Partial Differential Equations
.
.
Study Guide for
APM3701
Partial Differential Equations
.
Dr Justin MW Munganga
Department of Mathematical Sciences
.
UNIVERSITY OF SOUTH AFRICA
Study Guide for
.
APM3701
© 2021 University of South Africa
All rights reserved
Printed and Published by the
University of South Africa
Muckleneuk, Pretoria
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iv
v
CONTENTS
Contents
PREFACE
xi
PREREQUISITES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xi
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
xi
REMARK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii
CONVENTIONS FOR REFERENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xiii
1 PRELIMINARIES
1
1.1
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
LEARNING OUTCOMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3
DEFINITIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3.1
Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3.2
Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3.3
Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.3.4
Order of Partial Differential Equation . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3.5
Solution of a Partial Differential Equation . . . . . . . . . . . . . . . . . . . . . . . .
4
LINEAR OPERATORS AND LINEAR PARTIAL DIFFERENTIAL EQUATIONS . . . . .
6
1.4.1
Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4.2
Linear Partial Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
THE PRINCIPLE OF SUPERPOSITION . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5.1
The Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.5.2
Failure of the principle of superposition . . . . . . . . . . . . . . . . . . . . . . . . .
9
INITIAL–BOUNDARY–VALUE PROBLEM . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.4
1.5
1.6
1.7
1.6.1
Types of boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.6.2
Well–posedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
SOME IMPORTANT LINEAR PARTIAL DIFFERENTIAL EQUATIONS . . . . . . . . . 12
1.7.1
Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Exercises 1.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 GENERAL SOLUTION AND CLASSIFICATION OF PARTIAL DIFFERENTIAL
EQUATIONS
15
2.1
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2
LEARNING OUTCOMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
vi
2.3
CLASSIFICATION OF SECOND ORDER LINEAR PARTIAL DIFFERENTIAL EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.3.1
Classification of Partial Differential Equation: Elliptic – Parabolic – Hyperbolic
Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
Exercises 2.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.4
GENERAL SOLUTION OF LINEAR PARTIAL DIFFERENTIAL EQUATIONS . . . . . 18
2.4.1
Integrable equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.4.2
Reduction of Partial Differential Equations to canonical forms . . . . . . . . . . . . . 20
2.4.3
The General form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Exercises 2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.5
D’ALEMBERT SOLUTION OF THE WAVE EQUATION . . . . . . . . . . . . . . . . . . 28
Exercises 2.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.6
METHOD OF CHARACTERISTIC CURVES FOR PARTIAL DIFFERENTIAL EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.6.1
First order linear equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.6.2
Second order linear equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Exercises 2.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3 FOURIER SERIES ORTHOGONALITY AND STURM–LIOUVILLE THEORY
49
3.1
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.2
THE WAY FORWARD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.3
PRESCRIBED READING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.4
ORTHOGONALITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.4.1
Definition: orthogonal functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.4.2
Definition: orthogonal sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
3.4.3
Properties of orthogonal systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
Exercises 3.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
3.5
FOURIER SERIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.5.2
Definition: Fourier coefficients
3.5.3
Convergence of Fourier series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
3.7.1
Cosine and Sine Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
3.7.2
Complex Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Exercises 3.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.8
THE STURM-LIOUVILLE THEORY AND EIGENVALUE PROBLEM . . . . . . . . . . . 74
3.8.1
Sturm-Liouville Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.8.2
Regularity of the Sturm-Liouville equation
3.8.3
Eigenvalues and Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
3.8.4
Sturm-Liouville Eigenvalue Problems - Examples . . . . . . . . . . . . . . . . . . . . 77
3.8.5
Summary of Various Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . 85
3.8.6
Properties of eigenvalues and eigenfunctions of regular Sturm-Liouville problem . . . 87
. . . . . . . . . . . . . . . . . . . . . . . 75
Exercises 3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
vii
APM3701/1
4 PARTIAL DIFFERENTIAL EQUATIONS IN RECTANGULAR COORDINATES
89
4.1
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.2
LEARNING OUTCOME . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
4.3
MATHEMATICAL MODELLING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.4
4.3.1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
4.3.2
Mathematical Modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
THE HEAT EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.4.1
Derivation of the heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
4.4.2
Initial conditions for the Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . 93
4.4.3
Boundary conditions for the Heat Equation . . . . . . . . . . . . . . . . . . . . . . . 93
4.4.4
A Heat Transfer Problem with Dynamical Boundary conditions . . . . . . . . . . . . 95
Exercises 4.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95
4.5
SOLUTION OF THE ONE DIMENSIONAL HEAT EQUATION: METHOD OF SEPARATION OF VARIABLES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.5.1
Method of Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
4.5.2
The Heat equation with zero temperatures at finite ends . . . . . . . . . . . . . . . . 96
4.5.3
The Heat equation with insulated end(s) . . . . . . . . . . . . . . . . . . . . . . . . . 100
4.5.4
Heat Equation with Radiating End (or boundary conditions of the third kind) . . . 103
4.5.5
Heat conduction in a thin circular ring: Periodic Boundary Conditions . . . . . . . . 105
4.5.6
Steady state temperature distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 106
4.5.7
The Heat equation with non–homogeneous boundary conditions. . . . . . . . . . . . 108
Exercises 4.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
4.6
THE WAVE EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
4.6.1
Derivation of the wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
4.6.2
Initial and boundary conditions for the wave equation . . . . . . . . . . . . . . . . . 113
4.6.3
A Wave Problem with Dynamical Boundary Conditions . . . . . . . . . . . . . . . . 115
Exercises 4.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
4.7
SOLUTION OF THE WAVE EQUATION – METHOD OF SEPARATION OF VARIABLES116
4.7.1
Vibrating string with fixed ends
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
Exercises 4.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.8
THE LAPLACE EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.8.1
Derivation of the Laplace Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.8.2
Solution of the Laplace’s Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
Exercises 4.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
4.9
QUICK GUIDE FOR EIGENVALUES FOR SEPARATION OF VARIABLES . . . . . . . 121
4.10 LAWS OF CONSERVATION AND THE ENERGY METHOD . . . . . . . . . . . . . . . . 121
4.10.1 Basic Laws of Conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
4.10.2 Coupling Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
4.10.3 Derivation of the Heat Equation by the Law of Conservation . . . . . . . . . . . . . 126
4.10.4 Derivation of Burgers equation in one dimension . . . . . . . . . . . . . . . . . . . . 127
4.10.5 Derivation of the elastic beam equation in one dimension . . . . . . . . . . . . . . . 128
Exercises 4.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128
viii
5 QUALITATIVE QUESTIONS: MAXIMUM–MINIMUM PRINCIPLE UNIQUENESS
AND STABILITY RESULTS
131
5.1
INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
5.2
LEARNING OUTCOMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.3
DEFINITION: WELL-POSED PROBLEM . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
5.4
UNIQUENESS AND STABILITY OF SOLUTIONS . . . . . . . . . . . . . . . . . . . . . . 134
5.4.1
An Integral Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
5.4.2
An Energy Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
5.4.3
Maximum-Minimum Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
Exercises 5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
5.5
5.6
UNIQUENESS AND THE MAXIMUM PRINCIPLE FOR THE HEAT EQUATION . . . 141
5.5.1
Physical interpretation of the Maximum Principle Theorem . . . . . . . . . . . . . . 142
5.5.2
Stability of solution of the heat equation . . . . . . . . . . . . . . . . . . . . . . . . . 143
5.5.3
Interpretation of Theorem 5.5.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
A POSTERIORI EXISTENCE, UNIQUENESS AND STABILITY . . . . . . . . . . . . . . 144
6 FOURIER TRANSFORMS
149
6.1
MOTIVATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.2
LEARNING OUTCOMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.3
PRESCRIBED READING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.4
DEFINITION: FOURIER TRANSFORMS . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
6.6
DEFINITION: FOURIER TRANSFORMS - INVERSION THEOREM . . . . . . . . . . . 152
6.6.1
Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Exercises 6.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
6.7
FOURIER SINE AND COSINE TRANSFORMS . . . . . . . . . . . . . . . . . . . . . . . . 156
6.7.1
Fourier sine transforms
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
6.7.2
Fourier cosine transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
Exercises 6.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
6.8
PROPERTIES OF THE FOURIER TRANSFORM
. . . . . . . . . . . . . . . . . . . . . . 160
6.8.1
Linearity and shifting properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
6.8.2
Scaling property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
6.8.3
Differentiation of Fourier Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
6.8.4
Differentiation of Fourier cosine transform . . . . . . . . . . . . . . . . . . . . . . . . 161
6.8.5
Differentiation of the Fourier sine transform . . . . . . . . . . . . . . . . . . . . . . . 162
Exercises 6.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
6.9
CONVOLUTION THEOREM
6.9.1
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162
Properties of convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
Exercises 6.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
6.10 TABLES OF FOURIER TRANSFORMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
6.10.1 Table of Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
6.10.2 Table of Fourier Cosine Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
6.10.3 Table of Fourier sine transforms
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
ix
APM3701/1
6.11 SOLUTION OF BOUNDARY-VALUE PROBLEMS BY MEANS OF FOURIER TRANSFORMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166
6.11.1 Table of suitable Fourier transform for given boundary conditions and interval . . . 167
Exercises 6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
7 BESSEL SERIES. BOUNDARY–VALUE PROBLEMS IN OTHER COORDINATE
SYSTEMS
171
7.1
BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
7.2
LEARNING OUTCOMES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
7.3
ASSUMED KNOWLEDGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
7.4
PRESCRIBED READING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
7.5
BESSEL SERIES EXPANSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
7.5.1
Bessel’s Equation and Bessel Functions . . . . . . . . . . . . . . . . . . . . . . . . . 172
7.5.2
General Solution of the Bessel’s Equation . . . . . . . . . . . . . . . . . . . . . . . . 176
7.5.3
Parametric Bessel equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
Exercises 7.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
7.6
BESSEL SERIES EXPANSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
7.6.1
Differentiation of Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
7.6.2
Orthogonality of Bessel functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
7.6.3
Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
7.6.4
Bessel Series and Bessel-Fourier coefficients . . . . . . . . . . . . . . . . . . . . . . . 185
7.6.5
Very Important . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
Exercises 7.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
7.7
PARTIAL DIFFERENTIAL EQUATIONS IN POLAR AND CYLINDRICAL COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
7.7.1
The Laplacian in various coordinate systems . . . . . . . . . . . . . . . . . . . . . . 188
7.7.2
Steady Temperature in a circular domains . . . . . . . . . . . . . . . . . . . . . . . . 191
7.7.3
Laplace Equation in a cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196
7.7.4
Heat Distribution in an infinitely long cylinder . . . . . . . . . . . . . . . . . . . . . 198
7.7.5
Radial Vibration of a circular membrane . . . . . . . . . . . . . . . . . . . . . . . . . 199
Exercises 7.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202
A THE GAME AND BETA FUNCTIONS. THE ERROR FUNCTION
Appendix
203
203
A.1 THE GAMMA FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
A.1.1 Definition: Improper Integrals of The Second Type . . . . . . . . . . . . . . . . . . . 203
A.1.2 Definition: The Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
A.1.3 Recursion Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
A.1.4 Definition: The Gamma Function for p < 0. . . . . . . . . . . . . . . . . . . . . . . . 204
A.1.5 Graph of the Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
A.2 THE BETA FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
A.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
x
A.2.2 Definition The Beta Function . . . . . . . . . . . . . . . . . . . . . . .
A.2.3 Alternative forms for the Beta Function . . . . . . . . . . . . . . . . . .
A.3 RELATIONSHIP BETWEEN THE GAMMA AND THE BETA FUNCTIONS
A.4 APPLICATION OF GAMMA AND BETA FUNCTIONS. . . . . . . . . . . . .
A.5 THE ERROR FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A.5.1 Definition: The Error Function erf . . . . . . . . . . . . . . . . . . . . .
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xi
CONTENTS
PREFACE
PREREQUISITES
In this module, we will be studying and solving partial differential equations. We will make extensive use
of the following notions:
1. (a) solutions of ordinary differential equations,
(b) integrations of function of one variable,
(c) trigonometric identities,
(d) differentiation of functions of one and several variables,
(e) orthogonality of functions,
(f) eigenvalues and eigenvectors,
(g) series and
(h) properties of solutions of the equation of second degree of the form ax2 + bx + c = 0.
It is therefore of most importance for your success in this module that you revise these subjects. If you
are not familiar with these notions, I advise you not to continue with this module.
Before you start studying this module make sure that you have the following prerequisites:
1. Ordinary differential equations (APM2611);
2. Multivariable calculus (MAT2615);
3. Real analysis (MAT2613).
4. Linear algebra (MAT1503);
5. Mathematical modelling (APM2614)
BACKGROUND
The module APM3701 (formely APM301W) deals with Partial Differential Equations, one of the most
important and comprehensive branches of modern Applied Mathematics. In mathematical modelling
– real life situations (physical, biological, financial, economical, ...) – are often expressed in terms of
differential equations or a system of differential equations complemented by a system of constraints often
called boundary conditions and/or initial conditions. Usually the model consists of relations between an
xii
unknown function and its derivatives, which are based on underlying physical principles. Partial differential
equations not only accurately express these principles, but also help predict the behaviour of a system
from an initial state and from given external influences.
Often analysis of a mathematical model entails the investigation of the properties of the unknown function(s), without solving the equation(s). In some instances the equation(s) can be solved by some technique
for finding the solution. Even then the solution may be in a form which is so intractable that it satisfies
the particular equation.
This module will concentrate above all on the partial differential equations of Mathematical Physics,
amongst others the Laplace Equation, the Wave Equation and the Heat or Diffusion Equation. We consider
the Partial Differential Equations, not only in rectangular coordinate systems, but also in polar and
cylindrical coordinate systems. In the solution of boundary–value problems in any of the aforementioned
coordinate systems, Fourier, Bessel and Legendre equations play a vital role.
This Study Guide is produced as self–contained and complete as possible. Nevertheless you will need to
use it in conjunction with the Prescribed Book
C. Henry Edwards, David E. Penney and David Calvis
Differential Equations and Boundary Value Problems, Computing and Modelling,
Fifth Edition, Pearson global Edition, Edinburggh, 2016
ISBN 10: 1-292-10877-0
ISBN 13: 978-1-292-10877-3
as certain sections and chapters of this book are prescribed for self–study and we have taken most of the
exercises from it. You are advised to follow the order of chapters in the Study Guide and trace if necessary
the references to the Prescribed Book. The Prescribed Book contains an extensive solutions on Ordinary
Differential Equations and tables of Fourier Transforms.
I strongly recommend that you read the Prescribed Book, to revise the background materials related to
this module.
Chapter 5 and Appendix A of this Study Guide consist of revised sections 2.2–2.4, 2.6 and Appendix D
of the APM3701 Study Guide 1999 Edition by Professor M Grobbelaar.
This Study Guide contains a large number of exercises. The idea is not that you should do all the exercises,
but only as many as you find necessary to master the tutorial matter.
In compiling this Guide, we have apart from the Prescribed Book, made use of the following books:
1. Asmar N.H., Partial Differential Equations with Fourier and Boundary Value Problems. Pearson
Hall, Upper Saddle River, second edition, 2005.
2. Haberman, R., Elementary Applied Partial Differential Equations, 4nd Ed., Prentice–Hall, New
Jersey, 2004.
3. Myint-u T., Partial Differential Equations of Mathematical Physics, Elsevier Science & Technology,
Oxford, 1980.
4. Stephenson G., Partial Differential Equations for Scientists and Engineers, Longman, London, 1980.
xiii
APM3701/1
5. Sauer, N., Unpublished notes on Partial Differential Equations, 2001.
6. Pinchover Y. and Rubinstein J., An Introduction to Partial Differential Equations, Cambridge University Press, Cambridge, 2005.
REMARK
You are advised to follow the order of chapters and sections in the Study Guide and trace if necessary
the references to the Prescribed Book. The Prescribed Book also contains a very rich notes on Ordinary
Differential Equations and tables of Fourier Transforms.
CONVENTIONS FOR REFERENCES
References to equations, chapters, sections, Theorems,... will be followed by the source from where they
were taken from. Wherever there is no reference to the source, consider the source to be this guide. The
prescribed book always means
C. Henry Edwards, David E. Penney and David Calvis
Differential Equations and Boundary Value Problems, Computing and Modelling,
Fifth Edition, Pearson global Edition, Edinburggh, 2016
ISBN 10: 1-292-10877-0
ISBN 13: 978-1-292-10877-3
ACKNOWLEDGEMENT
I would like to thank the following students who brought to my intention typos and different ways of their
understanding of the problems. I really appreciate your contributions.
1. Mr Stephen “Charlie” Martindale (2009)
2. Mr Melusi Magele (2009)
3. Mr Sandile Mkhungo (2012/S2)
4. Miss Nchedi Portia Molekoa (2013/S1)
5. Miss Khotso Matlou (2013/S1–S2)
6. Mr Richard Leeman (2014/S2)
7. Mrs Keryn Barnes (2021/S1)
xiv
1
CONTENTS
Chapter 1
PRELIMINARIES
1.1
BACKGROUND
This chapter forms a link between ordinary differential equations and partial differential equations. Partial
derivatives, partial differential equations, linear operator are reviewed. Initial-boundary-value problem
and some important linear partial differential equations are given. The material in this chapter is mainly
background material.
1.2
LEARNING OUTCOMES
At the end of this chapter you will able to:
• define partial differential equation, its order and determine its linearity.
• define an initial-boundary value problem.
• recognise the three important linear Partial Differential Equations: heat, wave and Laplace equation.
1.3
DEFINITIONS
1.3.1
Partial Derivatives
Let u be a function of two (or three) variables e.g. u = u (x, y) [or u = u (x, y, z)], the partial derivative
∂u
∂x (also noted ux ) of u with respect to the variable x, is the ordinary differentiation of u with respect to
x, treating the other variables as constant. We will use the following notations
ux =
uxx =
∂u
∂u
∂u
, uy =
, ut =
,...
∂x
∂y
∂t
∂2u
∂ ∂u
∂ ∂u
∂2u
=
,
u
=
=
,...
xy
∂x2
∂x ∂x
∂y ∂x
∂y∂x
Example 1.3.1
• If f (x, y) = x2 y 2 + sin xy, then
•
∂f
∂
=
x2 y + sin (xy) = 2xy + y cos (xy) ,
∂x
∂x
∂f
∂
=
x2 y + sin (xy) = x2 + x cos (xy) ,
∂y
∂y
2
•
∂2
∂2f
2 y 2 + sin (xy) =
=
x
∂x2
∂x2
= 2y 2 − y 2 sin (xy) .
∂
∂x
∂
∂x
x2 y 2 + sin (xy)
∂
∂x
=
2xy 2 + y cos (xy)
∂2u ∂2u ∂2u
2
2
,
, if u(x, y) = 2ex +y + 4x2 y 4
,
2
2
∂x∂y ∂y ∂x
∂ ∂2u
∂ ∂u
2
2
=
∴
=
4xex +y + 8xy 4
∂x∂y
∂y ∂x
∂y
• Find
2
2
∂2u
∂y 2
= 8xyex +y + 32xy 3 .
∂ x2 +y2
∂ ∂u
2 3
=
=
4ye
+ 16x y
∂y ∂y
∂y
∂2u
∂x2
= 4ex +y + 8y 2 ex +y + 48x2 y 2 .
∂ ∂u
∂ 2
2
=
=
4xex +y + 8xy 4
∂x ∂x
∂x
2
= 4ex
1.3.1.1
2
2 +y 2
2
+ 8x2 ex
2
2 +y 2
+ 8y 4 .
Order of a Partial Derivative
The order of a partial derivative is the number of times that the function has been differentiated with
respect to any of its variables. By convention the function itself is a partial derivative of order 0.
Example 1.3.2
1. ux =
∂u
∂u
∂f
, uy =
, ft =
are partial derivatives of order 1.
∂x
∂y
∂t
2. fyy =
∂2u
∂2f
∂2f
,
u
=
,
f
,
=
are partial derivatives of order 2
xx
xt
∂y 2
∂x2
∂t∂x
3. ftxx =
1.3.1.2
∂3f
∂3u
∂3u
,
u
=
,
u
=
are partial derivatives of order 3.
yyy
xyz
∂x2 ∂t
∂y 3
∂z∂y∂x
Chain rule for Partial Derivatives
If f is a function of x and y, and x and y are functions of t, then
df
∂f dx ∂f ∂y
∂f
=
=
+
.
∂t
dt
∂x dt
∂y ∂t
Note that
df
is used because f depends only on t.
dt
Example 1.3.3
df
Find
, where f (x, y) = x2 + y 2 + x, and x = a cos t and y = b sin t (a and b constant).
dt
Solution
df
dt
∂f
∂x
∂f
∂y
=
∂f dx ∂f dy
+
∂x dt
∂y dt
= 2x + 1 = 2a cos t + 1,
= 2y = 2b sin t;
dx
= −a sin t
dt
dy
= b cos t.
dt
3
APM3701/1
Thus
df
dt
= − (2a cos t + 1) a sin t + 2b2 sin t cos t
= 2 b2 − a2 cos t sin t − a sin t
■
If f is a function of x and y, and x and y are functions of t and s, then
∂f
∂t
∂f
∂s
=
=
∂f ∂x ∂f ∂y
+
∂x ∂t
∂y ∂t
∂f ∂x ∂f ∂y
+
.
∂x ∂s
∂y ∂s
(1.1)
(1.2)
Example 1.3.4
∂f ∂f
Find
,
, if f (x, y) = x2 + y 2 + x, where x = s cos t and y = s sin t.
∂s
∂t
Solution
∂f
∂t
∂f
∂s
∂f
∂x
∂y
∂t
∂f ∂x ∂f ∂y
+
∂x ∂t
∂y ∂t
∂f ∂x ∂f ∂y
=
+
∂x ∂s
∂y ∂s
∂x
= 2x + 1,
= −s sin t,
∂t
∂x
= s cos t,
= cos t,
∂s
=
∂f
= 2y
∂y
∂y
= sin t.
∂s
Thus
∂f
∂t
= − (2x + 1) s sin t + 2ys cos t
= −s (2s cos t + 1) sin t + 2s2 sin t cos t
∂f
∂s
= −s sin t.
= (2x + 1) cos t + 2y sin t
= (2s cos t + 1) cos t + 2s sin2 t
= 2s + cos t.
■
1.3.2
Differential Equation
A differential equation is a relation between the derivatives of an unknown function. When the unknown
function depends on a single variable the differential equation is referred to as an ordinary differential
equation. Equations in which the unknown function depends on more than one variable are called
Partial Differential Equations. In this case the equation is a relationship between partial derivatives.
1.3.3
Partial Differential Equations
A Partial Differential Equation is a relation between the partial derivatives of an unknown function.
The order of a partial differential equation is the greatest order of the partial derivative of the function
appearing in the equation.
4
Example 1.3.5
1. The equations ut = c2 uxx , utt = 1c uxx and uxx + uyy = 0 are 2nd order partial differential equations
2. The equations ux + 2uy + u = 0, and ux + u uy = 0 are 1st order partial differential equations.
1.3.4
Order of Partial Differential Equation
A Partial Differential Equation is said to be of order k > 0, if k is the greatest order of the partial
derivative appearing in the equation.
Example 1.3.6
The following equations
(1) ut = c2 uxx ,
(2) ux uxx + uyy = 0,
(3) utt = c2 uxx
are Partial Differential Equations of order 2 (or simply 2nd order Partial Differential Equations) and
(4) uut + uxt uxxx = f (x, t) and (5) uxy + ux − uxyy = 0
are third order Partial Differential Equations.
1.3.5
■
Solution of a Partial Differential Equation
A solution of a Partial Differential Equation, on a specified region D, is any smooth function
defined on D such that the Partial Differential Equation is satisfied at all interior points of D. If no region
is specified, then a solution of a Partial Differential Equation is a smooth function defined on at least some
nonempty open region where it satisfies the Partial Differential Equation.
NOTE: By smooth function, we understand that, if a Partial Differential Equation is of order k, its
solution must be at least a C k –function (i.e. a function which is k–times continuously differentiable).
In practice, the unknown function u might be any one of the following quantities: temperature, pressure, mass density, concentration of a certain chemical, monetary flow, gravitational potential energy,
electrostatic potential energy, etc.
Example 1.3.7
Show that the function u (x, y) = x + cos x − ey is a solution of the Partial Differential Equation
∂2u
= 0.
∂x∂y
Solution
∂u
= 1 − sin x
∂x
∂2u
∂
and
=
∂x∂y
∂y
∂u
∂x
=
∂
(1 − sin x) = 0.
∂y
(1.3)
■
Example 1.3.8 Show that in general, if u (x, y) = H (x) + G (y) where H and G are arbitrary functions
twice differentiable in x and y respectively, then u satisfies (1.3).
Solution
∂2u
∂
=
∂x∂y
∂y
∂u
∂x
=
∂
H ′ (x) = 0, (Explain why?)
∂y
■
5
APM3701/1
Example 1.3.9 Show that u (x, y) = F (x + y) + G (x − y) satisfies the following partial differential equation
uxx − uyy = 0
where F and G are arbitrary C 2 –functions of their arguments.
Solution
ux = F ′ (x + y) + G′ (x − y)
(1.4)
where F ′ (x + y) and G′ (x − y) are derivatives with respect to x + y and x − y respectively.
uxx = F ′′ (x + y) + G′′ (x − y)
(1.5)
uy = F (x + y) − G (x − y)
(1.6)
′
′
uyy = F ′′ (x + y) + G′′ (x − y)
(1.7)
Therefore uxx − uyy = F ′′ (x + y) + G′′ (x − y) − F ′′ (x + y) − G′′ (x − y) = 0.
■
Example 1.3.10 Show that the function u (x, t) = sin (x + at) is a solution of the partial differential
equation utt − a2 uxx = 0.
Solution
uxx = − sin (x + at)
ut = a cos (x + at)
u (x, t) = sin (x + at)
ux = cos (x + at)
utt = −a2 sin (x + at).
Thus
utt − a2 uxx = −a2 sin (x + at) − a2 (− sin (x + at))
= −a2 sin (x + at) + a2 sin (x + at) = 0.
■
Example 1.3.11 Show that any function of the form u (x, t) = f (x + at) + g (x − at), (where f and
g are, arbitrary twice differentiable functions of their arguments), is a solution of the wave equation
utt − a2 uxx = 0.
Solution
ux = f ′ (x + at) + g ′ (x − at)
uxx = f ′′ (x + at) + g ′′ (x − at)
ut = af ′ (x + at) − ag ′ (x − at)
utt = a2 f ′′ (x + at) + a2 g ′′ (x − at) ;
where the prime’s are derivatives with respect to the argument x + at and x − at.
Thus
utt − a2 uxx = a2 f ′′ (x + at) + a2 g ′′ (x − at) − a2 f ′′ (x + at) − a2 g ′′ (x − at) = 0.
■
6
Example 1.3.12 Show that
u (x, y) =
1
f (x) + G (x) e−xy
x
is a solution of the partial differential equation
uyy + xuy = 0.
Solution
uy = −xG (x) e−xy , uyy = x2 G (x) e−xy .
So,
uyy + xuy = x2 G (x) e−xy + x −xG (x) e−xy
= x2 G (x) e−xy − x2 G (x) e−xy = 0.
■
1.4
LINEAR OPERATORS AND LINEAR PARTIAL DIFFERENTIAL EQUATIONS
1.4.1
Linear Operators
An operator is a mathematical rule which when applied to a function produces another function.
Example 1.4.1 L and M below are called differential operators.
∂u
∂2u
+ f (x)
−u
∂t
∂x∂t
2
∂2u
2∂ u
−
c
∂t2
∂x2
∂
∂2
+ f (x)
−1
∂t
∂x∂t
2
∂2
2 ∂
−
c
∂t2
∂x2
L[u] =
M [u] =
∴L =
and
=
M
There exist other types of operators, such as
Z
I [u] =
1
u (x, t) r (t, y) dt
0
H [u] = uxx (0, t) + ut (0, t)
The operator I is called an integral operator and H transforms the function u of two variables x and t
into another function H [u] of one variable t.
■
♦Note: In this module, we will only be concerned by differential operators.
1.4.2
1.4.2.1
Linear Partial Differential Equations
Definition: Linear Differential Operator
A differential operator is said to be linear if it satisfies the following properties:
1. L [cu] = cL [u]
for any constant c and any smooth function u.
7
APM3701/1
2. L [u + v] = L [u] + L [v] for u and v any smooth functions.
Properties (1) and (2) may be combined to express
L [au + bv] = aL [u] + bL [v]
(1.8)
where a and b are constants.
A Partial Differential Equation is said to be linear if it can be put in the form
L [u] = g (x, y, z, . . .)
(1.9)
where L is a linear differential operator and g is a continuous function in the independent variables
x, y, z, . . . , otherwise it is said to be nonlinear.
• IMPORTANT
Basically, we say that a Partial Differential Equation is linear if the function u and its derivatives
appear only in the first degree and if no product of u and its partial derivatives occurs.
Example 1.4.2
1. The equations
(a) ut = c2 uxx
(c) x2 + y uxy + ux − uxyy = y + 1
(b) utt = c2 uxx
are all linear.
2. One can verify that any 2nd order linear Partial Differential Equation in two independent variables
x and y can be put in the form:
A (x, y) uxx + 2B (x, y) uxy + C (x, y) uyy + D (x, y) ux + E (x, y) uy + F (x, y) u − G (x, y)
(1.10)
where A, B, . . . , G are functions of x and y and A, B and C do not vanish simultaneously. Equation
(1.10) will be studied in detail in chapter 2.
3. Equations
(a) ux uxx + uyy = 0
(c) ux + u2t = 0
(b) uut + ux uxxx + f (x, t) = 0
are all nonlinear (i.e. are not linear). Please explain!!!
Hint: For example, 3.(a) is not linear because there is a product of ux and uxx
1.4.2.2
■
Definition: Homogeneous Partial Differential Equations
A linear Partial Differential Equation is called homogeneous if it can be written in the form
Lu = 0
where L is a linear differential operator.
In other words, a linear Partial Differential Equation is homogeneous if each term contains only
the unknown function or one of its derivatives. Otherwise it is said to be non–homogeneous (or inhomogeneous). For instance, in Example 1.4.2, 1.(a) and 1.(b), are homogeneous equations. But 1.(c) is
not [because of the presence y + 1 on the right–hand side].
8
1.5 THE PRINCIPLE OF SUPERPOSITION
See also [1, Theorem 1, section 3.1]
The Principe of Superposition is a very important principle in the study of linear partial differential
equations. We know from section 1.4 and equation (1.9) that a linear partial differential equation can be
written in the form L [u] = g. Now, assume that u1 and u2 are solutions of L [u] = g1 and L [u] = g2
respectively. We have L [u1 ] = g1 and L [u2 ] = g2 . On the other hand we have
L [c1 u1 + c2 u2 ] = c1 L [u1 ] + c2 L [u2 ] = c1 g1 + c2 g2
i.e. c1 u1 + c2 u2 is a solution of L [u] = c1 g1 + c2 g2 . More generally, we state the following principle:
1.5.1
The Principle of Superposition
Theorem 1.5.1 Let u1 be a solution of L [u] = g1 and u2 be a solution of L [u] = g2 . Then for any
constants c1 and c2 , c1 u1 + c2 u2 is a solution of L[u] = c1 g1 + c2 g2 .
In other words
L [c1 u1 + c2 u2 ] = c1 g1 + c2 g2 .
(1.11)
In particular, if g1 = 0 and g2 = 0 (i.e. u1 and u2 are solutions of the homogeneous equation L [u] = 0),
then c1 u1 + c2 u2 is also a solution of the same homogeneous equation L [u] = 0.
More generally, if u1 , u2 , . . . , un are solutions of the homogeneous equation L [u] = 0 then for constants
c1 , c2 , . . . , cn , c1 u1 + c2 u2 + . . . + cn un is also a solutions of L [u] = 0.
■
Example 1.5.1 Verify that u1 (x, y) = y 3 is a solution of uxx + uy = 3y 2 and u2 = x3 + y 3 is a solution
of uxx + uy = 6x + 3y 2 . Find a solution of uxx + uy = 6x.
Solution
Let L [u] = uxx + uy , g1 (x, y) = 3y 2 and g2 (x, y) = 6x + 3y 2 . Since 6x = −g1 (x, y) + g2 (x, y) , choose
c1 = −1 and c2 = 1. We deduce that c1 u1 + c2 u2 = −y 3 + x3 + y 3 = x3 is a solution of L(u) = c1 g1 + c2 g2
i.e. uxx + uy = 6x.
■
Example 1.5.2 Verify that for every k, 0 < k ≤ n < ∞, uk = eky sin kx + eky cos kx, is a solution of
uxx + uyy = 0.
(1.12)
Find a more general solution of (1.12).
Solution
We have
(uk )x = k eky sin kx − eky cos kx , (uk )xx = k 2 −eky cos kx − eky sin kx ,
(uk )y = k eky sin kx + eky cos kx , (uk )yy = k 2 eky sin kx + eky cos kx ,
Hence, (uk )xx + (uk )yy = k 2 −eky cos kx − eky sin kx + k 2 eky sin kx + eky cos kx = 0.
Thus, for 0 < k ≤ n, uk is a solution of (1.12), therefore by principle of superposition,
n
P
ak eky sin kx + bk eky cos kx is also a solution of (1.12).
k=1
■
9
1.5.2
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Failure of the principle of superposition
The principle of superposition will generally fails for nonlinear partial differential equations.
Example 1.5.3 Consider the nonlinear equation
ux uy + u2 − u (ux + uy ) = 0
Show that ex and ey are solutions of the above equation, but ex + 2ey is not a solution. Thus the principle
of superposition does not hold.
Solution
The equation is nonlinear because of the terms ux uy , u2 and the product u (ux + uy ).
Let u1 = ex , u2 = ey and u3 = ex + 2ey . We have,
(u1 )x = ex , (u1 )y = 0, and (u1 )x (u1 )y + (u1 )2 − (u1 ) (u1 )x + (u1 )y = 0.
(u2 )x = 0 (u2 )y = ey and (u2 )x (u2 )y + (u2 )2 − (u2 ) (u2 )x + (u2 )y = 0
However, (u3 )x = ex
(u3 )y = 2ey
and
(u3 )x (u3 )y + (u3 )2 − (u3 ) (u3 )x + (u3 )y
=
2ex+y + (ex + 2ey )2 − (ex + 2ey ) (ex + 2ey )
= 2ex+y ̸= 0.
Thus, u3 = ex + 2ey is not a solution of the equation ux uy + u2 − u (ux + uy ) = 0. Therefore the principle
of superposition fails.
■
1.6
INITIAL–BOUNDARY–VALUE PROBLEM
Without any additional condition, generally a Partial Differential Equation admits an infinity many solutions. This is not useful if one wants to determine or predict the behaviour of a system. Usually Partial
Differential Equations are complimented by a system of constraints often called boundary and/or initial
conditions. These constraints are the topic of this section.
In this section we discuss initial and boundary conditions. Solutions of initial and boundary value
problems will be discussed in later sections (see section 2.4, and chapters 4, 6 and 7).
• A mathematical problem consists of finding an unknown function satisfying a Partial Differential
Equation as well as appropriate supplementary conditions. These conditions may be initial and/or
boundary conditions.
Initial conditions are usually prescribed for the unknown function at a certain time t = t0 (often
t = 0), and boundary conditions are prescribed for the unknown function at some points on the
boundary, or on all points of the boundary.
If both initial and boundary conditions are prescribed for the unknown function, the problem is
called Initial and Boundary-Value Problem (IBVP). More often the problem is autonomous
(i.e. independent of time), in this instance only the boundary conditions are prescribed: the problem
will then be called Boundary-Value Problem (BVP).
In many cases, the solution can be determined uniquely by prescribing initial conditions due to the
unboundedness of the domain. In this case, the problem is called Initial Value Problem (IVP).
See section 1.6.2
10
An example of an Initial–Boundary–Value Problem is given below:
Example 1.6.1
1
uxx = 0,
c2
u (x, 0) = f (x) ,
ut −
0 < x < L, t > 0,
(1.13)
0 ≤ x ≤ L,
(1.14)
u (0, t) = g1 (t) ,
t ≥ 0,
(1.15)
u (L, t) = g2 (t) ,
t ≥ 0.
(1.16)
Equations (1.13) – (1.16) describes the heat conduction in a rod of length L (see section 4.4.1, for more
details). Equation (1.14) is an initial condition, it prescribes the temperature u (x, t) throughout the given
rod at time t = 0. Equations (1.15) and (1.16) are boundary conditions, they describe the temperature at
the end–points of the rod x = 0 and x = L, which may be the outside temperature at the ends of the rod.
■
1.6.1
Types of boundary conditions
There are three main types of boundary conditions which arise frequently in the description of physical
problem:
(a) Dirichlet conditions: where the unknown function u is specified at each point of the boundary.
(b) Neumann conditions: where values of the normal derivative
∂u
of the unknown function are
∂η
prescribed on the boundary.
(c) Cauchy conditions: If one of the variables is t (time, say), and the values of both the unknown
∂u
on a boundary at t = t0 are given, i.e. the initial values of u
function u and its time derivative
∂t
∂u
and
, then the boundary conditions are of Cauchy type with respect to the variable t.
∂t
1.6.2
Well–posedness
The notion of well-posedness is a fundamental theoretical question which was first formulated by the
French mathematician Jacques Hadamard (1865–1963) [6].
A initial and/or boundary value problem is an appropriate mathematical model of real life (physical)
situation (or simply we will say that an (initial–) boundary value problem is properly posed (or well
posed).
Hadamard defined a well-posed problem as a problem which satisfies the following three requirements are
met:
1. Existence: there exists a solution to the problem
2. Uniqueness: the solution is unique (no more than one solution exists).
3. Stability or continuous dependence on the data: the solution is depends continuously on the
data. In other words, a small variation of the given data leads to a small change in the solution.
11
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Requirements (1.) and (2.) are logical conditions. It is not evident that the mathematical problem has a
solution just because the physical problem has one, and the physical problem may have a unique solution
but the mathematical problem may have more than one solution, or even does not have a solution at all.
Requirement (3.) is necessary due to small errors occurring in the measurement process.
A mathematical problem represents a physical phenomenon if it is well–posed (requirements (1.), (2.) and
(3.) are met). Otherwise the mathematical problem could not be associated with a physical (real life)
phenomenon, it is said to be ill–posed. We may be solving pretty mathematical equations which do not
have any physical application.
Example 1.6.2 – Ill-posed problems
1. The heat equation
ut =
1
uxx ,
c2
(1.17)
without any boundary condition, is ill–posed since it has an infinite family of solutions, for example
any function of the form u (x, t) = Kx, where K ∈ R is a solution of (1.17).
2. A second example of an ill–posed problem is the famous backwards heat equation (you may be
confused at this stage by this example; if it the case for you, please come back to it after studying
section 4.5.1).
ut = −k 2 uxx
0 < x < L,
t>0
(1.18)
subject to
u (0, t) = u (L, t) = 0
t>0
(1.19)
u (x, 0) = f (x)
0 < x < L.
(1.20)
In fact, we will show in section 4.5 that
u (x, t) =
∞
X
nπ 2
an e ( L )
kt
n=1
sin
nπx L
(1.21)
where
2
an =
L
Z
L
f (x) sin
0
nπx
dx
L
(1.22)
is a unique solution to the problem (1.18) – (1.20). Assume now that the data are changed by a
small amount,
nπx
1
f (x) → f (x) + sin
(1.23)
n
L
for large values of n. If we set
f1 (x) = f (x) +
the additional term
1
nπx
sin
,
n
L
1
nπx
sin
is very small, for large value of n.
n
L
12
However, the solution associated with the new initial condition is
u1 (x, y) =
∞
X
nπ 2
an e( L )
n=1
∞ X
kt
sin
nπx
L
L
nπ 2
nπx
nπx
f1 (x) sin
=
dx e( L ) kt sin
L
L
0
n=1
∞ Z L
X
nπ 2
2
nπx
nπx
=
f (x) sin
dx e( L ) kt sin
L 0
L
L
n=1
∞ Z L
X
nπ 2
1
2
nπx
2 nπx
sin
dx e( L ) kt sin
.
+
L 0 n
L
L
n=1
|
{z
}
extra term
In the solution, the extra term is
2
L
Z
∞
X
1 ( nπ )2 kt
nπx
sin
e L
n
L
n=1
which tend to infinity as t tends to infinity. Thus u (x, t) differs drastically when f (x) is slightly
disturbed.
The given problem is unstable, therefore ill–posed.
1.7
■
SOME IMPORTANT LINEAR PARTIAL DIFFERENTIAL EQUATIONS
In this section we list some simple equations which are of importance in mathematical physics [1, 3, 8, 9].
Some of them will be derived and analysed in Chapter 4.
ut + ux = 0
ut + uux = 0
1
ut = 2 uxx = 0
c
1
utt = 2 uxx
c
uxx + uyy = 0
uxx + uyy = f (x, y)
uxx + uyy + uzz = 0
1
uxxxx = − 2 utt
c
1
uxxxx + uyyyy = − 2 utt
c
uxx + uyy + α [E − V (x, y)] u = 0
uxx − c2 utt + λ2 u = 0
uxx + uyy = c2 utt + λ2 u = 0
1.7.1
Transport equation
(1.24)
Inviscid Burger’s equation
(1.25)
One–dimensional heat equation
(1.26)
One–dimensional wave equation
(1.27)
Two–dimensional Laplace equation
(1.28)
Two–dimensional Poisson equation
(1.29)
Three–dimensional Laplace equation
(1.30)
One–dimensional biharmonic equation
(1.31)
Two–dimensional biharmonic equation
(1.32)
Two–dimensional Schrödinger’s equation
(1.33)
One–dimensional Klein–Gordon equation
(1.34)
Two–dimensional Klein–Gordon equation.
(1.35)
Remarks
1. In this module, the methods of solving Partial Differential Equations will be particularly directed
at Partial Differential Equations (1.26) – (1.28) together with certain boundary conditions. Some
13
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non–homogeneous Partial Differential Equations of these kinds will also be considered.
2. A Partial Differential Equation in which no partial derivatives with respect to time appear, is called
stationary. Otherwise it is called non–stationary.
Exercises 1.7
1. For each of the following, state whether the Partial Differential Equation is linear, or non linear and
give its order. If it is linear, state whether it is homogeneous or nonhomogeneous. And if it is not,
state the reason.
(a) yuxy + uy = xy
(e) Auxx + Buxy + Cuyy + Euy + F u = G (x, y)
(b) 3xux − uuy = 0
(f ) uxx = cos x
(c) uuxx + u2y = 0
(g) uxxx + ln u = 0
(d) uxxxx + 2uxxyy + uyyyy = 0
(h) auxx + buxy + cuyy + u = cos u
2. Use the change of variables
α = px + qy and β = px − py,
where p and q are constants, to find solutions of the following differential equations:
(a) 1 + x2 ux + uy = 0.
(b) ut + 4ux = 0,
subject to u (x, 0) = cos (x + π) .
(c)
√
1 − x2 ux + xuy = 0,
subject to u (0, y) = 2y 2 .
2
(d) ex ux + xuy = 0,
subject to u (0, y) = ln y.
■
14
15
CONTENTS
Chapter 2
GENERAL SOLUTION AND
CLASSIFICATION OF PARTIAL
DIFFERENTIAL EQUATIONS
2.1
BACKGROUND
For purpose of classification of partial differential equations, this chapter will be restricted to partial
differential equations of order 1 and 2, where the unknown function u contains only two independent
variables.
Second order linear partial differential equations will be classified as elliptic, parabolic or hyperbolic
according to the formula B 2 − AC < 0, = 0 or > 0, respectively. General solution of linear partial
differential equation are found by direct integration and by reduction to canonical forms. For the wave
equation, d’Alembert solution is obtained.
2.2
LEARNING OUTCOMES
On completion of this chapter, the following outcomes will be achieved:
• Classify second order linear differential equations as elliptic, parabolic, or hyperbolic according to
the formula B 2 − AC < 0, = 0 or > 0, respectively.
• Reduce the second order linear partial differential equation to canonical forms using the method of
characteristic curves.
• Obtain the general solution of second order linear differential equation by reduction to canonical
forms .
• Obtain the solution of the wave equation using d’Alembert solution for the wave equation.
16
2.3
CLASSIFICATION OF SECOND ORDER LINEAR PARTIAL DIFFERENTIAL
EQUATIONS
We will be concerned mainly with linear 2nd order Partial Differential Equations. The classification of
Partial Differential Equations is of great importance, since conclusion about the well–posedness (see section
1.6.2) of a particular boundary–value problem can be drawn from the class to which a Partial Differential
Equation belongs. The class to which a Partial Differential Equation belongs also largely determines the
form of its general solution, as we will see in section 2.4.
For the purposes of classification of Partial Differential Equations, we shall restrict ourselves to Partial
Differential Equations of order 2, where the unknown function u contains only two independent variables.
2.3.1
Classification of Partial Differential Equation: Elliptic – Parabolic – Hyperbolic Equations
One can easily verify that any linear 2nd–order partial differential equation can be put in the form
A (x, y) uxx + 2B (x, y) uxy + C (x, y) uyy + D (x, y) ux + E (x, y) uy + F (x, y) u = G (x, y)
(2.1)
where A, B, C, D, E, F and G are differentiable functions. Most Partial Differential Equations mentioned
in section 1.7 are of this form.
Equation (2.1) will be called the General linear partial differential equation of order 2.
Note that the form (2.1) bears resemblance to the form of the equation of the general conic section
Ax2 + 2Bxy + Cy 2 + 2Dx + 2Ey + F = 0
(2.2)
and recall that (2.2) is an ellipse, parabola or hyperbola when B 2 − AC < 0, = 0 or > 0 respectively.
Accordingly, the partial differential equation (2.1) is classified as [1–3, 7, 9]:


2


elliptic

 B − AC < 0
if
parabolic
B 2 − AC = 0



 2
hyperbolic
B − AC > 0.
(2.3)
Consequently,
• the wave equation
utt =
1
uxx
c2
1
is hyperbolic, since B 2 − AC = 2 > 0. (Write the partial differential equation in the form utt −
c
1
uxx = 0, A = 1, B = 0, C = − c12 , by replacing y by t in (2.1))
c2
• The Laplace Equation in two variables
uxx + uyy = 0
is elliptic, since B 2 − AC = −1 < 0. A = 1, B = 0 and C = 1.
17
• The Heat Equation
APM3701/1
ut = k 2 uxx
is parabolic, since B 2 − AC = 0. (Write the partial differential equation in the form k 2 uxx − ut =
0, A = k 2 , B = 0, C = 0, by replacing y by t in (2.1)).
Remark 2.3.1
1. More often the coefficients A, B and C are functions of x and y, the class to which a Partial
Differential Equation of the form (2.1) belongs to, depends upon the region of the xy–plane.
2. The classification of linear Partial Differential Equations in three or more dimensions is similar, but
more complicated. Although the term elliptic, parabolic and hyperbolic should in this case be replace
by their 3–dimensional analogues, it is customary to call, for example, the heat equation
ut = k 2 ∇2 u
where ∇2 u = ux1 x1 + ux2 x2 + ux3 x3
parabolic, even if u = u (x1 , x2 , x3 ).
■
Example 2.3.1 Classify the following Partial Differential Equations:
1. x2 uxx + 2xyuxy + y 2 uyy = 0
2. yuxx + 2xuxy + yuyy = 0
Solution
1. A = x2
c = y 2 . So, B 2 − AC = x2 y 2 − x2 y 2 = 0.
B = xy
Thus the Partial Differential Equation is parabolic.
2. A = y
B=x
C = y, B 2 − AC = x2 − y 2
The Partial Differential Equation is elliptic in the region x2 − y 2 < 0, parabolic along the lines
x + y = 0 and x − y = 0, and hyperbolic in the region x2 − y 2 > 0 (see figure 2.1)
■
Example 2.3.2 Classify the following equations:
1. utt =
1
u
c2 xx
3. 2uxx + 3uxy − 4uyy + 5uy = cos x
2. uxx + 2uxy + uyy = 0
Solution
1. utt =
1
u ,
c2 xx
4. uxx − uxy + 2uyy = 0
write the partial differential equation in the form − c12 uxx + utt = 0,
A = − c12 , B = 0, C = 1, by replacing y by t.
We have B 2 − AC =
1
c2
> 0. Thus the equation is hyperbolic.
2. uxx + 2uxy + uyy = 0; A = 1, B = 1, C = 1
parabolic.
∴
B 2 − AC = 1 − 1 = 0. Thus the equation is
18
y
P
ar
ab
o
lic
lic
o
ab
ar
P
Elliptic
Hyperbolic
Hyperbolic
x
lic
P
ar
ab
o
lic
o
ab
ar
P
Elliptic
Figure 2.1: Classification of the equation yuxx + 2xuxy + yuyy .
3. 2uxx + 3uxy − 4uyy + 5uy = cos x;
A = 2; B = 23 , C = −4. B 2 − AC =
9
4
+8=
41
4
> 0.
Thus the equation is hyperbolic.
4. uxx − uxy + 2uyy = 0;
B 2 − AC =
1
4
A = 1, B = − 12 . C = 2
− 2 = − 74 < 0. Thus the equation is elliptic.
■
Exercises 2.3
1. Determine the region in which the given equation is elliptic, parabolic or hyperbolic.
2.4
(a) uxx + yuyy = 0
(b) 1 + x2 uxx + y 2 uyy = 0
(d) xuxx − uxy + xyuy − ex u = 0.
(c) uxx sin2 x + uxy sin 2x + uyy cos2 x = 0
(e) 3uxx + 4uxy − uyy = 0
■
GENERAL SOLUTION OF LINEAR PARTIAL DIFFERENTIAL EQUATIONS
Ideally, one would like to have a general technique that could be used to find all of the solutions of an
arbitrary Partial Differential Equation, or at least a relevant solution that satisfies certain initial/boundary
conditions. Unfortunately such a general technique for solving partial differential equations does not exist.
It is easy to find Partial Differential Equations for which there is no known method which will yield a
unique solution. Fortunately, the Partial Differential Equations which arise in practice, such as those we
will study in this module, are not completely arbitrary. Indeed, there are few different kinds of Partial
Differential Equations, or systems of Partial Differential Equations which regularly appear in applications.
Although there are some procedures, that apply to more than one relevant equation, we will prefer to
handle each relevant equation separately and develop the required procedure for solving it. Unlike the
theory of ordinary differential equations, the methods for solving Partial Differential Equations often
19
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depend more on the form of the impose boundary conditions than on the Partial Differential Equation
itself.
This makes it even more difficult to develop a unified theoretical edifice. In this section, we discuss some
elementary techniques, which will allow us to obtain general solutions of some simple equations directly.
2.4.1
Integrable equations
We start with an example on first–order Partial Differential Equation, to illustrate how a Partial Differential Equation can be integrated directly. (Not all Partial Differential Equations can be solved in this way,
more advance techniques of solutions are presented in chapter 4).
Example 2.4.1 Find the general solution of the following equation:
(
ux (x, y) = yexy ,
u (0, y) = y 2 + 1.
Solution
We integrate with respect to x (holding y constant).
We obtain u (x, y) = exy + f (y) . The condition u (0, y) = y 2 + 1 implies that 1 + f (y) = y 2 + 1.
Thus f (y) = y 2 . Therefore u (x, y) = exy + y 2 .
■
♦ Please take note of the following:
Whenever integrating with respect to one variable, remember to add
an arbitrary function of the other variables.
Example 2.4.2 Find the general solution of the following equations:
1. uxx = 0.



2. uxy = 0.
uyy = 0
3.
uy (x, 0) = ex


u (x, 0) = sin 2x.
Solution
1. uxx = 0 , we integrate once with respect to x we obtain ux = f (y) .
A second integration with respect to x gives u (x, y) = xf (y) + g (y) .
2. uxy = 0. Integration with respect to x yields uy = f (y) .
Next, we integrate with respect to y, we obtain u (x, y) = F (y) + G (x), where F (y) =


uyy = 0

3.
uy (x, 0) = ex


u (x, 0) = sin 2x.
R
f (y) dy.
A first integration of the Partial Differential Equation with respect to y yields uy (x, y) = f (x). The
condition uy (x, 0) = ex yields ex = f (x).
20
Thus uy (x, y) = ex .
A second integration with respect to y gives u (x, y) = yex + g (x) .
The condition u (x, 0) = sin 2x implies that 0 + g (x) = sin 2x.
Thus, g (x) = sin 2x. Therefore u (x, y) = yex + sin 2x.
2.4.2
■
Reduction of Partial Differential Equations to canonical forms
In this section we will study how to reduce the linear second order equation
Auxx + 2Buxy + Cuyy + Dux + Euy + F u = G
(2.4)
to standard or canonical forms. Before we tackle equation (2.4), let us first examine a special case of
equation (2.4), the Euler’s Equations.
2.4.2.1
Euler’s Equation
The equation
Auxx + 2Buxy + Cuyy = 0
(2.5)
where A, B and C are constants, with A, B or C not zero, is a special case of the general equation (2.4).
If one of the constants A or C is zero, the general solution of (2.5) may be obtained by direct integration
(see section 2.4.1). If A ̸= 0 and C ̸= 0 the general solution of equation (2.5) may be obtained as follows
(please note that a different method is proposed in Section 2.6.2):
1. First, we define two new variables α and β by (the aim is to reduce equation (2.5) to one of the
integrable equations studied in section 2.4.1).
(
α = px + qy
(2.6)
β = rx + sy
where p, q, r and s are arbitrary constants.
2. Next, using the chain rule for partial derivatives we have:
∂u
∂x
∂u
∂y
∂2u
∂x2
=
=
=
=
∂2u
∂y 2
=
=
∂2u
∂x∂y
=
=
∂u ∂α ∂u ∂β
∂u
∂u
+
=p
+r
∂α ∂x ∂β ∂x
∂α
∂β
∂u ∂α ∂u ∂β
∂u
∂u
+
=q
+s
∂α ∂y
∂β ∂y
∂α
∂β
∂ ∂u
∂
∂
∂u
∂u
= p
+r
p
+r
(from (2.7))
∂x ∂x
∂α
∂β
∂α
∂β
∂2u
∂2u
∂2u
p2 2 + 2pr
+ r2 2
∂α
∂α∂β
∂β
∂ ∂u
∂
∂
∂u
∂u
= q
+s
q
+s
(from (2.8))
∂y ∂y
∂α
∂β
∂α
∂β
∂2u
∂2u
∂2u
+ s2 2
q 2 2 + 2sq
∂α
∂α∂β
∂β
∂ ∂u
∂
∂
∂u
∂u
= p
+r
q
+s
(from (2.7) and (2.8))
∂x ∂y
∂α
∂β
∂α
∂β
∂2u
∂2u
∂2u
pq 2 + (ps + rq)
+ rs 2 .
∂α
∂α∂β
∂β
(2.7)
(2.8)
(2.9)
(2.10)
(2.11)
21
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Substitution of (2.7) – (2.11) into (2.5) yields
∂2u
∂2u
+
2
Apr
+
Csq
+
B
(rq
+
sp)
+
∂α2
∂α∂β
∂2u
= 0.
+ Ar2 + 2Brs + Cs2
∂β 2
Ap2 + 2Bpq + Cq 2
(2.12)
3. Then we choose arbitrary constants p, q, r and s such that p = r = 1 and such that q and s are the
roots λ1 and λ2 of the equation:
A + 2Bλ + Cλ2 = 0;
(2.13)
so that the coefficients of
∂2u
∂2u
and
become zero, consequently equation (2.12) becomes
2
∂α
∂β 2
[A + B (λ1 + λ2 ) + Cλ1 λ2 ]
Since
λ1 + λ2 = −
and
λ1 λ2 =
∂2u
= 0.
∂α∂β
2B
C
A
.
C
(2.14)
(2.15)
(2.16)
Equation (2.14) can be written as
∂2u
2
B 2 − AC
= 0.
C
∂α∂β
(2.17)
Hence, provided that equation (2.5) is not parabolic (i.e. B 2 − AC ̸= 0), equation (2.17) becomes
∂2u
= 0.
∂α∂β
(2.18)
Equation (2.18) may be integrated to give: (see example 2.4.2.(2.))
u = F (α) + G (β)
(2.19)
where F and G are arbitrary functions twice differentiable. Substitution of (2.6) into (2.19) yields
the solution
u (x, y) = F (x + λ1 y) + G (x + λ2 y)
(2.20)
where λ1 and λ2 are the roots of (2.13).
4. In a similar way, if we choose arbitrary constants p, q, r and s such that q = s = 1 and such that p
and r are the roots λ1 and λ2 of the equation (A ̸= 0).
Aλ2 + 2Bλ + C = 0.
So that the coefficients of
(2.21)
∂2u
∂2u
and
become zero, following the steps above ( equation (2.14) –
∂α2
∂β 2
(2.19)), we obtain
u = F (α) + G(β).
(2.22)
22
where F and G are arbitrary functions twice differentiable. Substitution of (2.6) into (2.22) yields
the solution
u (x, y) = F (λ1 x + y) + G (λ2 x + y)
(2.23)
where λ1 and λ2 are the roots of (2.21).
♦ In summary
When solving equation (2.5) with B 2 − AC ̸= 0, there are two possibilities:
Case 1: B 2 − AC > 0 (Hyperbolic equation)
The quadratic equation (choose one) has
A + 2Bλ + Cλ2 = 0
Solve for λ :
√
−B + B 2 − AC
λ1 =
and
√C
2
−B − B − AC
λ2 =
C
u (x, y) = F (x + λ1 y) + G (x + λ2 y)
F and G arbitrary C 2 functions
Aλ2 + 2Bλ + C = 0
Solve for λ :
√
−B + B 2 − AC
λ1 =
and
√A
2
−B − B − AC
λ2 =
A
u (x, y) = F (y + λ1 x) + G (y + λ2 x)
F and G arbitrary C 2 functions
or
Example 2.4.3 Find the general solution of the following equations:
1. uxx −
1
u
c2 tt
2. uxx + uxy − 2uyy = 0
=0
Solution
1. Quadratic equation
1 − c12 λ2 = 0
λ = ±c. Thus
u (x, y) = F (x + ct) + G (x − ct)
or
λ2 − c12 = 0
λ = ± 1c . Thus
u (x, y) = F t + 1c x + G t − 1c x
2. Quadratic equation
1 + λ − 2λ2 = 0
−1 + 3
1
λ1 =
=−
−4
2
−1 − 3
λ2 =
=1
−4
u (x, y) = F x − 21 y + G (x + y)
or
λ2 + λ − 2 = 0
−1 + 3
λ1 =
=1
2
−1 − 3
λ2 =
= −2
2
u (x, y) = F (y + x) + G (y − 2x)
■
Case 2: B 2 − AC < 0 (Elliptic equation)
The quadratic equation A + 2Bλ + Cλ2 = 0 or Aλ2 + 2Bλ + C = 0 has two distinct imaginary roots
λ1 and λ2 = λ1 .
23
APM3701/1
In this case we have
For A + 2Bλ + Cλ2 = 0
u (x, y) = F (x + λ1 y) + G x + λ1 y
where F and G are arbitrary C 2
functions, and λ1 and λ2 are
solutions of the equation
A + 2Bλ + Cλ2 = 0.
For Aλ2 + 2Bλ + C = 0
u (x, y) = F (y + λ1 x) + G y + λ1 x
where F and G are C 2 arbitrary
functions, and λ1 and λ2 are
solutions of the equation
Aλ2 + 2Bλ + C = 0.
or
Example 2.4.4 Find the general solution of the following equations.
1. uxx + uyy = 0
2. 2uxx + uxy + 4uyy = 0
Solution
1. Quadratic equation
1 + λ2 = 0 yields
λ = ±i. Thus
F (x, y) = F (x + iy) + G (x − iy)
or
λ2 + 1 = 0 yields
λ = ±i. Thus
u (x, y) = F (y + ix) + G (y − ix) .
2. Quadratic equation
2 + λ + 4λ2 = 0
or
√
λ1,2 =
−1±i 31
8
Thus u (x, y) = F x +
√
+G x − 1+i8 31 y
√
−1+i 31
y
8
2λ2 + λ + 4 = 0
λ1,2 =
√
−1±i 31
4 u (x, y) = F y + −1+i4
√
+G x − 1+i4 31 x
√
31
x
■
Case 3: B 2 − AC = 0 (Parabolic equations)
In deriving (2.19) and (2.22) we assumed that the equation (2.4) was not parabolic. In the case where
(2.4) is parabolic, equation (2.17) is satisfied identically, since B 2 − AC = 0.
Under these circumstances we go back to equation (2.12) and as before we set p = 1, q, r and s arbitrary.
We obtain
∂2u
∂2u
+
2
[Ar
+
Csq
+
B
(rq
+
s)]
+
∂α2
∂α∂β
∂2u
Ar2 + Cs2 + 2Brs
= 0.
∂β 2
A + 2Bq + Cq 2
(2.24)
If we choose q such that it is the solution of
A + 2Bq + Cq 2 = 0
(2.25)
then, since B 2 − AC = 0 by assumption
q=−
B
(twice).
C
(2.26)
24
Therefore, the first term of (2.24) is zero by virtue of (2.25). Using (2.26), the second term of (2.24) is
also zero, since
Br
Ar + Csq + B (rq + s) = Ar − Bs + B −
+s
(using(2.26))
C
r
= − B 2 − AC
= 0.
C
Consequently, provided that r and s are not both zero, equation (2.24) becomes
∂2u
= 0.
∂β 2
(2.27)
Integrating directly we obtain (see example 2.4.2(1.))
u = F (α) + βG (α)
where F and G are arbitrary C 2 -functions. From (2.6), since p = 1, q = −
(
(2.28)
B
≡ λ, we have
C
α = x + λy
β = rx + sy
where r and s are arbitrary (but not both zero). For simplicity, we choose r = 0, s = 1 we have
(
α = x + λy
β = y.
(2.29)
Thus,
u (x, y) = F (x + λy) + yG (x + λy) .
(2.30)
Similarly, if we choose s = 1, p, q and r arbitrary, we obtain
Ap2 + 2Bpq + Cq 2
∂2u
∂2u
∂2u
2
+
2
[Apr
+
Cq
+
B
(rq
+
p)]
+
Ar
+
2Br
+
C
= 0.
∂α2
∂α∂β
∂β 2
If we choose r such that it is the solution of
Ar2 + 2Br + C = 0
(2.31)
then, since B 2 − AC = 0 by assumption,
r=−
B
(twice).
A
(2.32)
Therefore, the first term of (2.31) is zero by virtue of (2.31). Using (2.32), the second term of (2.31) is
also zero, since
q
= 0. (Using(2.32))
Apr + Cq + B (rq + p) = − B 2 − AC
A
Consequently, provided that q and p are not both zero, equation (2.31) becomes
∂2u
= 0.
∂α2
Integrating directly we obtain
u = F (β) + αG (β)
25
APM3701/1
where F and G are arbitrary C 2 − functions. From (2.6), since s = 1, r = −
(
B
≡ λ, we have
A
α=x
β = λx + y
Thus,
u (x, y) = F (λx + y) + xG (λx + y) .
where λ is the double root of
Aλ2 + 2Bλ + C = 0.
(2.33)
Example 2.4.5 Find the general solution of the equation 4uxx + 4uxy + uyy = 0.
Solution
Quadratic equation
4 + 4λ + λ2 = 0
λ = − 24 = −2. Thus
u (x, y) = F (x − 2y) + yG (x − 2y)
or
4λ2 + 4λ + 1 = 0
λ = − 48 = − 12 . Thus
u (x, y) = F y − 21 x + xG y − 12 x
■
2.4.3
The General form
The general form (2.4) can be reduced to standard or canonical forms by performing suitable transformations. If this were possible and the canonical forms were closely related to the two–dimensional
forms of the standard equations of Mathematical Physics, a study of the latter equations would, to a great
extend, be sufficient in determining the properties of the more general equations and their solutions. More
precisely:
The general equation (2.4) with constant coefficients can be reduced to one of the following so–called
canonical forms:
∂2v
∂2v
+
+ γv + f1 (x, y) = 0 Elliptic type
(2.34)
∂α2 ∂β 2
or

∂2v
∂2v

−
+ γv + f1 (x, y) = 0 

∂α2 ∂β 2
Hyperbolic type
(2.35)
∂2v



+ γv + f1 (x, y) = 0
∂α∂β
or

∂2v
∂v

+ b2
+ f1 (x, y) = 0 

∂α2
∂β
Parabolic type
(2.36)
∂2v
∂v


+ b2
+ f1 (x, y) = 0 
∂α
∂β 2
where γ and b2 are constants depending on A, B, C, D, E and F. The transformations to be used are of
the form
(
α = x + λ1 y
(2.37)
β = x + λ2 y
26
with λ1 and λ2 are solutions of A + 2λB + Cλ2 = 0, for Elliptic and Hyperbolic type, and
(
α = x + λ̄y
β = y or β = x
(2.38)
where λ̄ is the double root of A + 2λB + Cλ2 = 0, for parabolic type. We introduce the new variable v
defined by
u = eτ α+µβ v (α, β)
(2.39)
where τ and µ are constants which have to be determined.
Example 2.4.6 Obtain the general solution of the equation
uxx − uyy + 3ux − 2uy + u = 0.
(2.40)
Solution
The equation is hyperbolic, B 2 − AC = 1 > 0. A + 2Bλ + Cλ2 = 0 becomes 1 − λ2 = 0 gives λ1 = 1 and
λ2 = −1. Thus, we use the transformation α = x + y and β = x − y. We have
ux = uα αx + uβ β x = uα + uβ
uy = uα αy + uβ β y = uα − uβ
uxx = (ux )x = (uα + uβ )x = (uα + uβ )α + (uα + uβ )β = uαα + 2uαβ + uββ
uyy = (uy )y = (uα − uβ )y = (uα − uβ )α − (uα − uβ )β = uαα − 2uαβ + uββ .
Now equation (2.40) becomes
4uαβ + uα + 5uβ + u = 0.
(2.41)
To get rid of first derivatives we introduce the following transformation
u = veαµ+ηβ
so,
uα = vα eµα+ηβ + µveµα+ηβ
uβ = vβ eµα+ηβ + ηveµα+ηβ
uαβ = vαβ eµα+ηβ + ηvα eµα+ηβ + µvβ eµα+ηβ + µηveµα+ηβ .
Equation (2.41) becomes
[4vαβ + (4η + 1) vα + (4µ + 5) vβ + (4µη + µ + 5η + 1) v] eµα+ηβ = 0
∴ 4vαβ + (4η + 1) vα + (4µ + 5) vβ + (4µη + µ + 5η + 1) v = 0.
In order to get rid of the first derivatives we must have
1
4
4η + 1 = 0
=⇒
η=−
4µ + 5 = 0
=⇒
5
µ=− .
4
and
(2.42)
27
APM3701/1
Thus, equation (2.42) becomes
1
4vαβ − v = 0.
4
We let v (α, β) = X (α) Y (β) . Equation (2.43) becomes
(2.43)
1
4X ′ Y ′ − XY = 0
4
dividing through by XY ′ , we obtain:
4
X′
1Y
=
X
4Y′
or
4
X′
1Y
=
=λ,
X
4Y′
λ ≡ constant (why? see sections on separation of variables in chapter 4)
which yields the solution
X = exp
λα
4
and
Hence
v (α, β) = exp(
Y = exp
1
β .
4λ
λα
1
+
β).
4
4λ
Consequently
u (α, β) = exp
λα
1
1
+
β exp − (5α + β) .
4
4λ
4
Finally
u (x, y) = exp
λ
1
3
+
−
4 4λ 2
x+
λ
1
−
−1 y ,
4 4λ
λ ≡ constant.
■
Exercises 2.4 .
1. Find the general solution of each of the following equations by direct integration:
(a) ux (x, y) = x3 + 3y 2
(c) uxyz (x, y, z) = ex + xy
(b) uxy (x, y) = xy 3
(d) uxtt (x, t) = exp(2x + 3t)
(e) uyy (x, y) − x2 u = 0
2. Find the general solution to the following equation.



(a) uxx + yux = 0
uxy + ux
(c)
uy (0, y) + u (0, y)


u (x, 0)
(b) xuxy + uy = ex+y
=
=
=
0
y + ey
x2
3. Obtain the general solution of the following equations:
(a) uxx − 4uxy − 3uyy = 0
(c) uxx − 4uxy + 3uyy = 0
(b) uxx + 4uyy = 0
(d) 2uxx + uyy = 0
(e) 4uxx + 4uxy + uyy = 0.
4. Obtain the general solution of
rutt − c2 rurr − 2c2 ur = 0
[Hint: Let v = ru].
(c = constant)
28
5. A function u (r, t) satisfies the equation
1 ∂
r2 ∂r
1 ∂2u
2 ∂u
r
= 2 2,
∂r
c ∂t
where c is constant. By introducing the new dependent variable v (r, t) = ru (r, t). Obtain the general
solution of the Partial Differential Equation.
6. Determine the general solution of
uxxxx − 2uxxyy + uyyyy = 0
[Hint: Let z = x + iy].
7. Use transformations of the form
(
α = x + λ1 y
β = x + λ2 y
and
v = ue−(aα+bβ)
a, b to be determined, to reduce the following equations to the form vαβ = cv, (c = constant) and
then obtain the general solution by separation of variables (see example 2.4.6).
(a) uxx − uyy − 5ux + 2uy + 3u = 0
2.5
(b) 3uxx + 7uxy + 2uyy + uy + u = 0.
■
D’ALEMBERT SOLUTION OF THE WAVE EQUATION
Consider the one–dimensional wave equation (see section 4.6)
utt = c2 uxx ,
−∞ < t, x < ∞
(2.44)
where u = u (x, t), and c is a constant called the wave speed, subject to the Cauchy initial conditions
u (x, 0) = f (x)
(2.45)
ut (x, 0) = g (x) .
(2.46)
The general solution of (2.44) (see example 2.4.3(a)) is:
u (x, t) = F (x + ct) + G (x − ct)
(2.47)
where F and G are arbitrary twice differentiable functions of their arguments.
From (2.45) and (2.46), we have (take t = 0 in (2.47))
where the primes refer to derivatives.
F (x) + G (x) = f (x)
(2.48)
cF ′ (x) − cG′ (x) = g (x)
(2.49)
29
APM3701/1
Integration of (2.49) gives
1
F (x) − G (x) =
c
x
Z
g (τ ) dτ
(2.50)
a
where the constant of integration has been incorporated in the lower limit by introducing an arbitrary
constant a. We add (2.48) and (2.50), we obtain
Z
1
1 x
F (x) = f (x) +
g (τ ) dτ
2
2c a
and subtract (2.50) from (2.48), we obtain
1
1
G (x) = f (x) −
2
2c
Thus
and
x
Z
g (τ ) dτ .
a
1
1
F (x + ct) = f (x + ct) +
2
2c
Z
1
1
G (x − ct) = f (x − ct) −
2
2c
Z
x+ct
g (τ ) dτ
(2.51)
g (τ ) dτ .
(2.52)
a
x−ct
a
Therefore, adding (2.51) and (2.52) we obtain the general solution of (2.44)–(2.46)in the form:
u (x, t) =
1
1
[f (x + ct) + f (x − ct)] +
2
2c
Z
x+ct
g (τ ) dτ .
(2.53)
x−ct
Formula (2.53) is known as d’Alembert solution of the wave equation.
Example 2.5.1 Use d’Alembert’s solution of the wave equation to find the solution of
4uxx = utt
u (x, 0) = 0
1
∂u
=
.
∂t t=0
1+x
Solution
Here c = 2, f (x) = 0, g (x) =
u (x, t) =
u (x, t) =
1
. Applying (2.53) we obtain
1+x
Z
1 x+2t 1
1
1
[0 + 0] +
dτ = ln (τ + 1)
2
4 x−2t 1 + τ
4
1 x + 2t + 1
ln
.
4 x − 2t + 1
x+2t
=
x−2t
1
ln (x + 2t + 1) − ln (x − 2t + 1)
4
■
Exercises 2.5
1. Use d’Alembert’s solution of the wave equation to determine the solution of the following initial–value
problems:
30
(a)
(b)
(c)
uxx −
1
u
c2 tt
=0
u (x, 0) = 0
ut (x, 0) = 1
uxx = 2uyy
u (0, y) = y 2
ux (0, y) = y
utt = 4uxx
u (x, 0) = ln 1 + x
2
(d)
utt − uxx = x + t
(e)
utt − c2 uxx = 2
(f )
ut (x, 0) = 2
u (x, 0) = x
ut (x, 0) = sin x
u (x, 0) = x2
ut (x, 0) = cos x
utt − c2 uxx = xex
u (x, 0) = sin x
ut (x, 0) = 0.
2. The vibration of a string of unit π, is described by u (x, t) , use d’Alembert’s solution of the wave
equation to solve the following equations with the given wave speed c, and subjet to the given initial
position and velocity.
(f ) u (x, 0) = π, ut (x, 0) = cos πx, c = 1.
(a) u (x, 0) = cos πx, ut (x, 0) = π, c = 1.
(b) u (x, 0) = sin πx cos π, ut (x, 0) = 0, c =
1
.
π
(c) u (x, 0) = sin πx + π sin 3πx,
1
ut (x, 0) = 0, c = .
π
(d) u (x, 0) = 0, ut (x, 0) = cos x, c = π.
(g) u (x, 0) = 0, ut (x, 0) = sin πx cos π, c =
(h) u (x, 0) = 0,
ut (x, 0) = sin πx + π sin 3πx, c =
(i) u (x, 0) = −π, ut (x, 0) = 0, c = π.
(e) u (x, 0) = 0, ut (x, 0) = −π, c = π.
(j) u (x, 0) = 0, ut (x, 0) = 1, c = 1.

1


2x
if 0 ≤ x ≤ ,


2
1
(k) u (x, 0) =
<
x
≤
1, ut (x, 0) = 0, c = 1.
2
(1
−
x)
if


2


0
if 1 < x ≤ π,
3. Use the solution obtained in Exercise 2. (k) above, and answer the following questions:
1
(a) plot the string at times t = 0, , 1.
2
1
(b) For t = , identify the points on the string that are still in rest position.
2
4. Show that if the initial velocity is zero, the solution u (x, t) of the wave equation
utt = c2 uxx ,
x ∈ [0, L] , −∞ < t < ∞
u (0, t) = 0 = u (L, t) ,
u (x, 0) = f (x) , ut (x, 0) = 0,
is given by
∞
u (x, t) =
nπx
1X nπx
an sin
(x − xt) + sin
(x + ct) .
2
L
L
n=1
nπx
.
L
Hint: Using the Fourier series of f (x) equation (3.8), (see section 3.5).
where an =
RL
0
f (x) sin
1
.
π
1
.
π
31
2.6
2.6.1
APM3701/1
METHOD OF CHARACTERISTIC CURVES FOR PARTIAL DIFFERENTIAL
EQUATIONS
First order linear equation
The method of characteristics [1] can be used to solve first order linear differential equations with nonconstant coefficients of the form
∂u
∂u
p (x, y)
+ q (x, y)
= r (x, y)
(2.54)
∂x
∂y
where p (x, y) , q (x, y) and r (x, y) are function of x and y, by reducing it to a solution of an ordinary
differential equation.
We note that, if u (x, y) is a solution of (2.54), then
(p (x, y) , q (x, y) , r (x, y)) · (ux (x, y) , uy (x, y) , −1) = 0.
(2.55)
From second year multi-variable calculus, we know that the normal to the surface S = {(x, y, u (x, y)) |x, y ∈ R}
at the point (x, y, u (x, y)) is given by ⃗n (x, y) = (ux (x, y) , uy (x, y) , −1) . Hence, from equation (2.55),
we see that the vector (p (x, y) , q (x, y) , r (x, y)) is perpendicular to the normal (ux (x, y) , uy (x, y) , −1) at
each point (x, y, u (x, y)) . Thus, the vector (p (x, y) , q (x, y) , r (x, y)) lies in the tangent plan to S.
To construct a such surface we follow the following steps:
1. first, we find a curve C which lies in S at each point (x, y, u (x, y)) ;
2. next, at each point on the curve C, we use a parametrisation s for C such that the vector
(p (x (s) , y (s)) , q (x (s) , y (s)) , r (x (s) , y (s))) is tangent to C.
n
o
3. then, we consider the particular curve C = x (s) , y (s) , z (s) , which satisfies the system
dx
ds
dy
ds
dz
ds
= p (x (s) , y (s))
(2.56)
= q (x (s) , y (s))
(2.57)
= r (x (s) , y (s)) .
(2.58)
The curve C sastifying (2.56)–(2.58) is know as an integral curve for the vector field (p (x, y) , q (x, y) .r (x, y))
The integral curves (2.56)–(2.58), associated with the partial differential equation (2.54) are known
as the characteristic equations for the equation (2.54).
4. By solving the the system (2.56)–(2.58), at each point (x, y, z) , the surface S is determined such
that the vector field V = (p (x, y) , q (x, y) , r (x, y)) lies in the tangent plane to S. The surface S is
known as an integral surface for V.
5. Finally, the solution of the PDE (2.55) is obtained by putting together the characteristic curve
obtained in step (4.) above.
Example 2.6.1 Find a solution of the equation
∂u
∂u
+ q (x, y)
= 0.
∂x
∂y
(2.59)
32
Solution
The characteristics are given by
dx
ds
dy
ds
dz
ds
= 1
= q (x, y)
= 0,
R
with solution x (s) = s + C1 , y (s) = η (x (s) , y (s)) + C2 , z (s) = C, where η (x (s) , y (s)) = q (x, y) ds.
Eliminating s, and yields y − η (x, y) = C2 or ϕ (x, y) = C, and z = C3 .
If S is the integral surface formed by these characteristic curves, we note that z (x, y) is constant along
ϕ (x, y) = C. Thus z (x, y) = F (ϕ (x, y)) , letting u (x, y) = z (x, y) = F (C) , where F is an arbitrary C 1
function. Therefore the solution of (2.59) is given by
u (x, y) = F (ϕ (x, y)) .
(2.60)
■
Remark 2.6.1 We note that since x = s + C1 , ϕ (x, y) is simple the solution of the characteristic
dy
= q (x, y) .
dx
(2.61)
Hence, in this case, the method of characteristic curves reduces the partial differential equation (2.59)
to an ordinary differential equation (2.61). Thus, to find the solution of the partial differential equation
(2.59) all we need to do is to solve the system of ordinary differential equations (2.61).
Example 2.6.2 Solve the following partial differential equations by method of characteristic curves:
∂u
∂u
∂u
∂u
+x
= 0. (b) x
+ y2
= 0.
(a) 2
∂x
∂y
∂x
∂y
Solution
∂u
x ∂u
x
(a) We rewrite the equation in the form
+
= 0. So, we have q (x, y) = . We need to solve
∂x
2 ∂y
2
dy
x
2
x2
x2
= , which yields the solution y = 4 + C, or y − 4 = C. Thus ϕ (x, y) = y − x4 , therefore the
dx
2
2
solution is u (x, y) = F y − x4 . Please verify by substituting in the equation.
∂u y 2 ∂u
y2
dy
y2
dy
dx
+
= 0, q (x, y) = . Solving
=
=⇒ 2 =
.
∂x
x ∂y
x
dx
x
y
x
Integration gives −y −1 = ln x + C, or y −1 + ln x = −C. Thus ϕ (x, y) = y −1 + ln x, and the solution
is u (x, y) = F y −1 + ln x . Please verify by substituting in the equation.
(b) We rewrite the equation in the form
■
Example 2.6.3 Find the solution of initial (boundary) problem
(
ux + q (x, y) uy = 0
u (0, y) = φ (y) .
(2.62)
33
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Solution
We know from example 2.6.1 and equation (2.60), that the solution of
ux + q (x, y) uy = 0
dy
given by the solution of the characteristic
= q (x, y) , which has solution y = η (x, y) + C. Thus the
dx
solution is given by u (x, y) = F (ϕ (x, y)) , where ϕ (x, y) = y − η (x, y) .
The initial/boundary condition u (0, y) = φ (y) = F (ϕ (0, y)) . Thus
u (x, y) = φ (ϕ (x, y)) .
(2.63)
■
Example 2.6.4 Solve the initial (boundary) problem
(
ux + (4x − 1) uy = 0
u (0, y) = ln y.
Solution
dy
= 4x − 1 is y = 2x2 − x + C. Thus
Here φ (y) = ln y, q (x, y) = 4x − 1. The solution of the characteristic
dx
ϕ (x, y) = y − 2x2 + x. Therefore (from Example 2.6.3), we have, u (x, y) = φ (ϕ (x, y)) = ln y − 2x2 + x .
■
Example 2.6.5 Solve the following partial differential equations by method of characteristic curves:
1. 2
∂u
∂u
+x
= 0.
∂x
∂y
2. x
∂u
∂u
+ y2
= 0.
∂x
∂y
Solution
∂u
x ∂u
x
1. We rewrite the equation in the form
+
= 0. So, we have q (x, y) = . We need to solve
∂x
2 ∂y
2
dy
x
2
x2
x2
= , which yields the solution y = 4 + C, or y − 4 = C. Thus ϕ (x, y) = y − x2 , therefore the
dx
2
2
solution is u (x, y) = F y − x2 . Please verify by substituting in the equation.
∂u y 2 ∂u
y2
dy
y2
dy
dx
+
= 0, q (x, y) = . Solving
=
=⇒ 2 =
.
∂x
x ∂y
x
dx
x
y
x
Integration gives −y −1 = ln x + C, or y −1 + ln x = −C. Thus ϕ (x, y) = y −1 + ln x, and the solution
is u (x, y) = F y −1 + ln x . Please verify by substituting in the equation.
2. We rewrite the equation in the form
■
Example 2.6.6 Find the solution of initial (boundary) problem
(
ux + q (x, y) uy = 0
u (0, y) = φ (y) .
Solution
We know from example 2.6.1 and equation (2.60), that the solution of
ux + q (x, y) uy = 0
(2.64)
34
dy
given by the solution of the characteristic
= q (x, y) , which has solution y = η (x, y) + C is given by
dx
u (x, y) = F (ϕ (x, y)) , where ϕ (x, y) = y − η (x, y) . The initial/boundary condition u (0, y) = φ (y) =
F (ϕ (0, y)) . Thus
u (x, y) = φ (ϕ (x, y)) .
(2.65)
■
Example 2.6.7 Solve the initial (boundary) problem
(
ux + (4x − 1) uy = 0
u (0, y) = ln y.
Solution
dy
= 4x − 1 is y = 2x2 − x + C.
Here φ (y) = ln y, q (x, y) = 4x − 1. The solution of the characteristic
dx
Thus ϕ (x, y) = y − 2x2 + x. Therefore u (x, y) = φ (ϕ (x, y)) = ln y − 2x2 + x .
■
Example 2.6.8
2.6.2
Second order linear equation
Consider the Euler equation (2.5), with variable coefficients
A (x, y) uxx + 2B (x, y) uxy + C (x, y) uyy = 0.
(2.66)
By using the change of variables
α = α (x, y) and β = (x, y) ,
(2.67)
we have
ux = αx uα + β x uβ
uy = αy uα + β y uβ
uxx = (ux )x = (αx uα + β x uβ )x = αxx uα + αx (uα )x + β xx uβ + β x (uβ )x
= αxx uα + αx (αx uαα + β x uαβ ) + β xx uβ + β x (αx uαβ + β x uββ )
= αxx uα + α2x uαα + 2αx β x uαβ + β xx uβ + β 2x uββ
uxy = (ux )y = (αx uα + β x uβ )y = αx (uα )y + β x (uβ )y + αxy uα + β xy uβ
uyy
= αxy uα + αx αy uαα + αx β y uαβ + β xy uβ + αy β x uαβ + β x β y uββ
= αx αy uαα + αx β y + αy β x uαβ + β x β y uββ + αxy uα + β xy uβ
= (uy )y = αy uα + β y uβ y = αy (uα )y + β y (uβ )y + αyy uα + β yy uβ
= α2y uαα + 2αy β y uαβ + β 2y uββ + αyy uα + β yy uβ
Substitution into equation (2.66) yields
Āuαα + 2B̄uαβ + C̄uββ + K̄ (α, β, u, uα , uβ ) = 0,
(2.68)
35
APM3701/1
where
Ā = Aα2x + 2Bαx αy + Cα2y ,
(2.69)
B̄ = Aαx β x + B αx β y + αy β x + Cαy β y ,
(2.70)
C̄ = Aβ 2x + 2Bβ x β y + Cβ 2y ,
(2.71)
K̄ (α, β, u, uα , uβ ) = A (αxx uα + β xx uβ ) + 2B αxy uα + β xy uβ + C αyy uα + β yy uβ
(2.72)
The discriminant of the equation (2.68) is given by
¯ = B̄ 2 − 4ĀC̄.
∆
(2.73)
It can be proven [4] that if the change of variables (2.67) is nondegenerate, i.e Jacobian J of the system
(2.67) is not zero (J ̸= 0), then the type of the equation (2.66) remains unchanged, according to the
classification (2.3).
One observes that the coefficients Ā and C̄ in equations (2.69) and (2.71) respectively, have the same form,
except (2.69) is an equation in α, whereas (2.71) is an equation in β. In order to integrate equation (2.68),
we can choose a change of variables such that the coefficients Ā and C̄ in (2.68), both vanish.
Equations (2.69) and (2.71) are then equivalent to the equation
Aη 2x + 2Bη x η y + Cη 2y = 0,
(2.74)
where η stands for α or β. The solutions of the equation (2.74) are called the characteristic curves of
the Euler equation (2.66). By dividing both sides of (2.74) by η 2y , we get
A
ηx
ηy
2
+ 2B
ηx
ηy
+ C = 0.
To obtain the solution of (2.66), we consider the following cases:
Case 1:
Constant coefficients A, B and C.
1. ∆ = B 2 − AC ̸= 0 (Hyperbolic and elliptic equations)
A general solution of (2.66) is obtained by method (a) or (b) below:
(2.75)
36
curves are given by
(a) We assume that A ̸= 0, but if A = 0 and
√
C ̸= 0, we can proceed in a similar way,
dy
B ± B 2 − AC
=
.
(2.77)
but this time dividing both sides of (2.74)
dx
A
ηy
2
instead
by η x and considering the ratio
ηx
(b) Alternatively, if we assume that C ̸= 0,
(see point 1.(b) and equation (2.78) bethe solution of equation (2.68) can be oblow). In the case where both A = C = 0,
tained as follow. We divide both sides of
then the equation (2.66) is already in the
(2.74) by η 2x , we get
reduced form, the equation is parabolic,
2
ηy
ηy
see case 1.(b) below. We note that the
+C
= 0. (2.78)
A + 2B
ηx
ηx
characteristic curves we are interested in
are the curves η (x, y) =constant, for exThe differentiation along those curves,
ample, we can take α (x, y) = α0 . Differgives
entiating along those curves, we get
dη = η dx + η dy = 0.
x
dη = η x dx + η y dy = 0.
which gives the slope of the characteristics
curve
η
dy
= − x.
dx
ηy
Substitution in equation (2.75) yields the
equation for the slope of the characteristic
curve
2
dy
dy
A
− 2B
+ C = 0. (2.76)
dx
dx
Depending on the sign of the discriminant
∆ = B 2 − AC, the quadratic equation
(2.76) has two, one, or no real solutions.
Thus, the solutions of the characteristic
or
y
ηy
dx
=− .
dy
ηx
Substitution in equation (2.78) yields the
equation for the slope of the characteristic
curve
2
dx
dx
A − 2B
+C
= 0.
dy
dy
Depending on the sign of the discriminant
∆ = B 2 − AC, the quadratic equation
(2.78) has two, one, or no real solutions.
Thus, the solutions of the characteristic
curves are given by
√
B ± B 2 − AC
dx
=
.
(2.79)
dy
C
37
• Similarly, if we use equation (2.79), we
have
√
B ± B 2 − AC
y − x = K,
(2.84)
C
• If we use equation (2.77), we have
√
B ± B 2 − AC
x + K,
y=
A
or
B±
√
B 2 − AC
x − y = K.
A
APM3701/1
(2.80)
and we let
α=
We let
α=
B+
and
β=
B−
√
√
B 2 − AC
x−y
A
(2.81)
(2.82)
Substitution into equation (2.68) gives
2
B − AC
−2
uαβ = 0.
A
Integration of the above equation yields
the solution
√
B 2 − AC
y−x
C
(2.85)
B 2 − AC
y − x.
C
(2.86)
and
β=
B 2 − AC
x − y.
A
B+
B−
√
Substitution into equation (2.68) gives
−2
B 2 − AC
C
uαβ = 0.
Integration of the above equation yields
the solution
u (α, β) = F (α) + G (β) .
u (α, β) = F (α) + G (β) .
Thus
Thus
u (x, y) = F (λ1 x − y) + G (λ2 x − y) ,
(2.83)
√
2
B + B − AC
,
where λ1 =
A
√
B − B 2 − AC
λ2 =
, F and G are C 2
A
functions of their arguments.
u (x, y) = F (λ1 y − x) + G (λ2 y − x) ,
(2.87)
√
2
B + B − AC
where λ1 =
,
C
√
B − B 2 − AC
λ2 =
, F and G are C 2
C
functions of their arguments.
Remark 1 Note that the solutions (2.83) and (2.87) are similar to the solutions obtained in
(2.20) and (2.23) with the difference that the solutions of (2.83) and (2.87) is obtained from
the equations
Aλ2 − 2B + C = 0 and A − 2B + Cλ2 = 0,
respectively, whereas the solutions of (2.20) and (2.23) are obtained from the equations
Aλ2 + 2Bλ + C = 0 and A + 2Bλ + Cλ2 = 0,
respectively. Beware that the difference is in the change in the sign of ”2B”
Example 2.6.9 Find a general solution of the following PDEs:
38
(a) 2uxx − 6uxy + 4uyy = 0
(c) uxx + 4uxy + 3uyy = cuy + dux + f (x, y) .
(b) uxx + 2uxy + 5uyy = 0
Solution
(a) A = 2, B = −3, C = 4 =⇒ B 2 − AC = 1 > 0, the equation is hyperbolic.
Alternatively, using formula (2.87),
 we get
√
1
B ± B 2 − AC
−3 ± 1  −
λ1,2 =
=
2
 −1,
C
4
Formula (2.83) gives
(
√
−1
B ± B 2 − AC
−3 ± 1
λ1,2 =
=
A
2
−2.
Thus,
giving the solution
u (x, y) = F (λ1 x − y) + G (λ2 x − y)
= F (−x − y) + G (−2x − y) .
u (x, y) = F (λ1 y − x) + G (λ2 y − x)
1
= F − y − x + G (−y − x) .
2
(b) A = 1, B = 1, C = 5 =⇒ B 2 − AC = −4 > 0, the equation is elliptic.
Alternatively, using formula (2.87), we get
Formula (2.83) gives
λ1,2
√
B 2 − AC
1 ± 2i
=
=
A
1
(
1 + 2i
=
1 − 2i.
B±
λ1,2 =
=





Thus,
√
1 ± 2i
B 2 − AC
=
C
5
1 + 2i
5
1 − 2i
,
5
B±
yields the solution
u (x, y) = F (λ1 x − y) + G (λ2 x − y)
= F ((1 + 2i) x − y)
+G ((1 − 2i) x − y) .
u (x, y) = F (λ1 x − y) + G (λ2 x − y)
1 + 2i
y−x
= F
5
1 − 2i
+G
y−x .
5
(c) uxx + 4uxy + 3uyy = uy + ux .
A = 1, B = 2, C = 3 =⇒ B 2 − AC = 1 > 0, the equation is hyperbolic. Since the
equation is not Euler, we define the characteristic curves by the formulae (2.85)–(2.86)
√
B + B 2 − AC
x−y
α=
A
and
β=
B−
√
B 2 − AC
x − y,
A
we get
α = 3x − y
β = x − y.
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APM3701/1
We have
αx = 3, αy = −1, β x = 1, β y = −1
ux = αx uα + β x uβ = 3uα + uβ
uy = αy uα + β y uβ = −uα − uβ
uxx = (ux )x = (3uα + uβ )x = 3 (3uα + uβ )α + (3uα + uβ )β
= 9uαα + 6uαβ + uββ
uxy = (ux )y = (3uα + uβ )y = − (3uα + uβ )α − (3uα + uβ )β
= −3uαα − 4uαβ − uββ
uyy = (uy )y = (−uα − uβ )y = − (−uα − uβ )α − (−uα − uβ )β
= uαα + 2uαβ + uββ .
Substituting into the PDE yields
−4uαβ = 2uα .
Which gives (integrating twice and using Integrating Factor)
u = e−1/2β (f (α) + g (β)) . [Workout the details.]
Thus,
u (x, y) = e−1/2(x−y) (f (3x − y) + g (x − y)) ,
where f and g are C 2 functions.
As an exercise, use formulae (2.85)–(2.86) to determine a general solution.
■
2. ∆ = B 2 − AC = 0 (Parabolic equation)
In this case, the equation (2.75) has only one solution, thus, the equation (2.68) has one set of
characteristic curves, so there will be no change of variables, if Ā = C̄ = 0. To get a solution of
( 2.68), we choose α to be the unique solution of (2.75). We choose β arbitrary as long as, the
Jacobian J of (2.67) is not zero. We can set β = x, thus we have the change of variables

 α = Bx − y
(2.88)
A

β = y.
One can verify easily that the Jacobian J of the system (2.88) is not zero.
Substitution into (2.68), yields
Cuββ = 0.
Integration gives,
u (α, β) = F (α) + βG (α) .
Thus
u (x, y) = F
B
B
x − y + yG
x−y .
A
A
(2.89)
40
Example 2.6.10 Use the method of characteristic, to find the general solution of the PDE
uxx − 4uxy + 4uyy = 0
Solution
A = 1, B = −2, C = 4, ∆ = B 2 − AC = 0, the equation is parabolic.
B
B
x − y + yG
x−y
The solution is give by (2.89), u (x, y) = F
A
A
or u (x, y) = F (−2x − y) + yG (−2x − y)
■
Case 2: A, B and C are not constants.
Suppose that A = A (x, y) , B = B (x, y) and C = C (x, y) , the change of variables (2.81) and (2.82)
[or (2.85) and (2.86)] can still be used as illustrated in the examples below 2.6.11
Example 2.6.11 Use the method of characteristic curves to solve the following PDEs:
1. uxx + 2 (x − y) uxy − 4xyuyy −
2. uxx + x2 uyy −
−2y + 4x2
2x + 1
ux −
uy = 0, x > 0, y > 0.
x+y
x+y
1
ux = 0, x > 0.
x
3. x2 uxx + 2xyuxy + y 2 uyy + 4xux = 0, x > 0, y > 0.
Solution
1. uxx + 2 (x − y) uxy − 4xyuyy −
F =−
2x + 1
−2y + 4x2
ux −
uy = 0, x > 0, y > 0
x+y
x+y
−2y + 4x2
2x + 1
,G = −
x+y
x+y
A = 1, B = x − y, C = −4xy, ∆ = B 2 − AC = (x + y)2 > 0. The equation is hyperbolic.
The characteristic curves are given by

√

 dy − 2x = 0
2
dy B ± B − AC
dx
−
= 0 =⇒
[Calculate the details]
dy

dx
A

+ 2y = 0.
dx
Solving the above equations gives
y − x2 = c1 and ln y + 2x = c2 .
Thus we let α = y − x2 and β = ln y + 2x.
αx = −2x, αy = 1, αxx = −2, αxy = 0, αyy = 0,
β x = 2, β y =
1
1
, β xx = 0, β xy = 0, β yy = − 2
y
y
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ux = αx uα + β x uβ
= −2xuα + 2uβ
uy = αy uα + β y uβ
1
= uα + uβ
y
uxx = αxx uα + α2x uαα + 2αx β x uαβ + β xx uβ + β 2x uββ
uxy
= −2uα + 4x2 uαα − 8xuαβ + 4uββ
= αx αy uαα + αx β y + αy β x uαβ + β x β y uββ + αxy uα + β xy uβ
2
2x
+ 2 uαβ + uββ
= −2xuαα + −
y
y
uyy = α2y uαα + 2αy β y uαβ + β 2y uββ + αyy uα + β yy uβ
1
1
2
= uαα + uαβ + 2 uββ − 2 uβ
y
y
y
Substituting in the PDE yields
uxx + 2 (x − y) uxy − 4xyuyy −
−2y + 4x2
2x + 1
ux −
uy
x+y
x+y
= −2uα + 4x2 uαα − 8xuαβ + 4uββ
2x
2
+2 (x − y) −2xuαα + −
+ 2 uαβ + uββ
y
y
2
1
1
−4xy uαα + uαβ + 2 uββ − 2 uβ
y
y
y
2x + 1
−2y + 4x2
1
−
[−2xuα + 2uβ ] −
uα + uβ
x+y
x+y
y
4
= − (x + y)2 uαβ = 0
y
Or uαβ = 0. Integrating twice gives u (α, β) = F (α) + G (β) .
The solution is then given by
u (x, y) = F y − x2 + G (ln y + 2x) .
where F and G are twice functions differentiable of their arguments.
Verify that u satisfies the PDE.
1
2. uxx + x2 uyy − ux = 0, x > 0.
x
A = 1, B = 0, C = x2 , B 2 − AC = −x2 < 0. Thus the equation is elliptic.
The characteristic curves are given
 by
√

 dy − xi = 0
dy B ± B 2 − AC
dx
−
= 0 =⇒
dy

dx
A

+ xi = 0.
dx
Solving the above equations yields
1
1
y − ix2 = c1 and y + ix2 = c2 .
2
2
42
1
1
We let α = y − ix2 and β = y + ix2
2
2
αx = −ix, αy = 1, αxx = −i, αxy = 0, αyy = 0,
β x = ix, β y = 1 , β xx = i, β xy = 0, β yy = 0
ux = αx uα + β x uβ
= −ixuα + ixuβ
uy = αy uα + β y uβ
= uα + uβ
uxx = αxx uα + α2x uαα + 2αx β x uαβ + β xx uβ + β 2x uββ
uxy
= −iuα − x2 uαα + 2x2 uαβ + iuβ − x2 uββ
= αx αy uαα + αx β y + αy β x uαβ + β x β y uββ + αxy uα + β xy uβ
= −ixuαα + (−ix + ix) uαβ − x2 uββ
uyy = α2y uαα + 2αy β y uαβ + β 2y uββ + αyy uα + β yy uβ
= uαα + 2uαβ + uββ
Substituting in the PDE yields
uxx + x2 uyy −
1
ux
x
= −iuα − x2 uαα + 2x2 uαβ + iuβ − x2 uββ + x2 [uαα + 2uαβ + uββ ] −
1
[−ixuα + ixuβ ]
x
= 4x2 uαβ = 0
Or uαβ = 0. Integrating twice gives u (α, β) = F (α) + G (β) .
The solution is then given by
1 2
1 2
u (x, y) = F y − ix + G y + ix ,
2
2
where F and G are twice functions differentiable of their arguments. Verify that u satisfies the
PDE.
3. x2 uxx + 2xyuxy + y 2 uyy + 4xux = 0, x > 0, y > 0
A = x2 , B = xy, C = y 2 , ∆ = B 2 − AC = (xy)2 − x2 y 2 = 0. The equation is parabolic.
The characteristic curves are given by
dy
y
dy B
−
= 0 =⇒
− = 0.
dx A
dx x
Solving the above equation, we get
y
= c1 .
x
y
Thus we let α = and choose β arbitrary provided the Jacobian J =
x
take β = y.
1
2y
1
y
αx = − 2 , αy = , αxx = 3 , αxy = 2 , αyy = 0,
x
x
x
x
β x = 0, β y = 1, β xx = 0, β xy = 0, β yy = 0
αx αy
βx βy
!
̸= 0. We can
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APM3701/1
ux = αx uα + β x uβ
y
= − 2 uα
x
uy = αy uα + β y uβ
1
uα + uβ
=
x
uxx = αxx uα + α2x uαα + 2αx β x uαβ + β xx uβ + β 2x uββ
=
uxy =
=
uyy =
=
y2
2y
u
+
uαα
α
x3
x4
αx αy uαα + αx β y + αy β x uαβ + β x β y uββ + αxy uα + β xy uβ
y
y
1
− 3 uαα − 2 uαβ + 2 uα
x
x
x
α2y uαα + 2αy β y uαβ + β 2y uββ + αyy uα + β yy uβ
1
2
uαα + uαβ + uββ
x2
x
Substituting in the PDE yields
x2 uxx + 2xyuxy + y 2 uyy + 4xux
hy
i
y2
y
1
2
y
1
2 2y
2
= x
u
+
u
u
−
u
+
u
u
+
u
+
u
−
4x
u
+
2xy
−
+
y
α
αα
αα
α
αα
α
αβ
αβ
ββ
x3
x4
x3
x2
x2
x2
x
x2
= y 2 uββ = 0
Or uββ = 0. Integrating twice gives u (α, β) = F (α) + βG (α) .
The solution is then given by
u (x, y) = F
y
x
+ βG
y
x
,
where F and G are twice functions differentiable of their arguments..
Verify that u satisfies the PDE.
■
Remark 2 It is important to note that the solutions obtained above are not unique. The change of
variables (2.81), (2.82), (2.85), (2.86), (2.88) determine which solution you will obtain.
2.6.2.1
General form
The change of variables obtained from equations (2.68)–(2.72) can be generalised to solve the general form
of the second order PDE (2.1) by following the steps below:
1. Obtain the for the characteristic curves by solving the equations
√
√
dy
B ± B 2 − AC
dx
B ± B 2 − AC
=
or
=
dx
A
dy
C
(2.90)
44
2. Express the integration of the equation (2.90) in the form ϕ (x, y) = c1 and φ (x, y) = c2 if B 2 −AC ̸=
0, or ϕ (x, y) = c if B 2 − AC = 0.
3. If B 2 − AC, let
α = ϕ (x, y) and β = φ (x, y) ,
(2.91)
α = ϕ (x, y) and β = x.
(2.92)
or, if B 2 − Ac = 0, let
4. Differentiate using the formula (2.93)–(2.97)
ux = αx uα + β x uβ
(2.93)
uy = αy uα + β y uβ
(2.94)
uxx = α2x uαα + 2αx β x uαβ + β 2x uββ + αxx uα + β xx uβ
uxy = αx αy uαα + αx β y + αy β x uαβ + β x β y uββ + αxy uα + β xy uβ
(2.95)
uyy = α2y uαα + 2αy β y uαβ + β 2y uββ + αyy uα + β yy uβ
(2.97)
(2.96)
5. Reduce the PDE in canonical form by substituting equations (2.93)–(2.97) into the PDE, simplify
futher, if necessary, and integrate to determine the solution.
Example 2.6.12 Use the method of characteristics to solve the following PDE
1. 4uxx + 12uxy + 9uyy − 9u − 9 = 0,
2. uxx + uxy − 2uyy − 3ux − 6uy = 9 (2x − y),
3. x2 uxx + 2xyuxy + y 2 uyy + xyux + y 2 uy = 0.
Solution
1. A = 4, B = 6, C = 9, =⇒ B 2 − AC = 0, the equation is parabolic.
The characteristics are given by
B
3
3
3
dy
=
= , or dy = dx. Which gives y = x + c.
dx
A
2
2
2
3
We let α = x − y and β = x.
2
3
Differentiation gives: αx = , αy = −1, β x = 1, β y = 0, αxx = αxy = αyy = 0, and β xx = β xy =
2
β yy = 0.
Using equations (2.93)–(2.97) and substituing into the PDE, we have:
4uββ − 9u = 9,
or,
9
9
uββ − u = .
4
4
We get the general solution
3
3
u (α, β) = f (α) e 2 β + g (β) e− 2 β − 1.
45
APM3701/1
Thus,
3
3
u (x, y) = f (3x − 2y) e 2 x + g (3x − 2y) e− 2 x − 1.
Verify that u satisfies the PDE.
1
9
2. A = 1, B = , C = −2 =⇒ B 2 − AC = > 0, the equation is hyperbolic.
2
4
The characteristics are given by
(
√
2
dy
B ± B 2 − AC
=
=
, or dy = 2dx and dy = −dx. Which gives y = 2x + c1 and
dx
A
−1
y = −x + c2 .
We let α = 2x − y and β = −x − y.
Differentiation gives: αx = 2, αy = −1, β x = −1, β y = −1, αxx = αxy = αyy = 0, and β xx = β xy =
β yy = 0.
Using equations (2.93)–(2.97) and substituing into the PDE yields
−9uαβ + 9uβ = 9α
or,
(uα − u)β = −α.
Integration with respect to β gives
uα − u = −αβ + f (α) ,
a second integration with respect to β (using integration factor, see p. 47 of the Prescribed book),
gives
(ueα )α = −αβeα + f (α) ea ,
or
ueα = −βeα (α − 1) + f (α) + g (β) .
Thus,
u (α, β) = −β (α − 1) + f (α) + e−α g (β) .
Therefore
u (x, y) = (x + y) (2x − y − 1) + f (2x − y) + e−2x+y g (−x − y) .
Verify that u satisfies the PDE.
3. A = x2 , B = xy, C = y 2 =⇒ B 2 − AC = 0, the equation is parabolic.
The characteristics are given by
B
xy
dy
dx
dy
=
= 2 , or
=
dx
A
x
y
x
. Which gives
We let α =
y
= c.
x
y
and β = x.
x
46
1
1
y
2y
Differentiation gives: αx = − 2 , αy = , β x = 1, β y = 0, αxx = 3 , αxy = − 2 , αyy = 0, and
x
x
x
x
β xx = β xy = β yy = 0.
Using equations (2.93)–(2.97) and substituing into the PDE, we have:
y
1
uα + uβ , uy = uα
2
x
x
y2
2y
2y
=
uαα − 2 uαβ + uββ + 3 uα
4
x
x
x
1
y
1
= − 3 uαα + uαβ − 2 uα
x
x
x
1
=
uαα
x2
ux = −
uαα
uαβ
uββ
Subtitution into the PDE yields
2
y
2y
2y
1
1
2 y
uαα − 2 uαβ + uββ + 3 uα + 2xy − 3 uαα + uαβ − 2 uα
x
x4
x
x
x
x
x
2
2
y
y
y
+ 2 uαα + xy − 2 uα + uβ + uα
x
x
x
2
= x uββ + xyuβ = 0
or
uββ +
y
uβ = 0 =⇒ uββ + αuβ = 0
x
uββ + αuβ = 0.
First integration with respect to β yields
uβ + αu = f (α) .
A second integration gives (using integrating factor, see p. 47 of the Prescribed book)
(ueα )β = f1 (α) eαβ =⇒ ueα = f1 (α)
eαβ
= g (α) .
α
Thus,
u (α, β) = f (α) + g (α) e−αβ .
Therefore,
u (x, y) = f
y
x
+g
y
x
e−y .
Verify that u satisfies the PDE.
■
Exercises 2.6
1. Solve the following transport equations by the method of characteristics, compare your answers with
the solutions of exercises 1.7(2.):
47
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√
(a) 1 + x2 ux + uy = 0. Plot some of the characteristic curves.
(c)
(b) ut + 4ux =
cos (x + π) .
(d) ex ux + xuy = 0, subject to u (0, y) = ln y.
0, subject to u (x, 0)
=
1 − x2 ux + xuy = 0, subject to
u (0, y) = 2y 2 .
2
(e) rutt − c2 rurr − 2c2 ur = 0
2. Use the method of characteristics to solve the problems in Exercises 2.4.(1.),(3.)-(4.).
3. Use the method of characteristics to solve the problems in Exercises 2.5.(1.)-(2.).
■
48
49
CONTENTS
Chapter 3
FOURIER SERIES ORTHOGONALITY
AND STURM–LIOUVILLE THEORY
3.1
BACKGROUND
Fourier theory of trigonometric series is of great practical importance, certain types of discontinuous
functions which cannot be expanded in power series can be expanded in Fourier series. More importantly,
a wide class of problems in physics and engineering possess periodic phenomena, and as a consequence,
Fourier’s trigonometric series become an indispensable tool in the analysis of these problems. Although
the analysis can be extended, the theory presented here will be concise, but sufficient for its applications
to many problems of mathematics physics.
This chapter is independent of the previous ones. Basic knowledge of calculus is required for the material
contained in this chapter. Section 3.8 on Sturm-Liouville theory involve ordinary differential equations
with constant coefficients.
The purpose of this chapter is to derive the basic properties of Fourier series, preparing the way for the
important applications of the following chapters. This chapter is essential for the understanding of the
whole module, particular attention must be given to it.
3.2
THE WAY FORWARD
Fourier series, as presented in this chapter are essential for all that follows. As we will see in the following
chapters, Fourier series are the most suitable expansions for solving certain classical problem in Applied
Mathematics. The are fundamental to the description of important physical phenomena from diverse
fields, such as mechanical and acoustic, heat transfer, planetary motion etc.
3.3
PRESCRIBED READING
Sections 2.1–2.8, 6.1 of [1].
♦ Note: This chapter is essential for this module, make sure that you have mastered its content before
moving on to the next chapter.
50
3.4
ORTHOGONALITY
3.4.1
Definition: orthogonal functions
Two functions f and g are said to be orthogonal with respect to the weight function q (x) on the
interval [a, b] if they are continuous on [a, b] and
Z b
f (x) g (x) q (x) dx = 0.
(3.1)
a
If q (x) = 1, then f and g are said to be orthogonal on the interval [a, b] , i.e.
Z b
f (x) g (x) dx = 0.
(3.2)
a
• If f = g, the number
∥f ∥ =
Z
b
12
2
f (x) q (x) dx
(3.3)
a
is called the norm of f with respect to the weight function q (x) .
If q (x) = 1, the number
∥f ∥ =
Z
b
2
12
(3.4)
f (x) dx
a
is simply called the norm of f.
Example 3.4.1
Show that the functions f (x) = sin x and g (x) = sin 2x are orthogonal on [−π, π] .
Solution
π
1
1
1
1
=0
cos x − cos 3x dx =
sin x − sin 3x
sin x sin 2xdx =
2
2
6
−π
−π 2
−π
Z π
1 Z π 1 π 1
2
2
2
√
1 1
1
1
2
∥sin x∥ =
sin xdx =
− cos 2x dx =
x − sin 2x
= π
2
2
4
−π
−π 2
−π
Z
π
Z
π
and
∥sin 2x∥ =
Z
π
2
sin 2xdx
−π
1
Z
2
π
=
−π
1 2
1 1
1
1
− cos 4x dx =
x − sin 2x
2 2
2
8
π
1
2
=
√
π.
−π
■
♦ Note: In general if n and m are positive integers n ̸= m, then sin nx and sin mx are orthogonal on
[−π, π] , since for n ̸= m
Z π
Z π
1
1
sin nx sin mxdx =
cos (n − m) x − cos (n + m) x dx
2
−π
−π 2
π
1
1
1
=
sin (m − n) x −
sin (m + n) x
2 m−n
n+m
−π
= 0
51
APM3701/1
and
Z
π
Z
2
π
sin nπxdx =
−π
−π
1 1
− cos 2nπx dx
2 2
1
(nx − cos nx sin nx)
2n
=
Therefore
∥sin nπx∥ =
√
π
= π.
−π
π ∀n = 1, 2, 3, . . .
Example 3.4.2
Show that the functions f (x) = cos nx and g (x) = cos mx; n, m = 0, 1, 2, . . . (n ̸= m) are orthogonal on
[−π, π] . Note that if n or m equals 0, cos 0 = 1.
Solution
Z
π
cos nx cos mxdx =
−π
=
Z
1 π
[cos (nx + mx) + cos (mx − nx)] dx
2 −π
π
1
1
1
sin (m − n) x +
sin (n + m) x
=0
2 m−n
n+m
−π
and
Z
π
Z
1 π
(1 + cos 2nx) dx
2 −π
π
1
(nx + cos nx sin x)
=π
2n
−π
2
cos nxdx =
−π
=
Thus ∥cos nx∥ =
And for n = 0,
√
π for all n = 1, 2, . . .
∥cos 0x∥ = ∥1∥ =
Z
1
π
2
dx
= x
−π
π
1
2
1
= (2π) 2 =
−π
√
2π
■
3.4.2
Definition: orthogonal sequence
A sequence of functions {ϕn (x)}n is said to be orthogonal with respect to the weight function q (x)
on the interval [a, b] if
Z b
ϕn (x) ϕm (x) q (x) dx = 0
for n ̸= m
(3.5)
a
The number
∥ϕn ∥ =
Z
b
ϕ2n (x) q (x) dx
a
21
≥0
(3.6)
is called the norm of the orthogonal system {ϕn } with respect to the weight function q (x) .
Example 3.4.3 Show that the sequence of functions
an orthogonal system on [0, 1] .
n
o∞
sin nπx
n=1
= {sin πx, sin 2πx, sin 3πx, . . .} form
52
Solution
Similar to example 3.4.1
Z
1
sin nπx sin mπxdx = 0
0
n ̸= m.
■
Example 3.4.4
n
o∞
Show that the sequence of functions cos nx
n=0
of function on [−π, π] .
= {1, cos x, cos 2x, cos 3x, . . .} form an orthogonal system
Solution
See solution of example 3.4.2.
■
∞
2πx
3πx
= sin πx
Example 3.4.5 Show that the sequence of functions sin nπx
L
L , sin L , sin L , . . . , L > 0
n=1
form an orthogonal system on [0, L] .
Solution
For n ̸= m, we have
Z Z L
mπx
1 L
(n − m) π
(n + m) π
nπx
sin
dx =
cos
x − cos
x dx
sin
L
L
2 0
L
L
0
L
1
L
(n − m) π
L
(n + m) π
=
sin
x−
sin
x
2 π (n − m)
L
(n + m) π
L
0
= 0.
■
Example 3.4.6
Show that the sequence of functions
n
2πx
3πx
πx
nπx o∞
= 1, cos
, cos
, cos
,...
cos
L n=0
L
L
L
form an orthogonal system on [0, L] , L > 0.
Solution
For n ̸= m
Z L
Z nπx
mπx
1 L
(nπ − mπ)
(nπ + mπ)
cos
cos
dx =
cos
x + cos
x dx
L
L
2 0
L
L
0
1
L
(nπ − mπ)
L
(nπ + mπ)
=
sin
x+
sin
x
2 (m − n) π
L
(n + m) π
L
= 0.
L
0
■
Example 3.4.7 Caution: Look alike sequences.
53
1. Show that the sequence of functions
where αn is solution of tan αn =
n
αn x o∞
sin
L n=1
L
do not form an orthogonal system on [0, L] .
αn
2. Show however that the sequence
where αn is solution of tan αn =
APM3701/1
n
αn x o∞
cos
L n=1
L
form an orthogonal system on [0, L] .
αn
Solution
1. For n ̸= m, we have
Z L
αn x
αm x
L
sin
sin
dx = − 2
(αn cos αn sin αm − αm sin αn cos αm )
L
L
α
−
α2m
0
n
= −
L2 cos αn
sin αn
= −
cos αn sin αm −
α2n
L2
−
L2 cos αm
sin αm
α2m
sin αn cos αm
cos2 αn sin2 αm − cos2 αm sin2 αn
(α2n − α2m ) sin αn sin αm
L2 cos αm αn
=
(cos αn − cos αm ) (cos αm + cos αn )
(α2n − α2m ) sin an sin αm
̸= 0 if αn ̸= αm + 2kπ or αn ̸= αm + (2k + 1) π.
2. For m ̸= n, we have:
Z L
αn x
αn x
cos
dx =
cos
L
L
0
L
(αn sin αn cos αm − αm cos αn sin αm )
− α2m
L
L cos αn
L cos αm
=
sin αn cos αm −
cos αn sin αm
α2n − α2m
sin αn
sin αm
L2 cos αm cos αn
(sin αn sin αm − sin αm sin αn )
=
(α2n − α2m ) sin αn sin αm
= 0.
α2n
■
3.4.3
Properties of orthogonal systems
The most important property of orthogonal systems is that every orthogonal system is linearly independent,
see exercise 3.4.(2.)
Exercises 3.4
1. Show that the polynomials 1, x,
1
3x2 − 1 are orthogonal to each other in the interval [−1, 1] .
2
2. Prove that every orthogonal set is linearly independent.
3. Show that the following sequences of functions are orthogonal systems on the prescribed interval.
54
n
nπx o∞
sin
on [−L, L]
L n=1
Find the norm of each functions. Deduce that sin nπ
[−π, π] .
∞
(b) cos nπx
on [−L, L].
L
n=1
(a)
∞
n=1
form an orthogonal system on
(i) Find the norm of each function.
h π πi
form
an
orthogonal
system
on
− ,
(ii) Deduce that {cos 2nπ}∞
.
n=1
2 2
4. Show that the sequence
πx
πx
2πx
2πx
3πx
3πx
, cos
, sin
, cos
, sin
, cos
,...
1, sin
L
L
L
L
L
L
is orthogonal on [0, L] and on [−L, L] .
5. Show that the following set of functions are orthogonal on the given interval, with respect to the given
w weight function [1]:
(a) Determine the constants a and b so that the functions 1, x, and a + bx + x2 become orthogonal
on the interval [−1, 1] .
(b) 1, sin πx, cos πx, sin 2πx, cos 2πx, sin 3πx, cos 3πx, ...; w (x) = 1, on [0, 2] .
1
on [−1, 1] . (Chebyshev polynomials of the first kind.)
(c) 1, x, −1 + 2x2 ; w (x) = √
1 − x2
1
on [−1, 1] .
(d) −3x + 4x3 , 1 − 8x2 + 8x4 ; w (x) = √
1 − x2
√
(e) 1, 2x, −1 + 4x2 ; w (x) = 1 − x2 on [−1, 1] . (Chebyshev polynomials of the second kind.)
2 − 4x + x2
(f ) 1, 1 − x,
; w (x) e−x on [0, ∞) . (Laguerre polynomials.)
2
2
(g) 1, 2x, −2 + 4x2 ; w (x) = e−x on [0, ∞) . (Hermite polynomials.)
2 − 4x + x2
(h)
, −12x + 8x3 ; W (x) = e−x on [0, ∞) .
2
(i) f (x) is an even function, g (x) is an odd function; w (x) = 1 on any symmetric interval about
0.
■
3.5
3.5.1
FOURIER SERIES
Introduction
We have seen in section 3.4.2 and in Exercise 3.4(4.) that the functions
1, cos
πx
πx
2πx
2πx
, sin
, cos
, sin
,...
L
L
L
L
(3.7)
are mutually orthogonal to each other on the interval [−L, L] and are linearly independent by Exercise
3.4.(2.). We form a series representing f (x) by assuming that f (x) can be written in the form
∞ X
kπx
kπx
+ bn sin
(3.8)
f (x) ∼ a0 +
an cos
L
L
k=1
55
APM3701/1
where ∼ indicates an association of a0 , ak and bk to f in some unique manner, often ” ∼ ” is replaced by
” = ”. Think of (3.7) as a base of a vector space.
The infinite sum on the right hand side of (3.8) may or may not be equal to f (x) at x. Whenever it is,
we will say that the sum converges to f (x) at x.
Let f (x) be integrable and defined on [−L, L] . We define the nth partial sum
n X
kπx
kπx
(3.9)
+ bk sin
Sn (x) = a0 +
ak cos
L
L
k=1
which is to define f (x) on [−L, L] . We seek the coefficient a0 , ak , and bk such that Sn (x) represents the
best approximation to f (x) in the sense of least squares, that is, we seek to minimise the function
Z L
2
F (a0 , ak , bk ) =
(3.10)
f (x) − Sn (x) dx.
−L
A necessary condition on a0 , ak , bk , in order for F to have a minimum is that the first derivatives of F with
respect to a0 , ak , bk must vanish (see second year Multivariable calculus). Substituting (3.9) into (3.10)
and differentiating with respect to a0, ak and bk yields


Z L
n X
∂F
jπx
jπx 
f (x) − a0 −
= −
aj cos
+ bj sin
dx
(3.11)
∂a0
L
L
−L
j=1


Z L
n X
∂F
jπx 
kπx
jπx
f (x) − a0 −
= −2
+ bj sin
cos
dx
(3.12)
aj cos
∂ak
L
L
L
−L
j=1


Z L
n X
∂F
jπx 
kπx
jπx
f (x) − a0 −
= −2
+ bj sin
sin
dx.
(3.13)
aj cos
∂bk
L
L
L
−L
j=1
Using the orthogonality of the sequence in (3.7) and observing that
Z L
Z L
jπx
jπx
cos
dx =
sin
dx = 0, for any integer j.
L
L
−L
−L
Equations (3.11), (3.12) and (3.13) become
Z L
Z L
∂F
= a0
dx −
f (x) dx
∂a0
−L
−L
Z L
Z L
∂F
kπx
kπx
cos2
= 2ak
dx − 2
f (x) cos
dx
∂ak
L
L
−L
−L
Z L
Z L
∂F
kπx
kπx
= 2bk
sin2
dx − 2
f (x) sin
dx
∂bk
L
L
−L
−L
which must vanish for F to have an extremal value. We have
Z L
Z L
∂F
= a0
dx −
f (x) dx = 0
∂a0
−L
−L
Z L
Z L
∂F
kπx
kπx
= 2ak
cos2
dx − 2
f (x) cos
dx = 0
∂ak
L
L
−L
−L
Z L
Z L
∂F
kπx
kπx
= 2bk
sin2
dx − 2
f (x) sin
dx = 0.
∂bk
L
L
−L
−L
(3.14)
(3.15)
(3.16)
(3.17)
(3.18)
(3.19)
(3.20)
56
Solving for a0 , ak and bk we obtain:
RL
a0 =
ak =
and
Z L
1
−L f (x) dx
f (x) dx
=
RL
2L −L
−L dx
RL
Z
kπx
1 L
−L f (x) cos L dx
=
f (x) cos
RL
2 kπx dx
L −L
L
−L cos
RL
bk =
kπx
−L f (x) sin L dx
RL
2 kπx
L dx
−L sin
1
=
L
Z
L
f (x) sin
−L
(3.21)
kπx
dx
L
kπx
dx.
L
(3.22)
(3.23)
It follows from (3.18), (3.19) and (3.20) that
∂2F
∂a20
=
∂2F
∂a2k
=
∂2F
∂b2k
=
Z
L
dx = 2L > 0
−L
Z L
−L
Z L
(3.24)
cos2
kπx
dx = 2L > 0
L
(3.25)
sin2
kπx
dx = 2L > 0.
L
(3.26)
−L
By expanding F in Taylor series about
(a0 , a1 , . . . , an , b1 , . . . , bn )
we obtain
F (a0 + ∆a0 , . . . , an + ∆an , . . . , bn + ∆bn ) = F (a0 , . . . , bn ) + ∆F
where ∆F is the remaining terms. We have
"
#
n 2
X
∂ F
1 ∂2F
∂2F
2
2
2 2
∆F =
∆a0 +
(∆ak ) +
(∆bk )
2! ∂a20
∂a2k
∂b2k
k=1
since all remaining higher derivatives vanish. By virtue of (3.24), (3.25) and (3.26), ∆F is positive. Hence,
for F to have a minimum value, the coefficients a0 , ak , bk must be given by equations (3.21), (3.22) and
(3.23) respectively.
3.5.2
Definition: Fourier coefficients
If the coefficients an (n = 0, 1, . . .) and bn (n = 1, 2, . . .) given by (3.18), (3.19) and (3.20) respectively,
exist and are all finite, then the Fourier series of f on [−L, L] is given by the expression
f (x) ∼ a0 +
∞ X
n=1
an cos
nπx
nπx + bn sin
.
L
L
(3.27)
The coefficients a0 , an , bn (n = 1, 2, 3, . . .) defined below
RL
a0 =
ak =
Z L
1
−L f (x) dx
=
f (x) dx
RL
2L −L
−L dx
RL
Z
kπx
1 L
−L f (x) cos L dx
=
f (x) cos
RL
2 kπx dx
L −L
cos
L
−L
(3.28)
kπx
dx
L
(3.29)
57
and
RL
bk =
kπx
−L f (x) sin L dx
RL
2 kπx
L dx
−L sin
=
APM3701/1
1
L
Z
L
f (x) sin
−L
kπx
dx.
L
(3.30)
are called the Fourier coefficients of f on [−L, L] .
Remark 3.5.1 The symbol ∼ is used to make a difference between the function f and its Fourier series,
most books, including the Prescribed Book, do not make clear this distinction. Surely, for sufficiently
smooth functions, (see theorem 3.7.1 in section 3.5.3), there is equality. In this Guide, we will use
either sign ” ∼ ” or ” = ”, please feel free to use any of the above, however you must bear in mind that
the function f (x) and its Fourier Series are two different functions, which are equal only when theorem
3.7.1 in section 3.5.3 is satisfied.
■
Example 3.5.1 Find the Fourier series of the function f (x) = x in [−L, L] . Solution
∞
∞
P
P
nπx
nπx
+
bn sin
. We have
Let f (x) = a0 +
an cos
L
L
n=1
n=1
a0 =
an =
=
bn =
=
=
Z L
Z L
1
1 1 2 L
1
f (x) dx =
xdx =
=0
x
2L −L
2L −L
2 2
−L
Z
Z L
h x
1 L
nπx
nπx iL
1
nπx
x cos
dx =
sin
−
sin
dx
L −L
L
nπ
L −L nπ −L
L
L
nπx L
0+
=0
n = 1, 2, 3, . . .
cos
L −L
(nπ)2
Z
Z L
1 L
−x
nπx
nπx L
1
nπx
x sin
dx =
cos
+
cos
dx
L −L
L
nπ
L −L nπ −L
L
2L
L
nπx L
−2L
=−
cos nπ +
sin
cos nπ
2
nπ
L −L
nπ
(nπ)
2L
(−1)n+1 ,
nπ
n = 1, 2, 3, . . .
since cos nπ = (−1)n .
Thus the Fourier series of f (x) = x is
x∼
∞
X
∞
n+1
(−1)
nπx
2L X (−1)n+1
nπx
2L
sin
=
sin
.
nπ
L
π
n
L
(3.31)
n=1
n=1
Please note that the Fourier series of f (x) and the function f are equal to zero at x = 0.
However, at x = L, f (L) = L and the Fourier series of f is
∞
2L X (−1)n+1
sin nπ = 0,
π
n
n=1
justifying the sign ” ∼ ” used in (3.31).
3.5.3
■
Convergence of Fourier series
A very good knowledge of calculus (limits, continuity, differentiations and integrals) is required for this
section.
58
3.5.3.1
Definitions
1. A function f (x) defined on [a, b] , is said to be piecewise continuous if and only if, there exists a
partition {x1 , x2 , · · · , xn } of [a, b] such that
1.1. f (x) is continuous on [a, b] except eventually at the points xi ,
1.2. f (xi +) = lim f (x) and f (xi −) = lim f (x) exist at the points xi , 1 ≤ i ≤ n.
x→xi +
x→xi −
Basically, we say that a function f (x) is piecewise continuous on [a, b] , if the interval [a, b] can
be broken up into a finite number of pieces such that f (x) is continuous in each piece, and the right
limit and left limit of f (x) at the points of discontinuity exist. The function f (x) does not have to be
continuous on [a, b], the only kind of discontinuity allowed is a finite number of jump discontinuity.
2. A function f (x) has a jump discontinuity at x = a if the limit from the left f (a−) and the limit
from the right f (a+) both exist but are not equal.
3. A function f (x) is said to be piecewise smooth if and only if f (x) and its derivatives are piecewise
continuous.
Figure 3.1: Jump discontinuity at x = a
Figure 3.2: Piecewise smooth function
1
Figure 3.3: Example of a function which is not no piecewise smooth: f (x) = x 3 .
59
APM3701/1
Figure 3.1 illustrates a jump discontinuity and Figure 3.2 illustrates a piecewise smooth function.
1
Whereas Figure 3.3 shows the graph of f (x) = x 3 , which is not piecewise smooth on any interval
containing x = 0; because
df
1 2
= x− 3
dx
3
and
df
(0) = ∞
dx
Therefore, any region including x = 0 cannot be broken up into pieces such that
the subinterval containing x = 0.
df
dx
is continuous in
4. A Fourier series of a function on [−L, L] is 2L-periodic. Thus, the Fourier series of f (x) on [−L, L]
is 2L–periodic. The function f (x) does not need to be periodic itself. We need the periodic
extension of f (x). Figure 3.4 shows the periodic extension of f (x).
Figure 3.4: Periodic extension of f (x)
To sketch the periodic extension of f (x), simply sketch f (x) on [−L, L] and then repeat continually
the same pattern with period 2L by translating the original sketch. We sketch the periodic extension
of f (x) = x in figure 3.5
Figure 3.5: Periodic extension of f (x) = x.
60
3.5.3.2
Convergence Theorem for Fourier series
Most of the results in this section are presented for 2π-period functions, they can be easily extended to
2L-periodic functions. Before we state and prove the fundamental theorem of convergence of Fourier series,
we prove some preliminary results.
Lemma 3.6 If f and f ′ are piecewise continuous on [a, b] , then
Z b
f (x) sin (kx) dx = 0,
lim
(3.32)
k→∞ a
and
b
Z
f (x) cos (kx) dx = 0.
lim
k→∞ a
(3.33)
Proof
Let {x1 , x2 , · · · , xn } be the points where f has jump discontinuity. We know that
Z
b
f (x) sin (kx) dx =
a
n−1
X Z xi+1
i=1
f (x) sin (kx) dx.
xi
To prove (3.32), it is enough to show that for i, 1 ≤ i ≤ n − 1
Z xi+1
lim
f (x) sin (kx) dx = 0.
k→∞ xi
Integration by parts gives
Z xi+1
Z
−f (x) cos (kx) xi+1 1 xi+1 ′
f (x) sin (kx) dx =
f (x) cos (kx) dx.
+
k
k xi
xi
xi
Since f and f ′ are bounded on [xi , xi+1 ] , we have
−f (x) cos (kx) xi+1
lim
= 0, and
k→∞
k
xi
1
lim
k→∞ k
Z
xi+1
f ′ (x) cos (kx) dx = 0 ,
xi
which complete the proof.
The proof of (3.33) is similar.
■
Lemma 3.7 If
1
KN (k) =
π
then for k ̸= 0, ±2π, ±4π, · · ·
KN (k) =
Furthermore KN (x) is continuous, even and
Z
N
1 X
+
cos (kx)
2
(3.34)
n=1
sin N +
2π sin
2π
N -periodic,
1
2
k
2
k
.
(3.35)
and
0
KN (x) dx =
−π
!
1
2
(3.36)
61
and
APM3701/1
π
Z
KN (x) dx =
0
1
2
(3.37)
Proof
KN (k) sin k =
=
N
1
π
X
1
sin k +
cos nk sin k
2
1
π
1
sin k +
2
n=1
N
X
1
2
h
n=1
N
Xh
!
sin ((n + 1) k) − sin ((n − 1) k)
!
i
!
=
1
2π
=
1 sin k + sin 2k − sin 0k + sin 3k − sin k + sin 4k − sin 2k+
2π
=
sin k +
sin ((n + 1) k) − sin ((n − 1) k)
n=1
i
+ sin 5k − sin 3k + · · · + sin ((N + 1) k) − sin ((N − 1) k)
1
1 sin kN + sin ((N + 1) k) = sin N + 21 k cos k2 ,
2π
π
where we have used the trigonometric identities:
sin x cos y =
1
[sin (x + y) − sin (y − x)] .
2
Since sin 2x = 2 sin x cos x, we have
KN (k) =
sin
N+
1
2
k cos k2
2π sin k2
.
From (3.34) we have:
0
1
KN (x) dx =
π
−π
Z
Z
0
−π
And
Z
0
π
1
KN (x) dx =
π
Z
0
π
!
"
#0
N
N
1 X
1 x 1X
+
cos nx dx =
+
sin nx
2
π 2 n
1
= .
2
!
"
#0
N
N
1 x 1X
1 X
+
cos nx dx =
+
sin nx
2
π 2 n
1
= .
2
n=1
n=1
n=1
n=1
−π
−π
■
Theorem 3.7.1 (Convergence Theorem for Fourier series) If f (x) is a piecewise smooth function on [−π, π] , then the Fourier series of f (x) converges to the following values:
1. to the periodic extension of f (x), where the periodic extension is continuous;
h
i
2. to the average of the two limits, usually 12 f (x+) + f (x−) , where the periodic extension has a
jump discontinuity.
62
In other words, if fp is periodic extension of f , then
(
∞ f (x+)+f (x−)
X
nπx nπx
2
a0 +
+ bn sin
=
an cos
L
L
f
(x)
n=1
if f has a jump discontinuity at x,
if fp is continuous at x
(3.38)
where
a0 =
1
2π
Z
π
f (x) dx,
1
π
an =
−π
Z
π
f (x) cos nxdx,
bn =
and
−π
1
π
Z
π
f (x) sin nxdx.
(3.39)
−π
Proof
Substitution of (3.39) into the left hand side of (3.38) gives
∞ X
nπx
nπx a0 +
an cos
+ bn sin
L
L
n=1
Z π
Z π
Z π
∞
1X
1
f (t) dt +
f (t) cos ntdt cos nx +
f (t) sin ntdt sin nx
=
2π −π
π
−π
−π
n=1
Z π
∞ Z π
1
1X
=
f (t) (cos nx cos nt + sin nx sin nt) dt
f (t) dt +
2π −π
π
−π
n=1
Z π
Z
∞
π
1X
1
f (t) dt +
f (t) cos n (t − x) dt.
=
2π −π
π
−π
n=1
Let consider the partial sum
fN (x)
1
2π
=
=⇒ fN
N Z
1X π
f (t) dt +
f (t) cos n (t − x) dt.
π
−π
n=1 −π
"
#
Z
N
1 π
1 X
=
+
cos n (t − x) dt .
f (t)
π −π
2
Z
π
(3.40)
n=1
Equation (3.38) is proven if
(
f (x+)+f (x−)
2
lim fN (x) =
if f has a jump discontinuity at x,
if fp is continuous at x.
fp (x)
N →∞
(3.41)
Without loss of generality, we assume that f (x) is 2π − periodic on R (take f equals to its periodic
extension). So that the function KN (t − x) f (t) and f (t + x) KN (t) are 2π− periodic in t. We substitute
(3.34) into (3.40), we obtain
Z
π
fN (x) =
−π
We let y = t − x, we have
Z
f (t) KN (t − x) dt.
π−x
fN (x) =
f (y + x) KN (y) dy
−π+x
Z
π
=⇒ fN (x) =
f (y + x) KN (y) dy
−π
(since KN (y) f (x + y) is 2π − periodic in y.)
Since KN is even, we have
Z
π
fN (x) =
−π
= −
Z
Z
f (t) KN (t − x) dt =
Z
π
−π
f (t) KN (x − t) dt
x−π
x+π
x+π
=
x−π
f (x − y) KN (y) dy
f (x − y) KN (y) dy
(Let y = x − t)
(3.42)
63
Z
APM3701/1
π
=⇒ fN (x) =
−π
f (x − y) KN (y) dy
(since KN (y) f (x − y) is 2π-periodic in y.)
(3.43)
Adding (3.42) and (3.43) yields
Z
π
fN (x) =
−π
f (x + y) + f (x − y)
KN (y) dy
2
Since the integrant is even, we have
Z
fN (x) =
0
π
f (x + y) + f (x − y) KN (y) dy.
We use (3.37) to obtain
Z
Z π
[f (x+) + f (x−)] KN (y) dy = [f (x+) + f (x−)]
π
KN (y) dy =
0
0
f (x+) + f (x−)
=⇒
=
2
Z
(3.44)
f (x+) + f (x−)
.
2
π
[f (x+) + f (x−)] KN (y) dy
(3.45)
0
Thus subtracting (3.44 ) from (3.45) gives
Z πh
i
f (x+) + f (x−)
− fN (x) =
f (x+) − f (x + y) + f (x−) − f (x − y) KN (y) dy.
2
0
f (x+) + f (x−)
We need to prove that lim
− fN (y) = 0. If
N →∞
2
Z π
lim
[f (x+) − f (x + y)] KN (y) dy = 0
(3.46)
(3.47)
N →∞ 0
and
Z
lim
N →∞ 0
then lim
N →∞
π
[f (x−) − f (x − y)] KN (y) dy = 0
(3.48)
f (x+) + f (x−)
− fN (x) = 0, and consequently (3.41) is satisfied. Let us prove (3.47), the
2
proof of (3.48) is similar. To prove (3.47), we note that
Z π
Z π
f (x+) − f (x + y)
y
lim
(f (x+) − f (x + y)) KN (y) dy = lim
sin
N →∞ 0
N →∞ 0
y
2π sin y2
N+
1
2
y dy
(3.49)
where we have used (3.35).
f (x+) − f (x + y)
y
′
If ϕ (y) =
y and ϕ (y) are piecewise continuous on [0, π] , lemma 3.6 implies that
y
2π
sin
2
Rπ
lim 0 (f (x+) − f (x + y)) KN (y) dy = 0 (by substituting N + 12 by N )
N →∞
f (x+) − f (x + y)
y
and u (y) =
, to prove that ϕ and ϕ′ are piecewise continuous on
y
2π sin y2
[0, π] , it is sufficient to prove that g, g ′ , u and u′ are piecewise continuous on [0, π] .
Let g (y) =
1. It is easy to verify that g, g ′ , u and u′ are continuous on (0, π] .
2. Let us show that g (0+) , g ′ (0+) , u (0+) and u′ (0+) exist.
64
2.1. g (0+) = lim
y→0+
f (x+) − f (x + y)
= −f ′ (x+) .
y
2.2. For 0 < y ≤ π,
−yf ′ (x + y) − y [f (x+) − f (x + y)]
y2
′
−y [f (x + y) − f ′ (x+)] − y [f ′ (x+) + f (x+) − f (x + y)]
=
y2
f (x + y) − f (x+)
f ′ (x+) −
′
′
[f (x + y) − f (x+)]
y
= −
−
.
y
y
g ′ (y) =
When y → 0, the second term in the above equation is zero, since
f (x + y) − f (x+)
′
lim
= f (x+) .
y→0+
y
Thus, we have
g ′ (0+) = lim g ′ (y) = lim −
y→0+
2.3. u (0+) = lim u (y) = lim
y→0+
y→0+
2.4. For 0 < y ≤ π, we have
u′ (0+) =
=
Thus lim
Rπ
N →∞ 0
y→0+
f ′ (x + y) − f (x+)
= −f ′′ (x+) .
y
1
y
= lim
y (by applying Hospital rule).
y→0+ π cos
2π sin y2
2
u′ (y)
lim u′ (y) =
y→0+
y
y
1 y cos 2 − 2 sin 2
=
.
4π
sin2 y2
y cos y2 − 2 sin y2
cos y2 − 21 y sin y2 − cos y2
1
1
=
lim
lim
4π y→0+
4π y→0+
2 sin y2 cos y2
sin2 y2
− 12 y sin y2
− 21 y
1
1
=
=0
lim
lim
4π y→0+ 2 sin y2 cos y2
4π y→0+ 2 cos y2
(by applying Hospital).
(f (x+) − f (x + y)) KN (y) dy = 0.
f (x+) + f (x−)
Note that if f is continuous, f (x+) = f (x−) =⇒
= f (x) . Thus (3.46) becomes
2
Rπ
f (x) − fN (x) = 0 [f (x+) − f (x + y) + f (x−) − f (x − y)] KN (y) dy.
Which completes the proof of the theorem.
Example 3.7.1 Consider the function
(
f (x) =
0 x<0
x x>0
(a) Determine the Fourier series of f (x) on [−L, L] .
(b) Use the result in (a) to show that
∞
X
π2
1
1
1
= 1 + 2 + 2 + ··· =
.
8
3
5
(2n + 1)2
n=0
Solution
■
65
APM3701/1
Figure 3.6: Sketch of f (x).
Figure 3.7: Sketch of the periodic extension f (x)
(a) We begin by sketching f (x) for all x (figure 3.6) and the periodic extension of f (x) (figure 3.7).
In figure 3.7 we marked by (×) the average values at the points of jump discontinuity. Thus
f (x) ∼ a0 +
a0 =
=
an =
=
=
bn =
=
∞
X
n=1
∞
an cos
nπx
nπx X
+
bn sin
L
L
(3.50)
n=1
Z L
1
f (x) dx =
xdx (since f (x) = 0 for − L < x < 0)
2L 0
−L
L
1 x2
L
=
2L 2 0
4
"
#
Z L
Z
Z L
1
nπx
1 L
nπx
1
L
nπx L
1
nπx
f (x) cos
dx =
x cos
dx =
x
sin
−
sin
dx
L −L
L
L 0
L
L
nπ
L 0
nπ 0
L
L nπx nπx
nπx L
cos
+
sin
n2 π 2
L
L
L
0
L
[(−1)n − 1]
n2 π 2
"
#
Z
Z
Z L
1 L
nπx
1 L
nπx
1
L
x L
L
x
−
f (x) sin
dx =
x sin
dx =
−x
cos nπ
cos nπ dx
L −L
L
L 0
L
L
nπ
L 0
nπ 0
L
"
#
L2
L
nπx L
L
L
− (−1)n −
sin
=−
(−1)n =
(−1)n+1 .
2
2
nπ
n π
L 0
nπ
nπ
1
2L
Z
L
Thus
∞
L X
f (x) = +
4
n=1
L
nπx
L
nπx
((−1)n − 1) cos
+
(−1)n+1 sin
n2 π 2
L
nπ
L
.
(3.51)
66
(b) Let x = 0, in equation (3.51), we have
∞
L X L
((−1)n − 1)
f (0) = 0 = +
2
2
4
n π
n=1
∞
1
1
1 X 1
1
2
1
1
n
∴ 0= + 2
((−1) − 1) = − 2 1 + 2 + 2 + 2 + . . . .
4 π
n2
4 π
3
5
7
n=1
Thus
∞
X
π2
1
1
1
.
= 1 + 2 + 2 + ... =
8
3
5
(2n + 1)2
n=0
■
Example 3.7.2 Let f (x) = x + x2 .
(a) Determine the Fourier series of f (x) on [−π, π] . (Plot f (x) and its periodic extension, determine
the value of the Fourier series at the points of jump discontinuity.)
(b) Compute the Fourier coefficients of f (x) on [−π, π] .
(c) Show that
∞
π2 X 1
=
6
n2
∞
and
n=1
π 2 X (−1)n+1
=
.
12
n2
n=1
Solution
(a) Since L = π, we have
f (x) = a0 +
∞
X
(an cos nx + bn sin nx)
(3.52)
n=1
(b)
a0 =
an =
=
=
=
Z π
Z π
π2
1
1 1 3 1 π
1
2
f (x) dx =
x + x dx =
x + x
=
2π −π
2π −π
2π 3
2 −π
3
Z π
Z π
1
1
f (x) cos nx dx =
x2 + x cos nx dx
π −π
x −π
Z π
Z π
π
2x sin nx
1 x sin nx π
sin nx
1 x2 sin nx
−
−
dx +
dx
π
n
n
π
n
n
−π
−π
−π
−π
Z π
cos nx
2
cos nx π
−
−
+
dx
nπ
n −π
n
−π
4
4
cos nπ = 2 (−1)n ,
n = 1, 2, 3, . . .
n2
n
Z
Z
1 π 2
1 π
f (x) sin nx dx =
x + x sin nx dx
π −π
π −π
Z π
1
cos nx π
cos nx
cos nx
2
=
−
+
dx +
−x2
π
n −π
n
π
n
−π
Z π
π
2
2 x sin nx
sin nx
= − cos nπ +
−
dx
n
nπ
n
n
−π
−π
2
2
= − cos π = − (−1)n ,
n = 1, 2, 3, . . .
n
n
bn =
π
−π
Z
π
+
−π
2x cos nx
dx
n
67
APM3701/1
Figure 3.8: Sketch of f (x) = x2 + x.
Thus
∞
π2 X
f (x) ∼
+
3
n=1
4
2
n
n
(−1) cos nx − (−1) sin nx .
n2
n
(3.53)
(c) Substituting x = π into (3.53) yields
∞
1 +
π2 X
(f (π) + f − (π)) = π 2 =
+
2
3
n=1
∴
4
(−1)n cos nπ
n2
∞
=
π2 X 4
+
(−1)n (−1)n
3
n2
n=1
∞
X
1
=4
3
n2
2π 2
n=1
∴
∞
π X 1
=
.
6
n2
n=1
For x = 0, we have
∞
1 +
π2 X 4
(f (0) + f − (0)) = 0 =
+
(−1)n ,
2
3
n2
n=1
Thus,
∞
∴
X (−1)n
π2
=−
.
12
n2
n=1
∞
π 2 X (−1)n+1
=
.
12
n2
n=1
■
68
Figure 3.9: Sketch of periodic extension of f (x) = x2 + x.
3.7.1
Cosine and Sine Series
Please read section 2.3 of [1] on even and odd function.
3.7.1.1
Cosine Fourier Series
nπx
nπx
is an even function and sin
is odd,
Let f (x) be an even function defined on [−L, L] . Since cos
L
L
nπx
nπx
the function f (x) cos
is an even function and f (x) sin
is odd. Thus the Fourier coefficients of
L
L
f (x) are given by:
an =
bn =
1
L
1
L
Z
L
f (x) cos
−L
Z L
f (x) sin
−L
nπx
2
dx =
L
L
nπx
dx = 0,
L
and
a0 =
1
2L
Z
L
Z
f (x) cos
0
nπx
dx,
L
n = 1, 2, 3 . . .
n = 1, 2, 3 . . . (because f (x) sin
L
f (x) dx =
−L
1
L
nπx
is odd)
L
(3.54)
(3.55)
L
Z
f (x) dx.
(3.56)
0
Hence, the Fourier series of an even function f (x) is a cosine series:
f (x) ∼ a0 +
∞
X
an cos
n=1
where ak (k = 0, 1, 2, . . .) are given by (3.54) and (3.56).
nπx
.
L
(3.57)
69
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Example 3.7.3 Find the Fourier expansion of f (x) = x2 on [−π, π] .
Solution
f (x) = x2 is even. Thus
f (x) = a0 +
a0 =
an =
=
=
=
=
∞
X
an cos nx
n=1
π
π2
x2 dx =
(L = π)
3
0
Z0 π
Z π
2
f (x) cos nxdx =
x2 cos nxdx
π 0
0
π
Z π
1 2
2
x sin nx dx
x sin nx −
n
nπ 0
0
iπ 1 Z π
2 h
x
−
cos nxdx
− cos nx +
nπ
n
n 0
0
iπ π
2
1 h
− cos πn + 2 sin nx
−
n
n
n
0
2
(−1)n .
n2
1
π
2
π
2
π
Z
1
f (x) dx =
π
Thus
Z
π
∞
f (x) =
X (−1)n
π2
+2
cos nx.
3
n2
n=1
■
Remark 3.7.1 (Even Extension) Sometimes we are required to find a cosine series expansion for
a function f (x) defined in [0, L] . The function f (x) does not need to be even on [−L, L] , we must first
define the even extension of f over the interval [−L, L] .
The even extension Fe (f (x)) of a function f (x) defined on [0, L] is defined by
(
Fe (f (x)) =
f (−x) if
f (x) if
−L<x<0
0 < x < L.
(3.58)
■
Let us illustrate this by an example.
Example 3.7.4
Find and plot the even extension of f (x) = x on [0, L] .
Solution
(
Fe (f (x)) =
−x −L < x < 0
x
0<x<L
Figure 3.10 shows the graph of the even extension of the f (x) = x.
■
70
Figure 3.10: Even extension of f (x) = x on [−L, L]
We then define cosine Fourier series of a function f (x) defined on [0, L] by
f (x) = a0 +
∞
X
an cos
n=1
nπx
L
(3.59)
where the coefficients an (n = 0, 1, 2, . . .) are given by
a0 =
and
an =
Z
1 L
f (x) dx,
L 0
Z
2 L
nπx
f (x) cos
dx.
L 0
L
(3.60)
(3.61)
Note that the even extension of f is an even function, thus its Fourier Series is a cosine series and the
coefficients of its Fourier series are given by (3.54) and (3.56).
Example 3.7.5
Find the cosine Fourier expansion of f (x) = x on [0, π] .
Solution
f (x) ∼ a0 +
a0 =
an =
∞
X
an cos nx
n=1
π
Z
Z
1 π
π
1
f (x) dx =
x dx =
π 0
π 0
2
Z π
Z π
2
2
2
f (x) cos nx =
x cos nx dx = 2 [(−1)n − 1]
π 0
π 0
n π
Thus
f (x) =
∞
2X 1 π
n
+
(−1)
−
1
cos nx
2 π
n2
n=1
■
71
3.7.1.2
APM3701/1
Sine Fourier Series
nπx
nπx
is an odd function and the function f (x) sin
is an
If a function is odd, the function f (x) cos
L
L
even function. Consequently,
Z
1 L
nπx
an =
f (x) cos
dx = 0,
n = 0, 1, 2, 3, . . .
L −L
L
Z
Z
1 L
nπx
2 L
nπx
bn =
dx =
dx, n = 1, 2, 3, . . .
(3.62)
f (x) sin
f (x) sin
L −L
L
L 0
L
Remark 3.7.2 Only occasionally are we given an odd function and asked to find its Fourier series. The
sine series frequently arise in real life situations, and the solution is defined only for values of x ∈ [0, L] .
If f is only given in [0, L] , then it can be extended as an odd function (see figure 3.11). The odd extension
Figure 3.11: Odd extension of f .
Fo (f ) of f is defined in [−L, L] by
(
Fo (f ) (x) =
−f (−x) if
f (x)
if
−L<x<0
0<x<L
(3.63)
The sine Fourier series of f is then defined by
∞
X
f (x) =
n=1
where
2
bn =
L
Z
bn sin
nπx
L
L
f (x) sin
0
nπx
.
L
Example 3.7.6 Let f (x) = x2 , 0 ≤ x ≤ π.
1. Determine and plot its odd extension in the interval [−3π, 3π] .
2. Determine the sine Fourier series of f.
Solution
(
1. F0 (f ) (x) =
2. f (x) =
∞
P
−x2 −π < x < 0
x2
0<x<π
bn sin nx.
n=1
2
bn =
π
Z
0
π
x2 sin nx dx = −
2 2 2
n
n+1
n
π
(−1)
+
2
(−1)
+
2
.
n3 π
(3.64)
(3.65)
72
Figure 3.12: Odd extension of f (x) = x2 .
Thus
∞
2X 1 2 2
n
n+1
f (x) =
n
π
(−1)
+
2
(−1)
+
2
sin nx.
π
n3
n=1
■
3.7.2
Complex Fourier Series
Sometimes, it is convenient to represent a function by an expansion in complex form.
∞
P
nπx
Let f (x) = a0 +
an cos nπx
+
b
sin
.
n
L
L
n=1
Observing that
sin x =
eix − e−ix
eix + e−ix
and cos x =
.
2i
2
We have
f (x) = a0 +
= a0 +
∞
X
"
n=1
∞ X
n=1
= c0 +
an
∞ X
n=1
e
nπx
i
L
+ e−
2
nπx
i
L
+ bn
e
nπxi
L
− e−
2i
nπx
i
L
#
nπx
nπx
1
1
(an − ibn ) ei L + (an + ibn ) e−i L
2
2
cn ei
nπx
L
+ c−n e−i
nπx
L
(3.66)
73
APM3701/1
where
c0
1
= a0 =
2L
L
Z
f (x) dx
−L
Z L
an − ibn
nπx
1
nπx f (x) cos
=
− i sin
dx
2
2L −L
L
L
Z L
nπx
1
f (x) e−i L dx,
2L −L
Z L
an + ibn
nπx
1
nπx f (x) cos
=
+ i sin
dx
2
2L −L
L
L
Z L
nπx
1
f (x) ei L dx.
2L −L
cn =
=
c−n =
=
Equation (3.66) can be written in the form
∞
X
f (x) =
cn ei
n=−∞
where
1
2L
cn =
Z
L
nπx
L
−L<x<L
f (x) e−i
nπx
L
dx.
(3.67)
(3.68)
−L
Example 3.7.7
Find the complex Fourier series expansion of the function f (x) = ex in the interval [−π, π] .
Solution
We have
∞
X
cn einπx
−π <x<π
f (x) = ex =
n=−∞
1
cn =
2π
Z
π
f (x) e
−inx
dx =
−π
=
=
=
=
Thus
ex =
Z π
1
ex e−inx dx
2π −π
Z π
1
e(1−in)x dx
2π −π
1 h (1−in)x i π
1
e
2π 1 − in
−π
1
1 (1−in)π
e
− e−(1−in)π
2π 1 − in
(1 + in)
sinh (1 − in) π.
π (n2 + 1)
∞
X
(1 + in) inxπ
e
sinh (1 − in) π.
π (n2 + 1)
n=−∞
■
74
Exercises 3.7
(
1. Let
f (x) =
−π if −π < x < 0
x
if 0 < x < π.
(a) Plot the periodic extension of f for −3π ≤ x ≤ 3π.
(b) Find the Fourier series of f and evaluate it at each point in the interval [−π, π]
2. Find the complex Fourier series expansion of the following functions:
(a) f (x) = e−x
(
(b) f (x) =
1
cos x
−π <x<π
if −π < x < 0
if
0<x<π
(c) f (x) = x, −1 < x < 1.
3. Exercises 2.3 of [1], all problems.
4. Exercises 2.4 of [1], all problems.
5. Exercises 2.6 of [1], all problems.
3.8
3.8.1
■
THE STURM-LIOUVILLE THEORY AND EIGENVALUE PROBLEM
Sturm-Liouville Equation
As we will see in section 4.5, the method of separation of variables transforms a second order homogeneous
partial equation into two ordinary differential equations (4.22) and (4.23), section 4.5 of this Guide. Those
two equations are special cases of more general differential equation
a (x)
dX
d2 X
+ b (x)
+ (c (x) + λ) X = 0.
dx2
dx
(3.69)
We introduce the following functions (think of integrating factor for ordinary differential equation)
R
p (x) = e
b
dx
a
,
q (x) =
c
p,
a
s (x) =
1
p.
a
(3.70)
Equation (3.69) becomes:
d
dx
dX
p
dx
+ (q + λs) X = 0
(3.71)
which is known as the Sturm-Liouville equation. In the Sturm-Liouville equation (3.71) λ is a parameter
independent of x, functions p (x) , q (x) and s (x) are given function of x in some interval, say [a, b] .
Solutions to the Sturm-Liouville equation exists, when q (x) and s (x) are continuous and p (x) is continuously differentiable in a closed finite interval, say [a, b] .
Some well-known special cases of the Sturm-Liouville equation is firstly the “Bessel equation”, see section
4.7 of [1], and section 7.5 of this guide.
1 d
dX
n2
x
+ λ− 2 X =0
(3.72)
x dx
dx
x
75
or
APM3701/1
′ n2
xX ′ − X + λxX = 0
x
where p (x) = x, q (x) =
(3.73)
−n2
and s (x) = x.
x
This equation describes the characteristic oscillations of a circular membrane with fixed radius.
A second special case of the Sturm-Liouville equation is the “Legendre equation”, see section 5.5 of [1].
′
1 − x2 X ′ + λX = 0
(3.74)
where p (x) = 1 − x2 , q (x) = 0 and s (x) = 1, which describes the characteristic oscillations of a sphere.
3.8.2
Regularity of the Sturm-Liouville equation
The Sturm-Liouville equation (3.71) is said to be “regular” in the interval [a, b] if the functions p (x) and
s (x) are positive in [a, b] . And it is said to be “singular”, when the interval is semi-infinite or infinite or
when p (x) or s (x) vanishes at one or both ends of a finite interval.
For example the Bessel Equation will be regular if and only if 0 < a < b < ∞, and the Legendre equation
will be regular in any subinterval of (−1, 1) .
3.8.3
Eigenvalues and Eigenfunctions
As we will see shortly, when subject to the boundary conditions of a particular type, the Sturm-Liouville
equation (3.71) will in general have non-trivial (non zero) solutions only for some particular values of λ.
Such solutions are called eigenfunctions of the Sturm-Liouville equation, and the corresponding values of
λ are called the eigenvalues.
Let us assume that Xn (x) and Xk (x) are two different eigenfunctions of (3.71) corresponding to two
different eigenvalues λn and λk respectively. Then, by definition, we have
dXn
d
p
+ (q + λn s) Xn = 0
(3.75)
dx
dx
d
dXk
p
+ (q + λk s) Xk = 0
(3.76)
dx
dx
We multiply (3.75) by Xk and (3.76) by Xn and subtract the resulting equations, we obtain
d
dXk
d
dXk
d
dXn
(λn − λk ) sXn Xk = Xn
p
− Xk
p
− Xk
p
dx
dx
dx
dx
dx
dx
d
pXn Xk′ − pXk Xn′
dx
=
(3.77)
where the primes refer to derivatives.
We integrate (3.77) with respect to x on (a, b) (arbitrary limits), we get,
(λn − λk )
Z
a
b
h
ib
s(x)Xn (x)Xk (x)dx = pXn Xk′ − pXk Xn′ .
a
(3.78)
76
If the eigenfunctions Xn and Xk are orthogonal with respect to the weight function s (see section 3.4.1 of
this Study Guide), then
h
ib
pXn Xk′ − pXk Xn′
= p (b) Xn (b) Xk′ (b) − p (b) Xk (b) Xn′ (b)
a
−p (a) Xn (a) Xk′ (a) − p (a) Xk (a) Xn′ (a) = 0.
(3.79)
Equation (3.79) can be satisfied in a number of ways by imposing certain types of boundary conditions on
X (x) . Well known examples of boundary conditions which satisfy equation (3.79) are:
1.
Xn (a) = Xn (b) = 0
and
Xk (a) = Xk (b) = 0.
(3.80)
Xn′ (a) = Xn′ (b) = 0
and
Xk′ (a) = Xk′ (b) = 0.
(3.81)
2.
3.
Xn (a) Xk′ (a) − Xk (a) Xn′ (a) = Xn (b) Xk′ (b) − Xk (b) Xn′ (b) = 0.
If the boundary conditions are such that
(
and
(
(3.82)
α1 Xn (a) + α2 Xn′ (a) = 0
α1 Xk (a) + α2 Xk′ (a) = 0
(3.83)
β 1 Xn (b) + β 2 Xn′ (b) = 0
β 1 Xk (b) + β 2 Xk′ (b) = 0
(3.84)
where α1 , α2 , β 1 , β 2 are real constants, then the simultaneous equations (3.83) in α1 and α2 may take
non-zero values only if Wronskian determinant
Xn (a) Xk (a)
Xn′ (a) Xk′ (a)
= 0.
(3.85)
Likewise the system (3.84) in β 1 and β 2 may take non-zero values only if
Xn (b) Xk (b)
Xn′ (b) Xk′ (b)
= 0.
(3.86)
Expressions (3.85) and (3.86) are respectively
Xn (a) Xk′ (a) = Xk (a) Xn′ (a) = 0.
(3.87)
Xn (b) Xk′ (b) − Xk (b) Xn′ (b) = 0.
(3.88)
and
Equations (3.87) and (3.88) together show that boundary conditions (3.82) are satisfied.
Consequently boundary conditions of type (3.83) and (3.84) which are linear combinations of eigenfunctions
and their derivatives at the end points of the interval (a, b) also give rise to eigenfunctions of the SturmLiouville equation which are orthogonal with respect to the weight function s (x) on the interval (a, b) .
All this boundary conditions are homogeneous in the sense that if X is substituted by cX, where c is
constant, then the boundary conditions are unaltered.
This is not the case with non-homogeneous boundary conditions, e.g. if Xn (a) = K, where K ̸= 0,
CXn (a) = CK ̸= K if C ̸= 1. In this case, the eigenfunctions of the Sturm-Liouville equation will not, in
general, form an orthogonal set.
77
3.8.4
APM3701/1
Sturm-Liouville Eigenvalue Problems - Examples
The Sturm-Liouville equation
d
dx
dX
p
+ (q − λs) X = 0
dx
(3.89)
together with the homogeneous boundary conditions of the form
α1 X (a) + α2 X ′ (a) = 0
β 1 X (b) + β 2 X ′ (b) = 0
(3.90)
where X ′ denotes the differentiation and the constants α1 , α2 , β 1 and β 2 are real numbers, is called the
Sturm-Liouville problem or Sturm-Liouville system. As mentioned in section 3.8.3 above, the values of λ for
which the Sturm-Liouville problem has a nontrivial solution are called “eigenvalues”, and the corresponding
solutions are called “eigenfunctions”. The “spectrum” of the system is the set of all eigenvalues of a regular
Sturm-Liouville system.
Example 3.8.1
Determine the eigenvalues and the corresponding eigenfunctions of the following problems
(a)
X ′′ + λX = 0, 0 ≤ x ≤ L
X (0) = 0, X (L) = 0.
)
X ′′ + λX = 0, a ≤ x ≤ b
X (a) = 0, X (b) = 0.
)
(3.91)
(b)
Solution
Here p = 1, q = 0 and s = 1.
The solution of the system (3.91) depends on the sign of λ:
(a) (a.1) If λ = 0, the problem (3.91) has solution
X (x) = Ax + B.
X (0) = 0 gives B = 0, and in a similar way X (L) = 0 gives A = 0. Thus X(x) ≡ 0
Hence λ = 0 is not an eigenvalue.
(a.2) If λ < 0, we set λ = −µ2 , the problem (3.91) has solution
X (x) = Aeµx + be−µx
or
X (x) = A cosh µx + B sinh µx.
The boundary conditions give:
X (0) = 0 implies that A = 0. Thus X(x) = B sinh µx, and for the same reason X (L) =
0 gives B sinh µL = 0, implies that B = 0. Thus X (x) ≡ 0. Hence λ < 0 is not an eigenvalue.
78
(a.3) If λ > 0, λ = µ2 , the solution of the problem (3.91) is
X (x) = A cos µx + b sin µx.
The end condition X (0) = 0 implies that A = 0. The condition X (L) = 0 yields
B sin µL = 0.
No trivial solutions to the Sturm-Liouville exist only when
sin µL = 0, B ̸= 0
or
µL = nπ, =⇒ µ =
Thus the eigenvalues are
λn =
nπ 2
L
and the corresponding eigenfunctions are
(b)
X ′′ + λX = 0, a ≤ x ≤ b
X (a) = 0, X (b) = 0.
n = 1, 2, 3, . . .
,
X(x) = Xn (x) = sin
nπ
.
L
nπ
x,
L
(3.92)
n = 1, 2, 3, . . . .
(3.93)
)
X ′′ + λX = 0, 0 ≤ x̄ ≤ b − a
Let x̄ = x − a, the above system becomes
X (0) = 0, X (b − a) = 0.
)
.
From (a) above, we know that (let L = b − a) the eigenvalues and eigenfunctions are λn =
and Xn (x̄) = sin
nπx̄
respectively. Thus
b−a
λn =
nπ
b−a
2
and Xn (x̄) = sin
nπ (x − a)
,
b−a
n = 1, 2, 3, . . .
nπ
b−a
2
(3.94)
■
Example 3.8.2 Consider the Sturm-Liouville equation
X ′′ + λX = 0.
Determine the eigenvalues and eigenfunctions, given the end conditions:
(a) X (0) = 0 and X ′ (L) = 0
(b) X ′ (0) = 0 and X ′ (L) = 0
(c) X (a) = 0 and X ′ (b) = 0
a < b.
(d) X ′ (a) = 0 and X ′ (b) = 0
a < b.
79
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Solution
(a)
(
X ′′ + λX = 0
0<x<L
′
X (0) = 0 = X (L)
(a.1) If λ = 0, the solution of the ordinary differential equation (3.95) is
X (x) = c1 x + c2 .
X (0) = 0 implies c2 = 0, and
X ′ (x) = c1 , thus X ′ (L) = 0 yields c1 = 0.
Therefore X (x) = 0. Thus λ = 0 is not an eigenvalue.
(a.2) If λ < 0, let λ = −µ2 , the solution of (3.95) is
X (x) = c1 cosh µx + c2 sinh µx.
X (0) = 0 yields c1 = 0. ⇒ X(x) = c2 sinh µx.
X ′ (x) = c2 µ cosh µx.
The end condition X ′ (L) = 0 gives
c2 µ cosh µL = 0 ⇒ c2 = 0 (since cosh x ̸= 0 for all x).
We have X (x) = 0, and consequently λ < 0 is not an eigenvalue.
(a.3) If λ > 0, let λ = µ2 , the Sturm-Liouville problem (3.95) has solution:
X (x) = c1 cos µx + c2 sin µx.
We apply the end condition X (0) = 0, we obtain c1 = 0. The condition X ′ (L) = 0 gives
c2 µ cos µL = 0
since λ = 0 is not an eigenvalue, the Sturm-Liouville problem has nontrivial solutions if
cos µL = 0, c2 ̸= 0 ⇒ µL =
(2n − 1)
π, n = 1, 2, 3, . . .
2
Thus the eigenvalues are
λn =
2n − 1
π
2L
2
,
n = 1, 2, 3 . . .
and the corresponding eigenfunctions are
Xn (x) = sin
(2n − 1)
πx,
2L
n = 1, 2, . . .
(3.95)
80
Therefore the eigenvalue of the problem are:
λn =
2n − 1
π
2L
2
, n = 1, 2, . . .
and the corresponding eigenfunctions are:
Xn (x) = sin
2n − 1
πx
2L
(b)
(
X ′′ + λX = 0
0<x<L
′
′
X (0) = 0 = X (L)
(3.96)
(b.1) If λ = 0, as before, the solution of (3.96) is
X (x) = Ax + B.
The conditions X ′ (0) = 0 and X ′ (L) = 0 give A = 0,. Thus X(x) = B, B arbitrary. λ = 0 is
eigenvalue with X(x) = B (take B = 1) as eigenfunction.
(b.2) If λ < 0, let λ = −µ2 , the solution of the Sturm-Liouville problem (3.96) is
X (x) = A cosh µ + B sinh µx
or X (x) = Aeµx + Be−µx .
The condition X ′ (0) = 0 implies that B = 0 and X ′ (L) = 0 gives Aµ sinh µL = 0.
Since λ ̸= 0 and L ̸= 0, we have A = 0.
Hence
X (x) ≡ 0,
λ < 0 is not an eigenvalue.
(b.3) If λ > 0, let λ = µ2 , as in (a) the Sturm-Liouville problem (3.96) has solution
X (x) = A cos µx + B sin µx.
The condition X ′ (0) = 0 yields B = 0.
X ′ (L) = 0 gives −Aµ sin µL = 0 ⇒ sin µL = 0 ⇒ µL = nπ, n = 1, 2, 3, . . .
Hence the eigenvalues are
λn =
nπ 2
L
,
n = 0, 1, 2, 3, . . .
(3.97)
and the corresponding eigenfunctions are
Xn (x) = cos
nπx
,
L
n = 0, 1, 2, 3, . . .
(3.98)
81
(
(c)
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X ′′ + λX = 0
a<x<b
′
X (a) = 0 = X (b)
Let x̄ = x − a. the problem becomes
(
X ′′ + λX = 0
0 < x̄ < b − a
′
X (0) = 0 = X (b − a)
From (a) above, we known that the eigenvalues and eigenfunctions are given by (take L = b − a)
(2n − 1) 2
2n − 1
λn =
π
and
Xn (x̄) = sin
πx̄, n = 1, 2, 3, . . . respectively. Thus the
2 (b − a)
2 (b − a)
eigenvalues and eigenfunctions are
λn =
(2n − 1)
π
2 (b − a)
2
and
Xn (x) = sin
2n − 1
π (x − a) , n = 1, 2, 3, . . .
2 (b − a)
respectively.
(
X ′′ + λX = 0
a<x<b
(d)
Let x̄ = x − a, the problem becomes
X ′ (a) = 0 = X ′ (b) .
(
X ′′ + λX = 0
0 < x̄ < b − a
′
′
X (0) = 0 = X (b − a)
From (b) above, we obtain the eigenvalue and eigenfunctions (take L = b − a): λn =
Xn (x̄) = cos
nπx̄
,
b−a
2
, and
n = 0, 1, 2, 3, . . . respectively. Thus, the eigenvalues and eigenfunctions are
λn =
nπ
b−a
nπ
b−a
2
,
Xn (x) = cos
and
respectively.
nπ (x − a)
, n = 0, 1, 2, 3, . . .
b−a
■
Example 3.8.3 Solve the eigenvalue problem
X ′′ + λX = 0.
With the following boundary conditions:
1. X (−L) = X (L) , and X ′ (−L) = X ′ (L) .
2. X (a) = X (b) , and X ′ (a) = X ′ (b) .
3. X (a) = X (a + 2L) , and X ′ (a) = X ′ (a + 2L) .
♦ Note: The above boundary conditions are known as periodic boundary conditions, see section 4.5.5 of
this Guide.
82
Solution
1.
X ′′ + λX = 0.
−L≤x≤L
X (−L) = X (L) , and
′
(3.99)
′
X (−L) = X (L) .
(a) If λ = 0, the solution of the Sturm-Liouville problem (3.99) is
X (x) = c1 x + c2
X (−L) = X (L) implies that −c1 L = c1 L ⇒ c1 = 0 and c2 arbitrary (take c2 = 1 as eigenfunction for the eigenvalue λ = 0). X ′ (−L) = X ′ (L) gives no useful information.
Thus λ0 = 0 is an eigenvalue and X0 (x) = 1 is the corresponding eigenfunction.
(b) If λ < 0, let λ = −µ2 then X (x) = A cosh µx + B sinh µx
X (−L) = X (L) implies A cosh (µL) − B sinh (µL) = A cosh (µL) + B sinh (µL)
⇒ −B sinh (µL) = B sinh (µL) = 0,
=⇒ B = 0.
X ′ (−L) = X ′ (L) =⇒ −Aµ sinh µL+µB cosh µL = Aµ sinh µL+Bµ cosh µL =⇒ A sinh (µL) .
⇒ A = 0 since sinh (µL) ̸= 0. Thus X (x) = 0, and λ < 0 is not an eigenvalue.
(c) If λ > 0, let λ = µ2 , the solution of the Sturm-Liouville problem (3.99) is:
X (x) = A cos µx + B sin µx.
X(−L) = X(L) =⇒ A cos (µL) − B sin (µL) = A cos (µL) + B sin (µL) , or
A (cos (µL) − cos (µL)) + B (sin (µL) + sin (µL)) = 0 =⇒ B sin (µL) = 0 =⇒ µL = nπ
=⇒ λn =
nπ 2
.
L
Therefore the eigenvalues and eigenfunctions of the Sturm-Liouville problem (3.99) are:
nπ 2
nπx
nπx
λ0 = 0, λn =
and X0 = 1, Xn (x) = An cos
+ Bn sin
L
L
L
n = 0, 1, 2, 3, . . . respectively.
(3.100)
(
2.
X ′′ + λX = 0
X (a) = X (b) , and
Let x̄ = x −
a≤x≤b
X ′ (a) = X ′ (b) .
b+a
, the problem becomes
2

b−a
b−a

 X ′′ + λX = 0,
−
≤x≤
2
2 b−a
b−a
b−a
b−a

′
′
 X −
=X
, and X −
=X
.
2
2
2
2
2nπ 2
b−a
From (a) above, the eigenvalues and eigenfunctions are (take L =
), λn =
and
2
b−a
2nπx̄
2nπx̄
Xn (x) = An cos
+ Bn sin
, n = 0, 1, 2, 3, . . . respectively. Thus the eigenvalues and
b−a
b−a
eigenfunctions of the problem are
2nπ x − b+a
2nπ x − b+a
2nπ 2
2
2
λn =
and Xn (x) = An cos
+ Bn sin
,
b−a
b−a
b−a
n = 0, 1, 2, 3, . . . respectively.
83
(
3.
X ′′ + λX = 0,
X (a) = X (a + 2L) , and
APM3701/1
a ≤ x ≤ a + 2L
X ′ (a) = X ′ (a + 2L) .
Let x̄ = x − a − L, the problem becomes
X ′′ + λX = 0.
−L≤x≤L
X (−L) = X (L) , and
X ′ (−L) = X ′ (L) .
From (a) above, the eigenvalues and eigenfunctions are
nπx̄
nπx̄
+ Bn sin
, n = 0, 1, 2, 3, . . . respectively.
Xn (x) = An cos
L
L
Thus the eigenvalues and eigenfunctions of the problem are
nπ 2
nπ
nπ
λn =
and Xn (x) = An cos
(x − a − L) + Bn sin
(x − a − L) ,
L
L
L
n = 0, 1, 2, 3, . . . respectively.
■
Example 3.8.4
1. Determine the eigenvalues and eigenfunctions of the Cauchy equation
x2 X ′′ + xX ′ + λX = 0,
a≤x≤b
with boundary conditions
X (a) = 0 and X (b) = 0.
2. In particular solve the following eigenvalue problem
x2 X ′′ + xX ′ + λX = 0,
1≤x≤e
X (1) = 0, X (e) = 0.
Solution
1. We let y = ln xa ⇒ x = aey
X′ =
X ′′ =
=
dX
dX dy
1 dX
1
dX
=
·
=
= e−y
(Chain rule)
dx
dy dx
x dy
a
dy
d2 X
d dX
d dX
dy
=
·
(Chain rule)
=
2
dx
dx dx
dy dx
dx
1 d 1 −y dX
1 −2y dX
d2 X
−y
e
e = − 2e
−
.
a dy a
dy
a
dy
dy 2
Substitution into the differential equation yields
d2 X
1
dX
a2 2y −2y dX
− 2e e
−
+ aey e−y
+ λX = 0
2
a
dy
dy
a
dy
d2 X
⇒
− λX = 0.
dy 2
84
The boundary conditions become:
x = a ⇒ y = ln
and
a
=0
a
b
x = b ⇒ y = ln .
a
The differential equation becomes
(
d2 X
dy 2
0 ≤ y ≤ ln ab
X (0) = 0 = X ln ab .
+ λX = 0
The eigenvalues and eigenfunctions of the above system are (see example 3.8.1, let L = ln ab ):
λn =
nπ
ln ab
!2
and
Xn (y) = sin
,
nπy
ln ab
n = 1, 2, 3, . . .
n = 1, 2, 3, . . .
since
x
y = ln , we have the eigenfunctions
a
nπ ln xa
Xn (x) = sin
, n = 1, 2, 3, . . . .
ln ab
corresponding to the eigenvalues:
λn =
nπ
ln ab
!2
,
n = 1, 2, 3, . . . .
(b) a = 1, b = e. Thus
λn = (nπ)2
and Xn (x) = sin (nπ ln x)
n = 1, 2, 3, . . . .
■
Example 3.8.5
Find the eigenvalues and eigenfunctions of the following Bessel parametric equation:
dR
1 d
r
= −λ2 , 0 < r < a
rR dr
dr
R (a) = 0.
♦ Note: This eigenvalue problem is encountered while solving partial differential equation involving Laplacian in polar coordinates, by method of separation of variables, is the parametric Bessel equation of order
0 (see section 7.5.1 of this guide and section 4.7 of [1]).
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Solution
The above ordinary differential equation can be rewritten by in the form
r
d2 R dR
+
+ λ2 rR = 0, 0 < r < a
dr2
dr
R (a) = 0.
Or
r2
dR
d2 R
+r
+ λ2 r2 R = 0, 0 < r < a
2
dr
dr
R (a) = 0.
The system is short of one boundary condition, this comes in form of boundedness at r = 0, i.e. |R (0)| < ∞.
The above differential equation has solution (see section 7.5.3)
R (r) = c1 J0 (λr) + c2 Y0 (λr)
where J0 and Y0 are Bessel functions of first and second kind.
Due to the boundedness at r = 0, we have c2 = 0, (see Bessel functions in section 7.5.1 of this guide).
Thus
R (r) = CJ0 (λr) .
The condition R (a) = 0 implies
J0 (aλ) = 0.
Thus the eigenvalue λn =
eigenfunctions are
αn
a ,
where αn are the zeros of the Bessel function J0 (x) . And the corresponding
J0 (λn r) , n = 1, 2, 3, . . .
(3.101)
■
Example 3.8.6 Example 4, section 6.2 of [1].
Example 3.8.7 Example 5, section 6.2 of [1].
3.8.5
Summary of Various Eigenvalue Problems
1. We summarise in Table 3.1 the eigenvalues and eigenfunctions of eigenvalue problem 3.102 with the
indicated boundary conditions
X ′′ + λX = 0
(3.102)
2. Table 3.2, give eigenvalues and eigenfunctions of the eigenvalue problem (3.103) with the indicated
boundary conditions
x2 X ′′ + xX ′ + λX = 0
(3.103)
3. We summarise in Table 3.3 the the eigenvalues and eigenfunctions of eigenvalue problem (3.104) with
the indicated boundary conditions
X ′′′′ − λX = 0
(3.104)
86
Boundary conditions
X (0) = X (L) = 0
X (0) = X ′ (L) = 0
X ′ (0) = X (L) = 0
X ′ (0) = X ′ (L) = 0
X (0) = X (L) = 0,
X ′ (0) = X ′ (L) = 0
X (−L) = X (L) ,
X ′ (−L) = X ′ (L)
Eigenvalues λ n
nπ 2
L
(2n − 1) π 2
2L
(2n − 1) π 2
2L
nπ 2
L 2
2nπ
L
nπ 2
L
nπ 2
b−a
(2n − 1) π 2
2 (b − a)
nπ 2
b−a
X (a) = X (b) = 0
X (a) = X ′ (b) = 0
X ′ (a) = X ′ (b) = 0
X (a) = X (b) ,
X ′ (a) = X ′ (b)
X (a) = X (a + 2L) ,
X ′ (a) = X ′ (a + 2L)
2nπ
b−a
2
nπ 2
L
Eigenfunctions Xn
nπx
sin
L
n − 12 πx
sin
L
n − 12 πx
cos
L
nπx
cos
L
2nπx
sin
L
nπx
sin
L
nπx
cos
L
nπ (x − a)
sin
b−a
(2n − 1) π (x − a)
sin
2 (b − a)
nπ (x − a)
sin
b−a 2nπ x − b+a
2
sin
b−a
2nπ x − b+a
2
cos
b−a
nπ (x − a − L)
sin
L
nπ (x − a − L)
cos
L
n = 1, 2, 3, ...
n = 1, 2, 3, ...
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
n = 0, 1, 2, 3, ...
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
n = 1, 2, 3, ...
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
Table 3.1: Eigenvalues and eigenfunctions of eigenvalue problem (3.102) with the indicated boundary
conditions
Boundary conditions Eigenvalues λ n Eigenfunctions Xn
!2
nπ ln xa
nπ
X (a) = X (b) = 0
sin
n = 1, 2, 3, ...
ln ab
ln ab
Table 3.2: Eigenvalues and eigenfunctions of eigenvalue problem (3.103) with the indicated boundary
conditions
87
Boundary conditions
X (0) = X (L) = 0, X ′′ (0) = X ′′ (L) = 0
X ′ (0) = X ′ (L) = 0, X ′′′ (0) = X ′′′ (L) = 0
APM3701/1
Eigenvalues λ n
nπ 4
L
nπ
4
L
Eigenfunctions Xn
nπx
sin
L
nπx
cos
L
n = 1, 2, 3, ...
n = 0, 1, 2, 3, ...
Table 3.3: Eigenvalues and eigenfunctions of eigenvalue problem (3.104) with the indicated boundary
conditions
3.8.6
Properties of eigenvalues and eigenfunctions of regular Sturm-Liouville problem
1. The eigenvalue of a regular Sturm-Liouville problem are real and form an increasing sequence λ1 <
λ2 < λ3 < ... <
λn −→ ∞ as n −→ ∞.
2. Each eigenvalue of a regular Sturm-Liouville problem has only one linearly independent eigenfunction
corresponding to it.
3. Eigenfunctions corresponding to different eigenvalues of a regular Sturm-Liouville problem are orthogonal with respect to weight function r.
For the proof see Theorem 2, in section 6.2 of [1].
Exercises 3.8
1. Show that every second ordinary differential equation of the form
p (x) X ′′ (x) + q (x) X ′ (x) + r (x) X (x) = 0
can be transformed into an equation of the form:
P (x) X ′′ + QX ′ + RX = 0
′
or P (x) X ′ + RX = 0.
where P ′ = Q.
Z q − p′
Hint: multiply (3.105) by exp
dx .
p
2. Without solving (explicitly) the eigenvalue problem
X ′′ − λX = 0
X (0) = 0,
0≤x≤1
X (1) + X ′ (1) = 0,
show that all eigenvalues are positive.
[Hint: Multiply the differential equation by X and integrate by parts on (0, 1), solve for λ.]
3. Exercises 6.2 of [1], all problems.
(3.105)
88
89
CONTENTS
Chapter 4
PARTIAL DIFFERENTIAL
EQUATIONS IN RECTANGULAR
COORDINATES
4.1
BACKGROUND
This chapter is the core of this module, it is mainly devoted to the derivation and the solution of the three
basic types of second order partial differential equations in rectangular coordinates, namely:
1. The heat Equation
ut − kuxx = 0,
2. The wave equation
utt − c2 uxx = 0,
3. The Laplace equation
uxx + uyy = 0.
This chapter is also devoted to the derivation of some well-known partial differential equations of mathematical physics from laws of conservation.
The method of solution for solving these partial differential equations by method of separation of variables
and using the Fourier series is discussed in details. We will append dynamic boundary conditions, i.e.
boundary conditions which contain time derivatives to the heat equation and wave equation.
4.2
LEARNING OUTCOME
The following outcome will be achieved at the end of this chapter:
• Derive the heat, wave and Laplace equation in rectangular coordinates.
• Apply Fourier series to obtain solution of boundary-value problems with various initial and boundary
conditions by method of separation of variables.
• Derive partial differential equations of mathematical physics by laws of conservation and by energy
method.
90
4.3
MATHEMATICAL MODELLING
Since, when we describe a physical situation as a partial differential equation together with a set of
boundary conditions, we are constructing a particular kind of mathematical model, we consider it as
appropriate to include in this introductory section a subsection on mathematical modelling in general.
Before proceeding to this, we need the following definitions:
4.3.1
Definition
The mathematical models, consisting of a Partial Differential Equation together with a set of boundary
conditions, which we are going to derive in this chapter are classified as so-called deterministic models.
This type of model may be described as the opposite of a probabilistic model, i.e. a model where an
element of chance is involved.
4.3.2
Mathematical Modelling
Following Haberman [5], we shall describe mathematical modelling as the process whereby the evolution or the state of a real-life system is represented by a set of mathematical relations, after proper
approximations and idealisations.
In general this process can be divided into the following steps: (we acknowledge this section in [5]):
4.3.2.1
Objective
In this step the real-life system to be modelled is defined. It is often convenient to cut the system down
to its essential features, so that a prototype for it can be constructed. This enables one to concentrate
on the essential processes taking place in the system under consideration.
4.3.2.2
Background
In this step the pertinent laws and data about the system have to be examined. In deterministic models
it is of importance to make sure that the problem is not over-specified (too many data) or under-specified
(few data), since then the problem will turn out to be ill-posed, (see also section 1.6.2).
4.3.2.3
Approximations and idealisations
Even when constructing a prototype model, some approximations and idealisations of reality must be
made. Thus all mathematical models are to an extent approximations of reality. These approximations
should, however, be kept within reasonable limits, so that the model will still correlate with the actual
behaviour of the system.
4.3.2.4
Modelling
The mathematical relations governing the system are now derived. In the case of a deterministic model
we now arrive at the boundary-value problem, i.e. the Partial Differential Equation with accompanying
boundary conditions.
91
4.3.2.5
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Model validation
Methods must now be developed to solve the model. The conclusions that may be drawn from the
solutions (model predictions), must be compared with the actual data about the system. If there are
large deviations between the model predictions and the data, the model must be re-examined and modified
accordingly.
This aspect is of utmost importance and the adequacy of the model should be tested even
if an exact solution cannot be constructed. (We will return to this point in Section 5)
4.3.2.6
Compounding
This entails the modification of the model in order to take into account aspects of the system that were
neglected earlier in the process of simplification.
In the following sections, we will derive the heat and the wave equations in one dimension, the Laplace
Equation in two dimensions, Burgers’s equation and the equation of the elastic beam in one dimension.
4.4
4.4.1
THE HEAT EQUATION
Derivation of the heat equation
Based on the Appendix at the end of Section 3.5 of [1].
In Section 4.10, we will derive the Heat Equation using another variant of law of conservation.
Heat may be transferred by conduction, convection and radiation. In conduction, the heat (molecular
motion or vibration) is transferred locally by impacts of molecules with adjacent molecules. With convection, heat is carried from one region to another by a current flow, and heat radiation occurs via
infrared electromagnetic waves. Here, we will limit ourselves to heat transferred by conduction. In this
section, we consider the case of one–dimensional heat flow.
Let us consider a thin straight rod of uniform cross section made of some heat–conduction substance, in
which thermal energy is transferred because of temperature differences in the rod and external sources of
thermal energy. We model the rod as an interval I = [0, L] and denote by u (x, t) the temperature at the
point x ∈ I at time t ≥ 0.
We also assume that the temperature u (x, t) is constant on each cross section located at x at a certain
time t. We introduce the following quantities:
ρ (x, t) = density of rod (mass per unit volume at point x and at time t),
c (x, t) = specific heat of the rod (amount of thermal energy needed to increase the temperature
of a unit mass of the material at point x and at time t by one degree),
L=
length of the rod,
A(x) =
area of cross section at the point x ∈ I,
E (t) =
total heat energy at time t.
Consider a slab of the rod of small thickness ∆x at x = x0 (see Figure 4.1). The mass of the slab is
ρA (x) ∆x. Hence the energy required to raise the temperature of the slab from a reference temperature
say, 0 to its current temperature u (x, t) is
c (x, t) u (x0 , t) ρ (x, t) A (x) ∆x.
92
Figure 4.1: Temperature distribution in an uniform rod of length L.
If u (x0 , t) < 0, then the energy must be extracted to lower the temperature. Adding up the energies of all
the slabs between x = a and x = b, letting ∆x → 0 (for accuracy), we arrive at the following expression
for the heat energy of the portion of the rod from x = a to x = b at time t.
b
Z
E (t) =
c (x, t) ρ (x, t) A (x) u (x, t) dx.
(4.1)
a
It is known (by experience) that heat energy flows from hotter regions to cooler regions, and that the rate
of heat flow is proportional to the temperature difference, divided by the distance between the regions.
In other words, the rate at which heat energy passes through the cross section at x = a in the positive
direction is
∂u
or − λ (x) A (x) ux (a, t) ,
(4.2)
−λ (x) A
∂x x=a
where λ (x) > 0, is called the thermal conductivity of the material of the rod.
If ux (a, t) < 0, then the temperature on the left hand side of x = a is greater than the temperature on
the right hand side of x = a. That is, the temperature decreases as x increases at x = a, the heat energy
flows to the right (positive), hence (4.2) must be positive, this explains the minus sign in (4.2). Fourier’s
law of heat conduction reads
∂u
ϕ(x, t) = −λ(x) .
∂x
Where ϕ(x, t) is the heat flux.
We assume that the rod is insulated on the outside, the only way that heat energy can enter the part of
the rod between x = a and x = b is through the cross sections at x = a and x = b.
We also assume that there are no thermal sources of heat such as chemical reactions or radioactivity.
Thus the net rate at which heat energy enters this part of the rod is the rate at which it enters the end
at x = a, minus the rate at which it leaves through the end at x = b. So we get
E ′ (t) = −λ (x) A (x) ux (a, t) − (−λ (x) A (x) ux (b, t))
Z b
b
∂
= λ (x) A (x) ux (x, t) =
(λ (x) A (x) ux (x, t)) dx.
a
a ∂x
(4.3)
Here we assumed that u is C 2 . By differentiating (4.1), we compute on the other hand E ′ (t) ,
d
E (t) =
dt
′
Z
b
c (x, t) ρ (x, t) A (x) ut (x, t) dx.
a
(4.4)
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Combining (4.3) and (4.4), we obtain
Z
a
or
b
Z b
i
∂ ∂ h
c (x, t) ρ (x, t) A (x) u (x, t) dx =
λ (x) A (x) ux (x, t) dx
∂t
a ∂x
Z b
a
∂ ∂ c (x, t) ρ (x, t) A (x) u (x, t) −
λ (x) A (x) ux (x, t) dx = 0.
∂t
∂x
Where [a, b] is an arbitrary subinterval of [0, L] , it follows that
∂ ∂ c (x, t) ρ (x, t) A (x) u (x, t) =
λ (x) A (x) ux (x, t) .
∂t
∂x
(4.5)
If c and ρ do not dependent on t, equation (4.5) becomes:
c (x) ρ (x) A (x) ut (x, t) =
∂ λ (x) A (x) ux (x, t)
∂x
(4.6)
λ
, where k is called the diffusivity
cρ
of heat or thermal diffusivity for the material in the rod, we obtain:
If λ, A and c are constant, we divide through by cρA and define k =
ut (x, t) − kuxx (x, t) = 0.
(4.7)
We assumed that the rod is insulated on the outside, except possibly over the ends at x = 0 and x = L.
Equation (4.7) which we have already encountered in the Section 1.7 is known as the heat equation.
4.4.2
Initial conditions for the Heat Equation
Equation (4.7) describes the flow of heat energy, it has one time derivative, we must be given one initial
condition (usually at t = 0). In many instances the initial temperature is not constant, but a function of
x:
u (x, 0) = f (x)
describing the temperature at x at initial time t = 0.
4.4.3
Boundary conditions for the Heat Equation
Two conditions are needed, as the second spatial derivatives present in (4.7) demand, usually one condition
at each end point. The appropriate condition depends on the physical mechanism in effect at each end
point, and often the condition at the boundary depends on both the material inside and outside the rod.
To simplify the mathematical problem, we will assume that the outside environment is known, we present
the following type of boundary conditions:
4.4.3.1
Prescribed temperature: Boundary conditions of first kind
The temperature at the end points of the rod, x = 0 or x = L, may be approximated by a prescribed
temperature
u (0, t) = up (t)
(4.8)
94
Often the temperature at the end points is constant, and equal to the temperature of the substance (bath)
in which the rod is submerged.
Well known example of prescribed temperature is the zero temperature at finite ends,
u(0, t) = 0 = u(L, t).
We will solve the heat equation with zero temperature at finite ends in section 4.5.2
4.4.3.2
Insulated boundary: Boundary of second kind
In this case, the heat flow is prescribed rather than the temperature,
−ux (0, t) = g (t)
(4.9)
where g (t) is given.
A well–known example of the prescribed heat flow boundary condition is when an end point is insulated
(both end points may be insulated). In this case there is no heat flow at the boundary, for instance
ux (0, t) = 0.
(4.10)
We will solve the heat equation with insulated ends in section 4.5.3.
♦ Remark: Equation (4.9) and (4.10) cannot be integrated because the slope ux is known only at one
value of x, i.e x = 0.
4.4.3.3
Newton’s law of cooling: Boundary of third kind
Imagine a very warm rod in contact with cooler moving air at one end. Heat will leave the rod, heating
up the air, at this end the rod loses heat to the surrounding medium. The air will then carry the heat
away, the air being however hotter near the rod. This process of heat transfer is called convection.
This is a complicated problem: the air temperature will vary with distance from the rod.
Experiments show that, as a good approximation, the heat flow leaving the rod is proportional to the
temperature difference between the rod and the prescribed external temperature. This boundary condition
is called Newton’s law of cooling or boundary condition of third kind. If it is valid at x = 0, then
−λAux (0, t) = −H u (0, t) − uB (t)
(4.11)
where λ is the thermal conductivity, A is the cross section and H is the proportionality constant called
heat transfer constant (or the convection coefficient). Attention must be given to the minus sign
in front of the proportionality constant. If the rod is hotter than the bath in which it is submerged
(u (0, t) > uB (t)) , then usually heat flows out of the rod at x = 0. Thus, heat is flowing to the left, and
in this case the heat flow would be negative, it is why we introduce a minus sign (H > 0) .
A similar derivation may be made at the other end x = L.
The coefficient H which is experimentally determined, depends on properties of the rod as well as the
bath properties. If H is very small, then very little heat energy flows across the boundary. In the limit
as H → 0, Newton’s law of cooling approaches the insulated boundary condition.
Heat equation with boundary of third kind is solved in section 4.5.4.
95
4.4.4
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A Heat Transfer Problem with Dynamical Boundary conditions
In this section we append dynamic boundary conditions, i.e. boundary conditions containing time
derivatives, which arise from the physics of the particular problem, to Partial Differential Equations.
Suppose that one end of a one-dimensional rod with unit length is kept at zero temperature, while at the
other end there is a convective exchange of heat between the rod and the surrounding medium, which may
be a fluid or a gas. If it is assumed that the change in heat is directly proportional to the heat flow over
the boundary (assume that the constant of proportionality is 1) and by prescribing the initial state, the
mathematical model consists of the heat equation (as before) with a dynamical boundary conditions at
one end:
∂u
∂t
u(0, t)
∂u
∂t x=1
u(x, 0)
∂2u
,
∂t2
= 0,
∂u
=
,
∂x x=1
= u0 (x)
= κ
0 < x < 1, t > 0,
t > 0,
t > 0,
0 < x < 1.
Note: A notation of the type u(x, 0) = u0 (x) or equivalently u
= u0 (x) always implies that the limit
t=0
lim u exists (as it must for any physically realistic problem) and equals to u0 .
t→0
Exercises 4.4
1. Consider a thin rod whose lateral surface is insulated. Assume that there is an internal source of
thermal heat Q (x, t): heat energy per unit volume generated per unit time. Show that equation (4.3)
becomes
Z b
Z b
∂
Q (x, t) dx.
(4.12)
[λAux (x, t)] dx +
E ′ (t) =
a
a ∂x
Show that if λ and A are constant then the heat equation with source of thermal energy is
ut = kuxx + Q (x, t) .
2. Assume that in (4.1) c, ρ and λ are not constant, but depend on x. Show that the heat equation is in
the form (A is constant)
∂u
∂
∂u
cρ
=
λ
(4.13)
∂t
∂x
∂x
[If c, ρ and λ are constant equation (4.13) becomes (4.7).]
3. If u (x, t) is known, give an expression for the total thermal energy contained in a bar (0 < x < L) .
RL
[Ans: E (t) = 0 cρAu(x, t)dx].
4. (a) Assume that the insulation around a rod is faulty in that the heat in each slab leaks through
the insulation at a rate proportional to the temperature of the slab. Show that the temperature
u(x, t) obeys the equation ut = kuxx − hu for some constant h > 0.
(b) Show that if w(x, t) solve wt = kwxx then u(x, t) = e−ht w(x, t) solves ut = kuxx − hu.
96
5. Consider a one dimensional rod, 0 ≤ x ≤ L. Assume that the heat energy flowing out of the rod at
the right end (x = L) is proportional to the temperature difference between the end temperature of
the bar and the known external temperature. Derive the boundary condition at x = L.
6. Two one–dimensional rods of different materials joined at x = a are said to be in perfect thermal
contact if the temperature is continuous at x = a, i.e.
u a+ , t = u a− , t
and no heat energy is lost at x = a (i.e. the heat energy flowing out of one rod flows into the other).
(a) What mathematical equation represent the latter condition at x = a?
(b) Under what special condition is ux continuous at x = a?
■
4.5
4.5.1
SOLUTION OF THE ONE DIMENSIONAL HEAT EQUATION: METHOD OF SEPARATION OF VARIABLES
Method of Separation of Variables
One of the most important solution techniques for boundary–value problems is the method of separation
of variables. This technique was first introduced by Bernoulli in 1755 and used by D’Alembert
in 1760 to solve the vibrating string problem. The method is heavily dependent on the concept of
Fourier Series (see Chapter 3). In the method of separation of variables, we make use of the superposition
principle (see section 1.5.1). The method of separation is used to find those solutions (if any) of a Partial
Differential Equation which are the sum of products of functions, each of which depends on just one of
the independent variables. In this chapter we will solve the heat, wave and Laplace using this technique.
4.5.2
The Heat equation with zero temperatures at finite ends
This section is base on Section 3.5 of [1].
We propose to study the following problem
∂2u
∂u
= k 2
∂t
∂x
u (0, t) = 0
0 < x < L, t > 0
(4.14)
(4.15)
u (L, t) = 0
(4.16)
u (x, 0) = f (x) .
(4.17)
The problem consists of finding the temperature in a rod of length L with prescribed temperatures
at both ends. We assume that the temperature u (x, t) is in the form of a product (section 3.5 of the
Prescribed Book is user-friendly and very lucid, please read the above section if you run into trouble here)
u (x, t) = X (x) T (t)
(4.18)
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where X is a function of x alone, and T is a function of t alone.
We substitute equation (4.18) into equation (4.14), we obtain:
∂u
∂t
∂2u
∂x2
dT
dt
2
d X
= T (t)
.
dx2
= X (x)
Equation (4.14) becomes
d2 X
dT
(4.19)
= k 2 T (t) .
dt
dx
We can separate variables by dividing both sides of (4.19) by kX (x) T (t) such that only functions depending on t appear on the left hand side, and only functions depending on x appear on the right hand
side:
(4.20)
1 dT
1 d2 X
=
dx2}
|kT{zdt}
|X {z
X (x)
function of t alone
function of x alone
The only way that a function of x can be equal to a function of t is for both functions to be equal to the
same constant, say −λ:
1 d2 X
1 dT
=
= −λ
(4.21)
kT dt
X dx2
where λ is arbitrary constant known as the separation constant. The meaning of the minus sign will
be explained later, in this section (see Remark 4.5.1).
Equation (4.21) yields two ordinary differential equation:
d2 X
= −λX
dx2
(4.22)
and
dT
= −λkT.
(4.23)
dt
The product solution u (x, t) = X (x) T (t) , must also satisfy the two homogeneous boundary conditions
(4.15) and (4.16). So,
u (0, t) = 0 implies X (0) T (t) = 0
∴ X (0) = 0 or T (t) = 0 for all t.
If T (t) = 0, from (4.18), the assumed product solution is identically zero, i.e. u (x, t) = 0. (u (x, t)) = 0, is
called the trivial solution, and it satisfies automatically any homogeneous Partial Differential Equation
with any homogeneous boundary condition.)
Thus, for nontrivial solutions, we must have
X (0) = 0.
In a similar way, u (L, t) = 0 implies that X (L) T (t) = 0,
∴ X (L) = 0 or T (t) = 0 imply that
X (L) = 0.
(4.24)
(4.25)
98
The time dependent problem (4.23) does not have additional conditions.
dT
−λkt .
dt = −λkT yields the solution T (t) = ce
Now we turn our attention to the x–dependent (spatial dependent) problem (4.22) with boundary
conditions (4.24) and (4.25):
d2 X
= −λX, 0 < x < L
dx2
X (0) = 0 and X (L) = 0
(4.26)
The problem (4.26) is a Sturm–Liouville equation was solved in example 3.8.1(a). and has eigenvalues
and eigenfunctions
λn =
nπ 2
L
and Xn (x) = sin
nπx
x,
L
n = 1, 2, 3, . . .
(4.27)
respectively.
Substitution of (4.23) and (4.27) into (4.18) yields the production solutions:
nπ 2
nπx
un (x, t) = e−k( L ) t sin
, n = 1, 2, 3, . . .
(4.28)
L
By the extended principle of superposition (see section 1.5), it follows that the solution of the heat equation
(4.14) – (4.17) will be the infinite linear combination of the eigenfunctions in (4.28):
u (x, t) =
∞
X
nπ 2
bn e−k( L ) t sin
n=1
nπx
.
L
(4.29)
Finally, we apply the initial condition (4.17), by letting t = 0 in (4.29), we obtain
f (x) =
∞
X
bn sin
n=1
which is the Fourier series of f.
The coefficients bn are determined as by equation (3.65).
Z
2 L
nπx
bn =
f (x) sin
dx,
L 0
L
nπx
L
n = 1, 2, 3, . . .
(4.30)
(4.31)
Remark 4.5.1
The fact that all the λ’s are positive is not surprising, the time–dependent solution (4.23) decays exponentially as t increases (k > 0) . Since we are solving a heat conduction problem and the temperature u (x, t) is
proportional to T (t) , it was not expected for the solution to grow in time.
Note also that (see figure 4.2)
πx
X1 (x) = sin
has no zeros for 0 < x < L,
L
2πx
X2 (x) = sin
has one zero in (0, L)
L
3πx
X3 (x) = sin
has 2 zeros in (0, L)
L
We may claim that
nπx
Xn (x) = sin
has (n − 1) zeros in (0, L)
L
■
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Figure 4.2: Zeros of eigenfunctions sin
nπx
L
♦ VERY IMPORTANT
1. While solving the Heat Equation


 ut = kuxx
u (0, t) = 0 = u (L, t)


u (x, 0) = f (x)
by the method of separation of variables, obtain the product solutions before applying the principle
of superposition and do not apply the initial condition u (x, 0) = f (x) until after the principle of
superposition of solutions.
2. The solution (4.29) with bn given by (4.31) is unique, stable i.e. is bounded dependent continuously
with the data for all x0 < x < L and all t > 0, see section 5.6 for the proof.
Example 4.5.1 Solve the problem
∂2u
∂u
= k 2,
t > 0, 0 < x < 1
∂t
∂x
u (0, t) = 0 = u (L, t) , t > 0
u (x, 0) = 50,
0 < x < L.
Solution
From (4.29), we know that
u (x, t) =
∞
X
nπ 2
bn e−k( L ) t sin
n=1
nπx
L
where L = 1 and
bn
2
L
Z
L
nπx
=
f (x) sin
dx = 2
L
0
1
1
cos nπx
= 100 −
nπ
0
100
n
=
(1 − (−1) ) .
nπ
Z
1
50 sin nπxdx
0
100
Thus
∞
100 X 1 1 − (−1)n sin nπx
π
n
u (x, t) =
100
π
=
n=1
∞
X
k=0
2
sin (2k + 1) πx.
2k + 1
■
Example 4.5.2 See example 1, section 3.5 page 138 of [1]. A thin bar of length π units is placed in a
hot medium (at A0 C). The bar reaches A0 C throughout, the bar is removed from the boiling water. With
the lateral sides kept insulated, suddenly, at time t = 0, the ends are immersed in a medium with constant
freezing temperature 00 C. Find the temperature u (x, t) for t > 0, we take c = 1.
Solution
The boundary value problem to solve is
∂2u
∂u
=
, 0 < x < π, t > 0
∂t
∂x2
u (0, t) = 0, u (π, t) = 0, t > 0,
u (x, 0) = A, 0 < x < π, (A constant).
The solution given by (4.29)-(4.31)
u (x, t) =
∞
X
2
an e−n t sin nx
n=1
with
an =
2
π
Z
0
π

 = 0, if n is even
2A
2A
n
A sin nxdx =
(1 − cos nπ) =
(1 − (−1) ) =
.
 = 4A , if n is odd.
nπ
nπ
nπ
Or
an =
Thus
∞
u (x, t) =
4A
, n = 1, 3, 5, 7, ....
nπ
2
∞
2
2A X en t
2A X e(2k+1) t
sin nx =
sin (2k + 1) x.
π
n
π
2k + 1
n=1
k=0
■
4.5.3
The Heat equation with insulated end(s)
Example 4.5.3 Solve the heat equation with insulated ends, and with initial temperature f (x) , 0 ≤
x ≤ L.
Solution
We need to solve the problem


0 ≤ x ≤ L,
t>0
 ut = kuxx ,
ux (0, t) = 0 and ux (L, t) = 0, t > 0


u (x, 0) = f (x) ,
0 < x < L.
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We assume that
u (x, t) = X (x) T (t) .
The Partial Differential Equation as before splits into two ordinary differential equations:
d2 X
dx2
dT
dt
= −λX
= −λkT
where λ is the separation constant. Again
T (t) = c e−λkt .
The boundary conditions imply that
X ′ (0) = 0 and X ′ (L) = 0.
Thus, we need to solve the Sturm–Liouville problem
 2
 d X
= −λX
dx2
 X ′ (0) = 0 and X ′ (L) = 0.
(4.32)
The eigenvalues and the eigenfunctions of (4.32) are given by (3.97) and (3.98), (see example 3.8.2(b) for
details.)
nπ 2
λ0 = 0, λn =
, n = 1, 2, 3, . . .
L
and the corresponding eigenfunctions are
X0 (x) = 1, Xn (x) = cos
nπx
,
L
n = 1, 2, 3, . . .
The resulting product solutions of the Partial Differential Equation are
nπ 2
u0 (x) = 1, and un (x, t) = e−( L )
kt
cos
nπx
,
L
n = 1, 2, 3, . . .
We use the principle of superposition, taking all product solutions, we have
u (x, t) = a0 +
∞
X
nπ 2
an e−( L )
kt
cos
n=1
=
∞
X
nπ 2
an e−( L )
kt
cos
n=0
nπx
L
nπx
.(Why?)
L
The initial condition u (x, 0) = f (x) is satisfied if
f (x) = a0 +
∞
X
n=1
an cos
nπx
.
L
(4.33)
Equation (4.33) represents the Fourier cosine series of f (x) in [0, L] , they are determined by (3.54) and
(3.56)(see example 3.7.5).
102
Therefore, we have
a0 =
2
an =
L
Z
1
L
L
Z
f (x)dx
0
L
nπx
dx.
L
f (x) cos
0
■
Example 4.5.4 Solve the problem


0 ≤ x ≤ L,
 ut = kuxx ,
ux (0, t) = 0,
u (L, t) = 0


u (x, 0) = f (x) .
t>0
Solution
We repeat all the above steps for a better understanding, you don’t have to it, if you understand how to
separate variables.
Let u (x, t) = X (x) T (t) . The boundary conditions yield
dX
(0) = 0 and X (L) = 0.
dx
The Partial Differential Equation splits in two ordinary differential equations
 2
d X



 dx2 = −λX



 dX (0) = 0
dx
and
X (L) = 0
dT
= −λkT.
dt
Equation (4.34) has solution
X (x) = a cos
√
(4.35)
√
λx + b sin λx.
Applying the boundary conditions, we have
√
√
√
√
dX
(x) = −a λ sin λx + b λ cos λx
dx
dX
(0) = 0 implies that b = 0
dx
and
X (L) = 0 implies that cos
Thus
λ = λn =
2n + 1
π
2L
√
λL = 0.
2
,
n = 0, 1, 2, 3, . . .
The solution of (4.35) is T (t) = Tn (t) = e−λn kt . Product solutions are
un (x, t) = e−(
2
2n+1
π kt
2L
)
cos
(4.34)
2n + 1
π,
2L
n = 0, 1, 2, . . .
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The principle of superposition gives
u (x, t) = a0 +
∞
X
an e−(
2
2n+1
π kt
2L
)
cos
n=1
(2n + 1)πx
.
2L
Applying the initial condition we obtain
f (x) = a0 +
∞
X
an cos
n=1
where
1
a0 =
L
and
an =
2
L
Z
L
f (x) cos
0
Z
(2n + 1)πx
2L
L
f (x) dx
0
(2n + 1)πx
dx,
2L
n = 1, 2, 3, . . .
■
4.5.4
Heat Equation with Radiating End (or boundary conditions of the third kind)
The boundary conditions of the third kind was introduced on page 94, equation (4.11). We will solve the
heat equation with zero prescribed temperature at x = 0, and with Newton’s law of cooling (radiating
end) applied at x = L (see page 94). We need to solve the problem:


ut = uxx



 u (0, t) = 0
(4.36)
 ux (L, t) + H u (L, t) = 0



 u (x, 0) = f (x)
As usually, we separate the variables u = XT where
T = e−λt
and X satisfies:

′′

 X = −λX, 0 ≤ x ≤ L,
X (0) = 0

 ′
X (L) + HX (L) = 0
t>0
(4.37)
The general solution of (4.37), has the form (as previously, λ > 0, let λ = µ2 )
X (x) = a cos µx + b sin µx
X (0) = 0 implies that a = 0, therefore X (x) = b sin µx. The second boundary condition implies that
µb cos µL + Hb sin µL = 0.
(4.38)
Since X (x) ̸= 0, we have b ̸= 0. Thus (4.38) yields
−
µ
= tan µL.
H
(4.39)
104
Since (4.39) is a transcendental equation, a solution for µ cannot be found exactly. We can, however, see
that there exist solutions by graphic approach, that is, by finding the points of intersection of the two
curves (see figure 4.3)
µ
y = tan µL and y = − .
H
This is done in Figure 4.3. Here we see that there exist infinitely many solutions λn .
Figure 4.3: Solutions of −
µ
= tan (µL)
H
The eigenfunctions are sin µn x, where µn are solutions of
tan µn L = −
µn
.
H
• Is the set of eigenfunction {sin µn x}∞
n=1 orthogonal?
For n ̸= m, we have
Z L
0
sin µn x sin µm xdx =
=
since
tan µn L = −
Z
1 L
[cos (µn − µm ) x − cos (µn + µm ) x] dx
2 0
µn cos µn L sin µm L − µm sin µn L cos µm L
µ2m − µ2n
µn
sin µn L
, ∴ µn = −H tan µn L = −H
.
H
cos µn L
Thus (4.40) becomes
1
µ2m − µ2n
sin µn L
sin µm L
−
cos µn L sin µm L + H
sin µn L cos µm L = 0,
cos µn L
cos µm L
(4.40)
105
APM3701/1
and
Z
0
L
Z
L sin (2µn L)
1 L
(1 − cos (2µn x)) dx = −
2 0
2
4µn
L 2 sin (µn L) cos (µn L)
L sin (µn L) cos (µn L)
−
= −
2
4µn
2
−2H tan (µn L)
2
sin µn xdx =
=
L
Z
=⇒
0
sin2 µn xdx =
L cos2 (µn L)
−
2
−2H
(4.41)
The principle of superposition yields
u (x, t) =
∞
X
2
bn e−µn t sin µn x
n=1
where µn ’s are solutions of
tan µn L = −
µn
.
H
Finally, we apply the initial condition u(x, 0) = f (x), we obtain
f (x) =
∞
X
bn sin µn x
n=1
where
RL
f (x) sin µn x dx
(work out the details)
RL 2
0 sin µn x dx
−1 Z L
L cos2 µn
=
+
f (x) sin µn x dx
2
2H
0
bn =
0
where we have used (4.41).
4.5.5
Heat conduction in a thin circular ring: Periodic Boundary Conditions
Suppose that a thin wire (with lateral sides insulated) is bent in the shape of a circle. Let 2L (for
convenience) be the circumference and r the radius of the circle formed by the wire. We have
2L = 2πr ∴ r =
2L
L
=
2π
π
We measure (for convenience) the arc length x, ranging from −L to L instead of 0 to 2L. Assume that
the wire is very tightly connected to itself at its ends (see exercise 4.4(6.)), we have
u (−L, t) = u (L, t) .
That is, u is continuous. Also, since the thermal conductivity is constant everywhere, the derivation of
the temperature is continuous, that is:
ux (−L, t) = ux (L, x)
106
With the initial temperature u (x, 0) = f (x), we have the following problem for the heat conduction in a
thin circular ring.


t>0
 ut = k uxx , −L ≤ x ≤ L,


 u (−L, t) = u (L, t) , t > 0
(4.42)

u
(−L,
t)
=
u
(L,
t)
,
t
>
0
x
x



 u (x, 0) = f (x)
Solution
Again we let u (x, t) = X (x) T (t) , with T (t) = e−λkt , as previously.
And the corresponding Sturm-Liouville problem is
d2 X
= −λX
dx2
X (−L) = X (L)
(4.43)
X ′ (−L) = X ′ (L) .
(4.45)
(4.44)
The boundary conditions (4.44) and (4.45) are referred to as periodic boundary conditions.
The solution of (4.43)-(4.45) is given by (3.100), (see example 3.8.3(a))
nπx
nπx
X0 = 1, Xn = An cos
+ Bn sin
, n = 1, 2, 3, . . .
L
L
We use the principle of superposition before applying the initial condition. We obtain:
u (x, t) = a0 +
∞
X
nπ 2
an e−( L )
∞
t
cos
nπ 2
nπx X
nπx
+
bn e−( L ) t sin
.
L
L
n=1
n=1
u (x, 0) = f (x) is satisfied if
f (x) = a0 +
∞ X
an cos
n=1
nπx
nπx + bn sin
L
L
which is the Fourier series of f in [−L, L] . Thus
a0 =
an =
bn =
1
L
2
L
2
L
Z
L
f (x) dx
−L
Z L
f (x) cos
nπx
dx,
L
n = 1, 2, 3, 4, . . .
f (x) sin
nπx
dx,
L
n = 1, 2, 3, 4, . . .
−L
Z L
−L
■
4.5.6
Steady state temperature distribution
Consider the heat equation in a rod 0 ≤ x ≤ L with non zero prescribed temperature at its ends:


 ut = k uxx , t ≥ 0, 0 ≤ x ≤ L
u (0, t) = A, u (L, t) = B, t > 0


u (x, 0) = f (x) , 0 ≤ x ≤ L.
107
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In the above problem, there are no source/sink to add/substract heat to the rod. Also the boundary
conditions are fixed temperatures and so they cannot change with time and we have not prescribed a
heat flux on the boundaries to continually add/absorb heat energy, which means that there is no forcing
of heat energy into or out of the rod and so while some heat energy may well naturally flow into or out
of the rod at the end points as the temperature changes, eventually the temperature distribution in the
rod should stabilize out and no longer depend on time. Intuitively, it makes sense that we expect, as
t → ∞, the temperature distribution, u (x, t) should behave as, lim u (x, t) = us (x) , where us (x) is
x→∞
called ”steady-state solution” or ”equilibrium temperature”. Note that the steady state solution
should also satisfy the heat equation and the boundary conditions, but not the initial condition since it is
the temperature distribution as t → ∞ whereas the initial condition is at t = 0. Because us is independent
dus
∂us
∂ 2 us
∂us
=
= 0, us satisfies the heat equation we have 0 =
=
of t, we have
∂t
dt
∂t
∂x2
d2 us
= 0,
us (0) = A
and us (L) = B.
(4.46)
=⇒
dx2
The solution of (4.46) is given by
B−A
us (x) = A +
x.
(4.47)
L
Formula (4.47) should not be memorized, instead you should consider a solution of the form Kx + G, and
find the constants K and G using the boundary conditions.
Example 4.5.5 The steady state solution of the problem
∂u
∂2u
= k 2
∂t
∂x
ux (0, t) = T1 and
0 < x < L,
u (L, t) = T2
t>0
(4.48)
u (x, 0) = f (x)
is given by us (x) = T1 (x − L) + T2 . (why?)
♦ REMARK
Note that for some types of boundary condition, the steady state solution does not exist. In fact the
steady state solution exists only when the solution of the nonhomogeneous solution approaches the steady
state solution as t → ∞, i.e. lim u (x, t) = lim (us (x) + v (x, t)) = us (x) , where v (x, t) is the solution of
t→∞
t→∞
the related homogeneous problem which has limit 0 as t → ∞.
For instance, when the heat is being drained out at one end at a rate different from which it is flowing in
at the other end. Let us consider the following problem


 ut = k uxx , t ≥ 0, 0 ≤ x ≤ L
ux (0, t) = T1 , ux (L, t) = T2 , t > 0


u (x, 0) = f (x) , 0 ≤ x ≤ L.
At the end x = 0 the heat is being drained at a rate ux (0, t) = T1 and at the end x = L, the heat is
being added to the rod at a rate ux (L, t) = T2 . If T1 > 0, then the heat is being added to the rod at a
constant rate T1 , if T2 < T1 , then the rod loses heat at constant rate T1 − T2 . Thus a steady state solution
does not exist, unless T1 = T2 . This can be proven by assuming that the steady state solution exists:
us (x) = Ax + B. So, ux (0, t) = T1 and ux (L, t) = T2 implies that B = T1 and B = T2 , which is impossible
unless T1 = T2 . The above problem will be solved in section 4.5.7.2.
108
4.5.7
4.5.7.1
The Heat equation with non–homogeneous boundary conditions.
The Method of homogenisation – Non zero boundary conditions.
Consider the heat equation in a rod 0 ≤ x ≤ L with non zero prescribed temperature at its ends:


 ut = k uxx , t ≥ 0, 0 ≤ x ≤ L
u (0, t) = A, u (L, t) = B, t > 0


u (x, 0) = f (x) , 0 ≤ x ≤ L.
(4.49)
The method of homogenisation involves the introduction of a new variable, by means of which the
non–homogeneous boundary–value problem is reduced to one with homogeneous boundary conditions.
We solve (4.49). by letting
u (x, t) = v (x) + w (x, t)
(4.50)
where v is a function of x only, and satisfies (4.49), that is
d2 v
= 0, 0 ≤ x ≤ L
dx2
v (0) = A and v (L) = B.
(4.51)


 wt = k wxx , t > 0, 0 ≤ x ≤ L
w (0, t) = 0 and w (L, t) = 0, t > 0


w (x, 0) = f (x) − v (x) , 0 ≤ x ≤ L.
(4.52)
And w (x, t) satisfies
Verify that u(x, t) = v(x) + w(x, t) satisfies (4.49). Equation (4.51) admits the solution
v (x) = A +
B−A
x
L
The solution of (4.52) is easily determined (see section 4.5.2):
w (x, t) =
∞
X
nπ 2
bn e−k( L ) t sin
n=1
where
2
bn =
L
Z
0
Thus
L
[f (x) − v (x)] sin
nπx
L
nπx
dx.
L
∞
X
nπ 2
B−A
nπx
x+
bn e−k( L ) t sin
u (x, t) = v(x) + w(x, t) = A +
L
L
n=1
where
2
bn =
L
Z
o
L
B−A
nπx
f (x) − A −
x sin
dx.
L
L
Example 4.5.6 Find the temperature distribution in the example 4.5.2, if after bringing the temperature
of the bar to 1000 , the end at x = 0, is frozen at 00 C, and the other end at x = π, is kept at 1000 C.
109
APM3701/1
Solution
The boundary value problem to solve is
∂u
∂2u
, 0 < x < π, t > 0
=
∂t
∂x2
u (0, t) = 0, u (π, t) = 100, t > 0,
u (x, 0) = 100, 0 < x < π.
Here k = 1, A = 0, B = 100, L = π, f (x) = 100. Thus,
nπ 2
∞
X
t
−
B−A
nπx
L
u (x, t) = A +
sin
x+
bn e
L
L
n=1
with
2
bn =
π
π
Z
0
B−A
nπx
f (x) − A −
x sin
dx.
L
L
We obtain the solution
∞
u (x, t) =
X
100
2
x+
bn e−n t sin nx
π
n=1
and
2
bn =
π
Z
0
π
100
200
100 −
x sin nxdx =
.
π
nπ
Therefore
∞
u (x, t) =
X 200
100
2
x+
e−n t sin nx
π
nπ
n=1
■
4.5.7.2
Constant heat flux at the boundary
Consider the following problem


 ut = k uxx , t ≥ 0, 0 ≤ x ≤ L
ux (0, t) = T1 and ux (L, t) = T2


u (x, 0) = f (x) .
(4.53)
To solve (4.53), we set
u (x, t) = up (x, t) + v (x, t)
where up (x, t) is a particular solution and v (x, t) is the solution of the related homogeneous problem. The
simplest form for a particular solution that reflects the fact that the heat energy is changing at a constant
rate is
up (x, t) = at + h (x) .
up is determined from (4.53) as follows:
∂up
∂ 2 up
∂u
∂2u
d2 h
a
a 2
=
=⇒
=
=⇒
a
=
k
=⇒ h′′ (x) =
=⇒ h (x) =
x + bx + c
2
2
2
∂t
∂x
∂t
∂x
dx
k
2k
(4.54)
110
The boundary conditions imply that:
ux (0, t)
=
and ux (L, t)
=
T1 =⇒ h′ (0) = b = T1
a
(T2 − T1 ) k
T2 =⇒ h′ (L) = L + T1 = T2 =⇒ a =
k
L
T2 − T1 2
h (x) =
kx + T1 x + c.
L
=⇒
Thus
up (x, t) =
=
(T2 − T1 )
(T2 − T1 ) 2
kt +
x + T1 x + c
L
2L
(T2 − T1 )
1
kt + x2 + T1 x + c,
L
2
where c is a constant.
The related homogeneous problem is


vt = k vxx , t ≥ 0, 0 ≤ x ≤ L



vx (0, t) = 0 and vx (L, t) = 0

(T
−
T
)

2
1
2

x + T1 x + c
 v (x, 0) = f (x) −
2L
which has solution (see section 4.5.3)
v (x, t) =
∞
X
nπ 2
An e−k( L ) t cos
n=0
nπx
L
where
A0 =
and
An =
L
(T2 − T1 ) 2
f (x) −
x + T1 x + c dx
2L
0
Z (T2 − T1 ) 2
nπx
2 L
f (x) −
x + T1 x + c cos
dx.
L 0
2L
L
1
L
Z
Thus
(T2 − T1 )
u (x, t) =
L
∞
X
nπ 2
1 2
nπx
kt + x + T1 x + c +
An e−k( L ) t cos
.
2
L
n=0
Note that the above solution is not bounded when t → ∞.
Exercises 4.5
1. Obtain the solution of the following initial boundary value problems:
(a) ut = k uxx ,
u (x, 0) =
x2 (L
0 ≤ x ≤ L,
t≥0
(b) ut = k uxx ,
− x) ,
u (0, t) = A,
u (0, t) = 0
u (L, t) = B
u (L, t) = 0
u (x, 0) = sin
0 ≤ x ≤ L,
t≥0
πx
.
L
2. Exercises 3.5. of [1], all problems, except problem 16.
3. Exercises 3.6. of [1], all problems.
■
111
4.6
APM3701/1
THE WAVE EQUATION
4.6.1
4.6.1.1
Derivation of the wave equation
Vertically vibrating string
Consider a horizontally stretched string with both ends tied down in a way that the tightly nature of the
string is maintained. Think of stringed musical instruments as example. Let x be the x–coordinate of
Figure 4.4: Vertically Vibrating string
a particle when the string is in the horizontal equilibrium position. In time as the string vibrates it is
located somewhere at (x, y) at a certain time(see Figure 4.4).
We assume that the slope of the string is small, in which case the horizontal displacement v can be
neglected. In this situation the vertical displacement u depends on x and t; y = u (x, t) .
Figure 4.5: Forces acting on a small portion of the string
We consider a tiny segment of the string between x and x + ∆x (see Figure 4.5). We assume that the
mass density ρ0 (x) of the unperturbed string is known.
112
For the tiny segment, the total mass is approximately
ρ0 (x) ∆x.
(4.55)
The forces acting on the tiny segment are:
• Body forces, acting vertically.
• Tension (forces acting on the end points of the segment of the string). We denote its magnitude by
T (x, t) . The force due to the tension pulls at both ends in the direction of the tangent, trying to stretch
the tiny segment of string. The angle depends on both the position x and time t. And the slope of the
string may be represented as tan θ. So we have the following:
tan (θ(x, t)) =
∂u
.
∂x
(4.56)
The horizontal component of Newton’s law describes the horizontal motion, which we assumed small and
can be neglected.
The vertical equation of motion states that the mass times the vertical component of
2 ∂ u
equals the vertical component of the tension plus the vertical component of the body
acceleration
∂t2
forces; using (4.55) we have:
ρ0 (x) ∆x
∂2u
∂t2
= T (x + ∆x, t) sin (θ (x + ∆x, t))
−T (x, t) sin (θ (x, t)) + ρ0 (x) ∆xQ (x, t)
(4.57)
where T (x, t) is the magnitude of the tension and Q(x, t) is the vertical component of the body force per
unit mass.
We divide (4.57) by ∆x and take the limit:
ρ0 (x)
i
∂2u
∂ h
=
T
(x,
t)
sin
θ
(x,
t)
+ ρ0 Q (x, t) .
∂t2
∂x
(4.58)
For θ is small we know that
∂u
sin θ
= tan θ =
≈ sin θ.
∂x
cos θ
Therefore equation (4.58) becomes
∂2u
∂
ρ0 (x) 2 =
∂t
∂x
∂u
T
+ ρ0 (x) Q (x, t) .
∂x
(4.59)
We will also assume that the string is perfectly elastic, meaning that the magnitude of the tension T (x, t)
depends only on the local stretching of the string. Since θ is small, the stretching of the string is nearly
the same as for the unperturbed string, where the tension is constant T0 (say).
So, for small vibrations of a highly stretched string, we have
ρ0 (x)
∂2u
∂2u
=
T
+ Q (x, t) ρ0 (x) .
0
∂t2
∂x2
(4.60)
113
4.6.1.2
APM3701/1
One dimensional wave equation – Free vibrations of the string
If we assume that the only body force per unit mass is gravity, then Q (x, t) = −g, in (4.60), this force is
very small
∂2u
ρ0 g <<< T0 2
∂x
and can be neglected. We also assume that the string is uniform in which case ρ0 (x) is constant ≡ ρ for
all 0 ≤ x ≤ L.
Equation (4.60) reduces to
2
∂2u
2∂ u
=
c
(4.61)
∂t2
∂x2
T0
where c2 =
, c is called the wave speed. Equation (4.61) is known as the one–dimensional wave
ρ
equation.
4.6.1.3
One dimensional wave equation – Forced vibrations of the string
See Case of Forced Vibration on page 111 of [1].
4.6.2
Initial and boundary conditions for the wave equation
Equation (4.61) has a second–order derivative in t and x. Two boundary conditions in x, and two initial
conditions in t.
4.6.2.1
Fixed end with zero displacement
The simplest boundary condition is that of a fixed end, usually fixed with zero displacement. For example
if a string is fixed (with zero displacement) at x = 0, then we have
u (0, t) = 0.
4.6.2.2
(4.62)
Varying end
Or, we might vary an end of the string in a prescribed way (for the end x = L) :
u (L, t) = f (t) .
(4.63)
The boundary condition (4.62) is homogeneous whereas (4.63) is nonhomogeneous.
4.6.2.3
Spring-mass system boundary condition
Another boundary condition is derived, where the string is attached to a spring–mass system. We
suppose that the left end point, x = 0, of a string is attached to a spring–mass system. The spring is
attached to the mass in a such way that if the position of the mass is y (t) then the position of the left
end point is:
u (0, t) = y (t) ,
(4.64)
y (t) is unknown and satisfies an ordinary differential equation determined by Newton’s laws. We
assume that at rest the spring has length ℓ and obeys Hooke’s law with constant k. The spring moves
in a prescribed way, ys (t) .
114
Figure 4.6: String attached to a spring–mass system.
The stretching of the spring is
y (t) − ys (t) − ℓ.
(4.65)
Newton’s law gives
d2 y
= −k [y(t) − ys (t) − ℓ] + external forces
dt2
where we have applied Hooke’s law with spring constant k.
Another vertical force acting on the mass m is the vertical component of the tension
m
T (0, t) sin θ (0, t) .
(4.66)
(4.67)
We denote by q (t) any other external forces on the mass.
We have assumed that θ is small, so that T = T0 is constant, and the vertical component is approximately
T (0, t) sin (θ(0, t)) ≈ T0
Since
∂u
(0, t) .
∂x
(4.68)
∂u
sin (θ(0, t))
(0, t) tan (θ(0, t)) ≈
.
∂x
cos (θ(0, t))
Thus, the boundary condition at x = 0 is
m
∂2
∂u
u (0, t) = −k [u (0, t) − ys (t) − ℓ] + T0
(0, t) + q (t)
2
∂t
∂x
(4.69)
where ys (t) is the variable support of the spring, and q (t) external forces.
4.6.2.4
The free end boundary conditions
The free end boundary condition is where the end is free to move up and down as before but with no
spring–mass, and no external forces. So in equation (4.69) we take the limit as k → 0, (q (t) = 0) , we
obtain
∂u
(0, t) = 0.
(4.70)
∂x
115
APM3701/1
Equation (4.70) means that the vertical component of the tension must vanish at the end since there is
no other vertical forces at that end.
4.6.2.5
Initial Conditions
The initial conditions are often given in the form
u (x, 0) = f (x)
∂u
(x, 0) = g (x)
∂t
4.6.3
(4.71)
(4.72)
A Wave Problem with Dynamical Boundary Conditions
Suppose that the left end of a vibrating string of density ρ is fixed. We assume that the right end, say at
x = L, is free to move in a vertical direction with a mass m which is attached to the end. If gravitational
force is not taken into account, we have the following dynamical boundary condition:
m
∂2u
∂u
(L, t) = −T
(L, t).
2
∂t
∂x
∂u
The latter equation is Newton’s law for the attached mass, with −T
(L, t) (T is the stress) the vertical
∂x
component of the stress at the end x = L. By imposing the initial conditions
∂u
(x, 0) = g(x),
∂t
u(x, 0) = f (x),
and setting c2 =
T
, we get the boundary-value problem
ρ
∂2u
∂t2
u(0, t)
2
∂ u
m 2 (L, t)
∂t
u(x, 0)
∂u
(x, 0)
∂t
= c2
∂2u
,
∂x2
0 < x < L, t > 0,
= 0,
t > 0,
∂u
(T, t),
∂x
= f (x),
= −T
t > 0,
= g(x).
Exercises 4.6
1. Suppose u (x, t) solves utt = c2 uxx , (c ̸= 0) .
(a) Let α, β, γ, and η be constants, with α ̸= 0. Show that the function
v (x, t) = u (αx + γ, βt + η) ,
satisfies
vtt =
βc
α
2
vxx .
116
(b) For any constant ω, let X = cosh ωx + c sinh ωt and T = c−1 sinh ωx + cosh ωt.
Show that x = cosh ωX − c sinh ωT and t = −c−1 sinh ωX + cosh ωT .
(a) Exercise 3.2, p. 111 of [1], all problems.
■
4.7
SOLUTION OF THE WAVE EQUATION – METHOD OF SEPARATION OF VARIABLES
4.7.1
Vibrating string with fixed ends
I strongly recommend that you read section 3.2 of [1], the method of separation of variables is clearly and
user friendly presented in this section.
Consider the one–dimensional wave equation, which represents a uniform vibrating string without external forces and with fixed ends. We have the equation

2

 utt = c uxx , 0 ≤ x ≤ L, t ≥ 0
(4.73)
u (0, t) = 0, u (L, t) = 0, t ≥ 0


u (x, 0) = f (x) , ut (x, 0) = g (x) , 0 ≤ x ≤ L
We let
u (x, t) = X (x) T (t) .
(4.74)
Substituting (4.74) into (4.73)1 yields (see section 4.5.1 for details calculations on separation of variables)
X
d2 X
d2 T
= c2 T
2
dt
dx2
(4.75)
Separating the variable (we also introduce the separation constant −λ)
1 1 d2 T
1 d2 X
=
= −λ
c2 T dt2
X dx2
(4.76)
Equation (4.76) splits in two ordinary differential equations
d2 T
= −λc2 T
dt2
(4.77)
and
d2 X
= −λX.
dx2
The homogeneous conditions (4.73)2 imply that
X (0) = X (L) = 0.
(4.78)
(4.79)
If λ > 0, the time–dependent problem (4.77) has solution
T (t) = a cos µct + b sin µct.
(4.80)
If λ = 0, T (t) = c1 + c2 t, and if λ < 0, T is a linear combination of exponential functions (growing and
decaying in time). Since we are solving a vibrating string, we expect the solution to oscillate with time.
Verify that λ ≤ 0 is not an eigenvalue.
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The x–dependent problem is the boundary–value problem
 2
d X


= −λX

dx2



X (0) = 0 and X (L) = 0
nπ 2
,
L
As before, we obtain the eigenvalues (see example 3.8.1) λn =
nπx
.
eigenfunctions sin
L
Using (4.80) we obtain two families of product solutions:
sin
n = 1, 2, 3, . . .with the corresponding
nπx
nπct
nπx
nπct
cos
and sin
sin
.
L
L
L
L
Or
nπx
nπct
nπx
nπct
cos
+ bn sin
sin
.
L
L
L
L
We apply the principle of superposition, we obtain:
un (x, t) = an sin
u (x, t) =
∞ X
n=1
nπx
nπct
nπx
nπct
an sin
cos
+ bn sin
sin
L
L
L
L
.
The initial conditions (4.73)3 , yield
∞
X
f (x) =
n=1
∞
X
g (x) =
an sin
bn
n=1
nπx
L
nπc
nπx
sin
L
L
which are the sine series of f and g respectively. Please note that the sine series of g has coefficients (see
section 3.7.1.2)
bn
2
nπc
=
L
L
g (x) sin
0
i.e
bn =
2
nπc
L
Z
Z
L
g (x) sin
0
nπx
,
L
nπx
,
L
and that of f has coefficients
2
an =
L
Z
L
f (x) sin
0
nπx
.
L
Example 4.7.1
A stretched string of unit length has it ends fixed at x = 0 and x = 1. The string vibrates from rest, when
released from rest from an initial position (as indicated in Figure 4.7). Determine the subsequent motion
1
of the string, if c = .
π
Solution
(
2x
if 0 ≤ x ≤ 21
The initial position u (x, 0) = f (x) =
1 − 2x if 12 < x ≤ 1.
118
u
u(x, 0) = f (x)
1
0
1
4
1
x
Figure 4.7: Example 4.7.1
Thus an = 2
or an =
1
R1
2π 2 n2
0
f (x) sin nπx
L dx = 4
nπ
2
6 sin
R
1
2
x sin nπxdx + 2
− 6nπ cos nπ + nπ cos nπ
2
0
R1
1
2
sin nπxdx − 4
R1
1
2
x sin nπxdx
Since the initial velocity is ut (x, 0) = g (x) = 0, we have bn = 0.
The solution is thus
u (x, t) =
∞
X
an sin nπx,
n=1
with
an =
nπ
nπ 1 6
sin
−
6nπ
cos
nπ
+
nπ
cos
2π 2 n2
2
2
■
Exercises 4.7
1. Exercise 3.2, p. 111 of [1], all problems.
(a) Derive the vibrations of a stretched homogeneous string in the following conditions:
(i) The only force acting on the string is the tension, with linear density of 0.002kg/m, and
τ = 200N.
(ii) When tension and weight are taken into account, with linear density of 0.002kg/m, and
τ = 200N.
(iii) No other force than the tension is taken into account. The string measures 1m and has a
mass of 10−2 kg, with the tension τ = 10N.
(iv) The tension and the weight of the string is taken into account. The string measures 1m
and has a mass of 10−2 kg, with the tension τ = 10N.
4.8
4.8.1
THE LAPLACE EQUATION
Derivation of the Laplace Equation
It is possible that a body reaches a state where the temperature u no longer varies with time. In such
case, heat can still flow in the body, but the flow itself no longer varies with time. The flow is thus the
same in each point at all times. In such a case we speak of stationary heat flow.
It follows that the heat equation in two–dimensions reduces to the Partial Differential Equation
∇2 u = 0.
Equation (4.81) is known as the Laplace equation.
(4.81)
119
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In two dimensions equation (4.81) becomes
∂2u ∂2u
+ 2 =0
∂x2
∂y
(4.82)
and in three dimensional
∂2u ∂2u ∂2u
+ 2 + 2 =0
(4.83)
∂x2
∂y
∂z
The Laplace equation is encountered in potential theory.
Note: Any C 2 –function which satisfies the Laplace equation is called a harmonic function. For example,
the following functions are harmonic (verify it):
(a) u (x, y) = ax + by + c,
4.8.2
4.8.2.1
(b) u (x, y) = f (x) + g (y), where f and g are arbitrary C 2 –functions.
Solution of the Laplace’s Equation
Laplace’s equation inside a rectangle
Consider the problem
uxx + uyy = 0,
0 < x < L, 0 < y < H
u (x, 0) = 0
u (x, H) = 0
(4.84)
u (0, y) = f (y)
u (L, y) = 0.
We propose to solve this problem by the method of separation of variables. We ignore momentarily the
non homogeneous boundary condition. We let
u = X (x) Y (y) .
(4.85)
The three homogeneous BC, give
X (L) = 0,
Y (0) = 0,
Y (L) = 0.
The Partial Differential Equation (4.84)1 , splits in two ordinary differential equations
and









d2 X
= λX
dx2
X (L) = 0
(4.86)
d2 Y
= −λY
dy 2
Y (0) = 0
Y (H) = 0.
(4.87)
The y–dependent problem (4.87) is a Sturm–Liouville problem, thus all its eigenvalues are positive, λ > 0,
and equal to
nπ 2
λn =
, n = 1, 2, 3, . . .
H
120
with corresponding eigenfunctions
nπy
,
H
Yn = sin
n = 1, 2, 3. . . .
The solution of the x–dependent problem (4.86) is a combination of exponentials or hyperbolic functions.
For convenience, we write the general solution in the form
Xn (x) = an cosh
nπ
nπ
(x − L) + bn sinh
(x − L)
L
L
(4.88)
Since Xn (L) = 0, we deduce that an = 0. Thus
Xn (x) = bn sinh
nπ
(x − L) ,
L
n = 1, 2, 3, . . .
(4.89)
Note that (4.88) is a simple translation of cosh nπx
and sinh nπx
L
L .
Translations of solutions of differential equations is not recommended if the coefficients of
the differential equation are not constant, or when the differential equation changes with a
translation.
Product solutions are
nπ
nπy
sinh
(x − L) .
un (x, y) = sin
H
L
We apply the principle of superposition to obtain
u (x, y) =
∞
X
an sin
n=1
nπy
nπ
sinh
(x − L) .
H
L
The non–homogeneous u (0, y) = f (y) yields
f (y) =
∞
X
n=1
an sin
nπy
sinh(−nπ)
H
which represents the Fourier sine series of f with coefficients
Z
nπL
2 H
nπy
an sinh
=
f (y) sin
dy.
H
L 0
H
Z H
2
nπy
∴ an =
f (y) sin
dy.
nπL
L
0
L sinh
H
Example 4.8.1
See Examples 2 section 3.8 pp 165–168 of [1].
■
Exercises 4.8
1. Solve the Laplace’s equation inside the rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H, with the following boundary
conditions:
(a) u (0, y) = f (y) ,
(b) u (0, y) = 0,
u (L, y) = 0,
u (L, y) = 0,
uy (x, 0) = 0,
u (x, H) = 0.
u (x, 0) − uy (x, 0) = 0,
u (x, H) = 0.
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y
u(x, b) = f2 (x)
b
∇2 u = f (x, y)
u(0, y) = g1 (y)
u(a, y) = g2 (y)
u(x, 0) = f1 (x)
u(x, 0) = f1 (x)
0
x
a
Figure 4.8: Exercise 4.8.(2.)
2. Solve the Dirichlet problem defined in Figure 4.8, with the following data
(a) a = 1, b = 2, f2 (x) = x, f1 = g1 = g2 = 0.
(b) a = b = 1, f1 = 0, f2 (x) = 100, g1 = 0, g2 (x) = 100.
(c) a = 3, b = 1, f1 (x) = 100, f2 = g2 = 0, g2 (y) = 100 (1 − y)
(d) a = b = 1, f1 (x) = 1 − x, f2 (x) = x, g1 = g2 = 0
(e) a = b = 1, f1 (x) = sin 3πx, f2 (x) = sin 2πx, g1 (y) = sin 2πy, g2 (y) = sin 4πy.
■
4.9
QUICK GUIDE FOR EIGENVALUES FOR SEPARATION OF VARIABLES
Name of equation
Equation
Heat Equation
∂u
∂t
Wave Equation
∂2u
∂t2
Laplace Equation
4.10
∂2u
∂x2
Separation equations
2
= k ∂∂xu2
=
+
2
c2 ∂∂xu2
∂2u
∂y 2
=0
1 dT
kT dt
1 1 d2 T
c2 T dt2
1 d2 X
X dx2
=
=
=
1 d2 X
X dx2
1 d2 X
X dx2
2
− Y1 ddyY2
= −λ
= −λ
Separated Equations
d2 X
= −λX
dx2
dT
dt = −λkT
d2 T
= −λc2 T
dt2
2
d X
= −λX
dx2
d2 X
dx2
d2 Y
dy 2
= −λX
= λY .
LAWS OF CONSERVATION AND THE ENERGY METHOD
Before proceeding to the derivation of deterministic models for the phenomena of e.g. heat transfer, we
first discuss the Basic Laws of Conservation.
4.10.1
Basic Laws of Conservation
You have undoubtedly already encountered laws of conservation. Two very simple examples of laws of
conservation are the following:
122
4.10.1.1
The law of constant momentum
This law reads as follows:
If the mass of a particle is constant and no external forces are applied to the particle, then the momentum
of the particle remains unchanged.
This law can easily be derived from Newton’s second law, viz
m
d
v
= (mv) = f,
dt
dt
(4.90)
dx
where m is the constant mass of a point x, f the (constant) force that is applied to it and v(t) =
the
dt
velocity of the point at time t. If f = 0, it namely follows from (4.90) that mv is constant, which means
that the momentum is preserved.
4.10.1.2
The law of conservation of Energy
The law of conservation of energy can be derived from Newton’s law. By multiplying the left and the
dx
right hand sides of (4.90) by v =
, and recalling that
dt
dv
1 d 2
=
v
dt
2 dt
we obtain
d 1
2
mv − f x = 0
(4.91)
dt 2
1
From (4.91) it follows that the quantity E = mv 2 − f x is constant, which is precisely the law of Conser2
vation of Energy.
When we assume that in a physical process in which certain quantities, for example mass, momentum, heat energy, are involved, there are no sources or sinks which release or absorb these quantities,
i.e. the quantity/quantities concerned are conserved, we obtain, in the simplest language, the following
law of conservation: The tempo with which a quantity increases in the region equals the nett tempo with
which it flows into the region, where the nett tempo of inflow is equal to the tempo of inflow minus the
tempo of outflow. In what follows, we will refer to this law as Law of Conservation.
This law can be mathematically formulated, depending on the physical situation. We illustrate this with
the aid of a few examples, from which it also becomes clear that a law of conservation gives rise to an
ordinary differential equation or a Partial Differential Equation.
Example 4.10.1
The contents of a tank are so thoroughly stirred that if any solution enters the tank, instantaneous mixing
takes place. Suppose a solution with concentration (density) k gram/litre is pumped in at a tempo of α
litre/second, while the mixed concentrate is pumped out at a tempo of β litre/second. Assume that the
concentration of dissolved substance at the moment t is c(t) gram/litre and that the volume of the solution
is V (t) litres. We assume that in the whole process no mass or volume is created or destroyed. The tempo
at which the dissolved substance flows in, is now αk gram/second, while the tempo at which it flows out,
is βc(t) gram/second at the moment t. The nett tempo of inflow is thus
αk − βc(t)
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at time t. Application of Law of Conservation on the mass of the dissolved substance yields
d c(t)V (t) = αk − βc(t).
dt
(4.92)
Equation (4.92) is thus the Conservation Law (with respect to the mass) as a formula, or in mathematical
language, a differential equation, in which V is, however, still unknown. By applying the principle in Law
of Conservation to the volume, we get
dV
= α − β,
(4.93)
dt
i.e. Conservation Law for the volume. Integration of (4.93) gives V (t) = V (0) + (α − β)t. By substituting
into (4.92), we get
d
((c(t)[V (0) + (α − β)t]) = αk − βc(t).
dt
By solving this ordinary differential equation for c(t), the concentration of the liquid can be determined at
any time.
■
Example 4.10.2
(This example may be compared with the case where we do not assume instantaneous mixing.)
Suppose a liquid is in rectilinear motion, with a constant velocity v. A quantity of dye is placed in the
stream of liquid and is carried downstream by the motion of the liquid. In order to describe the motion
of the dye, we observe the concentration of the dye at each point. Suppose that the concentration of dye
at the point x at time t is c(x, t). We neglect the diffusion phenomenon. Once again we assume that the
mass remains unchanged, i.e. we assume Law of Conservation for the mass.
The mass of dye over an arbitrary section of the stream, if we take this section as an interval [a, b], is
Z b
c(x, t)dx.
a
From Law of Conservation, applied to the mass, we have
Z
d b
c(x, t)dx = tempo of inflow - tempo of outflow
dt a
= c(a, t)v − c(b, t)v for all [a, b]
Z b
∂
= −v
c(x, t)dx
a ∂x
from the Fundamental Theorem of Calculus.
Z b
If we assume that the integral
c(x, t)dx converges uniformly in t (interchange of the time derivative and
a
the integral is now permissible), we have
Z b
Z b
∂
∂
c(x, t) dx = −v
c(x, t)dx for all[a, b]
∂t
∂x
a
a
Z b
∂
∂
c(x, t) + v c(x, t) dx = 0
for all [a, b].
∂t
∂x
a
By assuming that the integrand is continuous, it now follows that
∂c
∂c
+v
∂t
∂x
Thus Conservation Law gives rise to a Partial Differential Equation for the concentration c(x, t).
■
124
4.10.1.3
Abstract form of the Law of Conservation
In general the Law of Conservation can be put in mathematical language as follows:
Suppose that u(x, t) is the density function for the quantity that must be ”conserved” in a region G of the
N -dimensional space (N ≥ 1). Suppose ϕ(x, t) is a given flux vector, i.e. a vector in terms of which the
flux of the quantity of the boundary of the region can be described. Then we have:
(i) If N = 1, G = [a, b] :
d
dt
Z
b
u(x, t)dx = ϕ(a, t) − ϕ(b, t)
a
by keeping in mind that if u(x, t) is integrated over a region G, the resulting integral gives the mass
of the body that is in G at time t.
(ii) If N ≥ 2, ∂G the boundary of G :
d
dt
Z
G
u(x, t)dx = −
where n is the unit vector of the outward normal.
Z
∂G
ϕ · nds
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Remark 4.10.1
1. The integral over G is an N -tuple integral and that over ∂G is an (N − 1)-tuple integral.
2. The conservation law in (ii) is, put into words, the well-known principle that, if there are no sources
or sinks present within a region, then the speed with which a physical quantity increases in the region
is equal to the flux of
Z that quantity over the boundary of the region. Compare [Tikhonov & Samarskii;
p 28]. The term −
4.10.2
∂G
ϕ · nds, however, represents also the inflow into the region.
■
Coupling Equations
Suppose that u(x, t) is a density function for a quantity that must be conserved in a region G. Let ϕ(x, t)
denote the flux. Between the two functions u and ϕ there exists a definite relationship such that ϕ is in
some way or another dependent on u. We write
ϕ(x, t) = f (u(x, t)),
(4.94)
where f is a certain vector-valued functional, i.e. a mapping of a function onto a function. Examples of
this are:
f (u) =
f (u) = (u, |u|, u),
∂u ∂u ∂u
,
,
≡ grad u
∂x1 ∂x2 ∂x3
where x = (x1 , x2 , x3 ). (f (u) = grad u is thus indeed a vector.) Thus
(f1 , f2 , f3 )(u) =
∂
∂
∂
,
,
∂x1 ∂x2 ∂x3
(u).
Equation (4.94) is called the coupling equation, because it establishes a link between the functions u
and ϕ. As remarked earlier, a conservation law, together with a coupling equation, will often lead to a
Partial Differential Equation. We will show this in the next section, when we deduce, among others, the
Heat Equation.
In the case of heat conduction, the coupling equation, which links the flux vector ϕ(x, t) and the temperature of the substance through which heat flows, has the following form:
∂u
if x is a one-dimensional vector;
∂x
= −Kgrad u
∂u
∂u
,...
if x = (x1 , . . . , xn ),
≡ −K
∂x1
∂xn
ϕ = −K
(4.95)
(4.96)
(4.97)
where the positive constant K is the conductivity of the substance.
Equation (4.95)–(4.97) are known as Newton’s cooling law. The negative sign denotes that heat always
flows from a higher to a lower temperature, while the fact that K is constant means that the conduction
of heat is homogeneous, i.e. it is the same in all directions and at any position and any time. If the heat
conduction were to vary according to time and position, K would have to be replaced with a function, say
p = p(x, t).
126
4.10.3
4.10.3.1
Derivation of the Heat Equation by the Law of Conservation
In one dimension
We have already derived the equation of the Heat Equation in section 4.4 by direct physical approach.
Suppose heat flows through a thin rod (which we thus regard as a one-dimensional medium) with end
points a and b. Let ρ(x, t) be the mass density, c the specific heat or heat capacity of the rod, u(x, t)
the temperature of the rod at a point x at time t and ϕ(x, t) the flux.
Z b
The heat energy in (a, b) is then equal to
cρudx by recalling that the specific heat equals the amount
a
of heat energy required to allow the temperature of one unit mass of the substance to rise by 1o C. Assume
that there are no sources or sinks which release or absorb heat, i.e. heat flows in and out only at the end
points of the rod. According to Conservation Law we now have
d
dt
Z
a
b
cρudx = ϕ(a, t) − ϕ(b, t) = −
Z
a
b
∂ϕ
dx
∂x
Z
from the Fundamental Theorem of Calculus. By assuming that the integral
b
cρudx converges uniformly
a
in t, we may write
Z b
a
∂
∂ϕ
(cρu) +
dx = 0.
∂t
∂x
Since this relationship holds for arbitrary (a, b), we obtain the Partial Differential Equation
∂ϕ
∂ϕ
(cρu) +
= 0,
∂t
∂x
by furthermore assuming that the integrand is continuous. The conservation law thus leads to a relationship
between partial derivatives. The coupling equation for ϕ now leads to a Partial Differential Equation in
∂u
u. If, according to (4.95), we substitute ϕ = −K
in the equation above, we obtain
∂x
∂
∂
∂u
(cρu) +
−K
=0
∂t
∂x
∂x
If ρ, c are constant, we obtain
∂2u
∂u
=κ 2
∂t
∂x
with κ =
K
.
cρ
This equation is the standard form of the Heat Equation in one dimension.
Note that we have tacitly assumed that the length of the rod remains unchanged, in spite of the change in
temperature. This assumption may be regarded as an idealisation or an approximation. To illustrate the
process of compounding in mathematical modelling, we now present the derivation of the Heat Equation
in higher dimensions.
4.10.3.2
In N - dimensional space, N > 1
In this case heat flows through a region G ⫅ RN with boundary ∂G. Let c, k, ρ, u(x, t) and ϕ(x, t) be as in
section 4.10.3.1, where in this case
x = (x1 , . . . , xN ),
N > 1.
127
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By assuming that the heat energy is only dependent on the outflow (and inflow) of heat over the boundary,
we have, according to Conservation Law (ii) of section 4.10.1.3
Z
Z
d
cρu(x, t)dx = −
ϕ · nds
dt G
∂G
Z
divϕds
= −
G
according to Gauss’ Divergence theorem (see MAT215–W/APM212-W), i.e.
∂
(cρu) + divϕ = 0
∂t
similarly as in 4.10.3.1. Substitution of the coupling equation (4.96) yields
∂
(cρu) = Kdiv(grad u),
∂t
= K∇2 u,
∂u
K
= κ∇2 u where κ = .
∂t
cρ
This Partial Differential Equation is the standard form of the Heat Equation in higher dimensions.
4.10.4
Derivation of Burgers equation in one dimension
The Burgers equation
∂u
∂u
+u
= 0,
∂t
∂x
is a non-linear Partial Differential Equation, the solution of which is studied by making use of, amongst
others, the method of characteristics. Under special conditions, one of them being that viscosity is ignored,
this equation describes the one-dimensional movement of gas.
The equation is a special case of the Partial Differential Equation
∂u
∂u
+ a (u)
= 0.
∂t
∂x
The conservation law and coupling equation that in this case apply to the velocity u(x, t) and the flux
ϕ(x, t) are
Z
d b
u(x, t)dx = [ϕ(b, t) − ϕ(a, t)]
∀(a, b),
dt a
dg
with ϕ(x, t) = g(u(x, t)) where g satisfies g ′ (u) =
= c(u).
du
Under suitable assumptions (formulate them), we obtain
Z b
Z b
∂u
∂ϕ
dx +
dx = 0.
a ∂t
a ∂x
From
∂ϕ
∂u
= g ′ (u) , by again making suitable continuity assumptions, it follows that
∂x
∂x
∂u
∂u
+ c (u)
=0
∂x
∂x
By putting c(u) = u, we obtain the Burgers equation
∂u
∂u
+u
=0
∂x
∂x
128
4.10.5
Derivation of the elastic beam equation in one dimension
In this section we derive the boundary-value problem for the transversal vibrations of a (thin) cantilevered beam which is embedded at its left end x = 0 and free at its right end x = L. We assume that
the total energy is sufficiently described by the integral
Z " 2 2 2 #
∂ u
1 L
∂u
E(t) =
+
dx,
2 0
∂x2
∂t
with u representing the displacement of the beam. For energy conservation we need E constant, i.e.
Z L 2
dE
∂ u ∂3u
∂u ∂ 2 u
0 =
dx,
=
·
+
dt
∂x2 ∂t∂x2
∂t ∂t2
0
Z L
L
L
∂u ∂ 2 u ∂ 4 u
∂2u ∂2u
∂ 3 u ∂u
·
·
+ 4
dx
=
−
+
∂x2 ∂t∂x 0
∂x3 ∂t 0
∂t ∂t2
∂x
0
by applying integration by parts twice. Now the assumption that no energy is lost or gained within the
beam clearly leads to the Partial Differential Equation
∂2u ∂4u
+ 4 =0
∂t2
∂x
in (0, L). A variety of boundary conditions serve to annihilate the boundary terms. For the cantilever beam
∂u
∂2u
∂3u
the appropriate boundary conditions are u(0, t) = 0,
(0, t) = 0,
(L,
t)
=
0,
(L, 0) = 0. Thus
∂x
∂x2
∂x3
if we prescribe the initial displacement and the initial velocity, we obtain the following boundary-value
problem
∂2u ∂4u
+ 4
∂t2
∂x
= 0,
u(0, t) = 0,
∂2u
(L, t) = 0,
∂x2
u(x, 0) = u0 (x),
0 < x < L,
0 < t,
∂u
(0, t) = 0,
∂x
∂3u
(L, 0) = 0,
∂x3
∂u
(x, 0) = u1 (x).
∂t
Exercises 4.10
∂ρ
1. Deduce the continuity equation
+div(ρv) = 0. This equation occurs in gas dynamics, and describes
∂t
the movement of a gas with density ρ(x, t), velocity c(x, t) and pressure p = p(ρ, t), if viscosity and
heat conduction are ignored and no external forces are present.
2. Derive the boundary-value problem for the damped transversal vibrations of a cantilever beam which
is embedded at the end x = 0 and free at the end x = L. Introduce
∂5u
(i) A damping force which is proportional to the deformation rate,
, of the beam (this is
∂x4 ∂t
the case of damping of Kelvin-Voigt type);
∂3u
(ii) A damping force which is proportional to the bending rate, 2
of the beam. (This is the
∂ ∂t
case of Structural Damping).
129
APM3701/1
Hint: Apply integration by parts to
dE
+
dt
Z
dE
+
dt
Z
L
0
∂3u
∂x2 ∂t
2
∂5u
∂x4 ∂t
2
and to
L
0
dx
for Structural Damping
dx
for Kelvin-Voigt Damping.
dE
Bear in mind that, in contrast to the undamped (conservative) case where
= 0 is required, one
dt
dE
needs
< 0 to ensure energy dissipation. Assume as above that the energy is given by
dt
Z " 2 2 2 #
∂ u
∂u
1 L
+
dx.
E(t) =
2
2 0
∂x
∂t
Question: Suppose one were to test the adequacy of any of the models which incorporates energy
dissipation. What property should the solution exhibit as time tends to infinity?
3. Use Energy Methods to derive the boundary-value problem for the transversal vibrations of a beam
which has its end x = 0 rigidly fixed, while a load is attached to its other end x = L. Assume that
the energy is sufficiently described by the expression
1
E(t) =
2
"Z
0
L
∂2u
∂x2
2
+
∂u
∂t
2 !
∂u
dx +
(L, t)
∂t
2
#
.
Note that the last term represents the kinetic energy at x = L.
4. (Only for the ambitious student!) Repeat the above exercise, but introduce
(i) Kelvin-Voigt Damping;
(ii) Structural Damping.
You should find that in order to obtain energy dissipation, one has to introduce a damping term
not only in the Partial Differential Equation, but also into the dynamical boundary condition.
Hint: For the case of Structural Damping, assume that the energy is given by the expression
"Z
#
2 !
2
2
L 2 2
∂u
∂ u
∂u
1
∂u
+
dx +
(L, t) +
(L, t)
.
E(t) =
2 0
∂x2
∂t
∂t
∂x
(Here the last term may be regarded as the energy of deformation at the end x = L).
■
130
131
CONTENTS
Chapter 5
QUALITATIVE QUESTIONS:
MAXIMUM–MINIMUM PRINCIPLE
UNIQUENESS AND STABILITY
RESULTS
5.1
INTRODUCTION
As stated in the preface, a great deal of emphasis is laid in this course on developing techniques to solve
boundary-value problems. It is, however, important not to be satisfied with only solving the mathematical
models and interpreting their solutions, since in the case of nonlinear partial differential equations it is
often not even possible to construct an exact solution.
As we have already seen in Chapter 4, the logical progression in predicting the behaviour of a physical
system, begins with a description of the physical situation, which is made as concise as possible by making
idealisations and approximations. The second step is the construction of a mathematical model. Next the
testing of the mathematical model should follow. Unfortunately this aspect is completely overlooked in a
great majority of textbooks on partial differential equations!
This chapter discusses the qualitative mathematical concepts of existence, uniqueness and stability of
solutions. These properties when satisfied render a boundary-value problem well-posed. The maximum
and minimum principle is demonstrated. This chapter also presents uniqueness and stability of results for
Dirichlet problem for the heat Equation, wave and Laplace equation. A posteriori existence, uniqueness
and stability of solution will be examined.
A mathematical model can only be described as feasible or adequate if the following properties are validated:
(i) There exists one and only one solution, i.e. there exists a unique solution.
This is a naturally important requirement, since our experience in nature is such that a given set of
circumstances leads to only one outcome.
(ii) The solution is stable or depends continuously on the prescribed data, i.e. a small change in the
given data produces a correspondingly small change in the solution.
132
The stability property is vital, since when the boundary conditions are obtained experimentally,
certain small observational errors in their values will always exist. Stability ensures that such errors
will not lead to large changes in the solution.
5.2
LEARNING OUTCOMES
This chapter will achieve the following outcomes:
• Formulate the maximum–minimum principle for the heat and Laplace equation.
• Prove uniqueness and stability results for the heat, wave and Laplace equations.
• Establish a posterio existence, uniqueness and stability of solution.
5.3
DEFINITION: WELL-POSED PROBLEM
A boundary-value problem possessing a unique, stable solution will be called a well -posed problem. A
problem which is not well-posed, is ill-posed (see also section 1.6.2).
Remark 5.3.1
1. The following famous example of an ill-posed boundary-value problem is due to Hadamard (p.50):
Consider the following boundary-value problem for the Laplace equation:
∇2 u = 0
in R2 ,
u(x, 0) = 0,
1
∂u
(x, 0) =
sin kx
∂y
k
where k > 0 is an integer. It is easily verified that
u(x, y) =
1
sin kx · sinh ky
k2
solves this problem. Now since
1
1
sin kx < ,
k
k
∂u
(x, 0) = 0, in
∂y
which case the solution to the given boundary-value problem is u = 0. However, the solution
it follows that when k is large the boundary conditions are close to u(x, 0) = 0,
u(x, y) =
1
sin kx · sinh ky
k2
becomes unbounded as k → ∞ even for small values of |y| (why?). This shows that small changes in
the boundary conditions on u lead to large changes in the solution. Thus the solution is unstable.
2. Ill-posedness may also occur when a problem is over-specified, i.e when too many boundary conditions are prescribed, in which case a solution may not exist at all. A problem that has too few
boundary conditions will ”always” turn out to be ill-posed, since in this case the solution is not
unique.
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Clearly the establishment of well-posedness for a particular problem, enhances the confidence in the
underlying mathematical model. For the purposes of this module, the establishment of well-posedness
results will be divided into two steps:
(a) Showing existence and stability of a solution to a boundary-value problem from analysis of a
specific solution which is obtained by analytical methods. This is the a posteriori (“afterwards”)
approach and entails that one verifies that for instance a particular series solution you have
obtained, is convergent, satisfies the conditions of the model, and is continuous with respect to
small changes in the data.
(b) Demonstration of uniqueness and stability of solutions in a general manner without explicit
knowledge of the solution. This is the a priori approach and entails that we assume a priori
(beforehand) that a solution exists and then show that it is unique and stable.
In this chapter we will devote ourselves to establishing uniqueness and stability results by assuming
a priori that a boundary-value problem has a solution.
We shall restrict ourselves to continuous, bounded solutions with at least continuous and bounded
data. Although this seems to be a strong restriction, it enables us to gain the essence of the methods
without having to deal with the complications of discontinuities. The ground rules obtained for
the simple linear models also serve as an essential, though unfortunately only heuristic guide to
establishing well-posedness for nonlinear problems which are solved by e.g. Galerkin approximations.
Before proceeding to methods of proving uniqueness and stability, and presenting the maximum and
minimum principles, we conclude with the following remark which may give you an idea of the inherent
beauty of advanced study in the field of qualitative aspects of partial differential equations. I quote from
[Street; p. 173]: “Establishing the adequacy of mathematical models is one of the most interesting and
challenging subjects of applied mathematics. Over the years a number of sophisticated techniques have
been derived, particularly to prove the existence in general domains of solutions to general equations. The
history of the development of a proper existence proof for the Dirichlet problem for the Laplace equation
outlined by [Kellogg], illustrates the difficulties to be encountered. Green inferred the possibility of solving
the Dirichlet problem in 1828, but this was not a proof. Gauss put forth another argument in 1840, but
he did not properly restrict the region of the problem. Lord Kelvin in 1847 and Dirichlet in the 1850s
advanced techniques based on minimisation of an integral (a functional of the variational calculus), but
Weierstrass showed in 1870 that the result was not reliable. Not until 1899 did Hilbert sort out the proper
conditions for a reliable result.”
The problem of finding suitable models to describe a physical situation and then establishing well-posedness
for the corresponding boundary-value problem, seems to be inexhaustible. As long as the world around
us changes and new phenomena occur, or questions related to “old” phenomena remain incompletely
answered, researchers will carry on with the task of formulating models and establishing well-posedness
results under the best possible conditions. possible restrictions on the prescribed data.)
A few contributions of the author of this Guide in this field are the following papers:
1. Solutions in Lebesgue spaces of the Navier-Stokes equations with dynamic boundary conditions,
Proc. Roy. Soc. Edinburgh 123A (1993), 745-761. (Co-author: Niko Sauer).
134
2. On fractional powers of a closed pair of operators and a damped wave equation with dynamic
boundary conditions, Appl. Anal. 53 (1994), 41-54.
3. On the initial-boundary-value problem for the extensible beam with attached load, (Mathematical
Methods in the Applied Sciences), 1996.
4. Existence and uniqueness of classical solutions of the equations of motion for second grade fluids,
Arch. Rational Mech. Anal. 124, (1992), 221-237. (Co-authors: Giovanni P. Galdi & Niko Sauer).
5. Existence and uniqueness of classical solutions of the equations of motion for a fluid of second grade
with non-homogeneous boundary conditions, Int.J. Nonlinear Mechanics 30, (1995), 701-709. (Coauthors: Giovanni P. Galdi & Niko Sauer).
In all these papers existence and uniqueness results are established with the aid of functional analytic
methods. (The results can, however, be complemented and elucidated with the aid of methods from
Numerical Analysis.) The functional analytic approach falls outside the scope of this module. However,
should you after completion of this module, wish to proceed to advanced study in the field of partial
differential equations, Functional Analysis will be an important course for you, while a good knowledge of
Distribution Theory is indispensable.
5.4
UNIQUENESS AND STABILITY OF SOLUTIONS
We now proceed to methods of proving uniqueness and stability of solutions for boundary-value problems
a priori, i.e. without having constructed a solution to a particular problem, we assume that a solution
exists and then show that it is unique and stable.
There are three basic techniques for showing uniqueness and stability which we shall consider, viz. an
Integral Method, an Energy Method and Maximum-Minimum Principles. (The first two are used only in
uniqueness proofs.)
5.4.1
An Integral Method
This method is based on a general integral relation which is valid for any sufficiently differentiable function.
(This is why we have to impose smoothness assumptions on the solution we assume to exist!) In three
dimensions, the analysis is based on Gauss’ Divergence Theorem
Z
Z
v · nds =
divvdx
(5.1)
Ω
∂Ω
and the Green’s Identity
Z
∂g
ds =
f
∂Ω ∂n
Z
Ω
2
f ∇ gdx +
Z
Ω
(∇f · ∇g) dx
(5.2)
which is derived from (5.1).1
These results are valid under the following assumptions:
(i) Ω is a bounded simply connected region in Rn and n is the outward unit normal vector on the closure
∂Ω of the boundary ∂Ω which is assumed to be piecewise smooth to ensure amongst others that n
1
See Exercises 5.4 (1.)
135
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R
varies continuously over any smooth portion of the boundary ∂Ω. Thus Ω denotes an integral over
R
three-dimensional space, while ∂Ω denotes the integral over the two-dimensional boundary ∂Ω;
(ii) In (5.1) v is a continuous function of (x, y, z) in Ω ∪ ∂Ω = Ω, while in (5.2) it is required that f
and g be continuous on Ω ∪ ∂Ω = Ω and twice continuously differentiable in Ω - we use the notation
f, g ∈ C 2 (Ω).
We now prove the following theorem:
Theorem 5.4.1 Uniqueness for the Dirichlet Problem for the Laplace Equation
Consider the problem
∇2 u = 0
in Ω
u = f
on ∂Ω.
(5.3)
Suppose that a solution u exists for (5.3) and that u ∈ C 2 with respect to (x, y, z). Suppose additionally
that f ∈ C on ∂Ω. Then the solution of (5.3) is unique.
Proof
According to Green’s formula (5.2) we have
Z 2
2
u∇ u + (∇u)
Z
dX =
Ω
u
∂Ω
∂u
ds.
∂n
(5.4)
Suppose u1 and u2 are two non-identical solutions of the problem ∇2 v = 0 in Ω, v = v0 (x) on ∂Ω. Then,
as above U = u1 − u2 satisfies ∇2 U = 0 within Ω, and U = 0 on ∂Ω. Thus (5.4) becomes
Z
(∇U )2 dX = 0.
Ω
Consequently ∇U ≡ 0, so that U is a constant. Because U = 0 on ∂Ω, it follows that U ≡ 0, i.e. u1 = u2 .
This proves that the solution of the Dirichlet problem for the Laplace equation is unique. Indeed the
solution is uniquely determined by its values on the boundary.
■
We now employ an Integral Method to prove uniqueness for the solution of the Heat Equation in three
dimensions:
Theorem 5.4.2 Uniqueness for the Dirichlet problem for the Heat Equation
Consider the heat transfer problem
∂u
= κ∇2 u
∂t
∂u
= f
∂n
u(x, 0) = u0 (x).
in Ω
on ∂Ω
(5.5)
Suppose that a solution u exists for (5.5) and that u is C 2 with respect to (x, y, z), and C 1 with respect to
t in Ω. Then the solution u is unique.
136
Proof
Let U = u1 − u2 with u1 , u2 two non-identical solutions of (5.5). Clearly U satisfies
∂U
= κ∇2 U
∂t
∂U
= 0
∂n
U (x, 0) = 0.
in Ω
(5.6)
on ∂Ω
In order to make use of Green’s identity (5.2), we consider the functional
Z
2 2
U 2 dx ≥ 0
Q(t) = ρ c
2
(5.7)
Ω
which is at least C 1 with respect to t and represents the square of the internal energy in Ω. Clearly by
differentiation
Z
Z
Z
∂U
′
2 2
2
U
(∇U )2 ≤ 0
(5.8)
Q (t) = 2ρ c
U ∇ U dx = −2k
dx = 2k
∂t
Ω
Ω
Ω
where k = κρ2 c2 . For 0 < t by using Green’s Identity (5.2) and the fact that U vanishes on ∂Ω. Clearly
Q(0) = 0. From (5.8) we get
Z t
Z t
Q′ (τ )dτ + Q(0) =
Q′ (τ )dτ ≤ 0.
Q(t) =
0
0
This is in contradiction with (5.7) unless Q(t) ≡ 0. Again from (5.7) it follows that U must be zero. Thus
u1 = u2 and the solution u to (5.6) is unique. This completes the proof.
■
5.4.2
An Energy Method
Energy methods are based on consideration of the behaviour of the total energy of a physical system
as a function of time. We shall apply energy methods in proving the uniqueness of the solution of the
one-dimensional wave equation.
It can be shown that the total energy E(t) of a thin string undergoing small transverse vibrations while
both its ends are fixed, is given by
E(t) = KE + P E
Z " 2 2 #
1
∂u
∂u
=
c2
+
dx ≥ 0
2 Ω
∂x
∂t
with KE and P E representing respectively the kinetic and the potential energy. We now prove uniqueness
for the wave equation:
Theorem 5.4.3 Uniqueness for the Dirichlet Problem for the Wave Equation.
Consider the problem
2
∂2u
2∂ u
=
c
,
∂t2
∂x2
u(0, t) = u(L, t) = 0,
0 < x < L,
t > 0,
u(x, 0) = u0 (x),
∂u
(x, 0) = u1 (x).
∂t
2
The functional Q is a function of u. Since u is a function of t, the functional Q varies with time t, i.e. Q = Q(t).
(5.9)
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Suppose that there exists a solution u ∈ C 2 (0, L) to the problem (5.9). Assume u0 (x) ∈ C[0, L] with respect
to x and u1 (x) ∈ C 1 [0, L] with respect to t. Then the solution u is unique.
Proof
Assume that two non-identical solutions u1 and u2 satisfying the problem (5.9).
Their difference U = u1 − u2 satisfies
2
∂2U
2∂ U
=
c
,
∂t2
∂x2
U (0, t) = U (L, t) = 0,
0 < x < L,
t > 0,
U (x, 0) = 0,
∂U
(x, 0) = 0.
∂t
Clearly E(0) = 0. Also by differentiation
Z
′
L
E (t) =
2
c
0
∂U ∂ 2 U
∂x ∂x∂t
∂U ∂ 2 U
dx
+
∂t ∂t2
Now
L
Z
0
∂U ∂u
(Why does the term
∂x ∂t
Thus
L
0
Z L
∂U ∂U L
∂U ∂ 2 U
−
dx
∂x ∂t 0
∂t ∂x2
0
Z L
∂U ∂ 2 U
= −
.
∂t ∂x2
0
∂U ∂ 2 U
dx =
∂x ∂x∂t
disappear?)
′
Z
∂2U
+
−c
∂x2
∂t2
L E (t) =
0
2∂
It follows that
Z
E(t) =
t
2U
∂U
dx = 0.
∂t
E ′ (τ )dτ + E(0) = 0.
0
∂u
∂u
This implies that
=
within (0, L). Accordingly U must be constant.
∂x
∂t
As U = 0 at x = 0 and at x = L, we have U = 0 in [0, L].
This implies that u1 = u2 . The solution is therefore unique and the proof is complete.
5.4.3
■
Maximum-Minimum Principles
In this section we will show that a continuous solution in a region Ω for the Laplace Equation attains
both its maximum and minimum values on the boundary ∂Ω of Ω, while a continuous solution of the Heat
Equation attains its maximum and minimum values at the boundary or initially, i.e. at time t = 0. These
results form the basis for simple proofs of uniqueness and stability results.
We will establish two maximum-minimum theorems using a two-dimensional region Ω in the xy –plane.
The technique in both proofs are basically the same. We first prove the following theorem:
138
Theorem 5.4.4 Maximum–Minimum Principle for the Laplace Equation
Let the harmonic function u(x, y) be continuous on the closure Ω of the bounded, closed and simply connected region Ω in R2 . Then u(x, y) cannot be neither larger than its maximum on the boundary ∂Ω nor
smaller than its minimum on ∂Ω, i.e. maximum and minimum values are attained at the boundary.
Proof
We proceed by contradiction. We define
1
φ(x, y) = ϵ(x2 + y 2 ),
4
ϵ ≥ 0.
1
Let the origin of the xy–plane lie in Ω so that 0 ≤ φ(x, y) ≤ ϵρ2 , where ρ is the diameter of the smallest
4
circle enclosing Ω with centner at (0,0). Now ρ is less than or equal to the maximum distance across Ω.
Let M be the maximum value and m the minimum value of the harmonic function u on the boundary ∂Ω.
First ∇2 φ = ϵ ≥ 0. Therefore, since ∇2 u = 0 and in view of the positivity of φ, we have for U + = u + φ
∇2 U + = ∇2 φ ≥ 0,
U+ ≥ u
in
Ω
(5.10)
∇2 U − = −∇2 φ ≤ 0,
U− ≤ u
in
Ω.
(5.11)
while we have for U − = u − φ
Suppose now that U + takes its maximum value at some point in Ω, but not on ∂Ω. (Since we are using the
method of contradiction, we must assume that maximum and minimum values are not attained on ∂Ω.)
∂U +
∂U +
∂2U +
∂2U +
From calculus we know that at a maximum point
=
= 0 while also
≤
0
and
≤ 0.
∂x
∂y
∂x2
∂y 2
∂2U + ∂2U +
+
at the point of maximum. This is in contradiction with (5.10).
This implies that ∇2 U + =
∂x2
∂y 2
Consequently U + can only attain its maximum value on ∂Ω where
1
U + ≤ M + ϵρ2 .
4
Suppose now that U − takes its minimum value at some point in Ω but not on ∂Ω. By once more using
calculus rules, we obtain a contradiction with (5.11) and arrive at
1
U − ≥ m − ϵρ2 .
4
From (5.10) and (5.11) we have in Ω
−ϵ = ∇2 U − ≤ ∇2 u ≤ ∇2 U + = ϵ
1
1
m − ϵρ2 ≤ u ≤ M + ϵρ2 .
4
4
By letting ϵ → 0 we obtain for the harmonic function u the relation m ≤ u ≤ M . Therefore u can neither
be larger than its maximum on ∂Ω nor smaller than its minimum on ∂Ω. The proof is complete.
■
We now prove a maximum-minimum result for the Heat Equation:
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Theorem 5.4.5 (Maximum–Minimum Principle for the Heat Equation.) A continuous solution
u(x, y, t) of the Heat Equation
2
∂u
∂ u ∂2u
+
in Ω,
= κ
∂t
∂x2
∂y 2
u = f
on ∂Ω,
(5.12)
u(x, 0) = u0 (x)
in the finite space-time cylinder
n
ΩT = (x, y, t) : (x, y) ∈ Ω,
0<t≤T <∞
o
.
The solution u can neither be larger than its maximum nor smaller than its minimum on the surface
composed of the bounded cylinder S = {(x, y, t) : (x, y) ∈ ∂Ω, 0 < t ≤ T } and the base Ω × {0}. The
continuous solution u(x, y, t) attains its maximum and minimum values on the boundaries or initially.
Proof
The proof is analogous to the proof of the Theorem 5.4.4. Let φ be defined as before. Let M be the
maximum and m the minimum value of u(x, y, t) on S. By again defining U + = u + φ and U − = u − φ
and reasoning as before we obtain
1
U + ≤ M + ϵρ2
4
and
1
U − ≥ m − ϵρ2
4
on
S
while
U− ≤ u ≤ U+
generally.
Suppose now that U + attains its maximum value at some point (x0 , y0 , t0 ) in ΩT but not on S. Now
∂2U +
∂2U +
∂2U +
∂U +
∂U +
for a maximum
≤ 0,
≤ 0 and
≤ 0 while also
=
= 0 at (x0 , y0 , t0 ),
2
2
2
∂t
∂x
∂y
∂x
∂y
∂U +
= 0. If t0 = T , it is possible
the point of the supposed interior maximum. If t0 < T , we have
∂t
+
∂U
that U + (x0 , y0 , t0 − τ ) < U + (x0 , y0 , t0 ) for small τ so that
≥ 0 can be assumed. It follows that at
∂t
(x0 , y0 , t0 )
∂U +
− κ∇2 U + ≥ 0.
(5.13)
∂t
∂u
In view of
− κ∇2 u = 0 and ∇2 φ = ϵ > 0, the definition of U + = u + φ yields
∂t
∂U +
− κ∇2 U + ≤ 0.
∂t
(5.14)
This is in contradiction with (5.13). Thus an interior maximum is impossible and
1
U + ≤ M + ϵρ2 .
4
Supposing that U − takes its minimum inside ΩT at say (x1 , y1 , t1 ), yields analogously
1
U − ≥ m − ϵρ2 .
4
1
1
Consequently m − ϵρ2 ≤ u ≤ M + ϵρ2 in Ω. By letting ϵ → 0 we obtain m ≤ u ≤ M . The proof is
4
4
complete.
■
140
We can now establish uniqueness and stability results for the solutions to the Laplace equation and the
Heat Equation in two dimensions. (Since the Maximum-Minimum Principles can be extended to threedimensional space, these results can also be obtained in three-dimensional space.)
Theorem 5.4.6 (Uniqueness and Stability Result for the Laplace Equation.) Consider the problem
∇2 u = 0
u = f
in
on
Ω
∂Ω.
(5.15)
Suppose that there exists a solution u of (5.15) and that u ∈ C 2 with respect to (x, y). Suppose additionally
that f ∈ C on ∂Ω. Then the solution to Problem (5.15) is unique and stable.
Proof
Let U = u1 − u2 with u1 and u2 two non-identical solutions to (5.15). Then U satisfies
∇2 U
= 0
in
Ω
U
= 0
on
∂Ω.
Since U is a harmonic function, the values of U in Ω must be bounded by the maximum and minimum
values of U on ∂Ω. Therefore U ≡ 0 in Ω. Therefore u1 = u2 and the solution u is unique.
We now prove that the solution u is also stable. Let u1 be the unique solution to (5.15) and suppose u2 is
the solution to the so-called perturbed model, we have:
∇2 u2 = 0
u2 = f + δ
in Ω
on ∂Ω.
where δ(x, y) ∈ C and |δ(x, y)| ≤ ϵ > 0. Now U = u1 − u2 satisfies
∇2 U = 0
U =δ
in Ω
on ∂Ω.
By the result of Theorem 5.4.4, U is bounded by the maximum and minimum values of δ. Thus |U | ≤ δ ≤ ϵ
or
|u1 − u2 | ≤ ϵ.
Thus we have shown that a small change of order ϵ in the boundary function f produces an equally small
change in the solution. The stability of the solution u is therefore established.
■
141
APM3701/1
Exercises 5.4
1. Derive Green’s formula (5.2) from Gauss’ Divergence Theorem.
Hint: If u and v are scalar fields, then u∇v is a vector field. Apply the Gauss’ Theorem to u∇v.
Use also the vector identities ∇ · (v∇u) = v∇2 u + ∇v · ∇u.
2. Show that a necessary condition for the Neumann Problem
∇2 u = 0
∂u
=g
∂n
in Ω
on ∂Ω
to have any solution is that the compatibility condition
Z
gds = 0
∂Ω
is satisfied. From what physical considerations does this condition arise?
3. Use Gauss’ Divergence Theorem to show that the solution to the Neumann problem
∇2 u = 0
∂u
=g
∂n
in Ω
on ∂Ω
is unique only up to an additive constant , i.e. if u1 and u2 are two solutions, then u1 − u2 = k.
4. Use the Maximum-Minimum Principle for the Heat Equation to show that the continuous solution
to the heat transfer problem
∂u
= κ∇2 u
∂t
u=f
in Ω,
on ∂Ω,
u(x, t) = u0 (x).
is unique and stable.
5.5
■
UNIQUENESS AND THE MAXIMUM PRINCIPLE FOR THE HEAT EQUATION
We know from section 1.6.2 (Well–posedness) that the heat equation with its boundary conditions will
have no physical use if it does not admit a unique solution or if the solution exists but it is unstable.
We examine the methods for proving uniqueness. The first method, is a direct one, the second which is
based on the maximum principle also establishes stability.
Theorem 5.5.1 (Uniqueness theorem) If u1 (x, t) and u2 (x, t) are C 2 –solutions of the following problem, where a (t) , b (t) and f (x) are given C 2 –functions
ut = kuxx
0 ≤ x ≤ L,
u (0, t) = a (t) ,
u (x, 0) = f (x) ,
t>0
(5.16)
u (L, t) = b (t)
(5.17)
0 ≤ x ≤ L.
(5.18)
142
Then u1 (x, t) = u2 (x, t) for all 0 ≤ x ≤ L and t > 0.
Proof
See Example 2 in section 3.11 page 189 of [1].
■
Theorem 5.5.2 (The maximum principle for the heat equation.) Assume that a, b and f are given
C 2 –functions. Consider T > 0 a fixed future time. Define M to be the maximum value of the initial temperature, and let A and B be the maximum temperatures at the end points x = 0 and x = L during the
time intervals t = 0 to t = T, i.e.
A = max {a (t)} ,
0≤t≤T
B = max {b (t)}
0≤t≤T
and
M = max {f (x)} .
0≤x≤L
Let M = max {A, B, M } .
If u (x, t) is a C 2 –solution of
0≤x≤L
ut = kuxx
u (0, t) = a (t)
t≥0
(5.19)
u (L, t) = b (t)
(5.20)
u (x, 0) = f (x) .
(5.21)
Then
u (x, t) ≤ M
for all x and t with 0 ≤ x ≤ L, 0 ≤ t ≤ T.
Proof
See Appendix to section 3.11 page 190 of [1].
5.5.1
(5.22)
■
Physical interpretation of the Maximum Principle Theorem
In physical terms, the Maximum Principle Theorem states, that the temperature, at any point x inside the
rod at any time t (0 ≤ t ≤ T ) , is always less than the maximum of the initial temperature distribution or
the maximum of the temperatures prescribed at the ends during the time interval [0, T ] . This is expected,
since one does not in normal conditions expect heat to concentrate itself inside the rod, but rather to
dissipate.
Heat avoids itself, preferring to flow to colder regions. By the same token, cold avoids itself.
There is a Minimum Principle that follows from the Maximum Principle.
Corollary 5.5.3 The Minimum Principle for the heat equation
Let u (x, t) be a C 2 –solution of the Initial Boundary value problem (5.19) – (5.21) and let
a =
b =
min {a (t)} ,
0≤t≤T
min {b (t)}
0≤t≤T
143
APM3701/1
and
m = min {f (x)} .
0≤x≤L
Defining
m = min {a, b, m}
we have m ≤ u (x, t) for all x and t, 0 ≤ x ≤ L, 0 ≤ t ≤ T. .
■
Example 5.5.1
Use the maximum and minimum principles to deduce that the solution u(x, t) of
ut
=
9uxx ,
0 ≤ x ≤ 3, t > 0
u(0, t) = 0 = u(3, t)
πx
u(x, 0)
=
6 sin
+ 2 sin πx
3
√
satisfies the inequalities 0 ≤ u(x, t) ≤ 4 3
Solution
πx
We have k = 9, L = 3 and the initial temperature f (x) = 6 sin
+ 2 sin πx.
3
We know from the maximum that u (x, t) cannot exceed the maximum on the edges x = 0, x = 3 and t = 0
in the xt− plane.
For x = 0 and x = 3, u (0, t) = 0 = u (3, t) .
πx
To find the maximum and minimum of u (x, 0) , we differentiate u (x, 0) = f (x) = 6 sin
+ 2 sin πx :
3
πx
2πx
πx
f ′ (x) = 2π cos
+ cos πx = 4π cos
cos
.
3
3
3
3 3 9
3
9
′
Thus, in the interval [0, 3] , f (x) = 0, when x = , , . Thus, the maximum of f are f
=f
=
2 4
4
4
4 √
3
3
= 4. Thus the maximum of u (x, t) on the borders
4 2. At x = , f has a local minimum with f
2
4
√
√
x = 0, x = 3 and t = 0 is 4 2 and the minimum is 0. Thus 0 ≤ u (x, t) ≤ 4 2.
■
5.5.2
Stability of solution of the heat equation
Theorem 5.5.4 (Stability of Solution of the heat equation) Let u1 (x, t) and u2 (x, t) be C 2 –solutions
of the two respective problems (0 ≤ x ≤ L, t ≥ 0)




u
=
ku
ut = kuxx


t
xx




 u (0, t) = a (t)
 u (0, t) = a (t)
1
2
and
(5.21)


u
(L,
t)
=
a
(t)
u
(L,
t)
=
a
(t)
1
2






 u (x, 0) = f (x)
 u (x, 0) = f (x)
1
2
If, for some ε > 0, we have
|f1 (x) − f2 (x)| < ε
for 0 ≤ x ≤ L
and
|a1 (t) − a2 (t)| < ε
144
and
|b2 (t) − b2 (t)| < ε
for 0 ≤ t ≤ T
Then
|u1 (x, t) − u2 (x, t)| < ε
for 0 ≤ x ≤ L,
0 ≤ t ≤ T,
where T is arbitrary.
Proof
Let v (x, t) = u1 (x, t) − u2 (x, t) . Then v satisfies the following relations:
vt = kvxx
|v (x, 0)| = |a1 (t) − a2 (t)| ≤ ε
|v (L, t)| = |b1 (t) − b2 (t)| ≤ ε,
and ∥v(x, 0)∥ = ∥f1 (x) − f2 (x)∥ ≤ ε,
0≤t≤T
0 ≤ x ≤ L.
The maximum of v on the borders t = 0 (0 ≤ x ≤ L) and x = 0, x = L (0 ≤ t ≤ T ) is not greater than ε.
Thus the maximum and minimum principles yield
−ε ≤ v (x, t) ≤ ε
or
|u1 (x, t) − u2 (x, t)| < ε,
∀ε > 0.
■
5.5.3
Interpretation of Theorem 5.5.4
Theorem 5.5.4 states that if the initial and boundary conditions are slightly changed , then any solutions
will change slightly. This allows to approximate initial and boundary conditions by more manageable
functions, without disturbing the solution significantly. Physically it makes sense, since in practice initial
conditions can never be exactly known (measurement errors,....).
5.6
A POSTERIORI EXISTENCE, UNIQUENESS AND STABILITY
The qualitative properties of existence, uniqueness and stability were introduced in Section 2.6, where we
also presented well-posedness results by following the a priori approach.
As we are now acquainted with the Method of Separation of Variables, we are in a position to present an
existence and stability result by following the a posteriori approach, as promised in Section 2.6.1. For this
purpose we choose to analyse the series solution to the Dirichlet Problem for the Heat Equation, i.e. the
problem
∂2u
∂u
= k 2,
0 < x < L, t > 0
∂t
∂x
u(0, t) = 0,
t > 0,
(5.22)
u(L, t) = 0,
t > 0,
u(x, 0) = f (x),
0 < x < L.
By the result of section 4.5.2, the solution of problem (5.22) is given by
∞ Z L
2 2
X
2
nπτ
nπx
−k n π2 t
L
u(x, t) =
f (τ ) sin
dτ e
sin
.
L 0
L
L
n=1
(5.23)
145
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We now prove (see [Street; p 61])
Theorem 5.6.1 (Existence and stability result for the Dirichlet problem for the Heat Equation.)
The function u = u(x, t) given by (5.23) satisfies all the conditions of (5.22). The solution u(x, t) of (5.22)
is stable.
Proof
It is an elementary exercise to verify by differentiation (do so!) that the solution (5.23) satisfies the partial
differential equation in (5.22) as well as the boundary conditions u(0, t) = u(L, t) = 0.
We now verify that u(x, 0) = f (x) holds. Indeed we have from (5.23)
u(x, 0) =
∞ Z
X
2
n=1
L
L
0
i.e.
u(x, 0) =
nπτ
f (τ ) sin
dτ
L
∞
X
Cn sin
n=1
nπx
sin
L
nπx
L
with
2
Cn =
L
Z
L
f (τ ) sin
0
nπτ
dτ .
L
(5.24)
From what we have learnt about Fourier series, this is precisely the Fourier sine series expansion for f (x)
so that we have shown that u(x, 0) = f (x). Thus our series solution (5.23) satisfies all the conditions of
(5.22). Once we have shown that the series solution is also uniformly convergent, we will have accomplished
existence a posteriori. Since only a single set of Fourier coefficients have been found, the series solution
(5.23) is also unique.
We now show that the solution is continuous with respect to small changes in the data. For this purpose
we introduce a small error term ∆(x) into the initial data and employ a short-hand ∆-notation to denote
the terms involving the small error. With this notation (5.24) becomes
Cn∆
Z
2 L
nπτ
(f (τ ) + ∆τ ) sin
dτ
=
L 0
L
Z
Z
2 L
nπτ
2 L
nπτ
=
f (τ ) sin
dτ +
∆τ sin
dτ
L 0
L
L 0
L
Z
2 L
nπτ
∆τ sin
dτ
= Cn +
L 0
L
∴ Cn∆ = Cn + ∆Cn
so
∆
u (x, t) = u(x, t) +
∞
X
n=1
∆Cn e−k
(5.25)
(5.26)
(5.27)
n2 π
t
L2
sin
nπx
L
(5.28)
for t ≥ 0. From (5.26) and (5.27), we find that if max{∆(x)} denotes the largest absolute value of ∆(x)
on the interval, then
Z L
2
2
|∆Cn | ≤ max {∆ (x)}
dx = ∆m
(5.29)
L
L
0
146
because
Z
L
∆m = max {∆(x)}
dx
0
Z L
nπx
∆(x) sin
≥
dx
L
0
in view of
sin
nπx
≤ 1.
L
Therefore in (5.28)
u∆ (x, t) − u (x, t) ≤
∞
X
n2 π 2
nπx
2
∆m
e−k L2 t sin
L
L
(5.30)
n=1
2
by removing the constant term ∆m from each term of the series. Next, by virtue of the ratio test for
L
infinite series, the series
Sc (t) =
∞
X
e−k
n2 π 2
t
L2
n=1
is absolutely convergent for any fixed t > 0. Accordingly the Weierstrass M -test shows that the series
Sf (x, t) =
∞
X
e−k
n2 π 2
t
L2
sin
n=1
nπx
≤ Sc
L
(5.31)
is uniformly convergent. Clearly
∞
X
e−k
n2 π 2
t
L2
sin
n=1
nπx
≤ Sf
L
is always less than some finite number M because the convergent series Sc is bounded from above, whence
u∆ (x, t) − u (x, t) ≤ 2M max {∆(x)}
(5.32)
Thus the solution (5.23) is continuous with respect to small changes in the initial data, since small changes
in the data give rise to small changes in the solution - the change in u is namely of the order of the
maximum data change. We finally show that the series solution (5.23) is uniformly convergent for t > 0.
From (5.24)
2
|Cn | ≤ |f (x)|max
L
Z
0
L
dτ = 2 |f (x)|max .
Accordingly our solution (5.23) which we write as
∞
X
Cn e−k
n2 π 2
t
L2
n=1
can be written as
sin
nπx
L
∞
X
n2 π 2
nπx
u(x, t)
=
αn e−k L2 t sin
2 |f (x)|max
L
n=1
where
|αn | =
|Cn |
≤ 1.
2 |f (x)|max
147
APM3701/1
However, the absolute value of each term in this series is less than or equal to the corresponding term
of the uniformly convergent series Sf (x, t) (see (5.31)). It follows that (5.23) is a uniformly convergent
series for t > 0. We have now fully established existence and stability a posteriori.
■
148
149
CONTENTS
Chapter 6
FOURIER TRANSFORMS
6.1
MOTIVATION
In Chapter 4, the method of separation of variables leads to eigenvalue problems in finite intervals or
regions. When the intervals or regions are no longer finite, we no longer deal with a discrete set of
nπx
nπ
, Xn = sin
, n = 1, 2, 3, . . .), but with continuous set
eigenvalues and eigenfunctions (e.g. λn =
L
L
of eigenvalues λω, λ ∈ R, therefore the given boundary value problem will no longer be solved by means
of Fourier Series (see Chapter 4 and 5 of this guide). By recalling that an integral is the limit of a sum, it
is therefore natural to replace the series (sum) by an integral, and consequently the Fourier Series will be
replaced by a Fourier Integral. The Fourier integral may also be used to describe non-periodic functions.
Fourier transforms will be used to solve boundary value problems on unbounded regions, such as lines,
half lines, planes and half planes.
6.2
LEARNING OUTCOMES
The following outcomes are expected on completion of this chapter:
• Calculate Fourier transform as it is used in solving standard partial differential equations.
• Express the solution of boundary–value problems for infinite domain as convolutions involving the
boundary value problems or initial data.
• Apply Fourier cosine and sine transforms on semi–infinite regions.
6.3
PRESCRIBED READING
Chapter 7 of [1].
6.4
DEFINITION: FOURIER TRANSFORMS
Lemma 6.5 [8] If f is piecewise smooth function in the interval [0, b] , then, for b > 0,
Z
lim
x→0 0
b
f (x)
π
sin λx
dx = f (0+) .
x
2
150
Proof
Z
0
b
sin λx
f (x)
dx =
x
Z
b
sin λx
dx
x
f (0+)
0
b
f (x) − f (0+)
sin λxdx
x
0
Z b
Z λb
f (x) − f (0+)
sin t
dt +
sin λxdx.
(Let λx = t) = f (0+)
t
x
0
0
Z
+
Since f is piecewise smooth,
f (x) − f (0+)
sin λx < ∞ as λ → ∞,
x
and thus by the Riemann-Lebesgue Lemma,
Z
0
b
f (x) − f (0+)
sin λx → 0 as λ → ∞.
x
Therefore
Z
lim
λ→∞ 0
b
Z λb
sin t
sin λx
dx = f (0+)
dt
f (x)
x
t
0
π
=
f (0+) ,
2
since
Z
0
∞
sin t
π
dt = .
t
2
■
Theorem 6.5.1 (Fourier Integral Theorem [8].) If f is piecewise smooth in every finite interval, and
absolutely integrable on (−∞, ∞) , then
Z Z ∞
1
1 ∞
[f (x+) + f (x−)] =
f (t) cos α (t − x) dt dα.
(6.1)
2
π 0
−∞
Proof
First note that |cos α (t − x)| ≤ 1, and since f is absolutely integrable on (−∞, ∞) by hypothesis i.e.:
Z ∞
|f (x)| dx < ∞,
−∞
R∞
the integral −∞ f (t) cos α (t − x) dt converges independently of α and x. Therefore the order in the double
i
R λ hR ∞
integral 0 −∞ f (t) cos α (t − x) dt dα may be interchanged. We have
Z
∞
−∞
Z
0
λ
f (t) cos α (t − x) dα dt =
∞
sin λ (t − x)
dt
f (t)
t−x
−∞
Z
Z
−K
=
Z
x
+
−∞
Z
+
−K
K
Z
∞
+
x
f (t)
K
sin λ (t − x)
dt.
t−x
151
APM3701/1
If we let u = t − x, we have
K
Z
x
sin λ (t − x)
f (t)
dt =
t−x
K−x
Z
f (u + x)
0
sin λu
du,
u
and by Lemma 6.5,
K
Z
lim
λ→∞ x
and
Z
f (t)
π
sin λ (t − x)
dt = f (x+)
t−x
2
f (t)
sin λ (t − x)
π
dt = f (x−) .
t−x
2
x
lim
λ→∞ −K
If for ε > 0, we choose K sufficiently large, so that
Z
−K
f (t)
sin λ (t − x)
dt
(t − x)
<
ε
2
f (t)
sin λ (t − x)
dt
(t − x)
<
ε
.
2
−∞
Z
∞
K
and
Thus,
Z
−K
lim
λ→∞
Z
x
+
−∞
Z
+
K
Z
∞
+
−K
x
K
sin λ (t − x)
dt =
f (t)
(t − x)
=
Therefore
Z
0
∞ Z ∞
−∞
f (t) cos α (t − x) dt dα =
Z
x
lim
λ→∞
Z
K
+
−K
f (t)
x
sin λ (t − x)
dt
t−x
π
[f (x+) + f (x−)] .
2
π
[f (x) + f (x−)] .
2
■
♦ Notes:
1. If f is continuous at x, then
f (x+) = f (x−) = f (x) ,
thus (6.1) reduces to
1
f (x) =
π
Z
0
∞ Z ∞
−∞
f (t) cos α (t − x) dt dα.
Equation (6.2) is known as the Fourier integral of f.
2. Recalling that
cos α (t − x) =
i
1 h iα(t−x)
e
+ e−iα(t−x) ,
2
the Fourier integral (6.2) may be expressed in the complex form:
Z ∞Z ∞
Z ∞Z ∞
1
1
f (t) eiα(t−x) dtdα +
f (t) e−iα(t−x) dtdα.
f (x) =
2π 0
2π
−∞
0
−∞
(6.2)
152
If we change the integration variable from α to −α in the second integral, the above equation
becomes:
Z ∞
Z ∞
1
1
√
f (t) eiαt dt e−iαx dα,
(6.3)
f (x) = √
2π −∞
2π −∞
or
Z
∞
f (x) =
−∞
1
2π
Z
∞
f (t) e
iαt
dt e−iαx dα.
(6.4)
−∞
3. The constant √12π in (6.3) is a matter of choice, not of necessity. Many authors will attach the entire
1
2π factor in equation (6.3) to one equation like in (6.7) and (6.8) below. Whatever constant it is used
in one formula, the constant in the other formula is chosen such that the product of both constants
1
.
is 2π
Theorem 6.5.2 [8]
Let f (x) be continuous, piecewise smooth, and absolutely integrable. If
Z ∞
1
F (α) = √
f (t) e−iαt dt
2π −∞
then, for all x
1
f (x) = √
2π
(6.5)
∞
Z
F (α) eiαx dα.
(6.6)
f (t) eiαx dt
(6.7)
−∞
Alternatively, from the note 3 above, if
1
F (α) =
2π
then, for all x
Z
Z
∞
−∞
∞
F (α) eiαt dα.
f (x) =
(6.8)
−∞
6.6
DEFINITION: FOURIER TRANSFORMS - INVERSION THEOREM
Let f (x) be continuous, piecewise smooth, and absolutely integrable. The function
Z ∞
1
F (α) = √
f (t) eiαt dt
2π −∞
is called the Fourier Transform of f (x) , and the function
Z ∞
1
√
f (x) =
F (α) e−iαx dα
2π −∞
is called the inverse Fourier transform of F (α) , or alternately
Z ∞
1
F (α) =
f (t) e−iαt dt
2π −∞
is called the Fourier Transform of f (x) , and the function
Z ∞
f (x) =
F (α) eiαx dα
−∞
is called the inverse Fourier transform of F (α) .
(6.9)
(6.10)
(6.11)
(6.12)
153
6.6.1
APM3701/1
Notations
The Fourier transform of f is denoted by F (f ) , fˆ or simply by F.
Remark 6.6.1
1. Note that, in both formula (6.9) and (6.10) or [(6.11) and (6.12)] the product of the constants must
1
.
be 2π
2. Very often, you will find the term eiαt in (6.9) [or in (6.11)] replaced by e−iαt , and as a consequence,
e−iαx replaced by eiαx in (6.10) [or in (6.12)] (see example 6.6.2).
Example 6.6.1
(a) Find the Fourier transform of the function f (x) = e−|x| .
(b) Then show that
∞
cos αx
π
dα = e−|x| .
2
1+α
2
Z
∞
Z
0
(c) And prove in particular that
0
Solution
(a) Using the fact that e−|x| =
π
cos x
dx = .
2
1+x
2e
(
e−x if x > 0
,
ex
if x < 0
Z ∞
1
= √
F e−|x|
e−|t| eiαt dt
2π −∞
Z 0
Z ∞
1
= √
et(1+αi) dt +
e−t(1−α) dt
2π −∞
0
1
1
1
= √
+
2π 1 + iα 1 − αi
r
=
2 1
.
π 1 + α2
(b) Taking the inverse Fourier Transform of F (α) we have:
!
Z ∞ r
1
2 1
−|x|
f (x) = e
=√
e−iαx dα
π 1 + α2
2π −∞
Z
1 ∞ cos αx − i sin αx
=
dα since e−iαx = cos αx − i sin αx
2
π −∞
1+α
Z
1 ∞ cos αx
sin αx
=
dα
since
is
odd
π −∞ 1 + α2
1 + α2
Z
2 ∞ cos αx
cos αx
=
dα (since
is even).
2
π 0 1+α
1 + α2
Thus
π −|x|
e
=
2
Z
0
∞
cos αx
dα.
1 + α2
(6.13)
154
(c) In (6.13) we let x = 1, we obtain
π −1
e =
2
Substituting α by x we obtain:
Z
∞
0
Z
0
∞
cos α
dα.
1 + α2
cos x
π
dx = .
1 + x2
2e
■
♦ Notes: Show that you get to the same outcome by using the formulae
Z ∞
Z ∞
1
iαt
F (α) =
f (t) e dt
and
f (x) =
F (α) e−iαx dα.
2π
−∞
−∞
Example 6.6.2
(a) Determine the Fourier transform of
(
f (x) =
sin x, 0 < x < π
0,
otherwise.
(b) Show that
1
π
Z
∞
0
cos αx + cos α (π − x)
dα =
1 − α2
(
sin x, 0 < x < π
0,
otherwise.
Solution
(a) Note that
eix − e−ix
, and e±πi = cos (±π) + i sin (±π) = −1.
2i
Z ∞
Z π
1
1
−iαx
√
f (x) e
dx =
sin x e−iαx dx
2π 0
2π −∞
Z π
Z π
1
1
√
eix − e−ix e−iαx dx = √
eix(1−α) − e−ix(1+α) dx
2i 2π 0
2π2i 0
"
#π
1
eix(1−α)
e−ix(1+α)
√
+
i (1 + α)
2i 2π i (1 − α)
0
"
#
1
2
eπi(1−α) e−πi(1+α)
√
+
−
1−α
1+α
1 − α2
2π2i2
sin x =
F (α) =
=
=
=
1
(1 + α) e−πiα + (1 − α) e−πiα + 2
= − √
−
1 − α2
2 2π
1
−2e−πiα − 2
√
= −
1 − α2
2 2π
=
1 1 + e−πiα
√
.
2π 1 − α2
155
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(b) We take the inverse Fourier Transform,
Z ∞
1
1 + e−πiw iwx
e dw
f (x) =
2π −∞ 1 − w2
Z ∞
1
(1 + cos πw − i sin πw) (cos wx + i sin wx)
=
dw
2π −∞
1 − w2
Z ∞
cos wx + cos πw cos wx + sin πx sin wx
1
dw
=
2π −∞
1 − w2
Z
1 ∞ cos wx + cos w (π − x)
=
π 0
1 − w2
(
sin x, 0 < x < π
=
0,
otherwise.
■
Exercises 6.6
1. (a) Determine the Fourier transform of the function
(
1 |x| ≤ a
f (x) =
0 |x| > 0.
(b) Show that
Z
∞
0



sin αa cos αx
dα =

α

(c) Show in particular that
Z
0
∞
π
2
π
4
0
|x| < a
x=a .
|x| > a.
sin α
π
dα = .
α
2
2. Prove that
F −1 (F (f )) (t) = 2πf (−t) .
3. Exercises 7.1 of [1], all problems.
4. Exercises 7.2 of [1], problems 1–16, 21–46.
5. Find the Fourier transform of the given functions
(
(a) f (x) =
1
0


 −L
(b) f (x) =
1


0
if − L < x < L,
otherwise.
if − L < x < 0,
if 0 < x < 1,
otherwise.
(
1 − cos x
0
(
1 − |x|
0
(c) f (x) =
(d) f (x) =
(e) f (x) = e−|x| .
if − π2 < x < π2 ,
otherwise.
if − 1 < x < 1,
otherwise.
156
(
(f ) f (x) =
1 − x2
0


−2 − x



 x
(g) f (x) =

2−x



 0
if − 1 < x < 1,
otherwise.
if − 2 < x < −1,
if −1 < x < 1,
if 1 < x < 2,
otherwise.
6. If
(
f (x) =
1
0
if |x| ≤ 1,
otherwise.
Show that the Fourier transform of f is given by
2
f (x) =
π
Z
0
∞


 1
sin α cos αx
dα =

α

1
2
0
if |x| < 1,
if |x| = 1,
if |x| > 1.
7. Use Exercise 6.6 (6.) to show that
∞
Z
0
sin α cos α
π
dα = .
α
4
8. Use integration by parts and Exercise 6.6 (7.) to show that
Z
∞
0
sin2 α
π
dα = .
2
α
2
9. Use the identity sin2 α + cos2 α = 1 and Exercise 6.6 (8.) to show that
Z
0
∞
π
sin4 α
dα = .
2
α
4
10. Use Exercise 6.6 (6.) to show that
(a)
R∞
0
sin α cos 2α
dα
α
=0
(b)
R∞
0
1−cos α
dα
α2
= π2 .
Hint: sin2 α = sin4 α + cos2 α sin2 α = sin4 α + 14 sin2 2α
6.7
6.7.1
FOURIER SINE AND COSINE TRANSFORMS
Fourier sine transforms
Let f (x) be defined for 0 ≤ x < ∞, and let f (x) be extended as an odd function in (−∞, ∞) satisfying
the condition of Theorem 6.5.1. If, at the point of continuity
r Z ∞
2
Fs (α) =
f (t) sin αtdt
π 0
then
r Z ∞
2
f (x) =
Fs (α) sin αxdx.
π 0
157
APM3701/1
The function Fs (α) is called the Fourier sine transform of f (x) , and f (x) is the inverse Fourier sine
transform of Fs (α) .
♦ Remark: q
The constant π2 in the above formulae is a matter of choice, not of necessity. Many authors will attach
the entire π2 factor to one of either equations. Whatever constant it is used in one formula, the constant
in the other formula is chosen such that the product of both constants is π2 .
Example 6.7.1
(a) Determine the Fourier sine transform of


 −1 if −1 < x < 0
f (x) =
1 if 0 < x < 1


0 if
|x| > 1.
(b) Show that
∞
Z
0
Solution
(a) We have



(1 − cos α) sin αx
dα =

α

π
2
π
4
0
if
0<x<1
if
x=1
if |x| > 1, x = 0.
r Z ∞
2
f (x) sin αxdx
Fs (α) =
π 0
r Z 1
r
2
1
2
=
sin αxdx = −
cos αx
π 0
α π
=
1
α
r
x=1
x=0
2
(1 − cos α) .
π
(b) By taking the inverse Fourier since transform:
r Z ∞r
Z
2
21
2 ∞1
f (x) =
(1 − cos α) sin αxdα =
(1 − cos α) sin αxdα.
π 0
πα
π 0 α
For 0 < x < 1, f (x) = 1, (points of continuity) thus
Z
Z ∞
(1 − cos α) sin αx
2 ∞ (1 − cos α) sin αx
π
1=
dα ⇒
dα = .
π 0
α
α
2
0
1
For x = 1 (point of discontinuity): mean value = 1+0
2 = 2 , thus
Z
Z ∞
1
2 ∞ (1 − cos α) sin αx
(1 − cos α) sin αxdα
π
=
⇒
= .
2
π 0
α
α
4
0
For |x| > 1, f (x) = 0. And for x = 0 (point of discontinuity): mean value = 1−1
2 = 0. Thus
Z ∞
Z ∞
2
(1 − cos α) sin αx
(1 − cos α) sin αx
0=
⇒
dα = 0.
π 0
α
α
0
Finally, we have
Z
0
∞



(1 − cos α) sin αx
dα =

α

π
2
π
4
0
if
0<x<1
if
x=1
if |x| > 1, x = 0.
158
■
6.7.2
Fourier cosine transform
Let f (x) be defined for 0 ≤ x < ∞. Let f (x) be extended as an even function in (−∞, ∞) satisfying the
conditions of Theorem 6.5.1. If that the points of continuity
r Z ∞
2
f (t) cos αtdt
Fc (α) =
π 0
then
r Z ∞
2
f (x) =
Fc (α) cos αxdα.
π 0
♦ Remark: q
The constant π2 in the above formulae is a matter of choice, not of necessity. Many authors will attach
the entire π2 factor to one of either equations. Whatever constant it is used in one formula, the constant
in the other formula is chosen such that the product of both constants is π2 .
Example 6.7.2
(a) Determine the Fourier cosine transform of the function
(
1 − x2 |x| ≤ 1
f (x) =
0
|x| > 0.
(b) Use the result in part (a) to evaluate
Z ∞
0
x
x cos x − sin x
cos dx.
x3
2
Solution
(a) We have
Fc (α) =
=
=
=
r Z ∞
r Z 1
2
2
f (x) cos αxdx =
1 − x2 cos αxdx
π 0
π 0
#
r "
1
Z 1
2
sin
αx
x
sin
αx
1 − x2
+2
dx
π
α
α
0
0
#
r "
Z 1
2 −x cos αx 1
cos αx
2
+
dx
π
α2
α2
0
0
#
r "
2 − cos α
sin αx 1
2
+
π
α2
α3 0
r 2 α cos α − sin α
.
= −2
π
α3
(b) By taking the inverse Fourier cosine transform:
Z 4 ∞ α cos α − sin α
f (x) = −
cos αxdα.
π 0
α3
159
APM3701/1
We take x = 12 , we have
Z 1
3
α
4 ∞ α cos α − sin α
f
= =−
cos dα.
3
2
4
π 0
α
2
Thus replacing α by x, we have
Z
∞
0
x
x cos x − sin x
3π
cos dx = − .
3
x
2
16
■
Exercises 6.7
1. Determine the Fourier cosine transform of the function f (x) = e−|x| .
Use your result to show that
Z
0
∞
cos αx
π
dα = e−|x| .
2
α +1
2
2. Find the Fourier sine and cosine transforms of f (x) = e−ax (a > 0) .
3. (a) Let f (x) be defined for 0 ≤ x < ∞. Let f (x) be extended as an odd function in (−∞, ∞)
satisfying the conditions of Theorem 1. If at the points of continuity,
r Z ∞
2
f (x) sin αxdx,
Fs (α) =
π 0
then show that
r Z ∞
2
Fs (α) sin αxdα.
f (x) =
π 0
[Hint: Use equation (6.2).]
(b) Let f (x) be defined for 0 ≤ x < ∞. Let f (x) be extended as an even function in (−∞, ∞)
satisfying the conditions of Theorem 1. If at the points of continuity
r Z ∞
2
f (x) cos αxdx,
Fc (α) =
π 0
then show that
r Z ∞
2
f (x) =
Fc (α) cos αxdα.
π 0
[Hint: Use equation (6.2).]
4. Find the Fourier cosine transform of



x
0<x<1
f (x) =
2−x 1<x<2


0
x > 2.
[Ans:
r Z ∞
2
2 cos α − cos 2α − 1
cos αxdx .]
π 0
α2
5. Exercises 7.6 of [1], All problems.
■
160
6.8
6.8.1
PROPERTIES OF THE FOURIER TRANSFORM
Linearity and shifting properties
Theorem 6.8.1 Linearity and shifting properties
1. The Fourier transform is a linear transformation, i.e.
F [af + bg] = aF [f ] + bF [g] .
2. If F [f ] in a Fourier transform. Then
F [f (x − c)] = eiαc F [f (x)] ,
or equivalently
F [f (x − c)] = e−iαc F [f (x)] (see Remark 6.6.1 in Section 6.6).
where c is a real number (constant).
Proof
1. Linearity of the Fourier transform is trivial.
Z ∞
1
2. We have
F [f (x − c)] =
f (x − c) eiαx dx
2π −∞
Z ∞
1
= √
f (k) eiα(k+c) dk [k = x − c]
2π −∞
Z
Z
1 iαc ∞
1 iαc ∞
iαk
= √ e
f (k) e dk = √ e
f (x) eiαx dx
2π
2π
−∞
−∞
= eiαc F [f (x)] .
■
6.8.2
Scaling property
Theorem 6.8.2 Scaling property
If F [f ] is the Fourier transform of f, then
1 α
F [f (cx)] = F
|c|
c
where c ̸= 0 is a real constant, and F (α) is the Fourier transform of f (x) .
Proof
If c ̸= 0, we have
Z ∞
1
F [f (cx)] = √
f (cx) eiαx dx
2π −∞
Z ∞
α
1 1
√
=
f (k) ei c k dk [k = cx]
|c| 2π −∞
=
1 α
F
.
|c|
c
■
161
6.8.3
APM3701/1
Differentiation of Fourier Transform
Theorem 6.8.3 Differentiation of Fourier Transform
Let f be continuous and piecewise smooth in (−∞, ∞) and lim f (x) = 0. If f and f ′ are absolutely
|x|→∞
integrable, then
F f ′ = −iαF [f ] .
Proof
′
F f (x)
Z ∞
1
f ′ (x) eiαx dx
= √
2π −∞
Z ∞
∞
1
iαx
√ f (x) e
=
f (x) eiαx dx
− iα
2π
−∞
−∞
= −iαF [f (x)]
since lim f (x) = 0
[integration by parts]
|x|→∞
■
Remark 6.8.1
• This property is suitable to solve boundary-value problems when one variable, say x, varies from −∞
to ∞ and lim u (x, y) = 0. (See example 6.11.1 below.)
|x|→∞
• In general, if f and its first (n − 1) derivatives are continuous, and if f (n) is piecewise continuous,
f (k) is absolutely integrable (k = 0, 1, . . . , n) and
lim f (k) = 0, for k = 0, 1, . . . , n − 1
|x|→∞
then
h
i
F f (n) (x) = (−iα)n F [f (x)] , n = 0, 1, 2, . . . .
6.8.4
Differentiation of Fourier cosine transform
Theorem 6.8.4 Differentiation of Fourier cosine transform
If f and f ′ are continuous and lim f (x) = 0 = lim f ′ (x) , then
x→∞
Fc
x→∞
r
′′
2 ′
d2 f
2
(x) Fc f (x) = −α Fc [f (x)] −
f (0) .
dx2
π
162
Fc f ′′ (x) =
r Z ∞
2
f ′′ (x) cos αxdx
π 0
r 2
=
f ′ (x) cos αx
π
∞
0
Z
∞
+α
f (x) sin αxdx
′
r 2
=
−f ′ (0) + α (f (x) sin αx)
π
∞
0
−α
r r
2
π
′
2
=
−f (0) − α
Fc [f (x)]
π
2
r
2
= −α Fc [f (x)] −
[integration by parts]
0
2
Z
∞
f (x) cos αxdx
[2nd integration by parts]
0
(since f (x) → 0 as x → ∞)
2 ′
f (0) .
π
Remark 6.8.2
This property is suitable for use in boundary value problems where one variable, say x, varies from 0 to
∞, and lim u (x, t) = 0, and ux (0, t) is known. (See example 3.7.2 below.)
x→∞
6.8.5
Differentiation of the Fourier sine transform
Theorem 6.8.5 Differentiation of the Fourier sine transform
If f and f ′ are continuous and
lim f (x) = 0 = lim f ′ (x) ,
x→∞
x→∞
then
Fs
r
′′
2
d2 f
(x) = Fs f (x) =
αf (0) − α2 Fs [f (x)] .
dx2
π
Proof
Similar to Theorem 6.8.4.
■
Remark 6.8.3
Fourier sine transform are useful when solving boundary-value problems in u (x, t) in case where one
variable, say x, varies from 0 to ∞, lim u (x, t) = 0 and u (0, t) is known. (See example 3.7.3.)
x→∞
Exercises 6.8
1. Exercises 7.2 of [1], problems 17–20.
6.9
CONVOLUTION THEOREM
The function
1
(f ∗ g) (x) = √
2π
Z
∞
−∞
f (x − α) g (α) dα
is called the convolution of the functions f and g over the interval (−∞, ∞) .
163
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Theorem 6.9.1
If the Fourier transform of f and g exist, then the Fourier transform of f ∗ g exists, and
F [f ∗ g] (α) = F (α) · G (α)
(6.14)
where F and G are the Fourier transforms of f and g respectively.
Equivalently
F −1 (F · G) (x) = f ∗ g.
Proof
(See similar proof in [1], theorem 4 section 7.2)
We prove (6.14):
Z ∞
Z ∞
1
−iαx
−iαy
f (x) e
dx
g (y) e
dy
F (α) · G (α) = −
2π −∞
−∞
Z ∞Z ∞
1
f (x) g (y) e−iα(x+y) dxdy
= −
2π −∞ −∞
(6.15)
We introduce new variables in the double integral
u = x + y and v = y.
From calculus, we know that
dxdy =
∂x
∂u
∂ (x, y)
=
∂ (u, v)
∂x
∂v
1 −1
0
1
=
∂y
∂u
∂y
∂v
= 1.
Thus (6.15) becomes
F (α) · G (α) =
=
1
2π
Z
1
√
2π
∞
Z
−∞
Z
∞
−∞
∞
e
f (u − v) g (v) e−iαv du dv
−iαu
−∞
1
√
2π
Z
∞
−∞
f (u − v) g (v) dv du
= F (f ∗ g) (α) .
■
6.9.1
Properties of convolutions
1. f ∗ g = g ∗ f
2. f ∗ (g ∗ h) = (f ∗ g) ∗ h
(3) f ∗ (g + h) = f ∗ g + f ∗ h
Exercises 6.9
1. Exercises 7.2 of [1], problems 55–60.
164
6.10
TABLES OF FOURIER TRANSFORMS
See also Appendix B pp A66–A72 of [1], Examples 3, 4, 5, 6 of section 7.2 of [1].
6.10.1
Table of Fourier Transforms
Function
1
f (x) = √
2π
Fourier Transform
Z
∞
F
−∞
(α) eixα dα
1
F (x) = √
2π
Z
∞
f (x) e−ixα dx
−∞
1.
f (x)
F (α)
2.
f (ax − b)
iαb
αb
α
e− a
ei a α F
or
F
|a|
a
|a|
a
3.
f ∗ g (convolution)
F (α) · G (α)
r
2
a
π a2 + α2
4.
e−a|x|
r
5.
1
2
x + a2
π e−aα
2 a
r
6.
b−1 sin aπ
b
πa
+
cos
cosh πx
b
b
2 sinh aα
π sinh bα
7.
e−b
8.
f (n) (x)
in αn F (α) or (−i)n αn F (α)
9.
xn f (x)
in
10.
f
a>0
2 x2
(a) eibx
(see Theorem 5, section 7.2
of [1])
α2
1
√ e− 4b2
2b
n
dn F
n d F
or
(−i)
dαn
dαn
1
F
|a|
α−b
a
165
6.10.2
APM3701/1
Table of Fourier Cosine Transforms
Function
Fourier cosine Transform
r Z ∞
2
Fc (α) cos αxdα
f (x) =
π 0
r Z ∞
2
Fc (α) =
f (x) cos αxdx
π 0
0<x<∞
1.
f (x)
Fc (α)
r
2.
1
2
x + a2
3.
e−ax
4.
e−ax
5.
x
− 21
sin ax2
1
π
6.
1
π
7.
cos ax2
8.
d2 f (x)
dx2
−α2 F
(x > 0)
2
(a > 0)
2
π
2 e−aα
π |a|
r
2
a
2
π (a + α2 )
α2
1
√ e− 4a
π a
r
1
α
α2
α2
− sin
4a
4a
α2
α2
cos
+ sin
4a
4a
cos
r
c (α)
−
2 ′
f (0) .
π
0<α<∞
166
6.10.3
Table of Fourier sine transforms
Function
Fourier sine Transform
r Z ∞
2
Fs (α) sin αxdα
f (x) =
π 0
r Z ∞
2
Fs (α) =
f (x) sin αxdx
π 0
1.
f (x)
2.
x−1
3.
x
2
x + a2
4.
0<α<∞
Fs (α)
e−ax (x > 0)
xe
6.
x− 2
7.
tan−1
8.
d2 f (x)
dx2
r
2
π
r
2 −aα
e
π
2
π
r
√
α
2
π (α2 + a2 )
α2
2 αe− 4a
2π a 32
−ax2
5.
6.11
0<x<∞
1
√
α
1
x
r
2 e−aα
π α
r
2
αf (0) − α2 Fs (α) .
π
a
SOLUTION OF BOUNDARY-VALUE PROBLEMS BY MEANS OF FOURIER
TRANSFORMS
Based on section 7.3 of [1].
One can solve a boundary-value problem by using Fourier transform, by following these three steps:
1. Choose a suitable Fourier transform to reduce the partial differential equation to an ordinary differential equation. The choice of a suitable Fourier transform depends on the boundary conditions and
on the interval on which the solution must be found (see Table 6.11.1).
2. Solve the ordinary differential equation, and use the boundary conditions to determine the values of
arbitrary constants.
3. Take the inverse Fourier transform to obtain the solution of the given boundary-value problem.
The following table describes the suitable Fourier transform for given boundary conditions, and interval.
167
6.11.1
APM3701/1
Table of suitable Fourier transform for given boundary conditions and interval
Interval
Boundary conditions
−∞ < x < ∞
|x|→∞
lim u (x, y) = 0
Suitable Fourier
transform
Transformation
Fourier transform
uxx (x, y) → −α2 U (α, y)
u (x, 0) = f (x) → U (α, 0) = F (α)
u (x, 0) = f (x)
r
0<x<∞
lim u (x, y) = 0
x→∞
ux (0, y) = f (y)
0<x<∞
uxx (x, y) →
Cosine Fourier
−α2 U
c (α, y)
−
2
f (y)
π
transform
lim u (x, y) = 0
x→∞
u (0, y) = f (y)
Example 6.11.1
Solve the following heat equation:
Sine Fourier
uxx (x, y) → −α2 Us (α, y) + α
transform
u (0, y) = f (y) → Us (0, y)
∂u
∂t
=
q
2
πf
(y)
∂2u
, t > 0, |x| < ∞
∂x2
lim u (x, t) = 0, t > 0
|x|→∞
2
u (x, 0) = e−x .
Solution
Note that there is a boundary condition which is not given explicitly: u (x, y) < ∞ as y → ∞. The
boundary conditions and the interval suggest that we should use a Fourier transform; so we take the
Fourier transform with respect to x, since
Z ∞
Z ∞
∂
−iαx
ut (x, t) e
dx =
u (x, t) e−ixα dx,
∂t
−∞
−∞
and
F (uxx ) = −α2 U (α, t) , where U is the Fourier transform of u with respect to x.
The partial differential equation becomes
∂U
= −α2 U.
∂t
(6.16)
Equation (6.16) has the solution
2
U (α, t) = c (α) e−α t .
The Fourier transform of the initial condition:
u (x, 0) = e−x
168
transforms into
α2
1
2
U (α, 0) = F e−x = √ e− 4
2
so that
(see the Table 6.10.1(7) with b = 1)
α2
1
c (α) = √ e− 4 .
2
Thus
1
2
1
U (α, t) = √ e−α (t+ 4 ) .
2
By taking the inverse Fourier transform, and using
2 2
α2
1
F e−a x = √ e− 4a2
2a
we obtain
(see the Table 6.10.1(7)) ,
1
with a = √
.
4t + 1
x2
1
e− 4t+1
u (x, t) = √
4t + 1
■
Example 6.11.2 Solve the Laplace equation:
∂2u ∂2u
+ 2
∂x2
∂y
= 0, x > 0, y > 0
lim u (x, y) = 0,
x→∞
lim u (x, y) = 0,
y→∞
ux (0, y) = 0
u (x, 0) = e−x .
Solution
The type of Boundary condition and interval suggest that we can either use Fourier cosine transform with
respect to x or Fourier sine transform with respect to y.
Let us use the Fourier cosine.
We take the Fourier cosine transform with respect to x, we have:
∂ 2 Uc
− α2 Uc = 0,
∂y 2
(since Ux (0, y) = 0)
which yields the solution Uc = A(α)e−αy + B(α)eαy .
The boundary condition lim u = 0 implies that B = 0, thus
y→∞
Uc = A (α) e−αy .
169
APM3701/1
The boundary condition u (x, 0) = e−x , is transformed into
r Z ∞
2
Uc (α, 0) =
e−x cos αxdx
π 0
=
2
π
r
1
2
(see table 6.10.2(3)) .
π (1 + α2 )
Consequently
r
r
2 2 1
2 2 e−αy
A=
and Uc =
.
π π 1 + α2
π π 1 + α2
Taking the inverse Fourier cosine transforms yields:
Z
2 ∞ e−αy
u (x, y) =
cos αxdα.
π 0 1 + α2
■
Example 6.11.3 Solve
ut = c2 uxx ,
1
u (x, 0) =
x
u (0, t) = 0
0 < x < ∞,
t>0
lim u (x, t) = 0.
x→∞
Solution
We take the Fourier sine transforms with respect to x, we have:
∂Us
= −α2 c2 Us (α, t) ,
∂t
which yields the solution
2 2
Us (α, t) = A (α) e−c α t .
r
1
2
The boundary condition u (x, 0) = transforms into
x
π
r
2
A (α) =
(See Table 6.10.3(2)).
π
Thus
r
2 −c2 α2 t
e
.
π
By taking the inverse Fourier sine transforms, we obtain:
r Z ∞r
Z
2 ∞ −c2 α2 t
2
2 −c2 α2 t
u (x, t) =
e
sin αxdα =
e
sin αxdα.
π 0
π
π 0
Us (α, t) =
■
170
Exercises 6.11
1. Determine the movement of a semi-infinite rod x ≥ 0, when the displacement at the endpoint x = 0
is given and the rod is freely hinged. Initially, before the hinge is brought into motion, the rod is at
rest in its equilibrium position (i.e. along the x-axis), by solving the following problem:
4
∂u
2∂ u
+
α
∂t2
∂x4
= 0,
x>0
u (0, t) = f (t)
t > a,
t > 0,
uxx (0, t) = 0
t > 0,
u (x, 0) = 0
x>0
ut (x, 0) = 0
x>0
lim u = 0.
x→∞
[uxx = 0 means that the bending moment (curvature) at the endpoint is zero, i.e. the beam is hinged.]
[Hint: use the fact that
r
Fs [uxxxx ] = −
2 3
α u (0, t) + α4 Us (α, t)
π
provided that
uxx (0, t) = 0. ]
2. Determine the solution u (x, y) of the Laplace’s Equation in 0 < y < α, −∞ < x < ∞ subject to:
u (x, 0) = 0
u (x, a) = f (x)
lim u = 0.
|x|→∞
3. Solve the Laplace equation in x > 0 and y > 0, subject to the boundary conditions:
lim u =
x→∞
lim u = 0,
y→∞
u (0, y) = 0,
(
uy = (x, 0) =
4. Exercises 7.3 of [1], all problems.
1 0≤y≤1
0 otherwise.
171
CONTENTS
Chapter 7
BESSEL SERIES.
BOUNDARY–VALUE PROBLEMS IN
OTHER COORDINATE SYSTEMS
7.1
BACKGROUND
The aim of this final chapter is to solve boundary–value problems in other coordinate systems. More
precisely, we will solve problems in polar, spherical and cylindrical coordinate systems by using once more
the method of separation of variables.
This chapter presents the Laplace equation in polar, spherical and cylindrical coordinates, the vibrating
circular membrane, and the steady–state temperature in a circular plate.
This chapter requires the knowledge of Bessel functions and series. The study of these functions and their
properties is in itself interesting, and any third year student in Sciences or Engineering should be familiar
with them. For this reason, they are treated briefly, but sufficiently for their application to problems
solved in this module. Bessel functions and series are also very clearly presented in sections 4.7 and 4.8
of [1].
7.2
LEARNING OUTCOMES
The outcomes of this chapter are the following:
• Solve the Laplace equation in polar, cylindrical and spherical coordinates by separation of variables.
• Solve the vibrating membrane using Bessel series.
• Solve the steady temperature distribution in a circular plate.
7.3
ASSUMED KNOWLEDGE
The knowledge of series solutions of linear differential equations is required to understand Bessel functions.
Review of materials in APM2611 (Ordinary Differential Equations) on this topic is essential. Other special
functions which are encountered in the study of Bessel functions, are the Gamma function, the Beta
function and the Error function. They are treated briefly in Appendix A.
172
7.4
PRESCRIBED READING
I strongly recommend that you read Chapter 4 and 5, pp 237–261 of [1] and section 6.1–6.3, 11.5 of the
prescribed book of APM2611, Differential equations with Boundary–Value problems 5th Edition, by Denis
G. Zill and Michael R. Cullen.
7.5
BESSEL SERIES EXPANSIONS
In this chapter we will solve boundary–value problems over circular, cylindrical and spherical domains.
The solutions of these problems will entail new series expansions; such that the Bessel series. These series
look quite complicated at first, but with time, you will be able to see that they behave very much like
Fourier series.
7.5.1
Bessel’s Equation and Bessel Functions
7.5.1.1
Definition
The equation
x2 y ′′ + xy ′ + x2 − p2 y = 0,
x>0
(7.1)
is called Bessel’s Equation of order p ≥ 0. This equation arises when solving partial differential
equations involving the Laplacian in polar and cylindrical coordinates.
7.5.1.2
Solution of Bessel’s Equation
It is easy to show that x = 0 is a regular singular point of the Bessel’s equation (7.1), i.e. when, equation
(7.1) is rewritten in the standard form
y ′′ + P (x) y ′ + Q (x) y = 0,
(7.2)
the function f (x) = xQ (x) and q (x) = x2 Q (x) are both analytic at x = 0.
Since x = 0 is a regular singular point of Bessel’s equation, we know that there exists at least one solution
of the form (see section 12.5 of [1] on analytic functions)
y=
∞
X
cn xn+r .
(7.3)
n=0
Substitution of (7.3) into (7.1) gives:
2 ′′
′
2
x y +xy + x − p
2
y=
∞
X
n=0
cn (n + r) (n + r − 1) x
n+r
+
∞
X
n=0
cn (n + r) x
n+r
+
∞
X
n=0
cn x
n+r+2
−p
2
∞
X
cn xn+r .
n=0
After grouping the terms corresponding to n = 0 and n = 1 separating and simplifying, we obtain:
2
c0 r − p
2
∞ h
i
h
i
X
2
2
r+1
+
cn (r + n)2 − p2 + cn−2 xr+n = 0.
x + c1 (r + 1) − p x
r
n=2
173
APM3701/1
Equating coefficients of the series to zero gives:
c0 r2 − p2
= 0
(7.4)
c1 (r + 1)2 − p2 = 0
h
i
cn (r + n)2 − p2 + cn−2 = 0,
(7.5)
h
i
n≥0
(7.6)
As it will become clear, c0 must be ̸= 0 (see equation (7.11), we get:
r2 − p2 = (r − p) (r + p) = 0
(7.7)
with roots r = p and r = −p.
7.5.1.3
Bessel function of the first kind
We let r = p in (7.6), we obtain the recurrence relation
cn =
−1
cn−2 .
n (n + 2p)
n≥2
(7.8)
The relation is a two-step recurrence, the even and odd-indexed terms are determined separately. Let us
first determine the odd-indexed terms.
Substitute r = p in (7.5) we get
i
h
(7.9)
c1 (p + 1)2 − p2 = 0.
Since p ≥ 0 by assumption, (p + 1)2 − p2 = 2p + 1 > 0, thus the above equation implies that c1 = 0. Thus,
substituting in (7.8), we obtain c3 = c5 = c7 = · · · = 0.
Now, the even-indexed terms are determined by rewriting the recurrence relation by letting n = 2k :
c2k =
−1
c
,
+ p) 2(k−1)
22 k (k
k ≥ 1.
(7.10)
Or
−1
c0,
+ p)
−1
1
=
c2 = 4
c0
2
2 2 (2 + p)
2 2! (1 + p) (2 + p)
−1
−1
c4 = 6
c0
=
2
2 3 (3 + p)
2 3! (1 + p) (2 + p) (3 + p)
..
.
(−1)k
=
c0
22k k! (1 + p) (2 + p) · · · (k + p)
c2 =
c4
c6
c2k
22 (1
(7.11)
Substituting these coefficients into (7.3) gives one solution to Bessel’s equation
y = c0
∞
X
k=0
(−1)k
x2k+p .
22k k! (1 + p) (2 + p) · · · (k + p)
Equation (7.11) shows that c0 must ̸= 0, otherwise y ≡ 0.
(7.12)
174
We use the Gamma function (see Appendix A), we set
c0 =
1
,
2p Γ (p + 1)
(7.13)
we reduce the terms in the series by using the basic property of the Gamma function
Γ (p + 1) = pΓ (p)
Γ (1 + p + 1) = (1 + p) Γ (1 + p)
Γ (1 + p + 2) = (2 + p) Γ (2 + p) = (2 + p) (1 + p) Γ (1 + p)
and
Γ (1 + p) [(1 + p) (2 + p) · · · (k + p)] = Γ (2 + p) [(2 + p) · · · (k + p)]
= Γ (3 + p) [(3 + p) · · · (k + p)]
= · · · = Γ (k + p) (k + p)
= Γ (k + p + 1) .
(7.14)
We substitute the simplification (7.14) into (7.12), recalling (7.13), we obtain the first solution of the
Bessel’s equation, denoted by Jp and called ”Bessel function of the first kind of order p”,
Jp (x) =
∞
X
k=0
x 2k+p
(−1)k
.
k!Γ (k + p + 1) 2
(7.15)
When p is an integer n (p = n) , we have (see Section A.1.3 in Appendix A of this guide and pp 242-244
of [1])
Γ (k + p + 1) = (k + p)!,
thus the Bessel function of order p (n integer) is
Jp (x) =
∞
X
k=0
(−1)k x 2k+p
.
k! (k + p)! 2
(7.16)
The graphs of Jp (x) for p = 0, 1/2, 1, 2, 3 are shown in Figure 7.1.
Note that Jp (p ≥ 0) converges at least on the interval [0, ∞] , and is bounded at 0.
We now consider the second root of (7.7), n = −p. We let r = −p in (7.3), we obtain in exactly the same
fashion
∞
x 2k−p
X
(−1)k
J−p (x) =
.
(7.17)
k!Γ (k − p + 1) 2
k=0
J−p is not bounded at 0. When p is an integer J−p (x) = (−1)p Jp (x) p integer ≥ 0. So that Jp and J−p
are linearly dependent. However when n is not an integer Jp and J−p are linearly independent (see Denis
G. Zill and Michael R. Cullen, section 6.2 and 6.3), thus the general solution of (7.1) on (0, ∞) can be
written as linear combination of Jp and J−p i.e
y (x) = a1 Jp (x) + a2 J−p (x)
p ̸= integer.
(7.18)
175
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Figure 7.1: Bessel Functions of first kind
Example 7.5.1 Show that the general solution of the equation
1
x y + xy + x −
4
2 ′′
′
2
y=0
is
r
y (x) = A1
2
sin x + A2
πx
r
2
cos x.
πx
[Hint: Note that the above equation is the Bessel’s Equation of order p = 12 find its solution then show
r
r
2
2
that J1/2 (x) =
sin x and J−1/2 (x) =
cos x.]
πx
πx
Solution
We note that the above equation is the Bessel’s Equation of order p = 12 . The general solution is given
r
2
by (7.18), y(x) = A1 J1/2 (x) + A2 J−1/2 (x) . Now let us show that J1/2 (x) =
sin x and J−1/2 (x) =
πx
r
2
cos x .
πx
1
(2n)! √
1
First we use the following property of the gamma function: Γ n +
= 2n
π and Γ n + + 1 =
2
2 n!
2
(2n + 1)! √
π, we have (see Example A.4.3 page 208) we have:
22n+1 n!
1
∞
1 X (−1)k 22k+1 k! x 2k+
2
J1/2 (x) = √
k! (2k + 1)!
2
π
k=0
r
r
∞
2 X (−1)k 2k+1
2
=
x
=
sin x.
πx
(2k + 1)!
πx
k=0
176
By substituting p by −1/2 into (7.17), we obtain in a similar way to find
r
r
∞
2 X 1 2k
2
J−1/2 (x) =
x =
cos x.
πx
(2k)!
πx
k=0
■
7.5.1.4
Bessel function of second kind
For n ̸= integer, we defined the linear combination
Yp (x) =
Jp (x) cos pπ − J−p (x)
,
sin pπ
(7.19)
since Jp and J−p (p ̸= integer) are linearly independent solutions of Bessel’s equation (7.1), it follows that
Yp is also a solution of Bessel’s equation (see principle of superposition section 1.5). The function Yp is
called ”Bessel function of second kind of order p”. Figure 7.2 shows the graphs of Y0 , Y1/2 , Y1 , Y2 , Y3 .
It is noticeable from the graph that
lim Yp (x) = −∞,
(7.20)
x→0+
so that the Bessel’s functions of the second kind are not bounded at 0.
Figure 7.2: Bessel Functions of second kind
7.5.2
General Solution of the Bessel’s Equation
Summarising section 7.5.1, we conclude that the general solution of the Bessel’s equation can be written
in the form:
y (x) = A1 Jp (x) + A2 Yp (x) ,
(7.21)
177
APM3701/1
where Jp (x) is the Bessel function of first kind given by (7.15) which is bounded at 0, and Yp (x) is the
Bessel function of second kind given by (7.19) which is unbounded at 0.
7.5.3
Parametric Bessel equation
By replacing x by λx in (7.1) and using the chain rule, we obtain an alternative form of the Bessel’s
equation, known as the ”Parametric Bessel’s equation of order p”:
x2 y ′′ + xy ′ + λ2 x2 − p2 y = 0.
(7.22)
In fact, we set x̄ = λx, (7.22) becomes (use the chain rule):
x̄2 Jp′′ (x̄) + x̄Jp′ (x̄) + x̄2 − p2 Jp (x̄) = 0
(7.23)
Equation (7.23) is Bessel equation, its general solution is given by y (x̄) = A1 Jp (x̄) + A2 Yp (x̄) . Thus, the
general solution of (7.22) is
y (x) = A1 Jp (λx) + A2 Yp (λx) .
(7.24)
Example 7.5.2 Find the solution of the following equation
4x2 y ′′ + 4xy ′ + 3x2 − 1 = 0
Solution
Dividing through by 4, we obtain
2 ′′
′
x y + xy +
3 2 1
x −
4
4
y = 0,
√
1
3
which is the parametric Bessel’s equation of order with λ =
. The solution is given by (7.24):
2
2
√ !
√ !
3
3
y (x) = A1 J1/2
x + A2 Y1/2
x .
2
2
■
Exercises 7.5
1. Exercises 4.7 of [1] pp 244-248, all problems.
7.6
BESSEL SERIES EXPANSION
This section explores some properties of the bessel functions used in solving boundary value problems
occurring in this chapter. These properties are similar to properties of Fourier series: orthogonality and
expansion of function in Bessel series.
178
7.6.1
Differentiation of Bessel functions
Theorem 7.6.1 For p ≥ 0, we have the following basic identities
d p
[x Jp (x)] = xp Jp−1 (x)
dx
d −p
x Jp (x) = −x−p Jp+1 (x) .
dx
(7.25)
(7.26)
Proof
For (7.25), we use (7.15)
d p
[x Jp (x)] =
dx
d
dx
x
p
∞
X
k=0
x 2k+p
(−1)k
k!Γ (k + p + 1) 2
=
∞
x 2k+p
d X
(−1)k
xp
dx
k!Γ (k + p + 1) 2
=
∞
x 2k+2p
d X
(−1)k 2p
dx
k!Γ (k + p + 1) 2
!
k=0
k=0
=
∞
X
(−1)k 2p (k + p) x 2k+2p−1
k=0
= xp
k!Γ (k + p + 1)
∞
X
k=0
2
(−1)k x 2k+p−1
k!Γ (k + p) 2
p
= x Jp−1 (x) .
Now, we prove (7.26),
d −p
x Jp (x) =
dx
d
dx
x
−p
∞
X
k=0
x 2k+p
(−1)k
k!Γ (k + p + 1) 2
=
∞
x 2k+p
d X
(−1)k
x−p
dx
k!Γ (k + p + 1) 2
=
∞
d X
(−1)k
x2k
dx
k!Γ (k + p + 1) 22k+p
!
k=0
k=0
=
∞
X
k=1
∞
X
=
k=1
∞
X
k=0
(−1)k 2k
x2k−1
k!Γ (k + p + 1) 22k+p
(−1)k k
x2k−1
k!Γ (k + p + 1) 22k+p−1
(−1)k+1 (k + 1)
x2k+1
(k + 1)!Γ (k + p + 2) 22k+p+1
= −x−p
∞
X
k=0
Substituting k by k + 1, to start the sum with 0.
x 2k+p+1
(−1)k
k!Γ (k + p + 2) 2
= −x−p Jp+1 (x) .
Note that
d
[J0 (x)] = −J1 (x) .
dx
(Let p = 0 in either relation (7.25) or (7.26)).
■
179
APM3701/1
Properties (7.27) to (7.30) below can be easily proven by expanding (7.25) or (7.26) and using the product
rule (see also [1] section 4.8):
xJp′ (x) + pJp (x) = xJp−1 (x)
(7.27)
xJp′ (x) − pJp (x) = −xJp+1 (x)
(7.28)
Jp−1 (x) − Jp+1 (x) = 2Jp′ (x)
2p
Jp (x) .
Jp−1 (x) + Jp+1 (x) =
x
From (7.25) and (7.26)we deduce easily that
Z
J0 (x) dx = −J1 (x) + C
Z
xp+1 Jp (x) dx = xp+1 Jp+1 (x) + C
Z
x−p+1 Jp (x) dx = −x−p+1 Jp−1 (x) + C.
(7.29)
(7.30)
(7.31)
(7.32)
(7.33)
Hint for the proofs of (7.27)–(7.33):
To prove (7.27), expand the left hand side of (7.25), use the product rule to get:
xp Jp′ (x) + pxp−1 Jp (x) = xp Jp−1 (x) .
Dividing through by xp−1 gives (7.27).
For (7.28), we expand the left hand side of (7.26) using the product rule to obtain:
x−p Jp′ (x) − px−p−1 Jp (x) = −x−p Jp+1 (x) .
Dividing through by x−p−1 gives (7.28). Equation (7.29) is obtained by adding (7.27) and (7.28) and
dividing through by x. However subtracting (7.28) from (7.27) and dividing through by x yields (7.30).
Letting p → p + 1 into (7.25) and p → p − 1 into (7.26) and integrating yield (7.32) and (7.33) respectively.
Remark 7.6.1 .
Formulas (7.25)—(7.33) hold even when the Jp ’s are replaced by the Ys ’s. i.e
d p
[x Yp (x)] = xp
dx
d −p
x Yp (x) = −x−p Yp+1 (x)
dx
(7.34)
(7.35)
(7.36)
xYp′ (x) + pYp (x) = xYp−1 (x)
(7.37)
xYp′ (x) − pYp′ (x) = −xYp+1 (x)
(7.38)
Yp−1 (x) − Yp+1 (x) =
Yp−1 (x) + Yp+1 (x) =
2Yp′ (x)
(7.39)
2p
Yp (x) .
x
(7.40)
180
Z
Z
Z
7.6.2
Y0 (x) dx = −Y1 (x) + C
(7.41)
xp+1 Yp (x) dx = xp+1 Yp+1 (x) + C
(7.42)
x−p+1 Yp (x) dx = −x−p+1 Yp−1 (x) + C.
(7.43)
Orthogonality of Bessel functions
We recall the familiar example of the functions sin nπx, n = 1, 2, 3, · · · which are orthogonal on [0, 1] (see
Example 3.4.3), i.e
Z 1
sin nπx sin mπxdx = 0, if n ̸= m.
0
Please take note that the functions sin nπx are constructed from a single function sin x and its positive
zeros nπ, n = 1, 2, 3, · · ·
Likewise, we construct a system of orthogonal Bessel functions from a single Bessel function and its zeros.
For a fixed p ≥ 0, we consider the graph of Jp (x) for x > 0 (see Figure 7.3).
Figure 7.3: Zeros of Jp (x)
As we can see from Figure 7.3, the Bessel function Jp has infinitely many zeros for x > 0, (note the
similarity with sin x).
We denote the zeros of Jp (x) in ascending order
0 < αp1 < αp2 < αp3 < · · · < αpn < · · ·
so that αpn denotes the nth positive zero for Jp . Unlike the case of the sine function, where the zeros are
easily determined by nπ, the zero of the Bessel functions are determined numerically and they are found
181
APM3701/1
in most mathematical tables and computer programs. For instance, in Mathematica, the zeros of Bessel
functions can be found by using the command BesselJZero[n, p]. Table 7.1 lists some zeros of Jp (x): αpn
for n = 1, 2, 3, 4, 5, and p = 0, 1, 2, 3, 4, 5.
n
J0 (x)
J1 (x)
J2 (x)
J3 (x)
J4 (x)
J5 (x)
1
2.4048
3.8317
5.1356
6.3802
7.5883
8.7715
2
5.5201
7.0156
8.4172
9.7610
11.0647
12.3386
3
8.6537
10.1735
11.6198
13.0152
14.3725
15.7002
4
11.7915
13.3237
14.7960
16.2235
17.6160
18.9801
5
14.9309
16.4706
17.9598
19.4094
20.8269
22.2178
Table 7.1: Zeros of Jp (x), n = 1, 2, 3, 4, 5 and p = 0, 1, 2, 3, 4, 5
To generate orthogonal functions on the interval [0, a] , a > 0 from Jp , we use similar approach to the sine
nπx
functions on the interval [0, L] , sin
, n = 1, 2, 3, · · · We get the functions
L
α
pn
x , n = 1, 2, 3, · · ·
(7.44)
Jp
a
Figure 7.4 and 7.5 show the first five functions corresponding to p = 0 and p = 1 respectively. For the
seek of simplicity we denote by
αnp
, n = 1, 2, 3, · · ·
(7.45)
λpn =
a
So that λpn is the nth positive zero of Jp scaled by a factor 1/a.
7.6.3
Notation
αpn and λpn denote the zeros and the scaled zeros of the Bessel functions of order p, it is cumbersome to
work with, when it is clear which order we are dealing with, and there is no risk of confusion, we will drop
the index p and denotes αn and λn instead of αpn and λpn .
We are now in the position to state and prove the orthogonality property of Bessel functions.
Theorem 7.6.2
For a fixed p ≥ 0 and a > 0. Let αj , j = 1, 2, 3, · · · be the zeros of either Jp , Jp−1 or Jp+1 , on the interval
αj
, j = 1, 2, 3, · · · are as in (7.44) the zeros of either Jp , Jp−1 or Jp+1 on [0, a] , then we
[0, 1] . If λj =
a
have
Z 1
xJp (αn x) Jp (αm ) dx = 0
n ̸= m,
(7.46)
0
Z 1
1 2
xJp2 (αn x) dx =
J
(αn )
for n = 1, 2, 3, · · ·
(7.47)
2 p+1
0
Z a
xJp (λn x) Jp (λm x) dx = 0
n ̸= m,
(7.48)
0
Z a
a2 2
J
(λn a)
for n = 1, 2, 3, · · ·
(7.49)
xJp2 (λn x) dx =
2 p+1
0
Proof
One way of proving (7.46) is done in [1], theorem 3, section 4.8, p 255. An alternative method, illustrated
182
Figure 7.4: Orthogonal functions generated with J0 (x) : J0 (λ01 x), J0 (λ02 x), J0 (λ03 x), J0 (λ04 x), J0 (λ05 x)
below uses the relations (7.25) and (7.26). We rewrite the integrand of (7.46) as
xJp (αn x) Jp (αm x) = x−p Jp (αn x) xp+1 Jp (αm x)
n ̸= m.
Let u = x−p Jp (αn x)
du = −αn x−p Jp+1 (αn x) dx by (7.26)
1 p+1
dv = xp+1 Jp (αm x) dx
v=
x Jp+1 (αm x) by (7.25).
αm
Integration by parts, yields (we have assumed that αn are zero of Jp )
1
Z 1
xp+1
−p
Jp+1 (αm x)
xJp (αn x) Jp (αm x) dx = x Jp (αn x)
αm
0
0
Z 1
1
−p
p+1
−
−αn x Jp+1 (αn x)
x Jp+1 (αm x) dx .
αm
0
(7.50)
Or
Z
0
1
Z
αn 1
xJp (αn x) Jp (αm x) dx =
xJp+1 (αn x) Jp+1 (αm x) dx.
αm 0
[Since αn and αm are zeros of Jp (x) .]
Interchanging αn and αm in equation (7.51), we obtain
Z 1
Z
αm 1
xJp (αm x) Jp (αn x) dx =
xJp+1 (αn x) Jp+1 (αm x) dx.
αn 0
0
Subtracting (7.51) from (7.52), we get
α2 − α2m
0= n
αn αm
Z
1
xJp+1 (αn x) Jp+1 (αm x) dx.
0
(7.51)
(7.52)
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Figure 7.5: Orthogonal functions generated with J1 (x) : J1 (λ11 x), J1 (λ12 x), J1 (λ13 x), J1 (λ14 x), J1 (λ15 x)
Since αn ̸= αm , αn > 0 and αm > 0, we deduce that
Z
1
xJp+1 (αn x) Jp+1 (αm x) dx = 0.
(7.53)
0
Thus
Z
1
xJp (αn x) Jp (αm x) dx = 0.
(7.54)
0
Therefore the functions Jp (αn x) and Jp (αm x) are orthogonal with respect to the weight function x on
(0, 1) .
• If αn and αm are zero of Jp+1 , equation (7.50) implies that (7.53) is also satisfied, showing that
Jp (αn x) and Jp (αm x) are also orthogonal with respect to x on (0, 1) .
If we rewrite xJp (αn x) Jp (αm x) in the following form
xJp (αn x) Jp (αm x) = [xp Jp (αn x)] x−p+1 Jp (αm x) .
• Integration by parts, as above, yields
Z
1
1
1
xJp (αn x)
Jp−1 (αm x)
αm
0
Z 1
αn
xJp−1 (αn x) Jp−1 (αm x) dx.
αm 0
xJp (αn x) Jp (αm x) dx =
0
+
184
If αn and αm are zeros of Jp−1 (x), the above equation becomes
Z
0
1
αn
xJp (αn x) Jp (αm x) dx =
αm
Z
1
xJp−1 (αn x) Jp−1 (αm x) dx.
0
Interchanging αn and αm in the above equation we obtain as above:
1
Z
n ̸= m
xJp (αn x) Jp (αm x) dx = 0.
0
Thus Jp (αn x) and Jp (αm x) are orthogonal with respect to the weight function x when αn and αm are
zeros of Jp−1 .
Let us now prove (7.47). We know that Jp (x) satisfies the Bessel equation
x2 Jp′′ (x) + xJp′ (x) + x2 − p2 Jp (x) = 0
with
Jp (αn ) = 0.
We rewrite the above equation in the form:
p2
+ x−
x
′
xJp′
Jp = 0,
Jp (αn ) = 0
We multiply both sides of the equation by xJp′ and we put it in the form
i′
′ 2
xJp
h
+ x2 − p 2
Jp2
′
= 0.
We integrate on 0 < x < αn (integrate the 2nd term by parts), we obtain:
Z αn
2
2
′
2
2
2 2
αn Jp (αn ) + αn − p Jp (αn ) + p Jp (0) − 2
xJp2 (x) dx = 0.
0
We know that Jp (0) = 0 for p > 0, and obviously pJp (0) = 0 for p = 0, and since Jp (αn ) = 0, the above
equation reduces to
Z αn
2
′
αn Jp (αn ) − 2
xJp2 (x) dx = 0.
(7.55)
0
The use of (7.28) yields
αn Jp′ (αn ) − pJp (αn ) = −αn Jp+1 (αn ) .
Thus αn Jp′ (αn ) = −αn Jp+1 (αn ) , since Jp (αn ) = 0. Therefore (7.55) becomes
Z
0
αn
xJp2 (x) dx =
α2n 2
J
(αn ) .
2 p+1
(7.47) follows by replacing x by αn x.
Identities (7.48) and (7.49) follow by substituting x by ax into (7.46) and (7.47) respectively.
■
185
7.6.4
APM3701/1
Bessel Series and Bessel-Fourier coefficients
nπx
to expand functions
We follow an approach similar to Fourier sine series with the use of the function sin
L
in sine Fourier series on the interval [0, L] , we suppose that, for a given function f on the interval [0, a] ,
can be expanded in a series as:
∞
X
f (x) =
Ak Jp (λk x) ,
(7.56)
k=1
that is called ”Bessel series of order p” of f. The question of convergence of (7.56) will be dealt with
later (see theorem 3.7.1), assume for now that (7.56) is valid, similar to Fourier series, let us find the
coefficients in the series. Multiplying both side of (7.56) by xJp (λn x) and integrate term by term on the
interval [0, a, ] gives :
a
Z
xf (x) Jp (λn x) dx =
0
∞
X
Z
0
k=0
|
Z
= An
0
Thus
An = R a
0
1
xJp2 (λn x) dx
a
Ak
Z
a
xJp (λk x) Jp (λn x) dx
{z
}
=0 except when k=n.
xJp2 (λn x) dx.
(7.57)
a
xf (x) Jp (λn x) dx.
(7.58)
0
The number An defined by (7.58) is called the ”nth Bessel-Fourier coefficient”, or simply the ”Bessel
coefficient” of the function f. The values of these coefficients depend on how the eigenvalue λn are
defined. We will consider three cases:
Case 1: Jp (λn a) = 0, n = 1, 2, 3, · · ·
Case 2: hJp (λn a) + λn aJp′ (λn a) = 0, h ≥ 0, n = 1, 2, 3, · · ·
Case 3: J0′ (λn a) = 0,
n = 1, 2, 3, · · ·
We know that Jp (λx), p ≥ 0, satisfies the parametric Bessel equation (see section 7.5.3)
x2 y ′′ + xy ′ + λ2 x2 − p2 y = 0.
We set x̄ = λx, the above equation becomes (use the chain rule):
x̄2 Jp′′ (x̄) + x̄Jp′ (x̄) + x̄2 − p2 Jp (x̄) = 0.
Or
d ′
p2
x̄Jp + x̄ −
Jp = 0.
dx̄
x̄
We multiple the above equation by 2x̄Jp′ (x̄) and simplify, we obtain:
d ′ 2
x̄Jp + 2 x̄2 − p2 Jp′ Jp = 0
dx
186
Integration on [0, λa] gives (integrate the second term by parts)
Z
λa
2
0
x̄Jp2 dx̄ =
h
x̄Jp′
2
ia
+ x̄2 − p2 Jp2 .
0
Substituting x̄ by λx, we have
Z a
h
ia
2
2
xJp2 (λx) dx = λ2 a2 Jp′ (λx) + λ2 x2 − p2 Jp2 (λx)
2λ
0
0
Since Jp (0) = 0 and pJp (0) = 0 for p ≥ 0, (show this!!!), we have
2λ
2
Z
0
a
2
xJp2 (λx) dx = λ2 a2 Jp′ (λa) + λ2 a2 − p2 Jp2 (λa) .
(7.59)
We deduce the following from (7.59):
Case 1: If Jp (λa) = 0, we use (7.28) to deduce that Jp′ (λa) = −Jp+1 (λa),
Ra
2
2λ2 0 xJp2 (λx) dx = λ2 a2 Jp+1
(λa) . Thus
a
Z
0
xJp2 (λx) dx =
(7.59) becomes
a2 2
J
(λa) .
2 p+1
h
Jp (λa) substitution into (7.59) gives
λa
2
Z a
h
2
2
2 2
2λ
xJp (λx) dx = λ a
Jp2 (λa) + λ2 a2 − p2 Jp2 (λa)
λa
0
Z a
2 2
λ a − p2 + h2 2
=⇒
xJp2 (λa) dx =
Jp (λa) .
2λ2
0
Case 2: If hJp (λa) + λaJp′ (λa) = 0 =⇒ Jp′ (λa) = −
Case 3: If J0′ (λa) = 0 i.e. p = 0 and h = 0, (7.59) becomes
Z a
2
xJ02 (λx) dx = a2 J02 (λa) .
0
In this case λ = 0 is an eigenvalue. So for λ = 0, (J0 (0) = 1) we have
Z a
Z a
a2
2
xJ0 (λx) dx =
xdx = .
2
0
0
For λ ̸= 0
Z
a
0
7.6.4.1
xJ02 (λx) dx =
a2 2
J (λa) .
2 0
Bessel Coefficients
The Bessel series of f on (0, a) is defined by
f (x) =
∞
X
An Jp (λn x)
n=1
and using the three cases above, the Bessel coefficients are given by:
187
APM3701/1
(1) When λn are defined by Jp (λn a) = 0,
An =
2
2
(λn a)
a2 Jp+1
Z
a
xf (x) Jp (λn x) dx
0
n = 1, 2, 3, · · ·
(7.60)
(2) When λn are defined by hJp (λn a) + λn aJp′ (λn a) = 0,
2λ2n
An = 2 2
λn a − p2 + h2 Jp2 (λn a)
Z
a
xf (x) Jp (λn x) dx
0
n = 1, 2, 3, · · · .
(7.61)
(3) When λn are defined by J0′ (λn a) = 0,
2
A1 = 2
a
7.6.5
Z
0
a
2
xf (x) dx, and An = 2 2
a J0 (λn a)
Z
a
xJ0 (λn x) f (x) dx
0
n = 2, 3, 4, · · ·
(7.62)
Very Important
1. Coefficients of the Bessel series given by (7.60)–(7.62) are particular cases, the more general coefficients are given by
Z a
1
xf (x) Jp (λn x) dx, n = 1, 2, 3, · · · .
(7.63)
An =
2
λ2 a2 Jp′ (λa) + λ2 a2 − p2 Jp2 (λa) 0
2. We state in the next theorem, the condition under which the Bessel series expansion (7.56) with
coefficients given by (7.63) is valid.
Theorem 7.6.3 Convergence of Bessel series of order p
If f is a piecewise smooth function on [0, a] , then f has a Bessel series expansion of order p on the interval
(0, a) given by
∞
X
f (x) =
An Jp (λn x) ,
(7.64)
n=1
where An are defined by (7.60)–(7.62).
α1
, λ2 = αa2 , · · · are the scale
a
(x−)
positive zeros of Jp (x) given by (7.45). Furthermore, the series (7.64) converges to the average f (x+)+f
2
at the points of discontinuity (recall the similarities with the convergence of Fourier series, theorem 7.6.3).
α1 , α2 , · · · are the positive zeros of the Bessel function Jp (x), and λ1 =
The proof of this theorem is given in [10].
Exercises 7.6
1. Exercises 4.8 of [1] pp 257–261, all problems.
188
7.7
PARTIAL DIFFERENTIAL EQUATIONS IN POLAR AND CYLINDRICAL COORDINATES
Boundary-value problems considered up to now, have been expressed in terms of rectangular coordinate
system. If the problem under consideration are studied in circular, cylindrical or spherical domains, naturally the description of the problem in polar coordinates, cylindrical coordinates or spherical coordinates
respectively makes much sense. It is essential that we express the Laplacian ∇2 u = ∆u in terms of these
different coordinates systems. In the next sections, we will deal with the steady-state temperature in a
circular plate, heat distribution in an infinitely long cylinder, the vibration of a circular membrane, and
the state–state temperature in a sphere.
7.7.1
7.7.1.1
The Laplacian in various coordinate systems
The Laplacian in Polar Coordinates
Figure 7.6: Polar Coordinates
The relationship between polar coordinates and rectangular coordinates is given by
x = r cos θ, y = r sin θ,
r 2 = x2 + y 2 ,
tan θ =
y
.
x
For a function u = u (x, y) , the Laplacian ∇2 u of is given by
∇2 u =
∂2u ∂2u
+ 2.
∂x2
∂y
Let’s express ∇2 u in polar coordinates. We differentiate r2 = x2 + y 2 with respect to x and y, we obtain
2r
∂r
= 2x
∂x
or
∂r
x
∂r
= =⇒
= cos θ
∂x
r
∂x
2r
∂r
= 2y
∂y
or
∂r
y
∂r
= =⇒
= sin θ,
∂y
r
∂y
and
189
∂2r
∂x2
=
∂2r
∂y 2
=
∂θ
∂x
=
∂θ
∂y
=
∂2θ
∂x2
=
∂2θ
∂y 2
=
∂r
∂ x r − x ∂x
r − r cos2 θ
sin2 θ
=
=
=
=
,
∂x x
r2
r2
r
∂r
r−y
∂ y
∂ ∂r
r − r sin2 θ
cos2 θ
∂y
=
=
=
=
,
∂y ∂y
∂y x
r2
r2
r
y
− 2
y
sin θ
∂ −1 y x
=− 2 =−
tan
=
,
∂x
x
r
r
y2
1+ 2
x
1
∂ −1 y x
cos θ
x
tan
=
= 2 =
,
2
∂y
x
r
r
y
1+ 2
x
∂r
∂ y 2ry ∂x
2r2 sin θ cos θ
2 sin θ cos θ
−
=
=
=
,
2
2
4
4
∂x
r
r
r
r2
∂r
−2rx
∂ x
−2r sin θ
2 sin θ cos θ
∂y
=
=
=−
.
2
2
4
4
∂y r
r
r
r2
∂
∂x
∂r
∂x
The chain rule for partial differential equation gives:
∂u
∂x
∂2u
∂x2
=
=
=
=
=
=
∂u
∂y
∂2u
∂y 2
=
=
=
∂u ∂r
∂u ∂θ
∂u
∂u sin θ
+
=
cos θ −
,
∂r ∂x
∂θ r ∂θ ∂x ∂r
∂ ∂u
∂ ∂u
∂u sin θ
=
cos θ −
∂x ∂x
∂x ∂r
∂θ r
∂ ∂u
∂u sin θ
∂ ∂u
∂u sin θ sin θ
cos θ −
cos θ −
cos θ −
∂r ∂r
∂θ r
∂θ ∂r
∂θ r
r
2
2
∂ u sin θ ∂u sin θ
∂ u
cos θ −
+
cos θ
∂r2
∂r∂θ r
∂θ r2
2
∂ u
∂u
∂ 2 u sin θ ∂u cos θ sin θ
−
cos θ −
sin θ − 2
−
∂r∂θ
∂r
∂θ r
r
∂θ r
2
2
2
∂ u
∂ u sin θ cos θ ∂u sin θ cos θ
∂ u cos θ sin θ
cos2 θ −
+
−
2
2
∂r
∂r∂θ
r
∂θ
r
∂r∂θ
r
∂u sin2 θ ∂ 2 u sin2 θ ∂u cos θ sin θ
+
+ 2 2 +
∂r r
∂θ
r2
∂θ r
2
2
∂ u
∂ u sin θ cos θ
∂u sin θ cos θ
cos2 θ − 2
+2
2
∂r
∂r∂θ
r
∂θ
r2
2
2
2
∂u sin θ ∂ u sin θ
+
+ 2 2 ,
∂r r
∂θ r
∂u ∂r ∂u ∂θ
∂u
∂u cos θ
+
=
sin θ +
,
∂r ∂y
∂θ ∂y
∂r
∂θ r
∂ ∂u
∂ ∂u
∂u cos θ
=
sin θ +
∂y ∂y
∂y ∂r
∂θ r
∂ ∂u
∂u cos θ
∂ ∂u
∂u cos θ cos θ
sin θ +
sin θ +
sin θ +
∂r ∂r
∂θ r
∂θ ∂r
∂θ r
r
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190
∂2u
∂ 2 u cos θ ∂u cos θ
sin θ
=
sin θ +
−
∂r2
∂r∂θ r
∂θ r2
2
∂ u
∂u
∂ 2 u cos θ ∂u sin θ cos θ
+
sin θ +
cos θ − 2
−
∂r∂θ
∂r
∂θ r
r
∂θ r
2
2
2
∂ u
∂ u sin θ cos θ ∂u sin θ cos θ
∂ u cos θ sin θ
=
sin2 θ +
−
+
∂r2
∂r∂θ
r
∂θ
r2
∂r∂θ
r
2
2
2
∂u cos θ ∂ u cos θ ∂u cos θ sin θ
−
+
+ 2
∂r r
∂θ
r2
∂θ r2
∂2u
∂ 2 u sin θ cos θ
∂u sin θ cos θ
2
=
sin
θ
+
2
−2
∂r2
∂r∂θ
r
∂θ
r2
2
2
2
∂u cos θ ∂ u cos θ
+
+ 2
.
∂r r
∂θ r2
Thus
7.7.1.2
∂ 2 u 1 ∂u
∂2u ∂2u
1 ∂2u
+
=
+
+
∂x2
∂y 2
∂r2
r ∂r
r2 ∂θ2
(7.65)
Laplacian in cylindrical coordinates
Figure 7.7: Cylindrical Coordinates
If u = u (x, y, z) , the Laplacian is
∆u = ∇2 u =
∂2u ∂2u ∂2u
+ 2 + 2.
∂x2
∂y
∂z
We recall the relationship between rectangular coordinates and cylindrical coordinates:
x = ρ cos ϕ,
We have
∇2 u =
y = ρ sin ϕ,
z = z.
∂2u ∂2u ∂2u
∂ 2 u 1 ∂u
1 ∂2u ∂2u
+
+
=
+
+
+ 2
∂x2
∂y 2
∂z 2
∂r2
r ∂r
r2 ∂θ2
∂z
[From (7.65)]
(7.66)
191
APM3701/1
Figure 7.8: Spherical Coordinates
7.7.1.3
Laplacian in Spherical Coordinates
∂2u
∂2u
∂2u
+
+
. Using the chain rule as we did in section 7.7.1.1, we
∂x2
∂y 2
∂z 2
obtain the Laplacian in spherical coordinates
If u = u (x, y, z) , then ∇2 u =
∇2 u =
7.7.2
∂ 2 u 2 ∂u
∂2u
1
1 ∂ 2 u cot θ ∂u
+
+
.
+
+ 2
∂r2
r ∂r
r ∂θ
r2 sin2 θ ∂ϕ2 r2 ∂θ2
(7.67)
Steady Temperature in a circular domains
We saw in section 4.8 that it was possible for a body to reach a state where the temperature no longer
change with time. Heat still flows in the body, but the flow itself no longer varies with time.
In this section we are required to find the temperature distribution u in a circular plate of radius a, if
the temperature at the edge of the plate is known. We know that in this case the temperature does not
depend on time, we can formulate the problem as follow:
7.7.2.1
Problem: Steady–state temperature in a circular plate.
Find the steady–state temperature u (r, θ) in a circular plate of radius a if the temperature of the circumference edge is given by u (a, θ) = f (θ) , 0 < θ < 2π.
Solution
We need to solve the Laplace equation in polar coordinates
∂ 2 u 1 ∂u
1 ∂2u
+
+
= 0, 0 < θ < 2π, 0 < r < a.
∂r2
r ∂r
r2 ∂θ2
u (a, θ) = f (θ)
0 < θ < 2π.
u (r, 0) = u (r, 2π) and u′ (r, 0) = u′ (r, 2π)
We let u (r, θ) = R (r) Θ (θ) , and substitute into (7.68)
1
1
r2 Θ + R′ Θ + 2 RΘ′′ = 0.
r
r
(7.68)
[Periodic boundary conditions]
192
Figure 7.9: Temperature Distribution in Circular Plate
Separation of variables yields
r2 R′′ + rR′
Θ′′
=−
= λ,
R
Θ
or
r2 R′′ + rR′ − λR = 0
and
(7.69)
Θ′′ + λΘ = 0
(7.70)
and Θ′ (0) = Θ′ (2π)
Θ (0) = Θ (2π)
(7.71)
From example 3.8.3(b) (a = 0 and b = 2π) the solution of (7.70)–(7.71) is
λ n = n2
and
Θn = c1 cos nθ + c2 sin nθ,
n = 0, 1, 2, 3, · · ·
(7.72)
And the solution of (7.69) is given by
r −n
+ c4
,
n = 1, 2, 3, · · ·
a
a
r
and
R0 (r) = c3 + c4 ln .
a
r −n
r
Since the solution is bounded at r = 0, we have c4 = 0, because
and ln are not bounded at r = 0.
a
a
We have the product solutions
r n
u0 (r, θ) = A0 and un (r, θ) =
(An cos nθ + Bn sin nθ) ,
n = 1, 2, 3, · · ·
a
Rn = c3
r n
Superposition of these solutions yields
u (r, θ) = A0 +
∞ X
r n
n=1
a
(An cos nθ + Bn sin nθ) .
(7.73)
193
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The coefficients An and Bn are the Fourier coefficients of f (θ) on (0, 2π), given by (3.28)–(3.30):
A0 =
1
2π
Z
2π
f (θ) dθ
and An =
0
and
Bn =
1
π
Z
1
π
Z
2π
f (θ) cos nθdθ,
(7.74)
0
2π
f (θ) sin nθdθ,
0
n = 1, 2, 3, · · ·
(7.75)
■
Example 7.7.1 Find the steady-state temperature distribution in a disc of radius 1, if the upper half of
the circumference is into contact with a body which is at a constant temperature K and the lower half is
into contract with ice at temperature 0.
Solution
The problem to solve is the Laplace equation in a circular plate of radius 1, if the temperature of the
circumference is given by
(
K if 0 < θ < π
u (1, θ) = f (θ) =
0 if π < θ < 2π.
The solution is given by (7.73)–(7.75)
u (r, θ) = A0 +
∞
X
rn (An cos nθ + Bn sin nθ)
n=1
A0
and
Thus
Bn
Z 2π
Z π
1
K
=
Kdθ = , An =
K cos nθdθ = 0
2π 0
2
0
Z π
K
=
K sin nθdθ =
[1 − (−1)n ] .
nπ
0
∞
K
KX1
u (r, θ) =
+
[1 − (−1)n ] rn sin nθ.
2
π
n
n=1
■
7.7.2.2
Steady–state temperature in a circular sector (wedge)
Figure 7.10: Temperature Distribution in a Wedge
194
Find the steady–state temperature over the wedge in figure 7.10, supplied with the boundary conditions:
u (r, 0) = 0 = u (r, α) ,
∂
h u (a, θ) + ηu (a, θ) = f (θ) .
∂r
0<r<a
The latter condition means that the wedge is exchanging heat along its circular boundary at a rate given
by
∂
η
1
u (a, θ) = − ua, θ + f (θ) .
∂r
h
h
Solution
Note first that the wedge is defined by 0 < θ < α. The boundary–value problem to solve is
∂ 2 u 1 ∂u
1 ∂2u
+
+ 2 2 = 0, 0 < θ < α, 0 < r < a
∂r
r ∂r
r ∂θ
u (r, 0) = 0,
u (r, α) = 0,
∂
h u (a, θ) + ηu (a, θ) = f (θ) .
∂r
(7.76)
Separation of variables gives
Θ′′ + λΘ = 0,
2
Θ (0) = 0, Θ (α) = 0,
(7.77)
′′
r R + rR − λR = 0
(7.78)
nπθ
nπ
and Θn (θ) = sin
, n = 1, 2, 3, · · ·
Solution of (7.77) is given by λn =
α
α
The solution of (7.78) is given by (since r is bounded at 0)
r nπ
α .
Rn (r) = C
a
Superposition of solution gives
u (r, θ) =
∞
X
n=1
r nπ
nπθ
α sin
An
,
a
α
o < r < a, 0 < θ < α.
∂
u (a, θ) + ηu (a, θ) = f (θ) :
∂r


nπ
∞
∞
r nπ
X
X
r
∂ 
nπθ 
nπθ
α
α sin
An
sin
+η
An
h
∂r
a
α
a
α
The boundary condition h
n=1
r=a
=⇒ f (θ) =
∞
X
An
n=1
The coefficients An
n=1
hnπ
nπθ
+ η sin
,
aα
α
= f (θ)
r=a
0 < θ < α.
hnπ
+ η are determined as sine Fourier coefficients of f (see equation (3.62))
aα
Z
hnπ
2 α
nπθ
An
+η =
f (θ) sin
dθ, n = 1, 2, 3, · · ·
aα
α 0
α
2a
=⇒ An =
nhπ + aαη
Z
α
f (θ) sin
0
nπθ
dθ,
α
n = 1, 2, 3, · · ·
(7.79)
195
Thus
u (r, θ) =
∞
X
n=1
APM3701/1
r nπ
nπθ
α sin
An
,
a
α
where An are given by (7.79).
(7.80)
■
Example 7.7.2 Steady–state temperature in a semicircular plate
Find the steady–state temperature in the semicircular plate (see figure 7.11)
Figure 7.11: Temperature Distribution in a Semicircular Plate
Solution
The boundary–value problem to solve is:
1 ∂2u
∂ 2 u 1 ∂u
+
+
= 0, 0 < θ < π, 0 < r < a
∂r2
r ∂r
r2 ∂θ2
u (a, θ) = u0 ≡ constant, 0 < θ < π,
u (r, 0) = 0 = u (r, π) ,
0 < r < a.
This is the case of state-steady temperature in a wedge with α = π, h = 0, η = 1, and f (θ) = u0 . The
solution is given by (7.79)–(7.80),(with α = π, h = 0, η = 1, f (x) = u0 ):
Z
2u0
2a π
An =
u0 sin nθdθ = −
(1 − (−1)n ) ,
aπ 0
nπ
and
u (r, θ) =
∞
X
n=1
−
r n 2u
0
(1 − (−1)n ) sin nθ.
a
nπ
■
Example 7.7.3 A Robin condition on a wedge.
Find the steady–state temperature over a circular sector with central angle α, when the straight edges of
the boundary are in contact with ice at temperature 0 and the wedge exchanges the heat along its circular
196
boundary with the surrounding at a rate equals to −u (a, θ) − θ.
Solution
The problem to solve is:
1 ∂2u
∂ 2 u 1 ∂u
= 0,
0 < r < a,
0 < θ < α.
+
+
∂r2
r ∂r
r2 ∂θ2
u (r, 0) = 0 = u (r, α) ,
0 < r < a,
∂u
(a, θ) = −u (a, θ) − θ,
0 < θ < α.
∂r
This is a particular case of (7.76), with h = 1, η = 1, f (θ) = −θ. The solution is given by (7.79)-(7.80),
(h = 1, η = 1, f (θ) = −θ):
2a
An =
nπ + aα
α
Z
0
−θ sin
Thus
u (r, θ) =
nπθ
2a
α (−1)n
dθ =
, n = 1, 2, 3, . . . .
α
nπ + aα nπ
∞ X
r n
n=1
a
2a
α (−1)n
sin nθ.
nπ + aα nπ
■
7.7.3
7.7.3.1
Laplace Equation in a cylinder
Zero lateral and bottom surface temperature
We consider the steady–state temperature distribution inside a cylinder with lateral surface and bottom
kept at zero temperature and with radially symmetric temperature distribution at the top, i.e. independent
of θ, (see figure 7.12)
Solution
The boundary–value problem to solve is
∂ 2 u 1 ∂u ∂ 2 u
+
+ 2 = 0,
∂ρ2
ρ ∂ρ
∂z
u (ρ, 0) = 0
0 < ρ < a,
∇2 u =
u (a, z) = 0
0 < z < h,
u (ρ, h) = f (ρ)
0 < ρ < a.
Separation of variables u (ρ, z) = R (ρ) Z (z) , leads to the following equations:
ρ2 R′′ + ρR′ + ρ2 λ2 R = 0
′′
2
Z −λ R = 0
R (a) = 0
Z (0) = 0.
Solution of (7.81) is given by (3.101) (see example 3.8.5)
Rn (ρ) = J0 (λn ρ) ,
where λn are given by J0 (aλn ) = 0.
n = 1, 2, 3, · · ·
(7.81)
(7.82)
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Figure 7.12: Zero lateral and bottom surface temperature distribution in a cylinder
Solution of (7.82) is given by
n = 1, 2, 3, · · ·
Z (z) = sinh (λn z) ,
Superposing the product solutions yields
u (ρ, z) =
∞
X
An J0 (λn ρ) sinh (λn z) .
n=1
The boundary condition u (ρ, h) = f (ρ) yields
f (ρ) =
∞
X
An J0 (λn ρ) sinh (λn h) .
n=1
The coefficients An sinh (λn h) are given by (7.60)
Thus
2
An sinh (λn h) = 2 2
a J1 (λn a)
Z
2
An = 2 2
a J1 (λn a) sinh (λn h)
Z
a
f (ρ) J0 (λn ρ) ρdρ
0
a
f (ρ) J0 (λn ρ) ρdρ.
0
■
198
7.7.4
Heat Distribution in an infinitely long cylinder
Consider the heat transfer from the lateral surface of an infinitely long cylinder of radius a into a surrounding medium, the temperature inside the rod depends upon the time t and the distance r from its
axis (i.e. the axis through the centre and parallel to the lateral side). We assume that the surrounding
medium is at temperature zero and the initial temperature is constant at every point.
The problem to solve is the heat equation in polar coordinates:
2
∂ u 1 ∂u
∂u
=k
+
0 < r < a,
t > 0.
∂t
∂r2
r ∂r
Newton’s law of cooling at r = a:
∂u
(a, t) = −h (u (a, t) − UE )
∂r
where UE is the temperature of the surrounding medium. =⇒ UE = 0.
Thus
ur (a, t) = −hu (a, t) ,
t > 0,
Initial temperature is constant ⇒ u (r, 0) = C.
Boundedness: |u (0, t)| < ∞.
Thus the initial boundary value problem is:
2
∂ u 1 ∂u
∂u
= k
+
∂t
∂r2
r ∂r
ur (a, t) = −hu (a, t)
u (r, 0) = C,
0 < r < a,
t>0
|u (0, t)| < ∞.
Let u (r, t) = R (r) T (t), the problem splits into two ordinary differential equations:
rR′′ + R′ + λ2 rR = 0,
and
T ′ + kλ2 T
R′ (a) + hR (a) = 0.
= 0
with solutions
R = C1 J0 (λr) + C2 Y0 (λr)
T
2
= C3 e−kλ t .
Boundedness at r = 0 requires that C2 = 0, so that we have the product solutions:
2
uλ (r, t) = Ae−kλ t J0 (λr) .
The boundary condition ur (a, t) + hu (a, t) = 0 becomes
λJ0′ (λa) + hJ0 (λa) = 0,
or
kJ0 (λa) + λaJ0′ (λa) = 0.
(k = ah)
and hence λ is the positive root of the above equation. Superposing solutions corresponding to λn yields
u (r, t) =
∞
X
n=1
2
An e−kλn t J0 (λn r) .
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The initial condition u (r, 0) = C yields
C=
∞
X
An J0 (λn r) .
n=1
The coefficients An are the Bessel coefficients of C determined by (7.61), p = 0, h = k, f (x) ≡ C:
Z a
2λ2n C
rJ0 (λn r) dr
An =
λ2n a2 + k 2 J02 (λn a) 0
Z aλn
x
2λ2n C
2
=
J0 (x) dx (x = λn r)
2
2
2
a λn + k J0 (aλn ) 0
λ2n
Z aλn
d
2C
=
(xJ1 (x)) dx (by 7.25)
dx
a2 λ2n + h2 J02 (aλn ) 0
=
7.7.5
2Cλn J1 (aλn )
a λ2n + h2 J02 (aλn )
(k = ah) .
Radial Vibration of a circular membrane
The aim of this section is to study the vibrations of a thin circular membrane with uniform mass density,
clamped along its circumference. If we place the center of the membrane at the origin, the vibrations of
the membrane are described by the two–dimensional wave equation in polar coordinates:
2
1 ∂u
1 ∂2u
∂2u
2 ∂ u
=k
+
+ 2 2 .
(7.83)
∂t2
∂r2
r ∂r
r ∂θ
The initial position and initial velocity of the membrane are modeled by the functions f (r, θ) and g (r, θ)
respectively.
If the functions f and g are radially symmetric or asymmetric i.e f and g depend only on r: f = f (r)
and g = g (r) , in this case, it makes physical sense to assume that the solution also does not depend on θ.
∂u
= 0, and we formulate the problem as follow:
Consequently
∂θ
Find the displacement u (r, t) of a circular membrane of radius a clamped along its circumference, if initial
shape is f (r) and its initial velocity is g (r) .
Solution
The boundary–value problem to solve is:
2
∂2u
1 ∂u
2 ∂ u
= k
+
,
0 < r < a, t > 0,
(7.84)
∂t2
∂r2
r ∂r
u (a, t) = 0, t ≥ 0 [Membrane clamped at its circumference]
(7.85)
u (r, 0) = f (r) ,
ut (r, 0) = g (r)
0 < r < a [Initial shape of the membrane]
0 < r < a [Initial velocity].
(7.86)
(7.87)
Let u (r, t) = R (t) T (t) into the partial differential equation and separating the variable gives
T ′′
1
1 ′
′′
=
R + R = −λ2 ,
k2 T
R
r
and
rR′′ + R′ + λ2 rR = 0,
T ′′ + k 2 λ2 T
= 0.
R (a) = 0
(7.88)
(7.89)
200
Equation (7.88) is a parametric Bessel equation of order p = 0. Its general solution is
R (r) = c1 J0 (λr) + c2 Y0 (λr) , R (λa) = 0.
Since R (r) is bounded at 0, we have c2 = 0.
Thus
R (r) = CJ0 (λr) ,
J0 (λa) = 0.
Thus λ is the scaled positive zeros of J0 , λ1 , λ2 , λ3 , . . .
The general solution of (7.89) is
T = c3 cos kλt + c4 sin kλt.
The product solution yields
un (r, t) = (An cos kλn t + Bn sin kλn t) J0 (λn r) .
Superposing the product solution yields
u (r, t) =
∞
X
(An cos kλn t + Bn sin kλn t) J0 (λn r) .
(7.90)
n=1
The coefficients An and Bn are determined as follow: the condition u (r, 0) = f (r) gives
f (r) =
∞
X
An J0 (λn r) ,
0 < r < a, J0 (λn a) = 0
n=1
Bessel coefficients are given by (7.60)
An =
2
a2 J12 (aλn )
a
Z
rf (r) J0 (λn r) dr.
(7.91)
0
The second boundary condition ut (r, 0) = g (r) yields
g (r) =
∞
X
kλn Bn J0 (λn r) .
n=1
Thus kλn Bn are determined as above:
2
a2 J12 (aλn )
Z
2
2
a kλn J12 (aλn )
Z
kλn Bn =
∴
Bn =
a
rg (r) J0 (λn r) dr.
0
a
rg (r) J0 (λn r) dr.
(7.92)
0
■
Remark 7.7.1 The general case of Vibration of a circular membrane is studied in section 3.8 of [1]
Example 7.7.4 A circular membrane with clamped edges is vibrated by an explosion. Find the vibrations
of the membrane in the following cases:
201
APM3701/1
(a) The explosion impacts a uniform constant initial velocity, we assume that the initial shape of the
membrane is flat.
(b) The membrane starts from rest from the initial shape described by the function a2 − r2 .
Solution
(a) The boundary–value problem is the same as (7.84)–(7.87), with f (r) = 0 and g (r) = v0 . The solution
is given by (7.90)–(7.92). Thus
u (r, t) =
∞
X
(An cos kλn t + Bn sin kλn t) J0 (λn r)
n=1
where An = 0 and
Bn =
=
=
=
=
2
2
a kλn J12 (λn a)
Z
2v0
2
a kλn J12 (λn a)
Z
a
v0 J0 (λn r) rdr
0
aλn
0
x
J0 (x) dx
λ2n
(x = λn r)
Z aλn
d
2v0
(xJ1 (x)) dx
3 2
2
dx
a kλn J1 (λn a) 0
2v0
aλn J1 (aλn )
3 2
2
a kλn J1 (λn a)
2v0
.
2
akλn J1 (λn a)
Thus
u (r, t) =
∞
X
2v0
J0 (λn r) sin kλn t.
2
akλn J1 (λn a)
n=1
(b) The boundary conditions are u (r, 0) = f (r) = a2 − r2 , ut (r, 0) = 0.
As above the solution is given by
u (r, t) =
∞
X
(An cos kλn t + Bn sin kλn t) J0 (λn r)
n=1
with
An =
=
2
a2 λ2 J12 (λn a)
2
4 2
2
a λn J1 (λn a)
Integration by parts u = a2 λ2n − x2
dv = xJ0 (x) dx
a
Z
0
Z
0
a2 − r2 J0 (λn r) rdr
aλn
a2 λ2n − x2 J0 (x) xdx
du = −2xdx
v = xJ0 (x) .
(x = λn r)
(7.93)
202
We have
An =
=
=
=
=
2
a2 λ2n
a2 λ4 J12 (λn a)
4
Z
a2 λ4 J12 (λn a)
4
aλn
0
−x
2
xJ1 (x)
aλn
0
Z
aλn
2
x J1 (x) dx
+2
0
d 2
x J2 (x) dx
dx
aλn
x2 J2 (x)
a2 λ4 J12 (λn a)
0
4
J2 (aλn )
λ2 J12 (λn a)
8
(form (7.30))
3 2
λ J1 (λn a)
And Bn = 0.
Thus
u (r, t) =
∞
X
8
λ3 J12 (λn a)
n=1
J0 (λn r) cos kλn t.
■
Exercises 7.7
1. Exercises 4.2 of [1] pp 205–207, Problems 1–12.
2. Exercises 4.3 of [1] pp 214–216, Problems 1–7, 10–13.
3. Exercises 4.4 of [1] pp 223–228, All problems.
4. Exercises 4.5 of [1] pp 230–231, All problems.
203
CONTENTS
Appendix A
THE GAME AND BETA FUNCTIONS.
THE ERROR FUNCTION
A.1
A.1.1
THE GAMMA FUNCTION
Definition: Improper Integrals of The Second Type
Suppose that the function f is Riemann–integrable over [a, b] for each b in [a, ∞]. If the limit
Z b
lim
f (x)dx exists, then f is said to be improperly Riemann-integrable over [a, ∞). We write
b→∞ a
Z ∞
this limit as
f (x)dx and call the integral a convergent improper integral of the second type.
a
A.1.2
Definition: The Gamma Function
For positive p the Gamma Function is defined by
Z ∞
Γ(p) =
xp−1 e−x dx,
p > 0.
(A.1)
0
In this expression the integral is convergent, so that for p > 0 the Gamma Function is a convergent
improper integral of the second type.
We show that
Z ∞
xn e−x dx = n!
for n = 1, 2, 3, · · ·
0
Using integration by parts, we obtain
Z ∞
xn e−x dx = −xn e−x
0
=
From
lim −
x→∞
∞
0
xn
ex
Z
+n
∞
xn−1 e−x dx
0
− nx
∞
n−1 −x
e
0
+ n(n − 1)
Z
0
xn
=0
x→∞ ex
lim
and by repeating the process, it follows that
Z ∞
Z
n −x
x e = n!
0
0
∞
e−x dx = n!e−x
∞
0
= n!.
∞
xn−2 e−x dx.
204
R∞
The Gamma function Γ(p) = 0 xp−1 e−x dx is thus an extension of the factorial function, and for p an
integer, p ≥ 1 the relationship Γ(p) = (p − 1)! holds.
For non-integer p, 1 < p < 2, the value of Γ(p) is obtain from tables.
A.1.3
Recursion Formula
Theorem A.1.1
The following recursion formula holds:
Γ(p + 1) = pΓ(p).
Proof:
Z
∞
Γ(p + 1) =
xp e−x dx
0
p −x
∞
Z
−x e
−
0
0
Z ∞
= p
xp−1 e−x dx
=
∞
−e−x pxp−1 dx
0
= pΓ(p − 1)
p > 0.
.
A.1.4
■
Definition: The Gamma Function for p < 0.
1
Γ(p) = Γ(p + 1),
p
p < 0.
205
A.1.5
APM3701/1
Graph of the Gamma Function
Theorem A.1.2
√
1
Γ
= π
2
Proof:
Z ∞
Z ∞ 1
−
1
1
−x
2
√ e−x dx.
Γ
=2
x e dx =
2
x
0
0
Put x = y 2 . Then dx = 2ydy and
Z ∞
Z ∞
1
2
2
Γ
=2
e−y dy = 2
e−x dx.
2
0
0
(A.2)
Consequently
2
Z ∞
2
2
1
=4
Γ
e−(x +y ) dxdy.
2
0
Put x = r cos θ, y = r sin θ. Then
2
Z πZ ∞
1
2
Γ
= 4 2
e−r rdrdθ
2
0
0
2
=
i.e.
.
4π e−r
2 −2
∞
= π,
0
√
1
Γ
= π.
2
■
206
A.2
THE BETA FUNCTION
A.2.1
Definition
Suppose that f is Riemann-integrable over [a + ε, b] for each ε > 0, while f is unbounded over (a, b]. If
Rb
lim a+ε f (x)dx exists, this limit is called the improper Riemann integral of the first type of f, and we
ε→0
Rb
denote the integral by a f (x)dx.
A.2.2
Definition The Beta Function
For p, q ∈ (0, 1) the Beta Function B is defined by
Z 1
B(p, q) =
xp−1 (1 − x)q−1 dx,
0 < p < 1,
0 < q < 1.
0
With these choices of p and q the Beta Function is a convergent improper integral of the first type.
A.2.3
Alternative forms for the Beta Function
A.2.3.1
The trigonometrical form
Z π
B(p, q) = 2 2 (cos θ)2p−1 (cos θ)2q−1 dθ.
0
This formula is deduced from the definition by putting x = sin2 θ.
A.2.3.2
Improper integral of the second type
Z ∞
B(p, q) =
0
xp−1
dx.
(1 + x)p+q
This formula is obtained from the definition by substituting x =
A.3
y
.
1+y
RELATIONSHIP BETWEEN THE GAMMA AND THE BETA FUNCTIONS
Theorem A.3.1
B(p, q) =
Proof
R∞
Put t = y 2 in Γ(p) = 0 tp−1 e−t dt.
dt
Then
= 2y and
dy
Z
∞
Γ(p) = 2
Γ(p)Γ(q)
Γ(p + q)
2
y 2p−1 e−y dy.
0
In a similar way
Z
Γ(q) = 2
∞
2
y 2q−1 e−x dx.
0
Consequently
Z
∞Z ∞
Γ(p)Γ(q) = 4
0
0
x2p−1 y 2q−1 e−(x
2 −y 2
) dxdy.
207
APM3701/1
By using polar coordinates, we obtain
Z πZ
Γ(p)Γ(q) = 4 2
0
Z
= 2
∞
2
(r cos θ)2q−1 (r sin θ)2p−1 e−r rdrdθ
0
∞
r
2(p+q−1)
−r2
re
0
Z π
dr · 2 2 (cos θ)2q−1 (sin θ)2p−1 dθ
0
= Γ(p + q)B(p, q)
by putting x = r2 .
A.4
■
APPLICATION OF GAMMA AND BETA FUNCTIONS.
Many integrals can be calculated by using the Gamma and Beta Functions.
Example A.4.1
Evaluate
Z
0
1
√
Solution
Put x2 = y. Then
I =
=
=
=
=
=
x4
dx = I.
1 − x2
3
1
Z
−
1 1
2
y (1 − y) 2 dy
2 0
1
5 1
B
,
2
2 2
1
5
Γ
Γ
1
2
2
2
Γ(3)
3 √
3
Γ
π
12
2
2
2
31
1 √
Γ
π
82
2
3π
.
16
♦ Note: Evaluate I by using first year calculus:
Hint: Use the substitution x = sin2 θ. The integral is the reduced to
1
4
Z π
2 1 − 2 cos 2θ + 1 + cos 4θ dθ =
2
0
=
π
1 3θ
sin 4θ 2
− sin 2θ +
4 2
8
0
3π
.
16
■
208
Example A.4.2
Evaluate
Z π
2 sin4 θ cos5 θdθ = I.
0
Solution
By recalling the trigonometrical form of the Beta function, we have
5
Γ
Γ(3)
1
5
1
2
B
,3 =
11
2
2
2
Γ
2
1 3 5 √
· · · π
2 2 2
9 7 5 3 1√
· · · ·
π
2 2 2 2 2
8
.
315
I =
=
=
■
Example A.4.3
1
(2n)! √
1
(2n + 1)! √
Show that Γ n +
= n
π and Γ n + + 1 = n+1
π.
2
2 n!
2
2
n!
Solution
From (A.1) we have
1
Γ n+
2
∞
Z
=
0
Z
= 2
tn−1/2 e−t dt,
∞
let x2 = t, we obtain
2
x2n e−x dx.
0
We integrate by parts, and obtain
1
Γ n+
2
=
x2n−1 e−x
2
∞
+ 2 (2n − 1)
0
1
= 2 (2n − 1)
2
Z
∞
x
0
2n−2 −x2
e
1
2
Z
dx.
0
∞
2
x2n−2 e−x dx
209
APM3701/1
2
Since lim x2n−1 e−x = 0, repeating the process, we obtain
x→∞
Z
1 ∞ 2n−2 −x2
1
= 2 (2n − 1)
x
e dx
Γ n+
2
2 0
Z
1 ∞ 2n−4 −x2
= 2 (2n − 1) (2n − 3) 2
x
e dx
2 0
Z
1 ∞ 2n−6 −x2
= 2 (2n − 1) (2n − 3) (2n − 5) 3
x
e dx
2 0
Z ∞
1
2
x2n−8 e−x dx
= 2 (2n − 1) (2n − 3) (2n − 5) (2n − 7) 4
2 0
..
.
Z ∞
1
2
x2n−2k e−x dx
= 2 (2n − 1) (2n − 3) (2n − 5) (2n − 7) · · · (2n − (2k − 1)) k
2 0
..
.
Z ∞
1
2
= 2 (2n − 1) (2n − 3) (2n − 5) (2n − 7) · · · (2n − (2k − 1)) · · · (2n − (2n − 1)) n
x2n−2n e−x dx
2 0
Z ∞
1
2
= 2 (2n − 1) (2n − 3) (2n − 5) (2n − 7) · · · 3 × 1 × n
e−x dx
2 0
√
1
=
(2n − 1) (2n − 3) (2n − 5) (2n − 7) · · · 3 × 1 × π
n
2
√
2n (2n − 1) (2n − 2) (2n − 3) (2n − 4) (2n − 5) (2n − 6) (2n − 7) · · · 4 × 3 × 1 π
=
2n (2n − 2) (2n − 4) (2n − 6) · · · 4 × 2
2n
√
π
(2n)!
=
n
n
2 n (n − 1) (n − 2) (n − 3) · · · 2 × 1 2
(2n)! √
=
π.
22n n!
And
1
Γ n+ +1
2
1
1
=
n+
Γ n+
2
2
2n + 1 (2n)! √
=
π
2
22n n!
(2n + 1)! √
π.
=
2n+1 n!
■
A.5
THE ERROR FUNCTION
This function is often used in Probability Theory.
A.5.1
Definition: The Error Function erf
The Error Function erf is defined for 0 ≤ x ≤ ∞ by
2
erf(x) = √
π
Z
x
2
e−t dt.
0
2
The Error Function is the area under a section of the curve y = e−t , which has graph
210
The value of erf are obtained from tables.
Theorem A.5.1
erf(∞) = 1
Proof
erf(∞) =
=
Z ∞
2
2
√
e−t dt
π
√ 0
π
√ = 1.
π
■
211
BIBLIOGRAPHY
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