is written in this grand book, the universe, which stands
open to our gaze. But the book cannot be understood unless one
leams to comprehend the language and read the letters in which it
composed. It is written in the language of mathematics, and its
are triangles, circles and other geometric figures without which
is humanly impossible to understand a word of it; without these one
wanders nbout in a dark labyrinth.
GeI-neo Gnr-rnr
-
words of Galileo are just as true today as when he wrote them in the lTth century.
serious student of physics must first learn the language of mathematics. This is a more
task for today's student than it was for Galileo. The characters include integrals,
goups, tensors, and other concepts unheard of in Galileo's time. A physics major
also faced with a list of physics courses that may include not only the mechanics that
studied, but also electromagnetic theory, quantum mechanics, and relativity. With
push to include more physics in the undergraduate major, there is even less time to study
We usually attempt to remedy this by putting a lot of mathematics instruction
physics
the
classes, introducing the mathematical tools as they are needed.
graduate
A beginning
student usually needs to gather together tools that have been accrued
and develop a deeper understanding of their use. She or he also needs to review
reinforce material learned in undergraduate mathematics classes and become more
in using mathematical tools. That is the purpose of this book. A student who has
the material in this book should be prepared for graduate classes in mechanics,
mechanics, or electricity and magnetism.
The text is an outgrowth of a course at San Francisco State University that is taken by
in their final year of a physics BS or, even more commonly, by incoming graduate
These students have taken the usual undergraduate courses in mathematics: calcuvector calculus, differential equations, and linear algebra. They have also already taken
courses in mechanics and electricity and magnetism. While these students have
some of the material in the first few chapters of this book, they are rarely comfortable,
and accurate using these tools. At San Francisco State, incoming graduate stuwho already have a degree in physics take a placement test, which shows that most
them cannot perform vector calculations correctly, especially when asked to abandon
ian components! Thus, the early chapters include some basic material as a review
vtl
vilt
PREFACE
and for completeness. At the same time, this material can now be approached in a more
sophisticated manner.
There are many books that discuss the mathematics that graduate students in physics need
to know. Some of them are extensive reference works in two volumes. Every student needs
one of these books, but not as a textbook. Most textbooks cover too much material for a
one-semester course, and most of them are too mathematical in their approach for today's
students, who usually do not have an extensive undergraduate preparation in mathematics.
The goal of this book is to offer a one-semester course umed at physics students-a book
that provides students with the tools that they need and shows them how to use those tools
in physics problems. Because these students have some experience in physics, I draw on
that knowledge in examples and exercises.
The book is organized as follows. The eight chapters cover the basic material that I
have taught every year. Then five optional topics provide additional material, from which
instructors may select to round out the semester or which students may read on their own.
Some of the chapters include material normally covered somewhere in the undergraduate
curriculum, but pushed to more depth and/or breadth. Instructors may wish to skip some
of this material or assign it as required reading. Chapter 1 (on vectors and matrices) and
Chapter 3 (Differential Equations) fall into this class. Some chapters cover material not
normally explored in detail until the graduate program. Chapter 6 (on generalized functions)
and Chapter 7 (Fourier Transforms) are in this category. The optional topics are at a more
advanced level than the material in the chapters. Within each chapter, the material becomes
progressively more difficult. Instructors who want a lower-level presentation can simply
omit the final sections of each chapter and the optional topics. Instructors with more wellprepared students may want to use the early sections of each chapter as assigned reading
and concentrate on the more difficult material in the later sections.
Material introduced in one chapter is reinforced in later chapters. Thus, Chapter 2 (Complex Variables) uses material on vectors (Chapter 1); the discussion of inverse transforms
(Chapters 5 and 7) makes extensive use of contour integration (Chapter 2); and Chapter
7 (Fourier Transforms) uses material on delta functions (Chapter 6). Material on special
functions (Chapter 8) draws on series solutions of differential equations (Chapter 3) and
prior experience with Fourier series (Chapter 4) and Fourier transforms (Chapter 7). After
a tool such as the Fourier transform has been introduced, I immediately give examples of
how that tool is used in the solution of physics problems. Even though there is no single
chapter labeled "Partial differential equations," many techniques and examples involving
the solution of PDEs are presented throughout the text.
Mathematical proofs are kept to a minimum in the text, but many have been placed in the
appendices. Students who plan a career in theoretical work will need to study the proofs
in more detail at some time. My goal here is to show students how to use the mathematics
to get results (see, for example, the sections on using contour integration in Chapter 2).
Similarly, many of the examples are applications in physics.
I debated whether to include numerical methods in this text. I finally decided to omit
computer applications almost entirely. (There is a brief discussion of numerical solution of
differential equations in Chapter 3 and some discussion of computer methods in solving
matrix equations in Chapter l.) Of course, in the 21st century every physics student needs
competence in using a computer, and many of the problems suggest using a computer for
PREFACE
IX
graphics or numerical answers. But the student also needs to understand what the computer
is doing, and why. And that same student still needs to have the basic mathematical tools
and to understand why and how they work. The computer material belongs in a different
course, and in a different book.
A Student Solutions Manual, containing solutions to about 25 percent of the problems in
the book, is available for sale to students. A box around the problem number in the text indicates a problem solved inthe Student Solutions Manual. An Instructor's Solutions Manual,
containing solutions to all of the problems, is available to instructors. For further information about either of these ancillaries, instructors should contact their Cengage Learning
sales representative.
ACKNOWLEDGMENTS
The material has been extensively class
tested-with
a
favorable response-at San Francisco
University, and I thank the many graduate students who have taken this class and have
shaped my vision of what this book should be. I am also grateful to my colleagues at SFSU
who read the numerous drafts of the text. Special thanks go to Dr. John Burke (Physics and
Astronomy Department) and Dr. David Ellis (Department of Mathematics). My department
chair, Dr. James Lockhart, also provided useful criticism of the manuscript and supported
my efforts. I would also like to thank the following reviewers for their helpful comments
on various drafts of the manuscript:
State
Albert Altman,
U niv
e
r s ity
of M as s achu s ett s, Low
e
ll
Giles Auchmtty, University of Houston
Thomas Beatty, Florida Gulf Coast University
Paul L. DeVries, Miami University
Nevin Daniel Gibson, Denison University
Porter W. Johnson, Illinois Institute of Technology
David Kastor, University of Massachusetts, Amherst
Igor Ko goutio uk, M inne s o t a S tat e U niv e rs ity -M ankat o
Daniel
P.
Lathrop, University of Maryland
Romulo Ochoa, The College of New Jersey
Daniel Phillips, Ohio University
George R. Plitnik, Frostburg State Universiry
Asok K. Ray, University of Texas at Arlington
John F. Reading, Texas
A&M University
Peter S. Riseborough, Temple University
Sergei Shandarin, University of Kansas
Charles Stanton, CSU San Bemardino
Krzysztof Szalewicz, University of Delaware
Robert L.Zimmerman, University of Oregon
his love and suPPort'
Contents
Preface
1
vii
Describing the Universe
I
1.1 A Universal Language
1.2 Scalar and Vector Fields 17
1.3 Curvilinear Coordinates 28
1
'1.4 The Helmholtz Theorem
1.5 Vector Spaces 37
1.6 Matrices 41
Problems
2
37
65
Complex Variables 75
2.1 AllAbout Numbers 75
2.2 Functions of Complex Variables 89
2.3 Complex Series 101
2.4 Complex Numbers and Laplace's Equation 112
2.5 Poles and Zeros 117
2.6 The Residue Theorem 122
2.7 Using the Residue Theorem 127
2.8 Conformal Mapping 150
2.9 The Gamma Function 156
Problems
3
160
Differential Equations 169
3.1 Some Definitions 169
3.2 Common Differential Equations Arising in Physics 170
3.3 Solution of Linear, Ordinary Differential Equations 174
3.4 NumericalMethods 197
3.5 Partial Differential Equations: Separation of Variables 208
Problems
4
Fourier
4.1
211
Series
Fourier's
219
Theorem
219
XI
xll
CONTENTS
4.2
4.9
Finding the Coefficients 22o
Fourier Sine and Cosine Series 227
4'4UseofFourierSeriestoSolveDifferentialEquations234
4.5
Convergence of Fourier
Problems
Laplace
Series
24o
243
T[ansforms
251
5.1 Definition of the Laplace Transform 251
5.2 Some Basic Properties of the Transform 253
5.3 Use of the Laplace Transform to solve a Differential
5.4 Some Additional UsefulTricks 261
5.5 Convolution 265
5.6 The General lnversion Procedure 269
5.7 Some More PhYsics 273
Problems
Equation
279
Generalized Functions in
Physics
287
6.1 The Delta Function 287
6.2 Developing a Theory of Distributions
6.3 Properties of Distributions 307
6.4 Sequences and Series 310
6.5 Distributions in N Dimensions 313
305
6'6DescribingPhysicalQuantitiesUsingDeltaFunctions315
6.7
The Green's
Function
Problems
318
317
Fourier Tbansforms 323
7,1 Definition of the FourierTransform 323
7.2 Some ExamPles 325
7.3 Properties of the Fourier Transform 329
7.4
7.5
7.6
7.7
Causality
Problems
Sturm-Liouville
8.1
8.2
334
Use of FourierTransforms in the Solution
of Partial Differential Equations 336
Fourier Transforms and Power Spectra 341
Sine and Cosine Transforms 343
351
TheorY
The Sturm-Liouville
3s7
Problem
357
Use of Sturm-Liouville Theory in
Physics
363
E.3ProblemswithSphericalsymmetry:SphericalHarmonics36T
8'4ProblemswithCylindricalsymmetry:BesselFunctions39T
CONTENTS XIII
8.5 Spherical Bessel Functions 419
8.6 The ClassicalOrthogonal Polynomials 425
Problems
428
Topics 439
Tensors 439
.
Optional
A
A.1 Cartesian Tensors 439
A.2 lnner and Outer Products 444
A.3 Pseudo-tensors and Cross Products
A.4 GeneralTensor Calculus 447
A.5 The Metric Tensor 451
A.6 Contraction 452
A.7 Basis Vectors and Basis Forms 453
A.8 Derivatives 455
Problems
B
Group
B.1
445
460
Theory
465
Definition of a Group 465
Examples of Groups 466
8.2
8.3 Classes 470
8.4 Subgroups 471
8.5 Cyclic Groups 472
8.6 Factor Groups and Direct Product Groups
8.7 lsomorphism 473
8.8 Representations 474
8.9 Generators of Groups 484
8.10 Lie Algebras
Problems
C
Green's
487
490
Functions
495
C.1 Division-of-Region Method 496
C.2 Expansion in Eigenfunctions 499
C.3 Transform Methods 503
C.4 Extension to N Dimensions 504
C.5 lnhomogeneous Boundary Conditions 510
C.6 Green's Theorem 512
C.7 The Green's Function for Poisson's Equation
in a Bounded Region 514
Problems
D
Approximate Evaluation of
D.1
D.2
Integrals
The Method of Steepest Descent 531
The Method of Stationary Phase 535
Problems
L-_
526
537
s31
472
CONTENTS
E
Calculus of
Variations
541
E.1 lntegral Principles in Physics 541
E.2 The Euler Equation 544
E.3 Variation Subject to Constraints 548
E.4 Extension to Functions of More Than One Variable
Problems
551
552
Appendices
I
ll
lll
lV
V
sss
Transformation Properties of the Vector Cross Product 555
Proof of the Helmholtz Theorem 556
Proof by lnduction: The Cauchy Formula 559
The Mean Value Theorem for lntegrals 560
The Gibbs Phenomenon 562
The Laplace Transform and Convolution 565
vl
Proof That
Vl
ri@):
(-1)m(1
- *zynffiPrfu)
Vlll Proof of the Relation [f, oJ^(kdJ^(k'p)
lX The Error Function 571
X
Xl
:
1
;6(k
- k)
56e
Classification of Partial Differential Equations 574
TheTangent Function:A Detailed lnvestigation of Series Expansions
Bibliography
Index
dp
568
s8s
Mathematics for Physicists
i
t
i
t
CHAPTER
1
Describing the Universe
1.1. A UNIVERSAL LANGUAGE
The wonder of physics is that its laws are universal. So far as we can tell, the same laws
describe the behavior of things everywhere in the universe. We need a language for these
laws that is equally universal; that language is mathematics. The laws of physics are written
using mathematical terms that are independent of the reference frame that we use; that is, the
laws shouldbe independentof the coordinate systemthatwe choose, and also of any uniform
motion of the reference frame. Ensuring that all physical laws obey the velocity rule was
Einstein's inspiration for the theory of special relativity. Here, we'll begin by considering
the first constraint-that the physical laws be stated in a form that is independent of the
coordinate system used.
1
1.1.1. Common Coordinate Systems
Coordinate systems are used to describe the position and orientation ofobjects in space. The
three coordinate systems in most common use are the Cartesian, cylindrical, and spherical
coordinate systems.
Cartesian Coordinates
The Cartesian coordinate system is the familiar x, ), z system (Figure 1.1a). The coordinate
axes are mutually perpendicular and are generally chosen to be right-handed. That is, ifyou
pick a point to be the origin and then choose the orientation of two of the axes, the third
axis is determined by the right-handed convention.
I
L-_
See
Optional Topic A for more on the mathematics of special relativity.
CHAPTEF 1 DESCRIBING THE UNIVERSE
a point is described by its perpendicular distances from three mutually orthogonal planes, as shown here'
FIGURE 1.1a. Cartesian coordinates: the position of
The distance ds between two neighboring points with coordinates
y I dy, z + dz is given by Pythagoras'theorem (Figure 1'1b):
ds2:dx2+dy2+dz2
I,
y, z and x
I
dx,
(1.1)
using the
FIGURE 1..1b. The distance between two neighboring points (the line element) is found
PYthagorean theorem.
system'
The absence of cross terms like dx dy is characteristic of an orthogonal coordinate
give
We'll
perpendicular'
mutually
are
axes
the
because
The Cartesian system is orthogonal
1'3'1'
in
Section
a more careful definition of orthogonality
Cylindrical Coordinates
Thecylindricalcoordinatesarep,thedistancefromthez-axis; Q,ananglemeasuredcoun(Figure 1.2a).
terclockwise from a reference line (usually the positive x-axis); and z
1.1 A UNIVERSAL LANGUAGE
FIGURE 1.2a. Cylindrical coordinates:
the position
of a point is described by its perpendicular distance p from one coordinate axis, its perpendicular distance
z from a plane perpendicular to that
axis, and an angle. To obtain tbe an-
gle
reference
line
{,
we drop a perpendicular from
: 0, draw a line
from the origin to the foot of the perpendicular, and measure the angle @
between that line and a reference line
in the plane. Traditionally angles are
measured counterclockwise from the
reference line.
P to the plane z
Zp
........-...-'....-y
x-pcosQ, y-psinQ
(r.2)
p: \F+ y', Q:tan-ryx
(1.3)
and, conversely,
This is an orthogonal coordinate system because the unit vectors p (pointing outward from
the e-axis, parallel to the x-y plane), $ (perpendicular to p, pointing in the direction in which
{ increases), andh are mutually perpendicular. Most of the coordinate systems commonly
used in physics are orthogonal coordinate systems.
The distance ds between two neighboring points with coordinates p , Q , z afi p -f dp , Q *
dQ, z* dz may be found by recognizing that the distance along an arc of a circle with radius
p and angle dQ is p dQ (Figure 1.2b). Then, from Pythagoras'theorem,
h2
:
dp2
+
p2 dOz
+
dz2
(1.4)
FIGURE 1.2b. The distance between two neighboring points is found using the Pythagorean theorem.
The distance corresponding to the coordinate difference dQ is p dQ.
CHAPTER
1 DESCRIBING THE UNIVERSE
Spherical Coordinates
the origin; g' the angle
a point P arc r,the distance from
an angle measured counterusually the positive z-axis); and @'
1'3a)'
(Figure
Hne (fositive x-axis)
FIGURE1.3a.
polar axis
drical coordinates'The polar angle 0
to
is measured from the polar (z) axis
the Position vector of the Point'
reference
line
x
ln terms of Cartesian coordinates'
0
and
:
:
CoS
tZ
'-'
r
r sin0 sind,
Q:tarr-rl- z: r cos0
(1.s)
(1.6)
of radius r sin 0
dQ' the poinl P moves around a circle
When we increase the angle Q by
and so
*"o"gtt a distance r s1n0 dQ (Figure 1'3b)'
(1.7)
d'2 : dr2 + 12 doz + 12 sin2 e dO2
FIGUREl.3b.ThedistancebetweentwoneighboringpointsisfoundusingthePythagoreantheorem.
differences d0 and d'Q are r d0 and
co'ordinate
The distances conesponding to the
r sin9 dQ'
1.1 A UNIVERSAL LANGUAGT
1.1.2. Representing Physical Laws2
Physical laws are represented by equations containing mathematical quantities that are independent of the reference frame or coordinate choice we use to describe them. We can write
each physical law in a coordinate-independent way. For example, we may write Newton's
second law without reference to specific coordinates:
i : ^d.: di
dt
(
1.8)
The quantities appearing in this equation are scalars, vectors, and differential operators. No
coordinates appear explicitly in the equation. (In special relativity theory, we also demand
independence of inertial reference frame, and the time / becomes a fourth coordinate. Then
we have to write Newton's second law somewhat differently. For now we are considering
only coordinate independence and three spatial coordinates.)
Scalars
Scalars are mathematical quantities represented by a single number that is independent of
the coordinate system used. Examples are mass of a particle, electric charge of a particle,
and distance between two points. The quantity m in equation (1.8) is a scalan
Vectors
Vectors are geometrical objects (arrows) with magnitude and direction. Thus, equation (1.8)
is a relation between two arrows F and d. These two vectors must have the same direction,
and their magnitudes are related by the scalar ru. Vectors are added geometrically by placing
them head to tail. The sum has its tail at the tail of the first vector in the sum, and its head
at the head of the last vector in the sum.
Some problems can be solved by using the vector representation (1.8). But often we
find it easier to set up a coordinate system and work with the individual components of
the equation. In any particular coordinate system, a vector is represented by three numbersthe three components of the vector. The vector may be written
i : (ur, uy , ur)
or, using index notation, ut, where i : 1,2,3 represents the -r, y, or z component of the
vector. In this representation, we regard the vector i as the sum of three vectors, each
parallel to one of the coordinate axes. The magnitude of each of these vectors is given by
the magnitude of the corresponding component, while the sign of the component indicates
the direction (in the direction of increasing or decreasing the coordinate).
When we change the coordinate system, these three numbers (ur,uy, and u.) change.
But they must change in a specific way in order for equation (1.8) to remain true. In
order to see how the components change, let's restrict attention to Cartesian coordinate
2In this section
it is assumed that the reader has had some previous experience with matrices. Readers who have
not should read Sections L6.1-1.6.3 now.
L
CHAPTER
1 DESCRIBING THE UNIVERSE
systems.3 The two ways to change the coordinates and still maintain a rectangular system
are to (a) move the origin and (b) rotate the axes. Change of origin affects the position vector
but does not affect vectors like and d. Thr s, the important changes are rotations of the
i
i
coordinate axes.
Suppose that the system of coordinates xt , J' , z/ is obtained from the original system
x, y, zby rotating counterclockwise about the z-axis through an angle 0. Futher, suppose
that a vector i has components (ur, uy, uz) in the original coordinate system. Let's find the
components in the new system.
The
x- and y-components may be constructed geometrically by projecting the vector
onto the x-y plane and dropping perpendiculars from the end of the projected vector to the
coordinate axes, as shown in Figure 1.4. Thus,
Ur,:OAIAB
: -1D. * (uu cos d
: fa(r
cosd'
_
u, tand) sinO
sin2 o)
*
u, sin d
:urcos9*ursind
FIGURE 1,4. Rotation of coordinate
(1.e)
axes. Here the prime axes are obtained from the unprime axes
about the z-axis. OC is the projection
by rotating counterclockwise through an angle
of the vector onto the x-y Plane.
I
i
Similarly,
ur'
:
BC
(uy
-
u, tan0) cos0
--uxsing*urcos0
(1.10)
3In any coordinate system we need three numbers to completely describe a vector, but in a non-Cartesian system
the description is more complicated. See, for example, Problem 10 and Optional Topic A.
1.1 A UNIVERSAL LANGUAGE
uz'
Each new component is a linear
are
:
uz
combination of the old components. The two representations
related by a set of nine numbers Arj:
3
,!:DAijuj
(1.il)
i-l
where
:
cos
Aij
:
0,
At2
: sin9,
Azt
: - sin?,
A22
: cos9,
Azz
:
I
O.We may write these relations using matrix multiplicationa:
:(-r:i'"ifiJ;""13):(
,g il? lX,,
) ,,2,
i' :,Ai
(1.
l3)
(1.14)
the
rotation matrix that describes the coordinate transformation. This matrix allows us to
the components of any vector in the new coordinate system.
1,2,3. Similarly, the components
We write the vector components 4s ui,
ffix .A are the numbers A;7, where i labels the rows and labels the columns5 :
i :
j
Atz
Ars
Azz
Azz
x:1
of the
\
(1.1s)
)
Section 1.6.1. To perform the matrix multiplication, take the dot product of the nth row of the matrix with
vector expressed in the original components to obtain the nth new component.
is the most commonly used convention
for labeling the matrix components.
CHAPTER 1 DESCRIBING THE UNIVERSE
This matrix has several nice properties:
as a vector, is orthogonal to every other
1. It is orthogonal; that is, each row, considered
row:
t
A;iApi
I
1.i1'+;1,:
:lip
(1.16)
j
The same is true of the columns:
ty
(1.17)
T
In these expressions,
6;7.
is the Kronecker delta, which equals 1 if
I : k and zero
otherwise.
2. The determinant6 of the matrix is *1.
3. Because of properties (1) and (2),the inverseT of matrix.A equals its transpose:
/
cos?
A_,_n':(;"0
- sind
cos0
0
0
(
1.18)
01
Matrices that share the properties (1) and (3) but with determinant - I represent a transformation from a right-handed coordinate system to a left-handed system; that is, they
represent a combination of a rotation and a reflection.
The matrix multiplication may also be written using index notation, as in equation (1.11).
!.
Whenever
a commonly used shorthand calledthe summation convention, we drop the
an index is repeated once, we understand that we are to sum over that index. That is,
In
3
u'i: Aijuj
means
u'r:lli1u1
j:r
The index I in this expression is not summed; it is called a free index, and it must appear
once on each side of the equation. When using the summation convention, it is important to
remember that, in any one term, an index may not be wittenmore thantwice. For example,
A j ju j, with j repeated three times, is meaningless.
The identity matrix has components
':(i:l)
6See Section 1.6.2.
TSee Section 1.6.3.
(1.19)
1.1 A UNIVERSAL LANGUAGI
and is written in index form as E;7. Multiplication by the identity matrix leaves a vector
unchanged. In index notation, multiplication by the Kronecker delta 6;; simply replaces the
index label
I with l:
8;iu1
:
(1.20)
1t;
We may express the matrix ^A in terms of the angles between the old and new axes. Let
04 be the angle between the lth new axis and the 7th original axis. Then
Aij : cosoij
For the example above (rotation through 0 about the z-axis), 0n :
gn r 12 * 0, 022 : 0, and 0n : 0. The angles 9zj : n 12 fot j :
--
(r.2t)
e,0n : rl2 l, ).
0,
Invariance of Vector Equations
Now let's look at Newton's law again. In the first coordinate system, we may write each
component of the equation as
Fi
:
(1.22)
mai
Each of the vectors may now be transformed to the new system. The new components are
F!
: AijFj
(r.23)
and
a'i: Aiiai
of these relations to obtain
tion (1.23) on the left by the matrix .A-1:
We may invert each
(A-1)*i Fi
Then use the results
:
F; in terms of F/. First multiply
(1.24)
equa-
(A-1)p;A;; F;
: Ar and that A-1.4 is the unit matrix:
(Ar)rifi : (A-r)7.;A;1F1 :3p1F1
that,A-l
and so
A;pF{
: lp
(t.2s)
Equation (1.25) is the inverse of relation (1.23) and transforms the components in the
primed system to the original system. We can use this relation to transform both vectors in
equation (1.22):
AliFl
h--
:
mAiiati
10
cHAprER 1 DEScRIBING THE UNIvERSE
A1i(F1'-ma',):O
Multiply on the left
bY A:
Ap;A1i(F!
-ma',):O
Use equation (1.16):
Sri@i
-
ma',)
:0
Then use property (1.20):
Fi,
- ma|r:
O
if it is true in the original system.
(1.23) is a necessary consequence of the
law
transformation
the
that
see
we
Thus,
of coorgeometrical nature of vectors and the fact that vector relations are independent
a
vector'
of
dinate rotations. The transformation law (i.23) becomes part of the definition
So the equation is true in the primed system
by a set
A vector in three-dimensional space is represented in Cartesian coordinates
(1.23) when
to
equation
according
transform
that
(its
components)
of three numbers
the coordinate axes are rotated.
BA I,AB
and then
ComPare
t the new
Example
about the
it with th
order)'
z-axis (that is, the same rotations but performed in the reverse
: 033' )zt
Using rule (1.2I), we have for the first rotation 0n : 0
1ij
: r
l2'Thtts'
A_
(ril)
rder in which the matrices
the center: (B'N)kj : Bkt
Placement of the indices'
matrix multiPlication' we
repeated indices next to each
other
A
i
i Bp;
:
Bp; Ai
i = (lEA)1;'
: z' all other
1.1 A UNIVERSAL
For the second rotation about the new x-axis, 0zz
9ij
:
: n /2.5o
/t
u:lB
O
LANGUAGE
: 0t,0i2 : n, all other
o o\
-?
I)
The two successive rotations are represented by the matrix product:
u^:/l
o o\/ o
I o\ /o I o\
\o -? lJ(-; s ?J:l? s;J
A unit vector along the x-axis with components (1,0,0) in the original system
has
the following components in the rotated system:
(rii)(i):(l)
That is, it lies along the new z"-axis. Similarly, the y-axis has become the x"-axis,
and the z-axis has become the y//-axis (Figure 1.5).
FIGURE 1.5. The
set of rotations in'the first part of Example 1.1.
Doing the rotations in the opposite order, we get
*:(-; ii)(s
L--
?;):(l li)
11
12
cHAprER 1 DEScRtBtNGTHE UNIvERSE
This set of rotations sends .r to
-y't , y to -2"
, and z to
x" (Figure
1.6).
v
x'
FIGURE 1.6. In the second part of Example
1. 1,
the same rotations performed in the opposite order
give a different result.
However, two rotations about the same axis do commute. Thus, if we rotate about
the z-axis through an angle 01 and then again about the z-axis through dz, the result
is a rotation through 0t
]B.A:
-f
02.
cos91
sin91
sin
cos
91
(
00
(
I
r
(
sin(02 * 9r)
cos(dz * dr)
0
d1
cos 02 sin01
cos0z cos0r
t)
i)
*
sinOz
-
sin02sin91
cos9r 0
?
\
)
_AB
Multiplying Vectors
Product with a scalar A vector may be multiplied by a scalar. The result is another vector.
The product md in Newton's second law is an example of this kind of multiplication. The
new vector has the same direction as the original vector, but its magnitude is changed. The
ith component of the new vector equals the i th component of the original vector multiplied
by the scalar.
Scalar product of two
vectors Two vectors may be multiplied together to give either
scalar or a vector.9
9St
i.tly
speaking, this product is a pseudo-vector, as discussed below.
a
1.1 A UNIVERSAL
LANGUAGE
1
3
The scalar or dot producl oftwo vectors is
i.il:
(r.26)
uucos9
where u ardu arethe magnitudes of the two vectors
between them.
i
I
and fr, respectively, and
is the angle
Equivalently, we may write the product in terms of the Cartesian vector components:
v a:
(1.27)
uiui
where, as usual, the repeated index means that we sum over the index i. That is,
v. u :
s-3
Lr,r, :
Dlttl
I
u2u2
I utul
i:1
To show that this product is a scalar, let's start with the product in the prime frame and
transform to the unprime frame:
i.fr:
l)tpt;
:
AiiuiA;pu1,
Notice that we needed to be careful not to write the index -/ more than twice, so in transforming fr we called the summed index k. Now
AijAit : ,+!,Ait : A jrr Ai*:6 jr
and so
i.d:
Since the product
ulu.t;: A;1A;puiu1,: Sjrvjuk: t)kuk:
is the
same
in the two coordinate
?'d
systems,
it is a scalar. This
demonstration actually shows that the product is the same in any two coordinate systems, since the transformation represented by the matrix .A is an arbitrary rotation.
The dot product is also commutative:
i.il : i.i
The magnitude of a vector
i
may be written in terms of the dot product of
i
with itself:
u:lil :Jii
Examples of dot products in phy-sics include the definition of work
(dW
:
F ' aO' ttte
definitionofelectricflux(dO:fr,-ab,andtheexpressionsP:F'i:t'riforpower.
t--
14
CHAPTER
1 DESCRIBING THE UNIVERSE
or cross prodzct of two vectors is another vector
crossproduct or fuo vectors The vector
whose magnitude is given bY
l? x
il :
(1.28)
uu sin9
Thedirectionoftheproductisgivenbytheright-handrule,asshowninFigurel.Ta.
products. curl the flngers of the right hand from
FIGURE 1.7a. The right-hand rule for cross
towarJfr and the thumb gives the direction
i
i x i'
Its comPonents are given bY
(ixi)r:u2tt3-u3u2
(?xi)z:u3ttr-utu3
and
(ixfi)l:utuT-uzur
using the Levi-Civita symbol e;;1:
We may express the result in index form by
t-i
if i, j, k = an even Permutation of 1,2,3
it i, j, k= an odd Permutation of t, 2, 3
if any two of i, j, k are equal
(r.29)
sequence of numbers ijk --> kij ->
(To get an even pennutation' you may rotate the
--> ikj, obtained by
jki, but you may not interchange two of them. The pirmutation iik
Then
interchanging j and k'is an odd permutation')
(i x fr); :
Eijku juk
(1.30)
1.1 A UNIVERSAL LANGUAGI
15
Strictly speaking, the result is not a true vector. It transforms as a vector under coordinate
rotations but does not transform properly under reflections. (To see what this means, notice
that your right hand looks like a right hand however you turn it, but in a mirror it looks
like a left hand, as shown in Figure 1.7b.) The cross product is called a pseudo-vector. (See
Appendix I and Optional Topic A for the transformation law.)
FIGURE 1.7b. When viewed in a minor, the right hand becomes
a
left hand.
represents the parallelogram formed by the vectors i and fr. The
product
equals the area of the parallelogram, and the direction is
magnitude of the cross
parallelogram.
Examples of the cross product in physics include
plane
of the
normal to the
(t
:
i t Fl, angularrnot"nturn of a panicle (i : i x p)' and the Biot-Savart
torque
law for magnetic field (dB : /rgj x idV l4trr).
The cross product
i xi
vectors We may form a triple scalar product of three vectors by taking
the dot product of a third vector fr with the cross product fr x i. It is sometimes written
Prodacts of three
with square brackets as
td, n, fr'l
:
(fr x
i) 'fr': fr. (i x fr) :
(fr' x
i)
.i
(1.31)
why these three products are equal, note that each is the volume ofthe parallelepiped
with edges given by the three vectors fr, i, and fr lFigure 1.8). We may also derive the result
algebraically:
To see
tfr,n,*l
:
fr '
(i x fr) :
eijkuit)jwk
:
ekijuiujwk:
(fr x
i)' *
tt.32)
similarly for the third relation
For example, a particle in circular motion is acted on by a force F. The particle's angular
velocity is 6, and its linear velocity is i : 6 x i, where the vector i is the position vector
of the particle with respect to the center of the circle. The power delivered by the force F is
and
p
L-_
: F.i :F'
(dr
x i)
:
dr' (i
"
F;
:,;. t
CHAPTER
1 DESCRIBING THE UNIVERSE
of the parallelepiped with the three vectors
FIGURE 1,8. The triple scalar product equals the volume
along its edges.
of dot products:
The tripte vector producl may be expressed in terms
(i x i) x fr:
e;;7.(i x
i)itt':
eiikeihutumwk
(1.33)
To evaluate expression (1'33), we may use a nifty relation:
e
iti€ itm
:
6P13;a
- 5*^3it
(1.34)
index 7 in the same position in
To write relation (1.34) correctly, we first put the repeated
Then we pair up matching
both Levi-Civita symbols by a sequ"nc" oi "u"n permutations.
gives the positive term 6u3i^. Then we pair
indices in order (ft with I and i witt, -). This
negative term'
up the nonrepeated indices the other way to get the
all 3a possible combinations of
check
to
have
don't
(1.34).
We
equation
Let,s verify
k,i,l,andmseparate|ybecausewecanusethepropertiesoftheeand6symbolstoshow. if
First note that the e symbol is zero
that the relation is true for certain groups of values.
: i ot I : m' all three terms in the sum on
any two of its indices equal each oih"t, to \f k
: i or I : m,bothterms on the right-hand side are
the left-hand side are zero. Bot if k
identicalandtheirdifferenceiszero.Therelationistrueinthiscase.
Iftheindicesk,i'I,andruincludeallthevaluesl,2,and3,bothsidesalsovanish'side
k : I, i : 2, I : 2' and m : 3' Then on the left-hand
Suppose, for example, that
we have
eizeiZZ:0
: t323 :0' A quick check shows that in this
since e;rz : 0 unless i :3,but then ei23
on the right-hand side is also zero. So we
term
case one of the I(ronecker deltas in each
all cases except k -- I' i : m ot k : m'
zero-in
have verified the relation-both sides are
: The first of these possibilities (k : I, i : rn) gives
i
l.
e
jki€jrm:l elmeltm: *I
J
1.2 SCALARANDVECTOR
FIELDS 17
Note. In the middle term, we use the summation sign and do not use the summation
convention; that is, we do not sum over I and m.
Ontheright-handsideof equation(1.34),thefirstcombinationof
the second is zero. The other possibility is
e jki€ jtm
k
: s)
:
m,
i : l, which
eimteitm
6sis
1x 1: l,while
leads to
: -1
J
This time it is the second set of deltas on the right-hand side that equals 1, so the right-hand
side is also -1, as required, and relation (1.34) is proved.
Returning to the triple cross product (1.33), we have
(il x
i) x fr:
:
:
(i x i)
x fr
eijk€ jtmutl)muk
eikieilmulDmuk
(5u6i-
ukuiwk -
6p-6i1)u1uaw11
uiukwk
: ntd.m) - fr(n. n,)
(1.3s)
This result is sometimes known as the "bac-cab rule":
dxdxi;:6ta O-itd.6l
right, always start with the middte vector (i or 6 in the examples
above) times the dot product of the other two vectors. The second term has the other vector
To be sure you get the signs
inside the original parentheses (fr or i) times a dot product.
For example, the torque exerted by the magnetic force on a charged particle is
t: i
xF
: i,. q(i x E) : qli(i.il - iff ' tll
Tensors
Some physical laws cannot be represented using only vectors and scalars-they require an
object that is represented by more than three numbers. These objects are called tensors (see
Optional Topic A).
1.2. SCALAR ANDVECTOR FIELDS
in space. For example, a weather
map shows the air temperature in different cities. In principle, we could measure the
Some physical quantities are associated with points
temperature at each point
in space. Since temperature is a scalar quantity, the values of
18
cHAPTEB 1 DEScRIBING THE UNIvERSE
can represent the field visually by draw-
we shall make that assumption in
most cases, the derivatives are also continuous, and
what follows, except when explicitly noted otherwise'
we have a vector field'
when a vector quantity is associated with each point in space,
magnetic field'
m another' For example' the wind speed is
ity vector'
E
a
1.2.1. The Gradient
one vector field we can derive from a scalar field
coordinates, the gradient operator is written
o(i)
is the gradient
io.
tn Cartesian
/a a a\
u' ar)
v- = (a''
- /ao
vo=
(a"
The change in
o
ao
(1.36)
dy
when we move through an arbitrary displacement
4O
ao ao .
- -dx * no, aY4r:
a,
Vo.ad
4i : (dx, dy, d'z) is
(t.37)
: o. Thus, the gradient vector is perpendiculartant@ateachpoint.Inorderto-obtainthemaximum
changed@|d3|,dsmustbeparalleltoVo.Thus'thedirection
in which <D changes most rapidly at that point (Figof V<D at a
o is obtained by moving perpendicular to the lines of
ure 1.g). Th
If
d3 is tangent to the constant-(D contour, dQ
constant
(D.
Thed,irectionalderivativeofOinadirectiondescribedbyaunitvectoriisi'iO'
1.2 SCAI-AR AND VECTOR FIELDS
19
FIGURE 1.9. Contours of constant @. In this diagram, @ is increasing inward. The gradient of
a scalar field at a point is the steepest slope of the scalar field at that point. VO,
indicated by the arrows, is perpendicular to the constant-@ contours.
1.2.2. Properties of Vector Fields
The Divergence
In Cartesian coordinates, the divergence of a vector field is
i .t:
aux
0z
0y *9!u
0x *U"
(1.38)
In index notation, it is
V
.t: 9a
dxi
(1.3e)
The divergence is a scalar that indicates how much the vectors spread apart, or diverge,
from each other around a point. The electric field due to a positive point charge is a good
visual example of a vector field with positive divergence at the position of the charge.
A vector field whose divergence is zero is called solenoidal.
The Curl
In Cartesian coordinates, the curl of a vector field fr is given by
Du" 0u, 0r, au" \
i x n: (Yu
-Yt.
\ ay 0z' 0z 0x' Ar- Ay)
L-
(1.40)
20
cHAprER 1 DESoRIBING THE UNIvERSE
In index notation, it is given by
(ixil)i:€ijk#
(1.41)
The curl, like all cross products, is a pseudo-vector. The curl is nonzero at a point when the
vectors circulate around, or curl around, the point. The magnetic field due to a long straight
wire is a good visual example of a vector field with nonzero curl at the location of the wire.
A vector field whose curl is zero is called irrotational.
1.2.3. IntegralProperties of Vector Fields
Line Integrals and Circulation
A line integral of the form
t: ir.dl
(1.42)
occurs frequently in physics.l0 The integral is the sum of the contributions il ' dl for each
differentiai displacementll di ulong the path from A to B. For example, the work done by
a force F between A and B is
pB-
W(A
-->
B): I
Jt
F
'dl
In general, the result depends on the path taken between A and B (Figure 1.10).
FIGURE 1.10. Tlvo different paths between points A and B. Going from A to B along path
returning to A along path 2, we form a closed curve C.
10See,
II
for example, Lea and Burke, Section 7.1.4.
In Cartesian coordinates, di has components (dx, dy, dz).
1 and
1.2 SCALAR AND VECTOB
However, when
the path between
i : iO
A
arrd
I^
FIELDS
21
for some scalar field O, the integral (1.42) is independent of
B:
u
,r:
:
,".
l"^
dx
*
*
uy dy
u, dz)
/Aa * ao * ao \
(*"
Jo
,o' Eo')
rB
: ft ot:
Jn
o(B)
-
o(A)
(1.43)
When the integral is independent of path, the integral around the closed loop
A
-->
B
--> A
is zero:
f-18-1A
$n at: JA.pathl
1 ir.dt+ JB.path2
I n.dt
J
18-pB
- It. t i.dl- /t. patz n.dl:o
J
parn
J
:
The integral around the closed loop C
A --> B --> A is called the circulation of the vector
field d around the curve C. Thus, we have this result:
When a vector field fr is the gradient of a scalar field O in a region R, the circulation
ofil around any curve C in R is zero.
i
Conversely, if the circulation of fr around every crrrye C is zero, then is the gradient of
some scalar field O, and O may be determined (up to a constant) by equation (1.43).
The language that we use to describe vector fields comes from fluid theory, because
fluid velocity was one of the first vector fields studied. Water flowing out the drain in a sink
usually swirls around the drain. This is an example of a vector field with nonzero circulation
around any curve that surrounds the drain.
Flux
The flux of a vector field
i
through a surface S is
o,: JsI i.fiidA
where ff at any point on the surface is a unit vector normal to the surface at that point. For
fluids, with d equal to the fluid velocity, the flux describes the rate at which fluid flows
through the surface.
Although flux is a scalar, it has a sign that is associated with the direction of the vector
field d. Flux is positive if the field i points generally in the direction of ff and negative if fr
points generally opposite ff.
\-
22
CHAPTER 1 DESCRIBING THE UNIVEBSE
The Divergence Theorem
The divergence theoreml2 relates the flux through a closed surface S to the divergence
of the vector field in the enclosed volume. When integrating over a closed surface, it is
conventional to choose the normal vector ff to point outward' Then
du
o
oo:
l,i.itav
(r.44)
To prove this theorem, we cut the volume V up into a very large number of tiny cubes
(Figure 1.11).
FIGURE 1.11. Tlvo neighboring cubes inside volume V. On touching sides, the normals outward
from the two cubes are opposite each other.
The volume integral is the sum of the integrals over all the cubes. Notice that the normal
vectors on touching sides of two neighboring cubes are exactly opposite each other, and
so the contributions to the surface integrals f, n ' ff dA for the two cubes from these two
sides exactly cancel. Thus, the sum of the surface integrals for all the cubes reduces to
the integral over those surfaces that have no touching neighbors-that is, over the original
closed surface. So if we can prove the result for one cube (Figure l.l2),we will have proved
it for an arbitrary volume.
I'
differential cube with sides of length dx, dy, and' dz, respectively.
The normals to each side are in the directions +f,, * i, and * 2.
FIGURE 1.12. Integrating over
a
12This theorem is also called Gauss' theorem.
23
1.2 SCALAR AND VECTOR FIELDS
The volume integral on the right-hand side of equation (1.44) is
To evaluate the surface
sides at
x
and at
x
*
o. o' o'
(1.4s)
integral on the left-hand side of ( I .44), we first consider the pair of
dx. For the side at x, ff
: -i
I O OdA:
ls
where z,
.W.ff)
v idv : (Y
l,u**u"oo*
i
(x) is the r-component of
and
-ux(x\dy dz
evaluated at x, while for the side at x
*
dx, ff
: *i
and
/o o dA:ux(x-tdxldydz
Js
Using the definition of the partial derivative 0u, f 3 x , we find that the sum of the two terms is
[u,(x
*
dx)
-
u"(x)ldy dz
: !a*
dx
a, a,
y + dy arrd z, z + dz. Thus, the
left-hand side of equation (I.44) also equals the right-hand side of (1.45), and the theorem
We get similar contributions from the pairs of sides at y,
is proved.
To understand the meaning of the theorem, think of the electric field due to a positive
point charge. If we surround the charge with a volume V , then the electric field is outward
at each point of the surface S, and so the flux through the surface is positive. Similarly, the
divergence of the electric field is positive at the position of the charge, as required by the
theorem.
1.2. Compute the divergence of the vector field i : ki, where i is the
position vector and k is a constant. Find the flux of i through a cube of side a centered
at the origin and show that it equals / V i dV over the volume of the cube.
The divergence is
Exampte
i .i:
kV
.('l
+ yi
+
zi:
:1,
(
!.
\d"r
* !,
dy
: ,o
*3.)
dz /
Since the divergence is a constant, we can immediately compute the volume integral
as
3k times the volume, or 3kaj
The flux is
.
6
kk*-r yj + zit
.ffr a
e
Jsurface of cube
On each face of the cube, the normal is
atx:
Ai, *i,
or
*2,
so the flux through the face
a12is
t *!;t t;dA:klaz-oot
2
2
Jh."2
24
cHAprER 1 DESoRIBING THE UNIvERSE
while at x
: -a/2
we have
l,^""0
(-1)i
(-*) at
: kla2 :
+
The result is the same for each of the six faces, so the total flux is 3ka3 and equals
the volume integral of the divergence.
Stokes'Theorem
Stokes'theorem relates the line integral of a vector field il around a closed curve C to the
curl of the vector field. First we define a surface that spans the curve C to be any surface
whose edge is the curve C. Think of blowing bubbles, using the curve C as the wire. Dip
the wire in the soap solution. If curve C lies in a plane, you will get a planar film of soap
solution attached to the wire (the curve C) at its edges. This is the simplest surface that
spans C. Now as you blow the bubble, the surface expands but remains attached to the curve
C attheedges. You obtain a sequence of surfaces, each of which spans the curve C. Stokes'
theorem applies to all such surfaces.
We define the normal to the surface using a right-hand rule. Curl your fingers in the
direction of di and the thumb gives the direction of fi (Figure 1.13). Then
t';
ai:
l,ru
xi'y n ae
(1.46)
FIGURE1.13. DefinitionofthenormalvectorusedinStokes'theorem.Curlthefingersofyourright
hand in the direction
ofdi,
and your thumb points in the direction
offi'
To prove Stokes' theorem, we use a method similar to the one that we used for the
divergence theorem. Divide the surface up into a mesh of little rectangles. The surface
integral is just the sum of the integrals over all the rectangles. Summing the line integrals over all the rectangles, again we note that the contributions fr'dl from touching
sides cancel, since the vectors di have opposite directions. Thus, the sum of the line
integrals over all the rectangles reduces to the integral over the curve C, as shown in
Figure 1.14.
I.2
SCALAR AND VECTOB FIELDS
FIGURE 1.14. The surface divided into differential rectangles.
Let's look at one of the rectangles that is in a single plane, and choose that plane to be
the.r-y plane (Figure 1.15).
D(x,y + dy)
I
C(x
dx,y + dy)
oz
B(x + dx,y)
A(x, y)
FIGURE 1.f5. A single differential rectangle. We choose x- and y-axes parallel to the sides of this
rectangle.
The integral on the right-hand side of equation (1.46) is
f_
9J!\ a, a,
I-/differentialrecrangle
tvxi.l .ffdA:(Vxdl .zdxdy:(Y'
av)
\ 3x
while on the left we have
t/J AB+BC+CD+DA
il - dl
:
ux(x, y)dx
-
:
-l ur(x * dx, y)dy
u,(x, y * dy)dx
luy(x -f dx, y)
0u,, dv
- Ex'0y
'dx -
-
0u-
-
ur(x,
t)dl
ur(x, y)ldy
'dv
-
lu*(x, y
* dy) -
u*(x, y)ldx
dx
which equals the right-hand side. Thus, the theorem is proved.
For the special case of a curve lying entirely in the x-y plane, Stokes' theorem reduces to
f u, a* r u, dy : I,(* - Y) o. o,
26
cHAprER 1
DEScRTBTNG THE
Since any functions
f
(x ,
uNtvERsE
y) and g (x , y) may replace the x- and y-components of fr in this
proof
f ,r o.*sdy): l,(#-K)0,0,
(1.47)
In this form, the theorem is known as Green's theorem.
Example 1.3. Find the circulation of the vector field i : x2y3 (i *-i) around a
squareofside ainthex-yplane,centeredattheorigin.Alsocompute/(ixi;.fiaa
over the surface ofthe square, and show that the result is equal to the circulation.
We compute the circulation by calculating the line integral around each of the four
sides of the square, starting at x : a 12. We have
": I:;,(;)'"
We obtain the same result for the side at
o':
x
:
(;)'
'i1,"'-,,:o
-a12. At y
-
a/2,
,,: I"-r'' (;)' * o" : (;)' +,::,' : - (;)' ? (;)' : -? (;)'
whileaty:-a12,
,o:
a/2
lr"(_;),
*2 d*
rat.3 2 r
:
: _ (;), *!T _",,: - (;)'
; (;)' -? (;)'
Thus, the circulation is
:
f a ai:o-?(|)' *r-'r(;)' -tG)'
Now we calculate the curl. The normal to the square is in the 2 direction, so we only
need the z-component:
1i x i;.
:* -W - 2*y3 -3r'y'
1.2 SCALAR AND VECTOR FIELDS
27
Then the surface integral is
I (u "t)
naa
:
I::,11,,1"" -
:
l:;,(+
=,
Now if curl
i :
a'
- *"')1"'j.,,0.
1",(-* (;)') "
4
: -, (;)'
as we obtained
z'2v2) av
3
(;)'
from the circulation.
0 everywhere on S, then
thusl3 we can express
i : id
$"n'
ai:
0 for any curve C
lyingin S, and
on S. Thus,
ixi:o=+fi:id
Similarly, it is easy to showl4 from the expression for curl that
i:id+ixi:o
Notice that the divergence theorem and Stokes'theorem relate a locally defined property
ofthe vector field (divergence or curl) to a globally defined property (flux or circulation).
"Local" means that the property is defined at each point of the space, while "global" means
the property is defined over a region of the space (on a surface for the flux or around
complete curve for the circulation).
Numerous variants of these theorems exist, and they are all proved in a similar fashion.
example,l5
I,u'o':
frontdA
Repeated Operations with V
can define additional properties ofvector fields using second derivatives. For example,
have already observed that
curl(gradO):Vx(VO)=0
Section 1.2.3.
Problem 19.
Problem 30. Problem 31 gives an additional variant
(1.48)
28
cHAprERl
DEScRIBINGTHEUNIvERSE
Similarly, we can show that
div(curlil;:V'(Vxil):0
for any vector field
i.
(1.49)
Let's evaluate the expression in Cartesian coordinates:
*w-H.*(v-H.*(*-T)
E2u,
02ur: a*ay- ur*--
A2uy
+
A2uv
02u,
+
azar' ayaz-
U'u,"
ozoY
:O
-oxoz'
terms cancel in
Since the order in which we take the partial derivatives is irrelevant, the
parrs.
The second derivatives (1.48) and (1.49) are identically zero, but many others are not'
The combination
div(grado)
: i . tiol - v2o :# -#.#
(l.s0)
appears frequently in physics, as does the combination
curl(curlfr):Vx(ixfr)
We can evaluate this product using the bac-cab rule, being careful that each
i
operates on
everything to its right:
ix(ixi;:Vfi.tl-v2i
(1.s 1)
This expression effectively d'efines the quantity V2fr.
In Cartesian coordinates, the scalar operator V2, the lnplacian operator, is given by
^a2a2a2
v--a?*i7*rt
The same is not
and the same expression holds whether V2 operates on a scalar or a vector.
true in other coordinate systems, as we shall see below'
1.3. CURVILINEAR COORDINATES
is
When solving a physics problem, we always want to choose a coordinate system that
satela
orbit
of
the
well suited to the geometry of the problem. Thus, when studying
lite about the Earth, we would probably choose spherical coordinates with the origin at
1.3 CURVILINEAR COORDINATES
29
the center of the Earth. When studying the magnetic effects of a long current-carrying
wire, we would choose cylindrical coordinates with the z-axis along the wire. And for
studying the electric field due to a conducting disk, the best coordinates are spheroidal
coordinates. l6
1.3.1. UnitVectors
u , and u. The unit vector ff at any
point indicates the direction in which z increases while u and w are held fixed. The vectors
t and fr are defined similarly. We almost always luse an orthogonal coordinate system, in
which ff, i, and fr form a right-handed orthogonal set-that is, the triple scalar product
In general, we may call the three curvilinear coordin ates u ,
[fi,
i' ff1 :
11
We can relate these vectors to the Cartesian unit vectors i, i, and 2 geometrically, using
diagrams such as Figure 1.16, or we can find their components analytically. To see how,
first write the new coordinates u, u, and u in terms of x, y and z and, conversely, express
x, y, and z as functions of u, u, and w : x : x (u, u, w), y : y (u, u, w), and z : z(u, u, w).
Let point P be described by the position vector i and have coordinates uo, u0, and u.'s.
A neighboring point Q has coordinates zs * du, uo, and ru6. Then the vector from P to Q
is in the direction of ff:
di:dx?+dyy+aziq.dutr
(1.s2)
0z^
^ 0x^ 0y^+-z
uo(-x
0u 0u" 0u
and thus
FIGURE 1.16. Unit vectors in cylindrical coordinates. As the coordinate @ changes, so do the
directions of p and S.^We may express these vectors in Cartesian components:
0 : *cosd f isind, $ : -isin@ * $cos@. The third unit vector in this system
is 2, the same as in Cartesian coordinates. Note that the three vectors are mutually
perpendicular.
l6See Chapter
L-_
2, Problem 13, for a definition of these coordinates.
30
oHAPTER 1 DESoRIBING THE UNIvERSE
Since ff is a unit vector, the
ux:
l-,
y-, and z-components of fi are
0x
h1 0u
Jo*ta. T aylou)2 + (Ezlou)2
0y
:
"y-
:-- l0z
0z/0u
Jruta8i
(1.s3)
1ov
l0u
h1 0u
-
uz:
l0x
l0u
rWtor)2 + (ozlou)z
h1 0u
(1.54)
(1.s5)
where the constant of proportionality in equation (1'52) is
ht:
Similarly, we can find
i
and
(#)'. (#)' . (#)'
(1.s6)
fr:
^ I (0* 0y Az\
': h, \ar' ar' ar/
h2:
(#)'. (#)'. (#)'
(1.s7)
(1.s8)
and
^ I /0x
tn:
/r, \a,
(1.s9)
where
h3:
(x)'.
(#*)'
. (u*)'
(1.60)
1.3 CURVILINEAR COORDINATES
31
In a general non-Cartesian coordinate system, the directions of the unit vectors change
with position. This means that we have to be very careful when taking derivatives or integrals
of vectors expressed in curvilinear components.
1.3.2. Metric Coefficients
The distance between two
may be expressed in terms
we may
neighboring points described by the position vectors i andi * di
of differential changes in the coordin ates u , u, and u.r. In general,
write
di : h
and so
du tt + h2 du
i
-f ht dw itr
for an orthogonal system
dsz
: di
.
di : yl
au2 +
tfi av2 + nl awz
(1.61)
(compare with equations 1.1,1.4, and 1.7). The coefficients h; are called metric coefficients.
They relate the length of a geometrical line segment to the differential coordinate element
and can usually be obtained from geometry (for example, Figure l.2b), or they can be
calculated using equations (1.56), (1.58), and (1.60). We can see that such coefficients are
dimensional grounds. For example, in cylindrical coordinates, the coordinate
differential d@ is just a number and does not have the dimension of
length. The correspondinEhz : p (compare equations 1.4 and 1.61) does have the required
dimension of length.
Let's check the expressionfor h2 using equation (1.58). First, from equations (1.2) we
necessary on
@ is an angle, and so the
have
Ex
a0
and z is independent
-p
sin
@;
Y:
o"orr
,r:m
of@, so
:@:o
as
obtained from the geometry in Section
1.
1.1.
1.3.3. Div, Grad, and Curl in Curvilinear Coordinate Systems
Gradient
The
differential change in a scalar field O over a displacement
di
is (Section I.2.1, eqaa-
tion 1.37)
de
: i o . di : i o . (h du i + hz du i * fu dw fi)
32
CHAPTER
.I DESCRIBING THE UNIVERSE
But we can also derive the differential from the fact that O is a function of u, u, and w,
using the rules of calculus:
4q :
ao -f ao
ao
,du * *dw
-du
Comparing these two expressions, we find that the gradient is
tao. lao^ lao^
VO---riI--yI--w
h2 0u
ht 0u
Again we see that the
y'ls
hj
(1.62)
0w
are needed on dimensional grounds'
Divergence
In curvilinear coordinates, the divergenc e is definedby the divergence theorem ( I .44). Thus,
the divergence of a vector field f at a point P is
fv.r:,lryrvI fst-fidA
where the point P is inside the volume V. We begin by evaluating the integral on the righthand side. We can choose our volume V to be a curved "cube" with sides of length du, dv,
and dw and each face lying along a coordinate surface (Figure I .17 a). Then the normals
to the faces are *fi, ti, and *fr. From the pair of faces at u, u + du we get a contribution
- fufu,
where
u, w) h2h3 du dw
+ f"(u
I
du, u, w) hzhz du dw
fu is the u-component of the vector i.
FIGURE l.l7a. A differential volume used to calculate V . f in coordinates u, u, and u.r. The faces
ofthe volume lie along surfaces of constant u, u, arrd rl, respectively, and the edges
have lengths hy du, h2 du, and h3 dw.
33
1.3 CURVILINEAR COORDINATES
We get similar contributions from the other two pairs of faces. Dividing by the volume
hft2h3 du du dw and using the definition of each partial derivative, we have
v
.i:
lim
I- fu(u, u, w) h2h3 I fufu
I
du, u, w) hzhzl
hft2h3 du
du+O
I la(h2h3f,), oththtf,t,
;v'r-hthrkl
;a, - a, -
*
two similar terms
oth1h2f,1l
a,
(1.63)
I
Notice that the fts appear inside the derivatives, since they may be functions of the coordinates il, u, and u and so differ on opposite faces ofthe "cube."
In cylindrical coordinates (Figure l.l7b), with h1 : I, hz - p,andhz:1, we obtain
;, ;_ t la(ofil ,|fo , a(pil1 _1}(pfp) ,10f6 ,0f2
pl, Ep aQ 0z ) p 0p pA0 0z
(1.64)
Notice that the result is dimensionally correct.
FIGURE 1.17b. A differential volume in cylindrical coordinates.
1.4. Compute the divergence of the position vector, i . i, itr cylindrical
coordinates, and compare with the result in Cartesian coordinates (Example 1.2).
In cylindrical coordinates, the position vectorlT is i : p0 * zi. Thus, from equation (1.64), the divergence is
Exampte
:
,
v .i
.
I
I
II
tfrir
ly
: la?'zt dz: !po.
a 1 :z
p dp + pdQ +y
p
dz p Ap +y
!a(e,i
is the same result that we obtained in Example 1.2.
lTThere is no
@-component. Do you see why? See Problem 10.
i
I
t
E
I
II
:
t-
34
cHAprER 1 DESoRIBING THE UNIvERSE
Curl
In curvilinear coordinates, the curl of a vector
i
at a point P is defined via Stokes'theo-
rem (1.46):
6.i 'dl
tV x f) .ff: Iim '"
A+0
,
A
where A is the area of a surface S spanning the curve C, the point P lies on S, and i is the
normal to that surface.
Again we begin by evaluating the integral. We may choose a curve that lies in a constant
, ,rrrfu"" (Figure 1.18). Then the normal is fi : 0, and we will obtain the a-component of
the curl.
u,w
I
alda,wldw
dw
.P
u
a,w
*
du,w
FIGURE 1.18. Calculating the curl using a "rectangle" surrounding point P.
The line integral is
fxh2
dul. * f*hz
dwlu+au
-
frhz dulu+a.
- f*h3 dwl,
aa
dw
: -;(f,hz)du
dw +
*(f,h)du
while the arcais dA : hzhz du dw, and so the il-component of the curl is
1i
x
ir n: hlft,t,r,, - fi<t,r,tf
The other two components may be derived similarly:
a . l"
I ta
vxf : nrrln(f-hiar(f,h))n+
.#1ft,t.r,,- fr,r,',,f*
I
h,h,
t
l#,rt,r,) - *rr,r,r]
(1.6s)
1.3 CURVILINEAR COOROINATES
We may also express
this result using a determinant:
httr hzl
Vxf:
aaa
hft2h3
hsfi
(l.66)
0u 0u 0w
hf" hzfu htf,
When using this expression, take care to keep the terms in the correct order. The operators in the second row operate on the functions in the bottom row, including the metric
coefficients ft;.
In cylindrical coordinates, we obtain
i " i : il# - fr,,r,lu. l* - #16 * i 1ft,,n, - #1,
(t.67)
: (;# - *) ,. (* - #)o . i
lh*o, - *e!"lu
Example
1.5.
The magnetic field in a region of space is given by the function
fr: Bo@la3)(o-p)20,0 < p < a,andB:0forp >
a.Findthecurrent
density in the region.
From Maxwell's equations,
- lj:-VxB
Po
Since fr has only a @-component, we have
i,. n : (-#)o+
ror0
- Bo#lh@,"-
j:
"o
po
*
)1ft<,'alu
r,)],
l(a
i
pa' lrru - d2 -zp2 - D]
L
2Bo
: ffir"
- dtu - 2p)i
p
z-directionforal2<p<a
understand this result. While al
Notice that the curl is in the
L--
al2but
(Figure
ockwise
36
cHAprER 1 DEScRtBtNG THE UNIvERSE
gri
aiis positive when di
runs clockwise around a small curve C
ataf2 < p <
a
because the vectors get shorter as p increases in this region.
,
!\
)a r/ts
,
It
,
t/
/
/
tl
\
/j'
I /t
\
,
pla
a
i\
t\ t
1,
\
t
/t
/r
t/
I
\\
lt
I
I
J
--t./
i
in Example 1.5. The curl is in the positive z-direction ngar t!rc
negative
origin but is in the
z-direction for p > a /2.Note that the integrat fg ' ai ls
positive when di istaken cloclaMse around a small closed curve C placed at p > a /2,
FIGURE 1.19. The vector field
since the length of the vectors
i
i
is decreasing as we go away from tbe origin in this
reglon.
The Laplacian
With O a scalar function, the expressionl8 for V2O may be derived from the above expressions for gradient and divergence:
v2o:
V2O:
l8To evaluate
1
i.tior: u (*#u*
la
t^
hthzhz Ldu
V2i, with i
i
u,ryu.
*#,
ff#) .*(TT). a, (T#)l
a vector
a
function, you must use equation (1.51)
at the end
of Section 1.2.4.
(1.68)
37
1.5 VECTOR SPACES
In cylindrical coordinates, this is
"
V-o
a/lao\ * a/ao\l
: llalao\
; lu, \'a )* a (; uo ) u\'e ))
ralao\
ra2o
l^-lI--I- PoP\"aP)'
P2aO2'
a2o
(1.69)
az2
This operator occurs frequently in physics, and we shall need to refer to result (1.68).
1.4. THE HELMHOLTZ THEOREM
Any vector field
i
may be expressed as the sum of two terms:
F:d+i
fr: io
for some scalar field O and
i:Vx.4,
for some vector field A.
tne proof of this theoremle may
be found in Appendix IL Then
V.i:f.i:v2o
and
i"F:i*fixii
Inparticular,
andif V x F :
i :
i,
0, then we may choose O = constant and express F : i
"
0, then we may choose A : 0 and express F : VO, as we showed in
if i
.
Section 1.2.3.
1.5. VECTOR SPACES
1.5.1. Properties of Vector Spaces
In physics we regard vectors both as geometrical objects (anows) and also as algebraic
objects (an ordered set of numbers: the components of the vector). It is important to have
both descriptions, since each description proves most useful in certain classes ofproblems.
- Here we shall discuss aspects of the algebraic theory of vector spaces.
l
l9See also Morse and Feshbach, Methods of Theoretical Physlcs, Section 1.5.
38
cHAprER 1
DEScRTBTNG THE UNTvERSE
The number of components of a vector equals the dimension of the vector space. In
most applications in this book, we shall be concerned with three-dimensional vector spaces,
corresponding to the three dimensions ofphysical space. Infinite-dimensional vector spaces
importance-in quantum mechanics, for example-but we shall not discuss these
here.
spaces
We shall also restrict attention, for the most part, to spaces in which the vectors
purely
real.2o
are
A vector space consists ofa set ofobjects (the vectors) with the following properties2l:
are also of
1. The sum oftwo vectors is another vector in the
space:
d+6:i
2. Addition is commutative:
d+6:B+d
3. Addition is associative:
G+6)+i:a+6+i)
4. There is a zero vector, d, which
has the property that
d+d:d:d+d
5. Each vector d has a corresponding negative vector -d with the property that
d+(-a):d
6. Vectors
can be multiplied by scalars. The result is another vector:
td
:6
This product is also associative:
k(cd)
:
(kc)d
7. Multiplication by scalars satisfies two distributive laws:
(k
* c)d:
kd + cd
and
c(d+6):cd+ci
8. Multiplication by the scalar unity
20See Optional Topic B
leaves a vector unchanged:
for additional material.
21The first five properties establish the vectors as elements of a group; see Optional Topic C.
39
1.5 VECTOR SPACES
1.5.2. Basis Vectors
Properties (1) and (6) together imply that the sum
f,,0,
i:1
In particular, we can make this sum equal to the zero vector by appropriate choice
If the only possible choice of values that makes the sum equal to zero
:
is ci
0 for every number c;, then the set of vectors i;, i : 1 to n, is said to be linearly
independent. But if there is a set of numbers, at least two of which are not zero, such that
is a vector.
of the numbers c;.
f
,,i,:i
i:l
ofvectors i; is linearly dependent.
In each vector space, there is some number N with the following property: there
then the set
somevectorsi; suchthatDLr
cii; :0onlyif
eachc;
:0,butif
will
be
weaddonemorevector,
c;i1 - 0 with more than
any vector, then we can find a set of constants such that tllt
one of the ci nonzero. The number N is the dimension of the vector space. The original
N vectors are linearly independent and are said to span the space. They may be chosen
as basis vectors i; for the space. Then any vector in the space may be written as a linear
combination of the basis vectors:
N
f:!u;i;
t:l
The numbers ui a.re the components of the vector i. Thus, we may write the vector
algebraically as an ordered set ofnumbers, the components u;.
In physics applications, we define an additional operation called the inner product22 of
two vectors with respect to the basis23:
N
d*6:D",U,:6*d
(1.70)
i:l
Ifthe resulting value is zero, the two vectors are said to be orthogonal. The inner product
of two basis vectors is
N
i76
*i; : I6;1d;; :6;p
t:1
and so the basis vectors are
orthogonal by definition.
inner product may be defined in many different ways. This version is most useful for our purposes here
inner product becomes a more important and useful concept when it is independent ofthe basis chosen. This
to be the case when the operation that changes the basis is orthogonal, See Section 1.6.5 and Problem 50.
40
cHAprER 1 DEScRtBtNG THE UNIvERSE
It is most convenient to have the inner product defined by equation (1.70) correspond to
the dot product defined in Section 1 . 1.2 (equation I .26). We can do this if we choose a set of
basis vectors that are mutually orthogonal in the geometrical sense of being at right angles.
From any initial choice ofbasis vectors, we can construct a set such that the inner product
with respect to this basis is the geometrical dot product. Suppose we have one set of basis
vectors
i;
that are not mutually orthogonal; that is,
ii.6i 4o
+ j. Then we can use a procedure called Gram-Schmidt orthogonalization to
set f;. First we normalize each vector by dividing by the square root
orthogonal
an
create
product
itself:
with
its
dot
of
for some i
ei:
Then
6;'6;:1
These vectors are called unit vectors. Now we start with the vector
iz:62 -
ir :
6t
and define
(62'ir)Gr
so that
i2.ir:
(Gz
.or)
- Gz.6r)(6r .6r) :
o
and
iz .iz
:
:
- Gz. Gr)Grl . 16z - (62 ' 6r)Grl
62. 6z - Gz' 6)2 : | - (62. 6,)2
16z
Then we normalize to get
f2:
l=-"lfr.fo
We may continue in this manner to construct each succeeding f;:
i-r
i:oi-I,n.6)i*: i, :-L
k:t
The vectors
i;
/t, t
are both unit vectors and mutually orthogonal; they are called an ortho-
normal set.
With orthonormal basis vectors, the inner product defined by process (1.70) is identical
to the geometrical dot product defined in Section I.l.2.In what follows, we shall assume
1.6 MATRICES
that the basis vectors are
and the
41
orthonormal, and we shall not distinguish between the dot product
inner product.
In representing a vector algebraically by its components, we write the set of numbers
as
an array, either in a column such as
/o,
d: I
a2
\o,
called a column vector, or in a row such as
i:
(at, az, az)
called a row vector.24 Clearly both represent the same vector.
We may now write properties (1)-(8) of vectors in terms of the vector components in an
orthogonal basis. For example,
1. Two vectors are added by adding the corresponding components:
ci:ai*bi
4. The components ofthe zero vector are all zeros.
5. The components of B : -d are the negatives of the components of d:
bi
6. The components of f,
To
: ci
are
bi
:
: -ai
cai.
work efficiently with vectors written in terms of components, we shall need to develop
additional mathematical tools-the mathematics of matrices.
some
1.6. MATRICES
1.6.1. Basic Properties
ofproperties that defines a vector space includes several procedures for creating one
vector from another-for example, multiplication by a scalar. If the procedure is linear, each
component of the new vector i is a linear combination of the components of the original
vectol5 il:
The list
N
vi:lait<ut<
k:t
24sri.t1y speaking, ifthe column vectoris a, therow vectoris its transpose, ar. See Section 1.6.1.
25In this section we shall not use the summation convention introduced in Section l I.2. When summing over an
.index, we shall show the sum explicitly.
42
CHAPTER 1 DESCRIBINGTHE UNIVERSE
Thus, any linear operation that maps one vector in the space to another may be expressed
as an affay A of N x N numbers called a square matrix, whose elements are the set of
numbers ai1r.The simple operation of multiplying the components of a vector by a number
c is represented by the array
^:(:;l)
.TT
4tr
-
where II is the unit matrix
': (i :?)
This example illustrates the rule:
To multiply a matrix by a number c, multiply each element of the matrix by c.
The elements of the unit matrix may be represented by the Kronecker delta, d;y, whose
components arezerofor i 17 and I when i : ,/.
The usual convention26 for writing matrices with indices, as in a;p, is that the first index
(here l) labels the row while the second (here k) represents the column. Thus, a21 is the
element in the second row and the first column:
/ ott aD orr \
I@a22anl
oy ajz ott
\
/
An important example of a linear transformation is a rotation of the basis vectors (compare
with Section 1.1.2). Other examples are the angular momentum of a rigid body, L : lRri,
where the operator IR is represented by a matrix containing the moments of inertia of the
body and <i is the angular velocity vector.
A symmetric matrix is invariant under reflection about the diagonal that runs from upper
left to lower right; that is, aii : a;i. An antisymmetric matrix has the property that ai; :
-a;y. Thus, an antisymmetric matrix has zeros along the diagonal.
Two matrices are equal only if all of the corresponding elements are equal: .A : lE if and
only if a;i : bi j for each i and j. The elements of the sum of two matrices are the sums of
the corresponding elements: if C : ,A + ts, then c;i : aij * bij.
When two transformations are applied one after the other, we can describe the result with
matrix multiplication. If i : A* and il : lEi, then
i:
B(,Afr)
:
(lEA)fr
The operations in a matrix product are performed in order, right to left. ts.A means first do
A, then do lE.
26Any convention may be used, provided that you make it really clear.
1.6
i:
MATRICES 43
Let's investigate how to perform this matrix multiplication. The components of the vector
Afr
are
u;:l
N
j:r
a;;wi
Similarly,
un:f
b1,;u;
:f
E
Thus, the elements of the product
b1,;a;iwi:
C
:
(
E *rr,",,).,
:f
,0,r,
lB.A are computed as
N
'*i:lbrrioii
i:l
The element
ctj_ is the inner
product27
of
the row vector bp;
/ati\
column vector 4rl - I or', | . fnir row and column intersect
"
\"t"i /
indices
kj
:
(b6,bxz,brz) and the
at the
position labeled by the
(Figure 1.20).
/bn bn brs\
(
^l
,,: )
btz
atz
, \
/ ,tt
,):{t'
";;
a32
1/
Gi
\c:r
ct2 cl3 \
c22
El
ca2 q3 /
FIGURE 1.20. Matrix multiplication: C : lE,A. Element c;; is the dot product of the i th row of IB
with the jth column of A,. This row and column intersect at the position of element
c;y. In this diagram, i :2 and j :3.
Matrix multiplication is not, in general, commutative; that is, B^N + AB.
/r 2 3\
/o l l\
LetA:f I 0 I landB:l 2 I o l.co-putethe
\o t2)
\r 3t)
products .AIB and
Examplel.6.
lB.A,.
?\/g
l;\
^u:/r3
z/\
r
r
3
t)
\o
x2l3xI
x2*IxI
x2l2xl
x 1*2x 1*3x3 1x *2x0*3x
1x1*0x 1*1x3 1x .l0x0*1x
0x1*1x l]-2x3 0x *1x0*2x
1
I
,
s sense
iplied.
only
if
the row and column have the same length. Matrices that do not
44
CHAPTEH 1 DESCRIBINGTHE UNIVERSE
On the other hand,
]8.4,:
(iii)(i i)
:(
I l3\
i : z)
Clearly these products are not the same.
The transpose .Ar of a matrix A with elements a;7 is formed by interchanging rows for
columns. The components of the transpose are (Ar ),, : a j i. Inthe event that the elements
matrix are complex numbers, we can form the complex conjugate A* of the matrix A by
taking the complex conjugate28 of each element. The matrix formed by performing both of
these operations (complex conjugation and transposition) is called the hermitian conjugate
or adjoint matrix, A,t. A matrix is described as hermitian if it equals its adjoint: .A, : .At.
The trace of a matrix is the sum of the elements on the diagonal:
of
a
Tr(A):an+azzl"'*aNN
The trace is a linear operation: Tr (.4,
For a product, we find
Tr(A,JE):
ttIJ
*
lB)
:
Tr (A) + Tr (B).
a;ibii: tt
biiaij:Tr(lEA)
lr
This result is true even when the matrix product is nol commutative: .AlE + BA.
1.6.2. Determinants
A very important and useful property of a square matrix is its determinant. The determinant of a 2 x 2 matix is a number formed by multiplying along the diagonals and
subtracting:
otl an\ : orru,
\-.'./ A)) I
a"t (
\ A)t
ar2a21
The products a\pzz and a12a2r are called elementary products. In an N x N determinant,
each elementary product is a product of N elements, with no two being in the same row or
the same column. We attach a sign to each product of the form atja2k. . .dNm according
to whetherthe set tj,k, ...,ml is an even (*) or odd (-) permutation29 of 11,2,..., N\.
The determinant of an N x N matrix is the sum of all the signed elementary products. For
28See Chapter 2, Section 2.
l.
29To determine whether a permutation is even or odd, count the number of interchanges of two numbers that must
be performed to execute the permutation. An even permutation needs an even number of interchanges, and an odd
one needs an odd number.
45
1.6 MATRICES
a3 x 3 determinant, this result may be written using the Levi-Civita symbol (Section L 1.2,
equation 1.29):
3
A- t
rj
t
ijkali a2j
(1.71)
a3k
'k:1
The determinant of a 3 x 3 matrix .4, may also be formed as follows. Take any row of
the matrix-say, the top row. For each element a;i in that row, take the determinant of the
2 x 2 matix o;; obtained by removing the row and column that contain the chosen element
(Figure 1.21). Then the determinant is
det (A)
:
:
The quantity
an at2
ct2t 422
a3t 432
att(azza33
(-l)t+i
det
-
aI3
: !{-t)t+i
423
433
a;1det
(a;)
J
ana32)
-
ap(a21ay
-
ana3t)
*
an(azpzz
-
azzaz)
(o;;) -- A;1 is called the cofactor of the element a;;.
('"?dan\
( ozr
:
(
Z::)
",,
\i:', i:: i:: ) "''
FIGURE 1.21. Determining the elements of the submatrix
cv;y whose signed determinant
(- t;i+;
A;; -
det (a;; ) is the cofactor of the element a;; . This figure shows how to find
a12. Eliminate the row and column that contain the chosen element. The remaining
suhmatrix is a;7.
We can now write a rule for finding the determinant of any matrix:
The determinant of a matrix A is the sum of the elements of any row or column of the
matrix, each multiplied by its cofactor:
det (A)
:
lAl
: Do,iAti :Do,iA,i
0.72)
lt
Starting with the rule for a 2 x 2 determinant, we may use equation (1.72) to compute
thedeterminantofa3x3matrix,thena4x4matrix,andsoon.ThisiscalledtheLaplace
development.
These properties of determinants now follow30:
. If an N x N matrix is multiplied by a constant c, its determinant is multiplied
cN.
30See
Problems 37 through 39
and4l.
by
46
a
a
CHAPTER
1
DESCRIBING THE UNIVERSE
A matrix and its transpose have the same determinant.
If two rows or two columns of a maffix are identical, its determinant is zero.
The determinant proves very useful in understanding the properties of matrix products.
We will need the following theorem3l:
Product theorem for determinants. The determinant of the matrix C
:
AIE is rfte
product of the determinants of A andB.
Proof (for3x3matrices). MatrixChaselementsci;
is thus (equation 1.71)
-
D*aitbtj,andthedeterminant
3
det(C)
:
i,
D eijkcric2jcak
j,k:r
3
: D t,toDor*u^,Dornb,,\azrbpn
i,j,k:lmnP
: tt
\o1^o2,o3r(
m n p
f,
,,10u^,u4uro\
\,, j.k:l
/
if ffi,fl, p are an even permutation of 1,2,3, then the term in parentheses equals
det (JE), whereas if m,n, p aire al odd permutation of 1,2,3, then we get its negative. If
any two of m, n, p are equal, the term is zero, since it equals the determinant of a matrix
with two equal rows. Thus, we may write det (C) as
Now
det (C)
:
det (AlE)
: t
arma2na3pem,rp det (lB)
:
det (A) det (lB)
(r.73)
m,n,p
as asserted.
A matrix with determinant zero is said to be slngular Such matrices have some interesting
properties. For example, if such a matrix is multiplied by any other matrix, the determinant
of the resulting matrix is also zero. This fact allows us to understand why it is possible for
two matrices, each nonzero, to have a product that equals the zero matrix32:
,AIE
: 0
does not
imply A:0
or
IB
:0
As a consequence, in the mathematics of matrices, the square root of zero exists and is not
necessarily zero.
31
For additional proofs of this theorem, see, for example, Halmos, Section 53 or Tircker, p. 92.
42 and 43.
32See Problems
47
1.6 MATRICES
1.6.3. The Inverse of a Matrix
:
If the product of two matrices is the unit matrix, .A1B
ll, then lB is the inverse of .A, lE
:
.A.-
I
.
From the rule for computing the determinant of a matrix product (1.73), we conclude that
the determinant of A-l is l/det (A). Thus, the inverse exists only if the determinant of the
matrix is not zero (the matrix is nonsingular).
Using the basic rule for matrix multiplication, we have
N
\aiibir:64
j:r
This relation is actually N x N equations for the elements b1r.T\e solution may be found
by a process of successive elimination,33 with the result that the inverse matrix equals the
transpose of the matrix of cofactors of the elements of .4., divided by the determinant of .A:
u,r:K
(1.74)
Let's check this result. The product of a matrix and its inverse must be the identity matrix.
The product has elements
(AlE)4
:D,,,ff
:\a4bi*
l
l
flrst diagonal element is
D1ar1A11
tAt
=
lAl
1
tAt
we used equation (1.72) for the determinant with
2 and 3. The off-diagonal elements are of the form
i
:
1.
The result is the same for
t ot;ff
A't;
(1.7s)
j
numerator is the determinant of a matrix with two identical rows (here the first and
rows) and so is zero.
Example
See
L.7.
Find the inverse of the matrix
below for an example of how to do this.
(i:!)
48
cHAprER 1 DEScRtBtNG THE UNIvEFSE
The determinant of the matrix is D : 2, so the inverse exists. The cofactors are
At : 2, A12 : -4, An : 0, A2t : O, Azz :2, AZZ : 0, AZt : 0, AzZ: 0, and
Azl
:
L Thus, the inverse is
:(i: l):(-i :r)
We can check the result by multiplying the original matrix and the inverse we have
found:
(a: lX
i: i):(il
l)
as required.
A matrix whose transpose equals its inverse is called orthogonal.34 A matrix whose
adjoint equals its inverse is called unitary.
Computer mathematics packages such as Mathematica and Maple have matrix inversion
routines. Even spreadsheets can invert matrices. Formula (1.75) for the elements of the
inverse matrix is not usually the best computational tool.35 Procedures such as GaussJordan inversion are more efficient. To see how these methods work, first we note that
we may devise a set of matrices, called elementary matrices, whose effect is to perform
simplifying operations on a matrix. For example, a matrix with a 1 in each row and in each
column and zeros elsewhere permutes the entries in the matrix:
(r s;
:
X ;ii :,: z,) Qli
Multiplication by this matrix has moved row 2 to row
2.We can also add a multiple of one row to another:
( ; ; ? X ;ii xa: i:):
1,
row
1
z1:
ii)
to row 3, and row 3 to row
(",,7,':',, "-7,?"' "^"t",?',,)
Multiplication by this matrix subtracts two times row 1 from row 2. By putting the number
x on the diagonal of the elementary matrix, we can multiply a row by any number x:
(:
:
; ? X ;ii zi: z,) ftii'
.iai)
xa:
34You should convince yourself that this definition is equivalent to the one given in Section 1.1.2.
35see Press et a1., Chapter 2.
1.6 MATRICES
49
We can thus perform any of these operations on a matrix .A until we get the unit matrix.
Then the product of the elementary matrices that we used must be the inverse of .4.:
EnBn-r.
.
.
ts:EzErA
:
ll
* ErBr-t'
.' lE:lEzlEr
:
A,-l
We may evaluate this product of elementary matrices as a single matrix by applying the
same operations to the unit matrix:
IE,IE,-r
..
.lDrlEzErll
: .A-l[ : A-l
Once we have established this result, we do not actually need to compute the elementary
matrices; we can simply perform the operations on the matrix rows. The allowed operations
are
1. Interchange two rows.
2. Multiply a row by any number.
3. Subtract
a (positive or negative) number times one row from another row.
/z 3 4\
32I
{ I 2
3/
\0
First we check that the determinant is not zero, and thus that an inverse exists:
Example
1.8.
Invert the matrix
/z 3 4\
a"tl I 3 zl:s
\o 2 3)
Next we perform simplifying operations.
1. Subtract row 2 from row
2. Subtract row I from row
1:
(;
i)
/r
0 2\
2:
{o
\o ) \)
3.
Subtract 213 times row 2 from row 3:
(:
:i)
CHAPTER 1 DESCRIBINGTHE UNIVERSE
Subtract 213 times row 3 from row 1:
(r;l)
(r:l)
5. Divide rows 2 and 3 by 3:
Now we perform the exact same operations on the unit matrix.
1. Subtract row 2 from row l:
(i:i)
2.
Subtract row
I
from row 2:
/ r-t o
l-r 2 o
\o o
r
3.
Subtract 2/3 times row 2 from row 3:
/ t
l-r
-1
2
\ 3-t t)
4.
Subtract 2/3 times row 3 from row
l:
/ 5_L _2\
i
?-t
l_?
\
5. Divide rows 2 and 3 by 3:
/ t_t
I
t)
"
_2\
', I
\-i1-l +)
f
'+
Those 9s in the denominator are reassuring, since 9
result by multiplying:
as required.
:
det (,N). Finally, we check our
ei r)ti;i):(i,?)
1.6
MATRICES
51
1.6.4. Matrices and Linear Equations
A set of
N
equations
in N unknowns may be written in the language of matrices. The
equations
atxt+anx2+ ...lawxN aztxtlazzxzl "'*azNxN aNtxt + aNzxz* ':' + oyy*n1 :
bt
bz
6n1
may be written as
.N*:6
where the vector
i
contains the unknowns .rr as its components and the vector f, contains the
multiplying both sides by the inverse matrix .A,- I :
constants b; . The solution may be found by
(1.76)
where A.1i is the cofactor of element
aii. Ttrc numerator is the determinant of the maffix
formed by removing the ith column of .N and replacing it with the vector 6. Result (1.76) is
called Cramer's rule. It provides a formal solution to the equations, but it is rarely the best
computational procedure.
In the Gaussian elimination method,36 we use the first equation to eliminate 11 from
all of the remaining equations. Then we use the second equation to eliminate x2 from all
the equations below it, and so on. The final equation is then solved for x1y, and we then
substitute back up the sequence of equations to find all the unknowns. This algorithm is
easily programmed and is well suited to the characteristics of modern computers.
The homogeneous set of equations
Ai:0
a
nontrivial solution for the vector
are
i if and only if the determinant of .A, is zero. One
ion of the set of equations is that the row vectors with components a1i, a2i,
each perpendicular to the vector i. Thus, all three vectors must lie in the plane
to i. This means that the triple scalar product3T of the three vectors must be
or, equivalently, that the determinant lAl
0.
:
Change of Basis
Tfansformations
elements of a matrix that represents a linear operator in a vector space depend on the
vectors chosen for the space. If we change the basis vectors, then the elements of the
for example, Golub and van Loan, Chapter 4,
Section 1.1.2. Compare equation (1.32) with equation (1.71).
52
cHAprER 1
DEScRTBTNG THE UNTvERSE
matrix change, even though the operator is unchanged. The new basis vectors are linear
combinations ofthe old ones, so the operation ofchanging the basis is also represented by
a matrix. Let's call this matrix C. Thus, if a vector has components .{r in the original basis,
its components in the new basis will be x',, where
,';:l
N
j:r
r;1*1
0.77)
Now suppose an operator represented by A maps the vector
* to a new
vector
]:
N
ti:lai1x1
j=r
In the new basis, the operator is represented by .A/ with components at,, and
N
y',:\a'i1x'1
J:I
Thus,
/
"t-\-k:t
rt
,
:\-L
k:l
N
:\-Lxj
j:r
or, in matrix notation,38
i' :
Now the components
)c
j
CAi
are related to the components
xl
by the inverse of C,
x; :
Di(C-t)1^xl,so
i' : cAC-li' :.A,'i'
Thus,
,A':
C.A,C-l
(1.78)
38In this notation, i and i' represent the same vector, but the corresponding sets of components are expressed
with respect to different bases.
53
1.6 MATRICES
Equation (1.78) describes a similarity transformation. It is sometimes written in terms
of the matrix 'lf
: C-l:
.A.': T-1.47
(r.19)
The elements of the matrices C and lf are the components of the new basis vectors in the
original system. If a vector i with components x; is to be the first basis vector in the new
system, then its prime components are x\ : I, x'i : O if i I l. Thus,
rr:tt;ixt,-1;,
l
ri
Thus, the column vector
is the first column of the matrix
r, of the 7th basis vector form the jth column of 'lf
: C-l.
11.
Similarly, the components
We often want to find the basis that makes the matrix .A corresponding to a given operator
simple as possible. The simplest matrix is one with nonzero elements along the diagonal
and all other elements zero. Then if
61, the components of are simple multiples of
as
i: yi :
/:
Lixi.In particular, for
f
basis vector 6; that has only one nonzero
component, A'6i
),;6;. This relationship among vectors must remain true in any basis,
so there exist N vectors
such that,Ai;
l.iii. These N vectors are the eigenvectors of
the matrix A, and the constants ).; are the corresponding eigenvalues. The matrix.A/ has the
(suitably normalized) are chosen as the
values ),; along the diagonal when the vectors
the components of
-
a
:
i;
i;
basis vectors.
To diagonalize the matrix, we must solve the equation
(.4-).ll)i:0
where ll is the unit matrix. This equation has a nontrivial solution for the vector
det (.4
- lll) :
g
i
only
if9
(1.80)
This equation is called the characteristic equation. If the matrix is a2 x 2 matix, the characteristic equation is a quadratic equation; if the matrix is a 3 x 3 matrix, the characteristic
equation is a cubic; and so on. Thus, there are at most N real distinct eigenvalues. There
may be fewer than N if the eigenvalues are not distinct or if the roots are complex.
The eigenvalues of a real symmetric matrix are real. Let's see why. If l. is an eigenvalue,
then there exists a vector * such that
A*:
).i
both sides and take the complex conjugate4o:
i.A,*
: iA :.1,*i
See Section 1.6.4.
that we are working with vector spaces in which all the vectors are real.
54
cHAprER 1
DEScRTBTNG THE UNTvERSE
(The process of transposing does not change the right-hand side, except that the column
vector becomes a row vector.) Now take the inner product with i in both equations:
iAi : ),*'i: ).*i'i
Thus,
(),
i .i
Since
cannot be zero unless
i :
-
0,
:
),*)i .*
l :
o
).+ and the eigenvalues are real.
If
if
two eigenvalues are distinct, then the corresponding eigenvectors are orthogonal.
For
Ai1
:
'11frt
and
A*2:7'1'
then
*1A*2
: )'z*t'*'z
i2.Ai1
:
and
.l'riz
'ir
Transposing the latter relation and subtracting from the former, we have
0
:
(),2
Since the two eigenvalues are not equal,
Example
1.9.
-
).r)
i1 .*z
ir . iz
:
O
and the eigenvectors are orthogonal.
. (/t 2\
i -
Diagonalize the matrix
o)'
I
The characteristic equation is
Il-r-,r. 2
-'- - l:0
I
| 2 -)'l
-i.(3-r,)-4:0
L2 -?,).
- 4:o
So the eigenvalues are
,_3*JqT6_3+5_
:
^:
2
1A
2 --r'-
1.6 MATRICES
The diagonalized matrix is thus
, /-' n)
A:(
o 4/
We have solved the problem, but we can learn more from this example.
The two eigenvalues are real and distinct, so the eigenvectors are orthogonal. Their
components are found from the equation
2xy-)"x2:Q
Inserting the values of .],, we have
2x1
So,
with
- -x2
or
xr :2x2
xt: -1,
*t : (-1,2)
and, with
xz: ll,
*.2
:
(2,1)
These vectors may be normalized to provide unit vectors as a basis. Let's verify the
relation ,\*
i,i for these two vectors:
:
(1 3)(-i):(-\o):(_;
):-'(-;
)
()3)( ? ): ('|', ): ( I ): ^(?)
required.
We diagonalize the matrix by changing to a new basis with an operator (matrix)
C. Then ,\' : C.NC-I (equation 1.78), and the elements on the diagonal of .A/ are
the eigenvalues of .4. The eigenvectors of .4, are the new basis vectors. Then from
equation (I.77), the components of the kth eigenvector are related by
as
,,{k) _ \-1c-t),ir$t,:
--r
,,
!{c-r)i j3j*:
(c-r);r
J
Thus, the elements in the ftth column of C-l equal the components of the kth eigenvector. If the matrix is orthogonal, then these are the elements of the ftth row of C.
Here, the matrix that effects the transformation is orthogonal and has elements
':+(-" ?)
55
56
cHAprER 1 DEScRtBtNG THE UNIvERSE
Let's verify that this does the job:
cAC_
:+(_" ?)() 3)+c:
:i(-j ?)(-i t)
?)
o\
-(-r
4)
0
\
which is the matrix A/.
In this example, C turns out to be orthogonal. In fact, since the diagonal matrix A/ is
symmetric and an orthogonal transformation maps a symmetric matrix to another symmetric
matrix,4l any symmetric matrix ^A is diagonalized by an orthogonal transformation.
If a matrix has two or more equal eigenvalues, then we cannot prove that the eigenvectors
are orthogonal, but they may be.
Example
1.10.
Find the eigenvalues and eigenvectors of
tr,..utri* ( 3
2-)"
0
-1
-1
0
2-).
0 3-),
0 l:s-
Thus, the eigenvalues are I : 1 and tr
Next we seek the eigenvectors.
:
t5),.+7)"2-1.3:-(),-1)()'-3)2
3, the latter being a repeated root'
(ii sX,):^(l)
('-:ti,) :^(l
First we take the case of i.
:
1:
2x-z:x+x:z
3y:y*/:0
-xl2z:z+z:x
Thus, this vector
41See Problem 48.
i1
-;
)
\-t o 2)
First we find the eigenvalues:
0-
S
has components (x, 0,
x) for any x.
)
16 MATRICES 57
Then, with.l.
:
3, we have
2x-z:3xlx:-z
3Y :3Y t Y is arbitrary
-xl2z:3zlz:-x
Thus, these eigenvectors have components (x, y,-x) for any values of y and x.
Notice first that each of these vectors is orthogonal to i1. Second, we can find two
such vectors that are orthogonal to each other:
(x,y, -x)' (r,u, -u) : xu I yu I xu : O : 2xu -l yu
Thus, with x,u,and y chosen, we simply pick u : -2xuf y. We can always
this, provided that y is not zero. For example, we might pick
i3 : (1, -2, -l). Let's check that i2 is an eigenvector:
/zo
l-li
as
i)(_i
required. You should check that
):(
i2 :
(1, 1,
-1)
do
and
:,(_i)
i)
i3 is also an eigenvector.
If two matricas commute, then if their eigenvalues are all distinct, they have the same
eigenvectors. For example, suppose that AIE
lE,A and.Ai
l,*. Then
:
lB.A,i
:
: lB),i:
).lBi
But
BAi:
AlEi
AiBi:
l.lEi
Thus,
Thus, iEi is also an eigenvector of A with eigenvalue ).. If the eigenvectors of ,A are distinct,
then IB* must be a constant times i, and thus is also an eigenvector of IB.
i
Congruent Tbansformations
A
set of linear equations results when we consider small oscillations of a physical system about its equilibrium. We can often analyze such a system most efficiently using
Lagrangian mechanics. The Lagrangian of a system contains a kinetic energy term42 K
which is quadratic in the velocities and a potential energy term V which, in the case of
42In Lagrangian mechanics,
the symbol
the
transformation matrix, we'l1 use
? is often used for kinetic
K for the kinetic energy.
energy. But since we have already used
lf for
58
CHAPTER 1 DESCRIBING THE UNIVERSE
small displacements from equilibrium, may be expanded in a Taylor series with no linear
term. (This is the equilibrium condition.) Thus, the potential energy term is a quadratic
function of the coordinates. Each of these quadratic forms may be expressed in terms of
matrices:
r-dxT*dx:
v:xrvx
dt dt
where the vector xr on the left is a row vector and the vector x on the right is a column
vector. Furthermore, both matrices are symmetric. The motion of the system looks simplest
when we choose our coordinates to correspond to the normal modes of the system. But we
don't usually know what those modes are until we have solved the problem.
The normal modes of the system can be described by vectors x(N). Thus, to simplify
the system, we need to change our basis to make these vectors our basis Yectors. Let's see
what happens to a quadratic form when we change the basis. The energy is a scalar under
coordinate transformations, so
: x'zVx/: xrVx:V
V/
Butx:'lfx',so
x'zv'x'
:
(Tx')?Vllx'
:
x'z'llz\y'Tx'
and thus
V':
Tzv1f
(1.8 1)
This transformation is almost a similarity transformation (equation 1.79), but here the
transpose of 1f appears instead of the inverse. It is called a congruent transformation.
1l-1 and the two transformations are identical.
transformation is orthogonal, 'lfz
If the
:
Example 1.11. A system has two identical pendulums connected by a spring. Each
pendulum has a point object of mass rn on the end of a stiff but massless rod of length
l.Eachpendulum is attached to a pivot on the roof. The two pivots are a distance ss
apart. The two point objects are connected by a massless spring with spring constant
k and relaxed length s6. Find the possible motions of this system.
This system is most easily analyzed using Lagrangian methods.43 In equilibrium,
both rods hang straight down, and the spring exerts no horizontal force. Figure 1.22
shows the system when displaced from equilibrium. The kinetic energy is
K_
43See Optional Topic E.
:-rle)'.(#)')
1.6 MATRICES
FIGURE 1.22. A system of coupled pendulums (Example
l
I
l). The two objects are connected by
a spring.
The system has both gravitational potential energy and elastic potential energy. To
compute the gravitational potential energy, we put the reference level at the level of
the two pivots. Then
Vs
:
-mgl(cos 0t
*
cos 92)
The length of the spring in Figure 1.22 is a complicated function of the angles, but
we can obtain a much simpler expression if the two angles 01 and 02 remun small.
(This is the small oscillation approximation.) Then with d1 ( I and 0z 11 l, the
two objects are separated by a distance so * {(02 - dr) and the elastic potential
energy is
l"
V,:U**P2-0)2
Now applying the same constraint (0 < l) to the expression for the gravitational
potential energy, we obtain the total potential energy:
v
I
: -mB(I,/ - 70?- 7e"2\+ rke@z
- o)2
^
)
:
-2ms( +)telgt-t mg) +
)ulWt * ^s) -
kr?uzlt
Then the Lagrangian is
L:K-V-
:-el(#)'.(#)'l *2mgl
-)telo,t-t *d - )te:l.*t I
mil + kt20z0r
60
CHAPTER 1 DESCRIBINGTHE UNIVERSE
and Lagrange's equations are
d
aL
dtnF,d2o,
oL _n
^
mt"---J
-t(ktI*g)il*kt20z:0
(l.82)
dzg,
- t (kt * ^g) 0z * klzil :0
(1.83)
dtz
^
ml."---J
dtz
These coupled linear equations show that the motion of one pendulum influences the
motion of the other.
We shall now formulate this problem in the language of matrices. Note that both the
kinetic and potential energies may be expressed as products of a matrix and a vector
containing the velocities or the coordinates of the two pendulums.
| ^ / dor do'>'
y:)mtzt=:,=)(I
2
\dt'dt.
/
det
\
?)t l;,
(1.84)
I
\ dt /
and
V
-t mg) +
: -2mgl *
*
)telO,t
)tel<*t
: -2mgl I
)^t2re,", (
mg)
-
* *ox
:,i r)
kt20z0r
t;;
I
(1.8s)
Here the coupling is indicated by the off-diagonal terms in the matrix. Notice that we
have divided the symmetric termin9fi2 into two equal parts, making the off-diagonal
terms equal and the matrix symmetrical.
To find the normal modes of the system, we should change the basis to obtain
a diagonal matrix, indicating that the oscillations in the new coordinates are uncoupled.
Since the potential energy matrix is symmetric, we know it can be diagonalized by
an orthogonal transformation. A transformation of the type
V:lfrVlf
must also be applied to the matrix in the expression for kinetic energy, but since that
is the unit matrix, it remains unchanged if the transformation is orthogonal:
K'
:'lfrllr :
lfr1f
:'lf-l'll:
ll
1.6 MATRICES
The characteristic equation rs
kp
-+-ml
_L
m
-0
/k e
l-+:-t (.
\rn
:Q
__r k
m
and so the eigenvalues are
I *rL
x:1,
lLm
The eigenvectors are given by
rk c
(;*7
_.L\
Ir
\-;
o !,lG):^(":)
;-v/
(
",,):iG:)
In the first mode,
\t
tr):
,v/
Ltt,-gz):o or
m
iG)
ot:oz
In this mode, the two pendulums oscillate in phase, and the distance between them
stays fixed. The spring does not affect the system.
To find the other mode, we use the second eigenvalue:
(!:,: :l; . i;,,) : r' .,*) (z:)
k
--m
(0t
-10) :
Q
61
62
cHAprER 1 DEScRtBtNG THE UNIvERSE
Here 91 : -02,the two pendulums are 180' out of phase, and the spring is alternately
stretched and compressed.
We want to find a solution in which the system oscillates, so each angular displacement may be written in the form
xi : Ai cos(ait I
Qi)
:
-co! xi , and equations 1.82 and 1.83 take the following form:
Even mode: For 02 - 0y,
Then
d2
x; I dt2
-,?or
-(h*X)"+Ler:o
-1302
+Loz:o
-(L+9)e'
(/
m
\,fl
The equations are identical, as expected, and the solution forthe angular frequency is
0)e:
This is the oscillation frequency for a single pendulum oflength
Odd mode: For the second mode,02 : -01,
-azot
I, as expected'
* 9) o, Le, :o
- (!
\m t./ - m
This mode has a higher frequency because the spring provides an additional restoring
force.
If we initially displace the two pendulums by unequal amounts 0s and 9s, then we
must decompose the displacements into two parts, one corresponding to each normal
mode:
91
(o)
o2(o)
: to :
:
,u
:
ti#
to
*
;tu
to;lu
to
;to
The first term is the even mode, and the second is the odd mode. Thus, the system
evolves as
o1e)
: (r+)
"o,
| f,, .
(r+) "." l rL3,,
1.6 MATRICES
E /eL!!\ro,f;" *t,
oz(t):(eo+et\
2
\
)cos1/7t-l-z / u m t
We close by showing how the Lagrangian may be written in the new coordinates. Let
and
4
+ @t i- o).Then
: | {il - o)
L
:
)^
P
^h
:
lf**)'
-)tfO
+
,.lr)2(rrt
*
-t mg)
70,,
;,r,,)'l *,^,,
- )tfO - ,h)2(t t-t ^g) + trz1q + D@ - th)
: mt2lr*l' . (#)' *,x - o'x - r', (,*;)l
The absence of cross terms
nates. In matrix language,
L:m(2(#
Qlt shows that the motion is decoupled in these coordi-
/do
\
\dt
/
#)G Dll;l
* m(z1q.r,
(t,*o * fll
X)
: *t2(Q,,r, (X -r,,
,*
*oX _
.t2mgt
,,) (rr) * r*rn
The eigenvalues ,1. of the potential energy matrix are in fact the squares of the frequencies of the normal modes.
The transformation matrix has the components of the normalized eigenvectors as
its columns,
"-z\r _l)
rf
-
| (l
64
and
cHAprER 1 DEScRtBtNG THE UNIvERSE
it is orthogonal. Thus, the transformed potential energy matrix is
kgk
V':1f
?\
--l11
mLm
I -l )(
kk
mm -+
g
.(
g
I
11
:( 1 -l )(
2kl*gm
;)(1
_l)
mL
2k[.
*
Sm
m(.
)
0\
(:
2k(+
sm
m(.
/
tI
and it is diagonal, as expected. The diagonal elements are the eigenvalues of the
matrix.
The Lagrangian contains two matrices, and we have succeeded in making them both
diagonal simultaneously. The question then arises: Can we always do this? We can if both
matrices are rcal symmetric matrices and at least one of them is positive definite. A matrix
is positive definite if
x?.Ax
r0
(1.86)
for an arbitrary vector x. In the example above, one of our maffices was the unit matrix,
and the eigenvalues were found from the equation
det (V
- lll) :0
More generally, the equation for the eigenvalues in a Lagrangian oscillation problem is
det (V
where
K is the kinetic
- i.K) :
0
(1.87)
energy matrix. This matrix is always positive definite, and both
matrices are symmetric, so simultaneous diagonalization is possible in this case. (See Problem 53.) The resulting eigenvalues are called generalized eigerwalues. The eigenvectors are
solutions of the matrix equation
Vx:,i,Kx
The transformation in this case need not be, and usually is not, orthogonal.
PROBLEMS
65
1.6.6. Matrices and Quantum Mechanics
One very important application of matrices in physics is in quantum mechanics. The state
of a system is represented by a vector in a finite or infinite-dimensional vector space. Any
physically measurable property of the system may be obtained by allowing a matrix to
operate on the state. The matrix represents the physical operator. The eigenvalues of the
matrix represent the possible values of the physical measurement. The fact that matrices
do not commute, in general, leads to some interesting physical consequences, such as the
uncertainty principle.
PROBLEMS
: fl
i
,
I
I
|
Circular motion. Answer this question without using Cartesian components. A particle
it moving around a circle with angular velocity <ir. Wrlte its velocity vector i as a vector
product of 6 and the position vector i with respect to the center of the circle. Justify
yout expression using geometrical arguments. Differentiate your relation for i, and
hence derive the angular form of Newton's second law
(equation 1.8).
(i :
1d) from the standard form
the vector i : (1, 2,2) and perpendicular to
cts. Determine the transformation matrix .A
ordinate system with x'-axis along fr and yt-
: (-20, g, 12),zurd fr : (0, -4,3) are mutu-
ne the transformation matrix that transforms
stem to a system with x'-axis along il, y/-axis
ansformation to find components of the vectors
I, -2) in the prime system. Discuss the result
lectric and magnetic fields
i
and
i.
Show that
r\u,
)" "u
n
locity
i : io *
e6, where 6 is a unit vector
1)
and
fl : Bo(I, -2,1), set up a
{o(1,-t,
g E x B and y'-axis along E. Determine the
etermine the components of i(r) and i(r) in
e a criterion for "short time."
y <i. Using cylindrical coordinates with z-axis
of the velocity vector i at an arbitrary-point
in cylindrical coordinates to evaluate V x i.
66
CHAPTEB 1 DESCRIBINGTHE UNIVERSE
conservation of mass in a fixed volume V, use the divergence theorem to
derive the continuity equation for fluid flow:
6. Starting from
0p
At
7.
+i.(pi):o
where p is the fluid density and i its velocity.
Find the matrix that represents the transformation obtained by (a) rotating about the xaxis by 45o counterclockwise and then (b) rotating about the y'-axis by 30' clockwise.
What are the components of a unit vector along the original z-axis in the new (doubleprime) system?
[8] ooes the matrix
/ cosl
sind 0 \
I sin0 -cosd 0
o t)
\o
9.
I
represent a rotation of the coordinate axes? If not, what transformation does it represent?
Draw a diagram showing the old and new coordinate axes, and comment.
Represent the following transformation using matrices: (a) a rotation about the z-axis
through an angle z/3, followed by (b) a reflection in the line through the origin and in
the x-y plane, at an angle 2n 13 to the original x-axis, where both angles are measured
counterclockwise from the positive x-axis. Express your answer as a single matrix. You
should be able to recognize the matrix either as a rotation about the z-axis through an
angle a or as a reflection in a line through the origin at an angle cv to the -r-axis' Decide
whether this transformation is a reflection or a rotation, and give the value of a' (Note:
For the pulposes of this problem, reflection in a line in the .r-y plane leaves the z-axis
unchanged.)
10. Solve this problem without using Cartesian components. Using polar coordinates, write
the components of the position vectors of two points in a plane: P1, with coordinates
11 and 01, and P2, with coordin ales 12 and 02. (That is, write each vector in the form
i : uri + uo0.) What are the coordinates 13 and 93 of the point P3 whose position
vector is
iz:il
+iz?
Hint: Startby drawing the position vectors.
11. A skew (nonorthogonal) coordinate system in a plane has x'-axis along the x-axis and
y'-axis at an angle g to the x-axis, where 0 < ft 12.
(a) Write the transformation matrix that transforms vector components from the Carte-
x-y system to the skew system'
(b) Write an expression for the distance between two neighboring points in the skew
sian
system. Comment on the differences between your expression and the standard
Cartesian exPression.
(c) Write the equation for a circle of radius a, with center at the origin, in the skew
system.
PROBLEMS
67
E2l Prove the Jacobi identity:
dx
1.3.
$ ri)+6 x @ x d)+i x (d " B1 :o
Evaluate the vector product
(dxilx(ixd)
rn terms of triple scalar products. What is the result if all four vectors lie in a single
glane? What is the result if a, b, and i are mutually perpendicular? What is the result if
b:d?
14. Evaluate the product (d
x 6) . G x dl in terms of dot
products of d, 6,
i, and d.
15. Use the vector cross product to express the area of a triangle in three different ways.
Hence prove the sine rule (see figure):
sino_sinf_sinlz
ABC
C
PROBLEM
E6.lUse the dot product fd
15
- 6l . td - il to prove the cosine rule for a triangle:
,2 : o2 +bz -2ab cosy
17. A tetrahedron has its apex at the origin and its edges defined by the vectors d, f,, and
i, each of which has its tail at the origin (see the figure on the next page). Defining the
normal to each face to be outward from the interior of the tetrahedron, determine the total
vector area4 ofthe four faces ofthe tetrahedron. Find the volume ofthe tetrahedron.
4A vector area is represented by a vector
l-
cross product as in Section 1.1.2,
68
CHAPTER 1 DESCRIBINGTHE UNIVERSE
o
PROBLEM t7
18. A sphere ofunit radius is centered at the origin. Points U, V, and W on the surface of
the sphere have position vectors i, i, and fr. Show that points P and Q on the sphere,
located on a diameter perpendicular to the plane containing the points U, V, and W,
have position vectors given by
i:*
frxi+ixfr+*xd
li, i, fr'l
where I is the angle between the vectors i and i.
cos 0
L9. Show that
ix(io):o
for any scalar field O.
[zO.l rma an expression for
21. Prove the identity
ira.[t :
i x G x t)
(d.
in terms of derivatives of d and 6.
i)6+ 6. V)a+d x (i " B) +6 x (i
Hint: Start with the last two terms on the right-hand side.
22. Compfie i (a ,. 6) in t"r-, of curl d and curl 6.
23. Obtain an expression for i x (di), and hence show that i
@f,tne
x (Oi il
x
d)
:
O.
equation of motion for a fluid may be written
, (#+ 6 vrt) :-ip+pd
where
i
is the fluid velocity at a point, p its density, and P the pressure. The acceleration
due to gravity is E. Use the result of Problem 21 to show that for fluid flow that is
incompressible (p : constant) and steady (0 I 0t = 0), Bernoulli's law holds:
f + \Ou2 + pCh:
constant along a streamline
69
PROBLEMS
Hint: Express the statement "constant along a streamline" as a directional derivative
being equal to zero.
Under what conditions is P
throughout the fluid?
*
lOr2 + pgh equalto an absolute constant, the same
25. Evaluate the integral
$"
at
JC
where
(a) C is the unit circle in the x-y
plane and centered at the origin and
il:x2y* -*y25,
(b) C is a semicircle of radius a in the x-y plane with the flat side along the x-axis, the
center ofthe circle at the origin, and
i:
xy23 + y*2j,
(c) C is a 3-4-5 right-angled triangle with the sides of length 3 and 4 along the
),-axes, respectively, and
and
i^
:.x-x
+ xyy
u
@
x-
C is a semicircle of radius a inlhex-y plane with the flat side along the r-axis, the
center ofthe circle at the origin, and
i:
(2x
-
y3)i
-
(3y2 +
"')y
26, Evaluate the integral
Ir.aA
"/s
where
(a)
S
is a sphere ofradius 2 centered on the origin and
i: x3i +3yz2y -f3y2zi
(b)
S is a hemisphere of radius 1, with the center of the sphere at the origin, the flat side
in the x-y plane, and
i:
x2yz(! -12)
Show that the vector
i:xi-fyy-2zi
70
cHAprER 1
DESCRTBTNG THE
has zero divergence
function @ such that
uNrvEFsE
(it is solenoidal) and zero curl (it is inotational). Find a scalar
i:
and a vector
i.
Vd
such that
i:VxA
[2&l Strow that the vector
n:i
12
has zero divergence (it is solenoidal) and zero curl (it is irrotational) for
a scalar function @ such that
i:
rI
0. Find
Vd
and a vector d. such that
i:ixi
29. A surface S is bounded by a curve C. The solid angle subtended by the surface S at
a point P, where P is in the vicinity of but not on the curve, is given by
[4+
a:
""_ J,
R,
Here
*
i'
-
dal
is an element of area of the loop projected perpendicular to the vector f,. :
chosen origin O, and
*', * is the position vector of the point P with respect to some
is a vector that labels an arbitrary point on the surface or the curve. Now let the curve
rigidly displaced by a small amount d3. Express the resulting change in solid angle
dQ as an integral around the curve. Hence, show that
be
Vo:-i *{oi
JcR
30. Prove the theorems
(a)
l,u,
o,
:
f,ar
dA
(b)
it dv
IrU "
:
fla
'
i,t oo
71
PROBLEMS
31.
Prove
@
f-f-
6adt:/frxYedA
lc
Js
(b)
f
r/(fixvlxidA:$dtxi
Js
Jc
32. Derive the expressions for gradient, divergence, curl, and the Laplacian
coordinates.
in spherical
33. In polar coordinates in a plane, the unit vectors i and 6 are functions ofposition. Draw
a diagram showing the vectors i at two neighboring points with angular coordinates I
and 0 * d0.Use your diagram to find the difference Ai and hence find the derivative
ai I ae.
@
fne vector operator
I
L_ -FxV
i
appears in physics as the angular momentum operator. (Here
position vector.) Prove the identity
ifi' fl : i+ i(i
.
d) +
I : J=
and
i
is
the
;(i x i)
for an arbitrary vector d.
35. Can you express the vector d : (1, 2,3) as a linear combination of the v_ectors d1 :
(1, 1, 1), i2 : (1, 0,
-l), and i3 : Q,1,0)? Can you express the vector b : (1,3,2)
as a linear combination of the vectors il1 , iz , and d: ? Explain your answers geometrically.
36. Show that an antisymmetric 3 x 3 matrix has only three independent elements. How
many independent elements does a symmetric 3 x 3 matrix have? Extend these results
toanNxNmatrix.
37. Show that if any two rows of a matrix are equal, its determinant is zero.
b&lP.ove that a matrix with one row of zeros has a determinant equal to zero. Also show
that if an N x N matrix is multiplied by a constant c, its determinant is multiplied
by rN.
39. Prove that a matrix and its transpose have the same determinant.
i
40. Prove that the trace of a matrix is invariant under change of
basis-that is,
CAC-I) : Tr (A)
is invariant under change of basis-that is,
determinant of
a
real symmetric matrix equals
72
CHAPTER
1
DESCRIBING THE UNIVERSE
the product of two matrices is zero, it is not necessary that either one be zero. In
particular, show that a2 x 2 matrix whose square is zero may be written in terms of two
parameters a and b, and find the general form of the matrix. Verify that its determinant
@Jlt
1S ZerO.
43. If the product of the matrix ,\
: f:
\c
Or\
and another nonzero matrix lB is zero, find
d)
the elements of B. You may find it necessary to impose some conditions on matrix ,4.
so, state what they are.
If
44. Diagonalize the matrix
(iii)
45. Show that a real symmetric matrix with one or more eigenvalues equal to zero has no
inverse (it is singular).
i46. Diagonalize the
orthogonal?
(
? ) -O
^uti* \3! 4/
find the eigenvectors. Are the eigenvectors
If not, why not?
47. What condition must be imposed on the matrix ,A in order that.AlB
:
AC with B
If
B:(l s)
and
+
C?
a:(; l)
find a matrix ,A such that.AlE : AC.
48. Showthatif .AisarealsymmetricmatrixandCisorthogonal,then.A'/:
C.AC-I isalso
symmetric.
,A,lE : IEA if both A and lB are diagonal matrices.
that
the
inner product is invariant with respect to change ofbasis under orthogonal
Prove
[50.l
49. Show that
transformations.
51. A quadratic expression of the form
cv
x2 + 2frxy
I
yy2
:
1
represents a curve in the
x-y
plane.
(a) Write this expression in matrix form.
(b) Diagonalize
the matrix, and hence identify the form of the curve and find its symmetry axes. Determine how the shape of the curve depends on the values of a, p,
and y. Draw the curve in the case u : I :2,y :3.
52. Two small objects, each of mass m, are joined by a spring of relaxed Tength l.Identical
springs hold each mass to a wall. The walls are separated by a distance 31. Write the
Lagrangian for the system, and find the normal modes and the oscillation frequency for
each mode.
a jointed pendulum system. Two point objects, each of mass
linked by stiff but massless rods, each of length l. The upper rod is attached to
a pivot. The system is in equilibrium when both rods hang vertically below the pivot.
The figure shows the system when displaced from equilibrium.
53. Find the normal modes of
tn,
are
PROBLEMS 73
Write the Lagrangian for the system in matrix form. Simultaneously diagonalize the
kinetic energy and potential energy matrices, and find the generalized eigenvalues. Determine the eigenvectors. Find the matrix that effects the transformation, and verify that
both matrices are diagonalized.
PROBLEM
53
CHAPTER 2
Complex Variables
2.1. ALL ABOUT NUMBERS
Numbers are classified emotionally! The Greeks liked integers and rationals (a rational
number is the ratio of two integers).
Integers:
1.2,3.4,...
Rationals:
l3to7
2' 4' 436" "
But they did not like irrationals (irrationals are numbers that cannot be expressed as the
ratio of two integers). Yet irrationals show up in lots of common circumstances, like the
ratio of the circumference of a circle to its diameter.
Irrationals:
e, rr,
4,]rr-2,
..
.
In spite of the name, there is nothing wrong with these numbers, as we all know, and we
have no trouble computing with them.
What happens when we try to solve the following quadratic equation?
f(z):22+z+5=:g
z
It doesn't factor, so we use the formula:
1,J4
:_r=
:.: -r+/T-to2
2
This result does not exist in the list of numbers written above. We can see this geoqnetrically
by plotting the function f (z)
z2 + z +
{figo." 2.1) andnoticing that the function does
:
t
not cross the z-axis. The solution to the dilemma is to add a dimension to our number
system. We define the quantiryl^/J : i so that the solution to the equation is
I
--: -1*
2-ti:x*iy
i
lEngineers often use the notation
j = nE1.
75
'
L
76
CHAPTER
2
COMPLEX VARIABLES
f(x)
-2
FIGURE2.l.
-1
0
(x):
Thecurve f
12+r+512doesnotintersectthe.r-axis,sotheequation
has no real solutions.
f (x):O
We now need two real numbers, x and y, to describe the complex number z. The number
y multiplying the i is called the imaginary part,2 while.x is the real part. Now the original
expression becomes a function of the two real numbers x and y:
z2+z+
Since i2
5
5
t :(x*iy)2+x-liy1: , :
x2 -l2ixy
I i'y'
+x
I
iy
5
I ,
: -1, this expression becomes
f(z)
:
fi(x, y) r ifz@, !)
:
x2
-
y2
+*+
5
t -t iy(2x * t)
To plot this expression we would have to plot the two functions -fi :
and f2 - y(2x + 1). To obtain solutions of /(z) : 0, we require thar
*2 - y2 + x + ]
f1 and f2 be zero
simultaneously. By allowing the extra dimension, we find that both figures intersect the
plane f : 0 at two common points x : -+, y : ++..
By allowing ourselves to use complex numbers with both real and imaginary parts, we
are able to solve an enormous number of equations that do not have real solutions, notjust
quadratics of the type we considered here.
Complex numbers satisfy all the usual algebraic rules, so long as we remember that
i2
:
-1. Then, for example,
Addition:
u,,',i
2**'
: l' l,',1',,i ;] l',1 : ;,
5+t
2The name is unfortunate; these numbers are not "imaginary" in the normal sense of the word. In fact, they prove
extraordinarily useful in solving problems in the rea.l world of physics.
77
2.1 ALLABOUTNUMBERS
Subtraction:
(a'fib) -
(c
* id):
+ si) -
(3
- 4i) :
(2
(a
(2-
c) + i(b
3)
+
(s
-
d)
+ 4)i
: -l +9i
Multiplication:
: ac * i(bc 'f ad) + izaa
: ac - bd -f i(bc t ad)
(2+5i) x (3 - 4i):26*7i
(a
* ib) x (c *
id)
(2.1)
Division is trickier if we want to write the result in the standard x * iy form. As a preamble,
noticefromequation (2.1)that,ifwetakea - c,theimaginarypartoftheproductiszerofor
d: -b.Thus, theproductof anumberz : a *ib with its complex conjugate z* : a -ib
is purely real.
The complex conjugate z* of any complex number z is formed by changing the sign
of its imaginary part.
The product of a complex number and its complex conjugate is purely real and nonnegatrve:
zz*
: (a:i
ib)(a
-
ib)
:
a2
+
(2.2)
b2
We can make use of this result when dividing two complex numbers. We multiply the top
and bottom of the ratio by the complex conjugate of the denominator:
1!:#i: (T.*)(=#)
xtxz
I
yryz + i(y1x2
*l+fi
-
xrx2l yty2 . .(ytxz -
x1y2)
*]+f, "
xtYz)
*]+fi
So we may write the result of the division in the standard form:
(z+si) _ (2+si)
(3
- 4t)
(3
-
(3
4i1 G
+4i)
+
: _!25*T,
:,l!+T,i
2s
(3') + 4'z)
4i)
2.1.1. The Argand Diagram
The Argand diagram (or complex plane) is a diagram in which we plot complex numbers
as points with x and y coordinates equal to the real and imaginary parts of the number,
L-
78
CHAPTER
2
COMPLEX VARIABLES
respectively (Figure 2.2). Since we may equally well describe a point in the plane using
polar coordinates (r, 0), we can express the number as
z:x+iy:r(cos9*isin0)
(2.3)
where r : J*4 y'is the length of the line from the origin to P and is called the
amplitude (or absolute value) of z: r : lzl. The angle measured counterclockwise from the
x-axis, 0 : tan-r(ylx),is called the argument (or phase) of z:0 : arg (z). As x -+ 0,
y/x
-->
-|ooforpositivey,and9
-+ -n/2asx -+ 0.
< 2r or
--> +TTl2.Similarly,fornegativey,0
By convention, we choose a range of
-t <0 <n.
2r
for the angle 0. Usual choices are 0 < 0
v' Im (z)
-r, Re (z)
01/2
FIGURE 2.2. The Argand diagram allows
is described by Cartesian
usf
plot complex numbers as points in a plane. The point
or polar (z : ,"ie) coordinates.
Q,L
/ x + iy)
There is a nifty way to express z in terms of the polar coordinates
0z
ae
- r(-
sind
* i cos0) :
r
and 0 . Notice that
i7
We can integrate this differential equation to get
z
-
g(r)eio
at 0 : 0 and compare with the original expression (2.3) in terms
and sin 9, we see that we should take g(r) : r. Thus,
If we evaluate this result
of cos
I
z:
rei9
(2.4)
2.1 ALLABOUT
NUMBEHS 79
which is the polarform for z. In this form, multiplication and division are really easy:
:
ZIZ2
r1r2si(0t*02)
and
zl _ rt ^i(ot_oz)
z2 12
Comparing relations (2.4) and (2.3), we see that we may write
eio:cos1 f isin0
a
(2.s)
relation called Euler's formula. Changing the sign of the angle, we find
e-io
:
cosd
i sin0
-
Then, adding the two relations, we have
cos 0
eiq
+ e-iq
--
(2.6)
2
and subtracting, we have
sin0:
eiq _ e-iq
(2.7)
2i
These expressions are similar to the definitions of the hyperbolic functions;
coshx
: 'x + e-x and
22
sinhx
: e' - e-'
t
Thus, we have the relation
cosh
: cosd
(l0)
(2.8)
and, conver'bely,
cos
:
(l9)
cosh
d
(2.9)
Similarly,
sinh (i 9)
: I sinp
(2.10)
I sinhO
(2.1t)
and
sin (i 0)
:
80
CHAPTER
2
COMPLEX VARIABLES
a complex number is not necessarily a real
number between - 1 and I . When using complex numbers, we must recognize that functions
may behave differently than they do when we work with numbers that are purely real.
Using these relations, we note that the cosine of
Example
2.1.
sin (2i
*
Evaluate sin(2i
n
/4)
:
,i
-l r /4).
(2i -fn I 4)
_
2i
e-2(cosn f4
:
i
(2i+n I 4)
"-
*
rt (, sinh 2 *
;
I sin ir l4)
-
e2(cosn l4
-
i
sinn /4)
2i
cosh 2)
:2.6603 + 2.5646i
Notice that both the real part and the absolute value of this number are greater than
1.
Algebraic operations on complex numbers can be regarded as geometric operations in
the complex plane (Figure 2.3, a,b, c). For example,
Addition of complex numbers <----> Vector addition (Figure 2.3a)
Im (z)
Re (z)
012
FIGURE 2.3a. Addition of complex numbers is equivalent to addition of position vectors in the
Plane. Here
zZ:
zt
I
zZ.
Using this diagram, it is straightforward to prove the inequalities
lzr -f zzl
and
lzt-zzl
'
< lzrl * lzzl
(2.t2)
llzrl -lzzll
(2.r3)
2.1 ALLABOUTNUMBERS
81
Complex conjugate <---+ Reflection in the real axis (Figure 2.3b)
Im (z)
Re (z)
FIGURE 2.3b. The complex conjugate is formed by reflection in the real axis.
Multiplication of two numbers <----+ Magnification plus rotation
For example, in Figure 2.3c we show the product
zt x
z2
-
2"ir
14
t
3"in 16 - 6"i5tt lr2
Im (z)
Re (z)
FIGURE 2.3c. Multiplication is equivalent to magnification plus rotation. Here multiplication by
3ein /6 implies magnification times 3 and rotation through art angle r /6.
82
CHAPTER
2
COMPLEX VARIABLES
2.1.2. Roots
Since e2oi : cos (2n) * i sin (2n) : l, we can add2n to the argument of any complex
number without changing its value. That is,
z:
reio
:
rri(o+2n1
Now let's see what happens when we take the nth root of this number:
_ (reiolt/"
_ rl/nri9/n
zr/n
This is the primary root. Now add2tt to the aryument of z and take the root again:
zrln
- ,tl" e*p(ito +Znlln)
_
,1
In
,i0
/n
,2ri
In
This root is distinct from the primary root. We can continue to add
2n (n - l)
times
before we repeat a root. Thus, there are n distinct nth roots of any nonzero number. They
are given by
,1/n
<'
:-' rl/noi(0*2nm)/n
w
eJ4)
r
:rt/nl"o"(t *'"^) *,r,n (t *'"^\1,
n
n
L \
/
Example 2.2. Find the cube roots of
First write
1
:
lexp
\
m
: l:
l/3
eo
1t/3r2ti/3
o
<m
<n
(2.t5)
1.
(2rim), m:0,1,2
Then the roots are exp (2n i m I 3), with m
m:0:
/J
:
0, 7, and 2.
-,7
: cos(2r/3)*isin
(2n13)
| .Jj
: --+r22
and
m:2:
l/3"4ti/3 : cos(4r13)* j sin @n/3): -)-,f
Tbking m > 2 does not give any new roots-we simply repeat the values we have
already found. The roots fall at the vertices of an equilateral triangle in the complex
plane (Figure 2.4).
2.1
ALLABOUTNUMBERS 83
Im (z)
Z2
I
0.5
-0
Zl
0.
-t
Re (z)
05
-0.5.
I
-1
FIGURE 2.4. The cube roots of unity lie at the vertices of an equilateral triangle. One vertex lies on
the real axis at the point z
:
l.
The nth roots of a number always fall at the vertices of a symmetrical n-sided figure
the complex plane. For example, the fifth roots fall at the vertices of a pentagon.
in
2.1.3. Complex Functions
Mappings
A complex function u(z) takes the numb er z : x *iy and generates a new complex number
l.D -- u * iu. Since a plane is needed to plot a single qrmber, complex functions cannot
be represented in a single two- or three-dimensional dialram. We would need 2 x 2 : 4
dimensions. Thus, instead, we use two diagrams, and we think of a complex function as
a mapping of the complex plane onto another copy of itself.
Example
2.3.
Describe the mapping
W:In terms of the coordinates
w
.r ,
I
z
y or r, 0 , we have
:u +i, : -] . :'r].i!. :l
:
x+ty x'+y'
r "-,t
peiQ
We can map out in the u.,-plane the image of a curye in the z-plane. For example, the
unit circle in the z-plane (r : l) mapsto a circle (p : 1) in the u.r-plane, but traversed
in the opposite sense (Q : -0). The inside of the circle (r < 1) in the z-plane maps
to the outside of the circle (p > l) in the u.r-plane, and vice versa (Figure 2.5). This
function gives us the complete u-plane from the complete z-plane, one to one.
84
CHAPTER
2 COMPLEXVARIABLES
FIGURE2.5. Thefunction(mapping)
u:
l/zmapstheinteriorof thecircle lel
: l totheexterior
of the circle u I : 1. Traveling around the unit circle in the z-plane counterclockwise
corresponds to traveling around the unit circle in the u.,-plane clockwise.
I
Multivalued Functions and Branch Cuts
Some functions don't work that way. The log function exhibits some of the possible curious
behaviors:
w
:
lnz
:ln
(reie)
: lnr * i0 :
u
I
iu
If we take all possible points in the z-plane with 0 < 0 < 2n, we get only one strip of the
ur-plane,0<u<2n.Thisisthefirstbranch,orprincipalbranch,ofthefunction'Toget
the whole w-plane,we have to go around the z-plane more than once; that is, we have to add
l2rn
to the argument of z. (We can imagine the plane as a pad of paper containing many
I increases
also. (We move to the next sheet in the pad and to a
new branch of the function.) The positive x-axis forms the branch cut of the cornplex plane
for this function. As we cross the positive r-axis, we move from one sheet to the next, or
to a new copy of the plane. The branch cut-here, the positive x-axis-acts as a ramp that
allows us to move from one sheet to the next. With the log function, we have to go around
infinitely many times to generate the whole ra-plane. Thus, this funption has infinitely many
branches. (Our pad of paper has infinitely many sheeti.) With n : 0, we getthe principal
sheets, each labeled by the integer n.) Every time we cross the positive.r-axis,
by 2n, and thus u increases by
2t
branch of the function.
If we choose to define 0 such that -r < e < z, then the branch cut lies along the
negative x-axis at 0 : n. Thus, we can phoose the position ofthe branch cut by choosing
the range of values for 0. The choice tha[ we make affects the value of f (z) . For example,
consider the point zo that lies on the negative imaginary axis at a distance lzol : 2 from
the origin (Figure 2.6, a,b). With the branch cut along the positive real axis, and hence
0 < 0 < 2n, zohas argument 0 :3n 12, and so for the principal branch
ln z0
lbranch cut on positive real
axis: ln 2 *
3n
,.
t
However, with the branch cut along the negative real axis, and hence
-n < 0 < 1r, zohas
2.1 ALLABOUTNUMBERS
FIGURE 2.6a. Branch cut on the
positive
FIGURE
x-axis. The poirit z6 (x : 0,
| : -2) has argtmett3tr f 2.
argument 0
: -r
85
2.6b. Branch cut on the negative
-x-axis. The point zg has argu-
mert
-n /2.
12, and so for the principal branch
ln zo
luranctr cut on negative real
axis:
,r
h 2 - -t
2
Since the two values are different, so are the two functions.
The cube root function has three branches, since adding more multiples of 2n to the
argument simply reproduces the same roots that we already found with n : O, 1,2.
For both these functions, the branch cut ends at the origin, which is the branch point for
each function. Think of the branch cut as a tadpole with a fixed head (the branch point) and
a movable tail (the branch cut).
2.1.4. An Example from Physics
To demonstrate the use of complex variables in physics, let's look at an electromagnetic
wave. The general expression for a plane wave may be written in the form
y
:
yocos
(f. *. - at i
Q)
*uu"; f , the wave vector, gives the direction of propagation
and also the wavelength, ,1. : 2r /k| o is the angular frequency; and the wave speed is
u : @lk. The phase constant, /, is the wave phase at i : 0, / : 0. For an electromagnetic
wave, the electric wave amplitude is a vector:
where y6 is the amplitude of the
i:
We can
with d
i - c,.tt -f Q)
simplify the expression by choosing our r-axis along
kx - cot + d', we can write
:
E
where
frocos (R'
:
:
io :io"iQ
Eo Re (exp
Re
io
"iQ
(ilkx
e*p
-
(ilkx
f.
Then, using equation (2.5)
ot -t QD)
-
@tl)
:
Re
io
is the complex amplitude of the wave.
exp
(ilkx
-
\
cotl)
(2.16)
86
oHAPTER2ooMPLEXVARIABLES
This mathematical description makes it easier to investigate the propagation of a wave.
Let's start with Maxwell's equations:
Y
.D: pf
(2.17)
i.f,:o
(2.18)
ai
VxE
(2.19)
At
VxH:itt-aD
u
where
D : eE is the electric
displacement, the magnetic induction
(2.20)
i :-
pfr, and H is
the magnetic field. Now assume a solution of theform(2.16) forbothE andB, and substitute
into the equations. For waves in a region where the free charge and current densities are
pf : 0 and j/ : 0, Maxwell's equations take the following form:
Gauss' law (equation 2.17):
zero,
ikeE*
-
ikB"
-
O
Gauss'law for B (equation 2.18):
Q
: O;both i and B are perpendicular to the direction of propagation.
further
by choosing our y-axis along the direction of E. Then from
Now let's simplify
(equation
2.19)
we get
Faraday's law
Thus,
E, :
0 and Bx
aEy
0x
ikE,
_
_aBz
-
iaB,
0t
(2.21)
while from Ampere's law (equation 2.20) we get
(2.22)
- -iaeEy
-ikBJ
LL
Combining equations (2.21) and (2.22), we get
kEy:r\
Since we want a solution with
E, f
/aue \
Er)
*
0, we must have
a1
-:kJw
whichis the wavephase speed. Inavacuum, relation (2.23)becomesl^lf k
(2.23)
:
l/J1t'om = c'
87
2.1 ALLABOUTNUMBERS
Now let's see what happens if we allow the wave to propagate through a conducting
medium with conductivity o, but e and p, real. There will be a current given by
j:
"i,
which must be included in Ampere's law. Equation (2.22)becomes
-ikBJ
u
:
o Et.
-
nr9-
irotE,.* B. -
-i29!
-tK
Now when we combine this equation wirh (2.21), we get
kEr-iif"-iae)p.E,
Again we may cancel the factor En to get a relation between
k2
:
irott(o
a
and k:
- iae): ialto I ro2p,e
(2.24)
We expect the frequency to be a real number,3 but then the squarg of the wave number is
complex! This means that k itself must be complex, so let's write it as
k:rc*iy
(2.2s)
Before proceeding, we should ask whether this idea makes sense. If we put our expression
for k back into equation (2.16), we find
E
:
:
Re Eo exp
(t[(r * iy)x -
Re i, exp (i[rcx ' - io cos (rx ot I
-
rot])
rlrtD exp
(-yx)
(2.26)
Q)e-Y'
thus, the rlave amplitude decreases exponentially as the wave propagates. The wave energy
is converted to kinetic energy of electrons as the electric field drives an oscillating current
in the conducting medium.
OK; let's solve for this complex k. With
k2
:
rc2
-
y2 +2ircy
equation (2.24) becomes
*2
-
y2 -f 2ircy
:
iap,o
I
a2
pte
Equating real and imaginary parts, we get two equations for rc and o:
'2-Y2:'2
l"'
(2.27)
3Thi, ir th" usual convention, but it is sometimes useful to make the opposite assumption, choosing k rcal
complex.
and
a
88
CHAPTER
2
COMPLEX VARIABLES
and
2rcy
- olto
(2.28)
Then, from equation (2.28),
(Dl.LO
r-
(2.2e)
2r
Substituting this into equation (2.27) gives
*4-r21,e*2-(T)':o
We can solve this quadratic in rc2
to get
*r:t!9(r+
2\
Since
r
is real by definition, 12 must be positive, so we take the
*' :"{t
(, *
.\
*
sign:
(*)'
(2.30)
which reduces to our previous result (2.23) when o -+ 0, as required.
Now let's solve for y. We substitute our solution (2.30) for r into equation (2.29):
'2r
@l.Lo
2er+\n+@lr$
Now in the limit of low conductivity (o <<
ra.€), these
results become
which is only slightly different from the result for a nonconducting medium, and
p
^._o
, -ty;
which is small and directly proportional to o, as we might expect.
In the limit of high conductivity (o )) oe), we get
atJ@
T-
A Y*
'l-:
(pl.Ld
2
(2.3r)
89
2.2 FUNCTIONS OF COMPLEX VARIABLES
and
@polz
FO
Y: opo
2*: 2 Urw:Y ,. :*
Now what does this all mean? According to equation (2.26), when the conductivity is
low, the amplitude decreases slowly (y {1 rc), but when the conductivity is high, y : rc
and the wave travels only a short distance into the medium, as shown inFigure 2.7 .
Ey
Es
08
06
04
02
-02
FIGURE 2.7. The magnitude of the electric field vector decreases rapidly with distance inside
conducting medium. Here we show
(the high conductivity limit).
E,
as a
function of .x at a fixed time t , with K
:
a
y
2.2. FUNCTIONS OF COMPLEX VARIABLES
We have already discussed the notion of a complex function as a rnapping of the complex
plane onto itself. It is now time to explore some properties of these mappings.
2.2.1. Continuity
A real function f (x) is continuous at x : a if, for any positive e. we can choose
6 such that whenever lhl < 6, then also
a
positive
lf@+h)-f(a)l<e
/ (x) is close to / (a) whenever x is close to a (for example, see Stewart,
Section 2.5). We can extend this definition to complex functions simply by replacing x with
z and interpreting the absolute value as the absolute value of a complex number. Then
a function is continuous If f (z + ft) is close to f (z) in the mapped plane.
or, loosely speaking,
90
oHAPTER
Example
2 ooMPLEX vARIABLES
2.4.
Showthatthefunction
f (z) : l/z: (l/r)e-id
iscontinuousexcept
at the origin.
Leth
:
n
+
i6.Then, for z
I
0,
I
f(z+n): f@+iy+n+i6): x *rt * t(y +6)
x*n-i(y-1l)
(x -t D2 + (y
*6)2
Thus, neglecting squares of the small quantities 4 and 6, we have
f(z+
h1
i'=
x -f n - i(Y -f 6)
- -'=- fQ) - '(x-lD2+(y*6)2
x2+v2
(rl+,d)(-r -iy)2
(*2 + y2
*
(x2 +2xr1
2xr7 12yD@2
+ y2)
h(x - iy)2
+ y2 +2y6)(x2 + y2)
-+0
asft-+0
Thus, the function is continuous. But the function is not defined at the origin, where
x : ! :0, and so / cannotbe continuous there.
But look at the cube root function (Section 2.1.2)wirh branch cut along the positive
real axis (0 . I < 2n).If we consider only the first, or principal, branch, the function
is not continuous. Let zo : reis and the neighboring point zr - vsi(2r'6') (Figure 2.8a).
Then
: fr(zo) :
wt : h(z) :
wo
but
rr/3 exp (i6/3)
rt/3
exp
(ilZn
-
6]/3)
FIGURE 2.8a. The principal branch ofthe cube root function is not continuous across the branch
cut, here chosen to be along the positive x-axis. The two points zg and z1 are very
close together, but on opposite sides of the branch cut'
The point ur1 is not close to u.,s (Figure 2.8b).
However, there are two more branches of this function:
fz(z)
:
rrl3
exp
(i I2r + el
fi)
2.2 FUNCTIONS OF COMPLEX VARIABLES
91
and
fzk)
: ,i't p (i Vn + el F)
"
: f2(zt) : rr/3 expfi( n - S)/31 and w3 : ftk) :
:
rr/3expti(6n -6)/31
rr/3exp(-is/3) in the ur-plane also (Figure 2.8c). Then we
Let's plot the points w2
that /3(21) lies close to fi(zd. That is, the function is continuous if we are willing to
switch from one branch to another. Points that are on opposite sides of the branch cut are
not close for a single branch of the function. We have to change from one branch to another
see
as we cross the branch cut.
: fi(e6) and
ul : ft(zt) are not close
FIGURE 2.8b. The points u;6
FIGURE2.8c. Thepoint u3
to u;6.
-
-fl(zr)
is close
together.
Functions with More Than One Branch
Point
r
The function f (z) : .tnT? has two branch points, at z : *:i. Thus, there are also two
branch cuts, one starting at each ofthe branch points. We can understand the behavior of
-f i). Let's first choose each branch cut
this function if we factor it as f (z) :
^re= Dk
to run from its branch point upward along the imaginary axis (Figure 2.9a).Then, in polar
coordinates with origin at the point z : i, we write
Z-i:preiLt,
With origin at the point z
wrtn|>Qr
3tr
2
: -i,
: pz"ih,
f2 , Qz,
'- -34
2
Then, for points on the right side of the imaginary axis above z :
Qz: n12, pr: y - 1, and p2: y * l, so
Zl-i
f (z) : J npre@r+D
with
i
, we have @1
: J0 - D(y + t)"in/2 - i 17=
:
n 12,
92
CHAPTER
2
COMPLEX VARIABLES
FIGURE 2.9a. Here we have chosen both branch cuts to run upward. We describe
distances pr and p2 from the points
@2, as shown here.
I and -i
For points on the left side of the imaginary axis above z
:
a
point z by its
and the conesponding angles
i,
Qt
@1
and
: -3n /2 and Qz : -hu /2,
SO
- lxy + D r-i3tr/2 : i{y' - |
and the function is continuoo, u".orl the imaginary axis. Below z : -i, there a.re no branch
cuts and the function is continuous. However, for points between -i and *i, we have
f(z)
:/y
On the right:
tT
f k): t/ ptr-'"12Pzsi'r/2
_y2
- JPtn:
On the left:
f (z):
J;i-E7rer;t ;It : Je
Te
_
i0 + i"-i" _ _u[ _ yz
The function is not continuous across the imaginary axis. This choice for the two branch cuts
is equivalent to having a single branch cut that runs from one point to the other (Figure 2.9b).
i to run upward while that from
On the other hand, if we choose the branch cut from z
>
(Figure
>
then
r
2.9c),
runs downward
z
12 < Qz < 3n 12.
l2but
12 Qr
: -i
Above z
:
:
-3r
j, we have
-r
On the right:
f (z) :
J;A;n;;;n - "io JTn : i
/2
y2_
On the left:
.f (z)
: J;i=tf;/r;O;n -
,-itt /2 4Qy pz :
-i \f y2 -
and the function is discontinuous. A similar situation holds below
there are no branch cuts and the function is continuous.
-i.
1
Between
i
and
-i,
93
2.2 FUNCTIONS OF COMPLEX VARIABLES
FIGURE 2.9b. The situation in Figure 2.9a is
equivalent to having a single
branch cut that runs from one
FIGURE 2.9c, Here we have chosen
one
branch cut to run upward and
the other to run downward.
branch point to the other.
2.2.2. Di|erentiabi I ity
A real function
f (x) is differentiable
f(a*h\- "'f(al
lim"'
h+0
hi
at
x
:
a if, for h >
0,
existsandequals lim
'
i:0
(for example, see Stewart, Section 2.8). Thus, the function
differentiable everywhere, while the function
f
f(a)- f(a-h)
h
f(x) :
x2 (Figure 2.10a)
ifx<0
ifx>0
|. -.r
Q;: {'
f(x)
25
20
l5
10
5
FIGURE 2.10a.
24
The function f (x) :
x2 is differentiable.
ls
94
cHAPTER2ooMPLEXVARIABLES
(Figure 2.10b) is not differentiable at x
:
0 because
-h
lim_-
h+0
h
lim
h-->o
-l : -l
while
-/(o-r) :
i:ohh-ohn-o
l.m /(o)
tin]
o-,(-fr)
:
lim I
:
I
The two limits are not equal, and thus the derivative does not exist.
f(x)
FIGURE 2.10b. This function has an abrupt change of slope at the origin: limr-;g1 df /dx
but limr-6- df /dx : *1. The function is not differentiable at the origin.
: -l
Now with complex functions, we need to let one point approach the other not just along
the real axis, in either direction, but along any line in the complex plane (Figure 2.ll).ln
particular, we must get the same limit for the derivative if we approach z from a neighboring
point along either the real or the imaginary axisa:
.. .f(z+dz)-f(z)l
rrrrr
dz-0
That is, with
----------
/(z)
lim
:
u(x
.. f(z+dz)-f(dl
rrrrr
----l
laz:ax dz-O
dZ
clz
laz:idy
u(x, y) + iu(x, y),
*
dx, y) + iu(x
* dx, y) -
u(x, y)
-
dv)
u(x' v)
iu(x, y)
dx
dx--+O
:
lim
dy-->o
u(x, y
I
dy) + iu(x,
!*
idv
4This condition tums out to be not only necessary but also sufficient.
-
-
iu(x, v)
95
2.2 FUNCTIONS OF COMPLEX VARIABLES
FIGURE 2.11. For
a function to be differentiable at z, we must get the same
the point z along any line in the complex plane.
limit when we approach
Taking the limits, we obtain
0u
u+i
Du l/0u
u:
Eu\ : 0u Du
*'
;(u, u) -' ur* u,
Equating real and imaginary parts, we get the Cauchy-Riemann relations:
0u
0x
0u
0y
(2.32)
and
0u
6x
0u
0y
(2.33)
These relations must be satisfied by any function that is differentiable.s
We may now write the derivative as
df 0u .0u 0u .0u
dz 0x "ax- ay '0y
(2.34)
Conversely, if the Cauchy-Riemann conditions are satisfied and the partial derivatives are
continuous, then the function is differentiable. Consider a small displacement 6z : 6x*i 5y.
Then
ur:(#.,#)u,*(X+tfi)at
5See Problem 17
for an alternative statement ofthe differentiability condition.
96
oHAPTER2coMPLEXVARIABLES
Applying equations (2.32) and (2.33), we find
ur:
(y.,#)
(dx*i uD:
(#+tfr)u
and thus
d.f
-i: ,. 3f : (0, +i.0u\
ol!,* [* u )
for any 62.
Example
2.5'
Show that the function
First note that z2
of the function are
-
(x
*
iy)z
:
f
(z)
:
*2 + zixy
z2 is differentiable'
y2, so the real and imaginary parts
-
u(x, Y) : x2 -
Y2
and
u(x, y)
-
2xy
Then
0u^Eu
Ex
Dy
and
0u
0u
- -1r'
-0y-
0x
so the Cauchy-Riemann relations are satisfied and the partial derivatives are continuous. The derivative is
df
3u .0u I(0u
.0u\
;:i*';:;(ur*'u)
:2(x *
iy)
:22
which is exactly the expression we might expect. In fact, complex derivatives obey
all the usual rules of real derivatives, such as the product rule.
As is the case with real functions, complex functions that "blow up" are not differentiable,
and functions that are not continuous are not differentiable. Thus, functions that have branch
cuts are not differentiable at the cut. However, it is possible for a complex function to be
continuous everywhere but differentiable nowhere. The function f (z) : 7* : r - ly is
one such function. With 0u l0x -- -0u l0y : 1, the Cauchy-Riemann conditions are never
satisfied.
lI
2.2 FUNoToNS oF coMpLEX vARTABLES 97
r
2.2.3. Analyticity
A function / that is differentiable tt I : a andwithin aneighborhood6of z: a is said
to be analytic vt 7 : 6. Thus, a function analytic at a satisfies the Cauchy-Riemann relations within a small but finite region around z : a. The requirement that the function be
differentiable within a neighborhood ofa guarantees the existence ofhigher-order derivatives as well. A function that is analytic in the whole complex plane (except perhaps at
infinity) is called an entire function. Analyticity is a powerful constraint on a function, as
we shall see.
2.2.4. lntegrals
Since a complex function
integral
/(z)
represents a mapping of the z-plane onto the u-plane, any
fz2
I
Jzt
f(oa,
is an integralT along a path that must be specified between the points 7y and zz. Expanding
the integrand, we find
lr','
f {r) o,
:
f,'r'
@
*
iu)(dx
*
i dy)
Defining the vectors d, with components (u,
write the integral as
l,','
f k)or:
f,','
:
f,'r'
-u),
i
u
and
ai+ i
dx
i,
-
u
dy
*
i(u dy -f u dx)
with components (u, z), we may
l,','I
ai
(2.3s)
The Cauchy Theorem
Next we shall proves an extremely powerful theorem about integrals of analytic functions.
Consider the integral from 41 to z2 alonlgpath C1 and returning from zz to zr along path C2
(Figure 2.12).The combined path forms a closed curve C in the complex plane, and result
(2.35) applied to this path gives
f"rrtor:
fri.ai+
i
frt ai
6Aneighborhoodofz:aisasmallregioninthecomplexplanethatcontainsthepointz:astrictlywithin
example, the circle defined by lz - al < e for any e >
it-for
TSee Section 1.2.3 for information on path integrals.
SThe proof presented here has the advantage of being
relatively straightforward, but it requires continuity of the
partial derivatives. An altemative proof due to Goursat that does not require the use of Green's theorem (see, for
example, Jeffreys and Jeffreys, Section I L052) requires only that /(3) be differentiable and C have finite length.
98
CHAPTER
2
COMPLEX VARIABLES
Im (z)
Re (z)
f <O dzis apathintegral in the complex plane. Here the two paths
C1 and C2 canbe combined to form a closed curve C from 31 to z2 and returning
FIGURE 2.12. The integral
Ili
to {1.
Then we may use Green's theorem (equation I .47) to write each integral over C as a double
integral over the surface S spanning C:
t',
Now
if
.f (z) dz
:
l,(#
- x)
dx
dv,'
l,(# - #) o. no
the function is analytic everywhere on S, then both integrands are zero, by the
Cauchy-Riemann relations, and so we find
If f (z) is analytic in and on C, then
$ rrr;dz:o
Jc"
a
(2.36)
result known as the Cauchy theorem.
Cauchy's theorem applies to curves defined in a simply connected region. A curve in
simply connected region can be shrunk continuously to a point while remaining inside the
region. If the region has a hole in it (Figure 2.I3) and the curve surrounds the hole, then the
region is not simply connected and the theorem does not hold. We will see regions like this
in the following sections, and we will also find some ways to get around the restrictions.
An importantcorollary to Cauchy's theorem is thatthe integral of a function /(z) between
two points zr and 72 is path independent if the function / is analytic everywhere in the
region containing the paths of interest. This is important because it means you can often
use a path that hugely simplifies the integration.
a
2.2 FUNCTIONS OF COMPLEX VARIABLES
99
FIGURE 2.13. This white region has a hole (shaded circle), and so it is not simply connected. As
the curve C shrinks, it surrounds the hole and can never shrink to a point.
2.2.5. The Cauchy Formula
Careful application of the Cauchy theorem leads to many interesting results. For example,
consider the integralg
I:Q t'l
JcZ_a
-dz
where C is aclosed curve in the complexplane. The function f (z) : l/(z - a) is analytic
except at the single point z - a. Thus, by the Cauchy theorem, the integral is zero if the
curve C does not enclose the point z : a. If the curve includes z : e, we consider the
integral
I
,':fo _dz
z-a
where the curye C' : C + Br * f f 82 (Figure 2.14).The segments 81 and 82 are very
close together. Note that this new curve C/ does not enclose the point z : e, and thus the
function / is analytic everywhere in and on C'. Thus, I' is zero by the Cauchy theorem.
Since the function / is analytic everywhere except at z - a, it is continuous along 81 and
82 and so the contributions to the integral I' along .B1 and 82 cancel. (The integrands are
equal, but the path is traversed in opposite directions. Equivalently, the upper limit of the
9The symbol
$a as used here means an integral counterclockwise around the closed path C. When you traverse
such a curve, the interior of the curve is always to your left.
100
CHAPTER
2
COMPLEX VARIABLES
Im (z)
:
a from its interior. (When
as to exclude the point z
you travel counterclockwise around the curve, the interior, shown shaded, is on your
FIGURE 2.14. Thecurve C/ is constructed so
left.)
first integral is the lower limit of the second, and vice versa.)
| d,
I
or+6
o:{
'
- Jc+Br*riB2Z-adr:$
- Ic'z-a' or.:f
Jf.clockwiseZ-d
J6z-a
and so
ftfl
fr, - "o':
./r,.ro.r.* rr",
- od'
The integral around f is taken clockwise, the reverse of the usual direction. We may change
the direction if we also change the sign in front ofthe integral:
flfl
frr-"d':frr-ooNowlo on
f, z :
a
I
peiq,so d7
f\f2'l
ipeiq de andthen
l' ' - 'o': Jo fuo''' i ag
t2n
:, I
do
:Zri
Jo
lORemember: Addition ofcomplex numbers is equivalent to addition ofvectors in tbe complex plane (Figure 2.3).
2.3 COMPLEX SERIES
101
Thus, we obtain the first Cauchy formula:
I | ,__ lZni
Icz-o"\ o
if z:alieswithinC
We can easily modify the proof to show that for any function
$ fk, or:[ t'rlll\ar:
Jcz-'
/
that is analytic in and on C,
*+#aYi\ide
fo'"
(2.37)
otherwise
if z:arieswitbinc
otherwise
[o
Then we 7et p --> 0, to obtain
(
6 !''')^ar: 1
Jc1-
if
r2n
ta)
[ 0
"
Jo
d0
:2tif (a) if z :
a lies within
c
e3B)
otherwise
We may extend this result to obtain the general Cauchy formula:
t'rj- "9r": ffi
f('-r'{ztl,-o ir z:
a lies within
c
(2.3e)
The proof of this more general result is in Appendix III. This result verifies the claim
made in Section 2.2.3 that analytic functions possess derivatives of all orders.
Example2.6. Evaluate the integral
/
J circle
lzl:l
'ot'O,
Zt
around the circle of unit
radius centered at the origin.
The function f (z) : cos e is analytic within and on the circle lzl
Cauchy formula with n : 3, we have
cosz
-f
I
./.n.,.
,.,:,
a'
:
7. Using the
d2
: ;2tri fu
t'1.:o : zi (- cos zl':o) : -ti
""t
I
2.3. COMPLEX SERIES
2.3.1. Real Sequences and Series-A Review
Sequences
A sequence of numbers an is an ordered set with a rule for computing lhe nth element in the
set. The number a, is the nth number in the sequence and can often be written as a function
of n. Some examples are
l--
102
oHAPTER2coMPLEXVARTABLES
for which en
:
n, 2;rd
s2
:
{1,
-1,
1,
-1,...}
for which
a,
: (-I)n-r
unbounded if, for any real number M > 0, there is at least one member of the
sequence dn such thatlarl > M.The sequence ,Sr above is unbounded, since we need only
take M - N, any integer, and infinitely mar'y an are greater than N, no matter how large
N is. Conversely, if we can choose an M such that all members of the sequence S satisfy
larl < M, then the sequence is bounded. Sequence 52 is bounded-we could take M :2.
A sequence S is corwergent if therc is a number s such that, given any positive number
r, we can choose anm so that for every n > m
A sequence
S is
lan-sl<e
in which case the limit of the sequence is s. We write
n\on - t
or, more loosely,
an-->s ast -+@
Neither of the sequences .tl or 52 is convergent, but the sequence
s,
r ,;,
r
: (t,r,t t.r 4,
t
)
j
converges to zero. (Take m to be any integer greater than 1/e.)
A sequence that is bounded but not convergent-for example, sequence
oscillate finitely. Another example is
S2-is
said to
s*: {1* t-,,'}
L,?
)
Other sequences may be unbounded and not convergent-for example, sequence
51
above, but also series such as
55
: {/ncos (zn)}
which is oscillating but unbounded (infinitely oscillating).
Further properties of sequences can be found in texts on calculus (for example, see
Stewart, Chapter 11) or on real analysis.
Series
A
series is formed by summing the terms of a sequence. Using sequence 51, we form the
series
i': i,
n:1
n:1
e.4o)
2.3 COMPLEX
SERIES
103
A second sequence is formed from the partial sums of the series:
,+^:io,
n:l
lfthis sequence ofpartial sums is coryergent, then we say that the series is convergent. One
requirement for convergence is for the successive terrns to approach zero;that is, a, -+ 0.
Conversely, the series diverges
if
n\o" 7 o
This includes the case in which the limit does not exist at all.
Clearly, series (2.40) is not convergent, since limr-- a, does not exist. But what about
the
following series?
$t
Ln
(2.41)
n:l
: Iln
do approach zero, but not very fast.
Here the answer is not so clear. The terms dn
It is possible for a series to diverge even though an -+ O. We'l1 need to /esl this series for
convergence (see below).
If the series in which each an is replaced by its absolute value,
it',t
n:I
is said to be absolutely convergent.
Series that converge, but not absolutely, are said to be conditionally convergent. These
series do not have a well-defined sum, since by rearranging the terms we can achieve any
sum that we want (for example, see Stewart, p.736;Arfken and Weber, Section 5.4). Thus,
also converges, then the series
they must be used with caution. In contrast, an absolutely convergent series has a welldefined sum. In addition, absolutely convergent series may be multiplied together to form
a double sum that converges to the product of the original two sums.
There are numerous tests for convergence of a series. A number are listed in Gradshteyn
and Ryzhik, Section 0.22. (See also Stewart, Section 71.2.) They include the following.
The root test. If
]\lo*l'tk
andq
<
1, the series converges absolutely,
fails.
The ratio test.
but
:
q
if4 > l, the series diverges. Ifq :
test
If
r.,, lon*tl:o'
r-ol
ar I
1,11e
104
cHAPTEB
2 ooMPLEXVARTABLES
andq<l,theseriesconvergesabsolutely,butifq>l,theseriesdiverges.If4:1,fl[s
test fails, unless the ratio remains greater than 1 as the limit is approached, in which case
the series also diverges.
Theintegraltest. If
ap: f (k), where /(-r) isdefinedforx > q > l, thentheseries
converges or diverges according to whether the integral
r@
I
Jq
fGsax
converges or diverges.
The comparison test.
Iflanl < bn and ! b, converges, then ! a, converges absolutely.
The alternating series test. If the series is altemating (that is, successive terms in the
series alternate in sign), then the series converges provided
thatlalrlll
.
la*l and
la7.
|
-+
0.
i 1. Oru.rr"r.
Let's apply the integral test. Since oo : tii'i! aefined for k > l, we look at
Example2.7. Show that the series (2.41),
: lnrlf
l,* )0,
which diverges. Thus, the series also diverges.
If the series elements an are notjust numbers but functions of some variable, say x,
then convergence of the series may depend on the value of .r chosen. For any particular
value of -r, we can apply any of the tests listed above. The result determines the pointwise
convergence ofthe series at that value ofx. For example, the series
r"
converges if lxl < I but diverges otherwise. A particularly valuable series is one that converges independent of the value of x and converges equally well for all x. Such series are
said to be unifurmly convergent.
is uniformly convergent to / (x) in an interval 1 if, for any arbitrarily
small positive number €, we can find a number N such that
A sequence
f"(x)
lfo@)-f@)l.e
for all p > N, where N is independent of x in 1.
If
the uniformly convergent sequence f"(x) is the sequence of partial sums of a series
f,(x) : Dk:ou*(x), then that series is uniformly convergent in 1.
One of the major advantages of uniformly convergent series is that they may be integrated
and differentiated term by term, provided that the individual terms are continuous (for
integration) or differentiable (for differentiation).
2.3 COMPLEX
Tests for uniform convergence include theWeierstrass
SERIES
105
M test:
If there is a sequence of numbers M, suchthat lu"(x)l < M, for all x in the interval
la,bf and the series D|rM" converges, then the series
uniformly and absolutely in the intewal [a, b].
f[,
ur(x)
converges
Uniform convergence and absolute convergence are independent properties of a series.
Thus, the most useful series are those that converge both uniformly and absolutely.
2.3.2. Complex Series
A complex sequence is
a sequence
of complex numbers and thus is actually two sequences
of real numbers:
Zn:xrliYn
f
The sequence zn converges if and only
both the sequences
x,
and ),,? converge. Similarly,
the complex series
Drn:D*,*i\t"
if and only if both real series I x, and I y, converge.
We can define absolute convergence for a complex series as we did for a real series:
A series converges absolutely if the series of absolute values converges. But for complex
series, we are now talking about convergence of athird seies,
converges
f l.,l:f
x]+fi
which is different from the previous two. Thus, absolute convergence is a more powerful
constraint for complex series than for real series. If a series converges absolutely for z in
a range lz - zol < R, then R is called the radius of convergence for the series.
Let's look at an important example of a complex series: the geometric series
oo
fon
.L'
n:0
The
patial sums are
Sxr(a)
Now
if lzl <
:
* "' *zN
1 _zN*l
_1-zN+l_
I-z
l-z
l-z
I
*
z
-f
12
+
z3
1, then
t-N+1 I
r-rN+l
ll - z I
11
lt
l-lzl
-
zl
*0
asN--+rc
(2.42)
106
CHAPTER
2
COMPLEX VARIABLES
and so
oo
\-,n
/-'
I
-
n:o
I-z
iflzl <1
(2.43)
The geometric series is uniformly convergent for lzl < r arrd
r (here 1) is the radius of convergence for the series.
r <
1. The limiting
magnitude of
2.3.3. TheTaylor Series
Suppose that a function
z
:
f (z) is analytic in a region R: lz - al < p, centered on the point
f (z) as a series in powers of (z - a):
a. Then we may express
-f(d
: f h) * k - a) f' (at * tJ
t"(o) + "'
This series is uniformly convergent within the circle lz
*"
;i'' Hl,:"*
(2.44)
- al < p, where p is the radius of
convergence.
To prove this result, we'll use the Cauchy formula (equation 2.38) and the geometric
series (2.43). First we construct a closed curve f that surrounds both the points z and a
(Figure 2.15). Then we may wite f (z) as an integral around the curve f :
Next we write
where
lz.-al
I
1.4.1
l€-olsincezisinsideI-,andthusclosertoathanf,foranyfonf.(Wearefreetoconstruct
f
to make sure this is true.) Thus, we can use the geometric series (2.43) to expand the
second fraction:
+:(+) ('.=.(=)'.
t \$/.-aV
:f
:\s-")k\€-
")
)
2.3 COMPLEX
SERIES
1O7
Im (z)
Re (z)
FIGURE 2.15. The curve f for evaluating the coefficients in the Taylor series.
within the region R and encloses the point I : o,.
The geometric series is uniformly convergent on
respect to
f
f
is entirely enclosed
, so we may integrate term by term with
f:
r(z):+in:o e-ayn Ae*rru
27tt
(2.4s)
-
Then we may use the Cauchy formula (equation 2.39) to express the integral in terms of
derivatives of /:
a^f
f(z):i(z-d"
nt dzn lr=o
(2.46)
I
-
as
we set out to prove.
Now suppose there exists some other series for
/
about the same point a:
f (z):Dr,e - o)'
n--0
What can we say about the coefficients
times. Then
!clz :i
nr,(z
n=l
-
c, in this
series? Let's differentiate the series la
ayn-l
@
t2c
u
.I : )4^ n(, - l)cnk dz'
a)n-2
n:z
o: I^
dz'n
:
mtc^
+ i
n:m+I
nfn
-
1).. . (n
-
m
*
t)cnk
-
a)n-m
108
oHAPTER2 ooMPLEXVARTABLES
Evaluating tt 7
:
6, we have
dn
"f |
I
dz^ lr:o
So
if
-
ml.c.
any such series exists, its coefficients are given by equation (2.44).
Example
2.8.
Find the Taylor series about the origin for the exponential function
f (z) :
ez
:
ex-fiy
:
ete'J
:
er cos y
I
iex stny
:
0) is
This function is analytic everywhere:
3u: ,,
cos), :
A*
and
"^
Eu
Ay
3u r.
0u
A*:'-'smY:-ay
The derivatives are
d'f _
dzn
",
for every n, and so the Taylor series about the origin (a
-z2zn
e.:t*r+;+...+1t
This series is exactly what we might expect, since e" has the same form when x is
real.
2.3.4. The Laurent Series
Suppose that the function / is analytic in an annular region R centered ort I : ai pt <
lz - al < p2. Then we may still expand / in a series of powers of (z - a), but the series
will include negative as well as positive powers. This series is called a Laurent series. To
find the coefficients formally, we apply the same techniques that we used for the Taylor
series.
As before, we construct a curve f that surrounds the point z and lies entirely within
the region R. This time the curve cannot also surround a, but we can distort it to form
a composite curve having four parfs (Figure 2.16):
1. The original curve
f
with two snips in it
just inside the inner border of the annulus at lz
,
2. A circular curve C1
- al
3. A circular cuwe C2just inside the outer border of the annulus at lz - al
4. A set of cross cuts linking f to C1 and f to C2
:
:
pt
p2
t
-
e
6
2
3
COMPLEX SERIES
109
Im (z)
Re (z)
FIGURE 2.16. A curve for finding the Laurent series lies within the annulus but excludes the point
2.. C : C1 -l CZ I | - cross cuts.
Now we proceed as before, writing
/
as an
integral around the curve
f
:
t f f(€)dE
I(z):fifrq_,
But, when we traverse our composite curve as shown in the diagram, the integrand is analytic
everywhere inside and on the constructed composite curve, and thus the integral is zero by
the Cauchy theor m.
f,#o' : t'r,*ou * -{..",*0,',,"#.ou *
+$
{(€)-os :o
Jcrosscuts
t'r,.c,ockwise
*ot
I - (
The sum ofthe integrals along the cross cuts is zero, since the function is analytic within
R, and hence continuous at the cross cuts. We can use a minus sign to change the direction
ofthe clockwise integrals to the usual counterclockwise sense. Thus,
{ lLot: JcrE-z
$ {'€' oE - Jcrt-:
$ !"' oe
Jr€-z
Now we proceed as we did with the Taylor series.
For f on Cz, lz - al < l€ - al, and so we expand the denominator as we did before,
obtaining the same result (equation2.45 but not equation 2.46).However, for f on C1,
lz - al . l€ - al, so we have to expand as follows:
I
t-z
-1
(z-a)
1
-1 s/6-oY
(,-Lg\ - z-qk\r-')
\
z-o)
110
cHAPTER2 coMPLEXVARTABLES
and thus
-t t
IIc, €f@dt-$
-a)' d€
z
fr,f(€)(€
7:ok-o)'*a
which gives a series in negative powers of z
$
-
Pou:_i Q-o;n $
JLt 5
.
p
a. Writing
Jcr
,:_,
Here we cannot easily express the coefficents of (z
could for the Taylor series.
Putting together the two parts, we get
- - (n'f l), we get
-
f
(€) (€
-
a)p in terms of derivatives,
r(z):*A*ou: f
o-e-t
dE
as we
c,(z-a)n
where
,,: * t'r4*rtt
ror -oo <n <@
(2.47)
Once we have the result, we can evaluate the integrals over a cornmon circle C centered on a
and lying within the annulus, since the integrands in equation(2.47) are analytic everywhere
in the annulus.
This expression (2.41) for cn amply demonstrates the existence of the Laurent series, but
it is not generally very useful for finding a particular Laurent series. We can resort to a set
of "tricks" that work better. The next example illustrates the geometric series method.
. : t ,o, the function ^f+it;,t*" cannot
rhe function is not analytic at the r*o oo,ni, . : ;;;;:-1.
draw a circular region centered at z : 1 within which the function is analytic. So,
there is no Thylor series for this function about the point z : L However, there are
two annuli centered atz: l:0 . lz - ll . 2,with the point z: -l outside the
annulus and z - 1 inside an infinitesimal hole in the center, and2 < lz - 1l < oo,
Example
2.9.
Find a series about the point
with both points in the inner hole. Thus, we can find two Laurent series centered at
1: oneintheregion0 . lz
In the inner region, we have
z:
- ll
<2andasecondfor lz - Il > 2(Figtxe2.l7).
t/ 1
(z-1)(z+l)
tl
-t\.J-r+t)
I \
2.3 ooMPLEX
sERTES
111
Im (z)
2
Re (z)
4
FIGURE2.17. ThefunctioninExarnple2.9isnotanalyticatz:*l.ThereisaLaurentseriesvalid
in each of the annuli 0
<
lz
- ll < 2 and lz - ll > 2.
Now we can expand the second fraction using the geometric series. In the inner
annulus, lz
z
- Il < 2, so
11loo
+
t
- z - t +, -, (,.
\z/
and thus we have
:
+J :n(-"'(?)
*::l* :n'-"'(=)']
:1S
1-'in*r (t-1\'
\ z )
4'/'t't
/
'n:-1
which is a Laurent series with only one negative power: (z - g-t .
However, in the outer annulus, lz - 1l > 2, so we expand the fraction the other
way:
;-:;;041 :*i'-'r(=)'
and
+.:il-!-i,-D'=nnl
z2-t zlr-l f,:o \4-,, I
:- li
2?
-J-!(z-l)'+r
a series with infinitely many negative powers of z
e.4B)
- I and no positive powers.
112
CHAPTER
2 ooMPLEX
vARTABLES
2.4. COMPLEX NUMBERS AND LAPLACE'S EQUATION
2.4.1. The Equations of lncompressible Fluid Flow
i(i),
giving the velocity of each fluid element,
a vector field
and a scalar field p (i), giving the density at each point. Since the mass within a volume can
change only by flow of fluid into or out of the element,ll we have
A fluid may be described by
dff
; J,'dv:-
Jron'n'ae
where the minus sign arises because the unit vector ff normal to S points out of the volume
V. Now if we apply this relation to a fixed volume V, we can then apply the divergence
theoreml2 and rearrange to get
l,e+n
rpi))
dv:o
Then since this relation must hold for any fixed volume V, we may conclude that
3o
(2.4e)
;*r'(pi):o
which is known as the continuity equation.
Liquids have the property that their density remains almost constant under a very wide
range of applied pressure; they are incompressible. (Gases can also behave incompressibly
under some circumstances.) For incompressible fluids, equation (2.49) simplifies considerably, since the time and space derivatives of the density are identically zeroi
i.i:o
(2.50)
Certain classes of fluid flow are also irrotational,
ixi:O
(2.51)
which means that there are no swirling motions in the flow (no turbulence).
2.4.2. The Velocity Potential
A vector field that satisfies equation (2.51) may be described
function
@, since for any such
function
Vx(V@):Q
In our fluid flow problem, the function
@
as the gradient
of a scalar
i
is called the velocity potential. It is usual to
introduce a minus sign, so that
i:
1 1
See Chapter 1, Problem 6.
12See
Chapter 1, Section 1.2.3
u"i -f ury :
-fi6
l
(2.s2)
2,4 COMPLEX NUMBERS AND LAPLACE'S
EOUATION 113
Then we can put this result into equation (2.50) to get
Y'O
:0
which is Laplace's equation. In this respect, the solutions to irrotational, incompressible
fluid flow problems will resemble the solutions to problems in electrostatics, since the
governing differential equation is the same.
2.4.3. Analytic Functions as Solutions of Laplace's Equation
An analytic function w(z)
and2.33):
:
*
u
iu satisfies the Cauchy-Riemann relations (equations 2.32
0u 0u
Ex 0y
0u
0y
anl
0u
0x
Differentiating again, we have
02u02uA/02\
a*r: a*ay: O (-O/
and thus
02u
02u
ffi+#:y2u:o
(2.s3)
That is, u satisfies Laplace's equation in two dimensions. Such functions are said to be
harmonic. Similarly, we can show that u is also harmonic. Thus, the real part or the imaginary
part of any analytic function will be the solution of an incompressible, irrotational fluid flow
problem in two dimensions.
Let's define an analytic, complex velocity potential function
a:Q*ilr
The fluid velocity is given by equation (2.52).In particular, if we use the real part
as our solution for the real velocity potential, then the velocity components are
a0
a0
U*: anq ,y: 0X
Ay:
al'
A*
and so we can form the complex number
.
:Ux*iry-Then
/a0 .a,/\
(,r-,*)/,
.:-(#.,y):-#
@
of O
114
oHAPTER2coMPLEXVARIABLES
Thus, the velocity components are found from the real and imaginary parts of the derivative
of O; that is,
ux:
/ da\
-*'(a/
and
uy
/ da\
: t- (*/
(2.s4)
Now any physics problem is described mathematically by one or more differential equations plus a set of boundary conditions. For example, fluid cannot flow across a nonporous
boundary, such as a solid wall. On such a boundary we have the boundary condition
i.fi:-n.i4:o
(2.ss)
Thus, to solve such a problem we need only find an analytic function Q : Q * i ry' whose
real part satisfies the boundary condition (2.55).The real part @ then satisfies both Laplace's
equation and the boundary conditions, and the problem is solved.
If the real part of @ is the velocity potential, what is the imaginary part? Note that
3 :
=
V6'V{t
AlL+ A0 ArL :
0y Dy
0x 0x --
A0
--L---:-
u!(_y\ *ua:o
a'\ oy) oyox
(2.s6)
Thus, surfaces of constant ry' are perpendicular to surfaces of constant d. Ghis is the same
relation as that between equipotential surfaces and electric field lines in electrostatics.) The
constant ry' surfaces are thus the streamlines of the flow, and from equation (2.55), a solid
boundary surface must be a surface of constant ry'.
2.4.4. Steady lrrotational Flow Around an lnfinitely Long Cylinder
Suppose that fluid flows from a great distance with velocity i : Voi toward an infinitely
long solid cylinder of radius a (Figure 2.18). We want to find the flow velocity around
the cylinder. To solve this problem, we will use cylindrical coordinates with origin on the
cylinder axis. It is a two-dimensional problem with the boundary conditions
i -+ V6i
as
r -+ oo
(2.57)
and since the flow has to be al,ong the surface of the cylinder, streamlines follow the surface
a.
constant on the circle r
at r
a and
:
:
{ :
Y0
FIGURE 2.18. Fluid flows toward the cylinder (radius a) from infinity with velocity i
:
Voi.
2.4 COMPLEX NUMBERS AND LAPLACE'S
EQUATION 115
First we reformulate the problem in terms of complex functions. The solution for @ is the
real part of a complex function Q
Q * I ry' that is analytic everywhere outside the cylinder
and that satisfies the two boundary conditions. Since O is analytic everywhere outside the
cylinder (that is, in the region r > a), it can be written as a Laurent seriesl3 centered at the
:
origin:
*oo
:
(2.58)
n\'n'"
The real part of this function satisfies the differential equation for our problem. Now we
also need to satisfy the boundary conditions. At infinity, we have
,,: -# - vo:-* (f )
Thus, as
lzl -+ oo, Q : -Voz. The coefficients
cn,
and
n2
uy
:0
2, of the positive powers in
equation (2.58) are all zero, and the coefficient of z must be c1 : - Vo. The other coefficients
cn, n 1 0, are not yet determined, since all the negative powers of z go to zero at infinity.
The second boundary condition is f - constant on r : a, and we may choose that
constant to be zero. (As with electrostatic problems, we may add an arbitrary constant to
the potential without changing the values of its gradient, here the velocity field.) Now the
imaginary part of a complex function may be written as
lb:E(a_a*)
1
and we want this expression to be identically zerc for lxl : a. Inserting the series (2.58)
for O, with both z and c, in polar form (z : aeiq a\d cn : lcnl ei|'), we find
./
o
oo
\
: +(
,-'0 ) +f ,-,o-n r-in' - ,\no-n eine I
z'\ -vsa(eie n:t
/
o:
-voa
sind
:
-voa
sind
- i"n:7
c-nlsin (no - 3-,)
(9
E-,)
- .EJ
a rin - d-r) -to-"lr-nlsin(n0 -
n:z
We can14 satisfy this equation by taking
lc-nl
:0
for n
>
1 and
lc-r I : a2Vo, 6-t
Then
/
a2\
(D:-Volz*-l
\ z/ -
-vo(,.#) - -vo(,.#)
13See Section 2.3.4.
14In Chapter 4, we will show that this is the only possible solution.
-
n.
116
cHAPTER2 ooMPLEXVARTABLES
and on the curve lzl
:
a,
Q:-VoQ*z*):6*
as we require. Thus, our
Laurent series has only two terms, with powers
zl
and z-l
.
The velocity components are given by
ll)
*dQ
:
--
dz
:
vo(r
:
Vol I
-5)
/a2
\r' :ux-iuy
-;€ -rtt)
So
ux
:
vo(t
- 5"o, za)
;
vv
: -Voa2., sin20
The equipotential surfaces are
:
constant
: -Vol/ r - :a2\I sind :
r/
\
constant
0- -ro(, .
+)cosg
and the streamlines are given by
tb
The streamlines are shown in Figure 2.19.
o
FIGURE 2.19. Streamlines of flow around the cylinder. The lines are given by
t :
constant.
2.5 POLES AND
ZEROS 117
2.5. POLES AND ZEROS
2.5.1. Analytic Continuation
a function is analytic in a region R, pt < lz - al < p2, we may describe it using
a Laurent series valid in R. We may also be able to find a second Laurent series in a
different, but overlapping, region. (See, for example, the discussion of series for the tangent
function in Appendix XI.) The second series is an analytic continuation of / into the
second region.
For example, consider the function
If
srn z
frk) z-l
defined in the region
lzl < I (Figure 2.20).We may
evaluate the function with a Taylor
series:
/
-t
\.
-s
fzk):- (.- ;..;..
:
-z
- "-z -e"
-2; -
)(r
5,4
6<.
-
+. +22 +23 +24 +zs +...)
lol -t -.....
l2O. -"'
where f2Q) is valid only in the region lzl < l To evaluate the functionls outside this
region, we can expand about another point in the original region-for example, zt : -r /4.
Im (z)
Re (z)
FIGURE 2.20. The function
'' -- -*t4.
rtl
fzk)
may be analytically continued by expanding in a series about
.I
l5You might think that this would be unnecessary, since we already have the closed-form expression
fi
(3). But
we cannot actually compute sin z except through its series expansion. See Morse and Feshbach, p. 377, for a good
discussion of this point.
118
oHAPTER2 ooMPLEXVARTABLES
For convenience, we define a new variable u)
:
z
- zr :
z
-f
n 14.
Then we obtain the
series
ftk):
sin (z
/sinrl-cosu.,\
t r/a - n/4) sin(w - n l4) I t_t
-lnla)
(r
t)
u
r14/Z \ w-t.78s4 )
w - w3
+..' - (t - w212t+..')
r.78s4(t - w/1.7854)
(z+n1+
t
- -n
131
l/*2w3\
:138s4J-2(.t-'*i*
o*
/ww2w3\
xl1*
" \' ' :.zB54'I
:
0. 39605(1
-
)
r
' 1.78543 ' )
-r--r...
0.4399w * 0.25361w2 + 0.3087t1r,3 + ..' )
1.78542
< l-ln l4,or lz * T l4l < l-lr l4 : 1.7854'
Thus, we have extended the function fzk) into the shaded region shown inFigure2.2l.
andthis series /s(z) is validintheregion lu.rl
We can now pick a point in the new region and continue to extend the function indefinitely. The expressions ftQ), fzk), and fzk) are three different expressions for the same
function.
Im
FIGURE 2.21. The function
fzk)
(z)
has been continued into the shaded region as
fzk).
In some circumstances, we can simply use the original series outside of R as well, but
it does not necessarily describe the original function in the new region. The new function
is also an analytic continuation of /. In this case, we are redefining the function to be the
2.5 POLES AND
series. This result
at the function
is
ZEROS
119
called permanence of algebraic form. For example, suppose we look
111
f(x):i+z*r*:"r*"'
x is on the real axis with lxl > 1. (This series represents the real function
lnfxl@ - l)].) We can use this series everywhere in the annulus lzl > 1 in the complex plane, where it represents a complex function (which we might call the principal
where
lk - 1)]) that is identical with the original function on the real axis. It is an
analytic continuation of the original function.
Notice that we talk about "an" analytic continuation, not "the" analytic continuation.
There is more than one way to continue a given function. In any specific application, we
must choose the continuation that is best for that particular purpose.
branch of ln[z
2.5.2. Zeros
Thepoint
expand
/
z: aisazeroof
in
a
afunction
f it f (a):0. If /
- al < p:
is analytic ata,then wecan
Taylor series in some region lz
f (z)
:i',r,
-
n:0
oY
f , then c0 must be zero. If c1 10, then a is a simple zero of /. If both c6
andclarezeroandrzf0,thenaisazeroofordertwo.Ingeneral,lfcg,cl,...,cn-lare
If a is a zero of
/ of order n. Essentially, the order of the zero of a
6 is a statement about how fast the function goes to zero as z approaches a.
all zero but cn is not, then a rs a zero of
function ?t I
Example
:
2.10.
The function
If we expand in a Thylor
f (z) :
sin (22) has
a
zero at z
:
O.
What is its order?
series, we get
sin (22)
:.2 -
.6
i
+
In this series, both c6 and c1 are zero but c2 is not, so 2
of f .
What is the order of the zero z
:
0 for the function g
:
:
0 is a second-order zero
sin z2
-
z2?
2.5.3. Singularities
Isolated Singularities
It f(z) is analyticintheneighborhoodofapoint z: abut notactually dt 7:6,
then a is anisolated singularity of
f.
12O
oHAPTER2coMPLEXVARIABLES
There are three possible cases.
l. lf Q)l -+ oo as z
--+
a.An example of this
case is the functio n
singularities are called poles.
2. f (z)is
bounded as z
-+ a. An example of this
t
- -. *"r"
f (z): z-a
tot
/ : z-r12 . Then,
(11
case is the function
using I'Hospital's ru1e,16 we have
(z) .. - sinz
,!T1z-7
-,r12: ,!Tp I
-.
cos
: -t
These functions present no problem, since we can redefine the function as follows:
(
cos
f(z):1
(z)
z_n12
[ -l
* r/2
lfz:r/2
if z
The new function is analytic everywhere. This kind of singularity is called removable.
3. f (z)
oscillates. For example, consider the function
f (z):*p f 1)
\z/
Along the real axis, z
: r, we have
f (z) --> a
f(z) --> 0
But along the imaginary axis, 4
trk)t: l*'
as
-r
-+
as
x -+
-
ly,
0 through positive values
0 through negative values
(;)l : l*n (;)l : '
ror
al
varues
orv
Note that the Laurent series
/1\
1
I
".n(;/:|*;+7,7+"'
which is valid up ro the singularity at z : 0, has infinitely many negative powers. This
is a characteristic of this type of singularity, which is called an essential singularity.
(A Laurent series with infinitely many negative powers that is valid in an annulus with
afinite inner radius does not necessarily indicate an essential singularity. An example of
suchaseriesisinSection2.3.4,equation 2.48:theseriesfor
This function has a pole at z
: l.)
l6We could get the same result by expanding cos z in
a
Taylor series about z
+-validfor
zt-l
:
rr/2.
lz
- ll > 2.
2.5 POLES AND
ZEROS 121
Next we shall investigate singularities of the first type, the poles. The function / is
analytic in an annular region 0 < lz - al < p and may be expanded in a Laurent series
centereda;tT:qi
f (z):
-
,t*,,r,
oY
If the coefficientc-1 l0but allc-,-:0 for ffi t l, thenthepole is of order 1 (thepole
is simple).If c-2 is nolzerobut c-^: 0 for ffi > 2, the pole is of order 2 (whether or not
c-1 is zero). In general, if the series may be written as
f (z) :
or'
,t^,,r, -
with c-. # 0, the pole is of order m.lAn essential singularity (case 3 above) is a pole of
infinite order.l
We can test for the order of a pole without finding the Laurent series. The limit
I:!"(, -
a)P
f (z) :
ll{'k - ro ,t^cn(z - a)n
: ,_o
14 i
c'(z
-
a)n+P
p
:
m,and will not exist for
n_^
will
be zero for
p
) ffi,will
be a constant
(c-^)
for
p<
m.
Thus,
The order of a pole at z - a is the lowest integer p for which the
limr-o(z - a)P f (z) exists.
A function that has well-separated poles
limit
as its only singularities is described as mero-
morphic.
Other Kinds of Singularities
Functions may not be analytic for worse reasons than those cited above. For example:
.
t-
The function may have a branch poinr. A branch point is a point from which a branch cut
emerges (the head of the tadpole, see Sections 2.1.3 and2.2.l).To determine where the
branch points zo lie, note that we obtain successive branches ofthe function by increasing
the argument of z - zpby 2n . The function 7 : JVhas a branch point at the origin. The
principal branch ofthe function is not continuous across the branch cut, and therefore the
function is not analytic anywhere along the branch cut. Thus, the singularity at z : 0 is
not isolated.
122
.
cHAPTER2ooMPLEXVARIABLES
We may have an infinite set of singularities that converge to a limit point, so that any
neighborhood of that point contains infinitely many singularities. The function / :
tan (llz) has this property. The singularities are at lf z : nnf2,where n is an odd
integer, or
,n: L
This sequence converges to z : 0. No matter how small a neighborhood of the origin we
pick, say lzl . e, there are infinitely many zn in this neighborhood. To see this, let N be
an integer such that N > 1/e. Then
,*.4.?r.,
Nrt
7t
is inside the neighborhood, and so are all the zm, m
singularity of
f
> N.
Thus, z
:
O
is not an isolated
.
2.6. THE RESIDUE THEOREM
2.6.1. Definition of the Residue
If f(z)
is analytic in a neighborhood
the residue of
f
at a is
tl
,"1
ofe
:
a
axcePt perhaps
ata,then
f, f (z) dz, where the closed curve t "n"tot"t L.r,
By the Cauchy theorem, the residue of f at a is zero if the function / is analytic at a. But
the reverse is not necessarily true: A zero residue at a does not imply that the function is
analytic at a.
Comparison with the expression for the coefficients in the Laurent series (eqtanon2.47)
shows that
the residue of a function f at a isthe c-r coefficient
oI the Laurent series centered I a.
Example
2.11.
Find the residue of the function
f : Y
The function may be written as a series:
1/.2.4\
f (z):;
(t- i.* A+
-22Zl.'4t'
)
at z
:
Q'60)
0.
2.6 THE
The function has a second-order pole at z
residue there is zero.
Example
2.12.
Show that the function
:
O,
RESIDUETHEOREM 123
and since the coefficient c-1
f (z) :
sinz ez
-2-
:
Q, *1s
has a pole at the origin,
find its order, and find the residue there.
We expand the sine and the exponential in Taylor series:
r:Y -r']
:)(,-*.*:(,*.+!+{+
)
I
7- 1_z
<,
--i-.-
Thus,
/(z)
1
.
6.-11" -
3
_3
401"
-.-.
has a simple pole at the origin, and the residue there is
-1.
2.6.2. The ResidueTheorem
If a function
/ is analytic in a simply connected domain D except for afinite number
of isolated singularities and if curve C is within D, then
N
d f Or:2rilRes/(2,)
Jc
7l
wherc zn are the singularities of
/
(2.6t)
contained within C.
To prove this theorem, we deform the curve C so as to exclude each of the singularities
zn, as shown in Figure 2.22.The new curye C/ equals the original curve C, plus a set of
cross cuts leading to, and a small circle f, around, each singularity. We can do this only if
the singularities are isolated, since the function has to be analytic on the cross cuts and the
circles.
Then by the Cauchy theorem,
f
f.f
dz:o
since we have constructed Ct so that / is analytic everywhere in and on C'. Since the
function is analytic and thus continuous along the cross cuts, they do not contribute to the
integral.lT Thus,
^N
d tor+fd/f,,"ro"r*i..
Jc
il'i
lTWe used this argument previously in Section 2.2.5.
fdz:o
124
cHAPTER2ooMPLEXVARIABLES
Im (z)
C
/
Re (z)
FIGURE 2.22. Contolr for proving the residue theorem. The contour excludes the singularities at
zn.Two such points are shown in this diagram.
If the integral around C is in the standard counterclockwise direction, then each of the
integrals in the sum is taken clockwise around its circle f, . Now we move the sum to the other
side of the equation and use the resulting minus sign to change the direction of integration
to the counterclockwise direction. Using the definition of the residue (equation 2.59), we
have
^Nr
to,
{-ra,:Diln:lJl,
JL
N
:ZnilRes/(2,)
n:l
and the theorem is proved.
2.6.3. Finding Residues
There are several different methods for finding residues.
1. Find the Laurent series and pick out the coefficient c-1. This is what we did in
Section 2.6.1.
2. For a simple pole, the residue
at a is
Res/(a):I1E"k-a)f
(z)
(2.62)
2.6 THE
(z
_lim_
z+a
- a) f
(z)
:
RESTDUETHEoREM 125
@
- ol D
lim (e
z+a
cn(z
n:-l
-
a)n
@
: [m f
zta
cn(z-a)n+l
']t
All
the terms with positive powers of (z
Example
2.13.
a) go to zero,leaving c-
-
Find the residue of the functio n
f (z) :
tun'
r,
which is the residue.
at the origin.
First we find the order of the pole. Note that the limit
tanz
lio,.tu1z : lim
z-0 - Zz z+O Z
: li* ttl" :
z--+O I
I
exists, and thus the pole is simple. The above limit also gives the residue as 1.
3. For a pole of order ln, the residue is
Res/(a)
: !:+6+fr#tk - a)^ y
12y1
(2.63)
Again we use the Laurent series to demonstrate the result:
@+iffiu
I:g
-
a)^
rk): lg
- a)^,i^,,,, - o,'
^+fr#(z
Since the Laurent series is uniformly convergent, we may differentiate it term by term.
All terms with nr i n < m - 1 (that is, n < -l) differentiate to zero.
oo
sm_l
tim
*
z+q 42.," '
@
I
.,(. -a)n+^: z+a
lim I
-
-
oru,n
powers or (z
term: ",;;;,""
ld^-1/l\
lim _____:_ :-
7+a (m
as required.
-
1)l dzm-t
cn(nrm)(n*m-l).'.(n-12)(z-a)n+1
- ,;;trzero
in the limit, leaving only the n
_ l)(m _ 2)... I ) ._,
k'- _ a)* f (z) : { ,* l)! @
'
\(tn )
: C_l
: -r
126
oHAPTER2coMPLEXVARIABLES
Exampte
2.14.
The function
there.
cosh z
f : 7
has a pole at the origin. Find the residue
First we find the order of the pole. We find that
coshz: -. coshz
-.
llm
llm.:z-0 Zz z-O Z
-
does not exist. But
-. ,coshz:
..
lim
Itgot --
:
cosh z
I
exists, and so the pole is of second order. The residue is
Resf (0)
:
9
tir.
z-O dZ
coshz
-
lim sinhz
z-0
:0
We can check this result by finding the Laurent senes:
r(z):t-!ri*t
There is no
c-t
:\
*:.*.
term, and thus the residue is zero, as we found using method 3.
4. For a function of the form
J.Q)
where h(z) has a simple zero at z :
dt 7 : 6, and the residue is given by
a
sk)
h(z)
and g(z) is analytic at a,
Res/(a)
f (z) has a simple pole
: z+a
li- n,\Z)
,8,(,tl
(2.64)
To understand this result, first write h(z) in a Taylor series centered at z
h(z):f
r,r, _
:
a:
of
n:l
There is no c0 tenn because h(z) has a simple zero at z
given by (equation 2.44)
,,
l
nt
d"hl
dzn tr:o
I
:
a, and the coefficient c, is
2.7 USING THE RESIDUE
THEOREM 127
Now apply method (2):
Ir:trk
-
a)f (z)
: Irg"(, - ",ffi
:lg(._
',r#=*
gk)
Iim
- i:lr
D*
s@)
c1
tcn(z
_ q)n_l
s@)
h'(a)
as required.
Example2.15. The function f (z)
r 12. Find the residue there.
:Ianz - sinz/cosz has a simple pole at z :
By method 4, the residue is
*.'/ (;) : ,!T,r-"n j- :
5. Finally,
as a last resort, we can
-1
actually evaluate the integral in (2.59).
2.16. Evaluate the residue of the function f (z) : l lz at z
We choose C to be a circle of radius p centered at the origin. Then
Example
:
O.
I f
| ['" | ^,^i0..^ | [2",^
l - :1
t peta' d0: 2ri I ae: - 2ni
*A f k) dz 2triJo
2riJc"
2ti
--1 Jo
-pte
Of course, we could also have obtained this result trivially using method L
2.7. USING THE RESIDUE THEOREM
2.7.1. Evaluating the Integral of a Complex Function
The residue theorem is an extremely powerful tool for evaluating integrals. When using the
residue theorem to evaluate an integral of the form
f
dz
t, f {z\
I suggest that you always use the steps below
1. Draw a diagram showing
the contour C in the complex plane. Also mark on your diagram
any poles or other singularities of the integrand, f (z). If the function has a branch cut,
be sure to show it on your diagram too.
128
oHAPTER2coMPLEXVARIABLES
have to deform the contour C so that the entire branch
cut is excluded from the contour. The contour and the branch cut may not intersect at
any point!
2. If there is a branch cut, you will
3. Note which poles are inside the contour C.
4. Evaluate the residue of f ateach ofthe poles that are inside the contour'
5. Apply the residue theorem.
6. If the contour runs around a branch cut, you will have to evaluate the integral explicitly
along both sides of the cut.
Example
2.17.
Evaluate
/
sinz
f,-d'
where C is a square with comers at the points
Step 1: See Figure 2.23.
(-i),
(2
- i), (2 *
i), and (+i).
Im (z)
FIGURE 2.23. ^fhe contour for Example 2.77. The integrand has
a
pole at q
:
l. This pole is inside
the contour.
Steps 2 and 3: The integrand has a simple pole at Z
There are no other poles or branch cuts.
Step 4: Using method l, we have
Res/(l)
:.lTl
(.
:
- t)
: \.It is inside the contour.
sin(l)
=
Step 5:
"tn' o, :2ni
[ z|
Jc
(sinl)
:
5.287]i
2.7 USING THE
RESIDUETHEOREM 129
2.7.2. Using the ResidueTheorem to Calculate Real lntegrals
Integrals of Tiigonometric Functions
Integrals of the form
f2n
Jo /(sin
9,
cos9)d0
(2.6s)
frequently arise in the solution of physical problems. For example, with 0 : arl, the integral
gives 2t times the ayerage value of the function over one period. We may evaluate such
integrals around the unit circle in the complex plane. On the unit circle,
z:reio:eio,
dz:ieiod1+de:4
tz
e.66)
and
cos6:
eie+e-ie:i(..:)
22
(2.61)
Similarly,
sine:
eio-e-io:+G-:)
2i2
(2.68)
These relations allow us to convert the integral over g to an integral over z.
Example
2.18.
Evaluate
fltl
J, i#ade
This integral is not in the form (2.65) because the limits are 0 to r rather than 0 to
: (- sin g)2 : sin2 d] and so
2n.But notice that the integrand is even [sin2 (-O)
["#o':l:,;fu*'*lo" ffio'
:z [" 3--]-at
Jo + sin'P
130
oHAPTER2coMPLEXVARIABLES
Then we convert to an integral over the unit circle, using relations (2.66) and (2.68):
L"
#o'
: : I:"
u.*L.,o'
:
I
:lni,"i."r"
d.z
,.1+(,-:))'"
dz
I
I f
I
iXnitrirrtr7 \-,2 4' 422 '
:2i6 "zr
..ifunir circle z4 - l4zz +
1
Now we can apply the general method for evaluating contour integrals'
Step 1: SeeFigure2.24.
-QZ
Im (z)
Re (z)
FIGURE 2.24. The contour for Example 2.18 is the unit circle. There are four poles, but only two
are inside the contour.
Step 2 is not needed here.
Step 3: The integrand has four poles, given by
zl-t+zl*1:o
,
t4+
JTP= :7 r4J5:7
2
All four roots are real. The values
+6.9282
are
zr,2: +\nT
69282
:
+3.732r
and
zz,+:
LJT-
e9282: +0.26796
131
2.7 USING THE RESIDUE THEOREM
Of these four poles, only the last two are inside the unit circle (Figare 2.24).
Step 4: The relevant residues are
Res/(z:):lim(z-d#
z+23 -' (z
- z)(z - z)(z - z)(z -
where we used the fact that z2
zq)
(z-z+)#
z--,24'
(z - z)(z - z)(z _ zik _ z+)
D
:
45
k?
-
zl)kz
-
z+)
: -zr, and
Res/(z+): lim
Thus, since Z4
-
z4
-
, a
a.
,
\z'4-zi)k+-zz)
-23, we have
23z4l
'
_:______._____________T-_
residues inside
tzl
e
-
z,fi(zz
-
z+)
{zl
-
zl)k+
-
n)
k?
-
z?)
- -Jj
24
-4Jr-0 +4Jr-
Step 5: Applying the residue theorem, we find
I
..
f"
I 3+sin'e
Jo
Let's
see
_
( /3\
Jj
-- 24/I ::. 6
\
I
n
how an integral of this type arises from a physics problem.
-d0:2tri(2i)
Example 2.19. A circular wire loop of resistance R and radius a has its center at a
distance ds > a from a long straight wire. The loop is oriented so that a diameter of
the loop points directly at the long wire, as shown in Figure 2.25.The current in the
long wire is increasing atarate a : dI ldt. What is the current in the loop?
FIGURE 2.25. The physical system in Example 2.19 has a long straight wire carrying current 1 and
a circular wire loop whose center is distance dg from the wire.
132
CHAPTER
2
COMPLEX VARIABLES
According to Faraday's law, the emf induced in the loop has magnitude
:l!ldtJ[n dool
- ldtl
":1491
I
where O is the magnetic flux through the loop. The magnetic field produced by the
wire forms circles centered on the wire, and at a radial distance d from the wire,
D_,'
"o -
unI
27t(t
Now at a point inside the loop,
d:do*rcosd
and
dA:rdrd0
so the flux is
,: ['" fo2n (do,*o'
.,rdrdo
*rcosd)
Jo lo
The integral over
I
is of the form (2.65), so we may convert it to an integral over the
unit circle:
fzolfldz
_)O
Jo @o* rcosd)""t
/^
/unitcircle
:-2i f
,./unit
do+;(r*)),,
I
circre 2doz
+ r(* + Daz
The poles of the integrand are given by the roots of the quadratic in the denominator:
-2do
i. 4al -
+rz
2r
Now since do
> a > r,dsf r >
1. So only one of thetwo poles,
is inside the unit circle (Figure 2.26).T\e relevant residue is
2.7 USING THE RESIDUE
THEOREM 133
and thus
tf2n_)n:
Jo
I
(do*rcos0)
Im (z)
Re (z)
FIGURE 2.26. The integration contour for Example 2.19. Only one of the two poles is inside the
contour.
Then the flux is
6
: ltol ['
'D:
zir
2o
- ,.. _ Fol [d|-o' -du
I, prctr:
z Jot A
wherez:fi-r2.So
ur7 (3-o'
a:-+#l:;,
--ro1 (
un1
4-o' - d.)
(2.69)
and since 1 is the only time-dependent quantity in this expression, the current in the
loop is
: !:11491 - r'ooao
'-R-RltulR [' r
L'
We can check the result by looking at the limit
approximately
a 11 do.We
e = FoI oo, _ ltol oz
2nd0
2do
expect the flux to be
134
oHAPTER2coMPLEXVARIABLES
Now if we expand the square root in equation (2.69), we get
a:
uorao(t
-
ft) =
pordolt
I d2\l
I
- (t _ __
24))
I
/la2\
uol "
: psrds\ZA):;doo
as expected.
2.7.3. lntegrals Along the Entire RealAxis
Closing the Contour
IiS f U> dx may be converted to an integral in the complex plane
The idea is to "close the contour" by adding additional pieces
some
circumstances.
under
zero or some multiple of the original integral along the
is
either
the
integral
which
along
real axis.
The most common way to close' the contour is to add a large semicircle at infinity
(Figure 2.27).For example, the integral
An integral of the form
r*oo
I
I ----i-d*
' - J-*
7r'+o'y"^
r-
may be intepreted as an integral along the real axis in the complex plane:
r
: [
---]--------=az
Jreal axis (22
+ az)'
-RR
FIGURE 2.27. Closing the contour with
a
big semicircle of radius R
-+
oo in the upper half-plane.
Next we note that the integral along a large semicircle at infinity is zero. First look at the
absolute value of the integrand:
llrl
tt_
|'
)''' l -
kt-
l,
orl'
l1+?l
2.7 usrNc rHE RESIDUE
where we used the result (2.13) lzt f- zzl > I Izr I
big enough that lal I lzl < I /J2, and then
4
kl
-
THEoREM 135
lzzl l. We can make the large semicircle
on the semicircle
Then
lr
tt
r
I
;dzl (length of curve)(max
l,/se'icirct" (22 + az)2"* =
value
44n
:(zR)Ro:Rr*0
of linte$andl
on curve)
asR-+oo
Thus,
flftf
C
.d::
J (zz + az)'
I
+
Jrear axis (22
^d:*
oz1'
I
Jsemicircre
: Ifl ----: ^dz*o
Jreal axis (22 + oz)"
Thus, the integral around the closed contour composed of (a) a straight line
fte{qil
axis and (b) a large semicircle at infinity equals the
'g-tt\'
We may evaluate the integral around the closed con
d the
result equals the integral along the real axis.
The integrand has poles at z : X.ia. Each pole is of order 2. Only the pole at z : lia
is inside the contour (Figure 2.27).Using method 3, we find that the residue there is.
Res/(i
' a)
-
lim
z_ia
Idz,, -
ia12
--J--------=
+ az)"
(22
: ,'Yl"adt
k+;"j
-2
:- lim -2
;;" k + ia13 (2t612 Thus, the value of the integral is
and hence
4ia3
136
cHAPTER2 coMPLEXVARIABLES
Integrals of the Form
I jS "'k' f @) a*
Integrals of the form [lS ,ik' 71*) dx, where k is a real number, may be evaluated by
closing the contour with a semicircle at infinity. Fourier transforms (Chapter 7) are important
examples of this class of integrals.
ES fiOx, where k is real and fr > 0.
The first step is to close the contour using a large semicircle, as discussed above.
Onthelarge semicircle atinfinity, z: R(cos0 * I sin0), with0 < d < n. Thus,
Exampte
2.20.
Evaluate
exp
(ikz)
:
exp (jkR cos 9) exp
(-kR
0)
sin
(2.70)
and so
lexp
(ikz)l
-
lexp (iftR cos d) | exp
sincesin0 ispositivefor0 < 0
(-kR
sin
0)
:
exp
(-kR
sin
9) < I
<tr.
Thus, we have
l,["*"on"
ry#
(ength or path) max lintegrandl
"1=
I I
:rRn.Lax lexp fftz)llF
+
< tt
2
Rl=
R4
for R
2n
:Rr-0
The poles of the integrand are the fourth rootsls
exp(in14linr/2),n:0,1,2,3.
I
tl
> 2t/a
asR-+oo
of -1, exp(it14l2nni14)
Of these poles, only two, n
:0
and n
:
inside our contour (Figure 2.28).The residues are (by method 4)
/ik
nes/(er"r+)
: #1.:",",* :'SiI
\
/k
-
:tz"-'
\
*i!'
and
Res/ (e'3"/a)
18See Section 2.1.2
:
'*r(#r'-"
4e9iit /4
1/2exp
/-k
(VZ
(1
4(r + i)
+
\
t)/
:
1, are
- D)
2.7 usrNc THE REsTDUE THEoREM
137
FIGURE 2.28. Contour for the integral in Example 2.20. We close with a contour in the upper
half-plane, enclosing two of the four poles.
Thus,
E "!-o * : f #* -'tf
*, (-
+)(#
*#)
:,,*^,(-+x###.ffix)
:,,*"-(_+)
i
("r,
tA + r-ki /&) + t (r*ira - r-ki/a)
"(
a
k r\
:7r A*p (Jz*\r
(*'
2
[\- 2 /
a*"" a)
We include real integrals of the form /1S sin /c-r f (x) dx una fS cos kx f (x) dx in
this class of integrals, since each trigonometric function is a linear combination o1
"ikx
*6
. The sine and cosine must be treated as combinations of exponentials because
"-ikx
./semicircre/(.r)cos (kx)dx doesnotvanish. If youare surethattheresultoftheintegration
is a finite real value, then you may write
ll* ""0, f
(x)
dx:
*",[: ,ik*
y 1*1
d*
v
and
,fI
,t"
o' f (x) dx:
t-,ff
"ik'
7 7*1
d*
/
138
oHAPTER
2 ooMPLEXVARTABLES
and evaluate the integral as in Example 2.2o.However, if there is any chance that the integral
may have an imaginary part (and this does happen in physics problems; see the section below
on integrals with poles on the real axis), then you must be more careful and evaluate
f
+oo
''-"o,
/
J-Exampre
r
kx
2.21.
.f
i
r+m
a+m
(x)a*: ! I ,'k* f (*)a* + )2J-*
I ,-ik'7{*\d,
2J-*
Evaluate
l::
.ti#dx,whereft
is real.
Since the cosine is an even function, we may assume that the real number k is
positive. We expect this integral to have a real value, but let's check this assumption
by evaluating
coskx ,..
/+-
_
: 1 /*-
(ik{).t.- I /*- exp(-ikx)
d*
+,
p
fudx
"*p
J-- ;rraa' i J-*
J-* *o,
The first step is to close the contour. Look at the first of the two integrals. On the large
semicircle atinfinity, z
:
R(cos0
*i
sind), with0 < 0 <
r.
Using equation(2.70),
we have
lexp (lkz) I
Thus, for R
.
I
-
lexp (i kR cos d) | exp
olJ2,
(-kR
sin
0)
:
exp
(-kR
sin
0) < I
we have
t
lJsemicircle
o rl < (lengrh
'*! \'o'r)
z' I a' |
:
of path) max lintegrandl
zR max lexp (ikz)l
22n
toRn,
l-]--lz'+ a'l
I
0
asR-+oo
Thus,
(ikxl
I exp(ikz) oz
:
J-* ;riFo* fr* z2 + a2
/+-
exp
The integrand has two poles, at z
,
,
: !ia, but only the one at z - *ia
is inside the
contour (Figure 2.29).T\e pole is simple, and the residue is
Res/(ia)
: )y,(z - ia) #
:,ry,# : #
Using the residue theorem, we have
{
Ic+
or:zni(r=!"\:,a ,_oo
+at \2ia /
"*!r,n1)
zz
(2.jt)
I
2.7 USING THE RESIDUE
_R
THEOREM 139
R
-ra
C-
FIGURE 2.29.
lf
f (x)eikx, with k > 0, we close the contour in the
(C+). If & < 0, we close in the lower half-plane (C- ).
the integrand is of the form
upper half-plane
Now we turn to the second integral. Evaluating exp
(-ikz)
over the big semicircle,
we find
exp
(-ikz) :
exp
(-ikR
cos
9) exp (kR sin 9)
and the real exponential becomes arbitrarily largelg as R -+ m. (This is why we
cannot evaluate the integral ofthe cosine directly, but instead have to write the cosine
as a combination of exponentials.) However, if instead we close the contour downward
(Figure 2.29), where n < 0 < 2r, thensin 0 < 0 and the real exponential is bounded
by unity everywhere on this semicircle. Then we can show that
tr
(_ikz)
, -- 0
----; ---;-azl
I
a'
z'
+
lJsemicircle downward
exp
I
I
-+
as R
oo
I
Thus,
/+-
J-*
Now the pole at
-la
(-ikx) ,
I
-7 *7-o' :
e*p
is inside the contour
Res/(-ia) :
exp(-ikz)
fr- ,z a p
,
az
C-. The residue is
e-ikz
,-ikz
ia) ., , , : lim.
*
az
z+-ia
z--+-ra
zz I
z - ia
-
lim.
(z
e-ko
-2ia
lgAlthough we cannot show that the integral on the upper semicircle is zero, it is not necessarily infinite. Once we
for the integral along the real axis using the lower contour, we can use the upper contour
to calculate the (finite) integral along the upper semicircle. While possible, this is rarely useful.
have obtained the result
140
oHAPTER2ooMPLEXVARIABLES
As we apply the residue theorem, we have to remember that it applies to contours
traversed counterclockwise. This time we are going clockwise, so we have to add a
minus sign:
{
h_
exp(-ik.z)
q4oz: -2riRes
-J"r\vo/\ /(-ra)
z2+az
: -2ni( +\ \-zta/
o
a "-oo
which is the same result that we got for the first integral.
Finally, we add the two results to obtain
[.* r:2:#o*
J--
: ].(inr" *Tu*) :I,-o'
The result is a real number. Notice that we could also have obtained this result by
taking the realpart of equation (2.7I).
Jordan's Lemma
Integrals of the type considered above may be converted to complex contour integrals
providedthatthefunction
f (x) --> 0sufficientlyfastasx -+ oo.Theproofthattheintegral
along the semicircle is zero is facilitated by the use of Jordan's lemma:
It f (z)
converges uniformly to zeto whenever 4
rim
t
4*oo JCn
.f
-+
k)eik' dz
oo, then
:
o
where ft is any positive real number and Cp is the upper half of the circle lzl
Note. /(z) converges uniformly to zero if, given any e, there
f Q)l < e whenever lzl > M, no matter what the argument of z.
exists an
:
M
R.
such that
I
To prove the lemma, choose
R(cos0 * I sind).
R > M. Then on Cn, lf k)l <
e and
z:
Reiq
:
arl ,l [" .*r(i kR cos d) exp (-kR sine1 ni ei| ael
t f k)eik' --l
-= l/o
l"/c*"
I rn/2
exp (-kR sine) del
rtol{o
=
I
I
I
Since we need only an upper bound to the integral on the right, we note that sin d > 20 /n
sin e <
/n . This latter
throughout the range of integration (Figure 2.30) and thut ,-kR
"-2kR0
function may be integrated easily to give
llr -
r,rr,'o' o,l
='#
r, -,-oo t
=
T
Since e may be chpsen as small as we like, the integral goes to zero and the lemma is proved.
2.7 USINGTHE RESIDUETHEOREM 141
f(0)
0
o2
o4
08
l0
t.2
1.4
l6
l8
FIGURE 2.30. The graph ofthe two functions shows that sin 0 (curve) is greater than 20 /n (straight
line) throughout the range 0 < I < tr /2. Since the two functions have the same
valueatd :0 and 0 :n/2butthederivativeofsind iscosd andequals lat0:O
while the slope of the straight line equals 2 / n : 0 . 63662, the curve rises above the
straight line as 0 increases from zero, before falling back to the line at 0 : n /2.
Closing the Contour with a Rectangle
Integrands that contain hyperbolic functions do not lend themselves to the methods used
above, since the integrand does not go to zero on the contour Ca, whether closed upward or
downward. However, closing the contour with a rectangle may work. We choose the height
ofthe rectangle to make the integral along the top side ofthe rectangle equal to a constant
multiple20 of the integral along the real axis.
Example
2.22.
Evaluate the integral
/+- I
/-- .**""
If we try to close the contour with
a semicircle, we find that on CR,
,kRcos0 rikRsin0
1
6,-kRcos0 r-ikRsin0
- ,kRcos9 f 9-kRcos0
and the lower bound equals
I
on the imaginary axis (0
:
r l2).
So we cannot show
that the integral along the semicircle is zero. However, notice that on the line
20See Problem
31(c)
for
a slight variant on this theme
142
CHAPTER
2
COMPLEX VARIABLES
z:x+inlk,
coshkz: coshk
(, *rI) :!(ro*r'o a r-*xr-n):
-coshkx
and so the integral along this line from +oo to -oo, parallel to the real axis, equals
the original integral. Thus, we are led to consider the rectangular contour shown in
Figure 2.31.
FIGURE 2.31. Closing the contour with
a rectangle.
In Example 2.22,the upper side is at Im (3)
ir/k.
: i(2nll)n l2k. Only one pole, the one atz
(Figure
2.3I).
On the vertical side x : R, we have
is inside the contour
The integrandhas poles atz
It
lf,o"
r o'l:
I lf'1(/k
I
;kR;tF4;=Fa;-6i
.'rr*.
lJo
:
ni l2k,
I
dYl
r
:tlr'o
,kR lJo ;ir4;-rFn;:6dYl
r
ln
S-k-tO
A similar argument holds for the side at x
f
I
/+-
asR-+oo
: -R.
I
Thus,
4",*r," ainro': J-* *i*o*t
f-a+in/k I
J+*+inrt *rnoro'
f+ooI
.t_dx
- ' J-*
.o.tttr"^
Using method 4 (Section 2.6.3),we find that the residue at the pole is
I :
/ir\
Rtt/ (.;,/ :
,li,Ttrol.tt"h(/.2)
, \:l
I
:!(
n
krinh (ir/2) k \isin
"/2):
-
2.7 USING THE RESIDUE THEOREM
143
and thus the integral is
[**-l xo:!$
"/-cosh
|-o,:
1
/l\
,(2ni)
(nJ:
2 Jreqangle cosh kz
k
it
Z
Integrals with Poles on the Real Axis
Integrals such as
/+@
sin ftx
J-* * -rd*
in physics applications. The integrand is unbounded at x : 2, within the range
of integration, so we must calefully state exactly what we mean by this mathematical
occur
expression.
In our first definition, we simply remove the offending point from the range of integration.
The principal value of the integral is defined to be
/+m
PI
J_*
kx
/ fz-t sin kx -f /"+m sin k-x \
lim I I
"' I
|
-dx: ;-o
\/_m x -2-dx Jz+, x -2-dx)
-2
sin
x
(2.72)
provided that the limit exists.
When evaluating the integral using the residue theorem, we are not allowed to have any
singularities on the contour, so we must deform the contour so as to avoid the pole. For
example, we could put a small semicircle of radius e over the pole (Figure 2.32). Next we
split the integral up into two exponential integrals:
/+-
I
J-*
sinftx
x -2
I / f+*
,
'
2i \/--
exp(ikx)
x -2
_AX_
,
exp(-iftx) , \
f+* _AIl
I
J-*
x
-2
/
-Ar:-1
FIGURE 2.32. When there is
pole.
a
pole on the real axis, the contour must be deformed to go around the
144
cHAPTEB
2 coMPLEXVARIABLES
(In this discussion, we shall take k to be a positive real number.) To do the first integral,
we close the contour upward, as shown in Figure 2.32. (Compare with Example2.20.)The
integral along the big semicircle is zero, by Jordan's lemma, since the function I lk - 2)
goes to zero uniformly on the semicircle as R -> oo. There are no poles within the contour,
and so
f
exoikz
A #dz:o
JCPy <- z
For the second term, we have to close downward (Figure 2.33) so that the integral along the
big semicircle contributes zero.2r
FIGURE 2.33. The contour for the integral of e-ik' 11x - 2) is formed by closing with a semicircle
in the lower half-plane. The path along the real axis remains unchanged.
Now the pole at z
:
2 is inside the contour. The residue is
Res/(2)
:
linl
,-)''
(z
-
2rexPFilcz)
' z-2
:
exp
'
(-2ki)
Thus,
$
JCn-
exp_(-i!z)
1- L
oz: _2rie_2ki
Finally, we have to evaluate the integral around the little semicircle at the pole. On this
curve, z :2 * eei9 , where I varies from z to 0, so
!:*1,"fi0,:]To +t
: ){"t'o - e-izk) ,t
I:
:
2l Refe. to
i sin
(2k)(-r)
th" discussion in Example 2.21.
i
eieiede
2.7 USING THE RESIDUE
THEOREM 145
The principal value is the integral along the real axis, up to the beginning of the semicircle
and continuing from the end of the semicircle, so finally
.
sin k-r
p Ir*@ _dx
_zisin2k: I/+6 sin kx dx
J-* x-2
J-* x-2
:
1
(-2nie-2ki)l
E[o -
Thus,
+oo
sinkx :
n'
vs-'^t + :!(e2ki x -2
2i'
-dx
::G2ki + e-2ki)
2'
: n cos2k
e-2ki
I
In physics problems, however, the physics usually determines the path of integration
around the pole. We do not always want the principal value. More often we need the integral
along a continuous path from -oo to *oo. The pole settles onto the real axis as a result of
some approximation in the modeling process (zero friction,22 for example). In such cases,
a better interpretation of the integral may be
/n*-*tt
sin ftx
lTo/_-*,; * -rdx
where the path of integration passes over the pole,23 as shown inFigure 2.34.
FIGURE 2.34. Integration path that passes above
the pole.
22We
shall have to wait until future chapters for specific examples of this phenomenon. In Chapter 7, Example ?.4,
one pole moves onto the real axis as the damping parameter y approaches zero. The simpler equation may be
integrated directly to show that the correct answer is obtained
if
the path
ofintegration is
a straight line that passes
over the pole at the origin.
23This path is equivalent to a path along the real axis with the pole pushed slightly downward off the real axis.
146
CHAPTER
2
COMPLEX VARIABLES
We may evaluate this integral as we did above, but there is no integral around the little
semicircle. Thus,
1+oo*ie rin1*
}To"/--*,, * -ra*
: tI / r+oo+it exp(ikx)
O,
Uro/--*,, x -2
(-iftx)
/*-*'" expx-2 r"\
;-b/-oo+r'
/
: lto - (-2nie-2ki)f : nr-zki : o "o, 2k - ni sin2k
2i'
_ ti*
which differs from the previous result by having an additional imaginary part -ni sin2k.
On the other hand, the path may pass below the pole (Figure 2.35). In this case, the
integral around the lower curve is zero, but around the upper curve we have
$ Y*or:2nie2ki
1- L
JCn*
and so
-. 1*a-ie sinkx
I / ..
f+"o-i'
exp
(ikx)
Jro/---,. ,-ra': t Urr J-*-,, -=;dx
,
exp (-ikx)r"\
/*--t"
e+oJ-*-;" x-2
/
: !(zoir'ot - 0) : trr2ki : n cos2k * in sin2k
_ ti,,'
2i'
-R
R
FIGURE 2.35. lntegration path that passes below the pole.
Again the imaginary part of this result differs from both of the previous integrals. Thus, the
correct choice of path is crucial if we are to obtain physically meaningful results. We'll see
examples of these different choices in future chapters.24
d.
24See also Problem 36.
2.7 USING THE RESIDUE
THEOREM 147
Integrals of Multivalued Functions
We can also evaluate integrals of the form
fn*
*" f (") d*
where cv is not an integer. The function zo has a branch point at the origin.
branch cut along the positive real axis (that is, 0 < I < 2n),then
zo
-
Ifwe
choose the
lreie;o :Yariao
and so our real integral is an integral along the top of the branch cut where d : 0 and
,iuq - 1. We construct a closed contour, as shown in Figure 2.36, so that the branch point
at the origin and the entire branch cut are excluded from the interior of the contour.
FIGURE 2.36. A contour that excludes the branch cut along the positive real axis.
Example
2.23.
Evaluate the integral
"
, : [* -{-a*.
Jo xz*I
We choose the principal branch of the square root, as above, so 1 is the integral
along the top of the branch cut. Note that the integrand is analytic everywhere inside
the keyhole-shaped contour shown in Figure 2.36, except for poles at the two points
z : *.i. The residues at the poles are
Res
/(*i) :
{tlr
2i
:
and
Res
,Fi
/(-i ) : j;; :
-zt
148
oHAPTER2ooMPLEXVARIABLES
(Remember:
-i :
"3tti1z.,
0 < 0 < 2n for our chosen
branch of the square root function, so
Then, using the residue theorem, we find
f,ftro':zoifz:
oJ2
Along the big circle of radius R, we have
/F
|"fr l- ntr;Vie
l?, +--rl=
.
lnzl -
1
4
s tN/2
R3/2 (l - r/O
>
ror R
2
Thus,
-tgi l.{^ *a.15
*h
2'
R#: *F- \h :
o
Next we investigate the integral around the small circle at the origin. On this circle of
radius e, we set z : eeio i
:
n0
ie3/z
^i3g/2
-->
J",fudo
o
as e
-+
o
Finally, look at the integral along the bottom of the branch cut, where 0
: 2r:
l:ffidr : - L* ffi,, : fi,, :,
fo*
Thus, we have
fr*":
I,oo*"o,* lr-*l",,o,"or"u,
*
Ir":21
:nJ2
and so
fn Ji
':Jo iua':
,J2
z
2.7.4. Dispersion Relations
We have already noted how powerful the property of analyticity can be. In the 1920s,
H. A. Kramers and R. de L. Kronig discovered how to use this properfy to relate the real and
imaginary parts of the dielectric constant of a material, thus deriving relations that relate
the dispersive and absorptive properties of a material in its interaction with electromagnetic
waves. (See, for example, Jackson, Section 7.10.) The Kramers-Kronig relations arc a
specific example of a more general class of relations called dispersion relations. In recent
2.7 USINGTHE RESIDUETHEOREM
149
years, these relations have proved important in other branches of physics as well- for
example, in particle physics.
Suppose a function (z) is analytic everywhere in the upper half-plane. Then the first
"f
Cauchy formula (2.38) allows us to express the value of the function at a point z6 in terms
of an integral around a curye C that surrounds zo:
f
(zo):
(z)
:$ f o,
znt JcZ_ZO
where both zo and C are in the upper half-plane. Now if I f k)l --+ 0 as z -+ oo, we may
choose C to be a large semicircle with its flat side along the real axis. The integral along
the curved part is zero, and so we find
f
I
(zd: 2"i
/+oo f (x\
J-* ;-;a*
Now we let the point z0 approach the x-axis from above. The path of integration must
remain below the pole, so we put a small semicircle under the pole and obtain
r(xo)
: *1,
l:: *dx * ryl:"{k+P\i,," aef
t
I I I/"+oo f (xt
o*+ tzl(xo)l
-_tP
2ni I J-* x-xo
I
Thus,
f (xil:
From this it is clear that
[**
-L,
7t J-a
f (x) o*
X-xO
f (x) has both real and imaginary
parts, and they are related by
(x)l
I f+oo Im t'"'"dx
f
Re[/(xo)]--P
I
7t J_a .r-x0
(2.73)
and
Im [/(x6)]
If+oo
: --P
I
7t J--
(2.74)
Example2.24. The permittivity of any material approaches ee at very high
frequencies,
e(a\
€0
t + fko)
150
cHAPTER2coMPLEXVARIABLES
(a) is analytic in the upper half-plane. lf f : u * i u,where25 f * (a) : f (- a)
and f (o) -+ 0 as a) --> @, express the dispersion relations over the (physically
ard
e
meaningful) positive half of the co-axis.
Since e(a,l) does not approach zero at high frequencies, we work instead with the
function f Qo) : e(a) les - 1. The given condition on / shows that
u(a)
-
iu@;)
: u(-a) * iu(-a)
That is, u is an even function while u is odd. Then, from equation (2.73),
I r+o
u(as):-p
1
Ir J-a
o u(a\
or:lrl[o
'@) a,,]
'(') dr* J0
[** a-@o
I
n lJ-a@-aj
@-ao
I I roo u(-to\
'-'do*l /t+- ufui\dalI
::-Pll
7r LJO -o-@O
JO @-aO I
au(a) ,..
_lo t[*, uQo) [/ I + I \ldo:-P2 |f@-;---:nda
--'T Jo
n Jo @'-@6
\@+oo ,-roo/
and, conversely, from equation (2.74),
u(rr.,6):
f*
-!,7t J-o
'@) dr-
(D-aO
f
-l7t ,l LJO
uGa)
-a-@0
do* [** @-aO
'r'') arf
JO
J
- If* -l- --P
n Jo
a'-(D6"da
2
@ou(o)
These are the Kramers-Kronig relations. Measurements of the absorption properties
of a material, u(ro), determine the dispersion, and vice versa.
2.8. CONFORMAL MAPPING
2.8.1. Definition of a Conformal Transformation
In Section 2.I.3, we noted that complex functions are mappings of the complex plane onto
itself. If the function is analytic, then the mapping has some particularly nice properties.
If f (z) is analytic
at zo
ffid,f'(zo) does not vanish, then the mapping x --> f (a) is
conformal at zg.
Conformal mappings have the following properties:
1. Conformal mappings preserve angles.
25We shall see
in Chapter 7 that this condition arises from the fact that
function of time.
e(a-r) is the Fourier transform
of a real
2.8 CONFOHMAL
2. The magnification is the same for all curves passing through
3. Infinitesimal circles map to infinitesimal circles.
MAPPING 151
zo.
see how these properties follow from the definition. Consider two sets of curves:
constant and h(z) : constant (Figure 2.37a). Under the mapping w - f (z), we
get two new sets of curves in the rl-plane (Figure 2.37b).If we move along the original
curves from z6 by infinitesimal amounts dzt and dzz in the original plane, this movement
Let's
BQ) :
corresponds to the displacements dwy and dw2 along the mapped curves. (Remember that
addition of complex numbers corresponds to vector addition in the complex plane.)
8v-r(ol
FIGURE 2.37b. The mapped curves in
FIGURE 2.37a. Small displacements away
from 26 along the curves
g(z) : constant and h(zl :
the
ur-plane. If the mapping is
conformal, then?t ,the angle
between
constant in the z-plane.
dwt and dw2,
equals d, the angle between
dz1 and dz2.
Then
a*r: {ldz dzt and dwz
lzo
So
for
j :1,2,
df
dz.
dzz
-
the absolute value of dru; is
lawll:
df
dz
ldril
zo
and the argument is
ary(aw1):
+ arg(dz)
152
oHAPTER2ooMPLEXVARIABLES
Arg (dwi)differs
from
arg (d.21)by
the constant
^rr*
between the two curves in the u.r_plane is
0t
:
TE @w)
-
arg
*r( ll ). **,
\d'l'o/
(dw) : aq (dz) -
arg
and so the mapping preseryes angles (property 1)'
If a small circle about zo has radius , : ldzl, then each
a distance
(dz) :
the angle
0
dw on the mapped curve is
rt:ldwl:r
from
u.r6
: f kO,
and so the mapped curve is also a circle (property 3), with magnification
ldf ldzlrol (nroperty 2). Note that this works only
mapped "circle" reduces to a point.
if
df ldzlro is not zero; otherwise,
the
2.8.2. Some Examples of MaPPings
TVo trivial cases
f(z)
: z*
This is
a
a
translation of the whole plane by an amount a (Figure 2.38)'
:
z * a is a translation of the plane through the vector d, whose
components are the real and imaginary parts of the complex number a.
FIGURE 2.38. The mapping w
f (z):
a4
:
Petdl
This is a rotation by an angle a plus a magnification by p (Figwe2,39).
These two examples are not very useful because they do not change the shape of figures
in the plane.
j
2.8 CONFORMAL
MAPPING 153
FIGURE 2.39. The mapping u) : az conesponds to rotation plus magnification.
More interesting and useful maps
w:lnz
Writing z in polar form, z
:
reio , we find
tD:u+iu:lnr*i0
This function maps circles in the z-plane (constant r) to segments of lines in the u.r-plane
(constant u,0 I u < 2r) and maps radial lines from the origin (0 : constant) to lines
parallel to the a-axis (u : constant) (Figure 2.40).
w-Plane
z-Plarle
FIGURE 2.40. The function u = lnz maps circles centered at the origin to straight lines.
2a
z
w
2a
2a
: ---e-Iv
- -rr
(cosd
-
i sind)
:
u
+ iu
154
cHAPTER2coMPLEXVARIABLES
What kind of curves in the z-plane map to straight lines in the ur-plane? The points that
map to the line u : uo satisfy
2a
2a
-lsind-uo+r---sind
ru0
Thus, we have
2a
x:rcose:--sindcos9: -Lu0
sin20
U0
and
y:rsino --2o
sin2g: -Le-cos2o)
UO
UO
Then
:
r
(+)' . (++ t)2:
t
sin2
20 + cos2 2o
^ / *
x'+
(r
These curves are circles centered at
e-plane: circle centered at
FIGURE 2.41. The function
y
:
-a
-
y
a
:
-qf
o\2
^)
/a\2
: (;/
us with radius R
:
laluol figure 2.41).
w-plane: the circle maps to
the straight line u : a6
^
w :2a/z also maps circles
to straight lines. The circle centered at
/ u0 with u6 negative maps to the straight line u
:
u0.
2.8 CONFORMAL
MAPPING 155
2.25. To see how to use this transformation in a physics problem, let the
circle represent a metal cylinder ofradius R at electric potential V, separated by an
insulating strip from a metal plane on the x-axis at zero potential. Find the potential
everywhere outside the cylinder.
We use the transformation w : 2R2 /z.Then the cylinder maps to the line u - -R.
The plane at y - 0 (0 : 0 and 0 : n) maps to
Example
v
:2Rz
rr
sin (o)
: Q and u :2Rz
sin
(n)
:
Q
which is the real axis in the u-plane. Notice, though, that the origin maps to infinity
and the points at ) : *oo map to the origin. The region outside the cylinder is
described by circles of radius lR2/u6 | greater than R. Each such circle maps to a line
u : -'u0, with u6 < R. Similarly, the region inside the cylinder maps to points with
uo > R. The region below the plane (y < 0) maps to u > 0. The physical region of
interest (outside the cylinder and above the plane) maps to the region of the u-plane
with 0 > u > -R. Thus, in the ur-plane, the potential @ equals 0 at u : 0 and @
equals V at u : -R-a parallel plate capacitor! The potential between the plates in
the u-plane is given by the function Q : -V (u/R). We want this to be the real part
of an analytic function (see Section 2.4.3), so the complex function we need is
<D(u):
-vw:iL*
Rt
R
Mapping back to the z-plane, we find that the potential is
V
O(z):,8
2R2 2RV
,:i-s-tv
2RV
The physical potential is the real part of this expression, or
,
Q:2RV 'ine
And the equipotential surfaces (Figure 2.42) are described by
r:
2RV
0-sind
The field lines are described by the function
2RV
COSP
:
Constant
-
Clearly, the first step in using a conformal transformation in a physics problem is to find
the right transformation. This is a nontrivial task! The Schwarz-Christoffel transformation
maps the interior of a polygon with N sides to the upper half-plane (Morse and Feshbach,
Section 4.7). Circular arcs may be transformed to straight lines using the transformation
w : a2 l(z - zp),where zo is a point on the circular arc. A few additional cases are
explored in the problem set.
1s6
CHAPTER
2 COMPLEXVARIABLES
FIGURE 2.42. Equipotential surfaces around a conducting cylinder on a grounded plane (Example2.25).
2.9. THE GAMMA FUNCTION
The gamma function provides an example of a function that is defined in terms of a complex
integral. We will have an opportunity to see how to extend the definition of the function
from one with a purely real argument to one with a complex argument-another case of
analytic continuation.
The original definition of the gamma function (due to Euler) for x real and
f (x) :
fo*
"-'t'-t
x>
0 is
(2.7s)
at
From this definition we can determine the value of the function for integer arguments. First,
f(1)
: lfoo r-' dt :
Jo
-e-tl;"
:
I
Integrating the definition (2.75) by parts yields a recursion relation between
f(x -
1):
r(x)
:
lo*
'-'t'-r at
,'-r1-"-r;l-
-
Io*
o - t)(-e-t)t"-2
dt
f (x)
and
2.9 THE GAMMA
f(x):(x-
l)f(x- l)
FUNCTION 157
(x > l)
(2.76)
Then, when x is an integer,
f(n *
1)
: nl(n) :
n(n
- l)f(n -
1)
: "' :
n(n
- l)(n -2) "'
l(n]-l):nr.
1f(1)
(2.77)
For integer arguments, the gamma function is just a factorial.
An important special case of a noninteger argument is
r fl\ : [* e-trt/2
\2) Jo
To evaluate the integral, we change variables. Let
zau:
. (;) :
lo* "-u'
t:
l*
u2
dt
.Tlten dt
:
2u du
:
1_
e-u-
du: Jr
2tr/2 du. So
(2.78)
Appendix IX for the evaluation of the integral.)
We'd like to extend the definition of the gamma function to all real and complex arguments. The definition (2.75) remains valid for complex arguments z provided that Re (z) > 0
to ensure convergence of the integral, but it fails for.r < 0 because the integral diverges
at the lower limit. There is another difficulty in that the complex integrand lz-t,-t aut
a branch point at the origin if z is not an integer. We can deal with both problems if we
redefine the function as a contour integral with a well-chosen contour. The new definition
must coincide witir the original definition when z is a real number greater than zero. To
achieve this, we put the branch cut along the positive real axis and create a contour C in
the complex /-plane that runs along the top of the cut, around the branch point at the origin,
and back to infinity along the bottom of the branch cut (Figure 2.43).
(See
Im (t)
Re (t)
FIGURE 2.43. Contour for evaluating the gamma-function integral.
158
cHAPTER 2 coMPLEXVARIABLES
Let's evaluate the integral along this contour. The integrand is
f (2,t1:f-tr-r :
7z-t1r-t:
exp (ln
exp [(z
- 1)lnr -
r]
i0,where / : reiq. Onthe tophalf of contour C, the argument0
equals 0, while on the bottom half , 0 : 2n .The integrand differs by exp f2ni(z - l)l :
exp (2niz) on the two sides. Around the little circle lrl : 6,
Recallthatln/
:
ln
r
*
,'-' e-' dt : ['" Gr''1'-'
[
lc.
exp
{-eeio)ierio de
Jo
r2n
- ,' loLi'ei de
-
^2.
c-1"2niz
-
1)
as
* 0
e --+ o
as e
-+ 0 for Re (e) >
0
Thus,
lr*-'r-'
for
x > 0. Then
dt
:
(e2oi*
-
tl
l,
,x-tr-t dt :
(e2ni*
- l)f (x)
we define the gamma function through the integral expression:
f(z):
#-
r(z): -
tf
(2.79)
frt'-tr-'at
or, equivalently,
";e
Jr{-t)'-t
,-' at
(2.80)
so that the old definition and the new definition are consistent where they are both valid.
The contour C is the one shown in Figure 2.43'
The complex function f(z) defined by equation (2.79) has a singularity where
,2niz - 1 : 0, ot l : n, where n is any positive or negative integer' For positive n,
we must use a limiting process, since the numerator is also zero. Let Z : n * 6. Then
e2ntz-t
Jc
tu: a;=lJ;'+,-t"-'at
*
i(n't6) rn-t6-'
ln*,2tr
"-'
d.t)
: u:"',',n:":,- t [* ,,+t-rr-t dt: [ ,n+6-tr-t 4,
e2ftr\n+6t_lJo
Jo
-+f(n) as6-+0
2.9 THE GAMMA
FUNCTION 159
as required. So these singularities are removable. The singularities at negative integer values
of z are not removable; they are simple poles.
Properties of the gamma function are summarized in Gradshteyn and Ryzhik, Section 8.3.
One useful relation satisfied by the gamma function is
f (x)f (l - x):
,o
SIN
JT
-r
from which it follows that
rf1\r(t-1\:
'
\2) \ 2/ sin(r12)
f
/
a
\ 12
(;)l :7'
L.
/l \
:
'-(rr Ji
as we
found above by doing the integral. Also,
.
(j).
-(
'
l\_
('* :) :
*f_\:
\ 2/
-"
So
-7r
\-r)-;/{
\2/
-ir
t_
- _tr6
---vJ.
.>
2 \2/
where we used relation (2.76).
To compute the gamma function for large real arguments, we may use an asymptotic
series due to Stirling26:
rnlf
(x+l)l:ln(x!): )rnrn*(.* j)
'"'-.***...
Finally, we note that the incomplete gammafunctions are defined by the indefinite integrals
l@. x)
f
(a.
tx
: | ,-t ra-t tr,
Jo
A:
f6
J*
,-t ru-t 7,
where Re (cv) t 0.
The gamma function has numerous applications in physics and in statistics.
26This expression may be derived using the method of steepest descent (Optional Topic D).
160
cHAPTER2coMPLEXVARIABLES
PROBLEMS
fl
ff zr
:
5
+ 2i
and z2
:
3
4i, find zr/ zz and zt x
-
zz.
@)Use the polar representation of z to write two expressions for z3 in terms of
your result to express cos 30 and sin 30 in terms of cos and sin 0.
r and 0. Use
I
3. Prove De Moivre's theorem:
(cos0
* i sind)' :
eosn)
I i sinn?
4. The equation (y - yd2 : 4q(x - xs) describes a parabola. Write this equation in terms
of 7 : a I iy. Hint: Use the geometric definition of the parabola.
[-
Strow that the equation
lz-cl+lz-dl:q
represents an ellipse in the complex plane, where c and d are complex constants and cv
is a real constant. Use geometrical arguments to determine the position of the center of
the ellipse and its semi-major and semi-minor axes.
6.
Show that the equation
z:
aeiQ
*
be-iQ
represents an ellipse in the complex plane, where a and b are complex constants and @
is a real variable. Determine the position of the center of the ellipse and its semi-major
and semi-minor axes. Hint: Use the geometric interpretation of arithmetic operations in
, Section 2.1.I to determine the location of the axes.
7t-find c// solutions of the equations below Show your solutions in a plot of the complex
\-,/plane.
(a) zs
: -1
(b) za
:
16
E. Find all solutions of the equations below.
(a)
cosz:
100
(b)
sinz:6
: -5.
S
10. Find all complex numbers z such that z : ln (-5).
11. Investigate the function w : 1/ JZ.Find the functions u(r,0) and v(r,0), where rr) :
u I iu. How many branches does this function have? Find the image of the unit circle
fina all solutions of the equation
cosh z
under this mapping.
12. T\e function w(z) : zr/a . Find the functions u(r,0) and u(r, 0), where w : u * iu.
How many branches does this function have? Find the image under this mapping of
a square of side 1 centered at the origin.
E}l OUtate spheroidal coordinates u, u, w are defined in terms of cylindrical coordinates
p, Q, z by the relations
p
*iz:
ccosh (u
-liu), w:0
Show that the surfaces of constant u and constant u are ellipsoids and hyperboloids,
respectively. What values of u and u correspond to the z-axis and the z : 0 plane?
I
l
PROBLEMS 161
14. AnAC circuit contains acapacitor C in series with a coil with resistance R and inductance
L.The circuit is driven by an AC power supply with emf e : eo cos ot.
(a) Use Kirchhoff's rules to write equations for the steady-state current in the circuit.
(b) Using the fact that cos arl : Re (ei't),findthe cunent through the power supply in
the form
1
and show that 16
-
es
f
Z interms of l, R, C,
:
Re (Ioei.r)
Z, where Z is the complex impedance of the circuit. Express
and a.
(c) Use the result of (b) to find the amplitude and phase shift of the current.
(d) How much power is provided by the power supply? Show that the time-averaged
power is given by
p
with e
:
eseiat and 1
15. Small amplitude waves in
-
:
(esf
1ne (e/*)
22
- 1*" (r.r)
Z)ei.t.
a plasma are described
by the relations
0nA (nsu):g
^ *^dx
dt
AE
to- : -"n
and
^!At
: -eE -
mvu
where n6, e,m)v,and e6 are constants. The constant u is the collision frequency. Assume
that n, E, and u are all proportional to exp (ikx - iart). Solve the equations for nonzero
n, E , ard u to show that ar satisfies the equation
,2 + ir, :'o"'
:
m€o
,'n
where at, is the plasma frequency. Solve this equation to find the frequency ar and hence
---, show that collisions damp the waves.
[re.)wi."therealandimaginarypartSUanduofthecomplexfunctions.1
\,/
(a)
.f
: z2 sinz
(b)
/ : ;!
l+z
.t M )'
n 1,'^:-*,7 uL{
.I'tr4'/!)
)'i'(''';'"'f
t,
/\
:-r' +t "/
In each case, show that z and u obey the Cauchy-Riemann relations. First find the
derivative d.f ldz in terms of .r and y, and then express the answer in terms of z. Is the
result what you expected?
)
162
cHAPTER
2 ooMPLEX vARIABLES
The variables x andy in acomplex number
and its complex conjugate z*:
Z: x +iy
may be expressedinterms of
Z
1
16:;(z*z*)
z
|
,-.
Y:
- rlz-z*)
Show that the Cauchy-Riemann relations are equivalent to the condition
0f _n
0z*
.,<\
x3 ^ is thereal part of an analytic
^
xy2
On. of the functioos u1 : 2(x - flz anduz:
Q!
T function w(z) : u I iu. Which is it? Find the function u(x, y) and write ro as a function
of z.
19. A very long cylinder of radius a has potential V on one half and
The potential inside the cylinder may be written as a series:
-V
on the other half.
sir[2(2n + 1)0]
2n -11
Express each term in the sum as the imaginaly part of a complex number, and hence
sum the series. Show that the result may be expressed in terms of an inverse tangent.
20. The function -f
:
sin (22)
{t"" Example 2.10) also has a zero
a1a
: Ji.
What is its
order?
series for the following functions about the point specified. In each case,
determine the radius ofconvergence ofthe series.
21. Find the Taylor
ka)lzcoszaboutz:0
(b) ln (1 * z) about z :
(c)
0
sln z
raboutz:Tl2
z
(d) ''. about;:2
z"-r
1
22. Determine the Thylor or Laurent series for each of the following functions in the
immediate neighborhood of the point specified. In each case, determine the radius
of convergence of the series.
cos z-
(b)
z-l
,)
sln z-
aboutz
-
I
z
(c)
ez
-aboutz-0
about z. : ir
z-in
I
PROBLEMS 163
(d)
lnz
z- I.aboutz:l
(e) tan-l (z) about z :0
23. Determine allTaylor or Laurent series about the specified point for each of the following
functions.
eZ
(b)
(c)
-aboutz:3
-:
zz -9
(d)
" zz* +9
1
A.
about the orisin
zz+l
- I
'=-aboutz:i
zz+l
about the origin
Find all the singularities of each of the following functions and describe each of them
completely.
I
(a)
-ez --sincos z
sln z
(b)
477
.
tanhz
. . -(c)
z
(d) h(1+22)
25. Incompressible fluid flows over a thin sheet from
a
distance Xs into a corner, as shown in
:
i :
the diagram. The angle between the barriers is n f 3, andat x
X0,
Voi. Assuming
that the flow is as simple as possible, determine the streamlines and plot them. What is
the velocity
atr
:
Xo13,0
:
n16?
PROBLEM
25
E6.l Prove the Schwarz reflection principle: If a function
the real axis and /(x) is real when x is real, then
f*(z):
f(z*)
/ (z) is analytic in a region including
164
oHAPTER2ooMPLEXVARIABLES
Show that the result may be extended to functions that possess a Laurent series about
the origin with real coefficients.
(a) Verify the result for the function f(z) : cosz.
(b) Verify the result for the function f (z) : tan-l(z).
(c) Show that the result does not hold for all z if f(z)
:
lnz (the principal branch is
assumed).
27. Find the residue of each of the following functions at the point specified.
=1
atz:r
til #
<. -r
/1 I\ atz:O
\z- - /I
(b) exp |
(c)sin"z
at the ongrn
COS Z
(d)
'l/2
atz:116
-srnz
Evaluate the following integrals.
-[ 2Ydr,where C is a circle of radius 2 centeredat the origin
6 ) Jc
z
l0 ,{ fior,where c is a square of side 4 centered at the origin
q ,{ f:}ar,where C is a circle of radius 1 centered at the origin
@
,4
ff
uor,where
c
is a square of side 1 centered at the point z
:
(l + i) 14
29. Evaluate the following integrals.
@
(o)
f2n I + cos7
Jo z_rr*-do
f1T
sinz 0
Jo 1 +
f2n
(c)
" I
cosz
ede
I
lo l+sin'e
t(ol /" sinz, e de
Jo
-d0
30.-Evaluate each of the following
/+- I
,u)
J__ x4rdx
n(bl f+oo x
I
v P J-*
x5 ll
, -1 /n*- cos-dx
arx
G)
xzandx
J__
A
/+oo
_r
sinx
(g /_- xz +2x +2dx
integrals.
l
l
PROBLEMS 165
6/Ur" a rectangular contour to evaluate the integrals.
.1 r+- eo*
(91
where b is real and 0 < Re (a) < b
J-* t a ru,-dx'
t2- /+- sinh a-x
(V
sintr4o*dx
!l
J-*
1+m
xz
R /. *rt,*d'
32. Ev)luate the integrals
(")
/+/
x3/z
,rjdx
f@ xlll
(b) I
Jo x"+I
-dx
b3.'l Evaluate the integral
byintegratingoverapie-slicecontourwithsidesatd:0and
atf - r/N,0 < r < oo.
34. Evaluate the integral
fo*
""
a"
along the positive real axis by making the change of variable u2 : -ix2. Take care to
discuss the path of integration for the u-integral. Use the Cauchy theorem to show that
the resulting a-integral may be reduced to a known integral along the real axis. Hence
show that
(The result has numerous applications in
physics-for example, in signal propagation.)
35. The power radiated per unit solid angle by a charge undergoing simple harmonic motion is
dP
_-Ksinz0
dA
Q+pcosdsina-l/)s
where the constant K : e2cfla l4na2 and F : aatlc is the speed amplitude/c (see, for
example, Jackson, p. 701). Using methods from Section 2.7.2,performthe time average
over one period to show that
/dP\\
/ _
\ao/
K ,^
4+B2cos20
8 ""'" (t-prrorro),/,
_ _-i^Lo_
166
oHAPTER2 ooMPLEXVARTABLES
36. Langmuir waves. Waves in a plasma may be described by a wave form n :
n1exp(ikx - iri;) (compare with Section 2.I.4), where the relation between ar and
k (the dispersion relation) is given by
o:
4 [**
k J-*
I -F
3f (u)/ou
a-ku
o,
where roo is the plasma frequency ne2 f esm and/(u) is the one-dimensional Maxwellian
f (u) : tl
I m
*r;,
exe
/
mu2\
\- 2W )
Notice that the integrand has a singularity at u : co I k, which is on the real axis, if ar and
k are both real. Landau showed that the integral is to be regarded as an integral along
the real axis in the complex u-plane, and that the correct integration path passes around
and under the pole.
(a) Show that the integral may
be expressed as
(ut
If+@ af" "/au dt,:P
J-* a-ku
I
l
1
l
(Compare with Section 2.7.3,page I43.)
(b) Evaluate the principal value approximately, assuming a/k
)) v7 :
l
JEpT/m.
a series in powers
of ku la. Neglect the small effect of the pole, and find the frequency a-l as a function
of k. Now include the imaginary part due to the pole at o I k. Show that the wave is
damped. The result has been confirmed experimentally.
Hint:Firstintegrate by parts, and then expand the denominator in
(c) How would
the result change if the path of integration were to pass over, rather than
under, the pole?
Fil
ts the mapping
, :
z2
conformal? Find the image in the u.r-plane of the circle lz -
il :
I
in the z-plane, and plot it. Comment.
38. Is the mapping w
(a) the
x-axis
:
z
*
(1/z) conformal? Find the image in the u-plane of
(b) the
y-axis
(c) the unit circle in the z-plane
a cylindrical bump of radius 4 on it. The second plate is a distance
a away. One plate is maintained at potential V , and the other is grounded. Find the
potential everywhere between the plates.
A capacitor plate has
d
))
39. Show that the mapping z : tD * eu is conformal except at a finite set of points in the
z-plane.Aparallelplatecapacitorhasplatesthatextendfromx: -l tox : -oo.Find
an appropriate scaling that allows you to place the plates aty : ls . Show that the given
transformation maps the plates to the lines u : ln . Solve for the potential between the
plates in the ru-plane, map to the z-plane, and hence find the equipotential surfaces at
the ends ofthe capacitor. Sketch the field lines. This is the so-called fringing field.
PROBLEMS 167
@ f*o
conducting cylinders, each of radiu s a, ate touching. An insulating strip lies along
the line at which they touch. One cylinder is grounded, and the other is at potential V.
Use one of the mappings from the chapter to solve for the potential outside the cylinders.
: I I Q - 2) maps to straight line segments
(a) lz - 4l :2 with endpoints at z - 3 + Jii
(b) lz - (2+i)l: l withendpoints atz:3 *l and z: I*i
42. Show that f(x) < 0for -1 < x < 0.
41. Show that the mapping w
the arcs
f (z) is analytic and bounded in a region R, lz - zol = R
< M on the circle lz - zol: r, then the coefficients in the Taylor series
E3j Prove Cauchy's inequality:If
and if l/(z)l
expansion of
/
about z6 (equation 2.44) satisfy the inequality
M
," 1V
Hence prove Liouville's theorem:
If
f (z) is analytic and bounded
in the entire complex plane, then it is a constant.
44. A function f (z) is analytic except for well-separated simple poles at z
and zn
0. Show that the function may be expanded in a series
*
N
f(z):/(o)+;",(l+
\z'
,1
f
Zn,n
: I-
N,
t
Z-zn/)
where a, is the residue of at zr. Is the result valid for
H int : Ev aluate the integral
Ju
:
N -+
oo? Why or why not?
(r)_4.
2tti Jr^ a, - rl
: t t
f
where C1s is a circle of radius R1,' about the origin that contains the N poles. You may
assume that I f (z)l < eRlr on C1s for e a small positive constant.
CHAPTER 3
Differential Equations
Any physical situation can be described by a mathematical problem-usually one involving
the solution ofa differential equation subject to a set ofboundary conditions. Thus, solving
a physics problem frequently reduces to solving a differential equation. Because we are
concerned with a real physical situation, we expect a solution to exist. In this chapter, we
shall review some methods for solving differential equations that occur in physics. Much of
the rest ofthe book is devoted to developing additional techniques. A complete discussion of
this topic would require more space than is available here. For important theorems that prove
the existence and uniqueness of solutions, see, for example, the texts by Ince or Murray and
Miller. Interested students should refer to additional texts listed in the bibliography.
3.1. SOME DEFINITIONS
A differential equation is an equation for a function / ofone or more variables that includes
derivatives of / with respect to one or more of those variables. The order of a differential
equation is the order ofthe highest derivative that appears in the equation. Thus, an equation
of the form
d2t,
dl
+3--:- +2,'t :6
--+
dxt
dx
(3.1)
is a second-order differential equation for the function y(x).
Equation (3. l) is also a linear differential equation, because each term in the equation is
a
linear function of y or its derivatives. On the other hand,
dy _a
'dx
(3.2)
is a nonlinear differential equation, since it has a term with a product of y and y'.
These two equations are ordinary differential equations, which means that the solution
y is a function of the one variable x. When we have a function of more than one variable,
such as y (x , t) , and the equation involves partial derivatives with respect to both x and t ,
169
170
cHAprER 3
then we have
a
DTFFERENTTAL EeuATroNS
partial dffirential equation. These equations will be discussed later. (See
also Appendix X.)
Equations (3.1) and (3.2) are also inhomogeneous, since each equation has a term on
the right-hand side that is independent of y and all its derivatives. On the other hand, the
equation
d2v
.dv
-=1+3x'-il2y:o
dx'
clx
(3.3)
is a homogeneous equation. It is also linear. However, unlike equation (3.1), which
has
constant coefficients, equation (3.3) has a nonconstant coefficient, with x2 multiplying the
term in dy ldx.
3.2. COMMON DIFFERENTIAL EQUATIONS ARISING
IN PHYSICS
Some of the common differential equations that arise in physics are described below.
3.2.1. Newton's Second Law
Newton's second law, i : md, is a second-order differential equation for the position of
a particle. If the force is constant, such as the gravitational force on a particle near the
Earth's surface, then the equation is very simple. With the y-axis chosen vertically upward,
we have
m
d2y
or2
:
-m8'
This equation is quite easy to solve, so we may use it to draw some general conclusions.
We may integrate twice to obtain the solution:
l)
y:y0+-:dv t- t8'dto
(3.4)
There are two integration constants, 1lo and dy /dtls : u0, that must be specified to complete the solution.
The functions yr?) : I and yz1) : / are solutions of the homogeneous equation y// : Q.
Thus, the general solution is a linear combination of these two functions pllus a particular
integral-a function that satisfies the inhomogeneous equation. Here the particular inte-
gralis -!rgt2. Oft"n we obtain the values of y6 and dyldtls from the specified initial
conditions-the values of position and velocity at t : 0.
- Nowletussupposethatthereisairresistancethatisproportionaltotheparticle'svelocity:
Fa, : -ai. Then we get
d2v
m--:;:-m8-q,
dt'
dv
dt
(3.s)
This is a second-ordeq linear, inhomogeneous differential equation with constant coefficients. We shall solve this equation in Example 3.2.
3.2 COMMON DIFFERENTIAL EOUATIONS ABISING IN PHYSICS
171
3.2.2. Simple Harmonic Motion
If the force applied to a particle is given by Hooke's law and if y describes the displacement
of the particle from its equilibrium position, then we have
d2y
m7V
: -kt
(3'6)
which is a second-order, linear, homogeneous differential equation with constant coefficients.
If the system also has a damping force proportional to velocity, then the equation becomes
d2v
fll-:
dtz
d2y
dP
where
2q
: y lm *rd ,4 :
dv
'
dt
+2oQ *of,y:g
(3.7)
k/m. This is the equation for a damped harmonic oscillator
3.2.3. Bending of a Beam
A beam will bend when a load is placed on it. Let the coordinate x run along the undisplaced
beam, and let y(x) be the downward deflection of the beam's center line. The elastic properties of the beam are described by the Young's modulus E of its material. Its response to
applied loads is determined by the shape of the cross section, as measured by what engineers
call the moment of inertia 1 of the cross section about the long axis.l Then, with origin at
the left end of the beam, the deflection
y(x) may be computed
as
follows:
1. The beam bends as a result of the net torque acting on it. In equilibrium, torque due to
the internal stresses balances the external torques. Torque balance is expressed by the
differential equation2
d2v
I
:
--+
dx2 --m(x\
EI
(3.8)
where rn(x) is the net counterclockwise torque of all forces acting to the right of point
r. The minus sign arises because y increases downward.
2. The torque is due to the shearing force:
dm
_-f(x)
(3.e)
dx
where r (x) is the sum of all vertical (downward) components of forces acting to the right
of point x.
lLet x' , y' z/ be coordinates with origin at the centroid, with y/ (vertical) and z/ (horizontal) lying in the beam
,
I:
cross section. Then
[(y')2dy' dz'.
2This is the Bemoulli-Euler law See, for example, Long, pp.
9l-94.
172
CHAPTER
3
DIFFERENTIAL EQUATIONS
3. The shearing force results from the load per unit length:
dt
'
clx
Here the minus sign appears because t is
integral.
4. Putting these equations
: -q(x)
fi
q(x') dx' ,where x is
(3.10)
the
lower limit of the
together gives us the differential equation satisfied by the beam:
d4t,
dro:
1
(3.1 1)
EI(t@)
where q(x) is the load per unit length along the beam. This is one of the few equations
in physics that have order greater than2.
3.2.4. Electric Circuits
Applying Kirchhoff's laws to an electric circuit gives an equation for the charge on a capacitor or the current in the circuit. For a single-loop L RC circuit (Figure 3.1) with appropriate
choice of the circuit variables for charge q and current,3 we find
r#*o#*[:e
(3.12)
where t is the applied emf. The circuit is a driven, damped harmonic oscillator. Compare
equation (3.12) with equation (3.7).
C
FIGURE 3.1. Circuit diagram that leads to equation (3.12). The algebraic variables
i
and
q
are
defined here.
Multiloop circuits give rise to a set of coupled linear differential equations. In all such
problems, particular care must be taken in the definition of the variables and the consequent
relations among them.
3See, for example, Lea and Burke, p. 990.
3.2 COMMON DIFFERENTIAL EOUATIONS ARISING IN
PHYSICS 173
3.2.5. Diffusion
Processes involving many random interactions ("collisions") are usually described by
a diffusion equation. Heat conduction is an example of this kind of process. Heat transfer along a rod of cross-sectional area A, for example, is described by the heat flux
H : -kA 0T l0x, where k is the thermal conductivity of the rod's material. The minus
sign indicates that heat flows toward lower temperature. The rate of change of temperature
in a segment of length dx depends on the heat capacity of the segment and the imbalance
of heat flow into and out of that segment. Heat flow out of a segment of length dx reduces
that segment's temperature:
H(x -f dx)
- H(x) : -*r 4*!At
where m is the mass per unit length and c is the specific heat. Dividingby dx, we have the
differential equation
AT
AH kA_
A2T
'-'^
0x
0x2
AT KA A2T_n A2T
"'- 3t
dl
a-^a
mc dx'
dx'
(3.13)
(3.r4)
:
D kA lmc is called the diffusion coefficient. This is a partial differential equation,
since it contains partial derivatives with respect to r and r. The solution is a function of the
two variables x and t.
where
3.2.6. Waves
The wave equation relates the second time derivative of a function to its space derivative:
32y ,o2y
atz-" uz
where u is the phase speed of the wave. If we assume a solution of the form
Re ei't y1x\ (compare with Chapter 2, Section 2.1.4),then we find
(3.1s)
y(x,t) :
"
"d2v
u'-i
-@'Y(x):
dx'
d2y
dr, lk2y:g
(3.
r6)
where k2 : @2 /u2. We have reduced our partial differential equation to an ordinary differential equation known as the Helmholtz equation.
L
174
cHAprER 3
DTFFERENTTAL EeuATtoNS
3.3. SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS
A linear, homogeneous, ordinary differential equation of n th order has n linearly independent4 solutions. The general solution is a linear combination of the n independent solutions.
Thus, to completely specify the solution, we require n pieces of information (boundary
conditionss). For a second-order differential equation, we may know the value of both y
and yt ar one boundary (say x : 0, in which case we have an initial value problem) or
perhaps we may know the values of y at two boundaries (say x - 0 and x : l). In this
section, we shall focus on finding the two linearly independent solutions of a second-order
equation.
3.3.1. Equations with Gonstant Coefficients
The easiest class ofdifferential equations to solve is linear, ordinary differential equations
with constant coefficients, so we shall start with those.
Homogeneous Equations
A linear, homogeneous differential equation with constant coefficients, no matter what the
order, is solved by an exponential function y : et' , where s is a real or complex constant.
Simply substitute this solution into the differential equation to obtain an algebraic equation
for s. Each distinct solution for s corresponds to a linearly independent solution of the
original differential equation. Since a second-order equation gives a quadratic for s, a thirdorder equation gives a cubic equation, and so on, we can expect to find n solutions to an
nth-order equation. (The case ofrepeated roots is discussed below.)
Solve equation(3.12) with t : 0 and with the initial conditions 4
Q att :0 and the current I : dqldt : 0 at / : 0.
Substituting the trial solution est into the equation, we find
Example
3.1.
[.s2sst
a
Pssst
:
*Tett -0
,
1
and so
Ls2
+
Rs -l-
1
-C -0
(3.t7)
which has two solutions:
: -- R +
-2L
.S-L
4A set of
: -a tia
N functions ya(.r) is linearly independent if there are no nonzero coefficients an that make
DI:t onyn: 0 for all values of x. See Chapter 1, Section 1.5.2 for a related concept applied to vectors.
5In the particular case of the variable t (time) with conditions specifi ed att :0, the boundary conditions are called
initial conditions.
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 175
where
R
2L
and
.
-:
.,2 _
r I/R\2
_
LC- \n)
I
it Rl2L > IlJIrC, and both solutions decrease
with time. Butif Rl2L < llJLC, then the roots form a complex conjugate pair
with a negative real part. Since the complex exponential may be expressed in terms
Both roots are real and negative
of sines and cosines, these solutions are damped oscillations.
The general solution is a linear combination of the solutions e"+' and et-t
q
:
exp
(-at)[Aexp (iot)
f
B exp
:
(-iat)]
where the values of the constants A and B are given by the initial conditions. At
we have
/
:
0,
q(0):Q:A+n
and
I:+ldt
l,:o
:o:(-cu *ia)A+(-d
-iro)B
So
B:.4-oli'
ulia
and
/
-l- iro\
2io +-4: O r
\
O:-41 ll- -a
l--4=ll-i-l
u*ia
u*ito
2\
o/
\
/
cY
Then
B:e-A-f,Q+f)
The solution is then
n
:
3,-"'lQ - ,z),''' * (t *,!),-'''f
:
Qe-at
(.orrt +Y"inrt)
A special case arises lf RlzL
: I/J
LC. The
square root equals zero, andthere is
- R /2L. Thus, we find only one
only one solution to the quadratic equation for s: s
:
176
cHAprER 3
DTFFERENTTAL
EeuAroNS
:
:
e't .We look for a second solution to the differential equation,6
u(t)e't , and substitute into the original equation to obtain a new
differential equation for the function u:
solution for q, q1
ofthe form q2G)
dqz
dt
: utett *
st)ett
and
#
:
t)"e't +2su'e't + s2ue't
Substituting into the differential equation gives
Le't (u"
+
2su/
+
szu)
+
e't R1u'* su)
1
|
l\
:
o
Lu"+u'(2sL+n+(rs2
'+Rs+e)r:O
\
The coefficient of u is zero (equation 3.17). Then, using our solution for s, we have
Lu"+rlt(#) r+Rl ro:o
u":o
Thus, the solution for u is v : A + Bt. So the second, independent solution for 4
is te't . We may readily extend this method to the case where the equation for s has
a root repeated n times (see Problem 2).
Inhomogeneous Equations
To find the general solution to a linear, inhomogeneous differential equation, we first find
any solution of the inhomogeneous equation [the particular integral I p@)). This is not as
hard as it might seem, because we can often find the particular integral we need by setting
one or more of the derivatives to zero. Next we
ofthe homogeneous equation. The general sol
combination of y1 ard y2:
y(x):!p(x)|-ct
We can adjust the constants c1 and c2 to satisfy
3.2. Find the solution to equati
resistance) with the initial conditions y - y
First we find a solution yr (/) to the inhomo
tive to zero. (This means that the particle r
Example
6This is a general method that we discuss in more detail in
]
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 177
gravitational force is balanced by air resistance and the acceleration is zero.)
0:
-ms
-
"+dtq + yr(t): -^8 ,
Since the homogeneous equation is linear and has constant coefficients, its solution
is of the form e", where s satisfies the equation
*r2 +os :0 f
Thus, the general solution
s
:0,
m
yn(t) to the homogeneous
equation is
yHQ):A+8.*p(-"t)
\ m/
Next we add y7 to yp to obtain the solution to the complete equation:
!(t) : tn t
Y1
-
A+n
"*P
Finally, we make use of the initial conditions ),
the constants
A
and B:
(-o t) - I€t
\ m,/ a
-
),0 and dy
ldt :
us to determine
y0:y(0):A+B
and
dv
+
dt
a
mp
---B----::uolB:-[-l
"
m
d
t:o
rmr2
m
g--uo
\a/
a
Then
t mt,2
m
A:!o+(;,)
B*;uo
Thus, the solution is
y(t)
: * * (:)" l' - *n (-;')] *T @[r - "*n (-I,)] -',)
Check the limit as cv -+ 0 and convince yourself that we get back the solution (3.4).
What happens as a becomes very large?
3.3.2. Linear Equations with Nonconstant Coefficients
When the coefficients in a linear equation are not constant, the solution is no longer a simple
exponential. Before proceeding to a general method, we need to look at some properties of
these equations.
178
cHAprER3DTFFERENIALEeuATtoNS
Singular Points
The behavior of the solutions to a homogeneous differential equation is easiest to analyze
when the equation is written in standardform, where the coefficient of the highest derivative
is unity. For example, the standard form of a second-order equation is
dz,t
dv
dx
d)('
-+f(x)-i*g(x)y:o
Then the values of
y
and dy I
dx at any point determine the value of the
d2v
ji:
-f
A)
dv
dx
(3.
l8)
second derivative
- sk)v
By differentiating this expression, we may obtain the value of all higher derivatives that
exist.
The singular points of the differential equation are those points that are singularitiesT
of the function f (x) or the function S(r).
All other values of x
y
and dy /
are regular points of the differential equation. At a regular point,
dx may take on any values, and the resulting value of the second derivative will be
finite. We may then use the differential equation to compute the higher-order derivatives, and
hence find a Taylor series for y about the regular point. Providing the series has a nonzero
radius ofconvergence, the solution exists.
In contrast, at a singular point, y and dy/dx may take on only a set of special values.
The set may be empty (that is, no values exist). Thus, if we are given values of y and dy /dx
at the singular point and they are not in the set of special values, the Taylor series will not
exist.
The equation
d2v
*7?*2Y:o
has a singular point at x : 0. [In standard form, /(x) : 0 and 8(r) : 2/x,whichhas
a pole at x : 0.1 In order for the second derivative to exist at x :0, y(0) must be zero.
In fact, y must go to zero as r -+ 0 at least as fast as x. The value y(0) : 1 (or any other
finite nonzero constant) is not in the set of special values for this equation, and so there is
no analytic solutions with y(0) : 1.
If the equation is linear, the singular points are fixed. But the singular points of nonlinear
equations also depend on the value of y (and perhaps its derivatives). For example, the
TSee
Chapter 2. Section 2.5.3.
8See Problem 5.
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 179
equation
,#*(#)':,
has a singular point where
):
(3.19)
0. The solutione to this equation, with initial conditions
+ yil, and thus the singular point occurs at x : -yo/2y'o,
a value that depends on the values of y and yt at x : 0. This variability of the singular
),0 and y6, is y : {iQfi
points accounts for much of the challenge in solving nonlinear equations.
The Wronskian
If y1 and y2 are both solutions of
a linear, homogeneous differential equation, then any
By2
is
also
a solution.
|
TheWronskian of the two solutions is
linear combination Ay1
w(yt,
y) : lDL -
vzvl
(3.20)
The two solutions )1 and y2 arelineaily dependent if there are nonzero coefficients A and
B such rhat Ay1 * Byz : 0 for all values of x. In this case, the Wronskian is identically
zeto.
Now observe that
dW,
;:YtYi-YzY't'
We may evaluate this derivative using the differential equation (3.18). Then
# : ytGfyL - sJz) - vzGfvi- Blr)
:-.fw
(3.21)
This is a linear, first-order differential equation for W. To find an expression for 17, first
rewrite equation (3.21) as
1dw
Wdx
d
*(lnw): -f
and then integrate to obtain the solution:
w:w(xo)*n(\ou
l'"feoe)
can verify this solution by direct substitution into the equation. See also Problem 12.
(3.22)
180
cHAprER
s
DTFFERENTTAL
EeuATroNs
Since the exponential function is never zero for real arguments, V[ is zero for all x if it is
zero at a{.r : .r0; otherwise, it is never zero. When the Wronskian is not zero, the two
functions 1lr and y2 are linearly independent.
Equation (3.22) shows that I4z is a specific function of x. We can evaluate it by using any
form of the solutions 1,r and y2 or by performing the integration in equation (3.22).
Once one solution yr (.r) of the original differential equation is known, we can use the
Wronskian to find a second, linearly independent solution. Note that
d / vr\:
d. \n )
v'vl - vlvz
:7,w
---T-
Thus,
y2: yt [ \oJyi
(3.23)
While equation (3.23) provides a formal solution, it is not always a useful one, because the
integration may be difficult. It does prove, however, that a second-order differential equation
posseses two solutions.
3.3. One solution to the equation y" - 4y' * 4y - O is eb (compare with
Section 3.3.1). Use equation (3.23) to show that the second solution is xeb .
For this equation, the function f (x) = -4, and hence, from equation (3.22), the
Wronskian is
Example
w : w(o)"*n
(lr'
+ a6)
:
woe+'
Then the second solution is
rz',):
r' I
ffidx
: e2'wo
I
o'
: wsxe2*
and thus the second solution is xe2' . The constant W6 is arbitrary.
Method of Variation of Parameters
If one solution yr (x) of the differential equation has been found, we may search for a second
solution of the formlo yz(x) : u(x)yt@). This yields a second, and we hope simpler,
differential equation for the function z(.r).
low" did thi. in Example
3.1.
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
Example
3.4.
One solution to the Legendre equation
(I
is y(x)
EOUATIONS 181
. d2v
- *')li -
r*
dv
*
t2y
: Pr(x) : x. (Verify this by substituting
: xu(x).
:0
into the differential equation.)
Search for a second solution yz@)
Differentiating, we find
dYz
;:u*xu'
and
d'y, ^t,
+xu ,
dx2 -zu
Now substitute into the differential equation:
1l
-
x2112u'
* xu") x1r
2x(u
I
xu'I
- xzSu" + 2(l -
l2xu :
o
:
o
2x27u'
which is artr$-otdet equation for u/:
u" _ _, l -2x2_
-x(l-xz1
ttt
This equation may be integrated as follows. Let x2
:
u. Then 2x dx
:
d.w and
l-2x? dx:- [ r-2w ,.
rnr':-2 [ x(l-xz)
J
J G-w)w
:
- Jt (\u!- .l-)
l-uu/ dw -lnr,u -ln(l - u,)
:,"(a+;5)
Tbking the exponential ofboth sidesll gives
.
I
u':
x\t-x\:
t/t
Z (.P
1 \
t rl I
l \
+;7):2r2
-a (r -' * 1+.)
Integrating again, we get
,:-L+1mfti)
2x' 4"'\t-"/
1
lwe have omitted
y(x).o yz@).
the integration constant. You should
verify that including it adds a multiple of our first solution
182
cHAprER 3
DTFFERENTTAL EeuATtoNS
Finally,
yz:xu:-1*+,nf'*'\
2 4 \l-rl:1Qt@)
I
where in the last step we inserted the usual definition of the Legendrel2 function Q I .
With minor modifications, this method works for inhomogeneous equations as well. If y1
and y2 arc solutions of the corresponding homogeneous equation lh(x) :01, the solution
to the inhomogeneous equation
y" + f (x)y' -t s@)y
:
h(x)
may be written as
y(x)
:
ur@)yr@) -l uz@)yz@)
The differential equation leads to one equation for the two functions
free to impose an additional constraint. The first derivative is
y'
(x)
:
u'1
@)
I t @) *
u r @) y',
ul
and u2, so we are
(r) + u'2@) y2@) + u z@) yL@)
One convenient constraint is
u\(x)m@)*u'2@)yz@):O
(3.24)
in which case the derivative simplifies:
y'
(x)
:
u(x)yl(x) +
uz@)yL@)
The second derivative is then
y" (x)
:
u'r1x;y'r1x;
+ a@)y{(x) + u'r1x;y'r1x)
*
uz@)yi@)
and the differential equation becomes
h
(x)
:
u'r1x\ y'r1x1
+ u (x) y( (x) +
+ f lu@)vl(x)
:
:
[u(x)
*
't
l'2(x) +
uz @)
uz@)vL@)l-t slut@)vr(x)
u2@)ltyi +
u\(x)yl(x) +
u'2@)
yi @)
*
uz@)vz@)l
fyl + 8y1] + u\(x)t\@) + ui@)vL(x)
uL@)yL@)
Using equation(3.24) to eliminate u'2,wehave
h(x)
: r'r@ ly'r(*) - f,or,.,f : -u r@ff
l2The Legendre functions are discussed in some detail in Chapter 8.
Q.2s)
3
3
SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 183
where W(x) is the Wronskian (3.20) of the two solutions )1 and y2. Thus,
f
I
ut(x\:-
J
h(x\v'>(x\
'"
w(x)
d.r
(3.26)
and, similarly,
u.t(xt: I
I
h(x)YtG)
w(x)
o*
(3.21)
Note the similarity between equations (3.23) and (3.27).
Methods of Solution
We have discussed several methods for finding a second solution to a differential equation,
but how do we find the first one? Methods for solving an ordinary differential equation
include the following:
1. Guess
1
Form
the form of the solution.
a
power series-type solution.
3. Find an asymptotic solution.
4. Recognize the equation, or use a change of variable to transform the equation to a rec-
ognizable form.
5. Integrate numerically.
Guessing is not abad option. A given differential equation with a specified set ofboundary
conditions has a unique solution (we will not prove this theorem here), so if we can guess the
solution, we have the unique solution. Physical intuition and experience are useful guides
to the guess.
Later in the text we will identify some equations whose solutions are "named" functions
with properties that are well known. Legendre's equation (Example 3.7; Chapter 8, Section 8.3.1) and Bessel's equation (Example 3.9; Chapter 8, Section 8.4.1) are examples.
Once we recognize the equation, we may simply write down the solution. Some references
(for example, Polyanin andZaitsev, Murphy) list differential equations and their solutions.
These references allow us to look up the solution in the same way that integral tables allow
us to find the value of an integral.
Numerical integration gives a numerical solution for a specified set of boundary conditions, but the solution must be repeated for each new set of conditions. Thus, an analytic
solution is preferable if one can be found. Numerical methods can be very useful for nonlinear equations, however.
Additional methods for solving linear equations, such as expansion in eigenfunctions and
transform methods, will be developed in the chapters that follow. Here we shall begin by
studying power series solutions.
184
CHAPTER
3
DIFFERENTIAL EQUATIONS
3.3.3. Power Series Solutions
SolutionAbout a Regular Point
The solution to a linear, homogeneous differential equation can be expressed as a Taylor
0,
series about a regular point. Frequently (but not always) we want a solution about -r
so that the solution is a power series in .r. The method is to assume a solution of the form
:
I -lanx"
(3.28)
n:0
and substitute into the differential equation. Since a Taylor series is uniformly convergent
within its radius of convergence,13 we may differentiate term by term. Thus,
,oo
dY
' - \-) narxn-l
dx
(3.29)
?:o
Notice that the n : 0 term disappears because of the factor n multiplying ar. Then the
second derivative is
,')
o.-l
dxz
oo
:5-n(r -
(3.30)
l)anxn-2
n:0
where now both the n : 0 and n : 1 terms are zero.
For example, consider the Helmholtz equation (3.16)
d2v
7?+k"Y:o
(We can guess the solution to this equationi ) : sin kx or ) : cos k-r. Thus, we can check
our result.) This equation has no singular points, so .r : 0 is a regular point. Substitute in
the series (3.28) and (3.30):
anx' : o
irr, - r)anx'-z + k2f
n:o
n:o
The algebraic equation that results can be satisfied only if the coefficient of each power
of x is separately equal to zero. We always start with the lowest power that appears in the
equation.
"0:
We obtain this power by taking
Thus, we get
2. I.
l3See Chapter 2. Section 2.3.3.
a2
n
:
I
k2ag:
2 in the first term and
0
+ o,'2: -k'2
n
:
O
in the second term.
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 185
xl:
:3
We take n
in the first term and n
3. 2.
a3
: I in the second. Then
I k2at:
0
+
a3
: -ftzlt 3.2
until we come to the lowest power of x for which every sum
in the algebraic equation contributes to that power. Then we can draw a general conclusion
valid for all higher powers of x. Notice that both sums here contribute to the coefficient of
x I , so we are ready for a general statement. Let's look at the coeffic ient of x^-2 .
We have to consider special cases
x^-2i
We need n
:
m in the first term and n
m(m
:
m
- l)a^ + k2o*-r:
0
-
2 in the second.
* a^ : -k2-!!-2
rn\m - I)
a--2 interms of a*-4,
We can now use the same relation again to express
"'
-m-__--
-Pz
m(m
-
3)
etc.
(-k')'o^-o
-k2o^-4
r) (m - 2)(m -
(3.31)
m(m
-
l)(m
-
2)(m
-
3)
Continuing in this way, if m is even, we get
a^: (-I)m/'#oo
(3.32)
a^: (-I)@-t>PYlat
(3.33)
whlle if m is odd, we get
Equation (3.31) is the recursion relation forthe series. It relates each coefficient in the series
to the preceding one. In this case, the coefficents skip one: a* isrelatedto am-z rather than
a^-l.Eventually we can relate each coefficient to ag or al. But we have no relations that
determine the first two coefficients as and a1. These are the two arbitrary constants in the
general solution. The two solutions are
(kxta
yt:ao(t/ -; tkx\2+i
+ \ :aocoskx
)
and
y2:at
(
["-
*,3
k4x5
* _+...
,
5!
\
o,
l:_sinkx
)k
in agreement with our guess. Since (kx)z is positive, the series is alternating, and each term
decreases toward zero for m > lkxl * 1. Thus, these series converge everywhere.
186
cHAprER 3
Example
3.5.
DTFFERENTTAL EeuATroNS
Use a power series method to solve Hermite's equation:
dzv
dv
-2x
A?
* +2ay -
0
This equation arises in the quantum mechanical treatment of a harmonic oscillator.
The point x : 0 is a regular point of this equation, so we expect to be able to find
a Taylor series solution about x : 0. Substitute in series (3.28) and its derivatives
(3.29) and (3.30):
+ 2u\anx' :
irr, - r)anx'-2 - z*Dn&nxn-r
n:0
n:o
o
n--0
oo
D"@ -
l)anx'-2
@
- 2Dna,x' +2"Da,xn
n:O
n:0
:o
n:O
Now consider the coefficient of each power of x, starting with the lowest power.
r0:
The second term does not contribute .Take n
:
2 in the first term and n
:0
in the
third:
2' |'
a2
I
2aao
:
0
t
a2
: -aao
xl:
All terms contribute:
3' 2'
a3
-
2or +2aa1
- 0 + a3 : -To,
For all larger powers, all the terms in the equation contribute. So let's look at x^-2:
l
I
l
m(m
- l)a* -
2(m
-
2)a--z
*
2aa*-z
-
0
and thus
4^:
(m
^
ZA^-2-
Since the recursion relation relates
a-
-2) -a
to a*-2, one solution contains only
even
powers of x while the other contains odd powers. The two series give two independent
solutions of Hermite's equation. As in our previous example, the differential operator
in Hermite's equation, D(x) : d2 /dx2 - 2xd /dx I 2q, : D(-x), is even, and thus
the solution must be purely even or purely odd. The general solution is found by
choosing values of the two constants as arrd at; that is, it is a linear combination of
the two series.
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 187
For the even solution, we find
am
(m-2\-a
:2--- .--: ^ x 2 (m-4)-a Qm-4
(m-2)(m-3)
mlm - I)
q," (a - m -l 2)
: (-2ym/2d(a - 2)(a ao
ml
Notice that if a
:
2p for
some integer
a2p+2:
p, then we have
zoro,r!ffin:
o
Then a2ra4, a2pt6, and so on are all zero also. Thus, the series terminates with the
coefficient a2o; it is a polynomial of order 2p. These solutions are called Hermite
polynomials, Hzp@). Since the solution is a polynomial and not an infinite series,
the solution exists everywhere.
In this case (a :2p), the second solution to the differential equation does not
terminate. With
z
odd,
Q2n-ll
:2azn-t
(2n-t)-2p
(2n
*
l)(2n)
: (-2)n I2(p - n) - tll2(p
-
n)
and the numerator never vanishes. Indeed,
a2ny1
(2n
- 3l'.' (2p -
-l t)l
t)
al
Solution About a Singular Point
In the region surrounding a singular point, a Taylor series may not exist, so we must modify
our approach. If the singularity of the solution y (x) is isolated, we may express it in terms
of a Laurent series
't*o'"
- xo)n
valid for 0 . lx - xo I = p. However, the singularity may not be isolated; for example,
it may be a branch point. Thus, we may have to allow for noninteger powers of -r. [It is
also possible that, even though x6 is a singular point of the differential equation, one of the
solutions y(x) may be analytic at x6.l
We can allow for all these possibilities by choosing a series of the form
y(x)
:
(x
-
xdP
Do,@ -
xo)n
(3.34)
n:o
where p may be any number, positive or negative, integer or noninteger, real or complex.
This is called the Frobenius method.
188
cHAprER 3
DTFFEHENTTAL
EeuATroNs
If the solution is a Taylor or Laurent
series,
A < lx - r0l < p2, and we may differentiate
,'
it is uniformly convergent within a regionr4
term by term. If p is not an integer, then
: lfta-'o)o] (8" -.,') + (r - (n"",r" - "or'-')
^,'
The derivativeof (x - xs)p exists everywhere except on the branch cut. We may differentiate
everywhere in a region that is keyhole-shaped-that is, an annulus with a channel cut out
at the branch cut. The derivatives are
,oo
d\l
: )-(, + p)an@ i{
dx
(3.3s)
xo)n+P-l
n:o
and
d2,t
3
u.'- :f(n
dx' n:0
Example
3.6.
+ p)(n + p
-
l)an@
-
(3.36)
xs)n+n-z
Solve the hypergeometric equation
r*,-*tfi*(r.-)*+]r:o
Obtain a solution as a series in powers of x.
The equation has singularpoints wherex2 - x : O-that is, at r : 0 and atx : 1.
We are looking for a solution about x : 0, so we choose a series of the form (3.34).
Substituting into the equation, we find
oo@
9 : f{, * p)(n * p - t)anxn+e -D(u. t
n=O
n:O
p)(n + p
-
dr@
l)an76n*n-t
oo
+zfrn * p)a,xn+p - )D, * p)a,x'+P-t a ! lon*n*o
n:o
z=0
n:0
The lowest power that appears in this equation is
and fourth terms), and its coefficient is
1) + Iol"o
-loroZJ
L
Since we want a6
I
a
Taylor series, p1
-
:,
0 for a nontrivial solution, we must have
/
o(o
l4For
xP-l (with
r\
- ,):o
0 and the condition becomes lx
-
xol
.
pz.
n
:
0 in the second
EQUATIONS 189
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
This equation is called the indicial equation. It has the solutions p : 0 and p : I /2.
Looking atthe coefficientof rm +pgives us therecursionrelation forthecoefficients:
- l)a* - (m -f p -f l)(m *
l1
I .am-O
-;(mJ-P]-I)a^a1
z+
(m
*
p)(m
*
p
an.:.
:
p)a^+r
*
2(m
I p)a*
l- p)(m * p * 1) -f I /4
(m-rp-tl)(m+p+ll2)
(m
In this case, the two independent solutions arise from the two different possible values
of p. The recursion relation relates each a^a1to the preceding a^.
Forp-0,
am*t
+ rl4
+ wam
* tl2)2 :
: (m 111^
1 ry21a^
@+
r^ *
I /2m+l\ /2n-l\
:o\^at)\
:
m(m
* r)
r)(m
2m
2(m
*|
+ta^
* )^-'
-t l)ll
2m+t(m + l)! "
(2m
--erl
and the first solution is
S
yt:aoLn:0
- r)!! .
2^;r,:r"
(zn
This series is well behaved at the origin and converges for
The second solution has p - l/2.Then
am*t
lxl <
1.
(m+I/2)(m+312)+114^ _ m2+2m+t
am :
^
1* a3J2y^ trga^
@+3/r)@ + D
2(m * l)
2m+tQn I l)l
- 2m + 3 *^ - (2m -13)ll uu
:
:_ *m+r
+3424^
and
v,
-
Jiaoi-z:!--"
,_n r2n * l)!!
Once again the series converges for lxl < 1, but this function has a branch point at
the origin. The constant as in each solution may be chosen freely; it need not have
the same value in the two solutions.
The Frobenius method may fail to provide two independent solutions if the functions
and/or
l-
/(x)
g(x) in the standard form (3.18) are too badly behaved. In particular, the method
190
cHAprER 3
DTFFERENTTAL EeuATroNS
may fail to provide two solutions about x : r0 unless (x - xd f @) and (x - xs)z g(x) are
analyticls at x : xs, in which case the singular points are called regular singular points.
This result is known as Fuch's theorem. However, one solution may sometimes be obtained
about an irregular singular point, or two solutions may be obtained by choosing a different
central point that is a regular singular point.
The Frobenius method gives two independent solutions of equation (3. 18) about a regular
singularpointprovided thatthe indicial equation has two roots that do not differby an integer.
If the equation has a repeated root or two roots that differ by an integer,l6 the method may
provide only one solution to the differential equation. This problem arises because the
Frobenius method does not provide functions with logarithmic-type singularities. In such
cases, the second solution is of the forml7
oo
y2:
yt lnx +
!
(3.37)
anxn*P
n:0
where y1 (x) is the first solution and the coefficients on are to be determined.
3.7. Legendre's equation arises in the solution of Laplace's equation in
spherical coordinates (Chapter 8). It has the form
Example
(t
dv
^ d2v r*E+/(/
- x")7-
+ l)y
:0
This equation has singular points at x : tl. Let's look at the special case / : 0.
Then one solution is a constant and is valid everywhere. The second solution may
be found as a series in powers of x, but the series converges only for lxl < 1. Find
a solution valid for
x > l.
We can look for a solution about the singular point-that is, a solution in powers
of (x - 1). First we change variables to r, : r - l. Then we rewrite the equation in
terms of u.r:
dv
^ dzv 2x -!
: g
- x')---:dx' - dx
d2v
dv
(u +2)wA# *2(w * 1)A;:o
(1
The equation has singularities at u.; : -2 ard aI u) : 0, as expected. We assume
a solution of the Frobenius type in powers of u.r, and substituting into the differential
15See,
for example, Ince, Section 15.3, where this result is also extended to equations of higher order.
l6See Problem
1
l. Also
see
Problem 28(a) for
a case
the roots of the indicial equation differ by an integer.
lTSee Problem 10.
in which two independent solutions are obtained even though
l
l
I
I
EOUATIONS 191
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
equation, we get
0: t(n
-f p)(n
n:0
t p-
r)a,wn+p
*2f
fn + p)(n + p
-
r)anpn*t-r
n--0
+zl@ i
p)a,wn+P
+2\(n *
n:O
p)an6n+P-l
n:O
We can simplify by adding like powers:
0
: t(n
-t p)(n -t p
*
l)ansntn +
zl@ I
n:0
p)2anry"-rt-l
n:0
The lowest power that appears is
wP-r.Its coefficient is
2P2ag
:
and so the only possible nontrivial solution is
The coefficienr of w*tP : u- is
Q
p
-
0, a repeated root.
a*m(mf l) + 2a.ay(m*
1)2
:
g
of
I
l
N
t
r
,
i
i
am*l
:
an
-m
4^ t, l)
However, notice that for m :0, we get at : 0. There is no series! This is just the
constant solution mentioned above.
The second solution is found by introducing the logarithm:
I
@
I
y2:
where y1 is the first solution,
yt ln r.u *
wPlanw"
i:o
lr : I in this case. The derivatives
are
m
dtnl"f':
aww- +ftn
n:0
*p)anwn-+P-r
and
,1
d'v';
3ctw' :-
-m
I
"
+ftn
n:lJ
+ p)(nI p - l)anl2'+n-z
192
cHAprER 3
DTFFERENTTAL EouATtoNS
The differential equation becomes
o:
(u
it, * p)(n t p - t)anw'*o-')
-++
t''
/
7:o
\
(
+2),
-t 2(w
*
r, ( ! +
\r,ft/ir,
* p)a,w'+e-r\
Rearranging, this becomes
-yi2*rg=)
\rru/7:o
(
-12(w+ rl
i<,
n:o
0
+ (r, +
zlia * p)(n * p - t)ansn+p-l
* p)a,w'+P-l -
: I * D@ + p)(n -f p I
0
r)anwn+n
+
n:O
zl{" * p)2a,s'+n-t
n:O
The first term is a constant (u.,0 term). Thus, at least one term in the two series must
also contribute an equal and opposite constant. Therefore, p carrnot be a fraction,
since in that case n + p + 0 for any n.lf p were a negative number, P : -s, then
fot n :0 in the last term we would have
O
:2s2aow-t-1
requiring as = 0. Similarly, we would find ar: 0 for eachn < s. The first nonzero
term in the series would be erut-t : asrDo. Thus, we may as well take p : 0 and
start the series with a6. Then the x0 term is
I l2at:0 4 aI: -!2
The xm term has coefficient
m(m
or
I l)a* +2(m -t l)2a-a1- 0,
m
>I
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 193
The constant term as is not determined, since it does not appear in the differential
equation. It may be combined with the first solution, yl : constant. Thus, the second
solution is
@-oo
rn n,
*
i,-t,'*' (+)^.' ;n
m:l
:rn, - tt-tl' (:)^.'
#
The series may be identified as the series for
rn
(r +
\
:tn (t * *)
1)
2/
2 /
\
t)
:* f"l
\ 2 )
Thus, the second solution is
yz@)
:ln
- 1) - ln (r *
:n(r-1\ *,r,,
\x+l/
(x
1)
f
ln2
We may combine the constant ln 2 with the first solution if we wish, as we already
argued for the constant a6. The usual definition of the Legendre function Qo@) for
x>1is
I /x+1\
QoG):2"("_,i
which differs from our )2 only by an inconsequential overall factor of
-l
/2.
Indicial Equation with Complex Roots
The indicial equation may have complex roots.
Example
3.8.
Apply the Frobenius method to the Euler differential equation:
*2y"+xy'+y:g
:
Inserting the series (3.34) with r0
0, we find
@@oo
\f"
n:O
+ p)(n
* p-
l)a,xn+e
+l{" I
p)anxn+P
+lanx"+p :
n:0
o
n:O
The indicial equation is
P(P-1)*Pf1:o
:
which has the solutions p
general recursion relation is
[(m
*'i. This problem
-l P)(m i
P
-
has a second curiosity as well; the
:
l(m*i)2*lfa^-0
l) +
(m
-l P) + lfam
1m2
O
+2im1a*:g
194
cHAprEB 3 DTFFERENIAL EouATtoNs
and the only solution for m
of the Euler equation are
>
O
is
a* :
0' There is no series. Thus, the two solutions
!l: x'
and
!2: x-'
We may rewrite the solutions as follows:
yl :
exp
(lnx'; :
exp (i
lnx);
yz
:
exp
(-i lnx)
Thus, by taking appropriate linear combinations, we find that for real positive
have the real solutions cos (lnx) and sin (lnx).
x
we
Asymptotic Methods
Sometimes the equation simplifies for large values of x, allowing us to find the limiting
form ofthe solution.
Example
3.9.
The modified Bessel equation has the form
dzt
ldv
71+; d,-
/
(l
+
m2\
=,J
):o
(3'38)
Find the form of the solutions (the modified Bessel functions) for large argument.
For large x, the terms with powers of llx become very small, and the equation
simplifies to
d'Y*
t a
" -^
)'oo:u
with the solutions
!*:
ett
The two solutions have exponential behavior at infinity.
We can then find the complete solution by finding the function u(x), where
y
(x)
: u(r)y-(x) :
u
(x)eL'
Then
d! _
,rr+* +ue*,
dx
and
ue*'
#:It"eL* +2u'et' +
3.3 SOLUTION OF LINEAR, ORDINARY DIFFERENTIAL
EQUATIONS 195
Substituting this into the full differential equation (3.38), we have
It,,et, +2u,e*, +ue*,
*!
:O
(r *
5) r"*,
,,, + ( !+z),, *! (+t- 4),:o
x
1r,r*, +ve+*)_
\x
/
\
(3.3e)
"/
Once again we can simplify. Look at the order of the terms in the equation when x is
large:
Term
Order of
n"
u lx2
terml8
t)'
/x
lx
/x2
u'
u
u
u
u
/x
lx
u
u
lx2
lx2
Thus, we need to keep only the two terms of order uf x. Then, dividingby 2u,we
have
,)'
I
u2x
I
lnu:--ln-x
2
1
Jx
Thus, the general solution for large x is ofthe form
']::
e-)c
_ Of V:
'/x
e'
_
(3.40)
Jx
These are the asymptotic forms of the modified Bessel functions.
Solution About the "Point at
But what
values of
Infinity"
if we want more terms in our solution? Since we want a solution valid for large
x, we change variables to tt : l/x, so as to obtain a series in powers of l/x.
A solution valid in a neighborhood of a
:
0 is a solution valid about the "point at infinity"
in x.
Example
ers
3.10.
Obtain
a series
expansion for the modified Bessel functions in pow-
of If x.
We start with equation (3.39) in Example 3.9 and change variables to
The derivatives are
du
dx
du du
du dx
#(:,)
,du
du
u
: Ilx.
(3.4t)
l8We approximate hereby noting thatdifferentiating reduces each powerin a series expansion by 1-the same result
dividing by.x. We can (and should) check the approximation once we obtain the solution for u.
we obtain by
196
cHAprER o
DTFFERENTTAL EeuATtoNS
and
# : *(#) # : -,' (-,# - ;tt*) :t#
* 2,.ilr
Q
42)
Substituting in, we get
^d2u "du
"d
"o# -t 2f? - tu *ztu2fi*
u(*l -
.d2u
du
,' -fr + (u'^+2u)i+
(+1
-
m2u)u
:o
.
m'u)u
:o
This equation has a singular point at Lt : 0, so we look for a series solution of the
form (3.34) and substjtute in:
0
: D
an(n
*
p)(n + p
n:O
-
l)un+o+l
+l
a,1n
*
p)un+n+l
n--0
oo@@
+2Da,(n *
p)u"+p
!.1
anu"+P
n:O
n:0
- m2lonun*o*'
n:O
The lowest power that appears is up, and its coefficient is
ao(+2P
+ 1):0
The indicial equation gives p : I l2fotboth the positive and the negative exponential
terms. This is the leading term (3.40) that we found in Example 3.9.
Now look at the power uk-tp-tr .Its coefficient is
atl&* p)(k+ p -l) + (k+ p) -mzlLa1,a[l -2(k+ 1+p;1 :9
which gives the recursion relation:
ak+t
t : +*t
: +*gJl48(ft + l)
2(k + t)
-S*!
Thus, the two solutions are
tl\^t:
e'' /, (4m2 -1)
utJ--F I r -F ------;-'
yz@):
oo
8x
vr \'
-1-
(4m2 -9)(4m2
- l)
2(8x)2
-'-'
\
/t
and
e-r*
6
/'
[t
-
(4m2
- g4m2 - rl + \
- l\ + ---2'fi-,"
)
Am2
-
3.4 NUMERICAL
METHODS 197
The leading term is independent of the value of m, as we found above. One solution
is exponentially decaying, and one is exponentially growing. These are the functions
K^(x) and I*(x) (see also Chapter 8, Section 8.4.7).
3.4. NUMERICAL METHODS
The computer is an increasingly useful tool for the solution of physical problems. Many
useful pieces of software come with the computer's operating system or may be purchased
at a reasonable cost. Spreadsheet programs may be used to solve differential equations
numerically. Mathematical packages such as Mathematica and Maple have more sophisticated routines for solving differential equations. For more complex or unique problems,
you may find it necessary to write your own program.
3.4.1. Dimensionless Variables
To prepare a problem for numerical solution or even to plot up the results of an analytic
solution, the first step is to write the equation in dimensionless variables. For example, when
solving equation (3.11), we would use the variable u : xlL. The solution obtained can
then be applied to a beam of any length. Then
dud ld
dxdu Ldu
d
dx
Rewriting the equation, we have
I day q(u)
- EI
Scaling the deflection y similarly, we define p : y f L.T}.ren
L4 du4
I dau q(u)
dU4 EI
dau L3q(u)
du!: EI
L3
The source term q is load, or force per unit length, and
footnote 1), so the right-hand side has dimensions
,zM I
L
Tz
---
and since
lEf
:
forcelarea:
M
lLTz,
t
has dimensions of (length)a (see
I
L4lEl
the right-hand side is also dimensionless.
198
cHAprERoDTFFERENTTALEeuAloNs
Example
3.11.
Convert equation (3.5) (motion with air resistance) into dimension-
less form.
We would like to divide the space variable y by a characteristic length and the time
variable t by a characteristic time. Notice that each term in the equation is a force, so
the dimensions of the constant d, may be found from
I dvl
ML
MLT M
+lal:
,zT:T
P;l: 12
This suggests that we choose our characteristic time tobe m f a and the dimensionless
time variable to be r : tq/m. Then the equation becomes
ot dy
o2 d2y
*--lng-C(" mdt
mzdtt
q2 d2y
. o? dy
-'|
aia.=- - aiE
lll
1
Thus, we choose our characteristic length tobe m2 g f a2 and take the dimensionless
space variable to be u - ya2 1m2g. The equation is then
t!:-r-du
dtt
dr
Let's check the dimensions of the characteristic length:
lmzsl
| "t M2L T2 -r
aAl -frw-"
as required.
In what follows, we shall assume that the equations have been put into dimensionless form.
3.4.2. Difference Equations
The basic idea behind all numerical methods of solving differential equations is to replace
the differentials with finite differences. Recall the definition of derivative:
d.f .. f(x+h)-f(x)
E: i:\
h
With finite differences, we do not take the limit but allow ft to be small but finite. Thus,
a differential equation is replaced by a difference equation.
3.4 NUMERICAL METHODS
If a function is differentiable
at the point
x
:
199
r0, then the derivative may be calculated
equally well by taking the limits:
dfl :
dx l*n
-l
llm
h-o
f (xd - f @o -
h)
:
lim
f(xo+hlz)- f(xo-hl2)
h+O
Similarly, we may form finite differences in three different ways:
.
Forward dffirence:
Lf(x):f(x+h)-f(x)
.
Central dffirence:
/ h\
/
h\
d/(x):/(x+t)-f\,-Z)
,
Bachuard dffirence:
v
f(x)
:
f(x)
- f(x - h)
Higher-order differences may be formed similarly. Thus, the second forward difference is
^"
-irj:,; :';i.:'
The
differenceratio
Lf I h is an approximation to the derivative.
=?,i\:i
-
r (x,t
Inaccuracies occurbecause
ofrounding error in the numerical calculation, and because ofmistakes
in one or more of the values. These inaccuracies increase in opposite directions; making
h smaller improves the approximation to the derivative but increases the rounding error.
Thus, obtaining an accurate numerical approximation to a derivative is very difficult, and
we should attempt it only if it is absolutely unavoidable.
ft is not zero, because
3.4.3. Numerical Solution of a First-Order Differential Equation
Suppose we want to find a solution to a first-order equation
y'
: f(x,y)
that we have been unable to solve analytically. The equation need not be (and usually is not)
linear. The output of any numerical method is a set of values for the function y at a set of
y(rg) (a boundary condition) from
values x; for x. We need at least one given value )0
:
which to start.
200
cHAprER 3
DTFFERENTTAL EouATroNS
If the function f(x)
can be differentiated analytically, we can use the Thylor series to
step away from our given value:
yr
:
y(xo + h)
:
yo
: yo I
+ hf (xo, yd -l
hy' (xO
+
f;t"
@il +
"'
h2 d.f
t d- ro*"
Then
d.f
y2:yr*hf(x1,y)a h2
2dx,,
(3.43)
where x1 : .{o * ft, and so on.
As a first approximation, we drop the terms in powers of h greater than 1, effectively
expressing the derivative d y ldx using a forward difference. This is called the Euler method.
It is said to be first-order accurate, because we kept only the first power in ft.
A better approximation is obtained by taking more terms in the Taylor series expansion
(3.43). Thenumber of terms needed at each step is determined by the rate at which the series
converges, which usually depends on the value of h as well as the particular form of the
solution y. If a large number of terms is needed, this method can be inconvenient because
of the amount of analytic work that is required before beginning to compute.
Example
3.12.
Solve the differential equatron
Y'
:3xY - I/3
on the range 0 < x < 1, subject to the boundary condition y(0)
The function f :3xy - 1/3 has derivatives
:
1.
df
^.^dv
:'3y
-.*'3x dx
- :3[y * x(3xy - l/3)]:3y(1* 3x2S - x
dx
d2f
. dv
l8xy *3(l -l3x\!! - -l + lSxy * 3(1+3x2)(3xy +
-l
d*r!:
: -2 + 3x(9y *
9x2y
-
l13)
x)
At this point the labor is getting noticeable, so let's see if we can get away with
only these two derivatives. Let's take h :0.1, one-tenth of our interval. We can use a
spreadsheet to calculate the various terms and add them together. The spreadsheet is
shown in Table 3.1 ; the numerical values appear in Table 3.2.We can check the result
by choosing a value of ft equal to one-half of the previous value and repeating the
calculation. The maximum difference between the two results is 0.3Vo (Figure 3.2).
3.4 NUMERICAL
METHODS 2O1
TABLE 3.1. Spreadsheet for Example 3.12
D
I
Example 3.12
2
h
0t
3
x
0
I
4
5
0.1
6
0.2
7
0.3
8
0.4
--F7
9
0.5
=F8
l0
0.6
:F9
l1
0.7
=
12
0.8
l3
0.9
14
I
f= 3xy- U3
f'=
- ll3
- 1/3
- tl3
:3*A7*87 - ll3
:3+A8+B8 - l/3
: 3* A9*89 - | /3
: 3*Al0+Bl0 - 1/3
:3'All*B11- 1/3
: 3r Ar2*Bt2 - | 13
= 3*A13*B13 - 1/3
: 3* Ar4*Bt4 - | 13
- A4
- A5
- A6
3+B7* (l + 3* A7^2) - A7
3+88*(l + 3*A8^2) - A8
:3+B9+(l +3+A9 2) - A9
: 3+B10*(l + 3*A10^2) - A10
= 3*B1l+(1 + 3*A11^2) - A11
: 3*B12*(l +3+A12 2) - Al2
: 3+B13+(l + 3*A13^2) - A13
: 3*814*(l +3*Al4^2) - AI4
:
:
:
:
:
:3*
A4*84
= 3*A5+B5
:34 A6*86
1
:F4
:F5
:F6
F10
: Fl1
:FI2
: F13
-x
3y(1 + 3x^2)
3*B4* (1 + 3+ A4 2)
3+85*(1 + 3*A5^2)
3*86*(1 + 3*A6^2)
I
2
3f'
4
5
6
7
8
9
l0
l1
12
13
t4
y(n +
:
=
_2 1
(g*84 +
:
:
:
:
A4 2+84 - A4)
+9rA5^2*B5
- A5)
-2*3*A5+(9*B5
3+ A41
9*
- -2a3*A6r(9*B6+9'A6^2+86 : -2 + 3*A7i(9+B7 +9+A7^2*87 : -213+A8t(9*B8 +9*A8^2*B8 : -213*A9+(9*B9 +9*A9^2+B9 -
A6)
A7)
A8)
A9)
=
:
= -2 *3+Al0+(9+Bl0+9*Al0^2*810 -A10)
:-2*3*Al1*(9*Bl1+9*All^2+B11-A11)
= -2 * 3* Al2a (9*Bl2 + 9* Al2 2*Bl2 - Al2)
: -2*3*Al3*(9*Bl3+9+A13^2*B13-A13)
= -2*3+Al4+(9*Bl4+9*Al4^2*Bl4 - Al4)
l)
84 + $B$2'C4 + $B$2 2/2*D4 + $B$2^3/6*E4
85 + $B$2*C5 + $B$2^2/2*D5 +
86 * $B$2*C6 + $B$2^2/2+D6 +
B7 + $B$2*C7 + $B$2 2/2*D7 +
B8
$B$2+C8 + $B$2^2/2*D8 +
89 + $B$2*C9 + $B$2 2/2+D9 +
I
= Bl0 *
:
$B$2*C10
+
Bl 1 + $B$2+C11 +
= B12 * $B$2+Ct2 +
: Bl3 + $B$2*C13 +
: B14 + $B$2*Cl4 +
$B$2^3/6*E5
$B$2^3/6*E6
$B$2^3/6*E7
$8$2^3/6*88
$B$2"3/6+E9
$B$2^2/2*Dl0 + $B$2^3/6*E10
$B$2^2/2"D1 1 + $B$2^3/6+El I
$B$2 2/2*Dt2 +$B$2 3/6*Et2
$B$2 2/2"D13 + $B$2^3/6*El3
$B$2^2/2+Dl4 + $B$2^3/6*E14
TABLE 3.2. Numerical Values for Example 3.12
I
I
t
i
L
1
Example 3.12
0.1
h
c
D
f=3ry-1/3
f'= ly(l+ 3x2)
3
2.93232
0.646096
0.992209283
-x
f'
y(n + 1)
3
x
v
4
0
1
5
0.1
0.98
-0.3333
-0.0389
6
o.2
0.99
0.26199
3.t3382319
3.452247331
r.o34652997
7
0.3
1.03
0.59785
3.64202792
6.86495 1309
1.t1379273r
8
o.4
1.1
I
4.5452397f
tr.47359534
r.23875299
9
0.5
1.24
t.oo322
r.5248
6.OO34532
18.15395671
1.42427s53r
10
0.6
1.42
2.23036
8.28747931
28.2996385
r.693465796
1l
t2
o.7
1.69
3.22294
I 1.8485815
44.2t969028
2.082373136
0.8
2.O8
4.66436
17.M15887
69.84598597
2.647658296
13
0.9
2.65
6.81534
26.3444039
112.0219549
14
I
3.48
10.1054
40.7550206
182.89'75926
3.479585048
4.724385265
0.98 1333333
202
CHAPTER
3
DIFFERENTIAL EOUATIONS
+
+
h:0.05
h:0.1
\J)
/
1
0L
l0
o2
0
FIGURE 3.2. Numerical solution of the equation y/ - 3xy - 1/3 with boundary condition ) (0) : 1
(Example 3.12). The values computed with /r : 0.05 and h :0.1 do not differ
noticeably in this plot.
3.4.4. The Runge-Kutta Method
From Example 3.12, we observe that the derivative of
/
is
df af dy af af "af
dx- ox' d*ay- ax'r ay
Thus, the Taylor series is
y(xo
*
h)
:
y(xs) -l hf (xs, yO a
+(#.,X) +"'-!o*11
where all the terms on the right are evaluated ?t xo, yo. Thus, the increment 17 in the solution
y depends on values of / and all its derivatives at the starting point. We can improve the
accuracy of the truncated series if we can use values of f and its derivatives throughout the
range.r0 to xo * h. We can proceed by a process of successive approximations.
The first guess is to use
nr hf(xo, yo)
:
as the increment.
This takes us to the point P1 with coordinates (xo + h, )o
*
41). Now we
stepbacktothemidpointat(x6 +h/2,W*r11/2).Weevaluatetheslope yt
point and use this new value to compute an increment
nz
h
/
nr\ :
: hf(t*
Z' vo+ ; )
: hr (^. i. r, * T)
: hr(xo.
th,yo *
vo)
+
h
: f
atthis
where
/hAf
hf(xo,til -t h(Za*
This takes us to the point P2 with coordinates (x6
process to find a third increment:
qt
r12,
*
* ;nr0f\
u, )
ryz).
We can now repeat the
(r#. +#) .
3.4 NUMERICAL METHODS
203
Now we can evaluate the fourth increment by using the values at our last endpoint P3 with
coordinates (xo + h, Jo I rt):
ry+
: hf (xo * h, yo t ry) : hf (xo, yO + h(r# . ,tK) *
The best result is obtained by taking an appropriate weighted average of all these estimates
of the true increment 4. That average is
4:
1
(3.44)
O(4r*2nz+2qz'fn+)
This increment makes the expression for y(x * lr) correct to fourth order in h, and thus the
method is called the fourth-order Runge-Kutta method. In practical use, we can calculate the
numerical values we need fromthe function / without evaluating any 4erivatives analytically.
Example 3.13. Use the Runge-Kutta method to solve the equation y' :3xy - 713
with boundary condition y(0) : 1.
Again we can use a spreadsheet to calculate each of the 4s and sum them to
compute the increment in y. Our spreadsheet has columns containing the values
y0 + ry/2,q2,y2: yo*qz/2'xr: x0+h,
:
y4
y0
yo I ry, n, and finally
* 4, which becomes the y6 for the next value x1
of x. The beginning of the spreadsheet is shown in Table 3.3; the numerical solution
appears in Table 3.4. This time we find that the values computed with h :0.05 differ
from those with h :0.1 by at most 0.0037o (Figure 3.3).
x0,yo,ttr,xtl2: r0 * hl2,yt:
y3 :
h:
+
0.05
h:0.1
\))
/
I
0L
0
o.2
08
04
l0
x
FIGURE 3.3. Runge-Kutta solution of the equation
solutions for h
\.
:0.05
and h
:
yt : 3xy - ll3
(Example 3.13). Again the
O.l are not distinguishable.
204
cHAprER 3
DTFFEBENTTAL EouATroNS
TABLE 3.3. Spreadsheet for Example 3.L3
BC
Exsmple 3.13
hnx
0.100
l
2
= c3 +A$3
:c4+A$3
yl
I
v
xU2
I
=c3+A$3/2 =A$3*(3+C3*D3-l/3)
:c4+ A$3/2 = A$3-(3.C4.D4 - l/3)
:N3
:N4
eta
el^2
:
:
=D3+F3/2
=D4+F4/2
:c5+A$3/2 =A$3*(3*C5*D5-l/3) :Ds+Fs/2
A$3.(3-83*G3
A$3+(3*E4*G4
= A$3'(3+85*cs
- l/3)
- l/3)
- l/3)
TABLE 3.4. Numerical Solution for Example 3.13
Example 3 13
htr
0l
9
l0
1l
12
l3
0
I
202
3
4
5
6
1
8
909
l0
v
xl.l2
etal
0
I
-0
01
0 981,145
0.05
0.15
099
0.25
0.026213
0.05982
0.100393
1.005545
0 152618
3159E7
38
03
1.035032
04
1.114386
I 2396't8
I 425735
1 69581
2.086218
2.6541
3 490613
05
06
07
08
I
035
045
055
0.65
0.75
085
095
r.05
033333
-0.00389
o 223299
0.322',t8'l
0.46'1359
o.683274
I 01385
y1
0.983333
o.9795
1.(]f,4941
164583
53'1384
857203
2319897
2.995',737
3.99'7538
et^2
-0.018583
0.010744
0.042083
0.078486
0 123885
0 183805
0.266451
0.38453?
0.558241
o 820452
1.225891
The Runge-Kutta method, or indeed any numerical algorithm, works best if the step size
chosen is the largest possible one that will still achieve the desired level of accuracy. Press
et al. show how to write routines that adjust the step size automatically to obtain maximum
efficiency while retaining accuracy.
3.4.5. Higher-Order Equations
Higher-order differential equations may be reduced to systems of first-order equations
through the introduction of additional variables. For example, the second-order equation
'
y" +2y'2
-6:o
(3.4s)
may be reduced to two first-order equations. First introduce the new variable
U:
y'
(3.46)
Then the original differential equation becomes
u':6-2u2
(3.47)
Each of these equations may now be tackled using one of the standard methods.
3.14. Solve equation (3.45) subject to the initial conditions y(0) : 0.5
and y/(0) : 0. Use the Runge-Kutta method, and compute values for 0 < x < 1.
The differential equations are
Example
y': f(x,!,u):u
205
3.4 NUMERICAL METHODS
TABLE 3.3. (Continued)
!2
:D3+H3/2
=D4+H4/2
=D5+H5/2
eta3
y3
eta
eta4
:
:
:A$3*(3*(E3*r3-ll3) :D3+J3
:A$3*(3.(M*I4-l13) :D4+14
:A$3-(3.(Es.rs-l/3) =D5*J5
A$3.(3*C4-K3
A$3.(3-Cs-K4
= A$3*(3-C6-K5
- l/3)
- l/3)
- 1/3)
=
=
=
y4
(F3 + L3 + 2*(H3 +
(F4 +LA + 2*(H4+
(F5 + L5 + 2*(H5 +
I3))/6
=
rq)/6 :
J5))/6 :
D3 + M3
D4*M4
D5
*
M5
TABLE 3.4. (Continued)
I
t
I
I
y2
eta3
y3
eta4
eta
y4
i
0.990708
-0.018473
o 981527
0.986817
0.01 1073
0 992518
-0.003888
0 0262t8
0.981445
o.992438
1.01348
o 042618
o.o'79465
0.1254',n
1.035 l
t.074274
1.176329
1.331581
Q.186377
1.558963
o2'7M64
1.888079
2365338
0.391484
0.569828
3 064326
084
4.103558
I
l6
0 059827
I
114497
0 100406
018555
0.010994
o.042593
0 079355
I
23985',t
o.152645
o.\25292
1
426056
t.696399
2 087294
2 656046
3.4941
4;1499
o 22335',1
1.425',t35
0.683',199
0.186057
0.2'70015
0 390408
0.567882
I 014897
0.8365
I
259288
-0
o 322911
0 46',761',7
-0
033333
l2
0.9918 12
1.035032
1 I 14386
I
239678
69581
2086218
2 6541
3.490613
4 482425
and
: B(x,!,u):6 -2u2
We step forward from the boundary at x : 0, using an increment h in x. Then we
ut
compute the increments in y and u using
:
my :
rlz :
mz :
Qt
: hus
hg(xs, Y0, u0) : n$ hf (xo,
Ys, us)
hf (xo + h /2, yo -l
hg(xo
+ hl2, lo
*
2ul)
nr 12, vo
-f mt 12)
qr/2,
*
uo
mt12)
and so on. Then we use the increment 4 given by equation (3.44) with a similar
expression for m.
We set up a spreadsheet with columns for x,
x + h/2, and r * h; q; andm;,i :
1,2,3,4; q and m; and the new yalues ),(-r0 + h) : y(xo) * q and u(xo + h) :
u(xo) * m.The beginning of the spreadsheet is shown in Table 3.5; the numerical
solution appears in Table 3.6. The results for h :0.1 and ft : 0.05 are shown in
Figure 3.4. Splitting h inhalf gives results that differ by at most 0.004Vo.
The examples above illustrate the basic ideas involved in finding numerical solutions of
differential equations with given initial conditions. The same ideas are used in problems
with specified boundary values at two points, but the solution is more difficult and usually
involves repeated iteration to converge on a solution. Any one method may prove unsuitable
in a given case, and another method will have to be tried. Important issues of stability and
convergence have not been discussed here. Details can be found in the references in the
bibliography.
2OG
cHAprER 3
DTFFERENTTAL EeuATtoNS
TABLE 3.5. Spreadsheet for Example 3.14
I
C
D
: V4
:V5
ml
etal
: $B$3*C4 : $B$3+(6 -2*C4"2)
: $BS3*C5 : $B$3.(6 -2*C5'2)
: $B$3.C6 : $8$3*(6 - 2*C6'2)
G
Example 3.14
h
0.1
xyv
00.50
0.1 --T4
0.2 :T5
2
3
4
5
6
M
o
y3
eta3
: $B$3*I-,1 : B4 + M4
: $B$3*L5 : 85 + M5
: $B$3*L6 : 86 + M6
m3
yl
xl
: A4 + $B$3/2 :84
: As + $B$3/2 :Bs
: A6 + $B$3/2 :86
+D4/2
+Dsl2
+D6/2
a
1
2
3
4
5
6
:
:
:
$B$3+(6
-
elz4
m4
v3
: C4-l 04 : $B$3*P4 : $8$3+(6 - 2*P4 2)
: C5 + 05 : $B$3*P5 : $8$3*(6 -2+P5 2)
: C6 + 06 : $B$3*P6 : $B$3*(6 - 2*P6 2)
2*LA Z)
$8$3*(6 -2+L5^2)
$B$3* (6
-
2*L6 2)
TABLE 3.6. Numerical Solution to Example 3.14
c
I
2
Example 3.14
h
0.1
3
x
v
4
0
0.5
5
0.1
6
7
o.2
0.3
o.4
0.5
0.6
o.7
0.8
0.9
o.5294
0.6115
o.732
0.8766
8
9
t0
l1
t2
t3
t4
1
1.0348
0
o.5770
1.0386
1.3464
r.52'75
1.6264
t.2004
1.6783
1.3697
1.541
t.7049
t.1184
1.7t32
1.7252
1.88s9
t.7286
etal
ml
0.6
o.5334
0.3843
0.2375
0.1333
0.0709
0.0367
0.0187
0.0094
0.0047
x1
0.0s
0.15
0.25
0.35
0.45
0.55
0.65
0.75
0.85
0.95
yl
0.0000
0.0577
0.1039
0.1346
0.1528
0.1626
0.1678
0.1705
0.1718
0.1725
0.1729
0.00.24
1.05
1.9723
0.5000
0.5583
0.6634
o.7993
0.9530
r.tr62
t.2843
1.4550
t.6269
t.7994
N
o
eta3
y3
m3
v3
eta{
m4
0.o29t
0.s291
0 6100
0.7302
0.8752
0.5831
0.4701
0.3181
0.1900
0.1049
0.0553
0.0285
0.0145
0.0073
0.0037
0.0018
0.5831
0.0583
1.047 t
1.3568
0.to47
0.5320
0.3807
6
0.1 187
7
0.t432
8
1.0339
l1
0.1573
0.1650
0. l 690
0.1711
12
0.172t
t.7r3r
13
'14
o.1127
1.8859
o.r729
2.0588
l0
G
M
0.0806
9
D
t.t999
t.3694
1.5408
a
1.6818
1.7068
0.1357
0.1536
0.1632
0.1682
o.1707
t.7r94
0.r719
1.7257
0.t726
t;7289
o.1729
0.1730
t.5364
1.6324
1.7305
0.23t8
0.t279
0.0670
0.0343
0.o174
0.0088
0.0044
o.0022
0.0011
3.4 NUMERICAL METHODS
207
TABLE 3.5. (Continued)
:
:
:
:C4+84/2
:
:
y2
etaz
v1
c5 +Es/2
c6 +86/2
:
$B$3+H4
m2
$8$3*(6
$B$3*(6
$8$3*(6
:
:
:
84 +14/2
:Bs +r5/z
$B$3*H5
:86 +t6/2
$B$3+H6
v2
-
2*H4 2)
2)
2*H6 2)
-2*H5
-
:
:
c4 +K412
cs + Ks/2
:c6+K6/2
U
eta
(D4
(D5
= (D6
:
:
y(n+
m
1)
:
:
:
:84+54
:B5+55
:86+56
+ 2+t4 + 2*M4 + Q4) 16
+ 2*r5 + 2*M5 + Q5)/6
* 2x16 + 2*M6 + Q6)/6
(E4+2*K4+2*O4+R4)/6
(E5
(E6
+ 2+K5 + 2*O5 + R5)/6
+ 2xK6 + 2*06 + R6)/6
TABLE 3.6. (Continued)
K
v1
ela2
y2
m2
0.3
0.03
o.0843727
0.12307497
0.582
o.45762
o.29705
1.59418
o.t4650973
o.r594r777
0.5150
0.5716
0.6730
0.8052
0.9563
t.66t9t
0.16619128
l.rt79
t.69663
o.t6966329
0.t7142303
r.2852
0.17231005
r.627t
0.02429
0.01228
0.00618
0.t7275595
t.7996
0.0031
I
1.7267
0.r729798t
1.9724
0.00156
t.7294
0.84373
1.23075
1.4651
r.7t423
r.723r
t.72756
1.7298
1.4554
e!tr
y(n+
0.0294
0.5770
0.t205
0.t446
0.5294
0.6115
o.7320
o.8766
0.1582
m
v2
0.2910
0.8058
1.187
r
0.r707
r.4317
o.09172
r.5734
0.M76r
1.6502
1.6904
1.7110
t;12t5
0.46t6
v(n + 1)
o.5770
1.0386
o.3077
t.3464
l
1.5275
1.0348
0.0989
0.1655
0.1693
1.2004
0.05 19
r.6264
t.6783
r.369't
0.1712
1.5410
0.1722
r.7132
0.t'127
0.1730
1.8859
o.0266
0.0135
0.0068
0.0034
0.0017
0.0821
1)
2.0588
fr,.'11
0.181
1.7049
r.7r84
r.7252
1.7286
1.1303
v(n + 1)
C4 +U4
:
:c5+u5
: C6 +U6
208
cHAprER
s
DTFFERENTTAL
EeuAloNS
+
+
h:0.05
h:
O.I
\))
-{
F
4
F
I
-{
n-
0
0.2
0.6
0.4
0.8
1.0
x
FIGURE 3.4. Runge-Kutta solution of the second-order differential equation y" + 0.2y'2
with y(0) : 0.5 and )/(0) : 0 (Example 3.14).
-
6
:
0
3.5. PARTIAL DIFFERENTIAL EQUATIONS: SEPARATION
OF VARIABLES
Many of the differential equations that describe physical processes are partial dffirential
equations. The solution is a function of three space variables and one time variable, and
the equation relates the partial derivatives with respect to the four variables. The diffusion
equation (3r13) is an example. Maxwell's equations are also of this form. If we introduce the
potentials A and O, Maxwell's equations reduce to wave equations for these two potentials.
In SI units,
V"AIn the special case where
j
t a2L :
AM -tlo:
(3.48)
is independent of time, the equation simplifies to
V2A: _poj
A similar equation holds for O:
v2a: _L
t0
e.4g)
This equation is called Poisson's equation.
The mathematics of partial differential equations is extensive and complicated, and we
shall not delve into a general treatment here. Many of the partial differential equations in
physics may be solved by the method of separation of variables.lg In this method, we aim to
lgAdditional methods will be discussed in later chapters.
See,
for example, Chapter 7, Section 7.5.
3.5 PARTIAL DIFFERENTIAL EQUATIONS: SEPARATION OF VARIABLES
209
replace the partial differential equation with a set ofcoupled ordinary differential equations,
one in each of the variables. Then we can make use of any of the techniques we have already
discussed for the solution of ordinary differential equations. In particular, series solutions
are often used.
For example, let's look at equation (3.49) in the special case p
0. This simpler homo-
geneous equation
operator2o is
-
is called Laplace's equation. In Cartesian coordinates, the Laplacian
. a2o a2o* a2o
V-@:
art - af
u7
We look for a solution in which O is the product of three functions, each a function of one
(and only one) of the three variables:
a:
X@)Y(y)Z(z)
First we substitute into the differential equation:
a2o+ a2o+ :-i
a2o
: x"yz + xy"z * XyZ" :0
3
3
ctx' dy' OZ'
where the prime means differentiation with respect to the argument of the function.
Now for a nontrivial solution (O not identically zero),2r we may divide through by @
to get
Xtl ytt
_l_r_
X,Y,Z
Ztl
-0
At this point our equation has separated; each term is a function of only one of the three
variables
x,y,andz. If wenowvariedx, leaving yandz constant,wecouldchangethe
value of the first term without changing the second two. The equation would no longer
be satisfied. Thus, the only way we can satisfy this equation for all values of x , y , and z is
ifeach term separately equals a constant:
Xtt
yt,
XY
--4.
--8.
ztl
Z
and
A+B*C:0
Thus, the method of separation of variables leads to a set of coupled ordinary differential
equations. The constants A, B , and C are called separation constants.
20See Chapter 1, Section 1.2.4, equation (1.50). Solutions in curvilinear coordinate systems will be explored in
Chapter 8.
2lTechnically, we need the function O(i) to be nonzero except on a set ofmeasure zero or, roughly speaking,
within a region of zero volume.
210
CHAPTER
3
DIFFERENTIAL EOUATIONS
In this example, each function is an exponential. Let
-u2-p2.ThesolutionsareoftheformX
:
eLo",Y
-
A:
a2 and
:
e+Pv,andZ
B:
exp
F2.Tlten
C:
(uJdTP)
The boundary conditions must be invoked to complete the solution.
/ x w x hhas all its walls
(Thin rubber strips
potential.
is
at
a
nonzero
grounded except for the top, which
general
for the potential
solution
Find
the
separate the top from the other walls.)
Example
3.15. A rectangular
copper box of dimensions
everywhere inside the box.
The solution follows a set of steps that
we'll want to use for all problems of this
typ".'2
l.
Choose
an appropriate coordinate
system. Here
we
coordinateswithoriginatonecornerof theboxand0 <
choose rectangular
x < 1,0 < y .
w,
0<z<h.
2.
Separate the equation and determine the sets of solutions that exist. Here the solutions are real and complex exponentials in all three coordinates, as we determined
above.
3.
Choose one set of coordinates that has null boundary conditions. Choose the
function that satisfies the zero boundary condition at both boundaries. Here we
need
X(0) : 0 and X(l) :
0. We can combine the complex exponentials to
give a sine function, which is the solution we want. We reject the hyperbolic sine
because it has only one zero, at x : 0. We need a second zeto at x : /. Thus, we
must choose A to be negative: A : -k2 and X(x) : sin ftx.
4.
Choose the value of the separation constqnt to make the function zero at the second
boundary. Here we need
:0
srnk/
which will be satisfied if
kl
:
wT, or
k-
NT
T
This set of constants is the set of eigenvalues for the problem. The functions
X':sin
nrf
x
,
are called eigenfunctions.
5. Repeat the procedure for
the second set of coordinates:
Ym
22See also Chapter 8, Section 8.2.
-
mIf \)
sin ------'
w
PROBLEMS
211
For the last set of coordinates, the separation constant is already determined by
the two values of A and B that we have chosen in steps 4 and 5. Thus,
rmn:,2
c:-A-t:\i)tnn12+(;J
and is positive.
6.
Choose the final function to satisfy the one remaining zero boundary condition,
given the value of C. In this case, with real exponentials as the possible solutions,
we must choose the hyperbolic sine as our solution:
Z:
rinn
(T)' .)
(
Thus, the solution contains terms of the form
. wfx
mTy
Sln-SlIl-STDII
7. Write the general solution
.)
,
as a linear combination of such terms:
(T)'
This answers the question posed, but there is one final step:
8.
Use the final nonzero boundary condition to determine the cofficients
is the subject ofthe next chapter.
an^.This
PROBLEMS
f,
A vehicle moves under the influence of a constant force F and air resistance proportional
to velocity (compare with equation 3.5, with i replacing the gravitational force). Find
the speed of the vehicle as a function of time if it starts from rest at t : 0.
2. Find the general solution to the differential equation
y"'-3y"*3y'-y-o
3.
Hint: ExIend the result for a double root from Section 3.1. l.
A capacitor C, inductor L, and resistor R are connected in series with a switch (see
the figure). The capacitor is charged by connecting it across a battery with emf t. The
battery is disconnected, and then the switch is closed. Find the current in the circuit as
a
function of time after the switch is closed.
212
CHAPTER
3
DIFFERENTIAL EQUATIONS
PROBLEM
3
4. The Airy differential equation is
- xy:o
y"
Find the two solutions of this equation as power series in x.
S
Solve the equation xyl/
*2y :0
(Section 3.3.2) using the Frobenius method. Show that
as discussed in Section 3.3.2.
y(0) cannot equal any nonzero constant,
6. Find
a solution of Laguerre's differential equation
xy" + (l
that is regular at the origin. Show that if
of degree k.
7.
- x)y'*
cv
ay
:6
is an integer k, then this solution is a polynomial
Solve the Bessel equation
4x2y" +4xy'
as a Frobenius series in powers
ofx.
I
(4x2
- l)y :
o
Sum the series to obtain closed-form expressions
for the two solutions.
8.
Solve the hypergeometric equation
(*2
- *)y" + (3x -
112)yt
t
y
:
0
as a series
(a) in powers of
x
(b) in powers of x
-
1
9. Find two solutions of the Bessel equation
/
o\
x2y"+*v'+(*2-t-)v:o
' \
4/"
as series
in.r. Verify that your solutions
agree with the standard forms
fT ,"ot*
V;( "
\
*sinx/
PROBLEMS 213
EOl Consider
a
linear differential equation of the form
*2y"+xfy'+g!:o
Expand the function
f (x) in a power series
f(x):
of the form
fo* ftx
I
fzx2
+...
and expand the function g(.r) similarly. Find the indicial equation. What is the condition
on /6 and 96 if there is only one root? What is the value of the root in that case? Use the
method of variation of parameters to show that the second solution of the differential
equation is given by equation (3.37).Ifint.' Show that the equation for u may be reduced
to the form
dl
*(lnu'}:--+h(x)
where ft(x) is a series of positive powers of x. Integrate this equation twice to obtain
equation (3.37).
[iL
linear differential equation of the form x2 y" + xf y' + g/ : 0, the indicial equation
h(p) : 0, where ft (p) is a quadratic function. Show that in determining
the recursion relation, the coefficient of the cn teffi is h(p -l n). Hence argue that the
method fails to provide two solutions if the solutions of the equation h(p) : 0 differ by
an integer. Are there any exceptions?
For
a
may be writte n as
12. Solve the equation
:
* (*)'
,#
0 (equation 3.19) by writing it in the form
y"
y'
yty
and integrating twice.
13. Find the two solutions of the equation
y"-y'*I:o
x
as series
in powers of x.
14. Determine a solution of the equation
(l
I
x)y" +
(3
+ 2x)y'
* (2* x)y :o
x. Hence determine the solutions for all .r.
li5.l Oete.-lne the large argument expansion of the Legendre function
at large
tion of the equation
(l - *2)y" - 2*y' -l2y :
as a series
in powers of I lx.
o
Q1 by finding a solu-
i
214
cHAprER 3
DTFFERENTTAL EeuATroNS
L6. Solve the equation
*2y"-4xy'|_(6+x2)y:o
17. Solve the equation
xy"
-
y' -l 4x3y
-
O
18. The conical functions are Legendre functions with / : -L, + iX.
(a) Starting from the Legendre equation (Example 3.7), find the differential equation
satisfied by the conical functions P-;+ixt) and Q_y*,^(x).
(b)
Show that one solution is analytic at the point
r:
1
,
and determine a series expansion
fortheconicalfunction P_11ir(cos0) inpowers of sin(0 12). Henceshowthatthis
conical function is real.
) : 0 in standard form, and use Fuch's theorem to show that
the Frobenius method may not give two series-type solutions about x : 0. Change to
the new variable u : Ilx (as in Example 3.10) and show that the new equation can be
solved by the Frobenius method. Obtain the two solutions.
19. Write the equation *4y" +
[20.] Sotve the equation
Y"+Ycosh'r:0
Hint: Firstexpand the hyperbolic cosine in
a series, and then use a
power series method.
21. The Stark effect describes the energy shift of atomic energy levels due to applied electric
fields. The differential equation describing this effect may be written
xy"
+ y'*
(o
- f,r*' +T - *t), :
o
where the term Exz 14 is the perturbation due to the electric field. Obtain a power series
solution for y and obtain explicit expressions for the first four nonzero terms. How many
terms are needed before any effect of the electric field is included?
22. Show that the indicial equation for the Bessel equation
*Q#) *xy-0
has a repeated root. Show that this root leads to only one solution. Find the second
solution using equation (3.37). Try to get at least the first three terms in the series.
23. Attempt to solve the equation
*2y" +
!' :o
using the Frobenius method. Show that the resulting series does not converge for any
value of x.
PROBLEMS 215
[24.]
W"b"t's equation is
I x2\
l"+(* +___lv:0
2 4 /'
Show that the substitution 1l - exp(--x2/4)u(x) simplifies this equation. Find two
solutions for u(x) as power series in x.
25. The Schrcidinger equation in one dimension has the form
h2 d2il.,
2^ip
Develop a series solution for
of two nucleons:
lt
+ (E - V) tL :0
in the case where V is the potential due to the interaction
,,y
--L -'-o'
Obtain at least the first three nonzero terms.
26. The Kompaneets equation describes the evolution of the photon spectrum in
a
scattering
atmosphere:
!dt : n"orr+
\ !vo@, + n + n2)l
mcL x' dx
Here n is the photon number density, x is the dimensionless frequency, and o7 is the
Thomson scattering cross section. We may find a steady-state solution (A lAt :0) when
photons are produced by a source q(x) and subsequently escape from the cloud. When
n remains (( 1, the Kompaneets equation becomes a linear equation:
\lvorn'
x'dx
where
*n )l +
y is the Compton "y" parameter,
scattering)
x
q.")
equal
-on
y
:o
to (fractional energy change
(mean number of scatterings). Assume that
q(x)
-
0 except for
x(
per
1.
(a) Show that for x )) 1 the solution is an exponential. This is the Wien law.
(b) Show that in the special case ), - 1, with q(x) = 0, the solution is a power law in .r.
(c) Verify your answers to (a) and (b) by letting n : e-x D and finding a power series
solution for u. Obtain a solution for general y, and then let y -+ 1 to verify your
answer to (b).
27. Aparticle falls
d under gravity. Air resistance is proportional to the square of
.Wittethe differential equation that describes the particle's
position as a function of time. Choose dimensionless variables, as in Section 3.4.1, and
show that the equation may be put into the form
a distance
the particle's speed:
F
:
kuz
y"+oy'2-F:o
Use the Runge-Kutta method to solve this equation with cv : 0.1. Assume the particle
starts from rest. How long does it take for the particle to fall 10 m?
216
cHAprEB
s
DTFFERENTTAL EeuATtoNS
[ZAl tn astrophysics, the Lane-Emden equation describes the structure of a star with equation
the equation of hydrostatic equilibrium
of state P - Kp(n+t)ln.If we define p -
^.0",
becomes
i*e#) *Q':0
where x is a dimensionless distance variable with
the Lane-Emden equation.
x
:
0 at the center of the star. This is
(a) Find a series solution for @ in the case n : l.
(b) Find the first three nonzero terms in a series solution for @ for arbitrary n. Verify that
your result agrees with the result of part (a) when n : l. Hint: Begin by arguing
that the solution contains only even powers of
r.
(c) Solve the equation numerically for n :2, QQ) : I, and d'(0) : 0. At what value
of x does @ (x) first equal zero? (This corresponds to the surface of the star.)
29. Investigate the effect ofair resistance on the range ofa projectile launched with speed us.
Assumethatairresistanceisproportionaltovelocity:Fres: -ai.Writetheequationsfor
the x- and y-coordinates in dimensionless form. Scale the coordinates with the maximum
range
R
:
,E/5. What is the dimensionless air resistance parameter? Determine the
dimensionless range for values of the air resistance parameter equal to 0, 0.1, 0.2, 0.4,
and 0.5. Determine how the maximum range changes, and also determine how the launch
angle for maximum range changes as air resistance increases. Hint: lf there is no air
resistance, you can obtain exact expressions for the increments in position and velocity
in a time interval Ar. Use the same expressions when s I 0, but with acceleration
computed from the value of i at the beginning of your time interval.
30. The equation that describes the motion of a pendulum is
,r8.
I - -VsnY
(a) When y remains small, the equation may be reduced to the harmonic oscillator
equation. Solve this equation to obtain the solution
(b) With the initial conditions y(0)
: rl3,
yt(0)
y(t).
:
0, solve the nonlinear equation
period.
does
the period differ from your
By how much
numerically to obtain the
result in (a)?
31. Bessel's equation
oforder
u has the form
d2v
ldv
/
v2\
ii+;d*+(t-p,/
):0
Show that the differential equation
d2f
77 lz'7:g
may be converted to Bessel's equation through the relations
f:JZv
l
PROBLEMS 217
and
2
,l*r/,
--r+2'
resulting Bessel's equation? (The solutions are given in Chap-
plus sign is changed to a minus sign, so that the equation is
u"+u'+\u:o
has a solution of the form
u:1Qs-x/'*,
(;)
and find the order v of the modified Bessel function.
i
I
I
:
I
:
CHAPTER 4
Fourier Series
4.1. FOURIER'S THEOREM
The principle of superposition for waves states that when two or more wave disturbances
are present at the same time in a system, the total disturbance is the sum of the individual
disturbances. This physical principle remains valid so long as the linearity of the system is
preserved; that means that neither any of the individual disturbances nor their sum is too
large.l Mathematically, this principle is due to the linearity of the governing differential
equation (3.15). The functional form of the sum of the disturbances need not be a simple
sine function-it can be quite complicated and irregular.
This principle is also the idea behind Fourier's theorem, which says that any moderately
well-behaved2 function /(x) defined in a domain 0 < x < 2r may be expressed as a sum
of sines and cosines:
f (x) :\,{o"sinnl
*
bncosnx)
(4.1)
n:0
Equivalently, we may write the series as a sum of complex exponentials:
f (x):
(4.2)
n\r,dn*
llfthe equation is written in terms of suitable dimensionless variables
(see Chapter 3, Section 3.4.1), the necessary
conditionisy(1.
2we'll give specifics of how well-behaved the function must be in Section 4.5. For now, you may assume that any
function that is the solution to a physics problem is well enough behaved.
219
i
I
t
l
i
220
CHAPTER
4
FOURIER SEBIES
Here we are using a physical principle (the principle of superposition for waves) to
motivate the mathematical result that a function f (x) may be expressed as a superposition
of wavelike functions (the sines and cosines) no matter what use we may make of the
function /. Result (4.1) or (4.2) can be used in a variety of different physical systems, as
we shall see.
Since the sines and cosines are periodic functions, the Fourier series is also periodic. Thus,
f (x t2mr) : f (x), where m is an integer. We can find a Fourier series for a function that
is not periodic, but it will represent the original function only in a finite range. Outside of
that range, the original function and its Fourier series representation will differ; the series
gives a periodic extension of the function in the original range. In many physical situations,
we want a solution valid in a bounded region of space (for example, inside a box of side
L), and a Fourier series with period I works well, since the solution does not even exist
outside of the range 0 to Z.
4.2. FINDING THE COEFFICIENTS
4.2.1.The RealSeries
Let's assume that a series of the form (4.1) exists. Then how do we find the coefficients a,
and bn? First we note that the sine and cosine functions form a set of orthogonal functions
ontheinterval 0<x<2n(or,infact,onanyintervaloflength2r).Thefunctionsare
orthogonal3 in the sense that
fo'" "rnn,
sinmx dx
:o
(4.3)
l
and
j
fo'" "o"n*
unless
z:
cosmx dx
:o
cosmx dx
:o
(4.4)
nin both cases, and also
fo'" "rnr*
(4.s)
for all integers m and n. Multiplying two trigonometric functions and integrating over one
period is analogous to taking the dot product oftwo basis vectors.
3We shall give a more precise definition of the terrn orthogonal in Chapter 8.
4.2 FINDING THE COEFFICIENTS 221
To prove relations (4.3)-(4.5),
we'll
sincvcosp
:
p
:
sina sinB
:
cos
cy
cos
need the following relations:
jtrtn
)f"o"
jt"o,
-
p)l
cos (cv
-
p)l
cos (cv
*
P)l
(a +
fl)*
sin(o
(a +
fl)*
- fl)-
(a
Thus, we may write the integral in (4.5) as
t?n
/
Jo
sin
nx cosmx a*
If m : n, the integrand
: *I /12, trtn (n * m)x * sin (r LJo
m)xldx
\121t
I
I
:ll-| /
cos(n-d*ll
cos(n*m)x2 \ n*m
n-m
/lo
- 0 form fn
is j sin2mx and the integral is still zero.
Next we shall prove result (4.3):
fzosin nx sin mx dx : I f2" - (n
[cos
Z Jo
J,
m)x
: ^I I I sin(r zLn-m
-0 formfn
I
If n
:
-
cos (n
*
m)x1dx
tt2n
I
m)x ------:- sin (n +rn)xl
n+m
I
llo
m, the integral is
['"
Jo
"rn'
nx dx
:: ['" (l 2Jo
cos 2nx)
dx
: :( .- I ,in zr*\l'" : o
2\ 2n
/lo
if n : m :0, in which case the result is trivially zero.
Finally, the integral ofthe cosines is
except
r2n
J,
cosnx cosmx O.
: iI f2o [cos (n - m)x * cos (n -l m)x]dx
Jo
I I
I
l-sin(n-- 2ln-m
:0 formfn
and again the result is
r
1f
m
- nf
0. However, for m
r2t
I dx:2n
Jo
tt2n
I
mrx*n*m
sin(n+zlxll
'Jlo
: n:
O, we
now get
(4.6)
222
CHAPTER
4
FOURIER SERIES
Expressing a function /(x) as a Fourier series is akin to expressing a vector in terms
of its components. To find a component of a vector, we take the dot product of the vector
with the appropriate basis vector. Similarly, here we expect to find the coefficients an in
the Fourier series (4.1) by taking the "dot product" with the basis function sinkx. That is,
we multiply the function /(x) and its series (a.1) by sin kx for some integer k and integrate
from 0 to 2z:
r2n
l-"
Jo
.f1.xlsinkx dx
\
r2r/*
: l-. I Io,sinnx * b,cosnx I sinkxdx
Jo
/
\,=
If the series is uniformly convergent,4 we can interchange the sum and the integral. (We'll
test this hypothesis later.) Then
['" f ,rrsinkxdx:i.(",
['"
Jo
?:o\"Jo
:
*b,"Jo
['" cosnxsinkxdx)
,rnn*sinkxdx
/
AkTt
since all the other terms in the series are zero by relations (4.3) and (4.5). Thus,
ak:
: 1,," .f (x) sinkx dx,
kZ
l
(4.7)
1
Similarly, multiplying by cos kx gives the result
r
t2n
b1,:!1T I .fg;coskxdx, k>l
Jo
(4.8)
Whilefork:0,
uo:
I f2,
*
f
Jn
(x) dx
(4.e)
which is the average value of the function'over the interval.
Example
4.1.
Find the Fourier series for the step function
and/(x):1forr<x<2n.
f (x) :0
for 0 <
First we define the series to be
:i
,or: {l
U
n:O
"lo,=..,1",
4In Chapter 6, we show how to relax this restriction.
r,
sinnr
* bncosnx)
x<z
4.2 FINDING THE COEFFICIENTS
223
Then, using results (4.7) and (4.8), we obtain the coefficients:
an:
: t;"
f (x) sinnx dx
1l (-cosnx)
"' fin
:l
|
0
ifn
l-21
nn
:;1J,
12,
I
nT
-l-1
sinnx dx
+ (-1)'l
is even
if n is odd
and
br:-
tf 2n
I
nJo
/(x)cosrx O* :;
I f2"
J,
cosnxdx
:11
(sinnx)l2f :0
fin
We have to evaluate b0 separately:
bo:
lf2"1f2nl
2"J, fktdx:2oJ, o*:t
As expected, b6 is the average value ofthe function over the interval 0 to 2n. Thus,
the series is
r,"r
: {f ;l:=.i 1or, : :\
-
:i't:*
P:rJ
Figure 4.1 shows the first four partial sums5 of this series. Notice that the value of the
series at x
0 and at x T iq I/2, not 0 or 1. When a function has a discontinuity, the
Fourier series always gives a value equal to the midpoint of the jump. The series is a good
:
-
representation ofthe function, but it is not identically equal to the function at every point;
there can be significant differences at a finite number of points.
As we include more terms, the sum more closely represents the step function. Just beyond
the discontinuity at x : r,the series overshoots the value y : 1. Increasing the number of
terms moves the peak of the overshoot closer to x : n but does not reduce its magnitudean effect called the Gibbs phenomenon. The value of the series at the peak 0f the overshoot
is 1.179. (SeeAppendixV.) The Gibbs phenomenon occurs at every jump of adiscontinuous
function.
The series we have found actually represents a square wave. Figure 4.1 could be repeated
indefinitely to both the left and the right. Thus, the jump from 0 to 1 visible at x : n in
Figure 4.1 also occurs at every odd multiple of n, and, in reverse, the jump from I to 0
visible at x : 0 occurs at x : 2mn.Therefore, the series takes on the value L : LfO + I
when x equals any integer multiple of n.
5See
Chapter 2, Section 2.3.1 for the definition of partial sum.
224
CHAPTER
4
FOURIER SERIES
N:3
f(x)
1.0
0.8
0.6
0.4
o.2
0
FIGURE 4.1. Fourier series for the step function (heavy line). Shown are the partial
t'n!t',*,t)'
1 - tt
^
2 .un:O
2n I I
:
(solid line), and
N
for N :
0 (dashed
line). N : l
(dottedtine),
sums
N:
2
3 (dot-dashed line) terms. The series overshoots the step by
187o. As the number ofterms is increased, the overshoot does not decrease, but moves
closer to the step. This is the Gibbs phenomenon (see Appendix V).
If the function is defined over a range of values that does not span a range of 2n , then
we can proceed by changing to a new variable. For example, if the step function is
f
then we define the variable u
the series is
:
(o
(x\: { ifo<x<0.5
l1
\- if0.5<x<1
2nx
I
Itw\:
so that
2
S
\
0 < u < 2z corresponds to 0 <
2 nu^
n:U
sinl(2n
_
In general, if the function is defined for 0 <
2tt x I
L
so that u ranges from 0 to
f (x):E
2r
[,,
(4.10)
x<
-f l)2trxl
1. Then
(4.11)
2nll
x I L, then we choose the variable a :
anddefine the series
as
"r"(r) *b,cos(4r))
With this argument, the orthogonality relation for these sine functions is
L
1L /2nnx\ /2mnx\
/2"
sin(nr'r) sin(mu);du:
t'" (--) tt" ( r o*:
J,
/
Jo
L
,3^,
(4'12)
4.2 FINDING THE COEFFICIENTS
22s
and thus the expression for the coefficient an is
o": r2 fL /(x) sin / 2nnx\
Jo
\ , )0.
(4.13)
The factor in front of the integral is one divided by (half of the range of integration).
Similarly,
b,: ? lo' ro""" (rf1o.,
n>o
(4.14)
For bo, we get
bo:
;t;
f (x) dx
(4.1s)
Again we find that bs is the average value of the function over the interval.
For a function defined on the interval -L < x I L, we choose the variable u : ltx/L,
whichrangesinvaluefrom-n to1n,againarange of 2n.Thecoeffrcientsof thesinesin
the Fourier series are given by
f+L
o,: i,I J_r,or'^(T)
The coefficients
o-
(4.t6)
b, of the cosines are found similarly.
4.2.2. The Complex Series
Complex exponentials also form a complete set of orthogonal functions. In the onhogonality
relation for complex functions, we must multiply one function by the complex conjugate
of another:
r2tt
J,
y,yi,d*:
r2r
Jo
i(n "inx"-imx4*: --=
: o,
lo
m),itu-m)xl'_"
wlileifn:m,
f2n
J"
,'n'e-'n" dx
:
f2n
Jo
I dx
:2tr
m+n
226
cHAPTER
4
FouRrER SERTES
This expression is valid for n
:
m
:0
also. Thus, the coefficients in the senes
+oo
f(x): D
,nr'"'
(4.17)
are given by
Ln
-
+ l,'" f (x) e-'"* dx
(4.18)
In this case, the number in front ofthe integral is one divided by (the range ofintegration).
Ifthe function f(x) is real, then
\*
r
/ I f2, (x)e-inxdx) :
,;: l*
+ J,
l, f
p2n
f (x)ei"* dx:
c-,
(4.1e)
Let's evaluate the complex series for the step function (4.10):
(o
u ifo<x<o- : +m
:
/(')
D'nn',n',o*
iii.=:,':
; n:-6
\
1
Notice that we have chosen the variable 2n x as the argument of the exponential, since this
variable's value goes from 0 to2r as.r ranges from 0 to 1. Then
e
u^-Jrp'
c,: I fG)r-i2n,,a*:
[' ,-i2nnx4r:f
^L.)r-iznnxlllrP
Jo
\-2nri1
: !-k-2'ni - r-nni1: .Ltt - (-l)'l
znr
znTt
Like the an and bn we obtained previously, this coefficient is also zero unless n is odd, in
which case the term in parentheses is 2. We must evaluate the integral differently if n :0,
since our expression has a zero in the denominator in that case:
co:
Thus. the series is
r1
I o*:)
Jrlz
4
Now notice that the coefficient for n
n:
N:
: -N,
3
FOURIER SINE AND COSINE
SERIES 227
c_N, equals the negative of the coefficient for
c-N : -cN
Thus, we may rewrite the series as
f (x):
1lgi
;* i
t
n:1,
;ru,rn",
n odd
-
r-i2nrx1
1 2 S
sin(2nnx)
\_
_::____.
,
n
2 T ,:ioaa
A.2l\
where the sum is over odd values of n. The result is the same series (4.I 1) that we obtained
using sines and cosines from the start. One advantage ofusing the exponential series is that
the integrals are often easier to do.
A function /(x) may be represented by aFourier series of sines and cosines (equation 4.I
and Section 4.2.1) or of complex exponentials (equation 4.2). Since the two series represent
the same function,
ir,sinnx*
Expanding the sines
""0:"lr**
@ (
ii",
;'ot
ut
(Chapter
,inx - n-inx
-
b,cosnxl:
'2i
,,
f
,n"''*
"rrurton"l.i^tur.rr,we
i-* \
get
+m
,inx
+6--l!l:|',''n'
2 ) n?*
oo
b,
! b, .-in*] : f
iL) [-ia,_+
,inx
lian
'
'
2
z
t n:-6
n:O '
'
-
.,",,"
Thus, the relations between the coefficients are
c--:unI'o',
L-ncr:b-!--J!,
12
2
n>l
't-'
(4.22)
and
co:
bo
4.3. FOURIER SINE AND COSINE SERIES
The function shown in Figure 4.2 is obtained from equation (a.10) by shifting the origin
vertically and multiplying by - 1. The resulting step function,
(-tn
-L<x<o
S.(x)-< '-.- ififo<x<*z
l+t1z
(4.23)
CHAPTER
4
FOURIER SERIES
FIGURE 4.2. The step function 56(-r) is an odd function on the interval
(-L,
L).
is clearly odd in x over the range -L < x < ,L and so should be represented by a series
of sines, which are also odd in x. We did find that series (4.2I) for our previous step
function/square wave contained only sines. The origin shift in y is represented by the
constant 1/2 in equation (4.21). Thus,
g
(nr
S.(x)--T2 ) sin n x /L)
n:Goaa
(4.24)
When the function is odd on the range - L < x < L, the argument of the sines is n x I L
because the period is 2L. [In the previous section, we used the step function (equations 4.20
and 4.2I) with
I l2.l However, we can express the coefficients as an integral over half
the period, 0 < x < Z. The coefficients in a sine series ofthe form
-
I :
f (x):i,o,"inff
n:l
are given by
on
I fL
/nrx\
J-'f r*l ( J
'
fL
/nnx\ \
/nftx\
: ! (f
r\/-rl(x)sin(. )o'*Jo "f(x)sin\ t )d')
-- T
sln
ax
1 / fo
/ nTrxr
/lL
,nnrt, \
:T\Jrf
t-x)sin(- r )d(-x)+Jo /(r)sin\, )dr)
Then, since
an
/(x)
is odd,
I /
: L\-
t
/nlrx\1
1L
,r11)a*\
fo
o* *
/(x)sin
(,
-fa)
ll
\-r, /
l-sin
Jr
Jn
4.3 FOURIER SINE AND COSINE SERIES
an:
Conversely,
sented
/nlfx\
?t: /(x)sin \
if a function is even on the range
by a series of cosines with argument
nxI
f (x) :bo + i
' n:l
-Z
(4.2s)
, )d*
<
x<
229
L, the function will be repre-
L:
a,
nr )c
"o, L
The coefficients are given by
br: t2fL
Jo
/(x)
cos
/nfix\!)o
d*
\t
(4.26)
)
and
bo:
; t,' f (x) dx
We can make the odd square wave (4.23) into an
along the x-axis by L12 (Figve 4.3).
(4.27)
:n fu nction by shifting it to the left
f(x)
x
L
:
FIGURE 4.3. By shifting the origin to the right, we can create a step function Ss(-r) that is an even
function. This function is represented by a cosine senes.
230
cHAPTER
4 FouRrER
SERTES
Then series (4.24) becomes
oo
S"(x)
: -2 \lJ
1f
n:l,n
2
7t
2
7f
1
/x
-sinnzl-+
,?
\L
odd
.xwtxwf
-L2L2
sin nn
@
\./2
n:I,n
@
-l- cos nz
-
sln
-
n
odd
el)(n-r)/2
nL
odd
\,/-
n:l,n
)
nrx
(4.28)
which is a cosine series, as expected.
As with the complete series, if a function is defined only in the range 0 < x < L
(Figure 4.4a), then the Fourier sine (or cosine) series will represent a periodic extension of
the function outside ofthe original range (Figure 4.4b). The sine series gives an odd periodic
extension, while the cosine series gives an even periodic extension (Figure 4.4, c, d). Each
of these series has peiod2L.
f(x)
FIGURE 4.4a. The function /(.r)
is defined on the
range0<x<L.
f(x)
_LOL2L
FIGURE 4.4b. The complete Fourier series gives
period is L.
a
periodic repetition of the original function. The
4.3 FouRrER srNE AND cosrNE
sERrEs
231
f(x)
slne senes
FIGURE 4.4c. The Fourier sine series is a periodic repetition of the odd extension of the original
function. The period is 22.
f(x)
even extension
coslne senes
-LOL2L
FIGURE 4.4d. The Fourier cosine series is
a
periodic repetition of the even extension of
/(x).
The
period is 2.L.
Thus, in principle, we can find three different series for a function
rangeO<x<L:
1. The full series
f (x):
with coefficierts cn
: !
L
.f"t
f Al
,f_rn",rnn,,"
,_2innx/L
dx
andperiod
I.
/(x)
defined on the
-t
CHAPTER
4
FOURIER SERIES
2. The sine series
J$l:
with coefficients an
3. The cosine
a
z
r- anstnnTX
)
L
n:l
rL
I
. nJLI
: ; I /(x) sin
.
LJo
L
-
dx and period 2Z
series
3
nrx
f(x):bo+Lb"cos ,
n:l
: 2 fL / (x) cosnrx d * and b6 equal to the average value of the
L Jo
L
over the range 0 to I. This series also has period 22.
with coefficient" b"
function
Different applications require the use of different series,
as
we shall see in the next section.
Example 4.2. Find the Fourier sine series and the Fourier cosine series for the
function f(x) :x, 0 < x < L.
The sine series /(r) : DansinnnxfL represents an odd function on the range
-L
< x < L (Figure 4.5). The coefficients
o'
2 fL
/nlrx\
: T/
xsin(-l
o'
are (equation 4.25)
2/
L
: r[-';cos,
nr* L
fL L
nTrx \
o* Jo -cosi"
)
2 / L2
/ L\2
,rz*rr\ :
:- L\
I -1 cosn1+ (
l sin--:l
L to/I -ZLeD'
nr
ntT
\ro/
-
FIGURE 4.5. The first six terms in the sine series of the function
f (x) :
-r defined for 0
< x < Z.
233
4.3 FOURIER SINE AND COSINE SERIES
Thus,
-t:--
S) (-l)'
2r
o?
nrx
n
L
-
range
-L
<x<
/(x) : Dbncosnrxf L represents
an even function on the
(equations
(Figure
4.26 and 4.27)
are
L
4.6). The coefficients
The cosine series
bo:
IfL
Jo
i
*o': iL
and
L
,or.\
nrLl -dx
I
f(x)
x
L
FIGURE4.6.Thefirstfourtermsinthecosineseriesofthefunction/(x)=.rdefinedfor0<x<L.
Compare this graph with Figure 4.5. A few terms of the cosine series represent the
function more closely than does the sine series.
Thus,
is
b, :
0 for even n , and b,
: -4L / (ntr)2
for odd n. Therefore, the cosine series
I
L 4tS
z_t^ (2p * l)z
z nz P:O'
'
-
Qptt)nx
L
234
4
CHAPTER
FOURIER SERIES
Compare Figures 4.5 and4.6; the cosine series converges much faster than the sine series
(the terms decrease as If n2 rather than 1/n). This happens because the odd extension is
a discontinuous function with a jump at x : lL, whereas the even extension is a continuous
function. In Problem 2, you will be asked to look at the full series for this function.
4.4. USE OF FOURIER SERIESTO SOLVE DIFFERENTIAL
EQUATIONS
Fourier series may be used to solve differential equations under some circumstances. In this
section, we shall look at two of the most common applications.
4.4.1. An Inhomogeneous Linear Equation Has a Periodic Driving Term
When the source term in an inhomogeneous linear equation is periodic, we expect that the
solution will also be periodic. One example is an electric circuit connected to a power supply
that supplies a periodic emf.
Example
4.3.
A circuit has a 10 O resistor, a 2 mH inductor, and a 0.3 pF capacitor
connected in series with a power supply that generates a square wave emf t(r) with
:
a period of z
0.3 ms (Figure 4.7). The applied voltage varies between zero and
*1 V. What is the current in the circuit?
FIGURE 4.7. The series ZRC circuit in Example 4.3 with
current and charge variables
(I
a
periodic (square wave) emf. With the
and Q) as defined in this diagram,
I:
dQ/dt.
The charge and current variables I and Q are defined in Figure 4.7. Kirchhoff's
rules applied to the circuit give the differential equation
L-dI +RI+
dt
f,:erO
where 1
:
d Q / dt .
If we write the equation in standard form (equation
d2o
R do
- -),
dt2'Ldt----1
t-
ot
' I.C
I.
3. 18),
i
4.4 USE OF FOURIER SERIESTO SOLVE DIFFERENTIAL EQUATIONS
the coefficients of
dQ/dt
235
and Q arc
R 10s2
2a: lL 2mH
and
,3=
I
1
LC
(2 mH)(0.3 rr.F)
1010
6
s-2
Then
l0s
@o:
ftradls
:4 x lOa radls
We have already found the series for the square wave emf
t(t):
t
+oo
^(:.:
\
n:-
where z is the period of 0.3 ms and t6 :
series with period
t
(equation 4.20):
('':))
"ooi"'o
I V. Next, we write the solution
as a
Fourier
r:
Qft):
,t*r,*r(,'+)
(Note: It is much easier to use the series in exponential form here, because the equation
has both first and second derivatives. When we differentiate the series, each term in
the equation contains simple multiples of the original exponential terms. In contrast,
the odd-order derivatives mix sines and cosines: dsinxldx: cos-tr. Instead of one
equation for each coefficient cr, we would have two equations to solve simultaneously
for the coefficients of the sines and cosines.)
Now we substitute the series for t(t) and QG) into the differential equation:
i
i - ('+)'q,ri2nnt/t +2' n:-6
i22o nn"'"nt/'+@?0
:?(:.: r
\
n:-oo.
,,
f
nn"'"n'/'
n:-@
n:-@
Lr,r,,,t,\
odd
/
Next we make use of the orthogonality of the exponentials by multiplying the whole
equation by exp [-i(2mnt lt)] and integrating over one period. Only the terms with
n : m survive. Since this is true for any integer m, we aar' equate the coefficients of
each exponential separately.
The constant (n : 0) term gives
)
a54o:
c
ol)
-
+qo
toC
2
CHAPTER
4
FOURIER SERIES
The other terms are given by
| /zno\2
a,;-\;)*
4otinr "'l h
. +a6):ii
i
i
tstz
Qn:
I
L n l-(2wr)2 + 4qrinru + c'Sr2l
tsr2 4atr +[r3r'- (2nr)2]i/n
r L lr|t, - (2nn)212 I (4arnr)2
n
Notice that the real part of q, is even in n, while the imaginary part is odd. This is
exactly what we expect if the resulting series is to be real. The real parts combine to
give cosines, while the imaginary parts combine to give sines:
i
g(t\:t:9*t:!
2
nL
4wr + la;3r2 -
(2nn)2li
ln
^_^
(
,2nrt\
/
fl@t^o\'r
2ntrt I ^^ (2nr)'lsin
^ 2nnt
4cYrz cos
toC
r ;lafir'
_
- ,L $
- 2 -2tot2
k$t2-(2nr72J2+{4atnrS2
2
-@,n
Finally, we can differentiate to get 1. Then
4Est
I(t):
L
t
4narn sin
2nnt
@3t'
n:1
-
^
(2nr)'l
+ la'ot'
-
(2nn)212
a (4utnr)2
2nnt
The constant in front ofthe sum is
4tst
4(l V)(0.3
L
(2 mH)
ms)
:0.6 V.s :0.6A:10
-H
Thus,
I7t'1: -t0.6A;
p
L
4narn
,in2n!! + b3r2 - (2nr)2l"or'no'
Notice that if the natural frequency ar6 of the circuit times the period z of the emf
is very close to 2nn for some integer n, then the current will be very large if cv is
small; there is
a
resonance at frequency @n
aNf n
:
(4.I x
:
Znn
10a s-1;10.3
lr
.
In our example,
x l0-3 s)/r :3.9
4.4 USE OF FOURIER SERIES TO SOLVE DIFFERENTIAL EQUATIONS
237
which is very close to 4. The solution is shown in Figure 4.8. The current is dominated
by the n : 2 term, which has twice the frequency of the square wave emf.
1(A)
0.02
r (ms)
0
-0.02
-0.04
t--\
\n:2
-0.06
FIGURE 4.8. The first three terms of the solution for the current in Example 4.3. The thick solid line
is the sum. The thin solid line is the n : I term, the dashed line is the n : 2 term,
and the dotted line is the r : 3 term. The solution is dominated by the n : 2 term.
The square wave emf is shown schematically by a thin solid line. The square wave
indicates the phase (but not the amplitude) of the applied emf. The vertical axis does
not apply to this curve.
4.4.2. The Solution We Want Equals Zero at Two Boundaries, .r = 0
attdx=L
If a physical quantity is described by a function /(x) with "f (0) : f (L) : 0 and exists
only in the interval 0 < x < Z, we can make an odd extension of the function /(x) to the
range -L < x < L and represent it as a Fourier sine series. The sine series is always zero
at the endpoints r : 0 and x : I and thus automatically satisfies the boundary conditions.
For example, consider the problem of waves on a string of length L.The differential
equation for the string deflection y (x , r) is (equation 3. I 5)
.,32
u-
y
o2y
*z:
6P
We can solve the equation using separation of variables.6 Let
u2x"T
Dividing through by y
I
I
t
t
t
XT , we get
.X"
XT
t
i
:
: xr"
6See Chapter 3, Section 3.5.
T"
y(x,t): X(r)f
(/). Then
238
CHAPTER
4
FOURIER SERIES
Since each side of this equation is a function of only one variable, each side must equal
a constant if the equation is to remain true for all values of both x arrd t. Since we want the
stringdisplacementltoequal
X to be a sine,
zeroatx:0andx : Lfor alltimes/,wewantthefunction
so we choose the separation constant to be negative:
X,,
,-z
positive integer. Then the equation for Z
T"
T
:
n:l
(T)
k:
nnf L, where n is
t nn:,2 ^
t-2,.2 ___l_l
uz
sin (nn ut I
y(x.t):!sin
0; that is,
IS
"
and the solutions are
be written
sin(ft-r)
sin(kl) :
Next, we choose the constant ft to make
a
) X:
L)
\t )
and cos (nn ut I L) . Thus, the general solution may
fo,'rn
(t:)
* bn cos (tf)]
(4.2e)
which, for fixed r, is a Fourier sine series in x, as expected. The amplitude of each Fourier
component is oscillatory with angular frequency @n : nft u / L. The initial conditions determine the constants an and br.
4.4. Suppose we pull the string up at the middle so that it forms a triangle
of height h (Figure 4.9) and then let go. Find the subsequent motion of the string.
Example
L
FIGURE 4.9. The initial shape of the string (t
The solution has the form (4.29). At
v(x. 0)
:
/:
and the string velocit! at
t
:0
0) in Example 4.4.
0, we have
(xQh/L)
ifO<x.L/2
l(t - x)(2hlt) if Ll2 < x < L
{
:
:i0,"^(ry)
is zero:
Dvl
/nirx\
'l
:0:) Srzu
0rlr:o ?- -sinl-la_
L \ L/
4.4 USE OF FOURIER SEBIES TO SOLVE DIFFERENTIAL EQUATIONS
Thus, all of the coefficients an arezero, and, from equation (4.25), the coefficients bn
are given by
2 fL
bn: i
Jo
o)
nrx
sin
dx
"*'
'
: # (lr''' * sinffa' + l',rrt-
x) sin
#0.)
To do the first integral, we integrate by parts:
L ,
no*rlL/2 fr/z L
l-cos-llL /lo + I nr
nn\
Jo
72 /
wtr /L\2
wv
:rno
lL/2 nrx
I xsin-dx:xL
Jo
wtx
,
----=-dx
L
(-cosi)*\;)sin7
The cosine term is zero unless n is even, and the sine term is zero unless n is odd.
The second integral is
1L nftx
L2 r
nnxt.tL :- L2 t nnr
Ll
-Jrp- si11!!24*:1(-.or"',^)l
2 -,-tl')
L /lt/z nn\IcosjLnn\
and the last integral is
wrx
1L
nrx
L,
nnxr, lL
fL L
I xsin'-'---dx:y{cos )ltp
ll - I
--nn\--- Jtp"L
L Jtpntr
-cos-dxL
:
# (rr-rr' -*"7). (h)' sin)
Adding the three terms, we get the result
b,:4h
z -(-l)'
2
Z nTr 2 +cosf
wrLl-l""rT*Lsinf
*)(r,-r^ -*"7) * *'^T)
:- th
(nr)t
wr
2
which is zero unless n is odd, in which case the result is
u,
where we wrote n
:
ffir-g@-rtp
:2p *
1. Then the
v(x,t):Y\#{",in
P:l
: 12ol*1t-9n
solution for y is
(rzz
+ DT)"o, (tro
. r+)
240
CHAPTER
4
FOURIER SERIES
Figure 4. l0 shows the first four terms of this series. The plot shows the dimensionless variables7 y f h versus x
attimes ut f L :0,I14,I12, and l.
lL
v
n
./
--l---+---F--l---+-
\
t-l,",- - f - - -f ---t--
!
L
FIGURE4.10. Thestringdisplacementattimes/: L/utimesO(thicksolidline),714(dashedline),
1/2 (short dashes), and 1 (thin solid line). The first four terms of the series were used
to make this plot.
4.5. CONVERGENCE OF FOURIER SERIES
Now that we understand why Fourier series are useful, we should worry a little bit about
how well they converge.
4.5.1. Pointwise Convergence
A function is very smooth if at least the first two derivatives of the function exist and the
second derivative is continuous. It is piecewise very smooth if we can divide the interval
(-L, L) into a finite number of subintervals and the derivatives exist within each of the
pieces. The step function (4.23) satisfies this requirement, since the derivatives exist within
each of the two subintervals (-1, 0) and (0, I), even though the derivatives do not exist at
x
:0.
The Fourier series of a function f (x) Ihat is piecewise very smooth 6n an interval
-L < x < I converges pointwise to )U{, * 0) + f(x -0)l for each x in the
interval.
TSee
Chapter 3, Section 3.4.1
4.5 CONVERGENCE OF FOURIER
At the endpoints, we identify
f(L -
x : -L
and
SERIES 241
x : Z to obtain the limit ittef
+ 0) +
o)1.
Fourier series are not, in general, uniformly convergent. They are uniformly convergent
only in closed subintervals where /(x) is continuous. But Fourier series can be integrated
and differentiated term by term. We'll see why in Chapter 6.
4.5.2. Convergence in the Mean
Fourier series converge to the corresponding function
f (x) in the mean.
That is, the square
deviation
**
: J-rl
ft lru, - f ,nr,no,,,l' a* --o as N -+ oo
,7x
I
(This idea should remind you of a least-squares fit of a model to a set of data points.)
The function and the series may differ substantially at a finite number of isolated points
and still converge in the mean. We have already seen this happen with the step function
series (Figure 4.1), where the series always takes on the value at the middle of each step,
independent of the value of the function at that point.
The Fourier series of any piecewise continuous function converges in the mean. This
condition is sufficient but not necessary; that is, there are functions that behave more badly
yet whose series also converge in the mean. As physicists, we do not need to concern
ourselves with such functions because they do not arise in the solution of physics problems.
Now let's ask if we can do better than the series we have. Is there any set of coefficients
cn for which the remainder R1,, is smaller than that given by the Fourier coefficients?
Let
rLN2
RN: | ,f(xlJ
cnei""/L
|
-r.
(4.30)
dx
where the cn are unknown. Then, to minimize R1g, we choose the coefficients cp so that
ff:o
* l:,(t,', - ,t*'n''nn*") (r.u, -,i
Ifthesumff;=
-N
cneinrx/L isreal,thenc[
[ (-
kn x / L
r't
:
a 2'i ktr x / L
"
rL
J-r-f
''0"'tL
dx +
i
---N
c-m,and,if
^t- *'
-
che-i^'.rr) dx :0
f (x)isalsoreal,thederivativeis
*'-' ^"'')
IL ,;*nr/rr-^s-imnx/L
J-L
x
:
o
dx
:
o
d
242
cHAPTER
4 FouRtER SERIES
In the second term, every term in the sum integrates to zero except the one withm
:
k.
Thus, we have
rL
I ytik"r/tdx :2Lc_r
J-r
and so the value of cp that minimizes R1y is
ck:
r
;
rL
,-,0"*rtd*
J_rf
(4.31)
which is the usual Fourier coefficient (equation 4.18). Thus, the Fourier coefficients are
optimum in the sense that they make the square deviation R,ntr a minimum.
We can use the result that Rlr -+ 0 to prove another important result about the Fourier
coefficients. We begin by expanding expression (4.30):
rL/
N
N
\/
Rr: l- ( frrl- t
/-'\'
n:-N
cneino*/L
) (f,rr.- t
m:-N
/ \
\
cf,e-i^"'/Lldx
)
rLrLN
: I lfulP a* - l, \ lft*)*r, r
J -L
- -- n:_N
f')c\nleinttx/14*
rLNN
+ I \) ,nutnn'/' t
J_r,?r
cfle-i^n*/Ldx
m:_N
The last term may be rearranged to give
t
"i -t:-,,,i, I
:
and the integral is zero unless m
x/
L'-imn
x/L
r'iwt
4*
n, when it equals 21. Thus, the last term
rs
NN
2L
I
D c,ctr-zrn:-N
lc,lz
n:-N
We can also simplify the middle term, since
TLNNTL
I c\nrin"'tLdx : L r\, J|-L- tr*r rinnx/1tr,
l--t tr-l n?u
n:Jp
J
NN
: It
*
c* n2Lc-n
: Zf I
n:-N
lc,l2
PROBLEMS 243
and, similarly,
TL/VNN
>, f(x)*c,ei"ft'/Ldr:2L n:-N
D cncfi:ztn:-N
I - n:-N
\
J-L
b,l2
Combining the last two terms, we have
tLN
nx
: |
-L
1ye)12
dx
J
Since Rp
t
-2L \
n:-N
lr,t2
0 by its definition,
rIt
N
tL
\t"t2
;J-,trG)t24*' n=-N
which implies that the series on the right converges. Thus, for a series that converges in the
mean, R1g
-+
0 as
N ->
oo, and so
rIt
@
tL
\
; l_, f(*)2dx:n:-@
(4.32)
cn2
This is one version of Parseval's theorem.8
We can understand this result by thinking about the energy stored in the capacitor in the
circuit of Example 4.3. Averaged over one period, the energy stored in the capacitor is
'
:
I f, QG)2 _
: I \-,^,2
; Jo 1;o'
zc
Lb^r
From this we conclude that the time-averaged energy equals the sum of the energies in each
of the individual Fourier components.
PROBLEMS
I
Snow that the Fourier series (equation 4.1) for a function
f (x) :i
n,
n:O
"o,
/(x)
may be written
(nx -t Q,)
and find expressions for kn andQr.
8See Problem
variant.
2l for Parseval's theorem for
the sine and cosine coefficients, and see Problem 22 for an additional
244
2.
CHAPTER
4
FOURIER SERIES
f(x) : x
(a) overtherange0<x<1
(b) overtherange -l <.r < 1
(c) Make a plot showing the original function and the sum of the first three nonzero
Develop the full Fourier series for the function
terms in each series. Comment on the similarities and differences between the two
series.
3. Develop the Fourier series for the function f (x) : x2 over the range 0 < -r < 1 .
@ An odd function /(.r) on the range (-L, L) has the additional property that f (x + L)
:
- f(x).
(a) Make a sketch showing the important features of this function.
(b) Which kind of Fourier series (sine series, cosine series, or full series) represents this
function on the range -L < x < L?
(c) Show that the series has only terms of odd order (n :2m * l), and find a formula
for the coefficients as an integral over the range 0 < x < L 12.
(d) How does your answer change if f (x -f L) : t f (x)?
(e) How do your answers change if the function is even, but /(x + L) : - f (x)?
5. Which series, the sine series or the cosine series, do you expect will converge more
rapidly to the function f (x) : x3 on the range 0 < x < 1? Give reasons for your
answer. Evaluate the first four nonzero terms in the optimum series. How large is the
fractional deviation ls4
I {G) | u, :
"
lf@)l
sum of the first four terms.
0.5 and
x : l? In this expression,
Sa
is the
6. Find the Fourier series on the range 0 < x < 2n for the function f (x) : sin cvx, where
a is not an integer. Check your result by evaluating the limit a, --+ n. With the value
a : 0.7, plot the original function and the first three terms of your series on the range
0<x<2rr.Comment.
7. Find an exponential Fourier series for the function sinh cvx on the range 0 < x < 2n .
S
9.
By combining terms, rewrite your answer as a series in sines and cosines.
finO the first four nonzero terms in a Fourier series for the function tan r on the range
-n/4<x<n/4.
Use numerical integration to find the first ten terms in a Fourier series for the function
sin x2 on the range 0 < x < n. Discuss the percent error between your series and the
function sinx2 over the given range.
10. Find
a
Fourier series for the ramp function
.f
(x):
{r
ifO<x<1
ifl<x<2
ontheintervalO<x<2.
11. An electric circuit contains a 3 mH inductor, a 50 pF capacitor, and a 200 Q resistor
in series with a power supply that supplies a rectified sine wave voltage (see the figure)
with amplitude 110 V and period 2 ms. Determine the capacitor voltage as a Fourier
series.
system is driven by a periodic driving force with period
\') r",.,,_lo,
7:
ifO<r.T/2
\a(r-r) ifT/2<t
Find the response of the system.r(r) as a Fourier series.
<T
246
CHAPTER
4
FOURIER SERIES
14. A simply supported beam of length L bears aload W that is uniformly distributed over
the first 1/4 of its length. Determine the deflection of the beam as a Fourier series. Make
plots showing the first one, two, and three terms of your answer. How many terms are
needed to obtain a result accurate to l%o? (The differential equation satisfied by the beam
deflection is equation 3. I I , and the displacement is zero at the two ends.)
@e beam rests on supports at its ends, x : 0 and x : L. The load q(x) varies linearly
along the beam: 4 : ax.What are the boundary conditions? Find the displacement of
the beam as a Fourier series. Plot your results, and comment.
16. A guitar string of length L :65 cm is plucked by pulling it to the shape
v(r, o)
ifo<x.L/3
: Io*'
\{"/+'ttt-r12
if Ll3
<x <L
and then letting go. Determine the subsequent motion of the string. Which harmonics
are excited? Plot the string displacement as a function of x for t :0,0.4, and 0.8 times
I/u. Also plot the original string shape for comparison. Comment.
17. A violin string is plucked to the shape of a triangle with apex one-quarter of the way
along the string, as shown in the figure, and then let go. Find the displacement of the
stringatlatertimes.Plotyourresult(uptothen:10term)for/:
Lf2u,wherc
v : ltlJ
1t
L/lou,Ll5u,allid
is the wave speed. Are all harmonics excited?
I
L
0.010
0.008
0.006
0.004
0.002
0
o.4 0.6
PROBLEM
0.8
x
L
17
18. ApianostringoflengthZishitbyahammeroflengthl: tr/l0.Thehammeriscentered
at x : Ll4, and the impulse it imparts is 1. Assume that the impulse is uniformly
distributed over the hammer's length. Determine the subsequent displacement of the
string as a function of x and /. Which harmonics are excited? Plot the string displacement
as a function of x for t :0.1,0.5, and 0.75 times Llu.Use the first five nonzero terms
of the series to make the plot. Comment on your results.
PROBLEMS 247
19. Fourier series may be used to evaluate certain series of functions of integers.To illustrate
the method, develop the Fourier series for the function x2 on the rarrge -7r to z. Set
x
:0
and hence evaluate
I
St-l)'-r
\
2
n2
Which sum do you obtain by setting x
:
I
4's'
-t-.*-*...
n? Finally, use Parseval's theorem to evaluate
$r
Ln4
n:1
EO.'l
Us" the Fourier series for the step function to evaluate the sum
S
t-t.r'
L^2* *
1
m:o
(See Problem 19 for the method.) Use Parseval's theorem applied to the same series to
obtain the sum
$r
z-^
12m
21. A function
-f l)2
f (x) is represented by the Fourier senes
3 r
nrx
f G) : L lo, sin , *
n:O
nr(x\
bncos -----
)
I
on the range (-L, L). Derive a form of Parseval's theorem (equation 4.32) applicable
to this series; that is, express [j! f t*f a* in terms of the coefficients a, and bn.
22.
If f(x) is represented by the series D .fne'"' over the interval 0 < x < 2n,
: D grri'" over the same range, prove the genenlizedParseval theorem:
and
g(x)
1,2noom
L .f,t-,: n:_@
\ f,sl
- I f(xtg(xtdx: n:_@
ztf Jo
where the second expression applies when the function g(x) is real.
23. The capacitor shown in the figure on the next page is charged by the battery and discharges
through the bulb when the potential across it equals 0.9V. (See, for example, Lea and
Burke, Example 31.3,p.993.) Assuming that the capacitor discharges very rapidly, show
that the potential across the capacitor as a function of time is
vc:v(l
-"-,,*r1,
-\)
and repeats periodically with period
that represents this function.
T
:
o < r < RCtn lo
RC ln 10. Find a Fourier series with period ?
248
CHAPTER
4
FOURIER SERIES
PROBLEM
@l n
rectangular box of dimensions
y
23
a x b x a has conducting walls. All the walls
:
are
b. This wall is separated from the others by a thin
insulating strip, and it is at potential V. Using the method illustrated in Chapter 3,
Example 3.15, find the potential everywhere inside the box.
grounded except for the one at
x b x c has all its walls at temperature T1 except for
s, which is held at temperature 72. When the box comes to equilibrium,
the temperature function T(x,y,z) satisfies equation 3.14 (extended to three spatial
dimensions):
25. A rectangular box measuring a
the one dt 7
:
AT
u -
Dv2T
with the time derivative on the left equal to zero. Using the method of Chapter 3, Example 3.15, find the temperature T in the box in the form
T(x,y,z):Ttlt(x,y,z)
where z is expressed in a Fourier series
t (x , y , ,,
Find the function
f (z) nd
:
na*
sinlla
snff f fr>
the coefficients amn.
26. An infinitely long conducting tube with circular cross section of radius a is divided into
four equal pieces by insulating strips running along its length. One of the four pieces
is at potential V, and the other three are grounded. Solve Laplace's equation in two
dimensions, using the method of Example 3.15. Evaluate the solution at p - 6 411
show that the result is a Fourier series. Determine the coefficients, and hence find the
potential everywhere inside the tube.
Wl n Fourier series of the form
f (x):Dr,r'n*
PROBLEMS 249
may be expressed as a power series
f (x):D"nrn
where z : lirnr+t re'r ar'dr < L The function f (x) may be identifiedby summing
the power series. Use this technique to sum the Fourier series
@
s\_ Slfl/l-T
where 0
<x<
you found.
=n
rr . Check your result by evaluating the
Fourier sine series of the function
CHAPTER 5
Laplace Transforms
5.1. DEFINITION OFTHE LAPLACETRANSFORM
An integral transform allows us to convert an inhomogeneous linear differential equation to
an algebraic equation that is easier to solve. We shall look in detail at two such transforms:
the Laplace transform and the Fourier transform (Chapter 7).
The Laplace transform is an integral operator applied to a function f(t) that is defined
for0 < / <
oo:
L(f):F(s)
:
lo*
rale-'t
(s.1)
dt
The factor e-'t in the integrand causes the integral to converge for a very large class of
functions /.
A function
f
is of exponential order o6 ifthere exists a real positive constant
M
such
that
le-"ot y1tll <
for all
r,0 < t <
u
(s.2)
oo, and og is real.
The Laplace transform exists for all piecewise continuous functions of exponential order.
The transform is defined for Re (s) > os.
Let's look at some examples.
1. Exponential functions: f (t) : eot .The function eot is of exponential order cu-we
take o6 : a and M : I in equation (5.2). The transformis
can
|
F(s): [* ,o'r-"4t:
lo
cv-s "tc-s)rl€
Jo
251
252
cHAPTERs LAPLAoETRANSFoRMS
The transform exists for Re (s)
> q, in which
case
F(r):*
2. Powers: f (t)
:
.The Taylor series for the exponential function,
tm
-, :1+
e''
st
e^ t^
+ e2t2
^ * ...* *t +..'
shows that
,^ am!,"
em
and so
Itm e-€t I .
rrEmem
4rr,
"-r,
:
^t'
for any positive real number e. Thus, these functions are of exponential order e. With
m
:0,
the transform is
L(t): fo* "-"at: -+1, :
For m
>
1, we use
:
[6
Jo
te-'t dt
irRe (s) > o
: l,
integration by parts. For m
LO
:
: ---te-t, l* * ;I /l, /,
:
: -+ls'lo +
irRe (s) > o
foo
o-st l@
_,
e-'t dt
Then, in general,
L(t^): I t^e-'tdt
Jo
:
LL(t^-t)
:
1m
1 r@
s Ilo+:sJoI
ry
LG^-2)
:
.
mt^-re-'tL1
..
:
#
More powerfully, we can use the definition of the gamma function (equation 2.75). Make
a change of variable to u : s/. Then
Lltny
-
fo*
tne-,,d,
=#
which is valid for Re (s) > 0 and
:
lo*
(1)'
,"7
fo*uo"-"a":\fiP
p > 0. The
power p does not have to be an integer.
5.2 SOME BASIC PROPERTIES OF THE TRANSFORM
3.
253
Sines and cosines. These functions are easily taken care of, since they can be written
linear combinations of exponentials:
4(cosa.,t)
:
lo*
as
rye-'tdt
_1( I , I \_
- Z \s - ro- - t + i, ) - ;, +,Dt
.s
which is valid for Re (s)
>
0.
We can now begin to compile a table of transforms (Table 5.1). Each transform exists
when the real part of s exceeds some minimum value sg.
TABLE 5.1. Laplace Tbansforms
.f
(t)
f (t)
F(s)
I
J-0
eot
COS(r)T
1
srn @/
,t
r(p +
1P
F(s)
J
szlaz
a
-s2 to2
+
1)
Jp+1
This table gives transforms of the most common functions, but we will need methods
/(r) conesponding to any transform F(s). This process
is called inverting the transform. Since the Laplace transform is an integral operator, we
can change the value of the function f (t) at a finite number of points without changing the
value of the integral (5.1). Thus, the inverted transform cannot be unique. This does not
usually present any difficulties in the solution of physics problems, as we shall see. Usually
we choose a continuous function as the appropriate inverse.
that allow us to find the function
5.2. SOME BASIC PROPERTIES OFTHETRANSFORM
1. The Laplace transform is a linear operotor. This means that
L(af)
:
lo*
o, U, e-'t dt
:o
lo*
f (t)
e-''t dt
:
aL(f
)
where a is a constant, and
L(fi+f): JoIf6 (fi*f)e -" a, :
: L(f) + L(fz)
Io*
fr(t) e -', at +
lo
fzG) e-'t dt
254
cHAPTER
s
LAPLAoETRANSFoRMS
2, Thetransformof thederivativedf ldt maybefoundusingintegrationbyparts:
If / is of exponential
Re (s)
>
order s6, the integrated term aPproaches zero at infinity for
s6. Then
:
'(#)
-"f
(o)
*
(s.3)
sF(s)
Then the transform of any higher derivative may be found by iteration:
'(#) :'l* (#)l: - #1,. "(#)
:-#lo-"to'+s2r1s;
and, in general,
L(4:J\:
\dt^ /
su
F(s)
-t
d^-"
f I ^n-t
(s.4)
dtr'-^ lo"
n:I
3. The attenuation propertyi If a function / (r) is attenuated (or reduced) by
the factot e-at
,
then
L(r-o, f )
:
lo*
e-o'
f (t)
L(e-"'
e-'t dt
f):
F(s
:
*
lo*
71t1 e_-G+o)t
a)
dt
(s.5)
An inversion may be simplified if we can recognize the transform as a simpler function
of (s
*
a) for some constant d.
Exampte
5.1.
Invert the transform F(s)
: ll$2
-l2as + a2).
We recognize the denominator as (s + a)2.The function F(o) : lf o2 inverts
to give f (t) : t (Table 5.1: powers), and so the inverse of F(s) iste-at '
4. The shifting proper4r is closely related to the attenuation property. Suppose we shift the
function f (t) along the t-axis by an amount /6. Since the original function is defined
5
only for
3
USE OF THE LAPLACE TRANSFORM TO SOLVE A DIFFERENTIAL EQUATION
/ > 0, we must cut the shifted
function off at
/:
255
/o (Figure 5. 1) or, equivalently,
-to:0. (Theshiftedfunctionequals zerofort < /s.)Wecandothisbymultiplying
the shifted function by a step function S(l - rs):
att
shifted
/ :
S(/
-
td
f (t -
to)
f (t)
a function. The shifted function is zero for / < t6, but f (t
must cut off the dashed portion with the step function S(t
rO).
FIGURE 5.1. Shifting
-
- td is not, so we
Then the transform ofthe shifted function is given by
CtS(r- totf\t-rs)l:
Now change variables
/ s(r-r0)/0 -tde-'tdt:
Jo
to Lt. : / - /6. The limits become u :
f,fs(r- b\f(t-to\l:
LIS(I
[* ftt-to)e-'tdt
Jro
0 and oo.
I f(u\e-"'+'otdu:r-"o lo[* ffu)e-"du
lo
-tdf Q - ro)l:e-"'oF(s), /o > 0
(s.6)
Thus, atransformthatcontains an exponential factoris thetransformof a shiftedfunction.
In this result, the parameter /0 must be positive.
5.3. USE OFTHE LAPLACETRANSFORMTO SOLVE
A DIFFERENTIAL EQUATION
If we have a linear differential equation of the forml
d2v dv
oji*ti*cy:f(t)
lThe method is not limited to second-order equations (see Example 5.3). A second-order equation is used here for
illustration.
256
CHAPTER
5
LAPLACE TRANSFORMS
we may apply the Laplace transform operator to the entire equation. Then we can use the
Iinearity property to write
^'(#)
* u'(*sr) + CL(Y):
F(s)
and then use the derivative property:
+l
\ - dtlo-
A(
rytol + r2v(rl)
"/ + Bt-y(o)
* sy(s)l * cY(s):
This algebraic equation is easily solved for the transform
r(s)
:
F(s) + By(0) + A[dy
F(s)
I(s):
/dtls*
sy(O)]
(s.7)
As2+Bs+C
The problem of solving the original differential equation is thus reduced to the problem of
inverting the transform I(s). In many cases, it is not too difficult to invert the transform,
and this is the advantage of the transform method.
Linear differential
Laplace transform
equation for y(t)
direct
solution
for the transform
algebraic
I
J
Algebraic equation
solution
(aimcutt)
I(s)
I
J
(easy;
<- inverse transform <Because of the initial values of y and its derivatives that appear in expression (5.7) for the
transform Y, the Laplace transform is well suited to the solution of initial value problems.
5.2. A series ZRC circuit has a constantemf E and a switch. The switch
0. What is the current in the
has been open for a long time and is then closed at /
Example
:
circuitfort>0?
If the switch has been open for a long time, we are to asSume that the current has
dropped to zero and the capacitor is uncharged with q(0) : 0. Then we can define
the current variable i(l) so that
.dq
dt
(s.8)
shown in Figure 5.2. (It is very important that you take care in defining the algebraic
variables corresponding to the physical quantities ofcharge and current. It is easy to
make sign errors that will cause your solution to be incorrect. See, for example, Lea
and Burke, p. 990.) Then Kirchhoff's loop rule applied to the circuit after the switch
is closed gives the differential equation
as
dioL-+Ri*!:E
dtC
(s.e)
5.3 USE
OF THE LAPLACE TRANSFORM TO SOLVE A DIFFERENTIAL
EOUATION 257
FIGURE5.2. Definitionofcharge(i)andcurrent(4)variablesinExample5.l.Theswitchisclosed
at
I
:0.
Now we transform both equations (5.8) and (5.9):
,=t(#):se-q(o)
and
ot
LIsI-t(O)l+RI+;:;
Use the initial conditions 4(0)
:
0 and
t(0)
:
0, and substitute for Q(s):
I /1\ :; e
Lst*RI+r(;,/
Solve for 1:
t/s
t: l,s*R+(l/Cs)
(s.10)
With the usual definitions q
=
1(s
have
258
CHAPTER
5 LAPLACETRANSFORMS
There are several possible ways to invert the transform. We could factor the denominator and then use partial fractions. The roots of the denominator are
-.1 :
s+:-o*
,:
where we have defined
1@
-
I
s/
r/^\
_ | _
r\J''
-- t
-a !.i
ot."y6"n
\I
-
_-
)--
\(t-t+Xt-tJ
,3-o2 - -aLia
( t _ t \
t
LG+--) \Ft*
s-s-)
From Table 5.1, we can invert each fraction to get an exponential:
iO:
:
go
,hL(e'*'
t
,L,
"'
- e'-'): ,fr7e-"t{e+iot - r-i<t:t1
Sln (@f )
As / increases from 0, immediately after the switch is closed, the current is positive;
that is, it is charging the capacitor.
An alternative method uses the attenuation property (equation 5.5). First we complete the square in the denominator, which may be expressed in terms of s * o:
\
a
*
\l- t- I/
I
\(s +cv)'z +,.,3 -@z ) ofl \tt +a)2 +'2 )
I
r-- t/ I
Now, from Table 5.1, we recognize o/(s2 -f ri.21 as the transform of sin (art). Our
transform has (s * a)2 rather than s2, which indicates that the inverse is the sine
. Thus, the solution is
function multiplied by
"-o'
I
i
(t\ : :;e-dt
sin (ror)
as before.
Even if we do not have values of the function and all its derivatives att : 0, we can sometimes leave the unknown values in the solution and use the remaining boundary conditions
to eliminate the unknowns at the end.
5.3. A simply supported
beam of length Z is one that rests on a support
(Figure 5.3). If the beam supports a load Mg that is uniformly distributed
over a distanc e I at lhe center of the beam, find the displacement of the beam'
We set up a coordinate system with x-axis along the beam and origin at one end.
Then the given boundary conditions are
Example
at each end
y(0)
: 0;
v(L)
:
o
5.3 USE
OF THE LAPI..ACE TRANSFORM TO SOLVE A DIFFERENTIAL
EOUATION 259
FIGURE 5.3. A simply supported beam of length L carrying aload Mg.
The differential equation satisfied by the beam is equation (3.1 1):
dav 1
dr!: EIs@)
(s.1 1)
With a fourth-order equation, we need four boundary conditions. We can find the
remaining two boundary conditions that we need from equations (3.8)-(3.10). In
equilibrium, the torque2 about any origin must be zero, and thus m(0) :0. Also,
:0 since there are no forces to the right of the right end. Thus, we know two
quantities at 0, y and y", and two at x : l.
The load function is
m(L)
q(x)
tr (L -.t)12 < x < (L + t)12
: {yrtt
l0
otherwise
Now we transform equation (5.11):
sar
-
s3y10)
-
r2y'(0)
-sy"(0)
-
y"'(o)
: fiOAl
where
Msl ( L+t\s)-exv(/ L-r\1
-"^dx:i
e(s):[#9r-"'
,
2
, '/l
r+
Lexe(:
At
sl "-t'1''2
sinh
t
2
Put in the known initial conditions:
,aY
-
r(s)
:
s2
y'
10)
-
y"' (o)
- !E-r-t' tzz sinh?
Thus,
9"-u1z2sinh 4 + Y'!o) + Y"'\o)
sr EI [.
2
.sz
.s4
2Recall that m(.r) is the counterclockwise torque due to all forces to the right
ofx
260
oHAPTERsLAPLAoETRANSFoRMS
in the first term suggest that we apply the shifting proPerty
(equation 5.6) after inverting 1/ss. The powers of s invert to powers of .r (Table 5.1).
Thus,
The exponentials
:
up / r\ I /
riz \n )Lr
Y(x)
-'+)-
- +)^s('
(" -
+)^s(' -'+))
+ y'(0)r +
y"'Q)*
Now we need to apply the remaining boundary conditions to find the unknowns y/(0)
and y/'/(0). Evaluating the solution at x : L, we have
y(L)
:
ffil(+)^ - (+)^l *,',0,,
: ffil)ttrt'+ h)+
y'(o)r +
+ v"' (uLi
v"'Q1l
:s
and
_ Ms l('_:t\'_('=\'f
t2l
dr2l.:r-2rrtl\ z ) \ z )
:
ffitt
+ y,,,(o)L
*n,,,,0,,
\v
l''
:o
From this result we obtain
y"'(o) =
Then from
y(L)
:0
y'(o)
-#
it follows that
: -ffirr'
+
e) -
r"'e)* :
#9I;9
So the solution is
v(x)
:
#h{;l (" - +)^'('
-'+)-
(" -
+)^'(' - +))
.
e;r)._,uj
5.4 SOME ADDITIONAL USEFULTRICKS
The displacement of the beam in the case L
:
04
0.2
261
L12 is plotted in Figure 5.4.
06
x
i
YEI
MgL3
FIGURE 5.4. Deflection of th ebeamy EI / MgL3 versts x f L (Example 5.3). Notice the very different
scales on the two axes.
5.4. SOME ADDITIONAL USEFULTRICKS
In this section, we shall compile a list of tricks that may help us to find or to invert
a transform.
5.4.1. The Derivative of the Transform
Since the transform F(s) is a smooth function of s, it may be differentiated:
dFdf@_-fcod
ds
cls
:-
I
JO
ln*
.f (t)
e-" dt
: I
JO
f Q1
. e-"'dt
Cls
tf (t)e-'tdt
Thus,
dF
L(tf ) : --
(s.12)
We can repeat this process n times to obtain
L(t"
f): (-l)^#
(5.13)
262
cHAPTER5 LAPLAoETRANSFoRMS
Example
5.4.
From Table
5.
Find the transform of / sin a,l/.
l, the transform of sin art is o I $2
4(rsin
ar2). Thus,
(r+r)
,,1: -ft
:
+
2so
64 upy
5.4.2. The lntegral of theTransform
The transform may also be integrated:
Now do the integration over o first:
:
l,*
f !t)
Io*
,rolo"
"-"0'
: t(+)
(s.14)
It is essential to this derivation that the upper limit of the integral over o be infinite, so that
the exponential goes to zero at the upper limit. It is also necessary that the transform of //t
exist-that is, that the integral converge. For example, we cannot apply this result to the
function f = I, since the integral [f, (e-'t / t) dt does not exist.
Example
5.5.
Find the transform of the functio n
f (t) : l-f{
From Table 5.1 and the linearity of the transform, the transform of the function
1-cos/is
1s
s
s2+a2
5.4 SOME ADDITIONAL USEFULTRICKS
Then
f@/l
o
\
\ , ):J, \;-;taa)d"
4(L-cost1
:
(rno
\ - l2 +
:limrn-!-m-!
S+oo
rn (o2
J 52 +
:ln1-ttt-l
^/s2
.,r2))
')1,
ot2
I
Js2 +
a]
+ a2
/ o,2\
: lnJ;rT;z : t (.t
s
tln *;t/
5.4.3. Periodic Functions
Suppose the function
it
"f
(l) is periodic with period T (Figure 5.5a). Then we can represent
as
and so on, where
g(t)
is
f(t):g(t
n
gure 5.5b). Equivalently,
)
-2T)+...
0123
FIGURE 5.5a. A periodic function
/(r).
(5.15)
264
CHAPTER
5
LAPLACE TRANSFOBMS
0123
FIGURE 5.5b. The function g(t) is nonzero only in the first period
Now we can apply the shifting property to find the transform. First we transform g:
G(s)
: [' sfrle-'tdt
Jo
The upper limit is I, because the function g is zero for / > Z. Next we use the shifting
property (equation 5.6) to evaluate the transform of the second period of the function:
LIS(I
-
T)sQ
- T)f : e-'r G
and so on. Thus, from equation (5.15),
F(s): (Ile-'r *r-2sr +...)G
F(s):
G
(s. r6)
| - e-st
- series (equation2.43). Thus, a denominator
where we used the known sum of the geometric
of the form I - e-'r in the transform F(s) is the signature of aperiodic function /(r) with
period Z.
Example
period Z.
5.6.
The function
G(s) is
Find the Laplace transform of a square wave of amplitude 1 and
g(t)
:
o:
1
for 0
fo'''
< t < T l2and
e-'tdt:
is zero otherwise. Thus, the transform
+l'o''
:'- ':''''
5.5 CONVOLUTION
Note that the upper limit of the integral is
T
/2 . t < T. Then, using equation
T /2, since the function
(5.16), we have
265
g(t) : 0 for
F(s)::(T#):^;-,
The denominator I - e-'7 is the signature of a'periodic function, but it can be masked
by the form of G, as we see here for the square wave. The same thing happens with sines
and cosines-the prototypical periodic functions-for which G o<. | - e-'r , as you should
verify.
5.4.4. Discontinuous Functions
:
has a discontinuity at t
tr. Then we can find the transform F in
Suppose the function
the usual way. But we have to be more careful with the transform of the derivative, since
the derivative does not exist at /1. Thus, we divide the range of integration into two pieces
and integrate each piece by parts:
/
,(#):
:
:
:
Iou
.f
#u$dt
e-"1:;
+
*, l,
1,,*
#n"0,
ye-'tdt + f
"-,, T
Qr) - "f (tr+)l - /(0) *
- "f(0) - e-'tth
s-st1f,f
sF(s)
*, f,,* f e-,tdt
sF(s)
where ft : f (h+) - f (n-) is the jump in the value of the function at
of higher derivatives may be evaluated similarly.
/1 .
The transforms
5.5. CONVOLUTION
5.5.1. Systems with Memory
t
Certain physical systems have memory; that is, the behavior of the system at time / depends
on what has happened to the system at times less than /. The series LRC circuit is an
example of such a system, as we have seen in prior examples. The current in the circuit at
time r depends not only on the emf t applied to the circuit at time r, but also on the past
history of the circuit (when the switch was closed, for example). The current in the circuit
at any time / is determined both by the structure of the circuit and by the emf t(t) applied
to the circuit over time.
If we choose our current and charge variables as in Example 5.2, so that i : 'fdq / dt ,
Kirchhoff's loop rule results in the differential equation
t#*Ri+3:tG)
266
cHAPTER5LAPLAcETRANSFoRMS
Now we transform the equations to get
I:sQ-s(O)
and
LLst
Let's simplify by choosing
i(0)
-t(0)l + RI + 9:
L
:
q(O)
:0.
E(s)
Then, combining the equations gives
,2gt+sRe*9:zg)
or
E(.s)
Q:-#:E(s)R(s)
szl+sR+l/C
where
:i(r#T*)
R(s):Zi:,r\
t(sr+
tt+ tc1
The transform of 4 is the product of the two transforms E(s) and R(s), one being the
transform of the input function t(r) and the otherbeing the transform of the systemresponse
function r(l). The inverse of the transform E(s)R(s) is the convolution of the functions
€(t) andr(t):
t-t€n :
E*
r
: I to)rtt - t)dt
Jo
(5.17)
This result is called the convolution theorem. (See Appendix VI for the proof.)
Physically, this integral represents the input to the system at time r times the response at
timer(thatis,atimet-tlater)tothatinput,summedoveralltimesrfrom0to/.The
convolution may also be written
as
rt
t +r : r xt : I r(r)e(t - r)dr
Jo
that is, time measured backwards from the present (z
where r represents "time ago"
: t).
(t
to the time the clock was started
Thus, the charge on the capacitor in the circuit problem is
ft
q(t): I eG)rQ-t)dt
Jo
:
0)
5.5
where
r(/)
coNVoLUroN 267
is the function whose transform is R(s). We found this function in Example 5.2:
I
r(t): fie-"t
sinia,t
The circuit response is a damped oscillation.
To illustrate the use of the convolution, consider an AC circuit with
t(z) :
t0 cos Oz.
We have
q(t)
: f€o
(t
;cos or sin ar - r) exp [-a(t - r)l dr
J,
to -_o, tt
: -=-e
cos Or sin @ (t La Jo/
:
r)
edr
dr
['6rnyrr+ (o - o))t)+sin[arl - (a + a)rl]eo'dt
=t=o "_-o' Jo'
2La
The integral is most easily evaluated by writing the sines in terms of exponentials. The result
is
Y
\'/
Eo / a[2aosin or
Z, \
[cv2
+
-
(Q
sin a-lt]
a 7,t21
"-ut
(o
- d2][d2 + + @)2]
, ola2 - Q2 + a.,211cos {2t - e-at
(a2
+
Q2
-)
cos t,lf I
You should check that this solution satisfies the initial conditions q(0) :0 and i (0)
The long-time solution is an AC current. First note that for t )) | /a,
q(tt
to (294la sin Qt + <o[cY2 -- o2 + ar2] cos Qt \
- LrD\
)
and therefore
cos Ar + All - ro3l a2l sin Ar
de_
to
\
irt):E--5,2-\m)
^ nz /2a
: )€t
rocos Qr
* x sin or)
where Z is the impedance,3 given by
22
:-
R2
r x2 = n' + (ar- :)'
\ ac,) :
R2
r-'' )'1
: rrl+o, +\r-\r---@-))
:u
+ a2 ( t- "'
lu
: L2fa2 + (o - d2lwz + (o + d2llsz2
3See,
for example, Lea and Burke, p.1022.
See also Problem 2.14.
t
(aD2e - r3/a\2
:
\
0.
268
oHAPTER
5 t-nPr-AcE TRANSFoRMS
This is the expected long-time behavior. The other term in 4(r),
Es a(a2 + Q2 + a2) sinrirt + a\a2
la2 + (o
-
raD2llo2
+
-
Q2
(sz
+
+
or21
cos
@t
-ot
@)21
is an initial transient that goes to zero with a time scale I I a
:
2L I R.
5.5.2. TheTransform of an Integral
We may apply the convolution theorem in the special case that one of the functions is
g(t) = I with transform G(s) : l/s. Then we have
t(lr' r(t)s(t - do,):
,(1,'
a result
F(s)G(s)
rt"dr) _ F(s)
s
(5.18)
known as the primitive function theorem.
Example
5.7.
Find the Laplace transform of the sine integral
Si
(r)
:
/nt sin
Jn
z
;0"
First we use result (5.14):
,(Y) : I,* L(sint)do
: [* -J-oo:
t--lol*ls
J, o2+l
:;
-tan-r,
Then, using the primitive function theorem, we have
ltsi(r)l:;-itl 1t*-r"
or, since tan (n
/2
-
0)
:
cot 0, we may write the result in the more compact form
.CtSi
(r)l
:
cot_l '
.t
-
.s
5.6 THE GENERAL INVERSION PROCEDURE
269
5.6. THE GENERAL INVERSION PROCEDURE
We have so far amassed a collection of tricks for inverting a given transform F(s) to find
the function /(r). We'd like to have a single technique that we can use every time and that
does not require recognizing a trick that will work. This procedure is the Mellin inversion
integral:
f(t):
_t| 1r+ia F (s) e't ds
2ni Jr-;*
(s.19)
The integral is along a line parallel to the imaginary axis in the complex s-plane. The line
must be positioned so that all of the singularities of the function F(s) are to the left of the
line, as shown in Figure 5.6. This line is also called the Bromwich contour.
Im (s)
Re (s)
FIGURE 5.6. Contour for the inversion integral. The vertical line must be placed to the right of all
the singularities ofthe transform F(s).
To prove the result (5.19), we work backwards and take the transform of the function
/(t) defined by equation (5.19).
.
Ltf(t)l: [-* r-" zfit f*t* F(o)eotdodt
JO
"*,]- Jy_jso
Interchange the order of integration and perform the integration over /:
Ll,f(t)l::
:
['*'* ro>' Jo[*
2ri Jr-;*
1
:----.
'2nt
e-st+otdtdoi
1v+ia
e-$-ov
F(o)
o -,5
I
Jr-1*
-l
l@
lO
do
270
CHAPTER
5
LAPLACE TRANSFORMS
Now, provided that Re (s)
we have
> Re (o) : y, the exponential approaches zero as / -+
oo, and
1 fY+i@ F(o\
' do
Llf G)l: :---- I
/.|tt Jy-i6 S-o
Note that the function F(o) is analytic everywhere to the right of the path of integration,
so we may close the contour to the right, as shown in Figure 5.7.
Im (o)
Re (o)
FIGURE 5.7. Contour for proving that equation (5.19) correctly inverts the transform F(s). The
contour is closed to the right, enclosing the pole at o : r.
Then there is a single pole ofthe integrand inside the contour,4 at the point o : s. The
integrand along the big semicircle is zerg, provided that Max l^F(o)l approaches zero on
the semicircle at least as fast as R-6, where e > 0. Notice that we are traversing the contour
in the clockwise direction, rather than the standard counterclockwise direction, so we must
introduce a minus sign when applying the residue theorem. Then the result is
Llf (t)): -2ni ^L.lim (o zttto+s
:
s)I9
s-o
F(s)
/ is F, then / is the inverse transform of F.
When we use the Mellin inversion theorem to evaluate /, we cannot close the contour to
the right because the exponential e't causes the integrand to blow up on the large semicircle
at infinity. Thus, we extend the function F into the region to the left of the contour, using
analytic continuation, and close the contour to the left, thereby enclosing all the poles of F.
If the transform of
5.8.
Invert the transform F (s)
The denominator factors:
Example
t2 -2t - 3:
4Recall that we have already assumed Re (s)
r
z.
:
3s
I (s2
-
2s
(s
+
1)(s
-
3)
-
3)
.
271
5.6 THE GENERAL INVERSION PROCEDURE
Thus, the transform has simple poles at
inversion integral, we get
s : 3 and at s - -1. Using the Mellin
f (r): :2ni ['*'*
Jr-i*
-,r,
, *.tlr')'
l)(s - 3)
(s
y > 3. We close the contour with a big semicircle to the left (Figure 5.8).
Since F(s) goes uniformly to zero as R -+ oo, we can invoke Jordan's lemmas to
show that the integral along the semicircle goes to zero. The integral along each of
the small arcs to the right of the imaginary axis also goes to zero. Let the real part of
s on this piece of the contour be x, where 0 < x < y .Then
where
ltl
I I l<
J
I
arcl
llengthof path)(Maxof lintegrandl onpath)
: v3ert
' ,,,* ll(x+1
:
v3ert
, --, o
^J@
x2-2x-3\2 4@-t)2
o,V('
^u*
lr
) *-i
as R
-+
oo
Im (s)
Re (s)
FIGURE 5.8. Contour for actually performing the inversion (5.19). The contour is closed to the left,
thus enclosing all the singularities of F(s). Here the singularities are at s : -1 and
s
:
3, so we must choose
5To use the lemma, first make the change
y>
3.
of variable s
:
ico.
272
CHAPTER
5
LAPLACE TRANSFORMS
Then applying the residue theorem gives
/
\
_ 1rr"r,
f(r):r(. -e-t
_o * o )= 4.-_+ e_r),
3e3r
5.9. Invert the transform F (s) : 11 46.
This function has a branch point at the origin. We must choose the branch cut along
the negative real axis so that it does not cross the path of integration, and we must
deform our closed contour to avoid the branch cut, as shown in Figure 5.9. Applying
the Mellin inversion integral, we have
Example
f(t):
7 fY+i@ es'
I
2ri Jr-;* -ds
"/s
Im (s)
Re (s)
FIGURE 5.9. To invert the transform in Example 5.9, the contour must pass to the right of the branch
point at the origin and must exclude the entire branch cut. Here 7 > 0.
The integral around the closed contour is zero, since the integrand is analytic
everywhere inside. Thus,
(lr'-:: * Ir-*,/opo*.-"t *,* lr"*lo,,o,"oro.-.n.",)
#":
o
and so
t*['
['*'*!0,:-(\./a,,
Jr-,* ..6"" -
../ropof branch.u,
+[+[
' /a" '
-/bouo,nofbr-rt
)4r,
6*'
"ur)
The integral along Ciq goes to zero as R -+ oo, by Jordan's lemma, since 1/"6 goes
to zero uniformly as s -+ oo on Cp. The integral along the small arcs goes to zero
s.7 soME MORE
as R
-+ m
(and we can also let
y
PHYSTCS 273
--> 0 here). The integral around the small circle at
the origin is
[ lor: [_" exp(ee,_oj) '-rr"ieo,
Jc' s*" Jn JEeie/z
- r$ I,
exp(eei|t'1ei0/2de --r
0
as e --+ 0
We are left with the two integrals along the two sides of the branch cut. On the top
side,s: ret ,andso
et' ,
-/top or uranctr cu , 'rTot
f
t_
exp(-rt)
a' : - If' -----1---7-6r,
t\/r
Ja
f'exp(retftt)
:
J*-fij;''"
(-rt)
:J"f@ -ff4'
exp
and on the bottom side, s
t
./bottomofbranctrcur
:
Io,
/s
re-to
, so
: Jt[* "*'!'11":)
e-iod,: - [* "'!-!0 o'
Jt
Jrs-'n1z
-iJr
/-
:l-Ql
exp
(-rt)
,
t\/r
Je
Thus,
_tI
fY+ia ett
2ni
Jr-;* -ds
Js
:
-l
2ni
-2
I /rJo
Now change variables to u
f
(t):
h
Curiously, the function
tive constant).
:
exp
(-rt)
,
Jr
r/. Then
lr*
,-1/2"-u4r:
f (t) : t-r/2
J=r(;) :
I
fit
is its own transform (except for a multiplica-
5.7. SOME MORE PHYSICS
5.7.1. Modeling Problems
A difficulty that shows up in the mathematical solution of a problem often indicates that
we have chosen an inadequate model for the physical system. When the Laplace transform
274
cHAPTERsLAPLAoETRANSFoBMS
method is applied to the solution of a differential equation, the initial conditions must be
included in the solution. In some circumstances, the solution has a different value at t : 0,
indicating a physical inconsistency. This often happens in circuits containing inductance.
Example 5.L0. The two circuits in Figure 5.10 are coupled by mutual inductance.
The switch in the circuit on the right is closed at / : 0. Find the current in each circuit
fort > 0.
FIGURE 5.10. These circuits, coupled by mutual induction, exhibit some of the difficulties involved
with using an inadequate model for a physical system (Example 5.10). The current
variables i1 and i2 are defined in this diagram.
As usual, we begin by applying Kirchhoff's loop rule to each circuit. The current
variables i1 and i2 are defined as shown in the figure. For the circuit on the right, we
obtain
dir
V:itR+Li*r;
di't
and for the circuit on the left, we obtain
di'>
O:izR*t;**;
dit
with initial conditions that both currents are zeto at /
:
0.
Now we Laplace transform both equations:
v
-:
/rR
0:
IzR
*sLIt*sMI2
and
I
sLIz
*
sMIl
To solve the equations we use a trick. Adding the two equations, we get
V
-:XR*sLX]-sMX
.J
s.7 SOME MORE
where
X : It 112. Similarly, by subtracting and definingY :
11
PHYSTCS 275
- I2,we get
V:YR*sLY-sMY
Now we can solve:
(s.20)
t \
x:vr
s \s(L+M)+R)
1
cv
(L-fM)sIRis L+M
-J-. "-rt, where
--Then, from the primitive function theorem (equation 5.18),
-
From Table 5. r, the inverse of
r(t)
:
[' "-o' at :\(rR' =:
L*M- Jo
:
R
/ (L
* M).
r-o')
Similarly,
Y:yr
' \
s \s(t-M)+R/
y(t):f,<t-"-u'1
where B : R I (L - M). (P is positive since Z2
principles of.electromagnetic theory.) Now we can solve for the two currents:
h
lv
:
: 1@*y)
: Lr,
2R' -
I
> M2 is required by fundamental
e-o' + | - e-fr')
U<t -
e-ot
- "-ft)
while
iz: ){, - r) : {; -
:
e-ot
- (r -
v
- 2R',"-p'-e-ot\
i,-Y
.R
as expected.
and i,t-+0
e-frt),1
276
oHAPTER5LAPT-AoETRANSFoRMS
Now suppose we make this system by winding the wires for both circuits together
: M, and fl + m. We go back to equation (5.20) and
into a coil. Then we have L
notice that it has simplified:
V
J
:YRlsLY -sMY:YR
Thus,
v
v_
and, consequently,
Y(t)
:
v
R
which is a constant. Then the equation for X becomes
v
-:XR*s2LX
,s
v/
v
I
l \
_l
\2rr + R/
From Table 5.1 and the primitive function theorem (equation 5.18), we obtain
X:
vf'
t e-Ytdt
-2L Jo
: v - e-Y')
2Lv'
--(l
: Itt
e-v')
R' whercy = Rl2L.Thus,
lv
it=;(x*y):+(2-e-Y')
and
I.
iz: i@
-
y)
V
: --e-r'
Both solutions still have the correct behavioi for long times, but now neither satisfies
the corect initial conditions! At the closing of the switch, the currents have instantaV /2R and iz
neously jumped from zero to i1
-V /2R. Of course, this cannot
happen. The model fails6 because it is impossible for ^L to be exactly equal to M.
:
:
6This is, in fact, the exact solution to a somewhat different problem. Imagine combining the circuits by replacing
all the inductances with a single inductor L. The second circuit then forms a parallel combination. The solution
we have obtained would have zero cunent through the inductor at t : 0, and this is physically possible.
5.7 SOME MORE PHYSICS
277
Figure 5.11 shows how i2 behaves as M approaches Z. The absolute value of the
current increases more rapidly from zero as I - M approaches zero.
2R.
v'
t.,
-
t (s)
0
-0.2
-0.6
0.2
-0.8
FIGURE 5.11. Current i2 as a function of time for (L
s
- M)/R:
0.05, 0.1, and 0.2 s. The diagram
also shows the limit L
M (dashed line). As L --+ M, the cunent increases more
and more rapidly from zero. In the limit, the current makes an instantaneous jump to
V /2R at the moment the switch is closed. This is the correct mathematical limit, but
this solution is physically prohibited.
:
5.7.2. Nuclear Reaction Networks
A common application of the Laplace transform is to the solution of nuclear reaction problems. For example, in nuclear decay,T the number of nuclei of one species decreases, at the
same time increasing the number of nuclei of one or more additional daughter species. The
daughter nuclei may then in turn decay. For example, in a three-species chain, the decay
may be represented as
N1
-+ N2 -+
N3
The result is a set ofcoupled differential equations ofthe form
:-rrNr, ry
:
ry
dt
dt
* :xrr,
-LzNz*r1N1, dt
which we can more easily solve by converting them to a set of coupled algebraic equations
through the use of the Laplace transform:
sNz(s) sNr(s)
:
Nz(O) :
Nr(0)
-i.rNr(s)
-),zNzG)
sN:(s)-Nl(0):)'zNzG)
?See,
for e*a-ple, Lea and Burke, pp. 1 161-1 165.
* lrNr(s)
278
cHAPTERsLAPLAcETRANSFoRMS
Part of the natural radioactive sequence for the decay of 237Np involves the cv decay of2r7 At into 213Bi, which then B decays into 213Po98Eo of the time
209p6 by either an
and o decays into 209T1 2Vo of thetime. Both species then decay 1o
20eBi,
which is the stable endpoint
a decay or a p decay. The lead then B decays to
of the chain. We may write a set of equations for this reaction chain as follows.
At decays at rate ),1:
Example
5.11.
dNx :
-f
-ltN'q't
213Bi, produced by the decay of At, decays in two different ways:
dNsi
:.1"r Nr,
dt_/!IrtAt
-
o.oD,2Nsi- o.9g),sNsi
where ,tr2 is the decay rate of 213Bi into Tl and 13 is the decay rate o1 2133i into Po.
The other relations are obtained similarly:
:o'98lrNsi -
+
l+NPo
dN.t
-i:o'o2i"zNsi -lsNrt
)
dN"+.
-i
:
dNpl
_f
:
trsNr
*
)"+NPn
-'leNPr
1
I
LeN*,
(where B2 refers to 209Bi;. These relations transform to the much nicer looking set
of equations for the transforms:
sNe.t
sNsi
-
Net(0)
:
NR,
-i.tNe,t
- -0.0}'zNsi -
0.98),sNsi
t
-
0.98)":Nei
- l+Npo +
rNpo
-
NBi:
llNag
NPo:
Net(0)
sfl.r
ItNet
s*0.021.2*0.98r"3
0.981.:Nsi
s*1.+
0.02),zNet
- lsNr + Nn: s*l.s
lsNn * l+Npo
sNpr - lsNrr * aNp6 - i,6Np5 + Npt s*lo
sNn
-
0.02)'zNH
,tr
and, finally,
.rNs2
- l6tvr, +
Ns2
:
loNPu
.f
PROBLEMS 279
The transforms may be inverted by several of the methods we have discussed. For
example, if we want to find the abundance of polonium, we first solve for its transform:
0.98).gNsi 0.98).g
* l+
0.98).3
(s * r+) G *
(s
s
*
lt
*
r.+) (s
1.1
0.0D"2
+
N,q.t
0.0D"2
*
0.98)'3)
Net(0)
0.98).3) (s
*.r'r)
We may use the Mellin inversion integral to invert the transform. The integrand has
three simple poles, each giving an exponential term. The result is
e-Lqt
(0.0D,2 + 0.98).3
Npo
(r)
:
0.98,].:).r Net(0)
-
)'+) ()"r
-
I+)
,-(0'OD'z-lo'98\)t
+
Q"q
-
0.98).3) ()q
-
O.jD"z
e-Lt't
(0.0D,2 * 0.98i.3
-
).1)
0.0D"2
+ (l+
- lr)
-
-
0.98,r'g)
PROBLEMS
1. Show that
the following functions are of exponential order, and find their Laplace trans-
forms.
(t):
@ f ftl :
(a) f
sinha/
tanhat (Hint: Change variables
1o
,-Zat
: u and expand the integrand in
a series.)
r
(c) f(t): sinJat
(d) ,f (r) : s-dt2
(e) f(t): te{
(f) f (t):
sin(art
*
do)
(g) The ramp function f (t)
: {f, ifo<r<te
.Iro ift > 16
2. Using the shifting prop( rty or otherwise, find the Laplace transform of the function
\
/)
S
fo
f(r):1'
''"'-
itt <2
ltt -2)3s-at ift > 2
ni"a the Laplace transform of the function / cosh
4. Find the Laplace transform of the functio n
cvt.
f 1t 1 : + "t
280
cHAPTER5LAPLAoETRANSFoRMS
5. Find the Laplace transform of
(a) the triangle wave function with period T:
(aQ-nT)
"f(t):{
(b)
-
[a[(n + l)T
-tl
ifnT<t<nT+T/2
if nT *T12 <t < (n1-t)T
the sawtooth function:
if nT
f(t):a(t-nT)
<t <(nal)T
6. Use the Mellin inversion integral to invert the following transforms:
s
(a) F(s) : -=-----:sz
*2s t3
1
(b) F(s) :';--;---1----7'
(s2
(c) F(s)
*
at)z
: e-.6
s
- ln(l + s) (Express the answer in terms of the exponential integral
,
Elrt'): Ei(x):
fr
I
J-a
eu
dw
w
7. Invert the transform F(s)
,wherex<0.)
: +
{s'-a'
(a) by expanding in a series
(b) by using the Mellin inversion integral with a branch cut running from -a to a along
the real axis (A change of variable to z, where
'
: Z('* i)
may prove useful.)
8. Use the convolution theorem (equation 5. 17) to evaluate the inverse transform of
F(s)
9.
:
s4''t- to4'
Use the integration rule (equation 5. 14) and Table
5
.
I to derive
the result of Example
5
.9
for the transform of I /,r,/t.
EO.l fne diagram shows a simplified version of an automobile spark coil circuit. The spark
plug itself acts like an open circuit until the potential across it reaches the breakdown
voltage for air. Thus, you may ignore that branch of the circuit until the end of the
problem (part e).The battery voltage V : I2Y, C :0.1pF, R : l0 g, and L :
l0 mH. Assume that the switch (points) have been closed for a long time prior to t : 0.
(a) How long a "long time" is necessary? Write down expressions for the charge on the
capacitor and the current through the coil att :0.
,t
PROBLEMS
281
spark plug
PROBLEM
10
t : 0, the points open. Qualitatively discuss the circuit behavior. What is the
expected long-time solution for the charge and current?
(b) At
(c)
Use a Laplace transform method to solve for the potential difference across the spark
plug as a function of time.
(d) Plot your solution. What is the maximum potential difference achieved?
(e) If the breakdown voltage of air is 3 MV/m, what spark plug gap would be required
with this circuit? Remember that you would like the engine to start even if the battery
is a bit low!
11. A beam is supported at one end, as shown in the diagram. A block of mass M and length
/ is placed on the beam, as shown. Write down the known conditions at x :0. Use the
Laplace transform to solve for the beam displacement. Plot your results for.rs : Q.Sl
and,l : O.2L.
PROBLEM
11
12. Technetium is used in medical procedures as a diagnostic tool. The technetium is obtained
as the decay product of 99Mo, which decays 1o99m1" with a half-life of 66.02 h. The
technetium in turn decays with a half-life of 6.02 h. A medical radiology department
receives a source containing 100 mCi of 99Mo at 9:00 a.m. on Monday morning. Find
the amount of technetium present in the sample as a function of time after 9:00 a.m.
When is the amount of 99-Tc a maximum?
282
CHAPTER
5 LAPLACETRANSFORMS
13. An overdamped harmonic oscillator satisfies the equation
*dtz*zo!dt +of,x: f (t)
where cu2 > a2o andthedriving force is a square wave of period f . Find the displacement
x(t) if the initial conditions are x(0) : dxldtlt=o: 0. Plot the result for a :2ao,
q,T : l, and0 < t < 3T12.(Hint:Use theresultof Example5.6andexpandthe
transform in a series.)
14.
A harmonic oscillator with resonant frequency as and no damping is driven by a
sinusoidal force F(t) - Fs sin a,lf . If the initial conditions are x(0) : O, dx /dt : O
at t : O, use the Laplace transform to find x(t). What happens 1f at : a>o?
(b) Solve the same problem with the initial conditions x(0) : 0, dx I dt : a at t : O.
@
15. The two circuits in diagrams (a) and (b) show how we might use a capacitor to prevent
sparks across a switch when the switch is opened. Assume that the switch has been
closed for a long time and is opened at t : 0. For each circuit, use Kirchhoff's rules to
solve for the current through the inductor and the charge on the capacitor as a function
of time after the switch is opened. Discuss the merits of each of the circuit designs.
PROBLEM
15a
PROBLEM
15b
PROBLEMS
283
switch has been in position A in the circuit shown in the diagram for a long time.
What are the charge on the capacitor and the current I through the inductor? At time
r = 0, the switch is moved to position B. What are the charge on the capacitor and the
current a long time later? Find the charge on the capacitor as a function of time for t > 0.
Give your answer in terms of a.b, where a;B : l/LC,a : RlL, ar.rd B :l/RC. You
You may assume
may also find it useful to define y : (a + p)/2and O : Jr"pthat I is real.
6. The
I
i
I
i
17. The switch in the circuit shown in the diagram has been closed for a long time, and
a constant current flows. What is the charge on the capacitor? At time t : 0, the switch
is opened. What are the charge on the capacitor and the current through the inductor
a long time later? Find the cunent through the inductor as a function of time for t > 0.
Give your answer in terms of a4 and o, where eE : I / LC and cv : R/2L.
PROBLEM
[&'l
17
The switch in the circuit shown in the diagram has been open for a long time. At t : 0, the
switch is closed. Find the current through the inductor and the charge on the capacitor
284
CHAPTER
5 ilAPLACE TRANSFORMS
as functions of time for t > 0. Give your answer in terms of ag,u, and p, where
a;B : l/LC,a : R/2L, and P : oft1+" : ll2RC.You may also find it useful to
defineo:JZ;A-a2-fl2.
PROBLEM
18
L9. In the figure shown, capacitor Cr has charge Q and capacitor C2 is uncharged. At
t : 0, the switch is closed. The two capacitances are equal. Find the voltage across each
capacitor as a function of time for t > 0.
PROBLEM
19
20. The switch in the circuit shown in the diagram has been closed for a long time and is
opened at
t:
0.
What are the currents in the circuit for
t < 0? Use the Laplace transform
C
fa
fD
t2l
PROBLEM
20
PROBLEMS
to find the currents in the circuit as a function of time for
the initial conditions?
285
t > 0. Does your answer satisfy
Rework the problem, leaving the initial value of the current iz in arm AB as an
unknown to be found. Find the solution for the current i1 through R1 and require that
it satisfy ir(0) : E/(h + Rz/2). What value of iz(0) is required? Give a physical
explanation of this result. If R1 : Rz : R and L1 : L2 :.L, plot both solutions. Plot
current in units of E/R versus time in units of L / R. How long is it before both solutions
give the same result to within 17o?
21. The radioactive series that begins with neptunium 93 contains the following decays:
Decay
237ryn
233pu
233pu
2339 -2339
--r 22976
2297A _- 22spu
22s16
*
22s4"
225;" --r 22tp,
22tp, --r 217 61
217
S1 _- 2t3gi
2t3gi --r 2r3Po 198vo1
2r3gi -* 20eTl (ZEo)
i
I
213po
20911
209p6
Half-life
Type
_-
2.14
p
27.0d
x
1.6
d
7340y
p
14.8 d
d,
10.0d
q,
4.8 min
a
0.032 s
47 min
p
4.2
20ep6
20e91
p
3.3 h
209p6
_*
_*
706
y
105y
d,
a
a
p
--
x
a
trt"s
2.2min
If
we regard any decay that takes less than one year to be essentially instantaneous,
then the chain simplifies to
trtNp _- 233g
i
__s 22eTh
_> 2oeBi
Write a series of differential equations that describes this decay chain. Apply the Laplace
transformto findthe fraction of the original 237Np thatis in the form of uranium, thorium,
and bismuth after 10s and 106 years. Make a plot showing the amounts of each element
as a function of time.
22. Find the Laplace transform of the function g(r) : t df ldt. Express the result in terms of
the transform F(s) of the function /(l). Use the result to solve the differential equation
tY'1-Y:s-t
23. Apply the Laplace transform to the differential equation
y"-t2y:t2
Does the Laplace transform offer any advantages in solving this equation? Using any
method of your choice, solve the original equation or the transformed equation subject
286
CHAPTER
5 LAPLACETRANSFORMS
to the initial conditions y(0) : 1 and y/(0)
may be expressed as a power series in 1/s.)
:
0, and comment. (Hint: The transform
@f,fak.e the Laplace transform of the Bessel equation of order zero,
..
y"+-y'*):0
1
and show that
(s2+t)r'(s)*sY(s):Q
Solve for Y(s) and hence find an integral expression for
for more information on Jo(x).]
y(.r)
:
.Io(x). [See Chapter
8
CHAPTER 6
Generahzed Functions in Physics
6.1. THE DELTA FUNCTION
The parlicle is an often used and very valuable physical model. We can describe an electron
as a particle because its size, as far as we know, is smaller than anything else we might care
about. Sometimes we describe something as large as an automobile as a particle because
its size and shape are not important to the question we are interested in. When something is
modeled as a particle, all ofits physical properties, such as mass and charge, are concentrated
at a single point. In a similar fashion, an impulsive force is modeled as a force that is
concentrated at a point in time. To describe the density of the particle or the force
a function of time, we need a function that has the following properties:
.
.
.
as
Its value is zero everywhere except where its argument is zero.
Its value is infinite where the argument is zero.
The integral ofthe function is 1.
This mathematical object was first used in physics by P. A. M. Dirac and is called the
delta function 6(x). Using this function, we write an impulsive force as
F(t) :1511;
The impulse delivered is
r*oo
r+oo
7: J-I ielat:ilJ-*
3(t) dt
: I
To write the density of a particle at the origin, we need a three-dimensional delta function:
p(i)
:
M
3(i)
:
M
3(x)6(y)3(z)
(6.1)
The particle's mass is
M
: JallI
p1)dv
space
r*oo /+oo /+oo
: M
I I
"^ I
J_* J_* J_*
6(x)d(y)d(z)
d.x d.y
dz:
M
287
288
CHAPTER
6 GENERALIZED FUNCTIONS
IN PHYSICS
In both these examples, we can see that the physical dimensionality of the delta function
is one over the dimension of its argument:
I**
uurdt
: t+ t6(r)r : # : TI
and, similarly,
1 1 1:
td(i)l:td(x)6(Y)6(z)l:
- Ixl lyl lzl -
1=
L3
The properties we have listed for the delta function do not describe any proper mathematical function. Thus, we must extend our ideas about functions to include these objects, which
are known as distributions or generalizedfunctions. To begin our study of distributions, we
shall study the delta function in one dimension in more detail.
5.1.1. Delta Sequences
The delta function is defined as a mathematical object that possesses a property known as
the sifiing property:
ll-rr.rur,)dx:
(6.2)
r(o)
That is, if we multiply a function f (x) by the delta function and integrate over the whole
real line, we sift out the value of the function at the origin. Let's see how the sifting property
follows from the properties listed above:
l*
rortr*)
d.x
:
f' ,f{*)o{*) o* because d(x) is zero except at x
:
f (xo)
l" ,u{r)d",
: f (xd [*
J-a
uG)
o*
where
-
e < xo <
:0
e
because 6(.r) is zero except at
= f (xd because the integral of the delta function
+"f(0) ase-+0
is
x
:
0
1
In the second step, we used the mean value theoreml for integrals.
Delta functions arise from modeling a physical quantity that is distributed over a small but
finite region as being concentrated at a point. Thus, we should expect that the mathematical
I
See Appendix
IV for
a
proof of this result. The application here can be made rigorous by prop€r use of limiting
procedures, as we show below.
6.1 THE DELTA FUNCTION
289
quantity d (x) would arise as the limit of a set of functions that become increasingly concenffated at a single point. We can construct a sequence of proper functions d, (-r) labeled by
apara;rrrctetn suchthatasr, -+ oothesequenceQ"@) approachesthedeltafunction.What
we mean by "Qn approaches the delta function" is that as n --> @ we obtain the sifting
property; that is,
lim [** Q*@)f (x)dx
n-co J_e
: f (o)
/(x). Such a sequence of functions is called a delta sequerce.
The simplest delta sequence is a sequence ofrectangular blocks with unit area:
for any continuous function
o'(x) :
("/2 it-l/n <x <lf
otherwise
[o
n
(6.3)
As n gets larger, the blocks get taller and narrower (Figure 6.1), and so @, "looks like" the
delta function as n -+ oo. Next we must test for the sifting property. We multiply QrQ) by
a function /(x) that is continuous at the origin and integrate:
f_l o.av@)dx:;
I:,,:,' r(x)dx
n2
::'r?lrol
1
,,r'
",,
::'
*t
I
=
6"(x)
-l
FIGURE 6.1. The "block" delta sequence
shown here.
0
-0.5
@"
0.5
I
(-r) given by equation (6.3). The first five firnctions are
290
cHAprER 6 GENERALTzED FUNoIoNS
rN
pHystcs
(We used the mean value theorem to evaluate the integral.) Since we obtain the sifting
property, S"@) is a delta sequence.
We can use this sequence to determine the sifting property of 6(x - a). We look at the
integral:
+oo
I O,U
f
J-*
*oo
: If
J-a
- a)f(x)dx
O,t"l;fu *a)du
1
I
<
: n2
where-;-.f(uola),
zn
n =uo +-n
--> f (a) asr? -+ oo
3(x - a) sifts out the value of the function at x : a.
Similarly, we can determine how to express 6(ax). First let a > 0. Then
So we conclude that
l+@
I
J-*
Qn@x).f
f+oo
rut, du
(I
4,tr)f
'"
\a/)-a
J-a
(x)dx :
I
-f (0)
-->
where we used the change of variables u
gives
l-l
o,r-rrrx)dx
:
ax . But
ast4
-+
oo
if a < 0, the
same change of variables
: l** o,rt
r*oo
: - J_*
0,,
>
1
--"f(0)
a
We can express both results as
8(ax):
I
(6.4)
lo16(*)
Sometimes we need a sequence of functions that are better behaved than the block functions @n (x). For example, to investigate the derivative of the delta function, we need a set
of functions that are each differentiable. The functions
Qn$):
I
n.T
sin2
nx
.
(6.5)
6.1 THE DELTA FUNCTION
291
infinitely differentiable and also get peakier around the origin as n increases (Figure 6.2).
(The first zero of this function occrus zt xg - tr f n, and .16 approaches zero as n --+ @.
The value of the function at the origin is nf n and increases linearly with n.) Again these
functions "look like" the delta function as n -+ oo, but we'll need to test for the sifting
property.
are
Q"@)
,
,
\,
n:3
t
I
l
I
t.
t.'
t..
t
I
t
t
.'l
.'l
:l
I
i
i^
Y.
l'.
l' .-
.//
/n:
Z
-5-4-3-2-1012345
FIGURE 6.2. The first three functions in the delta sequence @4(x) given by equation (6.5).
The derivative is
I ( 2nsin nx cos nx
dQ,
E:
n'T
:
This clearly exists for aIl x
sine and cosine for small x:
2
\-----7-
"inn: @x cosnx-
nr xJ
.
\
-'--#
)
sin2 nx
sin n.r)
10. It also exists for.r : 0, as we can see by expanding
/
'2'2\
il,:o:lTo,o,'1"*(t- ,
2nx
4!tl
I /
/-('"-
the
rz3x3\l
o1
:ri^(-?r"\:o
;+0\ 3 o/
This result (dQ, ldx
slope zero at x :0,
:
0 at
x
:
0) agrees with the graphs shown in Figure 6.2, which have
292
CHAPTER
6 GENERALIZED FUNCTIONS
IN PHYSICS
Now let's check for the sifting property. We write the sine in terms of complex exponentials:
:l::#(
"2inx
_2a"
-4
-2inx
),o,
o.
- ('^' ) /t', r,
: 4nn
.L J-*
[** ('
\ r' /"'
I /.+oo 11-"-2inx1
(x)dx
* 0,,
J_* (---;r-/ f
Now we evaluate each integral along the real axis in the complex plane and use the residue
theorem. Suppose the function (z) is analytic except for a set of isolated simple poles and
"f
approaches zero uniformly at infinity. Then the poles of the integrands are the poles of /(z)
and
of
9n,+(z):
There is
1
_
"L2inz
z2
a pole of gn,1(z) at the origin. Let's look at the Laurent series:
1
_
"I2inz
)
Bn.+k)
: +2i! +2n2 + "'
z
This pole is simple, and the residue of g",+(z) there is a2in (Chapter 2, Section 2.6.3,
method 1). We evaluate the first integral by closing the contour with a big semicircle in
the upper half-plane. The pole is on the real axis, so we shall evaluate the integral by moving the contour slightly downward so that the path of integration lies below the real axis2
(Figure 6.3). Then, if the poles of are simple, the integral3 is
/
2ni I
Ir: 4nn
l{-zin)t'lo)+
:
,f (0)
!Res/1 ,ols,,*frr>f
t _
1_^2inz.^
+ .f(0) as n -+ oo
. +)rtesf (z)::;:
p
2The result for the complete integral is independent ofthe method we choose to avoid the pole, provided that
/(z)
has no poles on the real axis, because the singularity of (sin2 nx) l xz is removable.
3There may not be a pole of the integrand at the origin if
/(3) is zero at the origin. But then the integral is zero in
limit and so still equals /(0). If the poles of / are not simple, the expression for the residue of the integrand
zp is more complicated, but the result 11 -+ /(0) as n --+ oo still holds. Note that zp has a positive imaginary
part, since the relevant poles are in the upper half-plane, and so the exponential approaches 0 as n --+ oo faster
tban any power of n. See Problem 30.
the
at
6.1 THE DELTA FUNCTION
293
Im (z)
Re (z)
FIGURE 6.3. To demonstrate the sifting property of sequence (6.5), we evaluate the integral using
a path that passes
below the real axis, under the pole at z
:
For the second integral, we have to close downward. The pole at z
this contour, and so the integral is zero in the limit. Thus,
/+oo I
sin2
0.
:
O
is excluded from
nx
,qL/-- ,"--;z-f Q)dx: f (o)
The sequence
/,
possesses the sifting property as n
-+
oo.
6.1.2. The Derivative of the Delta Function
Now we are ready to determine the sifting property of the derivative of the delta function.
Let's use the delta sequence4 (6.5) to evaluate the integral:
lll r;o,r,.,0.
We can do the integral by parts:
f_l
o"ort(x)dx:
Q,@)r(x)
_: -
/_
Qn@)rt(x)dx
4since the explicit form of
Qn@) is not needed, any differentiable delta sequence
will suffice.
294
cHAprER 6 GENERALTZED FUNCIoNS tN pHystcs
The integrated term is zero, provided that
take the limit as n --> @, we find
f (x) remains
bounded as .r
-+ *oo. Now if we
f+@
/+oo
f +oo
lim , q"G)f'7x7dx
t'tx)f tx)dx: n-6
lim I O"@)f- (x)dx: - n-@
I
J_oo
J _a
J_Then, using the sifting property of @n, we obtain
ll*
a'a>r@)dx: -f'(o)
(6.6)
which is the sifting property for the derivative of the delta function.
We can repeat the steps above to evaluate the sifting property for the nth derivative of the
delta function. The result is
.
I
ll*
l
u,' t", f (x) dx: 1- t;n Y(') 1s;
(6.7)
l
l
I
l
I
moment p : p*. Express the charge
density in terms of delta functions. Given that the electric potential is
Example
6.1. An ideal dipole has dipole
I t p(7') .a
o(x):4"^l
l*.-*tra"''
t
find the electric potential due to a dipole placed at the origin.
We begin by modeling the dipole as two point charges q ffid -q, separated by
a distance .( along the x-axis. Each point charge has a charge density represented by
a delta
function (see equation 6.1):
p(i) : -qd(x)s(y)s
:
Now we let
[
p(i)
-->
:
0
and
(z)
|
-q6(y)6(z)
*
q6(x
6(r)_
lvtl
-
q -+ oo such that the product
't-oLf
-ps(y)s(z)lim
6(x)
t)6(y)3(z)
j(r_
- {(r - l)l
(
)
21
ql :
:
p remains finite.s Then
-oa1r;a k)6'(x)
5Since the delta function is not a proper function, we should demonstrate that the
derivative. The result rnay be proved using a delta sequence; see Problem 23.
limit below behaves as the
I
I
6.1 THE DELTA FUNCTION
295
We evaluate the potential using the sifting property of the delta functions:
o(i):#',lffio,*'
p
/+oo /+oo /+oo:
4nes J -oa J -* J -* Je
reoJ--J--J--rM +oo
+oo
I-* I-*
4neo
J
J
6t(xt)3(yt)3(zt)
x')2 + (y - y')2 -l
8,(xt)a(y,)
l-G- lY + () - y,)2 + z2
P [**-Lo*'
-- 4reo
-a 1/1x - x')2 * y2 +
J
k-
4xt dy' dz'
z.'12
-
dx'dy'
z2
This integral is evaluated using the siftUg_ptgPg4y_glhe derivative of the delta
function (result 6.6) with /(x') : t11/1*- ,F+ y2T7t
.f)
o(i)
f' {x').,_01 : 4neo
l1a
: -f-F
+T
eO
px
4t
where 12
:
x2
*
r3
e11
y2
+
22.
6.1.3. The Delta Function of a Function
Sometimes the argument of the delta function is itself a function. We can use any of our
delta sequences to determine the sifting property of 6[S(x)]:
r*oo
akt")lf(x)dx
/__
lim [** o,rr@)lf @)dx
==nrooJ_oo
(6.8)
We make a change of variable to u : g(x). Then du : g'(x) dx. Next we divide the
range of integration into N pieces, where g/(x) I 0 within each piece (Figure 6.4). Thus,
g/(x) is either positive or negative within each piece and is zero only on the boundaries
at x : x;. Notice also that there is at most one zero of g(x) in each piece. We label the
values of r where g(x) : 0 as xs;, i : I to N' Then the integral on the right-hand side of
equation (6.8) is
N
I
fxi+1
D
Ixi
i:l
J
Q,ts@)tf(x)dx =
,
du
)- J8$,)
/Tskiattentu\f1r-trr)lih
-ro
Let's look at one of the integrals in the sum. If gt(x) is positive throughout the ranEe xi to
.r;11 then g(xi) is less than g@i+). As we take the limit n --+ @, Qn approaches the delta
296
CHAPTER
6 GENERALIZED FUNCTIONS
IN PHYSICS
g(x), g'(x)
FIGURE 6.4. A function g(x) and its derivative. The function is zero at the points x6;; the derivative
is zero at x2 and x3. Thus, g/(.r) is positive between -a and x2 and between x3 and
oo, and is negative between x2 and x3.
function and thus goes to zero outside any finite range containing u :0. We may expand
the range of integration for r.r without changing the value of the integral6:
p|Gi+r)
,J'!L Jr",,,,
Q,@)
r+oo
,!L J_* Q,{"t f 1g-t (u\ffi\l
f (xoi)
: ,rts-l(o)l g\*r)
,
du
f [e-' tuilV# tun:
du
t1t--:
where the particular zero xoi of g lies between x; and x;a1. If there is no zero of g in this
range, then the integral is zero.
Now if g/ (x) is negative throughout the range xi to x;a1, then g (xi ) is greater than g (x;+r )
and we get
^\
t*Gi+r)
du
-im [-*
'| - du
al7#@i
,,,
Qn@)f ls-t(D1{#@j:,1;i;
J**
Jr,,,,
^-rurfrp-t(r
,f
ts-l(o)l
:
8'rst{dr
6St i"tly, we must also redefine the function
to u
: g(xi+t).
fl7-l
f (xoi)
k(rCI)
@)l to be zero (or any constant) outside the
ran ge
u
:
g(x;)
297
6.1 THE DELTA FUNCTIOI.I
Adding the results for each of the pieces gives
ffurrr.' r@)dx:I#ffi
where the sum is over all the zeros of the function
also write the result in the form
(6.e)
:
g(x): 8(xo;)
0,
i-
6rg(x)l:If#
1
to N. We may
(6.10)
Equation (6.9) (or 6.10) has numerous important applications in physics. It is valid only
if the sum is over a finite number N of zeros and there are no repeated roots. [If there is
a repeated root, then g'@oi) : 0 for those roots, and our demonstration fails.]
As a specific example, consider the function g(x) : xz - 4. There are two roots at
x : !2, and g/(x) : 2x is zero at x : 0. Thus, we divide the range of integration as
follows:
f0
/+oo .
- .
te' -Dfe)dx:
IJ-a
IJ-a 6G2-Df@)dx+ JoIl+* 3@2-4)f(x)dx
Now we change variables to u
and so
x
:
x2
- -4Q 14.f!rus,
-
ll* ur., -gr(x)dx
:
I.:s@)
rGJu
+
4, du
:
2x dx .
o+=
In the first integral, x is negative,
a)#
*
6(D r du +
l*o*
by defining f eJn +4 : 0 for < -4.
Next we extend the range of integration
precise value chosen is unimport nt, since 6(u) is zero for u
I
0. Thus,
r+oo
I
J-*
6(x'
- 4) f (x) dx
f+oo
- t stu)f GJu + +)
-l_*o\w)J\-\4I1t
Then, using the sifting
property,
f
+I
+@
J_o
du
q\---:
t<utf <J,' +t-trlQtr4
e have
[**ou'-ofu)a*:ff+
which is in agreement with the g neral result (6.9).
'
The
298
CHAPTER
5 GENERALIZED FUNCTIONS
IN PHYSICS
6.2. A sheet of charge lies in the z : 0 plane. The surface charge density
is os. Express the volume charge density in spherical coordinates.
The charge is localized to one value of zi Z :0. Thus, we may express p with
a delta function in z:
Example
p(i) :
k6(z)
To determine k, we integrate over a cylindrical volume of cross-sectional area dA
extending from z : -oo to z : *oo (Figure 6.5). The cylinder acts like a cookie
cutter, cutting out an amount of charge og dA from the sheet. The amount of charge
inside the cylinder may also be computed from the charge density:
r+@
pi)dv: I
dq: If
Jcylinder
J
k6Q)dzdA:kdA
-a
andthusk:oo:
p(*)
:
oo6(z)
Now we change to spherical coordinates: z
p(i) :
:
r cos 0, and
so
o66(r cos0)
FIGURE 6.5. Cutting out an area dA from the charged sheet (Example 6.2).The charge inside the
cylinder is o6 dA.
To interpret the delta function, we first note that charge exists at all values of r,
0 < r < oo, so we cannot have a delta function in r. But charge exists at only one
value of 0, 0 : r /2. Thus, we must have a delta function in 0. Then we apply result
(6.4) with e: rl
p(x)
:
6
(cos d)
r
6.1 THE DELTA FUNCTION
299
Now we may use result (6.10) to evaluate 6(cos 0). There is only one zero of cos 0 in
the physical range of interest, 0 < 0 < n,and that is at 0 : n f2.Thus,
:t 12) :
: oor ,5(o
l- srnub:r/2
p(7)
!9xe - r /2)
r
We can check the result by calculating the amount of charge inside a spherical shell
of thickness dr (Figure 6.6). This shell intersects the sheet in a circle of thickness dr,
and so the charge inside is dq : oo dA - os2nr dr.lntegrating the volume density
and using the sifting property to evaluate the integral over the delta function, we get
on
:
I,n"upg)dv
:
Io'"
fi
:2roor sin;dr
z
tr
l:
"|urt - r/2)r2
sinl d0 dr
og2nr dr
as required.
FIGURE 6.6. A thin spherical shell ofradius
of inner radius
r
and thickness
r and thickness dr intersects
the plane in an annulus
dr.
6.1.4. The Integral of the Delta Function
of a delta function, let's
x < -l I n. Then
To interpret the integral
(equation 6.3). First let
[.
J-a
o,@) du
integrate our block delta sequence
:o
since @, (a) is zero throughout the range of integration. Now let
x>
nltl" :
fx
ftlnn
Q"@)du: I
-ul
-du
I
2 l;tn
Jt1n2
J-a'"
so
f'
J-*Q^fu)0":
(o ifx<-1/n
\L
if x >
*r/n
I
I
f n. Then
300
cHAprER 6 GENERALTZED FUNcroNs
rN
pHysrcs
As we let n --> @, we have
:
l'*o''
{0,
o'
ifx <0 :
ifr>0
(D(x)
(6.1
l)
where @(-r) is the step distribution.T It has the property that
/*J-a
",r,,r,
x) dx
: [**
Jo
f (r) o,
Since @(x) is the integral of d(.r), we may reasonably conclude that
The delta function is the derivative of the step distribution:
d@
_ :6(x)
(6.12)
dx
We can check this conclusion by determiningthat d@
/dx
possesses the sifting property:
f** 499\
.vf (*)4v: f (x)@t.r)l1S - J-*
/
ax
J-a
'
/^+oo
:,Tlo fG)-o-
f'(x1@@7dx
r*m
J,
f'(x)dx
:,l*f(x)-/(x)13
:,l* f @) -,$"rt')
: (o)
+ /(o)
"f
as required.
6.1.5. The Fourier Series of a Delta Function
We may find the Fourier series of a delta function by finding the Fourier series of a delta
sequence and taking the limit as n --> oo. Since the Fourier series is periodic, we will
actually get a periodic repetition of the delta function. The easiest sequence to handle is
the sequence ofblocks (6.3). The Fourier series ofthis delta sequence Qn@) on the range
-L<x<Lmaybewritten
*oo
Qn@):
^D*r-r'*"''t
TPreviously we used the notation S(-r) for the step function. The notation @(.r) distinguishes the distribution from
the function.
6.1 THE DELTA
FUNCTION 301
where (equation 4.31)
r rL
c^: )= I O,@)r-'mnx/L 4,
2L
J-L
L f""
2L
n
J-ttn
:,
r-imnx/L 4*
(e-imn/nL-,imn/nL\:
4mr \
lnL +
Now as n --> @, mn
first term, we find
2
,in^o
nL
'
2mr
/
-i
0. Expanding the sine in a Taylor series and keeping only the
c*
nrnfi1
- zmn nL: i
Therefore, the Fourier series of the delta function is
. +oo
6(x): a |Lt
2L
(6.13)
"imnxlL
which is a very odd-looking series! Since the coefficients do not decrease as m increases,
the series does not converge in the usual sense (pointwise or in the mean). That should not
surprise us unduly, since 3(x) is not a proper limit. However, we can check for the sifting
property:
+oo t
*oo
r*L ,
I
'
J-, 2L I
,imnx/L y1x1
^L_*
r*L
dx: \- ' I
l-r
^?*2L
a-^
The integral on the right gives the coefficient
"nimnx/L f e)dx
in the Fourier series
*oo
f (x): I
p:-6
for the function
/(x)
[,' +
to
*L.
on the range
-Z
f
f(x)dx
m:-@
",-o","
orrtoo'/t
Then
:f
m:-@
o-^
:f
m:-@
"*:,f(o)
and so the series (6.13) possesses the sifting property. Thus, equation (6.13) represents the
delta function on the range (-L, L).
We should perhaps worry about changing the order of the sum and the integral, since the
series is not uniformly convergent. In fact, it does not converge at all! The validity of this
procedure will be discussed in Section 6.3.
Fourier sine and cosine series may be found sirnilarly.
302
CHAPTER
Example
x:
L
6
6.3.
GENEFIALIZED FUNCTIONS IN PHYSICS
An initially stationary string of length L is hit by a hammer blow at
Determine the subsequent motion of the string. (The speed of waves on the
:
1t, where Z is the tension and pc is the mass per unit length. See
^1tlJ
4,
also Chapter
Section 4,4.2.)
We may model the hammer blow as an impulse 10 occurring at t : O and at x :
L/3.The impulse per unit length is
13.
string is y
{4!-ts6(x-Lt3t
dx
Using the impulse-momentum theorem,8 we can determine the initial velocity of the
string at / : 0+. The change in momentum of a string element of length dx and mass
dm
-
ptdx is
0*)
d I,(x) .
Uclx: * clx
*
Oy(x,0*) 16 .;-_:6(x_Ll3)
0y(x,
which, together with y(x, 0) : 0, gives the initial conditions for the string. We expect
a solution of the form (equation 4.29)
v(x. 1)
: .!)
d.
srn
nrx
nrut
frLL
which satisfies the initial condition of zero string displacement at r
Ey
0t
@
/nzx\ nTu
: s-)
L'' d.srnl-l\L/
L
n:l
ovl
nnD
'l
:) .a
orl,:o fr,,L
:
0. Then
nTrut
L
wrx
L
Applying the second initial condition, we have
bu,*
p
- L/3\:?L"'Lo-n" ,inno'
L
Now we find the coefficients in the usual way:
2Ig nr
Io ^.
wrx
nftu 2 fL -:3(x-Ll3)sin
dx: jsrn---a,-.:=
I
.
" L
p
L
L1t 3
LJo
(6.14)
where we used the sifting property. As we expect for a delta function series, the term
on the right-hand side does not decrease with increasing n. (Compare equation 6.14
8see Lea and Burke, p. 200.
303
6.1 THE DELTA FUNCTION
with equation 6.13.) However, the coefficients a, that describe the string displacement
do decrease:
,
a,:
L
2
sin)
nnu Lpt" 3 ?'o
nnu
Io
p,srn)3
and the solution for the string displacement is
t): i'
b rirT
L "inY!
L
3 "in\!
*t':rnnu\
Notice that all the terms with n : 3m and m an integer are zero, since we have
ajm q. sinmr : 0 and so every third harmonic is missing. The first few terms are
v(x,
nut I
y: Jitn / nx
o'\ttn7srn-+tsrn
4zut
2trx 2trut I 4nxsrn-+"')
srn
L
4ttn L
t
\
Figure 6.7 shows the sum of the series for the string displacement, up to the n : lO
term, at times utf L: l18,I14,and 1/2. Notice the peak atx : L13 atthe earliest
time.
apy
111
\
\
\
\
\1
\a
t
t
I
\
x
L
FIGURE 6.7. The string displacement (Example 6.3) at times ut/L : ll8, l/4, and ll2.The
vertical axis shows the dimensionless quantity ulty/10. The dimensionless variable
x / L is plotted on the horizontal axis. The hammer blow at x : L /3 causes the peak
in the displacement at L 13 at the earliest time. The displacement spreads along the
string.
6.1.6. lntegral Representations of the Delta Function
We can
find an integral representation of the delta function by finding its Laplace transform.
for / > 0. We can use a set of blocks
To use this transform, we need a delta sequence defined
304
cHAprER 6 GENERALTzED FUNoIoNS tN pHystcs
defined as follows:
(o if r <o
t0^(t):\n if0<t<lln
(0
(6.1s)
lln
if t >
This sequence offunctions has essentially the same behavior as our previous block sequence
(6.3), and the proof that it is a delta sequence is almost identical. The Laplace transform is
op(s)
:
fo*
o,r-"0, : Iot'^ ne-'tdt
:,
::r,
e-'/n)
=lii"
Now we take the limit as n --> oo:
Ll6(t)l:
-
i {'-
.}(:)'. ]}
l'-;
t-t+...-+1
asn-+oo
Thus, the Laplace transform ofthe delta function is 1.
We could also obtain this result from the relatione 6(/)
Laplace transform ofthe step function is l/s. Then
LI6(t)l:
s.C[@(/)]
-
@(0)
:
:
d@
s/s
: I
ldt
and the fact that the
If we invert the transform using the Mellin inversion integral (equation 5.19), we obtain
I
r+ioo
6(1): ' I
2ri J-i*
ettds
where we may integrate along the imaginary axis since the integrand has no poles.
Now make the change of variable s : iro. We find
r
.+m
I /+oo
,i''
ei''ida: 2"
J-J_-
' I
d(r): 2ni
d,
(6.16)
This expression is extremely important and finds many applications in physics. Notice the
similarity to equation (6.13). The integral is the extension ofthe Fourier series representation
to the case of a continuous, rather than a discrete, frequency spectrum.l0
9This demonstration appears less satisfactory because we have to invoke a value of the step distribution ar the
origin. Using the theory developed in Section 6.2, we can show that the initial condition may be dropped. See
Problem 24.
l0Demonstration ofthe sifting property may be accomplished in several ways. For example, expression (6.16)
may be written as limn;*(1/22) /]S exp (- ,2 14n2 + itot) da;. The integral is easily evaluated (Chapter 7,
Example '/ .2), and the result is the delta sequence (6.18)
305
6.2 DEVELOPING ATHEORY OF DISTRIBUTIONS
6.2. DEVELOPING ATHEORY OF DISTRIBUTIONS
The delta function (together with its derivatives) is a very useful mathematical object that
often simplifies calculations in physical problems. Because it is not a properly defined
function, we used sequences of proper functions-the delta sequences-to understand the
properties of the delta function. The delta function is mathematically defined by the sifting
property (6.2).ln this section, we shall begin to formalize and generalize these ideas.
To develop a more rigorous theory of distributions,l 1 we start with the definitions of two
classes of functions: corefunctions andtestfunctions. The properties of these functions determine the paticular distribution theory. The example below is illustrative of the techniques
but should not be regarded as a description ofthe only class ofdistributions that exists.
Test functions. For our illustrative theory we choose the test functions /(x) to be infinitely differentiable and to go to zero very fast as -{ -+ m. Taking ,f : 0 outside some
finite range will certainly be fast enough. We'll explore this feature further as we develop
the theory.
Core functions. For our theory we choose the core functions g(.r) to be infinitely differentiable.
We'll also need to define
a new
form of convergence, called weak convergence.
Weak convergence. Let g"(x) be a sequence of core functions. Then the sequence is
weakly convergent
if
lim I^+oo sn(x) f (x) dx
n+a J-*
exists for any test function
/(x).
The definition of a distribution as the weak limit of a sequence of core functions now
follows.
.
Definition of a distribution.
(x) is a distribution if there exists a sequence of core functions g, (x) that converges
weakly to @(.r), in which case
@
,,g/:
for any test function
gn@)f(x)r,:
l::
Q@)f
(x)dx
(6.17)
/(x).
llsee Lighthill's book for more on this topic. The test functions here are akin to his "good" functions and the
core functions to his "fairly good" functions.
306
cHAprER 6 GENERALTzED FUNcrloNS lN PHYslcs
Many different sequences of core functions may converge to the same distribution.
The sequence of core functions
gnT):
n
\/ lt
,-n"'
(6.18)
illustrates some of these concepts. The sequence does not converge pointwise, since
gr(x) --> 0 as n -+ oo for x I 0, but gr(x) diverges as t? -+ oo for x : 0. However, the
sequence does converge weakly. To see which distribution
it converges to, we evaluate the
integral (6.17):
lim
le-""y1r1d*
1+*
n*@ J_oo
\/iT
:.uu (l_:" .
l:,':: . l:,;,) ft,-""f
:
-f Iz
nBL Qt
*
{,)o*
Iz)
/(x) is infinitely differentiable and goes to zero as t -+ too.
Let l/(.{)l < Ml for 0 < x < oo. Then
2=r-"
2?r-n"'d*
:
Now the test function
/(x)
is bounded.
,,
where
* f*
a11uQ
= 2
o(.fr)
J
;1
: +. [:
Ji
tt
,t
\/1T
is the error function,l2 which approaches
<D(.r):
I
as
du
- aQ-n)l
*rt
2
n -+
oo.
Thus,
+
fr r-''dt:t-+':!x +"'
,/tr JO
41T
[The asymptotic expression for O(x) may be found in Appendix IX, equation (29).] Thus,
13 -+ 0 as r? -+ oo. Similarly, we can show that 11 -+ 0. However, for 12, we have
rJi
: fc)ii't Jo
e-,'du
by the meanvalue theorem, where
-ll^fr
=
€<
+llJn andu:
nx. Thus'
12: f G)aQfr)
and so
(*)dx -- rim 12: f e): [**
lim [**
+r-,'*'.f
n+@
J
J _*
Jn
n-cr,
l2The error function is also called erf (-r). See Appendix IX.
-*
Q@)f (x)dx
6.3 PROPERTIES OF DISTRIBUTIONS
307
The distribution Q@), defined as the weak limit of sequence (6.18), exhibits the sifting
property, and so it is the delta function. We may write
ft"-n"'
-+ 6('t)
weaklY
(6.1e)
Some sequences of core functions may have pointwise limits that are ordinary functions
and may also have weak limits that are distributions. It is even possible that gr(x) -+ g(x)
pointwise as n -->
q,
gn(x)
l**
--> Q@) weakly, but
oarr@)dx
t ll) rr,rrr,ro.
This kind of bizarre sequence is rarely of any importance in physics.
In physics, pointwise convergence is not always important, since a physical quantity
cannot be measured at a point, but only over some small but finite region. Thus, weak
convergence is often all we need.
There is a close relationship between functions and distributions.
The Smudging Theorem. For every continuous function h(x), we can construct a sequence gn(x) of core functions such that
,$
ls,(")
-
h(x)l <
e
for any e > 0 and for all x in any finite interval.r3
According to this theorem, we can "smooth" the function h(x) over the interval
- lln,x -f l/n) to create an infinitely differentiable g"(x), and the two functions
differ negligibly. Since the sequence of core functions converges to a distribution weakly,
this means we can replace the function with a distribution.
(x
For every continuous function, an equivalent distribution exists.
We cannot make the reverse statement; there is not necessarily a continuous function
corresponding to every distribution. The delta function is an example of a distribution for
which there is no corresponding continuous function.
The class of distributions is an important addition to our mathematical arsenal. In the
next section, we shall learn more about how to use them.
6.3. PROPERTIES OF DISTRIBUTIONS
.
Distributions may be added, subtracted, multiplied by constants, or multiplied by infinitely
differentiable functions. These assertions may be easily proved using the definition of
l3For the proof of tbis theorem, see Butkov, p. 243, or Lighthill, p.
2 1.
308
cHAprEB 6 cENERALIZED FUNcrloNs lN PHYslcs
a distribution as a weak limit of a sequence of core functions. For example, let h(x) be
an infinitely differentiable function, and let d(-r) be the weak limit of the sequence gr.
Then
+@
f+oo
I
J-*
hfOO(x)lf
- @) dx
f
: n+mJ-m
lim I
lt (r)s"@)lf (x) dx
/"+oo
,\%
/-""
s'@)Ih(x)
f (x)) dx
f*m
I
J-a
Q@ltntof @)ldx
h(x)/(r) is a test function if / is. So we can define a new distribution ry'(x)
h(x)Q@). Thus, for example, rd(x) is a distribution. Let's see what it does.
since
lllwt<.ttr',)dx
Thus,
.
.
.
x6(x)
:
:
l-:
d('r)txl(x)l dx
:0x
/(0)
:
:
Q
0, the zero distribution.l4
Distributions may NOT be multiplied or divided by other distributions depending on the
same variable. Forexample,t6(x)12 is meaningless. [However, d(i) : 6(x)d(y)d(z) does
make sense becauSe the arguments x , y , and z of the three delta functions are independent
variables.l
lf Q@) is a distribution, so are Q@ - a) and QGx).
Distributions are infinitely differentiable, because the core functions are. Moreover, if
gn(x) --> d (x) weakly, then gl (x) -> Qt (x):
,$"/J
s'^@)f
(x)d,: n\(tror^r*>t
=
-,[:
g^@)f'(idx)
The integrated term is zero by the properties ofthe test functions, and so
lll
o'rnr(x)dx: -
I:: Q@)rt(x)dx
(6.20)
This is a property we have already established for the delta function, and it serves to
define the derivative
.
QI
(x). Higher-order derivatives may be computed similarly.
Since disffibutions can be differentiated, they can be the solutions of differential equations.
Example
6.4.
Find the solution of the differential equation
d,t
@-D**y:0
14we'll need this result in the next chapter.
6.3 PROPERTIES OF DISTRIBUTIONS 309
We can integrate this equation directly to obtain one solution:
dy
dx
y
x-r
lny--ln(x-l)+C
A
t- *-1
:
This solution is validforx > I orforx < I but not at x
L
Now consider the distribution y
6 (x
1) . If this satisfies the differential equation,
-
:
then we must have
/^+oo
I
J_q
x*
-
l)6/(.r
- 1) + s(x - l)lf @) dx :
o
Let's evaluate the first term using the sifting property ofd/ (equation 6.6):
f+*
I eJ_a
l)6'(x -r)f(x)dx
d
: - If+@a@- tlilfx
- t)f(x)ldx
ctx
J_a
r*oo
: - II
J_m
d(_r
- t)t/(x) * (r - t)ft(x)ldx
: -f(l)
The second term is
/n+oo
/
J-a
a("
-
l),f(x) dx
: f (Il
so the differential equation is satisfied. We must choose this delta function solution
the point x
I is within our domain of interest.
:
if
Note that since the differential equation is first order, we expect only one solution. What
we learn here is that the character of that solution depends on the domain of validity.
Some differential equations contain a distribution. The solutions of such equations are
also distributions.
Example
6.5.
Find the solution of Poisson's equation for the electric field due to
p : oo6(x - a):
a sheet ofcharge with charge density
dEt oo^.
E-j6(x-a)
From equation (6.12), the solution is
Oi
E,:A*j@(x-a)
€0
310
cHAprER 6 GENERALTzED FUNcrtoNs lN PHYSlcs
where @ is the step distribution (see Section 6.1.4). Since the electric field points
directly away from positive charge, we can find the integration constant A by invoking
symmetry about the sheet (Figure 6.8):
E1(x <
a):
>a)
A_
#)
and so
.oo
2eo
Thus,
E*
:
o)
(-; .
otx
-
a))
FIGURE 6.8. Electric field due to a sheet of charge (Example 6.5).
6.4. SEQUENCES AND SERIES
We can extend the concept of weak convergence to sequences and series. For example,
the function e'" is a core function and therefore may also be regarded as a distribution.
This means that any Fourier series may be regarded as a series of distributions. The series
converges weakly if the sequence of partial sums converges weakly. Once we make this
identification, some very useful results follow.
The sequence of functions etn' has no proper limit. But let's see if there's a weak limit:
6.4 SEQUENCES AND
SERIES
31
1
The integrated term is zero, by the properties of the test functions [/(x) is zero outside
some finite range a < x < bf . The absolute value of the remaining term is
l_J
# t'(x) dx : : I"' ,in' y'1*\ d*
=: l,' l,'^' f'utl o,
by the mean value theorem, where
a < f, < b
[a, b]. Thus,
e'"
: Il'.'"u r'c>l@ - a)
and lei"€ 7'(6) |
ir
-->
0
as
n -+
oo
bounded in the interval
--> 0 weakly
If a sequence of distributions Q,(x) converges weakly to @ (x),Q @)
:
lirnn+oo Q"@),
then the sequence of derivatives QL@) converges to @/(x):
Q'@):
6.21)
,\QL{*l
To see why, note that
lim [**
n-@J_a
O',r*,
f (x) dx
"
: - n-@J_m
lim [**
"
0,@)
f
t(x)
dx
(result 6.20)
Q@)f'(x)dx
0'@)
f @) ax
A convergent series of distributions can be differentiated term by term.
To prove this statement, let S^,(x) : If:, Qn@).Because the sum has a finite number
of terms, we may differentiate to obtain Sir(x) : DItQ'"@). Then, taking the limit, we
have S(x) : limN-oo Sr,,r(x), and, by theorem (6.21) above, S'(") : limru-oo Sir(x).
This is a powerful result: Series of distributions possess some of the same properties as
uniformly convergent series.
If we start with a convergent Fourier series S(x), we can differentiate term by term, thus
multiplying each coefficientby in. We can do this m times, so the series with coefficients
(in)* a" also converges to dm S f dxm. We have thus shown that every Fourier series, considered as a series of distributions, converges (weakly), even when the coefficients do not
decrease with increasing n. As a consequence, every Fourier series may be integrated and
differentiated term by term.l5
15We made
use
of this result in Chapter 4.
312
cHAprER 6 GENEBALTZED FUNcrloNS lN PHYSlcs
Example 6.6. Find a Fourier series for a delta function by differentiating the Fourier
series for the triangle function
f(x):{
on the range (0,
L)
(x(2h/L)
ifO<x.L/2
l(L - x)(2hlL) if L/2 < x < L
(see Example 4.4):
f(x):Yn##,,"s+r!
This series is nicely convergent.
(compare with equation 4.28):
f,(*):
If we differentiate once, we
{'!#,,
{:-
get a step distribution
@(x
- L/D}
(2ntI)nx
sh s (-l)',
:i1rz"*tlcos
"lorir.='*!":T
awhich also converges, but more slowly, and
function:
if we differentiate
f"(*): -?uu
- L/2):+?.i,-,,'*'
t' I'-)
L
"t
again, we get a delta
gl:l!
6.22)
Actually it is a negative delta function, since the step function steps down at x :
L 12. Compare this result with equation (6.14). Here we have no even terms, because
sinl(nn/L)(L/2)l : sin(nnl2) is zero whenever n is even. The series (6.22) is
a weakly convergent series of distributions. The sum of the first twenty terms of the
series is plotted in Figure 6.9.
The process of finding the coefficients in a Fourier series of a distribution such as 6(x)
requiresaminormodificationtoourtheory.First,theintegralsthatgivethecoefficientsare
notoverthecorrectrange(-oo,+oo),andsecond,thefunctions
e'n'(orsinnxorcosnx)
are not appropriate test functions as we have defined them. We can solve both problems by
differentiable "tudge function" u(x) such
;;l;;G;il
,ii
+
'ora
that
L: "in'u(*)o@)0,: #, ll"' "'"'rr*ro'
(6'23)
l
I
6,5 DISTRIBUTIONS IN N DIMENSIONS 313
L2f,
8h
x
L
FIGURE 6.9. The first twenty terms in the series (6.22) for the second derivative of the triangle
function. The variables are L2 7" 18h versus
delta function at x
/L :0.5.
x/I.
This series represents a negative
Then etnx u(x) is a test function, and the integrals can be evaluated in the usual way.16 Once
we know that such a procedure exists, we can just use the normal expression for the Fourier
coeffrcients-that is, the integral on the right-hand side of equation (6.23). Of course, the
resulting series gives a periodic repetition of the original distribution.
6.5. DISTRIBUTIONS IN N DIMENSIONS
The theory of distributions can be extended to N dimensions. In physics, we are frequently
concerned with problems in three space dimensions and one time dimension. Sometimes we
work in a six-dimensional phase space with three position coordinates and three components
of momentum. The modifications are straightforward.
1. The test functions should possess partial derivatives of all orders and go to zero sufficiently fast outside a finite volume of the space.
2. T\e core functions should
possess partial derivatives of all orders.
3. The integrals that define the distributions become volume integrals over the Ndimensional space.
l6Examples of such functions a can be found in Butkou p. 255, and Lighthill, Section 5.2.
314
cHAprER 6 GENERALTZED FUNcrtoNs lN PHYSlcs
Then all the properties of distributions extend as expected' For example,
:
L#,,o0, l(orrr= - I:: o{a.)a,a,
:- Jv[ oTav
dx
The delta function in three dimensions is readily expressed in Cartesian coordinates.
is nonzero only at the origin, so
6(i) :6(x)6(y)6(z)
We may write this as a
It
(6.24)
limit of the core functions (6.18):
:,\%
fim !--e-n"'4"-n'Y' '
6(i): n+a
t/It
\/iT "-""
JiT
(e)' '*r
(6'2s)
A particularly useful relation in three dimensions is
:
"'(i)
-4zd(i)
(6.26)
First we note that the left-hand side of equation (6.26) has the sort of behavior we expect
for a delta function. If r I 0, then
/a2 * a2 a2\
(P i7*
a?
)ffi:
a /-x\ * a /-y\. *
a" \;t/ r, \;t/
/ 3y
-3 / 3x\ - r\-F
'\-Fl
I
)
-3 r2
:7+3-:o
but as
r
--> O, I /
r -+ oo and its derivatives
cannot be computed; the left-hand side of (6.26)
is undefined.
Next we must test for the sifting property. Since we have already shown that V21t 1 r1
for r 10, the volume integral
./,,,n*.
o'
"'(:) f G)dv: ,/nn"."or.u*o." (i) "o,o'
:
O
I
6.6 DESCRIBING PHYSICAL QUANTITIES USING DELTA FUNCTIONS
315
reduces to an integral over a small spherical volume surrounding the origin. We write the
integrand in terms of a divergence so that we can use the divergence theorem:
./pr,... or,uaiu.,
t' (i)
f G) dv
:./nn",.o,.u0,",,
vl v rfav
lu (rclvl) -
(rrrrvl) .aoo- [
: Jsurfaceofsphere\
[
r/
(rrn])
:[
# aa + [
' ' t'
\"
/
:I
Yo,on-[
0r
-].iyav
Jsphereofradiusel'o
,./sphereofradius,
./surfaceof sphere
/sphereofradiuse
_f
- t
J solid angle of sphere
:
Jsurfaceofsphere-
0r
a, aa
rl)da
lfli=6-ft)lr:elda
f
-"f(o)
\!,,
rz
J"nno,on: -4trf
(o)
Thus, since we have obtained the sifting property, relation (6.26) is proved.
A similar relation holds in two dimensions:
vr'(h
L):znqil
where Vr2 is the Laplacian operator in two dimensions (Vr2
constant length, and p
(6.27)
:
A2
/Ax2
+
A2
/Ay\,a
is any
- \FT|
6.6. DESCRIBING PHYSICAL QUANTITIES USING
DELTA FUNCTIONS
We often want to use delta functions to describe the mass or charge density of a system that
is confined to an infinitesimally thin sheet or line. Let's investigate how to describe such
physical quantities using delta functions.
Example 6.7. A ring of charge of radius a lies in the x-y plane. It has a variable
line charge density ),e cos {. What is the volume charge density? Use (a) cylindrical
coordinates and (b) spherical coordinates.
316
cHAprER 6 cENERALIZED FUNcrtoNs tN PHYslcs
(a) First we note that the charge exists only
at z
:0
and at
p
:
4, so we must have
two delta functions:
p(i):C6(z)6(p-a)
To find C (which may be a function of the coordinates), we integrate over a wedge of
space with -oo < z < +@,0 < p < oo, and angular extent dQ (see Figure 6.10a)'
(This volume is chosen by using the entire range of each coordinate in which we have
a delta function, but only a differential range in the third coordinate.) The amount
charge d4 in this volume is )'adQ. Thus, using the sifting proPerty, we have
il.scosQ dQ
- dq :
-
and so C
:
lo
cos
f*"o""rurrr6(p
d:
cos
FIGURE 6.10a.
To find an expression for the charge density
in Example 6.7, we integrate over a wedge
of an infinite cylinder with
<p<
a) p dp dz dQ
Cad0
p(*) :.r"e
O
-
of
-oo < z <
oo, and angular extent d@.
@,
@
6(z)6(p
-
a)
FIGURE 6.10b.
To find an expression for the charge density
in spherical coordinates, we integrate over
the wedge of angular width d@ that extends
throughout the full range of the coordinates
0 andr:0 < 0 < z and0 < r < oo.
6.7 THE GREEN'S
(b) In spherical
coordinates, the x-y plane is at 0
p(i)
:
FUNCTION 317
n 12, so
: ca(e -|) x, - "t
Now weintegrateoveran"orange-wedge" shapewith0 < 0 < 7T,0 <
extent dQ in the azimuthal angle (Figure 6.10b):
a),scosQ dQ
:
:
Thus, C
: lo (cosd)/a,
f*.or"ca(e
Ca2^n
sin
r < oo, and
- Z) uU - a)r2 sin0 dr d0 dQ
,aO:
CazdQ
and
p(i):
l,6cos@d
(t -;)t+
Check the dimensions of the result!
This example illustrates a general method for finding densities of linear or planar charge
or mass distributions in terms of delta functions in curvilinear coordinates.
First, determine which coordinates take on only a single value, and find the delta functions
in these coordinates. (In the example above, charge exists only at 0 : r /2 and r : a.)
Second, determine the function that multiplies the delta functions by integrating over
a region of space that extends over the full range of the coordinates in the delta functions,
but over only a differential range in the other coordinate or coordinates. The value ofyour
integral should be equal to the value determined by elementary methods. (In our example, we
integratedoverawedgethatextendsfrom@ toQ
of the ring and thus contains an amount d e
:
+ dQ.Tltis wedgeslicesoffalengthadQ
)ra dQ of charge.) This method guarantees that
you will not lose information about the dependence of the coefficient C on the coordinates.
Integrating over all space does not work because you will lose the detailed information that
you need.
6.7. THE GREEN'S FUNCTION
An important application of distribution theory is the Green's function. The Green's function
is actually a distribution that describes the response of a physical system to a unit delta
function input (a "point" source). We have already seen an example of this in Example 6.3,
where we applied a point impulse to a string. Other examples are the electric field due to
a point charge, a point mass on a beam, and a voltage impulse applied to an electric circuit.
If the physical system is linear (and thus is described by a linear differential equation), the
response of the system to a sum of inputs is just the sum of the responses to the individual
inputs. Then ifwe can model any input as the sum (integral) ofa set ofpoint (delta function)
inputs, we can use the Green's function to compute the response.
318
cHAprER 6 GENERALTzED FUNcrloNS lN PHYSlcs
If the impulse per unit length applied to the string is 1(x), we can write it
7L
I(x): JoI
If
t1x'p1x-x'1dx'
the response of the system to the delta function is
y(x)
:
fo"
as
,
G(x,
{*')o{*,
x'7, the response to 1(.r) is
(6.28)
x'; dx'
The Green's function is described in more detail in Optional Topic C, where we show
how to find an appropriate Green's function for several different systems and in different
geometries.
PROBLEMS
1. Show that the following
liail a. r*t
-
n
sequences offunctions are delta sequences:
e-nlrl
: L ( -+-)
/r \l+n'x'/
1 - cosnx
(c) d,(x) : ---------1
nTr x'
(b) d,(x)
2. Find a Fourier
t*nr, Use contour integration.)
series representation of the delta function
6
(x) in the range
(-
L,
*L)
in
two ways.
(a) Start with the Fourier series for a step function (equation 4.24, for example) and
differentiate.
(b)
Start with the block functions (equation 6.3) and form the Fourier series. Take the
limit
as
n
--> @.
Are the results the same?
If not, why
not? Give a quantitative as well as a qualitative
account of any discrepancy.
I at a distance
end. Find the displacement of the beam
kail if tne beam is supported at the left end, as in Problem 5. 1 1
3. A point load Mg is placed on a beam of length
(b) if
Ll3
from the left-hand
the beam rests on supports at each end, as in Example 5.3
Ignore the weight of the beam itself.
4. A damped harmonic oscillator (compare with equation 3.7 andProblem 4.13) has initial
conditionsx(0) : x6 and dxldtl,:s: u0.Animpulse 1is applied att : to' Findthe
motion of the oscillator for t > 0.
5. Distributions may be multiplied by infinitely differentiable functions. Do you expect the
product
yt\ t
-
6(x
-
a)
PROBLEMS 319
to be a valid distribution? Why or why not? Investigate the properties of this quantity by
evaluating the integral
f@ 6-(x - a\
(x)dx
J-*-;;f
where Q"@) is a delta sequence of your choice and f (x) is
determine the result for functions that have the property
distribution in this case? Can you identify it?
a
test function. In particular,
f (a) :0.
Is ry'(x) a valid
6. Evaluate
@ /a s-txt51*2 -l2x - 3) dx
o) /]S e-" 3(x2 -t x - 6) dx
7. A string of length Z, with tension Z and mass per unit length p,, is hit simultaneously
at t : 0 at the two points x : L I 3 and x : 2L I 3. The impulse delivered at each point
@
is 1. Find the subsequent displacement of the string.
Using a general curvilinear coordinate system (Chapter 1, Section 1.3) with coordinates
u, u, and u.r, find the charge density due to a point charge q placed at the point u : t!0,
1) : uo, tr : tDo. Hint: Start with the delta sequence (6.25) and note that as n -->
m, only a differential line element ds2 is needed in the exponent. Then make use of
equation (1.61).
9.
Show that the sequence of functions
nf*o
fn7): t;G J_, ,*r[-r2(* - *')2]d*'
converges weakly to the distribution that gives the average value of any test function on
the interval -a to a.
10. Show that the sequence of functions
f'(x):
n
2coshz nx
converges weakly to the delta function. Hint: Use a method similar to the one used in
this chapter for sequence (6. I 8).
1L. According to the properties of distributions in Section 6.3,
Which distribution is it?
12. Starting with the integral (6.16), show that
6(x)
E
:
e-'6t(x) is a distribution.
: I,* coskx dk
gV expanding the cosines in exponentials or otherwise, show that, for
positive real numbers,
3(x
-
a)
_2 [*
-rJn
cos
kx
cos ka
dk
x
and a both
320
cHAprER 6 GENERALIzED FUNcrloNS lN PHYslcs
and obtain a similar expression as an integral over sines. Are these results consistent
with Problem 12? Discuss.
14. Find the Laplace transform
of 6(r - a).
Express the inverse as an integral using
equation (5.19) and demonstrate that this integral possesses the sifting property.
15. A uniform disk of radius a and mass M lies in the .r-y plane. Express the density in
terms of delta functions
(a) in rectangular Canesian coordinates
(b) in cylindrical coordinates
(c) in spherical coordinates
16. A uniform rod of length I and mass M lies along the x-axis with one end at the origin.
Express the density in terms of delta functions
(a) in rectangular Cartesian coordinates
(b) in cylindrical coordinates
(c) in spherical coordinates
E7J a. me of charge with uniform line charge density ), lies along the z-axis. Find the volume
charge density
(a) in cylindrical coordinates (b) in spherical coordinates
18. A disk ofcharge with radius a and surface charge density o(r)
-
osrfalies in the r-y
plane with center at the origin. Find the volume charge density
(a) in cylindrical coordinates (b) in spherical coordinates
19. Current 1 flows in a circular loop of radius a lying in the x-y plane with its center
at the
origin. Find an expression for the current density
(a) in cylindrical coordinates (b) in spherical coordinates
20. Prove the relation (equation 6.27)
v2ktg:zrs(i)
a
where p is the radial coordinate in a cylindrical coordinate system, a is a constant length,
and p is the position vector in a plane. Use the result to find the potential due to a line
charge
I
running parallel to the z-axis at x
:
Q,
!:
b.
21. A circuit contains a resistor, a capacitor, and a square wave power supply with period Z.
Use Kirchhoff's loop rule to write an equation for the current in the circuit in terms of
delta functions, and solve it to find the current as a function of time.
Ezl
starting with the result
D.r
O: :-r3
for the electric potential due to a dipole placed at the origin (see Example 6.1), calculate
the electric field everywhere, including ar the origin. llint: Show that the electric field
contains a delta function term. Use a method similar to that used in Section 6.5 to prove
relation (6.26).
PROBLEMS 321
23. Using
a delta sequence
of your choice, show that the limit
..
16(r)
-
6(x
libl z
-
t)1
l
exhibits the sifting property of d/(r).
24. Use the derivative property (6.20) to show that, for distributions, the Laplace transform
of the derivative Qt(x) equals s times the Laplace transform of @. Show that the Laplace
transform of lnr is -(y + lns)/s, where 7 is Euler's constant, - If, r-'lnxdx :
0.5'772. Hence show that the Laplace transform
tion, is
ln s. (See Zemanian, Chapter 8.)
of I / t (t > 0), considered
as a
distribu-
-y -
25. Stating from equation (6.16), show that
sin Rx
*9- "r :6(r)
Confirm your result by demonstrating the sifting property. Use contour integration to do
the integral.
Similarly, show that
if the integral
is taken to be the principal value.
Demonstrate the plausibility of the results by evaluating
2
sin Nxrx
I, :
and h' : [*
/@
n J,
x
J,
numerically for a set ofvalues ofe ( I and
proaches unity and 12 decreases toward zero.
N )) l.
cos
Nx
x
Show that as
o*
N increases,
E6.l Show that
dx (x) :261*r,
sizn
*
where sign
(t) :
l"l
27. Show that
**Urnrrrr:
*v'
IO
If-t;'1,t71n -
Hint: Use proof by induction (Appendix III).
if n < m
m)116("--)1v1 ifn > m
11 ap-
322
cHAprER 6 GENERALTZED FUNcrtoNS tN PHYSlcs
28. The integral If, *" f (r) dx : I:S xd @(x) f (x) dr, where o < 0, may be integrated
if xo is intetpreted as a distribution. First show that
xoo(x) :
where a
t" + rlt"
t --.
+zl'.'
4n_[x"+,@(x)]
(cv
* n > O and n is an integer. Use the result to evaluate the integral
[* *-:l'r-*
lo
@l
* n) dx'4*
absorbs light at frequency u; due to an atomic transition. The imaginary part
of the dielectric constant may be approximated as a66(u u1,). Use the Kramers-Kronig
lmaterial
-
relations (Chapter 2, Example 2.24) to determine the behavior of the refractive index
n
JPalt/eS as a function of frequency. Comment.
:
30. Demonstrate the sifting property of the delta
sequence (6.5),
I sin2 nx
""--Fin the case that f (x) has a second-order pole 4t z : 7o in the upper half-plane. Can you
hnE):
extend the result to a pole of order m?
CHAPTER 7
Fourier Transforms
7.1. DEFINITION OFTHE FOURIERTRANSFORM
The Fourier transform is the second integral transform that we shall study. Like the Laplace
transform, it provides a convenient way to solve an inhomogeneous differential equation.
But the Fourier transform proves to have many other useful applications in physics, as we
shall see.
Recall that a function f (x) may be represented as a Fourier series over any finite range
-L<x<lL:
f (x):
*oo
,D_on"'"o''t
Outside of the selected range, we obtain a periodic extension of the original function. The
set of coefficients a, forms a unique representation of the function /(x) over this finite
range. Each an may.be calculated using the integral (equation 4.31)
o^: i,1
f+L
J_, trnr-inrx/14,
Now we'd like to increase the range (-L, L) until / may be represented over the whole
real line. But as I -+ oo, nn I L --> 0 for any finite n, so we cannot use equation (4.31) as
it stands. We begin by defining
k:no
ard an
:
L
a(k)
so that
f
(x)
:
*oo
t
L-4lL-
a(k) eik'
^
and
a(k):
+ I_"" f (x)e-ik, dx
323
324
cHAPTERTFoURIERTRANSFoRMS
We multiply this expressionby
Here we may let
L
-->
a
LJTIi
and define
to obtain
F(k):
# l::
(7.1)
r@)e-ik'dx
provided that the integral converges.
Now the difference between two neighboring values of ft is
ui:-
(n*l\n wr r
L - L:T
Thus, we may rewrite the original series in terms of
,f(")
:D
as
2
7f
: -+l- )
Jzn
-
7
Then, as we let
obtain
F(k) and Ak
F(k) ,ikx
61l'
l, -+ oo and consequently A,k --> 0, the sum becomes an integral and we
I r+f(x): _t.,8; I F(k)eik'dk
-*
(7.2)
Equations (7.1) and (7.2) define the Fourier transform and its inverse. The factor of I / J2n
appears in both expressions-this isthe symmetric Fourier transform. The symmetric transform obeys the following symmetry relation:
If F(k)
is the symmetric Fourier transform of
f (x), then /(-k)
is the transform of
F(x).
An alternative definition of the transform has a factor I l2r in front of one integral and
unity in front of the other. The product of both factors must be ll2n.It is also possible to
define the transform with a plus sign in the exponential in equation (7.1) and a minus sign
in the exponential in equation (7.2);infact, this is usually done when the variable is
variable.l
lThe difference is related to the metric of space-time in special relativily.
See Section 7.3.7
a
time
for an example.
7.2 SOME
EXAMPLES 325
The Fourier transform and the Laplace transform are closely related. To see how, let
ift
:
s in equation (7.1). Then we have
Jzo
In the special case that
transform of
/.
f (x) :
reir: [**
J-*
0 for
.r <
f (x) e-sx dx
0, the righrhand side becomes the Laplace
The inverse (equation 7.2) is
I 1*;oo
f(x):GJ
:FGis),'.0:
I
:
/+'oo
J-,* "o'(-is)
^
i
i
I
e"
ds
which is the Mellin inversion integral (equation 5.19) with y --> 0.
An important difference between the two transforms is that the Fourier transform may be
applied to functions defined over the whole range -oo < r < +oo. However, the function
must approach zero as x -+ *oo for the transform to exist.2 In contrast, the Laplace
transform exists for functions that diverge as r -+ oo, provided that they do not diverge
faster than an exponential (see Chapter 5, Section 5.1). The Laplace transform is restricted
to functions defined for positive values of the argument.
7.2. SOME EXAMPLES
Example7.1. Find the Fourier transform of the function
where cv > 0.
/(.r) - exp(-o lxl),
The transform is given by equation (7.1):
F(k):
:
I
/+oo
h J_*
h
:&
,-utxt
"-ikx
(l_*u'.ik*
7,
dx
+
fo**
"-'*r-'o'
or)
_ I (ro_ltt,1o _r-ro+ilr"l+-\
\"-,-l---
t / r
"+," l, )
-l \
___
- Jr; I\cv-ik
a+ik)
_.1
_ td
Y;aTF
I
(7.3)
2This restriction can be lifted if we are willing to consider the function
/(x) to be a generulized function.
Lighthill's book for an extensive discussion. See also Example 7.3 and Problem 28.
See
326
cHAPTERTFoURIERTRANSFoRMS
Now let's go backwards and find the function that corresponds to this transform. From
equation (7.2),
: ;l/+-a
dk
J-* ;+ tceik'
This integral is of the type that can be evaluated using the residue theorem (Section 2.7.3,
Example 2.20). For x > 0, we close the contour upward with a big semicircle (Figure 7.1).
Im (k)
i
j
I
Re (k)
FIGURE 7.1. To invert the transform (7.3), we complete the contour in the upper half-plane when
x>0.
The integral along the semicircle is zero, by Jordan's lemma. The integrand has two
: Lia. Only one of the poles is inside the contour. Thus, the integral is
simple poles at k
l_-
*,ik'dk: f ,+r,ik'dk
/a\
: 2r i (;i)'i(ia)x
-
7T
e-o'
and so
(x): lbre-"'):
1
.f
s-dx
7V
For
x < 0, we must close the contour downward. The pole at k : -ia
contour, and we go around the contour clockwise (Figute7.2).
is inside the
7.2 SOME EXAMPLES
Im (k)
Re (k)
FIGURE 7.2. For x < 0, we close the contour in the lower half-plane.
Thus, the integral is
[_ *,ikx41,: t' -+r,ikx7p
: -2ri (-:)
,i(-iutx :
\-2ia /
andhence
f (x):
eax.We maycombineourresultsforx
f (x) :
ireux
< 0andx > 0toobtain
exP (-cv lx l)
as required.
Exampte 7.2. Find the Fourier transform of the Gaussian function
The transform is
F(k)
: 1 /+oo
G J"
7yr-o2*2r-ik* 7*
We do this integral by completing the square:
-o*2
-
ikx
j1\
: -ot2
- (\^*, *' 4*q,2n 4oa ) ik\2 _
"/
:_q"\.**)
Now we change variables to u
F(k): N
t*
-
4q2
p2
4A
: olx + ikl2a2):
/ e_\
Jzr'*P\-m)
[+@+ikl2d e
._,zdu
J-a+irpa
-
f (x) :
N
e-o"'
328
CHAPTER
7
FOURIER TRANSFORMS
The path of integration has been moved offthe real axis by the amount k f 2a.However,
since the integrand has no poles, the value of the integral is not changed. To see why,
constructarectangularcontourfromRe (u) : -RtoRe (u) : *R, withheightkl2a
(Figure 7.3). The integral along the two vertical pieces at the ends of the rectangle
goes to zero as R -+ oo:
ll,,o.u*
o"l= *max
*^
le-t+n+;rr'l
le-n'?+ziw+f|
*"*r(*)'"-R'--' 0
asR-+oo
Im (z)
Re (a)
FIGURE 7.3. The transform of the Gaussian is computed using an integral along the upper side
of the rectangle. Since there are no poles inside the closed contour and there is no
contribution from the short sides, the integral is the same along both long sides.
The integral around the entire rectangle is zero, by the Cauchy theorem, so the integral
along the upper path is the same as the integral along the real axis. We know the value
(Appendix IX). Thus,
of this integral-it is
"fi
F(k)
:
N
/ k2\
oJr"*p \)
^d
Therefore, the transform of a Gaussian is also a Gaussian, but their widths are inversely
related. The original function reaches one half its maximum value at x : 0.83/a; the
transform reaches one half its maximum value at k : 0.83(2q).
This result is closely related to the uncertainty principle in quantum mechanics. The
relation between momentum and wave number k is
P:hk
A particle is described by a wave packet, and the uncertainty in its position is determined
by the width of the packet---essentially llu.T\e width of the corresponding pulse in
momentum space is h A,k
:2ha.Thus,
the product of the two uncertainties is
L.x A,p
I
:2h
- -2ha
d
7.3 PROPERTIESOFTHEFOURIERTRANSFORM
329
for a Gaussian wave packet. The quantum mechanical relation is
t*tp2h '2
Thus, a particle that is well localized in space is poorly located in momentum, and vice
versa.
Example
7.3.
Find the Fourier transform of the function
/(.r) : l.
The transform is
F(k\: L
'.,/2n
[** ,-ikx 4,
J-a
We can recognize this integral as the delta function (equation 6.16):
F(k): Jn
ag,>
Then, by the sifting property, the inverse is
f
(x): I
f+oo
h J_* J2n 6(k)"ik'dk: t
[Note that etkr does not approach zero as x approaches infinity, so it is not a proper test
function as we defined them in Chapter 6. We must n€urow the class of distributions
by putting the requirement g(x) + 0 as.r -+ oo onto the core functions rather than
the test functions. This narrower class still includes the delta function as well as many
other useful distributions. l
Examples 7.1-:7.3 demonstrate the most commonly used methods for evaluating trans-
forms and their inverses:
1.
2.
3.
4.
Integrate an exponential with linear argument.
Complete the contour with a large semicircle and use the residue theorem.
Complete the square.
Identify the delta function or use the sifting property.
7.3. PROPERTIES OFTHE FOURIER TRANSFORM
Because the Fourier transform is similar to the Laplace transform,
properties that we found in Chapter 5.
7.3.1. Linearity
The Fourier transform operator
f
is a linear operator:
T(f+il:fU)+F(s)
f
(af)
: aF(f)
it
shares many of the
330
oHAPTERTFoURTERTRANSFoRMS
7.3.2. Complex Conjugate
Ifthe function f(x) is real, then
F*(k\
f+oo
: I1
l:
I
f (*)"-ik" orf. :
lJ2r l_a
# I:: f (r)e+ik'dx: F(-k)
(7.4)
7.3.3. Differentiation
We evaluate the transform of a derivative using integration by parts:
'(#):hI IIJ#''*0.t
s.-ikxl*-"'l-- -- ,D; ,'
/*--
h J-* -ik-se-ik* dx
:
'(#) h ll* ,"-'o' dx - ikF(k)
(7.s)
The integrated term vanishes, since the transform exists only for functions /(x) that approach zero as x -+ *oo. Notice that in the case of the Fourier transform, unlike that
of the Laplace transform, no initial conditions appear. Extension to higher derivatives is
straightforward:
,(#):
(ik)'F(k)
(7.6)
The sign that we use in the exponential when forming the transform (equation 7.1) is
reflected in the sign that appears in the derivative rule (equations 7.5 and 7.6). If the transform
is defined as
F(o):
I
f+- (t)e+i't
dt
f
h J_*
then the first derivative has transform
'(#)
: -iaFko)
(7.7)
i
7.3 PROPERTIES
TRANSFORM 331
OF THE FOURIER
7.3.4. Attenuation and Shifting
Let g(x)
:
eo'
f (x) . Then the transform of g is
G(k):
:
If g(x)
: f (x -
I
/+oo
e"''
h J_* f
(x)r-ik'r 4*
h l** ror"_-i(ia-tk)x 4*
:
F(k -l
G(k)
:
I
: x-
a, we have
ia)
(7.8)
a), then
Changing variables to u
G(k):
:
/+oo
(x
h J_* f - a) r-ikx 4*
I
h
/+- (u)e-iktu+o)
J_* f
4u
e-ik" F (k)
(7
.9)
Thdse relations are analogous to (5.5) and (5.6) for the Laplace transform.
7.3.5. Parseval's Theorem
The integral of the product of two functions may be related to the integral of the product of
their transforms.
r
a+m
a+oo
f+*
.,._ f+*
d*l
a*rQe)eik'I
IJ-- fr*le(x)dx:+l
"
2r J-J-*
J--
:
l::
F&)dk
:
l_:
F&)dk
:
f +oo
J_*
doGlo)ei'r
lll crao,* f+* "io'+'t'd*
l_l
cr,t 6(k-ta)da
F(k)GeDdk
If the function g is real, then we may
use equation (7 .4) to write the result as
l** to, g@)dx
F&) G*&) dk
(7.l0)
332
cHAPTERTFoURIERTRANSFoRMS
This result is called Parseval's theorem.3 An important special case occurs when
/ :
g.
Then
r+oo
r*oo
I Lfe)l'dx:I
J-a
J-*
The absolute value signs are necessary since even
(7.tt)
F&)2dk
if f (x) is real, F(k)
may not be.a
7.3.6. Convolution
Suppose the transform
H(k)
:
F(k) G(k) is the product of the two transforms F and G.
Then the inverse is
h(x):
:
I
/^+-
h
J_*
I
/t+oo
H(k)eik'dk
h J-* 'rorG(k)eik'dk
Now we write the transform G(k) in terms of the function g:
h(x):
+
l:: ,n ll.: s(u)e-iku aul,ik'an
We then combine the exponentials and interchange the order of integration:
h(x):
I
f+oo
h
J
r
*
r+oo
I
s{")-L16
J_*
.+m
I
-L
Jzr J-*
s@).f(x-u)du
which is the convolution5 of the functions g and
h(x)
,rorrik(x-u) 4P4,
f
.
Equivalently, we may write
:
the inverse transform of
:
-l-*
F(k) G(k)
(7.12)
l** ra g@ - u) du
3compare with equation (4.32) for Founer senes.
4See Example 7.4.
5This convolution differs from the one defined in Chapter 5 only by the numerical factor
of the limits of integration.
l/J2tr
and the values
7.3 PROPERTIESOFTHEFOURIERTRANSFORM
333
7.3.7. Extension to N Dimensions
We may extend the theory of Fourier transforms to as many dimensions as we wish.
Frequently, in physics, N : 4: three space dimensions and one time dimension. If / is
a function of the three Cartesian coordinates x, y, z (N : 3), we simply transform with
respect to each of the variables, one at a time.
FG) : F(kx,ky,kz)
:
I:: l-: l-- ,r'!,
#
: : lf
/
/til
(J2r)z Jdl
,pu"" "
exp
z) e-ik"x dx e-ikvv dv e-ik'z
(-ik .i) dv
It is usual to define the time transform with the opposite signs, so that for N
F(i.a): ^: [*" I
\z1r )' J -a Jdl
d.
f G,r)exp
[-i([ .i-
:
4,
att)ld3idt
space
Then the function
,f
lf+@l
(i, r) :
I I
-l\zn ). _a Jail i
J
Fd<, alexp
space
[i(f .i -
c,.,r)]
a3iaat
is a sum of plane waves.6
The transform of a derivative may be calculated using a method similar to the one we
used with the one-dimensional derivatives in Section 7.3.3. For example,
-lt-
F(v
f
)
(t/2n\3
/
Jall
tv./(i)lexp (-ik .*) dv
space
lf
------:= /
tVtf t*) exp (-tf ' *)l - /(DV exp (-tf .h\ dv
- \\/'27r)'
Jall
lf
:: (J2tt)3/surfaceu,-"''
I
fr*)exp(-if .i)fids
space
lf
- ("/'2T)' JallI
-----
Now provided thatT
/
space
-irtt(i)
exp
(-ii.;i dv
approaches zero faster than I / r2 as
r ->
oo, the surface integral
(-ii
.*) dv
is zero, and
rri n :
0
+ irt
ah
l^r"o
*/(*)
exp
6Co.p*" with
Chapter 2, Section 2.1.4
TAgain, this condition is also required for the existence of the transform of
/.
334
cHAPTER
7 FouFtER TRANSFoRMS
F(V
f):
(7.t3)
ikF(k)
Similarly, we may show that
TN2fl:-nzr1i<1
(7.14)
f(ixi):;f"i
(7.1s)
and so on.
7.4. CAUSALITY
In an initial value problem, the initial conditions appear explicitly in Laplace transform
theory. In Fourier transform theory they appear more subtly. Let's look at an example to
see how this occurs.
7.4. An electron, initially at rest, is acted upon by an electric fielA fr1r; :
io"-"|for t > 0, where o is real and positive. The electron is also subject to a damping
Example
: -yi. Find the subsequent motion of the electron.
The equation satisfied by the electron's velocity is
force F4
d_
^fit+yi:-ei,1*..t)
First notice that the motion will be one dimensional. Then we take the Fourier transform of the entire equation with respect to time. Define the transform by the relations
i(r)
:
I
/+-. i{a1e-i't da
h J-*
and
ilar;
: I r+- i(r)ei't dt
h J_*
Then the transform of the electric field is
E1r;
: ErhI
:E-:- "
/"+-
J,
L-at
"-at
uiat 4,
ri<ot 16
J2r -a *
-11
: f^-,:
-l
"J2ra-ia
ialn
335
7.4 CAUSALITY
The range ofintegration reduces to positive values of /, since E
:
0 for
/ < 0.
Transforming the differential equation and making use of relation (7.7) for the
transform of the derivative, we find
-imai
I yi : -rfi,o!
E@-ia)
1
and thus
ilar;
-1
: eEs
Jz"
-
e
@
-
ico)(ima
1
Eo
^""
-
y)
1
(7.16)
J2n @*ia)(a-fiy/m7
Now we invert the transform:
i1r;:
hl_:
e)
' En I'
nx " Jzr
1
ko
* ia)(o *
iy lm) "-iat
7.
We can use the residue theorem to evaluate the integral if we close the contour with
a big semicircle.8 For / > 0 we close the semicircle downward, while for r < 0
we must close upward. But for / < 0 the electron has felt no force, and so i must
remain zero. Thus, causality9 requires that the transform have no poles in the upper
half-planel The poles of our integrand are at (D : -ia and o : -iy /m, and both
are in the lower half-plane (Figure 7.4). Notice that one of these poles is contributed
by the differential equation (-iy lm), and one by the driving force (-icv). It is at this
point that the initial conditions enter our solution.
Im
(rrr)
Re (rrr)
FIGURE 7.4. Contour for inverting the transform (7.16) when r > 0. Both poles are in the lower
half-plane and thus are inside the contour.
8see Chapter 2, Section 2.7.3, especially Examples 2.20
and2.2l.
9This argument proves important in deciding where to put the integration path when there is a pole on the real
axis. See Chapter 2, Section 2.7.3.
336
CHAPTER
7
FOURIER TRANSFORMS
For / > 0 we integrate around the lower contour. The contour is traversed clockwise,
so we introduce a minus sign. The solution for r > 0 is
i(r)
/
: !!ior-ro,,
2rm
/m)t
e-i(-ia)t
\r-n+ota*
"-i(-iy
(-iy
/m -f ia)
:
'"o (r-ot - r-hty
am-y
It is easy to verify that i : O at t :0, and also that i
-+ 0 as t -> oo.
7.5. USE OF FOURIERTRANSFORMS INTHE SOLUTION
OF PARTIAL DIFFERENTIAL EQUATIONS
The solution of a partial differential equation will be a function of two or more variables. If
the equation is linear, we can apply the transform operator in one or more of the variables
to obtain either an ordinary'differential equation or an algebraic equation. In this section,
we show how to apply these techniques to two common partial differential equations.
7.5.1. The Wave Equation
J.5. Suppose an infinitely long cable is pulled up at .r : 0, so that its
shape is described by the function f (x),: \s-lxl/a, and then let go. What is the
subsequent motion of the cable?
The equation of motion for the cable is the wave equation (3.15):
Exampte
o2y: ui*'
,ozy
aP
Let's transform the equation with respect to space and time. We define the transform
as
follows:
i(t<,a):
*rE
y(x, t)
"-ikx-tiat
4*
49
I
l
l
and, conversely,
y(x,t): !2n [** f* ,rn,11"ik*-i't
J-* J-* "Then the transformed equation is
a-
i
a
a-
-(t)'Y : -U'k'Y
ko-uk)(a*uk)j:g
dkao
IN THE SOLUTION OF PARTIAL DIFFEBENTIAL
7.5 USE OF FOURIER TRANSFORMS
EQUATTONS 337
One solution to this equation is the trivial solution i : 0, but this is not the solution
we want. We might be tempted to say that the equation is solved by taking o - *.uk,
but we want a solution for the transform i@, k). The transform is zero except where
a : fsft-v
property that is reminiscent of delta functions. The nonzero solution
that we need is
j :
A(k)6(a
-
uk)
+ B(k)8@ * uk)
since we have already established that x6(x) is the zero distribution (Chapter 6,
Section 6.3). The equation becomes
+ B(k)(ot - uk)(co * uk)6(to'f uk)
: A(k)(to* uk) x O * B(k)(c,t- uft) x 0 : 0
A(k)(a
*
uk)(a
-
uk)6(a
-
uk)
as required.
Then the solution for y(.r,
y(x, t)
t) is
: I /+oo /"+oo tattlo
; J _* ;/__
(o
-
uk)
*
*
B(k)s(a
uk)l ,ikx-iat 4p 4,
: ! [** ,ik'[A(k)"-iukt * B(k)e+i'ktf dk
2n
J-*
This expression shows that the displacement of the cable is a sum of rightward-moving
and leftward-moving waves.
0y
To solve for the remaining unknowns, we must use the initial conditions that
/0t :0 and y - 11"-lxl/a at t : O. Thus,
y(r,0)
:*
I::
,ik'tA(k)+ B(ft)l dk
-
he-txt/a
and
#1,:,:
* I::
iukeik*1-A1k) +
B(k))dk:
o
From the second of these relations we conclude that A : B, while from the first we
may make use of relation (7.3) with a : l/a to write the transform of the initial
condition on y:
t, ha
Y
:
1-...
alL\
zl+kzaz tE-""-'
-'l
A(K):
ha
TTFT
So, finally, our solution is
t- nu
r-Foo
^ikx
I
:----(e-ivt't + e+iukt)dk
y(x,t\ "- 2o J-* lrk2al
338
7
CHAPTER
FOUFIER TRANSFORMS
To do this integral, we use the residue theorem:
hf+*1
^
y(x,t)--L,
ztra J-* &+i/a)(k-ilat
tlsikx-ivkt+eikx+iukt)dk
There are two simple poles on the imaginary axis at k : Ii /a.
lst term: Case I: x - ut < 0. We must close the contour downward, thus enclosing
the pole at k : -i / a. Then the integral along the big semicircle is zero, by Jordan's
lemma. We traverse the contour clockwise. Thus,
2ni ei(-i)(x-ut)/a
:;i ,k-ut)/a
it:_*ii
z
Ji
Case
k
: li
II: x
- ut > 0. We must close the contour
upward, enclosing the pole at
/a. The result is
tr:ih2ni
2nd term: Case III: x
*
ut <
2n
"i(i)(x-ut)/a
x
:h "-(x-vt)/a
2
0. We close downward and get
i
ex+Dt
tz---=-,.-- "i(-i)(x1-ur)/a
-,.
Lft
-ztCase [V:
x
I
ut >
/a
z^
0. We close upward. The result is
iz:
2ni ei(i)(x*utl/a
*ii
,
:;i
e-@+Dt)/a
z
Finally, we put all this together:
If x <
-ut
for
/ > 0, then also x <
y(x,t) If
-ut
h "(x-ut)/a
u/, so we have Case I and Case III:
I
"@)-vt)/a
:
he"/o"orh
A
a
< x < ut, then we have Case I and Case IV:
Y(x't) - oe(x-ut)/a
+2e-G+u')/a
-
|rr-ut/a cosh 1
a
If x > ut, then we have Case I[ and Case IV:
4
Y\x,t): h"-(x-rt)la 2-"-G*ut)/a -
he-xlo
{
"orh a
This solution is shown in Figure 7.5. The initial peak at x : 0 propagates both
rightward and leftward along the string and decreases in magnitude with time. The
7.5 USE OF FOURIER TBANSFORMS
339
IN THE SOLUTION OF PARTIAL DIFFERENTIAL EOUATIONS
displacement is never negative, so the string does not oscillate as we might have
expected. This is a consequence of the infinite length of the string and the fact that
the wave is propagating in one dimension (along a line-the string).
Y(x, t)
I
I
0.F
0.6
..'a't.
\
--::.....
!a
-5-4-3-2-1012345
FIGURE 7.5. Shape of the string in Example 1.5 at
ut
/a:
0, l, and 2.
7.5.2. The Diffusion Equation
7.6. At t : 0, a small amount of mud of mass rn is introduced at the point
:
x { into an infinitely long pipe of cross-sectional area A, containing fresh water.
Example
Determine the distribution of mud in the pipe at times t > 0.
The appropriate differential equation is the diffusion equation (equation
3.
l4):
Dp
. _n _o2p
E-"ar2
where D is the diffusion coefficient for the problem. The boundary conditions are
p:0forallxfort < 0,andp -+ 0asx + oo. Att:0,allthemudisconcentrated
at one point. We may model the density as a delta function:
p(x,O):YX, - t)
Since the differential equation is first order in time, we may be successful by taking
the Fourier transform with respect to x only:
F&,t):
I
h
/^+-
J_*
or*,t1e-ik* dx
Transforming the whole equation with respect to.r, we get the first-order equation
ai :
At
-k2
Di
340
oHAPTERTFoURTERTRANsFoRMs
which has the solution
i :
The initial conditions determine
io
fu"-k2Dt
/s:
: v(k,q :
h lll ,r,,01 e-ik'
dx
1 /.+oo
::+m
I tk - t\e'ik, dx
AJ2n J-*
:7*e-ikl
m
Thus, we have
v&,t):|ff"-|"'
and transforming back gives
p(x,t):ft+,f*"^o
(ikx
-
ikt
-
k2D\dk
To do the integral, we complete the square:
k2
Dt + ik((.
-o
:
(*Jot
.,#)'
*
la =!f
Thus,
p(x,t):T+^,(-i+) /_ *, (r-*,?)'foo
lNow let u
:
kJ Dt +
p(x,,)
i(t - x)/2.,,fii,
: T*
"-,
m I
'72"trDt
and thus du
: tDt
(-#) t::::H
dk. Then we have
e-"'
:h
( (x-l)2\
"^o
\- o- /
Thus, the mud distribution is a Gaussian that spreads with time while the maximum
density decreases (Figure 7.6).
7.6 FOURIER TRANSFORMS AND POWER SPECTRA
341
4 p(x)
m
(''... /'2
S-4.,
-5-4-3-2-1012345
FIGURE 7.6. Distributionof mudin Example 7.6for
and 2 m (dotted line) and I : 0.
x(m)
JDt : l12m(solidline), I m (dashedline),
7.6. FOURIER TRANSFORMS AND POWER SPECTRA
The Fourier transform relates the time dependence of a function f(t) to the frequency
dependence ofthe transforn F(a). Thus, it is a useful tool for analyzing the frequency
components present in a physical system. One of the most common applications is to the
electromagnetic power radiated by a system.
An accelerated charge radiates. The power radiated per unit area of wavefront is given
by the Poynting flux:
S=
ExB
Po
Far from the source,
i
is perpendicular to both fi and fr:
k(E.E)
Using
dA
:
- E(k.E)
r2d{2, we find that the power radiated per unit solid angle is
4! :rr13,1-"t'
dA
LLoc
342
CHAPTER
7
FOURIER TRANSFORMS
where E and hence P are functions of time. Then the total energy radiated per unit solid
angle is
ry:
[** Lfr'ro.fr,(Dat
dA J-Foc
Using Parseval's theorem (equation 7.71), we may write this in terms of an integral over
the transform of E:
ry:
[** Li.td'fr:@)aa
dQ J-tLoc
Thus, the energy radiated per unit solid angle per unit angular frequency is
+:da
Li,@).i.(,)
dQ
(7.r7)
LLoc
which is the power spectrum.
The electric field due to an accelerated charge in nonrelativistic motion has two parts: the
Coulombfieldthatdecreases aslf r2 andtheradiationfieldthatdecreases asl/r.Thus,at
large distances from the charge, the electric field is dominated by the radiation field, which
takes the form (Jackson, Chapter 14)
*E: skx(txd)
4nw+
:
d2*/ dt2 is the acceleration of the charge and fi is a unit vector along the direction
ofpropagation. Therefore, the energy radiated per unit solid angle per unit angular frequency
is independent of the distance r to the source, and d2 W / dQ do may be expressed in terms
of the time transform of the charge's position, since d(ar) : -ro2i.(to) (by the derivative
property, Section'7 .3.3):
where d
d2w :
'" ^
:
q2
iN&;AFaa
)
^,1-,,
cr (4n)zes
where d is the angle between d and [.
Integrating over the angles, with pc
dw :
;
q2
1*'1a112 sin2 e
:
,a
lir(r)12
sin2
(7.18)
e
cos 9, we find
r*l
4,0
l*(,)12 (r",
J_, -
Fffi,a
r,2)drt
: r!3 cs(4n)zeo
,,1'.^ ,a 1fi.(r)lz
: -{ra
bft c" eo
Bko)12
('t.ts)
343
7.7 SINE AND COSINETRANSFORMS
7.7. Find the power spectrum radiated by the electron whose motion we
considered in Example 7.4.
Example
Here
it is more
convenient to write d1a-l; ln terms of
i(ro) so that we can use
equation (7.16) forthe transform ofthe electron velocity. The electron radiates energy
per unit angular frequency:
I _l'
I
e2 .ledw _ q2 .,2e , :
lv(all' l-En
---:-tD- t'r-tt
6o1310* li"t J2" @*irr)@+iy/m)l
d, 6113^,2
,4 E3
-0)z
(7.20)
-
12n2*2r3 eo (a2
*
a2)la2 + 0 /m)21
The power spectrum (equation 7 .20) is shown in Figure -7 .7 . The spectrum peaks at
o-
Juffi.
dw
dt
I
L
I
t
I
i
45
Id
6'7
FIGURE 7.7. Power spectrum radiated by the electron in Example 7.7 for y /m
and y
lm - 2a (dashed line).
The frequency variable is
:
0.5o (solid line)
o/u.
7.7. SINE AND COSINE TRANSFORMS
In Chapter 4, we saw that the Fourier series of an even function contains only cosines while
the series of an odd function contains only sines. Similarly, we can construct sine and cosine
transforms for odd and even functions.
Sine and cosine transforns are used when either
1. the function f (x) is known to be even or odd or
2. the function is defined only forx > 0.
344
7
.7.1
Let
cHAPTEFTFoURTERTRANSFoBMS
. The Cosine Transform
f (x) be defined for x >
range
0, and let us make an even extension of the function to the
.r < 0. Then the Fourier transform of the extended function is
F(k):
# I::
ysle-ikr
dx
: -* (lt
(x)
*ral "-ikx 4, * Io** f e-ik' dx)
:
(i
# (- /: f G,u) "ik' du * lo** f e-tk' dx)
:
# lo** fr*rrrikx + e-ikx) dlx
,
F(k)
: ,l:
lr.* /(x)
cos kx
dx = F"(k)
/(r).
which is the Fourier cosine transform of the original function
transform is an even function of /c:
Fc(-k) :
?
7f
lo**
/(x)
Io** /(x)
cos
(7.21)
Note that the cosine
(-kx) dx
cos kx
dx
:
Fc(k)
The inverse transform is given by
6
.f
r+o
(x): rl1 I
Y 7t Jo
F"(k)coskxdk
To verify this expression, we expand out the cosine:
f(x):
:
h
(1.-
F"(k) eik' dk
+ Io** r"Q<) e-ik' ate)
(7.22)
l
l
I
7.7 SINE AND COSINE
Now let rc --
-k
in the second term, and use the fact that F"(ft) is an even function of k:
f (x) :
:
Thus, equatiot
7.7
.2. The
TRANSFORMS 345
(7
Si ne
h (lr*
#
I**
F"1k1eik' oo
'"n'
"ik'
-
Io-*
F.(-r) ,i-' a*)
dk
.22) corresponds to the usual inversion relation (7.2).
Transform
The sine transform is defined similarly by making an odd extension of
/.
The Fourier
transform of the extended function is
F(k):
:
:
# I_: r@)e-ik'dx
h (l'*ro, "-ikx 6* *
h(-
/:
f ?u)
"iku
du
lo**
*
Io**
f (i e-tk- dx)
d*
lo** fr*rr"-ikx - "ik';
r+*
: -i tl lT
I
/(x) sin k-r dx : -i
UTJo
=
f (x) e-ik' dx)
h
F"(k)
where the sine transform is defined by
F.(k) =
(7.23)
which is an odd function of fr. Again the inverse is
f (x)
: ,l 1 lr.* F,(k) sinkx dk
(7.24)
346
cHAPTERzFoURTERTRANSFoRMS
Let's verify equation (7.24):
tl
lT r+* Fs(k) sinkx dk :
; J,
tl:*/*-
nror< eikx
: ,11+(1.*
: # I**
-
e-ikx)dk
'!!,' "'o' oo
oro,
"ik'
dk
* lo-* Fl--*) "'.' d*)
: f (x)
as required.
7.7.3. Use of the Sine and GosineTransforms
Like the Laplace and Fourier transforms, the sine and cosine transforms prove to be useful
tools for solving differential equations. Thus, wti: will need the transforms of derivatives of
a function in terms of the transform of the original function. Let's start by finding the cosine
transform ofdfldx:
t;
: _V;f
ro1+ftrs(k)
The transform of the derivative brings in the initial condition /(0), as is the case with the
Laplace transform, but it also mixes up the sine and cosine transforms. The Fourier cosine
transform of df I dx is k times the Fourier sine transform of / (x). A similar mixing occurs
with the sine transform:
^(#):,11 l,** ffsinkxdx
,11(tsinkxlf -olo**
:
_kF"(k)
This time no initial conditions appear.
f
cosrcxdx)
7.7 SINE AND COSINETRANSFORMS
347
We get back the transform we started with (sine or cosine) when we look at the second
derivative:
#.:o.r^(#)
4!^
,:o+
df
dx ,:o
kl-kFc&)l
-
k2
p.(k)
(7.2s)
and
'"(#)
: _r^(#) :
:
(#)
^
r,
-/. (-
|]
17 r <ot -
k2
rror+
k*(k))
r"(k)
(7.26)
These results suggest that the sine and cosine transforms will be most useful in solving
differential equations that have only even derivatives or only odd derivatives. The choice
of transform will be determined by the initial conditions that are given. For second-order
equations:
. If the value of the function is known at x : 0, use the sine transform.
. If the value of the first derivative of the function is known at x :
0, use the cosine
transform.
If we were to use the Laplace transform method, we would nee d both initial conditions . It
appears that we can get away with only one condition if we use the sine or cosine transform.
But it is implicit in the Fourier transform (sine or cosine) that the function /(x) approach
zero as r --> oo, and this provides our second boundary condition. No such constraint is
needed with the Laplace transform.
Example
7.8.
Use a Fourier ffansform to solve the equation
d2v
77-d"Y:o
with the boundary conditions y(0)
:
y0 and
y -+ 0 as r
+
oo.
348
oHAPTERTFoURTERTRANSFoRMS
Since we are given y(0), we use the sine transform:
"IT
-k2f,(v) + r1f
;n - o2F,1v1 = s
The solution for the transform is
l,
,rvn
F,(y): l;eld
Now the inverse is
y(x):
,l: I,* li;+rsinkxdk
The integrand is even in rk, so we can extend the range of integration:
I /Poo kvn ,ikx - ,-ikx
t\x):;J__Ffr
, ar
We use the residue theorem to evaluate the integral. The integand has poles at k :
*ia. For the first term, we close the contour upward (remember t t 0), and the
integral around the big semicircle is zero, by Jordan's lemma. Only the pole at k :
*ic is inside the contour (see Figure 7.1), and we get
1 [*
L_
,!!e
t ro
dk _
,i{ro1, _
2ia
2ri J -* & * ia)(k - iq)"ik,
!.o
2
"-ax
Similarly, for the second term we close downward, enclosing the pole at k
Iz:
I
/Poo
fr J--G *
kyo
ia)(k
- ia)"-ikx
41,
: -ia:
- _-ioro "-i(-ia)x - -l!2-,-ox
-2ia
where the extra minus sign accounts for the fact that we go around the contour
clockwise. Combining the two terms, we find
)(x): h-Iz:loe-ax
which is the expected result.
Let's
x
:0,
see
how the solution goes using the Laplace transform. We don't k'now dy
so let's call
it b.
r2Y(r)-slo- b-a2Y:o
-i------ru-slo*b
- s' d,'
-
/dx
at
TRANSFORMS 349
7.7 SINE AND COSINE
Inverting,lo we get
:
y(x)
yo cosh cr.r
+4
a
sinh
cx
Now the second boundary condition requires that the positive exponential terms in this
result sum to zero, so we need
yo+!:o+
b:-ato
a
Then the solution is
y(x):
yg(coshcux
-
sinhcux)
:
yoe-dx
as we obtained before.
If we can always use the Laplace transform in this way, why should we bother with sine
and cosine transforms? The answer is that we cannot always use the Laplace transform.
7.9. Suppose the system in Example 7.8 is. now driven by a function /(x),
where the exact functional form is for the moment unknown. Find y(x) in terms of
Example
f
(x).
In this case, the Laplace transform becomes
b
F
rz-qz- s2-.,2
svn -l-
i:
To invert the second term, we use the
convolution theorem (equation
5
.
17). The inverse
is
y(x)
:y0
coshcvr
+
I
sinhar
aJoq
* ['
,rrrsirtha(x
-
E)
o,
Until we evaluate the convolution integral, we cannot make use of the boundary
condition at infinity. However, if we use the sine transform, we get
.fs())
l, kyo- F,(f)
: y;F
*7 trTA
To invert this expression, we need to work out the convolution theorem for the sine
transform:
r;ttt"(f)a(s)l
: rl:
Ir*
F"(f)F,(d
sinkx dk
lZ /^oo I/ lZ
- foo g)sinkt d€ \
F,@
| ; J,
f
ll ; J,
)
sinkx dk
:?ltJo[* f @ Jo
[* r,(s)]tcosr(x
-f) -cosk(r +€)tdkd€
z
louse partial fractions and Table 5.1, or use the Mellin inversion integral.
3s0
CHAPTER
7
FOURIER TRANSFORMS
F;t[F"(f)F,(g)]
:
r[*
Jzn
Jo
f G)tE@ -
5)
-
E@
+ €)ld€
(7.27)
where
E@): F;rtF,(dl
is the inverse cosine tansform of .Fr(g).
To solve our problem, we need the inverse of G(t) : ll(k2 + q.2). This function
is even, but fortunately we need its inverse cosine transform. The method closely
follows what we did in Example 7.8, the only difference being the lack of the factor
k in the numerator. (See also Example 7.1.)
i@):
r;' (Fld)
:
:
I
:
f* 'ikx 1 e-ik' "
& l-* 4*'ao4"*
li ,-"'
2tri f ,itir)* ,-i(-ia)x\
ft
(td6 - t-,6
): ''li;
(x > o)
greater than zero, x - f is not, because f ranges from zero
since
we took the inverse cosine transform, the function f(x)
to infinity. However,
this
example,
must be even. In
E@) : Ji-f2(s-ulxl/cv). Then the solution of our
is
differential equation
Although
x is always
v(x)
:
roe-o,
*
*
Ir*
f G)@-"1,-E1 _ r-u(x+€)y 6q
: loe-dx * * (lr. f G) s-a(x-1) aE + l,*
-*
Ir*
rro e"@-il d€)
f (ile-d(x+€) d€
: e-ax (*. *
d€
lo' rcleo€ - *
*# l,* f c)e-"€
Ir*
rlsve-"e as)
aq
which clearly obeys the condition y(x) + 0 as .r
/, provided that its Fourier transform exists.
-+
oo no matter what the function
PROBLEMS 351
We conclude this discussion by finding the convolution theorem for the cosine transform:
r;tlr"$)F"(s)l
:
rlZ
f
U 7T Jo
F"u)F"(s)coskx dk
: l? ,"G) (,li
coskxdk
l,*
lo* rc,cos/,6d6)
:1
F;tlF"(f)F"(s)l
lr*
:
r"Glf @)tcosk(x -
llooI
Jn
Jo
f G)tsM
-
5) + cosk(x + €)ldkd€
6l) + s(x + E)ld€
(7.28)
This time the functions are not mismatched with the transform, but we must take the even
extension
ofg
when evaluating S@
-
€) for
f > r.
PROBLEMS
1. Find the Fourier transform of the following functions, and verify your results by computing the inverse transform.
" _x2-l 4x-ll3
(a)
1
(b) e-"'2
cos px
(c) I {x)
@ coshax
1"1
x<
otr,.r*ir.
: *
[o
l(t) :
if 0 <
(
(rr-o'
[o
if
I
t>o
otherwise
(0 --+---t
x'+a'
2. Invert the following
(a)
F(ft):
transforms to find the corresponding functions:
t-2ik
l+4k2
I
l(b)lr(k):11,.1r
(c) F(ft) :
1
i sinh ak
-
352
CHAPTER
7
FOURIER TRANSFORMS
istheFouriertransform of f(x), showthat idF/dk isthetransform ofxf(x).
What conditions must F(ft) satisfy for this result to hold?
4. Verify Parseval's theorem in the form of equation (7.10) by evaluating the transforms
of the functions /(x) : cos pr and g(.r) : s-ax2 and evaluating the two integrals in
3.
IfF(t)
equation (7.10).
S
Verify Parseval's theorem in the form ofequation (7.11) by evaluating the transform of
(t if -1<x<l
[o otherwise
/(x) :
f (x)2 and F (t)
F(/c) isthetransform of f (x),then(lla)F(k/a)isthetransform
2
and evaluating the integral s of
6. Provethatif
7.
I
|
.
of
f
(ax).
Show that the result is consistent with Parseval's theorem.
Find the Fourier transform of the function
:
r(t)
(Acosri,nt if
-T<t<T
[o
otherwise
2
that represents a finite train of data. Plot the Fourier power spectrum I F (ar) | as a function
I andaoT 10. Comment. What happens as 7 increases
of arT forthe two cases ar6Z
:
:
toward infinity?
8. Find the Fourier transform of
if -T<t<T
r,-\ : [l-vtt,
\o
'l trr
otherwise
Hence find the transform of the function
(t
I
g(r):(-1
[o
S
if -T<r<o
ifO<r<I
otherwise
Strow that the square deviation between two functions,
o: [**
J-a
V@)
-
g(x)12 dx
equals the square deviation between the transforms,
o:
r+oo
J_* tF(ft) -
G(DP dk
10. A spring-and-dashpot system satisfies the equation
d2x
dx
.
AV+2aV+ofix:f(t)
I
I
l
PROBLEMS
with
@0
>
cy.
The driving force per unit mass
f (t) :
for
11.
t > 0. Find x(l)
for
"f
(t) is zero for
353
/ < 0 and
e-ot sin Q/
t > 0, and verify thatyour
method gives
x
:
0 for
/ < 0.
An electron in an atom may be modeled classically as a damped harmonic oscillator
(compare with Problem l0 above). The electron is driven by an incoming EM wave with
electric field E(t) : Eo(sin At)/fu for -oo < t < oo. What is the appropriate /(l)
for this problem? Solve for the transform .r(ar) of the electron's position.
Use the results of Section 7.6 to determine the power spectrum of the radiated energy.
Plotyourresultsinthecase q
-
10,9:2on.
ogf
Comment.
12. The electric displacement D is related to the electric field B by the dielectric constant €.
In general, e is a function of frequency, so the relationship is one between the Fourier
transforms of D and E:
rt@,
(a)
r'>)
: e(tDfr,.;, a)
Show that the relationship between f,1x,
D(x, r)
:
E(r, t)
r; and i(-r, t) is
/n
OO
+ I G(z)E(x, t J-*
t) dt
and determine an expression for G(r) in terms of e(co).
(b) Find G(t) for the one-resonance model
e(0))-r*a#_*
where ap, ar6, and y are real positive constants and y < ag.
(c) Discuss the physical meaning of your result. Be specific.
An electron in an atom may be represented by a damped harmonic
trl quency
i
&D ang damping rate f . (Compare with Problem 10 with 2a
electric field E0) acts on the electron. Find the Fourier transform i
position as a function of time. If the electron loses energy at a rate P
use Parseval's theorem to show that the total energy loss is
i
LU:-
I
e2
m
/^oo
I
J-* @?o-rt)t+@:212
Note that the integmnd is sharply peaked at @ - a,l6, while E(ar) is a slowly varying
function, and so the integral may be approximated as
LU
:
-2
LtfrtuDP
.oo
a2l
I _*6=#9**0,
Evaluate the integral by contour integration to show that AU is independent of f , and
AU. (In this expression, @s and f are real positive constants, and cr.ro > f .)
hence find
354
CHAPTER
7
FOURIER TRANSFORMS
14. The radon problem. Radon diffuses from the ground into the atmosphere at a rate
r (atoms/m2.s). Model the atmosphere as a semi-infinite medium with boundary (the
ground) at y : 0. Then the density p(y,t) of atmospheric radon is described by the
equation
Do
02o
u-"aF' -)'P
where D is the appropriate diffusion coefficient and
boundary condition at the ground is
apl
d lr-
:
constant
:
I
is the decay rate for radon. The
-cY
What is the boundary condition at y --> oo? Use the Fourier cosine transform in y to
derive an integral expression for p(y, r) in the case that p(y,O): 0. Evaluate 0pl1t at
/ : 0, and hence determine a in terms of r and D.
Extra credit: Obtain expressions for p(0, t) and p(y, oo), and obtain p(y, r) as an
integral over /.
: I cm2 is initially at 15oC. At time / :
15. A long copper rod of cross-sectional area A
one end (at x
0) is placed into a vat of hot
-
0,
oil at 300'C.
(a) Refer to Chapter 3, Section 3.2.5. Write the equation lhat describes the change of
temperature at position x along the rod at time /.
(b) Write an expression for the temperature I(x) of the rod immediately after the end
is placed in the oil.
(c) Discuss
the use of Fourier and/or Laplace transforms in solving this equation. What
determines the best choice of transform for this problem?
(d) Find
the temperature of the rod as a function of position and time for
/ > 0.
(e) Given the following data for copper, plot the temperature along the first 5 m of the
/ : 0.5, 1.5, 3.0, and 6.0 s. Thermal conductivity: 400
heat: 385 J/kg.K; density: 8.96 kg/m3.
rod at times
16.
Wm.K; specific
A long beam is resting on an elastic foundation. The equation satisfied by the beam
displacement is
ur
d4v
lF
:
q@)
- ay(x)
4(x) is the load and q is a constant describing the elastic properties ofthe foundation. If the load is concentrated toward the center of the beam, then we may assume
that y -+ 0 as x -+ too. Transform the equation, and find Y(k) in terms of Q(k). Solve
for the beam displacement if
where
(a) q(x)
(b) q(x)
: Mg6(x - a)
: (MglL)[S(x - a + L/2) -
S(x
- a - L/2)l
PROBLEMS 355
e-t sinx, find
(a) the Fourier sine transform
(b) the Fouriercosine transform
18. For the function x€-o', where c is a real positive constant, find
17. For the function
@
(b)
m" Fourier sine transform
the Fourier cosine transform
thefunction
19. ShowthattheFouriercosinetransformof
lz t
xp-r for0 < p < 1is
D,t
,,1;*cos7r(P)
Hence show that the function 1 / Ji is its own cosine transform. Obtain similar results
for the sine transform. (The results of Chapter 2, Section 2.9 may prove useful.)
[Z0l Oetermine the form of Parseval's theorem (equation 7.10 and 7.ll) that applies to the
cosine transform.
21. The magnetic field in a conducting medium diffrrses away according to the equation
TH(x,t)
02H(x,t):
lto
ox2
at
Solve this equation by taking the Fourier transform in space. Find I1(.r,
netic field att :0 is a step function:
t) if
the mag-
(no it-d/2<x<d/2
|.0 otherwise
II(x.0) : {
Express your answer in terms of the error function (Appendix IX).
22. At t : 0, the distribution of chlorine in a pipe of water is given by
p(x,
O)
:
pos-'2/o2
t > 0 in terms
of the diffusion coefficient D.
E3.l Develop a three-dimensional version of the convolution theorem. Use the result to obtain
the solution of Poisson's equation
Find the distribution of chlorine for
V2(D
: -P(i)
60
f,
as an integral over space. Hint: lJse spherical coordinates to do the integration ou",
and use the principal value in the integral over k. Evaluate the resulting integral explicitly
if p(i)
:
46(i).
24. Find the three-dimensional Fourier transform of the charge distribution
p(i):
ooI!r
+7t
This transform is called the form factor of the charged particle.
356
25.
CHAPTER
Tiake the
7
FOURIER TRANSFORMS
Fourier transform of the three-dimensional wave equation
02s
ift-u'Y"s:f(*'t)
and solve for the transform S([, a.r). Show that the introduction of a damping force
(through the addition of a term u 0s /0t on the left-hand side) moves the poles off the
real axis. Invert the transform in the case cv -+ 0 for the cases
(a) (i, t) : s-rla61l), where r is distance from the origin
"f
(b) ,f (i, r) : 6(*)d(r)
26. At / : 0, the distribution of salt in a pipe of fresh water is given by
1\
p(x,o):pol/ sinax+ 4)
,*
Solve the diffusion equation (for example, equation 3.14 and Section 7.5.2) to find the
salt distribution at
t > 0 in terms of the diffirsion coefficient D. Hint: The result of
Problem 5 may prove useful.
27. Sum the
series
2p -11
i,-'," x2+(zp+t)2
p:0
by taking the Fourier transform of each term, summing the series in the transform space,
and then transforming back.
[2&l Use the derivative rule (7.6) and the symmetry property of Fourier transforms to evaluate the tralsform of x'. Check your result by ,inverting the transform.
29. The differential equation that describes the evolution of neutron density p is
opp
DY2P: S6(i)6(r)
Dt't -
-!_
r
is the neutron lifetime and D is the diffusion coefficient. The source is a point
source S at i : 0, / : 0, and the initial condition is p(i, 0) : 0.
Using any method of your choice, find the density p(i, t) of neutrons for t > 0.
where
CHAPTER 8
Sturm-Liouville Theory
8.1. THE STURM-LIOUVILLE PROBLEM
Many differential equations describing physical systems can be reduced to one or more
linear ordinary differential equations of the form
d/
dr\
@)E
dx \f
)
where u.r(x) > 0 on the range
conditions of the form
-
e@)Y
*
)"u(x)Y
:
(8.1)
0
o < x < b and the solution y(x)
qt
-l
dv
0 atx:
Ari
dx -
also obeys boundary
a
(8.2)
and
qzY
dv
* flz-i
dx -0
atx:b
We want to determine values of the constant ). for which there are nontrivial solutions
This is called the Sturm-Liouville problem. If
dyldx:
0 (Neumannconditions),otif
p
:
a :0
y(r).
the boundary condition simplifies to
0theconditionis y
:
0(Dirichletconditions).
The constants o and p cannot both be zero.
An example of a Sturm-Liouville problem that we have already seen (Chapter 4, Section 4.4.2) is the problem of waves on a string with fixed ends. Using separation of variables,
we reduced the partial differential equation to two ordinary differential equations
s2v
!+
+k2x:o
dxz
(8.3)
357
358
CHAPTER
8
STURM-LIOUVILLE THEORY
and
tI*k2u2T:o
dtz
with the boundary conditions in x:
X(0):X(L):O'
The differential equation for X is of the form (8.1) with /(x) = 1, g(x) : 0, and w(x) : 1.
Also, the boundary conditions have p; : 0, so they are Dirichlet conditions. The solutions
to this problem are a set of sine functions sin/cr with a specified set of values for k:
k : nr lL. These are the eigenvalues for the problem, and the resulting functions X"(x) :
sinnn x f L are the eigenfunctions. We also found that these functions have a property called
orthogonality (for example, equations 4.3 and 4.12):
- :
fL sinnirxsinmrx dx
a
,
J,
L
^
26^n
We shall see that all problems of Sturm-Liouville type have a similar structure-the general
solution is a linear combination of orthogonal eigenfunctions.
8.1.1. Orthogonalaty of the Eigenfunctions
First, let us show that solutions corresponding to different eigenvalues are orthogonal on
the interval (a,b) with respect to the weight function u.r(x). Suppose that there is a set
of solutions y^(x) to equation (8.1) called eigenfunctions, with corresponding eigenvalues
).., and that these solutions also satisfy the boundary conditions (8.2). Then for one solution
labeled
y.,
- s<aw t
* ('o'*)
)'^w(x)Y*
:
(8.4)
Q
while for a second solution labeled yr,
*(,*,*)
- g(x)y" l')'nw(x)y, -
(8.s)
Q
Now we multiply equation (8.a) by yr, multiply equation (8.5) by
term g(x)y-yn cancels, and we are left with
y^,
and subtract. The
,"ft(ratff) - ,** Q atff) t (),^ - )'n)w(x)v^vn : e
Next we integrate the whole equation over the range of interest, -{
:
a to
x
:
b:
(ratff)ldx-()"n-)"^) f,b '{*)Y^Y' a'
(8.6)
8.1 THE STURM-LIOUVILLE
PROBLEM 359
On the left-hand side, we integrate by parts. The first term is
o
,,* (ror*) dx: vnr(xr*u"-
b
dY'
,r*rdY^ o,
dx"'
dx
Clearly the integrated term vanishes if we have Dirichlet ()" : 0) or Neumann (dy^ ldx =
0) conditions at each of the two boundaries. In the general case, we may use the boundary
conditions (8.2) to eliminate the derivative yt at each boundary, giving
r(b)y,(b)(-;r-*r) - r(a)y,(a)(-;r-rr) - l"'
= -firrot,@)y*(b)
All three parts of this expression
+frf
are
<o)r,(a)ym@)
symmetricinm
- I,'
rofrffo.
,or*ffa-
and n. We get an
identical contribution
from the second term on the left-hand side of (8.6), and they subtract to give zero. Thus,
the left-hand side reduces to zero, and so the right-hand side must be zero too. There are
two possibilities:
o either
Lm: Ln and m:
whichl
means that
n
lm = !n,
oOf
w(x)y^y, dx
:
O
(8.7)
Equation (8.7) is the orthogonality integral that we set out to find. Notice that the eigenfunctions y*(x) are weighted by the function u(x). [In the case of the string problem
(equation 8.3), the weighting function is identically equal to 1.1
There are two other important cases in which the left-hand side of equation (8.6) reduces
to zeto. If the function f (x) has the value zero at x : a and x : b,then the integrated
term is zero no matter what the boundary conditions on y(x), provided that y(.r) remains
finite. We shall see that this is the case for Legendre's equation.2 Finally, if yy' f has period
(b - a), then the integrated term vanishes.
I
See Section 8. 1.3 for the degenerate case )"w
2Section 8.3.1.
:
Ln with m
I
n
l
l
l
360
cHAprER
I
sruRM-LrouvrLLE rHEoRy
When equation (8.7) is satisfied, the set of eigenfunctions yn(x) forms a complete
orthogonal set on the interval [a, b]. This means that any reasonably well-behaved function
/(-r) defined for a < x < b can be expanded in a series of eigenfunctions:
f (x):Donrn@)
(8.8)
n=O
where the coefficients an may be found, as we did with the Fourier series (Chapter 4), by
multiplying both sides of relation (8.8) by w(x)y^(x) and integrating over the range a to
b. Only the one term in the sum withm : /, survives the integration, and
um-
I! f @)v^(*)w(x)dx
(8.e)
Illv^@)lzw(x) dx
As we found with the Fourier series in Chapter 4, the sum converges to the function in
the mean; that is,
*ry,*
l,u [ro, -
L^r,or)
w(x)dx
:o
We can also obtain a useful relation, called the completeness relation, by inserting the
expression for a" (8.9) into equation (8.8):
f (x) :
fn:o I!
f t'')Y'(!')w(x')
dx'
In
,,rr,
where
7b
In: I
Ja
lyn(x)l2w(x)dx
Interchanging the sum and the integral, we have
yn@)yn(xt)w(x')
f(x): I,u ro'ri
-ax
In
,
,
l
361
8.1 THE STURM.LIOUVILLE PROBLEM
Since the quantity multiplying f (x') in the integrand exhibits the sifting property (Chapter 6,
equation 6.2),we may conclude that
i
yn(x)yn@')w(x')
:
In
n=O
6(x
-
x')
(8.10)
Equation (8.10) is the completeness relation for the set of eigenfunctions
yr(r).
8.1.2. Reality of the Eigenvalues
Even if the functions f (x), g(x), and u.r(x) are real, the eigenfunctions may be complex.
i" a solution to
Again equation (8.3) offers an example, since the complex function
"ikx
this equation. But, provided that the weight function is real and w(x) > 0 on (4, D), the
eigenvalues are real even if the eigenfunctions a.re not. To prove this result and demonstrate
the orthogonality relation, we take the complex conjugate of equation (8.1):
ft (r<aff) - sr'rf * Llw(x)vi:s
(8.1 1)
Using a method similar to that used in proving orthogonality, we multiply equation (8.1)
by yfi and equation (8.11) by y^ and subtract. Again the term in g(x) subtracts away, and
we have
,^ft (r<.t#) - *ft (ra>*) .
Now we integrate over the rarrge x
two terms by parts. We then have
(t-ratff
:
e to
- v;f @*)1,
x:
(^tr
- x^)w(x)v1v^:0
b, as we did before, integrating the first
: u, -
^D
I,
w@)vf,v^ dx
We can make the left-hand side zero with appropriate boundary conditions, as we did in the
case
ofreal eigenfunctions (Section 8.1.1). Then, if m : n,
0:
The integral is
(ln
-
),D
pb
I
Ja
> 0 if w(x) >
w|)y|yn dx :
0, since lynl2
)'n
and the eigenvalues are real.
t
,0
pb
(),n
-
),tr)
I
Ja
w@)lynl2 dx
no matter what the form of y,. Thus,
: )tl'
362
CHAPTER
8
STURM-LIOUVILLE THEORY
We also obtain the orthogonality integral for complex functions:
I"u
*'*""'n
: o'
dx
n+
(8.12)
m
8.1.3. Degeneracy
The above proofs fail if ),m - 1., for some m + n. Then we cannot conclude that the corresponding eigenfunctions y. and y, are orthogonal (athough in some cases they are). Ifthere
are N eigenfunctions that have the same eigenvalue, then we have an N-fold degeneracy.
In the case of double degeneracy, we can always construct two linear combinations of the
eigenfunctions so that these new functions are orthogonal. Degeneracy reflects a symmetry
of the underlying physical system.
Suppose that we are looking for solutions of equation (8.3) with periodic boundary conditions: y(0) : y(L) and ]'(0) : y/(Z). (Recall that this was the final case we considered
in establishing orthogonality in Section 8.1.1.) There are two eigenfunctions, sinknx and
eos knx , with the same eigenvalue'. k, : 2nn I L. In this case of double degeneracy, these
two eigenfunctions are orthogonal on the range x : 0 to x : l, as we established in
Chapter 4.
Homogeneity of space allows us to place the origin anywhere that we like. If we shift the
origin by an amount x6, then, for example, a new eigenfunction is
2nnxg
2nr(x-xo):sin 2nttxcos,j-cos
,
,
.
sin
2nnx 2nrxg
sin
,
L
which is a linear combination of the previous two functions. The eigenvalues remain the
ofthe system has not changed.
Example 3.15 (potential in a box) shows what can happen in two dimensions. In that
example, the equation for the functions of x and y was found to be
same, as they must, since the physical behavior
a2.f
. a2 f
ar'2+ uV+k'f
:o
The eigenfunctions are fnm : sin(nr x I I) sin(mr y / w), with corresponding eigenvalues
4- : 1n2 1t2 + m2 1w21t 2.If the box is square, I : u), the eigenfunctions f,^ nd f-n
rtt-" ttr" same eigenvalue, reflecting the symmetry of the square.3
8.1.4. The Sturm-Liouville Operator as a Self-Adioint Operator
We may write equation (8.1) in terms of an operator .C, where
Ly=
d/
E\f
dv\
rr>--=,
-
s@)y
3There are additional degeneracies here, since in some cases we can obtain the same value
values of m and n. See Problem 3 for another example.
^2
+ n2 with
several
8.2 USE OF STURM-LIOUVILLE THEOBY IN PHYSICS
so equation
363
(8.1) is
LY
-l )"w(x)Y(r) :
0
Then the left-hand side of equation (8.6) may be written
t;
We have already shown that
!nL!^ o*
-
I"u
y^Lyn dx
if the eigenfunctions y satisfy appropriate boundary conditions,
then this difference is zero, or
f"u
,'r,^ o, :
I"u
y*Ly, dx
The operator,C is then said to be self-adjoint.
8.2. USE OF STURM.LIOUVILLE THEORY IN PHYSICS
The Sturm-Liouville problem often arises in physics problems as a result of separating
variables in a partial differential equation. Thus, there are usually two or more differential
equations, each in a different variable, that are linked by a relation between the eigenvalues.
The general method4 of solution is as follows:
set of functions that satisfy each differential equation for arbitrary l.
which
coordinates have homogeneous boundary conditions (the function or
2. Determine
is
zero)
at both ends ofthe range. These conditions are necessary ifwe are
its derivative
to have a Sturm-Liouville problem.
3. Choose the eigenfunction that satisfies one ofthe zero boundary conditions'
1. Determine the
4. Choose the eigenvalue to satisfy the second zero boundary condition.
5. Repeat steps 3 and 4 for any other coordinates with two zero boundary conditions.
6. For the remaining coordinate, choose a solution that satisfies a zero boundary condition
at one of the boundaries. The remaining condition must be a nonzero condition if the
solution is nontrivial.
7. Write the general solution as a linear combination of the eigenfunctions you have found.
8. Use the orthogonality of the eigenfunctions to determine the unknown constants in the
linear combination.
Let's see how this plan works in an example.
8.1. Find the electrostatic potential inside an infinitely long rectangular
wave guide with conducting walls. The guide measures a x b. One of the sides of
length a is held at potential V; the other sides are grounded.
Example
4see also Example 3.15.
t
364
CHAPTER
8
STURM-LIOUVILLE THEORY
First we choose a coordinate system that fits the problem. With a rectangular system,
we choose Cartesian coordinates and put the origin at one corner. Then the interior
oftheguideisdefinedby0. x < a and0 < y < b (Figure8.1).Thepotential
is independent of z. The differential equation satisfied by the potential is Laplace's
equation:
:0
V2o
I
b
l_
'I
FIGURE E.1. The task in Example 8.1 is to determine the potential in the region O < x < a,
0 < y < b. The boundary conditions are that @ is zero on the sides at x : O, x : a,
y
:
b has potential V.
Separating variables, we let @
:
X(x)Y (y),
and
y
:
0. The side at
and the differential equation becomes
a2e a2o a2x
a2Y
-r
:
-r
: v
;---^;--T
dx' ;--.dy' ;--t,
dx'
dy'
Dividing through by
@
: XI,
we have
r
a2x
I
a2Y
xa7*Tarr:'
where the first term is a function of -r only and the second is a function of y only.
Thus, we can satisfy the equation for all values of x and y only if both terms are
constants:
I A2X
X 0x2 - -),
and
I
AzY
---A
Y 8,tt -1
l
8.2 USE OF STURM.LIOUVILLE THEORY IN PHYSICS
36s
We have chosen the same constant i, with the opposite sign in the second equation so
that the two terms sum to zero, as required.
Each of these equations is of Sturm-Liouville type with
w(x) : 1. The boundary conditions in.r are
f (x) = I, g(x) :
0, and
X(0):X(a):Q
making the problem in x (differential equation plus boundary conditons) a SturmLiouville problem. But the boundary conditions in y are
Y(0)
:0;
Y(b):
V
and thus are not of the correct Sturm-Liouville form.
Now let's follow the steps of the general method.
. By choosing the
Step 1. The solutions for X are exponentials of the form
"+t"fi,x
constant ). to be positive, we obtain complex exponentials, or sines and cosines that
are linear combinations of the complex exponentials. The y-equation is of the same
form and has real exponentials (or sinh r/-i,x, cosh JT'D as its solutions.
Step 2.We have O : 0 at two values of x, so we look at the functions of .t first.
Step 3. To satisfy the constraint X(0) : 0, we need a solution of the form sin kr,
where ). : k2 has been chosen to be positive.
Step 4. To obtain X(a) : 0, we must choose k so that ta is one of the zeros of
the sine function-thatis, ka : wr, ot k : nn /a, for some integer n. These are the
eigenvalues for the problem. Then the solutions for X are Xn : sinnnxf a.Tltese
are the eigenfunctions in the variable x.
Step 5 is not needed here.
Step 6. Now that the separation constant I has been determined-it is (nn la)2-the
equation for Y is
a2Y r nn t,2
Ar2: \;) '
and the solutions are real exponentials. The function Y must be zero at
appropriate linear combination that we need is the hyperbolic sine:
Y
:
sinh
/nfl
\
\;,
)
y
-
0, so the
Notice thatthe function I(y) contains the lengtha of the wave guide inthex-direction.
This happens because the equations are coupled through the separation constant that
was chosen to fit the boundary conditions in x.
Step 7. The general solution is a linear combination of the eigenfunctions we have
found:
O(x, y)
oo
nrfx
nTv
: ) .cn sin
sinh '
aa
n:I
-
366
CHAPTER
8
STURM-LIOUVILLE THEORY
Since each eigenfunction sin (nn x / a) sirr}r (nn y / a) is zero at the three boundaries
: O, x : a, and y : 0, so is the linear combination.
Step 8. The one remaining boundary condition is the value of O at ! : b:
x
V
:
oo
<D(x,
D
:rrn"inry
rinhno-b
n:laa
which is a Fourier sine series in x . We find the coefficients cn by using the orthogonality
of the sines. We multiply both sides by sin (mt x la) and integrate:
fa
mfix
/ V sin o
Jo
-dx
a /
mnx\p :
V- l-cos-ll
mft \
a
'/tO
/4 3
nfix nnb mnx
I ) cn sin a sinh a sin a
Jo7:,
-dx
3
nrb fa nrx mrx
) .cnsinh- / sin-sin-dr
a JO
The integral on the right-hand side is zero unless
:
14
o
3
,#f-cosmn
+ 1) : Icn
_lu
,.. : a
_ (_l)",1
;cm
m7fzQ
sinh
sinh
a
m,in which
n:l
V
a
case iteqaals a12:
nnbra
T (Zu,^)
mnb
_
Thus,
Cm: V
- (-t)-l
2lt
mn sinh(mnbla)
and the complete solution is
2vs.
O(x,y):-)1Tu
N:T
[1
- (-l)',]
n sinh (nnb
Since only odd values ofn contribute [1
for n odd], we may write n :2m * 7:
o(x.r,)-_4v3
)
2^2m
71 m:(l
/a)
- (-l)' :
sin
nftx
nftv
_
sinh '
AQ
0 for n even and 1
- (-l)" :2
(2m*l)ny
I
* |
|
(2m
I
.,
+ l;zx'l srnn
"
"
.-:OtT7W
| srnh
-
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL HARMONICS
367
. Let's see how this solution looks. Figure 8.2 shows the sum of the first ten terms
at various values of y, with b : 2a. Notice how the solution approaches the value
Q I V : 1 as y approaches b, while maintaining the value zero at the boundaries in x.
q
v
FIGURE 8.2. The solution for the potential with b :2a.The plot shows the first ten terms in the
series for @ versus x for several values of y. The dashed line represents Q(x , y) at
y : 0.5a; the dotted line, y : a; the dot-dash line, y : l.Sa;the heavy solid line,
) : l.8a; and the thin solid line, y - 1.9a.
8.3. PROBLEMS WITH SPHERICAL SYMMETRY:
SPHERICAL HARMONICS
Suppose our potential problem has spherical boundaries. Then we would like to solve the
problem in spherical coordinates. Let's look at Laplace's equation again:
^
vzo
la /"ao\ * r a / ao\+ I
: p;1";
) 1"* aa ('ine- ) 7R
a2a
e aor
:0
(8'r3)
We apply the same techniques that we used in the rectangular problem; only the details
change. We are looking for a solution of the form
E:
R(r)P(o)w(Q)
368
8
CHAPTEH
STUBM-LIOUVILLE THEORY
Substituting into the differential equation and dividing by O, we get
t a /"aR\
I
a/
aP\l
I
a2w
a,zu\'"*) + ,rriner, (ttn'r )A- wZRo aQ, =
To separate out an equation for @, we multiply the whole equation by 12 sinz 0:
sin2
o
o
R0r (r#)*sinB-L
I
ta2w
(""r#) -P + --:0
WA6t
Now the last term is a function of @ only, while the sum of the first two terms is a function
ofr and 0 only. Thus, ifthe solution is to satisfy the differential equation for a/l values of
r,0, and Q, each of these two pieces must equal a constant.
Often the region of interest is the inside or outside of a complete sphere. In such a region,
an increase of / by any integer multiple of 2n corresponds to the same physical point.
Thus, the function O must have the same value for Q - Qr and Q : Qr * 2n; that is, the
function W must be periodic with period 2r.We may achieve this behavior if we choose
the separation constant so that
I A2W
--:
W aqz
'l
--2
1
l
with m equal to an integer. Then the solutions are the periodic functions
( sinmd
W:1
or
w : eti^Q
lcosln@
The equation in
r
and d then becomes
a/ aP\l
e a /^aR\
l+sind-lsine-lR0r lr"
arl
a0\
ae/P -m2:o
\
sin2
Next, to separate the
r
and d dependences, we divide through by sin20 to get
| _
^' _n
.##('',#)
P sin2o-"
+*(r#)
The first term in this equation is a function of r only, while the sum of the last two terms is
a function of0 only. Again both pieces must be constant. The equation has separated:
t a I/.aR\
__
ROr\rz_0r/| -
k
(8.14)
369
8.3 PROBLEMS WITH SPHERICAL SYMMETRY SPHERICAL HARMONICS
and
I a/lsinp_l___*k:0
aP\l
m2
P
Ae
sinz
0
\
/
(8.
sin9 00
When working in spherical coordinates, changing variables to 1t
trick. Then dp, : - sin9 d0, and the theta equation becomes
d /
A
[(t -
m2
"dP\
u")
ou)
-
cos
I
is often a useful
*kP:o
r_ irP
rs)
(8.16)
Equation (8.16) is known as the associated Legendre equation. Let's begin our study ofthis
equation by tackling a special case.
8.3.1. Problems with Axisymmetry:The Legendre Polynomials
If the problem
a
has rotational symmetry about the polar axis, then the function W must be
constant (O is independent of @) and so m : O. Then equation (8.16) simplifies:
h(, - t,#)
*kP
:o
(8.17)
We can solve this equation by looking for a power series solution.5 The singular points of
the equation are at p,: *1, so we should be able to find a solution about /.r : 0 of the form
v:fo,p'
n:0
Substituting into the equation, we have
@@@@
D"@
n--0
- I)o,t n-2 -D"@ - l)a,p' - zDnanFn -l kl
n:0
where each power of
n:0
anp."
:
g
n:0
p must separately equal zero. The constant term in the equation is
2az*kao:0 * a2- -Loo
and the first power of trl has coefficient
3
x
2a3
-
2at + ka1 -
O
)
az
- or'J
'3 x2
For all higher powers, every term in the equation contributes. Looking at
p -f 2 in the first term and n - p in the rest, we find
(p + 2)(p
5See
Chapter 3, Section 3.3.3.
I
l)ara2
-
p@
-
l)ap
-
2pap
I
kao
-
0
p"P
, setttng n
37O
cHAprER 8 sruRM-LtouvtLLE THEoRY
and so the recursion relation is
Apl2:
Ap
p(p
-
l) +zp
-
p(p-tt)-k
t<
(p+2)(p+t)
o
(8.1 8)
(p +2)(p + t)
The first two relations we obtained can also be described by this formula with p : 0 and
p : l, respectivelY.
The solution we have obtained is valid for - 1 < l-L < 1, but the series does not converge
for p, - *1. This is a concern, since p - fl corresponds to I : 0 and & : -l to 0 : r.
These points are on the z-axis, where usually we do not expect the potential to blow up.
Thus, we need a solution that remains valid up to and including these points. We can solve
the problem by choosing the separation constant k so that the series terminates after a finite
number of terms. In particular, if we choose k to have the value
k:l(l+D
for some integer /, then according to the recursion relation (8.18),
aI+2:
and so every succeeding ao for
takes the form
al
t(t+t)-t(t+D :0
(t+2)(t+t)
p > I +2
d/
dp
\'
-tn-
is also zero. The differential equation (8.17) now
t,#) +l(l+l)P:0
(8.1e)
and the corresponding solutions are the Legendre polynomials
choose a6 (for even /) or al (for odd /) so that
hj-t).By
convention,
l
P;(1)
=
(8.20)
I
I
The recursion relation becomes
Ap!2:
p(p*1)-/(/+1)
Clp
(8.21)
(p+2)(p+t)
The first few polynomials are found as follows.
I : 0: The only nonzero coefficient is 46, which must equal
Ps(p)
:
l
I
I to make Ps(1)
:
1, 5q
(8.22)
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHEBICAL
HARMONICS 371
/ : 1: The only nonzero coefficient is at, and again we must take at
&(1) : 1. Thus,
h[-r):
I :2:There
tt
= |
to make
(8.23)
are two nonzero terms,
q2:ao(/-2x3\ --3ao
, ,/
and the subsequent an are all zero. Then
PzQD:as(l-3P2)
Evaluating this at pc : 1, we find
P2Q): a0e2):
1
* oo: -).
Thus,
pzQ-r):)ft*'-t>
(8.24)
Notice the pattern. We use the recursion relation to determine the nonzero coefficients as
multiples of the leading coefficient (as or a1). Then we evaluate the resulting polynomial
at p, : 1 and set the result equal to 1, thus determining the value ofthe leading coefficient.
Let's do one more.
I : 3: Applying the recursion relation (8.21) with / : 3, we find
/
hIl):ot(pr+
Evaluating at
p,:1
lx2-3x4,\
t"
3"2
)
gives
P3(r)
/ s\ :, ] a1- 3
: o' (.t
--;)
and so
PtQ.t):!rS*,-Z>
(8.2s)
372
cHAprERssruRM-LtouvtLLETHEoRY
The first four polynomials are shown in Figure 8.3.
P(tt")
l
j
j
1
l
I
!
FIGURE E.3. The first four Legendre polynomials. P6(g.) and P2Qt) are even tunctions; P1 (g.) and
fuQ.t) arc odd functions. All the functions are chosen to have the value I at p. - 1
(top right).
8.3.2. Solution for the Potential
Now that we have the function of d, let's return to the potential problem and solve for the
function of r. With the separation constant determined, equation (8.14) becomes
L (,r9!\:
a'\'ar)Solutions to this equation are powers of
r(/
{
r)R
r: rP, where
* (rT) : # (,'o,o-') : p@ + t)rp : t(t * t)rp
p - /; there is a second solution with p
/0 + 1), as required. We have
One solution has
and
p(p + 1) :
R:rtor+
: -(/ + 1). Then p I I :
-1,
(8.26)
8.3 PROBLEMS WITH SPHERICAL SYMMETRY SPHERICAL
HARMONICS 373
Thus, an axisymmetric potential may be expressed as
o(r. d)
:
f
(o,,, +
1)
r,ot
(8.27)
where the constants A1 and 8; must be determined by the boundary conditions in r.
8.3.3. Legendre Functions of the Second Kind
Noticethatbychoosingtheseparationconstantk: l(l+l)weforceoneofthetwosolutions
of equation (8.19) to terminate. The second solution does not terminate. This solution is
called the Legendre function of the second kind: QtQ-t). For example, with / : 0, the second
solution begins with a1 and contains only odd powers of trc. The recursion relation (8.21) is
Qp*2:ao
p
p(p+l)-O
1r*2yoag:api+2:ap-z
p-2
p-2 p
p p+2:ae-2 p+2
al
p+2
and none ofthe
a, is zero unless a1 is. Thus, the solution is
/ujus\
QoQt)--arlp*+++ +"'I
\35/
at. /l+p\
:ttn\t-r/
In this function, at is taken to be 1 so that
I /1+u\
eoo.t):r^\r_*)
(8.28)
This expression is valid for -1 < 1t, < l.
These functions Qt are not used in the solution of spherical potential problems, but they
do find uses in other situations. For example, they are used when solving potential problems
in spheroidal coordinates.6 Unlike the Legendre polynomials, which diverge at large values
6See Problem 13 inChapter 2 and Problem 9
in this chapter.
;
374
cHAprER 8 sruRM-LtouvtLLE THEoRY
of the argument, the Legendre functions of the second kind approach zero as the argument
tends to infinity. To obtain the limiting form for Qt(x), we solve the differential equation
in inverse powers of x. We find7
I
QrG) --> 't"-fo+1)
q;S
as
e*t
r
--+ oo
(8.2e)
8.3.4. Orthogonality of the Legendre Functions
The Legendre equation (8.19) is of the Sturm-Liouville form (8.1) with
fQD=t-t"2
SQr)
=
0
w(p) :1
and
The eigenvalue is ,1, : l(l + l). Even without any boundary conditions specified, except that
the function remain finite, the Legendre functions must be orthogonal on the range t- 1, ll
because
f (l): f(-l):0.
I*rt,,r*rr,,et)
dp.
:o
for t
I/
(8.30)
To make use of this relation in forming series expansions in Legendre polynomials (that is,
to use equation 8.9), we will need to find the value of the integral for I : l/. In the next few
sections, we shall collect some useful tools that will allow us to do that integral.
8.3.5. Properties of Legendre Polynomials
The Generating Function
Suppose we put a point charge 4 on the polar axis at a distance s from the origin (Figure 8.4).
Then the potentials at point P is
-^-
r
4T es
Q
D --
I
q
4T es
J?TF - zrs cose
TSee Problem I 1
8See, for example, Lea and Burke, Chapter 25, equation 25.9.
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL
HARMONICS 375
FIGURE 8.4. The potential at P due to a point chatge q on the e-axis at z - s may be expressed
in terms of Legendre polynomials using equation (8.27). It may also be written as
q
/4n esD.
whichwecanalsoexpressintheform(8.27).Weletx:rlsfotconvenience,andthenfor
r < s we can expand the denominator to get
-ct1q1
-
,fr
-
4ness
f-.7
J
4neos
Vt*"2-2;u
uf -
21611a
162
q (,_(*2-z*p)_Let/2\(-3/2\.., - "-...\
:_ G*
-2xtt)'* )
z * --T(x"
(t -
: #*('
: *t*
* "u -
*r, - 3r,2). )
(r +"r, (y) + x2 pzur)
which has the form (8.27) with A1
+...)
: q l(4n e6sl+1)
for each I and Bt
:
O.
Thus, we have
the identity
-oo
t-
\-
1/l-2xp,*xz
-
-./ n 2..'
(8.31)
-r:o
The function
G(x,
tr):
\fr
-z.nTp
(8.32)
376
cHAprERBsruRM-LtouvtLLETHEoRY
is called the generating function for the Legendre polynomials. We can use it to determine
several useful properties of the polynomials.
The Orthogonality Integral
We can obtain the integral we want by integrating the square of the generating function:
Il,'o,r,: l::
F;G?r.: l::f,
l=0
,,r*>i*,'' rurDo,
lt--0
: ii",*,' [*,t ptet)Py(p,) dp.
/=0
1/=0
The integral ofthe Pls is zero unless
G2 by achange of variable to u : I
oo
r+l
n*'
For
x<
I
J_,
/' (equation 8.30). Thus, evaluating
2x 1t
*
;lnl-":;lrt+
the integral
of
x2, we have
1
*
ll-x)2 6,
Ju*,j, ?
(l*x)2 1. 1*.r
:;Inl.
- -tn 1-x
O4
logarithm:
x2t+t
*3 xs
*T+"'+zt+t*
s
2/
:r(,
-
:
ProDPtQ.t)ap:
1, we may expand the
l+x
/
*t* h*
:
\
)
) ,2." I*,'
,,,rrr,e,)d*
Both sides of this equation contain only even powers of x. Equating the coefficients of each
power gives
hQt)hQt") dtt
(8.33)
which is the desired result.
Recursion Relations
Our next task is to find a set of relations between different Legendre polynomials. These
relations are known as recursion relations. We begin by finding a relation among polynomials with successive values of l: the pure recursion relation.9 First we differentiate the
9The name "pure" arises from the fact that this relation contains the eigenfunctions only and no derivatives appear.
HARMONICS 377
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL
generating function with respect to x:
AG:
ax
-@ - tt)
-zxpT6n
0
:
-(x 0
-
tt)
zr'tt +
x\t'
Rearranging, we obtain
(t-2xP'+*\Y:Qt-x)G
dx
We proceed by inserting the expansion of G in Legendre polynomials into this expression,
{1
\r
-
)t6
p*
"rl i/:o
txt-t ptet")
:
Q.t
- *)ir,
r1q"y
t:o
and gathering up in powers of .r:
Da *
r)xl+t pteD
The coefficient ofeach power
ulru
-
t:o
ofx
lht1l) -
pt1.r)
*\xtp1et1+Dtxt-r
t:0
:0
/=0
must separately equal zero; so, for
(21+ I)tthQr) + (/ + l)Pr+r(p)
Equation (8.34), the pure recursion relation, relates each
P1
/>
1,
:0
(8.34)
to the ones above and below
it in the sequence. It may be used to determine the P;s once the first two are known, and it
thus provides an alternative to using relation (8.21).
Next we obtain a relation involving the derivatives of P;. Again we start by differentiating
the generating function, this time with respect to trr,:
AGxx
W - Q-
zxlL +
t)'/' - (l -'rxlt
+
t)
Rearranging and inserting the expansion of G gives
(r
i/:0r'*' ri @l - i/:0"'*t
1
Pr
Q-t)
-
-t
2
2xp.
+ x2)\xl rioD
r, Pl QDI
+
:
PitQD
-
zt't'Pi Q.r)
/=0
il=0 *t r; q4 : o
Setting the coefflcient of each power of x to zero,we have, for
P:QD
: xD*' ,,(r)
/:0
*
/>
Pi+Jr")
1,
(8.3s)
378
cHAprER 8 sruRM-LtouvtLLE THEoRy
If we differentiate the pure recursion relation (8.34), we obtain
t
for
I>
ri q@)
1.
- (2t + r) Pt@) - (2t + r) rr Pi QD + 0 + t) Pi*rQtl :
(8.36)
s
Using this relation to eliminate P/*tfrom equation (8.35), we find
PrQ.t)
:
-
Pi_JtD
-
2pPlQD
t
o,
: 1+I I Pi-t}r)
- PtPiQt) t
and so, for
-
Pi-r@)
1
(2t
+
r)
hQD
l+l
- Qt + t)u'Pi0t)
2l+1
* H r',(r)
+I
I > l,
thQ.D: 1t Pifu)
- Pi_Jt')
(8.37)
Similarly, by eliminating Pi-r, we obtain
I Pt
0.0
:
(2t
+
r) Pt1D
a Qt *
(t
I
r)
t t,
Pi QD
t)hQt)
:
-
0
+
Pl+rQr)
- zltl Pi @) +
- pP{(p)
D Pi +rQD
which is valid for / > 0.
Then, adding equations (8.37) and (8.38), we can express the polynomial
terms of derivatives:
(2t
+
I)hQtl: riu@) -
P7
I
Pi{@)
(8.38)
entirely in
(8.3e)
Pi_rQ")
which is valid for I > 1. This expression proves useful in evaluating integrals ofthe P1s.
Next we can combine (8.37) with (8.38) to eliminate fi-t@).Letl --+ / - 1 in (8.38) to
obtain
t
pt_r1r)
:
ri@)
-
:
pi@)
*
(t !)
t-tpl_t1*)
h-tQ.D:
rrPrQ.D
-
p.l.pPi1D
-
I hQDl
(8.40)
'la>
Equation (8.40) is valid for I > l. This relation is called a ladder operator because it
allows us to step down the series in /. If, instead, we eliminate fiu@), we obtain the
step-up ladder operator:
(l
*
r) P11yQt)
: pPl+rjr) -
Pi @)
: trl(t + t) P1QD + p.PlQ.|l -
piQ")
S.3 PROBLEMS WITH SPHERICAL SYMMETRY SPHERICAL
h+rQ.t):
r.tpre.t)
HABMONICS 379
r,*,
- (#)
(8.41)
These relations prove especially useful in quantum mechanics.
The Rodrigues Formula
The Legendre polynomials may be expressed as
P1(x):
h#o' - r'
(8.42)
a relation known as the Rodrigues formula. To demonstrate the validity of equation (8.42),
let's evaluate the derivatives. First we expand the quantity (r2 - l)t using the binomial
expansion:
(r2
-
t)t
:
(-r)r (t - u,
=
(-
*l(l - t) *++... + t-rl, rr,)
lr
r)r
D{-t), ---J!-.*to
After differentiating / times, only those terms with2p
fiu'-
l)/
:,-,,'
r:,# r,rrr-rr'
: (-1)/
i
2 / remain:
#
r.#r"
1-t1,-SE!-{l*',-'
P)! dxr-t
P=l\t+1t/21 Pl(l I
: (-r)/ t 1-tyo2Jg!-2!)#,r,-,
p:t(t+r)/21
where [(/
+
1)
12] means the integer pan of
{r*'dxt '
r)'
(l + l) 12. Continuing in this way, we find
: (-r)1/! i
1-110*zr-t
**
Qp - l)lpl(l - p)l^p:t(t+t)/zt
(8.43)
Notice that if / is even, this polynomi alhas I 12 even powers of x, while if / is odd, it contains
(t + I) /2 odd powers of x. The ratio of the coefficient of xk+2 to the coefficient of xft is
380
CHAPTER
8
STURM-LIOUVILLE THEORY
found by first setting 2p
ctk+2
*:
- I:
k
*
2 and then setting 2p
- I:
k:
&+z+l)t
lI-k -\
&+Dt(L+/+2\
t
\
)t( , -t)l
(/+r+
0+r)t
2)(t+k.t,(?)
/k+l+2\
(k+2)(k+tt(
z
-k2 +t-k
: - t2(k+
l)(k +2):
/
k(k+1) -t(t+t)
(k+t)(k+2)
which is the recursion relation (8.21) for the Legendre polynomials. The numerical
factor in front of the derivatives ensures the correct normalization. We can easily check
itlo by looking at the first few polynomials (/ : 0 is trivial).
I _ L.1.
I
-
I dt
UIIN@"
ld
.
rd*@"
- 1)' :
- l) :x
which is correct.
, _4.
L-L.
**u'dxt'
r)r:
2tlt
:
h#o^ -zx2 +r): * fte*'-o,t
!B*, 2'
r\
which is also correct.
As a by-product, we can use equations (8.42) and (8.43) to obtain an explicit formula
for
P1:
nfur):
Or, setting
k:
p
(-1)r
-f,
- l/2for
t
I
p=l(t+1 )/2t
(2p)t
(I1n *2r-t
(2p-t)tpt(t-p)l
/ even gives
ho.+):+fr-ttor'o
act
losee also Problem ?.
(8.44)
(2k + t)t
(rc.
i),(i-
(8.4s)
r)'
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL
and setting k
:
p
-
(l
-
l)
/2for / odd
HARMONICS 381
gives
(+t)12
Pto4:--T f- <-rlkw, -'
-1-Y{t_r)/z
,
(8.46)
We can also use these expressions to evaluate &(0). For I odd, the polynomial has no
constant term, and so Pl(O) : 0. For I even, the constant term (ft : 0) in equation (8.45) is
&(o)
:
+fu^,?:
(-Dt/2#:
eDt/2(t
-
r)tt
(8.47)
8.3.6. Solving a Potential Problem: Fourier-Legendre Series
Example 8.2. A hollow copper sphere of radius a is divided into two halves at the
equator by a thin insulating strip (Figure 8.5). The top half of the sphere is held at
potential V, and the bottom is grounded. What is the potential everywhere inside?
FIGURE 8.5. The potential on the top half of the sphere in Example 8.2 is V. The potential on the
bottom half is zero.
Since V2O : 0 inside the sphere and the system has axisymmetry, the solution
must be of the form (8.27):
a(r,0)
:
f (o''' * ft) nra
982
cHAprERssruRM-LtouvtLLETHEoRy
We must set all the coefficients 81 Io zero, because nothing is at the origin to cause
a divergent potential. We find the coefficients A7 by evaluating the potential at the
surface r : ai
a@,0)
:io,o'
1:o
r,Q")
0
: {Y T1
"
ifo>pr>-l
.
to
Next we make use of the orthogonality of the P;s by multiplying both sides of the
equation by Py (t t) and integrating from - 1 to *I :
r*l oo
rl
I-t D,
1:o
PtQt)Pr'oD dp
Arat
J
: I
Jo
v P,Qt) dp'
On the left-hand side, only the term with I : // survives the integration. The integral
is given by equation (8.33). Then, dropping the primes, we get
nl
, 2
:,
t,"'h
Jo
Pr1Ddu
Now we need to evaluate the integral on the right-hand side. We can make use of
equation (8.39), which is valid for I > 0:
rtlfl
Jo
r'toap: ai Joei*'tul - Pillr)ldLL
Here the right-hand side can be integrated immediately:
fl
Jo
Since
r'ul
aP
I
: a?
[Pt+(t't)
- &-r(P)]lA
&(l) : I for all values of l,
I
fl
h@) d r' : - zt t[&+r (O) - &-r (0)]
+
Jo
: -fi-''+rror (r . +) - - Pr+l(o)
where we used the pure recursion relation (8.34) to express Pr-r(0) in terms of
Pl+r(0). Then
v2l+l
Ar:-2-;-Pr+r(0),
For the remaining case,
/:
I>1
0, we can easily do the integral, since P6(9,) - |
(equation 8.22):
Sl
I
Jo
pottDap:
nl
I tdp:r
Jo
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL
HARMONICS 383
oo:I
and
e(r.ot:il,
E'+Pr+r(o)
(L)'
Po"r)
The sum reduces to a sum over odd /, since Pl+r (0) : 0 if / * 1 is odd. Then,
use equation (8.47) for Pl+r (0) and set I : 2n * 1, the potential is
a(r
o)
: : ('.
:,-
u.
;ffi
#+,# (:)'.'
if
we
P,.*,u,))
The first few terms are
a(r.d)
'7 rr:,3
3r
: tVz\ (/ I + icosl cos6(5cos2d - 3;
;l)
+ll6\a)fl)t
"ol'(urcos4e -70cos2
e
)
+' tJ'ts)+
t
)
Figure 8.6 shows the first four terms in the series for the potential versus angle for
rf a:0.I,0.2,and0.5.NoticethatthepotentialequalsV/2attheequator
(0
:
r12)
for all values of r. (This should remind you of a similar property of the Fourier series:
The series for a discontinuous function always converges to the midpoint of the jump.)
polar axis
--I
I
,/
/q
,
r
- :0.5
^^ \
\
,t
\
\r
08
08
equator
FIGURE 8.6. The potential inside the sphere in Example 8.2 at several values of r: r/a :0.1,0.2,
and 0.5. The value of the potential @ / V at a specific angle 0 is given by the distance
of the curve from the origin.
384
CHAPTER
8
STURM-LIOUVILLE THEORY
8.3.7. The Associated Legendre Functions and Spherical Harmonics
Orthogonality of Associated Legendre Functions
When a problem does not have symmetry about the polar axis, we need a set of eigenfunctions conesponding to nonzero values of the separation constant m.Thenthe equation
for the function of theta is equation (8.16), where we keep the value k l(l + 1) for that
:
separation constant:
,,
" dP\
d/
,^- ^P+l(l*l)P:0
a(.(' -*")or)- l-p"
(8.48)
Equation (8.48) is of Sturm-Liouville form with /(lr.) : 1 - t-t2, g(l.t) : *2 /(1 - p\,
w(t-t) : l, and ). : l(I * 1). (Note that z is the eigenvalue for the phi equation.)
The solutions of this equation are the associated Legendre functions P1' (l). They satisfy
the orthogonality relation
l*r'
,r
r*r
r,
d
t','
:o
unless
/
: //
(8.4e)
z
is the same in both functions.
Alternatively, if we disregard the physical origin of this equation, we-may identify it as
a Sturm-Liouville equation with S(p) : -l(t * l), w(p') : ll0 - u'2), and eigenvalue
zr2. In this case, the orthogonality relation is
where the value of
'lY\dp:o
untessm:m'
(8.50)
where the value of / is the same in both functions.ll
Form of the Associated Legendre Functions
To obtain an expression for the function P( , we could solve in a series as we did for the
Legendre polynomials. But it is more efficient to relate the solutions P,n to the polynornials
we have already found. We expand out the differential operator in Legendre's equation (8. l9)
II
See Problem 19
for the value of the integral when m
:
mt
.
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL HARMONICS
385
and then differentiate the whole equation m times:
I-
u'>ffi -
r*#
+ Kt +r)P,
:o
,'tffi -r,t*ffi -r4u* +t(t + ufi:o
e - p\# - z, zwffi+ r/(/ + t) - 2 x ttffi :o
s-
Continuing in this way, we get
^ d-*2P,
(l - lt")
a,^+z -2(m *
1n'tt p,
dn pt
r)uiifr
+U(t + t) - m(m + r)lAi :o
(8.s1)
so the mth derivative of the Legendre polynomial P; satisfies the equation
(r
-
p2)y',i,
-
2(m
*
t)pyk +l/0 + t)
InAppendixVll, we show that the substitution
(r
-
t2)2" -2pz'
- m(m* r)ly, :
0
(8.s2)
l^ : (l - p2) ^/22(p) yields the equation
+,ltu+
1)
- {r1:,
and thus z(9,) satisfies the associated Legendre equation (8.48). Thus, we have shown that
Cz(p) : fifu) for any constant C. In physics applications, it is common to choose
f, : (-l)n. Then
P\QD: (-r)-(1 - p\m/2#Pt1')
(8.s3)
Clearly Pf : 0 for m > /, since the highest power of pc that appears in Pl is tr^r/. The
function & (p) is even if / is even and odd if / is odd. The quantity (t - t"2)^12 is an even
function of pr, and after m derivatives, the highest power in (dm /d lt^) hQ-t) is p/-'. Thus,
fi @) is even if I -l m is even and odd if I * m is odd.
Also, since the associated Legendre equation contains m2, the eigenvalue -la leads to
the same differential equation. Thus, Pfn must be a constant times Pln.It is convenient
to define
rl^@):
r-rrffiP(e")
(8.s4)
386
CHAPTER
8
STURM-LIOUVILLE THEORY
the appropriate solution corresponding to the eigenvalue
shown in Figure 8.7 and Table 8.1.
as
-m. A few of the functions
are
P(p)
l:2,m:2
o*,. l:3, m:2
--f
-------I:3, m:
I
\,:,
FIGURE 8.7. The first few associated Legendre functions
Figure 8.3. Notice that
Pf eD :
TABLE 8.1. Associated Legendre Functions
0 for m
Pf
+
P{.
For functions with
Qt)
m:3
m:7
l:0
l:l
t _a
I
-L
l:3
m : 0, see
O.
1
1^
l.L
1Qw',
u
-
1)
,(5w' -3)
2_
-irt*'-t)1/r-*z
3(l l51t
t,.2)
(I -
p.2)
-15(1
- ,zf/z
The Orthogonality Integral
Next we need to find the value of the orthogonality integral when
/
:
//. Using the definition
(8.53), we have
,,-: I_: pf o,t)pt ar: l*r'{t - t"\^{#{#
o*
8.3 PFOBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL
HARMONICS 387
We integrate by parts:
a^
fi^: (t - t)'2)^1^dp*
!'9) dpm-t
'1ltrll+l
l-r
f'1dp l,tL' - J-t
d*,'!t(,t"t
dp^ ) dp^-' o,
p2)^d':tYL)l
The integrated term is zero, and the derivative in the integrand is
!apl"lo - Pp\^u
!'!P1 :
' dpI
-2mtt(t
- -'
-
u2y-ta^ dl.r^
!r!t) +r\r(r
P4!)
t dpm+l
- *ury{i'
We can simplify this expression by using equation (8.51) with m --> m
- l:
dt-l Pt
.- p")1 d-+l Pt ^ d^ Pt:
(ll) m(m- l)laU^*, -2^p aU^ -II(I+ dt;_l
Thus,
11^
: lt(t+ l) - men
' '- /*' (r
' -l)l
Jt
:(l+m)(l-m*l)11,*-1
rr2)^-t
loii'il'
LdP^-t I
o
u
Now we may step down to get
I1*:
:
:
-lm)(I - ml l)(l -lm - l)(l - m -12)h,m-z
(/ + m)(I -f m - t)(l * m - 2)(I - m + 1)0 - m -l 2)(l - m I
(/ + m)(l * tn - 1) . . . (l + l)l .' . (l - m I 2)(l - rn -t 7) h,o
(I
:
3)
h,^-z
(t*m)t 2
(l - m)l2l -t l
where we used equation (8.33) for 4,s. Thus,
l*r' 'i'{D'1t'o':
(8.ss)
388
8
CHAPTER
STURM-LIOUVILLE THEORY
Spherical Harmonics
The general solution to Laplace's equation in spherical coordinates may be written as
e(r,o,r,:
(8.s6)
* E,(*^,,* #)rrrtoei^Q
[The previous result (equation8.27) is a special case of equation (8.56) witham
: 0.1 Next we define the combination
:
bI^
:
0
unless m
2l*l(l-m)l P{(p.)ei^o
=Yt^(o,Q)
4n (I * m)l
where the constant has been chosen to make the functions
l:r' lr*
Y1a(0.Q)Yf
^,(e.o)dQdtt
-
61v6^^,
I1.
(8.s7)
orthonormal; that is,
: I Y14@'Q)Yf
^,@,0)de
JsPhere
(g.5g)
The functions Yt*@, Q) ate called spherical harmonics (see Figure 8.8 and Table 8.2). They
find application not only in potential problems but also in the quantum mechanics of atoms,
wave mechanics, oscillations of spheres (for example, the sun), and the structure of the
cosmic microwave background.
TABLE 8.2. Spherical Harmonics
m:2
m:0
t:0
l=1
I
=3
m=3
I
J4"
,f ! "o.,
Y4n
- i)
'lE (1"",',
fT
/5 " 3\
1f *coso 11"o"e - z)
-^17
V8z
sins.i6
-1,/E"o.B.;oe"io
4Y n
-!l{<s-"te
-1)sinoeil
!,f3!
8V
! , fT
8V r
r
"in2s.zi6
.o"s
"in2
s"2io
,fT
tlr
"in3
e"til
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL HARMONICS
The defining equation for each hgure is
{Retyi(G,
il)}z
m:0, l:0
m:l,I:I
m:0, l:l
m: O,l:2
m:2, l:2
J
m:2, l:3
The firr
firstt few
plotted here.
ics, with
/
:
O,1,2,3; m
:0, l, 2, 3. The real part is
390
CHAPTER
8
STURM-LIOUVILLE THEORY
Recursion Relations for the
P{
Here we shall derive a relation between functions with the same I but differentm. Let's start
with equation (8.51) and multiply by (-1)n(1 - Lt2)*12:
0
:
(-t)z(1
-
p2)tm+2t/2{2
t
d*-*,
-
Pl
t)*(-t)*(t
\' - *zr^nd!+t
r /
"
dpm+l
2(m -tI t)ft\
L\t't
P:
1ll(t + t) - m(m+ 1)l(-1)-( t - uzr^ndi
' du,^
Now we use the definition (8.53) to get
0:
p{+2 -t2(m-r
Dl+FPi'+l
+ U(/ +
1)
-
m(m
*
t)lPin
(8.se)
Numerous similar relations12 may be obtained by differentiating the relations for the P1.
The Addition Theorem for Spherical Harmonics
Suppose y is the angle between two vectors * and i', as shown in Figure 8.9. The addition
theorem allows us to express the Legendre polynomial P1(cosy) in terms of spherical
harmonics in the angles e, Q, e' , and Qt that describe the vectors i and i/.
polar axis
FIGURE 8.9. The angle y is the angle between the vectors
l2Gradshteyn and Ryzhik has an extensive list. See also Problems
l5 and l6'
i
and
i/
8
3
391
PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL HARMONICS
First we note that rt P1(cosl) is a solution of Laplace's equation. Thus, using a ubai'
coordinate system with polar axis along the vector i', we have
l(l + ll
-)
Vir&(cos y) : -:;z:
Pr(cos Z)
Vfi, is the angul ar part of the Laplacian operator (the second two terms in
equation 8.13). Since P1@osy) is a function of the angles only, the first term in the
Laplacian (with derivatives in r) is zero, so we may add it back in to obtain
where
V2Pr("o*
D
:
(l+!&(cos
-l
rz)
Since V2 is a scalar operator, we may express it in any coordinate system. Returning to our
original system, we note that the solutions of
vrf@.0):-t!i!f(e,o)
are the spherical
harmonics Y1* with-l <
m< l. Thus, P;(cos y) must
be expressible as
asumofthese Y;r:
I
P1@osy):
A.(g',|')Yt*@,0)
D
(8.60)
m:-l
In principle, we can find the coefficients
o^ :
I
A-
inthe usual way,
p1@os
y)yf^@,
O)
da
(8.6r)
but this integral is not easy to do!
Instead, we turn things around and regard Yk(e, il as a function of the angles y and p
i in a coordinate system with polar axis along i'. We can expand Yf*(e, fi in
spherical harmonics in y and B, again with a single value of l:
that describe
I
Yk@,O:
t
B^,Y1^,(y,P)
(8.62)
mt---l
If we let
y
--> O,
i
moves onto the polar axis and we have axisymmetry, so only the m'
term remains.l3 Thus, as
d -+ 0t
hm Yh(e.
y+v
and
il :
Q
--> Ql
, wehave
Yf. (e' ,O')
.
47t
"Yr-(O,d)
:0
unless ln
:0,
because
Pf
0 :0,
:
BoYrcQ,
hQ)
Bs
as you should verify.
fl)
:
0
392
cHAprER 8 sruRM-LtouvtLLE THEoRY
(Of course
Y76
is independent of p.) Therefore,
uo:
lEurYf^(e"Q')
Alternatively, we can find the coefficients
B,
(8.63)
using orthogonality of the Y1^ in the (y, B)
coordinate system:
:
B*,
I
yk@, Q)Yh,(y, fl) dszy
In particular,
no
:
J
Yi^@.
ilYfo(y, p) dQy
'f
: Jr v|^re,ottfltt)-t
rt@osv)de,
l2t+t
: tl'#A.(0',0')
U 4/t
,
(8.64)
where we used equation (8.61) in the last step. Since we integrate over the whole sphere,
does not matter whether we use the coordinates 0, Q or y, fl as the integration variables.
Finally, we combine equations (8.63), (8.64), and (8.60) to get
&(cos z)
: -l
4n
F_,fi1Yr^<e'
ilYh (e'' Q')
it
(8.6s)
This is the addition theorem for spherical harmomcs.
Note the expected symmetry in the primed and unprimed coordinates. [It does not matter
which function has the complex conjugate, since the sum is over positive and negative values
of m, andYt,-^ : (-DmYk.lWe can also check the result by letting 0/ approach zero,in
which case the left-hand side is just Pl(cos 9). Verify that the right-hand side also reduces
to P;(cos 9) in this case.
In the special case y - 0, we obtain the sum rule for spherical harmonics:
I
I vm{e,il:2
m---l
2l +1
4n
(8.66)
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHEBICAL HARMONICS
393
The addition theorem is useful for performing rotations of the coordinate axes. When
combined with the generating function for the Legendre polynomials, it also allows us to
express the inverse distance I / l* - i'I in terms of spherical harmonics.l4 This is a particularly valuable result, because this inverse distance appears in the integral expressions for
electric scalar potential, magnetic vector potential, and Newtonian gravitational potential.
The orthogonality of the spherical harmonics often makes the integrals more tractable.
Boundary Value Problems Using Spherical Harmonics
Example
8.3.
Find the potential inside a conducting sphere of radius a with poten-
V onthehalf 0 < Q. r andzero onthehalf z . Q.2n.
The potential may be expressed in terms of spherical harmonics, as in equation (8.56). Since there is nothing at the origin to cause a divergent potential, we
tial
O:
exclude the negative powers of r. Thus,
ool
Q(r,0,d):
tl:0 m---l
I A6rtY1^@,Q)
To find the constants A1^,we evaluate the potential on the surface and set it equal to
the given expression:
-
e(a,0,0):I I
A6aty1^@
t:om:-t
ifo<Q<tr
ifr<Q<2n
^r:[v [0
To evaluate the coefficients, we use the orthogonality of the Y1-.We multiply both
@) and integrate over the sphere:
sides by Yf
-,@,
i o,^o' l"nnu"y1a(o,Q)yf^,@,o)da:,
i m:_l
Jsphere
I:O
Only one term on the left-hand side is nonzero. With I
l:r'
lo"
rr^,rr,Q)dudQ
: lt and m :
mt , the
equals 1. Then
A1,^,ql'
-
Y
rii' (D o*
lo "-i-'Q
Now we may drop the primes:
tA,
ttm
l4See Problem 20.
v
-
7
Pf0r)rr(#)
dO
integral
394
oHApTERBSTURM-LrouvtLLETHEoRy
The result is zero unless m is odd. Then the integral over LL is zero unless /
also odd, for then f( @) is an even function. Let's call this integral l1^.Then
v-
AI*:
The case I
ot
: m:0
1S
+ | (l - m)l L* 2 for/ odd andm odd
^l2I
4r
\l
e + m\1"* im
is special, for
f+l ft
I
I I Y6s : t/4tr
Ji
J -t Jo
-(2)(n):
grvmg
Am:
If m :
O
I t'
but
v
2
0, the result is zero because
f+t f+t I PoQ)hjt)dry:0
I nU"SdP:
J-l
J-r
(equation 8.30). So the potential is
e(r,o.Q):Yz."i
: Y, *,
We can demonstrate
combining the terms
ffi
,,
-
tt'",
bf:y
2l-ft(l-m)l 16. 2 Y1p(0,Q)
4n (l -f m)l tm
h- ? pi
G)"+
.
r,
b
^ry
t
-":;;
,"t,
Pf{
1p'vsi ^ao
:
ffi
:
ffi"'fie''
,,
*
h
+
) ei
-o
for a real physical quantity, by
Using equation (8.54), we have
as expected
withm: M andm: -M.
3
ot
ffi', * h'; * o")'-i
-
r,, trt@'*Q _ e-i MQ)
rutsinMQ
M Q
8.3 PROBLEMS WITH SPHERICAL SYMMETRY: SPHERICAL HARMONICS
Thus,
(D(r, 0,
a)
:
:. ; i
The result is clearly real and
e(r, 0. il
::.:
{It
t,
> r:l
+ffi;-ri
a;""r"rally
T'-tsin
d
@)
sinmQ
correct. The first few terms are
*, (:)' llt#tu,tu' -
1's 1/
t
-
fi
sinq
*ljf,"ttsin3esinro. I)
r3 sinosin@
:, (t ;o
* rze\ot
' "-\t[:1s.or'e-Dsindsin@
r.- --\r+
*5sin3 osin3Q. ])
We can use the addition theorem to show that this result agrees with the result of Example 8.2. Let's compare the two pictures (Figure 8.10).
FIGURE 8.10. Example 8.3 (on the right)
is
just Example 8.2 (on the left) rotated through 90o.
The result of Example 8.2 is
o::{'- I
where
;
y is the angle between the position vector i and the z-axis in the picture on the left
or, equivalently, between the position vector and the y-axis in the picture on the right. To
along the y-axis, so
relate the two results, we may use the addition theorem with vector
i 0':Q':r/2.
l
'#r,*,(o) (l)1arc",r,r}
i'
396
cHAprER
I
sruRM-LrouvtLLE THEoRy
Let's look at the first few terms:
P1(cos/)
:
cos
4nl3
t' :
rr
i L* "*
:!la,"ine)2cos
3 Lgo '"^"v/:lvr
,cos?
* l8trrin L2 "i11gpi{t-n/z) q r-i(Q-tr/zt1l')
1)l :
\Y' - 2/ )
/a
sin0 sind
and
P3(cos
7
/): +olt
t lGrz@)rt(o)+2G
But Pj"(0)
p3 (cos
:
0 unless 3
/)
:
*
g
^=L-ffiPiQ-t)pi(o)"o''(d
m is even, so there is no contribution from m
2l]" (;)'
G
r.r2 -1) sin
-;)l
e
(-
1) sin
d
* fi rra'
sin3
:2.Thus,
B
(-,
"
3d)]
: -(+(s"or', - t) sind sin@ + fr sin3 e sin&)
and the potential in Example 8.2, rotated into the new coordinates, is
a:I['- [
'+Pr+r(o) 1r)'ar"",nr]
: :{' -' (-;) (1)'i"e'i'o
-:tE Gf
::{t.tuG)sind
l-'
(*1tt
cos2
P't'
e -1) sing sin @ *
*1 "n'
3d)
])
sin@
. *(:)t[3(5cos2 e -
Dsin0sin@
r5sin30sin3{].
}
which agrees with Example 8.3.
Now let's try to get
a
general result. We insert equation (8.65) into the result of Example 8.2
and get
e (r, o, a)
:
:- " I
zff r+r10, (:)'
: : -,
n+&+r
(0)
(:)'
2,hr*@,
p,ffi
,i'
o)Yh
(;, +)
r*;ei^a ry {o)s-int'l 1z
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY BESSEL
FUNCTIONS 397
Again we note that Pi'(0) :0 unless I * m is even. But we also need I * I to be even so
/2 : i-m : 1-9@-t) lz 1 i:
that P711 (0) I 0. Thus, I and m are both odd, and
"-imr
A(r,0,0)
(0) (:)' t F_#rr rryei-a r( {0r#
- v2 -u fu,_,zr 2l+ r pr+r''"'\a/
Lt,(ll
m odd
'*o
.-
zl+r_I
-'
L
-, z l,_rS$
r'
I
-'#',*,(o)
|
L
1
^ (1- m)tpy(ot(_1,,(m-t\/2(r_:t
-Q)eD@-t\/'? (L)'
\a/ r((Dsinrn{
(t +
(t+m):
^).Pf
J
|
-Etfr,
rodd u odd
I
Compare this with the result of Example 8.3:
(b(r, 0,
a)
The results agree
if
::.:
(I)' rr rr, sinmQ
D P- +ffi,,^
,,-: llr' Pi@)dw
:' Pi'Q't)dw
(8.67)
lo'
:
r
Pr
+t (o)
Pi
e)l
;
g<^+tt
t
z
, l,modd
Thus, we have obtained a useful expression for the integral on the left. In Problem 8.21,
you will be asked to verify this result.
8.4. PROBLEMS WITH CYLINDRICAL SYMMETRY:
BESSEL FUNCTIONS
8.4.1. Bessel Functions
Bessel functions arise as solutions of potential problems in cylindrical coordinates. Laplace's
equation in cylindrical coordinates is (equation 1.69)
lalao\
la2(D
a2Q
iar\oao-)*FeF* urz:o
398
cHAprER
I
sruRM-LtouvtLLE THEoRY
To separate the variables, we let
<D
:
R(p)W(Q)Z(z).Then we find
I a / aR\
Rp ap
*
\oa )
1 A2W rA2Z
*o'w * z ar' :t)
The last term is a function of z only, while the sum of the first two terms is a function of p
and @ only. Thus, we take each part to be a constantls called k2.Then
a2z
arr:k
Z
and the solutions are
Z
: etk'
The remaining equation is
lalaR\tazw.
Rpapl'u=)*W uFtk':o
(8'68)
Next we multiply through by p2:
o3 /aR\
..
la2W
iA\'A)*0"0"* * uF:'
Here the last term is a function of @ only, and the first two terms are functions of p only. As
with spherical coordinates, we often want a solution that is periodic in @ with peiod2r,
so we choose a negative separation constantl6:
A2W
ffi:-^2w+w:eri-Q
Finally, we have the equation for the function R of p:
a/aR\
,*lr;)+k2p2n-*2R:o
To see that this equation is of Sturm-Liouville form, we divide through by p:
a/AR\
^
m2
^ lP^dp/l+k'Pn--R-0
p
dp\
(8.6e)
15Here we choose a positive constant (real k). The solution is rather different if the constant is negative, as we
shall see in Section 8.4.7.
16In this application, m is usually an integer. Noninteger values are also of interest; see, for example, Section 8.5.
In that case, the separation constant is usually written as -u2.
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL FUNCTIONS 399
Now we have a Sturm-Liouville equation (8.1) with f (p) : p, C@) : m2 /p, eigenvalue
)' : k2 , and weighting function w(p) : p. Equation (8.69) is Bessel's equation.
It is simpler and more elegant to solve this equation if we change to the dimensionless
variable
x
:
kp.Then
A / aR\ ^2
k_IkD_l+k"oR_2_4:0
akp V akp)
kp
d/dR\
m2
_lx_l*xR__R:0
dx\ dx)
x
The equation has a singular point at
x
:
(8.70)
0, so we look for a series solution ofthe Frobenius
type (refer to Chapter 3, Section 3.3.2):
R: xPio,*'
n:O
Then the equation becomes
oo
!t,
n--0
+ p)(n -f p - r)anxn+o-t
+D@ I
p)anvn-tP-r
n:0
+ io,"'+p+r
n:0
-
m2fanx'+n-r
:o
n:O
The indicial equation is
p(p-l)+p-^2:O+p:aa
Thus, one of the solutions (with p : m) is analytic at x : 0, and one (with
not. To find the recursion relation, we look at the k * p - I power of x:
(k
+ p)(k * p - I)a*+ (ft + p)ax + ak-z
-
-
m2a1,
and so
oo------o!-2-:(k*p)2-m2
ak-z
oo-'=
=
k2+zkp+ p2-^2
=
ak-z
-F**:-*1*+zny
O
p : -m)
is
400
CHAPTER
8
STURM-LIOUVILLE THEORY
p
will always
Let's look first at the solution with
the series with as, then ft
at1,
-l
*
-
2n(2n
-
23n(n
-
: *m. We can step down to find each ap.If
:2n, and
-l
2m) (2n
-2)(2n -2-l2m)
(- 1)3
-
r)(n
-
2)23(n
2n nl.
2n(n
*
m)(n
(-r)'
: AA"
we start
be even, k
* m)(n I
a2n-4
-
1)(n
(m
-l l)
m
* m-
2)
a2n-6
1
*
m
- l).
..
The usual convention is to take
(8.7r)
Then
(8.72)
and the solution is the Bessel function:
oo
/
1\n
J^(x)
r)
(;)-.*
(8.73)
The function J^(x) has only even powers if m is an even integer and only odd powers if m
is an odd integer. The series converges for all values of x. Although we have assumed m to
be an integer thus far, expression (8.73) is also valid when m is not an integer.
Let's see what the second solution looks like. with p : -m' the recursion relation is
Qk:
ak-2
k(k
-
(8.74)
2m)
where again k :2n. Now if m is an integer, we will not be able to determine a2^,because
the recursion relation blows up. One solution to this dilemma is to start the series with a2^.
Then we can find the succeeding coefficients a2@+m):
azu+m): -420-lt+zm
721naDn
-42@-z)+2*
24@+m)(n*m-l)n(n-l)
:
(-l)uf(z*1)
nlllna**1122nu'^
:
lm. (ComThese coefficients generate the same series that we had before in the case P
pare the equation above with equation 8.72.) Thus, we do not get a linearly independent
solution this way. This dilemma does not arise if the separation constant is taken tobe
-v2
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL FUNCTIONS 401
with v noninteger. In that case, the second recursion relation provides a series
is linearly independent of the first.
With u : m, an integer, we find
l-
^
(x
)
: i"
####
:S
(-t)'r(m -f l)2m
n:U
a2*
\ "tr1, 'r^ *,
and
if
762(n
t
m)
-
J-r(.r) that
m
/x\m+2n
"'* \r)
we choose
a2m:
(-D^
r@ + Dz^
then
J-^(x) : (-I)m J^(x)
(8.7s)
With this choice, J, (x) is a continuous function of u. (Notice that we can also express the
series using equation 8.72 for the coefficients, with ffi ) -ffi, n --> k * m, and a2n
=O
for n < m.)
We still have to determine the second linearly independent solution of the Bessel equation
when m is an integer. We can find it by taking the limit as v --> m of a linear combination
of ./, and J-, known as the Neumann function, Nr(x):
Nr(x):
Jr(x)cos vtr
sln
-
J-u(x)
vz
Thus,
-. J,(x) cos urr - J-r(x)
N^(x):limN,(x):lim
v+m
v+m
sln uT
e+0
:
lim
e+0
J^+e(x)
(-l)n sinen
- (-l)^ J-6a"1@)
efi
(8.76)
4O2
cHAprER 8 sruRM-LrouvrLLE THEoRy
where we expanded the trigonometric functions to first order in e. Next we expand the
Bessel functions in a Taylor series in u to first order in e and use relation (8.75):
N,(x)
: JTo *lt. +, *1"=,N-(x)
-
I
*'Tl,:,)]
_(_t)*Tl":_l
It_dJ"
" ldD
<-o^ (t-^
(8.77)
,--^
The derivative has a logarithmic terml7:
L (.,s
t-rl, r1)r'\
:
r,.liT
dv d, \ "-o
+ U \il )
"
dJv
: ,"'
$
dv L^
/x\2, )_r,4 i
L^
\;)
d, n:U
n:O
(-l)'
/xv2n
ntt(n+ , + rl (t,)
and
dx'
dv
d
dv
,ulnx
:lnxeuln* : x, lnx
has a term containing Jvlnx. This term diverges as t -+ 0, provided that
0 for u f 0,
zero (that is, for v : 0). The function Ny (.r) also diverges as x
because it contains negative powers of x. (The series for ,I-, starts with a term x-u.) But
Nr(.r) does not diverge as ,r -+ oo, because "/, -+ 0 faster than the logarithm approaches
and so
dlrldv
/, (0) is not
infinity.
TWo additional functions called Hankel functions are defined as linear combinations
of
,/ and N:
HP @) : J*(x) -f iN^(x)
(8.78)
H#)@): J^(x) - iN-(x)
(8.7e)
and
17This result should not be surprising. We could also find the second solution using the methods of Chapter 3,
Section 3.3.3, particularly in the form of equation (3.37). See, for example, Chapter 3, Problem22.
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL FUNCTIONS
403
8.4.2. "Properties of the Bessel Functions
The Bessel functions (Js) are well behaved both at the origin and as x + oo. They have
infinitely many zeros. All of the J^, except Js, arc zero at,r : 0. The first few functions
are shown in Figure 8.11.
f(x)
1.0
0.8
0.6
0.4
0.2
-0.2
-0.4
FIGURE 8.11. The first three Bessel functions. All the functions except
equal zero at -r : 0,
"16(-r)
and all of them approach zero as r -+ oo. All of the J^ oscillate with decreasing
amplitude.
For small values of the argument, we may approximate the function with the first term
in the series:
J^(x)
x
"- -,
(;)^
for
x <( I
(8.80)
The Neumann functions are not well behaved at x : O. N6 has a logarithmic singularity,
> O, N. diverges as an inverse power of x:
and for m
2
11" for x <<
(m-l\l /2\^
N^(x)ry--l_l
rr \x,/
N6(x) = T
(8.E1)
1
forx((
(8.82)
4O4
cHAprER E sruRM-LrouvlLLE THEoBy
For large values of the argument, both .I and N oscillate. They are like damped cosine or
sine functions:
TT / mn z\
J^(x)xV;"ot\., - o)
lT / rnn z\
N.(x)-r,l
"*sin(x- 2 4)
for
x
)) l,m
(8.83)
for
x
)) l,m
(8.84)
ror x
)) t,m
(8.8s)
Thus, the Hankel functions are like complex exponentialsl8:
se,2)
o
rl*" r[*,(, - ry -il
Noticethatifm>l,thelargeargumentexpansionsapplyforxlmrutherthantheusual
x)1.
8.4.3. Relations Between the Functions
As we found with the Legendre functions, we can determine a set of recursion relations that
relate successive J^(x). Starting with the series representation (8.73), we divide by (x /2)^
and then differentiate:
d
?r)nn (!\r"-,
\ :$
xm / ?:onlf (n*m*l) \2/
(2^J^(x)
dr \
t-tl'
(!\"-'
-$
L.@-l)!r(n*m*l)\2/
n=l
Now let
k:
n
- l:
d (2^J^(x) \ : _$
t-tlo
(L\ro*,
\2/
m*r*t)
dr \ xm /
futlr(k+
om @
(-l)k
x^ t-0 1rty11r 1* a
:_o sL_
: l8Refer to Chapter 2, equation (2.5).
2m
_Jm+t(x)
1x12k+m+l
a g \i)
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL
FUNCTIONS 405
and thus
d (J^(x)\_
J.+tQ)
E\,^ 1-which is valid for m
> 0.In particular, with m :
(8.86)
,-
0 we obtain
J1@): -J6@)
(8.87)
Similarly,
*'^
:
2^
(;)'^
ftk^, ^{,)t
*D"frffir,
n=O
* n)
?Dn(m
:2,
-' S
1x12m+2n-l
4"tf(r+^+D\r)
:,'i
(-l)'
(x\m+2n-r
\2/
l+l)
lnlf(n*mn=U
:
ftu^ r^{*)f
which is valid for m
x^
J^-{x)
(8.88)
> l.
Combining relations (8.86) and (8.88), we get
r^+r
r
Im-t
: -.^ * (#)
-,-
(-*h.
*
i
*o^
h)
+
r*(x)l
I (**^-'r^
+ *^
rl,)
:*b-th+^**Jh:2#U
Jm+t
I Jm-r:
2m
x
-T
(8.89)
406
cHAprERssruRM-LrouvtLLETHEoRy
which is the pure recursion relation for the Bessel functions. Similarly, by subtracting instead
of adding, we find
-dJ^
Jm+t-Jm-t:-2-;ax
The same relations hold for the Ns and the
(8.e0)
.F1s.
8.4.4. Bessel Functions as Solutions of the Helmholtz Equation:
The Generating Function
Like Laplace's equation, the Helmholtz equation (equation 3.16 in Chapter 3 extended to
two or three dimensions)
(v2+t2)o:o
may be written in cylindrical coordinates and separated. Suppose we look for a solution in
two dimensions, with @ independent of z. Then the equation becomes
1 a / aR\ I AzW ,.
Rpap\oA)* *7 aF+k':tr
which is the same equation (8.68) that we obtained from Laplace's equation after separating
out the z-dependence. Thus, the solutions are of the same form:
+oo
':
*E*o*J^(kP)ei^o
We may exclude the functions N^(kp) if the origin is within our region of interest.
As with the Legendre functions, we may exploit a simple physical situation-here a plane
wave-to obtain a generating function for the Bessel functions. A plane wave propagating
along the y-direction has the form
eiky
-
uikpsinQ
which must therefore be expressible in Bessel functions:
*oo
,ikpsrnQ
- t
a_, J^,(kp)ei^'Q
mt:-@
Now we make use of the orthogonality of the ei^Q . We multiply both sides by e-i*Q and,
integrate over the range 0 to 2r . Only the one term with mt : m survives on the right-hand
side:
lo2"
ei*o "ino-i*Q d.Q
:
2r a- J*(kp)
407
8.4 PFOBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL FUNCTIONS
:
0, we can evaluate both sides for p - 0 to obtain ao : l.If m t' 0, we expand
the function ,ikpsinQ using the exponential series and expand the Bessel function on the
right using its series (8.73):
With m
* I'",-'*o (r .|n#-),r
n*m.,
- "^ (ry)-
(T)"
I 0, the first term in the integral on the left-hand side integrates to zero. Now for
small ftp, we keep only the leading term on the right-hand side. Then we have
With m
*>=(Y)"
Look first at the case m
,+.
lr'"
iQ
'-imQf
-e-i')n
dQ:a-.#
: l. Keeping only the leading (n :
(+)^
1) term on the left-hand side,
we have
# Ir'" e-iQ@iQ -
e-iQ)a6:
fi<rn-
0)
: "rT +
at:
r
> 1, the terms with n < m ontheleft-hand side are allzerobecause leiQ - s-iQln :
(l - ,-2ib1n has no ei-Q lrormif n < m. (This must be true for consistency, because the
For m
ei"Q
lowest power of p in the Bessel function series on the right is p- .) Thus, the first nonzero
term has n : m.In the factor leiQ - ,-iQ1m, only the term ei-Q survives the integration
over @. We h e
(ikp)* |
ml(2i)m
['o
2n J11"-,-ar,^e
d6 =
sY
-
Thus, we have a generating function for the
,ikpsinQ
_
o_ t-f
l@ l)
qm
e)^ )a*-l
J*,
+oo
t
J^(kp)ei^Q
(8.e1)
and the integral representation,
(8.92)
408
cHAprERBsruRM-LrouvrLLErHEoRy
Next we write sin @ : (eiQ for the generating function:
e-i\ /Zt and let eiQ :
*p^12
l+ (,\ +)l
tll
8.4.5. Orthogonality of the
t. Then we have an alternative form
:f
l^(kr)tm
(8.e3)
m:-6
"I,z
Since the Bessel equation is of Sturm-Liouville form, the Bessel functions are orthogonal
if we demand that they satisfy boundary conditions of the form (8.2). In particular, suppose
the region ofinterest is p - 0 to p : a, and the boundary conditions are that Jm(ka) : O
and J.(0) is finite. We do not need a more specific boundary condition at
the function f(p) : p iszerc there. Then the eigenvalues are
p
-
0 because
dmn
.
nmn--
a
of J-. (The zeros are tabulated in standard references such
and
as Abramowitz
Stegun. Programs such as Mathematica and Maple can also compute
(8.6) becomes
Then
equation
them.)
where a^n is the nth zero
r * (k*^,
d p!!!#@1"
o
-, ^ o^, o, ofu#@|,',
:
(4n,-
t?*,)
fo oh&^,,p)J^(k*np)dp
The boundary conditions make the left-hand side zero, and thus,
fo
o
h&^,,
p)
J^(kanp) dp
if n
I
(8.94)
n',
:o
(8.es)
n : n' , we replace kmn, with an arbitrary value
(8.94) becomes
the
eigenvalues.
Then
equation
of k, not one of
To determine the value of the integral when
J*(ka)a
ffro^,Alr:o:
,o'
- &,t fo oh&*,p)
J^(kp) dp
l
Now we differentiate this expression with respect to ft:
dJ*(ka) dJ'-11r^,p11 :2k
a-o;ak^n
irk^np ro:o Jo["
+&2
I
pJ^(k^np) J^(kp) dp
- u^, Ir' pr^(k^,p)4!#O
dp
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY BESSEL FUNCTIONS
Next let
k
409
--> k^n. The second term on the right vanishes, and we have
k*nlall(k^na)12
:ro^,
fo
pU*(k^nil12 dp
and so
fo ou^&^ndl2
dp
: lvh{t^,o)l'
(8.e6)
which is the integral we need.le
8.4.6. Solving a Potential Problem
Example 8.4. A cylinder of radius a and height /r has its curved surface and its
bottom grounded. The top surface has potential V (Figure 8.12). What is the potential
inside the cylinder?
FIGURE E.12. The top end of the cylinder in Example 8.4 is at potential V. All the other surfaces
are grounded.
The potential has no dependence on @, and so only eigenfunctions with m :
0 contribute. The potential is zero at p : a, so the solution we need is Jo(/cp)
with eigenvalues chosen to make Jo(ka) : 0. Thus, the eigenvalues are given by
19See Problem
29
for the integral with Neumann conditions.
410
cHAprER
I
sruRM-LtouvtLLE THEoRy
kgna : agn, where don are the zeros of the function "16. The remaining function
must be zeto at z : 0, so we choose the hyperbolic sine. Thus, the potential is
a@, z)
:lc,Jo(ks,p)
:
Q(p, D
z
sinh (k6nz)
:
To find the coefficients cu, we evaluate this expression at z
V
of
:DcnJs(ks,p)
hi
sinh (ft0nft)
Next we make use of the orthogonality lf ,h" S"rr"t fonctions. We multiply both sides
by pJo(ko,D and integrate from 0 to a. Only the one term in the sum with n : r
survives the integration:
V
ra
fa
I oJoko, d dp : I
JoJon
pJo&o,dlcnJs(ks,p)
sinh(konh) dp
: c, Ifq p lJo&0, dl2 dp sinh (ko,h)
Jo
a2
: q7U
6ko,a)12 slrr'h (ko, h)
To evaluate the left-hand side, we use equation (8.88) with m
: l:
!' oJokddp: lfad
E J, ooo[kpJr(kp)]dp
Jo
: 1-a
i pJr&p)16:
So
,,:#Jr&o,^ffi
-
v2
k
,"
h(korq)sinh (ko,h)
J6: -h.
where we used the result from equation (8.87) that
,
cD
:
^,,
zv
Finally, our solution is
^-.S Jo(konp) sinh(ko,z)
: 2v
2 kr*tt&r*)tt"h (ftrlr)
The first two zeros of Js are aot
two terms in the potential are
\v
-JtGa)
:
2.4048 and
cu62
:
5.5201, and thus the first
( JoQ.a048pla) sinh(2.404821a)
\rAo48JtQn4q
{rnb QA048Ua) -
Jo(5.5201p/a) sinh (5.52Dtzla)\
(tsrolv") )
55201fi55201)
"1"h
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL
FUNGTIONS 411
These terms are plotted in Figure 8.13 for h/o - 2 and z/a: 0.5, 1, 1.5, and 1.8.
For zf a: 1.8, the first two terms do not represent the potential well. The three-term
result is also shown for Z /a : 1.5 and 1.8. While the third term makes little difference
at zf a : 1.5, it is vbry significant at zf a : 1.8.
o
n
two terms
three terms
p
a
FIGURE 8.13. First two terms in the series for the potential inside the cylinder versus radius for
h/a :2 and zf a: 0.5, l, 1.5, and 1.8. As 2/a increases, the series converges more
:
slowly and tbese first two terms do not represent the potential well at zf a
1.8hence the dip near the axis. The grey lines show the potential when a third term is
added to the series. At the smaller values of z, the extra term makes a negligible
difference in this diagram.
8.4.7. Modified Bessel Functions
Suppose we change the potential problem in Example 8.4 so that the top and bottom
:
of
the cylinder are grounded but the outer wall at p
a has a known potential V (Q, z).Then
we need to choose a negative separation constant so that the solutions ofthe z-equation are
trigonometric functions
:
a2z
-k2Z
02.2
:
:
+ Z:
asinkz
*
bcoskz
: 0. We also need Z(h) : Q,
nn /h.
This change in sign ofthe separation constant also affects equation (8.69) for the function
R(p) because the sign of the k2 term changes here too. The equation becomes
At z
O,
Z(z)
0, so we need the sine and therefore s€t b
so we choose the eigenvalue
k
:
a/aR\
^
^ lp^dp/l-k"pRdp\
^2 R-0
p
412
cHAprERssruRM-LtouvtLLETHEoRy
or, with variables changed to
x
:
kp,
m2
x
* ("#) -xR--R-0
(8.e7)
which is called the modified Bessel equation. The solutions to this equation are J^(ikp).It
is usual to define the modified Bessel function I^(x) by the relation
I
I*(x): ;J^(ix)
(8.e8)
so that the function f. is always real (whether or not m is an integer). Using equation (8.73),
we can write a series expansion for 1.:
ra(x)
t\4)
('!\^*'"
::
- im i4nr.f.=l:11* + l) \2 )
(n
n=U
m
m
I
I-(-r\: )
4nlf(n*m
N:U
As with the Js, if
z
is an integer,
1-'
I-^(x)
*l) (;)^.^
is not independent of
:- i* J-^(ix) :
I*;
(8.ee)
in fact,
I*(x)
The second independent solution is usually chosen to be
K^(x)
:
Lim+t P$) 6x)
(8.100)
Then these functions have the limiting forms for small argument:
r^(x)
x
i.,
(;)^
forr
(
I
(8.10r)
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL
FUNCTIONS 413
and
Ks(x)x-0.5772-t"t forx <
l(m\ /2\n
K^(x)*i
l;),
(8.102)
1
m>0, forx(
I
(8.103)
At large x , x >> I , m, the asymptotic forms are
+f,
r^(x)
x
K^(x)
x
(8.104)
and
7f
_e
2x
(8.105)
(Compare with Chapter 3, Examples 3.9 and 3.10.) These functions, like the real exponentials, do not have multiple zeros and are not orthogonal functions. Note that the 1s are well
behaved at the origin but diverge at infinity. For the Ks, the reverse is true. They diverge at
the origin but are well behaved at infinity (Figure 8.14).
Is
I1
I2
Ks
K1
K2
0
FIGURE 8.14. The first three modified Bessel functions. The functions Kn (.x) diverge
and the functions
1n
(-r) diverge as
r
--)
oo.
at the origin,
414
cHAprERBsruRM-LrouvrLLETHEoRy
The recursion relations satisfied by the modified Bessel functions are similar to, but not
identical to, the relations satisfied by the Js. For the 1s, again we can start with the series:
d (2*I^(x)\_S
1xlzn-t
d"\ xm )-2ntt(ntm+D\z)
Now let
k:
n
- I:
I
d (2mr^(x)\:s
(!\ro*t
d"\ xm / Aklr(ft+m*r-rt)\2/
d (I^(x)\ /,+r(x)
)cm
dx\ x- )-
(8.106)
Similarly,
:
ft{** r^l
xm
Im-r
(8.107)
Expanding out and combining, we get
I*(x)-mb:I^+t
x
4++:tm-l
,
Adding, we obtain
2Ik:
Im+r
I
Im-t
(8.108)
while subtracting, we get
2m
-x
Im
- Im-t -
Im+l
(8.109)
8
4
PROBLEMS WITH CYLINDRICAL SYMMETRY: BESSEL
FUNCTIONS 415
For the Ks, the relations are
fr{*^
*^)
-
-x* K*-l
and
d
dx
/ K.(x)\
K,+r(x)
*-
\"')
(8.1 10)
and, consequently,
2m
K^-l - Km+t:--Km
x
K.-t * Km+t:-2Kk
(8.1 1 1)
(8.1
l2)
8.4.8. Combining Functions
When solving a physics problem, we start with a partial differential equation and a set of
boundary conditions. Separation ofvariables produces asetof coupled ordtnary differential
equations in the various coordinates. The standard solution method (Section 8.2) requires
that we choose the separation constants by fitting the zero boundary conditions first. In
a standard three-dimensional problem, once we have chosen the two separation constants
we have no more freedom, and the third function is determined.
When we solve Laplace's equation in cylindrical coordinates, the functions couple as
follows.
Zero boundary conditions
J. (o^n
I)
where
J^(a*r) :
in
p.
?,"
The eigenfunctions are of the form
sinh a-n1
*
B^n
0. The set of functions J^(a^npf
set on the constant z surfaces that bound the
Zero boundary conditions in
"*
z.
u
a)eri'd
^,1) "+;*o
forms a complete orthogonal
region.
The eigenfunctions are of the form
(#) lA*, r^ ff)
The set of functions sin (nt zl
surface p : constant.
"osh
+ B*n K -
(T)1,.'-'
h)eti'Q forms a complete orthogonal
set on the boundary
Thus, in solutions of Laplace's equation, the ,Is in p always couple with the hyperbolic
p always couple with
sines and cosines (or real exponentials) in z, while the 1s and Ks in
the sines and cosines (or complex exponentials) in z.
416
cHAprER
I
sruRM-LtouvtLLE THEoRY
8.5. Find the potential inside a cylinder of height /r and radius a when the
top and bottom are grounded and the curved walls have a potential V (Q, z) : Vo for
Example
0<d<
n and-Vs forn < Q.2n.
As we discussed in Section 8.4.7 , the z-functions are of the form sin (nn z/ h) so
that the potential equals zero at z : 0 and at z : h.The Potential should be finite on
the axis at p - 0, so we exclude the K functions. Thus, the potential is of the form
Q(p, z, o
.oo+oo
:D
D o^^ "^ (T),*(T o) "'^'
n:lm:'6
:
Now we evaluate the potential at p
e(a,
z,t,
:
{l'*
41
'rlorlrr'.oro
::.f:*
An^
sin
(?
n(ffo)
,t^o
This is a Fourier sine series, and we find the coefficients in the usual way:
t.^f;r^(7")r":ro
loo
"^(T)0,(1," - f)e-i^,do
h
2t1 - (-l),,11
: Vo-(l
-cosrln)-f
lm
wf
]h
(
: ( v,wrmt,
[o
We must evaluate the terms with
A,0 :0' Thus,
:T
m
:
for n odd and rn odd
otherwise
0 separately. The integral over
i i'"
l=-la
(#)' ^
@
^:
\.i-o)'' *'
n I^(Tt)
1,=".?
Combining positive and negative m terms, we have
. /n!3\
Q(p.:.0\:ryi
.-
n=t,
Figure 8.15 shows the
i aP'r
m:t.
solution;";t"
a:
h/2.
r (wr ^\
sinmolm\ P)
'
m ,^(Tt)
is zero,
so
8.4 PROBLEMS WITH CYLINDRICAL SYMMETRY BESSEL
FUNCTIONS 417
o
V,
h
4
p
0
0.5
n
: h/2. This plot shows @(p)/ V6 versus pf h
The
dotted
line
represents
z : h/8; the dashed line, z - h/4; the solid
/2.
h / 2. The first three values of m and n are included in the sum.
FIGURE 8.15. The solution to Example 8.5 with a
at
0
:
Iine, z
tr
:
8.4.9. Continuous Set of Eigenvalues: The Fourier-BesselTransform
In Chapter 7, we approached the Fourier transform by letting the length of the domain in
a Fourier series problem become infinite. The orthogonality relation for the exponential
functions
+ [,^, (,7)
*p
(-iT)
ax
: a^n
becomes
r[*
2n
J-*
,ikx"-ik'x dx : 6(k _ k,)
That is, the Kronecker delta becomes a delta function, and the countable set of eigenvalues
nn f L becones a continuous set of values -oo < k < oo.
The same thing happens with Bessel functions. With a finite domain in p, say 0 < p < a,
we can determine a countable set ofeigenvalues from the set ofzeros ofthe Bessel functions
J., as we did in Example 8.4. If our domain in p becomes infinite, then we cannot determine
the eigenvalues, and instead we have a continuous set. The orthogonality relation
Io"
,^
("^:) n (,*e)
p dp
:
f;ul{o*,)t't,o
418
cHAprER
I
sruRM-LrouvrLLE THEoRy
becomes
(8. r
l3)
(The proof of this relation is in Appendix VIII.) Then the solution to the physics problem
is determined as an integral over ft. For example, a solution of Laplace's equation may be
written
as
e(p,Q,z) : I
,'^o [* A*(k)f (kz)J^(kp)dk
7Jo
where
/(kz)
exponentials
depends on the boundary conditions
,-kz
un6
in z. It will be a combination of the
"*kz.
The potential on a plane at z : O is given by the function
Find the potential above the plane at z > 0.
sin(pla).
Vo@/p)
Example
8.6.
Theappropriatefunction of zise-k',chosensothatO
V(p)
:
+ 0asz -+ oo(along
way from the plane). Then the solution is of the form
Q(p, Q, r,
:
+oc
nN
*\:*r'-' Jo
Evaluating O on the plane at z
:
A^(k)e-kz J*(kp) dk
0 gives
+oo
V(p,il: o(p,d,0) : l.
*a*
roo
"i'Q JoI
l,-(t)t^(kp)dk
Now we can make use of the orthogonality of the ei*Q. We multiply both sides by
e-i^'o and integrate over the range 0 to 2n. On the right-hand side, only the term
with
m:
nrl suryives the integration, and we get
Io'"
,rr,Q)s-i-'q dQ:2n
fo*
o^,{o)r^,(kp)dk
which is a Fourier-Bessel transform. Next2o we multiply both sides by pJ^(k'p),
integrate from 0 to oo in p, and use equation (8.1 13) to get
/^oo f2o
J, J,
v @, Q)s-ima d0 J^(ktp)p dp
: zo
: r,
20We also drop the primes onthe mt for convenience.
fo*
[@
Jo
fo*
o^{o)r^(kp)J^(k' p)p
3(k - k'\
e^&)*:dk
dp dk
A^(k'\
:2r:T
8.5 SPHERICAL BESSEL
FUNCTIONS 419
which determines the coefficient A^(k') in terms of the known function V (p , Q).
In our example, V (p) is independent of @, so only m : 0 survives the integration
over S,leaving
Ao(k):
ifka>t
' f*--a
'lallT
vs: sin Lro&doap:rovo{o
'
a
rt
Jo
-G8 k:a . r
k
(The integral is Gradshteyn and Ryzhik formula 6.671#7.) Thus,
a@,2):
voa2
:ro
[''" -L]o(kp)e-kz
Jo
1/l - (ka)z
f1
dk
x/
Jn'#n(*2)"-"/'a*
The solution for O(p, z) on the z-axis and at p
-
2a is shown in Figure 8.16.
o
Vs
1.0
0.8
0.6
0.4
o.2
z
a
0123
FIGURE 8.16. This plot of the solution to Example 8.6 shows
p
:2a.
@
(p, z) on the z-axis (P
:
0) and at
The integration was done numerically.
8.5. SPHERICAL BESSEL FUNCTIONS
8.5.1. The Wave Equation in Spherical Coordinates
In Section 8.4.1, we showed how Bessel functions arise as solutions of Laplace's equation
in cylindrical coordinates. Closely related functions-the spherical Bessel functions-arise
as solutions ofthe wave equation in spherical coordinates.
The wave equation (3.15)
I a2F _n
" _ __
V.F
,2 at2 -"
420
CHAPTER
8
STURM-LIOUVILLE THEORY
may be solved by first Fourier transforming in time. The wave equation is then transformed
to the Helmholtz equation (3.16) with P : 02 /c2 ,where a.r is the Fourier transform variable
(angular frequency ofthe wave) and c is the wave phase speed. For waves from a point source
or in some other spherical geometry it makes sense to write the V2 operator in spherical
coordinates:
I a /"aF\ *,,sinl L(rineiF) + r A2F +k'F :0
l;1";
) =l 00\ ap ) 7suzt aF
(8'l14)
Next we separate variables, as we have done before. First we write F : R(r)@(d)O(@),
multiply the whole equation by r2 sin2 9, and divide by F, thus isolating the @ dependence:
sin2
e
a
R0r
(r#).##("",#)
The last term must therefore be a constant, and if our region of interest is 0 < Q < 2r , then
we must choose the constant tobe -m2 so that the solutions are periodic with period 22.
Then
Q
: sinmQ,
cosmQ
and, after dividing by sin2 d, equation (8.114) becomes
ta
Eu
("#)
tk2r2.
#*# ('"#) - k
:o
Now we have separated the equation completely, since the first two terms are functions
of r only while the last two are functions of 0 only. The last two terms are set equal to
us the Legendre equation (refer to Section 8.3, equations 8.15 and 8.48)
with solutions Pf (cos 0) and Ql @os 0). The remaining terms give the equation for R:
-l(l + 1), giving
*# t'#) lk2r2-I(l+1):o
We proceed by letting
(8. r 1s)
n = Z /"fi.Then
dR
dr -
IZ
213/2
'
Z,
rt/2
-J--
and
d
d,
So the equation
/"dR\ : rz
ll; ) -;;E
+ rt/2 z' + 13/22"
for Z is
-ii,
1vt/2zt
+
r3/22" + k2r3/22
-
r(t
+
t's
,,,1-
:o
8
5
421
SPHERICAL BESSEL FUNCTIONS
Dividing by rr/2 and gathering terms, we have
d
/
t\22
*(rz't*k'Zr-(,*ri --0
which is Bessel's equation (8.69) of order /
that we need is
+
I12. Thus, the solution of equation (8.115)
: !*''lo
R(r)
This function is called a spherical Besselfunction.The usual normalization is
irk):
T7
(8.1
\f 2*Jr+rn@)
l6)
Thus, the full solution to the wave equation in spherical coordinates is of the form
F
(r, 0, Q,
t)
:
j1(kr) Pi
@os 7Seimf
r-itcct
(8.
l 17)
We may define a similar "spherical" analog of each Bessel function. For example, the
spherical Neumann function is
n1(x): tf
fiN+tp1,1
In the large
argument
(8.1 I 8)
limit, the spherical Hankel functions (refer to Section
8.4.1,
equations 8.78, 8.79, and 8.8 i) have the form
rtlr){rrr)
=
rEry -'#
: (;,1+t{
asr ->
oo
(8.'e)
Thus, solutions of the form
F (r, 0 , Q
'
t)
:
n[t)
1t
r)
r1' 6os g)ei-Q e-ikct o ,ik(r -ct)
correspond to outgoing waves. Since the wave intensity is proportional to lF l2 , and I F | l/r forlarge r, this result corresponds to the usual inverse square law. The second Hankel
function 7@ (kr) describes incoming waves.
Properties of the spherical Bessel functions follow in a straightforward manner from the
properties of the Bessel functions. For example, the recursion relations (8.89) and (8.90)
422
cHAprER
I
sruHM-LrouvrLLE THEoRy
become
*
it+r@)
yr-r (x)
:
2t+1
+t/2)..
2(t
.
(8.120)
xx
-/t(x)
-j,(x):
and
jr+r(x)
-
jr-t@)
: - 2d
: - 2d
a*Jt*tpG)
.F *1,/x7@)1
6
- -"djr$)
dxx-
jr?)
We can simplify this relation by using equation (8.120):
(21* t)ljt+r@)
(l-t
jr-t(x)l
:
-2(2t + gditr?)
ax
I) jr+r@)
-
tjt-r@)
-
:
-
-(21*
Ur+r(x)
*
"r)-r(x)l
DT
(8.
r21)
We may also write the spherical Bessel functions in series expansions. For example, using
the series (8.73) for h+t/2, we find
i)'
1x1t+!+2"
: ,- ---J-*
irGT:,Fi
\r)
" V k;rT;+T+T
jt@)
In particular, for
I
:
t
_Ji L^nlf
2
N:U
(-l)'
(n-tl*
\
1r
f
(8.122)
0,
io@):+fi##t
Let's simplify the
(;)'.^
(;)^
function:
.(,*'u):(,.j) .(,.;) :(,*;) (,-;) .("-:)
: (. * :)("-;)("-) j.(;)
:s#!r':W#:W#
(8 123)
8.5 SPHERICAL BESSEL FUNCTIONS
423
and so
rrr (n +
7l
\/ -(2n
:):
-l l)l
22n+l
Thus,
sln.t
,,b(x):L^#{r,^
(8.124)
x
In fact, each spherical Bessel function may be written in terms of sines and cosines. For
example,
jt@):
jz(x)
sin
x
----;-
-
cos
.x
(8.125)
x
l\: (/3
lsinx\r'--- x/
3
(8.126)
1 cosx
(See Figure 8.17.) As / increases, the expressions become more and more complicated.
f
(x)
1.0
0.8
0.6
0.4
o.2
0
-/
16
-o.2
FIGURE 8.17. The first few spherical Bessel functions j6(x),71 (x), and 72(x). Compare with
Figure
8.
I
l.
424
cHAprER
I
sruRM-LrouvrLLE THEoRy
For large arguments, the expressions become
simplell:
1 /
T\
j"(x) - -sm(\.r-"r)
for
x
)) I,n
(8.r27)
8.5.2. Orthogonality
Equation(8.115)isof Sturm-Liouvilleformwith
the weighting
function w(r) :
f(r):-12,g(r):l(l+ l),f.:
k2,and
12. Thus, the orthogonality relation is of the form
f"' ,' 1,{t r)
11(k'|r)
dx
:o
(8.128)
unless fr : t', provided that boundary conditions of the form (8.2) are satisfied. As with
the regular Bessel functions, since /(0) : 0, we have
fb
,2
Jo
.irtt
r) jr\k'r) dr :
o
provided that the boundary conditions (8.2) are satisfied at r : b,regardless ofthe boundary
conditions, if any, that apply at r : 0. (As always, we do require that the function remain
finite when its argument is zero.)
For k : kt : qt+tl2,pf b,where dr+r/z,p is the pth zero of J+t/2, and therefore also of
7), we may use relation (8.96) to obtain
pb ^
rz1r{trl1r(kr)dr
Jo
rb2
: ;T fbrJ1,412(kr)J1+rp(kr)dr : ;;tJ|*rp@t+,tz.)12
J,
But
.
i'(x)
: tlfnZeJ' J,:
lit
,,1
J
Zt xy2
,lT (''.
I
: Tir
ur' - ;i
,,1
*)
Thus, here we have
7b
Jo
,,i,ro,,i,(kr)dr
:
:
2lCompare with equation (8.83).
n b22kb
;7'i
jt@+tp,il12
*;##)
liir",*'/2.il
1.,,
f;tiiro,*r12,))2
(8. l 29)
425
8.6 THE CLASSICAL ORTHOGONAL POLYNOMIALS
where we used the fact that dr+r /2, p is a zero of n. We can write the result in terms of the
functions themselves rather than the derivatives22 by using the recursion relations (8.120)
and (8.121):
fn'
,,
i,{*ii,&r)
dr
:
(8.130)
f;tj,*r{o+t/2,))2
Spherical Bessel functions find particular application in the study of electromagnetic
waves (see Jackson, Chapter 9) and also in the quantum mechanics ofa particle in a spherical
cavity (Problem 55).
8.6. THE CLASSICAL ORTHOGONAL POLYNOMIALS
The functions we have studied in this chapter have several common features:
.
They are orthogonal on an interval
example, relation 8.30).
.
.
.
They satisfy a set ofrecursion relations (for example, equations 8.34 through 8.41).
[a,b]with
respect to a weight function ur(x) (for
They may be computed from a generating function (for example, equation 8.32).
They may be computed from a Rodrigues{ype formula (for example, equation 8.42).
These properties are coflrmon to a larger class of orthogonal polynomials that arise in
physics problems. They are defined by the generalized Rodrigues formula:
cp(x)
11dn
- K" r@ dr"lw(x)snl
(8.1
3
1)
u(x), the weight function, is real, positive, and integrable on the interyalla,bl;
s(x) is a polynomial of degree < 2 withreal roots; and the product rus satisfies the boundary
where
conditions
w(a)s(a): u(b)s(b) :0
In addition, the first polynomial Cr(x) is a first-degree polynomial in x. The constant K,
serves to normalize the polynomials. [Notice that the Legendre polynomials satisfy these
conditionswith
22See Problem 48.
?r(r):1,s(.t) :
x2
-l,a:
-l,b:1,
Kn -2nnl,andC1
(x):1.]
426
CHAPTER
8
STURM-LIOUVILLE THEORY
We shall begin by showing that the function
defined by equation (8. 13 I ) is a poly-
C
"(x)
nomial of degree n. To construct the proof, we use the symbol
pr(x) to represent
any
polynomial of degree < k. Then
:
ffi@r" rD
(8.132)
u)sn-m pk+m
where ppy^ is another polynomial of degree < k I m. To prove this result, we start with
the definition of Cr from equation (8.131) to obtain
wK1C1:,# *
whereKrCr
that
r# + ,# :.(*rr, - #)
-dsldx isapolynomialof degree <
1.
(Remember:Wehavealreadyspecified
Cl is a first-degree polynomial.) Then
pil : d,' ,n po + nw!s'-t
!@r"
dx
dx
clx
rr, + wsndlk
ax
:,(*rr, - #)u-'rx + nwfrs"-'po +.u*
=
u,sn-t
l("'t'
+ (n -
"#)rr
*'#f
Sinces(x)isapolynomialof degree <2,dsf dx isof degree< l,anddppldx isof
degree
< k - l, the term in brackets on the right-hand side is a polynomial of degree < k +
d
fttws'P1r'1
:
1:
wsn-t Pk+l
[Given specific forms for the functions u(x), s(x), and a polynomial pp, we could use this
relation to construct a specific form for the polynomial p*+t, but we don't need to do that.l
We can continue to differentiate until we obtain the result (8.132).
Setting n : m and k : 0 in equation (8.132) and using the result in equation (8.131),
we obtain
Cn(x):
Pn
that is, Cn is a (yet unspecified) polynomial of degree < n. This polynomial may be written
as a polynomial of degree < n - l, with the possible addition of a term in xn:
Cn
:
pn-r
I Ax"
Next we shall show that A is not zero, and thus Cn is
a
polynomial of degree n.
(8.133)
427
8.6 THE CLASSICAL ORTHOGONAL POLYNOMIALS
To show that A is not zero, and at the same time establish the orthogonality property
the polynomials Cr, we shall first show that
f"u
n.{*lc^(x)u.'(x) dx
:o
for m
<
n
of
(8.134)
for any p*(x). We begin with the defining relation (8.131) and integrate by parts:
rb
I
J"
p^@)Cn(x)w(x) dx
:
lfbdn
i
J" P^(*)*;l*(x\snldx
tb
rh
tn-l
tn-l
\
| /
: nlo^t*>fr,-tt*t')"11,
- J" o^-,@h*J G)s\dx)
where we used the result that the derivative of a polynomial of degree < m is another
polynomialof degree < m-7.Result(8.132) showsthatallthederivatives(d- f dx^)(us')
withm < n vanishattheendpointsr : a andx: b of theintervalbecause u.rs does, and
thus the integrated term vanishes. We may now continue to integrate by parts, reducing the
order of the polynomial in the integrand by one each time, until we obtain
b
p^(x)C"(x)w(x) dx
:
< n, and the next integration shows that the integral (8.134) is zero.
Now we multiply equation (8.133) by C"w and integrate:
for m
l"u
{r,)'w(x)dx:
f"u
o,-rc,wdx *
:o*A
f"u
t
fob
*nc,*d,
*'c^rdx:1,
where we used result (8.134) with m : n - 1 to eliminate the first term. The left-hand
side is > 0, since the integrand is positive throughout the range of integration, so A must
be greater than zerc, and Cn is thus a polynomial of order n. Finally, we set pm : C* in
relation (8.134) to obtain the orthogonality relation we seek:
I,'
r-orr,(x)w(x)dx
:o
form < n
Properties of some of the polynomials used in physics are listed in Table 8.3.
(8.
l3s)
428
cHAprERosruRM-LrouvrLLETHEoRy
TABLE 8.3. Classical Orthogonal Polynomials
u(x)s(r)abKn
-|
-1
e-"
I
-oo +m
xue-"
r
Irgendre
I
Hermite
Laguene
x2
o
|
!t-i-
I
nl
+oo
+l
-l
2n
(-l)'
1l
Ji2"nl
f(n*u*l)
nl
(-2\nnl ,ffi
Jacobi (l-.r)v(la.r)r l-x2 -1 +l
Tchebichef
2
2nnl
+l
(-l\nr
(Zn\l
2nn!
tt
2
-
PROBLEMS
fl
fina the eigenfunctions for the one-dimensional Helmholtz equation
d2v
7j
+ k'Y
:o
subject to the boundary conditions
)=0
atx
:0
and
!' :o
at
x
:
L. (This corresponds to a vibrating string with one fixed end and one free end.)
2. Find the eigenfunctions for the Helmholtz equation
d2v
7fi
+ k'Y
:o
subject to the boundary conditions
atx
:0
al *
bYt
:Q
*
Fy'
:0
and
ay
atx:
L.
429
PROBLEMS
Determine specific forms for the functions, and obtain the eigenvalues for the following
CASES:
(i) a:aandb:F
(ii) aL:bandF:2aL
You should be able to obtain numerical values for the product
kl.
3. The displacement of a square vibrating membrane of side L satisfies
the two-dimensional
Helmholtz equation
02s 02s
arr+ arr*k's:o
where
k : olu,
rr.r
is the frequency, and u is the speed of waves on the membrane.
x : 0, L and y : O, L. Separate variables
and solve for the eigenfunctions s(x, y). Show that the system exhibits degeneracythat is, that there is more than one eigenfunction corresponding to a given eigenvalue
k2. In particular, show that there are two eigenfunctions s1 and s2 that correspond to the
eigenvalue k2 : 5n2 / L2 .What symmetry of the physical system causes this degeneracy?
(Hints: Where are the nodal lines for the two modes? What happens if one side of
the membrane is slightly shorter, equal to L - e?) Any linear combination of the two
eigenfunctions is also a solution. Find some of the nodal lines for combinations of the
modes-for example, .rt + s2. How do these modes reflect the symmetry of the system?
Can you find an eigenvalue that has threefold degeneracy? If so, what do those modes
look like?
Suppose the membrane is fixed at its edges at
set of eigenfunctions y,(x) satisfies the Sturm-Liouville equation (8.1) with boundary
conditions (8.2). The function g = 0. Show that the derivatives un(x) : y'n@) are
also orthogonal functions. Determine the weighting function for these functions. What
boundary conditions are required for orthogonality? Apply your results to the Legendre
equation to determine the orthogonality of the derivatives fi@).
@A
5.
Use the recursion relations in Section 8.3.5 to show that the derivatives Pi
(f)
of the
Legendre polynomials are orthogonal on the range (-1, 1) with weighting function
0 - 1"2), in agreement with the results of Problem 4.
6. To obtain Fourier-Legendre series, we often need to evaluate integrals ofthe form
,f
:
lo'
tr"
h1t)
dpt
(a) Start by evaluating tf , tl, I[, and Ii.
(b) Next use the recursion relations for the polynomials in Section 8.3.5 to determine
recursion relations for the integrals f . Multiply equation (8.37) by ptn andintegrate
by parts to obtain
(c) Use
a
relation between
ti nd li-rt
.
these results to step down until you can use your results from (a) to obtain an
430
CHAPTER
8
STURM-LIOUVILLE THEORY
explicit expression for
If.
Show that
nl
(n-l)ll(n+/+1)!l
t-n-t (l-n-2\ll
(-l)---rn!'
(n*/*l)!!
0
if n > I
If n <
I
andl
-n is odd
ifn<landl-niseven
7. Yeify that the Rodrigues formula gives the correct normalization for every Legendre
polynomial. Hint: Wite (r2 - t)t : (x - l)l(-x + 1)', differentiate, and evaluate the
resultatx:I.
@
Evaluate the integral
f+l
t
PtG)
J-r Jl - xz
dv
for the function I l\n-'.
and hence obtain a Fourier-Legendre series
cursion relations in Section 8.3.5 may prove useful.
Hint: Tlte re-
9. Write Laplace's equation in oblate spheroidal coordinates (refer to Chapter 2, Problem I 3), separate variables, and hence show that the solution requires Legendre functions
in both the coordinates z and u. Argue that the solution exterior to an oblate spheroidal
boundary requires the use of the Legendre function of the second kind, Q.
10. Expand the Legendre function Qo@) for large values of the argument, and show that
your result agrees with the asymptotic form in equation (8.29), modulo a constant.
11. Rewrite the Legendre equation
d
d. (o -
l1|
*tff) *,0 + t)et :
o
in terms of the variable u : I I x, and obtain a solution as a series in u. Show that for
latge x, Q7(r) goes to zero as If xt+t. Show that for / : 0, the solution Qo@) may be
written as in equation (8.28) but with x - I in the denominator instead of 1 - x.
Cunent flow in aconducting sheetis describ-edby the relations fr : -VO ana j: ofr.
Use the charge conservation law (0pl0t + V .j : 0) to show that, in a steady state, O
satisifes Laplace's equation. Find the eigenfunctions for current flow in a circular copper
plate.
Current 1 flows into and out of a plate of thickness t and conductivity o through
r: a extendingfromd :7t -yl2ton 1-y12 andfrom I : -y12to
*y /2.Determine the potential, and plot the current flow lines.
electrodes at
13. A solid sphere of radius a is immersed in a vat of fluid at temperature 16. Heat is conducted
into the sphere according to equation(3.14).If the temperature at the boundary is fixed at
Zo and the initial temperature of the sphere is Ir, find the temperature within the sphere
as a
function of time.
14. Use the Cauchy formula together with the Rodrigues formula to write hQt) as a contour
integral in the complex plane. Take the contour to be a circle of radius \62=, and
PROBLEMS 431
hence obtain the integral expression
I fr r
r1
hA): - Jo lx + t/x2 -
tcosQ)
dQ
15. Starting from the relations in Section 8.3.5, derive the following recursion relations for
the associated Legendre functions:
:
(a) (/ - * + Dt/TlEPl-t
Pi\1 - wPi
(b) (2t + D\/T=EP{-t Pilr - Piir
:
Hint: Start with the pure recursion relation and differentiate.
(c) From relations (a) and (b), derive the following:
(2t
+ I)p.Pf @) : (t -
m
* r)Pi\+
0 + m)PLt
16. Starting from the definition (8.53), obtain the m.rusing recursion relation for the associated Legendre functions:
-^-+Pi1n
- JTl: dP eitu)
t/l - lt'
r;"+t{u):
Combine this result with equation (8.59) to obtain the m-loweingrelation
(t
*m)(t -m
t
t)Pi'-'
: ,f,4hPi' - mJ+EPf
17. Use the results of Problem 15 to show that, for I + m even,
+ m - r)lt'
Pf Q) : eD(t+D/2(l
(t - m)tt
lE.
Show by direct substitution into equation (8.15) that P#(e) o< sin- d. Use the value
the orthogonality integral (8.55), together with the result
rn /2
["''
Jo
,inz^+t
o
dg
:
of
Qm)ll
Qm 1- l)ll
(for example, Gradshteyn and Ryzhik formula 3.621#4), to show that
P#@)
: (-t1m94
2mml,
"inm
s
This relation, together with the z-lowering relation (Problem 16), may be used to generatethe Pf .
19. Verify the result
If+t
Jt
I (t r m)l
' Ut)12
dLr:-l-p' = ' m(l -m)l
Lpl"
432
CHAPTER
8
STURM-LIOUVILLE THEORY
for the second orthogonality integral in Section 8.3.7 in the cases
(c)l:m
(a) l:m:l
(b)l:2,m-l
(d) Stepping down in m, vse proof by induction (Appendix III) to show that the result
is true in general.
[ZOl Using the generating function G(x, tD (equation 8.32) and the addition theorem (8.65),
derive the expansion
-ml
#u : [,t, h#'^''
ilYfi^ (e"
6'7
where r= and r, are the lesser and the larger ofr and r/, respectively.
Hence find the magnetic vector potential due to a circular loop of wire of radius a that
is carrying current 1. (Refer also to Problem 6.19.)
21. Verify the result (8.67)
I,'
where
/
Pi
@)d w
: (- 1)("+r),'tT r,*rfo) Pf Q)
and m are both odd.
(a) First evaluate the Legendre functions to show that
l-2)lt(l*m-t)tt
_
f+t
It^:
I PiQt)du:
, ,i,, +,(l-m)ll
\r'' ' -mn )i
t
(+t)tl
J-t
(b)
I,m
odd
Use the expression for Pff (0) from Problem 18 to show that the result is true for
m : I . Hint: Use contour integration.
(c) Show that the result is true for m : I and / equal to any odd integer.
(d) Use proof by induction (Appendix III) to show that the result is true for all m, | <
m < I, with both I andm odd. Hint:Usethe result from part (b) and step down in m.
22. Find the electrostatic potential inside a hemisphere of radius a with potential O
the flat side and Q : V on the curved part.
:
0 on
23. Quantum mechanical treatment of the harmonic oscillator results in the Hermite differential equation
y,,-2*y,*ly:0
Write this equation in standard Sturm-Liouville form.
If
the boundary conditions are
y(x) -+ 0asx -+ foo,showthatthesolutionsareorthogonalontherange(-oo, *oo),
and find the weight function u.r(x). Solve the equation to find a series expansion for the
Hermite functions. What value of the eigenvalue l, is required for the functions to remain
bounded throughout the interval, including x -+ Iec? (Hint: Experience with Legendre
functions should prove useful.) Normalize the solutions by choosing the coefficient of
the highest power x' to be 2n , and hence determine the first three eigenfunctions.
PROBLEMS 433
[Z+]fne generating function for Hermite polynomials is
G(x.t): '-t2+2'r: i l"r,t,
7=on!
Use this generating function to establish a pure recursion relation for Hermite polynomials (analogous to equation 8.34 for Legendre polynomials). Also obtain the derivative
dH"ldx in terms of the Hn (analogous to equations 8.40 and 8.41).
Differentiate G(x, t) with respect to t a total of n times to obtain the Rodrigues-type
formula
Hn@):
(-l)'r*'#{"
25. By using
the Rodrigues formula (Problem 24) for the Hermite polynomials or otherwise,
obtain the normalization integral:
/n+oo
I ,-"lHn(x)12
J--
dx
- 2"nt.Ji
26. Starting with relation (8.86), derive the Rodrigues-type formula for Bessel functions:
Jn(x)
/ I dY
- *' (\ -:ilxdx/
Jo(x)
27. A drumhead is a circular membrane of radius a. When it is struck, waves propagate
across the drumhead. The membrane vibrates with displacement f , where f (r,
ry1r,0)e-i't and ryQ, d) satisfies the Helmholtz equation
0,t) :
la (-an\-7flzn
;u\'u)+7aertk"q-s
: ,2 lu2 and u is the speed with which waves propagate across the drumhead.
(The speed u depends on the tension in the drumhead, among other things.) The boundary condition is that 4 :0 at r : a. Separate variables, and find the eigenfunctions.
Determine the first three allowable frequencies co in terms of the drum parameters u
where k2
and a.
28. Sound waves propagating through a tube may be described by a velocity potential (refer
to Chapter 2, Section 2.4) that satisfies the Helmholtz equation
/
a.,2\
"
(o'*7)*:o
where c, is the sound speed in the tube. Assume that for propagation along the length of
the tube (in the *z-direction), the potential may be written
Q : Qteikz
434
cHAprEB 8 sruRM-LrouvrLLE THEoRy
(D1 is a function of the transverse coordinates (x and y or r and 0). Because the
air cannot move perpendicular to the walls of the tube, the boundary condition is
where
-ao
fr.VO : =- :0
0n
on the boundary surface
Write the differential equation and boundary conditions satisfied by
the eigenvalues and the set ofallowed frequencies a-t if
<D1
and hence find
[a)] tne tube has a rectangular cross section measuring a x b
(b) the tube has a circular cross section ofradius a
In each case, show that there is a minimum frequency for waves that propagate along
the tube with <D1 not constant.
29. If y.n is the nth zero of Jh@), show that the Bessel functions satisfy the orthogonality
relation:
F
Jo"
t^
/ o\
(^,*) l.
/ o\
o2 /
(y*ol) o ao : T
lt
^2\
- ;)
u^ (y.)t23,t
30. Use the generating function (8.93) to show that
(a) sinx
(b)
@
:2D
n:o
Jz,+r(x)
1:,/o(x)+zitz,@)
n:l
31. Use the generating function (8.91) to show that
+co
Jn(x*y): t
m:-@
J*@)Jn-^(y)
,
and hence show that
Jo(2x)
: t& @)+ z ir-rl
^ J3@)
m:l
bi]
(Hint: Use the orthogonality of the complex exponentials.)
Starting from the Bessel differential equation, show that
[@ J^(x)Jn@)
. 2 sinl(m -n)r/2] fornr *rr > 0
'
' dx:
I
---#
X
It n7'-n'
JO
and that this result equals ll2n when ltl : n. Does this result constitute
a second
orthogonality relation for the Bessel functions? Why or why not?
33. At time / : 0, the surface of the water in a pond has the form s(p, Q,0) : AJo@d.
By taking the Fourier transform of the wave equation with two spatial dimensions, find
the displacement s (p, 0 , t) at later times.
PBOBLEMS 435
34. (a) Show that
[@l
I e-ox Jn(x\dx ::
Jo
Ja2+r
Hint: Use the integral expression (8.92) for Jo(x) and perform the integral over x
(b)
first. Do the integral over the angle using methods from Chapter 2.
Use the same technique to evaluate the integral
e-o*
fo*
J^(*) d*
35. Show that
*t roro*, a,
fot
: (t - 1) rro
if, instead, J1@) : Ql
(other
than zero) of the Bessel function J^(x), xr, 1 , is an
36. Show that the first zero
increasing function of m-that is,
If Jo@)
:
0. What is the value of the integral
I0,1 <xl,l<XZ,l<Xj,I
and so on. Hint: Use relations (8.86)-(8.90).
@ e pendulum has steadily increasing length /(r) :
/o
*
cvt. Show that the equation that
describes small oscillations of this pendulum is
10"
+2a0'
*
g0
:0
Change variables to tt : (1 -f at I lslr/2 and y : u0, and hence show that the general
solution may be expressed in terms of Bessel functions. Find the solution if the pendulum
is released from rest at an angle 0g at t :0 (that is, when I : ld.
38. The equation that describes the angular displacement of
a vertical pole or column is
d2o
trl7
* P,xO :o
where x increases downward from the top of the pole, E is theYoung's modulus, 1 is the
moment of inertia (see also Chapter 3, Section 3.2.3), and L is the mass per unit length.
Make a change of variables to
2fil"^
':;,,1fr*'' v : Ji
e
and hence show that the solution may be expressed in terms of Bessel functions. Show
that there is no solution that fits the boundary conditions 0 (L) : 0 and d/(0) : 0 unless
the pole has a minimum length 1161. Find an expression for L,;n in terms of the physical
parameters of the pole.
436
cHAprERssruRM-LrouvrLLETHEoRy
39. Establish the addition theorem for Bessel functions:
.Io(kR)
:
J*(kp)J^(kpt)
^t*r'^,
where
n
: Jfi
@,)2
-
2ppt cosQ
Hint: Begin by expressing I/l* - i'l as an expansion in Bessel functions, as in
Section 8.4.9, and set 3 : z' : O and 0' : 0. Evaluate the constant A.(k) using the
result of Problem 34.
FO.l Starting from the definitions (8.100), (8.76), and (8.78), show that
Ku(x)
n l-u(x) - Ir(x)
- ,
,i"r"
where u is not an integer. Hence, show that
41. The potential on a plane is V6 for p < a
K-u@)
and zero for
:
Kr(x).
r > a. Find an integral expression
for the potential everywhere. Evaluate the integral with
p:
0 to find the potential on
the z-axis.
42. A cylinder of height h and radius a has the top and bottom grounded. The potential on
the wall at p - a is Vs. Find the potential inside the cylinder.
43. A cylinder of height ft and radius a is grounded except for its base at z : 0. On the base
the potential is Vs f \/ I - p2 / a2 . Find the potential inside the cylinder.
E4.l
(al
Use the series for "Io(;) to show that its Laplace transform is
LtJod.)l:
(b)
-L
Jl*sz
Then use the recursion relation (8.87) to find the Laplace transform of Jr (x). Extend
the result to show that
LIJ-(x)l: 1uffi_gm
'JP
+l
Compare with Problem 34.
(c) Use the convolution theorem to establish the relation
['
JO
,or,
-
u) Jo@)
du
:
sinx
45. Obtainexpression(8.126)for jz@) fromtheexpressionsfor jsand j1 andtherecursion
relation (8.120).
46. Starting with the recursion relations (8.86) and (8.88), derive the relations
a (it@)\_
7+t@)
E\-;-)----7-
PROBLEMS 437
AIId
d
t+l
ftt
ir{il:
xt+t
it-r@)
lZTl Use proof by induction (Appendix III) to establish the Rodrigues-type formula
jn(x)
:
(_t)n xn
(: *)" (Y)
48. Use the recursion relations to show that the orthogonality relation (8.130) is equivalent
to (8.129).
49. Starting from the definition (8.76), show that
n1
(x)
: (- 1)l+l j-rl+ri (")
Hence show that
ns(x)
- -cos'r
x
50. Starting from relations (8.111) and (8.112), establish the recursion relations for the
modified spherical Bessel tunctions kt@)
: JT/iiKlayp@):
h-r-kt+r:-('zJlf\k,
\r
/
and
(t
i
t)k1a1-t tk1-1
:
-(2t + lft*t<x1
[it--] rne Fresnel integrals
s(x)
: ,E
I'
sint2 dt
and
may be expressed as series of spherical Bessel functions. First show that
I f*2
s(x):hJ,
Jiio@)au
and obtain a similar expression for
C(x). Use the recursion relation to do
and hence establish the result
6oo
S(x)
: \l:.D
I J'
n:l
Determine a similar expression for C(x).
jzn-t@2)
the integration
438
CHAPTER
8
STURM-LIOUVILLE THEORY
52. Sound waves in a spherical cavity satisfy the differential equation (V2 + t<2) F : 0 for
r < R with 0F l0r : 0 at r : R. Find the eigenvalues k, for the problem and hence
find the allowed frequencies @n : knu for sound waves inside.the cavity.
53. Electromagnetic waves in a spherical cavity may be described by a mathematical problem
similar to that described in Problem 52, with r) : c, the speed of light. The boundary
conditions depend on the polarization. Find the allowed frequencies if the boundary
condition is F(R) : 0.
54. The modified spherical Bessel functions are defined23 as
i1e):
and k1@): rlZK+trzi)
,l!t,*r,r(r)
Y |Tx
U 2x
Using expression (8.99) and the result of Problem'8.40, verify the expressions for the
modified spherical Bessel functions 16 (x) : sinh x I x and ko (r) : e-x / x. Use proof by
induction to show that
h@)
:1_7)txt(:*)'+
55. We may model the force between particles in an atornic nucleus with
a
three-dimensional
square-well potential
v(r\:[-vo
|.0
forr<R
forr >
R
Schrodinger's equation for this system takes the form
m
(o, - zftvr,t) v : -2-Eill
hz
Write the differential operator in spherical coordinates and show that the solution may be
writtenintermsof sphericalBesselfunctions.Withcy :2(m/h2)VoR2 ande: -ElVo
with E < 0, show that the energy levels are determined by the equation
Jr -irt("J;) i+r("Jr [iCl
:
') J;i,(",h - ')r,t@1r61
Find the energy ofthe lowest energy level for / : 0, cy : 10.
fn" density of neutrons in uranium is described by the equation
!:pyzrqo,
At
where D (the diffusion coefficient) and a (the net production rate) may be taken to be
constants in space and time. Solve the equation using separation ofvariables. Look for
a solution with spherical symmetry that satisfies the boundary condition n : 0 atr : R.
Show that the density increases exponentially if R exceeds a critical value Rs;1, and
determine that value in terms of
D
23Some authors choose different normalization.
and a.
OPTIONAL TOPIC A
Tensors
A.1. CARTESIAN TENSORS
In Chapter 1, we showed how physical laws, such as Newton's second law (equation 1.8),
may be-represented as mathematical relations between vectors. In equation (1.8), the force
vector F and the acceleration vector d are parallel. It is sometimes the case that one vector
is linearly related to another, as in equation (1.8), but the directions of the vectors are not
the same. An example from mechanics is the angular momentum L of a rigid body, which
is determined by the body's angular velocity <i but is not parallel to 6 utless the body has
sufficient symmetry.1 In electricity and magnetism, the current density j in a magnetized
plasma is not necessarily parallel to the electric field E that drives it, because the magnetic
force causes particles to gyrate. In these cases, the vectors are related by a linear operator
that mixes the components. For example, the angular momentum is
i :
ll<ir
where ll is the inertia tensor.2 The vector components are related by a matrix. In index
notation,3
L;:
of
I;ia1
(A.1)
The matrix ll has 3 x 3 : 9 components.4 It is an example of a rank-two tensor. The rank
a tensor is indicated by the number of indices needed to describe its components.
TABLE 4.1. Tensors in Three-Dimensional Space
Object
Notation
Number of Components
Rank of Tensor
scalar
m
Ui
1:30
3:31
9 :32
0
vector
rank-two tensor
I ij
I
2
I
See, for example, Lea and Burke, Chapter 9; Marion and Thornton; Goldstein, Chapter 5.
2This is not the identity matrix. Here II stands for inertia. In Chapter 1, we used the symbol IR for this operator.
3Throughout this topic we shall use the summation convention discussed in Chapter 1, Section 1.1.2.
4The number 3 here represents the three dimensions of space.
439
44O
oPTroNALToPrc A TENSoRS
As with Newton's second law, it is essential that a physical relationship such as (A.1)
remain true when we change the coordinate system (refer to Section 1.1.1). In the new
system,
L'i: Ilj@'j
(A'2)
We already know how the vector components transform under coordinate rotations, so let's
fransform them:
L'i: AitLt
where A;p is the rotation matrix (equation 1.21). To go in the reverse direction, we multiply
on the left by {-t : A7:
A[,L',
:
:
Al;A;pL1,
6nkLk
:
Ln
so (equation 1.25)
Lr:
AinLti
rom:
Aim@}
Similarly,
so equation (A.1) becomes
A;nLt;: In6Aiaoj
Now we multiply on the left by,A,, and we have
Apn
A[, L',
:
Akn
Ir^ A j^tD'j
6P; Lti
:
L'k
:
Ak,
A
i^In^r'j
Comparing with equation (A.2), we find the transformation rule for the tensor
It,
:
AprAi^In*
ly:
(A.3)
Notice how the indices match up in equation (A.3); the repeated indices indicate that we
must sum over both n and m.
l
A.1 CARTESIANTENSORS
441
We can use matrix multiplication to perform the calculation indicated by the indices if
we first put the summed indices next to each other.5 To do this, we have to transpose one
of the matrices:
Ilri:
Alrnlr^Afi1
or, in matrix notation,
['
:
^AJl.Ar
(A.4)
The index notation (A.3) is more general and more powerful. It extends easily to tensors of
rank greater than two, whereas the matrix notation does not.
The transformation laws for vectors, scalars, and tensors are compared in Table A.2.
TABLE A.2. TFansformation Laws
Object
Transformation Law
scalar
ml:m
u'i: Aiju j
It^ : ApnAyli Ini
vector
rank-two tensor
We need one transformation matrix for each index of the tensor. Extending the rule we
have found, we get the transformation law for a rank-three tensor:
Tlip:
Example
A.1.
A;nAi-AkpTn^p
(A.s)
The inertia tensor has components
ti,:.1{r2ti1 -xix)dm
(A.6)
A uniform
square plate of side s is rotating about an axis perpendicular to the plane
of the square and through its center. The angular momentum of the square about its
center is
L:'t'
6
io
Compute the components of the inertia tensor in a coordinate system with origin at
the center of the square ard x- and y-axes parallel to the sides of the square, and
verify the expression for the angular momentum. Next, compute the components of
5See footnote 8 in Chapter 1.
-.!
442
oPTroNALToPrc A TENSoBS
the tensor in a primed coordinate system with axes rotated by 45" about the original
x-axis. Hence find the angular momentum about its center when the square rotates
about an axis at 45o to the plane ofthe square.
FIGURE
A.1.
The square plate in Example I rotates about (a) the z-axis and (b) the z/-axis.
Since the square is a planar object and all mass elements are at
we may immediately conclude that
I;3-13;-Q
unless i also equals 3. Then, with dm : o dA :
(M /s2) dx
dy
we have
tr,
: \ [''' f '' @2 + v2) dx dv
S' J
J
-s12
-t /2
Pl2 113
:5M l_;,(;.
:5 I::,(*
'
's/2
*,')l_,,,0,
. ,,,)
a,
M /s3
u31rs/2
: ;r
(i, *'?)l_",,
M s4
-s26-
Ms2
6
The other components are
,,,
: 5 l::,1::,,,
+ v,
- *27 d. dv
yt f /z l'''
Ms2
:M
,, Tl_,,r,1_,,r: T
z:
0 (Figure A.1),
and 12
:
x2
+
y2,
A.1 CARTESIANTENSORS
443
and
r,,
: 5 I::, I::,-xy dx o, : -5
Similar calculations give 122: 1rr and
lzt : In.
+l:::,,,*1,'_,,,:
o
Thus, in this system,
tr: ,''/'1S\
o ;;)
lB
Then, when the system rotates about the z-axis,
L;:r;iai:#(i:
g
)
(I
:+( I
)
)
i is parallel to <i in this case.
Now we change to a coordinate system rotated by 45o about the x-axis (Figure A. I ).
The transformation matrix is
Thus,
'/& -;??\i )
^:z\3
The new components of the inertia tensor are found from the transformation rule
(A.3):
I!,:
A;nIn^A[,
::(f, ii)#(s:!Xf
:+(f ii)(f i i)
i i)
:*"24 (3 St ?\
3)
\o
In the new coordinate system, the angular velocity is along the z/-axis and has only
one component, so the angular momentum is
L,i:r,@,j:#
(i?
i)(I)
:+( i )
OPTIONALTOPIC A TENSORS
andisnot parallel to dr.
The angular velocity has components @(A/D(O, -1, l) in the original coordinate system. We may check our result by computing the components of L in this
frame and transforming the result into the prime frame.
',:ry(i: l) ,+(-i ): M,"*( ;)
Transforming the resulting vector to the new frame, we have
i i)(-l ): +( i)
L,i:AiLi:u,,,fit"(f,
which agrees with the result from the ffansformed inertia tensor.
A.2. INNER AND OUTER PRODUCTS
The components of the inertia tensor (4.6) are made up of components of the position
vectors of all the elements of the body. Such definitions are common. In fact, the products
a;b i of vectorcomponents are the components of a tensor called an outer product of the two
vectors d and b. All such products are tensors. In an outer product, the rank ofthe resulting
tensor is the sum of the ranks of the tensors making up the product.
When we sum over two or more indices, as in the dot product a;b; or in the product of
a tensor and a vector to give another vector l;itoi, the result is an inner product. All inner
products are also tensors. The process of reducing the rank of a tensor by summing over
a pair ofindices is called contraction. Every contraction reduces the rank ofthe tensor by
two.
It is fairly straightforward to show that an inner or outer product oftwo tensors obeys the
ffansformation law for a tensor of the appropriate rank.6 In fact, we may prove a yet more
powerful result called the quotient theorem.
If a tensor b is the inner or outer product of c and d
and
if c is a tensor with arbitrary
components, then the set of components d is also a tensor.
For example, let
b;i
Then
:
c;PdPi
if b is a tensor, we can transform the components using equation (A.3):
bL^:
6See Problems 4 and 5.
An;Aaib;;
:
An;Aaic;pdpi
A.sPSEUDO-TENSORSANDCROSSPRODUCTS
445
But from the definition of b in the prime frame and the fact that c is also a tensor,
bL^:
c'^od'p-
-
An;Apkcrkd'p*
Setting the two expressions for b'n^ equal, we have
Ari A^i c;pd11i
Now we multiplyT on the left Ay
enl :
Anq
A^icqtd*i
Next we multiply on the left bV
A,l :
(6,jd*j
(d1r,
-
-
Ani A plrc itd'p^
and use
the result An/ Ani
:
6qi
to obtain
Aorcqtdtpm
Amri
: O
A4rApnd'o_)cq* :0
A^, Appdlo^)cq*
Because the components c4ft are arbitrary, the term in parentheses must equal zero, and so
we must have
dk,
Thus, the components
: ApnA^rdp^: e;) e;)alo^
ofd transform
as a tensor, as required.
A similar proof works for tensors of any rank.
A.3. PSEUDO.TENSORS AND CROSS PRODUCTS
In Chapter 1, we saw that the vector cross product is not a true vector because it does not
possess the proper behavior under reflection of the coordinate axes. Also recall that we may
express the cross product using the Levi-Civita symbol (see Chapter 1, equation 1.30):
(fr x
i), :
eijku
jvk
i are both vectors yet fr x i is not, the symbol e;;1 cannot be a tensor. It represents
pseudo-tensor,
a
or a tensor density. The transformation law for a tensor density is similar
to that for a tensor (Thble A.2) except that we must also multiply by the determinant of the
transformation matrix.8 Thus, for the Levi-Civita tensor density,
Since fr and
eli p
:
det (A) A; o A I q A p, e pqr
From this result and the fact that det (A)
: tl,
we may easily see that
1. The product of two tensor densities is a tensor.
2. The product ofa tensor and a tensor density is another tensor density.
7We cannot divide out the "factors"
8See Problem I 1 and Appendix I.
Ar; andcit
because we are summing over
i
and k.
446
oPTroNALToPrc A TENSoRS
We may avoid the difficulties associated with the cross product if instead we represent
the components of the cross product as components of an antisymmetric tensor. A ranktwo antisymmetric tensor has only three independent components, and these will be the
components of the cross product il x i if we choose
Tij:uiuj-uiuj
In matrix form, the components are
0
/
uruz
- u2ut
utu3
- urrt \
u2u3 - u3uz
T: I u2Dt - utuz 0
0
/
\ ,rr' - utui uyu2 - u2u3
(dxi;, -(frxi)r\
o
/
:l -lilxi)j
(ixi)1 |
o
\ tdxn)z -1frxi)1 o I
I
Then we may also define the vector density:
di
:
which is known as the dual of the tensor
,,
at
:r(Tzl -
az
:
1
1
1Qt -
1
(A.7)
t€ijkTjk
T;1 .
The components of d; are
Tn)
:
T23
:
u2u3
-
u31)2
:
(fr x i;1
Tn)
-
T3r
:
u3ut
-
u1u3
:
(fr x
and similarly for d3. Thus, the vector density d
The inverse dual relationship is
T;.i
:
i)z
ir ttr" cross product i x
i.
(A.8)
e;ipdp
We may demonstrate this relationship using equation (1.34) from Chapter 1:
r;1
:
:
r,1t
)rkrmTrm
1
1(T;1
-
:
){t,,ai^
Tii)
Since ?l; is antisymmetric, the result follows.
-
t;^6i)Tm
l
A.4 GENERALTENSOR
CALCULUS 447
Example A.2. Express the angular velocity and angular momentum of a particle in
circular motion as tensors.
We choose to put the origin at the center of the circle. The angular velocity of the
particle is given by
i:rixi
Since
i
and
i
are vectors,
6 is a pseudo-vector. We can find an expression for dl by
ri is perpendicular to i:
taking anothei cross product and using the fact that
ixi:ix(6xi;-drr2
Thus,
- 1,
@-;rXv
_
We may express these components using the antisymmetric tensor:
to;i:
"Giu1
-u;ri)= pn,t
The angular momentum is
i:ix
i:mixi
We may also define an antisymmetric tensor with components equal to the components
of L:
L;i : m(r;ui -
uir j)
- *?ii
Thus, returning to the pseudo-vector representation, we can relate the components
L to the components of 6 by
of
Li : m'2ai
A.4. GENERAL TENSOR CALCULUS
a non-Cartesian coordinate system, we need to identify two different classes of vector-like
entities that obey different transformation laws. First let's write the transformation matrix
for vectors in terms of the old (u') and new (D') components. A differential displacement
vector d* has components dTi that are functions of the original coordinates x'. Thus, we
may use the usual rules of partial differentiation to express dit as
In
dii :
oT',
0xl
d.*j
448
OPTIONALTOPIC A TENSORS
Since all vectors transform the same way, we obtain the transformation law for any vector
i:
ATt
-iut:rjur:A,jul
where the transformation matrix has components
A,j
ATI
(A.e)
flxJ
You should check that this expression is consistent with our previous result (equations 1. 1 1
and I.L2 in Chapter 1) in terms of rotation angles.
Notice that we are now writing the index on the vector component as an upper, rather than
a lower, index. The transformation matrix is written with one upper index and one lower
index. The lower index corresponds to the vector in the denominator on the right-hand side
of equation (A.9). We'll discuss this convention further below.
Now let's look at the gradient of a scalar function O. The gradient has components
- lao ao ao\
v<D:l_._._l
\0x' 0y' 0z /
3;O:
'
ao
]xI
-
(A.10)
The lower index on the left-hand side reminds us that the upper index vector component is
in the denominator ofthis expression. To transform the gradient to a new coordinate system
(labeled -'), we use the chain rule for derivatives. Since <D is a scalar, O : <D.
aO |xt
3;O:
'
0Q
aT' -07'}xJ
-:
and thus the ffansformation matrix for the gradient has components
,
B,t:,
DxJ
AT'
(A. r l)
The gradient operator is written with a lower index to indicate that it transforms with the
matrix IB. Upper index and lower index quantities transform differently, in general. The upper
index quantities are called c ontravariant tensors, while the lower index quantities are called
covariant tensors, or forms. Thus, the differential displacement vector is a contravariant
vector (or just a vector), while the gradient is a covariant vector (or one-form). If the
geometrical image for a vector is an arrow with a specified length and direction, then the
image for a one-form is a set of level surfaces with a given orientation and separation
(Figure A.2).
A.4GENERALTENSORCALCULUS 449
FIGURE A.2. Level surfaces offer a geometric image for a one-form. The dot product d3. i Q : d.@ .
When the vector dd is laid across the level surfaces representing iO, aO ir ttr"
difference in the value of O between the head and the tail of the vector.
In the case of Cartesian coordinates, where the allowed set of transformations is rotations,
these distinctions are not necessary, because the two transformation matrices are the same.
But let's look at what happens when we transform from Cartesian coordinates to cylindrical
coordinates. The relations between the coordinates are
x-pcosQ; y-psinQ; z:Z
(A.r2)
and the inverse relations are
p: ,rF + t';
Q
: t*-l I; Z: z
x
(A.13)
We may differentiate tan@ to obtain an expression for 6Q/0x,
y
06 :- tand 1
v
. .06
tanQ:'-+sec'Q^
,cos'Q:
^
pcosQ
dx --= x' + dx
x
sin@
p
and hence find the components of the transformation matrices.
At,1ox,0o,0o
- ax- p.:COS@; A'Z:-:Sind;
"Idy
Ad
sind
Ad cosd
AL" _
AL" _
'0xp"0yp"0Z
A\:
^ -Q
-- az-"
Ad _ N
AZ" _
'
and
-E-g;
'0x-dyJ1z
A3,
A3r:E:gi
,fr:!:1
while
urt:H-cos@; 4:H:sind:
Bl:93:s
,0x..Eyr]z
---: - -psinQ;
B.r':
'a0-dQ"a0
Bz": * - pcosQ:
Bj: ^ -Q
lv
g]:!:-:g1
B1J:-:l
and
_r
0x ^
"azazaz
B1r:-:0.
^
_z
0z
450
oPTroNALToPrc A rENsoRS
Thus, the two matrices9 are
@
-sinQlp
cos
00
sin
@
cosQ/p
(A.14)
?)
and
cos@
-p sinQ
00
sin@
p cosQ
(A.1s)
t)
which are clearly different. In fact,
m.T
- n-l
(A.16)
Let's verify this result:
f cosQ -p sin
lEr.A:lsind pcosQ
0
@
\o
:(
o
?)
0
1
cos@sin@ - sin@cos@
cos2 4 + siri24
sin@cos/ - cos QsinQ cos2 4 +sin2 6
00
)
i):(i:?)
Relation (A.16) is true in general. (See Problem 15.)
Since the components of the transformation matrix are not constants, in general, but
functions of position, we cannot add vectors defined at different points of space because the
sum would not have a well-defined transformation law For example, we cannot subtract
velocity vector_s at two different points of space to determine an average acceleration. A finite
displacement D43 from point A to point B is no longer a true vector. However, operations
such as adding the electric field contributions E1 gndES at qpoint P due to two different
charge distributions to give the total electric field E : Er * E2 remain valid.
Example
A.3.
The electric field due to a long straight wire may be written in Carte-
I
j
sian coordinates as
i
-),/xv\
E:2ir€o
(7;7';rrP'o)
Transform the electric field vector into the cylindrical coordinate system.
i labels the rows, and j labels the columns. Even though one index is up and one is down, we shall still
write one index to the left ofthe other so that we can maintain the conventions introduced in Chapter I for labeling
rows and columns.
9Here,
1
I
A.5 THE METRIC
TENSOR
451
The electric field is a contravariant vector. Thus, in the new coordinate system, the
components are
E,:AE
(; )
:( -iiif,,
)=-=
"J!"f,,
o
0 rf
3
\
zneoo'
\o/
/ xcos@*Ysin@ \
^ 1l ,-rsind*OycosQllp I
zneop\
/
pcos2q*psin2f
: ^ I t-'cos@ sin@ * P sin Acoso)/o \
2orw2
)
\
:
x
:#G(
f
):h(
i)
The electric field has only a radial component. This is the expected result.
A.5. THE METRIC TENSOR
The line element (Chapter 1, Sections 1.1.1 and 1.3.2) is an important example of a scalar
that is formed as an inner product of a tensor, called the metric tensor, and the differential
vector di:
ds2
:
gii dxidxi
(A.17)
Since the lefrhand side is a scalar and the vector d*. is a vector whose components may
take on any values, we may apply the quotient theorem to conclude that g;1 is a tensor. In
Cartesian coordinates, gii : 6;i.In any orthogonal coordinate system, g;; is diagonal. For
example, in the cylindrical coordinate system (equation 1.4),
dsz
:
dp2
a p2 dq2 +
dz2
and so
/r o o\
,,,:186, ?)
(A.r8)
452
oPTroNALToPlcA TENSoRS
We may obtain these components from the Cartesian components by applying the transformation law in the usual way. The lower indices imply that g;; is a covariant tensor, and so
we transform with the matrix IE:
8,: EsBr: (
_,$tr ,"xf,
?
) (i
: ? ) (qr l;il, I)
: (-'.;tr, ,.'::, (:t, l#,
?)
l)
: ('*'d+sin2d p2sin2elor"o"rQ
l): (i (
?)
in agreement with the result above (equation A.18).
A.6. CONTRACTION
To form an inner product of two tensors of ranks m and n, we must contract the tensors by
summing over (at least) two of the indices. The result is a tensor of lower order (by two for
each pair of summed indices) than the sum of the ranks of the two tensors in the product,
and thus it must transform as a tensor of rank m -l n - 2. As an example, let's consider the
contraction of two vectors to form a dot product-a scalar. If we try to contract by summing
over two upper indices, we find that the result is not a scalar.
d-u'
:
Airuj Aiouk
But the product A'i A'o + 67r except for rotations of a Cartesian coordinate system, and so
this product is not a scalar in general. However, if we contract an upper index with a lower
index, then
-u'u;:
and the product Ai, B ,k
:
6f
Airui
B1ku1'
(eOuation A. 16), so
U'ui
-
L
U"uk
and this inner product is a scalar. Thus, we have the following rule:
In general tensor algebra, an upper index may be contracted with a lower index, but
not with another upper index.
'
i
i
I
A.7 BASIS VECTORS AND BASIS FORMS
4s3
In order to form an inner product of two vectors, we must first form the lower-index
covariant vector that corresponds to the upper-index contravariant vector. The line element
(equation A.17) shows us how to do this. Writing the line element as a dot product of di
with itself,
:
ds2
dx;
:
dxi dx;
:
Bi; d.xi
dxi
we observe that
gii dxl
In this way, the metric tensor maps any contravariant vector to a corresponding covariant
vector (or one-form):
u;:
g;iul
(A.1e)
Similarly, the contravariant metric tensor g'J, whose components form the inverse of the
matrix of components gij, maps a covariant vector to a contravariant vector:
u' : g'luj
(A.20)
Sometimes we say that the metric tensor g'J is used to raise an index, while the tensor grj
is used to lower an index.
These expressions show that there is a covariant vector corresponding to every contravariant vector, and vice versa. The metric tensor is an isomorphism that maps one space to
the other, its "dual" space.
When an inner product is formed, it does not matter which index of a contracted pair is
up and which is down. For example,
lt'L);
:u'gijuJ:ujul
A.7. BASIS VECTORS AND BASIS FORMS
in any coordinate system are defined to have components (1,0,0),
(0, 1,0), and (0,0, 1). However, in a non-Cartesian coordinate system, the basis vectors
are not necessaily unit vectors. We may find the magnitude of a vector from its dot product
The basis vectors
OPTIONALTOPIC A TENSORS
with itself:
fil:
Ju'ui
Thus, for the nth basis vector, we have
t-
16,,,1:
i and j
where (n) labels the basis vector and
ei6siiett,t
1f
eln',
label the components. But
:
6',
and so
li<,rl
: {r,,tfil:
^E;
(A.21)
where in these expressions we do not sum10 over n. Thus, in cylindrical coordinates, the
second basis vector has magnitude ,En : p, and the basis vectors in this system are p, pQ,
and2.
The basis forms are defined similarly. So, for example, the first basis form has components
(1, 0, 0), and its magnitude i" t/if .
A.4. Starting from the components of the velocity vector in Cartesian
coordinates, ffansform to cylindrical coordinates to find the components of i in the
new system, and hence write the velocity vector in cylindrical coordinates.
First we transform the velocity vector:
Example
,
l
)
1
-u' :
A'juJ
sin
@
cos
@
p
0
0xcosd 3v
+ *sind
*
3x
sin@
0y cos@
r__
M p'at
p
0z
u
l0There is no inconsistency here: n is a label, not an index.
I
455
A.8 DERIVATIVES
Next we write the derivatives in terms of the new coordinates, using equations (A.12):
"':
cos
@
sin
@
l)
(ff*'r-o
(ff*"r-o
""a(ysin@*0,",r#) )
ryesind* r""'r#)
I
)
as expected from the definition. To write the vector, we must multiply each component
by the corresponding basis vector. Then
^ 0z^z
- 0p"l) + AQ^ * 32" : Ap^0t. AQPQ-t
o:
^
^
^
* ",
t",r,
dt
dt
dt
which is the standard result.
"e(2)
A.8. DERIVATIVES
We have already shown that the gradient of a scalar is a covariant vector. We must be
more careful when taking the derivative of tensors of rank one or higher. For the moment,
let's discuss vectors. In taking the derivative, we must compare the values of the vector
components at neighboring points. But if the coordinates are not Cartesian, then even if
the vector is unchanged (its magnitude and direction remain the same), its components at
the two neighboring points are not the same (see Figure A.3). Thus, before we can subtract
FIGURE A.3. Parallel displacement. When
the
vector is moved in a non-Cartesian
system, its components change
'
even if the magnitude and direction do not. Here a vector that is in
the r direction at one point has both
r and 0 components when moved
to a second nearby point.
456
OPTIONAL TOPIC A TENSORS
the two vectors, we must first displace the vector parallel to itself and evaluate its new
components. This process is called parallel displacement.
It is customary to start with a covariant vector. Since we are considering differential
displacements, the change in the component u; under parallel displacement is proportional
1
to the displacement. It should also be proportional to the vector components,l so we demand
that
6v;:li.nr'4*t'
(A.22)
where the components f ,{o are called an affinity that specifies the affine connection of the
space. These components are not a tensor. Now, since the dot product u'v; is a scalar, we
require that it not change under parallel displacement,rz 61ui u;1 : 0. Thus,
u; 5u'
* u' 6t)i :0 :
ui
3ui
+ uilirnu i dxk
Since this result must be true for any covariant vector u;, we may conclude that
: -lj*uj
6ui
dxk
(4.23)
Now if the change du' inayector's components is equal to the change 6z' due to parallel
displacement, then the vector has not actually changed, and so its derivative is zero. Thus,
we define the covariant derivative z'., in terms of the difference between the actual change
in components and the change (A.23) due to parallel transport:
du'
-
3u'
:
u';.i dxJ
(A.24)
And so we obtain an expression for the covariant derivative in terms of the affinity:
,t.i
:#
tripiuk
(4.2s)
Working from equation (A.22), we obtain the resultl3
ui:j:
0u;
a*l -l^ij'*
(4.26)
Similarly, by looking at invariant combinations, we may write the covariant derivative of
any rank tensor. It turns out that we need one term in the affinity f for each index of the
llThis is actually a consequence ofrequiring that the parallel displacement in general be consistent with the
familiar definition in Euclidean space. See Lawden, p. 98.
l2Thir i, clearly true in Euclidean space. Once again we are requiring that the new ideas be consistent with the
old. This is a constraint on the kinds of space we are willing to consider as physically interesting.
l3See also Problem 28.
457
A.8 DERIVATIVES
tensor. It will be added, as in equation (A.25), for an upper index, and subtracted, as in
equation (A.26), for a lower index.
The covariant derivative of a vector (equation A.25) is a rank-two tensor. This result
follows from the definition (A.24), since the quantity on the left-hand side is the difference
between two vectors defined at the same point of space and thus is also a vector. Since
the components of dxi are arbitrary, we may apply the quotient theorem to show that the
covariant derivative is a tensor. Similarly, the covariant derivative of any tensor is also
a tensor.
It remains to find the components of the affinity. 14 The result we obtain should be valid in
any space, including ordinary Euclidean space. But if the space is Euclidean,15 we may set up
aCartesiancoordinatesystemxtattheoriginalpointandcomputetheCartesiancomponents
tr
of the covariant vector rr,. These components do not change under parallel displacement:
6ui - Q. We may obtain the components ui from ui using the usual transformation matrix:
a7l
ui:
axiuj
Then
6u;
/a]j\
a7l
- 3\e; )o, + *\ui
:
^r-;
L'!!-d*kv1+o
Dxt Exk
: a2tj
*o*
)-L
,-ox^
A-*j
r^
and thus, comparing with equation (A.22), we find
nm
I i*:
azti
go;
6**
0x^
67
(
.27)
This expression shows that the affinity is symmetric in its lower two indices. However, it is
not very convenient for calculating f.
To obtain a more convenient expression, note that we can obtain the covariant derivative
u;;i in two ways:
1.
Tiake the
covariant derivative u'., and then lower indices.
or
2. Lowq indices and then take the derivative.
l4ln their Section 1.5, Morse and Feshbach give a different derivation of this result and provide an expression for
the affinity in terms of the metric coefficients h;, discussed in ChaPter 1, Section 1.3.2.
l5This requirement is more restrictive than necessary, but allows us to present a more transparent argument. In
fact, even in non-Euclidean space we can set up a local Caftesian system, provided that det (S) * 0, and then
show that the parallel displacement of the components in this system is zero to first order in tbe displacement dx.
458
oPTroNALToPrc A TENSoRS
These two methods must give the same result, and so
gi-u\j : uiij :
(gi*uk);j
:
gik;juk
*
git uk;j
In the second step, we assumed that the product rule applies to the covariant derivative in
the same way as for ordinary derivatives.l6 Then it must be true that
qi*;i
:0
(A.28)
That is, the covariant derivative of the metric tensor is zero, a pleasing and intuitive result.
This requirement actually determines the f . We write it explicitly as
#-tf,1si^-tfis^t,:o
We may permute the indices in this expression to obtain two more similar expressions:
#-rlisi^-t!*E*i:o
and
Y-tfil*^-rfrs^i:o
Now we add the first two, subtract the last, and use the symmetry of the
f
s and the
g;; to
get
#
-2rf,1ti^*H
-ffi:o
Thus,
rf,1si*::(#*H-Y)
Multiplying by g'n, we find
8" ( |gir , }gii asu \
-n
'kj-- 2 \arl - a'l'- a.,)
(A.29)
Expression (A.29) is called the metric affinity. It is also called the Cbristoffel symbol of the
second kind.
We now have the mathematical apparatus that we need to describe physical systems in
any coordinate frame.
l6See Problem 29.
A.8 DERIVATIVES
A.5.
Example
459
The covariant divergence of a vector u'. t is a scalar. The electric field
inside a uniform cylinder of charge is E : poil2eo, where p6 is the charge density.
Evaluate the divergence ofthis vector field in Cartesian coordinates (d : xi + yi).
Evaluate the necessary components of the metric affinity in cylindrical coordinates,
and hence calculate the divergence of the electric field in cylindrical coordinates.
Verify that the results are the same.
In Cartesian coordinates, the
i .i:
f
Ei;
are zero, so
:
Po 0E'
2es 0TI
(0x
Po^
,0Y\
--%r\t-ay)
2eo
Po
p0
so
which is the expected result from Gauss'law.
In cylindrical coordinates, the only derivative of the metric tensor components that
is not zero is
ogzz
ffi:20
The divergence is
. AEi
Eii:
U, +l'kiE"
:9!'
* t(W *V+0x' 2 3x' \xk
\
|
:
pn /2e0, E2
:
Eq\ro
1xn
/
:
-
po l2eo.
O,and E3
0. Thus, in this system, A Ei I axi
The metric tensor is diagonal, so in the second term we get contributions from the
n. With the sums put in explicitly, the term becomes
sum over i only when i
where E
:
lilrrEk
8n'
- \L2
n,k
W*W-#)*
The terms in gr1, arc nonzero only if n
eachn. Thus, the only nonzero term is
t
n,k
gnn ogrn
2 ]xk
:
k, but 1gnrf 0x" (not summed) is zero for
,o - B" ogzz ,t - | .^,po
'tp\zot
2 1xl
:
^0
and so
V.E:
as we
ui,
:#+r',,iEk
found in Cartesian coordinates.
p0a_-_ p0 po
2€o' 2eg- eg
po
6
460
OPTIONAL TOPIC
A TENSORS
Although we have used flat space in our examples here, the maximum benefit of general
tensor mathematics accrues in applications such as special and general relativity, where the
metric cannot be reduced to the simple form 3;; with +1 along the diagonal. Some examples
are given in the problem set.
PROBLEMS
fl
oete.mine the velocity of an electron driven by an electric field E : tor--t't in the
presence of a constant uniform magnetic field Bo. Choose the z-axis along 86, but do
not make any assumptions about the direction of Eo. If there are n electrons per unit
volume, write the current density in the form ji : oij E j and determine the components
of o;i.
2. Starting with the expression ?
: 6xi
expression (,4'.6) given in Example
3.
A.l for the inertia
tensor.
Compute the inertia tensor (equation ,4..6) for a uniform cylinder of radius R and height
ft. Hence find its angular momentum when it rotates with angular speed a,l about
(a)
(b)
(c)
4.
for a particle in circular motion, derive the
an axis through its center and along its length
an axis through its center and along a diameter
an axis through its center making
Show that the outer product aib
j
a45"
angle with each of the axes in (a) and (b)
of two vectors obeys the transformation law for
a
rank-
two tensor.
S
Strow that the inner product a;i1rb1, of a rank-three tensor and a vector obeys the transformation law for a rank-two tensor.
6.
Show that the Kroneker delta tensor 6;7 has the same components in every coordinate
frame.
7.
Show that if a tensor b;y is symmetric in one frame (thatis, bi.i
:
bji
), then it is
in every frame. Similarly, show that the property of antisymmetry (bii
preserved under coordinate transformations.
8. (a) The following
symmetric
: -bj;)
is
set of components is defined in two-dimensional Cartesian space:
a,;:
(/ -y'i,
:J-ry \)
Does this set of components transfonn as a tensor? Why or why not? Hint: Try to
express the components in terms of inner and outer products of known vectors and
tensors. Alternatively, form inner or outer products with vectors and use the quotient
theorem.
Repeat the problem for the following sets of components:
x2\
y- -xy/
(b) Bi; : (' -*l.t
"
\
I
PROBLEMS
(c)
cii:
G, ?)
(d) In three-dimensional
space,
Dij
E fn" magnetic
(j : i)
moment tensor has components
Mik:
where
(a)
(b)
j
461
|
,;lrav
is the current density and the integral is over all space.
Show that Mip is antisymmetric for any steady, finite current distribution.
Show that the corresponding cross product (the dual vector, equation A.7) reduces
to the usual magnetic moment vector fr : I Aff in the case of a planar current loop.
10. The electric quadrupole tensor is given by
g,t
: f (3x;xi J
r2lii)P(*) av
Calculate the quadrupole tensor for a set of four point charges, two of charge q and two of
charge -e, at the corners of a square of side a.Tlte charges alternate in sign, so charges
of equal sign are at opposite ends of the diagonals of the square. Use a coordinate system
with 7- and y-axes along the diagonals of the square.
The force on a quadrupole charge distribution placed in an external electric field is
I
E2E,r
Fi: =Qik
b a& a*k
where the derivatives are evaluated at the origin. Find the force on the square when it
is placed in an external electric field i : o(2*y, -i2 , x2 + y2). The square's normal
lies in the x-z plane at angle I to the z-axis, and its center is at the origin.
11. Show that the components of the Levi-Civita symbol
e;;7.
transform as a tensor density
under coordinate rotations and reflections.
12. Starting from the components of the velocity vector in Cartesian coordinates, transform
to spherical coordinates to find the components of i in the new system, and hence write
the velocity vector in spherical coordinates. (Refer to Example A.4.)
Ell
tn
a
region of space, the electric scalar potential has the form
<D
:
-EoZ.
(a) Working in Cartesian coordinates, compute the gradient to obtain the electric field
components. Transform to a spherical coordinate system, using the appropriate ffansformation law from Section A.4.
462
OPTIONAL TOPIC
A TENSORS
(b) Write the potential in spherical coordinates, and compute the gradient using the
operator 3; (equation A. 10).
Confirm that both methods give the same electric field.
14. Write the components of the gradient form in (a) cylindrical coordinates and (b) spherical
coordinates. Use the metric tensor to raise indices, thus mapping to the corresponding
vector. Finally, multiply by the basis vectors to obtain the conventional expression for
VO, as derived in Chapter I (equation 1.62).
15. Use equations (A.9) and (A.11) in Section A.4 to show that the relation lBr
true in general.
Aii - 4l', show that Ai 1 : Aji ardthat
Can you find a relation between Ai i and A/;? Why or why not?
16. If the tensor Ai"/ is symmetric,
:
.A,-l is
A;i - Aji.
17. Which of the following relations between tensor components could possibly be true?
Say what is wrong with the incorrect expressions.
(a) Vi
(b) fii
(c) V'
(d) y'
: eiik(Jr
: Xikyrj
: Xik(Jr, + Wi
: eiikIJiWlXl,Yi
[f8..l fn special relativity, space-time is described by four-component vectors u". (It is con-
ventional to use Greek letters to label the indices 0,1,2, and 3 in four-dimensional
space-time.) The coordinates of an event in space-time are xo : (ct, x, y, z), and the
metriclT is
/t o o o\
lo-r o ol
'*:lB B-; -i)
The Lorentz transformation matrix relating two coordinate systems moving with relative
speed u along the x-axis is
I
l
I
/ v -vP o o\
nt:l -YP ,n o, Sl
\ o o o t)
where
y
: ll \fr}
and
B
:
y1s.
The electromagnetic potential is described by a four-vectorrvith components Al' :
(O, a", Ar, Ar), where <D is the electric scalar potential and A is the magnetic vector
potential. The electromagnetic field tensor has components
Fl"u
_
AP
Av
_ Av AtL
lTThir i, the sign convention used by Jackson. There are others.
I
l
j
l
PROBLEMS 463
Show that, in the Gaussian unit system, the components of the field tensor in terms
ofE
and B are
Fllv
-
(?'
i,"; fr)
Two particles, each with charge 4 and mass z, are moving along lines parallel to the
x- axis and a distance d apart. Each particle moves with speed u << c. Stan in a reference
frame in which the two particles are at rest. Compute the components of the field tensor
in this reference frame, and hence find the force acting on each particle. Now transform
the field tensor to the lab frame, and again compute the force on each particle. Verify
your result to first order in the small quantity 0 : u lc by computing E and B in the lab
frame using Coulomb's law and the Biot-Savart law. (In the Biot-Savart law, the source
1dl should be identified with a point source qi atthe position of one of the charges.)
19. Using the metric of Lorentz space-time and the electromagnetic field tensor (see P-roblem
18 above), verify that an electromagnetic wave has the same field structure (E I B,
E : B) (in cgs Gaussian units) in any inertial frame.
20. What invariants can you form from a tensor Zpv? Compute these invariants for the
electromagnetic field tensor in Lorentz space-time (see Problem 18).
the wave four-vector has components ktr : (alc,k',ky,kr).
Use the Lorcntz transformation matrix Ap, (see Problem 18) to find the components
of the wave vector in a second frame moving-with velocity i - ui with respect to the
first. What is the result if (a) [ : ki and O) f : ki? Compare with the nonrelativistic
Doppler shift formula, and comment.
[Zil fn Lorcntz space-time,
22. Use the metric gpy for Lorentz space-time (see Problem 18) to compute the line element
ds2 : Bp, dxp dx' . The proper time r is defined by the relation dt : ds f c. Compfie
the proper
timeintewaldt between two neighboring points
movingatspeedu.(Letthepointshavecoordinates
on the world line of a particle
ct,x,y,zandc(t*dt),xldx,y*
z*dz,where dx : Dx dt and dy and dz are defined similarly.) Express your result in
terms of the time interval dt, P : u f c, and y : I I \/ I - p2. Compute the components
of thefour-velocityuP : dxpldt of aparticleandcomputetheinvariantprodtctupuu.
Comment.
dy ,
23. The set of components
*,
I -l
[0
(
t"fv6 :
if a py 6 :
if apy 3 :
0123 or an even permutation of this
1023 or an even permutation of this
otherwise
maybeusedtoformthetensorFof
-
je"fr6 Frt dualtoF(comparewithequationA.T).
(a) Show that the components of edfv6 ffansform as a tensor in Minkowski space-time.
(b) Find the invariant F"0 Fop if F"P is the electromagnetic field tensor defined in
Problem 18. Comment.
464
oPTroNALToPrc A TENSoRs
24. IJse Gauss' law in integral form to find the electric field inside a uniformly charged
sphere. Compute the necessary components of fir, and hence find the divergence of
this electric field in spherical coordinates. Show that the divergence equals the (uniform)
charge density divided by e6.
[2il
In two-dimensional flat space described with polar coordinates, a vector d is in the radial
direction. Displace the vector to a neighboring point (refer to Figure A.3), and compute
the new components in terms of the displacement (6p, d0). Compare with relation (A.23)
in the text and hence compute the components f 1;; of the affrnity. Perform the same
operations with a vector in the I direction to find the remaining components of l. Hint:
Remember that the basis vectors a.re not unit vectors in this system.
:
ff;
26. In three-dimensional
(equation A.29) has 33
space, the affinity
27 components.
Show that the affinity for Euclidean space with cylindrical coordinates has only three
nonzero components, and compute them.
Use equations A.14, A.15, and A.27 to obtain the same result. (Remember that in
equation A.27 i are the Cartesian coordinates.)
27
.
The tensor density etift is defined in Cartesian coordinates as in Chapter I (Section l. 1.2,
equation 1.29). Transform to flat space with cylindrical coordinates, and determine the
p.Compute
components of e'lft . Lower indices to fi nd the components of e, ik, E
rk, andEi i
:
the cross product 6 x i
d for a particle in uniform circular motion with <il : ari.
r
Hint:
16 x
i;t : eiikajuk:
eijkroi uk
28. By lowering indices, show that the covariant derivative of a covariant vector may
be
]
written
ui:j :
as
0u;
a*,] -l^iju*
in equation (4.26). Obtain the same result from equation (4.22).
l
i
[2q.lVe.ify the product rule
l
I
ui.i : (Tikulr);j : ukTik;j + Tikut, j
l
j
I
I
j
.t
I
I
I
OPTIONAL TOPIC B
Group Theory
In modern physics, group theory is becoming an increasingly important mathematical tool.
As we develop the theory we shall see that groups form the natural mathematical description
of symmetries of physical systems. These symmetries are in turn related to the conservation
laws of physics, and thus the mathematics of groups allows us to investigate these laws in
a very general way. Space constraints here do not allow us to do more than touch on the
basics of group theory. Students who plan to pursue an interest in topics for which group
theory is especially important, such as particle physics, will need to consult more advanced
texts to supplement the material here.
B.1. DEFINITION OF A GROUP
A group G is
a set of elements {a}, together with an operation
*, that obeys the following
set of rules:
l.
If a
and
b
are elements
of the group, then so is a x b.
2, T\eoperation * is associative: (a * b) * c - a * (b * c).
3. Thereis aunitelement I in G suchthat 1+ a: a*l: afor everyelement ainG.
4. EveryelementainGpossessesaninverseelementa-t,alsoinG,whichhastheproperty
that
I
-l
a*a':a'*a:I
where
1
is the identity element (see rule 3).
In the special case where the operation is commutative (a + b
:
b * a),the group is said
tobe abelian.
The number of elements in the group is called the order of the group. The order may be
finite or infinite.
465
466
OPTIONAL TOPIC
B GROUP THEORY
8.2. EXAMPLES OF GROUPS
The group of order one The most trivial group is of order one and consists of the single
identity element. Convince yourself that this set obeys all of the rules listed above.
two
The group of order
There is exactly one group of order two. It consists of the identity
1. This group represents the
and one other element a that is its own inverse, so that a r( e
physical symmetry of reflection in a plane or of rotation through an angle z about a single
:
fixed axis.
The group of order three There is exactly one group of order three. It contains the identity
and two other elements a and b, where b is the inverse of a. We may write a multiplication
table for this group, as shown in Table B.l.
Since b : a2, wealso have the relation a * b : e * a2 : aJ : l.This group and the
previous one are examples of cyclic groups. A cyclic group is generated by a single element.
Every element of the group is obtained as a power of this single element, and in a group of
order n, the n th power gives the identity. The cyclic group of ordet n is called Cn. Every
cyclic group is abelian.
TABLE B.1. Multiplication Table
for the Group of Order Three
The cyclic group C, represents a symmetry of a polygon with n sides. We can think
the element a as a rotation through 2n ln that rotates the polygon into an identical copy
of
of
itself.
The set of integers under the operation of addition The set of (positive and negative)
integers forms a group of infinite order. The operation x is addition:
mtrn=mln
Then the sum of any two integers is another integer. The operation is associative:
n*(m*p):(n*m)*p
The identity element for this group is the integer zero:
nl0:0ln:n
for any integer n. The inverse of an element n is the negative of n:
-n.
Then
n*(-n):(-n)ln:O
1
8.2 EXAMPLES OF GROUPS
467
This group is abelian, since the order of the integers in the sum does not matter: n+m : mIn
for all integers m andn.
The set of integers does not form a group under ordinary multiplication, because the
inverse of n,lf n, is not in the set of integers.
The set of all rational numbers m ln under the operatian of multiplication, where n,
m+0
* *P :*
nqn
p
q
X 1-:
mxp
nxq
:"e":) : (T"")":
Theidentityelementis
and has infinite order.
l:1/l,andtheinverse
of m/nisn/m.Thisgroupisalsoabelian
The rotation group Spatial rotations may be represented by rotation matrices (see, for
example, Chapter l). The identity is a rotation through 0 radians (represented by the identity
matrix), and the inverse of a rotation through 0 is a rotation through -9 about the same
axis. This group is not abelian. The set of rotations in a plane is given the name SO(2).
The rotations are represented by orthogonal2 x 2 matrices with determinant +1. The O in
SO(2) refers to the orthogonality of the matrices, the 2 refers to the fact that they are 2 x 2
matrices, and the S stands for special and refers to the condition that the determinant is +1.
This group is abelian.
Notice that in this group the elements themselves are "operations" (rotations), represented
mathematically by matrices, while the gro up operation is "follow one rotation with another"
and is represented mathematically by matrix multiplication.
Permutqtion groaps In these groups, also, the group elements are things we do ("operations"). Tlte group operation is "follow one permutation with another." As an example,
consider the set of permutations of three objects, 53. One element of this group is the
permutation {interchange first two elements}, which we'll label Gp.
Gn: (qbc) --> (bac)
The identity element is {do nothing}.-
l:
(abc) -'+ (abc)
Consider the two elements GB : {interchange lst and 3rd} and G23
and 3rd ) . The product of these two elements may be written
GzzGn:
{@bc) --> (cba)} followedby {(cba)
:
{interchange 2nd
-+ (cab)}
468
OPTIONAL TOPIC
B GROUP THEORY
The result is the group element Gsrz, which means {take the last element and put it first}.
(In standard notation, the operations are done in order from right to left. This is consistent
with the notation for matrix multiplication.) Now if we multiply by Gr: again, we find
GpGlGs :
{@bc)
+
(cba)l {(cba) --> (cab)} l(cab) --> (bac)}
- Gn
where G12 is the group element {interchange lstand 2nd}. Eachofthesethree "interchange"
elements is its own inverse, since interchanging two elements twice gets us back to where
we started. Thus, we can also write this relation as
GnGnGr]
-
(B.l)
Gn
We say that the element G12is conjugate to Gzt.
The sixth element is Gnz, which means {take the first element and put
it last},
or
l+3-->2--+L.
GB2:
{@bc) -+ (bca)}
Using these symbols for the group elements, we can set up a multiplication tablel listing
all of the 3l : 6 group elements for the group 53. (See Thble B.2.) Note that this group is
not abelian, since, for example,2
GrczGn:
{@bc) --> (bac) --+ (acb)l
GnGnz:
l@bc)
:
G23
while
-->
(bca)
-->
(cba)\
- Gn
TABLE B.2. Multiplication Thble for the Group 53
Gn
Gn
I
Go
Gn
Gzz
Gzt
Gnz
Gnz
Gzn
1
Gvz
Gn
Gn
Gzt
Gnz
Gzn
Gn
I
Gtz
Gnz
Gzt
Gnz
Gvz
Grrz
Gn
Gzz
Gzn
Gnz
1
Gnz
Gzz
Gn
Gzn
1
Gn
Grc
Gzz
Gn
Gs
Gzz
Gn
Gn
Gztz
I
I
Gnz
The elements of this group may also be written using the cycle structure, in which G13 is
written (13)(2): atwo-cycle (13) in whichthe element 1 goes to 3 and 3 goes to 1, and a onecycle (2), showing that element 2 goes to itself. The one-cycle is sometimes omitted, since
it is redundant. The element G132 is written (132). The sequence of numbers in the cycle is
I
Since this group is not abelian, the table must be read as follows: Multiply the element in the column on the far
left by the element in the top row to obtain each table entry.
2Remember:We do the operation on the right first.
8.2 EXAMPLES OF GROUPS
469
important, but it does not matter which one we write first, since 2 -+ 3 -> I -+ 2 --> 3.In
this notation, Gyp is written as the three-cycle (3I2) : (123).
We can understand how the operations in this permutation group are related to physical
symmetries by labeling the vertices of an equilateral triangle 1,2, and 3, as shown in
Figure B.1. Then each ofthe two-cycles corresponds to rotation by n about an axis in the
plane of the triangle and through one apex. The cycle (12) cornesponds to rotation about
axis A in the figure. The three-cycles correspond to rotation by 2n 13 or 4n /3 about an
axis through the center of the triangle and perpendicular to the plane of the figure. Each
of these operations leaves the triangle unchanged, except for the positions of the labels
on the vertices. Once we make this identification, it is easy to see why (I23) + (123) :
GyzGzo: Gnz: G32) and (123) * (132) : l.
FIGURE B.1. The group
,S3
describes the symmetry of this equilateral triangle.
The unitary groups Another set of groups that is important in physics is the set of unitary
groups. The unitary n x n matices form the group U(n). A matrix is unitary if its adjoint
equals its inverse. The adjoint is formed by taking the complex conjugate and then the
transpose:
adj (A)
:
G*.)
If A is real, then the condition of unitarity becomes tr : A-r;that is, the matrix is orthogonal. The special group SU(n) is the group of n x n unitary matrices with determinant *1.
The general form of a group element of SU(2) is
(-i. :.
)
with aa*
*
bb*
For example,
l/
eiq
eif \
fi\_r-w ,-'t )
would satisfy these constraints.
:
t
(8.2)
47O
oProNALToPrc B GRoUPTHEoRY
B.3. CLASSES
If a, b, and c
are three elements
of a group and
.
b:
_l
CnC '
(B.3)
then b is said to be conjugate to a (compare with equation B.1).
1. An element is always conjugate to itself, since we can take c to be the identity element.
Then
,or-l : lal:
a
2. If a is conjugate to b, then b is conjugate to a:
a: cbc-l
We multiply on the right by c and on the left by
c-l.
Then
,-1o, : c-lcbc-lc : b: fof-l
where
/ : c-r.
3. If a is conjugate to b and c, then b and c are conjugate to each
Ifa
is conjugate to both b and c, then there are elements
f
other.
and S such that
o:fb.f-t:gcg-l
Thus, if f -r g : h, then h-r - U-r il-r : g-r
on the left by f -r and on the right by /, we find
f .Multiplying
the above expressions
7-t767-17:b:hch-l
and so b is conjugate to c.
The complete set of elements C : at, a2,. . . that are conjugate to one another is called
an equivalence class, or simply a class, of the group. The identity element forms a class
by itself, since ala-t : I for each element a of the group. To find the elements that are
conjugate to a, calculate gag-rfor each element g of the group. The results may not all be
distinct.
For example, in the permutation group S:, the elements conjugate to G n are computed
in Thble B.3. The elements in the class ne {Gp,Gs,Gzzl. This seems sensible, since
these three elements are the "interchange" elements; that is, they do the same kind of thing.
Similarly, the two "cycling" elements form a second class for this group, as you should
check.
I
SUBGROUPS 471
B.4
TABLE B.3. Classof Elements
Conjugate to G12
IGpl-l :GpGoGt) -Gn
GpGpGsr:GzztGn-Gzz
G4G2G2]:GnzGzz-Gn
GyyGpGytl: GnGnz - Gzz
Gp2GpGl]2: GztGzzr - Gn
8.4. SUBGROUPS
Sometimes we can find a subset of the original group that obeys all the rules for a group;
this is called a subgroup. For example, the subset ll, Gn| is a subgroup of 53 because
(Gn)2 : 1. A second example is the set of all even integers. This set forms a subgroup
of the group of integers, since the sum of any two even integers is also even, the subset
includes the identity (zero), and each element has its own inverse within the subgroup. This
subgroup also has infinitely many elements. The group of all rotations by multiples of n 12
around the x-axis is a subgroup of the rotation group. This group has only four elements:
rotations by n 12 : -3r 12, T = -n,3n /2 : -T 12, and2n : 0. This last element is the
identity element. This subgroup is abelian, while the group of all rotations is not.
If a group G has a subgroup S and element a is not in S, then the sets of elements as;
and s;a are the left and ight cosets of the subgroup S with respect to a. Note that none
of the elements in the cosets are in the subgroup if a is not. We can then write the group
elements as
G:Sfa1S*a251...
Thus, the order of the subgroup S is a divisor of the order of the group G.
The set of conjugate elements
aSa-l
forms a subgroup S'. To see this, note the following:
1. The product of two elements is in the
set.
If
s1 and ,t2 are elements
of S, then
(asp-l)(as2a-l) : as1(a-r a)s2a-1 : asrs2a | : as3a-l
where s3
:
sls2 is also in S, and so aqa-r is in S/.
2. The associative property is preserved:
(as1a-l
*
as2a-11
* orjo-l : asra-l * (as2a-l *
3. The identity element is in S/. If S is a subgroup, it contains
the element ala-r : aa-7 : l.
aq,a-l
)
the identity. Thus, S/ contains
472
oPTToNAL ToPrc
B GRoUP THEoRY
4. The inverse of asa-l is as-la-l:
(asa-t)(as-r o-r)
-
alss-r1a-r
: aa-r :
1as-r a-t)(asa.-r)
-
a(s-r s)a-r
: ee-r : I
I
and
If the new subgroup S/ is the same as the original subgroup S, that subgroup is called
invariant, normal, or self-conjugate.
Note that the subgroup of even integers is an invariant subgroup, since
n
l2m -t (-n) :2m
for any integer n. However, the subgroup {l,Gnl of 53 is not invariant,
GpGpGpr - Gzt (see Table B.3) and Gzl is not in the subgroup.
because
B.5. CYCLIC GROUPS
a group G, then a' is also an element, for any integer n. By the first
rule of groups,at( a: a.2 : b must alsobeinthe group. Thena +b: a3 must also be in
the group, and so on. If the group has finite order, then an : I for some n.
If a is an element of
If N is the smallest integer for which aN : l,
then
N is called
the order of the
element a.
The set {o, o2 , . . . , aN : 1} is called the period of a.Itis a group called a cyclic group
Clv. If the order of an element is less than the order of the group, then its period forms a
cyclic subgroup of G. In the permutation group of three objects, ,S3, each of the interchange
elements has order two, while the other two elements (Gzy ard G3p) have order three.
Then, for example, the period [Gn, GLz: 1] forms a cyclic subgroup of order two. The
symmetry of cyclic groups was discussed in Section B.2.
8.6. FACTOR GROUPS AND DIRECT PRODUCT GROUPS
If S is an invariant
subgroup of a group G, then we can identify another group, called the
factor group G/S, whose elements are the cosets aS. The group operation for the factor
group is defined by the rule
(aS)*(bS):(ab)S
where the product ab is computed using the group operation of the original group G. The
invariant subgroup itself acts as the identity element. The inverse of aS is a-l S:
(aS)
* (a-lS; : 1aa-r)S :
S
j
8.7 ISOMORPHISM
473
The group structure of the factor group follows directly from the group structure of G itself.
If the subgroup has m elements, then the order of the group is mn for some integer n, and
the factor group has order n.
A group may be written as a product of subgroups under some circumstances.
G is a direct product of subgroups 51 and Sz, G
:
51
x 52, if
(i) all the elements of 51 commute with all the elements
(ii) every element in G may be written as
8
:
of 52 and
sl't2
where s1 is a member of the subgroup 51 and s2 is a member of 52.
Then the product of two group elements is
gi
I
i
:'t1'is2't't1'is2'i
: sr,isr,jsz,is2,j by property (i)
: sl,ks2,k since 51 and 52 are subgroups
:8k
and is
ofthe correct form (ii).
B.7. ISOMORPHISM
We can relate the elements of two groups to one another by a mapping. The mapping may be
one-to-one, in which one element of Gt maps to exactly one element of Gz; many-to-one,
in which two or more elements of G1 map to a single element of Gz; or one-to-many, in
which a single element of Gt maps to more than one element of G2. A mapping is said to be
"onto"if themappingmapsallof Gr ontoallof G2.Forexample,wecanmaptheelements
of the permutation group of three elements, 53, to the group of integers by assigning each
permutation a number, I through 6.
I -+l
Gp --> 2
Gp --> 3
Gv --> 4
Gp2 --> 5
G3e -+ 6
This mapping is one-to-one but not onto.
We can also construct a mapping from the integers to the permutation group 53. We divide
each integer by 6 and note the remainder. That element maps to the permutation group as
474
oPTroNALToPrc B GRoUPTHEoRY
follows:
Remainder maps to permutation group element
0
1-+1
2
3-+Gn
4
5
-+
Gzrz
-->
Gn
-->
-+
Gzz
Gnz
This mapping is onto but not one-to-one, since, for example, both 6 and 12mapto G312.It
is many-to-one.
In an important class of mappings called homomorphisrns, the mapping also "preserves
that maps group G to group
the operation." To see what this means, consider a mapping
f1. Then we can write
/(g; : h.If /
/
is
a
homomorphism, then
fQr*s):f$ixf(sz)
(B.4)
For example, the mapping that maps the rational fractions to the integers via
n
-m
--+ nm
is a homomorphism, since
, (h.'a) : , (#)
:,'P'?'Q
: (nm)(pq): f (;) - ,G)
This mapping is onto (since every integer is also a rational fraction of the form n I l), but
not one-to-one (l and f both map to 6).
A homomorphism that is also one-to-one and onto is called an isomorphism. The group
of rotations in a plane is isomorphic to the group of 2 x 2 orthogonal matrices, since each
rotation maps to one matrix, and vice versa.
8.8. REPRESENTATIONS3
B.8.1. Reducible and lrreducible Representations
are isomorphic to one another are the same in a fundamental sense. An
isomorphism allows us to represent the elements of an abstract group with the elements of
another group. This is a useful idea because the elements of the isomorphic group can be
mathematical structures such as the matrices that represent rotations. The group of order
two can be represented by the numbers 1 and -l through the isomorphism /(1) : I and
All groups that
f (a): -1.
3In this section, we choose to draw conclusions from examples, ratber than by explicit proof. For proofs of the
results presented here, the reader should refer to the bibliography.
,
I
l
I
475
8.8 REPRESENTATIONS
We can use a homomorphism to construct a representation of an abstract group as follows.
We allow the group elements to operate on a vector space of dimension n. For example,
to represent the permutation group Sr, we can label the points of a triangle with x and y
coordinates, thus allowing the group elements to operate on two-component vectors. Then
A representation of an abstract group G is defined as a homomorphism /: G
where M is a group of n x n matrices. This representation has dimension n.
-->
M,
If the mapping is an isomorphism, the representation is said to be faithful.
Every group has a trivial one-dimensional representation in which every element of the
group is represented by the number 1. This mapping clearly preserves the group operation,
but we have lost a good deal of information about the group structure; the representation is
not faithful.
Another representation, of dimension equal to the order of the group, may be formed
using the group multiplication table. We form n-dimensional vectors whose components
are the elements of the group, in some order, We label the group elements gt, 92, . . . , Bn.
Then we can represent any element g; with an n x n matrix M; that has only one nonzero
element in each row and each column. If
gigj
:8k
then (M;) 1a : 6mk. This matrix represents the multiplication properties of element g;. For
example, the cyclic group C3 of order three whose multiplication table is given in Table B.l
may be represented by the following matrices:
Mt:
(r l
l) M2:(l i ;) M3:(: ? i)
where M2 represents element
a
and M3 represents b. Then
*(i) :(li:X',):( i)
which correctly reproduces the second row of the multiplication table. Note also that
M2M2
-
(li3X?il):(:ls)
-Mz
M2M3
-
(ii:Xris):(i:i)
-Mt
and
(B.s)
476
OPTIONAL TOPIC
B GROUP THEORY
The matrices themselves satisfy the same multiplication table, as required for a representation. This representation is called the regular representation.
A one-dimensional representation of the same group may be formed from the complex
numbers
fi(1) = 1, fr@) : ,2ni/3, f1(b) :
safti/3
(B.6)
Recall from Chapter 2, Section 2.1.1 that multiplication by a complex number represents a
rotation. So each of these elements may be visualized as a rotation in the complex plane.
Extending this idea, we may form a two-dimensional representation using the 2 x 2
rotation matrices:
(t?)
We may also use the
r/-r
-t
z
\ -./r
J1
-1
) :G-f)
(B.7)
3x3 matrices:
Ml:
ML:
(-f,, f,i: i)
-Jt/2
-r/2
0
i)
(8.8)
These 3 x 3 matrices operate on three-component vectors, but they leave the third component
unchanged. Labeling the components .r, y, and z in the conventional way, we may write
any vector in the space as the sum of a vector u in the x-y plane plus a vector w parallel to
the z-axis: v : u * w. The group elements leave w unchanged.
We may generalize this result. If an n -dimensional space V may be written as the sum of
an m-dimensional space and an (n - m)-dimensional space, (V : U I W) and if, under
the group operations, vectors in 14/ remain in I4z, then the subspace 17 is invariant under the
group. As we saw in Chapter 1, we can change the basis vectors of the vector space. The
entries in the matrix will also change. The matrices with respect to the two bases are related
by a similarity transformation (equation 1.78):
M'
:
SMS-r
(B.e)
The corresponding representations M and Mt are said to be equivalenl. But in the primed
basis, the invariance ofthe subspace W is less obvious.
If we reverse this argument, we can see that invariance of a subspace under a group means
that there will be a basis for which the group representation makes the invariance obvious.
The representation is said to be reducible.
8.8 REPRESENTATIONS
477
A representation is reducible if g; is represented with respect to some basis by a matrix
Mi thattakes the form
Mi:
(
f' ";,)
where A;, B;, and D; arc submatrices of dimensions rn x m, m x p, and p x p, respectively, and n : m * p is the order of the representation. Then the group element g1g; is
represented by
(
f' '; )(t'
where the submatrices A*,
u;,): (
^'f,
Bu
o''b!,u,'',
)
: ( f- ';r)
and Dp are of the specified form and
A,u * Biur \
(f,3; )(L): /(r,,)
so that the subspace I4z is invariant under the group.
completely reducible and is written4
If B
:
0, then the representation M is
M:A@D
This corresponds to the case in which both subspaces U andW are invariant under the group,
as is true for the representation (B.8). For finite groups, every reducible representation is
equivalent to a completely reducible representation.
If the submatrices A and/or D are reducible, we may continue this decomposition until
M is diagonal or the remaining submatrices are not reducible. The resulting representation
is then said to be ireducible.
In the representation (B.8), the matrix D is a I x I matrix-that is, just a number. The set
of matrices D; is the trivial representation of the group C3; each Di : 1. We can continue to
reduce the representation (B.8) by diagonalizing5 the matrices Ai. We can do this because
41 is already diagonal and A2 and A3 give rise to the same equation (equation 1.80) for the
eigenvalues:
), +Jit2 :0
I +Jitz -) - x
| -+ -
4The symbol O does not imply addition in the usual sense. It cannot, because the matrices A and D may not have
the same dimension. They are to be "added" by forming a larger matrix with A and D along the diagonal.
5See Chapter
l, Section
1.6.5.
478
oProNALroPrcBGRoUPTHEoRY
So
| . J,.
.
),: -;*;,
- ,2ni/3 o, "4ni/3
This solution is just the representation (8.6) that we found above. There is, in fact,
a
second
representation here:
: l,
f21;.)
fz@)
: u4tri/3,
fz(b)
:
r8ni/3
- ,2tri/3
G.10)
Thus, the irreducible representations are the trivial representation, (8.6) and (B.10). The
regular representation and (B.8) may be reduced to the form
/t o o\
lt o o
r
o
M;:l
M',':l
o
l, ' s ,2ni/3 o
'
\o o r)'
\o o
/t o o \
o I
M;:l s ,4nit3
g
,zni/3
\o
\
I,
,+"iF)
(B.u)
f
The representation has now been fully reduced and may be written
M:T@\@Fz
@.12)
where 7 is the trivial representation and F1 and F2 are as given in (8.6) and (B.10).
In a direct sum ofinequivalent, irreducible representations (irreps), as in equation (B.12),
a given representation may appear more than once. The representations in the direct sum
need not, and usually do not, all have the same dimension.
B.8.2. Orthogonality of the lrreps
Using the representation (B.11), we can form a number of three-component vectors by
selecting one component from each of the matrices that represent the group elements.
These vectors are orthogonal:
3
/
..\*
)-.(ui)/P rrrt;n Ei.
K6rq
K is a constant. Here the different irreps are labeled by
group elements. For example, with p : 1 and Q :2, we have
where
I x 1* l r
r2tri/3
+t
x e4oi/3 : |
p and q,
and k labels the
- ++ |tJl - l- jiJi :o
while with p : q : l, we find K : 3.
Now we want to generalize this result to a group of order m in which one or more of the
irreps have dimension greater than one. Let us label the distinct, inequivalent, irreducible
l
l
479
8.8 REPRESENTATIONS
representatio\s M1, M2, .. .. Within representation i, with dimension ni, a gtoup element
g is represented by the ni x ni matrix MiG). The representations satisfy an orthogonality
relation that we may write as
:
{ill rn[M i G-\],,
\tu,
K 3 ij 3 p,
I
where
6;;
-+
K
depends on 4 and
r. We can determine the value of K by setting
; : j.
Then
1, and
{il) rnlMi (s-r)1,,
\tu,
-
K 6 p,
c
Now we take the trace; that is, we set p
matrix product:
\tu,
p
1
{s- )l, r[u
Since the trace
of
i {B))
6os
:
q
p
:
lM i G-
:
1
)
s and sum from
M i (E))r q
:
I to n;. On the left, we have
1
lM i G- g)1, q
:
lM i (l)1, n
:
the
3, o
ni,
Du,n -,n6ra - Kni
and
u
:
#u,n
Thus, the orthogonality relation is
\
I
ru, {dl rn lM 1 G-r)),,
When the M; arc unitary, we may write this relation
:
L
6,n
;i
E
p
(B.13)
as
{il) onlM G\!, : L lij
l1u,
I
1
3
6
p,
Dq,
(B.14)
We can use this relation to determine a limit on the number and dimensionality of the
irreps. As we did with the irreps of C3, we can think of this relation as the dot product of
zz-dimensional vectors, wherc m is the order of the group. Here the vectors are labeled by
two indices, p arrd q, each of which runs over ni values. There are nf such vectors, and
they are all orthogonal. Similarly, s and r each run over n j values. Furthermore, these n2,
480
oPTroNALToPrc B GRoUPTHEoRY
vectors are orthogonal to the first nl vectors. In total, there are
vectors in the m-dimensional vector space. Thus,
!,
nf mutually orthogonal
t n?.^
i
In fact, it can be shown that the equality holds:
(B.rs)
\"?:*
We found three irreps for the group C3, each of dimension one, and so the left-hand side
of
equation (B.15) equals 3, the order ofthe group, as required.
B.8.3. Character
Equivalent representations are essentially the same. It would be nice to have a convenient
way to characteize representations that labels all equivalent representations the same way.
Again we can review the properties of matrices (Chapter 1) to find the one that we need: It
is the trace,6 since
Tr (SMS-r)
:Tr
(M)
The character of a representation is the set {1; } of traces of the matrices that represent
the group elements.
The traces are not all distinct. In fact, every element
elements g; and gi are representedby M; and Mi,then
TrlM(sisi7it)l
:
Tr
in a class has the same fface. If
(M1M;M,\:rr 1u)
(8.16)
The character of the representation (B.8) of the cyclic group C3 is the set
{3, 0, 0}
and the character of the regular representation (B.5) is the same. Since the characters are
the same, the two representations must be equivalent.
It is most instructive to consider the characters of the ineducible representations. In
the case of C3, or indeed any abelian group, the irreducible representations are all onedimensional,T and the traces are just the numbers themselves. The characters of the three
ineducible representations that we have found m
6See Chapter 1, Problem 40.
TSee Problem ?.
8.8 REPRESENTATIONS
481
vectors, where m is the order ofthe group (equal to 3 for C3). These vectors are mutually
orthogonal in the following sense8:
lfr
:l.xf
<x!)* :6ob
(B.17)
i:l
The index i in this expression labels the elements of the group, and a andb label the different
characters.
We can obtain a second relation by noting that all elements in the same equivalence class
are represented by matrices with the same trace (equation B.16). Thus, equation (B.17)
becomes
*Lo,^;rx!)*
:6ob
(8.18)
where N is the number of distinct classes in the group and p j is the number of elements in
the 7th class. This constitutes a second orthogonality relation for the N-dimensional vectors
J p14. Since there can be no more than N of these, we find that the number of inequivalent,
irreducible representations is less than or equal to the number of classes in the group. Once
again it turns out that the equality holds.
Let's check this result for the group C3 . There are three distinct classes, since each element
forms its own class. Thus, there are three inequivalent, irreducible representations, as we
have found.
We can use the results found so far to determine the character table for any finite group.
Let's find the character table for the permutation group9 53. There are three classes,
and so there are three irreps. One is the trivial representation of dimension 1. Then, by
equation (8.15), the other two irreps have dimensions that satisfy
nl+ n2r:5
Thus, n2 : 1 and n3 : 2. With a one-dimensional representation, the multiplication
table for the characters is the same as the group multiplication table. Therefore, for the
"interchange" class, we must have Xl : I and thus 1; : *1. For the "cycling" class,
x? : l. We also have x1y" - Xi , so Xi : -l and X" : 1. For the last representation, the
character of the identity equals the dimension of the representation, here 2. We have now
obtained the results shown in Table B.4.
8Forone-dimensionalrepresentations,equation(B.14)withp:q:r:r:landn;:lreducesto
I has been replaced by i, and i and j have been replaced by a and b.
equation (B.17). In equation (B.17), the label
9Recall that this group has order six.
482
oPTToNAL ToPrc B GRoUP THEoRY
TABLE 8.4. Character Table for
53
Identity
Class
Three-cycles
Interchange
Number of elements in class
Trivial rep I-dimension 1
1
J
2
1
1
1
2ndrep M2-dimension 1
3rd rep M3-dimension 2
1
-1
1
p
a
2
Then the orthogonality relation (B.18) gives us two equations for
u
and
p:
2-3a-l2P:g
2+3ot-l2B:g
Thus,
0:-l
and cv:0
We can understand these results by referring back to the symmetries in Figure B.1. Let
the triangle be in the x-y plane. Then the second representation has basis 2; the cycling
elements correspond to rotations about the z-axis and so leave the z-axis invariant. The
interchange elements correspond to rotation about an axis in the plane; these operations
invert the z-axis. The third representation must have two basis vectors in the plane of the
triangle. With axis A labeled as the x-axis, the interchange elements are represented by the
reflection matriceslo
",,
:(;
-?
)
GB
: (-'fr,, -{;/'
),
G23:
( -!^fr,, {i')
while the cycling elements are represented by rotation matrices representing rotations of
2r I 3 and 4n f 3, rcspectively :
G23t:
(-_lJir,,
_{;/') -o Gt3z: (-'fr,, _{i')
Notice that these matrices correctly reproduce the character table. Check that they also
satisfy the multiplication table for the group.
The character of a fully reduced representation is the sum of the characters of the irreducible representations that compose it. This result can be helpful in determining the
decomposition of any particular representation.
B.8.4. Representations and Physics
Here we give a brief example to illustrate how group representations can simplify our
understanding of physical systems. The ammonia molecule is a pyramid with the nitrogen
loRefer to Chapter 1, Problem 8.
8.8 REPRESENTATIONS
483
atom at the top and three hydrogen atoms in an equilateral triangle below (see Figure 8.2).
The group 53 is the symmetry group for this system, because we can permute the hydrogen
atoms but we must leave the nitrogen molecule where it is.
FIGURE 8.2. An ammonia molecule is shaped like a tetrahedron with the nitrogen molecule
at
the top.
We can obtain a physical representation of the symmetry group by using real threedimensional space as the vector space ofthe representation and asking how a vector in threedimensional space transforms under these symmetries. This representation is reducible. To
see how it decomposes, we can compute the character. The trace of the representation of the
identity is, as always, the dimension of the rep, here 3. We can take one of the interchanges
to be reflection in the x-z plane, represented by the matrix
'*:(3 ;l)
with trace
* L The cycles
(B.re)
are rotations about the z-axis. One of these is represented by
M{,":
('f;,, -f/'
I)
with trace zero. Thus, the character is
:
^v
Comparing with Table B.4, we see that Xv
13,
:
r,oy
Xt
*
Xr and thus
Mv :T @Mz
484
OPTIONALTOPIC B GROUPTHEORY
In particular, the trivial representation is included. This means that we can find a vector that
remains invariant under the symmetry group and hence, for example, the molecule can have
an electric dipole moment.
A magnetic dipole moment, however, is a pseudo-vectorll that transforms differently
under reflection. The representation (B.19) changes to
/-t o o\
,,f,": (, .;
_i )
3
with trace - 1. The character of the new rep is {3, - 1, 0}, and the decomposition is Mz@ Mt.
The trivial representation is not included. Thus, the molecule cannot have a permanent
magnetic moment. Notice that we did not need to know any of the physical properties of
the system----other than its geometrical structure-to draw these conclusions.
8.9. GENERATORS OF GROUPS
The groups we considered in Section B.8 were primarily finite groups. But many of the important groups in physics have infinite order. In groups such as the rotation group SO(3),'the
group elements depend on a set of continuous parameters-the rotation angles in SO(3)and we can express the group elements in terms of these parameters. The group elements may
be expressed in terms of generators that relate each group element to the set of parameters.
It is convenient to relate the identity element of the group to the parametet zeto.
We can understand the idea behind generators by considering the group of rotations in a
plane-that is, SO(2). Each group element may be represented by a matrix of the form
R(o): / cos?
[_sind
d\
)
parameter is the angle 0, and with 0 : 0 we obtain the identity
sin
cose
element,
Here the relevant
as required. We can express the cosines and sines in exponential form and then use the Pauli
matrices as a basis. These are
"t:(? I
) *:(?
1)
and
",:(i
-? )
Thus,
R@):"*, ( ;
?
).'i.o ( -? I )
lcos0 *iozsin0
For very small rotations
so we may write
d
((
I (that is, "near" the identity element), cos d ry l and sin 0 ry 0,
R(9)
t1
See Appendix
:
exP(io20)
I and Optional Topic A, especially Problem 9.
(8.20)
B.gGENERATORSOFGROUPS 485
At this point, you might ask what it means to have
The meaning is clear
a
matrix
if we define the exponential by its
as the argument of an exponential.
series expansion:
eM:t+MI+1****...
2
Note that, in general,
,A,B:
(t*^+jn-A,+..
-rr
- . -A+lB
+.A.*lE
I
Xt.n+jm-u*
* rA*
^4,
*B
*
I
I
r1-r
7, -rA+lB +lB x.A* 2B xA*A + rB
:
+lE
*'..
+A
*...
1
1,\*lE
*,lE
)
eBea
Expanding the exponential in the rotation matrix, we have
R(g):
t*
io20
-!o?e'-lto"tt
2-z- 6--2- + loteo
24-. + *ioles
120 " -
But
"?:(l
f-.ofeu
720 "
+..'
tXl -t):(l ?)
so we may simplify:
R(g): t*io20 -!e'-lioze3
2 6' + !.eo
24 a liozes
120'
- *e6+...
720
:, -)r' * *t^ - fit'+...+ o(e -it' * *t'.
)
:cosd lio2sin9
This suggests that the representation
R:
exp(ioz9)
holds for all values of 0, not just small ones. The matrix
S:io":(
-
o I\
\-t
0)
is the generatorl2 for the rotation matrices.
We can extend this idea to get the generators for the full set of three-dimensional rotations.
We need the three matrices
s,
/o o o\
/ o o l\
:lo o -l I, sr:l o o o l,
\0, o)
\-r oo)
and
o\
/o
s::11 -lo o
\o oo)
l2Whether S or a2 is taken as the generator is primarily a matter of taste. ackson uses S.
I
486
B GRoUP THEoRY
oPTToNAL ToPrc
Then any rotation may be written as
R
:
(B.2r)
exp (-<o . S)
where the dot product is
ro
If the vector
<o
.S
:
arrSr
*
azSz
*
@3S3
:
I\-rt ;'
/
o
-(D1, ar \
o'-;;
at o
(B.22)
I
)
has only one component o3, we get back our previous result for rotations
about the z-axis. The other components generalize the result for a rotation about an arbitrary
axis described by the vector c). Note that since the rotation matrix has determinant +1, the
(MI)1.
. S is traceless
ldet (eM) "Tr
It is useful to note the relations between the matrices S;:
matrix
<,r
SrSz
:(l? ;X r li):(lls)
and
SzSr
:(-l
So
SrSz
li)(sl :):(lil)
(:sl)
-
S2S1
:
-Sa
Similar relations hold for the other products. Since the gener4tors do not commute, neither
do the rotations they generate.
The rotation group is a subgroup of the Lorentz group (Jackson, Section I 1.7;Arfken and
Weber, Section 4.5). The full group has six generators: three for the three components of the
relative velocity of the frames and three for the three parameters of the rotation (direction
ofrotation axis and angle ofrotation).
The unitary group SU(2) is generated by the three Pauli matrices and is closely related
to the rotation group SO(3). They are homomorphic, not isomorphic. That is, we can find a
mapping that maps two elements of SU(2) to one element of SO(3).13
Now let's generalize. Suppose that elements near the identity in a group G can be described in terms of N real parameters ay. and that the identity element corresponds to ap - 0
for each k. The identity element is represented by the identity matrix:
M\a)lo'_.o
1
3See
Arfken and Weber, Section 4.2, pp.232-236.
:1
B.1O LIE
ALGEBRAS 487
where we have written the N parameters as a vector cr. Since the parameters describe the
matrices continuously, we may use a Taylor series to write elements near the identity asla
M(du)
: M(0).
#|":ooon+
. - M(0) + t (-,
#1":,)
:l*iXtdo&
,*
(B.23)
where the repeated index ft implies summation. The quantities
xn: -i
Yl
dqr la:o
are the generators of the group. If cr has only one component 6u, we can generate the group
element represented by M (a) by repeating the operation represented by M (a I fl) a total of
p
B times. (You can understand this by thinking about rotations-rotate through an angle 0 I
about the same axis B times.)
M(a)
:
I
l,
/
a\1f
\e ))
This element must be a member of the group if M(ql F) is. Now we let B --> a, af p
becomes differential, we obtain M(ulfl) from equation (B.23), and we generate M(a)
through the limit:
M(u)
: nl,(;)]' : /11 (, .,+)'
Expanding the expression on the right, we have
(,.,+| -, * u'+ . P+1,+)' *
M(a)
* iax**)fioxol' +...
-+
1
-
si"xr
asB -+ oo
(8.24)
Thus, the structure of the group is defined by the elements near the identity.
B.10. LIE ALGEBRAS
Suppose a group G with elements R; has generators
equations B2l and 8.24)
$. Then we can write (compare with
R : et'S
l4The introduction of
i
here is to make the resulting generators Hermitian.
488
oPTroNALToPrc B cRoUPTHEoRY
In particulaq there is a set of elements near the identity for which
R;
with e;
(
1.
-
asisi
1
*eiSi +:t?t
+
For such an element,
Rlt :
Now
:
e-eisi
:l -eisi +)t?s? +
if R is unitary, then
det (R;)
and so Tr
(&)
:
RtrR'Ri
:
exp [ei Tr (S;)]
:
1
0. Let R7 be defined similarly. Then
-R;'(l +r,s, +:r?s?* )o,
-rr
-' r
_rrr
-,
:
eiR; rS;R;
Ei
*
/
(l -e;s;
I ^
tt?Rjtsr2n;
+
I *eiSi *e;e;(SiS;
+...
1..\ /
I ^ ^\ I " "
+
yisi )$(t +e;s; * r'iti
) ;isf
-
S;S,)
+:t.S?
where we have dropped terms of third order in the small parameters ei and
compute
RlrR;lnin, : (t - e;s; *
)'f *)(r
:l+e,€j(&Si-SiS,)
+ r,s,
r
e;e;(s;s;
-
srs;)
e
j.
Finally, we
+;'?S?)
to second order in the es. Now this element of the group can also be expressed in terms of
the generators, so we must have
1
*
e;ej(s;si
- si$) :
s.'ii's
: t +lrilr;rs*
k
In particular, if we write
,!, :
(SiS;
t;t 1o!r, then we must have
-
57S;)
= tsi, S;l :1"!,Sn
(B.2s)
k
This is a fundamental property of the generators
Rl andR; thatwepick"d.il;';;ff
,S;,
inde
particular
elements
"i;;;;lledthe #:i,H;l:fft:
i
j
I
B.1O LIE
ALGEBRAS
489
Because of the antisymmetry of the commutator [S;, Sy1 (equation B.25),
u!,
: -aji
(B.26)
We can now compute the double commutator:
rsi' rs''
s'lrr:;;fJ
,l:;3
_l]:ll,-i#
tl
ui1,s^l :\ufi,lsi,
lt,,I
Lm)*mn
s-l
: D"ToDoi^s,
Then the Jacobi identity
ls;, [s;, sr]l + [s;, [sr, si]l
* [S*, [si, s;JJ :0
(8.27)
which follows from the definition of the commutator, gives another constraint on the structure constants:
u!^s, +t t afiaf,^sn : o
tnrnD ul1,a'!^sn +DD"t
nm
nm
tnmt s, (afi,ui* + uf;a|- + affai^) : o
\
("Tr"f* + afraj
^
+ uffai*)
:o
(8.28)
For the rotation group SO(3), we found
/o -r o\
0 0 l:sl
\o o ol
SrSz-SzSr:l I
so
of, -
1.
Then, from equation (8.26), we have
for this group are the Levi-Civita symbols
a), : -l.In
ofj : tij*
fact, the structure constants
@.29)
i
490
oProNALroPrcB GBoUPTHEoRY
Although we used a specific representation to calculate the structure constants, they are
independent of the particular representation.
We can check that relation (B.28) is also satisfied:
D ("yo"y^ + af,aj* + ufiai^) :l
:l
(ei*^enim
: j,6*i 6
*
6ii6*n
Eki*trr*
-f
6p$;1
-
(ei**eimn
I
e1a;*ei^n
I
e;i-e1,*n)
* rrr^iru^)
3;r3it a
3;n31*
-
6i"6*i
_n
-u
If we define the commutator as the multiplication operator for the generators, the vector
space of generators, with both addition and multiplication defined, becomes a field or
algebra called the Lie algebra for the group G.
A Lie algebra is a vector spacels of elements u; with an operation (the bracket or commutator) that satisfies the following rules.
1. The bracket is linear:
faur +
bu2,4l : alur,
uzl
I
bfuz, uzf
and
lu1. au2
-f bql
: afut, uzl I
b[u1,
4l
2. The bracket is antisymmetric:
lut,uzl
- -[uz,ui
3. The Jacobi relation is satisfied:
lq,luz,uzlf *fuz, [u:, ur]l *
[us, [ur, uzff
:O
A
group whose generators form a Lie algebra is called a Lie group. Lie groups are
tremendously important in physics. The group structure is determined by the Lie algebra
of the generators, as expressed by the structure constants. The rotation $roup, the Lorentz
group (which includes the rotation group as a subgroupl6), and the symmetry groups of
particle physics are all Lie groups.
PROBLEMS
I
Strow that the set of permutations of two elements is a group. What is its order? Write the
multiplication table for this group. How many classes are there? Is the group abelian?
15SeeChapter
1.
l6See Problem 15.
PROBLEMS
491
2. The symmetry group of a square contains those operations that
leave the square unchanged. Show that this group has eight elements, and write the multiplication table.
What are the classes? Are there any subgroups?
3.
4.
S
6.
7.
Show that the set {1, -1, l, -l} forms a group under algebraic multiplication. Write the
multiplication table. How many classes are there? Show that this group is isomorphic to
a group of rotations that preserve the symmetry of a square (compare with Problem 2
and Problem 10).
Show that there are two goups of order four, and determine their multiplication tables.
Strow that any group of order n is isomorphic to a subgroup of S,.
Show that unitary matrices of the form (B.2) form a group under matrix multiplication.
Show that in any abelian group, each element forms its own class. Hence show that all
of the irreps of an abelian group are one-dimensional.
8. Showthatthe setof 2x2matricesof theform
/ * y\
(-';)
forms a group under the operations of (a) addition and (b) matrix multiplication. Show
that the group of complex numbers x I iy is isomorphic with this group of matrices in
each of the two cases.
9. The quaternions are four-dimensional complex numbers of the form Q : a + bi +
cj * dk, where a, b, c, and d are real numbers and the quantities i, i , and k obey the
multiplication rules
i2:j2:k2:-7
and
i.i
:k;
ji:-k
(a) Show that the set {+1, +i, +j, tk} forms a group under this multiplication.
(b) Show that i, j , and k may be represented by the matrices
':(;lil)':(i:;i)
.:(,il')
Determine the classes of this group.
(c) Determine the number and dimension of irreps of this group, and find the character
table.
(d) Is the representation in (b) reducible? If so, how?
492
OPTIONAL TOPIC
B GROUP THEORY
[tO.] Strow that the integers I through 4 form a group under the operation of multiplication
mod 5. Write the multiplication table. What is the identity element? How many classes
are there? Is the group abelian? Another group with four elements is a group of rotations
that preserve the symmetry of a square-that is, rotations by multiples of t 12. Are these
groups isomorphic? Why or why not?
11. Consider the mapping
/
that maps the group of rationals to the group of integers by
/m\
r \;):m*n
Is the mapping a homomorphism? Why or why not?
12. Show that all elements in the same class have the same period.
13. Show that if a set of elements {e; } forms a class of a group G, then the
[f+l
I
set {e, } of tne
inverses of {e;} is also a class.
fne center Z of a group G is the set of elements that commute with every element in
the group. Show that the center is an abelian subgroup of G. Is it possible for a group to
have no center?
15. A homomorphism / maps group A to group B. The kernel of the homomorphism is the
set of all elements of A that map to the identity element of B. Show that the kernel is an
invariant subgroup of A.
16. A one-dimensional translation operator f, translates a function along the x-axis by an
amount nd , where d is a fixed step length: Tn f @) : f (x -l nd) .
(a) Show that the set of operators 7l, forms a group that may be represented by the
complex numbers Tn
=
e-ik'd . What
are the conesponding basis functions?
(b) Work out the orthogonality relation (B.14) for this representation, and comment.
You will have to make some changes to account for the infinite order of the group.
(c) Nowlettheoperatortranslatebyanarbitraryamount xt:T(xt)f (x1 : f @*xt).
What are the generators of this group?
17. The water molecule is an isosceles triangle with the oxygen atom at one vertex and the
hydrogen atoms symmetrically located on either side. Determine the symmetry group
for this molecule. May this molecule possess a pefinanent electric dipole moment? What
about a magnetic moment?
[ft.l fne molecule SbSs is square-pyramidal. Four
S atoms form a square base with the Sb
atom at the center. The fifth S atom sits at the top of the pyramid. Determine the symmetry
group for this system. What is the order of the group? Work out the multiplication table.
How many classes are there? Determine the character table. May this molecule possess
a permanent electric dipole moment?
19. The Lorentz group has generators that are 4 x 4 matices with mostly zero elements.
The matrices K; are given by
/01 o o\
Kt:
l'Isl,)
I
PROBLEMS
493
and so on. (The nonzero elements of K; are the ith elements in the top row and the first
column, where the first element is labeled with 0, not 1.) Similarly, the generators S; are
given by
/o o o o\
St=
[,'sl;,J
(The nonzero elements of S; are given by a1*6) : sijk, whete ai* is the submatrix
formed by removing the top row and left-most column and j,k : 1,2,3') Find the
group element generated by K1 and also by 51 .
Compute the product of the two group elements eoKr and ebKz. Hence show that the
elements generated by K; do not form a subgroup. Do the elements formed by the $
form a subgroup? Are there any other subgroups? If so, what are they?
: ax * b (where x' , x, a, and b are real numbers and
two-dimensional representation of this group that acts on
20. Show that the transformations .r'
a+
form a group. Form
the vectors (.r, 1).
O)
a
21. A homomorphism / maps a group G to a group FI. Show that the image f (G) in H
is isomorphic to the factor group G / K, where K is the kernel of the homomorphism
(Problem 15).
@l
Xgroup G has an invariant subgroup S. If element a of group G has period N, where N
is prime, and a is not a member of the subgroup S, show that element aS of the factor
group G/S also has period N.
OPTIONAL TOPIC C
Green's Functions
Many physical systems are linear and consequently are described by a linear differential
equation. An important example is the electromagnetic field, and we shall treat this system
in detail in Section C.5. Such systems obey a superposition principle: The response of
the system to a sum of inputs is just the sum of the responses to the individual inputs. As
aresult, these systems can often be analyzed using a Green's function. The Green's function
is actually a distribution (Chapter 6) that describes the response of a physical system to a unit
delta function input.
In general, any physical system in the region 0 < .x < I is described by a differential
equation for some quantity, say y(x), plus a set of boundary conditions. The boundary
conditions provide a sort of shorthand for representing additional sources that exist outside
the region of interest. The system is given an input 1(x), and we want to find the response.
The Green's function is the solution to a similar physical problem, but with the source term
replaced by a delta function (a "point source") and with zero (or, at worst, constant) values
on the boundary. The Green's function problem is thus easier to solve than the original
mathematical problem, which is the advantage of the Green's function method.
We shall begin by discussing one-dimensional problems for which the solution itself is
zero at the boundaries. A physical system exists in a region defined by 0 < x < L.Ifthe
input to the system is 1(r), we can write it as
I (x)
:
Io'
I (x')3(x
-
x'])
dx'
If the response of the system at-x to the delta function at x/ is G(x, r'), then, by the principle
of superposition for linear systems, the response to the input 1(x) is
y(x)
:
fot
,
{*')o{*,
x'; dx'
(c.1)
The Green's function is symmetric in the two variables:
G(x, x')
:
G(x' , x)
(c.2)
495
496
oproNALToprc
c
GBEEN's FUNcToNS
If the variable is a time variable, then
G(t, t')
: G(-t' , -t)
(c.3)
This sign change is necessary to preserve causality-the response cannot precede the event
that caused it. (See Morse and Feshbach, Section 7 .3, for a more detailed discussion of this
point.)
Because the differential equation satisfied by the Green's function contains a delta function, strictly speaking G is not a function, but a distribution (Chapter 6). Indeed, the physical
solution is obtained as an integral over G, ad expected if G is a distribution. The method is
advantageous provided that the integral (C.l) is relatively easy to do.
The methods used to find the Green's function for a given problem may be classified into
three groups: (a) divide the region of interest into two pieces with the point source on the
boundary between them, (b) expand the Green's function as a series of eigenfunctions, or
(c) use an integral transform.
C.l.
DIVISION.OF-REGION METHOD
First we write the differential equation and boundary conditions that describe the system
in the absence of sources. (The boundaries may be in space, time, or both.) With no sources,
the differential equation will be a homogeneous equation. We determine the solutions of
this homogeneous equation.
Next we write the differential equation for the Green's function problem. The source is
nowadeltafunction[forexample, S(x-xl)linwhichthepositionx/ofthesourceisforthe
moment consideredfxed.This source may be imagined as dividing space into two regions,
x < x' and x > x'. We write down the appropriate solutions of the homogeneous equation
that match the boundary condition or conditions in the two regions. There will be one or
more unknown constants in these solutions.
We also have to consider the boundary conditions that apply at the intermediate boundary
that we have constructed at x : x' .We need enough boundary conditions to determine the
unknown constants in G.
Once G(x, -r') has been determined, we can use it to find the response of the system to
any input, using the integral (C.1). In evaluating this integral, we considerx/ to be variable
and x to be fixed.
To illustrate the method, consider a particle moving under the influence of a driving force
F(r) with damping proportional to velocity. The equationl describing the system is
*Edu *au:
lCompare with equation (3.5), with rng replaced by
F(t)
,
F(l).
C.1 DIVISION-OF-BEGION METHOD
The boundary conditions in time are u
function is
-+
0 as
t -+ too.
The equation for the Green's
dG
m . -fqG:6(t-t')
tlt
For
tI
tt
497
(c.4)
, the delta function is zero and the equation simplifies to
dG
^E IaG:O
which has the solution (Chapter 3, Section 3.3.1)
G: A*r(_;,)
This function approaches zero as / -> *oo, butblows up as / -+ -@. Since the appropriate
solution for / < // must approach zero as t --> -@, it must be identically zero. The system
is set in motion by the impulse applied att : tt. Thus, we can write the solution for t < tl
andfor
t > tt:
. : (o
G(t. t'\'
Since G is a function of both
t
ift<t'
1
[A
exp
(-at
lm)
if t > t'
(c.5)
and tt , we expect that the "constant" A is actually a function
of t/. Our next task is to find this function.
We imagine the delta function input at t : t/ dividing the system into two regions, with
the point t : tt on the boundary between them. To find an expression for A, we integrate
the differential equation (C.4) across the (imaginary) boundary at t : tt:
dc) o' :
[,]"' (*#+
1,,',:,"
s(t
- /) dt :7
where we used the sifting property to evaluate the right-hand side. On the left-hand side,
the first term integrates easily:
mct:|,!:,
*, 1,,',:"' G dt :
t
The remaining integral is bounded:
I',1,'
oarl s
'u*
tct(2e)
<2elAl
Provided G remains finite, this integral approaches zero as e -+ 0. Now we insert our
solution (C.5) for G in the two regions into the resulting equation. At the lower limit, t < tl
498
and
oploNALToprc
G:0;
c
GREEN's FUNoIoNS
attheupperlimit,
t > ttandG:
^lt"*p
?Ir)
-
Aexp
o]
:
(-at/m).Thus,
as
e
-+
0,
'
I zo .r
A--exp[-t')
m\m,/
and so
(o
G(t,t'):{
L'\"I'
- t
ift </
/
l;'-o(-;
l' -''l) it'
(c.6)
'''
A piecewise-defined solution for G always results from the division-of-region method.
Notice that the reciprocity relation (C.3) is satisfied by the function (C.6).
Example
times r :
C.1.
Find u(r) if the input F(t) is a force that is a constant F6 between
0 and
t
: T andzero otherwise.
The resulting velocity is (compare with equation C.1)
ift'<O
r+*/0
f+oo
ro
ifo<t'<T \lc1t,fiat'
u(r): I re)c(t.t')d/: I I
J_a \o
J__
ift,>T
I
Since the force becomes zero for
t/ < 0 and for tt > 7, the limits
of the integral are
0tof.Thus,
ift<0
u(t;:0
andforr > 0,
u(t)
:
1T
Fo
Jo
G(t,t') dt'
If 0 < r < Z, since Gis zerofort < // (equationC.6),theupperlimitof theintegral
is t:
u(r)
:
,o
F6
m
Fs
a
/l\ at'
[' ! "*p-\(-!1,
m - /
JOm
C.2 EXPANSION IN EIGENFUNCTIONS
On the other hand,
499
if t > T,
u(t)
fTl
/ d
: Fo"Jo
t']\ at'
I -exp
m '\ [ --V
m' - '/
/d,
--[t
Fs exp [\m
m
- t'l)
T
qlm
0
l
(-!v
zl\ "-- /-9,\
- &d f"^p
\ m - ) - "^'\
t
tn')l
& exp^\(-gt) f.^p^\m
(" r)/ - rll
d,
m/L
Once the force ceases, the speed decreases exponentially (see Figure C.l). Notice that
both expressions give the same result for u(t) at t : T .
ua
F11
t
T
as a function of time for the particle in Example C.l
velocity decreases exponentially at large times.
FIGURE C.1. Velocity
withaT/m: l.The
C.2. EXPANSION IN EIGENFUNCTIONS
Instead of dividing our region into two pieces and obtaining a piecewise-defined Green's
function, we can sometimes expand the Green's function in a series of eigenfunctions.
If the governing differential equation is of the Sturm-Liouville type (equation 8.1) and
the boundary conditions are of the form (8.2), then there exists a set of eigenfunctions ), (r)
for the problem, with associated eigenvalues l"n. We will assume that the eigenfunctions
have been normalized so that
l"'
r rrrr^rx)yn@)
d,x
: 3-n
(c.7)
500
oproNALToprc
c
cREEN's FUNcroNs
We require that the differential equation for our Green's function problem be of the same
form but with a constant I that is not equal to any of the eigenvalues ).r. The source is
a delta function:
d/
dc\
n \f al::- ) - s{x)c * )"w(x)G: -4n6(x - x')
(c.8)
The factor of -4n is raditional-it arises from the use of this method in potential theory.2
Then the Green's function may be expanded in eigenfunctions:
G(x, x')
:Dy,
n:0
(c.e)
@') l^@)
For the moment we consider the value x/ tobefixed.We substitute this assumed form (C.9)
into equation (C.8):
i
r^<.'
t
lft (t o ft/,
(,)) -
g (x)
*'l,
v,(x)
(x)v, (x)]
: -4r6(x -
xt)
(c.10)
Now we know that each eigenfunction satisfies an equation of the form
- sr'rr' r
* ('o'*)
)'lnw(x)Yn
-
Q
so we use this to simplify equation (C.10):
oo
D
y,
@'
)
[-
)'nw (x)
yn
(x)
*
)'w (x) y n(x
)]
:
-
41 6 (x
-
xt )
n=O
Next we make use of the orthogonality of the eigenfunctions by multiplying the whole
equation by
y^(x)
I' 2"
and integrating over the ran9e
-
x
)')w(x)yn(x')vn@)v^(x) dx
:
& to
: -0,
iU - ),)vn@) f"b ,{r)v^{r)v^(x)dx:
x
:
I"u
b:
6(x
-
xt)v^(x) dx
-4nv^(xt)
()'- )'^)y^(x') : -4ny^(x')
where we used the sifting property to evaluate the integral on the right and equation (C.7)
to evaluate the integral on the left. Then
y^(x') =
4o!-92
lum
lt
-
2See Section C.6 below.
c.2 EXpANsloN
rN
ETGENFUNoToNS 501
and the Green's function is
(c.ll)
Note the symmetry in
x
and
r/,
by equation (C.2).
as required
Example C.2. Find the Green's function forthe one-dimensional Helmholtz equation in the region 0 < x < I with y(x) :0 atx : 0 and at x : L.Hence find y(x)
it f(x) equals I fot L14 < x < 3L/4 and equals 0 outside this range.
The equation for the Green's function is
:
7fi + k'y
d2,v
which is of the form (C.8) with ).
replaced
:
6(*
-
*')
k2 and the factor
(c.12)
-4n
on the right-hand side
by 1. Thus, the equation for the eigenfunctions is of the Sturm-Liouville
form
dzY'
a;
+
r,)
^
:0
with 1", I k2.The solutions to the eigenfunction equation that satisfy the boundary
condition atx:0are
ln :
To satisfy the second condition at
x
cnsin
:
(r/l''.r)
L, we need
1/fi,L:
nn
and so the eigenvalues are
tnnt2
^,--\i)
We still need to normalize the functions. We choose the constant cn so that
1L
I
Jo
,1
lo'
Yn@)Yn@)
,in'
dx =
1
(74 dx:,]L- :1
-,-y ItL
5O2
oproNAlToprc c
GREEN's FUNcIoNS
So the normalized eigenfunctions are
!n(x):
6
tli"^(T-)
and the Green's function is (compare with equation C.11 with
4n replacedby -1)
too
G(x, x')
: '\rlif k :
nn lL for any n. Physically, this
corresponds to driving the system at a resonant frequency. The displacement becomes
arbitrarily large in the absence of damping, and there is no solution.
Notice that the expression is not valid
Thenif
f(x) : I for L14 < x < 3Lf4,
1L
13L/4
I cG,x')f(*')dx': Jr1+
I
-y(x): Jo
:?L
J
GQ,x')dx'
["'o Y-"tnT"nT
or'
rtq ? tQ - (nn lL)z ---. wrx
a
Slh
n
SlIl-
nT x,
:1\- = - L = I13Llasin-'77
L
L?kt-(nrlL\zJ11+
:
2
L
. wfx
\-
L
,
cos
I
2
iwr - cos iwt
n,
+F - <r,rtUt'
nT nft
. nrfx 2sin-5inoo sinL
2
4
-o\- L 1n, 1712 1r2
-2,
wr
n:l
fnrxlnr
-4
i odd(-1v{n.1121#F)
L'
I
n:1.
-
n
The sum is over odd values of n, because
Figure C.2.
'3
sin(nnl2):0 if n is even.
C,3 TRANSFORM METHODS
o
FIGURE
0.2
1.0
0.6
0.8
:
:2/L.
5
are
Terms up to n
C.2. The result of Example C.2 with k
Including tems up to n :9 does not change the plot noticeably.
503
*t
0.4
plotted here
C.3. TRANSFORM METHODS
We have already found the Fourier (Chapter 7, Example 7.3) and Laplace (Chapter 6,
Section 6. 1.6) transforms of the delta function. Thus, we can solve for the Green's function
by transforming its defining equation.
The Green's function for the damped harmonic oscillator is given by
d2cft.
i;
ac (t. t'\
t'\ 12qi--trrJ
+ aloc1t,t,):6(t
-
t,)
(c.13)
We may Fourier transform this equation to get
-r2c1r, tt) - 2iodc(to, t') -l ,20c1r,1'; - -J-ri'"
Jnr
Then the transform of G is
1
"iatl
G(o, t') t/2r afi - 2iao
-
o2
Inverting, we get
G(t, t,)
:*
I_: d-4, _-,-ia,
4,
(c.14)
We may evaluate the integral using the residue theorem. The integrand has two simple poles,
where the denominator is zero. These are
aP: -ia t
,2o
-
o2
504
OPTIONALTOPIC C GREEN'S FUNCTIONS
Both poles are in the lower half-plane, as we expect from causality (Chapter 7, Section 7.4).
For / < //, we must close the contour upward. There are no poles inside the contour, and
the result is zero:
fort <tl
G(t,t'I -g
For
t>
//, we close downward, enclosing both poles. Then, writing
have
: J-e-o{t-tl ("-ia<t-t) - eial-/))
- tt) for t > tl
G(t, /) "-a1t-tl,sit1{2(ta
(c.ls)
{o}* "'=
S), we
(c.16)
The result is of the same form that we would get from the division-of-region method
(Section C.l).
C.4. EXTENSION TO N DIMENSIONS
Problems defined in a region with two or more dimensions are handled in a similar manner.
The Green's function is defined in the same region of space (and/or time) as the original
function /, but both the differential equation and the boundary conditions are simpler.
1. The source term on the right-hand side ofthe original differential equation is replaced
by a delta function representing a point source.
2. The boundary conditions are replaced by
(a) Dirichlet conditions: G(*, i/) : 0 on S
or
(b)
Neumann conditions: n '
VG(i, i/)
:
constant3 on S
where S is the surface that bounds the volume of interest, V.
In case (a), we have the Dirichlet Green's function; in case (b), we have the Neumann
Green's function.
The method used to find the Green's function will be a combination of the methods used
in the one-dimensional case.
We'll begin with an example in two dimensions with homogeneous boundary conditions.
We'll consider inhomogeneous boundary conditions later.
A potential problem is defined in a two-dimensional rectangular region of dimensions
x
a b.The boundaries are conducting and are grounded. We'll choose a coordinate system
3The constajrt may be zero in some cases. See Section C.6 below.
i
1
I
C.4 EXTENSION TO
If
DIMENSIONS
s05
with x- and y-axes parallel to the sides of the region and the origin in one corner. Then the
potential satisfies Poisson's equation:
a2o+ a2o :
A*,
Ay,
-4no(x'
Y)
where the term on the right-hand side describes the sources for the potential (the charge
distribution in the region).
The Green's function is the solution to the similar, but simpler, equation
a2c_(ii')
dx'
where for the moment we consider
+
a'c;!l_.
dy'
*'
i')
:
_4n
6(i _ i,)
(c.17)
tobe fixed.
Method. I Our goal is to reduce equation (C.17) to a one-dimensional equation so that we
may integrate across the discontinuity, as we did in Section C.1. We do this by expanding in
eigenfunctions. This is equivalent to expressing the delta function in one ofthe coordinates
using the completeness relation (8.10).
We imagine the delta function dividing the rectangular region into two pieces. We can
choose the interior boundary through the point *' to be parallel to the x-axis or parallel
to the y-axis. In this example, we'll choose to divide in y (Figure C.3). Then within each
region there are no sources, and the differential equation for G is simpler:
a2G6,1') , a2c(i(, *') _ .,
------;----;.----;--;-_dx'
d!'
(c.18)
0e
0
FIGURE C.3. Dividing
into two regions: region I with y < y/ and region
y/) is on the boundary between the two regions.
a two-dimensional space
II with y > y/. The source
at (x/,
506
oproNAlToptc c
GREEN's FUNcrloNS
We begin by solving this equation by separation of variables, subject to the boundary
conditions at x : 0 and x : a. Let G : X (-t) I(y). Then
X,,Y
We divide through by G
:
+XY,,:O
XY:
X,,
Y,I
i*T:o
To satisfy the equation at aII vahes of x and y, we must have
X,,
-k2
x
and
Y,,
Y -Lz
where we chose the separation constant to be negative so that the solution for X is
a function with more than one zero. (Compare with Chapter 8, Section 8.2.) Here we need
X : sinkx. Then to satisfy the boundary condition at x : a, we need to choose the
eigenvalue kn
:
nr la
so that
sinkna:
0. The equation for Y becomes
.. tnn12
Y
,,,:\;)
with solutions4
Y
In region I, 0
: sirrhnftl and Y - ssshlll
. y { }', we must choose the hyperbolic
sine function so that
y(0)
:
0.
Then our solution has the form
Gr(i,
i') :f
,,{*' ,
y')
sirlla sinn!!)
for
v < v'
n:l
InregionII,b> y > y',wemustchooseasolutionthatiszero ary: b.Theappropriate
choice is sinh
lnn(b
G u(*,
-
y)/al.Then we have
i') :
in:l D,(x' , y') sinlla ,in1",Y9-2
for
v > v'
Now G represents the potential due to a unit "point" charges placed at */, and so the
potential
it produces must be continuous everywhere in our region. This gives our first
4Alternatively, we could choose the functions exp (*nrry/a), but the hyperbolic functions are more useful here.
5The word "point" here refers to a source located at a point in the two-dimensional space. Equivalently, this
solution also applies to a region that is infinite in the third (z) dimension, with no z-dependence, in which case
our source would be a line charge.
1
507
C.4 EXTENSION TO N DIMENSIONS
boundary condition at the internal boundary at
G1(x, yt:
y
:
yt:
,' , y'), Gg(x, y': x' , y')
oo
,, r,, y, \ sin !!! rinh'o;!'
fn--tQa7=taa
r
:i
r, {r,,
Thus, by orthogonality ofthe eigenfunctions
)
r-,
"in
(C.19)
!!a
r'
nn
nn (b
-
v' )
sin(nnxfa),
. - nr(b-y')
,.sinh
. nTTv'
' : Dr(x' , y'), sinh
Cn@' , y')
Defining the new "constant" En(x', y'), we have
n
-
Dn(x' , y')
Cn(x' , y')
sinh
[m(b - y)'laf -
sinh
(nny'la)
(The "constants" are functions of x' and y/, but remember that for the moment we are
considering these values fixed.) Then the Green's function becomes
o,
: i
En(*'
,
y')
,
y')
n:laaa
*
"inno
sinh
'o) ,' *nn
(b
-
yt)
(c.20)
and
oo
cr,
: i
En(x'
n:laaa
rin"* "'*nt(b -
Y)
(c.2t)
,inh'o-J'
We need one more condition to find the unknowns Er. We obtain it by returning to the differential equationfor G (equationC.lT). Writing G intheform D S"O, x', y') sin (nn x /a),
we have
(a\'
i {\ \a/
n:l
-l
+ f!: ,innn' }
""snsin!!!
a
lYt
a )
:
-+o01.,
- x,) 6(y -
y,)
We begin by reducing this relation to an equation in the one coordinate y that we chose to
divide the region. To make use of the orthogonality of the sines, we multiply both sides by
sin (n' n x l a) and integrate from 0 to a. Only the one term in the sum with n : rzl survives
the integration. Then, dropping the prime on n, we have
;lfff '^*#l- -o't'inn'-''
t1v
-v'1
(c.22)
Thus,E, cxsin(nnx'fa).Wecanincorporatethisresultbywriting En: Frsin(nnxtfa)
and, equivaleafly, gn : yn(y,y')sin(nnxt/a). Our expression for G now has complete
508
oproNAlToptc c cREEN's FUNcloNs
symmetry in the primed and unprimed coordinates
equation (C.22) across the boundary at y : yt:
I,'il""
if Ii
is a constant. Next we integrate
;l- ff)' ".#)dY : -4n I,''*"' u' -
Y')dY
The first term goes to zero as we let e --+ 0, since G(*, *'), and hence yn, is continuous
y : y' . We evaluate the right-hand side using the sifting property. Then we have
!
^ tt/+6
oY,
f :
2 oy ly,_, -4n
At the upper limit, y
:
y' + €
)
J',so G
:
(c.23)
G1 and, using result (C.21),we have
I oYnl : -!!! rl" cosh nn(b a
2a
2 |lly,+
At
the lower
limit, y
:
y'
- I < !',so G :
at
Y')
a
"i'hno!'
G1 (equation C.20), and so
l ovnl ::o,sinhnn(b
z 0y ly,- 2- " ------ a
Y')
(Y"ornn")')
ct /
\a
Thus, equation (C.23) becomes
nr /
l)rinh no)' *
-;Fn(cosh-"o9:a
a
sinh
nn(b
a
y')
no)') :
-4tr
"orh a /
The term in parentheses simplifies, leaving
Fn-
nsinh (nnbla)
and so the Green's function (equations C.20 and C.21) is
G(*,
i,)
: i -'=j
,inno''
4nsinh(nnb/a) a
n:l
-
a
"innn'rinh
,'*nn(b 'o)
a
&
y')
G.z4)
fory<y/and
G(i, *,)
: i -.=9 '^ ,inn'*'
a
a "in'o*
n:l
-nsrnn\nrb/a)
for y > y/.
a
"inhno!'
,'nnnn(b
a
- y)
G.z5)
Here again we note the symmetry of the function: G(i, i') : G(*', i). The symmetry
may be made more obvious by writing the function as a single expression. We define
y. :
min (y, y/)
\
C.4 EXTENSION TO N DIMENSIONS
509
and
y, :
max (y, y/)
(compare with Jackson, Classical Electrodynamics) so that we may write
@
G(i,i')' :t-.
.j . ^ rinno*'
z-/.nsinh(nnb/a)
a
a "in"'rinhno)'
a
e r'*nz(b-y')
(c.26)
n=l
Once again we find that division of the region leads to a piecewise-defined Green's
function, but this time each piece is expressed as a series ofeigenfunctions in the undivided
coordinate.
2
An altemative method allows us to expand the Green's function in a double
x and y. But eigenfunctions of which equation? Observe
(C.18)
that the differential equation
is a special case ofthe equation
Method
sum, using eigenfunctions in both
a2c_(ij
dx'
i')
+
a2c_(i_,
dy'
i')
I
),a2G
:
O
(c.27)
with the constant ), : 0. Thus, the eigenfunctions we need are solutions of equation (C.27)
with 1., not equal to zero. This is the Helmholtz equation (compare with equation 3.16 or
Chapter 8, Section 8.4.4). The solutions of this equation are again found by separation of
variables:
X,,
Y,,
V+Tl)'^n:O
Again we choose the separation constant for the X equation to be negative and then choose
the constant tobe -(nn/a)2 so as to obtain the eigenfunctions sin (nnx/a). Then the
equation for Y is
Y"
t nt:,2
v tL^n- \;) :Q
This time we also want eigenfunctions in y that ate zeto at both boundaries, so we choose
Then
and
51
0
oproNAL roptc c cREEN's FUNcrtoNS
Thus, the eigenfunctions are
XY
:
Cmn
. wrxsin mTy
sin'--'
-;:
(C.28)
AD
where we must choose the coefficient C*n to normalize the functions-that is,
lr'
Iru
(c-,"in?!a
dx dy
"inry)'
:
I
Thus,
2
wmn
-
-----
4ab
The corresponding eigenvalue is
r nn..,2 (*o
,
\2
L^,:\;)
*\
r/
Now we make use of the general result (C.l
c(i. *')
l) with
^ "in'TX
: o"F_#u
. nrx
.1,
:
,in^o!
rotlt,
(c.2e)
0 to get
"in"*'
+
,in*T!'
<^,llrr,
mntyttnnfix'
b
mry'
o= Itn a
o "n 4
-= YIy'ttn
ob
fur/a)2 -l (mrlb)2
^L^
(c.30)
Here again the symmetry in the primed and unprimed variables is obvious.
C.5. TNHOMOGENEOUS BOUNDARY CONDITIONS
In the examples we have seen so far, we have imposed the value zero on our solution at the
boundaries. This is not always the case. As an example, let's look at a problem involving
diffusion in one dimension. Suppose heat is conducted along a rod that extends from x : 0
to r : oo. The value of the temperature at x : O is a specified function of time,
T(0,t): t(t)
and T
(x,0)
:
0 for
x > 0. The differential equation for the problem is equation
AT
(3.14):
A2T
i - D ax,:o
We can convert this problem into a one-dimensional problem by using a transform. The sine
transform applied to the space variable is useful because T (x, /) is defined only for positive
CONDITIONS 51 1
C,5 INHOMOGENEOUS BOUNDARY
values of r, our equation has only second derivatives in space, and we know the value of
our function T ar x :0. (Refer to Chapter 7, Section 7.7.3.) When we apply the transform,
T(k,t):
our equation becomes
(E
ai
-ur) :0
(r)
D
at - (kV -u
a7-tk'DT : Dkl;t(t)
lT
*
(c.31)
Equation (C.31) is a one-dimensional inhomogeneous differential equation that we can
solve using methods already discussed.
From Section C.l, with the correspondence
Green's function is (equation C.6)
G(k'
t
.
t')
'
:
Thus, the solution for the transform
u
--+ T, uf
(o
lexr [-kz De
i
(t , t)
- t)l
m -->
k2
D,
the appropriate
ift<t'
ir
t>
(c.32)
t,
i"
: o*rl7
yTJo[' "*p1-t'o(t -
t')lr(t') dt'
Transforming back, we have
T (x ,
t)
:
?
J'
, /JO
o ,rn
or [' ,*,l-k2 D(t - t)lr (t)
JO
dt'|
dk
(c.33)
The expression (C.33) is most easily evaluated by doing the k-integration first and noting
that k sin kx : - (d I dx) cos kx:
T
(x, t)
:
- n?
* lr' , tt' t
cos kx exp
fo*
[-k2 D(t
-
t)f
dk dt,
We write the cosine in terms of exponentials and complete the square to obtain
: Jo[' --i:!L
O(t
J4tr
-
t')3"*n'
*' \ r,,
4D(t
\
- t') /
(
(c.34)
512
oproNALroptc c cREEN's FUNcrloNS
This last expression6 has the form that we expect for a Green's function-type solution, but
now it is the boundary condition rather than the source function that appears in the integral.
In the next section, we generalize this result to show how both the boundary conditions and
the sources can contribute to a solution.
C.6. GREEN'S THEOREM
Here we shall develop a general theorem applicable to solutions of Poisson's equationT in
a region where the solution is not necessarily zero on the boundaries.
Let O and V be scalar functions of position defined in a volume V. Then
i .roivl :9o. iv + ov2{/
and also
V
: Vv . io + q/v2o
.tvVol
Now we subtract and integrate over the volume V and use the divergence theorem (Chapter 1, equation 1.44):
f
/Jv-tV- .toVw) - v .({rvo)l-dv : Js| (ov\v r
\vvo) .ffds
where S is the surface of the volume V and fi is the outward normal. Then
lr(ro'* - vv2o) o,
:
Ir(onv -
vio)
.nas
(c.35)
This relation is known as Green's theorem. It is valid for all scalar functions with derivatives
that exist within the volume V and on the surface S.
Now we choose <D to be the required solution of the Poisson equation
YzcD:
PG)
€0
in the volume V. We choose V to be the Green's function that satisfies the related equation
v2c(i,
*'y: -4n6(i - i')
(c.36)
where * and i/ are within V. Physically, the Green's function G(*, i/) is a constant times
the potential at i due to a unit point charge at i'. In SI units,8 the constant is 4ne6. Then
: 0 in this expression. This is an artifact ofour use ofthe sine transform. The resulting
function is necessarily odd and thus equals zero at x : 0. To obtain the correct boundary condition, we must take
the limit as -r -+ 0 for positive values of x. See Problem 10'
TSee also Problem 5.
8In Gaussian units, the constant is 1, a more pleasing result.
6It appe-, that Z(0, r)
C.6
GREEN'STHEOREM 513
equation (C.35) becomes
t^^f
I @v2c Jv
cv2o) dv
: Jsl(ovc - cvo).ffds
And using the differential equations satisfied by O and G, we get
i')r
," : l,raic- cvo) . nds
l,la-o"u(i - - " (-*)]
The integrals are over the variable
have
-o(i')
.#
i,
i/
with
lroru.i')p(i)
held fixed. Thus, using the sifting property, we
: * l;ovc-
cvo)'nds
o, - G tovc /
GVo) .fids
o'
giving the formal solution for O:
,lflf
o(i') :
4"^ Jrc(i,
*')p(i)
The boundary conditions specify either O or iO ' ff on the surface, but not both. (See
Appendix X.)
(a) If O is specified on the surface (Dirichlet conditions), we choose Go(i, i') : 0 on S
and obtain
o(*')
: J[ Gn(i,i')p(*)dv - !+lt JS[ rvco.ffds
47t€g Jy
(c.37)
This solution9 is consistent with our previous expressions in the special case O : 0 on S.
(b) If iO . ff is specified on the surface (Ne rmann conditions), we choose iCrq(i, *') . n
: constant: K on S and obtain
o(i'):
+#f
hl,G^r(i,*')p(*)dv
In general, we cannot choose
K:
o-u..nds-
0, because
f vrcdv :
J,
I u<n-;(.')dv
-4tr rv
-
-4n
But according to the divergence theorem,
Iro'o
gco-p-"
o,
:
lrfc(i,
this result with equation (C.34). See Problem 18.
ir'y
.ffds
: KA
[
lrr**1ot
51
4
oproNAL Toprc c cREEN's FUNcIoNS
where A is the arealo of surface S. Thus,
K : -4nlA
and so the solution for (D becomes
o(*'):
h
lrGr.r(i,i')p(*)dv
o(*'):
#
l,GN(i,i/)p(i)dv
+
[
+
f / c.vo
frc*vr'fids+
.
nds +
]
lrras
(o)
(c.38)
where (O) is the average value of <D on the surface S. This additional constant does not
affect the physics, which is governed by the gradient of O, and for most purposes can be
removed by redefining the reference point for O.
C.7. THE GREEN'S FUNCTION FOR POISSON'S EQUATION
IN A BOUNDED REGION
C.7.1. Symmetry
The Dirichlet Green's function is symmetric in the primed and unprimed coordinates. To
see why, we begin with equation (C.35) with O : G(i, i) and V : G(fr, i';. Then
Irloru,i;v,2c1d,
:
*')
-
c(d,
*)vlcln,ll]
a3n
*;i,c1t, i') - c(i, i')i,c(i, *l]
,{ [t,u,
.nas,
The Dirichlet Green's function is zero on S, so the right-hand side is zero. On the left, we
use the defining differential equation (C.36):
-+, lv[
[c(fr, *)6(il
-
*')
-
c(d, i/)E(i
-
*)] d3fr
:0
Thus, using the sifting property, we have
G(*"
i) : G(i, i')
which is the symmetry relation we need (compare with equation C.2).
For the Neumann case, the right-hand side becomes
-!
f;ort
i) - c(i, *)tdsu
lONote that A is the area of the total bounding surface of the volume V.
pieces, A is the sum of the areas of all the pieces.
If the surface comprises several separated
C.7 THE GREEN'S FUNCTION FOR POISSON'S EQUATION IN A BOUNDED
REGION 515
Thus, symmetry may be proved only for infinitely large bounding surfaces (A -+ oo).
However, symmetry can always be imposed as an additional condition on the Neumann
Green's function.
C.7.2. The Division-of-Region Method
Because G satisfies equation (C.36), the solution is of the form
c(i.*'):-]-+f(i.i')
lx x'l
-
where V2(1/li - i'l) : -416(* - i/) (equation6.26) andY2lr: 0. The function ry'
is required for G to satisfy the boundary conditions. This form of solution is not always
useful, however, because it can be hard to integrate. Instead we can expand the. Green's
function in a set ofeigenfunctions appropriate to the region ofinterest and make use ofthe
orthogonality of the eigenfunctions when performing the integrations in equations (C.37)
and (C.38).
Here we outline the divison-of-region method for calculating the Green's function in
a bounded region of space. The method works for either the Dirichlet or the Neumann
Green's function. In the Neumann case, it produces the symmetric Green's function. Remember that in calculating the Green's function we take the coordinates of i' to be fixed.
1. Draw the volume under consideration.
2. Choose a suitable coordinate system. (We'll call
the coordinates u, u, w.) Each boundary
of your volume should correspond to a constant value of one of the coordinates.
3. Place a point charge of magnitude 1 at an arbitrary point i' within the volume.
4. Determine the solutions of Laplace's equation V2O : 0 in your coordinate system.
Note which of the functions are orthogonal functions.
5.
Choose one of the coordinates, say u, ar'd use it to divide your volume into two regions,
with the point charge on the boundary between those two regions-that is,
Regionl:u <u'
Regionll:u>u'
It is important that the functions in the other two
coordinates (u, u') be orthogonal
if you are using spherical coordinates, you may only dlide
space in r, since the only functions of r in the solution of V2O : 0 are powers and
functions. This means that
the
are not orthogonal functions. With other systems, you have some choice.
6. The equation for the Green's function is equation (C.36). Thus, in either of your two
regions (butnot on the boundary between them), V2G : 0. Thus, you may expand G in
the eigenfunctions identified in step 4. Use the orthogonal functions in u and u. You must
choose the eigenvalues and eigenfunctions so as to satisfy the boundary conditions on
G at the edges of your volume. Remember: For the Dirichlet Green's function, Go : 0
on the boundary, and for the Neumann problem, \Gyl0n : -4r lA on the boundary.
51
6
oproNAL Toprc
c
GREEN's
FUNcloNs
You will have two different linear combinations of the (nonorthogonal) eigenfunctions
in u: one valid in region I (u < u') and one valid in region II (u > ut). At this point,
your function will have two sets of unknown constants.
7. Use continuity of the potential across the inner boundary at u
G(u: u'-,u,w): G(u:
:
uti
ut+,t),u))
Equivalently, you may invoke the symmetry of the Green's function: G(i,
i') :
G(i', i). This condition will allow you to express one set of unknown constants in terms
of the other. Now you should have only one set of unknown constants remaining.
8. Substitute your expression into the differential equation for G and use orthogonality to
reduce to an equation in u.
9. Use discontinuity ofthe field at i : i' by integrating the differential equation across the
boundary atu : ut. The last set ofconstants is thereby found.
10. Check for symmetry, correct dimensions, and so forth.
C.7.3. Dirichlet Green's Function for the Region Outside a Sphere
Let's find the Dirichlet Green's function for the region outside a sphere of radius a.
Step 1. The diagram is shown in Figure C.4.
FIGURE C.4. The sphere's boundary is at r
:
a. The region outside the sphere is divided into two
regions,r<rtandr>rt.
Step 2. In spherical coordinates, the inner boundary of the region is at
r:
a, and the
outerboundary is r -+
oo.
Step 3. We put a point charge at
i/
with coordinates r'
, e'
,0' . Then we
need to solve the
equation
v2 G(*.,
*')
:
-4n 6(i
- i')
(c.3e)
with the boundary condition
G(*,i'):0
forr:a andforr -> oo
(c.40)
C.7 THE GREEN'S FUNCTION FOR POISSON'S EQUATION IN A BOUNDED
Note: At this point,
i
HEGION 517
is the variable and*t isfixed.
Step 4. Within each region, we have the simpler equation
vzG(*.,i')
:o
with solutions (equation 8.56)
c(i, i,)
:Y
(
"l- \
e,^r,* r'-'/
?3)
ym@,o)
in each region.
Step 5. The functions Y7^(0, @) are orthogonal functions, but the powers of
we divide the space in r;
Regionl:a<r<rl
Regionll:r/<r<oo
Step 6. Next we apply the boundary conditions atr :
a and
In region II, as r -> oo, G must approach zero, so AI^ :0,
Gr(i. i/)
r
are
not, so
r -+ oo.
:Dfut,^@,Q)
(c.41)
I,M
while at
r:
a, equation (C.40) requires
,
A1^at
Br,
I ffi:o
and so
BI^ : -o2ltl AlThus,
Gr(i, *')
:lt,^
(, - #)
Step 7. Next we look at the boundary at
Ym(o. Q)
r : r'. The potential
(c.42)
(G) must be continuous
across this boundary, so
o,-
F-
(t'' t' -
#)
Yt
^(0,
Q)
:
n r-!ft
h^ (0, Q)
Nowweusetheorthogonalityofthe Y1-.WemultiplybothsidesbyYf
{) andintegrate
^,(0,
must be separately
over the whole range of 0 and Q, to show that the coefficients of each I;.
equal. Then
Bt^: At*I(rrlzl*l -
o2l+r1
51
I
oproNALToprc
c
GREEN's FUNoIoNS
Inserting this result into equation (C.41), we have
Grr(*.
i')
: D !4!Yh.(0,
Q)
G
while G1 is given by G.aD. We can make this look more symmetric by writing
At* :
a1^f (r')t+l .Tlten
Gr(i,
g2t+r i') : L,.l'1,,,
ur^ (7t)r+tYt+t'
Grr(i.
i') :
a2t+t)Ym(e, Q)
DY4!#J#\Y^@,
\r'), ''r
Q)
-t_
which exhibits the symmetry of G in r and r' .
Step 8. Now we have one set of constants left to find,the a1*, and one remaining boundary
condition atr : rt . To see what it is, we go back to the differential equation (C.39). Writing
C : D ghYh and expressing the delta function in spherical coordinates (see Chapter 6,
especially Problem 8), we have
nG+YY*
+Y2uoeYt^r,^)
: -a'6(';!)srpt -
1't'st1q
-
q'1
t
l.m
Again we multiply both sides by Yi-,and integrate over the full range of 0 and d. On the
left-hand side, we get zero, except when / : lt and m : mt . We use the sifting property to
evaluate the integral on the right. Then, dropping the primes on / and m, we get
:+y -,ffr,^
,
-aos(';!)yk@,,Q,)
Now we can relabel again: Let gtm : f6Yf^(0',Q') and, correspondingly, arm:
PhYk(0' , //). Then the Green's function is symmetric in all three coordinates if pl.
is a constant, independent of all the coordinates. With this relabeling,
6(r - r')
102rf1. /( + tl -+n
,z
12 ''^
r 0r2
Step 9. Now we multiply this equation by
f :
ft:
r
and then integrate across the boundary at
C.7 THE GREEN'S FUNCTION FOR POISSON'S EOUATION IN A BOUNDED
REGION
51
9
On the righChand side, we use the sifting property to evaluate the integral:
arf,*l'*" _ [''*' /(L+l) fi^dr:
1,,-,-
J,,-, ,
f
4n
-7
Next we let e -+ 0. We have already ensured that f1- is continuous at r : rt , and r is also
continuous, so the second term on the left-hand side goes to zero. Then we have
ar!t^1,,+
: _y
r'
e.43)
3r 1,,At the upper limit,
r : rtl,
lrft^l
we are in region II, and so
a pr, le')2t+t
-
azt+lf
|
F,- lv)''*t - a2t+tf
-' , ---CryT7iiil-
u *: fi ----Gi+Gr- 1,,*:
: -rn-utr:!!!:!!I: -ttu^ () - #;)
1,,
At the lower limit, we are in region I, and so
+1,
*' t(r') +'#)
-: *fft(''*' - +)1,,*: #ft(u
: frm\/l+l,,
+l
d21+l
\
,ryo )
Then equation (C.43) becomes
/l
d2l+l \
/l+l
-ttu^\l - 6W ) - tu^\
,,
d21+l \
+t77p
)
-ft^
The r/ cancels, which it must, since
coordinates. Then
p6
2l+1
4n
r,
4n
1
should be a constant, not a function of either set of
4n
flm: 24
1
and so finally we have
Gr(i,
i') := 5-
4' 9!!'1]'*t'
t-Girrr-Ym@' ilYi'(e'' O')
|,, *
o''*'l
Grr(i. *') := \- .1t , k'f'll-l r, * t--irni-Yr^@'
QtYh(e'' o'\
52O
In
a
oproNAL roprc c cREEN's FUNoIoNS
more compact form,
(c.44)
where r. and r, are the lesser and the greater of r and r/, respectively.
Step 10. Since G is 4neo times the potential due to a unit point charge, it should have
dimensions of l/length, as our expression does. This series should converge nicely fot r > a.
As an example of using this result, let's find the potential due to a ring of charge of radius
b > a and uniform linear charge density )., lying in the equatorial plane of a grounded
sphere of radius a. Using equation (C.37), we have
o1i;
: -t
I Gp(i.i';p1*')dv' -l4tr Js/or*'lV'Gp'ffds'
4treoJv
When we are doing the integration, the coordinates * are fixed while */ is the integration
variable. With O : 0 on S, the second integral is zero:
o(i):
I
4treo
[ )\6(r' -b)8(P')
b
Jv
" rLt =2l11
+ |,
:
I
i.s
tr2!+r-r,.' 'r,'
4----------------r+t
'
',
h^@, Q)yh(0' . Q')o')2 dr' d p' d4'
.02j+t-o2t+t1
2l
+l
4n
^*24'Yt^(o'olu-ffi
ffir;"r', l,
"-imQ
4q'l
where now r. and r, are the lesser and the greater ofr and b, respectively.
The integral over Qtyields zero unless lrt :0. Then
rb \o(*): zto
1r2l+1
-
ozl-1r,,
? ffih(o)hIt)
Now we check dimensions: 2n),b is the total charge on the ring, and the term in r has
dimensions of l/length, so the answer is of order charge/(es x length), which is correct for
potential.
Note that & (0)
:
0 for / odd, so only even / appear in the sum. The first few terms for
0<r<bare
o(i)
-:t
(' :2
4 rrbz
4reob\ r " -!"
3 19 -a9
+' +
.:-(35
64 rrba
r3cos2e
cosa
g
-
-
ts
3o cos2 e
+ 3) +
521
C.7 THE GREEN'S FUNCTION FOR POISSON'S EQUATION IN A BOUNDED REGION
whileforr >
b,
O t'q1*' -o2t-tt1 n(o)Pr(P)
o(i): 4neobl
/+rbt'+r
o (r-1\
- 4zeor
/
\
b
"
o
4T esb
l(r, i)Q:*1
-*'=#(35cosae-30cos2
0+3).
]
The first term is the potential due to a point charge, and this term dominates for large r. The
magnitude of the point charge is the sum of the charge on the ring plus the charge induced
on the surface of the sphere. Thus, the total induced charge on the sphere is Qind : - Qa I b.
The solution in the equator and on the polar axis is shown in Figure C.5.
r
a
@ times 4n esb /Q in the equatorial plane (0 : n /2, solid line) and on the
polar axis (0 : O, dashed line) versus r/a. The first three terms in the series are shown
FIGURE C.5. Potential
here, with b
:2a.
C.7.4. Dirichlet Green's Function for the lnterior of a Cylinder
Again we follow the steps of the standard method. Our region is an infinitely long cylindrical
tube ofradius a.
Step 1. The volume is shown in Figure C.6.
Step 2. We use cylindrical coordinates. The volume is bounded by the surfaca p : a.
Step 3.We place a point charge at i/ with coordinates p' , Q', z'.
Step 4.The solutions oflaplace's equation in cylindrical coordinates are given in Chapter 8, Section 8.4.8. We may choose to divide the space in any one of the three coordinates.
522
OPTIONALTOPIC C GREEN'S FUNCTIONS
region
II
region
I
I
l_
FIGURE C.6. The interior of the tube is divided
Step 5. Here
dt
I:
zt
.
we'll divide the volume in z:
Regionll-z<z'
Region
ll'.2 > z'
Step 6. The eigenfunctions in p that take the value zero at p : e are the functions
J-(x*"p/a),where x^n is the nth zero of J^.We exclude the Ns because they are not
well behaved on the cylinder axis at p - 0. The appropriate functions it z are the real
exponentials,
"-x^nzfa
Gr(i,
for large positive Z
i') :l
vnd.
elx^"2/o for large negative z. Thus, we have
A^nJ^(.*":)
ex^nz/arimt for z < z'
m,n
and
Gn(i, i') :\B^,J^(.^":)
r-xmnzfa"imQ for z > z'
Step 7. Setting the two functions equal at the inner boundary, Z
DA^nJ^(-.,:)
m,n
ex^,z'/arimQ
:I
B^nJ^(-*,:)
m.n
:
Zt, we have
r-xnnztf arimQ
REGION 523
C.7 THE GBEEN'S FUNCTION FOB POISSON'S EOUATION IN A BOUNDED
Since the functions J-(x^np la)ei^Q form an orthogonal set, we may equate the coefficients
of each term in the sum separately:
A^rgx^nz' la
:
fi-ng-xmnZ' lo
= d^,
Thus,
Gr(i, i') :
,-o^,
Gu(*,
t^ (.-,:)
exnntz-z'
*') : D o^, J^ (r^":)
)
ta
eimQ
-z) /a rimQ
"x^n(z'
m,n
Step 8. Next we substitute into the differential equation for G, which we now write as
c
: D s^,(z)J^(r^^I) ,'^r
with s^nQ)
:
c"mnQXP
(TU.- .'))
(c.45)
z. and z. equal to the lesser and greater of z and 4/, respectively. Then, with the delta
function expressed in cylindrical coordinates, we have
with
:
td'ef\@]
)-.r. (*^^L\
- p,)a(Q - 6')6{z \ d/ r,^r l-fur^,k)
dz' ) -!ur,
P
L
z')
We multiply both sides by pJ^, (x-,n, pf a) e-i^'0 and integrate over p and @. On the left,
we use the orthogonality of the eigenfunctions and equation (8.96); on the right, we use the
sifting property.
!
(zD
V ko^,r1'
l-* B*nk) .
e#) -
-4n
r^(r^"+)
,-i*Q'5(z
Now we rewrite:
E^,(z)
-
ymnexp
(:u.-.,))
t- (.-"+)
Step 9. The next step is to integrate across the boundary at 7
a2 lt'^1x.,11'
"
I: "' l-*
_
r.,,*
(z' - z' |
*
-
e-i^o'
7t to find the y^n:
t-#)
rz'+e
: -4n I tk - z') dz'
2.,_t
J
As e
-+
0, the first term on the left vanishes, and the equation simplifies to
o2l${r^,)]2
dymneryQ'-z'\
dz.
-_L
o
r'
-
z')
524
oproNAlToprc c cREEN's FUNoIoNS
At the upper limit, z > z', so the derivative is
*" r(Tu' -.,)1.:., : -ry
while at the lower limit, z < z',
so
xmn
d
/xmn. ,.tl
l.;(t
)1,=,,:
o
E"*P
"
Thus,
a2
1fi 1x ^n112 r*nlY g21 : -4
and so
l^n:
,^noLi^r*^rll,
Inserting this value into the expression (C.45) for G gives
c (i,
i')
:
i2 ;frh;rt r ^ (*^n l) t - (, ^, +)
ei ^(Q
-o'
)
"
v
(! <r.- ., ))
Step 10. The result exhibits the necessary symmetry in * and i' and has dimensions of
inverse length, as required.
What is the potential if the cylinder has a band with potential V6 between z : -a and
z: *a and is grounded everywhere else?
This time only the second integal in equation (C.37) is nonzero. Remember that fr is the
outward normal, here p.
a@.Q,i: -GI f2n f+o,o AG
Jo J_" fr\,,=,adQ'dz'
I
: -#
I,* l:," ?Xnni;mr^(.*':)
rh(x^)ei^(Q-o')
" "*o (fu<r.
- ,))
a dO' dz'
: O terms survive the integration over @/, as expected for a system with
azimuthal symmetry. Thus,
Only the m
e(p, z)
: -++
W
ll."
"*o
(Yu- - z)) az'
C.7 THE GREEN'S FUNCTION FOR POISSON'S EQUATION IN A BOUNDED REGION
For e
525
> a, this becomes
a@' z)
On the other hand, for
: 24!HP ll"' "*, (Tu' - D) az'
: voD !ffiP *sinh x6, "*v (-*o,l)
-a < z
14,
Q(p,z):VoDffi# ll,*' " , (Tu - '\) a'' + f',"*v (Tu' - o) atl
: vo"ut
Jo@9'plo)
lero^11-s-,o, + e''o,t)
*onJr(xor) L
:zvoDJ-q(xo'Pla)
"
?
*orJt(xor)
l1
L
n-'o'"*h
-
I
e-xon*
k'o^I-
e-'0,)l'l
(r*1)]
This solution is plotted in Figure C.7, where we show <D/ Vs versus p /o. A few terms in the
series represent the solution well for z > a, but many terms are required for lzl < a, and
even then the expression
failsll
at p
:
a.
o
Vg
p
0.4
0.6
0.8
1.0
a
FIGURE C.7. Solution for the potential O(p) inside the tube. The solid line shows the potential at
z - a /2 (42 terms). The dashed line shows tbe potential at z : 2a (10 terms). For
z > a, fewer tenns are needed in the sum to obtain an accurate result. The additional
terms decrease exponentially.
C.7.5. Expansion in Eigenfunctions
An alternative method of solution produces a single expression for G(i, i/) throughout the
region. We write G as an expansion in eigenfunctions, as in Sections C.2 andC.4, method 2.
The penalty we pay for having a single expression is that we have a triple sum rather than
llW" hun" seen this phenomenon before in Section C.5. Here our eigenfunctions are zero at p : a, and so the
solution, a superposition of the eigenfunctions, is also zero on the boundary. The solution has the correct limit as
p -+ a from below In Figure C.7, the solid curve @(p) approaches Vg as p --+ a, but dives towatdzero aI p : a.
s26
OPTIONALTOPIC C GREEN'S FUNCTIONS
the double sum we obtained using the division-of-region method. The eigenfunctions we
need are solutions of the Helmholtz equation:
(v2 +
k\f (i; :
(c.46)
o
To demonstrate the method, let's look at the Dirichlet problem in the interior of a cylinder of
radius a and height ft. The solutions of equation (C.46) that satisfy the boundary conditions
h are (Chapter 8, Section 8.4.4)
G 0 tt p a and
:
z:
-
.f
: J^(r*,:)
sinPlrz si^o
,2
/ x.n\2
k-^np:\;)*\l,/
I PT \2
(*)
and the eigenvalues are
where x^n is the n th zero of J^(x).We must normalizethe eigenfunctions, using equations
(8.96) and (4.I2). Then, from equation (C.11) extended to three dimensions, with ). :
k2
:0,
c(*, *')
: +n \
m,n,p
r^(,-,*) '*
1^("^,+)
,rnP!Z' ,-i^o'
! [ri,,*-,,)'f;r,l(+)' . (+)')
- Ih \./'L
L \-\^+mooo
a:-a
Plz ,i-o
n:l p:l
r^(.*,:) r^(,-,#) '* ffsinPTz' ,i-1q-Q')
r * +t (x * )t2
* (T)')
t
l.r,
The dimensions of this expression are l/length, as required.
PROBLEMS
fl
Use the division-of-region method to find the Green's function for a damped harmonic
oscillator. Hence find the response of the oscillator to the input /(r) : 1 - t lT for
0 < / < T andzerootherwise.
2. Find the Green's function for a beam supported at one end (refer to Chapter 5, Problem l1) using a Laplace transform method.
Find
the Green's function for a wave on a string of length I,
3.
A2G
"a2G
,;, - "# :6k -
x')a(t
-
t')
by taking the Fourier transform of the equation in time and using the divison-of-region
method for the space dependence. Give your answer as an integral over the Fourier
transform variable ar.
I
!
,!
I
I
j
PROBLEMS 527
4. Find the Green's function for the diffusion equation
AG
AzG
rp:3(x- x')8(t-t')
;-r
by taking the Fourier transform in space and using the division-of-region method in time.
Take -oo < r < ooand0 < / < oo.
Use the result to compute the function f (x, t) that satisfies the diffusion equation
af
a2f
S(x, r)
-
;-oup:S(x,,)
with the source function
S
,-*2 /o' 3(t)
Strow that we can use Green's theorem (Section C.6) to obtain a solution for O, where
O satisfies the Helmholtz equation
(v2
+t<z)o(i): s(i)
with a source function S(*). Determine the solution for O in terms of the Green's function
when O satisfies the Dirichlet boundary conditions O (i) : f (i), a known function, on
the boundary surface S.
6. Find the Green's function for the one-dimensional Poisson equation
d2e
dx2
with boundary conditions O (x)
when
:
0 at
x
: -P(x)
:0
and
x
:
a. Hence find the solution for O
(a) P (-t) : sit (n x la)
(b) p(r) : *2(o2 - *2)
7. Use a division-of-region method to find the Green's function for the one-dimensional
Helmholtz equation
d2v
d*r+k-Y:f(x)
y(x):0atx:0and.x :
a. FindtheFouriersineseries
intheregion0 < x < a,with
for G(x, x/) and hence show that your result agrees with the result of Example C.2.
8. Sometimes we may expand the Green's function as a series of eigenfunctions even if
the differential equation is not of the Sturm-Liouville form. The governing differential
equation for the displacement of a beam is equation (3. I 1):
dav q(x)
dx4 EI
528
oproNALToprc
c
GREEN's FUNoToNS
A beam of length Z rests on a support at each end so that the boundary conditions are
y(0) : !(L) : 0. Show that the Green's function may be expanded in a series of
f,
eigenfunctions, and determine the form of the Green's function. Use it to find the beam
displacement when it is subjected to a load q(x) : ax lL. Compare with Chapter 4,
Problem 15.
fina the Green's function for heat transfer along a rod with insulated ends. The relevant
differential equation is
+At - r#:
0x
e@,t)
and the boundary conditions are 0 G / 0 x : O at x : 0 and at x : L; G(.r, 0) : 0. Treat
the problem as a two-dimensional problem, and use method 1 in Section C.4, dividing
the region in time.
10. Verify that equation (C.34) gives the correct result z(t) in the limit x -+ 0 from above.
Hint: Expand t (t') in a Taylor series about t/ : t.
11. Find the Green's function for the wave equation in three space dimensions, using spherical coordinates. The wave equation is
u2vzc
62
- atrc:
6(i
-
i')6(r
-
r')
G(i, i' , t, t'| is the displacement and V2 is the Laplacian operator in three dimensions. Transform the equation in space and time, and solve for G. Hence find the
displacement f (i, t) if the source is h 3 (t) e- o2". Co*p*" with Example 7.5.
where
12. Find the Dirichlet Green's function for the two-dimensional Helmholtz equation in
a circular region ofradius a. Obtain the result as a double sum over appropriate eigen-
ltil
functions.
pma the Neumann Green's function for the one-dimensional Poisson equation
d2Q p(x)
dxz
60
withboundary conditions dGldx:0 atx :0 and x: L. Express youransweras
series of eigenfunctions. Hence find the potential O if
lporlL ifO<x.Llz
Dlxl:
and
dQ/dx
:
0 at
x
:
0 and
x
a
<
|.0
:
otherwise
L.
14. Use a Fourier transform method and cylindrical coordinates to find the Green's function
for the wave equation in two dimensions.
15. Using the division-of-space method, find (a) the Dirichlet and (b) the Neumann Green's
function for Poisson's equation in the interior of a sphere of radius a.
16. (a) Find the Dirichlet Green's function for Poisson's equation in the half-space z > 0
using Canesian coordinates.
PROBLEMS
(b)
529
Evaluate the integral to show that G may be interpreted as the potential due to a point
charge and its image in the plane z :0. Hint: Use polar coordinates to evaluate
the integral, and use contour integration to evaluate the integral over the angular
variable, as in Chapter 2, Section2.T .2.
17. Find the Dirichlet Green's function for Poisson's equation in the interior of
a sphere
of
radius a as a triple sum over appropriate eigenfunctions.
Eal
OUtain a relation analogous to equation (C.37) for the diffusion equation
af
At
-
Dv2.f
:
S(i, r)
Define the Green's function through the equation
AG
E - Dv2G: Dr(i - *')6(t - /)
Ffint.'Notethatsince G(t,t'): G(t-t'),0Gl0t' : -0G/0t.Integrateoverbothspace
and time, and let G : 0 on the bounding surface.
Apply your result to the example in Section C.5. Apply the sine transform in space to
the equation
AG
A2G
; - o up -
D6(x
-x')6(r
-
r')
and obtain the Green's function. Show that the solution (C.34) may be expressed in terms
0Gl\xt and that this solution is consistent with the general result you found above.
of
19. Find the Dirichlet Green's function for Poisson's equation in the interior of
of radius a.
a
hemisphere
(a) Choose O < 0 <z and0 < 0 < tr.
(b) Choose0 < P < n/2 and0 < Q < 2r.
(c) Using one of the two Green's functions, (a) or (b), evaluate the potential inside the
hemisphere if O : 0 on the spherical surface and O(r) : Vo(l - r/a) on the flat
face.
20. Obtain the Green's function inside a cylindrical tube (Section C.7 .4) by dividing space
tn p.
OPTIONAL TOPIC D
Approximate Evaluation of Integrals
D.1. THE METHOD OF STEEPEST DESCENT
Often functions are defined in terms of integrals in the complex plane. The gamma function
(Chapter 2) is one example that we have seen. Another example is the Hankel functionl
nJ,)(6)
:: l,*, l; (,_:))f"
where the contour C runs from the origin to
-oo,
as shown
(D.1)
in Figure D.1.
Im (z)
Re (z)
FIGURE D.1. Path C for evaluating the first Hankel function U{t) @).The path starts at the origin
and goes to
-oo
in the upper half-plane.
Such integrals may be written in the form
I(€): II sk)expl€f (z)ldz
Jc
lThis function is introduced in Chapter 8 from
a different
(D.2)
point of view.
531
532
OPTIONALTOPIC D APPROXIMATE EVALUATION OF INTEGRALS
where f is real, (z) is an analytic function, and C is a specified contour in the complex
"f
plane. We wish to evaluate the function 1(6) in the limit that f becomes large. In this case,
the exponent
Ef
:
e)
€lu(x, y) + iv(x, y)l
has a large absolute value. The imaginary part causes the integrand to oscillate wildly, contributing very little to the integral, except where u is approximately constant. The integrand
is maximized where the real part u is maximum.
We know that we can deform the contour C without changing the value of the integral,
so long as we do not cross any singularities of the integrand.2 We can evaluate the integral
most easily if we deform the contour so that it passes through a point z0 such that
1. u is approximately constant near z6 and
2. u is a maximum at zo md decreases rapidly along
the deformed contour C' as we move
away from zg.
Then the integral will be dominated by a small region of the path very near zo.
To make u a maximum, we need
{:o
clz
atz-
(D.3)
zo
since 0u f 0x and 0u l0y are then both zero at zo. (Refer to Chapter 2, Section 2.2.2.) Now
if the function f (z) is analytic, then the function u(x, y) is harmonic; that is,
^ 02u
Y'r,t:ip+
E2u
Ayr:O
if a is a maximum with respect to x (02u/0x2 = 0), it must be a minimum with
respect to y (02u I Ey2 r 0). Thus, the point z6 that we seek is a saddle point. In order that
the contribution of the integrand to the total integral be concentrated in the vicinity of zo,
we must choose our path C/ so that it crosses the saddle in such a way that u decreases from
its value at zo at the maximum rate. (Ct looks]ike the rider's legs on the saddle.)
We also know that for analytic functions Va is perpendicular to Vu (Chapter 2, Section2.4.3).Thus, by choosing the path that maximizes the change in a, we are also choosing
the path along which the imaginary part u remains almost constant. This is exactly what we
Thus,
want.
Now
if
the function g(z) varies relatively slowly in the neighborhood of 26, we can
approximate the integral as follows:
1(6)
: If sk)expl€f(z)ldz
Jc
- skil
:
fonnn*.0 "*o
gkilexp [5,f (zo)]
lf (rt.ol
*)u - ro>'f"ko))) dz
lru,nn"*,0*o
[e
2See Chapter 2, Section 2.2.4.We may not cross a branch cut either
if.
-
zo\2
7" {zs1fdz
(D.4)
D.1 THEMETHODOFSTEEPESTDESCENT
533
where we used the Taylor series to express f (z) near eo -d truncated after the third
term. The second term [involving f'(z)l is zero. To choose the path, we set z - zs =
reiQ and f"(zd : aeia, where a and cv are real constants. Then the Taylor series
becomes
f (z) : f (zo) +
u
:
u(xo, to)
+
u
:
u(xo, to)
+
)r2
"2ia
orio
(2Q
+ q)
sin (2@
* cv)
)or'cos
and
1
1ar2
To keep u approximately constant, we choose the path so that 2Q
*u:
nn; that is,
an
Q:Qo:-t*t,
(D.s)
This path is a straight line in the neighborhood of eo. Then
u
: u(xo,ld +
)or2
cos (nn)
:
u(xo, ys)
+ (-D')ar2
Now we war,ttu to decrease as we move away from 20, so we choose n
u
:
u(xo,
yil
: *1. Then
- !or,
2
The choice of n : *l or - I is made by requiring that r go from a negative value3 before
passing through 46 to a positive value after. The integral (D.4) is then reduced to an integral
over r, with the value of r increasing throughout the range of integration:
[
(-+\z
I (E) =8(zo) exp lEf {zd)
"*p \
./pathnearzg
:
lkd
exP tE
f Qo)te'^
/,: ; #:
a, ,'oo
/
".'
(-+)
o'
3In the usual polar representation of a complex number, the amplitude r is a positive real number, and negative
real parts are described by phase angles between r /2 and 3tt 12. But here we want to describe the straight line
path with a constant phase angle {g, and so we must allow r to take negative values.
i
534
oproNALToprc D AppRoxtMATE EVALUAIoN oF
TNTEGRALS
because f is very large. Thus, we may
The exponent larz 12 becomes very large for r
extend the limits in r to aoo without appreciably changing the value of the integrala:
l0
r
(e =g(zo)
I
(il :
exp t€f kdt,'^
l::
o,
2n
k)leih
g(zo) exp tEf
^, (-+)
(D.6)
.
q,a
The result (D.6) is the asymptotic form of the integral (D.2).
Let's use this method to evaluate the asymptotic form of the Hankel function (D.l).
Comparing equation (D.1) with the standard form (D.2), we find that the function "f (z) is
t/
l\
f(z):l\'-;)
and is analytic except at z
:
O.
Similarly,
1
sQ): ,,+t
We want the path of integration to pass through 20, where
dfl
r\
t/
;1,:,,:t(t*d:o+Zo:ti
Of these choices, only
pointz6
:
*i
lies near the original path C. So we deform C to pass through the
i. Then
..
I
f"(zd: - -l : -t :
- tt
I
7J
So
c: I
4, :
and cv
r /4 -f r 12
:
:
-n
le-i1t/2
|,
12. Thus, the new path C/ must pass through zo at an angle
-
3n /4 (equation D.5). Before zo, the difference z
zo has a positive
so r is negative, as
real part on this path (see Figure D.2) and ,3it/4
+
- -+A
required.
Then
r(zo):f(i):;(t-i) :,
4See Appendix
IX for the evaluation of the integral.
\iJ1,
D.2 THE METHOD OF STATIONARY PHASE
s35
Im (z)
Re (z)
FIGURE D.2. PathCtfor evaluating the first Hankel function n{t) @).The path passes through the
point z
: i at45o to the imaginary
axis.
Thus, applying the result (D.6), we have
n{')G) =
,Eu*'r(e
1f
+
3n
4
z\
-o+2)r)
z\
|
-u
2 - -4/
which is the expression given in Chapter 8, Section 8.4.2 (equation 8.85).
D.2. THE METHOD OF STATIONARY PHASE
A similar method applies to integrals of the form
I (x)
s(x) exp [iQ@)]dx
(D.7)
when the function d(x) (the phase) is real-valued and g(x) is a slowly varying function.
Integrals of this type arise through the use of Fourier transforms, for example, and are of
importance in determining the signal that arrives after propagation through a dispersive
medium (see, for example, Jackson, Chapter 7). The major contribution to the integral
comes from the neighborhood of the point (or points) where @ (x) is stationary-hence the
name of the method. Away from the stationary points, the integrand oscillates wildly and
there is very little contribution to the integral.5
Again we expand the phase in a Taylor series in the vicinity of the stationary point .rr:
Q@):d(x,) +
5lt.
)a - x,)2 Q"(x,)
(D.8)
-o." precise statement may be made in the language of generalized functions. See Chapter 6, especially
Section 6.4.
536
oproNAL Toptc D APPRoxIMATE
EVALUATIoN
oF INTEGRALS
Then, since g(x) varies much more slowly than eif
I(x)
,
- B(xJeiQGr l)1,' "-, (Uo - *,;2q"6"1) ax
Here also we can extend the range of integration to (-oo, *oo) with negligible change in
the value of the integral.
To evaluate the integral, we make a change of variables to
tu:(x-r,)tltO"@,)
The limits of the integral also change correspondingly. Let's assume for the moment that
Q" @r) is positive6 and equals 2A2.Then
Q"
u-(x-xt)
(xt)
-ir/4_A@_x")e-io/4
2
and the integral becomes
I (x)
-
g1x";si\GJeila
f""-" au
Im (z)
Re (u)
FIGURE D.3. The.contour for the u-integral runs from -oo to +oo but at a 45o angle to the real
The path of integration C for u is at 45o to the real axis (see Figure D.3). However,
since the integrand has no poles and the contribution to the integral ftom u : *oo is
zero, the value of the Gaussian integral (equal to Ji when evaluated along the real axis)
6lf
6't 1x"1 is negative, the path is rotated to cross the real axis at an angle of +T
is the same.
f4 ftther
than
-t
/4. The result
PROBLEMS
537
is not changed by this path shift. Thus,
(x)
-
I (x)
-
r
g1x;siQl)eila
ll* ,-"'a,
(D.e)
g(xJsiQ@J.i'/4
If there is more than one stationary point, then the integral is the sum of the contributions
from each ofthe stationary points.
PROBLEMS
f,
Use the method of steepest descent to evaluate the asymptotic form of the gamma function
f (t):
[* tE-tr-'4t = tE6E-rtzr-e
Jo
where the path of integration is along the real axis. This is Stirling's formula.
2.
The modified Bessel function
Kr(f ) has an integral representation
K,(E):
I roo _t ( "_.1\l d'
"*p L-z \.'*;/l ;|
1Jo
with path of integration along the real axis. Use the method of steepest descent to find
the asymptotic form of Kr(f ) as f -+ m.
3. The Bessel function may be represented by an integral
J,(E)
lf
: 2"
Jrexp
(-if
sin z
*
ivz) dz
where the contour C is as shown in the figure. Use the method of steepest descent or
stationary phase, as appropriate, to derive the asymptotic form (8.83).
PROBLEM
3
538
@
OPTIONALTOPIC D APPROXIMATE EVALUATION OF INTEGRALS
Expand the function g(x) in equation (D.7) in a Taylor series about the stationary point.
Show that there is no contribution to the integral from the second (linear) term in the series
if the expansion of the phase @ is truncated at the quadratic term, as in equation (D.8).
5. The function n
j2) 1xl has the
integral expression
nJ,)c):
(,-:)]f"
* l,*'[l
where the path of integration goes from -oo to zero along a path in the lower half-plane
that is the mirror image of the path in Figure D. 1. Verify the asymptotic form (8.85) for
this function.
6. An alternative integral expression for the Bessel functions is
Fu@)
:
O
l"exp
(ivu
- ix sinu) du
where
(a) for I1(1) use contour C1 and k : I lt
(b) for I1(2) use contour C2 andk : lln
(c) for ,/ use contour C3 and k : I l2n
and the contours are as shown in the figure. Evaluate the integrals for large values of
and verify the asymptotic forms in Chapter 8.
PROBLEM
6
7. The Airy integral
Ai(x)
:
+"
I:""n; (r' *+)0,
-x
PROBLEMS
539
arises in the study of diffraction. The path of integration lies slightly abovethercalaxis.T
Use the method of stationary phase to show that
Ai(x)
*, (-'r. '')
- #*
for large, positive x
E tr"
amplitude of a signal arriving from a distant source after propagation through a dispersive medium may be written as a Fourier integral of the form
s(x, r)
:
I*
orr',explik(a)x
-
iorldo.t
where k(ar) is the dispersion relation for the medium (see, for example, Jackson, Chapter 7). Use the method of stationary phase to show that, at time r, the largest amplitude
signal is contributed by frequencies with group speed dro ldk : D I t , where D is the distance from source to receiver. Find an approximate expression for the amplitude at time
t. Obtain
an explicit form for the solution
TSee Jeffreys and Jeffreys, Section 17.07.
if ck(a)
:
F4,
and a, is a constant.
OPTIONAL TOPIC E
Calculus of Variations
E.l.
INTEGRAL PRINCIPLES IN PHYSICS
Physical systems are often characterized by being an extremum (maximum, minimum, or
point of inflection) of some physical property. I For example, a system in equilibrium is at an
extremum of the potential energy. For stable equilibrium, the potential energy is a minimum.
According to Fermat's principle, light rays follow the path of minimum time. The physical
quantity being minimized in these examples (energy or time) is often expressed as an integral
over the system. For example, the time for light to travel a distance dt is dt : dl / u, where
u is the light speed, and thus the total time for light to travel from point A to point B is
fB d(.
t:
I
Je
(E.1)
u
where the integral is taken along the path from A to B. The path from A to B is described
by one or more parameters, and the path actually followed by the light between A and B
will cause the value of the integral (E.l) to be a minimum.
If we want to find the true path between points A and B, we must adjust the parameters
so as to make the integral (E.1) an extremum. To see how to do this, first recall that if a
function /(x) has an extremum at the point rs, then df /dxlro: 0, and the change in /
due to a small change dx
in.r is
6f= df
dx
- 0
0x
*
terms of order (dr)2
xo
to first order in 6x
if the integral (E.l) is an extremum, its value will not change when we make
small changes in the parameters. That is, the integral along paths that lie very close to the
true path will be the same as the integral along the true path, to first order in changes in
the parameters that describe the path. The difference 61 between the value of the integral 1
along the ffue path and that along a neighboring path is called the variation in the integral 1.
Similarly,
I
Note the language carefully; "a" maximum is not the same
as
"the" maximum. We
are interested in local extrema.
541
542
OPTIONAL TOPIC
E CALCULUS OF VARIATIONS
E.L. As a simple first example, consider the path taken by light moving
from point A in a medium with refractive index n1 to point B in a medium with
refractive index n2 (see Figure 8.1). For now, let's take it for granted that the light
travels in a straight line within each medium. Then the path meets the boundary
surface at point P, and we can parameterize the path by the distance s of the point
P along the boundary surface, as shown. Since the speed of light in each medium is
u : c /n, the total time taken for light to travel from A to B is
Example
,
:
lo'
!!dt
n1
+
n2
I" -dt
s2+tfi+
c
.^
c
n2
c
(w
-
s)2
where in this case the integrals are trivial. Now we compute the change 3r in r due to
a small change 6s in s:
6/
:
At
-3s
0s
,,I
t
A
FIGURE E.1. The path of a light ray from point A in medium
I
to point B in medium 2 passes
through point P on the boundary between the two materials. We use Fermat's principle
to determine the location of point P and also to show that line A P is straight.
This change must be zero for any 6s when the value of s corresponds to the true
path. Thus, the value of s we need is given by
nl
S
n2
U-.t
-0
543
E.1 INTEGRAL PRINCIPLES IN PHYSICS
or, in a more familiar form,
n1
sinfi:
nzsin)z
which is Snells'law.
Now let's ask whether the light does travel in a straight line from A to P. Let the
path be described by the function y(r), where y(0) : 0 at point A and y(s) : ftt is
point P. A differential piece of the path has length
d(.:
\/dx2
+
dy2
: 1[* (!oj
o-
(8.2)
and the total time taken for light to travel from A to P is
,:T l,
(#)' dx
Here the integrand is a function of the derivative y/ of y. Our goal is to find the
function y(x) that minimizes the time r, subject to the constraints that the endpoints
y(0) and y(s) are fixed. The quantity / is a functional; that is, it is a function of the
function y(x).
Now as the function y(x) changes, so does its derivative y'(x).If the derivative
changes by a small amount 6y/(.r) (the variation in y'), then the integral changes by
an amount 6r, where
tt6t: T
l,'
T
l,' t/l +
13yt)2 dx
(y')\
zyny'
a,
to first order in 6y/
:T l,'JTrory('.ffi)'.
-+-!-
nl tfs
v'6v'
gT6,y", lo ---J==--:-)-
When y(x) describes the true path, 3t
by parts:
'' :!)
-;
6t
:
0. To obtain 6t in terms of 3y, we integrate
u, "'"
o*
- !-291!:l
-Ll' * 1 Jot' l-L
Jt+ryflo',
t\frTory 2 0 +o))3/2)"r
With the endpoints fixed, the variation 6y must equal zero at x :0 and x : s. Thus,
the integrated term is zero. The remaining integral must equal zero no matter what
the variation 6y(x) if y(x) describes the true path. Thus, the term in square brackets
544
oproNAL Toprc E oALcULUS oF vARrAroNs
must be zero. This term simplifies to
t),1
11+(Y')zlztz-"
and thus the second derivative J" : 0. This means that the slope of the line, y/(x),
must be a constant, and thus the path is a straight line.
E.2. THE EULER EOUATION
Let's review the general procedure that we used in Example E.l. We have ai integral
r (y)
: I"u , o, y, y')
(E.3)
d.x
evaluated along some path between the points a and b. That path is described by the function
y(.r). The endpoints a andb are fixed, and thus the function y(x) has fixed values l(a)
Yr
and y(b)
Y2 at the endpoints. Our goal is to find the function y(x) such that the integral
:
:
1(y) is an extremum.
We begin by expressing the variation 61 in the integral
function y and its derivative y/:
u,:
tb / af
J"
I
in terms of the variation in the
{un,)
[rrrr. ,r, ) ,*
_
To express the second term in terms of the variation 6y, we integrate by parts:
I' (#r') a* : ffiul,'"- I' * (#)',
The integrated term2 is zero because
6r:-
6/
:
0 at
x
:
e and .r
:
r.
b. Thus,
1"'l*ffi -#)ro.
I
We want the variation 61 to be zero for arbitrary3 small variations 8y(x) from the true
function y(x), and so the term in square brackets must be zero:
d (y\_Ef :o
dx \|yt)
2In th"
andx
: "ase
b.
(E.4)
Ay
that the value of y is not fixed at the endpoints, we obtain a second conditi on: 3f / Tyt
3Some restrictions apply, most importartly that 3y be differentiable.
:
0 at
x
:
a
i
1
I
E.2 THE EULER
EQUATION 545
The arbitrariness of 6y is essential to this argument. We can perhaps find a specially contrived
Ey to make the integral zero, but if 6y must be arbitrary, then we are forced to make the
square bracket zero. Then equation (E.4) is the desired equation for the function y(x). This
equation is known as the Euler-Lagrange equation.
An alternative form of equation (E.4) may be derived by noting that the total derivative
d.f a.f af dy 0f dzy
d*: a** u**87?
Solving tor 0f l0y and substituting into equation (8.4), we obtain
L
(
y\ -af : L ( y\ - ! (4L -af -914\
y'
oy - dx
ox
ax \ay'
)
\ay'
)
\dx
ayt dx2
:o
)-"
and so
df _af _ Ln,, _ r,L (91\
'
dx 0x 0y''
dx \Dy'/
:o
We may simplify by noting that
d (.,,a/\ _:' .r,Ef .__.,,d
*'E (af \
a"
V
8
ar,)
\av,)
So finally we have
*(, -r#) -#:o
This version is particularly useful when
/
does not depend explicity on x
since then we may integrate once to obtain
, - ,'#:
constant
."s)
(0fl3x =
O),
(E.6)
E.2.1. Application to Mechanics
The Lagrangian of a mechanical system is
L:T -V
where
Z is the kinetic
V the potential energy of the system. These quantities are
qi that describe the position and velocity
parts of the system. For a system of particles, the coordinates qi arejust the
and
expressed in terms ofthe generalized coordinates
of all of the
546
OPTIONAL TOPIC
E CALCULUS OF VARIATIONS
ordinary coordinates of the particles in the system. Hamilton's principlea says that the
motion of the system will be such as to minimize the integral
Itz
I:
I L(qi,qi,t)dt
Jt,
Thus, we can determine the motion of the system [that is, determine the
calculus of variations.
qi(t)l
using the
E.2. A particle of mass lzl slides down a wedge whose sloping side of
L makes an angle 0 with the horizontal (see Figure E.2). The wedge of mass
Example
Tength
M isfree to slide horizontally. How long
does the particle take to reach the ground?
FIGURE E.2. A particle of mass la slides down a wedge of mass M that is itself free to slide
horizontally.
It is convenient to choose our generalized coordinates to be the distance s that the
particle has slid along the wedge surface and the distance r that the wedge has slld
over the ground. With these coordinates, the kinetic energy of the wedge rs lU*2.
The velocity of the particle with respect to the ground is
- ldx ds \^
n:
(, - Vcoso )*-
ds
Esnav
and thus the total kinetic energy is
,:l* (#)' .;^l(# -ff*",)' *
:
I
r(M
+*)
/ dx\2
dx
ds
\i ) -*iicoss*
I /
ds\2
r^ \A )
With reference level at the top of the wedge, the potential energy is
V
:
4See, for example, Goldstein, Chapter 7, Section 5.
-mgs sin9
E.2 THE EULER
EOUATION 547
Thus, the Lagrangian is
1
/ dx\2 dx ds
| / ds\2 jmgssinl
L:r -v - r(M
+^)
-*iicoso*
r*\i )
\i )
The Euler-Lagrange equations for this system are
!dr (9!\ -oL
:o
0x
\0i /
/dx\ d'-"ore'l
lla,L(M**)\A)-mA
-o:o
I
(8.7)
and
!dr (9!\ -oL :o
\3.i
o (-**cosg*-+)
""dt)
dr\ "'
)
0s
-mgsino:o
(E.8)
Equation (E.7) may be integrated immediately to get
(M + m)
(#) - mfrcose : c1
@.e)
which may be recognized as conservation of momentum in the .r-direction. Thus, if
the system starts from rest, the integration constant Cr : 0. Integrating again, we
find
(M
and again, if
x
:
s
:
0 at
t:
*
m)x
-
0, then Cz
ms cos9
:
:
C2
O.Equation (E.8) may also be integrated
to get
/ dx
ds\
(-tcoso* i)- grsind:o
where again we used the information that the system starts from rest. We may now
dx/dt using equation (E.9) and integrate again with the initial condition
eliminate
s(0)
:0:
!!(tdt\
m
M*m "or'e):ptsino
/ "
/ M+m \gr2
|
" - \,vl +msin20)
2 ""'"
-Slnd
-
548
OPTIONAL TOPIC
E CALCULUS OF VARIATIONS
The particle reaches the bottom of the wedge (s
Tn
t:V'*"\
: I)
in
a
time
/ M +,,,sin2o\
M+^
)
+
oo (fixed wedge) to obtai n
letting M
the correct result for a particle accelerating at g sin 9.
We may check our result by
t
-
JTLI
9FF'|0
,
E.3. VARIATION SUBJECT TO CONSTRAINTS
Sometimes the variations cannot be completely arbitrary because of additional constraints
If we want to find an extremum of an integral 1 (equation E.3) with
integral J : I: g(x, y , y') dx temainconstant, we proceed by
a
second
that
the constraint
(but
yet unknown) multiplier.tr" to form the combination
as
a
constant
introducing
on the physical system.
K:I*)'J
Then
if
1 is stationary and "I is constant,
K
must also be stationary. Proceeding
as
Section E.2, we obtain the Euler-Lagrange equation (E.4 or E.6) but with the function
replaced by the functio11 11 : f | )'9.
E.3. A uniform cable of length L hangs between two points at the same
height and a distance D apart. Find the shape of the curve described by the cable. The
curve is called a catenary.
In this problem, the relevantphysical quantity is the gravitational potential energy of
the cable, which will be a minimum when the cable is in its equilibrium configuration.
If we choose the reference level at the height of the two endpoints of the cable and let
y(x) describe the distance of the cable below this level at coordinate x, 0 < x < D,
the potential energy is
Example
u(y):
rD
|
-t@)s a^ = _u.c JoI y\/t + (y,)2
dx
(8.10)
where g, is the mass per unit length of the cable, dm : It dt, and we express the
length element dt of the cable as in Example E.1 (equation E.2). At the endpoints,
y(0)
: y(D) :0.
However, the length of the cable is a constant, so we may not vary
completely arbitrary manner. We must have
L: I
dL:
loD
G/t) dx:
Ioo
,/T+
1y,y
a,
y(x) in
a
(E.1r)
[Here we choose to call the integrand G(yt) to avoid confusion with the acceleration
due to gravity, g.l None of the integrands contain x explicitly, so we may use equation (E.6), including the constraint. We may absorb the constants p and g into the
in
f
E.3 VARIATION SUBJECT TO CONSTRAINTS
constant multiplier l,:
f
+ ),G - y'a$
!!G) :
0y'
+ ))Jt + 0'Y :
W - r' !t,
dy'
o + xlJ*
o+DJt+ry'Y(r-#h)
constant
c
:.
y+^:
stfrT(Y'Y
Solving for the derivative, we find
dv
ll
]:trl
dx
-.(y*x)'-l
As the cable hangs, the slope dy/dx is positive on the first half of the cable and
negative on the second. Now we may integrate to relate the coordinates (X, Y) of
apointonthecable. For0 < x < Df2,
tX
:Jodx
1,,
Let y
* ), :
C cosh0, dy
:
40
+r)2
-t
C sinhd dd. Then
lcosn-rtr+r)/c C sirn0 dg
J"ort-'(t
/c)
c l"osr'-r fI+)
C
L
\
: x
'inha
:t
CJ
/ -cosh-r1l
Rearranging, we can solve equation (8.12) for
!:
C cosh
I
for0 <
x < Dlz
on the first half of the cables:
/x
(a *cr"n-tl\-l'
C/
The slope of the cable is then
4 :
dx
sinh
(1
\C
+
(E.12)
"ortr-t
1)
Cl
sHaving completed the integration, we no longer need to distinguish between Y and y.
(E.13)
550
oproNAL roprc E cALcuLUs oF vARtATtoNS
At the midpoint, x
:
D 12, the slope is zero:
o:
sinh
1\
/D
l#.""tn-' ;)
Thus,
It is convenient
D
,I
cosh-'2CC
to choose C to be negative, C : -Y, and then
),
:
-y
"o"n
fi
Our solution (E.13) becomes
t
D
-\l
Y:zfcoshr-cos,n(Dl2\ y /l
with slope
dy .. /Dl2-x\
d":stnn\
, )
Next we apply the length constraint (E.
the first half of the cable.
11
) to find y . Again it is easiest to work with
LfDl2fD/2
::zJoJo
| ,/t+(t,)zdx: I
1*sinh2
( Dl2 - *) ,, :
: [''' "orn
---y
lo
\ v )
sinh
(r7)r.
( D/2
-
\ v
x\lD/'z
/lo
D
-ysinh'2y
which is a transcendental equation for y that can be solved numerically. If the cable
length t is twice the distance D between the supports, then
DD
_
: sinh
with solution
Dly :4.355,
shown in Figure E.3.
and hence
2y
y :0.2296D.
The shape of the cable is
551
E.4 EXTENSION TO FUNCTIONS OF MORE THAN ONE VARIABLE
0 0.2 0.4 0.6 0.8 1.0
0
x
D
0.2
0.4
0.6
0.8
v
D
FIGURE E.3. The cable's shape is such as to minimize its potential energy. Here we
length Z : 2D. Remember that y increases downward.
see a cable
with
E.4. EXTENSION TO FUNCTIONS OF MORE THAN ONE
VARIABLE
The theory of calculus of variations may easily be extended to integrals over more than one
variable. The integral is of the form
r: JI f $i;u)d,*
x;, i -- I - n, arc the independent variables and a is a function of the variables -r;.
Upon taking the variation, we have
where
u,:
| (#'".D#rn,(#))"-
Once again, we integrate by parts to convert the variation in each derivative to a variation
in a. We obtain the Euler equation:
Z*ffi^)-{:o
(E.14)
In this expression, the derivative E l0x; is apartial derivative in the sense that we are holding
all the other coordinates fixed, but it is a total derivative in the sense that we must include
the implicit as well as the explicit dependence upon r. For example, if n :3,
aFll-l
-tax
aFl
tT
f
I"on.ty.z dx l"onr, u.y.z.ux.uy,uz
I-
aF 0 (0ul0y)
' A Gu/Ay) 0x
552
OPTIONALTOPIC E CALCULUS OF VARIATIONS
If we already knew the function u(x, y, z), we could obtain the same result by explicitly
writing out the coordinate-dependence of the function u and all its derivatives to obtain
F (x , y , z) md then evaluating the usual partial derivative 0 F I 0 x . But in cases where we
want to use equation (E.14), the function u is the solution that we seek, and so the explicit
dependence onr is notknown apriori.
Example E.4. _ The electrostatic field in a volume V may be described as the gradient
of a potential, i : -id, where the potential Q@, y, z) takes on specified values on
the bounding surface S. Show that the function @ that minimizes the electric energy
in the volume V satisfies Laplace's equation: Y'Q :0.
The electric energy is given by the integral
u:'; lun'av:') lrr'ofav
:7 l,l(*)'. (#)' . (y)'),,
The integrand
tion (8.14) is
f :
(0Q/0x)2
+
a (0Ql0z)2, and so the Euler equa-
@Q1AD2
*(#).*(#) .*(#) :Y2Q:O
PROBLEMS
fl
u : us(l * y). what is the path of a
ray? This model describes the propagation of seismic waves through the Earth's outer
rne
speed of waves in a medium varies with
)
8S
layers.Ifthewavesstartat):0,.tr:0,findthevalueofratwhichthewavesreturn
to the surface as a function of the initial slope of the ray dyldx : m : tan0' Also
determine the total time of travel for each ray.
2. Find the path of a light ray through a medium whose refractive index increases linearly
with depth: n : I *x/a. Assume that the path starts from the origin, I : y : 0, and
dyldx:latx:0.
3. Rework Problem 2, reversing the roles of the labels x and y and describing the time as
an integral over the new r. Is the result the same? Which method is easier?
4. Repeat Problem 2 with refractive index function n(x) - nsex '
A(0, 0) and
S fne brachistochrone. A smooth wire runs between two fixed pointsfriction
on the
y).
without
particle
sliding
a
Find the shape of the wire such that
B(X,
point
A
from
particle
starts
the
that
wire reaches point B in minimum time. Assume
that
of
0.
Show
y
functions
as
x
and
obtain
with speed u6. Hint: Set y/ - tand and
the coordinates of the two fixed points are sufficient to determine the initial and final
values of 0 and the two integration constants. Determine an explicit solution in the case
u0 : 0, X : Y : -1. Plot the shape of the wire in this case.
PRoBLEMS 553
6. Show that the Sturm-Liouville equation (8.1) anses from the problem of finding the
exffemum of the integral
t
:
l"ou{r,),
+
eyrldr: I,u Fdx
subject to the constraint
,:
l"u
,y2 d,
:
l"u
odx
: constant
7. Show that the catenary (Example E.3) is symmetric about the midpoint-that
is,
y(D-x):y(x).
8.
f,
Show that the curve that encloses the greatest area with a fixed perimeter is a circle.
Investigate the problem of finding an extremum of the integral
r:l{H{*dx
subject to the constraint
J:lty'ty'*d.x:t
where 11 is the Hamiltonian operator
rr2 d2
ttu --2^o*'
_ -lV(x)
and V(x) is a known function. Show that the resulting differential equation is the
Schrcidinger equation.
Hint: First integrate by parts to eliminate the second derivative.
10. Using polar coordinates, write the Lagrangian for a particle moving in the potential
V
: - G M m f r, and form the Euler-Lagrange equations.
Show that the equation in the
angular coordinate indicates conservation of angular momentum.
Ll.
A spherical pendulum is a mass free to move on the end of a string of fixed length l. Write
the Lagrangian in terms of the spherical angles 0 and Q, and hence find the equations of
motion. Show that one possible motion is the conical pendulum with constant 9. What
is the value of dQ I dt in this case?
12. Consider the one-dimensional motion of
a particle with potential energy V(.r) that is
independent of time. Show that the Euler-Lagrange equations may be written in the form
of equation (E.6). Give a physical interpretation of this equation.
[if]fne
Lagrangian for a vibrating string may be written
,: lo'l:-(#)' -:, (#)'1,,
554
OPTIONAL TOPIC
E CALCULUS OF VARIATIONS
where y(x, t) is the displacement of the string, p is the mass per unit length, and Z is
the tension. The first term in the integrand is the kinetic energy density, and the second
is the potential energy density. Determine the Euler-Lagrange equations for the system,
and comment.
14. As an alternative approach to Problem 13, we may expand the displacement y(x, /) as a
Fourier series in x (refer to Chapter 4, Section 4.2):
y(x, t)
:ia^(t) sinff
n:o
Write the Lagrangian as a function of the generalized coordinates ar(t) and the time /.
What are the Euler-Lagrange equations now?
1.5. A volume V is formed by rotating a curve y(x) defined for -a < .r < a around the
x-axis. Given that the curve is symmetric about the y-axis, y(a) : )(-a) : 0, and
y'(0) : 0, show that the curve that gives the maximum volume for a given surface area
is a circle and the corresponding volume is a sphere.
16. The Lagrangian for a panicle moving under the influence of electromagnetic fields is
g: !^r2 - qQ + qi-i
2
Eil
where 4! and A are given functions of position and time. Find the equations of motion, and
hence show that the force acting on the particle is the Lorentz force F : q(E * i x B).
Strow that the shortest distance between two points on the surface of a sphere is a great
circle. Hint: You may place the polar axis through one of the points.
APPENDICES
I. TRANSFORMATION PROPERTIES OF THE VECTOR
CROSS PRODUCT
In Chapter 1, we noted that the cross product oftwo vectors does not transform as a vector
(Section I.1.2).Here we shall investigate the transformation properties of the cross product.l In a second, primed coordinate system, the definition of the cross product in terms
of components (Chapter 1, equation 1.30) is unchanged. We can then express the vector
components in the primed system in terms of the original (unprimed) components, using
the transformation law for vectors:
(il' x i'), : eij*u'juL:
eijkA
jutAkmum
(1)
We want to compare this expression with the usual transformation law (1.23) for vectors:
A;o(it x i) p
:
AipE pjku
juk
These expressions are not the same.
As a first step toward finding the correct transformation law, we write the determinant2
of the transformation matrix using the Levi-Civita symbol:
An An
det.A:
:
:
An
Azr Azz Azt
Att Azz Azz
An(AzzAtz
-
A1i e iu AzkA3r
AzzAz)
-
Ap(A21A3z
-
AzzAy)
i An(AuAzz -
AzzAz)
: - A2;e;i pAli A3
Similarresultsareobtainedforallpermutationsoftheindices
and r :3, the first expression for det A gives directly
epnr detA,
:
eluApiAq*Art
l,2,and3.Forp
- l,e:2,
(2)
l See also Optional Topic A, Section A.3.
2See also Chapter l, Section 1.6.2.
555
556
APPENDICES
This relation is confirmed fot p - 2, Q : 1, and r : 3 using the second expression for
det A and the index name changes i -+ i, i --> k, k --+ l. Similarly, we may verify the
relation for each set of values of p, q, and r .
Now we can make use of this relation to see how the cross product transforms. We start
with the definition ( I .30), apply the transformation matrix, and then write the result in terms
of the components in the primed system:
A;p(it
xi)o :
A;pepikuiuk
: Aipepi*Ailtuie;)'h
Since the inverse of the transformation matrix A equals its transpose, we have
Ai
r(it x i),
:
e
oi* Aip Ari
A*pulufi
Now we use equation (2):
A1r(fr x
i)r:
ei1^ut1ul
detA,: (fr' x i'),
detA'
(3)
The usual transformation law for vectors does not lead to the correct expression (1) for the
cross product, so we multiply both sides of equation (3) by det A and use the result that
det.A,
: tl
to obtain
(d' x
i')i :
(detA)A;p(d x
i)o
(4)
which is the transformation law we seek. Because of the determinant in equation (4),
(il'xi'):A(frxi)
for rotations, but
(i'xi';:-A1fixi)
for reflections.3
II. PROOF OFTHE HELMHOUTZTHEOREM
The Helmholtz theorem states that any vector field F may be decomposed into the sum
V x A.
two vectors:
+ i, where fr = VO and
i:
i:i
of
The proof takes the form of a demonstration of how to construct the two functions
A(r, y, z). First we construct the vector function:
@(x, y, z) and
fi(x, y, d:
[ ..space
Jdt
3For additional information, see Optional Topic A.
,u'o:,]';z')
dx, dy,dz,
(5)
II PROOFOFTHEHELMHOLTZTHEOREM 557
where
R:li-i,l
and the integral is over all space. This function satisfies the equation
v2fr': -i
(6)
Let's demonsffate this. Since the operator I differentiates with respect to r, y, e but not
x', y' , z', we may move it inside the integral, where it operates only on the function 1/R:
v2fr
: oz
:
I
fi'ul]t
t>
1
dx,
*r*', y', z')l-6(i.
dy, dz,
:
- *')ldv'
I
:
,r*,, t,,
z)v2J-
dx,
dy,
dz,
-fr(x, y, z)
where we used equation (6.26) andthe sifting property. Thus, we have verified equation (6).
But from equation (1.51),
v2fr:ifi.*l-i"(ixfr)
and so
F:ix(ix*l-Vfi.*l
Then we define the scalar function O by
o
and the vector function
i
: -i .fr'
(7)
Uy
i:ixfr
(8)
so that
i:Vr,.i+io
as required
(9)
by the theorem.
Next we can find explicit expressions for
<D
and
.i. L"t't
start with @ (equations 7
and 5):
cb(x,
y,.)
: -i
.*
: -i *o;l;,,
I
dx,
: - [ U .l(x'-,Y'az') dx, dy, dz,
4nR
J
: - I i(r'. v'. z') .i=l dx' dy'dz.'
4nR
J
fl
dy,
dz,
558
APPENDToES
Then we use the result
(z-z')o
(x-x')^ (y-y')^
: V-------=-2:-\:
-Rr
Rr "
R3
:l17
'R
where
i'
--I-
3,1
R
is the grad operator with respect to the primed coordinates. Thus,
:
a
fF(x"
J
_l',, z',)
.\-'' I
4tr R
dx'dv'dz.'
Now we move the grad operator back through the vector F again, using the result
/V
d+n. ir2 withi :
Fand qr
:
i
'
1,2f1
:
I/R:
o: llr, (t'a.l.tt) #l dx,dy,dz,
(10)
We can rewrite the first term using the divergence theorem (equation 1.44):
q: I i.x"Y"z')'fidA'[n,'!or'
4rR
J 4rR
./sThe surface integral is zero provided that
i +
0 on the surface at infinity at least as fast as
.p-(1+e).11ur,
,:- lY
4tr R
(11)
V,
We can obtain an expression for A similarly:
:i
A :-Yi x *vr-'^l
dx' dv' dz'
* [ i(r''+"nY''z') u^srs<
First we move the curl inside the integral, where it operates only on the function 1/R:
v|' t') dx'dy'az, :
L: Jfv " Ft*"4rR
[(v]=)
J\4rR/
"
F1*',
y'.2')dx'dy'dz'
Next we convert to a derivative with respect to the prime variables and use the result of
Problem 1.23:
A:- Ji (v'-!)
\ 4rR/
:-
"F1'',
y'.2')dx'dy'dz'
llu'" (o*t) - #u'"'l "' dv'dz'
lll
PROOF BY INDUCTION:THE CAUCHY
FOBMULA 559
Now we use a variant ofthe divergence theorem ( Problem 1.30b) to convert the first integral
to a surface integral:
A:- Js[ ,i'' f-L\ ae'+ I[i'"i4nR
\+nn)
dx'dv'dz'
i
i: .lI i','
o*' dy'dz'
4rR
(r2)
again provided that F -+ 0 on the surface at infinity at least as fast as iq-(l+e;.
Equations (11) and (12) allow us to calculate the vectors [ : iO and i : V ,. A.
lll.
PROOF BY INDUCTION:THE CAUCHY FORMULA
To prove the Cauchy formula (Chapter 2,equation2.39),
f,j+r,:#rr@-'){ol,_"
(13)
we use the idea of proof by induction. First we assume that the result is true for some value
of n, say n : m. Then we shall show that the result must also be true for n : m * 1. Since
we have already shown that the result holds for n : I (Chapter 2, equation2.38), it must
be true for all values of n > l.
We start with n : ml1 and express the derivative on the right-hand side of equation (13)
in terms of the (m - l)st derivative:
f '^' k)|,--o: lgb
i (r'*- ",rrl,:"*^ - f '^-D <al,-")
Now we use the assumed result for n
:
m to express the (m
-
l)st derivatives:
1,)t ( {
ri^tu{ f@ orl
'-'lr:o : h-o
- Jck-a1^
---Jo - ar.-)
"f,,,etl
2nih \/clz-(a*h\)n
t
:1-('-l)l
''\k-a-h)'
h+0 2nih f r,rr(
I \
- tr-oy)d'
Jc"
k-a)--(z-a-h)*
_1. fu-l)l f
,:\
2"ih
frf{')ffid'
560
APPENDIcES
We expand the second term
in the numerator of the fraction in the integrand,
using
the binomial series:
(z- a)* - (z - a'- h)^
(z-a-h)*(z-a)*
(z-a)^-lQ-a1^
(z-a-h)m(z-a)m
-
mh(z
a)m-r
-
m(m
-
2
l)
t'r,
-
q'
oY-27r2
a
"'
(z-a-h)^(z-a)^
Then, since
z-
a
+0
anywhere on C, wemay
divideoutafactorof (z -
a)--r toget
(m-l\
I-""; "(z-a)-th+"'
L
---l-
Then the
(z-a-h)^(z-a)
nth derivative becomes
.f'-'e)l z:a
I
which shows that the result (13) is also true for n
:
m
I l. Hence the general result (13)
is proven.
IV. THE MEAN VALUETHEOREM FOR INTEGRALS
If the functions
/(x)
and
g(x)
are continuous on the closed
interval la, bland g(x)
>
O
on the open interval (a, b),then
pb
1b
lf@)s?)dx=f(c)ls@)dx
- Jo
J"
for some c in the interval (a, b).
(14)
INTEGRALS 561
IV THE MEANVALUETHEOREM FOR
The proof uses the mean value theorem for derivativesa applied to the function
F(x)
:
fx
J" f ltls{tl
o,
Afx
- i J" sQ) dt
(ls)
where
o:
I"'f(x)g(x)dx
is the left-hand side ofequation (14) and
u
:
1b
J, sk)dx
is the integral on the right-hand side of equation (14). Notice that
continuous, then
F is differentiable. Then there exists
F'(c): -'F(b) -_
;
But F(D) :
F(a):
0, so F'(c)
equation (15):
-
if
some c between
and g are
and b such that
both
a
/
F(a)
"
0. Next we evaluate the derivative directly from
F'(x)- f(x)s@)-1s';,)
6
Setting
x
:
c, we obtain
A
f (c)s@) - ,sG)
Then, since g(x)
> 0 for all x in (a, b),
:o
we may divide out the factor g(c) to obtain
f(c)B:A
or
f (d
l"u
g(x)dx
which is the result we set out to prove.
In the special case g(x) : 1, then B
:
b
:
-
l"u
f {*)r{*)o*
a, andwe obtain the simpler result
pb
I ff*)dx :(b -a)f(c),
Ja
4See, for example, Stewart, p. 289
wherea < c
<b
(16)
562
APPENDICES
The area under the curve
width b
-
/(x)
a andheight /(c);
between
see
Figure
x : a andx :
l.
b equals the area of a rectangle of
f(x)
0
FIGURE 1. In the simpler version (16) of the mean value theorem, the area of the rectangle equals
the area under the curve /(-r). For this function with a : 0 andb : 4, there are two
possible values ofc, as shown.
V. THE GIBBS PHENOMENON
The Fourier series of a function that has a discontinuity always converges to the midpoint
of the discontinuity, and it overshoots as it makes the jump. To investigate these phenomena
further, we will concentrate our attention on the step function.
We will denote the sum of the first N terms of a Fourier series by Sls:
N
S1v
: !(a,
sinnx
*
bncosnx)
n--0
where (equation 4.7)
t
r2n
an:! JoI
7T
.f<*>sinnxdx
and bn is defined similarly (equation 4.8). Then we can write
an sirtnx
r b, cosnx : !tr ['" f (u)(sinnusin nx *
Jo
r
t2n
: ! I
7T Jo
for n
>
1, and (equation
/(r,r) cos n(u
- x) du
4.9)
bo:
I
f20
z" J,
f
(u) du
cos t?u cos nx) du
V THE GIBBS PHENOMENON
We'll also need the combination
an cosnx
-
In this case, the term with n
:
T
b,
sin
: -I If2n f (u) sinn(u nJo
nx
x) du
0 is identically zero. The series
(x)
: D
@"cos r?r
n:0
-
bn sinnx)
is called the allied series. The sum of its first N terms is denoted by ft,,. Then
t r2r
rt
Srs*iT,v-I "f@)1
" trJo
'tZ * $
u_t,i'tu-'t\ 7u
)
where the first term in the braces (l/2) is the n
Chapter 2, Section 2.3.2, equation 2.42 with ,
(
I f2,
t
Sr,r*iZr:_l
ftu)\-;*
"
rJo
t 2
:
-
0 term. We can evaluate the sum (refer to
"i(u-x))
6
get
1-ri(N+l)(a-.r))
I-et\u-x) )ldu
I f2o
,-itu-xt/2 _ ,i(N*l/2)(u-x)
:iJo
/(I/)t@
(
;\,"
The quantity in braces may be rewritten as follows:
cos
(f) - i sin (5:)
_
-cos f(n + *) (u - x)
(x+l
-t
1
(?) - cos [(ru + j) (z - xt] * sin [(ru + j) (rz - xt]
zsln (f )
2sin (f )
-cos
'
Thus, the real part of this equation is
S1g(x):
1
^
zlr
f2o . sin [(N + +)(u - .{)]
1
Jo .f @S---frdu
,ini r-"'l
\2 /
^
,
(17)
Now let's applythisresulttothe stepfunction f (u):1for0 < u < ir and f (u):0for
rr < u < 2n . For this function, the upper limit of the integral in equation ( 17) becomes n :
S1v(x)
:
I
fJT
2" J,
du
564
APPENDICES
Now let N
+
+
=
r.,, arrrd
(u
-
x)4
-
P:
1 p@-x)n sin u du
S1s(xl:;'2n '
J-,n sin(u/2D n
1
I fxn sin u
:2ort
-J--r*ffid'+ z-
1@-x)n sin u
sinlr/2il4v
J,,
For0<xlT,thefirsttermismuchlargerthanthesecond,sincesinu/2r1
isnotcloseto
zero in the second range of integration. In the first integral, the integrand is even, and so
srv(x): f- ['n ,sin!^ du foro < x < z
srnu/24
n4
Jo
Thus, as we let N, and hence 4, become large, sin
u
/2n
-
u
/2r7 and we have
2 fxn sinu
for0<x<z
Srs(x)=- /
TTJo u
-du
The integral is zero5 for x : 0 and increases steadily as x increases until 4x :
(18)
z,
because
the integrand is positive throughout this range. For u > n, the integrand becomes negative
and the integral begins to decrease. Further cycles of the sine cannot increase the integral
rr because of the larger denominator. As 4 -+ m, we find6
back to its value dt x4
:
s(,r)- ?
:,
[*tin'dr:?1
12
u
nJo
as expected. The peak
foro<
x<n
of the overshoot is thus greater than one and is given by7
? [" "-:7r:
T|JO u
?si Qr):1.179
1r
(1e)
The peak occurs at
'
and gets closer to
5wh"n
x
:
r = 0, the second
7f
N+t/2
0 as N increases (see Figure 2).
integral becomes
* ff' "|a"
-- |
as
n
--+
@,so we retrieve the expected result
that the series converges to the midpoint of the jump (Chapter 4, Section 4.2' 1).
6See Chapter 2, Section 2.7.3, for the evaluation of the integral.
TThe integral may be evaluated numerically. See also Abramowitz and Stegun, Table 5'1.
VI THE LAPLACE TRANSFORM AND CONVOLUTION
565
r.2
r1
tl ^
1.0
S7,,'(x) o.s
0.6
ru
o4ot
I=5
0.1
S1y (.r) (equation l8) for N : 5, 10, 20, 40, and 80. The peak moves closer
0 as N increases, while the height of the peak remains approximately constant
FIGURE 2. The function
to
x
:
for large N.
VI. THE LAPLACE TRANSFORM AND CONVOLUTION
Toprove the convolution theorem (5.17) forLaplace transforms, let's calculate the transform
of the convolution of the functions f and g:
L(f
*il:
lo*
,-"
Io'
f G)e(t -r)dtdt:
,ti!*lr' lr'
e-'t f
(r)s(t -r)drdt
The integral is taken over the half-space in the /-r plane bounded by the t-axis and the line
r : t.T\e integral is evaluated by summing over vertical strips, as shown in Figure 3a.
FIGURE 3a. The Laplace transform of the convolution is an integral over the triangular
the diagonal line
r:
r, summed over the vertical strips shown.
area below
APPENDICES
OT
FIGURE
3b.
We can equally well compute the area as a sum of the horizontal strips shown here.
We can equally well evaluate the integral by summing over horizontal strips, as shown
in Figure 3b. Then the integral becomes
L(f
x
il:
Now we change variables
L(f
*
il:
:
.gn- lr' l,'
to u : t - ri
,lg- lo'
o,
lo'-'
e-'t f
(t)g(t
ou
e-s(u+r)
rT / rT-t
rrr*Jo (/
stu)e-'u
-i
dt
dr
f G)sfu)
\
au)tttle-"
dr
The integral is over a triangular region of the t -u plane, as shown in Figure 4.
As Z + oo, the integral over the square gives
du [''' f ,rre-" dr
lim [''' ,(u)e-'u
T-a Js
JO
:
_[*
Jo
sfu)e-"
du
f@
ft)e-" dt :
Jo f
G(s)F(s)
We can show that the integrals over the two smaller triangles go to zero in the limit:
I
Both
/
ltop ,,i-er. I
:l [' f ft) e-,, o, Jo
['-'
lJrp
g1u)
e-"
dul
I
and g are of exponential order, so, from equation (5.2) in Chapter 5, max lg(u) I
Mlsotu ontherange 0 < u <T12,arrdasimilarrelationholdsfor
'
f: lf G)l 1M2eo2t
VI THE LAPLACE TRANSFORM AND CONVOLUTION
567
FIGURE 4. In the u-z plane, we integrate over the area to the left of the diagonal line. We may divide
this area into the square and the two smaller triangles shown here.
forT/2 < z < T.Thus,
l4op
ri*gr"l <
MrMzllr',r"''-u' dt
: MtMz [' ,'o'-"'
du
1-
Jrtz
:
#llr',r7u--')t
-
r(or--s)r
-
e@,-')') dr]
- e@z-or)T12
02-01
n{or-rlTe@2-ot)T
-ozT
-e -sT'
which approaches zero as I -+ oo provided thats Re (s) > max (ot,o). We can use a
similar argument to show that the integral over the lower triangle also goes to zero. Then
we have
L(f * d:
F(s)G(s)
and, equivalently,
c-t@c): f * I
8The argument rnust be modified
slightly
ifal -
o2. The result is unchanged.
568
vl.
APPENDTcES
PRooF THAT Pf QD= (-1)m(1
- p\^/2#PrQ.r\
In Chapter 8, we showed that the mth derivative of the Legendre polynomial Pr satisfies the
equation (8.52):
(t
Now let
y^(lD
- t"2)yk-2(m'rDpyk+Il(l
+1) -m(m
* l)lv,:0
(20)
: (l - lr2)'z(t ). We will be able to show that z(l.t) satisfies the associated
y^:
Legendre equation with a suitable choice of the power r. We begin by differentiating
yk
: 0-
p'2)' z'
-f r(-2p')(r
-,2;r-r,
and
yk:
-4rlt(l - *27r-t/ -2r(l - 2rpl-2(r - 1)pl(1 - ,2;r-2,
Q
-
1"2)'2"
: (l - ,27r-rl{, - *'rr" - 4rpz' - {*l
1"2)'-rz
+ t2
-t*'af
Substituting this into the differential equation (20), we get
0:6 - p\(t -,zy-tlr,- u\r" -4r1tz'- {p<,+
-
2(m
*
+
U(l
+ I)
1)pt(1
-
-
m(m
p,2)'z'
+
We can divide out a factor of (1
o
:
-
-
1)l (1
-
2r
-
-
-r*'D)
t'r2)'-1 zf
p2)'z
1t2)' to get
+ rr2 -
- 4r1tz' - {ro
+ll(l+t)-m(m+1)lz
frr
p'(l
t'2
p2)2"
rr'.)- 2@ r lp(r' - ;+tr)
Gathering up terms, we have
0:
p.2)2"
-2p.(2r +m*l)zl
(1
-
*
rl,u+ t) - m(mtt>
- , --"rrr+
pt2
-2p2, -2(m-tDt"\)
QD
We want this equation to become equation (8.48). We'll get the correct coefficient of zt
we choose r : -m12.Inserting this value into equation (21), we get
0: (r - p.2)2" -2112' t
1,,,*
t) -m(m+
: (l - p2)2,, -21t2, + rlru * r, - +)
l)+
lVrt
-
p2
- u'nf
if
Vlll PRooF oF THE REI-ATION
If;
and we have the equation we seek. Thus, we have shown that Cz :
C. In physics applications, it is usual to choose C : (-1)^ and then
piQ"):
as
(-1),?(l
: !!i*'t
pJm(kp'tJ^(k' pl ap
Pl
569
for any constant
- ,zynffirt@)
in equation (8.53).
Vlll. PROOF
OF THE RELRTION J[*n
J*(kp) J^&' p) d p =tS;!)
To prove the delta function relation above (equation 8.113), we start with the Bessel differ-
ential equation (8.69):
* ('ry)
With
a
+
k2 o
t^1r,pt
-
( n<r,o't :
o
different eigenvalue k/, the equation is
0 / 0J_&,o\\
,.
&,p)_^2J^1k,py_g
p
uo\o-i )+&')'or^
Following the usual procedure for proving orthogonality (Section 8.1.1), we multiply the
first equation by J^(ktp), multiply the second by J^(kp), and subtract:
l - (k' D
h
Qry) - r
^
(kd
+
(r*P)
+ w'
-
&' )2t
p
t
^
(rc p)
l
^
(kt
p)
:e
Now we integrate from 0 to oo. We integrate the first two terms on the left-hand side by
parts, and the integrals cancel, leaving only the integrated terms:
olt^ro'off - r^(kd9J!P]f : t&)2 -o', Io
pr^(kp)r^(k'p)dp
The integrated terms vanish at the lower limit p : 0. To evaluate these terms ur,n" ffi
limit, we use the asymptotic form (8.83) for J*(kp) and the recursion relation (8.90). The
first term is
;*T
t^n'
d
lJ^-tQrp)
- J^+(kp)]
z\
: lim kP 2
\
2-4)
P-a2npJkk' "o"(k'o-^'
" f"", (oo -'- -t'"
-l
-"o, (oo
-gy-
;)]
570
APPENDTcES
Using the identity
cos A cos
f:
jt"o,
(A
+ B)*
cos (A
-
B)l
we find the first term to be
,:_, I lk I
cos[(ft +k')p-mirl*cos[(k -k')p+nl2]
I
Vft'I -cos[(k +k')p-mlt -z]-cos l(k- k')p-"/zl I
tTk
: p-*nY
lim -^l ; leD^ cos (k t k')p - sin (k - k')n]
k
i*n
The second integrated term is found by interchanging k andk' in this expression. Thus, the
difference of the two integrated terms is
tTk
lim lr/ _
p-ar
Y kl
[t-t)-
cos (k
*
k')p
-
sin(k
-
kt)p]
tlu
k')p-sin(k'-k1pl
- -\l
ftu Ik' ff-rl'cos(k*
:
!lt.- - *' r-r),cos
lim
p+6
7r
L
\/kk'
(k
+ k,)p +k+k'
sin(/<,
'/kk'
-k)p'lI
Inserting this result into equation (22), we have
tf(-l)'+l
,:*
llm
-l
p-@rv
"ot(!J!)t- k+k,
L Jtctc,
I sin(k'-k\o1
' Jklt ;;):
f@
J,
pJ'(kp)J*(k'p)dp
(23)
Now we already know9 from Chapter 6 (equation 6.16) that
rim I [*
n-o2n
J_p "'t'dr:6(k)
1 sin/cR
:6(ft)
,i^
"'oo - "-'n^ : lim
"
R+m 2ikn
R*oo z k
and thus the sine term on the left-hand side of equation (23) is
r ^., .. 3(k'-k)
@6&', -k):The cosine term is zero in the limit. (See Chapter 6, Problem 25.) We can understand this
result by looking at plots of the two functions; see Figure 5.
Thus, we have the desired result:
fo*
gSee
also
Lighthill, p.29.
ot^{tro)J^(k' p) ap
:
ltgr' -
t1
IX THE ERROR FUNCTION
571
f(k)
FIGURE 5. The functio",
{F#
(heavy lines)
""0
for Rkt :
a
anffi24
(thin lines) versus k/k/
20 (solid line) and l0 (dashed line). While the function with the sine has
prominent peak, the function with the cosine has no peak for ft and k/ positive.
IX. THE ERROR FUNCTION
The Gaussian function
*'(-+)
(24)
occurs frequently in statistics. For example, it describes the distribution of a set of measurements around the mean at x : xo, provided that the number of data points is large and
the errors are random. As a consequence, it also appears in physics-for example, in the
Maxwellian velocity distribution, where, for each component,
f (u,)
/ m 1/2 /
*p
- \r"* ',)
The function (24) has a maximum value of
value where
t;4:
-rn
I
mu?\
\-rkr )
at
x
(;) :
:
x6. It reaches one-half its maximum
".2:o6e3t5
or at
x
:
xo
I aArA:
xo
* 0.83255a
572
APPENDICES
The spread x - xo : 0.83255a is called the half-width half-maximum, while the spread
2 x 0.83255a : l.665la is called the full-width half-maximum (Figure 6).
f
-4-2024
FIGURE 6. The Gaussian
function
e-"
(x)
.Thedashed line shows the half-width half-maximum.
Frequently we need values of the area under the curve of this function. The function
o(x)
2fx
- Ji
-I
(2s)
lo "-u'd,
is called the error function or the probability integral. It is also sometimes wriften as ed (x).
The normalization arises from the value of the integral
fo*
"-,'au::
I:: "-u'du : +
(26)
To evaluate this integral (26), we begin with its square:
,': l_: "-"d* Il* ,-nor: I_:
ll*
"-"'*v\dxdy
This is anintegral overthe wholex-y plane. Wemayrewritethis integralinpolarcoordinates.
The area element is
dxdy-dA:rdrd0
and
*2+y2:v2
rx rHE ERRoR
FUNcroN
573
Thus,
,,: Io fr'" ,-,'rdrdo:r, lo* "-*+
where we have made the change of variable w
12
:
12. Then
: n(_ e_,1f,) : n
giving
,
/^+m )
e-'' dx :
:
J_*
Ji
The complementary error function
erfc (.r)
:
1
-
o(x)
:
+
f,*
represents the area under the Gaussian curve from the point
normalized.
For small
r,
(27)
"-u'du
r
out to infinity, appropriately
we may usefully approximate the error function using the series expansion:
@
o
z
i_
,O(-t): -: ) (-l)r-
J" ?o'
A-zk+l
Qk
(28)
+ t)kl
which may be easily derived by integrating the series expansion for
ofx, the function approaches the value I exponentially,
o(x)
I
..2/
-, - {r*r-^
(.t
e-"' . For large
I
3
15
\
- ;t * o*o- 816 + "' )
values
(2e)
and the complementary function likewise approaches zero:
erfc(.r)
=hr'(t-#.
Values of the function erf (-r) are shown in Thble
I
)
and Figure 7.
(30)
574
APPENDTcES
o(x)
FIGURE
7.
The error function erf (x).
TABLE 1. Values of the Error Function
erf (x)
erf (x)
0.7
0.8
0.9
0.6'n8
0.3286
0.4284
1.0
0.5205
0.6039
2.0
0.8427
0.9661
0.9953
2.5
0.9996
0
0
0.1
o.tt25
0.2
0.3
0.4
0.2227
0.5
0.6
1.5
0.7421
0.7969
X. CLASSIFICATION OF PARTIAL DIFFERENTIAL EQUATIONS
Here we shall give a brief summa.ry of the types of partial differential equations most
frequently encountered in physics. For a more extensive discussion, see, for example, Morse
and Feshbach, Chapters 2 and 6. The discussion here is presented in two dimensions, but
the extension to higher dimensions is relatively straightforward.
Suppose we have a function f (x, y) that satisfies a linear partial differential equation of
the form
-u'{ + ctl
et4
'- 0x2 +' r.u
-- 0x
0y - 6yz
:'("'
''''Y'H)
(31)
where A, B, and C are each functions of the coordinates. The function / exists in a specified
region with boundaries defined parametrically by the curyes x : EG), y : 0(r). The type
ofboundary conditions that we need in order to find a solution depends on the form ofthe
X CLASSIFICATION OF PARTIAL DIFFERENTIAL EOUATIONS
575
differential equation, as specified by the relative values of the functions A, B, and C, and
also on the form of the boundary. The boundary conditions are classified as follows:
1. Dirichlet conditions: The value of the function / is specified on the boundary.
2. Neumann conditions: The value of the normal derivative of / is specified on the boundary. For example, if one of the boundaries is at x : a,we would specify 0f l0y at x : a.
3. Cauchy conditions: Both / and its derivative are specified on the boundary.
The boundary is closed if it forms a single closed curve, part of which may be at infinity,
and values are specified everywhere on the boundary, including at infinity. It is open if it
extends to infinity in some region and no values are specified on the part at infinity.
The differential equation (31) is classified according to the values of the functions A, B,
and C, as shown in Table 2. This table also shows examples of each class from the text,
along with specific values for A, B, and C for each example.
IABLE 2. Classification
of PDEs
Name
Hyperbolic
Definition 82 > AC everywhere
Example
Variables
Wave equation (3.15)
B
:0,
x,t
AC
:
Elliptic
Parabolic
82
:
AC everywhere
Diffusion equation
(3. 14)
x,t
A:D,B:C:O
-u2
82 < AC everywhere
Poisson's equation (3.49)
x,y
A:C:1,ll:0
To determine what kind of boundary conditions we need for a solution (Table 3), we step
away from the boundary using a Taylor series. We find that this is possible given Cauchy
TABLE 3. Existence of Solutions for Given Boundary Conditions
Boundary Boundary
Condition Tlpe
Ilyperbolic
Dirichlet
insufficient
to determine
open
Parabolic
Equation
Equation
closed
solution not
Neumann
open
insufficient
to determine
Neumann
closed
Cauchy
open
Cauchy
closed
solution
in one direction
to determine solution
(Example 7.6)
no solution;
overdetermined
unique solution exists
unique
(Examples 8.1, 8.2, 8.3)
unique solution exists insufficient
in one direction
to determine solution
(Problem 7. 14)
no solution;
overdetermined
unique solution exists no solution;
(Examples4.4,'1.5) overdetermined
no solution;
no solution;
overdetermined overdetermined
solution not
Equation
uniquesolutionexists insufficient
solution
unique
Dirichlet
Elliptic
unique solution exists
(Problem 8.12)
unstable,
unphysical solution
no solution;
overdetermined
I
576
APPENDICES
conditions, provided that the boundary does not coincide with one of the characteristic
curves for the equation, which are specified by
Ady:(a+Jnr_ ec)a,
B
(32)
To understand this result, consider the wave equation (3.15). Here we have A:
: O, and C : -1, and the characteristics satisfy
dx
dt
u2,
*u
with solutions
x
Lut:
constant
These are the wave fronts. It is possible to find a solution for a wave on a sffing, for example,
ut t't.
and Ef
at t
O (see Example 4.4). But the line x
if we know both
constant is a wave front. Any function of x ut satisfies the differential equation. Infinitely
:
l|t
/(x)
:
-
-
:
many functions f(u) have a specified value and a specified derivative at one given point,
and so the solution is not determined if the boundary is a line x - ut : constant.
The directionality of the parabolic equations is well illustrated by our example: the
diffusion equation. We can integrate forward in time, but not backward. Physically, this
is related to the increase in entropy with time. The solutions smooth out as time increases,
and it is not possible, in general, to determine the initial state from a given final state.
xl. THETANGENT
FUNCTION: A DETAILED INVESTIGATION OF
EXPANSIONS
SERIES
The tangent function offers us the opportunity to investigate Taylor and Laurent series in
some detail. While the Taylor series for tan x is familiar and easily found in reference books,
the Laurent series for the function tan z are rarely seen. They do have uses, however. We'll
use this function to discuss the relation ofthese less familiar series to the Taylor series and
to discuss some methods for finding them.
First let's see whether the function tan z is analytic by checking the Cauchy-Riemann
relations. We start by finding the functions u and u:
tartz:
z.
srn z
e'z
- e-'z
2i
cos zz
e'
stz
2
I s-tz
_
"-i(x+iy)
si(x-liy) a r-i(x+iy)
-i- "i(x-liy)
e-2y +r2ix _r-2ix -,2y
e-2y + r2ix L,
+ g2l
"-2ix
* i sinh2y :uliu
cos2x I cosh2y
sin2x
e' ,g v
-t x ey
_T *e
v
x
1 ei ; ey
,"
,"
,
,
,
-t
et
+ e a Xe
+ e' a
9
,_
_,
2i sin2x - 2sinh2y
2cos2x | 2cosh2y
Xl THE TANGENT FUNCTION: A DETAILED INVESTIGATION OF SERIES
EXPANSIONS 577
Thus, for this function,
u(x, y)
:
u(x, y)
-
sin2x
cos2x
*
cosh2y
and
sinh 2y
cos2x
I
cosh 2y
You might want to check that these expressions are correct in the case 1l : 0.
Now let's compute the partial derivatives to check the Cauchy-Riemann conditions:
2cos2x
-2sin2x
--" (cos 2x
* cosh 2y-sin2xI cosh2y)z
I * cos 2x cosh 2y
_ 2 cos 2x (cos 2x I cosh 2y) + 2 sinz 2x _,
(cos2x+cosh2y)2 = " 1"o"2* 1*"h2ry
0u
0x
-:
"^^^
cos 2x
-
and
0u
0y
--
2sinh2y
2cosh2y
'
-sinh2 (cos2x * cosh2y)2
cos2x * cosh2y
2 cosh
2y(cos 2x
I
(cos2x
-
cosh2y)
*
cosh
^- I + cos 2x cosh 2y
2sinhz 2y
(cos 2x * cosh2y)2 -
2y)2
0u
A*
Similarly,
0u
0y
-2
sin2x sinh 2y
(cos 2x
I
cosh2y)2
and
Du -2(- sin2x) sinh 2y
Ex - (cos 2x I cosh2y)2 -
2sin2x sinh 2y
0u
(cos2x * cosh 2y)2 -
So the Cauchy-Riemann relations are satisfied.
Where, if ever, does this fail? It fails if the numerator blows up
denominator goes to zero. The denominator is zero where
cos2x
which can only happen
(n
+ l)n.
if y :
*
cosh 2y
0 and cos2x
:
: -1,
0y
(y -+
oo) or
if
the
Q
or 2x
:
(2n
* l)n;that
is,
x:
We already know that the tan function approaches infinity (for real arguments)
when x equals an odd number times n 12. So this is consistent.
578
APPENDToES
Now let's find some series representations for tanz and look at the relations between
them.
Series
1
We can find a Taylor series about the origin in the region
tanz
lzl < r
12 (see Figure 8):
-io,r'
n:o
FIGURE 8. The region within which we can find
a
Taylor series for the tangent function: lzl
The coefficients are given by (equation2.44)
tdn rtanz)l
"r- aazr,
lo
I
Thus,
ao:ttn(0):0
at
!: 1rd
rdz tunzllo :
I
|
or: i1d2
O*ru"rlo: i,
1
r""2 (0)
:
I
secz(secztanz)lo
:0
Before proceeding, it helps to simplify this derivative:
d2
42tan
z
:2^ sinz
cos3 z
Then
I t d sinz. l | /cosz -sinz(-sinz)\l
3---*J, )1,
ot: tI d3
1rttan.lo: j A*J;lo:3 (;e; -
I cos2z-l3sin2zl I l+2sin2zl:- I
3 cos4z lo 3 cosaz le 3
<
n 12.
Xl THE TANGENT FUNCTION: A DETAILED INVESTIGATION OF SERIES EXPANSIONS 579
oo:
t dl+zsin2zl
:
o * E--;;{, l,
4,. rttan.lo
I d4
1
*",
I
4sinzcos2z
-
4(l +2.inz.;1-sinz)
""r=
I
ro
1l
303
srn z
(z+
cos5
Iz n2z )
:Q
0
and
t""1.:* adt2sinz+sin3z
cos5 z
0
.)
5x3
(2cosz13cosz sln- e)cosz-5(- sin x)(2sinz 1sin3 z I
cos5 z
0
2
l5
and so on. Thus, we have
l, * 2.
rrz'*.'.
tanz- z*12'
Let's have the computer package Maple check this. Asked for the series expansion of
tan z,
it outputs
tanz-
z+lf * trr'* #r'
+
ffif + o(zro)
So the program agrees with our calculations (and gives us some more terms, as well).
Series
2
z
/2. The radius p
:
at z
n
- n l2l < z that surrounds the singularity at
by the position of the neighboring singularities
3n /2 (see Figure 9). To minimize confusion, we change to a new
Now let's look in the annulus lz
: tt is determined
: -n /2 and at z :
: z - n 12. Then
variable u)
J^n'
sin
-
cos
z
z
sin (w
-:
cos
* n 12)
(u -l
n
cos u,
12) - sin u;
580
APPENDToES
Now we use the previous series for tan
u.r:
l1
tanuJ w+lu3 +?srt +#.7
,(r + lwz + ?s.o + #*6
+
+
&.n
fu.t
+...1
*o(u.,10)
, (, - *,, - h.o - #,u +... + (-t)(-2) (!., * ?=,0 *#,u* f&*t * )' \
- -;
+ elt?@ (*., * ?r*o * #.u *#3.,'*. )'* .
)
\
:-; l' -!.' - tr.* -#.u* *(f,' **.u.
=
)- *,'l*
!,oo
*...)
-!u\ ( t - !,,2
3 - 45 -aru
945
/
lll.2.
---+-u+-uJ+-uJ+,,,
u345945
I
I / ft\
| r zr3
2 r rrr5
:-e-n/r)*r(t- z)+ qs\'-z)
*sos(t-r)
+'
which is the series we set out to find. It is a Laurent series with only one negative power
and is valid for 0 < lz - r /21 < r.
FIGURE 9. Both Series
I
and Series 2 for tanz are valid in the shaded region, so they must be
identical there.
Now in the region of overlap, the two series are supposed to be identical. Let's check. The
overlap region is 0 < lzl < n/2 (see Figure 9). In this region, we can expand the second
series. We begin by writing the first term as a function of 2zf n, whose absolute value is
Xl THE TANGENT FUNCTION: A DETAILED INVESTIGATION OF SERIES EXPANSIONS 581
less than
I in the overlap
tanz- -
1
region. Then
1
/
*
e-,tlz) 5(t -
: ;2 ,, 1
-rrr*
I
,t -
,t
r
I z
rr\3
t)+ E\'- r)
I
l,
+
ir' *23
+
2 r z\5
%s\'- Z) +"'
1"
1 n
- i"" + ftzn2 -
I
360-o'
| ,rot* | ,oo- | ot*...
2 ,t- r ,or*J-rtor*- g4i"
l5l2o"
r89" " ' r89" "
378" " ' r5r2'"
g " 76,
I
I
2/ rl-lz*
2 4, *l-zt
\I lo
: 1l
+llzo
+.
^zz
=)=o3
z\ tr- rr2- 7rr- n+/ 6 360
_ 1 ot*''ar+ I ^ 7 / - I 1 - | o 3+!rt
t5t20
60ro'* 1gV1zn* *z'n ,Uz"n- * +Sr'
'
1."
I
2
* *z'tr' - lggr"o -l *rz' *...
I . 4
2 r
I .
6o
-
360o"
I
I
n
o2
*#ro
I
8n
n
- t|nor' + F, + az + 60ro' -l ,rzur- + orr'
- *r'n - *,,.r'o' * #r' * *r'
:6.65 x t0-3 *0.972*7.1x
+
fif
lo-222 *0.2423
- firoo +..'
*8.8 x l0-2za +..'
Each term in the expanded series is itself a series, of which we have evaluated only the first
few terms. We can compare with the first series:
l.
2.
tanz-z*12' *,rz-*...
The a6 term should be zero; with four terms, we have 0.006 and are decreasing toward
zero. The er tem' should be 1; we have 0.97. If we could add enough terms, we should be
able to get back the first seies emctly.
Series 3 Next we'll look at the series in the annulus n /2 < lzl < 3r /2. This region is
centered on the origin, but excludes the singularities at z : *n /2 (see Figure 10). We have
a
Laurent series with coefficients (equation2.47)
I f
an: 2ni
tanz
frVa
o'
The contour C must lie within the annulus. Thus, it contains three poles, at
Using the residue theorem, we have
3
an:DRes
p=l
(zp)
I-
Q,
*.n 12.
582
APPENDTcES
v
2T
\
-7f
_7f
FIGURE 10. Annulus
tt
/2 < lzl <
3n 12
within which Series 3 for tanz is valid.
The residue at z : 0 is a'n, the nth coefficient of the Taylor series valid in the immediate
neighborhood of z - 0. To see this, note that near z : 0,
tanz D*o'^z^ \L
Zn+l
Zn+l
-:
a'^
,n*l-m
The residue is the coefficient of the z-1 term in this series (m : n)-that is, the coefficient atr.
To evaluate the residues at the other poles, we use the series valid about each of them.
For example, neur z - n f2, we found
tanz---
I
+!('-!t-
e- ft/D* :(' t)
The integrand has a simple pole at z
:
+
1r
zr3
n\'- z) +"'
r 12, so the residue there, by method
t *!(,_1)+!(,_,,t
n.r/1):
r.^(z-nl2)(_
\2) 7+n/2
V.l-\-(r-"/r)+:\t-t)+u\'-Z)
:
-l
/ 2\'+l
1o121*': - (;/
Intheneighborhood of
-rf2,letu: z- (-n12): zln12.Then
sinz _ sin(a - 1rl2) _ -cos&
tut4a"*-
*t("-"/D
-
*-
1, is
+ \)
Xl THE TANGENT FUNCTION: A DETAILED INVESTIGATION OF SERIES EXPANSIONS 583
andthus the series fortanzrra?t 7
The residue at z : -r 12 is then
n",
"*
- -n12is Series 2, with r.u replacedby, - z*n12.
: (- ,rn (z\'+r
: -' ,, : - (\.-?\'*'
\ z):;*r*r:
tt /
Y" )
(-1'\
These results are valid for both positive and negative values of n. Thus, the integral that
gives the Laurent coefficients is
o^:
$ Ho=
" ^L.
2ni Jc
z'*t
fu,:, *", (zi): a',* (?)".' t-l + (-1)'l
0 for negative n. The term - 1 + (- 1)' is zero for even values of n and equals
odd values ofn (positive or negative). Thus, Series 3 is
atn:
where
-2for
tan
T: -'(+)^ + -'Gf ; -:..('- #)
..' - + a'). ul+ - : (1)').
[l
'-
-
12.
r124.9348
z - - +z(0. 18943)
-
I'76-
+23(4.8219
This is
lzl <
a
x
10-3) + z5(t.9zee
x t0-4) +...
Laurent series with infinitely many positive and negative powers, valid for ir 12 <
31 12.
:
n and compare with the known value tan zr
Let's evaluate this series 4t 7
sum of the six terms listed above gives
tanrr
:
0. The
:
-+
rr 3 tr 15
23n 2r r
13651,2
: "' - ff
* fr"t+ "' - -3'2x t0-2
+r +
3tr3
7T
The next two terms are
( ?\' 1 * f ?)'.,.l * 11 ,, :2.s246x ro-2 at z:
-rl
"L\")
"l':ts
,''\''/
Thus, the sum
of eight terms of the series is -3. 1981
-6.735 x l0-3. The series
n
x l0-2 + 2.5246 x I0-2 -
is approaching the correct value, but rather slowly.
584
APPENDToES
At z :3n 14,the value should be -1. With six terms, we get
,un3n
--'4
:...-
. (?)'
:
/ 4 \3 I
1\t -4.g3,*t(;/
-i+
\3n/
t2.t76f
,o.rrrnx 10-3).
(?)'
3n
4(o'r8e43)
(1.s266x r0-4)
... _ 0.97 +...
These series clearly converge more slowly than the standard ones we are used to using,
but they do have their uses. Notice how we used the first two series to obtain the coefficients
for the third; this is a good demonstration of the idea of analytic continuation.
The region of validity for the third series overlaps with that for the second. Again, the
two series are identical in the region of overlap.
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Arfken, George B., and Weber, Hans
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Cook, David. Computation and Problem Solving in Undergraduate Physics. Brooks/Cole,
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586
BTBLToGRAPHY
Halmos, Paul R. Finite Dimensional Vector Spaces. Springer-Verlag, New York, 1974.
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Index
The main reference or definition is given in boldface'
Abelian grouP, 465
Absolute convergence, 103' 105
Absolute value, of complex number, 78
Absorption, 150,322
AC current, 161,267
Addition
of complex numbers, 76
as vector addition, 80
of matrices, 42
of vectors, 5, 41
Addition theorem
for Bessel functions, 436
for spherical harmonics, 392
Adjoint matrix,44' 469
Affine connection,456
Affinity,456
Air resistance, 17 O' 2ll, 215, 216
Airy equation,2l2
Airy integral, 538
Algebra, Lie,49O
Algebraic form, Permanence of, 119
Allied series, 563
Altemating series, 104
Ammonia,483
Ampere's law, 35, 86
Amplitude
of complex number, 78
ot of , 44'l
71
oPerator,
Angular momentum
tensor rePresentati
Angular velocitY, 15
Antisymmetric matix, 42
Antisymmetric tensor, 446
Aperiodic function,
Fourier series for, 220
Argand dragram,TT
Argument, of comPlex number, 78
and branch cuts, 84, 121
Associated Legendre equation, 369, 384,
38s, 420, 568
Associated Legendre functions, 385' 569
integral of,387,431
with negative m, 385
orthogonalitY of' 384
table of, 386
Associative ProPerty, 3E, 465, 47 L
Asymptotic form, for modified Bessel
functions, 196
Asymptotic methods
for differential equations, 183' 194
for integrals, 531
Atom, 353,389
Attenuation, and Fourier transform, 331
Attenuation ProPertY, 254
AxisymmetrY, 369, 409, 419
ofwave,85
Analytic continuation, ll7 ' 156,21O
bac-cab rule, 17
Analyticity,97
Basis, change of,51, 58,476
Basis forms, 454
Basis vectors, 39, 220, 453' 47 6
change of, 53
and tangent function, 584
Angles, between curves, 1 5 I
Angular frequencY, 85
Angular momentum, 42, 439, 441
ofparticle,
Backward difference, 199
15
587
588
Branch point, 85,
Beam
bendingof, 17l
on elastic foundation, 354
l2l,
147,157,187,272
more than one, 91
Bromwich contour, 269
simply supporte d, 258-261
Bernoulli-Euler laq 171
Bernoulli's laq 68
Cable, hanging,548
Bessel functions
Calculus of variations, 541
with more than one variable, 551
Capacitor
energy stored in, 243
parallel plate, 155, 166
Cartesian coordinates, I
asymptotic form of, 195, 404, 413, 424
generating function
fol
407
integral of, 409, 418, 434
integral representation of , 286,
407,531,538
Laplace transform of , 436
with negative index, 401
orthogonality of
in finite domain,408
in infinite domain, 418,434,569
relations between, 405406
series, 400
small argument approximation of, 403
zeros of, 41O,522
Bessel's equation, 212, 214, 398, 399
Laplace transform of, 286
modified, 194,217,412
series solution of, 400
Biot-Savart law, 15, 463
Boundary
closed, 575
open, 575
Boundary conditions
cylindrical, 409, 415, 418
for fluid flow problems, I 14
and Fourier series, 237
and Green's functions, 495, 510, 513
for Laplace's equation, 210
and numerical solution of differential
equations, 199
and ordinary differential equations, 169,
174,176
and partial differential equations, 575
spherical, 381, 393
and Sturm Liouville theory 357, 363
Boundary value problem, 381, 393, 409,
415,418,5t2-516
Brachistochrone, 552
Bracket,490
Branch, offunction,84
Branch cut, 84, 90,91,96,121, 188
and contour integrals, 128, 147 , 157 ,272
curl in, 19
divergence
gradientin,
in,
19
18
Green's function in, 501
Laplacian in, 28
potential in, 364
vectors in,5, 10
Cartesian tensor, 439
Catenary, 548
Cauchy conditions, 575
Cauchy formula, 99, 106, 107, 149, 559
Cauchy-Riemann relations, 95, 162
Cauchy theorem, 97,109, 122
Cauchy's inequality, 167
Causality
and Fourier transform, 334
and Green's functions, 496,504
Center,
of agrotry,492
Central difference, 199
Chain, of nuclear reactions, 277
Change of basis, 53
Character,480
orthogonality of,48l
Character table,482
Characteristic curve, 5 76
Characteristic equation, 5 3
Charge, 172,234
induced, 521
Charge density
dipole,294
expressed in terms of delta
functions, 317
ring,315
sheet, 298
Choice
of coordinates ,28,210
of transform, 347
INDEX
Christoffel symbol of 2nd kind, 458
Circle
in complex plane, 153, 154
great,554
lunit, 129
Circuit, LRC, 172, 17 4, 234, 256, 265
Circular motion, 15, 447
Circulation, 21
Class,470
and number of irreps, 48 I
Classical orthogonal polynomials, 425428
Closing the contour, 134
Coefficient, in Fourier seies, 222, 225,
226,227,229,242
Cofactor,45
Collisions,
16l,l73
Column vector, 4l
Commutative property
ofaddition,38
of multiplication, 10, 13,43
Commutatot 488,490
Comparison test, 104
Complementary error function, 573
Completeness relation, 361, 505
Complex conjugate, 44, 7 7
and differentiab ility, 9 6, 162
of Fourier transform, 330
offunction,163
and orthogonality, 225, 362, 388
as reflection in real axis, 8l
Complex exponential, 79
weak
limit
oi
311
Complex impedance, 161
Complex integral, path independence of, 98
Complex number, polar representation
ol
78
Complex plane,77
Complex potential, I 13, 155
Complex roots, of indicial equation, 193
Components, of vector, 5, 37, 39, 41, 439
Compton scattering, 215
Conditional convergence, 103
Conductivity, 87,460
Conformal mapping, 150
Congruent transformation, 5 8
Conical functions, 214
Conjugate elements, 468, 470
Connection, affine,456
Conservation of momentum, 547
Constants
separation, 209
structure, 488
Constrained variation, 548
Continuity, 89
of Green's function, 506, 5 16
Continuity equation, 66, ll2
Continuous function, 89
Continuous set of eigenvalues, 417
Contour
Bromwich, 269
closing, 134, 136, 138, 270
deforming, 100, 108, 124,532
for gamma function integral, 157
for Hankel function integral, 531
keyhole-shaped, 147
rectangular, l4l
semi-circular, 134
Contour integral, basic method for
evaluating,127
Contraction, 444,452
Contravariant tensor, 448
Convention
for labeling matrix components, 7, 42
summation, 8
Convergence
absolute, 103, 105
of Fourier series, 240
in the mean, 241,360
pointwise, 104
of Fourier series, 240
radius of, 105, 106, 184
of series, 103
uniform
of function, 140,241
ofseries, l04,24l
weak,305
Convergence tests, 103-104
Convolution
cosine transform,35l
Fourier transform, 332
Laplace transform, 265, 565
sine transform, 350
Coordinate rotations, 6, 476
Coordinates
Cartesian, 1
curl in, 19
divergence
gradientin,
in,
18
19
589
590
Coordinates (continueA
Green's function in, 501
Laplacian in, 28
potential in, 364
vectors in,5, l0
choosing,28,2lO
cylindrical, 2
charge density in. 3 I 6
curl in, 35
divergence in,33,459
gradientin,32
Green's function in, 521
Laplace's equation in, 397
Laplacianin,3T
transformation matrix into, 449
unit vectors in, 29
velocity in, 454
general curvilinear, 28
curl in, 34
delta function in, 3 19
divergence in, 33
gradient in, 32
Laplacian in, 36
generalized,546
skew, 66
spherical, 4, 28, 298, 314, 315-316
Green's function in, 516
Laplace's equation in, 367
wave equation in, 419
spheroidal, 29, 160, 430
Core function, 305, 313, 329
Coset,471
Cosine function
Laplace transform of, 253
series
transformation law for, 445,556
Cube root function, 90
Cube roots, 83
Curl
in Cartesian coordinates, 19
in curvilinear coordinates, 34
in cylindrical coordinates, 35
expressed with determinant, 35
Fourier transform of , 334
Current
in circuit, l3l, 172, l7 4, 234, 256, 265
in conducting sheet, 430
Current density, 87
Curvilinear coordinates, 28
curl in, 34
delta function in, 319
divergence in, 33
gradient
in,32
Laplacian in, 36
Cycle,468
Cyclic group, 466,472
Cylindrical coordinates, 2
charge density
in,316
curl in, 35
divergence in, 33
covariant, 459
grad and curl in, 31
gradient in,32
Green's function in, 521
Laplace's equation in, 397
Laplacian in, 37
transformation matrix into, 449
unit vectors in, 29
velocity in, 454
for, 185
Cosine rule, 67
Cosine series,229,232
Cosine transfomt,344
of derivative, 346
Cosines
Damped harmonic oscillator, 171
product of,22l
sum of, 221
Coupled pendulums, 58
Covariant derivative, 45 6
of metric tensor,458
Covariant tensor,448
Cramer's rule, 51
Cross product, 14, 445, 555
and antisymmetric tensors, 446
Deflection, of beam,261
Deforming integration contour, 532
Degeneracy,362
Damping,87,17l
Daughter species,277
De Moivre's theorem, 160
Decay, mclear,277
Del squared operator, 28
Delta function, 287 -304
derivative of, 293
expressed as sum of eigenfunctions, 361
Fourier series of, 300
Fourier transform of, 329, 503
INDEX
of
a function, 295
and Green's functions, 495, 505, 512
integral of,299
integral representation of, 303-304, 319,569
Laplace transform of, 304
physical dimension oi 288
as weak
limit, 307
Delta sequence, 288, 3 1 8
Densities, expressed with
delta functions, 317
Derivative
of complex function, 94, 96
covariant, 456
directional, l8
of distribution, 308
expressed as line integral, 101, 559
Fourier transform oi 330, 333
Laplace transform of, 254
of Laplace transform, 261
numerical, 199
Determinants,44, 5l
and curl, 35
and Levi-Civita symbol, 45, 555
product theorem for, 46
Deviation, squ'are, 241, 352
Diagonal, of mafrx,42
Diagonalize a matrix, 53,47'7
Dielectric constant, 86, 148, 353
Difference, backward, central, and
forward, 199
Difference equation, 198
Differentiability, 93
Differential area,34
Differential equations
with constant coefficients, 1 74
with distributions, 308, 495
Fourier series solution of,234
homogeneous, 169, 174
inhomogeneous, l7 O, l7 6
Laplace transform solution oi 255
linear, 169, 17 4, 177, 251, 255,
334,336
numerical solution of, 199
orderof,169
ordinary, 169
partial, 169,173,363
classification of, 574
existence of solutions for, 575
and Fourier transforms, 336-340
separation ofvariables in, 208
power series solution of, 184-196
Differential operator, even, 186
Differential volume, 32, 33
Differentiate term by term, 104, 311
Diffusion, 173
of magnetic field, 355
Diffusion coefficient, 173
Diffusion equation, 173, 57 5
and Fourier transforms, 339
Green's function for, 5lO, 529
Dimension
of group representation, 475
of vector space, 38, 39,475
Dimensionless variable, 1 97
Dipole,294
electric field due to, 320
Dipole moment and molecules,484
Dirac, P. 4.M,287
Direct product grotp, 473
Direct sum of irreps, 478
Directional derivative, 1 8
Directionality, in solution of differential
equations, 576
Dirichlet conditions, 357, 513, 57 5
for Green's functions, 504
Dirichlet Green's function
cylindrical, 524,526
rectangular, 505-5 l0
spherical, 516, 529
Discontinuity, in electric field, 516
Discontinuous functions, 240
Fourier series for, 223
and Laplace transform, 265
Dispersion relations, 148
Displacement, 450
electric, 353
parallel, 456
of string, 303
Distributions, 305
derivatives of, 308
general theory of, 305
in N dimensions, 313
product of, 308
properties of, 307
as solutions of differential equations,
308-309
Distributive law, 38
591
592
Divergence, 19
covariant,459
in curvilinear coordinates, 32
Divergence theorem, 22, 32
Dividing space, 505
Division-of-region method, for Green's
function, 496,515
Dot product, 12-13, 40, 220, 444, 452
Drum,433
Dual, of tensor, 446
Dual space,453
Eigenfunctions , 2lO, 358, 499-502, 505,
509,515,525
complex,361-362
and delta function, 361, 417, 418, 569
and Green's functions, 499
orthogonality of, 359
Eigenvalues, 210, 358
of associated Legendre equation, 384
choosing, 363,409
of commuting matrices, 57
continuous set,4l7
generalized,64
of Legendre equation, 370
of matrix,53
reality of, 53
in Sturm-Liouville theory, 361
Eigenvectors, 53
of commuting matrices, 57
orthogonality of, 54, 56
Electromagnetic waves, in conducting
Equipotentials, I 16, 156
Equivalence class, 470
Equivalent representations, 476
Erf function, 572
Erfc function, 573
Error function, 306, 57 2
asymptotic expansion for, 573
series expansion
values of, 574
ol
573
Essential singularity, 120
Euler differential equation, 193
Euler equation, 544
Euler-Lagrange equation, 544
Euler method, 200
Euler's constant, 321
Euler's definition of gamma function, 156
Euler's formula, T9
Even differential operator, 186
Even function, 129
Existence of solutions, for partial differential
equations, 575
Exponential, of matrix, 485
Exponential function, 79, 85
as asymptotic form, 194
and attentuation, 254
Laplace transform of, 25 1
series for, 108
as solution
of differential equation, 174
as solution of Laplace's equation, 210
Exponential integral, 280
Exponential order, 25 I
Extremum,541
medium,87-89
Elementary matrices,48
Elementary product, 44
Ellipse, 160
Elliptic differential equation, 575
Energy
electrostatic, 552
gravitational potential, 548
Entire function, 97
Entropy,576
Equality of matrices, 42
Equation of state, 216
Equations
that reduce to Bessel's equation,
216,2t7
transcendental, 550
Equilibrium,58
Factor grotp,472
Factorial, 157
Fairly good functions, 305
Faithful representation, 475
Faraday's law,86, 132
Fermat's principle, 541
Fieldlines,114
Fluid,21, 66,68,112
Flux,21, ll2,l32
electric, l3
magnetic, 132
Form,448
basis,453
Form factor, 355
Forward difference, 199
Fourier-Bessel transform, 4l 8
INDEX
Fourier cosine series, 229,23I
Fourier-Legendre series, 3 8 I
Fourier series,2ll, 219 etseq,323
for aperiodic function, 220
multivalued, 84, 147
odd,512
periodic, Laplace transform
263
Functional,543
convergence of , 240-243
of delta function, 300
Gamma function, 156
as distribution, 311
Fourier sine series, 228,231
Fourier transform
of delta function, 329,503
of derivative, 330, 333
as distribution, 337
evaluating, 329
asymptotic form of, 159,537
incomplete, 159
and Laplace transform of powers, 252
Gauss-Jordan method, 48
Gauss'law, 86,459
Gauss'theorem, 22
Gaussian elimination, 5l
Gaussian function, 340, 57 I
as delta sequence, 306
Fourier transform of, 327
Gaussian integral, 157, 572
along path off real axis, 328, 534,536
Generalized eigenvalues, 64
Generalized function, 287
Generalized Rodrigues formula, 425
ofGaussian,32T
of grad, curl, and Laplaciary 334
and Green's function, 503
inverse, 324
and Laplace transform, 325
in N dimensions, 333
symmetric, 324
Fourier's theorem, 219
Fourthroots, 136
Generating function
for Bessel functions, 407,408
for Herrnite polynomials, 433
for Legendre polynomials, 375
Generator, of group, 485, 487
Geometric series, 105, 106, 110
Gibbs phenomenon, 223, 562
Fractions, partial, 258
Free index, 8
Frequency
nafirzl,236
normal mode, 63
resonant, 236,502
Fresnel integrals,437
Fringing field, 167
Frobenius method, 187
Fuch's theorem, 190
Full-width half-maximum, 572
Function
analytic,9T
with branch point, integral of, 147
complex,83
continuous, 89
and distribution, 307
Global2'l
Good functions, 305
Goursat, 97
Gradient, 18,31,448
Gram-Schmidt orthogonalization, 40
Great circle, 554
Green's function, 317, 495
for beam, 526,527
for diffusion equation, 5lO,529
division-of-region method for, 496,
505,515
core, 305
expansion of, in eigenfunctions, 501,
delta,287
discontinuous, 223, 240, 265
508,510, 519,526
for Helmholtz equation, 50I,527
in N dimensions, 512
for Poisson's equation, 505, 5 12
entire,97
even,129
garnma, 156,537
generalized, 288
meromorphic,
oi
test, 305
l2l
with more than one branch point,
with multiple zeros, 210
91
symmetry of,495,514
transform methods for, 503,
Green's theorem, 26, 512
5 10
593
594
as solutions ofLaplace's equation, 211
Hypergeometric equation, 188, 212
Group,465
abelian, 465
cyclic,466,472
direct product, 473
factor,472
generator
Identity element, 465, 466
character of, 481
of,484
for factor grotp,472
Lie,490
Lorentz,486,492
multiplication table for, 466,468
operation,465
permutation, 467,482
rotation, 467,484
unitary,469
Group speed, 539
Guitar,246
Half-width half-maximum, 572
Hamiltonian, 553
Hamilton's principle, 546
Hankel functions, 402, 421
contour for evaluating, 531, 535
Harmonic functions, ll3, 532
Harmonic oscillator, l7 l, 186, 282,
352-353,432
damped, 171
Heat conduction, 173,
5
l0-5 I I
Heat flux, 173,354,430
in semi-infinite rod, 510
Helmholtz equation, 173, 433
in cylindrical coordinates, 406
eigenfunctions of, and Green's
function, 509
Green's function for, 5Ol, 527
in spherical coordinates, 420
Taylor series solution of, 184
Helmholtz theorem, 37, 556
Hermite polynomials, 187, 428, 432, 433
Hermite's equation, 186, 432
Hermitian matix,44
Ha@\402
Homogeneity of space, 362
Homogeneous boundary conditions, 504
Homogeneous differential equation, 169, l7 4
Homomorphism, 474
Hooke's law, 171
Hyperbolic differential equation, 575
Hyperbolic functions, 79, l4l
and catenary, 549
and generators, 484. 486
Identity matrix, 8
ra@),412
Imaginary part,76
Impedance,267
complex, 161
Impulse, 287
applied to string, 302
Incomplete gamma function, 159
Incompressible flow, 68, 112
Indices, raising and lowering, 453
Indicial equation, 189, 190
complex roots of, 193
repeated roots of, 190, 213,214
roots of that differ by an integer, 190, 213
Inductance, 16l, 172, 234,256
Induction, proof by, 559
Inequalities, for complex numbers, 80
Inertia tensor, 42, 439, 441, 460
Infinity, point at, 195
Inhomogeneous boundary conditions,
510,513
Inhomogeneous differential equation, 170,
176,334
choice oftransform for, 349
Fourier series solution of,234
general solution of, 183
Laplace transform solution of, 255,265
Initial conditions, 170, 174, l'77
and Fourier transform, 334
and Laplace transform, 254, 256, 27 4
and sine and cosine transforms, 346-347
for vibrating string, 238
Initial value problem, 256
Inner product, 39, 444, 452
Integers, 75
group of,466
Integral representation, of delta function, 304
Integral test, 104
Integral
Airy,538
asymptotic form of, 534
INDEX
complex,9T
contour, 127-148,269
general method for,127
of delta function, 299
exponential, 280
ofGaussian,328
of Laplace transform, 262
mean value theorem for, 560
of multivalued functions, 147
orthogonality, 359
path,2O
path independence of, 98
of P1-,38'7,431,432
Ladder operatoq 378
Lagrange's equations, 60, 54'l
Lagrangian, 57,59,545
with EM fields, 554
for vibrating string, 553, 554
Laguerre polynomials, 428
Laguerre's eqtation, 212
Landau damping, 166
Lane-Emden equation. 2 I 6
Langmuir waves, 165
Laplace development, 45
Laplace transforms, 251, 325, 347, 348, 436
ofBessel function,436
probability,572
and convolution, 265, 565
along real axis, 134-148
of delta function, 304, 503
of a derivative, 254
of a discontinuous function, 265
sine, 143,564
of trigonometric functions
over one peiod, 129-134, 220-221
along real axis, 137
Interior, of curve, 100
Invariance,9,
13
Invariant subgroup, 472
Invariant subspace, 476
Inverse, of matix,8,47
Inverse element,465
Inverse square law, 342,421
Inverting a transform, 253, 269, 324, 344-345
Irrationals, 75
Irreducible representations, 477
Irreps, 478
number of, 480, 481
Irrotational flow, 112
Irrotational vector field, 20, 70
Isolated singularity, ll9, 123
Isomorphism,4T4
Isotherm, 18
Jacobi identity, 67, 489
Jacobi polynomi als, 428
Ja@),400
Jordan's lemma, 140
Kernel,492
Kinetic energy, 57-58
Kirchhoff's laws, 161, 172,234,256
KmG),4t2
Kompaneets equation,
21 5
Kramers-Kronig relations, 150, 322
Kronecker delta, 8, 39, 42
595
and inversion, 256, 258, 269
of a periodic function, 263
ofa square wave,265
table of,253
Laplace's equation, 367, 552
and analytic functions, I 13
in cylindrical coordinates, 397
solution of
in Cartesian coordinates, 209,364
in cylindrical coordinates, 415, 418
in spherical coordinates, 373, 388
in spherical coordinates, 367
use of in finding Green's function, 515
Laplacian,28,36
and delta function
in three dimensions, 314
in two dimensions, 315
Fourier transform of, 334
Laurent series, 1.08-11.1, 1 15
and differential equations, 187
and residues, 122,124
and singularities, l2O, l2l
for tangent function, 580, 583
Legendre functions
with complex l, 214
of second kind, 193, 373,430
Legendre polynomials, 37V37 2, 425, 428
derivatives of, 385
explicit expression for, 381
integral of, 376
orthogonality of,374
value with zero argument, 381
lrgendre's equation, 214, 370, 420
and Laplace's equation, 367-369
second solution oi 181, 190
series solution of, 190, 369
Levi-Civita symbols, 14, 16,45
in four dimensions, 463
product of, 16
as tensor density, 445, 464
Lie algebra,490
Lie group,490
Limit point, of singularities, 122
Line charge, 506
Line element, 31,319, 451
Line irtegral,20,97
path independence of ,
2l
Mathernatica,4S, 197
Matrices, 7, 41, 469, 47 4480
indices of, convention for, 42
inverse of, 8,47
multiplication of , 10,
4L4!
441, 467, 479
orthogonal,48
positive definite, 64
sum of,42
transpose of, 44
unitary,48
Maxwellian velocity distribution, 166, 57 1
Maxwell's equations, 86, 131, 208
Mean value theorem for integrals, 288, 560
Mellin inversion integral, 269
Memory systems with, 265
Linear differential equation, 169
Linear independence
of functions, 174, 179
of vectors, 39
Linear operator, 253, 329
Linearity, of Fourier transform, 329
Liouville's theorem, 167
Load, on bearn,172
Meromorphic function, 121
Metric coefficients, 3l
Metric tensor, 451
Modeling, 145
problems in, 274
Modified Bessel equation, 194,217 , 412
Modified Bessel functions, 412
asymptotic form of, 196,413
Local,27
integral representation of, 537
recursion relations for, 414415
series representation of, 412
Moment of inertia, 17 l, 439, 441
Log function
branches of, 84
as
conformal mapping, 153
for, ll9, 193,376
Lorcntz force,554
Lorentz grotp,486,492
Momentum,328
Motion
with air re-sistance, 176
in E and B fields, 65,554
Multiplication table, of group, 466, 468, 47 5
Multivalued tunctions, 84,90, 147
Mutual inductance, 274
Lorentz transformation, 462
Lowering indices,453
Neighborhood,97
cosine and sine of, 194
and delta function, 315
and second solution of differential
equation, 190
series
ZRC circuit, 172,256
with periodic driving emf,234,267
Magnetic field, 15, 35,86,132,432
diffusion of, 355
Magnification, 81, 151
Magnitude
of complex number, 78 (see alsoAmplitude)
ofvector,
13
Network, nuclear reaction, 277
Neumann conditions, 357, 513, 528,
for Green's function, 504
Neumann function, 4Ol, 421
Neutron density, evolution of, 356
in uranium,438
Newton's second law, 5,9,170
57 5
N^(x),402
Nonsingular matix,47
Map\e,48,197,579
Normal modes, 58,60,72
Mapping,473
Normal subgroup,472
conformal, 150
Normal to a surface, 24
function as, 83
Normal vector, and Stokes'theorem, 24
INDEX
Nuclear reactions, 27 7, 28I, 285, 438
Nucleon interaction, 215
Numerical solution
of lst order differential equation, 199
of 2nd order differential equation, 204
Oblate spheroidal coordinates, 160
Laplace's equation in, 430
One-to-one, 83,473
Onto,473
Operation
group,465
preservation of,474
Operator
ladder, 378
self-adjoint, 362
translation,492
Order
of differential equation, 169
of element,472
of group, 465,471,479
of numerical integration method, 2OO,2O3
ofpole, l2l
ofzero,119
Ordinary differential equations,
coupled, 209
Orthogonal coordinates, 1,29, 451
Orthogonal functions, 220, 35V359, 384,
388,408, 418,424,425
Orthogonal matrix, 8, 48,467,484
Orthogonal polynomials, 37 0, 425428, 432
table of,428
Orthogonal transformation, 56
Orthogonal vectors, 39
orthogonality, 220, 35u359
of associated Legendre functions, 384
of Bessel functions
in finite domain, 408,434
in infinite domain, 4I7,569
of characters, 481
of complex eigenfunctions, 362
of complex exponentials, 225
of derivatives of eigenfunctions, 429
of eigenfunctions, proof of, 358-359
of eigenvectors, 54
of eigenvectors with equal eigenvalues, 56
of group representations, 479
in infinite domain, 417
of Legendre functions, 374
of sine functions,224
of spherical Bessel functiots,424
of spherical harmonics, 388
of trigonometric functions, 220
ofvectors,39
Orthogonality integral, 359
for associated Legendre functions, 387
for Bessel functions, 409
for Legendre polynomials, 376
for spherical Bessel functions,425
Orthonormal basis vectors, 40
Orthonormal functions, 388, 499
Oscillations
damped, 175
small,57
Outer product,444
Parabola, 160
Parabolic differential equation, 575
Parallel displacement, 456
Parameters, variation ol 180
Parseval's theorem
for Fourier seies, 243,247
for Fourier transform, 33 l, 332
Partial differential equations
classification of,574
and Fourier transforms, 336
Partial fractions, 258
Partial sums, 103
of Fourier seies, 224, 241, 562
Particular integral, 17 0, 17 6
Path independence, of integrals, 2 I, 98
Path integral, 20, 97, 531, 541
Pauli marices,484
Pendulum,216
conical, 553
coupled, 58
with increasing length, 435
Period
ofgroup element,472
integral over, 129, 220-221
Periodic function, Laplace
transform of, 263
Permanence of algebraic form, 119
Permittivity, 149
Permutation group, 467,
characters of, 481
Permutations, 467
even or odd,14,44
47
0
597
598
INDEX
Phase
of a complex number,
78
535
85
constant,533
85
l'73
Physical dimension
ofdelta function, 288
of Green's function, 520
Physical law,5
Physical model,I45,273
Piano,246
Piecewise smooth, 240
Plane wave, 85, 333
in cylindrical coordinates, 406
Plasma waves, 161, 165
Point at infinity, 195
Point charge, 287
Point source, 317,495
Poisson's equation, 208
as elliptic equation, 575
and Fourier transform, 355
Green's function for,5l4
in one dimension, 309
in two dimensions, 505
Polar angle, 4
Polar axis, 4, 395
symmetry about, 369
Polar form, of complex number, 78-79
Pole, 120
order of, 121
onrealaxis, 143, 149
stationary,
a wave,
Phase angle,
Phase constant,
Phase speed, 86,
of
in Cartesian coordinates, 363-367
in cylindrical coordinates, 409,415
due to dipole, 295
due to point charge,375
in spherical coordinates, 388
vector,208
velocity,
ll2
Potential energy, 58, 545,548
Power, 13, 15
in AC circuit, 16l
radiated, 165,341
Power series, 105-lll, 184-196
Power spectrum, 342
Poynting flux, 341
Preservation, of operation, 474
Primitive function theorem, 268
Principal branch, 84
Principal value, of integral,t43
Principle, uncertainty, 328
Probability integral,5l2
Product
inner and ofier,444
of Levi-Civita symbols, 16
Product rule for covariant derivatives,
458,464
Product theorem for determinants, 46
Projectile with air resistance,2l6
Proofby induction, 559
Pseudo-tensor,445
Pseudo-vector, 15,445
Pure recursion relation for Legendre
polynomials, 377
Pythagoreantheorem,2,3,4
Pole expansion, 167
Polygon
83
symmetry of,466
Polynomials
classical orthogonal,425
in complex plane,
Qo@),193,373
QtG),182,213
Quadratic form, 58
Quaternions,49l
Quotient theorem,444
Hermite, 187, 428, 432, 433
Laguene,428
Legendre,369
Position vectors, 23
angle between, 390
in polar coordinates, 33, 66
Positive definite matrix, 64
Potential
axisymmetric, 373
complex, 113
Radiation due to accelerating charge, 341
Radius ofconvergence, l05, 184, 190
Radon, 354
Raising indices,453
Range, ofprojectile with air resistance, 216
Rank, of tensor, 439
Ratio test, 103
Rationals, T5
group of,467
599
Recursion relations
for associated Legendre functions, 390,431
for Bessel functions, 405
for Legendre polynomials, 377 -3'1 8
for modified Bessel functions,474
for power series, 185
Reducibility, of group representations, 477
Reflection
ofcoordinate axes, 8, 15,66,556
in real axis, 81, 163
Reflection matrices, 8, 66, 483
Regular point of a differential equation, 78
solution about, 184
Regular representation, 476
Regular singular point, 190
Relativity, special, l, 5, 462, 463
Remainder,24l
Removable singularity, 120
Repeated roots, 17 6, 297
of indicial equation, l9O,2l4
Representations
dimensions of, 475
equivalent,4T6
faithful, 475
of group, 475
regular,476
trivial,4T'l
Residue, 122
methods for finding, 124-126
Residue theorem, 123
use of, 127-150, 269-27 3
Resonance, 236, 282, 502
Response function, 266
Right-hand rule, 14
Right-handed coordinates, l, 29
Ring of charge, potential due to, 520
Rodrigues formula, 379
generalized,425
Rodrigues-type formula
for Bessel functions, 433
for Hermite polynomials, 433
for spherical Bessel functions, 437
Root test, 103
Roots,82, 136
Rotation matrix, 7 -9, 467, 484
Rotations
in complex plane, 81
of coordinate axes, 6, 395,467, 556
group of, 467,484
repeated,
l0
Rounding error, 199
Row vector,4l
Runge-Kutta method, 202
Saddle point, 532
Scalar field, 18
Scalar product, 12, 40, 452
Scalars, 5, 12,31, 439
Schrodinger eqtation, 215, 438, 553
Schwarz-Christoffel transformation, 1 55
Schwarz reflection principle, 163
Second derivatives, ofvectors, 27
Seismic waves, 552
Self-adjoint
oper ator,
3
62
Self-conjugate subgroup, 472
Semicircle at infinity, 134, 149
Separation constants, 209
for Helmholtz equation, 420
for Laplace's equation, 365, 368,
37 O,
398, 41 I
for waves on string, 238
Separation of variables, 208
and Green's functions, 509
and Helmholtz equation, 420
and Laplace's equation, 363, 367, 398
and waves on string, 237
Sequences, 101
delra, 288
of disributions,3l0
of partial sums, 103
radioactive,2TS
Series, 102, 184, 356
allied, 563
complex, 105, 106, 108
complex Fourier, 219, 225
convergence of, 102, 105
of distributions, 3 I I
Fourier, 219
Fourier-Legendre,382
geometric, 105
Laurent,108, 187
for tangent function, 580, 583
for log tunction, ll9, 193,376
power, 105-l I I, 184-196
for tangent function, 576
Taylor,106, 184
Series solutions, of differential equations, I 84
Shearing force,172
I
I
I
I
600
INDEX
Shifting property
of Fourier transform, 331
oflaplace transform, 254
Si (.r), 268
Sifting property,288,361
Similarity transformation, 53
Simple harmonic motion, 165,l7l
Simple pole, 121
Simple zero, 119,126
Simply connected region, 98
Simply supported beam, 258-261
diagonalization,64
function
Laplace transform of, 253
series for, 185
Sine integral, 268,564
Sine rule, 67
Sine series, 228,232
Sine transform,345, 510
of a derivative, 346
Sines
ofcomplex numbers, T9-80
sum of, 221
Singular matrix, 46
Singular point
of differential equation, 178
solution about, 187
variability of, 179
regular, 190
Singularities, 119-122
essential, 120
of gammafunction, 159
isolated, I 19
of Laplace transform, 269
removable, 120
Skew coordinate system, 66
Small oscillations, 57
Smooth function, 240
Smudging theorem, 307
Snell's law,543
5C(2),467
Solenoidal vector, 19, 70
Solid angle, 70
Span, 39
Specific heat, l'73
Spherical Bessel functions, 421
integrals of, 425
orthogonality of ,424
series representation of, 422
Simultaneous
Sine
Spherical coordinates,4,28,298,
Green's function in, 516
Laplace's equation in, 367
wave equation in, 419
3
14, 3 15-3 16
Spherical harmonics, 388-389
addition theorem for,392
sum rule for,392
table of, 388
Spheroidal coordinates, 29, 160,430
Spreadsheet, 201,204,205,206,207
Square deviation,24l,352
Square wave,223,234
Laplace transform oi 264
Standard form, of differential equation, 178
Star, structure of, 216
Stark effect, 214
Stationary phase, method of, 535
Steepest descent, method of, 531
Step function
eve\229
Fourier series for, 224,226,228,230,563
odd,228
and shifting property, 255
and step distribution, 300
Step size, in numerical solution, 204
Stirling's approximation,159,53'l
Stokes' theorem,24
Streamlines, 114,116
Structure constants,4S8
ofSO(3),489
Sturm-Liouville differential equation,
357,499,553
Sturm-Liouville problem, 357
Sturm-Liouville theory, general method for
use of,210, 363
SU(n),469
Subgroup, 471
Submatrix,45
Subspace,
invaiant,476
Sumrule for spherical harmonics,392
Sumrnation convention, 8
Superposition, principle of, 219
Surface spanning ctwe,24,98
Symmetric matix, 42,56,60,64
Symmetry
and degeneracy, 362
of Fourier transform, 324
of Green's function, 495,496,514
and groups, 465, 466, 483
INDEX
function,576
106
Tangent
Taylor series, 58,
and approximation ofintegrals, 532,
for cross product, 556
for tensors,44l
535
Transient, 268
anddifferentialequations,lTS,lS4,lST Translation,l52,492
200,202
579
Tchebichef polynomials,428
Technetium,28l
Temperature,inrod,5lO
Tensor density, 445
Tensors, 17
Cartesian, 439
and numerical solutions,
for tangent function,
ofmatrix, 8, 44
Triangle function, 238,312
Trigonometric function, integral of
one period, 129-134,220-221
Transpose,
alongrealaxis, I37
Triple scalar product, 15, 29, 5l
Triple vector product, 17
Trivial represeniation, 475,482,484
contravariant and covariant, 448
general,441
metric,45l
Terminal velocity, 176
Testfunction,305,313,329
Tests for convergence, 103
Theorem
addition, for spherical harmonics, 390
Cauchy, 97
divergerce,22
Uncertainty principle, 328
Uniform convergence, 104, 106, 140
Unit circle, 84
integral around,l29
Unit vectors, 3, 29,30, 40, 453
Unitary group,469
Unitary matrix, 4E,469,486
Upper and lower indices, 448
Uranium,438
Green's, 26,512
Helmholtz,7,556
Liouville's,167
560
function,268
product, for determinants, 46
quotient,444
smudging, 307
Stokes', 24
Thermalconductivity, lT3
Torque, 15, l7 , 259
on beam, 171
Trace, of matrix,44,71,480
Transcendental equation, 550
Transform
choice of, 347
Fouier,323
Fourier-Bessel,4l8
of integral, 268
Laplace,25l
Transform method, for Green's function,
503,510
Transformation
mean value, for integrals,
primitive
conformal, 150
congruent, 58
orthogonal, 55
Transformation law
Variables, separation of,208
Variation
constrained, 548
of integral,54l
of parameters, 1 80
Vector area, 15, 67
Vector field, 17, 19
Vector potential,208,554
Vectorproduct, 14
Vector space, 37 , 47 5
Vectors, 5, 38
basis, 39, 53,220, 453,476
column,4l
component of, 5, 39,4l
definition of, l0
row,4l
sum
oi
5,
38,4l
Eansformation law for, 7,9, 441
wit,29
Velocity potential, 112
Yiolin,246
Volume, of parallelepiped, 15
Water molecule, 492
Wave equation
in cylindrical coordinates,406
601
602
INDEX
Wave equation
(c ont inue
d)
for finite sting,237
and Fourier transforms, 336,356
Green's function for, 528
as hyperbolic equation, 575
for infinite string, 336
in one dimension, 173
in spherical coordinates, 419,438
Wave front, 576
Wave number, 328
Wave packet, 328
Wave vector, 85
sound, in cavity, 438
on string, 237, 302, 501, 553, 554
in three dimensions, 356,419421
Weak convergence, 305
Weber's equation, 215
Weighting function, 358, 425
Weirstrass
M
test, 105
Wien law,215
Wronskian, 179
Y6,388
Young's modulus,
17
l,
435
Wavelength, 85
Waves, 173
and complex variables, 85
electromagnetic, 85-89, 34 I
on infinite string, 336-339
plane, 85,406
seismic, 552
Zero distribution, 308, 337
Zeromatix,46
Zero vector,38,41
Zeros, ll9
of Bessel functions, 410,522
Gradient
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Second derivatives
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#NMFR;
ITIONIREA?}IBIilRIE$
rdxBtxi=B(i.i)-dd i)
ti ,. 6) .ri " i) : (d.06.dl - ra.al<[,tt
(i x B) x G x dy :
6ta,
i,dt -
dtf,, a, dl
Cylindrical coordinates
- - ao + lao" -t- ao
p0Q' -2
0p' --d
0z
-o
V.i: pta@fp)
p0Q*of,
0p *lafa
0z
VO
Vxf= /t0f,
\p ad
*)u. (*
v2o:
- #)o * i lftr,ra - W),
i#Q*\*i*4.#
Spherical coordinates
ao lao^
I ao^
vo-_i+__0+__6
0r r00 rsin90Q'
I o(sind/6), I afo
;,;_
v'r-7 lav2fr),
-rti"e
a,
ao -rti"eaO
vx
r:
t /a sino[o
^r"B
(m
t/a
+ ;\*rfe
"
v'o
a \ t/ r
- uok )i * ; (*" O' - *,o)u
a \^
- *f,)+
| a / ao\
l a20
: 71aa, /"ao\
* ;rr'r** (ttntm + ;;rR eW
a,
\'" )
)
The first term rnay be written in the alternative forms
r a2
a2e+-,. zaa
--(rO):
r 'drz '
0rz
r 0r
I
a