Advanced Engineering Mathematics by Stroud K.A. Booth Dexter J(1)

Advanced Engineering Mathematics K. A. Stroud Formerly Principal Lecturer Department of Mathematics, Coventry University with additions by Dexter J. Booth Formerly Principal Lecturer School of Computing and Engineering, University of Huddersfield FIFTH EDITION Review Board for the fifth edition: Professor Jem Hebden, University College London, UK Dr Colin Steele, University of Manchester, UK Professor Pete Peterson, John Tyler Community College, Virginia, USA Dr Alan McCall, University of Hertfordshire, UK Dr Daphne O’Doherty, Cardiff University, UK INDUSTRIAL PRESS, INC. NEW YORK # K.A. Stroud 1986, 1990, 1996 # K.A Stroud and Dexter J. Booth 2003 & 2011 All rights reserved. No reproduction, copy or transmission of this publication may be made without written permission. No portion of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Designs and Patents Act 1988, or under the terms of any licence permitting limited copying issued by the Copyright Licensing Agency, Saffron House, 6–10 Kirby Street, London EC1N 8TS. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. Published in North America under license from Palgrave Macmillan Ltd., Houndmills, Basingstoke, Hants RG21 6XS Industrial Press Inc. 989 Avenue of the Americas New York, NY 10018, USA Palgrave1 and Macmillan1 are registered trademarks in the United States, the United Kingdom, Europe and other countries ISBN 978–0–8311–3449–5 paperback This book is printed on paper suitable for recycling and made from fully managed and sustained forest sources. Logging, pulping and manufacturing processes are expected to conform to the environmental regulations of the country of origin. A catalogue record for this book is available from the British Library. A catalog record for this book is available from the Library of Congress. Printed in China Summary of contents Preface to the first edition Preface to the second edition Preface to the third edition Preface to the fifth edition Hints on using the book Useful background information xviii xix xix xx xxi xxii 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 1 46 92 123 155 193 236 267 297 334 357 378 398 439 482 519 563 593 642 691 735 771 818 869 895 935 983 1014 Numerical solutions of equations and interpolation Laplace transforms 1 Laplace transforms 2 Laplace transforms 3 Difference equations and the Z transform Introduction to invariant linear systems Fourier series 1 Fourier series 2 Introduction to the Fourier transform Power series solutions of ordinary differential equations 1 Power series solutions of ordinary differential equations 2 Power series solutions of ordinary differential equations 3 Numerical solutions of ordinary differential equations Partial differentiation Partial differential equations Matrix algebra Systems of ordinary differential equations Numerical solutions of partial differential equations Multiple integration 1 Multiple integration 2 Integral functions Vector analysis 1 Vector analysis 2 Vector analysis 3 Complex analysis 1 Complex analysis 2 Complex analysis 3 Optimization and linear programming Appendix Answers Index 1063 1072 1105 Contents Preface to the first edition xviii Preface to the second edition xix Preface to the third edition xix Preface to the fifth edition xx Hints on using the book xxi Useful background information xxii Programme 1 Numerical solutions of equations and interpolation 1 Learning outcomes Introduction The Fundamental Theorem of Algebra Relations between the coefficients and the roots of a polynomial equation Cubic equations Transforming a cubic to reduced form Tartaglia’s solution for a real root Numerical methods Bisection Numerical solution of equations by iteration Using a spreadsheet Relative addresses Newton–Raphson iterative method Tabular display of results Modified Newton–Raphson method Interpolation Linear interpolation Graphical interpolation Gregory–Newton interpolation formula using forward finite differences Central differences Gregory–Newton backward differences Lagrange interpolation Revision summary 1 Can you? Checklist 1 Test exercise 1 Further problems 1 1 2 2 v 4 7 8 8 10 10 12 12 14 15 16 21 24 25 25 25 31 33 34 38 41 42 43 vi Contents Programme 2 Laplace transforms 1 46 Learning outcomes Introduction Laplace transforms Theorem 1 The first shift theorem Theorem 2 Multiplying by t and t n Theorem 3 Dividing by t Inverse transforms Rules of partial fractions The ‘cover up’ rule Table of inverse transforms Solution of differential equations by Laplace transforms Transforms of derivatives Solution of first-order differential equations Solution of second-order differential equations Simultaneous differential equations Revision summary 2 Can you? Checklist 2 Test exercise 2 Further problems 2 46 47 47 54 55 57 60 61 67 68 69 70 72 74 81 86 89 90 90 Programme 3 Laplace transforms 2 92 Learning outcomes Introduction Heaviside unit step function Unit step at the origin Effect of the unit step function Laplace transform of uðt cÞ Laplace transform of uðt cÞ f ðt cÞ (the second shift theorem) Differential equations involving the unit step function Convolution The convolution theorem Revision summary 3 Can you? Checklist 3 Test exercise 3 Further problems 3 92 93 93 94 94 97 98 108 112 117 118 120 120 121 Programme 4 Laplace transforms 3 123 Learning outcomes Laplace transforms of periodic functions Periodic functions Inverse transforms The Dirac delta function – the unit impulse Graphical representation Laplace transform of ðt aÞ The derivative of the unit step function 123 124 124 130 134 135 136 139 vii Contents Differential equations involving the unit impulse Harmonic oscillators Damped motion Forced harmonic motion with damping Resonance Revision summary 4 Can you? Checklist 4 Test exercise 4 Further problems 4 140 142 144 146 149 151 152 153 154 Programme 5 Difference equations and the Z transform 155 Learning outcomes Introduction Sequences Difference equations Solving difference equations Solution by inspection The particular solution The Z transform Table of Z transforms Properties of Z transforms Linearity First shift theorem (shifting to the left) Second shift theorem (shifting to the right) Translation Final value theorem The initial value theorem The derivative of the transform Inverse transforms Solving difference equations Sampling Revision summary 5 Can you? Checklist 5 Test exercise 5 Further problems 5 155 156 156 158 160 160 163 166 171 171 171 172 173 174 175 176 176 177 180 183 186 189 190 190 Programme 6 Introduction to invariant linear systems 193 Learning outcomes Invariant linear systems Systems Input-response relationships Linear systems Time-invariance of a continuous system Shift-invariance of a discrete system Differential equations The general nth-order equation 193 194 194 195 196 199 201 202 202 viii Contents Zero-input response and zero-state response Zero-input, zero-response Time-invariance Responses of a continuous system Impulse response Arbitrary input Exponential response The transfer function Differential equations Responses of a discrete system The discrete unit impulse Arbitrary input Exponential response Transfer function Difference equations Revision summary 6 Can you? Checklist 6 Test exercise 6 Further problems 6 203 206 208 209 209 209 213 215 217 220 220 221 223 224 225 229 232 233 234 Programme 7 Fourier series 1 236 Learning outcomes Introduction Periodic functions Graphs of y ¼ A sin nx Harmonics Non-sinusoidal periodic functions Analytic description of a periodic function Integrals of periodic functions Orthogonal functions Fourier series Dirichlet conditions Effect of harmonics Gibbs’ phenomenon Sum of a Fourier series at a point of discontinuity Revision summary 7 Can you? Checklist 7 Test exercise 7 Further problems 7 236 237 237 237 238 239 239 243 247 247 250 257 258 259 261 262 263 264 Programme 8 Fourier series 2 267 Learning outcomes Functions with periods other than 2 Function with period T Fourier coefficients Odd and even functions Products of odd and even functions 267 268 268 269 272 275 ix Contents Half-range series Series containing only odd harmonics or only even harmonics Significance of the constant term 12 a0 Half-range series with arbitrary period Revision summary 8 Can you? Checklist 8 Test exercise 8 Further problems 8 282 286 288 289 292 293 294 295 Programme 9 Introduction to the Fourier transform 297 Learning outcomes Complex Fourier series Introduction Complex exponentials Complex spectra The two domains Continuous spectra Fourier’s integral theorem Some special functions and their transforms Even functions Odd functions Top-hat function The Dirac delta The triangle function Alternative forms Properties of the Fourier transform Linearity Time shifting Frequency shifting Time scaling Symmetry Differentiation The Heaviside unit step function Convolution The convolution theorem Fourier cosine and sine transforms Table of transforms Revision summary 9 Can you? Checklist 9 Test exercise 9 Further problems 9 297 298 298 298 303 304 305 307 310 310 310 312 314 316 316 317 317 318 318 318 319 320 321 322 323 325 327 327 330 331 332 Programme 10 334 Power series solutions of ordinary differential equations 1 Learning outcomes Higher derivatives Leibnitz theorem 334 335 338 x Contents Choice of functions for u and v Power series solutions Leibnitz–Maclaurin method Cauchy-Euler equi-dimensional equations Revision summary 10 Can you? Checklist 10 Test exercise 10 Further problems 10 340 341 342 349 353 354 355 355 Programme 11 357 Power series solutions of ordinary differential equations 2 Learning outcomes Introduction Solution of differential equations by the method of Frobenius Indicial equation Revision summary 11 Can you? Checklist 11 Test exercise 11 Further problems 11 357 358 358 360 376 377 377 377 Programme 12 378 Power series solutions of ordinary differential equations 3 Learning outcomes Introduction Bessel functions Graphs of Bessel functions J0 ðxÞ and J1 ðxÞ Legendre’s equation Legendre polynomials Rodrigue’s formula and the generating function Sturm–Liouville systems Orthogonality Legendre’s equation revisited Polynomials as a finite series of Legendre polynomials Revision summary 12 Can you? Checklist 12 Test exercise 12 Further problems 12 378 379 380 385 385 386 386 388 390 391 392 393 395 396 396 Programme 13 398 Numerical solutions of ordinary differential equations Learning outcomes Introduction Taylor’s series Function increment First-order differential equations Euler’s method 398 399 399 400 401 401 xi Contents The exact value and the errors Graphical interpretation of Euler’s method The Euler–Cauchy method – or the improved Euler method Euler–Cauchy calculations Runge–Kutta method Second-order differential equations Euler second-order method Runge–Kutta method for second-order differential equations Predictor–corrector methods Revision summary 13 Can you? Checklist 13 Test exercise 13 Further problems 13 410 414 416 417 422 425 425 427 432 434 436 436 437 Programme 14 439 Partial differentiation Learning outcomes Small increments Taylor’s theorem for one independent variable Taylor’s theorem for two independent variables Small increments Rates of change Implicit functions Change of variables Inverse functions General case Stationary values of a function Maximum and minimum values Saddle point Lagrange undetermined multipliers Functions with three independent variables Revision summary 14 Can you? Checklist 14 Test exercise 14 Further problems 14 439 440 440 440 442 444 445 446 450 452 458 459 465 470 473 476 478 479 480 Programme 15 482 Partial differential equations Learning outcomes Introduction Partial differential equations Solution by direct integration Initial conditions and boundary conditions The wave equation Solution of the wave equation Solution by separating the variables The heat conduction equation for a uniform finite bar Solutions of the heat conduction equation 482 483 484 484 485 486 487 487 496 497 xii Contents Laplace’s equation Solution of the Laplace equation Laplace’s equation in plane polar coordinates The problem Separating the variables The n ¼ 0 case Revision summary 15 Can you? Checklist 15 Test exercise 15 Further problems 15 502 502 507 508 508 511 514 515 516 517 Programme 16 519 Matrix algebra Learning outcomes Singular and non-singular matrices Rank of a matrix Elementary operations and equivalent matrices Consistency of a set of equations Uniqueness of solutions Solution of sets of equations Inverse method Row transformation method Gaussian elimination method Triangular decomposition method Using an electronic spreadsheet Comparison of methods Matrix transformation Rotation of axes Revision summary 16 Can you? Checklist 16 Test exercise 16 Further problems 16 519 520 521 522 526 527 531 531 535 539 542 548 552 553 555 557 559 560 561 Programme 17 563 Systems of ordinary differential equations Learning outcomes Eigenvalues and eigenvectors Introduction Cayley–Hamilton theorem Systems of first-order ordinary differential equations Diagonalisation of a matrix Systems of second-order differential equations Revision summary 17 Can you? Checklist 17 Test exercise 17 Further problems 17 563 564 564 571 572 577 582 589 590 591 591 xiii Contents Programme 18 Numerical solutions of partial differential equations 593 Learning outcomes Introduction Numerical approximation to derivatives Functions of two real variables Grid values Computational molecules Summary of procedures Derivative boundary conditions Second-order partial differential equations Second partial derivatives Time-dependent equations The Crank–Nicolson procedure Dimensional analysis Revision summary 18 Can you? Checklist 18 Test exercise 18 Further problems 18 593 594 594 597 598 601 605 608 612 615 619 624 631 632 635 636 637 Programme 19 642 Multiple integration 1 Learning outcomes Introduction Differentials Exact differential Integration of exact differentials Area enclosed by a closed curve Line integrals Alternative form of a line integral Properties of line integrals Regions enclosed by closed curves Line integrals round a closed curve Line integral with respect to arc length Parametric equations Dependence of the line integral on the path of integration Exact differentials in three independent variables Green’s theorem Revision summary 19 Can you? Checklist 19 Test exercise 19 Further problems 19 642 643 650 653 655 657 660 661 664 666 667 671 672 673 678 679 686 688 689 690 Programme 20 691 Learning outcomes Double integrals Surface integrals Multiple integration 2 691 692 697 xiv Contents Space coordinate systems Volume integrals Change of variables in multiple integrals Curvilinear coordinates Transformation in three dimensions Revision summary 20 Can you? Checklist 20 Test exercise 20 Further problems 20 703 708 717 719 727 729 731 732 732 Programme 21 735 Integral functions Learning outcomes Integral functions The gamma function The beta function Relation between the gamma and beta functions Application of gamma and beta functions Duplication formula for gamma functions The error function The graph of erf ðxÞ The complementary error function erfc ðxÞ Elliptic functions Standard forms of elliptic functions Complete elliptic functions Alternative forms of elliptic functions Revision summary 21 Can you? Checklist 21 Test exercise 21 Further problems 21 735 736 736 745 749 750 754 755 756 756 758 759 759 763 766 768 769 769 Programme 22 771 Vector analysis 1 Learning outcomes Introduction Triple products Properties of scalar triple products Coplanar vectors Vector triple products of three vectors Differentiation of vectors Differentiation of sums and products of vectors Unit tangent vectors Partial differentiation of vectors Integration of vector functions Scalar and vector fields Grad (gradient of a scalar field) Directional derivatives Unit normal vectors 771 772 777 778 779 781 784 789 789 792 792 795 795 798 801 xv Contents Grad of sums and products of scalars Div (divergence of a vector function) Curl (curl of a vector function) Summary of grad, div and curl Multiple operations Revision summary 22 Can you? Checklist 22 Test exercise 22 Further problems 22 803 805 806 807 809 812 814 815 815 Programme 23 818 Vector analysis 2 Learning outcomes Line integrals Scalar field Vector field Volume integrals Surface integrals Scalar fields Vector fields Conservative vector fields Divergence theorem (Gauss’ theorem) Stokes’ theorem Direction of unit normal vectors to a surface S Green’s theorem Revision summary 23 Can you? Checklist 23 Test exercise 23 Further problems 23 818 819 819 822 826 830 831 834 839 844 850 853 859 862 864 865 866 Programme 24 869 Vector analysis 3 Learning outcomes Curvilinear coordinates Orthogonal curvilinear coordinates Orthogonal coordinate systems in space Scale factors Scale factors for coordinate systems General curvilinear coordinate system ðu, v, wÞ Transformation equations Element of arc ds and element of volume dV in orthogonal curvilinear coordinates Grad, div and curl in orthogonal curvilinear coordinates Particular orthogonal systems Revision summary 24 Can you? Checklist 24 Test exercise 24 Further problems 24 869 870 874 875 879 880 882 883 884 885 888 890 892 893 894 xvi Contents Programme 25 Complex analysis 1 895 Learning outcomes Functions of a complex variable Complex mapping Mapping of a straight line in the z-plane onto the w-plane under the transformation w ¼ f ðzÞ Types of transformation of the form w ¼ az þ b Non-linear transformations Mapping of regions Revision summary 25 Can you? Checklist 25 Test exercise 25 Further problems 25 899 903 912 917 931 932 932 933 Programme 26 935 Complex analysis 2 895 896 897 Learning outcomes Differentiation of a complex function Regular function Cauchy–Riemann equations Harmonic functions Complex integration Contour integration – line integrals in the z-plane Cauchy’s theorem Deformation of contours at singularities Conformal transformation (conformal mapping) Conditions for conformal transformation Critical points Schwarz–Christoffel transformation Open polygons Revision summary 26 Can you? Checklist 26 Test exercise 26 Further problems 26 935 936 937 939 941 946 946 949 954 963 963 964 967 972 978 979 980 981 Programme 27 983 Complex analysis 3 Learning outcomes Maclaurin series Radius of convergence Singular points Poles Removable singularities Circle of convergence Taylor’s series Laurent’s series 983 984 988 989 989 990 990 991 993 xvii Contents Residues Calculating residues Integrals of real functions Revision summary 27 Can you? Checklist 27 Test exercise 27 Further problems 27 997 999 1000 1007 1009 1010 1011 Programme 28 1014 Optimization and linear programming Learning outcomes Optimization Linear programming (or linear optimization) Linear inequalities Graphical representation of linear inequalities The simplex method Setting up the simplex tableau Computation of the simplex Simplex with three problem variables Artificial variables Minimisation Applications Revision summary 28 Can you? Checklist 28 Test exercise 28 Further problems 28 1014 1015 1015 1016 1016 1022 1022 1024 1032 1036 1047 1051 1055 1056 1057 1058 Appendix 1063 Answers 1072 Index 1105 Preface to the first edition The purpose of this book is essentially to provide a sound second year course in Mathematics appropriate to studies leading to B.Sc. Engineering Degrees and other qualifications of a comparable level. The emphasis throughout is on techniques and applications, supported by sufficient formal proofs to warrant the methods being employed. The structure of the text and the techniques used follow closely those of the author’s first year book, Engineering Mathematics – Programmes and Problems, to which this further book is a companion volume and a continuation of the highly successful learning strategies devised. As with the previous work, the text is based on a series of self-instructional programmes arising from extensive research and rigid evaluation in a variety of relevant courses and, once again, the individualised nature of the development makes the book eminently suitable both for general class use and for personal study. Each of the course programmes guides the student through the development of a particular topic, with numerous worked examples to demonstrate the techniques and with increased responsibility passing to the student as mastery is achieved. Revision exercises are provided where appropriate and each programme terminates with a Revision Summary of the main points covered, a Test Exercise based directly on the work of the programme and a set of Further Problems which provides opportunity for the additional practice that is essential for ensured success. The ability to work at one’s own pace throughout is of utmost importance in maintaining motivation and in achieving mastery. In several instances, the topic of a programme is a direct extension of basic work covered in Engineering Mathematics and where this is so, the title page of the programme carries a brief reference to the relevant programme in the first year treatment. This clearly directs the student to worthwhile revision of the prerequisites assumed in the further development of the subject matter. A complete set of Answers to all problems and a detailed Index are provided at the end of the book. Grateful acknowledgement is made of the constructive suggestions and cooperation received from many quarters both in the development of the original programmes and in the final preparation of the text. Recognition must also be made of the many sources from which examples have been gleaned over the years and which contribute in no small measure to the success of the work. Finally my sincere appreciation is due to the publishers for their patience, advice and ready cooperation in the preparation of the text for publication. K.A. Stroud xviii Preface to the second edition Since the first publication of Further Engineering Mathematics as core material for a typical second year engineering degree course, requests have been received from time to time for the inclusion of further topics to cover the particular requirements of individual syllabuses. Some limit, inevitably, has to be placed on the physical size of the text, but it has been possible at least to include a programme on Linear Optimisation (Linear Programming) which was one of the subjects most frequently required. The treatment of the additional material follows the structure of the rest of the book and the emphasis is largely on the practical use of the simplex method for the solution of both maximisation and minimisation problems. The opportunity has also been taken to amend and clarify a number of minor points in the existing text and my thanks are due to those correspondents who have undertaken to write with constructive comment. Such feedback is always welcome. K.A.S. Preface to the third edition With the new edition of Further Engineering Mathematics, the opportunity has been taken to incorporate a number of minor revisions and amendments to the previous text. The format of the pages has been changed and the publishers have undertaken the complete resetting of the text to result in a more open presentation of the material and to facilitate the learning process still further. Once again, my sincere thanks are due to all those correspondents who have kindly written with constructive comment concerning the book and to the publishers for their continued support, advice and cooperation throughout the preparation, production and marketing of the work. K.A.S. xix Preface to the fifth edition It is now over 40 years since Ken Stroud first developed his approach to personalised learning with his classic text Engineering Mathematics, now in its sixth edition and having sold over half a million copies. Some 15 years later he followed this with Further Engineering Mathematics which was restyled as Advanced Engineering Mathematics for its fourth edition. As in all earlier editions his unique and hugely successful programmed learning style is continued in this fifth edition of Advanced Engineering Mathematics, and I am delighted to have been asked to contribute to it. As with the fourth edition, I have endeavoured to retain the very essence of the style, particularly the timetested Stroud format with its close attention to technique development throughout. This methodology has, over the years, contributed significantly to the mathematical abilities of so many students all over the world. New to this edition To cater for continual changes in engineering mathematics the work of this edition builds upon material that was introduced in the previous edition. Notably, the new programme, Introduction to invariant linear systems, builds upon the new-styled Z transforms, now called Difference equations and the Z transform. It was also felt that the work on convolution was incorrectly located so it has been moved to Laplace transforms 2 and significantly expanded to further clarify the concept. The programme on Fourier series has been split into two parts; the first dealing with periodic functions with period 2 and the second dealing with general periodic functions as well as half-range series. The programme, Power series solutions of ordinary differential equations, was very large and unwieldy and has been split into three. The first part deals with the Leibnitz–Maclaurin method with the addition of Cauchy–Euler equations, the second part deals with the Frobenius method and the third part deals with special functions. Finally, the programme on Matrix algebra has been split into two with the second part concentrating on systems of differential equations. Acknowledgements This is a further opportunity that I have had to work on the Stroud books. It is as ever a challenge and an honour to be able to work with Ken Stroud’s material. Ken had an understanding of his students and their learning and thinking processes which was second to none, and this is reflected in every page of this book. As always, my thanks go to the Stroud family for their continuing support for and encouragement of new projects and ideas which are allowing Ken’s hugely successful teaching methodology to be offered to a whole new range of students. I should also like to express my thanks and appreciation for the valuable feedback that has been provided by all the reviewers during the writing of this new edition and of previous editions upon xx xxi Preface to the fifth edition and Hints on using the book which this one builds. In particular I should like to mention Professor Pete Peterson of John Tyler Community College, Virginia, USA. Engineering mathematics is not a static universe and it is always a challenge to best determine how a new edition is to be developed. Pete’s encouraging comments and sympathetic treatment of the new material was greatly appreciated. Finally, I should like to thank the entire production team at Palgrave Macmillan for all their care and principally my editor Helen Bugler whose enthusiasm and professionalism I greatly value and admire. Huddersfield March 2011 Dexter J Booth Hints on using the book This book contains twenty-eight Programmes, each of which has been written in such a way as to make learning more effective and more interesting. It is almost like having a personal tutor, for you proceed at your own rate of learning and any difficulties you may have are cleared before you have the chance to practise incorrect ideas or techniques. You will find that each Programme is divided into sections called frames. When you start a Programme, begin at Frame 1. Read each frame carefully and carry out any instructions or exercise which you are asked to do. In almost every frame, you are required to make a response of some kind, testing your understanding of the information in the frame, and you can immediately compare your answer with the correct answer given in the next frame. To obtain the greatest benefit, you are strongly advised to cover up the following frame, where necessary, until you have made your response. When a series of dots occurs, you are expected to supply the missing word, phrase, or number. At every stage, you will be guided along the right path. There is no need to hurry: read the frames carefully and follow the directions exactly. In this way, you must learn. At the end of each Programme, you will find a Revision summary and a Can you? checklist that matches the Learning outcomes given at the beginning of the Programme. Read these carefully to make sure you have not missed anything. Next you will find a short Test exercise. This is set directly on what you have learned in the Programme: the questions are straightforward and contain no tricks. When you have completed these, return to the Can you? checklist as a final reminder of the contents of the Programme. To provide you with the necessary practice, a set of Further problems is also included. Remember that in mathematics, as in many other situations, practice makes perfect – or more nearly so. Even if you feel you have done some of the topics before, work steadily through each Programme: it will serve as useful revision and fill in any gaps in your knowledge that you may have. Useful background information 1 Algebraic identities ða þ bÞ2 ¼ a2 þ 2ab þ b2 ða þ bÞ3 ¼ a3 þ 3a2 b þ 3ab2 þ b3 ða bÞ2 ¼ a2 2ab þ b2 ða bÞ3 ¼ a3 3a2 b þ 3ab2 b3 ða þ bÞ4 ¼ a4 þ 4a3 b þ 6a2 b2 þ 4ab3 þ b4 ða bÞ4 ¼ a4 4a3 b þ 6a2 b2 4ab3 þ b4 a2 b2 ¼ ða bÞða þ bÞ a3 þ b3 ¼ ða þ bÞða2 ab þ b2 Þ a3 b3 ¼ ða bÞða2 þ ab þ b2 Þ 2 Trigonometrical identities (1) sin2 þ cos2 ¼ 1; sec2 ¼ 1 þ tan2 ; cosec2 ¼ 1 þ cot2 (2) sinðA þ BÞ ¼ sin A cos B þ cos A sin B sinðA BÞ ¼ sin A cos B cos A sin B cosðA þ BÞ ¼ cos A cos B sin A sin B cosðA BÞ ¼ cos A cos B þ sin A sin B tan A þ tan B 1 tan A tan B tan A tan B tanðA BÞ ¼ 1 þ tan A tan B (3) Let A ¼ B ¼ ; sin 2 ¼ 2 sin cos tanðA þ BÞ ¼ cos 2 ¼ cos2 sin2 ¼ 1 2 sin2 ¼ 2 cos2 1 2 tan 1 tan2 ; sin ¼ 2 sin cos 2 2 cos ¼ cos2 sin2 2 2 ¼ 1 2 sin2 ¼ 2 cos2 1 2 2 tan 2 ¼ (4) Let ¼ 2 xxii xxiii Useful background information 2 tan tan ¼ 2 2 CþD CD cos (5) sin C þ sin D ¼ 2 sin 2 2 CþD CD sin sin C sin D ¼ 2 cos 2 2 CþD CD cos cos C þ cos D ¼ 2 cos 2 2 CþD CD cos D cos C ¼ 2 sin sin 2 2 (6) 2 sin A cos B ¼ sinðA þ BÞ þ sinðA BÞ 1 tan2 2 cos A sin B ¼ sinðA þ BÞ sinðA BÞ 2 cos A cos B ¼ cosðA þ BÞ þ cosðA BÞ 2 sin A sin B ¼ cosðA BÞ cosðA þ BÞ (7) Negative angles: sinðÞ ¼ sin cosðÞ ¼ cos tanðÞ ¼ tan (8) Angles having the same trigonometrical ratios: (a) Same sine: and ð1808 Þ (b) Same cosine: and ð3608 Þ, i.e. ðÞ (c) Same tangent: and ð1808 þ Þ (9) a sin þ b cos ¼ A sinð þ Þ a sin b cos ¼ A sinð Þ a cos þ b sin ¼ A cosð Þ a cos b sin ¼ A cosð þ Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 8 2 2 > <A ¼ a þ b where > : ¼ tan1 b ð08 < < 908Þ a 3 Standard curves (a) Straight line dy y2 y1 Slope, m ¼ ¼ dx x2 x1 Angle between two lines, tan ¼ m2 m1 1 þ m1 m2 For parallel lines, m2 ¼ m1 For perpendicular lines, m1 m2 ¼ 1 xxiv Useful background information Equation of a straight line (slope ¼ m) (1) Intercept c on real y-axis: y ¼ mx þ c (2) Passing through ðx1 , y1 Þ: y y1 ¼ mðx x1 Þ y y1 x x1 ¼ (3) Joining ðx1 , y1 Þ and ðx2 ; y2 Þ: y2 y1 x 2 x 1 (b) Circle Centre at origin, radius r: x2 þ y 2 ¼ r 2 Centre ðh; kÞ, radius r: ðx hÞ2 þ ðy kÞ2 ¼ r 2 x2 þ y 2 þ 2gx þ 2fy þ c ¼ 0 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi with centre ðg; f Þ: radius ¼ g 2 þ f 2 c General equation: Parametric equations: x ¼ r cos , y ¼ r sin (c) Parabola Vertex at origin, focus ða, 0Þ: y 2 ¼ 4ax Parametric equations: x ¼ at 2 , y ¼ 2at (d) Ellipse pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 y 2 Centre at origin, foci a2 þ b2 ; 0 : 2 þ 2 ¼ 1 a b where a ¼ semi-major axis, b ¼ semi-minor axis Parametric equations: x ¼ a cos , y ¼ b sin (e) Hyperbola pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 y 2 Centre at origin, foci a2 þ b2 ; 0 : 2 2 ¼ 1 a b Parametric equations: x ¼ a sec , y ¼ b tan Rectangular hyperbola: a a a2 ¼ c2 Centre at origin, vertex pffiffiffi ; pffiffiffi : xy ¼ 2 2 2 a where c ¼ pffiffiffi i.e. xy ¼ c2 2 Parametric equations: x ¼ ct, y ¼ c=t 4 Laws of mathematics (a) Associative laws – for addition and multiplication a þ ðb þ cÞ ¼ ða þ bÞ þ c aðbcÞ ¼ ðabÞc (b) Commutative laws – for addition and multiplication aþb¼bþa ab ¼ ba (c) Distributive laws – for multiplication and division aðb þ cÞ ¼ ab þ ac bþc b c ¼ þ (provided a 6¼ 0Þ a a a Programme 1 Frames 1 to 87 Numerical solutions of equations and interpolation Learning outcomes When you have completed this Programme you will be able to: . Appreciate the Fundamental Theorem of Algebra . Find the two roots of a quadratic equation and recognise that for polynomial equations with real coefficients complex roots exist in complex conjugate pairs . Use the relationships between the coefficients and the roots of a polynomial equation to find the roots of the polynomial . Transform a cubic equation to its reduced form . Use Tartaglia’s solution to find the roots of a cubic equation . Find the solution of the equation f ðxÞ ¼ 0 by the method of bisection . Solve equations involving a single real variable by iteration and use a spreadsheet for efficiency . Solve equations using the Newton–Raphson iterative method . Use the modified Newton–Raphson method to find the first approximation when the derivative is small . Understand the meaning of interpolation and use simple linear and graphical interpolation . Use the Gregory–Newton interpolation formula with forward and backward differences for equally spaced domain points . Use the Gauss interpolation formulas using central differences for equally spaced domain points . Use Lagrange interpolation when the domain points are not equally spaced 1 2 Programme 1 Introduction 1 In this Programme we shall be looking at analytic and numerical methods of solving the general equation in a single variable, f ðxÞ ¼ 0. In addition, a functional relationship can be exhibited in the form of a collection of ordered pairs rather than in the form of an algebraic expression. We shall be looking at interpolation methods of estimating values of f ðxÞ for intermediate values of x between those listed among the ordered pairs. First we shall look at the Fundamental Theorem of Algebra, which deals with the factorisation of polynomials. The Fundamental Theorem of Algebra 2 The Fundamental Theorem of Algebra can be stated as follows: Every polynomial expression f (x) = an xn + an1 xn1 + + a1 x + a0 can be written as a product of n linear factors in the form f (x) = an (x r1 )(x r2 )( )(x rn ) As an immediate consequence of this we can see that there are n values of x that satisfy the polynomial equation f ðxÞ ¼ 0, namely x ¼ r1 , x ¼ r2 , . . . , x ¼ rn . We call these values the roots of the polynomial, but be aware that they may not all be distinct. Furthermore, the polynomial coefficients ai and the polynomial roots ri may be real, imaginary or complex. For example the quadratic equation x2 þ 5x þ 6 ¼ 0 can be written ðx þ 2Þðx þ 3Þ ¼ 0 so it has the two distinct roots x ¼ 2 and x ¼ 3 x2 4x þ 4 ¼ 0 can be written as ðx 2Þðx 2Þ ¼ 0 so it has the two coincident roots x ¼ 2 and x ¼ 2 x2 þ x þ 1 ¼ 0 can be written as ðx þ aÞðx þ bÞ ¼ 0 so it has the two roots x ¼ a and x ¼ b To find the numerical values of a and b we need to use the formula for finding the roots of a general quadratic equation. Can you recall what it is? If not, then refer to Frame 14 of Programme F.6 in Engineering Mathematics, Sixth Edition. The solution to the quadratic equation ax2 þ bx þ c ¼ 0 is . . . . . . . . . . . . The answer is in the next frame 3 Numerical solutions of equations and interpolation x¼ b pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b2 4ac 2a 3 So the roots of x2 þ x þ 1 ¼ 0 are . . . . . . . . . . . . Next frame pffiffiffi 1 3 x¼ j 2 2 4 Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffi b2 4ac 1 1 4 ¼ a ¼ b ¼ c ¼ 1 and so x ¼ 2a 2 pffiffiffi 1 3 ¼ j 2 2 b This quadratic equation has two distinct complex roots. Notice that the two roots form a complex conjugate pair – each is the complex conjugate of the other. Whenever a polynomial with real coefficients ai has a complex root it also has the complex conjugate as another root. pffiffiffi So given that x ¼ 2 þ j 5 is one root of a quadratic equation with real coefficients then the other root is . . . . . . . . . . . . pffiffiffi x ¼ 2 j 5 Because pffiffiffi pffiffiffi The complex conjugate of x ¼ 2 þ j 5 is x ¼ 2 j 5 and complex roots of a polynomial equation with real coefficients always appear as conjugate pairs. The quadratic equation with these two roots is . . . . . . . . . . . . 5 4 Programme 1 6 x2 þ 4x þ 9 ¼ 0 Because If x ¼ a and x ¼ b are the two roots of a quadratic equation then ðx aÞðx bÞ ¼ 0 gives the quadratic equation. That is ðx aÞðx bÞ ¼ x2 ða þ bÞx þ ab ¼ 0. pffiffiffi pffiffiffi Here, the two roots are x ¼ 2 þ j 5 and x ¼ 2 j 5 so that h h pffiffiffii pffiffiffii x 2 þ j 5 x 2 j 5 ¼ 0 h pffiffiffi pffiffiffii h pffiffiffiih pffiffiffii That is x2 x 2 þ j 5 2 j 5 þ 2 þ j 5 2 j 5 ¼ 0. So x2 þ 4x þ 9 ¼ 0. Notice that the coefficients are . . . . . . . . . . . . 7 Real Relations between the coefficients and the roots of a polynomial equation Let , , be the roots of px3 þ qx2 þ rx þ s ¼ 0. Then, writing the expression px3 þ qx2 þ rx þ s in terms of , , gives q r s px3 þ qx2 þ rx þ s ¼ p x3 þ x2 þ x þ p p p ¼ ............ 8 pðx Þðx Þðx Þ Therefore q r s px3 þ qx2 þ rx þ s ¼ p x3 þ x2 þ x þ p p p ¼ pðx Þðx Þðx Þ ¼ p x2 ½ þ x þ ðx Þ 3 ¼ p x ½ þ x2 þ x x2 þ ½ þ x ¼ p x3 ½ þ þ x2 þ ½ þ þ x Þ Therefore, equating coefficients (a) þ þ ¼ . . . . . . . . . . . . (b) þ þ ¼ . . . . . . . . . . . . (c) ¼ . . . . . . . . . . . . 5 Numerical solutions of equations and interpolation q (a) ; p (b) r ; p (c) 9 s p This, of course, applies to a cubic equation. Let us extend this to a more general equation. So on to the next frame In general, if 1 , 2 , 3 . . . n are roots of the equation then 10 p0 xn þ p1 xn1 þ p2 xn2 þ . . . þ pn1 x þ pn ¼0 sum of the roots ¼ sum of products of the roots, two at a time ¼ p1 p0 p2 p0 sum of products of the roots, three at a time ¼ sum of products of the roots, n at a time ðp0 6¼ 0Þ p3 p0 ¼ ð1Þn : pn p0 So for the equation 3x4 þ 2x3 þ 5x2 þ 7x 4 ¼ 0, if , , , are the four roots, then (a) þ þ þ ¼ . . . . . . . . . . . . (b) þ þ þ þ þ ¼ . . . . . . . . . . . . (c) þ þ þ ¼ . . . . . . . . . . . . (d) ¼ . . . . . . . . . . . . 2 (a) ; 3 (b) 5 ; 3 7 (c) ; 3 (d) 4 3 Now for a problem or two on the same topic. Example 1 Solve the equation x3 8x2 þ 9x þ 18 ¼ 0 given that the sum of two of the roots is 5. Using the same approach as before, if , , are the roots, then (a) þ þ ¼ . . . . . . . . . . . . (b) þ þ ¼ . . . . . . . . . . . . (c) ¼ . . . . . . . . . . . . 11 6 Programme 1 12 (a) 8; So we have þ þ ¼ 8 ;5þ ¼8 Also (b) 9; (c) 18 Let þ ¼ 5 ;¼3 ¼ 18 ð3Þ ¼ 18 ; ¼ 6 þ ¼ 5 ; ¼ 5 ; ð5 Þ ¼ 6 2 5 6 ¼ 0 ; ð 6Þð þ 1Þ ¼ 0 ; ¼ 1 or 6 ; ¼ 6 or 1 Roots are x ¼ 1, 3, 6 13 Example 2 Solve the equation 2x3 þ 3x2 11x 6 ¼ 0 given that the three roots form an arithmetic sequence. Let us represent the roots by ða kÞ, a, ða þ kÞ Then the sum of the roots ¼ 3a ¼ . . . . . . . . . . . . and the product of the roots ¼ aða kÞða þ kÞ ¼ . . . . . . . . . . . . 14 3 3a ¼ ; 2 ;a¼ 1 2 aða þ kÞða kÞ ¼ 1 1 k2 ¼ 3 2 4 ;k¼ 6 ¼3 2 5 2 1 a ¼ ; a k ¼ 3; a þ k ¼ 2 2 5 1 If k ¼ a ¼ ; a k ¼ 2; a þ k ¼ 3 2 2 If k ¼ 5 2 1 ; required roots are 3, , 2 2 Here is a similar one. Example 3 Solve the equation x3 þ 3x2 6x 8 ¼ 0 given that the three roots are in geometric sequence. a This time, let the roots be , a, ak k a a Then ¼ a þ ak ¼ . . . . . . . . . . . . and ðaÞðakÞ ¼ . . . . . . . . . . . . k k 7 Numerical solutions of equations and interpolation sum of roots ¼ 3; product of roots ¼ 8 15 It then follows that the roots are . . . . . . . . . . . ., . . . . . . . . . . . ., . . . . . . . . . . . . 4, 16 2, 1 The working rests on the relationships between the roots and the coefficients, i.e. if , , are the roots of the cubic equation ax3 þ bx2 þ cx þ d ¼ 0 then (a) þ þ ¼ ............ (b) þ þ ¼ . . . . . . . . . . . . (c) ¼ . . . . . . . . . . . . b (a) ; a (b) c ; a (c) d a 17 In each of the three examples reconstruct the cubic to confirm that they are correct. Now on to the next stage Cubic equations The Fundamental Theorem of Algebra tells us that every cubic expression f ðxÞ ¼ ax3 þ bx2 þ cx þ d can be written as a product of three linear factors f ðxÞ ¼ aðx r1 Þðx r2 Þðx r3 Þ Consequently, every cubic equation f ðxÞ ¼ aðx r1 Þðx r2 Þðx r3 Þ ¼ 0 has three roots which may be distinct or coincident and which may be real or complex. However, because complex roots of a polynomial with real coefficients always appear in complex conjugate pairs we can say that every such cubic equation has at least one . . . . . . . . . . . . 18 8 Programme 1 19 at least one real root To find the value of this real root we can employ a formula equivalent to the formula used to find the two roots of the general quadratic. This is called Tartaglia’s method but before we can proceed to look at that we must first consider how to transform the general cubic to its reduced form. Next frame 20 Transforming a cubic to reduced form In every case, an equation of the form x3 þ ax2 þ bx þ c ¼ 0 can be converted into the reduced form y 3 þ py þ q ¼ 0 by the substitution a x¼y . 3 The example will demonstrate the method. Example 4 Express f ðxÞ ¼ x3 þ 6x2 4x þ 5 ¼ 0 in reduced form. a 6 Substitute x ¼ y , i.e. x ¼ y ¼ y 2. Put x ¼ y 2. 3 3 The equation then becomes ðy 2Þ3 þ 6ðy 2Þ2 4ðy 2Þ þ 5 ¼ 0 ðy 3 3y 2 2 þ 3y4 8Þ þ 6ðy 2 4y þ 4Þ 4ðy 2Þ þ 5 ¼ 0 which simplifies to . . . . . . . . . . . . 21 y 3 16y þ 29 ¼ 0 Tartaglia’s solution for a real root In the sixteenth century, Tartaglia discovered that a root of the cubic equation x3 þ ax þ b ¼ 0, where a > 0, is given by ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 b a3 b2 þ þ þ x¼ þ 2 2 27 4 27 4 That looks pretty formidable, but it is a good deal easier than it appears. Notice rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b a3 b2 that and þ occur twice and it is convenient to evaluate these first 2 27 4 and then substitute the results in the main expression for x. 9 Numerical solutions of equations and interpolation Example 5 Find a real root of x3 þ 2x þ 5 ¼ 0. b Here, a ¼ 2, b ¼ 5 ; ¼ 2:5 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ffi a3 b2 8 25 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 6:5463 ¼ 2:5586 þ þ ¼ 27 4 27 4 : : Then x ¼ ð2 5 þ 2 5586Þ1=3 þ ð2:5 2:5586Þ1=3 ¼ 0:3884 1:7166 ¼ 1:3282 x ¼ 1:328 Once we have a real root, the equation can be reduced to a quadratic and the remaining two roots determined. They are x ¼ 0:664 þ j 1:823 and x ¼ 0:664 j 1:823 (see Engineering Mathematics, Sixth Edition, Programme F.6). Example 6 Determine a real root of 2x3 þ 3x 4 ¼ 0. This is first written x3 þ 1:5x 2 ¼ 0 ; a ¼ 1:5, b ¼ 2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi b a3 b2 þ Now you can evaluate and and so determine 2 27 4 x ¼ ............ 22 0.8796 Because ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 þ ¼ f2:06066g1=3 ¼ 1:2725 and þ 2 27 4 ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 þ ¼ f0:6066g1=3 ¼ 0:3929, 2 27 4 therefore x ¼ 1:2725 0:3929 ¼ 0:8796 Note: If you wish to find the real root of a cubic of the form x3 þ ax þ b ¼ 0 where a < 0 then it is best that you resort to numerical methods. Read on. Next frame 10 Programme 1 Numerical methods 23 The methods that we have used so far to solve quadratic equations and to find the real root of a cubic equation are called analytic methods. These analytic methods used straightforward algebraic techniques to develop a formula for the answer. The numerical value of the answer can then be found by simple substitution of numbers for the variables in the formula. Unfortunately, general polynomial equations of order five or higher cannot by solved by analytic methods. Instead, we must resort to what are termed numerical methods. The simplest method of finding the solution to the equation f ðxÞ ¼ 0 is the bisection method. Bisection The bisection method of finding a solution to the equation f ðxÞ ¼ 0 consists of Finding a value of x, say x ¼ a, such that f ðaÞ < 0 Finding a value of x, say x ¼ b, such that f ðbÞ > 0 The solution to the equation f ðxÞ ¼ 0 must then lie between a and b. Furthermore, it must lie either in the first half of the interval between a and b or in the second half. f (x) f (b) > 0 f ([a + b]/2) a f (a) < 0 a+b 2 b x Find the value of f ð½a þ b=2Þ – that is halfway between a and b. If f ð½a þ b=2Þ > 0 then the solution lies in the first half and if f ð½a þ b=2Þ < 0 then it lies in the second half. This procedure is repeated, narrowing down the width of the interval by a half each time. An example should clarify all this. Example 7 Find the positive value of x that satisfies the equation x2 2 ¼ 0. Firstly we note that if x ¼ 1 then x2 2 < 0, and that if x ¼ 2 then x2 2 > 0, so the solution that we seek must lie between 1 and 2. We look for the . . . . . . . . . . . . 11 Numerical solutions of equations and interpolation 24 The mid-point between 1 and 2 which is 1.5 Now, when x ¼ 1:5, x2 2 ¼ 0:25 > 0 so the solution must lie between . . . . . . . . . . . . 25 1 and 1.5 The mid-point between 1 and 1.5 is 1.25. When x ¼ 1:25, x2 2 ¼ 0:4375 < 0 so the solution must lie between . . . . . . . . . . . . 26 1.25 and 1.5 The mid-point between 1.25 and 1.5 is 1.375. We now evaluate x2 2 at this point and determine in which half interval the solution lies. This process is repeated and the following table displays the results. In each block of six numbers the first column lists the end points of the interval and the midpoint. The second column contains the respective values f ðxÞ ¼ x2 2. Construct the table as follows. (a) For each block of six numbers copy the last number in the first column into the second place of the first column of the following block. This represents the centre point of the previous interval. (b) For each block of six numbers copy the number that represents the other end point of the new interval from the first column into the first place of the first column of the following block. Look at the signs in the second column of the first block to decide which is the appropriate number. a b ða þ bÞ=2 a b ða þ bÞ=2 a b ða þ bÞ=2 a b ða þ bÞ=2 1.0000 1.0000 2.0000 2.0000 1.5000 0.2500 1.0000 1.0000 1.5000 0.2500 1.2500 0.4375 1.5000 0.2500 1.2500 0.4375 1.3750 0.1094 1.5000 0.2500 1.3750 0.1094 1.4375 0.0664 1.3750 0.1094 1.4375 0.0664 1.4063 0.0225 1.4375 0.0664 1.4063 0.0225 1.4219 0.0217 1.4063 0.0225 1.4219 0.0217 1.4141 0.0004 1.4219 0.0217 1.4141 0.0004 1.4180 0.0106 1.4141 0.0004 1.4180 0.0106 1.4160 0.0051 1.4141 0.0004 1.4160 0.0051 1.4150 0.0023 1.4141 0.0004 1.4150 0.0023 1.4146 0.0010 1.4141 0.0004 1.4146 0.0010 1.4143 0.0003 1.4141 0.0004 1.4143 0.0003 1.4142 0.0001 1.4143 0.0003 1.4142 0.0001 1.4142 0.0001 1.4142 0.0001 1.4142 0.0001 1.4142 0.0000 The final result to four decimal places is x ¼ 1:4142 which is the correct answer to that level of accuracy – but it has taken a lot of activity to produce it. A much faster way of solving this equation is to use an iteration formula that was first devised by Newton. Next frame 12 Programme 1 Numerical solution of equations by iteration 27 The process of finding the numerical solution to the equation f ðxÞ ¼ 0 by iteration is performed by first finding an approximate solution and then using this approximate solution to find a more accurate solution. This process is repeated until a solution is found to the required level of accuracy. For example, Newton showed that the square root of a number a can be found from the iteration equation 1 a xi þ , i ¼ 0, 1, 2, . . . xiþ1 ¼ 2 xi where x0 is the approximation that starts pffiffiffi the iteration off. So, to find a succession of approximate values of 2, each of increasing accuracy, we proceed as follows. Let x0 ¼ 1:5 – found by the first stage of the bisection method. Then 1 a x0 þ ¼ 0:5ð1:5 þ 2=1:5Þ ¼ 1:4166 . . . x1 ¼ 2 x0 This value is then used to find x2 . By rounding x1 to 1.4167, the value of x2 is found to be . . . . . . . . . . . . 28 x2 ¼ 1:4142 Because x2 ¼ 1 a x1 þ ¼ 0:5ð1:4167 þ 2=1:4167Þ ¼ 1:4142 . . . 2 x1 This has achieved the same level of accuracy as the bisection method in just two steps. Using a spreadsheet This simple iteration procedure is more efficiently performed using a spreadsheet. If the use of a spreadsheet is a totally new experience for you then you are referred to Programme 4 of Engineering Mathematics, Sixth Edition where the spreadsheet is introduced as a tool for constructing graphs of functions. If you have a limited knowledge then you will be able to follow the text from here. The spreadsheet we shall be using here is Microsoft Excel, though all commercial spreadsheets possess the equivalent functionality. Open your spreadsheet and in cell A1 enter n and press Enter. In this first column we are going to enter the iteration numbers. In cell A2 enter the number 0 and press Enter. Place the cell highlight in cell A2 and highlight the block of cells A2 to A7 by holding down the mouse button and wiping the highlight down to cell A7. Click the Edit command on the Command bar and Numerical solutions of equations and interpolation 13 point at Fill from the drop-down menu. Select Series from the next dropdown menu and accept the default Step value of 1 by clicking OK in the Series window. The cells A3 to A7 fill with . . . . . . . . . . . . the numbers 1 to 5 29 In cell B1 enter the letter x – this column is going to contain the successive xvalues obtained by iteration. In cell B2 enter the value of x0 , namely 1.5. In cell B3 enter the formula ¼ 0.5*(B2+2/B2) The number that appears in cell B3 is then . . . . . . . . . . . . 1.416667 30 Place the cell highlight in cell B3, click the command Edit on the Command bar and select Copy from the drop-down menu. You have now copied the formula in cell B3 onto the Clipboard. Highlight the cells B4 to B7 and then click the Edit command again but this time select Paste from the drop-down menu. The cells B4 to B7 fill with numbers to provide the display ............ n 0 1 2 3 4 5 x 1.5 1.416667 1.414216 1.414214 1.414214 1.414214 By using the various formatting facilities provided by the spreadsheet the display can be amended to provide the following n 0 1 2 3 4 5 x 1.500000000000000 1.416666666666670 1.414215686274510 1.414213562374690 1.414213562373090 1.414213562373090 The number of decimal places here is 15, which is far greater than is normally required but it does demonstrate how effective a spreadsheet can be. In future we shall restrict the displays to 6 decimal places. 31 14 Programme 1 Notice that to find a value accurate to a given number of decimal places or significant figures it is sufficient to repeat the iterations until there is no change in the result from one iteration to the next. Save your spreadsheet under some suitable name such as Newton because you may wish to use it again. Now we shall look at this spreadsheet a little more closely 32 Relative addresses Place the cell highlight in cell B3 and the formula that it contains is ¼ 0.5*(B2+2/B2). Now place the cell highlight in cell B4 and the formula there is ¼ 0.5*(B3+2/B3). Why the difference? When you enter the cell address B2 in the formula in B3 the spreadsheet understands that to mean the contents of the cell immediately above. It is this meaning that is copied into cell B4 where the cell immediately above is B3. If you wish to refer to a specific cell in a formula then you must use an absolute address. Place the cell highlight in cell C1 and enter the number 2. Now place the cell highlight in cell B3 and re-enter the formula ¼ 0.5*(B2+$C$1/B2) and copy this into cells B4 to B7. The numbers in the second column have not changed but the formulas have because in cells B3 to B7 the same reference is made to cell C1. The use of the dollar signs has indicated an absolute address. So why would we do this? Change the number in cell C1 to 3 to obtain the display . . . . . . . . . . . . 33 n 0 1 2 3 4 5 x 1.500000000000000 1.750000000000000 1.732142857142860 1.732050810014730 1.732050807568880 1.732050807568880 pffiffiffi These are the iterated values of 3 – the square root of the contents of cell C1. We can now use the same spreadsheet to find the square root of any positive number. Newton’s iterative procedure to find the square root of a positive number is a special case of the Newton–Raphson procedure to find the solution of the general equation f ðxÞ ¼ 0, and we shall look at this in the next frame. 15 Numerical solutions of equations and interpolation Newton–Raphson iterative method Consider the graph of y ¼ f ðxÞ as shown. Then the x-value at the point A, where the graph crosses the x-axis, gives a solution of the equation f ðxÞ ¼ 0. y If P is a point on the curve near to A, then x ¼ x0 is an approximate value of the root of f ðxÞ ¼ 0, the error of the approximation being given by AB. 34 x Let PQ be the tangent to the curve as P, crossing the x-axis at Q ðx1 ; 0Þ. Then x ¼ x1 is a better approximation to the required root. From the diagram, PB dy ¼ QB dx the point P, x ¼ x0 . i.e. the value of the derivative of y at P PB ¼ f 0 ðx0 Þ and PB ¼ f ðx0 Þ QB PB f ðx0 Þ ; QB ¼ 0 ¼ 0 ¼ h (say) f ðx0 Þ f ðx0 Þ ; x1 ¼ x0 h ; x1 ¼ x0 f ðx0 Þ f 0 ðx0 Þ If we begin, therefore, with an approximate value (x0 ) of the root, we can determine a better approximation (x1 ). Naturally, the process can be repeated to improve the result still further. Let us see this in operation. On to the next frame Example 1 35 The equation x3 3x 4 ¼ 0 is of the form f ðxÞ ¼ 0 where f ð1Þ < 0 and f ð3Þ > 0 so there is a solution to the equation between 1 and 3. We shall take this to be 2, by bisection. Find a better approximation to the root. We have f ðxÞ ¼ x3 3x 4 ; f 0 ðxÞ ¼ 3x2 3 If the first approximation is x0 ¼ 2, then f ðx0 Þ ¼ f ð2Þ ¼ 2 and f 0 ðx0 Þ ¼ f 0 ð2Þ ¼ 9 A better approximation x1 is given by f ðx0 Þ x0 3 3x0 4 ¼ x 0 f 0 ðx0 Þ 3x0 2 3 ð2Þ ¼ 2:22 x1 ¼ 2 9 ; x0 ¼ 2; x1 ¼ 2:22 x1 ¼ x0 16 Programme 1 If we now start from x1 we can get a better approximation still by repeating the process. x2 ¼ x1 f ðx1 Þ x1 3 3x1 4 ¼ x 1 f 0 ðx1 Þ 3x1 2 3 Here x1 ¼ 2:22 36 f ðx1 Þ ¼ . . . . . . . . . . . . ; f ðx1 Þ ¼ 0:281; f 0 ðx1 Þ ¼ . . . . . . . . . . . . f 0 ðx1 Þ ¼ 11:785 Then x2 ¼ . . . . . . . . . . . . 37 x2 ¼ 2:196 Because 0:281 x2 ¼ 2:22 : ¼ 2:196 11 79 Using x2 ¼ 2:196 as a starter value, we can continue the process until successive results agree to the desired degree of accuracy. x3 ¼ . . . . . . . . . . . . 38 x3 ¼ 2:196 Because f 0 ðx2 Þ ¼ f 0 ð2:196Þ ¼ 11:467 f ðx2 Þ ¼ f ð2:196Þ ¼ 0:002026; f ðx2 Þ 0:00203 ; x3 ¼ x2 0 ¼ 2:196 (to 4 sig. fig.) ¼ 2:196 f ðx2 Þ 11:467 The process is simple but effective and can be repeated again and again. Each repetition, or iteration, usually gives a result nearer to the required root x ¼ xA . In general xnþ1 ¼ . . . . . . . . . . . . 39 xnþ1 ¼ xn f ðxn Þ f 0 ðxn Þ Tabular display of results Open your spreadsheet and in cells A1 to D1 enter the headings n, x, f ðxÞ and f 0 ðxÞ Fill cells A2 to A6 with the numbers 0 to 4 In cell B2 enter the value for x0 , namely 2 In cell C2 enter the formula for f ðx0 Þ, namely ¼ B2^3 – 3*B2 – 4 and copy into cells C3 to C6 17 Numerical solutions of equations and interpolation In cell D2 enter the formula for f 0 ðx0 Þ, namely ¼ 3*B2^2 – 3 and copy into cells D3 to D6 In cell B3 enter the formula for x1 , namely ¼ B2 – C2/D2 and copy into cells B4 to B6. The final display is . . . . . . . . . . . . n 0 1 2 3 4 x f ðxÞ 2 2.222222 2.196215 2.195823 2.195823 2 0.30727 0.004492 1.01E-06 5.15E-14 f 0 ðxÞ 9 11.81481 11.47008 11.46492 11.46492 40 As soon as the number in the second column is repeated then we know that we have arrived at that particular level of accuracy. The required root is therefore x ¼ 2:195823 to 6 dp. Save the spreadsheet so that it can be used as a template for other such problems. Now let us have another example. Next frame Example 2 41 The equation x3 þ 2x2 5x 1 ¼ 0 is of the form f ðxÞ ¼ 0 where f ð1Þ < 0 and f ð2Þ > 0 so there is a solution to the equation between 1 and 2. We shall take this to be x ¼ 1:5. Use the Newton–Raphson method to find the root to six decimal places. Use the previous spreadsheet as a template and make the following amendments In cell B2 enter the number . . . . . . . . . . . . 1.5 42 Because That is the value of x0 that is used to start the iteration In cell C2 enter the formula . . . . . . . . . . . . ¼ B2^3 + 2*B2^2 – 5*B2 – 1 Because That is the value of f ðx0 Þ ¼ x0 3 þ 2x0 2 5x0 1. Copy the contents of cell C2 into cells C3 to C5. In cell D2 enter the formula . . . . . . . . . . . . 43 18 Programme 1 44 ¼ 3*B2^2 + 4*B2 – 5 Because That is the value of f 0 ðx0 Þ ¼ 3x0 2 þ 4x0 5. Copy the contents of cell D2 into cells D3 to D5. In cell B2 the formula remains the same as . . . . . . . . . . . . 45 ¼ B2 – C2/D2 The final display is then . . . . . . . . . . . . 46 n 0 1 2 3 x 1.5 1.580645 1.575792 1.575773 f ðxÞ 0.625 0.042798 0.000159 2.21E-09 f 0 ðxÞ . 7 75 8.817898 8.752524 8.75228 We cannot be sure that the value 1.575773 is accurate to the sixth decimal place so we must extend the table. Highlight cells A5 to D5, click Edit on the Command bar and select Copy from the drop-down menu. Place the cell highlight in cell A6, click Edit and then Paste. The seventh row of the spreadsheet then fills to produce the display n 0 1 2 3 4 x 1.5 1.580645 1.575792 1.575773 1.575773 f ðxÞ 0.625 0.042798 0.000159 2.21E-09 8.9E-16 f 0 ðxÞ . 7 75 8.817898 8.752524 8.75228 8.75228 And the repetition of the x-value ensures that the solution x ¼ 1:575773 is indeed accurate to 6 dp. Now do one completely on your own. Next frame 47 Example 3 The equation 2x3 7x2 x þ 12 ¼ 0 has a root near to x ¼ 1:5. Use the Newton–Raphson method to find the root to six decimal places. The spreadsheet solution produces . . . . . . . . . . . . 19 Numerical solutions of equations and interpolation 48 x ¼ 1:686141 to 6 dp Because Fill cells A2 to A6 with the numbers 0 to 4 In cell B2 enter the value for x0 , namely 1.5 In cell C2 enter the formula for f ðx0 Þ, namely ¼ 2*B2^3 – 7*B2^2 – B2 + 12 and copy into cells C3 to C6 In cell D2 enter the formula for f 0 ðx0 Þ, namely ¼ 6*B2^2 – 14*B2 – 1 and copy into cells D3 to D6 In cell B3 enter the formula for x1 , namely ¼ B2 – C2/D2 and copy into cells B4 to B6. The final display is . . . . . . . . . . . . n 0 1 2 3 4 x 1.5 1.676471 1.686103 1.686141 1.686141 f ðxÞ 1.5 0.073275 0.000286 4.46E-09 0 49 f 0 ðxÞ 8.5 7.60727 7.54778 7.54755 7.54755 As soon as the number in the second column is repeated then we know that we have arrived at that particular level of accuracy. The required root is therefore x ¼ 1:686141 to 6 dp. First approximations The whole process hinges on knowing a ‘starter’ value as first approximation. If we are not given a hint, this information can be found by either (a) applying the remainder theorem if the function is a polynomial (b) drawing a sketch graph of the function. Example 4 Find the real root of the equation x3 þ 5x2 3x 4 ¼ 0 correct to six significant figures. Application of the remainder theorem involves substituting x ¼ 0, x ¼ 1, x ¼ 2, etc. until two adjacent values give a change in sign. f ðxÞ ¼ x3 þ 5x2 3x 4 f ð0Þ ¼ 4; f ð1Þ ¼ 1; f ð1Þ ¼ 3 The sign changes from f ð0Þ to f ð1Þ. There is thus a root between x ¼ 0 and x ¼ 1. Therefore choose x ¼ 0:5 as the first approximation and then proceed as before. Complete the table and obtain the root x ¼ ............ f(x) –1 3 0 –4 x 20 Programme 1 50 x ¼ 0:675527 The final spreadsheet display is n 0 1 2 3 4 51 f ðxÞ 1.375 0.11907 0.000582 1.43E-08 0 x 0.5 0.689655 0.675597 0.675527 0.675527 f 0 ðxÞ 7.25 8.469679 8.386675 8.386262 8.386262 Example 5 Solve the equation ex þ x 2 ¼ 0 giving the root to 6 significant figures. It is sometimes more convenient to obtain a first approximation to the required root from a sketch graph of the function, or by some other graphical means. In this case, the equation can be rewritten as ex ¼ 2 x and we therefore sketch graphs of y ¼ ex and y ¼ 2 x. x 0.2 0.4 0.6 0.8 1 x e 1.22 1.49 1.82 2.23 2.72 2x 1.8 1.6 1.4 1.2 1 y y = ex y=2–x x It can be seen that the two curves cross over between x ¼ 0:4 and x ¼ 0:6. Approximate root x ¼ 0:4 f ðxÞ ¼ ex þ x 2 f 0 ðxÞ ¼ ex þ 1 x ¼ ............ Finish it off 21 Numerical solutions of equations and interpolation 52 x ¼ 0:442854 The final spreadsheet display is n 0 1 2 3 x 0.4 0.443412 0.442854 0.442854 f ðxÞ 0.10818 0.001426 2.42E-07 7.11E-15 f 0 ðxÞ 2.491825 2.558014 2.557146 2.557146 Note: There are times when the normal application of the Newton–Raphson method fails to converge to the required root. This is particularly so when f 0 ðx0 Þ is very small, so before we leave this section let us consider this difficulty. 53 Modified Newton–Raphson method If the slope of the curve at x ¼ x0 is small, the value of the second approximation x ¼ x1 may be further from the exact root at A than the first approximation. f (x) x1 x0 x If x ¼ x0 is an approximate solution of f ðxÞ ¼ 0 and x ¼ x0 h is the exact solution then f ðx0 hÞ ¼ 0. By Taylor’s series f ðx0 hÞ ¼ f ðx0 Þ hf 0 ðx0 Þ þ h2 00 f ðx0 Þ . . . ¼ 0 2! (a) If we assume that h is small enough to neglect terms of the order h2 and higher then this equation can be written as f ðx0 hÞ f ðx0 Þ hf 0 ðx0 Þ, that is f ðx0 Þ hf 0 ðx0 Þ 0 and so f ðx0 Þ f ðx0 Þ giving x1 ¼ x0 0 as a better approximation h 0 f ðx0 Þ f ðx0 Þ to the solution of f ðxÞ ¼ 0. This is, of course, the relationship we have been using and which may fail when f 0 ðxÞ is small. Notice: h is positive unless the sign of f ðx0 Þ is the opposite of the sign of f 0 ðx0 Þ. 22 Programme 1 (b) If we consider the first three terms then h2 00 f ðx0 Þ 0, that is 2! 0 2 00 2f ðx0 Þ 2hf ðx0 Þ þ h f ðx0 Þ 0 f ðx0 hÞ f ðx0 Þ hf 0 ðx0 Þ þ Since f 0 ðx0 Þ is small we shall assume that we can neglect it so sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ unless the signs of f ðx0 Þ and f 0 ðx0 Þ are different when That is h ¼ f 00 ðx0 Þ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ . We use this result only when f 0 ðx0 Þ is found to be it is h ¼ f 00 ðx0 Þ very small. Having found x1 from x0 we then revert to the normal f ðx0 Þ for subsequent iterations. relationship xnþ1 ¼ xn 0 f ðx0 Þ Note this 54 Example 6 The equation x3 1:3x2 þ 0:4x 0:03 ¼ 0 is known to have a root near x ¼ 0:7. Determine the root to 6 significant figures. We start off in the usual way. f ðxÞ ¼ x3 1:3x2 þ 0:4x 0:03 f 0 ðxÞ ¼ 3x2 2:6x þ 0:4 and complete the first line of the normal table. n xn 0 0.7 f ðxn Þ f 0 ðxn Þ h¼ f ðxn Þ f 0 ðxn Þ xnþ1 ¼ xn h h¼ f ðxn Þ f 0 ðxn Þ xnþ1 ¼ xn h Complete just the first line of values. 55 We have n xn f ðxn Þ f 0 ðxn Þ 0 0.7 0.044 0.05 0.88 1.58 We notice at once that (a) The value of x1 is well away from the approximate value (0.7) of the root. (b) The value of f 0 ðx0 Þ is small, i.e. 0.05. To obtain x1 we therefore make a fresh start, using the modified relationship x1 ¼ . . . . . . . . . . . . 23 Numerical solutions of equations and interpolation sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ x1 ¼ x0 f 00 ðx0 Þ 56 f ðxÞ ¼ x3 1:3x2 þ 0:4x 0:03 ¼ ½ðx 1:3Þx þ 0:4x 0:03 ¼ ð3x 2:6Þx þ 0:4 f 0 ðxÞ ¼ 3x2 2:6x þ 0:4 f 00 ðxÞ ¼ 6x 2:6 n x0 f ðx0 Þ 0 0.7 0.044 00 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ 00 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ h¼ f 00 ðx0 Þ f ðx0 Þ x1 ¼ x0 h Complete the line. n x0 f ðx0 Þ f ðx0 Þ 0 0.7 0.044 1.6 x1 ¼ x0 h 0.2345 57 0.9345 Note that in the expression x1 ¼ x0 h, we chose the positive sign since at x0 ¼ 0:7, f ðx0 Þ is negative and the slope f 0 ðx0 Þ is positive. 0.7 x x1 – 0.044 Having established that x1 ¼ 0:9345, we now revert to the usual xnþ1 ¼ xn f ðxn Þ for the rest of the calculation. Complete the table therefore and f 0 ðxn Þ obtain the required root. The final spreadsheet display is n 0 1 2 3 4 5 x 0.7 0.934521 0.892801 0.887387 0.887298 0.887298 f ðxÞ 0.044 0.024625 0.002544 4.02E-05 1.06E-08 9.16E-16 58 0 f ðxÞ 0.05 0.590233 0.469997 0.45516 0.454919 0.454919 00 f ðxÞ 1.6 Therefore to six decimal places the required root is x ¼ 0:887298. Note that we only used the modified method to find x1 . After that the normal relationship is used. 24 Programme 1 And now . . . To date our task has been to find a value of x that satisfies an explicit equation f ðxÞ ¼ 0. This is quite general because any equation in x can be written in this form. For example, the equation sin x ¼ x e3x can always be written as sin x x þ e3x ¼ 0 and then approached by one of the methods that we have discussed so far. What we want to do now is to work the other way – given a value of x, to find the corresponding value of f ðxÞ. If f ðxÞ is given explicitly then this is no problem, it is just a matter of substituting the value of x in the formula and working it out. However, many times a function exists but it is not given explicitly, as in the case of a set of readings compiled as a result of an experiment or practical test. We shall consider this problem in the following frames. Next frame Interpolation 59 When a function is defined by a well-understood expression such as f ðxÞ ¼ 4x3 3x2 þ 7 or f ðxÞ ¼ 5 sinðexp½xÞ the values of the dependent variable f ðxÞ corresponding to given values of the independent variable x can be found by direct substitution. Sometimes, however, a function is not defined in this way but by a collection of ordered pairs of numbers. Example 1 A function can be defined by the following set of data: x f ðxÞ 1 4 2 14 3 40 4 88 5 164 6 274 Intermediate values, for example, x ¼ 2:5, can be estimated by a process called interpolation. The value of f ð2:5Þ will clearly lie between 14 and 40, the function values for x ¼ 2 and x ¼ 3. Purely as an estimate, f ð2:5Þ ¼ . . . . . . . . . . . . What do you suggest? 25 Numerical solutions of equations and interpolation 60 27 1 Linear interpolation If you gave the result as 27, you no doubt agreed that x ¼ 2:5 is midway between x ¼ 2 and x ¼ 3, and that therefore f ð2:5Þ would be midway between 14 and 40, i.e. 27. This is the simplest form of interpolation, but there is no evidence that there is a linear relationship between x and f ðxÞ, and the result is therefore suspect. Of course, we could have estimated the function value at x ¼ 2:5 by other means, such as ............ by drawing the graph of f ðxÞ against x 61 2 Graphical interpolation We could, indeed, plot the graph of f ðxÞ against x and, from it, estimate the value of f ðxÞ at x ¼ 2:5. f(x) This method is also approximate and can be time consuming. f ð2:5Þ 26 x In what follows we shall look at interpolation using finite differences, which work well and quickly when the values of x are equally spaced. When the values of x are not equally spaced we need to resort to the more involved algebraic method called Lagrangian interpolation (which could also be used for equally spaced points). Next frame 3 Gregory–Newton interpolation formula using forward finite differences x .. . f ðxÞ .. . x0 x1 .. . f ðx0 Þ f ðx1 Þ .. . f0 ¼ f ðx1 Þ f ðx0 Þ We assume that x0 , x1 , . . . are distinct, equally spaced apart, and x0 < x1 < . . . 62 26 Programme 1 For each pair of consecutive function values, f ðx0 Þ and f ðx1 Þ, in the table, the forward difference f0 is calculated by subtracting f ðx0 Þ from f ðx1 Þ. This difference is written in a third column of the table, midway between the lines carrying f ðx0 Þ and f ðx1 Þ. x f ðxÞ 1 4 2 14 3 .. . 40 .. . f 10 26 63 Complete the table for the data given in Frame 59 which then becomes . . . . . . . . . . . . x f ðxÞ 1 4 2 14 3 40 4 88 5 164 6 274 f 10 26 48 76 110 We now form a fourth column, the forward differences of the values of f , denoted by 2 f , and again written midway between the lines of f . These are the second forward differences of f ðxÞ. So the table then becomes . . . . . . . . . . . . 64 x f ðxÞ 1 4 2 14 3 40 4 88 5 164 6 274 f 2 f 10 26 48 76 110 16 22 28 34 A further column can now be added in like manner, giving the third differences, denoted by 3 f , so that we then have . . . . . . . . . . . . 27 Numerical solutions of equations and interpolation x f ðxÞ 1 4 2 14 3 40 4 88 5 164 6 274 f 2 f 65 3 f 10 16 26 22 48 28 76 6 6 6 34 110 Notice that the table has now been completed, for the third differences are constant and all subsequent differences would be zero. Now we shall see how to use the table. So move on To find f ð2:5Þ x0 h x1 66 xp x f ðxÞ 1 4 2 14 3 40 4 88 5 164 6 274 f 10 26 48 76 110 2 f 3 f 16 22 28 34 6 6 6 We have to find f ð2:5Þ. Therefore denote x ¼ 2 as x0 x ¼ 2:5 as xp x ¼ 3 as x1 Let h ¼ the constant range between successive values of x, i.e. h ¼ x1 x0 xp x0 , 0<p<1 Express ðxp x0 Þ as a fraction of h, i.e. p ¼ h : 2 5 2:0 Therefore, in the case above, h ¼ 1 and p ¼ ¼ 0:5. 1 All we now use from the table is the set of values underlined by the broken line drawn diagonally from f ðx0 Þ. So we have p ¼ ............; 2 f0 ¼ . . . . . . . . . . . . ; f0 ¼ . . . . . . . . . . . . ; 3 f0 ¼ . . . . . . . . . . . . ; f0 ¼ . . . . . . . . . . . . 28 Programme 1 67 p ¼ 0:5 f0 ¼ 14; 2 f0 ¼ 22; f0 ¼ 26; 3 f0 ¼ 6 Now we are ready to deal with the Gregory–Newton forward difference interpolation formula pðp 1Þ 2 pðp 1Þðp 2Þ 3 f0 þ f0 þ . . . fp ¼ f0 þ pf0 þ 12 123 This is sometimes written in operator form pðp 1Þ 2 pðp 1Þðp 2Þ 3 fp ¼ 1 þ p þ þ þ . . . f0 2! 3! which you no doubt recognise as the binomial expansion of fp ¼ ð1 þ Þp f0 Substituting the values in the above example gives f ð2:5Þ ¼ fp ¼ . . . . . . . . . . . . 68 24.625 Because 0:5ð0:5Þ 0:5ð0:5Þð1:5Þ ð22Þ þ ð6Þ fp ¼ 14 þ 0:5ð26Þ þ 12 123 ¼ 14 þ 13 2:75 þ 0:375 ¼ 27:375 2:75 ¼ 24:625 Comparing the results of the three methods we have discussed (a) Linear interpolation f ð2:5Þ ¼ 27 (b) Graphical interpolation f ð2:5Þ ¼ 26 (c) Gregory–Newton formula f ð2:5Þ ¼ 24:625 – the true value Example 2 x 69 f ðxÞ It is required to determine the value of f ðxÞ at x ¼ 5:5. 2 14 In this case 4 88 x0 ¼ ............ x1 ¼ . . . . . . . . . . . . 6 274 h ¼ ............ p ¼ ............ 8 620 10 1174 x0 ¼ 4; x1 ¼ 6; h ¼ 2; p ¼ 0:75 Because h ¼ x1 x0 ¼ 6 4 ¼ 2 xp x0 5:5 4 1:5 ¼ ¼ 0:75 ¼ p¼ 2 2 h First compile the table of forward differences . . . . . . . . . . . . 29 Numerical solutions of equations and interpolation x f ðxÞ 2 14 4 88 6 274 8 620 10 1174 x0 x1 f 74 186 346 554 2 f 112 160 208 70 3 f 48 48 The Gregory–Newton forward difference interpolation formula is fp ¼ ð1 þ Þp f0 i.e. fp ¼ . . . . . . . . . . . . 71 pðp 1Þ 2 pðp 1Þðp 2Þ 3 þ þ . . . f0 2! 3! pðp 1Þ 2 pðp 1Þðp 2Þ 3 f0 þ f0 þ . . . ¼ f0 þ pf0 þ 2! 3! fp ¼ 1 þ p þ So, substituting the relevant values from the table, gives f ð5:5Þ ¼ fp ¼ . . . . . . . . . . . . 72 214.4 Because xp x0 x1 x f ðxÞ 2 14 4 88 6 274 8 620 10 1174 f 74 186 346 554 2 f 112 160 208 0:75ð0:25Þ ð160Þ 12 0:75ð0:25Þð1:25Þ ð48Þ þ 123 ¼ 88 þ 139:5 15 þ 1:875 ¼ 214:375 ; f ð5:5Þ ¼ 214:4 f ð5:5Þ ¼ fp ¼ 88 þ 0:75ð186Þ þ Finally, one more. 3 f 48 48 30 Programme 1 Example 3 Determine the value of f ð1Þ from the set of function values. x 4 2 0 2 4 6 8 f ðxÞ 541 55 1 53 155 31 1225 Complete the working and then check with the next frame. 73 f ð1Þ ¼ 10 Here is the working; method as before. xp x0 ¼ 2; x0 x1 x1 ¼ 0; x f ðxÞ 4 541 2 55 0 486 54 1 2 53 4 155 6 31 8 1225 xp ¼ 1; f 54 102 186 1194 ; h ¼ 2; 2 f 432 0 48 288 1008 3 f 432 48 336 720 4 f 384 384 384 p ¼ 12 pðp 1Þ 2 pðp 1Þðp 2Þ 3 f0 þ f0 12 123 pðp 1Þðp 2Þðp 3Þ 4 þ f0 1234 1 3 1 1 1 1 2 2 2 2 2 ð0Þ þ ð48Þ ¼ 55 þ ð54Þ þ 2 12 123 1 12 32 52 ð384Þ þ2 1234 fp ¼ f0 þ pf0 þ ¼ 55 27 þ 0 3 15 ¼ 10 ; fp ¼ f ð1Þ ¼ 10 This table of data does have its restrictions. For example, if we had wanted to find f ð2:5Þ from the table we would have run out of data because there is no 4 f entry available. In such a case we can resort to a zig-zag path through the table using central differences. Next frame Numerical solutions of equations and interpolation 31 Central differences The central difference operator is defined by its action on the expression f ðxÞ as f ðxÞ ¼ f ðx þ h=2Þ f ðx h=2Þ and using this operator the interpolated value of f ðxÞ near to the given value of f0 is defined by the Gauss forward formula as pðp 1Þ 2 ðp þ 1Þpðp 1Þ 3 f0 þ f0þ1 2 2! 3! ðp þ 1Þpðp 1Þðp 2Þ 4 f0 þ . . . þ 4! fp ¼ f0 þ pf0þ1 þ 2 or by the Gauss backward formula as ðp þ 1Þp 2 ðp þ 1Þpðp 1Þ 3 f0 þ f01 2 2! 3! ðp þ 2Þðp þ 1Þpðp 1Þ 4 f0 þ . . . þ 4! fp ¼ f0 þ pf01 þ 2 There are no tabulated values at the half-interval values x0 þ h=2 and x0 h=2 and so these are taken to be the differences evaluated at mid-interval as given in the forward difference table. This means that the tables for the Gregory– Newton forward differences and the central differences are identical (apart, that is, from the column headings); the method of tracing through the table, however, is different. For example, to find f ð2:5Þ for the example given in Frame 59 x f ðxÞ 1 4 2 14 f ðxÞ 2 f ðxÞ 3 f ðxÞ 10 16 26 3 40 4 88 5 164 6 274 6 22 48 6 28 76 6 34 110 Here x0 ¼ 2, f0 ¼ 14, f0þ1 ¼ 26, 2 f0 ¼ 16, 3 f0þ1 ¼ 6, 4 f0 ¼ 0 and p ¼ 0:5. 2 2 Thus ð0:5Þð0:5Þ ð0:5Þð0:5Þð1:5Þ fp ¼ 14 þ ð0:5Þ26 þ 16 þ 6 2 6 ¼ 14 þ 13 2 0:375 ¼ 24:625 which agrees with the value found using the Gregory–Newton forward difference formula. 74 32 Programme 1 Try one for yourself. The given tabulated values are x f ðxÞ 0 5 1 2 f ðxÞ 2 f ðxÞ 3 f ðxÞ 3 6 9 2 7 3 34 4 91 12 18 27 12 30 57 Using the Gauss forward difference formula, the interpolated value of f ð2:2Þ ¼ . . . . . . . . . . . . Next frame 75 10.576 Because pðp 1Þ 2 pðp 1Þðp þ 1Þ 3 f0 þ f0þ12 þ . . . and 2! 3! following the solid line through the table where x0 ¼ 2, f0 ¼ 7, f0þ1 ¼ 27, 2 f0 ¼ 18, 3 f0þ1 ¼ 12 and p ¼ 0:2, Using fp ¼ f0 þ pf0þ12 þ 2 2 ð0:2Þð0:8Þ ð0:2Þð0:8Þð1:2Þ then fp ¼ 7 þ ð0:2Þ27 þ 18 þ 12 2 6 ¼ 7 þ 5:4 1:44 0:384 ¼ 10:576 Using the Gauss backward difference formula (following the broken line) fp ¼ f0 þ pf012 þ pðp þ 1Þ 2 pðp 1Þðp þ 1Þ 3 f0 þ f012 þ . . . 2! 3! where here f012 ¼ 9 and 3 f012 ¼ 12 and so ð0:2Þð1:2Þ ð0:2Þð1:2Þð0:8Þ fp ¼ 7 þ ð0:2Þ9 þ 18 þ 12 2 6 ¼ 7 þ 1:8 þ 2:16 0:384 ¼ 10:576 as found with the Gauss forward difference formula. Next frame Numerical solutions of equations and interpolation 33 Gregory–Newton backward differences We have seen that the Gregory–Newton forward difference procedure loses terms if the interpolation is for points sufficiently forward in the table. We have also seen how this difficulty can be avoided by using central differences. However, even with central differences we can run out of data before completing a full traverse of the table. In such a situation we resort to the Gregory–Newton backward difference formula fp ¼ f0 þ pf1 þ 76 pðp þ 1Þ 2 pðp þ 1Þðp þ 2Þ 3 f2 þ f3 þ . . . 2! 3! As an example, consider the table of Frame 74. x f ðxÞ 1 4 2 14 f 2 f 3 f 10 16 26 3 40 4 88 5 164 6 274 6 22 48 6 28 76 6 34 110 Using this table we can calculate f ð5:5Þ by tracing back through the table (see broken line) as ð0:5Þð1:5Þ 2 ð0:5Þð1:5Þð2:5Þ 3 f2 þ f3 f ð5:5Þ ¼ f0 þ ð0:5Þf1 þ 2 6 ð0:5Þð1:5Þ28 ð0:5Þð1:5Þð2:5Þ6 þ ¼ 164 þ ð0:5Þ76 þ 2 6 : ¼ 214 375 In each of the examples that we have looked at so far the tabular display of differences eventually results in a column of zeros and this determines the number of terms in an interpolation calculation. The zeros have arisen because all the examples have been derived from polynomials. The following example deals with a tabular display of differences which does not result in a column of zeros. In this case the number of terms used in the interpolation calculation determines confidence in the accuracy of the result. 77 34 Programme 1 Example Use the Gregory–Newton forward difference method to find f ð0:15Þ to 4 decimal places from the following finite difference table. x f ðxÞ 0 0.000000 f 2 f 3 f 0.099833 0. 1 0.099833 0.000998 0.098836 0. 2 0.198669 0. 3 0.295520 0.000988 0.001985 0.096851 0.000968 0.002953 0.093898 0. 4 0.389418 0.000938 0.003891 0.090007 0. 5 0.479426 Here x0 ¼ 0:1, x1 ¼ 0:2, xp ¼ 0:15 and therefore p ¼ 0:5, and pðp 1Þ 2 pðp 1Þðp 2Þ 3 f0 þ f0 þ . . . 2! 3! 1 1 1 ð0:001985Þ=2 ¼ 0:099833 þ ð0:098836Þ þ 2 2 2 1 1 3 ð0:000969Þ=6 þ . . . þ 2 2 2 : : ¼ 0 099833 þ 0 049418 þ 0:000248 0:000061 þ . . . fp ¼ f0 þ pf0 þ ¼ 0:1494 to 4 dp As you can see, the calculation can continue indefinitely and termination is dictated by the number of decimal places required in the final answer. Lagrange interpolation 78 If the straight line pðxÞ ¼ a0 þ a1 x passes through the two points ðx0 , f ðx0 ÞÞ and ðx1 , f ðx1 ÞÞ, where a0 and a1 are constants, then the equation for this line can also be written as x x1 x x0 pðxÞ ¼ f ðx0 Þ þ f ðx1 Þ x0 x1 x1 x0 For example, the straight line pðxÞ ¼ 3 þ 2x passes through the two points ð1, 5Þ and ð2, 7Þ. Substituting the values for the variables in the above equation demonstrates this alternative form for the equation Numerical solutions of equations and interpolation pðxÞ ¼ 35 x2 x1 5þ 7 ¼ 10 5x þ 7x 7 ¼ 3 þ 2x 12 21 So, given the two data points from Frame 59, ð2, 14Þ and ð3, 40Þ, using linear interpolation f ð2:5Þ pð2:5Þ ¼ . . . . . . . . . . . . 79 27 Because x x1 x x0 f ðx0 Þ þ f ðx1 Þ x0 x1 x1 x0 x3 x2 14 þ 40 ¼ 26x 38 ¼ 23 32 pðxÞ ¼ and so f ð2:5Þ pðxÞ ¼ 26ð2:5Þ 38 ¼ 27 The principle of Lagrange interpolation is that a function f ðxÞ whose values are given at a collection of points is assumed to be approximately represented by a polynomial pðxÞ that passes through each and every point. The polynomial is called the interpolation polynomial and it is of degree one less than the number of points given. For two data points the interpolating polynomial is taken to be a linear polynomial, as you have just seen in the last example. For three data points the interpolating polynomial is taken to be a quadratic, for four data points the interpolation polynomial is taken to be a cubic, and so on. In the same manner as before it can be shown that the quadratic pðxÞ ¼ a0 þ a1 x þ a2 x2 that passes through the three points ðx0 , f ðx0 ÞÞ, ðx1 , f ðx1 ÞÞ and ðx2 , f ðx2 ÞÞ can be written as pðxÞ ¼ ðx x1 Þðx x2 Þ ðx x0 Þðx x2 Þ f ðx0 Þ þ f ðx1 Þ ðx0 x1 Þðx0 x2 Þ ðx1 x0 Þðx1 x2 Þ þ ðx x0 Þðx x1 Þ f ðx2 Þ ðx2 x0 Þðx2 x1 Þ So let’s try one. Given the collection of values x f ðxÞ 1.5 0.405 2.1 0.742 3 1.099 by Lagrangian interpolation, f ð1:8Þ . . . . . . . . . . . . to 2 decimal places. 80 36 Programme 1 81 0.58 Because ðx x1 Þðx x2 Þ ðx x0 Þðx x2 Þ f ðx0 Þ þ f ðx1 Þ ðx0 x1 Þðx0 x2 Þ ðx1 x0 Þðx1 x2 Þ ðx x0 Þðx x1 Þ f ðx2 Þ þ ðx2 x0 Þðx2 x1 Þ : ðx 2:1Þðx 3Þ :405 þ ðx 1 5Þðx 3Þ 0:742 0 ¼ : ð1 5 2:1Þð1:5 3Þ ð2:1 1:5Þð2:1 3Þ : : ðx 1 5Þðx 2 1Þ : þ 1 099 ð3 1:5Þð3 2:1Þ ðx2 5:1x þ 6:3Þ : ðx2 4:5x þ 4:5Þ : 0 405 þ 0 742 ¼ 0:9 ð0:54Þ pðxÞ ¼ ðx2 3:6x þ 3:15Þ : 1 099 1:35 2 ¼ 0:11x þ 0:958x 0:784 þ So that f ð1:8Þ pð1:8Þ ¼ 0:58 to 2 decimal places. By carefully considering the interpolating polynomials for two and three data points you should be able to see a pattern. Write down what you think the interpolating polynomial should be for four data points: ............ 82 pðxÞ ¼ ðx x1 Þðx x2 Þðx x3 Þ ðx x0 Þðx x2 Þðx x3 Þ f ðx0 Þ þ f ðx1 Þ ðx0 x1 Þðx0 x2 Þðx0 x3 Þ ðx1 x0 Þðx1 x2 Þðx1 x3 Þ ðx x0 Þðx x1 Þðx x3 Þ ðx x0 Þðx x1 Þðx x2 Þ f ðx2 Þ þ f ðx3 Þ þ ðx2 x0 Þðx2 x1 Þðx2 x3 Þ ðx3 x0 Þðx3 x1 Þðx3 x2 Þ Use this interpolating polynomial for the data points x 1 1.2 1.3 1.5 f ðxÞ 0.368 0.301 0.273 0.223 To 2 decimal places, f ð1:4Þ . . . . . . . . . . . . Numerical solutions of equations and interpolation 0.25 Because pðxÞ ðx x1 Þðx x2 Þðx x3 Þ ðx x0 Þðx x2 Þðx x3 Þ f ðx0 Þ þ f ðx1 Þ ¼ ðx0 x1 Þðx0 x2 Þðx0 x3 Þ ðx1 x0 Þðx1 x2 Þðx1 x3 Þ ðx x0 Þðx x1 Þðx x3 Þ ðx x0 Þðx x1 Þðx x2 Þ f ðx2 Þ þ f ðx3 Þ þ ðx2 x0 Þðx2 x1 Þðx2 x3 Þ ðx3 x0 Þðx3 x1 Þðx3 x2 Þ ðx 1:2Þðx 1:3Þðx 1:5Þ : ðx 1Þðx 1:3Þðx 1:5Þ ¼ 0 368 þ : 0:301 : : : ð1 1 2Þð1 1 3Þð1 1 5Þ ð1 2 1Þð1:2 1:3Þð1:2 1:5Þ ðx 1Þðx 1:2Þðx 1:5Þ ðx 1Þðx 1:2Þðx 1:3Þ 0:273 þ : 0:223 þ : : : : : ð1 3 1Þð1 3 1 2Þð1 3 1 5Þ ð1 5 1Þð1:5 1:2Þð1:5 1:3Þ ðx3 4x2 þ 5:31x 2:34Þ : ðx3 3:8x2 þ 4:75x 1:95Þ : 0 368 þ 0 301 ¼ ð0:03Þ 0:006 ðx3 3:7x2 þ 4:5x 1:8Þ : ðx3 3:5x2 þ 4:06x 1:56Þ : 0 273 þ 0 223 ð0:006Þ 0:03 ¼ 0:167x3 þ 0:767x2 1:415x þ 1:183 þ So that f ð1:4Þ pð1:4Þ ¼ 0:25 to 2 decimal places The general Lagrange interpolation polynomial for n þ 1 data points at x0 , x1 , . . . , xn is pðxÞ ¼ ðx x1 Þðx x2 Þð Þðx xn Þ f ðx0 Þ ðx0 x1 Þðx0 x2 Þð Þðx0 xn Þ ðx x0 Þðx x2 Þð Þðx xn Þ f ðx1 Þ þ þ ðx1 x0 Þðx1 x2 Þð Þðx1 xn Þ ðx x0 Þðx x1 Þð Þðx xn1 Þ þ f ðxn Þ ðxn x0 Þðxn x1 Þð Þðxn xn1 Þ This now completes the work of this Programme. What follows is a Revision summary and a Can you? checklist. Read the summary carefully and respond to the questions in the checklist. When you feel sure that you are happy with the content of this Programme, try the Test exercise. Take your time, there is no need to hurry. Finally, a collection of Further problems provides valuable additional practice. 37 83 38 Programme 1 Revision summary 1 84 1 The Fundamental Theorem of Algebra can be stated as follows: Every polynomial expression f ðxÞ ¼ an xn þ an1 xn1 þ þ a1 x þ a0 can be written as a product of n linear factors in the form f ðxÞ ¼ an ðx r1 Þðx r2 Þð Þðx rn Þ 2 Relations between the coefficients and the roots of a polynomial equation Whenever a polynomial with real coefficients ai has a complex root it also has the complex conjugate as another root. If , , , . . . are the roots of the equation p0 xn þ p1 xn1 þ p2 xn2 þ þ pn1 x þ pn ¼ 0 then, provided p0 6¼ 0 sum of roots ¼ p1 p0 sum of the product of the roots, taken two at a time ¼ p2 p0 p3 p0 n pn sum of the product of the roots, taken n at a time ¼ ð1Þ . p0 sum of the product of the roots, taken three at a time ¼ 3 Cubic equations Reduced form Every cubic equation of the form x3 þ ax2 þ bx þ c ¼ 0 can be written in a reduced form y 3 þ py þ q ¼ 0 by using the transformation x ¼ y . 3 Tartaglia’s solution Every cubic equation with real coefficients has at least one real root that may be found analytically using Tartaglia’s method. The real root of x3 þ ax þ b ¼ 0 when a > 0 is ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 ( rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi)1=3 b a3 b2 b a3 b2 þ þ þ x¼ þ 2 2 27 4 27 4 4 Numerical methods Bisection The bisection method of finding a solution to the equation f ðxÞ ¼ 0 consists of If a < 0 it is best to resort to numerical methods. Finding a value of x such that f ðxÞ < 0, say x ¼ a Finding a value of x such that f ðxÞ > 0, say x ¼ b. The solution to the equation f ðxÞ ¼ 0 must then lie between a and b. Furthermore, it must lie either in the first half of the interval between a and b or in the second half. 39 Numerical solutions of equations and interpolation 5 Numerical solution of equations by iteration The process of finding the numerical solution to the equation f ðxÞ ¼ 0 by iteration is performed by first finding an approximate solution and then using this approximate solution to find a more accurate solution. This process is repeated until a solution is found to the required level of accuracy. 6 Using a spreadsheet Iteration procedures are more efficiently performed using a spreadsheet. 7 Newton–Raphson iteration method If x ¼ x0 is an approximate solution to the equation f ðxÞ ¼ 0, a better approximation x ¼ x1 is given by x1 ¼ x0 8 f ðx0 Þ f ðxn Þ , and in general xnþ1 ¼ xn 0 f 0 ðx0 Þ f ðxn Þ Modified Newton–Raphson iteration method If, in the Newton–Raphson procedure f 0 ðx0 Þ is sufficiently small enough to cause the value of x1 to be a worse approximation to the solution than x0 , then x1 is obtained from the relationship sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2f ðx0 Þ x1 ¼ x0 f 00 ðx0 Þ Subsequent iterations then use xnþ1 ¼ xn 9 10 Interpolation Linear Graphical Gregory–Newton interpolation formulas using forward finite differences fp ¼ f0 þ pf0 þ 11 f ðxn Þ . f 0 ðxn Þ pðp 1Þ 2 pðp 1Þðp 2Þ 3 f0 þ fo þ 2! 3! Gauss interpolation formulas using central finite differences Gauss forward formula pðp 1Þ 2 ðp þ 1Þpðp 1Þ 3 f0 þ f0þ12 2! 3! ðp þ 1Þpðp 1Þðp 2Þ 4 þ f0 þ 4! Gauss backward formula fp ¼ f0 þ pf0þ12 þ ðp þ 1Þp 2 ðp þ 1Þpðp 1Þ 3 f0 þ f01 2 2! 3! ðp þ 2Þðp þ 1Þpðp 1Þ 4 f0 þ þ 4! fp ¼ f0 þ pf01 þ 2 40 Programme 1 12 Gregory–Newton interpolation formula using backward finite differences fp ¼ f0 þ pf1 þ 13 pðp þ 1Þ 2 pðp þ 1Þðp þ 2Þ 3 f2 þ f3 þ 2! 3! Lagrange interpolation If the straight line pðxÞ ¼ a0 þ a1 x passes through the two points ðx0 , f ðx0 ÞÞ and ðx1 , f ðx1 ÞÞ, where a0 and a1 are constants, then the interpolation polynomial (straight line) for this line can be written as x x1 x x0 pðxÞ ¼ f ðx0 Þ þ f ðx1 Þ x0 x1 x1 x0 The quadratic interpolating polynomial that passes through the three points ðx0 , f ðx0 ÞÞ, ðx1 , f ðx1 ÞÞ and ðx2 , f ðx2 ÞÞ can be written as pðxÞ ¼ ðx x1 Þðx x2 Þ ðx x0 Þðx x2 Þ f ðx0 Þ þ f ðx1 Þ ðx0 x1 Þðx0 x2 Þ ðx1 x0 Þðx1 x2 Þ ðx x0 Þðx x1 Þ f ðx2 Þ þ ðx2 x0 Þðx2 x1 Þ The cubic interpolating polynomial that passes through the four data points ðx0 , f ðx0 ÞÞ, ðx1 , f ðx1 ÞÞ, ðx2 , f ðx2 ÞÞ and ðx3 , f ðx3 ÞÞ can be written as pðxÞ ¼ ðx x1 Þðx x2 Þðx x3 Þ f ðx0 Þ ðx0 x1 Þðx0 x2 Þðx0 x3 Þ ðx x0 Þðx x2 Þðx x3 Þ f ðx1 Þ þ ðx1 x0 Þðx1 x2 Þðx1 x3 Þ ðx x0 Þðx x1 Þðx x3 Þ f ðx2 Þ þ ðx2 x0 Þðx2 x1 Þðx2 x3 Þ ðx x0 Þðx x1 Þðx x2 Þ f ðx3 Þ þ ðx3 x0 Þðx3 x1 Þðx3 x2 Þ The interpolating polynomial that passes through n þ 1 data points is pðxÞ ¼ ðx x1 Þðx x2 Þð Þðx xn Þ f ðx0 Þ ðx0 x1 Þðx0 x2 Þð Þðx0 xn Þ ðx x0 Þðx x2 Þð Þðx xn Þ f ðx1 Þ þ þ ðx1 x0 Þðx1 x2 Þð Þðx1 xn Þ ðx x0 Þðx x1 Þð Þðx xn1 Þ f ðxn Þ þ ðxn x0 Þðxn x1 Þð Þðxn xn1 Þ 41 Numerical solutions of equations and interpolation Can you? 85 Checklist 1 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Appreciate the Fundamental Theorem of Algebra? Yes Frames 1 to 3 4 to 6 7 to 17 18 to 20 21 and 22 . Find the solution of the equation f ðxÞ ¼ 0 by the method of bisection? Yes No 23 to 26 . Solve equations involving a single real variable by iteration and use a spreadsheet for efficiency? Yes No 27 to 33 34 to 52 53 to 58 59 to 61 62 to 73 74 to 77 78 to 83 No . Find the two roots of a quadratic equation and recognise that for polynomial equations with real coefficients complex roots exist in complex conjugate pairs? Yes No . Use the relationships between the coefficients and the roots of a polynomial equation to find the roots of the polynomial? Yes No . Transform a cubic equation to reduced form? Yes No . Use Tartaglia’s solution to find the real root of a cubic equation? Yes No . Solve equations using the Newton–Raphson iterative method? Yes No . Use the modified Newton–Raphson method to find the first approximation when the derivative is small? Yes No . Understand the meaning of interpolation and use simple linear and graphical interpolation? Yes No . Use the Gregory–Newton interpolation formula using forward and backward differences for equally spaced domain points? Yes No . Use the Gauss interpolation formulas using central differences for equally spaced domain points? Yes No . Use Lagrange interpolation when the domain points are not equally spaced? Yes No 42 Programme 1 Test exercise 1 86 1 pffiffiffi Given that x ¼ 1 þ j 3 is one root of a quadratic equation with real coefficients, find the other root and hence the quadratic equation. 2 Solve the cubic equation 2x3 7x2 42x þ 72 ¼ 0. 3 Write the cubic 3x3 þ 5x2 þ 3x þ 5 in reduced form and use Tartaglia’s method to find the real root. 4 Use the method of bisection to find a solution to x3 5 ¼ 0 correct to 4 significant figures. 5 Use the Newton–Raphson method to find a positive solution of the following equation, correct to 6 decimal places: cos 3x ¼ x2 6 Use the modified Newton–Raphson method to find the solution correct to 6 decimal places near to x ¼ 2 of the equation x3 6x2 þ 13x 9 ¼ 0 7 Given the table of values x f ðxÞ 1 2 3 4 5 6 0 19 70 171 340 595 estimate (a) f ð2:5Þ using the Gregory–Newton forward difference formula (b) f ð3:4Þ using the Gauss central difference formula (c) f ð5:6Þ using the Gregory–Newton backward difference formula. 8 Given the table of values x f ðxÞ 1 4 2 9 5 108 use Lagrangian interpolation to estimate the value of f ð2:2Þ. Numerical solutions of equations and interpolation 43 Further problems 1 1 pffiffiffi 1 j 3 1 þ j and x ¼ pffiffiffi are two roots of a quartic equation 2 2 with real coefficients, find the other two roots and hence the quartic equation. Given that x ¼ 2 Solve the equation x3 5x2 8x þ 12 ¼ 0, given that the sum of two of the roots is 7. 3 Find the values of the constants p and q such that the function f ðxÞ ¼ 2x3 þ px2 þ qx þ 6 may be exactly divisible by ðx 2Þðx þ 1Þ. 4 If f ðxÞ ¼ 4x4 þ px3 23x2 þ qx þ 11 and when f ðxÞ is divided by 2x2 þ 7x þ 3 the remainder is 3x þ 2, determine the values of p and q. 5 If one root of the equation x3 2x2 9x þ 18 ¼ 0 is the negative of another, determine the three roots. 6 Solve the equation x3 7x2 21x þ 27 ¼ 0, given that the roots form a geometric sequence. 7 Form the equation whose roots are those of the equation x3 þ x2 þ 9x þ 9 ¼ 0 each increased by 2. 8 Form the equation whose roots exceed by 3 the roots of the equation x3 4x2 þ x þ 6 ¼ 0. 9 If the equation 4x3 4x2 5x þ 3 ¼ 0 is known to have two roots whose sum is 2, solve the equation. 10 Solve the equation x3 10x2 þ 8x þ 64 ¼ 0, given that the product of two of the roots is the negative of the third. 11 Form the equation whose roots exceed by 2 those of the equation 2x3 3x2 11x þ 6 ¼ 0. 12 If , , are the roots of the equation x3 þ px2 þ qx þ r ¼ 0, prove that 2 þ 2 þ 2 ¼ p2 2q. 13 Using Tartaglia’s solution, find the real root of the equation 2x3 þ 4x 5 ¼ 0 giving the result to 4 significant figures. 14 Solve the equation x3 6x 4 ¼ 0. 15 Rewrite the equation x3 þ 6x2 þ 9x þ 4 ¼ 0 in reduced form and hence determine the three roots. 16 Show that the equation x3 þ 3x2 4x 6 ¼ 0 has a root between x ¼ 1 and x ¼ 2, and use the Newton–Raphson iterative method to evaluate this root to 4 significant figures. 17 Find the real root of the equations: (a) x3 þ 4x þ 3 ¼ 0 (b) 5x3 þ 2x 1 ¼ 0. 87 44 Programme 1 18 Solve the following equations: (a) x3 5x þ 1 ¼ 0 (b) x3 þ 2x 3 ¼ 0 3 (c) x 4x þ 1 ¼ 0. 19 Express the following in reduced form and determine the roots: (a) x3 þ 6x2 þ 9x þ 5 ¼ 0 (b) 8x3 þ 20x2 þ 6x 9 ¼ 0 (c) 4x3 9x2 þ 42x 10 ¼ 0. 20 Use the Newton–Raphson iterative method to solve the following. (a) Show that a root of the equation x3 þ 3x2 þ 5x þ 9 ¼ 0 occurs between x ¼ 2 and x ¼ 3. Evaluate the root to four significant figures. (b) Show graphically that the equation e2x ¼ 25x 10 has two real roots and find the larger root correct to four significant figures. (c) Verify that the equation x cos x ¼ 0 has a root near to x ¼ 0:8 and determine the root correct to three significant figures. (d) Obtain graphically an approximate root of the equation 2 ln x ¼ 3 x. Evaluate the root correct to four significant figures. (e) Verify that the equation x4 þ 5x 20 ¼ 0 has a root at approximately x ¼ 1:8. Determine the root correct to five significant figures. (f) Show that the equation x þ 3 sin x ¼ 2 has a root between x ¼ 0:4 and x ¼ 0:6. Evaluate the root correct to five significant figures. (g) The equation 2 cos x ¼ ex 1 has a real root between x ¼ 0:8 and x ¼ 0:9. Evaluate the root correct to four significant figures. (h) The equation 20x3 22x2 þ 5x 1 ¼ 0 has a root at approximately x ¼ 0:6. Determine the value of the root correct to four significant figures. 21 A polynomial function is defined by the following set of function values x 2 4 6 8 10 y ¼ f ðxÞ 7.00 9.00 97.0 305 681 Find (a) f ð4:8Þ using the Gregory–Newton forward difference formula (b) f ð7:2Þ using the Gauss central difference formula (c) f ð8:5Þ using the Gregory–Newton backward difference formula. 22 For the function f ðxÞ x 4 5 6 7 8 9 10 f ðxÞ 10 12 56 128 234 380 572 Find (a) f ð4:5Þ and f ð6:4Þ using the Gregory–Newton forward difference formula (b) f ð7:1Þ and f ð8:9Þ using the Gregory–Newton backward difference formula. 45 Numerical solutions of equations and interpolation 23 x 2 4 6 8 10 12 f ðxÞ 9 35 231 675 1463 2691 For the function defined in the table above, evaluate (a) f ð2:6Þ and (b) f ð7:2Þ. 24 A function f ðxÞ is defined by the following table x 4 2 0 2 4 6 8 f ðxÞ 277 51 1 17 147 533 1319 Find (a) f ð3Þ and f ð1:6Þ using the Gregory–Newton forward difference formula (b) f ð0:2Þ and f ð3:1Þ using the Gauss central difference formula (c) f ð4:4Þ and f ð7Þ using the Gregory–Newton backward difference formula. 25 Given the table of values x f ðxÞ 1 3 2.71828 0.04979 5 0.00674 use Lagrangian interpolation to find the value of f ð3:4Þ. 26 Given the table of values x f ðxÞ 6 7.2 0.801153 0.82236 9 0.73922 0.994808 13 use Lagrangian interpolation to find the value of f ð8Þ. 27 Given the table of values x f ðxÞ 2 2.63906 2.48491 0 5 6 1.94591 1.79176 use Lagrangian interpolation to find the values of (a) f ð0:8Þ (b) f ð0:8Þ (c) f ð5:5Þ. Programme 2 Frames 1 to 94 Laplace transforms 1 Learning outcomes When you have completed this Programme you will be able to: . Obtain the Laplace transforms of simple standard expressions . Use the first shift theorem to find the Laplace transform of a simple expression multiplied by an exponential . Find the Laplace transform of a simple expression multiplied or divided by a variable . Use partial fractions to find the inverse Laplace transform . Use the ‘cover up’ rule . Use the Laplace transforms of derivatives to solve differential equations . Use the Laplace transform to solve simultaneous differential equations Prerequisite: Engineering Mathematics (Sixth Edition) Programme 26 Introduction to Laplace transforms 46 47 Laplace transforms 1 Introduction The solution of a linear, ordinary differential equation with constant coefficients such as the second-order equation af 00 ðtÞ þ bf 0 ðtÞ þ cf ðtÞ ¼ gðtÞ can be solved by first obtaining the general form for the expression f ðtÞ. This general form will contain a number of integration constants whose values can be found by applying the appropriate boundary conditions (see Engineering Mathematics, Sixth Edition, Programme 25). A more systematic way of solving such equations is to use the Laplace transform which converts the differential equation into an algebraic equation and has the added advantage of incorporating the boundary conditions from the beginning. Furthermore, in situations where f ðtÞ represents a function with discontinuities, the Laplace transform method can succeed where other methods fail. Laplace transform techniques also provide powerful tools in numerous fields of technology such as Control Theory where a knowledge of the system transfer function is essential and where the Laplace transform comes into its own. Let us see what it is all about. (For a more detailed introduction see Engineering Mathematics, Sixth Edition, Programme 26.) Laplace transforms The Laplace transform of an expression f ðtÞ is denoted by Lff ðtÞg and is defined as the semi-infinite integral ð1 f ðtÞest dt ð1Þ Lff ðtÞg ¼ t¼0 The parameter s is assumed to be positive and large enough to ensure that the integral converges. In more advanced applications s may be complex and in such cases the real part of s must be positive and large enough to ensure convergence. In determining the transform of an expression, you will appreciate that the limits of the integral are substituted for t, so that the result will be an expression in s. Therefore ð1 f ðtÞest dt ¼ FðsÞ Lff ðtÞg ¼ t¼0 Make a note of this general definition: then we can apply it 1 48 2 Programme 2 So we have Lff ðtÞg ¼ ð1 f ðtÞest dt ¼ FðsÞ 0 Example 1 To find the Laplace transform of f ðtÞ ¼ a (constant). st 1 ð1 1 e a st Lfag ¼ ae dt ¼ a ¼ est 0 s s 0 0 a a ¼ f0 1g ¼ s s a ðs > 0Þ ; Lfag ¼ s ð2Þ Example 2 To find the Laplace transform of f ðtÞ ¼ eat (a constant). As with all cases, we multiply f ðtÞ by est and integrate between t ¼ 0 and t ¼ 1. ð1 ð1 ; Lfeat g ¼ eat est dt ¼ eðsaÞt dt 0 0 ¼ ............ Finish it off. 3 Lfeat g ¼ 1 sa Because at Lfe g ¼ ð1 at st e e dt ¼ 0 ð1 e ðsaÞt 0 1 eðsaÞt dt ¼ ðs aÞ 0 1 1 f0 1g ¼ sa sa 1 ðs > aÞ ; Lfeat g ¼ sa ¼ ð3Þ So we already have two standard transforms Lfag ¼ a s and ; Lfeat g ¼ 1 sa Lf4g ¼ . . . . . . . . . . . . ; Lf5g ¼ . . . . . . . . . . . . ; Lfe4t g ¼ . . . . . . . . . . . . Lfe2t g ¼ . . . . . . . . . . . . 49 Laplace transforms 1 4 ; s 5 Lf5g ¼ ; s 4 1 s4 1 Lfe2t g ¼ sþ2 Lfe4t g ¼ Lf4g ¼ Note that, as we said earlier, the Laplace transform is always an expression in s. Now for some more examples Example 3 5 To find the Laplace transform of f ðtÞ ¼ sin at. We could, of course, apply the definition and evaluate ð1 Lfsin atg ¼ sin at est dt 0 using integration by parts. However, it is much shorter if we use the fact that e j ¼ cos þ j sin so that sin is the imaginary part of e j , written iðe j Þ. The function sin at can therefore be written iðe jat Þ so that ð1 ð1 ejat est dt ¼ i eðsjaÞt dt Lfsin atg ¼ Lfiðe jat Þg ¼ i 0 0 ( 1 ) eðsjaÞt 1 ¼i ½0 1 ¼i ðs jaÞ ðs jaÞ 0 1 ¼i s ja We can rationalise the denominator by multiplying top and bottom by ............ 6 s þ ja s þ ja a ; Lfsin atg ¼ i 2 ¼ 2 s þ a2 s þ a2 a ; Lfsin atg ¼ 2 s þ a2 ð4Þ We can use the same method to determine Lfcos atg since cos at is the real part of e jat , written <ðe jat Þ. Then Lfcos atg ¼ . . . . . . . . . . . . 50 Programme 2 7 Lfcos atg ¼ Because s þ ja Lfcos atg ¼ < 2 s þ a2 Recapping then: ¼ s2 s þ a2 s s2 þ a2 Lf1g ¼ . . . . . . . . . . . . ; Lfsin 2tg ¼ . . . . . . . . . . . . ; 8 Lf1g ¼ Lfsin 2tg ¼ (5) 1 ; s Lfe3t g ¼ . . . . . . . . . . . . Lfcos 4tg ¼ . . . . . . . . . . . . 1 s3 s Lfcos 4tg ¼ 2 s þ 16 Lfe3t g ¼ 2 ; s2 þ 4 Example 4 To find the transform of f ðtÞ ¼ t n where n is a positive integer. ð1 n By the definition Lft g ¼ t n est dt. 0 Integrating by parts st 1 ð e n 1 st n1 þ e t dt Lft n g ¼ t n s 0 s 0 1 ð 1 n 1 n1 st ¼ t n est þ t e dt s s 0 0 We said earlier that in a product such as t n est the numerical value of s is large enough to make the product converge to zero as t ! 1 1 ; t n est ¼00¼0 0 ð n 1 n1 st t e dt ð6Þ ; Lft n g ¼ s 0 ð1 ð1 You will notice that t n1 est dt is identical to t n est dt except that 0 0 n is replaced by ðn 1Þ. ð1 ð1 n st ; If In ¼ t e dt, then In1 ¼ t n1 est dt 0 0 n and the result (6) becomes In ¼ :In1 s This is a reduction formula, and if we now replace n by ðn 1Þ we get In1 ¼ . . . . . . . . . . . . ð7Þ 51 Laplace transforms 1 In1 ¼ 9 n1 :In2 s If we replace n by ðn 1Þ again in this last result, we have n2 :In3 s ð1 n So In ¼ t n est dt ¼ :In1 s 0 n n1 ¼ : :In2 s s n n1 n2 : :In3 etc. ¼ : s s s ¼ . . . . . . . . . . . . (next line) In2 ¼ In ¼ 10 n n1 n2 n3 : : : :In4 s s s s So finally, we have n n 1 n 2 n 3 n ðn 1Þ : : ... :I0 In ¼ : s s s s s But 1 s nðn 1Þðn 2Þðn 3Þ . . . ð3Þð2Þð1Þ n! ; In ¼ ¼ nþ1 nþ1 s s n! n ; Lft g ¼ nþ1 s I0 ¼ Lft 0 g ¼ Lf1g ¼ ; Lftg ¼ 1 ; s2 Lft 2 g ¼ 2 ; s3 Lft 3 g ¼ ð8Þ 6 s4 and with n ¼ 0, since 0! ¼ 1, the general result includes Lf1g ¼ which we have already established. 1 s Example 5 Laplace transforms of f ðtÞ ¼ sinh at and f ðtÞ ¼ cosh at. Starting from the exponential definitions of sinh at and cosh at, i.e. sinh at ¼ 12 ðeat eat Þ and cosh at ¼ 12 ðeat þ eat Þ we proceed as follows, recalling that the Laplace transform is a linear transformation so that: Lfaf ðtÞ þ bgðtÞg ¼ aLff ðtÞg þ bLfgðtÞg where a and b are constants [Refer: Engineering Mathematics, Sixth Edition, Programme 26, Frame 11.] 52 Programme 2 (a) f ðtÞ ¼ sinh t. 1 at 1 at e e 2 2 1 1 at L eat ¼ L e 2 2 ¼ ............ Lfsinh tg ¼ L Complete it 11 Lfsinh tg ¼ a s2 a2 Because 1 1 1 1 1 1 L eat L eat ¼ 2 2 2s a 2s þa 1 ðs þ aÞ ðs aÞ ¼ 2 s2 a 2 a ¼ 2 s a2 ð9Þ (b) f ðtÞ ¼ cosh t. Proceeding in the same way Lfcosh tg ¼ . . . . . . . . . . . . Next frame 12 Lfcosh tg ¼ s2 s a2 1 at 1 at e þ e 2 2 1 1 at þ L eat ¼ L e 2 2 1 1 1 1 þ ¼ 2s a 2s þ a 1 ðs þ aÞ þ ðs aÞ ¼ 2 s2 a2 s ¼ 2 s a2 Lfcosh tg ¼ L ð10Þ So we have accumulated several standard results: a Lfag ¼ ; s Lfeat g ¼ a ; s2 þ a2 a Lfsinh atg ¼ 2 ; s a2 Lfsin atg ¼ 1 ; sa Lft n g ¼ n! snþ1 s s2 þ a 2 s Lfcosh atg ¼ 2 s a2 Lfcos atg ¼ Make a note of this list if you have not already done so: it forms the basis of much that is to follow. 53 Laplace transforms 1 The Laplace transform is a linear transform, by which is meant that: 13 (1) The transform of a sum (or difference) of expressions is the sum (or difference) of the individual transforms. That is Lff ðtÞ gðtÞg ¼ Lff ðtÞg LfgðtÞg (2) The transform of an expression that is multiplied by a constant is the constant multiplied by the transform of the expression. That is Lfkf ðtÞg ¼ kLff ðtÞg Note: Two transforms must not be multiplied together to form the transform of a product of expressions – we shall see later that the product of two transforms is the transform of the convolution of two expressions. Example 6 (a) L 2et þ t ¼ L 2et þ Lft g ¼ 2L et þ Lftg ¼ 2 1 2s2 þ s þ 1 þ 2¼ 2 sþ1 s s ðs þ 1Þ (b) Lf2 sin 3t þ cos 3tg ¼ 2Lfsin 3tg þ Lfcos 3tg 3 s sþ6 þ ¼ ¼ 2: 2 s þ 9 s2 þ 9 s2 þ 9 (c) L 4e2t þ 3 cosh 4t ¼ 4L e2t þ 3Lfcosh 4tg 1 s 4 3s ¼ 4: þ 3: 2 þ ¼ s2 s 16 s 2 s2 16 7s2 6s 64 ¼ ðs 2Þðs2 16Þ So 1. Lf2 sin 3t þ 4 sinh 3tg ¼ . . . . . . . . . . . . 2. Lf5e4t þ cosh 2tg ¼ . . . . . . . . . . . . 3. L t 3 þ 2t 2 4t þ 1 ¼ . . . . . . . . . . . . 1. 18ðs2 þ 3Þ ; s4 81 2. 6s2 4s 20 ; ðs 4Þðs2 4Þ 3. 1 3 s 4s2 þ 4s þ 6 s4 The working is straightforward. 3 3 þ 4: 2 s2 þ 9 s 9 6 12 18ðs2 þ 3Þ ¼ 2 þ 2 ¼ 4 s þ9 s 9 s 81 1. Lf2 sin 3t þ 4 sinh 3tg ¼ 2: 2. L 5e4t þ cosh 2t ¼ 5 s 6s2 4s 20 þ 2 ¼ s 4 s 4 ðs 4Þðs2 4Þ 14 54 Programme 2 3. 3! 2! 1! 1 þ 2: 3 4: 2 þ 4 s s s s 1 3 2 ¼ 4 s 4s þ 4s þ 6 s L t 3 þ 2t 2 4t þ 1 ¼ We have been building up a list of standard transforms of simple expressions. Before we leave this part of the work, there are three important and useful theorems which enable us to deal with rather more complicated expressions. 15 Theorem 1 The first shift theorem The first shift theorem states that if Lff ðtÞg ¼ FðsÞ then L eat f ðtÞ ¼ Fðs þ aÞ ð1 ð1 Because L eat f ðtÞ ¼ eat f ðtÞest dt ¼ f ðtÞeðsþaÞt dt ¼ Fðs þ aÞ t¼0 t¼0 That is L eat f ðtÞ ¼ Fðs þ aÞ The transform Lfeat f ðtÞg is thus the same as Lff ðtÞg with s everywhere in the result replaced by ðs þ aÞ. 2 For example Lfsin 2tg ¼ 2 s þ4 then L e3t sin 2t ¼ 2 ðs þ 3Þ2 þ 4 Similarly, L t 2 ¼ 16 ¼ 2 s2 þ 6s þ 13 2 s3 ; L t 2 e4t ¼ . . . . . . . . . . . . 2 ðs 4Þ3 Because L t 2 ¼ ; L t 2 e4t ¼ 2 . ; L t 2 e4t s3 2 is the same with s replaced by ðs 4Þ. ðs 4Þ3 Here is a short exercise by way of practice. Exercise Determine the following. 1. L e2t cosh 3t 4. L e2t cos t 2. L 2e3t sin 3t 5. L e3t sinh 2t 3. t 6. L t 3 e4t Lf4te g Complete all six and then check with the results in the next frame 55 Laplace transforms 1 Here they are. 17 s s2 9 1. Lfcosh 3t g ¼ 2. Lfsin 3tg ¼ 3. Lf4tg ¼ 4: 1 s2 4. Lfcos tg ¼ s s2 þ 1 5. Lfsinh 2tg ¼ 6. L t3 ¼ sþ2 ; L e2t cosh 3t ¼ 3 s2 þ 9 ðs þ 2Þ2 9 sþ2 ¼ 2 s þ 4s 5 6 ; L 2e3t sin 3t ¼ ðs 3Þ2 þ 9 6 ¼ 2 s 6s þ 18 ; L 4tet ¼ ; L e2t cos t ¼ 4 ðs þ 1Þ2 s2 ðs 2Þ2 þ 1 s2 ¼ 2 s 4s þ 5 2 s2 4 2 ; L e3t sinh 2t ¼ 3! s4 ðs 3Þ2 4 2 ¼ 2 s 6s þ 5 ; L t 3 e4t ¼ 6 ðs þ 4Þ4 Now let us deal with the next theorem 18 Theorem 2 Multiplying by t and t n If Lff ðtÞg ¼ FðsÞ then Lftf ðtÞg ¼ F0 ðsÞ ð1 ð1 dest st dt Because Lftf ðtÞg ¼ tf ðtÞe dt ¼ f ðtÞ ds t¼0 t¼0 ð d 1 ¼ f ðtÞest dt ¼ F 0 ðsÞ ds t¼0 That is Lftf ðtÞg ¼ F0 ðsÞ For example, and similarly, 2 s2 þ 4 d 2 ; Lft sin 2tg ¼ 2 ds s þ 4 Lfsin 2tg ¼ Lft cosh 3tg ¼ . . . . . . . . . . . . ¼ 4s ðs2 þ 4Þ2 56 Programme 2 19 s2 þ 9 ðs2 9Þ2 Because Lft cosh 3tg ¼ d s ðs2 9Þ sð2sÞ s2 þ 9 ¼ ¼ 2 2 2 ds s 9 ðs 9Þ ðs2 9Þ2 We could, if necessary, take this a stage further and find L t 2 cosh 3t ( ) d s2 þ 9 2 L t cosh 3t ¼ Lftðt cosh 3tÞg ¼ ds ðs2 9Þ2 ¼ 2sðs2 þ 27Þ ðs2 9Þ3 Likewise, starting with Lfsin 4tg ¼ 4 s2 þ 16 Lft sin 4tg ¼ . . . . . . . . . . . . and 20 8s ; ðs2 þ 16Þ2 L t 2 sin 4t ¼ . . . . . . . . . . . . 8ð3s2 16Þ ðs2 þ 16Þ3 d fFðsÞg in each case. ds Theorem 2 obviously extends the range of functions that we can deal with. So, in general, if Lff ðtÞg ¼ FðsÞ, then applying Lftf ðtÞg ¼ Lft n f ðtÞg ¼ ð1Þn dn fFðsÞg dsn Make a note of this in your record book Laplace transforms 1 Theorem 3 Dividing by t 57 21 ð1 f ðtÞ ¼ If Lff ðtÞg ¼ FðsÞ then L FðÞ d t ¼s f ðtÞ provided Lim exists. To demonstrate this we start from the right-hand t t!0 side of the result ð 1 ð 1 ð1 Notice the dummy variable . t FðÞ d ¼ f ðtÞe dt d The end result is an expression ¼s ¼s t¼0 in s which comes from the ð1 ð1 t lower limit of the integral so ¼ f ðtÞe d dt the variable of integration, t¼0 ¼s ð 1 ð1 which is absorbed during the ¼ f ðtÞ et d dt process of integration, is chant¼0 ¼s ged to . Notice also that we ð1 est interchange the order of intedt ¼ f ðtÞ t t¼0 gration. f ðtÞ ¼L t f ðtÞ This rule is somewhat restricted in use, since it is applicable only if Lim t t!0 exists. In indeterminate cases, we use L’Hôpital’s rule to find out. Let’s try a couple of examples. Example 1 sin at Determine L t sin at 0 0 ¼ which gives the indeterminate form of . So, First we test Lim t 0 0 t!0 by L’Hôpital’s rule, we differentiate top and bottom separately and substitute t ¼ 0 in the result to ascertain the limit of the new expression. sin at a cos at ¼ Lim ¼ a, that is, the limit exists and the theorem Lim t 1 t!0 t!0 can therefore be applied. So Lfsin atg ¼ ð1 a sin at a ¼ , therefore L d 2 2 2 2 s þa t s þa h i1 ¼ arctan a s s ¼ arctan 2 a a ¼ arctan s 22 58 Programme 2 Notice that arctan a s þ arctan ¼ , as can be seen s a 2 from the figure s a Example 2 1 cos 2t Determine L t First we test whether Lim t!0 23 1 cos 2t t exists. Result? . . . . . . . . . . . . the limit exists 1 cos 2t 11 0 ¼ ¼ ¼? t 0 0 t!0 1 cos 2t 2 sin 2t 0 ¼ Lim ¼ ¼0 Lim t 1 1 t!0 t!0 Lim Lf1 cos 2t g ¼ ; limit exists. 1 s s s2 þ 4 Then, by Theorem 3 ð1 1 cos 2t 1 ¼ 2 L d t þ4 ¼s 2 1 1 1 2 1 ¼ ln ln þ 4 ln 2 ¼ 2 2 þ 4 ¼s ¼s 2 ! ln 1 ¼ 0 When ! 1, ln 2 þ4 1 cos 2t ¼ ............ Therefore, L t Complete it 24 ln Because 2 1 cos 2t 1 s L ¼ ln 2 t 2 s þ4 rffiffiffiffiffiffiffiffiffiffiffiffiffi s2 þ 4 ¼ ln s2 rffiffiffiffiffiffiffiffiffiffiffiffiffi s2 þ 4 s2 2 s ¼ ln 2 s þ4 1=2 Let us pause here for a while and take stock, for we have met a number of results important in the future work. 59 Laplace transforms 1 1 Standard transforms f ðtÞ a eat sin at cos at sinh at cosh at tn 2 Theorem 1 Lff ðtÞg ¼ FðsÞ a s 1 sa a s2 þ a2 s s2 þ a2 a s2 a2 s s2 a2 n! snþ1 (n a positive integer) The first shift theorem If Lff ðtÞg ¼ FðsÞ, then Lfeat f ðtÞg ¼ Fðs þ aÞ 3 Theorem 2 Multiplying by t If Lff ðtÞg ¼ FðsÞ, then Lftf ðtÞg ¼ 4 d fFðsÞg ds Theorem 3 Dividing by t ð1 f ðtÞ If Lff ðtÞg ¼ FðsÞ, then L ¼ FðÞ d t ¼s f ðtÞ exists. provided Lim t t!0 Now let us work through a short revision exercise, so move on Exercise 25 Determine the Laplace transforms of the following expressions. 1 sin 3t 6 t cosh 4t 2 cos 2t 7 3 e4t 8 4 6t 2 t 2 3t þ 4 e3t 1 t 3t e cos 4t 5 sinh 3t 9 10 t 2 sin t Complete the whole set and then check results with the next frame 60 26 Programme 2 Here are the results. 1 2 3 4 5 s2 3 þ9 6 s s2 þ 4 1 s4 12 s3 3 s2 9 7 8 9 10 s2 þ 16 2 ðs2 16Þ 1 2 4s 3s þ 2 3 s s ln s3 s3 s2 6s þ 25 6s2 2 3 ð s2 þ 1 Þ It is just a case of applying the standard tranforms and the three theorems. Now on to the next piece of work Inverse transforms 27 Here we have the reverse process, i.e. given a Laplace transform, we have to find the function of t to which it belongs. a For example, we know that 2 is the Laplace transform of sin at, so we s þ a2 a can now write L1 2 ¼ sin at, the symbol L1 indicating the inverse s þ a2 transform and not a reciprocal. 1 4 ¼ . . . . . . . . . . . . ; (c) L1 ¼ ............ ; (a) L1 s2 s s 12 (b) L1 2 . . . . . . . . . . . . ; (d) L1 2 ¼ ............ s þ 25 s 9 28 1 ¼ e2t ; s2 s 1 ¼ cos 5t; (b) L s2 þ 25 (a) L1 4 ¼4 s 12 1 (d) L ¼ 4 sinh 3t s2 9 (c) L1 Therefore, given a transform, we can write down the corresponding expression in t, provided we can recognise it from our table of transforms. 61 Laplace transforms 1 But what about L1 3s þ 1 ? This certainly did not appear in our list of s2 s 6 standard transforms. 3s þ 1 3s þ 1 , it happens that we can write 2 as s2 s 6 s s6 1 2 the sum of two simpler functions þ which, of course, makes all the sþ2 s3 difference, since we can now proceed 3s þ 1 1 2 1 1 ¼L L þ s2 s 6 sþ2 s3 In considering L1 which we immediately recognise as . . . . . . . . . . . . e2t þ 2e3t The two simpler expressions 1 2 and are called the partial fractions of sþ2 s3 3s þ 1 , and the ability to represent a complicated algebraic fraction in s2 s 6 terms of its partial fractions is the key to much of this work. Let us take a closer look at the rules. Rules of partial fractions 1 The numerator must be of lower degree than the denominator. This is usually the case in Laplace transforms. If it is not, then we first divide out. 2 Factorise the denominator into its prime factors. These determine the shapes of the partial fractions. A where A is a constant A linear factor ðs þ aÞ gives a partial fraction sþa to be determined. 3 A B þ . ðs þ aÞ ðs þ aÞ2 4 A repeated factor ðs þ aÞ2 gives 5 Similarly ðs þ aÞ3 gives 6 A quadratic factor s2 þ ps þ q gives 7 A B C þ . þ ðs þ aÞ ðs þ aÞ2 ðs þ aÞ3 Ps þ Q . s2 þ ps þ q 2 Repeated quadratic factors s2 þ ps þ q give Ps þ Q Rs þ T þ . s2 þ ps þ q ðs2 þ ps þ qÞ2 So s 19 has partial fractions of the form . . . . . . . . . . . . ðs þ 2Þðs 5Þ 29 62 Programme 2 30 A B þ sþ2 s5 and 3s2 4s þ 11 ðs þ 3Þðs 2Þ2 has partial fractions of the form . . . . . . . . . . . . Be careful of the repeated factor. 31 A B C þ þ s þ 3 ðs 2Þ ðs 2Þ2 Let us work through the various steps with an example. Example 1 To determine L1 5s þ 1 . s2 s 12 (a) First we check that the numerator is of lower degree than the denominator. In fact, this is so. (b) Factorise the denominator 5s þ 1 5s þ 1 . ¼ s2 s 12 ðs 4Þðs þ 3Þ (c) Then the partial fractions are of the form . . . . . . . . . . . . 32 A B þ s4 sþ3 We therefore have the identity s2 5s þ 1 A B þ s 12 s 4 s þ 3 If we multiply through both sides by the denominator s2 s 12 ðs 4Þðs þ 3Þ we have 5s þ 1 Aðs þ 3Þ þ Bðs 4Þ This is also an identity and true for any value of s we care to substitute – our job is now to find the values of A and B. We now substitute convenient values for s (a) Let ðs 4Þ ¼ 0, i.e. s ¼ 4 (b) ; 21 ¼ Að7Þ þ Bð0Þ ; A¼3 Let ðs þ 3Þ ¼ 0, i.e. s ¼ 3 and we get . . . . . . . . . . . . 63 Laplace transforms 1 B¼2 33 5s þ 1 3 2 þ ; 2 s s 12 s 4 s þ 3 5s þ 1 1 ¼ ............ ; L s2 s 12 3e4t þ 2e3t 34 Example 2 Determine L1 9s 8 . s2 2s Working as before, f ðtÞ ¼ . . . . . . . . . . . . 4 þ 5e2t Because Lff ðtÞg ¼ 9s 8 . s2 2s (a) Numerator of first degree; denonominator of second degree. Therefore rule satisfied. (b) 9s 8 A B þ . sðs 2Þ s s 2 (c) Multiply by sðs 2Þ. ; 9s 8 ¼ Aðs 2Þ þ Bs. (d) Put s ¼ 0. 8 ¼ Að2Þ þ Bð0Þ ; A ¼ 4. (e) Put s 2 ¼ 0, i.e. s ¼ 2. 10 ¼ Að0Þ þ Bð2Þ ; B ¼ 5. 4 5 ¼ 4 þ 5e2t ; f ðtÞ ¼ L1 þ s s2 Example 3 Express FðsÞ ¼ s2 15s þ 41 ðs þ 2Þðs 3Þ2 inverse transform. s2 15s þ 41 ðs þ 2Þðs 3Þ2 in partial fractions and hence determine its has partial fractions of the form . . . . . . . . . . . . 35 64 Programme 2 36 A B C þ þ s þ 2 s 3 ðs 3Þ2 Now we multiply throughout by ðs þ 2Þðs 3Þ2 and get s2 15s þ 41 Aðs 3Þ2 þ Bðs þ 2Þðs 3Þ þ Cðs þ 2Þ Putting ðs 3Þ ¼ 0 and then ðs þ 2Þ ¼ 0 we obtain . . . . . . . . . . . . 37 A ¼ 3 and C ¼ 1 Now that we have run out of ‘crafty’ substitutions, we equate coefficients of the highest power of s on each side, i.e. the coefficients of s2 . This gives ............ 38 1¼AþB s2 15s þ 41 ; 1¼3þB ; B ¼ 2 3 2 1 þ s þ 2 s 3 ðs 3Þ2 ðs þ 2Þðs 3Þ 3 2 ¼ . . . . . . . . . . . . and L1 ¼ ............ Now L1 sþ2 s3 So 2 ¼ 39 3e2t and 2e3t ( 1 1 ) ? ðs 3Þ2 1 We remember that L1 2 ¼ . . . . . . . . . . . . s But what about L 40 t and that by Theorem 1, if Lff ðtÞg ¼ FðsÞ then Lfeat f ðtÞg ¼ Fðs þ aÞ. 1 1 ; is like 2 with s replaced by ðs 3Þ i.e. a ¼ 3. 2 s ðs 3Þ ( ) 1 1 ¼ te3t ; L ðs 3Þ2 ( ) 2 1 s 15s þ 41 ¼ 3e2t þ 2e3t þ te3t ; L ðs þ 2Þðs 3Þ2 65 Laplace transforms 1 Example 4 Determine L1 4s2 5s þ 6 . ðs þ 1Þðs2 þ 4Þ Notice that this time we have a quadratic factor in the denominator 4s2 5s þ 6 A Bs þ C þ ðs þ 1Þðs2 þ 4Þ s þ 1 s2 þ 4 ; 4s2 5s þ 6 Aðs2 þ 4Þ þ ðBs þ CÞðs þ 1Þ. (a) Putting ðs þ 1Þ ¼ 0, i.e. s ¼ 1, 15 ¼ 5A ; A ¼ 3 (b) Equate coefficients of highest power, i.e. s2 4¼AþB ; 4¼3þB ; B¼1 (c) We now equate the lowest power on each side, i.e. the constant term 6 ¼ 4A þ C ; 6 ¼ 12 þ C ; C ¼ 6 Now you can finish it off. f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ 3et þ cos 2t 3 sin 2t Because 3 s 6 þ s þ 1 s2 þ 4 s2 þ 4 ; f ðtÞ ¼ 3et þ cos 2t 3 sin 2t Lff ðtÞg ¼ Example 5 1 Determine L s2 9s 7 ðs2 þ 2s þ 2Þðs 1Þ Again we have a quadratic factor in the denominator that does not have simple factors s2 9s 7 A Bs þ C þ ðs2 þ 2s þ 2Þðs 1Þ s 1 s2 þ 2s þ 2 ; s2 9s 7 Aðs2 þ 2s þ 2Þ þ ðBs þ CÞðs 1Þ (a) Putting s 1 ¼ 0, that is s ¼ 1, 15 ¼ 5A ; A ¼ 3 (b) Equate coefficients of highest power, that is s2 1¼AþB ; 1 ¼ 3 þ B ; B¼4 (c) We now equate the lowest power on each side, that is the constant term 7 ¼ 2A C So that Lff ðtÞg ¼ ; 7 ¼ 6 C 4s þ 1 3 . s2 þ 2s þ 2 s 1 ; C¼1 41 66 Programme 2 Here we have a denominator with no simple factors. We therefore complete the square and use the First Shift Theorem [Refer: Frames 15–17]. 4s þ 1 3 s2 þ 2s þ 2 s 1 4s þ 1 3 ¼ ðs þ 1Þ2 þ 1 s 1 Lff ðtÞg ¼ ¼ ¼ 4ðs þ 1Þ 3 ðs þ 1Þ2 þ 1 4ðs þ 1Þ ðs þ 1Þ2 þ 1 3 s1 3 ðs þ 1Þ2 þ 1 3 s1 By the First Shift Theorem f ðtÞ ¼ 4et cos t 3et sin t 3et Now you try one. Given Lff ðtÞg ¼ ðs2 2s2 þ s 3 then f ðtÞ ¼ . . . . . . . . . . . . þ 4s þ 5Þðs þ 1Þ Next frame 42 f ðtÞ ¼ 3e2t cos t 4e2t sin t et Because 2s2 þ s 3 A Bs þ C þ ðs2 þ 4s þ 5Þðs þ 1Þ s þ 1 s2 þ 4s þ 5 ; 2s2 þ s 3 Aðs2 þ 4s þ 5Þ þ ðBs þ CÞðs þ 1Þ (a) Putting s þ 1 ¼ 0, that is s ¼ 1, 2 ¼ 2A ; A ¼ 1 (b) Equate coefficients of highest power, that is s2 2¼AþB ; 2 ¼ 1 þ B B¼3 (c) We now equate the lowest power on each side, that is the constant term 3 ¼ 5A þ C ; 3 ¼ 5 þ C ; C¼2 3s þ 2 1 þ 4s þ 5 s þ 1 3s þ 2 1 ¼ 2 ðs þ 2Þ þ 1 s þ 1 So that Lff ðtÞg ¼ ¼ ¼ s2 3ðs þ 2Þ 4 2 2 ðs þ 2Þ þ 1 3ðs þ 2Þ ðs þ 2Þ þ 1 1 sþ1 4 2 ðs þ 2Þ þ 1 1 sþ1 By the First Shift Theorem f ðtÞ ¼ 3e2t cos t 4e2t sin t et 67 Laplace transforms 1 The ‘cover up’ rule While we can always find A, B, C, etc., there are many cases where we can use the ‘cover up’ methods and write down the values of the constant coefficients almost on sight. However, this method only works when the denominator of the original fraction has non-repeated, linear factors. The following examples illustrate the method. 43 Example 1 9s 8 A B has partial fractions of the form þ . By the sðs 2Þ s s2 1 ‘cover up’ rule, the constant A, that is the coefficient of , is found by s temporarily covering up the factor s in the denominator of FðsÞ and finding the limiting value of what remains when s (the factor covered up) tends to zero. 1 9s 8 Therefore A ¼ coefficient of ¼ Lim ¼ 4. That is A ¼ 4. s s2 s!0 We know that FðsÞ ¼ 1 , is obtained by covering up the factor s2 ðs 2Þ in the denominator of FðsÞ and finding the limiting value of what remains when ðs 2Þ ! 0, that is s ! 2. 1 9s 8 Therefore B ¼ coefficient of ¼ Lim ¼ 5. That is B ¼ 5. So that s 2 s!2 s Similarly, B, the coefficient of 9s 8 4 5 ¼ þ sðs 2Þ s s 2 Another example Example 2 44 s þ 17 A B C FðsÞ ¼ þ þ . ðs 1Þðs þ 2Þðs 3Þ s 1 s þ 2 s 3 A: cover up ðs 1Þ in FðsÞ and find s þ 17 18 ¼ ; A ¼ 3 Lim ðs þ 2Þðs 3Þ 6 s!1 Similarly B: . . . . . . . . . . . . ; B ¼ ............ C: . . . . . . . . . . . . ; C ¼ ............ 68 45 Programme 2 s þ 17 15 ¼1 B ¼ Lim ¼ ðs 1Þðs 3Þ ð3Þð5Þ s!2 s þ 17 20 C ¼ Lim ¼ ¼2 ðs 1Þðs þ 2Þ ð2Þð5Þ s!3 ; B¼1 ; C¼2 1 2 3 þ sþ2 s3 s1 So f ðtÞ ¼ e2t þ 2e3t 3et ; FðsÞ ¼ Every entry in our table of standard transforms gives rise to a corresponding entry in a similar table of inverse transforms. Let us tabulate such a list. 46 Table of inverse transforms FðsÞ a s 1 sþa n! snþ1 1 sn a s2 þ a2 s 2 s þ a2 a s2 a2 s 2 s a2 f ðtÞ a eat tn (n a positive integer) t n1 ðn 1Þ! (n a positive integer) sin at cos at sinh at cosh at Theorem 1 The first shift theorem can be stated as follows. If FðsÞ is the Laplace transform of f ðtÞ then Fðs þ aÞ is the Laplace transform of eat f ðtÞ. Here is a short revision exercise. 69 Laplace transforms 1 Exercise 1 Find the inverse transforms of (a) 2 (b) 5 3 ðs 4Þ ; (c) 3s þ 4 . s2 þ 9 Express in partial fractions (a) 3 1 ; 2s 3 22s þ 16 ; ðs þ 1Þðs 2Þðs þ 3Þ (b) Determine 2 4s 17s 24 ; (b) (a) L1 sðs þ 3Þðs 4Þ 4s2 21s þ 30 (c) L1 . ðs2 6s þ 13Þðs 4Þ 1 3t=2 5 2 4t e ; t e ; (b) 2 2 1 4 5 þ ; sþ1 s2 sþ3 s2 11s þ 6 ðs þ 1Þðs 2Þ2 L1 . 5s2 4s 7 ; ðs 3Þðs2 þ 4Þ 4 3 cos 3t þ sin 3t 3 2 1 4 (b) s þ 1 s 2 ðs 2Þ2 1 (a) 2 (b) 3 (a) 2 þ 3e3t e4t ; (c) 5 2e4t þ 2e3t cos 2t þ e3t sin 2t 2 (c) (b) 5 2e3t þ 3 cos 2t þ sin 2t 2 Solution of differential equations by Laplace transforms To solve a differential equation by Laplace transforms, we go through four distinct stages (a) Rewrite the equation in terms of Laplace transforms. (b) Insert the given initial conditions. (c) Rearrange the equation algebraically to give the transform of the solution. (d) Determine the inverse transform to obtain the particular solution. We have spent some time finding the transforms of a variety of functions of t and the inverse transforms of functions of s, i.e. we have largely covered steps (a) and (d) of the above list. However, to write a differential equation in Laplace transforms, we must obtain the transforms of the first and second derivatives of f ðtÞ, that is the transforms of f 0 ðtÞ and f 00 ðtÞ. 47 70 Programme 2 Transforms of derivatives Lff 0 ðtÞg ¼ By definition ð1 est f 0 ðtÞ dt. 0 Integrating by parts 1 ð 1 Lff 0 ðtÞg ¼ est f ðtÞ f ðtÞ sest dt 0 st When t ! 1, e 0 f ðtÞ ! . . . . . . . . . . . . 48 0 Because s is positive and large enough to ensure that est decays faster than any possible growth of f ðtÞ. ; Lff 0 ðtÞg ¼ f ð0Þ þ sLff ðtÞg Replacing f ðtÞ by f 0 ðtÞ gives Lff 00 ðtÞg ¼ . . . . . . . . . . . . 49 Lff 00 ðtÞg ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ Because Lff 0 ðtÞg ¼ f ð0Þ þ sLff ðtÞg so Writing Lff 00 ðtÞg ¼ f 0 ð0Þ þ sLff 0 ðtÞg ¼ f 0 ð0Þ þ sðf ð0Þ þ sLff ðtÞgÞ Lff ðtÞg ¼ FðsÞ as usual, we have Lff ðtÞg ¼ FðsÞ Lff 0 ðtÞg ¼ sFðsÞ f ð0Þ Lff 00 ðtÞg ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ We can see a pattern emerging Lff 000 ðtÞg ¼ . . . . . . . . . . . . 71 Laplace transforms 1 Lff 000 ðtÞg ¼ s3 FðsÞ s2 f ð0Þ sf 0 0Þ f 00 ð0Þ 50 Alternative notation We make the working neater by adopting the following notation. Let x ¼ f ðtÞ and at t ¼ 0, we write x ¼ x0 dx ¼ x1 dt 2 d x ¼ x2 dt 2 dn x ¼ xn ; dt n i.e. f ð0Þ ¼ x0 i.e. f 0 ð0Þ ¼ x1 i.e. f 00 ð0Þ ¼ x2 etc. i.e. f n ð0Þ ¼ xn Also we denote the Laplace transform of x by x, i.e. x ¼ Lfxg ¼ Lff ðtÞg ¼ FðsÞ. So, using the ‘dot’ notation for derivatives, the previous results can be written . . . . . . . . . . . . Lfxg ¼ x _ ¼ sx x0 Lfxg 51 Lfx€g ¼ s2 x sx0 x1 Lf_€ xg ¼ s3 x s2 x0 sx1 x2 In each case, the subscript indicates the order of the derivative, dn x i.e. xn ¼ the value of n at t ¼ 0. dt Notice the pattern of the results. Lf_€ x_g ¼ . . . . . . . . . . . . Lf_€ x_g ¼ s4 x s3 x0 s2 x1 sx2 x3 Now, at long last, we can start solving differential equations. 52 72 Programme 2 Solution of first-order differential equations Example 1 dx 2x ¼ 4 given that at t ¼ 0, x ¼ 1. dt We go through the four stages. Solve the equation (a) Rewrite the equation in Laplace transforms, using the last notation Lfxg ¼ x; 53 _ ¼ ............ Lfxg Lf4g ¼ . . . . . . . . . . . . _ ¼ sx x0 ; Lfxg Lf4g ¼ 4 s 4 s (b) Insert the initial condition that at t ¼ 0, x ¼ 1, i.e. x0 ¼ 1 ðsx x0 Þ 2x ¼ Then the equation becomes ; sx 1 2x ¼ 4 s (c) Now we rearrange this to give an expression for x x ¼ . . . . . . . . . . . . 54 x ¼ sþ4 sðs 2Þ (d) Finally, we take inverse transforms to obtain x. sþ4 in partial fractions gives . . . . . . . . . . . . sðs 2Þ 55 3 2 s2 s Because sþ4 A B þ sðs 2Þ s s 2 (1) Put ðs 2Þ ¼ 0, i.e. s ¼ 2 (2) Put s ¼ 0 ; x ¼ ; s þ 4 ¼ Aðs 2Þ þ Bs ; 6 ¼ Bð2Þ ; B¼3 ; 4 ¼ Að2Þ ; A ¼ 2 sþ4 3 2 ¼ sðs 2Þ s 2 s Therefore, taking inverse transforms sþ4 3 2 1 1 x¼L ¼L ¼ ............ sðs 2Þ s2 s 73 Laplace transforms 1 56 x ¼ 3e2t 2 This solution should now be substituted back into the differential equation to verify that it is, indeed, correct. Example 2 Solve the equation dx þ 2x ¼ 10e3t given that at t ¼ 0, x ¼ 6. dt (a) Convert the equations to Laplace transforms, i.e. ............ ðsx x0 Þ þ 2x ¼ 57 10 s3 (b) Insert the initial condition, x0 ¼ 6 sx 6 þ 2x ¼ 10 s3 (c) Rearrange to obtain x ¼ . . . . . . . . . . . . x ¼ 58 6s 8 ðs þ 2Þðs 3Þ (d) Taking inverse transforms to obtain x 6s 8 ¼ ............ x ¼ L1 ðs þ 2Þðs 3Þ Complete the solution 59 x ¼ 4e2t þ 2e3t Because 6s 8 A B þ ðs þ 2Þðs 3Þ s þ 2 s 3 ; 6s 8 ¼ Aðs 3Þ þ Bðs þ 2Þ (1) Put ðs 3Þ ¼ 0, i.e. s ¼ 3 ; 10 ¼ Bð5Þ ; B¼2 (2) Put ðs þ 2Þ ¼ 0, i.e. s ¼ 2. ; 20 ¼ Að5Þ ; A ¼ 4 6s 8 4 2 ¼ þ ðs þ 2Þðs 3Þ s þ 2 s 3 4 2 ; x ¼ L1 þ ¼ 4e2t þ 2e3t sþ2 s3 ; x ¼ 74 Programme 2 Example 3 dx 4x ¼ 2e2t þ e4t , given that at t ¼ 0, x ¼ 0. dt Work this through the four steps in the same way as before and complete it on your own. Solve the equation x ¼ ............ 60 x ¼ e4t e2t þ te4t The working is quite standard. dx 4x ¼ 2e2t þ e4t dt 2 1 þ s2 s4 2 1 þ (b) x0 ¼ 0 ; sx 4x ¼ s2 s4 2 1 þ (c) ; x ¼ ðs 2Þðs 4Þ ðs 4Þ2 (a) ðsx x0 Þ 4x ¼ (d) 2 A B þ ; 2 ¼ Aðs 4Þ þ Bðs 2Þ ðs 2Þðs 4Þ s 2 s 4 Putting ðs 2Þ ¼ 0, i.e. s ¼ 2 ; 2 ¼ Að2Þ ; A ¼ 1 Putting ðs 4Þ ¼ 0, i.e. s ¼ 4 ; 2 ¼ Bð2Þ 1 1 1 þ ; x ¼ s 4 s 2 ðs 4Þ2 ; B¼1 ; x ¼ e4t e2t þ te4t Now on to the next frame 61 Solution of second-order differential equations The method is, in effect, the same as before, going through the same four distinct stages. Example 1 Solve the equation d2 x dx þ 2x ¼ 2e3t , given that at t ¼ 0, x ¼ 5 and 3 dt 2 dt dx ¼ 7. dt (a) We rewrite the equation in terms of its transforms, remembering that Lfxg ¼ x _ ¼ sx x0 Lfxg Lfx€g ¼ s2 x sx0 x1 The equation becomes . . . . . . . . . . . . 75 Laplace transforms 1 s2 x sx0 x1 3ðsx x0 Þ þ 2x ¼ 2 s3 62 (b) Insert the initial conditions. In this case x0 ¼ 5 and x1 ¼ 7 ; s2 x 5s 7 3ðsx 5Þ þ 2x ¼ 2 s3 (c) Rearrange to obtain x ¼ . . . . . . . . . . . . x ¼ 63 5s2 23s þ 26 ðs 1Þðs 2Þðs 3Þ Because s2 x 5s 7 3sx þ 15 þ 2x ¼ s2 3s þ 2 x 5s þ 8 ¼ 2 s3 2 s3 2 2 þ 5s2 23s þ 24 þ 5s 8 ¼ s3 s3 5s2 23s þ 26 ; x ¼ ðs 1Þðs 2Þðs 3Þ ðs 1Þðs 2Þx ¼ (d) Now for partial fractions 5s2 23s þ 26 A B C ¼ þ þ ðs 1Þðs 2Þðs 3Þ s 1 s 2 s 3 ; 5s2 23s þ 26 ¼ Aðs 2Þðs 3Þ þ Bðs 1Þðs 3Þ þ Cðs 1Þðs 2Þ So that A ¼ . . . . . . . . . . . . ; B ¼ ............; A ¼ 4; B ¼ 0; C ¼ ............ C¼1 64 4 1 þ s1 s3 ; x ¼ ............ ; x ¼ x ¼ 4et þ e3t As you see, the Laplace transform method can be considerably shorter than the classical method which requires (a) determination of the complementary function (b) determination of a particular integral (c) obtaining the general solution, before (d) arriving at the particular solution by substitution of the initial conditions in the general solution. 65 76 Programme 2 Here is another example. Example 2 Solve d2 x dx ¼ 4. 4x ¼ 24 cos 2t given that at t ¼ 0, x ¼ 3 and dt 2 dt (a) In Laplace transforms . . . . . . . . . . . . 66 24s s2 x sx0 x1 4x ¼ 2 s þ4 (b) Insert initial condition, i.e. x0 ¼ 3; x1 ¼ 4 s2 x 3s 4 4x ¼ 24s s2 þ 4 24s ; s2 4 x ¼ 3s þ 4 þ 2 s þ4 3s3 þ 4s2 þ 36s þ 16 ¼ s2 þ 4 (c) x ¼ 3s3 þ 4s2 þ 36s þ 16 ðs2 þ 4Þðs 2Þðs þ 2Þ Expressed in partial fractions, this becomes ............ 67 3s3 þ 4s2 þ 36s þ 16 As þ B C D þ 2 þ 2 ðs þ 4Þðs 2Þðs þ 2Þ s þ 4 s 2 s þ 2 ; 3s3 þ 4s2 þ 36s þ 16 ðAs þ BÞðs 2Þðs þ 2Þ þ Cðs2 þ 4Þðs þ 2Þ þ Dðs2 þ 4Þðs 2Þ Putting ðs 2Þ ¼ 0, i.e. s ¼ 2, gives C¼4 Putting ðs þ 2Þ ¼ 0, i.e. s ¼ 2, gives D¼2 3 Equating coefficients of s and also the constant terms gives A ¼ 3 and B ¼ 0. ; x ¼ 3s3 þ 4s2 þ 36s þ 16 4 2 3s þ ¼ ðs2 þ 4Þðs 2Þðs þ 2Þ s 2 s þ 2 s2 þ 4 ; x ¼ ............ 68 x ¼ 4e2t þ 2e2t 3 cos 2t Now let us solve another equation, this time using the ‘cover up’ rule. 77 Laplace transforms 1 Example 3 Solve x€ þ 5x_ þ 6x ¼ 4t, given that at t ¼ 0, x ¼ 0 and x_ ¼ 0. 4 As usual we begin s2 x sx0 x1 þ 5ðsx x0 Þ þ 6x ¼ 2 s 4 x0 ¼ 0; x1 ¼ 0 ; s2 þ 5s þ 6 x ¼ 2 s 4 ; x ¼ 2 s ðs þ 2Þðs þ 3Þ The s2 in the denominator can be awkward, so we introduce a useful trick and detach one factor s outside the main expression, thus 1 4 1 A B C þ þ x ¼ ¼ s sðs þ 2Þðs þ 3Þ s s sþ2 sþ3 Applying the ‘cover up’ rule to the expressions within the brackets 1 4 1 2 4 1 : þ : x ¼ s 6 s ðs þ 2Þ 3 s þ 3 Now we bring the external x ¼ 1 back into the fold s 2 1 2 4 1 : þ : 3 s2 sðs þ 2Þ 3 sðs þ 3Þ and the second and third terms can be expressed in simple partial fractions so that x ¼ . . . . . . . . . . . . x ¼ 2 1 1 1 4 1 4 1 : þ þ : : 3 s2 s s þ 2 9 s 9 s þ 3 69 which can now be simplified into 2 1 5 1 1 4 1 : : þ : 3 s2 9 s s þ 2 9 s þ 3 ; x ¼ ............ x ¼ x¼ 2 5 4 t þ e2t e3t 3 9 9 There are times when a quadratic coefficient of x cannot be expressed in simple linear factors. In that case, we merely complete the square converting the expression into ðs kÞ2 a2 . Let us see such an example. 70 78 Programme 2 Example 4 Solve x€ 2x_ þ 10x ¼ e2t , given that at t ¼ 0, x ¼ 0 and x_ ¼ 1. We find the expression for x as before. x ¼ . . . . . . . . . . . . 71 x ¼ s1 ðs 2Þðs2 2s þ 10Þ Because s2 x sx0 x1 2ðsx x0 Þ þ 10x ¼ x0 ¼ 0; x1 ¼ 1 1 s2 1 s2 1 s1 ¼ ; s2 2s þ 10 x ¼ 1 þ s2 s2 s1 ; x ¼ ðs 2Þðs2 2s þ 10Þ ; s2 x 1 2sx þ 10x ¼ Expressing this in partial fractions x ¼ . . . . . . . . . . . . 72 Evaluate the coefficients. x ¼ 1 1 s 10 2 10 s 2 s 2s þ 10 Because s1 A Bs þ C þ ðs 2Þðs2 2s þ 10Þ ðs 2Þ s2 2s þ 10 ; s 1 ¼ A s2 2s þ 10 þ ðs 2ÞðBs þ CÞ 1 10 Put ðs 2Þ ¼ 0, i.e. s ¼ 2 1 ¼ Að4 4 þ 10Þ ; A¼ 2 s 0¼AþB ; B¼ ½CT 1 10 1 ¼ 10A 2C ; 2C ¼ 2 ; C¼1 1 1 s 10 ; x¼ 10 s 2 s2 2s þ 10 Now we have to find the inverse transforms to obtain x. The first 1 s 10 term is easy enough, but what of 2 ? The denominator will not s2 s 2s þ 10 factorise into simple linear factors; therefore we complete the square in the denominator and write it as s2 s 10 s 10 ¼ 2s þ 10 ðs 1Þ2 þ 9 79 Laplace transforms 1 and then we improve this still further and write it in the form s1 are quite happy with this, for 2 is merely s2 ðs 1Þ 9 . We ðs 1Þ2 þ 9 s with s replaced by þ9 ðs 1Þ þ 9 ðs 1Þ, which indicates an extra factor et in the final function of t (Theorem 1). ( ) 1 1 s1 9 þ So x ¼ 10 s 2 ðs 1Þ2 þ 9 ðs 1Þ2 þ 9 ; x ¼ ............ x¼ 73 1 2t e et cos 3t þ 3et sin 3t 10 Just try one more like this one Example 5 74 t Solve x€ þ x_ þ x ¼ e given that at t ¼ 0, x ¼ 0 and x_ ¼ 1. We find the expression for x as before. x ¼ . . . . . . . . . . . . x ¼ Because s2 x sx0 x1 þ ðsx x0 Þ þ x ¼ s2 x 1 þ sx þ x ¼ 75 sþ2 ðs þ 1Þðs2 þ s þ 1Þ 1 where x0 ¼ 0 and x1 ¼ 1 so that sþ1 1 sþ1 therefore x s2 þ s þ 1 ¼ 1 þ 1 sþ2 ¼ sþ1 sþ1 giving x ¼ sþ2 ðs þ 1Þðs2 þ s þ 1Þ Expressing this in partial fractions x ¼ . . . . . . . . . . . . Evaluate the coefficients 80 Programme 2 76 x ¼ 1 s1 s þ 1 s2 þ s þ 1 Because x ¼ sþ2 A Bs þ C þ ¼ ðs þ 1Þðs2 þ s þ 1Þ s þ 1 s2 þ s þ 1 so that s þ 2 ¼ A s2 þ s þ 1 þ ðBs þ CÞðs þ 1Þ Put s þ 1 ¼ 0, that is s ¼ 1 then 1 ¼ Að1 1 þ 1Þ so that A ¼ 1 2 s 0¼AþB so that B ¼ 1 ½CT 2 ¼ A þ C so that C ¼ 1 Therefore x ¼ 1 s1 s þ 1 s2 þ s þ 1 Completing the squares in the second term gives s1 ¼ ............ s2 þ s þ 1 77 s1 ¼ s2 þ s þ 1 s þ 12 2 pffiffi s þ 12 þ 23 pffiffiffi pffiffi3 3 2 2 2 pffiffi s þ 12 þ 23 Because s1 s1 ¼ 2 s2 þ s þ 1 s þ 1 þ3 2 4 s þ 12 32 ¼ 2 pffiffi s þ 12 þ 23 s þ 12 ¼ 2 pffiffi s þ 12 þ 23 so that x ¼ . . . . . . . . . . . . 2 pffiffiffi pffiffi3 3 2 2 2 pffiffi s þ 12 þ 23 2 2 81 Laplace transforms 1 s þ 12 1 x ¼ 2 pffiffi sþ1 s þ 12 þ 23 pffiffiffi pffiffi3 3 2 þ 2 2 pffiffi s þ 12 þ 23 78 2 and so x ¼ . . . . . . . . . . . . x¼e t e t=2 pffiffiffi pffiffiffi 3t pffiffiffi t=2 3t þ 3e cos sin 2 2 79 Before we leave this topic, the same general approach can be employed for solving simultaneous differential equations. Let us see an example in the next frame 80 Simultaneous differential equations Example 1 Solve the pair of simultaneous equations y_ x ¼ et x_ þ y ¼ et given that at t ¼ 0, x ¼ 0 and y ¼ 0. (a) We first express both equations in Laplace transforms. 1 s1 1 ðsx x0 Þ þ y ¼ sþ1 ðsy y0 Þ x ¼ (b) Then we insert the initial conditions, x0 ¼ 0 and y0 ¼ 0. 9 1 > > ; sy x ¼ = s1 1 > > ; sx þ y ¼ sþ1 ð1Þ (c) We now solve these for x and y by the normal algebraic method. Eliminating y we have 1 s1 s sy þ s2 x ¼ sþ1 2 1 s2 2s 1 ¼ ; s2 þ 1 x ¼ s þ 1 s 1 ðs þ 1Þðs 1Þ sy x ¼ ; x ¼ s2 2s 1 ðs 1Þðs þ 1Þðs2 þ 1Þ Representing this in partial fractions gives . . . . . . . . . . . . 82 Programme 2 81 1 1 1 1 s 1 : þ þ x ¼ : 2 s 1 2 s þ 1 s2 þ 1 s2 þ 1 Because s2 2s 1 A B Cs þ D þ þ ðs 1Þðs þ 1Þðs2 þ 1Þ s 1 s þ 1 s2 þ 1 ; s2 2s 1 ¼ Aðs þ 1Þ s2 þ 1 þ Bðs 1Þ s2 þ 1 x ¼ þ ðs 1Þðs þ 1ÞðCs þ DÞ Putting s ¼ 1 and s ¼ 1 gives A ¼ 12 and B ¼ 12. Comparing coefficients of s3 and the constant terms gives C ¼ 1 and D ¼ 1. 1 1 1 1 sþ1 : : þ 2 2 s1 2 sþ1 s þ1 ; x ¼ ............ ; x ¼ 82 x ¼ 12 et 12 et þ cos t þ sin t We now revert to equations (1) and eliminate x to obtain y and hence y, in the same way. Do this on your own. y ¼ ............ 83 y ¼ 12 et þ 12 et cos t þ sin t Here is the working. 9 s > > s2 y sx ¼ s 1= 1 > > ; y þ sx ¼ sþ1 s 1 s2 þ 2s 1 þ ¼ ; s2 þ 1 y ¼ s 1 s þ 1 ðs 1Þðs þ 1Þ s2 þ 2s 1 A B Cs þ D þ þ ðs 1Þðs þ 1Þðs2 þ 1Þ s 1 s þ 1 s2 þ 1 ; s2 þ 2s 1 ¼ Aðs þ 1Þ s2 þ 1 þ Bðs 1Þ s2 þ 1 ; y ¼ þ ðs 1Þðs þ 1ÞðCs þ DÞ Putting s ¼ 1 and s ¼ 1 gives A ¼ 12 and B ¼ 12. Equating coefficients of s3 and the constant terms gives C ¼ 1 and D ¼ 1. 1 1 1 1 s 1 : þ : þ 2 s 1 2 s þ 1 s2 þ 1 s2 þ 1 1 1 ; y ¼ et þ et cos t þ sin t 2 2 ; y ¼ 83 Laplace transforms 1 So the results are 1 t e þ et þ sin t þ cos t ¼ sin t þ cos t cosh t 2 1 t y ¼ e þ et þ sin t cos t ¼ sin t cos t þ cosh t 2 ; x ¼ sin t þ cos t cosh t; y ¼ sin t cos t þ cosh t x¼ Simultaneous equations are all solved in much the same way. Here is another. Example 2 Solve the equations 2y_ 6y þ 3x ¼ 0 3x_ 3x 2y ¼ 0 given that at t ¼ 0, x ¼ 1 and y ¼ 3. Expressing these in Laplace transforms, we have ............ ............ 2ðsy y0 Þ 6y þ 3x ¼ 0 3ðsx x0 Þ 3x 2y ¼ 0 84 Then we insert the initial conditions and simplify, obtaining ............ ............ 3x þ ð2s 6Þy ¼ 6 ð3s 3Þx 2y ¼ 3 (1) (2) (a) To find x 3x þ ð2s 6Þy ¼ 6 ðs 3Þð3s 3Þx ð2s 6Þy ¼ 3ðs 3Þ ½ðs 3Þð3s 3Þ þ 3x ¼ 3s 9 þ 6 ; 3s2 12s þ 12 x ¼ 3s 3 2 s 4s þ 4 x ¼ s 1 (1) (2) ðs 3Þ Adding, ; x ¼ s1 ðs 2Þ2 A B Aðs 2Þ þ B þ ¼ s 2 ðs 2Þ2 ðs 2Þ2 ; s 1 ¼ Aðs 2Þ þ B ; x ¼ 1 1 þ s 2 ðs 2Þ2 giving A ¼ 1 and B ¼ 1 ; x ¼ e2t þ te2t (b) Going back to equations (1) and (2), we can find y. y ¼ ............ 85 84 Programme 2 86 y ¼ 12 6e2t þ 3te2t Because, eliminating x we get ( ) ( ) 6s 9 1 A B 1 Aðs 2Þ þ B ¼ þ y ¼ 2 2ðs 2Þ2 2 s 2 ðs 2Þ2 ðs 2Þ2 ; 6s 9 ¼ Aðs 2Þ þ B ( ) 1 6 3 þ ; y ¼ 2 s 2 ðs 2Þ2 ; A ¼ 6; B¼3 ; y ¼ 12 6e2t þ 3te2t Simultaneous second-order equations are solved in like manner. Again, with all these solutions it is a worthwhile exercise to substitute the solution back into the differential equation to verify that the solution is correct. 87 Example 3 If x and y are functions of t, solve the equations x€ þ 2x y ¼ 0 y€ þ 2y x ¼ 0 given that at t ¼ 0, x0 ¼ 4; y0 ¼ 2; x1 ¼ 0; y1 ¼ 0. 2 We start off as usual with s x sx0 x1 þ 2x y ¼ 0 2 s y sy0 y1 þ 2y x ¼ 0 and Inserting the initial conditions, we have s2 x 4s þ 2x y ¼ 0 s2 y 2s þ 2y x ¼ 0 Simplifying these we can eliminate y to obtain x and hence x. x ¼ ............ 88 x ¼ 3 cos t þ cos pffiffiffi 3t Because 2 s þ 2 x y ¼ 4s x þ s2 þ 2 y ¼ 2s Eliminating y and simplifying gives x ¼ 4s3 þ 10s ðs2 þ 1Þðs2 þ 3Þ 4s3 þ 10s As þ B Cs þ D þ 2 s þ3 þ 1Þðs2 þ 3Þ s2 þ 1 3 2 2 ; 4s þ 10s ¼ s þ 3 ðAs þ BÞ þ s þ 1 ðCs þ DÞ ; x ¼ ðs2 ð1Þ ð2Þ 85 Laplace transforms 1 Equating coefficients of like powers of s 3 4¼AþC ; AþC¼4 s ½CT 0 ¼ 3B þ D ; 3B þ D ¼ 0 Putting s ¼ 1, 14 ¼ 4A þ 4B þ 2C ¼ 2D Putting s ¼ 1 14 ¼ 4A þ 4B 2C þ 2D ; 2A þ 2B þ C þ D ¼ 7 ; 2A 2B þ C D ¼ 7 Putting C ¼ 4 A and D ¼ 3B in the last two leads to A ¼ ............; C ¼ ............; B ¼ ............; D ¼ ............ A ¼ 3; B ¼ 0; C ¼ 1; D ¼ 0 89 3s s þ s2 þ 1 s2 þ 3 ; x ¼ ............ ; x ¼ x ¼ 3 cos t þ cos pffiffiffi 3t To find y we could return to equations (1) and (2) and repeat the process, eliminating x so as to obtain y and hence y. But always keep an eye on the original equations, the first of which is x€ þ 2x y ¼ 0 Therefore, in this particular case, y ¼ x€ þ 2x. So all we have to do is to differentiate x twice and substitute pffiffiffi x ¼ 3 cos t þ cos 3t pffiffiffi pffiffiffi x_ ¼ 3 sin t 3 sin 3t pffiffiffi x€ ¼ 3 cos t 3 cos 3t pffiffiffi pffiffiffi ; y ¼ 3 cos t 3 cos 3t þ 6 cos t þ 2 cos 3t pffiffiffi ; y ¼ 3 cos t cos 3t which is a good deal quicker. So, as we have seen, the method of solving differential equations by Laplace transforms follows a general routine. (a) Express the equation in Laplace transforms (b) Insert the initial conditions (c) Simplify to obtain the transform of the solution (d) Rewrite the final transform in partial fractions (e) Determine the inverse transforms and, by now, you are fully aware of the importance of partial fractions! 90 86 Programme 2 That brings us to the end of this particular Programme. We shall continue our study of Laplace transforms in the next Programme. Meanwhile, be sure you are familiar with the items listed in the Revision summary that follows, and respond to the questions in the Can you? checklist. You will then have no difficulty with the Test exercise and the Further problems provide additional practice. Revision summary 2 91 1 Laplace transform Lff ðtÞg ¼ ð1 f ðtÞest dt ¼ FðsÞ. 0 2 Table of transforms f ðtÞ a eat sin at cos at sinh at cosh at tn 3 Lff ðtÞg ¼ FðsÞ a s 1 sa a 2 s þ a2 s s2 þ a2 a 2 s a2 s s2 a2 n! snþ1 (n a positive integer) Linearity of the Laplace transform (a) The transform of a sum (or difference) of expressions is the sum (or difference) of the individual transforms. That is Lff ðtÞ gðtÞg ¼ Lff ðtÞg LfgðtÞg (b) The transform of an expression that is multiplied by a constant is the constant multiplied by the transform of the expression. That is Lfkf ðtÞg ¼ kLff ðtÞg 4 Theorem 1 First shift theorem If Lff ðtÞg ¼ FðsÞ, then L eat f ðtÞ ¼ Fðs þ aÞ. 87 Laplace transforms 1 5 Theorem 2 Multiplying by t If Lff ðtÞg ¼ FðsÞ, then Lftf ðtÞg ¼ 6 Theorem 3 7 Inverse transform d fFðsÞg. ds Dividing by t ð1 f ðtÞ If Lff ðtÞg ¼ FðsÞ, then L ¼ FðÞ d t s f ðtÞ provided that Lim exists. t t!0 If Lff ðtÞg ¼ FðsÞ, then L1 fFðsÞg ¼ f ðtÞ. 8 Rules of partial fractions (a) The numerator must be of lower degree than the denominator. If not, divide out. (b) Factorise the denominator into its prime factors. A where A is a constant (c) A linear factor ðs þ aÞ gives a partial fraction sþa to be determined. A B (d) A repeated factor ðs þ aÞ2 gives þ . s þ a ðs þ aÞ2 A B C þ þ . (e) Similarly ðs þ aÞ3 gives s þ a ðs þ aÞ2 ðs þ aÞ3 Ps þ Q (f) A quadratic factor s2 þ ps þ q gives 2 . s þ ps þ q 2 (g) A repeated quadratic factor s2 þ ps þ q gives s2 9 Ps þ Q Rs þ T . þ þ ps þ q ðs2 þ ps þ qÞ2 The ‘cover up’ rule The ‘cover up’ rule often enables the values of the constant coefficients to be written down almost on sight. However, this method only works when the denominator of the original fraction has non-repeated, linear factors. 88 Programme 2 10 Table of inverse transforms f ðtÞ FðsÞ a s 1 sþa n! snþ1 1 sn a s2 þ a 2 s s2 þ a 2 a s2 a 2 s s2 a 2 a eat tn (n a positive integer) t n1 ðn 1Þ! sin at cos at sinh at cosh at By the first shift theorem If FðsÞ is the Laplace transform of f ðtÞ then Fðs þ aÞ is the Laplace transform of eat f ðtÞ. 11 Laplace transforms of derivatives Lfxg ¼ x dx _ ¼ sx_ x0 ¼ Lfxg L dt ( ) d2 x ¼ Lfx€g ¼ sx sx0 x1 etc. L dt 2 where x0 ¼ value of x at t ¼ 0 x1 ¼ value of 12 dx at t ¼ 0, etc. dt Solution of differential equations (a) Rewrite the equation in terms of Laplace transforms. (b) Insert the given initial conditions. (c) Rearrange the equation algebraically to give the transform of the solution. (d) Express the transform in standard forms by partial fractions. (e) Determine the inverse transforms to obtain the particular solution. 89 Laplace transforms 1 13 Simultaneous differential equations Convert the simultaneous differential equations into simultaneous algebraic equations by taking the Laplace transform of each equation in turn. Insert the initial values. Solve the simultaneous algebraic equations in the usual manner and take the inverse Laplace transform of the algebraic solutions to find the solutions to the simultaneous differential equations. Can you? 92 Checklist 2 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Obtain the Laplace transforms of simple standard expressions? Yes No Frames 1 to 14 . Use the first shift theorem to find the Laplace transform of a simple expression multiplied by an exponential? Yes No 15 to 17 . Find the Laplace transform of a simple expression multiplied or divided by a variable? Yes No 18 to 26 . Use partial fractions to find the inverse Laplace transform? Yes No 27 to 42 . Use the ‘cover up’ rule? Yes 43 to 46 . Use the Laplace transforms of derivatives to solve differential equations? Yes No 47 to 79 . Use the Laplace transform to solve simultaneous differential equations? Yes No 80 to 90 No 90 Programme 2 Test exercise 2 93 1 Determine the Laplace transforms of the following functions. (a) 3e4t 5e4t (b) sin 4t þ cos 4t (c) t 3 þ 2t 2 t þ 4 et e2t . (d) e2t cos 5t (e) t sin 3t (f) t 2 Determine the inverse transforms of the following. s5 s2 þ 3s 7 (b) (a) ðs 3Þðs 4Þ ðs 1Þðs2 þ 2Þ (c) 3 s2 3s 4 2 ðs 3Þðs 1Þ 2s2 6s 1 . ðs 3Þðs2 2s þ 5Þ Solve the following equations by Laplace transforms. dx þ 3x ¼ e2t (a) given that x ¼ 2 when t ¼ 0 dt (b) 3x_ 6x ¼ sin 2t given that x ¼ 1 when t ¼ 0 (c) x€ 7x_ þ 12x ¼ 2 given that at t ¼ 0, x ¼ 1 and x_ ¼ 5 (d) x€ 2x_ þ x ¼ tet 4 (d) given that at t ¼ 0, x ¼ 1 and x_ ¼ 0. Solve the following pair of simultaneous equations where x and y are functions of t and given that at t ¼ 0, x ¼ 4 and y ¼ 1. x_ þ y_ þ x þ 2y ¼ e3t x_ þ 3x þ 5y ¼ 5e2t Further problems 2 94 1 Determine the Laplace transforms of the following functions. (a) e4t cos 2t (b) t sin 2t (c) t 3 þ 4t 2 þ 5 sinh 2t . (d) e3t t 2 þ 4 (e) t 2 cos t (f) t 2 Determine the inverse transforms of the following. 2s 6 5s 8 s2 2s þ 3 (b) (c) (a) ðs 2Þðs 4Þ sðs 4Þ ðs 2Þ3 2 11s s s5 (e) 2 (f) 2 . (d) 2 2 ðs 2Þðs þ 2s þ 2Þ s þ 4s þ 20 ðs þ 1Þðs þ 4Þ 91 Laplace transforms 1 In Questions 3 to 11, solve the equations by Laplace transforms. 3 x_ 4x ¼ 8 at t ¼ 0, x ¼ 2. 4 3x_ 4x ¼ sin 2t at t ¼ 0, x ¼ 13 . 5 x€ 2x_ þ x ¼ 2ðt þ sin tÞ at t ¼ 0, x ¼ 6, x_ ¼ 5. 6 7 8 9 10 11 x€ 6x_ þ 8x ¼ e3t x€ þ 9x ¼ cos 2t at t ¼ 0, x ¼ 0, x_ ¼ 2. at t ¼ 0, x ¼ 1, x_ ¼ 3. x€ 2x_ þ 5x ¼ e2t x€ þ 4x_ þ 4x ¼ t 2 þ e2t at t ¼ 0, x ¼ 0, x_ ¼ 1. at t ¼ 0, x ¼ 12 , x_ ¼ 0. x€ þ 8x_ þ 32x ¼ 32 sin 4t at t ¼ 0, x ¼ x_ ¼ 0. x€ þ 25x ¼ 10ðcos 5t 2 sin 5tÞ at t ¼ 0, x ¼ 1, x_ ¼ 2. In Questions 12 to 17, solve the pairs of simultaneous equations by Laplace transforms. ) 12 y_ þ 3x ¼ e2t at t ¼ 0, x ¼ y ¼ 0. x_ 3y ¼ e2t 13 4x_ 2y_ þ 10x 5y ¼ 0 at t ¼ 0, y ¼ 4, x ¼ 2. y_ 18x þ 15y ¼ 10 14 x_ 2y_ 3x þ 6y ¼ 12 at t ¼ 0, x ¼ 12, y ¼ 8. 3y_ þ 5x þ 2y ¼ 16 15 2x_ þ 3y_ þ 7x ¼ 14t þ 7 at t ¼ 0, x ¼ y ¼ 0. 5x_ 3y_ þ 4x þ 6y ¼ 14t 14 16 2x_ þ 2x þ 3y_ þ 6y ¼ 56et 3et at t ¼ 0, x ¼ 8, y ¼ 3. x_ 2x y_ 3y ¼ 21et 7et ) 17 x€ y€ þ x y ¼ 5e2t at t ¼ 0, x ¼ 1, y ¼ 2, x_ ¼ 0. 2x_ y_ þ y ¼ 0 18 Find an expression for x in terms of t, given that y€ x_ þ 2x ¼ 10 sin 2t y_ þ 2y þ x ¼ 0 and when t ¼ 0, x ¼ y ¼ 0. 19 If x€ þ 8x þ 2y ¼ 24 cos 4t and 4y€ þ 2x þ 5y ¼ 0 and at t ¼ 0, x ¼ y ¼ 0, x_ ¼ 1, y_ ¼ 2, determine an expression for y in terms of t. 20 Solve completely, the pair of simultaneous equations 5x€ þ 12y€ þ 6x ¼ 0 5x€ þ 16y€ þ 6y ¼ 0 given that, at t ¼ 0, x ¼ 74, y ¼ 1, x_ ¼ 0, y_ ¼ 0. Programme 3 Frames 1 to 50 Laplace transforms 2 Learning outcomes When you have completed this Programme you will be able to: . Use the Heaviside unit step function to ‘switch’ expressions on and off . Obtain the Laplace transform of expressions involving the Heaviside unit step function . Solve linear, constant coefficient ordinary differential equations with piecewise continuous right-hand sides . Understand what is meant by the convolution of two functions and use the convolution theorem to find the inverse transform of a product of transforms 92 93 Laplace transforms 2 Introduction In the previous Programme, we dealt with the Laplace transforms of continuous functions of t. In practical applications, it is convenient to have a function which, in effect, ‘switches on’ or ‘switches off’ a given term at predescribed values of t. This we can do with the Heaviside unit step function. 1 Heaviside unit step function Consider a function that maintains a zero value for all values of t up to t ¼ c and a unit value for t ¼ c and all values of t c. f(t) c f ðtÞ ¼ 0 for t < c f ðtÞ ¼ 1 for t c t This function is the Heaviside unit step function and is denoted by f ðtÞ ¼ uðt cÞ where the c indicates the value of t at which the function changes from a value of 0 to a value of 1. Thus, the function f (t) t is denoted by f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ uðt 4Þ Similarly, the graph of f ðtÞ ¼ 2uðt 3Þ is ............ 2 94 Programme 3 3 f (t) t So uðt cÞ has just two values for t < c, uðt cÞ ¼ . . . . . . . . . . . . for t c, uðt cÞ ¼ . . . . . . . . . . . . 4 t < c, uðt cÞ ¼ 0; t c, uðt cÞ ¼ 1 Unit step at the origin u(t) If the unit step occurs at the origin, then c ¼ 0 and f ðtÞ ¼ uðt cÞ becomes f ðtÞ ¼ uðtÞ t i.e. uðtÞ ¼ 0 for t < 0 uðtÞ ¼ 1 for t 0. Effect of the unit step function The graph of f ðtÞ ¼ t 2 is, of course, as shown. f (t) t Remembering the definition of uðt cÞ, the graph of f ðtÞ ¼ uðt 2Þ t 2 is ............ 95 Laplace transforms 2 5 f (t) t For t < 2, uðt 2Þ ¼ 0 ; uðt 2Þ t 2 ¼ 0 t 2 ¼ 0 t 2, uðt 2Þ ¼ 1 ; uðt 2Þ t 2 ¼ 1 t 2 ¼ t 2 So the function uðt 2Þ suppresses the function t 2 for all values of t up to t ¼ 2 and ‘switches on’ the function t 2 at t ¼ 2. Now we can sketch the graphs of the following functions. (a) f ðtÞ ¼ sin t for 0 < t < 2 (b) f ðtÞ ¼ uðt =4Þ sin t for 0 < t < 2. These give . . . . . . . . . . . . and . . . . . . . . . . . . 6 f(t) t – f(t) – t 4 That is, the graph of f ðtÞ ¼ uðt =4Þ sin t is the graph of f ðtÞ ¼ sin t but suppressed for all values prior to t ¼ =4. If we sketch the graph of f ðtÞ ¼ sinðt =4Þ we have f(t) t 4 Since uðt cÞ has the effect of suppressing a function for t < c, then the graph of f ðtÞ ¼ uðt =4Þ sinðt =4Þ is ............ 96 Programme 3 7 f (t) t 4 That is, the graph of f ðtÞ ¼ uðt =4Þ sinðt =4Þ is the graph of f ðtÞ ¼ sin t ðt > 0Þ, shifted =4 units along the t-axis. In general, the graph of f ðtÞ ¼ uðt cÞ sinðt cÞ is the graph of f ðtÞ ¼ sin t ðt > 0Þ, shifted along the t-axis through an interval of c units. Similarly, for t > 0, sketch the graphs of (a) f ðtÞ ¼ et (b) f ðtÞ ¼ uðt cÞ et (c) f ðtÞ ¼ uðt cÞ eðtcÞ (d) f ðtÞ ¼ et fuðt 1Þ uðt 2Þg. Arrange the graphs under each other to show the important differences. 8 (a) 1 f(t) f (t) = e –t 0 (b) t 1 f(t) 0 (c) f (t) = u (t – c).e –t c t 1 f (t) = u (t – c).e –(t – c) f(t) 0 (d) c t 1 f (t) = e –t {u (t – 1) – u (t – 2)} f(t) 0 In In In In 1 2 (a), we have the graph of f ðtÞ ¼ et (b), the same graph is suppressed prior to t ¼ c (c), the graph of f ðtÞ ¼ et is shifted c units along the t-axis (d), the graph of f ðtÞ ¼ et is turned on at t ¼ 1 and off at t ¼ 2. t 97 Laplace transforms 2 Laplace transform of u (t – c ) Lfuðt cÞg ¼ Because Lfuðt cÞg ¼ ecs s ð1 est uðt cÞ dt 0 but est uðt cÞ ¼ so that Lfuðt cÞg ¼ 0 for 0 < t < c est for t c ð1 est uðt cÞ dt ¼ 0 est ¼ s ð1 est dt c 1 ¼ c esc s Therefore, the Laplace transform of the unit step at the origin is LfuðtÞg ¼ . . . . . . . . . . . . 1 s 9 Because c ¼ 0. ecs s 1 LfuðtÞg ¼ . s Lfuðt cÞg ¼ So and Also from the definition of uðtÞ: Lð1Þ ¼ Lf1 uðtÞg LðtÞ ¼ Lft uðtÞg Lff ðtÞg ¼ Lff ðtÞ uðtÞg Make a note of these results: we shall be using them As we have seen, the unit step function uðt cÞ is often combined with other functions of t, so we now consider the Laplace transform of uðt cÞ f ðt cÞ. 10 98 Programme 3 Laplace transform of u ( t – c ) . f ( t – c ) (the second shift theorem) Lfuðt cÞ ðf ðt cÞg ¼ ecs Lff ðtÞg ¼ ecs FðsÞ Because ð1 est uðt cÞ f ðt cÞ dt Lfuðt cÞ f ðt cÞg ¼ 0 0 for 0 < t < c but est uðt cÞ ¼ est for t c so that Lfuðt cÞ f ðt cÞg ¼ ð1 est f ðt cÞ dt c We now make the substitution t c ¼ v so that t ¼ c þ v and dt ¼ dv. Also for the limits, when t ¼ c, v ¼ 0 and when t ! 1, v ! 1. Therefore ð1 esðcþvÞ f ðvÞ dv Lfuðt cÞ f ðt cÞg ¼ 0 ð1 cs ¼e esv f ðvÞ dv ð1 Now 0 esv f ðvÞ dv has exactly the same value as 0 ð1 est f ðtÞ dt which 0 is, of course, the Laplace transform of f ðtÞ. Therefore Lfuðt cÞ f ðt cÞg ¼ ecs Lff ðtÞg ¼ ecs FðsÞ 11 Lfuðt cÞ f ðt cÞg ¼ ecs FðsÞ where FðsÞ ¼ Lff ðtÞg n o So L uðt 4Þ ðt 4Þ2 ¼ e4s FðsÞ where FðsÞ ¼ L t 2 2! 2e4s ¼ e4s 3 ¼ 3 s s Note that FðsÞ is the transform of t 2 and not of ðt 4Þ2 . In the same way: Lfuðt 3Þ sinðt 3Þg ¼ . . . . . . . . . . . . 12 e3s þ1 s2 Because Lfuðt 3Þ sinðt 3Þg ¼ e3s FðsÞ 1 3s ; Lfuðt 3Þ sinðt 3Þg ¼ e s2 þ 1 where FðsÞ ¼ Lfsin tg ¼ 1 s2 þ 1 99 Laplace transforms 2 So now do these in the same way. n o (a) L uðt 2Þ ðt 2Þ3 ¼ ............ (b) Lfuðt 1Þ sin 3ðt 1Þg (c) L uðt 5Þ eðt5Þ ¼ ............ ¼ ............ (d) Lfuðt =2Þ cos 2ðt =2Þg ¼ . . . . . . . . . . . . Here they are n o (a) L uðt 2Þ ðt 2Þ3 ¼ e2s FðsÞ where FðsÞ ¼ L t 3 3! 6e2s ¼ e2s 4 ¼ 4 s s 13 (b) Lfuðt 1Þ sin 3ðt 1Þg ¼ es FðsÞ where FðsÞ ¼ Lfsin 3tg 3 3es s ¼ 2 ¼e 2 s þ9 s þ9 n o (c) L uðt 5Þ eðt5Þ ¼ e5s FðsÞ where FðsÞ ¼ L et 1 e5s 5s ¼ ¼e s1 s1 (d) Lfuðt =2Þ cos 2ðt =2Þg ¼ es=2 FðsÞ where FðsÞ ¼ Lfcos 2tg s s es=2 ¼ 2 ¼ es=2 2 s þ4 s þ4 So Lfuðt cÞ f ðt cÞg ¼ ecs FðsÞ where FðsÞ ¼ Lff ðtÞg. Written in reverse, this becomes If FðsÞ ¼ Lff ðtÞg, then ecs FðsÞ ¼ Lfuðt cÞ f ðt cÞg where c is real and positive. This is known as the second shift theorem. Make a note of it: then we will use it If FðsÞ ¼ Lff ðtÞg, then ecs FðsÞ ¼ Lfuðt cÞ f ðt cÞg This is useful in finding inverse transforms, as we shall now see. 14 100 Programme 3 Example 1 Find the function whose transform is The numerator corresponds to ecs uðt 4Þ. 1 Then 2 ¼ FðsÞ ¼ Lftg ; f ðtÞ ¼ t. s 4s e ¼ uðt 4Þ ðt 4Þ ; L1 s2 e4s . s2 where c ¼ 4 and therefore indicates Remember that in writing the final result, f ðtÞ is replaced by ............ 15 f ðt cÞ Example 2 1 Determine L 6e2s . s2 þ 4 The numerator contains e2s and therefore indicates . . . . . . . . . . . . 16 uðt 2Þ The remainder of the transform, i.e. ; 17 6 2 , can be written as 3 s2 þ 4 s2 þ 4 6 ¼ FðsÞ ¼ Lf. . . . . . . . . . . .g s2 þ 4 Lf3 sin 2tg ; L1 18 6e2s s2 þ 4 ¼ ............ 3uðt 2Þ sin 2ðt 2Þ Because 2s 6e L1 2 s þ4 ¼ uðt 2Þ f ðt 2Þ where f ðtÞ ¼ L1 s2 6 þ4 ¼ uðt 2Þ 3 sin 2ðt 2Þ Example 3 Determine L1 s es . s2 þ 9 This, in similar manner, is . . . . . . . . . . . . 101 Laplace transforms 2 19 uðt 1Þ cos 3ðt 1Þ Because the numerator contains es which indicates uðt 1Þ. s ¼ FðsÞ ¼ Lfcos 3tg Also 2 s þ9 ; f ðtÞ ¼ cos 3t ; f ðt 1Þ ¼ cos 3ðt 1Þ. s es ; L1 2 ¼ uðt 1Þ cos 3ðt 1Þ s þ9 Remember that, having obtained f ðtÞ, the result contains f ðt cÞ. Here is a short exercise by way of practice. Exercise Determine the inverse transforms of the following. 2e5s s3 3e2s (b) 2 s 1 8e4s (c) 2 s þ4 2s e3s s2 16 5es (e) s s es=2 (f) s2 þ 2 (a) (d) Results – all very straightforward. 20 (a) uðt 5Þ ðt 5Þ2 (b) 3uðt 2Þ sinhðt 2Þ (c) 4uðt 4Þ sin 2ðt 4Þ (d) 2uðt 3Þ cosh 4ðt 3Þ (e) 5uðt 1Þ pffiffiffi (f) uðt 1=2Þ cos 2ðt 1=2Þ. Before looking at a more interesting example, let us collect our results together as far as we have gone. The main points are (a) uðt cÞ ¼ 0 0<t<c 21 ) ¼1 tc 9 ecs > > (b) Lfuðt cÞg ¼ = s > 1 > ; LfuðtÞg ¼ s (c) Lfuðt cÞ f ðt cÞg ¼ ecs FðsÞ where FðsÞ ¼ Lff ðtÞg (d) If FðsÞ ¼ Lff ðtÞg, then e cs FðsÞ ¼ Lfuðt cÞg f ðt cÞg Now let us apply these to some further examples. ð1Þ ð2Þ ð3Þ ð4Þ 102 Programme 3 Example 1 Determine the expression f ðtÞ for which Lff ðtÞg ¼ 3 4es 5e2s 2 þ 2 s s s We take each term in turn and find its inverse transform. 3 1 (a) L1 ¼ 3L1 ¼ 3 i.e. 3uðtÞ s s s 4e ¼ uðt 1Þ 4ðt 1Þ (b) L1 s2 2s 5e ¼ ............ (c) L1 s2 22 uðt 2Þ 5ðt 2Þ 3 So we have L s s 4e L1 s2 2s 5e L1 s2 1 ¼ 3uðtÞ ¼ uðt 1Þ 4ðt 1Þ ¼ uðt 2Þ 5ðt 2Þ ; FðtÞ ¼ 3uðtÞ uðt 1Þ 4ðt 1Þ þ uðt 2Þ 5ðt 2Þ To sketch the graph of f ðtÞ we consider the values of the function within the three sections 0 < t < 1, 1 < t < 2, and 2 < t. Between t ¼ 0 and t ¼ 1, f ðtÞ ¼ . . . . . . . . . . . . 23 f ðtÞ ¼ 3 Because in this interval, uðtÞ ¼ 1, but uðt 1Þ ¼ 0 and uðt 2Þ ¼ 0. In the same way, between t ¼ 1 and t ¼ 2, f ðtÞ ¼ . . . . . . . . . . . . 24 f ðtÞ ¼ 7 4t Because between t ¼ 1 and t ¼ 2, uðtÞ ¼ 1, uðt 1Þ ¼ 1, but uðt 2Þ ¼ 0. ; f ðtÞ ¼ 3 4ðt 1Þ þ 0 ¼ 3 4t þ 4 ¼ 7 4t Similarly, for t > 2, f ðtÞ ¼ . . . . . . . . . . . . 103 Laplace transforms 2 25 f ðtÞ ¼ t 3 Because for t > 2, uðtÞ ¼ 1, uðt 1Þ ¼ 1 and uðt 2Þ ¼ 1 ; f ðtÞ ¼ 3 4ðt 1Þ þ 5ðt 2Þ ¼ 3 4t þ 4 þ 5t 10 ¼ t 3 So, collecting the results together, we have for 0 < t < 1, f ðtÞ ¼ 3 1 < t < 2, f ðtÞ ¼ 7 4t f ðtÞ ¼ t 3 2 < t, ðt ¼ 1, f ðtÞ ¼ 3; t ¼ 2, f ðtÞ ¼ 1Þ ðt ¼ 2, f ðtÞ ¼ 1; t ¼ 3, f ðtÞ ¼ 0Þ Using these facts we can sketch the graph of f ðtÞ, which is . . . . . . . . . . . . 26 f(t) t Here is another. Example 2 Determine the expression f ðtÞ ¼ L1 2 3es 3e3s þ 2 2 s s s and sketch the graph of f ðtÞ. First we express the inverse transform of each term in terms of the unit step function. This gives . . . . . . . . . . . . L1 2 s ¼ 2uðtÞ; s 3e L1 s2 3s 3e L1 s2 ¼ uðt 1Þ 3ðt 1Þ ¼ uðt 3Þ 3ðt 3Þ ; f ðtÞ ¼ 2uðtÞ þ uðt 1Þ 3ðt 1Þ uðt 3Þ 3ðt 3Þ So there are ‘break points’, i.e. changes of function, at t ¼ 1 and t ¼ 3, and we investigate f ðtÞ within the three intervals. 0<t<1 f ðtÞ ¼ . . . . . . . . . . . . 1<t<3 f ðtÞ ¼ . . . . . . . . . . . . 3<t f ðtÞ ¼ . . . . . . . . . . . . 27 104 28 Programme 3 0 < t < 1, f ðtÞ ¼ 2; 1 < t < 3, f ðtÞ ¼ 3t 1; 3 < t, f ðtÞ ¼ 8 Because with uðtÞ ¼ 1, but uðt 1Þ ¼ uðt 3Þ ¼ 0 0 < t < 1, ; f ðtÞ ¼ 2 uðtÞ ¼ 1, uðt 1Þ ¼ 1, but uðt 3Þ ¼ 0 1 < t < 3, ; f ðtÞ ¼ 3t 1 ; f ðtÞ ¼ 2 þ 3ðt 1Þ ¼ 3t 1 uðtÞ ¼ 1, uðt 1Þ ¼ 1, uðt 3Þ ¼ 1 3 < t, ; f ðtÞ ¼ 2 þ 3t 3 3t þ 9 ; f ðtÞ ¼ 8 Therefore, the graph of f ðtÞ is . . . . . . . . . . . . 29 f(t) t Between the break points, f ðtÞ ¼ 3t 1 t ¼ 1, f ðtÞ ¼ 2 t ¼ 3, f ðtÞ ¼ 8 Now move on for the next example 30 Example 3 If f ðtÞ ¼ L1 1 e2s 1 þ e4s s2 , determine f ðtÞ and sketch the graph of the function. Although at first sight this looks more complicated, we simply multiply out the numerator and proceed as before. 2s þ e4s e6s 1 1 e f ðtÞ ¼ L s2 2s 1 e e4s e6s ¼ L1 2 2 þ 2 2 s s s s We now write down the inverse transform of each term in terms of the unit function, so that f ðtÞ ¼ . . . . . . . . . . . . 105 Laplace transforms 2 f ðtÞ ¼ uðtÞ t uðt 2Þ ðt 2Þ þ uðt 4Þ ðt 4Þ uðt 6Þ ðt 6Þ 31 and we can see there are break points at t ¼ 2, t ¼ 4, t ¼ 6. For 0 < t < 2, f ðtÞ ¼ t 0 þ 0 0 f ðtÞ ¼ t 2 < t < 4, f ðtÞ ¼ t ðt 2Þ þ 0 0 f ðtÞ ¼ 2 4 < t < 6, f ðtÞ ¼ t ðt 2Þ þ ðt 4Þ 0 f ðtÞ ¼ t 2 6 < t, f ðtÞ ¼ t ðt 2Þ þ ðt 4Þ ðt 6Þ f ðtÞ ¼ 4 The second and fourth components are constant, but before sketching the graph of the function, we check the values of f ðtÞ ¼ t and f ðtÞ ¼ t 2 at the relevant break points. f ðtÞ ¼ t. At t ¼ 0, f ðtÞ ¼ 0; at t ¼ 2, f ðtÞ ¼ 2 f ðtÞ ¼ t 2. At t ¼ 4, f ðtÞ ¼ 2; at t ¼ 6, f ðtÞ ¼ 4. So the graph of the function is . . . . . . . . . . . . 32 f(t) It is always wise to calculate the function values at break points, since discontinuities, or jumps, sometimes occur. On to the next frame Now for one in reverse. Example 4 A function f ðtÞ is defined by f ðtÞ ¼ 4 for 0 < t < 2 ¼ 2t 3 for 2 < t. Sketch the graph of the function and determine its Laplace transform. We see that for t ¼ 0 to t ¼ 2, f ðtÞ ¼ 4. 33 106 Programme 3 34 f Notice the discontinuity at t ¼ 2. Expressing the function in unit step form: f ðtÞ ¼ 4uðtÞ 4uðt 2Þ þ uðt 2Þ ð2t 3Þ Note that the second term cancels f ðtÞ ¼ 4 at t ¼ 2 and that the third switches on f ðtÞ ¼ 2t 3 at t ¼ 2. Before we can express this in Laplace transforms, ð2t 3Þ in the third term must be written as a function of ðt 2Þ to correspond to uðt 2Þ. Therefore, we write 2t 3 as 2ðt 2Þ þ 1. Then f ðtÞ ¼ 4uðtÞ 4uðt 2Þ þ uðt 2Þ f2ðt 2Þ þ 1g ¼ 4uðtÞ 4uðt 2Þ þ uðt 2Þ 2ðt 2Þ þ uðt 2Þ ¼ 4uðtÞ 3uðt 2Þ þ uðt 2Þ 2ðt 2Þ ; Lff ðtÞg ¼ . . . . . . . . . . . . 35 Lff ðtÞg ¼ 4 3e2s 2e2s þ 2 s s s Here is one for you to work through in much the same way. Example 5 A function is defined by f ðtÞ ¼ 6 ¼ 8 2t ¼4 0<t<1 1<t<3 3 < t. Sketch the graph and find the Laplace transform of the function. 107 Laplace transforms 2 36 f Expressing this in unit step form we have f ðtÞ ¼ 6uðtÞ 6uðt 1Þ þ uðt 1Þ ð8 2tÞ uðt 3Þ ð8 2tÞ þ uðt 3Þ 4 where the second term switches off the first function f ðtÞ ¼ 6 at t ¼ 1 and the third term switches on the second function f ðtÞ ¼ 8 2t, which in turn is switched off by the fourth term at t ¼ 3 and replaced by f ðtÞ ¼ 4 in the fifth term. Before we can write down the transforms of the third and fourth terms, we must express f ðtÞ ¼ 8 2t in terms of ðt 1Þ and ðt 3Þ respectively. 8 2t ¼ 6 þ 2 2t ¼ 6 2ðt 1Þ 8 2t ¼ 2 þ 6 2t ¼ 2 2ðt 3Þ ; f ðtÞ ¼ 6uðtÞ 6uðt 1Þ þ uðt 1Þ f6 2ðt 1Þg uðt 3Þ f2 2ðt 3Þg þ 4uðt 3Þ ¼ 6uðtÞ 6uðt 1Þ þ 6uðt 1Þ uðt 1Þ 2ðt 1Þ 2uðt 3Þ þ uðt 3Þ 2ðt 3Þ þ 4uðt 3Þ which simplifies finally to f ðtÞ ¼ . . . . . . . . . . . . 37 f ðtÞ ¼ 6uðtÞ uðt 1Þ 2ðt 1Þ þ uðt 3Þ 2ðt 3Þ þ 2uðt 3Þ from which Lff ðtÞg ¼ . . . . . . . . . . . . Lff ðtÞg ¼ 38 6 2es 2e3s 2e3s 2 þ 2 þ s s s s Note that, in building up the function in unit step form (a) to ‘switch on’ a function f ðtÞ at t ¼ c, we add the term uðt cÞ f ðt cÞ (b) to ‘switch off’ a function f ðtÞ at t ¼ c, we subtract uðt cÞ f ðt cÞ. Next we shall look at some differential equations that use what we have done so far in the Programme. Next frame 108 Programme 3 Differential equations involving the unit step function 39 We can now use the work on the unit step function to solve constant coefficient differential equations with a piecewise continuous right-hand side. To solve the differential equation: f 0 ðtÞ þ 3f ðtÞ ¼ uðt 1Þ where f ð0Þ ¼ 0 we start by taking the Laplace transform of both sides to find that Lff 0 ðtÞ þ 3f ðtÞg ¼ Lfuðt 1Þg that is sFðsÞ f ð0Þ þ 3FðsÞ ¼ es so that s es es giving FðsÞ ¼ s sðs þ 3Þ es 1 1 ¼ 3 s sþ3 s e es L1 Therefore f ðtÞ ¼ L1 3s 3ðs þ 3Þ s s 1 1 e e L L1 ¼ 3 s sþ3 i 1h ¼ uðt 1Þ uðt 1Þe3ðt1Þ 3 uðt 1Þ 1 e3ðt1Þ ¼ 3 ðs þ 3ÞFðsÞ ¼ 1.2 input u(t – 1) 1.0 0.8 0.6 response f(t) 0.4 0.2 0 –0.2 0 2 4 6 t You try one. The solution of the equation 5f 0 ðtÞ f ðtÞ ¼ uðt 4Þ where f ð0Þ ¼ 0 is: f ðtÞ ¼ . . . . . . . . . . . . Next frame 109 Laplace transforms 2 ðt 4Þ 1 uðt 4Þ exp 5 40 Because Taking the Laplace transform of both sides we find that Lf5f 0 ðtÞ f ðtÞg ¼ Lfuðt 4Þg that is 5sFðsÞ f ð0Þ FðsÞ ¼ e4s so that s e4s e4s giving FðsÞ ¼ s sð5s 1Þ 5 1 4s ¼e 5s 1 s 4s 4s 1 5e 1 e Therefore f ðtÞ ¼ L L 5s 1 s " ( ) # 4s 4s 1 e 1 e ¼ L L s s 15 h i ¼ uðt 4Þeðt4Þ=5 uðt 4Þ expðt 4Þ ¼ uðt 4Þ 1 5 ð5s 1ÞFðsÞ ¼ 1.2 input u(t – 4) 1.0 0.8 0.6 0.4 response f(t) 0.2 0 –0.2 0 2 4 6 t And another. The solution of the equation f 00 ðtÞ þ 5f 0 ðtÞ þ 6f ðtÞ ¼ uðt 2Þ sinðt 2Þ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 is: f ðtÞ ¼ . . . . . . . . . . . . Next frame 110 41 Programme 3 f ðtÞ ¼ i uðt 2Þ h sinðt 2Þ cosðt 2Þ þ 2e2ðt2Þ e3ðt2Þ 10 Because Taking the Laplace transform of both sides we find that Lff 00 ðtÞ þ 5f 0 ðtÞ þ 6f ðtÞg ¼ Lfuðt 2Þ sinðt 2Þg that is s2 FðsÞ sf ð0Þ f 0 ð0Þ þ 5FðsÞ 5f ð0Þ þ 6FðsÞ ¼ so that ðs2 þ 5s þ 6ÞFðsÞ ¼ giving FðsÞ ¼ ðs2 e2s þ1 s2 e2s s2 þ 1 e2s þ 1Þðs þ 2Þðs þ 3Þ 1 e2s 1 se2s 1 e2s 1 e2s þ þ 10 s2 þ 1 10 s2 þ 1 5 s þ 2 10 s þ 3 i uðt 2Þ h Therefore f ðtÞ ¼ sinðt 2Þ þ cosðt 2Þ þ 2e2ðt2Þ e3ðt2Þ 10 ¼ 1.5 input 1.0 response 0.5 0 –0.5 0 2 4 6 t –1.0 –1.5 And just one more to make sure. The differential equation f 00 ðtÞ þ f 0 ðtÞ þ f ðtÞ ¼ gðtÞ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 and where the graph of gðtÞ is as shown 1 1 2 Next frame 111 Laplace transforms 2 ! pffiffiffi pffiffiffi 1 3ðt 1Þ 3ðt 1Þ pffiffiffi sin f ðtÞ ¼ uðt 1Þe 1 cos 2 2 3 ! pffiffiffi pffiffiffi 1 3ðt 2Þ 3ðt 2Þ ðt2Þ=2 pffiffiffi sin 1 cos uðt 2Þe 2 2 3 ðt1Þ=2 Because The graph is a single square pulse of width 1 between t ¼ 1 and t ¼ 2 This is described algebraically as the unit step turned on at t ¼ 1 and then turned off at t ¼ 2 That is gðtÞ ¼ uðt 1Þ uðt 2Þ the differential equation then becomes f 00 ðtÞ þ f 0 ðtÞ þ f ðtÞ ¼ uðt 1Þ uðt 2Þ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 Taking the Laplace transform of both sides we find that Lff 00 ðtÞ þ f 0 ðtÞ þ f ðtÞg ¼ Lfuðt 1Þ uðt 2Þg that is s2 FðsÞ sf ð0Þ f 0 ð0Þ þ FðsÞ f ð0Þ þ FðsÞ ¼ ðs2 þ s þ 1ÞFðsÞ ¼ es e2s es e2s so that s s 1 1 and so FðsÞ ¼ es e2s s sðs2 þ s þ 1Þ Separating the factors in the denominator we find that 1 sþ1 FðsÞ ¼ es e2s s s2 þ s þ 1 8 9 > > <1 = sþ1 s 2s ¼ e e pffiffi2 > > 2 :s s þ 12 þ 23 ; 9 8 > > = <1 1 1 sþ2 2 ¼ es e2s pffiffi2 pffiffi2 > > 2 2 :s s þ 12 þ 23 s þ 12 þ 23 ; 8 9 pffiffi > > <1 = 3 1 sþ2 1 2 p ffiffiffi ¼ es e2s pffiffi2 pffiffi2 > 2 3 s þ 1 2þ 3 > :s ; s þ 12 þ 23 2 2 Taking the inverse Laplace transforms we see that ! pffiffiffi pffiffiffi 1 3ðt 1Þ 3ðt 1Þ ðt1Þ=2 pffiffiffi sin f ðtÞ ¼ uðt 1Þe 1 cos 2 2 3 ! pffiffiffi pffiffiffi 1 3ðt 2Þ 3ðt 2Þ ðt2Þ=2 pffiffiffi sin 1 cos uðt 2Þe 2 2 3 Graph overleaf 42 112 Programme 3 1.2 input 1.0 0.8 0.6 0.4 response 0.2 0 0 –0.2 2 4 6 t And now a slight digression. In Frame 13 of the previous Programme it was stated that two Laplace transforms must not be multiplied together to form the transform of a product of expressions. This is because the product of two transformations is not the transform of a product of expressions but rather, the transform of the convolution of the two expressions. We shall now see what is meant by convolution. Move to the next frame Convolution 43 The convolution of the two functions f ðtÞ and gðtÞ is defined as: ð1 cðtÞ ¼ f ðtÞ gðtÞ ¼ f ðxÞgðt xÞ dx x¼1 where the denotes the operation of convolution. You will notice that this is a function t as denoted by cðtÞ and exactly what is happening here does require an explanation. We wish to end up with a function depending on the variable t so we set the variable of integration to be x. Next, against the same set of coordinate axes we draw the graphs of the functions f ðxÞ and gðxÞ where the graph of gðxÞ is simply the graph of gðxÞ reflected in the vertical axis. If this reflected graph were to be moved along the horizontal axis to the point where its leading edge was at x ¼ t then in its new position it would be the graph of gðt xÞ. As an example consider the rectangular function: ( f ðtÞ ¼ 2 0t3 0 otherwise 2 with graph 3 113 Laplace transforms 2 and the rectangular function: ( gðtÞ ¼ 1 0t2 with graph 0 otherwise 1 2 The second function reflected in the vertical is: ( gðtÞ ¼ 1 2 t 0 with graph 0 otherwise 1 –2 The convolution integral is then: ð1 f ðxÞgðt xÞdx cðtÞ ¼ f ðtÞ gðtÞ ¼ x¼1 with the following graphical configuration: f(x) g(t – x) x=t x In this position t < 0 and f ðxÞgðt xÞ ¼ 0 for all values of x. This means that the value of the convolution integral is also zero. c(t) t As t becomes positive the non-zero parts of the graphs of gðt xÞ and f ðxÞ overlap and so the integrand f ðxÞgðt xÞ becomes non-zero as does the resulting convolution integral. As t increases further so the range of values of x for which the integrand is non-zero also increases. c (t) f (x) g(t – x) x=t Here, cðtÞ ¼ f ðtÞ gðtÞ ðt 2 1 dx ¼ x¼0 ¼ 2t x provided 0 t 2 t 114 Programme 3 Eventually the range of values of x for which the integrand is non-zero reaches its maximum and remains there as t increases further. During this stage the convolution integral assumes a constant value. Here, cðtÞ ¼ f ðtÞ gðtÞ ðt 2 1 dx ¼ provided 2 t 3 x¼t2 ¼ 2t 2ðt 2Þ ¼4 c (t) f (x) g (t – x) x=t x t At some point the range of values of x for which the integrand is non-zero decreases. Here, cðtÞ ¼ f ðtÞ gðtÞ ð3 2 1 dx ¼ provided 3 t x¼t2 ¼ 6 2ðt 2Þ ¼ 10 2t c(t) f (x) g(t – x) x=t x t Finally, as t increases further there comes a point where f ðxÞgðt xÞ ¼ 0 for all values of x greater than t. c(t) f (x) g (t – x) x=t In conclusion: 8 2t > > > <4 cðtÞ ¼ > 10 2t > > : 0 0t<2 2t<3 3t5 otherwise x t 115 Laplace transforms 2 Now you try one. Given the rectangular function 1 1 t 1 f ðtÞ ¼ 0 otherwise and the truncated exponential t e 0t1 gðtÞ ¼ 0 otherwise g (t) the convolution of these two functions is t cðtÞ ¼ . . . . . . . . . . . . The answer is in the next frame 8 1 e1t > > > < 1 e1 cðtÞ ¼ > e1t e1 > > : 0 44 1 t < 0 0t<1 1t2 otherwise Because: cðtÞ ¼ f ðtÞ gðtÞ ð1 ¼ f ðxÞgðt xÞ dx x¼1 ¼ ð1 f ðxÞgðt xÞ dx x¼1 The evaluation of this integral is separated into three parts and it is always advisable to draw small sketches of the configurations to help in deciding the limits of the integrals. (a) 1 t < 0: cðtÞ ¼ f ðtÞ gðtÞ ð1 f ðxÞgðt xÞ dx ¼ x¼1 ðt 1 eðtxÞ dx ¼ x¼1 ðt ¼ ext dx f(x) g (t – x) x¼1 h it ¼ ext 0 1 1t ¼e e 1t ¼1e x=t x 116 Programme 3 (b) 0 t < 1: cðtÞ ¼ f ðtÞ gðtÞ ð1 ¼ f ðxÞgðt xÞ dx x¼1 ðt 1 eðtxÞ dx ¼ x¼t1 ðt ¼ ext dx f (x) g(t – x) x¼t1 it h ¼ ext t1 ¼ e0 et1t ¼ 1 e1 x x=t (c) 1 t < 2: cðtÞ ¼ f ðtÞ gðtÞ ð1 ¼ f ðxÞgðt xÞ dx f (x) g (t – x) x¼1 ¼ ¼ ð1 ðtxÞ 1e x¼t1 ð1 ext dx x¼t1 i1 h xt ¼ e dx t1 ¼ e1t e1 8 1 e1t > > > < 1 e1 Giving cðtÞ ¼ > e1t e1 > > : 0 x=t x 1 t < 0 0t<1 1t2 otherwise with the following graph. c (t) 0.6 0.5 0.4 0.3 0.2 0.1 0 –1 –0.5 0 0.5 1 1.5 2 t We can now return to our main theme, namely the properties of the Laplace transform Next frame 117 Laplace transforms 2 The convolution theorem The convolution theorem concerns the product of a pair of Laplace transforms. Given two functions f ðtÞ and gðtÞ where f ðtÞ ¼ 0 and gðtÞ ¼ 0 when t < 0 and their respective Laplace transforms FðsÞ and GðsÞ then the Laplace transform of the convolution of f ðtÞ and gðtÞ is equal to the product of their Laplace transforms. That is: Lff ðtÞ gðtÞg ¼ Lff ðtÞgLfgðtÞg Because f ðtÞ and gðtÞ are interchangeable in this equation, that is: Lff ðtÞ gðtÞg ¼ Lff ðtÞgLfgðtÞg ¼ LfgðtÞgLff ðtÞg ¼ LfgðtÞ f ðtÞg we see that convolution is a commutative operation; f ðtÞ gðtÞ ¼ gðtÞ f ðtÞ. Furthermore, because f ðtÞ ¼ gðtÞ ¼ 0 when t < 0 this means that ð t ð t FðsÞGðsÞ ¼ L f ðt xÞgðxÞ dx ¼ L gðt xÞf ðxÞ dx 0 0 Notice that the upper limit is t because f ðt xÞ ¼ 0 and gðt xÞ ¼ 0 for t x < 0, that is for x > t. For example using the convolution theorem to 1 1 , we see that if evaluate L s2 ðs 3Þ 1 FðsÞ ¼ 2 ¼ Lff ðtÞg then f ðtÞ ¼ t and s 1 ¼ LfgðtÞg then gðtÞ ¼ e3t . GðsÞ ¼ s3 As a result FðsÞGðsÞ ¼ Lff ðtÞ gðtÞg so that 1 1 ¼ f ðtÞ gðtÞ L s2 ðs 3Þ ð1 ¼ f ðxÞgðt xÞ dx 1 Since convolution is a commutative operation the choice is made of which expression is represented by f ðxÞ and which by gðt xÞ so as to result in the simplest integral. Therefore ðt 1 1 ¼ xe3ðtxÞ dx L s2 ðs 3Þ 0 ðt t 1 e3t 3t xe3x dx ¼ þ ¼e 3 9 9 0 which is in agreement with the partial fraction procedure. Now you try one. ( ) s 1 ¼ ............ By the convolution theorem L ðs2 þ 1Þ2 The answer is in the next frame 45 118 Programme 3 46 t sin t 4 Because: s ðs2 2 þ 1Þ ¼ 1 s ðs2 þ 1Þ ðs2 þ 1Þ ¼ FðsÞ GðsÞ so that f ðtÞ ¼ sin t and gðtÞ ¼ cos t L1 fFðsÞGðsÞg ¼ f ðtÞ gðtÞ ðt ¼ sinðt xÞ cos x dx 0 ðt ¼ fðsin t cos x sin x cos tÞ cos xg dx 0 ðt ðt ¼ sin t cos2 x dx cos t sin x cos x dx 0 0 ðt ðt cos 2x þ 1 sin 2x dx cos t dx ¼ sin t 2 2 0 0 t sin t sin 2x cos t cos 2x t ¼ þx 2 2 2 2 0 0 1 ¼ fsin t sin 2t þ t sin t þ cos t cos 2t cos t g 4 t sin t ¼ 4 You have now reached the end of this Programme and this brings you to the Revision summary and the Can you? checklist. Following that is the Test exercise. Work through this at your own pace. A set of Further problems provides additional valuable practice. Revision summary 3 47 1 Heaviside unit step function: uðt cÞ 1 f ðtÞ ¼ 0 ¼1 f(t) 0 c t 0<t<c c<t 119 Laplace transforms 2 2 Suppression and shift u(t – c) . f (t) f(t) a a t t c u(t – c) . f (t – c) a c 3 Laplace transform of uðt cÞ Lfuðt cÞg ¼ 4 t ecs ; s LfuðtÞg ¼ 1 . s Laplace transform of uðt cÞ f ðt cÞ Lfuðt cÞ f ðt cÞg ¼ ecs FðsÞ where FðsÞ ¼ Lff ðtÞg. 5 Second shift theorem If FðsÞ ¼ Lff ðtÞg, then ecs FðsÞ ¼ Lfuðt cÞ f ðt cÞg where c is real and positive. 6 Convolution theorem The convolution of two expressions f ðtÞ and gðtÞ is denoted as f ðtÞ gðtÞ and is defined as the definite integral ðt f ðt xÞgðxÞ dx f ðtÞ gðtÞ ¼ x¼1 Also, convolution is a commutative operation. That is ð1 ð1 f ðt xÞgðxÞ dx ¼ gðtÞ f ðtÞ ¼ gðt xÞf ðxÞ dx f ðtÞ gðtÞ ¼ x¼1 x¼1 The convolution theorem states that if Lff ðtÞg ¼ FðsÞ and LfgðtÞg ¼ GðsÞ then Lff ðtÞ gðtÞg ¼ Lff ðtÞgLfgðtÞg ¼ FðsÞGðsÞ so that L1 fFðsÞGðsÞg ¼ f ðtÞ gðtÞ 120 Programme 3 Can you? 48 Checklist 3 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Use the Heaviside unit step function to ‘switch’ expressions on and off? Yes No 1 to 8 . Obtain the Laplace transform of expressions involving the Heaviside unit step function? Yes No 8 to 38 . Solve linear, constant coefficient ordinary differential equations with piecewise continuous right-hand sides Yes No 39 to 42 43 to 46 . Understand what is meant by the convolution of two functions and use the convolution theorem to find the inverse transform of a product of transforms Yes No Test exercise 3 49 1 In each of the following cases, sketch the graph of the function and find its Laplace transform. (a) f ðtÞ ¼ 3t 0t<2 ¼6 (b) f ðtÞ ¼ e2t ¼0 (c) f ðtÞ ¼ t 2 0t<3 3t 0t<2 ¼2 2t<3 ¼4 3t (d) f ðtÞ ¼ sin 2t ¼0 2 2t 0t< t. Determine the function f ðtÞ whose transform FðsÞ is 1 2 5es þ 8e3s . FðsÞ ¼ s Sketch the graph of the function between t ¼ 0 and t ¼ 4. 121 Laplace transforms 2 3 If f ðtÞ ¼ L 1 1 þ 3e2s 1 e3s s2 , determine f ðtÞ and sketch the graph of the function. 4 Determine the function f ðtÞ for which 2ð1 es Þ f ðtÞ ¼ L1 . sð1 e3s Þ Sketch the waveform and express the function in analytical form. 5 6 Solve the differential equation f 0 ðtÞ f ðtÞ ¼ uðtÞet uðt 1Þeðt1Þ where f ð0Þ ¼ 0 ( ) 2s 1 where f ð0Þ ¼ 0. Use the convolution theorem to find L ðs2 16Þ2 Further problems 3 1 2 3 1 3s þ 2e2s 2e5s , determine f ðtÞ. s2 ð1 es Þ 1 þ e2s 1 , find f ðtÞ in terms of the unit step function. If f ðtÞ ¼ L s2 If Lff ðtÞg ¼ A function f ðtÞ is defined by f ðtÞ ¼ 4 0t<3 ¼ 2t þ 1 3 t. Sketch the graph of the function and determine its Laplace transform. 4 Express in terms of the Heaviside unit step function (a) f ðtÞ ¼ t 2 0t<3 ¼ 5t 3 t. (b) f ðtÞ ¼ cos t ¼ cos 2t ¼ cos 3t 0t< t < 2 2 t. 5 A function f ðtÞ is defined by f ðtÞ ¼ 0 0t<2 ¼tþ1 2t<3 ¼0 3 t. Determine Lff ðtÞg. 6 A function f ðtÞ is defined by f ðtÞ ¼ t 2 0t<2 ¼4 2t<5 ¼0 5 t. Determine (a) the function in terms of the unit step function (b) the Laplace transform of f ðtÞ. 50 122 Programme 3 7 Solve the differential equations (a) f 0 ðtÞ þ 2f ðtÞ ¼ tuðtÞ ðt 1Þuðt 1Þ where f ð0Þ ¼ 0 (b) f 00 ðtÞ 4f 0 ðtÞ þ 4f ðtÞ ¼ uðtÞ uðt 2Þ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 (c) f 00 ðtÞ f ðtÞ ¼ uðtÞ sin 3t ðt 4Þ sin 3ðt 4Þ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 (d) f 00 ðtÞ þ f 0 ðtÞ þ f ðtÞ ¼ ðt 1Þuðt 1Þ where f 0 ð0Þ ¼ f ð0Þ ¼ 0 8 Determine the inverse Laplace transforms of each of the following 1 1 1 (a) (b) 3 2 (c) 2 2 ðs þ 1Þðs þ 1Þ s ðs 2Þ ð3s 4Þð2s2 1Þ 9 Show that (a) uðtÞ uðtÞ ¼ tuðtÞ (b) tuðtÞ et uðtÞ ¼ ðet t 1Þ where uðtÞ is the Heaviside unit step function. Programme 4 Frames 1 to 70 Laplace transforms 3 Learning outcomes When you have completed this Programme you will be able to: . Find the Laplace transforms of periodic functions . Obtain the inverse Laplace transforms of transforms of periodic functions . Describe and use the unit impulse to evaluate integrals . Obtain the Laplace transform of the unit impulse . Use the Laplace transform to solve differential equations involving the unit impulse . Solve the equation and describe the behaviour of an harmonic oscillator 123 124 Programme 4 Laplace transforms of periodic functions 1 Periodic functions Let f ðtÞ represent a periodic function with period T so that f ðt þ nTÞ ¼ f ðtÞ with a graph of the following form f(t) f (t1+T)=f(t1) t1 0 T t1 +T 2T 3T t If we describe the first cycle by fðtÞ then f ðtÞ for 0 t < T f ðtÞ ¼ 0 otherwise The second cycle is identical to the first cycle except that it is shifted by T units of time along the t-axis. Therefore the second cycle can be described in terms of the Heaviside unit step function as fðt TÞuðt TÞ. That is f ðtÞ for T t < 2T fðt TÞuðt TÞ ¼ 0 otherwise By this reasoning the periodic function f ðtÞ is represented by f ðtÞ ¼ fðtÞuðtÞ þ . . . . . . . . . . . . 2 f ðtÞ ¼ fðtÞuðtÞ þ fðt TÞuðt TÞ þ fðt 2TÞuðt 2TÞ þ Because uðtÞ switches on fðtÞ at time t ¼ 0, uðt TÞ switches on fðt TÞ at time t ¼ T and uðt 2TÞ switches on fðt 2TÞ at time t ¼ 2T, etc. Consider now the Laplace transform of fðtÞ. By definition ðT ð1 est fðtÞ dt ¼ est f ðtÞ dt ¼ FðsÞ LffðtÞg ¼ 0 0 because for t > T, fðtÞ ¼ 0 and so the semi-infinite integral becomes an integral just over the period of f ðtÞ. Using the second shift theorem (see Frame 10 of Programme 3), the Laplace transform of f ðtÞ is Lff ðtÞg ¼ LffðtÞuðtÞg þ L fðt TÞuðt TÞ þ L fðt 2TÞuðt 2TÞ þ That is Lff ðtÞg ¼ . . . . . . . . . . . . 125 Laplace transforms 3 þ esT FðsÞ þ e2sT FðsÞ þ Lff ðtÞg ¼ FðsÞ 3 Because L fðtÞuðt cÞ ¼ esc L fðtÞ by the second shift theorem. and write Lff ðtÞg as We can factor out FðsÞ Lff ðtÞg ¼ 1 þ esT þ e2sT þ . . . FðsÞ Now, do you remember the series 1 þ x þ x2 þ x3 þ . . .? This can be written in closed form as 1 þ x þ x2 þ x3 þ . . . ¼ . . . . . . . . . . . . 1 þ x þ x2 þ x3 þ . . . ¼ 4 1 1x Because 1 ¼ ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ . . . 1x either by the binomial theorem or by performing the long division. So, if we let x ¼ esT then 1 þ esT þ e2sT þ . . . ¼ . . . . . . . . . . . . 1 þ esT þ e2sT þ . . . ¼ 5 1 1 esT And so the Laplace transform of f ðtÞ is given as ¼ . . . . . . . . . . . . where FðsÞ ¼ ............ Lff ðtÞg ¼ 1 þ esT þ e2sT þ . . . FðsÞ Lff ðtÞg ¼ 1 where FðsÞ ¼ FðsÞ ð1 esT Þ ðT est f ðtÞ dt 0 Note that we integrate est f ðtÞ over one cycle, that is from t ¼ 0 to t ¼ T, and not from t ¼ 0 to t ¼ 1 as we did previously. This is an important result. Make a note of it – then we shall apply it 6 126 Programme 4 Example 1 Find the Laplace transform of the function f ðtÞ defined by f ðtÞ ¼ 3 0 < t < 2 f ðt þ 4Þ ¼ f ðtÞ ¼0 2<t<4 f(t) t The expression for Lff ðtÞg is ............ 7 Lff ðtÞg ¼ (do not evaluate it yet) 1 1 e4s ð4 est f ðtÞ dt 0 Because the period ¼ 4, i.e. T ¼ 4. The function f ðtÞ ¼ 3 for 0 < t < 2 and f ðtÞ ¼ 0 for 2 < t < 4. ð2 1 ; Lff ðtÞg ¼ est 3 dt ¼ . . . . . . . . . . . . 1 e4s 0 8 Lff ðtÞg ¼ 3 sð1 þ e2s Þ Because 2s st 2 3 e 3 e 1 ¼ 1 e4s s 0 1 e4s s s 2s 3 1e 3 ¼ ¼ 1 e4s sð1 þ e2s Þ s Lff ðtÞg ¼ That is all there is to it. Now for another, so move on 9 Example 2 Find the Laplace transform of the periodic function defined by f ðtÞ ¼ t=2 f ðt þ 3Þ ¼ f ðtÞ 0<t<3 f (t) t 127 Laplace transforms 3 Because in this case, period ¼ 3, i.e. T ¼ 3. ðT 1 est f ðtÞ dt ; Lff ðtÞg ¼ 1 eTs 0 ð3 1 t dt est ¼ 1 e3s 0 2 ð3 ; 2 1 e3s Lff ðtÞg ¼ t est dt 0 Integrating by parts and simplifying the result gives Lff ðtÞg ¼ . . . . . . . . . . . . Lff ðtÞg ¼ 1 3s 1 2s2 e3s 1 Because 2 1 e3s Lff ðtÞg ¼ ð3 test dt 0 est ¼ t s 3 þ 0 1 s ð3 est dt 0 3e3s 1 est þ ¼ s s s 3 0 3e3s e3s 1 2 þ 2 ¼ s s s 1 3se3s ; Lff ðtÞg ¼ 2 1 2s 1 e3s 1 3s ¼ 2 1 3s 2s e 1 Example 3 Sketch the graph of the function f ðtÞ ¼ et f ðt þ 5Þ ¼ f ðtÞ 0<t<5 and determine its Laplace transform. First we sketch the graph of f ðtÞ, which is . . . . . . . . . . . . 10 128 11 Programme 4 f(t) t Clearly, period ¼ 5 ; T¼5 ðT 1 est f ðtÞ dt Lff ðtÞg ¼ 1 eTs 0 Lff ðtÞg ¼ . . . . . . . . . . . . gives Complete the working 12 Lff ðtÞg ¼ 1 e5ðs1Þ ðs 1Þð1 e5s Þ Because ð5 1 est et dt 1 e5s 0 ð5 ; 1 e5s Lff ðtÞg ¼ eðs1Þt dt Lff ðtÞg ¼ 0 ¼ ; Lff ðtÞg¼ o eðs1Þt 1 n ¼ 1 e5ðs1Þ ðs 1Þ 0 s 1 5 1 e5ðs1Þ ðs 1Þð1 e5s Þ All very straightforward. Example 4 Determine the Laplace transform of the half-wave rectifier output waveform defined by f ðtÞ ¼ 8 sin t 0 < t < f ðt þ 2Þ ¼ f ðtÞ ¼0 < t < 2 f(t) t Here the period is 2 i.e. T ¼ 2. In general, for a periodic function of period T Lff ðtÞg ¼ . . . . . . . . . . . . 129 Laplace transforms 3 Lff ðtÞg ¼ 1 1 eTs ðT est f ðtÞ dt 13 0 So, for this example ð 2 1 est f ðtÞ dt 1 e2s 0 ð ; 1 e2s Lff ðtÞg ¼ est 8 sin t dt Lff ðtÞg ¼ 0 Writing sin t as the imaginary part of e jt , i.e. sin t ie jt , ð 1 e2s Lff ðtÞg ¼ 8i est e jt dt ð0 ¼ 8i eðsjÞt dt 0 and this you can finish off in the usual manner, giving Lff ðtÞg ¼ . . . . . . . . . . . . Lff ðtÞg ¼ 8 ðs2 þ 1Þð1 es Þ Because ð 1 e2s Lff ðtÞg ¼ 8 i eðsjÞt dt 0 ðsjÞt e ¼8i ðs jÞ 0 i 8 h ðsjÞ ¼i 1 e sj 1 1 es e j ¼8i sj But e j ¼ cos þ j sin ¼ 1. 1 ð1 þ es Þ sj sþj 1 þ es ¼8i 2 ð1 þ es Þ ¼ 8 s þ1 s2 þ 1 1 1 þ es ; Lff ðtÞg ¼ 8 2s 1e s2 þ 1 8 ¼ ð1 es Þðs2 þ 1Þ ; 1 e2s Lff ðtÞg ¼ 8 i Now let us consider the corresponding inverse transforms when periodic functions are involved. 14 130 15 Programme 4 Inverse transforms Finding inverse transforms of functions of s which are transforms of periodic functions is not as straightforward as in earlier examples, for the transforms result from integration over one cycle and not from t ¼ 0 to t ¼ 1. Hence we have no simple table of inverse transforms upon which to draw. However, all difficulties can be surmounted and an example will show how we deal with this particular problem. Example 1 Determine the inverse transform 2 þ e2s 3es L1 sð1 e2s Þ 1 e2s in the denominator, 1 which suggests a periodic function of period 2 units, i.e. where 1 eTs T ¼ 2. The key to the solution is to write 1 e2s in the denominator as 1 1 e2s in the numerator and to expand this as a binomial series. The first thing we see is the factor We remember that ð1 xÞ1 ¼ . . . . . . . . . . . . 16 ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ . . . 1 2 3 ¼ 1 þ e2s þ e2s þ e2s þ . . . ; 1 e2s ¼ 1 þ e2s þ e4s þ e6s þ . . . 1 2 þ e2s 3es 1 ¼ 2 þ e2s 3es 1 e2s ; Lff ðtÞg ¼ 2s s sð1 e Þ ¼ 1 2 þ e2s 3es 1 þ e2s þ e4s þ e6s þ e8s þ . . . s We now multiply the second series by each term of the first in collect up like terms, giving 8 þ2e4s þ2e6s þ2e2s >2 1< Lff ðtÞg¼ þ e4s þ e6s þ e2s s> : s 3s 5s 3e 3e 3e ¼ ............ turn and 9 ...> = ... > ; ... 131 Laplace transforms 3 Lff ðtÞg ¼ 17 1 2 3es þ 3e2s 3e3s þ 3e4s 3e5s þ . . . s Each term is of the form ecs , so, expressing f ðtÞ in unit step form, we have s f ðtÞ ¼ . . . . . . . . . . . . 18 f ðtÞ ¼ 2uðtÞ 3uðt 1Þ þ 3uðt 2Þ 3uðt 3Þ þ 3uðt 4Þ . . . and from this we can sketch the waveform, which is therefore ............ 19 2 f (t) 0 –1 1 2 3 4 5 6 t We can finally define this periodic function in analytical terms. f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ 2 0<t<1 ¼ 1 1 < t < 2 f ðt þ 2Þ ¼ f ðtÞ 20 The key to the whole process is thus to . . . . . . . . . . . . express 1 eTs in the denominator as 1 1 eTs in the numerator and to expand this as a binomial series. We do this by making use of the basic series ð1 xÞ1 ¼ . . . . . . . . . . . . 21 132 Programme 4 22 ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ x4 þ . . . Example 2 1 Determine L 3ð1 es Þ sð1 e3s Þ and sketch the resulting waveform of f ðtÞ. 1 3 ð1 es Þ 1 e3s s ¼ ............ Lff ðtÞg ¼ 23 Lff ðtÞg ¼ (next step) 3 ð1 es Þ 1 þ e3s þ e6s þ e9s þ . . . s which multiplied out gives 3 1 es þ e3s e4s þ e6s e7s þ . . . s 3 3es 3e3s 3e4s 3e6s ¼ þ þ ... s s s s s Lff ðtÞg ¼ And in unit step form, this gives f ðtÞ ¼ . . . . . . . . . . . . 24 f ðtÞ ¼ 3uðtÞ 3uðt 1Þ þ 3uðt 3Þ 3uðt 4Þ þ . . . The waveform is thus . . . . . . . . . . . . 25 f(t) t f ðtÞ ¼ 3 0<t<1 f ðtÞ ¼ 0 1<t<3 f ðt þ 3Þ ¼ f ðtÞ And now, one more. They are all done in the same way 133 Laplace transforms 3 Example 3 26 1 2e4s , determine f ðtÞ and sketch the waveform. If Lff ðtÞg ¼ 2 2s sð1 e4s Þ 1 t The first term is easy enough. In unit step form L1 ¼ uðtÞ 2s2 2 From the second term 1 o 2e4s 2 n 4s e ¼ 1 e4s 4s sð 1 e Þ s 2 4s e ¼ 1 þ e4s þ e8s þ e12s þ . . . s 2e4s 2e8s 2e12s 2e16s þ þ þ þ ... ¼ s s s s ; f ðtÞ ¼ . . . . . . . . . . . . (in unit step form) f ðtÞ ¼ 27 t uðtÞ 2uðt 4Þ 2uðt 8Þ 2uðt 12Þ . . . 2 Now we have to draw the waveform. Consider the function terms up to each break point in turn. 0<t<4 4<t<8 8 < t < 12 t 2 t f ðtÞ ¼ 2 2 t f ðtÞ ¼ 2 2 2 f ðtÞ ¼ f ð0Þ ¼ 0; f ð4Þ ¼ 2 f ð4Þ ¼ 0; f ð8Þ ¼ 2 f ð8Þ ¼ 0; f ð12Þ ¼ 2 etc. So the waveform is . . . . . . . . . . . . 28 f(t) t Expressed analytically, we finally have f ðtÞ ¼ t 2 0 < t < 4, f ðt þ 4Þ ¼ f ðtÞ 134 Programme 4 The Dirac delta – the unit impulse 29 So far we have dealt with a number of standard Laplace transforms and then the Heaviside unit step function with some of its applications. We now come to consider an entity that is different from any of the functions we have used before because it is not a proper function. Rather than being defined by its inputs and corresponding outputs it is defined by its effect on other functions. If f ðtÞ represents a function then the Dirac delta ðtÞ is defined by the integral ð1 f ðtÞðt aÞ dt ¼ f ðaÞ 1 ðtÞ is often referred to as the Dirac delta function even though it is not a function in the conventional sense of being completely defined in terms of its outputs for the corresponding inputs. The nearest that can be achieved in defining it in function terms is 0 t 6¼ 0 ðtÞ ¼ undefined t ¼ 0 From the definition, if f ðtÞ ¼ 1 then ð1 ðt aÞ dt ¼ . . . . . . . . . . . . 1 ð1 30 ðt aÞ dt ¼ 1 1 Because ð1 f ðtÞðt aÞ dt ¼ f ðaÞ and f ðtÞ ¼ 1 so f ðaÞ ¼ 1, therefore ð1 1 ðt aÞ dt ¼ 1 hence the name unit impulse. 1 Also, if p < a < q then ðq p ðt aÞ dt ¼ . . . . . . . . . . . . 135 Laplace transforms 3 ðq 31 ðt aÞ dt ¼ 1 p Because ð1 ðt aÞ dt ¼ 1 ðp ðt aÞ dt þ 1 ¼0þ So that ðt aÞ dt þ p ðq ðt aÞ dt þ 0 p ðq ðq ð1 ðt aÞ dt q since ðt aÞ ¼ 0 for 1 < t p and q t < 1 ¼1 ðt aÞ dt ¼ 1 p Graphical representation Graphically the Dirac delta or unit impulse ðt aÞ is represented by the horizontal axis with a vertical line of infinite length at t ¼ a and where the infinite nature of the line is indicated by an arrow-head. f(t) a t So far, then, we have ðq (a) ðt aÞ dt ¼ 1 p (b) ðq f ðtÞ ðt aÞ dt ¼ f ðaÞ p provided, in each case, that p < a < q. Example 1 To evaluate ð3 2 t þ 4 ðt 2Þ dt. 1 The factor ðt 2Þ shows that the impulse occurs at t ¼ 2, i.e. a ¼ 2. ; f ðaÞ ¼ f ð2Þ ¼ 4 þ 4 ¼ 8 f ðtÞ ¼ t 2 þ 4 ð3 2 ; t þ 4 ðt 2Þ dt ¼ f ð2Þ ¼ 8 1 32 136 Programme 4 Example 2 To evaluate ð cos 6t ðt =2Þ dt. 0 ð cos 6t ðt =2Þ dt ¼ f ð=2Þ ¼ cos 3 ¼ 1 0 and in the same way ð6 5 ðt 3Þ dt ¼ . . . . . . . . . . . . (a) 0 ð5 (b) e2t ðt 4Þ dt ¼ . . . . . . . . . . . . 2 ð1 3t 2 4t þ 5 ðt 2Þ dt ¼ . . . . . . . . . . . . (c) 0 33 (a) ð6 0 ð5 (b) (c) 5 ðt 3Þ dt ¼ 5 1 ¼ 5 e2t ðt 4Þ dt ¼ f ð4Þ ¼ e2t ð21 t¼4 ¼ e8 3t 2 4t þ 5 ðt 2Þ dt ¼ 12 8 þ 5 ¼ 9 0 Nothing could be easier. It all rests on the fact that, provided p < a < q ðq f ðtÞ ðt aÞ dt ¼ . . . . . . . . . . . . p 34 f ðaÞ Now let us consider the Laplace transform of ðt aÞ. On then to the next frame 35 Laplace transform of (t – a) We have already shown that ðq f ðtÞ ðt aÞ dt ¼ f ðaÞ p<a<q p Therefore, if p ¼ 0 and q ¼ 1 ð1 f ðtÞ ðt aÞ dt ¼ f ðaÞ 0 Hence, if f ðtÞ ¼ est , this becomes ð1 est ðt aÞ dt ¼ Lfðt aÞg ¼ . . . . . . . . . . . . 0 137 Laplace transforms 3 eas 36 i.e. the value of f ðtÞ, i.e. est , at t ¼ a. Lfðt aÞg ¼ eas It follows from this that the Laplace transform of the impulse function at the origin is . . . . . . . . . . . . 37 1 Because For a ¼ 0, Lfðt aÞg ¼ LfðtÞg ¼ e0 ¼ 1 ; LfðtÞg ¼ 1 Finally, let us deal with the more general case of Lff ðtÞ ðt aÞg. We have ð1 est f ðtÞ ðt aÞ dt. Lff ðtÞ ðt aÞg ¼ 0 Now the integrand est f ðtÞ ðt aÞ ¼ 0 for all values of t except at t ¼ a at which point est ¼ eas , and f ðtÞ ¼ f ðaÞ. ð1 ðt aÞ dt ; Lff ðtÞ ðt aÞg ¼ f ðaÞ eas 0 ¼ f ðaÞ eas ð1Þ ; Lff ðtÞ ðt aÞg ¼ f ðaÞeas Another important result to note. Then let us deal with some examples We have Lff ðtÞ ðt aÞg ¼ f ðaÞ eas 38 Therefore (a) Lf6 ðt 4Þg a ¼ 4, ; Lf6 ðt 4Þg ¼ 6e4s (b) Lft ðt 2Þg a ¼ 2, ; Lft 3 ðt 2Þg ¼ 8e2s 3 Similarly (c) Lfsin 3t ðt =2Þg ¼ . . . . . . . . . . . . es=2 Because Lfsin 3t ðt =2Þg ¼ sin 3t t¼=2 and (d) Lfcosh 2t ðtÞg ¼ . . . . . . . . . . . . es=2 ¼ es=2 39 138 Programme 4 40 1 Because Lfcosh 2t ðtÞg ¼ cosh 2t t¼0 e0 ¼ cosh 0 ð1Þ ¼ 1 So our main conclusions so far are as follows. ðq (1Þ ðt aÞ dt ¼ . . . . . . . . . . . . provided . . . . . . . . . . . . p ðq (2Þ f ðtÞ ðt aÞ dt ¼ . . . . . . . . . . . . provided . . . . . . . . . . . . p (3) Lfðt aÞg ¼ . . . . . . . . . . . . (4) LfðtÞg ¼ . . . . . . . . . . . . (5) Lff ðtÞ ðt aÞg ¼ . . . . . . . . . . . . ðq 41 (1) ðt aÞ dt ¼ 1 provided p < a < q p (2Þ ðq f ðtÞ ðt aÞ dt ¼ f ðaÞ provided p < a < q p (3) Lfðt aÞg ¼ eas (4) LfðtÞg ¼ 1 (5) Lff ðtÞ ðt aÞg ¼ f ðaÞ eas Just check that you have noted this important list – the basis of all work on the Dirac delta function. Now for one further example on this section Example Impulses of 1, 4, 7 units occur at t ¼ 1, t ¼ 3 and t ¼ 4 respectively, in the directions shown. f(t) t Write down an expression for f ðtÞ and determine its Laplace transform. We have f ðtÞ ¼ 1 ðt 1Þ 4 ðt 3Þ þ 7 ðt 4Þ. Then Lff ðtÞg ¼ . . . . . . . . . . . . 139 Laplace transforms 3 Lff ðtÞg ¼ es 4e3s þ 7e4s and that is all there is to that. The derivative of the unit step function One further consideration is interesting. Consider some function f ðtÞ that is zero outside some finite interval ½a, b of the real line. That is, f ðtÞ ¼ 0 for t < a and t > b, then ð1 ½uðtÞf ðtÞ0 dt ¼ ½uðtÞf ðtÞ1 1 ¼ 0 1 where uðtÞ is the unit step function and f ðtÞ is zero at the limits. Now ð1 ð1 ð1 ½uðtÞf ðtÞ0 dt ¼ u0 ðtÞf ðtÞ dt þ uðtÞf 0 ðtÞ dt 1 1 1 and so ð1 ð1 u0 ðtÞf ðtÞ dt ¼ uðtÞf 0 ðtÞ dt 1 1 This means that ð1 ð1 u0 ðtÞf ðtÞ dt ¼ uðtÞf 0 ðtÞ dt 1 1 ð1 f 0 ðtÞ dt ¼ 0 h i1 ¼ f ðtÞ Because the unit step is zero for negative t 0 ¼ f ð1Þ þ f ð0Þ Because f ð1Þ ¼ 0 by definition ¼ f ð0Þ ¼ ð1 1 ðtÞf ðtÞ dt By the definition of the Dirac delta and so u0 ðtÞ ¼ ðtÞ – the unit impulse is equal to the derivative of the unit step function. 42 140 Programme 4 Differential equations involving the unit impulse 43 Example 1 A system has the equation of motion x€ þ 6x_ þ 8x ¼ gðtÞ where gðtÞ is an impulse of 4 units applied at t ¼ 5. At t ¼ 0, x ¼ 0 and x_ ¼ 3. Determine an expression for the displacement x in terms of t. The impulse of 4 units is applied at t ¼ 5. ; gðtÞ ¼ 4 ðt 5Þ. ; x€ þ 6x_ þ 8x ¼ 4 ðt 5Þ At t ¼ 0, x ¼ 0, x_ ¼ 3. Taking Laplace transforms this differential equation becomes ............ 44 s2 x sx0 x1 þ 6ðsx x0 Þ þ 8x ¼ 4e5s Now x0 ¼ 0; x1 ¼ 3 ; s2 x 3 þ 6sx þ 8x ¼ 4e5s ; s2 þ 6s þ 8 x ¼ 3 þ 4e5s ; x ¼ 3 þ 4e5s Writing 1 ðs þ 2Þðs þ 4Þ 1 in partial fractions, we get ðs þ 2Þðs þ 4Þ x ¼ . . . . . . . . . . . . 1 1 1 1 x ¼ 3 þ 4e5s 2 sþ2 2 sþ4 45 ; x ¼ 5s 3 1 1 e e5s þ2 2 sþ2 sþ4 sþ2 sþ4 Taking inverse transforms n o 3 2t x¼ e e4t þ 2 e2ðt5Þ uðt 5Þ e4ðt5Þ uðt 5Þ 2 3 2t e e4t þ 2 e2t e10 uðt 5Þ e4t e20 uðt 5Þ ¼ 2 which simplifies to x ¼ . . . . . . . . . . . . 46 x ¼ e2t 3 3 þ 2e10 uðt 5Þ e4t þ 2e20 uðt 5Þ 2 2 141 Laplace transforms 3 Example 2 Solve the equation x€ þ 4x_ þ 13x ¼ 2 ðtÞ where, at t ¼ 0, x ¼ 2 and x_ ¼ 0. x€ þ 4x_ þ 13x ¼ 2 ðtÞ x0 ¼ 2; x1 ¼ 0 Expressing in Laplace transforms, we have ............ s2 x sx0 x1 þ 4ðsx x0 Þ þ 13x ¼ 2 ð1Þ 47 Inserting the initial conditions and simplifying, x ¼ . . . . . . . . . . . . x ¼ ð2s þ 10Þ 48 1 s2 þ 4s þ 13 Rearranging the denominator by completing the square, this can be written 1 x ¼ ð2s þ 10Þ ðs þ 2Þ2 þ 9 ; x ¼ ............ 49 x ¼ 2e2t fcos 3t þ sin 3t g Because x ¼ 2ðs þ 2Þ 2 ðs þ 2Þ þ 9 þ 6 ðs þ 2Þ2 þ 9 2t cos 3t þ 2e2t sin 3t ; x¼ 2e2t cos 3t þ sin 3t ; x ¼ 2e Now for one further example for you to work through on your own. So move on Example 3 The equation of motion of a system is x€ þ 5x_ þ 4x ¼ gðtÞ where gðtÞ ¼ 3 ðt 2Þ. At t ¼ 0, x ¼ 2 and x_ ¼ 2. Determine an expression for the displacement x in terms of t. We have x€ þ 5x_ þ 4x ¼ 3 ðt 2Þ with x0 ¼ 2 and x1 ¼ 2. As before, you can express this in Laplace transforms, substitute the initial conditions, simplify to obtain an expression for x and finally take inverse transforms to determine the required expression for x. Work right through it carefully. It is good revision and there are no snags. x ¼ ............ 50 142 Programme 4 x ¼ et 2 þ e2 uðt 2Þ e8 e4t uðt 2Þ 51 Here is the working for you to check. x€ þ 5x_ þ 4x ¼ 3 ðt 2Þ with x0 ¼ 2 and x1 ¼ 2 2 s x sx0 x1 þ 5ðsx x0 Þ þ 4x ¼ 3e2s s2 x 2s þ 2 þ 5sx 10 þ 4x ¼ 3e2s 2 s þ 5s þ 4 x 2s 8 ¼ 3e2s ; ðs þ 1Þðs þ 4Þx ¼ 2s þ 8 þ 3e2s 2ðs þ 4Þ 3 þ e2s ; x ¼ ðs þ 1Þðs þ 4Þ ðs þ 1Þðs þ 4Þ 2 1 1 þ e2s ¼ sþ1 sþ1 sþ4 ; x ¼ 2 e2s e2s þ sþ1 sþ1 sþ4 ; x ¼ 2et þ uðt 2Þ eðt2Þ uðt 2Þ e4ðt2Þ ¼ 2et þ uðt 2Þ e2 et uðt 2Þ e8 e4t x ¼ et 2 þ e2 uðt 2Þ e8 e4t uðt 2Þ Harmonic oscillators 52 If the position of a system at time t is described by the expression f ðtÞ where f ðtÞ satisfies the differential equation af 00 ðtÞ þ bf ðtÞ ¼ 0, f ð0Þ ¼ and f 0 ð0Þ ¼ (and where a and b have the same sign) then, taking Laplace transforms of both sides gives Lfaf 00 ðtÞ þ bf ðtÞg ¼ Lf0g That is a s2 FðsÞ s þ b FðsÞ ¼ 0 Collecting like terms gives 2 as þ b FðsÞ ¼ aðs þ Þ giving FðsÞ ¼ aðs þ Þ as2 þ b 143 Laplace transforms 3 s þ 2 and so þ ðb=aÞ s þ ðb=aÞ rffiffiffi rffiffiffi rffiffiffi b a b t þ sin t f ðtÞ ¼ cos a b a Therefore FðsÞ ¼ s2 The natural frequency rffiffiffi system executes simple harmonic, oscillatory motion with rffiffiffi b 2 a radians per unit of time and with period pffiffiffiffiffiffiffiffi ¼ 2 . It is called an a b b=a harmonic oscillator. Let’s try some examples. Example 1 Find the solution to the harmonic oscillator f 00 ðtÞ þ 16f ðtÞ ¼ 0 where f ð0Þ ¼ 1 and f 0 ð0Þ ¼ 0 Taking Laplace transforms gives FðsÞ ¼ . . . . . . . . . . . . FðsÞ ¼ s2 s þ 16 53 Because Taking Laplace transforms Lff 00 ðtÞ þ 16f ðtÞg ¼ Lf0g. That is s2 FðsÞ s þ 16FðsÞ ¼ 0 and so s FðsÞ ¼ 2 s þ 16 This means that f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ cos 4t Because s s ¼ 2 so f ðtÞ ¼ cos 4t from the Table of Laplace þ 16 s þ 42 transforms on page 68. FðsÞ ¼ s2 The motion of this system is then periodic with frequency 4 radians per unit of time and with period 2=4 ¼ =2 units of time. 54 144 Programme 4 Example 2 The frequency and period of the harmonic oscillator whose position f ðtÞ satisfies the differential equation 5f 00 ðtÞ þ 10f ðtÞ ¼ 0 where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 4 is given as frequency . . . . . . . . . . . . radians per unit of time and period . . . . . . . . . . . . units of time 55 frequency pffiffiffi pffiffiffi 2 and period 2 Because Taking Laplace transforms gives Lf5f 00 ðtÞ þ 10f ðtÞg ¼ Lf0g that is 5s2 FðsÞ 4 þ 10FðsÞ ¼ 0 so that 4 4=5 ¼ FðsÞ ¼ 2 5s þ 10 s2 þ 2 and from the Table of Laplace transforms on page 68 pffiffiffi pffiffiffi 2 2 sin 2t f ðtÞ ¼ 5 pffiffiffi This is periodic with frequency 2 radians per unit of time and period pffiffiffi pffiffiffi 2= 2 ¼ 2 units of time. pffiffiffi 2 2 . Notice that the amplitude of the motion is 5 56 Damped motion Consider the equation 5f 00 ðtÞ þ 5f 0 ðtÞ þ 10f ðtÞ ¼ 0 where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 4 This is the same as the last equation in Frame 54 with an extra term added, namely 5f 0 ðtÞ. This term describes a particular effect on the system as you will see from the solution. Solving the differential equation gives f ðtÞ ¼ . . . . . . . . . . . . 145 Laplace transforms 3 pffiffiffi 8 f ðtÞ ¼ pffiffiffi et=2 sin 7t=2 5 7 57 Because Taking Laplace transforms gives Lf5f 00 ðtÞ þ 5f 0 ðtÞ þ 10f ðtÞg ¼ Lf0g that is 5 s2 FðsÞ 4 þ 5sFðsÞ þ 10FðsÞ ¼ 0 so that FðsÞ ¼ 20 4 4 ¼ ¼ 2 5s2 þ 5s þ 10 s2 þ s þ 2 ðs þ 1=2Þ2 þ pffiffiffi 7=2 and from the Table of Laplace transforms on page 68 pffiffiffi 8 f ðtÞ ¼ pffiffiffi et=2 sin 7t=2 7 This is periodic with frequency 1 radian per unit of time and period 2 units of time but with an amplitude that is decreasing with time. The graph of this function is as follows 2 1.5 f (t) 1 0.5 t 0 –0.5 5 10 15 –1 The effect of the 5f 0 ðtÞ in the differential equation is to introduce damping into the oscillatory motion so causing the oscillations to decay. Let’s try another example. Example 3 Consider the equation 5f 00 ðtÞ þ f 0 ðtÞ þ 10f ðtÞ ¼ 0 where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 4 This equation is again similar to the previous equation but with a smaller damping term of f 0 ðtÞ instead of 5f 0 ðtÞ. Then here f ðtÞ ¼ . . . . . . . . . . . . 146 Programme 4 58 pffiffiffiffiffiffiffiffiffiffi 4 f ðtÞ ¼ pffiffiffiffiffiffiffiffiffiffi e0:1t sin 1:99t 1:99 Because Taking Laplace transforms gives Lf5f 00 ðtÞ þ f 0 ðtÞ þ 10f ðtÞg ¼ Lf0g that is 5 s2 FðsÞ 4 þ sFðsÞ þ 10FðsÞ ¼ 0 so that FðsÞ ¼ 20 4 4 ¼ ¼ 5s2 þ 1s þ 10 s2 þ 0:2s þ 2 ðs þ 0:1Þ2 þ 1:99 and from the Table of Laplace transforms on page 68 pffiffiffiffiffiffiffiffiffiffi 4 f ðtÞ ¼ pffiffiffiffiffiffiffiffiffiffi e0:1t sin 1:99t 1:99 pffiffiffiffiffiffiffiffiffiffi This is periodic with frequency 1:99 radians per unit of time and period pffiffiffiffiffiffiffiffiffiffi 2= 1:99 units of time and with an amplitude that is decreasing with time. The graph of this function is as follows 3 2 1 f(t) t 0 –1 5 10 15 20 25 30 –2 –3 Again, the effect of the f 0 ðtÞ in the differential equation is to introduce damping into the oscillatory motion so causing it to decay. Also because the coefficient of f 0 ðtÞ is smaller in this example, the damping is less severe. Forced harmonic motion with damping 59 The equation f 00 ðtÞ þ f 0 ðtÞ þ f ðtÞ ¼ et where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 0 we know would represent damped harmonic motion were it not for the exponential on the right-hand side. To see the effect of the exponential we solve the equation. Taking Laplace transforms we see that FðsÞ ¼ . . . . . . . . . . . . 147 Laplace transforms 3 FðsÞ ¼ 60 1 ðs 1Þðs2 þ s þ 1Þ Because Lff 00 ðtÞ þ f 0 ðtÞ þ f ðtÞg ¼ L et that is s2 þ s þ 1 FðsÞ ¼ FðsÞ ¼ 1 so s1 1 ðs 1Þðs2 þ s þ 1Þ Separating into partial fractions gives FðsÞ ¼ . . . . . . . . . . . . FðsÞ ¼ 1 sþ2 3ðs 1Þ 3ðs2 þ s þ 1Þ 61 Because 1 A Bs þ C þ 2 ¼ 2 ðs 1Þðs þ s þ 1Þ ðs 1Þ ðs þ s þ 1Þ A s2 þ s þ 1 þ ðBs þ CÞðs 1Þ ¼ ðs 1Þðs2 þ s þ 1Þ Equating numerators and then comparing coefficients of powers of s gives 1 ¼ A s2 þ s þ 1 þ ðBs þ CÞðs 1Þ [s2 ]: 0¼AþB ð1Þ [s]: 0 ¼ A B þ C ð2Þ [CT]: 1¼AC So ð2Þ þ ð3Þ: 1 ¼ 2A B 2 ð1Þ: ð3Þ 0 ¼ 2A þ 2B 1 ¼ 3B Therefore: so B ¼ 1=3 ¼ A and C ¼ 2=3 1 1 sþ2 ¼ Thus FðsÞ ¼ ðs 1Þðs2 þ s þ 1Þ 3ðs 1Þ 3ðs2 þ s þ 1Þ Consequently f ðtÞ ¼ . . . . . . . . . . . . pffiffiffi pffiffiffi ! pffiffiffi et 1 t=2 3 3 f ðtÞ ¼ e cos t þ 3 sin t 3 3 2 2 Because FðsÞ ¼ ¼ 1 sþ2 3ðs 1Þ 3ðs2 þ s þ 1Þ 3 s þ 12 1 2 2 2 3ðs 1Þ 3 s þ 1 þ 3 3 s þ 1 þ3 2 4 2 4 So pffiffi pffiffi pffiffiffi et 1 t=2 cos 23 t þ 3 sin 23 t e 3 3 from the Table of Laplace transforms on page 68. f ðtÞ ¼ 62 148 Programme 4 8000 7000 6000 5000 f (t) 4000 3000 2000 1000 0 t 2 4 6 8 10 12 pffiffi pffiffi pffiffiffi 1 t=2 e cos 23 t þ 3 sin 23 t represents damped harmo3 et nic motion and is called the transient term whereas the term represents a 3 steady-state term, so called because as the transient term decays the steadystate term remains the dominant part of the solution. The steady-state solution is a direct consequence of the term on the right-hand side of the differential equation. Try another one for yourself. The transient and steady-state terms of the system described by the differential equation Notice that the term f 00 ðtÞ þ 2f 0 ðtÞ þ 5f ðtÞ ¼ e2t where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 1 are 63 Transient term . . . . . . . . . . . . Steady-state term . . . . . . . . . . . . 1 t 5 t 1 2t e cos 2t þ e sin 2t, e 13 13 13 Because Taking Laplace transforms, Lff 00 ðtÞ þ 2f 0 ðtÞ þ 5f ðtÞg ¼ L e2t . That is 1 , that is s2 2 1 s1 ¼ s þ 2s þ 5 FðsÞ ¼ 1 þ s2 s2 s1 A Bs þ C So that FðsÞ ¼ þ 2 ¼ . Hence 2 ðs 2Þðs þ 2s þ 5Þ s 2 s þ 2s þ 5 2 s 1 ¼ A s þ 2s þ 5 þ ðBs þ CÞðs 2Þ. Equating powers of s gives s2 FðsÞ 1 þ 2sFðsÞ þ 5FðsÞ ¼ [s2 ]: [s]: [CT]: 0¼AþB 1 ¼ 2A 2B þ C 1 ¼ 5A 2C 149 Laplace transforms 3 Solving these three equations gives A ¼ 1=13, B ¼ 1=13 and C ¼ 9=13 so that 1 s9 13ðs 2Þ 13ðs2 þ 2s þ 5Þ 1 s9 . That is ¼ 13ðs 2Þ 13 ðs þ 1Þ2 þ 22 FðsÞ ¼ FðsÞ ¼ 1 sþ1 10 þ 2 13ðs 2Þ 13 ðs þ 1Þ þ 22 13 ðs þ 1Þ2 þ 22 Therefore f ðtÞ ¼ 1 2t 1 t 5 t e e cos 2t þ e sin 2t 13 13 13 60 50 40 f(t) 30 20 10 0 t 1 2 3 4 Next frame 64 Resonance These differential equations with a function on the right-hand side are called inhomogeneous differential equations. They represent systems whose behaviour f ðtÞ is dictated by the structure of the left-hand side and the forcing function on the right-hand side. If an external force is applied to an undamped harmonic oscillator with a vibrational frequency equal to the oscillator’s natural frequency [Frame 52] the oscillator will be set in motion and vibrate in sympathy at its natural frequency. This is called resonance. If the applied force is maintained unabated the oscillator will continue to resonate but with an increasing amplitude. An example will illustrate this. The differential equation f 00 ðtÞ þ f ðtÞ ¼ 0 where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 1 represents an undamped, unforced system with behaviour f ðtÞ ¼ . . . . . . . . . . . . 150 Programme 4 65 f ðtÞ ¼ sin t Because Taking the Laplace transform of both sides of the equation gives Lff 00 ðtÞ þ f ðtÞg ¼ Lf0g that is s2 FðsÞ 1 þ FðsÞ ¼ 0 so that 1 giving f ðtÞ ¼ sin t FðsÞ ¼ 2 s þ1 If the forcing term 2 sin t is applied to the right-hand side of the equation it has the same period as the natural frequency of the system being forced and so resonance will set in. The differential equation to solve is then f 00 ðtÞ þ f ðtÞ ¼ 2 sin t where f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 1 This has the solution f ðtÞ ¼ . . . . . . . . . . . . 66 f ðtÞ ¼ t cos t Because Taking the Laplace transform of both sides of the equation gives Lff 00 ðtÞ þ f ðtÞg ¼ Lf2 sin tg that is s2 FðsÞ 1 þ FðsÞ ¼ so that FðsÞ ¼ s2 2 þ1 1 2 s2 1 giving FðsÞ ¼ . Now, the 2 2 þ 1 ðs2 þ 1Þ ðs2 þ 1Þ Laplace transform of cos t is Therefore s2 s and s2 þ 1 0 s s2 1 ¼ . 2 s2 þ 1 ðs2 þ 1Þ f ðtÞ ¼ t cos t 40 20 f(t) t 0 10 20 30 40 50 –20 –40 The system undergoes periodic behaviour with an increasing amplitude. You have now reached the end of this Programme and this brings you to the Revision summary and the Can you? checklist. Following that is the Test exercise. Work through this at your own pace. A set of Further problems provides additional valuable practice. 151 Laplace transforms 3 Revision summary 4 1 Periodic functions 67 f ðtÞ ¼ f ðt þ nTÞ n ¼ 1, 2, 3, . . . Period ¼ T. 2 Laplace transform of a periodic function with period T ðT 1 est f ðtÞ dt. Lff ðtÞg ¼ 1 eTs 0 3 Inverse transforms involving periodic functions 1 þ 2e3s 3e2s e.g. L1 sð1 e3s Þ 1 Expand 1 e3s as a binomial series, like ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ . . . Multiply out and take inverse transforms of each term in turn. 4 Dirac delta function or unit impulse function δ(t – a) f(t) ðt aÞ ¼ 0 ¼1 a 0 5 t ¼ a. t Delta function at the origin f(t) a¼0 ; ðtÞ ¼ 0 ¼1 δ(t) t 6¼ 0 t ¼ 0. t 0 6 t 6¼ a Area of pulse = 1 ðq f(t) ; δ(t – a) ðt aÞ dt ¼ 1 p p<a<q 0 p a q t 152 Programme 4 7 Integration of the impulse function ðq f ðtÞ ðt aÞ dt ¼ f ðaÞ p<a<q p 8 Laplace transform of ðt aÞ Lfðt aÞg ¼ eas LfðtÞg ¼ 1 because a ¼ 0 Lff ðtÞ ðt aÞg ¼ f ðaÞ eas . 9 Harmonic oscillators The equation of af 00 ðtÞ þ bf ðtÞ ¼ 0, f ð0Þ ¼ and f 0 ð0Þ ¼ , where a and b are of the same sign, represents a system undergoing simple harmonic motion and is referred to as an harmonic oscillator. The system rffiffiffi b radians per unit of time and with period oscillates with a frequency of a rffiffiffi 2 a pffiffiffiffiffiffiffiffi ¼ 2 units of time. If a first derivative term is added to the b b=a left-hand side of the equation then, provided all three coefficients have the same sign, the system will undergo damped harmonic motion. 10 Forced harmonic motion Forced harmonic motion is achieved by the existence of a term on the right-hand side of the equation giving rise to transient and steady-state parts of the solution. 11 Resonance If an external force is applied to an undamped harmonic oscillator with a vibrational frequency equal to the oscillator’s natural frequency the oscillator will be set in motion and vibrate in sympathy at its natural frequency. This is called resonance. If the applied force is maintained unabated the oscillator will continue to resonate but with an increasing amplitude. Can you? 68 Checklist 4 Check this list before and after you try the end of Programme test. On a scale of 1 to 5, how confident are you that you can: . Find the Laplace transforms of periodic functions? Yes No . Obtain the inverse Laplace transforms of transforms of periodic functions? Yes No Frames 1 to 14 15 to 28 153 Laplace transforms 3 . Describe and use the unit impulse to evaluate integrals? Yes No 29 to 34 . Obtain the Laplace transform of the unit impulse? Yes No 35 to 42 . Use the Laplace transform to solve differential equations involving the unit impulse? Yes No 43 to 51 . Solve the equation and describe the behaviour of an harmonic oscillator? Yes No 52 to 66 Test exercise 4 1 Determine the Laplace transform of the periodic function shown. 69 f(t) t 2 Evaluate ð4 (a) e3t ðt 2Þ dt (b) ð1 0 0 sin 3t ðt Þ dt (c) (b) L e3t ðt 2Þ . ð3 2 2t þ 3 ðt 2Þ dt. 1 3 Determine (a) Lf4 ðt 3Þg, 4 Sketch the graph of f ðtÞ ¼ 3 ðtÞ þ 4 ðt 2Þ 3 ðt 4Þ and determine its Laplace transform. 5 Solve the equation x€ þ 6x_ þ 10x ¼ 7 ðtÞ given that, at t ¼ 0, x ¼ 1 and x_ ¼ 0. 6 The equation of motion of a system is x€ þ 3x_ þ 2x ¼ 3 ðt 4Þ. At t ¼ 0, x ¼ 2 and x_ ¼ 4. Determine an expression for the displacement x in terms of t. 7 Find the frequency, periodic time and solution for each of the following harmonic oscillators. (a) f 00 ðtÞ þ f ðtÞ ¼ 0 given that f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 1 (b) 6f 00 ðtÞ þ 2f 0 ðtÞ þ 9f ðtÞ ¼ 0 given that f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 3. 8 Find the transient and steady-state solutions of the forced harmonic oscillator f 00 ðtÞ þ 2f 0 ðtÞ þ 3f ðtÞ ¼ 4e5t given that f ð0Þ ¼ 2 and f 0 ð0Þ ¼ 6. 154 Programme 4 Further problems 4 70 0t< f ðt þ 2Þ ¼ f ðtÞ, t < 2 a prove that Lff ðtÞg ¼ 2 . ðs þ 1Þð1 es Þ 1 If f ðtÞ ¼ a sin t ¼0 2 If f ðtÞ ¼ a sin t 3 Find the Laplace transforms of the following periodic functions. (a) f ðtÞ ¼ t 0t<T f ðt þ TÞ ¼ f ðtÞ t 0 t < 2 f ðt þ 2Þ ¼ f ðtÞ (b) f ðtÞ ¼ e (c) f ðtÞ ¼ t 0t<1 f ðt þ 2Þ ¼ f ðtÞ ¼0 1t<2 0t<2 (d) f ðtÞ ¼ t 2 f ðt þ 3Þ ¼ f ðtÞ 2t<3 ¼4 4 A mass M is attached to a spring of stiffness !2 M and is set in motion at t ¼ 0 by an impulsive force P. The equation of motion is M x€ þ M!2 x ¼ P ðtÞ. 0t< f ðt þ Þ ¼ f ðtÞ, determine Lff ðtÞg. Obtain an expression for x in terms of t. 5 An impulsive voltage E is applied at t ¼ 0 to a series circuit containing inductance L and capacitance C. Initially, the current and charge are zero. The current i at time t is given by di q L þ ¼ E ðtÞ dt C dq where q is the instantaneous value of the charge on the capacitor. Since i ¼ , dt determine an expression for the current i in the circuit at time t. 6 A system has the equation of motion x€ þ 5x_ þ 6x ¼ FðtÞ where, at t ¼ 0, x ¼ 0 and x_ ¼ 2. If FðtÞ is an impulse of 20 units applied at t ¼ 4, determine an expression for x in terms of t. 7 Find the frequency, periodic time and solution for each of the following harmonic oscillators. (a) 12f 00 ðtÞ þ f ðtÞ ¼ 0 given that f ð0Þ ¼ 1 and f 0 ð0Þ ¼ 2 (b) f 00 ðtÞ þ 12f ðtÞ ¼ 0 given that f ð0Þ ¼ 2 and f 0 ð0Þ ¼ 1. 8 Solve for each of the following harmonic oscillators. (a) 4:6f 00 ðtÞ þ 2:2f ðtÞ ¼ 0 given that f ð0Þ ¼ 1:6 and f 0 ð0Þ ¼ 3:1 pffiffiffi 00 pffiffiffi (b) 2f ðtÞ þ 3f ðtÞ ¼ 0 given that f ð0Þ ¼ 0 and f 0 ð0Þ ¼ . 9 Find the transient and steady-state solutions of the forced harmonic oscillator 4f 00 ðtÞ þ 3f 0 ðtÞ þ 2f ðtÞ ¼ et given that f ð0Þ ¼ 0 and f 0 ð0Þ ¼ 6. Programme 5 Frames 1 to 53 Difference equations and the Z transform Learning outcomes When you have completed this Programme you will be able to: . Convert the descriptive prescription of the output form of a sequence into a recursive description and recognise the importance of initial terms . Recognise a difference equation, determine its order and generate its terms from a recursive description . Obtain the solution to a difference equation as a sum of the homogeneous solution and the particular solution . Define the Z transform of a sequence and derive transforms of specified sequences . Make reference to a table of standard Z transforms . Recognise the Z transform as being a linear transform and so obtain the transform of linear combinations of standard sequences . Apply the first and second shift theorems, the translation theorem, the initial and final value theorems and the derivative theorem . Use partial fractions to derive the inverse transforms . Use the Z transform to solve linear, constant coefficient difference equations . Create a sequence by sampling a continuous function and demonstrate the relationship between the Laplace and the Z transform 155 156 Programme 5 Introduction 1 The Laplace transform deals with continuous functions and can be used to solve many differential equations that arise in science and engineering. There are occasions, however, when we have to deal with discrete functions – sequences – and their associated difference equations. For example, the central processing unit of your computer can only handle information in the form of pulses of electricity. This information transmission is called digital transmission. There are, however, times when information is fed into the computer in the form of a continuously varying signal called an analogue signal. For instance, a mouse can be moved about the flat surface of your desk in a continuous manner but the central processing unit will only recognise position on the screen to the nearest pixel. The analogue signal coming from the mouse needs to be converted into a digital signal for recognition by the computer’s central processing unit. This conversion of a signal from analogue to digital is achieved by a device called a demodulator that samples the analogue signal at regular intervals of time and outputs the sampled values as the digital signal – as a sequence of numbers. The Z transform, which is allied to the Laplace transform, deals with such sequences and the recurrence relations – or difference equations – that arise. Sequences 2 Any function of a single real variable f whose input is restricted to integer values n has an output f ðnÞ in the form of a discrete sequence of numbers. Accordingly, such a function is called a sequence. For example, the function defined by the prescription f ðnÞ ¼ 5n 2 where n is an integer 1 is a sequence. The first three output values corresponding to the successive input values 1, 2 and 3 are f ð1Þ ¼ 5 1 2 ¼ 3 f ð2Þ ¼ 5 2 2 ¼ 8 f ð3Þ ¼ 5 3 2 ¼ 13 Each output value f ðnÞ of the sequence is called a term of the sequence. An alternative way of describing the terms of this sequence can be found from the following consideration: f ð1Þ ¼ 5 1 2 ¼ 3 f ð2Þ ¼ 5 2 2 ¼ 3 þ 5 f ð3Þ ¼ 5 3 2 ¼ 8 þ 5 The value of any term is the value of the previous term plus 5 and provided we know that the first term is 3 we can compute any other term. A process such as Difference equations and the Z transform 157 this one that repeatedly uses known values to compute an unknown value is called a recursive process. That is: f ðn þ 1Þ ¼ 5ðn þ 1Þ 2 ¼ ½5n 2 þ 5 ¼ f ðnÞ þ 5 so that f ðn þ 1Þ ¼ f ðnÞ þ 5 where f ð1Þ ¼ 3. This description, in which each term of the sequence is seen to depend upon another term of the same sequence, is called a recursive description and can make the computing of the terms of the sequence more efficient and very amenable to a spreadsheet implementation. In this particular example we simply start with 3 and just add 5 to each preceding term to get: 3, 8, 13, 18, 23, 28, 33, 38, . . . Notice that without the initial term f ð1Þ ¼ 3 the recursive description would be of little worth because we would not know how to start the sequence. Find the recursive description and compute the first four terms of each of the following sequences: (a) f ðnÞ ¼ 7n þ 4 where n is an integer 1 (b) f ðnÞ ¼ 8 2n where n is an integer 0 (c) f ðnÞ ¼ 4n where n is an integer 3 [slightly different involving multiplication rather than addition] The answers are in the following frame 3 f ðn þ 1Þ ¼ f ðnÞ þ 7 where f ð1Þ ¼ 11: 11, 18, 25, 32 f ðn þ 1Þ ¼ f ðnÞ 2 where f ð0Þ ¼ 8: 8, 6, 2, 4 1 1 1 1 f ðn þ 1Þ ¼ 4f ðnÞ where f ð3Þ ¼ : , , ,1 64 64 16 4 Because: (a) f ðn þ 1Þ ¼ 7ðn þ 1Þ þ 4 ¼ ½7n þ 4 þ 7 ¼ f ðnÞ þ 7 so f ðn þ 1Þ ¼ f ðnÞ þ 7 where f ð1Þ ¼ 11 giving the first four terms as 11, 18, 25, 32 (b) f ðn þ 1Þ ¼ 8 2ðn þ 1Þ ¼ ½8 2n 2 ¼ f ðnÞ 2 so f ðn þ 1Þ ¼ f ðnÞ 2 where f ð0Þ ¼ 8 giving the first four terms as 8, 6, 4, 2 (c) f ðn þ 1Þ ¼ 4nþ1 ¼ 4½4n ¼ 4f ðnÞ so f ðn þ 1Þ ¼ 4f ðnÞ where f ð3Þ ¼ 43 ¼ giving the first four terms as 1 1 1 , , ,1 64 16 4 1 64 Next frame 158 Programme 5 Difference equations 4 The recursive equation f ðn þ 1Þ ¼ f ðnÞ þ 5 can also be written as f ðn þ 1Þ f ðnÞ ¼ 5 and is an example of a first order, constant coefficient, linear difference equation or linear recurrence relation. It is linear because there are no products of terms such as f ðnÞ f ðmÞ and it is first order because f ðn þ 1Þ is just one term away from f ðnÞ. The order of a difference equation is taken from the maximum number of terms between any pair of terms so that, for example: (a) f ðn þ 2Þ þ 2f ðnÞ ¼ 3n4 þ 2 is a second order difference equation because f ðn þ 2Þ is two terms away from f ðnÞ and (b) 3f ðn þ 3Þ f ðn þ 2Þ þ 5f ðn 1Þ þ 4f ðn 2Þ ¼ 6n2 cos n is a fifth order difference because f ðn þ 3Þ is five terms away from f ðn 2Þ So the order of the difference equation 89f ðn 3Þ þ 17f ðn þ 1Þ 3f ðn 2Þ þ 5f ðn þ 5Þ ¼ 13n2 2n4 is . . . . . . . . . . . . The answers are in the following frame 5 8 Because: f ðn þ 5Þ is 8 terms away from f ðn 3Þ. In order to generate the terms of the sequence from the recursive description it is necessary to have as many initial terms as the order of the difference equation. For example if we are given the second order difference equation with a single initial term: f ðn þ 2Þ þ 2f ðnÞ ¼ 3n þ 2 where f ð1Þ ¼ 1 then by substituting into the difference equation we see that: f ð1 þ 2Þ þ 2f ð1Þ ¼ 3 1 þ 2 that is f ð3Þ ¼ 5 2f ð1Þ ¼ 3 and f ð2 þ 2Þ þ 2f ð2Þ ¼ 3 2 þ 2 that is f ð4Þ ¼ 8 2f ð2Þ ¼ ? and f ð3 þ 2Þ þ 2f ð3Þ ¼ 3 3 þ 2 that is f ð5Þ ¼ 11 2f ð3Þ ¼ 5 and f ð4 þ 2Þ þ 2f ð4Þ ¼ 3 4 þ 2 that is f ð6Þ ¼ 16 2f ð4Þ ¼ ? With the single initial term given we can find all those terms of the sequence that correspond to an odd value of n but unless we are given the value of a term that corresponds to an even value of n, a second initial term, we cannot find any of the other terms of the sequence. The order and the number of initial conditions necessary to generate the terms of the sequence f ðnÞ from the difference equation: f ðn þ 4Þ 3f ðn þ 2Þ þ 5f ðn 3Þ ¼ 2nuðnÞ are . . . . . . . . . . . . and . . . . . . . . . . . . The answers are in the following frame Difference equations and the Z transform 159 6 7 and 7 Because: f ðn þ 4Þ is 7 terms away from f ðn 3Þ and the number of initial conditions required to recover the terms of the sequence from a recursive description is equal to the order of the equation. For example, if a sequence has terms that satisfy the second order difference equation f ðn þ 2Þ 3f ðn þ 1Þ þ 2f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 then the first five terms of the sequence are 0, 1, . . . . . . . . . . . ., . . . . . . . . . . . ., . . . . . . . . . . . . Next frame 7 0, 1, 4, 11, 26 Because: Since f ðn þ 2Þ 3f ðn þ 1Þ þ 2f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 then f ð2Þ 3f ð1Þ þ 2f ð0Þ ¼ 1 that is f ð2Þ 3 1 þ 2 0 ¼ 1 and so f ð2Þ ¼ 4 f ð3Þ 3f ð2Þ þ 2f ð1Þ ¼ 1 that is f ð3Þ 3 4 þ 2 1 ¼ 1 and so f ð3Þ ¼ 11 f ð4Þ 3f ð3Þ þ 2f ð2Þ ¼ 1 that is f ð4Þ 3 11 þ 2 4 ¼ 1 and so f ð4Þ ¼ 26 Try another yourself. The first six terms of the sequence that satisfies the second-order difference equation f ðn þ 2Þ f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 are 0, 1, . . . . . . . . . . . . , . . . . . . . . . . . . , . . . . . . . . . . . . , . . . . . . . . . . . . Next frame 8 0, 1, 1, 0, 2, 1 Because: Since f ðn þ 2Þ f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 then f ð2Þ f ð0Þ ¼ 1 that is f ð2Þ 0 ¼ 1 and so f ð2Þ ¼ 1 f ð3Þ f ð1Þ ¼ 1 that is f ð3Þ f ð1Þ ¼ 1 and so f ð2Þ ¼ 0 f ð4Þ f ð2Þ ¼ 1 that is f ð4Þ 1 ¼ 1 and so f ð4Þ ¼ 2 f ð5Þ f ð3Þ ¼ 1 that is f ð5Þ 0 ¼ 1 and so f ð5Þ ¼ 1 They are all done the same way. Move on to the next frame 160 Programme 5 Solving difference equations 9 We have seen how the prescription for a sequence such as: f ðnÞ ¼ 5n 2 where n is an integer 1 can be manipulated to create the difference equation f ðn þ 1Þ f ðnÞ ¼ 5 where f ð1Þ ¼ 3 What we wish to be able to do now is to reverse this process. That is, given the difference equation we wish to find the prescription for the sequence which is the solution to the difference equation. In their most general form these linear difference equations can be written as: an f ðnÞ þ an1 f ðn 1Þ þ . . . þ ank f ðn kÞ ¼ bm gðmÞ þ bm1 gðm 1Þ þ . . . þ bml gðm lÞ where the sequence f on the left is unknown and the sequence g on the right along with all the a and b coefficients are known. It is not dissimilar in structure to an ordinary differential equation and we shall find that the process of finding the solution is similar as well [Ref: Engineering Mathematics, Sixth Edition]. Read on. Move on to the next frame 10 Solution by inspection The solution to a constant coefficient, linear recursive difference equation is analogous to the solution of a constant coefficient, linear differential equation. It is of the form f ðnÞ ¼ fh ðnÞ þ fp ðnÞ where fh ðnÞ is the solution to the homogeneous equation: an f ðnÞ þ an1 f ðn 1Þ þ . . . þ ank f ðn kÞ ¼ 0 and fp ðnÞ is a particular solution of the inhomogeneous difference equation. Also, for a complete solution, an nth order difference equation must be accompanied by n initial terms. For example to solve the second order difference equation: f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ ¼ 1 for n 0 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 we first consider the homogeneous difference equation f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ ¼ 0 and assume a solution of the form fh ðnÞ ¼ Kwn so that the equation becomes: ............ ¼ 0 Next frame Difference equations and the Z transform 161 Kwn w2 7w þ 12 ¼ 0 11 Because: Substitution of f ðnÞ ¼ Kwn into f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ ¼ 0 yields Kwnþ2 7Kwnþ1 þ 12Kwn ¼ K wnþ2 7wnþ1 þ 12wn ¼ Kwn w2 7w þ 12 ¼0 This is called the characteristic equation of the difference equation and it has roots given from: w2 7w þ 12 ¼ ðw 3Þðw 4Þ ¼ 0. That is w ¼ 3 or w ¼ 4 therefore: fh ðnÞ ¼ A 3n þ B 4n where A and B are constants To find fp ðnÞ we assume a form of fp ðnÞ ¼ C1 n þ C2 where C1 and C2 are constants. Substitution yields: C1 ¼ . . . . . . . . . . . . and C2 ¼ . . . . . . . . . . . . Next frame C1 ¼ 0 and C2 ¼ 12 1 6 Because: Substituting f ðnÞ ¼ C1 n þ C2 into f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ ¼ 1 yields ðC1 ðn þ 2Þ þ C2 Þ 7ðC1 ðn þ 1Þ þ C2 Þ þ 12ðC1 n þ C2 Þ ¼ 1, that is 1 6C1 n 5C1 þ 6C2 ¼ 1 so that C1 ¼ 0 and C2 ¼ . 6 1 Therefore fp ðnÞ ¼ 6 The complete solution is then: f ðnÞ ¼ A 3n þ B 4n þ 1 6 From the initial terms we have: f ð0Þ ¼ 0: f ð1Þ ¼ 1: 1 1 ¼ 0 that is A þ B ¼ 6 6 1 5 1 1 A 3 þ B 4 þ ¼ 1 that is 3A þ 4B ¼ 6 6 A 30 þ B 40 þ From these two equations we see that: A ¼ . . . . . . . . . . . . and B ¼ . . . . . . . . . . . . and hence f ðnÞ ¼ . . . . . . . . . . . . Next frame 162 Programme 5 13 3 4 1 A ¼ , B ¼ , f ðnÞ ¼ 3nþ2 2 4nþ1 1 2 3 6 Because: 5 1 3 4 and A þ B ¼ so that 3A þ 3B ¼ and 4A þ 4B ¼ 6 6 6 6 3 4 then A ¼ and B ¼ and hence 2 3 Since 3A þ 4B ¼ 3 4 1 f ðnÞ ¼ 3n þ 4n þ 2 3 6 3nþ1 4nþ1 1 ¼ þ þ 6 2 3 nþ2 3 2 4nþ1 1 þ þ ¼ 6 6 6 1 nþ2 ¼ 3 2 4nþ1 1 6 Just to re-cap to make sure you are clear of what we have done here. The sequence f with terms f ðnÞ ¼ 1 nþ2 3 2 4nþ1 1 6 is the solution of the difference equation: f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ ¼ 1 where f ð0Þ ¼ 0, f ð1Þ ¼ 1 As a check: f ðn þ 2Þ 7f ðn þ 1Þ þ 12f ðnÞ 1 1 ¼ 3nþ4 2 4nþ3 1 7 3nþ3 2 4nþ2 1 6 6 1 nþ2 nþ1 24 1 þ 12 3 6 1 7 ¼ ð81 3n 128 4n 1Þ þ ð27 3n 32 4n 1Þ 6 6 2ð9 3n 8 4n 1Þ 81 189 128 224 1 7 þ 18 þ 4n þ 16 þ þ2 ¼ 3n 6 6 6 6 6 6 ¼1 Now you try one. If you follow the route given here you will find that it is quite straightforward The difference equation: f ðn þ 2Þ þ 3f ðn þ 1Þ 10f ðnÞ ¼ 4 where f ð0Þ ¼ 1 and f ð1Þ ¼ 0 has solution f ðnÞ ¼ . . . . . . . . . . . . The answer is in the following frame Difference equations and the Z transform 163 14 1 3 2 ð5Þn þ 33 2n 14 21 Because: f ðnÞ ¼ fh ðnÞ þ fp ðnÞa and assuming fh ðnÞ ¼ Kwn we arrive at the characteristic equation: w2 þ 3w 10 ¼ ðw þ 5Þðw 2Þ with roots w ¼ 5 and w ¼ 2 therefore: fh ðnÞ ¼ A ð5Þn þ B 2n To find fp ðnÞ we assume a form of fp ðnÞ ¼ C1 n þ C2 then substitution yields: ðC1 ðn þ 2Þ þ C2 Þ þ 3ðC1 ðn þ 1Þ þ C2 Þ 10ðC1 n þ C2 Þ ¼ 4 that is, 2 6C1 n þ 5C1 6C2 ¼ 4 so that C1 ¼ 0 and C2 ¼ ¼ fp ðnÞ 3 The complete solution is then: f ðnÞ ¼ A ð5Þn þ B 2n 2 3 From the initial terms we find that: 2 5 ¼ 1 that is A þ B ¼ 3 3 2 2 1 1 f ð1Þ ¼ A ð5Þ þ B 2 ¼ 0 that is 5A þ 2B ¼ 3 3 f ð0Þ ¼ A ð5Þ0 þ B 20 From these two equations we find that: 8 9 and B ¼ therefore: 21 7 8 9 2 ð5Þn þ 2n f ðnÞ ¼ 21 7 3 1 3 n ¼ 2 ð5Þ þ 33 2n 14 21 A¼ Move on to the next frame 15 The particular solution To find the particular solution of a difference equation we assumption about a certain form for the solution and apply difference equation. The form assumed depends upon the form of hand side of the equation and a sample of these are listed in the table: make an it to the the rightfollowing gðnÞ Particular solution Polynomial term nm Cmþ1 nmþ1 þ Cm nm þ Cm1 N m1 þ . . . þ C0 Exponential an Can an cos bn, an sin bn an ðC1 cos bn þ C2 sin bnÞ 164 Programme 5 For example, the solution to the difference equation f ðn þ 1Þ 3f ðnÞ ¼ gðnÞ where f ð0Þ ¼ 1 and (a) gðnÞ ¼ n2 (b) gðnÞ ¼ 2n (c) gðnÞ ¼ cos 2n is . . . . . . . . . . . . The answers are in the next frame 16 3nþ1 ðn2 þ n þ 1Þ 2 (b) f ðnÞ ¼ 2ð3n 2n1 Þ cos 2 3 sin 2 ½cos 2n 3n þ sin 2n (c) f ðnÞ ¼ 3n þ 10 6 cos 2 10 6 cos 2 (a) f ðnÞ ¼ Because The homogeneous equation f ðn þ 1Þ 3f ðnÞ ¼ 0 has solution fh ðnÞ ¼ Kwn : giving the characteristic equation Kwn ðw 3Þ ¼ 0 so that w ¼ 3 and fh ðnÞ ¼ K 3n . The particular solution is: (a) fp ðnÞ ¼ Cn2 þ Dn þ e. Substitution into the inhomogeneous equation yields Cðn þ 1Þ2 þ Dðn þ 1Þ þ E 3ðCn2 þ Dn þ EÞ ¼ n2 so that Cðn2 þ 2n þ 1Þ þ Dðn þ 1Þ þ E 3ðCn2 þ Dn þ EÞ ¼ n2 ð2CÞ þ nð2C 2DÞ þ ðC þ D 2EÞ ¼ n2 Hence C¼D¼E¼ 1 2 therefore fp ðnÞ ¼ n2 þ n þ 1 n2 þ n þ 1 and f ðnÞ ¼ K 3n . 2 2 Applying the boundary condition f ð0Þ ¼ 1 we see that f ð0Þ ¼ K 1=2 ¼ 1 giving K ¼ 3=2 so that: f ðnÞ ¼ 3nþ1 ðn2 þ n þ 1Þ 2 Check: f ð0Þ ¼ 31 1 ¼1 2 Difference equations and the Z transform (b) fp ðnÞ ¼ C 2n . Substitution into the inhomogeneous equation yields C 2nþ1 3C 2n ¼ 2n so that C 2nþ1 3C 2n ¼ ð2C 3CÞ 2n ¼ C 2n ¼ 2n Hence C ¼ 1 therefore fp ðnÞ ¼ 2n and f ðnÞ ¼ K 3n 2n . Applying the boundary condition f ð0Þ ¼ 1 we see that f ð0Þ ¼ K 1 ¼ 1 giving K ¼ 2 so that: f ðnÞ ¼ 2ð3n 2n1 Þ Check: f ð0Þ ¼ 2ð1 21 Þ ¼ 1 (c) fp ðnÞ ¼ A cos 2n þ B sin 2n. Substitution into the inhomogeneous equation yields A cos 2½n þ 1 þ B sin 2½n þ 1 3A cos 2n 3B sin 2n ¼ cos 2n so that Aðcos 2n cos 2 sin 2n sin 2Þ þ ðB sin 2n cos 2 þ sin 2 cos 2nÞ 3A cos 2n 3B sin 2n ¼ cos 2n. Hence fA cos 2 þ B sin 2 3Ag cos 2n þ fA sin 2 þ B cos 2 3Bg sin 2n ¼ cos 2n so that Aðcos 2 3Þ þ B sin 2 ¼ 1 A sin 2 þ Bðcos 2 3Þ ¼ 0 Multiplying the first equation by sin 2 and the second by cos 2 3 gives Aðcos 2 3Þ sin 2 þ B sin2 2 ¼ sin 2 Aðcos 2 3Þ sin 2 þ Bðcos 2 3Þ2 ¼ 0 and multiplying the second equation by sin 2 and the first by cos 2 3 gives Aðcos 2 3Þ2 þ B sin 2ðcos 2 3Þ ¼ cos 2 3 A sin2 2 þ B sin 2ðcos 2 3Þ ¼ 0 so that n o A ðcos 2 3Þ2 þ sin2 2 ¼ A cos2 2 þ sin2 2 6 cos 2 þ 9 ¼ cos 2 3 and n o B ðcos 2 3Þ2 þ sin2 2 ¼ B cos2 2 þ sin2 2 6 cos 2 þ 9 ¼ sin 2 therefore cos 2 3 sin 2 A¼ and B ¼ 10 6 cos 2 10 6 cos 2 Therefore cos 2 3 sin 2 cos 2n þ sin 2n and fp ðnÞ ¼ 10 6 cos 2 10 6 cos 2 cos 2 3 sin 2 f ðnÞ ¼ K 3n þ cos 2n þ sin 2n. 10 62 10 6 cos 2 165 166 Programme 5 Applying the boundary condition f ð0Þ ¼ 1 we see that f ð0Þ ¼ K þ cos 2 3 cos 2 3 ¼ 1 giving K ¼ 1 so that: 10 6 cos 2 10 6 cos 2 f ðnÞ ¼ 3n þ cos 2 3 sin 2 ½cos 2n 3n þ sin 2n 10 6 cos 2 10 6 cos 2 Check: f ð0Þ ¼ 30 þ cos 2 3 sin 2 ½cos 0 30 þ sin 0 ¼ 1 10 6 cos 2 10 6 cos 2 Solving linear, constant coefficient difference equations in this way is quite straightforward and quite analogous to the method used for solving linear, constant coefficient differential equations for particularly simple equations. As soon as the inhomogeneous equation becomes in any way complicated the algebraic manipulation becomes very labour intensive. Fortunately, there is a simpler way out. Just as we can solve constant coefficient linear differential equations by using the Laplace transform and simultaneously incorporating the boundary conditions so we can solve constant coefficient difference equations using a transform called the Z transform again, simultaneously incorporating the initial conditions. Read on. Next frame The Z transform 17 We have already seen that the Laplace transform of the piecewise continuous function f ðtÞ is given as ð1 f ðtÞest dt Lff ðtÞg ¼ t¼0 ð1 f ðtÞ ¼ dt st t¼0 e This is, in fact, a single-sided Laplace transform and is a special case of what is called the bilateral Laplace transform where the integration ranges from minus infinity to plus infinity: ð1 f ðtÞest dt Lff ðtÞg ¼ t¼1 ð1 f ðtÞ ¼ dt st t¼1 e The bilateral transform is identical to the familiar single-sided transform when f ðtÞ ¼ 0 for t < 0 The equivalent transform for a function that is not piecewise continuous but discrete is: Zff ðnÞg ¼ 1 X f ðnÞ zn n¼1 ¼ FðzÞ where n is an integer Difference equations and the Z transform 167 This is called the Z transform of the sequence. For example, the sequence . . . , 32 , 31 , 30 , 31 , 32 , . . . has a general term of the form f ðnÞ ¼ 3n and its Z transform is: Zff ðnÞg ¼ ¼ 1 X f ðnÞ zn n¼1 1 X 3n zn n¼1 1 X 3 z n¼1 3 ¼ ... þ z ¼ n 1 3 þ z 0 3 þ z 1 þ 3 z 2 þ... Using this definition the Z transform of the sequence f ðnÞ ¼ 1 is . . . . . . . . . . . . The answer is in the following frame . . . þ z2 þ z þ 1 þ 1 1 þ þ ... z z2 Because: Zff ðnÞg ¼ ¼ 1 X f ðnÞ zn n¼1 1 X 1 n¼1 zn 1 1 1 1 1 þ þ þ þ þ ... z2 z1 z0 z1 z2 1 1 ¼ . . . þ z2 þ z þ 1 þ þ 2 þ . . . z z ¼ ... þ It is noticeable that this sum does not converge for any value of z because 1 only if jzj < 0 and diverges if jzj 1 . . . þ z2 þ z þ 1 converges to 1z 1 1 1 1 1 1 1 1 only if and þ 2 þ 3 þ . . . ¼ 1 þ 2 þ 3 þ . . . converges to z z z z z z z 1 1=z 1 1, that is jzj > 1 and diverges if jzj 1. Since jzj 1 or jz > 1 the sum z must necessarily diverge. 18 168 Programme 5 For a Z transform to have any worth it must converge and we need to know the values of z that ensures this. As a first step we shall avoid doubly infinite sequences and only concern ourselves with those sequences for which f ðnÞ ¼ 0 for n < 0. For example, the Z transform FðzÞ of the discrete unit step function uðnÞ where: ( 1 n0 uðnÞ ¼ is 0 n<0 FðzÞ ¼ ZfuðnÞg 1 X uðnÞ ¼ zn n¼1 1 X 1 ¼ zn n¼0 1 1 1 1 1 þ 1 þ 2 þ 3 þ 4 þ ... 0 z z z z z 1 1 1 1 ¼ 1 þ þ 2 þ 3 þ 4 þ ... z z z z 1 recall that ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ . . . provided jxj < 1 ¼ 1 1 z z 1 provided ¼ < 1, that isjzj > 1 z1 z ¼ Therefore, the sequence f ðnÞ ¼ an 0 n f ðnÞ ¼ a uðnÞ has the Z transform n0 n<0 , which can be written as FðzÞ ¼ . . . . . . . . . . . . provided jzj > . . . . . . . . . . . . The answer is in the following frame Difference equations and the Z transform 169 z provided jzj > jaj za 19 Because: Since f ðnÞ ¼ an uðnÞ then FðzÞ ¼ Zff ðnÞg 1 X f ðnÞ ¼ zn n¼1 1 X an uðnÞ ¼ zn n¼1 1 X an ¼ zn n¼0 ¼ ¼ 1 X a z n¼0 a z ¼1þ 0 þ n a z 1 þ a a þ z z a z 2 2 þ þ a z a z 3 3 þ þ a z a z 4 4 þ... þ... a provided <1 a z 1 z z provided jzj > jaj ¼ za Let’s try another. The sequence f ðnÞ ¼ nuðnÞ has the Z transform ¼ 1 FðzÞ ¼ . . . . . . . . . . . . The answer is in the following frame 1 2 3 4 þ þ þ þ ... z z2 z3 z4 Because: FðzÞ ¼ Zff ðnÞg 1 X f ðnÞ ¼ zn n¼1 1 X nuðnÞ ¼ zn n¼1 1 Xn ¼ zn n¼0 0 1 2 3 4 þ þ þ þ þ ... z0 z1 z2 z3 z4 1 2 3 4 ¼ þ 2 þ 3 þ 4 þ ... z z z z ¼ 20 170 Programme 5 By comparing this sequence with the derivative of the series representation of ð1 xÞ1 , this sequence can be written as a rational expression in z as: FðzÞ ¼ . . . . . . . . . . . . provided jzj . . . . . . . . . . . . The answer is in the following frame 21 z ðz 1Þ2 provided jzj > 1 Because: ð1 xÞ1 ¼ 1 þ x þ x2 þ x3 þ x4 . . . provided jxj < 1 and by differentiating both sides 1 ð1 xÞ2 ¼ 1 þ 2x þ 3x2 þ 4x3 þ . . . provided jxj < 1. Comparing this with 1 2 3 4 þ þ þ þ ... z z2 z3 z4 1 3 4 5 1 þ þ 2 þ 3 þ 4 þ ... ¼ z z z z z " # 1 1 1 < 1 that is provided jzj > 1 provided ¼ z ð1 1=zÞ2 z FðzÞ ¼ So, multiplying numerator and denominator by z2 z FðzÞ ¼ provided jzj > 1 ðz 1Þ2 And another example. The Z transform of the discrete unit impulse: ðnÞ ¼ 1 n¼0 0 otherwise is FðzÞ ¼ . . . . . . . . . . . . The answer is in the following frame 22 1 Because: FðzÞ ¼ ZfðnÞg 1 X ðnÞ ¼ zn n¼0 1 0 0 þ þ þ ... z0 z1 z2 ¼1 ¼ Next frame Difference equations and the Z transform 171 Table of Z transforms We list the results that we have obtained so far as well as some additional ones for future reference. Sequence Transform F(z) Permitted values of z ðnÞ ¼ f1, 0, 0, . . .g 1 All values of z z z1 z uðnÞ ¼ f1, 1, 1, . . .g n uðnÞ ¼ f0, 1, 2, 3, . . .g ðz 1Þ2 zðz þ 1Þ n2 uðnÞ ¼ f0, 1, 4, 9, . . .g ðz 1Þ3 z z2 þ 4z þ 1 n3 uðnÞ ¼ f0, 1, 8, 27, . . .g ðz 1Þ4 z ðz aÞ az an uðnÞ ¼ 1, a, a2 , a3 , . . . n an uðnÞ ¼ 0, a, 2a2 , 3a3 , . . . ðz aÞ2 23 jzj > 1 jzj > 1 jzj > 1 jzj > 1 jzj > jaj jzj > jaj Next frame Properties of Z transforms 24 1 Linearity The Z transform is a linear transform. That is, if a and b are constants then Zðaf ðnÞ þ bgðnÞÞ ¼ aZff ðnÞg þ bZfgðnÞg For example, the Z transform of the sequence n uðnÞ is Zfn uðnÞg ¼ . . . . . . . . . . . . and the Z transform of the sequence e2n uðnÞ is Zfe2n uðnÞg ¼ . . . . . . . . . . . . Zfn uðnÞg ¼ z ðz 1Þ 2 and Z e2n uðnÞ ¼ z z e2 Because Zfn uðnÞg ¼ z from the table and, also from the table, ðz 1Þ2 z so when a ¼ e2 , Zfan uðnÞg ¼ za z Z e2n uðnÞ ¼ z e2 Consequently, the Z transform of ð3n 5e2n ÞuðnÞ is . . . . . . . . . . . . 25 172 Programme 5 5z3 þ 13z2 z 3e2 þ 5 26 ðz 1Þ2 ðz e2 Þ Because Z ð3n 5e2n ÞuðnÞ ¼ 3Zfn uðnÞg 5Z e2n uðnÞ 3z 5z ¼ 2 z e2 Þ ð ðz 1Þ 3z z e2 5zðz 1Þ2 ¼ ðz 1Þ2 ðz e2 Þ ¼ ¼ 3z2 3ze2 5z3 þ 10z2 5z ðz 1Þ2 ðz e2 Þ 5z3 þ 13z2 z 3e2 þ 5 ðz 1Þ2 ðz e2 Þ 2 First shift theorem (shifting to the left) If Zff ðnÞg ¼ FðzÞ then Zff ðn þ mÞg ¼ zm FðzÞ zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm 1Þ is the Z transform of the sequence that has been shifted by m places to the left. For example Zff ðn þ 1Þg ¼ zFðzÞ zf ð0Þ Zff ðn þ 2Þg ¼ z2 FðzÞ z2 f ð0Þ zf ð1Þ These will be used later when solving difference equations. Note the similarity between these results and the Laplace transforms for the first and second derivatives for continuous functions. z then For example, given that Zf4n uðnÞg ¼ z4 Z 4nþ3 uðnÞ ¼ . . . . . . . . . . . . Difference equations and the Z transform 173 27 64z z4 Because Zff ðn þ mÞg ¼ zm FðzÞ zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm 1Þ so Z 4nþ3 uðnÞ ¼ z3 Zf4n uðnÞg z3 40 þ z2 41 þ z42 where Zf4n uðnÞg ¼ z z4 z z3 þ 4z2 þ 16z z4 z4 ¼ z3 þ 4z2 þ 16z z4 z4 z3 þ 4z2 þ 16z ðz 4Þ ¼ z4 4 4 z z 64z ¼ z4 64z ¼ z4 In this way we have derived the Z transform of the sequence f64, 256, 1024, . . .g by shifting the sequence f1, 4, 16, 64, 256, . . .g three places to the left and losing the first three terms. ¼ z3 Try another. Given that Zfn uðnÞg ¼ z ðz 1Þ2 then Zfðn þ 1ÞuðnÞg ¼ . . . . . . . . . . . . 28 z2 2 ðz 1Þ Because Zfxkþm g ¼ zm FðzÞ zm x0 þ zm1 x1 þ . . . þ zxm1 so z Z fk þ 1g ¼ z ½z 0 ðz 1Þ2 ¼ z2 ðz 1Þ2 3 Second shift theorem (shifting to the right) If Zff ðnÞg ¼ FðzÞ then Zff ðn mÞg ¼ zm FðzÞ the Z transform of the sequence that has been shifted by m places to the right. z then For example, given that Zff ðnÞuðnÞg ¼ z1 Zff ðn 3Þuðn 3Þg ¼ . . . . . . . . . . . . 174 Programme 5 29 1 z2 ðz 1Þ Because Zff ðn mÞg ¼ zm FðzÞ so z z1 1 ¼ 2 z ðz 1Þ Zff ðn 3Þuðn 3Þg ¼ z3 In this way we have derived the Z transform of the sequence f0, 0, 0, 1, 1, 1, . . .g by shifting the sequence f1, 1, 1, 1, . . .g three places to the right and defining the first three terms as zeros. Try this one. The sequence ff ðnÞuðnÞg with Z transform Zff ðnÞuðnÞg ¼ 1 , where a is a constant, is f. . . . . . . . . . . .g ðz aÞ 30 f ðnÞ ¼ an1 uðn 1Þ Because From the table of transforms the nearest transform to the one in question is z which is the Z transform of fan uðnÞg. Now ðz aÞ 1 1 z ¼ ðz aÞ z ðz aÞ ¼ z1 FðzÞ where FðzÞ ¼ Zfan uðnÞg and so 1 ¼ Z an1 uðn 1Þ ðz aÞ which is the Z transform of an uðnÞ, shifted one place to the right. 4 Translation If the sequence f ðnÞ has the Z transform Zff ðnÞg ¼ FðzÞ then the sequence an f ðnÞ has the Z transform Zfan f ðnÞg ¼ F a1 z . For example, Zfn uðnÞg ¼ z ðz 1Þ2 so that Zf2n n uðnÞg ¼ . . . . . . . . . . . . Difference equations and the Z transform 175 31 2z 2 ðz 2Þ Because z Since Zfn uðnÞg ¼ ðz 1Þ2 Zf2n n uðnÞg ¼ F 21 z ¼ ¼ ¼ FðzÞ then by the translation property 21 z ð21 z 1Þ 2 2z ðz 2Þ2 5 Final value theorem For the sequence f ðnÞ with Z transform FðzÞ z1 FðzÞ provided that Lim f ðnÞ exists. Lim f ðnÞ ¼ Lim z n!1 z!1 n!1 1n For example, the sequence f ðnÞ ¼ 2 uðnÞ has the Z transform z 2z . ¼ FðzÞ ¼ 1 z 2 2z 1 Now Lim z!1 and Lim n!1 z1 2ðz 1Þ FðzÞ ¼ Lim ¼0 z 2z 1 z!1 1 2 n uðnÞ ¼ 0 which confirms the final value theorem. Using the final value theorem the final value of the sequence with the Z transform FðzÞ ¼ 10z2 þ 2z ðz 1Þð5z 1Þ2 is . . . . . . . . . . . . 0.75 Because Lim z!1 ( ) z1 z1 10z2 þ 2z FðzÞ ¼ Lim z z z!1 ðz 1Þð5z 1Þ2 ( ) 10z þ 2 ¼ Lim z!1 ð5z 1Þ2 12 ¼ 16 ¼ 0:75 32 176 Programme 5 6 The initial value theorem For the sequence f ðnÞ with Z transform FðzÞ f ð0Þ ¼ Lim fFðzÞg z!1 For example, the sequence f ðnÞ ¼ an uðnÞ has the Z transform FðzÞ ¼ z 1 ¼ Lim ¼ 1 z a z!1 z!1 1 z and za Lim FðzÞ ¼ Lim z!1 by L’Hôpital’s rule. Furthermore f ð0Þ ¼ a0 ¼ 1, so demonstrating the validity of the theorem. 7 The derivative of the transform If Zff ðnÞg ¼ FðzÞ then zF 0 ðzÞ ¼ Zfn f ðnÞg This is easily proved. 1 1 1 X X 1X FðzÞ ¼ f ðnÞzn and so F 0 ðzÞ ¼ f ðnÞðnÞzn1 ¼ f ðnÞn zn z n¼0 n¼0 n¼0 1 ¼ Zfn f ðnÞg z and so zF 0 ðzÞ ¼ Zfn f ðnÞg For example, the sequence f ðnÞ ¼ an uðnÞ has the Z transform FðzÞ ¼ z and za so the sequence n an uðnÞ has Z transform Zfn an uðnÞg ¼ zF 0 ðzÞ ¼ . . . . . . . . . . . . 33 Zfn an uðnÞg ¼ az ðz aÞ2 Because z 0 zaz zF ðzÞ ¼ z ¼ z za ðz aÞ2 0 ! ¼ az ðz aÞ2 Notice that this is in agreement with the Table of transforms in Frame 23. Next frame 34 Summary We now summarise the properties that we have just discussed. Linearity Zfaf ðnÞ þ bgðnÞg ¼ aZff ðnÞg þ bZfgðnÞg Difference equations and the Z transform 177 If Zff ðnÞg ¼ FðzÞ then: Shifting to the left Zff ðn þ mÞg ¼ zm FðzÞ zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm 1Þ Shifting to the right Zff ðn mÞg ¼ zm FðzÞ Translation Zfan f ðnÞg ¼ F a1 z Final value theorem z1 FðzÞ provided Lim f ðnÞ exists Lim f ðnÞ ¼ Lim z n!1 z!1 n!1 Initial value theorem f ð0Þ ¼ Lim fFðzÞg z!1 Derivative of the transform zF 0 ðzÞ ¼ Zfnf ðnÞg Inverse transforms If the sequence f ðnÞ has Z transform Zff ðnÞg ¼ FðzÞ, the inverse transform is defined as Z1 FðzÞ ¼ f ðnÞ There are many times when, given the Z transform of a sequence, it is not possible to immediately read off the sequence from the Table of transforms. Instead some manipulation may be required and, as with Laplace transforms, very often this involves using partial fractions. Example z . To find the inverse z2 5z þ 6 transform, and hence the sequence, we recognise that the denominator can be factorised and separated into partial fractions as The sequence f ðnÞ has Z transform FðzÞ ¼ FðzÞ ¼ . . . . . . . . . . . . 35 178 Programme 5 36 FðzÞ ¼ 3 2 z3 z2 Because z z2 5z þ 6 z ¼ ðz 2Þðz 3Þ A B þ ¼ z2 z3 Aðz 3Þ þ Bðz 2Þ ¼ ðz 2Þðz 3Þ FðzÞ ¼ Equating numerators gives z ¼ Aðz 3Þ þ Bðz 2Þ, giving A þ B ¼ 1 and 3A 2B ¼ 0. From these two equations we find that A ¼ 2 and B ¼ 3. So FðzÞ ¼ 3 2 z3 z2 The nearest Z transform in the table to either of these two partial fractions z . Therefore if we write is Zfan uðnÞg ¼ za 3 2 FðzÞ ¼ z3 z2 3 z 2 z ¼ z z3 z z2 so Z1 FðzÞ ¼ . . . . . . . . . . . . 37 Z1 FðzÞ ¼ ð3n 2n ÞuðnÞ Because 3 z 2 z z z3 z z2 ¼ 3 z1 Zf3n uðnÞg 2 z1 Zf2n uðnÞg FðzÞ ¼ and so Z1 FðzÞ ¼ 3 3n1 uðn 1Þ 2 2n1 uðn 1Þ by the second shift theorem ¼ 3n uðnÞ 2n uðnÞ So f ðnÞ ¼ ð3n 2n ÞuðnÞ. There is a simpler way of doing this without employing the second shift theorem. Recognising that z appears in the numerator of FðzÞ, we consider FðzÞ instead the partial fraction breakdown of z FðzÞ ¼ ............ z Difference equations and the Z transform 1 1 z3 z2 Because FðzÞ 1 z ¼ 2 z z z 5z þ 6 1 ¼ 2 z 5z þ 6 1 ¼ ðz 2Þðz 3Þ A B ¼ þ z2 z3 Aðz 3Þ þ Bðz 2Þ ¼ ðz 2Þðz 3Þ Equating numerators gives 1 ¼ Aðz 3Þ þ Bðz 2Þ, giving [z]: [CT]: FðzÞ z FðzÞ AþB¼0 3A 2B ¼ 1 with solution A ¼ 1 and B ¼ 1. So that 1 1 that is z3 z2 z z ¼ z3 z2 ¼ Zf3n uðnÞg Zf2n uðnÞg and so ¼ Z1 FðzÞ ¼ 3n uðnÞ 2n uðnÞ ¼ ð3n 2n ÞuðnÞ Thus the use of the second shift theorem is avoided. So try one yourself. The sequence f ðnÞ has Z transform FðzÞ ¼ 5z ðz2 4z þ 4Þðz þ 2Þ therefore f ðnÞ ¼ . . . . . . . . . . . . 179 38 180 Programme 5 39 5 ð2n 1Þ2n þ ð2Þn uðnÞ 16 Because FðzÞ 1 5z ¼ 2 z z ðz 4z þ 4Þðz þ 2Þ 5 ¼ ðz 2Þ2 ðz þ 2Þ A B C þ þ ¼ ðz 2Þ2 z 2 z þ 2 ¼ Aðz þ 2Þ þ Bðz 2Þðz þ 2Þ þ Cðz 2Þ2 ðz 2Þ2 ðz þ 2Þ Equating numerators gives 5 ¼ Aðz þ 2Þ þ B z2 4 þ C z2 4z þ 4 , giving [z2 ]: [z]: [CT]: BþC¼0 A 4C ¼ 0 2A 4B þ 4C ¼ 5 with solution A ¼ 5=4, B ¼ 5=16 and C ¼ 5=16, so FðzÞ 5=4 5=16 5=16 ¼ þ giving 2 z z 2 zþ2 ðz 2Þ FðzÞ ¼ 5 2z 5 z 5 z þ and so 8 ðz 2Þ2 16 z 2 16 z þ 2 5 5 5 n2n uðnÞ 2n uðnÞ þ ð2Þn uðnÞ 8 16 16 5 ¼ ð2n 1Þ2n þ ð2Þn uðnÞ 16 Move on to the next frame Z1 FðzÞ ¼ Solving difference equations 40 If a sequence satisfies a difference equation with given initial terms then the general term of the sequence can be found by using the Z transform. For example, to solve the difference equation f ðn þ 2Þ 5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 we begin by taking the Z transform of both sides of the equation to give: Zff ðn þ 2Þ 5f ðn þ 1Þ þ 6f ðnÞg ¼ Zf1g that is Zff ðn þ 2g 5Zff ðn þ 1Þg þ 6Zff ðnÞg ¼ Zf1g Using the first shift theorem where Zff ðnÞg ¼ FðzÞ this then becomes 2 z z FðzÞ z2 f ð0Þ zf ð1Þ ð5zFðzÞ zf ð0ÞÞ þ 6FðzÞ ¼ z1 Difference equations and the Z transform 181 Collecting like terms and substituting for the initial terms f ð0Þ ¼ 0 and f ð1Þ ¼ 1 gives 2 z 5z þ 6 FðzÞ z ¼ z z z2 so z2 5z þ 6 FðzÞ ¼ z þ ¼ that is z1 z1 z1 2 2 z z FðzÞ ¼ ¼ and so 2 ðz 1Þðz 5z þ 6Þ ðz 1Þðz 2Þðz 3Þ FðzÞ z ¼ z ðz 1Þðz 2Þðz 3Þ This has the partial fraction breakdown FðzÞ . . . . . . . . . . . . . . . . . . ¼ þ þ z z1 z2 z3 The answer is in the next frame FðzÞ ¼ 1=2 4 9=2 þ z1 z2 z3 41 Because: Letting z A B C ¼ þ þ ðz 1Þðz 2Þðz 3Þ z 1 z 2 z 3 Aðz 2Þðz 3Þ þ Bðz 1Þðz 3Þ þ Cðz 1Þðz 2Þ ðz 1Þðz 2Þðz 3Þ and so z ¼ Aðz 2Þðz 3Þ þ Bðz 1Þðz 3Þ þ Cðz 1Þðz 2Þ. ¼ Taking z ¼ 1, 2 and 3 in turn yields A ¼ 1=2, B ¼ 2 and C ¼ 3=2. Consequently, FðzÞ ¼ 1 z z 3 z 2 þ and so f ðnÞ ¼ . . . . . . . . . . . . 2 z1 z2 2 z3 The answer is in the next frame f ðnÞ ¼ 1 3nþ1 2nþ1 þ uðnÞ 2 2 Because: f ðnÞ ¼ Z1 fFðzÞg 1 z z 3 z 2 þ 2 z1 z2 2 z3 n z o 3 n z o 1 1 n z o 1 2Z þ Z1 ¼ Z 2 z1 z2 2 z3 1 3 n n ¼ uðnÞ 2 2 uðnÞ þ 3 uðnÞ 2 2 nþ1 1 3 2nþ1 þ ¼ uðnÞ 2 2 ¼ Z1 42 182 Programme 5 Try one yourself. The solution of the second order difference equation f ðn þ 2Þ f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 is f ðnÞ ¼ . . . . . . . . . . . . Next frame 43 f ðnÞ ¼ 1 3 ð2n 3Þ þ ð1Þn uðnÞ 4 4 Because: Taking the Z transform of the difference equation gives Zff ðn þ 2Þ f ðnÞg ¼ Zf1g. That is Zff ðn þ 2Þg Zff ðnÞg ¼ Zf1g so that 2 z . z FðzÞ z2 f ð0Þ zf ð1Þ FðzÞ ¼ z1 Substituting f ð0Þ ¼ 0 and f ð1Þ ¼ 1 gives FðzÞ ¼ . . . . . . . . . . . . Next frame 44 FðzÞ ¼ z2 þ 2z ðz þ 1Þðz 1Þ2 Because: 2 z becomes z FðzÞ z2 f ð0Þ zf ð1Þ FðzÞ ¼ z1 2 z z FðzÞ þ z FðzÞ ¼ so that z1 z z2 þ 2z ðz2 1ÞFðzÞ ¼ z þ ¼ and so z1 z1 z2 þ 2z z2 þ 2z FðzÞ ¼ 2 ¼ ðz 1Þðz 1Þ ðz þ 1Þðz 1Þ2 Therefore FðzÞ ...... ...... ...... þ ¼ þ 2 z z1 zþ1 ðz 1Þ Next frame Difference equations and the Z transform 183 45 FðzÞ 1=2 3=4 3=4 ¼ þ 2 z z 1 z þ1 ðz 1Þ Because: FðzÞ z þ 2 ¼ z ðz þ 1Þðz 1Þ2 ¼ ¼ A ðz 1Þ2 þ B C þ z1 zþ1 Aðz þ 1Þ þ Bðz þ 1Þðz 1Þ þ Cðz 1Þ2 ðz þ 1Þðz 1Þ2 giving z þ 2 ¼ Aðz þ 1Þ þ Bðz þ 1Þðz 1Þ þ Cðz 1Þ2 1 3 3 and hence A ¼ , B ¼ and C ¼ 2 4 4 From this we conclude that: f ðnÞ ¼ Z1 fFðzÞg ( ) 1 1 z 3 1 n z o 3 1 z Z Z ¼ Z þ 2 4 z1 4 zþ1 ðz 1Þ2 1 3 3 n uðnÞ uðnÞ þ ð1Þn uðnÞ 2 4 4 1 3 ð2n 3Þ þ ð1Þn uðnÞ ¼ 4 4 ¼ Move on to the next frame Sampling If a continuous function f ðtÞ of time t progresses from t ¼ 0 onwards and is measured at every time interval T then what will result is the sequence of values ff ðkTÞg ¼ ff ð0Þ, f ðTÞ, f ð2TÞ, f ð3TÞ; . . .g A new, piecewise continuous function f ðtÞ can then be created from the sequence of sampled values such that f ðtÞ ¼ f ðkTÞ 0 if t ¼ kT otherwise 46 184 Programme 5 The graph of this new function consists of a series of spikes at the regular intervals t ¼ kT f *(t) T 2T 3T nT This function can alternatively be described in terms of the delta function ðtÞ as f ðtÞ ¼ f ð0ÞðtÞ þ f ðTÞðt TÞ þ f ð2TÞðt 2TÞ þ f ð3TÞðt 3TÞ þ . . . 1 X f ðkTÞðt kTÞ ¼ k¼0 The Laplace transform of f ðtÞ is then given as F ðsÞ ¼ Lff ðtÞg ð1 ¼ ff ð0ÞðtÞ þ f ðTÞðt TÞ þ f ð2TÞðt 2TÞ þ . . .gest dt 0 ¼ f ð0Þ þ f ðTÞesT þ f ð2TÞe2sT þ f ð3TÞe3sT þ . . . 1 X f ðkTÞeksT ¼ k¼0 Define a new variable z ¼ esT and we see that 1 1 X X f ðkTÞ Lff ðtÞg ¼ f ðkTÞzk ¼ zk k¼0 k¼0 which is the Z transform of the sequenceff ðkTÞg. Example 1 The function f ðtÞ ¼ eat is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . . 47 FðzÞ ¼ z z eaT Because Defining f ðtÞ ¼ P1 k¼0 f ðkTÞðt kTÞ ¼ transform of f ðtÞ is given as F ðsÞ ¼ 1 X k¼0 ekaT eksT P1 k¼0 eakT ðt kTÞ then the Laplace Difference equations and the Z transform 185 This means that the Z transform of ff ðkTÞg is FðzÞ ¼ 1 kaT X e k¼0 zk ¼ 1 z ¼ aT z eaT e 1 z Notice that this agrees with the Z transform of the sequence bn uðnÞ z when b is replaced by eaT . which is zb Try another. Example 2 The function f ðtÞ ¼ t is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . . FðzÞ ¼ Because The Z transform of ff ðkTÞg is FðzÞ ¼ 1 X f ðkTÞ k¼0 FðzÞ ¼ 48 Tz ðz 1Þ2 zk . Here f ðkTÞ ¼ kT and so 1 X kT k¼0 ¼T zk 1 2 3 þ 2 þ 3 þ ... z z z T 1 þ 2z1 þ 3z2 þ 4z3 þ . . . z d 1 þ z1 þ z2 þ z3 þ . . . ¼ Tz dz d 1 1 T 1 2 Tz ¼ Tz 1 1 ¼ ¼ dz z z z ðz 1Þ2 ¼ Example 3 The function f ðtÞ ¼ cos t is sampled every interval of T. The Z transform of the sampled function is then . . . . . . . . . . . . 186 Programme 5 49 FðzÞ ¼ zðz cos TÞ z2 2 cos T þ 1 Because f ðTÞ ¼ cos T ¼ FðzÞ ¼ e jT þ ejT and the Z transform of ekaT is 2 z . z eaT e jT þ ejT is Therefore the Z transform of 2 jT þ z z ejT 1 z z 1 z ze þ ¼ 2 z ejT z e jT 2 ðz ejT Þðz e jT Þ 2 1 2z z e jT þ ejT ¼ 2 z2 ½e jT þ ejT z þ 1 zðz cos TÞ ¼ 2 z 2z cos T þ 1 And that is the end of the Programme on Z transforms. All that remain are the Revision summary and the Can you? checklist. Read through these closely and make sure that you understand all the workings of this Programme. Then try the Test exercise; there is no need to hurry, take your time and work through the questions carefully. The Further problems then provide a valuable collection of additional exercises for you to try. Revision summary 5 50 1 Sequences Any function f whose input is restricted to integer values n has an output f ðnÞ in the form of a discrete sequence of numbers. A sequence can be defined by a prescription for the nth term. Alternatively, it can be defined recursively where terms are defined by the values of previous terms. A recursively defined sequence requires one or more initial terms to start the process of evaluating successive terms. 2 Difference equations The equation that recursively defines a sequence is called a difference equation. A linear, constant coefficient difference equation consists of a sum of general terms of the sequence, each multiplied by a constant. The order of a difference equation is the maximum number of terms between any pair of terms in the equation. Difference equations and the Z transform 187 3 Solving difference equations In analogy with linear constant coefficient inhomogeneous differential equations, a linear constant coefficient inhomogeneous difference equation can be solved by first finding the inhomogeneous solution in terms of unknown constants, adding this to the particular solution and then applying the initial terms to find the values of the unknown constants. 4 Z transform The Z transform of the sequence f ðnÞ is 1 X f ðnÞ ¼ FðzÞ where the value of z is chosen to zn n¼1 ensure that the sum converges. f ðnÞ and Zff ðnÞg form a Z transform pair. Zff ðnÞg ¼ 5 Table of Z transforms Sequence ðnÞ ¼ f1, 0, 0, . . .g uðnÞ ¼ f1, 1, 1, . . .g n uðnÞ ¼ f0, 1, 2, 3, . . .g n2 uðnÞ ¼ f0, 1, 4, 9, . . .g n3 uðnÞ ¼ f0, 1, 8, 27, . . .g an uðnÞ ¼ 1, a, a2 , a3 , . . . n an uðnÞ ¼ 0, a, 2a2 , 3a3 , . . . 6 Transform F(z) 1 z z1 z ðz 1Þ2 zðz þ 1Þ ðz 1Þ3 z z2 þ 4z þ 1 ðz 1Þ4 z ðz aÞ az ðz aÞ2 Permitted values of z All values of z jzj > 1 jzj > 1 jzj > 1 jzj > 1 jzj > jaj jzj > jaj Linearity The Z transform is a linear transform. That is, if a and b are constants then Zfaf ðnÞ þ bgðnÞg ¼ aZff ðnÞg þ bZfgðnÞg. 7 First shift theorem (shifting to the left) If Zff ðnÞg ¼ FðzÞ then Zff ðn þ mÞg ¼ zm FðzÞ zm f ð0Þ þ zm1 f ð1Þ þ . . . þ zf ðm 1Þ the Z transform of the sequence that has been shifted by m places to the left. 8 Second shift theorem (shifting to the right) If Zff ðnÞg ¼ FðzÞ then Zfn mg ¼ zm FðzÞ the Z transform of the sequence that has been shifted by m places to the right. 188 Programme 5 9 Translation If the sequence f ðnÞ has the Z transform Zff ðnÞg ¼ FðzÞ then the sequence an f ðnÞ has the Z transform Zfan f ðnÞg ¼ F a1 z . 10 Final value theorem For the sequence f ðnÞ with Z transform FðzÞ z1 FðzÞ provided that Lim f ðnÞ exists. Lim f ðnÞ ¼ Lim z n!1 z!1 n!1 11 The initial value theorem For the sequence f ðnÞ with Z transform FðzÞ f ð0Þ ¼ Lim fFðzÞg. z!1 12 The derivative of the transform If Zff ðnÞg ¼ FðzÞ then zF 0 ðzÞ ¼ Zfnf ðnÞg. 13 Inverse transformations If the sequence f ðnÞ has Z transform Zff ðnÞg ¼ FðzÞ, the inverse transform is defined as Z1 FðzÞ ¼ f ðnÞ. 15 Solving difference equations If a sequence f ðnÞ satisfies a difference equation with given initial conditions then the general term of the sequence can be found by using the Z transform where Zff ðnÞg ¼ FðzÞ. This is referred to as solving the difference equation. 15 Sampling If a continuous function f ðtÞ is sampled at equal intervals, the resulting sequence has a Z transform that is related to the Laplace transform of the piecewise function created from the sequence of sample values. Lff ðtÞg ¼ 1 X k¼0 f ðkTÞzk ¼ 1 X f ðkTÞ k¼0 zk ¼ Zff ðkTÞg where ff ðkTÞg ¼ ff ð0Þ, f ðTÞ, f ð2TÞ, f ð3TÞ, . . .g, f ðkTÞ if t ¼ k f ðtÞ ¼ 0 otherwise and z ¼ esT . Difference equations and the Z transform 189 Can you? 51 Checklist 5 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Convert the descriptive prescription of the output form of a sequence into a recursive description and recognise the importance of initial terms? Yes No Frames 1 to 3 . Recognise a difference equation, determine its order and generate its terms from a recursive description? Yes No 4 to 8 . Obtain the solution to a difference equation as a sum of the homogeneous solution and the particular solution? Yes No 9 to 16 . Define the Z transform of a sequence and derive transforms of specified sequences? Yes No 17 to 22 . Make reference to a table of standard Z transforms? Yes No . Recognise the Z transform as being a linear transform and so obtain the transform of linear combinations of standard sequences? Yes No . Apply the first and second shift theorems, the translation theorem, the initial and final value theorems and the derivative theorem? Yes No . Use partial fractions to derive the inverse transforms Yes No . Use the Z transform to solve linear, constant coefficient difference equations? Yes No . Create a sequence by sampling a continuous function and demonstrate the relationship between the Laplace and the Z transform? Yes No 23 24 to 26 26 to 34 35 to 39 40 to 45 46 to 49 190 Programme 5 Test exercise 5 52 1 Find a recursive description corresponding to each of the following prescriptions for the output of a sequence: (a) f ðnÞ ¼ 5n 9 where n is an integer 1 (b) f ðnÞ ¼ 23 4n where n is an integer 0 (c) f ðnÞ ¼ 3n where n is an integer 2 2 Determine the order and find the first six terms of each of the following sequences: (a) f ðn þ 3Þ f ðnÞ ¼ 5n where f ð0Þ ¼ 1, f ð1Þ ¼ 1 and f ð2Þ ¼ 3 (b) f ðn þ 1Þ 5f ðnÞ þ 6f ðn 1Þ ¼ 2n where f ð1Þ ¼ 0 and f ð0Þ ¼ 1 (c) f ðn þ 2Þ f ðn þ 1Þ þ 12f ðnÞ ¼ 3uðnÞ where f ð0Þ ¼ 2 and f ð1Þ ¼ 5 3 Obtain the solution to the following difference equation in the form of a sum of homogeneous and particular solutions: f ðn þ 1Þ 5f ðnÞ þ 6f ðn 1Þ ¼ 2n where f ð1Þ ¼ 0 and f ð0Þ ¼ 1 Check that your answer is in agreement with the answer to 2(b). 4 Find the Z transform of each of the sequences with output: (a) f ðnÞ ¼ ð1Þn uðnÞ (b) f ðnÞ ¼ ð4n 2an ÞuðnÞ (c) f ðnÞ ¼ ðn 3ÞuðnÞ (d) f ðnÞ ¼ 5nþ2 uðnÞ 5 Find the inverse Z transform of z2 ðz 3Þ FðzÞ ¼ 2 . ðz 2z þ 1Þðz 2Þ 6 Solve the difference equation f ðn þ 2Þ 4f ðn þ 1Þ þ 4f ðnÞ ¼ 3 where f ð0Þ ¼ 1 and f ð1Þ ¼ 0. 7 The function f ðtÞ ¼ sin t is sampled at equal intervals of t ¼ T. Find the Z transform of the resulting sequence of values. Further problems 5 53 1 Find the Z transform of f ðnÞ ¼ ðaÞn where a > 0. 2 Solve each of the following difference equations in the form of the homogeneous solution plus the particular solution: (a) f ðn þ 2Þ þ 5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 (b) 3f ðn þ 2Þ 7f ðn þ 1Þ þ 2f ðnÞ ¼ n where f ð0Þ ¼ 1 and f ð1Þ ¼ 0 (c) f ðn þ 2Þ 9f ðnÞ ¼ 2n2 where f ð0Þ ¼ 1 and f ð1Þ ¼ 1. Difference equations and the Z transform 3 Given that aðn þ 1Þ ¼ bðnÞ and that bðn þ 1Þ ¼ cðnÞ where cðnÞ ¼ f ðnÞ gðnÞ, show that f ðn þ 2Þ þ f ðnÞ ¼ gðnÞ and solve for f ðnÞ when gðnÞ ¼ ðnÞ, the unit impulse sequence where f ð0Þ ¼ 0 and f ð1Þ ¼ 1. 4 If pðn þ 1Þ ¼ qðnÞ qðn þ 1Þ ¼ rðnÞ rðnÞ ¼ f ðnÞ qðnÞ pðnÞ where and are constants, show that pðn þ 2Þ þ pðn þ 1Þ þ pðnÞ ¼ f ðnÞ Solve this recurrence relation when f ð0Þ ¼ 1, f ð1Þ ¼ 0 for (a) ¼ 4, ¼ 4 and f ðnÞ ¼ ðnÞ, the unit impulse sequence (b) ¼ 4, ¼ 4 and f ðnÞ ¼ uðnÞ the unit step sequence. 5 Find the Z transform of each of the following sequences. (a) f1, 0, 1, 0, 1, 0, . . .g (b) f0, 1, 0, 1, 0, 1, . . .g (c) f1, 0, 1, 1, 0, 0, 0, 1g (d) f1, 1, 1, 0, 0, 0, 1, 1g (e) f0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1g (f) f1, 1, 0, 0, 0, 1, 1g Note that the last four of these are finite sequences. 6 Find the inverse transform of z (a) FðzÞ ¼ ðz þ 1Þðz þ 2Þðz þ 3Þ z2 ðz þ 1Þðz þ 2Þðz þ 3Þ zð3z þ 1Þ (c) FðzÞ ¼ ðz 2Þðz 3Þ (b) FðzÞ ¼ (d) FðzÞ ¼ 7 z2 . 2 3z þ z2 Given 3z2 z2 z þ 1 show that FðzÞ ¼ Z1 FðzÞ ¼ f3, 3, 3, 3, . . .g. Hint: Use long division on FðzÞ. 8 Given FðzÞ ¼ 1þ 2 z 3 show that Z1 FðzÞ ¼ f1, 6, 24, 48, . . .g. Hint: Use the binomial theorem on FðzÞ. 191 192 Programme 5 9 Find the final value of the sequence f ðnÞ with Z transform 4z2 z . FðzÞ ¼ 2 2z 3z þ 1 10 What is the initial value of the sequence whose Z transform is given by 2z2 z þ 1 FðzÞ ¼ ? 5 3z 7z2 11 Given the sequence of n terms f ðkÞ for 0 k n 1 with Z transform Fn ðzÞ, show that the Z transform of the sequence formed by continually repeating the terms f ðkÞ is given as Fn ðzÞ . FðzÞ ¼ 1 zn 12 Using the result of Question 11, show that the Z transform of the sequence obtained by continually repeating the three term sequence f1, 0, 1g is z2 . FðzÞ ¼ 2 z þ1 13 Find the Z transforms of the sequence of values obtained when f ðtÞ is sampled at regular intervals of t ¼ T where (a) f ðtÞ ¼ sinh t (b) f ðtÞ ¼ cosh at (c) f ðtÞ ¼ eat cosh bt. 14 Solve each of the following difference equations using the Z tranform (a) f ðn þ 2Þ þ 5f ðn þ 1Þ þ 6f ðnÞ ¼ 1 where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 (b) f ðn þ 2Þ 7f ðn þ 1Þ þ 2f ðnÞ ¼ n where f ð0Þ ¼ 1 and f ð1Þ ¼ 0 (c) 3f ðn þ 2Þ 9f ðnÞ ¼ 2 where f ð0Þ ¼ 1 and f ð1Þ ¼ 1 (d) f ðn þ 2Þ þ 2f ðn þ 1Þ 15f ðnÞ ¼ 4n where f ð0Þ ¼ 0 and f ð1Þ ¼ 1 15 If f ðn þ 1Þ ¼ 3ðn þ 1Þf ðnÞ show that f ðn þ 1Þ ¼ 3nþ1 ðn þ 1Þ!f ð0Þ 16 Show that the difference equation gðn þ 2Þ gðn þ 1Þ 6gðnÞ ¼ 0 can be derived from the coupled difference equation f ðn þ 1Þ ¼ gðnÞ gðn þ 1Þ ¼ gðnÞ þ 6f ðnÞ Find f ðnÞ and gðnÞ given that f ð1Þ ¼ 0 and gð1Þ ¼ 1. 17 Show that f ðnÞ ¼ n!uðnÞ satisfies the difference equation f ðn þ 1Þ ðn þ 1Þf ðnÞ ¼ ðn þ 1Þ. 18 Use the derivative property to find the Z transform of f ðnÞ ¼ 3n n uðn 3Þ. 19 Solve the equation for the Fibonacci sequence: f ðn þ 2Þ ¼ f ðn þ 1Þ þ f ðnÞ where f ð0Þ ¼ 0, f ð1Þ ¼ 1 and n 0 Programme 6 Frames 1 to 46 Introduction to invariant linear systems Learning outcomes When you have completed this Programme you will be able to: . Recognise a system as a process whereby an input (either continuous or discrete) is converted to an output, also called the response of the system . Distinguish between linear and non-linear systems and recognise time-invariant and shift-invariant systems . Determine the zero-input response and the zero-state response . Appreciate why zero valued boundary conditions give rise to a time-invariant system . Demonstrate that the response of a continuous, linear, time-invariant system to an arbitrary input is the convolution of the input with response of the system to a unit impulse . Understand the role of the exponential function with respect to a linear, time-invariant system . Use the convolution theorem to find the response of a continuous, linear, time-invariant system to an arbitrary input . Derive the system transfer function of a constant coefficient linear differential equation and use it to solve the equation . Demonstrate that the response of a discrete, linear, shift-invariant system to an arbitrary input is the convolution sum of the input with response of the system to a unit impulse . Understand the role of the exponential function with respect to a discrete linear, shift-invariant system . Derive the system transfer function of a constant coefficient linear difference equation and use it to solve the equation . Derive the constant coefficient difference equation from knowledge of its unit impulse response. Prerequisites: Advanced Engineering Mathematics (Fifth Edition) Programme 3 Laplace transforms 2, Programme 4 Laplace transforms 3 and Programme 5 Difference equations and the Z transform 193 194 Programme 6 Invariant linear systems 1 Systems A system is a process that is capable of accepting an input, processing the input and producing an output, also called the response of the system. In Engineering Mathematics, Sixth Edition a function was described as an example of a system where the input was a number x which was processed by the function f to produce a number output f ðxÞ as shown in the box diagram: x f f (x) For example the function f with input x and output f ðxÞ ¼ sin x can be represented as: x f(x) = sin x f This system description simply links the input number to the output number via the function. How the function performs the process of evaluating the sine of the input number is not accounted for in this description it is just accepted that the function f can do it. In this Programme we are going to extend this application of a system to one that will accept an expression as input, process the expression and produce another expression as output. For continuous inputs the box diagram for this system will be: x(t) L y(t) that is yðtÞ ¼ LfxðtÞg or LfxðtÞg ¼ yðtÞ Here the input and output expressions involve the parameter t. In what follows we shall take this to represent the variable time but it can represent whatever variable is appropriate to the problem in hand. For a discrete system the box diagram is: x [n] L y[n] that is y½n ¼ Lfx½ng or Lfx½ng ¼ y½n Here the input and output expressions involve the discrete integer parameter n. For the purposes of this Programme the integer parameter is placed within square braces to indicate the discrete nature as opposed to round braces used to indicate a continuous nature. That is: x1 , x2 , x3 , . . . ;xn , . . . or x½1, x½2, x½3, . . . , x½n, . . . Just as a system can be used to describe the input–output relationship linking two numbers so a system can be used to describe the input–output relationship linking two expressions. What we need to look for now are 195 Introduction to invariant linear systems input–output relationships linking two expressions that can be described by a system. Further, just as there are many different types of relationship there are many different types of system. The specific type of system we shall be interested in is an invariant linear system (but we get ahead of ourselves). Move to the next frame 2 Input-response relationships Many physical situations in science and engineering can be described by a linear, constant coefficient, ordinary differential equation of the type met in the previous Programmes. Their method of solution may differ depending on the structure of the differential equation but the desire to obtain the solution is common to all. Take for instance the particularly simple first order differential equation dy1 ðtÞ ¼ 2t where y 1 ð0Þ ¼ 0 dt By integrating this equation: ð ð ð dy1 ðtÞ t2 dt ¼ 2t dt that is dy1 ðtÞ ¼ y1 ðtÞ ¼ 2 þ C ¼ t 2 þ C dt 2 and applying the boundary condition y1 ð0Þ ¼ 0 ¼ 0 þ C we arrive at the solution y1 ðtÞ ¼ t 2 . The same equation, but with a different right-hand side, dy2 ðtÞ ¼ 4t 3 where y2 ð0Þ ¼ 0 has solution . . . . . . . . . . . . dt The answer is in the next frame t4 Because: By integrating this equation: ð ð ð dy2 ðtÞ t4 dt ¼ 4t 3 dt that is dy2 ðtÞ ¼ y2 ðtÞ ¼ 4 þ C0 ¼ t 4 þ C0 dt 4 and applying the boundary condition y2 ð0Þ ¼ 0 ¼ 0 þ C0 we arrive at the solution y2 ðtÞ ¼ t 4 . The general form of this simple equation can be given as: dyðtÞ ¼ xðtÞ where yð0Þ ¼ 0 dt and in both cases we insert the specific expression xðtÞ in the right-hand side and then manipulate the equation to obtain the solution yðtÞ. It is this commonality of procedure that merits further study. In each case, the method used to find the solution can be represented by a system where the differential equation specifies the relationship between the 3 196 Programme 6 input and the output. The process is that of integration and evaluating the integration constant, the input is the term on the right-hand side and the output or the system response is what we are trying to find, the solution to the differential equation. We can use a box diagram to represent the system: 2t L 4t 3 t2 L t4 In the first box diagram 2t is input and t 2 is the response and in the second box diagram 4t 3 is input and t 4 is the response The process L is the same for each differential equation; what differs are the respective inputs and their corresponding responses. The response of the differential equation where yð0Þ ¼ 0 is dyðtÞ ¼ xðtÞ to the input xðtÞ ¼ sin t dt yðtÞ ¼ . . . . . . . . . . . . The answer is in the next frame 4 cos t þ 1 Because: In the differential equation dyðtÞ ¼ xðtÞ, yðtÞ is the response to the input dt xðtÞ ¼ sin t so that ð ð dyðtÞ dt ¼ sin t dt. dt That is ð dyðtÞ ¼ yðtÞ ¼ cos t þ A where A is the integration constant. and applying the boundary condition yð0Þ ¼ 0 ¼ 1 þ A we arrive at the solution yðtÞ ¼ cos t þ 1. Move to the next frame 5 Linear systems Systems that are linear are of particular interest because many problems in science and engineering can be posed as linear systems. A system yðtÞ ¼ LfxðtÞg is linear if sums and scalar multiples are preserved, that is if Lfx1 ðtÞ þ x2 ðtÞg ¼ Lfx1 ðtÞg þ Lfx2 ðtÞg and LfxðtÞg ¼ LfxðtÞg where is a constant. In particular, by choosing ¼ 0 then Lf0g ¼ 0 which shows that if nothing is put into a linear system nothing will come out – zero input yields zero output. 197 Introduction to invariant linear systems These two properties can be combined. yðtÞ ¼ LfxðtÞg is a linear system if: Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg where a and b are constants. For the discrete case, the system is linear if: Lfax1 ½n þ bx2 ½ng ¼ aLfx1 ½ng þ bLfx2 ½ng where a and b are constants. For example, consider the system in which the output is 5 times the input. That is: yðtÞ ¼ LfxðtÞg ¼ 5xðtÞ To show that this is a linear system we consider two distinct inputs x1 ðtÞ and x2 ðtÞ and their respective responses y1 ðtÞ ¼ 5x1 ðtÞ and y2 ðtÞ ¼ 5x2 ðtÞ. We also consider the linear combination of the inputs xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ where a and b are constants and where yðtÞ is the corresponding response. Then: yðtÞ ¼ LfxðtÞg ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ 5½ax1 ðtÞ þ bx2 ðtÞ ¼ 5ax1 ðtÞ þ 5bx2 ðtÞ the response is 5 times the input ¼ ay1 ðtÞ þ by2 ðtÞ ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg that is Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg Therefore the system is linear. On the other hand, the discrete system whose output is the square of the input, that is: y½n ¼ Lfx½ng ¼ x2 ½n is not linear because, if y1 ½n ¼ x1 2 ½n, y2 ½n ¼ x2 2 ½n and x½n ¼ ax1 ½n þ bx2 ½n where a and b are constants and where y½n is the corresponding response, then: y½n ¼ Lfx½ng ¼ Lfax1 ½n þ bx2 ½ng ¼ ½ax1 ½n þ bx2 ½n2 the response is the square of the input ¼ a x1 ½n þ 2abx1 ½nx2 ½n þ b2 x2 2 ½n 2 2 However, aLfx1 ½ng þ bLfx2 ½ng ¼ ax1 2 ½n þ bx2 2 ½n therefore Lfax1 ½n þ bx2 ½ng 6¼ aLfx1 ½ng þ bLfx2 ½ng and so the system is not linear. A system that is not linear is called a non-linear system. So, which of the following represent a linear system and which a non-linear system? (a) yðtÞ ¼ LfxðtÞg ¼ xðtÞ sin pt (b) yðtÞ ¼ LfxðtÞg ¼ exðtÞ (c) y½n ¼ Lfx½ng ¼ x½n þ 4x½n 1 (d) y½n ¼ Lfx½ng ¼ cosðx½nÞ The answers are in the following frame 198 Programme 6 6 (a) linear (b) non-linear (c) linear (d) non-linear Because: (a) yðtÞ ¼ LfxðtÞg ¼ xðtÞ sin pt so if y1 ðtÞ ¼ Lfx1 ðtÞg ¼ x1 ðtÞ sin pt, y2 ðtÞ ¼ Lfx2 ðtÞg ¼ x2 ðtÞ sin pt and xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ where a and b are constants and where yðtÞ is the corresponding response. Then: yðtÞ ¼ LfxðtÞg ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ ½ax1 ðtÞ þ bx2 ðtÞ sin pt the response is the input times sin pt ¼ ax1 ðtÞ sin pt þ bx2 ðtÞ sin pt ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg and so sums and scalar products are preserved The system is linear. (b) yðtÞ ¼ LfxðtÞg ¼ exðtÞ so if y1 ðtÞ ¼ Lfx1 ðtÞg ¼ ex1 ðtÞ , y2 ðtÞ ¼ Lfx2 ðtÞg ¼ ex2 ðtÞ and xðtÞ ¼ ax1 ðtÞ þ bx2 ðtÞ where a and b are constants and where yðtÞ is the corresponding response. Then: yðtÞ ¼ LfxðtÞg ¼ Lfax1 ðtÞ þ bx2 ðtÞg ¼ eax1 ðtÞþbx2 ðtÞ ax1 ðtÞ the response is e to the power of the input bx2 ðtÞ e ¼e ¼ Lfax1 ðtÞg Lfbx2 ðtÞg 6¼ Lfax1 ðtÞg þ Lfbx2 ðtÞg Therefore yðtÞ ¼ LfxðtÞg ¼ exðtÞ is not a linear system – it is a non-linear system (c) y½n ¼ Lfx½ng ¼ x½n þ 4x½n 1 so if y1 ½n ¼ Lfx1 ½ng ¼ x1 ½n þ 4x1 ½n 1, y2 ½n ¼ fx2 ½ng ¼ x1 ½n þ 4x2 ½n 1 and x½n ¼ ax1 ½n þ bx2 ½n where a and b are constants and where y½n is the corresponding response. Then: y½n ¼ Lfx½ng ¼ Lfax1 ½n þ bx2 ½ng ¼ ðax1 ½n þ bx2 ½nÞ þ 4ðax1 ½n 1 þ bx2 ½n 1Þ ¼ aðx1 ½n þ 4x1 ½n 1Þ þ bðx2 ½n þ 4x2 ½nÞ ¼ aLfx1 ½ng þ bLfx2 ½ng The system is linear. and so sums and scalar products are preserved 199 Introduction to invariant linear systems (d) y½n ¼ Lfx½ng ¼ cosðx½nÞ so if y1 ½n ¼ Lfx1 ½ng ¼ cosðx1 ½nÞ, y1 ½n ¼ Lfx1 ½ng ¼ cosðx2 ½nÞ and x½n ¼ ax1 ½n þ bx2 ½n where a and b are constants and where y½n is the corresponding response. Then: y½n ¼ Lfx½ng ¼ Lfax1 ½n þ bx2 ½ng ¼ cosðax1 ½n þ bx2 ½nÞ 6¼ Lfax1 ½ng þ Lfbx2 ½ng because Lfax1 ½ng þ Lfbx2 ½ng ¼ cosðax1 ½nÞ þ cosðbx2 ½nÞ. The system is non-linear. Move to the next frame 7 Time-invariance of a continuous system Consider the plot of the input to and the corresponding response of an arbitrary continuous system x (t) y(t) t t If this response pattern is retained but shifted wholesale through t0 when the input is similarly shifted through t0 then the system is said to be timeinvariant. In other words it does not matter when we activate the system, we always get the same response for the same input; the response will be the same on Tuesday as it was on Monday. x (t) t0 x(t – t0) t y(t) t0 That is, a system is said to be time-invariant if: LfxðtÞg ¼ yðtÞ and Lfxðt t0 Þg ¼ yðt t0 Þ where t0 is a constant y(t – t0) t 200 Programme 6 For example, if LfxðtÞg ¼ yðtÞ ¼ ðt xðÞ d then if x1 ðtÞ ¼ xðt t0 Þ 1 Lfxðt t0 Þg ¼ Lfx1 ðtÞg ðt x1 ðÞ d ¼ 1 ¼ ðt xð t0 Þ d because x1 ðtÞ ¼ xðt t0 Þ xðsÞ ds where s ¼ t0 so that ds ¼ d and when t ¼ 1, s ¼ t t0 1 ¼ ð tt0 1 ¼ yðt t0 Þ Therefore: LfxðtÞg ¼ yðtÞ and Lfxðt t0 Þg ¼ yðt t0 Þ so the system is time-invariant. Which of the following are time-invariant? (a) LfxðtÞg ¼ yðtÞ ¼ xðtÞ sin pt (b) LfxðtÞg ¼ yðtÞ ¼ exðtÞ The answers are in the next frame 8 (a) not time-invariant (b) time-invariant Because: (a) LfxðtÞg ¼ yðtÞ ¼ xðtÞ sin pt so if x1 ðtÞ ¼ xðt t0 Þ then Lfxðt t0 Þg ¼ Lfx1 ðtÞg ¼ x1 ðtÞ sin pt ¼ xðt t0 Þ sin pt because x1 ðtÞ ¼ xðt t0 Þ However, yðt t0 Þ ¼ xðt t0 Þ sin pðt t0 Þ 6¼ Lfxðt t0 Þg so that L is not time-invariant. (b) LfxðtÞg ¼ yðtÞ ¼ exðtÞ so let x1 ðtÞ ¼ xðt t0 Þ then Lfxðt t0 Þg ¼ Lfx1 ðtÞg ¼ ex1 ðtÞ ¼ exðtt0 Þ ¼ yðt t0 Þ Therefore LfxðtÞg ¼ yðtÞ and Lfxðt t0 Þg ¼ yðt t0 Þ so that L is timeinvariant. Next frame Introduction to invariant linear systems 201 Shift-invariance of a discrete system 9 A discrete system is said to be shift-invariant if: Lfx½ng ¼ y½n and Lfx½n n0 g ¼ y½n n0 For example, if Lfx½ng ¼ y½n ¼ x½n þ 3 then if x1 ½m ¼ x½m n0 we see that Lfx½n n0 g ¼ Lfx1 ½ng ¼ x1 ½n þ 3 ¼ x½n þ 3 n0 here m ¼ n þ 3 ¼ x½n n0 þ 3 ¼ y½n n0 Therefore Lfx½ng ¼ y½n and Lfx½n n0 g ¼ y½n n0 so the system is shiftinvariant. Which of the following are shift-invariant: (a) Lfx½ng ¼ y½n ¼ x½n 1 þ x½n 2 (b) Lfx½ng ¼ y½n ¼ nx½n The answers are in the next frame (a) shift-invariant (b) not shift-invariant Because: (a) Lfx½ng ¼ y½n ¼ x½n 1 þ x½n 2 so if x1 ½m ¼ x½m n0 then Lfx½n n0 g ¼ Lfx1 ½ng ¼ x1 ½n 1 þ x1 ½n 2 ¼ x½n 1 n0 þ x½n 2 n0 here m ¼ n 1 ¼ x½n n0 1 þ x½n n0 2 ¼ y½n n0 Therefore Lfx½ng ¼ y½n and Lfx½n n0 g ¼ y½n n0 and so L is shiftinvariant (b) Lfx½ng ¼ y½n ¼ nx½n so if x1 ½m ¼ x½m n0 then Lfx½n n0 g ¼ Lfx1 ½ng here m ¼ n ¼ nx1 ½n ¼ nx½n n0 ¼ y1 ½n However, y½n n0 ¼ ðn n0 Þx½n n0 6¼ y1 ½n and so L is not shiftinvariant. Move to the next frame 10 202 Programme 6 Differential equations 11 The general nth-order equation Linear, constant coefficient differential equations define a linear system so let us refresh our memory. The general nth-order, linear, constant coefficient, inhomogeneous differential equation: an dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ n1 dt n dt n1 m m1 d xðtÞ d xðtÞ ¼ bm þ bm1 þ . . . þ b0 xðtÞ m m1 dt dt coupled with the values of the n boundary conditions: dn yðtÞ dn1 yðtÞ , , . . . , yðt0 Þ dt n t¼t0 dt n1 t¼t0 describes the input-response relationship of a continuous linear system with input xðtÞ and response yðtÞ. Such an equation has a solution in the form yðtÞ ¼ yh ðtÞ þ yp ðtÞ where yh ðtÞ is the complementary function solution to the homogeneous equation an dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ ¼ 0 n1 dt n dt n1 and yp ðtÞ is a particular integral or particular solution to the inhomogeneous equation. [Refer to Page 1101, Frame 23ff of Engineering Mathematics, Sixth Edition.] The procedure for solving such an equation is: (i) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (ii) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (iii) Apply the boundary conditions to find the values of the unknown integration constants in yh ðtÞ. The solution can alternatively be written as yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ where yzi ðtÞ is called the zero-input response and yzs ðtÞ is called the zero-state response. The zero-input response of the equation depends only on the initial conditions and is independent of the input. It is obtained by solving the homogeneous equation and applying the boundary conditions. The zero-state response depends only on the input and is independent of the initial conditions, it is obtained by solving the inhomogeneous equation but with all the boundary conditions equated to zero. Here the procedure is: (i) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (ii) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ Introduction to invariant linear systems 203 (iii) Equate the boundary conditions to zero and then find the values of the unknown integration constants in yðtÞ. This is then the zero-state response yzs ðtÞ (iv) Apply the original boundary conditions to find the values of the unknown integration constants in yh ðtÞ. This is then the zero-input solution yzi ðtÞ. Move to the next frame for an example Zero-input response and zero-state response Consider the first-order, linear, constant coefficient, inhomogeneous differential equation: dyðtÞ þ 4yðtÞ ¼ tuðtÞ dt where yð0Þ ¼ y0 is the boundary condition and yðtÞ ¼ 0 for t < 0. Homogeneous solution yh ðtÞ dyh ðtÞ þ 4yh ðtÞ ¼ 0 the auxiliary equation is m þ 4 ¼ 0 and so m ¼ 4 dt and yh ðtÞ ¼ Ae4t uðtÞ. Note the use of the Heaviside unit step function which acts as a switch to ensure that yh ðtÞ ¼ 0 for t < 0 [Refer to Frame 4 of Programme 3]. From Particular solution yp ðtÞ Based on the form of the right-hand side the form of the particular solution is yp ðtÞ ¼ Ct þ D (C, D are constants). Substituting yp ðtÞ ¼ Ct þ D into dyðtÞ þ 4yðtÞ ¼ tuðtÞ gives: dt C þ 4C þ 4D ¼ t for t 0 from which we can see that 4C ¼ 1 and C þ 4D ¼ 0. Therefore C ¼ 1=4 and D ¼ 1=16 so that t 1 t 1 uðtÞ and yðtÞ ¼ Ae4t þ uðtÞ yp ðtÞ ¼ 4 16 4 16 Again, the Heaviside unit step function ensures that yp ðtÞ ¼ 0 and yðtÞ ¼ 0 for t < 0. Zero-state solution To obtain the zero-state response we now equate the boundary condition to zero and find the value of the integration constant A. That is, we write yð0Þ ¼ y0 ¼ 0 and find that: yð0Þ ¼ A 1 1 1 4t ¼ 0 so that A ¼ and yzs ðtÞ ¼ e þ 4t 1 uðtÞ 16 16 16 12 204 Programme 6 Zero-input solution To obtain the zero-input response we apply the original boundary condition yð0Þ ¼ y0 to yh ðtÞ ¼ Ae4t uðtÞ to give: yzi ðtÞ ¼ y0 e4t uðtÞ The complete solution is the sum of the zero-state and zero-input responses. That is: yðtÞ ¼ 1 ½16y0 þ 1e4t þ 4t 1 uðtÞ 16 So, the zero-input and zero-state responses of each of the differential equations: dyðtÞ yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 and yðtÞ ¼ 0 for t < 0 dt d2 yðtÞ dyðtÞ dyðtÞ þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ 1 and (b) 5 ¼ 1 and dt 2 dt dt t¼0 (a) yðtÞ ¼ 0 for t < 0 are . . . . . . . . . . . . The answers are in the next frame 13 (a) yzi ðtÞ ¼ et uðtÞ yzs ðtÞ ¼ uðtÞ sinh t (b) yzi ðtÞ ¼ 4e2t 3e3t uðtÞ uðtÞ yzs ðtÞ ¼ 9e2t þ 4e3t þ 6t þ 5 36 Because: (a) dyðtÞ yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 and yðtÞ ¼ 0 for t < 0 dt Homogeneous solution yh ðtÞ dyh ðtÞ yh ðtÞ ¼ 0 the auxiliary equation is m 1 ¼ 0 and so m ¼ 1 and dt yh ðtÞ ¼ Aet uðtÞ. From Particular solution yp ðtÞ Based on the form of the right-hand side the form of the particular solution is dyðtÞ yp ðtÞ ¼ Bet (B constant). Substituting yp ðtÞ ¼ Bet into yðtÞ ¼ et uðtÞ dt gives: 1 Bet Bet ¼ et for t 0 from which we can see that B ¼ . 2 1 t 1 Therefore yp ðtÞ ¼ e uðtÞ and yðtÞ ¼ Aet et uðtÞ 2 2 Introduction to invariant linear systems Zero-state solution To obtain the zero-state response we now equate the boundary condition to zero and find the value of the integration constant A. That is, we write yð0Þ ¼ 0 and find that: 1 1 1 t 1 t e e uðtÞ ¼ uðtÞ sinh t yð0Þ ¼ A ¼ 0 so that A ¼ and yzs ðtÞ ¼ 2 2 2 2 Zero-input solution To obtain the zero-input response we apply the original boundary condition yð0Þ ¼ 1 to yh ðtÞ ¼ Aet uðtÞ to give: yzi ðtÞ ¼ et uðtÞ The complete solution is the sum of the zero-state and zero-input responses. That is: yðtÞ ¼ ðet þ sinh tÞuðtÞ (b) d2 y dyðtÞ dy þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ 1 and 5 ¼ 1 dt 2 dt dt t¼0 Homogeneous solution yh ðtÞ From d2 yðtÞ dyðtÞ þ 6yðtÞ ¼ 0 the auxiliary equation is 5 dt 2 dt m2 5m þ 6 ¼ ðm 2Þðm 3Þ ¼ 0 and so m ¼ 2, 3 and yh ðtÞ ¼ Ae2t þ Be3t uðtÞ. Particular solution yp ðtÞ Based on the form of the right-hand side the form of the particular solution is yp ðtÞ ¼ Ct þ D (C, D constants). Substituting yp ðtÞ ¼ Ct þ D into the differential equation d2 yðtÞ dyðtÞ þ 6yðtÞ ¼ tuðtÞ gives: 5 dt 2 dt 5C þ 6ðCt þ DÞ ¼ t for t 0 from which we can see that 6C ¼ 1 and 5C þ 6D ¼ 0 Therefore t 5 þ uðtÞ giving C ¼ 1=6 and D ¼ 5=36 so that yp ðtÞ ¼ 6 36 t 5 1 yðtÞ ¼ Ae2t þ Be3t þ þ uðtÞ and y 0 ðtÞ ¼ 2Ae2t þ 3Be3t þ 6 36 6 205 206 Programme 6 Zero-state solution To obtain the zero-state response we now equate the boundary conditions to zero and find the value of the integration constants A and B. That is, we write yð0Þ ¼ 0 and y 0 ð0Þ ¼ 0 to find that: yð0Þ ¼ A þ B þ 5 1 9 4 and y 0 ð0Þ ¼ 2A þ 3B þ ¼ 0 giving A ¼ and B ¼ 36 6 36 36 so that: uðtÞ yzs ðtÞ ¼ 9e2t þ 4e3t þ 6t þ 5 36 Zero-input solution To obtain the zero-input response we apply the original boundary conditions yð0Þ ¼ 1 and y 0 ð0Þ ¼ 1 to yh ðtÞ ¼ Ae2t þ Be3t uðtÞ to give: yzi ðtÞ ¼ 4e2t 3e3t uðtÞ The complete solution is the sum of the zero-state and zero-input responses. That is: uðtÞ yðtÞ ¼ 135e2t 104e3t þ 6t þ 5 36 Move to the next frame 14 Zero-input, zero-response We have already seen that a system yðtÞ ¼ LfxðtÞg is linear if sums and scalar multiples are preserved, that is if Lfx1 ðtÞ þ x2 ðtÞg ¼ Lfx1 ðtÞg þ Lfx2 ðtÞg and LfxðtÞg ¼ LfxðtÞg where is a constant. In particular, by choosing ¼ 0 then Lf0g ¼ 0 which shows that if nothing is put into a linear system nothing will come out – zero input yields zero output. In the previous frame we considered two systems described by the differential equations: dyðtÞ yðtÞ ¼ et uðtÞ where yð0Þ ¼ 1 dt d2 yðtÞ dyðtÞ dyðtÞ þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ 1 and (b) 5 ¼ 1 dt 2 dt dt t¼0 (a) Do these equations give rise to linear or non-linear systems? (a) . . . . . . . . . . . . (b) . . . . . . . . . . . . The answers are in the following frame 207 Introduction to invariant linear systems (a) Non-linear (b) Non-linear 15 Because: (a) yzi ðtÞ ¼ et uðtÞ so that the zero-input response is not zero. Therefore this differential equation does not give rise to a linear system but to a nonlinear system. (b) yzi ðtÞ ¼ 4e2t 3e3t uðtÞ so that the zero-input response is not zero. Therefore this differential equation does not give rise to a linear system but to a non-linear system. Is it possible for a differential equation to give rise to a linear system? To answer this question we now re-cast these two differential equations with general boundary conditions. Firstly, dyðtÞ yðtÞ ¼ et uðtÞ where yð0Þ ¼ A dt Here, for t < 0 the equation becomes, with zero-input: dyðtÞ yðtÞ ¼ 0 where yð0Þ ¼ A dt With solution being the zero-input response yzi ðtÞ ¼ Aet and this can only be zero if A ¼ 0. The differential equation only gives rise to a linear system if the value of the boundary condition is zero. Now you try. The conditions that the differential equation: d2 yðtÞ dy dyðtÞ þ 6yðtÞ ¼ tuðtÞ where yð0Þ ¼ K 5 and ¼ K2 1 dt 2 dt dt t¼0 gives rise to a linear system are . . . . . . . . . . . . The answer is in the next frame K1 ¼ 0 and K2 ¼ 0 Because: Here, for t < 0 the equation becomes, with zero-input: d2 yðtÞ dyðtÞ dyðtÞ 5 ¼ K2 þ 6yðtÞ ¼ 0 where yð0Þ ¼ K1 and dt 2 dt dt t¼0 with solution being the zero-input response yzi ðtÞ ¼ Ae2t þ Be3t . Applying the boundary conditions: yð0Þ ¼ K1 : dyðtÞ ¼ K2 : dt A þ B ¼ K1 2A þ 3B ¼ K2 t¼0 with solution A ¼ 3K1 K2 and B ¼ K2 2K1 giving yzi ðtÞ ¼ ð3K1 K2 Þe2t þ ðK2 2K1 Þe3t 16 208 Programme 6 This can only be zero if: 3K1 K2 ¼ 0 2K1 þ K2 ¼ 0 that is if K1 ¼ 0 and K2 ¼ 0 The differential equation only gives rise to a linear system if the values of the boundary conditions are zero. This is a general property – a constant coefficient, linear differential equation only gives rise to a linear system if the values of all the boundary conditions are zero. This is an important fact to be remembered. Next we look at these differential equations and time-invariance. Move to the next frame 17 Time-invariance Consider the differential equation in Frame 12, namely: dyðtÞ þ 4yðtÞ ¼ tuðtÞ where yð0Þ ¼ y0 and yðtÞ ¼ 0 for t < 0 dt with solution yðtÞ ¼ 1 ½16y0 þ 1e4t þ 4t 1 uðtÞ. 16 If we re-visit this equation but this time change the boundary condition to yð0Þ ¼ 0 the solution is the zero-state solution: yzs ðtÞ ¼ 1 4t e þ 4t 1 uðtÞ 16 If we now delay the input by 3 units so that the input becomes ðt 3Þuðt 3Þ the differential equation becomes: dyðtÞ þ 4yðtÞ ¼ ðt 3Þuðt 3Þ where yð3Þ ¼ 0 and yðtÞ ¼ 0 for t < 3 dt The homogeneous solution is again yh ðtÞ ¼ Ae4t uðtÞ and the particular solution has the form yp ðtÞ ¼ Ct þ D. However, substituting into the differential equation we now find that C þ 4Ct þ 4D ¼ t 3 for t 3 from which we find that 4C ¼ 1 and C þ 4D ¼ 3. Therefore: 13 12 1 3 1 ¼ ¼ and the particular solution is C ¼ 1=4, D ¼ 16 16 16 4 16 t 3 1 t 3 1 4t uðt 3Þ so that yðtÞ ¼ Ae þ uðt 3Þ. yp ðtÞ ¼ 4 16 4 16 Introduction to invariant linear systems 209 Applying the boundary condition yð3Þ ¼ 0: yð3Þ ¼ Ae12 1 1 2e12 ¼ that is A ¼ giving the solution to 16 16 16 dyðtÞ þ 4yðtÞ ¼ ðt 3Þuðt 3Þ where yð3Þ ¼ 0 as dt 1 4ðt3Þ yðtÞ ¼ þ 4ðt 3Þ 1 uðt 3Þ 2e 16 This is the same solution but delayed by the same amount as the input. Consequently, the system is not only linear but it is also time-invariant. Indeed, the zero values of the boundary conditions ensure that the general constant coefficient, linear differential equation gives rise to a system that is not only linear but also time-invariant. Move to the next frame Responses of a continuous system Impulse response We shall soon see that any continuous, linear, time-invariant system has the important property that its response to any input can be found from knowing its response to the unit impulse ðtÞ – a property that can be exploited to solve such differential equations as considered here [refer to Programme 4]. When the input to a linear, time-invariant system is the unit impulse ðtÞ the response is denoted by hðtÞ and is referred to as the impulse response. That is: hðtÞ ¼ LfðtÞg and, because the system is time-invariant hðt t0 Þ ¼ Lfðt t0 Þg Arbitrary input From the properties of the unit impulse ðtÞ we can express an arbitrary input xðtÞ in terms of the unit impulse ðtÞ as: ð1 xðÞðt Þ d xðtÞ ¼ 1 so that if the response to this arbitrary input is yðtÞ then: yðtÞ ¼ LfxðtÞg ð1 xðÞðt Þ d ¼L 1 18 210 Programme 6 Because the variable inside the integral is the variable of integration and the operator L is acting on t and not on the operator can be moved inside the integral. (Recall that an integral is a limit of a sum and, for linear systems, sums are preserved.) Therefore: ð1 ð1 xðÞðt Þ d ¼ xðÞLfðt Þgd L 1 1 ð1 ¼ xðÞhðt Þd 1 because the system is given as time invariant. This is a remarkable result so we shall look at it very closely. We have just found that: ð1 xðÞhðt Þd. LfxðtÞg ¼ 1 You have seen integrals like this one before, can you recall? This integral is the . . . . . . . . . . . . between the input to the system and the impulse response of the system. The answer is in the next frame 19 convolution Because: The convolution between xðtÞ and hðtÞ is obtained by first reversing hðtÞ to form hðtÞ, changing the variable to the dummy variable of the integral to form xðÞ and hðÞ, advancing hðÞ by t to form hð þ tÞ ¼ hðt Þ, taking the product of this with xðÞ and integrating with respect to to form [refer to page 112, Frame 43]: ð1 xðÞhðt Þd ¼ xðtÞ hðtÞ yðtÞ ¼ 1 Aside: It is also worth remembering that convolution is a commutative operation. That is: ð1 xðÞhðt Þd and yðtÞ ¼ xðtÞ hðtÞ ¼ ð1 1 yðtÞ ¼ hðtÞ xðtÞ ¼ hðÞxðt Þd 1 Bear this fact in mind as we shall make use of it little later on. This is a most important result because it tells us that if we know the impulse response of a continuous, linear, time-invariant system then we can find the response of the system to any input simply by evaluating the convolution of the input with the system impulse response. 211 Introduction to invariant linear systems As an example consider the response to a unit step input uðtÞ at time t ¼ 0 of a system that has an impulse response: hðtÞ ¼ uðtÞekt k>0 The graphs of the input and the impulse response are: x(t) = u(t) h (t) = u(t)e–kt 1 1 t t To evaluate the convolution yðtÞ ¼ xðtÞ hðtÞ we first need to change t to the variable of integration to form xðÞ and hðÞ. We then require hðÞ to be advanced by t to form hð tÞ where t > 0. Next we flip hð tÞ about the vertical to form hð½ tÞ ¼ hðt Þ to overlap with the unit step. h(t – τ) 1 1 t τ t τ The only non-zero overlap inside the convolution integral is then between the values ¼ 0 and ¼ t provided t > 0. If t < 0 then there is no overlap at all with the unit step function so that: yðtÞ ¼ xðtÞ hðtÞ ð1 ¼ xðÞhðt Þd 1 ðt ð1 xðÞhðt Þd þ xðÞhðt Þd þ xðÞhðt Þd 1 0 t ðt ¼ 0 þ xðÞhðt Þd þ 0 xðÞ ¼ 0 for < 0 and hðt Þ ¼ 0 for > t ¼ ð0 0 Now xðÞ ¼ uðÞ and hðt Þ ¼ uðt ÞekðtÞ so that ðt yðtÞ ¼ uðÞuðt ÞekðtÞ d 0 ðt ¼ uðt ÞekðtÞ d 0 ¼ ð0 t uðxÞekðxÞ dðxÞ where x ¼ t so x ¼ t when ¼ 0 and x ¼ 0 when ¼ t 212 Programme 6 Furthermore, ¼ t x so d ¼ dðxÞ ¼ dx so that ð0 yðtÞ ¼ uðxÞekx dx t ðt ¼ uðxÞekx dx 0 ðt ¼ ekx dx 0 ¼ ekx k t ¼ 0 1 1 eðtÞ k The response can then be written as yðtÞ ¼ 1 1 ekt uðtÞ k You try one. The response to the input xðtÞ ¼ uðt aÞ of a system with impulse response hðtÞ ¼ uðtÞebt where a > 0 and b > 0 is . . . . . . . . . . . . The answer is in the next frame 20 yðtÞ ¼ 1 1 ebðtaÞ uðt aÞ b Because: The graphs of the input and the impulse response are: x(t) = u (t – a) h (t) = u(t)e–bt 1 1 t t To evaluate the convolution yðtÞ ¼ xðtÞ hðtÞ we first need to change t to the variable of integration to form xðÞ and hðÞ. We then require hðÞ to be advanced by t to form hðt Þ where t > 0. Next we flip hðt Þ about the vertical to form hð½ tÞ to overlap with the unit step. If t < a there is no overlap with the unit step function. h(t – τ) 1 1 t τ t a τ 213 Introduction to invariant linear systems Provided t > a the only non-zero overlap inside the convolution integral is over the interval a to t so that: yðtÞ ¼ xðtÞ hðtÞ ð1 ¼ xðÞhðt Þ d 1 ðt ð1 ða xðÞhðt Þ d þ xðÞhðt Þ d þ xðÞhðt Þ d ¼ 1 a t ðt xðÞ ¼ 0 for < a and hðt Þ ¼ 0 for ¼ 0 þ xðÞhðt Þ d þ 0 a >t ðt ¼ uð aÞuðt ÞebðtÞ d a ðt ¼ uðt ÞebðtÞ d a ¼ ¼ ð0 uðxÞebðxÞ dðxÞ ta ð ta where x ¼ t uðxÞebx dx 0 ta ebx b 0 1 1 ebðtaÞ for t > a ¼ b ¼ The response can then be written as yðtÞ ¼ 1 1 ebðtaÞ uðt aÞ b Next frame 21 Exponential response When the input to a continuous, linear, time-invariant system with impulse response hðtÞ is the exponential xðtÞ ¼ Aest (A and s being constants) the system response is given as: yðtÞ ¼ Aest hðtÞ or, because convolution is commutative, yðtÞ ¼ hðtÞ Aest That is: yðtÞ ¼ A ð1 es hðt Þ d or yðtÞ ¼ A 1 ð1 hðÞesðtÞ d 1 Taking the latter form of the convolution of the input with the impulse response we see that: ð1 yðtÞ ¼ A hðÞesðtÞ d 1 ð1 st ¼ Ae hðÞes d 1 ¼ Aest HðsÞ ð1 hðÞest d. where HðsÞ ¼ 1 214 Programme 6 Notice that if hðtÞ ¼ 0 for t < 0 then: ð1 HðsÞ ¼ hðÞes d so that HðsÞ is the . . . . . . . . . . . . transform of hðtÞ 0 The answer is in the next frame 22 Laplace Because: Recalling from Frame 1 on page 47 the Laplace transform FðsÞ of function f ðtÞ is defined as: ð1 FðsÞ ¼ f ðtÞest dt 0 The expression HðsÞ is called the system’s transfer function. Furthermore, because the system response is simply a scaled version of the input, the scaling factor being HðsÞ we can say that: LfAest g ¼ HðsÞ Aest which tells us that Aest is an eigenfunction of the operator L with corresponding eigenvalue HðsÞ (refer to Engineering Mathematics, Sixth Edition, page 578). The converse is also true. The eigenfunctions of a linear, time-invariant system are exponential functions, which places the exponential function in a special position with respect to linear time-invariant systems as we shall appreciate later. For example, the transfer function corresponding to the input 5est of the continuous, linear, time-invariant system with impulse response: ( t0 e6t hðtÞ ¼ 0 t<0 is: HðsÞ ¼ 5 ¼5 ð1 ð01 e6t es d eð6þsÞt d 0 1 eð6þsÞt ð6 þ sÞ 0 1 ¼5 0 ð6 þ sÞ 5 provided s > 6 ¼ sþ6 ¼5 If s 6 then s þ 6 0 and the integral diverges. So the transfer function corresponding to the input 3est uðt 2Þ of the continuous, linear, time-invariant system with impulse response hðtÞ ¼ et uðtÞ is . . . . . . . . . . . . The answer is in the next frame 215 Introduction to invariant linear systems 23 3e2ð1þsÞ provided s > 1 sþ1 Because: HðsÞ ¼ 3 ¼3 ð1 ð01 est uðt 2ÞuðtÞe d eð1þsÞt d 2 1 eð1þsÞt ð1 þ sÞ 2 eð1þsÞ2 ¼3 0 ð1 þ sÞ ¼3 ¼ 3e2ð1þsÞ sþ1 provided s > 1 If s 1 then s þ 1 0 and the integral diverges. Move to the next frame 24 The transfer function We have seen that the response yðtÞ of a continuous, linear, time-invariant system to an input xðtÞ is given in terms of the system’s unit impulse response hðtÞ as the convolution: yðtÞ ¼ xðtÞ hðtÞ We have also seen that provided hðtÞ ¼ 0 for t < 0 then the system’s transfer function HðsÞ is the Laplace transform of hðtÞ. That is: ð1 hðtÞest dt HðsÞ ¼ 0 Referring now to Frame 45 of page 117 we see that the convolution theorem states that the Laplace transform of a convolution of two functions is equal to the product of their respective Laplace transforms. Therefore, if ð1 ð1 yðtÞest dt and XðsÞ ¼ xðtÞest dt then YðsÞ ¼ XðsÞHðsÞ YðsÞ ¼ 0 0 For example to find the response of a time-invariant linear system with impulse response hðtÞ ¼ uðtÞet to an input xðtÞ ¼ uðtÞ uðt 1Þ all we need do is: (a) Find the Laplace transforms of hðtÞ ¼ uðtÞet and xðtÞ ¼ uðtÞ uðt 1Þ These are HðsÞ ¼ es 1 es and XðsÞ ¼ s sþ1 s 216 Programme 6 (b) Obtain YðsÞ ¼ XðsÞHðsÞ es e2s sðs þ 1Þ sðs þ 1Þ 1 1 ¼ es e2s s sþ1 This is YðsÞ ¼ (c) Take the inverse Laplace transform es es e2s e2s s sþ1 s sþ1 ðt1Þ so yðtÞ ¼ uðt 1Þ uðt 1Þe uðt 2Þ uðt 2Þeðt2Þ ¼ uðt 1Þ 1 eðt1Þ uðt 2Þ 1 eðt2Þ YðsÞ ¼ You try one. The response of a time-invariant linear system with impulse response hðtÞ ¼ uðt 1Þ to an input xðtÞ ¼ uðtÞ sin t uðt 1Þ sinðt 1Þ is yðtÞ ¼ . . . . . . . . . . . . The answer is in the next frame 25 uðt 1Þð1 cosðt 1ÞÞ uðt 2Þð1 cosðt 2ÞÞ Because: (a) The Laplace transforms of hðtÞ ¼ uðt 1Þ and xðtÞ ¼ uðtÞ sin t uðt 1Þ sinðt 1Þ are: HðsÞ ¼ es 1 es and XðsÞ ¼ 2 2 s þ1 s þ1 s (b) The Laplace transform of yðtÞ is: YðsÞ ¼ XðsÞHðsÞ es e2s sðs2 þ 1Þ sðs2 þ 1Þ 1 s ¼ es e2s s s2 þ 1 ¼ (c) Then the inverse Laplace transform is l1 fYðsÞg ¼ l1 es se2s e2s se2s 2 l1 2 s s þ1 s s þ1 so yðtÞ ¼ ðuðt 1Þ uðtÞ cosðt 1ÞÞ ðuðt 2Þ uðt 2Þ cosðt 2ÞÞ ¼ uðt 1Þð1 cosðt 1ÞÞ uðt 2Þð1 cosðt 2ÞÞ Introduction to invariant linear systems 217 In summary, to find the response of a continuous, linear, time-invariant system all we need to do is take the inverse Laplace transform of the product of the Laplace transform of the input and the transfer function – the transfer function being the Laplace transform of the unit impulse response. yðtÞ ¼ l1 fXðsÞHðsÞg where HðsÞ ¼ lfhðtÞg So, given the input all we need do to proceed is to determine the impulse response. However, for those differential equations that give rise to continuous, linear, time-invariant systems we have a much simpler way of determining the transfer function. Next frame Differential equations To solve the equation y 00 ðtÞ 5y 0 ðtÞ þ 6yðtÞ ¼ xðtÞ where yð0Þ ¼ 0, y 0 ð0Þ ¼ 0 and y 00 ð0Þ ¼ 0 we note that the differential equation gives rise to a continuous, linear, timeinvariant system. That is yðtÞ ¼ LfxðtÞg Accordingly, the exponential function Aest is an eigenfunction of the system whose corresponding eigenvalue is the system’s transfer function HðsÞ. That is: if xðtÞ ¼ Aest then yðtÞ ¼ Aest HðsÞ Substituting these into the differential equation, we then see that: st 00 0 Ae HðsÞ 5 Aest HðsÞ þ6 Aest HðsÞ ¼ Aest s2 5s þ 6 HðsÞAest ¼ Aest 1 HðsÞ ¼ 2 s 5s þ 6 HðsÞ is the Laplace transform of the left-hand side of the differential equation and now you see the importance of all the boundary conditions having a value of zero. If they did not then their non-zero values would be automatically incorporated into the Laplace transform so giving a different expression to this one here. Now if, for example, the input is xðtÞ ¼ e5t uðtÞ then its Laplace transform is 1 XðsÞ ¼ giving the Laplace transform of the system’s response yðtÞ as: sþ5 YðsÞ ¼ XðsÞHðsÞ 1 ðs þ 5Þðs 2Þðs 3Þ P Q R þ þ ¼ sþ5 s2 s3 ¼ 26 218 Programme 6 Taking inverse Laplace transforms we see that yðtÞ ¼ Pe5t þ Qe2t þ Re3t where the values of P, Q and R can be found – the usual partial fractions procedure giving the solution to the differential equation as: e5t e2t e3t þ 56 7 8 Therefore, the solution to yðtÞ ¼ y 00 ðtÞ þ 3y 0 ðtÞ 28yðtÞ ¼ et uðtÞ where yð0Þ ¼ 0, y 0 ð0Þ ¼ 0 and y 00 ð0Þ ¼ 0 is yðtÞ ¼ . . . . . . . . . . . . The answer is in the next frame 27 yðtÞ et e7t e4t þ þ 30 66 55 Because: 1 s2 þ 3s 28 1 The Laplace transform of the input xðtÞ ¼ et uðtÞ is XðsÞ ¼ sþ1 The Laplace transform of the response yðtÞ is then 1 1 2 YðsÞ ¼ s þ 1 s þ 3s 28 1 ¼ ðs þ 1Þðs þ 7Þðs 4Þ The auxiliary equation is s2 þ 3s 28 ¼ 0 so that HðsÞ ¼ ¼ P Q R þ þ sþ1 sþ7 s4 which gives the response as yðtÞ ¼ Pet þ Qe7t þ Re4t . Now 1 P Q R ¼ þ þ ðs þ 1Þðs þ 7Þðs 4Þ s þ 1 s þ 7 s 4 ¼ Pðs þ 7Þðs 4Þ þ Qðs þ 1Þðs 4Þ þ Rðs þ 1Þðs þ 7Þ ðs þ 1Þðs þ 7Þðs 4Þ ¼ ðP þ Q þ RÞs2 þ ð3P 3Q þ 8RÞs þ ð28P 4Q þ 7RÞ ðs þ 1Þðs þ 7Þðs 4Þ Therefore 0 10 1 0 1 PþQþR¼0 1 1 1 P 0 B CB C B C 3P 3Q þ 8R ¼ 0 that is @ 3 3 8 A@ Q A ¼ @ 0 A 28P 4Q þ 7R ¼ 1 28 4 7 R 1 1 0 1 0 1=30 P B C B C giving @ Q A ¼ @ 1=66 A 1=55 R Therefore yðtÞ ¼ et e7t e4t þ þ 30 66 55 219 Introduction to invariant linear systems And just one more. To find the solution to y 00 ðtÞ 4yðtÞ ¼ ½uðtÞ uðt 1Þt where yð0Þ ¼ 0, y 0 ð0Þ ¼ 0 and y 00 ð0Þ ¼ 0 we first need to arrange the input into a form that is Laplace transformable. In other words, we need to convert uðt 1Þt into a form involving uðt 1Þðt 1Þ that is: ½uðtÞ uðt 1Þt ¼ . . . . . . . . . . . . The answer is in the next frame ½uðtÞ uðt 1Þt ¼ uðtÞt uðt 1Þðt 1Þ uðt 1Þ 28 Because uðt 1Þðt 1Þ ¼ uðt 1Þt uðt 1Þ therefore ½uðtÞ uðt 1Þt ¼ uðtÞt uðt 1Þðt 1Þ uðt 1Þ The differential equation then becomes: y 00 ðtÞ 4yðtÞ ¼ uðtÞt uðt 1Þðt 1Þ uðt 1Þ where yð0Þ ¼ 0, y 0 ð0Þ ¼ 0 and y 00 ð0Þ ¼ 0. The solution is then yðtÞ ¼ . . . . . . . . . . . . The answer is in the next frame o uðt 1Þ n uðtÞ 4t þ e2t e2t 4t þ 3e2ðt1Þ þ e2ðt1Þ 16 16 Because Taking the Laplace transform of the left-hand side tells us that HðsÞ ¼ 1 1 1 1 ¼ s2 4 4 s 2 s þ 2 Taking the Laplace transform of the right hand side we get XðsÞ ¼ 1 es es s2 s2 s Therefore 1 es es 1 1 1 YðsÞ ¼ 2 2 s 4 s2 sþ2 s s 29 220 Programme 6 Breaking into partial fractions: 1 1 2 1 1 ¼ þ 2 2 s ðs þ 2Þ 4 s s sþ2 1 1 1 1 ¼ sðs þ 2Þ 2 s s þ 2 1 1 2 1 1 2 þ ¼ s2 ðs 2Þ 4 s s s2 1 1 1 1 ¼ sðs 2Þ 2 s 2 s Therefore 1 es es 1 1 1 YðsÞ ¼ 2 2 s 4 s2 sþ2 s s ð1 es Þ 4 1 1 es 2 1 1 2þ þ þ ¼ 16 s s2 sþ2 s s2 sþ2 8 Giving o uðt 1Þ n uðtÞ 4t þ e2t e2t 4ðt 1Þ þ e2ðt1Þ e2ðt1Þ 16 16 o uðt 1Þ n 2 þ e2ðt1Þ þ e2ðt1Þ 8 o uðt 1Þ n uðtÞ 4t þ e2t e2t 4t þ 3e2ðt1Þ þ e2ðt1Þ ¼ 16 16 This completes our work on continuous linear systems. Now we shall move on to consider discrete linear systems. Read on. yðtÞ ¼ Responses of a discrete system 30 The discrete unit impulse The value of the unit impulse ðtÞ in the study of continuous linear systems cannot be overestimated. It permits the system response to any input to be found once the system’s response to the unit impulse is known. In the case of discrete systems the equivalent is called the discrete unit impulse ½n which is defined as: ½n ¼ 1 n¼0 0 n 6¼ 0 where n is an integer. Associated with the discrete unit impulse is the shifted discrete unit impulse: ½n k ¼ 1 n¼k 0 n 6¼ 0 which enables us to select a particular component of an expression x½n via the equation: x½k ¼ x½n½n k 221 Introduction to invariant linear systems This is because the right-hand side x½n½n k ¼ 0 unless n ¼ k. Indeed, any sequence x½n can be considered as consisting of a collection of scaled and shifted discrete unit impulses. For example, the geometric sequence: x½n ¼ 3n has values . . . , 32 , 31 , 1, 3, 32 , 33 , . . . and can be alternatively written as the sum x½n ¼ . . . þ 32 ½n ð2Þ þ 31 ½n ð1Þ þ 1½n 0 þ 3½n 1 þ 32 ½n 2 þ . . . From this sum any term of the sequence can be selected. For instance: x½2 ¼ . . . þ 32 ½2 ð2Þ þ 31 ½2 ð1Þ þ 1½2 0 þ 3½2 1 þ 32 ½2 2 þ . . . ¼ . . . þ 32 ½4 þ 31 ½3 þ 1½2 þ 3½1 þ 32 ½0 þ . . . ¼ . . . þ 32 0 þ 31 0 þ 1 0 þ 3 0 þ 32 1 þ . . . ¼ 32 For this reason the discrete unit impulse is also referred to as the unit sample and it can be used to decompose any sequence into a sum of weighted and shifted unit samples. For example: x½n ¼ . . . þ x½2½n ð2Þ þ x½1½n ð1Þ þ x½0½n 0 þ x½1½n 1 þ x½2½n 2 þ . . . 1 X x½k½n k ¼ k¼1 Note the analogy with the continuous case: ð1 f ðsÞðt sÞ ds f ðtÞ ¼ 1 When the input to a linear, shift-invariant system is the discrete unit impulse ½n the response is denoted by h½n and is referred to as the discrete unit impulse response. That is: h½n ¼ Lf½ng and, because it is shift-invariant, h½n n0 ¼ Lf½n n0 g Next frame 31 Arbitrary input Just like the continuous system a discrete, linear, shift-invariant system has the important property that its response to any input can be found from its response to the discrete unit impulse ½n. Recalling the discrete decomposition of a sequence as x½n ¼ 1 X k¼1 x½k½n k 222 Programme 6 then the response of a linear system to this input is y½n where: y½n ¼ Lfx½ng ( ) 1 X x½k½n k ¼L k¼1 ¼ ¼ ¼ 1 X k¼1 1 X k¼1 1 X Lfx½k½n kg because L is linear and so sums are preserved x½kLf½n kg because L is linear and so scalar multiples are preserved x½kh½n k because L is shift-invariant k¼1 1 X That is y½n ¼ x½kh½n k which is referred to as the convolution sum of k¼1 x½n and h½n, alternatively written as: x½n h½n (also h½n x½n since the convolution sum is commutative) So, by direct analogy with a continuous system, the response of a discrete linear system can be obtained from the convolution sum of the input with the system’s unit impulse response. For example, a discrete, linear, shift-invariant system has the unit impulse response: h½n ¼ u½n 1 the discrete unit step function where u½n ¼ 1 n0 0 n<0 To find the response y½n to the input x½n ¼ 2n u½n we see that: y½n ¼ x½n h½n 1 X x½kh½n k ¼ ¼ ¼ k¼1 1 X 2k u½ku½n k 1 k¼1 1 X 2k u½n k 1 since u½k ¼ 0 for k < 0 k¼0 ¼ n1 X 2k since u½n k 1 ¼ 0 for n k 1 < 0 ie k > n 1 k¼0 ¼ 2n 1 sum of the first n terms of a geometric series with common ratio 2 Try one yourself. A discrete linear shift-invariant system has a unit impulse response h½n ¼ u½n 4 and a response to the input x½n ¼ nu½n of : y½n ¼ . . . . . . . . . . . . The answer is in the next frame 223 Introduction to invariant linear systems y½n ¼ 32 1 ðn 4Þðn 3Þ 2 Because: y½n ¼ x½n h½n 1 X ¼ x½kh½n k ¼ ¼ k¼1 1 X ku½ku½n k 4 k¼1 1 X ku½n k 4 since ku½k ¼ 0 for k 0 k since u½n k 4 ¼ 0 for k > n 4 k¼0 ¼ n4 X k¼0 ¼ ðn 4Þðn 5Þ 2 sum of the first n 4 integers Move to the next frame Exponential response When the input to a discrete, linear, shift-invariant system with impulse response h½n is the exponential x½n ¼ Azn (A being constant) the system response is given as: y½n ¼ h½n x½n 1 X h½kx½n k ¼ ¼ k¼1 1 X h½kAznk k¼1 1 X ¼ Azn h½kzk k¼1 ¼ Azn H½z where H½z ¼ 1 X h½kzk which you will recognise as the Z transform of h½k k¼1 [refer to Programme 5]. 33 224 Programme 6 As in the continuous case we call H½z the system transfer function. For example, the transfer function H½z of a discrete, linear, shift-invariant system with impulse response: ( n 1 0n4 5 h½n ¼ 0 otherwise is HðzÞ ¼ 1 X h½kzk k¼1 ¼ 4 X 1 k 5 zk k¼0 1 2 3 4 z0 þ 15 z1 þ 15 z2 þ 15 z3 þ 15 z4 1 1 1 1 þ ¼1þ þ þ 5z 25z2 125z3 625z4 So, the transfer function HðzÞ of a discrete, linear, shift-invariant system with impulse response: nu½n 0 n 3 h½n ¼ 0 otherwise ¼ 10 5 is . . . . . . . . . . . . The answer is in the next frame 34 HðzÞ ¼ 1 2 3 þ þ z z2 z3 Because: HðzÞ ¼ 1 X h½kzk k¼1 ¼ 3 X ku½kzk k¼0 ¼ 0u½0z0 þ 1u½1z1 þ 2u½2z2 þ 3u½3z3 1 2 3 ¼ þ 2þ 3 z z z Move to the next frame 35 Transfer function We have seen in Frame 31 that the response y½n to the input x½n to a discrete, linear, shift-invariant system with impulse response h½n is given as the convolution sum: y½n ¼ h½n x½n 1 X h½kx½n k k¼1 225 Introduction to invariant linear systems So the Z transform of the response YðzÞ is given as: 1 X y½nzn YðzÞ ¼ n¼1 ¼ ¼ ¼ ! 1 X 1 X n¼1 k¼1 1 X k¼1 1 X h½k h½kx½n k zn 1 X ! x½n kz n interchanging sums n¼1 h½kzk XðzÞ by the first shift property of the Z k¼1 ¼ HðzÞXðzÞ where the transfer function HðzÞ ¼ 1 X h½kzk is the Z transform of the k¼1 discrete impulse response h½k. Consequently, for discrete, linear, shiftinvariant systems the transfer function (as in the continuous case) completely characterises the system and permits the response to any input to be obtained. Move to the next frame 36 Difference equations Given a linear, constant coefficient difference equation it is possible to derive its transfer function and from that the impulse response. For example, consider the difference equation: y½n ¼ 4y½n 2 þ x½n Taking the Z transform of both sides where Zfx½ng ¼ XðzÞ and Zfy½ng ¼ YðzÞ we see that: YðzÞ ¼ ð. . . . . . . . . . . .ÞXðzÞ Next frame YðzÞ ¼ 37 1 XðzÞ ð1 4z1 Þ Because Zfy½ng ¼ Zf4y½n 1 þ x½ng ¼ 4Zfy½n 1g þ Zfx½ng That is: YðzÞ ¼ 4z1 YðzÞ þ XðzÞ so that YðzÞ 1 4z1 ¼ XðzÞ and so: YðzÞ ¼ 1 XðzÞ ð1 4z1 Þ This means that the transfer function is: HðzÞ ¼ . . . . . . . . . . . . Next frame 226 Programme 6 38 z z4 Because YðzÞ ¼ 1 XðzÞ ð1 4z1 Þ ¼ HðzÞXðzÞ Giving: 1 ð1 4z1 Þ z ¼ z4 HðzÞ ¼ From this we can now determine the impulse response: h½n ¼ . . . . . . . . . . . . Next frame 39 4n u½n Because h½n ¼ Z1 fHðzÞg n z o ¼ Z1 z4 ¼ 4n u½n Now you try one. The transfer function and hence the impulse response of the difference equation: y½n ¼ 2y½n 1 þ 3y½n 2 þ x½n 1 2x½n 2 are given as: HðzÞ ¼ . . . . . . . . . . . . h½n ¼ . . . . . . . . . . . . Next frame 40 3=4 1=4 þ ðz þ 1Þ ðz 3Þ n o 1 3 ð1Þn1 þ 3n1 u½n h½n ¼ 4 4 HðzÞ ¼ 227 Introduction to invariant linear systems Because Taking the Z transform of both sides we see that: Zfy½ng ¼ 2Zfy½n 1g þ 3Zfy½n 2g þ Zfx½n 1g 2Zfx½n 2g that is: YðzÞ ¼ 2z1 YðzÞ þ 3z2 YðzÞ þ z1 XðzÞ 2z2 XðzÞ so that: YðzÞ 1 2z1 3z2 ¼ XðzÞ z1 2z2 and so: 1 z 2z2 YðzÞ ¼ XðzÞ ¼ HðzÞXðzÞ giving: ð1 2z1 3z2 Þ 1 z 2z2 HðzÞ ¼ ð1 2z1 3z2 Þ ¼ ðz 2Þ ðz2 2z 3Þ ¼ ðz 2Þ ðz þ 1Þðz 3Þ ¼ 3=4 1=4 þ ðz þ 1Þ ðz 3Þ From this we can now determine the impulse response: h½n ¼ Z1 fHðzÞg 3 1 1 z 1 z Z þ Z1 z1 z 4 ðz þ 1Þ 4 ðz 3Þ 3 1 ¼ ð1Þn1 u½n þ 3n1 u½n 4 4 ¼ Next frame This procedure can be reversed. For example to find the linear, constant coefficient difference equation whose impulse response is h½n ¼ 0:5 ð5Þnþ2 u½n we proceed as follows. Since the transfer function is the Z transform of the unit impulse then: HðzÞ ¼ Zfh½ng n o ¼ Z 0:5 ð5Þnþ2 u½n o 1 n ¼ Z ð5Þnþ2 u½n 2 1 z ¼ z2 2 ðz þ 5Þ ¼ 1 z3 2 ðz þ 5Þ 41 228 Programme 6 From this we can deduce the input-output relationship: YðzÞ HðzÞ ¼ XðzÞ ¼ 1 z3 2 ðz þ 5Þ so that: 1 z3 XðzÞ 2 ðz þ 5Þ 1=2 XðzÞ ¼ 2 ðz þ 5z3 Þ YðzÞ ¼ therefore: 1 XðzÞ 2 and so the resulting difference equation is: y½n 2 þ 5y½n 3 ¼ 0:5x½n or y½n þ 1 þ 5y½n ¼ 0:5x½n þ 3 ðz2 þ 5z3 ÞYðzÞ ¼ You try one now. The difference equation whose unit impulse response is given as: h½n ¼ 3 2n2 u½n is . . . . . . . . . . . . Next frame 42 y½n þ 1 2y½n ¼ 3x½n 1 Because: HðzÞ ¼ Zfh½ng ¼ Z 3 2n2 u½n ¼ 3Z 2n2 u½n z ¼ 3 z2 ðz 2Þ ¼3 z1 ðz 2Þ From this we can deduce the input-output relationship: YðzÞ HðzÞ ¼ XðzÞ ¼3 z1 ðz 2Þ so that: YðzÞ ¼ 3 z1 XðzÞ ðz 2Þ therefore: ðz 2ÞYðzÞ ¼ 3z1 XðzÞ and so the resulting difference equation is: y½n þ 1 2y½n ¼ 3x½n 1 Introduction to invariant linear systems 229 And that is the end of the Programme on invariant linear systems. All that remain are the Revision summary and the Can you? checklist. Read through these thoroughly and make sure you understand all the workings of this Programme. Then try the Test exercise; there is no need to hurry, take your time and work through the questions carefully. The Further problems then provide a valuable collection of additional exercises for you to try. Revision summary 6 1 Systems A system L is a process capable of accepting an input xðtÞ and processing the input to produce an output yðtÞ, also called the response of the system. This is written as yðtÞ ¼ LfxðtÞg. 2 Linear systems A linear system preserves sums and scalar products. If yðtÞ ¼ LfxðtÞg then L is linear if Lfax1 ðtÞ þ bx2 ðtÞg ¼ aLfx1 ðtÞg þ bLfx2 ðtÞg 3 Time-invariance A continuous linear system is time-invariant if yðtÞ ¼ LfxðtÞg and yðt t0 Þ ¼ Lfxðt t0 Þg 4 Shift-invariance A discrete linear system is shift-invariant if y½n ¼ Lfx½ng and y½n n0 ¼ Lfx½n n0 g 5 Differential equations The general nth-order, linear, constant coefficient, inhomogeneous differential equation: dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ n1 dt n dt n1 dm xðtÞ dm1 xðtÞ ¼ bm þ bm1 þ . . . þ b0 xðtÞ m dt dt m1 an coupled with the values of the n boundary conditions dn yðtÞ dn1 yðtÞ , , . . . , yðt0 Þ dt n t¼t0 dt n1 t¼t0 describes the input-response relationship of a continuous linear system with input xðtÞ and response yðtÞ. Such an equation has a solution in the form yðtÞ ¼ yh ðtÞ þ yp ðtÞ where yh ðtÞ is complementary function solution to the homogeneous equation an dn yðtÞ dn1 yðtÞ þ a þ . . . þ a0 yðtÞ ¼ 0 n1 dt n dt n1 and yp ðtÞ is a particular integral or particular solution to the inhomogeneous equation. The procedure for solving such an equation is: 43 230 Programme 6 (i) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (ii) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (iii) Apply the boundary conditions to find the values of the unknown integration constants in yh ðtÞ. 6 Zero-input and zero-state The solution of the general nth-order, linear, constant coefficient, inhomogeneous differential equation can alternatively be written as yðtÞ ¼ yzi ðtÞ þ yzs ðtÞ where yzi ðtÞ is called the zero-input response and yzs ðtÞ is called the zero-state response. The zero-input response of the equation depends only on the initial conditions and is independent of the input. It is obtained by solving the homogeneous equation and applying the boundary conditions. The zero-state response depends only on the input and is independent of the initial conditions. It is obtained by solving the inhomogeneous equation but with all the boundary conditions equated to zero. Here the procedure is: (i) Find the homogeneous solution yh ðtÞ in terms of unknown integration constants (ii) Find the particular solution yp ðtÞ and form the complete solution yðtÞ ¼ yh ðtÞ þ yp ðtÞ (iii) Equate the boundary conditions to zero and then find the values of the unknown integration constants in yðtÞ. This is then the zerostate response yzs ðtÞ. (iv) Apply the original boundary conditions to find the values of the unknown integration constants in yh ðtÞ. This is then the zero-input solution yzi ðtÞ. 7 Zero input and zero response For a linear time-invariant system zero input yields zero response. This is equivalent to all the boundary conditions having a zero value. 8 Arbitrary input If hðtÞ is the response of a continuous linear time-invariant system to the unit impulse ðtÞ, that is hðtÞ ¼ LfðtÞg then the response to an arbitrary input xðtÞ is the convolution of the input with the unit impulse response. That is: ð1 xðÞhðt Þ d ¼ xðtÞ hðtÞ LfxðtÞg ¼ 1 9 Exponential response The response of a linear, time-invariant system to an exponential input is a scaled exponential. That is LfAest g ¼ HðsÞðAest Þ. Therefore the exponential is an eigenfunction of the system and the scaling factor HðsÞ is the eigenvalue. This eigenvalue HðsÞ is referred to as the system transfer function. Introduction to invariant linear systems 10 Transfer function The transfer function HðsÞ of a linear, time-invariant system is the Laplace transform of the unit impulse response. That is: ð1 hðtÞest dt HðsÞ ¼ 1 11 Convolution theorem The fact that the response of a continuous linear time-invariant system is the convolution of the input with the unit impulse response enables the use of the convolution theorem as it applies to the Laplace transform: yðtÞ ¼ xðtÞ hðtÞ and so lfyðtÞg ¼ lfxðtÞ hðtÞg ¼ lfxðtÞglfhðtÞg That is YðsÞ ¼ XðsÞHðsÞ, the Laplace transform of the response, is equal to the product of the Laplace transform of the input and the system’s transfer function. 12 Arbitrary input to a discrete system If h½n is the response of a discrete linear shift-invariant system to the discrete unit impulse ½n, that is h½n ¼ Lf½ng then the response to an arbitrary input x½n is the convolution sum of the input with the unit impulse response. That is: Lfx½ng ¼ 1 X x½kh½n k ¼ x½n h½n k¼1 13 Exponential response The response of a discrete linear, time-invariant system to an exponential input is a scaled exponential. That is LfAzn g ¼ HðzÞðAzn Þ. Therefore the exponential is an eigenfunction of the system and the scaling factor HðzÞ is the eigenvalue. This eigenvalue HðzÞ is referred to as the system transfer function. 14 Transfer function The transfer function HðzÞ of a discrete linear, shift-invariant system is the Z transform of the unit impulse response. That is: HðzÞ ¼ 1 X h½kzk k¼1 15 Difference equations The transfer function HðzÞ of a discrete linear system described by a difference equation can be derived by taking the Z transform of the equation. By taking the inverse Z transform of the transfer function the unit impulse response can be found. Alternatively, given the impulse response of a discrete system the corresponding difference equation can be derived. 231 232 Programme 6 Can you? 44 Checklist 6 Check this list before and after you try the end of Programme test On a scale of 1 to 5 how confident are you that you can: . Recognise a system as a process whereby an input (either continuous or discrete) is converted to an output, also called the response of the system? Yes No Frames 1 to 4 . Distinguish between linear and non-linear systems and recognise time-invariant and shift-invariant systems? Yes No 5 to 10 . Determine the zero-input response and the zero-state response? Yes No 11 to 13 . Appreciate why zero valued boundary conditions give rise to a time-invariant system? Yes No 14 to 17 18 to 20 21 to 23 24 to 25 26 to 29 30 to 32 33 to 35 . Demonstrate that the response of a continuous, linear, time-invariant system to an arbitrary input is the convolution of the input with response of the system to a unit impulse? Yes No . Understand the role of the exponential function with respect to a linear, time-invariant system? Yes No . Use the convolution theorem to find the response of a continuous, linear, time-invariant system to an arbitrary input? Yes No . Derive the system transfer function of a constant coefficient linear differential equation and use it to solve the equation? Yes No . Demonstrate that the response of a discrete, linear, shift-invariant system to an arbitrary input is the convolution sum of the input with response of the system to a unit impulse? Yes No . Understand the role of the exponential function with respect to a discrete linear, shift-invariant system? Yes No 233 Introduction to invariant linear systems . Derive the system transfer function of a constant coefficient linear difference equation and use it to solve the equation? Yes No 36 to 40 . Derive the constant coefficient difference equation from knowledge of its unit impulse response? Yes No 41 to 42 Test exercise 6 1 Which of the following are linear, non-linear, time-invariant and shift-invariant: (a) yðtÞ ¼ LfxðtÞg ¼ 3xðtÞ (e) yðtÞ ¼ LfxðtÞg ¼ txðtÞ (b) y½n ¼ Lfx½ng ¼ 2x½n4 (c) yðtÞ ¼ LfxðtÞg ¼ e2t sin xðtÞ (d) y½n ¼ Lfx½ng ¼ 2x½n cos x½n 2 (f) y½n ¼ Lfx½ng ¼ x½n½n 4 xðtÞ (g) yðtÞ ¼ LfxðtÞg ¼ 4 (h) y½n ¼ Lfx½ng ¼ Lfx½ng ¼ 4n x½n Find the zero-input response and the zero-state response for each of the following and determine which are time-invariant: (a) y 0 ðtÞ 3yðtÞ ¼ t 2 uðtÞ : yð0Þ ¼ 2 (b) y 00 ðtÞ 5y 0 ðtÞ þ 4yðtÞ ¼ uðtÞ sin t : y 0 ð0Þ ¼ 4, yð0Þ ¼ 0 (c) 5y 0 ðtÞ þ 4yðtÞ ¼ et uðtÞ : yð0Þ ¼ 0 (d) y 00 ðtÞ þ 2y 0 ðtÞ þ yðtÞ ¼ uðtÞ : y 0 ð0Þ ¼ 0, yð0Þ ¼ 0 3 A linear, time-invariant system has the impulse response hðtÞ ¼ e3t uðtÞ find the system response to the input xðtÞ ¼ uðtÞ uðt 3Þ. 4 A linear, time-invariant system has the impulse response hðtÞ ¼ tuðt 1Þ find the transfer function HðsÞ and use it to find the response to the input xðtÞ ¼ uðtÞ 2uðt 1Þ þ uðt 2Þ 5 Given the differential equation y 00 ðtÞ þ 3y 0 ðtÞ 4yðtÞ ¼ 30e2t : y 0 ð0Þ ¼ 0, yð0Þ ¼ 0 find the transfer function and solve the equation. 6 A linear, shift-invariant system has the impulse response h½n ¼ nu½n find the system response to the input x½n ¼ 4n u½n. 7 Find the impulse response of the difference equation y½n þ 1 3y½n þ 2y½n 1 ¼ x½n þ 1 x½n. 8 A linear, shift-invariant system has the impulse response h½n ¼ nu½n, find the difference equation. 45 234 Programme 6 Further problems 46 1 For what values of is the system y½n ¼ Lfx½ng ¼ x½n shift-invariant? 2 For what values of a and b is the system yðtÞ ¼ LfxðtÞg ¼ axðtÞ þ b linear? 3 Is the system yðtÞ ¼ LfxðtÞg ¼ 1 X xðtÞðt nt0 Þ linear and time-invariant? n¼0 4 A linear, time-invariant system has the impulse response hðtÞ ¼ e3t uðtÞ find the system response to the input xðtÞ ¼ e3t uðtÞ. 5 Is the system y½n ¼ Lfx½ng ¼ x½n x½n linear and shift-invariant? Is the system y½n ¼ Lfx½ng ¼ x n3 linear and shift-invariant? 6 7 Show that the 8 2 > > > <4 x½n ¼ > 6 > > : 0 sequence: n¼0 n¼1 n¼2 otherwise can be represented as x½n ¼ 2½n þ 4½n 1 þ 6½n 2 or as x½n ¼ 2ðu½n þ u½n 1 þ u½n 2 3u½n 3Þ 8 The sign function sgnðxÞ (called the signum function to avoid confusion with the sine function) is defined as: 8 8 x<0 n<0 > > < 1 < 1 sgnðxÞ ¼ 0 x ¼ 0 the discrete form being sgn½n ¼ 0 n¼0 > > : : 1 x>0 1 n>0 The signum function is essentially the sign of a number so that x ¼ jxjsgnðxÞ because if x < 0 then x ¼ jxj and if x > 0 then x ¼ jxj. Show that if x½n ¼ an u½n then the even part of x½n is: 1 jnj xe ½n ¼ a þ ½n 2 and the odd part is 1 xo ½n ¼ ajnj sgn½n. 2 9 10 11 Show that the convolution sum of a½n ¼ nu½n 1 and b½n ¼ n2 u½n is n2 ðn2 1Þ . 12 1 anþ1 Show that if x½n ¼ an u½n (a 6¼ 1) and y½n ¼ u½n then x½n y½n ¼ u½n. 1a Show that if p½n 6¼ 0 only for m1 n m2 and q½n 6¼ 0 only for M1 n M2 then p½n q½n 6¼ 0 only for m1 þ M1 n m2 þ M2 . Introduction to invariant linear systems 12 The cross-correlation of two sequences a½n and b½n is defined as: 1 X a½n+ b½n ¼ a½kb½n þ k k¼1 Show that if x½n ¼ an u½n then ajnj 1 a2 [the cross-correlation of a sequence with itself is called the autocorrelation of the sequence]. n A linear, shift-invariant system has the impulse response h½n ¼ 14 u½n find the system response to the complex input x½n ¼ e jn!0 u½n. x½n+ x½n ¼ 13 14 Solve the differential equation y 00 ðtÞ þ 2y 0 ðtÞ þ yðtÞ ¼ et uðtÞ: yð0Þ ¼ 0, y 0 ð0Þ ¼ 0. 15 Find the impulse response of the differential equation 1 1 y 0 ðtÞ þ yðtÞ ¼ xðtÞ : yð0Þ ¼ 0 a a 16 Solve the differential equation 1 G y 0 ðtÞ ¼ yðtÞ þ uðtÞ : yð0Þ ¼ 0 T T 17 Solve the difference equation y½n ¼ y½n 1 þ ð1 Þu½n : y½0 ¼ 0. 18 Solve the difference equation 7 y½n þ 1 y½n ¼ ðy½n 20Þ : y½0 ¼ 160. 100 19 Solve the difference equation 2y½n ¼ y½n 1 þ y½n þ 1 : y½0 ¼ 0, y½1 ¼ 8. 20 A continuous, linear, time-invariant system has output yðtÞ ¼ tuðtÞ when the input is xðtÞ ¼ uðtÞ. Find the impulse response of the system and the output when the input is xðtÞ ¼ uðt 1Þ. 21 A discrete, linear, shift-invariant system has output y½n ¼ nu½n when the input is x½n ¼ 2n u½n. Find the impulse response of the system and the output when the input is x½n ¼ 3n u½n. 235 Programme 7 Frames 1 to 41 Fourier series 1 Learning outcomes When you have completed this Programme you will be able to: . Determine the period and amplitude of a periodic function . Write down the harmonics of a periodic trigonometric function . Give an analytic description of a non-sinusoidal periodic function . Evaluate integrals with periodic integrands . Demonstrate the orthogonality of the trigonometric functions sin nx and cos nx for n ¼ 0, 1, 2, . . . . Describe a periodic function as a Fourier series subject to Dirichlet conditions . Obtain the Fourier coefficients and hence the Fourier series of a periodic function . Describe the effects of the harmonics in the construction of the Fourier series . Find the value of the Fourier series at a point of discontinuity of the periodic function Prerequisite: Engineering Mathematics (Sixth Edition) Programmes 15 Integration 1 and 17 Reduction formulas 236 237 Fourier series 1 Introduction We have seen earlier that many functions can be expressed in the form of infinite series. Problems involving various forms of oscillations are common in fields of modern technology and Fourier series, with which we shall now be concerned, enable us to represent a periodic function as an infinite trigonometrical series in sine and cosine terms. One important advantage of a Fourier series is that it can represent a function containing discontinuities, whereas Maclaurin’s and Taylor’s series require the function to be continuous throughout. Periodic functions A function f ðxÞ is said to be periodic if its function values repeat at regular intervals of the independent variable. The regular interval between repetitions is the period of the oscillations. y f (x) 0 x1 x x1+p Graphs of y ¼ A sin nx (a) y ¼ sin x The obvious example of a periodic function is y ¼ sin x, which goes through its complete range of values while x increases from 08 to 3608. The period is therefore 3608 or 2 radians and the amplitude, the maximum displacement from the position of rest, is 1. y x (b) y ¼ 5 sin 2x The amplitude is 5. The period is 1808 and there are thus 2 complete cycles in 3608. y f (x) x 1 238 Programme 7 (c) y ¼ A sin nx Thinking along the same lines, the function y ¼ A sin nx has amplitude . . . . . . . . . . . .; period . . . . . . . . . . . .; and will have . . . . . . . . . . . . complete cycles in 3608. 2 amplitude ¼ A; period ¼ 3608 2 ¼ ; n cycles in 3608 n n Graphs of y ¼ A cos nx have the same characteristics. By way of revising earlier work, then, complete the following short exercise. Exercise In each of the following, state (a) the amplitude and (b) the period. 1 y ¼ 3 sin 5x 5 y ¼ 5 cos 4x 2 y ¼ 2 cos 3x x y ¼ sin 2 6 y ¼ 2 sin x 7 y ¼ 3 cos 6x y ¼ 4 sin 2x 8 y ¼ 6 sin 3 4 2x 3 Deal with all eight. They will not take much time. 3 No. Amplitude Period No. Amplitude Period 1 3 2=5 5 5 =2 2 2 2=3 6 2 3 1 4 7 3 =3 4 4 8 6 3 Harmonics A function f ðxÞ is sometimes expressed as a series of a number of different sine components. The component with the largest period is the first harmonic, or fundamental of f ðxÞ. y ¼ A1 sin x is the first harmonic or fundamental y ¼ A2 sin 2x is the second harmonic y ¼ A3 sin 3x is the third harmonic, etc. and in general y ¼ An sin nx is the . . . . . . . . . . . . harmonic, with amplitude . . . . . . . . . . . . and period . . . . . . . . . . . . 239 Fourier series 1 nth harmonic; amplitude An ; period ¼ 2 n 4 Non-sinusoidal periodic functions Although we introduced the concept of a periodic function via a sine curve, a function can be periodic without being obviously sinusoidal in appearance. Example In the following cases, the x-axis carries a scale of t in milliseconds. (a) y x (b) period ¼ 8 ms y x (c) period ¼ . . . . . . . . . y period ¼ . . . . . . . . . x (b) period ¼ 6 ms; (c) period ¼ 5 ms Analytic description of a periodic function A periodic function can be defined analytically in many cases. Example 1 y f (x) x (a) Between x ¼ 0 and x ¼ 4; y ¼ 3, i.e. f ðxÞ ¼ 3 0<x<4 (b) Between x ¼ 4 and x ¼ 6, y ¼ 0, i.e. f ðxÞ ¼ 0 4<x<6 5 240 Programme 7 So we could define the function by 3 0<x<4 f ðxÞ ¼ 0 4<x<6 f ðx þ 6Þ ¼ f ðxÞ the last line indicating that . . . . . . . . . . . . 6 the function is periodic with period 6 units Example 2 y f (x) x In this case (a) Between x ¼ 0 and x ¼ 2, y ¼ x i.e. f ðxÞ ¼ x 0<x<2 x x 2<x<6 (b) Between x ¼ 2 and x ¼ 6, y ¼ þ 3, i.e. f ðxÞ ¼ 3 2 2 (c) The period is 6 units i.e. f ðx þ 6Þ ¼ f ðxÞ. So we have ( x x 3 2 f ðx þ 6Þ ¼ f ðxÞ. f ðxÞ ¼ 0<x<2 2<x<6 Example 3 y In this case f(x) x 7 5x 8 f ðx þ 8Þ ¼ f ðxÞ f ðxÞ ¼ 0<x<8 ............ ............ 241 Fourier series 1 Here is a short exercise. Exercise Define analytically the periodic functions shown. y 1 f(x) x 2 y f(x) x y 3 f(x) x y 4 f(x) x – y 5 f(x) x Finish all five and then check the results. Here are the details. 2x 0<x<3 1 f ðxÞ ¼ 1 3<x<5 f ðx þ 5Þ ¼ f ðxÞ. 8 0<x<4 2 <3 f ðxÞ ¼ 5 4<x<7 : 0 7 < x < 10 f ðx þ 10Þ ¼ f ðxÞ. 8 242 Programme 7 3 4 5 8 0<x<4 <x f ðxÞ ¼ 4 4<x<7 : 0 7<x<9 f ðx þ 9Þ ¼ f ðxÞ. 8 3x > > 0<x<4 < 4 f ðxÞ ¼ 7 x 4 < x < 10 > > : 3 10 < x < 13 f ðx þ 13Þ ¼ f ðxÞ. 8 0<x<2 < 1 f ðxÞ ¼ 3 2<x<5 : 1 5<x<7 f ðx þ 7Þ ¼ f ðxÞ. Now we have the same thing in reverse. Exercise Sketch the graphs of the following, inserting relevant values. 4 0<x<5 1 f ðxÞ ¼ 0 5<x<8 f ðx þ 8Þ ¼ f ðxÞ. 2 3 4 5 9 f ðxÞ ¼ 3x x2 0<x<3 f ðx þ 3Þ ¼ f ðxÞ. 2 sin x 0<x< f ðxÞ ¼ 0 < x < 2 f ðx þ 2Þ ¼ f ðxÞ. 8 x > < 0<x< f ðxÞ ¼ 2 x > : < x < 2 2 f ðx þ 2Þ ¼ f ðxÞ. 8 2 x > > < 0<x<4 4 f ðxÞ ¼ 4 4<x<6 > > : 0 6<x<8 f ðx þ 8Þ ¼ f ðxÞ. Here they are: check carefully. y 1 f(x) x 243 Fourier series 1 y 2 f(x) x y 3 f(x) x y 4 f(x) x y 5 f(x) x All this is in preparation for what is to come, so let us now consider Fourier series. Move on then to the next frame Integrals of periodic functions Before we proceed we need to consider some specific integrals involving integers m and n. These are integrals over a single period of periodic integrands. You will already know some of these and the others you will easily be able to work out. The integrals that we are concerned with are those of sines, cosines and their combinations where the integration is over a single period from to . First, though, we list the integral of the unit constant over the period. ð h i dx ¼ x ¼ 2 1 ð sin nx cos nx dx ¼ ðn 6¼ 0Þ 2 n ¼ 3 ð sin n sinðnÞ n n ¼0 because sin n ¼ 0 sin nx dx ¼ . . . . . . . . . . . . ðn 6¼ 0Þ 10 244 Programme 7 ð 11 sin nx dx ¼ 0 Because ð h cos nxi sin nx dx ¼ ðn 6¼ 0Þ n cos n cosðnÞ þ ¼ n n ¼ 0 because cosðxÞ ¼ cos x ð ð cos 2nx þ 1 2 dx because cos 2A ¼ 2 cos2 A 1 4 cos nx dx ¼ 2 sin 2nx x ðn 6¼ 0Þ ¼ þ 4n 2 ¼ sin 2n sinð2nÞ ðÞ þ 4n 2 4n 2 ¼ ð sin2 nx dx ¼ . . . . . . . . . . . . 5 ðn 6¼ 0Þ ð 12 sin2 nx dx ¼ Because ð ð 1 cos 2nx dx because cos 2A ¼ 1 2 sin2 A sin2 nx dx ¼ 2 x sin 2nx ðn 6¼ 0Þ ¼ 2 4n sin 2n ðÞ sinð2nÞ þ ¼ 2 4n 2 4n ¼ ð 6 cos mx cos nx dx 1 ¼ 2 ð ½cosðm þ nÞx þ cosðm nÞx dx because 2 cos A cos B ¼ cosðA þ BÞ þ cosðA BÞ sinðm þ nÞx sinðm nÞx ¼ ðm 6¼ nÞ þ mþn mn ¼ 7 sinðm þ nÞ sinðm nÞ sinðm þ nÞðÞ sinðm nÞðÞ þ mþn mn mþn mn ¼0 ð sin mx sin nx dx ¼ . . . . . . . . . . . . ðm 6¼ nÞ 245 Fourier series 1 ð sin mx sin nx dx ¼ 0, m 6¼ n Because ð sin mx sin nx dx ð 1 ¼ ½cosðm nÞx cosðm þ nÞx dx 2 because 2 sin A sin B ¼ cosðA BÞ cosðA þ BÞ sinðm nÞx sinðm þ nÞx ¼ ðm 6¼ nÞ mn mþn sinðm nÞ sinðm þ nÞ sinðm nÞðÞ sinðm þ nÞðÞ þ ¼ mn mþn mn mþn ¼0 ð 8 cos mx sin nx dx ðm 6¼ nÞ ð 1 ¼ ½sinðm þ nÞx sinðm nÞx dx 2 because 2 cos A sin B ¼ sinðA þ BÞ sinðA BÞ 1 cosðm þ nÞx cosðm nÞx ¼ ðm 6¼ nÞ þ 2 mþn mn 1 cosðm þ nÞ cosðm nÞ ¼ þ 2 mþn mn cosðm þ nÞðÞ cosðm nÞðÞ þ mþn mn ¼0 because cosðxÞ ¼ cos x And finally, when m ¼ n ð cos mx sin mx dx ¼ . . . . . . . . . . . . 9 13 246 Programme 7 ð 14 cos mx sin mx dx ¼ 0 ¼ Because ð cos mx sin mx dx ð 1 ¼ sin 2mx dx because sin 2A ¼ 2 sin A cos A 2 1 cos 2mx ðm 6¼ 0Þ ¼ 2 2m 1 cos 2m cos 2mðÞ þ ¼ 2 2m 2m ¼0 15 because cosðxÞ ¼ cos x Summary ð 1 2 ð h i dx ¼ x ¼ 2 cos nx dx ¼ 0 ð 3 sin nx dx ¼ 0 4 ð ( cos mx cos nx dx ¼ mn 5 ð where mn ¼ 1 if m ¼ n 0 if m 6¼ n ðmn is called the Kronecker deltaÞ sin mx sin nx dx ¼ mn 6 ð cos mx sin nx dx ¼ 0 Note that the same results are obtained no matter what the end points of the integrals are, provided that the interval between them is one period. So, for example ð kþ2 sin nx kþ2 cos nx dx ¼ ðn 6¼ 0Þ n k k sinðnk þ 2nÞ sin nk ¼ n n ¼ 0 because sinðx þ 2nÞ ¼ sin x Now to put all these integrals to practical use 247 Fourier series 1 16 Orthogonal functions If two different functions f ðxÞ and gðxÞ are defined on the interval a x b ðb f ðxÞgðxÞ dx ¼ 0 and a then we say that the two functions are orthogonal to each other on the interval a x b. In the previous frames we have seen that the trigonometric functions sin nx and cos nx where n ¼ 0, 1, 2, . . . form an infinite collection of periodic functions that are mutually orthogonal on the interval x , indeed on any interval of width 2. That is ð cos mx cos nx dx ¼ 0 for m 6¼ n ð sin mx sin nx dx ¼ 0 for m 6¼ n and ð cos mx sin nx dx ¼ 0 Fourier series Given that certain conditions are satisfied then it is possible to write a periodic function of period 2 as a series expansion of the orthogonal periodic functions just discussed. That is, if f ðxÞ is defined on the interval x where f ðx þ 2nÞ ¼ f ðxÞ then f ðxÞ ¼ 1 a0 X þ ðan cos nx þ bn sin nxÞ 2 n¼1 This is the Fourier series expansion of f ðxÞ where the an and bn are constants called the Fourier coefficients. But how do we find the values of these constants? Quite easily. We make use of the mutual orthogonality of the trigonometric functions in the expansion. 17 248 Programme 7 For example, to find a10 we multiply f ðxÞ by cos 10x and integrate over a period. That is ð f ðxÞ cos 10x dx ! ð 1 a0 X þ ¼ ðan cos nx þ bn sin nxÞ cos 10x dx 2 n¼1 ð ð 1 X a0 ¼ cos 10x dx þ an cos nx cos 10x dx 2 n¼1 ð 1 X bn sin nx cos 10x dx þ n¼1 ¼ 1 X 1 X a0 0þ an n;10 þ bn 0 2 n¼1 n¼1 ¼ a0 0 þ a1 0 þ . . . þ a9 0 þ a10 1 þ a11 0 þ . . . ¼ a10 So that a10 1 ¼ ð f ðxÞ cos 10x dx ð f ðxÞ cos mx dx ¼ . . . . . . . . . . . . In just the same way ð 18 f ðxÞ cos mx dx ¼ am Because ð f ðxÞ cos mx dx ! 1 a0 X þ ¼ ðan cos nx þ bn sin nxÞ cos mx dx 2 n¼1 ð ð 1 X a0 ¼ cos mx dx þ an cos nx cos mx dx 2 n¼1 ð 1 X bn sin nx cos mx dx þ ð n¼1 ¼ 1 X a0 an n;m þ bn 0 0þ 2 n¼1 n¼1 1 X ¼ am and so am ¼ 1 ð f ðxÞ cos mx dx Finally ð f ðxÞ sin mx dx ¼ . . . . . . . . . . . . 249 Fourier series 1 ð f ðxÞ sin mx dx ¼ bm 19 Because ð f ðxÞ sin mx dx ! 1 a0 X þ ¼ ðan cos nx þ bn sin nxÞ sin mx dx 2 n¼1 ð ð 1 X a0 sin mx dx þ an cos nx sin mx dx ¼ 2 n¼1 ð 1 X þ bn sin nx sin mx dx ð n¼1 1 1 X X a0 0þ ¼ an 0 þ bn n;m 2 n¼1 n¼1 ¼ bm and so bm ¼ 1 ð f ðxÞ sin mx dx 20 Summary Given that certain conditions are satisfied, if f ðxÞ is defined on the interval x and where f ðx þ 2nÞ ¼ f ðxÞ then f ðxÞ ¼ 1 a0 X þ ðan cos nx þ bn sin nxÞ 2 n¼1 This is the Fourier series expansion of f ðxÞ where the an and bn are constants called the Fourier coefficients and where ð ð 1 1 f ðxÞ cos nx dx and bn ¼ f ðxÞ sin nx dx, n ¼ 0, 1, 2, . . . an ¼ Look in particular at the constant function f ðxÞ ¼ c which can be considered as a periodic function with any period we wish to choose. Choosing the period to be 2 then ð ð 1 c c cos nx dx ¼ cos nx dx ¼ 2cn;0 . an ¼ a0 as expected. That is a0 ¼ 2c so c ¼ 2 Also 1 bn ¼ ð c c sin nx dx ¼ ð sin nx dx ¼ 0 250 Programme 7 From this we see that we have two choices to represent the Fourier series. We can either write f ðxÞ ¼ where an ¼ 1 1 a0 X þ fan cos nx þ bn sin nxg 2 n¼1 ð f ðxÞ cos nx dx and bn ¼ 1 ð f ðxÞ sin nx dx or we can write f ðxÞ ¼ 1 X fan cos nx þ bn sin nxg n¼0 where ð 1 f ðxÞ dx, an ¼ f ðxÞ cos nx dx ðn 6¼ 0Þ ð 1 and bn ¼ f ðxÞ sin nx dx. a0 ¼ 1 2 ð We choose the former and so avoid having a separate integral for a0 . Dirichlet conditions 21 If a function f ðxÞ is such that (a) f ðxÞ is defined, single-valued and periodic with period 2 (b) f ðxÞ and f 0 ðxÞ have at most a finite number of finite discontinuities over a single period – that is they are piecewise continuous then the series 1 a0 X þ fan cos nx þ bn sin nxg 2 n¼1 ð ð 1 1 f ðxÞ cos nx dx and bn ¼ f ðxÞ sin nx dx converges to f ðxÞ when ðx, f ðxÞÞ is a point of continuity. The Dirichlet conditions are sufficient for the Fourier series to represent f ðxÞ not only at a point of continuity but, with a slight modification, also at a point of discontinuity, as we shall see later in Frame 36. Also the periodicity of the function need not be restricted to 2, as we shall see in Programme 8. Note that these conditions, while being sufficient, are not necessary because there are functions that do not satisfy these conditions which still possess a convergent Fourier series. However, the cases met in science and engineering do generally meet these conditions. where an ¼ 251 Fourier series 1 Exercise If the following functions are defined over the interval < x < and f ðx þ 2Þ ¼ f ðxÞ, state whether or not each function can be represented by a Fourier series. 1 f ðxÞ ¼ x3 4 2 f ðxÞ ¼ 4x 5 2 f ðxÞ ¼ x 5 1 x5 f ðxÞ ¼ tan x 6 f ðxÞ ¼ y 3 1 2 Yes Yes 3 No: infinite discontinuity at x ¼ 0 f ðxÞ ¼ where x2 þ y 2 ¼ 2 22 4 Yes 5 No: infinite discontinuity at x ¼ =2 6 No: two valued On then Example 1 23 y Find the Fourier series for the function shown. Consider one cycle between x ¼ and x ¼ . x 8 > 0 > > > < The function can be defined by f ðxÞ ¼ 4 > > > > :0 <x< f ðx þ 2Þ ¼ f ðxÞ. (a) As before f ðxÞ ¼ 1 X 1 a0 þ fan cos nx þ bn sin nxg 2 n¼1 The expression for a0 is . . . . . . . . . . . . <x< 2 2 <x< 2 2 252 Programme 7 24 1 a0 ¼ ð f ðxÞ dx This gives (ð ) ð =2 ð =2 1 0 dx þ 4 dx þ 0 dx a0 ¼ =2 =2 =2 1 4x ; a0 ¼ 4 ¼ =2 (b) To find an an ¼ 1 1 ; an ¼ ð f ðxÞ cos nx dx ( ð =2 ð0Þ cos nx dx þ ð =2 4 cos nx dx þ =2 ð =2 ; an ¼ . . . . . . . . . . . . 25 an ¼ 8 n sin n 2 Then considering different integer values of n, we have If n is even If n ¼ 1, 5, 9, . . . If n ¼ 3, 7, 11, . . . an ¼ 0 8 an ¼ n an ¼ 8 n We keep these in mind while we find bn . (c) To find bn ð 1 bn ¼ f ðxÞ sin nx dx ¼ . . . . . . . . . . . . ) ð0Þ cos nx dx 253 Fourier series 1 26 bn ¼ 0 Because we have (ð ) ð =2 ð =2 1 ð0Þ sin nx dx þ 4 sin nx dx þ ð0Þ sin nx dx bn ¼ =2 =2 4 ¼ 4 cos nx =2 sin nx dx ¼ n =2 =2 ð =2 no 4 n n cos cos ¼0 ; bn ¼ 0 n 2 2 So with a0 ¼ 4; an as stated above; bn ¼ 0; the Fourier series is 8 1 1 1 cos x cos 3x þ cos 5x cos 7x þ . . . f ðxÞ ¼ 2 þ 3 5 7 ¼ In this particular example, there are, in fact, no sine terms. Example 2 y f(x) x Determine the Fourier series to represent the periodic function shown. It is more convenient here to take the limits as 0 to 2. y The function can be defined as x f ðxÞ ¼ 0 < x < 2 2 y = x/2 f(x) x f ðx þ 2Þ ¼ f ðxÞ i.e. period ¼ 2. Now to find the coefficients. 2 ð ð 1 2 1 2 x 1 2 dx ¼ x (a) a0 ¼ f ðxÞ dx ¼ 0 0 2 4 0 ¼ 1 (b) an ¼ ; a0 ¼ ð 2 0 1 f ðxÞ cos nx dx ¼ ð 2 x cos nx dx 2 0 ð 1 2 x cos nx dx 2 0 ¼ . . . . . . . . . . . . (integrating by parts) ¼ 254 Programme 7 27 an ¼ 0 Because 1 an ¼ 2 ð 2 1 x cos nx dx ¼ 2 0 1 1 ð0 0Þ ð0Þ ¼ 2 n 1 (c) bn ¼ ð 2 ( ) ð x sin nx 2 1 2 sin nx dx n n 0 0 ¼0 ; an ¼ 0 f ðxÞ sin nx dx ¼ . . . . . . . . . . . . 0 28 bn ¼ 1 n Straightforward integration by parts, as for an , gives the result stated. So we now have a0 ¼ . . . . . . . . . . . . ; 29 a0 ¼ ; an ¼ . . . . . . . . . . . . ; an ¼ 0; bn ¼ bn ¼ . . . . . . . . . . . . 1 n Now the general expression for a Fourier series is ............ 30 f ðxÞ ¼ 1 X 1 a0 þ fan cos nx þ bn sin nxg 2 n¼1 Therefore in this case 1 X þ fbn sin nxg because an ¼ 0 2 n¼1 1 1 1 ¼ þ sin x sin 2x sin 3x . . . 2 1 2 3 1 1 ; f ðxÞ ¼ sin x þ sin 2x þ sin 3x þ . . . 2 2 3 f ðxÞ ¼ Note that in this example, the series contains a constant term and sine terms only. 255 Fourier series 1 Example 3 Find the Fourier series for the function defined by y f ðxÞ ¼ x f ðxÞ ¼ 0 < x < 0 0<x< f ðx þ 2Þ ¼ f ðxÞ. x The general expressions for a0 , an , bn are a0 ¼ . . . . . . . . . . . . an ¼ . . . . . . . . . . . . bn ¼ . . . . . . . . . . . . a0 ¼ an ¼ bn ¼ 1 1 1 ð f ðxÞ dx 31 ð f ðxÞ cos nx dx ð f ðxÞ sin nx dx With that reminder, in this example a0 ¼ . . . . . . . . . . . . a0 ¼ 2 Because 1 (a) a0 ¼ ð 1 f ðxÞ dx ¼ 0 1 x2 ðxÞ dx ¼ ¼ 2 2 ð0 (b) To find an ð 1 an ¼ f ðxÞ cos nx dx ¼ . . . . . . . . . . . . 32 256 Programme 7 33 an ¼ 2 n2 ðn oddÞ; ðn evenÞ 0 Because an ¼ 1 ð f ðxÞ cos nx dx ¼ ¼ 1 1 ¼ ¼ ð0 1 ðxÞ cos nx dx ð0 x cos nx dx ( sin nx x n 0 1 n ) ð0 sin nx dx 1 1 h cos nxi0 ð0 0Þ n n 1 f1 cos ng n2 ¼ But cos n ¼ 1 (n even) or 1 (n odd) ; an ¼ 2 n2 ðn odd) or 0 ðn even) (c) Now to find bn . Working as for an , we obtain bn ¼ . . . . . . . . . . . . 34 bn ¼ 1 n ðn even) 1 n or ðn oddÞ Because ð 1 0 f ðxÞ sin nx dx ¼ ðxÞ sin nx dx ð0 1 ¼ x sin nx dx ( ) ð 1 cos nx 0 1 0 x ¼ þ cos nx dx n n ( ) 1 cos n 1 sin nx 0 cos n þ ¼ ¼ n n n n 1 bn ¼ ; bn ¼ So we have ð 1 n 1 n ðn oddÞ an ¼ 0 ðn evenÞ ðn evenÞ; a0 ¼ ; 2 bn ¼ 1 n or ðn evenÞ or 2 ðn oddÞ n2 1 ðn oddÞ n ; f ðxÞ ¼ . . . . . . . . . . . . Complete the series 257 Fourier series 1 2 1 1 cos x þ cos 3x þ cos 5x þ . . . f ðxÞ ¼ 4 9 25 1 1 1 þ sin x sin 2x þ sin 3x sin 4x þ . . . 2 3 4 35 It is just a case of substituting n ¼ 1, 2, 3, etc. In this particular example, we have a constant term and both sine and cosine terms. Effect of harmonics It is interesting to see just how accurately the Fourier series represents the function with which it is associated. The complete representation requires an infinite number of terms, but we can, at least, see the effect of including the first few terms of the series. Let us consider the waveform shown. We established earlier in Frames 23–26 that the function y x 8 > 0 > > > > < f ðxÞ ¼ 4 > > > > > :0 <x< 2 <x< 2 2 <x< 2 f ðx þ 2Þ ¼ f ðxÞ gives the Fourier series 8 1 1 1 f ðxÞ ¼ 2 þ cos x cos 3x þ cos 5x cos 7x þ . . . 3 5 7 If we start with just one cosine term, we can then see the effect of including subsequent harmonics. Let us restrict our attention to just the right-hand half of the symmetrical waveform. Detailed plotting of points gives the development on the next page. 258 Programme 7 1 8 f ðxÞ ¼ 2 þ cos x y f (x) x 2 f ðxÞ ¼ 2 þ y 8 1 cos x cos 3x 3 f (x) x 3 8 1 1 cos x cos 3x þ cos 5x f ðxÞ ¼ 2 þ 3 5 y f (x) x 4 f ðxÞ ¼ 2 þ y 8 1 1 1 cos x cos 3x þ cos 5x cos 7x 3 5 7 f (x) x As the number of terms is increased, the graph gradually approaches the shape of the original square waveform. The ripples increase in number and, apart from the one nearest to the step, decrease in amplitude. A perfectly square waveform is unattainable in practice. For practical purposes, the first few terms normally suffice to give an accuracy of acceptable level. Gibbs’ phenomenon You will notice from the previous diagrams that near the discontinuity, as more terms are taken into acount, the series tends to overshoot on one side and undershoot on the other. This over and undershooting on either side of the discontinuity does not go away as the number of terms in the Fourier series that are taken into account is increased, rather it tends to two spikes on either side of the discontinuity. This effect is called the Gibbs’ phenomenon. 259 Fourier series 1 36 Sum of a Fourier series at a point of discontinuity f ðxÞ ¼ 12 a0 þ 1 X fan cos nx þ bn sin nxg n¼1 If f ðxÞ is continuous at x ¼ x1 , the series converges to the value f ðx1 Þ as the number of terms included increases to infinity. A particular point of interest occurs at a point of finite discontinuity or ‘jump’ of the function y ¼ f ðxÞ. y As x ! x1 , the expression f ðxÞ approaches y1 or y2 depending on the direction of approach. y1 y2 x x1 y If we approach x ¼ x1 from below that value, the limiting value of f ðxÞ is y1 . y1 x1 x y If we approach x ¼ x1 from above that value, the limiting value of f ðxÞ is y2 . y2 x1 x To distinguish between these two values we write y1 ¼ f ðx1 0Þ denoting immediately before x ¼ x1 y2 ¼ f ðx1 þ 0Þ denoting immediately after x ¼ x1 . In fact, if we substitute x ¼ x1 in the Fourier series for f ðxÞ, it can be shown that the series converges to the value 12 ff ðx1 0Þ þ f ðx1 þ 0Þg i.e. 12 ðy1 þ y2 Þ, the average of y1 and y2 . Example Consider the function 0 < x < f ðxÞ ¼ a < x < 2 f ðx þ 2Þ ¼ f ðxÞ. y a x First of all, determine the Fourier series to represent the function. There are no snags. f ðxÞ ¼ . . . . . . . . . . . . 260 Programme 7 37 f ðxÞ ¼ a 2a þ fsin x þ 13 sin 3x þ 15 sin 5x þ . . .g 2 Check the working ð ð 1 1 1 f ðxÞ dx ¼ a dx ¼ ; a0 ¼ a (a) a0 ¼ ax ¼ a 0 0 ð ð 1 1 f ðxÞ cos nx dx ¼ a cos nx dx (b) an ¼ 0 a sin nx ¼0 ; an ¼ 0 ¼ n 0 ð ð 1 1 f ðxÞ sin nx dx ¼ a sin nx dx (c) bn ¼ 0 a cos nx a a ð1 cos nÞ ¼ 1 ð1Þn ¼ ¼ n n n 0 and because cos n ¼ 1 ðn evenÞ and 1 ðn oddÞ 2a ðn oddÞ bn ¼ 0 ðn evenÞ; n 1 X 1 ; f ðxÞ ¼ a0 þ bn sin nx 2 n¼1 a 2a 1 1 ; f ðxÞ ¼ þ sin x þ sin 3x þ sin 5x þ . . . 2 3 5 A finite discontinuity, or ‘jump’, occurs at x ¼ 0. If we substitute x ¼ 0 in the series obtained, all the sine terms vanish and we get f ðxÞ ¼ a=2, which is, in fact, the average of the two function values at x ¼ 0. y Note also that at x ¼ , another finite discontinuity occurs and substituting x ¼ in the series gives the same result. a a x The Revision summary and Can you? checklist now follow, after which you will have no trouble with the Test exercise. The Further problems provide additional practice. 261 Fourier series 1 Revision summary 7 1 Graphs of y ¼ A sin nx and A cos nx 38 3608 2 Amplitude ¼ A; period ¼ ¼ radians n n 2 Harmonics y ¼ A1 sin x is the first harmonic or fundamental y ¼ An sin nx is the nth harmonic. 3 Periodic function f ðx þ PÞ ¼ f ðxÞ 4 P ¼ period Fourier series – functions of period 2 f ðxÞ ¼ 12 a0 þ a1 cos x þ a2 cos 2x þ a3 cos 3x þ . . . þ an cos nx . . . þ b1 sin x þ b2 sin 2x þ b3 sin 3x þ . . . þ bn sin nx . . . 1 X fan cos nx þ bn sin nxg ¼ 12 a0 þ n¼1 5 Dirichlet conditions (a) The function f ðxÞ must be defined, single-valued and periodic. (b) f ðxÞ and f 0 ðxÞ must be piecewise continuous in the periodic interval. 6 Fourier coefficients ð 1 f ðxÞ dx a0 ¼ ð 1 an ¼ f ðxÞ cos nx dx ð 1 bn ¼ f ðxÞ sin nx dx where, in each case, n ¼ 1, 2, 3, . . . 7 Sum of Fourier series at a finite discontinuity y At x ¼ x1 , series for f ðxÞ converges to the value y1 1 2 ff ðx1 y2 x1 x 0Þ þ f ðx1 þ 0Þg ¼ 12 ðy1 þ y2 Þ 262 Programme 7 Can you? 39 Checklist 7 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Determine the period and amplitude of a periodic function? Yes No Frames 1 . Write down the harmonics of a periodic trigonometric function? Yes No . Give an analytic description of a non-sinusoidal periodic function? Yes No and 2 3 4 to 9 10 to 15 . Demonstrate the orthogonality of the trigonometric functions sin nx and cos nx for n ¼ 0, 1, 2, . . . ? Yes No 16 to 20 . Describe a periodic function as a Fourier series subject to the Dirichlet conditions? Yes No 21 and 22 . Obtain the Fourier coefficients and hence the Fourier series of a periodic function? Yes No 23 to 25 . Evaluate integrals with periodic integrands? Yes No . Describe the effects of the harmonics in the construction of the Fourier series? Yes No . Find the value of the Fourier series at a point of discontinuity of the periodic function? Yes No 35 36 and 37 263 Fourier series 1 Test exercise 7 1 What is the amplitude and the period of the function with output pffiffiffi 3x f ðxÞ ¼ 2 cos ? 4 2 Give an analytic description of the function with the following graph: f(x) 2 3 Draw the graph of: 8 > <x f ðxÞ ¼ 1 > : ðx 42 Þ f ðx þ 4Þ ¼ f ðxÞ 4 3 4 6 7 x 0x<1 1x<3 3x<4 for 0 x 8 If f ðxÞ is defined in the interval x < and f ðx þ 2Þ ¼ f ðxÞ, state whether or not each of the following functions can be represented by a Fourier series. (a) f ðxÞ ¼ x4 (d) f ðxÞ ¼ e2x (b) f ðxÞ ¼ 3 2x 1 (c) f ðxÞ ¼ x (e) f ðxÞ ¼ cosec x pffiffiffiffiffiffi (f) f ðxÞ ¼ 4x 5 Determine the Fourier series for the function defined by f ðxÞ ¼ 2x 0 x 2 f ðx þ 2Þ ¼ f ðxÞ 6 What is the value at x ¼ 4 of the Fourier series for the function defined by 8 0x<2 > < 2 f ðxÞ ¼ 4 2x<4 > : 2 4 x < 2 f ðx þ 2Þ ¼ f ðxÞ 40 264 Programme 7 Further problems 7 41 1 For each of the following graphs give the analytical description of the function drawn. f(x) 2 (a) 2 6 4 x 8 f(x) 3 9 3 (b) x 12 6 –3 f(x) 2 (c) –4 –3 2 Draw the graph of sin x (a) ðxÞ ¼ cos x f ðx þ Þ ¼ f ðxÞ for cos x (b) f ðxÞ ¼ sin x f ðx þ Þ ¼ f ðxÞ for ( x2 (c) f ðxÞ ¼ 6x (d) x –1 0 x < =2 =2 x < 2 x 2 0 x < =2 =2 x < 2 x 2 0x<2 2 x < 10 f ðx þ 10Þ ¼ f ðxÞ for 20 x 20 8 3 0x<2 > <x f ðxÞ ¼ 8 2x<3 > : 3 3x<5 ð5 xÞ f ðx þ 5Þf ðxÞ for 10 x 10 265 Fourier series 1 ( (e) f ðxÞ ¼ ðx þ 4Þ2 4 2x f ðx þ 4Þ ¼ f ðxÞ 3 4 x < 2 2 x < 0 for 8 x 8 A periodic function f ðxÞ is defined by x 0 x < 2 f ðxÞ ¼ 1 f ðx þ 2Þ ¼ f ðxÞ Determine the Fourier series up to and including the third harmonic. 4 A function is defined by þx x < 0 f ðxÞ ¼ x 0x< f ðx þ 2Þ ¼ f ðxÞ Obtain the Fourier series. 5 A periodic function is defined by A sin x 0x< f ðxÞ ¼ A sin x x < 2 f ðx þ 2Þ ¼ f ðxÞ Obtain the Fourier series up to and including the fourth harmonic. 6 A function is defined by 0 x < 0 f ðxÞ ¼ x 0x< f ðx þ 2Þ ¼ f ðxÞ Obtain the Fourier series. 7 A function is defined by cos x x < 0 f ðxÞ ¼ 0 0x< f ðx þ 2Þ ¼ f ðxÞ Obtain the Fourier series. 8 A function is defined by f ðxÞ ¼ x2 f ðx þ 2Þ ¼ f ðxÞ x< Obtain the Fourier series 9 A function is defined by 3x f ðxÞ ¼ 7 f ðx þ 2Þ ¼ f ðxÞ x< Obtain the Fourier series up to the fourth harmonic. 266 Programme 7 10 A function is defined by 8 þx > < x < 0 2 f ðxÞ ¼ x > : 0x< 2 f ðx þ 2Þ ¼ f ðxÞ Obtain the Fourier series. 11 A function is defined by f ðxÞ ¼ x2 f ðx þ 2Þ ¼ f ðxÞ 0 x < 2 Obtain the Fourier series. 12 Given a periodic function f ðxÞ with period 2 and Fourier series a0 X þ f ðxÞ ¼ fan cos nx þ bn sin nxg 2 n¼1 show that ð 1 n o 1 a0 2 X ½f ðxÞ2 dx ¼ þ an 2 þ bn 2 2 n¼1 13 Given two periodic functions f ðxÞ and gðxÞ each with period 2 and Fourier series 1 1 a0 X p0 X þ f ðxÞ ¼ fan cos nx þ bn sin nxg and gðxÞ ¼ þ fpn cos nx þ qn sin nxg 2 n¼1 2 n¼1 show that ð 1 1 a0 p0 X þ f ðxÞgðxÞ dx ¼ fan pn þ bn qn g 2 n¼1 14 Given two periodic functions f ðxÞ and gðxÞ ¼ ð xÞ each with period 2 and Fourier series 1 1 a0 X p0 X þ f ðxÞ ¼ fan cos nx þ bn sin nxg and gðxÞ ¼ þ fpn cos nx þ qn sin nxg 2 n¼1 2 n¼1 show that ð 1 X 1 2 bn f ðxÞð xÞ dx ¼ 2 0 n n¼1 Programme 8 Frames 1 to 52 Fourier series 2 Learning outcomes When you have completed this Programme you will be able to: . Obtain the Fourier coefficients of a function with arbitrary period T . Recognise even and odd functions and their products . Derive the Fourier series of even and odd functions . Derive half-range Fourier series . Recognise the conditions for the Fourier series to contain only odd or only even harmonics . Explain the geometric significance of the constant term a0 =2 . Derive half-range Fourier series with arbitrary period 267 268 Programme 8 Functions with periods other than 2 1 In Programme 7 we considered functions f ðxÞ with period 2. In practice, we often encounter functions defined over periodic intervals other than 2, T T e.g. from 0 to T, to , etc. 2 2 Functions with period T T T to , i.e. has a period T, we can convert 2 2 this to an interval of 2 by changing the units of the independent variable. In many practical cases involving physical oscillations, the independent variable is time (t) and the periodic interval is normally denoted by T, i.e. If y ¼ f ðxÞ is defined in the range f ðt þ TÞ ¼ f ðtÞ y f (t) a a+T t (s) x period = T Each cycle is therefore completed in T seconds and the frequency f hertz 1 (oscillations per second) of the periodic function is therefore given by f ¼ . T If the angular velocity, ! radians per second, is defined by ! ¼ 2f , then !¼ 2 T and T ¼ 2 . ! The angle, x radians, at any time t is therefore x ¼ !t and the Fourier series to represent the function can be expressed as 1 a0 X þ fan cos n!t þ bn sin n!t g 2 n¼1 1 a0 X 2nt 2nt þ bn sin ¼ þ an cos T T 2 n¼1 f ðtÞ ¼ Fourier series 2 269 Fourier coefficients 2 With the new variable, the Fourier coefficients become f ðtÞ ¼ 12 a0 þ 1 X fan cos n!t þ bn sin n!tg n¼1 a0 ¼ an ¼ bn ¼ 2 T 2 T 2 T ðT f ðtÞ dt ¼ 0 ðT f ðtÞ cos n!t dt ¼ 0 ðT f ðtÞ sin n!t dt ¼ 0 ! ! ! ð 2=! f ðtÞ dt 0 ð 2=! f ðtÞ cos n!t dt 0 ð 2=! f ðtÞ sin n!t dt. 0 We can see that there is very little difference between these expressions and T T those that have gone before. The limits can, of course, be 0 to T, to , 2 2 2 etc. as is convenient, so long as they cover a complete period. to , 0 to ! ! ! Example Determine the Fourier series for a periodic function defined by ( 2ð1 þ tÞ 1 < t < 0 f ðtÞ ¼ 0 0<t<1 f ðt þ 2Þ ¼ f ðtÞ The first step is to sketch the waveform which is . . . . . . . . . . . . 270 Programme 8 3 y y = 2(1 + t) x We have 1 a0 X 2nt 2nt þ bn sin an cos f ðtÞ ¼ þ T T 2 n¼1 ¼ 1 a0 X þ fan cos nt þ bn sin ntg 2 n¼1 because T ¼ 2 Therefore ð ð1 ð0 ð1 2 T=2 f ðtÞ dt ¼ f ðtÞ dt ¼ 2ð1 þ tÞ dt þ ð0Þ dt a0 ¼ T T=2 1 1 0 0 ¼1 ¼ 2t þ t 2 1 and ð ð1 2 T=2 f ðtÞ cos nt dt ¼ f ðtÞ cos nt dt T T=2 1 ð0 ¼ 2ð1 þ tÞ cos nt dt ¼ . . . . . . . . . . . . an ¼ 1 4 an ¼ 0 ðn evenÞ; an ¼ 4 ðn oddÞ n 2 2 Because ð0 an ¼ 2ð1 þ tÞ cos nt dt 1 ( ) ð sin nt 0 1 0 ¼ 2 ð1 þ tÞ sin nt dt n 1 n 1 ( ) 1 cos nt 0 2 ¼ 2 2 ð1 cos nÞ ¼ 2 ð0 0Þ n n n 1 ¼ 2 1 ð1Þn n2 2 so that an ¼ 0 ðn evenÞ, an ¼ Now for bn 2 bn ¼ T 4 n2 2 ð T=2 T=2 ðn oddÞ f ðtÞ sin 2nt dt ¼ . . . . . . . . . . . . T 271 Fourier series 2 bn ¼ 5 2 n Because ð0 bn ¼ 2ð1 þ tÞ sin nt dt 1 ( ) ð cos nt 0 1 0 ¼ 2 ð1 þ tÞ þ cos nt dt n 1 n 1 ( ) 1 sin nt 0 2 2 2 þ þ 2 2 ðsin nÞ ¼ ¼ ¼2 n n 1 n n n So the first few terms of the series give f ðtÞ ¼ . . . . . . . . . . . . f ðtÞ ¼ 1 4 1 1 þ 2 cos t þ cos 3t þ cos 5t þ . . . 2 9 25 2 1 1 1 sin t þ sin 2t þ sin 3t þ sin 4t þ . . . 2 3 4 6 The Fourier series 1 a0 X f ðtÞ ¼ þ fan cos n!t þ bn sin n!t g 2 n¼1 can also be written in the form 1 A0 X þ f ðtÞ ¼ Bn sinðn!t þ n Þ 2 n¼1 Comparing these two expressions we see that A0 ¼ a0 , Bn sin n ¼ an and Bn cos n ¼ bn . From this it follows that Bn ¼ . . . . . . . . . . . . and n ¼ . . . . . . . . . . . . Bn ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2n þ b2n ; n ¼ arctan an bn So B1 sinð!t þ 1 Þ is the first harmonic or fundamental (lowest frequency) B2 sinð2!t þ 2 Þ is the second harmonic (frequency twice that of the fundamental) Bn sinðn!t þ n Þ is the nth harmonic (frequency n times that of the fundamental). And for the series to converge, the values of Bn must eventually decrease with higher-order harmonics, i.e. Bn ! 0 as n ! 1. 7 272 8 Programme 8 Odd and even functions (a) Even functions A function f ðxÞ is said to be even if f ðxÞ ¼ f ðxÞ i.e. the function value for a particular negative value of x is the same as that for the corresponding positive value of x. The graph of an even function is therefore reflection symmetrical about the y-axis. y y ¼ f ðxÞ ¼ x2 is an even function because a2 f ð2Þ ¼ 4 ¼ f ð2Þ f ð3Þ ¼ 9 ¼ f ð3Þ –a a etc. x y y ¼ f ðxÞ ¼ cos x is an even function because a –a cosðxÞ ¼ cos x f ðaÞ ¼ cos a ¼ f ðaÞ. x (b) Odd functions A function f ðxÞ is said to be odd if f ðxÞ ¼ f ðxÞ i.e. the function value for a particular negative value of x is numerically equal to that for the corresponding positive value of x but opposite in sign. If the graph of an odd function is rotated about the origin through 1808 it coincides with the original graph. We say it is symmetrical about the origin. y y =x3 x y ¼ f ðxÞ ¼ x3 is an odd function because f ð2Þ ¼ 8 ¼ f ð2Þ f ð5Þ ¼ 125 ¼ f ð5Þ y etc. y ¼ f ðxÞ ¼ sin x is an odd function because x sinðxÞ ¼ sin x f ðaÞ ¼ f ðaÞ. So, for an even function f ðxÞ ¼ f ðxÞ, symmetrical about the y-axis for an odd function f ðxÞ ¼ f ðxÞ, symmetrical about the origin. 273 Fourier series 2 Example 1 y x f ðxÞ shown by the waveform is therefore an . . . . . . . . . . . . function because it is . . . . . . . . . . . . odd; symmetrical about the origin, i.e. f ðxÞ ¼ f ðxÞ 9 Example 2 y x Hence the waveform of y ¼ f ðxÞ depicts an . . . . . . . . . . . . function, because it is ............ even; symmetrical about the y-axis, i.e. f ðxÞ ¼ f ðxÞ 10 Example 3 y x In this case, the waveform shows a function that is . . . . . . . . . . . . because . . . . . . . . . . . . neither even nor odd; not symmetrical about either the y-axis or the origin 11 274 Programme 8 Exercise State whether each of the following functions is odd, even, or neither. y 1 x y 2 x y 3 x y 4 x y 5 x y 6 – 12 – x 1 4 odd neither 2 5 odd even 3 6 even odd We shall shortly see that a knowledge of odd and even functions can save a lot of unnecessary calculation. First, however, let us consider products of odd and even functions in the next frame 275 Fourier series 2 13 Products of odd and even functions The rules closely resemble the elementary rules of signs. ðevenÞ ðevenÞ ¼ ðevenÞ like ðþÞ ðþÞ ¼ ðþÞ ðoddÞ ðoddÞ ¼ ðevenÞ ðÞ ðÞ ¼ ðþÞ ðoddÞ ðevenÞ ¼ ðoddÞ ðÞ ðþÞ ¼ ðÞ. The results can easily be proved. (a) Two even functions Let FðxÞ ¼ f ðxÞgðxÞ where f ðxÞ and gðxÞ are even functions. Then FðxÞ ¼ f ðxÞgðxÞ ¼ f ðxÞgðxÞ since f ðxÞ and gðxÞ are even ; FðxÞ ¼ FðxÞ i.e. FðxÞ is even (b) Two odd functions Let FðxÞ ¼ f ðxÞgðxÞ where f ðxÞ and gðxÞ are odd functions. Then FðxÞ ¼ f ðxÞgðxÞ ¼ ff ðxÞgfgðxÞg since f ðxÞ and gðxÞ are odd ¼ f ðxÞgðxÞ ¼ FðxÞ ; FðxÞ ¼ FðxÞ i.e. FðxÞ is even Finally (c) One odd and one even function Let FðxÞ ¼ f ðxÞgðxÞ where f ðxÞ is odd and gðxÞ even. Then FðxÞ ¼ f ðxÞgðxÞ ¼ f ðxÞgðxÞ ¼ FðxÞ ; FðxÞ ¼ FðxÞ i.e. FðxÞ is odd So if f ðxÞ and gðxÞ are both even, then f ðxÞgðxÞ is even and if f ðxÞ and gðxÞ are both odd, then f ðxÞgðxÞ is even but if either f ðxÞ or gðxÞ is even and the other odd, then f ðxÞgðxÞ is odd. Now for a short exercise, so move on Exercise 14 State whether each of the following products is odd, even, or neither. 1 x2 sin 2x 6 ð2x þ 3Þ sin 4x 2 x cos x 7 sin2 x cos 3x 3 cos 2x cos 3x 8 x3 ex 4 x sin nx 9 5 3 sin x cos 4x ðx4 þ 4Þ sin 2x 1 cosh x xþ2 3 10 Finish all ten and then check with the next frame 276 Programme 8 15 1 2 3 4 5 odd (E)(O) odd (O)(E) even (E)(E) even (O)(O) odd (O)(E) ¼ (O) ¼ (O) ¼ (E) ¼ (E) ¼ (O) 6 7 8 9 10 neither (N)(O) even (E)(E) neither (O)(N) odd (E)(O) neither (N)(E) ¼ (N) ¼ (E) ¼ (N) ¼ (O) ¼ (N) Two useful facts emerge from odd and even functions. Thinking in terms of areas under the graphs (a) y For an even function ða ða f ðxÞ dx ¼ 2 f ðxÞ dx –a a x a 0 (b) y –a a x For an odd function ða f ðxÞdx ¼ 0 a We can now look at two important theorems concerning odd and even functions. Theorem 1 If f ðxÞ is defined over the interval < x < and f ðxÞ is even, then the Fourier series for f ðxÞ contains cosine terms only. Included in this is a0 which may be regarded as an cos nx with n ¼ 0. ð ð0 f ðxÞ dx ¼ f ðxÞ dx Proof : Since f ðxÞ is even, 0 ð ð ð 1 2 2 (a) a0 ¼ f ðxÞ dx ¼ f ðxÞ dx ; a0 ¼ f ðxÞ dx 0 0 ð 1 (b) an ¼ f ðxÞ cos nx dx. But f ðxÞ cos nx is the product of two even functions and therefore itself even. ; an ¼ . . . . . . . . . . . . 277 Fourier series 2 2 an ¼ ð f ðxÞ cos nx dx 16 0 Because as the integrand is even, ð ð 1 2 an ¼ f ðxÞ cos nx dx ¼ f ðxÞ cos nx dx. 0 ð 1 f ðxÞ sin nx dx (c) bn ¼ Arguing along similar lines, this gives bn ¼ . . . . . . . . . . . . 17 bn ¼ 0 Because, since f ðxÞ sin nx is the product of an even function and an odd function, it is itself odd. ð 1 ; bn ¼ f ðxÞ sin nx dx ¼ 0: ; bn ¼ 0 Therefore, there are no sine terms in the Fourier series for f ðxÞ. Now for an example. Example The waveform shown is symmetrical about the y-axis. The function is therefore even and there will be no sine terms in the series. y x ; f ðxÞ ¼ 1 X 1 a0 þ an cos nx 2 n¼1 =2 ð ð ð 1 2 2 =2 2 4x f ðxÞ dx ¼ f ðxÞ dx ¼ 4 dx ¼ ¼4 0 0 0 ð ð 1 2 f ðxÞ cos nx dx ¼ f ðxÞ cos nx dx (b) an ¼ 0 (a) a0 ¼ ¼ ............ Finish the integration. 278 Programme 8 18 an ¼ 0 an ¼ ðn evenÞ; 8 n an ¼ 8 n ðn ¼ 1, 5, 9, . . .Þ; ðn ¼ 3, 7, 11, . . .Þ Because an ¼ 2 ð f ðxÞ cos nx dx ¼ 0 2 ð =2 4 cos nx dx 0 8 sin nx =2 8 n sin ¼ ¼ n n 2 0 But sin n ¼0 2 for n even ¼1 for n ¼ 1, 5, 9, . . . ¼ 1 for n ¼ 3, 7, 11, . . . Hence the result stated. (c) We know that bn ¼ 0, because f ðxÞ is an even function. Therefore, the required series is f ðxÞ ¼ . . . . . . . . . . . . 19 f ðxÞ ¼ 2 þ 8 1 1 1 cos x cos 3x þ cos 5x cos 7x þ . . . 3 5 7 If you care to look back to Example 1 in Frame 23 of Programme 7, you will see how much time and effort we have saved by not having to evaluate bn . A similar theorem applies to odd functions. Theorem 2 If f ðxÞ is an odd function defined over the interval < x < , then the Fourier series for f ðxÞ contains sine terms only. ð ð0 f ðxÞ dx ¼ f ðxÞ dx. Proof : Since f ðxÞ is an odd function, 0 ð 1 (a) a0 ¼ f ðxÞ dx. But f ðxÞ is odd ; a0 ¼ 0 ð 1 (b) an ¼ f ðxÞ cos nx dx Remembering that f ðxÞ is odd and cos nx is even, the product f ðxÞ cos nx is ............ 279 Fourier series 2 20 odd 1 ; an ¼ ; an ¼ 0 ð 1 f ðxÞ cos nx dx ¼ ðincluding a0 ¼ 0Þ ð (odd function) dx ¼ 0 Now for bn we have ð 1 f ðxÞ sin nx dx and because f ðxÞ and sin nx are each odd, (c) bn ¼ the product f ðxÞ sin nx is . . . . . . . . . . . . 21 even Then bn ¼ 1 2 ¼ ð f ðxÞ sin nx dx ¼ ð 2 ; bn ¼ 1 ð (even function) dx f ðxÞ sin nx dx 0 ð f ðxÞ sin nx dx 0 So, if f ðxÞ is odd, a0 ¼ 0; an ¼ 0; bn ¼ 2 ð f ðxÞ sin nx dx 0 i.e. the Fourier series contains sine terms only. Example Consider the function shown. y f ðxÞ ¼ 6 x f ðxÞ ¼ 6 f ðx þ 2Þ ¼ f ðxÞ. <x<0 0<x< Before we do any evaluation, we can see that this is . . . . . . . . . . . . and therefore ............ an odd function; sine terms only, i.e. a0 ¼ 0 and an ¼ 0 ð 1 f ðxÞ sin nx is a product of two odd bn ¼ f ðxÞ sin nx dx. functions and is therefore even. ð 2 ; bn ¼ f ðxÞ sin nx dx ¼ . . . . . . . . . . . . 0 22 280 Programme 8 23 bn ¼ 0 Because 2 bn ¼ ðn evenÞ or 24 n ðn oddÞ 12 cos nx 12 ð1 cos nÞ: 6 sin nx dx ¼ ¼ n n 0 0 ð Hence the result stated above. So the series is f ðxÞ ¼ . . . . . . . . . . . . 24 f ðxÞ ¼ 24 1 1 sin x þ sin 3x þ sin 5x þ . . . 3 5 Because cos n ¼ ð1Þn . Of course, if f ðxÞ is neither an odd nor an even function, then we must obtain expressions for a0 , an and bn in full. One more example Example Determine the Fourier series for the function shown. y y y = 2x y =2 x x This is neither odd nor even. Therefore we must find a0 , an and bn . f ðxÞ ¼ 1 X 1 a0 þ fan cos nx þ bn sin nxg 2 n¼1 ð ð 2 1 2 f ðxÞ dx ¼ 2dx xdx þ 0 0 ( ) 2 1 x2 1 ; a0 ¼ 3 ¼ f þ 4 2g ¼ 3 þ 2x ¼ 0 ð ð ð 2 1 2 1 2 x cos nx dx þ (b) an ¼ f ðxÞ cos nx dx ¼ 2 cos nx dx 0 0 ð ð 2 2 1 x sin nx 1 sin nx dx þ cos nx dx ¼ n 0 n 0 1 (a) a0 ¼ ð 2 ¼ ............ Finish it off 281 Fourier series 2 an ¼ 0 ðn evenÞ; an ¼ 4 2 n 2 ðn oddÞ 25 Because ( ) 2 1 1 cos nx sin nx 2 ð0 0Þ þ an ¼ n n n 0 2 1 2 ð1Þn þ 1 þ ð0 0Þ ¼ n 2 ¼ 2 2 1 ð1Þn n and so an ¼ 0 ðn evenÞ and an ¼ 4 2 n 2 ðn oddÞ (c) To find bn , we proceed in the same general manner bn ¼ . . . . . . . . . . . . Complete it on your own bn ¼ 2 n Here is the working. ð ð ð 2 1 2 1 2 x sin nx dx þ bn ¼ f ðxÞ sin nx dx ¼ 2 sin nx dx 0 0 ð ð 2 2 1 x cos nx 1 þ cos nx dx þ sin nx dx n 0 n 0 ( ) 2 1 1 sin nx cos nx 2 ð cos nÞ þ þ ¼ n n n n 0 ¼ ¼ 2 1 1 cos n þ ð0 0Þ ðcos 2n cos nÞ n n ¼ 2 1 2 cos 2n ¼ cos 2n n n But cos 2n ¼ 1: ; bn ¼ 2 n So the required series is f ðxÞ ¼ . . . . . . . . . . . . 26 282 Programme 8 3 4 1 1 cos 5x þ . . . f ðxÞ ¼ 2 cos x þ cos 3x þ 2 9 25 2 1 1 1 sin x þ sin 2x þ sin 3x þ sin 4x . . . 2 3 4 27 At this stage, let us take stock of our findings so far. If a function f ðxÞ is defined over the range to , or any other periodic interval of 2, then the Fourier series for f ðxÞ is of the form 1 X 1 a0 þ fan cos nx þ bn sin nxg 2 n¼1 ð 1 f ðxÞ dx a0 ¼ ð 1 an ¼ f ðxÞ cos nx dx ð 1 bn ¼ f ðxÞ sin nx dx f ðxÞ ¼ where We also know that (a) if f ðxÞ is an even function, the series will contain no sine terms (b) if f ðxÞ is an odd function, the series will contain only sine terms (c) if f ðxÞ is neither odd nor even, the series will, in general, contain a constant term, cosine terms and sine terms. 28 Half-range series Sometimes a function of period 2 is defined over the range 0 to , instead of the normal to , or 0 to 2. We then have a choice of how to proceed. For example, if we are told that between x ¼ 0 and x ¼ , f ðxÞ ¼ 2x, then, since the period is 2, we have no evidence of how the function behaves between x ¼ and x ¼ 0. y x (a) y x If the waveform were as shown in (a), the function would be an even function, symmetrical about the yaxis and the series would have only cosine terms (including possibly a0 ). 283 Fourier series 2 y (b) On the other hand, if the waveform were as shown in (b), the function would be odd, being symmetrical about the origin and the series would have only sine terms. x y (c) x Of course, if we choose something quite different for the waveform between x ¼ and x ¼ 0; then f ðxÞ will be neither odd nor even and the series will then contain ............ both sine and cosine terms (including a0 ) 29 In each case, we are making an assumption on how the function behaves between x ¼ and x ¼ 0, and the resulting Fourier series will therefore apply only to f ðxÞ between x ¼ 0 and x ¼ for which it is defined. For this reason, such series are called half-range series. Example 1 y A function f ðxÞ is defined by x f ðxÞ ¼ 2x f ðx þ 2Þ ¼ f ðxÞ. 0<x< Obtain a half-range cosine series to represent the function. To obtain a cosine series, i.e. a series with no sine terms, we need an . . . . . . . . . . . . function. even y Therefore, we assume the waveform between x ¼ and x ¼ 0 to be as shown, making the total graph symmetrical about the y-axis. y=2x x Now we can find expressions for the Fourier coefficients as usual. a0 ¼ . . . . . . . . . . . . 30 284 Programme 8 31 a0 ¼ 2 Because a0 ¼ 2 ð f ðxÞ dx ¼ 0 2 ð 2x dx ¼ 0 2 2 x ¼ 2 0 ; a0 ¼ 2 Then we need an which is . . . . . . . . . . . . 32 an ¼ 0 Because 2 an ¼ 4 ¼ ð ðn evenÞ 4 ð 2x cos nx dx ¼ 0 x sin nx n 0 1 n ¼ 8 n2 ðn oddÞ ð x cos nx dx 0 sin nx dx 0 4 1 cos nx 4 ð0 0Þ ¼ 2 ðcos n 1Þ ¼ n n n 0 cos n ¼ 1 ; an ¼ 0 ðn evenÞ ðn evenÞ and ¼ 1 ðn oddÞ 8 an ¼ 2 ðn odd) n All that now remains is bn which is . . . . . . . . . . . . 33 zero, since f ðxÞ is an even function, i.e. bn ¼ 0 So a0 ¼ 2; 34 an ¼ 0 8 ðn oddÞ, bn ¼ 0. n2 Therefore f ðxÞ ¼ . . . . . . . . . . . . ðn evenÞ or 8 1 1 f ðxÞ ¼ cos x þ cos 3x þ cos 5x þ . . . 9 25 Let us look at a further example, so move on to the next frame 285 Fourier series 2 Example 2 35 y Determine a half-range sine series to represent the function f ðxÞ defined by f ðxÞ ¼ 1 þ x f ðx þ 2Þ ¼ f ðxÞ. x y – x – 0<x< We choose the waveform between x ¼ and x ¼ 0 so that the graph is symmetrical about the origin. The function is then an odd function and the series will contain only sine terms. ; a0 ¼ 0 and an ¼ 0 bn can now easily be determined and the required series obtained. f ðxÞ ¼ . . . . . . . . . . . . f ðxÞ ¼ 4 1 1 þ 2 sin x þ sin 3x þ sin 5x þ . . . 3 5 1 1 1 sin 2x þ sin 4x þ sin 6x þ . . . 2 2 4 6 Check the working. ð ð 2 2 cos nx 1 ð1 þ xÞ bn ¼ ð1 þ xÞ sin nx dx ¼ þ cos nx dx 0 n 0 n 0 2 1þ 1 1 sin nx cos n þ þ ¼ n n n n 0 2 1 1þ 2 cos n ¼ f1 ð1 þ Þ cos ng ¼ n n n cos n ¼ 1 ðn evenÞ ¼ 1 ðn oddÞ 2 4 þ 2 ; bn ¼ ðn evenÞ ¼ ðn oddÞ n n 1 X bn sin nx we have Substituting in the general expression f ðxÞ ¼ n ¼ 1 4 þ 2 1 1 f ðxÞ ¼ sin x þ sin 3x þ sin 5x þ . . . 3 5 1 1 1 2 sin 2x þ sin 4x þ sin 6x þ . . . 2 4 6 So a knowledge of odd and even functions and of half-range series saves a deal of unnecessary work on occasions. Now let us consider the presence of odd or even harmonics, so move on 36 286 37 Programme 8 Series containing only odd harmonics or only even harmonics f ðxÞ ¼ 12 a0 þ a1 cos x þ a2 cos 2x þ a3 cos 3x þ . . . þ b1 sin x þ b2 sin 2x þ b3 sin 3x þ . . . If we replace x by ðx þ Þ, this becomes f ðx þ Þ ¼ 12 a0 þ 1 X fan cos nðx þ Þ þ bn sin nðx þ Þg n¼1 Now cosðnx þ nÞ ¼ cos nx cos n sin nx sin n: But for n ¼ 1, 2, 3, . . . sin n ¼ 0 ; cos nðx þ Þ ¼ cos nx cos n Also for n ¼ 1, 2, 3, . . . cos n ¼ 1 ; cos nðx þ Þ ¼ cos nx ðn evenÞ ðn evenÞ ¼ 1 ðn oddÞ. ¼ cos nx ðn oddÞ ð1Þ Similarly, sinðnx þ nÞ ¼ sin nx cos n þ cos nx sin n. Therefore, as before sin nðx þ Þ ¼ sin nx ; f ðx þ Þ ¼ 1 2 a0 ðn evenÞ ¼ sin nx ðn oddÞ ð2Þ a1 cos x þ a2 cos 2x a3 cos 3x þ . . . b1 sin x þ b2 sin 2x b3 sin 3x þ . . . But f ðxÞ ¼ 1 2 a0 þ a1 cos x þ a2 cos 2x þ a3 cos 3x þ . . . þ b1 sin x þ b2 sin 2x þ b3 sin 3x þ . . . If f ðxÞ ¼ f ðx þ Þ, these two series are equal and the odd harmonics that you see differ in sign must be zero. ; f ðxÞ ¼ f ðx þ Þ ¼ 12 a0 þ a2 cos 2x þ a4 cos 4x þ . . . þ b2 sin 2x þ b4 sin 4x þ . . . ; If f ðxÞ ¼ f ðx þ Þ, the Fourier series for f ðxÞ contains even harmonics only. Similarly, from the same two series above if f ðxÞ ¼ f ðx þ Þ, the Fourier series for f ðxÞ contains odd harmonics only. ; f ðxÞ ¼ a1 cos x þ a3 cos 3x þ . . . þ b1 sin x þ b3 sin 3x þ . . . Make a note of these two results: you will find them useful 38 Example 1 y Here f ðxÞ ¼ f ðx þ Þ Therefore, the series contains ............ x x x 287 Fourier series 2 39 even harmonics only Example 2 y Here we see that f ðxÞ ¼ f ðx þ Þ. x x Therefore, the series contains ............ x+ 40 odd harmonics only Now we can apply our knowledge to date to the following exercise. Exercise From each of the following waveforms, we can describe the nature of the terms in the relevant Fourier series. 1 2 y y y=sinx x x 3 4 y y x x 5 6 y a a x y – x 288 41 Programme 8 1 cosine terms ( þ a0 ) only; even harmonics only 2 3 4 sine terms only; odd harmonics only sine terms only; all harmonics cosine terms ( þ a0 ) only; odd harmonics only 5 6 cosine terms ( þ a0 ) only; all harmonics a0 ; sine and cosine terms; even harmonics only. On we go 42 Significance of the constant term 12 a0 We might, at this point, note that the effect of the constant term 12 a0 is to raise, or lower, the whole waveform on the y-axis. y y 1 a 2 0 x x In electrical applications to alternating currents, the constant term 12 a0 of the Fourier series indicates the d.c. component. For example, from Frames 58–61 we found that the odd square wave 6 < x < 0 f ðxÞ ¼ f ðx þ 2Þ ¼ f ðxÞ 6 0<x< f (x) 6 –π π 2π 3π x –6 has the Fourier series expansion 24 1 1 sin x þ sin 3x þ sin 5x þ . . . f ðxÞ ¼ 3 5 The function gðxÞ ¼ 2 þ f ðxÞ has the Fourier series expansion 24 1 1 gðxÞ ¼ 2 þ sin x þ sin 3x þ sin 5x þ . . . 3 5 g (x) a0 8 –π 4 π 2π 3π x –4 Here a0 =2 ¼ 2 – the amount by which the graph of the original function has been raised. 289 Fourier series 2 43 Half-range series with arbitrary period We now extend the work on half-range sine and cosine series to functions with arbitrary period. (a) Even function Half-range cosine series y y=f (t) f(t) y ¼ f ðtÞ 0<t< T 2 f ðt þ TÞ ¼ f ðtÞ –T 2 T 2 x symmetrical about the y-axis. With an even function, we know that bn ¼ 0 1 X 1 a0 þ an cos n!t 2 n¼1 ð 4 T=2 a0 ¼ f ðtÞ dt T 0 ð T=2 4 an ¼ f ðtÞ cos n!t dt T 0 ; f ðtÞ ¼ where and (b) Odd function Half-range sine series y y ¼ f ðtÞ f(t) –T 2 T 2 0<t< T 2 f ðt þ TÞ ¼ f ðtÞ t x symmetrical about the origin. ; a0 ¼ 0 and an ¼ 0 Then and f ðtÞ ¼ . . . . . . . . . . . . bn ¼ . . . . . . . . . . . . f ðtÞ ¼ 1 X bn sin n!t; bn ¼ n¼1 4 T ð T=2 44 f ðtÞ sin n!t dt 0 Now for an example or two. So move on 290 45 Programme 8 Example 1 A function f ðtÞ is defined by f ðtÞ ¼ 4 t; 0 < t < 4: y We have to form a half-range cosine series to represent the function in this interval. f(t) t x y First we form an even function, i.e. symmetrical about the y-axis. x t y t Now for a useful little trick. If we lower the waveform 2 units, i.e. to its ‘average’ position, balanced above and below the xaxis, then in this new position 12 a0 ¼ 0 and we have been saved one set of calculations. x y The function is now y ¼ f1 ðtÞ ¼ 2 t and, for the moment 12 a0 ¼ 0. Also, being an even function bn ¼ 0. All we need to do is to evaluate an . y=2 – t t So an ¼ 4 T ð T=2 x ð 4 4 ð2 tÞ cos n!t dt 8 0 ¼ ............ f1 ðtÞ cos n!t dt ¼ 0 46 an ¼ 0 ðn evenÞ ¼ 1 n2 !2 ðn oddÞ Simple integration by parts gives 1 2 sin 4n! 1 an ¼ 2 2 ðcos 4n! 1Þ 2 n! n ! 2 2 ¼ ¼ T 8 4 1 2 sin n 1 2 2 ðcos n 1Þ an ¼ 2 n! n ! But ! ¼ sin n ¼ 0; ; an ¼ 0 cos n ¼ 1 ðn evenÞ ; f1 ðtÞ ¼ . . . . . . . . . . . . and n ¼ 1, 2, 3, . . . ðn evenÞ; cos n ¼ 1 1 an ¼ 2 2 ðn oddÞ n ! ðn oddÞ 291 Fourier series 2 1 1 1 cos 5!t þ . . . f1 ðtÞ ¼ 2 cos !t þ cos 3!t þ ! 9 25 Now if we finally lift the waveform back to its original position by restoring the 2 units (i.e. 12 a0 ¼ 2), the original function is regained with f ðtÞ ¼ f1 ðtÞ þ 2. 1 1 1 cos 5!t þ . . . ; f ðtÞ ¼ 2 þ 2 cos !t þ cos 3!t þ ! 9 25 where ! ¼ : 4 Example 2 A function f ðtÞ is defined by f ðtÞ ¼ 3 þ t 0<t<2 f ðt þ 4Þ ¼ f ðtÞ. f ( t) Obtain the half-range sine series for the function in this range. t f (t) Sine series required. Therefore, we form an odd function, symmetrical about the origin a0 ¼ 0; an ¼ 0; T ¼ 4 t f ðtÞ ¼ 1 X bn sin n!t n¼1 ; bn ¼ 4 T ð2 0 f ðtÞ sin n!t dt ¼ ð2 ð3 þ tÞ sin n!t dt 0 This you can easily evaluate and then, putting n ¼ 1, 2, 3, . . . obtain the series f ðtÞ ¼ . . . . . . . . . . . . 47 292 Programme 8 2 1 4 1 4 sin !t sin 2!t þ sin 3!t sin 4!t . . . f ðtÞ ¼ ! 2 3 4 48 Because Straightforward integration by parts gives 1 1 bn ¼ ð3 5 cos 2n!Þ þ 2 2 ðsin 2n!Þ n! n ! 2 2 ; !¼ ¼ But T ¼ ! T 2 8 2 > > < ðn evenÞ 1 1 n! ð3 5 cos nÞ þ 2 2 sin n ¼ ; bn ¼ > 8 n! n ! > : ðn oddÞ n! Therefore 2 1 4 1 f ðtÞ ¼ 4 sin !t sin 2!t þ sin 3!t sin 4!t . . . ! 2 3 4 And that just about brings this particular Programme to an end. Fourier series have wide applications so it is very worthwhile paying considerable attention to them. The Revison summary and Can you? checklist now follow, after which you will have no trouble with the Test exercise. The Further problems provide additional practice. Revision summary 8 49 1 Functions with period T 1 X fan cos n!t þ bn sin n!tg f ðtÞ ¼ 12 a0 þ n¼1 a0 ¼ an ¼ 2 T 2 T 2 bn ¼ T ðT f ðtÞ dt 0 ðT f ðtÞ cos n!t dt ¼ 0 ðT 0 ! ! ! f ðtÞ sin n!t dt ¼ where ! ¼ 2 ¼ ð 2=! f ðtÞ dt 0 ð 2=! f ðtÞ cos n!t dt 0 ð 2=! f ðtÞ sin n!t dt. 0 2 2 so that T ¼ T ! Odd and even functions (a) Even function: f ðxÞ ¼ f ðxÞ; symmetrical about the y-axis. (b) Odd function: f ðxÞ ¼ f ðxÞ; symmetrical about the origin. 293 Fourier series 2 Product of odd and even functions ðevenÞ ðevenÞ ¼ ðevenÞ ðoddÞ ðoddÞ ¼ ðevenÞ ðoddÞ ðevenÞ ¼ ðoddÞ 3 Sine series and cosine series If f ðxÞ is even, the series contains cosine terms only (including a0 ) If f ðxÞ is odd, the series contains sine terms only. 4 Half-range series A function defined over the domain 0 x can be extended into either an odd function or an even function with period 2. 5 Odd and even harmonics If f ðx þ Þ ¼ f ðxÞ, the Fourier series for f ðxÞ contains even harmonics only If f ðx þ Þ ¼ f ðxÞ, the Fourier series for f ðxÞ contains odd harmonics only 6 Significance of the constant term The effect of the constant term a0 =2 is to raise or lower the waveform on the vertical axis. 7 Half-range series with arbitrary period T Even function 1 X 1 a0 þ fan cos n!t g 2 n¼1 ð 4 T f ðtÞ dt a0 ¼ T 0 ð 4 T an ¼ f ðtÞ cos n!t dt T 0 f ðtÞ ¼ where ! ¼ Odd function 1 X f ðtÞ ¼ fbb sin n!t g n¼1 bn ¼ 4 T ðT f ðtÞ sin n!t dt 0 2 2 that is T ¼ T ! Can you? 50 Checklist 8 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Obtain the Fourier coefficients of a function with arbitrary period T? Yes No . Recognise even and odd functions and their products? Yes No Frames 1 to 7 8 to 15 294 Programme 8 . Derive the Fourier series of even and odd functions? Yes No 15 to 27 . Derive half-range Fourier series? Yes 28 to 36 37 to 41 No . Recognise the conditions for the Fourier series to contain only odd or only even harmonics? Yes No . Explain the geometric significance of the constant term a0 =2? Yes No 42 . Derive half-range Fourier series with arbitrary period? Yes No 43 to 48 Test exercise 8 51 1 Given the function f ðtÞ ¼ t 2 0t<2 f ðt þ 2Þ ¼ f ðtÞ obtain the Fourier series and determine the value of the series when t ¼ 2. 2 3 State whether each of the following products is odd, even, or neither. (a) x3 cos 2x (d) x2 e2x (b) x2 sin 3x (e) ðx þ 5Þ cos 2x (c) sin 2x sin 3x (f) sin2 x cos x. A function f ðxÞ is defined by f ðxÞ ¼ x 0<x< f ðx þ 2Þ ¼ f ðxÞ. Express the function (a) as a half-range cosine series (b) as a half-range sine series. 4 Comment on the nature of the terms in the Fourier series for the following functions. (a) (b) y x y x 295 Fourier series 2 (c) (d) y y x x 5 A function f ðtÞ is defined by 0 2 < t < 0 f ðtÞ ¼ t 0<t<2 f ðt þ 4Þ ¼ f ðtÞ. Determine its Fourier series. Further problems 8 1 Determine the Fourier series representation of the function f ðtÞ defined by 3 2 < t < 0 f ðtÞ ¼ 5 0<t<2 f ðt þ 4Þ ¼ f ðtÞ. 2 Determine the half-range cosine series for the function f ðxÞ ¼ sin x defined in the range 0 < x < . 3 Determine the Fourier series to represent a half-wave rectifier output current, i amperes, defined by 8 T > < A sin !t 0<t< 2 i ¼ f ðtÞ ¼ > T :0 <t<T 2 f ðt þ TÞ ¼ f ðtÞ. 4 A function f ðxÞ is defined by 8 > a 0<x< > > 3 > > < 2 0 <x< f ðxÞ ¼ 3 3 > > > > 2 > : a <x< 3 f ðx þ Þ ¼ f ðxÞ. Obtain the Fourier series to represent the function. 5 If f ðxÞ is defined by f ðxÞ ¼ xð xÞ 0 < x < , express the function as (a) a half-range cosine series (b) a half-range sine series. 52 296 Programme 8 6 Determine the Fourier cosine series to represent the function f ðxÞ where 8 > < cos x 0<x< 2 f ðxÞ ¼ > :0 <x< 2 f ðx þ 2Þ ¼ f ðxÞ. 7 If f ðxÞ ¼ 8 > <0 0<x< 2 f ðx þ 2Þ ¼ f ðxÞ, <x< 2 obtain the Fourier cosine series for f ðxÞ in the range x ¼ 0 to x ¼ . > : cos x 8 A function f ðxÞ is defined over the interval 0 < x < by 8 <x 0<x< 2 f ðxÞ ¼ : x <x< 2 For the range x ¼ 0 to x ¼ , determine the Fourier sine series. 9 A function f ðtÞ is defined by 1 1 < t < 0 f ðtÞ ¼ 2t 0<t<1 f ðt þ 2Þ ¼ f ðtÞ. Obtain the Fourier series up to and including the third harmonic. 10 A function f ðtÞ is defined by f ðtÞ ¼ 1 t 2 f ðt þ 2Þ ¼ f ðtÞ. 1<t <1 Determine its Fourier series. 11 Determine the Fourier series 8 2 < t > < 1 f ðtÞ ¼ 0 1 < t > : 1 1<t for a periodic function such that < 1 <1 <2 f ðt þ 4Þ ¼ f ðtÞ. 12 Determine the Fourier series for the function f ðtÞ defined by ( 0 2 < t < 0 f ðtÞ ¼ 3t 0<t<4 4 f ðt þ 6Þ ¼ f ðtÞ. Programme 9 Frames 1 to 53 Introduction to the Fourier transform Learning outcomes When you have completed this Programme you will be able to: . Convert a trigonometric Fourier series into a doubly infinite sum of complex exponentials . Derive the complex Fourier series of a function that satisfies Dirichlet’s conditions . Recognise the function sinc ðtÞ . Separate a discrete complex spectrum into an amplitude spectrum and a phase spectrum . State Fourier’s integral theorem in terms of complex exponentials . Define and derive the Fourier transform of a function satisfying Dirichlet’s conditions . Separate a continuous complex spectrum into an amplitude spectrum and a phase spectrum . Recognise the functions a ðtÞ and a ðtÞ and derive their Fourier transforms along with those of the Dirac delta and the Heaviside unit step . Recognise alternative forms of the function–transform pair . Reproduce a collection of properties of the Fourier transform . Evaluate the convolution of two functions and describe its Fourier transform . Derive the Fourier sine and cosine transformations 297 298 Programme 9 Complex Fourier series 1 Introduction In the previous Programme we saw how a periodic function can be represented by an infinite sum of periodic, trigonometric harmonics. Each harmonic has a definite frequency which is an integer multiple of the fundamental frequency. A non-periodic function can be similarly represented, not as a sum but as an integral over a continuous range of frequencies. Before we do this, however, we shall convert the infinite Fourier series in terms of sines and cosines into a doubly infinite series involving complex exponentials. Complex exponentials Recall the exponential form of a complex number and its relationship to the polar form, namely z ¼ rðcos þ j sin Þ ¼ re j From this equation we can see that cos þ j sin ¼ e j and so cosðÞ þ j sinðÞ ¼ ej ¼ cos j sin Using these two equations we can find the complex exponential form of the trigonometric functions as cos ¼ . . . . . . . . . . . . 2 cos ¼ e j þ ej 2 and sin ¼ . . . . . . . . . . . . and sin ¼ e j ej 2j Because cos þ j sin ¼ e j and cos j sin ¼ ej so adding these two equations gives 2 cos ¼ e j þ ej that is cos ¼ e j þ ej 2 ð1Þ and subtracting the two equations gives 2j sin ¼ e j ej that is sin ¼ e j ej 2j ð2Þ These two equations permit us to develop an alternative representation of a Fourier series. 299 Introduction to the Fourier transform In the previous Programme we found that the Fourier series of the piecewise continuous function f ðtÞ with piecewise continuous derivative and where f ðt þ TÞ ¼ f ðtÞ is given as 1 a0 X þ ðan cos n!0 t þ bn sin n!0 t Þ 2 n¼1 ð 2 2 T=2 and where an ¼ where !0 ¼ f ðtÞ cos n!0 t dt T T T=2 ð 2 T=2 f ðtÞ sin n!o t dt and bn ¼ T T=2 f ðtÞ ¼ ð3Þ Now, if we substitute the right-hand sides of equations (1) and (2) into equation (3) we obtain 1 n o n o a0 X . . . . . . . . . . . . e jn!0 t þ . . . . . . . . . . . . ejn!0 t f ðtÞ ¼ þ 2 n¼1 1 a0 X an jbn jn!0 t an þ jbn jn!0 t þ e e f ðtÞ ¼ þ 2 n¼1 2 2 Because 1 a0 X þ ðan cos n!0 t þ bn sin n!0 tÞ 2 n¼1 1 a0 X e jn!0 t þ ejn!0 t e jn!0 t ejn!0 t þ bn an ¼ þ 2 n¼1 2 2j 1 a0 X an þ bn =j jn!0 t an bn =j jn!0 t e e ¼ þ þ 2 2 2 n¼1 1 a0 X an jbn jn!0 t an þ jbn jn!0 t e e ¼ þ þ 2 n¼1 2 2 f ðtÞ ¼ In the next frame we shall make some notational changes to simplify this expression 3 300 4 Programme 9 If we now define cn ¼ an jbn so that the complex conjugate of cn is 2 an þ jbn we can write this sum as 2 1 X f ðtÞ ¼ c0 þ cn e jn!0 t þ cn ejn!0 t cn ¼ Note that we have taken b0 ¼ 0. There is no problem about this. There is no term sin 0!0 t in the Fourier series and so b0 ¼ 0 n¼1 ¼ c0 þ 1 X cn e jn!0 t þ n¼1 ¼ c0 þ 1 X 1 X cn e jn!0 t þ ¼ ¼ n¼1 1 X 1 X For notational convenience we denote cn by cn . This means that an ¼ an and bn ¼ bn jn!0 t cn e n¼1 cn e jn!0 t þ n¼1 1 X cn ejn!0 t n¼1 n¼1 ¼ c0 þ 1 X 1 X As n ranges from 1 to 1 so n ranges from 1 to 1 cn e jn!0 t n¼1 cn e jn!0 t þ c0 þ 1 X Notice the reversed order of summation in the first sum cn e jn!0 t n¼1 cn e Combining all three terms into the doubly infinite sum jn!0 t n¼1 ð an jbn 2 T=2 ¼ where cn ¼ f ðtÞðcos n!0 t j sin n!0 t Þ dt. That is 2T T=2 2 ð T=2 1 cn ¼ f ðtÞejn!0 t dt. T T=2 In the next frame we shall look at some examples 5 Example 1 To find the complex Fourier series for the function 8 T=2 < t < a=2 <0 f ðtÞ ¼ 1 a=2 < t < a=2 where f ðt þ TÞ ¼ f ðtÞ : 0 a=2 < t < T=2 f (t) –T 2 we proceed as on the next page. –a 2 1 0 a 2 T 2 t 301 Introduction to the Fourier transform 1 X f ðtÞ ¼ cn ¼ ¼ cn e jn!0 t n¼1 ð T=2 1 T 1 T where !0 ¼ 2 and T f ðtÞejn!0 t T=2 ð a=2 ejn!0 t dt Because f ðtÞ ¼ 1 for a=2 < t < a=2 a=2 a=2 1 ejn!0 t T jn!0 a=2 jn!0 a=2 e e jn!0 a=2 ¼ j2n ¼ sin n!0 a=2 n sin na=T ¼ n a sin na=T ¼ T na=T ¼ Provided n 6¼ 0 Since !0 ¼ 2 T Recall that sin ¼ Since !0 ¼ e j ej 2j 2 T Provided n 6¼ 0 When n ¼ 0 ð ð 1 T=2 1 a=2 a c0 ¼ f ðtÞ dt ¼ dt ¼ T T=2 T a=2 T Therefore f ðtÞ ¼ 1 a X a sin na=T jn!0 t e þ T n¼1 T na=T n6¼0 In the next frame we shall look at the same function retarded by half the width of the peak Example 2 6 To find the complex Fourier series for the function 1 0<t<a f ðtÞ ¼ where f ðt þ TÞ ¼ f ðtÞ 0 a<t<T f (t) –T 0 a T We find that, for n 6¼ 0, cn ¼ . . . . . . . . . . . . t 302 Programme 9 7 jna=T cn ¼ e a sin na=T T na=T Because 1 X f ðtÞ ¼ cn ¼ ¼ cn e jn!0 t n¼1 ð T=2 1 T 1 T where ! ¼ 2 and T f ðtÞejn!0 t dt T=2 ða jn!0 t e dt 0 a 1 ejn!0 t T jn!0 0 jn!0 a e 1 ¼ j2n jn!0 a=2 e jn!0 a=2 jn!0 a=2 e ¼e j2n a sin na=T ¼ ejna=T T na=T ¼ Provided n 6¼ 0 Provided n 6¼ 0 To finish c0 ¼ . . . . . . . . . . . . 8 c0 ¼ a T Because c0 ¼ ¼ 1 T 1 T ð T=2 f ðtÞ dt T=2 ða dt ¼ 0 a T Therefore 1 a X jna=T a sin na=T f ðtÞ ¼ þ e e jn!0 t T n¼1 T na=T n6¼0 Next frame 303 Introduction to the Fourier transform sin na=T that occurs in both of na=T sin x these examples. This is an example of a commonly occurring expression x which has the special name sinc ðxÞ. Notice that sinc ð0Þ is not defined. sin x ¼ 1 we define sinc ð0Þ ¼ 1. However, because Lim sinc ðxÞ ¼ Lim x x!0 x!0 Before we move on, consider the expression 9 sinc (x) 1 –12 –9 –6 –3 0 3 6 9 12 x This means that c0 can be incorporated into the summations so the solutions to Examples 1 and 2 become f ðtÞ ¼ f ðtÞ ¼ 1 X n¼1 1 X ða=TÞ sinc ðna=TÞe jn!0 t ða=TÞejna=T sinc ðna=TÞe jn!0 t respectively. n¼1 Now let’s compare these two results Complex spectra The coefficients cn in Example 1 of Frames 5 and 9 are real numbers, namely a cn ¼ sincðna=TÞ T whereas in Example 2 or Frames 8 and 9 they are complex numbers, namely, a cn ¼ sincðna=TÞejnaT T In general, the cn are complex numbers and can be written as cn ¼ jcn je jn where, in Example 2 jcn j ¼ a a sin na=T a sincðna=TÞ ¼ for n 6¼ 0 and c0 ¼ and T T na=T T n ¼ na=T. These complex coefficients constitute a discrete complex spectrum where cn represents the spectral coefficient of the nth harmonic. Each spectral coefficient couples an amplitude spectrum value jcn j and a phase spectrum value n . The amplitude spectrum tells us the magnitude of each of the harmonic components and has, for both examples, the graph shown on the next page. 10 304 Programme 9 |cn| 0 πa T t The phase spectrum n ¼ na=T tells us the phase of each harmonic relative to the fundamental harmonic frequency !0 . ϕn 1 –3 –2 2 3 –1 0 n The phase spectrum of the first example is zero for all n and tells us that each harmonic is in phase with the fundamental harmonic. The phase spectrum of the second example, which is a retarded form of the first example, tells us that the nth harmonic is shifted out of phase from the fundamental harmonic by n!0 . Next frame The two domains 11 A periodic waveform and its spectrum are described in different terms. The waveform is described in terms of behaviour in time whereas the spectrum is described in terms of behaviour relative to frequency. Thus time and frequency form two domains of definition of our functions and whatever information can be gleaned from within one domain can equally be gleaned from within the other. For example, the power content of a periodic function f ðtÞ of period T is defined in the time domain as the mean square value of f ðtÞ ð 1 T=2 ðf ðtÞÞ2 dt T T=2 Within the frequency domain the power content is given as ............ 305 Introduction to the Fourier transform 1 X 12 jcn j2 n¼1 Because ð ð 1 T=2 1 T=2 2 ðf ðtÞÞ dt ¼ T T=2 T T=2 ¼ ¼ ¼ ¼ 1 X n¼1 1 X n¼1 1 X n¼1 1 X cn cn ! 1 X cn e jn!0 t f ðtÞ dt n¼1 1 T 1 T ð T=2 f ðtÞe jn!0 t dt T=2 ð T=2 f ðtÞejðnÞ!0 t dt T=2 cn cn ¼ 1 X cn cn n¼1 jcn j 2 n¼1 So the power content can be obtained from either domain. Next frame Continuous spectra Of interest in the analysis of periodic functions is the behaviour of the Fourier series as the period increases without limit. Consider Example 1 from Frame 5 8 T=2 < t < a=2 <0 f ðtÞ ¼ 1 a=2 < t < a=2 where f ðt þ TÞ ¼ f ðtÞ : 0 a=2 < t < T=2 f (t) –T 2 –a 2 1 0 a 2 T 2 t which has the Fourier series f ðtÞ ¼ 1 X n¼1 cn e jn!0 t na sin 2 a T and where cn ¼ where !0 ¼ na T T T As the period increases the separation between the pulses increases and in the limit as T ! 1 only a . . . . . . . . . . . . remains and the resulting function is no longer . . . . . . . . . . . . 13 306 Programme 9 14 only a single pulse remains and the resulting function is no longer periodic f(t) –T 2 f (t) 1 –a 2 a 2 –a 2 T 2 a 2 t In the Fourier series the distance between neighbouring harmonics in the 2 the complex spectra is the fundamental frequency !0 ¼ and, in the limit as T T ! 1, so !0 ! 0. This means that as the period increases the space between lines in the spectrum decreases so the spectrum lines come closer together and in the limit merge into a continuous spectrum. That is, for large T n!0 ¼ n! and as T ! 1 so n! ! ! where ! is the continuous frequency variable. To see the effect of this on the general form of the Fourier series we start with f ðtÞ ¼ 1 X cn e jn!0 t where !0 ¼ n¼1 1 and where cn ¼ T ð T=2 2 T f ðtÞejn!0 t dt T=2 Substituting the integral form of cn into the sum gives " ð # 1 X 1 T=2 jn!0 u f ðtÞ ¼ f ðuÞe du e jn!0 t T T=2 n¼1 where u is a dummy variable in place of the variable t. 2 and so T " ð # 1 X 1 T=2 jn!0 u f ðtÞ ¼ f ðuÞe du !0 e jn!0 t 2 T=2 n¼1 Now, !0 ¼ If T is large then !0 ¼ ! and " ð # 1 X 1 T=2 jn!u f ðtÞ ¼ f ðuÞe du e jn!t ! 2 T=2 n¼1 307 Introduction to the Fourier transform In the limit as T ! 1 so n! ! !, the sum becomes an integral and ! becomes the differential d! giving ð ð1 1 1 f ðtÞ ¼ f ðuÞej!u du e j!t d! !¼1 2 u¼1 ð ð 1 1 1 1 pffiffiffiffiffiffi f ðuÞej!u du e j!t d! ¼ pffiffiffiffiffiffi 2 !¼1 2 u¼1 ð ð 1 1 1 1 j!t p ffiffiffiffiffi ffi p ffiffiffiffiffi ffi ¼ Fð!Þe d! where Fð!Þ ¼ f ðuÞej!u du 2 !¼1 2 u¼1 These two integrals form the conclusion of Fourier’s integral theorem. Next frame Fourier’s integral theorem Given function f ðtÞ with derivative f 0 ðtÞ where 15 (a) f ðtÞ and f 0 ðtÞ are piecewise continuous in every finite interval ð1 (b) f ðtÞ is absolutely integrable in ð1, 1Þ, that is jf ðtÞj dt is finite 1 then 1 f ðtÞ ¼ pffiffiffiffiffiffi 2 ð1 1 Fð!Þe j!t d! where Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð1 f ðtÞej!t dt 1 The discrete harmonic values n!0 of the periodic function are now replaced by the continuous harmonic variable ! and the discrete spectra cn ¼ jcn je jn are replaced by the continuous spectra Fð!Þ ¼ jFð!Þje jð!Þ . Fð!Þ is referred to as the Fourier transform of f ðtÞ and can also be written as fðf ðtÞÞ. Deriving the Fourier transform of a function is then a matter of applying the second of these two integrals. The expressions f ðtÞ and Fð!Þ form a Fourier transform pair where f ðtÞ can be referred to as the inverse Fourier transform of Fð!Þ. That is, f ðtÞ ¼ f1 ½Fð!Þ. Next frame 308 16 Programme 9 Example 3 Find the Fourier transform of 8 t < a=2 <0 f ðtÞ ¼ 1 a=2 < t < a=2 : 0 a=2 < t 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 ð a=2 f (t) –a 2 ej!t dt a=2 t a 2 1 hej!t ia=2 ¼ pffiffiffiffiffiffi 2 j! a=2 j!a=2 1 e e j!a=2 ¼ pffiffiffiffiffiffi j! 2 rffiffiffi j!a=2 2 e e j!a=2 ¼ 2j! rffiffiffi 2 sin !a=2 ¼ ! a sin !a=2 ¼ pffiffiffiffiffiffi 2 !a=2 a ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 A plot of Fð!Þ produces the continuous amplitude spectrum of f ðtÞ F(ω) ω Notice the similarity between the plots of Fð!Þ and the discrete spectrum of Frame 10. The lines in the discrete spectrum have merged to form a continuous spectrum while retaining the envelope of the discrete spectrum. Now you try one 17 Example 4 The function of the previous example time delayed by t ¼ a=2 units is 1 0<t<a f ðtÞ ¼ 0 otherwise And has the Fourier transform Fð!Þ ¼ . . . . . . . . . . . . f (t) 1 0 a t 309 Introduction to the Fourier transform 18 aej!a=2 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 Because 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 ða ej!t dt 0 a 1 ej!t ¼ pffiffiffiffiffiffi 2 j! 0 j!a 1 e 1 ¼ pffiffiffiffiffiffi j! 2 j!a=2 2 e e j!a=2 ¼ pffiffiffiffiffiffi ej!a=2 2j! 2 2 j!a=2 sin !a=2 ¼ pffiffiffiffiffiffi e ! 2 a sin !a=2 ¼ pffiffiffiffiffiffi ej!a=2 !a=2 2 aej!a=2 ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 Here Fð!Þ is a complex function so we write Fð!Þ ¼ jFð!Þje jð!Þ where p ffiffiffiffiffiffi jFð!Þj ¼ a= 2 sinc ð!a=2Þj is the continuous amplitude spectrum and ð!Þ ¼ a!=2 is the continuous phase spectrum. |F(ω)| ϕ(ω) ω ω Again, notice the similarity between the plots of ð!Þ and the discrete phase spectrum of Frame 10. The lines in the discrete spectrum have merged to form a continuous spectrum while retaining the envelope of the discrete spectrum. Next frame 310 Programme 9 Some special functions and their transforms 19 Even functions If f ðtÞ is an even function then ð 1 1 Fð!Þe j!t d! f ðtÞ ¼ f ðtÞ and f ðtÞ ¼ pffiffiffiffiffiffi 2 1 where Fð!Þ ¼ . . . . . . . . . . . . ð1 f ðtÞ . . . . . . . . . . . . dt 0 20 Fð!Þ ¼ rffiffiffi ð 2 1 f ðtÞ cos !t dt 0 Because ð 1 1 f ðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð ð 1 0 1 1 ¼ pffiffiffiffiffiffi f ðtÞej!t dt þ pffiffiffiffiffiffi f ðtÞej!t dt 2 1 2 0 ð ð 1 1 1 1 f ðtÞej!t dt þ pffiffiffiffiffiffi f ðtÞej!t dt ¼ pffiffiffiffiffiffi 2 0 2 0 reversing the limits on the first integral ð ð 1 1 1 1 j!t f ðtÞe dðtÞ þ pffiffiffiffiffiffi f ðtÞej!t dt ¼ pffiffiffiffiffiffi 2 0 2 0 changing the variable of integration in the first integral from t to t ð 1 1 f ðtÞ e j!t þ ej!t dðtÞ ¼ pffiffiffiffiffiffi 2 0 rffiffiffi ð ð 2 1 2 1 ¼ pffiffiffiffiffiffi f ðtÞ cos !t dt ¼ f ðtÞ cos !t dt 0 2 0 Notice that if f ðtÞ is even then Fð!Þ is real. Odd functions If f ðtÞ is an odd function then 1 f ðtÞ ¼ f ðtÞ and f ðtÞ ¼ pffiffiffiffiffiffi 2 ð1 Fð!Þe j!t d! 1 where Fð!Þ ¼ . . . . . . . . . . . . ð1 0 f ðtÞ . . . . . . . . . . . . dt Introduction to the Fourier transform rffiffiffi ð 2 1 Fð!Þ ¼ j f ðtÞ sin !t dt 0 Because 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 ð1 f ðtÞej!t dt 1 ð 1 1 f ðtÞej!t dt þ pffiffiffiffiffiffi f ðtÞej!t dt 2 0 1 ð ð 1 1 1 1 ¼ pffiffiffiffiffiffi f ðtÞej!t dt þ pffiffiffiffiffiffi f ðtÞej!t dt 2 0 2 0 1 ¼ pffiffiffiffiffiffi 2 ð0 reversing the limits on the first integral ð ð 1 1 1 1 ¼ pffiffiffiffiffiffi f ðtÞe j!t dðtÞ þ pffiffiffiffiffiffi f ðtÞej!t dt 2 0 2 0 changing the variable of integration in the first integral from t to t ð 1 1 ¼ pffiffiffiffiffiffi f ðtÞ e j!t þ ej!t dt 2 0 ð 2j 1 f ðtÞ sin !t dt ¼ pffiffiffiffiffiffi 2 0 rffiffiffi ð 2 1 ¼ j f ðtÞ sin !t dt 0 Notice that if f ðtÞ is odd then Fð!Þ is imaginary. An example will show the converse of these two results. Example Given that ff ðtÞ ¼ Fð!Þ ¼ Að!Þ þ jBð!Þ where Að!Þ and Bð!Þ are real functions of !, then if (a) Að!Þ 6¼ 0 and Bð!Þ ¼ 0 then f ðtÞ is an . . . . . . . . . . . . function (b) Að!Þ ¼ 0 and Bð!Þ 6¼ 0 then f ðtÞ is an . . . . . . . . . . . . function 311 21 312 Programme 9 22 (a) Að!Þ 6¼ 0 and Bð!Þ ¼ 0 then f ðtÞ is an even function (b) Að!Þ ¼ 0 and Bð!Þ 6¼ 0 then f ðtÞ is an odd function Because The Fourier transform is given as ð 1 1 f ðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 f ðtÞ½cos !t j sin !t dt ¼ pffiffiffiffiffiffi 2 1 ð1 ð 1 1 1 ¼ pffiffiffiffiffiffi f ðtÞ cos !t dt j pffiffiffiffiffiffi f ðtÞ sin !t dt 2 1 2 1 ¼ Að!Þ þ jBð!Þ ð1 (a) If f ðtÞ sin !t dt ¼ 0 then f ðtÞ sin !t is odd. But sin !t is odd, so 1 f ðtÞ must be even. ð1 f ðtÞ cos !t dt ¼ 0 then f ðtÞ cos !t is odd. But cos !t is even, so (b) If 1 f ðtÞ must be odd. Top-hat function This function is a special form of the function met in Example 3 in Frame 16, and is defined by 8 t < a=2 < 0 f ðtÞ ¼ 1=a a=2 < t < a=2 : 0 a=2 < t Πa (t) –a 2 1 a a 2 t It is, because of its shape, referred to as the top-hat function and is denoted by the symbol a ðtÞ. It is a special form of the function in Example 3 because it has a unit area – width height ¼ a ð1=aÞ ¼ 1, or ð1 ð a=2 t a=2 a ðtÞ dt ¼ ð1=aÞ dt ¼ ¼1 a a=2 1 a=2 The Fourier transform of the top-hat function is Fð!Þ ¼ . . . . . . . . . . . . 313 Introduction to the Fourier transform 23 1 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 Because ð 1 1 a ðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 a=2 ð1=aÞej!t dt ¼ pffiffiffiffiffiffi 2 a=2 ð a=2 1 ej!t dt ¼ pffiffiffiffiffiffi a 2 a=2 1 ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 This function is useful in that it can be used to select any segment of any function, so acting as a filter. For example ðtÞ sin t selects the segment of sin t between =2 and reduces the rest to zero. πΠπ(t) sin (t) 1 –π 2 π 2 t –1 So ðt Þ cos t selects the segment of cos t between . . . . . . . . . . . . and . . . . . . . . . . . . 314 Programme 9 24 =2 and 3=2 Because 8 < 0 ðt Þ ¼ 1= : 0 t < =2 =2 < t < =2 =2 < t that is 8 < 0 ðt Þ ¼ 1= : 0 and so t < =2 =2 < t < 3=2 =2 < t ðt Þ cos t ¼ cos t 0 =2 < t < 3=2 otherwise selects the segment of cos t between =2 and 3=2. πΠπ(t – π)cos t π 2 3π 2 t The Dirac delta (refer to Programme 4, Frames 29ff ) In science and technology we often require to use the notion of a force that acts for a very brief interval of time. To simulate this mathematically we can use the unit-area pulse – the top-hat function. If we take the duration of this pulse to decrease while at the same time retaining a unit-area then in the limit we are led to the notion of the Dirac delta. That is ð1 fa ðtÞg dt ¼ Lim 1 ¼ 1 Lim a!0 1 a!0 Here as a ! 0 the width of the top-hat decreases as the height increases but all the while retaining the area beneath the top-hat as unity. It is this limit that we can use to justify the integral definition of the Dirac delta because ð1 ð1 ð1 ðtÞ dt ¼ 1 fa ðtÞg dt ¼ Lim fa ðtÞg dt ¼ Lim a!0 1 1 a!0 1 and it is also in this sense that we accept the validity of the integral ð1 f ðtÞðt t0 Þ dt ¼ f ðt0 Þ 1 because, like the top-hat function, it selects only that part of f ðtÞ over which it is non-zero, namely at t ¼ t0 . So if f ðtÞ ¼ ðtÞ then Fð!Þ ¼ . . . . . . . . . . . . 315 Introduction to the Fourier transform 25 1 pffiffiffiffiffiffi 2 Because 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 ej!0 ¼ pffiffiffiffiffiffi 2 1 ¼ pffiffiffiffiffiffi 2 ð1 ðtÞej!t dt 1 because ðtÞ ¼ ðt 0Þ Try another. f (t) 1 O t The truncated exponential function at e t>0 f ðtÞ ¼ 0 t<0 where a > 0 can be also expressed in the form f ðtÞ ¼ eat uðtÞ and has the Fourier transform Fð!Þ ¼ . . . . . . . . . . . . 1 Fð!Þ ¼ pffiffiffiffiffiffi 2ða þ j!Þ Because ð 1 1 at p ffiffiffiffiffi ffi e uðtÞej!t dt Fð!Þ ¼ 2 1 ð 1 1 ðaþj!Þt ¼ pffiffiffiffiffiffi e dt 2 0 1 ¼ pffiffiffiffiffiffi 2ða þ j!Þ 26 316 Programme 9 The triangle function 8 < ða þ tÞ=a2 a ðtÞ ¼ ða tÞ=a2 : 0 Λa (t) a < t < 0 0<t<a jtj > a Notice that this also has unit area 1 a The Fourier transform of 1 ðtÞ –a a t 27 is Fð!Þ ¼ . . . . . . . . . . . . 1 Fð!Þ ¼ pffiffiffiffiffiffi sinc 2 ð!=2Þ 2 Because ð 1 1 Fð!Þ ¼ pffiffiffiffiffiffi 1 ðtÞej!t dt 2 1 ð ð 1 0 1 1 ¼ pffiffiffiffiffiffi ð1 þ tÞej!t dt þ pffiffiffiffiffiffi ð1 tÞej!t dt 2 1 2 0 ð ð1 1 0 1 ¼ pffiffiffiffiffiffi ð1 tÞe j!t dt þ pffiffiffiffiffiffi ð1 tÞej!t dt 2 1 2 0 changing the variable of integration in the first integral from t to t ð 2 1 and integration by parts yields ¼ pffiffiffiffiffiffi ð1 tÞ cos !t dt 2 0 ! 2 1 sin2 ð!=2Þ ¼ pffiffiffiffiffiffi 2 2 ð!=2Þ2 1 ¼ pffiffiffiffiffiffi sinc 2 ð!=2Þ 2 Alternative forms 28 It should be noted that there are a number of alternative forms for the Fourier transform – each dealing with a different location for the constant 2. Other forms are ð ð1 1 1 Fð!Þe j!t d! where Fð!Þ ¼ f ðtÞej!t dt f ðtÞ ¼ 2 1 1 or f ðtÞ ¼ 1 2 ð1 1 Fð!Þe j!t d! where Fð!Þ ¼ ð1 1 f ðtÞej!t dt 317 Introduction to the Fourier transform or, by absorbing the 2 in the exponential by defining ! ¼ 2 ð1 ð1 f ðtÞ ¼ FðÞe j2t d where FðÞ ¼ f ðtÞej2t dt 1 1 We shall remain with our original form because it has the simplest exponential factor and we do not need to remember which integral has the constant in front of it and which does not. Next frame Properties of the Fourier transform We now list a number of properties of the Fourier transform that are useful in their manipulation. 29 Linearity If the Fourier transforms fðf1 ðtÞÞ ¼ F1 ð!Þ and fðf2 ðtÞÞ ¼ F2 ð!Þ then fð1 f1 ðtÞ þ 2 f2 ðtÞÞ ¼ 1 fðf1 ðtÞÞ þ 2 fðf2 ðtÞÞ ¼ 1 F1 ð!Þ þ 2 F2 ð!Þ where 1 and 2 are constants. Example The Fourier transform of f ðtÞ ¼ 22 ðtÞ 62 ðtÞ is Fð!Þ ¼ . . . . . . . . . . . . rffiffiffi 2 sinc ð!Þð1 3sinc ð!ÞÞ Because 1 If f ðtÞ ¼ 22 ðtÞ then Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!Þ and if f ðtÞ ¼ 2 ðtÞ then 2 1 2 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!Þ. Since f ðtÞ ¼ 22 ðtÞ 62 ðtÞ then 2 2 6 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!Þ pffiffiffiffiffiffi sinc2 ð!Þ 2 2 rffiffiffi 2 sinc ð!Þð1 3sinc ð!ÞÞ ¼ 30 318 Programme 9 Time shifting If fðf ðtÞÞ ¼ Fð!Þ then fðf ðt t0 ÞÞ ¼ e j!t0 Fð!Þ Example 1 The Fourier transform of 2 ðtÞ is pffiffiffiffiffiffi sinc ð!Þ so, by the time shifting property, 2 the Fourier transform of 2 ðt 5Þ is . . . . . . . . . . . . 31 and of 2 ðt þ 3Þ is . . . . . . . . . . . . e j5! pffiffiffiffiffiffi sinc ð!Þ and 2 ej3! pffiffiffiffiffiffi sinc ð!Þ 2 Frequency shifting If fðf ðtÞÞ ¼ Fð!Þ then f f ðtÞe j!0 t ¼ Fð! !0 Þ Example If the Fourier transform of f ðtÞ is Fð!Þ then the transform of f ðtÞ cos 4t is ............ 32 1 ðFð! þ 4Þ þ Fð! 4ÞÞ 2 Because e j4t þ ej4t 2 1 1 j4t ¼ f ðtÞe þ f ðtÞej4t 2 2 1 ¼ f ðtÞe j4t þ f ðtÞej4t 2 f ðtÞ cos 4t ¼ f ðtÞ 1 ðFð! 4Þ þ Fð! þ 4ÞÞ by the linearity 2 and the frequency shifting properties. and so the Fourier transform is Time scaling If fðf ðtÞÞ ¼ Fð!Þ then 1 ! fðf ðktÞÞ ¼ F jkj k So, for example, given f ðtÞ ¼ a ðtÞ with Fourier transform Fð!Þ, if f ðtÞ is shrunk to half its width then Fð!Þ is stretched to twice its width but shrunk to half its height. 319 Introduction to the Fourier transform Example If Fð!Þ is the Fourier transform of f ðtÞ then the Fourier transform of f ðtÞ is ............ Fð!Þ Because 1 jkj Fð!=kÞ ¼ pffiffiffiffiffiffi 2 1 ð1 33 f ðktÞej!t dt and when k ¼ 1 then 1 1 j 1j1 Fð!=½1Þ ¼ pffiffiffiffiffiffi 2 ð1 f ðtÞej!t dt ¼ Fð!Þ 1 Symmetry If fðf ðtÞÞ ¼ Fð!Þ then fðFðtÞÞ ¼ f ð!Þ Example 1 The Fourier transform of f ðtÞ ¼ 2 ðtÞ is Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!Þ, so the Fourier 2 transform of 1 FðtÞ ¼ pffiffiffiffiffiffi sinc ðtÞ is . . . . . . . . . . . . 2 f ð!Þ ¼ 2 ð!Þ 34 Because 1 The Fourier transform of FðtÞ ¼ pffiffiffiffiffiffi sinc ðtÞ 2 is f ð!Þ ¼ 2 ð!Þ ¼ 2 ð!Þ Try one yourself. Example The Fourier transform of the unit constant function f ðtÞ ¼ 1 is f½1 ¼ . . . . . . . . . . . . pffiffiffiffiffiffi 2ð!Þ Because pffiffiffiffiffiffi 1 1 f½ðtÞ ¼ pffiffiffiffiffiffi so f pffiffiffiffiffiffi ¼ ð!Þ, therefore f½1 ¼ 2ð!Þ 2 2 35 320 Programme 9 Differentiation If f ðtÞ ! 0 as t ! 1 and if fðf ðtÞÞ ¼ Fð!Þ then fðf 0 ðtÞÞ ¼ . . . . . . . . . . . . 36 j!Fð!Þ Because ð 1 1 0 f½f 0 ðtÞ ¼ pffiffiffiffiffiffi f ðtÞej!t dt 2 1 ð 1 1 j! 1 ¼ pffiffiffiffiffiffi f ðtÞej!t 1 þ pffiffiffiffiffiffi f ðtÞej!t dt 2 2 1 ¼ 0 þ j!Fð!Þ In general, if f ðtÞ ! 0 as t ! 1 and if fðf ðtÞÞ ¼ Fð!Þ then If fðf ðtÞÞ ¼ Fð!Þ then f f ðnÞ ðtÞ ¼ ðj!Þn Fð!Þ where the superscript ðnÞ indicates the nth derivative. Example The differential equation for unforced and undamped harmonic motion is of the form mf 00 ðtÞ þ kf ðtÞ ¼ 0. If we take the Fourier transform of this equation we immediately find that the permitted frequencies of oscillation are ! ¼ ............ rffiffiffiffiffi k m 37 !¼ Because If Fð!Þ is the Fourier transform of f ðtÞ then taking the Fourier transform of both sides of the equation mf 00 ðtÞ þ kf ðtÞ ¼ 0 gives by the differentiation property mðj!Þ2 Fð!Þ þ kFð!Þ ¼ ðm!2 þ kÞFð!Þ ¼ 0 so if Fð!Þ 6¼ 0 then m!2 ¼ k and so the permitted frequencies are rffiffiffiffiffi k . !¼ m Next frame 321 Introduction to the Fourier transform The Heaviside unit step function The Heaviside unit step function is defined as uðtÞ where 0 t<0 uðtÞ ¼ 1 t>0 38 If we follow the definition of the Fourier transform we find that ð 1 1 uðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 So that Fð!Þ ¼ . . . . . . . . . . . . 1 Fð!Þ ¼ pffiffiffiffiffiffi 1 Lim ej!t 2j! t!1 39 Because ð 1 1 uðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 j!t ¼ pffiffiffiffiffiffi e dt 2 0 1 ¼ pffiffiffiffiffiffi 1 Lim ej!t 2j! t!1 Because ej!t ¼ cos !t j sin !t we cannot say what happens to the exponential as t ! 1. So how do we resolve the problem? Next frame Let fuðtÞ ¼ Fð!Þ and so, by the scaling property,fuðtÞ ¼ Fð!Þ. Now, uðtÞ þ uðtÞ ¼ 1, therefore f½uðtÞ þ fu½ðtÞ ¼ f½1. That is, from Frame 35 pffiffiffiffiffiffi Fð!Þ þ Fð!Þ ¼ 2ð!Þ We now assume that Fð!Þ consists of a combination of the Dirac delta and an arbitrary function Gð!Þ Fð!Þ ¼ ð!Þ þ Gð!Þ so that Fð!Þ þ Fð!Þ ¼ ð!Þ þ Gð!Þ þ ð!Þ þ Gð!Þ ¼ 2ð!Þ þ Gð!Þ þ Gð!Þ since ð!Þ ¼ ð!Þ pffiffiffiffiffiffi ¼ 2ð!Þ rffiffiffi Therefore ¼ and Gð!Þ þ Gð!Þ ¼ 0. That is, Gð!Þ ¼ Gð!Þ. 2 rffiffiffi ð!Þ þ Gð!Þ. Consequently f½uðtÞ ¼ Fð!Þ ¼ 2 40 322 Programme 9 rffiffiffi ð!Þ þ Gð!Þ and since u0 ðtÞ ¼ ðtÞ then 2 rffiffiffi 1 1 f½u0 ðtÞ ¼ f½ðtÞ ¼ pffiffiffiffiffiffi giving j! ð!Þ þ Gð!Þ ¼ pffiffiffiffiffiffi 2 2 2 1 1 Since !ð!Þ ¼ 0, then j!Gð!Þ ¼ pffiffiffiffiffiffi and so Gð!Þ ¼ pffiffiffiffiffiffi thereby giving 2 j! 2 1 1 f½uðtÞ ¼ pffiffiffiffiffiffi ð!Þ þ j! 2 Now, f½u0 ðtÞ ¼ j!Fð!Þ ¼ j! The next property deals with the Fourier transform of a product of functions but before we go any further we need to recap what is meant by the convolution of two functions. Next frame Convolution 41 You will recall that from Programme 3, Frame 43 onwards we defined the convolution of two functions f ðtÞ and gðtÞ as ð1 f ðxÞgðt xÞ dx ¼ hðtÞ f ðtÞ gðtÞ ¼ 1 where denotes the operation of convolution. As a refresher consider the convolution f ðtÞ gðtÞ where ( sec2 t jtj < =4 f ðtÞ ¼ uðtÞ and gðtÞ ¼ 0 otherwise where uðtÞ is the Heaviside function then hðtÞ ¼ f ðtÞ gðtÞ ¼ . . . . . . . . . . . . 42 1 þ tan2 t 1 þ tan t Because ð1 ð1 f ðxÞgðt xÞ dx ¼ uðxÞgðt xÞ dx 1 1 ¼ ð =4 sec2 ðt xÞ dx 0 because uðtÞ ¼ 0 for t < 0 and gðtÞ ¼ 0 for t > =4 =4 ¼ tanðt xÞ 0 ¼ f tanðt =4Þ þ tan t g ¼ tan t 1 1 þ tan2 t þ tan t ¼ 1 þ tan t 1 þ tan t Next frame Introduction to the Fourier transform 323 The convolution theorem If Fð!Þ and Gð!Þ are the Fourier transforms of f ðtÞ and gðtÞ respectively then 43 (a) The Fourier transform of the convolution of f ðtÞ and gðtÞ is equal to the product of the individual Fourier transforms. That is pffiffiffiffiffiffi f½f ðtÞ gðtÞ ¼ 2Fð!ÞGð!Þ and so 1 f1 ½Fð!ÞGð!Þ ¼ pffiffiffiffiffiffi ½f ðtÞ gðtÞ 2 (b) The Fourier transform of the product f ðtÞgðtÞ is equal to the convolution of the individual Fourier transforms. That is 1 f½f ðtÞgðtÞ ¼ pffiffiffiffiffiffi Fð!Þ Gð!Þ and so 2 pffiffiffiffiffiffi f1 ½Fð!Þ Gð!Þ ¼ 2f ðtÞgðtÞ These provide useful methods of finding inverse transforms. Example To find the inverse transform of Fð!Þ ¼ 1 2ða þ j!Þ2 1 1 ¼ pffiffiffiffiffiffi pffiffiffiffiffiffi where a > 0 2ða þ j!Þ 2ða þ j!Þ 1 then from Frame 26 we note that if F1 ð!Þ ¼ pffiffiffiffiffiffi 2ða þ j!Þ f1 ðtÞ ¼ f1 ½F1 ð!Þ ¼ . . . . . . . . . . . . f1 ðtÞ ¼ eat uðtÞ Now, because Fð!Þ ¼ F1 ð!ÞF1 ð!Þ then 1 f ðtÞ ¼ f1 ½Fð!Þ ¼ f1 ½F1 ð!ÞF1 ð!Þ ¼ pffiffiffiffiffiffi ½f1 ðtÞ f1 ðtÞ 2 ð 1 1 ¼ pffiffiffiffiffiffi f1 ðxÞf1 ðt xÞ dx 2 1 ð 1 1 ax e uðxÞeaðtxÞ uðt xÞ dx ¼ pffiffiffiffiffiffi 2 1 ð eat 1 ax e uðxÞeax uðt xÞ dx ¼ pffiffiffiffiffiffi 2 1 ð eat 1 uðxÞuðt xÞ dx ¼ pffiffiffiffiffiffi 2 1 44 324 Programme 9 Now, uðxÞuðt xÞ ¼ 0 when x < 0 or when t x < 0, that is when x > t. 1 if 0 < x < t Therefore, uðxÞuðt xÞ ¼ so 0 otherwise ð eat t f ðtÞ ¼ pffiffiffiffiffiffi dx 2 0 8 at > < te pffiffiffiffiffiffi if t > 0 teat that is, f ðtÞ ¼ pffiffiffiffiffiffi uðtÞ ¼ 2 > 2 : 0 if t < 0 Now you try one. The inverse Fourier transform of Fð!Þ ¼ 5 is 6 þ 5j! !2 f ðtÞ ¼ . . . . . . . . . . . . 45 f ðtÞ ¼ pffiffiffiffiffiffiffiffiffi 2t 50 e e3t uðtÞ Because Fð!Þ ¼ 5 6 þ 5j! !2 5 ð2 þ j!Þð3 þ j!Þ 1 Let F1 ð!Þ ¼ pffiffiffiffiffiffi so that f1 ðtÞ ¼ e2t uðtÞ and 2ð2 þ j!Þ ¼ 1 F2 ð!Þ ¼ pffiffiffiffiffiffi so that f2 ðtÞ ¼ e3t uðtÞ so that 2ð3 þ j!Þ Fð!Þ ¼ 10½F1 ð!ÞF2 ð!Þ By the convolution theorem 10 f ðtÞ ¼ pffiffiffiffiffiffi ½f1 ðtÞ f2 ðtÞ 2 pffiffiffiffiffiffiffiffiffi ð 1 f1 ðxÞf2 ðt xÞ dx ¼ 50 1 ð 1 pffiffiffiffiffiffiffiffiffi ¼ 50 e2x uðxÞe3ðtxÞ uðt xÞ dx 1 ð1 pffiffiffiffiffiffiffiffiffi ¼ 50e3t ex uðxÞuðt xÞ dx 1 ðt pffiffiffiffiffiffiffiffiffi 1 if 0 < x < t ¼ 50e3t ex dx since uðxÞuðt xÞ ¼ 0 otherwise 0 t ðt i pffiffiffiffiffiffiffiffiffi 3t h t 1 if t > 0 e ex dx ¼ ¼ 50e e 1 uðtÞ since 0 if t < 0 0 i pffiffiffiffiffiffiffiffiffih 2t 3t uðtÞ ¼ 50 e e Move to the next frame 325 Introduction to the Fourier transform Fourier cosine and sine transforms Given that 46 ð 1 1 Fð!Þe j!t d! where f ðtÞ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 f ðtÞej!t dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 f ðtÞðcos !t þ j sin !tÞ dt ¼ pffiffiffiffiffiffi 2 1 if f ðtÞ is an even function so that f ðtÞ ¼ f ðtÞ then ð 1 1 f ðtÞðcos !t þ j sin !tÞ dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð1 1 f ðtÞ cos !t dt since f ðtÞ sin !t is odd ¼ pffiffiffiffiffiffi 2 1 rffiffiffi ð 2 1 f ðtÞ cos !t dt ¼ 0 This is referred to as the Fourier cosine transformation and is denoted by Fc ð!Þ. That is rffiffiffi ð 2 1 Fc ð!Þ ¼ f ðtÞ cos !t dt 0 Similarly if f ðtÞ is an odd function so that f ðtÞ ¼ f ðtÞ then ð 1 1 f ðtÞðcos !t þ j sin !tÞ dt Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð1 j ¼ pffiffiffiffiffiffi f ðtÞ sin !t dt since f ðtÞ cos !t is odd 2 1 rffiffiffi ð 2 1 f ðtÞ sin !t dt ¼j 0 This gives rise to the Fourier sine transformation, denoted by Fs ð!Þ where rffiffiffi ð 2 1 Fs ð!Þ ¼ f ðtÞ sin !t dt 0 Example 1 The Fourier cosine transformation of f ðtÞ ¼ 1 if 0 < t < a is 0 if t a Fc ð!Þ ¼ . . . . . . . . . . . . 326 Programme 9 rffiffiffi 2 a sinc ð!aÞ Fc ð!Þ ¼ 47 Because rffiffiffi ð 2 1 Fc ð!Þ ¼ f ðtÞ cos !t dt rffiffiffi ð0 2 a ¼ cos !t dt rffiffiffi 0 2 sin !t a ¼ ! 0 rffiffiffi rffiffiffi 2 sin !a 2 ¼ asinc ð!aÞ ¼ ! Example 2 The Fourier sine transformation of f ðtÞ ¼ 1 if 0 < t < a is 0 if t a Fs ð!Þ ¼ . . . . . . . . . . . . rffiffiffi 2 2 2a ! sinc2 ð!aÞ Fs ð!Þ ¼ 48 Because rffiffiffi ð 2 1 Fs ð!Þ ¼ f ðtÞ sin !t dt rffiffiffi ð0 2 a ¼ sin !t dt rffiffiffi 0 2 cos !t a ¼ ! 0 rffiffiffi rffiffiffi rffiffiffi 2 1 cos !a 2 2 sin2 !a 2 2 ¼ ¼ 2a ! sinc 2 ð!aÞ ¼ ! ! The Fourier cosine and sine transforms are useful when f ðtÞ is only defined for t 0 and where an extension can be added to f ðtÞ for t < 0 that makes the extended f ðtÞ into an even or odd function respectively. 327 Introduction to the Fourier transform Table of transforms f ðtÞ ¼ 1 0 1 0 f ðtÞ ¼ a ðtÞ ¼ if a=2 < t < a=2 otherwise a Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 if 0 < t < a otherwise aej!a=2 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 1 Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ 2 1 1 Fð!Þ ¼ pffiffiffiffiffiffi ð!Þ þ j! 2 1 1 Fð!Þ ¼ pffiffiffiffiffiffi ð! þ aÞ þ j! 2 1=a if a=2 < t < a=2 0 otherwise f ðtÞ ¼ uðtÞ f ðtÞ ¼ eat uðtÞ f ðtÞ ¼ teat uðtÞ 1 Fð!Þ ¼ pffiffiffiffiffiffi 2ða þ j!Þ2 f ðtÞ ¼ ðtÞ 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 49 The main points of the Programme are listed in the Revision summary that follows. Read it in conjunction with the Can you? checklist and refer back to the relevant parts of the Programme, if necessary. You will then have no trouble with the Test exercise and the Further problems provide valuable additional practice. Revision summary 9 1 Complex Fourier series The Fourier series of the piecewise continuous function f ðtÞ with piecewise continuous derivative and where f ðt þ TÞ ¼ f ðtÞ is given as f ðtÞ ¼ 1 X cn e jn!0 t n¼1 where cn ¼ 2 1 T ð T=2 f ðtÞejn!0 t dt. T=2 Discrete complex spectra The cn are complex numbers and can be written as cn ¼ jcn je jn These complex coefficients constitute a discrete complex spectrum where cn represents the spectral coefficient of the nth harmonic. Each spectral coefficient couples an amplitude spectrum value jcn j and a phase spectrum value n . 50 328 Programme 9 3 Fourier’s integral theorem If (a) f ðtÞ and f 0 ðtÞ are piecewise continuous in every finite interval ð1 jf ðtÞj dt is (b) f ðtÞ is absolutely integrable in ð1, 1Þ, that is 1 finite then 1 f ðtÞ ¼ pffiffiffiffiffiffi 2 4 ð1 1 Fð!Þe j!t d! where Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð1 f ðtÞej!t dt. 1 Continuous complex spectra The Fourier transform Fð!Þ is a complex function so we write pffiffiffiffiffiffi Fð!Þ ¼ jFð!Þje jð!Þ where jFð!Þj ¼ a= 2 sinc ð!a=2Þ is the continuous amplitude spectrum and ð!Þ ¼ a!=2 is the continuous phase spectrum. 5 Transforms of special functions Top-hat function 1=a a=2 < t < a=2 a ðtÞ ¼ 0 otherwise 1 with Fourier transform Fð!Þ ¼ pffiffiffiffiffiffi sinc ð!a=2Þ. 2 The Dirac delta 1 If f ðtÞ ¼ ðtÞ then Fð!Þ ¼ pffiffiffiffiffiffi . 2 The Heaviside unit step function 0 t<0 uðtÞ ¼ has the Fourier transform 1 t>0 1 1 . Fð!Þ ¼ pffiffiffiffiffiffi ð!Þ þ j! 2 The triangle 0 ðtÞ ¼ 1 1 Fð!Þ ¼ pffiffiffiffiffiffi 2 6 function jtj > 1 jtj < 1 has the Fourier transform sinc 2 ð!=2Þ. Alternative forms There are a number of alternative forms for the Fourier transform – each dealing with a different location for the constant 2. Other forms are ð ð1 1 1 Fð!Þe j!t d! where Fð!Þ ¼ f ðtÞej!t dt or f ðtÞ ¼ 2 1 1 ð ð1 1 1 f ðtÞ ¼ Fð!Þe j!t d! where Fð!Þ ¼ f ðtÞej!t dt or 2 1 ð1 ð 11 j2!t f ðtÞ ¼ Fð!Þe d! where Fð!Þ ¼ f ðtÞej2!t dt. 1 1 Introduction to the Fourier transform 7 Properties of the Fourier transform Time shifting ð 1 1 f ðtÞej!t dt then If Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 f ðt t0 Þej!t dt. e j!t0 Fð!Þ ¼ pffiffiffiffiffiffi 2 1 Linearity If F1 ð!Þ and F2 ð!Þ are the Fourier transforms of f1 ðtÞ and f2 ðtÞ respectively then 1 F1 ð!Þ þ 2 F2 ð!Þ is the Fourier transform of 1 f1 ðtÞ þ 2 f2 ðtÞ where 1 and 2 are constants. Frequency shifting If Fð!Þ is the Fourier transform of f ðtÞ then the Fourier transform of f ðtÞej!0 t is Fð! !0 Þ. Time scaling ð 1 1 f ðtÞej!t dt then If Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 f ðktÞej!t dt. jkj1 Fð!=kÞ ¼ pffiffiffiffiffiffi 2 1 Symmetry If Fð!Þ is the Fourier transform of f ðtÞ then the Fourier transform of FðtÞ is f ð!Þ. Differentiation ð 1 1 f ðtÞej!t dt then If Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 ðnÞ f ðtÞej!t dt and ðj!Þn Fð!Þ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 F ðnÞ ð!Þ ¼ pffiffiffiffiffiffi ðj!Þn f ðtÞej!t dt. 2 1 8 Convolution The convolution of two functions f ðtÞ and gðtÞ is defined as ð1 f ðtÞ gðtÞ ¼ f ðxÞgðt xÞ dx ¼ hðtÞ. 1 The convolution theorem If Fð!Þ and Gð!Þ are the Fourier transforms of f ðtÞ and gðtÞ respectively then (a) The Fourier transform of the convolution of f ðtÞ and gðtÞ is equal to the product of the individual Fourier transforms. That is pffiffiffiffiffiffi f½f ðtÞ gðtÞ ¼ 2Fð!ÞGð!Þ. (b) The Fourier transform of the product f ðtÞgðtÞ is equal to the convolution of the individual Fourier transforms. That is 1 f½f ðtÞgðtÞ ¼ pffiffiffiffiffiffi Fð!Þ Gð!Þ. 2 329 330 Programme 9 9 Fourier cosine and sine transforms ð 1 1 Fð!Þe j!t dt where Given that f ðtÞ ¼ pffiffiffiffiffiffi 2 1 ð 1 1 Fð!Þ ¼ pffiffiffiffiffiffi f ðtÞej!t dt 2 1 where f ðtÞ is even then rffiffiffi ð rffiffiffi ð 2 1 2 1 f ðtÞ ¼ Fc ð!Þ cos !t d! where Fc ð!Þ ¼ f ðtÞ cos !t dt 0 0 and where Fc ð!Þ is called the Fourier cosine transformation. This transformation is useful when f ðtÞ is defined only for t 0 and where an extension can be added to f ðtÞ for t < 0 that makes the extended f ðtÞ into an even function. If f ðtÞ is odd then rffiffiffi ð rffiffiffi ð 2 1 2 1 f ðtÞ ¼ Fs ð!Þ sin !t d! where Fs ð!Þ ¼ f ðtÞ sin !t dt 0 0 and where Fs ð!Þ is called the Fourier sine transformation. This transformation is useful when f ðtÞ is defined only for t 0 and where an extension can be added to f ðtÞ for t < 0 that makes the extended f ðtÞ into an odd function. Can you? 51 Checklist 9 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Convert a trigonometric Fourier series into a doubly infinite sum of complex exponentials? Yes No . Derive the complex Fourier series of a function that satisfies Dirichlet’s conditions? Yes No . Recognise the function sinc ðtÞ? Yes Frames 1 to 4 5 to 8 9 No . Separate a discrete complex spectrum into an amplitude spectrum and a phase spectrum? Yes No 10 to 12 331 Introduction to the Fourier transform . State Fourier’s integral theorem in terms of complex exponentials? Yes No 13 to 15 . Define and derive the Fourier transform of a function satisfying Dirichlet’s conditions? Yes No 16 and 17 . Separate a continuous complex spectrum into an amplitude spectrum and a phase spectrum? Yes No . Recognise the functions a ðtÞ and a ðtÞ and derive their Fourier transforms along with those of the Dirac delta and the Heaviside unit step? Yes No 18 19 . Recognise alternative forms of the function–transform pair? Yes No . Reproduce a collection of properties of the Fourier transform? Yes No . Evaluate the convolution of two functions and describe its Fourier transform? Yes No . Derive the Fourier sine and cosine transformations? Yes No to 27 28 29 to 40 41 to 45 46 to 48 Test exercise 9 1 Find the complex Fourier series of the sawtooth wave f ðtÞ ¼ t, 0 < t < 1 and where f ðt þ 1Þ ¼ f ðtÞ. 2 Find the Fourier transform of at e jtj < 1 f ðtÞ ¼ a>0 0 otherwise 3 4 1 Given that the Dirac delta ðtÞ has the Fourier transform Fð!Þ ¼ pffiffiffiffiffiffi, show, by 2 considering the inverse Fourier transform, that ð ð 1 1 j!t 1 1 ðtÞ ¼ e d! ¼ cos !t d!. 2 1 0 If f ðtÞ and Fð!Þ form a Fourier transform pair, find the Fourier transform of f ðtÞ sin !0 t where !0 is a constant. 52 332 Programme 9 5 Find the inverse transform of Fð!Þ ¼ 6 Show that (a) uðtÞ uðtÞ ¼ t uðtÞ (b) t uðtÞ et uðtÞ ¼ et t 1 uðtÞ 6 . !2 þ 5j! 4 where uðtÞ is the unit step function. 7 Find the Fourier sine and cosine transformations of f ðtÞ ¼ ekt for t > 0 and k > 0. Further problems 9 53 1 By comparing the trigonometric Fourier series of a periodic function with its complex exponential counterpart show that qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 bn 2 2 an þ bn and n ¼ arctan where cn ¼ jcn je jn . jcn j ¼ 2 an 2 Prove Parseval’s identity for the periodic function with period T ð 1 1 X a2 1 X 1 T=2 jcn j2 ¼ 0 þ a2 þ b2n ff ðtÞg2 dt ¼ T T=2 4 2 n¼1 n n¼1 and show that 1 X 1 2 . ¼ n2 6 n¼1 3 Draw the graph and find the complex Fourier series of the rectified sine wave f ðtÞ ¼ sin t, 0 < t < 1 where f ðt þ 1Þ ¼ f ðtÞ. 4 Draw the graph and find the complex Fourier series of the rectified cosine wave f ðtÞ ¼ cos t, 1=2 < t < 1=2 where f ðt þ 1Þ ¼ f ðtÞ. 5 Draw the graph and find the complex Fourier series of f ðtÞ ¼ et , 0<t<2 where f ðt þ 2Þ ¼ f ðtÞ. 6 Draw the graph and find the complex Fourier series of the sawtooth wave t 1 f ðtÞ ¼ þ , 0 < t < T where f ðt þ TÞ ¼ f ðtÞ. T 2 7 If f1 ðtÞ ¼ 1 X cn e jn!0 t and f2 ðtÞ ¼ n¼1 n¼1 convolution f1 ðtÞ f2 ðtÞ ¼ 1 X cn dn e jn!0 t . n¼1 8 1 X Find the Fourier transform of cosh t for jtj < 1 f ðtÞ ¼ 0 for jtj > 1 dn e jn!0 t where !0 ¼ 2=T, show that the Introduction to the Fourier transform 9 Find the Fourier transform of sinh t for jtj < 1 f ðtÞ ¼ 0 for jtj > 1. 10 Find the Fourier transform of sin t for 0 < t < 1 f ðtÞ ¼ 0 otherwise. 11 Find the Fourier transform of cos t for jtj < 1=2 f ðtÞ ¼ 0 otherwise. 12 Draw the graph and find the Fourier transform of f ðtÞ ¼ eajtj , 13 a > 0. Given that 8 for 1 < t < 0 <1 f ðtÞ ¼ 1 for 0 < t < 1 : 0 otherwise (a) Draw the graph of f ðtÞ (b) Express f ðtÞ in terms of the Heaviside unit step function (c) Find the Fourier transform of f ðtÞ. 14 Draw the graph and find the Fourier transform of f ðtÞ ¼ ðuðtÞ uðt ÞÞ cos kt. 15 Show that if f ðtÞ is real then the corresponding Fourier transform Fð!Þ ¼ jFð!Þje jð!Þ is such that jFð!Þj is even and ð!Þ is odd. 16 Show that if the Fourier transform of a real function is real then f ðtÞ is even, and if the Fourier transform of a real function is imaginary then f ðtÞ is odd. 17 Defining the squared modulus of the Fourier transform jFð!Þj2 ¼ Fð!ÞF ð!Þ where F ð!Þ is the complex conjugate of Fð!Þ, prove Parseval’s theorem ð1 ð1 2 ½f ðtÞ dt ¼ jFð!Þj2 dt. 1 1 18 Show that the convolution of a top-hat function with itself is the triangle function. That is a ðtÞ a ðtÞ ¼ a ðtÞ. 19 Show that sinc ðtÞ sinc ðtÞ ¼ sinc ðtÞ. 20 Find the Fourier sine and cosine transforms of at e for jtj < 1 f ðtÞ ¼ 0 otherwise. 21 Find the Fourier sine and cosine transforms of cosh t for jtj < 1 f ðtÞ ¼ 0 otherwise. 333 Programme 10 Frames 1 to 45 Power series solutions of ordinary differential equations 1 Learning outcomes When you have completed this Programme you will be able to: . Obtain the nth derivative of the exponential, circular and hyperbolic functions . Apply the Leibnitz theorem to derive the nth derivative of a product of expressions . Use the Leibnitz–Maclaurin method of obtaining a series solution to a second-order homogeneous differential equation with constant coefficients . Solve Cauchy–Euler equi-dimensional equations Prerequisite: Engineering Mathematics (Fifth Edition) Programmes 13 Series 1, 14 Series 2 and 25 Second-order differential equations 334 Power series solutions of ordinary differential equations 1 335 Higher derivatives dy ¼ cos x ¼ sin x þ dx 2 d2 y 2 ¼ sin x ¼ sinðx þ Þ ¼ sin x þ dx2 2 3 d y 3 etc: ¼ cos x ¼ sin x þ dx3 2 dn y n We see a pattern developing. In general ¼ sin x þ : Before we go dxn 2 further, we introduce a shorthand notation for the nth derivative of y as dn y y ðnÞ ¼ n . Note, however, we still use the ‘prime’ notation y 0 , y 00 and y 000 to dx represent the first, second and third derivatives respectively. The results above can therefore be written If y ¼ sin x ; y 0 ¼ cos x ¼ sin x þ 2 2 00 y ¼ sin x ¼ sin x þ 2 3 000 y ¼ cos x ¼ sin x þ 2 n and, in general, y ðnÞ ¼ sin x þ 2 It is therefore possible to write down any particular derivative of sin x without calculating all the previous derivatives. For example d7 y 7 ð7Þ ¼ cos x ¼ y ¼ sin x þ dx7 2 If y ¼ sin x 1 Similarly, starting with y ¼ cos x, we can determine an expression for the nth derivative of y which is . . . . . . . . . . . . n y ðnÞ ¼ cos x þ 2 Because ; y 0 ¼ sin x ¼ cos x þ 2 2 y 00 ¼ cos x ¼ cos x þ 2 3 etc: y 000 ¼ sin x ¼ cos x þ 2 n ; y ðnÞ ¼ cos x þ 2 Many of the standard functions can be treated in a similar manner. y ¼ cos x For example, if y ¼ eax , then y ðnÞ ¼ . . . . . . . . . . . . 2 336 Programme 10 3 y ðnÞ ¼ an eax Because y ¼ eax , y 0 ¼ aeax , y 00 ¼ a2 eax , y 000 ¼ a3 eax , etc. In general, y ðnÞ ¼ an eax . With no great effort, we can now write down expressions for the following If y ¼ sin ax; y ðnÞ ¼ . . . . . . . . . . . . If y ¼ cos ax; y ðnÞ ¼ . . . . . . . . . . . . n y ðnÞ ¼ an sin ax þ 2 n y ¼ cos ax, y ðnÞ ¼ an cos ax þ 2 4 y ¼ sin ax, Now one more. If y ¼ ln x; 5 y ðnÞ ¼ . . . . . . . . . . . . y ðnÞ ¼ ð1Þn1 ðn 1Þ! xn Because y ¼ ln x ; y0 ¼ 1 x y 00 ¼ y 000 ¼ 1 x2 2 x3 y ð4Þ ¼ 3! x4 ; y ðnÞ ¼ ð1Þn1 ðn 1Þ! xn dy 1 ¼ y 0 ¼ ¼ x1 . dx x Therefore, if the result obtained for y ðnÞ is to be valid for n ¼ 1, then We already know that, if y ¼ ln x; y 0 ¼ ð1Þ0 But y 0 ¼ x1 0! 0! ¼ x x ; 0! ¼ . . . . . . . . . . . . 337 Power series solutions of ordinary differential equations 1 6 0! ¼ 1 Now let us consider the derivatives of sinh ax and cosh ax. Next frame If y ¼ sinh ax, y 0 ¼ a cosh ax 7 y 00 ¼ a2 sinh ax y 000 ¼ a3 cosh ax etc: Because sinh ax is not periodic, we cannot proceed as we did with sin ax. We need to find a general statement for y ðnÞ containing terms in sinh ax and in cosh ax, such that, when n is even, the term in cosh ax disappears and, when n is odd, the term in sinh ax disappears. This we can do by writing y ðnÞ in the form y ðnÞ ¼ an ½1 þ ð1Þn sinh ax þ ½1 ð1Þn cosh ax 2 In very much the same way, we can determine the nth derivative of y ¼ cosh ax as . . . . . . . . . . . . y ðnÞ ¼ an ½1 ð1Þn sinh ax þ ½1 þ ð1Þn cosh ax 2 Finally, let us deal with y ¼ xa . y ¼ xa ; y 0 ¼ axa1 y 00 ¼ aða 1Þ xa2 y 000 ¼ aða 1Þða 2Þ xa3 ............ ðnÞ ; y ; y ðnÞ ¼ aða 1Þða 2Þ . . . ða n þ 1Þ xan a! xan ¼ (a is a positive integer) ða nÞ! So, collecting our results together, we have a! y ðnÞ ¼ xan y ¼ xa ða nÞ! y ¼ eax y ðnÞ ¼ an eax n yðnÞ ¼ an sin ax þ 2 n y ¼ cos ax yðnÞ ¼ an cos ax þ 2 an ðnÞ ½1 þ ð1Þn sinh ax þ ½1 ð1Þn cosh ax y ¼ sinh ax y ¼ 2 an ðnÞ ½1 ð1Þn sinh ax þ ½1 þ ð1Þn cosh ax y ¼ cosh ax y ¼ 2 Make a note of these, as a set, and then move on to the next frame y ¼ sin ax 8 338 9 Programme 10 Exercise Determine the following derivatives 1 y ¼ sin 4x y ð5Þ ¼ . . . . . . . . . . . . 2 y ¼ ex=2 y ð8Þ ¼ . . . . . . . . . . . . 3 y ð12Þ ¼ . . . . . . . . . . . . 4 y ¼ cosh 3x pffiffiffi y ¼ cosðx 2Þ 5 y ¼ x8 y ð6Þ ¼ . . . . . . . . . . . . 6 y ¼ sinh 2x y ð7Þ ¼ . . . . . . . . . . . . yð10Þ ¼ . . . . . . . . . . . . Finish them all; then check with the next frame 10 Here are the solutions 5 1 y ð5Þ ¼ 45 sin 4x þ ¼ 1024 sin 4x þ ¼ 1024 cos 4x 2 2 8 1 1 x=2 e ¼ ex=2 =256 ex=2 ¼ 2 y ð8Þ ¼ 2 256 312 f0 sinh 3x þ 2 cosh 3xg ¼ 312 cosh 3x 2 pffiffiffi pffiffiffi 10 4 y ð10Þ ¼ ð 2Þ10 cos x 2 þ 2 pffiffiffi pffiffiffi ¼ 32 cosðx 2 þ 5Þ ¼ 32 cosðx 2Þ 8! 5 yð6Þ ¼ x2 ¼ 20 160 x2 2! o 27 n 6 y ð7Þ ¼ ½1 þ ð1Þ7 sinh 2x þ ½1 ð1Þ7 cosh 2x 2 3 y ð12Þ ¼ ¼ 27 cosh 2x Leibnitz theorem – n th derivative of a product of two functions If y ¼ uv, where u and v are functions of x, then dv du and u0 ¼ dx dx y 00 ¼ uv00 þ v 0 u0 þ vu00 þ u0 v 0 ¼ u00 v þ 2u0 v 0 þ uv00 y 0 ¼ uv0 þ vu0 and where v0 ¼ If we differentiate the last result and collect like terms, we obtain y 000 ¼ . . . . . . . . . . . . Power series solutions of ordinary differential equations 1 y 000 ¼ u000 v þ 3u00 v 0 þ 3u0 v 00 þ uv000 A further stage of differentiation would give y ð4Þ ¼ uð4Þ v þ 4uð3Þ v ð1Þ þ 6uð2Þ v ð2Þ þ 4uð1Þ v ð3Þ þ uv ð4Þ These results can therefore be written y ¼ uv y 0 ¼ u0 v þ uv0 y 00 ¼ u00 v þ 2u0 v 0 þ uv00 y 000 ¼ u000 v þ 3u00 v 0 þ 3u0 v 00 þ uv000 y ð4Þ ¼ uð4Þ v þ 4uð3Þ v ð1Þ þ 6uð2Þ v ð2Þ þ 4uð1Þ v ð3Þ þ uvð4Þ Notice that in each case (a) the superscript of u decreases regularly by 1 (b) the superscript of v increases regularly by 1 (c) the numerical coefficients are the normal binomial coefficients. Indeed, ðuvÞðnÞ can be obtained by expanding ðu þ vÞðnÞ using the binomial theorem where the ‘powers’ are interpreted as derivatives. So the expression for the nth derivative can therefore be written as y ðnÞ ¼ uðnÞ v þ nuðn1Þ v ð1Þ þ nðn 1Þ ðn2Þ ð2Þ u v 12 nðn 1Þðn 2Þ ðn3Þ ð3Þ u v þ ... 123 nðn 1Þ ðn2Þ ð2Þ u v ¼ uðnÞ v þ nuðn1Þ v ð1Þ þ 2! nðn 1Þðn 2Þ ðn3Þ ð3Þ u þ v þ ... 3! ¼ uðnÞ v þ n C1 uðn1Þ v ð1Þ þ n C2 uðn2Þ v ð2Þ þ . . . þ i.e. y ðnÞ þ n Cn1 uð1Þ v ðn1Þ þ uvðnÞ where n Cr ¼ If y ¼ uv n! r!ðn rÞ! y ðnÞ ¼ n X n Cr uðnrÞ v ðrÞ where uð0Þ u r¼0 This is the Leibnitz theorem. We shall certainly be using it often in the work ahead, so make a note of it for future reference. Then we can see it in use. 339 11 340 12 Programme 10 Choice of function for u and v For the product y ¼ uv the function taken as (a) u is the one whose nth derivative can readily be obtained (b) v is the one whose derivatives reduce to zero after a small number of stages of differentiation. Example 1 To find y ðnÞ when y ¼ x3 e2x . Here we choose v ¼ x3 whose fourth derivative is zero u ¼ e2x because we know that the nth derivative uðnÞ ¼ . . . . . . . . . . . . 13 uðnÞ ¼ 2n e2x Using the Leibnitz theorem: nðn 1Þ ðn2Þ ð2Þ u v 2! nðn 1Þðn 2Þ ðn3Þ ð3Þ u þ v þ ... 3! y ðnÞ ¼ uðnÞ v þ nuðn1Þ v ð1Þ þ 14 v ¼ x3 ; v ð1Þ ¼ 3x2 ; u ¼ e2x ; uðnÞ ¼ 2n e2x v ð2Þ ¼ 6x; v ð3Þ ¼ 6; v ð4Þ ¼ 0 ; y ðnÞ ¼ . . . . . . . . . . . . y ðnÞ ¼ e2x 2n3 8x3 þ 12nx2 þ nðn 1Þ 6x þ nðn 1Þðn 2Þ Example 2 If x2 y 00 þ xy 0 þ y ¼ 0, show that x2 y ðnþ2Þ þ ð2n þ 1Þ xy ðnþ1Þ þ ðn2 þ 1Þ y ðnÞ ¼ 0. We take the given equation x2 y 00 þ xy 0 þ y ¼ 0 and differentiate n times, treating each term in turn. If w ¼ x2 y 00 wðnÞ ¼ . . . . . . . . . . . . If w ¼ xy 0 wðnÞ ¼ . . . . . . . . . . . . If w ¼ y wðnÞ ¼ . . . . . . . . . . . . Power series solutions of ordinary differential equations 1 nðn 1Þ ðnÞ y 2 þ 0... 2! ¼ y ðnþ1Þ x þ ny ðnÞ 1 þ 0 þ . . . w ¼ x2 y 00 ; wðnÞ ¼ y ðnþ2Þ x2 þ ny ðnþ1Þ 2x þ w ¼ xy 0 ; wðnÞ w¼y ; wðnÞ ¼ y ðnÞ Then x2 y 00 þ xy 0 þ y ðnÞ ¼0 341 15 becomes ............ x2 y ðnþ2Þ þ ð2n þ 1Þxyðnþ1Þ þ ðn2 þ 1Þy ðnÞ ¼ 0 16 which is what we had to show. Example 3 Differentiate n times ð1 þ x2 Þy 00 þ 2xy 0 5y ¼ 0. The result . . . . . . . . . . . . ð1 þ x2 Þ y ðnþ2Þ þ 2ðn þ 1Þ xy ðnþ1Þ þ ðn2 þ n 5Þ y ðnÞ ¼ 0 17 Because, by the Leibnitz theorem nðn 1Þ ðnÞ y 2 2! n o þ 2 xy ðnþ1Þ þ nyðnÞ 1 5y ðnÞ ¼ 0 y ðnþ2Þ ð1 þ x2 Þ þ nyðnþ1Þ 2x þ ð1 þ x2 Þ y ðnþ2Þ þ 2ðn þ 1Þ xy ðnþ1Þ þ fnðn 1Þ þ 2n 5g y ðnÞ ¼ 0 ð1 þ x2 Þ y ðnþ2Þ þ 2ðn þ 1Þ xy ðnþ1Þ þ ðn2 þ n 5Þ y ðnÞ ¼ 0 We shall be using the Leibnitz theorem in the rest of this Programme, so let us move on to see some of its applications. Power series solutions Second-order linear differential equations with constant coefficients of d2 y dy the form a þb þ cy ¼ 0 can be solved by algebraic methods dx2 dx giving solutions in terms of the normal elementary functions such as exponentials, trigonometric and polynomial functions. 18 342 Programme 10 d2 y dy þ Q ðxÞ y ¼ 0, where þ P ðxÞ 2 dx dx P ðxÞ and Q ðxÞ are functions of x, cannot be solved in this way. However, it is often possible to obtain solutions in the form of infinite series of powers of x – and the next section of work investigates one of the methods which make this possible. In general, equations of the form Leibnitz–Maclaurin method As the title suggests, for this we need to be familiar with the Leibnitz theorem and with Maclaurin’s series. The Leibnitz theorem states that, if y ¼ uv, where u and v are functions of x, then y ðnÞ ¼ . . . . . . . . . . . . 19 nðn 1Þ ðn2Þ ð2Þ v þ ... u 2! nðn 1Þ . . . ðn r þ 1Þ ðnrÞ ðrÞ u þ v þ . . . þ uvðnÞ r! y ðnÞ ¼ uðnÞ v þ nuðn1Þ v ð1Þ þ where uðrÞ and v ðrÞ denote d ru d rv and respectively. dx r dx r Maclaurin’s series for y ¼ f ðxÞ can be stated as y ¼ ............ 20 y ¼ ðyÞ0 þ xðy 0 Þ0 þ where y ðnÞ 0 x2 00 xn ðnÞ ðy Þ0 þ . . . þ y 0 þ ... 2! n! denotes the value of the nth derivative of y at x ¼ 0. On to the next frame 21 Example 1 Find the power series solution of the equation x d2 y dy þ þ xy ¼ 1. dx2 dx The equation can be written xy 00 þ y 0 þ xy ¼ 1 In the first product term xy 00 , treat y 00 as u and x as v. Then, differentiating the equation n times by the Leibnitz theorem, gives ............ Power series solutions of ordinary differential equations 1 xy ðnþ2Þ þ n 1 y ðnþ1Þ þ y ðnþ1Þ þ xy ðnÞ þ n 1 y ðn1Þ ¼ 0 i.e. 343 22 xy ðnþ2Þ þ ðn þ 1Þy ðnþ1Þ þ xy ðnÞ þ ny ðn1Þ ¼ 0 At x ¼ 0, this becomes ðn þ 1Þ yðnþ1Þ 0 þ n y ðn1Þ 0 ¼ 0 n n1 ; y ðnþ1Þ 0 ¼ y ðn1Þ 0 nþ1 This relationship is called a recurrence relation. We can now substitute n ¼ 1, 2, 3, . . . and get a set of relationships between the various coefficients. n¼1 ðy 00 Þ0 ¼ 12 ðyÞ0 n¼2 ðy 000 Þ0 ¼ 23 ðy 0 Þ0 n¼3 ðy ð4Þ Þ0 ¼ 34 ðy 00 Þ0 ¼ 34 12 ðyÞ0 Continuing in the same way, ðy ð5Þ Þ0 ¼ . . . . . . . . . . . . ðy ð6Þ Þ0 ¼ . . . . . . . . . . . . ðy ð7Þ Þ0 ¼ . . . . . . . . . . . . ðy ð8Þ Þ0 ¼ . . . . . . . . . . . . n¼4 n¼5 n¼6 n¼7 ðy ð5Þ Þ0 ¼ 45 ðy ð3Þ Þ0 ¼ 45 23 ðy ð1Þ Þ0 ðy ð6Þ Þ0 ¼ 56 ðy ð4Þ Þ0 ¼ 56 34 12 ðyÞ0 ðy ð7Þ Þ0 ¼ 67 ðy ð5Þ Þ0 ¼ 67 45 23 ðy ð1Þ Þ0 ðy ð8Þ Þ0 ¼ 78 ðy ð6Þ Þ0 ¼ 78 56 34 12 ðyÞ0 Notice that, by this means, the values of all the derivatives at x ¼ 0 can be expressed in terms of ðyÞ0 and ðy 0 Þ0 . If we now substitute these values for ðy ðrÞ Þ0 in the Maclaurin series y ¼ ðyÞ0 þ xðy 0 Þ0 þ we obtain . . . . . . . . . . . . x2 00 x3 xr ðy Þ0 þ ðy 000 Þ0 þ . . . þ ðy ðrÞ Þ0 þ . . . 2! 3! r! 23 344 24 Programme 10 x2 1 x3 2 ðyÞ0 þ ðy 0 Þ0 2 3 2! 3! 4 5 x 3 1 x 4 2 ðyÞ0 þ ðy 0 Þ0 þ 4 2 5 3 4! 5! 6 x 5 3 1 ðyÞ0 þ . . . . . . . . . . . . þ 6 4 2 6! y ¼ðyÞ0 þ xðy 0 Þ0 þ Simplifying, this gives y ¼ ðyÞ0 1 x2 x4 x6 þ þ ... 22 22 42 22 42 62 þ ðy 0 Þ0 x x3 x5 þ 2 þ ... 2 3 3 52 The values of ðyÞ0 and ðy 0 Þ0 provide the two arbitrary constants for the secondorder equation and are obtained from the given initial conditions. dy ¼ 1; then the relevant particular For example, if at x ¼ 0, y ¼ 2 and dx solution is . . . . . . . . . . . . 25 y ¼2 1 x2 x4 x6 þ 2 2 þ ... 2 2 2 2 4 2 42 62 þ x Because at x ¼ 0, y ¼2 dy ¼1 dx x3 x5 þ 2 þ ... 2 3 3 52 i.e. ðyÞ0 ¼ 2 i.e. ðy 0 Þ0 ¼ 1. To be a valid solution, the series obtained must converge. Application of the ratio test will normally indicate any restrictions on the values that x may have. The Leibnitz–Maclaurin (power series) method therefore involves the following main steps: (a) Differentiate the given equation n times, using the Leibnitz theorem. (b) Rearrange the result to obtain the recurrence relation at x ¼ 0: (c) Determine the values of the derivatives at x ¼ 0, usually in terms of ðyÞ0 and ðy 0 Þ0 : (d) Substitute in the Maclaurin expansion for y ¼ f ðxÞ. (e) Simplify the result where possible and apply boundary conditions if provided. That is all there is to it. Let us go through the various steps with another example. 345 Power series solutions of ordinary differential equations 1 Example 2 Determine a series solution of the equation d2 y dy þ y ¼ 0. þx dx2 dx The equation can be written y 00 þ xy 0 þ y ¼ 0 (a) Differentiate n times using the Leibnitz theorem, which gives ............ 26 y ðnþ2Þ þ xy ðnþ1Þ þ ðn þ 1Þy ðnÞ ¼ 0 Because y 00 þ xy 0 þ y ¼ 0 n o ; y ðnþ2Þ þ xy ðnþ1Þ þ n 1 y ðnÞ þ y ðnÞ ¼ 0 ; y ðnþ2Þ þ xyðnþ1Þ þ ðn þ 1Þy ðnÞ ¼ 0. (b) Determine the recurrence relation at x ¼ 0, which is ............ 27 y ðnþ2Þ ¼ ðn þ 1Þ y ðnÞ (c) Now taking n ¼ 0, 1, 2, 3, 4, 5, determine the derivatives at x ¼ 0 in terms of ðyÞ0 and ðy 0 Þ0 . List them, as we did before, in table form. n¼0 ðy 00 Þ0 ¼ ðyÞ0 000 28 ¼ ðyÞ0 0 0 1 ðy Þ0 ¼ 2 ðy Þ0 ¼ 2 ðy Þ0 2 ðy ð4Þ Þ0 ¼ 3 ðy 00 Þ0 ¼ ð3Þ½ðyÞ0 ¼ 3 ðyÞ0 3 ðy ð5Þ Þ0 ¼ 4 ðy 000 Þ0 ¼ ð4Þ½2ðy 0 Þ0 ¼ 2 4 ðy 0 Þ0 4 ðy ð6Þ Þ0 ¼ 5 ðy ð4Þ Þ0 ¼ ð5Þ½3ðy 00 Þ0 ¼ 3 5 ðyÞ0 5 ðy ð7Þ Þ0 ¼ 6 ðy ð5Þ Þ0 ¼ ð6Þ½4ðy 000 Þ0 ¼ 2 4 6 ðy 0 Þ0 (d) Substitute these expressions for the derivatives in terms of ðyÞ0 and ðy 0 Þ0 in Maclaurin’s expansion y ¼ ðyÞ0 þ x ðy 0 Þ0 þ Then x2 00 x3 x4 ðy Þ0 þ ðy 000 Þ0 þ ðy ð4Þ Þ0 þ . . . 2! 3! 4! y ¼ ............ 346 29 Programme 10 x2 x3 x4 x5 ðyÞ0 þ ð2y 0 Þ0 þ ð3yÞ0 þ ð8y 0 Þ0 2! 3! 4! 5! x6 x7 0 þ ð15yÞ0 þ ð48y Þ0 þ . . . 6! 7! y ¼ ðyÞ0 þ xðy 0 Þ0 þ Collecting now the terms in ðyÞ0 and ðy 0 Þ0 , we finally obtain y ¼ðyÞ0 1 x2 x4 x6 þ þ ... 2 24 246 þ ðy 0 Þ0 x x3 x5 x7 þ þ ... 3 35 357 They are all done in very much the same way. Here is another. Example 3 Solve the equation d2 y dy þ þ 2xy ¼ 0 given that at x ¼ 0, y ¼ 0 and dx2 dx dy ¼ 1. dx First write the equation as y 00 þ y 0 þ 2xy ¼ 0, differentiate n times by the Leibnitz theorem and obtain the recurrence relation at x ¼ 0, which is ............ n o y ðnþ2Þ ¼ y ðnþ1Þ þ 2nyðn1Þ 30 n1 Because y 00 þ y 0 þ 2xy ¼ 0 ; y ðnþ2Þ þ y ðnþ1Þ þ 2xy ðnÞ þ n2y ðn1Þ ¼ 0 At x ¼ 0, y ðnþ2Þ þ y ðnþ1Þ þ 2nyðn1Þ ¼ 0 n o ; y ðnþ2Þ ¼ y ðnþ1Þ þ 2ny ðn1Þ Since we have a term in y ðn1Þ , then n must start at 1 to give ðyÞ0 . Therefore the recurrence relation applies for n 1. We now take n ¼ 1, 2, 3, . . . to obtain the relationships between the coefficients up to ðyð6Þ Þ0 . Complete the table and check with the next frame. 347 Power series solutions of ordinary differential equations 1 ðy ð3Þ Þ0 ¼ ðy ð2Þ Þ0 þ 2ðyÞ0 ðy ð4Þ Þ0 ¼ ðy ð3Þ Þ0 þ 4ðy 0 Þ0 ðy ð5Þ Þ0 ¼ ðy ð4Þ Þ0 þ 6ðy ð2Þ Þ0 ðy ð6Þ Þ0 ¼ ðy ð5Þ Þ0 þ 8ðy ð3Þ Þ0 n¼1 n¼2 n¼3 n¼4 31 We therefore have expressions for ðy 000 Þ0 ; ðy ð4Þ Þ0 ; ðy ð5Þ Þ0 ; ðy ð6Þ Þ0 ; but what about ðy 00 Þ0 ? If we refer to the initial conditions, we know that at x ¼ 0, y ¼ 0 and y 0 ¼ 1. ; ðyÞ0 ¼ 0 and ðy 0 Þ0 ¼ 1. We can find ðy 00 Þ0 by reference to the given equation itself, because y 00 þ y 0 þ 2xy ¼ 0 Therefore, at x ¼ 0; ðy 00 Þ0 þ ðy 0 Þ0 ¼ 0 ; ðy 00 Þ0 ¼ ðy 0 Þ0 ¼ 1. So now we have ðyÞ0 ¼ 0 ðy 0 Þ0 ¼ 1 ðy 00 Þ0 ¼ 1 ðy 000 Þ0 ¼ ðy 00 Þ0 þ 2ðyÞ0 ðy ð4Þ Þ0 ¼ ðy 000 Þ0 þ 4ðy 0 Þ0 ðy ð5Þ Þ0 ¼ ðy ð4Þ Þ0 þ 6ðy 00 Þ0 ðy ð6Þ Þ0 ¼ ðy ð5Þ Þ0 þ 8ðy 000 Þ0 ¼ fð1Þ þ 0g ¼ 1 ¼ f1 þ 4g ¼ 5 ¼ fð5Þ 6g ¼ 11 ¼ f11 þ 8g ¼ 19 The required series solution is therefore y ¼ ............ y ¼x x2 x3 5x4 11x5 19x6 þ þ þ ... 2! 3! 4! 5! 6! 32 Because x2 00 x3 x4 ðy Þ0 þ ðy 000 Þ0 þ ðy ð4Þ Þ0 þ . . . 2! 3! 4! x2 x3 x4 x5 x6 ¼ 0 þ xð1Þ þ ð1Þ þ ð1Þ þ ð5Þ þ ð11Þ þ ð19Þ 2! 3! 4! 5! 6! x2 x3 5x4 11x5 19x6 þ þ ... ; y¼x þ 2! 3! 4! 5! 6! y ¼ ðyÞ0 þ xðy 0 Þ0 þ One more of the same kind. Example 4 Determine the general series solution of the equation ðx2 þ 1Þy 00 þ xy 0 4y ¼ 0 As usual, establish the recurrence relation at x ¼ 0, which is ............ 33 348 Programme 10 34 y ðnþ2Þ ¼ ð4 n2 Þy ðnÞ Because x2 þ 1 y 00 þ xy 0 4y ¼ 0 therefore n o nðn 1Þ ðnÞ y x2 þ 1 y ðnþ2Þ þ 2xnyðnþ1Þ þ 2 þ xy ðnþ1Þ þ ny ðnÞ 4y ðnÞ ¼ 0 2! At x ¼ 0, this becomes y ðnþ2Þ þ nðn 1Þy ðnÞ þ ny ðnÞ 4y ðnÞ ¼ 0 that is y ðnþ2Þ ¼ 4 n2 y ðnÞ Then, starting with n ¼ 0, determine expressions for ðy ðnÞ Þ0 as far as n ¼ 7. They are . . . . . . . . . . . . 35 n¼0 ðy 00 Þ0 ¼ 4ðyÞ0 ¼ 4ðyÞ0 n¼1 ðy 000 Þ0 ¼ 3ðy 0 Þ0 ¼ 3ðy 0 Þ0 n¼2 ðy ð4Þ Þ0 ¼ 0 ¼0 n¼3 ðyð5Þ Þ0 ¼ 5ðy 000 Þ0 ¼ 15ðy 0 Þ0 n¼4 ðy ð6Þ Þ0 ¼ 12ðy ð4Þ Þ0 ¼ 0 n¼5 ðyð7Þ Þ0 ¼ 21ðy ð5Þ Þ0 ¼ ð21Þð15Þðy 0 Þ0 Now substitute in Maclaurin’s expansion and simplify the result. y ¼ ............ 36 y ¼ Að1 þ 2x2 Þ þ B x þ x3 x5 x7 þ þ ... 2 8 16 Because x2 00 x3 x4 ðy Þ0 þ ðy 000 Þ0 þ ðy ð4Þ Þ0 þ . . . 2! 3! 4! x2 x3 x4 x5 0 0 ¼ ðyÞ0 þ xðy Þ0 þ 4ðyÞ0 þ 3ðy Þ0 þ ð0Þ þ ð15Þðy 0 Þ0 þ etc. 2! 3! 4! 5! x3 x5 x7 2 0 ¼ ðyÞ0 f1 þ 2x g þ ðy Þ0 x þ þ þ ... 2 8 16 y ¼ ðyÞ0 þ xðy 0 Þ0 þ Putting ðyÞ0 ¼ A and ðy 0 Þ0 ¼ B, we have the result stated. Now to something slightly different Power series solutions of ordinary differential equations 1 Cauchy–Euler equi-dimensional equations Closely allied to some of the equations that we have been looking at are the Cauchy–Euler equi-dimensional differential equations. The general nth order equation is an inhomogeneous equation with the structure an xn y ðnÞ ðxÞ þ an1 xn1 y ðn1Þ ðxÞ þ . . . þ a1 xy 0 ðxÞ þ a0 yðxÞ ¼ gðxÞ where a0 , . . . , an are constants and where the nth derivative has a coefficient containing the nth power of x – hence the name equi-dimensional. To simplify matters here we shall only deal with second order equations but the method employed can be quite easily extended to higher orders. We shall just look at Cauchy–Euler equations of the form: ax2 y 00 ðxÞ þ bxy0 ðxÞ þ cyðxÞ ¼ gðxÞ where x > 0. Solutions Just like the solution to a linear, constant coefficient ordinary differential equation, the solution to ax2 y 00 ðxÞ þ bxy0 ðxÞ þ cyðxÞ ¼ gðxÞ is in two parts: yðxÞ ¼ yh ðxÞ þ yp ðxÞ -- the homogeneous solution yh ðxÞ where ax2 yh00 ðxÞ þ bxyh0 ðxÞ þ cyh ðxÞ ¼ 0 -- the particular solution yp ðxÞ whose form depends on the form of gðxÞ. We shall proceed by example. Example 1 To solve the Cauchy–Euler differential equation x2 y 00 ðxÞ 4xy 0 ðxÞ þ 6yðxÞ ¼ 0 where yð1Þ ¼ 1 and yð2Þ ¼ 0 we first solve the homogeneous equation. The homogeneous solution x2 yh00 ðxÞ 4xyh0 ðxÞ þ 6yh ðxÞ ¼ 0. We assume a solution of the form: yh ðxÞ ¼ Kxn so that yh0 ðxÞ ¼ nKxn1 and yh00 ðxÞ ¼ nðn 1ÞKxn2 . Substituting into the homogeneous equation gives: x2 yh00 ðxÞ 4xyh0 ðxÞ þ 6yh ðxÞ ¼ Kxn ðnðn 1Þ 4n þ 6Þ ¼ Kxn n2 5n þ 6 ¼ Kxn ðn 3Þðn 2Þ ¼ 0 Therefore n ¼ 3 or n ¼ 2 so that yh ðxÞ ¼ Ax3 þ Bx2 (A, B constants). 349 37 350 Programme 10 The particular solution x2 y 00 ðxÞ 4xy 0 ðxÞ þ 6yðxÞ ¼ x. Since the right-hand side of the inhomogeneous equation is gðxÞ ¼ x we assume a form for the inhomogeneous solution of yp ðxÞ ¼ Cx þ D (C, D constants) just as we did for the linear constant coefficient case. This means that that yp0 ðxÞ ¼ C and yp00 ðxÞ ¼ 0. Substituting into the inhomogeneous equation gives: 4Cx þ 6ðCx þ DÞ ¼ 2Cx þ 6D ¼ x x Then C ¼ 1=2 and D ¼ 0 giving yp ðxÞ ¼ . 2 The complete solution Adding the homogeneous solution to the particular solution gives x yðxÞ ¼ Ax3 þ Bx2 þ . 2 Finally, since yð1Þ ¼ 1 and yð2Þ ¼ 0 we see that: yð1Þ ¼ A þ B þ 1 ¼1 2 AþB ¼ so that yð2Þ ¼ 8A þ 4B þ 1 ¼ 0 so that 1 2 8A þ 4B ¼ 1 so A ¼ 3 5 and B ¼ 4 4 giving 2x þ 5x2 3x3 4 2þ53 4 þ 20 24 ¼ 1 and yð2Þ ¼ ¼0 Check: yð1Þ ¼ 4 4 yðxÞ ¼ Now you try one. Example 2 Given the Cauchy–Euler equation x2 y 00 ðxÞ 8xy 0 ðxÞ þ 20yðxÞ ¼ 3x where yð1Þ ¼ 0 and yð3Þ ¼ 102 we first consider the homogeneous equation x2 yh00 ðxÞ 8xyh0 ðxÞ þ 20yh ðxÞ ¼ 0. Assuming a solution of the form yh ðxÞ ¼ Kxn we find that: yh ðxÞ ¼ . . . . . . . . . . . . 38 yh ðxÞ ¼ Ax5 þ Bx4 Because Since yh ðxÞ ¼ Kxn then yh0 ðxÞ ¼ nKxn1 and yh00 ðxÞ ¼ nðn 1ÞKxn2 so substituting into x2 yh00 ðxÞ 8xyh0 ðxÞ þ 20yh ðxÞ ¼ 0 gives: nðn 1Þ 8n þ 20 ¼ n2 9n þ 20 ¼ ðn 5Þðn 4Þ ¼0 so that n ¼ 5 or n ¼ 4. That is yh ðxÞ ¼ Ax5 þ Bx4 (A, B constants). Power series solutions of ordinary differential equations 1 351 The particular solution yp ðxÞ satisfies the equation x2 y 00 ðxÞ 8xy 0 ðxÞ þ 20yðxÞ ¼ 3x so that, assuming a form yp ðxÞ ¼ Cx þ D, yp ðxÞ ¼ . . . . . . . . . . . . and so yðxÞ ¼ . . . . . . . . . . . . yp ðxÞ ¼ x x and yðxÞ ¼ Ax5 þ Bx4 þ 4 4 39 Because Assuming a form yp ðxÞ ¼ Cx þ D so that yp0 ðxÞ ¼ C and yp00 ðxÞ ¼ 0 and substituting into the inhomogeneous equation gives the equation 8Cx þ 20ðCx þ DÞ ¼ 12Cx þ 20D ¼ 3x so that C ¼ 1=4 and D ¼ 0 therefore yp ðxÞ ¼ x x and yðxÞ ¼ Ax5 þ Bx4 þ . 4 4 Applying the boundary conditions yð1Þ ¼ 0 and yð3Þ ¼ 102 we find the complete solution to be: yðxÞ ¼ . . . . . . . . . . . . yðxÞ ¼ 3x5 4x4 þ x 4 Because 1 1 ¼ 0 that is A þ B ¼ 4 4 3 5 yð3Þ ¼ 102 so that yð3Þ ¼ 243A þ 81B þ ¼ 102 that is 3A þ B ¼ 4 4 3 therefore A ¼ and B ¼ 1 so that 4 5 3x 4x4 þ x yðxÞ ¼ 4 yð1Þ ¼ 0 so that yð1Þ ¼ A þ B þ Check: yð1Þ ¼ 729 324 þ 3 34þ1 ¼ 102 and yð3Þ ¼ 4 4 And just to make sure try another, but this time try and obtain the complete solution yourself. Example 3 The Cauchy–Euler equation x2 y 00 ðxÞ þ xy 0 ðxÞ yðxÞ ¼ 6x2 þ 8x3 , where yð1Þ ¼ 3, yð2Þ ¼ 40 and yp ðxÞ is of the form Cx2 þ Dx3 , has solution yðxÞ ¼ . . . . . . . . . . . . 40 352 Programme 10 41 yðxÞ ¼ 1 4 x þ 2x3 þ 16x2 16 x Because We first consider the homogeneous equation x2 yh00 ðxÞ þ xyh0 ðxÞ yh ðxÞ ¼ 0. Assuming a solution of the form yh ðxÞ ¼ Kxn we find that nðn 1Þ þ n 1 ¼ n2 1 ¼ ðn þ 1Þðn 1Þ ¼0 so that n ¼ 1 or n ¼ 1. 1 That is yh ðxÞ ¼ Ax þ Bx (A, B constants). Since the particular solution is of the form Cx2 þ Dx3 then yp ðxÞ ¼ Cx2 þ Dx3 , yp0 ðxÞ ¼ 2Cx þ 3Dx2 and yp00 ðxÞ ¼ 2C þ 6Dx. Substituting in the equation x2 y 00 ðxÞ þ xy 0 ðxÞ yðxÞ ¼ 6x2 þ 8x3 results in the equation: x2 ð2C þ 6DxÞ þ xð2Cx þ 3Dx2 Þ ðCx2 þ Dx3 Þ ¼ 6x2 þ 8x3 . That is: x2 ½3C þ x3 ½8D ¼ 6x2 þ 8x3 so that C ¼ 2, D ¼ 1, yp ðxÞ ¼ 2x2 þ x3 therefore yðxÞ ¼ Ax þ Bx1 þ 2x2 þ x3 . Applying the boundary conditions yð1Þ ¼ 3, yð2Þ ¼ 40 yð1Þ ¼ A þ B þ 2 þ 1 ¼ 3 B yð2Þ ¼ 2A þ þ 8 þ 8 ¼ 40 2 so that A þ B ¼ 0 B so that 2A þ ¼ 24 giving A ¼ 16, B ¼ 16 2 Therefore yðxÞ ¼ 16x 16x1 þ 2x2 þ x3 1 ¼ ðx4 þ 2x3 þ 16x2 16Þ x Check: yð1Þ ¼ 1ð1 þ 2 þ 16 16Þ ¼ 3 1 yð2Þ ¼ ð16 þ 16 þ 64 16Þ ¼ 40 2 And that completes the work of this Programme. The main points that we have covered in this Programme are listed in the Revision summary that follows. Read this in conjunction with the Can you? check list and note any sections that may need further attention: refer back to the relevant parts of the Programme, if necessary. There will then be no trouble with the Test exercise. The set of Further problems provides an opportunity for further practice. Power series solutions of ordinary differential equations 1 353 Revision summary 10 1 Higher derivatives y a! xan ða nÞ! eax an eax cos ax sinh ax cosh ax 2 y xa sin ax 42 ðnÞ n an sin ax þ 2 n an cos ax þ 2 an ½1 þ ð1Þn sinh ax þ ½1 ð1Þn cosh ax 2 an ½1 ð1Þn sinh ax þ ½1 þ ð1Þn cosh ax 2 Leibnitz theorem — nth derivative of a product of functions. If y ¼ uv y ðnÞ ¼uðnÞ v þ nuðn1Þ v ð1Þ þ nðn 1Þ ðn2Þ ð2Þ u v 2! nðn 1Þðn 2Þ ðn3Þ ð3Þ v þ ... u 3! nðn 1Þðn 2Þ . . . ðn r þ 1Þ ðnrÞ ðrÞ u v þ ... ... þ r! 1 X n ¼ Cr uðnrÞ v ðrÞ . þ i.e. y ðnÞ r¼0 ðnÞ ðuvÞ can be obtained by expanding ðu þ vÞðnÞ using the binomial theorem where the ‘powers’ are interpreted as derivatives. 3 Power series solution of second-order differential equations Leibnitz–Maclaurin method (1) Differentiate the equation n times by the Leibnitz theorem. (2) Put x ¼ 0 to establish a recurrence relation. (3) Substitute n ¼ 1, 2, 3, . . . to obtain y 0 , y 00 , y 000 , . . . at x ¼ 0. (4) Substitute in Maclaurin’s series and simplify where possible. 354 Programme 10 4 Cauchy–Euler equi-dimensional equations The second order Cauchy–Euler equi-dimensional equation has the structure ax2 y 00 ðxÞ þ bxy0 ðxÞ þ cyðxÞ ¼ gðxÞ where the coefficient of the nth derivative contains xn . The solution consists of the sum of a homogeneous solution yh ðxÞ and a particular solution yp ðxÞ. The homogeneous solution is assumed to be of the form yh ðxÞ ¼ Kxn and substitution into the homogeneous equation results in as many values of n as the degree of the equation. The form of the particular solution depends upon the form of the right-hand side of the equation gðxÞ. Can you? 43 Checklist 10 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Obtain the nth derivative of the exponential and circular and hyperbolic functions? Yes No 1 to 9 . Apply the Leibnitz theorem to derive the nth derivative of a product of expressions? Yes No 10 to 17 18 to 36 37 to 43 . Apply the Leibnitz–Maclaurin method of obtaining a series solution to a second-order homogeneous differential equation with constant coefficients? Yes No . Solve Cauchy–Euler equi-dimensional equations? Yes No Power series solutions of ordinary differential equations 1 355 Test exercise 10 1 If y ¼ ex y 2 ðnþ2Þ 2 þx , show that y 00 ¼ y 0 ð2x þ 1Þ þ 2y and hence prove that ¼ ð2x þ 1Þy ðnþ1Þ þ 2ðn þ 1Þy ðnÞ . 44 Obtain a power series solution of the equation ð1 þ x2 Þy 00 3xy 0 5y ¼ 0 up to and including the term in x6 . 3 Solve each of the following (a) x2 y 00 ðxÞ þ 2xy 0 ðxÞ 2yðxÞ ¼ 0 (b) 2x2 y 00 ðxÞ þ 5xy 0 ðxÞ 9yðxÞ ¼ x2 (c) x2 y 00 ðxÞ xy 0 ðxÞ þ yðxÞ ¼ 3x2 2x3 where yð1Þ ¼ yð2Þ ¼ 4 Further problems 10 (a) Use the Leibnitz theorem for the following. 1 If y ¼ x3 e4x , determine y ð5Þ . 2 Find the nth derivative of y ¼ x3 ex for n > 3. 3 If y ¼ x3 ð2x þ 1Þ2 , find y ð4Þ . 4 Find the 6th derivative of y ¼ x4 cos x. 5 If y ¼ ex sin x, obtain an expression for y ð4Þ . 6 Determine y ð3Þ when y ¼ x4 ln x. 7 If x2 y 00 þ xy 0 þ y ¼ 0; show that 8 9 x2 y ðnþ2Þ þ ð2n þ 1Þxy ðnþ1Þ þ ðn2 þ 1Þy ðnÞ ¼ 0. x If y ¼ ð2x Þ4 sin , evaluate y ð6Þ when x ¼ =2. 2 If y ¼ ex cos x, show that y ð4Þ þ 4y ¼ 0: 10 Find the ð2nÞth derivative of (a) y ¼ x2 sinh x (b) y ¼ x3 cosh x. 11 If y ¼ ðx3 þ 3x2 Þe2x , determine an expression for y ð6Þ . 12 Find the nth derivative of y ¼ eax cos ax and hence determine y ð3Þ . 13 If y ¼ sin x , show that 1 x2 (a) ð1 x2 Þy 00 4xy 0 ð1 þ x2 Þy ¼ 0 (b) y ðnþ2Þ ðn2 þ 3n þ 1Þy ðnÞ nðn 1Þy ðn2Þ ¼ 0 at x ¼ 0. 45 356 Programme 10 (b) Use the Leibnitz–Maclaurin method to determine series solutions for the following. 14 ð1 þ x2 Þy 00 þ xy 0 9y ¼ 0. 15 ðx þ 1Þy 00 þ ðx 1Þy 0 2y ¼ 0. 16 ð1 x2 Þy 00 7xy 0 9y ¼ 0. 17 ð1 x2 Þy 00 2xy 0 þ 2y ¼ 0. 18 xy 00 þ y 0 þ 2xy ¼ 0. (c) Solve the following Cauchy–Euler equi-dimensional equations. 20 x2 y 00 ðxÞ 5xy 0 ðxÞ þ 8yðxÞ ¼ x3 where yð1Þ ¼ 3 and yð2Þ ¼ 4 pffiffiffi 6x2 y 00 ðxÞ þ 19xy 0 ðxÞ þ 6yðxÞ ¼ 99x3 56x2 where yð1Þ ¼ 1 4 2 and yð8Þ ¼ 448 21 x2 y 00 ðxÞ 2yðxÞ ¼ 4x3 where yð1Þ ¼ 2 and yð2Þ ¼ 2 22 x2 y 00 ðxÞ þ xyðxÞ yðxÞ ¼ 3x2 where yð1Þ ¼ 1 and yð3Þ ¼ 19 17 3 Programme 11 Frames 1 to 54 Power series solutions of ordinary differential equations 2 Learning outcomes When you have completed this Programme you will be able to: . Apply Frobenius’ method of obtaining a series solution to a second-order homogeneous ordinary differential equation by first differentiating the assumed series several times . Substitute into the differential equation and equate coefficients of corresponding powers . Derive the indicial equation . Distinguish between the possible four distinct outcomes arising from the indicial equation 357 358 Programme 11 Introduction 1 In the previous Programme we established the solutions of second order ordinary differential equations as power series in integer powers of x. Such solutions are not always possible and a more general method is to assume a trial solution of the form y ¼ xc fa0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .g where a0 is the first coefficient that is not zero. The type of equation that can be solved by this method is of the form y 00 þ Py 0 þ Qy ¼ 0 where P and Q are functions of x. However, certain conditions have to be satisfied. (a) If the functions P and Q are such that both are finite when x is put equal to zero, x ¼ 0 is called an ordinary point of the equation. (b) If xP and x2 Q remain finite at x ¼ 0, then x ¼ 0 is called a regular singular point of the equation. In both of these cases, the method of Frobenius can be applied. (c) If, however, P and Q do not satisfy either of these conditions stated in (a) or (b), then x ¼ 0 is called an irregular singular point of the equation and the method of Frobenius cannot be applied. Solution of differential equations by the method of Frobenius To solve a given equation, we have to find the coefficients a0 ; a1 ; a2 ; . . . and also the index c in the trial solution. Basically, the steps in the method are as follows (a) Differentiate the trial series as required. (b) Substitute the results in the given differential equation. (c) Equate coefficients of corresponding powers of x on each side of the equation. The following examples will demonstrate the method – so move on 2 Example 1 Find a series solution for the equation d2 y dy þ y ¼ 0. þ dx2 dx The equation can be written as 2xy 00 þ y 0 þ y ¼ 0. 2x Assume a solution of the form y ¼ xc fa0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .g c ; y ¼ a0 x þ a1 x cþ1 þ a2 x cþ2 þ . . . þ ar x cþr þ ... Differentiating term by term, we get y0 ¼ ............ a0 6¼ 0: 359 Power series solutions of ordinary differential equations 2 3 y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxcþ1 þ . . . þ ar ðc þ rÞxcþr1 þ . . . Repeating the process one stage further, we have y 00 ¼ . . . . . . . . . . . . (give yourself plenty of room) 4 y 00 ¼ a0 cðc 1Þxc2 þ a1 cðc þ 1Þxc1 þ a2 ðc þ 1Þðc þ 2Þxc þ . . . þ ar ðc þ r 1Þðc þ rÞxcþr2 þ . . . So far, we have 2xy 00 þ y 0 þ y ¼ 0 y ¼ a0 xc þ a1 xc þ 1 þ a2 xc þ 2 þ . . . þ ar xc þ r þ . . . y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxc þ 1 þ . . . þ ar ðc þ rÞxc þ r1 þ . . . y 00 ¼ a0 cðc 1Þxc2 þ a1 cðc þ 1Þxc1 þ a2 ðc þ 1Þðc þ 2Þxc þ . . . þ ar ðc þ r 1Þðc þ rÞxc þ r2 þ . . . Considering each term of the equation in turn 2xy 00 ¼ 2a0 cðc 1Þxc1 þ 2a1 cðc þ 1Þxc þ 2a2 ðc þ 1Þðc þ 2Þxc þ 1 þ . . . þ 2ar ðc þ r 1Þðc þ rÞxc þ r1 þ . . . y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxc þ 1 þ . . . þ ar ðc þ rÞxc þ r1 þ . . . y ¼ a0 xc þ a1 xc þ 1 þ . . . þ ar xc þ r þ . . . Adding these three lines to form the left-hand side of the equation, we can equate the total coefficient of each power of x to zero, since the right-hand side is zero. c1 gives . . . . . . . . . . . . x ½xc1 : 5 2a0 cðc 1Þ þ a0 c ¼ 0 ; a0 cð2c 1Þ ¼ 0 a0 cð2c 1Þ ¼ 0 So, ½xc1 gives c Similarly, ½x gives . . . . . . . . . . . . ð1Þ 360 Programme 11 6 2a1 c ðc þ 1Þ þ a1 ðc þ 1Þ þ a0 ¼ 0 Simplifying, this becomes a1 ð2c2 þ 3c þ 1Þ þ a0 ¼ 0 i.e. a1 ðc þ 1Þð2c þ 1Þ þ a0 ¼ 0 cþ1 gives . . . . . . . . . . . . Also x 7 ð2Þ 2a2 ðc þ 1Þðc þ 2Þ þ a2 ðc þ 2Þ þ a1 ¼ 0 and this simplifies straight away to a2 ðc þ 2Þð2c þ 3Þ þ a1 ¼ 0 ð3Þ c Note that the coefficient of x involves all three lines of the expressions and, from then on, a general relationship can be obtained for xc þ r , r 0. In the expression for 2xy 00 and y 0 we have terms in xc þ r1 . If we replace r by ðr þ 1Þ, we shall obtain the corresponding terms in xc þ r . In the series for 2xy 00 , this is 2arþ1 ðc þ rÞðc þ r þ 1Þxc þ r In the series for y 0 , this is arþ1 ðc þ r þ 1Þxc þ r In the series for y, this is a r xc þ r Therefore, equating the total coefficient of xc þ r to zero, we have ............ 8 2arþ1 ðc þ rÞðc þ r þ 1Þ þ arþ1 ðc þ r þ 1Þ þ ar ¼ 0 and this tidies up to arþ1 fðc þ r þ 1Þð2c þ 2r þ 1Þg þ ar ¼ 0 ð4Þ Make a note of results (1), (2), (3) and (4): we shall return to them in due course. Then move on 9 Indicial equation Equation (1), formed from the coefficient of the lowest power of x, that is xc1 , is called the indicial equation from which the values of c can be obtained. In the present example a0 cð2c 1Þ ¼ 0 ; c ¼ ............ 361 Power series solutions of ordinary differential equations 2 c ¼ 0 or 10 1 , since a0 6¼ 0, by definition 2 Both values of c are valid, so that we have two possible solutions of the given equation. We will consider each in turn. (a) Using c ¼ 0 (2) gives a1 ð1Þð1Þ þ a0 ¼ 0 ; a1 ¼ a0 Similarly (3) gives . . . . . . . . . . . . 11 a2 ð2Þð3Þ þ a1 ¼ 0 a1 ¼ a0 and and from (4) a1 a0 ¼ 23 23 ar ¼ ðr þ 1Þð2r þ 1Þ a2 ¼ arþ1 r0 From the combined series, the term in xc and all subsequent terms involve all three lines and the coefficient of the general term can be used. ar So we have a1 ¼ a0 and arþ1 ¼ for r ¼ 0, 1, 2, . . . ðr þ 1Þð2r þ 1Þ a1 a0 ¼ ; a2 ¼ 23 23 a2 a0 a2 ¼ ¼ 3 5 ð2 3Þð3 5Þ a3 a0 ¼ etc. a4 ¼ 4 7 ð2 3 4Þð3 5 7Þ a0 a0 0 2 3 ; y ¼ x a0 a0 x þ x x þ ... ð2 3Þ ð2 3Þð3 5Þ x2 x3 x4 ; y ¼ a0 1 x þ þ þ ... ð2Þð3Þ ð2 3Þð3 5Þ ð2 3 4Þð3 5 7Þ Now we go through the same steps using our second value for c, i.e. c ¼ 12. Next frame (b) Using c ¼ 12 12 Our equations relating the coefficients were a0 cð2c 1Þ ¼ 0 which gave c ¼ 0 or c ¼ 12 ð1Þ a1 ðc þ 1Þð2c þ 1Þ þ a0 ¼ 0 ð2Þ a2 ðc þ 2Þð2c þ 3Þ þ a1 ¼ 0 ð3Þ arþ1 ðc þ r þ 1Þð2c þ 2r þ 1Þ þ ar ¼ 0 ð4Þ Putting c ¼ 12 in (2) gives . . . . . . . . . . . . 362 Programme 11 13 a1 ¼ Similarly (3) gives a2 ¼ a0 3 a1 a0 ¼ 10 3 10 and from the general relationship, (4), we have . . . . . . . . . . . . 14 arþ1 ¼ So i.e. 15 i.e. ar ðr þ 1Þð2r þ 3Þ a0 3 a1 a0 ¼ a2 ¼ 2 5 ð1 2Þð3 5Þ a2 a0 ¼ a3 ¼ 3 7 ð1 2 3Þð3 5 7Þ a3 a0 ¼ a4 ¼ 4 9 ð1 2 3 4Þð3 5 7 9Þ a1 ¼ etc. y ¼ xc fa0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .g y ¼ ............ a0 a0 a0 1 x2 x3 þ . . . y ¼ x2 a0 x þ 3 ð1 2Þð3 5Þ ð1 2 3Þð3 5 7Þ x x2 x3 1 þ þ ... y ¼ a0 x2 1 ð1 3Þ ð1 2Þð3 5Þ ð1 2 3Þð3 5 7Þ Since a0 is an arbitrary (non-zero) constant in each solution, its values may well be different, A and B say. If we denote the first solution by uðxÞ and the second by vðxÞ, then x2 x3 x4 þ þ ... u¼A 1xþ ð2 3Þ ð2 3Þð3 5Þ ð2 3 4Þð3 5 7Þ and 1 v ¼ B x2 1 x x2 x3 þ þ ... ð1 3Þ ð1 2Þð3 5Þ ð1 2 3Þð3 5 7Þ The general solution y ¼ u þ v is therefore . . . . . . . . . . . . 16 y ¼A 1xþ x2 x3 x 1 þ . . . þ B x2 1 ð1 3Þ ð2 3Þ ð2 3Þð3 5Þ x2 x3 þ þ ... ð1 2Þð3 5Þ ð1 2 3Þð3 5 7Þ Power series solutions of ordinary differential equations 2 363 The method may seem somewhat lengthy, but we have set it out in detail. It is a straightforward routine. Here is another example with the same steps. Example 2 Find the series solution for the equation 3x2 y 00 xy 0 þ y xy ¼ 0: We proceed in just the same way as in the previous example. Assume y ¼ xc fa0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .g i.e. y ¼ a0 xc þ a1 xc þ 1 þ a2 xc þ 2 þ . . . þ ar xc þ r þ . . . ; y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxcþ1 þ . . . þ ar ðc þ rÞxc þ r1 þ . . . y ¼ ............ 00 and y 00 ¼ a0 cðc 1Þxc2 þ a1 ðc þ 1Þcxc1 þ a2 ðc þ 2Þðc þ 1Þxc þ . . . 17 þ ar ðc þ rÞðc þ r 1Þxcþr2 þ . . . Now we build up the terms in the given equation. 3x2 y 00 ¼ 3a0 cðc 1Þxc þ 3a1 ðc þ 1Þcxc þ 1 þ 3a2 ðc þ 2Þðc þ 1Þxc þ 2 þ . . . þ 3ar ðc þ rÞðc þ r 1Þxc þ r þ . . . xy 0 ¼ a0 cxc a1 ðc þ 1Þxcþ1 a2 ðc þ 2Þxcþ2 . . . ar ðc þ rÞxc þ r . . . y ¼ a0 xc þ a1 xc þ 1 þ a2 xc þ 2 þ . . . þ ar xc þ r þ . . . xy ¼ a0 xc þ 1 a1 xc þ 2 . . . ar xc þ r þ 1 . . . The indicial equation, i.e. equating the coefficient of the lowest power of x to zero, gives the values of c. Thus, in this case c ¼ ............ c ¼ 1 or 1 3 Because the lowest power is xc and the coefficient of xc equated to zero gives 3a0 cðc 1Þ a0 c þ a0 ¼ 0 ; a0 ð3c2 4c þ 1Þ ¼ 0 ; c ¼ 1 or ; ð3c 1Þðc 1Þ ¼ 0 since a0 6¼ 0 1 3 The coefficient of the general term, i.e. xc þ r gives 3ar ðc þ rÞðc þ r 1Þ ar ðc þ rÞ þ ar ar1 ¼ 0 ; ar ¼ . . . . . . . . . . . . 18 364 19 Programme 11 ar ¼ ar1 2 3ðc þ rÞ 4ðc þ rÞ þ 1 ¼ ar1 ðc þ r 1Þð3c þ 3r 1Þ (a) Using c ¼ 1 the recurrence relation becomes ar1 ar ¼ rð3r þ 2Þ a0 ; r¼1 a1 ¼ 15 a1 a0 r¼2 a2 ¼ ¼ 2 8 ð1 2Þð5 8Þ a2 a0 ¼ r¼3 a3 ¼ 3 11 ð1 2 3Þð5 8 11Þ Our first solution is therefore y ¼ ............ 20 a0 x a0 x2 a 0 x3 þ þ þ ... y ¼ x a0 þ ð1 5Þ ð1 2Þð5 8Þ ð1 2 3Þð5 8 11Þ x x2 x3 þ þ þ ... ; y ¼ Ax 1 þ ð1 5Þ ð1 2Þð5 8Þ ð1 2 3Þð5 8 11Þ 1 (b) For the second solution, we put c ¼ 13 : The recurrence relation then becomes ar ¼ . . . . . . . . . . . . 21 ar ¼ ar1 rð3r 2Þ Therefore we can now determine the coefficients for r ¼ 1, 2, 3, . . . and complete the second solution. y ¼ ............ 365 Power series solutions of ordinary differential equations 2 1 y ¼ Bx3 1 þ x þ 22 x2 x3 þ ð2 4Þ ð2 3Þð4 7Þ x4 þ þ ... ð2 3 4Þð4 7 10Þ Because a0 a1 a0 ; a2 ¼ ¼ 11 24 ð1 2Þð2 4Þ a2 a0 ¼ a3 ¼ 3 7 ð2 3Þð4 7Þ a3 a0 ¼ a4 ¼ 4 10 ð2 3 4Þð4 7 10Þ x2 x3 1 ; y ¼ a0 x3 1 þ x þ þ ð2 4Þ ð2 3Þð4 7Þ a1 ¼ x4 þ ... þ ð2 3 4Þð4 7 10Þ Therefore, the general solution is y ¼ ............ x x2 x3 þ þ þ ... y ¼ Ax 1 þ ð1 5Þ ð1 2Þð5 8Þ ð1 2 3Þð5 8 11Þ x2 x3 x4 1 þ Bx3 1 þ x þ þ þ þ ... ð2 4Þ ð2 3Þð4 7Þ ð2 3 4Þð4 7 10Þ Example 3 Find the series solution for the equation d2 y y ¼0 dx2 i.e. y 00 y ¼ 0. As usual, we start off with the assumed solution y ¼ xc fa0 þ a1 x þ a2 x2 þ . . . þ ar xr þ . . .g i.e. y ¼ a0 xc þ a1 xcþ1 þ a2 xcþ2 þ . . . þ ar xcþr þ . . . ; y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxcþ1 þ . . . þ ar ðc þ rÞxcþr1 þ . . . y 00 ¼ a0 cðc 1Þxc2 þ a1 ðc þ 1Þcxc1 þ a2 ðc þ 2Þðc þ 1Þxc þ . . . þ ar ðc þ rÞðc þ r 1Þxcþr2 þ . . . These three expansions are required regularly, so make a note of them 23 366 24 Programme 11 Now we build up the terms in the left-hand side of the equation. y 00 ¼ a0 cðc 1Þxc2 þ a1 ðc þ 1Þcxc1 þ a2 ðc þ 2Þðc þ 1Þxc þ . . . þ ar ðc þ rÞðc þ r 1Þxcþr2 þ . . . y ¼ a0 xc þ a1 xcþ1 þ . . . þ ar xcþr þ . . . The term in xcþr in the first of these expansions is ............ 25 arþ2 ðc þ r þ 2Þðc þ r þ 1Þxcþr Because replacing r by ðr þ 2Þ in ar ðc þ rÞðc þ r þ 1Þxcþr2 gives this result. Then y 00 y ¼ . . . . . . . . . . . . 26 y 00 y ¼ a0 cðc 1Þxc2 þ a1 ðc þ 1Þcxc1 þ ½a2 ðc þ 2Þðc þ 1Þ a0 xc þ . . . þ ½arþ2 ðc þ r þ 2Þðc þ r þ 1Þ ar xcþr þ . . . We now equate each coefficient in turn to zero, since the right-hand side of the equation is zero. The coefficient of the lowest power of x gives the indicial equation from which we obtain the values of c. So, in this case, c ¼ ............ 27 c¼0 or 1 For the term in xc1 ; we have [xc1 ]: a1 ðc þ 1Þc ¼ 0. With c ¼ 1; a1 ¼ 0. But with c ¼ 0; a1 is indeterminate, because any value of a1 combined with the zero value of c would make the product zero. a0 a2 ðc þ 2Þðc þ 1Þ a0 ¼ 0 ; a2 ¼ [xc ]: ðc þ 1Þðc þ 2Þ For the general term [xcþr ]: ............ Power series solutions of ordinary differential equations 2 arþ2 ¼ ar ðc þ r þ 1Þðc þ r þ 2Þ 367 28 Because arþ2 ðc þ r þ 2Þðc þ r þ 1Þ ar ¼ 0. Hence the result above. From the indicial equation, c ¼ 0 or c ¼ 1. (a) When c ¼ 0 In general r¼1 r¼2 a1 is indeterminate a0 a2 ¼ 2 ar arþ2 ¼ ðr þ 1Þðr þ 2Þ a1 ; a3 ¼ 23 a2 a0 ¼ a4 ¼ 3 4 4! Therefore, one solution is . . . . . . . . . . . . n o a0 a1 a0 y ¼ x0 a0 þ a1 x þ x2 þ x3 þ x4 . . . 2! 3! 4! x2 x4 x3 x5 y ¼ a0 1 þ þ þ . . . þ a1 x þ þ þ . . . 2! 4! 3! 5! a0 and a1 are arbitrary constants depending on the boundary conditions. x2 x4 x3 x5 ; y ¼ A 1 þ þ þ ... þ B x þ þ þ ... 2! 4! 3! 5! i.e. Notice that these two series are the Maclaurin series expansions of the hyperbolic functions, so that y ¼ A cosh x þ B sinh x It is not very often the case that the series solution is so easily expressible in terms of known functions. (b) Similarly, when c ¼ 1 a1 ¼ 0 a0 23 ¼ ............ a2 ¼ arþ2 29 368 Programme 11 30 arþ2 ¼ r¼1 r¼2 r¼3 ar ðr þ 2Þðr þ 3Þ ; a1 ¼ 0 a0 a2 ¼ 3! a1 ¼0 a3 ¼ 34 a2 a0 a4 ¼ ¼ 4 5 5! a3 ¼ 0 etc. a5 ¼ 56 A second solution with c ¼ 1 is therefore y ¼ ............ 31 x3 x5 y ¼ a0 x þ þ þ . . . 3! 5! and, because a0 is an arbitrary constant x3 x5 x7 y ¼ C x þ þ þ þ ... 3! 5! 7! Note: This is not, in fact, a separate solution, since it already forms the second series in the solution for c ¼ 0 obtained previously. Therefore, the first solution, with its two arbitrary constants, A and B, gives the general solution. This happens when the two values of c differ by an integer. Make a note of the following: If the two values of c, i.e. c1 and c2 , differ by an integer, and if c ¼ c1 results in a1 being indeterminate, then this value of c gives the general solution. The solution resulting from c ¼ c2 is then merely a multiple of one of the series forming the first solution. Our last problem was an example of this. So far, we have met two distinct cases concerning the two roots c ¼ c1 and c ¼ c2 of the indicial equation. (a) If c1 and c2 differ by a quantity NOT an integer then two independent solutions, y ¼ uðxÞ and y ¼ vðxÞ, are obtained. The general solution is then y ¼ Au þ Bv. (b) If c1 and c2 differ by an integer, i.e. c2 ¼ c1 þ n, and if one coefficient ðar Þ is indeterminate when c ¼ c1 , the complete general solution is given by using this value of c. Using c ¼ c1 þ n gives a series which is a simple multiple of one of the series in the first solution. Make a note of these two points in your record book. Then move on Power series solutions of ordinary differential equations 2 There is a third category to be added to (a) and (b) above. 369 32 (c) If the roots c ¼ c1 and c ¼ c1 þ n of the indicial equation differ by an integer and one coefficient ðar Þ becomes infinite when c ¼ c1 , the series is rewritten with a0 replaced by kðc c1 Þ. Putting c ¼ c1 in the rewritten series and that of its derivative with respect to c gives two independent solutions. Add this to the previous two. Then we will see how it works in practice Example 4 33 Find the series solution of the equation xy 00 þ ð2 þ xÞy 0 2y ¼ 0. Using y ¼ xc ða0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .Þ and its first two derivatives, the expansions for xy 00 ¼ . . . . . . . . . . . . 2y 0 ¼ . . . . . . . . . . . . xy 0 ¼ . . . . . . . . . . . . 2y ¼ . . . . . . . . . . . . Method as before. xy 00 ¼ a0 cðc 1Þxc1 þ a1 ðc þ 1Þcxc þ a2 ðc þ 2Þðc þ 1Þxcþ1 þ . . . þ ar ðc þ rÞðc þ r 1Þx 0 c1 2y ¼ 2a0 cx c cþr1 þ 2a1 ðc þ 1Þx þ 2a2 ðc þ 2Þx 34 þ ... cþ1 þ 2a3 ðc þ 3Þxcþ2 þ . . . þ 2ar ðc þ rÞxcþr1 þ . . . xy 0 ¼ a0 cxc þ a1 ðc þ 1Þxcþ1 þ a2 ðc þ 2Þccþ2 þ . . . þ ar ðc þ rÞxcþr þ . . . 2y ¼ 2a0 xc 2a1 xcþ1 2a2 xcþ2 2a3 xcþ3 . . . 2ar xcþr . . . From which, the indicial equation is . . . . . . . . . . . . a0 ðc2 þ cÞ ¼ 0 i.e. equating the coefficient of the lowest power of x, ðxc1 Þ, to zero. a0 6¼ 0 ; c ¼ 0 or 1 Also, from the expansions, the total coefficient of xc gives a1 ¼ . . . . . . . . . . . . 35 370 Programme 11 36 a1 ¼ a0 ðc 2Þ ðc þ 1Þðc þ 2Þ From the terms in xc , all four expansions are involved, so we can form the recurrence relation from the coefficient of xcþr . arþ1 ¼ . . . . . . . . . . . . 37 arþ1 ¼ ar ðc þ r 2Þ ðc þ r þ 1Þðc þ r þ 2Þ Because arþ1 ðc þ r þ 1Þðc þ rÞ þ 2arþ1 ðc þ r þ 1Þ þ ar ðc þ rÞ 2ar ¼ 0 arþ1 ðc þ r þ 1Þðc þ r þ 2Þ þ ar ðc þ r 2Þ ¼ 0 ; arþ1 ¼ ar ðc þ r 2Þ ðc þ r þ 1Þðc þ r þ 2Þ r0 ; a2 ¼ . . . . . . . . . . . . 38 a2 ¼ a0 ðc 1Þðc 2Þ ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ and, from the recurrence relation, when r ¼ 2 a3 ¼ . . . . . . . . . . . . 39 a3 ¼ ( ; y ¼ a0 x c a0 cðc 1Þðc 2Þ ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ c2 ðc 1Þðc 2Þ xþ x2 ðc þ 1Þðc þ 2Þ ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ ) cðc 1Þðc 2Þ 3 x þ ... ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ 1 From the indicial equation above, the values of c are 0 and 1. Putting c = 0, we have one solution y ¼ u ¼ ............ 371 Power series solutions of ordinary differential equations 2 x2 y ¼ u ¼ a0 1 þ x þ 6 40 Note that coefficients after the x2 term are zero, because of the factor c in the numerator. Putting c ¼ 1, we soon find that . . . . . . . . . . . . 41 coefficients become infinite, because of the factor ðc þ 1Þ in the denominator. Therefore, we substitute a0 ¼ kðc c1 Þ ¼ kðc ½1Þ ¼ kðc þ 1Þ. ( c2 ðc 1Þðc 2Þ c xþ x2 ; y ¼ kðc þ 1Þx 1 ðc þ 1Þðc þ 2Þ ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ ( c ¼ kx ðc þ 1Þ cðc 1Þðc 2Þ ðc þ 1Þðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ c2 ðc 1Þðc 2Þ 2 x xþ cþ2 ðc þ 2Þ2 ðc þ 3Þ cðc 1Þðc 2Þ ðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ ) 3 x þ ... ) 3 x þ ... Now, putting c ¼ 1: y ¼ ............ x3 y ¼ kx1 3x þ 3x2 þ 2 All subsequent terms are zero, since the numerators all contain a factor ðc þ 1Þ. x2 ; y ¼ v ¼ 3 þ 3x þ 2 is a solution. 42 372 Programme 11 A solution is also given by So, starting from ( y ¼ kxc ðc þ 1Þ @y ¼ kxc ln x @c ( @y ¼ 0: @c c2 ðc 1Þðc 2Þ 2 xþ x cþ2 ðc þ 2Þ2 ðc þ 3Þ cðc 1Þðc 2Þ ðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ ) 3 x þ ... c2 ðc 1Þðc 2Þ 2 xþ x cþ2 ðc þ 2Þ2 ðc þ 3Þ ) cðc 1Þðc 2Þ 3 x þ ... ðc þ 2Þ2 ðc þ 3Þ2 ðc þ 4Þ ( ) c2 ðc 1Þðc 2Þ 2 c @ ðc þ 1Þ xþ x ... þ kx @c cþ2 ðc þ 2Þ2 ðc þ 3Þ ðc þ 1Þ We now have to determine the partial derivative of each term. @ ðc þ 1Þ ¼ 1 @c @ c2 ¼ ............ @c c þ 2 @ c2 4 ¼ @c c þ 2 ðc þ 2Þ2 43 Now we have to differentiate Let t ¼ ðc 1Þðc 2Þ ðc þ 2Þ2 ðc þ 3Þ ðc 1Þðc 2Þ ðc þ 2Þ2 ðc þ 3Þ ; ln t ¼ lnðc 1Þ þ lnðc 2Þ 2 lnðc þ 2Þ lnðc þ 3Þ 1 @t 1 1 2 1 ¼ þ ; t @c c 1 c 2 c þ 2 c þ 3 @t ðc 1Þðc 2Þ 1 1 2 1 ¼ þ ; @c ðc þ 2Þ2 ðc þ 3Þ c 1 c 2 c þ 2 c þ 3 @ ðc þ 1Þ ¼ 1 @c @ c2 ¼4 @c c þ 2 ( ) @ ðc 1Þðc 2Þ ¼ ............ @c ðc þ 2Þ2 ðc þ 3Þ ; when c ¼ 1; Power series solutions of ordinary differential equations 2 10 373 44 Therefore, when c ¼ 1: @y x3 ¼ kx1 ln x 0 þ 3x þ 3x2 þ þ . . . @c 2 þ kx1 f1 4x 10x2 þ . . .g ; Another solution is x2 y ¼ w ¼ C ln x 3 þ 3x þ þ . . . þx1 ð1 4x 10x2 þ . . .Þ 2 Now we have a problem, for we seem to have three separate series solutions for a second-order differential equation. x2 (a) y ¼ u ¼ A 1 þ x þ 6 x2 (b) y ¼ v ¼ B 3 þ 3x þ 2 x2 (c) y ¼ w ¼ C ln x 3 þ 3x þ þ . . . þ x1 ð1 4x 10x3 þ . . .Þ 2 But (b) is clearly a simple multiple of (a) and thus not a distinct solution. So finally, we have just (a) and (c). x2 i.e. y ¼u¼A 1þxþ 6 x2 y ¼ w ¼ B ln x 3 þ 3x þ þ . . . þ x1 ð1 4x 10x3 þ . . .Þ and 2 The complete solution is then y ¼uþw In general if c1 c2 ¼ n where n is a non-zero integer the solution is of the form: y ¼ ð1 þ k ln xÞxc1 a0 þ a1 x þ a2 x2 þ . . . þ xc2 b0 þ b1 x þ b2 x2 þ . . . Finally we have just one more variation to the list in Frames 31 and 32, so move on Example 5 45 Solve the equation xy 00 þ y 0 xy ¼ 0. Start off as before and build up expansions for the terms in the left-hand side of the equation. xy 00 ¼ . . . . . . . . . . . . y0 ¼ ............ xy ¼ . . . . . . . . . . . . 374 Programme 11 46 xy 00 ¼ a0 cðc 1Þxc1 þ a1 ðc þ 1Þcxc þ a2 ðc þ 2Þðc þ 1Þxcþ1 þ . . . þ ar ðc þ rÞðc þ r 1Þxcþr1 þ . . . y 0 ¼ a0 cxc1 þ a1 ðc þ 1Þxc þ a2 ðc þ 2Þxcþ1 þ . . . þ ar ðc þ rÞxcþr1 þ . . . a0 xcþ1 a1 xcþ2 . . . xy ¼ ar xcþrþ1 . . . The indicial equation, therefore, gives c ¼ . . . . . . . . . . . . 47 c¼0 Because a0 fcðc 1Þ þ cg ¼ 0 (twice) a0 6¼ 0 ; c2 ¼ 0 ; c ¼ 0 (twice) Coefficient of xc gives . . . . . . . . . . . . 48 a1 ¼ 0 [xc ]: [x cþ1 ; a1 ðc þ 1Þ2 ¼ 0 a1 ðc2 þ c þ c þ 1Þ ¼ 0 ; a1 ¼ 0 ]: [xcþr1 ]: ar fðc þ rÞðc þ r 1Þ þ ðc þ rÞg ar2 ¼ 0 ar2 ; ar ðc þ rÞ2 ¼ ar2 ; ar ¼ ðc þ rÞ2 ; y ¼ ............ ( 49 y¼x c ( i.e. y ¼ a0 x c 1þ a0 þ a0 ðc þ 2Þ2 x2 ðc þ 2Þ2 þ x þ 4 ðc þ 2Þ2 ðc þ 4Þ2 x4 ðc þ 2Þ2 ðc þ 4Þ2 ; When c ¼ 0 x2 x4 y ¼u¼A 1þ 2þ 2 þ . . . 2 2 42 ) a0 2 x þ ... ) þ ... ð1Þ 375 Power series solutions of ordinary differential equations 2 This is one solution. Another is given by v ¼ @y @c ( ) @y x2 x4 c ¼ a0 x ln x 1 þ þ þ ... @c ðc þ 2Þ2 ðc þ 2Þ2 ðc þ 4Þ2 ( ) @ x2 x4 1þ þ a0 x þ þ ... @c ðc þ 2Þ2 ðc þ 2Þ2 ðc þ 4Þ2 c @ ð1Þ ¼ 0; @c Now Let t ¼ ( ) @ 1 2 ¼ @c ðc þ 2Þ2 ðc þ 2Þ3 1 ðc þ 2Þ2 ðc þ 4Þ2 ; ln t ¼ 2 lnðc þ 2Þ 2 lnðc þ 4Þ 1 @t 2 2 @t 2 1 1 ; ¼ ; ¼ þ t @c c þ 2 c þ 4 @c ðc þ 2Þ2 ðc þ 4Þ2 c þ 2 c þ 4 ( ) @y x2 x4 c ¼ a0 x ln x 1 þ þ þ ... ; @c ðc þ 2Þ2 ðc þ 2Þ2 ðc þ 4Þ2 ( þ a0 x c 0 2x2 ðc þ 2Þ3 4x4 ðc þ 3Þ ðc þ 2Þ3 ðc þ 4Þ3 ) þ ... ; When c ¼ 0 y ¼ v ¼ ............ x2 x4 x2 3x4 y ¼ v ¼ B ln x 1 þ 2 þ 2 þ . . . þ . . . 2 2 42 22 23 42 (2) So our two solutions are y ¼ u (at 1) and y ¼ v (at 2). The complete solution is therefore y ¼ u þ v. In general if c1 ¼ c2 ¼ c the solution is of the form y ¼ ð1 þ k ln xÞxc a0 þ a1 x þ a2 x2 þ . . . þ xc b1 x þ b2 x2 þ . . . The main points that we have covered in this Programme are listed in the Revision summary that follows. Read this in conjunction with the Can you? checklist and note any sections that may need further attention: refer back to the relevant parts of the Programme, if necessary. There will then be no trouble with the Test exercise. The set of Further examples provides an opportunity for further practice 50 376 Programme 11 Revision summary 11 51 Frobenius’ method Assume a series solution of the form y ¼ xc fa0 þ a1 x þ a2 x2 þ . . . þ ar xr þ . . .g a0 6¼ 0 (1) Differentiate the assumed series to find y 0 and y 00 . (2) Substitute in the equation. (3) Equate coefficients of corresponding powers of x on each side of the equation – usually written with zero on the right-hand side. (4) Coefficient of the lowest power of x gives the indicial equation from which values of c are obtained, c ¼ c1 and c ¼ c2 . Case 1: c1 and c2 differ by a quantity not an integer. Substitute c ¼ c1 and c ¼ c2 in the series for y. Case 2: c1 and c2 differ by an integer and make a coefficient indeterminate when c ¼ c1 . Substitute c ¼ c1 to obtain the complete solution. Case 3: c1 and c2 ðc1 < c2 Þ differ by an integer and make a coefficient infinite when c ¼ c1 . Replace a0 by kðc c1 Þ. Two independent solutions are then @y obtained by putting c ¼ c1 in the new series for y and for . @c In general if c1 c2 ¼ n where n is a non-zero integer, the solution is of the form y ¼ ð1 þ k ln xÞxc1 a0 þ a1 x þ a2 x2 þ . . . þ xc2 b0 þ b1 x þ b2 x2 þ . . . Case 4: c1 and c2 are equal. Substitute c ¼ c1 in the series for y and for the substitution after differentiating. In general if c1 ¼ c2 ¼ c, the solution is of the form y ¼ ð1 þ k ln xÞxc a0 þ a1 x þ a2 x2 þ . . . þ xc b1 x þ b2 x2 þ . . . @y . Make @c 377 Power series solutions of ordinary differential equations 2 Can you? 52 Checklist 11 Check this item before and after you try the end of Programme test On a scale of 1 to 5 how confident are you that you can: . Apply Frobenius’ method of obtaining a series solution to a second-order homogeneous ordinary differential equation by first differentiating the assumed series several times? Yes No . Substitute into the differential equation and equate coefficients of corresponding powers? Yes No . Derive the indicial equation? Yes Frames 1 to 4 5 to 8 9 No . Distinguish between the possible four distinct outcomes arising from the indicial equation? Yes No 10 to 50 Test exercise 11 1 Determine a series solution for each of the following (a) 3xy 00 þ 2y 0 þ y ¼ 0 53 (b) y 00 þ x2 y ¼ 0 (c) xy 00 þ 3y 0 y ¼ 0 Further problems 11 1 Use the method of Frobenius to obtain a series solution for each of the following (a) 3xy 00 þ y 0 y ¼ 0 (b) y 00 þ y ¼ 0 (c) y 00 xy ¼ 0 (d) 3xy 00 þ 4y 0 þ y ¼ 0 (e) y 00 xy 0 þ y ¼ 0 (f) xy 00 3y 0 þ y ¼ 0 (g) xy 00 þ y 0 3y ¼ 0 54 Programme 12 Frames 1 to 44 Power series solutions of ordinary differential equations 3 Learning outcomes When you have completed this Programme you will be able to: . Apply Frobenius’ method to Bessel’s equation to derive Bessel functions of the first kind . Apply Frobenius’ method to Legendre’s equation to derive Legendre polynomials . Use Rodrigue’s formula to derive Legendre polynomials and the generating function to obtain some of their properties . Recognise a Sturm–Liouville system and the orthogonality properties of its eigenfunctions . Write a polynomial in x as a finite series of Legendre polynomials 378 Power series solutions of ordinary differential equations 3 379 Introduction A common feature of certain differential equations is that they appear in a multiplicity of guises in the application of mathematics to problems in physics and engineering. For example, Bessel’s equation appears in the study of electromagnetic radiation, heat conduction, vibrational modes of a membrane and signal processing to name but a few. Many of these equations have solutions (called special functions) in the form of infinite series that are accessible by the method of Frobenius and in this Programme we shall consider two of these equations, namely Bessel’s equation and Legendre’s equation. Move to the next frame 1 A second-order differential equation that occurs frequently in branches of technology is of the form 2 x2 y 00 þ xy 0 þ ðx2 v 2 Þy ¼ 0 where v is a real constant. Starting with y ¼ xc ða0 þ a1 x þ a2 x2 þ a3 x3 þ . . . þ ar xr þ . . .Þ and proceeding as before with the Frobenius method of Programme 11, we obtain c ¼ v and a1 ¼ 0 ar ¼ The recurrence relation is ar2 v 2 ðc þ rÞ2 for r 2. It follows that a1 ¼ a3 ¼ a5 ¼ a7 ¼ . . . ¼ 0 and that a2 ¼ . . . . . . . . . . . . ; a2 ¼ a0 v2 a6 ¼ h ðc þ 2Þ 2 ; a4 ¼ . . . . . . . . . . . . ; a4 ¼ h v2 a6 ¼ . . . . . . . . . . . . a i 0h i; v 2 ðc þ 4Þ2 ðc þ 2Þ 2 a0 ih ih i v 2 ðc þ 4Þ2 v 2 ðc þ 6Þ2 v 2 ðc þ 2Þ ; When c ¼ þv 2 a2 ¼ . . . . . . . . . . . . ; a6 ¼ . . . . . . . . . . . . ; a4 ¼ . . . . . . . . . . . . ar ¼ . . . . . . . . . . . . 3 380 4 Programme 12 a2 ¼ a0 ; þ 1Þ 22 ðv a4 ¼ 24 a0 2ðv þ 1Þðv þ 2Þ a6 ¼ a0 26 3!ðv þ 1Þðv þ 2Þðv þ 3Þ ar ¼ ð1Þr=2 a0 for r even 2r ðr=2Þ!ðv þ 1Þðv þ 2Þ . . . ðv þ r=2Þ The resulting series solution is therefore y ¼ u ¼ ............ 5 y ¼ u ¼ Axv 1 x2 x4 þ 22 ðv þ 1Þ 24 2!ðv þ 1Þðv þ 2Þ x6 6 þ ... 2 3!ðv þ 1Þðv þ 2Þðv þ 3Þ This is valid provided v is not a negative integer. Similarly, when c ¼ v x2 x4 y ¼ w ¼ Bxv 1 þ 2 þ 4 2 ðv 1Þ 2 2!ðv 1Þðv 2Þ x6 þ ... þ 6 2 3!ðv 1Þðv 2Þðv 3Þ This is valid provided v is not a positive integer. Except for these two restrictions, the complete solution of Bessel’s equation is therefore y ¼ u þ w with the two arbitrary constants A and B. 6 Bessel functions It is convenient to present the two results obtained above in terms of the gamma function ðxÞ where the letter is the capital Greek gamma. We shall deal with gamma functions in more detail in Programme 16 but for now we only require two simple properties of gamma functions, namely For x > 0, ðx þ 1Þ ¼ xðxÞ and ð1Þ ¼ 1 What happens when x 0 or what ðxÞ looks like in terms of a general variable x does not matter for now. What is important is that for x > 0 these simple properties give rise to the following equations: ðx þ 1Þ ¼ xðxÞ ðx þ 2Þ ¼ ðx þ 1Þðx þ 1Þ ¼ ðx þ 1ÞxðxÞ ðx þ 3Þ ¼ ðx þ 2Þðx þ 2Þ ¼ ðx þ 2Þðx þ 1ÞxðxÞ, etc. 381 Power series solutions of ordinary differential equations 3 Then if x ¼ 1 ð1 þ 1Þ ¼ 1 þ ð1Þ ¼ 1 ð1 þ 2Þ ¼ ð1 þ 1Þð1 þ 1Þ ¼ 2 1 ð1Þ ¼ 2 1 ð1 þ 3Þ ¼ ð1 þ 2Þð1 þ 2Þ ¼ 3 2 1 ð1 þ 4Þ ¼ . . . . . . . . . . . . Next frame 7 4 3 2 1 ¼ 4! Because: ð1 þ 4Þ ¼ ð1 þ 3Þð1 þ 3Þ ¼ 4 3 2 1 ¼ 4! And so, if x ¼ 1 and n is a positive integer ð1 þ nÞ ¼ . . . . . . . . . . . . The answer is in the next frame 8 n ðn 1Þ ð. . .Þ 2 1 ¼ n! Because: ð1 þ nÞ ¼ ð1 þ ½n 1Þ ð1 þ ½n 1Þ ¼ n ð1 þ ½n 1Þ ¼ n ð1 þ ½n 2Þ ð1 þ ½n 2Þ ¼ n ðn 1Þ ð1 þ ½n 2Þ ¼ ............ ¼ n ðn 1Þ ð. . .Þ ð1 þ ½n nÞ ¼ n ðn 1Þ ð. . .Þ ð1Þ ¼ n ðn 1Þ ð. . .Þ 1 ¼ n! If now, in Frame 3, we assign to the arbitrary constant a0 the value 1 2v ðv then we have, for c ¼ v a2 ¼ a0 v2 2 ¼ a0 a0 ¼ ðv c 2Þðv þ c þ 2Þ 2ð2v þ 2Þ ðc þ 2Þ 1 1 1 ¼ ¼ 2 2 ðv þ 1Þ 2v ðv þ 1Þ 2vþ2 ð1!Þðv þ 2Þ Similarly a4 ¼ . . . . . . . . . . . . , þ 1Þ 382 Programme 12 9 a4 ¼ 1 2vþ4 ð2!Þðv þ 3Þ Because a2 a2 a2 ¼ ¼ v 2 ðc þ 4Þ2 ðv c 4Þðv þ c þ 4Þ 4ð2v þ 4Þ 1 1 1 ¼ ¼ 3 2 ðv þ 2Þ 2vþ2 ð1!Þðv þ 2Þ 2vþ4 ð2!Þðv þ 3Þ a4 ¼ and a6 ¼ . . . . . . . . . . . . 10 a6 ¼ 1 2vþ6 ð3!Þðv þ 4Þ We can see the pattern taking shape. ar ¼ ð1Þr=2 r for r even: r 2vþr ! v þ þ 1 2 2 ; Put r ¼ 2k The result then becomes a2k ¼ . . . . . . . . . . . . 11 a2k ¼ ð1Þk 2vþ2k ðk!Þðv þ k þ 1Þ k ¼ 1, 2, 3, . . . Therefore, we can write the new form of the series for y as 1 x2 x4 y ¼ xv v vþ2 þ vþ4 ... 2 ðv þ 1Þ 2 ð1!Þðv þ 2Þ 2 ð2!Þðv þ 3Þ This is called the Bessel function of the first kind of order v and is denoted by Jv ðxÞ. xv 1 x2 x4 þ ... ; Jv ðxÞ ¼ 2 ðv þ 1Þ 22 ð1!Þðv þ 2Þ 24 ð2!Þðv þ 3Þ This is valid provided v is not . . . . . . . . . . . . 12 a negative integer – otherwise some of the terms would become infinite. If we take the other value for c, i.e. c ¼ v, the corresponding result becomes Jv ðxÞ ¼ . . . . . . . . . . . . Power series solutions of ordinary differential equations 3 Jv ðxÞ ¼ v x 1 x2 x4 þ 2 ... 2 ð1 vÞ 2ð1!Þð2 vÞ 2 ð2!Þð3 vÞ 383 13 provided that v is not a positive integer. In general terms xv X 1 ð1Þk x2k Jv ðxÞ ¼ 2 k¼0 22k ðk!Þðv þ k þ 1Þ Jv ðxÞ ¼ 1 xv X 2 k¼0 ð1Þk x2k v þ 1Þ 22k ðk!Þðk The convergence of the series for all values of x can be established by the normal ratio test. Jv ðxÞ and Jv ðxÞ are two independent solutions of the original equation. Hence, the complete solution is y ¼ AJv ðxÞ þ BJv ðxÞ where A and B are constants. Make a note of the expressions for Jv ðxÞ and Jv ðxÞ. Then on to the next frame Some Bessel functions are commonly used and are worthy of special mention. This arises when v is a positive integer, denoted by n. x n X 1 ð1Þk x2k ; Jn ðxÞ ¼ 2 k¼0 22k ðk!Þðn þ k þ 1Þ 14 From our work on gamma functions, ðk þ 1Þ ¼ k! for k ¼ 0, 1, 2, . . . ; ðn þ k þ 1Þ ¼ ðn þ kÞ! and the result above then becomes Jn ðxÞ ¼ . . . . . . . . . . . . Jn ðxÞ ¼ x n X 1 2 k¼0 ð1Þk x2k 22k ðk!Þðn þ kÞ! We have seen that Jv ðxÞ and Jv ðxÞ are two solutions of Bessel’s equation. When v and v are not integers, the two solutions are independent of each other. Then y ¼ AJv ðxÞ þ BJv ðxÞ. When, however, v ¼ n (integer), then Jn ðxÞ and Jn ðxÞ are not independent, but are related by Jn ðxÞ ¼ ð1Þn Jn ðxÞ. This can be shown by referring once again to our knowledge of gamma functions. ðx þ 1Þ x and for negative integral values of x, or zero, ðxÞ is infinite. ðx þ 1Þ ¼ xðxÞ ; ðxÞ ¼ 15 384 Programme 12 From the previous result: Jv ðxÞ ¼ xv X 1 2 k¼0 ð1Þk x2k 22k ðk!Þðk v þ 1Þ k ¼ 0, 1, 2, . . . Let us consider the gamma function ðk v þ 1Þ in the denominator and let v approach closely to a positive integer n. Then ðk v þ 1Þ ! ðk n þ 1Þ. When k n þ 1 0, i.e. when k ðn 1Þ, then ðk n þ 1Þ is infinite. The first finite value of ðk n þ 1Þ occurs for k ¼ n. When values of ðk v þ 1Þ are infinite the coefficients of Jv ðxÞ are ............ 16 zero The series, therefore, starts at k ¼ n n X 1 x ð1Þk x2k ; Jn ðxÞ ¼ 2k 2 2 ðk!Þðk n þ 1Þ k¼n ¼ 1 X k¼n ¼ 1 X p¼0 ð1Þk x2kn n þ 1Þ 22kn ðk!Þðk Put k ¼ p þ n ð1Þpþn x2pþn 22pþn ðk!Þðk nÞ! 1 X ¼ ð1Þn p¼0 ð1Þp x2pþn 22pþn ðp!Þðp þ nÞ! x n X 1 ð1Þp x2p ¼ ð1Þn 2 p¼0 22p ðp!Þðp þ nÞ! ¼ ð1Þn x n X 1 2 k¼0 ð1Þk x2k 22k ðk!Þðk þ nÞ! n ; Jn ðxÞ¼ ð1Þ Jn ðxÞ So, after all that, the series for Jn ðxÞ ¼ . . . . . . . . . . . . 17 Jn ðxÞ ¼ xn 1 2 x2 x4 1 1 þ ............ n! ðn þ 1Þ! 2 ð2!Þðn þ 2Þ! 2 From this we obtain two commonly used functions J0 ðxÞ ¼ . . . . . . . . . . . . 385 Power series solutions of ordinary differential equations 3 J0 ðxÞ ¼ 1 18 1 x 2 1 x 4 1 x 6 þ þ... ð1!Þ2 2 ð2!Þ2 2 ð3!Þ2 2 J1 ðxÞ ¼ . . . . . . . . . . . . and J1 ðxÞ ¼ x 1 x 2 1 x4 1 þ þ... 2 ð1!Þð2!Þ 2 ð2!Þð3!Þ 2 19 Bessel functions for a range of values of n and x are tabulated in published lists of mathematical data. Of these, J0 ðxÞ and J1 ðxÞ are most commonly used. 20 Graphs of Bessel functions J0(x) and J1(x) y y= J0 (x) y= J1(x) x Legendre’s equation Another equation of special interest in engineering applications is Legendre’s equation of the form ð1 x2 Þy 00 2xy 0 þ kðk þ 1Þy ¼ 0 where k is a real constant. This may be solved by the Frobenius method as before. In this case, the indicial equation gives c = 0 and c = 1, and the two corresponding solutions are kðk þ 1Þ 2 kðk 2Þðk þ 1Þðk þ 3Þ 4 x þ x ... (a) c ¼ 0: y ¼ a0 1 2! 4! ðk 1Þðk þ 2Þ 3 x (b) c ¼ 1: y ¼ a1 x 3! ðk 1Þðk 3Þðk þ 2Þðk þ 4Þ 5 x ... þ 5! where a0 and a1 are the usual arbitrary constants 21 386 Programme 12 Legendre polynomials When k is an integer (n), one of the solution series terminates after a finite number of terms. The resulting polynomial in x, denoted by Pn ðxÞ, is called a Legendre polynomial, with a0 or a1 being chosen so that the polynomial has unit value when x ¼ 1. P2 ðxÞ ¼ . . . . . . . . . . . . For example 22 P2 ðxÞ ¼ 1 ð3x2 1Þ 2 Because, in P2 ðxÞ, n ¼ k ¼ 2 23 2 x þ 0 þ 0 þ ... ; y ¼ a0 1 2! ¼ a0 f1 3x2 g The constant a0 is then chosen to make y ¼ 1 when x ¼ 1 i.e. 1 ¼ a0 ð1 3Þ ; a0 ¼ 12 ; P2 ðxÞ ¼ 12 ð1 3x2 Þ ¼ 12 ð3x2 1Þ P3 ðxÞ ¼ . . . . . . . . . . . . Similarly 23 P3 ðxÞ ¼ 1 ð5x3 3xÞ 2 Here n ¼ k ¼ 3 25 3 x þ 0 þ 0 þ ... ; y ¼ a1 x 3! 5x3 ¼ a1 x 3 5 3 ¼ 1 ; a1 ¼ y ¼ 1 when x ¼ 1 ; a1 1 3 2 3 3 5x 1 x ; P3 ðxÞ ¼ ¼ ð5x3 3xÞ 2 2 3 24 Rodrigue’s formula and the generating function Legendre polynomials can be derived by using Rodrigue’s formula Pn ðxÞ ¼ 1 dn 2 x 1 2n n! dxn n so using this formula P4 ðxÞ ¼ . . . . . . . . . . . . Power series solutions of ordinary differential equations 3 P4 ðxÞ ¼ 1 35x4 30x2 þ 3 8 387 25 Because 1 d4 2 4 x 1 24 4! dx4 1 d4 8 ¼ x 4x6 þ 6x4 4x2 þ 1 384 dx4 1 d3 7 8x 24x5 þ 24x3 8x ¼ 384 dx3 1 d2 56x6 120x4 þ 72x2 8 ¼ 384 dx2 1 d 336x5 480x3 þ 144x ¼ 384 dx 1 1680x4 1440x2 þ 144 ¼ 384 1 ¼ 35x4 30x2 þ 3 8 P4 ðxÞ ¼ In addition to Rodrigue’s formula, the function 1 X 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Pn ðxÞt n , jtj < 1 1 2xt þ t 2 n¼0 is called the generating function for Legendre polynomials and can be used to obtain some of their properties. For example using this generating function we find that Pn ð1Þ ¼ . . . . . . . . . . . . Pn ð1Þ ¼ 1 Because When x ¼ 1 the generating function becomes 1 X 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Pn ð1Þt n , jtj < 1 1 2t þ t 2 n¼0 1 1 1 ¼ ð1 tÞ1 , the left-hand side Noting that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 1t 2 1 2t þ t ð1 tÞ is expanded by the binomial theorem to give 1 X ð1 tÞ1 ¼ 1 þ t þ t 2 þ t 3 þ . . . ¼ tn. n¼0 Therefore 1 X tn ¼ n¼0 1 X Pn ð1Þt n and so Pn ð1Þ ¼ 1 n¼0 By a similar reasoning Pn ð1Þ ¼ . . . . . . . . . . . . 26 388 Programme 12 27 Pn ð1Þ ¼ ð1Þn Because When x ¼ 1 the generating function becomes 1 X 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Pn ð1Þt n 1 þ 2t þ t 2 n¼0 1 1 1 ¼ ð1 þ tÞ1 , the left-hand side Noting that pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 1þt 2 1 þ 2t þ t ð1 þ tÞ is expanded by the binomial theorem to give ð1 þ tÞ1 ¼ 1 t þ t 2 t 3 þ . . . ¼ 1 X ð1Þn t n . Therefore n¼0 1 1 X X ð1Þn t n ¼ Pn ð1Þt n and so Pn ð1Þ ¼ ð1Þn n¼0 n¼0 Legendre’s equation, whose solutions are expressed in terms of Legendre polynomials, is an example of a particular class of differential equations referred to as Sturm–Liouville systems. In the following frames we shall look at such systems more closely. So on to the next frame Sturm–Liouville systems 28 A boundary value problem that is described by a differential equation of the general form 0 ðpðxÞy 0 Þ þðqðxÞ þ rðxÞÞy ¼ 0 for a x b and rðxÞ > 0 where the boundary conditions can be written in the form 1 yðaÞ þ 2 y 0 ðaÞ ¼ 0 and 1 yðbÞ þ 2 y 0 ðbÞ ¼ 0 is called a Sturm–Liouville system. Solutions of such a system are in the form of an infinite sequence of eigenfunctions yn , each corresponding to an eigenvalue n of the system for n ¼ 0, 1, 2, . . .. For example, consider the differential equation y 00 þ y ¼ 0 for 0 x 5 where here, a ¼ 0 and b ¼ 5. Also yð0Þ ¼ 0 and yð5Þ ¼ 0 By comparing this equation with the general form given above we can see that pðxÞ ¼ . . . . . . . . . . . . ; qðxÞ ¼ . . . . . . . . . . . . ; rðxÞ ¼ . . . . . . . . . . . . ; 2 ¼ . . . . . . . . . . . . ; 2 ¼ . . . . . . . . . . . . 389 Power series solutions of ordinary differential equations 3 pðxÞ ¼ 1; qðxÞ ¼ 0; rðxÞ ¼ 1; 2 ¼ 0; 2 ¼ 0 29 Because By performing the differentiation on the left-hand term of 0 ðpðxÞy 0 Þ þðqðxÞ þ rðxÞÞy ¼ 0 we find that the differential equation can be written as pðxÞy 00 þ p0 ðxÞy 0 þ ðqðxÞ þ rðxÞÞy ¼ 0 By inspection, comparing this form with the differential equation y 00 þ y ¼ 0 it is easily seen that pðxÞ ¼ 1, qðxÞ ¼ 0, rðxÞ ¼ 1 and comparing boundary conditions gives 2 ¼ 0 and 2 ¼ 0. To solve the equation y 00 þ y ¼ 0 we use the auxiliary equation m2 þ ¼ 0 pffiffiffi which has solutions m ¼ j (refer to Engineering Mathematics (Sixth Edition), page 1093, Frame 5ff). This means that the solution can be written in the form y ¼ A sin . . . . . . . . . . . . þ B cos . . . . . . . . . . . . pffiffiffi pffiffiffi y ¼ A sin x þ B cos x 30 Because When the solutions to the auxiliary equation are of the form m ¼ j the solution to the differential equation is of the form pffiffiffi y ¼ ex ðA sin x þ B cos xÞ and here ¼ 0 and ¼ Applying the boundary condition yð0Þ ¼ 0 then B ¼ . . . . . . . . . . . . B¼0 31 Because pffiffiffi pffiffiffi y ¼ A sin p ffiffiffix þ B cos x and so yð0Þ ¼ A sin 0 þ B cos 0 ¼ B ¼ 0. Therefore y ¼ A sin x Applying the boundary condition yð5Þ ¼ 0 then ¼ ............ ¼ Because n2 2 25 pffiffiffi pffiffiffi y ¼ A sin x therefore yð5Þ ¼ A sin 5 ¼ 0. If A ¼ 0 the solution reduces to pffiffiffi the trivial solution y ¼ 0. For a non-trivial solution sin 5 ¼ 0 and so pffiffiffi 5 ¼ n, n ¼ 0, 1, 2, 3, . . .. This means that pffiffiffi n ¼ 5 and so ¼ n2 2 25 32 390 Programme 12 There is an infinity of eigenvalues, the nth eigenvalue being denoted by n n2 2 and to each eigenvalue there is an eigenvector solution where n ¼ 25 nx . yn ¼ An sin 5 33 Orthogonality If two different functions f ðxÞ and gðxÞ are defined on the interval a x b and ðb f ðxÞgðxÞ dx ¼ 0 a then we say that the two functions are mutually orthogonal. If, on the other hand, a third function wðxÞ > 0 exists such that ðb f ðxÞgðxÞwðxÞ dx ¼ 0 a then we say that f ðxÞ and gðxÞ are mutually orthogonal with respect to the weight function wðxÞ. One important property of the solutions to a Sturm–Liouville system is that the solutions are all mutually orthogonal with respect to the weight function rðxÞ. For instance, in the previous example the individual solutions were given as nx where rðxÞ ¼ 1 yn ¼ An sin 5 and so if m 6¼ n ð5 ym ðxÞyn ðxÞrðxÞ dx ¼ . . . . . . . . . . . . 0 34 ð5 ym ðxÞyn ðxÞrðxÞ dx ¼ 0 0 Because ð5 ð5 mx nx An sin dx where rðxÞ ¼ 1 ym ðxÞyn ðxÞrðxÞ dx ¼ Am sin 5 5 0 0 ð5 mx nx ¼ Am An sin sin dx 5 5 0 ð Am An 5 ðm nÞx ðm þ nÞx dx cos ¼ cos 5 5 2 0 Am An 5 ðm nÞx ¼ sin ðm nÞ 5 2 þ ¼0 5 ðm þ nÞx sin ðm þ nÞ 5 5 provided m 6¼ n 0 391 Power series solutions of ordinary differential equations 3 35 Summary 1 A Sturm–Liouville system is a differential equation of the form pðxÞy 00 þ p 0 ðxÞy 0 þ ðqðxÞ þ rðxÞÞy ¼ 0 a x b and rðxÞ > 0 for where the boundary conditions can be written in the form 1 yðaÞ þ 2 y 0 ðaÞ ¼ 0 and 1 yðbÞ þ 2 y 0 ðbÞ ¼ 0 2 Solutions yn to a Sturm–Liouville system are called eigenvectors, each corresponding to an eigenvalue n for n ¼ 0, 1, 2, . . . 3 The solutions yn are mutually orthogonal with respect to the weighting rðxÞ. That is ðb ym ðxÞyn ðxÞrðxÞ dx ¼ 0 ðm 6¼ nÞ a Keep going 36 Legendre’s equation revisited The equation 1 x2 y 00 2xy 0 þ nðn þ 1Þy ¼ 0 is Legendre’s equation and has Legendre polynomials as solutions. That is yn ¼ Pn ðxÞ where Pn ð1Þ ¼ 1 and Pn ð1Þ ¼ ð1Þn This equation is an example of a Sturm–Liouville system pðxÞy 00 þ p 0 ðxÞy 0 þ ðqðxÞ þ rðxÞÞy ¼ 0 with boundary conditions 1 yðaÞ þ 2 y 0 ðaÞ ¼ 0 and 1 yðbÞ þ 2 y 0 ðbÞ ¼ 0 where pðxÞ ¼ . . . . . . . . . . . . ; qðxÞ ¼ . . . . . . . . . . . . ; rðxÞ ¼ . . . . . . . . . . . . ; 1 , 2 ¼ . . . . . . . . . . . . ; pðxÞ ¼ 1 x2 ; qðxÞ ¼ 0; rðxÞ ¼ 1; 1 , 2 ¼ . . . . . . . . . . . . 1 , 2 ¼ 1, 0; 1 , 2 ¼ 1, 0 37 Consequently, Legendre polynomials are mutually orthogonal. That is, if m 6¼ n ð1 Pm ðxÞPn ðxÞ dx ¼ . . . . . . . . . . . . 1 ð1 1 Pm ðxÞPn ðxÞ dx ¼ 0 38 392 Programme 12 Polynomials as a finite series of Legendre polynomials Many differential equations cannot be solved by the normal analytical means and solution by power series provides a powerful tool in many situations. Furthermore, any polynomial can be written as a finite series of Legendre polynomials. Example 1 Show that f ðxÞ ¼ x2 can be written as a series of Legendre polynomials. Assume that f ðxÞ ¼ x2 ¼ 1 X an Pn ðxÞ, then n¼0 x2 ¼ a0 P0 ðxÞ þ a1 P1 ðxÞ þ a2 P2 ðxÞ þ . . . ¼ a0 ð1Þ þ a1 ðxÞ þ a2 3x2 1 5x3 3x þ a3 þ ... 2 2 Since the left-hand side is a polynomial of degree 2 then any Legendre polynomial on the right-hand side containing powers of x greater than 2 must be excluded. Therefore a3 ¼ a4 ¼ . . . ¼ 0, so that x2 ¼ a0 a2 3 þ a1 x þ a2 x2 2 2 giving a2 ¼ 2 a2 , a1 ¼ 0, a0 ¼ 0 3 2 1 therefore a0 ¼ , and 3 x2 ¼ 1 2 P0 ðxÞ þ P2 ðxÞ 3 3 Now you try one 39 Example 2 The polynomial 1 þ x þ x3 can be written as a series of Legendre polynomials in the form 1 þ x þ x3 ¼ . . . . . . . . . . . . 40 8 2 1 þ x þ x3 ¼ P0 ðxÞ þ P1 ðxÞ þ P3 ðxÞ 5 5 Because 1 þ x þ x3 ¼ a0 P0 ðxÞ þ a1 P1 ðxÞ þ a2 P2 ðxÞ þ . . . ¼ a0 ð1Þ þ a1 ðxÞ þ a2 3x2 1 5x3 3x þ a3 þ ... 2 2 Power series solutions of ordinary differential equations 3 393 Since the left-hand side is a polynomial of degree 3 then any Legendre polynomial on the right-hand side containing powers of x greater than 3 must be excluded. Therefore a4 ¼ a5 ¼ . . . ¼ 0, so that a2 3 3 5 1 þ x þ x3 ¼ a0 þ a1 a3 x þ a2 x2 þ a3 x3 2 2 2 2 2 3 a2 This gives a3 ¼ , a2 ¼ 0, a1 a3 ¼ 1, a0 ¼ 1 therefore a0 ¼ 1, 5 2 2 8 and a1 ¼ so 5 8 2 1 þ x þ x3 ¼ P0 ðxÞ þ P1 ðxÞ þ P3 ðxÞ 5 5 As usual, the main points that we have covered in this Programme are listed in the Revision summary that follows. Read this in conjunction with the Can you? checklist and note any sections that may need further attention: refer back to the relevant parts of the Programme, if necessary. There will then be no trouble with the Test exercise. The set of Further problems provides an opportunity for further practice. Revision summary 12 1 Bessel’s equation x2 y 00 þ xy 0 þ ðx2 v 2 Þ ¼ 0 where v is a real constant. Bessel functions: Express the two solutions obtained in terms of gamma functions. x v 1 x2 x4 þ ... Jv ðxÞ ¼ 2 ðv þ 1Þ 22 ð1!Þðv þ 2Þ 24 ð2!Þðv þ 3Þ This is the Bessel function of the first kind of order v – valid for v not a negative integer. x v 1 x2 x4 Also Jv ðxÞ ¼ þ 2 ... 2 ð1 vÞ 2ð1!Þð2 vÞ 2 ð2!Þð3 vÞ provided that v is not a positive integer. Complete solution is therefore y ¼ AJv ðxÞ þ BJv ðxÞ. 41 394 Programme 12 When v ¼ n ðan integerÞ Jn ðxÞ ¼ ð1Þn Jn ðxÞ x n 1 x 2 x4 1 1 Jn ðxÞ ¼ þ 2 n! ðn þ 1Þ! 2 ð2!Þðn þ 2Þ! 2 x 6 1 þ... ð3!Þðn þ 3Þ! 2 In particular J0 ðxÞ ¼ 1 1 x 2 1 x 4 1 x6 þ þ... ð1!Þ2 2 ð2!Þ2 2 ð3!Þ2 2 and x 1 x2 1 x 4 1 x6 1 J1 ðxÞ ¼ þ þ... 2 ð1!Þð2!Þ 2 ð2!Þð3!Þ 2 ð3!Þð4!Þ 2 2 Legendre’s equation ð1 x2 Þy 00 2xy 0 þ kðk þ 1Þy ¼ 0 where k is a real constant. Solution by Frobenius gives kðk þ 1Þ 2 kðk 2Þðk þ 1Þðk þ 3Þ 4 x þ x ... c ¼ 0: y ¼ a0 1 2! 4! ðk 1Þðk þ 2Þ 3 x c ¼ 1: y ¼ a1 x 3! ðk 1Þðk 3Þðk þ 2Þðk þ 4Þ 5 x ... þ 5! When k is an integer, one series terminates. The resulting polynomial in x, Pn ðxÞ, is a Legendre polynomial, with a0 or a1 being chosen so that the polynomial has unit value when x ¼ 1. 3 Rodrigue’s formula 1 dn 2 x 1 2n n! dxn Generating function Pn ðxÞ ¼ n 1 X 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Pn ðxÞt n 1 2xt þ t 2 n¼0 4 Sturm–Liouville systems 0 ðpðxÞy 0 Þ þðqðxÞ þ rðxÞÞy ¼ 0 for a x b and rðxÞ > 0 with boundary conditions 1 yðaÞ þ 2 y 0 ðaÞ ¼ 0 and 1 yðbÞ þ 2 yðbÞ ¼ 0 Solutions yn to a Sturm–Liouville system are called eigenvectors, each corresponding to an eigenvalue n for n ¼ 0, 1, 2, . . . 395 Power series solutions of ordinary differential equations 3 5 Orthogonality If two different functions f ðxÞ and gðxÞ are defined on the interval a x b and ðb f ðxÞgðxÞ dx ¼ 0 a then the two functions are orthogonal to each other. If a function wðxÞ > 0 exists such that ðb f ðxÞgðxÞwðxÞ dx ¼ 0 a then f ðxÞ and gðxÞ are orthogonal to each other with respect to the weight function wðxÞ. The solutions of a Sturm–Liouville system yn are mutually orthogonal with respect to the weighting rðxÞ. That is ðb ym ðxÞyn ðxÞrðxÞ dx ¼ 0 ðm 6¼ nÞ a 6 Legendre polynomials are mutually orthogonal If m 6¼ n then ð1 Pm ðxÞPn ðxÞ dx ¼ 0 1 The orthogonality of the Legendre polynomials permits any polynomial to be written as a finite series of Legendre polynomials. Can you? 42 Checklist 12 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Apply Frobenius’ method to Bessel’s equation to derive Bessel functions of the first kind? Yes No 1 to 20 . Apply Frobenius’ method to Legendre’s equation to derive Legendre polynomials? Yes No 21 to 23 . Use Rodrigue’s formula to derive Legendre polynomials and the generating function to obtain some of their properties? Yes No 24 to 27 396 Programme 12 . Recognise a Sturm–Liouville system and the orthogonality properties of its eigenfunctions? Yes No 28 to 37 . Write a polynomial in x as a finite series of Legendre polynomials? Yes No 38 to 40 Test exercise 12 43 1 Use Rodrigue’s formula Pn ðxÞ ¼ 1 dn 2n n! dxn x2 1 n to derive the Legendre polynomials P2 ðxÞ and P3 ðxÞ, and show that P2 ðxÞ and P3 ðxÞ are orthogonal on ð1, 1Þ. 2 Write f ðxÞ ¼ 1 2x2 as a series of Legendre polynomials. Further problems 12 44 1 Verify that y 00 þ y ¼ 0 where y 0 ð0Þ ¼ 0 and yð2Þ ¼ 0 is a Sturm–Liouville system. Find the eigenvalues and eigenfunctions of the system and prove that they are orthogonal in ð0, 2Þ. 2 Series solutions of the equation y 00 2xy 0 þ 2ny ¼ 0 are known as Hermite polynomials, Hn ðxÞ, where n 2 d x2 e Hn ðxÞ ¼ ð1Þn ex dxn Derive the first four Hermite polynomials and show that they are orthogonal 2 with respect to the weight ex in ð1, 1Þ. 3 Series solutions of the equation xy 00 þ ð1 xÞy 0 þ ny ¼ 0 are known as Laguerre polynomials, Ln ðxÞ, where dn Ln ðxÞ ¼ ex n ðxn ex Þ dx Derive the first four Laguerre polynomials and show that they are orthogonal with respect to the weight ex in ð0, 1Þ. 4 Given the generating function for Laguerre polynomials Ln ðxÞ as 1 ext=ð1tÞ X Ln ðxÞ n t ¼ n! 1t n¼0 show that Ln ð0Þ ¼ n! Power series solutions of ordinary differential equations 3 5 Given the generating function for Hermite polynomials Hn ðxÞ as 1 X Hn ðxÞ n 2 t e2txt ¼ n! n¼0 show that H2nþ1 ð0Þ ¼ 0. 6 Given the generating function for Legendre polynomials Pn ðxÞ as 1 X 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ Pn ðxÞt n 1 2xt þ t 2 n¼0 show that P2nþ1 ð0Þ ¼ 0. 397 Programme 13 Frames 1 to 69 Numerical solutions of ordinary differential equations Learning outcomes When you have completed this Programme you will be able to: . Derive a form of Taylor’s series from Maclaurin’s series and from it describe a function increment as a series of first and higher-order derivatives of the function . Describe and apply by means of a spreadsheet the Euler method, the Euler–Cauchy method and the Runge–Kutta method for first-order differential equations . Describe and apply by means of a spreadsheet the Euler second-order method and the Runge–Kutta method for second-order ordinary differential equations . Describe and apply by means of a spreadsheet a simple predictor– corrector method. Prerequisite: Engineering Mathematics (Sixth Edition) Programme F.4 (Using a spreadsheet) 398 399 Numerical solutions of ordinary differential equations Introduction The range of differential equations that can be solved by straightforward analytical methods is relatively restricted. Even solution in series may not always be satisfactory, either because of the slow convergence of the resulting series or because of the involved manipulation in repeated stages of differentiation. In such cases, where a differential equation and known boundary conditions are given, an approximate solution is often obtainable by the application of numerical methods, where a numerical solution is obtained at discrete values of the independent variable. The solution of differential equations by numerical methods is a wide subject. The present Programme introduces some of the simpler methods, which nevertheless are of practical use. Taylor’s series Let us start off by briefly revising the fundamentals of Maclaurin’s and Taylor’s series. y y = f (x) f(h) f (0) O h x Maclaurin’s series for f ðxÞ is f ðxÞ ¼ f ð0Þ þ xf 0 ð0Þ þ x2 00 xn f ð0Þ þ . . . þ f n ð0Þ þ . . . 2! n! ð1Þ and expresses the function f ðxÞ in terms of its successive derivatives at x ¼ 0, i.e. at the point K. Therefore, at P, f ðhÞ ¼ . . . . . . . . . . . . 1 400 Programme 13 2 h2 hn f ðhÞ ¼ f ð0Þ þ hf ð0Þ þ f 00 ð0Þ þ . . . þ f n ð0Þ þ . . . 2! n! (2) 0 y y = f (a+x) f (a+h) f (a) O a x h If the y-axis and origin are moved a units to the left, the equation of the same curve relative to the new axes becomes y ¼ f ða þ xÞ and the function value at K is f ðaÞ. At P, f ða þ hÞ ¼ f ðaÞ þ hf 0 ðaÞ þ h2 00 hn f ðaÞ þ . . . þ f n ðaÞ þ . . . 2! n! This is one common form of Taylor’s series. Make a note of it and then move on 3 Function increment y y = f (x) y f(a+h) f(a) O a h x If we know the function value f ðaÞ at A, i.e. at x ¼ a, we can apply Taylor’s series to determine the function value at a neighbouring point B, i.e. at x ¼ a þ h. f ða þ hÞ ¼ f ðaÞ þ hf 0 ðaÞ þ h3 h2 00 f ðaÞ þ f 000 ðaÞ þ . . . 3! 2! The function increment from A to B ¼ y ¼ f ða þ hÞ f ðaÞ i.e. f ða þ hÞ ¼ f ðaÞ þ y where y ¼ hf 0 ðaÞ þ h2 00 h3 f ðaÞ þ f 000 ðaÞ þ . . . 2! 3! ð3Þ 401 Numerical solutions of ordinary differential equations This entails evaluation of an infinite number of derivatives at x ¼ a: in practice an approximation is accepted by restricting the number of terms that are used in the series. This approximation of Taylor’s series forms the basis of several numerical methods, some of which we shall now introduce. It should be noted that these early examples have been selected because exact solutions can also be found. The purpose of this is to enable a comparison between the results obtained by a particular method with those obtained from an exact solution, and so to demonstrate the accuracy of the method. On then to the next frame First-order differential equations Numerical solution of y ¼ y0 : dy ¼ f ðx; yÞ with the initial condition that, at x ¼ x0 , dx Euler’s method The simplest of the numerical methods for solving first-order differential equations is Euler’s method, in which the Taylor’s series f ða þ hÞ ¼ f ðaÞ þ hf 0 ðaÞ þ h2 00 h3 f ðaÞ þ f 000 ðaÞ þ . . . 2! 3! is truncated after the second term to give f ða þ hÞ f ðaÞ þ hf 0 ðaÞ ð4Þ This is a severe approximation, but in practice the ‘approximately equals’ sign is replaced by the normal ‘equals’ sign, in the knowledge that the result we obtain will necessarily differ to some extent from the function value we seek. With this in mind, we write f ða þ hÞ ¼ f ðaÞ þ hf 0 ðaÞ y C A y1 y0 O y=f (x) B a h a+h If h is the interval between two near ordinates and if we denote f ðaÞ by y0 , then the relationship f ða þ hÞ ¼ f ðaÞ þ hf 0 ðaÞ becomes x y1 ¼ y0 þ hðy 0 Þ0 ð5Þ 0 Hence, knowing y0 , h and ðy Þ0 , we can compute y1 , an approximate value for the function value at B. Make a note of result (5): we shall be using it quite a lot. Then move on for an example 4 402 5 Programme 13 Example 1 dy ¼ 2ð1 þ xÞ y with the initial condition that at x ¼ 2, y ¼ 5, we dx can find an approximate value of y at x ¼ 2:2, as follows. Given that We have y 0 ¼ 2ð1 þ xÞ y x0 ¼ 2; y0 ¼ 5 with ; ðy 0 Þ0 ¼ . . . . . . . . . . . . 6 ðy 0 Þ0 ¼ 1 We obtain this by substituting x0 and y0 in the given equation: ðy 0 Þ0 ¼ 2ð1 þ x0 Þ y0 ¼ 2ð1 þ 2Þ 5 ; ðy 0 Þ0 ¼ 1 So we have x0 ¼ 2; y0 ¼ 5; ðy 0 Þ0 ¼ 1; x1 ¼ 2:2; h ¼ 0:2: By Euler’s relationship: y1 ¼ y0 þ hðy 0 Þ0 7 ; y1 ¼ . . . . . . . . . . . . y1 ¼ 5:2 Because y1 ¼ y0 þ hðy 0 Þ0 ¼ 5 þ ð0:2Þ1 ¼ 5:2 y B At B, x1 ¼ 2:2; y1 ¼ 5:2; and C A ðy 0 Þ1 ¼ . . . . . . . . . . . . y1 y0 O x x0 h x1 403 Numerical solutions of ordinary differential equations 8 ðy 0 Þ1 ¼ 1:2 ðy 0 Þ1 ¼ 2ð1 þ x1 Þ y1 ¼ 2ð1 þ 2:2Þ 5:2 ¼ 1:2 y y1 y0 O x x0 h x1 0 If we take the values of x, y and y that we have just found for the point B and treat these as new starter values x0 , y0 , ðy 0 Þ0 , we can repeat the process and find values corresponding to the point C. At B, x0 ¼ 2:2; y0 ¼ 5:2; ðy 0 Þ0 ¼ 1:2; x1 ¼ 2:4. Then at C: y1 ¼ . . . . . . . . . . . .; y1 ¼ 5:44; ðy 0 Þ1 ¼ . . . . . . . . . . . . 9 ðy 0 Þ1 ¼ 1:36 y1 ¼ y0 þ hðy 0 Þ0 ¼ 5:2 þ ð0:2Þ1:2 ¼ 5:44 ðy 0 Þ ¼ 2ð1 þ x1 Þ y1 ¼ 2ð1 þ 2:4Þ 5:44 ¼ 1:36 1 So we could continue in a step-by-step method. At each stage, the determined values of x1 , y1 and ðy 0 Þ1 become the new starter values x0 , y0 and ðy 0 Þ0 for the next stage. Our results so far can be tabulated thus x0 y0 ðy 0 Þ0 x1 y1 ðy 0 Þ1 2.0 5.0 1.0 2.2 5.2 1.2 2.2 5.2 1.2 2.4 5.44 1.36 2.4 5.44 1.36 Continue the table with a constant interval of h ¼ 0:2. The third row can be completed to give x1 ¼ . . . . . . . . . . . . ; y1 ¼ . . . . . . . . . . . . ; ðy 0 Þ1 ¼ . . . . . . . . . . . . 404 Programme 13 10 x1 ¼ 2:6; y1 ¼ 5:712; ðy 0 Þ1 ¼ 1:488 Because x1 ¼ x0 þ h ¼ 2:4 þ 0:2 ¼ 2:6 y1 ¼ y0 þ hðy 0 Þ0 ¼ 5:44 þ ð0:2Þ1:36 ¼ 5:712 ðy 0 Þ ¼ 2ð1 þ x1 Þ y1 ¼ 2ð1 þ 2:6Þ 5:712 ¼ 1:488 1 Now you can continue in the same way and complete the table for x ¼ 2.0, 2.2, 2.4, 2.6, 2.8, 3.0 Finish it off and compare results with the next frame 11 Here is the result. x0 y0 ðy 0 Þ0 x1 y1 ðy 0 Þ1 2.0 5.0 1.0 2.2 5.2 1.2 2.2 5.2 1.2 2.4 5.44 1.36 2.4 5.44 1.36 2.6 5.712 1.488 2.6 5.712 1.488 2.8 6.009 6 1.590 4 2.8 6.009 6 1.590 4 3.0 6.327 68 1.672 32 3.0 6.327 68 1.672 32 In practice, we do not, in fact, enter the values in the right-hand half of the table, but write them in directly as new starter values in the left-hand section of the table. x0 y0 ðy 0 Þ0 2.0 5.0 1.0 2.2 5.2 1.2 2.4 5.44 1.36 2.6 5.712 1.488 2.8 6.009 6 1.590 4 3.0 6.327 68 1.672 32 The particular solution is given by the values of y against x and a graph of the function can be drawn. Draw the graph of the function carefully on graph paper. 405 Numerical solutions of ordinary differential equations Graph of the solution of dy ¼ 2ð1 þ xÞ y with y ¼ 5 at x ¼ 2. dx 12 y x It is an advantage to plot the points step-by-step as the results are built up. In that way, one can check that there is a smooth progression and that no apparent errors in the calculations occur at any one stage. dy ¼ 2ð1 þ xÞ y can be solved by the integration The differential equation dx factor method (see Engineering Mathematics, Sixth Edition, Programme 24) to give the solution y ¼ 2x þ e2x and in the following table we compare our results with the actual values to determine the errors. x y (Euler) y (actual) Absolute error 2.0 5.0 5.0 0 2.2 5.2 5.218 731 0.018 731 2.4 5.44 5.470 320 0.030 320 2.6 5.712 5.748 812 0.036 812 2.8 6.009 6 6.049 329 0.039 729 3.0 6.327 68 6.367 879 0.040 199 The errors involved in the process are shown. These errors are due mainly to .................................... 13 406 14 Programme 13 the fact that Taylor’s series was truncated after the second term By now you will appreciate the amount of arithmetic manipulation involved in solving these differential equations – a large amount of which is repetitive. To avoid the tedium and to make the computations more efficient we shall resort to the use of a spreadsheet. If the use of a spreadsheet is a totally new experience to you then you are referred to Programme F.4 of Engineering Mathematics, Sixth Edition, where the spreadsheet is introduced as a tool for constructing graphs of functions. If you have a limited knowledge then you will be able to follow the text from here. The spreadsheet we shall be using here is Microsoft Excel, though all commercial spreadsheets possess the equivalent functionality. Alternatively, an iteration process can be used in any computer algebra package such as Derive, Maple or Mathematica. Open your spreadsheet and in cell A1 enter the letter n and press Enter. In this first column we are going to enter the iteration numbers. In cell A2 enter the number 0 and press Enter. Place the cell highlight in cell A2 and highlight the block of cells A2 to A12 by holding down the mouse button and wiping the highlight down to cell A12. Click the Edit command on the Command bar and point at Fill from the drop-down menu. Select Series from the next drop-down menu and accept the default Step value of 1 by clicking OK in the Series window. The cells A3 to A12 fill with . . . . . . . . . . . . 15 The numbers 1 to 10 In cell B1 enter the letter x – this column is going to contain the successive xvalues for which the y-value is going to be enumerated. In cell B2 enter the number 2 – the initial x-value. We now could fill the column in much the same way as we filled the first column, but we have a better way. Place the cell highlight in cell F1 and enter the number 0.2 – this is the value of h, the increment in x. Now place the cell highlight in cell B3 and enter the formula = B2 + $F$1 followed by Enter (uppercase or lowercase, it does not matter) The number 2.2 appears in cell B3. Place the cell highlight in cell B3, click the Edit command and select Copy from the drop-down menu. You have now copied the contents of cell B3 to the clipboard. Now place the cell highlight in cell B4 and highlight the block of cells from B4 to B12. Click the Edit command again but this time select Paste from the drop-down menu. The cells B4 to B12 fill with the numbers . . . . . . . . . . . . Numerical solutions of ordinary differential equations The numbers 2.4 to 4.0 in intervals of 0.2 How has this happened? When you typed in the cell reference B2 into the formula in cell B3, the spreadsheet understood this to mean the contents of the cell immediately above current cell B3. When the formula is copied into cell B4 it means the contents of the cell immediately above current cell B4. Entered in this way the address B2 is a relative address. On the other hand, when you typed in $F$1 the spreadsheet understood this to mean the contents of cell F1 and that meaning remains when it is copied – the dollar signs indicate an absolute address. So as you move down the column the contents of a cell contain the contents of the cell immediately above it plus the contents of cell F1. You will shortly see the advantages of all this. For now, place the cell highlight in cell C1 and enter the letter y – this column is going to contain the computed y-values against the corresponding x-values in column B. Place the cell highlight in cell C2 and enter the number 5 – the initial y-value. Before we can compute the y-values in column C we need to be able to tabulate the values of y 0 – the derivatives of y. Place the cell highlight in cell D1 and enter y 0 – this column will contain the values of the derivatives of y against the corresponding x-values. Cell D2 will contain the initial value of y 0 which can be computed from the equation y 0 ¼ 2ð1 þ xÞ y When x ¼ x0 ¼ 2 and y ¼ y0 ¼ 5 then y00 ¼ 2ð1 þ x0 Þ y0 ¼ 2ð1 þ 2Þ 5 ¼ 1 so place the cell highlight in cell D2 and enter the formula = 2 (1 + B2) – C2 (B2 contains x0 and C2 contains y0 ) The number 1 appears in cell D2. We need to copy this formula down the y 0 column. Place the cell highlight in cell D2, click Edit and select Copy. Now place the cell highlight in cell D3 and highlight the block of cells D3 to D12. Click the Edit command again and select Paste. The cells D3 to D12 fill with . . . . . . . . . . . . 407 16 408 Programme 13 17 The numbers 6.4 to 10.0 in intervals of 0.4 Because the cells in the C2 column are currently empty, these values are just 2 (1 + B2) – 0. Now, to compute the y-values we use the equation y1 ¼ y0 þ hðy 0 Þ0 . Place the cell highlight in cell C3 and enter the formula = C2 + $F$1 D2 (C2 contains y0 , F1 contains h and D2 contains ðy 0 Þ0 ) and the number 5.2 appears. That is, y1 ¼ 5 þ ð0:2Þð1Þ ¼ 5:2. This now completes the sequence of operations required to find y1 . To find the values of y2 ¼ yðx2 Þ ¼ yð2:4Þ this sequence is repeated and, to ensure this, all that remains is to copy the formula in cell C3 into cells C4 to C12. So do this to reveal the following display n 0 1 2 3 4 5 6 7 8 9 10 x 2 2.2 2.4 2.6 2.8 3 3.2 3.4 3.6 3.8 4 y y’ 5 5.2 5.44 5.712 . 6 0096 6.32768 6.662144 . 7 0097152 7.36777216 7.734217728 8.107374182 1 1.2 1.36 . 1 488 1.5904 1.67232 . 1 737856 1.7902848 1.83222784 . 1 865782272 1.892625818 0.2 Now that was a lot easier than all that arithmetic manipulation by hand, wasn’t it? We can tidy this display up by using the Format command and by using the various options on the tool bars to change the column widths and to display the numbers in a regular format of 10 decimal places to produce a display that is easier to read. Next frame 409 Numerical solutions of ordinary differential equations n 0 1 2 3 4 5 6 7 8 9 10 x 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 y 5.0000000000 5.2000000000 5.4400000000 5.7120000000 6.0096000000 6.3276800000 6.6621440000 7.0097152000 7.3677721600 7.7342177280 8.1073741824 y’ 1.0000000000 1.2000000000 1.3600000000 1.4880000000 1.5904000000 1.6723200000 1.7378560000 1.7902848000 1.8322278400 1.8657822720 1.8926258176 h = 0.2 Notice that we have added h= in cell E1 and justified it to the right and then justified the number 0.2 in F2 to the left so that together they read as an equation. The advantage of isolating the step value 0.2 in cell F1, as we have done, is that we can change the value and immediately see the effects on the calculations. For example, if the contents of F1 are changed to 0.1 the display changes automatically to n 0 1 2 3 4 5 6 7 8 9 10 x 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 y 5.0000000000 5.1000000000 5.2100000000 5.3290000000 5.4561000000 5.5904900000 5.7314410000 5.8782969000 6.0304672100 6.1874204890 6.3486784401 y’ 1.0000000000 1.1000000000 1.1900000000 1.2710000000 1.3439000000 1.4095100000 1.4685590000 1.5217031000 1.5695327900 1.6125795110 1.6513215599 h = 0.1 Notice that the different values of h produce different corresponding values in the tables. For example, for h ¼ 0:2 we find that yð3:0Þ ¼ 6:327 680 0000 whereas for h ¼ 0:1 we have yð3:0Þ ¼ 6:348 678 4401. The smaller the value of h then, the smaller the errors in the calculation – we shall see this demonstrated explicitly in the next frame. Go to the next frame 18 410 19 Programme 13 The exact value and the errors The differential equation y 0 ¼ 2ð1 þ xÞ y can be solved using the integration factor method (see Engineering Mathematics, Sixth Edition, Programme 24) to give the solution y ¼ 2x þ e2x We can programme this into the spreadsheet to compare the exact solution with the solution obtained numerically and compute the actual errors. Place the cell highlight in cell E1 and highlight cells E1 and F1. Click Insert on the Command bar and select Columns. Immediately two new columns appear. Notice that the numbers in the display do not change despite the fact that the h-value of 0.2 has moved from F1 to H1 – all the formulas in the spreadsheet will have automatically adjusted themselves. You can check this by highlighting a cell with a formula in it to see the change. In cell E1 enter the word Exact and in cell F1 enter Errors (%). In cell E2 enter the right-hand side of the equation y ¼ 2x þ e2x by using the formula = 2 B2 + EXP(2 – B2) (the EXP stands for the exponential function) and copy this into the block of cells E3 to E12. In cell F2 enter the formula for the error = (E2 – C2) 100/E2 (the error as a percentage of the exact value) and copy this into the block of cells F3 to F12 to produce the following display n x 0 1 2 3 4 5 6 7 8 9 10 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 y 5.0000000000 5.2000000000 5.4400000000 5.7120000000 6.0096000000 6.3276800000 6.6621440000 7.0097152000 7.3677721600 7.7342177280 8.1073741824 y’ 1.0000000000 1.2000000000 1.3600000000 1.4880000000 1.5904000000 1.6723200000 1.7378560000 1.7902848000 1.8322278400 1.8657822720 1.8926258176 Exact 5.0000000000 5.2187307531 5.4703200460 5.7488116361 6.0493289641 6.3678794412 6.7011942119 7.0465969639 7.4018965180 7.7652988882 8.1353352832 Errors (%) 0.00 0.36 0.55 0.64 0.66 0.63 0.58 0.52 0.46 0.40 0.34 h=0.2 411 Numerical solutions of ordinary differential equations Change the value of h to 0.1 and produce the following display n x 0 1 2 3 4 5 6 7 8 9 10 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 y y’ 5.0000000000 5.1000000000 5.2100000000 5.3290000000 5.4561000000 5.5904900000 5.7314410000 5.8782969000 6.0304672100 6.1874204890 6.3486784401 1.0000000000 1.1000000000 1.1900000000 1.2710000000 1.3439000000 1.4095100000 1.4685590000 1.5217031000 1.5695327900 1.6125795110 1.6513215599 Exact 5.0000000000 5.1048374180 5.2187307531 5.3408182207 5.4703200460 5.6065306597 5.7488116361 5.8965853038 6.0493289641 6.2065696597 6.3678794412 Errors h = 0.1 (%) 0.00 0.09 0.17 0.22 0.26 0.29 0.30 0.31 0.31 0.31 0.30 When h ¼ 0:2 the error in yð3:0Þ is 0.63% whereas when h ¼ 0:1 the error in yð3:0Þ is 0.30%. The smaller the value of h the . . . . . . . . . . . . smaller the error Having completed your first spreadsheet you can now use it as a template for similar problems. To avoid losing the work that you have already done, save your spreadsheet under some suitable name. When that is complete, highlight all the cells from A1 to G12 and copy them onto the clipboard using the Edit-Copy sequence of commands. Now click the Sheet 2 tab at the bottom of your spreadsheet to reveal a blank worksheet. Place the cell highlight in cell A1, click Edit and select Paste. The entire contents of Sheet 1 are now copied to Sheet 2 in readiness for editing to accommodate a new problem. So let’s look at another example. Example 2 Obtain a numerical solution of the equation dy ¼1þxy dx with the initial condition that y ¼ 2 at x ¼ 1, for the range x ¼ 1:0ð0:2Þ3:0, that is from x ¼ 1:0 to x ¼ 3:0 with step length x ¼ 0:2. As initial conditions, we have x0 ¼ . . . . . . . . . . . . and y0 ¼ . . . . . . . . . . . . 20 412 Programme 13 21 x0 ¼ 1, y0 ¼ 2 Because x0 ¼ 1 and y0 ¼ 2 are given initial conditions. These values can now be inserted into the spreadsheet in cells . . . . . . . . . . . . 22 x1 ¼ 1 in B2, y0 ¼ 2 in C2 Notice how the numbers in column B have changed to accommodate the new sequence of x-values. The contents of the cells in column C do not need to be changed as they refer to the equation y1 ¼ y0 þ hðy 0 Þ0 which is the same in this spreadsheet as it was in the previous spreadsheet. The contents of column D do have to be changed because they currently refer to the equation to be solved in the previous problem. The equation to be solved here is y0 ¼ 1 þ x y so in cell D3 the contents need to be changed to . . . . . . . . . . . . 23 = 1 + B2 – C2 This formula must then be copied into cells C3 to C12. Finally, the Exact column needs to be amended to reflect the exact solution to this equation, which is again found by using the integration factor method as y ¼ x þ e1x So, in E2, enter the formula . . . . . . . . . . . . 413 Numerical solutions of ordinary differential equations 24 = B2 + EXP(1 – B2) This formula needs to be copied into cells E3 to E12. This completes the editing of the spreadsheet to reflect the new problem to give the display n x y y’ Exact Errors h=0.2 (%) 0 1.0 2.0000000000 0.0000000000 2.0000000000 0.00 1 1.2 2.0000000000 0.2000000000 2.0187307531 0.93 . . . . 2 1 4 2 0400000000 0 3600000000 2 0703200460 1.46 . . . . 3 1 6 2 1120000000 0 4880000000 2 1488116361 1.71 . . . . 4 1 8 2 2096000000 0 5904000000 2 2493289641 1.77 . . . . 5 2 0 2 3276800000 0 6723200000 2 3678794412 1.70 . . . . 6 2 2 2 4621440000 0 7378560000 2 5011942119 1.56 . . . . 7 2 4 2 6097152000 0 7902848000 2 6465969639 1.39 . . . . 1.22 8 2 6 2 7677721600 0 8322278400 2 8018965180 . . . . 9 2 8 2 9342177280 0 8657822720 2 9652988882 1.05 . . . . 10 3 0 3 1073741824 0 8926258176 3 1353352832 0.89 A plot of the graph of y against x for both the computed value and the exact value looks as follows y y = x + e1 – x 3.5 3.0 2.5 2.0 1.5 approximate exact 1.0 0.5 0 x 0 1 2 3 4 414 25 Programme 13 Graphical interpretation of Euler’s method If AT is the tangent to the curve at A, NT dy then ¼ ðy 0 Þ0 ¼ AN dx x¼x0 NT ¼ ðy 0 Þ0 ; NT ¼ hðy 0 Þ0 h ; At x ¼ x1 , MT ¼ y0 þ hðy 0 Þ0 y y1 y0 O x0 h By Euler’s relationship, x1 x y1 ¼ y0 þ hðy 0 Þ0 i.e. MT. The difference between the calculated value of y, i.e. MT, and the actual value of the function y, i.e. MB, at x ¼ x1 , is indicated by TB. This error can be considerable, depending on the curvature of the graph and the size of the interval h. It is inherent to the method and corresponds to the truncation of the Taylor’s series after the second term. Euler’s method, then (a) is simple in procedure (b) is lacking in accuracy, especially away from the starter values of the initial conditions (c) is of use only for very small values of the interval h. In spite of its practical limitations, it is the foundation of several more sophisticated methods and hence it is worthy of note. Here is one more example to work on your own. Example 3 dy ¼ x þ y with the initial condition that y ¼ 1 at x ¼ 0, dx : : for the range x ¼ 0ð0 1Þ0 5. Obtain the solution of By using a previously constructed spreadsheet as a template, the solution is ............ The function values are given in the next frame 415 Numerical solutions of ordinary differential equations n x 0 1 2 3 4 5 6 7 8 9 10 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y y’ 1.0000000000 1.1000000000 1.2200000000 1.3620000000 1.5282000000 1.7210200000 1.9431220000 2.1974342000 2.4871776200 2.8158953820 3.1874849202 Exact 1.0000000000 1.2000000000 1.4200000000 1.6620000000 1.9282000000 2.2210200000 2.5431220000 2.8974342000 3.2871776200 3.7158953820 4.1874849202 1.0000000000 1.1103418362 1.2428055163 1.3997176152 1.5836493953 1.7974425414 2.0442376008 2.3275054149 2.6510818570 3.0192062223 3.4365636569 h = 0.1 Errors (%) 0.00 0.93 1.84 2.69 3.50 4.25 4.95 5.59 6.18 6.73 7.25 Because The initial conditions are entered as 0 in cell B2 1 in cell C2 0.1 in cell H1 (the initial x-value) (the initial y-value) (the x step length) The formulas are entered as = B2 + C2 in cell D2, copied into cells D3 to D12 (the successive y 0 -values) = C2 + $H$1 D2 in cell C3 copied into cells C4 to C12 (the successive y-values) The exact solution found by using the integration factor method is y ¼ 2ex x 1 and so = 2 EXP(B2) – B2 – 1 is entered into cell E2 and copied into cells E3 to E12 Notice how the errrors here are significant, which is very evident from the graphs of the computed values and the exact values. y . 40 y = 2ex – x – 1 3.5 3.0 2.5 2.0 1.5 calculated 1.0 exact 0.5 0 0 0.2 0.4 0.6 0.8 1.0 1.2 x 26 416 27 Programme 13 The Euler–Cauchy method – or the improved Euler method y y1 y0 O x0 x1 x h In Euler’s method, we use the slope ðy 0 Þ0 at A ðx0 , y0 Þ across the whole interval h to obtain an approximate value of y1 at B. TB is the resulting error in the result. y y1 y0 O x0 x0 + 12 h x1 x In the Euler–Cauchy method, we use the slope at A ðx0 , y0 Þ across half the interval and then continue with a line whose slope approximates to the slope of the curve at x1 . Let y 1 be the y-value of the point at T. The error (MB) in the result is now considerably less than the error (TB) associated with the basic Euler method and the calculated results will accordingly be of greater accuracy. Numerical solutions of ordinary differential equations Euler–Cauchy calculations 417 28 The steps in the Euler–Cauchy method are as follows. 1 2 3 We start with the given equation y 0 ¼ f ðx, yÞ with the initial condition that at x ¼ x0 , y ¼ y0 . We have to determine function values for x ¼ x0 ðhÞxn . From the equation and the initial condition we obtain ðy 0 Þ0 ¼ f ðx0 ; y0 Þ. Knowing x0 , y0 , ðy 0 Þ0 and h, we then evaluate (a) x1 ¼ x0 þ h (b) the auxiliary value of y, denoted by y where y 1 ¼ y0 þ hðy 0 Þ0 .This is the same step as in Euler’s method. (c) Then y1 ¼ y0 þ 12 hfðy 0 Þ0 þ f ðx1 ; y 1 Þg Note that f ðx1 , y1 Þ is the right-hand side of the given equation with x and y replaced by the calculated values of x1 and y 1 . (d) Finally ðy 0 Þ1 ¼ f ðx1 ; y1 Þ. We have thus evaluated x1 , y1 and ðy 0 Þ1 . The whole process is then repeated, the calculated values of x1 , y1 and ðy 0 Þ1 becoming the starter values x0 , y0 , ðy 0 Þ0 for the next stage. Make a note of the relationships above. We shall be using them quite often. Then on to the next frame for an example of their use Example 1 29 Apply the Euler–Cauchy method to solve the equation y0 ¼ x þ y with the initial condition that at x ¼ 0, y ¼ 1, for the range x ¼ 0ð0:1Þ1:0. We proceed as before by copying our template solution to a new worksheet. Before we continue we need to decide what the entries are going to be in our spreadsheet. 1 We are going to have to enter new initial conditions, so Enter 0 in cell B2 Enter 1 in cell C2 Enter 0.1 in cell H1 2 that is x0 ¼ 0 that is y0 ¼ 1 this is the x step length The equation to be solved is y 0 ¼ x þ y, so enter the formula = B2 + C2 in cell D2 and copy the contents of D2 into cells D3 to D12 3 The Euler–Cauchy method tells us that 1 y1 ¼ y0 þ h ðy 0 Þ0 þf x, y 1 2 where y 1 ¼ y0 þ hðy 0 Þ0 so that f x1 , y 1 ¼ x1 þ y 1 ¼ x1 þ y0 þ hðy 0 Þ0 Therefore y1 ¼ . . . . . . . . . . . . 418 Programme 13 30 1 y1 ¼ y0 þ h x1 þ y0 þ ð1 þ hÞðy 0 Þ0 2 Because By replacing f x1 , y 1 with x1 þ y0 þ hðy 0 Þ0 in the expression y1 ¼ y0 þ 12 h ðy 0 Þ0 þf x1 , y 1 we find that y1 ¼ y0 þ 12 h ðy 0 Þ0 þx1 þ y0 þ hðy 0 Þ0 ¼ y0 þ 12 h x1 þ y0 þ ð1 þ hÞðy 0 Þ0 In cell C3 enter the formula . . . . . . . . . . . . 31 = C2 + (0.5) $H$1 (B3 + C2 + (1 + $H$1) D2) Because y0 is in cell C2, h is in cell H1, x1 is in cell B3 and ðy 0 Þ0 is in cell D2. Copy the contents of cell C3 into cells C4 to C12. 4 Finally, for comparison purposes, the exact solution of this equation is y ¼ 2ex x 1 and this is entered into E2 by the formula . . . . . . . . . . . . and copied into cells . . . . . . . . . . . . 32 = 2 EXP(B2) – B2 – 1 and copied into cells E3 to E12 The resulting display looks as follows n x 0 1 2 3 4 5 6 7 8 9 10 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y 1.0000000000 1.1100000000 1.2420500000 1.3984652500 1.5818041013 1.7948935319 2.0408573527 2.3231473748 2.6455778491 3.0123635233 3.4281616932 y’ 1.0000000000 1.2100000000 1.4420500000 1.6984652500 1.9818041013 2.2948935319 2.6408573527 3.0231473748 3.4455778491 3.9123635233 4.4281616932 Exact 1.0000000000 1.1103418362 1.2428055163 1.3997176152 1.5836493953 1.7974425414 2.0442376008 2.3275054149 2.6510818570 3.0192062223 3.4365636569 Errors (%) 0.00 0.03 0.06 0.09 0.12 0.14 0.17 0.19 0.21 0.23 0.24 h = 0.1 419 Numerical solutions of ordinary differential equations Comparing these results with the same equation being solved by the Euler method demonstrates how much more accurate the Euler–Cauchy method is, as can be seen from the following table of comparative errors x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Euler 0.00 0.93 1.84 2.69 3.50 4.25 4.95 5.59 6.18 6.73 7.25 Euler–Cauchy 0.00 0.03 0.06 0.09 0.12 0.14 0.17 0.19 0.21 0.23 0.24 Next frame Now for another example, but before that, complete the following without reference to your notes – if possible. In the Euler–Cauchy method the relevant relationships are 33 x1 ¼ . . . . . . . . . . . . y1 ¼ . . . . . . . . . . . . y1 ¼ . . . . . . . . . . . . ðy 0 Þ1 ¼ . . . . . . . . . . . . Next frame 34 x1 ¼ x0 þ h 0 y 1 ¼ y0 þ hðy Þ0 1 y1 ¼ y0 þ h ðy 0 Þ0 þf x1 , y 1 2 ðy 0 Þ1 ¼ f ðx1 , y1 Þ Example 2 Determine a numerical solution of the equation y 0 ¼ 2ð1 þ xÞ y with the initial condition that y ¼ 5 when x ¼ 2, for the range 2:0ð0:2Þ4:0. Try this one yourself. The exact solution is given as y ¼ 2x þ e2x and the final display of results is . . . . . . . . . . . . 420 35 Programme 13 n x 0 1 2 3 4 5 6 7 8 9 10 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 y 5.0000000000 5.2200000000 5.4724000000 5.7513680000 6.0521217600 6.3707398432 6.7040066714 7.0492854706 7.4044140859 7.7676195504 8.1374480313 y’ 1.0000000000 1.1800000000 1.3276000000 1.4486320000 1.5478782400 1.6292601568 1.6959933286 1.7507145294 1.7955859141 1.8323804496 1.8625519687 Exact 5.0000000000 5.2187307531 5.4703200460 5.7488116361 6.0493289641 6.3678794412 6.7011942119 7.0465969639 7.4018965180 7.7652988882 8.1353352832 Errors (%) 0.00 0.02 0.04 0.04 0.05 0.04 0.04 0.04 0.03 0.03 0.03 h = 0.2 Because 1 The initial conditions are entered as Enter 2 in cell B2 (that is x0 ¼ 2); enter 5 in cell C2 (that is y0 ¼ 5) Enter 0.2 in cell H1 (this is the x step length) 2 The equation to be solved is y 0 ¼ 2ð1 þ xÞ y, so enter the formula = 2 (1 + B2) – C2 in cell D2 and copy the contents of D2 into cells D3 to D12 3 The Euler–Cauchy method tells us that 1 y1 ¼ y0 þ h ðy 0 Þ0 þf x1 , y 1 where y 1 ¼ y0 þ hðy 0 Þ0 so that 2 f x1 , y 1 ¼ 2ð1 þ x1 Þ y 1 ¼ 2ð1 þ x1 Þ y0 hðy 0 Þ0 therefore 1 y1 ¼ y0 þ h ðy 0 Þ0 þ2ð1 þ x1 Þ y0 hðy 0 Þ0 that is 2 1 y1 ¼ y0 þ h 2ð1 þ x1 Þ y0 þ ð1 hÞðy 0 Þ0 2 This is accommodated by the formula in C3 (copied into cells C4 to C12) = C2 + (0.5) $H$1 (2 (1 + B3) – C2 + (1 – $H$1) D2) 4 Finally the exact solution y ¼ 2x þ e2x is entered into cell E2 as = 2 B2 + EXP(–2 B2) and copied into cells E3 to E12. Refer to Frame 19 for a comparison of errors between this method and the Euler method. Then another example for you to try just to make sure you are clear about the processes involved. Next frame 421 Numerical solutions of ordinary differential equations Example 3 36 0 2 Solve the equation y ¼ y þ xy with initial condition that at x ¼ 1, y ¼ 1, for the range x ¼ 1:0ð0:1Þ1:7. Use the Euler–Cauchy method and work to 6 places of decimals. The solution is . . . . . . . . . . . . n 0 1 2 3 4 5 6 7 x 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 y . 1 000000 1.238000 1.591023 2.152410 3.145846 5.251007 11.595613 57.704110 y’ . 2 000000 2.894444 4.440583 7.431004 14.300528 35.449581 153.011211 3427.861242 h = 0.1 Because 1 The initial conditions are entered as Enter 1 in cell B2 (that is x0 ¼ 1); enter 1 in cell C2 (that is y0 ¼ 1) Enter 0.1 in cell H1 (this is the x step length) 2 The equation to be solved is y 0 ¼ y 2 þ xy, so Enter the formula = C2^2 + B2 C2 in cell D2 and copy the contents of D2 into cells D3 to D9. Note that C2^2 = C2 C2 – the ‘hat’ indicates raising to a power. 3 The Euler-Cauchey method tell us that 1 where y 1 ¼ y0 þ hðy 0 Þ0 so that y1 ¼ y0 þ h ðy 0 Þ0 þf x1 , y 1 2 2 2 f x1 , y 1 ¼ y 1 þ x1 y 1 ¼ y0 þ hðy 0 Þ0 þx1 y0 þ hðy 0 Þ0 therefore 2 o 1 n y1 ¼ y0 þ h ðy 0 Þ0 þ y0 þ hðy 0 Þ0 þx1 y0 þ hðy 0 Þ0 2 This is accommodated by the formula in C3 (copied into cells C4 to C9) = C2 + (0.5) $F$1 (D2 + (C2 + $F$1 D2))^2 + B3 (C2 + $F$1 D2)) The table shows that as x increases, the computed values of y and its derivative increase dramatically. This is an indication that the exact solution increases without bound near to the larger values of x considered, so bringing the accuracy of these computed values into question. This emphasises the importance of checking every method against a known solution so as to form some idea of the method’s accuracy. However, all numerical methods produce significant accuracies whenever the exact solution diverges in this way. 37 422 38 Programme 13 Runge–Kutta method The Runge–Kutta method for solving first-order differential equations is widely used and affords a high degree of accuracy. It is a further step-by-step process where a table of function values for a range of values of x is accumulated. Several intermediate calculations are required at each stage, but these are straightforward and present little difficulty. In general terms, the method is as follows. To solve y 0 ¼ f ðx; yÞ with initial condition y ¼ y0 at x ¼ x0 , for a range of values of x ¼ x0 ðhÞxn . Starting as usual with x ¼ x0 , y ¼ y0 , y 0 ¼ ðy 0 Þ0 and h, we have x1 ¼ x0 þ h Finding y1 requires four intermediate calculations k1 ¼ h f ðx0 ; y0 Þ ¼ hðy 0 Þ0 k2 ¼ h f ðx0 þ 12 h; y0 þ 12 k1 Þ k3 ¼ h f ðx0 þ 12 h; y0 þ 12 k2 Þ k4 ¼ h f ðx0 þ h; y0 þ k3 Þ The increment y0 in the y-values from x ¼ x0 to x ¼ x1 is then y0 ¼ 16 fk1 þ 2k2 þ 2k3 þ k4 g y1 ¼ y0 þ y0 . and finally We shall be using these repeatedly, so make a note of them for future reference. Then let us see an example 39 Example 1 Find the numerical solution of y 0 ¼ x þ y using the Runge–Kutta method with y ¼ 1 and x ¼ 0 for values in the range x ¼ 0ð0:1Þ1:0. We shall proceed with the solution of this differential equation using a spreadsheet in much the same manner as before. However, we are going to require a different structure in order to accommodate the four variables ki for i ¼ 1, 2, 3, 4. The structure we shall use is headed by 1 A n B x C k1 D k2 E k3 F k4 G y H y’ I h= where the value of h is held in cell J1. 1 2 3 4 Enter the values 0 to 10 in column A from A2 to A12 using the Edit-FillSeries sequence of commands. These are the iteration numbers. Enter the x step value of 0.1 in cell J1. Enter the initial value of x in cell B2 as 0 and in B3 enter the formula = B2 + $J$1. Now copy the contents of B3 into cells B4 to B12. Enter the initial value of y in cell G2 as 1. 423 Numerical solutions of ordinary differential equations We can now progressively enter the table of values from the left. k1 ¼ hf ðx0 , y0 Þ ¼ hðy 0 Þ0 – the y 0 -values are in column H, so in cell C2 enter the formula = $J$1 H2. Copy the contents of C2 into cells C3 to C12. k2 ¼ hf x0 þ 12 h, y0 þ 12 k1 ¼ h x0 þ 12 h þ y0 þ 12 k1 , so in cell D2 enter the formula = $J$1 (B2 + 0.5 $J$1 + G2 + 0.5 C2). Copy the contents of D2 into cells D3 to D12. k3 ¼ hf ðx0 þ 12 h, y0 þ 12 k2 Þ ¼ hðx0 þ 12 h þ y0 þ 12 k2 Þ, so in cell E2 enter the formula = $J$1 (B2 + 0.5 $J$1 + G2 + 0.5 D2). Copy the contents of E2 into cells E3 to E12. k4 ¼ hf ðx0 þ h, y0 þ k3 Þ ¼ hðx0 þ h þ y0 þ k3 Þ, so in cell F2 enter the formula = $J$1 (B2 + $J$1 + G2 + E2). Copy the contents of F2 into cells F3 to F12. y1 ¼ y0 þ 16 fk1 þ 2k2 þ 2k3 þ k4 g, so in cell G3 enter the formula = G2+(1/6) (C2 + 2 D2 + 2 E2 + F2). Copy the contents of G3 into cells G4 to G12. y 0 ¼ x þ y, so in H2 enter the formula = B2 + G2. Copy the contents of H2 into cells H3 to H12. The results are displayed in the next frame 5 6 7 8 9 10 n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 k1 0.1000000 0.1210342 0.1442805 0.1699717 0.1983648 0.2297441 0.2644236 0.3027503 0.3451079 0.3919203 0.4436559 k2 0.1100000 0.1320859 0.1564945 0.1834703 0.2132831 0.2462313 0.2826448 0.3228878 0.3673633 0.4165163 0.4708387 k3 0.1105000 0.1326385 0.1571052 0.1841452 0.2140290 0.2470557 0.2835558 0.3238947 0.3684761 0.4177461 0.4721979 k4 0.1210500 0.1442980 0.1699910 0.1983862 0.2297677 0.2644497 0.3027792 0.3451398 0.3919555 0.4436949 0.5008757 y 1.0000000 1.1103417 1.2428051 1.3997170 1.5836485 1.7974413 2.0442359 2.3275033 2.6510791 3.0192028 3.4365595 y’ h = 0.1 1.0000000 1.2103417 1.4428051 1.6997170 1.9836485 2.2974413 2.6442359 3.0275033 3.4510791 3.9192028 4.4365595 with the following errors n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Exact 1.0000000 1.1103418 1.2428055 1.3997176 1.5836494 1.7974425 2.0442376 2.3275054 2.6510819 3.0192062 3.4365637 Error (%) 0.0000000 0.0000153 0.0000301 0.0000444 0.0000578 0.0000703 0.0000820 0.0000929 0.0001030 0.0001124 0.0001213 Error (%) 0.00 0.93 1.84 2.69 3.50 4.25 4.95 5.59 6.18 6.73 7.25 The column to the far right contains the errors using the Euler method and, as you can see, the Runge–Kutta method provides a significant improvement in accuracy. 40 424 Programme 13 Now, without reference to your notes, complete the following expressions for k1 ¼ . . . . . . . . . . . . k2 ¼ . . . . . . . . . . . . k3 ¼ . . . . . . . . . . . . k4 ¼ . . . . . . . . . . . . y0 ¼ . . . . . . . . . . . . y1 ¼ . . . . . . . . . . . . It speeds up your working if you can remember them. 41 k1 ¼ hðy 0 Þ0 k2 ¼ hf x0 þ 12 h, y0 þ 12 k1 k3 ¼ hf x0 þ 12 h, y0 þ 12 k2 k4 ¼ hf ðx0 þ h, y0 þ k3 Þ y0 ¼ 16 ðk1 þ 2k2 þ 2k3 þ k4 Þ y1 ¼ y0 þ y0 With those in mind, let us move on to a further example. Next frame Example 2 42 Solve y 0 ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y for x ¼ 0ð0:2Þ2:0 given that at x ¼ 0, y ¼ 0:8. Using the spreadsheet for the previous example as a template for this example. The solution is . . . . . . . . . . . . 43 n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 k1 0.1788854 0.2030063 0.2339473 0.2698011 0.3091304 0.3509358 0.3945381 0.4394732 0.4854190 0.5321472 0.5794925 k2 0.1896779 0.2174206 0.2510185 0.2887709 0.3294604 0.3722562 0.4165946 0.4620889 0.5084682 0.5555390 0.6031595 k3 0.1902460 0.2180825 0.2516977 0.2894271 0.3300769 0.3728285 0.4171237 0.4625781 0.5089213 0.5559599 0.6035518 k4 0.2030021 0.2339548 0.2698134 0.3091435 0.3509482 0.3945492 0.4394829 0.4854274 0.5321545 0.5794989 0.6273385 y 0.8000000 0.9902892 1.2082838 1.4598160 1.7490394 2.0788983 2.4515074 2.8684170 3.3307894 3.8395148 4.3952888 y’ h = 0.2 0.8944272 1.0150316 1.1697366 1.3490055 1.5456518 1.7546790 1.9726904 2.1973659 2.4270948 2.6607358 2.8974625 Because 1 2 The initial conditions are entered as x0 ¼ 0 and y0 ¼ 0:8. The x step length is entered as 0.2 The formula for the variable k1 remains the same as = $J$1 H2 425 Numerical solutions of ordinary differential equations 3 4 5 6 7 The formula for the variable k2 is changed to = $J$1 (((B2 + 0.5 $J$1)^2 + G2 + 0.5 C2)^0.5) The formula for the variable k3 is changed to = $J$1 (((B2 + 0.5 $J$1)^2 + G2 + 0.5 D2)^0.5) The formula for the variable k4 is changed to = $J$1 (((B2 + $J$1)^2 + G2 + E2)^0.5) The formula for y remains the same as = G2+{1/6) (C2 + 2 D2 + 2 E2 + F2) The formula for y 0 is changed to = (B2^2+G2)^0.5 That is it. Now move on to the next frame where we make a new start and apply similar methods to the solution of second-order differential equations by numerical methods. Second-order differential equations 44 Euler second-order method The first method we will deal with is really an extension of the Euler method for the first-order equations and is a direct application of a truncated form of Taylor’s series. We anticipate, therefore, that the method will be relatively easy, but the results will not be accurate to a high degree. Taylor’s series: h2 00 h3 f ðxÞ þ f 000 ðxÞ þ . . . 2! 3! Differentiating term by term with respect to x, we obtain f ðx þ hÞ ¼ f ðxÞ þ h f 0 ðxÞ þ h2 000 h3 f ðxÞ þ f 0000 ðxÞ þ . . . 2! 3! If we neglect terms in f 000 ðxÞ and subsequent terms in each of these two series, we have the approximations f 0 ðx þ hÞ ¼ f 0 ðxÞ þ h f 00 ðxÞ þ f ðx þ hÞ . . . . . . . . . . . . f 0 ðx þ hÞ . . . . . . . . . . . . f ðx þ hÞ f ðxÞ þ h f 0 ðxÞ þ h2 00 f ðxÞ 2! f 0 ðx þ hÞ f 0 ðxÞ þ h f 00 ðxÞ Although these are approximations, in practice we tend to write them with the ‘equals’ sign. Therefore, at x ¼ a, these become and ........................ ........................ 45 426 Programme 13 46 f ða þ hÞ ¼ f ðaÞ þ h f 0 ðaÞ þ h2 00 f ðaÞ 2! f 0 ða þ hÞ ¼ f 0 ðaÞ þ h f 00 ðaÞ and these, with the notation we have previously used, can be written y1 ¼ y0 þ hðy 0 Þ0 þ h2 00 ðy Þ0 2! ðy 0 Þ1 ¼ ðy 0 Þ0 þ hðy 00 Þ0 Thus, if x0 , y0 , ðy 0 Þ0 and ðy 00 Þ0 are known, we can find an approximate value of y1 at x1 ¼ x0 þ h: Make a note of these two relationships: then we can apply them. 47 Example Solve the equation y 00 ¼ xy 0 þ y for x ¼ 0ð0:2Þ2:0 given that at x ¼ 0, y ¼ 1 and y 0 ¼ 0. We shall set about finding the numerical solution to this equation as we have done previously by using a spreadsheet. The headings for the sheet will be 1 A n B x C y D y’ E y’’ F Exact G Errors (%) H h= The entries will then be 1 2 3 4 Column A contains the iteration number from 0 in A2 to 10 in A12. Cell I1 contains the x step length which is 0.2. Column B contains the successive x-values from 0.0 to 2.0 in steps of 0.2. The initial value of x0 ¼ 0 is entered into cell B2 and the formula = B2 + $I$1 is entered into cell B3 and copied into cells B4 to B12. Column C contains the computed y-values. The initial value of y0 ¼ 1 is entered into cell C2 and the equation y1 ¼ y0 þ hðy 0 Þ0 þ h2 00 ðy Þ0 2! is represented in cell C3 by the formula = C2 + $I$1 D2 + ($I$1^2) E2/2 5 copied into cells C4 to C12. Column D contains the computed y 0 -values. The initial value of ðy 0 Þ0 ¼ 0 is entered into cell D2 and the equation ðy 0 Þ1 ¼ ðy 0 Þ0 þhðy 00 Þ0 6 is represented in cell D3 by the formula = D2 + $I$1 E2 copied into cells D4 to D12. Column E contains the y 00 -values which are obtained from the equation y 00 ¼ xy 0 þ y which is represented in cell E2 by the formula = B2 D2 + C2 copied into cells E3 to E12. 427 Numerical solutions of ordinary differential equations 7 Column F contains the values obtained from the exact solution which can 2 be shown to be y ¼ ex =2 . This is represented in cell F2 by the formula = EXP((B2^2)/2) copied into cells F3 to F12. Column G contains the percentage errors. In cell G2 enter the formula = (F2 – C2) 100/F2 copied into cells G3 to G12. 8 Your spreadsheet should now look like the one on the next page (with the appropriate formatting to make it easier to read). n x 0 1 2 3 4 5 6 7 8 9 10 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2 .0 y 1.0000000 1.0200000 1.0812000 1.1885200 1.3524648 1.5908157 1.9317893 2.4194277 3.1226113 4.1501667 5.6764580 y’ 0.0000000 0.2000000 0.4120000 0.6612000 0.9782480 1.4052606 2.0044759 2.8719080 4.1599278 6.1156269 9.1472859 y’’ 1.0000000 1.0600000 1.2460000 1.5852400 2.1350632 2.9960763 4.3371604 6.4400989 9.7784957 15.1582952 23.9710299 Exact 1.0000000 1.0202013 1.0832871 1.1972174 1.3771278 1.6487213 2.0544332 2.6644562 3.5966397 5.0530903 7.3890561 Errors (%) 0.00 0.02 0.19 0.73 1.79 3.51 5.97 9.20 13.18 17.87 23.18 h = 0.2 You will notice that the errors are significant and grow dramatically as the value of x increases. The main cause of errors is . . . . . . . . . . . . the truncation of the Taylor’s series on which the method is based A greater degree of accuracy can be obtained by using the Runge–Kutta method for second-order differential equations, which is an extension of the method we have already used for first-order equations. As before, more intermediate calculations are required, but the reliability of results reflects the extra work involved. Runge–Kutta method for second-order differential equations Starting with the given equation y 00 ¼ f ðx; y; y 0 Þ and initial conditions that at x ¼ x0 , y ¼ y0 and y 0 ¼ ðy 0 Þ0 , we can obtain the value of y1 at x1 ¼ x0 þ h as follows. (a) We evaluate k1 ¼ 12 h2 f fx0 ; y0 ; ðy 0 Þ0 g ¼ 12 h2 ðy 00 Þ0 k1 h k 2 k3 ¼ 12 h2 f x0 þ 12 h; y0 þ 12 hðy 0 Þ0 þ 14 k1 ; ðy 0 Þ0 þ h 2k3 k4 ¼ 12 h2 f x0 þ h; y0 þ hðy 0 Þ0 þ k3 ; ðy 0 Þ0 þ h k2 ¼ 12 h2 f x0 þ 12 h; y0 þ 12 hðy 0 Þ0 þ 14 k1 ; ðy 0 Þ0 þ 48 428 Programme 13 (b) From these results, we then determine P ¼ 13 fk1 þ k2 þ k3 g Q ¼ 13 fk1 þ 2k2 þ 2k3 þ k4 g (c) Finally, we have x1 ¼ x0 þ h y1 ¼ y0 þ hðy 0 Þ0 þ P Q ðy 0 Þ1 ¼ ðy 0 Þ0 þ h It is not as complicated as it looks at first sight. Copy down this list of relationships for reference when dealing with some examples that follow. Then move on 49 Note the following 1 Four evaluations for k are required to determine a single new point on the solution curve. The method is self-starting in that no preliminary calculations are required. The equation and initial conditions are sufficient to provide the next point on the curve. As with the Runge–Kutta method for first-order equations, the method contains no self-correcting element or indication of any error involved. 2 3 Example 1 Use the Runge–Kutta method to solve the equation y 00 ¼ xy 0 þ y for x ¼ 0:0ð0:2Þ2:0 given that at x ¼ 0, y ¼ 1 and y 0 ¼ 0. This is the same problem that we have just encountered and in due course we shall compare results. As expected, we shall use a spreadsheet to derive the solution. The headings for the sheet this time will be 1 A n B x C k1 D k2 E k3 F k4 G P H Q I y J y’ K y’’ L h= The entries will then be 1 2 3 4 Column A contains the iteration number from 0 in A2 to 10 in A12. Cell M1 contains the x step length which is 0.2. Column B contains the successive x-values from 0.0 to 2.0 in steps of 0.2. The initial value of x0 ¼ 0 is entered into cell B2 and the formula = B2 + $M$1 is entered into cell B3 and copied into cells B4 to B12. Column C contains the computed k1 -values and the equation k1 ¼ 12 h2 ðy 00 Þ0 is represented in cell C2 by the formula . . . . . . . . . . . . Numerical solutions of ordinary differential equations = (0.5) ($M$1^2) K2 429 50 The contents of cell C2 are then copied into cells C3 to C12. 5 Column D contains the computed k2 -values and the equation k2 ¼ 12 h2 f x0 þ 12 h, y0 þ 12 hðy 0 Þ0 þ 14 k1 , ðy 0 Þ0 þk1 =h ¼ 12 h2 ðx0 þ 12 hÞ ðy 0 Þ0 þk1 =h þ y0 þ 12 hðy 0 Þ0 þ 14 k1 is represented in cell D2 by the formula . . . . . . . . . . . . =(0.5) ($M$1^2) ((B2 + 0.5 $M$1) (J2 + C2/$M$1) + I2 + 0.5 $M$1 J2+0.25 C2) 51 The contents of cell D2 are then copied into cells D3 to D12. 6 Column E contains the computed k3 -values and the equation k3 ¼ 12 h2 f x0 þ 12 h, y0 þ 12 hðy 0 Þ0 þ 14 k1 , ðy 0 Þ0 þk2 =h ¼ 12 h2 ðx0 þ 12 hÞ ðy 0 Þ0 þk2 =h þ y0 þ 12 hðy 0 Þ0 þ 14 k1 is represented in cell E2 by the formula . . . . . . . . . . . . =(0.5) ($M$1^2) ((B2 + 0.5 $M$1) (J2 + D2/$M$1) + I2 + 0.5 $M$1 J2+0.25 C2) 52 The contents of cell E2 are then copied into cells E3 to E12. 7 Column F contains the computed k4 -values and the equation k4 ¼ 12 h2 f x0 þ h, y0 þ hðy 0 Þ0 þk3 , ðy 0 Þ0 þ2k3 =h ¼ 12 h2 ðx0 þ hÞ ðy 0 Þ0 þ2k3 =h þ y0 þ hðy 0 Þ0 þk3 is represented in cell F2 by the formula . . . . . . . . . . . . =(0.5) ($M$1^2) ((B2 + $M$1) (J2 + 2 E2/$M$1) + I2 + $M$1 J2 + E2) The contents of cell F2 are then copied into cells F3 to F12. 8 Column G contains the computed P-values and the equation P ¼ 13 ðk1 þ k2 þ k3 Þ is represented in cell G2 by the formula . . . . . . . . . . . . 53 430 Programme 13 54 =(1/3) (C2 + D2 + E2) The contents of cell G2 are then copied into cells G3 to G12. 9 Column H contains the computed Q-values and the equation Q ¼ 13 ðk1 þ 2k2 þ 2k3 þ k4 Þ is represented in cell H2 by the formula ............ 55 =(1/3) (C2 + 2 D2 + 2 E2 + F2) The contents of cell H2 are then copied into cells H3 to H12. 10 Column I contains the computed y-values. The initial value of y0 ¼ 1 is entered into cell I2 and the equation y1 ¼ y0 þ hðy 0 Þ0 þP is represented in cell I3 by the formula . . . . . . . . . . . . 56 = I2 + $M$1 J2 + G2 The contents of cell I3 are then copied into cells I4 to I12. 11 Column J contains the computed y 0 -values. The initial value of ðy 0 Þ0 ¼ 0 is entered into cell J2 and the equation ðy 0 Þ1 ¼ ðy 0 Þ0 þQ=h is represented in cell J3 by the formula . . . . . . . . . . . . 57 = J2 + H2/$M$1 The contents of cell J3 are then copied into cells J4 to J12. 12 Column K contains the y 00 -values which are obtained from the equation y 00 ¼ xy 0 þ y which is represented in cell K2 by the formula . . . . . . . . . . . . 58 = B2 J2 + I2 The contents of cell K2 are then copied into cells K3 to K12 and the final spreadsheet looks like the following n 0 1 2 3 4 5 6 7 8 9 10 x 0 .0 0 .2 0 .4 0 .6 0 .8 1 .0 1 .2 1 .4 1 .6 1 .8 2 .0 k1 0.0200000 0.0212202 0.0251322 0.0325641 0.0451694 0.0659480 0.1002542 0.1577302 0.2560654 0.4284592 0.7387844 k2 0.0203000 0.0227790 0.0282477 0.0378798 0.0538673 0.0801269 0.1236497 0.1970991 0.3238945 0.5483881 0.9567270 k3 0.0203030 0.0228258 0.0284035 0.0382519 0.0546501 0.0816865 0.1266912 0.2030044 0.3354254 0.5711745 1.0024949 k4 0.0212182 0.0251351 0.0325752 0.0451961 0.0660061 0.1003762 0.1579840 0.2565931 0.4295622 0.7411112 1.3181400 P 0.0202010 0.0222750 0.0272612 0.0362319 0.0512289 0.0759205 0.1168650 0.1859446 0.3051284 0.5160073 0.8993354 Q 0.0408081 0.0458550 0.0570033 0.0766745 0.1094035 0.1633170 0.2529733 0.4048434 0.6680891 1.1362318 1.9917894 y y’ y’’ h = 0.2 1.0000000 0.0000000 1.0000000 1.0202010 0.2040403 1.0610091 1.0832841 0.4333153 1.2566102 1.1972083 0.7183318 1.6282074 1.3771066 1.1017045 2.2584702 1.6486764 1.6487218 3.2973982 2.0543413 2.4653068 5.0127095 2.6642677 3.7301734 7.8865105 3.5962469 5.7543906 12.8032719 5.0522535 9.0948361 21.4229585 7.3872279 14.7759954 36.9392186 431 Numerical solutions of ordinary differential equations The errors have been dramatically reduced, as can be seen from the following table in comparison with those in Frame 47. n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 Exact 1.0000000 1.0202013 1.0832871 1.1972174 1.3771278 1.6487213 2.0544332 2.6644562 3.5966397 5.0530903 7.3890561 Error (%) 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.01 0.02 0.02 Next frame Now here is one for you to do entirely on your own. The method is exactly the same as before and there are no snags. Use the spreadsheet that you created for the previous example as a template for this one. 59 Example 2 Solve the equation y 00 ¼ x y 2 for x ¼ 0:0ð0:2Þ2:0 where at x ¼ 0, y ¼ 0 and y 0 ¼ 0. When you have finished, check the results with the next frame n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 k1 0.0000000 0.0040000 0.0079977 0.0119741 0.0158546 0.0194477 0.0223654 0.0239482 0.0232395 0.0190914 0.0104978 k2 0.0020000 0.0059996 0.0099915 0.0139351 0.0177065 0.0210264 0.0233782 0.0239504 0.0216639 0.0153806 0.0043750 k3 k4 0.0020000 0.0039999 0.0059996 0.0079974 0.0099915 0.0119731 0.0139351 0.0158524 0.0177065 0.0194436 0.0210264 0.0223594 0.0233782 0.0239421 0.0239504 0.0232394 0.0216639 0.0191107 0.0153806 0.0105578 0.0043750 0.0026809 P 0.0013333 0.0053331 0.0093269 0.0132814 0.0170892 0.0205002 0.0230406 0.0239497 0.0221891 0.0166175 0.0064159 Q 0.0040000 0.0119986 0.0199789 0.0278556 0.0353748 0.0419709 0.0466068 0.0476631 0.0430019 0.0303905 0.0084390 y 0.0000000 0.0013333 0.0106664 0.0359919 0.0852508 0.1661731 0.2858812 0.4501006 0.6618359 0.9194737 1.2145418 y’ 0.0000000 0.0199999 0.0799930 0.1798875 0.3191655 0.4960396 0.7058940 0.9389280 1.1772436 1.3922530 1.5442053 y’’ h=0.2 0.0000000 0.1999982 0.3998862 0.5987046 0.7927323 0.9723865 1.1182719 1.1974094 1.1619732 0.9545681 0.5248882 Because The only items that need amending from the previous spreadsheet are the references to the actual differential equation. Consequently The formula in D2 for k2 now reads as = (0.5) ($M$1^2) (B2 + 0.5 $M$1 – (I2 + 0.5 $M$1 J2 + 0.25 C2)^2) The formula in E2 for k3 now reads as = 0.5 ($M$1^2) (B2 + 0.5 $M$1 – (I2 + 0.5 $M$1 J2 + 0.25 C2)^2) 60 432 Programme 13 The formula in F2 for k4 now reads as = 0.5 ($M$1^2) (B2+$M$1 – (I2 + $M$1 J2 + E2)^2) The formula in K2 for y 00 now reads as = B2 – I2^2 Predictor–corrector methods 61 So far, all the methods that we have used for the numerical solution of differential equations have been single-step methods. By this is meant that, given the differential equation y 0 ¼ f ðx, yÞ, a set of starting values (x0 and y0 ) and a step length (h), we can then find the value of y1 . The values of x1 and y1 become the starting values for the next iteration and so the procedure goes on, one step at a time. More accurate methods employ a multi-step procedure where, instead of starting with just a single set of initial values, we use a collection of previously calculated values. A very simple multi-step method is given by the equations y 1 ¼ y0 þ hf ðx0 , y0 Þ y1 ¼ y0 þ 12 h f ðx0 , y0 Þ þ f ðx1 , y 1 Þ Here we calculate y 1 first from the given initial conditions x0 and y0 . We call this equation the predictor because it gives y 1 as a first estimate of y1 . Using y 1 in the second equation then gives a more accurate value for y1 . We call this equation the corrector. An even better pair of predictor–corrector equations is given by y iþ1 ¼ yi þ 12 hð3f ðxi , yi Þ f ðxi1 , yi1 ÞÞ yiþ1 ¼ yi þ 12 h f ðxi , yi Þ þ f ðxiþ1 , y iþ1 Þ for i ¼ 0, 1, 2, 3, . . . Here, in order to use the predictor for the first time when i ¼ 0 we need to know the value of f ðx01 , y01 Þ ¼ f ðx1 , y1 Þ, which we do not. Instead we shall use the equation y 1 ¼ y0 þ hf ðx0 , y0 Þ when i ¼ 0. In the next frame we shall look at an example 62 Example Solve the equation y 0 ¼ x þ y for x ¼ 0:0ð0:1Þ1:0 where y ¼ 1 when x ¼ 0. We have solved this equation before in Frame 32 using the Euler–Cauchy method and have viewed the accuracy of this method when compared with the exact solution. Here we shall see that this predictor–corrector method is even more accurate. Set up the following heading on your spreadsheet 1 A n B x C y* D y E Exact F Errors (%) G h= 433 Numerical solutions of ordinary differential equations As usual, column A contains the iteration numbers 0 to 10 in cells A2 to A12 and column B contains the x-values stepped according to the step length h ¼ 0:1 which is in cell H1. The initial value of y ¼ 1 must be entered into cell D2. Column C contains the predictor values given by the equations y 1 ¼ y0 þ hf ðx0 , y0 Þ y iþ1 ¼ yi þ 12 hð3f ðxi , yi Þ f ðxi1 , yi1 ÞÞ for i > 0 To accommodate these equations in cell C3 enter the formula . . . . . . . . . . . . 63 = D2 + $H$1 (B2 + D2) And in cell C4 enter the formula . . . . . . . . . . . . 64 = D3 + 0.5 $H$1 (3 B3 + 3 D3 – B2 – D2) And copy into cells C5 to C12. Column D contains the corrector values given by the equation yiþ1 ¼ yi þ 12 h f ðxi , yi Þ þ f ðxiþ1 , y iþ1 Þ To accommodate this equation in cell D3 enter the formula . . . . . . . . . . . . 65 = D2 + 0.5 $H$1 (B2 + D2 + B3 + C3) And copy into cells D4 to D12. We have seen that the exact solution to this equation is 2ex x 1, so this can be programmed into the sheet entering the formula = 2 EXP(B2) – B2 – 1 in cell E2 and then copying it into cells E3 to E12. The final table looks as follows n 0 1 2 3 4 5 6 7 8 9 10 x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 y* 1.1000000 1.2415000 1.3984613 1.5824421 1.7963085 2.0432055 2.3266093 2.6503618 3.0187092 3.4363445 y 1.0000000 1.1100000 1.2425750 1.3996268 1.5837303 1.7977322 2.0447791 2.3283485 2.6522840 3.0208337 3.4386926 Exact 1.0000000 1.1103418 1.2428055 1.3997176 1.5836494 1.7974425 2.0442376 2.3275054 2.6510819 3.0192062 3.4365637 Error (%) 0.00 0.03 0.02 0.01 0.01 0.02 0.03 0.04 0.05 0.05 0.06 h = 0.1 434 Programme 13 Here the errors are significantly reduced, as seen from the comparisons below. 1 0.00 0.93 1.84 2.69 3.50 4.25 4.95 5.59 6.18 6.73 7.25 2 0.00 0.03 0.06 0.09 0.12 0.14 0.17 0.19 0.21 0.23 0.24 3 0.00 0.03 0.02 0.01 0.01 0.02 0.03 0.04 0.05 0.05 0.06 Here 1 refers to Euler, 2 refers to Euler–Cauchy and 3 refers to the predictor– corrector method just used. And that is it. There are many other more sophisticated methods for the solution of ordinary differential equations by numerical methods and a detailed study of these is a course in itself. The methods we have used give an introduction to the processes and are practical in application. The Revision summary and Can You? checklist now follow as usual. Check them carefully and refer back to the Programme for any points that may need further brushing up. Then you will be ready for the Test exercise, and the Further problems provide further practice. Revision summary 13 66 1 Taylor’s series f ða þ hÞ ¼ f ðaÞ þ h f 0 ðaÞ þ 2 h 00 h f ðaÞ þ f 000 ðaÞ þ ::: 2! 3! Solution of first-order differential equations Equation y 0 ¼ f ðx; yÞ with y ¼ y0 at x ¼ x0 for x0 ðhÞxn : (a) Euler’s method y1 ¼ y0 þ hðy 0 Þ0 . (b) Euler–Cauchy method x1 ¼ x0 þ h y 1 ¼ y0 þ hðy 0 Þ0 y1 ¼ y0 þ 12 hfðy 0 Þ0 þ f ðx1 ; y 1 Þg ðy 0 Þ1 ¼ f ðx1 ; y1 Þ. 435 Numerical solutions of ordinary differential equations (c) Runge–Kutta method x1 ¼ x0 þ h k1 ¼ h f ðx0 ; y0 Þ ¼ hðy 0 Þ0 k2 ¼ h f ðx0 þ 12 h; y0 þ 12 k1 Þ k3 ¼ h f ðx0 þ 12 h; y0 þ 12 k2 Þ k4 ¼ h f ðx0 þ h; y0 þ k3 Þ y0 ¼ 16 ðk1 þ 2k2 þ 2k3 þ k4 Þ y1 ¼ y0 þ y0 ðy 0 Þ1 ¼ f ðx1 ; y1 Þ: 3 Solution of second-order differential equations Equation y 00 ¼ f ðx; y; y 0 Þ with y ¼ y0 and y 0 ¼ ðy 0 Þ0 at x ¼ x0 for x ¼ x0 ðhÞxn . (a) Euler’s second-order method h2 00 ðy Þ0 2! 0 0 00 ðy Þ1 ¼ ðy Þ0 þ hðy Þ0 . y1 ¼ y0 þ hðy 0 Þ0 þ (b) Runge–Kutta method x1 ¼ x0 þ h k1 ¼ 12 h2 f fx0 ; y0 ; ðy 0 Þ0 g ¼ 12 h2 ðy 00 Þ0 k1 h k 2 k3 ¼ 12 h2 f x0 þ 12 h; y0 þ 12 hðy 0 Þ0 þ 14 k1 ; ðy 0 Þ0 þ h 2k3 k4 ¼ 12 h2 f x0 þ h; y0 þ hðy 0 Þ0 þ k3 ; ðy 0 Þ0 þ h k2 ¼ 12 h2 f x0 þ 12 h; y0 þ 12 hðy 0 Þ0 þ 14 k1 ; ðy 0 Þ0 þ P ¼ 13 ðk1 þ k2 þ k3 Þ Q ¼ 13 ðk1 þ 2k2 þ 2k3 þ k4 Þ y1 ¼ y0 þ hðy 0 Þ0 þ P Q ðy 0 Þ1 ¼ ðy 0 Þ0 þ h ðy 00 Þ1 ¼ f fx1 ; y1 ; ðy 0 Þ1 g. 4 Predictor–corrector Equation y 0 ¼ f ðx, yÞ with y ¼ y0 and y 0 ¼ ðy 0 Þ0 at x ¼ x0 for x ¼ x0 ðhÞxn , then Predictor y iþ1 ¼ yi þ 12 hð3f ðxi , yi Þ f ðxi1 , yi1 ÞÞ y 1 ¼ y0 þ hf ðx0 , y0 Þ for i ¼ 1, 2, 3, . . . for i ¼ 0 Corrector yiþ1 ¼ yi þ 12 h f ðxi , yi Þ þ f ðxiþ1 , y iþ1 Þ for i ¼ 0, 1, 2, 3, . . . 436 Programme 13 Can you? 67 Checklist 13 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Derive a form of Taylor’s series from Maclaurin’s series and from it describe a function increment as a series of first and higher-order derivatives of the function? Yes No . Describe and apply by means of a spreadsheet the Euler method, the Euler–Cauchy method and the Runge–Kutta method for first-order differential equations? Yes No . Describe and apply by means of a spreadsheet the Euler second-order method and the Runge–Kutta method for second-order ordinary differential equations? Yes No . Describe and apply by means of a spreadsheet a simple predictor–corrector method? Yes No Frames 1 to 3 4 to 43 44 to 60 61 to 65 Test exercise 13 68 1 2 Apply Euler’s method to solve the equation dy ¼ 1 þ xy for x ¼ 0ð0:1Þ0:5 dx given that at x ¼ 0; y ¼ 1. dy ¼ x2 2y is subject to the initial condition y ¼ 0 at x ¼ 1. Use dx the Euler–Cauchy method to obtain function values for x ¼ 1:0ð0:2Þ2:0: The equation 3 Using the Runge–Kutta method, solve the equation dy ¼ 1 þ y x for x ¼ 0ð0:1Þ0:5 dx given that y ¼ 1 when x ¼ 0. 4 Apply Euler’s second-order method to solve the equation y 00 ¼ y x for x ¼ 2:0ð0:1Þ2:5 given that at x ¼ 2; y ¼ 3 and y 0 ¼ 0: Numerical solutions of ordinary differential equations 5 Use the Runge–Kutta method to solve the equation y 00 ¼ ðy 0 =xÞ þ y for x ¼ 1:0ð0:1Þ1:5 given the initial conditions that at x ¼ 1:0; y ¼ 0 and y 0 ¼ 1:0. 6 Use the predictor–corrector method in the text to solve the equation y 0 ¼ 1 þ xy for x ¼ 0ð0:1Þ1 437 given that x ¼ 0 when y ¼ 0. Further problems 13 Solve the following differential equations by the methods indicated. Euler’s method 1 y 0 ¼ 2x y x ¼ 0, y ¼ 1 x ¼ 0ð0:2Þ1:0 2 y 0 ¼ 2x þ y 2 x ¼ 0, y ¼ 1:4 x ¼ 0ð0:1Þ0:5 x ¼ 1, y ¼ 2 x ¼ 1:0ð0:2Þ2:0 x ¼ 0, y ¼ 0:5 x ¼ 0ð0:1Þ0:5 x ¼ 0, y ¼ 1 x ¼ 0ð0:1Þ0:5 xþy xy x ¼ 1, y ¼ 1 x ¼ 1:0ð0:1Þ1:5 7 y 0 ¼ y sin x þ cos x x ¼ 0, y ¼ 0 x ¼ 0ð0:1Þ0:5 8 y 0 ¼ 2x y x ¼ 0, y ¼ 1 x ¼ 0ð0:2Þ1:0 9 y 0 ¼ x y2 x ¼ 0, y ¼ 1 x ¼ 0ð0:1Þ0:5 10 y 0 ¼ y 2 xy pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 11 y 0 ¼ 2x þ y x ¼ 0, y ¼ 0:4 x ¼ 0ð0:2Þ1:0 x ¼ 1, y ¼ 2 x ¼ 1:0ð0:2Þ2:0 12 y 0 ¼ 1 x3 =y x ¼ 0, y ¼ 1 x ¼ 0ð0:2Þ1:0 x ¼ 0, y ¼ 1 x ¼ 0ð0:2Þ1:0 14 y 00 ¼ ðx þ 1Þy 0 þ y x ¼ 0, y ¼ 1, y 0 ¼ 1 x ¼ 0ð0:1Þ0:5 15 y 00 ¼ 2ðxy 0 4yÞ x ¼ 0, y ¼ 3, y 0 ¼ 0 x ¼ 0ð0:1Þ0:5 Euler–Cauchy method 3 y 0 ¼ 2 y=x 4 y 0 ¼ x2 2x þ y 1 2 5 y 0 ¼ ðy x2 Þ 6 y0 ¼ Runge Kutta method 13 y 0 ¼ yx yþx Euler second-order method 69 438 Programme 13 Runge–Kutta second-order method 16 y 00 ¼ x y xy 0 x ¼ 0, y ¼ 0, y 0 ¼ 1 x ¼ 0ð0:2Þ1:0 17 y 00 ¼ ð1 xÞy 0 y x ¼ 0, y ¼ 1, y 0 ¼ 1 x ¼ 0ð0:2Þ1:0 18 y 00 ¼ 1 þ x y 2 x ¼ 0, y ¼ 2, y 0 ¼ 1 x ¼ 0ð0:1Þ0:5 19 y 00 ¼ ðx þ 2Þy 2y 0 x ¼ 0, y ¼ 1, y 0 ¼ 0 x ¼ 0ð0:2Þ1:0 x ¼ 1, y ¼ 0, y 0 ¼ 1 x ¼ 1:0ð0:2Þ2:0 21 y 0 ¼ 2 y=x x ¼ 1, y ¼ 2 x ¼ 1:0ð0:2Þ2:0 22 y 0 ¼ 2x y pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 23 y 0 ¼ 2x þ y x ¼ 0, y ¼ 1 x ¼ 0:0ð0:2Þ1:0 x ¼ 1, y ¼ 2 x ¼ 1:0ð0:2Þ2:0 20 y 00 ¼ y xy 0 x2 Predictor–corrector Programme 14 Frames 1 to 81 Partial differentiation Learning outcomes When you have completed this Programme you will be able to: . Derive the expression for a small increment in an expression of two real variables using Taylor’s theorem . Apply the notion of small increments in expressions in two and three real variables to a variety of problems . Determine the rate of change with respect to time of an expression involving two or three real variables . Differentiate implicit functions . Determine first and second derivatives involving change of variables in expressions of two real variables . Use the Jacobian to obtain the derivatives of inverse functions of two real variables . Locate and identify maxima, minima and saddle points of functions of two real variables . Solve problems where the independent variables are constrained by using the method of Lagrange undetermined multipliers for functions of two and three real variables Prerequisite: Engineering Mathematics (Sixth Edition) Programmes 9 Differentiation applications 2, 10 Partial differentiation 1 and 11 Partial differentiation 2 439 440 Programme 14 Small increments 1 Taylor’s theorem for one independent variable Taylor’s theorem expands f ðx þ hÞ in terms of f ðxÞ, powers of h and successive derivatives of f ðxÞ, and can be stated as f ðx þ hÞ ¼ f ðxÞ þ hf 0 ðxÞ þ h2 00 hn f ðxÞ þ . . . þ f ðnÞ ðxÞ þ . . . 2! n! where f ðnÞ ðxÞ denotes the nth derivative of f ðxÞ. You will also, no doubt, remember that, by putting x ¼ 0 in the result and then letting h ¼ x, we obtain Maclaurin’s series f ðhÞ ¼ f ð0Þ þ hf 0 ð0Þ þ h2 00 hn f ð0Þ þ . . . þ f ðnÞ ð0Þ þ . . . 2! n! Taylor0 s theorem for two independent variables If we consider z ¼ f ðx, yÞ where z is a function of two independent variables x and y, then, in general, increases in x and y will produce a combined increase in z. So, if z ¼ f ðx, yÞ then z þ z ¼ f ðx þ h, y þ kÞ (x + h , y + k ) y h ¼ increase in x k k ¼ increase in y. (x, y) h (x + h , y ) x O For R: f ðx þ h, yÞ ¼ f ðx, yÞ þ hfx ðx, yÞ þ where fx ðx, yÞ denotes From R to Q: h2 fxx ðx, yÞ þ . . . 2! ð1Þ @ @2 f ðx, yÞ; fxx ðx, yÞ denotes 2 f ðx, yÞ etc. @x @x ðx þ hÞ is constant; y changes to ðy þ kÞ ; f ðx þ h, y þ kÞ ¼ f ðx þ h, yÞ þ kfy ðx þ h, yÞ þ k2 fyy ðx þ h, yÞ þ . . . 2! ð2Þ To express (2) in terms of f ðx, yÞ we can substitute result (1) for the first term f ðx þ h, yÞ and similar expressions which we shall obtain for fy ðx þ h, yÞ, fyy ðx þ h, yÞ and so on. If we differentiate (1) with respect to y, we have fy ðx þ h, yÞ ¼ . . . . . . . . . . . . 441 Partial differentiation fy ðx þ h, yÞ ¼ fy ðx, yÞ þ hfyx ðx, yÞ þ h2 fyxx ðx, yÞ þ . . . 2! 2 If we now differentiate this result again with respect to y fyy ðx þ h, yÞ ¼ . . . . . . . . . . . . fyy ðx þ h; yÞ ¼ fyy ðx; yÞ þ hfyyx ðx; yÞ þ h2 fyyxx ðx; yÞ þ . . . 2! 3 Then our previous expansion (2), i.e. f ðx þ h; y þ kÞ ¼ f ðx þ h; yÞ þ kfy ðx þ h; yÞ þ k2 fyy ðx þ h; yÞ þ . . . 2! now becomes h2 f ðx þ h, y þ kÞ ¼ f ðx, yÞ þ hfx ðx, yÞ þ fxx ðx, yÞ þ . . . 2! h2 þ k fy ðx, yÞ þ hfyx ðx, yÞ þ fyxx ðx, yÞ þ . . . 2! 2 k h2 fyy ðx, yÞ þ hfyyx ðx, yÞ þ fyyxx ðx, yÞ þ . . . þ 2! 2! þ ... Rearranging the terms by collecting together all the first derivatives, and then all the second derivatives, and so on, we get f ðx þ h, y þ kÞ ¼ . . . . . . . . . . . . f ðx þ h; y þ kÞ ¼ f ðx; yÞ þ fhfx ðx; yÞ þ kfy ðx; yÞg 1 þ fh2 fxx ðx; yÞ þ 2hkfxy ðx; yÞ þ k2 fyy ðx; yÞg þ . . . 2! This is Taylor’s theorem for two independent variables. 4 442 Programme 14 Small increments If z ¼ f ðx, yÞ, h ¼ x, k ¼ y, then Taylor’s theorem can be written as @z @z 1 @2z @2z @2z þ h2 2 þ 2hk þ k2 2 þ . . . z þ z ¼ z þ h þ k @x @y 2! @x @y @x @y Subtracting z from each side: @z @z 1 @2z @2z @2z 2 2 z ¼ x þ y þ ðx yÞ þ þ ... ðxÞ þ 2 ðyÞ @x @y 2! @x2 @y @x @y 2 Since x and y are small, the expression in the brackets is of the next order of smallness and can be discarded for our purposes. Therefore, we arrive at the result If z ¼ f ðx; yÞ then z ¼ @z @z x þ y @x @y As already explained above, this result is, in fact, an approximation since the smaller terms in the series have been neglected. For practical purposes, however, the result can be used as stated. Be sure to make a note of the result, for it is the foundation of much that follows. 5 z ¼ f ðx, yÞ; z ¼ @z @z x þ y @x @y The following diagram illustrates the result. z y z δy y δz = z δx + z δy x y δy (x, y) δx O @z is the slope of PN @x @z is the slope of PL @y QT ¼ QM þ MT z δx x x @z x ¼ QM @x @z ; SL ¼ y ¼ MT @y ; RN ¼ ; z ¼ @z @z x þ y @x @y This is the total increment of z ¼ f ðx, yÞ from P to Q. 443 Partial differentiation It is worth noting at this stage that the result can be extended to the case of three independent variables, i.e. if u ¼ f ðx, y, zÞ u ¼ @u @u @u x þ y þ z @x @y @z One or two straightforward applications will lay the foundations for future development. Example A rectangular box has sides measured as 30 mm, 40 mm and 60 mm. If these measurements are liable to be in error by 0:5 mm, 0:8 mm and 1:0 mm respectively, calculate the length of the diagonal of the box and the maximum possible error in the result. First build up an expression for the diagonal d in terms of the sides, a, b d and c. a d¼ A d¼ C b c pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2 þ c 2 6 Because d2 ¼ a2 þ AC2 ¼ a2 þ b2 þ c2 and so d ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 þ b2 þ c2 @d @d @d a þ b þ c @a @b @c We now determine the partial differential coefficients and obtain an expression for d, but all in terms of a, b and c. Do not yet insert numerical values. Then d ¼ d ¼ . . . . . . . . . . . . 1 d ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi faa þ bb þ ccg 2 a þ b2 þ c 2 Now, substituting the values a ¼ 30, b ¼ 40, c ¼ 60 a ¼ 0:5, b ¼ 0:8, c ¼ 1:0 the calculated length of the diagonal ¼ . . . . . . . . . . . . the maximum possible error ¼ . . . . . . . . . . . . 7 444 Programme 14 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi diagonal ¼ a2 þ b2 þ c2 ¼ 78:10 mm maximum error ¼ 1:37 mm 8 Because d ¼ 1 f30ð 0:5Þ þ 40ð 0:8Þ þ 60ð 1:0Þg 78:10 Greatest error when the signs are the same ; d ¼ 1 f ð15 þ 32 þ 60Þg ¼ 1:37 mm 78:10 Rates of change If z ¼ f ðx; yÞ, then we have seen that z ¼ @z @z x þ y @x @y z @z x @z y ¼ þ t @x t @y t dz @z dx @z dy ¼ þ dt @x dt @y dt Dividing through by t: Then if t ! 0: Note the result. Then on to an example. Example The base radius r of a right circular cone is increasing at the rate of 1.5 mm/s while the perpendicular height is decreasing at 6.0 mm/s. Determine the rate at which the volume V is changing when r ¼ 12 mm and h ¼ 24 mm. Find an expression for 9 dV in terms of r and h which is . . . . . . . . . . . . dt dV 2rh dr r 2 dh ¼ þ dt 3 dt 3 dt V¼ 1 2 r h; 3 @V 2rh ¼ ; @r 3 ; dV @V dr @V dh ¼ þ dt @r dt @h dt @V r 2 ¼ @h 3 dV 2rh dr r 2 dh ¼ þ dt 3 dt 3 dt 445 Partial differentiation Finally, we insert the numerical values: r ¼ 12; h ¼ 24; dr ¼ 1:5; dt dh ¼ 6:0 dt (h is decreasing) dV ¼ 288 288 ¼ 0 dt ; At the instant when r ¼ 12 mm and h ¼ 24 mm, the volume is unchanging: Implicit functions @z @z x þ y enables us to determine the @x @y derivative of an implicit function f ðx, yÞ ¼ 0, i.e. in a case where y is not defined explicitly in terms of x. If f ðx, yÞ ¼ 0 is an implicit function, we let z ¼ f ðx, yÞ. The same initial result, z ¼ Then, as before: Dividing through by x: Then, if x ! 0: But z ¼ 0 ; dz ¼0 dx So, if x2 xy y 2 ¼ 0, @z @z x þ y @x @y z @z @z y ¼ þ x @x @y x dz @z @z dy ¼ þ dx @x @y dx @z @z dy ; þ ¼0 @x @y dx dy @z @z ¼ ; dx @x @y z ¼ dy ¼ ............ dx 10 dy 2x y ¼ dx x þ 2y @z ¼ 2x y @x The rest follows immediately. Putting z ¼ x2 xy y 2 , and @z ¼ x 2y @y Now on to the next frame The work so far, important though it is, is largely by way of revision of the more basic ideas of partial differentiation. We now extend these same ideas to further applications. 11 446 Programme 14 Change of variables If z ¼ f ðx, yÞ and x and y are themselves functions of two new independent @z @z and . variables, u and v, then we need expressions for @u @v Yet again, we start from the result we established at the beginning of this Programme. If z ¼ f ðx, yÞ z ¼ then @z @z x þ y @x @y Dividing in turn by u and v: z @z x @z y ¼ þ u @x u @y u z @z x @z y ¼ þ v @x v @y v Then, as u ! 0 and v ! 0, these become @z @z @x @z @y ¼ þ @u @x @u @y @u @z @z @x @z @y ¼ þ @v @x @v @y @v Example 1 If z ¼ x2 y 2 and x ¼ r cos and y ¼ r sin , then and @z @z @x @z @y ¼ þ @r @x @r @y @r @z @z @x @z @y ¼ þ @ @x @ @y @ We now need the various partial derivatives @z ¼ ............; @x @z ¼ ............; @y 12 @z ¼ 2x; @x @z ¼ 2y; @y @x ¼ ............; @r @x ¼ ............; @ @x ¼ cos ; @r @x ¼ r sin ; @ @y ¼ ............ @r @y ¼ ............ @ @y ¼ sin @r @y ¼ r cos @ Substituting in the two equations and simplifying: @z ¼ ............; @r @z ¼ ............ @ 447 Partial differentiation @z ¼ 2x cos 2y sin ; @r @z ¼ ð2xr sin þ 2yr cos Þ @ 13 Finally, we can express x and y in terms of r and as given, so that, after tidying up, we obtain @z ¼ ............; @r @z ¼ 2r ðcos2 sin2 Þ; @r @z ¼ ............ @ @z ¼ 4r 2 sin cos @ 14 Of course, we could express these as @z @z ¼ 2r cos 2 and ¼ 2r 2 sin 2 @r @ From these results, we can, if necessary, find the second partial derivatives in the normal manner. @2z @ @z @ ¼ ð2r cos 2Þ ¼ 2 cos 2 ¼ @r 2 @r @r @r Similarly @2z ¼ ............ @2 and @2z ¼ 4r 2 cos 2; @2 Because and @2z ¼ ............ @r@ @2z ¼ 4r sin 2 @r@ @2z @ @z @ ¼ ð2r 2 sin 2Þ ¼ 4r 2 cos 2 ¼ @2 @ @ @ @2z @ @z @ ¼ ð2r 2 sin 2Þ ¼ 4r sin 2 ¼ @r@ @r @ @r Example 2 If z ¼ f ðx, yÞ and x ¼ 12 ðu2 v 2 Þ and y ¼ uv, show that @z @z @z @z u v ¼2 x y @v @u @y @x Although this is much the same as the previous example, there is, at least, one difference. In this case, we are not told the precise nature of f ðx, yÞ. We must remember that z is a function of x and y and, therefore, of u and v. With that in mind, we set off with the usual two equations. @z ¼ ............ @u @z ¼ ............ @v 15 448 Programme 14 16 @z @z @x @z @y ¼ þ @u @x @u @y @u @z @z @x @z @y ¼ þ @v @x @v @y @v From the given information: @x ¼ ............; @u @x ¼ ............; @v 17 @x ¼ u; @u @x ¼ v; @v Whereupon @y ¼ ............ @u @y ¼ ............ @v @y ¼v @u @y ¼u @v @z ¼ ............ @u @z ¼ ............ @v 18 @z @z @z ¼u þv @u @x @y @z @z @z ¼ v þu @v @x @y If we now multiply the first of these by (v) and the second by u and add the two equations, we get the desired result. @z @z @z ¼ uv v2 @u @x @y @z @z @z ¼ uv þ u2 u @v @x @y @z @z @z @z Adding u v ¼ 2uv þ ðu2 v 2 Þ @v @u @x @y @z @z þ 2x ¼ 2y @x @y @z @z @z @z ¼2 x y ; u v @v @u @y @x v 449 Partial differentiation With the same given data, i.e. 1 2 ðu v 2 Þ and y ¼ uv 2 2 @2z @2z @ z @2z 2 2 we can now show that 2 þ 2 ¼ ðu þ v Þ . þ @u @v @x2 @y 2 In determining the second partial derivatives, keep in mind that z is a @z @z and . function of u and v and that both of these variables also occur in @x @y @2z ¼ ............ @u2 z ¼ f ðx; yÞ with x ¼ 19 2 2 @ 2 z @z @2z 2@ z 2@ z ¼ þ 2uv þ u þ v @u2 @x @x2 @x@y @y 2 Because @ @ @ @ @ u þv and ¼ v þu @x @y @v @x @y 2 @ z @ @z @z @z @ @z @ @z ¼ þ v ; ¼ u þ v þ u @u2 @u @x @y @x @u @x @u @y @z @ @ @z @ @ @z þu u þv þv u þv ¼ @x @x @y @x @x @y @y @ ¼ @u ¼ @z @2z @2z @2z @2z þ u2 2 þ uv þ uv þ v2 2 @x @x @x@y @x@y @y 2 2 @ 2 z @z @2z 2@ z 2@ z þ u þ v ¼ þ 2uv @u2 @x @x2 @x@y @y 2 2 @ z @ @z @ @z @z ¼ v þu ¼ Likewise, @v 2 @v @v @v @x @y ; ð1Þ ¼ ............ 20 @ 2 z @z @2z @2z @2z þ v 2 2 2uv þ u2 2 ¼ 2 @v @x @x @x@y @y Because @2z @ @z @z @z @ @z @ @z v þ u ¼ þ þ u ¼ @v 2 @v @x @y @x @v @x @v @y @z @ @ @z @ @ @z v v þu þ u v þu ¼ @x @x @y @x @x @y @y ¼ ; @z @2z @2z @2z @2z þ v 2 2 uv uv þ u2 2 @x @x @x@y @x@y @y @ 2 z @z @2z @2z @2z þ v 2 2 2uv þ u2 2 ¼ 2 @v @x @x @x@y @y Adding together results (1) and (2), we get . . . . . . . . . . . . ð2Þ 450 Programme 14 2 @2z @2z @ z @2z 2 2 þ ¼ ðu þ v Þ þ @u2 @v 2 @x2 @y 2 21 and that is it. Now, for something slightly different, move on to the next frame Inverse functions 22 If z ¼ f ðx, yÞ and x and y are functions of two independent variables u and v defined by u ¼ gðx, yÞ and v ¼ hðx, yÞ, we can theoretically solve these two equations to obtain x and y in terms of u and v. Hence we can determine @x @x @y @y @z @z , , ; and then and as required. @u @v @u @v @x @y In practice, however, the solution of u ¼ gðx, yÞ and v ¼ hðx, yÞ may well be difficult or even impossible by normal means. The following example shows how we can get over this difficulty. Example 1 If z ¼ f ðx, yÞ and u ¼ ex cos y and v ¼ ex sin y, we have to find @x @x @y @y , ; , . @u @v @u @v We start off once again with our standard relationships @u @u x þ y @x @y @v @v x þ y v ¼ @x @y ð1Þ u ¼ ð2Þ Now u ¼ ex cos y and v ¼ ex sin y So @u ¼ ............; @x @v ¼ ............; @x @u ¼ ............ @y @v ¼ ............ @y 451 Partial differentiation @u ¼ ex cos y; @x @v ¼ ex sin y; @x 23 @u ¼ ex sin y @y @v ¼ ex cos y @y Substituting in equations (1) and (2) above, we have u ¼ ex cos y x ex sin y y ð3Þ v ¼ ex sin y x þ ex cos y y ð4Þ Eliminating y from (3) and (4), we get x ¼ . . . . . . . . . . . . x ¼ ex cos y ex sin y u þ v cos 2y cos 2y 24 Because ex cos y u ¼ cos2 y x sin y cos y y ð3Þ ex cos y: ex sin y v ¼ sin2 y x þ sin y cos y y ð4Þ ex sin y: Adding: ex cos y u þ ex sin y v ¼ ðcos2 y sin2 yÞ x ex cos y ex sin y u þ v cos 2y cos 2y @x @x u þ v But x ¼ @u @v @x ex cos y @x ex sin y ; ¼ and ¼ @u cos 2y @v cos 2y ; x ¼ which are, of course, two of the expressions we have to find. Starting again with equations (3) and (4), we can obtain y ¼ . . . . . . . . . . . . y ¼ ex sin y ex cos y u þ v cos 2y cos 2y Because ð3Þ ex sin y: ð4Þ ex cos y: Adding: ; y ¼ ex sin y u ¼ sin y cos y x sin2 y y ex cos y v ¼ sin y cos y x þ cos2 y y ex sin y u þ ex cos y v ¼ ðcos2 y sin2 yÞ y ex sin y ex cos y u þ v cos 2y cos 2y But, y ¼ . . . . . . . . . . . . Finish it off. 25 452 Programme 14 26 @y @y u þ v @u @v @y ex sin y @y ex cos y ; ¼ and ¼ @u cos 2y @v cos 2y y ¼ So, collecting our four results together: @x ex cos y ¼ ; @u cos 2y @y ex sin y ¼ ; @u cos 2y @x ex sin y ¼ @v cos 2y @y ex cos y ¼ @v cos 2y We can tackle most similar problems in the same way, but it is more efficient to investigate a general case and to streamline the results. Let us do that. 27 General case If z ¼ f ðx, yÞ with u ¼ gðx, yÞ and v ¼ hðx, yÞ, then we have @u @u x þ y @x @y @v @v x þ y v ¼ @x @y ð1Þ u ¼ ð2Þ We now solve these for x and y: Eliminating y; we have @v : @y @u ð2Þ : @y ð1Þ Subtracting: @v @v @u @v @u u ¼ x þ y @y @y @x @y @y @u @u @v @u @v v ¼ x þ y @y @y @x @y @y @v @u @u @v @v @u u v ¼ : : x @y @y @x @y @x @y @v @u u v @y @y ; x ¼ @u @v @v @u @x @y @x @y Starting afresh from (1) and (2) and eliminating x, we have y ¼ . . . . . . . . . . . . 453 Partial differentiation y ¼ @u @v v u @x @x 28 @u @v @v @u @x @y @x @y The two results so far are therefore @v @u u v @y @y x ¼ @u @v @v @u @x @y @x @y and @u @v v u @x y ¼ @x @u @v @v @u @x @y @x @y You will notice that the denominator is the same in each case and that it can be expressed in determinant form @u @v @u @v @v @u @x @x ¼ @x @y @x @y @u @v @y @y This determinant is called the Jacobian of u, v with respect to x, y and is denoted by the symbol J: @u @v @x @x and is often written as @ðu; vÞ J ¼ i.e. @u @v @ðx; yÞ @y @y @u @v @ðu; vÞ @x @x So ¼ J¼ @ðx; yÞ @u @v @y @y Our last two results can therefore be written x ¼ . . . . . . . . . . . . ; u @u @v @u u v @y @y @y ¼ x ¼ @u J @x @u @y v @v @y , @v @x @v @y y ¼ . . . . . . . . . . . . @u @v @x @x @u @v v u u v @x @x y ¼ ¼ @u @v J @x @x @u @v @y @y 29 454 Programme 14 We can now get a number of useful relationships. (a) If v is kept constant, v ¼ 0 ; x Dividing by u and letting u ! 0 Similarly (b) If u is kept constant, u ¼ 0 ; x Dividing by v and letting v ! 0 Similarly @x @u @y @u @x @v @y @v @v u J ¼ @y @v ¼ J @y @v ¼ J @x @u ¼ v J @y @u ¼ J @y @u ¼ J @x So, at this stage, we had better summarise the results. Summary If z ¼ f ðx, yÞ and u ¼ gðx, yÞ and @x @v @x @u ¼ ¼ J J @u @y @v @y @y @v @y @u J J ¼ ¼ @u @x @v @x where, in each case @u @ðu; vÞ @x ¼ J¼ @ðx; yÞ @u @y v ¼ hðx, yÞ then @v @x @v @y Let us put this into practice by doing again the same example that we started with (Example 1 in Frame 22), but by the new method. First of all, however, make a note of the important summary listed above for future reference. 455 Partial differentiation Example 1 30 @x @x If z ¼ f ðx, yÞ and u ¼ ex cos y and v ¼ ex sin y, find the derivatives , , @u @v @y @y , . @u @v u ¼ ex cos y @u ¼ ex cos y @x @u ¼ ex sin y @y @u @ðu; vÞ @x ¼ J¼ @ðx; yÞ @u @y v ¼ ex sin y @v ¼ ex sin y @x @v ¼ ex cos y @y @v x @x e cos y ¼ @v ex sin y @y ex sin y ex cos y ¼ ðex cos yÞðex cos yÞ ðex sin yÞðex sin yÞ ¼ cos2 y sin2 y Then ¼ cos 2y @x @v ex cos y ¼ J¼ ; @u @y cos 2y @y @v ex sin y ¼ J¼ ; @u @x cos 2y @x @u ex sin y ¼ J¼ @v @y cos 2y @y @u ex cos y ¼ J¼ @v @x cos 2y which is a lot shorter than our first approach. Move on for a further example Example 2 31 @x @x @y @y If z ¼ f ðx; yÞ with u ¼ x2 y 2 and v ¼ xy, find expressions for , , , : @u @v @u @v First we need @u @u @v @v ¼ ............; ¼ ............; ¼ ............; ¼ ............ @x @y @x @y @u ¼ 2x; @x @u ¼ 2y; @y @v ¼ y; @x @v ¼x @y Then we calculate J which, in this case, is . . . . . . . . . . . . 32 456 Programme 14 33 J ¼ 2ðx2 þ y 2 Þ Because @u @v @ðu, vÞ @x @x 2x ¼ J¼ ¼ @ðx, yÞ @u @v 2y @y @y y ¼ 2x2 þ 2y 2 x Finally, we have the four relationships @x @v @x @u ¼ ¼ J ¼ ............; J ¼ ............ @u @y @v @y @y @v @y @u ¼ J ¼ ............; ¼ J ¼ ............ @u @x @v @x 34 @x x ¼ ; 2 @u 2ðx þ y 2 Þ @y y ¼ ; @u 2ðx2 þ y 2 Þ @x y ¼ 2 @v x þ y 2 @y x ¼ @v x2 þ y 2 And that is all there is to it. If we know the details of the function z ¼ f ðx; yÞ then we can go one @x @x @y @y @z @z , , , to find and . stage further and use the results @u @v @u @v @u @v Let us see this in a further example. Example 3 If z ¼ 2x2 þ 3xy þ 4y 2 and u ¼ x2 þ y 2 and v ¼ x þ 2y, determine @x @x @y @y @z @z , , , (b) and . @u @v @u @v @u @v Section (a) is just like the previous example. Complete that on your own. (a) 457 Partial differentiation @x 1 ¼ ; @u 2x y @x y ¼ ; @v 2x y @y 1 ¼ ; @u 2ð2x yÞ @y x ¼ @v 2x y 35 Because if u ¼ x2 þ y 2 and v ¼ x þ 2y @u @u @v @v ¼ 2x; ¼ 2y; ¼ 1; ¼2 @x @y @x @y @u @v @ðu; vÞ @x @x 2x 1 ¼ J¼ ¼ ¼ 4x 2y ¼ 2ð2x yÞ @ðx; yÞ @u @v 2y 2 @y @y @x @v 1 Then ¼ J¼ 2 2ð2x yÞ ¼ @u @y 2x y @x @u y ¼ J ¼ 2y 2ð2x yÞ ¼ @v @y 2x y @y @v 1 ¼ J ¼ 1 2ð2x yÞ ¼ @u @x 2ð2x yÞ @y @u x ¼ J ¼ 2x 2ð2x yÞ ¼ @v @x 2x y ; @x 1 ¼ ; @u 2x y @x y ¼ ; @v 2x y @y 1 ¼ ; @u 2ð2x yÞ @y x ¼ @v 2x y Now for part (b). Since z is also a function of u and v, the expressions for @z @z and are @u @v @z ¼ ............ @u @z ¼ ............ @v @z @z @x @z @y ¼ þ @u @x @u @y @u @z @z @x @z @y ¼ þ @v @x @v @y @v The only remaining items of information we need are the expressions for @z @z and which we obtain from z ¼ 2x2 þ 3xy þ 4y 2 @x @y @z @z ¼ 4x þ 3y and ¼ 3x þ 8y @x @y Using these and the previous set of derivatives, we now get @z @z ¼ ............; ¼ ............ @u @v 36 458 Programme 14 37 @z 5x 2y ¼ ; @u 2ð2x yÞ @z 3x2 þ 4xy 3y 2 ¼ @v 2x y Because @z @z @x @z @y ¼ þ @u @x @u @y @u @z 1 1 ; ¼ ð4x þ 3yÞ þ ð3x þ 8yÞ @u 2x y 2ð2x yÞ 5x 2y @z 5x 2y ; ¼ ¼ 2ð2x yÞ @u 2ð2x yÞ @z @z @x @z @y ¼ þ @v @x @v @y @v @z y x ¼ ð4x þ 3yÞ þ ð3x þ 8yÞ ; @v 2x y 2x y and ¼ 3x2 þ 4xy 3y 2 2x y ; @z 3x2 þ 4xy 3y 2 ¼ @v 2x y They are all done in the same general way. Now on to the next topic Stationary values of a function 38 You will doubtless remember that in earlier work you established the characteristics of stationary points on a plane curve and derived the conditions that enable these critical points to be calculated. y y = f(x) O At A and B For maximum For minimum x1 x2 dy ¼0 dx d2 y is negative ðx ¼ x1 Þ dx2 2 d y is positive ðx ¼ x2 Þ dx2 x 459 Partial differentiation We now progress to the application of these same considerations to three dimensions, where z ¼ f ðx, yÞ. The function is now represented by a surface and stationary values of the function z ¼ f ðx, yÞ occur when the tangent plane to the surface at a point P (a, b) is parallel to the plane z ¼ 0, i.e. to the x–y plane. Let us take a closer look at this. 39 Maximum and minimum values z P (a,b) Q (a+h,b+k) b y O a x A function z ¼ f ðx, yÞ is said to have maximum value at P (a, b) if f ða, bÞ is greater than the value at a near-by point Q (a þ h, b þ k) for all values of h and k however small, positive or negative, i.e. in all directions from P. z Q (a+h,b+k) P (a, b) b O y a x Similarly, z ¼ f ðx, yÞ is said to have a minimum value at P (a, b) if f ða, bÞ is less than the value at a neighbouring point Q (a þ h, b þ k) in any direction from P. To establish maximum and minimum values, we must therefore investigate the sign of the value of f ða þ h, b þ kÞ f ða, bÞ. If f ða þ h, b þ kÞ f ða, bÞ < 0 we have a maximum value at P ða, bÞ. If f ða þ h, b þ kÞ f ða, bÞ > 0 we have a minimum value at P ða, bÞ. 460 Programme 14 To pursue this further we turn to the total differential df ðx, yÞ ¼ @f @f dx þ dy @x @y The total differential measures the rise or fall in the tangent plane from the point of its contact with the surface at ðx, yÞ to the point ðx þ dx, y þ dyÞ. tangent plane tangential at P P dx df dy y x surface f (x,y) If the point of contact is a maximum or a minimum then @f ¼ ............ @x 40 @f ¼0 @x and @f ¼ ............ @y @f ¼0 @y and Because The tangent plane is parallel with the x–y plane and so the tangent plane neither rises nor falls, so that df ðx, yÞ ¼ @f @f dx þ dy ¼ 0 @x @y Also because dx 6¼ 0 and dy 6¼ 0 then @f @f ¼ 0 and ¼ 0. @x @y @f Notice the logic here. If there is a maximum or a minimum, then ¼ 0 and @x @f @f @f ¼ 0. However, just because ¼ 0 and ¼ 0 at a point does not imply that @y @x @y a maximum or a minimum exists at that point. What we can say is that a stationary point exists at that point and, as we shall see later, not all stationary points are maxima or minima. Example 1 Determine the values of x and y at which the stationary values of f ðx, yÞ ¼ x2 þ xy þ y 2 þ 5x 5y þ 3 occur. @f @f and , equate each to zero @x @y and then solve the pair of simultaneous equations so obtained. In which case All we need to do is to obtain expressions for x ¼ ............ and y ¼ ............ 461 Partial differentiation 41 x ¼ 5 and y ¼ 5 Because @f @f ¼ 2x þ y þ 5 and ¼ x þ 2y 5 giving the pair of simultaneous @x @y equations 2x þ y þ 5 ¼ 0 x þ 2y 5 ¼ 0 ð1Þ ð2Þ Adding ð1Þ þ ð2Þ gives 3x þ 3y ¼ 0, that is y ¼ x Substitution in ð1Þ gives x ¼ 5 and so y ¼ 5 Although a stationary value occurs at ð5, 5Þ we have no evidence as to whether it is a maximum or a minimum value. Let us investigate further. From the previous definitions f ða, bÞ will be a maximum value if f ða þ h, b þ kÞ f ða, bÞ < 0 f ða, bÞ will be a minimum value if f ða þ h, b þ kÞ f ða, bÞ > 0 Now, from Taylor’s theorem @f @f þk @x @y 2 1 @ f @2f @2f h2 2 þ 2hk þ k2 2 þ . . . þ 2! @x @x@y @y f ða þ h, b þ kÞ ¼ f ða, bÞ þ h and we have already seen that at a stationary value @f @f ¼ 0 and ¼ 0. So, at a @x @y stationary point, Taylor’s theorem becomes 2 2 1 @2f 2@ f 2@ f h þk f ða þ h, b þ kÞ f ða, bÞ ¼ þ... þ 2hk 2! @x2 @x@y @y 2 where subsequent terms are of higher orders of h and k and are neglected. The expression in the brackets on the right-hand side can be written as ( !) 2 2 2 2 1 @2f @2f @2f 2 @ f @ f h 2þk þk : @x @x@y @x2 @y 2 @x@y @2f @x2 Take a moment and expand the brackets and confirm that this is so. 462 42 Programme 14 @2f @2f @2f þ k2 2 þ 2hk 2 @x @x@y @y ( !) 2 2 2 2 2 2 2 1 @ f @ f @ f @ f @ f ¼ 2 h 2þk þ k2 @x @x@y @x2 @y 2 @x@y @ f 2 @x 2 2 @ f @2f Now h 2 þ k , being a square, is always positive and if @x @x@y So h2 @2f @2f @2f 2> 2 @x @y @x@y 2 the second term will also be positive. In that case the sign of the whole expression is given by that of front. @2f at the @x2 2 2 2 2 @2f @2f @ f @2f @2f @ f > , i.e. – > 0, this can be 2 2 2 2 @x @y @x@y @x @y @x@y @2f and have the same sign. Therefore, @y 2 Furthermore, if so only if @2f @x2 for f ða; bÞ to be a maximum, @2f @2f and are both negative 2 @x @y 2 and for f ða; bÞ to be a minimum, @2f @2f and are both positive. @x2 @y 2 So, to determine whether a known stationary value is a maximum or a @2f @2f @2f . minimum value, we must find the second derivatives 2 , 2 and @x @y @x@y Then 2 2 @2f @2f @ f > 0, the stationary value is a true maximum @x2 @y 2 @x@y or minimum value. (a) If (b) In that case @2f @x2 @2f (2) if 2 @x (1) if and and @2f are both negative, f ða, bÞ is a maximum @y 2 2 @ f are both positive, f ða, bÞ is a minimum. @y 2 Make a careful note of the conclusions (a) and (b): then let us apply them. 43 Example 2 Investigate further the stationary value of the function z ¼ x2 þ xy þ y 2 þ 5x 5y þ 3 We have already seen that this function has a stationary point at x ¼ ............; y ¼ ............ 463 Partial differentiation x ¼ 5; 44 y¼5 2 2 @2z @2z @ z . If this is greater @x2 @y 2 @x@y than zero at ð5, 5Þ, then either a maximum or a minimum occurs at that point. Check whether this is so. Next, we investigate the value of Yes. @2z @x2 45 2 2 @2z @ z >0 @y 2 @x@y Because @2z ¼ 2; @x2 @2z ¼ 2; @y 2 @2z ¼ 1. @x@y This confirms that ð5, 5Þ is either a maximum or a minimum. To decide which it is, we note that @2z @2z and are both positive. 2 @x @y 2 ; at ð5, 5Þ, z is a . . . . . . . . . . . . 46 minimum Of course to find the actual minimum value of z we substitute x ¼ 5 and y ¼ 5 into the expression for z. That is really all there is to it. Another example. Example 3 Determine the stationary values, if any, of the function z ¼ x3 6xy þ y 3 The four steps in the routine are: @z @z @z and and solve the equations ¼ 0 and @x @y @x 2 2 2 2 @ z @ z @ z > 0. (b) Determine whether @x2 @y 2 @x@y (a) Find (c) If so, note the sign of @z ¼ 0. @y @2z @2z and 2 to distinguish between max. and min. 2 @x @y (d) Evaluate the maximum or minimum value of z. In this example, stationary values occur at . . . . . . . . . . . . 464 Programme 14 47 z ¼ 0 at ð0, 0Þ and z ¼ 8 at ð2, 2Þ Because z ¼ x3 6xy þ y 3 ; @z ¼ 3x2 6y @x @z ¼ 6x þ 3y 2 @y @z @z ¼ 0 and ¼ 0 ; x2 2y ¼ 0 and 2x þ y 2 ¼ 0 @x @y A possible stationary point exists when x2 2y ¼ 0 and 2x þ y 2 ¼ 0. From the first equation y ¼ x2 =2 and substitution into 2x þ y 2 ¼ 0 gives 2x þ x4 =4 ¼ 0. That is x4 8x ¼ x x3 8 ¼ 0 and so x ¼ 0 or x ¼ 2. When x ¼ 0 then y ¼ 0 and when x ¼ 2 then y ¼ 2. ; There are stationary values at (0, 0) and (2, 2) Next we determine whether @2z @x2 2 2 @2z @ z >0 2 @y @x@y Result . . . . . . . . . . . . 48 No max. or min. at ð0, 0Þ; Either max. or min. at ð2, 2Þ Because @z ¼ 3x2 6y @x @z ¼ 6x þ 3y 2 @y ; at ð0, 0Þ At ð2, 2Þ @2z ¼ 6x @2z @x2 ¼ 6 2 @x@y @ z ; ¼ 6y @y 2 2 2 2 2 @ z @ z @ z ¼ ð0Þð0Þ 36 < 0 2 2 @x @y @x@y ; ; No max. or min. at ð0, 0Þ 2 2 2 2 @ z @ z @ z ¼ ð12Þð12Þ 36 > 0 @x2 @y 2 @x@y ; Either max. or min. at ð2, 2Þ @2z @2z and are positive. Therefore the 2 @x @y 2 stationary value at ð2, 2Þ is a . . . . . . . . . . . . We see that at ð2, 2Þ both 49 minimum Finally, the minimum value of z is . . . . . . . . . . . . 465 Partial differentiation 8 Therefore, 50 zmin ¼ 8 and occurs at ð2, 2Þ Before doing a further example, let us consider one other aspect of stationary values. On to a new frame Saddle point 51 In the previous example, when we substituted the coordinates ð0, 0Þ in the 2 2 2 2 @ z @ z @ z we found that this did not satisfy the expression @x2 @y 2 @x@y condition that for a maximum or minimum value 2 2 2 2 @ z @ z @ z >0 2 2 @x @y @x@y @z @z ¼ 0 and ¼0 @x @y 2 2 2 2 @ z @ z @ z <0 @x2 @y 2 @x@y In fact, if and this is an indication of a form of stationary value described as a saddle point, as shown at P below. A saddle point is, in effect, a combined maximum and minimum (a,b) configuration in different directions. Its name is obvious from the shape. Add this then to the list of conditions for stationary values that we have built up. At this stage, one naturally asks, what is implied if 2 2 2 2 @ z @ z @ z ¼0 @x2 @y 2 @x@y In such a case, further detailed study of the function is necessary. Now for an example to see it all in practice. Example 4 Determine the stationary values of z ¼ 5xy 4x2 y 2 2x y þ 5. Stationary values (or turning points) occur where @z @z ¼ 0 and ¼ 0, i.e. at . . . . . . . . . . . . @x @y 52 466 Programme 14 53 x ¼ 1, y ¼ 2 Because @z @z ¼ 5y 8x 2 ¼ 5x 2y 1 @x @y ) ; 8x 5y þ 2 ¼ 0 gives x ¼ 1; y ¼ 2 5x 2y 1 ¼ 0 Therefore, the only stationary value occurs at ð1, 2Þ. Next we substitute these x and y values in 2 2 2 2 @ z @ z @ z and find . . . . . . . . . . . . @x2 @y 2 @x@y 54 @2z @x2 2 2 @2z @ z <0 @y 2 @x@y Because @2z @2z @2z ¼5 ¼ 8; ¼ 2; 2 2 @x @y @x@y 2 2 2 2 @ z @ z @ z ; ¼ ð8Þð2Þ 25 ¼ 9 i:e: < 0 @x2 @y 2 @x@y The stationary value at ð1, 2Þ is therefore a . . . . . . . . . . . . 55 saddle point Example 5 Determine stationary values of z ¼ x3 3x þ xy 2 and their nature. We go through the same routine as before. First find @z @z @z @z and and solve ¼ 0 and ¼ 0. @x @y @x @y Possible stationary values therefore occur at . . . . . . . . . . . . 467 Partial differentiation pffiffiffi x ¼ 0, y ¼ 3; 56 x ¼ 1, y ¼ 0 Because @z ¼ 3x2 3 þ y 2 @x @z ¼ 2xy @y If x ¼ 0, y2 ¼ 3 If y ¼ 0, 3x2 ¼ 3 ; x ¼ 0 or y ¼ 0 pffiffiffi pffiffiffi ; y¼ 3 x ¼ 0, y ¼ 3 ; x ¼ 1 x ¼ 1, y ¼ 0: Now we need the second derivatives and the usual tests. Finish if off. The nature of the stationary values: pffiffiffi pffiffiffi ð0, 3Þ . . . . . . . . . . . . ; ð0, 3Þ . . . . . . . . . . . . ð1, 0Þ . . . . . . . . . . . . ; ð0, pffiffiffi 3Þ saddle point; ð1, 0Þ minimum; ð1, 0Þ . . . . . . . . . . . . ð0, pffiffiffi 3Þ saddle point 57 ð1, 0Þ maximum Because @2z @2z @2z ¼ 2y ¼ 6x; ¼ 2x; @x2 @y 2 @x@y 2 2 2 2 @ z @ z @ z @x2 @y 2 @x@y pffiffiffi ð0, 3Þ ð0Þð0Þ 12 i.e. < 0 ; saddle point pffiffiffi ð0, 3Þ ð0Þð0Þ 12 i.e. < 0 ; saddle point ð1, 0Þ ð6Þð2Þ 0 i.e. > 0 ; minimum ð1, 0Þ ð6Þð2Þ 0 i.e. > 0 ; maximum and that just about does everything. Substitution of ð1, 0Þ and ð1, 0Þ in z ¼ x3 3x þ xy2 gives the minimum and maximum values of z. zmin ¼ 2; zmax ¼ 2. The value of z at each of the saddle points is zero. Let’s now look at some examples where the second derivative test fails Example 6 58 2 2 Determine the stationary values of z ¼ x 6xy þ 9y . Here we see that @z @z ¼ 2x 6y, ¼ 6x þ 18y and so these two derivatives @x @y vanish when ............ 468 Programme 14 59 y ¼ x=3 Because @z ¼ 2x 6y ¼ 0 when 2x ¼ 6y, that is when y ¼ x=3 and @x @z ¼ 6x þ 18y ¼ 0 when 6x ¼ 18y, that is when y ¼ x=3, and so there is an @y infinity of stationary points lying along the line y ¼ x=3. Now 60 @2z @x2 2 2 @2z @ z ¼ ............ @y 2 @x@y 0 Because @2z @2z @2z ¼ 6 so that ¼ 2, ¼ 18 and @x2 @y 2 @x@y 2 2 2 2 @ z @ z @ z ¼ 18 2 36 ¼ 0 2 2 @x @y @x@y So the second derivative test does not apply and we must look elsewhere to decide the nature of the stationary points. Since x2 6xy þ 9y 2 ¼ ðx 3yÞ2 then z 0 for all values of x and y. Therefore the stationary points are minima – there is an infinity of minimum points along the line y ¼ x=3. 469 Partial differentiation Example 7 Find the stationary points of z ¼ x4 y 3 . @z @z ¼ 4x3 , ¼ 3y 2 and so these two derivatives @x @y vanish when x ¼ . . . . . . . . . . . ., y ¼ . . . . . . . . . . . . Here we see that x ¼ 0, y ¼ 0 61 Because @z @z ¼ 4x3 ¼ 0 when x ¼ 0 and ¼ 2y 2 ¼ 0 when y ¼ 0, so there is @x @y just one stationary point at ð0, 0Þ. Now, at the stationary point 2 2 2 2 @ z @ z @ z ¼ ............ @x2 @y 2 @x@y 0 Because @2z @2z ¼ 12x2 , ¼ 4y and 2 @x @y 2 2 2 2 2 @ z @ z @ z ¼0 @x2 @y 2 @x@y @2z ¼ 0 so that at ð0, 0Þ: @x@y So the second derivative test does not apply. However, in the z–x plane y ¼ 0 and so z ¼ x4 . This means that the line of intersection of the surface with the z–x plane has a minimum at the origin. In the z–y plane x ¼ 0 and so z ¼ y3 . This means that the line of intersection of the surface with the z–y plane has a point of inflexion at the origin. 62 470 Programme 14 Lagrange undetermined multipliers 63 Closely allied to the problem of locating the stationary points of some function u ¼ f ðx, yÞ is the problem of locating points where u ¼ f ðx, yÞ attains its greatest or its least value (an extremal value) subject to the condition that x and y are related to each other via the equation ðx, yÞ ¼ 0 ð1Þ The problem can be clarified if we consider it graphically. The graph of u ¼ f ðx, yÞ is a surface within the ðx, y, uÞ coordinate system. Selecting a plane parallel to the x–y plane on which the value of u is constant, uk , we see that the surface intersects the plane in a curve given by the equation f ðx, yÞ ¼ uk . u f (x, y) = uk uk y x y This line of intersection can now be projected onto the x–y plane to form what is known as a un level curve. Different values of uk determine different planes (all parallel to the x–y plane), u3 u2 different lines of intersection and hence u1 different level curves. Accordingly, an alterx native graphical description of u ¼ f ðx, yÞ is that of a family of level curves in the x–y plane with each member of the family being associated with a particular value of uk , where we assume that u1 < u2 < u3 < . . . < un or u1 > u2 > u3 > . . . > un . We now superimpose onto this family of level curves the graph of the constraint equation ðx, yÞ ¼ 0. Clearly, in the figure alongside, u3 is the y extremal value of f ðx, yÞ that coincides with ðx, yÞ ¼ 0, and at the point P where they meet un they share the same tangent line dy=dx. Now, P since ðx, yÞ ¼ 0 we see that u3 u2 u1 dy ¼ ............ dx ϕ (x, y) = 0 x 471 Partial differentiation 64 dy @=@x ¼ dx @=@y Because d ¼ @ @ dy @=@x dx þ dy ¼ 0 so that ¼ @x @y dx @=@y The same tangent can be found from du ¼ @f @f dx þ dy @x @y by equating the differential du ¼ 0. Therefore dy @f =@x @=@x ¼ ¼ dx @f =@y @=@y The latter two fractions are equivalent fractions which means that the two numerators and the two denominators each differ by the same multiplicative factor K, enabling us to say that @f @ @f @ ¼K and ¼K so that @x @x @y @y @f @ þ ¼0 @x @x @f @ þ ¼0 @y @y ð2Þ ð3Þ ¼ K is called a Lagrange multiplier and equations (2) and (3), coupled with the constraint equation ðx, yÞ ¼ 0, give us three relationships from which the values of x and y at the extremal points – and also the value of if required – can be found. Quite often the value of is not important. Let us see how it works in a simple example. Example 1 65 Find the stationary points of the function u ¼ x2 þ y 2 subject to the constraint x2 þ y 2 þ 2x 2y þ 1 ¼ 0. In this case, u ¼ x2 þ y 2 ¼ x2 þ y 2 þ 2x 2y þ 1 We need to know @u ¼ ............; @x @ ¼ ............; @x @u ¼ ............ @y @ ¼ ............ @y 472 Programme 14 66 @u ¼ 2x; @x @u ¼ 2y; @y @ ¼ 2x þ 2; @x @ ¼ 2y 2 @y @u @ þ ¼0 @x @x @u @ þ ¼0 @y @y Then we form and solve ¼ x2 þ y 2 þ 2x 2y þ 1 ¼ 0 together with x ¼ ............; y ¼ ............; ¼ ............ which gives pffiffiffi pffiffiffi pffiffiffi 2 2 ; y ¼1 ; ¼ 21 x ¼ 1 2 2 67 @u @ þ ¼ 0 ; 2x þ ð2x þ 2Þ ¼ 0 @x @x @u @ þ ¼ 0 ; 2y þ ð2x 2Þ ¼ 0 @y @y ; x ðx þ 1Þ ¼ y ðy 1Þ ; xy x ¼ xy þ y ; x þ ðx þ 1Þ ¼ 0 ; y þ ðy 1Þ ¼ 0 ; y ¼ x Substituting this in x2 þ x2 þ 2x þ 2x þ 1 ¼ 0 2x2 þ 4x þ 1 ¼ 0 pffiffiffi 2 ; x ¼ 1 2 pffiffiffi 2 But y ¼ x ; y ¼1 2 pffiffiffi To find , we have x þ ðx þ 1Þ ¼ 0 ; ¼ 2 1 As we have already said, we do not really need to find the value of . On to the next 473 Partial differentiation 68 Functions with three independent variables The argument is very much the same as before. To find stationary points of the function subject to the constraint u ¼ f ðx, y, zÞ ðx, y, zÞ ¼ 0 ð1Þ ð2Þ Again we have, at stationary points @u @u @u x þ y þ z ¼ 0 @x @y @z ð3Þ and since ðx; y; zÞ ¼ 0 then @ @ @ x þ y þ z ¼ 0 @x @y @z ð4Þ Multiplying each term in (4) by and adding (4) to (3), we have ............ @u @ @u @ @u @ x þ y þ z ¼ 0 þ þ þ @x @x @y @y @z @z from which @u @ þ ¼0 @x @x @u @ þ ¼0 @y @y @u @ þ ¼0 @z @z 69 ð5Þ ð6Þ ð7Þ Equations (5), (6), (7), together with the constraint (2), provide all the information to determine x, y, z, and, if necessary, . Example 2 70 To find the stationary points of the function u ¼ x2 þ 2y 2 þ z subject to the constraint So @u ¼ ............; @x @ ¼ ............; @x ðx; zÞ ¼ x2 z2 2 ¼ 0. @u ¼ ............; @y @ ¼ ............; @y @u ¼ ............ @z @ ¼ ............ @z 474 Programme 14 71 @u ¼ 2x; @x @ ¼ 2x; @x @u ¼ 4y; @y @ ¼ 0; @y @u ¼1 @z @ ¼ 2z @z Now compile the equations @u @ þ ¼ 0; @x @x @u @ þ ¼ 0; @y @y @u @ þ ¼0 @z @z and, together with the constraint ¼ x2 z2 2 ¼ 0, establish that stationary points occur at . . . . . . . . . . . . 72 ð32 ; 0; 12Þ and ð 32 ; 0; 12Þ Because @u @ þ ¼0 @x @x @u @ þ ¼0 @y @y @u @ þ ¼0 @z @z ; 2x þ 2x ¼ 0 ; ¼ 1 4y þ ð0Þ ¼ 0 ; y ¼ 0 ¼ x 2 z2 2 ¼ 0 1 2z ¼ 0 ; z¼ ; x2 14 2 ¼ 0 Therefore, stationary points at 3 2; 1 1 ¼ 2 2 ; x ¼ 32. 0; 12 and 32 ; 0; 12 . The method of Lagrange multipliers does not lend itself easily to give a distinction between the various types of stationary points. In many practical applications, however, whether a result is a maximum or a minimum value will be apparent from the physical consideration of the problem. Let us finish with one further example. So move on 73 Example 3 A hot water storage tank is a vertical cylinder surmounted by a hemispherical top of the same diameter. The tank is designed to hold 400 m3 of liquid. Determine the total height and the diameter of the tank if the surface heat loss is to be a minimum. We first write down the function for the total surface area, A. A ¼ ............ 475 Partial differentiation 74 A ¼ 3r 2 þ 2rh Because The surface area of the hemisphere is 2r 2 , the area of the base of the tank is r 2 and the area of the cylindrical side is 2rh, giving a total area of 3r 2 þ 2rh. This is the function which has to be a minimum. The constraint in this problem is that . . . . . . . . . . . . 75 the volume is 400 m3 So we have constraint So let We now want A ¼ 3r 2 þ 2rh 2 V ¼ r 2 h þ r 3 ¼ 400 3 2 2 ¼ r h þ r 3 400 ¼ 0 3 @A ¼ ............; @r @ ¼ ............; @r @A ¼ 6r þ 2h; @r @A ¼ 2r; @h Now we form and ð1Þ ð2Þ @A ¼ ............ @h @ ¼ ............ @h @ ¼ 2rh þ 2r 2 @r @ ¼ r 2 @h @A @ þ ¼0 @r @r @A @ þ ¼0 @h @h 2 and, with the constraint, ¼ r 2 h þ r 3 400 ¼ 0, 3 we eventually obtain r ¼ . . . . . . . . . . . . and h ¼ . . . . . . . . . . . . Finish it off and hence find the total height and the diameter. 76 476 Programme 14 77 r ¼ 4:243 m; h ¼ 4:243 m Check the working: @A @ þ ¼ 0 ; 6r þ 2h þ ð2rh þ 2r 2 Þ ¼ 0 @r @r @A @ þ ¼ 0 ; 2r þ r 2 ¼ 0 @h @h 2 Substitute this in (3) From (4): ¼ r 2 6r þ 2h ð2rh þ 2r 2 Þ ¼ 0 r ; 6r þ 2h 4h 4r ¼ 0 ; h¼r 2 5 Also r 2 h þ r 3 ¼ 400 ; r 3 ¼ 400 ; r ¼ 4:243 3 3 : ; Total height ¼ h þ r ¼ 8 49 m; Diameter ¼ 8:49 m ð3Þ ð4Þ That brings us to the end of this particular Programme and to the usual Revision summary that follows. Check through the Can you? checklist and be sure to revise any section should you feel that is necessary. Then you will find the Test exercise straightforward – no tricks. The Further problems provide valuable additional practice. Revision summary 14 78 1 Small increments @z @z x þ y @x @y @u @u @u u ¼ f ðx; y; zÞ u ¼ x þ y þ z @x @y @z z ¼ f ðx; yÞ z ¼ 2 Rates of change dz @z dx @z dy ¼ : þ : z ¼ f ðx; yÞ dt @x dt @y dt 3 Implicit functions z ¼ f ðx; yÞ ¼ 0 4 dy @z @z ¼ dx @x @y Change of variables z ¼ f ðx; yÞ x and y are functions of u and v @z @z @x @z @y ¼ þ @u @x @u @y @u @z @z @x @z @y ¼ þ @v @x @v @y @v 477 Partial differentiation 5 Inverse functions z ¼ f ðx; yÞ u ¼ g ðx; yÞ v ¼ hðx; yÞ @x @v @x @u ¼ J; ¼ J @u @y @v @y @y @v @y @u ¼ J; ¼ J @u @x @v @x @u @v @ðu; vÞ @x @x where J ¼ ¼ @u @v @ðx; yÞ @y @y 6 Stationary points z ¼ f ðx; yÞ @z @z ¼ 0 and ¼0 @x @y 2 2 2 2 @ z @ z @ z >0 (b) @x2 @y 2 @x@y 2 2 2 2 @ z @ z @ z <0 @x2 @y 2 @x@y (a) @2z @x2 (c) 7 2 2 @2z @ z ¼0 @y 2 @x@y with with no decision without further information and @2z both negative for maximum @y 2 @2z @x2 and @2z both positive for minimum. @y 2 @u @ þ ¼0 @x @x @u @ þ ¼0 @y @y ðx; yÞ ¼ 0. Three independent variables u ¼ f ðx; y; zÞ with constraint ðx; y; zÞ ¼ 0 Solve for saddle point @2z @x2 Lagrange multipliers Two independent variables u ¼ f ðx; yÞ with constraint ðx; yÞ ¼ 0 Solve for max. or min. @u @ þ @x @x @u @ þ @y @y @u @ þ @z @z ðx; y; zÞ ¼0 ¼0 ¼0 ¼ 0. 478 Programme 14 Can you? 79 Checklist 14 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Derive the expression for a small increment in an expression of two real variables using Taylor’s theorem? Yes No 1 to 4 . Apply the notion of small increments in expressions in two and three real variables to a variety of problems? Yes No 5 to 7 . Determine the rate of change with respect to time of an expression involving two or three real variables? Yes No 8 and 9 9 and 10 . Determine first and second derivatives involving change of variables in expressions of two real variables? Yes No 11 to 21 . Use the Jacobian to obtain the derivatives of inverse functions of two real variables? Yes No 22 to 37 . Locate and identify maxima, minima and saddle points of functions of two real variables? Yes No 38 to 57 58 to 77 . Differentiate implicit functions? Yes No . Solve problems where the independent variables are constrained by using the method of Lagrange undetermined multipliers for functions of two and three real variables? Yes No 479 Partial differentiation Test exercise 14 1 xy , show that If z ¼ xy @z @z (a) x þ y ¼z @x @y (b) x2 (c) z 2 3 80 @2z @2z y2 2 ¼ 0 2 @x @y @2z @z @z ¼2 : . @x@y @x @y Two sides of a triangular plate are measured as 125 mm and 160 mm, each to the nearest millimetre. The included angle is quoted as 608 18. Calculate the length of the remaining side and the maximum possible error in the result. 1=2 If z ¼ x2 y 2 and x is increasing at 3.5 m/s, determine at what rate y must change in order that z shall be neither increasing nor decreasing at the instant when x ¼ 5 m and y ¼ 3 m. dy d2 y and 2 . dx dx 4 If 2x2 þ 4xy þ 3y 2 ¼ 1, obtain expressions for 5 If u ¼ x2 þ y 2 and v ¼ 4xy, determine @x @x @y @y , , , . @u @v @u @v 6 Determine the position and nature of the stationary points of the functions: (a) z ¼ 2x2 y 2 þ 4xy 2 4y 3 þ 16y þ 5 (b) z ¼ 4 25x2 þ 20xy 4y 2 . 7 A rectangular storage tank is to have a capacity of 1:0 m3 . If the tank is closed and the top is made of metal half as thick as the sides and base, use Lagrange’s method of undetermined multipliers to determine the dimensions of the tank for the total amount of metal used in its construction to be a minimum. 8 Use Lagrange’s method of undetermined multipliers to obtain the stationary values of u ¼ x2 þ y 2 þ z2 subject to the constraint ¼ 3x 2y þ z 4. 480 Programme 14 Further problems 14 81 1 2 If z ¼ 2x2 3y with u ¼ x2 sin y and v ¼ 2y cos x, determine expressions for @z @z and . @u @v @x @x @y @y If u ¼ x2 þ e3y and v ¼ 2x þ e3y , determine , , , . @u @v @u @v 3 If z ¼ f ðx; yÞ where x ¼ uv and y ¼ u2 v 2 , show that @z @z @z @z ¼u þv (a) 2x þ 2y @x @y @u @v @z 1 @z @z ¼ v . (b) 2 u @y u2 þ v 2 @u @v 4 If V ¼ f ðx, yÞ and x ¼ r cos and y ¼ r sin , show that @ 2 V @ 2 V @ 2 V 1 @V 1 @ 2 V þ þ 2 ¼ 2 þ . @x2 @y @r r @r r 2 @2 5 If z ¼ cosh 2x sin 3y and u ¼ ex ð1 þ y 2 Þ and v ¼ 2yex , determine expressions for @x @x @y @y @z @z , , , , and hence find and . @u @v @u @v @u @v 6 If z ¼ f ðu; vÞ where u ¼ 12 ðx2 y 2 Þ and v ¼ xy, prove that 2 @2z @2z @ z @2z @2z @z þ2 . þ4v ¼ 2u 2 2 2 2 @x @y @u @v @u@v @u 7 Locate the stationary points of the following functions. Determine the nature of the points and calculate the critical function values. (a) z ¼ y 2 þ xy þ x2 þ 4y 4x þ 5 (b) z ¼ y 2 þ xy þ 2x þ 3y þ 6 (c) z ¼ 3xy 6y 2 3x2 þ 6y þ 6x þ 7. 8 Find the stationary points of the function z ¼ ðx2 þ y 2 Þ2 8ðx2 y 2 Þ and determine their nature. 9 Verify that the function z ¼ ðx þ y 1Þ=ðx2 þ 2y 2 þ 2Þ has stationary values at ð2, 1Þ and 23 , 13 and determine their nature. 10 Locate stationary points of the function z ¼ 4x2 þ 10xy þ 4y 2 x2 y 2 and determine their nature. 11 Find the stationary points of the following functions and determine their nature. (a) z ¼ xðx2 3Þ þ 3yðx 1Þ2 þ 18y 2 ð2y 3Þ (b) z ¼ x2 y 2 x2 y 2 . 481 Partial differentiation 12 Find the stationary points of the following functions and determine their nature. (a) z ¼ ðx yÞ x2 þ xy þ y 2 (b) z ¼ 6 x2 þ 8xy 16y 2 (c) z ¼ cos x2 þ y 2 . 13 A metal channel is formed by turning up the sides of width x of a rectangular sheet of metal through an angle . If the sheet is 200 mm wide, determine the values of x and for which the cross-section of the channel will be a maximum. 14 A container is in the form of a right circular cylinder of length l and diameter d, with equal conical ends of the same diameter and height h. If V is the fixed volume of the container, find the dimensions l, h and d for minimum surface area. 15 A solid consists of a cylinder of length l and diameter d, surmounted at one end by a cone of vertex angle 2 and base diameter d, and at the other end by a hemisphere of the same diameter. If the volume V of the solid is 50 cm3 , determine the dimensions l, d and so that the total surface area shall be a minimum. 16 A rectangular solid of maximum volume is to be cut from a solid sphere of radius r. Determine the dimensions of the solid so formed and its volume. 17 Use Lagrange’s method of undetermined multipliers to obtain the stationary values of the following functions u, subject in each case to the constraint . (a) u ¼ x2 y 2 z2 ¼ x 2 þ y 2 þ z2 4 ¼ 0 (b) u ¼ x2 þ y 2 ¼ 4x2 þ 6xy þ 4y 2 ¼ 9. Programme 15 Frames 1 to 66 Partial differential equations Learning outcomes When you have completed this Programme you will be able to: . Summarise the introductory methods of solving ordinary differential equations . Solve partial differential equations that are amenable to solution by direct integration . Apply initial and boundary conditions . Solve the one-dimensional wave and heat equations by separating the variables and obtaining eigenfunctions and corresponding eigenvalues . Solve the two-dimensional Laplace equation in Cartesian coordinates . Recognise the need for alternative coordinate systems and solve the two-dimensional Laplace equation in plane polar coordinates Prerequisite: Engineering Mathematics (Sixth Edition) Programmes 24 First-order differential equations and 25 Second-order differential equations 482 483 Partial differential equations Introduction The formation of ordinary linear differential equations and their solution by various methods were covered in some detail in Programmes 24 and 25 of Engineering Mathematics (Sixth Edition), and reference to these before undertaking the new work of this Programme could be beneficial – especially Programme 25 which dealt with second-order equations. Working through the Test exercise of that Programme would provide worthwhile revision. The main results obtained are listed here for convenience and easy reference. 1 Equations of the form a d2 y dy þ cy ¼ 0 þb dx2 dx Auxiliary equation am2 þ bm þ c ¼ 0: Solutions depend on the roots of this equation. (a) Real and different roots: m ¼ m1 and m ¼ m2 Solution y ¼ Aem1 x þ Bem2 x ð1Þ (b) Real and equal roots: m ¼ m1 (twice) Solution y ¼ em1 x ðA þ BxÞ ð2Þ (c) Complex roots: m ¼ j Solution y ¼ ex ðA cos x þ B sin xÞ 2 Equations of the form (a) d2 y þ n2 y ¼ 0 dx2 ð3Þ d2 y n2 y ¼ 0 dx2 ; m2 þ n2 ¼ 0 ; m2 ¼ n2 ; m ¼ jn Solution y ¼ A cos nx þ B sin nx (b) d2 y n2 y ¼ 0 dx2 ; m2 n2 ¼ 0 ; m2 ¼ n2 9 Solution y ¼ A cosh nx þ B sinh nx > = or y ¼ Aenx þ Benx > ; or y ¼ A sinh nðx þ Þ ð4Þ ; m ¼ n ð5Þ In each case, A and B are arbitrary constants depending on the initial conditions, and in the last form is an arbitrary constant. 1 484 Programme 15 Partial differential equations 2 A partial differential equation is a relationship between a dependent variable u and two or more independent variables ðx, y, t, . . .Þ and partial derivatives of u with respect to these independent variables. The solution is therefore of the form u ¼ f ðx, y, t, . . .Þ. Solution by direct integration The simplest form of partial differential equation is such that a solution can be determined by direct partial integration. Example 1 Solve the equation @2u ¼ 12x2 ðt þ 1Þ given that at x = 0, u ¼ cos 2t and @x2 @u ¼ sin t. Notice that the boundary conditions are functions of t and not just @x @2u constants. 2 ¼ 12x2 ðt þ 1Þ. Integrating partially with respect to x, we have @x @u ¼ 4x3 ðt þ 1Þ þ ðtÞ where the arbitrary function ðtÞ takes the place of the @x normal arbitrary constant in ordinary integration. Integrating partially again with respect to x gives u ¼ ............ 3 u ¼ x4 ðt þ 1Þ þ xðtÞ þ ðtÞ where ðtÞ is a second arbitrary function. To find the two arbitrary functions ðtÞ and ðtÞ, we apply the given initial @u conditions that at x ¼ 0, ¼ sin t and u ¼ cos 2t. Substituting these in the @x relevant equations gives ðtÞ ¼ . . . . . . . . . . . . ; ðtÞ ¼ . . . . . . . . . . . . 4 ðtÞ ¼ sin t; ðtÞ ¼ cos 2t Because u ¼ x4 ðt þ 1Þ þ x sin t þ cos 2t Example 2 Solve the equation @2u @u ¼ sin ðx þ yÞ, given that at y = 0, ¼ 1 and at @x@y @x x ¼ 0; u ¼ ðy 1Þ2 : In just the same way as before, u ¼ . . . . . . . . . . . . 485 Partial differential equations 5 u ¼ sinðx þ yÞ þ x þ sin x þ ðy 1Þ2 Because @2u @u ¼ sinðx þ yÞ ; ¼ cosðx þ yÞ þ ðxÞ. @x@y @x @u ¼ 1 ; 1 ¼ cos x þ ðxÞ ; ðxÞ ¼ 1 þ cos x At y ¼ 0; @x @u ; ¼ cosðx þ yÞ þ 1 þ cos x @x Integrating again partially, this time with respect to x, we have u ¼ sinðx þ yÞ þ x þ sin x þ ðyÞ But at x ¼ 0; u ¼ ðy 1Þ2 : ; ðy 1Þ2 ¼ sin y þ ðyÞ ; ðyÞ ¼ ðy 1Þ2 þ sin y ; u ¼ sinðx þ yÞ þ x þ sin x þ sin y þ ðy 1Þ2 Initial conditions and boundary conditions As with any differential equation, the arbitrary constants or arbitrary functions in any particular case are determined from the additional information given concerning the variables of the equation. These extra facts are called the initial conditions or, more generally, the boundary conditions since they do not always refer to zero values of the independent variables. Example 3 Solve the equation @2u ¼ sin x cos y, subject to the boundary conditions that @x@y @u , ¼ 2x and at x ¼ ; u ¼ 2 sin y. 2 @x Work through it: it is easy enough. u ¼ . . . . . . . . . . . . at y ¼ 6 u ¼ x2 þ cos xð1 sin yÞ þ sin y þ 1 2 Because @2u @u ¼ sin x cos y ; ¼ sin x sin y þ ðxÞ @x@y @x @u But ¼ 2x at y ¼ ; ðxÞ ¼ 2x sin x @x 2 @u ¼ 2x sin xð1 sin yÞ ; u ¼ x2 þ cos xð1 sin yÞ þ ðyÞ ; @x But u ¼ 2 sin y at x ¼ ; ðyÞ ¼ 1 2 þ sin y u ¼ x2 þ cos xð1 sin yÞ þ sin y þ 1 2 On to the next frame 486 Programme 15 7 Before we take a closer look at some of the more important partial differential equations occurring in branches of technology, let us recall the fact that if u ¼ u1 , u ¼ u2 , u ¼ u3 , . . . are different solutions of a linear partial differential equation, so also is the linear combination u ¼ c1 u1 þ c2 u2 þ c3 u3 þ . . . where c1 , c2 , c3 , . . . are arbitrary constants. There are many types of partial differential equations, some requiring special treatment in their solution. In this Programme we are concerned with a restricted number of such equations that occur in branches of science and technology, which can be solved by the method of separating the variables, and which also link up with the work we have done on Fourier series techniques. Let us make a new start 8 The wave equation u =f (x,t) u (x, t) O x l x Consider a perfectly flexible elastic string stretched between two points at x ¼ 0 and x ¼ l with uniform tension T. If the string is displaced slightly from its initial position of rest and released, with the end points remaining fixed, then the string will vibrate. The position of any point P in the string will then depend on its distance from one end and on the instant in time. Its displacement u at any time t can thus be expressed as u ¼ f ðx, tÞ where x is its distance from the left-hand end. @2u 1 @2u T in ¼ , where c2 ¼ The equation of motion is given by @x2 c2 @t 2 which T is the tension in the string and the mass per unit length of the string. The displacement of the string is regarded as small so that T and remain constant. Now let us deal with the solution of this equation. On to the next frame Partial differential equations 487 Solution of the wave equation 9 The new equation @2u 1 @2u ¼ has a solution uðx, tÞ. @x2 c2 @t 2 Boundary conditions: (a) The string is fixed at both ends, i.e. at x ¼ 0 and at x ¼ l for all values of time t. Therefore u ðx; tÞ becomes ) uð0, tÞ ¼ 0 for all values of t 0 uðl, tÞ ¼ 0 Initial conditions: (b) If the initial deflection of P at t ¼ 0 is denoted by f ðxÞ, then uðx, 0Þ ¼ f ðxÞ (c) Let the initial velocity of P be gðxÞ, then @u ¼ gðxÞ @t t¼0 So now we have listed all the information available from the question. Next we turn to solving the equation. Solution by separating the variables We assume a trial solution of the form uðx, tÞ ¼ XðxÞTðtÞ where XðxÞ is a function of x only TðtÞ is a function of t only. If we simplify the symbols to u ¼ XT and denote derivatives with respect to their own independent variables by primes, we have u ¼ XT ; @u ¼ X0 T @x @u ¼ XT 0 @t The wave equation and and @2u ¼ X00 T @x2 @2u ¼ XT 00 @t 2 @2u 1 @2u ¼ can then be written as @x2 c2 @t 2 ............ 488 Programme 15 10 X00 T ¼ 1 XT 00 c2 X00 1 T 00 ¼ 2 c T X Notice that the left-hand side expression involves functions of x only and that the right-hand side expression involves functions of t only. Therefore, if these two expressions are to be equal for all values of the separate variables, then both expressions must be equal to ............ and this can be transposed into 11 a constant Denote this arbitrary constant by k. Then we have X00 1 T 00 ¼ k and 2 ¼k c T X ; X00 kX ¼ 0 and T 00 c2 kT ¼ 0 Let us consider the first of these two equations for different values of k. (1) If k ¼ 0, X00 ¼ 0 ; X0 ¼ a ; X ¼ ax þ b. ) But X ¼ 0 at x ¼ 0 ; b ¼ 0 ; X ¼ ax ; a¼b¼0 and X ¼ 0 at x ¼ l ; a ¼ 0 ; X ¼ 0 which is not oscillatory as the problem requires it to be. (2) If k is positive, let k ¼ p2 ; X00 p2 X ¼ 0. The auxiliary equation is therefore m2 p2 ¼ 0 ; m 2 ¼ p2 m ¼ p ; X ¼ Ae px px þ Be But X ¼ 0 at x ¼ 0 ; 0¼AþB and X ¼ 0 at x ¼ l ; 0 ¼ Ae Aepl pl ; A¼0 ; B ¼ A ; 0 ¼ Aðepl epl Þ ; A¼B¼0 Here again X ¼ 0 which is not oscillatory. (3) If k is negative, let k ¼ p2 ; X00 þ p2 X ¼ 0: This is one of the standard equations listed at the beginning of the Programme and gives a solution X ¼ A cos px þ B sin px which fits the requirements. The second equation T 00 c2 kT ¼ 0 therefore now becomes ............ ð1Þ 489 Partial differential equations 12 T 00 þ c2 p2 T ¼ 0 because the same value for k must apply. This equation is of the same form as before and gives the solution T ¼ C cos cpt þ D sin cpt ð2Þ So our suggested solution u ¼ XT now becomes uðx; tÞ ¼ ðA cos px þ B sin pxÞ ðC cos cpt þ D sin cptÞ ; p ¼ , this becomes c uðx; tÞ ¼ A cos x þ B sin x ðC cos t þ D sin tÞ c c and, if we put cp ¼ ð3Þ where A, B, C, D are arbitrary constants. The result, of course, must satisfy the set of boundary conditions which we now turn to. (a) u ¼ 0 when x ¼ 0 for all values of t. From this, we get ............ A¼0 Because, substituting u ¼ 0 and x ¼ 0 in result (3) above 0 ¼ AðC cos t þ D sin tÞ for all t ; A ¼ 0 ; uðx; tÞ ¼ B sin xðC cos t þ D sin tÞ c l (b) u ¼ 0 when x ¼ l for all t ; 0 ¼ B sin ðC cos t þ D sin tÞ c l Now B 6¼ 0 or uðx; tÞ would be identically zero. ; sin ¼ 0. c l nc ¼ n where n ¼ 1; 2, 3, . . . ; ¼ for n ¼ 1, 2, 3, . . . ; c l Note that we exclude n ¼ 0 since this would also make uðx, tÞ identically zero. 13 490 Programme 15 As we can see, there is an infinite set of values of and each separate value gives a particular solution for uðx; tÞ. The values of are called the eigenvalues and each corresponding solution the eigenfunction. Putting n ¼ 1; 2, 3, . . . we therefore have Eigenvalues n 1 nc l c 1 ¼ l ¼ 2 2 ¼ 2c l 3 3 ¼ 3c l .. . .. . r r ¼ Eigenfunctions x fC cos t þ D sin tg uðx; tÞ ¼ B sin c x ct ct u1 ¼ sin C1 cos þ D1 sin l l l 2x 2ct 2ct u2 ¼ sin C2 cos þ D2 sin l l l 3x 3ct 3ct u3 ¼ sin C3 cos þ D3 sin l l l .. . rc l ur ¼ sin rx rct rct Cr cos þ Dr sin l l l Note that the constant B has been absorbed into the constants C and D so that BC ¼ Cn and BD ¼ Dn , where C1 ; C2 , C3 , . . . and D1 ; D2 , D3 , . . . are arbitrary constants. Since the original wave equation is linear in form, we have already noted that if u ¼ u1 , u ¼ u2 , u ¼ u3 . . . are particular solutions, a more general solution is ............ 14 u ¼ u1 þ u2 þ u3 þ . . . The more general solution is therefore 1 1 X X rx rct rct Cr cos þ Dr sin ur ¼ sin uðx; tÞ ¼ l l l r¼1 r¼1 ð4Þ We still have to find Cr and Dr and for this we use the initial conditions which we have not yet taken into account. (c) At t ¼ 0; uðx, 0Þ ¼ f ðxÞ for 0 x l Therefore from ð4Þ, uðx, 0Þ ¼ f ðxÞ ¼ 1 X Cr sin r¼1 (d) Also at t ¼ 0; rx . l @u ¼ gðxÞ for 0 x l @t t¼0 We therefore differentiate (4) with respect to t and put t ¼ 0, which gives ............ 491 Partial differential equations gðxÞ ¼ 15 1 c X rx Dr r sin l r¼1 l Because 1 @u X rx rc rct rc rct ¼ Cr sin þ Dr cos sin @t l l l l l r¼1 ; With t ¼ 0, 1 X @u rc rx Dr ¼ gðxÞ ¼ sin @t l l r¼1 ; gðxÞ ¼ 1 c X rx Dr r sin l r¼1 l Finally we can draw on our knowledge of Fourier series techniques to determine the coefficients Cr and Dr . rx between x ¼ 0 and x ¼ l Cr ¼ 2 mean value of f ðxÞ sin l ð 2 l rx ; Cr ¼ dx r ¼ 1; 2, 3, . . . f ðxÞ sin l 0 l rc rx and Dr ¼ 2 mean value of gðxÞ sin between x ¼ 0 and x ¼ l l l ð 2 l rx dx r ¼ 1; 2, 3, . . . ; Dr ¼ gðxÞ sin rc 0 l The general solution (4) then becomes 1 ð l X 2 rw rct rx f ðwÞ sin uðx, tÞ ¼ dw cos sin l l l l 0 r¼1 ðl 2 rw rct rx þ gðwÞ sin dw sin sin rc 0 l l l ð5Þ Notice that the variable of integration has been changed from x to w because we wish to use the variable x in the final expression for uðx; tÞ. The value of a definite integral depends only on the limit points of the integral and we are free to use any symbol that we desire for the variable of integration – we call such a variable a dummy variable. At first sight, the solution seems very involved, but it can be analysed into a definite sequence of logical steps. Given the equation and relevant initial and boundary conditions, we go through the following stages. (a) Assume a solution of the form u ¼ XT and express the equation in terms of X and T and their derivatives. (b) Transpose the equation by separation of the variables and equate each side to a constant, so obtaining two separate equations, one in x and the other in t. (c) Choose k ¼ p2 to give an oscillatory solution. 492 Programme 15 (d) The two solutions are of the form X ¼ A cos px þ B sin px T ¼ C cos cpt þ D sin cpt Then uðx; tÞ ¼ fA cos px þ B sin pxgfC cos cpt þ D sin cptg. (e) Putting cp ¼ , i.e. p ¼ , this becomes c uðx; tÞ ¼ A cos x þ B sin x fC cos t þ D sin tg: c c (f) Apply boundary conditions to determine A and B. (g) List the eigenvalues and eigenfunctions for n ¼ 1; 2, 3, . . . and determine the general solution as an infinite sum. (h) Apply the remaining initial or boundary conditions. (i) Determine the coefficients Cr and Dr by Fourier series techniques. Make a list of these steps: then we can follow them with an example. 16 Example u (x,0) A stretched string of length 20 cm is set oscillating by displacing its mid-point a u =f(x) distance 1 cm from its rest position and releasing it with zero initial velocity. @2u 1 @2u x ¼ Solve the wave equation @x2 c2 @t 2 where c2 ¼ 1 to determine the resulting motion, uðx; tÞ. First we make a list of the boundary conditions from the data given in the question. uð0, tÞ ¼ . . . . . . . . . . . . ; uð20, tÞ ¼ . . . . . . . . . . . . uðx, 0Þ ¼ . . . . . . . . . . . . ............ @u ¼............ @t t ¼ 0 17 uð0; tÞ ¼ 0; uð20; tÞ ¼ 0 ðfixed end pointsÞ x uðx; 0Þ ¼ f ðxÞ ¼ 0 x 10 10 20 x ¼ 10 x 20 10 @u ¼0 ðzero initial velocityÞ @t t ¼ 0 493 Partial differential equations u(x,0) u=f(x) Now we can apply our sequence of operations which we listed. u= x 10 u =2– x 10 x So move on (a) Assume a solution u ¼ XT where X is a function of x only and T is a 2 function of t only. Then the equation 18 2 @ u @ u ¼ (since c2 ¼ 1) becomes @x2 @t 2 ............ X00 T ¼ XT 00 19 Because u ¼ XT and @2u @2u ¼ @x2 @t 2 ; @u ¼ X0 T @x @u ¼ XT 0 @t @2u ¼ X00 T @x2 @2u ¼ XT 00 @t 2 ; X00 T ¼ XT 00 (b) Next we rearrange the equation to separate the variables, giving ............ X00 T 00 ¼ X T 20 (c) Since the two sides are equal for all values of the variables, each must be equal to a constant k and to give an oscillatory solution we put k ¼ p2 . The two separate equations then are written . . . . . . . . . . . . and . . . . . . . . . . . . X00 þ p2 X ¼ 0 and T 00 þ p2 T ¼ 0 (d) These have solution X ¼ . . . . . . . . . . . . T ¼ ............ so that uðx, tÞ ¼ . . . . . . . . . . . . 21 494 Programme 15 22 X ¼ A cos px þ B sin px; T ¼ C cos pt þ D sin pt ; uðx, tÞ ¼ fA cos px þ B sin pxgfC cos pt þ D sin ptg (e) We normally now put cp ¼ , but in this case c ¼ 1 ; p ¼ and uðx, tÞ ¼ . . . . . . . . . . . . 23 uðx, tÞ ¼ fA cos x þ B sin xgfC cos t þ D sin tg (f) Now we determine A and B from the boundary conditions. (1) uð0, tÞ ¼ 0 ; 0 ¼ AðC cos t þ D sin tÞ ; A¼0 ; uðx, tÞ ¼ B sin xðC cos t þ D sin tÞ (2) uð20, tÞ ¼ 0 ; 0 ¼ B sin 20ðC cos t þ D sin tÞ B 6¼ 0 or u would be identically zero. ; sin 20 ¼ 0: n ; 20 ¼ n ; ¼ 20 n n n ; uðx, tÞ ¼ sin x P cos t þ Q sin t 20 20 20 where P ¼ B C and Q ¼ B D: (g) The next step is to list the eigenvalues and eigenfunctions. Eigenvalues n ¼ n 20 1 1 ¼ 20 2 2 ¼ 2 20 3 3 ¼ 3 20 .. . .. . r r ¼ r 20 Eigenfunctions uðx; tÞ ¼ sin xfP cos t þ Q sin tg t t P1 cos þ Q1 sin 20 20 2x 2t 2t u2 ¼ sin P2 cos þ Q2 sin 20 20 20 3x 3t 3t u3 ¼ sin P3 cos þ Q3 sin 20 20 20 .. . rx rt rt ur ¼ sin Pr cos þ Qr sin 20 20 20 u1 ¼ sin u ¼ u1 þ u2 þ u3 þ . . . ; uðx, tÞ ¼ x 20 1 X r¼1 sin rx rt rt Pr cos þ Qr sin 20 20 20 495 Partial differential equations (h) Now we apply the remaining initial conditions x 0 x 10 ð1Þ uðx; 0Þ ¼ f ðxÞ ¼ 10 20 x ¼ 10 x 20 10 Also uðx; 0Þ ¼ . . . . . . . . . . . . uðx, 0Þ ¼ 1 X r¼1 Pr sin rx 20 24 rx Then Pr ¼ 2 mean value of f ðxÞ sin between x ¼ 0 and x ¼ 20 20 ð 20 2 rx dx f ðxÞ sin ¼ 20 0 20 ð 10 ð 20 x rx 20 x rx ; 10Pr ¼ sin dx þ sin dx 10 20 10 20 0 10 þ I2 ¼ I1 ð 10 x rx I1 ¼ sin dx ¼ . . . . . . . . . . . . 20 0 10 I1 ¼ 20 r 40 r cos þ 2 2 sin r 2 r 2 25 Using integration by parts ð 20 20 x rx I2 ¼ sin dx ¼ . . . . . . . . . . . . 10 20 10 I2 ¼ Then 10 Pr ¼ 20 r 40 r cos 2 2 sin r sin r 2 r 2 20 r 40 r 20 r 40 r cos þ 2 2 sin þ cos 2 2 sin r sin r 2 r 2 r 2 r 2 8 r ; For r ¼ 1, 2, 3, . . . Pr ¼ 2 2 sin r 2 1 X rx 8 r rt rt ; uðx; tÞ ¼ cos þ Q sin sin sin r 20 r 2 2 2 20 20 r¼1 ð2Þ Also at t ¼ 0, @u ¼ 0. @t @u ¼ ............ @t 26 496 27 Programme 15 1 @uðx; tÞ X rx ¼ sin @t 20 r¼1 ; At t ¼ 0; 0¼ 1 X sin r¼1 So finally we have 28 8 r sin 2 2 r 2 r rt sin 20 20 r rt þ Qr cos 20 20 rx r Qr 20 20 ; Qr ¼ 0 uðx; tÞ ¼ . . . . . . . . . . . . uðx; tÞ ¼ 1 8X 1 rx r rt sin cos sin 2 r¼1 r 2 20 2 20 And that is it. Now let us turn to a slightly different equation, but one for which the method of solution is very much along the same lines. The heat conduction equation for a uniform finite bar The conduction of heat in a uniform bar depends on the initial distribution of temperature and on the physical properties of the bar, i.e. the thermal conductivity and specific heat of the material, and the mass per unit length of the bar. With a uniform bar insulated except at its ends, any heat flow is along the bar and, at any instant, the temperature u at a point P is a function of its distance x from one end and of the time t. y u =f (x ,t ) u(x, t ) O x x The one-dimensional heat equation is then of the form @ 2 u 1 @u ¼ @x2 c2 @t ð1Þ k in which k ¼ thermal conductivity of the material; ¼ specific heat of the material; ¼ mass per unit length of the bar. where c2 ¼ 497 Partial differential equations You will already have noticed that the heat equation differs from the wave equation only in the fact that the right-hand side contains the first partial derivative instead of the second. It is not surprising therefore that the method of solution is very much like that of our previous examples. Solutions of the heat conduction equation Consider the case where (a) the bar extends from x ¼ 0 to x ¼ l (b) the temperature of the ends of the bar is maintained at zero (c) the initial temperature distribution along the bar is defined by f ðxÞ. u = f (x, t ) u(x , t ) O x l x The boundary conditions can be expressed as . . . . . . . . . . . . uð0, tÞ ¼ 0 and uðl, tÞ ¼ 0 for all t 0 29 uðx, 0Þ ¼ f ðxÞ for 0 x l As before, we assume a solution of the form uðx, tÞ ¼ XðxÞTðtÞ where X is a function of x only T is a function of t only. Then, starting with u ¼ XT we can write the equation @ 2 u 1 @u in terms ¼ @x2 c2 @t of X and T, and separating the variables, we obtain ............ X00 1 T0 ¼ 2 c T X Arguing as before, since the left-hand side is a function of x only and the righthand side a function of t only, for these to be equal each side must equal the same constant. Let this be ðp2 Þ as before. X00 ¼ p2 ; X00 þ p2 X ¼ 0 giving X ¼ A cos px þ B sin px X 1 T0 and 2 ¼ p2 ; T 0 þ p2 c2 T ¼ 0 giving T ¼ . . . . . . . . . . . . c T ; 30 498 Programme 15 31 T ¼ Cep 2 2 c t Because T0 2 2 ¼ p2 c2 ; ln T ¼ p2 c2 t þ c1 ; T ¼ Cep c t T 2 2 uðx; tÞ ¼ XT ¼ fA cos px þ B sin pxg Cep c t ; uðx; tÞ ¼ fP cos px þ Q sin pxgep 2 2 c t where P ¼ AC and Q ¼ BC Now put pc ¼ ; p ¼ c 2 ; uðx; tÞ ¼ P cos x þ Q sin x e t c c Applying the boundary condition uð0; tÞ ¼ 0 gives . . . . . . . . . . . . and . . . . . . . . . . . . 32 2 P ¼ 0 and uðx; tÞ ¼ Qe t sin x c Also uðl; tÞ ¼ 0 and from this we get ............ 33 ¼ nc for n ¼ 1, 2, 3, . . . l Because if u ¼ 0 when x ¼ l, 2 0 ¼ Qe t sin l c Q 6¼ 0 or uðx, tÞ would be identically zero ; sin ; l ¼ n c ; ¼ nc l n ¼ 1; 2, 3, . . . l ¼0 c 499 Partial differential equations Now we can compile the table of eigenfunctions. nc l n ¼ 1 1 ¼ 2 2 ¼ 3 .. . r 2 uðx; tÞ ¼ Qe t sin c l 2 u1 ¼ Q1 e1 t sin 2c l 3c 3 ¼ l nx l x l 2x l 3x 2 u3 ¼ Q3 e3 t sin l 2 u2 ¼ Q2 e2 t sin .. . .. . rc r ¼ l 2 ur ¼ Qr er t sin rx l u ¼ u1 þ u2 þ u3 þ . . . 1 X rx 2r t Qr e sin ; uðx, tÞ ¼ l r¼1 The remaining boundary condition still to be applied is that when t ¼ 0, uðx; 0Þ ¼ f ðxÞ 0xl This gives f ðxÞ ¼ . . . . . . . . . . . . f ðxÞ ¼ 1 n X Qr sin r¼1 rxo l 34 and from our knowledge of Fourier series techniques: Qr ¼ . . . . . . . . . . . . Qr ¼ 2 mean value of f ðxÞ sin ; Qr ¼ 2 l ðl rx dx and the final solution becomes l 1 ð l 2 X rw rx 2r t dw e uðx; tÞ ¼ f ðwÞ sin sin l r¼1 l l 0 f ðxÞ sin 0 where r ¼ rx from x ¼ 0 to x ¼ l l rc l r ¼ 1, 2, 3; . . . Now on to the next frame for an example 35 500 36 Programme 15 Example A bar of length 2 m is fully insulated along its sides. It is initially at a uniform temperature of 108C and at t ¼ 0 the ends are plunged into ice and maintained at a temperature of 08C. Determine an expression for the temperature at a point P a distance x from one end at any subsequent time t seconds after t ¼ 0. 10 u(x, 0) 0 x 2m 10 P u (x, t ) u (x ,t ) 0 We have the heat equation x 2m x @ 2 u 1 @u with the boundary conditions ¼ @x2 c2 @t . . . . . . . . . . . . ; . . . . . . . . . . . . ; and . . . . . . . . . . . . 37 uð0, tÞ ¼ 0; uð2, tÞ ¼ 0; uðx, 0Þ ¼ 10 Assuming a solution of the form u ¼ XT, we know that this gives for this equation X ¼ A cos px þ B sin px and T ¼ Cep 2 2 c t so that the general solution is uðx; tÞ ¼ fP cos px þ Q sin pxgep 2 2 c t and the solution becomes c 2 uðx; tÞ ¼ P cos x þ Q sin x e t c c If we now write pc ¼ , p¼ Applying the first two of the boundary conditions gives us ............ 501 Partial differential equations n nxo 2 t P ¼ 0 and uðx; tÞ ¼ Q sin e 2 38 Because 2 uð0, tÞ ¼ 0 ; 0 ¼ Pe t ; P ¼ 0 2 ; uðx, tÞ ¼ Q sin x e t c 2 2 t e Also uð2, tÞ ¼ 0 ; 0 ¼ Q sin c 2 2 nc Q 6¼ 0 ; sin ¼0 ; ¼ n ; ¼ c c 2 nx 2 t e ; uðx, tÞ ¼ Q sin 2 n ¼ 1, 2, 3, . . . There is, of course, an infinite number of such solutions with different values of n. We can write the solution so far therefore as uðx, tÞ ¼ . . . . . . . . . . . . uðx; tÞ ¼ 1 X Qr sin r¼1 rx 2r t e 2 39 Finally, there is the remaining initial condition that at t ¼ 0, u ¼ 10. ; uðx; 0Þ ¼ f ðxÞ ¼ 10 ; 10 ¼ 1 X Qr sin r¼1 rx 2 rx from x ¼ 0 to x ¼ 2. 2 ; Qr ¼ . . . . . . . . . . . . where Qr ¼ 2 mean value of 10 sin 0 ðr evenÞ; 40 ðr oddÞ r Because ð2 ð2 rx rx dx ¼ 10 sin dx 2 2 0 0 20 h rxi2 20 ¼ cos f1 cos rg ¼ r 2 0 r 40 ðr oddÞ ¼ 0 ðr evenÞ and r Qr ¼ 2 2 10 sin Therefore the required solution is uðx, tÞ ¼ . . . . . . . . . . . . 40 502 Programme 15 41 uðx, tÞ ¼ 1 40 X 1 rx 2r t sin e r ðoddÞ¼1 r 2 where r ¼ r ¼ 1; 3; 5; . . . rc 2 By now you will appreciate that the approach to all these problems is very much the same, as indeed it still is with the next important equation. Laplace’s equation The Laplace equation concerns the distribution of a field, e.g. temperature, potential, etc., over a plane area subject to certain boundary conditions. z y z=u (x, y) P O x The potential at a point P in a plane can be indicated by an ordinate axis and is a function of its position, i.e. z ¼ uðx, yÞ where uðx, yÞ is the solution of the @2u @2u Laplace two-dimensional equation 2 þ 2 ¼ 0. @x @y Let us consider the situation in the next frame 42 Solution of the Laplace equation y z b y u (x, y) O a x O x @2u @2u þ ¼ 0 for the @x2 @y 2 rectangle bounded by the lines x ¼ 0, y ¼ 0, x ¼ a, y ¼ b, subject to the following boundary conditions We are required to determine a solution of the equation u¼0 when x ¼ 0 0yb u¼0 u¼0 when x ¼ a when y ¼ b 0yb 0xa u ¼ f ðxÞ when y ¼ 0 0 x a i.e. uð0, yÞ ¼ 0 and uða, yÞ ¼ 0 for 0 y b and uðx, bÞ ¼ 0 and uðx, 0Þ ¼ f ðxÞ for 0 x a. 503 Partial differential equations The solution z ¼ uðx, yÞ will give the potential at any point within the rectangle OPQR. We start off, as usual, by assuming a solution of the form uðx, yÞ ¼ XðxÞYðyÞ where X is a function of x only and Y is a function of y only. We now express the equation in terms of X and Y and separate the variables to give ............ 43 X00 Y 00 ¼ X Y Because u ¼ XY ; @u ¼ X0 Y @x @u ¼ XY 0 @y @2u ¼ X00 Y @x2 @2u ¼ XY 00 @y 2 and and The equation is then X00 Y ¼ XY 00 ; X00 Y 00 ¼ X Y Putting each side equal to a constant ðp2 Þ gives two equations X00 þ p2 X ¼ 0 and Y 00 p2 Y ¼ 0 X00 þ p2 X ¼ 0 has a solution X ¼ . . . . . . . . . . . . X ¼ A cos px þ B sin px In the introduction to this Programme we said that the equation Y 00 p2 Y ¼ 0 has a solution of the form Y ¼ C cosh py þ D sinh py which can also be expressed as Y ¼ E sinh pðy þ Þ. ; uðx, yÞ ¼ fA cos px þ B sin pxgE sinh pðy þ Þ ; uðx, yÞ ¼ fP cos px þ Q sin pxg sinh pðy þ Þ Now we apply the first of the boundary conditions. uð0, yÞ ¼ 0 ; 0 ¼ P sinh pðy þ Þ ; P ¼ 0 ; uðx, yÞ ¼ Q sin px sinh pðy þ Þ From the second boundary condition, we have uða, yÞ ¼ 0 ; 0 ¼ Q sin pa sinh pðy þ Þ ; sin pa ¼ 0 ; pa ¼ n for n ¼ 1; 2, 3, . . . n and uðx, yÞ ¼ Q sin x sinh ðy þ Þ a Now from the third condition If we write ¼ p then ¼ uðx, bÞ ¼ 0 from which we have . . . . . . . . . . . . 44 504 Programme 15 45 uðx, yÞ ¼ Q sin x sinh ðb yÞ Because 0 ¼ Q sin x sinh ðb þ Þ ; sinh ðb þ Þ ¼ 0 ; ¼ b. ; uðx, yÞ ¼ Q sin x sinh ðy bÞ sinh ðy bÞ ¼ sinh ðb yÞ ; uðx, yÞ ¼ Q sin x sinh ðb yÞ, the minus sign being absorbed in the symbol Q whose value has yet to be n with n ¼ 1, 2, 3, . . . and there is therefore an infinite found. Now ¼ a number of values for and hence an infinite number of solutions for uðx, yÞ. Therefore, again using u ¼ u1 þ u2 þ u3 þ . . . we have uðx, yÞ ¼ . . . . . . . . . . . . 46 uðx, yÞ ¼ 1 X Qr sin r x sinh r ðb yÞ r¼1 Now there remains the fourth boundary condition to be applied. uðx, 0Þ ¼ f ðxÞ ; f ðxÞ ¼ 1 X Qr sin r x sinh r b r¼1 ; Qr sinh r b ¼ 2 mean value of f ðxÞ sin r x from x ¼ 0 to x ¼ a ð 2 a ¼ f ðxÞ sin r x dx a 0 ð 2 a rx dx ¼ f ðxÞ sin a 0 a from which the coefficients Qr can be found. Let us work through an example with numerical values. Example Determine a solution uðx, yÞ of the Laplace equation @2u @2u þ ¼ 0 subject @x2 @y 2 to the following boundary conditions u ¼ 0 when x ¼ 0; u ! 0 when y ! 1; u ¼ 0 when x ¼ u ¼ 3 when y ¼ 0 As always, we begin with uðx, yÞ ¼ XðxÞYðyÞ, rewrite the equation in terms of X and Y and separate the variables. The equation then becomes ............ 505 Partial differential equations 47 X00 Y 00 ¼ X Y Equating each side to p2 , we have X00 þ p2 X ¼ 0 and Y 00 p2 Y ¼ 0: X00 þ p2 X ¼ 0 has a solution . . . . . . . . . . . . 48 X ¼ A cos px þ B sin px The solution of Y 00 p2 Y ¼ 0 can be stated in three different forms Y ¼ C cosh py þ D sinh py; Y ¼ Cepy þ Depy ; Y ¼ C sinh pðy þ Þ On this occasion, we will use the second one Y ¼ Cepy þ Depy Then uðx, yÞ ¼ fA cos px þ B sin pxgfCepy þ Depy g Application of the first boundary condition uð0, yÞ ¼ 0 gives . . . . . . . . . . . . and . . . . . . . . . . . . A ¼ 0 and uðx, yÞ ¼ sin pxfPepy þ Qepy g 49 Because 0 ¼ AfCepy þ Depy g ; A¼0 and uðx, yÞ ¼ B sin pxfCepy þ Depy g ¼ sin pxfPepy þ Qepy g. The second boundary condition uð, yÞ ¼ 0 then gives ............ uðx, yÞ ¼ sin nx fPeny þ Qeny g n ¼ 1, 2, 3, . . . Because u ¼ 0 when x ¼ ; 0 ¼ sin pfPepy þ Qepy g ; sin p ¼ 0 ; p ¼ n ; p ¼ n n ¼ 1, 2, 3, . . . ; uðx, yÞ ¼ sin nx fPeny þ Qeny g The third condition is that u ! 0 as y ! 1. 50 506 Programme 15 Because eny ! 0 as y ! 1 then 0 ¼ sin nxfPeny g, so that P ¼ 0 ; uðx, yÞ ¼ Qeny sin nx But n can have an infinite number of values giving an infinite number of solutions u1 ¼ Q1 ey sin x u2 ¼ Q2 e2y sin 2x u3 ¼ Q3 e3y sin 3x .. .. . . ur ¼ Qr ery sin rx So the solution at this stage can be written as uðx, yÞ ¼ . . . . . . . . . . . . 51 uðx, yÞ ¼ 1 X Qr ery sin rx r¼1 Now we turn to the final boundary condition that u ¼ 3 when y ¼ 0. ; 3¼ 1 X Qr sin rx from which we obtain r¼1 Qr ¼ . . . . . . . . . . . . 52 Qr ¼ 0 ðr evenÞ; Qr ¼ 12 ðr oddÞ r Because Qr ¼ 2 mean value of 3 sin rx between x ¼ 0 and x ¼ ð 2 6 cos rx 6 3 sin rx dx ¼ ¼ ð1 cos rÞ ¼ 0 r r 0 12 ðr oddÞ ; Qr ¼ 0 ðr evenÞ and r 1 X 12 ry ; uðx, yÞ ¼ r ¼ 1, 3, 5, . . . e sin rx r r ðoddÞ¼1 12 y 1 1 ; uðx, yÞ ¼ e sin x þ e3y sin 3x þ e5y sin 5x þ . . . 3 5 507 Partial differential equations Laplace’s equation in plane polar coordinates Laplace’s equation 53 @ 2 uðx, yÞ @ 2 uðx, yÞ þ ¼0 @x2 @y 2 is often referred to as the potential equation because such physical entities as the electrostatic and gravitational potentials can be shown to satisfy it. It is an equation that is commonly met in science and engineering. Solving this equation inside a region of the x–y plane subject to some specified condition applied to uðx, yÞ on the boundary of the region is known as a Dirichlet problem. To solve this Dirichlet problem we proceed, as we have seen, by separating the variables to find the general solution and then matching up the general solution to the boundary conditions to find the specific solution. However, the process of finding the specific solution from the general solution is very sensitive to the shape of the boundary, and difficulties can arise if the symmetries of the boundary do not match the symmetries of the coordinate system used. For example, if the region under consideration is bounded by the circle x2 þ y 2 ¼ a2 employing Cartesian coordinates will create difficulties when we come to match up the general solution in Cartesians to the boundary conditions on the circular boundary. To avoid such difficulties we choose a coordinate system that has the same symmetries as the boundary where the coordinate symmetries are exhibited when we let one variable vary while keeping all the others constant. The Cartesian coordinate system ðx, yÞ produces straight lines x ¼ constant as y varies and y ¼ constant as x varies. The plane polar coordinate system ðr, Þ, on the other hand, produces circles r ¼ constant when varies and so is suitable for dealing with circular boundaries in the plane. y r r θ O θ x r = constant x = r cosθ y = r sinθ 508 Programme 15 Before we attempt to find the solution we must pose the problem from the beginning in terms of the coordinates that are appropriate to the boundary conditions. This means, of course, that Laplace’s equation must also be given in the same coordinates. To convert Laplace’s equation from its current form in Cartesians ðx, y) to a new form in plane polar coordinates ðr, Þ where x ¼ r cos and y ¼ r sin requires manipulations using Frame 11 onwards of Programme 14. We shall not go into this here, suffice it to say that in plane polar coordinates Laplace’s equation is @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ ¼0 þ 2 @r 2 r @r r @2 where vðr, Þ is the expression obtained by changing the coordinates in uðx, yÞ using x ¼ r cos and y ¼ r sin . We shall now pose the problem anew in the next frame 54 The problem Find the solution to @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ ¼0 þ 2 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ a2 (that is, for 0 r a) of the plane where 1 vðr, Þ is finite for 0 r a and for all 2 vða, Þ ¼ f ðÞ – the condition on the boundary of the circular region 3 is unbounded but vðr, þ 2Þ ¼ vðr, Þ for 0 r a. That is, though can take any finite value, the value of vðr, Þ repeats itself as winds round every 2. Separating the variables The variables are r and and we assume they are separable and write vðr, Þ ¼ RðrÞðÞ. This form is then substituted into Laplace’s equation and r2 the entire equation multiplied by to obtain RðrÞðÞ ............ ¼ 0 509 Partial differential equations 55 r 2 d2 RðrÞ r dRðrÞ 1 d2 ðÞ þ þ ¼0 RðrÞ dr ðÞ d2 RðrÞ dr 2 Because Substituting RðrÞðÞ for vðr, Þ gives @ 2 RðrÞðÞ 1 @RðrÞðÞ 1 @ 2 RðrÞðÞ þ 2 þ ¼0 @r 2 r @r r @2 That is ðÞ d2 RðrÞ ðÞ dRðrÞ RðrÞ d2 ðÞ þ 2 þ ¼0 dr 2 r dr r d2 Multiplying the entire equation by r2 then gives RðrÞðÞ r 2 d2 RðrÞ r dRðrÞ 1 d2 ðÞ þ þ ¼0 2 RðrÞ dr ðÞ d2 RðrÞ dr From this result we can say that r 2 d2 RðrÞ r dRðrÞ 1 d2 ðÞ ¼ þ ¼k 2 RðrÞ dr ðÞ d2 RðrÞ dr which gives rise to the two uncoupled, second-order ordinary differential equations r 2 d2 RðrÞ r dRðrÞ ¼ k so that þ RðrÞ dr RðrÞ dr 2 r2 d2 RðrÞ dRðrÞ ¼ kRðrÞ þr dr 2 dr ð1Þ and 1 d2 ðÞ d2 ðÞ ¼ k so that ¼ kðÞ ðÞ d2 d2 ð2Þ The general solution to equation (2) for k > 0 is ............ n ðÞ ¼ an cos n þ bn sin n where n ¼ 1, 2, . . . Because d2 ðÞ d2 ðÞ ¼ kðÞ, that is þ kðÞ ¼ 0 we use the auxiliary d2 d2 pffiffiffi equation m2 þ k ¼ 0 with solutions m ¼ j k. This gives the solution, periodic with period 2 as pffiffiffi pffiffiffi ðÞ ¼ A cos k þ B sin k ð3Þ To solve 56 510 Programme 15 provided k > 0 so that m is pure imaginary. If k < 0 then non-periodic solutions would result which would be physically incorrect. To ensure periodicity, that is to ensure that k > 0 write k ¼ n2 , n ¼ 1, 2, . . .. n ðÞ ¼ an cos n þ bn sin n is a solution to equation ð2Þ. We shall look at the case n ¼ 0 later. Substituting k ¼ n2 into equation (1) then gives r2 d2 RðrÞ dRðrÞ ¼ n2 RðrÞ þr dr 2 dr ð4Þ As a trial solution to equation (4) let RðrÞ ¼ pr q . Substitution into (4) gives q ¼ ............ 57 q ¼ n where n ¼ 1, 2, . . . Because dRðrÞ dðpr q Þ d2 RðrÞ ¼r ¼ rpqr q1 ¼ pqr q . Similarly r 2 ¼ pqðq 1Þr q . dr dr dr 2 d2 RðrÞ dRðrÞ ¼ n2 RðrÞ gives þr Therefore, substitution into r 2 dr 2 dr r ½qðq 1Þ þ qpr q ¼ n2 pr q and so q2 n2 pr q ¼ 0 giving q ¼ n where n ¼ 1, 2, . . . Therefore, a solution to equation (4) is Rn ðrÞ ¼ cn r n þ dn r n provided n 6¼ 0. The case n ¼ 0 is special. 58 Summary To summarise the results so far, we have started to solve Laplace’s equation @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ a2 (that is, for 0 r aÞ of the plane where 1 2 3 vðr, Þ is finite for 0 r a and for all vða, Þ ¼ f ðÞ is unbounded but vðr, þ 2Þ ¼ vðr, Þ for 0 r a. We have found that, assuming vðr, Þ ¼ RðrÞðÞ then, provided n 6¼ 0 n ðÞ ¼ an cos n þ bn sin n Rn ðrÞ ¼ cn r n þ dn r n So that vn ðr, Þ ¼ Rn ðrÞn ðÞ ¼ ðcn r n þ dn r n Þðan cos n þ bn sin nÞ If we now apply the boundary condition 1 we find that dn ¼ . . . . . . . . . . . . 511 Partial differential equations 59 dn ¼ 0 Because vðr, Þ is finite for 0 r a. In particular, the solution is finite when r ¼ 0 and so we cannot have a term of the form r n . Accordingly dn ¼ 0, so omitting the r n term the solution then becomes vn ðr, Þ ¼ cn r n ðan cos n þ bn sin nÞ There is an infinite number of such solutions (eigenfunctions), one for each eigenvalue n. The complete solution to Laplace’s equation is then a linear combination of all these eigenfunctions. That is vðr, Þ ¼ 1 X en vn ðr, Þ ¼ n¼1 1 X r n ðAn cos n þ Bn sin nÞ n¼1 And now for the n ¼ 0 case 60 The n = 0 case When n ¼ 0 then k ¼ 0 and equation (1) becomes r2 d2 RðrÞ dRðrÞ ¼0 þr dr 2 dr dRðrÞ then this equation becomes dr dSðrÞ 2 dSðrÞ þ rSðrÞ ¼ 0, that is r r þ SðrÞ ¼ 0 and so r dr dr dSðrÞ d½rSðrÞ þ SðrÞ ¼ ¼0 r dr dr and if we let SðrÞ ¼ This has the solution rSðrÞ ¼ (constant) and so SðrÞ ¼ dRðrÞ ¼ dr r giving RðrÞ ¼ ln r þ ð5Þ When n ¼ 0 then k ¼ 0 and equation (2) becomes d2 ðÞ ¼ 0 with solution ðÞ ¼ þ d2 ð6Þ Applying the boundary conditions to the solutions (5) and (6) gives ¼ . . . . . . . . . . . . and ¼ ............ 512 Programme 15 61 ¼ 0 and ¼0 Because (a) vðr, Þ is finite for 0 r a, in particular when r ¼ 0, and so ¼ 0 (b) vðr, þ 2Þ ¼ vðr, Þ. That is, though can take any finite value, the value of vðr, Þ repeats itself as winds round every 2 and this means that ¼ 0. So, when n ¼ 0 the solution is v0 ðr, Þ ¼ constant. We therefore write the complete solution as vðr, Þ ¼ 1 A0 X þ r n ðAn cos n þ Bn sin nÞ 2 n¼1 A0 . 2 Applying the condition on the boundary where vða, Þ ¼ f ðÞ we see that where the constant is taken to be in the form f ðÞ ¼ 1 A0 X þ an ðAn cos n þ Bn sin nÞ 2 n¼1 which is a Fourier series and hence the form of the constant term being taken A0 as . 2 The Fourier coefficients are then ð ð 1 2 1 2 An ¼ f ðÞ cos n d and Bn ¼ f ðÞ sin n d 2an 0 2an 0 Example Solve Laplace’s equation @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ a2 of the plane where 1 2 3 vðr, Þ is finite for 0 r a and for all vða, Þ ¼ sin vðr, þ 2Þ ¼ vðr, Þ for 0 r a. The solution, as we have seen, is vðr, Þ ¼ 1 A0 X þ r n ðAn cos n þ Bn sin nÞ where 2 n¼1 An ¼ . . . . . . . . . . . . and Bn ¼ . . . . . . . . . . . . 513 Partial differential equations An ¼ 0 and Bn ¼ 1 2an 62 1; n Because An ¼ 1 2an ð 2 0 ð 2 f ðÞ cos n d ¼ 1 2an ð 2 sin cos n d ¼ 0 and 0 ð 2 1 1 1 f ðÞ sin n d ¼ sin sin n d ¼ 2an 0 2an 0 2an where 1; n is the Kronecker delta Bn ¼ 1; n 1 That is, B1 ¼ , Bn ¼ 0 for n ¼ 2, 3, . . .. The complete solution is then 2a r vðr, Þ ¼ sin a Notice that all three conditions in Frame 61 are satisfied by this solution, that is 1 2 3 r vðr, Þ ¼ sin is finite for 0 r a and for all a a vða, Þ ¼ sin ¼ sin a r r vðr, þ 2Þ ¼ sinð þ 2Þ ¼ sin ¼ vðr, Þ for 0 r a. a a That covers the main steps in the method of solving linear, second-order partial differential equations applied specifically to the wave equation, the heat conduction equation and Laplace’s equation. The same approach can be made with other similar equations. The Revision summary and the Can you? checklist now follow, then the Test exercise with problems like those we have considered. Although the solutions take rather more steps than with other forms of equations, the method is straightforward and follows a clear pattern. The Further problems give additional practice. 514 Programme 15 Revision summary 15 63 1 Ordinary second-order linear differential equations (a) Equation of the form a d2 y dy þb þ cy ¼ 0 2 dx dx Auxiliary equation am2 þ bm þ c ¼ 0 (1) Real and different roots: m ¼ m1 and m ¼ m2 y ¼ Aem1 x þ Bem2 x (2) Real and equal roots: m ¼ m1 (twice) y ¼ em1 x ðA þ BxÞ (3) Complex roots: m ¼ j y ¼ ex fA cos x þ B sin xg. (b) Equations of the form ð1Þ ð2Þ d2 y þ n2 y ¼ 0; dx2 d2 y n2 y ¼ 0; dx2 or or 2 d2 y n2 y ¼ 0 dx2 y ¼ A cos nx þ B sin nx y ¼ A cosh nx þ B sinh nx y ¼ Aenx þ Benx y ¼ A sinh nðx þ Þ. Partial differential equations Solution u ¼ f ðx, y, t, . . .Þ Linear equations: If u ¼ u1 , u ¼ u2 ; u ¼ u3 ; . . . are solutions, so also is 1 X u ¼ u1 þ u2 þ u3 þ . . . þ ur þ . . . ¼ ur : r¼1 (a) Wave equation – transverse vibrations of an elastic string @2u 1 @2u ¼ @x2 c2 @t 2 (b) Heat conduction equation – heat flow in uniform finite bar @ 2 u 1 @u k where c2 ¼ ¼ @x2 c2 @t k ¼ thermal conductivity of material ¼ specific heat of the material ¼ mass per unit length of bar. (c) Laplace equation – distribution of a field over a plane area @2u @2u þ ¼ 0. @x2 @y 2 515 Partial differential equations 3 Separating the variables Let uðx, yÞ ¼ XðxÞYðyÞ where XðxÞ is a function of x only and YðyÞ is a function of y only. Then @u ¼ X0 Y; @x @u ¼ XY 0 ; @y @2u ¼ X00 Y @x2 @2u ¼ XY 00 @y 2 Substitute in the given partial differential equation and form separate differential equations to give XðxÞ and YðyÞ by introducing a common constant ðp2 Þ. Determine arbitrary functions by use of the initial and boundary conditions. 4 Laplace’s equation in plane polar coordinates @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 Separating the variables by vðr, Þ ¼ RðrÞðÞ produces two uncoupled, second-order ordinary differential equations d2 RðrÞ dRðrÞ ¼ kRðrÞ þr 2 dr dr d2 ðÞ and ¼ kðÞ d2 r2 These two ordinary differential equations can then be solved under the application of appropriate boundary conditions. Can you? 64 Checklist 15 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Summarise the introductory methods of solving ordinary differential equations? Yes No 1 . Solve partial differential equations that are amenable to solution by direct integration? Yes No . Apply initial and boundary conditions? Yes No 2 to 7 5 to 7 516 Programme 15 . Solve the one-dimensional wave and heat equations by separating the variables and obtaining eigenfunctions and corresponding eigenvalues? Yes No . Solve the two-dimensional Laplace equation in Cartesian coordinates? Yes No . Recognise the need for alternative coordinate systems and solve the two-dimensional Laplace equation in plane polar coordinates? Yes No 8 to 40 41 to 52 53 to 62 Test exercise 15 65 1 Solve the following equations @2u @u ¼ 4t: ¼ 24x2 ðt 2Þ, given that at x ¼ 0; u ¼ e2t and @x2 @x @2u @u ¼ 4ey cos 2x, given that at y ¼ 0; ¼ cos x (b) @x@y @x 2 and at x ¼ , u ¼ y . (a) 2 A perfectly elastic string is stretched between two points 10 cm apart. Its centre point is displaced 2 cm from its position of rest at right angles to the original direction of the string and then released with zero velocity. Applying the @2u 1 @2u equation 2 ¼ 2 2 with c2 ¼ 1, determine the subsequent motion uðx; tÞ. @x c @t 3 One end A of an insulated metal bar AB of length 2 m is kept at 08C while the other end B is maintained at 508C until a steady state of temperature along the bar is achieved. At t ¼ 0, the end B is suddenly reduced to 08C and kept at that @ 2 u 1 @u temperature. Using the heat conduction equation 2 ¼ 2 , determine an @x c @t expression for the temperature at any point in the bar distance x from A at any time t. 4 A square plate is bounded by the lines x ¼ 0, y ¼ 0; x ¼ 2; y ¼ 2: Apply the @2u @2u Laplace equation þ ¼ 0 to determine the potential distribution uðx, yÞ @x2 @y 2 over the plate, subject to the following boundary conditions. u ¼ 0 when x ¼ 0 0 y 2 u ¼ 0 when x ¼ 2 0 y 2 u ¼ 0 when y ¼ 0 0 x 2 u ¼ 5 when y ¼ 2 0 x 2. Partial differential equations 5 517 Solve Laplace’s equation in plane polar coordinates @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ 1 of the plane where (1) (2) (3) vðr, Þ is finite for 0 r 1 and for all vð1, Þ ¼ 5 cos 3 vðr, þ 2Þ ¼ vðr, Þ for 0 r 1. Further problems 15 @2u 1 @2u ¼ 0 is satisfied by @x2 c2 @t 2 u ¼ f ðx þ ctÞ þ Fðx ctÞ where f and F are arbitrary functions. 1 Show that the equation 2 If 3 The centre point of a perfectly elastic string stretched between two points A and B, 4 m apart, is deflected a distance 0.01 m from its position of rest perpendicular to AB and released initially with zero velocity. Apply the wave @2u 1 @2u equation 2 ¼ 2 2 where c ¼ 10 to determine the subsequent motion of a @x c @t point P distant x from A at time t. 4 An elastic string is stretched between two points 10 cm apart. A point P on the string 2 cm from the left-hand end, i.e. the origin, is drawn aside 1 cm from its position of rest and released with zero velocity. Solve the one-dimensional wave equation to determine the displacement of any point at any instant. 5 An insulated uniform metal bar, 10 units long, has the temperature of its ends maintained at 08C and at t ¼ 0 the temperature distribution f ðxÞ along the bar @ 2 u 1 @u is defined by f ðxÞ ¼ xð10 xÞ. Solve the heat conduction equation 2 ¼ 2 @x c @t with c2 ¼ 4 to determine the temperature u of any point in the bar at time t. 6 The ends of an insulated rod AB, 10 units long, are maintained at 08C. At t ¼ 0, the temperature within the rod rises uniformly from each end reaching 28C at the mid-point of AB. Determine an expression for the temperature uðx; tÞ at any point in the rod, distant x from the left-hand end at any subsequent time t. @2u 1 @2u ¼ and c ¼ 3, determine the solution u ¼ f ðx; tÞ subject to the @x2 c2 @t 2 boundary conditions uð0; tÞ ¼ 0 and uð2; tÞ ¼ 0 for t 0 @u uðx; 0Þ ¼ xð2 xÞ and ¼0 0 x 2. @t t¼0 66 518 Programme 15 7 A rectangular plate OPQR is bounded by the lines x ¼ 0, y ¼ 0, x ¼ 4, y ¼ 2. Determine the potential distribution uðx; yÞ over the rectangle using the @2u @2u Laplace equation 2 þ 2 ¼ 0, subject to the following boundary conditions @x @y uð0; yÞ ¼ 0 0y2 uð4; yÞ ¼ 0 0y2 uðx; 2Þ ¼ 0 0x4 uðx; 0Þ ¼ xð4 xÞ 0 x 4. 8 Two sides AB and AD of a rectangular plate ABCD lie along the x and y axes respectively. The remaining two sides are the lines x = 5 and y = 2. The sides BC, CD and DA are maintained at zero temperature. The temperature distribution along AB is defined by f ðxÞ ¼ xðx 5Þ. Determine an expression for the steadystate temperature at any point in the plate. 9 Solve Laplace’s equation in plane polar coordinates @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ 1 of the plane where (1) (2) (3) 10 vðr, Þ is finite for 0 r 1 and for all vð1, Þ ¼ sin 2 4 cos vðr, þ 2Þ ¼ vðr, Þ for 0 r 1. Solve Laplace’s equation in plane polar coordinates @ 2 vðr, Þ 1 @vðr, Þ 1 @ 2 vðr, Þ þ 2 þ ¼0 @r 2 r @r r @2 in the circular region x2 þ y 2 ¼ 1 of the plane where (1) (2) (3) vðr, Þ is finite for 0 r 1 and for all vð1, Þ ¼ 3 sin2 vðr, þ 2Þ ¼ vðr, Þ for 0 r 1. Programme 16 Frames 1 to 69 Matrix algebra Learning outcomes When you have completed this Programme you will be able to: . Determine whether a matrix is singular or non-singular . Determine the rank of a matrix . Determine the consistency of a set of linear equations and hence demonstrate the uniqueness of their solution . Obtain the solution of a set of simultaneous linear equations by using matrix inversion, by row transformation, by Gaussian elimination, by triangular decomposition and by using an electronic spreadsheet . Use matrices to represent transformations between coordinate systems Prerequisite: Engineering Mathematics (Sixth Edition) Programmes 4 Determinants and 5 Matrices 519 520 Programme 16 Singular and non-singular matrices 1 Every square matrix A has associated with it a number called the determinant of A and denoted by jAj. If jAj 6¼ 0 then A is called a non-singular matrix. Otherwise if jAj = 0, then A is called a singular matrix. Example 1 0 1 1 2 8 Is A ¼ @ 4 7 6 A singular or non-singular? 9 5 3 1 2 8 jAj ¼ 4 7 6 9 5 3 7 6 4 6 4 7 ¼ 1 2 þ 8 5 3 9 3 9 5 ¼ ð21 30Þ 2ð12 54Þ þ 8ð20 63Þ ¼ 9 þ 84 344 ¼ 269 Because jAj 6¼ 0 then A is non-singular. Example 2 0 3 9 Is A ¼ @ 1 5 2 7 1 2 6 A singular or non-singular? 4 A is . . . . . . . . . . . . 2 singular Because 3 jAj ¼ 1 2 9 5 7 2 6 4 ¼ 3ð20 42Þ 9ð4 12Þ þ 2ð7 10Þ ¼ 66 þ 72 6 ¼0 Because jAj ¼ 0 then jAj is singular. 521 Matrix algebra Exercise Determine whether each of the following is singular or non-singular. ! ! 3 4 4 5 2 jBj ¼ 1 jAj ¼ 6 8 2 3 0 1 0 1 4 1 2 3 2 4 B C B C C 3 jCj ¼ B 3C 4 jDj ¼ B @1 7 A @5 1 6A 5 8 1 2 0 3 1 3 non-singular singular 2 4 singular non-singular Because Straightforward evaluation of the relevant determinants gives 1 jAj = 2 2 jBj ¼ 0 3 jCj ¼ 0 4 jDj ¼ 5 Closely related to the notion of the singularity or otherwise of a square matrix is the notion of rank of a general n m matrix. Rank of a matrix The rank of an n m matrix A is the order of the largest square, non-singular sub-matrix. That is, the largest square sub-matrix whose determinant is nonzero. If n ¼ m, so making A itself square, then this sub-matrix could be the matrix A itself. Example 0 1 3 4 5 To find the rank of the matrix A ¼ @ 1 2 3 A we note that 4 5 6 3 4 5 jAj ¼ 1 2 3 ¼ . . . . . . . . . . . . 4 5 6 3 522 Programme 16 4 0 Because 3 jAj ¼ 1 4 4 2 5 5 3 6 ¼ 3ð12 15Þ 4ð6 12Þ þ 5ð5 8Þ ¼ 9 þ 24 15 ¼ 0 Therefore we can say that the rank of A is . . . . . . . . . . . . 5 not 3 Because jAj ¼ 0 and therefore A is singular. Now try a sub-matrix of order 2. 3 4 1 2 ¼ 6 4 ¼ 2 6¼ 0. Therefore the rank of A is . . . . . . . . . . . . 6 2 Because The largest square, non-singular sub-matrix of A has order 2 therefore A has rank 2. This method of finding the rank of a matrix can be a very hit and miss affair and a better, more systematic method is to use elementary operations and the notion of an equivalent matrix. Next frame Elementary operations and equivalent matrices 7 Each of the following row operations on matrix A produces a row equivalent matrix B, where the order and rank of B is the same as that of A. We write A B. 1 2 3 Interchanging two rows Multiplying each element of a row by the same non-zero scalar quantity Adding or subtracting corresponding elements from those of another row are operations called elementary row operations. There is a corresponding set of three elementary column operations that can be used to form column equivalent matrices. 523 Matrix algebra Example 1 0 1 3 4 5 Given A ¼ @ 1 2 3 A then 4 5 6 0 1 0 1 3 4 5 0 2 4 B C B C 2 3A @1 2 3A @1 4 5 6 4 5 6 0 1 0 2 4 B C @1 2 3A 0 0 3 6 1 0 3 6 B C @1 2 3A 0 3 6 0 1 0 3 6 B C @1 2 3A 0 0 0 by subtracting 3 times each element of row 2 from row 1 by subtracting 4 times each element of row 2 from row 3 by multiplying each element of row 1 by 3/2 by subtracting corresponding elements of row 1 from row 3 ¼B The row of zeros in matrix B means that its determinant is zero and so its rank is not 3. The largest sub-matrix with non-zero determinant has order 2 and so the rank of B is 2. Because matrix B is row equivalent to matrix A we can say that the rank of A is also 2. Example 2 0 1 2 Determine the rank of A ¼ @ 4 7 9 5 1 8 6A 3 By taking 4 times the elements of row 1 from row 2 we obtain the equivalent matrix . . . . . . . . . . . . 0 1 2 @ 0 1 9 5 1 8 26 A 3 8 By taking 9 times the elements of row 1 from row 3 we obtain the equivalent matrix . . . . . . . . . . . . 0 1 @0 0 1 2 8 1 26 A 13 69 By multiplying the elements of row 2 by 13 we obtain the equivalent matrix ............ 9 524 Programme 16 0 10 1 @0 0 1 2 8 13 338 A 13 69 By adding corresponding elements of row 2 to row 3 we obtain the equivalent matrix . . . . . . . . . . . . 0 11 1 @0 0 2 13 0 1 8 269 A 69 Because all the elements below the main diagonal of this matrix are zero we call the matrix an upper triangular matrix. By inspection we can see that the determinant of this triangular matrix is non-zero, being the product of its three diagonal elements 1 13 ð69Þ ¼ 897. Therefore its rank is 3 and so the rank of matrix A is also 3. Try another one for yourself. Example 3 0 1 2 6 A is . . . . . . . . . . . . 4 3 9 The rank of A ¼ @ 1 5 2 7 12 2 Because 0 3 B A ¼ @1 2 9 2 1 0 0 C B 5 6A @1 7 4 0 2 0 B @1 0 6 7 6 5 0 3 3 0 5 3 3 8 1 C 6A 5 0 B @1 0 0 0 B @1 0 0 1 B @0 16 Subtracting 3 times row 2 from row 1 4 1 16 C 6 A Subtracting 2 times row 2 from row 3 8 1 8 C 6A Multiplying row 1 by 1=2 8 1 C 5 6A 0 0 1 5 6 C 3 8A 0 0 Adding row 1 to row 3 Interchanging rows 1 and 2 525 Matrix algebra 1 and 0 0 5 6 3 8 ¼ 0. So the rank of this matrix is not 3. The largest 0 0 square sub-matrix of this matrix with non-zero determinant is, by inspection, of order 2 and so the rank of this matrix, and hence the rank of the equivalent matrix A is 2. Finally try a non-square matrix. Example 4 0 2 2 The rank of A ¼ @ 0 8 1 7 1 1 4 A is . . . . . . . . . . . . 2 3 2 3 13 3 Because 0 2 B A ¼ @0 1 2 3 1 1 0 0 12 3 8 2 8 2 C B 4A @0 7 3 2 0 3 1 7 3 0 8 2 2 B @0 8 2 1 C 4 A Subtracting 2 times row 3 from row 1 2 1 C 4A 1 7 3 2 0 1 0 0 0 2 B C @0 8 2 4A 1 7 3 2 Multiplying row 1 by 2=3 Adding row 2 to row 1 It is possible to find a 3 3 sub-matrix of this matrix that has non-zero determinant, namely 0 1 0 0 2 0 0 2 @ 8 2 4 A where 8 2 4 ¼ 2ð24 14Þ ¼ 20. 7 3 2 7 3 2 Consequently, this matrix and hence matrix A has rank 3. 526 Programme 16 Consistency of a set of equations 14 In solving sets of simultaneous equations, we can express the equations in matrix form. For example a11 x1 þ a12 x2 þ a13 x3 ¼ b1 a21 x1 þ a22 x2 þ a23 x3 ¼ b2 a31 x1 þ a32 x2 þ a33 x3 ¼ b3 can be written in the form 0 10 1 0 1 x1 b1 a11 a12 a13 B CB C B C a a a x ¼ b @ 21 @ 2A 22 23 A@ 2 A a31 i.e. a32 a33 x3 b3 Ax ¼ b The set of three equations is said to be consistent if solutions for x1 , x2 , x3 exist and inconsistent if no such solutions can be found. In practice, we can solve the equations by operating on the augmented coefficient matrix, i.e. we write the constant terms as a fourth column of the coefficient matrix to form Ab . 0 1 a11 a12 a13 b1 Ab ¼ @ a21 a22 a23 b2 A a31 a32 a33 b3 which, of course, is a ð3 4Þ matrix. The general test for consistency is then: A set of n simultaneous equations in n unknowns is consistent if the rank of the coefficient matrix A is equal to the rank of the augmented matrix Ab . If the rank of A is less than the rank of Ab , then the equations are inconsistent and have no solution. Make a note of this test. It can save time in working 15 Example 1 3 x1 4 If ¼ then 2 6 x2 5 1 3 4 1 3 and Ab ¼ A¼ 2 6 5 2 6 1 3 Rank of A: ¼66¼0 ; rank of A ¼ 1 2 6 1 3 ¼ 0 as before Rank of Ab : 2 6 3 4 ¼ 15 24 ¼ 9 ; rank of Ab ¼ 2 but 6 5 In this case, rank of A < rank of Ab ; ............ 527 Matrix algebra no solution exists 16 Remember that, for consistency, rank of A ¼ . . . . . . . . . . . . rank of Ab 17 Uniqueness of solutions 1 With a set of n equations in n unknowns, the equations are consistent if the coefficient matrix A and the augmented matrix Ab are each of rank n. There is then a unique solution for the n equations. Note that if the rank of A ¼ n then A is a non-singular sub-matrix of Ab and so the rank of Ab ¼ n also. Therefore there is no need to test for the rank of Ab in this case. If the rank of A and that of Ab is m, where m < n, then the matrix A is singular, i.e. jAj ¼ 0, and there will be an infinite number of solutions for the equations. As we have already seen, if the rank of A < the rank of Ab , then no solution exists. Copy these up in your record book; they are important 2 3 Writing the results in a slightly different way: With a set of n equations in n unknowns, checking the rank of the coefficient matrix A and that of the augmented matrix Ab enables us to see whether (a) a unique solution exists rank A ¼ rank Ab ¼ n (b) an infinite number of solutions exist rank A = rank Ab ¼ m < n (c) no solution exists rank A < rank Ab Example 4 5 8 10 x1 x2 ¼ 3 6 Finding the rank of A and of Ab leads us to the conclusion that ............ 18 528 Programme 16 19 there is an infinite number of solutions Because 4 A¼ 8 5 10 Rank of A: Rank of Ab : and Ab ¼ 4 8 4 8 4 5 8 10 3 6 5 ¼ 40 þ 40 ¼ 0 ; Rank of A ¼ 1 10 5 5 3 4 3 ¼ 0; ¼ 0; ¼0 10 10 6 8 6 ; Rank of Ab ¼ 1 ; Rank of A ¼ rank of Ab ¼ 1 But there are two equations in two unknowns, i.e. n ¼ 2 ; Rank of A ¼ rank of Ab ¼ 1 < n ; Infinite number of solutions. 4x1 3 . 5 You will recall that, for a unique solution of n equations in n unknowns The solutions can be written as x1 arbitrary and x2 ¼ ............ 20 rank A ¼ rank Ab ¼ n Now for some examples for you to try. In each of the following cases, apply the rank tests to determine the nature of the solutions. Do not solve the sets of equations. Example 1 0 1 10 1 0 1 2 1 x1 C CB C B 4 2 A@ x2 A ¼ @ 2 A x3 3 1 4 3 0 0 1 1 1 2 1 B B C A ¼ @3 4 2 A and Ab ¼ @ 3 1 1 4 3 1 B @3 1 2 1 1 C 4 2 2 A 4 3 3 Finish it off and we find that . . . . . . . . . . . . 529 Matrix algebra a unique solution exists 21 Because n ¼ 3; rank of A ¼ 3; rank of Ab ¼ 3. ; rank of A ¼ rank of Ab ¼ 3 ¼ n ; Solution unique And this one. Example 0 2 1 @4 2 3 1 2 10 1 0 1 2 7 x1 2 A@ x2 A ¼ @ 5 A x3 1 3 This time we find that . . . . . . . . . . . . no solution is possible Because 0 2 B A ¼ @4 1 2 3 1 n ¼ 3; 1 7 C 2 A; 0 2 B Ab ¼ @ 4 1 7 2 2 3 1 3 3 rank of A ¼ 2; 1 2 C 5A 1 rank of Ab ¼ 3 ; rank of A < rank of Ab ; No solution exists and finally Example 3 0 10 1 0 1 1 1 2 3 x1 @1 3 4 A@ x2 A ¼ @ 2 A x3 3 2 5 1 In this case, we find that . . . . . . . . . . . . 22 530 Programme 16 23 infinite number of solutions possible Because 0 1 2 A ¼ @1 3 2 5 0 1 1 2 3 4 A and Ab ¼ @ 1 3 2 5 1 Rank of A: 0 1 0 1 2 3 1 B C B A ¼ @1 3 4 A @0 2 5 1 2 0 1 B @0 0 0 1 B @0 0 2 1 5 2 1 1 2 1 0 1 3 C 7 A 1 1 3 C 7 A 7 1 3 C 7 A 0 1 3 1 4 2A 1 3 Subtracting row 1 from row 2 Subtracting 2 times row 1 from row 2 Subtracting row 2 from row 3 and so rank of A is 2 by inspection. Rank of Ab : 0 1 0 1 2 3 1 1 B C B Ab ¼ @ 1 3 4 2 A @ 0 2 5 1 3 2 0 1 B @0 0 0 1 B @0 2 1 5 2 1 1 2 1 0 0 3 1 1 C 1A 3 1 3 1 C 7 1A 7 1 1 3 1 C 7 1A 7 1 0 Subtracting row 1 from row 2 Subtracting 2 times row 1 from row 2 Subtracting row 2 from row 3 0 and so rank of Ab is 2 by inspection. Therefore rank of A ¼ rank of Ab ¼ 2 < n (that is 3), therefore there is an infinite number of solutions. Now let us move on to a new section of the work 531 Matrix algebra Solution of sets of equations 24 1 Inverse method Let us work through an example by way of explanation. Example 1 To solve 3x1 þ 2x2 x3 ¼ 4 2x1 x2 þ 2x3 ¼ 10 x1 3x2 4x3 ¼ 5: We first write this in matrix form, which is . . . . . . . . . . . . 0 3 @2 1 0 3 Then if A ¼ @ 2 1 2 1 3 10 1 0 1 2 1 x1 4 1 2 A@ x2 A ¼ @ 10 A x3 3 4 5 1 0 1 0 1 1 x1 4 1 2 A then @ x2 A ¼ A @ 10 A x3 4 5 where A1 is the inverse of A. To find A1 (a) Form the determinant of A and evaluate it. 3 2 1 jAj ¼ 2 1 2 ¼ 3ð4 þ 6Þ 2ð8 2Þ 1ð6 þ 1Þ ¼ 55 1 3 4 (b) Form a new matrix C consisting of the cofactors of the elements in A. The cofactor of any one element is its minor together with its ‘place sign’ 0 1 A11 A12 A13 i.e. C ¼ @ A21 A22 A23 A A31 A32 A33 where A11 is the cofactor of a11 in A. 1 2 2 2 ¼ 10; A12 ¼ ¼ 10; A11 ¼ 3 4 1 4 2 1 ¼ 5 A13 ¼ 1 3 2 1 3 1 ¼ 11; A22 ¼ ¼ 11; A21 ¼ 3 4 1 4 3 2 ¼ 11 A23 ¼ 1 3 A31 ¼ . . . . . . . . . . . . ; A32 ¼ . . . . . . . . . . . . ; A33 ¼ . . . . . . . . . . . . 25 532 26 Programme 16 A31 ¼ 2 1 1 3 ¼ 3; A32 ¼ 2 2 1 3 2 ¼ 8; A33 ¼ ¼ 7 2 2 1 0 1 10 10 5 So C ¼ @ 11 11 11 A 3 8 7 We now write the transpose of C, i.e. CT in which we write rows as columns and columns as rows. CT ¼ . . . . . . . . . . . . 0 27 10 11 CT ¼ @ 10 11 5 11 1 3 8 A 7 This is called the adjoint (adj) of the original matrix A i.e. adj A ¼ CT Then the inverse of A, i.e. A1 is given by 0 1 10 11 3 1 1 @ 1 T C ¼ A ¼ 10 11 8 A jAj 55 5 11 7 As a check that all the calculations have been done correctly and without error, the product of matrix A with its adjoint should be equal to the unit matrix multiplied by the determinant of A. That is A adj A ¼ det A I For this case 0 3 2 1 10 10 B CB A adj A ¼ @ 2 1 2 A@ 10 1 3 4 5 0 1 55 0 0 B C ¼ @ 0 55 0A 0 0 55 11 11 11 3 1 C 8 A 7 ¼ det A I Thus all is well. We can now continue to find the solution. 0 1 0 1 x1 4 B C B C So @ x2 A ¼ A1 @ 10 Abecomes 0 x3 1 0 5 10 11 x1 1 B B C @ 10 11 @ x2 A ¼ 55 5 11 x3 10 1 3 4 CB C 8 A@ 10 A ¼ . . . . . . . . . . . . 7 5 533 Matrix algebra 28 x1 ¼ 3; x2 ¼ 2; x3 ¼ 1 Because 0 10 1 0 1 10 11 3 4 x1 B CB C B C 1 B CB C B x2 C ¼ @ A 55 @ 10 11 8 A@ 10 A 5 11 7 5 x3 0 1 0 1 40 þ110 þ15 3 C B C 1 B B 40 110 40 C ¼ B 2 C ¼ @ A @ A 55 20 þ110 35 1 ; x1 ¼ 3; x2 ¼ 2; x3 ¼ 1 The method is the same every time. To solve Ax ¼ b x ¼ A1 b To find A1 (1) Evaluate jAj If jAj 6¼ 0 then proceed to (2) If jAj ¼ 0 then there is no inverse and hence no unique solution. Later we shall discover how to determine whether there is an infinity of solutions or none. (2) Form C, the matrix of cofactors of A (3) Write CT , the transpose of C 1 CT . jAj Now apply the method to Example 2. (4) Then A1 ¼ Example 2 4x1 þ 5x2 þ x3 ¼ 2 x1 2x2 3x3 ¼ 7 3x1 x2 2x3 ¼ 1. x1 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . ; x3 ¼ . . . . . . . . . . . . 534 Programme 16 29 x1 ¼ 2; x2 ¼ 3; x3 ¼ 5 Here is the complete working. 0 1 4 5 1 4 5 1 B C A ¼ @ 1 2 3 A; jAj ¼ 1 2 3 ¼ 26 3 1 2 3 1 2 0 1 A11 A12 A13 B C C ¼ @ A21 A22 A23 A A31 A32 A33 2 3 A11 ¼ ¼1 1 2 5 1 ¼9 A21 ¼ 1 2 5 1 ¼ 13 A31 ¼ 2 3 0 1 7 B ; C¼@ 9 11 13 3 ¼ 7 2 4 1 A22 ¼ ¼ 11 3 2 4 1 A32 ¼ ¼ 13 1 3 1 0 5 1 C B T 19 A ; C ¼ @ 7 A12 ¼ 1 3 2 ¼5 1 4 5 A23 ¼ ¼ 19 3 1 4 5 A33 ¼ ¼ 13 1 2 1 9 13 C 11 13 A A13 ¼ 1 3 13 5 19 13 0 1 1 9 13 1 1 B C 1 T C ¼ @ 7 11 13 A A ¼ jAj 26 5 19 13 0 10 1 0 1 0 1 1 9 13 2 x1 2 B CB C B C B C C B B x2 C ¼ A1 B 7 C ¼ 1 B 7 11 13 A@ 7 C A @ A @ A 26 @ 5 19 13 1 x3 1 0 1 2 þ63 13 C 1 B ¼ B 14 77 þ13 C A 26 @ 10 þ133 13 0 1 0 1 52 2 C B C 1 B B C 78 C ¼ B A ¼ @ 3 A 26 @ 130 5 ; x1 ¼ 2; 13 x2 ¼ 3; x3 ¼ 5 With a set of four equations with four unknowns, the method becomes somewhat tedious as there are then sixteen cofactors to be evaluated and each one is a third-order determinant! There are, however, other methods that can be applied – so let us see method 2. 535 Matrix algebra 30 2 Row transformation method Elementary row transformations that can be applied are as follows (a) Interchange any two rows. (b) Multiply (or divide) every element in a row by a non-zero scalar (constant) k. (c) Add to (or subtract from) all the elements of any row k times the corresponding elements of any other row. Equivalent matrices Two matrices, A and B, are said to be equivalent if B can be obtained from A by a sequence of elementary transformations. Solutions of equations The method is best described by working through a typical example. Example 1 Solve 2x1 þ x2 þ x3 ¼ x1 þ 3x2 þ 2x3 ¼ 5 1 3x1 2x2 4x3 ¼ 4: 1 0 10 1 0 5 2 1 1 x1 This can be written @ 1 3 2 A@ x2 A ¼ @ 1 A x3 4 3 2 4 and for convenience we introduce the 0 10 1 0 1 2 1 1 x1 @1 3 2 A@ x2 A ¼ @ 0 x3 0 3 2 4 0 1 1 0 0 where @ 0 1 0 Amay be regarded as 0 0 1 unit matrix 10 1 0 0 5 1 0 A@ 1 A 0 1 4 1 5 the coefficient of @ 1 A 4 We then form the combined coefficient 0 2 1 1 @1 3 2 3 2 4 0 matrix 1 0 0 1 0 0 1 0A 0 1 and work on this matrix from now on. On then to the next frame 536 31 Programme 16 The rest of the working is mainly concerned with applying row transformations to convert the left-hand half of the matrix to a unit matrix and the righthand side to the inverse, eventually obtaining 0 1 1 0 0 a b c @0 1 0 d e f A 0 0 1 g h i with a, b, c, . . . g, h, i being evaluated in the process. The following notation will be helpful to denote the transformation used: ð1Þ ð2Þ denotes ‘interchange rows 1 and 2’ ð3Þ 2ð1Þ denotes ‘subtract twice row 1 from row 3’, etc. So off we go. ð1Þ ð2Þ ð2Þ 2ð1Þ 0 3 2 0 1 0 B @2 1 1 1 C 0 0A 0 3 2 4 0 1 3 2 B ð3Þ 3ð1Þ @ 0 5 3 0 11 10 ð3Þ 2ð2Þ 0 ð2Þ ð3Þ 0 ð3Þ 5ð2Þ 1 1 1 3 2 B 0 5 3 @ 0 1 0 1 C 1 2 0 A 0 3 1 1 0 1 0 C 1 2 0 A 0 1 4 2 1 3 B @0 1 1 0 1 1 0 1 1 2 0 4 C 2 1 1 A 0 5 3 1 1 3 2 0 B @0 1 4 C 2 1 1 A 0 2 0 1 0 1 0 0 17 11 1 0 10 6 4 3 1 B ð3Þ ð17Þ @ 0 1 4 2 1 1 C A 7=17 5=17 1 1=17 C 3=17 A ð1Þ 3ð2Þ ð1Þ þ 10ð3Þ 0 0 0 0 1 0 B ð2Þ 4ð3Þ @ 0 1 0 0 1 11=17 7 5 8=17 2=17 0 10=17 11=17 1 7=17 0 11=17 5=17 We now have 0 10 1 0 10 1 8 2 1 5 1 0 0 x1 @ 0 1 0 A@ x2 A ¼ 1 @ 10 11 3 A@ 1 A 17 x3 11 7 5 4 0 0 1 ; x1 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . ; x3 ¼ . . . . . . . . . . . . 537 Matrix algebra x1 ¼ 2; 0 1 0 x1 40 1 B C B @ x2 A ¼ @ 50 17 x3 55 x1 ¼ 2; x2 ¼ 3; x2 ¼ 3; 2 þ11 7 x3 ¼ 4 32 1 0 1 0 1 4 34 2 1 C B C B C 12 A ¼ @ 51 A ¼ @ 3 A 17 þ20 68 4 x3 ¼ 4 Of course, there is no set pattern of how to carry out the row transformations. It depends on one’s ingenuity and every case is different. Here is a further example. Example 2 2x1 x2 3x3 ¼ 1 x1 þ 2x2 þ x3 ¼ 3 2x1 2x2 5x3 ¼ 2. First write the set of equations in matrix form – with the unit matrix included. This gives . . . . . . . . . . . . 0 2 @1 2 10 1 0 1 3 x1 1 0 2 1 A@ x2 A ¼ @ 0 1 x3 2 5 0 0 10 1 0 1 0 A@ 3 A 1 2 33 The combined coefficient matrix is now . . . . . . . . . . . . 0 2 1 3 @1 2 1 2 2 5 1 0 0 1 0 0 1 0 0A 1 If we start off by interchanging the top two rows, we obtain a 1 at the beginning of the top row which is a help. 0 1 ð1Þ ð2Þ 1 2 1 0 1 0 B C @ 2 1 3 1 0 0 A 2 2 5 0 0 1 Now, if we subtract 2 row 1 from row 2 and 2 row 1 from row 3, we get ............ 34 538 Programme 16 0 35 1 @0 0 2 1 0 5 5 1 6 7 0 1 2 2 1 0 0A 1 Continuing with the same line of reasoning, we then have 0 1 ð2Þ ð3Þ 1 2 1 0 1 0 B C 1 2 1 0 1 A @0 0 6 7 0 2 1 0 1 ð3Þ þ 6ð2Þ 1 2 1 0 1 0 B C 0 1 A @0 1 2 1 ð1Þ 2ð2Þ ð3Þ 5 0 1 B0 @ 0 0 0 5 6 2 5 1 0 3 2 1 2 1 2 1 0 1 C A 6 2 0 1 5 5 1 Notice the three diagonal 1s appearing at the left-hand end What do you suggest we should do now? ............ 36 Add three times row 3 to row 1 and subtract twice row 3 from row 2 Right. That gives 0 ð1Þ þ 3ð3Þ 1 0 B ð2Þ 3ð3Þ @ 0 1 0 1 ; @0 0 0 0 8 5 75 6 5 15 4 5 25 1 1 1C A 1 0 0 1 0 10 1 8 1 0 0 x1 1 4 1 0 A@ x2 A ¼ @ 7 5 6 2 0 1 x3 10 1 5 1 5 A@ 3 A 5 2 Now you can finish it off. x1 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . ; 37 x3 ¼ . . . . . . . . . . . . x1 ¼ 1; x2 ¼ 3; x3 ¼ 2 Because 0 1 0 1 0 1 0 1 x1 8 3 10 5 1 1 1 @ x2 A ¼ @ 7 þ 12 þ 10 A ¼ @ 15 A ¼ @ 3 A 5 5 x3 6 6 10 10 2 Let us now look at a somewhat similar method with rather fewer steps involved. So move on 539 Matrix algebra 38 3 Gaussian elimination method Once again we will demonstrate the method by a typical example. Example 1 2x1 3x2 þ 2x3 ¼ 9 3x1 þ 2x2 x3 ¼ 4 x1 4x2 þ 2x3 ¼ 6. We start off as usual 0 10 1 0 1 9 2 3 2 x1 @3 2 1 A@ x2 A ¼ @ 4 A x3 6 1 4 2 We then form the augmented coefficient matrix by including the constants as an extra column on the right-hand side of the matrix 0 1 9 2 3 2 @3 4A 2 1 1 4 2 6 Now we operate on the rows to convert the first three triangular matrix 0 1 0 ð1Þ ð3Þ 1 4 2 6 ð2Þ ð3Þ 1 B C B 2 1 4 A @3 @2 2 3 2 9 3 0 0 1 ð2Þ 2ð1Þ 1 4 2 6 ð2Þ 5 1 B B C ð3Þ 3ð1Þ @ 0 5 2 3 A ð3Þ 7 @ 0 0 0 1 ð3Þ 2ð2Þ B0 @ 0 14 7 14 1 4 2 6 1 25 35 C A 0 15 45 ð3Þ ð5Þ columns into an upper 4 3 2 4 1 0 2 1 B @0 0 4 0 1 0 2 6 1 C 2 9A 1 4 1 2 6 C 25 35 A 1 2 1 2 6 C 25 35 A 1 4 The first three columns now form an upper triangular matrix which has been our purpose. If we now detach the fourth column back to its original position on the right-hand side of the matrix equation, we have ............ 540 Programme 16 0 39 1 4 2 0 1 B @0 0 10 x1 1 0 6 1 CB C B C 1 25 A@ x2 A ¼ @ 35 A x3 4 Expanding from the bottom row, working upwards x3 ¼ 4 ; x3 ¼ 4 x2 25 x3 ¼ 35 x1 4x2 þ 2x3 ¼ ; x1 ¼ 2; ; x2 ¼ 35 þ 85 ¼ 1 ; x2 ¼ 1 ; x1 4 þ 8 ¼ 6 6 x2 ¼ 1; ; x1 ¼ 2 x3 ¼ 4 It is a very useful method and entails fewer tedious steps, and can be used to solve efficiently higher-order sets of equations and non-square systems. It can also solve a sequence of problems with the same coefficient matrix A by using the augmented matrix ðAb1 b2 . . . bn Þ. Example 2 x1 þ 3x2 2x3 þ x4 ¼ 1 2x1 2x2 þ x3 2x4 ¼ 1 x1 þ x2 3x3 þ x4 ¼ 6 3x1 x2 þ 2x3 x4 ¼ 3. First we write this in matrix form and compile the augmented matrix which is ............ 40 0 1 3 2 1 B 2 2 1 2 B @1 1 3 1 3 1 2 1 1 1 1C C 6A 3 Next we operate on rows to convert the left-hand side to an upper triangular matrix. There is no set way of doing this. Use any trickery to save yourself unnecessary work. So now you can go ahead and complete the transformations and obtain x1 ¼ . . . . . . . . . . . . ; x3 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . x4 ¼ . . . . . . . . . . . . 541 Matrix algebra x1 ¼ 2; x2 ¼ 3; x3 ¼ 1; 41 x4 ¼ 4 Here is one way. You may well have taken quite a different route. 0 1 1 1 3 2 1 B 2 2 1 2 1C B C @1 6A 1 3 1 3 1 2 1 3 ð2Þ 2ð1Þ ð3Þ ð1Þ ð4Þ ½ð1Þ þ ð2Þ ð2Þ 4ð4Þ ð3Þ ð4Þ ð2Þ ð4Þ ð3Þ 4 ð4Þ 7ð3Þ 0 1 B0 B @0 0 0 1 B0 B @0 0 0 1 B0 B @0 0 0 1 B0 B @0 0 3 2 1 8 5 4 2 1 0 2 3 0 1 1 3C C 7A 3 3 2 1 0 7 4 0 4 0 2 3 0 1 1 9 C C 4A 3 3 2 1 2 3 0 0 1 0 0 7 4 1 1 3C C 1A 9 3 2 1 2 3 0 0 1 0 0 0 4 1 1 3C C 1A 16 Returning the right-hand column to its original position 1 0 10 1 0 1 1 3 2 1 x1 B C B B 0 2 3C 3 0C C B CB x2 C ¼ B @0 1A 0 1 0 A@ x3 A @ x4 16 0 0 0 4 Expanding from the bottom row, we have 4x4 ¼ 16 ; x4 ¼ 4 x3 ¼ 1 2x2 þ 3x3 ¼ 3 ; x3 ¼ 1 ; 2x2 ¼ 6 x1 þ 3x2 2x3 þ x4 ¼ 1 ; x2 ¼ 3 ; x1 9 þ 2 þ 4 ¼ 1 ; x1 ¼ 2; x2 ¼ 3; x3 ¼ 1; ; x1 ¼ 2 x4 ¼ 4 542 42 Programme 16 We still have a further method for solving sets of simultaneous equations. 4 Triangular decomposition method A square matrix A can usually be written as a product of a lower-triangular matrix L and an upper-triangular matrix U, where A ¼ LU. 0 1 1 2 3 B C For example, if A ¼ @ 3 5 8 A, A can be expressed as 4 9 10 0 10 1 0 u11 u12 u13 l11 0 B CB C A ¼ LU ¼ @ l21 l22 0 A@ 0 u22 u23 A l31 l32 l33 0 0 u33 ðLÞ ðUÞ 0 1 l11 u13 l11 u11 l11 u12 B C ¼ @ l21 u11 l21 u12 þ l22 u22 l21 u13 þ l22 u23 A l31 u11 l31 u12 þ l32 u22 l31 u13 þ l32 u23 þ l33 u33 Note that, in L and U, elements occur in the major diagonal in each case. These are related in the product and whatever values we choose to put for u11 , u22 , u33 . . . then the corresponding values of l11 , l22 , l33 . . . will be determined – and vice versa. For convenience, we put u11 ¼ u22 ¼ u33 . . . ¼ 1 0 1 l11 l11 u12 l11 u13 B C Then A ¼ LU ¼ @ l21 l21 u12 þ l22 l21 u13 þ l22 u23 A l31 l31 u12 þ l32 l31 u13 þ l32 u23 þ l33 0 1 1 2 3 B C In our example, A ¼ @ 3 5 8A 4 9 10 l11 u12 ¼ 2 ; u12 ¼ 2; l11 u13 ¼ 3 ; u13 ¼ 3 ; l11 ¼ 1; l21 ¼ 3; Similarly l22 ¼ . . . . . . . . . . . . ; u23 ¼ . . . . . . . . . . . . l31 ¼ 4; l32 ¼ . . . . . . . . . . . . ; l33 ¼ . . . . . . . . . . . . 43 l22 ¼ 1; u23 ¼ 1; l32 ¼ 1; l33 ¼ 3 Because l21 u12 þ l22 u22 ¼ 5 that is 3 2 þ l22 1 ¼ 5 and so l22 ¼ 1 l21 u13 þ l22 u23 ¼ 8 that is 3 3 þ ð1Þ u23 ¼ 8 and so u23 ¼ 1 l31 u12 þ l32 u22 ¼ 9 that is 4 2 þ l32 1 ¼ 9 and so l32 ¼ 1 l31 u13 þ l32 u23 þ l33 u33 ¼ 10 that is 4 3 þ 1 1 þ l33 1 ¼ 10 and so l33 ¼ 3 Now we substitute all these values back into the upper and lower triangular matrices and obtain A ¼ LU ¼ . . . . . . . . . . . . 543 Matrix algebra 0 1 0 A ¼ LU ¼ @ 3 1 4 1 10 0 1 2 0 A@ 0 1 3 0 0 1 3 1A 1 44 We have thus expressed the given matrix A as the product of lower and upper triangular matrices. Let us now see how we use them. Example 1 x1 þ 2x2 þ 3x3 ¼ 16 3x1 þ 5x2 þ 8x3 ¼ 43 4x1 þ 9x2 þ 10x3 ¼ 57. 0 10 1 0 1 16 1 2 3 x1 i.e.@ 3 5 8 A@ x2 A ¼ @ 43 A x3 57 4 9 10 i.e. Ax ¼ b. We have seen above that A can be 0 10 1 0 0 1 A ¼ LU ¼ @ 3 1 0 A@ 0 4 1 3 0 written as LU where 1 2 3 1 1A 0 1 To solve Ax ¼ b, we have LUx ¼ b i.e. LðUxÞ ¼ b Putting Ux ¼ y, we solve Ly ¼ b to obtain y and then Ux ¼ y to obtain x. 0 10 1 0 1 1 0 0 y1 16 B CB C B C B C B C (a) Solving Ly ¼ b B 0C @ 3 1 A@ y2 A ¼ @ 43 A 4 1 3 y3 57 Expanding from the top y1 ¼ 16; 3y1 y2 ¼ 43 4y1 þ y2 3y3 ¼ 57 ; 64 þ 5 3y3 ¼ 57 ; y3 ¼ 4 0 1 0 1 16 y1 ; @ y2 A ¼ @ 5 A y3 4 0 (b) Solving Ux ¼ y 1 2 @0 1 0 0 ; y2 ¼ 5; and 10 1 0 1 3 x1 16 1 A@ x2 A ¼ @ 5 A 1 x3 4 Expanding from the bottom, we then have x1 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . ; x3 ¼ . . . . . . . . . . . . 544 Programme 16 45 x1 ¼ 2; x2 ¼ 1; x3 ¼ 4 Note: 1 If lii ¼ 0, then either decomposition is not possible, or, if A is singular, i.e. jAj ¼ 0, there is an infinite number of possible decompositions. 2 Instead of putting u11 ¼ u22 ¼ u33 . . . ¼ 1, we could have used the alternative substitution l11 ¼ l22 ¼ l33 . . . ¼ 1 and obtained values of u11 , u22 , u33 . . . etc. The working is as before. 3 One advantage of employing LU decomposition over Gaussian elimination is in the solution of a sequence of problems in which the same coefficient matrix occurs. Now for another example. 46 Example 2 x1 þ 3x2 þ 2x3 ¼ 19 2x1 þ x2 þ x3 ¼ 13 4x1 þ 2x2 þ 3x3 ¼ 31. 0 10 1 0 1 19 1 3 2 x1 B CB C B C ; @ 2 1 1 A@ x2 A ¼ @ 13 A i.e. Ax ¼ b x3 31 4 2 3 1 0 10 1 u12 u13 0 l11 0 B CB C CB C A ¼ LU ¼ B @ l21 l22 0 A@ 0 1 u23 A 0 0 1 l31 l32 l33 0 l11 u13 l11 l11 u12 B ¼B @ l21 l21 u12 þ l22 l21 u13 þ l22 u23 l31 u12 þ l32 1 1 3 2 B C C ¼B @2 1 1A 4 2 3 0 l31 l31 u13 þ l32 u23 þ l33 1 C C A 545 Matrix algebra Now we have to find the values of the various elements. The usual order of doing this is shown by the diagram. 1 0 C C C C C C C C C C C C A B B B B B B B B B B B B @ That is, first we can write down values for l11 , l21 , l31 from the left-hand column; then follow this by finding u12 ; u13 from the top row; and proceed for the others. So, completing the two triangular matrices, we have A ¼ LU ¼ . . . . . . . . . . . . 0 1 A ¼ LU ¼ @ 2 4 As we stated before: then 0 5 10 10 0 1 0 A@ 0 0 1 1 3 2 1 35 A 0 1 47 Ax ¼ b; LðUxÞ ¼ b. Put Ux ¼ y (a) solve Ly ¼ b to obtain y (b) solve Ux ¼ y to obtain x. 0 1 0 1 y1 ...: B C B C Solving Ly ¼ b gives @ y2 A ¼ @ . . . : A and y3 ...: 0 1 0 1 y1 19 @ y2 A ¼ @ 5 A y3 5 48 Because 0 10 1 0 1 19 1 0 0 y1 @2 5 0 A@ y2 A ¼ @ 13 A y3 31 4 10 1 Expanding from the top gives y1 ¼ 19; y2 ¼ 5; y3 ¼ 5. (b) Now solve Ux ¼ y from which x1 ¼ . . . . . . . . . . . .; x3 ¼ . . . . . . . . . . . . x2 ¼ . . . . . . . . . . . .; 546 Programme 16 49 x1 ¼ 3; x2 ¼ 2; x3 ¼ 5 Because we have 0 i.e. 1 B B0 @ 0 3 2 10 Ux ¼ y 1 0 x1 19 1 CB C B C CB x2 C ¼ B 5 C A@ A @ A 0 1 x3 5 1 3 5 3 Expanding from the bottom x3 ¼ 5; x2 þ x3 ¼ 5 5 and x1 þ 3x2 þ 2x3 ¼ 19 ; x1 þ 6 þ 10 ¼ 19 ; x1 ¼ 3; x2 ¼ 2; ; x2 ¼ 2 ; x1 ¼ 3 x3 ¼ 5 We can of course apply the same method to a set of four equations. Example 3 x1 þ 2x2 x3 þ 3x4 ¼ 9 2x1 x2 þ 3x3 þ 2x4 ¼ 23 3x1 þ 3x2 þ x3 þ x4 ¼ 5 4x1 þ 5x2 2x3 þ 2x4 ¼ 2. 0 10 1 0 1 1 2 1 3 9 x1 B C B B 2 1 C 3 2C B CB x2 C B 23 C i.e. B CB C ¼ B C @3 3 1 1 A@ x3 A @ 5 A i.e. Ax ¼ b x4 4 5 2 2 2 10 1 0 1 1 u12 u13 u14 1 2 1 3 l11 0 0 0 Bl CB C B 3 2C B 21 l22 0 0 CB 0 1 u23 u24 C B 2 1 C A¼LU¼ B CB C¼B C @ l31 l32 l33 0 A@ 0 0 1 u34 A @ 3 3 1 1A 0 0 l41 l42 l43 l44 l11 u13 l11 l11 u12 0 0 0 1 l11 u14 4 5 2 2 Bl l21 u14 þ l22 u24 B 21 l21 u12 þ l22 l21 u13 þ l22 u23 A¼ B @ l31 l31 u12 þ l32 l31 u13 þ l32 u23 þ l33 l31 u14 þ l32 u24 þ l33 u34 1 C C C A l41 l41 u12 þ l42 l41 u13 þ l42 u23 þ l43 l41 u14 þ l42 u24 þ l43 u34 þ l44 Now we have to find the values of the individual elements. It is easy enough if we follow the order indicated in the diagram earlier. So the two triangular matrices are A ¼ LU ¼ ð . . . . . . . . . . . . Þð . . . . . . . . . . . . Þ 547 Matrix algebra 0 1 B2 B A ¼ LU ¼ B @3 4 0 10 1 0 0 5 0 3 1 B 0C CB 0 CB 0 A@ 0 3 1 66 5 0 As usual Ax ¼ b; LðUxÞ ¼ b. Put Ux ¼ y 2 1 3 1 1 1 0 1 4C 5C C A 28 5 0 0 1 50 ; Ly ¼ b (a) Solving Ly ¼ b 0 1 0 0 B 2 5 0 B B @ 3 3 1 4 3 10 1 0 1 0 y1 9 B C B C 0C CB y2 C B 23 C CB C ¼ B C 0 A@ y3 A @ 5 A 1 66 y4 2 5 0 1 0 1 ... y1 By C B...C B 2C B C ; B C¼B C @ y3 A @ . . . A y4 ... 0 1 0 1 y1 9 B y2 C B 1 C B C¼B C @ y3 A @ 25 A y4 5 (b) Solving Ux ¼ y 0 10 1 0 1 1 2 1 3 9 x1 4 CB C B 1 C B 0 1 1 B B C 5 CB x2 C B C¼B B C C @ A @0 0 A @ A x 25 1 28 3 5 x4 5 0 0 0 1 which finally gives x1 ¼ . . . . . . . . . . . . ; x2 ¼ . . . . . . . . . . . . x3 ¼ . . . . . . . . . . . . ; x4 ¼ . . . . . . . . . . . . 51 548 Programme 16 52 x1 ¼ 1; x2 ¼ 2; x3 ¼ 3; x4 ¼ 5 5 Using an electronic spreadsheet The four methods just considered for multiplying and inverting matrices clearly demonstrate the effects of certain properties of matrices and their algebraic manipulation. For this reason alone these methods are invaluable for providing a facility and familiarity with matrix algebra. However, by far the most efficient way of proceeding to multiply and invert small matrices with a numerical content is to use an electronic spreadsheet. This not only provides a faster method but also provides a method that is not prone human arithmetic error! The spreadsheet used to demonstrate the method is the Excel 2010 spreadsheet provided by Microsoft. There are two functions in particular that we shall use, namely: MINVERSE(array) MMULT(array1, array2) for obtaining the inverse of a matrix for multiplying two matrices together Their use is quite straightforward. We start with an example we have done before in Frame 25 to solve the matrix equation: 0 10 1 0 1 4 3 2 1 x1 B CB C B C 2 A@ x2 A ¼ @ 10 A @ 2 1 1 3 4 x3 5 The solution is given as: 0 1 0 11 0 1 3 2 1 4 x1 B C B C B C 2 A @ 10 A @ x2 A ¼ @ 2 1 x3 1 3 4 5 0 0 11 3 3 2 1 B B C where @ 2 1 2 A is the inverse of matrix @ 2 1 1 3 4 2 1 3 1 1 C 2 A. 4 Open up a new blank worksheet and enter the elements of the matrix 0 1 3 2 1 B C 2 A into cells A1 to C3. @ 2 1 1 3 4 Now highlight the empty cells A5 to C7 this is an empty 3 3 array ready to take the inverse matrix. With these cells highlighted type in: =MINVERSE(A1:C3) and wait! You may be tempted just to press Enter but don’t; you must press Ctrl-ShiftEnter (hold down the Ctrl and Shift keys together and then press Enter). And there, in the allotted array appear the numbers . . . . . . . . . . . . Next frame 549 Matrix algebra 0:181818182 0:181818182 0:090909091 0:2 0:2 0:2 53 0:054545455 0:145454545 0:127272727 This is the inverse matrix. We now need the 3 1 matrix so in cells E1 to E3 enter the numbers: 4 10 5 Now place the cursor in cell A9 and highlight the three cells A9 to A11. With these three cells highlighted type in: =MMULT(A5:C7,E1:E3) and press Ctrl-Shift-Enter. The result is . . . . . . . . . . . . Next frame 54 3 2 1 Because: 0 1 0 x1 3 B C B @ x2 A ¼ @ 2 x3 0 2 1 3 1 11 0 1 1 4 C B C 2 A @ 10 A 4 0:1818 . . . 0:2 B ¼ @ 0:1818 . . . 0:2 0:0909 . . . 0:2 0 1 3 B C ¼ @ 2 A 5 10 1 0:0545 . . . 4 CB C 0:1454 . . . A@ 10 A 0:1272 . . . 5 1 Try one yourself. The solution of the set of equations: 2x1 x2 3x3 ¼ 1 x1 þ 2x2 þ x3 ¼ 3 2x1 2x2 5x3 ¼ 2 is x1 ¼ . . . . . . . . . . . . , x2 ¼ . . . . . . . . . . . . , x3 ¼ . . . . . . . . . . . . The answer is in the next frame 550 Programme 16 55 x1 ¼ 1, x2 ¼ 3, x3 ¼ 2 Because: 2x1 x2 3x3 ¼1 x1 þ 2x2 þ x3 2x1 2x2 5x3 ¼3 ¼2 can be written in matrix form as 0 10 1 0 1 2 1 3 x1 1 B CB C B C 1 2 1 x ¼ @ A@ 2 A @ 3 A 2 2 5 x3 2 Entering the 3 3 array on the left in cells A1 to C3 and the 3 1 array on the right in cells E1 to E3 we then highlight cells A5 to C7 and type in the instruction: =MINVERSE(A1:C3) and press Ctrl-Shift-Enter to reveal the display . . . . . . . . . . . . Next frame 56 1:6 0:2 1 1:4 0:8 1 1:2 0:4 1 This is the inverse matrix. Next, we multiply this array by the array in cells E1 to E3 so highlight the cells A9 to A11 and type in the formula =MMULT(A5:C7,E1:E3) Press Ctrl-Shift-Enter to reveal the result 1 3 giving the solution to the three equations as x1 ¼ 1, x2 ¼ 3, x3 ¼ 2 2 Now try this one and see how much easier the whole process is as the number of equations increases. The solution to the set of equations: 2x 3y þ z þ 4w ¼ 13 x þ 2y 3z þ w ¼ 25 3x y þ 4z 2w ¼ 34 xþyþzþw ¼6 is x ¼ ............, y ¼ ............, z ¼ ............, w ¼ ............ The answer is in the next frame 551 Matrix algebra 57 x ¼ 1, y ¼ 3, y ¼ 3, z ¼ 4, w ¼ 6 Because: 2x 3y þ z þ 4w ¼ 13 x þ 2y 3z þ w ¼ 25 3x y þ 4z 2w ¼ 34 xþyþzþw ¼6 can be written in matrix form as 1 0 10 1 0 x 13 2 3 1 4 B C B C B 1 2 3 1C B CB y C B 25 C C B CB C ¼ B @ 3 1 4 2 A@ z A @ 34 A w 6 1 1 1 1 Entering the 4 4 array on the left in cells A1 to D4 and the 4 1 array on the right in cells F1 to F4 we then highlight cells A6 to D9 and type in the instruction: =MINVERSE(A1:D4) and press Ctrl-Shift-Enter to reveal the display . . . . . . . . . . . . Next frame 0196078431 0:078431373 0:019607843 0:294117647 0:764705882 0:294117647 0:176470588 0:647058824 0:607843137 0:156862745 0:039215686 0:411764706 0:333333333 0:333333333 0:333333333 0 This is the inverse matrix. Next, we multiply this array by the array in cells F1 to F4 so highlight the cells A11 to A14 and type in the formula =MMULT(A6:D9,F1:F4) Press Ctrl-Shift-Enter to reveal the result 1 3 4 6 giving the solution to the three equations as x ¼ 1, y ¼ 3, z ¼ 1, w ¼ 6 58 552 Programme 16 We can even combine the two processes of taking the inverse and performing the multiplication into one formula. For example, to solve the matrix equation: 1 0 10 1 0 5 2 1 1 x1 @1 3 2 A@ x2 A ¼ @ 1 A x3 4 3 2 4 Enter the 3 3 array on the left in cells A1 to C3 and the 3 1 array on the right in cells E1 to E3. We then highlight cells A5 to A7 and type in the instruction: =MMULT(MINVERSE(A1:C3),E1:E3) and press Ctrl-Shift-Enter to reveal the display . . . . . . . . . . . . Next frame 59 2 3 4 giving the solution x1 ¼ 2, x2 ¼ 3, x3 ¼ 4. Comparison of methods Inverse method This is an elementary method but it is very inefficient when the number of equations to solve increases beyond three. Row transformation method An efficient method but each case is different and relies on ingenuity to see the way forward. Gaussian elimination method The most efficient method and should be used in most cases. It must be used when there is a singular or non-square system. Triangular decomposition method An alternative to Gaussian elimination in some cases and by far the most efficient method of all for very large matrices. Spreadsheet method Whilst a spreadsheet cannot be used for matrices with algebraic content, for numerical content it provides an efficient method for small matrices. Now let us proceed to something rather different, so move on to the next frame for a new start 553 Matrix algebra Matrix transformation y x–y plane v u–v plane Q (u, v) P (x, y) y O v x x O u u If for every point Q ðu, vÞ in the u–v plane there is a corresponding point P ðx, yÞ in the x–y plane, then there is a relationship between the two sets of coordinates. In the simple case of scaling the coordinate where u ¼ ax and v ¼ by we have a linear transformation and we can combine these in matrix form u a 0 x ¼ v 0 b y a 0 The matrix then provides the transformation between the vector 0 b x u in one set of coordinates and the vector in the other set of y v coordinates. Similarly, if we solve the two equations for x and y, we have 1 1 x ¼ u and y ¼ v b a x 1=a 0 u ; ¼ y 0 1=b v which allows us to transform back from the u–v plane coordinates to the x–y plane coordinates. Now for an example. 60 554 Programme 16 Example x 2 2 0 If X ¼ ¼ with the transformation T ¼ determine y 1 2 1 u U¼ ¼ TX and show the positions on the x–y and u–v planes. v In this case u 2 ¼ v 2 y 0 1 2 4 ¼ 1 5 u–v plane x–y plane v transforms into O x O u If T is non-singular and U ¼ TX then X ¼ T1 U and since 2 0 T¼ then T1 ¼ . . . . . . . . . . . . 2 1 61 T1 ¼ 1=2 0 1 1 There are several ways of finding the inverse of a matrix. One method is as follows. 2 0 T¼ 2 1 ! ! 2 0 j 1 0 2 0 j 1 0 2 1 j 0 1 0 1 j 1 1 ! 1 0 j 1=2 0 0 1 j 1 1 1=2 0 ; T1 ¼ 1 1 So we have U ¼ TX ; X ¼ T1 U x 1=2 0 u ; ¼ y 1 1 v 1 x Hence a vector in the u–v plane transforms into in the x–y 4 y x plane where ¼ ............ y 555 Matrix algebra x 1=2 ¼ y 5 x 1=2 ¼ y 1 0 1 62 1 1=2 ¼ 4 5 y v transforms into O u O x Rotation of axes A more interesting case occurs with a degree of rotation between the two sets of coordinate axes. v (x, y) (u, v) y y u O Let P be the point ðx; yÞ in the x–y plane and the point ðu; vÞ in the u–v plane. v u x x Let be the angle of rotation between the two systems. From the diagram we can see that ) x ¼ u cos v sin ð1Þ y ¼ u sin þ v cos In matrix form, this becomes x cos sin u ¼ y sin cos v which enables us to transform from the u–v plane coordinates to the corresponding x–y plane coordinates. Make a note of this and then move on 556 63 Programme 16 If we solve equations (1) for u and v, we have x sin ¼ u sin cos v sin2 y cos ¼ u sin cos þ v cos2 ; y cos x sin ¼ vðcos2 þ sin2 Þ ¼ v Also x cos ¼ u cos2 v sin cos y sin ¼ u sin2 þ v sin cos ; x cos þ y sin ¼ uðcos2 þ sin2 Þ ¼ u So u ¼ x cos þ y sin v ¼ x sin þ y cos and written in matrix form, this is . . . . . . . . . . . . u cos sin x ¼ v sin cos y 64 So we have and cos sin u x ¼ sin cos v y u cos sin x ¼ v sin cos y i.e. X ¼ TU and U ¼ T1 X where T is the matrix of transformation and the equations provide a linear transformation between the two sets of coordinates. Example If the u–v plane axes rotate through 308 in an anticlockwise manner from the x–y plane axes, determine the ðu, vÞ coordinates of a point whose ðx, yÞ coordinates are x ¼ 2, y ¼ 3 in the x–y plane. This is a straightforward application of the results above. u So ¼ ............ v 557 Matrix algebra u ¼ v Because u v ¼ ¼ ¼ cos sin ! pffiffiffi 3:23 3 þ 3=2 pffiffiffi ¼ 1:60 1 þ 3 3=2 2 cos ¼ 65 pffiffiffi 3=2 sin cos 3 sin ¼ 1=2 ! pffiffiffi 2 3=2 1=2 pffiffiffi 3 1=2 3=2 ! pffiffiffi 3:23 3 þ 3=2 pffiffiffi ¼ 1:60 1 þ 3 3=2 As usual, the Programme ends with the Revision summary, to be read in conjunction with the Can you? checklist. Go back to the relevant part of the Programme for any points on which you are unsure. The Test exercise should then be straightforward and the Further problems give valuable additional practice. Revision summary 16 1 Singular square matrix: jAj ¼ 0 Non-singular square matrix jAj ¼ 6 0. 2 Rank of a matrix – order of the largest non-zero determinant that can be formed from the elements of the matrix. 3 Elementary operations and equivalent matrices Each of the following row operations on matrix A produces a row equivalent matrix B where the order and rank of B are the same as those of A. We write A B. (1) Interchanging two rows (2) Multiplying each element of a row by the same non-zero scalar quantity (3) Adding or subtracting corresponding elements from those of another row. These operations are called elementary row operations. There is a corresponding set of three elementary column operations that can be used to form column equivalent matrices. 66 558 Programme 16 4 Consistency of a set of n equations in n unknowns with coefficient matrix A and augmented matrix Ab . (a) Consistent if rank of A ¼ rank of Ab (b) Inconsistent if rank of A < rank of Ab . 5 6 Uniqueness of solutions – n equations with n unknowns. (a) rank of A ¼ rank of Ab ¼ n unique solutions (b) rank of A ¼ rank of Ab ¼ m < n infinite number of solutions (c) rank of A < rank of Ab no solution Solution of sets of equations (a) Inverse matrix method Ax ¼ b; To find A x ¼ A1 b 1 (1) evaluate jAj (2) form C, the matrix of cofactors of A (3) write CT , the transpose of A (4) A1 ¼ 1 CT . jAj (b) Row transformation method Ax ¼ b; Ax ¼ Ib (1) form the combined coefficient matrix ½A I 1 (2) row transformations to convert to ½I A (3) then solve x ¼ A1 b. Ax ¼ b (c) Gaussian elimination method (1) form augmented matrix ½A b 0 (2) operate on rows to convert to ½U b where U is the uppertriangular matrix. (3) expand from bottom row to obtain x. (d) Triangular decomposition method Ax ¼ b Write A as the product of upper and lower triangular matrices. A ¼ LU; LðUxÞ ¼ b. Put Ux ¼ y ; Ly ¼ b (1) solve Ly ¼ b to obtain y (2) solve Ux ¼ y to obtain x. (e) Using an electronic spreadsheet The spreadsheet used to demonstrate the method is the Excel spreadsheet provided by Microsoft. The two functions used are: MINVERSE(array) MMULT(array1, array2) for obtaining the inverse of a matrix for multiplying two matrices together 559 Matrix algebra 7 Matrix transformation (a) U ¼ TX, where T is a transformation matrix, transforms a vector in the x–y plane to a corresponding vector in the u–v plane. Similarly, X ¼ T1 U converts a vector in the u–v plane to a corresponding vector in the x–y plane. (b) Rotation of axes y u cos sin x ¼ v sin cos y (x, y) (u, v) v x u O y ¼ cos sin sin cos u v x Can you? 67 Checklist 16 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Determine whether a matrix is singular or non-singular? Yes No 1 to 3 . Determine the rank of a matrix? Yes 3 to 13 14 to 23 24 to 59 60 to 65 No . Determine the consistency of a set of linear equations and hence demonstrate the uniqueness of their solution? Yes No . Obtain the solution of a set of simultaneous linear equations by using matrix inversion, by row transformation, by Gaussian elimination, by triangular decomposition and by using a spreadsheet? Yes No . Use matrices to represent transformations between coordinate systems? Yes No 560 Programme 16 Test exercise 16 68 1 Determine the rank of A and of Ab for the following sets of equations and hence determine the nature of the solutions. Do not solve the equations. (a) x1 þ 3x2 2x3 ¼ 6 (b) x1 þ 2x2 4x3 ¼ 3 4x1 þ 5x2 þ 2x3 ¼ 3 x1 þ 2x2 þ 3x3 ¼ 4 x1 þ 3x2 þ 4x3 ¼ 7 0 2 3 2 If Ax ¼ b where A ¼ @ 3 1 solve the set of equations. 2x1 þ 4x2 þ x3 ¼ 3. 1 0 1 3 2 4 5 4 A and b ¼ @ 10 A, determine A1 and hence 2 3 9 Given that 3x1 þ 2x2 þ x3 ¼ 1 x1 x2 þ 3x3 ¼ 5 2x1 þ 5x2 2x3 ¼ 0 apply the method of row transformation to obtain the value of x1 , x2 , x3 . 4 5 By the method of Gaussian elimination, solve the equations Ax ¼ b, where 0 1 0 1 1 2 4 3 A ¼ @2 1 3 A and b ¼ @ 4 A. 1 3 2 5 0 1 0 1 1 2 1 7 If Ax ¼ b where A ¼ @ 3 1 2 A and b ¼ @ 3 A, express A as the product 5 3 3 5 A ¼ LU where L and U are lower and upper-triangular matrices and hence determine the values of x1 , x2 , x3 . 6 Use an electronic spreadsheet to solve the set of equations: 3a 2b þ 4c d þ e ¼ 32 a þ 3b 2c þ 5d þ 3e ¼ 3 a b þ c d þ e ¼ 12 4a 6b þ 2c þ 8d e ¼ 20 a 5b 7c þ 2d 3e ¼ 40 7 (a) Determine the vector in the u–v plane formed by U ¼ TX, where the 2 1 3 transformation matrix is T ¼ and X ¼ is a vector in the 3 4 2 x–y plane. (b) The coordinate axes in the x–y plane and in the u–v plane have the same origin O, but OU is inclined to OX at an angle of 608 in an anticlockwise 4 manner. Transform a vector X ¼ in the x–y plane into the 6 corresponding vector in the u–v plane. 561 Matrix algebra Further problems 16 0 1 2 1 0 1 5 2 3 6 If Ax ¼ b where A ¼ @ 3 2 2 A and b ¼ @ 5 A, determine A1 and 4 3 1 5 hence solve the set of equations. Apply the method of row transformation to solve the following sets of equations. (a) x1 3x2 2x3 ¼ 8 (b) x1 3x2 þ 2x3 ¼ 8 2x1 þ 2x2 þ x3 ¼ 4 2x1 x2 þ x3 ¼ 9 3x1 4x2 þ 2x3 ¼ 3 3 3x1 þ 2x2 þ 3x3 ¼ 5: Solve the following sets of equations by Gaussian elimination. (a) x1 2x2 x3 þ 3x4 ¼ 4 2x1 þ x2 þ x3 4x4 ¼ 3 3x1 x2 2x3 þ 2x4 ¼ 6 x1 þ 3x2 x3 þ x4 ¼ 8 (b) 2x1 þ 3x2 2x3 þ 2x4 ¼ 2 4x1 þ 2x2 3x3 x4 ¼ 6 x1 x2 þ 4x3 2x4 ¼ 7 3x1 þ 2x2 þ x3 x4 ¼ 5 (c) x1 þ 2x2 þ 5x3 þ x4 ¼ 4 3x1 4x2 þ 3x3 2x4 ¼ 7 4x1 þ 3x2 þ 2x3 x4 ¼ 1 x1 2x2 4x3 x4 ¼ 2. 4 Using the method of triangular decomposition, solve the following sets of equations. 1 0 10 1 0 2 1 4 1 x1 (a) @ 4 2 3 A@ x2 A ¼ @ 1 A x3 18 7 3 2 1 0 10 1 0 2 1 2 3 x1 (b) @ 2 1 5 A@ x2 A ¼ @ 17 A x3 22 6 3 2 1 0 10 1 0 3 1 2 3 1 x1 B C B C B3 1 3 2C CB x2 C ¼ B 14 C. (c) B @5 3 2 3 A@ x3 A @ 21 A x4 10 2 4 2 4 5 Use an electronic spreadsheet to solve all the equations in questions 2, 3 and 4. [Hint: For all those sets of three equations you only need a single template just change the numbers. The same applies for all those sets of four equations. 69 562 Programme 16 0 6 1 8 10 7 Invert the matrix A ¼ @ 5 9 4 A and hence solve the equations 9 11 8 8I1 þ 10I2 þ 7I3 ¼ 0 5I1 þ 9I2 þ 4I3 ¼ 9 7 9I1 þ 11I2 þ 8I3 ¼ 1. 0 1 0 1 2 3 2 If A ¼ @ 4 6 7 A and B ¼ @ 1 5 8 9 2 6 6 2 1 4 5 A, verify that AB ¼ kI where I is a 2 unit matrix and k is a constant. Hence solve the equations x1 þ 2x2 þ 3x3 ¼ 2 4x1 þ 6x2 þ 7x3 ¼ 2 5x1 þ 8x2 þ 9x3 ¼ 3. Programme 17 Frames 1 to 54 Systems of ordinary differential equations Learning outcomes When you have completed this Programme you will be able to: . Obtain the eigenvalues and corresponding eigenvectors of a square matrix . Demonstrate the validity of the Cayley–Hamilton theorem . Solve systems of first-order ordinary differential equations using eigenvalue and eigenvector methods . Construct the modal matrix from the eigenvectors of a matrix and the spectral matrix from the eigenvalues . Solve systems of second-order ordinary differential equations using diagonalisation 563 564 Programme 17 Eigenvalues and eigenvectors 1 Introduction Matrices commonly appear in technological problems, for example those involving coupled oscillations and vibrations, and give rise to equations of the form Ax ¼ x where A ¼ ðaij Þ is a square matrix, x ¼ ðxi Þ is a column matrix and is a scalar quantity, that is a number. For non-trivial solutions, that is for x 6¼ 0, the values of are called the eigenvalues, characteristic values or latent roots of the matrix A and the corresponding solutions of the given equations Ax ¼ x are called the eigenvectors, or characteristic vectors of A (refer to Engineering Mathematics, Sixth Edition, pages 578ff). The set of equations 0 10 1 0 1 a11 a12 . . . a1n x1 x1 B a21 a22 . . . a2n CB x2 C B x2 C B CB C B C B .. .. CB .. C ¼ B .. C .. @ . @ . A . A@ . A . an1 an2 . . . ann xn then simplifies to 0 ða11 Þ a12 ... B a21 ða Þ . .. 22 B B .. .. @ . . an1 an2 xn a1n a2n .. . ðann Þ 10 1 0 1 x1 0 CB x2 C B 0 C CB C B C CB .. C ¼ @ .. A A@ . A . xn 0 That is, Ax ¼ x becomes Ax x ¼ 0 i.e. ðA IÞx ¼ 0 the unit matrix I being introduced since we can subtract only a matrix from another matrix. For this set of homogeneous linear equations (right-hand side constant terms all zero) to have non-trivial solutions jA Ij must be zero This is called the characteristic determinant of A and jA Ij ¼ 0 is the characteristic equation, the solution of which gives the values of , i.e. the eigenvalues of A. 565 Systems of ordinary differential equations Example 1 2 Find the eigenvalues and corresponding eigenvectors of 2 3 Ax ¼ x where A ¼ . 4 1 The characteristic equation is jA Ij ¼ 0 i.e. 2 3 ¼ 0, which, when expanded, gives 4 1 1 ¼ . . . . . . . . . . . . and 2 ¼ . . . . . . . . . . . . 1 ¼ 2 and 2 ¼ 5 3 Because ð2 Þð1 Þ 12 ¼ 0 ; 2 3 þ 2 12 ¼ 0 2 3 10 ¼ 0 ð 5Þð þ 2Þ ¼ 0 ; ¼ 2 or 5 Now we substitute each value of in turn in the equation ðA IÞx ¼ 0 With ¼ 2 2 3 1 ð2Þ 4 1 0 2 2 3 þ 0 4 1 0 x1 0 ¼ 1 x2 0 0 0 x1 ¼ 0 2 x2 0 4 3 x1 ¼ 0 4 3 x2 Multiplying out the left-hand side, we get ............ 4x1 þ 3x2 ¼ 0 from which we get x2 ¼ 43 x1 i.e. not specific values for x1 and x2 , but a relationship between them. Whatever value we assign to x1 we obtain a corresponding value of x2 . 3 6 9 x1 ¼ or or , etc. x1 ¼ 4 8 12 x2 The most convenient way to do this is to choose x1 ¼ 1 and then scale x1 to obtain integer elements. So here we find for x1 ¼ 1 then x2 ¼ 4=3 so x1 is of the form ! 1 4 3 4 566 Programme 17 This is now scaled up by multiplying by 3 to give 3 x1 ¼ where is a constant multiplier. 4 The simplest result, with ¼ 1, is the one normally quoted. 3 ; for 1 ¼ 2; x1 ¼ 4 Similarly, for 2 ¼ 5, the corresponding eigenvector is . . . . . . . . . . . . 5 x2 ¼ 1 1 Because, with 2 ¼ 5, ðA IÞx ¼ 0 becomes 0 2 3 1 0 x1 ¼ 5 x2 0 4 1 0 1 0 2 3 5 0 x1 ¼ x2 0 4 1 0 5 0 3 3 x1 ¼ x2 0 4 4 ; 3x1 þ 3x2 ¼ 0 i.e. x2 ¼ x1 1 ; with 2 ¼ 5, the corresponding eigenvector is x2 ¼ 1 1 Again, taking ¼ 1, for 2 ¼ 5; x2 ¼ 1 So the required eigenvectors are 3 x1 ¼ corresponding to 1 ¼ 2 4 1 x2 ¼ corresponding to 2 ¼ 5. 1 Example 2 Determine the eigenvalues and corresponding eigenvectors of 3 10 Ax ¼ x where A ¼ . 2 4 The characteristic equation is jA Ij ¼ 0, which in this case can be written as ............ 567 Systems of ordinary differential equations 3 2 6 10 ¼0 4 Expanding the determinant and solving the equation gives 1 ¼ . . . . . . . . . . . . ; 1 ¼ 1; 2 ¼ . . . . . . . . . . . . 7 2 ¼ 8 ; 2 7 8 ¼ 0 Because the equation is ð3 Þð4 Þ 20 ¼ 0 ; ð þ 1Þð 8Þ ¼ 0 ; ¼ 1 or 8 (a) With 1 ¼ 1, we solve ðA IÞx ¼ 0 to obtain an eigenvector, which is ............ x1 ¼ Because A¼ 3 10 2 4 ! ( ; 3 10 5 2 ! 8 1 ð1Þ 2 4 0 ( ! 3 10 1 þ 2 4 0 4 2 2 ; 4x1 þ 10x2 ¼ 0 ; x2 ¼ x1 5 0 !) 1 0 !) 1 10 ! x1 x2 x1 x2 x1 5 x2 5 ! ¼ ! ¼ ! ¼ 0 ! 0 0 ! 0 0 ! 0 x1 ¼ 2 5 ; with ¼ 1 1 ¼ 1 and x1 ¼ 2 (b) In the same way the corresponding eigenvector x2 for 2 ¼ 8 is ............ 568 Programme 17 2 x2 ¼ 1 9 Because ( ! 3 10 2 ( 4 3 10 2 4 8 ! 1 0 0 1 8 0 0 8 !) !) 5 10 2 4 ! x1 x2 x1 x2 x1 x2 ! ¼ ! ¼ ! ¼ 1 ; 5x1 þ 10x2 ¼ 0 ; x2 ¼ x1 2 ; with ¼ 1, 2 ¼ 8 and x2 ¼ 0 ! 0 0 ! 0 0 ! 0 2 x2 ¼ 1 2 1 The same basic method can similarly be applied to third-order sets of equations. Example 3 Determine 0 1 A ¼ @0 3 the eigenvalues and eigenvectors of Ax ¼ x where 1 0 4 2 0 A. 1 3 As before, we have ðA IÞx ¼ 0 with characteristic equation jA Ij ¼ 0. i.e. 1 0 3 0 4 2 0 ¼0 1 3 Expanding this we have 1 ¼ . . . . . . . . . . . . ; 2 ¼ . . . . . . . . . . . . ; 3 ¼ . . . . . . . . . . . . 569 Systems of ordinary differential equations 1 ¼ 2; 2 ¼ 3; 10 3 ¼ 5 Because ð1 Þfð2 Þð3 Þ 0g þ 4f0 3ð2 Þg ¼ 0 ð1 Þð2 Þð3 Þ 12ð2 Þ ¼ 0 ; ð2 Þfð1 Þð3 Þ 12g ¼ 0 ; ¼ 2 or 2 þ 2 15 ¼ 0 ; ð 3Þð þ 5Þ ¼ 0 ; ¼ 2, 3, or 5 (a) With 1 ¼ 2; ðA IÞx ¼ 0 becomes 80 1 0 190 1 0 1 4 1 0 0 > 0 > < 1 0 = x1 B C B C B C B C 0 A 2 @ 0 1 0 A @ x2 A ¼ @ 0 A @0 2 > > : ; 3 1 3 0 0 1 x3 0 80 1 0 190 1 0 1 x 1 0 4 2 0 0 0 > > 1 = < B C B C B C B C 0 A @ 0 2 0 A @ x2 A ¼ @ 0 A @0 2 > > ; : x3 0 3 1 3 0 0 2 0 10 1 0 1 1 0 4 x1 0 B CB C B C ; @ 0 0 0 A@ x2 A ¼ @ 0 A 3 1 5 x3 0 from which a corresponding eigenvector x1 is . . . . . . . . . . . . 0 1 4 x1 ¼ @ 7 A 1 11 Because we have x1 þ 4x3 ¼ 0 ; x3 ¼ 14 x1 3x1 þ x2 5x3 ¼ 0 ; 3x1 þ x2 54 x1 ¼ 0 ; x2 ¼ 74 x1 ; x1 , x2 , x3 are in the ratio 1 : (b) Similarly 80 < 1 @0 : 3 7 1 : i.e. 4 : 7 : 1 4 4 x2 ¼ . . . . . . . . . . . . 4 1 B C C ; x1 ¼ B @ 7 A 1 for 2 ¼ 3, ðA IÞx ¼ 0 1 0 190 1 0 1 0 4 1 0 0 = x1 0 2 0 A 3@ 0 1 0 A @ x2 A ¼ @ 0 A ; 1 3 0 0 1 x3 0 from which a corresponding eigenvector is 0 570 Programme 17 0 1 2 x2 ¼ @ 0 A 1 12 Because 80 1 0 1 0 4 3 > > <B C B B0 2 B 0C @ A @0 > > : 3 1 3 0 0 2 B B 0 @ 3 190 1 0 1 0 > > x1 C=B C B C B C B C 3 0C A>@ x2 A ¼ @ 0 A > ; 0 3 x3 0 10 1 0 1 0 4 x1 0 CB C B C C B C B 1 0 A@ x2 A ¼ @ 0 C A 1 6 x3 0 0 0 ; 2x1 þ 4x3 ¼ 0 ; x3 ¼ 12 x1 Also x2 ¼ 0 ; x2 ¼ 0 0 1 2 B C ; x2 ¼ @ 0 A 1 (c) All that now remains is 3 ¼ 5. A corresponding eigenvector x3 is x3 ¼ . . . . . . . . . . . . Finish it on your own. Method just the same as before. 0 1 2 x3 ¼ @ 0 A 3 13 Check the 0 1 B A ¼ @0 3 80 1 > > <B B0 @ > > : 3 working. 1 0 4 C 2 0 A and 3 ¼ 5 with ðA IÞx ¼ 0. 1 3 1 0 190 1 0 1 0 0 4 1 0 0 > > x1 C B C=B C B C B C B C B C 2 0C A þ 5@ 0 1 0 A>@ x2 A ¼ @ 0 A > ; x3 0 1 3 0 0 1 0 10 1 0 1 0 6 0 4 x1 B CB C B C B 0 7 0 CB x2 C ¼ B 0 C @ A@ A @ A x3 0 3 1 2 ; 6x1 þ 4x3 ¼ 0 ; x3 ¼ 32 x1 7x2 ¼ 0 ; x2 ¼ 0 0 2 1 B C ; x3 ¼ @ 0 A 3 571 Systems of ordinary differential equations Collecting the results together, we finally have 0 1 0 1 0 1 4 2 2 1 ¼ 2, x1 ¼ @ 7 A; 2 ¼ 3, x2 ¼ @ 0 A; 3 ¼ 5, x3 ¼ @ 0 A 1 1 3 Cayley–Hamilton theorem The Cayley–Hamilton theorem states that every square matrix satisfies its characteristic equation. For example the matrix 2 3 A¼ 4 1 of Frame 54 has the characteristic equation 2 3 10 ¼ 0 and so the Cayley–Hamilton theorem tells us that A2 3A 10I ¼ 0 To verify this we note that 2 3 2 3 16 9 2 A ¼ ¼ 4 1 4 1 12 13 ! 16 9 2 A2 3A 10I ¼ 3 12 13 4 ! 16 9 6 ¼ 12 13 12 so that ! 3 1 10 1 0 ! 9 10 3 0 You try one. Verify that the matrix A¼ 3 2 0 ! 1 0 10 10 4 ! ¼ 0 0 0 0 ! of Frame 5 with the characteristic equation 2 7 8 ¼ 0 satisfies the Cayley–Hamilton theorem, that is . . . . . . . . . . . . 14 572 Programme 17 15 A2 7A 8I ¼ 0 Because 3 10 3 10 29 70 ¼ so that A2 ¼ 2 4 2 4 14 36 ! ! ! 29 70 3 10 1 0 2 7 8 A 7A 8I ¼ 14 36 2 4 0 1 ! ! ! 8 0 21 70 29 70 ¼ ¼ 0 8 14 28 14 36 0 0 0 0 ! Now on to something different Systems of first-order ordinary differential equations 16 Matrix methods involving eigenvalues and their associated eigenvectors can be used to solve systems of coupled differential equations, though we shall only consider cases where the relevant eigenvalues are distinct. We proceed by example. Example 1 Consider the system of two coupled ordinary differential equations f10 ðxÞ ¼ 2f1 ðxÞ þ 3f2 ðxÞ f20 ðxÞ ¼ 4f1 ðxÞ þ f2 ðxÞ where f1 ð0Þ ¼ 2 and f2 ð0Þ ¼ 1 These can be written in matrix form as . . . . . . . . . . . . 17 f10 ðxÞ f20 ðxÞ ¼ 2 4 3 1 f1 ðxÞ f2 ðxÞ That is F0 ðxÞ ¼ AFðxÞ 0 f1 ðxÞ 2 3 f1 ðxÞ 0 and A ¼ and where where FðxÞ ¼ , F ðxÞ ¼ f20 ðxÞ 4 1 f2 ðxÞ f1 ð0Þ 2 Fð0Þ ¼ ¼ are the boundary conditions in matrix form. f2 ð0Þ 1 Systems of ordinary differential equations 573 The matrix differential equation F0 ðxÞ ¼ AFðxÞ is similar in form to the single differential equation f 0 ðxÞ ¼ af ðxÞ (a constant) which has solution f ðxÞ ¼ eax ( constant), so to solve the matrix equation we try a solution of the form FðxÞ ¼ Cekx where the number k and the constants c1 and c2 of the c1 are to be determined. matrix C ¼ c2 Substituting FðxÞ ¼ Cekx into the matrix equation F0 ðxÞ ¼ AFðxÞ gives ............ kCekx ¼ ACekx 18 Because FðxÞ ¼ Cekx so F0 ðxÞ ¼ kCekx . Since F0 ðxÞ ¼ AFðxÞ then kCekx ¼ ACekx Dividing both sides by ekx gives kC ¼ AC that is AC ¼ kC. So, from Frame 1, k is an eigenvalue of A and C is the corresponding eigenvector. Therefore, we must first find the eigenvalues of A and for this matrix these have been found earlier in Frames 2 to 5. They are 3 ¼ 2 (and so k ¼ 2) with corresponding eigenvector 4 1 ¼ 5 (and so k ¼ 5) with corresponding eigenvector 1 To each eigenvalue the matrix FðxÞ ¼ Cekx is a solution. The complete solution to F0 ¼ AF is then ... ... F1 ðxÞ ¼ a1 e... and F2 ðxÞ ¼ a e... ... ... 2 F1 ðxÞ ¼ 3 1 a1 e2x and F2 ðxÞ ¼ a e5x 4 1 2 Because FðxÞ ¼ Cekx is the solution corresponding to the eigenvalue k with associated eigenvector C. The complete solution to the equation F0 ðxÞ ¼ AFðxÞ is then a combination of these two solutions in the form 3 2x 1 5x FðxÞ ¼ A e þB e 4 1 Applying the boundary conditions gives Fð0Þ ¼ . . . . . . . . . . . . 19 574 Programme 17 20 Fð0Þ ¼ 3A þ B 4A þ B 2 ¼ 1 Because FðxÞ ¼ A 3 2x 1 5x 3 1 e þB e and so Fð0Þ ¼ A þB 4 1 4 1 3A þ B 2 ¼ ¼ 4A þ B 1 Therefore 3A þ B ¼ 2 4A þ B ¼ 1 with solution A ¼ 1=7 and B ¼ 11=7, giving the final solution as FðxÞ ¼ . . . . . . . . . . . . 21 FðxÞ ¼ 3=7 2x 11=7 5x e þ e 4=7 11=7 Summary To solve an equation of the form F0 ðxÞ ¼ AFðxÞ 1 2 3 Find the eigenvalues 1 , 2 , . . ., n of A (assuming they are all distinct) Find the associated eigenvectors C1 , C2 , . . ., Cn P Write the solution of the equation as FðxÞ ¼ nr¼1 Ar er x Cr and use the boundary conditions to find the values of ar for r ¼ 1, 2, . . . , n. Now you try one. Next frame 22 Example 2 The system of two coupled ordinary differential equations f10 ðxÞ ¼ 3f1 ðxÞ þ 10f2 ðxÞ f20 ðxÞ ¼ 2f1 ðxÞ þ 4f2 ðxÞ where f1 ð0Þ ¼ 0 and f2 ð0Þ ¼ 1 has the solution (refer to Frames 5 to 9) f1 ðxÞ ¼ . . . . . . . . . . . . f2 ðxÞ ¼ . . . . . . . . . . . . 575 Systems of ordinary differential equations 23 10 x 10 8x e þ e 9 9 4 5 f2 ðxÞ ¼ ex þ e8x 9 9 f1 ðxÞ ¼ Because f10 ðxÞ ¼ 3f1 ðxÞ þ 10f2 ðxÞ f20 ðxÞ ¼ 2f1 ðxÞ þ 4f2 ðxÞ can be written in matrix form as ............ f10 ðxÞ f20 ðxÞ ¼ 3 2 10 4 f1 ðxÞ f2 ðxÞ 24 That is F0 ðxÞ ¼ AFðxÞ 0 f1 ðxÞ 3 f1 ðxÞ 0 , F ðxÞ ¼ and A ¼ where FðxÞ ¼ 0 f ðxÞ f2 ðxÞ 2 2 0 Fð0Þ ¼ . 1 10 4 and where To solve the matrix equation we first need the eigenvalues and associated eigenvectors of the matrix A. These have already been found in Frames 5 to 9 and they are 5 ¼ 1 with corresponding eigenvector 2 2 ¼ 8 with corresponding eigenvector 1 The complete solution of F0 ¼ AF is then 2 8x 5 x e FðxÞ ¼ A e þB 1 2 That is f1 ðxÞ ¼ . . . . . . . . . . . . f2 ðxÞ ¼ . . . . . . . . . . . . 576 Programme 17 25 f1 ðxÞ ¼ 5Aex þ 2Be8x f2 ðxÞ ¼ 2Aex þ Be8x Because FðxÞ ¼ f1 ðxÞ f2 ðxÞ 2 8x 5 x ¼A e e þB 1 2 and so f1 ðxÞ ¼ 5Aex þ 2Be8x f2 ðxÞ ¼ 2Aex þ Be8x Applying the boundary conditions, we find 0 ...A þ ...B Fð0Þ ¼ ¼ 1 ...A þ ...B 5A þ 2B 0 f1 ð0Þ ¼ Fð0Þ ¼ ¼ 2A þ B 1 f2 ð0Þ 26 Because The boundary conditions are f1 ð0Þ ¼ 0 and f2 ð0Þ ¼ 1 therefore 5A þ 2B 0 f1 ð0Þ ¼ Fð0Þ ¼ ¼ 2A þ B f2 ð0Þ 1 This gives the pair of simultaneous equations 5A þ 2B ¼ 0 2A þ B ¼ 1 which have solution A ¼ . . . . . . . . . . . . and B ¼ . . . . . . . . . . . . 27 A ¼ 2=9 and B ¼ 5=9 This gives the complete solution as 10=9 x f1 ðxÞ 10=9 8x ¼ e þ FðxÞ ¼ e 4=9 f2 ðxÞ 5=9 10 10 8x f1 ðxÞ ¼ ex þ e 9 9 4 5 f2 ðxÞ ¼ ex þ e8x 9 9 577 Systems of ordinary differential equations Diagonalisation of a matrix Modal matrix 28 We have already discussed the eigenvalues and eigenvectors of a matrix A of order n. In this section we shall assume that all the eigenvalues are distinct. If the n eigenvectors xi are arranged as columns of a square matrix, the modal matrix of A, denoted by M, is formed i.e. M ¼ ðx1 , x2 , x3 , . . . , xn Þ For example, we have seen earlier that if 0 1 1 0 4 A ¼ @0 2 0 A then 1 ¼ 2, 2 ¼ 3, 3 ¼ 5 3 1 3 and the corresponding eigenvectors are 0 1 0 1 0 1 4 2 2 x1 ¼ @ 7 A, x2 ¼ @ 0 A, x3 ¼ @ 0 A 1 1 3 0 1 4 2 2 Then the modal matrix M ¼ @ 7 0 0A 1 1 3 Spectral matrix Also, we define the spectral matrix of A, i.e. S, as a diagonal matrix with the eigenvalues only on the main diagonal 0 1 1 0 0 . . . 0 B 0 2 0 . . . 0 C B C i.e. S ¼ B .. .. C .. .. @ . . A . . 0 0 0 ... n So, in the example above, S ¼ . . . . . . . . . . . . 0 2 0 S ¼ @0 3 0 0 1 0 0A 5 Note that the eigenvalues of S 0 5 6 B So, if A ¼ @ 1 1 3 0 and A are the same. 1 1 C 0 A has eigenvalues ¼ 1, 2, 4 and 1 0 1 0 1 0 1 0 1 3 B C B C B C corresponding eigenvectors @ 1 A, @ 1 A, @ 1 A 6 3 3 then M ¼ . . . . . . . . . . . . and S ¼ . . . . . . . . . . . . 29 578 Programme 17 0 30 0 1 M ¼ @1 1 6 3 1 3 1 A; 3 0 1 S ¼ @0 0 0 2 0 1 0 0A 4 Now how are these connected? Let us investigate. The eigenvectors x arranged in the modal matrix satisfy the original equation Ax ¼ x M ¼ ð x1 Also AM ¼ Að x1 ¼ ð Ax1 Then ¼ ð 1 x1 0 . . . xn Þ x2 x2 . . . xn Þ Ax2 . . . Axn Þ 2 x2 1 0 .. . 0 2 .. . ... ... 0 ; AM ¼ MS 0 . . . n B B Now S ¼ B B @ 0 0 .. . . . . n xn Þ 1 since Ax ¼ x C C C ; ð 1 x1 C A 2 x2 . . . n xn Þ ¼ MS If we now pre-multiply both sides by M1 we have M1 AM ¼ M1 MS But M1 M ¼ I ; M1 AM ¼ S Make a note of this result. Then we will consider an example 31 Example 1 From the results of a previous example in Frame 13, if 0 1 1 0 4 B C A ¼ @0 2 0 A then 1 ¼ 2, 2 ¼ 3, 3 ¼ 5 and 3 1 3 0 1 0 1 0 1 4 2 2 B C B C B C x1 ¼ @ 7 A; x2 ¼ @ 0 A; x3 ¼ @ 0 A. 1 1 3 0 1 4 2 2 B C Also M ¼ @ 7 0 0 A. 1 1 3 We can find M1 by any of the methods we have established previously. M1 ¼ . . . . . . . . . . . . 579 Systems of ordinary differential equations 0 M1 0 ¼ @ 3=8 1=8 1 1=7 0 1=4 1=4 A 1=28 1=4 32 Here is one way of determining the inverse. You may have done it by another. 0 1 0 1 7 0 0 1 0 0 0 1 0 4 2 2 B C B C 0 1 0 A @ 1 1 3 0 0 1A 0 @ 7 0 1 1 3 0 0 1 4 2 2 1 0 0 0 1 0 1 1 0 0 1 0 0 0 1=7 0 0 1=7 0 B C B C 0 1=7 1 A @ 0 1 3 0 1=7 1A @ 0 1 3 0 2 2 1 B 0 @ 0 0 1 B @0 0 0 1 0 0 3 1 0 0 1 0 0 1 0 0 0 B ; M ¼ @ 3=8 1=8 0 1 B So now A ¼ @ 0 1 1 0 0 1=8 4=7 1=7 1=7 1=28 1 B ; AM ¼ @ 0 0 2 3 10 4 4 2 CB 0 A@ 7 0 1 3 0 0 B 1 Then M AM ¼ B @ 3=8 1=8 3 0 0 1 0 C 1 A 1=4 1 0 1=7 0 C 3=8 7=28 1=4 A 1=8 1=28 1=4 1 1=7 0 C 1=4 1=4 A 1=28 1=4 1 0 0 4 4 C B 2 0 A and M ¼ @ 7 3 1 0 0 1 1=7 1=4 1=28 ¼ ............ 8 2 0 1 2=7 2 1 2 C 0A 1 1 3 1 0 1 2 8 6 10 C B C 0 A ¼ @ 14 0 0A 1 3 2 3 15 10 1 0 8 6 10 CB C B 1=4 C 0C A@ 14 0 A 1=4 2 3 15 580 Programme 17 0 33 2 0 M1 AM ¼ @ 0 3 0 0 1 0 0A 5 So we have transformed the original matrix A into a diagonal matrix and notice that the elements on the main diagonal are, in fact, the eigenvalues of A i.e. M1 AM ¼ S Therefore, let us list a few relevant facts 1 2 3 4 M1 AM transforms the square matrix A into a diagonal matrix S. A square matrix A of order n can be so transformed if the matrix has n independent eigenvectors. A matrix A always has n linearly independent eigenvectors if it has n distinct eigenvalues or if it is a symmetric matrix. If the matrix has repeated eigenvalues and is not symmetric, it may or may not have n linearly independent eigenvectors. Now here is one straightforward example with which to finish. Example 2 If A ¼ 6 5 ; 4 2 M ¼ ............; M1 ¼ . . . . . . . . . . . . ; and hence M1 AM ¼ . . . . . . . . . . . . Work through it entirely on your own: (1) Determine the eigenvalues and corresponding eigenvectors. (2) Hence form the matrix M. (3) Determine M1 , the inverse of M. (4) Finally form the matrix products AM and M1 ðAMÞ. 34 M¼ 1 2 5 1=6 ; M1 ¼ 2 1=6 5=12 ; 1=12 M1 AM ¼ Here is the working. See whether you agree. 6 5 6 5 ; ¼0 A¼ 4 2 4 2 ð6 Þð2 Þ 20 ¼ 0 ð 4Þð þ 8Þ ¼ 0 ; 2 þ 4 32 ¼ 0 ; ¼ 4 or 8 4 0 0 8 581 Systems of ordinary differential equations ( (a) 1 ¼ 4 6 5 ! 4 2 4 0 0 4 10 !) 5 ! 2 x1 x2 x1 ! 0 ¼ ! 0 ! 0 ¼ ! 0 1 ; 10x1 þ 5x2 ¼ 0 ; x2 ¼ 2x1 x1 ¼ 2 ! ! ( ! !) 0 6 5 8 0 x1 ¼ þ (b) 2 ¼ 8 0 4 2 0 8 x2 ! ! ! 0 2 5 x1 ¼ 0 4 10 x2 5 ; 2x1 þ 5x2 ¼ 0 ; x2 ¼ 25 x1 ; x2 ¼ 2 1 5 ; M¼ 2 2 1 5 j 1 0 To find M1 2 2 j 0 1 4 Operating on rows, we have 0 5 j 1 0 0 12 j 2 1 1 5 j 0 1 1 j 0 j 0 1 j 1=6 1=12 1=6 5=12 1=6 1=12 6 5 1 ¼ ¼ ; M1 ¼ x2 ; AM ¼ 1 0 1=6 1=12 1=6 5=12 5 ¼ 4 40 4 2 2 2 8 16 1=6 5=12 4 40 4 ¼ ; M1 AM ¼ 1=6 1=12 8 16 0 4 0 ; M1 AM ¼ 0 8 0 8 582 Programme 17 Systems of second-order differential equations 35 The process of uncoupling a system of differential equations to obtain their solution can be achieved by diagonalising the matrix of coefficients. For simplicity we shall only consider second-order equations and again, we proceed by example. Example 1 Consider the system of coupled second-order differential equations f100 ðxÞ ¼ 2f1 ðxÞ þ 3f2 ðxÞ f200 ðxÞ ¼ 4f1 ðxÞ þ f2 ðxÞ where f1 ð0Þ ¼ 2, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 4 and f20 ð0Þ ¼ 3 These can be written in matrix form as ............ 36 f100 ðxÞ f200 ðxÞ ¼ 2 3 4 1 f1 ðxÞ f2 ðxÞ That is F00 ðxÞ ¼ AFðxÞ 00 f1 ðxÞ 2 3 f1 ðxÞ 00 and where , F ðxÞ ¼ and A ¼ where FðxÞ ¼ f200 ðxÞ 4 1 f2 ðxÞ 0 f1 ð0Þ f1 ð0Þ 4 2 Fð0Þ ¼ ¼ are the boundary ¼ and F0 ð0Þ ¼ f20 ð0Þ 3 1 f2 ð0Þ conditions in matrix form. The matrix differential equation F00 ðxÞ ¼ AFðxÞ is similar in form to the single differential equation f 00 ðxÞ ¼ af ðxÞ (a constant) which has solution pffiffi pffiffi ax ax f ðxÞ ¼ e þ e (, constants), so to solve the matrix equation we try a solution of this form. We already know from Frames 2 to 5 that the eigenvalues and eigenvectors of matrix A are 3 ¼ 2 with corresponding eigenvector 4 1 ¼ 5 with corresponding eigenvector 1 The modal matrix of A is the matrix M and the spectral matrix of A is the matrix S where ... ... ... ... M¼ and S ¼ ... ... ... ... 583 Systems of ordinary differential equations M¼ 3 4 1 1 and S ¼ 2 0 0 5 37 Because The modal matrix is formed from the eigenvectors of A. That is 3 1 3 1 M¼ where the two eigenvectors are and 4 1 4 1 The spectral matrix is formed from the eigenvalues of A. That is 2 0 S¼ where the two eigenvalues are 2 and 5 0 5 If we now define the matrix GðxÞ by the equation FðxÞ ¼ MGðxÞ, then differentiating gives F00 ðxÞ ¼ ½MGðxÞ00 ¼ MG00 ðxÞ where F00 ðxÞ ¼ AFðxÞ ¼ AMGðxÞ and so, from Frame 85, M1 MG00 ðxÞ ¼ G00 ðxÞ ¼ M1 AMGðxÞ ¼ SGðxÞ. That is G00 ðxÞ ¼ SGðxÞ Therefore, in component terms 00 g1 ðxÞ 2 0 g1 ðxÞ 00 ¼ SGðxÞ ¼ G ðxÞ ¼ g200 ðxÞ 0 5 g2 ðxÞ and so g100 ðxÞ ¼ . . . g1 ðxÞ with solution g1 ðxÞ ¼ k11 e...x þ k12 e...x g200 ðxÞ ¼ . . . g2 ðxÞ with solution g2 ðxÞ ¼ k21 e...x þ k22 e...x pffiffi pffiffi g100 ðxÞ ¼ 2g1 ðxÞ with solution g1 ðxÞ ¼ k11 e j 2x þ k12 ej 2x pffiffi pffiffi g200 ðxÞ ¼ 5g2 ðxÞ with solution g2 ðxÞ ¼ k21 e 5x þ k22 e 5x Now, FðxÞ ¼ MGðxÞ so f1 ðxÞ ... FðxÞ ¼ ¼ f2 ðxÞ ... 38 584 Programme 17 39 FðxÞ ¼ f1 ðxÞ ¼ f2 ðxÞ 3k11 e j pffiffi 2x þ pffiffi j 2x 4k11 e pffiffi 2x þ pffiffi j 2x 3k12 ej 4k12 e Because FðxÞ ¼ f1 ðxÞ f2 ðxÞ and so f1 ðxÞ ¼ 3k11 e j ¼ MGðxÞ ¼ pffiffi 2x and f2 ðxÞ ¼ 4k11 e j þ 3k12 ej pffiffi 2x pffiffi 2x 4k12 ej 3 1 4 1 þ k21 ej pffiffi 2x pffiffi 5x þ k21 e k21 e pffiffi 5x þ pffiffi 5x þ k21 e pffiffi 5x pffiffi 5x k22 e þ k22 e ! pffiffi pffiffi k11 e jpffiffi2x þ k12 ejpffiffi2x k21 e 5x þ k22 e 5x þ k22 e pffiffi 5x pffiffi 5x þ k22 e pffiffi 5x This solution can be written in terms of circular and hyperbolic trigonometric expressions as ... ... P cos . . . x þ Q sin . . . x FðxÞ ¼ ... ... R cosh . . . x þ S sinh . . . x 40 FðxÞ ¼ Because 3 4 1 1 pffiffiffi pffiffiffi P cos p2ffiffiffix þ Q sin p 2ffiffiffi x R cosh 5x þ S sinh 5x pffiffi þ 3k12 ej 2x pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ 3k11 cos 2x þ j sin 2x þ 3k12 cos 2x j sin 2x pffiffiffi pffiffiffi ¼ P cos 2x þ Q sin 2x where P ¼ 3k11 þ 3k12 and Q ¼ ð3k11 3k12 Þj 3k11 e j and pffiffi 2x pffiffi þ k22 e 5x pffiffiffi pffiffiffi pffiffiffi pffiffiffi ¼ k21 cosh 5x þ sinh 5x þ k22 cosh 5x sinh 5x pffiffiffi pffiffiffi ¼ R cosh 5x þ S sinh 5x where R ¼ k21 þ k22 and S ¼ k21 k22 pffiffi 5x k21 e Therefore FðxÞ ¼ ... ... Systems of ordinary differential equations FðxÞ ¼ pffiffiffi pffiffiffi pffiffiffi pffiffiffi ! 3P cos 2x þ 3Q sin 2x þ R cosh 5x þ S sinh 5x pffiffiffi pffiffiffi pffiffiffi pffiffiffi 4P cos 2x 4Q sin 2x þ R cosh 5x þ S sinh 5x 585 41 That is f1 ðxÞ ¼ . . . . . . . . . . . . f2 ðxÞ ¼ . . . . . . . . . . . . pffiffiffi pffiffiffi pffiffiffi pffiffiffi f1 ðxÞ ¼ 3P cos 2x þ 3Q sin 2x þ R cosh 5x þ S sinh 5x pffiffiffi pffiffiffi pffiffiffi pffiffiffi f2 ðxÞ ¼ 4P cos 2x 4Q sin 2x þ R cosh 5x þ S sinh 5x Because FðxÞ ¼ f1 ðxÞ f2 ðxÞ 42 and so pffiffiffi pffiffiffi pffiffiffi pffiffiffi f1 ðxÞ ¼ 3P cos 2x þ 3Q sin 2x þ R cosh 5x þ S sinh 5x pffiffiffi pffiffiffi pffiffiffi pffiffiffi f2 ðxÞ ¼ 4P cos 2x 4Q sin 2x þ R cosh 5x þ S sinh 5x Applying the boundary conditions, we find 2 4 ...P þ ...R ...Q þ ...S Fð0Þ ¼ ¼ ¼ and F0 ð0Þ ¼ 1 3 ...Q þ ...S ...P þ ...R Fð0Þ ¼ pffiffiffi pffiffiffi 2 3P þ R 4 3 p2ffiffiffiQ þ p5ffiffiffiS ¼ and F0 ð0Þ ¼ ¼ 1 4P þ R 3 4 2Q þ 5S Because f1 ð0Þ ¼ 2, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 4 and f20 ð0Þ ¼ 3 and so f1 ð0Þ 3P þ R 2 ¼ and ¼ Fð0Þ ¼ 1 4P þ R f2 ð0Þ pffiffiffi pffiffiffi ! 0 f1 ð0Þ 4 3 2Q þ 5S 0 pffiffiffi F ð0Þ ¼ ¼ ¼ pffiffiffi f20 ð0Þ 3 4 2Q þ 5S This gives the two sets of simultaneous equations pffiffiffi pffiffiffi 3 2Q þ 5S ¼ 4 3P þ R ¼ 2 and which have solution pffiffiffi pffiffiffi 4P þ R ¼ 1 4 2Q þ 5S ¼ 3 P ¼ ............, R ¼ ............, Q ¼ . . . . . . . . . . . . and S ¼ . . . . . . . . . . . . 43 586 Programme 17 pffiffiffi pffiffiffi P ¼ 1=7, R ¼ 11=7, Q ¼ 1= 7 2 and S ¼ 25= 7 5 44 This gives the complete solution as pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 11 25 f1 ðxÞ ¼ cos 2x þ pffiffiffi sin 2x þ cosh 5x þ pffiffiffi sinh 5x 7 7 7 5 7 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi 4 4 11 25 cosh 5x þ pffiffiffi sinh 5x f2 ðxÞ ¼ cos 2x pffiffiffi sin 2x þ 7 7 7 5 7 2 This method is quite straightforwardly extended to three or more such coupled differential equations. Summary To solve the system of coupled second-order differential equations F00 ðxÞ ¼ AFðxÞ 1 2 3 Find the eigenvalues and eigenvectors of matrix A and construct the modal matrix M and the diagonal spectral matrix S Solve the equation G0 ðxÞ ¼ SGðxÞ (note that even though M1 is used there was no need to calculate it) Apply FðxÞ ¼ MGðxÞ to find FðxÞ. Try one yourself. Next frame 45 Example 2 The system of coupled second-order differential equations (refer to Frames 5 to 9) f100 ðxÞ ¼ 3f1 ðxÞ þ 10f2 ðxÞ f200 ðxÞ ¼ 2f1 ðxÞ þ 4f2 ðxÞ where f1 ð0Þ ¼ 0, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 1 and f20 ð0Þ ¼ 0 has the solution f1 ðxÞ ¼ . . . . . . . . . . . . f2 ðxÞ ¼ . . . . . . . . . . . . 587 Systems of ordinary differential equations pffiffiffi pffiffiffi 5 2 f1 ðxÞ ¼ 10 cos x þ sin x þ 10 cosh 2 2x þ pffiffiffi sinh 2 2x 9 9 2 pffiffiffi pffiffiffi 2 1 f2 ðxÞ ¼ 4 cos x sin x þ 5 cosh 2 2x þ pffiffiffi sinh 2 2x 9 9 2 46 Because f100 ðxÞ ¼ 3f1 ðxÞ þ 10f2 ðxÞ f200 ðxÞ ¼ 2f1 ðxÞ þ 4f2 ðxÞ can be written in matrix form as ............ f100 ðxÞ f200 ðxÞ ¼ 3 10 2 4 f1 ðxÞ f2 ðxÞ That is F00 ðxÞ ¼ AFðxÞ 00 f1 ðxÞ 3 10 f1 ðxÞ , F00 ðxÞ ¼ and A ¼ where FðxÞ ¼ f200 ðxÞ 2 4 f2 ðxÞ 0 ð0Þ f 0 1 1 and where Fð0Þ ¼ and F0 ð0Þ ¼ ¼ . f20 ð0Þ 1 0 To solve the matrix equation we first need the eigenvalues and associated eigenvectors of the matrix A. These have already been found in Frames 5 to 9 and they are 5 ¼ 1 with corresponding eigenvector 2 2 ¼ 8 with corresponding eigenvector 1 The complete solution of F00 ¼ AF is then pffiffiffi pffiffiffi 5 2 FðxÞ ¼ ðP cos x þ Q sin xÞ þ R cosh 2 2x þ S sinh 2 2x 2 1 pffiffiffi pffiffiffi ! 5P cos x þ 5Q sin x þ 2R cosh 2 2x þ 2S sinh 2 2x pffiffiffi pffiffiffi ¼ 2P cos x 2Q sin x þ R cosh 2 2x þ S sinh 2 2x That is f1 ðxÞ ¼ . . . . . . . . . . . . f2 ðxÞ ¼ . . . . . . . . . . . . 47 588 Programme 17 pffiffiffi pffiffiffi f1 ðxÞ ¼ 5P cos x þ 5Q sin x þ 2R cosh 2 2x þ 2S sinh 2 2x pffiffiffi pffiffiffi f2 ðxÞ ¼ 2P cos x 2Q sin x þ R cosh 2 2x þ S sinh 2 2x 48 Because FðxÞ ¼ f1 ðxÞ f2 ðxÞ and so pffiffiffi pffiffiffi f1 ðxÞ ¼ 5P cos x þ 5Q sin x þ 2R cosh 2 2x þ 2S sinh 2 2x pffiffiffi pffiffiffi f2 ðxÞ ¼ 2P cos x 2Q sin x þ R cosh 2 2x þ S sinh 2 2x Applying the boundary conditions, we find 1 ...Q þ ...S 0 ...P þ ...R 0 ¼ Fð0Þ ¼ ¼ and F ð0Þ ¼ 0 ...Q þ ...S 1 ...P þ ...R 49 pffiffiffi 5P þ 2R 0 1 5Q þ 4 p2ffiffiffiS 0 and F ð0Þ ¼ ¼ ¼ Fð0Þ ¼ 2P þ R 1 0 2Q þ 2 2S Because The boundary conditions are f1 ð0Þ ¼ 0, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 1 and f20 ð0Þ ¼ 0, therefore 5P þ 2R 0 f1 ð0Þ ¼ and Fð0Þ ¼ ¼ f2 ð0Þ 2P þ R 1 pffiffiffi ! 0 f1 ð0Þ 1 5Q þ 4 2S 0 pffiffiffi ¼ F ð0Þ ¼ ¼ f20 ð0Þ 0 2Q þ 2 2S This gives the two sets of simultaneous equations pffiffiffi 5Q þ 4 2S ¼ 1 5P þ 2R ¼ 0 and which have solution pffiffiffi 2P þ R ¼ 1 2Q þ 2 2S ¼ 0 P ¼ ............, R ¼ ............, Q ¼ . . . . . . . . . . . . and S ¼ . . . . . . . . . . . . 50 pffiffiffi P ¼ 2=9, R ¼ 5=9, Q ¼ 1=9 and S ¼ 1= 9 2 This gives the complete solution as pffiffiffi pffiffiffi 10 5 10 2 cos x þ sin x þ cosh 2 2x þ pffiffiffi sinh 2 2x 9 9 9 9 2 pffiffiffi pffiffiffi 4 2 5 1 f2 ðxÞ ¼ cos x sin x þ cosh 2 2x þ pffiffiffi sinh 2 2x 9 9 9 9 2 f1 ðxÞ ¼ Systems of ordinary differential equations 589 As usual, the Programme ends with the Revision summary, to be read in conjunction with the Can you? checklist. Go back to the relevant part of the Programme for any points on which you are unsure. The Test exercise should then be straightforward and the Further problems give valuable additional practice. Revision summary 17 1 Eigenvalues and eigenvectors Ax ¼ x Sets of equations of form Ax ¼ x, where A ¼ coefficient matrix, x ¼ column matrix, ¼ scalar quantity. Equations become ðA IÞx ¼ 0: For non-trivial solutions, jA Ij ¼ 0 is the characteristic equation and gives values of i.e. the eigenvalues. Substitution of each eigenvalue gives a corresponding eigenvector. 2 Cayley–Hamilton theorem Every square matrix satisfies its own characteristic equation. 3 Solving systems of first-order ordinary differential equations To solve the system of coupled first-order differential equations F0 ðxÞ ¼ AFðxÞ (a) Find the eigenvalues and eigenvectors of matrix A and construct the modal matrix M and the diagonal spectral matrix S (b) Solve the equation G0 ðxÞ ¼ SGðxÞ (c) Apply FðxÞ ¼ MGðxÞ to find FðxÞ. 4 Diagonalisation of a matrix Modal matrix of A If A has distinct eigenvalues M ¼ ðx1 , x2 , . . . , xn Þ, where x1 , x2 , . . ., xn are eigenvectors of A, then M1 AM ¼ S where S is the spectral matrix of A 1 0 1 0 . . . 0 B 0 2 . . . 0 C C B B : : : C C B and S ¼ B : : C C B : @ : : : A 0 0 . . . n 1 , 2 , . . . , n are the eigenvalues of A. 51 590 Programme 17 5 Solving systems of second-order ordinary differential equations To solve an equation of the form F00 ðxÞ ¼ AFðxÞ (a) Find the eigenvalues 1 , 2 , . . . , n of A (b) Assuming the eigenvectors are all distinct, find the associated eigenvectors C1 , C2 , . . . , Cn (c) Write the solution of the equation as n pffiffi pffiffiffiffi X ar e r x þ br e r x Cr FðxÞ ¼ r¼1 and use the boundary conditions to find the values of ar and br for r ¼ 1, 2, . . . , n. Can you? 52 Checklist 17 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Obtain the eigenvalues and corresponding eigenvectors of a square matrix? Yes No Frames 1 to 13 14 and 15 . Solve systems of first-order ordinary differential equations using eigenvalue and eigenvector methods? Yes No 16 to 27 . Construct the modal matrix from the eigenvectors of a matrix and the spectral matrix from the eigenvalues? Yes No 28 to 34 . Solve systems of second-order ordinary differential equations using diagonalisation? Yes No 35 to 50 . Demonstrate the validity of the Cayley–Hamilton theorem? Yes No Systems of ordinary differential equations 591 Test exercise 17 1 2 Determine the eigenvalues and corresponding eigenvectors of Ax ¼ x where 0 1 1 3 0 B C A¼@ 1 2 1 A. 2 1 1 3 2 determine If x1 and x2 are eigenvectors of Ax ¼ x where A ¼ 4 1 (a) M ¼ x 1 x2 53 (b) M1 (c) M1 AM. 3 Solve the system of first-order differential equations f10 ðxÞ ¼ 5f1 ðxÞ 2f2 ðxÞ where f1 ð0Þ ¼ 3 and f2 ð0Þ ¼ 2. f20 ðxÞ ¼ f1 ðxÞ þ 4f2 ðxÞ 4 Solve the system of second-order differential equations f100 ðxÞ ¼ f1 ðxÞ þ 6f2 ðxÞ where f1 ð0Þ ¼ 1, f2 ð0Þ ¼ 0, f10 ð0Þ ¼ 2, f20 ð0Þ ¼ 1. f200 ðxÞ ¼ 3f1 ðxÞ 2f2 ðxÞ Further problems 17 1 Solve each of the following systems of first-order differential equations. (a) f10 ðxÞ ¼ 2f1 ðxÞ 5f2 ðxÞ f20 ðxÞ ¼ f1 ðxÞ 4f2 ðxÞ where f1 ð0Þ ¼ 1 and f2 ð0Þ ¼ 0 (b) f10 ðxÞ ¼ 5f1 ðxÞ þ 9f2 ðxÞ f20 ðxÞ ¼ f1 ðxÞ þ 3f2 ðxÞ where f1 ð0Þ ¼ 0 and f2 ð0Þ ¼ 2 (c) f10 ðxÞ ¼ 5f1 ðxÞ 6f2 ðxÞ þ f3 ðxÞ f20 ðxÞ ¼ f1 ðxÞ þ f2 ðxÞ f30 ðxÞ ¼ 3f1 ðxÞ þ f3 ðxÞ where f1 ð0Þ ¼ 1, f2 ð0Þ ¼ 0 and f3 ð0Þ ¼ 2 (d) f10 ðxÞ ¼ 4f1 ðxÞ þ 10f2 ðxÞ 8f3 ðxÞ f20 ðxÞ ¼ f1 ðxÞ þ 2f2 ðxÞ þ f3 ðxÞ f30 ðxÞ ¼ f1 ðxÞ þ 2f2 ðxÞ þ 3f3 ðxÞ where f1 ð0Þ ¼ 4, f2 ð0Þ ¼ 2 and f3 ð0Þ ¼ 1. 54 592 Programme 17 0 2 1 0 3 A; determine the three eigenvalues 1 , 2 , 3 of A and 9 0 1 9 1 1 verify that if M ¼ @ 3 2 4 A then M1 AM ¼ S, where S is a diagonal 1 3 3 1 If A ¼ @ 3 0 3 10 3 matrix with elements 1 , 2 , 3 . 3 Solve each of the following systems of second-order differential equations. (a) f100 ðxÞ ¼ 4f1 ðxÞ þ 3f2 ðxÞ f200 ðxÞ ¼ 2f1 ðxÞ þ 5f2 ðxÞ where f1 ð0Þ ¼ 0, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 4 and f20 ð0Þ ¼ 1 (b) f100 ðxÞ ¼ 6f1 ðxÞ þ 5f2 ðxÞ f200 ðxÞ ¼ 4f1 ðxÞ þ 2f2 ðxÞ where f1 ð0Þ ¼ 0, f2 ð0Þ ¼ 1, f10 ð0Þ ¼ 1 and f20 ð0Þ ¼ 0 (c) f100 ðxÞ ¼ 2f1 ðxÞ þ 7f2 ðxÞ f200 ðxÞ ¼ f1 ðxÞ þ 3f2 ðxÞ þ f3 ðxÞ f300 ðxÞ ¼ 5f1 ðxÞ þ 8f3 ðxÞ where f1 ð0Þ ¼ 1, f2 ð0Þ ¼ 1, f3 ð0Þ ¼ 0, f10 ð0Þ ¼ 0, f20 ð0Þ ¼ 0 and f30 ð0Þ ¼ 1 (d) f100 ðxÞ ¼ 3f1 ðxÞ þ 6f3 ðxÞ f200 ðxÞ ¼ 4f1 ðxÞ þ 5f2 ðxÞ þ 3f3 ðxÞ f300 ðxÞ ¼ f1 ðxÞ þ 2f2 ðxÞ þ f3 ðxÞ where f1 ð0Þ ¼ 1, f2 ð0Þ ¼ 1, f3 ð0Þ ¼ 0, f10 ð0Þ ¼ 0, f20 ð0Þ ¼ 0, f30 ð0Þ ¼ 1. Programme 18 Frames 1 to 65 Numerical solutions of partial differential equations Learning outcomes When you have completed this Programme you will be able to: . Derive the finite difference formulas for the first partial derivatives of a function of two real variables and construct the central finite difference formula to represent a first-order partial differential equation . Draw a rectangular grid of points overlaid on the domain of a function of two real variables and evaluate the function at the boundary grid points . Construct the computational molecule for a first-order partial differential equation in two real variables and use the molecule to evaluate the solutions to the equation at the grid points interior to the boundary . Describe the solution as a set of simultaneous linear equations and use matrices to represent them . Invert the coefficient matrix and thereby represent the solution to the partial differential equation as a column matrix . Take account of a boundary condition in the form of the derivative normal to the boundary . Obtain the central finite difference formulas for the second derivatives of a function of two real variables and construct finite difference formulas for second-order partial differential equations . Use the forward difference formula for the first time derivatives in partial differential equations involving time and distance . Use the Crank–Nicolson procedure for a partial differential equation involving a first time derivative . Appreciate the use of dimensional analysis in the conversion of a partial differential equation modelling a physical system into a dimensionless equation 593 594 Programme 18 Introduction 1 The numerical solution of partial differential equations is a large subject and can form the content of a course in itself. Here we shall just introduce the subject by considering the basic methods of solving some first- and secondorder partial differential equations that involve functions of two real variables. The approach that is used is to construct finite difference formulas for the first and second partial derivatives and then to construct a finite difference formula that represents an approximation to the differential equation. However, before we move into the realm of functions of two real variables we shall derive the finite difference formulas for the ordinary first derivative of a function of a single real variable. Next frame Numerical approximation to derivatives 2 A function of one real variable f ðxÞ has the Taylor series expansion f ðx þ hÞ ¼ f ðxÞ þ hf 0 ðxÞ þ h2 00 h3 f ðxÞ þ f 000 ðxÞ þ . . . 2! 3! and, equally, replacing h byh f ðx hÞ ¼ f ðxÞ hf 0 ðxÞ þ h2 00 h3 f ðxÞ f 000 ðxÞ þ . . . 2! 3! From the first equation we can see that by dividing through by h, we have f ðx þ hÞ f ðxÞ h h2 ¼ f 0 ðxÞ þ f 00 ðxÞ þ f 000 ðxÞ þ . . . h 2! 3! and from the second equation f ðx hÞ f ðxÞ h h2 ¼ f 0 ðxÞ þ f 00 ðxÞ f 000 ðxÞ þ . . . h 2! 3! If we now neglect terms of the order two and higher we see that f 0 ðxÞ f ðx þ hÞ f ðxÞ h [this is the forward difference formula for the first derivative of f ðxÞ] f ðxÞ f ðx hÞ h [this is the backward difference formula for the first derivative of f ðxÞ] and f 0 ðxÞ Numerical solutions of partial differential equations 595 and both of these are accurate up to terms of order two. A more accurate estimate of the derivative can be obtained by subtracting the two Taylor series expansions from each other to get f 0 ðxÞ . . . . . . . . . . . . neglecting terms of the order of . . . . . . . . . . . . f ðx þ hÞ f ðx hÞ 2h neglecting terms of the order two and higher f 0 ðxÞ Because h2 h3 f ðxÞ þ hf 0 ðxÞ þ f 00 ðxÞ þ f 000 ðxÞ þ . . . 2! 3! 2 h h3 f ðxÞ hf 0 ðxÞ þ f 00 ðxÞ f 000 ðxÞ þ . . . 2! 3! h3 ¼ 2 hf 0 ðxÞ þ f 000 ðxÞ þ . . . 3! f ðx þ hÞ f ðx hÞ ¼ and so f ðx þ hÞ f ðx hÞ h2 ¼ f 0 ðxÞ þ f 000 ðxÞ þ . . . 2h 3! giving f ðx þ hÞ f ðx hÞ 2h neglecting terms of the order two and higher. f 0 ðxÞ The derivative at x is given as the difference between the two values either side of f(x) divided by 2h. This is called the central difference formula for the derivative of f ðxÞ and because it is the most accurate of the three for small h, it is the one that we shall use in the remainder of the Programme. Now we need to look at the second derivative. By adding the first two Taylor series expansions in Frame 2 we find that f 00 ðxÞ . . . . . . . . . . . . neglecting terms of the order . . . . . . . . . . . . 3 596 Programme 18 4 f ðx þ hÞ 2f ðxÞ þ f ðx hÞ h2 neglecting terms of the order two and higher f 00 ðxÞ Because f ðx þ hÞ þ f ðx hÞ ¼ f ðxÞ þ hf 0 ðxÞ þ h2 00 h3 f ðxÞ þ f 000 ðxÞ þ . . . 2! 3! h2 00 h3 000 þ f ðxÞ hf ðxÞ þ f ðxÞ f ðxÞ þ . . . 2! 3! h2 h4 ¼ 2 f ðxÞ þ f 00 ðxÞ þ f iv ðxÞ þ . . . 2! 4! 0 ¼ 2f ðxÞ þ h2 f 00 ðxÞ þ h4 iv f ðxÞ þ . . . 12 and so f ðx þ hÞ 2f ðxÞ þ f ðx hÞ h2 iv 00 f ðxÞ þ . . . ¼ f ðxÞ þ h2 12 Therefore f 00 ðxÞ f ðx þ hÞ 2f ðxÞ þ f ðx hÞ h2 neglecting terms of the order two and higher This is the central difference formula for the second derivative and, as you see, it possesses the same level of accuracy as the central difference formula for the first derivative. 597 Numerical solutions of partial differential equations Functions of two real variables A function of two real variables f ðx, yÞ is graphically represented as a surface in three-dimensional space. f (x, y) f (x0, y0) (x0, y0, f(x0, y0)) y (x0, y0) x If f ðx, yÞ is single-valued, then to every domain point ðx, yÞ there corresponds a single range point f ðx, yÞ and hence a single surface point ðx, y, f ðx, yÞÞ. If we know the exact form of f ðx, yÞ then we can compute its value at any domain point ðx, yÞ selected at random. If we do not know the exact form of f ðx, yÞ but we do know that it satisfies a given differential equation then to evaluate f ðx, yÞ numerically we have to be more systematic. What we do is to lay a rectangular grid over the domain and evaluate f ðx, yÞ at the grid points – the points of intersection of the lines parallel with the x-axis and the lines parallel with the y-axis. y x 5 598 Programme 18 In this Programme we shall be considering functions of two real variables that satisfy given differential equations and whose domains are restricted to being rectangular. This restriction avoids many of the problems that occur with arbitrary domain shapes where the grid lines can cross the domain boundary. Grid values 6 The rectangular domain of the function is overlaid by a grid whose mesh size is of h units in the x direction and k units in the y direction. We shall denote the value of f ðx, yÞ at the ij th grid point as fi; j f ðih, jkÞ The values of the expression f ðx, yÞ are required to be found at the grid points as shown: fi1; jþ1 fi1; j fi1; j1 fi; jþ1 fi; j fi; j1 fiþ1; jþ1 fiþ1; j fiþ1; j1 Notice as you move along the jth row of this table that the value of y is constant at yj ¼ y0 þ jk for all points on that row. Similarly, as you move up and down the i th column that the value of x is constant at xi ¼ x0 þ ih for all points in that column. These facts now enable us to define the central difference formulas for the partial derivatives of f ðx, yÞ. The first partial derivative of f ðx, yÞ with respect to the variable x is obtained by differentiating f ðx, yÞ with respect to x whilst keeping the value of the variable y constant. Therefore, as with the ordinary derivative @f ðx, yÞ @x ij is equal to the difference between the two adjacent values of f ðx, yÞ in the x-direction divided by twice the mesh size in the x-direction. That is fiþ1; j fi1; j @f ðx, yÞ ¼ @x ij 2h This is the central difference formula for the partial derivative with respect to x. Similarly, the central difference formula for the partial derivative with respect to y is @f ðx, yÞ ¼ ............ @y ij 599 Numerical solutions of partial differential equations fi; jþ1 fi; j1 @f ðx, yÞ ¼ @y ij 2k 7 Because @f ðx, yÞ @y ij is equal to the difference between the two adjacent values of f ðx, yÞ in the y-direction divided by twice the mesh size in the y-direction. That is @f ðx, yÞ fi; jþ1 fi; j1 ¼ @y ij 2k Let’s try an example so that we can put all this information together. Example 1 @f ðx, yÞ @f ðx, yÞ 4 ¼ 0, for 0 x 1 and 0 y 1 @x @y given that the boundary conditions are Find the solution to 3 f ðx, 0Þ ¼ 4x þ 4 f ðx, 1Þ ¼ 4x þ 7 f ð0, yÞ ¼ 3y þ 4 f ð1, yÞ ¼ 3y þ 8 for a mesh of size 1=4 in the x-direction and of size 1=3 in the y-direction. Next frame The first thing we must do is to make a reasonable drawing of the domain of the function with the grid overlaid. The domain of f ðx, yÞ is the square of side length 1 as shown in the diagram. y (7) 1 (6) 2/3 (5) 1/3 0 (4) (8) (9) (10) A B C D E F 1/ 4 (5) 1/ 2 (6) 3/ 4 (7) (11) (10) (9) 1 (8) x 8 600 Programme 18 Overlaid on the function domain in the x–y plane is a mesh of grid points. The values of f ðx, yÞ that we can compute directly from the boundary conditions are shown in brackets. For example, from f ðx, 0Þ ¼ 4x þ 4 we obtain f ð1=4, 0Þ ¼ 5, f ð1=2, 0Þ ¼ 6, f ð3=4, 0Þ ¼ 7 and f ð1, 0Þ ¼ 8. From f ð1, yÞ ¼ 3y þ 8 we obtain f ð1, 0Þ ¼ 8, f ð1, 1=3Þ ¼ 9, f ð1, 2=3Þ ¼ 10 and f ð1, 1Þ ¼ 11. N o t i c e t h a t t h e v a l u e f o u n d a t f ð1, 0Þ ¼ 8 u s i n g f ðx, 0Þ ¼ 4x þ 4 is the same as the value found using f ð1, yÞ ¼ 3y þ 8, as of course it must be. The values of f ðx, yÞ that we have to determine are labelled A to F. The second part of the procedure is to find the central difference formula that describes the differential equation: fiþ1; j fi1; j @f ðx, yÞ ¼ We have ¼ 2ðfiþ1; j fi1; j Þ because h ¼ 1=4 @x ij 2h fi; jþ1 fi; j1 @f ðx, yÞ ¼ 1:5ðfi; jþ1 fi; j1 Þ because k ¼ 1=3 ¼ @y ij 2k Therefore 3 9 @f ðx, yÞ @f ðx, yÞ 4 ¼ 0 becomes . . . . . . . . . . . . @x @y 6ðfiþ1; j fi1; j Þ 6ðfi; jþ1 fi; j1 Þ ¼ 0 Because @f ðx, yÞ @f ðx, yÞ 4 ¼ 0 evaluated at the ijth grid point is @x @y @f ðx, yÞ @f ðx, yÞ 4 ¼0 3 @x ij @y ij 3 which is 3 2ðfiþ1; j fi1; j Þ 4 1:5ðfi; jþ1 fi; j1 Þ ¼ 0, that is 6ðfiþ1; j fi1; j Þ 6ðfi; jþ1 fi; j1 Þ ¼ 0 601 Numerical solutions of partial differential equations Computational molecules The value of the first derivative with respect to x at the point ðxi , yj Þ on the grid overlaying the function domain is found by evaluating the right-hand side of the equation @f ðx, yÞ fiþ1; j fi1; j fi1; j þ fiþ1; j ¼ ¼ @x ij 2h 2h and this process is repeated for every grid point in the function domain. We can construct a graphic template to assist us in this process: –1 2h 1 2h 0 ij The three circles in a row are used to calculate the contribution of three adjacent row members to the equation. If the circle labelled ij is laid over the ijth grid point then the derivative at that point is given by multiplying the value of the function at the i 1; j grid point (one to the left) by 1=2h and adding the product of the value of the function at the i þ 1; j grid point (one to the right) by 1=2h. The number 0 in the centre circle means that we multiply fi; j by zero because it does not enter into the formula. This template is called a computational molecule. The horizontal structure reflects the fact that we are evaluating along a row. By a similar reasoning the first derivative with respect to y at the ijth grid point is @f ðx, yÞ fi; jþ1 fi; j1 ¼ @y ij 2k and this is represented by the computational molecule: 1 2k 0 ij –1 2k The vertical structure reflects the fact that we are evaluating up and down a column. 10 602 Programme 18 By combining such computational molecules we can construct a composite molecule that represents the entire differential equation. For example, the partial differential equation a @f ðx, yÞ @f ðx, yÞ þb ¼c @x @y evaluated at the ijth grid point is @f ðx, yÞ @f ðx, yÞ a þb ¼c @x ij @y ij and is represented by the central difference formula a b ðfiþ1; j fi1; j Þ þ fi; jþ1 fi; j1 Þ ¼ c 2h 2k which is in turn represented by the composite computational molecule: b 2k –a 2h a 2h 0 ij –b 2k @f ðx, yÞ @f ðx, yÞ 4 ¼ 0 which is represented by @x @y the finite difference formula So the equation 3 6ðfiþ1; j fi1; j Þ 6ðfi; jþ1 fi; j1 Þ ¼ 0 has the computational molecule . . . . . . . . . . . . 603 Numerical solutions of partial differential equations 11 –6 –6 6 0 =0 ij 6 We now place the centre of the molecule, in turn, on each of the grid points at which we need to find the value of f ðx, yÞ: On A 36 48 þ 6B þ 6D ¼ 0 On B 6A 54 þ 6C þ 6E ¼ 0 On C ............ On D ............ On E ............ On F ............ On A 36 48 þ 6B þ 6D ¼ 0 On B 6A 54 þ 6C þ 6E ¼ 0 On C 6B 60 þ 60 þ 6F ¼ 0 On D 30 6A þ 6E þ 30 ¼ 0 On E 6D 6B þ 6F þ 36 ¼ 0 On F 6E 6C þ 54 þ 42 ¼ 0 We now have six simultaneous linear equations in six unknowns. These can be written in matrix form as . . . . . . . . . . . . 12 604 Programme 18 0 13 0 6 B 6 0 B B 0 6 B B 6 0 B @ 0 6 0 0 That is: 0 6 0 0 0 6 6 0 0 0 6 0 0 6 0 6 0 6 1 10 1 0 84 A 0 C B C B 0C CB B C B 54 C C B C B 6 CB C C B 0 C C C B C¼B 0C CB D C B 0 C A @ A @ 36 A E 6 96 F 0 Ax ¼ b with solution x ¼ A1 b There are many ways to derive the inverse matrix A1 , many of them time consuming and prone to arithmetic error. An efficient method in terms of time and accuracy is to use a spreadsheet, provided of course that the spreadsheet has the appropriate functionality. Here we shall use the Microsoft Excel spreadsheet which possesses matrix functions. If your spreadsheet does not have these functions then you are referred to Programme 16, Matrix algebra. If you do possess the Microsoft Excel spreadsheet then follow the instructions in the next frame. Next frame 14 1 2 3 4 5 Open your spreadsheet. Place the cell highlight in cell A1 and then enter the values of matrix A into the cells A1 to F6. Place the cell highlight in cell H1 and then enter the values of matrix b into the cells H1 to H6. Place the cell highlight in cell A8 and drag the mouse to highlight the block of cells A8 to F13 – this is where the inverse of A is going to go. With this block of cells highlighted, type the function: =MINVERSE(A1 : F6) and then press the three keys Ctrl-Shift-Enter together As you type, the function is entered into cell A8 and when you press the Ctrl-Shift-Enter keys together the block of cells A8 to F13 fills with entries. This block of cells is the inverse matrix A1 . (Note: You must remember to press the three keys Ctrl-Shift-Enter together. If you just press Enter it will not work.) MINVERSE(array) is the Excel function that computes the inverse of the square matrix denoted by array. 6 7 Place the cell highlight in cell H8 and drag the mouse to highlight the block of cells H8 to H13 – this is where the solution x is going to go. With this block of cells highlighted type the function: =MMULT(A8 : F13, H8 : H13) and then press the three keys Ctrl-ShiftEnter together MMULT(array1,array2) is the Excel function that multiplies the two matrices denoted by array1 and array2. 605 Numerical solutions of partial differential equations As you type, the function is entered into cell H8 and when you press the Ctrl-Shift-Enter keys together the block of cells H8 to H13 fills with entries. This block of cells is the product matrix A1 b, that is, the solution matrix x. 0 1 0 1 A 7 B B C B8C B C B C BCC B9C B C¼B C BDC B6C B C B C @ E A @7A F 8 These values are identical to the values found from the exact solution which is f ðx, yÞ ¼ 4x þ 3y þ 4. Next frame Summary of procedures The procedure to solve a first-order partial differential equation requires a number of steps to be completed in a certain order, and the following list describes the sequence: 1 2 Draw the domain of the function with the grid overlaid. On the drawing enter the values of f ðx, yÞ that can be obtained from the boundary conditions. Put these values in brackets so that they will be easily distinguished from the x- and y-values on the axes. 3 4 5 6 7 8 Label the grid points at which f ðx, yÞ is to be evaluated with capital letters. Construct the central difference equation that represents the numerical approximation to the partial differential equation. Construct the computational molecule for this equation. Lay the centre of the molecule on each of the lettered grid points in turn and derive a set of simultaneous linear equations – the unknowns being represented by the letters at the grid points. Write the simultaneous linear equations in matrix form Ax ¼ b. Find the inverse matrix A1 and compute the solution x ¼ A1 b. Now try one yourself. Just follow the procedure in order and you should have no problems. 15 606 Programme 18 Example 2 The solution to x that @f ðx, yÞ @f ðx, yÞ y ¼ 0, for 0 x 1 and 0 y 1 given @x @y f ðx, 0Þ ¼ 2 f ðx, 1Þ ¼ x þ 2 f ð0, yÞ ¼ 2 f ð1, yÞ ¼ y þ 2 for a mesh of 1=4 in the x-direction and 1=3 in the y-direction is: . . . . . . . . . . . . 0 1 0 : 1 0 1 A 2 166 . . . 13=6 B B C B 2:33 . . . C B 7=3 C B C B C B C B C C B 2:5 C B 5=2 C B C¼B C¼B C B D C B 2:0833 . . . C B 25=12 C B C B C B C @ E A @ 2:166 . . . A @ 13=6 A F 2:25 9=4 16 Because The domain of the function f ðx, yÞ with the overlaid grid looks as follows: y (2) 1 (2) 2/3 (2) 1/3 0 (2) ( 9/4 ) ( 10/4 ) (11/4) A B C D E F 1/ 4 (2) 1/ 2 (2) 3/ 4 (2) ( 12/4 ) ( 8/3 ) ( 7/3 ) 1 (2) x where the numbers at the grid points in brackets are the values of f ðx, yÞ obtained by applying the boundary conditions and the letters A . . . F represent the values of f ðx, yÞ that we have yet to determine. 607 Numerical solutions of partial differential equations The central difference formulas for the two first partial derivatives of f ðx, yÞ are fiþ1; j fi1; j @f ðx, yÞ ¼ 2ðfiþ1; j fi1; j Þ because h ¼ 1=4 ¼ @x ij 2h fi; jþ1 fi; j1 @f ðx, yÞ ¼ 1:5ðfi; jþ1 fi; j1 Þ because k ¼ 1=3 ¼ @y ij 2k Therefore x @f ðx, yÞ @f ðx, yÞ y ¼ 0 becomes . . . . . . . . . . . . @x @y 17 2ðxi fiþ1; j xi fi1; j Þ 1:5ðyj fi; jþ1 yj fi; j1 Þ ¼ 0 Because x @f ðx, yÞ @f ðx, yÞ y ¼0 @x @y is written using the central difference formulas as 2xi ðfiþ1; j fi1; j Þ 1:5yj ðfi; jþ1 fi; j1 Þ ¼ 2ðxi fiþ1; j xi fi1; j Þ 1:5ðyj fi; jþ1 yj fi; j1 Þ ¼ 0 This has the following computational molecule: –1.5yj –2xi 0 2xi =0 ij 1.5yj Placing the centre of the molecule, in turn, on each of the grid points that we need to evaluate, we obtain the six simultaneous equations: 2 14 ð2Þ 32 23 94 þ 2 14 B þ 32 23 D ¼ 0 On A at 14 , 23 : 1 3 2 On B at 12 , 23 : 2 12 A 32 23 10 4 þ2 2 Cþ2 3 E¼0 38 3 2 2 34 B 32 23 11 On C at 34 , 23 : 4 þ2 4 3 þ2 3 F ¼0 On D at 14 , 13 : 2 14 ð2Þ 32 13 A þ 2 14 E þ 32 13 ð2Þ ¼ 0 On E at 12 , 13 : 2 12 D 32 13 B þ 2 12 F þ 32 13 ð2Þ ¼ 0 On F at 34 , 13 : 2 34 E 32 13 C þ 2 34 73 þ 32 13 ð2Þ ¼ 0 These six equations can be simplified as . . . . . . . . . . . . 608 Programme 18 18 On A B=2 þ D ¼ 13=4 On B A þ C þ E ¼ 10=4 On C 3B=2 þ F ¼ 5=4 On D A=2 þ E=2 ¼ 0 On E B=2 D þ F ¼ 1 On F C=2 3E=2 ¼ 9=2 These six simultaneous linear equations can be expressed in matrix form as ............ 0 19 0 B 1 B B 0 B B 0:5 B @ 0 0 0:5 0 1:5 0 0:5 0 0 1 0 0 0 0:5 1 0 0 0 1 0 0 1 0 0:5 0 1:5 10 1 0 1 0 A 13=4 B C B C 0C CB B C B 10=4 C C B C B 1 CB C C B 5=4 C C B C¼B C 0C CB D C B 0 C 1 A@ E A @ 1 A 0 F 9=2 That is Ax ¼ b with solution x ¼ A1 b Inverting the matrix A we find that 0 1 0 : 1 0 1 A 2 166 . . . 13=6 B B C B 2:3 . . . C B 7=3 C B C B C B C B C C B 2:5 C B 5=2 C B C¼B C¼B C B D C B 2:0833 . . . C B 25=12 C B C B C B C @ E A @ 2:166 . . . A @ 13=6 A F 2:25 9=4 which is identical to the values found from the exact solution f ðx, yÞ ¼ xy þ 2. Next frame Derivative boundary conditions 20 The process of solving a differential equation, either ordinary or partial, involves using indefinite integration and each time we integrate we produce an integration constant. For a differential equation to have a complete solution, where all the integration constants are evaluated, the differential equation must be accompanied by a set of conditions that are sufficient to do this. If the differential equation involves time t then it is natural for these conditions to give values of the function and its derivatives at time t ¼ 0. Such conditions are known as initial conditions and we have met these before when 609 Numerical solutions of partial differential equations we studied the Laplace transform, for example. Other conditions, like the conditions we met in the previous two examples, are called boundary conditions because they gave the values of the function on the boundary of the function domain. We now consider boundary conditions in the form of derivatives normal to the boundary and this we do in the following example. Example 3 @f ðx, yÞ @f ðx, yÞ þ2 ¼ 3, for 0 x 1 and 0 y 1 @x @y given that the boundary conditions are Find the solution to 4 and f ðx; 0Þ ¼ f ðx; 1Þ ¼ f ð0; yÞ ¼ 10 @f ðx, yÞ ¼2 @x x¼1 for a mesh of size 1=3 in both the x-direction and the y-direction. Next frame The domain of f ðx, yÞ is the square of side length 1 as shown in the diagram: y (10) (10) 1 (10) (10) f =2 x (10) 2/3 (10) 1/3 0 (10) A B C G D E F H 1/ 3 (10) 2/ 3 (10) 1 (10) 4/ 3 x Overlaid on the function domain in the x–y plane is a mesh of grid points. Because the boundary condition relating to the side x ¼ 1 is in the form of a derivative normal to the side, we extend the grid over the boundary of the function domain by adding two additional points outside the domain and distant 1=3 from it, as shown in the figure. The values of f ðx, yÞ that we can compute from the boundary conditions alone are shown in brackets. The values of f ðx, yÞ that we have to determine are labelled A to F and we shall need the two additional points G and H outside the domain of f ðx, yÞ to do this. 21 610 Programme 18 The second part of the procedure is to find the central difference formula that describes the differential equation: @f ðx, yÞ fiþ1; j fi1; j ¼ 1:5 fiþ1; j fi1; j ¼ We have @x ij 2h @f ðx, yÞ fi; jþ1 fi; j1 ¼ ¼ 1:5 fi; jþ1 fi; j1 @y 2k ij because both h and k ¼ 1=3 Therefore 4 @f ðx, yÞ @f ðx, yÞ þ2 ¼ 3 becomes . . . . . . . . . . . . @x @y 6 fiþ1; j fi1; j þ 3 fi; jþ1 fi; j1 Þ ¼ 3 22 Because @f ðx, yÞ @f ðx, yÞ þ2 ¼ 3 can be written as @x @y 4 1:5 fiþ1; j fi1; j þ 2 1:5 fi; jþ1 fi; j1 ¼ 3, that is 6 fiþ1; j fi1; j þ 3 fi; jþ1 fi; j1 ¼ 3 4 This has the computational molecule . . . . . . . . . . . . 23 3 6 0 –6 =3 ij –3 We now place the centre of the molecule, in turn, on each of the grid points that we need to evaluate: On A 60 þ 30 þ 6B 3E ¼ 3 On B 6A þ 30 þ 6C 3E ¼ 3 On C ............ On D ............ On E ............ On F ............ 611 Numerical solutions of partial differential equations On A 60 þ 30 þ 6B 3E ¼ 3 On B 6A þ 30 þ 6C 3E ¼ 3 On C 6B þ 30 þ 6G 3F ¼ 3 On D 60 þ 3A þ 6E 30 ¼ 3 On E 6D þ 3B þ 6F 30 ¼ 3 On F 6E þ 3C þ 6H 30 ¼ 3 24 @f ðx, yÞ ¼ 2 can be written At the boundary x ¼ 1 the boundary condition @x x¼1 using the central difference formula as fiþ1; j fi1; j @f ðx, yÞ ¼ 1:5ðfiþ1; j fi1; j Þ ¼ 2 ¼ @x x¼1 2h y¼yj which has the computational molecule: –1.5 1.5 0 ij We now place the centre of this molecule, in turn, on each of the grid points C and F to obtain On C 1:5B þ 1:5G ¼ 2 On F On C On F ............ 1:5B þ 1:5G ¼ 2 1:5E þ 1:5H ¼ 2 25 We can now use these last two equations either to eliminate the points G and H from the six equations in Frame 24 or to form an 8 8 system. We shall eliminate the points G and H to obtain the six equations, with the constant on the right-hand side as ............ On A 6B 3E ¼ 33 On B 6A þ 6C 3E ¼ 27 On C 3F ¼ 35 On D 3A þ 6E ¼ 93 On E 6D þ 3B þ 6F ¼ 33 On F 3C ¼ 25 These six simultaneous linear equations can be expressed in matrix form as ............ 26 612 Programme 18 0 27 0 B 6 B B 0 B B 3 B @ 0 0 6 0 0 0 3 0 0 6 0 0 0 3 0 0 0 0 6 0 1 10 1 0 33 A 3 0 C B C B 3 0 C CB B C B 27 C C B C B 0 3 CB C C B 35 C C C B C¼B 6 0 C CB D C B 93 C A @ A @ 33 A E 0 6 25 F 0 0 That is Ax ¼ b with solution x ¼ A1 b Inverting the matrix A1 we find that x ¼ . . . . . . . . . . . . 0 1 0 : 1 0 1 A 6 777 . . . 61=9 B B C B 11:555 . . . C B 104=9 C B C B C B C B C C B 8:333 . . . C B 25=3 C B C¼B C¼B C B D C B 11:9444 . . . C B 215=18 C B C B C B C @ E A @ 12:111 . . . A @ 109=9 A F 11:666 . . . 35=3 28 Next frame Second-order partial differential equations 29 The most general form of a second-order partial differential equation is aðx; yÞ @2f @2f @2f @f @f þ bðx; yÞ þ dðx; yÞ þ eðx; yÞ þ gðx; yÞ ¼ 0 þ cðx; yÞ @x2 @x@y @y 2 @x @y Three types of equation are of particular interest because they feature so prominently in engineering and science. Elliptic equations If b2 4ac < 0 the partial differential equation is called an elliptic equation. Such equations arise out of steady-state problems as occur in potential or flow theory. Two examples are Poisson’s equation @ 2 ðx; yÞ @ 2 ðx; yÞ þ ¼ gðx; yÞ @x2 @y 2 Laplace’s equation @ 2 ðx; yÞ @ 2 ðx; yÞ þ ¼0 @x2 @y 2 In both cases a ¼ 1, b ¼ 0 and c ¼ 1 and so b2 4ac < 0. 613 Numerical solutions of partial differential equations Hyperbolic equations If b2 4ac > 0 the partial differential equation is called an hyperbolic equation. Such equations arise out of vibrational and radiative problems as occur in wave mechanics. An example is The wave equation @ 2 ðx; tÞ 1 @ 2 ðx; tÞ ¼ 2 2 @x @t 2 Here a ¼ 1, b ¼ 0 and c ¼ 1 and so b2 4ac > 0. 2 Parabolic equations If b2 4ac ¼ 0 the partial differential equation is called a parabolic equation. Such equations arise out of transient flow problems as occur in conduction or consolidation. An example is The consolidation (or heat conduction) equation @ 2 ðx; tÞ 1 @ðx; tÞ ¼ @x2 @t Here a ¼ 1, b ¼ 0 and c ¼ 0 and so b2 4ac ¼ 0. In the equations above a, b and c are constant but in the general case they depend on x and y and so a given equation may change from one type to another within the same domain. Determine whether each of the following equations are elliptic, hyperbolic or parabolic: @ 2 ðx, yÞ @ 2 ðx, yÞ þ þ kðx, yÞ ¼ ðx, yÞ @x2 @y 2 2 @ðr, tÞ @ ðr, tÞ 2 @ðr, tÞ ¼a þ ðr, tÞ (b) þ @t @r 2 r @r (a) @ 2 ðx, tÞ @ðx, tÞ @ 2 ðx, xÞ ¼ p2 þk þ qðx, yÞ 2 @t @t @x2 @ðx, tÞ @ 2 ðx, tÞ @ðx, tÞ ¼ þ ðx, tÞ (d) @t @x2 @x 2 @2 @ðx, yÞ @ @ðx, yÞ n n (e) þ ¼ rðx, yÞ px py @x @y @x @y @ 2 ðx, tÞ @ @ðx, yÞ þ qm ðx, yÞ ¼p (f) n ðx, yÞ @t 2 @x @y (c) Next frame 614 Programme 18 30 (a) Elliptic (b) Parabolic (cÞ Hyperbolic (d) Parabolic (e) Elliptic (f) Hyperbolic Because: Comparing with the general equation aðx; yÞ (a) @2f @2f @2f @f @f þ cðx; yÞ þ bðx; yÞ þ dðx; yÞ þ eðx; yÞ þ gðx; yÞ ¼ 0 @x2 @x@y @y 2 @x @y @ 2 ðx, yÞ @ 2 ðx, yÞ þ þ kðx, yÞ ¼ ðx, yÞ @x2 @y 2 Here b ¼ 0 so b2 4ac < 0, Elliptic. The Helmholtz equation. 2 @ðr, tÞ @ ðr, tÞ 2 @ðr, tÞ þ ðr, tÞ (b) þ ¼a @t @r 2 r @r Here b ¼ c ¼ 0 so b2 4ac ¼ 0. Parabolic. The heat equation with central symmetry. (c) 2 @ 2 ðx, tÞ @ðx, tÞ 2 @ ðx, xÞ þ k þ qðx, yÞ ¼ p @t 2 @t @x2 Here b ¼ 0, a > 0 and c < 0 so b2 4ac > 0. Hyperbolic. The telegraph equation. (d) @ðx, tÞ @ 2 ðx, tÞ @ðx, tÞ ¼ þ ðx, tÞ @t @x2 @x Here b ¼ c ¼ 0 so b2 4ac ¼ 0. Parabolic. Burger’s equation. @2 @2 n @ðx, yÞ n @ðx, yÞ þ ¼ rðx, yÞ (e) px py @x @y @x @y Here b ¼ 0 so b2 4ac < 0. Elliptic. The anisotropic heat diffusion equation. @ 2 ðx, tÞ @ @ðx, yÞ n (f) ¼ p ðx, yÞ þ qm ðx, yÞ @t 2 @x @y Here b ¼ 0, a > 0 and c < 0 so b2 4ac > 0. Hyperbolic. Next frame Numerical solutions of partial differential equations 615 Second partial derivatives In Frame 4 we found that for a function of a single real variable f ðxÞ the central difference formula approximating the second derivative was f 00 ðxÞ 31 f ðx hÞ 2f ðxÞ þ f ðx þ hÞ h2 The second derivative at x is given as the sum of the two adjacent values less twice the value at the point, all divided by h2 . If we apply this to a function of two real variables f ðx, yÞ and use fi; j f ðih, jkÞ to represent the value of f ðx, yÞ at the point ðih; jkÞ then the central difference formulas for the second partial derivatives with respect to x and y are seen to be . . . . . . . . . . . . fi1; j 2fi; j þ fiþ1; j @ 2 f ðx, yÞ 2 @x h2 ij fi; j1 2fi; j þ fi; jþ1 @ 2 f ðx, yÞ 2 @y k2 ij Because The second derivative at xi is given as the sum of the two adjacent values on the jth row less twice the value at xi , all divided by the cell width squared – h2 , and so fi1; j 2fi; j þ fiþ1; j @ 2 f ðx, yÞ @x2 ij h2 The second derivative at yj is given as the sum of the two adjacent values in the jth column less twice the value at yj , all divided by the cell height squared – k2 , and so fi; j1 2fi; j þ fi; jþ1 @ 2 f ðx, yÞ 2 @y k2 ij We are now ready to consider the construction of central difference formulas for second-order partial differential equations. We shall proceed by example. 32 616 Programme 18 Example 4 Given a grid with mesh size h ¼ k ¼ 1=3, find a numerical solution to the equation @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ 0 for 0 x 1, 0 y 1, given that @x2 @y 2 f ðx; 0Þ ¼ f ðx; 1Þ ¼ 5 3x f ð0; yÞ ¼ 9y 2 9y þ 5 and @f ðx, yÞ ¼ 6 @x x¼1 The domain with the grid overlaid is . . . . . . . . . . . . 33 y (4) (5) 1 (3) (2) f = –6 x (3) 2/3 (3) 1/3 0 (5) A B C G D E F H 1/ 3 (4) 2/ 3 (3) 1 x (2) The solution is to be evaluated at the grid points A to F – the external grid points G and H are inserted to accommodate the derivative boundary condition. The numbers in brackets are the values of f ðx, yÞ as found from the boundary conditions. The central difference formula that represents the partial differential equation is . . . . . . . . . . . . 617 Numerical solutions of partial differential equations 9 fiþ1; j þ fi; jþ1 4fi; j þ fi1; j þ fi; j1 34 Because @ 2 f ðx, yÞ @ 2 f ðx, yÞ fiþ1; j 2fi; j þ fi1; j fi; jþ1 2fi; j þ fi; j1 þ þ @x2 ij @y 2 ij h2 k2 ¼ 9 fiþ1; j 2fi; j þ fi1; j þ 9 fi; jþ1 2fi; j þ fi; j1 ¼ 9 fiþ1; j þ fi; jþ1 4fi; j þ fi1; j þ fi; j1 From this we can construct the computational molecule for this differential equation as . . . . . . . . . . . . 35 9 9 –36 9 =0 ij 9 If we applied this computational molecule to the grid points A to F then the six simultaneous linear equations that result would all have a common factor of 9 arising from the 9 in the molecule. If we divided every equation by 9 to remove this common factor we would not change the overall validity of the equations. So, to make the computation simpler we divide each term in the computational molecule by 9 and use the resulting molecule: 1 1 1 –4 ij 1 =0 618 Programme 18 We now proceed as we have done before. Laying the centre of the computational molecule on each grid point in turn gives the six simultaneous linear equations: 36 On A 3 þ 4 þ B þ D 4A ¼ 0 On B ............ On C ............ On D ............ On E ............ On F ............ On A 3 þ 4 þ B þ D 4A ¼ 0 On B A þ 3 þ C þ E 4B ¼ 0 On C B þ 2 þ G þ F 4C ¼ 0 On D 3 þ A þ E þ 4 4D ¼ 0 On E D þ B þ F þ 3 4E ¼ 0 On F E þ C þ H þ 2 4F ¼ 0 We now apply the derivative boundary condition at the grid points C and F by using the computational molecule for the first partial derivative with respect to x fiþ1; j fi1; j 3 @f ðx, yÞ ¼ ðfiþ1; j fi1; j Þ ¼ 6 ¼ @x 2 2h x¼1 This gives . . . . . . . . . . . . 37 3 ðB þ GÞ ¼ 6 2 3 ðE þ HÞ ¼ 6 2 Because The computational molecule for the first partial derivative with respect to x is fi1; j þ fiþ1; j 3 @f ðx, yÞ ¼ ¼ fi1; j þ fiþ1; j because h ¼ 1=3 @x 2 2h ij Applying this molecule at the boundary points C and F gives the two equations 3 ðB þ GÞ ¼ 6 so G ¼ 4 þ B 2 3 ðE þ HÞ ¼ 6 so H ¼ 4 þ E 2 619 Numerical solutions of partial differential equations Substitution of these two equations into the first six eliminates the grid points G and H to produce the six equations in six unknowns. These are written in matrix form as . . . . . . . . . . . . 0 4 B 1 B B 0 B B 1 B @ 0 0 1 0 1 0 4 1 0 1 2 4 0 0 0 0 4 1 1 0 1 4 0 1 0 2 10 1 0 1 0 A 7 B C B C 0 C CB B C B 3 C BCC B 2 C 1 C CB C ¼ B C B C B C 0 C CB D C B 7 C A @ A @ 1 E 3 A 4 F 2 38 which has solution . . . . . . . . . . . . 0 1 0 1 A 28=9 B B C B 7=3 C B C B C B C C B 8=9 C B C¼B C B D C B 28=9 C B C B C @ E A @ 7=3 A F 8=9 39 Next frame Time-dependent equations Many physical systems have their behaviour modelled by a differential equation. For example, a long thin metal bar of length L, insulated along its length, has its ends maintained at a temperature of 08C and, at time t ¼ 0, the temperature distribution is given by Tðx, 0Þ ¼ x2 2xL þ L2 The future distribution of temperature Tðx, tÞ can then be found by solving the partial differential equation (the heat equation) @ 2 Tðx, tÞ 1 @Tðx, tÞ ¼ @x2 @t K is ! called the diffusivity constant where K is the thermal conductivity and ! is the specific heat per unit volume of the metal that constitutes the rod. Apart from the physical considerations that set up the equation in the first place, the dimensions of are ½L2 T1 and are necessary to balance the dimensions on either side of the equation. subject to the given boundary and initial conditions. The constant ¼ 40 620 Programme 18 If we wished to solve the heat equation numerically as it stands then we would need to know the value of , and this would vary depending upon the specific metal used for the bar. We can overcome this problem by absorbing using a process of dimensional analysis when we transform the equation into an equation of the form @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t where the variables x and t are now dimensionless – they are measured in numbers rather than units of distance and time respectively. How this is done we shall leave to the end of the Programme. For now we are interested in numerically solving such dimensionless equations over a rectangular domain of width 1, and as usual we shall proceed by example. Next frame 41 Example 5 Solve the partial differential equation @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 and t 0 where f ð0; tÞ ¼ 1 f ðx; 0Þ ¼ 1 þ x and @f ðx, tÞ ¼0 @x x¼1 We now have a change in procedure. Hitherto, the first thing we did was to draw the domain of the function with the grid overlaid. We could do this because we knew the step lengths in the x- and y-directions from the beginning. Here, the first thing we must do is to construct the finite difference formula that will represent the differential equation because its structure will dictate the step lengths. We can immediately write down the central difference formula for the second derivative on the left of this equation. It is . . . . . . . . . . . . 42 @ 2 f ðx, tÞ fi1; j 2fi; j þ fiþ1; j @x2 ij h2 To use a central difference formula for the derivative with respect to t would require a knowledge of f ðx, tÞ for values of t < 0 and this we do not possess. Consequently, for the derivative with respect to t we use the forward difference formula. Do you remember this one? It is . . . . . . . . . . . . Numerical solutions of partial differential equations fi; jþ1 fi; j @f ðx, tÞ @t ij k 621 43 Because For a function of a single real variable the forward difference formula is given as fi; jþ1 fi; j f ðx þ hÞ f ðxÞ @f ðx, tÞ 0 and so f ðxÞ h @t ij k Using these two finite difference formulas we can write down the finite difference representation of the partial differential equation. The finite difference representation is . . . . . . . . . . . . fi1; j 2fi; j þ fiþ1; j fi; jþ1 fi; j ¼ h2 k 44 That is k ðfi1; j 2fi; j þ fiþ1; j Þ h2 It can be shown that there will be no growth of rounding errors when k 1 evaluating this equation if 2 . h 2 In compliance with this condition we shall take h ¼ 0:2 and k ¼ 0:02 k 1 so that 2 ¼ . We shall also restrict ourselves to finding solutions for t h 2 ranging from 0 to 0.16. fi; jþ1 ¼ fi; j þ The finite difference equation then reduces to . . . . . . . . . . . . fi; jþ1 ¼ 1 fi1; j þ fiþ1; j 2 Because k ðfi1; j 2fi; j þ fiþ1; j Þ and so h2 1 1 fi; jþ1 ¼ fi; j þ ðfi1; j 2fi; j þ fiþ1; j Þ ¼ ðfi1; j þ fiþ1; j Þ 2 2 Notice that this is an equation for stepping forwards in time, so that given the solution is known at t ¼ 0 then the solution at t ¼ k can be found from this equation. We can use our spreadsheet to construct the solution from this equation. Open your spreadsheet and fi; jþ1 ¼ fi; j þ 1 2 Cell A1 enter t \ x to represent the fact that the first column will contain the t-values and the first row the x-values. In cells B1 to H1 enter the values of x from 0 to 1.2 in steps of 0.2. 45 622 Programme 18 The column headed 1.2 contains grid points outside the domain of f ðx, tÞ to accommodate the derivative boundary condition. 3 4 5 6 In cells A2 to A10 enter the values of t from 0 to 0.16 in steps of 0.02. In cells B2 to B10 enter the value 1 to represent the boundary condition f ð0, tÞ ¼ 1. In cell C2 enter the formula = 1 + C1 to represent the initial condition f ðx, 0Þ ¼ 1 þ x. Copy this formula into cells D2 to G2. In cell C3 enter the formula = 0.5(B2 + D2) to represent the finite difference equation fi; jþ1 ¼ 1 fi1; j þ fiþ1; j 2 7 Copy the contents of cell C3 into the block of cells C3 to G10. @f ðx, tÞ ¼ 0 is represented Because the derivative boundary condition @x x¼1 by the central difference formula fiþ1; j fi1; j ¼ 0, the values of f ðx, tÞ at the external grid points when x ¼ 1:2 are equal to the values at the internal grid points when x ¼ 0:8. 8 In cell H2 enter the formula = F2 and copy this into cells H3 to H10 to produce the following final display: t\x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.20000 1.20000 1.20000 1.20000 1.20000 1.18750 1.18750 1.17188 1.17188 1.40000 1.40000 1.40000 1.40000 1.37500 1.37500 1.34375 1.34375 1.31250 1.60000 1.60000 1.60000 1.55000 1.55000 1.50000 1.50000 1.45313 1.45313 1.80000 1.80000 1.70000 1.70000 1.62500 1.62500 1.56250 1.56250 1.50781 2.00000 1.80000 1.80000 1.70000 1.70000 1.62500 1.62500 1.56250 1.56250 1.80000 1.80000 1.70000 1.70000 1.62500 1.62500 1.56250 1.56250 1.50781 If the diffusion equation in Frame 40 to which this solution refers is taken to represent the temperature distribution along a heated rod then this tableau displays how the temperature is changing both in time and spatially along the rod. Notice how, as the heat diffuses through the rod, the temperature changes faster at points that are further away from the end that is maintained at constant temperature. Try one yourself. Next frame 623 Numerical solutions of partial differential equations Example 6 46 The solution of the partial differential equation @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 taken in steps of h ¼ 0:2 and 0 t 0:16 in steps of k ¼ 0:02 where @f ðx, tÞ ¼ 0:5 f ð0, tÞ ¼ 2, f ðx, 0Þ ¼ 2 þ x and @x x¼1 is . . . . . . . . . . . . t\x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.20000 2.20000 2.20000 2.20000 2.20000 2.19375 2.19375 2.18594 2.18594 2.40000 2.40000 2.40000 2.40000 2.38750 2.38750 2.37188 2.37188 2.35625 2.60000 2.60000 2.60000 2.57500 2.57500 2.55000 2.55000 2.52656 2.52656 2.80000 2.80000 2.75000 2.75000 2.71250 2.71250 2.68125 2.68125 2.65391 3.00000 2.90000 2.90000 2.85000 2.85000 2.81250 2.81250 2.78125 2.78125 3.00000 3.00000 2.95000 2.95000 2.91250 2.91250 2.88125 2.88125 2.85391 Because k fi1; j 2fi; j þ fiþ1; j and so h2 1 1 fi; jþ1 ¼ fi; j þ fi1; j 2fi; j þ fiþ1; j ¼ fi1; j þ fiþ1; j 2 2 We can use our spreadsheet to construct the solution from this equation. Open your spreadsheet and fi; jþ1 ¼ fi; j þ 1 In cell A1 enter t \ x to represent the fact that the first column will contain the t-values and the first row the x-values. 2 In cells B1 to H1 enter the values of x from 0 to 1.2 in steps of 0.2. The column headed 1.2 contains grid points outside the domain of f ðx, tÞ to accommodate the derivative boundary condition. 3 4 5 6 In cells A2 to A10 enter the values of t from 0 to 0.16 in steps of 0.02. In cells B2 to B10 enter the value 2 to represent the boundary condition f ð0, tÞ ¼ 2. In cell C2 enter the formula = 2 + C1 to represent the initial condition f ðx, 0Þ ¼ 2 þ x. Copy this formula into cells D2 to G2. In cell C3 enter the formula to represent the finite difference equation 1 fi; jþ1 ¼ fi1; j þ fiþ1; j 2 The formula is . . . . . . . . . . . . 47 624 Programme 18 48 = 0.5(B2 + D2) 7 Copy the contents of cell C3 into the block of cells C3 to G10. @f ðx, tÞ ¼ 0:5 is represented @x x¼1 by the central difference formula fiþ1; j fi1; j ¼ 0:2, the values of f ðx, tÞ at the external grid points when x ¼ 1:2 are equal to Because the derivative boundary condition The values at the internal grid points when x ¼ . . . . . . . . . . . . plus . . . . . . . . . . . . 49 x ¼ 0:8 plus 0.2 8 In cell H2 enter the formula = F2 + 0.2 and copy this into cells H3 to H10 to produce the following display: t\x 0.0 0.2 0.4 0.6 0.8 1.0 1.2 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.00000 2.20000 2.20000 2.20000 2.20000 2.20000 2.19375 2.19375 2.18594 2.18594 2.40000 2.40000 2.40000 2.40000 2.38750 2.38750 2.37188 2.37188 2.35625 2.60000 2.60000 2.60000 2.57500 2.57500 2.55000 2.55000 2.52656 2.52656 2.80000 2.80000 2.75000 2.75000 2.71250 2.71250 2.68125 2.68125 2.65391 3.00000 2.90000 2.90000 2.85000 2.85000 2.81250 2.81250 2.78125 2.78125 3.00000 3.00000 2.95000 2.95000 2.91250 2.91250 2.88125 2.88125 2.85391 The Crank–Nicolson procedure 50 The forward difference formula that we used for the derivative with respect to time is not as accurate as a central difference formula. However, because we do not possess information about f ðx, tÞ for t < 0 we were forced to adopt the forward difference formula. To overcome this the Crank–Nicolson procedure makes the assumption that the partial differential equation is satisfied not just at the grid points but also at points in time halfway between two grid points. That is @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 i; jþ1=2 @t i; jþ1=2 625 Numerical solutions of partial differential equations We can then derive a central finite difference formula for the time derivative based on this intermediate point fi; jþ1 fi; j fi; jþ1 fi; j @f ðx, tÞ ¼ ¼ @t i; jþ1=2 2ðk=2Þ k Here the two grid points either side of the i, j þ 1=2th point are the i, jth and the i, j þ 1th, each separated by half the grid step in the time direction. You will note that the outcome is identical to the forward difference taken from the i, jth grid point. However, the finite difference formula that represents the partial differential equation will not be the same. For the second derivative with respect to x on the left-hand side of the equation we use a finite difference formula that is the average of the central difference formulas for the i, jth grid point and the i, j þ 1th grid point. That is @ 2 f ðx, tÞ 1 fi1; j 2fi; j þ fiþ1; j fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 ¼ þ @x2 i; jþ1=2 2 h2 h2 The partial differential equation is then represented by the central difference formula . . . . . . . . . . . . fi; jþ1 fi; j 1 fi1; j 2fi; j þ fiþ1; j fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 ¼ þ 2 h2 h2 k 51 That is k fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 2 2h k ¼ fi; j 2 fi1; j 2fi; j þ fiþ1; j 2h fi; jþ1 þ k 2h2 and different choices of h and k will result in different difference k formulas. If we choose 2 ¼ 1 this difference formula becomes 2h Unlike the previous case there is now no restriction on the value of fi1; jþ1 3fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fi; j fiþ1; j So we have three unknown quantities on the left-hand side of this equation given in terms of three known quantities on the right. We shall do an example to see exactly how this procedure operates. Next frame 626 52 Programme 18 Example 7 Use the Crank–Nicolson procedure to solve the partial differential equation @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 taken in steps of h ¼ 0:25 and 0 t 0:5 in steps of k ¼ 0:125 where: f ð0, tÞ ¼ f ð1, tÞ ¼ 0 f ðx, 0Þ ¼ xð1 xÞ We can use our spreadsheet to construct the solution from this equation. Open your spreadsheet and 1 2 3 4 5 6 In cell A1 enter t \ x to represent the fact that the first column will contain the t-values and the first row the x-values. In cells B1 to F1 enter the values of x from 0 to 1 in steps of 0.25. In cells A2 to A6 enter the values of t from 0 to 0.5 in steps of 0.125. In cells B2 to B6 enter the value 0 to represent the boundary condition f ð0, tÞ ¼ 0. In cells F2 to F6 enter the value 0 to represent the boundary condition f ð1, tÞ ¼ 0. In cell C2 enter the formula = C1(1 – C1) to represent the boundary condition f ðx, 0Þ ¼ xð1 xÞ and copy into cells D2 to F2. We now want to know the values that are going to go into the block of cells C3 to E6. We shall work on one row at a time and consider cells C3, D3 and E3 – we shall call these values A, B and C respectively. Applying the central difference formula for the differential equation fi1; jþ1 3fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fi; j fiþ1; j we find that by working along rows 2 and 3 From columns B to D: From columns C to E: From columns D to F: 0 3A þ B ¼ 0 þ 0:1875 0:25, that is 3A þ B ¼ 0:0625 A 3B þ C ¼ 0:1875 þ 0:25 0:1875, that is A 3B þ C ¼ 0:125 B 3C þ 0 ¼ 0:25 þ 0:1875 0, that is B 3C ¼ 0:0625 These equations have solution A ¼ 0:044643, B ¼ 0:071429 and C ¼ 0:044643 Enter these values into cells C3 to E3 respectively and repeat the procedure to find the values in cells C4 to E4. These are C4: . . . . . . . . . . . ., D4: . . . . . . . . . . . . and E4: . . . . . . . . . . . . 627 Numerical solutions of partial differential equations 53 C4: 0:014031, D4: 0:015306 and E4: 0:014031 Because From columns C to E: 3A þ B ¼ 0:026786 A 3B þ C ¼ 0:017857 From columns D to F: B 3C ¼ 0:026786 From columns B to D: These equations have solution A ¼ 0:014031, B ¼ 0:015306 and C ¼ 0:014031 This process is repeated until all the appropriate values have been found, giving the following display: t\x 0.00 0.25 0.50 0.75 1.00 0.000 0.125 0.250 0.375 0.500 0.000000 0.000000 0.000000 0.000000 0.000000 0.187500 0.044643 0.014031 0.002369 0.001328 0.250000 0.071429 0.015306 0.005831 0.000521 0.187500 0.044643 0.014031 0.002369 0.001328 0.000000 0.000000 0.000000 0.000000 0.000000 Try one yourself. Next frame Example 8 54 Use the Crank–Nicolson procedure to solve the partial differential equation @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 taken in steps of h ¼ 0:2 and 0 t 0:2 in steps of k ¼ 0:04 where f ð0, tÞ ¼ 2 f ð1, tÞ ¼ 1 f ðx, 0Þ ¼ 2 x2 The very first thing we must do in solving this equation numerically is . . . . . . . . . . . . 628 Programme 18 55 Derive the finite difference equation to be used Because The Crank–Nicolson procedure tells us that k fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 2 2h k ¼ fi; j 2 fi1; j 2fi; j þ fiþ1; j 2h fi; jþ1 þ so for each different ratio k we have a different finite difference 2h2 formula. Here we choose h ¼ 0:2 and k ¼ 0:04 so that k 1 ¼ and the terms in 2h2 2 fi; j do not appear. This gives the finite difference formula . . . . . . . . . . . . fi1; jþ1 4fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fiþ1; j 56 Because k fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 2 2h k ¼ fi; j 2 fi1; j 2fi; j þ fiþ1; j 2h fi; jþ1 þ and so fi; jþ1 þ 1 1 fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 ¼ fi; j fi1; j 2fi; j þ fiþ1; j 2 2 that is 1 1 fi1; jþ1 4fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fiþ1; j 2 2 giving fi1; jþ1 4fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fiþ1; j The complete solution required is . . . . . . . . . . . . 629 Numerical solutions of partial differential equations t\x 0.00 0.20 0.40 0.60 0.80 1.00 0.000 0.040 0.080 0.120 0.160 0.200 2.000000 2.000000 2.000000 2.000000 2.000000 2.000000 1.960000 1.901818 1.870083 1.847483 1.832271 1.821919 1.840000 1.767273 1.713058 1.676875 1.65221 1.635467 1.640000 1.567273 1.513058 1.476875 1.45221 1.435467 1.360000 1.301818 1.270083 1.247483 1.232271 1.221919 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 Because Using your spreadsheet to construct the solution from this equation 1 2 3 4 5 6 In cell A1 enter t \ x to represent the fact that the first column will contain the t-values and the first row the x-values. In cells B1 to G1 enter the values of x from 0 to 1 in steps of 0.2. In cells A2 to A7 enter the values of t from 0 to 0.2 in steps of 0.04. In cells B2 to B7 enter the value 2 to represent the boundary condition f ð0, tÞ ¼ 2. In cells G2 to G7 enter the value 1 to represent the boundary condition f ð1, tÞ ¼ 1. In cell C2 enter the formula = 2 – C1^2 to represent the boundary condition f ðx; 0Þ ¼ 2 x2 and copy into cells D2 to F2. We now want to know the values that are going to go into the block of cells C3 to F7. We shall work on one row at a time and consider cells C3, D3, E3 and F3 – we shall call these values A, B, C and D respectively. Applying the central difference formula for the differential equation fi1; jþ1 4fi; jþ1 þ fiþ1; jþ1 ¼ fi1; j þ fiþ1; j Then by working along rows 2 and 3 From columns B to D: 2 4A þ B ¼ 2 1:6, that is 4A þ B ¼ 5:6 From columns C to E: ............ From columns D to F: ............ From columns E to G: ............ 57 630 Programme 18 58 From columns B to D: 4A þ B ¼ 5:6 From columns C to E: From columns D to F: A 4B þ C ¼ 3:2 B 4C þ D ¼ 2:8 From columns E to G: C 4D ¼ 3:4 Because From columns B to D: From columns C to E: From columns D to F: From columns E to G: 2 4A þ B ¼ 2 1:6, that is 4A þ B ¼ 5:6 A 4B þ C ¼ 1:8 1:4, that is A 4B þ C ¼ 3:2 B 4C þ D ¼ 1:6 1:2, that is B 4C þ D ¼ 2:8 C 4D þ 1 ¼ 1:4 1:0, that is C 4D ¼ 3:4 These equations have solution A ¼ ............, B ¼ ............, C ¼ . . . . . . . . . . . . and D ¼ . . . . . . . . . . . . 59 A ¼ 1:901818 B ¼ 1:767273 C ¼ 1:567273 D ¼ 1:301818 Enter these values into cells C3 to F3 respectively and repeat the procedure to find the values for cells C4 to F4. These are C4: . . . . . . . . . . . . , D4: . . . . . . . . . . . . , E4: . . . . . . . . . . . . and F4: . . . . . . . . . . . . 60 C4: 1:870083 D4: 1:713058 E4: 1:513058 F4: 1:270083 Continuing in this way we find the complete solution as: t\x 0.00 0.20 0.40 0.60 0.80 1.00 0.000 0.040 0.080 0.120 0.160 0.200 2.000000 2.000000 2.000000 2.000000 2.000000 2.000000 1.960000 1.901818 1.870083 1.847483 1.832271 1.821919 1.840000 1.767273 1.713058 1.676875 1.65221 1.635467 1.640000 1.567273 1.513058 1.476875 1.45221 1.435467 1.360000 1.301818 1.270083 1.247483 1.232271 1.221919 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 Next frame 631 Numerical solutions of partial differential equations Dimensional analysis The equation of Frame 40 @ 2 Tðx, tÞ 1 @Tðx, tÞ for 0 x L and t ¼ @x2 @t 61 0 models the temperature distribution Tðx, tÞ along a long thin metal bar of length L. Solutions of this equation will produce values for the temperature distant x along the rod ð0 x LÞ at time t. The dimensions of the left- and right-hand sides of this equation due to the derivatives are 2 @ @ 2 ½T 1 ½L and 2 @t @x To ensure that the dimensions of the left-hand side are the same as the dimensions of the right-hand side we find that the dimensions of 1 are 1 ½L2 T This then ensures that the equation compares quantities with the same dimension. To solve this equation numerically would require a knowledge of the value of which would be different for different problems. To avoid this we transform the equation into a dimensionless form, so ensuring that the variables are measured in numbers and not in any particular dimensional units. We do this as follows. x t Define new dimensionless variables as: X ¼ (so that 0 X 1), ¼ 2 and L L define UðX, Þ ¼ T ðx½X, t½Þ then @T d @U @U ¼ ¼ 2 and @t dt @ L @ @T dX @U 1 @U ¼ ¼ @x dx @X L @X @2T @ @T @ 1 @U dX 1 @ 2 U 1 @2U ¼ ¼ therefore ¼ ¼ 2 2 2 @x @x @x @x L @X dx L @X L @X2 This means that so @ 2 Tðx; tÞ 1 @Tðx; tÞ becomes ¼ 2 @x2 @t 1 @ 2 UðX; Þ 1 @UðX; Þ 1 @UðX; Þ ¼ ¼ 2 2 2 2 L @X L @ L @ @ 2 UðX; Þ @UðX; Þ ¼ @X2 @ is the required equation in dimensionless form. 632 Programme 18 This now completes the work for this Programme. Read through the Revision summary that follows and then check your understanding against the Can you? checklist. When you are satisified that you do understand the contents of the Programme, try the Test exercises. There are no tricks and you should find them quite straightforward. Finally there are some Further problems to give additional practice. Revision summary 18 62 1 Numerical approximation to derivatives of f ðxÞ The forward difference formula f 0 ðxÞ f ðx þ hÞ f ðxÞ neglecting terms of the order h h The backward difference formula f 0 ðxÞ f ðxÞ f ðx hÞ neglecting terms of the order h h The central difference formulas 2 f 0 ðxÞ f ðx þ hÞ f ðx hÞ neglecting terms of the order h2 2h f 00 ðxÞ f ðx þ hÞ 2f ðxÞ þ f ðx hÞ h2 neglecting terms of the order h2 . Functions of two real variables If f ðx, yÞ is single-valued, then to every domain point ðx; yÞ there corresponds a single range point f ðx, yÞ. Grid values The rectangular domain of the function is overlaid by a grid whose mesh size is of h units in the x-direction and k units in the y-direction. The value of f ðx, yÞ at the ijth grid point is denoted by fi; j f ðx0 þ ih; y0 þ jkÞ The values of the expression f ðx, yÞ are required to be found at the grid points fi1; jþ1 fi1; j fi1; j1 3 fi; jþ1 fi; j fi; j1 fiþ1; jþ1 fiþ1; j fiþ1; j1 Central difference formulas for partial derivatives @f ðx, yÞ fiþ1; j fi1; j @f ðx, yÞ fi; jþ1 fi; j1 ¼ ¼ and @x ij @y ij 2h 2k 633 Numerical solutions of partial differential equations 4 Computational molecules @f ðx, yÞ @f ðx, yÞ þb ¼ c, evaluated at The partial differential equation a @x @y @f ðx, yÞ @f ðx, yÞ the ijth grid point, is a þb ¼ c and is by the central @x ij @y ij difference formula b a fiþ1; j fi1; j þ fi; jþ1 fi; j1 ¼ c 2h 2k which is in turn represented by the composite computational molecule: b 2k –a 2h 0 ij a 2h =c –b 2k 5 Numerical solutions The solutions are in the form of simultaneous linear equations in that they can be written in matrix form as Ax ¼ b with solution x ¼ A1 b. Using the Microsoft Excel spreadsheet the two functions MINVERSE(array) and MMULT(array1, array2) are employed. 6 Derivative boundary conditions The grid is extended over the boundary of the function domain by adding additional points outside the domain. 7 Second-order partial differential equations The most general form of a second-order partial differential equation is aðx; yÞ @2f @2f @2f @f @f þ cðx; yÞ 2 þ dðx; yÞ þ eðx; yÞ þ gðx; yÞ ¼ 0 þ bðx; yÞ 2 @x @x@y @y @x @y Elliptic equations If b2 4ac < 0 then the partial differential equation is called an elliptic equation Hyperbolic equations If b2 4ac > 0 then the partial differential equation is called an hyperbolic equation Parabolic equations If b2 4ac ¼ 0 then the partial differential equation is called a parabolic equation. 634 Programme 18 8 Second partial derivatives – central difference formulas fi1; j 2fi; j þ fiþ1; j @ 2 f ðx, yÞ @x2 ij h2 fi; j1 2fi; j þ fi; jþ1 @ 2 f ðx, yÞ and 2 @y k2 ij 9 Time-dependent equations To use a central difference formula for the derivative with respect to t would require a knowledge of f ðx, tÞ for values of t < 0 and this we do not possess. Consequently, for the derivative with respect to t we use the forward difference formula fi; jþ1 fi; j @f ðx, tÞ @t k ij So the partial differential equation fi; jþ1 ¼ fi; j þ @ 2 f ðx, tÞ @f ðx, tÞ ¼ becomes @x2 @t k ðfi1; j 2fi; j þ fiþ1; j Þ h2 where it can be shown that there will be no growth of rounding errors when evaluating this equation if k 1 . h2 2 10 The Crank–Nicolson procedure The Crank–Nicolson procedure makes the assumption that the partial differential equation can be satisfied at points in time halfway between two grid points. That is @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 i; jþ1=2 @t i; jþ1=2 This gives fi; jþ1 fi; j fi; jþ1 fi; j @f ðx, tÞ ¼ ¼ @t 2ðk=2Þ k i; jþ1=2 2 @ f ðx, tÞ 1 fi1; j 2fi; j þ fiþ1; j fi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 ¼ þ @x2 i; jþ1=2 2 h2 h2 So that k ðfi1; jþ1 2fi; jþ1 þ fiþ1; jþ1 Þ 2h2 k ¼ fi; j 2 ðfi1; j 2fi; j þ fiþ1; j Þ 2h k with no restriction on the value of 2 . 2h fi; jþ1 þ 635 Numerical solutions of partial differential equations Can you? 63 Checklist 18 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Derive the finite difference formulas for the first partial derivatives of a function of two real variables and construct the central finite difference formula to represent a first-order partial differential equation? Yes No . Draw a rectangular grid of points overlaid on the domain of a function of two real variables and evaluate the function at the boundary grid points? Yes No . Construct the computational molecule for a first-order partial differential equation in two real variables and use the molecule to evaluate the solutions to the equation at the grid points interior to the boundary? Yes No . Describe the solution as a set of simultaneous linear equations and use matrices to represent them? Yes No . Invert the coefficient matrix and thereby represent the solution to the partial differential equation as a column matrix? Yes No . Take account of a boundary condition in the form of the derivative normal to the boundary? Yes No . Obtain the central finite difference formulas for the second derivatives of a function of two real variables and construct finite difference formulas for second-order partial differential equations? Yes No . Use the forward difference formula for the first time derivatives in partial differential equations involving time and distance? Yes No Frames 1 to 4 5 to 9 10 and 11 12 and 13 14 to 19 20 to 28 29 to 39 40 to 49 636 Programme 18 . Use the Crank–Nicolson procedure for a partial differential equation involving a first time derivative? Yes No . Appreciate the use of dimensional analysis in the conversion of a partial differential equation modelling a physical system into a dimensionless equation? Yes No 50 to 61 Test exercise 18 64 1 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ 4 ¼ 5 5 @x @y for 0 x 1 with a step length h ¼ 1=4 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 3x 4, f ðx, 1Þ ¼ 3x þ 1, f ð0, yÞ ¼ 5y 4 and f ð1, yÞ ¼ 5y 1 2 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ 10 þ8 ¼ 10 @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where @f ðx, yÞ f ðx, 0Þ ¼ 7x þ 5, f ðx, 1Þ ¼ 7x 5, f ð0, yÞ ¼ 5 10y and ¼7 @x x¼1 3 Name the type of equation in each of the following. @f ðx, yÞ @f ðx, yÞ 3y ¼ 4xy (a) 2 @x @y (b) @f ðx, yÞ @ 2 f ðx, yÞ @f ðx, yÞ x þ ¼ @x @x@y @y y @ 2 f ðx, yÞ @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ 2 ¼0 2 @x @x@y @y 2 @ @f ðx, yÞ @f ðx, yÞ x2 ¼ 3 (d) þ @x @x @y y (c) (e) 3 4 @ 2 f ðx, yÞ @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ 2 ¼ 3xy @x2 @x@y @y 2 Solve the following equation numerically. @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ 2 for 0 x 1 and 0 y 1 @x2 @y 2 with step lengths h ¼ k ¼ 1=3 where f ðx, 0Þ ¼ f ðx, 1Þ ¼ x 2, f ð0, yÞ ¼ y 2 y 2 and @f ðx, yÞ ¼1 @x x¼1 60 Numerical solutions of partial differential equations 5 Solve the following equation numerically using the forward difference approximation for the first derivative with respect to time. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:2 with step length k ¼ 0:02 where @f ðx, tÞ 2 ¼ 0:25 f ðx, 0Þ ¼ x , f ð0; tÞ ¼ 0 and @x x¼1 6 Solve the following equation numerically using the Crank–Nicolson procedure. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:2 with step length k ¼ 0:04 where 637 f ðx, 0Þ ¼ x2 x þ 1 and f ð0; tÞ ¼ f ð1; tÞ ¼ 1 Further problems 18 1 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ þ ¼0 2 @x @y for 0 x 1 with a step length h ¼ 1=4 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ x 3, f ðx, 1Þ ¼ x 1, f ð0, yÞ ¼ 2y 3 and f ð1, yÞ ¼ 2y 2 2 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ 9 7 ¼ 7 @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 7x þ 4, f ðx, 1Þ ¼ 7x þ 14, f ð0, yÞ ¼ 10y þ 4 and f ð1, yÞ ¼ 10y þ 11 3 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ x þ ðy þ 1Þ ¼0 @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ x 1, f ðx, 1Þ ¼ ðx 2Þ=2, f ð0, yÞ ¼ 1 and f ð1, yÞ ¼ y=ðy þ 1Þ 65 638 Programme 18 4 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ ¼ x2 þ y 2 @y @x for 0 x 1 with a step length h ¼ 1=4 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 0, f ðx, 1Þ ¼ xðx 1Þ, f ð0, yÞ ¼ 0 and f ð1, yÞ ¼ yð1 yÞ 5 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ 5 ¼ 4 3 @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 7x þ 15, f ðx, 1Þ ¼ 7x þ 20, f ð0, yÞ ¼ 5y þ 15 @f ðx, yÞ and ¼7 @x x¼1 6 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ þ 12 ¼ 19 11 @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 5x þ 21, f ðx, 1Þ ¼ 5x þ 18, f ð0, yÞ ¼ 21 3y @f ðx, yÞ and ¼5 @x x¼1 7 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ y ¼ 8x2 2x @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 2x2 þ 4, f ðx, 1Þ ¼ 2x2 3x þ 4, f ð0, yÞ ¼ 4 @f ðx, yÞ and ¼ 4 3y 2 @x x¼1 8 Solve the following equation numerically. @f ðx, yÞ @f ðx, yÞ þx ¼ x4 y 4 y @x @y for 0 x 1 with a step length h ¼ 1=3 and 0 y 1 with a step length k ¼ 1=3 where f ðx, 0Þ ¼ 0, f ðx, 1Þ ¼ xðx þ 1Þðx 1Þ, f ð0, yÞ ¼ 0 @f ðx, yÞ and ¼ yð3 y 2 Þ @x x¼1 Numerical solutions of partial differential equations 9 Solve the following equation numerically. @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ 4 @x2 @y 2 for 0 x 1 and 0 y 1 with step lengths h ¼ k ¼ 1=3 where @f ðx, yÞ f ðx, 0Þ ¼ 3x2 , f ðx, 1Þ ¼ 3x2 5, f ð0, yÞ ¼ 5y 2 and ¼6 @x x¼1 10 Solve the following equation numerically. @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ 2ðx þ yÞ @x2 @y 2 for 0 x 1 and 0 y 1 with step lengths h ¼ k ¼ 1=3 where f ðx, 0Þ ¼ 1, f ðx, 1Þ ¼ x2 þ 3x 1, f ð0, yÞ ¼ 1 @f ðx, yÞ and ¼ y 2 þ 4y @x x¼1 11 Solve the following equation numerically. @ 2 f ðx, yÞ @ 2 f ðx, yÞ þ ¼ ð2 x2 Þ cos y @x2 @y 2 for 0 x 1 and 0 y 1 with step lengths h ¼ k ¼ 1=3 where @f ðx, yÞ 2 2 : f ðx, 0Þ ¼ x , f ðx, 1Þ ¼ 0 540302x , f ð0, yÞ ¼ 0 and ¼ 2x cos y @x x¼1 12 Solve the following equation numerically. @ 2 f ðx, yÞ @ 2 f ðx, yÞ ¼ 4ðx yÞ @x2 @y 2 for 0 x 1 and 0 y 1 with step lengths h ¼ k ¼ 1=3 where f ðx, 0Þ ¼ x3 , f ðx, 1Þ ¼ ðx þ 1Þðx2 þ 1Þ, f ð0, yÞ ¼ y 3 @f ðx, yÞ and ¼ y 2 þ 2y þ 3 @x x¼1 13 Given the central difference formula @ 2 f ðx, yÞ 1 ¼ 2 fi1; j1 fiþ1; j1 fi1; jþ1 þ fiþ1; jþ1 @x@y ij 4h where the step length in both directions is h, construct the computational molecule for this formula. Solve the equation @ 2 f ðx, yÞ ¼1 @x@y for 0 x 1 and 0 y 1 with step lengths h ¼ 1=3 where @f ðx, yÞ f ðx, 0Þ ¼ 0, f ðx, 1Þ ¼ x, f ð0, yÞ ¼ 0 and ¼y @x x¼1 639 640 Programme 18 14 Given the central difference formula @ 2 f ðx, yÞ 1 ¼ 2 fi1; j1 fiþ1; j1 fi1; jþ1 þ fiþ1; jþ1 @x@y ij 4h where the step length in both directions is h, construct the computational molecule for this formula. Solve the equation @ 2 f ðx, yÞ ¼ 2ðx yÞ @x@y for 0 x 1 and 0 y 1 with step lengths h ¼ 1=3 where @f ðx, yÞ f ðx, 0Þ ¼ 0, f ðx, 1Þ ¼ xðx 1Þ, f ð0, yÞ ¼ 0 and ¼ yð2 yÞ @x x¼1 15 Solve the following equation numerically using the forward difference approximation for the first derivative with respect to time. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:2 with a step length k ¼ 0:02 where @f ðx, tÞ ¼1 f ðx, 0Þ ¼ xðx 1Þ, f ð0; tÞ ¼ 2t and @x x¼1 16 Solve the following equation numerically using the forward difference approximation for the first derivative with respect to time. @ 2 f ðx, tÞ 1 @f ðx, tÞ ¼ : @x2 0 1 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:2 with a step length k ¼ 0:02 where @f ðx, tÞ ¼ 0:54et=10 f ðx, 0Þ ¼ sin x, f ð0; tÞ ¼ 0 and @x x¼1 17 Solve the following equation numerically using the forward difference approximation for the first derivative with respect to time. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:2 with a step length k ¼ 0:02 where @f ðx, tÞ : ¼ 2:41e0 41t f ðx, 0Þ ¼ 3 sinð0:64xÞ, f ð0; tÞ ¼ 0 and @x x¼1 18 Solve the following equation numerically using the Crank–Nicolson procedure. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:2 and 0 t 0:6 with a step length k ¼ 0:04 where f ðx, 0Þ ¼ x2 þ x 1 and f ð0, tÞ ¼ 2t 1, f ð1; tÞ ¼ 1 þ 2t Numerical solutions of partial differential equations 19 Solve the following equation numerically using the Crank–Nicolson procedure. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:1 and 0 t 0:14 with a step length k ¼ 0:02 where f ðx, 0Þ ¼ 10xðx 1Þ and f ð0; tÞ ¼ f ð1; tÞ ¼ 20t 20 Solve the following equation numerically using the Crank–Nicolson procedure. @ 2 f ðx, tÞ @f ðx, tÞ ¼ @x2 @t for 0 x 1 with a step length h ¼ 0:1 and 0 t 0:6 with a step length k ¼ 0:04 where f ðx, 0Þ ¼ 100 sin x and f ð0; tÞ ¼ f ð1; tÞ ¼ 0 641 Programme 19 Frames 1 to 83 Multiple integration 1 Learning outcomes When you have completed this Programme you will be able to: . Evaluate double and triple integrals and apply them to the determination of the areas of plane figures and the volumes of solids . Understand the role of the differential of a function of two or more real variables . Determine exact differentials in two real variables and their integrals . Evaluate the area enclosed by a closed curve by contour integration . Evaluate line integrals and appreciate their properties . Evaluate line integrals around closed curves within a simply connected region . Link line integrals to integrals along the x-axis . Link line integrals to integrals along a contour given in parametric form . Discuss the dependence of a line integral between two points on the path of integration . Determine exact differentials in three real variables and their integrals . Demonstrate the validity and use of Green’s theorem Prerequisite: Engineering Mathematics (Sixth Edition) Programme 23 Multiple integrals 642 643 Multiple integration 1 Introduction The introductory work on double and triple integrals was covered in detail in Programme 23 of Engineering Mathematics (Sixth Edition) and another look at the main points before launching forth on the current development could well be worth while. You will no doubt recognise the following. 1 Double integrals ð y2 ð x2 f ðx, yÞ dx dy y1 x1 is a double integral and is evaluated from the inside outwards, i.e. ð y2 ð x 2 y1 f ðx, yÞ dx 1 2 dy x1 A double integral is sometimes expressed in the form ð x2 ð y2 dy f ðx, yÞ dx y1 x1 in which case, we evaluate from the right-hand end, i.e. ð y2 ð x2 dy y1 y1 2 1 x1 ð y2 ð x 2 then f ðx, yÞ dx 2 f ðx, yÞ dx dy x1 Triple integrals Triple integrals follow the same procedure. ð z2 ð y2 ð x2 f ðx, y, zÞ dx dy dz is evaluated in the order z1 y1 x1 ð z2 ð y2 ð x2 z1 y1 x1 f ðx; y; zÞ dx 1 3 2 dy dz 1 644 Programme 19 3 Applications (a) Areas of plane figures y y1 = f (x) P (x, y) y2 y2 = F(x) δy y1 O a x b δx x Area of element A ¼ xy y¼y X2 Area of strip xy y¼y1 Area of all such strips x¼b X (y¼y X2 x¼a y¼y1 If x ! 0 and y ! 0, A ¼ ð b ð y2 a ) xy dy dx y1 (b) Areas of plane figures bounded by a polar curve r ¼ f ðÞ and radius vectors at ¼ 1 and ¼ 2 δr y θ = θ2 r = f (θ) r δθ δθ O r δθ θ r δr Small arc of circle of radius r, subtending angle at centre. θ = θ1 ; Arc ¼ r x Area of element A r r Area of thin sector r¼f ðÞ X r r r¼0 ; Total area of all such sectors ðÞ X ¼ 2 r¼f X ¼1 ; If r ! 0 and ! 0, A ¼ ð 2 ð r¼f ðÞ 1 0 r¼0 r dr d ) r r 645 Multiple integration 1 (c) Volume of solids z x1 z z = f(x, y) O y1 y2 y x2 δz O x x Volume of element V ¼ x y z Volume of column z¼fX ðx, yÞ x y z z¼0 Volume of slice y¼y X2 (z¼f ðx, yÞ X y¼y1 ) x y z z¼0 ; Total volume V sum of all such slices i.e. V ðx, yÞ x¼x X2 z¼fX X2 y¼y x ¼ x1 y¼y1 x y z z¼0 Then, if x ! 0, y ! 0, z ! 0, ð x2 ð y2 ð z¼f ðx, yÞ V¼ dz dy dx x1 y1 0 If z ¼ f ðx, yÞ, this becomes ð x 2 ð y2 V¼ f ðx, yÞ dy dx x1 y1 δy δx y 646 2 Programme 19 4 Revision examples As a means of ‘warming up’, let us work through one or two straightforward examples on the previous work. Example 1 Find the area of the plane figure bounded by the curves y1 ¼ ðx 1Þ2 and y2 ¼ 4 ðx 3Þ2 : The first thing, as always, is to sketch the curves – each of which is a parabola – and to determine their points of intersection. y y1 = (x – 1)2 y2 y2 = 4 – (x – 3)2 y1 O x x δx Points of intersection: ðx 1Þ2 ¼ 4 ðx 3Þ2 x2 2x þ 1 ¼ 4 x2 þ 6x 9 i.e. x2 4x þ 3 ¼ 0 ; ðx 1Þðx 3Þ ¼ 0 ; x ¼ 1 or x ¼ 3. Now we have all the information to determine the required area, which is ............ 3 A ¼ 2 23 square units Because ð x¼3 ð y¼4ðx3Þ2 ð x¼3 ð y2 dy dx ¼ dy dx A¼ x¼1 ¼ ð3 y1 x¼1 y¼ðx1Þ2 f4 ðx 3Þ2 ðx 1Þ2 g dx ¼ 2 1 ð3 1 ðx2 4x þ 3Þ dx 3 3 x 2x2 þ 3x ¼ 2 23 square units ¼ 2 3 1 Now for another. 647 Multiple integration 1 Example 2 A rectangular plate is bounded by the x and y axes and the lines x ¼ 6 and y ¼ 4: The thickness t of the plate at any point is proportional to the square of the distance of the point from the origin. Determine the total volume of the plate. First of all draw the figure and build up the appropriate double integral. Do not evaluate it yet. The expression is therefore V ¼ ............ V¼ ð x¼6 ð y¼4 x¼0 kðx2 þ y 2 Þ dy dx 4 y¼0 y Thickness t of plate at P is t ¼ k OP2 ¼ kðx2 þ y 2 Þ y O Element of area ¼ y x x x ; Element of volume at P kðx2 þ y 2 Þ y x ð x¼6 ð y¼4 ; Total volume V ¼ kðx2 þ y 2 Þ dy dx x¼0 y¼0 Now we can evaluate the integral. We start from the inside with ð y¼4 kðx2 þ y 2 Þ dy, y¼2 remembering that for this integral (volume of the strip) x is constant. This gives . . . . . . . . . . . . k 64 4x2 þ 3 Because y¼4 y3 64 2 k ðx þ y Þ dy ¼ k x y þ ¼ k 4x þ 3 3 y¼0 0 ð6 64 dx ¼ . . . . . . . . . . . . V¼k 4x2 þ 3 0 ð4 Then 2 2 2 5 648 Programme 19 6 V ¼ 416 k cubic units That was easy enough. Notice that an alternative interpretation of this problem could be that of a uniform lamina with a variable density ¼ k x2 þ y 2 at any point ðx, yÞ. Now for one in polar coordinates. Example 3 Express as a double integral the area enclosed by one loop of the curve r ¼ 3 cos 2 and evaluate the integral (refer to Engineering Mathematics (Sixth Edition), Programme 23, Frame 10). θ=π 4 Consider the half loop shown. r = 3 cos2θ θ O r 3 First set up the double integral which is . . . . . . . . . . . . 7 A¼ ð ¼ =4 ð r¼3 cos 2 r dr d r¼0 ¼0 θ=π 4 r δθ r θ O δr 3 x Area of element ¼ r r ; Area of sector r¼3X cos 2 r r r¼0 ; Area of half loop ¼=4 cos 2 X r¼3X ¼0 r r r¼0 If r ! 0 and ! 0, ð ¼=4 ð r¼3 cos 2 A¼ r dr d ¼0 r¼0 Now finish it off to find the area of the whole loop, which is ............ x 649 Multiple integration 1 8 9 square units 8 Because A¼ ð ¼=4 ð r¼3 cos 2 r dr d r¼0 ¼0 ð =4 2 3 cos 2 r d 2 0 0 ð 9 =4 cos2 2 d ¼ 2 0 ð 9 =4 ¼ ð1 þ cos 4Þ d 4 0 9 sin 4 =4 þ ¼ 4 4 0 ¼ ¼ 9 16 This is the area of a half loop. Required area ¼ 9 square units 8 Now here is another. Example 4 Find the volume of the solid bounded by the planes z ¼ 0, x ¼ 1, x ¼ 3, y ¼ 1, y ¼ 2 and the surface z ¼ x2 y 2 . As always, we start off by sketching the figure. When you have done that, check the result with the next frame. 9 z y z = x2 y2 δx δz δy O x 650 Programme 19 We now build up the integral which will give us the volume of the solid. Element of volume V ¼ x y z Volume of column 2 2 z¼x Xy x y z z¼0 Volume of slice 8 2 2 y¼2 <z¼x Xy X 9 = x y z : z¼0 ; 8 9 2 2 y¼2 z¼x x¼3 <X = X Xy Volume of solid x y z : ; x¼1 y¼1 z¼0 y¼1 When x ! 0, y ! 0, z ! 0, ð x¼3 ð y¼2 ð z¼x2 y2 dz dy dx V¼ x¼1 y¼1 z¼0 V ¼ ............ Evaluating this, 10 V ¼ 20 29 cubic units Because, starting with the innermost integral ð3 ð2 ð x¼3 ð y¼2 x2 y2 z dy dx ¼ x2 y 2 dy dx V ¼ x¼1 ¼ ð3 1 y¼1 x2 y 3 3 y¼2 dx y¼1 1 0 ¼ 1 ð3 7x2 dx ¼ 20 29 1 3 Now that we have revised the basics, let us move on to something rather different Differentials 11 It is convenient in various branches of the calculus to denote small increases in value of a variable by the use of differentials. The method is particularly useful in dealing with the effects of small finite changes and shortens the writing of calculus expressions. 651 Multiple integration 1 We are already familiar with the diagram from which finite changes y and x in a function y ¼ f ðxÞ are depicted. y = f (x) y (x1, y1) (x0, y0) δy δx O x0 x0 + δx x The increase in y from P to Q ¼ MQ ¼ y ¼ f ðx0 þ xÞ f ðx0 Þ MT If PT is the tangent at P, then MQ ¼ MT þ TQ. Also ¼ f 0 ðx0 Þ x ; MT ¼ f 0 ðx0 Þx ; MQ ¼ y ¼ f 0 ðx0 Þ x þ TQ and, if Q is close to P, then y f 0 ðx0 Þx We define the differentials dy and dx as finite quantities such that dy ¼ f 0 ðx0 Þ dx y = f (x) y dy δy dx = δx O x0 x0 + δx x Note that the differentials dy and dx are finite quantities – not necessarily zero – and can therefore exist alone. Note too that dx ¼ x: From the diagram, we can see that y is the increase in y as we move from P to Q along the curve. dy is the increase in y as we move from P to T along the tangent. As Q approaches P, the difference between y and dy decreases to zero. The use of differentials simplifies the writing of many relationships and is based on the general statement dy ¼ f 0 ðxÞ dx: 652 Programme 19 For example (a) y ¼ x5 then dy ¼ 5x4 dx (b) y ¼ sin 3x then dy ¼ 3 cos 3x dx (c) y ¼ e 4x then dy ¼ 4 e4x dx (d) y ¼ cosh 2x then dy ¼ 2 sinh 2x dx Note that when the left-hand side is a differential dy the right-hand side must also contain a differential. Remember therefore to include the ‘dx’ on the right-hand side. The product and quotient rules can also be expressed in differentials. d dv du ðuvÞ ¼ u þv becomes dðuvÞ ¼ u dv þ v du dx dx dx du dv u v d u u v du u dv ¼ dx 2 dx becomes d ¼ dx v v v2 v So, if and if 12 y ¼ e2x sin 4x, cos 2t y¼ t2 dy ¼ . . . . . . . . . . . . dy ¼ . . . . . . . . . . . . y ¼ e2x sin 4x, cos 2t y¼ ; t2 dy ¼ 2e2x ð2 cos 4x þ sin 4xÞ dx 2 dy ¼ 3 ft sin 2t þ cos 2tg dt t That was easy enough. Let us now consider a function of two independent variables, z ¼ f ðx, yÞ: If z ¼ f ðx, yÞ then z þ z ¼ f ðx þ x; y þ yÞ ; z ¼ f ðx þ x, y þ yÞ f ðx, yÞ Expanding z in terms of x and y, gives z ¼ Ax þ By þ higher powers of x and y, where A and B are functions of x and y. If y remains constant, i.e. y = 0, then z ¼ A x þ higher powers of x ; z A x @z @x Similarly, if x remains constant, i.e. x ¼ 0, then z B z ¼ B y þ higher powers of y ; y @z ; If y ! 0, then B ¼ @y @z @z x þ y þ higher powers of small quantities ; z ¼ @x @y @z @z ; z ¼ x þ y @x @y ; If x ! 0, then A ¼ 653 Multiple integration 1 In terms of differentials, this result can be written @z @z dx þ dy If z ¼ f ðx, yÞ. then dz ¼ @x @y The result can be extended to functions of more than two independent variables. @z @z @z dx þ dy þ dw If z ¼ f ðx, y, wÞ, dz ¼ @x @y @w Make a note of these results in differential form as shown. Exercise Determine the differential dz for each of the following functions. 1 z ¼ x2 þ y 2 2 z ¼ x3 sin 2y 3 z ¼ ð2x 1Þ e 3y 4 z ¼ x2 þ 2y 2 þ 3w2 5 z ¼ x3 y 2 w. Finish all five and then check the results. 1 dz ¼ 2ðx dx þ y dyÞ 2 dz ¼ x2 ð3 sin 2y dx þ 2x cos 2y dyÞ 3 dz ¼ e3y f2 dx þ ð6x 3Þdyg 4 dz ¼ 2ðx dx þ 2y dy þ 3w dwÞ 5 dz ¼ x2 yð3yw dx þ 2xw dy þ xy dwÞ 13 Now move on Exact differential We have just established that if z ¼ f ðx, yÞ @z @z dx þ dy dz ¼ @x @y We now work in reverse. Any expression dz ¼ P dx þ Q dy, where P and Q are functions of x and y, is an exact differential if it can be integrated to determine z. @z @z and Q ¼ ; P¼ @x @y @P @2z @Q @2z @2z @2z ¼ and ¼ and we know that ¼ . @y @y @x @x @x @y @y @x @x @y @P @Q ¼ and this is the test Therefore, for dz to be an exact differential @y @x we apply. Now 14 654 Programme 19 Example 1 dz ¼ ð3x2 þ 4y 2 Þ dx þ 8xy dy. If we compare the right-hand side with P dx þ Q dy, then P ¼ 3x2 þ 4y 2 Q ¼ 8xy @P @Q ¼ @y @x @P ¼ 8y @y @Q ; ¼ 8y @x ; ; dz is an exact differential Similarly, we can test this one. Example 2 dz ¼ ð1 þ 8xyÞ dx þ 5x2 dy: From this we find . . . . . . . . . . . . 15 dz is not an exact differential Because dz ¼ ð1 þ 8xyÞ dx þ 5x2 dy ; P ¼ 1 þ 8xy Q ¼ 5x2 @P @Q 6¼ @y @x @P ¼ 8x @y @Q ¼ 10x ; @x ; ; dz is not an exact differential. Exercise Determine whether each of the following is an exact differential. 16 1 dz ¼ 4x3 y 3 dx þ 3x4 y 2 dy 2 dz ¼ ð4x3 y þ 2xy 3 Þ dx þ ðx4 þ 3x2 y 2 Þ dy 3 dz ¼ ð15y 2 e3x þ 2xy 2 Þ dx þ ð10ye3x þ x2 yÞ dy 4 dz ¼ ð3x2 e2y 2y 2 e3x Þ dx þ ð2x3 e2y 2ye3x Þ dy 5 dz ¼ ð4y 3 cos 4x þ 3x2 cos 2yÞ dx þ ð3y 2 sin 4x 2x3 sin 2yÞ dy. 1 Yes 2 Yes 3 No 4 No 5 Yes We have just tested whether certain expressions are, in fact, exact differentials – and we said previously that, by definition, an exact differential can be integrated. But how exactly do we go about it? The following examples will show. 655 Multiple integration 1 Integration of exact differentials @z @z and Q ¼ dz ¼ P dx þ Q dy where P ¼ @x @y ð ð ; z ¼ P dx and also z ¼ Q dy Example 1 dz ¼ ð2xy þ 6xÞ dx þ ðx2 þ 2y 3 Þ dy: ð @z ¼ 2xy þ 6x ; z ¼ ð2xy þ 6xÞ dx P¼ @x ; z ¼ x2 y þ 3x2 þ f ðyÞ where f ðyÞ is an arbitrary function of y only, and is akin to the constant of integration in a normal integral. ð @z 2 3 ¼ x þ 2y ; z ¼ ðx2 þ 2y 3 Þ dy Also Q ¼ @y ; z ¼ ............ z ¼ x2 y þ 17 y4 þ FðxÞ where FðxÞ is an arbitrary function of x only 2 So the two results tell us z ¼ x2 y þ 3x2 þ f ðyÞ and z ¼ x2 y þ ð1Þ 4 y þ FðxÞ 2 ð2Þ For these two expressions to represent the same function, then y4 already in ð2Þ 2 2 FðxÞ in ð2Þ must be 3x already in ð1Þ f ðyÞ in ð1Þ must be and ; z ¼ x2 y þ 3x2 þ y4 2 Example 2 Integrate dz ¼ ð8e4x þ 2xy 2 Þ dx þ ð4 cos 4y þ 2x2 yÞ dy. Argue through the working in just the same way, from which we obtain z ¼ ............ 656 Programme 19 18 z ¼ 2e4x þ x2 y 2 þ sin 4y Here it is. dz ¼ ð8e4x þ 2xy 2 Þ dx þ ð4 cos 4y þ 2x2 yÞ dy ð @z ¼ 8e4x þ 2xy 2 P¼ ; z ¼ ð8e4x þ 2xy 2 Þdx @x ; z ¼ 2e4x þ x2 y 2 þ f ðyÞ ð @z ¼ 4 cos 4y þ 2x2 y ; z ¼ ð4 cos 4y þ 2x2 yÞ dy Q¼ @y ; z ¼ sin 4y þ x2 y 2 þ FðxÞ For (1) and (2) to agree, f ðyÞ ¼ sin 4y and FðxÞ ¼ 2e ð1Þ ð2Þ 4x ; z ¼ 2 e4x þ x2 y 2 þ sin 4y They are all done in the same way, so you will have no difficulty with the short exercise that follows. On you go. Exercise Integrate the following exact differentials to obtain the function z. 1 dz ¼ ð6x2 þ 8xy 3 Þ dx þ ð12x2 y 2 þ 12y 3 Þ dy 2 dz ¼ ð3x2 þ 2xy þ y 2 Þ dx þ ðx2 þ 2xy þ 3y 2 Þ dy 3 dz ¼ 2ðy þ 1Þe2x dx þ ðe2x 2yÞ dy 4 dz ¼ ð3y 2 cos 3x 3 sin 3xÞ dx þ ð2y sin 3x þ 4Þ dy 5 dz ¼ ðsinh y þ y sinh xÞdx þ ðx cosh y þ cosh xÞ dy. Finish all five before checking with the next frame. 19 1 z ¼ 2x3 þ 4x2 y 3 þ 3y 4 2 z ¼ x3 þ x2 y þ xy 2 þ y 3 3 z ¼ e2x ð1 þ yÞ y 2 4 z ¼ y 2 sin 3x þ cos 3x þ 4y 5 z ¼ x sinh y þ y cosh x: In the last one, of course, we find that the two expressions for z agree without any further addition of f ðyÞ or FðxÞ. We shall be meeting exact differentials again later on, but for the moment let us deal with something different. On then to the next frame 657 Multiple integration 1 Area enclosed by a closed curve One of the earliest applications of integration is finding the area of a plane figure bounded by the x-axis, the curve y ¼ f ðxÞ and ordinates at x ¼ x1 and x ¼ x2 : y y = f (x) A1 ¼ A1 O ð x2 y dx ¼ ð x2 x1 x x2 x1 f ðxÞ dx x1 If points A and B are joined by another curve, y ¼ FðxÞ y y = F (x) A2 ¼ FðxÞ dx x1 A2 O ð x2 x1 x x2 Combining the two figures, we have y y = f (x) A ¼ A1 A2 ð x2 ð x2 ; A¼ f ðxÞ dx FðxÞ dx A y = F (x) O x1 x2 x1 x1 x It is convenient on occasions to arrange the limits so that the integration follows the path round the enclosed area in a regular order. y y = f (x) c2 c1 y = F(x) O x1 x2 x 20 658 Programme 19 For example ð x2 FðxÞ dx gives A2 as before, but integrating from B to A along c2 with x1 y ¼ f ðxÞ, i.e. ð x1 x2 ð x1 f ðxÞ dx, is the integral for A1 with the sign changed, i.e. x2 f ðxÞ dx ¼ ð x2 f ðxÞ dx x1 ; The result A ¼ A1 A2 ¼ ð x2 f ðxÞ dx ð x2 FðxÞ dx becomes x1 x1 A ¼ ............ 21 A¼ ð x2 FðxÞ dx ð x1 x1 i:e: A¼ f ðxÞ dx x2 ð x2 FðxÞ dx þ ð x1 x1 f ðxÞ dx x2 If we proceed round the boundary in an anticlockwise manner, the enclosed area is kept on the left-hand side and the resulting area is considered positive. If we proceed round the boundary in a clockwise manner, the enclosed area remains on the right-hand side and the resulting area is negative. The final result above can be written in the form þ A ¼ y dx þ where the symbol indicates that the integral is to be evaluated round the closed boundary in the positive (i.e. anticlockwise) direction ð x2 ð x1 þ FðxÞ dx þ f ðxÞ dx ; A ¼ y dx ¼ x1 x2 ðalong c1 Þ y ðalong c2 Þ y = f (x) c2 A c 1 y = F (x) O x1 Let us apply this result to a very simple case. x2 x 659 Multiple integration 1 Example 1 Determine the area enclosed by the graphs of y ¼ x3 and y ¼ 4x for x 0: First we need to know the points of intersection. These are ............ 22 x ¼ 0 and x ¼ 2 y y = x3 y= 4x c2 A c1 O x We integrate in an anticlockwise manner c1 : y ¼ x3 , c2 : y ¼ 4x, limits x ¼ 2 to x ¼ 0. þ A ¼ y dx ¼ . . . . . . . . . . . . limits x ¼ 0 to x ¼ 2 A ¼ 4 square units Because ð 2 þ ð0 A ¼ y dx ¼ x3 dx þ 4x dx (0 2 2 ) 0 x4 ¼4 þ 2x2 ¼ 4 0 2 Another example. Example 2 Find the area of the triangle with vertices ð0, 0Þ, ð5, 3Þ and ð2, 6Þ. y (2, 6) 6 The equation of OA is . . . . . . . . . . . . 3 O (5, 3) 2 5 x BA is . . . . . . . . . . . . OB is . . . . . . . . . . . . 23 660 Programme 19 24 OA is y ¼ 35 x BA is y ¼ 8 x OB is y ¼ 3x þ Then A ¼ y dx y 6 c2 c3 ¼ ............ A 3 Write down the component integrals with appropriate limits. c1 O 5 x 2 þ 25 A ¼ y dx ¼ ð 5 3 x dx þ 05 ð2 ð8 xÞ dx þ ð0 5 3x dx 2 The limits chosen must progress the integration round the boundary of the figure in an anticlockwise manner. Finishing off the integration, we have A ¼ ............ 26 A ¼ 12 square units The actual integration is easy enough. The work we have just done leads us on to consider line integrals, so let us make a fresh start in the next frame Line integrals 27 y y Ft δs Ft F O Fn x O x If a field exists in the x–y plane, producing a force F on a particle at K, then F can be resolved into two components Ft along the tangent to the curve AB at K Fn along the normal to the curve AB at K. 661 Multiple integration 1 The work done in moving the particle through a small distance s from K to L along the curve is then approximately Ft s. So the total work done in moving a particle along the curve from A to B is given by ............ Lim X ð Ft s ¼ Ft ds from A to B 28 s!0 ð Ft ds where A and B are the end points of This is normally written AB ð Ft ds where the curve c connecting A and B is the curve, or as c defined. Such an integral thus formed is called a line integral since integration is carried out along the path of the particular curve c joining A and B. ð ð ; I¼ Ft ds ¼ Ft ds AB c where c is the curve y ¼ f ðxÞ between A ðx1 , y1 Þ and B ðx2 , y2 Þ. There is in fact an alternative form of the integral which is often useful, so let us also consider that 29 Alternative form of a line integral It is often more convenient to integrate with respect to x or y than to take arc length as the variable. y δs δy δx O Ft If Ft has a component Q P in the x-direction Q in the y-direction P x ð Ft ds ¼ ; ð AB then the work done from K to L can be stated as P x þ Q y. ðP dx þ Q dyÞ AB where P and Q are functions of x and y. In general then, the line integral can be expressed as ð ð I ¼ Ft ds ¼ ðP dx þ Q dyÞ c c where c is the prescribed curve and F, or P and Q, are functions of x and y. Make a note of these results – then we will apply them to one or two examples 662 30 Programme 19 Example 1 ð Evaluate ðx þ 3yÞdx from A ð0, 1Þ to B ð2, 5Þ along the curve y ¼ 1 þ x2 . c The line integral is of the form ð ðP dx þ Q dyÞ y c c where, in this case, Q = 0 and c is the curve y ¼ 1 þ x2 . O x It can be converted at once into an ordinary integral by substituting for y and applying the appropriate limits of x. ð ð2 ð I ¼ ðP dx þ Q dyÞ ¼ ðx þ 3yÞ dx ¼ ðx þ 3 þ 3x2 Þ dx c c x2 þ 3x þ x3 ¼ 2 0 2 ¼ 16 0 Now for another, so move on 31 Example 2 Evaluate I ¼ ð ðx2 þ yÞ dx þ ðx y 2 Þ dy from A ð0, 2Þ to B ð3, 5Þ along c the curve y ¼ 2 þ x. ð I ¼ ðP dx þ Q dyÞ y c P ¼ x2 þ y ¼ x2 þ 2 þ x ¼ x2 þ x þ 2 c Q ¼ x y 2 ¼ x ð4 þ 4x þ x2 Þ ¼ ðx2 þ 3x þ 4Þ Also y ¼ 2 þ x O ; dy ¼ dx and the limits are x ¼ 0 to x ¼ 3. ; I ¼ ............ 32 I ¼ 15 Because ð3 I ¼ fðx2 þ x þ 2Þ dx ðx2 þ 3x þ 4Þ dxg 0 ð3 0 2 3 ð2x þ 2Þdx ¼ x 2x ¼ 15 0 x 663 Multiple integration 1 Here is another. Example 3 Evaluate I ¼ ð fðx2 þ 2yÞ dx þ xy dyg from O ð0, 0Þ to B ð1, 4Þ along the curve c y ¼ 4x2 . y In this case, c is the curve y ¼ 4x2 . ; dy ¼ 8x dx Substitute for y in the integral and apply the limits. c O Then I ¼ . . . . . . . . . . . . x Finish it off: it is quite straightforward. 33 I ¼ 9:4 Because ð I¼ ðx2 þ 2yÞ dx þ xy dy y ¼ 4x2 ; dy ¼ 8x dx c Also x2 þ 2y ¼ x2 þ 8x2 ¼ 9x2 ; xy ¼ 4x3 ð1 ð1 ; I ¼ f9x2 dx þ 32x4 dxg ¼ ð9x2 þ 32x4 Þ dx ¼ 9:4 0 0 They are all done in very much the same way. Move on for Example 4 Example 4 Evaluate I ¼ 34 ð ðx2 þ 2yÞ dx þ xy dy from O ð0, 0Þ to A ð1, 0Þ along line y ¼ 0 c and then from A ð1, 0Þ to B ð1, 4Þ along the line x ¼ 1. y c O c x (1) OA: c1 is the line y ¼ 0 ; dy ¼ 0: Substituting y ¼ 0 and dy ¼ 0 in the given integral gives 3 1 ð1 x 1 ¼ IOA ¼ x2 dx ¼ 3 0 3 0 (2) AB: Here c2 is the line x ¼ 1 ; dx ¼ 0 ; IAB ¼ . . . . . . . . . . . . 664 Programme 19 35 IAB ¼ 8 Because IAB ¼ ð4 fð1 þ 2yÞð0Þ þ y dyg 0 ¼ ð4 y dy 0 ¼ 2 4 y ¼8 2 0 Then I ¼ IOA þ IAB ¼ 13 þ 8 ¼ 8 13 ; I ¼ 8 13 If we now look back to Examples 3 and 4 just completed, we find that we have evaluated the same integral between the same two end points, but . . . . . . . . . . . . 36 along different paths of integration If we combine the two diagrams, we have y c O where c is the curve y ¼ 4x2 and c1 þ c2 are the lines y ¼ 0 and x ¼ 1: c The results obtained were x c Ic ¼ 9 25 and Ic1 þ c2 ¼ 8 13 Notice therefore that integration along two distinct paths joining the same two end points does not necessarily give the same results. 37 Let us pause here a moment and list the main properties of line integrals. Properties of line integrals ð Fds ¼ 1 ð c 2 AB 3 ð fP dx þ Q dyg ð Ð Ð Fds ¼ F ds and AB fP dx þ Q dyg ¼ BA fP dx þ Q dyg c BA i.e. the sign of a line integral is reversed when the direction of the integration along the path is reversed. (a) For a path of integration parallel to the y-axis, i.e. x ¼ k, ð ð P dx ¼ 0 ; Ic ¼ Q dy. dx ¼ 0: ; c c (b) For a path of integration parallel to the x-axis, i.e. y ¼ k, ð ð dy ¼ 0: ; Q dy ¼ 0 ; Ic ¼ P dx. c c 665 Multiple integration 1 4 If the path of integration c joining A to B is divided into two parts AK and KB, then Ic ¼ IAB ¼ IAK þ IKB : In all cases, the function y ¼ f ðxÞ that describes the path of integration involved must be continuous and single-valued – or dealt with as in item 6 below. If the function y ¼ f ðxÞ that y describes the path of integra(x2, y2) tion c is not single-valued for part of its extent, the path is y = f2(x) divided into two sections. (x3, y3) 5 6 y = f1(x) y ¼ f1 ðxÞ from A to K y ¼ f2 ðxÞ from K to B. (x1, y1) O x Make a note of this list for future reference and revision Example Evaluate I ¼ 38 ð ðx þ yÞ dx from A ð0, 1Þ to B ð0, 1Þ along the semi-circle c x2 þ y 2 ¼ 1 for x 0: y O x The first thing we notice is that ............ the function y ¼ f ðxÞ that describes the path of integration c is not single-valued pffiffiffiffiffiffiffiffiffiffiffiffiffiffi For any value of x, y = 1 x2 . Therefore, we divide c into two parts y c pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 from A to K pffiffiffiffiffiffiffiffiffiffiffiffiffiffi (2) y ¼ 1 x2 from K to B. ð As usual, I ¼ ðP dx þ Q dyÞ and in this (1) y ¼ O x c c particular case, Q = . . . . . . . . . . . . 39 666 Programme 19 40 Q¼0 ; I¼ ð P dx ¼ c ¼ ð1 ð1 xþ ð0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 dx þ ðx 1 x2 Þ dx 0 1 ð 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 x2 dx ðx þ 1 x x þ 1 x Þ dx ¼ 2 0 0 Now substitute x ¼ sin and finish it off. I ¼ ............ 41 I¼ Because ð 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi I¼2 1 x2 dx x ¼ sin 2 ; dx ¼ cos d 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 x2 ¼ cos Limits: x ¼ 0, ¼ 0; x ¼ 1, ¼ ; I¼2 ð =2 0 cos2 d ¼ ð =2 2 ð1 þ cos 2Þ d 0 sin 2 =2 ¼ þ 2 0 ¼ 2 Now let us extend this line of development a stage further. 42 Regions enclosed by closed curves A region is said to be simply connected if a path joining A and B can be deformed to coincide with any other line joining A and B without going outside the region. Another definition is that a region is simply connected if any closed path in the region can be contracted to a single point without leaving the region. 667 Multiple integration 1 Clearly, this would not be satisfied in the case where the region R contains one or more ‘holes’. The closed curves involved in problems in this Programme all relate to simply connected regions, so no difficulties will arise. 43 Line integrals round a closed curve We have already introduced the symbol þ to indicate that an integral is to be evaluated round a closed curve in the positive (anticlockwise) direction. y Positive direction (anticlockwise) line integral þ denoted by . c O x c y Negative direction (clockwise) line integral þ denoted by . O x With a closed curve, the y-values on the path c cannot be single-valued. Therefore, we divide the path into two or more parts and treat each separately. y y x1 y = f2(x) y = f1(x) L O M x2 ð1Þ Use y ¼ f1 ðxÞ for ALB x O x1 x2 x ð2Þ Use y ¼ f2 ðxÞ for BMA. Unless specially required otherwise, we always proceed round the closed curve in an . . . . . . . . . . . . 668 Programme 19 44 anticlockwise direction Example 1 Evaluate the line integral I ¼ þ ðx2 dx 2xy dyÞ where c comprises the three c sides of the triangle joining O ð0, 0Þ, A ð1, 0Þ and B ð0, 1Þ. First draw the diagram and mark in c1 , c2 and c3 , the proposed directions of integration. Do just that. y 45 c c O x c The three sections of the path of integration must be arranged in an anticlockwise manner round the figure. Now we deal with each part separately. (a) OA: c1 is the line y ¼ 0 ; dy ¼ 0. þ Then I ¼ ðx2 dx 2xy dyÞ for this part becomes 3 1 x 1 I1 ¼ x dx ¼ ¼ 3 3 0 0 ð1 2 ; I1 ¼ 1 3 (b) AB: c2 is the line y ¼ 1 x ; dy ¼ dx I2 ¼ . . . . . . . . . . . . (evaluate it) 46 I2 ¼ 23 Because c2 is the line y ¼ 1 x ; dy ¼ dx. ð0 ð0 I2 ¼ fx2 dx þ 2xð1 xÞ dxg ¼ ðx2 þ 2x 2x2 Þ dx 1 ¼ ð0 ð2x x2 Þdx ¼ x2 1 3 0 x 3 1 ¼ 1 2 3 ; I2 ¼ 2 3 Note that anticlockwise progression is obtained by arranging the limits in the appropriate order. Now we have to determine I3 for BO. (c) BO: c3 is the line x ¼ 0 I3 ¼ . . . . . . . . . . . . 669 Multiple integration 1 47 I3 ¼ 0 x¼0 Because for c3 , ; dx ¼ 0 ð ; I3 ¼ 0 dy ¼ 0 Finally, I ¼ I1 þ I2 þ I3 ¼ 13 23 þ 0 ¼ 13 ; I3 ¼ 0 ; I ¼ 13 Let us work through another example. Example 2 þ Evaluate y dx when c is the circle x2 þ y 2 ¼ 4. c y x2 þ y 2 ¼ 4 c R= ; I ¼ y is thus not single-valued. Therefore use pffiffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ 4 x2 for ALB between x ¼ 2 and pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ¼ 2 and y ¼ 4 x2 for BMA between x ¼ 2 and x ¼ 2. 2 O –2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; y ¼ 4 x2 x ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 x2 dx þ f 4 x2 g dx 2 2 ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ¼2 4 x dx ¼ 2 4 x2 dx 2 ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 4 4 x2 dx. 2 0 To evaluate this integral, substitute x ¼ 2 sin and finish it off. I ¼ ............ 48 I ¼ 4 Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; 4 x2 ¼ 2 cos limits: x ¼ 0, ¼ 0; x ¼ 2, ¼ 2 ð =2 ð =2 ; I ¼ 4 2 cos 2 cos d ¼ 16 cos2 d x ¼ 2 sin ; dx ¼ 2 cos d 0 ¼ 8 ð =2 0 Now for one more 0 sin 2 =2 ð1 þ cos 2Þ d ¼ 8 þ ¼ 4 2 0 670 Programme 19 Example 3 Evaluate I ¼ þ xy dx þ ð1 þ y 2 Þ dy where c is the boundary of the rectangle c joining A ð1, 0Þ, B ð3, 0Þ, C ð3, 2Þ and D ð1, 2Þ. First draw the diagram and insert c1 , c2 , c3 , c4 . That gives . . . . . . . . . . . . y 49 c c c c O x Now evaluate I1 for AB; I2 for BC; I3 for CD; I4 for DA; and finally I. Complete the working and then check with the next frame 50 I1 ¼ 0; I2 ¼ 4 23 ; I3 ¼ 8; I4 ¼ 4 23 ; I ¼ 8 Here is the complete working. þ I ¼ fxy dx þ ð1 þ y 2 Þ dyg c (a) AB: c1 is y ¼ 0 ; dy ¼ 0 ; I1 ¼ 0 (b) BC: c2 is x ¼ 3 ; dx ¼ 0 2 ð2 y3 ¼ 4 23 ; I2 ¼ ð1 þ y 2 Þdy ¼ y þ 3 0 0 (c) CD: c3 is y ¼ 2 ; dy ¼ 0 1 ð1 ; I3 ¼ 2x dx ¼ x2 ¼ 8 3 ; I2 ¼ 4 23 ; I3 ¼ 8 3 (d) DA: c4 is x ¼ 1 ; dx ¼ 0 0 ð0 y3 ¼ 4 23 ; I4 ¼ ð1 þ y 2 Þ dy ¼ y þ 3 2 2 ; I4 ¼ 4 23 671 Multiple integration 1 Finally I ¼ I1 þ I2 þ I3 þ I4 ¼ 0 þ 4 23 8 4 23 ¼ 8 ; I ¼ 8 Remember that, unless we are directed otherwise, we always proceed round the closed boundary in an anticlockwise manner. On now to the next piece of work Line integral with respect to arc length We have already established that ð ð I¼ Ft ds ¼ fP dx þ Q dyg AB AB where Ft denoted the tangential force along the curve c at the sample point K ðx, yÞ. The same kind of integral can, of course, relate to any function f ðx, yÞ which is a function of the position of a point on the stated curve, so that ð I¼ f ðx, yÞ ds. c This can readily be converted into an integral in terms of x. (Refer to Engineering Mathematics (Sixth Edition), Programme 19, Frame 30.) sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð ð ds ds dy ¼ 1þ I ¼ f ðx, yÞ ds ¼ f ðx, yÞ dx where dx dx dx c c sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð ð x2 dy ; f ðx, yÞ ds ¼ f ðx, yÞ 1 þ dx ð1Þ dx c x1 Example Evaluate I ¼ ð ð4x þ 3xyÞ ds where c is the straight line joining O ð0, 0Þ to c A ð1, 2Þ. y A dy c is the line y ¼ 2x ; ¼2 dx sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 pffiffiffi ds dy ¼ 1þ ¼ 5 ; dx dx c O x ; I¼ ð x¼1 x¼0 ð4x þ 3xyÞ ds ¼ ð1 pffiffiffi ð4x þ 3xyÞð 5Þ dx. But y ¼ 2x 0 ; I ¼ ............ 51 672 Programme 19 pffiffiffi I¼4 5 52 Because ð1 pffiffiffi pffiffiffi ð 1 pffiffiffi I ¼ ð4x þ 6x2 Þð 5Þ dx ¼ 2 5 ð2x þ 3x2 Þ dx ¼ 4 5 0 0 Try another. The path length of the parabola defined by y ¼ x2 betwen the values x ¼ 0 and x ¼ 2 is given by the integral ð I ¼ ds ¼ . . . . . . . . . . . . to 3 dp c 53 3.393 to 3 dp Because ð2 ð I ¼ ds ¼ c ¼ ð2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dy 1þ dx dx x¼0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 2x dx x¼0 Let u ¼ 1 þ 2x so that du ¼ 2dx and so ð5 du I¼ u1=2 2 u¼1 5 1 2 3=2 u ¼ 2 3 1 1 1=2 125 1 ¼ 3 ¼ 3:393 to 3 dp 54 Parametric equations When x and y are expressed in parametric form, e.g. x ¼ f ðtÞ, y ¼ gðtÞ, then sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 2 ds dx dy dx dy ¼ þ ; ds ¼ þ dt dt dt dt dt dt and result (1) above becomes sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð ð t2 dx dy I ¼ f ðx, yÞ ds ¼ dt f ðx, yÞ þ dt dt c t1 ð2Þ Make a note of results (1) and (2) for future use 673 Multiple integration 1 Example Evaluate I ¼ 55 þ 4xy ds where c is defined as the curve x ¼ sin t, y ¼ cos t c between t ¼ 0 and t ¼ We have x ¼ sin t y ¼ cos t : 4 dx ¼ cos t dt dy ¼ sin t ; dt ds ; ¼ ............ dt ; ds ¼1 dt Because sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dy þ ¼ cos2 t þ sin2 t ¼ 1 dt dt sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð =4 ð t2 dx dy f ðx, yÞ þ dt ¼ 4 sin t cos t dt ; I¼ dt dt 0 t1 ð =4 cos 2t =4 sin 2t dt ¼ 2 ¼1 ; I¼1 ¼2 2 0 0 ds ¼ dt Dependence of the line integral on the path of integration We saw earlier in the Programme that integration along two separate paths joining the same two end points does not necessarily give identical results. With this in mind, let us investigate the following problem. Example Evaluate I ¼ þ 3x2 y 2 dx þ 2x3 y dy between O ð0, 0Þ and A ð2, 4Þ c (a) along c1 i.e. y ¼ x2 (b) along c2 i.e. y ¼ 2x (c) along c3 i.e. x ¼ 0 from ð0, 0Þ to ð0, 4Þ and y ¼ 4 from ð0, 4Þ to ð2, 4Þ. Let us concentrate on section (a). First we draw the figure and insert relevant information. This gives . . . . . . . . . . . . 56 674 Programme 19 57 y y =x2 c O (a) I ¼ ð x f3x2 y 2 dx þ 2x3 y dyg c The path c1 is y ¼ x2 ; dy ¼ 2x dx ð2 ð2 ; I1 ¼ f3x2 x4 dx þ 2x3 x2 2x dxg ¼ ð3x6 þ 4x4 Þ dx 0 2 ¼ x7 ¼ 128 0 ; I1 ¼ 128 0 (b) In (b), the path of integration changes to c2 ; i.e. y ¼ 2x y So, in this case, y = 2x c I2 ¼ . . . . . . . . . . . . O x 58 I2 ¼ 128 y ¼ 2x ; dy ¼ 2 dx ð2 ð2 ; I2 ¼ f3x2 4x2 dx þ 2x3 2x2 dxg ¼ 20x4 dx Because with c2 , 0 2 ¼ 4 x5 ¼ 128 0 ; I2 ¼ 128 0 (c) In the third case, the path c3 is split x ¼ 0 from ð0, 0Þ to ð0; 4Þ y ¼ 4 from ð0, 4Þ to ð2, 4Þ Sketch the diagram and determine I3 . I3 ¼ . . . . . . . . . . . . 675 Multiple integration 1 59 I3 ¼ 128 y y=4 c c x=0 O x From ð0, 0Þ to ð0, 4Þ x ¼ 0 ; dx ¼ 0 ; I3a ¼ 0 From ð0, 4Þ to ð2, 4Þ y¼4 ; I3b ¼ 48 ; dy ¼ 0 ð2 x2 dx ¼ 128 0 ; I3 ¼ 128 On to the next frame In the example we have just worked through, we took three different paths and in each case, the line integral produced the same result. It appears, therefore, that in this case, the value of the integral is independent of the path of integration taken. y 60 c c How then does this integral perhaps differ from those of previous cases? c c O x We have been dealing with I ¼ Let us investigate ð f3x2 y 2 dx þ 2x3 y dyg c On reflection, we see that the integrand 3x2 y 2 dx þ 2x3 y dy is of the form P dx þ Q dy which we have met before and that it is, in fact, an exact differential of the function z ¼ x3 y 2 ; because @z @z ¼ 3x2 y 2 and ¼ 2x3 y @x @y Provided P, Q and their first partial derivatives are finite and continuous at all points inside and on any closed curve, this always happens. If the integrand of the given integral is seen to be an exact differential, then the value of the line integral is independent of the path taken and depends only on the coordinates of the two end points Make a note of this. It is important 61 676 62 Programme 19 y A c2 If I ¼ ð fP dx þ Q dyg and ðP dx þ Q dyÞ is an c exact differential, then c1 Ic1 ¼ Ic2 O x y c If we reverse the direction of c2 , then Ic1 ¼ Ic2 i.e. Ic1 þ Ic2 ¼ 0 c O x Hence, if ðP dx þ Q dyÞ is an exact differential, then the integration taken round a closed curve is zero. þ ; If ðP dx þ Q dyÞ is an exact differential, ðP dx þ Q dyÞ ¼ 0 63 Example 1 Evaluate I ¼ ð f3y dx þ ð3x þ 2yÞ dy g from A ð1, 2Þ to B ð3, 5Þ. c No path is given, so the integrand is probably an exact differential of some @P @Q function z ¼ f ðx, yÞ. In fact ¼3¼ . @y @x We have already dealt with the integration of exact differentials, so there ð is no difficulty. Compare with I ¼ fP dx þ Q dyg. @z ¼ 3y P¼ @x @z ¼ 3x þ 2y Q¼ @y ð c ; z ¼ 3y dx ¼ 3xy þ f ðyÞ ð ; z ¼ ð3x þ 2yÞ dy ¼ 3xy þ y 2 þ FðxÞ For (1) and (2) to agree f ðyÞ ¼ . . . . . . . . . . . . and FðxÞ ¼ . . . . . . . . . . . . ð1Þ ð2Þ 677 Multiple integration 1 f ðyÞ ¼ y 2 ; 64 FðxÞ ¼ 0 Hence z ¼ 3xy þ y 2 ð ð ð3, 5Þ ; I ¼ f3y dx þ ð3x þ 2yÞ dyg ¼ dð3xy þ y 2 Þ ð1, 2Þ c ð3, 5Þ ¼ 3xy þ y 2 ð1, 2Þ ¼ ð45 þ 25Þ ð6 þ 4Þ ¼ 60 Example 2 ð ðx2 þ yex Þ dx þ ðex þ yÞ dy between A ð0, 1Þ and B ð1, 2Þ. ð As before, compare with fP dx þ Q dyg: Evaluate I ¼ c c @z ¼ x2 þ yex P¼ @x @z ¼ ex þ y Q¼ @y ; z ¼ ............ ; z ¼ ............ Continue the working and complete the evaluation. When you have finished, check the result with the next frame 65 x3 þ yex þ f ðyÞ 3 y2 z ¼ yex þ þ FðxÞ 2 z¼ For these expressions to agree, Then I ¼ x3 y2 þ yex þ 3 2 ð1, f ðyÞ ¼ y2 ; 2 FðxÞ ¼ x3 3 2Þ ð0, 1Þ 5 ¼ þ 2e 6 So the main points are that, if ðP dx þ Q dyÞ is an exact differential ð (a) I ¼ ðP dx þ Q dyÞ is independent of the path of integration c ð (b) I ¼ ðP dx þ Q dyÞ is zero when c is a closed curve. c On to the next frame 678 66 Programme 19 Exact differentials in three independent variables A line integral in space naturally involves three independent variables, but the method is very much like that for two independent variables. dw ¼ Pdx þ Qdy þ R dz is an exact differential of w ¼ f ðx, y, zÞ @P @Q @P @R @R @Q ¼ ; ¼ ; ¼ if @y @x @z @x @y @z If the test is successful, then ð (a) ðP dx þ Q dy þ R dzÞ is independent of the path of integration þc (b) ðP dx þ Q dy þ R dzÞ is zero when c is a closed curve. c Example Verify that dw ¼ ð3x2 yz þ 6xÞdx þ ðx3 z 8yÞdy þ ðx3 y þ 1Þdz is an exact differeð ntial and hence evaluate dw from A ð1, 2, 4Þ to B ð2, 1, 3Þ. c First check that dw is an exact differential by finding the partial derivatives above, when P ¼ 3x2 yz þ 6x; Q ¼ x3 z 8y; and R ¼ x3 y þ 1. We have . . . . . . . . . . . . 67 @P ¼ 3x2 z; @y @Q ¼ 3x2 z @x @P ¼ 3x2 y; @z @R ¼ x3 ; @y @R ¼ 3x2 y @x @Q ¼ x3 @z ; @P @Q ¼ @y @x @P @R ¼ @z @x @R @Q ¼ ; @y @z ; ; dw is an exact differential Now to find w. ; P¼ @w ¼ 3x2 yz þ 6x @x @w ¼ x3 z 8y @y @w ¼ x3 y þ 1 @z @z ; @x @z @w ; R¼ @y @z ð ; w ¼ ð3x2 yz þ 6xÞdx Q¼ ¼ x3 yz þ 3x2 þ f ðy; zÞ ð ; w ¼ ðx3 z 8yÞ dy ¼ x3 zy 4y 2 þ Fðx; zÞ ð ; w ¼ ðx3 y þ 1Þ dz ¼ x3 yz þ z þ gðx, yÞ For these three expressions for z to agree f ðy; zÞ ¼ . . . . . . . . . . . . ; Fðx; zÞ ¼ . . . . . . . . . . . . ; gðx, yÞ ¼ . . . . . . . . . . . . 679 Multiple integration 1 f ðy; zÞ ¼ 4y 2 ; Fðx; zÞ ¼ z; 68 gðx, yÞ ¼ 3x2 ; w ¼ x3 yz þ 3x2 4y 2 þ z ð2; 1; 3Þ 3 2 2 ; I ¼ x yz þ 3x 4y þ z ð1; 2; 4Þ ¼ ............ 69 I ¼ 36 Because ð2; 1; 3Þ 3 2 2 I ¼ x yz þ 3x 4y þ z ð1; 2; 4Þ ¼ ð24 þ 12 4 þ 3Þ ð8 þ 3 16 þ 4Þ ¼ 36 The extension to line integrals in space is thus quite straightforward. Finally, we have a theorem that can be very helpful on occasions and which links up with the work we have been doing. It is important, so let us start a new section Green’s theorem Let P and Q be two functions of x and y that are, along with their first partial derivatives, finite and continuous inside and on the boundary c of a region R in the x–y plane. y 70 c O x If the first partial derivatives are continuous within the region and on the boundary, then Green’s theorem states that þ ð ð @P @Q dx dy ¼ ðP dx þ Q dyÞ @y @x R c That is, a double integral over the plane region R can be transformed into a line integral over the boundary c of the region – and the action is reversible. Let us see how it works. 680 Programme 19 Example 1 þ Evaluate I ¼ fð2x yÞ dx þ ð2y þ xÞ dy g around the boundary c of the ellipse c x2 þ 9y 2 ¼ 16: The integral is of the form I ¼ þ fP dx þ Q dyg where c P ¼ 2x y ; @P ¼ 1 @y @Q ¼ 1. and Q ¼ 2y þ x ; @x ð ð @P @Q ; I¼ dx dy @y @x R ð ð ð1 1Þ dx dy ¼ R ¼2 But ð ð ð ð dx dy R dx dy over any closed region gives . . . . . . . . . . . . R 71 the area of the figure In this case, then, I ¼ 2A where A is the area of the ellipse x2 þ 9y 2 ¼ 16 i.e. x2 9y 2 þ ¼1 16 16 4 3 16 ; A ¼ ab ¼ 3 32 ; I ¼ 2A ¼ 3 ; a ¼ 4; b ¼ To demonstrate the advantage of Green’s theorem, let us work through the next example (a) by the method of line integrals, and (b) by applying Green’s theorem. 681 Multiple integration 1 Example 2 Evaluate I ¼ þ fð2x þ yÞ dx þ ð3x 2yÞ dy g taken in anticlockwise manner c round the triangle with vertices at O ð0, 0Þ, A ð1, 0Þ and B ð1, 2Þ. y c O I¼ c þ fð2x þ yÞ dx þ ð3x 2yÞ dy g c x c (a) By the method of line integrals There are clearly three stages with c1 ; c2 ; c3 : Work through the complete evaluation to determine the value of I. It will be good revision. When you have finished, check the result with the solution in the next frame 72 I¼2 (a) (1) c1 is y ¼ 0 ; dy ¼ 0 1 ð1 ; I1 ¼ 2x dx ¼ x2 ¼ 1 0 ; I1 ¼ 1 0 (2) c2 is x ¼ 1 ; dx ¼ 0 2 ð2 ; I2 ¼ ð3 2yÞ dy ¼ 3y y 2 ¼ 2 0 ; I2 ¼ 2 0 (3) c3 is y ¼ 2x ; dy ¼ 2 dx ð0 ; I3 ¼ f4x dx þ ð3x 4xÞ2 dxg 1 ¼ ð0 1 0 2x dx ¼ x2 ¼ 1 ; I3 ¼ 1 1 I ¼ I1 þ I2 þ I3 ¼ 1 þ 2 þ ð1Þ ¼ 2 ; I¼2 Now we will do the same problem by applying Green’s theorem, so move on 682 73 Programme 19 (b) By Green’s theorem þ I ¼ fð2x þ yÞ dx þ ð3x 2yÞ dyg c @P ¼ 1; Q ¼ 3x 2y P ¼ 2x þ y ; @y ð ð @P @Q I¼ dx dy @y @x R ; @Q ¼3 @x Finish it off. I ¼ . . . . . . . . . . . . 74 I¼2 Because ð ð I¼ ð1 3Þ dx dy ð Rð dx dy ¼ 2A ¼2 R ¼ 2 the area of the triangle ¼21¼2 ; I ¼2 Application of Green’s theorem is not always the quickest method. It is useful, however, to have both methods available. If you have not already done so, make a note of Green’s theorem. þ ð ð @P @Q dx dy ¼ ðP dx þ Q dyÞ @y @x R c 75 Example 3 Evaluate the line integral I ¼ þ fxy dx þ ð2x yÞ dyg round the region c bounded by the curves y ¼ x2 and x ¼ y 2 by the use of Green’s theorem. y y= x Points of intersection are O ð0, 0Þ and A ð1, 1Þ. P and Q are known, so there is no difficulty. y = x2 O x Complete the working. I ¼ ............ 683 Multiple integration 1 I¼ 76 31 60 Here is the working. þ I ¼ fxy dx þ ð2x yÞ dyg c þ ð ð @P @Q dx dy fP dx þ Q dyg ¼ @y @x c R @P @Q P ¼ xy ; ¼ x; Q ¼ 2x y ; ¼2 @y @x ð ð y ðx 2Þ dx dy I¼ R y= x ¼ ð 1 ð y¼pffiffix 0 y = x2 O x ; I¼ ð1 ¼ ð1 pxffiffi ðx 2Þ y dx 0 x ðx 2Þ dy dx y¼x2 x2 pffiffiffi ðx 2Þð x x2 Þ dx 0 ¼ ð1 ¼ ðx3=2 x3 2x1=2 þ 2x2 Þ dx 0 2 5=2 1 4 4 3=2 2 3 x x x þ x 5 4 3 3 1 ¼ 0 31 60 Before we finally leave this section of the work, there is one more result to note. In the special case when P ¼ y and Q ¼ x @P ¼ 1 and @y @Q ¼ 1 @x Green’s theorem then states ð ð þ f1 ð1Þg dx dy ¼ ðPdx þ Q dyÞ R ð ð þc i.e. 2 dx dy ¼ ðy dx x dyÞ R þ c ¼ ðx dy y dxÞ c Therefore, the area of the closed region þ ð ð 1 A¼ dx dy ¼ ðx dy y dxÞ 2 c R Note this result in your record book. Then let us see an example 684 77 Programme 19 Example 1 Determine the area of the figure enclosed by y ¼ 3x2 and y ¼ 6x. y Points of intersection: c 3x2 ¼ 6x ; x ¼ 0 or 2 þ Area A ¼ 12 ðx dy y dxÞ y = 3x 2 y = 6x c c O x We evaluate the integral in two parts, i.e. OA along c1 and AO along c2 2A ¼ ð ðx dy y dxÞ þ ð c1 ðalong OAÞ I1 : ðx dy y dxÞ ¼ I1 þ I2 c2 ðalong AOÞ c1 is y ¼ 3x2 ; dy ¼ 6x dx 2 ð2 ð2 ; I1 ¼ ð6x2 dx 3x2 dxÞ ¼ 3x2 dx ¼ x3 ¼ 8 0 0 0 ; I1 ¼ 8 Similarly, I2 ¼ . . . . . . . . . . . . 78 I2 ¼ 0 Because c2 is y ¼ 6x ; dy ¼ 6 dx ð0 ; I2 ¼ ð6x dx 6x dxÞ ¼ 0 ; I2 ¼ 0 2 ; I ¼ I1 þ I2 ¼ 8 þ 0 ¼ 8 ; A ¼ 4 square units Finally, here is one for you to do entirely on your own. Example 2 Determine the area bounded by the curves y ¼ 2x3 , y ¼ x3 þ 1 and the axis x ¼ 0 for x 0. Complete the working and see if you agree with the working in the next frame 685 Multiple integration 1 Here it is. 79 y y ¼ 2x3 ; y ¼ x3 þ 1; x ¼ 0 y = x3 + 1 Point of intersection c 2x3 ¼ x3 þ 1 ; x3 ¼ 1 ; x ¼ 1 þ 1 Area A ¼ ðx dy y dxÞ 2 c þ ; 2A ¼ ðx dy y dxÞ y = 2x3 c c O x c c1 is y ¼ 2x3 ; dy ¼ 6x2 dx ð ð1 ðx dy y dxÞ ¼ ð6x3 dx 2x3 dxÞ ; I1 ¼ (a) OA: c1 ¼ ð1 0 1 3 4x dx ¼ x4 ¼ 1 0 ; I1 ¼ 1 0 c2 is y ¼ x3 þ 1 ; dy ¼ 3x2 dx ð0 ð0 ; I2 ¼ f3x3 dx ðx3 þ 1Þ dxg ¼ ð2x3 1Þ dx (b) AB: 1 ¼ (c) BO: 4 x x 2 0 1 1 ¼ ð12 1Þ ¼ 12 c3 is x ¼ 0 ; dx ¼ 0 ð y¼0 ðx dy y dxÞ ¼ 0 I3 ¼ ; I2 ¼ 12 ; I3 ¼ 0 y¼1 ; 2A ¼ I ¼ I1 þ I2 þ I3 ¼ 1 þ 12 þ 0 ¼ 1 12 ; A ¼ 34 square units And that brings this Programme to an end. We have covered some important topics, so check down the Revision summary and the Can you? checklist that follow and revise any part of the text if necessary, before working through the Test exercise. The Further problems provide an opportunity for additional practice. 686 Programme 19 Revision summary 19 80 1 Differentials dy and dx (a) y = f (x) y (x, y) dy δy dx= δx O x 0 dy ¼ f ðxÞ dx @z @z dx þ dy @x @y @z @z @z dz ¼ dx þ dy þ dw. @x @y @w (b) If z ¼ f ðx; yÞ, dz ¼ If z ¼ f ðx; y; wÞ; (c) dz ¼ P dx þ Q dy, where P and Q are functions of x and y, is an exact @P @Q differential if ¼ : @y @x 2 Line integrals – definition ð ð I ¼ f ðx, yÞ ds ¼ ðP dx þ Q dyÞ c 3 c Properties of line integrals (a) Sign of line integral is reversed when the direction of integration along the path is reversed. ð (b) Path of integration parallel to y-axis, dx ¼ 0 ; Ic ¼ Q dy. c ð Path of integration parallel to x-axis, dy ¼ 0 ; Ic ¼ P dx: c 4 (c) The y-values on the path of integration must be continuous and single-valued. þ Line of integral round a closed curve ð Positive direction ’ anticlockwise ð ð ð Negative direction @ clockwise; i.e. @ ¼ ’. 687 Multiple integration 1 5 Line integral related to arc length ð ð F ds ¼ ðP dx þ Q dyÞ I¼ AB AB sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð x2 dy ¼ f ðx, yÞ 1 þ dx dx x1 c With parametric equations, x and y in terms of t, sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð t2 ð dx dy f ðx, yÞ þ dt I ¼ f ðx, yÞ ds ¼ dt dt c t1 6 Dependence of line integral on path of integration In general, the value of the line integral depends on the particular path of integration. 7 Exact differential If P dx þ Q dy is an exact differential where P, Q and their first derivatives are finite and continuous inside the simply connected region R @P @Q ¼ @y @x ð (b) I ¼ ðP dx þ Q dyÞ (a) (c) I ¼ þ c is independent of the path of integration where c lies entirely within R ðP dx þ Q dyÞ is zero when c is a closed curve lying entirely within R. Exact differentials in three variables c 8 If P dx þ Q dy þ R dz is an exact differential where P, Q, R and their first partial derivatives are finite and continuous inside a simply connected region containing path c @P @Q @P @R @R @Q ¼ ; ¼ ; ¼ @y @x @z @x @y @z ð (b) ðP dx þ Q dy þ R dzÞ is independent of the path of (a) c integration þ ðP dx þ Q dy þ R dzÞ is zero when c is a closed curve. (c) c 9 Green’s theorem þ ð ð @P @Q dx dy ðP dx þ Q dyÞ ¼ @y @x c R and, for a simple closed curve þ ð ð dx dy ¼ 2A ðx dy y dxÞ ¼ 2 c R where A is the area of the enclosed figure. 688 Programme 19 Can you? 81 Checklist 19 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Evaluate double and triple integrals and apply them to the determination of the areas of plane figures and the volumes of solids? Yes No Frames 1 to 10 . Understand the role of the differential of a function of two or more real variables? Yes No 11 to 13 . Determine exact differentials in two real variables and their integrals? Yes No 14 to 19 . Evaluate the area enclosed by a closed curve by contour integration? Yes No 20 to 26 27 to 41 42 to 50 51 to 53 . Link line integrals to integrals along a contour given in parametric form? Yes No 54 to 56 . Discuss the dependence of a line integral between two points on the path of integration? Yes No 56 to #65 . Determine exact differentials in three real variables and their integrals? Yes No 66 to 69 70 to 79 . Evaluate line integrals and appreciate their properties? Yes No . Evaluate line integrals around closed curves within a simply connected region? Yes No . Link line integrals to integrals along the x-axis? Yes No . Demonstrate the validity and use of Green’s theorem? Yes No 689 Multiple integration 1 Test exercise 19 1 Determine the differential dz of each of the following. (a) z ¼ x4 cos 3y; 2 (b) z ¼ e2y sin 4x; (c) z ¼ x2 yw3 . Determine which of the following are exact differentials and integrate where appropriate to determine z. (a) dz ¼ ð3x2 y 4 þ 8xÞ dx þ ð4x3 y 3 15y 2 Þ dy (b) dz ¼ ð2x cos 4y 6 sin 3xÞdx 4ðx2 sin 4y 2yÞ dy (c) dz ¼ 3e3x ð1 yÞ dx þ ðe3x þ 3y 2 Þ dy. 3 Calculate the area of the triangle with vertices at O ð0, 0Þ, A ð4, 2Þ and B ð1, 5Þ. 4 Evaluate the following. Ð (a) I ¼ c fðx2 3yÞ dx þ xy 2 dyg from A ð1, 2Þ to B ð2, 8Þ along the curve y ¼ 2x2 : Ð (b) I ¼ c ð2x þ yÞ dx from A ð0, 1Þ to B ð0, 1Þ along the semicircle x2 þ y 2 ¼ 1 for x 0. Þ I ¼ c fð1 þ xyÞ dx þ ð1 þ x2 Þ dyg where c is the boundary of the rectangle joining A ð1, 0Þ, B ð4, 0Þ, C ð4, 3Þ and D ð1, 3Þ. Ð I ¼ c 2xy ds where c is defined by the parametric equations x ¼ 4 cos , y ¼ 4 sin between ¼ 0 and ¼ . 3 Ð I ¼ c fð8xy þ y 3 Þ dx þ ð4x2 þ 3xy 2 Þ dyg from A(1, 3) to B(2, 1). Þ I ¼ c fð3x þ yÞ dx þ ðy 2xÞ dyg round the boundary of the ellipse (c) (d) (e) (f) x2 þ 4y 2 ¼ 36. 5 6 Apply Green’s theorem to determine the area of the plane figure bounded by pffiffiffi the curves y ¼ x3 and y ¼ x. Verify that dw ¼ 2xyz þ 2z y 2 dx þ x2 z 2yx dy þ x2 y þ 2x dz is an exact differential and find the value of ð dw where c (a) c is the straight line joining ð0, 0, 0Þ to ð1, 1, 1Þ (b) c is the curve of intersection of the unit sphere centred on the origin and the plane x þ y þ z ¼ 1. 82 690 Programme 19 Further problems 19 83 1 2 3 Ð Show that I ¼ c fxy 2 w2 dx þ x2 yw2 dy þ x2 y 2 w dwg is independent of the path of integration c and evaluate the integral from A ð1, 3, 2Þ to B ð2, 4, 1Þ. Determine whether dz ¼ 3x2 ðx2 þ y 2 Þ dxÐ þ 2yðx3 þ y 4 Þ dy is an exact differential. If so, determine z and hence evaluate c dz from A ð1, 2Þ to B ð2, 1Þ. þ xdy ydx where c is the boundary of the Evaluate the line integral I ¼ 2 þ y2 þ 4 x c segment formed by the arc of the circle x2 þ y 2 ¼ 4 and the chord y ¼ 2 x for x 0: 4 5 6 7 8 9 Show that Ð I ¼ c fð3x2 sin y þ 2 sin 2x þ y 3 Þ dx þ ðx3 cos y þ 3xy 2 Þ dyg is independent of the path of integration and evaluate it from A ð0; 0Þ ; : to B 2 Ð Evaluate the integral I ¼ c xy ds where c is defined by the parametric equations x ¼ cos3 t, y ¼ sin3 t from t ¼ 0 to t ¼ . 2 xdx ydy 2 Verify that dz ¼ 2 for x > y 2 is an exact differential and evaluate x y 2 x2 y 2 z ¼ f ðx, yÞ from A ð3, 1Þ to B ð5, 3Þ. The parametric equations of a circle, centre (1, 0) and radius 1, can be expressed as x ¼ 2 cos2 , y ¼ 2 cos sin . Ð Evaluate I ¼ c fðx þ yÞ dx þ x2 dyg along the semicircle for which y 0 from O ð0, 0Þ to A ð2, 0Þ. Þ Evaluate c fx3 y 2 dx þ x2 y dyg where c is the boundary of the region enclosed by the curve y ¼ 1 x2 , x ¼ 0 and y ¼ 0 in the first quadrant. Use Green’s theorem to evaluate þ I ¼ fð4x þ yÞ dx þ ð3x 2yÞ dyg c 10 where c is the boundary of the trapezium with vertices A ð0, 1Þ, B ð5, 1Þ, C ð3, 3Þ and D ð1, 3Þ. Ð Evaluate I ¼ c fð3x2 y 2 þ 2 cos 2x 2xyÞ dx þ ð2x3 y þ 8y x2 Þ dyg (a) along the curve y ¼ x2 x from A ð0, 0Þ to B ð2, 2Þ 11 (b) round the boundary of the quadrilateral joining the points ð1, 0Þ, ð3, 1Þ, ð2, 3Þ and ð0, 3Þ y x xy Verify that dw ¼ dx þ dy 2 dz is an exact differential and find the value z z z ð dw where c is the straight line joining ð0, 0, 1Þ to ð1, 2, 3Þ for either region of c z > 0 or z < 0. Programme 20 Frames 1 to 77 Multiple integration 2 Learning outcomes When you have completed this Programme you will be able to: . Evaluate double integrals and surface integrals . Relate three-dimensional Cartesian coordinates to cylindrical and spherical polar forms . Evaluate volume integrals in Cartesian coordinates and in cylindrical and spherical polar coordinates . Use the Jacobian to convert integrals given in Cartesian coordinates into general curvilinear coordinates in two and three dimensions 691 692 Programme 20 Double integrals 1 Let us start off with an example with which we are already familiar. Example 1 A solid is enclosed by the planes z ¼ 0, y ¼ 1, y ¼ 2, x ¼ 0, x ¼ 3 and the surface z ¼ x þ y 2 . We have to determine the volume of the solid so formed. First take some care in sketching the figure, which is ............ 2 z y dy dx O x In the plane y ¼ 1, z ¼ x þ 1, i.e. a straight line joining ð0, 1, 1Þ and ð3, 1, 4Þ In the plane y ¼ 2, z ¼ x þ 4, i.e. a straight line joining ð0, 2, 4Þ and ð3, 2, 7Þ In the plane x ¼ 0, z ¼ y 2 , i.e. a parabola joining ð0, 1, 1Þ and ð0, 2, 4Þ In the plane x ¼ 3, z ¼ 3 þ y 2 , i.e. a parabola joining ð3, 1, 4Þ and ð3, 2, 7Þ. Consideration like this helps us to visualise the problem and the time involved is well spent. Now we can proceed. The element of volume v ¼ x y z ððð Then the total volume V ¼ dx dy dz between appropriate limits in each case. 693 Multiple integration 2 We could also have said that the element of area on the z ¼ 0 plane a ¼ y x and that the volume of the column vc ¼ z a ¼ z x y 2 Then, since z ¼ x þ y , this becomes vc ¼ ðx þ y 2 Þ x y Summing in the usual way then gives ð V ¼ z da ð ð ¼ ðx þ y 2 Þ dx dy R where R is the region bounded in the x–y plane. Now we insert the appropriate limits and complete the integration V ¼ ............ V ¼ 11:5 cubic units Because V¼ ð y¼2 ð x¼3 y¼1 ð2 ðx þ y 2 Þ dx dy x¼0 x¼3 x2 þ xy 2 ¼ dy 1 2 x¼0 ð2 9 þ 3y 2 dy ¼ 1 2 2 9 y þ y3 ¼ 2 1 ¼ 11:5 ; V ¼ 11:5 cubic units Although we have found a volume, this is, in fact, an example of a double integral since the expression for z was a function of position in the x–y plane within the closed region ð ð f ðx, yÞ da i.e. I ¼ ðR ð f ðx, yÞ dy dx ¼ R In this particular case, R is the region in the x–y plane bounded by x ¼ 0, x ¼ 3, y ¼ 1, y ¼ 2. 3 694 Programme 20 Example 2 A triangular thin plate has the dimensions shown and a variable density where ¼ 1 þ x þ xy. y We have to determine (a) the mass of the plate (b) the position of its centre of gravity G. O x (a) Consider an element of area at the point P ðx, yÞ in the plate y a ¼ x y The mass m of the element is then y = 2x m ¼ x y y O x x ; Total mass M ¼ ð ð dm ¼ R ð ð dx dy R Now we insert the limits and complete the integration, remembering that ¼ ð1 þ x þ xyÞ M ¼ ............ 4 M ¼ 17 1 3 Because we have ð x¼2 ð y¼2x ð ð dx dy ¼ ð1 þ x þ xyÞ dy dx M¼ R x¼0 y¼0 y¼2x ð2 xy 2 ¼ y þ xy þ dx 2 y¼0 0 ð2 ¼ f2x þ 2x2 þ 2x3 gdx 0 ¼ x2 þ 2x3 x4 þ 3 2 2 ¼ 17 0 1 3 (b) To find the position of the centre of gravity, we need to know ............ 695 Multiple integration 2 5 the sum of the moments of mass about OY and OX (1) To find x, we take moments about OY. y Moment of mass of element about OY ¼ x m ¼ xð1 þ x þ xyÞ x y y = 2x δm O x x ; Sum of first moments ¼ ð ð ðx þ x2 þ x2 yÞ dx dy R ¼ ............ 26 6 2 15 Because sum of first moments ¼ ð x¼2 ð y¼2x x¼0 ¼ ð2 0 ¼ ð2 ðx þ x2 þ x2 yÞ dy dx y¼0 x2 y 2 xy þ x y þ 2 2 y¼2x dx y¼0 f2x2 þ 2x3 þ 2x4 gdx 0 ¼2 ð2 ðx2 þ x3 þ x4 Þ dx 0 3 2 x x4 x5 2 þ þ ¼ 26 ¼2 15 3 4 5 0 Now Mx ¼ sum of moments ; x ¼ ............ 7 x ¼ 1:508 1 We found previously that M ¼ 17 3 33 which gives x ¼ 1 ¼ 1:508 65 ; 1 2 17 x ¼ 26 3 15 (2) To find y we proceed in just the same way, this time taking moments about OX. Work right through it on your own. y ¼ ............ 696 Programme 20 8 y ¼ 1:754 y Moment of element of mass m about OX ¼ y m ¼ yð1 þ x þ xyÞ x y y O x ; Sum of first moments about OX ¼ ¼ ð ð ðy þ xy þ xy 2 Þ dx dy R ð x¼2 ð y¼2x ðy þ xy þ xy 2 Þ dy dx x¼0 y¼0 y¼2x y xy 2 xy 3 dx þ þ 2 3 y¼0 0 2 ð2 8x4 2 3 dx 2x þ 2x þ ¼ 3 0 3 2 2x x4 8x5 ¼ þ þ 3 2 15 0 2 ¼ 30 5 ¼ ; My ¼ 30 2 5 ; y ¼ 30 2 5 17 ð2 2 1 ¼ 1:754 3 So we finally have: y O x Note that this again referred to a plane figure in the x–y plane. Now let us move on to something slightly different Multiple integration 2 697 Surface integrals 9 When the area over which we integrate is not restricted to the x–y plane, matters become rather more involved, but also more interesting. If S is a two-sided surface in space and R is its projection on the x–y plane, then the equation of S is of the form z ¼ f ðx, yÞ where f is a singlevalued function and continuous throughout R. Let A denote an element of R and S the corresponding element of area of S at the point Pðx; y; zÞ in S. γ z δs O x y δA = δx δy Let also ðx; y; zÞ be a function of position on S (e.g. potential) and let denote the angle between the outward normal PN to the surface at P and the positive z-axis. A ¼ A sec and Then A S cos i.e. S cos X ðx; y; zÞS is the total value of ðx; y; zÞ taken over the surface S. As S ! 0, this sum becomes the integral ð I ¼ ðx, y, zÞ dS S and, since S A sec , the result can be written ð ð ðx, y, zÞ sec dx dy < I¼ 2 R ^ k, where k is the unit vector in the z-direction and n ^ is Notice that cos ¼ n the unit normal to the surface at P. With limits inserted for x and y, the integral seems straightforward, except for the factor sec , which naturally varies over the surface S. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 @z @z þ We can, in fact, show that sec ¼ 1 þ @x @y (see Appendix, page 1065) Therefore, the surface integral of ðx, y, zÞ over the surface S is given by ð (a) I ¼ ðx, y, zÞ dS S sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð ð @z @z ðx, y, zÞ 1 þ þ dx dy or (b) I ¼ @x @y R where z ¼ f ðx, yÞ ð1Þ ð2Þ 698 Programme 20 Note that, when ðx; y; zÞ ¼ 1; then I ¼ ð dS gives the area of the surface S. S ; S¼ ð dS ¼ ð ð S R sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 @z @z 1þ dx dy þ @x @y ð3Þ Make a note of these three important results. Then we will apply them to a few examples. 10 Example 1 Find the area of the surface z ¼ x2 þ y 2 ¼ 1: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð ð @z @z 1þ þ dx dy S¼ @x @y R pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y 2 over the region bounded by @z @z and and determine So we now find @x @y sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 @z @z 1þ þ @x @y which is . . . . . . . . . . . . pffiffiffi 2 11 Because z ¼ ðx2 þ y 2 Þ1=2 ; @z 1 2 x ¼ ðx þ y 2 Þ1=2 2x ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 @x 2 x þ y2 @z 1 2 y ¼ ðx þ y 2 Þ1=2 2y ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 @y 2 x þ y2 2 2 @z @z x2 þ y 2 þ ¼1þ 2 ¼2 ; 1þ @x @y x þ y2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 pffiffiffi @z @z ¼ 2 ; 1þ þ @x @y pffiffiffi pffiffiffi ð ð dx dy ¼ 2 . . . . . . . . . . . . ; S¼ 2 R 699 Multiple integration 2 12 the area of the region R But R is bounded by x2 þ y 2 ¼ 1, i.e. a circle, centre the origin and radius 1. ; area ¼ pffiffiffi pffiffiffi ð ð dx dy ¼ 2 ; S¼ 2 R Example 2 Find the area of the surface S of the paraboloid z ¼ x2 þ y 2 cut off by pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the cone z ¼ 2 x2 þ y 2 . z z z = 2 x2 + y2 z = 2y z = y2 z = x2 + y2 y O O x y We can find the point of intersection A by considering the y–z plane, i.e. put x ¼ 0: Coordinates of A are . . . . . . . . . . . . 13 A ð2, 4Þ The projection of the surface S on the x–y plane is ............ 14 the circle x2 þ y 2 ¼ 4 z S¼ ð ð R S O R z = x2 + y2 y sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 @z @z 1þ dx dy þ @y @x For this we use the equation of the surface S. The information from the projection R on the x–y plane will later provide the limits of the two stages of integration. x For the time being, then, S ¼ . . . . . . . . . . . . 700 Programme 20 15 S¼ ð ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4x2 þ 4y 2 dx dy R y y= O i.e. 4 – x2 Using Cartesian coordinates, we could integrate with respect to y from y ¼ 0 to pffiffiffiffiffiffiffiffiffiffiffiffiffiffi y ¼ 4 x2 and then with respect to x from x ¼ 0 to x ¼ 2. Finally, we should multiply by four to cover all four quadrants. x x ð x¼2 ð y¼pffiffiffiffiffiffiffiffi 4x2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi S¼4 1 þ 4x2 þ 4y 2 dy dx x¼0 y¼0 But how do we carry out the actual integration? It becomes a lot easier if we use polar coordinates. The same integral in polar coordinates is . . . . . . . . . . . . 16 S¼ ð ¼2 ð r¼2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4r 2 r dr d ¼0 y r O θ x y x r¼0 x ¼ r cos ; y ¼ r sin x2 þ y 2 ¼ r 2 dx dy ¼ r dr d (refer to Frame 67) ð ¼2 ð r¼2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 4r 2 r dr d S¼ ¼0 r¼0 ; S ¼ ............ Finish it off. 701 Multiple integration 2 S ¼ 36:18 square units 17 Because 2 ð ¼2 ð r¼2 ð 2 1 S¼ ð1 þ 4r 2 Þ3=2 d ð1 þ 4r 2 Þ1=2 r dr d ¼ ¼0 r¼0 0 12 0 2 ð 2 1Ð ¼ f173=2 1g d ¼ 5:7577 ¼ 36:18 12 0 0 Now on to Example 3. Example 3 18 To determine the moment of inertia of a thin spherical shell of radius a about a diameter as axis. The mass per unit area of shell is . z δs x O x θ r y Equation of sphere x2 þ y 2 þ z2 ¼ a2 z y Mass of element ¼ m ¼ S δA I mr 2 Sr 2 Let us deal with the upper hemisphere ð ; IH ¼ r 2 dS S sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð ð @z @z 2 r 1 þ ¼ dx dy þ @x @y R Now determine the partial derivatives and simplify the integral as far as possible in Cartesian coordinates. IH ¼ . . . . . . . . . . . . IH ¼ ð ð R a r 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dy 2 a x2 y 2 In this particular example, R is, of course, the region bounded by the circle x2 þ y 2 ¼ a2 in the x–y plane. Converting to polar coordinates x ¼ r cos ; y ¼ r sin ; dx dy ¼ r dr d the integral becomes IH ¼ . . . . . . . . . . . . 19 702 Programme 20 20 IH ¼ a ð ¼2 ð r¼a ¼0 Because for x2 þ y 2 ¼ r 2 : ð ð r3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dr d 2 a r2 r¼0 limits of r: limits of : r ¼ 0 to r ¼ a ¼ 0 to ¼ 2 a r 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r dr d 2 a r2 R ð ¼2 ð r¼a r3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dr d ¼ a 2 a r2 ¼0 r¼0 IH ¼ First we have to evaluate ða r3 Ir ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dr a2 r 2 0 If we substitute u ¼ a2 r 2 then the integral is evaluated as Ir ¼ . . . . . . . . . . . . 21 Ir ¼ 2a3 3 Because When u ¼ a2 r 2 then du ¼ 2r dr so that r 2 ¼ a2 u and du r dr ¼ . Therefore 2 ða ða r3 r2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r dr Ir ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dr ¼ a2 r 2 0 r¼0 a2 r 2 ð0 2 a u du pffiffiffi ¼ 2 u 2 u¼a ð ð 2 0 a 1 0 1=2 u du þ u1=2 du ¼ 2 u¼a2 2 u¼a2 a2 h 1=2 i0 1 2 3=2 0 u ¼ 2u þ u¼a2 2 3 2 u¼a2 ¼ a3 ¼ a3 3 2a3 3 Now, to complete IH we have ð 2 2a3 d 3 0 ¼ ............ IH ¼ a 703 Multiple integration 2 IH ¼ 22 4a4 3 Because IH ¼ a ð 2 0 2 2a3 2a4 4a4 ¼ d ¼ 3 3 3 0 Therefore, the moment of inertia for the complete spherical shell is Is ¼ 8a4 3 The total mass of the shell M ¼ 4a2 ; I ¼ 2Ma2 3 Now let us turn our attention towards volume integrals and in preparation review systems of space coordinates. 23 Space coordinate systems 1 Cartesian coordinates ðx, y, zÞ – referred to three coordinate axes OX, OY, OZ at right angles to each other. These are arranged in a right-handed manner, i.e. turning from OX to OY gives a right-handed screw action in the positive direction of OZ. z z y–z plane (x = 0) z–x plane (y = 0) y O x O y x–y plane (z = 0) x The three coordinate planes, x ¼ 0, y ¼ 0, z ¼ 0, divide the space into eight sections called octants. The section containing x 0, y 0, z 0 is called the first octant. z x x P (x, y, z) z O For a point P ðx, y, zÞ y y L OL2 ¼ x2 þ y 2 OP2 ¼ x2 þ y 2 þ z2 Note that this is Pythagoras’ theorem in three dimensions. We are all familiar with this system of coordinates. 704 24 Programme 20 Cylindrical coordinates ðr, , zÞ are useful where an axis of symmetry occurs. 2 z Any point P is considered as having a position on a cylinder. If L is the projection of P on the x–y plane, then ðr; Þ are the usual polar coordinates of L. The cylindrical coordinates of P then merely require the addition of the z-coordinate. P (r, θ, z) z r0 O θ r y L (r, θ) x Relationship between Cartesian and cylindrical coordinates z If we consider a combined figure, we can easily relate the two systems. P (x, y, z) (r, θ, z) Expressing each of the following in terms of the alternative system, z x x 25 O θ y x ¼ ............ y ¼ ............ z ¼ ............ y r L pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 þ y 2 x ¼ r cos r¼ y ¼ r sin ¼ arctanðy=xÞ z¼z z¼z So, in cylindrical coordinates, the surface defined by ð1Þ r ¼ 5 is . . . . . . . . . . . . ð2Þ ¼ =6 is . . . . . . . . . . . . ð3Þ z ¼ 4 is . . . . . . . . . . . . r ¼ ............ ¼ ............ z ¼ ............ 705 Multiple integration 2 26 (1) r ¼ 5 is a right cylinder, radius 5, with OZ as axis. (2) ¼ =6 is a plane through OZ, making an angle =6 with OX. (3) z ¼ 4 is a plane parallel to the x–y plane cutting OZ at 4 units above the origin. So position P ð2, 3, 4Þ in Cartesian coordinates ¼ . . . . . . . . . . . . in cylindrical coordinates and position Q ð2:5, =3, 6Þ in cylindrical coordinates ¼ . . . . . . . . . . . . in Cartesian coordinates. pffiffiffiffiffiffi P ð2, 3, 4Þ ¼ ð 13, 0:983, 4Þ in cylindrical coordinates Q ð2:5, =3, 6Þ ¼ ð1:25, 2:165, 6Þ in Cartesian coordinates. Spherical coordinates ðr; ; Þ are appropriate where a centre of symmetry occurs. The position of a point is considered as being a point on a sphere. 3 z r is the distance of P from the origin and is always taken as positive. (x, y, z) (r, θ, ϕ) θ O is the angle between OP and the positive OZ axis r y ρ ϕ L is the projection of P on the x–y plane is the angle between OL and the OX axis. x Note that (a) may be regarded as the longitude of P from OX (b) may be regarded as the complement of the latitude of P. Relationship between Cartesian and spherical coordinates z r θ x x z O ϕ y The combined figure shows the connection between the two systems, so ρ y x ¼ ............ y ¼ ............ z ¼ ............ r ¼ ............ ¼ ............ ¼ ............ 27 706 Programme 20 28 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 2 þ y 2 þ z2 x ¼ r sin cos r¼ y ¼ r sin sin ¼ arccosðz=rÞ z ¼ r cos ¼ arctanðy=xÞ For the spherical coordinates of any point in space r 0; 0 ; 0 2 So, converting Cartesian coordinates ð2, 3, 4Þ to spherical coordinates gives ............ 29 P ðr, , Þ ¼ ð5:385, 0:734, 0:983Þ Because x ¼ 2, y ¼ 3, z ¼ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ; r ¼ x2 þ y 2 þ z2 ¼ 4 þ 9 þ 16 ¼ 29 ¼ 5:385 pffiffiffiffiffiffi ¼ arccosðz=rÞ ¼ arccosð4= 29Þ ¼ 0:734 ¼ arctanðy=xÞ ¼ arctan 1:5 ¼ 0:983 And, in reverse, spherical coordinates ð5, =4, =3Þ transform into Cartesian coordinates . . . . . . . . . . . . 30 P ðx, y, zÞ ¼ ð1:768, 3:061, 3:536Þ Because x ¼ r sin cos ¼ 5 sin cos ¼ 5ð0:707Þð0:5Þ ¼ 1:768 4 3 y ¼ r sin sin ¼ 5 sin sin ¼ 5ð0:707Þð0:866Þ ¼ 3:061 4 3 z ¼ r cos ¼ 5 cos ¼ 5ð0:707Þ ¼ 3:536. 4 One of the main uses of cylindrical and spherical coordinates occurs in integrals dealing with volumes of solids. In preparation for this, let us consider the next important section of the work. So move on 707 Multiple integration 2 Element of volume in space in the three coordinate systems 1 Cartesian coordinates z We have already used this many times. δz x O y y δx x 2 v ¼ x y z z δy Cylindrical coordinates r δθ z r δθ δz δz δr δr O δθ θ y r v ¼ r r z x 3 ; v ¼ r r z δr Spherical coordinates z δr r δθ r θ O ϕ x δθ rsinθ δϕ δϕ y ρ v ¼ r r r sin ; v ¼ r 2 sin r It is important to make a note of these results, since they are required when we change the variables in various types of integrals. We shall meet them again before long, so be sure of them now. 31 708 Programme 20 Volume integrals z 32 z2 = F(x, y) z2 z1 x1 x2 O y1 z1 = f(x, y) y2 y x A solid is enclosed by a lower surface z1 ¼ f ðx, yÞ and an upper surface z2 ¼ Fðx, yÞ: Then, in general, using Cartesian coordinates, the element of volume is v ¼ x y z: The approximate value of the total volume V is then found (a) by summing v from z ¼ z1 to z ¼ z2 to obtain the volume of the column (b) by summing all such columns from y ¼ y1 to y ¼ y2 to obtain the volume of the slice (c) by summing all such slices from x ¼ x1 to x ¼ x2 to obtain the total volume V. Then, when x ! 0, y ! 0, z ! 0; the summation becomes an integral ð x¼x2 ð y¼y2 ð z¼z2 V¼ dz dy dx x¼x1 y¼y1 z¼z1 Example 1 Find the volume of the solid bounded by the planes z ¼ 0, x ¼ 0, y ¼ 0, x2 þ y 2 ¼ 4 and z ¼ 6 xy for x 0, y 0, z 0. First sketch the figure, so that we can see what we are doing. Take your time over it. 709 Multiple integration 2 z O y x x2 + y2 = 4 v ¼ x y z Volume of column z¼6xy X x y z z¼0 Volume of slice Total volume pffiffiffiffiffiffiffiffi( 4x2 6xy X X y¼0 z¼0 pffiffiffiffiffiffiffiffi 2 6xy 2 4x X X X x¼0 y¼0 ) x y z x y z z¼0 If x ! 0, y ! 0, z ! 0, then ffiffiffiffiffiffiffiffi2 ð ð 2 ð p4x 6xy dz dy dx V¼ 0 0 0 Starting with the innermost integral 6xy ð 6xy dz ¼ z 0 Then 0 ¼ 6 xy ffiffiffiffiffiffiffiffi2 ð p4x ð6 xyÞ dy ¼ . . . . . . . . . . . . 0 33 z = 6 – xy 710 Programme 20 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x 6 4 x2 4 x2 2 34 Because ð pffiffiffiffiffiffiffiffi2 y¼pffiffiffiffiffiffiffiffi 4x2 xy 2 ð6 xyÞ dy ¼ 6y 2 y¼0 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ¼ 6 4 x2 ð4 x2 Þ 2 ð2 x3 Then finally V ¼ 6ð4 x2 Þ1=2 2x þ dx 2 0 ð Now we are faced with ð4 x2 Þ1=2 dx. You may remember that this is a ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi xo standard form a2 x2 dx ¼ 12 x a2 x2 þ a2 arcsin . a ð 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 x2 dx, put x ¼ 2 sin and proceed from If not, to evaluate 4x 0 there. Finish off the main integral, so that we have V ¼ ............ 35 V ¼ 6 2 16:8 cubic units Because we had ð2 x3 dx V¼ 6ð4 x2 Þ1=2 2x þ 2 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x 2 x4 ¼ 3 x 4 x2 þ 4 arcsin x2 2 0 8 0 ¼ 3f4 arcsin 1 4 arcsin 0g 4 þ 2 ¼ 3f2g 2 ¼ 6 2 16:8 711 Multiple integration 2 Alternative method We could, of course, have used cylindrical coordinates in this problem. z z = 6 – xy v ¼ r r z x ¼ r cos ; y ¼ r sin ; z ¼ 6 xy δv x O z r θ y ¼ 6 r 2 sin cos ¼6 y x r2 sin 2 2 x2 + y2 = 4 ; V¼ ¼ ð 2 ð =2 ð 6ðr 2 =2Þ sin 2 r dr d dz r¼0 ð =2 ¼0 ð2 ¼0 r¼0 z¼0 ð 6ðr 2 =2Þ sin 2 dz r dr d z¼0 ¼ ............ Finish it V ¼ 6 2 (as before) r2 sin 2 r dr d 2 ¼0 r¼0 ð =2 ð 2 r3 6r sin 2 dr d ¼ 2 ¼0 r¼0 r¼2 ð =2 4 r 2 ¼ 3r sin 2 d 8 0 r¼0 ð =2 ð12 2 sin 2Þ d ¼ V¼ ð =2 ð 2 6 0 =2 ¼ 12 þ cos 2 0 ¼ ð6 1Þ 1 ; V ¼ 6 2 In this case, the use of cylindrical coordinates facilitates the evaluation. Let us consider another example. 36 712 37 Programme 20 Example 2 To find the moment of inertia and radius of gyration of a thick hollow sphere about a diameter as axis. Outer radius ¼ a; inner radius ¼ b; density of material ¼ c. It is convenient to deal with one-eighth of the sphere in the first octant. a b b a O z θ ϕ ; Total mass of the solid M1 ¼ δv r a b y ρ M1 ¼ 1 4 ða3 b3 Þc ¼ ða3 b3 Þc 8 3 6 x Using spherical coordinates, the element of volume v ¼ . . . . . . . . . . . . 38 v ¼ r 2 sin r Also the element of mass m ¼ cv Second moment of mass of the element about OZ ¼ m2 ¼ mðr sin Þ2 ¼ c r 2 sin r r 2 sin2 ¼ c r 4 sin3 r ; Total second moment for the solid I1 =2 X =2 X a X 1 M 8 c r 4 r sin3 ¼0 ¼0 r¼b Then, as usual, if r ! 0, ! 0; ! 0; we finally obtain ð =2 ð =2 ð a I1 ¼ c r 4 dr sin3 d d ¼0 ¼0 r¼b which you can evaluate without any difficulty and obtain I1 ¼ . . . . . . . . . . . . 713 Multiple integration 2 I1 ¼ 39 5 ða b5 Þc 15 Because ð =2 ð =2 5 a r I1 ¼ c sin3 d d 5 b 0 0 ð =2 ð =2 c 5 ða b5 Þ sin3 d d ¼ 0 0 5 ð =2 ð =2 c ¼ ða5 b5 Þ ð1 cos2 Þ sin d d 5 0 0 =2 ð =2 c cos3 ¼ ða5 b5 Þ cos þ d 5 3 0 0 ð =2 c 5 1 5 d 1 ¼ ða b Þ 5 3 0 =2 2c 5 c 5 ¼ ða b5 Þ ða b5 Þ ¼ 15 15 0 Therefore, the moment of inertia for the whole sphere I is I ¼ 8I1 i.e. I¼ Radius of gyration ðkÞ 8 5 ða b5 Þc 15 Mk2 ¼ I ; k ¼ ............ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a5 b5 k¼ 5 a3 b3 40 We had already calculated the total mass M ¼ 8 5 ða b5 Þ then 15 4 3 8 5 ða b3 Þck2 ¼ ða b5 Þc 3 15 2 a5 b5 ; k2 ¼ 5 a3 b3 4 3 ða b3 Þc and since 3 I¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a5 b5 ; k¼ 5 a3 b3 We have set the working out in considerable detail, since spherical coordinates may be a new topic. Many of the statements can be streamlined when one is familiar with the system. Now move on for another example 714 41 Programme 20 Example 3 Find the total mass of a solid sphere of radius a, enclosed by the surface x2 þ y 2 þ z2 ¼ a2 and having variable density c where c ¼ 1 þ r z and r is the distance of any point from the origin. This is a case where spherical coordinates can clearly be used with advantage. z a O θ ϕ x a r a ρ y (a) . . . . . . . . . In the element of volume, the three dimensions are (b) . . . . . . . . . 42 (a) r (b) r (c) ¼ r sin so that 43 (c) . . . . . . . . . v ¼ . . . . . . . . . . . . v ¼ r 2 sin r Then the mass of the element ¼ c v ¼ ð1 þ r z Þ v and z ¼ r cos ; m ¼ c v ¼ ð1 þ r 2 cos Þ r 2 sin r Since the density uses j z j ¼ 1 we must only consider the region where cos 0 and so we consider the upper hemisphere only. The integral for the total mass M1 is M1 ¼ . . . . . . . . . . . . Write out the integral and insert the limits. 715 Multiple integration 2 M1 ¼ ð ¼2 ð ¼=2 ð r¼a ¼0 i.e. M1 ¼ ð 2 ð =2 ð a ¼0 ð1 þ r 2 cos Þr 2 sin dr d d 44 r¼0 fr 2 sin dr d d þ r 4 sin cos dr d dg ¼0 ¼0 r¼0 ¼ I1 ¼ ð 2 ð =2 ð a 0 0 þ I1 I2 r 2 sin dr d d gives . . . . . . . . . . . . 0 Do not work it out. You can doubtless recognise what the result would represent. The volume of the hemisphere 45 Because the integral is simply the summation of elements of volume throughout the region of the hemisphere. Thus, without more ado, I1 ¼ 2 3 a . 3 Now for I2 . ð 2 ð =2 ð a I2 ¼ r 4 sin cos dr d d 0 0 0 ¼ . . . . . . . . . . . . Evaluate the triple integral. I2 ¼ a5 5 Because ð 2 ð =2 a5 sin cos d d 5 0 0 =2 ð a5 2 sin2 d ¼ 2 0 5 0 ð a5 2 ¼ 1d 10 0 2 a5 a5 ¼ ¼ 10 5 0 I2 ¼ ; I2 ¼ a5 5 So now finish it off. For the complete sphere M ¼ ............ 46 716 Programme 20 47 M¼ 2a3 ð10 þ 3a2 Þ 15 Because M1 ¼ I1 þ I2 ¼ 2 3 a5 a3 a þ ¼ ð10 þ 3a2 Þ 3 5 15 2a3 ð10 þ 3a2 Þ 15 Each problem, then, is tackled in much the same way. Then, for the whole sphere, M ¼ 2M1 ¼ (a) (b) (c) (d) Draw a careful sketch diagram, inserting all relevant information. Decide on the most appropriate coordinate system to use. Build up the multiple integral and insert correct limits. Evaluate the integral. And now we can apply the general guide lines to a final problem. Example 4 Determine the volume of the solid bounded by the planes x ¼ 0, y ¼ 0, z ¼ x, z ¼ 2 and y ¼ 4 x2 in the first quadrant. First we sketch the diagram. z 48 z2 = 2 y = 4 – x2 z1 = x y O x There is no axis of symmetry and no spherical centre. We shall therefore use . . . . . . . . . . . . coordinates. 49 Cartesian So off you go on your own. There are no snags. V ¼ ............ 717 Multiple integration 2 V¼6 50 2 cubic units 3 Here is the complete solution. V 2 4x 2 X X2 X x y z x¼0 y¼0 z¼x ; V¼ ð 4x2 ð 2 ð2 dz dy dx x¼0 y¼0 ¼ ð 2 ð 4x2 0 ¼ ð2 0 ¼ ð2 z¼x ð2 xÞ dy dx 0 4x2 2y xy dx y¼0 f8 2x2 4x þ x3 gdx 0 2 2x3 x4 2x2 þ ¼ 8x 3 4 0 2 ¼6 3 And that is it. Now we move to the next section of work 51 Change of variables in multiple integrals In Cartesian coordinates, we use the variables ðx, y, zÞ; in cylindrical coordinates, we use the variables ðr, , zÞ; in spherical coordinates, we use the variables ðr, , Þ; and we have established relationships connecting these systems of variables, permitting us to transfer from one system to another. These relationships, you will remember, were obtained geometrically in Frames 23 to 30 of this Programme. There are occasions, however, when it is expedient to make other transformations beside those we have used and it is worth looking at the problem in a rather more general manner. This we will now do 718 52 Programme 20 First, however, let us revise a result from an earlier Programme on determinants to find the area of the triangle ABC. y (x3, y3) (x2, y2) (x1, y1) O x1 If we arrange the vertices x3 x2 x A ðx1 , y1 Þ B ðx2 , y2 Þ C ðx3 , y3 Þ in an anticlockwise manner then area triangle ABC ¼ trapezium AMPC þ trapezium CPNB trapezium AMNB ¼ 12 fðx3 x1 Þðy1 þ y3 Þ þ ðx2 x3 Þðy2 þ y3 Þ ðx2 x1 Þðy1 þ y2 Þg ¼ 12 fx3 y1 x1 y1 þ x3 y3 x1 y3 þ x2 y2 þ x2 y3 x3 y2 x3 y3 x2 y1 x2 y2 þ x1 y1 þ x1 y2 g ¼ 1 2 fðx2 y3 1 ¼ 12 x1 y1 x3 y2 Þ þ ðx3 y1 x1 y3 Þ þ ðx1 y2 x2 y1 Þg 1 1 x2 x3 y2 y3 The determinant is positive if the points A, B, C are taken in an anticlockwise manner. We shall need to use this result in a short while, so keep it in mind. On to the next frame 719 Multiple integration 2 Curvilinear coordinates ð ð Consider the double integral ðx, yÞdA where dA ¼ dxdy in Cartesian 53 R coordinates. Let u and v be two new independent variables defined by u ¼ Fðx, yÞ and v ¼ Gðx, yÞ where these equations can be simultaneously solved to obtain x ¼ f ðu, vÞ and y ¼ gðu, vÞ. Furthermore, these transformation equations are such that every point ðx, yÞ is mapped to a unique point ðu, vÞ and vice versa. Let us see where this leads us, so on to the next frame The equation u ¼ Fðx, yÞ will be a family of curves depending on the particular constant value given to u in each case. y Curves u ¼ Fðx; yÞ for different constant values of u. O u=1 u=2 u=3 x etc. u=4 Similarly, v ¼ Gðx; yÞ will be a family of curves depending on the particular constant value assigned to v in each case. y v=4 v=3 Curves v ¼ Gðx, yÞ for different constant values of v. v=2 v=1 x These two sets of curves will therefore cover the region R and form a network, and to any point P ðx0 , y0 Þ there will be a pair of curves u ¼ u0 (constant) and v ¼ v0 (constant) that intersect at that point. y R v0 + δv S δA1 P Q u0 u0 + δu v0 x 54 720 Programme 20 The u- and v-values relating to any particular point are known as its curvilinear coordinates and x ¼ f ðu, vÞ and y ¼ gðu, v) are the transformation equations between the two systems. In the Cartesian coordinates ðx, yÞ system, the element of area A ¼ xy and is the area bounded by the lines x ¼ x0 , x ¼ x0 þ x, y ¼ y0 , and y ¼ y0 þ y. In the new system of curvilinear coordinates (u, v) the element of area A1 can be taken as that of the figure P, Q, R, S, i.e. the area bounded by the curves u ¼ u0 , u ¼ u0 þ u, v ¼ v0 and v ¼ v0 þ v. Since A1 is small, PQRS may be regarded as a parallelogram i.e. A1 2 area of triangle PQS and this is where we make use of the result previously revised that the area of a triangle ABC with vertices ðx1 , y1 ), ðx2 , y2 Þ, ðx3 , y3 ) can be expressed in determinant form as Area ¼ . . . . . . . . . . . . 55 Area ¼ 1 1 x1 2 y1 1 x2 y2 1 x3 y3 Before we can apply this, we must find the Cartesian coordinates of P, Q and S in the diagram on page 646 where we omit the subscript 0 on the coordinates. If x ¼ f ðu, vÞ, then a small increase x in x is given by x ¼ . . . . . . . . . . . . 56 x ¼ i.e. x ¼ @f @f u þ v @u @v @x @x u þ v @u @v and, for y ¼ gðu; vÞ y ¼ . . . . . . . . . . . . 721 Multiple integration 2 y ¼ @y @y u þ v @u @v 57 Now (a) P is the point ðx, y) (b) Q corresponds to small changes from P. x ¼ @x @x u þ v @u @v and y ¼ But along PQ v is constant. @y @y u þ v @u @v ; v ¼ 0. @x @y u and y ¼ u @u @u @x @y u, y þ u . i.e. Q is the point x þ @u @u ; x ¼ (c) Similarly for S, since u is constant along PS u ¼ 0 and @x @y ; S is the point x þ v, y þ v @v @v So the Cartesian coordinates of P, Q, S are @x @y @x @y u, y þ u ; S x þ v, y þ v P ðx, yÞ; Q x þ @u @u @v @v ; The determinant for the area PQS is . . . . . . . . . . . . 1 Area ¼ 1 x 2 y 1 1 @x @x xþ u x þ v @u @v @y @y u y þ v yþ @u @v Subtracting column 1 from columns 2 and 3 gives 1 0 0 @x @x u v 1x @u @v Area ¼ 2 @y @y u v y @u @v which simplifies immediately to ............ 58 722 59 Programme 20 @x @x 1 @u u @v v Area ¼ @y 2 @y u v @u @v Then, taking out the factor u from the first column and the factor v from the second column, this becomes Area ¼ . . . . . . . . . . . . 60 @x @x 1 @u @v uv 2 @y @y @u @v The area of the approximate parallelogram is twice the area of the triangle. @x @x @u @v ; Area of parallelogram ¼ A1 ¼ u v @y @y @u @v Expressing this in differentials @x @x @u @v dA ¼ du dv @y @y @u @v and, for convenience, this is often written @ðx, yÞ du dv dA ¼ @ðu, vÞ @ðx, yÞ is called the Jacobian of the transformation from the Cartesian @ðu, vÞ coordinates (x, y) to the curvilinear coordinates ðu, vÞ. @x @x @ðx, yÞ @u @v ¼ ; Jðu; vÞ ¼ @ðu; vÞ @y @y @u @v So, if the transformation equations are x ¼ uðu þ vÞ and y ¼ uv2 Jðu, vÞ ¼ . . . . . . . . . . . . 723 Multiple integration 2 61 Jðu; vÞ ¼ uvð4u þ vÞ Because @x ¼ 2u þ v @u @y ¼ v2 @u @x ¼u @v @y ¼ 2uv @v 2u þ v u ; Jðu; vÞ ¼ v 2 ¼ 4u2 v þ 2uv2 uv2 2uv ¼ 4u2 v þ uv2 ¼ uvð4u þ vÞ Next frame Sometimes the transformation equations are given the other way round. That is, where u and v are given as expressions in x and y. In such a case Jðu, vÞ can be found using the fact that @ðx, yÞ 1 ¼ @ðu, vÞ @ðu, vÞ @ðx, yÞ 62 For example, if the transformation equations are given as u ¼ x2 þ y 2 and v ¼ 2xy then Jðu, vÞ ¼ . . . . . . . . . . . . 1 Jðu, vÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 u2 v 2 Because @u @u @ðu, vÞ 2x @x @y ¼ ¼ @v @v 2y @ðx, yÞ @x @y 2y ¼ 4x2 4y 2 2x and so Jðu, vÞ ¼ @ðx, yÞ 1 1 ¼ ¼ 2 @ðu, vÞ @ðu, v 4ðx y 2 Þ @ðx, yÞ Now u v ¼ x2 2xy þ y 2 ¼ ðx yÞ2 and u þ v ¼ x2 þ 2xy þ y 2 ¼ ðx þ yÞ2 pffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi and so x2 y 2 ¼ ðx yÞðx þ yÞ ¼ u v u þ v ¼ u2 v 2 giving 1 Jðu, vÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 u2 v 2 There is one further point to note in this piece of work, so move on 63 724 64 Programme 20 Note: In the transformation, it is possible for the order of the points P, Q, R, S to be reversed with the result that A may give a negative result when the determinant is evaluated. To ensure a positive element of area, the result is finally written dA ¼ @ðx, yÞ du dv @ðu, vÞ where the ‘modulus’ lines indicate the absolute value of the Jacobian. ð ð Therefore, to rewrite the integral Fðx, yÞ dx dy in terms of the new R variables, u and v, where x ¼ f ðu; v) and y ¼ gðu; v), we substitute for x and y @ðx, yÞ in Fðx, yÞ and replace dx dy with du dv: @ðu; vÞ The integral then becomes ð ð @ðx, yÞ Fff ðu, vÞ, gðu; vÞg du dv @ðu, vÞ R Make a note of this result 65 Example 1 Express I ¼ ð ð xy2 dx dy in polar coordinates, making the substitutions R x ¼ r cos , y ¼ r sin . @x ¼ cos @r @y ¼ sin @r @x ¼ r sin @ @y ¼ r cos @ ; Jðr; Þ ¼ . . . . . . . . . . . . 66 Jðr; Þ ¼ r cos r sin ¼ r cos2 þ r sin2 ¼ r sin r cos ð ð xy 2 dx dy becomes . . . . . . . . . . . . Then I ¼ Jðr; Þ ¼ R 725 Multiple integration 2 I¼ ð ð r 3 sin2 cos r dr d 67 R Because xy 2 ¼ r cos r 2 sin2 ¼ r 3 sin2 cos @ðx, yÞ dr d ¼ r dr d @ðr; Þ ð ð ð ð r 3 sin2 cos r dr d ¼ r 4 sin2 cos dr d ; I¼ R R Now this one. Example 2 Express I ¼ ð ð ðx2 þ y 2 Þdx dy in terms of u and v, given that x ¼ u2 v 2 R and y ¼ 2uv. First of all, the expression for @ðx, yÞ gives . . . . . . . . . . . . @ðu; vÞ 4 u2 þ v 2 Because @x @x ¼ 2u ¼ 2v @u @v @y @y y ¼ 2uv ; ¼ 2v ¼ 2u @u @v @x @y 2u 2v @ðx, yÞ @u @u ; ¼ ¼ 4ðu2 þ v 2 Þ ¼ @ðu; vÞ @x @y 2v 2u @v @v x ¼ u2 v 2 ; Also x2 þ y 2 ¼ ðu2 v 2 Þ2 þ ð2uvÞ2 ¼ u4 2u2 v 2 þ v 4 þ 4u2 v 2 ¼ u4 þ 2u2 v 2 þ v 4 ¼ ðu2 þ v 2 Þ2 Then I ¼ ðð ðx2 þ y 2 Þ dx dy becomes I ¼ . . . . . . . . . . . . R 68 726 Programme 20 69 I¼4 ð ð ðu2 þ v 2 Þ3 du dv R One more. Example 3 By substituting x ¼ 2uv and y ¼ uð1 v) where u > 0 and v > 0, express ð ð x2 y dx dy in terms of u and v. the integral I ¼ R I ¼ ............ Complete it: there are no snags. 70 I¼8 ð ð u4 v 2 ð1 vÞ du dv R Working: @x ¼ 2v @u @y y ¼ u uv ¼1v @u @x @y @u @u @ðx, yÞ ¼ ; Jðu; vÞ ¼ @ðu; vÞ @x @y x ¼ 2uv @v v v 1 x2 y ¼ 4u2 v 2 ðu uvÞ ¼ 4u3 v 2 ð1 vÞ ð ð 4u3 v 2 ð1 vÞ 2u du dv ; I¼ R I¼8 R 2u u ¼ 2u 1 0 @ðx, yÞ ¼ 2u @ðu; vÞ ð ð 1v 1 ¼ 2u 1 2v ¼ @v 1v ¼ 2u ; @x ¼ 2u @v @y ¼ u @v ; u4 v 2 ð1 vÞ du dv 727 Multiple integration 2 Transformation in three dimensions If we extend the previous results to convert variables ðx, y, zÞ to ðu, v, wÞ, we proceed in just the same way. If x ¼ f ðu, v, wÞ; y ¼ gðu, v, wÞ; z ¼ hðu; v; wÞ Then @x @u @ðx; y; zÞ @x ¼ Jðu; v; wÞ ¼ @ðu; v; wÞ @v @x @w @y @u @y @v @y @w @z @u @z @v @z @w and the element of volume dV ¼ dx dy dz becomes dV ¼ J ðu; v; wÞ jdu dv dw ððð Also Fðx; y; zÞ dx dy dz is transformed into ððð Gðu; v; wÞ @ðx; y; zÞ du dv dw @ðu; v; wÞ Now for an example, so move on Example 4 To transform a triple integral I ¼ ððð 71 Fðx; y; zÞ dx dy dz in Cartesian coordinates to spherical coordinates by the transformation equations x ¼ r sin cos y ¼ r sin sin z ¼ r cos : First we need the partial derivatives, from which to build up the Jacobian. These are . . . . . . . . . . . . 728 Programme 20 72 @x ¼ sin cos @r @x ¼ r cos cos @ @x ¼ r sin sin @ @y ¼ sin sin @r @y ¼ r cos sin @ @y ¼ r sin cos @ sin cos sin sin @z ¼ cos @r @z ¼ r sin @ @z ¼0 @ cos ; Jðr; ; Þ ¼ r cos cos r cos sin r sin r sin sin r sin cos 0 r cos cos r cos sin ¼ cos r sin sin r sin cos sin cos þ r sin r sin sin sin sin r sin cos ¼ ............ 73 r 2 sin Because Jðr, , Þ ¼ r 2 cos2 sin cos sin sin cos cos sin sin cos cos sin ¼ ðr 2 sin3 þ r 2 sin cos2 Þ sin cos þ r 2 sin3 ¼ r 2 sin ðsin2 þ cos2 Þðcos2 þ sin2 Þ ¼ r 2 sin ððð ; I¼ Gðu; v; wÞr 2 sin dr d d which agrees, of course, with the result we had previously obtained by a geometric consideration. And that is about it. Check carefully down the Revision summary and the Can you? checklist that now follow, before working through the Test exercise. The Further problems give additional practice. 729 Multiple integration 2 Revision summary 20 1 Surface integrals ð ð ð f ðx, yÞ dy dx I ¼ f ðx, yÞ da ¼ R 2 3 74 R Surface in space ð ð ð ðx; y; zÞ sec dx dy I ¼ ðx; y; zÞ dS ¼ S R ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ð ð @z @z ¼ ðx; y; zÞ 1 þ þ dx dy @x @y R ð < =2Þ Space coordinate systems (a) Cartesian coordinates ðx, y, zÞ z P (x, y, z) First octant: x 0; y 0; z 0 z x O y y x (b) Cylindrical coordinates ðr, , zÞ z P (x, y, z) (r, θ, z) x ¼ r cos y ¼ r sin z¼z z O r θ r0 y pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ x2 þ y 2 ¼ arctanðy=xÞ z¼z x (c) Spherical coordinates (r, ; Þ x ¼ r sin cos y ¼ r sin sin z ¼ r cos z θ r O ϕ x r0 ρ (x, y, z) (r, θ, ϕ) y pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ x 2 þ y 2 þ z2 ¼ arccosðz=rÞ ¼ arctanðy=xÞ 730 Programme 20 4 Elements of volume (a) Cartesian coordinates z δz v ¼ x y z z x y O y x δx δy r0 (b) Cylindrical coordinates r δθ z δz δz rδθ δr z O r θ δr y δθ x v ¼ r r z (c) Spherical coordinates δr z r θ r δθ δθ r sinθ δϕ O ϕ x ρ y δϕ ρ δϕ v ¼ r 2 sin r 5 Volume integrals ððð V¼ dz dy dx ððð I¼ f ðx, y, zÞ dz dy dx 731 Multiple integration 2 6 Change of variables in multiple integrals (a) Double integrals x ¼ f ðu, vÞ; y ¼ gðu, vÞ @x @y @ðx, yÞ @u @u ¼ Jðu; vÞ ¼ @ðu; vÞ @x @y @v @v ð ð ð ð @ðx, yÞ Fðx, yÞ dx dy ¼ F f f ðu, vÞ, gðu, vÞg I¼ du dv @ðu, vÞ R R @ðx, yÞ dA ¼ du dv; @ðu, vÞ (b) Triple integrals x ¼ f ðu, v, wÞ; y ¼ gðu, v, wÞ; z ¼ hðu, v, wÞ @x @u @ðx, y, zÞ @x ¼ Jðu, v, wÞ ¼ @ðu, v, wÞ @v @x @w ððð Then I ¼ ¼ @y @u @y @v @y @w @z @u @z @v @z @w Fðx, y, zÞ dx dy dz ððð Gðu, v, wÞ @ðx, y, zÞ du dv dw @ðu, v, wÞ Can you? 75 Checklist 20 Check this list before and after you try the end of Programme test. On a scale of 1 to 5, how confident are you that you can: . Evaluate double integrals and surface integrals? Yes Frames 1 to 22 . Relate three-dimensional Cartesian coordinates to cylindrical and spherical polar forms? Yes No 23 to 31 . Evaluate volume integrals in Cartesian coordinates and in cylindrical and spherical polar coordinates? Yes No 32 to 50 51 to 73 No . Use the Jacobian to convert integrals given in Cartesian coordinates into general curvilinear coordinates in two and three dimensions? Yes No 732 Programme 20 Test exercise 20 76 1 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Determine the area of the surface z ¼ x2 þ y 2 over the region bounded by x2 þ y 2 ¼ 4. ð 1 Evaluate the surface integral I ¼ dS where ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi over the surface of S x2 þ y 2 the sphere x2 þ y 2 þ z2 ¼ a2 in the first octant. 3 (a) Transform the Cartesian coordinates (1) (4, 2, 3) to cylindrical coordinates ðr, , zÞ (2) (3, 1, 5) to spherical coordinates ðr, , Þ. (b) Express in Cartesian coordinates ðx; y; zÞ (1) the cylindrical coordinates ð5; =4; 3) (2) the spherical coordinates ð4; =6; 2Þ. 4 Determine the volume of the solid bounded by the plane z ¼ 0 and the surfaces x2 þ y 2 ¼ 4 and z ¼ x2 þ y 2 þ 1. Determine the total mass of a solid hemisphere bounded by the plane z ¼ 0 and the surface x2 þ y 2 þ z2 ¼ a2 ðz 0Þ if the density at any point is given by ¼ 1 z ðz < aÞ. ð ð ðx yÞ dx dy in terms of u and v, where 6 (a) Express the integral I ¼ 5 R x ¼ uð1 þ vÞ and y ¼ u v. (b) Express the triple integral I ¼ ð ð ð xþz dx dy dz in terms of u, v, w using y the transformation equations x ¼ u þ v þ w; y ¼ v 2 w; z ¼ u w. Further problems 20 77 1 Evaluate the surface integral I ¼ ð ðx2 þ y 2 Þ dS over the surface of the cone S z2 ¼ 4ðx2 þ y 2 Þ between z ¼ 0 and z ¼ 4. 2 Find the position of the centre of gravity of that part of a thin spherical shell x2 þ y 2 þ z2 ¼ a2 which exists in the first octant. 3 Determine the surface area of the plane 6x þ 3y þ 4z ¼ 60 cut off by x ¼ 0, y ¼ 0, x ¼ 5, y ¼ 8. 4 Find the surface area of the plane 3x þ 2y þ 3z ¼ 12 cut off by the planes x ¼ 0, y ¼ 0, and the cylinder x2 þ y 2 ¼ 16 for x 0, y 0. 733 Multiple integration 2 5 Determine the area of the paraboloid z ¼ 2ðx2 þ y 2 Þ cut off by the cone pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi z ¼ x2 þ y 2 . 6 Find the area of the cone z2 ¼ 4ðx2 þ y 2 Þ which is inside the paraboloid z ¼ 2ðx2 þ y 2 ). 7 Cylinders x2 þ y 2 ¼ a2 and x2 þ z2 ¼ a2 intersect. Determine the total external surface area of the common portion. 8 Determine the surface area of the sphere x2 þ y 2 þ z2 ¼ a2 cut off by the cylinder x2 þ y 2 ¼ ax. 9 A cylinder of radius b, with the z-axis as its axis of symmetry, is removed from a sphere of radius a, a > b, with centre at the origin. Calculate the total curved surface area of the ring so formed, including the inner cylindrical surface. 10 Find the volume enclosed by the cylinder x2 þ y 2 ¼ 9 and the planes z ¼ 0 and z ¼ 5 x. 11 Determine the volume of the solid bounded by the surfaces y ¼ x2 , x ¼ y 2 , z ¼ 2 and x þ y þ z ¼ 4. 12 Find the volume of the solid bounded by the plane z = 0, the cylinder x2 þ y 2 ¼ a2 and the surface z ¼ x2 þ y 2 . 13 A solid is bounded by the planes x ¼ 0, y ¼ 0, z ¼ 2, z ¼ x and the surface x2 þ y 2 ¼ 4. Determine the volume of the solid. 14 Find the position of the centre of gravity of the part of the solid sphere x2 þ y 2 þ z2 ¼ a2 in the first octant. pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A solid is bounded by the cone z ¼ 2 x2 þ y 2 , z 0, and the sphere 2 x2 þ y 2 þ ðz aÞ ¼ 2a2 . Determine the volume of the solid so formed. 15 x 2 y 2 z2 þ þ ¼ 1. a2 b2 c2 16 Determine the volume enclosed by the ellipsoid 17 Find the volume of the solid in the first octant bounded by the planes x ¼ 0, y ¼ 0, z ¼ 0, z ¼ x þ y and the surface x2 þ y 2 ¼ a2 . ðð Express the integral ðx2 þ y 2 Þ dx dy in terms of u and v, using the 18 transformations u ¼ x þ y, v ¼ x y. 19 Determine an expression for the element of volume dx dy dz in terms of u, v, w using the transformations x ¼ uð1 vÞ, y ¼ uv, z ¼ uvw. 20 A solid sphere of radius a has variable density c at any point ðx, y, zÞ given by c ¼ kða zÞ where k is a constant. Determine the position of the centre of gravity of the sphere. ðð Calculate x2 y 2 dx dy over the triangular region in the x–y plane with vertices 21 ð0, 0Þ, ð1, 1Þ, ð1, 2Þ. 734 Programme 20 22 Evaluate the integral I ¼ ð 2 ð pffiffiffiffiffiffiffiffi 4y2 0 y pffiffiffiffiffiffiffiffiffiffiffi x2 þ y 2 dx dy by transforming to polar yð2yÞ coordinates. 23 Evaluate I ¼ ð1 ðy 0 xy 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx dy. x2 þ y 2 0 24 Find the volume bounded by the cylinder x2 þ y 2 ¼ a2 , the plane z ¼ 0 and the surface z ¼ x2 þ y 2 . Convert to polar coordinates and show that a4 V¼ 2. 25 By changing the order of integration in the integral ða ða 2 y dy dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I¼ x2 þ y 2 0 x pffiffiffi show that I ¼ 13 a3 lnð1 þ 2Þ. Programme 21 Frames 1 to 79 Integral functions Learning outcomes When you have completed this Programme you will be able to: . Derive the recurrence relation for the gamma function and evaluate the gamma function for certain rational arguments . Evaluate integrals that require the use of the gamma function in their solution . Identify the beta function and evaluate integrals that require the use of the beta function in their solution . Derive the relationship between the gamma function and the beta function . Use the duplication formula to evaluate the gamma function for half integer arguments . Recognise the error function and its relation to the Gaussian probability distribution . Recognise elliptic functions of the first and second kind . Evaluate integrals that require the use of elliptic functions in their solution . Use alternative forms of the elliptic functions Prerequisite: Engineering Mathematics (Sixth Edition) Programmes 15 Integration 1, 16 Integration 2 and 17 Reduction formulas 735 736 Programme 21 Integral functions 1 Some functions are most conveniently defined in the form of integrals and we shall deal with one or two of these in the present Programme. The gamma function The gamma function ðxÞ is defined by the integral ð1 ðxÞ ¼ t x1 et dt ð1Þ 0 and is convergent for x > 0: ð1 t x et dt From ð1Þ: ðx þ 1Þ ¼ 0 Integrating by parts t 1 ð1 e ðx þ 1Þ ¼ t x þx et t x1 dt 1 0 0 ¼ f0 0g þ xðxÞ ; ðx þ 1Þ ¼ xðxÞ ð2Þ This is a fundamental recurrence relation for gamma functions. It can also be written as ðxÞ ¼ ðx 1Þðx 1Þ With it we can derive a number of other results. For instance, when x ¼ n, a positive integer 1, then ðn þ 1Þ ¼ nðnÞ But ðnÞ ¼ ðn 1Þðn 1Þ ¼ nðn 1Þðn 1Þ ðn 1Þ ¼ ðn 2Þðn 2Þ ¼ nðn 1Þðn 2Þðn 2Þ ¼ nðn 1Þðn 2Þðn 3Þ . . . 1ð1Þ ¼ n!ð1Þ But, from the original definition 2 ð1Þ ¼ . . . . . . . . . . . . ð1Þ ¼ 1 Because ð1Þ ¼ ð1 t 0 et dt ¼ et 1 0 Therefore, we have and ¼0þ1¼1 0 ð1Þ ¼ 1 ðn þ 1Þ ¼ n! provided n is a positive integer. ; ð7Þ ¼ . . . . . . . . . . . . ð3Þ 737 Integral functions 3 ð7Þ ¼ 720 Because ð7Þ ¼ ð6 þ 1Þ ¼ 6! ¼ 720. Knowing ð7Þ ¼ 720, ð8Þ ¼ . . . . . . . . . . . . and ð9Þ ¼ . . . . . . . . . . . . ð8Þ ¼ 5040; 4 ð9Þ ¼ 40 320 Because ð8Þ ¼ ð7 þ 1Þ ¼ 7ð7Þ ¼ 7ð720Þ ¼ 5040 ð9Þ ¼ ð8 þ 1Þ ¼ 8ð8Þ ¼ 8ð5040Þ ¼ 40 320 We can also use the recurrence relation in reverse ðx þ 1Þ ¼ xðxÞ ðx þ 1Þ ; ðxÞ ¼ x ð4Þ For example, given that ð7Þ ¼ 720; we can determine ð6Þ ð6 þ 1Þ ð7Þ 720 ¼ ¼ ¼ 120 6 6 6 ð5Þ ¼ . . . . . . . . . . . . ð6Þ ¼ and then ð5Þ ¼ 24 ð5Þ ¼ ð5 þ 1Þ ð6Þ 120 ¼ ¼ ¼ 24: 5 5 5 So far, we have used the original definition ð1 ðxÞ ¼ t x1 et dt 0 for cases where x is a positive integer n. What happens when x ¼ 12? We will investigate. ð1 1 t 1=2 et dt 2 ¼ 0 Putting t ¼ u2 , dt ¼ 2u du, then 12 ¼ . . . . . . . . . . . . 5 738 Programme 21 6 12 ¼ 2 Because ð1 12 ¼ 2 u1 eu 2u du ¼ 2 0 ð1 ð1 2 eu du 0 2 eu du. 0 Unfortunately, ð1 2 eu du cannot easily be determined by normal means. 0 It is, however, important, so we have to find a way of getting round the difficulty. ð1 2 Evaluation of ex dx 0 ð1 2 2 ex dx; then also I ¼ ey dy 0 0 ð 1 ð 1 ð1 ð1 2 2 2 x2 y 2 ; I ¼ e dx e dy ¼ eðx þy Þ dx dy Let I ¼ ð1 0 0 0 0 a ¼ x y represents an element of area in the x–y plane and the integration with the stated limits covers the whole of the first quadrant. y δy δx y O x x Converting to polar coordinates, the element of area a ¼ r r. Also, x2 þ y 2 ¼ r 2 2 ; eðx þy 2 Þ ¼ er 2 For the integration to cover the same region as before, y δθ r δθ δr r O θ x the limits of r are r ¼ 0 to r ¼ 1 the limits of are ¼ 0 to ¼ =2. y x 739 Integral functions ð =2 " r2 #1 e ; I ¼ e r dr d ¼ d 2 0 0 0 0 =2 ð =2 1 ¼ ¼ d ¼ 2 2 4 0 0 pffiffiffi ; I¼ p2ffiffiffi ð1 2 ex dx ¼ ; 2 0 2 ð =2 ð 1 r 2 ð5Þ This result opens the way for others, so make a note of it and then move on to the next frame Before that diversion, we had established that ð1 2 eu du 12 ¼ 2 7 0 pffiffiffi pffiffiffi e du ¼ ; 12 ¼ We now know that 2 0 From this, using the recurrence relation ðx þ 1Þ ¼ xðxÞ, we can obtain the following pffiffiffi pffiffiffi 32 ¼ 12 ð12Þ ¼ 12 ð Þ ; ð32Þ ¼ 2 pffiffiffi pffiffiffi 3 ; 52 ¼ 52 ¼ 32 32 ¼ 32 2 4 72 ¼ . . . . . . . . . . . . ð1 u2 7 2 pffiffiffi 15 ¼ 8 8 Because pffiffiffi pffiffiffi 5 5 5 5 3 7 15 2 ¼ 2þ1 ¼2 2 ¼ ¼ 2 4 8 Using the recurrence relation in reverse, i.e. ðxÞ ¼ obtain ðx þ 1Þ , we can also x ð 12Þ ð12Þ pffiffiffi ¼ 43 ¼ 32 ¼ 3 3 1 2 ð 2Þð 2Þ Negative values of x ðx þ 1Þ ; then as x ! 0; ðxÞ ! 1 ; ð0Þ ¼ 1: x The same result occurs for all negative integral values of x – which does not follow from the original definition, but which is obtainable from the recurrence relation. Since ðxÞ ¼ 740 Programme 21 Because at ð0Þ ¼1 1 ð1Þ ð2Þ ¼ ¼ 1 etc: 2 ð1Þ pffiffiffi 12 ¼ 21 ¼ 2 2 x ¼ 1, ð1Þ ¼ x ¼ 2, Also, at x ¼ 12 , and at x ¼ 32 , ð 12Þ 4 pffiffiffi 32 ¼ ¼ 3 32 Similarly ð 52Þ ¼ . . . . . . . . . . . . and ð 72Þ ¼ . . . . . . . . . . . . 9 8 pffiffiffi ; 52 ¼ 15 16 pffiffiffi 72 ¼ 105 So we have (a) For n a positive integer ðn þ 1Þ ¼ nðnÞ ¼ n! ð1Þ ¼ 1; ð0Þ ¼ 1; pffiffiffi (b) 12 ¼ ; pffiffiffi ; 32 ¼ 2 pffiffiffi 3 ; 52 ¼ 4 pffiffiffi 15 72 ¼ ; 8 ðnÞ ¼ 1 pffiffiffi 12 ¼ 2 4 pffiffiffi 32 ¼ 3 8 pffiffiffi 52 ¼ 15 16 pffiffiffi 72 ¼ 105 This is quite a useful list. Make a note of it for future use 10 Graph of y ¼ ðxÞ Values of ðxÞ for a range of positive values of x are available in tabulated form in various sets of mathematical tables. These, together with the results established above, enable us to draw the graph of y ¼ ðxÞ. x 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 ðxÞ 1 1.772 1.000 0.886 1.000 1.329 2.000 3.323 6.000 x 0.5 1.5 2.5 3.5 ðxÞ 3.545 2.363 0.945 0.270 741 Integral functions y y = Γ(x) x pffiffiffiffiffiffiffiffiffi For large n it can be shown that ðn þ 1Þ 2n nn en which gives rise to Stirling’s formula for an approximation to the factorial of a large number pffiffiffiffiffiffiffiffiffi n! 2n nn en 11 Revision Let us now revise the main points before we move on to some examples. The definition of ðxÞ is that ðxÞ ¼ . . . . . . . . . . . . ðxÞ ¼ ð1 t x1 et dt 12 0 The recurrence relation states that ðx þ 1Þ ¼ . . . . . . . . . . . . ðx þ 1Þ ¼ x ðxÞ 13 When x is a positive integer, i.e. x ¼ n, then ðn þ 1Þ ¼ . . . . . . . . . . . . 14 ðn þ 1Þ ¼ n! Then we have a number of specific results ð1Þ ¼ . . . . . . . . . . . . ; ð0Þ ¼ . . . . . . . . . . . . ; 12 ¼ . . . . . . . . . . . . ð1Þ ¼ 1; ð0Þ ¼ 1; pffiffiffi 12 ¼ and finally, for all negative integral values of n ðnÞ ¼ . . . . . . . . . . . . 15 742 Programme 21 16 ðnÞ ¼ 1 Listing them together, we have ð1 ðxÞ ¼ t x1 et dt 0 ðx þ 1Þ ¼ xðxÞ ðn þ 1Þ ¼ n! ð1Þ ¼ 1; ð0Þ ¼ 1; ð12Þ for n a positive integer pffiffiffi ¼ ðnÞ ¼ 1 for n a negative integer. 17 Now for a few examples of evaluation of integrals. Example 1 ð1 x7 ex dx: Evaluate 0 We recognise this as the standard form of the gamma function ð1 ðxÞ ¼ t x1 et dt with the variables changed. 0 It is often convenient to write the gamma function as ð1 xv1 ex dx ðvÞ ¼ 0 Our example then becomes ð1 ð1 x7 ex dx ¼ xv1 ex dx I¼ 0 18 where v ¼ . . . . . . . . . . . . 0 v¼8 ; I ¼ ðvÞ ¼ ð8Þ ¼ . . . . . . . . . . . . 743 Integral functions ð8Þ ¼ 7! ¼ 5040 ð1 i.e. 19 x7 ex dx ¼ ð8Þ ¼ 7! ¼ 5040 0 Example 2 ð1 Evaluate x3 e4x dx. 0 If we compare this with ðvÞ ¼ ð1 xv1 ex dx, we must reduce the power of e 0 to a single variable, i.e. put y = 4x, and we use this substitution to convert the whole integral into the required form. y ¼ 4x ; dy ¼ 4 dx Limits remain unchanged. The integral now becomes . . . . . . . . . . . . I¼ ð1 0 1 ; I¼ 4 4 ð1 y 3 ey dy ¼ 0 y 4 1 ðvÞ 44 3 ey dy 4 where v ¼ . . . . . . . . . . . . v¼4 Because ð1 ð1 y v1 ey dy ¼ y 3 ey dy 0 20 21 ; v¼4 0 ; I¼ 1 ð4Þ ¼ . . . . . . . . . . . . 44 I¼ 3 128 Because 1 1 6 3 ð4Þ ¼ ð3!Þ ¼ ¼ I¼ 256 256 256 128 One more. Example 3 ð1 2 Evaluate x1=2 ex dx. 0 The substitution here is to put . . . . . . . . . . . . 22 744 Programme 21 23 y ¼ x2 Work through it as before. When you have completed it, check with the next frame. 24 Here is the working. y ¼ x2 ; dy ¼ 2x dx Limits x ¼ 0, y ¼ 0; x ¼ 1, y ¼ 1. x ¼ y 1=2 ; x1=2 ¼ y 1=4 ð1 ð 1 1=4 y y e dy ; I¼ y 1=4 ey dy=2x ¼ 2y 1=2 0 0 ð1 1 ¼ y 1=4 ey dy 2 0 ð 1 1 v1 y 3 1 3 ; I¼ ¼ y e dy where v ¼ 2 0 4 2 4 From tables, ð0:75Þ ¼ 1:2254 ; I ¼ 0:613 Here is part of a table that may be useful. x ðxÞ x ðxÞ 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 3.6256 1.7725 1.2254 1.0000 0.9064 0.8862 0.9191 1.0000 1.1330 1.3293 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00 1.6084 2.0000 2.5493 3.3234 4.4230 6.0000 8.2851 11.6318 16.5862 24.0000 2.50 Now we will move on to another set of functions closely related to gamma functions. Let us start a new frame 745 Integral functions 25 The beta function The beta function Bðm, nÞ, is defined by ð1 Bðm, nÞ ¼ xm1 ð1 xÞn1 dx ð1Þ 0 which converges for m > 0 and n > 0: Putting ð1 xÞ ¼ u ; x¼1u ; dx ¼ du Limits: when x ¼ 0; u ¼ 1; when x ¼ 1; u ¼ 0 ð0 ð1 ; Bðm, nÞ ¼ ð1 uÞm1 un1 du ¼ ð1 uÞm1 un1 du ¼ ð1 1 0 un1 ð1 uÞm1 du ¼ Bðn; mÞ 0 ; Bðm, nÞ ¼ Bðn, mÞ ð2Þ Alternative form of the beta function We had Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx 0 If we put x ¼ sin2 , the result then becomes . . . . . . . . . . . . Bðm, nÞ ¼ 2 ð =2 26 sin2m1 cos2n1 d 0 Because if x ¼ sin2 ; dx ¼ 2 sin cos d: When x ¼ 0, ¼ 0; when x ¼ 1, ¼ =2. 1 x ¼ 1 sin2 ¼ cos2 ð =2 ; Bðm, nÞ ¼ 2 sin2m2 cos2n2 sin cos d 0 ; Bðm, nÞ ¼ 2 ð =2 sin2m1 cos2n1 d 0 Make a note of this result. We shall need to use it later. ð3Þ 746 27 Programme 21 Reduction formulas In Programme 17 of Engineering Mathematics (Sixth Edition) we established useful reduction formulas relating to integrals of powers of sines and cosines, particularly when the integral limits are 0 and =2. ð ð =2 n 1 =2 n1 (a) sinn x dx ¼ sinn2 x dx i.e. Sn ¼ ð4Þ Sn2 n n 0 0 ð ð =2 n 1 =2 n1 n Cn2 (b) cos x dx ¼ cosn2 x dx i.e. Cn ¼ ð5Þ n n 0 0 A third reduction formula for products of powers of sines and cosines is ð ð =2 m 1 =2 m n sin x cos x dx ¼ sinm2 x cosn x dx (c) mþn 0 0 ð =2 If we denote sinm x cosn x dx by Im; n , the last result can be written 0 m1 Im2; n Im; n ¼ mþn ð =2 sinm x cosn x dx can be expressed as Alternatively, ð6Þ 0 ð n 1 =2 sinm x cosn2 x dx mþn 0 n1 Im; n2 i.e. Im; n ¼ ð7Þ mþn ð =2 Now Bðm, nÞ ¼ 2 sin2m1 cos2n1 d and if we apply (6) to the integral, 0 we have ð =2 sin2m1 cos2n1 d 0 ð =2 ð2m 1Þ 1 sin2m3 cos2n1 d ¼ ð2m 1Þ þ ð2n 1Þ 0 ð =2 m1 sin2m3 cos2n1 d ¼ mþn1 0 Now, using (7) with the right-hand integral ð =2 sin2m1 cos2n1 d 0 ð =2 m1 ð2n 1Þ 1 sin2m3 cos2n3 d m þ n 1 ð2m 3Þ þ ð2n 1Þ 0 ð =2 m1 n1 sin2m3 cos2n3 d ¼ mþn1 mþn2 0 ¼ 747 Integral functions ð =2 ðm 1Þðn 1Þ sin2m3 cos2n3 d 2 ðm þ n 1Þðm þ n 2Þ 0 ðm 1Þðn 1Þ Bðm 1; n 1Þ i.e. Bðm, nÞ ¼ ðm þ n 1Þðm þ n 2Þ ; Bðm, nÞ ¼ ð8Þ This is obviously a reduction formula for Bðm, nÞ and the process can be repeated as required. For example Bð4, 3Þ ¼ . . . . . . . . . . . . Bð4, 3Þ ¼ ð3Þð2Þ ð2Þð1Þ Bð2, 1Þ ð6Þð5Þ ð4Þð3Þ 28 Because, applying (8) Bð4, 3Þ ¼ ð3Þð2Þ ð3Þð2Þ ð2Þð1Þ Bð3, 2Þ ¼ Bð2, 1Þ ð6Þð5Þ ð6Þð5Þ ð4Þð3Þ Now we must evaluate Bð2, 1Þ for we can go no further in the reduction process, since, from the definition of Bðm, nÞ, m and n must be ............ 29 >0 But Bð2, 1Þ ¼ 2 ð =2 sin3 cos d ¼ 2 0 =2 sin4 1 ¼ 4 0 2 ð3Þð2Þ ð2Þð1Þ 1 ; Bð4, 3Þ ¼ ð6Þð5Þ ð4Þð3Þ 2 ð3Þð2Þð1Þ ð2Þð1Þ ð3!Þð2!Þ ¼ ¼ ð6Þð5Þð4Þð3Þð2Þð1Þ ð6!Þ Similarly, Bð5, 3Þ ¼ . . . . . . . . . . . . Bð5; 3Þ ¼ ð4!Þð2!Þ ð7!Þ Because ð4Þð2Þ ð4Þð2Þ ð3Þð1Þ Bð4; 2Þ ¼ Bð3; 1Þ ð7Þð6Þ ð7Þð6Þ ð5Þð4Þ 6 =2 ð =2 sin 1 5 Bð3; 1Þ ¼ 2 sin cos d ¼ 2 ¼ 6 3 0 0 ð4Þð2Þ ð3Þð1Þ 1 ð2Þ ð4!Þð2!Þ ; Bð5; 3Þ ¼ ¼ ð7Þð6Þ ð5Þð4Þ 3 ð2Þ ð7!Þ Bð5; 3Þ ¼ 30 748 Programme 21 ðm 1Þ!ðn 1Þ! ðm þ n 1Þ! ð =2 Bðk, 1Þ ¼ 2 sin2k1 cos d ð9Þ In general Bðm, nÞ ¼ Note that 0 ¼2 ð =2 sin2k1 dðsin Þ 0 " #=2 sin2k 1 ¼2 ¼ 2k k 0 1 ; Bðk, 1Þ ¼ k ; Bðk, 1Þ ¼ Bð1, kÞ ¼ 1 k We can also use the trigonometrical definition (3) to evaluate B B 12 ; 12 ¼ . . . . . . . . . . . . 1 2;2 B 12 ; 12 ¼ 31 Because Bðm, nÞ ¼ 2 ; B 1 1 2, 2 ¼2 ¼2 ð =2 0 ð =2 0 ð =2 sin2m1 cos2n1 d sin0 cos0 d =2 1 d ¼ 2 ¼ 0 Revision Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx ð11Þ 0 Now let us summarise our various results so far. 32 ð10Þ 1 Next frame m > 0; n > 0 0 Bðm, nÞ ¼ Bðn; mÞ ð =2 sin2m1 cos2n1 d Bðm, nÞ ¼ 2 0 ðm 1Þðn 1Þ Bðm 1; n 1Þ ðm þ n 1Þðm þ n 2Þ ðm 1Þ!ðn 1Þ! m and n positive integers Bðm, nÞ ¼ ðm þ n 1Þ! 1 ; Bð1; 1Þ ¼ 1 Bðk, 1Þ ¼ Bð1; kÞ ¼ k Bð12 , 12Þ ¼ Bðm, nÞ ¼ Be sure that you are familiar with all these. We shall be using them all in due course. 749 Integral functions 33 Relation between the gamma and beta functions If m and n are positive integers Bðm, nÞ ¼ ðm 1Þ!ðn 1Þ! ðm þ n 1Þ! Also, we have previously established that, for n a positive integer, n! ¼ ðn þ 1Þ ; ðm 1Þ! ¼ ðmÞ and ðn 1Þ! ¼ ðnÞ and also ðm þ n 1Þ! ¼ ðm þ nÞ ; Bðm, nÞ ¼ ðm 1Þ!ðn 1Þ! ðmÞðnÞ ¼ ðm þ n 1Þ! ðm þ nÞ The relation Bðm, nÞ ¼ ð12Þ ðmÞðnÞ holds good even when m and n are not ðm þ nÞ necessarily integers. We will prove this in the next frame, so move on Proof that Let ðmÞ ¼ Bðm, nÞ ¼ ð1 ðmÞðnÞ ðm þ nÞ 34 xm1 ex dx and ðnÞ ¼ 0 ; ðmÞðnÞ ¼ ¼ ð1 m1 ð01 x ð1 0 0 x e dx ð1 ð1 y n1 ey dy 0 y n1 ey dy 0 xm1 y n1 eðxþyÞ dx dy Note that the integration is carried out over the first quadrant of the x–y plane. Putting x ¼ u2 and y ¼ v 2 dx ¼ 2u du and dy ¼ 2v dv ð1 ð1 2 2 u2m2 v 2n2 eðu þv Þ uv du dv ; ðmÞðnÞ ¼ 4 ð01 ð01 2 2 ¼4 u2m1 v 2n1 eðu þv Þ du dv 0 y 0 O x 750 Programme 21 y If we now convert to polar coordinates, u ¼ r cos ; v ¼ r sin ; du dv ¼ r dr d u2 þ v 2 ¼ r 2 0<r<1 δθ 0 < < =2 δr r O ; ðmÞðnÞ ¼ 4 ¼4 ð =2 ð 1 0 ð =2 0 ð1 0 0 y θ x x 2 r 2m1 cos2m1 r 2n1 sin2n1 er r dr d 2 r 2mþ2n2 er cos2m1 sin2n1 r dr d 2 ; dw ¼ 2 r dr Then, writing w ¼ r ð1 ð =2 mþn1 w ðmÞðnÞ ¼ 2 w e dw sin2n1 cos2m1 d 0 0 ¼ ðm þ nÞ Bðm, nÞ ðmÞðnÞ ; Bðm, nÞ ¼ ðm þ nÞ ð13Þ So Bð32 , 12Þ ¼ . . . . . . . . . . . . 35 B 3 1 2, 2 ¼ 2 Because Bð32 , 12Þ ð3Þð1Þ ¼ 2 2 ¼ ð2Þ pffiffiffi pffiffiffi =2 ¼ 1 2 Now for some examples. 36 Application of gamma and beta functions The use of gamma and beta functions in the evaluation of definite integrals depends largely on the ability to change the variables to express the integral in ð1 the basic form of the beta function xm1 ð1 xÞn1 dx or its trigonometrical 0 ð =2 2m1 2n1 form 2 sin cos d. 0 Example 1 Evaluate I ¼ ð1 x5 ð1 xÞ4 dx: 0 Compare this with Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx 0 Then m 1 ¼ 5 ; m ¼ 6 and n 1 ¼ 4 ; n¼5 ; I ¼ Bð6, 5Þ ¼ . . . . . . . . . . . . 751 Integral functions 5! 4! 1 ¼ 10! 1260 I ¼ Bð6, 5Þ ¼ Example 2 Evaluate I ¼ ð1 37 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x4 1 x2 dx: 0 Comparing this with Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx 0 we see that we have x2 in the root, instead of a single x. Therefore, put x2 ¼ y 1 ; x ¼ y2 The limits remain unchanged. 1 dx ¼ 12 y 2 dy ; I ¼ ............ I ¼ 12 B 5 3 2;2 38 Because ð ð1 1 1 1 1 1 3 2 12 2 y 2 ð1 yÞ2 dy I ¼ y ð1 yÞ y dy ¼ 2 2 0 0 m 1 ¼ 32 ; m ¼ 52 and n 1 ¼ 12 ; n ¼ 32 ; I ¼ 12 B 52 ; 32 Expressing this in gamma functions I ¼ ............ I¼ 1 ð52Þð32Þ 2 ð4Þ 39 From our previous work on gamma functions pffiffiffi pffiffiffi 3 32 ¼ ; 52 ¼ ; ð4Þ ¼ 3! 2 4 ; I ¼ ............ I¼ 32 Because I¼ pffiffiffi pffiffiffi 1 ð3 =4Þð =2Þ ¼ . 2 3! 32 Now you can work through this one in much the same way. There are no tricks. 40 752 Programme 21 Example 3 Evaluate I ¼ ð3 x3 dx pffiffiffiffiffiffiffiffiffiffiffiffi. 3x 0 You need to compare this with Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx so bring every- 0 thing up on to the top line and then make the necessary change in the variables. Finish it off and then compare the results with the next frame. pffiffiffi 864 3 ¼ 42:76 I¼ 35 41 Here is the working; see whether you agree. ð3 3 ð3 ð3 x dx x 12 3 12 p ffiffiffiffiffiffiffiffiffiffiffi ffi I¼ x3 1 ¼ x ð3 xÞ dx ¼ 3 3 3x 0 0 0 x Put ¼ y; i.e. x ¼ 3y ; dx ¼ 3 dy 3 Limits: x ¼ 0; y ¼ 0; x ¼ 3; y ¼ 1 pffiffiffi ð 1 1 ; I ¼ 27 3 y 3 ð1 yÞ2 dy 12 0 pffiffiffi pffiffiffi ; I ¼ 27 3 B 4; 12 ¼ 27 3 dx m1¼3 ; m¼4 n 1 ¼ 12 ; n ¼ 12 ð4Þð12Þ ð9=2Þ pffiffiffi pffiffiffi 105 ; ð4Þ ¼ 3! Now 2 ¼ ; ð9=2Þ ¼ 16 pffiffiffi pffiffiffi pffiffiffi 16 864 3 pffiffiffi ¼ ; I ¼ 27 3 6 ¼ 42:76 105 35 1 Example 4 Evaluate I ¼ ð =2 sin5 cos4 d: 0 Bðm, nÞ ¼ 2 ð =2 sin2m1 cos2n1 d 0 ; 2m 1 ¼ 5 ; m ¼ 3; 2n 1 ¼ 4 ; I¼ 1 2 ; n ¼ 5=2 Bð3, 5=2Þ ¼ . . . . . . . . . . . . Finish it off 753 Integral functions I¼ 42 8 315 1 1 ð3Þð5=2Þ Bð3; 5=2Þ ¼ 2 2 ð11=2Þ pffiffiffi pffiffiffi 1 2!ð3 Þ=4 3 32 8 pffiffiffi pffiffiffi ¼ ¼ ¼ 2 ð945 Þ=32 4 945 315 I¼ Finally, one more. Example 5 Evaluate I ¼ ð =2 pffiffiffiffiffiffiffiffiffiffiffi tan d: 0 Somehow, we need to turn this into the form ð =2 Bðm, nÞ ¼ 2 sin2m1 cos2n1 d 0 So off you go; express the result in gamma functions I ¼ ............ I¼ 43 1 ð34Þð14Þ ð1Þ 2 Because I¼ ð =2 pffiffiffiffiffiffiffiffiffiffiffi ð =2 1 1 tan d ¼ sin2 cos2 d 0 0 1 3 1 ; m ¼ ; 2n 1 ¼ ; 2m 1 ¼ 2 4 2 1 3 1 1 ð34Þð14Þ ¼ ; I¼ B , ð1Þ 2 4 4 2 ; n¼ 1 4 and, unless we have appropriate tables to evaluate ð34Þ and ð14Þ, we cannot proceed much further. However, we do have such a table in Frame 24 so refer to it to evaluate the integral of our example. I ¼ ............ 754 Programme 21 44 I ¼ 2:2214 Because ð0:25Þ ¼ 3:6256 and ð0:75Þ ¼ 1:2254 1 ð1:2254Þð3:6256Þ ; I¼ ¼ 2:2214 2 1:0000 Duplication formula for gamma functions We already know that, when n is a positive integer ðnÞ ¼ ðn 1Þ! A useful formula enables us to calculate the gamma functions for values of n halfway between the integers. This is the duplication formula which can be stated as pffiffiffi ð2nÞ ðn þ 12Þ ¼ 2n1 ð14Þ 2 ðnÞ Thus, to find ð3:5Þ ðnÞ ¼ ð3Þ ¼ 2! ð2nÞ ¼ ð6Þ ¼ 5! pffiffiffi 5! ; ð3:5Þ ¼ ð3 þ 12Þ ¼ 5 ¼ 3:3234 2 2! The formula is quoted here without proof, but it is useful to have on occasions. So ð6:5Þ ¼ . . . . . . . . . . . . 45 ð6:5Þ ¼ 287:9 pffiffiffi ð12Þ 1 : ð6 5Þ ¼ ð6 þ 2Þ ¼ 11 2 ð6Þ ð6Þ ¼ 5!; ð12Þ ¼ 11!; 211 ¼ 2048 pffiffiffi 11! : ¼ 287:9 ; ð6 5Þ ¼ 2048 5! Now let us consider another function represented by an integral. On then to the next frame 755 Integral functions The error function The error function erf ðxÞ is defined as ð 2 x 2 erf ðxÞ ¼ pffiffiffi et dt 0 46 and occurs in statistics and various studies in physics and engineering. This integral, for arbitrary x, can only be evaluated numerically and values of erf ðxÞ for various values of x are obtained from tables. ðb 2 et dt are zero or 1, however, an exact result Where the limits of a ð1 2 et dt in Frame 6 is possible. We have already considered the integral I ¼ 0 when dealing with gamma functions and we established then that ð1 2 et dt ¼ 0 ð1 2 et dt ¼ 0 Consequently 2 Lim ðerf ðxÞÞ ¼ pffiffiffi x!1 ð1 pffiffiffi 1 1 2 ¼ 2 2 47 2 et dt ¼ 1 0 By representing the exponential function in the integral by its Maclaurin series we see that 1 2 X ............ erf ðxÞ ¼ pffiffiffi n¼0 1 2 X ð1Þn x2nþ1 erf ðxÞ ¼ pffiffiffi n¼0 n!ð2n þ 1Þ Because 2 erf ðxÞ ¼ pffiffiffi 2 ¼ pffiffiffi ðx 2 et dt 0 ! ðx X 1 ð1Þn t 2n 0 n¼0 1 ð x X 2 ¼ pffiffiffi n¼0 n! dt ð1Þn t 2n dt n! 0 1 2 X ð1Þn x2nþ1 ¼ pffiffiffi n¼0 n!ð2n þ 1Þ Consequently erf ðxÞ ¼ erf ðxÞ and so erf ðxÞ is an odd function. 48 756 Programme 21 The graph of erf (x ) erf (x) 2 1.5 1 0.5 –1.5 –2 –0.5 –1 –0.5 0.5 1 1.5 –1 –1.5 –2 The complementary error function erfc (x ) The complementary error function is defined as ð 2 1 2 erfc ðxÞ ¼ pffiffiffi et dt x which is related to the error function by the relation erfc ðxÞ ¼ . . . . . . . . . . . . 49 erfc ðxÞ ¼ 1 erf ðxÞ Because 2 erfc ðxÞ ¼ pffiffiffi 2 ¼ pffiffiffi ð1 2 et dt x ð 1 2 et dt 0 ðx 2 et dt 0 ¼ 1 erf ðxÞ Example 1 In terms of the complementary error function, for 0 < a < b ðb 2 et dt ¼ . . . . . . . . . . . . a 2 x 757 Integral functions pffiffiffi ½erfc ðaÞ erfc ðbÞ 2 50 Because ðb ðb ða 2 2 2 et dt ¼ et dt et dt a 0 p0ffiffiffi p ffiffiffi erf ðbÞ erf ðaÞ ¼ 2 2 pffiffiffi pffiffiffi ¼ ½1 erfc ðbÞ ½1 erfc ðaÞ 2 2 pffiffiffi ¼ ½erfc ðaÞ erfc ðbÞ 2 Example 2 In statistics the integral ð 1 x t 2 =2 ðxÞ ¼ pffiffiffiffiffiffi e dt 2 1 is the area beneath the Gaussian or normal probability distribution 1 2 pffiffiffi et =2 for the values 1 < t x. 2 1 e –x 2/2 2π 0.3 0.25 0.2 0.15 0.1 0.05 –2 –1.5 –1 –0.5 0.5 –0.075 1 The area beneath the complete Gaussian curve is then ð 1 1 t 2 =2 pffiffiffiffiffiffi e dt ¼ . . . . . . . . . . . . 2 1 1.5 2 x 758 Programme 21 51 1 Because ð1 ð 1 1 t 2 =2 1 t 2 =2 pffiffiffiffiffiffi e dt ¼ pffiffiffiffiffiffi 2 e dt 2 1 2 0 rffiffiffi ð 2 1 t 2 =2 ¼ e dt 0 rffiffiffi ð 2 pffiffiffi 1 u2 2 e du ¼ 0 ¼1 because the integrand is even pffiffiffi where u ¼ t= 2 from Frame 47 For positive x, ðxÞ is related to the error function ðxÞ ¼ . . . . . . . . . . . . 1 1 x ðxÞ ¼ þ erf pffiffiffi 2 2 2 52 Because ð 1 x t 2 =2 ðxÞ ¼ pffiffiffiffiffiffi e dt 2 1 ð ð 1 0 t 2 =2 1 x 2 ¼ pffiffiffiffiffiffi e dt þ pffiffiffiffiffiffi et =2 dt 2 1 2 0 ð x=pffiffi2 pffiffiffi p ffiffiffi 1 1 2 eu du where u ¼ t= 2 ¼ þ pffiffiffiffiffiffi 2 2 2 0 1 1 x ¼ þ erf pffiffiffi 2 2 2 Now let us consider a new set of integral funtions. Elliptic functions The use of elliptic functions provides a means of evaluating a further range of definite integrals, provided that the integrals can be converted by various appropriate substitutions into certain standard forms. pffiffiffiffiffiffiffiffiffiffi If an integrand is a rational function of x and of PðxÞ where PðxÞ is a polynomial in x of degree 3 or 4, then the integral is said to be elliptic. ð1 dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi is an elliptic integral. The name is For example, ð1 2x2 Þð4 3x2 Þ 0 derived from such an integral occurring in the determination of the arc length of part of an ellipse. 53 759 Integral functions Standard forms of elliptic functions (a) Of the first kind ð d Fðk; Þ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 k2 sin2 where 0 ð1Þ =2 and 0 < k < 1. (b) Of the second kind ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 d Eðk; Þ ¼ ð2Þ 0 and 0 < k < 1: 2 Make a careful note of these two standard forms: then we can apply them to some examples. where 0 Example 1 ð =2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Evaluate 4 sin2 d in terms of an elliptic function. 54 0 Taking out a factor 4 to reduce the first term to 1 ð =2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 sin2 d I¼2 4 0 The integral now agrees with the standard form, where k2 ¼ 14, i.e. k ¼ 12 and ¼ =2: ; I ¼ ............ I ¼ 2Eð12 , =2Þ Complete elliptic functions In each of the cases (1) and (2) listed above, if ¼ =2; the integral is said to be complete and then Fðk; =2Þ is denoted by KðkÞ and Eðk, =2Þ is denoted by EðkÞ. The method, then, rests on making suitable substitutions in a given integral to transform the integrand into one of the standard forms stated above. For various values of k and , values of the functions Fðk, Þ, Eðk, Þ, KðkÞ and EðkÞ are obtainable from published tables. These tables, which are quite extensive, are not reproduced here and so many required values will be given in the text. Incidentally, the result of Example 1 above, i.e. I ¼ 2Eð12 , =2Þ could also be written as I ¼ ............ 55 760 Programme 21 56 I ¼ 2E 1 2 because, in this case, ¼ =2. From tables, we find that Eð12Þ ¼ 1:4675 Example 2 ; I ¼ 2:935 ð =6 d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. 0 1 4 sin2 At first sight, this seems to be in standard form, but notice that the value of k2 is 4, i.e. k ¼ 2 – and this does not comply with the requirement that 0 < k < 1. We therefore proceed as follows. ð =6 d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I¼ 0 1 4 sin2 Evaluate I ¼ Put 4 sin2 ¼ sin2 i.e. 2 sin ¼ sin cos d 2 cos Also, for the new limits, when ¼ 0, ; 2 cos d ¼ cos d ; d ¼ when ¼ =6, and 57 ¼ 0, ; I¼ ð =2 0 ¼ ............ ¼ ............ ¼ 0; ¼ =6, ¼ =2 1 cos d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 cos 2 1 sin We now transform the cos sin ¼ 12 sin ; 1 cos2 ¼ 14 sin2 ð 1 =2 1 cos d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; I¼ 2 0 cos 1 1 sin2 ¼ 1 2 ð =2 0 ; cos ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 14 sin2 4 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi which is now in standard form 1 14 sin2 ; I ¼ ............ 761 Integral functions I ¼ 12 F 1 2 ; =2 ¼ 12 K 1 Kð12Þ ¼ 1:6858 From the appropriate tables, 58 2 ; I ¼ 0:8429 Now for another Example 3 ð =3 d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. 0 3 4 sin2 The first step is to . . . . . . . . . . . . Evaluate I ¼ take out a factor 3 to reduce the first term to 1 1 ; I ¼ pffiffiffi 3 ð =3 0 59 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 43 sin2 Next, we see that k2 > 1. Therefore, we put . . . . . . . . . . . . 2 4 3 sin 2 pffiffiffi sin ¼ sin 3 60 ¼ sin2 2 ; pffiffiffi cos d ¼ cos 3 pffiffiffi 3 cos d ; d ¼ 2 cos d Then, so far, we have I ¼ . . . . . . . . . . . . 1 I ¼ pffiffiffi 3 ð ¼ =3 ¼0 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 sin2 pffiffiffi 3 cos d 2 cos 2 pffiffiffi sin ¼ sin 3 ¼0 pffiffiffi 2 2 3 ¼ 1 ; ¼ =2 ¼ ; pffiffiffi sin ¼ pffiffiffi 3 2 3 3 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Also cos ¼ 1 sin2 ¼ 1 34 sin2 Limits: when ¼ 0, ; I ¼ ............ 61 762 Programme 21 62 I¼ 1 2 ð =2 0 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 34 sin2 which is now in standard form with k ¼ pffiffiffi 3 and ¼ =2 2 ! pffiffiffi pffiffiffi! 1 3 3 ; =2 ¼ K 2 2 2 pffiffiffi! 3 ¼ 2:1565 ; I ¼ 1:078 From tables K 2 1 ; I¼ F 2 Now, what about this one? Example 4 Evaluate I ¼ ð =2 0 d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. 1 þ 4 sin2 The trouble here is the plus sign in the denominator. Were it a minus sign as in Example 2, the integral could be converted into standard form and would present no difficulty. In this case, the key is to put ¼ =2 ; i.e. sin ¼ cos . Expressing the integral in terms of , we have I ¼ ............ 63 I¼ ð0 d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 4 sin2 =2 Because ¼ =2 ; d ¼ d 2 1 þ 4 sin ¼ 1 þ 4ð1 cos2 Þ ¼ 5 4 cos2 ¼ 5 4 sin2 Limits: when ¼ 0, ¼ =2; when ¼ =2, immediately follows. ¼ 0 and the expression above Move on 64 So we have I ¼ ð0 =2 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 4 sin2 The minus sign in the numerator can be absorbed by . . . . . . . . . . . . 763 Integral functions 65 changing the order of the limits ; I ¼ ð =2 0 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5 4 sin2 Finally, taking out a factor 5 from the denominator, the integral becomes ð 1 =2 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I ¼ pffiffiffi 5 0 1 45 sin2 and this can then be written . . . . . . . . . . . . 1 2 1 2 ¼ pffiffiffi K pffiffiffi I ¼ pffiffiffi F pffiffiffi , 2 5 5 5 5 2 From tables K pffiffiffi ¼ Kð0:8944Þ ¼ 2:2435 5 66 ; I ¼ 1:003 Alternative forms of elliptic functions (a) Of the first kind ðx du Fðk; xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð1 u Þð1 k2 u2 Þ 0 where 0 x 1 and 0 < k < 1. (b) Of the second kind ð x rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 u2 du Eðk; xÞ ¼ 1 u2 0 where 0 x ð3Þ ð4Þ 1 and 0 < k < 1. Note these two new forms and then we can deal with a few examples. As before, it is a case of transforming the given integrand into the required form by suitable substitutions. Example 1 ð 1=pffiffi2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi 4 3u Evaluate I ¼ du. 1 u2 0 Here we remove a factor 4 from the numerator to reduce the first term to 1. sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð 1=pffiffi2 1 34 u2 du I¼2 1 u2 0 This is now in standard form with k ¼ . . . . . . . . . . . . and x ¼ . . . . . . . . . . . . 67 764 Programme 21 pffiffiffi 3 ; k¼ 2 68 1 x ¼ pffiffiffi 2 ! pffiffiffi 3 1 , pffiffiffi ¼ 2ð0:7282Þ from tables ; I ¼ 2E 2 2 ; I ¼ 1:4564 Example 2 Evaluate I ¼ ð 1=2 0 du pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi. 5 6u2 þ u4 Factorising the denominator gives I ¼ . . . . . . . . . . . . 69 I¼ ð 1=2 0 du pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 u2 Þð5 u2 Þ Taking out a factor 5 ð 1 1=2 du qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi I ¼ pffiffiffi 5 0 2 ð1 u Þð1 15 u2 Þ pffiffiffi which is in standard form with k ¼ 1= 5 and x ¼ 1=2 ; I ¼ ............ 1 1 1 I ¼ pffiffiffi F pffiffiffi , 5 5 2 70 1 In some tables, k is quoted as sin , i.e. sin ¼ pffiffiffi 5 1 and x is quoted as sin , i.e. sin ¼ 2 pffiffiffi : Then Fð1= 5, 1=2Þ ¼ 0 528 ; ¼ 268 340 ; ¼ 308. ; I ¼ 0:236 Now move on for Example 3 765 Integral functions Example 3 71 ffi ð p3ffiffi=4 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 Evaluate I ¼ dx: 1 4x2 0 ð rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 u2 du, so first concentrate We have to convert this into the form 1 u2 on the denominator. Any suggestions? 72 Put 4x2 ¼ u2 i.e. 2x ¼ u 4x2 ¼ u2 ; 2x ¼ u ; 2 dx ¼ du Limits: when x ¼ 0, u ¼ 0 and when x ¼ Also pffiffiffi pffiffiffi 3=4, u ¼ 3=2 2 x2 ¼ 2 u2 =4 The integral now becomes . . . . . . . . . . . . ð p3ffiffi=2 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 u2 =4 du I¼ 1 u2 2 0 73 Finally, taking out the factor 2 in the numerator I ¼ ............ i.e. k2 ¼ 1 8 ð pffiffi3=2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 u2 =8 1 I ¼ pffiffiffi du 1 u2 2 0 pffiffiffi pffiffiffi 3 2 ; k¼ and x ¼ 2 4 So I ¼ . . . . . . . . . . . . 1 I ¼ pffiffiffi E 2 74 pffiffiffi pffiffiffi! 2 3 , 4 2 pffiffiffi pffiffiffi 3 2 0 Then sin ¼ ; ¼ 208 42 and sin ¼ 2 4 pffiffiffi pffiffiffi! 3 2 ¼ 1:029 ; I ¼ 0:728 , From tables, E 2 4 75 ; ¼ 608 So it is all just a question of manipulation to transform the given integral into the required standard forms, and then of reference to the appropriate tables. 766 Programme 21 The Revision summary follows, to be read in conjunction with the Can you? checklist, checking with the relevant parts of the Programme any points of which you are unsure. You will then find the Test exercise straightforward. Finally the Further problems provide additional practice. Revision summary 21 76 1 Gamma functions ð1 t x1 et dt (a) ðxÞ ¼ x>0 0 ðx þ 1Þ ¼ xðxÞ (b) If x ¼ n, a positive integer ðn þ 1Þ ¼ n! ð1Þ ¼ 1 ð0Þ ¼ 1 pffiffiffi ð1 2 (c) ex dx ¼ 2 0 ðnÞ ¼ 1 pffiffiffi 2 pffiffiffi 15 ð72Þ ¼ 8pffiffiffi p ffiffiffi 4 ð 12Þ ¼ 2 ð 32Þ ¼ 3 pffiffiffi ð2nÞ (e) Duplication formula n þ 12 ¼ 2n1 2 ðnÞ pffiffiffi pffiffiffi 3 ð52Þ ¼ 4 (d) ð12Þ ¼ 2 ð32Þ ¼ Beta functions (a) Bðm, nÞ ¼ ð1 xm1 ð1 xÞn1 dx m > 0; n > 0 0 Bðm, nÞ ¼ Bðn, mÞ ð =2 sin2m1 cos2n1 d Bðm, nÞ ¼ 2 0 767 Integral functions ðm 1Þðn 1Þ Bðm 1, n 1Þ ðm þ n 1Þðm þ n 2Þ 1 Bðk, 1Þ ¼ Bð1, kÞ ¼ k Bð1, 1Þ ¼ 1; Bð12 , 12Þ ¼ ðmÞ ðnÞ Bðm, nÞ ¼ ðm þ nÞ (c) m and n positive integers ðm 1Þ!ðn 1Þ! Bðm, nÞ ¼ ðm þ n 1Þ! (b) Bðm, nÞ ¼ 3 Error function ð 2 x 2 (a) erf ðxÞ ¼ pffiffiffi et dt 0 pffiffiffi ð1 2 x (b) e dx ¼ 2 ð01 pffiffiffi 2 ex dx ¼ ; 1 ð1 ex 2 =2 dx ¼ pffiffiffiffiffiffi 2 1 Complementary error function ð 2 1 t 2 erfc ðxÞ ¼ pffiffiffi e dt ¼ 1 erf ðxÞ x 4 Elliptic functions (a) Standard forms ð d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 k2 sin2 ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 sin2 d (2) of the second kind: Eðk; Þ ¼ (1) of the first kind: Fðk, Þ ¼ 0 In each case, 0 =2; (b) Complete elliptic integrals ¼ 2 E k; 2 F k; 0 < k < 1. 2 ¼ K ðkÞ ¼ E ðkÞ (c) Alternative forms of elliptic functions ðx du (1) of the first kind: Fðk; xÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 u2 Þð1 k2 u2 Þ 0 ð x rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 k2 u2 (2) of the second kind: Eðk; xÞ ¼ du 1 u2 0 In each case 0 x 1; 0 < k < 1. 768 Programme 21 Can you? 77 Checklist 21 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: Frames . Derive the recurrence relation for the gamma function and evaluate the gamma function for certain rational arguments? Yes No 1 to 16 . Evaluate integrals that require the use of the gamma function in their solution? Yes No 17 to 24 . Identify the beta function and evaluate integrals that require the use of the beta function in their solution? Yes No 25 to 32 . Derive the relationship between the gamma function and the beta function? Yes No 33 to 44 . Use the duplication formula to evaluate the gamma function for half integer arguments? Yes No 44 and 45 . Recognise the error function and its relation to the Gaussian probability distribution? Yes No 46 to 52 . Recognise elliptic functions of the first and second kind? Yes No . Evaluate integrals that require the use of elliptic functions in their solution? Yes No . Use alternative forms of the elliptic functions? Yes No 53 54 to 66 66 to 75 769 Integral functions Test exercise 21 1 ð 12Þ ð6Þ ð1:5Þ (b) (c) : 3ð4Þ ð2 5Þ ð12Þ ð1 ð1 2 (d) x5 ex dx (e) x6 e4x dx. 0 2 0 Determine ð1 x5 ð2 xÞ4 dx (a) ð =2 (b) 0 3 5 ð =8 sin7 cos3 d (c) 0 Show that ða pffiffiffi 2 et dt ¼ erf ðaÞ (a) ð1 (b) ek 2 2 t dt ¼ 0 Evaluate (a) erfc ð1Þ sin2 4 cos5 4 d. 0 a 4 78 Evaluate (a) pffiffiffi , 2k k > 0. (b) erfc ð0Þ. Express the following in elliptic functions. ð =4 ð pffiffi3=2 d du ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . p (a) (b) 2 4 5u2 þ u4 0 0 1 2 sin Further problems 21 1 ð12Þ ð5Þ ð2:5Þ ; (b) ; ; (c) 1 2ð3Þ ð3:5Þ ð 2Þ ð1 ð1 (d) x4 ex dx; (e) x8 e2x dx. 0 2 Determine (a) 4 ð1 ð01 0 3 x x e dx; (b) ð1 x4 e3x dx; ð01 pffiffiffi pffiffix dx. xe 0 0 ð1 n If m and n are positive constants, show that xm eax dx can be expressed in 0 1 mþ1 . the form n n aðmþ1Þ=n (c) 3 79 Evaluate (a) Evaluate the following. ð 1=2 x4 ð1 2xÞ3 dx (a) 0 ð =2 (d) 0 2 x2 e2x dx; pffiffiffiffiffiffiffiffiffiffiffiffiffi sin cos5 d (d) ð 1=pffiffi2 (b) 0 ð =4 (e) 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 2x2 dx ð =2 (c) sin5 cos4 d 0 sin3 2 cos6 2 d ð 1=3 (f) 0 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x2 1 9x2 dx. 770 Programme 21 d 2 2 erf ðxÞ ¼ pffiffiffi ex . dx 5 Show that 6 Show that the Laplace transform of the error function is given as ð1 2 es =4 s erfc FðsÞ ¼ erf ðtÞ est dt ¼ for s > 0. 2 s 0 7 The Fresnel integrals are defined as 2 2 ðx ðx t t dt and SðxÞ ¼ sin dt CðxÞ ¼ cos 2 2 0 0 Show that rffiffiffiffiffi! 1 j pffiffiffiffiffi erf x ¼ CðxÞ jSðxÞ 2 2j 8 Express the following in elliptic functions. ð =2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð =2 ð 1 rffiffiffiffiffiffiffiffiffiffiffiffiffi2ffi d 4x 2 pffiffiffiffiffiffiffiffiffiffiffi (a) 1 þ 4 sin d (b) dx (c) 1 x2 cos 0 0 0 ð2 ð2 dx dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (d) (e) 2 2 ð9 x Þð16 x Þ ð4 x2 Þð5 x2 Þ 0 0 ð =6 ð =3 d d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . (f) (g) 2 2 2 0 =4 sin þ 2 cos sin þ 2 cos2 9 Using the substitution x ¼ tan prove that the integral ð1 dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 þ x2 Þð1 þ 4x2 Þ 0 can be expressed in the form ð 1 =4 d qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 0 1 3 cos2 4 , evaluate the integral in terms of elliptic functions. 2 Evaluate the following. ð 0: 5 ð 1: 0 dx dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (a) (b) 2 4 3 4x þ x 3 4x2 þ x4 0 0:5 ð =2 ð =3 d d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi . (c) (d) 2 0 0 25 þ 9 sin 4 þ 3 sin2 Hence, using ¼ 10 Programme 22 Frames 1 to 95 Vector analysis 1 Learning outcomes When you have completed this Programme you will be able to: . Obtain the scalar and vector product of two vectors . Reproduce the relationships between the scalar and vector products of the Cartesian coordinate unit vectors . Obtain the scalar and vector triple products and appreciate their geometric significance . Differentiate a vector field and derive a unit vector tangential to the vector field at a point . Integrate a vector field . Obtain the gradient of a scalar field, the directional derivative and a unit normal to a surface . Obtain the divergence of a vector field and recognise a solenoidal vector field . Obtain the curl of a vector field . Obtain combinations of div, grad and curl acting on scalar and vector fields as appropriate Prerequisite: Engineering Mathematics (Sixth Edition) Programme 6 Vectors 771 772 Programme 22 Introduction 1 The initial work on vectors was covered in detail in Programme 6 of Engineering Mathematics (Sixth Edition) and, if you are in any doubt, spend some time reviewing that section of the work before proceeding further. The current Programmes on vector analysis build on these early foundations, so, for quick reference, the essential results of the previous work are summarised in the following list. Summary of prerequisites 1 2 3 A scalar quantity has magnitude only; a vector quantity has both magnitude and direction. The axes of reference, OX, OY, OZ, form a right-handed set. The symbols i, j, k denote unit vectors in the directions OX, OY, OZ, respectively. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If OP ¼ r ¼ ax i þ ay j þ az k then OP ¼ jrj ¼ a2x þ a2y þ a2z where jrj is the modulus of r. The direction cosines [l, m, n] are the cosines of the angles between the vector r and the axes OX, OY, OZ, respectively. For any vector r ¼ ax i þ ay j þ az k ay ax az l¼ ; m¼ ; n¼ jrj jrj jrj and 4 l 2 þ m2 þ n2 ¼ 1. Scalar product (‘dot product’) A B ¼ AB cos where is the angle between A and B and where A and B are the moduli of A and B. If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k then A B ¼ ax bx þ ay by þ az bz 5 and AB¼BA Vector product (‘cross product’) A B ¼ AB sin in a direction perpendicular to A and B so that A, B, ðA BÞ form a right-handed set. Therefore jA Bj ¼ AB sin i j k Also A B ¼ ax bx 6 ay by az where A B ¼ B A bz Angle between two vectors cos ¼ l1 l2 þ m1 m2 þ n1 n2 where l1 ; m1 ; n1 and l2 ; m2 ; n2 are the direction cosines of vectors r1 and r2 respectively. For perpendicular vectors l1 l2 þ m1 m2 þ n1 n2 ¼ 0 For parallel vectors l1 l2 þ m1 m2 þ n1 n2 ¼ 1. One or two examples will no doubt help to recall the main points. 773 Vector analysis 1 Example 1 Direction cosines z If i; j; k are unit vectors in the directions OX, OY, OZ, respectively, then any position vector OP ð¼ rÞ can be represented in the form az ax OP ¼ r ¼ ax i þ ay j þ az k. Then jrj ¼ . . . . . . . . . . . . qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2x þ a2y þ a2z 2 The direction of OP is denoted by stating the direction cosines of the angles made by OP and the three coordinate axes. OL ax ¼ OP jrj m ¼ cos ¼ OM ay ¼ OP jrj y x jrj ¼ l ¼ cos ¼ ay O z az ax ON az ¼ OP jrj ; l, m, n ¼ cos , cos , cos n ¼ cos ¼ γ β α ay O y x So, if P is the point ð3, 2, 6Þ, then jrj ¼ . . . . . . . . . . . . ; l ¼ ............; m ¼ ............; n ¼ ............ jrj ¼ 7; l ¼ 0:429; m ¼ 0:286; n ¼ 0:857 Because ðj r jÞ2 ¼ 9 þ 4 þ 36 ¼ 49 l ¼ cos ¼ 3 ¼ 0:4286 7 m ¼ cos ¼ 27 ¼ 0:2857 n ¼ cos ¼ 67 ¼ 0:8571. ; j r j¼ 7 3 774 Programme 22 Example 2 Angle between two vectors If the direction cosines of A are l1 ; m1 ; n1 and those of B are l2 ; m2 ; n2 , then the angle between the vectors is given by cos ¼ l1 l2 þ m1 m2 þ n1 n2 : ð1Þ If A ¼ 2i þ 3j þ 4k and B ¼ i 2j þ 3k, we can find the direction cosines of each and hence which is . . . . . . . . . . . . 4 ¼ 668 360 Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 4 þ 9 þ 16 ¼ 29 2 3 4 ; l1 ¼ pffiffiffiffiffiffi ; m1 ¼ pffiffiffiffiffiffi ; n1 ¼ pffiffiffiffiffiffi 29 29 29 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi j r2 j¼ 1 þ 4 þ 9 ¼ 14 1 2 3 ; l2 ¼ pffiffiffiffiffiffi ; m2 ¼ pffiffiffiffiffiffi ; n2 ¼ pffiffiffiffiffiffi 14 14 14 1 cos ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f2 6 þ 12g ¼ 0:3970 14 29 For A: j r1 j¼ For B: Then ; ¼ 668 360 Let us now look at the question of scalar and vector products. On to the next frame 5 Example 3 Scalar product If A and B are two vectors, the scalar product of A and B is defined as A B ¼ AB cos ð2Þ where is the angle between the two vectors. If A B ¼ 0 then A ? B. If we consider the scalar products of the unit vectors i; j; k, which are mutually perpendicular, then and i j ¼ ð1Þð1Þ cos 908 ¼ 0 i i ¼ ð1Þð1Þ cos 08 ¼ 1 ; ij¼jk¼ki¼0 ; i i ¼ j j ¼ k k ¼ 1: I n g e n e r a l , i f A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k A B ¼ ax bx þ ay by þ az bz which is, of course, a scalar quantity. then So, if A ¼ 2i 3j þ 4k and B ¼ i þ 2j þ 5k, then A B ¼ ............ 6 A B ¼ 2 6 þ 20 ¼ 16 Also, since A B ¼ AB cos ; we can determine the angle between the vectors. In this case ¼ . . . . . . . . . . . . 775 Vector analysis 1 7 ¼ 578 90 A ¼ 2i 3j þ 4k B ¼ i þ 2j þ 5k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ; A ¼j A j¼ 4 þ 9 þ 16 ¼ 29 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ; B ¼j B j¼ 1 þ 4 þ 25 ¼ 30 We have already found that A B ¼ 16 and A B ¼ AB cos pffiffiffiffiffiffi pffiffiffiffiffiffi ; 16 ¼ 29 30 cos ; cos ¼ 0:5425 ; ¼ 578 90 So, the scalar product of A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k is A B ¼ ax bx þ ay by þ az bz and A B ¼ AB cos where is the angle between the vectors. It can also be shown that (a) A B ¼ B A and (b) A ðB þ CÞ ¼ A B þ A C Make a note of these results Example 4 Vector product 8 If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k the vector product A B has magnitude jA Bj ¼ AB sin in the direction perpendicular to A and B such that A; B and (A B) form a right-handed set. We can write this as A B ¼ ðAB sin Þn ð3Þ where n is defined as a unit vector in the positive normal direction to the plane of A and B, i.e. forming a right-handed set. Also i A B ¼ ax bx j ay by k az bz If we consider the vector products of the unit vectors, i, j, k, then i j ¼ ð1Þð1Þ sin 908k ¼ k j k ¼ i, ki¼j Note that j i ¼ ði jÞ ¼ k k j ¼ i, i k ¼ j Also i i ¼ ð1Þð1Þ sin 08n ¼ 0 jj¼kk¼0 ð4Þ 776 Programme 22 It can also be shown that (a) A ðB þ CÞ ¼ A B þ A C ð5Þ (b) A B ¼ ðB AÞ and Make a note of these results (3), (4) and (5). Then, if A ¼ 3i 2j þ 4k and B ¼ 2i 3j 2k A B ¼ ............ 9 A B ¼ 16i þ 14j 5k We simply evaluate the determinant i j k AB¼ 3 2 2 3 4 2 ¼ ið4 þ 12Þ jð6 8Þ þ kð9 þ 4Þ ¼ 16i þ 14j 5k Move on to the next frame 10 We have seen therefore that the scalar product of two vectors is a scalar but that the vector product of two vectors is a vector. We know also that jA Bj ¼ AB sin Therefore, the angle between the vectors A and B given in Example 4 is ¼ ............ 11 ¼ 798 400 Because A ¼ 3i 2j þ 4k; B ¼ 2i 3j 2k; and A B ¼ 16i þ 14j 5k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffi ; j A B j ¼ 162 þ 142 þ 52 ¼ 477 ¼ 21:84 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼j A j¼ 32 þ 22 þ 42 ¼ 29 ¼ 5:385 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi B ¼j B j¼ 22 þ 32 þ 22 ¼ 17 ¼ 4:123 ; 21:84 ¼ ð5:385Þð4:123Þ sin ; sin ¼ 0:9838 ; ¼ 798 400 777 Vector analysis 1 So, to recapitulate: If A ¼ ax i þ ay j þ az k and B ¼ bx i þ by j þ bz k and is the angle between them (a) Scalar product ¼ A B ¼ ax bx þ ay by þ az bz ¼ AB cos i j (b) Vector product ¼ A B ¼ ax ay bx by k az bz j A B j ¼ AB sin . and Make a note of these fundamental results: we shall certainly need them. Then, in the next frame, we can set off on some new work Triple products We now deal with the various products that we form with three vectors. 12 Scalar triple product of three vectors If A; B; C are three vectors, the scalar formed by the product A ðB CÞ is called the scalar triple product. If A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k; C ¼ cx i þ cy j þ cz k; then i B C ¼ bx cx j by cy k bz cz i ; A ðB CÞ ¼ ðax i þ ay j þ az kÞ bx cx j by cy k bz cz Multiplying the top row by the external bracket and remembering that ij¼jk¼ki¼0 and i i ¼ j j ¼ k k ¼ 1 ax ay az we have A ðB CÞ ¼ bx cx by cy bz cz Example If A ¼ 2i 3j þ 4k; B ¼ i 2j 3k; C ¼ 2i þ j þ 2k; 2 3 then A ðB CÞ ¼ 1 2 4 3 2 1 2 ¼ ............ ð6Þ 778 Programme 22 13 A ðB CÞ ¼ 42 Because 2 3 4 A ðB CÞ ¼ 1 2 2 1 3 2 ¼ 2ð4 þ 3Þ þ 3ð2 þ 6Þ þ 4ð1 þ 4Þ ¼ 42 As simple as that. 14 Properties of scalar triple products (a) bx B ðC AÞ ¼ cx ax by cy ay bz ax cz ¼ cx az bx ay cy by az cz bz since interchanging two rows in a determinant reverses the sign. If we now interchange rows 2 and 3 and again change the sign, we have ax B ðC AÞ ¼ bx cx ay by cy az bz ¼ A ðB CÞ cz ð7Þ ; A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ i.e. the scalar triple product is unchanged by a cyclic change of the vectors involved. bx B ðA CÞ ¼ ax cx (b) by ay cy bz ax az ¼ bx cz cx ay by az bz cy cz ð8Þ ; B ðA CÞ ¼ A ðB CÞ i.e. a change of vectors not in cyclic order, changes the sign of the scalar triple product. (c) ax ay az A ðB AÞ ¼ bx ax by ay bz ¼ 0 since two rows are identical. az ; A ðB AÞ ¼ B ðC BÞ ¼ C ðA CÞ ¼ 0 Example If A ¼ i þ 2j þ 3k; B ¼ 2i 3j þ k; C ¼ 3i þ j 2k A ðB CÞ ¼ . . . . . . . . . . . . C ðB AÞ ¼ . . . . . . . . . . . . ð9Þ 779 Vector analysis 1 A ðB CÞ ¼ 52; C ðA BÞ ¼ 52 Because 1 2 3 A ðB CÞ ¼ 2 3 1 ¼ 1ð6 1Þ 2ð4 3Þ þ 3ð2 þ 9Þ ¼ 52 3 1 2 C ðB AÞ is not a cyclic change from the above. Therefore C ðB AÞ ¼ A ðB CÞ ¼ 52 Coplanar vectors The magnitude of the scalar triple product j A ðB CÞ j is equal to the volume of the parallelepiped with three adjacent sides defined by A, B and C. A n C ϕ θ B The scalar triple product A ðB CÞ ¼ AðBC sin nÞ ¼ ABC sin cos where n is a unit vector perpendicular to the plane containing B and C, is the angle between B and C and is the angle between A and n. Therefore j A ðB CÞ j ¼ ABC j sin cos j Notice that in the figure both and are drawn as acute but in the general case this may not be so. Now, BC j sin j is the area of the parallelogram defined by B and C. The altitude of the parallelepiped is A j cos j and so ABC j sin cos j is the volume of the parallelepiped with three adjacent sides defined by A, B and C. Consequently if A ðB CÞ ¼ 0 then the volume of the parallelepiped is zero and the three vectors A, B and C are coplanar. Example 1 Show that A ¼ i þ 2j 3k; B ¼ 2i j þ 2k; and C ¼ 3i þ j k are coplanar. We just evaluate A ðB CÞ ¼ . . . . . . . . . . . . and apply the test. 15 780 Programme 22 16 A ðB CÞ ¼ 0 Because 1 2 3 A ðB CÞ ¼ 2 1 2 ¼ 1ð1 2Þ 2ð2 6Þ 3ð2 þ 3Þ ¼ 0: 3 1 1 Therefore A, B, C are coplanar. Example 2 If A ¼ 2i j þ 3k; B ¼ 3i þ 2j þ k; C ¼ i þ pj þ 4k are coplanar, find the value of p. The method is clear enough. We merely set up and evaluate the determinant and solve the equation A ðB CÞ ¼ 0. p ¼ ............ 17 p ¼ 3 Because A ðB CÞ ¼ 0 2 1 3 2 1 ¼0 ; 3 1 p 4 ; 2 ð8 pÞ þ 1ð12 1Þ þ 3ð3p 2Þ ¼ 0 ; 7p ¼ 21 ; p ¼ 3 One more. Example 3 Determine whether the three vectors A ¼ 3i þ 2j k; B ¼ 2i j þ 3k; C ¼ i 2j þ 2k are coplanar. Work through it on your own. The result shows that ............ 18 A, B, C are not coplanar Because 3 2 in this case A ðB CÞ ¼ 2 1 1 2 ; A ðB CÞ 6¼ 0 1 3 ¼ 13 2 ; A, B, C are not coplanar. Now on to something different 781 Vector analysis 1 19 Vector triple products of three vectors If A; B and C are three vectors, then ) A ðB CÞ are called the vector triple products. and ðA BÞ C ð10Þ Consider A ðB CÞ where A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k and C ¼ cx i þ cy j þ cz k: Then ðB CÞ is a vector perpendicular to the plane of B and C and A ðB CÞ is a vector perpendicular to the plane containing A and ðB CÞ, i.e. coplanar with B and C. Note that, similarly, ðA BÞ C is coplanar with A and B and so in general A ðB CÞ 6¼ ðA BÞ C. Now i ðB CÞ ¼ bx cx Then j by k by bz ¼ i cy cz i cy A ðB CÞ ¼ ¼ bz bx j cz cx ax by bz cy cz bz cz þk j k ay az bx bz bx by cx cz cx cy i j k ax ay az by bz bz bx bx by cy cz cz cx cx cy bx cx by cy In symbolic form, further expansion of the determinant becomes somewhat tedious. However a numerical example will clarify the method. So make a note of the definition (10) above and then go on to the next frame Example 1 If A ¼ 2i 3j þ k; B ¼ i þ 2j k; triple product A ðB CÞ. 20 C ¼ 3i þ j þ 3k; determine the vector We start off with B C ¼ . . . . . . . . . . . . 782 Programme 22 21 B C ¼ 7i 6j 5k Because i j BC¼ 1 2 3 1 k ¼ ið6 þ 1Þ jð3 þ 3Þ þ kð1 6Þ 1 3 ¼ 7i 6j 5k Then A ðB CÞ ¼ . . . . . . . . . . . . 22 A ðB CÞ ¼ 21i þ 17j þ 9k Because i j A ðB CÞ ¼ 2 3 k 1 7 6 5 ¼ ið15 þ 6Þ jð10 7Þ þ kð12 þ 21Þ ¼ 21i þ 17j þ 9k That is fundamental enough. There is, however, an even easier way of determining a vector triple product. It can be proved that A ðB CÞ ¼ ðA CÞB ðA BÞC and ðA BÞ C ¼ ðC AÞB ðC BÞA ð11Þ The proof of this is given in the Appendix. For the moment, make a careful note of the expressions: then we will apply the method to the example we have just completed. 23 A ¼ 2i 3j þ k; B ¼ i þ 2j k; C ¼ 3i þ j þ 3k and we have A ðB CÞ ¼ ðA CÞB ðA BÞC ¼ ð6 3 þ 3Þði þ 2j kÞ ð2 6 1Þð3i þ j þ 3kÞ ¼ 6 ði þ 2j kÞ þ 5ð3i þ j þ 3kÞ ¼ 21i þ 17j þ 9k which is, of course, the result we achieved before. Here is another. Example 2 If A ¼ 3i þ 2j 2k; B ¼ 4i j þ 3k; C ¼ 2i 3j þ k determine ðA BÞ C using the relationship ðA BÞ C ¼ ðC AÞB ðC BÞA. ðA BÞ C ¼ . . . . . . . . . . . . 783 Vector analysis 1 24 50i 26j þ 22k Because ðA BÞ C ¼ ðC AÞB ðC BÞA ¼ ð6 6 2Þð4i j þ 3kÞ ð8 þ 3 þ 3Þð3i þ 2j 2kÞ ¼ 2ð4i j þ 3kÞ 14ð3i þ 2j 2kÞ ¼ 50i 26j þ 22k Now one more. Example 3 If A ¼ i þ 3j þ 2k; B ¼ 2i þ 5j k; C ¼ i þ 2j þ 3k A ðB CÞ ¼ . . . . . . . . . . . . ðA BÞ C ¼ . . . . . . . . . . . . Finish them both. 25 A ðB CÞ ¼ 11i þ 35j 58k ðA BÞ C ¼ 17i þ 38j 31k Because A ðB CÞ ¼ ðA CÞB ðA BÞC ¼ ð1 þ 6 þ 6Þð2i þ 5j kÞ ð2 þ 15 2Þði þ 2j þ 3kÞ ¼ 13ð2i þ 5j kÞ 15ði þ 2j þ 3kÞ ¼ 11i þ 35j 58k and ðA BÞ C ¼ ðC AÞB ðC BÞA ¼ ð1 þ 6 þ 6Þð2i þ 5j kÞ ð2 þ 10 3Þði þ 3j þ 2kÞ ¼ 13ð2i þ 5j kÞ 9ði þ 3j þ 2kÞ ¼ 17i þ 38j 31k These two results clearly confirm that A ðB CÞ 6¼ ðA BÞ C so beware! Before we proceed, note the following concerning the unit vectors. (a) ði jÞ ¼ k z ; i ði jÞ ¼ i k ¼ j ; i ði jÞ ¼ j (b) ði iÞ j ¼ ð0Þ j ¼ 0 ; ði iÞ j ¼ 0 x O y and once again, we see that i ði jÞ 6¼ ði iÞ j On to the next 784 26 Programme 22 Finally, by way of revision: Example 4 If A ¼ 5i 2j þ 3k; B ¼ 3i þ j 2k; C ¼ i 3j þ 4k; determine (a) the scalar triple product A ðB CÞ (b) the vector triple products (1) A ðB CÞ (2) ðA BÞ C. Finish all these and then check with the next frame 27 (a) A ðB CÞ ¼ 12 (b) ð1Þ A ðB CÞ ¼ 62i þ 44j 74k ð2Þ ðA BÞ C ¼ 109i þ 7j 22k Here is the working. 5 2 3 1 2 (a) A ðB CÞ ¼ 3 1 3 4 ¼ 5ð4 6Þ þ 2ð12 þ 2Þ þ 3ð9 1Þ ¼ 12 (b) (1) A ðB CÞ ¼ ðA CÞB ðA BÞC ¼ ð5 þ 6 þ 12Þð3i þ j 2kÞ ð15 2 6Þði 3j þ 4kÞ ¼ 23ð3i þ j 2kÞ 7ði 3j þ 4kÞ ¼ 62i þ 44j 74k (2) ðA BÞ C ¼ ðC AÞB ðC BÞA ¼ 23ð3i þ j 2kÞ ð8Þð5i 2j þ 3kÞ ¼ 109i þ 7j 22k Let us now move to the next topic 28 Differentiation of vectors In many practical problems, we often deal with vectors that change with time, e.g. velocity, acceleration, etc. If a vector A depends on a scalar variable t, then A can be represented as AðtÞ and A is then said to be a function of t. If A ¼ ax i þ ay j þ az k then ax ; ay ; az will also be dependent on the parameter t. i.e. AðtÞ ¼ ax ðtÞi þ ay ðtÞj þ az ðtÞk Differentiating with respect to t gives . . . . . . . . . . . . 785 Vector analysis 1 d d d d fAðtÞg ¼ i fax ðtÞg þ j fay ðtÞg þ k faz ðtÞg dt dt dt dt In short 29 day dA dax daz ¼i þj þk . dt dt dt dt The independent scalar variable is not, of course, restricted to t. In general, if u is the parameter, then dA ¼ ............ du day dA dax daz ¼i þj þk du du du du δA A (u + δu) A(u) O 30 If a position vector OP moves to OQ when u becomes u þ u, then as u ! 0, the direction of the chord PQ becomes that of the tangent to the dA curve at P, i.e. the direction of is along the du tangent to the locus of P. T = dA du O A(u) Example 1 If A ¼ ð3u2 þ 4Þi þ ð2u 5Þj þ 4u3 k, then dA ¼ ............ du dA ¼ 6ui þ 2j þ 12u2 k du If we differentiate this again, we get When u ¼ 2, Then d2 A ¼ 6i þ 24uk du2 d2 A dA ¼ 6i þ 48k ¼ 12i þ 2j þ 48k and du2 du dA ¼ ............ du and d2 A ¼ ............ du2 31 786 Programme 22 dA : du ¼ 49 52; 32 2 d A : du2 ¼ 48 37 Because and dA ¼ f122 þ 22 þ 482 g1=2 ¼ f2452g1=2 ¼ 49:52 du d2 A ¼ f62 þ 482 g1=2 ¼ f2340g1=2 ¼ 48:37 du2 Example 2 If F ¼ i sin 2t þ je3t þ kðt 3 4tÞ, then when t ¼ 1 dF ¼ ............; dt 33 d2 F ¼ ............ dt 2 dF ¼ 2 cos 2i þ 3e3 j k dt d2 F ¼ 4 sin 2i þ 9e3 j þ 6k dt 2 From these, we could if required find the magnitudes of dF ¼ ............; dt dF ¼ 60:27; dt 34 dF d2 F and 2 . dt dt d2 F ¼ ............ dt 2 2 d F : dt 2 ¼ 180 9 Because dF ¼ fð2 cos 2Þ2 þ 9e6 þ 1g1=2 dt ¼ f0:6927 þ 3631 þ 1g1=2 ¼ 60:27 and d2 F ¼ fð4 sin 2Þ2 þ 81e6 þ 36g1=2 dt 2 ¼ f13:23 þ 32 678 þ 36g1=2 ¼ 180:9 One more example. Example 3 If A ¼ ðu þ 3Þi ð2 þ u2 Þj þ 2u3 k, determine (a) dA du (b) d2 A du2 (c) dA du (d) d2 A du2 at u ¼ 3. Work through all sections and then check with the next frame 787 Vector analysis 1 Here is the working. (a) A ¼ ðu þ 3Þi ð2 þ u2 Þj þ 2u3 k dA ¼ i 2uj þ 6u2 k du At u ¼ 3, 35 dA ¼ i 6j þ 54k du d2 A d2 A ¼ 2j þ 12uk At u ¼ 3, ¼ 2j þ 36k 2 du du2 dA ¼ f1 þ 36 þ 2916g1=2 ¼ ð2953Þ1=2 ¼ 54:34 (c) du (b) (d) d2 A ¼ f4 þ 1296g1=2 ¼ ð1300Þ1=2 ¼ 36:06 du2 The next example is of a rather different kind, so move on Example 4 36 A particle moves in space so that at time t its position is stated as x ¼ 2t þ 3; y ¼ t 2 þ 3t; z ¼ t 3 þ 2t 2 . We are required to find the components of its velocity and acceleration in the direction of the vector 2i þ 3j þ 4k when t ¼ 1. First we can write the position as a vector r r ¼ ð2t þ 3Þi þ ðt 2 þ 3tÞj þ ðt 3 þ 2t 2 Þk Then, at t ¼ 1 d2 r ¼ ............ dt 2 dr ¼ ............; dt dr ¼ 2i þ 5j þ 7k; dt d2 r ¼ 2j þ 10k dt 2 Because and dr ¼ 2i þ ð2t þ 3Þj þ ð3t 2 þ 4tÞk dt dr ; At t ¼ 1; ¼ 2i þ 5j þ 7k dt 2 d r ¼ 2j þ ð6t þ 4Þk dt 2 d2 r ; At t ¼ 1; ¼ 2j þ 10k dt 2 Now, a unit vector parallel to 2i þ 3j þ 4k is . . . . . . . . . . . . 37 788 Programme 22 38 2i þ 3j þ 4k 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffi ð2i þ 3j þ 4kÞ 4 þ 9 þ 16 29 Denote this unit vector by I. Then dr the component of in the direction dt of I dr ¼ cos dt dr ¼ I dt 1 ¼ pffiffiffiffiffiffi ð2i þ 5j þ 7kÞ ð2i þ 3j þ 4kÞ 29 ¼ ............ 39 8:73 Because 1 1 pffiffiffiffiffiffi ð2i þ 5j þ 7kÞ ð2i þ 3j þ 4kÞ ¼ pffiffiffiffiffiffi ð4 þ 15 þ 28Þ 29 29 47 ¼ pffiffiffiffiffiffi 29 : ¼ 8 73 Similarly, the component of 40 d2 r in the direction of I is dt 2 ............ 8:54 Because d2 r d2 r cos ¼ I dt 2 dt 2 1 ¼ pffiffiffiffiffiffi ð2j þ 10kÞ ð2i þ 3j þ 4kÞ 29 1 ¼ pffiffiffiffiffiffi ð6 þ 40Þ 29 46 ¼ pffiffiffiffiffiffi 29 : ¼ 8 54 dr dt I 789 Vector analysis 1 Differentiation of sums and products of vectors If A ¼ AðuÞ and B ¼ BðuÞ, then d dA fcAg ¼ c du du d dA dB fA þ Bg ¼ þ (b) du du du d dB dA fA Bg ¼ A þ B (c) du du du d dB dA fA Bg ¼ A þ B. (d) du du du (a) These are very much like the normal rules of differentiation. However, if AðuÞ AðuÞ ¼ a2x þ a2y þ a2z ¼ jAj2 ¼ A2 is a constant then d d d fAðuÞ AðuÞg ¼ AðuÞ fAðuÞg þ AðuÞ fAðuÞg du du du d d ¼ 2AðuÞ fAðuÞg ¼ fA2 g ¼ 0 du du Assuming that AðuÞ 6¼ 0, then since AðuÞ that AðuÞ and d d fAðuÞg ¼ fA2 g ¼ 0 it follows du du d fAðuÞg are perpendicular vectors because du ............ 41 d d fAðuÞg ¼j AðuÞ j fAðuÞg cos ¼ 0 du du ; cos ¼ 0 ; ¼ 2 AðuÞ Now let us deal with unit tangent vectors. Unit tangent vectors We have already established in Frame 30 of this Programme that if OP is a position vector AðuÞ in space, then the direction of the vector d fAðuÞg is denoting du ............ O A (u) 790 Programme 22 42 parallel to the tangent to the curve at P Then the unit tangent vector T at P can be found from d fAðuÞg du T¼ d fAðuÞg du O A (u) In simpler notation, this becomes: If r ¼ ax i þ ay j þ az k then the unit tangent vector T is given by T¼ dr=du jdr=duj Example 1 Determine the unit tangent vector at the point ð2, 4, 7Þ for the curve with parametric equations x ¼ 2u; y ¼ u2 þ 3; z ¼ 2u2 þ 5. First we see that the point ð2, 4, 7Þ corresponds to u ¼ 1. The vector equation of the curve is r ¼ ax i þ ay j þ az k ¼ 2ui þ ðu2 þ 3Þj þ ð2u2 þ 5Þk ; 43 dr ¼ ............ du dr ¼ 2i þ 2uj þ 4uk du and at u ¼ 1; dr ¼ 2i þ 2j þ 4k du Hence dr ¼ ............ du pffiffiffi dr ¼ 2 6; du 44 1 T ¼ pffiffiffi fi þ j þ 2kg 6 Because pffiffiffi dr ¼ f4 þ 4 þ 16g1=2 ¼ 241=2 ¼ 2 6 du dr 2i þ 2j þ 4k 1 du pffiffiffi T¼ ¼ ¼ pffiffiffi fi þ j þ 2kg dr 2 6 6 du Let us do another. and T ¼ . . . . . . . . . . . . 791 Vector analysis 1 Example 2 Find the unit tangent vector at the point ð2; 0; Þ for the curve with parametric equations x ¼ 2 sin ; y ¼ 3 cos ; z ¼ 2. We see that the point ð2; 0; Þ corresponds to ¼ =2: Writing the curve in vector form r ¼ . . . . . . . . . . . . 45 r ¼ 2 sin i þ 3 cos j þ 2 k Then, at ¼ =2, dr ¼ ............ d dr ¼ ............ d T ¼ ............ Finish it off pffiffiffiffiffiffi dr dr ¼ 3j þ 2k; ¼ 13 d d 1 T ¼ pffiffiffiffiffiffi ð3j þ 2kÞ 13 46 And now Example 3 Determine the unit tangent vector for the curve x ¼ 3t; y ¼ 2t 2 ; z ¼ t2 þ t at the point ð6, 8, 6Þ. On your own. T ¼ . . . . . . . . . . . . 1 T ¼ pffiffiffiffiffiffi ð3i þ 8j þ 5kÞ 98 The point ð6, 8, 6Þ corresponds to t ¼ 2 r ¼ 3ti þ 2t 2 j þ ðt 2 þ tÞk dr ¼ 3i þ 4tj þ ð2t þ 1Þk ; dt dr ¼ 3i þ 8j þ 5k dt pffiffiffiffiffiffi ¼ 98 At t ¼ 2, r ¼ 6i þ 8j þ 6k and dr ¼ ð9 þ 64 þ 25Þ1=2 dt dr=dt 1 ; T¼ ¼ pffiffiffiffiffiffi ð3i þ 8j þ 5kÞ jdr=dtj 98 ; 47 792 Programme 22 Partial differentiation of vectors 48 If a vector F is a function of two independent variables u and v, then the rules of differentiation follow the usual pattern. If F ¼ xi þ yj þ zk then x, y, z will also be functions of u and v. Then @F @x @y @z ¼ iþ jþ k @u @u @u @u @F @x @y @z ¼ iþ jþ k @v @v @v @v @2F @2x @2y @2z ¼ 2iþ 2jþ 2k 2 @u @u @u @u @2F @2x @2y @2z ¼ iþ 2jþ 2k @v 2 @v 2 @v @v @2F @2x @2y @2z ¼ iþ jþ k @u@v @u@v @u@v @u@v and for small finite changes du and dv in u and v, we have dF ¼ @F @F du þ dv @u @v Example If F ¼ 2uvi þ ðu2 2vÞj þ ðu þ v 2 Þk @F ¼ ............; @u @2F ¼ ............; @u2 49 @F ¼ 2vi þ 2uj þ k; @u @2F ¼ 2j; @u2 @F ¼ ............ @v @2F ¼ ............ @u@v @F ¼ 2ui 2j þ 2vk @v @2F ¼ 2i @u@v This is straightforward enough. Integration of vector functions The process is the reverse of that for differentiation. If a vector F ¼ xi þ yj þ zk where F; x; y; z are expressed as functions of u, then ðb ðb ðb ðb F du ¼ i x du þ j y du þ k z du. a a a a 793 Vector analysis 1 Example 1 If F ¼ ð3t 2 þ 4tÞi þ ð2t 5Þj þ 4t 3 k, then ð3 ð3 ð3 ð3 F dt ¼ i ð3t 2 þ 4tÞ dt þ j ð2t 5Þ dt þ k 4t 3 dt ¼ . . . . . . . . . . . . 1 1 1 1 42i 2j þ 80k 50 Because 3 ð3 F dt ¼ iðt 3 þ 2t 2 Þ þ jðt 2 5tÞ þ kt 4 1 1 ¼ ð45i 6j þ 81kÞ ð3i 4j þ kÞ ¼ 42i 2j þ 80k Here is a slightly different one. Example 2 If F ¼ 3ui þ u2 j þ ðu þ 2Þk and V ¼ 2ui 3uj þ ðu 2Þk ð2 ðF VÞdu. evaluate 0 First we must determine F V in terms of u. F V ¼ ............ F V ¼ ðu3 þ u2 þ 6uÞi ðu2 10uÞj ð2u3 þ 9u2 Þk 51 Because i j k F V ¼ 3u u2 ðu þ 2Þ 2u 3u ðu 2Þ which gives the result above. ð2 ðF VÞ du ¼ . . . . . . . . . . . . Then 0 4 3 f14i Because ð ð2 ; 0 ðF VÞdu ¼ þ 13j 24kg 3 4 u4 u3 u u þ þ 3u2 i 5u2 j þ 3u3 k 4 3 3 2 ðF VÞdu ¼ ð4 þ 83 þ 12Þi ð83 20Þj ð8 þ 24Þk ¼ 43 f14i þ 13j 24kg 52 794 Programme 22 Example 3 If F ¼ A ðB CÞ where A ¼ 3t 2 i þ ð2t 3Þj þ 4tk B ¼ 2i þ 4tj þ 3ð1 tÞk C ¼ 2ti 3t 2 j 2tk ð1 determine Fdt. 0 First we need to find A ðB CÞ. The simplest way to do this is to use the relationship A ðB CÞ ¼ . . . . . . . . . . . . 53 A ðB CÞ ¼ ðA CÞB ðA BÞC So and 54 A C ¼ ............ A B ¼ ............ A C ¼ 6t 3 6t 3 þ 9t 2 8t 2 ¼ t 2 A B ¼ 6t 2 þ 8t 2 12t þ 12t 12t 2 ¼ 2t 2 Then F ¼ A ðB CÞ ¼ t 2 f2i þ 4tj þ 3ð1 tÞkg 2t 2 f2ti 3t 2 j 2tkg ð1 F dt ¼ . . . . . . . . . . . . ; 0 Finish off the simplification and complete the integration. 55 1 60 f20i þ 132j þ 75kg Because F ¼ A ðB CÞ ¼ ð2t 2 4t 3 Þi þ ð4t 3 þ 6t 4 Þj þ ð3t 2 þ t 3 Þk Integration with respect to t then gives the result stated above. Now let us move on to the next stage of our development 795 Vector analysis 1 Scalar and vector fields ϕ (x, y, z) P (x, y, z) z z y 56 y O x If every point P ðx; y; zÞ of a region R of space has associated with it a scalar quantity ðx; y; zÞ, then ðx; y; zÞ is a scalar function and a scalar field is said to exist in the region R. x Examples of scalar fields are temperature, potential, etc. Similarly, if every point P ðx; y; zÞ of a region R has associated with it a vector quantity Fðx; y; zÞ, then Fðx; y; zÞ is a vector function and a vector field is said to exist in the region R. F (x, y, z) z P (x, y, z) z y x O y x Examples of vector fields are force, velocity, acceleration, etc. Fðx; y; zÞ can be defined in terms of its components parallel to the coordinate axes, OX, OY, OZ. That is, Fðx; y; zÞ ¼ Fx i þ Fy j þ Fz k. Note these important definitions: we shall be making good use of them as we proceed 57 Grad (gradient of a scalar function) If a scalar function ðx; y; zÞ is continuously differentiable with respect to its variables x; y; z; throughout the region, then the gradient of , written grad , is defined as the vector grad ¼ @ @ @ iþ jþ k @x @y @z ð12Þ Note that, while is a scalar function, grad is a vector function. For example, if depends upon the position of P and is defined by ¼ 2x2 yz3 , then grad ¼ 4xyz3 i þ 2x2 z3 j þ 6x2 yz2 k 796 Programme 22 Notation The expression (12) above can be written @ @ @ grad ¼ i þ j þ k @x @y @z @ @ @ where i þ j þ k is called a vector differential operator and is denoted @x @y @z by the symbol r (pronounced ‘del’ or sometimes ‘nabla’) @ @ @ i.e. r i þj þk @x @y @z Beware! r cannot exist alone: it is an operator and must operate on a stated scalar function ðx, y, zÞ. If F is a vector function, rF has no meaning. So we have: @ @ @ þj þk @x @y @z @ @ @ ¼i þj þk @x @y @z r ¼ grad ¼ i ð13Þ Make a note of this definition and then let us see how to use it 58 Example 1 If ¼ x2 yz3 þ xy 2 z2 , determine grad at the point P ð1, 3, 2Þ. By the definition, grad ¼ r ¼ @ @ @ iþ jþ k. @x @y @z All we have to do then is to find the partial derivatives at x ¼ 1, y ¼ 3, z ¼ 2 and insert their values. ; r ¼ . . . . . . . . . . . . 59 4ð21i þ 8j þ 18kÞ Because ¼ x2 yz3 þ xy 2 z2 @ @y ¼ x2 z3 þ 2xyz2 Then, at ð1, 3, 2Þ ; @ ¼ 2xyz3 þ y 2 z2 @x @ ¼ 3x2 yz2 þ 2xy 2 z @z @ ¼ 48 þ 36 @x @ ¼ 8 þ 24 @y @ ¼ 36 þ 36 @z @ ¼ 84 @x @ ; ¼ 32 @y @ ; ¼ 72 @z ; ; grad ¼ r ¼ 84i þ 32j þ 72k ¼ 4ð21i þ 8j þ 18kÞ 797 Vector analysis 1 Example 2 If A ¼ x2 zi þ xyj þ y 2 zk and B ¼ yz2 i þ xzj þ x2 zk determine an expression for grad ðA BÞ: This we can soon do since we know that A B is a scalar function of x, y and z. First then, A B ¼ ............ 60 A B ¼ x2 yz3 þ x2 yz þ x2 y 2 z2 rðA BÞ ¼ . . . . . . . . . . . . Then 2xyzðz2 þ 1 þ yzÞi þ x2 zðz2 þ 1 þ 2yzÞj þ x2 yð3z2 þ 1 þ 2yzÞk Because if ¼ A B ¼ ðx2 zi þ xyj þ y 2 zkÞ ðyz2 i þ xzj þ x2 zkÞ ¼ x2 yz3 þ x2 yz þ x2 y 2 z2 @ ¼ 2xyz3 þ 2xyz þ 2xy2 z2 ¼ 2xyzðz2 þ 1 þ yzÞ @x @ ¼ x2 z3 þ x2 z þ 2x2 yz2 ¼ x2 zðz2 þ 1 þ 2yzÞ @y @ ¼ 3x2 yz2 þ x2 y þ 2x2 y 2 z ¼ x2 yð3z2 þ 1 þ 2yzÞ @z ; r ðA BÞ ¼ 2xyzðz2 þ 1 þ yzÞi þ x2 zðz2 þ 1 þ 2yzÞj þ x2 yð3z2 þ 1 þ 2yzÞk Now let us obtain another useful relationship. z dr dy r y x If OP is a position vector r where r ¼ xi þ yj þ zk and dr is a small displacement corresponding to changes dx, dy, dz in x, y, z respectively, then dz dx z O y dr ¼ dx i þ dy j þ dz k x If ðx; y; zÞ is a scalar function at P, we know that @ @ @ iþ jþ k @x @y @z grad dr ¼ . . . . . . . . . . . . grad ¼ r ¼ Then 61 798 Programme 22 62 grad dr ¼ @ @ @ dx þ dy þ dz @x @y @z Because @ @ @ grad dr ¼ iþ jþ k ðdx i þ dy j þ dz kÞ @x @y @z @ @ @ ¼ dx þ dy þ dz @x @y @z ¼ the total differential d of That is d ¼ dr grad ð14Þ This will certainly be useful, so make a note of it 63 Directional derivatives Q (r + dr) z We have just established that P (r) d ¼ dr grad If ds is the small element of arc between P ðrÞ and Q ðr þ dr) then ds ¼ jdrj dr dr ¼ ds jdrj dr dy ds O dz dx y x dr is thus a unit vector in the direction of dr. and ds ; d dr ¼ grad ds ds If we denote the unit vector dr ^ then the result becomes by a ds d ^ grad ¼a ds d ^ and is called the is thus the projection of grad on the unit vector a ds ^. It gives the rate of change of directional derivative of in the direction of a d ^ and ^ grad will be a with distance measured in the direction of a ¼a ds ^ and grad have the same direction, since then maximum when a ^ grad ¼j a ^ jj grad j cos and will be zero. a Thus the direction of grad gives the direction in which the maximum rate of change of occurs. 799 Vector analysis 1 Example 1 Find the directional derivative of the function ¼ x2 z þ 2xy 2 þ yz2 at the point (1, 2, –1) in the direction of the vector A ¼ 2i þ 3j 4k. We start off with ¼ x2 z þ 2xy 2 þ yz2 ; r ¼ . . . . . . . . . . . . r ¼ ð2xz þ 2y 2 Þi þ ð4xy þ z2 Þj þ ðx2 þ 2yzÞk 64 Because @ ¼ 2xz þ 2y 2 ; @x @ ¼ 4xy þ z2 ; @y @ ¼ x2 þ 2yz @z Then, at ð1, 2, 1Þ r ¼ ð2 þ 8Þi þ ð8 þ 1Þj þ ð1 4Þk ¼ 6i þ 9j 3k ^ where A ¼ 2i þ 3j 4k Next we have to find the unit vector a ^ ¼ ............ a 1 ^ ¼ pffiffiffiffiffiffi ð2i þ 3j 4kÞ a 29 65 Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼ 2i þ 3j 4k ; j A j¼ 4 þ 9 þ 16 ¼ 29 A 1 ^¼ ¼ pffiffiffiffiffiffi ð2i þ 3j 4kÞ a jAj 29 1 ^ ¼ pffiffiffiffiffiffi ð2i þ 3j 4kÞ So we have r ¼ 6i þ 9j 3k and a 29 d ^ r ¼a ; ds ¼ ............ d 51 ¼ pffiffiffiffiffiffi ¼ 9:47 ds 29 Because d 1 ^ r ¼ pffiffiffiffiffiffi ð2i þ 3j 4kÞ ð6i þ 9j 3kÞ ¼a ds 29 1 51 ¼ pffiffiffiffiffiffi ð12 þ 27 þ 12Þ ¼ pffiffiffiffiffiffi ¼ 9:47 29 29 66 800 Programme 22 That is all there is to it. (a) From the given scalar function , determine r. ^ in the direction of the given vector A. (b) Find the unit vector a (c) Then d ^ r. ¼a ds Example 2 Find the directional derivative of ¼ x2 y þ y 2 z þ z2 x at the point ð1, 1, 2Þ in the direction of the vector A ¼ 4i þ 2j 5k. Same as before. Work through it and check the result with the next frame 67 d 23 ¼ pffiffiffi ¼ 3:43 ds 3 5 Because ¼ x 2 y þ y 2 z þ z2 x ; r ¼ ð2xy þ z2 Þi þ ðx2 þ 2yzÞj þ ðy 2 þ 2zxÞk ; At ð1, 1, 2Þ, r ¼ 2i 3j þ 5k pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi A ¼ 4i þ 2j 5k ; j A j ¼ 16 þ 4 þ 25 ¼ 45 ¼ 3 5 1 ^ ¼ pffiffiffi ð4i þ 2j 5kÞ ; a 3 5 ; d 1 ^ r ¼ pffiffiffi ð4i þ 2j 5kÞ ð2i 3j þ 5kÞ ¼a ds 3 5 1 23 ¼ pffiffiffi ð8 6 25Þ ¼ pffiffiffi ¼ 3:43 3 5 3 5 Example 3 Find the direction from the point ð1, 1, 0Þ which gives the greatest rate of increase of the function ¼ ðx þ 3yÞ2 þ ð2y zÞ2 . This appears to be different, but it rests on the fact that the greatest rate of increase of with respect to distance is in ............ 68 the direction of r All we need then is to find the vector r, which is ............ 801 Vector analysis 1 69 r ¼ 4ð2i þ 8j kÞ Because ¼ ðx þ 3yÞ2 þ ð2y zÞ2 @ @ ¼ 2ðx þ 3yÞ; ¼ 6ðx þ 3yÞ þ 4ð2y zÞ; @x @y @ @ @ ¼ 8; ¼ 32; ¼ 4 ; At ð1, 1, 0Þ, @x @y @z ; @ ¼ 2ð2y zÞ @z ; r ¼ 8i þ 32j 4k ¼ 4ð2i þ 8j kÞ ; greatest rate of increase occurs in direction 2i þ 8j k So on we go 70 Unit normal vectors The equation of ðx; y; zÞ ¼ constant represents a surface in space. For example, 3x 4y þ 2z ¼ 1 is the equation of a plane and x2 þ y 2 þ z2 ¼ 4 represents a sphere centred on the origin and of radius 2. z dr If dr is a displacement in this surface, then d ¼ 0 since is constant over the surface. r y x O z y x Therefore our previous relationship dr grad ¼ d becomes dr grad ¼ 0 for all such small displacements dr in the surface. But dr grad ¼j dr jj grad j cos ¼ 0: ; grad is perpendicular to dr, i.e. grad is a vector perpendicular ; ¼ 2 to the surface at P, in the direction of maximum rate of change of . The magnitude of that maximum rate of change is given by j grad j. 802 Programme 22 The unit vector N in the direction of grad is called the unit normal vector at P. z ; Unit normal vector N¼ dr r y x r j r j ð15Þ z O y x Example 1 Find the unit normal vector to the surface x3 y þ 4xz2 þ xy 2 z þ 2 ¼ 0 at the point ð1; 3; 1Þ. Vector normal ¼ r ¼ . . . . . . . . . . . . 71 r ¼ ð3x2 y þ 4z2 þ y 2 zÞi þ ðx3 þ 2xyzÞj þ ð8xz þ xy2 Þk Then, at ð1, 3, 1Þ, r ¼ 4i 5j þ k and the unit normal at ð1, 3, 1Þ is . . . . . . . . . . . . 72 1 pffiffiffiffiffiffi ð4i 5j þ kÞ 42 Because pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi 16 þ 25 þ 1 ¼ 42 r 1 ¼ pffiffiffiffiffiffi ð4i 5j þ kÞ N¼ j r j 42 j r j ¼ and One more. Example 2 Determine the unit normal to the surface xyz þ x2 y 5yz 5 ¼ 0 at the point ð3, 1, 2Þ. All very straightforward. Complete it. 803 Vector analysis 1 1 Unit normal ¼ N ¼ pffiffiffiffiffiffi ð8i þ 5j 2kÞ 93 73 Because ¼ xyz þ x2 y 5yz 5 ; r ¼ ðyz þ 2xyÞi þ ðxz þ x2 5zÞj þ ðxy 5yÞk pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi At ð3, 1, 2Þ, r ¼ 8i þ 5j 2k; j r j¼ 64 þ 25 þ 4 ¼ 93 r 1 ; Unit normal ¼ N ¼ ¼ pffiffiffiffiffiffi ð8i þ 5j 2kÞ j r j 93 Collecting our results so far, we have, for ðx, y, zÞ a scalar function @ @ @ iþ jþ k @x @y @z @ @ @ (b) d ¼ dr grad where d ¼ dx þ dy þ dz @x @y @z d ^ grad ¼a (c) directional derivative ds r . (d) unit normal vector N ¼ j r j (a) grad ¼ r ¼ Copy out this brief summary for future reference. It will help Grad of sums and products of scalars @ @ @ (a) rðA þ BÞ ¼ i ðA þ BÞ þ j ðA þ BÞ þ k ðA þ BÞ @x @y @z @A @A @A @B @B @B ¼ iþ jþ k þ iþ jþ k @x @y @z @x @y @z ; rðA þ BÞ ¼ rA þ rB @ @ @ ðABÞ þ j ðABÞ þ k ðABÞ (b) rðABÞ ¼ i @x @y @z @B @A @B @A @B @A þB þB þB þj A þk A ¼i A @x @x @y @y @z @z @B @B @B @A @A @A iþA jþA k þ B iþB jþB k ¼ A @x @y @z @x @y @z @B @B @B @A @A @A iþ jþ k þB iþ jþ k ¼A @x @y @z @x @y @z ; rðABÞ ¼ AðrBÞ þ BðrAÞ Remember that in these results A and B are scalars. 74 804 75 Programme 22 Example If A ¼ x2 yz þ xz2 and B ¼ xy2 z z3 , evaluate rðABÞ at the point ð2, 1, 3Þ. We know that rðABÞ ¼ AðrBÞ þ BðrAÞ At ð2, 1, 3Þ, rB ¼ . . . . . . . . . . . . ; 76 rA ¼ . . . . . . . . . . . . rB ¼ 3i þ 12j 25k; rA ¼ 21i þ 12j þ 16k rB ¼ @B @B @B iþ jþ k ¼ y 2 zi þ 2xyzj þ ðxy 2 3z2 Þk @x @y @z ¼ 3i þ 12j 25k at ð2, 1, 3Þ rA ¼ @A @A @A iþ jþ k ¼ ð2xyz þ z2 Þi þ x2 zj þ ðx2 y þ 2xzÞk @x @y @z ¼ 21i þ 12j þ 16k at ð2, 1, 3Þ Now rðABÞ ¼ AðrBÞ þ BðrAÞ ¼ . . . . . . . . . . . . Finish it 77 rðABÞ ¼ 3ð117i þ 36j 362kÞ Because rðABÞ ¼ AðrBÞ þ BðrAÞ A ¼ x2 yz þ xz2 2 ; at ð2, 1, 3Þ, A ¼ 12 þ 18 ¼ 30 3 B ¼ xy z z ; at ð2, 1, 3Þ, B ¼ 6 27 ¼ 21 ; rðABÞ ¼ 30ð3i þ 12j 25kÞ 21ð21i þ 12j þ 16kÞ ¼ 351i þ 108j 1086k ¼ 3ð117i þ 36j 362kÞ So add these to the list of results. rðA þ BÞ ¼ rA þ rB rðABÞ ¼ AðrBÞ þ BðrAÞ where A and B are scalars. Now on to the next page 805 Vector analysis 1 Div (divergence of a vector function) 78 The operator r (notice the ‘dot’; it makes all the difference) can be applied to a vector function Aðx, y, zÞ to give the divergence of A, written in short as div A. If A ¼ ax i þ ay j þ az k @ @ @ div A ¼ r A ¼ i þ j þ k ax i þ ay j þ az k @x @y @z @ax @ay @az ; div A ¼ r A ¼ þ þ @x @y @z Note that (a) the grad operator r acts on a scalar and gives a vector (b) the div operation r acts on a vector and gives a scalar. Example 1 If A ¼ x2 yi xyzj þ yz2 k then div A ¼ r A ¼ . . . . . . . . . . . . div A ¼ r A ¼ 2xy xz þ 2yz 79 We simply take the appropriate partial derivatives of the coefficients of i, j and k. It could hardly be easier. Example 2 If A ¼ 2x2 yi 2ðxy2 þ y 3 zÞj þ 3y 2 z2 k, determine r A, i.e. div A. Complete it. r A ¼ ............ rA¼0 Because A ¼ 2x2 yi 2ðxy 2 þ y 3 zÞj þ 3y 2 z2 k @ax @ay @az þ þ rA¼ @x @y @z ¼ 4xy 2ð2xy þ 3y 2 zÞ þ 6y 2 z ¼ 4xy 4xy 6y 2 z þ 6y 2 z ¼ 0 Such a vector A for which r A ¼ 0 at all points, i.e. for all values of x, y, z, is called a solenoidal vector. It is rather a special case. 80 806 Programme 22 Curl (curl of a vector function) The curl operator denoted by r, acts on a vector and gives another vector as a result. If A ¼ ax i þ ay j þ az k, then curl A ¼ r A. @ @ @ ðax i þ ay j þ az kÞ i.e. curl A ¼ r A ¼ i þ j þ k @x @y @z i ¼ ; rA¼i @ @x ax @az @ay @y @z j k @ @ @y @z ay az @ay @ax @ax @az þj þk @z @x @x @y Curl A is thus a vector function. It is best remembered in its determinant form, so make a note of it. If r A ¼ 0 then A is said to be irrotational. Then on for an example 81 Example 1 If A ¼ ðy 4 x2 z2 Þi þ ðx2 þ y 2 Þj x2 yzk, determine curl A at the point ð1, 3, 2Þ. curl A ¼ r A ¼ i @ @x j @ @y k @ @z y 4 x 2 z2 x2 þ y 2 x2 yz Now we expand the determinant @ @ @ @ rA¼i ðx2 yzÞ ðx2 þ y 2 Þ j ðx2 yzÞ ðy 4 x2 z2 Þ @y @z @x @z @ 2 @ ðx þ y 2 Þ ðy 4 x2 z2 Þ þk @x @y All that now remains is to obtain the partial derivatives and substitute the values of x, y, z. ; r A ¼ ............ 82 2i 8j 106k r A ¼ ifx2 zg jf2xyz þ 2x2 zg þ kf2x 4y 3 g. ; At ð1, 3, 2Þ, r A ¼ ið2Þ jð12 4Þ þ kð2 108Þ ¼ 2i 8j 106k 807 Vector analysis 1 Example 2 Determine curl F at the point (2, 0, 3) given that F ¼ ze2xy i þ 2xz cos yj þ ðx þ 2yÞk. In determinant form, curl F ¼ r F ¼ . . . . . . . . . . . . i @ @x ze2xy j @ @y 2xz cos y x þ 2y k @ @z 83 Now expand the determinant and substitute the values of x, y and z, finally obtaining curl F ¼ . . . . . . . . . . . . curl F ¼ r F ¼ 2ði þ 3kÞ Because r F ¼ if2 2x cos yg jf1 e2xy g þ kf2z cos y 2xze2xy g ; At ð2, 0, 3Þ r F ¼ ið2 4Þ jð1 1Þ þ kð6 12Þ ¼ 2i 6k ¼ 2ði þ 3kÞ Every one is done in the same way. Summary of grad, div and curl (a) Grad operator r acts on a scalar field to give a vector field. (b) Div operator r acts on a vector field to give a scalar field. (c) Curl operator r acts on a vector field to give a vector field. (d) With a scalar function ðx; y; zÞ grad ¼ r ¼ @ @ @ iþ jþ k @x @y @z (e) With a vector function A ¼ ax i þ ay j þ az k @ax @ay @az þ þ @x @y @z i j k @ @ @ ð2Þ curl A = rA ¼ @x @y @z ax ay az ð1Þ div A ¼ r A ¼ Check through that list, just to make sure. We shall need them all 84 808 85 Programme 22 By way of revision, here is one further example. Example 3 If ¼ x2 y 2 þ x3 yz yz2 and F ¼ xy 2 i 2yzj þ xyzk determine for the point P ð1, 1, 2Þ, (a) r, (b) unit normal, (c) r F, (d) r F. Complete all four parts and then check the results with the next frame 86 Here is the working in full. ¼ x2 y 2 þ x3 yz yz2 (a) r ¼ @ @ @ iþ jþ k @x @y @z ¼ ð2xy 2 þ 3x2 yzÞi þ ð2x2 y þ x3 z z2 Þj þ ðx3 y 2yzÞk ; At ð1; 1; 2Þ r ¼ 4i 4j þ 3k pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi r j r j¼ 16 þ 16 þ 9 ¼ 41 j r j 1 ; N ¼ pffiffiffiffiffiffi ð4i þ 4j 3kÞ 41 (b) N ¼ (c) F ¼ xy 2 i 2yzj þ xyzk rF¼ @ax @ay @az þ þ @x @y @z ; r F ¼ y 2 2z þ xy ; At ð1, 1, 2Þ (d) i @ rF¼ @x xy 2 r F ¼ 1 4 1 ¼ 4 j @ @y k @ @z 2yz xyz ; r F ¼ 4 ; r F ¼ iðxz þ 2yÞ jðyz 0Þ þ kð0 2xyÞ ¼ ðxz þ 2yÞi yzj 2xyk ; At ð1, 1, 2Þ r F ¼ 2j þ 2k ; r F ¼ 2ðj þ kÞ Now let us combine some of these operations. 809 Vector analysis 1 87 Multiple operations We can combine the operators grad, div and curl in multiple operations, as in the examples that follow. Example 1 If A ¼ x2 yi þ yz3 j zx3 k @ @ @ i þ j þ k ðx2 yi þ yz3 j zx3 kÞ then div A ¼ r A ¼ @x @y @z Then ¼ 2xy þ z3 þ x3 ¼ say @ @ @ iþ jþ k grad (div AÞ ¼ rðr AÞ ¼ @x @y @z ¼ ð2y þ 3x2 Þi þ ð2xÞj þ ð3z2 Þk i.e. grad div A ¼ rðr AÞ ¼ ð2y þ 3x2 Þi þ 2xj þ 3z2 k Move on for the next example Example 2 88 2 2 2 If ¼ xyz 2y z þ x z , determine div grad at the point ð2, 4, 1Þ. First find grad and then the div of the result. At ð2, 4, 1Þ, div grad ¼ r ðrÞ ¼ . . . . . . . . . . . . div grad ¼ 6 Because we have ¼ xyz 2y 2 z þ x2 z2 grad ¼ r ¼ @ @ @ iþ jþ k @x @y @z ¼ ðyz þ 2xz2 Þi þ ðxz 4yzÞj þ ðxy 2y 2 þ 2x2 zÞk ; div grad ¼ r ðrÞ ¼ 2z2 4z þ 2x2 ; At ð2, 4, 1Þ, div grad ¼ r ðrÞ ¼ 2 4 þ 8 ¼ 6 Example 3 If F ¼ x2 yzi þ xyz2 j þ y 2 zk determine curl curl F at the point (2, 1, 1). Determine an expression for curl F in the usual way, which will be a vector, and then the curl of the result. Finally substitute values. curl curl F ¼ . . . . . . . . . . . . 89 810 Programme 22 90 curl curl F ¼ r ðr FÞ ¼ i þ 2j þ 6k Because i @ curl F ¼ @x j @ @y x2 yz xyz2 k @ @z y2 z ¼ ð2yz 2xyzÞi þ x2 yj þ ðyz2 x2 zÞk Then i @ @x j @ @y k @ @z 2yz 2xyz x2 y yz2 x2 z curl curl F ¼ ¼ z2 i ð2xz 2y þ 2xyÞj þ ð2xy 2z þ 2xzÞ k ; At (2, 1, 1), 91 curl curl F ¼ r ðr FÞ ¼ i þ 2j þ 6k Remember that grad, div and curl are operators and that they must act on a scalar or vector as appropriate. They cannot exist alone and must be followed by a function. One or two interesting general results appear. (a) Curl grad where is a scalar @ @ @ iþ jþ k @x @y @z i j k @ @ @ ; curl grad ¼ @x @y @z @ @ @ @x @y @z 2 2 @ @2 @ @2 ¼i j @y@z @z@y @z@x @x@z 2 @ @2 þk @x@y @y@x grad ¼ ¼0 ; curl grad ¼ r ðrÞ ¼ 0 811 Vector analysis 1 (b) Div curl A where A is a vector. A ¼ ax i þ ay j þ az k i j k @ @ @ curl A ¼ r A ¼ @x @y @z ax ay az @a @ay @ax @az @az @ax y j þk ¼i @y @z @x @z @x @y @ @ @ Then div curl A ¼ r ðr AÞ ¼ i þ j þ k ðr AÞ @x @y @z @ 2 az @ 2 ay @ 2 az @ 2 ax @ 2 ay @ 2 ax þ þ @x@y @z@x @x@y @y@z @z@x @y@z ¼0 ¼ ; div curl A ¼ r ðr AÞ ¼ 0 (c) Div grad where is a scalar grad ¼ @ @ @ iþ jþ k @x @y @z Then div grad ¼ r ðrÞ @ @ @ @ @ @ ¼ i þj þk iþ jþ k @x @y @z @x @y @z ¼ @2 @2 @2 þ þ @x2 @y 2 @z2 ; div grad ¼ r ðrÞ ¼ @2 @2 @2 þ þ @x2 @y 2 @z2 ¼ r2 , the Laplacian of The operator r2 is called the Laplacian. So these general results are (a) curl grad ¼ r ðrÞ ¼ 0 (b) div curl A ¼ r ðr AÞ ¼ 0 (c) div grad ¼ r ðrÞ ¼ @2 @2 @2 þ þ . @x2 @y 2 @z2 That brings us to the end of this particular Programme. We have covered quite a lot of new material, so check carefully through the Revision summary and Can you? checklist that follow: then you can deal with the Test exercise. The Further problems provide an opportunity for additional practice. 812 Programme 22 Revision summary 22 92 If A ¼ ax i þ ay j þ az k; B ¼ bx i þ by j þ bz k; C ¼ cx i þ cy j þ cz k; then we have the following relationships. 1 Scalar product (dot product) AB¼BA A B ¼ AB cos A ðB þ CÞ ¼ A B þ A C and If A B ¼ 0 and A, B 6¼ 0 then A ? B. 2 Vector product (cross product) A B ¼ ðAB sin Þn n = unit normal vector where A, B, n form a right-handed set. i A B ¼ ax bx j ay by k az bz A B ¼ ðB AÞ and A ðB þ CÞ ¼ A B þ A C 3 Unit vectors (a) i i ¼ j j ¼ k k ¼ 1 i j ¼ j k ¼ k i ¼ 0. (b) i i ¼ j j ¼ k k ¼ 0 i j ¼ k, j k ¼ i, k i ¼ j. 4 A ðB CÞ Scalar triple product ax A ðB CÞ ¼ bx cx ay by cy az bz cz A ðB CÞ ¼ B ðC AÞ ¼ C ðA BÞ Unchanged by cyclic change of vectors. Sign reversed by non-cyclic change of vectors. A ðB CÞ ¼ 0: 5 Coplanar vectors 6 Vector triple product A ðB CÞ and ðA BÞ C A ðB CÞ ¼ ðA CÞB ðA BÞC and 7 ðA BÞ C ¼ ðC AÞB ðC BÞA: Differentiation of vectors If A, ax , ay , az are functions of u day dA dax daz ¼ iþ jþ k du du du du 8 Unit tangent vector T dA T ¼ du dA du 813 Vector analysis 1 9 Integration of vectors ðb ðb ðb ðb A du ¼ i ax du þ j ay du þ k az du a 10 a a a Grad (gradient of a scalar function ) @ @ @ iþ jþ k @x @y @z @ @ @ ‘del’ ¼ operator r ¼ i þ j þ k @x @y @z d ^ grad ¼ a ^ r where a ^ is a unit (a) Directional derivative ¼a ds vector in a stated direction. Grad gives the direction for maximum rate of change of . (b) Unit normal vector N to surface ðx; y; zÞ ¼ constant. grad ¼ r ¼ N¼ 11 r j r j Div (divergence of a vector function A) div A ¼ r A ¼ @ax @ay @az þ þ @x @y @z If r A ¼ 0 for all points, A is a solenoidal vector. 12 Curl (curl of a vector function AÞ i @ curl A ¼ r A ¼ @x ax j @ @y ay k @ @z az If r A ¼ 0 then A is an irrotational vector. 13 Operators grad ðrÞ acts on a scalar and gives a vector div ðrÞ acts on a vector and gives a scalar curl ðrÞ acts on a vector and gives a vector. 14 Multiple operations (a) curl grad ¼ r ðrÞ ¼ 0 (b) div curl A ¼ r ðr AÞ ¼ 0 (c) div grad ¼ r ðrÞ ¼ @2 @2 @2 þ þ @x2 @y 2 @z2 ¼ r2 , the Laplacian of . 814 Programme 22 Can you? 93 Checklist 22 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Obtain the scalar and vector product of two vectors? Yes No Frames 1 to 4 . Reproduce the relationships between the scalar and vector products of the Cartesian coordinate unit vectors? Yes No 5 to 11 . Obtain the scalar and vector triple products and appreciate their geometric significance? Yes No 12 to 27 . Differentiate a vector field and derive a unit vector tangential to the vector field at a point? Yes No 28 to 48 49 to 55 . Obtain the gradient of a scalar field, the directional derivative and a unit normal to a surface? Yes No 56 to 77 . Obtain the divergence of a vector field and recognise a solenoidal vector field? Yes No 78 to 80 80 to 86 87 to 91 . Integrate a vector field? Yes . Obtain the curl of a vector field? Yes No No . Obtain combinations of div, grad and curl acting on scalar and vector fields as appropriate? Yes No 815 Vector analysis 1 Test exercise 22 1 Find (a) the scalar product and (b) the vector product of the vectors A ¼ 3i 2j þ 4k and B ¼ i þ 5j 2k: 2 If A ¼ 2i þ 3j 5k; B ¼ 3i þ j þ 2k; 94 C ¼ i j þ 3k; determine (a) the scalar triple product A ðB CÞ (b) the vector triple product A ðB CÞ. 3 Determine whether the three vectors A ¼ 2i þ 3j þ k; B ¼ i 2j þ 2k; C ¼ 3i þ j þ 3k are coplanar. 4 If A ¼ ðu2 þ 5Þi ðu2 þ 3Þj þ 2u3 k, determine dA d2 A dA ; (b) ; (c) ; all at u ¼ 2. du du2 du Determine the unit tangent vector at the point ð2, 4, 3Þ for the curve with parametric equations (a) 5 x ¼ 2u2 ; 6 y ¼ u þ 3; z ¼ 4u2 u. If F ¼ 2i þ 4uj þ u2 k and G ¼ u2 i 2uj þ 4k, determine ð2 ðF GÞdu: 0 7 Find the directional derivative of the function ¼ x2 y 2xz2 þ y 2 z at the point ð1, 3, 2Þ in the direction of the vector A ¼ 3i þ 2j k. 8 Find the unit normal to the surface ¼ 2x3 z þ x2 y 2 þ xyz 4 ¼ 0 at the point ð2, 1, 0Þ. 9 If A ¼ x2 yi þ ðxy þ yzÞj þ xz2 k; B ¼ yzi 3xzj þ 2xyk; and ¼ 3x2 y þ xyz 4y 2 z2 3; determine, at the point ð1, 2, 1Þ (a) r; (b) r A; (c) r B; (d) grad div A; (e) curl curl A. Further problems 22 1 If A ¼ 2i þ 3j 4k; B ¼ 3i þ 5j þ 2k; C ¼ i 2j þ 3k; determine A ðB CÞ. 2 If A ¼ 2i þ j 3k; B ¼ i 2j þ 2k; C ¼ 3i þ 2j k; find A ðB CÞ: 3 If A ¼ i 2j þ 3k; B ¼ 2i þ j 2k; C ¼ 3i þ 2j þ k; find (a) A ðB CÞ; 4 (b) ðA BÞ C. If F ¼ x2 i þ ð3x þ 2Þj þ sin xk, find (a) dF ; dx (b) d2 F ; dx2 (c) dF ; dx (d) d ðF FÞ at x ¼ 1. dx 95 816 Programme 22 5 If F ¼ ui þ ð1 uÞj þ 3uk and G ¼ 2i ð1 þ uÞj u2 k, determine d d d ðF GÞ; (b) ðF GÞ; (c) ðF þ GÞ. (a) du du du 6 Find the unit normal to the surface 4x2 y 2 3xz2 2y 2 z þ 4 ¼ 0 at the point ð2, 1, 2Þ. 7 Find the unit normal to the surface 2xy 2 þ y 2 z þ x2 z 11 ¼ 0 at the point ð2, 1, 3Þ. 8 Determine the unit vector normal to the surface xz2 þ 3xy 2yz2 þ 1 ¼ 0 at the point ð1, 2, 1Þ. 9 Find the unit normal to the surface x2 y 2yz2 þ y 2 z ¼ 3 at the point ð2, 3, 1Þ. 10 Determine the directional derivative of ¼ xey þ yz2 þ xyz at the point ð2, 0, 3Þ in the direction of A ¼ 3i 2j þ k. 11 Find the directional derivative of ¼ ðx þ 2y þ zÞ2 ðx y zÞ2 at the point (2, 1, 1) in the direction of A ¼ i 4j þ 2k. 12 Find the scalar triple product of (a) A ¼ i þ 2j 3k; B ¼ 2i j þ 4k; C ¼ 3i þ j 2k. (b) A ¼ 2i 3j þ k; B ¼ 3i þ j þ 2k; C ¼ i þ 4j 2k: (c) A ¼ 2i þ 3j 2k; 13 (a) A ¼ 3i þ j 2k; B ¼ 2i þ 4j þ 3k; C ¼ i 2j þ k. (b) A ¼ 2i j þ 3k; B ¼ i þ 4j 5k; C ¼ 3i 2j þ k: B ¼ 2i 3j þ 2k; C ¼ 3i 3j þ k: If F ¼ 4t 3 i 2t 2 j þ 4tk, determine when t ¼ 1 (a) 15 C ¼ 2i 5j þ k: Find the vector triple product A ðB CÞ of the following. (c) A ¼ 4i þ 2j 3k; 14 B ¼ 3i j þ 3k; dF ; dt (b) d2 F ; dt 2 (c) d ðF FÞ. dt If ¼ x2 sin z þ zey find, at the point ð1, 3, 2Þ, the values of (a) grad and (b) j grad j. 16 Given that ¼ xy 2 þ yz2 x2 , find the derivative of with respect to distance at the point ð1, 2, 1Þ, measured parallel to the vector 2i 3j þ 4k: 17 Find unit vectors normal to the surfaces x2 þ y 2 z2 þ 3 ¼ 0 and xy yz þ zx 10 ¼ 0 at the point (3, 2, 4) and hence find the angle between the two surfaces at that point. 18 If r ¼ ðt 2 þ 3tÞi 2 sin 3tj þ 3e2t k, determine (a) 19 dr ; dt (b) d2 r d2 r ; (c) the value of at t ¼ 0. dt 2 dt 2 (a) Show that curl ðyi þ xjÞ is a constant vector. (b) Show that the vector field ðyzi þ zxj þ xykÞ has zero divergence and zero curl. 817 Vector analysis 1 20 If A ¼ 2xz2 i xzj þ ðy þ zÞk, find curl curl A. 21 Determine grad where ¼ x2 cosð2yz 0:5Þ and obtain its value at the point ð1, 3, 1Þ. 22 Determine the value of p such that the three vectors A, B, C are coplanar when A ¼ 2i þ j þ 4k; B ¼ 3i þ 2j þ pk; C ¼ i þ 4j þ 2k: 23 If A ¼ pi 6j 3k; B ¼ 4i þ 3j k; C ¼ i 3j þ 2k (a) find the values of p for which (1) A and B are perpendicular to each other (2) A; B and C are coplanar. (b) determine a unit vector perpendicular to both A and B when p ¼ 2. Programme 23 Frames 1 to 87 Vector analysis 2 Learning outcomes When you have completed this Programme you will be able to: . Evaluate the line integral of a scalar and a vector field in Cartesian coordinates . Evaluate the volume integral of a vector field . Evaluate the surface integral of a scalar and a vector field . Determine whether or not a vector field is a conservative vector field . Apply Gauss’ divergence theorem . Apply Stokes’ theorem . Determine the direction of unit normal vectors to a surface . Apply Green’s theorem in the plane 818 819 Vector analysis 2 We dealt in some detail with line, surface and volume integrals in an earlier Programme, when we approached the subject analytically. In many practical problems, it is more convenient to express these integrals in vector form and the methods often lead to more concise working. Line integrals Let a point P on the curve c joining A and B be denoted by the position vector r with respect to a fixed origin O. If Q is a neighbouring point on the curve with position vector r þ dr, then PQ ¼ dr: The curve c can be divided up into many (n) such small arcs, approximating to dr1 , dr2 , dr3 . . . drp . . . so that AB ¼ n X dr (a) r + dr r O (b) drp dr3 dr2 dr1 drp p¼1 where drp is a vector representing the element of arc in both magnitude and direction. Scalar field If a scalar field V exists for all points on the curve, then n X V drp with dr ! 0, p¼1 defines the line integral of V along the curve c from A to B, ð i.e. line integral ¼ V dr c We can illustrate this integral by erecting a continuous ordinate proportional to V at each point of ð the curve. V dr is then repre- y z V c sented by the area of the curved surface between the ends A and B of the curve c. O r x To evaluate a line integral, the integrand is expressed in terms of x, y, z, with dr ¼ . . . . . . . . . . . . 1 820 Programme 23 2 dr ¼ i dx þ j dy þ k dz In practice, x, y and z are often expressed in terms of parametric equations of a fourth variable (say u), i.e. x ¼ xðuÞ; y ¼ yðuÞ; z ¼ zðuÞ. From these, dx, dy and dz can be written in terms of u and the integral evaluated in terms of this parameter u. The following examples will show the method. Example 1 If V ¼ xy 2 z, evaluate ð V dr along the curve c having parametric equations c x ¼ 3u; y ¼ 2u2 ; z ¼ u3 between A ð0, 0, 0Þ and B ð3, 2, 1Þ. V ¼ xy 2 z ¼ ð3uÞð4u4 Þðu3 Þ ¼ 12u8 dr ¼ i dx þ j dy þ k dz ¼ . . . . . . . . . . . . 3 dr ¼ i 3 du þ j 4u du þ k 3u2 du Because x ¼ 3u, y ¼ 2u2 , z ¼ u3 , Limits: ; dx ¼ 3 du ; dy ¼ 4u du ; dz ¼ 3u2 du A ð0, 0, 0Þ corresponds to u ¼ . . . . . . . . . . . . B ð3, 2, 1Þ corresponds to u ¼ . . . . . . . . . . . . 4 A ð0; 0; 0Þ u ¼ 0 ð V dr ¼ ; ð1 B ð3; 2; 1Þ u ¼ 1 12u8 ði 3 du þ j 4u du þ k 3u2 duÞ 0 c ¼ ............ Finish it off 5 4i þ 24 36 jþ k 5 11 Because ð ð1 V dr ¼ 12 ði 3u8 du þ j 4u9 du þ k 3u10 duÞ c 0 which integrates directly to give the result quoted above. Now for another example. 821 Vector analysis 2 Example 2 2 If V ¼ xy þ y z, evaluate 6 ð V dr along the curve c defined by c x ¼ t 2 ; y ¼ 2t; z ¼ t þ 5 between A ð0, 0, 5Þ and B ð4, 4, 7Þ. As before, expressing V and dr in terms of the parameter t we have V ¼ ............ V ¼ 6t 3 þ 20t 2 ; dr ¼ . . . . . . . . . . . . dr ¼ i 2t dt þ j 2 dt þ k dt 7 Because V ¼ xy þ y 2 z ¼ ðt 2 Þð2tÞ þ ð4t 2 Þðt þ 5Þ ¼ 6t 3 þ 20t 2 : 9 dx ¼ 2t dt > Also x ¼ t 2 = ; dr ¼ i dx þ j dy þ k dz y ¼ 2t dy ¼ 2 dt > ; z ¼tþ5 dz ¼ dt ¼ i 2t dt þ j 2 dt þ k dt ð ð ; V dr ¼ ð6t 3 þ 20t 2 Þði 2t þ j 2 þ k Þ dt c c A ð0, 0, 5Þ t ¼ . . . . . . . . . . . . Limits: B ð4, 4, 7Þ t ¼ . . . . . . . . . . . . A ð0; 0; 5Þ t ¼ 0; ð V dr ¼ ; c ð2 B ð4; 4; 7Þ t ¼ 2 8 ð6t 3 þ 20t 2 Þði 2t þ j 2 þ k Þ dt 0 ¼ . . . . . . . . . . . . Complete the integration. 8 ð444 i þ 290 j þ 145 kÞ 15 ð V dr ¼ 2 c ð2 fð6t 4 þ 20t 3 Þi þ ð6t 3 þ 20t 2 Þj þ ð3t 3 þ 10t 2 Þkg dt 0 The actual integration is simple enough and gives the result shown. All line integrals in scalar fields are done in the same way. 9 822 10 Programme 23 Vector field If a vector field F exists for all points of the curve c, then for each element of arc we can form the scalar product F dr. Summing these products for all elements n X F drp of arc, we have F dr r + dr c r O p¼1 Then, if drp ! 0, the sum becomes the integral ð F dr, c i.e. the line integral of F from A to B along the stated curve ð . ¼ F dr c In this case, since F dr is a scalar product, then the line integral is a scalar. To evaluate the line integral, F and dr are expressed in terms of x, y, z and the curve in parametric form. We have F ¼ Fx i þ Fy j þ Fz k dr ¼ i dx þ j dy þ k dz and F dr ¼ ðFx i þ Fy j þ Fz k Þ ði dx þ j dy þ k dzÞ ¼ Fx dx þ Fy dy þ Fz dz ð ð ð ð ; F dr¼ Fx dx þ Fy dy þ Fz dz Then c c c c Now for an example to show it in operation. Example 1 2 If F ¼ x yi þ xzj 2yzk, evaluate ð F dr between A ð0, 0, 0Þ and B ð4, 2, 1Þ c along the curve having parametric equations x ¼ 4t; y ¼ 2t 2 ; z ¼ t 3 : Expressing everything in terms of the parameter t, we have F ¼ ............ dx ¼ . . . . . . . . . . . . ; dy ¼ . . . . . . . . . . . . ; dz ¼ . . . . . . . . . . . . 823 Vector analysis 2 11 F ¼ 32t 4 i þ 4t 4 j 4t 5 k dx ¼ 4 dt; dy ¼ 4t dt; dz ¼ 3t 2 dt Because x2 y ¼ ð16t 2 Þð2t 2 Þ ¼ 32t 4 3 xz ¼ ð4tÞðt Þ ¼ 4t 4 x ¼ 4t y ¼ 2t ; dx ¼ 4 dt 2 ; dy ¼ 4t dt z ¼ t3 ; dz ¼ 3t 2 dt 2yz ¼ ð4t 2 Þðt 3 Þ ¼ 4t 5 ð ð Then F dr ¼ ð32t 4 i þ 4t 4 j 4t 5 k Þ ði 4 dt þ j 4t dt þ k 3t 2 dtÞ ð ¼ ð128t 4 þ 16t 5 12t 7 Þ dt A ð0, 0, 0Þ t ¼ . . . . . . . . . . . . ; Limits: A t ¼ 0; ð F dr ¼ ; c ð1 B ð4, 2, 1Þ t ¼ . . . . . . . . . . . . 12 Bt¼1 ð128t 4 þ 16t 5 12t 7 Þ dt ¼ . . . . . . . . . . . . 0 13 128 8 3 803 þ ¼ ¼ 26:77 5 3 2 30 ð F dr represents the If the vector field F is a force field, then the line integral c work done in moving a unit particle along the prescribed curve c from A to B. Now for another example. Example 2 If F ¼ x2 yi þ 2yzj þ 3z2 xk, evaluate (a) along the straight lines then and (b) along the straight line c1 c2 c3 c4 ð F dr between A ð0, 0, 0Þ and B ð1, 2, 3Þ c from ð0, 0, 0Þ to ð1, 0, 0Þ from ð1, 0, 0Þ to ð1, 2, 0Þ from ð1, 2, 0Þ to ð1, 2, 3Þ joining ð0, 0, 0Þ to ð1, 2, 3Þ. As before, we first obtain an expression for F dr which is ............ 824 Programme 23 14 F dr ¼ x2 y dx þ 2yz dy þ 3z2 x dz Because ð ; F dr ¼ ðx2 y i þ 2yz j þ 3z2 x k Þ ði dx þ j dy þ k dzÞ ð ð ð F dr ¼ x2 y dx þ 2yz dy þ 3z2 x dz z (a) Here the integration is made in three sections, along c1 , c2 and c3 . (1, 2, 3) x (1) c1 : 1 c1 (1, 0, 0) y ¼ 0, z ¼ 0, dy ¼ 0, dz ¼ 0 ð F dr ¼ 0 þ 0 þ 0 ¼ 0 ; c1 (2) c2 : The conditions along c2 are ............ 15 x ¼ 1, c2 : ð z ¼ 0, dx ¼ 0, dz ¼ 0 F dr ¼ 0 þ 0 þ 0 ¼ 0 ; c2 (3) c3 : x ¼ 1, y ¼ 2, dx ¼ 0, dy ¼ 0 ð F dr ¼ . . . . . . . . . . . . ; c3 16 27 Because ð ð3 F dr ¼ 0 þ 0 þ 3z2 dz ¼ 27 c3 0 Summing the three partial results ð ð1; 2; 3Þ ð F dr ¼ 0 þ 0 þ 27 ¼ 27 ; ð0; 0; 0Þ F dr ¼ 27 c1 þc2 þc3 c3 O c2 2 (1, 2, 0) y 825 Vector analysis 2 z (b) If t is taken as the parameter, the parametric equations of c are (1, 2, 3) c4 x ¼ ............ y ¼ ............ z ¼ ............ O x x ¼ t; y ¼ 2t: 1 2 y z ¼ 3t 17 and the limits of t are . . . . . . . . . . . . t ¼ 0 and t ¼ 1 18 As in Example 1, we now express everything in terms of t and complete the integral, finally getting ð F dr ¼ . . . . . . . . . . . . c4 ð F dr ¼ c4 115 ¼ 28:75 4 Because F ¼ 2t 3 i þ 12t 2 j þ 27t 3 k dr ¼ i dx þ j dy þ k dz ¼ i dt þ j 2 dt þ k 3 dt ð1 ð F dr ¼ ð2t 3 i þ 12t 2 j þ 27t 3 k Þ ði þ 2j þ 3k Þ dt ; 0 c4 ¼ ð1 ð2t 3 þ 24t 2 þ 81t 3 Þ dt ¼ 0 ð1 ð83t 3 þ 24t 2 Þ dt 0 1 t4 115 ¼ 28:75 ¼ 83 þ 8t 3 ¼ 4 4 0 So the value of the line integral depends on the path taken between the two end points A and B ð (a) F dr via c1 ; c2 and c3 ¼ 27 (b) ð F dr via c4 ¼ 28:75 We shall refer to this topic later. One further example on your own. The working is just the same as before. 19 826 Programme 23 Example 3 ð If F ¼ x2 y 2 i þ y 3 zj þ z2 k, evaluate F dr along the curve x ¼ 2u2 , y ¼ 3u, c z ¼ u3 between A ð2, 3, 1Þ and B ð2, 3, 1Þ. Proceed as before. You will have no difficulty. ð F dr ¼ . . . . . . . . . . . . c ð 20 F dr ¼ c 500 ¼ 23:8 21 Here is the working for you to check. x ¼ 2u2 y ¼ 3u z ¼ u3 x2 y 2 ¼ ð4u4 Þð9u2 Þ ¼ 36 u6 3 3 3 dx ¼ 4u du 6 y z ¼ ð27u Þðu Þ ¼ 27u 2 dy ¼ 3 du 6 dz ¼ 3u2 du z ¼u Limits: A ð2, 3, 1Þ corresponds to u ¼ 1 B ð2, 3, 1Þ corresponds to u ¼ 1 ð1 F dr ¼ ðx2 y 2 i þ y 3 zj þ z2 k Þ ði dx þ j dy þ k dzÞ ð ; 1 c ¼ ¼ ð1 1 ð1 ð36u6 i þ 27u6 j þ u6 k Þ ði 4u du þ j 3 du þ k 3u2 duÞ ð144u7 þ 81u6 þ 3u8 Þ du 1 81u7 u9 þ ¼ 18u þ 7 3 8 1 ¼ 1 500 ¼ 23:8 21 Now on to the next section Volume integrals 21 If V is a closed region bounded by a surface S and F is a vector field at each ð point of V and on its boundary surface S, then F dV is the volume integral of V F throughout the region. z dV = dxdy dz ð O x F dV ¼ V y ð x 2 ð y2 ð z2 F dz dy dx x1 y1 z1 827 Vector analysis 2 Example 1 ð Evaluate F dV where V is the region bounded by the planes x ¼ 0, x ¼ 2, V y ¼ 0, y ¼ 3, z ¼ 0, z ¼ 4, and F ¼ xyi þ zj x2 k. We start, as in most cases, by sketching the diagram, which is ............ 22 z dV O y x Then F ¼ xy i þ z j x2 k and dV ¼ dx dy dz ð4 ð3 ð2 ð F dV ¼ ðxyi þ zj x2 kÞ dx dy dz ; 0 V 0 ð4 ð3 0 x ¼ 2 x2 y x3 dy dz ¼ i þ xzj k 2 3 0 0 x¼0 ð4 ð3 8 ¼ 2yi þ 2zj k dy dz 3 0 0 ¼ . . . . . . . . . . . . Complete the integral. ð F dV ¼ 4ð9i þ 12j 8kÞ V Because y ¼ 3 ð ð4 8 F dV ¼ dz y 2 i þ 2yzj yk 3 0 V y¼0 ð4 ¼ ð9i þ 6zj 8kÞ dz 0 2 4 ¼ 9zi þ 3z j 8zk 0 ¼ 36i þ 48j 32k ¼ 4ð9i þ 12j 8kÞ Now another. 23 828 Programme 23 Example 2 ð Evaluate F dV where V is the region bounded by the planes x ¼ 0, y ¼ 0, V z ¼ 0 and 2x þ y þ z ¼ 2, and F ¼ 2zi þ yk . To sketch the surface 2x þ y þ z ¼ 2, note that when z ¼ 0, 2x þ y ¼ 2 i.e. y ¼ 2 2x when y ¼ 0, 2x þ z ¼ 2 i.e. z ¼ 2 2x when x ¼ 0, yþz¼2 i.e. z ¼ 2 y Inserting these in the planes x ¼ 0, y ¼ 0, z ¼ 0 will help. The diagram is therefore ............ 24 z 2 dV O x – x) y = 2(1 1 2 y So 2x þ y þ z ¼ 2 cuts the axes at A ð1, 0, 0Þ; B ð0, 2, 0Þ; C ð0, 0, 2Þ. Also F ¼ 2zi þ yk; z ¼ 2 2x y ¼ 2ð1 xÞ y ð ð 1 ð 2ð1xÞ ð 2ð1xÞy ; F dV ¼ ð2zi þ yk Þ dz dy dx 0 V ¼ ¼ 0 ð 1 ð 2ð1xÞ 0 ð1 0 0 ð 2ð1xÞ 0 z2 i þ yzk z ¼ 2ð1xÞy dy dx z¼0 f½4ð1 xÞ2 4ð1 xÞy þ y 2 i 0 þ ½ 2ð1 xÞy y 2 kg dy dx ð 1 y3 i ¼ 4ð1 xÞ2 y 2ð1 xÞy 2 þ 3 0 2ð1xÞ y3 2 k þ ð1 xÞy dx 3 y¼0 ¼ ............ Finish the last stage 829 Vector analysis 2 ð F dV ¼ V 25 1 ð2i þ kÞ 3 Because ð ð1 8 4 ð1 xÞ3 i þ ð1 xÞ3 k dx F dV ¼ 3 0 3 V 1 2 1 1 ¼ ð1 xÞ4 i ð1 xÞ4 k ¼ ð2i þ k Þ 3 3 3 0 And now one more, slightly different. Example 3 ð Evaluate F dV where F ¼ 2i þ 2zj þ yk and V is the region bounded by the V planes z ¼ 0, z ¼ 4 and the surface x2 þ y 2 ¼ 9. z It will be convenient to use cylindrical polar coordinates ð, , zÞ so the relevant transformations are dV z O ϕρ x ¼ ............; y ¼ ............ z ¼ ............; dV ¼ . . . . . . . . . . . . y x x ¼ cos ; z ¼ z; Then ð F dV ¼ ððð V y ¼ sin dV ¼ d d dz ð2i þ 2zj þ yk Þ dx dy dz: V Changing into cylindrical polar coordinates with appropriate change of limits this becomes ð ð 2 ð 3 ð 4 F dV ¼ ð2i þ 2zj þ sin k Þ dz d d ¼0 V ¼ ¼ ð 2 ¼0 ð3 z¼0 2zi þ z2 j þ sin zk ¼0 ¼0 ð 2 ð 3 4 d d z¼0 ð8i þ 16j þ 4 sin k Þ d d 0 ¼4 0 ð 2 ð 3 0 ð2i þ 4j þ 2 sin k Þ d d 0 Completing the working, we finally get ð F dV ¼ . . . . . . . . . . . . V 26 830 Programme 23 27 72ði þ 2jÞ Because 3 ð ð 2 3 2 2 F dV ¼ 4 i þ 2 j þ sin k d 3 0 V 0 ð 2 ¼4 ð9i þ 18j þ 9 sin k Þ d 0 ð 2 ¼ 36 ði þ 2j þ sin kÞ d 0 2 ¼ 36 i þ 2j cos k 0 ¼ 36fð2i þ 4j k Þ ðk Þg ¼ 72ði þ 2jÞ You will, of course, remember that in appropriate cases, the use of cylindrical polar coordinates or spherical polar coordinates often simplifies the subsequent calculations. So keep them in mind. Now let us turn to surface integrals – in the next frame Surface integrals 28 (A x B) The vector product of two vectors A and B has magnitude jA Bj ¼ AB sin at right angles to the plane of A and B to form a right-handed set. B A AB n B A I f ¼ , t h e n jA Bj ¼ AB i n t h e 2 ^ is direction of the normal. Therefore, if n a unit normal then ^ ¼ AB n ^ A B ¼ jAj jBjn 831 Vector analysis 2 z If P ðx, yÞ is a point in the x–y plane, the element of area dx dy has a vector area dS ¼ ði dxÞ ðj dyÞ: dx dyk O dx x y dS P dy i.e. dS ¼ dx dyði jÞ ¼ dx dy k i.e. a vector of magnitude dx dy acting in the direction of k and referred to as the vector area. dSn S For a general surface S in space, each element of surface dS has a vector area dS ^. such that dS ¼ dS n dS You will remember we established previously that for a surface S given by the ^ is given by equation ðx; y, zÞ ¼ constant, the unit normal n ^¼ n grad r ¼ jgrad j jrj Let us see how we can apply these results to the following examples. 29 Scalar fields Example 1 A scalar field V ¼ xyz exists over the curved surface S defined by x2 þ y 2 ¼ 4 ð between the planes z ¼ 0 and z ¼ 3 in the first octant. Evaluate V dS over S this surface. We have V ¼ xyz ^ dS dS ¼ n S: x2 þ y 2 4 ¼ 0, z ¼ 0 to z ¼ 3 ^¼ where n r jrj @ @ @ iþ jþ k ¼ 2xi þ 2yj and @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4x2 þ 4y 2 ¼ 2 x2 þ y 2 ¼ 2 4 ¼ 4 Now r ¼ Therefore r xi þ yj xi þ yj ^ dS ¼ ^¼ ¼ so that dS ¼ n dS n jr j 2 2 ð ð ^ dS V dS ¼ V n ; S S ð 1 ¼ xyzðxi þ yj Þ dS 2 S ð 1 ¼ ðx2 yzi þ xy 2 zj Þ dS 2 S ð1Þ 832 Programme 23 We have to evaluate this integral over the prescribed surface. Changing to cylindrical coordinates with ¼ 2 z dS x O ϕ dSn z x ¼ ............; y ¼ ............ z ¼ ............; dS ¼ . . . . . . . . . . . . y ρ 30 x ¼ 2 cos ; z ¼ z; y ¼ 2 sin dS ¼ 2 d dz ; x2 yz ¼ ð4 cos2 Þð2 sin ÞðzÞ ¼ 8 cos2 sin z xy 2 z ¼ ð2 cos Þð4 sin2 ÞðzÞ ¼ 8 cos sin2 z Then result (1) above becomes ð ð ð 1 =2 3 V dS ¼ ð8 cos2 sin zi þ 8 cos sin2 zj Þ 2 dz d 2 0 0 S ð =2 ð 3 ¼4 ðcos2 sin i þ cos sin2 j Þ 2z dz d 0 ¼4 ð =2 0 ðcos2 sin i þ cos sin2 j Þ 9 d 0 and this eventually gives ð V dS ¼ . . . . . . . . . . . . S 31 ð V dS ¼ 12ði þ jÞ S Because =2 ð cos3 sin3 iþ j V dS ¼ 36 ¼ 12ði þ jÞ 3 3 S 0 833 Vector analysis 2 Example 2 A scalar field V ¼ x þ y þ z exists over the surface S defined by 2x þ 2y þ z ¼ 2 bounded by x ¼ 0; y ¼ 0; z ¼ 0 in the first octant. ð Evaluate V dS over this surface. S z k n γ S: dS y R 2x þ 2y þ z ¼ 2 x¼0 z ¼ 2 2y y¼0 z ¼ 2 2x z¼0 y ¼1x dR x ^ dS dS ¼ n ^¼ where n r jrj @ @ @ iþ jþ k ¼ 2i þ 2j þ k and @x @y @z pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4 þ 4 þ 1 ¼ 9 ¼ 3 Now r ¼ Therefore ^¼ n r 2i þ 2j þ k 1 ^ dS ¼ ð2i þ 2j þ kÞ dS ¼ so that dS ¼ n jr j 3 3 If we now project dS onto the x–y plane, dR ¼ dS cos ^ k¼ cos ¼ n 1 1 ð2i þ 2j þ kÞ ðkÞ ¼ 3 3 1 ; dS ¼ 3dR ¼ 3 dxdy ; dR ¼ dS 3 ð ð ð ð 1 ^ dS ¼ ðx þ y þ zÞ ð2i þ 2j þ k Þ3 dx dy ; V dS ¼ V n 3 S S S But z ¼ 2 2x 2y ð ð 1 ð 1x ; V dS ¼ ð2 x yÞð2i þ 2j þ k Þ dy dx S x¼0 y¼0 ¼ ............ 834 Programme 23 32 2 ð2i þ 2j þ kÞ 3 Because 1x ð ð1 y2 V dS ¼ 2y xy ð2i þ 2j þ k Þ dx 2 0 S 0 1 3 x3 2 xx þ ð2i þ 2j þ kÞ ¼ 2 6 0 2 ¼ ð2i þ 2j þ kÞ 3 33 Vector fields Example 1 A vector field F ¼ yi þ 2j þ k exists over a surface S defined by x2 þ y 2 þ z2 ¼ 9 ð bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant. Evaluate F dS over the S surface indicated. ^ dS dS ¼ n ^¼ where n r where ¼ x2 þ y 2 þ z2 9 ¼ 0 jrj @ @ @ iþ jþ k ¼ 2x i þ 2y j þ 2z k and @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffi jr j ¼ 4x2 þ 4y 2 þ 4z2 ¼ 2 x2 þ y 2 þ z2 ¼ 2 9 ¼ 6 Now r ¼ z n 1 ð2xi þ 2yj þ 2zkÞ 6 1 ¼ ðxi þ yj þ zkÞ 3 ^¼ ; n θ O ϕ r y x ð ð 1 ^ dS ¼ ðyi þ 2j þ k Þ ðxi þ yj þ zk Þ dS Fn 3 S S ð 1 ¼ ðxy þ 2y þ zÞ dS 3 S F dS ¼ S ð Before integrating over the surface, we convert to spherical polar coordinates. x ¼ ............; y ¼ ............ z ¼ ............; dS ¼ . . . . . . . . . . . . 835 Vector analysis 2 x ¼ 3 sin cos ; z ¼ 3 cos ; 34 y ¼ 3 sin sin dS ¼ 9 sin d d Limits of and are ¼ 0 to ; ¼ 0 to : 2 2 ð ð ð 1 =2 =2 ; F dS ¼ ð9 sin2 sin cos þ 6 sin sin 3 0 0 S þ 3 cos Þ 9 sin d d ð =2 ð =2 ð3 sin3 sin cos þ 2 sin2 sin ¼9 0 0 þ sin cos Þ d d ¼ ............ Complete the integral ð 3 F dS ¼ 9 1 þ 4 S 35 Because ð ð =2 1 d F dS ¼ 9 2 sin cos þ sin þ 2 2 S 0 =2 3 ¼ 9 sin2 cos ¼9 1þ 2 2 0 4 Example 2 ð Evaluate F dS where F ¼ 2yj þ zk and S is the surface x2 þ y 2 ¼ 4 in the first S two octants bounded by the planes z ¼ 0, z ¼ 5 and y ¼ 0. : x2 þ y 2 4 ¼ 0 ^¼ n r jr j @ @ @ iþ jþ k ¼ 2xi þ 2yj @x @y @z qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; jr j ¼ 4x2 þ 4y 2 ¼ 2 x2 þ y 2 pffiffiffi ¼2 4¼4 r 2xi þ 2yj 1 ^¼ ; n ¼ ¼ ðxi þ yj Þ jr j 4 2 ð ð ^ dS ¼ . . . . . . . . . . . . F dS ¼ F n ; z dS r ¼ S S z O ϕ x ρ n y 836 Programme 23 ð 36 y 2 dS S Because ð ð 1 ^ dS ¼ ð2yj þ zk Þ ðxi þ yj Þ dS Fn 2 S S ð ð 1 ¼ ð2y 2 Þ dS ¼ y 2 dS 2 S S This is clearly a case for using cylindrical polar coordinates. x ¼ ............; y ¼ ............ z ¼ ............; 37 x ¼ 2 cos ; z ¼ z; ð ; F dS ¼ S Limits: ð y 2 dS ¼ ð ð S dS ¼ . . . . . . . . . . . . y ¼ 2 sin dS ¼ 2 d dz 4 sin2 2 d dz ¼ 8 S ¼ 0 to ¼ ; ð ð sin2 d dz S z ¼ 0 to z ¼ 5 ð ; F dS ¼ . . . . . . . . . . . . S 38 20 Because ð ð5 F dS ¼ 4 S ð ð1 cos 2Þ d dz z¼0 ¼0 ð5 sin 2 dz ¼4 2 0 0 ð5 5 ¼ 4 dz ¼ 4 z ¼ 20 0 0 Example 3 ð Evaluate F dS where F is the field x2 i yj þ 2zk and S is the surface S 2x þ y þ 2z ¼ 2 bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant. We can sketch the diagram by putting x ¼ 0; y ¼ 0; z ¼ 0 in turn in the equation for S. y When x¼0 y þ 2z ¼ 2 z¼1 2 y¼0 xþz¼1 z¼1x z¼0 2x þ y ¼ 2 y ¼ 2 2x So the diagram is . . . . . . . . . . . . 837 Vector analysis 2 z 39 k n γ x F ¼ x2 i yj þ 2zk; y R dR 2x þ y þ 2z 2 ¼ 0 : @ @ @ iþ jþ k ¼ 2i þ j þ 2k @x @y @z ð ð ^ dS F dS ¼ F n r ¼ S jr j ¼ 3 S ¼ . . . . . . . . . . . . ðnext stageÞ 1 3 ð 40 ð2x2 y þ 4zÞ dS S Because ð ð 1 ^ dS ¼ ðx2 i yj þ 2zk Þ ð2i þ j þ 2k Þ dS Fn 3 S S ð 1 ð2x2 y þ 4zÞ dS ¼ 3 S If we now project the element of surface dS onto the x–y plane dR ¼ dS cos ^ k cos ¼ n ^ k dS ; dR ¼ n ; dS ¼ 1 2 3 ð2i þ j þ 2k Þ ðk Þ ¼ ; dS ¼ dx dy 3 3 2 ð ð ^ dS Using these new relationships, F dS ¼ F n ^ k ¼ ; n S S ¼ ............ dx dy ^ k n 838 Programme 23 ð ð 41 R 1 ð2x2 y þ 4zÞ dx dy 2 Because ð ð 1 ^ dS ¼ Fn ð2x2 y þ 4zÞ dS 3 S S ð ð 1 3 ð2x2 y þ 4zÞ dx dy ¼ 3 R 2 ð ð 1 ð2x2 y þ 4zÞ dx dy ¼ 2 R Limits: y ¼ 0 to y ¼ 2 2x; x ¼ 0 to x ¼ 1 ð ð ð 1 1 22x ^ dS ¼ Fn ð2x2 y þ 4zÞ dy dx ; 2 0 0 S But 2x þ y þ 2z ¼ 2 1 ð2 2x yÞ 2 ð ^ dS ¼ . . . . . . . . . . . . ; Fn ; z¼ S Complete the integration 42 1 2 Here is the rest of the working. ð ð ð ð 1 1 22x ^ dS ¼ F dS ¼ F n ð2x2 y þ 4 4x 2yÞ dy dx 2 0 0 S S ð ð 1 1 22x ð2x2 4x þ 4 3yÞ dy dx ¼ 2 0 0 22x ð 1 1 3y 2 ð2x2 4x þ 4Þy dx ¼ 2 0 2 0 ð 1 1 ð4x2 8x þ 8 4x3 þ 8x2 8x 6 þ 12x 6x2 Þ dx ¼ 2 0 ð ð1 1 1 ð6x2 4x3 4x þ 2Þ dx ¼ ð3x2 2x3 2x þ 1Þ dx ¼ 2 0 0 1 4 x 1 ¼ x3 x2 þ x ¼ 2 2 0 While we are concerned with vector fields, let us move on to a further point of interest. 839 Vector analysis 2 Conservative vector fields In general, the value of the line integral ð F dr between two stated c points A and B depends on the particular path of integration followed. If, however, the line integral between A and B is independent of the path of integration between the two end points, then the vector field F is said to be conservative. þ It follows that, for a closed path in a conservative field, F dr ¼ 0: c c c c c2 ðBAÞ c c þ ; Because, if the field is conservative ð ð F dr ¼ F dr c ðABÞ c2 ðABÞ ð ð1 F dr ¼ F dr But c2 ðABÞ Hence, for the closed path ABc1 þ BAc2 þ ð ð F dr ¼ F dr þ F dr c1 ðABÞ c2 ðBAÞ ð ð F dr F dr ¼ c1 ðABÞ c2 ðABÞ ð ð F dr F dr ¼ 0 ¼ c1 ðABÞ c1 ðABÞ F dr ¼ 0 Note that this result holds good only for a closed curve and when the vector field is a conservative field. Now for an example. Example ð If F ¼ 2xyzi þ x2 zj þ x2 yk , evaluate the line integral F dr between A ð0, 0, 0Þ and B ð2, 4, 6Þ (a) along the curve c whose parametric equations are x ¼ u, y ¼ u2 , z ¼ 3u (b) along the three straight lines c1 : ð0, 0, 0Þ to ð2, 0, 0Þ; ð2, 4, 0Þ; c3 : ð2, 4, 0Þ to ð2, 4, 6Þ. Hence determine whether or not F is a conservative field. First draw the diagram ............ c2 : ð2, 0, 0Þ to 43 840 Programme 23 44 z B (2, 4, 6) c A c y c x c (a) F ¼ 2xyzi þ x2 zj þ x2 yk y ¼ u2 ; x ¼ u; ; dx ¼ du; z ¼ 3u dy ¼ 2u du; dz ¼ 3 du. F dr ¼ ð2xyzi þ x2 zj þ x2 yk Þ ði dx þ j dy þ k dzÞ ¼ 2xyz dx þ x2 z dy þ x2 y dz Using the transformations shown above, we can now express F dr in terms of u. F dr ¼ . . . . . . . . . . . . 45 15u4 du Because 2xyz dx ¼ ð2uÞðu2 Þð3uÞ du ¼ 6u4 du x2 z dy ¼ ðu2 Þð3uÞð2uÞ du ¼ 6u4 du x2 y dz ¼ ðu2 Þðu2 Þ3 du ¼ 3u4 du ; F dr ¼ 6u4 du þ 6u4 du þ 3u4 du ¼ 15u4 du The limits of integration in u are ............ 46 u ¼ 0 to u ¼ 2 ð F dr ¼ ; c ð2 0 4 15u du ¼ 2 3u5 0 ¼ 96 ð F dr ¼ 96 c 841 Vector analysis 2 (b) The diagram for (b) is as shown. We consider each straight line section in turn. z B (2, 4, 6) c3 c1 y c2 x ð ð F dr ¼ ð2xyz dx þ x2 z dy þ x2 y dzÞ c1 : ð0; 0; 0Þ to ð2; 0; 0Þ; y ¼ 0, z ¼ 0, dy ¼ 0, dz ¼ 0 ð F dr ¼ 0 þ 0 þ 0 ¼ 0 ; c1 In the same way, we evaluate the line integral along c2 and c3 : ð ð F dr ¼ . . . . . . . . . . . . ; F dr ¼ . . . . . . . . . . . . c2 c3 ð F dr ¼ 0; c2 Because we have ð ð 47 F dr ¼ 96 c3 ð F dr ¼ ð2xyz dx þ x2 z dy þ x2 y dzÞ c2 : ð2; 0; 0Þ to ð2; 4; 0Þ; x ¼ 2, z ¼ 0, ð F dr ¼ 0 þ 0 þ 0 ¼ 0 ; c ð 2 F dr ¼ 0 dx ¼ 0, dz ¼ 0 c3 : ð2; 4; 0Þ to ð2; 4; 6Þ; x ¼ 2, y ¼ 4, dx ¼ 0, dy ¼ 0 6 ð6 ð F dr ¼ 0 þ 0 þ 16 dz ¼ 16z ¼ 96 ; 0 c 0 ð 3 F dr ¼ 96 c2 c3 Collecting the three results together ð ð F dr ¼ 0 þ 0 þ 96 ; c1 þc2 þc3 F dr ¼ 96 c1 þc2 þc3 842 Programme 23 In this particular example, the value of the line integral is independent of the two paths we have used joining the same two end points and indicates that F may be a conservative field. It follows that ð ð þ F dr F dr ¼ 0 i.e. F dr ¼ 0 c1 þc2 þc3 c So, if F is a conservative field; þ F dr ¼ 0 Make a note of this for future use 48 Two tests can be applied to establish that a given vector field is conservative. If F is a conservative field (a) curl F = 0 (b) F can be expressed as grad V where V is a scalar field to be determined. For example, in the work we have just completed, we showed that F ¼ 2xyzi þ x2 zj þ x2 yk is a conservative field. (a) If we determine curl F in this case, we have curl F ¼ . . . . . . . . . . . . 49 curl F ¼ 0 Because i @ curl F ¼ @x 2xyz j @ @y k @ @z x2 z x2 y ¼ ðx2 x2 Þi ð2xy 2xyÞj þ ð2xz 2xzÞk ¼ 0 ; curl F ¼ 0 (b) We can attempt to express F as grad V where V is a scalar in x, y, z. If V ¼ f ðx; y; zÞ grad V ¼ @V @V @V iþ jþ k @x @y @z and we have F ¼ 2xyzi þ x2 zj þ x2 yk ; @V ¼ 2xyz @x @V ¼ x2 z @y @V ¼ x2 y @z ; V ¼ x2 yz þ f ðy, zÞ ; V ¼ ............ ; V ¼ ............ We therefore have to find a scalar function V that satisfies the three requirements. V ¼ ............ 843 Vector analysis 2 50 V ¼ x2 yz Because @V ¼ 2xyz @x @V ¼ x2 z @y ; V ¼ x2 yz þ f ðy, zÞ @V ¼ x2 y @z ; V ¼ x2 yz þ hðx, yÞ ; V ¼ x2 yz þ gðx, zÞ These three are satisfied if f ðy, zÞ ¼ gðz, xÞ ¼ hðx, yÞ ¼ 0 ; F ¼ grad V where V ¼ x2 yz So two tests can be applied to determine whether or not a vector field is conservative. They are (a) . . . . . . . . . . . . (b) . . . . . . . . . . . . 51 (a) curl F ¼ 0 (b) F ¼ grad V Any one of these conditions can be applied as is convenient. Now what about these? Exercise Determine which of the following vector fields are conservative. (a) F ¼ ðx þ yÞi þ ðy zÞj þ ðx þ y þ zÞk (b) F ¼ ð2xz þ yÞi þ ðz þ xÞj þ ðx2 þ yÞk (c) F ¼ y sin z i þ x sin z j þ ðxy cos z þ 2zÞk (d) F ¼ 2xyi þ ðx2 þ 4yzÞj þ 2y 2 zk (e) F ¼ y cos x cos z i þ sin x cos z j y sin x sin z k. Complete all five and check your findings with the next frame. (a) No (b) Yes (c) Yes (d) No (e) Yes 52 844 Programme 23 Divergence theorem (Gauss’ theorem) z For a closed surface S, enclosing a region V in a vector field F, ð ð div F dV ¼ F dS S V V O S y x In general, this means that the volume integral (triple integral) on the lefthand side can be expressed as a surface integral (double integral) on the righthand side. Let us work through one or two examples. Example 1 Verify the divergence theorem for the vector field F ¼ x2 i þ zj þ yk taken over the region bounded by the planes z ¼ 0, z ¼ 2, x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 3. Start off, as always, by sketching the relevant diagram, which is ............ 53 z dV ¼ dx dy dz dV O We have to show that ð ð div F dV ¼ F dS y V S x ð (a) To find div F dV @ @ @ iþ j k ðx2 i þ zj þ ykÞ @x @y @z @ 2 @ @ ¼ ðx Þ þ ðzÞ þ ðyÞ ¼ 2x þ 0 þ 0 ¼ 2x @x @y @z ð ððð div F dV ¼ 2x dV ¼ 2x dz dy dx V div F ¼ r F ¼ ð ; V V V Inserting the limits and completing the integration ð div F dV ¼ . . . . . . . . . . . . V 845 Vector analysis 2 ð 54 div F dV ¼ 6 V Because 2 ð1 ð3 ð2 ð1 ð3 ð 2xz dy dx div F dV ¼ 2x dz dy dx ¼ V 0 ¼ ð1 0 0 3 dx ¼ 4xy 0 0 12x dx ¼ 6x2 0 1 ¼6 0 F dS Now we have to find S ð 0 0 0 ð F dS i.e. (b) To find ð1 S ð ^ dS Fn S z O y x The enclosing surface S consists of six separate plane faces denoted as S1 , S2 , . . . , S6 as shown. We consider each face in turn. F ¼ x2 i þ zj þ yk (1) S1 (base): z ¼ 0; ð ; ^ ¼ k (outwards and downwards) n ; F ¼ x2 i þ yk dS1 ¼ dx dy ðð ^ dS ¼ Fn ðx2 i þ yk Þ ðk Þ dy dx S1 S1 ¼ ð1 ð3 0 ð1 ðyÞ dy dx 0 y2 ¼ 2 0 9 ¼ 2 (2) S2 (top): z ¼ 2; 3 dx 0 ^ ¼k n dS2 ¼ dx dy ð ^ dS ¼ . . . . . . . . . . . . ; Fn S2 846 Programme 23 55 9 2 Because ð ðð ^ dS ¼ Fn ðx2 i þ 2j þ yk Þ ðk Þ dy dx S2 S2 ¼ ð1 ð3 0 y dy dx ¼ 0 9 2 So we go on. y ¼ 3; (3) S3 (right-hand end): ^ ¼j n dS3 ¼ dx dz F ¼ x2 i þ zj þ yk ðð ð ^ dS ¼ Fn ðx2 i þ zj þ 3k Þ ðj Þ dz dx ; S3 S3 ¼ ð1 ð2 0 z dz dx 0 ð1 ð 1 2 2 z dx ¼ 2 dx ¼ 2 ¼ 0 2 0 0 ^ ¼ j (4) S4 (left-hand end): y ¼ 0; n dS4 ¼ dx dz ð ^ dS ¼ . . . . . . . . . . . . ; Fn S4 56 2 Because ð ðð ð1 ð2 ^ dS ¼ Fn ðx2 i þ zj þ yk Þ ðj Þ dz dx ¼ ðzÞ dz dx S4 S4 ¼ ð1 0 2 2 z 2 dx ¼ ð1 0 0 ð2Þ dx ¼ 2 0 0 Now for the remaining two sides S5 and S6 . Evaluate these in the same manner, obtaining ð ^ dS ¼ . . . . . . . . . . . . Fn ð S5 ^ dS ¼ . . . . . . . . . . . . Fn S6 847 Vector analysis 2 ð ð ^ dS ¼ 6; Fn S5 57 ^ dS ¼ 0 Fn S6 Check: ^ ¼i n dS5 ¼ dy dz (5) S5 (front): x ¼ 1; ðð ðð ð ^ F n dS ¼ ði þ zj þ yk Þ ðiÞ dy dz ¼ 1 dy dz ¼ 6 ; S5 S5 S5 ^ ¼ i n dS6 ¼ dy dz (6) S6 (back): x ¼ 0; ðð ðð ð ^ dS ¼ Fn ðzj þ yk Þ ði Þ dy dz ¼ 0 dy dz ¼ 0 ; S6 S6 S6 Now on to the next frame where we will collect our results together For the whole surface S we therefore have ð 9 9 F dS ¼ þ þ 2 2 þ 6 þ 0 ¼ 6 2 2 S and from our previous work in section (a) 58 ð div F dV ¼ 6 V We have therefore verified as required that, in this example ð ð div F dV ¼ F dS V S We have made rather a meal of this since we have set out the working in detail. In practice, the actual writing can often be considerably simplified. Let us move on to another example. Example 2 Verify the Gauss divergence theorem for the vector field F ¼ xi þ 2j þ z2 k taken over the region bounded by the planes z ¼ 0, z ¼ 4, x ¼ 0, y ¼ 0 and the surface x2 þ y 2 ¼ 4 in the first octant. z Divergence theorem ð ð div F dV ¼ F dS V O x y x 2 + y2 = 4 (a) div F ¼ r F ¼ @ @ @ iþ jþ k @x @y @z ¼ ............ S S consists of five surfaces S1 , S2 , . . . , S5 as shown. ðxi þ 2j þ z2 k Þ 848 Programme 23 59 1 þ 2z ð div F dV ¼ ; ð V r F dV ¼ ððð V ð1 þ 2zÞ dx dy dz V Changing to cylindrical polar coordinates ð, , zÞ x ¼ cos y ¼ sin z¼z dV ¼ d d dz Transforming the variables and inserting the appropriate limits, we then have ð div F dV ¼ . . . . . . . . . . . . V Finish it 60 20 Because ð ð =2 ð 2 ð 4 div F dV ¼ ð1 þ 2zÞ dz d d 0 V ¼ 0 ð =2 ð 2 0 ¼ ð =2 0 z þ z2 0 10 0 2 4 0 2 d ¼ 0 ð d d ¼ ð =2 ð =2 ð 2 20 d d 0 0 40 d ¼ 20 ð1Þ 0 F dS over the closed surface. (b) Now we evaluate S z k (S2) –i (S4) –j (S3) n (S5) O x y The unit normal vector for each surface is shown. F ¼ xi þ 2j þ z2 k –k (S1) ^ ¼ k z ¼ 0; n F ¼ xi þ 2j ð ð ^ dS ¼ ðxi þ 2j Þ ðk Þ dS ¼ 0 ; Fn (1) S1 : S1 S1 849 Vector analysis 2 ^ ¼k z ¼ 4; n F ¼ xi þ 2j þ 16k ð ð ð ^ dS ¼ ðxi þ 2j þ 16k Þ ðk Þ dS ¼ Fn 16 dS ; S2 S2 S2 4 ¼ 16 ¼ 16 4 ð ^ dS ¼ . . . . . . . . . . . . In the same way for S3 : Fn S ð3 ^ dS ¼ . . . . . . . . . . . . Fn and for S4 : (2) S2 : S4 ð ^ dS ¼ 16; Fn S3 ð ^ dS ¼ 0 Fn 61 S4 Because we have ^ ¼ j y ¼ 0; n F ¼ xi þ 2j þ z2 k ð ð ^ dS ¼ ðxi þ 2j þ z2 kÞ ðj Þ dS ; Fn S3 S ð3 ¼ ð2Þ dS ¼ 2ð8Þ ¼ 16 (3) S3 : S3 ^ ¼ i x ¼ 0; n F ¼ 2j þ z2 k ð ð ^ dS ¼ ð2j þ z2 k Þ ði Þ dS ¼ 0 ; Fn (4) S4 : S4 S4 Finally we have (5) S5 : x2 þ y 2 4 ¼ 0 ^ ¼ ............ n ^¼ n 1 ðxi þ yjÞ 2 Because rS 2xi þ 2yj xi þ yj ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 2 jrS j 2 4x þ 4y ð ð ð xi þ yj 1 ^ dS ¼ ðxi þ 2j þ z2 k Þ dS ¼ ; Fn ðx2 þ 2yÞ dS 2 2 S5 S5 S5 x2 þ y 2 4 ¼ 0 ^¼ n Converting to cylindrical polar coordinates, this gives ð ^ dS ¼ . . . . . . . . . . . . Fn S5 62 850 Programme 23 63 4 þ 16 Because we have ð ð 1 ^ dS ¼ Fn ðx2 þ 2yÞ dS 2 S5 S5 also x ¼ 2 cos ; y ¼ 2 sin z ¼ z; ð dS ¼ 2 d dz ^ dS ¼ Fn ; S5 1 2 ¼2 ¼2 ð 4 ð =2 0 ð4 0 ð =2 0 0 ð 4 ¼2 0 fð1 þ cos 2Þ þ 2 sin g d dz 0 ð4 ð4 cos2 þ 4 sin Þ 2 d dz =2 sin 2 2 cos dz 2 0 þ 2 dz ¼ 4 þ 16 2 Therefore, for the total surface S ð ^ dS ¼ 0 þ 16 16 þ 0 þ 4 þ 16 ¼ 20 Fn ð ð S div F dV ¼ F dS ¼ 20 ; V ð2Þ S Other examples are worked in much the same way. You will remember that, for a closed surface, the normal vectors at all points are drawn in an outward direction. Now we move on to a further important theorem. Stokes’ theorem n 64 dS c If F is a vector field existing over an open surface S and around its boundary, closed curve c, then ð þ curl F dS ¼ F dr S c This means that we can express a surface integral in terms of a line integral round the boundary curve. The proof of this theorem is rather lengthy and is to be found in the Appendix. Let us demonstrate its application in the following examples. 851 Vector analysis 2 Example 1 A hemisphere S is defined by x2 þ y 2 þ z2 ¼ 4 ðz 0Þ: A vector field F ¼ 2yi xj þ xzk exists over the surface and around its boundary c. þ ð curl F dS ¼ F dr: Verify Stokes’ theorem, that S k z γ c n S: x2 þ y 2 þ z2 4 ¼ 0 F ¼ 2yi xj þ xzk dS c is the circle x2 þ y 2 ¼ 4. O y x c þ (a) F dr ¼ c ¼ ð ð2yi xj þ xzk Þ ði dx þ j dy þ k dzÞ ðc ð2y dx x dy þ xz dzÞ c Converting to polar coordinates x ¼ 2 cos ; y ¼ 2 sin ; dx ¼ 2 sin d; z¼0 dy ¼ 2 cos d; Limits ¼ 0 to 2 Making the substitutions and completing the integral þ F dr ¼ . . . . . . . . . . . . c þ 65 F dr ¼ 12 c Because þ ð 2 F dr ¼ ð4 sin ½2 sin d 2 cos 2 cos dÞ c 0 ¼ 4 ¼ 4 ð 2 0 ð 2 ð2 sin2 þ cos2 Þ d 2 ð1 þ sin Þ d ¼ 2 0 ¼ 2 3 sin 2 2 ð 2 ð3 cos 2Þ d 0 2 ¼ 12 ð1Þ 0 On to the next frame 852 66 Programme 23 ð curl F dS (b) Now we determine ð S curl F dS ¼ ð ^ dS curl F n F ¼ 2yi xj þ xzk ; curl F ¼ . . . . . . . . . . . . 67 curl F ¼ zj 3k Because i @ curl F ¼ @x 2y j @ @y x k @ ¼ i ð0 0Þ j ðz 0Þ þ k ð1 2Þ ¼ zj 3k @z xz rS 2xi þ 2yj þ 2zk xi þ yj þ zk ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ jrS j 2 4x2 þ 4y 2 þ 4z2 ð ð xi þ yj þ zk ^ dS ¼ ðzj 3k Þ dS Then curl F n 2 S S ð 1 ¼ ðyz 3zÞ dS 2 S ^¼ Now n Expressing this in spherical polar coordinates and integrating, we get ð ^ dS ¼ . . . . . . . . . . . . curl F n S 68 12 Because x ¼ 2 sin cos ; y ¼ 2 sin sin ; z ¼ 2 cos ; dS ¼ 4 sin d d ð ð ð 1 ^ dS ¼ ð2 sin sin 2 cos 6 cos Þ4 sin d d curl F n ; 2 S S ð 2 ð =2 ð2 sin2 cos sin þ 3 sin cos Þ d d ¼ 4 0 0 =2 2 sin3 sin 3 sin2 þ d 3 2 0 0 ð 2 2 3 sin þ d ¼ 12 ¼ 4 3 2 0 ¼ 4 ð 2 ð2Þ So we have from our two results (1) and (2) ð þ curl F dS ¼ F dr S c Before we proceed with another example, let us clarify a point relating to the direction of unit normal vectors now that we are dealing with surfaces. So on to the next frame 853 Vector analysis 2 Direction of unit normal vectors to a surface S When we were dealing with the divergence theorem, the normal vectors were drawn in a direction outward from the enclosed region. With an open surface as we now have, there is in fact no inward or outward direction. With any general surface, a normal vector can be drawn in either of two opposite directions. To avoid confusion, a convention must therefore be agreed upon and the established rule is as follows. n c c n n c ^ is drawn perpendicular to the surface S at any point in the A unit normal n direction indicated by applying a right-handed screw sense to the direction of integration round the boundary c. Having noted that point, we can now deal with the next example. Example 2 A surface consists of five sections formed by the planes x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 3, z ¼ 2 in the first octant. If the vector field F ¼ yi þ z2 j þ xyk exists over the surface and around its boundary, verify Stokes’ theorem. n=k z n = –i n=j n = –j n=i c c c y c x If we progress round the boundary along c1 , c2 , c3 , c4 in an anti-clockwise manner, the normals to the surfaces will be as shown. ð þ We have to verify that curl F dS ¼ F dr S c þ F dr (a) We will start off by finding ð c F dr ¼ . . . . . . . . . . . . 69 854 Programme 23 ð 70 (1) Along c1 : ð F dr ¼ ðy dx þ z2 dy þ xy dzÞ y ¼ 0; z ¼ 0; dy ¼ 0; dz ¼ 0 ð ð F dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ; c1 (2) Along c2 : x ¼ 1; z ¼ 0; dx ¼ 0; dz ¼ 0 ð ð F dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ; c2 In the same way ð F dr ¼ . . . . . . . . . . . . and c3 F dr ¼ . . . . . . . . . . . . c4 ð 71 ð F dr ¼ 3; ð c3 F dr ¼ 0 c4 Because (3) Along c3 : y ¼ 3; z ¼ 0; dy ¼ 0; dz ¼ 0 0 ð0 ð F dr ¼ ð3 dx þ 0 þ 0Þ ¼ 3x ¼ 3 ; 1 c3 (4) Along c4 : x ¼ 0; z ¼ 0; dx ¼ 0; dz ¼ 0 ð ð F dr ¼ ð0 þ 0 þ 0Þ ¼ 0 ; c þ4 F dr ¼ 0 þ 0 3 þ 0 ¼ 3 ; þc F dr ¼ 3 c ð curl F dS: (b) Now we have to find S First we need an expression for curl F. F ¼ yi þ z2 j þ xyk ; curl F ¼ . . . . . . . . . . . . 1 ð1Þ 855 Vector analysis 2 curl F ¼ ðx 2zÞi yj k 72 Because i j @ @ curl F ¼ r F ¼ @x @y k @ @z y z2 xy ¼ i ðx 2zÞ j ðy 0Þ þ k ð0 1Þ ¼ ðx 2zÞi yj k ð ð ^ dS Then, for each section, we obtain curl F dS ¼ curl F n (1) S1 (top): ^ ¼k n ð ^ dS ¼ . . . . . . . . . . . . curl F n ; S1 3 Because ð ð ^ dS ¼ fðx 2zÞi yj k g ðk Þ dS curl F n S1 S ð1 ¼ ð1Þ dS ¼ ðarea of S1 Þ ¼ 3 S1 Then, likewise ^ ¼j (2) S2 (right-hand end): n ð ð ^ dS ¼ fðx 2zÞi yj k g ðj Þ dS curl F n ; S2 S ð2 ¼ ðyÞ dS S2 But y ¼ 3 for this section ð ð ^ dS ¼ ð3Þ dS ¼ ð3Þð2Þ ¼ 6 curl F n ; S2 S2 ^ ¼ j (3) S3 (left-hand end): n ð ^ dS ¼ . . . . . . . . . . . . ; curl F n S3 73 856 Programme 23 74 0 Because ð ð ^ dS ¼ fðx 2zÞi yj k g ðj Þ dS curl F n S3 S ð3 y dS ¼ S3 But y ¼ 0 over S3 ð ^ dS ¼ 0 ; curl F n S3 Working in the same way ð ^ dS ¼ . . . . . . . . . . . . ; curl F n ð S4 ð 75 ^ dS ¼ . . . . . . . . . . . . curl F n S5 ð ^ dS ¼ 6; curl F n S4 ^ dS ¼ 12 curl F n S5 Because ^ ¼i (4) S4 (front): n ð ð ^ dS ¼ fðx 2zÞi yj k g ðiÞ dS ; curl F n S4 S ð4 ¼ ðx 2zÞ dS S4 But x ¼ 1 over S4 2 ð ð3 ð2 ð3 2 ^ dS ¼ zz ; curl F n ð1 2zÞ dz dy ¼ dy S4 0 ¼ ð3 0 ð2Þ dy ¼ 3 2y 0 0 0 ¼ 6 0 ^ ¼ i with x ¼ 0 over S5 n ð Similar working to that above gives (5) S5 (back): ^ dS ¼ 12 curl F n S5 Finally, collecting the five results together gives ð ^ dS ¼ . . . . . . . . . . . . curl F n S 857 Vector analysis 2 ð ^ dS ¼ 3 6 þ 0 6 þ 12 ¼ 3 curl F n (2) 76 S So, referring back to our result for section (a) we see that ð þ curl F dS ¼ F dr S c Of course we can, on occasions, make use of Stokes’ theorem to lighten the working – as in the next example. Example 3 A surface S consists of that part of the cylinder x2 þ y 2 ¼ 9 between z ¼ 0 and z ¼ 4 for y 0 and the two semicircles of radius 3 in the planes z ¼ 0 and ð z ¼ 4. If F ¼ zi þ xyj þ xzk , evaluate curl F dS over the surface. S The surface S consists of three sections z n=k (a) the curved surface of the cylinder (b) the top and bottom semicircles. n We could therefore evaluate ð curl F dS S O y over each of these separately. However, we know by Stokes’ theorem that ð curl F dS ¼ . . . . . . . . . . . . x n = –k S þ F dr where c is the boundary of S c z F ¼ zi þ xyj þ xzk þ þ ; F dr ¼ ðzi þ xyj þ xzk Þ c c ði dx þ j dy þ k dzÞ þ ¼ ðz dx þ xy dy þ xz dzÞ c c c O c y c x Now we can work through this easily enough, taking c1 , c2 , c3 , c4 in turn, and summing the results, which gives ð þ curl F dS ¼ F dr ¼ . . . . . . . . . . . . S c 77 858 Programme 23 78 24 þ F dr ¼ Here is the working in detail. c þ ðz dx þ xy dy þ xz dzÞ c (1) c1 : y ¼ 0; z ¼ 0; dy ¼ 0; dz ¼ 0 ð ð F dr ¼ ð0 þ 0 þ 0Þ ¼ 0 c1 c1 (2) c2 : x ¼ 3; y ¼ 0; dx ¼ 0; dy ¼ 0 4 ð ð 3z2 F dr ¼ ð0 þ 0 3z dzÞ ¼ ¼ 24 2 0 c2 c2 (3) c3 : y ¼ 0; z ¼ 4; dy ¼ 0; dz ¼ 0 ð ð ð3 F dr ¼ ð4 dx þ 0 þ 0Þ ¼ 4 dx ¼ 24 c3 3 c3 (4) c4 : x ¼ 3; y ¼ 0; dx ¼ 0; dy ¼ 0 2 0 ð ð 3z F dr ¼ ð0 þ 0 þ 3z dzÞ ¼ ¼ 24 2 4 c4 c4 Totalling up these four results, we have þ F dr ¼ 0 24 þ 24 24 ¼ 24 c ð curl F dS ¼ But S þ F dr c ð curl F dS ¼ 24 ; S ð curl F dS over the three This working is a good deal easier than calculating S separate surfaces direct. So, if you have not already done so, make a note of Stokes’ theorem: ð þ curl F dS ¼ F dr S c Then on to the next section of the work 859 Vector analysis 2 Green’s theorem Green’s theorem enables an integral over a plane area to be expressed in terms of a line integral round its boundary curve. We showed in Programme 19 that, if P and Q are two single-valued functions of x and y, continuous over a plane surface S, and c is its boundary curve, then ðð þ @Q @P ðP dx þ Q dyÞ ¼ dx dy @y c S @x 79 where the line integral is taken round c in an anticlockwise manner. In vector terms, this becomes: y S is a two-dimensional space enclosed by a simple closed curve c. dS = dxdy dS ¼ dx dy c O ^ dS ¼ k dx dy dS ¼ n x If F ¼ Pi þ Qj where P ¼ Pðx; yÞ and Q ¼ Qðx; yÞ then curl F ¼ . . . . . . . . . . . . @Q @P k @x @y 80 Because i j k @ @ @ curl F ¼ @x @y @z P Q 0 @Q @P @Q @P j 0 þk ¼i 0 @z @z @x @y @Q @P @Q @P But in the xy plane, ¼ ¼ 0. ; curl F ¼ k @z @z @x @y ð ð ^ dS and in the xy plane, n ^ ¼k So curl F dS ¼ curl F n ðð @Q @P @Q @P ðk Þ dS ¼ dx dy curl F dS ¼ k ; @x @y @y S S S @x ð ðð @Q @P ; curl F dS ¼ dx dy @y S S @x ð ð Now by Stokes’ theorem . . . . . . . . . . . . ð1Þ 860 Programme 23 ð 81 curl F dS ¼ S and, in this case; þ c ¼ F dr ¼ ; c F dr c F dr ¼ þ þ þ ðPi þ Qj Þ ði dx þ j dy þ k dzÞ þc þc ðP dx þ Q dyÞ ðP dx þ Q dyÞ ð2Þ c Therefore from (1) and (2) ð þ Stokes’ theorem curl F dS ¼ F dr in two dimensions becomes S c þ ðð @Q @P dx dy ¼ ðP dx þ Q dyÞ Green’s theorem @y S @x c Example þ Verify Green’s theorem for the integral fðx2 þ y 2 Þ dx þ ðx þ 2yÞ dyg taken c round the boundary curve c defined by y 2 y¼0 0x2 2 0x2 0 y 2. x þy ¼4 x¼0 c2 c O Green’s theorem: In this case c x þ ðð @Q @P dx dy ¼ ðP dx þ Q dyÞ @y S @x c ðx2 þ y 2 Þ dx þ ðx þ 2yÞ dy ¼ P dx þ Q dy ; P ¼ x2 þ y 2 and Q ¼ x þ 2y We now take c1 , c2 , c3 in turn. (1) c1 : y ¼ 0; dy ¼ 0 3 2 ð ð2 x 8 ; ðP dx þ Q dyÞ ¼ x2 dx ¼ ¼ 3 0 3 c1 0 (2) c2 : x2 þ y 2 ¼ 4 ; y 2 ¼ 4 x2 ; y ¼ ð4 x2 Þ1=2 x þ 2y ¼ x þ 2ð4 x2 Þ1=2 1 x dy ¼ ð4 x2 Þ1=2 ð2xÞ dx ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 2 4 x2 ð ðP dx þ Q dyÞ ¼ . . . . . . . . . . . . ; c2 Make any necessary substitutions and evaluate the line integral for c2 . 861 Vector analysis 2 82 4 Because we have ð ð pffiffiffiffiffiffiffiffiffiffiffiffiffiffi x ðP dx þ Q dyÞ ¼ 4 þ ðx þ 2 4 x2 Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx 4 x2 c2 c2 ð x2 4 2x pffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ 4 x2 c2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi Putting x ¼ 2 sin , 4 x2 ¼ 2 cos dx ¼ 2 cos d Limits: x ¼ 2; ¼ ; x ¼ 0; ¼ 0. 2 ð ð0 4 sin2 ; 2 cos d ðP dx þ Q dyÞ ¼ 4 4 sin 2 cos c2 =2 1 sin 2 0 ¼ 4 2 sin sin2 2 2 =2 h i ¼4 21 ¼4 4 Finally x ¼ 0; dx ¼ 0 0 ð ð0 ; ðP dx þ Q dyÞ ¼ 2y dy ¼ y 2 ¼ 4 (3) c3 : c3 2 2 ; Collecting our three partial results þ 8 16 ðP dx þ Q dyÞ ¼ þ 4 4 ¼ 3 3 c That is one part done. Now we have to evaluate P ¼ x2 þ y 2 ; ð1Þ ðð @Q @P dx dy @y S @x @P ¼ 2y @y @Q Q ¼ x þ 2y ; ¼1 @x ðð ðð @Q @P dx dy ¼ ; ð1 2yÞ dy dx @y S @x S It will be more convenient to work in polar coordinates, so we make the substitutions x ¼ r cos ; y ¼ r sin ; dS ¼ dx dy ¼ r dr d ðð ð =2 ð 2 @Q @P ð1 2r sin Þr dr d ; dx dy ¼ @y S @x 0 0 ¼ ............ Complete it 862 Programme 23 83 16 3 Here it is: ðð ð =2 ð 2 @Q @P ðr 2r 2 sin Þ dr d dx dy ¼ @y S @x 0 0 2 ð =2 2 r 2r 3 sin d ¼ 2 3 0 0 ð =2 16 sin d ¼ 2 3 0 =2 16 16 ¼ 2 þ cos ¼ 3 3 0 ð2Þ So we have established once again that þ ðð @Q @P dx dy ðP dx þ Q dyÞ ¼ @y c S @x And that brings us to the end of this particular Programme. We have covered a number of important sections, so check carefully down the Revision summary and the Can you? checklist, and then work through the Test exercise that follows. The Further problems provide valuable additional practice. Revision summary 23 84 1 Line integrals ð V dr (a) Scalar field V: c The curve c is expressed in parametric form. dr ¼ i dx þ j dy þ k dz ð (b) Vector field F: F dr c F ¼ Fx i þ Fy j þ Fz k dr ¼ i dx þ j dy þ k dz F dr ¼ Fx dx þ Fy dy þ Fz dz 863 Vector analysis 2 2 Volume integrals F is a vector field; V a closed region with boundary surface S. ð ð x2 ð y2 ð z2 F dV ¼ F dz dy dx V 3 x1 y1 z1 Surface integrals (surface defined by ðx, y, zÞ ¼ constant) (a) Scalar field Vðx, y, zÞ: ð ð r grad ^ dS; ^¼ ¼ V dS ¼ V n n jr j jgrad j S S (b) Vector field F ¼ Fx i þ Fy j þ Fz k: ð ð r ^ dS; ^¼ F dS ¼ F n n jr j S S 4 Polar coordinates (a) Plane polar coordinates ðr, Þ y dθ dS x ¼ r cos ; r O y y ¼ r sin dS ¼ r dr d θ x x (b) Cylindrical polar coordinates ð, , zÞ z P (ρ, ϕ, z) x ¼ cos y ¼ sin z O ϕ y ρ z¼z dS ¼ d dz dV ¼ d d dz x (c) Spherical polar coordinates ðr, , Þ z O ϕ P (r, θ, ϕ) θ x ¼ r sin cos y ¼ r sin sin z ¼ r cos r y x 5 Conservative vector fields A vector field F is conservative if þ (a) F dr ¼ 0 for all closed curves c (b) curl F ¼ 0 (c) F ¼ grad V where V is a scalar. dS ¼ r 2 sin d d dV ¼ r 2 sin dr d d 864 Programme 23 6 Divergence theorem (Gauss’ theorem) z Closed surface S enclosing a region V in a vector field F. ð ð div F dV ¼ F dS S V V S y O x 7 Stokes’ theorem z n c dS c An open surface S bounded by a simple closed curve c, then ð þ curl F dS ¼ F dr S c y O x 8 Green’s theorem y The curve c is a simple closed curve enclosing a plane space S in the x–y plane. P and Q are functions of both x and y. c O x þ ðð @Q @P dx dy ¼ ðP dx þ Q dyÞ. Then @y S @x c Can you? 85 Checklist 23 Check this list before and after you try the end of Programme test. On a scale of 1 to 5 how confident are you that you can: . Evaluate the line integral of a scalar and a vector field in Cartesian coordinates? Yes No . Evaluate the volume integral of a vector field? Yes Frames 1 to 20 21 to 27 28 to 42 No . Evaluate the surface integral of a scalar and a vector field? Yes No 865 Vector analysis 2 . Determine whether or not a vector field is a conservative vector field? Yes No 43 to 52 52 to 63 64 to 68 . Determine the direction of unit normal vectors to a surface? Yes No 69 to 78 . Apply Green’s theorem in the plane? Yes 79 to 83 . Apply Gauss’ divergence theorem? Yes No . Apply Stokes’ theorem? Yes No No Test exercise 23 1 If V ¼ x3 y þ 2xy 2 þ yz, evaluate ð V dr between A ð0, 0, 0Þ and B ð2, 1, 3Þ c 2 along the curve with parametric equations x ¼ 2t, y ¼ t 2 , z ¼ 3t 3 . ð 2 3 2 2 If F ¼ x y i þ yz j þ zx k , evaluate F dr along the curve x ¼ 3u2 , y ¼ u, 3 z ¼ 2u3 between A ð3, 1, 2Þ and B ð3; 1; 2Þ. ð F dV where F ¼ 3i þ zj þ 2yk and V is the region bounded by the Evaluate 4 planes z ¼ 0, z ¼ 3 and the surface x2 þ y 2 ¼ 4. ð 2 If V is the scalar field V ¼ xyz , evaluate V dS over the surface S defined by 5 x2 þ y 2 ¼ 9 between z ¼ 0 and z ¼ 2 in the first octant. ð Evaluate F dS over the surface S defined by x2 þ y 2 þ z2 ¼ 4 for z 0 and c V S S bounded by x ¼ 0; y ¼ 0; z ¼ 0 in the first octant where F ¼ xi þ 2zj þ yk. 6 Determine which of the following vector fields are conservative. (a) F ¼ ð2xy þ zÞi þ ðx2 þ 2yzÞj þ ðx þ y 2 Þk (b) F ¼ ðyz þ 2yÞi þ ðxz þ 2xÞj þ ðxy þ 3Þk (c) F ¼ ðyz2 þ 3Þi þ ðxz2 þ 2Þj þ ð2xyz þ 4Þk. ð 7 F dS where By the use of the divergence theorem, determine S F ¼ xi þ xyj þ 2k, taken over the region bounded by the planes z ¼ 0, z ¼ 4, x ¼ 0, y ¼ 0 and the surface x2 þ y 2 ¼ 9 in the first octant. 86 866 Programme 23 8 A surface consists of parts of the planes x ¼ 0, x ¼ 2, y ¼ 0, y ¼ 2 and z ¼ 3 y ð curl F dS over the in the region z 0. Apply Stokes’ theorem to evaluate S surface where F ¼ 2xi þ xzj þ yzk where S lies in the z ¼ 0 plane. 9 Verify Green’s theorem in the plane for the integral þ 2 xy 2x dx þ x þ 2xy 2 dy c where c is the square with vertices at ð1, 1Þ, ð1, 1Þ, ð1, 1Þ and ð1, 1Þ. Further problems 23 87 1 If V ¼ x2 yz, evaluate ð V dr between A ð0, 0, 0Þ and B ð6, 2, 4Þ c (a) along the straight lines c1 : ð0, 0, 0Þ to ð6, 0, 0Þ c2 : ð6, 0, 0Þ to ð6, 2, 0Þ c3 : ð6, 2, 0Þ to ð6, 2, 4Þ 2 (b) along the path c4 having parametric equations x ¼ 3t, y ¼ t, z ¼ 2t. ð 2 If V ¼ xy þ yz, evaluate to one decimal place V dr along the curve c having 3 parametric equations x ¼ 2t 2 , y ¼ 4t, z ¼ 3t þ 5 between A ð0, 0, 5Þ and B ð8, 8, 11Þ. ð Evaluate to one decimal place the integral ðxyz þ 4x2 yÞ dr along the curve c 4 with parametric equations x ¼ 2u, y ¼ u2 , z ¼ 3u3 between A ð2, 1, 3Þ and B ð4, 4, 24Þ. ð If F ¼ xyi þ yzj þ 3xyzk , evaluate F dr between A ð0; 2; 0Þ and B ð3; 6; 1Þ 5 where c has the parametric equations x ¼ 3u, y ¼ 4u þ 2, z ¼ u2 . ð F ¼ x2 i 2xyj þ yzk . Evaluate F dr between A ð2, 1, 2Þ and B ð4, 4, 5Þ c c c c where c is the path with parametric equations x ¼ 2u, y ¼ u2 , z ¼ 3u 1: 6 7 A unit particle is moved in an anticlockwise manner round a circle with centre ð0, 0, 4Þ and radius 2 in the plane z ¼ 4 in a force field defined as F ¼ ðxy þ zÞi þ ð2x þ yÞj þ ðx þ y þ zÞk . Find the work done. ð Evaluate F dV where F ¼ i yj þ k and V is the region bounded by the plane V z ¼ 0 and the hemisphere x2 þ y 2 þ z2 ¼ 4; for z 0. 867 Vector analysis 2 8 V is the region bounded by the planes x ¼ 0, y ¼ 0, z ¼ 0 and the surfaces y ¼ 4 x2 ðz 0Þ and y ¼ 4 z2 ðy 0Þ. ð F dV throughout the region. If F ¼ 2i þ y 2 j k , evaluate V 9 If F ¼ 3i þ 2j 2xk , evaluate ð F dV where V is the region bounded by the V planes y ¼ 0, z ¼ 0, z ¼ 4 y ðz 0Þ and the surface x2 þ y 2 ¼ 16. 10 A scalar field V ¼ x þ y exists over a surface S defined by x2 þ y 2 þ z2 ¼ 9, bounded by the planes x ¼ 0; y ¼ 0; z ¼ 0 in the first octant. Evaluate ð V dS over the curved surface. S 11 A surface S is defined by y 2 þ z ¼ 4 and is bounded by the planes x ¼ 0, x ¼ 3, ð y ¼ 0, z ¼ 0 in the first octant. Evaluate V dS over this curved surface where S 12 V denotes the scalar field V ¼ x2 yz. ð curl F dS over the surface S defined by 2x þ 2y þ z ¼ 2 and Evaluate S bounded by x ¼ 0, y ¼ 0, z ¼ 0 in the first octant and where 13 F ¼ y 2 i þ 2yzj þ xyk. ð Evaluate F dS over the hemisphere defined by x2 þ y 2 þ z2 ¼ 25 with z 0, S where F ¼ ðx þ yÞi 2zj þ yk : 14 A vector field F ¼ 2xi þ zj þ yk exists over a surface S defined by x2 þ y 2 ðþ z2 ¼ 16, bounded by the planes z ¼ 0, z ¼ 3, x ¼ 0, y ¼ 0. F dS over the stated curved surface. Evaluate 15 Evaluate ð S F dS, where F is the vector field x2 i þ 2zj yk , over the curved S surface S defined by x2 þ y 2 ¼ 25 and bounded by z ¼ 0, z ¼ 6, y 3. 16 A region V is defined by the quartersphere x2 þ y 2 þ z2 ¼ 16, z 0, y 0 and the planes z ¼ 0, y ¼ 0. A vector field F ¼ xyi þ y 2 j þ k exists throughout and on the boundary of the region. Verify the Gauss divergence theorem for the region stated. 17 A surface consists of parts of the planes x ¼ 0, x ¼ 1, y ¼ 0, y ¼ 2, z ¼ 1 in the first octant. If F ¼ yi þ x2 zj þ xyk , verify Stokes’ theorem. 18 S is the surface z ¼ x2 þ y 2 bounded by the planes z ¼ 0 and z ¼ 4. Verify Stokes’ theorem for a vector field F ¼ xyi þ x3 j þ xzk . 868 Programme 23 19 A vector field F ¼ xyi þ z2 j þ xyzk exists over the surfaces x2 þ y 2 þ z2 ¼ a2 , x ¼ 0 and y ¼ 0 in the first octant. Verify Stokes’ theorem that ð þ curl F dS ¼ F dr. 20 A surface is defined by z2 ¼ 4ðx2 þ y 2 Þ where 0 z 6. If a vector field F ¼ zi þ xy 2 j þ x2 zk exists over the surface and on the boundary circle c, show ð þ curl F dS. that F dr ¼ S c c 21 S Verify Green’s theorem in the plane for the integral þ ðx yÞ dx ðy 2 þ xyÞ dy c where c is the circle with unit radius, centred on the origin. Programme 24 Frames 1 to 40 Vector analysis 3 Learning outcomes When you have completed this Programme you will be able to: . Derive the family of curves of constant coordinates for curvilinear coordinates . Derive unit base vectors and scale factors in orthogonal curvilinear coordinates . Obtain the element of arc ds and the element of volume dV in orthogonal curvilinear coordinates . Obtain expressions for the operators grad, div and curl in orthogonal curvilinear coordinates 869 870 Programme 24 1 This short Programme is an extension of the two previous ones and may not be required for all courses. It can well be bypassed without adversely affecting the rest of the work. Curvilinear coordinates Let us consider two variables u and v, each of which is a function of x and y i.e. u=a y u ¼ f ðx, yÞ v ¼ gðx, yÞ v=b If u and v are each assigned a constant value a and b, the equations will, in general, define two intersecting curves. O x If u and v are each given several such values, the equations define a network of curves covering the x–y plane. y a1 a2 u = constant a3 a4 b4 O b2 b1 x b3 v = constant y u = ar (x, y) (u, v) y v = br O x x A pair of curves u ¼ ar and v ¼ br pass through each point in the plane. Hence, any point in the plane can be expressed in rectangular coordinates ðx, yÞ or in curvilinear coordinates ðu, vÞ. Let us see how this works out in an example, so move on 871 Vector analysis 3 Example 1 2 2 Let us consider the case where u ¼ xy and v ¼ x y. 4 (a) With u ¼ xy, if we put u ¼ 4, then y ¼ and we can plot y against x x to obtain the relevant curve. Similarly, putting u ¼ 8; 16; 32; . . . we can build up a family of curves, all of the pattern u ¼ xy. 0.5 1.0 2.0 3.0 4.0 u¼ 4 8 4 2 1.0 u¼ 8 16 8 4 1.33 2.67 x y u ¼ 16 32 16 8 u ¼ 32 64 32 16 5.33 10.67 2 4 8 If we plot these on graph paper between x ¼ 0 and x ¼ 4 with a range of y from y ¼ 0 to y ¼ 20, we obtain ............ 3 y u = 32 u = 16 O u=8 u=4 x Note that each graph is labelled with its individual u-value. (b) With v ¼ x2 y, we proceed in just the same way. We rewrite the equation as y ¼ x2 v; assign values such as 8, 4, 0, 4, 8, 12, 16, . . . to v; and draw the relevant curve in each case. If we do that for x ¼ 0 to x ¼ 4 and limit the y-values to the range y ¼ 0 to y ¼ 20, we obtain the family of curves ............ 872 Programme 24 4 y v = –16 v = –8 v = –4 v=0 v=4 v=8 O x The table of function values is as follows. x y 0 1 2 3 4 v¼ 8 8 7 4 1 8 v¼ 4 4 3 0 5 12 v¼ 0 0 1 4 9 16 v¼ 4 4 5 8 13 20 v¼ 8 8 9 12 17 24 v ¼ 12 12 13 16 21 28 v ¼ 16 16 17 20 25 32 Note again that we label each graph with its own v-value. This again is a family of curves with the common pattern v ¼ x2 y, the members being distinguished from each other by the value assigned to v in each case. Now we draw both sets of curves on a common set of x–y axes, taking and the range of x from x ¼ 0 to x ¼ 4 the range of y from y ¼ 0 to y ¼ 20. It is worthwhile taking a little time over it – and good practice! When you have the complete picture, move on to the next frame 873 Vector analysis 3 y v = –8 v = –16 5 v = –4 v=0 v=4 v=8 u = 32 u = 16 u=8 u=4 O x The position of any point in the plane can now be stated in two ways. For example, the point P has Cartesian rectangular coordinates x ¼ 2, y ¼ 8. It can also be stated in curvilinear coordinates u ¼ 16, v ¼ 4, for it is at the point of intersection of the two curves corresponding to u ¼ 16 and v ¼ 4. Likewise, for the point Q, the position in rectangular coordinates is x ¼ 2:65, y ¼ 5:0 and for its position in curvilinear coordinates we must estimate it within the network. Approximate values are u ¼ 13, v ¼ 2. Similarly, the curvilinear coordinates of R ðx ¼ 1:8, y ¼ 14Þ are approximately u ¼ ............; u ¼ 26; v ¼ ............ 6 v ¼ 11 Their actual values are in fact u ¼ 25:2 and v ¼ 10:76. Now let us deal with another example. Example 2 2 7 2 1 2 ðu 2 If u ¼ x þ 2y and v ¼ y ðx þ 1Þ , these can be rewritten as y ¼ x Þ and y ¼ v þ ðx þ 1Þ2 . We can now plot the family of curves, say between x ¼ 0 and x ¼ 4, with u ¼ 5ð5Þ30 and v ¼ 20ð5Þ5, i.e. values of u from 5 to 30 at intervals of 5 units and values of v from 20 to 5 at intervals of 5 units. The resulting network is easily obtained and appears as ............ 874 Programme 24 8 y v = 5 v = 0 v = –5 v = –10 v = –15 u = 30 v = –20 u = 25 u = 20 u = 15 O x u=5 u = 10 For P, the rectangular coordinates are ðx ¼ 2:18, y ¼ 5:1Þ and the curvilinear coordinates are ðu ¼ 15, v ¼ 5Þ. For Q, the rectangular coordinates are . . . . . . . . . . . . and the curvilinear coordinates are . . . . . . . . . . . . 9 Q: ðx ¼ 3:5, y ¼ 3:0Þ; ðu ¼ 18:5, v ¼ 17Þ Orthogonal curvilinear coordinates If the coordinate curves for u and v forming the network cross at right angles, the system of coordinates is said to be orthogonal. The test for orthogonality is given by the dot product of the vectors formed from the partial derivatives. This is, if @u @v @u @v þ ¼ 0 then u and v are orthogonal. @x @x @y @y Example 3 Given the curvilinear coordinates u and v where u ¼ xy and v ¼ x2 y 2 then u and v form a coordinate system that is . . . . . . . . . . . . 875 Vector analysis 3 10 orthogonal Because @u @u @v @v ¼ y and ¼ x, v ¼ x2 y 2 so ¼ 2x and ¼ 2y. @x @y @x @y @u @v @u @v Then þ ¼ 2xy 2xy ¼ 0 and so u and v form a coordinate @x @x @y @y u ¼ xy so system that is orthogonal. Example 4 Given the curvilinear coordinates u and v where u ¼ x2 þ 2y and v ¼ y ðx þ 1Þ2 then u and v form a coordinate system that is . . . . . . . . . . . . 11 not orthogonal Because u ¼ x2 þ 2y so and @u @u @v ¼ 2x and ¼ 2, v ¼ y ðx þ 1Þ2 so ¼ 2ðx þ 1Þ @x @y @x @v ¼ 1. @y Then @u @v @u @v þ ¼ 4xðx þ 1Þ þ 2 6¼ 0 and so u and v form a coordinate @x @x @y @y system that is not orthogonal. Let us extend these ideas to three dimensions. Move on Orthogonal coordinate systems in space Any vector F can be expressed in terms of its components in three mutually perpendicular directions, which have normally been the directions of the coordinate axes, i.e. 12 z Fz k F k O F ¼ Fx i þ Fy j þ Fz k x i Fx i Fy j j where i, j, k are the unit vectors parallel to the x, y, z axes respectively. y 876 Programme 24 Situations can arise, however, where the directions of the unit vectors do not remain fixed, but vary from point to point in space according to prescribed conditions. Examples of this occur in cylindrical and spherical polar coordinates, with which we are already familiar. 1 z Cylindrical polar coordinates ð, , zÞ Let P be a point with cylindrical coordinates ð; ; zÞ as shown. The position of P is a function of the three variables ; ; z r x (a) If and z remain constant and varies, then P will move out along @r and the unit AP by an amount @ vector I in this direction will be given by @r @r I¼ @ @ z O ϕ ρ y z r I z O ϕ ρ y x (b) If, instead, and z remain constant and varies, P will move ............ 13 round the circle with AP as radius @r is therefore a vector along the tangent to @ the circle at P and the unit vector J at P will be given by @r @r J¼ @ @ z r x O ϕ ρ z y 877 Vector analysis 3 z r O ϕ (c) Finally, if and remain constant and @r will be z increases, the vector @z to the z-axis and the unit vector K in this direction will be given by @r @r K¼ @z @z z y ρ x Putting our three unit vectors on to one diagram, we have ............ 14 z r O ϕ ρ I z y x Note that I, J, K are mutually perpendicular and form a right-handed set. But note also that, unlike the unit vectors i, j, k in the Cartesian system, the unit vectors I, J, K, or base vectors as they are called, are not fixed in directions, but change as the position of P changes. So we have, for cylindrical polar coordinates @r @r I¼ @ @ @r @r J¼ @ @ @r @r K¼ @z @z If F is a vector associated with P, then F(r) ¼ F I þ F J þ Fz K where F , F , Fz are the components of F in the directions of the unit base vectors I, J, K. Now let us attend to spherical coordinates in the same way. 878 Programme 24 2 15 Spherical polar coordinates ðr, , Þ z P is a function of the three variables r, , . θ r O ϕ y x z I q O f r y x z O θ r y ϕ x (a) If and remain constant and r increases, P moves outwards in the @r is thus a vector direction OP. @r normal to the surface of the sphere at P and the unit vector I in that direction is therefore @r @r I¼ @r @r (b) If r and remain constant and increases, P will move along the @r is a ‘meridian’ through P, i.e. @ tangent vector to this circle at P and the unit vector J is given by @r @r J¼ @ @ (c) If r and remain constant and increases, P will move ............ 16 along the circle through P perpendicular to the z-axis z O ϕ x θ r y @r is therefore a tangent vector at P @ and the unit vector K in this direction is given by @r @r K¼ @ @ So, putting the three results on one diagram, we have . . . . . . . . . . . . 879 Vector analysis 3 z O ϕ 17 I θ r y x Once again, the three unit vectors at P (base vectors) are mutually perpendicular and form a right-handed set. Their directions in space, however, change as the position of P

Advanced Engineering Mathematics by Stroud K.A. Booth Dexter J(1)

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