FOURIER SERIES
AND
BOUNDARY VALUE PROBLEMS
Donated by
Mrs. Yemuna Bappn
to
The Indian Institute of Astrophysics
from the personal collection
of
Dr. M. K. V. Bappu
FOURIER SERIES
AND
BOUNDARY VALUE PROBLEMS
BY
RUEL V. CHURCHILL
Associate Professor of Mathematics
University of Michigan
First Edition
Seventh Impression
McGRAW-HILL BOOK COMPANY, Inc.
NEW YORK AND LONDON
1941
FOURIER SERIES AND BOUNDARY VALUE PROBLEMS
Copyright, 1941, by the
McGraw-Hill Book Company, Ino.
PRINTED IN THE UNITED STATES OF AMERICA
All rights reserved. This book , or
parts thereof , may not be reproduced
in any form without permission of
the publishers.
PREFACE
This is an introductory treatment of Fourier series and their
application to the solution of boundary value problems in the
partial differential equations of physics and engineering. It is
designed for students who have had an introductory course in
ordinary differential equations and one semester of advanced
calculus, or an equivalent preparation. The concepts from the
field of physics which are involved here are kept on an elementary
level. They are explained in the early part of the book, so that
no previous preparation in this direction need be assumed.
The first objective of this book is to introduce the reader to
the concept of orthogonal sets of functions and to the basic
ideas of the use of such functions in representing arbitrary
functions. The most prominent special case, that of represent-
ing an arbitrary function by its Fourier series, is given special
attention. The Fourier integral representation and the repre-
sentation of functions by scries of Bessel functions and Legendre
polynomials are also treated individually, but somewhat less
fully. The material covered is intended to prepare the reader
for the usual applications arising in the physical sciences and to
furnish a sound background for those who wish to pursue the
subject further.
The second objective is a thorough acquaintance with the
classical process of solving boundary value problems in partial
differential equations, with the aid of those expansions in series
of orthogonal functions. The boundary value problems treated
here consist of a variety of problems in heat conduction, vibra-
tion, and potential. Emphasis is placed on the formal method of
obtaining the solutions of such problems. But attention is also
given to the matters of fully establishing the results as solutions
and of investigating their uniqueness, for the process cannot be
properly presented without some consideration of these matters.
The book is intended to be both elementary and mathe-
matically sound. It has been the author's experience that
careful attention to the mathematical development, in contrast
PREFACE
vi
to more formal procedures, contributes much to the student's
interest as well as to his understanding of the subject, whether
he is a student of pure or of applied mathematics. The few
theorems that are stated here without proofs appear at the end
of the discussion of the topics concerned, so they do not reflect
upon the completeness of the earlier part of the development.
Illustrative examples are given whenever new processes are
involved.
The problems form an essential part of such a book. A rather
generous supply and wide variety will be found here. Answers
are given to all but a few of the problems.
The chapters on Bessel functions and Legendre polynomials
(Chaps. VIII and IX) are independent of each other, so that
they can be taken up in either order. The continuity of the
subject matter will not be interrupted by omitting the chapter
on the uniqueness of solutions of boundary value problems
(Chap. VII) or by omitting certain parts of other chapters.
This volume is a revision and extension of a planographed form
developed by the author in a course given for many years to
students of physics, engineering, and mathematics at the Uni-
versity of Michigan. It is to be followed soon by a more
advanced book on further methods of solving boundary value
problems.
The selection and presentation of the material for the present
volume have been influenced by the works of a large number of
authors, including Carslaw, Courant, Byerly, B6cher, Riemann
and Weber, Watson, Hobson, and several others.
To Dr. E. D. Rainville and Dr. R. C. F. Bartels the author
wishes to express his gratitude for valuable suggestions and for
their generous assistance with the reading of proof. In the
preparation of the manuscript he has been faithfully assisted by
his daughter, who did most of the typing, and by his wife and son.
Ann Arbor, Mich.,
January, 1941.
Ruel V. Churchill.
CONTENTS
Page
Preface v
Chapter I
Section INTRODUCTION
1. The Two Related Problems 1
2 . Linear Differential Equations 2
3. Infinite Series of Solutions 5
4. Boundary Value Problems 6
Chapter II
PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS
5. Gravitational Potential 10
6 . Laplace’s Equation 12
7. Cylindrical and Spherical Coordinates 13
8 . The Flux of Heat 15
9. The Heat Equation 17
10 . Other Cases of the Heat Equation 19
11. The Equation of the Vibrating String 21
12 . Other Equations. Types 23
13. A Problem in Vibrations of a String 24
14. Example. The Plucked String 28
15. The Fourier Sine Series 29
16. Imaginary Exponential Functions 31
Chapter III
ORTHOGONAL SETS OF FUNCTIONS
17. Inner Product of Two Vectors. Orthogonality 34
18. Orthonormal Sets of Vectors 35
19. Functions as Vectors. Orthogonality 37
20 . Generalized Fourier Scries 39
21 . Approximation in the Mean 40
22 . Closed and Complete Systems 42
23. Other Types of Orthogonality 44
24. Orthogonal Functions Generated by Differential Equations. ... 46
25. Orthogonality of the Characteristic Functions 49
Chapter IV
FOURIER SERIES
26. Definition 53
27. Periodicity of the Function. ^Example ’ ’ 55
28. Fourier Sine Scries. Cosine Series 57
vii
viii CONTENTS
Section Page
29. Illustration 59
30. Other Forms of Fourier Series 61
31. Sectionally Continuous Functions 64
32. Preliminary Theory 67
33. A Fourier Theorem 70
34. Discussion of the Theorem 72
35. The Orthonormal Trigonometric Functions 74
Chapter V
FURTHER PROPERTIES OF FOURIER SERIES;
FOURIER INTEGRALS
36. Differentiation of Fourier Series 78
37. Integration of Fourier Series 80
38. Uniform Convergence 82
39. Concerning More General Conditions 85
40. The Fourier Integral 88
41. Other Forms of the Fourier Integral 91
Chapter VI
SOLUTION OF BOUNDARY VALUE PROBLEMS BY THE USE
OF FOURIER SERIES AND INTEGRALS
42. Formal and Rigorous Solutions 94
43. The Vibrating String 95
44. Variations of the Problem 98
45. Temperatures in a Slab with Faces at Temperature Zero 102
46. The Above Solution Established. Uniqueness 105
47. Variations of the Problem of Temperatures in a Slab 108
48. Temperatures in a Sphere 112
49. Steady Temperatures in a Rectangular Plate 114
50. Displacements in a Membrane. Fourier Series in Two Variables 116
51. Temperatures in an Infinite Bar. Application of Fourier Integrals 120
52. Temperatures in a Semi-infinite Bar 122
53. Further Applications of the Series and Integrals 123
Chapter VII
UNIQUENESS OF SOLUTIONS
54. Introduction 127
55. Abel’s Test for Uniform Convergence of Series 127
56. Uniqueness Theorems for Temperature Problems 130
57. Example 133
58. Uniqueness of the Potential Function 134
59. An Application 137
Chapter VIII
BESSEL FUNCTIONS AND APPLICATIONS
60. Derivation of the Functions J n (x) 143
61. The Functions of Integral Orders 145
CONTENTS ix
Section Page
62. Differentiation and Recursion Formulas 148
63. Integral Forms of J n (x) 149
64. The Zeros of J n (x) 153
65. The Orthogonality of Bessel Functions 157
66. The Orthonormal Functions 161
67. Fourier-Bessel Expansions of Functions 162
68. Temperatures in an Infinite Cylinder 165
69. Radiation at the Surface of the Cylinder 168
70. The Vibration of a Circular Membrane 170
Chapter IX
LEGENDRE POLYNOMIALS AND APPLICATIONS
71. Derivation of the Legendre Polynomials 175
72. Other Legendre Functions 177
73. Generating Functions for P„ (s) 179
74. The Legendre Coefficients 181
75. The Orthogonality of P n (x). Norms 183
76. The Functions P n (x ) as a Complete Orthogonal Set 185
77. The Expansion of x m 187
78. Derivatives of the Polynomials 189
79. An Expansion Theorem 191
80. The Potential about a Spherical Surface 193
81. The Gravitational Potential Due to a Circular Plate 198
Index 203
FOURIER SERIES AND
BOUNDARY VALUE PROBLEMS
CHAPTER I
INTRODUCTION
1. The Two Related Problems. We shall be concerned here
with two general types of problems: (a) the expansion of an
arbitrarily given function in an infinite series whose terms are
certain prescribed functions and (6) boundary value problems
in the partial differential equations of physics and engineering.
These two problems are so closely related that there are many
advantages, especially to those interested in applied mathematics,
in an introductory treatment that deals with both of them
together.
In fact an acquaintance with the expansion theory is neces-
sary for the study of boundary value problems. The expansion
problem can be treated independently. It is an interesting
problem in pure mathematics, and its applications are not con-
fined to boundary value problems. But it gains in unity and
interest when presented as a problem arising in the solution of
partial differential equations.
The series in the problem type (a) is a Fourier series when its
terms are certain linear combinations of sines and cosines.
Fourier encountered this expansion problem, and made the first
extensive treatment of it, in his development of the mathe-
matical theory of the conduction of heat in solids.* Before
Fourier's work, however, the investigations of others, notably
D. Bernoulli and Euler, on the vibrations of strings, columns
of air, elastic rods, and membranes, and of Legendre and Laplace
on the theory of gravitational potential, had led to expansion
* Fourier, “Thdorie analytique do la (dm, lour,” 1822. A translation of
this book by Freeman appeared in 1878 under the title “The Analytical
Theory of Heat.”
1
2
FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 2
problems of the kind treated by Fourier as well as the related
problems of expanding functions in series of Bessel functions,
Legendre polynomials, and spherical harmonic functions.
These physical problems which led the early investigators
to the various expansions are all examples of boundary value
problems in partial differential equations. Our plan of pres-
entation here is in agreement with the historical development
of the subject.
The expansion problem as presented here will stress the
development of functions in Fourier series. But we shall also
consider the related generalized Fourier development of an arbi-
trary function in series of orthogonal functions, including the
important series of Bessel functions and Legendre polynomials.
2. Linear Differential Equations. An equation in a function
of two or more variables and its partial derivatives is called a
partial differential equation. The order of a partial differential
equation, as in the case of an ordinary differential equation, is
that of the highest ordered derivative appearing in it. Thus
the equation
d 2 u 0 du
a) + =
is one of the second order.
A partial differential equation is linear if it is of the first degree
in the unknown function and its derivatives. The equation
ro \ d 2 u 9 du .
f2) w +xy -d~y = 3x y
is linear; equation (1) is nonlinear. If the equation contains
only terms of the first degree in the function and its derivatives,
it is called a linear homogeneous equation. Equation (2) is
nonhomogeneous, but the equation
d 2 u
dx 2
+ xy'
du
dy
= 0
is linear and homogeneous.
Thus the general linear partial differential equation of the
second order, in two independent variables x and y, is
Sec. 21 INTRODUCTION 3
where the letters A, B, • m m , Q. represent functions of x and y.
If F is identically zero, the equation is homogeneous.
The following theorem is sometimes referred to as the principle
of superposition of solutions.
Theorem 1. Any linear combination of two solutions of a linear
homogeneous differential equation is again a solution .
The proof for the ordinary equation
(3) y" + Py ' + Qy = 0,
where P and Q may be functions of x , will show how the proof
can be written for any linear homogeneous differential equation,
ordinary or partial.
Let y = yi(x) and y = y%(x) be two solutions of equation (3).
Then
(4) y” + Py[ + Qyx = 0,
(5) 2/2 + Py'i + Qyi — 0-
It is to be shown that any linear combination of yi and y 2 —
namely, Ayi + By 2) where A and B are arbitrary constants — is
a solution of equation (3). By multiplying equations (4) by A
and (5) by B and adding, the equation
Ay" + By'f + P(Ay[ + By 2 ) + Q(Ayi + Byf) == 0
is obtained. This can be written
d 2 d
^2 (Ay i + Byf) + P — (Ayi + Byf) + Q(Ayi + By 2 ) = 0,
which is a statement that Ayi + By* is a solution of equation (3).
For an ordinary differential equation of order n, a solution
containing n arbitrary constants is known as the general solu-
tion. But a partial differential equation of order n has in
general a solution containing n arbitrary functions. These are
functions of k — 1 variables, where k represents the number
of independent variables in the equation. On those few occa-
sions here where we consider such solutions, we shall refer to
them as “general solutions” of the partial differential equations.
But the collection of all possible solutions of a partial differential
equation is not simple enough to be represented by just this
“general solution” alone.*
* See, for instance, Courant and Hilbert, “ Mcthoden der nmthermitiHohen.
Physik,” Vol. 2, Chap. I; or Forsyth, “Theory of Differential Equations,”
Vols. 5 and 6.
4 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 2
Consider, for example, the simple partial differential equation
in the function u{x, y ) :
^ = 0 .
dx
According to the definition of the partial derivative, the solution
IS
u = f{y),
where /(y) is an arbitrary function. Similarly, when the equation
dx 1
is written —It—) — 0, its general solution is seen to be
dx \dxj
u = xf(y) + g(y),
where f(y) and g(y) are arbitrary functions.
PROBLEMS
1. Prove Theorem 1 for Laplace's equation
d 2 ^ d 2 u dhx _
dx 2 + dy 2 + dz 2 “
2. Prove Theorem 1 for the heat equation
du _ / dhi dhi d 2 iA
dt ~ \d# 2 ”* d?/ 2 dz 2 y
Note that k may be a function of x } y, z , and t here.
3. Show by means of examples that the statement in Theorem 1 is
not always true when the differential equation is nonhomogeneous.
4. Show that y = f(x + at) and y = g(x — at) satisfy the simple
wave equation
d*y 0 d 2 y
dt 2 a dx 2 ’
where a is a constant and / and g are arbitrary functions, and hence that a
general solution of that equation is
V =f(x+ at) + g{x - at).
5. Show that e-» s ‘ sin nx is a solution of the simple heat equation
du dhi
Sec. 3]
INTRODUCTION
5
If A i, A 2 , • * * , Ax are constants, show that the function
N
u=%A n e~ nH sin nx
71 — 1
is a solution having the value zero at x = 0 and x = r, for all t.
3. Infinite Series of Solutions. Let u n (n = 1, 2, 3, • • • ) be
an infinite set of functions of any number of variables such that
the series
-f- U 2 + * * * + u n -f- ■ ■ •
converges to a function u. If the series of derivatives of u ny
with respect to one of the variables, converges to the same
derivative of u , then the first series is said to be termwise differ-
entiable with respect to that variable.
Theorem 2. If each of the functions u 2j • • • , u n , • ■ • , is a
solution of a linear homogeneous differential equation , the function
u
u n
is also a solution provided this infinite series converges and is
termwise differentiable as far as those derivatives which appear in
the differential equation are concerned.
Consider the proof for the differential equation
(i)
d 2 u d 2 u
dx 2 ^ dx dt
+ qu
0 ,
where p and q may be functions of x and t. Let each of the
functions u n (x, t) (n = 1, 2, * • • ) satisfy equation (1). The
series
00
is assumed to be convergent and termwise differentiable; hence
if u(x , t) represents its sum, then
00 00 00
du _ "^1 du n dhi _ d'hi 7l d 2 U __ d 2 U n
dx dx ’ dx 2 dr' 1 ’ dx dt dx dt
l l l
Substituting these, the left-hand number of equation (1) becomes
i l l
( 2 )
6 FOURIER SERIES AND BOUNDARY PROBLEMS I*
and if this quantity vanishes, the theorem is true. Now c.xpi » s
sion (2) can be written
2(^ +p Ss + s “”)’
since the series obtained by adding three convejmt series
term by term converges to the sum of the t ice ui , •
sented by those series. Since u n is a solution o cqi
dhc n
dx 2
+ V
d 2 Un
dx dt
+ qUn
= 0
(n « 1, 2, • • * h
and so expression (2) is equal to zero. Hence u(x , t) satisfies
equation (1). , ,
This proof depends only upon the fact that the diffcien la
equation is linear and homogeneous. It can clearly be app u (
to any such equation regardless of its order or number of vm in > <
4. Boundary Value Problems. In applied problems in dif-
ferential equations a solution which satisfies some specified con-
ditions for given values of the independent variables is usnnlh
sought. These conditions are known as the boundary eondit ions.
The differential equation together with these boundary con-
ditions constitutes a boundary value problem. The student is
familiar with such problems in ordinary differential equat ions.
Consider, for example, the following problem.
A body moves along the x-axis under a force of attraction
toward the origin proportional to its distance from the origin.
If it is initially in the position x = 0 and its position one. second
later is x = 1, find its position x(t) at every instant.
The displacement z(t) must satisfy the conditions
( 1 )
( 2 )
d 2 x
dt?
x = 0 when t = 0
= —k 2 x,
, x = 1 when t = 1,
where k is a constant. The boundary value problem here con-
sists of the equation (1) and the boundary conditions (2),
which assign values to the function x at the extremities (or on
the boundary) of the time interval from t = 0 to t = 1.
The general solution of equation (1) is
x = C\ cos kt + C% sin kt .
Sec. 4]
INTRODUCTION
7 ■
According to the conditions (2), Ci — 0 and C 2 = 1/sin so
the solution of the problem is
sin kt
x = - — r *
sin k
From this the initial velocity which makes x = 1 when t = 1
can be written
dx _ k
dt sin k
when t = 0.
This condition could have been used in place of either of the
conditions (2) to form another boundary value problem with
the same solution.
In general, the boundary conditions may contain conditions
on the derivatives of the unknown function as well as on the
function itself.
The method corresponding to the one just used can sometimes
be applied in partial differential equations. Consider, for
instance, the following boundary value problem in u(x, y ) :
(3)
(4)
w(0, y ) = y~, u( 1, y) = 1.
Here the values of u are prescribed on the boundary, consisting
of the lines x = 0 and x = 1, of the infinite strip in the ary-plane
between those lines.
The general solution of equation (3) is
u(x, y ) = xf(y) + g(y),
where /(y) and g(y) are arbitrary functions. Tho conditions (4)
require that
( 5 ) ff(y) = y 2 , f(y) + g(y) = 1,
so f(y) — 1 — y 2 , and the solution of the problem is
u(x y y) = x(l - ?/) + y\
But it is only in exceptional cases that problems in partial
differential equations can be solved by the above method. The
general solution of the partial differential equation usually
cannot be found in any practical form. But even when a gen-
8 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 4
eral solution is known, the functional equations, corresponding
to equations (5), which are given by the boundary conditions arc
often too difficult to solve. A more powerful method will bo
developed in the following chapters — a method of combining
particular solutions with the aid of Theorems 1 and 2. It is, of
course, limited to problems possessing a certain linear character.
The number and character of the boundary conditions which
completely determine a solution of a partial differential equation
depend upon the character of the equation. In the physical
applications, however, the interpretation of the problem will
indicate what boundary conditions are needed. If, after a
solution of the problem is established, it is shown that only one
solution is possible, the problem will have been shown to be com-
pletely stated as well as solved.
PROBLEMS
1. Solve the boundary value problem
dhc
dxdy = 0; u @,y)=y,
2. Solve the boundary value problem
dhb
tedy = 2z; 14(0, y ) = 0,
u(x , 0) = sin x .
Ans. u = y sin x.
u(x , 0) = x 2 .
o q * -rv n . Ans. u = xhy + x 2 .
the condition ° * W 6n the Second bou ndary condition is replaced by
du(x t 0)
dx
= z 2 .
4. By substituting the new independent variables U * V + ***‘
X = * + at, [x = x — at,
show that the wave equation d W = «*0 V«**) becomes
dt*
-23L _ n
d\ dn ~ °>
ion of the w*
problem
= V(x, 0) = F(x), 0 ) _ n
fit " u >
6. Solve thfLuXy S iT u r P rtbt e m WaVe eqUati ° n (Pr ° b ' 4 ’ ^ 2K
Sec. 4]
INTRODUCTION
9
where F(x) is a given function defined for all real x.
Ans. y = i[F(x + at) +• F(x — at)].
6. Solve Prob. 5 if the boundary conditions are replaced by
y{x, 0) = 0, = G(x).
= G(x).
Also show that the solution under the more general conditions
y(x, 0) = Fix), = G(x)
is obtained by adding the solution just found to the solution of Prob. 5.
Ans. V -(l /2a) f*** G(0 d!-.
CHAPTER II
PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS
5. Gravitational Potential. According to the universal law
of gravitation, the force of attraction exerted by a particle of
mass m at the point (re, y, z) upon a unit mass at (X, Y , Z ) is
directed along the line joining the two points, and its magnitude
and sense are given by the equation
where k is a positive constant and r is the distance between the
two masses:
r = V(X -xy + (F - y y + (Z =r sp.
The positive sense is taken from the point (x, y, z), called Q,
toward the point P (X, Y, Z).
The gravitational potential V at any point P due to the mass
m at Q is defined to be the function
y __ km
r
So the derivative of this function is the force:
dV __ km
dr r 2
Let Q be fixed and consider V as a function of X, F, and Z.
It will now be shown that the directional derivative of the
potential in any direction gives the projection of the force F
in that direction. .
First let the direction be parallel to the X-axis. Then
dV _ dV dr
dX “ Hr dX
kmX — x
r 2 r
= F cos a = F x ,
where cos a is the first direction cosine of the radius vector r,
10
Sec. 5] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS
and F x is the projection of F on the X-axis.
( 1 )
Similarly,
11
Now if s is the directed distance along any line through P hav-
ing the direction angles a', /3', -y' ; the directional derivative of V
can be written
(2) W = dV6X dVdY dV dZ
ds dX ds ^ dY ds dZ ds
= F x cos a' + F v cos /S' + F z cos y'.
The last expression is the projection of the force in the direction
of the line along which s is measured.
, extension to the potential and force due to a continuous
distribution of mass is quite direct. The potential function due
to a mass of density S(x, y, z) distributed throughout a volume r,
at a point P not occupied by mass, is defined to be
(3) F(X, Y.Z) — h f f f s .( x > Ui z ) dx dy dz
’ ) J J Jr [(* - *)■ + (F -yY + (Z -
2) 2 l i ‘
This integral can be differentiated with respect to X, Y or Z
inside the integral. Thus '
(4)
This is the total component F x of the gravitational forces exerted
by all the elements of mass in r upon a unit mass at P. Like-
wise the total, components F y and F z satisfy relations (1), so
that the directional derivative has the same form as in equation
( 2 ).
Hence the projection, along any direction, of the force exerted
by a mass distribution upon a unit mass at (X, Y, Z) is given
by the directional derivative, along that direction, of the poten-
tial function (3) ; that is,
A force which can be derived in this manner from a potential
function is known as a conservative force.
12 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 6
Let s be the arc length along any curve joining two points (Xi,
Y h Z x ) and (X 2 , 7 2 , 7 2 ), at which s = si and s = s 2 , respectively.
Then, according to formula (5),
f>,ds = F(X 2 , F 2j Zi) - V(X h Y i, Zx).
J8l
That is, the difference between the values of the potential V
at two points represents the work done by the gravitational
force upon a unit mass which is moved from one of these points
to the other. The amount of work depends upon the positions
of the points, but not upon the path along which the unit mass
moves.
6. Laplace’s Equation. The potential V(X, Y, Z) due to any
distribution of mass will now be seen to satisfy an important-
partial differential equation. Upon differentiating both members
of equation (4), Sec. 5, with respect to X, we find that
fx* - ~ k J/I 5 dxd V dz -
Likewise
'■-‘in
327
dZ'
I _ 3(7 - y) 2
pS y.5
1 _ 3(z - z y :
8 dx dy dz,
8 dx dy dz .
The sum of the terms inside the three brackets
is zero; so
&V dW d 2 V
dX 2 + dY 2 + dZ~ 2 = °-
This is Laplace’s equation. It is often written
V 2 F = 0,
T"*”- V ' squared,”
V s = ~ 1 3 I d 2
3X 2 ‘ r dY 2 + dZ ~ 2
^ -L V ZJ -
We have ^ust r ^ presen *’ n severa -l other important equations.
by the tlvitadon^ LaplaCe ’ S e ^ Uation is satisfied
gravitational potential at points in space not occupied
Sec. 7] PARTIAL DIFFERENTIAL EQ UA TIONS OF PHYSICS 13
by mass. It is satisfied as well by static electric or magnetic
potential at points free from electric charges or magnetic poles,
since the law of attraction or repulsion and the definition of the
potential function in these cases are the same, except for constant
factors, as in the case of gravitation.
Other important functions in the applications satisfy Laplace's
equation. One of them is the velocity potential of the irrota-
tional motion of an incompressible fluid, used in hydrodynamics
and aerodynamics. Another is the steady temperature at points
in a homogeneous solid; this will be shown further on in this
chapter.
The gravitational potential at points occupied by mass of
density 8 can be shown to satisfy Poisson’s equation:
V 2 V = -4t rS,
a nonhomogeneous equation. The equations of Laplace and
Poisson, like most of the important partial differential equations
of physics, are linear and of the second order.
7. Cylindrical and Spherical Coordinates. Since cylindrical
and spherical surfaces occur frequently in the boundary value
problems of physics, it is important to have expressions for
the Laplacian operator in terms of cylindrical and spherical
coordinates.
The cylindrical coordinates (r, <p, z) determine a point P
(Fig. 1) whose rectangular coordinates are
(1) x = r cos <p, y — r sin <p, z = z.
These relations can be written
(2) r = \A 2 + y 2 , tp = arctan -> z = z,
x
14 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 7
provided it is observed that the quadrant of the angle p is
determined by the signs of x and y , not by the ratio y/x alone.
Let u be a function of r , p } and z. In view of relations (2) it
is also a function of x, y , z, and according to the formula for
differentiating a composite function,
du __ du dr du dp
dx dr dx d<p dx
Therefore
du x
dr \/x 2 + V 2
du y
dip x 2 + y 2
d 2 u __ f x\ _ (v\ \ ^ y d ( eu\
dx 2 dr dx\r ) dcp dx \r 2 ) r dx \dr ) r 2 dx \d<p)
The last two indicated derivatives can be written
a_/3i
lA _ d 2 u X
d 2 u y
cte \di
* ) dr 2 r
dr dtp r 2 ’
u\ _ d 2 U X
d 2 u y
\d
\ pj dp dr r
Vr 1 '
Substituting and simplifying, we find that
d 2 u __ y 2 du , 2 xy du x 2 d 2 u 2 xy d 2 u y 2 d 2 u
dx 2 r z dr r 4 dip r 2 dr 2 r z dr dip r 4 dtp 2
Similarly, it is found that
d 2 u __ ^ du __ 2xy du y 2 d 2 u 2 xy d 2 u x 2 d 2 u
dy 2 r 3 dr r A dp ^ r 2 dr 2 ~r*' dr dip + r 4 dp 2 ’
so that the Laplacian of u in cylindrical coordinates is
(3)
\7 2 w = — 4- I — -L L ^ I dHt
dr 2 ^ r dr f r 2 <V dz 2 *
It is simpler to transform the right-hand member of equation
(3) into rectangular coordinates. This operation furnishes a
verification of equation (3).
The spherical coordinates (r, p, 6) of a point P (Fig. 2), also
called polar coordinates, are related to the rectangular coordinates
as follows:
(4) x = r sin 6 cos p,
y = r sin 6 sin p,
z = r cos 6 .
Sec. 8] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 15
The Lapladan of a function u in spherical coordinates is
(5) = 4 \r f- 2 ( ru ) + ~ ± ( s i n d —
' L dr 2 ; J sm S dd \ dd J
+
1 d 2 u
sm 2 6 d<p 2 J
The derivation or verification of this formula can be carried
out in the same manner as that of the corresponding formula
(3) for cylindrical coordinates. It is left as an exercise.
PROBLEMS
1. Derive the expression given above for dhi/dy- in cylindrical
coordinates, and thus complete the derivation of formula (3).
2. Verify formula (3) by transforming its right-hand member into
rectangular coordinates.
3. Verify formula (5) by transforming its right-hand member into
rectangular coordinates.
4. Write the formulas which give the spherical coordinates in tenm
of x, y, z.
5 Derive formula (5) for the Laplacian in terms of spherical coordi-
8. The Flux of Heat. Consider an infinite slab of homogene-
ous solid material bounded by the pianos x = 0 and x = L Let
the faces x = 0 and * = L be kept at fixed uniform temperatures
ui and u„ respectively. After the temperatures have become
steady, the amount of heat per unit time which flows from the
surface x = 0 to the surface x = L, per unit area, is
JT ^2 U 1
A j—,
where the constant K is known as the thermal conductivity
Ihis statement is essentially a definition of the conductivity K.
16 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 8
The time rate of flow of heat per unit area through a surface
is called the flux of heat. For the flux F through any isothermal
surface (a surface at uniform temperature), the natural extension
of the above definition is
Here u is the temperature as a function of position, n o is the
distance measured along a directed normal to the isotherm,
and the positive sense of the flux F is that of the normal. In
formula (1) the conductivity K may be variable, and the solid
nonhomogeneous.
Fig. 3.
To indicate the extension of this formula to the flux F n normal
to an arbitrary surface in a solid at a point P, let coordinate
axes be chosen with origin at P so that the xy - plane is tangent
to the isotherm through P (Fig. 3). Let X, y, v be the direction
cosines of the normal n of the given surface. Now let the
surface be displaced parallel to itself through a distance p, so
that its tangent plane and the coordinate planes bound an
elementary volume in the form of a tetrahedron.
If A A is the area of the face QRS made by the tangent plane,
vA A is the area of the face in the xy- plane. As p approaches
zero, the rate of flow of heat into the element through one of
these faces must approach the rate of flow out through the other:
F n AA = F t vAA,
where F z is the flux through the isotherm. The remaining two
faces are perpendicular to the isotherm, so that the flux of heat
through them is zero.
Sec. 9] PARTIAL DIFFERENTIAL EQ UA TIONS OF PHYSICS 17
According to formula (1), F z = —K du/dz , so that
But according to the formula for the directional derivative,
du_^du du du _
dn d# ^ <9y v 62
dw
d2 ?
since du/dx and du/dy are both zero, owing to the fact that
x and y are distances along the isothermal surface. It follows
that
( 2 )
jp _
Fn ~ ~ K Tn'
that is, the flux of heat through any surface in the direction of the
normal to that surface is 'proportional to the rate of change of
the temperature with respect to distance along that normal.
In the derivation of relation (2) it was assumed that there is
no source of heat in the neighborhood of the point P, and that
the derivatives of the temperature function u exist. Further-
more, our argument involved approximations, such as the use
of tangent planes in place of surfaces, the validity of which use
was not examined.
We shall not attempt to make the derivation of relation (2)
precise. In a rigorous development of the mathematical theory
of heat conduction, relation (2) can be postulated instead of
(1). The results which follow from (2) — in particular, the heat
equation derived in the next section — have long been known to
agree with experimental measurements.
It should be observed that the temperature u serves as the
potential function from which the flux of heat is obtained by
finding its directional derivative. In the case of the gravitational
or electrical potential, the directional derivative gives, respec-
tively, the gravitational force or the flux of electricity in that
direction; the flux of electricity is the current per unit area of
surface normal to the direction.
9. The Heat Equation. Let u(x, y , 2, t ) represent the tem-
perature at a point P (x y z) of a solid at time t, and let K be
the thermal conductivity of this solid, where K may be a func-
tion of x, y, z and t, or of u . Suppose that the point P is enclosed
18 FOURIER SERIES AND BOUNDARY PROBLEMS [Sac. 9
by any surface S lying entirely within the solid, and let n repre-
sent the outward-drawn normal to the closed surface S. Then
accor ing to the formula for the flux in Sec. 8, the time rate of
flow of heat into the volume V enclosed by S, through the surface
( 1 )
K d ^dS.
dn
Now if 6 is the density of the solid and c its specific heat, or
tile amount of heat required to raise the temperature of a unit
mass of the solid 1 degree, another expression for the rate of
increase of heat in the volume V is
(2)
If X, n, v are the direction cosines of the normal n, the integral
(1) can be written
/X( xk 5+**S + -*S)<»
This can be transformed, according to Green’s theorem, into
the volume integral
J JX [1 0 1) +!(*!)+ 1 (* £)] ^
which must be equal to the integral (2), so that
(3)
+
dz
( K ^)
C3 - ) are^rmt^™ 11 ^ f ^ ^ * erms * n *be brackets in equation
he int, eZT 0US ^ °f “ a nei ^borhood of P Since
the integral m equation (3) vanishes for every volume V its
SrTfnT* r iSi P ' For if were peak™
V tkL ZhiTT r°'‘ d o 5 ‘ 1U "' a small volume
l pSv! The T* \ “ d throush0l ' t «“> integrand
P toe ' The y »ould then be positive, in
Sec. 10] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 19
contradiction to equation (3). Similarly if the integrand were
negative. Therefore, at P,
fA\ d /^chA , <9 / v du\ d ( ^ du\
(4) , *Tt-Tx\ K irx) + Ty\ K T,) + Tii\ K Tz)
This is a general form of the equation of conduction of heat, or
the heat equation.
It should be noted that we have assumed in the derivation
that no sources of heat exist in the neighborhood of the point
(z, V> z).
10. Other Cases of the Heat Equation. If the conductivity
K is constant, or does not depend upon the coordinates, the
heat equation becomes
(1)
du _ j ( d 2 U d 2 U d 2 u \
Tt 3F/
where the coefficient k , called the diffusivity , is defined thus:
The equation appears most frequently in the form (1), or the
abbreviated form
(2) g = kVht.
The right-hand member can be expressed in terms of other
coordinates by using the results of Sec. 7.
The heat equation is also called the equation of diffusion.
It is satisfied by the concentration u of any substance which
penetrates a porous solid by diffusion.
It was shown above that the temperature u everywhere
within a solid satisfies the heat equation. To determine u as a
definite function of x , y } 2 , and /, it is of course necessary to
use, in addition to the heat equation, boundary conditions which
describe the thermal state of the surface of the solid and the
initial temperature. All these conditions make up the boundary
value problem in the conduction of heat.
There are several special cases and simple generalizations of
the heat equation which arc important. First there are the cases
in which the temperature is independent of one or more of the
four independent variables, which consist of the space coordinates
20 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 10
and time t. If the temperatures are “ steady ” — that is, if u
does not change with time — u satisfies Laplace's equation. This
is approximately the case, for example, if the temperature
distribution on the surface of a solid has been kept the same for
a long period of time.
If conditions are such that there can be no flow of heat in the
direction of the z-axis, the heat equation for “ two-dimensional
flow” applies:
du _ , ( SHt d 2 lA
at “ * \dx* + W
Similarly for one-dimensional flow.
Continuous sources of heat may exist within a solid. If at
each point (x, y, z) they supply heat at the rate of F(x, y , z, t)
units per unit time per unit volume, the heat equation becomes
nonhomogeneous. Tor the case of one-dimensional flow, where
the strength F of the source is a function of x and t , the equation
becomes
(3) ci t,-fX K f^)+ F ^ e >-
This follows readily from the derivation in Sec. 9. Equation (3)
may apply, for instance, to the temperature u in a wire which
carries an electric current.
PROBLEMS
1. The lateral surface of a homogeneous prism is insulated against
the flow of heat. The initial temperature is zero throughout, and the
end x = 0 is kept at temperature zero while the end x = L is kept at
To, a constant temperature. Write the heat equation for this case.
Ans. du/dt = k(d 2 u/dx 2 ).
2. Find the steady temperature in Prob. 1, after the conditions given
there have been maintained for a very long time. What is the flux
through one end during the steady state?
Ans. u = (T 0 /L)x; flux - KT 0 /L.
3. State a physical problem whose solution is represented by the
finite series in Prob. 5, Sec. 2.
4. Show that the temperature u in a uniform circular disk whose
entire surface is insulated, and whose initial temperature is a function
only of the distance r from the axis of the disk, satisfies the equation
du z / d 2 u 1 du\
dt ~ * \dr* + r TrJ
Sec. 11] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 21
5. If the initial temperature of a homogeneous sphere is a function
only of the distance r from the center, and the surface is insulated, show
that the temperature u of points inside satisfies the equation
du _ ( dhc 2 du\
Bt “ * \dr* + r dr)'
11. The Equation of the Vibrating String. The transverse
displacements of the points of a stretched string satisfy an
important partial differential equation. Let the string be
stretched between two fixed points on the a>axis and then given a
displacement or velocity parallel to the y- axis. Its subsequent
motion, with no external forces acting on it, is to be considered;
this is described by finding the displacement y as a function of
x and t.
It will be assumed that 5, the mass per unit length, is uniform
over the entire length of the string, and that the string is perfectly
flexible, so that it can transmit tension but not bending or
shearing forces. It will also be assumed that the displacements
are small enough so that t he square of the inclination dy/dx
can be neglected in comparison to 1; hence, if 8 is distance
measured along the string at any instant,
approximately. The length of each part of the string therefore
remains essentially unaltered, and hence the tension is approxi-
mately constant.
Consider the vertical components of the forces exerted by the
string upon any element A s of its length, lying between x and
x + Ax (Fig. 4). The ^-component of the tensile force P
exerted upon the element, at the end (x, y) is
_ p dy = __ P d JL d Jt =
ds dx ds bx
22 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 11
approximately. The corresponding force at the end whose
abscissa is x + Ax is
P^f + P^y Ax + R(Axy,
dx dx 2
where R is the usual factor in the remainder in Taylor’s formula.
Setting the sum of these forces equal to the product of the
mass of the element and the acceleration in the ^-direction, we
have
pg Ax + R(Ax)^5Ax-^-
By div iding by Ax and letting Ax approach zero, it follows that
d>y . d*v
(i) w = a a?
where
This is the equation of the vibrating string; it is also called the
simple wave equation, since it is a special case of the wave
equation of theoretical physics. . .
If an external force parallel to the y-axis acts along the string,
it is easily seen that the equation becomes
( 2 )
g-*3+n*o.
where SF(x, t) is the force per unit length of string. In case
the weight of the string is to be considered, for instance, the
function F becomes the constant g, the acceleration of gravity.
If the transverse displacements arc not confined to tlio ay-plane,
two equations of type (2) are found, one involving the y, the
other the z, of the points of the string, while the acceleration r
is replaced by the y and z components of the external acceleration
in those two equations, respectively.
Equation (1) is also satisfied by the longitudinal displacements
in a homogeneous elastic bar; y is then the displacement along
the bar of any point from its position of equilibrium. A column
of air may be substituted for the bar, and the equation becomes
one of importance in the theory of sound. The equation also
applies to the torsional displacements in a right circular cylinder.
Sec. 12] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 23
PROBLEMS
1. Derive equation (2) above.
2. State Prob. 5, Sec. 4, as a problem of displacements in a stretched
string of infinite length. Show that the motion given by the result of
that problem can be described as the sum of two displacements, obtained
by separating the initial displacement into two equal parts, one of which
moves to the left along the string with the velocity a, and the other to
the right with the same velocity.
3. If a damping force proportional to the velocity, such as air resist-
ance, acts upon the string, show that the equation of motion has the
form
d 2 y 0 d 2 y dy
dt 2 " a dx* dt’
where b is a positive constant.
12. Other Equations. Types. Some further partial differ-
ential equations of importance in the applications will be described
briefly at this point. For their derivation and complete descrip-
tion, the reader should refer to books on the subjects involved.
A natural generalization of equation (1) of the last section
is the equation of the vibrating membrane:
Here the position of equilibrium of the stretched membrane is
the rry-plane, so that z is the transverse displacement of any point
from that position. Assumptions similar to those in the case
of the string are necessary. The membrane is assumed to be
thin and perfectly flexible, with uniform mass 5 per unit area.
The tensile stress P, or tension per unit length across any line,
is assumed to be large, and the displacements small. The
constant a 2 is then the ratio P/8.
The telegraph equation ,
(2) ~ = KL g + (RK + SL) + RSv,
is satisfied by either the electric potential or the current in a
long slender wire with resistance P, the electrostatic capacity K ,
the leakage conductance S , and the self-inductance L, all per
24 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 13
unit length of wire. The simple wave equation is a special
case of this.
The transverse displacements y(x, t) of a uniform beam satisfy
the fourth-order equation
(3)
d*y
dt 2
+ c 5
d 4 y
dx 4
= o,
where the constant c 2 depends upon the stiffness and mass of
the beam.*
Airy’s stress function tp(x } y) , used in the theory of elasticity,
satisfies the fourth-order equation
(4)
, O dV , gv = 0
dx 4 ^ dx 2 dy 2 ^ dy 4
often written VV = 0. It serves in a sense as a potential
function from which shearing and normal stresses within an
elastic body can be derived. The form (4) assumes that no
deformations exist in the 2 -direction.
Linear partial differential equations of the second order with
two independent variables x ) y are classified into three types
in the theory of these equations. If the terms of second order,
when collected on one side of the equation, are
+ B
d 2 u
dx dy
, c d 2 U
+ L -r-T y 7
dy-
where A, B , C are constants, the equation is of elliptic , parabolic ,
or hyperbolic type according as (J5 2 — 4 AC) is negative, zero, or
positive. In the study of boundary value problems it will be
observed that these three types require different kinds of bound-
ary conditions to completely determine a solution.
Note that Laplace’s equation in x and y is elliptic, while the
heat equation and the simple wave equation in x and t arc
parabolic and hyperbolic, respectively. The telegraph equation
is also hyperbolic, if KL ^ 0.
13. A Problem in Vibrations of a String. When the differen-
tial equation is linear and the boundary conditions consist of
linear equations, the boundary value problem itself is called linear .
*See, for instance, Timoshenko, “Vibration Problems in Engineering,”
p. 221.
Sec. 13] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 25
A method which can be used to solve a large class of such prob-
lems will now be illustrated. It will be seen that the process
leads naturally to a problem in Fourier series. A formal solution
of the following problem will be given.
Find the transverse displacements y(x, t) in a string of length
L stretched between the points (0, 0) and (L, 0) if it is displaced
initially into a position y = f(x) and released from rest at this
position with no external forces acting.
The required function y is the solution of the following bound-
ary value problem :
(1) «>0,0<*<i),
(2) 2/(0, 0 = 0, y(L, <)=0 (< £ 0),
(3) y(x, 0) = fix) (o ^ X g L),
(4) =0 (0 g x g L).
ot
Our method consists of finding particular solutions of the
partial differential equation (1) which satisfy the homogeneous
boundary conditions (2) and (4), and then of determining a
linear combination of those solutions which satisfies the non-
homogeneous boundary condition (3).
Particular solutions of equation (1) of the type
( 5 ) y = XT,
where X is a function of x alone and T a function of t alone, can
easily he found by means of ordinary differential equations.
According to equation (5), dy/dx = X'T, dy/dt = XT', etc.,
where the prime denotes the ordinary derivative with respect
to the only independent variable involved in the function.
Substituting into equation (1), we find
XT" = a*X"T,
or, upon separating the variables by dividing by a 2 XT,
X"(x) T"(t)
X(xf a ir T(t ) "
Since the member on the left is a function of x alone, it cannot
vary with t; it is equal to a function of t alone, however, and
26 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 13
thus it cannot vary with x. Hence both members must be equal
to a constant, say y, so that
(6) X"(x) - yX(x) = 0,
(7) T”{t) - ya?T(t) = 0.
If our particular solution is to satisfy conditions (2), XT must
vanish when x = 0 and when x = L, for all values of t involved.
Therefore
(8) X(0) = 0, X{L) = 0.
Similarly, if it is to satisfy condition (4),
(9) T'( 0) = 0.
Equations (6) and (7) are linear homogeneous ordinary
differential equations with constant coefficients. The auxiliary
equation corresponding to (6), m 2 — 7 = 0, has the roots
m, = ± Vy. The general solution of equation (6) is therefore
X = C,e x ^ + C 2 e~*
where C 1 and C 2 are arbitrary constants. But if 7 is positive,
it is easily seen that there are no values of C\ and C 2 for which
this function X satisfies both of conditions (8).
Suppose 7 is negative, and write
7 = -£ 2 .
The general solution of equation (6) can then be written
X = A sin fix + B cos
where A and B arc arbitrary constants.
If X(0) = 0, the constant B must vanish. Then A must
be different from zero, since we are not interested in the trivial
solution X(x) = 0. So if X{L) = 0, we must have
sin fiL — 0.
Hence there is a discrete set of values of /3 } namely,
(» = 1 , 2 ,--*),
Sec. 13] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 27
for which the system consisting of equation (6) and conditions
(8) has solutions. These solutions are
X = A sin
mrx
Note that no new solutions are obtained when n = — 1, —2,
- 3 , • • ■ .
Substituting — nV 2 /L 2 for y in differential equation (7) and
applying condition (9), we find that
T = C cos
mrat
where C is an arbitrary constant.
Therefore all the functions
(10)
A n sin
mrat
mrx
T cos T
(» = 1 , 2 , • • • )
are solutions of our partial differential equation (1) and satisfy
the linear homogeneous conditions (2) and (4), when A J? A 2 , • • •
are arbitrary constants.
Any finite linear combination of these solutions will also
satisfy the same conditions (Theorem 1, Chap. I); but when
t = 0, it will reduce to a finite linear combination of the functions
sin ( mrx/L ). Thus condition (3) will not be satisfied unless the
given function f(x) has this particular character.
Consider an infinite series of functions (10),
oo
/11N . . mrx mrat
(11) V = An sin -j~ cos •
This satisfies equation (1) provided it converges and is termwise
differentiable (Theorem 2, Chap. I); it also satisfies conditions
(2) and (4). It will satisfy the nonhomogeneous condition (3)
provided the numbers A n can be so determined that
( 12 )
90
f(x) = 2 An
. mrx
sm - T - *
It will be shown in Sec. 15 that if such an expansion of f(x) is
possible, the numbers A n must have the values
28 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 14
Equation (11) with coefficients (13) is formally the solution
of the boundary value problem (l)-(4).
The series on the right of equation (12) with the coefficients
defined by (13) is called the Fourier sine series of the function
f(x). In a later chapter it will be shown that this series actually
converges to the function f(x) in the interval 0 x ^ L, pro-
vided /(x) satisfies certain moderate conditions — conditions which
are almost always satisfied by functions which arise in the
applications.
Other questions are left unsettled at this point in the treatment
of this problem. Series (11) has not been shown to be conver-
gent, or to represent a continuous function, or to be termwise
differentiable twice with respect to either x or t. It has not
been shown that series (11) is the only solution of the problem
(l)-(4). Questions of this character are to be treated later on.
14. Example. The Plucked String. As a special case of the
problem just treated, let the string be stretched between the
points (0, 0) and (2, 0), and suppose its mid-point is raised to a
height h above the rc-axis. The string is then released from rest
in this broken-line position (Fig. 5).
The function f(x) which describes the initial position can be
written, in this case,
f(x) = hx when 0 ^ x g 1,
= — hx + 27i when 1 ^ x ^ 2.
The coefficients in solution (11), Sec. 13, are, according to
formula (13), Sec. 13,
A n = f(x) sin dx
= h x sin V ^‘ dx + h ( — x + 2) sin dx.
Sec. 15] PARTIAL DIFFERENTIAL EQUA TIONS OF PHYSICS 29
After integrating and simplifying, we find that
A 8h . mr
An = -r-r, sin ;
7 r 2 n 2 2 7
so that the displacement y(x, t ) in this case of the plucked string
is given by the formula
V =
8h 1 . rnr . mrx
V>2jn> sm Y- sm - T cos
mrat
8 h(
7T 2 \
tx irat 1 . 2nrx
sm — cos g sin — cos
St at
~2~
, 1 ■ 5 tx 57 rat \
+ 23 sm — COS -g- - • • . y
Another form of this solution will be obtained later [formula
(4), Sec. 43].
15. The Fourier Sine Series. In the solution of the problem
of Sec. 13 it was necessary to determine the coefficients A „ so
that the series of sines would converge to f(x). Assuming that,
an expansion of the type needed there, namely,
(1) Six) = A i sin ^ + A« sin ~ x •
D L
, A . UirX
i A n sin - T -j-
Jj
is possible when 0 ^ i 1 1, and that the scries can be integrated
term by term after being multiplied by sin (mrx/L), it is easy to
see what values the coefficients must have.
It is necessary to recall that
sin sin «•* . i &.-»>» _ „„ (» +
oJ T1 2 m ' 7TJC 1 ( i 2ni7rx\
h,n L -2\ l -™»- L )>
and hence, when m and n arc integers,
( 2 )
r' J . nnr.r . riTX ,
Jo K1U ' // H,n '// dx = 0 if
L
2
in n,
if ni = ?i.
The functions sin (mrx/L) (a = 1, 2, • • • ) therefore form an
orthogonal system m the interval 0 < x < /,; that is, the integral
30 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 15
over that interval of the product of any two distinct functions
of the system is zero.
Now let all terms in equation (1) be multiplied by sin (rnrx/L)
and integrated between 0 and L. The first term on the right
becomes
This is zero unless n = 1, according to the orthogonality property
(2) . Likewise all terms on the right except the nth one become
zero; so the process gives
(* ft \ • '0/KX -j A ‘ 7 A-nfi
I f[x) sm -j— ax = A n I sin 2 -j- dx =
according to property (2). Hence the coefficients in equation (1)
must have the values
(3) A n = ^ f(x) sin ^ dx.
The Fourier sine series corresponding to f(x) can be written
(4)
where the sign ^ is used here to denote correspondence. It
is to be shown later on that the series does converge to f(x) in
general.
PROBLEMS
1. Show that the Fourier sine series corresponding to fix) = 1 in the
interval 0 < x < tt is
1 ^sin x + ^ sin Sx + i sin 5x -j- ■ • • j.
2. Show that the sine series for f(x) = x in the interval 0 < x < 1 is
x
sin mrx.
3. Find the solution of the problem of the string in Sec. 13 if the
initial displacement is f(x) = A sin (ttx/L). Discuss the motion.
Arts, y — A sin (tx/L) cos (tt at/L),
Sec. 16 ] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 31
16. Imaginary Exponential Functions. According to the
power series expansion of e z ,
e ix
( ix) n
n\
t 2
2! 1 4!
where i = V"— 1- So
— 1
” 1 2 ! ' 4 !
+ * (* -
, x*
3! + 6!
(1) e ix = cos x + i sin x.
This is usually taken as the definition of the exponential function
with imaginary exponents. Then
e -ix — cos x — { sin x f t
and by first eliminating cos x and then sin x between this equa-
tion and equation (1), we find that
(2) i sin x = ^ = si n h (ix),
nix I ix
(3) cos x = ^ = cosh (ix).
When the coefficients of a linear homogeneous differential
equation are constant, particular solutions in the form of exponen-
tial functions can be found. ^
To illustrate the use of exponential functions in partial differ-
ential equations, consider again the problem in Sec. 13. The
function
y = fOcx+01^
where a and 13 are constants, is clearly a solution of the equation
(4)
provided that
Hence the functions
dt 2 dx 1 ’
/3 2 = a 2 « 2 .
(\ ax (> i C~ ax C aat
are solutions.
Except for a constant factor, the difference between the two
products just written is the only linear combination which
32 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 16
vanishes at x = 0 [condition (2), Sec. 13]. Thus the functions
e±aoU si n h aX
satisfy that condition as well as equation (4). The linear com-
bination of these two functions which satisfies the condition that
dy/dt = 0 when t = 0 is the sum, or
(6) sinh ax cosh aat
But here a must be imaginary, since our function is to vanish
when x = L, because the hyperbolic sine of a real argument
vanishes only when the argument is zero. According to equations
(2) and (3), when a = in the function (5) can be written, except
for a constant factor, as
sin ixx cos j uat.
This vanishes when x = L if n == mr/L.
Thus we again have the particular solutions
, . utx meat
A n sm -j- cos -j—
of equation (4), which satisfy the homogeneous conditions in the
problem in Sec. 13. From this point on the procedure is the
same as in that section.
As another application of imaginary exponential functions,
llote that
N N
2(cos 0 + cos 20 + • • • + cos N9) = ^ e in9 + ^ e~ in9 .
Summing the finite geometric series on the right, this becomes
N
2 cos nO =
l
e i6 (l - e iN9 ) e~ i9 (l - e~ iN9 )
1 — e l9 1 — e~ 19
— JV4-$) _|_ g— \iQ g—iO(N+}f)
q— bid
This can be written at once in the form
which is known as Lagrange’s trigonometric identity . This
identity will be useful in the theory of Fourier series.
Sec. 16] PARTIAL DIFFERENTIAL EQUATIONS OF PHYSICS 33
PROBLEMS
1. Use exponential functions to determine particular solutions of the
simple heat equation
du d 2 w
dt = dF 2
which vanish when x = 0 and x = x. (Compare Prob. 5, Sec. 2.)
2. Use exponential functions to determine particular solutions of the
equation
d' l u dhi _
dx* dy 2, ~
such that u = 0 when y = 0, and du/dx = 0 when x = 0 and x = 1.
Ans. u — An cos mrx sinh mry ( n = 0, 1, 2, • ■
CHAPTER III
ORTHOGONAL SETS OF FUNCTIONS
17. Inner Product of Two Vectors. Orthogonality. The
concept of an orthogonal set of functions is a natural generaliza-
tion of that of an orthogonal set of vectors, that is, a set of
mutually perpendicular vectors. In fact, a function can be
, considered as a generalized vector, so that the fundamental
properties of the set of functions ar$ suggested by the analogous
properties of the set of vectors. In the following discussion
of simple vectors, the terminology and notation which apply
to the generalized case will be used whenever it seems advanta-
geous for the later generalizations.
Let either g or g(r) denote a vector in ordinary three-dimen-
sional space whose rectangular components are the three numbers
#(1), <7(2), and <7(3). It is the radius vector of the point having
these numbers as rectangular cartesian coordinates. The square
of the length of this vector, called its norm , will be written
N(g)] it is the sum of the squares of the components of g:
(1) N(g) = g\ 1) + g\ 2 ) + (7*0) = g\r).
r = 1
If N (g) = 1, g is a unit vector, also called a normed or a normal-
ized vector.
Let 6 be the angle between two vectors g x (r) and g 2 (r ). Since
the components £i(l), 0i(2), g x {Z) are proportional to the direc-
tion cosines of the vector g Xj and similarly for g 2 , the formula
from analytic geometry for cos 6 can be written
n = ffi(l)ff2(l) + gi(2)g 2 (2) + <7i(3)gr 2 (3)
The numerator on the right is called the inner product (or scalar
product) of the vectors g x and g 2) denoted by the symbol (g h g 2 ) ;
thus
34
Sec. 18]
ORTHOGONAL SETS OF FUNCTIONS
35
3
( 2 ) (01, gt ) = 5 ) 01M02M
T = 1
= VN( gi ) v%) cos 6.
When JV(0 2 ) = 1, (g 1, #2) is the projection of the vector <71 in the
direction of g 2 .
The condition that the vectors <71 and g 2 be orthogonal, or
perpendicular to each other, can be written
(3) (g h 02) = 0
or, in terms of components,
(4) X gi(r)gt(r) = 0.
r = 1
Note also that expression (1) for the norm of g can be written
N(g) = (0, 0).
18. Orthonormal Sets of Vectors. Given an orthogonal set
of three vectors g n (n = 1, 2, 3), a set of unit vectors <p n having
the same directions can be formed by dividing each component
of g n by the length of g n . The components of <p h for instance,
are <pi(r) = 0i(r)[JV(0i)]-* (r = 1, 2, 3). This set of mutually
perpendicular unit vectors <p n , obtained by normalizing the
mutually perpendicular vectors g n , is called an orihonormal set.
Such a set can be described by means of inner products by writing
(1) (<Pm, <pn) = & mn (w, U = 1, 2, 3),
where 5 mn , called Kronecker’s <5, is 0 or 1 according as m and n
are different or equal:
&rnn — 0 if 771 7^ 72,
= 1 if rn = n.
The condition (1) therefore requires that each vector of the
set <p h <p 2 , <ps is perpendicular to every other one in that set, and
that each has unit length.
The symbol {<?„} will be used to denote an orthonormal set
whose vectors arc <p h and <^ 3 . The simplest example of
such a set is that consisting of the unit vectors along the three
coordinate axes.
36 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 18
Every vector / in the space considered can be expressed as a
linear combination of the vectors (pi, (pi, and <ps. That is, three
numbers Ci, c 2 , c% can be found for which
(2) f(r) = Ci<pi(r) + c 2 (pi(r) + c z <pz(r) (r = 1, 2, 3),
when the components /( 1), /( 2), /( 3) are given. To find the
number Ci in a simple way, consider equation (2) as a vector
equation and take the inner product of both its members by (pi.
This gives
(/, ^l) — (pi) + Ci(<pi, (pi) + Cz((pz, <pl) = Cl,
since ((pi, <pi) = 1 and (<pi, (pi) = (<Pz, <pi) = 0, according to
condition (1). Similarly c 2 and c z are found by taking the inner
product of the members of equation (2) by (pi and (p z , respectively.
The coefficients are therefore
(3) c n = (/, <p n ) = X f(r)ip n (r) (n = 1, 2, 3).
r — 1
The representation (2) can then be written
(4) f(r) = (/, <pi)<pi(r) + (/, <Pi)<Pi{r) + (/,
= X
n = 1
The representation (2) or (4) may be called an “expansion”
of the arbitrary vector / in a finite series of the orthonormal
reference vectors <pi, (pi , and <p 3 . These orthogonal reference
vectors were assumed to be normalized only as a matter of
convenience, in order to obtain the simple formulas (3) for the
coefficients in the expansion. The normalization is not neces-
sary, of course.
The definitions and results just given can be extended immedi-
ately to vectors in a space of k dimensions. In this case the
index r , which indicates the component, has values from 1 to k,
instead of 1 to 3; similarly the indices m and n, which distinguish
the different vectors of an orthonormal set, run from 1 to k.
The definition of the inner product of the vectors g\ and g%
in this space, for instance, becomes
k
(g i, gt) = X s'iMffaW.
r = l
( 5 )
Sec. 19] ORTHOGONAL SETS OF FUNCTIONS 37
The formal extension to vectors in a space of a countably
infinite number of dimensions (k = °o ) is also possible. In this
case the numbers g(r) (r = 1, 2, • • * ) which define a vector g
would be so restricted that the infinite series involved, such as
the series in (5) with k infinite, would converge. The possibility
of the representation corresponding to (2) would have to be
examined, of course.
A generalization of another sort is also possible. The units
of length on the rectangular coordinate axes, with respect
to which the components of vectors are measured, may vary
from one axis to another. In such a case the scalar product of
two vectors gi and g 2 in three-dimensional space has the form
3
(?i, 0*) = X
r = 1
The “ weight numbers” p(l), p( 2), and p( 3) here depend upon
the units of length used along the three axes.
19. Functions as Vectors. Orthogonality. A vector g(r) in
three dimensions was described above by the numbers g( 1),
g(2) y g( 3). Any function g(r) which has real values when
r = 1, 2, 3 will represent a vector if it is agreed that these values
are the components of the vector. This function may not be
defined for any other values of r, in which case its graph would
consist only of three points.
The function g(r) will represent a vector in space of k dimen-
sions if it has real values when r = 1, 2, * • • , /c, which are
considered as the components of the vector. If g(r) is defined
only at these points, it is determined by the vector; graphically
it is represented by k points whose abscissas are r = 1, 2, * * • ,
kj and whose ordinates arc the corresponding components of the
vector.
Now let g(x) be a function defined for all values of x in an
interval a ^ x ^ b. To consider this function as a vector, the
components should consist of all the ordinates of its graph in
the interval. The argument x, which has replaced r here, has
as many values as there are points in the interval, so that the
number of components is not only infinite but uncountable. It
is therefore impossible to sum with respect to x as we do with
the index r. The natural process now is to sum by integration.
38 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 19
The norm of the function or vector g(x), or the sum of the squares
of its components, is therefore defined as the number
(1) N(g) = f*\g(x)]*dx.
The inner product of two functions g m (x) and g n (x) is defined
as the number
( 2 ) (i g m , g n ) = f b g m (x)g n (x) dx ,
Ja
in analogy to equation (5), Sec. 18. The condition that the two
functions be orthogonal is written
(Qm, gn) = 0 ,
or
(3) gm(x)g n (x) dx = 0.
Just as before, definition (1) can be written N{g) = ( g , g).
A set (or system) of functions {<7n(z)} (n = 1, 2, • ■ ■ ) is
orthogonal in the interval (a, b) if condition (3) is true when
m t* n for all functions of the set. The functions of the set
are normed by dividing each function g n (x) by [2V (£«)]*, thus
forming a set {<p n (a;)} (n = 1, 2, ■ • • ), which is normed and
orthogonal, or orthonormal . An orthonormal system in (a, b ) is
then characterized as follows:
(4) <Pn) = <5mn (^, XI == 1, 2, * * * );
where 5 m n is Kronecker's 5, defined in Sec. 18. Written in full,
equation (4) becomes
(5) r <Pm{x)ip n {x) dx = 0 if m n,
= 1 if m = ft, (m, n = 1, 2, • • • ).
The interval (a, b) over which the functions and their inner
products are defined is called the fundamental interval. Func-
tions for which the integrals representing the inner product and
the norm fail to exist must, of course, be excluded.
Throughout this book, only functions which are bounded and
integrable in the fundamental interval , and whose norms are not
zero , will be considered. The aggregate of all such functions for
the given interval makes up the function space being considered,
Sec. 20]
ORTHOGONAL SETS OF FUNCTIONS
39
in just the same way that the three-dimensional vector space con-
sists of all vectors with three components g(r) (r = 1, 2, 3).
An example of an orthogonal set of functions has already been
given in Sec. 15; namely, the functions
sm
mrx
(n = 1, 2, • • • ).
The fundamental interval is the interval (0, L). The norm of
all these functions is the same number, L/2, so the orthonormal
set consists of the functions
f 2 . mrx , %
Vi sin it (« = i, 2, • ■ •
The set {sin (nirx/L)} is also orthogonal in the interval
( — L, L ); the normalizing factor is easily seen to be l/\/L in
this case.
PROBLEMS
1. Show that the set of functions j cos nx} (n = 0, 1, 2, • • ■ )
is orthogonal in the interval (0, 7r). What is the corresponding ortho-
normal set? Ans. {l/V7r, V2/7T cos nx\ (n = 1, 2, • • • ).
2. Show that the set (sin x, sin 2x, sin 3$, ••■,], cos x 1 cos 2x,
• - * } is orthogonal in the interval (—7 r, t). Normalize this set.
20. Generalized Fourier Series. Given a countably infinite
orthonormal set of functions {<£>«,(#) 1 (n = 1, 2, • • ■ ), it may
be possible to represent an arbitrary function in the fundamental
interval as a linear combination of the functions <p, t (r),
(1) f(x) = Ci<pi(x) + C 2 <p*(x) + • • * + Cn<Pn(x) + * * *
(a < x < b).
This corresponds to representation (2), Sec. IS, of any vector
in terms of the vectors of an orthonormal set.
If the series in equation (1) converges and if, after being mul-
tiplied by <p„(:r), it can be integrated term by term over the funda-
mental interval (a, b), the coefficients c n can be found in the same
way as before. Writing tin 1 , inner product of both members of
equation (1) by <p n (x) — that is, multiplying (1) by <p n and integrat-
ing over (a, b) — we have
(/, <Pn) = ipn) + C 2 (^2, <Pn) + * * ' + C n (<p n , <Pn) + * * *
40 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 21
since ( <p m , , <p n ) — 8 mn . That is, c n is the projection of the vector
/ on the unit vector p*.
These numbers c n are called the Fourier constants of f(x)
corresponding to the orthonormal system { <p n (x) } ; they can be
written
(2) Cn = £ f(x)<p n (x) dx (n = 1, 2, • • • ).
The series in (1) with these coefficients is called the generalized
Fourier series corresponding to f(x), written
(3) f(x) ~ ^ C n (p n (x) = tp n {x) £ b f (£) <Pn(£) d%.
The above correspondence between J{x) and its series will
not always be an equality. This can be anticipated at once
by considering the case of vectors in three dimensions. In
that case if only two vectors <pi(r), ^ 2 (r), make up the orthonormal
system, any vector not in the plane of those two could not
be represented in the form Ci<pi(r) + c 2 <p 2 (r). The reference
system here is not complete, in the sense that there is a vector
in the three-dimensional space which is perpendicular to both
of its vectors <pi and <p 2 .
Likewise in formula (3), if f(x) is orthogonal to every member
<p n (x) of the system, every term in the series on the right is zero,
and so the series does not represent f(x).
If there is no function in the space considered which is orthog-
onal to every <p n (x), the system {vn(,x)\ is called complete . So
the system must necessarily be complete if all functions are
to be represented by their generalized Fourier series with respect
to that system.
PROBLEMS
1. Show that the set \\^2/L cos ( mrx/L )} {n = 1, 2, • • * ) is ortho-
normal in the interval (0, L), but not complete without the addition of a
function corresponding to n = 0.
2. Show that the system {sin nirx] (n = 1, 2, * • • ) is orthogonal
but not complete in the interval ( — 1, 1).
21. Approximation in the Mean. Let K m (x) represent a finite
linear combination of m functions of an orthonormal set { <Pn(x)\
(n = 1, 2, ■ • • ; a ^ x ^ 6); that is,
(1) K m {x) = ym(x) + 72 ^ 2 ^) + * * * + 7
Sec. 21] ORTHOGONAL SETS OF FUNCTIONS 41
The values of the constants y n can easily be found for which
Km(x) is the best approximation in the mean to any given function
f(x ) ; this means the best approximation in the sense that the
value of the integral
(2) J = £ [f(x) - K n (x)Y dx
is to be as small as possible; it is also the approximation in the
sense of least squares.
Writing c n for the Fourier constants of f(x) with respect to
<Pn( X )>
Cn = £ f(x)<p n (x) dx,
the integral J can be written
J — f a lf ( x ) ~ yi<pi(z) ~ 72^2(2) — • • • — r m<p?n(x)] 2 dx
= fa + Tl + 7i + ■ ' ' + jl
- 2 T ,Ci - 2 -y id — • • • - 27 m C m .
Completing the squares here by adding and subtracting rf,
cl, ' ' ' A, gives
( 3 ) J = f b [f(x)] 2 dx - c\ - 4 - • • • - d + ( Tl - o,)2
+ (72 — c 2 ) 2 + • • • + (7,,, - c m y-.
It is clear from (2) that J ^ 0, so it follows from equation (3)
that J has its least value when 71 = c lt y 2 = * * • y m = c m .
The result can be stated as follows:
Theorem 1. The Fourier constants of a function f(x) with
respect to the functions <p 1 (r), <p' 2 (x), • • • , (p m (x) of an ortho normal
set are those coefficients for which a linear combination K m (x)
of these functions is the best approximation in the mean to f(x), in
the fundamental interval (a, b).
Since / S 0, it follows from equation (3), by taking y n = c ni
that
W cl + 4 + ■■■ + d S £ [f(x)]* dx.
This is known as Bessel’s inequality. The number on the right
is independent of m\ so it follows that the scries of squares of th©
42 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 22
Fourier constants of any function
c? + d+ • • • + c* + • • • =X C *
1
always converges; and its sum is not greater than the norm of
/c*o,
(5) =
It follows that the Fourier constants of every function correspond-
ing to any orthonormal system {<p n ) approach zero as n tends to
infinity:
(6) lim c n = 0;
n— > oo
because a necessary condition for the convergence of the series
in (5) is that its general term c\ approaches zero as n becomes
infinite.
22. Closed and Complete Systems. Let S m (x) be the sum of
m terms of the generalized Fourier series corresponding to f(x),
with respect to an orthonormal set of functions {tp n ) (n = 1,
2, * • ■ ) ; that is,
m
(1) " S m (x) = 5} C n <Pn{x).
This is the sum K m (x) in the last section when y n = c n .
The sum S n (x) is said to converge in the mean to the function
m if
(2) lim p [ f{x ) — *S m (a;)] 2 dx = 0.
m — * so
This is also written
l.i.m. S m (x) = f(x),
m—> oo
where the abbreviation l.i.m. stands for limit in the mean .
If the relation (2) is true for each f(x) in the function space
considered, the system is said to be closed .* According
to Theorem 1, then, the system is closed if every function can be
* The definitions of the terms closed and complete (Sec. 20) given here
are those most commonly used today. Many German writers use the term
closed ( abgeschlossen ) to denote what we have called complete, and complete
(vollstandig) for our concept of closed.
Sec. 22] ORTHOGONAL SETS OF FUNCTIONS 43
approximated arbitrarily closely in the mean by some linear
combination of the functions <p n (x ).
By expanding the integrand in equation (2) and keeping the
definition of c n in mind, we have
i m m
I fa dX ” 2 2/ C « + X C *[ == °*
Hence for every closed systern it is true that
( 3 ) X c » = Jf r /(*)] 2 dx -
This is known as ParsevaVs theorem. When written in the
form
(4) X & = N ^’
T
it identifies the sum of the squares of the components of /, with
respect to the reference vectors <p n , with the* norm of /.
Suppose B(x) is a function which is orthogonal to every func-
tion of the closed set. Substituting it for / in equation (4)
gives N(8) = 0, so that 0(.r) cannot belong to the function
space; and thus it is shown that the set is complete (Sec. 20).
The following theorem is therefore established:
Theorem 2. If the set {(Pn(z)\ i* closed , it is complete.
It is an immediate consequence that if there is a function
which is orthogonal to every member of the set, the set cannot be
closed.
This is only a bare introduction to a general theory which
has been developed extensively in recent years. To carry it
further (even to prove the converse of Theorem 2), a broader
class of functions and the idea of the Lebesgue integral are needed.
But the term “closed” was defined hero with respect to con-
vergence in the mean, and this type of convergence does not
guarantee ordinary convergence at any point. That is, the
statement (2) is quite different from the statement of ordinary
convergence :
lim & m (x) = f(x) (a g x ^ b).
m— > oo
It is this ordinary convergence, and the concept of closed orthog-
44 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 23
onal sets with respect to it, which are usually needed in the
applications.
No general tests of a practical nature exist for showing that a
set of functions is closed. That is another reason for deserting
the general theory at this point.
23. Other Types of Orthogonality. Some of the important
extensions of the concept of orthogonal sets of functions should
be noted.
a. A set {0n(a;)} (n = 1, 2, • • • ) is orthogonal in an interval
(a, b ) with respect to a given weight function p{x) y where it is
usually supposed that p{x) ^ 0 in (a, b ), if
(1) p(x)g m (x)g n (x) dx = 0 whenra n (m, n = 1, 2, • - • ).
The integral on the left represents the inner product (g m , g n )
with respect to the weight function, a generalization of the inner
product of vectors in terms of components with respect to axes
along which different units of length are used (Sec. 18). The
norm of g n {x) in this case is, of course,
N(g n ) = ( g n , g n ) = JT p(x)[g n (x )] 2 dx (n = 1, 2, • • • ).
By multiplying each function g n of the set by the normalizing
factor [N(g n )]-* } the corresponding orthonormal set is obtained.
This type of orthogonality can be reduced at once to the ordi-
nary type having the weight function 1. It is only necessary
to use the products y/ p(x)g n (x) as the functions of the system;
then equation (1) shows that the system so formed has ordinary
orthogonality in the interval (a, b).
An important instance of orthogonality with respect to weight
functions will be seen in the study of Bessel functions later on.
The Tchebichef polynomials ,
(2) T n (x) = 2^1 cos ( n arccos x ), T 0 (x) = 1
(n = 1, 2, ■ - ■
also form a set of this type. This set is orthogonal in the interval
( — 1, 1) with respect to the weight function
p(x) = (1 — x 2 )~l.
Sjuc. 23] ORTHOGONAL SETS OF FUNCTIONS 45
This is easily verified by integration; thus,
J 1 dx 1 C r
T m (x)T n (x) = J o cos me cos nd de
= 0 if m ^ n.
6. Another extension of orthogonality applies to a system of
complex functions of a real variable x (a g x g 6). A system
consisting of the functions gJx), where
g n (x) = w„(a;) + iv n (x),
is said to be orthogonal in the Hermitian sense if
(3) fa ffm(x)g n (x) dx = 0 when m 9 ^ n,
where g n (x) = u n (x) — iv n (x), the conjugate of g n . The system
is normed if
J[ 6 9n(x)g n (x) dx = 1;
that is, if
£ Wl{x) + vl(x)] dx = 1
for every n.
When the functions are real, v n (x) = 0, and this type reduces
to the ordinary orthogonality.
Imaginary exponential functions furnish the most important
examples of such systems. For instance, the functions
(4) e inx — cos nx + i sin nx (n -= 0, ±1, ±2, • • • )
form a system which is orthogonal on the interval (— tt, tt) in the
above sense. The proof is left as a problem.
c. Extensions to cases in which the fundamental interval is
infinite in length arc obtained by replacing a by — 00 or & by 00 ,
or both.
d. For systems [g n (x, y)} (n = 1, 2, • • • ) of functions of
two variables, the fundamental interval is replaced by a region
in the £ 2 /-plane, and the integrations are carried out over this
region. Similar extensions apply when three or more variables
are present. Weight functions may be introduced in such cases
too, as well as in case c.
46 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 24
PROBLEMS
1. By using the binomial expansion and equating real parts in the
well-known formula
(cos 6 + i sin d) n = cos nd + i sin nd ,
obtain the identity
cos nd = cos n 6 —
cos n_2 6 sin 2 6 -f
cos n ” 4 6 sin 4 6 — * * • + A n ,
where
are the binomial coefficients, and A n
i n sin 71 6 if n is even,
A n = ni n ~ l cos 0 sin 71- * 1 6 if n is odd. Hence show that the functions
T n (x) defined by equations (2) above are actually polynomials in x of
degree n.
2. Prove that the system of exponential functions in equation (4) of
this section is orthogonal on the interval ( — t, t) in the Hermitian sense.
3. Prove that the system ^exp
/ 2 mrix\ \
[b-a)j
(n
0 , ± 1 , ± 2 , • • * ),
where exp(w) denotes e u , is orthogonal in the Hermitian sense on the
interval (a, b).
24. Orthogonal Functions Generated by Differential Equa-
tions. In solving the problem of displacements in a stretched
string in Sec. 13, we used particular solutions of the partial
differential equation of motion which vanished when x = 0 and
x — L. In order that y = X(x)T{t) be such a solution, it was
found that the function X{x) must satisfy the conditions
(1) X"(x) + \X(x) = 0,
(2) X(0) = 0, X(L) = 0,
for some constant value of X, denoted there by —7.
Equations (1) and (2) form a homogeneous boundary value
problem in ordinary differential equations containing \ as a
parameter. Since the solution of equation (1) that vanishes
when x = 0 is X = C sin \/Xx, the problem has solutions not
identically zero only if X satisfies the equation
(3) sin V^L = 0.
Therefore X = n 2 7r 2 /L 2 (n = 1 , 2, • • • ), and the corresponding
solutions of the problem (l)-(2) for these values of X are,
Sec. 24]
ORTHOGONAL SETS OF FUNCTIONS
47
except for a constant factor, sin ( mrx/L ). These functions were
shown to form an orthogonal set on the interval (0, L).
Corresponding results can be found in much more general
cases. When applied to a more general partial differential
equation, separation of variables will yield an equation in X(x)
of the type
X" +Mx)X' + [Mx) + \f*(x)]X = 0.
Here/ 1 ,/ 2 , and/ 3 are known functions involved in the coefficients
of the partial differential equation, and X is the constant which
arises upon separation of variables.
When the last equation is multiplied through by the factor
r(x ), where
r(x) = eJViWrf*
it takes the form
(4) Tx [ r{x) 2] + [«(*> + x ^ x = °>
known as the Sturm-Liouville equation.
The boundary conditions on X(x) may have the form
(5) aiX(a) + a 2 X\a) = 0, b.Xib) + b 2 X'(b) = 0,
where a Xj a 2 , b i, and b 2 arc constants.
The problem composed of the differential equation (4) and the
boundary conditions (5) is called a Sturm-Liouvitte problem or
system , in honor of the two mathematicians who made the first
extensive study of that problem. *
Under rather general conditions on the functions p , q, and
r, it can be shown that there is a discrete set of values Xj, X 2 , * * *
of the parameter X for which the system (4)-(5) has solutions
not identically zero. These numbers X n are called the character-
istic numbers of the system. In the above special case — equations
(1) and (2) — they are the numbers nV 2 /L 2 , the roots of the char-
acteristic equation (3).
The solutions X n (x) (n = 1, 2, • • • ), obtained when X = X n
in equation (4), are the characteristic functions of the Sturm-
Liouville problem. These are the functions sin (mrx/L) in the
special case.
* Papons by Liouvillo and Sturm on this problem will be found in the first
three volumes of Journal de matk&maliquc. , 183G-1838.
48 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 24
It will be showx in the following section that the set of func-
tions jZnOc)} (n = 1, 2, • • • ) is orthogonal on the interval
(a, b) with respect to the weight function p(x).
Moreover, it can be shown that any function f(x ), defined in
the interval (a, b ) and satisfying certain restrictions as to its
continuity and differentiability, is represented by its generalized
Fourier series corresponding to that orthogonal set of functions.
That is, if (p n (x ) is the function obtained by normalizing X n (x),
the series
2) C n <p n (x),
where
P (?&)f (**0 Vn (%) dx,
converges to the function f(x) in the interval (a, b). It should
be noted that the normalizing factor for X n here is the number
(£vxid X y.
When r(b ) = r(a), the statements made above are also true
when the boundary conditions (5) are replaced by the conditions
(7) X(a) = X(b), X'(a) = X'(b),
called the periodic boundary conditions. Conditions of this
sort frequently arise when x represents a coordinate such as the
angle 0 in polar coordinates, or cos 0.
The proof that series (6) converges to the function f(x) is
quite long and involved, as we may well expect in view of the
fact that the coefficients in differential equation (4) are arbitrary
functions of x. The proofs generally make use of the theory of
functions of complex variables, or the comparison of the expan-
sion with a Fourier series, or both. The development of a general
expansion theorem, along with other interesting and useful
results in the general theory of Sturm-Liouville systems, is
beyond the scope of the present volume.*
The expansions considered in the chapters to follow are all
special cases of the general theory. But in two of the important
cases, those of Bessel and Legendre functions, the equations are
* -A- treatment of these topics is included in a companion volume, now
being prepared by the author, on further methods of solving partial differen-
tial equations. Also see Ince, “Ordinary Differential Equations,” pp.
235#., 1927; and the references listed at the end of the present chapter.
Sec. 25]
ORTHOGONAL SETS OF FUNCTIONS
49
singular cases of equation (4) which must be treated separately
even in the general theory. Hence the plan of presentation
followed here is not especially inefficient.
Many of the important sets of orthogonal functions are gen-
erated in the above manner, as solutions of a homogeneous
differential system involving a parameter. The expansion
theorem shows that these sets are closed with respect to ordinary
convergence, rather than convergence in the mean, so that we
have an important advantage over the general theory discussed
in the preceding sections.
25. Orthogonality of the Characteristic Functions. Two
theorems from the general theory of Sturm-Liouvillc systems
can easily be established here. They will be useful in the
following chapters. The first shows the orthogonality of the
characteristic functions, and the second shows that the char-
acteristic numbers arc real. The existence of such functions
and numbers will be established in each case treated later on, of
course, by actually finding them.
Theorem 3. Let the coefficients p, q, and r in the Sturm-Liouville
problem be continuous in the interval a ^ x g b, and let X w , \ n be
any two distinct characteristic numbers , and X m (x), X n (x) the
corresponding characteristic functions , whose derivatives X' m (x) y
X' n (z ) are continuous. Then X m (x) and X n (x) are orthogonal on
the interval (a, b ), with respect to the weight function p(x).
Furthermore , in case r(a) = 0, the first of the conditions (5),
Sec. 24, can be dropped from the problem , and if r(b) = 0 the
second of those conditions can be dropped. If r(b) = r(a) f those
conditions can be replaced by the periodic conditions (7), Sec. 24.
Since X m and A r n arc solutions of equation (4), Sec. 24, when
X = A w and X = X n , respectively,
(rX'J + (q + \ m p)X m = 0,
~ (rXJ + (q + Kp)X n = 0.
Multiplying the first equation by X n and the second by X m , and
subtracting, gives
(X m - \ n )pX m X n = X, A (rX'J - X. (rXJ
= [(rXJX,„ - (rXJXJ.
50 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 25
Integrating both members over the interval (a, V),
(1) (Xm - X„) £ pX m X n dx = [r(X m X' n - XnX'J ]*.
In the special case when = b% = 0, the boundary conditions
(5), Sec. 24, become
( 2 ) X{a) = 0, X(b) = 0.
Since both X m (x) and X n (x) then satisfy these conditions, it is
evident that the right-hand member of equation (1) vanishes.
But X m — X n 0, so that
(3) £ p(x)X m (x)X n (x) dx = 0,
which is the statement of orthogonality between X m and X n .
In case r(a) = 0, it is clear that (3) follows from (1) without
the use of the first of the conditions (2). Similarly, if r(b) = 0,
the second condition is not needed.
The proof that equation (3) follows when the general boundary
conditions (5), Sec. 24, or the periodic boundary conditions, are
substituted for (2), will be left for the problems.
Theorem 4. If in addition to the conditions stated in Theorem 3
the coefficient p(x) does not change sign in the interval ( a , b), then
every characteristic number of the Sturm-Liouville problem is real . *
Suppose there is a complex characteristic number X, where
X = cx. + i^
Let
X(x) = u(x) + iv(x)
be the corresponding characteristic function. Substituting this
in the Sturm-Liouville equation, we have
d
fa ( ™ ' + irv ') + (q + ap + i$p) (u + iv) = 0.
Equating the real and imaginary parts to zero, separately,
fa (™0 + (? + ap)u — /Spy = 0,
d
fa ( rv ') + (Q + ap)v + Ppu = 0,
* The functions p , q , and r are assumed to be real.
Sec. 25] ORTHOGONAL SETS OF FUNCTIONS 51
and, upon multiplying the first of these equations by v and the
second by u , and subtracting, it follows that
— P(u 2 + v 2 )p = u ^ ( rv' ) — v ( ru ')
= ^ [(rv')u - ( ru')v ].
Consequently, an integration gives the relation
(4) — (u 2 + v 2 )p dx = ( uv ' —
Again let us complete the proof here when the boundary condi-
tions of the Sturm-Liouville problem have the special form (2).
Since our characteristic function u + iv satisfies (2), its real
and imaginary parts must each vanish when x = a and x = b.
The right-hand member of (4) therefore vanishes. But if the
function p(x) in the integral does not change sign in the interval
(a, b) 7 the integrand itself cannot change sign, and so the integral
cannot vanish. It follows that ft = 0, and therefore the char-
acteristic number X is real.
As before, if r(a) = 0, the first of conditions (2) is not needed,
and if r(b) — 0, the second can be dropped.
The argument is not essentially different when the more
general boundary conditions (5), Sec. 24, or t.he periodic con-
ditions are used. This matter is left for the problems.
PROBLEMS
1. Complete the proof of Theorem 3 when the boundary conditions
are (a) the conditions (5), Sec. 24; (/>) the periodic conditions (7), Sec. 24,
assuming that r(b) = r(a).
2. Complete the proof of Theorem 4 when the boundary conditions
are (a) the conditions (5), Sec. 24; (/>) the periodic conditions (7),
Sec. 24, assuming that r(b) = r(a).
Find the characteristic functions of each of the following special cases
of the Sturm-Liouville problem. Also note the interval and weight
function in the orthogonality relation ensured by Theorem 3, and find
the normalizing factors.
3. X" + XX = 0; X'(0) = 0, X f (L) = 0.
A ns. X n = cos ( mrx/L ) (n = 0, 1, 2, • • ■ ).
4. X" + XA r = 0; X ((»_= 0, X'(L) = 0.
Am. ip n = V2/Lsin [(2 n — \)ttx/( 2L)] (n = 1, 2, • * - ).
52 FOURIER SERIES AND BOUNDARY PROBLEMS (Sec. 25
5. X" + XX = 0; X(tt) - X(-tt), X'(tt) = X'(-tt).
Ans. {1, cos a, cos 2a;, • • • , sin a;, sin 2a;, • * * }.
6. X" + XX = 0; X(0) = 0, X'(l) + hX(l) = 0, where h is a con-
stant. Show in this case that X n = sin a n x, where a n represents the
positive roots of the equation tan a = —a/h, an equation whose roots
can be approximated graphically. Also show that X n is normalized by
multiplying it by V2 h/(h + cos 2 a n ).
7. (d/dx)(x*X r ) + \xX = 0; X(l) = 0, X(e) = 0. Note that the
equation here reduces to one of the Cauchy type after the indicated
differentiation is carried out.
Ans. (p n = (a/ 2 / x ) sin (mr log x) (n — 1, 2, ■ • • ).
REFERENCES
1. Courant, R., and D. Hilbert: “Methoden der mathematischen Physik,”
Vol. 1, 1931.
2. Mises, R. v.: “Die Differential- und Integralgleichungen der Mechanik
und Physik” (Riemann- Weber), Vol. 1, 192,5.
CHAPTER IV
FOURIER SERIES
26. Definition. The trigonometric series
(1) |a 0 + (ai cos x + b i sin x) + (a 2 cos 2x + b 2 sin 2x)
+ * • * + (a n cos nx + b n sin nx) + ■ • ■
is a Fourier series provided its coefficients are given by the
formulas
( 2 )
(In
b,
= 1 T /(*) e
If J — v
.-if
T J-7T
cos nx dx
f(x) sin nx dx
(n = 0, 1, 2,
(n = 1, 2,
),
),
where f(x) is some function defined in the interval (—7 r, 7r). In
particular, series (1) with the coefficients (2) is called the Fourier
series corresponding to f(x) in the interval (— t, 7r), written
(3) /(*)
+ 2^
(a n cos nx + b n sin nx)
(— 7T < X < 7r).
Formulas (2) for the coefficients arc special cases of those
for the generalized Fourier series in the chapter preceding. The
functions 1, cos x, sin x, cos 2x , sin 2x } • • * constitute an orthog-
onal (but not normalized) set in (— tt, 1 r). This was noted in
Prob. 5, Sec. 25; but we can easily show it here independently.
For if m, n = 0, 1 , 2, • • • , then
J* cos mx cos nx dx = 0,
J sin mj; sin nx dx = 0, if m 5* n f
and whether m and n are distinct or not,
cos mx sin nx dx = 0.
53
FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 26
When, m = n the first two integrals become
cos 2 nxdx = r
= 2 T
sin 2 nx dx = t.
if ft 0,
if ft = 0;
Considering (3) as an equality and multiplying first by
and integrating therefore gives formally
cos nx
fl T fi x ) cos nx dx = 7m„ (n = 0, 1, 2, • • ■ ).
Similarly, multiplying by sin nx and integrating gives
fl T fi x ) sin nx dx = irb n (» = 1, 2, • • • )•
These are formulas (2) for the Fourier coefficients.
Again, the corresponding orthonormal set of functions is
1 cos x sin x cos ‘lx sin 2x
V2t Vir Vir ’ Vir ’ ’
and the Fourier constants c„ of f(z) corresponding to these
functions are the integrals of the products, or the inner products,
? \ . eSe ^ unc ^ ons by f(x). So the Fourier series, with respect
to this set, corresponding to Six), is
/(*)
v&L.
'\/2t
f(x') dx'
cos nx' , , cos nx
V? vT
+ r kx') s ^dx' s ^ i
J-ir V 7T V 7T J
where x' is used for the variable of integration. This can be
written
(4) Six) ~ g- j_J(x') dx' + i 2 [cos nx J* Six') cos nx' dx'
+ sin nx L Six') sin nx' d*'l;
which is the same as series (3) with the coefficients (2).
Note that the constant term is the mean value of f(x) over the
interval (— x, vr).
27. Periodicity of the Function. Example. Every term in
the above series is periodic with the period 2i r. Consequently,
if the series converges to f(x) in the interval (-t, tt), it must
converge to a periodic function with period 27 r for all values of x.
Thus it would represent /(x) for every finite value of x , provided
Fig. (>.
the definition of /(x) is extended to include all values of x by
the periodicity relation
f(x + 2tt) = /(*).
Thus the Fourier scries may conceivably serve either of two
purposes: (a) to represent, a function defined in the interval
(— 7r, 7 r), for values of x in that interval, or (/;) to represent a
periodic function, with period 2tt, for all values of x. It clearly
cannot represent a function for all values of x if that function is
not periodic.
The particular interval (—t, tt) was introduced only as a
matter of convenience. We shall soon see that it is easy to
change to any other finite interval.
It is not necessary that f(x) be described by a single analytic
expression, or that it be continuous, in order to determine the
coefficients in its Fourier series. Of course the mere fact that
the series can be written does not ensure its convergence or, if
convergent, that its sum will be /(x). Conditions for this are
to be established in Sec. 33.
56 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 27
Example. Write the Fourier series corresponding to the func-
tion /(s) defined in the interval — r < x < * as follows:
f(x) = 0
= x
when —t<x^0,
when 0 g x < w.
The graph of this function is indicated by the heavy lines in
*lg. b. ihe Fourier coefficients are
do
dn
1 C T i p*
~ Z I f( x ) COS nxdx = - I
r J - T * JO
cos nx dx
cos nx + nx sin nx
cos mr
n
(- 1 )* .
sin nx
x cos nx dx
r ]o = (C0S nx - 1 )’
~ t I s ' n nx dx = - I x sin nx dx
J ~ T * Jo
= ^2 |^ s i n n x — nx cos nx j =
The series is therefore
(i) /(*)^; + >br(-i)--i
w ^2 cosnx-
= | + (sin x - 2 cos a;) - i s in 2x
+ (| sin 3a: - |- cos 3a:) - i sin 4a: + • • • .
If converges to /(*) when - r < x < T , it also converges
or all other values of x to the periodic function represented
by the dotted lines in the figure. Note that this periodic func-
sZelbvZ tmU0U \ at V ±T ’ ±3t ’ • * • ’ the vaIue ^pre-
sented by the senes at such points will be found later.
As an indication of the convergence of the series (1) to f(x)
it is instructive to sum a few terms of the series by composition
t o™ be ,0U " d - ,0r tata “ ce '
7T e 2
= j + sin x — - cos x
* r
1 • o
2 sm
Sec. 28]
FOURIER SERIES
57
is a wavy approximation to the curve shown in the figure.
The addition of more terms from the series generally improves
the approximation.
PROBLEMS
Write the Fourier series corresponding to each of the following func-
tions defined for In a few of the problems, sum a few
terms of the series graphically; also show graphically the periodic func-
tion which is represented by the series provided the series converges to
the given function.
1. fix) = x when —t < x <tt. (Also note the sum of the series
when x = ±i r.)
2. f(x) = e x when —7 r < x < ir.
2 sinh 7 r
A ns
2
n
sin nx.
A ns.
\ + 2 r+T 2 ( c,oB nx “ n sin
l
3. fix) = 1 when —tt<x<0; fix) = 2 when 0 < x < ir.
a 3 , 1 X? 1 r- ( — l)"' .
Am - 2 + t^J sinr
1
4. f{x) = 0 when —tt<x< 0; /(;r) = sin x when 0 < x < r.
00
1,1. 2 cos 2 nx
_ + - sm * - - 2j 4^-“T
1
A ns.
28. Fourier Sine Series. Cosine Series. When /( — #) =
is called an odd function; its graph is symmetric with
respect to the origin, and its integral from —w to ir is zero. When
f( — x) = f{x), the function is wen; its graph is symmetric to
the axis of ordinates, and
Jy /O) dx = 2 J['/(.r) dx.
As examples, the funetions x, x : \ and r- sin lex are odd, while
1, x 2 , cos lex, and x sin lex are even.
Although most funetions are neither even nor odd, every
function can bo written as the sum of an even and an odd one by
means of the identity
f(x) = i[f(x) +/(-*)! + *[/(*) - /(-*)]•
( 1 )
58 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 28
When f(x) is an odd function defined in ( — t, 7 r), formulas (2),
Sec. 26, for its Fourier coefficients become
a n = 0 (n = 0, 1, 2, • • • )»
2 f T
= - I f(x) sin nx dx (n = 1, 2, • • • ).
Hence its Fourier series reduces to
(2) /(a;) ~ ~ sin na: J* /(a; 7 ) sin ra;' da;'.
The series in (2) is known as the Fourier sine series. It can
clearly be written when fix) is any function defined in the
interval (0, t), provided the integrals representing its coefficients
exist. Furthermore, when f(x) is defined only in the interval
(0, 7r), an odd function exists in (—7 r, 7 r) which is identical with
f(x) in (0, 7r) . If that odd function is represented by its Fourier
series, so is fix) in (0, t). Thus the question of convergence
of the sine series to fix) in (0, t) depends directly upon the
conditions of convergence of the series in the last section.
Similarly, when fix) is an even function defined in the interval
(— 7r, 7 r), the coefficients in its Fourier series are
2 r x
a n = - I fix) cos nx dx (n = 0, 1, 2 • • ■ ),
t jo
b, = 0 (n = 1, 2, • • • ) ;
and the series becomes the Fourier cosine series
(3) /(s) ~ ^ f(x ') dx' + ^ cos nx J f(x') cos nx’ dx'.
When /Or) is defined only in (0, tt), this series can be written,
in general, and again the conditions under which it converges
to /(a;) will be known when the conditions are found for the more
general series in the last section.
For functions defined in the interval (0, if), then, both the sine
series and the cosine series representations can be considered.
As indicated earlier, these are the series corresponding to /(a-)
with respect to two different sets of functions, sin nx],
and {l/Vw, y/2/x cos nx) (n = 1, 2, • • • ), each of which is
Sec. 29 ]
FOURIER SERIES
59
orthonormal in (0, t). The series (2) and (3) can be written
more easily from this viewpoint; but in the theory of these series
it is important to consider them as special cases of the series in
Sec. 26.
Every term of the sine series is an odd function. So if the
series converges when 0 < x < w, it must converge to an odd
function with the period 2t for all values of x.
Similarly, if the cosine series converges, it must represent an
even periodic function with period 2t.
29. Illustration. Let us write («) the Fourier sine series, and
( b ) the Fourier cosine series, corresponding to the function /(x),
defined in the interval 0 < x < r as follows:
f(x) = x when 0 ^ x <
£j
7T
= 0 when - < x ^ w.
Z
a. The coefficients in the sine series are
b n — - I /Or) sin nx <lx = - I x sin nx dx
7T jo n Jo
1 . nir ht\
= r t l 2 Sill -X UT COS 7T 1;
7 Tti" \ Z Z /
so the series is
/W
2 / 2 . 'Hit it rnr\ .
t \n 2 KIU "2" ~ n <! ° 1 * * * S 2 ) sm UX
\ ( 7 f ^ /i
= - ( 2 sin a; + x sin 2a: — - sin 3a; — T sin 4a; +
7T \ Z
2
9
7T
4
>
If this sine series converges to our function f(x :), it must also
represent the odd periodic extension of f(x) shown in Fig. 7.
60 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 29
b. The coefficients in the cosine series are
2
CLq = —
T
X dx = - A y
4
= - I f(x) cos nx dx = - I x cos nx dx
* JO 7T JO
1 / 0 nir .nr 0 \
= iro* \ 2 C0S T + nir sm "2 - 2 /‘
/./ N 7T 1 / 2 n 7 r. 7 r.n 7 r 2 \
/W~8 + i2i^ cos T + ; m T-^j
Therefore
«■ , 1 \ ,
8 ir L T _
2) cos x — cos 2x
-(i+D
cos nx
cos Sx +
Assuming its convergence to f(x) } this cosine series would
converge for all x to the even periodic function shown in Fig. 8.
PROBLEMS
Find (a), the Fourier sine series, and ( b ) the Fourier cosine series,
corresponding to each of the following functions defined in the interval
(0, 7r). Assuming that each series represents its function within that
interval, show what function it represents outside the interval.
1. f(x) = x when 0 < x < it. (Compare Prob. 1, Sec. 27.)
00 J0
Ans. (a) 2 "S’ f-iu+i ^ nx - fh\ Z _ i ^ cos (2n - l)x
^ } n ’ w 2 T Zx ~(2n - 1)* "
2. f(x) = sin x when 0 < x < tt.
3. f(x) = cos x when 0
Ans.
(a) sin x; (b) - - -
7 r 7T
1
cos 2nx
4 ri l — 1
< X < 7T.
Ans.
n sin 2 nx „ ,
~ 4 n 2 _ 1 ; ( 0 ) cos 3 .
Sec. 30]
FOURIER SERIES
4 . = 7 i- — x when 0 < x < tt.
, „ „ sin nx , n ir , 4 ^ cos (2n - 1)*
Am. (a) 2 ^ — fr ~, + - 1)2 •
l i
5. f(x) = 1 when 0 < x < x/2, f(x) = 0 when x/2 < x < x.
. / \ 2 rwr\ sin Tia;
Ans. (a) - 2i y - cos t) ~r~ ;
1
6. fix) = x when 0 < x < ir/2, fix) = tt — x when tt/2 < x <tt.
, \ a ^A^ ,ia 8 ^ # cos (4?^ 2)s
(a) (Compare Sec. 14); (6) j ^ "
7. /(a;) = e* when 0 < x <ir.
, * 2 ^ ri , nN 1 u sin na;
s * ^ x ^ n 2 + 1 ;
l
... e* - — 1 2 "ST* ri ... .cos nx
(b)
8. Obtain series (4), Sec. 2(5, for any function in ( — 7r, tt) from series
(2) and (3), Sec. 28, for odd and even functions, respectively, using
identity (1), Sec. 28.
30. Other Forms of Fourier Series. The Fourier series cor-
responding to any function F(z), defined in the interval
7T < Z < 7T,
J* F{z') dz' + ~ J^coh nz J F{z r ) cos nz r dz r
+ sin nz l /(*') •
sin nz f dz r .
Substituting the new variable x and the new variable of inte-
gration x f throughout, where
Lz . Lz '
X = ; X = 7
7 T 7T
and writing fix) for F(irx/L) } ihe above correspondence becomes
62 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 30
( 1 ) /(*)■
, 1 "sh r nirz C L
+ r^L c “ s — J_,
fix') cos dx'
•L h
i • n irx C L . rnrx ' , ,1
+ sin —j~ J L f\ x ) sin — cfo' .
The series in (1) is the Fourier series on the interval ( —L, L )
corresponding to any function f(x) defined in that interval.
The same substitution changes the sine series to one cor-
responding to a function /(a;) defined in the interval (0, L), or an
odd function in (— L, L):
( 2 ) f(x) ~ sin J f{x r ) sin dx'.
It also changes the cosine series to the form
(3) f(x) ~ 1 £ f(x’) <b'+|2“ B T J[V) cos dx',
corresponding to a function f(x) defined in the interval (0, L), or
an even function in (-L, L). The substitution simply changes
the unit of length on the 3 -axis.
Of course the forms (1) to (3) can also be written by noting
the orthogonality of the sine and cosine functions involved there
in the interval (— L, L ) or (0, L). Let us obtain the series for
any interval (a, b) in this manner.
Upon integrating, it will be found that
f 6 ( 2mirix\ ( 2nirix \ ,
J. exp [b^J ,xp *
That is, the set of complex functions
/ (2mrix\\
(»- 0 , ± 1 , ± 2 , ■ • ■ ),
= 0 if m t* —n y
= b — a if m = ~n.
is orthogonal on the interval (a, &), in the Hermitian sense.
Sec. 30]
FOURIER SERIES
63
Assuming a series representation of f(x) in terms of those
functions,
/(*) = 2 C - exp (K) o<*< *>.
the coefficients C„ can be found formally by multiplying through
by exp [—2 mirix/{b — a)] and integrating. In view of the above
orthogonality property, this gives
J fix) exp dx = (b — a)C m .
Thus, the exponential form of the Fourier series corresponding
to a function fix) defined in the interval (a, b) is
Grouping the terms for which the indices n differ only in
sign, (4) takes the trigonometric form,
<«>
2mr(x' — x)
+
of the Fourier scries corresponding to jf(rr) in (a, b). This can
be obtained as well from the earlier form (5), Sec. 26, for the
interval (— 7 r, tt) by making a linear substitution in the variables
x and x
These additional forms of the series, therefore, arise from the
original form for the interval (— t, 1 r) by changing the origin
and the unit of length on the rr-axis. So it is only necessary to
develop the theory of convergence of the series for the interval
(-7T, tt); the results will then be evident for the other forms.
Form (5) contains the earlier forms as special cases. The
series represents a periodic function with period ( b - a), if
it converges. Therefore it can be considered as a possible
expansion of either a function which is periodic with period
(b - a), or a function which is defined only in the interval (a, b).
Both types of applications are important. In the second case,
64 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 31
however, there may be many Fourier series representations of
the function; for the function can be defined at pleasure in
any extension of the interval, and the new series in the extended
interval may still represent its function. It would then represent
the given function in (a, b ) .
# PROBLEMS
1. Write formula ( 1 ) in the form corresponding to ( 3 )-( 2 ), Sec. 26;
also in the form corresponding to (5), Sec. 26.
2 . Write formula (5) when a = 0 , b = 2 L, and compare it with for-
mula ( 1 ).
Write the Fourier series corresponding to each of the following
functions.
3. fix) = — 1 when —L < x < 0, fix) — 1 when 0 < x < L.
Ans
oo
4 X? 1
x 2 n — 1
• (^ n iy&
sin
4. f{x) = \x\ when — L < x < L; that is, fix) = —x in (— L, 0) and
(2 n — 1 )xx
_J (2n — l) 2 cos
l
f(x) = * in ( 0 , L). Ans. \ - § ^ ( 2 ^
5 . /(as) = x 2 when — L < x < L.
, L 2 , 4L 2 '^(-1)“ mrx
Ans - 1 + 1 ? 2i~^ cos ~T'
1
2 1 .
—5 cos nxx — - sm mrx
irn 2 n
6 . fix) = x + x 2 when — 1 < x < 1.
i x
7. /(x) = 0 when —2 < x < 1, /(x) = 1 when 1 < x. < 2.
, 1 1 XI if. 7VTT
Ans. -r > - sm it
4 x n *2
l L
■>
nxx ,
cos — +
/ nx\ . nxxl
l cos nx — cos ~2 1 sin *
8 . fix) = 1 when 0 < x < 1, fix) = 2 when 1 < x < 3, and
fix + 3) = fix) for all x.
A„.
3 x n
l L
2 nx
2 nxx , (
. 2mr\
. 2 nxx
sin
-TjT COS
— + {
l-COS-g-j
sui- 3 -
9. fix) = e* when 0 < x < 1 , using the exponential form of the
Fourier series.
31. Sectionally Continuous Functions. At this point let us
introduce some special classes of functions, the use of which will
Sbc. 31] FOURIER SERIES gg
keep the theory which is to follow on a fairly elementary level
These classes will include most of the functions which arise
in the applications; but they are rather old-fashioned classes.
As we shall point out from time to time, our principal results can
be obtained for a considerably broader class of functions by
using somewhat more advanced methods of modern analysis.
A function is sectionally continuous , or 'piecewise continuous, in
a finite interval if that interval can be subdivided into a finite
number of intervals in each of which the function is continuous
and has finite limits as the variable approaches either end point
from the interior. Any discontinuities of such a function are
of the type known as ordinary points of discontinuity. Every
such function is bounded and integrable over the interval, its
integral being the sum of a finite number of integrals of continu-
ous functions.
The symbol /(a* + 0) denotes the limit of f(x) as x approaches
x 0 from the right. For /(*# - 0) the approach is from the left
That is, if X is positive,
f(x 0 + 0) = lim f(x 0 + X),
X— 0
/(.To — 0) = lim f(x„ — X).
A— >0
We define the right-hand derivative, or derivative from the right,
of /(a:) at xo as the following limit:
lim ^ Xo Q ~ f( Xa + 0)
x-o X '
where X is positive, provided of course that this limit exists.
Similarly, the left-hand derivative is
lim /if l.~ . 0) ~ /fa ~ X)
x— ►o X '
where X is again a positive variable.
, ^ f°ll° WK c,f once that if f(x) has an ordinary derivative
f (x) at x 0 , then its derivatives from the right and left both exist
there and have the common value f'(x„). But a function may
ave one-sided derivatives without having an ordinary derivative.
For example, if
f(x) = x 2 when x ^ 0,
— sin x when x 0,
66 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 31
then /'( 0) does not exist, but at the point x = 0 the derivatives
from the right and left have the values 1 and 0, respectively.
Again, for the step function
J{x) — 0 when x < 0,
= 1 when x > 0,
/'( 0) does not exist, but its one-sided derivatives have the common
value zero.
All the functions described in the problems and examples
in this book are sectionally continuous and have one-sided
derivatives.
If two functions fix) and fix) have derivatives from the
right at a point x = x 0 , so does their product. For the right-
hand derivative of their product is the limit, as X approaches
zero through positive values, of the ratio
fijxo -j- \)fjXQ + X) — fijxp + 0)fjXQ + 0)
X
This can be written
Mx* + x) /^° + x ) . -/^° . + °)
A
+ + 0 )
fi(xo + X) — fijxo + 0)
The limit of fi(xo + X) exists, and the limits of the two frac-
tions exist, since they represent the right-hand derivatives of
fix) and fiix) at the point x Q . Hence the limit of the ratio
representing the right-hand derivative of the product / 1/2 exists.
In the same manner it can be seen that the left-hand derivative
of the product exists at each point where the two factors have
left-hand derivatives.
One further property will be useful, in connection with our
theorem on the differentiation of Fourier series (Chap. V).
Let f(x) be a function which is continuous in an interval
a ^ x ^ 6, and whose derivative f'{x) exists and is continuous
at all interior points of that interval. Also let the limits /'(a + 0)
and f r (b — 0) exist. Then the right-hand derivative of fix)
exists at x = a, and the left-hand derivative exists at x = 6, and
these have the values /'(a 0) and/'(b — 0), respectively.
Sec. 32]
FOURIER SERIES
67
Since f(x) is continuous, and differentiable when a < x < b,
the law of the mean applies. So, for every X (0 < X < b — a),
a number 0 (0 < 0 < 1) exists such that
/( a + X) - jf(fl)
X
= f'(a + 0X).
Since f'(a + 0) exists, the limit, as X approaches zero, of the
function on the right exists and has that same value. The
function on the left must have the same limit; that is, the deriva-
tive from the right at x = a has the value f (a + 0).
Similarly for the derivative from the left at x = b.
It follows at once that if f(x) and f{x) are sectionally continu-
ous , the one-sided derivatives of f(x) exist at every 'point.
32. Preliminary Theory. In order to establish conditions
under which a Fourier series converges to its function, a few
preliminary theorems, or lemmas, on limits of trigonometric
integrals are useful. The integrals involved in these lemmas
are known as Dirichlct\s integrals.
The lemmas here will be so formulated that they can also be
used in the theory of the Fourier integral (Chap. V). There
it is essential that the parameter fc used in the first lemma be
permitted to vary continuously rather than just through the
positive integers. In the latter case ( [k = n) the limit in Lemma 1
would follow quite easily from equation (6), Sec. 21.
Lemma 1. If F(x) is sectionally continuous in the interval
a ^ x ^ bj then
(1) lim f b F(x) sin kx dx = 0.
Jfe-> oo
Let the interval (a, b) be divided into a finite number of
parts in each of which F(x) is continuous, and let ( g , h) represent
any one of those parts. Then, if it is shown that
(2) lim f h F(x) sin kx dx = 0,
the lemma will be proved.
Divide the interval (g, h) into r equal parts by the points
z 0 = g, x h x 2 , • • * , x r = h. Then the integral in equation (2)
can be written
r ~ 1 2
J F(x) sin kx dx ,
60 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 32
or
r — 1
2 0 1 £‘ +l F{ ^ sin kx dx + r ^ '*) - F &)] sin kxdx\-
Carrying out the first integration and using the fact that the
absolute value of an integral is not greater than the integral
ot the absolute value of the integrand, we find that
( 3 )
I C K r -Z\ c i
| F{x) sin kx dx S I F(x x )
0 iZ o 1 I
cos kxi— cos kxi+i
k
+
J |[^ (z) — F( Xi )] sin kx\ dx
The oscillation of F(x) in the interval ( Xi , Xi¥l ) is the difference
between the greatest and least values of the function in that
interval. Let Vr be the greatest oscillation of F(x) in any of the r
intervals (xi, x^), so that \F(x) - F( Xi )\ g Vr in each interval.
Also let M be the greatest value of |F(s)| in the interval (g, h).
Then according to (3),
sin kx dx
2M, /
£ ~r Vr[Xi + i —
- 2M J + 1J r (h — g ).
Now let r be selected as the largest integer which does not
exceed y/k. Then
2M r z <L
k ~ y/k
and this approaches zero as k tends to infinity. But r tends to
m mty with k, and so Vr approaches zero too, because the
osci ation of a continuous function approaches zero uniformly
in all intervals of length (h - g)/ r as r becomes infinite. Hence
lim
£ F(x) sin kx dx = 0;
so relation (2) is true, and the lemma is established.
Lemma 2. If F(z) is sectionally continuous in the interval
u — x = b and has a right-hand derivative at x = 0, then
Sec. 32]
FOURIER SERF EE
69
(4) l™JV W ^*_I P(+0) .
The integral in (4) can ho writ, ten as the sum
(«) IV°> =r* * + X" *,
Consider the first of these integrals. We can write
lim J fl F(+ 0) ^ tf, = /'’(+0) lun = IFC+O),
A:
since
f>jnj* rfw *
Jo M 2
The function — /'’( +0 )]/.t in the second integral in (5)
is scctionally continuous in the interval (0, h) since F(x) itself
is, and since ' 1
lim W
r—> f-o .r
exists because F(r) has a right -hand derivative at * = 0. Lemma
1 therefore applies to the second integral in (5) ; giving
rtijiLT. «.+«!> „ hlt , rf ,
k~> °a J 0 .r
The limit of expression (5) is therefore F (- f 0)r/2; hence (4) is
true and the lemma is proved.
Lemma 3. If F(x) is scctionally continuous in the interval
(a, b) and has derivatives from the right and left at a point x = Xo
where a < x 0 < b, then
0)].
<*> & J> * - l Wr. + 0) + ffe -
I h(^ integral in ((}) can bo written as the sum
.TV) d.r + f /-■(,) dx _
Substituting x f = .r (l — .r in tin* first of those integrals, and
x = x — Xo in the s(*cond, wo can write* their sum as
70 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 33
a F(xo - x') dx' + X °F(. x" + Xo) Sin J X " dx".
Lemma 2 applies to each of the integrals here, and since
and
lim F(x o — x') = F(xq — 0),
x f — >+0
lim F(x" + a; 0 ) = F(x 0 + 0),
x n — ► +0
the limit of their sum is [F{x 0 — 0) + F(x 0 + 0)]t/2 . State-
ment (6) in the lemma is therefore true.
33. A Fourier Theorem. A theorem which gives conditions
under which a Fourier series corresponding to a function con-
verges to that function is called a Fourier theorem. One such
theorem will now be established. The conditions are only
sufficient for the representation; necessary and sufficient con-
ditions are not known.
It will be convenient to consider the function as periodic with
period 2w.
Theorem 1. Let fix) satisfy these conditions: ( a ) fix + 2t) =f(x)
for all values of and ( b ) fix) is sectionally continuous in the
interval (— 7r, tt). Then the Fourier series
(1)
i a ° + ( a n cos nx + b n sin nx) 7
where
i r
a n = - I fix) cos nx dx (n = 0, 1, 2, •
■ • ),
K J-7T
(2)
l c*
b n = - fix) sin nx dx ( n =1,2, •
TT J-t
• • ),
converges to the value
m* + o) +/(*- o)]
at every point where fix) has a right- and left-hand derivative .
Condition ( h ) ensures the existence of the Fourier coefficients
defined by equations (2), since the products fix) cos nx and
fix) sin nx are continuous by segments and therefore integrable.
It was pointed out in Sec. 26 that series (1) with coefficients
(2) can be put in the form
Sec. 33]
FOURIER SERIES
71
cos [m(x r — a;)] dx f .
The sum S n {x) of the first n + 1 terms of the series can there-
fore be written
s„0) = i J_ /(x') || + 2 cos NC®' - ®)]| dx '-
Applying Lagrange’s trigonometric identity (Sec. 16) to the
sum of cosines here, we have
S n (x) = i
TT
sin [(n + - s)] , ,
2 sin [|(x' — a:)]
The integrand here is a periodic function of x' with period 2t ;
hence its integral over every interval of length 2 tt is the same.
Let us integrate over the interval (a, a + 2i r), where the number
a has been selected so that the point x is in the interior of that
interval; that is, a < x < a + 2t.
Introducing the factor (x f - x) in both the numerator and
the denominator of the integrand, we have
1 si
(3) S n (x) = - F(x') -
K J a
sin [( n + i)(af ~ ®)] dx>
x — X
where
Now
(4)
Moreover, F{x') is written as the product of two functions
each of which is sectionally continuous in every interval and
has a derivative from the right and left at the point x f = x ,
This was assumed in the theorem for the first factor f{x') }
and it is easily verified for the second. Therefore, F{x f ) is
sectionally continuous, and, according to Sec. 31, its derivatives
from the right and left exist at x f = x.
Therefore F(x') satisfies the conditions of Lemma 3 in which
xq = x and k = n + •£. Applying that lemma to the integral
in equation (3), we have
72 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 34
lim S n (x) = W(x + 0) + F(x - 0 )].
71— ► «
But according to equation (4),
Fix + 0 ) = f(x + 0), F(x - 0 ) = fix - 0),
and therefore
lim S n ix ) = i[f(x + 0 ) + fix — 0 )].
71— > 80
This is the same as the statement in the theorem.
34. Discussion of the Theorem. At any point where the
periodic function f(x) is continuous,
fix + 0) = fix - 0) = fix ) ;
hence at such a point the mean value of the limits of the function,
from the right and left, is the value of the function. If the
one-sided derivatives of f(x) exist there, the Fourier series
converges to f(x).
Suppose f(x) is defined only in the interval (— 7 r, 7r). Then
it is the periodic extension of this function which is referred
to in Theorem 1 . Consequently, if j{x) is sectionally continuous,
its Fourier series converges to the value
«/(* + 0 ) +/(* “ 0 )]
at each interior point where both one-sided derivatives exist.
But at both the end 'points x = ±7 r the series converges to the value
M/fr — 0) +/(— 7r + 0)],
provided f(x) has a right-hand derivative at x = —t and a
left-hand derivative at x = t, because that is the mean value
of the periodic function at those points.
It follows that if the series is to converge to /(— t +0)when
x — —x, or to /(x — 0 ) when x = x, it is necessary that the
function have equal limiting values at the end points of its
interval; that is,
/( — x + 0) = /(x — 0).
It was pointed out that the other forms of Fourier series
(Sec. 30) arise from the form used in Theorem 1 by changing
the unit or the origin of the variable x . The sine series and
cosine series are special cases arising when f(x) is an odd or an
Sec. 34]
FOURIER SERIES
73
even function. Consequently the Fourier theorem applies to
these series at once with the quite obvious modifications neces-
sary because of the changes in the interval.
For the series corresponding to the interval (~L, L), for
example, the theorem becomes
Corollary 1. Let f(x + 2 L) = f(x) for all x , and let f{x) be
sectionally continuous in the interval (— L, L). Then at any point
where f(x) has a right - and left-hand derivative , it is true that
(1) |[/(* + 0 )+/(*- 0 )]
where
= 2 ao +
?(-
cos
nrx
+ b n sin
mrx\
J f(x) cos dx
J ^ f(x) sin ~~ dx
(n = 0, 1, 2, • • • ),
(n = 1, 2, ■ • • ).
It should be observed here, as well as in Theorem 1, that the
existence of the one-sided derivatives is not required at all points
of the interval, but only at those points where representation (1)
is used. The function \/x~ 2 in the interval (— L, L), for instance,
does not have one-sided derivatives at x = 0. But, according
to our expansion theorem, the Fourier series corresponding to
this function must converge to \Zx^ at all points for which
— L g x <0 or 0 < x S L. At x = 0 the convergence is not
ensured by our theorem.
Again, if f{x) is defined in the interval (0, L ) and is sectionally
continuous there, its Fourier sine series
00
/o\ V , • nirx
(2) >sm
where
b n = y f f(x) sin dx (n = 1, 2, • • • ),
U Jo U
converges to i-[f(x + 0) + /(: v — 0)] at each point, a: (0 < x < L)
where f(x ) lias one-sided derivatives. Series (2) obviously
always converges to zero when x = 0 and when x = L.
74 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 35
Under the same conditions f(x) is represented by its Fourier
cosine series in the interval (0, L ) :
(3) i[/(* + 0)+/(3-0)] = la 0 + 2 a " cos ^
(0 <x <L),
where
a n = j- f(x) cos dz (ft = 0, 1, 2, • • * ).
But in view of the even periodic function represented by the cosine
series, this series converges to /(+0) at the point x = 0 when the
derivative from, the right exists at that point. It converges to
f(L — 0) at the point x = L when fix) has a left-hand derivative
at that point.
Broader conditions than these, under which the Fourier series
converges to its function, will be stated in the next chapter.
35. The Orthonormal Trigonometric Functions. Let us denote
by 4i the aggregate or space of all functions defined in the
interval (— L, L) which are sectionally continuous there and
which possess right- and left-hand derivatives at all points,
except the end points, of the interval. At the end points let
the derivatives from the interior exist. Also let every function
of the class A L be defined at each point x of discontinuity to
have the value ■§[/(# + 0) +/(a; — 0)], and at the end points
x = ± L to have the value %[f(L — 0) +/(—L + 0)].
Then, according to Corollary 1, for every function /(x) belong-
ing to the class A h there is a series (the Fourier series) of the
functions sin (nirx/L), cos ( mrx/L ) which converges in the ordi-
nary sense to fix). This can be stated as follows in the termi-
nology of Chap. III.
Corollary 2. In the function space A L) the orthonormal set
consisting of all the functions
( 1 )
_1 1 _
V2l ; Vl
mrx 1 . nirx
C0S ~L ! VX 8m ~L~
(n
1 , 2 ,
),
is closed with respect to ordinary convergence. It is also complete.
The proof of completeness is left for the problems.
Similar statements can be made for functions defined in the
interval (0, L), with respect to either the set of sine functions
or the set of cosine functions.
Sec. 35]
FOURIER SERIES
75
Note that the last corollary is a statement about functions
whose one-sided derivatives exist at all points of the interval, a
condition which is not used in Corollary 1.
Let us observe finally how the conditions of our Fourier
theorem apply to our examples. The function in the example
treated in Sec. 27 ; namely,
fix) = 0 when — r < x ^ 0,
= x when 0 g x < t,
is continuous in the interval (— ir, r). It has one-sided deriva-
tives at all points. Series (1), Sec. 27, therefore converges to
fix) at all points in the interval — r < x < r, according to
Theorem 1. At the points x = ±r it converges to the value
7r/2, since /(— t + 0) =0 and /( r — 0) = r. The graph of the
periodic function shown there (Fig. 6) would be a complete
representation of the function represented by the series if the
points (±7r, 7r / 2) , (±37t, 7t/2), • • • were inserted.
In Sec. 29 the cosine and sine series were found for the function
fix ) = x when 0 ^ x <
= 0 when ^ < x g r.
This function is sectionally continuous in the interval (0, t),
and its one-sided derivatives exist there. The sine series there-
fore converges to fix) when 0 g x S tt, except »at x — 7t/2,
where it converges to 7t/4. At x = 0 and x = t it converges to
jT ( — | — 0) and fir — 0), since these are both zero. The cosine
series for this function converges in just the same manner in the
interval (0, r).
PROBLEMS
1. Show that each of the functions described in Probs. 1 to 4, Sec. 27,
satisfies the conditions under which the series found there converges to
the function, except possibly at certain points. What is the sum of the
series at those points?
Ans. Prob. 1: x = ±r; sum = 0;
Prob. 2: x = ±r; sum = cosh7r;
Prob. 3: x =0, ±w; sum — f.
2. Solve Prob. 1 above for each of the functions in Probs. 1 to 7,
Sec. 29.
76 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 35
3. Solve Prob. 1 above for each of the functions in Probs. 3 to 9,
Sec. 30.
4. If f(x) = 0 when -1 < x < 0, fix) = cos tx when 0 < x < 1,
/(0) = §■>/(!) = ~h, and fix + 2) — fix) for ! all x, show that
1 *
fix) = g cos tx + - ^ 4w2 ”_ l sin 2mrx
Ti
for all values of x.
6. If fix) = c/4 - x when OiiS c/2, fix) = x — 3c/4 when
c/2 S i S c, show that
COS
(4 n — 2 )ttx
for all x in the interval 0 ^ x ^ c.
6. If /( x) = x 2 when —1 < as ^ 0, f(x ) = 0 when 0 ^ a? < 1,
/(!•) = I? and J{x + 2) = f(x) for all x, find its Fourier series and show
that it converges to f(x) for all values of x.
7. Prove that the orthonormal set of functions in Corollary 2 is
complete in the function space A Lt (Compare Sec. 22; show that any
function in A L which is orthogonal to every member of the set must be
identically zero.)
8. State and prove the corollary, corresponding to Corollary 2, for
functions defined in the interval (0, L ), with respect to the orthonormal
set of functions { V2/Z sin (rnrx/L) } .
9. Show that the series
L . mrx \
f(x) sin -g- ax J
of squares of the coefficients in the Fourier sine series converges when-
ever f(x) is bounded and integrate on the interval (0, L), and that
2 - 1 J 0 t/wi 2 dx -
[See formula (5), Sec. 21.]
10. Show that the series
1
2 a
2
0
mrx _
cos — f- dx
)
involving the squares of the coefficients in the Fourier
cosine series,
Sec. 35]
FOURIER SERIES
77
converges whenever f(x) is bounded and integrable in the interval
(0, L), and that
2 a ° + - L I dx -
l c/O
(Compare Prob. 9.)
11. If f(x) is bounded and integrable in the interval (— L, L), show
that the series ’ ’
§«0 + 2 ) al +
where a n and b n are the coefficients in Fourier series (1), Sec. 34, con-
verges to a sum not greater than
X f [/(*)] 2 dx.
(Compare Prob. 9.)
12. For every function which is bounded and integrable in the interval
(-L, L), the Fourier coefficients a n and b n in series (1), Sec. 34, approach
zero as n tends to infinity. Show how this follows from Prob. 11.
When the function is sectionally continuous, show that the result for b n
follows also from Lemma 1.
13. The coefficients (i n and b, l} in Corollary 1, are those for which the
sum of any fixed finite number of terms of the series written there will be
the best approximation in the mean to f(x), in the interval (-L, L).
Show how this follows as a special case of Theorem 1, Chap. III.
14. Find the values of A lf A 2 , and A* such that the function
y = Ai sin y + A 2 sin + A [t sin
will be the best approximation in the mean to the function f(x) = 1,
over the interval (0, 2) (compare Prob. 13). Also draw the graph of y [
using the coefficients found, and compare it to the graph of f(x).
Ans. A , = 4/j r, A, = 0, A z = 4/(3tt).
16. Show that it follows from the expansion in Prob. 5, Sec. 30, by
setting x = L, that
Similarly, show that
CHAPTER V
FURTHER PROPERTIES OF FOURIER SERIES;
FOURIER INTEGRALS
36. Differentiation of Fourier Series. We have seen that the
Fourier series representation of the function f(x) = x is valid
in the interval —x < x < x; thus (Prob. 1, Sec. 29)
x = 2(sin x — i sin 2x + i sin 3x — * • • )
when — x < x < 7T. But the series obtained by differentiating
this series term by term, namely,
2(cos x — cos 2x + cos 3x — • • • ),
does not converge to the derivative of x in the interval (— x, x).
The term cos nx does not approach zero as n tends to infinity;
hence the series does not converge.
For all values of x, the above series for the function f(x) = x
represents a periodic function with discontinuities at the points
x = ±x, ±3x, • • * . We shall see that the continuity of the
periodic function is an important condition for the termwise
differentiation of a Fourier series. A complete set of sufficient
conditions can be stated as follows:
Theorem 1. Let f(x) be a continuous function in the interval
“X # = 7r such that /(x) = /(— x), and let its derivative f'(x )
be sectionally continuous in that interval. Then the One-sided
derivatives of f(x) exist (Sec. 31), and hence f(x) is represented by
its Fourier series
00
(1) f(x) — %a 0 + ^ (a n cos nx + b n sin nx) (— x ^ x ^ x),
where
1 i C*
(2) a» = - I f(x) cos nx dx , b n = - I f{x) sin nx dx ,
xJ_ T irj-f
and at each point where f{x) has a derivative that series can be
differentiated termwise; that is,
78
Sec. 36] FURTHER PROPERTIES OF FOURIER SERIES
79
(3) fix) = ^ n( — a n sin nx + b n cos nx) (— 7r < x < x).
Since fix) satisfies the conditions of our Fourier theorem, it
is represented by its Fourier series at each point where its deriva-
tive/"^) exists. At such a point fix) is continuous, so that
(4) f(x) = ^a' 0 + 2) i a h cos nx + b r n sin nx ),
where
if 7 " 1 r ,r
(5) a' - - I /'(a) cos no; ire, b' n = - I /'(x) sin nx da;.
T J-7T
These integrals can be integrated by parts, since fix) is con-
tinuous and fix) is scctionally continuous. Therefore
a' — -
(6)
fix) cos nx j + ~ J fix) sin nx dx
LfM ~ /( — tt)] + nb».
7T
cos n7r
This reduces to n& n because of our condition that/( 7 r) = /(— 7 r),
Furthermore, aj = 0. Likewise,
b^, = i £/(x) w ^ u j — - /(x) cos nx dx
= —na n .
Substituting these values of a' and b r n into equation (4), we
have
0O
fix) = ^ (nbn cos nx — na n sin nx).
This is the equation (3) which was obtained by differentiating
(1) term by term; hence the theorem is proved.
It is important to observe that, according to equation (6), the
Fourier scries for fix) docs not reduce to series (3) obtained by
termwise differentiation if the function fails to satisfy the condition
fW) = fi~i r).
This condition ensures the continuity of the periodic extension
of/(x) at the points x = ±7r, and therefore at all points, in view
of the continuity of fix) in the interval (— x, 7 r).
80 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 37
At a point where f(x) has a derivative from the right and
from the left, but no ordinary derivative, we can easily see
from the above proof that termwise differentiation is still valid
in the sense that
00
i[f(x + 0) + f'(x — 0)] = V n( — a n sin nx + b n cos nx).
i
Since this is true for the periodic extension of fix), the derived
series converges at the points x = ±tt to the value
«f(- 7T + 0) + /'(*•- 0)]
if fix) has a right-hand derivative at -tt and a left-hand deriva-
tive at t. We are assuming the continuity of the periodic
extension of f(x) at all points, of course.
Theorem 1 applies with the usual changes to the other forms
of Fourier series.
PROBLEMS
1. Show that the series in Prob. 4, Sec. 27, can be differentiated term
by term, and state what function is represented by the derived series.
(Compare Prob. 4, Sec. 35.)
2. In the problems, Sec. 29, obtain the series in Prob. 3a by differen-
tiating the series in Prob. 26. Note that this is permissible according
to Theorem 1; but we cannot reverse the process and*obtain the latter
series by differentiating the former.
3. In Probs. 1 to 7, Sec. 29, which of the series can be differentiated
termwise? _ Am. 1(6); 2(a), (6); 3(6); 4(6); 6(a), (6); 7(6).
4 . Show that in Probs. 4 and 5, Sec. 30, the series are termwise
differentiable.
5. Show that the Fourier coefficients a n and b n for the function f(x),
described in the first sentence of Theorem 1, satisfy the relations
lim na n = 0, lim n6„ = 0.
71 — ► oo 7 i — > oo
37. Integration of Fourier Series. Termwise integration of
a Fourier series is possible under much more general conditions
than those for differentiation. This is to be expected, because
an integration introduces a factor n in the denominator of the
general term. It will be shown in the following theorem that
it is not even essential that the original series converge to its
function, in order that the integrated series converge to the
Sec. 37] FURTHER PROPERTIES OF FOURIER SERIES 81
integral of the function. Of course, the integrated series is not a
Fourier series if a 0 ^ 0, for it contains a term a Q x/ 2.
Theorem 2. Let f(x) be sectionally continuous in the interval
(-“ir, tt). Then whether the Fourier series corresponding to f(x),
c©
(1) f(%) ~ ^ (a n cos nx + b n sin nx),
converges or not, the following equality is true : ,
( 2 ) /(*) dx = — (x + tt)
+ ^ - [a w sin nx — b n (cos nx — cos nx)],
7 n
when —ir^x^T. The latter series is obtained by integrating
the former one term by term .
Since /(x) is sectionally continuous, the function F(x), where
(3) F(x) = J* r f(&) dx — ioox,
is continuous; moreover
F'(x) = f(x) -
except at points where /(x) is discontinuous, and even there
F(x) has right- and left-hand derivatives. Also,
F( tt) = J^/(x) dx — = a 0 7r — -ja ( )7r = ^-aoTr,
and F(— 7r) = ^a ( )7r; hence /'’(t) = According to our
Fourier theorem then, for all x in the interval — 7 r 5^ x ^ 7 r, it is
true that
00
F(x) = ^ (A„ cos nx + B n sin nx),
where
\ C T i
A n = - I F(x) cos nx dx , B n = - I F(x) sin nx dx.
J -TT IT J -TT
Since F{x) is continuous and F'(x) is sectionally continuous,
the integrals for A n and B n can be integrated by parts. Thus if
n y* 0,
82 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 38
A n = -J- F(x) sin nx — ( F'(x) sin nx dx
nr L J-* nr J_ T
= f_ w [ f(x) ~ i \ sin dx = ~l b -
Similarly, B n = a n /n; hence
1 00 1
(4) F(x) = g A 0 + - (a n sin nx — 6* cos no;).
l
But since F(r ) = ^oott,
- a 0 7T = ^ A 0 — i b n cos n7r.
1
Substituting the value of A 0 given here in equation (4),
F( x ) 1“ + - [a n sin nx — b n ( cos nx — cos nir)].
l -
In view of equation (3), equation (2) follows at once.
The theorem can be written for the integral from x 0 to x, when
—ir ^ x Q ^ r and —it ^ x ^ tt, by noting that
j£7(*) = /_V^) dx - <f*.
The other forms of Fourier series can be integrated termwise
under like conditions, of course.
Still more general conditions under which the Fourier series
can be integrated term by term will be noted in Sec. 39.
PROBLEMS
1. By integrating the expansion found in Prob. 4, Sec. 27, from —tt
to x , obtain the expansion
N X 1 1 1
F(x) = - + 9 — 9 COS X
T L A 7 r
1 sin 2 nx
n4n 2 — 1 *
where and F(x) = 0 when —ir^x^O, F(x) = l — C os x
when 0 ^ x ^ x.
2. Integrate the series obtained in Probs. 1 and 3, Sec. 27, from 0 to x ,
and describe the functions represented by the new series.
38. Uniform Convergence. If A n and B n (n = 1, 2, • * • ; m)
represent any real numbers, the equation
Sec. 38] FURTHER PROPERTIES OF FOURIER SERIES 83
m m hi m
2, ( A » X + 5 ») 2 = X 2 X A l + 2* 2 A ' B » + 2 B * = 0
1 111
cannot have distinct real roots. In fact, if it has a real root
x = x Q , then A n x 0 + B n = 0 for all n, and the ratio B n /A n must
be independent of n. The discriminant of the quadratic equa-
tion in x is therefore negative or zero; that is,
( m \ 2 in m
X A -Bn) g X A l X Bl
With the help of this relation, known as Cauchy 1 s inequality ,
we can readily show that the convergence of the Fourier series
to the function f(x) described in Theorem 1 is absolute and
uniform.
Broader conditions for uniform convergence will be cited in
the next section. But it should be noted that a Fourier series
cannot converge uniformly in any interval containing a dis-
continuity of its function, since a uniformly convergent series of
continuous functions always converges to a continuous function.
Theorem 3. Let f(x) be a continuous function in the interval
— 7 r S x ^ t such that /(t) = /(— 7 r), and let its derivative f'(x)
be sectionally continuous in that interval. Then the Fourier series
for the function f{x) converges absolutely and uniformly in the
interval ( — 7 r , tt).
The theorem will be proved if we can show that for each
positive number e an integer m 0 , independent of x , can be found
such that
m'
^ | a n cos nx + b n sin nx\ < e
m
when m > m 0 , for all m' > m. The term between the absolute
value signs represents, of course, the general term in the Fourier
series corresponding to f(x). Since it can be written as
\fa\ + bl cos ( nx — 0) ^0 = arctan
it is clear that
|a n cos nx + b n sin nx | ^ +
84 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 38
so it will suffice to show that
( 2 )
m
X VoJ + b\ < €
( m > mo, ml > m).
In the proof of Theorem 1 we found that
( 3 ) «» = £>'„ = —na n ,
where a'„ and fc' are the Fourier coefficients of the function
Therefore
wt m'
2 = 2
Applying inequality (1) to the last sum, we have
, m* m’ v *
( 4 ) 2 = { 2^2 [(<)2 + w i] \ ■
Bessel's inequality (4), Sec. 21, applies to the bounded inte-
grate function f'(x) 3 with respect to the orthonormal set of
functions
(6> cos ^ sfa “} • • ■ >■
giving the relation
2 K <) 2 +
"l
2 dx
for every integer m'. Let M denote the member on the right
here ; then the second sum on the right of inequality (4) does not
exceed the number M.
Now the series
converges; so for any positive number e 2 /M an integer m 0 can
be found such that
m
Sec. 39] FURTHER PROPERTIES OF FOURIER SERIES 85
when m > mo, for all rr\I > m; and mo is clearly independent of
x. For this choice of mo, the right-hand member of inequality (4)
is less than €, so that inequality (2) is established and the theorem
is proved.
But in view of inequality (2) we have also shown that, under
the conditions in Theorem 3, the series
00
2) Val + bl
always converges . Consequently each of the following series
converges :
5) W, | \bn\.
It is of interest to note that the Parse val relation (3), Sec. 22,
applies to the class of functions described in Theorem 3 with
respect to the orthonormal set of trigonometric functions (5).
This follows by multiplying the Fourier series expansion of f(x)
by f(x), thus leaving it still uniformly convergent, and integrat-
ing, to obtain
[ f ( x )] 2 dx = -g-ao dx + ^a n f* v f(x) cos nx dx
f(x ) sin nx ctaj.
In view of the definitions of a n and h n , this can be written
(6) S-* dX= * [i a ° + X (“n + K) ] -
This is the Parse val relation.
PROBLEM
Show that if a class of functions satisfies the Parseval relation, the
orthonormal set is closed with respect to the limit in the mean (Sec. 22).
Hence deduce that set (5) is closed in that sense, for the class of all
functions satisfying the conditions in Theorem 3.
39. Concerning More General Conditions. The theory of
Fourier series developed above will be sufficient for our purposes.
Let us note at this point, however, a few of the many more
86 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 39
general results which are known. These will be stated without
proof, since our purpose is only to inform the reader of the
existence of such theorems. They cannot be stated in their
most general form, usually, without introducing Lebesgue
integrals in place of the Riemann integrals considered here.
a. Fourier Theorem . Let f{x) denote here a periodic func-
tion with period 2 x, and let f(x) dx exist. If the integral is
improper, let it be absolutely convergent. Then the Fourier
series corresponding to /(x) converges to the value
M* * * § + o) +/(*-o)]
at each point x which is interior to an interval in which /(x) is of
bounded variation.*
b. Uniform Convergence . If the periodic function /(x) described
under ( a ) is continuous and of bounded variation in some interval
(a, b) then its Fourier series converges to /(x) uniformly in any
interval interior to (a, b). f
We have noted earlier that the partial sums S n (x ) of a Fourier
series cannot approach the function f(x) uniformly over any
interval containing a point of discontinuity of /(x). The nature
of the deviation of S n (x ) from /(x) in such an interval is known
as the Gibbs phenomenon
c. Integration. The Parseval relation
(1) \ j [/(z)P dx = i a 2 0 + ^ ( a * + h n),
is true whenever jf(x) is bounded and integrable in the interval
( — x, 7r).§ That is, the series of squares of the Fourier coefficients
of /(x) on the right of equation (1) converges to the number on
the left.
Now let a n and ($ n be the Fourier coefficients of a function
<p(x), bounded and integrable in the interval (—7 r, tt). Then
(a n + <*n) and ( b n + £ n ) are the coefficients of the function
U + <?)> an d according to equation (1) we have
* See first the proof in Ref. 2 at the end of this chapter.
t For a proof, see Ref. 2.
t See Ref. 1.
§ See first the proof given in Ref. 2.
Sue. 39] FURTHER PROPERTIES OF FOURIER SERIES
87
l f [/(*) + <?(x)] 2 dx
T J ~T
GO
= 2 ( a ° a ° ) 2 ” t “ [C + <*») 2 +' (&» + j 8 ») 2 ].
I -
Likewise
<p(x)] 2 dx
to
= i (do — ao) 2 + 2 [(“» ~ a ") 2 + (&» ~ 0») 2 ],
and by adding the last two equations we find that
( 2 )
r/_
+ b n fi n ).
In form (2) of the Parseval formula suppose that
<p(x) = g(x) when — ir < x < t,
= 0 when t < x < ir (— 7r ^ t ^ 7r),
where {/(x) is bounded and integrable in the interval (— 7r, tt).
Then
1 f
#(x) cos nx dx, f$,
,-ir
7 T J-x
g(x) sin no; dx,
and form (2) becomes
(3)
J f(x)g(:r) dx = £a<> *7(x) dx
-T J IT
+
?[-f
< 7 (x) cos no: do: + 7)
"J-
< 7 (x) sin no: dx
So it follows from statement (c) that if the Fourier series cor-
responding to any bounded integrable function /(x) is multiplied
by any other function of the same class and then integrated
term by term, the resulting series converges to the integral of the
product f(x)g(x). When g(x) = 1, we have a general theorem
for the termwise integration of a Fourier series.
88 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 40
PROBLEM
Assuming statement (c), show that it follows that the set of func-
tions (5), Sec. 38, is closed, in the sense of convergence in the mean,
with respect to the class of bounded integrable functions in the interval
(— 7 r, 7r). (Compare the problem at the end of Sec. 38.)
40. The Fourier Integral. The Fourier series (Sec. 30) cor-
responding to fix) in the interval (— L, L) can be written
It converges to \ [fix + 0) + fix — 0)] when — L < x < L,
provided f(x) is sectionally continuous and has right- and left-
hand derivatives in the interval (— L, L). If fix) satisfies those
conditions in every finite interval, then L may be given any fixed
value, arbitrarily large but finite, in order that we may obtain
a representation of f(x) in a large interval. But this series
representation cannot be valid outside that interval unless fix)
is periodic with the period 2 L, since series (1) represents only
such functions.
To indicate a representation which may be valid for all real
x when fix) is not periodic, it is natural to try to extend series (1)
to the case L = oo . The first term would then vanish, assuming
that J f ^ fix) dx converges. Putting Aa = t/L, the remaining
terms can be written
£ 2 j C ° S [t ^ X ' ~ *)] dX '
60
= ^ A a j f(x') cos [nAa(V — a;)] dx'.
1 —L
OO
The last series has the form ^F(nAo:)Aa, where
Fia) = J^ L fi x ') cos [aix' — x)] dx';
hence when Aa is small, it may be expected to approximate the
integral F(a) da . (Note, however, that its limit as A a
3ec. 40] FURTHER PROPERTIES OF FOURIER SERIES
89
approaches zero is not the definition of this integral; further-
more, when A a. approaches zero, L becomes infinite, so F(a) itself
changes.) But if the process were sound, when L becomes
infinite series (1) would become
1 f* 00 f* 00
- I da I f(x') cos [a(x r — a;)] dx r .
* Jo J- «
This is the Fourier integral of f{x). Its convergence to f(x)
for all finite values is suggested but by no means established
by the above argument. It will now be shown that this repre-
sentation is valid whenf(x) satisfies the conditions in the follow-
ing Fourier integral theorem:
Theorem 4. Let f(x) be sectionally continuous in every finite
interval (a, b), and let * \f(x)\dx converge . Then at every 'point
x ( — °o < x < °o), where f{x) has a right- and left-hand deriva-
tive , f(x) is represented by its Fourier integral as follows:
(2) i[/(x + 0) +f(x - 0)]
= - I (la I fix' ) cos [a (x' — x)] dx'.
In every interval (a, b),f(x) satisfies the conditions of Lemma 3,
Sec. 32, so that
(3) | [f(x + 0) +f(x - 0)] = lim C f(x') a jllJ0xL - g)j dx >
at, any point x (a < x < b), where f(x) has a right- and left-hand
derivative. Now
(4)
fix')
A [<*(*' - *)] dx>
i:
Whenever a < x,
r
f(x') sin _
x — X
and the latter integral converges because |/(a;)| dx does.
90 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 40
Similarly for the last integral in equation (4), when b > x.
Hence for any e > 0 a positive number N can be found such that
^ a < and b > N, the first and last integrals on the right
of equation (4) will each be numerically less than e/3. The
second integral there can be made to differ from the value
+ 0) + f(x — 0)] by an amount numerically less than
<:/3 by taking a sufficiently large, according to equation (3).
Hence the integral (4) differs numerically from the above value
by an amount less than e for all a greater than some fixed number;
that is,
(6) M 8ln 5 - , - 1 w* + 0) +f(„ - 0)].
Writing the fraction in the integrand as an integral, and divid-
ing by 7 r, this becomes
2 f f( x + 0) + f(x — 0)]
= 1™ l J m fix') dx' cos [o' (a/ - *)] da'
= l l d “' /_ _ fix') COS [a'(x' - x)] dx'.
The inversion of order of integration in the last step is valid
because the integrand does not exceed \f(x')\ in absolute value, so
that the integral
f_ " fix ' ) cos W(x' - x)] dx'
converges uniformly for all «'.* The last equation is the same
as statement (2) in the theorem.
Fourier integral theorems with somewhat broader conditions
on f(x) are also known. The more modem theorems take
advantage of the use of Lebesgue integration.
PROBLEMS
1. Verify the Fourier integral theorem directly for the function
f(x) = 1 when -1 < * < 1, fix) = 0 when x < - 1 and when x > 1.
ihe following integration formula, usually established in advanced
calculus, will be useful:
* See, for instance, p. 199 of Ref. 1.
U FURTHER PROPERTIES OF FOURIER SERIES
91
f °° sinfcx dx =Z ii k > 0,
Jo * i 2
= 0 if k = 0,
= ~ ilk <0.
how that the function f(x) = 0 when x < 0, f(x) = e~~ x when
■ /(0) = is represented by its Fourier integral; hence show that
;egral
X
cos ax + a sin ax
1 + a 2
da
y value 0 if x < 0, tt/2 if x = 0, and we"* if a; > 0.
how that the Fourier integral of the function f(z) = 1 does not
ge.
Other Forms of the Fourier Integral. Let f(x) be an odd
:m which satisfies the conditions of Theorem 4. Then
(x') cos [a(x' — x)] dx f
f* OO /• 00
J ^ jf(x') cos [a (s' — *)] dx' +1 f( — y) cos [a(y + x)] dy
j /(x') cos [a(x' — x)] d,x' — J /(x') cos [a(x' + x)] dx'
f* oo
5 sin ax I f(x r ) sin ax' dx' .
Jo
the Fourier integral formula becomes
[f(x + 0) + f(x — 0)]
sin ax da
sin ax' dx'.
i is the Fourier sine integral, corresponding to the Fourier
M’ies. If f(x) is defined only when x > 0, formula (1)
<1 provided f(x) is piecewise continuous in each finite
i,l in x ^ 0 and has a right- and left-hand derivative
point x (x > 0), and provided £ |/(:r)| dx converges.
Iarly if f(x ) is an even function satisfying the conditions
orem 4, it is represented by its Fourier cosine integral:
92 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 41
(2) g[/(* + 0) +/(*- 0)]
= T Jo C0S aX Jo ft*') C0S aX> dx '-
Under the conditions just given for the sine integral, formula (2)
is also valid if f(x) is defined only when x > 0. Moreover, the
integral converges at x = 0 to /(+0) provided fix) has a right-
hand derivative there. “
By writing cos [a(x' — x)] in terms of imaginary exponential
functions, the integral formula of Theorem 4 can be reduced to
(3) 2 [/(* + 0) + fix — 0)]
“ ^ J m eiax da J- " e~ iax ' fix') dx'.
This is the exponential form of the Fourier integral of the func-
tion f{x) defined for all real values of x.
If gi.a) is a known function when a > 0, note that the integral
equation
(4)
fix') sin ax' dx' = gia),
can be solved easily for the unknown function fix) (x > 0),
provided that function is one of the class for which the Fourier
integral formula (1) is true. For by multiplying equation (4)
through by y/2 sin ax and integrating with respect to a over
the interval (0, oo) W e have, in view of formula (1),
(5)
fix) =
g(a) sin ax da
ix > 0).
Of course this formula would give the mean value of fix) at a
point of discontinuity. K
The integral in equation (4) is called the Fourier sine transform
of fix) Formula (5), which gives fix) in terms of its transform
Qi<x), has precisely the same form as equation (4).
In view of formula (2), the sine functions can clearly be
replaced by cosines in equations (4) and (5).
PROBLEMS
1. Show that the formula in Theorem 4 reduces to formula (2) when
fix) is an even function.
Sec. 41] FURTHER PROPERTIES OF FOURIER SERIES
93
2. Transform the formula in Theorem 4 to the exponential form (3).
3. Apply formula (2) to the function /(as) = 1 when 0 g* x < 1,
/(as) — 0 when x > 1, and hence show that
sin a cos ax , ir , A . ^ -
da = - when 0 ^ x < 1.
a 2
— | when a; — 1,
= 0 when a; > 1.
4. Apply formula (1) to the function /(as) = cos as, and thus show
that
I
15 sin ax , t _ ^ A
-r— — da = - e~ x cos as if xb > 0.
x 4 + 4 2
6. By applying formula (1) to the function /(as) = sin as when
0 £ x /(as) = 0 when x > t, show that
X
sin ax sin 7 ro :
1 — a 1
da
t .
= - sin x
= 0
if 0 ^ as ^ 7 r,
if as > 7 r.
6. Apply formula (2) to the function /(as) of Prob. 5 and obtain
another integration formula.
7. Show that the solution of the integral equation
sin ax dx = g(a),
where g{a) = 1 when 0 < a <7 r, g (a) =0 when a > ir, is
. 2 1 — cos7ras
m = - 5 —
8. Show that the integral equation
(x > 0).
has the solution
/(as) cos ax dx = e~ a
ft*) = l ttv*
(as > 0).
REFERENCES
1. Carslaw, H. S.: ‘‘Fourier’s Series and Integrals,” 1930.
2. Whittaker, E. T., and G. N. Watson: “ Modern Analysis,” Chap. 9, 1927.
3. Titchmarsh, E. C.: “Theory of Functions,” Chap, 13, 1939. This is
more advanced.
CHAPTER VI
SOLUTION OF BOUNDARY VALUE PROBLEMS BY THE
USE OF FOURIER SERIES AND INTEGRALS
42. Formal and Rigorous Solutions. In an introductory
treatment of boundary value problems in the partial differential
equations of physics, it seems best to follow to some extent
the plan used in introductory courses in ordinary differential
equations; that is, to stress the method of obtaining a solution
of the problem as stated, and give less attention to the precise
statement of the problem that would ensure that the solution
found is the only one possible. But it is important that the
student be aware of the shortcomings of this sort of treatment;
hence some discussion of the rigorous statement and solution of
problems will be given. The subject of boundary value problems
in partial differential equations is still under development; in
particular, the uniqueness of the solutions of some of the impor-
tant types of problems has not yet been satisfactorily investigated.
In ordinary differential equations, the solution for all x ^ 0
of the simple boundary value problem
y'{x) = 2, 2/(0) = 0,
would generally be given as y = 2x, because it is understood
that y(x) must be continuous. Without such an agreement,
however, the function y = 2x + c when x > 0, y = 0 when
x = 0, is a solution for every constant c; that is, the solution is
not unique. Even when the boundary condition is written
2/(+0) = 0, the solution could be written, for instance, as y = 2x
when 0^x^ a, y = 2x + c when x > a, unless y(x) is required
to be continuous for all 3 ^ 0.
Such tacit agreements necessary for the existence of just
one solution are not nearly so evident in partial differential
equations. Furthermore, if the result is found only in the
form of an infinite series or integral, it is sometimes quite difficult
94
Sec. 43] SOLUTION OF BOUNDARY VALUE PROBLEMS 95
to determine the precise conditions under which that series qr
integral converges and represents even one possible solution.
The treatment of an applied boundary value problem is only
a formal one unless it is shown (a) that the result found is actu-
ally a solution of the differential equation and satisfies all the
boundary conditions, and ( b ) that no other solution is possible.
The physical problem will require that there should be only
one solution; hence the mathematical statement of the problem
is not strictly complete unless the uniqueness condition ( b ) is
satisfied.
43. The Vibrating String. The formula for the displacements
y(x, t) in a string stretched between the points (0, 0) and (L, 0)
and given an initial displacement y = f{x) was found in Sec. 13
to be
00
.... . . mrx rnr at
(1) 2 / = > | A n sin -j- cos -£->
“i
where
. 2 f L \ . nirx ,
A n = j I f(x ) sin -j-dx.
The function fix) must of course be continuous in the interval
0 ^ x ^ L and vanish when x = 0 and x — L. In addition,
let, fix) be required to have a right- and left-hand derivative at
each point. Then the Fourier sine series obtained when t = 0
in formula (1) does converge to/(z); hence this initial condition
is actually satisfied. Thus an important improvement in the
formal solution is made possible by the theory of Fourier series.
The nature of the problem requires the solution y(x, t) to he
continuous with respect to x and t. Since y(x, t) is to satisfy
the equation of motion
( 2 )
d' l y „ d' 2 y
W ~ n ~ a * 5
it > 0, 0 < x < L),
and all the boundary conditions
2 /( 0 , t ) = 0, yiL, 0 = 0,
, o, o) - /<*>,
some conditions relative to the existence of its derivatives
must also be satisfied. We shall now examine the function
96 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 43
defined by formula (1) to see if it is actually a solution of our
problem.
The Solution Established. It is possible to sum the series in
formula (1); that is, to write the result in a closed, or finite,
form. This will make it much easier to examine the function
y(x, t).
Since
2 sin cos ^ = sin [*£ (* - at)] + sin [*£ (* + at)],
equation (1) can be written
(3) y = \ 2 An sin [t ~ a< ^]
+ j2 A ” sin [x (a: + a<) '
The two series here are those obtained by substituting (x — at)
and (x + at), respectively, for the variable x in the Fourier
sine series for f(x). Since the sine series represents an odd
periodic function, the last equation can be written
(4) y = $F(x - at) + F(x + at)],
where the function F(x') is defined for all real values of x ’ as the
odd periodic extension of f(x') ; that is,
F(x r ) = fix') if 0 ^ a ■/ £L,
F(-x') = -F{x'),
and
Fix' + 2 L) = Fix') for all x'.
The function jix) is continuous in the interval (0, L) and
vanishes at the end points; hence Fix') is continuous for all x .
According to our Fourier theorem, the two sine series in equation
(3) converge to the functions in equation (4) whenever fix) has
one-sided derivatives. The same function yix, t) is then repre-
sented by each of the three formulas (1), (3), and (4); moreover,
according to (4), yix, t) is a continuous function of x and t for
all values of these variables.
Sec. 43] SOLUTION OF BOUNDARY VALUE PROBLEMS
97
By differentiating equation (4) we can easily see that y(x, t )
satisfies differential equation (2) whenever the derivative F"(x')
exists. When it is observed that F'(x') and F"(x') are even
and odd functions, respectively, it can be seen that the second
derivative exists for all x' provided fix) has a second derivative
whenever 0 < x < L, and provided that the one-sided derivatives
of f'(x) at the end points x = 0 and x = L exist and have the
value zero.
Under these rather severe conditions on f(x), then, our func-
tion y(x f t) satisfies the equation of motion for all x and t } and
it is also evident from equation (4) that dy/dt is continuous
and vanishes when t = 0. The remaining boundary conditions
are clearly satisfied, in view of either equation (1) or (4); hence
y(Xj t) is established as a solution.
If we permit /'(rc) and f"(x) to be only sectionally continuous,
or if the one-sided second derivatives of fix) do not vanish at the
points x = 0 and x = L, then at each instant t there will be a
finite number of points x at which the second derivatives of
y(x, t) fail to exist. Except at these points, differential equation
(2) will still be satisfied. In this case we have a solution of our
problem in a broader sense.
In cither case an examination of the uniqueness of the solution
found would be necessary to make the treatment of the problem
complete.
An Approximate Solution. Except for the nonhomogeneous
boundary condition
(5) y(x, 0) = J{x) 9
our boundary value problem is satisfied by the sum of any
finite number of terms of the scries in equation (1), say
(6)
Vn —
N
mrx
L
cos
mrat
~77
where N is some integer. In place of condition (5) this function
satisfies the condition
N
, x , . UTX
(7) ysix, 0) = An sm ~7“‘
i
The function y„(r, t) has continuous derivatives of all orders.
98 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 44
The sum in condition (7) is that of the first N terms of the
Fourier sine series for f(x). According to Theorem 3, Chap. V,
that series converges uniformly to f(x) provided f(x) is con-
tinuous by segments. Hence, by taking N sufficiently large,
the sum can be made to approximate f{x) arbitrarily closely for
all values of x in the interval 0 ^ x ^ L.
The function y N {x , t) is therefore established as a solution of
the “approximating problem,” obtained by replacing condition
(5) of the original boundary problem by condition (7).
Similar approximations can be made to the problems to be
considered later on. But the remarkable feature in the present
case is that the approximating function y N (x, t) does not deviate
from the actual displacement y(x, t) by more than the maximum
deviation of ynix, 0) from/(x). This is true because ynix, t ) can
be written
Vn = 5 An sin (x - at) J + ^ A « sin [77 (* + «oJ|>
and each sum here consists of the first N terms of the sine series
for the odd periodic extension of /($), except for substitutions of
new variables. But the greatest deviation of the first sum from
F(x — at), or of the second from F(x + at), is the same as the
greatest deviation of ynix, 0) from jf(x).
PROBLEMS
1. Show that the motion of every point of the string in the above
problem is periodic in t with the period 2 L/a.
2 . The position of the string at any time t can be found by moving
the curve y = iF( x ) to the right with the velocity a and an identical
curve to the left at the same rate and adding the ordinates, in the
interval 0 ^ x ^ L, of the two curves so obtained at the instant t.
Show how this follows from formula (4).
3. Plot a few positions of the plucked string of Sec. 14, using the
method of Prob. 2 above.
44. Variations of the Problem. If each point of the string is
given an initial velocity in its position of equilibrium, the bound-
ary value problem in the displacement y(x, t) is the following:
Sec. 44] SOLUTION OF BOUNDARY VALUE PROBLEMS
99
y(0, t) = 0, y(L, t) = 0,
y(x, 0) = 0, = g(x).
As before, functions of the form y = X(x)T(t) which satisfy
the differential equation and all the homogeneous boundary
conditions can be found. Writing the series of these particular
solutions, we have
00
2 , . nirx . mrat
An sin -j- sm ~j ~ *
i
The final condition, that dy/dt = g(x) when t = 0, shows that
the numbers mraA n /L should be the Fourier sine coefficients of
g{x ) ; hence the solution of the problem becomes
a)
sin
mrx r
~L~
dx'.
By the method of the last section, dy/dt can be written here
in terms of the odd periodic extension G(x r ) of the function
g(x f ). This leads to the closed forms
( 2 )
y = 2 J 0 “ at ') + + a ^)] dt'
of solution (1). The details of these derivations are left for the
problems.
Superposition of Solutions. If the string is given both an
initial displacement and initial velocity, the last two boundary
conditions become
(3) y(x, 0) = f(x), = d( x )-
All the other conditions of the linear boundary problem are
homogeneous. They are satisfied by the solution of the problem
of the preceding section and by solution (2) above, and therefore
by the sum of those two functions, namely
1 1
(4) y = 2 l F ( x ~ at ) + F ( x + ®01 + 2a J ( dx '-
100 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 44
When t = 0, the first function of the sum becomes f(x) and the
second vanishes; hence the first of conditions (3) is satisfied.
Likewise it is seen that the second of those conditions is satisfied,
and therefore equation (4) is the required solution.
In general the solution of a linear problem containing more
than one nonhomogeneous boundary condition can be written
as the sum of solutions of problems each of which contains only
one nonhomogeneous condition. Of course we cannot always
find the solutions of the simpler problems which are to be super-
posed in this way.
Units . It is often possible and advantageous to select units
so that some of the constants in our problem become unity.
For example, if we write r for the product (at), the equation of
motion of the string reduces to
d^y = cPy
dr 2 dx 2
Such changes sometimes help to bring out reductions in compu-
tation, or general properties of the solution.
Since the boundary problem of the last section, for example,
does not involve the number a when the problem is written in
terms of r and x , its solution must be a function only of x, L, and
the product (at). This conclusion is possible without our
knowing the formula for the solution. But a 2 is proportional
to the tension in the string; hence if y i(x, t) and y 2 (x, t) are the
displacements when the tension has the values Pi and P 2 , respec-
tively, then
(5) yi(x, h) = y 2 (x, t 2 ) if k\/Pi = £ 2 a/P^
That is, the same set of instantaneous positions is assumed by
the string whether the tension is Pi or P 2 , but the tim es U a nd
t 2 required to reach any one position are in the ratio \/P 2 /Pi.
Nonhomogeneous Differential Equations. The substitution of a
new unknown function sometimes reduces a linear differential
equation which is not homogeneous to one which is homogeneous,
so that our method of solution can be employed.
To illustrate this, consider the problem of displacements in a
stretched string upon which an external force acts proportional
to the distance from one end. If the initial displacement and
velocity are zero, the units for t and x can be so selected that the
Sec. 44] SOLUTION OF BOUNDARY VALUE PROBLEMS
101
problem becomes
^ = H + Az (0 < * < 1, t > 0),
2/(0, t) = 0, 2/(1, t) = 0,
2 /(*, o) = o, = 0.
In terms of the new function F, where
2/(x, 0 = Y(x , 0 + ^(x),
and iA(x) is to be determined later, the differential equation
becomes
d 2 Y d 2 Y
= -^2 + ^" 0*0 + Ax (0 < x < 1 , £ > 0 ).
This will be homogeneous if
(6) ^"(x) = —Ax (0 < x < 1).
The first pair of boundary conditions on F are
m o + m = o, f(i, t) + ^d) = o;
hence these arc homogeneous if
(7) *(0) = 0, *(1) = 0.
In view of conditions (6) and (7),
(8) i (z) = 4 (* - .x 3 ) (0 < x < 1),
and with this choice of ^ the problem in F becomes a special
case of the problem in the preceding section; for the initial
conditions arc
Y{x, 0) = -i(x), = 0.
The solution of our problem in forced vibrations therefore can
be written ,
(9) y = yjy{x) - - t) + x + /)],
where ^(x') is the odd periodic extension of the function ^(x')
defined by equation (8) in the interval (0, 1).
102 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 45
PROBLEMS*
1. Carry out the details in the derivation of formula (1).
2. Write out the steps used in deriving formulas (2) from (1).
3. Show that relation (5) fails to hold between the displacements of a
given string under different tensions if the initial velocity is the same
in both cases and not zero. What change in initial velocity must accom-
pany an increase in tension to cause a more rapid vibration with the
same amplitude?
4. A string is stretched between the points (0, 0) and (1,0). If it is
initially at rest on the z-axis, find its displacements under a constant
external force proportional to sin ttx at each point. Verify your solu-
tion by showing that it satisfies the equation of motion and all boundary
conditions’. Ans. y = A/(j 2 a 2 ) sin7rx(l — cos 7 rat).
5. A wire stretched between two fixed points of a horizontal line is
released from rest while it lies on that line, its subsequent motion being
due to the force of gravity and the tension in the wire. Set up and
solve the boundary value problem for its displacements. Show that its
solution can be written in the form (9), if a = 1 and \p(x) — (: x 2 — Lx)g / 2
in the interval (0, L), where g is the acceleration of gravity.
45. Temperatures in a Slab with Faces at Temperature Zero.
Let a slab of homogeneous material bounded by the planes
x = 0 and x = ir have an initial temperature u = f(x), varying
only with the distance from the faces, and let its two faces be
kept at temperature zero. The formula for the temperature u
at every instant and at all points of the slab is to be determined.
In this problem it is clear that the temperature is a function
of the variables x and t only; hence at each interior point this
function u(x } t) must satisfy the heat equation for one-dimen-
sional flow,
/i\ du , d 2 u ^
(1) -jfi = k (0 < x < V, t > 0).
In addition, it must satisfy the boundary conditions
(2) t*(+0, t) = 0, w(tt - 0, 0 = 0 (t> 0),
(3) u(x, +0) = f(x) (0 < x < tt).
The boundary value problem (l)-(3) is also the problem
of temperatures in a right prism or cylinder whose length is ir
* Only formal solutions of the boundary value problems here and in the
sets of problems to follow are expected, unless it is expressly stated that the
solution is to be completely established.
Sec. 45] SOLUTION OF BOUNDARY VALUE PROBLEMS 103
(taken so for convenience in the computation), provided its
lateral surface is insulated. Its ends x = 0 and x = t are held at
temperature zero and its initial temperature is f(x ).
To find particular solutions of equation (1) that satisfy
conditions (2), we write u = X(x)T(t). When substituted in
equation (1), this gives XT' = kX"T , or
X" _ V
~x~W
Since the function on the left can vary only with x and the one
on the right only with t, they must both equal a constant a;
that is,
(4) X" - aX = 0, T' - akT = 0.
Moreover, if the function XT' is to satisfy conditions (2), then
(5) X(0) = 0, X(r) = 0,
provided X(x) is a continuous function.
The solution of the first of differential equations (4) that
satisfies the first of conditions (5) is X = C\ sinh X's/a, and this
can satisfy the second of conditions (5) only if
a = — ft 2 (ft = 1, 2, • • - ).
Then X = C 2 sin nx. The solution of the second of equations
(4) is, then, T — CV n2 K Hence the solutions of equations (1)
and (2) of the form u — XT are
(6) b n (~ n2kt sin nx (?i = 1, 2, ■ • • ),
where the constants b n are arbitrary.
Clearly no sum of a finite number of functions (6) can satisfy
the nonhomogencous condition (3) unless f(x) happens to be a
linear combination of sines of multiples of x. But the infinite
series of those functions,
0Q
(7) u(x, /.) = ^ b n c~ n * kt sin nx ,
does in general reduce to f(x) in (0, 7 r) when t = 0, provided the
coefficients b n are those of the Fourier sine series for /(a:); namely,
.-5 r
7T JO
f{x) sin nx dx (n = 1,2,
b
104 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 45
More precisely, if f(x) is sectionally continuous and has one-
sided derivatives at all points in (0, r), then
QO
u(x, o ) = 2} bn sin nx = i[f(x + 0) + fix — 0)] (0 < x < 7 r),
and this represents f(x) at each interior point where f(x) is
continuous.
With those mild restrictions on f(x) } then, the solution of the
problem is
(8) u(x } t) = ^ e~ n * kt sin nx J fix') sin nx 1 dx r ,
provided this series converges to a function u(x, t) such that
u(x, +0) = u(x , 0) when 0 < x < tt, ^(+0, t) = u( 0 , t) and
u(t — 0, t) = u(tt, t) when t > 0, and provided the series can
be differentiated termwise once with respect to t and twice
with respect to x when t > 0 and 0 < x < tt. It will be shown
in the next section that the series does satisfy those conditions.
PROBLEMS
1. Solve the above problem if the faces of the slab are the planes
x — 0 and x — L.
a / ,v 2 ( nVki\ . mrx C L . mrx'
Ans. u(x, t) = Tj Za exP ^ jJ~) sm irj 0 ^ sin ~TT dx> '
2. Find the formula for the temperatures in a slab of width L which
is initially at the uniform temperature u 0 , if its faces are kept at tem-
perature zero.
Ans ‘ u(x, 0-^r2 2^=1 ex P
(2 n - 1 )Vkt
IJ
sin
(2 n — 1)7 rx
L
3. The initial temperature in a bar with ends x = 0 and x = tt is
u = sin x. If the lateral surface is insulated and the ends are held at
zero, find the temperature u(x, t) . V erify your result completely. How
does the temperature distribution vary with time?
Ans. u = e~ kt sin x.
4. Write the solution of Prob. 1 if f(x) = A when 0 < x < L /2
fix) = 0 when L/2 < x < L.
Ans.
_ M sin 2 (mr/4) / nVkt\ . mrx
tt n exp L 2 J sm ~L~
Sec. 46] SOLUTION OF BOUNDARY VALUE PROBLEMS 105
5. Two slabs of iron, each 20 cm. thick, one at temperature 100°C.
and the other at temperature 0°C. throughout, are placed face to face
in perfect contact, and their outer faces are kept at 0°C. (compare
Prob. 4). Given that 7c = 0.15 c.g.s. (centimeter-gram-second) unit,
find to the nearest degree the temperature 10 min. after contact was
made, at a point on their common face and at points 10 cm. from it.
Am. 37°C.;33°C.; 19°C.
6. If the slabs in Prob. 5 are made of concrete with k = 0.005 c.g.s.
unit, how long after contact will it take the points to reach the same
temperatures found in the iron slabs after 10 min.? Am. 5 hr.
46. The Above Solution Established. Uniqueness. It is not
difficult to show that the series found in Sec. 45, namely
00 1
(1) ^ b n e~~ n ' kt sin nx ,
represents a function u(x, t) which satisfies all the conditions
of the boundary value problem, provided the initial temperature
function /(x) is sectionally continuous in the interval (0, x) and
has one-sided derivatives at all interior points of that interval.
For the sake of convenience, we define the value of J{x) to be
%[f(x + 0) +f(x — 0)] at each point x where the function is
discontinuous.
Since |/(x)| is bounded,
|6 n | =-\ \ f(x) sin nx dx ^ - I |/(x)| dx < M,
7T | Jo T Jo
where M is a fixed number independent of n. Consequently, for
each to > 0,
| b n e~ n2kt sin nx\ < Me- n * kto when t ^ t Q .
The series of the constant terms c- nHto converges; hence, accord-
ing to the Weierstrass M-test, series (1) converges uniformly
with respect to x and t when t ^ to, 0 ^ x ^ x. Also, the terms
of series (1) are continuous with respect to x and t, so that
the function u(x, t) represented by the series is continuous for
those values of x and t ; consequently, whenever t > 0,
u(+ 0, 0 = u( 0, t) = 0,
u(x — 0, l) = u(ir , t) = 0.
106 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 46
The terms of the series obtained by differentiating (1) with
respect to t satisfy the inequality
| —kb n n 2 e~- n2kt sin nx\ < kMn 2 e~ n2kt ° when t 5; to.
Since the series whose terms are n 2 e ~ nm ° also converges, accord-
ing to the ratio test, that differentiated series is uniformly
convergent for all t ^ to- Hence series (1) can be differentiated
termwise; that is,
oo
= 2 ^ ( b ^~ nakt sin nx ) <t > o).
"i
In just the same way it follows that the series can be differenti-
ated twice with respect to x whenever t > 0, and since each
term of series (1) satisfies the heat equation, the function u(x, t )
must do so whenever t > 0 (Theorem 2, Chap. I).
It only remains to show that u(x , t) satisfies the initial condition
(2) u{x j +0) = f{x) (0 < x < 7 r).
This can be shown with the aid of a test, essentially due to
Abel, for the uniform convergence of a series. At this time
let us show how the test applies to the present problem, and
defer the general statement of the test and its proof to the
following chapter (see Theorem 1, Chap. VII).
oo
For each fixed x (0 < x < ir), the series b n sin nx con-
verges to f(x). According to Abel’s test, the new series formed
by multiplying the terms of a convergent series by the cor-
responding members of a bounded sequence of functions of t,
such as e~ n * k \ whose functions never increase in value with n,
converges uniformly with respect to t. Series (1) therefore con-
verges uniformly with respect to t when 0 ^ t ^ h, 0 < x < w,
for every positive t\.
The terms of series (1) are continuous functions of t ; hence
the function u(x, t) represented by that series is continuous with
respect to t when t ^ 0 and 0 < x < r. Therefore
u(x, +0) = u(x, 0),
and condition (2) is satisfied because u(x, 0) = f(x) (0 < x <i r).
The function u{x, t ) is now completely established as a solution
of the boundary value problem (l)-(3), Sec. 45.
Sec. 46] SOLUTION OF BOUNDARY VALUE PROBLEMS
107
It is necessary to add to the statement of the problem some
further restrictions as to the properties of continuity of the
function sought, before we can prove that we have the only
solution possible. We illustrate this by stating one complete
form of the problem. For the sake of simplicity, we shall impose
rather severe conditions of regularity on the functions involved.
A Complete Statement of the Problem . Let the function u(x, t)
be required to satisfy the heat equation and boundary conditions
as given by equations (1) to (3), Sec. 45, in which the function
/( x) is now supposed continuous in the interval 0 ^ x g 7 r. We
also assume that /( 0) = f(r) = 0, and that f'(x) is sectionally
continuous in the interval (0, it) . In addition let it be required
that u(x , t) be continuous with respect to the two variables
x , t together when 0 ^ x ^ t, t ^ 0, and that the derivative
du/dt be continuous in the same manner whenever t > 0.
We can show that there is just one possible solution of this
problem, and that solution is the function represented by series
(1) *
It was shown above that that function satisfies the heat equa-
tion and boundary conditions; also, that the series for du/dt
converges uniformly with respect to x and t together when
0 g x g tt, t t o (/<o > 0). Since the terms of the derived
series arc continuous functions of x and t together, it follows that
du/dt is continuous with respect to both variables together
whenever t > 0,
The continuity of the function when 0 ^ x ^ tt and t ^ 0
follows again from our form of Abel’s test. For the conditions
00
on /(;r) ensure the uniform convergence of the series ^ 6 n sin nx.
In this case the introduction of the factors e~ n * kt into the terms
of that series produces a series which is uniformly convergent
with respect to x and t together, when 0 ^ x ^ t, 0 g t S h,
for every positive h. Hence scries (1) has this uniform con-
vergence, and the continuity follows as before.
The function defined by series (1) therefore satisfies all the
conditions of the problem. Of course, the derivative d 2 u/dx 2 is
continuous in the same sense as du/dt , since these two deriva-
tives differ only by the factor k.
* Concerning the continuity of a series with respect to more than one
variable, see the remarks preceding Theorem 1, Chap. VII.
108 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 47
It is not difficult to show that two distinct functions, satisfying
all the requirements made upon u( x, t) in the above statement
of the problem, cannot exist. The complete statement and
proof of this uniqueness theorem will be given later (Theorem 2
Chap. VII). If we accept this statement for the present, the
only possible solution of our problem has been found.
47. Variations of the Problem of Temperatures in a Slab.
With only slight modifications in the method, the temperature
distribution can be found for the slab of Sec. 45 when the faces
are subject to certain other conditions, or when the heat equation
is modified.
a. One Face at Temperature A. To find the temperature
u(x, t) in a slab with initial temperature f(x) when the face
x = 0 is held at zero and the face x = % at constant temperature
A, a simple transformation can be used to obtain the result from
that of Sec. 45.
Here u(x, t ) must be a solution of the boundary value problem
du _ , d 2 U
Ht ~ “dx*
m(+0, t) = 0, u(tt — 0, t )
U(x, +0) = f(x).
(0 < x < it, t > 0),
= A,
It follows that the function
(1) v(x, t) = u(x, t ) — — x,
7T
must satisfy the conditions
dv _ d 2 v
dt ~ & dx 2
(0 < z < r, t > 0),
*K+o» t) = o, v (x — 0, t) = 0,
v(x, +0) = fix) --X.
7T
This is the boundary value problem of Sec. 45 with fix) replaced
by fix) — Ax/t, so that its solution is
vix, t)
sin nx
Ax r
7T
sin nx r dx r .
Substituting this for v(x } t) in equation (1) and carrying out
part of the integration, we obtain the following solution of the
Sec. 47 ] SOLUTION OF BOUNDARY VALUE PROBLEMS 109
problem:
u(x, t) = — x
7 r
+ ^ 2 e ~ nik ‘ s * n nx j^( — 1 )’‘ ~ + J* f ( x 0 sin nx> cfo'l.
6. Insulated Faces. Find the temperature u(x, t) in a slab
with initial temperature fix) if the faces x = 0 and x = ir are
thermally insulated.
Since the flux of heat through those faces is proportional to
the values of du/dx there, the boundary value problem can be
written
( 2 )
du _ , d 2 u
~dt ~ W 2
(0 < x < T, t > 0),
(3)
(4)
drt(+0, t) _
dx
0 ,
du(£ — 0, f) _ 0
dx
u(x, +0) = f(x)
it > 0),
(0 < x < it).
Setting u = X{x)T(t), it is found that the functions
a n e~ ntkt cos nx (n = 0, 1, 2,
)
satisfy the homogeneous conditions (2) and (3). The infinite
series of those functions satisfies condition (4) as well, provided
the coefficients a n arc those in the Fourier cosine series correspond-
ing to fix). So if fix) satisfies the conditions of our Fourier
theorem, the solution of the problem is
(5) uix, t)
c. One Face Insulated. If the face x = 0 is held at tempera-
ture zero and the face x = x is insulated, the problem can be
reduced to one in which both faces are held at zero.
Let the slab be extended to x = 2ir with the face x = 2ir
held at temperature zero, and let the initial temperature of the
new slab be symmetric with respect to the plane x = x. Then,
when ir < x < 2ir, the. initial temperature is f(2ir — x), where
fix) is the initial temperature of the original slab. In the
110 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 47
physical problem the symmetry indicates clearly that no heat
will flow through the plane x = tt. When the solution is found,
it can be verified that du/ dx = 0 when x = x.
According to Prob. 1, Sec. 45, the temperature in the extended
slab is
,n . nx' , ,
x') sm dx' .
By substituting a new variable of integration in the second
integral, this can be reduced to
2 C*
i(x, t) = - ^ e -m\kt s i n mnX /(s') sin m n x' dx',
n = 1
where m n = (2n — l)/2. When 0 ^ x ^ t, this is the solution
of problem c.
d. The Radiating Wire . Suppose the diameter of a wire or
bar is small enough so that the variation of temperature over
every cross section can be neglected. If the lateral surface is
exposed to surroundings at temperature zero and loses or gains
heat according to Newton’s law, the heat equation takes the
form
( 6 )
du _ , d*u
~dt ” dx*
hu.
where x is the distance along the wire and h is a positive constant.
Newton’s law of surface heat transfer is an approximate law
of radiation and convection according to which the flux of
heat through the surface of a solid is proportional to the differ-
ence between the temperature of the surface and that of the
surroundings. It is generally valid only for small temperature
differences; but it has the advantage over the more exact laws
of being a linear relation. That the heat equation does take
the form (6) when such surface heat transfer is present can be
seen from the derivation of the heat equation (Sec. 9).
When the ends x = 0, x = 7r, of the radiating wire are kept
at temperature zero and the initial temperature is f(x), the
temperature function can be found by the method of Sec. 45.
Sec. 47] SOLUTION OF BOUNDARY VALUE PROBLEMS 111
The result is
(7) u(x , t) = e~ ht Ui(x, t),
where Ui(x, t) is the function u in equation (S), Sec. 45.
When the ends are insulated, the result is
(8) u(x, t) = e~ ht u 2 (x, t),
where u 2 (x, t) is the function u in equation (5) above.
PROBLEMS
1. Derive the solution of the problem in Sec. 476 above when the faces
are x — 0 and x = L.
A ns. u = i f(x') dx'
+
12'
nWJct
L‘i
COS
IT jc Kx ' ]
cos ■
UTX
dx',
2. Show that the result of Prob. 1 can be completely established as a
solution of the boundary value problem by the method of Sec. 46.
3. Solve the problem in Sec. 47c above for a slab of width L with the
face x — L insulated. It will be instructive to carry out the solution
directly by obtaining particular solutions u = XT, without using the
method of extension, noting the orthogonal functions generated by the
differential equation in A r and its boundary conditions (compare Sec. 25).
A ns. u
-12'
w = 1
P
>l n X
Jo
f(x') sin m n x' dx',
where m n — (n — l/2)ir/L.
4. Derive formula (7).
5. Derive formula (8).
6. Use the substitution v = uc ht to simplify equation (6) and, by
writing the boundary value problem in terms of v{x , t), obtain formulas
(7) and (S) from known results.
7. For a wire in which heat is being generated at a constant rate,
while the lateral surface is insulated, the heat equation takes the form
du _ dhi
dt ~ k Ox *
+ B,
where B is a positive constant. If the ends x = 0 and x = tt are kept
at temperature zero and the initial temperature is f(x), set up the
boundary value problem for u(x, t) and solve it. Note the result when
112 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 48
f(x) = Bx(i r — x)/(2 k). Suggestion : Apply the method used in Sec. 44
to reduce the nonhomogeneous differential equation to a homogeneous
one.
Ans . u = ^ (tt — x)
+
12
e -n*kt gi n nx
c* r Bx r , ,
Jo L» (I
■)+/(*')
sin nx' dx f .
is insulated, instead of being
8. Solve Prob. 7 when the end x = i r
kept at temperature zero.
9. A wire radiates heat into surroundings at temperature zero. The
ends x = 0 and x = t are kept at temperatures zero and A , respectively,
and the initial temperature is zero. Set up and solve the boundary
value problem for the temperature u(x, t). Suggestion: Substitute
v — u + yp{x) : then determine ^ so that hf/" — hf/ = 0 and \[/(fl) = 0,
^(tt) = -A.
Ans.
u = A sinh xVh/l + 2Ak e _ ht
sinh iry/h/k ir
n( — l) n
h + kn 2
e -n*kt s i n nx.
10. The face x = 0 of a slab is kept at temperature zero and heat is
supplied or extracted at a constant rate at the face x = tt, so that
du/dx = A when x - tt. If the initial temperature is zero, derive the
formula
u “ Ax + 2 (i -H)2 e (n i,! ‘ sin _ i)*.
for the temperatures in the slab, where the unit of time has been so
chosen that h = 1.
48. Temperatures in a Sphere. Let the initial temperature in
a homogeneous solid sphere of radius c be a function /(r) of the
distance from the center, and let the surface r = c be kept at
temperature zero. The temperature is then a function u{r , t),
of r and t only, and the heat equation in spherical coordinates
becomes
du _ h d 2 (ru)
dt r dr 2
The boundary conditions are
u(c — 0, t) = 0
utr, +0) = f(r)
(t > 0),
(0 < r < c).
Sec. 48] SOLUTION OF BOUNDARY VALUE PROBLEMS
113
If we set v(rj t) = ru(r , t ), the boundary value problem here
can be written
dv 7 d 2 v
di = k d?’
t) = 0, v(c — 0, t) — 0,
v (r, +0) = rf(r),
where the condition v(+0, t) = 0 is included because u(r, t )
must be bounded at r = 0. Except for the presence of r instead
of x and r/(r) instead of fix), this problem is that of the tempera-
tures in a slab of width c. Hence the temperature formula
for the sphere can be written at once (Prob. 1, Sec. 45). It is
( 1 )
n 2 r 2 kt
sin
mrr
J r'f(r') sin dr'.
PROBLEMS
1. Find the temperatures in a sphere if the initial temperature is zero
throughout and the surface r — c. is kept at constant temperature A.
Ans. u(r, t) — A +
2 Ac
TV
(-PV
n
nhr 2 kt
sin ■
2. Prove that the sum of the temperature function found in Prob. 1
and the function given by formula (1) above represents the temperature
in a sphere whose initial temperature is f(r) and whose surface is kept
at temperature A.
3. An iron sphere with radius 20 cm., initially at the temperature
100°C. throughout, is cooled by keeping its surface at 0°C. Find to
the nearest degree the temperature at its center 10 min. after the cooling
begins, taking k = 0.15 e.g.s. unit. Ans. 22°C.
4. Solve Prob. 3, assuming that the sphere is made of concrete with
k = 0.005 e.g.s. unit. Ans. 100°C.
5. The surfaces r = b and r — c of a solid in the form of a hollow
sphere are kept at temperature zero. The initial temperature of the
solid is f(r) {b < r < c). Derive the following formula for the tem-
peratures u(r , t) in the solid:
nVkt
(c - by
. mr(r — b)
sin — — i
c — b
mr(r — b)
r/(r) sin - —
dr.
where
114 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 49
6. Show that when the surface of a sphere is insulated, the solution
of the temperature problem no longer involves the expansion of rf(r) in
a Fourier series, but an expansion in a series of the functions sin a n r,
where a n are the roots of the mixed equation tan ac — ac . Show why
these f un ctions form an orthogonal set in the interval 0 < r < c
(Sec. 25).
49. Steady Temperatures in a Rectangular Plate. Let u{x, y)
be the steady temperature at points in a plate with insulated
faces, the edges of the plate being the lines (or planes) x = 0,
x = a, y = 0, and y — b. Let three of the edges be kept at
temperature zero and the fourth at a fixed temperature distribu-
tion. Then u(x , y) is the solution of the following problem:
(1) S + 0 =O (0<*<a,0 <y<b),
(2) w(+0, y) = 0, u(a - 0, y) = 0, (0 < y < b ),
(3) u(x, b — 0) = 0, u(x, +0) = f{x), (0 < x < a).
Since special case (1) of the heat equation is also a case of
Laplace’s equation, the function u(x, y) is also the potential
in the rectangular region when the potential on the edges is
prescribed by conditions (2) and (3). The region also may be
considered as an infinitely long rectangular prism, or the right
section of any prism in which the potential or steady temperature
depends only upon x and y.
Settings = X(x)Y(y), the functions
sin ^ sinh ^ (y - C) J (n = 1, 2, • • • )
are found as solutions of (1) which satisfy conditions (2), for
every constant C. If C = b, they also satisfy the first of con-
ditions (3), and the series
u = A n sin sinh (y — b) j
satisfies the nonhomogeneous condition in (3) provided
fix) = — 2 A n sinh sin (0 < x < a).
i
According to the Fourier sine series, this is true if the coeffi-
cients A n are determined so that
Sec. 49] SOLUTION OF BOUNDARY VALUE PROBLEMS 115
— A n sinh
mrb
a
. mcx
Sill —
a
dx .
So the formal solution of the problem can be written
“<*■*> - ; % - 1 * ” j>> * ¥ *■
Our result can be completely established as a solution of the
problem (l)-(3) by the method used in See. 46. But in this
case let us defer that part of the discussion, along with a com-
plete statement of the problem which ensures just one solution^
until a later time when the necessary tests have been derived
(Sec. 59).
PROBLEMS
1. Find the solution of the above problem if u(x, y) is zero on all edges
except x = a, and u(a, y) = g(y).
Am.
sinh (ft7 rx/b) . mr ; y
sinh (mra/b) SlU b
g{y') sin
2. When the temperature distributions on all four edges are given,
show how the formula for the steady temperatures in the plate can be
written by combining results already found.
3. What is the steady temperature at the center of a square plate
with insulated faces, (a) if three edges arc kept at ()°C. and the fourth
at 100°C.; (6) if two adjacent edges arc kept at 0°C. and the others at
100°C.? Suggestion: Superpose the solutions of like problems here to
obtain the obvious case in which all four edges are kept at 100°C.
Am. (a) 25°C.; ( b ) 50°C.
4. A square plate has its faces and its edge y — 0 insulated. Its
edges x - 0 and x — x are kept at temperature zero, and its edge y = x
at temperature fix). Derive the formula for its steady temperature.
CO
Am. u(x, y) - ^ ■ ° S J 1 tl]f sin nx I f K x') sin ?ix' dx'.
x cosh nx J ()
5. Derive the formula for the electric potential V{x, y) in the space
0 ^ x ^ L, y ^ 0, if the planes x = 0 and x — L are kept at zero
potential and the points of the plane y — 0 at the potential f(x), if
V(x , y) is to be bounded as y becomes infinite.
Ans. V(x, y) = j
-r2-
niry
" L
sin
r
. mrx'
fix') sin
dx',
116 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 50
6. Find the electric potential in Prob. 5 if the planes x = 0 and x = L,
instead of being kept at potential zero, are insulated, so that the electric
force normal to those planes is zero; that is, dV/dx = 0. Also state
this problem as a temperature problem.
Am. Vix^y) = ^ J fix') dx'
2 nTX f L ,\ nwx' ,
+ ^ e L cos J /(* ) cos — jj- ax .
7. Solve Prob. 5 if the electric potential is zero on the plane x = 0
and the electric force normal to the plane x = L is zero.
8. Find the steady temperatures in a semi-infinite strip whose faces
are insulated and whose edges x = 0 and x = ir are kept at temperature
zero, if the base y = 0 is kept at temperature 1 (Prob. 5).
Ans. u(x, y) = - ^e-*' sin x + | e~ Zy sin 3x + jr e~ hy sin ox + * ■ ■
9. In the power series expansion of [log (1 + z) — log (1 — z)]
(|s| < 1), set z = re l(p and equate imaginary parts to find the sum of the
series
S = r sin cp + \r z sin 3<p + ir 5 sin 5<p + • • • ;
also note that
log [p( cos d + i sin 6)] - log ( pe 1 6 ) = log p + i&,
and therefore show that
S =
1 , 2r sin <p
2 arctan ^ *
Thus show that the answer to Prob. 8 can be written in closed form as
follows:
u{x i y) — ~ arctan
sin x
sinh y
Verify the answer in this form. Also trace some of the isotherms,
u(x , y) ~ a constant.
50. Displacements in a Membrane. Fourier Series in Two
Variables. Let z represent the transverse displacement at each
point (x ) y) at time t in a membrane stretched across a rigid
rectangular frame in the zy-plane. Let the boundaries of the
rectangle be the lines x = 0, x = xo, y = 0, and y = yo. If the
initial displacement z is a given function f(x, y) } and the mem-
brane is released from rest after that displacement is made, the
Sec. 50] SOLUTION OF BOUNDARY VALUE PROBLEMS 117
boundary value problem in z(x, y, t ) is the following:
/ 1 \ d 2 z _ „ / d 2 z dh\
^ dt 2 a \dx 2 + dy V’
zfro, y, t ) = 0;
z{x, 2 / 0 , <) = 0;
= 0, z(x, y, 0) = f(x, y).
s(0, y, t ) = 0,
z(x, 0, t) = 0,
dzQc, y, Q)
dt
In order that the product z = X(x)Y{y)T(t) be a solution
of equation (1), its factors must satisfy the equation
TIL - TL _l XL
a 2 T X + 7*
All three terms in this equation must be constant, since they are
functions of x ) y , and t separately. Write
then
T" = - a\a 2 + /3 2 )7 7 .
The solutions of these three equations, for which z = XYT satis-
fies all the homogeneous boundary conditions, are
X = sin ax, Y = sin @y , T = cos (aV a 2 + j B 2 t),
where a = mw/x o, and (3 = mr/ijo (m, n = 1, 2, • • • ).
So the function
( 2 )
/ . fm 2 . n 2 \ • . n7r?/
cos I 7r at . — *• q — r ) sin sin — -
V \ y\) ^ yo
satisfies equation (1) and all the boundary conditions, formally,
provided the coefficients A mn can be determined so that z = f(x, y)
when t = 0; that is, provided
22 *
rn = 1 // = 1
. mirx . mry
sin sin
xq y o
(3) f(x, y)
m = 1 n = l
(0 ^ x ^ Xo, 0 ^ y ^ yo).
By formally grouping the terms of the series, equation (3)
can be written
< 4 )
rn*l 'n»l '
118 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 50
For each fixed y between zero and y 0 this series is the Fourier
sine series of the function /(x, y) of the variable x (0 g j
provided the coefficients of sin ( nnrx/xo ) are those of the Fourier
sine series. So equation (4) is true in general if
(5)
sm
ft = i
nr y
2/o
-5 J.
. m ttx , .
sm dx r .
x 0
Again, using the formula for the Fourier sine coefficients of the
functions F m (y) f where
r% pxo /
Fm(v) = - J o fix', y ) sin dx’ (0 ^ y ^ y 0 ),
expansion (5) is valid if
A mn = - C° F m (y’) sin VElI dy’ .
2/o Jo 2/o
The series in equation (3) is then a Fourier sine series in two
variables for /(x, y ) provided its coefficients have the values
( 6 )
A 4 p° . r*° . . . mrx . nry ,
A mn = — I dy f(x,y) sm sm — ^ dx.
Zo2/o Jo Jo x 0 2/0
The formal solution of the membrane problem is then given
by equation (2) with the coefficients defined by equation (6).
According to equation (2), the displacement z is not in general
periodic in t, since the numbers [(m 2 /xg) + (n 2 /?/§)]* do not
change by multiples of any fixed number as the integers m and n
change. Consequently the vibrating membrane, in contrast to
the vibrating string, will not generally give a musical note. It
can be made to do so, however, by giving it the proper initial
displacement.
If, for instance, for any fixed integers M and N,
z(x, y, 0) = A sin
Mrx
x 0
. Nry
sm —
2/o
then the displacement (2) is given by a single term :
/ \ / m ata
z \ x , V, 0 = A cos ( rat H 2 * ) sin ;
\ \ x o 2/o /
In this case z is periodic in t with the period
Mrx
Xo
sin
Nry
~y ~
(2/a)(Myxl + Ny v i)-y
Sec. 50] SOLUTION OF BOUNDARY VALUE PROBLEMS
119
PROBLEMS
1. Solve the above problem if the membrane starts from the position
of equilibrium, 3 = 0, with an initial velocity at each point; that is,
- "<*. »>•
2. The four edges of a plate tv units square are kept at temperature
zero and the faces are insulated. If the initial temperature is/(x, y ),
derive the following formula for the temperature u{x, y, t) :
SO 00
u — X X A mn exp [ — A:(m 2 + n 2 )t] sin mx sin ny,
m = 1 n = 1
where
4 C T f 6 7 "
A mn = — , I sin ny dy I f(x, y) sin mxdx.
T Jo Jo
3. When fix, y) = Ax, show that the solution of Prob. 2 is
u = 1/1(2, t)u 2 (y, t), where
Ml
.+1
— C
(-1)"
n
nZkt sin nXj
>-n*kt s i n n y '
Show that ?/i and u 2 represent temperatures in cases of one-dimensional
(low of heat with initial temperatures Ax and 1, respectively.
4. Solve Prob. 2 if, instead of being kept at temperature zero, the
edges arc insulated. Note the result when f(x, y) = 1.
5. If the faces x = 0, x = tt, y = 0, y = tt, z = 0, 2 = tt of a cube
are kept at temperature zero and the initial temperature is given at each
point u(x, y, z, 0) = fix, y, z), show that the temperature function is
u(x, //, 2, 0 = X XX Amnv Hitl mX Sln Uy Sin PZe * (w2+, * S+pl) h
m “ 1 7i—l /> = 1
where
I I I /(•£, y, z) sin ms sin ny sin ps ds dp dz.
7r ’ Jo Jo Jo
6. When f{x, ?/, 2) = 1, show that the solution of Prob. 5 reduces to
u = u 2 (x , 0^2(3/, £)m 2 (z, 0, where the function u 2 is defined in Prob. 3.
120 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 51
51. Temperatures in an Infinite Bar. Application of Fourier
Integrals. Let the length of a homogeneous cylinder or prism
be so great that it can be considered as extending the entire
length of the z-axis. If the lateral surface is insulated and the
initial temperature is given as a function f(x) of position along
the bar, the temperature u(x , t) is the solution of the following
boundary value problem:
(1) Tt =h M (-*<*<
(2) u(x, +0) = f(x) (—00 < X < oo).
Particular solutions of equation (1) which are bounded for all
x and t (t ^ 0) are found by the usual method to be
(3) e~ a2kt cos [a(x + C)],
where a and C are arbitrary constants. Any series of these
functions, formed in the usual manner by taking a as multiples
of a fixed number, would clearly reduce to a 'periodic function
of x when t = 0. But f(x) is not assumed periodic, and con-
dition (2) is to be satisfied for all values of x; hence it is natural
to try to use the Fourier integral here in place of the Fourier series.
Since function (3) is a solution of equation (1), so is the
function
- f{x r )e~ a * kt cos [a(x r — x)] 7
7 r
where the parameters x' and a are independent of x and t. The
integral of this with respect to these parameters,
1 f* o° f* oo
(4) u(x , t) = - I da I f(x')e- a * kt cos [a(x' — z)] dx f ,
* J0 J- oo
is then a solution of equation (1) provided this integral can be
differentiated twice with respect to x and once with respect to t
inside the integral signs.
When t = 0, the right-hand member of equation (4) becomes
the Fourier integral corresponding to fix). Hence if f(x ) satisfies
the conditions of the Fourier integral theorem, and if the function
u{x , t) defined by equation (4) is such that u(x , 0) == u(x, +0),
then
<x, +0) =*[/(* + 0)+ /(a -0)],
and this is condition (2) at each point where f(x) is continuous.
Sec. 51] SOLUTION OF BOUNDARY VALUE PROBLEMS 121
The solution of the problem is therefore given, at least formally,
by equation (4) . By inverting the order of integration and using
the integration formula
J 0 cos «(*' - x) da = exp [ - (x ~ f >* ] <t > 0),
equation (4) becomes
(6) <x ' 0 ” 5^7S( “ xp [“ t]* 1
This can be still further reduced by using a new variable of
integration £, where
x' — x
* -
this gives
(6) u( x, t) = -L
V? r
When fix) is bounded for all values of x and integrable in
every finite interval, it can be shown that the function defined
by equation (5) satisfies equation (1) and condition (2).* Under
those conditions, then, the required solution is given either by
equation (5) or by equation (6) .
a
PROBLEMS
J* fix + 2 Vkt £)<r-* s d£.
1. Derive the temperature function for the above bar if fix) is periodic
with period 2tt.
1 C*
Ans. u(x, t) — ^ I fix') dx'
00
+£2*
1
2. If fix) = 0 when x < 0, and fix) = 1 when x > 0, show that the
temperature formula for the infinite bar becomes, for t > 0,
,% 1C
fix') cos [nix' — x)] dx'.
2 y/kt
1 , _1 r a: x 1 2 3 _
2 s/ir 2 Vkt 3(2 Vkty 5 • 2!(2 \/kty
* For a proof, see p. 31 of Ref. 1 at the end of this chapter.
122 FOURIER SERIES AND BOUNDARY PROBLEMS (Sec. 52
3. One very thick layer of rock at 100°C. is placed upon another of
the same material at 0°C. If k = 0.01 c.g.s. unit, find the temperatures
to the nearest degree at points 60 cm. on each side of the plane of con-
tact, 100 hr. after contact is made. Am . 76°C.; 24°C.
52. Temperatures in a Semi-infinite Bar. If the bar of the
foregoing section extends only along the positive half of the
z-axis and the end x = 0 is kept at temperature zero, the bound-
ary value problem in the temperature function u(x, t ) becomes
the following:
(1)
du , d 2 u
dt~ dx 2
(x > 0 , t > 0),
(2)
«(+o, t) = 0
(t > 0),
(3)
u(x, +0) = f(x)
(x > 0).
The solution can be formed from the function e ~ a * kt sin ax ,
which satisfies conditions (1) and (2). Multiplying this by
(2/7 r)f(x l ) sin ax' and integrating with respect to the parameters
a and x', which are independent of x and t, the function
2 r 00 r 00
(4) u(x, t) = - I e~ aikt sin ax da I f(x') sin ax' dx'
n Jo Jo
is found. When t = 0, the integral on the right reduces to
the Fourier sine integral of f(x), which represents f(x) when
0 < x < 00.
If we write
2 sin ax sin ax' — cos [ot(x' — re)] — cos [a(x' + a?)],
the integration formula used in the foregoing section can be
applied to reduce formula (4) to the form
<5)
4 kt
jy-
-U
(x -
A
1 kt
when t > 0. This can be written
r (a? + s') 2 ]) j ,
(6) v.ix, () = --L f e~ p f(x + 2 f)
V* J -X
da
2"\/kt
-J
e~Pf(—x + 2 \, / Ft £) da
iVki
Sec. 53] SOLUTION OF BOUNDARY VALUE PROBLEMS 123
These results can also be found directly from those of the last
section by making f(x) there an odd function. Under the
conditions stated in the preceding section, function (5) then
satisfies all the conditions of the problem.
PROBLEMS
1. When f(x) = 1, prove that the temperature in the semi-infinite bar,
or in a semi-infinite solid x S 0, with its boundary x = 0 at zero, is
X
o r 2 kt
u(x, t) = —pz I <?-i 2 d£
Jo
2 r x x 3 x 5
~ Vv 1.2 Vkt ~ 3(2 Vfe) 3 + 5 - 2!(2 s/ktf ~
2. When the end x = 0 is kept at temperature A and the initial
temperature of the bar is zero, show that
X
3. Show that when a semi-infinite solid initially at a uniform temper-
ature throughout is cooled or heated by keeping its plane boundary at a
constant temperature, the times required for any two points to reach
the same temperature are proportional to the squares of their distances
from the boundary plane.
4. Show that the function
x ~ —
U\ — ^ e 4kt
satisfies all conditions of the boundary value problem consisting of
equations (1) to (3) when f(x) — 0. Hence this function can be multi-
plied by any constant and added to the solution obtained above, to
obtain as many solutions of that problem as we please. But also show
that U[ is not bourn led at x = t = 0; this can be seen by letting x
vanish while x 2 = t.
63. Further Applications of the Series and Integrals. Many
other boundary value problems, arising frequently as problems
in engineering or geology, can be solved by the methods of this
chapter. A few will be slated at this point. The derivation
of the results given here can be left as problems for the student.
a. Electric Potential between Parallel Planes. The plane y = 0
is kept at electric potential V = 0, and the plane y = b at the
124 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 53
potential V =f(x). Assuming that the space between those
planes is free of charges, the potential V(x , y) in the space is to
be determined.
It can be shown that
Problems of this type are idealizations of problems arising
in the design of vacuum tubes. They are also problems in steady
temperatures, or steady diffusion, in solids ; hence their applica-
tions are quite broad. The following problem is another of the
same type.
b. Potential in a Quadrant. A medium free of electric charges
has the planes x = 0 and y = 0 as its boundaries. If those
planes are kept at electric potential 7 = 0 and 7 = f(x),
respectively, and if the potential V(x, y) is bounded for all x
and y (x ^ 0, y ^ 0), the formula for V(x, y) is to be found.
The result can be written, when y > 0, as
(2) 7
-fJM
i
2 + (x' - x) 1 y 2 + O' + x)
2
dx r .
When f{x) = 1, this formula becomes
(3) V = — arctan — •
tv y
In this case the equipotential surfaces are the planes x = cy ,
where the constant c has the value tan (wV/2).
c . Angular Displacements in a Shaft. Let 6(x, t) be the angular
displacement or twist in a shaft of circular cross section with
its axis along the z-axis. If the ends x = 0 and x = L of the
shaft are free, the displacements 0(x, t) due to an initial twist
9 = f(x) must satisfy the boundary value problem
cPd
dt 2
M(0, 0 = 0
dx
0) _ n
dt
a 2
d 2 e
dx 2 ’
dJiL, t) = Q
dx
9(x, 0) = /(*),
where a is a constant.
Sec. 53] SOLUTION OF BOUNDARY VALUE PROBLEMS 125
The solution of this problem can be written
,.v - 1 . nirx rnr at
(4) 6 = 2^0 + ^ a n cos -j- cos —j-*
i
where a n (n = 0, 1, 2, • • • ) are the coefficients in the Fourier
cosine series for f(x) in the interval (0, L).
d. The Simply Supported Beam. The differential equation
for the transverse displacements y(x , t) in a homogeneous beam
or bar was given in Sec. 12. At an end which is simply supported
or hinged, so that both the displacement and the bending moment
are zero there, it can be shown that d*y/dx 2 must vanish as well
as y . The displacements are to be found in a beam of length L
with both ends simply supported, when the initial displacement
is y = /($), and the initial velocity is zero.
The result is
/kn 2 . nirx nVct C L /N . nirx r ,
(5) 2 / = ^ > i sm "TT cos ~U~ ^ x ' sm ~TT dx 7
where c is the constant appearing in the differential equation.
PROBLEMS
1. Write the boundary value problem in Sec. 53a above, and derive
solution (1).
2. Write the boundary value problem in Sec. 536, and derive solution
( 2 ).
3 . Obtain solution (3) from (2), and show that the function (3)
satisfies all the conditions of the boundary value problem when /(a) = 1.
4. Derive the solution (4) of Sec. 53c. Also show how this formula
can be written in closed form in terms of the even periodic extension of
the function f(x).
6. Set up the boundary value problem in Sec. 53d, and derive solution
(5).
6. Derive the formula for the temperatures u(x, t) in the semi-
infinite solid x ^ 0, if the initial temperature is f{x) and the boundary
x = 0 is kept insulated.
7 . Find the formula for the displacements y(x, t) in a string stretched
between the points (0, 0) and (t, 0), if the string starts from rest in the
position y = f(x) and is subject to air resistance proportional to the
velocity at each point. Let the unit of time be selected so that
the equation of motion becomes
&V _ &V _ o h d]L
dt 2 “ dx* m dt ’
126 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 53
where h is a positive constant.
A ns. y
= e~ ht b n (cos Kj + sin Kj^j si
sin nx , where
K n = Vn 2 - h 2 ,
and b n are the coefficients in the Fourier sine series for f(x) in the
interval (Q, t).
8 . Let V(r, p) be the electric potential in the space inside the cylin-
drical surface r = 1, when the potential on this surface is a given function
f(p) of (p alone. Note that V (r, p) must be periodic in (p with period 27 t;
it must also be a continuous function within the cylinder, since the space
is supposed free of charges. Derive the following formula for V (r, <p) :
V = ia Q + ^ r n (a n cos np +■ b n sin np),
where a n and b n are the Fourier coefficients of f(p) for the interval
(-7T, tt).
9. In Prob. 8, suppose /(<p) = —1 when —tt < p < 0, and f(p) = 1
when 0 < <p < and show in this case that the potential formula can
be written in the closed form
V =
2 , 2 r sin p
- arctan -= 5 -
with the aid of the result found in Prob. 9, Sec. 49.
10. From the infinite solid cylinder bounded by the surface r = c
a wedge is cut by the axial planes ip = 0 and p = <p 0 . Find the steady
temperatures u(r, cp) in this wedge if u — 0 on the surfaces p — 0 and
p = <po, and u — f(p) (0 < p < po) on the convex surface of the wedge.
oo mr
Ans. u — ^ b n (r/cY Q sin ( nirp/p 0 ), where b n are the coefficients in
the Fourier sine series for f{p) in the interval (0, p Q ).
11. If in Prob. 10, f(p) = A, where A is a constant, show that the
formula for u{r, p) can be written in closed form with the aid of the
result found in Prob. 9, Sec. 49.
REFERENCES
1 . Carslaw, H. S.: “ Mathematical Theory of the Conduction of Heat in
Solids/' 1921.
2. Ingersoll, L. R., and O. J. Zobel: “Mathematical Theory of Heat Conduc-
tion with Engineering and Geological Applications," 1913.
3. Timoshenko, S.: “Vibration Problems in Engineering," 1937.
CHAPTER VII
UNIQUENESS OF SOLUTIONS
54. Introduction. For the most part, our solutions of the
boundary value problems in the last chapter were formal, in that
we did not usually attempt to establish our result completely,
or to find conditions under which the formula obtained represents
the only possible solution. We shall develop a few theorems
here which will furnish the reader interested in such matters with
a mathematically complete treatment of many of our problems.
A multiplicity of solutions may actually arise when the problem
is incompletely stated. Also, it is generally not a simple matter
to transcribe a physical problem completely into its mathe-
matical form as a boundary value problem. Consequently, the
precise treatment of such problems is of practical as well as
theoretical interest.
Our first theorem (Abel’s test) enables us to establish the
continuity of many of our results obtained in the form of
series. The continuity property is useful both in demonstrating
that our result is actually a solution of the boundary value
problem, and in showing that it is the only solution.
The remaining theorems give conditions under which not
more than one solution is possible. It will be evident that they
can be applied only to specific types of problems. But no
“general” uniqueness theorem exists in the theory of boundary
value problems in partial differential equations, in the sense that
the same theorem applies to temperature problems, potential
problems, etc.
The uniqueness theorems given below arc again special in
that they require a high degree of regularity of the functions
involved. But. they will make possible a complete treatment of
many of the problems considered in this book.
55. Abel’s Test for Uniform Convergence of Series. We now
establish a test for the uniform convergence of infinite series
whose terms arc products of specified types of functions. Appli-
cations of this test have already been made in the foregoing
127
128 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 55
chapter, to establish the continuity of the solution of a boundary
value problem (Sec. 46).
The function represented by a uniformly convergent series
of continuous functions is continuous. This is true regardless of
the number of independent variables, as will be evident upon our
recalling the method of proof for a single variable.* It is to
be understood that the terms of the series are continuous with
respect to all the independent variables taken together, in some
region. The uniform convergence of the series in this region
then ensures the same type of continuity of the sum of the series.
A sequence of functions T n (t) (n = 1,2, - - - ) is said to be uni-
formly bounded for all values of t in an interval if a constant K,
independent of n, exists for which
( 1 ) \Tn(t)\<K
for every n and all values of t in the interval. The sequence is
monotone with respect to n if either
(2) ZWO ^ T n (t)
for every t in the interval and for every n, or else
( 3 ) Tn+i(t) T n (t)
for every t and n.
The following somewhat generalized form of a test due to
Abel shows that when the terms of a uniformly convergent
series are multiplied by functions T n (t) of the type just described,
the new series is uniformly convergent.
Theorem 1. The series
X n (x)T n (t)
converges uniformly with respect to the two variables x and t together ,
in a closed region R of the xt-plane, provided that ( a ) the series
00
X n (x ) converges uniformly with respect to x in R, and ( b ) for
v
all t in R the functions T n (t ) {n = 1, 2, ■ • • ) are uniformly
bounded and monotone with respect to n.
Let S n denote the partial sum of our series,
S n (x, t) = Xi(x)Ti (0 +X a (*)r,(0 + • • • + X n (x)T n (t).
* See, for instance, Sokolnikoff, “ Advanced Calculus,” p. 256, 1939.
Sec. 55]
UNIQUENESS OF SOLUTIONS
129
We are to prove that, given any positive number e, an integer
N independent of x and t can be found such that
| S m (x, t) - SnO, t) | < e if n > N,
for all integers m = n + l,n + 2, • * • , and for all t in the
region R.
If we write
Sn(x) = Xi(x) + X 2 (^) +■'*-+- X n (x),
then for every pair of integers m, n (m > n), we have
(4) S m - Sn
= Xn+lTn+l + Xn+^Tn+2 + * * * + X m T m
— (Sn-fl — 7 i-|-i H~ (Sn-4-2 S71+1) T * ' * H” (,^m Sm— l) Tm
= (S n+ 1 “ S n )(T n+ 1 — T n + 2 ) + (Sn+2 “ S n )(T n +2 — T n +z)
“b * * * "b (Sw— 1 $n)(T m -l "l - Sr^'Tm*
Suppose now that the functions T n arc nonincreasing, with
respect to n, so that they satisfy relation (2). Also let K be an
upper bound of their absolute values, so that condition (1) is
true. Then the factors (7 7 „ + i — 7 1 „+2), (3P»+2 — T n+3 ), * * • , in
equation (4) arc non-negative, and |7\»| < K. Since the series
oo
V X n (x) is uniformly convergent, an integer N can be found for
x
which
|s n+p - s„| < when n> N,
for all integers p, where e is any given positive number and N
is independent of x. For this choice of N it follows from equa-
tion (4) that
|/S„, — Sn I < [(jf’n-l-1 — Tn+i) + (Tn +2 — T n+ . s)
+ • • • + \T m II = 3 ^ [Tn +I - T m + \T m \\,
and therefore
|jS„ — <S„| < e, when n > N (m > n).
The proof of the theorem is similar when it is supposed that
the functions 7’» arc; of the nondecreasing type (3), with respect
to n.
130 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 56
When the variable x is kept fixed, or when the functions
X n (x) are constants, the theorem shows that the series with
terms X n T n is uniformly convergent with respect to L The only
requirement on the series in X n in this case is that the series
shall converge.
Extensions of the theorem to the case in which the functions
X n involve the variable t as well as x , or where X n and T n are
functions of any number of variables, become evident when it
is observed that our proof rests on the uniform convergence of
the Xn-series and the bounded monotone character, with respect
to n, of the functions T n .
56. Uniqueness Theorems for Temperature Problems. Let R
denote the region interior to a solid bounded by a closed surface
and let R + S denote the closed region consisting of the
points within the solid and upon its surface. If u(x, y, z, t)
represents the temperature at any point in the solid at time t, a
rather general problem in the distribution of temperatures
in an arbitrary solid is represented by the following boundary
value problem;
(!) + <p{x, y, z, t ) (t > 0),
at all points (z, y, z) in R;
( 2 ) » = /(*, y, z)
in R, when t = 0;
(3) u = g(x, y, z, t) (f > 0),
when ( x , y, z) is on S.
This is the problem of determining the temperatures u in a
solid with prescribed initial temperatures /(x, y , z) and surface
temperatures g(x, y, z, t). A continuous source of heat, whose
strength is proportional to <p(x } y ) z, t), may be present in the
solid.
Suppose there are two solutions
u = ui(x, y, z, t), u = u 2 (x, y , z, t),
of this problem, where both U\ and u 2 are continuous functions
of Xj y } z y t } together, in the region R + S when t ^ 0, while
dui/dty du 2 /dt, and all* the derivatives of u\ and u 2 once or twice
UNIQUENESS OF SOLUTIONS
Sec. 56]
131
with respect to either x, y , or z are continuous functions when
(x, y , 2 ) is in R + S and t > 0.
Since ui and u 2 satisfy each of the linear conditions (1) to (3), it
follows at once that their difference w,
w(x, y , 2 , t) = ui(x, y, z, t ) - u 2 (x, y , 2 , t),
satisfies the following linear homogeneous problem:
(4)
^ = kV*w
at
in R
(t >0);
(5)
w = 0
when t = 0,
in R ;
(6)
w = 0
on S
(t > 0).
Moreover, w and its derivatives appearing in equation (4) must
have the continuity properties required above of u y and u 2 and
their derivatives.
We shall show now that w must vanish at all points of R
for all t > 0, so that the two solutions Ui and u 2 are identical.
It follows that not more than one solution of the problem (1)-
(3) can exist if the solution is required to satisfy the continuity
conditions stated above.
Since the function w is continuous in R + S, the integral
where dV = dx dy dz , is a continuous function of t when t ^ 0.
According to condition (5),
J( 0) = 0.
I 11 view of the continuity of dw/dt when t > 0, we can write
-*///•
dw JTr
w — dV
R at
wV 2 w dV
(t > 0).
Since the second derivatives of w with respect to each of the
coordinates art; continuous functions in R + S when t > 0, we
can use Green’s theorem to write
■in.
dw
w — dS
dn
wV 2 w dV +
132 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 56
Here n is the outward-drawn normal to the surface S. But
according to condition (6), w = 0 on 8, and so
-- k Si X [(£)■ + (£)' + (£)'] ^ # > »>•
Since the integrand here is never negative,
• J'(t) g 0 when t > 0.
The mean-value theorem applies to J (t) to give
J(t) ~ J(0) = U'iff) (0 < h < t),
and since J (0) = 0, it follows that
J(t) ^ 0 whenever t > 0.
However, the definition of the integral J shows that
J(t) ^ 0 (t £ 0).
Therefore
J(t) = 0 (t ^ 0);
and so the integrand w 2 of the integral J cannot be positive
in R. Consequently
w(x, y, z,t) = 0
throughout R + S, when 0.
This completes the proof of the following uniqueness theorem:
Theorem, 2. Let u(x, y, z, t) satisfy these conditions of regu-
larity: (a) it is a continuous function of x, y, z, t, taken together,
when (x, y, z) is in the region R S and t S: 0; (6) those derivatives
of u which are present in the heat equation (1) exist in R and are
continuous in the same manner when t > 0. Then if u is a solu-
tion of the boundary value problem (l)-(3), it is the only possible
solution satisfying the conditions (a) and (6).
Our proof required only that the integral
in Green’s theorem be zero or negative. The integral vanished,
since w = 0 on S because of condition (3) j but it is never positive
if (3) is replaced by the condition
(8) ^ + hu = g(x, y, z, t ) on S,
where h is a non-negative constant or function. So our theorem
can be modified as follows :
Sec. 57]
UNIQUENESS OF SOLUTIONS
133
Theorem 3. The statement in Theorem 2 is true if boundary
condition (3) is replaced by condition (8), or if (3) is satisfied on
part of the surface S, and (8) on the remainder of S.
The condition that u be continuous when t = 0 makes our
uniqueness test somewhat limited. This condition is clearly
not satisfied, for instance, if the initial temperature function is
discontinuous in R + S, where the initial temperature on S is
taken as the surface temperature.
If the regularity conditions (a) and ( b ) in Theorem 2 are
added to the requirement that u must satisfy the heat equation
and boundary conditions, our temperature problem will be
completely stated provided it has a solution. For that will be
the only possible solution.
57. Example. In the problem of temperatures in a slab with
insulated faces and initial temperature f{x) (Sec. 475), suppose
f(x) is continuous when 0 ^ x g 7r, and f'(x) is sectionally
continuous in that interval. Then the Fourier cosine series for
f(x) converges uniformly in the interval.
Let u(x, t) denote the function defined by the series
(1) a n e~ nVit cos nx,
l
which was obtained in Sec. 47 as the formal solution, a n being
the coefficients in the Fourier cosine series for f{x).
Series (1) converges uniformly with respect to x and t together
when and t ^ 0, according to Theorem 1. In any
interval throughout which t > 0, the series obtained by differ-
entiating series (1) term by term, any number of times with
respect to either variable, is uniformly convergent according
to the Wcicrstrass M-tcst. It readily follows that u{x , t) not
only satisfies all the conditions of the boundary value problem
(compare Sec. 46), but that it is also continuous when 0 ^ x ^ tt,
t ^ 0, and its derivatives du/dt , d 2 u/dx 2 are continuous when
0 g x ^ tt, t > 0. That is, u(x, t) satisfies our conditions of
regularity.
The temperature problem for a slab is just the same as the
problem for a cylindrical bar with its lateral surface insulated
( du/dn — 0); hence the region R, can be considered here as a
finite cylinder. Theorem 3 therefore applies, showing that the
134 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 58
function defined by series (1) is the only possible solution which
satisfies the above regularity conditions.
PROBLEMS
1. In Prob. 7, Sec. 47, let fix) be continuous, and fix) sectionally
continuous, in the interval (0, t), and suppose /(0) = fiw) = 0. Show
that the solution found is the only one possessing the regularity prop-
erties stated above.
2. Make a complete statement of Prob. 8, Sec. 47, so that it has one
and only one solution.
3. Establish the solution of Prob. 10, Sec. 47, and show that it is the
only possible solution satisfying the regularity properties stated above.
58. Uniqueness of the Potential Function. A function of
x, y, z is said to be harmonic in a closed region R + S, where S
is a closed surface bounding a region R, if it is continuous in
R + S and if its second ordered derivatives with respect to
x, y , and z are continuous in R and satisfy Laplace's equation
there.
Let Uix , y, z) be a harmonic function whose derivatives of the
first order are continuous in R + S. Then since
( 1 ) V*U = 0
throughout R, Green's formula (7), Sec. 56, can be written as
follows:
This formula is valid for our function [/, even though we have
not required the continuity of the second ordered derivatives
of U in the closed region R + S. We shall not stop here to
prove that, since V 2 E/ = 0, this modification of the usual condi-
tions in Green's theorem is possible.*
If JJ = 0 at all points on S , the first integral in equation (2)
is zero, so the second integral must vanish. But the integrand
of the second integral is clearly non-negative. It is also con-
tinuous in R, So it must vanish at all points of R ; that is,
(3)
dU
dx
* The proof is not difficult,
of this chapter.
dU = dU
' dy dz
See, for instance, p. 119 of Ref. 1 at the end
UNIQUENESS OF SOLUTIONS
Sec. 58]
135
so that U is constant in R. But U is zero on 8 and continuous
in R + S, and therefore U = 0 throughout R + S.
Suppose that dU/dn, instead of U, vanishes on 8; or, to make
the condition more general, suppose
(4) + hU = 0 on S,
where h ^ 0, and h can be either a constant or function of x , y,
and z. Then on S>
= —hU 2 ^ 0,
cm ’
so that the integral on the left in equation (2) is not positive.
But the integral on the right is not negative. Both integrals
therefore vanish and again condition (3) follows, so that U is
constant throughout R.
Of course U may vanish over part of S and satisfy condition (4)
over the rest of the surface, and our argument still shows that U
is constant in R. In this case the constant must be zero.
Now suppose that the function V(x, y, z), together with its
derivatives of the first order, is continuous in R + S, and let
its derivatives of the second order be continuous in R. Also let
V(x, y, z) be required to satisfy these conditions:
(5) V 2 7 = /
when (x, y, z ) is in R ;
( 6 ) P^~+hV = g
when (x, y, z) is on the surface S. The prescribed quantities
f, p, h, and g may be functions of (x 7 y , z); but it is assumed that
p ^ 0 and h ^ 0.
We have made boundary condition (6) general enough to
include various cases of importance. When p = 0 on S, or on
part of S, the value of V is assigned there; and when h = 0, the
value of dV/dn is assigned. Of course p and h must not vanish
simultaneously.
If V = V i and V = V<> arc two solutions of this problem, then
their difference,
U = V x - V 2
136 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 58
satisfies Laplace’s equation (1) in R , and the condition
+ hU = 0
on S. Since U is harmonic and has continuous derivatives
of the first order in R + S, we have shown that U must be
constant throughout R + S. Moreover, if p = 0 at any point
of S , so that U vanishes there, then U = 0 throughout R •+ S.
We therefore have the following uniqueness theorem for
problems in potential or steady temperatures, and other prob-
lems in which the differential equation is that of Laplace or
Poisson.
Theorem 4. Let the function V ( x , y , z ) be required to satisfy
these conditions of regularity: (a) it is continuous , together with
its partial derivatives of the first order , in the region R + S; and
(6) its derivatives d 2 V/dx 2 , d 2 V/dy 2 , and d 2 V/dz 2 are continuous
functions in R. Then if V is a solution of the. boundary value
problem (5)-(6), it is the only possible solution satisfying the
conditions of regularity , except for an arbitrary additive constant .
If, in condition (6), p = 0 at any point of S , then the additive
constant is zero and the solution is unique.
It is possible to show that this theorem also applies when R
is the infinite region outside the closed surface S, provided V
satisfies the additional requirement that the absolute values of
pV,
} dV
J dy }
t dV
' dz
shall be bounded for all p greater than some fixed number,
where p is the distance from the point ( x , y, z) to any fixed point.*
Since V is required to approach zero as p becomes infinite, the
additive constant in this case is always zero. But note that even
here S is a closed surface, so that this extension of our unique-
ness theorem docs not apply, for instance, to the infinite region
between two planes or the infinite region inside a cylinder.
The regularity requirement (a) in Theorem 4 is quite severe.
It will not be satisfied, for instance, in problems in which V
is prescribed on the boundary as a discontinuous function, or as a
function with a discontinuous derivative of the first order.
* For the proof, see Ref. 1.
Sec. 59]
UNIQUENESS OF SOLUTIONS
137
For problems in which V is prescribed on the entire boundary
S [that is, p = 0 in condition (6)] of the finite region R, it is
possible to relax the conditions of regularity so as to require
only the continuity of V itself in R + 8. The derivatives of the
first and second order are only required to be continuous in R.
This follows directly from a remarkable theorem in potential
theory: that if a function is harmonic in R + S, and not constant,
its maximum and minimum values will be assumed at points on
Sj never in R.* But this uniqueness theorem is limited in its
applications to boundary value problems, because it does not
permit such a condition as d V/dn = 0 on any part of S, a condi-
tion which is often present or implied in the problem. This will
be illustrated in the example to follow.
59. An Application. To illustrate the use of the theorem in
the preceding section, consider the problem, in Sec. 49, of deter-
mining the steady temperature u(x, y ) in a rectangular plate
with three edges kept at temperature zero and with an assigned
temperature distribution on the fourth. The faces of the plate
are kept insulated. For the purpose of illustration it will be
sufficient to consider here only the case of the square plate with
edge 7 r units long. We also observe that as long as du/dn = 0
on the faces, the thickness of the plate does not affect the problem.
We may as well consider this as a problem in the potential
V(x , y) in the finite region R bounded by the planes x = 0,
x — 7T, y = 0, y = 7r, and any two planes z = Z\, z = z 2 . Then
our boundary value problem can be written
(1)
aw dW
dx 2 dy‘ l
(2)
V(0, y) = 0,
V(t, y ) = 0,
(3)
V(x, 0) = f(x),
o'
11
and of course, dV/dz = 0 on z = Z\ and z = z 2 .
The given function /(x) will be required here to be continuous,
together with its first derivative, in the interval (0, n r). It is
also supposed that/"(x) is seetionally continuous in that interval;
and finally, we require /(x) to satisfy the conditions
/( 0) = /( x) = 0.
* The proofs of those theorems will he found quite interesting, and not
difficult to follow. See Kefs. 1 and 2.
138 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59
Then, according to our theory of Fourier series, the sine series
for /(*),
ao _
(4) 2 bn sin nx L" = ^ Jo ^ sin nx dx ’
converges uniformly, and so does the cosine series for f(x),
oo
(5) ^ nb n cos nx,
obtained by differentiating the sine series termwise. In demon-
strating the uniform convergence of the Fourier series in Sec. 38,
however, we proved that the series of the constants + b\
converges. In the case of series (5), in which the sine coefficients
are zero and the cosine coefficients are nb n , this means that the
series
2) \ nb *\
is convergent. Since the absolute values of the terms of the
series
00
(6) V nb n sin nx
1
are not greater than \rib n \, it follows from the Weierstrass test
that the series (6) also converges uniformly.
In addition to the conditions (1) to (3), let the unknown
function V(x , y) be required to satisfy the regularity conditions
(a) and ( b ) of Theorem 4. That is, V, dV /dx, and dV/dy must
be continuous in the closed region 0 ^ x S 0 ^ y ^ tt, while
d 2 V/dx 2 and d 2 V/dy 2 are required to be continuous at all interior
points of the region. We shall call this a complete statement
of the problem of determining the function V(x, y). For accord-
ing to Theorem 4, this problem cannot have more than one
solution, and we shall now prove that it does have a solution.
The series derived in Sec. 49 as the formal solution of our
problem can be written here as
sinh n{j — y)
sinh 7i7T
sin nx .
(7)
Sec. 59 ]
UNIQUENESS OF SOLUTIONS
139
Let us show that this represents a function y ) which satis-
fies all the requirements made upon V in the complete statement
of our problem, so that V = \p (x, y ) is the unique solution of that
problem.
To examine the uniform convergence of series (7), let us
first show that the sequence of the functions
/on sinh u(t - y)
sinh mr ’
which appear as factors in the terms of that series, is mono-
tone nonincreasing as n increases, for every y in the interval
0 ^ y ^ t. This is evident when y = 0 and y = t. It is true
when 0 < y < 7r, provided that the function
T(t) =
sinh bt
sinh at
always decreases in value as t grows, when t > 0 and a > b > 0.
Now
2T'{t) sinh 2 at
= 2b sinh at cosh bt — 2 a sinh bt cosh at
= —(a — b) sinh ( a + b)t + (a + b) sinh ( a — b)t
[sinh ( a + b)t sinh (a — b)t~\
T^b
= -(a 2 - b>)
L a b
= -(a 2 - b-) 2 K« + b)°- n - (a -
£2n-H
( 2 »+ 1)1
The terms of this series are positive, so that
T'(t) < 0,
and T(t) decreases as t increases. Therefore functions (8) never
increase as n grows.
Likewise the functions
(9)
cosh n (it — y )
sinh mr
(0 g y ^ tt)
never increase in value when n grows; because the squares of
these functions can be written as the sum
sinh 2 n(ir — y )
sinh 2 mr ’
( 10 )
sinh 2 mr
140 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59
and as n grows the first term of the sum clearly decreases, while
the second was just shown to be nonincreasing.
Functions (8) are clearly positive and not greater than unity
for all values of y and n involved. Functions (9) are also uni-
formly bounded; this is evident from the expression (10) for
their squares.
Therefore the sequence of functions (8), or of functions (9),
can be used in our form of Abel’s test for uniform convergence.
So, from the uniform convergence of the sine and cosine series
(4), (5), and (6), when 0 ^ x ^ 7r, we conclude not only that
our series (7) converges uniformly with respect to x, y in the
region 0 ^ x ^ t, 0 ^ y S but also that this uniform con-
vergence holds true for the series
2 , sinh n(r — y)
nl> - smhn„
obtained by differentiating series (7) with respect to x } and for
the series
oo
-2
Tibn
cosh n(w — y)
sinh mr
sm nx,
obtained by differentiating series (7) with respect to y.
Consequently series (7) converges to a function i p(x, y) which,
together with its partial derivatives of the first order, is con-
tinuous in the closed region 0 ^ x ^ w, 0 y S v. The func-
tion \j/ clearly satisfies boundary conditions (2) and (3).
When differentiated twice with respect to either x or y, the
terms of series (7) have absolute values not greater than the
numbers
( 11 )
n 2 \b n \
sinh n(7r — y 0 )
sinh mr
for all x and y in the region 0 ^ x ^ tt, y 0 ^ y ^ x, where y 0 is
any positive number less than t. Since the series of the constants
(11) converges, the series of the second derivatives of the terms
of series (7) converges uniformly in the region specified. Hence
series (7) can be differentiated termwise in this respect whenever
0 < y < 7r; also the derivatives d^jdx 2 , d^/dy 2 are continuous
whenever Q^x^w, 0<y^T.
Sec. 59]
UNIQUENESS OF SOLUTIONS
141
Thus i p(x } y ) satisfies the regularity conditions. It only
remains to note that it satisfies Laplace's equation (1) in R .
This is true because the terms of series (7) satisfy that equation,
and series (7) is termwise differentiable twice with respect to x
and to y in R , so that Theorem 2, Chap. I, applies.
The only solution of our completely stated problem is therefore
V =
00
sinh n(ir — y)
sinh mr
sin nx.
In particular, note that we have shown that our complete
problem, which includes the condition that dV/dz = 0 on the
boundaries z = Zi and z = z 2 , has no solution which varies with z.
In the formal treatment of the problem given earlier, the absence
of the variable z was regarded as physically evident. In the
present section we have omitted the term d 2 V/dz 2 in Laplace's
equation, and at other times have neglected writing the variable
z, only as a matter of convenience.
PROBLEMS
1. Show that the formal solution found in Sec. 49 can be completely
established as one possible solution of the boundary value problem
written there, provided the function /(x) is sectionally continuous in the
interval (0, a) and has one-sided derivatives there, and fix) is defined to
have the value [fix + 0) + fix - 0)]/2 at each point x of discontinuity
(0 < # < a).
2. Make a complete statement of the boundary value problem for the
steady temperatures in a square plate with insulated faces, if the edges
x = 0, x = 7 r, and y = 0 arc insulated, and the edge y = t is kept at
the temperature u = fix). Assume that fix) is continuous when
0 ^ x Z 7 T, and that /'(()) = /'(x) = 0. Show that your problem has
the unique solution
cosh ny
cosh mr
cos nx
2 r
(In = ™
* Jo
fix) cos nx dx
3. Establish the result found in Prob. 5, See. 49, as a solution (but
not as the only possible one) of the boundary value problem, when the
function fix) there is represented by its Fourier sine series.
4. In Prob. 8, Sec. 53, let the infinite cylinder be replaced by a finite
cylinder bounded by the surfaces r — 1, z = z\, z = on the last two
of which dV/dz = 0. Also let the periodic function f(<p) have a con-
142 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 59
tinuous second derivative. Then show that the result found there is
actually a solution, and that it is the only possible solution of the
problem satisfying our conditions of regularity.
REFERENCES
1. Kellogg, 0. D .: “ Foundations of Potential Theory,” 1929.
2. Courant, R., and D. Hilbert: “Methoden der Mathematischen Physik,”
Vol. 2, 1937.
CHAPTER VIII
BESSEL FUNCTIONS AND APPLICATIONS
60. Derivation of the Functions Any solution of the
differential equation
(1) * 2 § + *i + (x2_n2)2/ = 0 ’
known as Bessel 1 s equation , is called a Bessel function or cylindrical
function. It will be shown later on how this equation arises
in the process of obtaining particular solutions of the partial
differential equations of physics, written in cylindrical coordi-
nates. We shall let the parameter n be any real number.
A particular solution of Bessel’s equation in the form of a
power series multiplied by x l \ where p is not necessarily an integer,
can always be found. Let a () be the coefficient of the first non-
vanishing term in such a series, so that a 0 ^ 0. Then our pro-
posed solution has the form
00 00
(2) y = x* ]£ ajX* = X ap***.
3=0 3=0
If the series here can be differentiated termwise, twice, the
coefficients aj can be determined so that the series is a solution
of equation (1). For upon differentiating and substituting in
equation (1), we obtain the equation
X + j) (p +j — 1) + (p + j) + (x 2 - n.*)]ajX*+i = 0.
3=0
Dividing through by :v p and collecting the coefficients of the
powers of x ) we can write the equation in the form
(p 2 — n 2 )a 0 + [(p + l) 2 - n 2 ]a x x
+ X ! Kp + j) 2 - n 2 ]a s + a,,-2)xf = 0.
J- 2
143
144 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 60
This is to be an identity in x , so that the coefficient of each
power of x must vanish. The constant term vanishes only if
p = ±n. The second term vanishes if ai = 0; and the coeffi-
cients of the second and higher powers of x all vanish if
[(p + j) 2 - + Oj-s = 0 (i = 2, 3, • • • );
that is, if
(p - n +j)(p + n + f)a, = -a,-_ 2 (j = 2, 3, • • • )•
This is a recursion formula for a iy giving each coefficient in
terms of one appearing earlier in the series.
Let us make the choice
V = n,
so that the recursion formula becomes
(3) j(2n +j)a i = -o,-_ 2 O' = 2, 3, • • • ).
Since a L = 0, it follows that a 3 = 0; hence a 5 = 0, etc.; that is,
(4) a 2fc _i = 0 (Jc = 1, 2, • • - ),
provided n is such that 2n + j ^ 0 in formula (3). But even
if 2 n + j does vanish for some integer j, coefficients (4) still
satisfy formula (3). Since this is all that is required to find a
solution, we can take all the coefficients a 2k - 1 as zero regardless
of the value of n.
Replacing j by 2 j in formula (3), we can write
a 23 ' = 2 2 j(n + J) a2j “ 2 ^ = ^
provided n is not a negative integer. Replacing j by j — 1 here,
we have
-1
a2l '~ 2 ~ 2*0' - 1)(» + 3 ~ 1) a2l '“ 4 ’
so that
(- 1) 2
a2i 200 - l)(n+i)(»+j - I) 023 '" 4 '
Continuing in this manner, it follows that
a 2 j = ( — l) fc a2 3 --2fc/[2 2fc j(j — 1) • ■ * (j — k + 1)
(n + j) (n +j - 1) • • • (n +j ~ fc + 1)];
Skc. 6 . 1 ] BESSEL FUNCTIONS AND APPLICATIONS 145
so that when k = j, we have the formula for a v in terms of a Q :
( — 1 ) 3 CLq .
0/27 ~ 2 23 j\{n +j)(n +7 - 1) ' * • (n + 1) ^ = 1 » 2 > ' ' * )•
The coefficient a 0 is left as an arbitrary constant. Let its
value be assigned as follows :
1
a ° 2T (n + 1)'
Recalling that the Gamma function has the factorial property
kT(k) = r(fc + 1 ),
it follows that
(n + j)(n + j - 1) • • • (n + 2 )(n + l)r(n + 1)
= r(n + i + 1).
Our formula for a 2? - can therefore be written
(5) a 2 j = + j _|_ i)2^+2/ 0’ = ‘ " ‘ )>
where j! = 1 if j = 0.
The function represented by series (2) with coefficients (4)
and (5) is called a Bessel f unction of the first kind of order n:
( 6 )
(- 0 '
j\T(n + j + 1)
_ r x 2
2 n T(n + 1) [_ 2(2 n + 2)
+
2 • 4(2 n -j- 2)(2n -|- 4)
The series in brackets is absolutely convergent for all values
of x ) according to the ratio test. It is a power series, so that
the termwise differentiation employed above is valid, and hence
function (6) is a solution of Bessel's equation. Of course, when
n is not a positive integer, J n (x) or its derivatives beyond a
certain order will not exist at x = 0, because of the factor x n .
61. The Functions of Integral Orders. When n = 0, the
important case of the function of order zero is obtained:
Jo(x) = 1
x 2 x A
22 + 2 2 • 4 2
2 2 • 4 2 * 6 2
a +
146 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 61
When n is a negative integer, the choice p = -n can be made
and the recursion formula (3) of the last section gives the coeffi-
cients just as before. If, in this case,
do
2 n
r(-n + 1)’
the solution of Bessel’s equation will be found to be
( 1 )
2 (-l) a '
_ v j!r(-n + j + 1)
J-n(x).
Now if we define l/r(j?) to be zero when p = 0, 1, 2 • • •
formula (6) of the last section can be used to define a function
J n {x) even when n is a negative integer. For if n = -m where
m is a positive integer, that formula becomes
J —m
Summing with
written
(- 1 )*
x \-m+2f
(x) 2jfr(-m+j + l) ^2,
respect to k, where k = -m + j, this can be
(-1)* /x\
f^(k + m)\T(k + 1) \2/
r
(k + m) !r(& + 1 )
-(-u- 2 ^
j^kWim + k + 1 )
But the last series represents J m (x); hence for functions of
integral order,
(2) J-m(x) = (-1 ) m J m (x) ( m = i, 2, 3, • ■ • ).
According to solution (1) now, the function y = ( — iw ( x )
and hence the function y = J n (x) is a solution of Bessel’s equa-
tion when .n is a negative integer; hence the junction dejined by
equation (6), Sec. 60, is a solution for every real n.
When n is neither a positive nor a negative integer, nor zero, it
can be shown that the particular solution /_ n (z) obtained by
aiang p - ms not a constant times the solution J n (x) ; hence
the general solution of Bessel’s equation in this case is
1/ ~ AJ n(x) -)- BJ-. n (x),
where A and B are arbitrary constants.
Sec. 61] BESSEL FUNCTIONS AND APPLICATIONS
147
When n is an integer, the general solution of Bessel’s equation is
V = AJ n (x) + BY n (x),
where F»( x) is a Bessel function of the second kind of order n.
These functions will not be used here. For their derivation
and properties, as well as for a more extensive treatment- of the
theory of Bessel functions of the first kind than we can give
here, the reader should consult the references at the end of this
chapter.
There are several other ways of defining the functions J n (x).
When n is zero or a positive or negative integer, the generating
function exp [£x(f — I/O], is often used for this purpose. By
multiplying the two series
it can be seen that
(3) exp | (i - D = 2 J &)t m
L n — — vo
= Jq(x) + Ji(x)t + Jv(x)t 2 + • • •
+ J-\(x)t~ l -b + * * * ,
for all values of x and t except t = 0. Hence J n (x) can be defined
as the coefficients in this expansion. It is on the basis of this
definition that the above choice of the constant a 0 was made.
PROBLEMS
1. Prove tlmt
J -if*)
2. Prove that
J\{x)
3. Derive solution (1) when n is a negative integer.
4. Carry out the derivation of formula (8).
5. Show that, for every n,
Jn(-X) = (-1 )»J n ( X ).
148 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 62
62. Differentiation and Recursion Formulas. By differenti-
ating the series
<» - I. W+j'+D
it follows that
(2) xJ' n (x)
2 (-iy(n + 2j) /x\ n+ii
00
= nJ n (x ) + X 2 Tfz:
(- 1 V
J-l
(j - iy.T(n+j + 1)
(I)
n+2j—l
That is,
(3)
TtJ nix) X
A!r(n + A + 2)
n+l+2fc
•oj n(^0 XiJ nix') xj n+l(x).
Similarly, if we write n + 2j = 2(n + j) — n in the sec-
ond member of equation (2), and replace r(n + j + 1) by
(n + j)T(n -+- j), we obtain the relation
xJ'nix) - — nJ„( x) + X
that is,
(4) xJ' n {x) = —nJ n {x) + zJ n-i{x).
Elimination of J n {x) between equations (3) and (4) gives the
formula
(®) ‘^'J'ni 3 ') = J n— \(x) / n+1 (z)j
and the elimination of J' n (x ) between the same two equations
gives the formula
(6) ~ J n ix) = J n _,ix) + J n+ i(x).
(xy~
,_ u i !r ( n +3) W
The recursion formula (6) gives the function Jn+i(x) of any order
in terms of the functions J n(x) and J n -\{x) of lower orders.
By multiplying equation (4) by x n ~\ and equation (3) by ar"- 1 ,
we can write these formulas, respectively, as
d
\% n Jn(%y\ = X n J n -l(x),
^[a :~ n J n (x)] = —x^Jn+iix).
Sec. 63] BESSEL FUNCTIONS AND APPLICATIONS 149
The following consequences of the above formulas should be
noted:
A(x) = -J i(x) =
(7) rJo(r) dr = xJi(x ).
PROBLEMS
1. Obtain formula (7) above.
2. Prove that
*V"(x) = [»(n - 1) - *V.(*) + xJ n+1 (x).
3. With the aid of Probs. 1 and 2, Sec. 61, prove that
- S(^- cosx )
4. When n is half an odd integer, show that J n (x) can always be
written in closed form in terms of sin x, cos x, and powers of \/\fx.
63. Integral Forms of J n (z). Let us first recall that the Beta
function is defined by the formula
B(n + i, j + i) = 2 £ sin 2n 0 cos 2 * d dd (n > — j > — £).
Let i be zero or a positive integer. Then
B(n + j + ■*) = sin 2n 0 cos 2 * 6 d6 (n > — ■£).
This function is given in terms of the Gamma function by the
formula
B
j
r (n + j)r(j + 1)
r(n + j + 1) '
and as a consequence we shall be able to write the general
term of our series for J n (x) in terms of trigonometric integrals.
Our formula for J n (x ) can be written
Jn(x)
Now
1
2 2j j‘!r(n + j + 1)
© n 'sh ( —
2j2*>jlT(n+j + 1)’
j = U
j • ^ • I • • • 0 ~ nrCD
(2j)!r(n +j + l)r(i)
r(j + 1) = £(«■ + j + j)
( 2 j!)r(n +i + l)r(i) (2i)!r(i)r(n + l)
150 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 63
Therefore
(1) Jn(x) = Cn 2 XU Jo sin2 " 6 C0s2) 9
de,
where
( 2 )
When n ^ 0, the series
(-i)
_ (s/2)»
” r(*)r<» + i)‘
3-0
m
x 23 sin 271 0 cos 23 0
converges uniformly with respect to 0 in the interval (0, it);
because the absolute values of its terms are not greater than the
corresponding terms of the convergent series
i- o
and the terms here are independent of 0. The first series can
therefore be integrated termwise with respect to 0 over the
interval (0, 7 r). In other words, the integral sign in formula
(1) can be written either before or after the summation sign.
Therefore,
sin 2n 0
3 =0
(-iy
m
(x cos 0) 23 dd.
Since the series in the integrand represents cos (x cos 0),
(3) J n (x) = C n sin 2n 0 cos ( x cos 0) dd,
where C n is defined by formula (2).
Formula (3) gives one of LommeVs integral forms of J n (x ).
Although the above derivation holds only for n ^ 0, form (3)
is valid when n > This can be shown by writing the first
term in series (1) separately and integrating the remaining terms
by parts to obtain the equation
Jn(x) = Cn
sin 2n 0 dd
( — l) 3 ’# 23 ' 2j - 1
(2 j ) ! 2n + 1
I si
sin 27l+2 0 cos 23 ’” 2 0 dd
f
Sec. 63] BESSEL FUNCTIONS AND APPLICATIONS
151
if n > —Jr- Here again the second integral sign can be written
before the summation sign, and the series in the integrand can
be seen to represent the function
sin 2n+ i 0 d / cos (x cos 0) — 1
2n + 1 dB \ cos 0
The details here are left to the reader. Integrating this by
parts and adding the first integral in brackets gives formula (3)
for n > — i*
When n = 0, formula (3) becomes
1 C T
J 0 (x) = - I cos ( x cos 0) d$.
T Jo
When ft = 0, 1, 2, • • ■ , the following integral form is valid:
i r
(4) J n (x) = - I cos {nd — x sin 0) dQ (n — 0, 1, 2, • ■ ■ ).
This is known as Bessel's integral form. By writing the integrand
as
cos nd cos (x sin '0) + sin nd sin ( x sin 0),
it can be seen that formula (4) reduces to
1 f *
(5) Jn(x) = — I cos nd cos (a: sin 0) dd if n = 0, 2, 4, • • • :
rr Jo
(6) Jn(x) = “ nd s * n ( x 0) dd if n = 1, 3, 5, * ■ • .
These forms can be obtained from formula (3), Sec. 61. By
substituting t — e i0 in that formula, we find that
(7) cos (x sin 0) + i sin ( x sin 0)
00 00
= J„(z) + 2 2) Jin(r) cos 2nd + 2 i V J in _^x) sin (2 n — 1)0.
n = 1 n = 1
Equating real parts and imaginary parts separately here, and
multiplying the resulting equations by sin nd or cos nd and
integrating, using the orthogonality of these functions in the
interval 0 < 0 < 7r, we get formulas (5) and (6). Formula (4)
follows by the addition of the right-hand members of formulas (5)
and (6). The details are left for the problems.
152 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 63
From formula (4) the important property of boundedness
\Jn(x)\ £ 1 (» = 0, 1 , 2, • • • ),
follows at once. It also follows from the same formula that
each derivative of J n (x ) is bounded for all x:
d k
Jn(%) = 1 (n = 0, 1 , 2, • * • j k = 1 , 2, • • • ).
According to formula (5),
2«7 2 n(^r) = a> 2 n (n — 1, 2, • • *
where a n denotes the coefficients in the Fourier cosine series,
with respect to 0, of the function cos (x sin 0). Similarly if b n
denotes the coefficients in the Fourier sine series of the function
sin ( x sin 0), formula (6) shows that
2J r 2»— i(x) = &2n— i (n = 1 , 2 , ■ • • ).
Since the Fourier coefficients of every bounded integrable
function tend to zero as n tends to infinity, it follows that for
every x the Bessel functions of integral orders have the property
lim J n (x) = 0.
n — > oo
As to the behavior of the functions J n (x) for large values of x, it
can be shown that
(8) lim J n (x) = 0 (n = 0, 1, 2, • • • ).
X — > 00
The proof is left to the problems.
PROBLEMS
1. Use the Lommel integral form of J n (x ) to prove that
2. Prove in different ways that
Jn(-x) = ( — 1 yj n (x) (n = 0 , 1 , 2, • • ■ ),
and hence that J n (x) is an even or odd function of x according as n is an
even or odd integer. Also deduce that
J 2n— 1(0) = 0
(n - 1, 2, ■ • •
8ec. 64] BESSEL FUNCTIONS AND APPLICATIONS
153
3. Prove in different ways that
/o(0) = 1.
4. Obtain formula (7) above by the method indicated there, and
follow the process outlined to derive Bessel's integral forms (5) and (6),
and thence (4).
6. Deduce from formula (5) that
7T
2 7*2
Jinix) = - I cos 2nd cos ( x sin 0) d6 (n = 0, 1, 2, • ■ • ).
X Jo
6. Deduce from formula (6) that
J 2n-l(x)
2
7 r
J~ 2 sin (2 n — 1)0 sin ( x sin 0) dQ
(n
1, 2, • • • ).
7. Write the integral in Prob. 5 as the sum of the integrals over the
intervals (0, tt/2 — rj) and (• tt/2 — 77, t/2), where rj > 0, and thus show
that
\JUx)\ s l
1 7*2 v cos 2n0
5 J 0 cos e
cos (x sin 0) d(x sin 0)
V-
By integration by parts, show that the absolute value of the integral
appearing here is not greater than a positive number M V1 independent
of x. Hence, given any small positive number e, by first selecting 17
sufficiently small and then x large, show that
|*/2»0c)( < € when x > x Q .
This establishes formula (S) when n = 0, 2, 4, • • ■ , there.
8. Apply the procedure of Prob. 7 to the formula in Prob. 6, and thus
complete the proof of formula (8).
9. Note that the functions cos ( x sin 0) and sin (x sin 0), of the vari-
able 0, satisfy the conditions in our theorem in Sec. 38; also, since they
are even and odd functions, respectively, the series of absolute values
of their Fourier coefficients converges. Deduce that the series
00
X •'»(*)
n =0
is absolutely convergent for every x.
64. The Zeros of J n (x). The following theorem gives further
information of importance in the applications of Bessel functions
to boundary value problems.
154 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 64
Theorem 1. For any given real n the equation J n (x ) = 0 has an
infinite number of real positive roots Xi, x 2 , * • * , x m , • * • ,
which become infinite with m.
This will first be proved when — i < n ^ The proof for
every real n will then follow from Rolled theorem. For if
J n (%) vanishes when x = xi and x ~ x 2) for any real n, then so
do x n J n(x) and x~ n J n (x), and hence their derivatives vanish
at least once between xi and x 2 . But it was shown in Sec. 62 that
these derivatives are x n J n ~i(x) and —zr n J lH _ 1 (x), respectively.
Therefore between two zeros of J n (x ) there is at least one
zero of Jn- i(x), and one of J n+1 (x). So if there is an infinite
number of zeros of J n (x) when — \ < n ^ the same is true
when n is diminished or increased by unity, and repetitions of
the argument show the same for any real n.
For the proof when — £ < n ^ ■£, we shall use the Lommel
formula derived in the last section; namely,
r , \ (a/2 Y r
(1) Jn(z) - r(^)r(n~+ J) J 0 sm2n 6 cos & cos °) de -
Now suppose that x is confined to the alternate intervals of
length 7r/2 on the positive axis; that is,
x = mir + g ty
where m = 0, 1, 2, * * • , and 0 S t S 1. Also let a new
variable of integration X, where
A * \
cos d = — X,
2x 1
be introduced into the integral in formula (1). Then the
integral becomes
2x
7T r 7T COS ( 7tX/2) d\
2 X 2 x [1 - (ir\/2xy 2 ]-"+i ;
and except for a factor which is always positive, this can be
written
( 2 )
X
2/n-j-i
cos (ttX/ 2) dX
[(2m + t) % — X 2 ]“ w+ *
Sec. 64] BESSEL FUNCTIONS AND APPLICATIONS
155
The sign of J n (x) is therefore the same as the sign of integral
(2) . That integral can be broken up into this sum of integrals:
(3) —I\ + I 2 — ^3 + * • * + (—1 ) m I m + ( — l) m H m ,
where
rzj
cos (ttX/2) d\
(i = 1 , 2 ,
H.
2 [(2m + ty - X 2 ]-“+i
2m+t cos (xX/2)
r/rfc„
2m
, m),
= (-1)” P
J 2 n
[(2m + ty — X 2 ]~ n+ *
Now let 7j be broken up into the sum of two integrals, one
over the interval (2 j — 2, 2j — 1) and the other over the interval
(2j — 1, 2 j). By substituting a new variable of integration /z
into these integrals, where
in the first, and
A = 2j - 1 - ju
A - 2j - 1 + ji
in the second, it will be found that
U = j Q l Fi(») sin ^dn,
where
FA?) = [(2m + ty - (2j - 1 + m) 2 ] 71 "*
- [(2m + 0 2 - (2 j - 1 - m) 2 ] 71 "*.
Since n — % = 0, the function F,(ix) is never negative. By
letting j assume continuous values and differentiating F 7 (/z) with
respect to j, wo find that this function always increases in value
with j.
The integral /,• is therefore a positive increasing function of j]
that is,
0 ~ I\ = I 2 = * ‘ * =
Furthermore H m is not negative; because the numerator
cos (t\/2)
in the integral there can be written as ( — l)"* cos (7171/2) when
X = 2m + /z, and cos (7171/2) is positive.
156 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 64
Now the sum (3) can be written
( — + (Irn — I m — l) + (Im-2 “ I m -z) +***],
where the final term in the brackets is (I 2 — /i) if m is even, and
1 1 if m is odd. The quantity in the brackets is therefore positive,
and consequently the sign of J n (mi r + irt/2) is that of ( — l) m ;
that is
J n (win + g ^ > 0 if m = 0, 2, 4, • • • ,
<0 if m = 1 , 3, 5, • • • .
Since J n (x) is a continuous function of x ) its graph therefore
crosses the rr-axis between x = ir/2 and x = t, and again between
x = 3?r/2 and x = 2tt, and so on, when — J- < n g That is,
J nix) vanishes at an infinite number of points x = x Xj x^ • ■ • ,
where x m tends to infinity with m. The theorem is therefore
proved.
Fig. 9.
It follows at once from Rolle’s theorem that the equation
J'nix ) = 0
also has an infinite number of positive roots x' m (m = 1 , 2, ■ • • ),
and x' m tends to infinity with m.
It should be observed that whenever x m is a zero of J n (x), the
number — x m is also a zero. This is true for any w, as is evident
from our series for J n (x ), Sec. 60.
The difference between successive roots of J n (x) = 0 can be
shown to approach i r as the roots become larger.
Tables of numerical values of J n (x ), and of the zeros of these
functions, will be found in the references at the end of this
chapter.* We list below the values, correct to four significant
* See Refs. 1, 3, and 4.
Sec. 65] BESSEL FUNCTIONS AND APPLICATIONS 157
figures, of the first five zeros of Jo(x), and the corresponding
values of Ji(x).
J b(Xvi) — 0
m
1
2
3
4
5
Xm
Jl(Xm)
2.405
0.5191
5.520
- 0.3403
8.654
0.2715
11.79
- 0.2325
14.93
0.2065
The graphs of the functions J Q (x) and Ji(x) are shown in
Fig. 9.
PROBLEMS
1. Draw the graph of (See Prob. 2, Sec. 61.)
2. Draw the graph of (See Prob. 1, Sec. 61.)
3. Draw the graph of / 2 (^) by composition of ordinates, using recur-
sion formula (6), Sec. 62, and the graphs of J Q (x) and Ji(x).
65. The Orthogonality of Bessel Functions. Since J n (r ) satis-
fies BessePs equation, we can write
+ rJ' n (r) + (r 2 - n 2 )J n (r) = 0.
Substituting the new variable x , where r = \x and X is a constant,
it follows that
d 2
dx 1
2 JnO^x) + x - Jr k (\x) + (\ 2 x 2 — n 2 )J n (\x ) = 0;
that is, J n (\x) satisfies BessePs equation in the form
( 1 )
d_
dx
X ~ J n (\x ) + (\*x - J n (\x) = 0.
For each fixed n this form is a special case of the Sturm-Liou-
ville equation
li. [ r< ^ ] + M*) + M*)]-* = °»
with the parameter written as X 2 instead of X (Sec. 24). The
function r(x) = x here; hence it vanishes v,hen x = 0. It
follows from Theorem 3, Sec. 25, that those solutions of equation
(1) in an interval 0 < x < c, which satisfy the boundary condition
Jn(Xc) = 0,
158 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 65
form an orthogonal set of functions on that interval, with
respect to the weight function p(x) = x . It will be observed
that in this case q(x) = —n 2 /x, so it is discontinuous when
x = 0, unless n = 0; but this does not affect the proof of the
theorem.
Now J n (\x) is a solution of equation (1), and it was shown in
the last section that J n (\c) — 0 if Xc = xj (j = 1, 2, • • • ),
where Xj are the positive zeros, infinite in number, of J n (x). Let
Xj denote the corresponding values of X, so that
Then the functions J n (\jx) ( j = 1, 2, • • • ), are orthogonal
provided their derivatives J„(\jx) are continuous. This is
true except possibly at x = 0, and if n ^ 0 an inspection of the
series for J n shows that xJ' n (\jx) vanishes at x = 0, which is all
that is necessary in the theorem. The result can be stated as
follows :
Theorem 2. Let X,- (j = 1, 2, • * • ) be the positive roots of the
equation
(2) J n (\c) = 0 ,
where n is fixed and n ^ 0. Then the functions J n (X ix), J n (\ 2 x)j
• • • form an orthogonal set in the interval (0, c) with respect to the
weight function x; that is ,
(3) r xJ n (Xjx)J n (\j c x) dx = 0 if k 7* j.
No new functions of the set are obtained by using the negative
roots of equation (2), for an inspection of the scries for J n shows
that
Jn( — \jX) = ( — 1 ) n J n (\jX).
It should be carefully noted that n is the same for all functions
of the set; hence an infinite number of sets has been determined
here , one for each n (n ^ 0).
Another boundary condition at x = c which gives still other
orthogonal systems can be seen by examining the proof of the
above theorem. For any two distinct real values Xy and of
the parameter X, the functions J n (\,x) and J n (M x ) satisfy
Sec. 65 ] BESSEL FUNCTIONS AND APPLICATIONS
159
equation (1 ) ; hence
d_
dx
A
dx
% dx ^
X ^ Jn(\kX)
+
+
(\%x - J »(X,<c) = 0,
(xfc - ^ J n (\ k x ) = 0.
Multiplying the first of these equations by J n (\ k x) and the
second by Jnfax), then subtracting and integrating, we find that
(X? - xi) f xJ J n(X/i;iC) dx
= J q |/„(X,a:) J„(Xifea;)j - J n (\nx) ^ jj*^ J„(X,-a:)j| da:
— ^ ~(lx 7i(Xj3/) n(X^3/) n(Xfc2/) «^"n(X j3/) dx.
When n ^ 0, both terms in the brackets in the last expression
vanish when x = 0; hence
(4) (A? - Afc) J^ C xJ n (\jx)J n (\ k x) dx
= cKkJ n(Xjc)*/ n (\&c) cXjtT n (X^c)t/" n (\jC),
where /^(Xc) denotes the value of ( d/dr)J n (r ) when r = Xc.
Since X? — Aj? ^ 0, the orthogonality (3) exists whenever the
right-hand member of equation (4) vanishes. This will be the
case when Ay and \ k are two distinct values of X which satisfy
the equation
(5) XcJ'(Xc) = hJ 7i (Xc) ,
where h is any constant, including zero. The result can be
written thus:
Theorem 3. For any fixed n (n ^ 0), the functions J n (\ jx)
(j = 1, 2, • • * ) form an orthogonal set in the interval (0, c)
with respect to the weight function x, when Ay are the non-negative
roots of equation (5).
Here again, for every root Xy there is a root — X 7 . This can be
seen by writing equation (5) in the form
(6) (n + h)J n (\c ) — Xc/n+i(Xc) = 0.
Consequently the negative roots introduce no new characteristic
functions. The details here can be left to the problems.
If n + h ^ 0, equation (6) has no purely imaginary roots.
This is easily seen by examining our series for J n (x ). From now
160 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 65
on let us assume that h ^ 0, as is usually the case in the applica-
tions; then n + h S 0.
Similarly, equation (2) has no purely imaginary roots.
Now equation (5) can be written
(7) c [J„(Xz)] + hJJ\x) = 0 when x = c (h 3: 0);
hence it is a boundary condition of the type introduced earlier.
It involves a linear combination of the dependent variable in
Bessel’s equation and the derivative of that variable.
Consider the Sturm-Liouville problem, consisting of Bessel’s
equation (1) and either one of the boundary conditions (2) or
(7). A boundary condition at x = 0 is not involved because
the function r(x ) in the general Sturm-Liouville equation is the
independent variable x in this case, and it vanishes at x = 0.
The characteristic functions here, J n (\ ; x), are continuous in the
interval (0, c), since ti^O. Likewise for their first ordered
derivatives, except possibly at the point x = 0; but the product
x(djdx) J n (\jx) is continuous and vanishes at the point x = 0,
which is all that matters. Finally, note that the function p(x)
is also x itself here, and therefore it does not change sign in the
interval (0, c). Hence according to Theorem 4, Sec. 25, the
characteristic numbers Xf are all real.
According to equation (6), X = 0 is a root of equation (5) only
if either /„(0) = 0 or n + h = 0. In the first case the char-
acteristic function J n (Xx) vanishes, so that the root X = 0 can
contribute a characteristic function only if n = h = 0. A root
X = 0 of equation (2) can never contribute a characteristic
function.
We state our results as follows :
Theorem 4. When n S: 0 and h ^ 0, equations (2) and (5)
have only real roots X ; . For either equation we use only the non-
negative roots, since no new characteristic Junctions correspond to
the negative roots. The root X = 0 is used only in the case of
equation (5) with n = h = 0.
PROBLEMS
1. Derive form (6) of equation (5).
2. Prove that when X,- is any root of equation (6), -X, is also a root
Sec. 66] BESSEL FUNCTIONS AND APPLICATIONS 161
66. The Orthonormal Functions. It can be shown that, when
n § 0 , the function J n (x ) is, except for a constant factor, the only
solution of BessePs equation that is bounded at x = 0. Hence
it follows from the results of the last section that the functions
J n (\x) (j = 1 , 2, • • • ) represent all the characteristic functions
of the Sturm-Liouville problem involved there, on the interval
(0, c). We can therefore anticipate an expansion of an arbitrary
function in series of the functions of this set.
It should be observed that the orthogonality here with respect
to the weight function x is the same as the ordinary orthogonality
of the set of functions
{VxJJX-x)} ( j = 1 , 2, • • • )•
Let us now find the value of the norm,
Nnj = x[J n (\jX)] 2 dX ,
of the functions J n (\-x)] these functions can then be normalized
by multiplying them by the factors (N n j)~K
If we multiply the terms in BessePs equation,
by the factor (2 xd/dx)J n (\z), we can write the equation as
i i-'-M 1 ’ - »■
Integrating, and using integration by parts in the second term, we
find that
js + ( X2x ' 2
Since
rJ’n(r)
it follows that
- n 2 )[J n (\x)} 2 C
_o
= nJ n (r) — rj n+ i(r ),
2X 2 jT° x[J n (^x)] 2 dx
= [{nJ.(X*) - \xJ„ +[ (\x)V> + (XV - n 2 ){/„(Xx)} 2 ]° c ;
162 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 67
or, since n ^ 0,
(!) J Q 4Jn(\x)]*dx = J {[/ n (Xc)] 2 + [/^(Xc)] 2 }
TIC
«L>(Xc)t/ n +i(Xc).
Hence, when X,- represents the roots of the equation J„(Xc) = 0,
(2) z[J*(X 3 -a:)] 2 da; = 1 [/„ +1 (X,c)j 2 (j = 1, 2, • • • )-
FAen X 3 - represents the roots of equation (5), £ec. 65, we have
seen that
\jCj n+l(XyC) = (n + A)t/n(XyC),
and hence formula (1) reduces to
(3) £ a, - - »■
(J - 1 , 2 , ■ • • ).
The normalized functions <p n ,(x) can now be written
Vni(x) =
'f n(XjZ)
VIKt
O' = 1, 2,
),
where the norms N nj are given for the two types of boundary
conditions by equations (2) and (3). The set of functions
Wnj{x)\ is orthonormal in the interval (0, c) with a; as a weight
function; that is, for each fixed n (n ^ 0),
f 0 x<p n} (x)<p nk (x) dx = 0 if j y£ k,
= 1 if j = k (j, k = 1, 2, • • • ).
67. Fourier-Bessel Expansions of Functions. Let c B/ be the
Fourier constants of a function /( x) with respect to the functions
<f> n ,{x) of our orthonormal set, where fix) is defined in the interval
(0, c). Then
°ni = x<p n ;(x')f(x) dx
(i = l, 2, • • • ),
Sec. 67] BESSEL FUNCTIONS AND APPLICATIONS 163
and the generalized Fourier series corresponding to f(x) can be
written here as
<W*) = 2 £ x ' J ^')f( x ') dx',
when n ^ 0, and 0 < x < c.
In view of the formulas given in the last section for N n j, this
series can be written
(1) X (* ^ 0),
i=i
where the coefficients A, are defined as follows:
< 2 > A < - dx '
when Xi, X 2 , • * • are the positive roots, in ascending order of
magnitude, of the equation
(3) J n(Xc) = 0;
but
2x? f c
(4) A, ~ = (Xfc 2 + ¥ - n 2 )[J„(XjC)]' 2 Jo xJ ^ X i x )f( x ) dx,
when Xi, X 2 , • * * arc the positive roots of the equation
(5) XcJ'(Xc) + hJ n (\c) =0 (h ^ 0, n ^ 0).
However, in the special case where h = n = 0, Xi is to be taken
as zero , and the first term of the scries is the constant
It can be shown that, when 0 < x < c, the series here does
converge to f(x) under the conditions given earlier for the repre-
sentation of this function by its Fourier series. Let us state one
such theorem here explicitly, and accept it without proof for the
purposes of the present volume.*
Theorem 5. Let f(x) he arty function defined in the interval
(0, c)j such that ^/x f(x) dx is absolutely convergent. Then at
each point x (0 < x < c) which is interior to an interval in which
* A proof, using contour integrals in the complex plane, will be found in
Ref. 1.
164 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 67
fix) is of bounded variation, series (1) converges to
#/(* + 0 ) +/(* - 0 )];
that is,
oo
(7) i lf(. x ”1" 0) + f( x 0)] ~ AjJ n (\jx) (0 <C x <C c),
i =i
where the coefficients Aj are defined by equation (2) or (4), and
n ^ 0, h ^ 0.
The theorem holds true for the special case h = n — 0, men-
tioned above, if A i is defined by formula (6).
It can be shown that all conditions here on fix) are satisfied
everywhere, so that formula (7) is true for every x (0 < x < c)
when f{x) and its derivative f' (x) are sectionally continuous in the
interval (0, c). These conditions are narrower, but perhaps more
practical for us, than those stated in the theorem.
Expansion (7) is usually called the Fourier-Bessel expansion;
but when X ? - represents the roots of equation (5), the expansion is
sometimes referred to as Dini’s.
Other expansion formulas in terms of the Bessel functions J n
are known. There is, for instance, an integral representation
of an arbitrary function which corresponds to the Fourier
integral representation.
Suppose the interval /0, c) is replaced by some interval (a, b)
in the Sturm-Liouville problem with Bessel's equation, where
(a, b) does not contain the point x = 0. Then a boundary
condition is required at each end point x = a and x = b, and
the problem is no longer a singular case, but an ordinary special
case, of the Sturm-Liouville problem. Hence the expansion in
this case will be another one in series of Bessel functions; but
here the functions of the second kind may be involved together
with the functions J n .
PROBLEMS
1. Expand the function fix) = 1, when 0 < x < c, in series of the
functions Jq(X/:c), where X,- are the positive roots of the equation
An,. <»<*<'>'
2. In the expansion of fix) = 1 (0 < x < c) in series of Jo(X/c),
where X,- are the non-negative roots of «/J(Xc) = 0, show that A,- = 0
when; = 2, 3, * • • , and Ai — 1.
Sec. 68] BESSEL FUNCTIONS AND APPLICATIONS
165
3. Expand the function fix) = 1 when 0 < x < 1, f(x) = 0 when
1 < x < 2, /Cl) = in series of J 0 (X,-a;) where X,- are the roots of
J o(2X) = 0. Ans.
^ 2 2 Xj[Ji(2X,)] 2 (° < x < 2 )-
3 = 1
4. Expand f(x) = x (0 < x < 1) in series of J i(K,<c), where X,- are the
positive roots of Ji(K) = 0. Also note the function represented by
the series in the interval —1 < x ^ 0.
Ans . x — 2 ^ ^i(X,a;)/[Xj/ 2 (X J -)] (“1 < a; < 1).
i = i
68. Temperatures in an Infinite Cylinder. Let the convex sur-
face r = c of an infinitely long solid cylinder, or a finite cylinder
with insulated bases, be kept at temperature zero. If the
initial temperature is a function /(r), of distance from the axis
only, the temperature at any time t will be a function u(r , t ).
This function is to be found.
The heat equation in cylindrical coordinates, and the boundary
conditions, are
/i\ du ( d 2 U 1 dlA ^
(1) = (0 = r < c ’ * > °)>
(2) u(c - 0, t) = 0 (< > 0),
( 3 ) + 0 ) = f(r) (0 < r < c ).
It will be supposed that /(?•) and f'(r) are sectionally continuous
in the interval (0, c) and, for convenience, that f{r) is defined
to have the value l[f(r -f- 0) +/(r — 0)] at each point r where
it is discontinuous.
Particular solutions of equation (1) can be found by separation
of variables. The function u = R(r)T(t) is a solution, provided
that is, if
Since the member on the left is a function of t alone, and that
on the right is a function of r alone, they must be equal to a
constant; say, —X 2 . Hence we have the equations
rR" + R' + XbR = 0 ,
T + k\*T = 0 .
166 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 68
The equation in R here is Bessel’s equation (Sec. 65), in which
n = 0. If the function RT is to satisfy the condition (2), then
R(r ) must satisfy the condition
S(c) = 0.
According to Theorem 4, there are only real values of the param-
eter X in Bessel’s equation for which a solution exists and
satisfies this condition, and the positive values alone yield all
possible solutions. We are supposing that the function R(r)
and its derivative of the first order are continuous functions
when 0 ^ r ^ c. It can be shown that the second fundamental
solution, R = Fo(Xr), of Bessel’s equation, or Bessel’s function
of the second kind, is infinite when r = 0. Therefore the only
functions R(r) which satisfy the required conditions are Jo(X,r),
where X, are the positive roots of the equation
(4) Jq(\c) = 0.
The only particular solutions u = RT of the heat equation (1)
which satisfy the homogeneous boundary condition (2) are
therefore (except for a constant factor)
u = Jo(\jr)e~ kXiH ,
where X, are the positive roots of equation (4).
A series of these solutions,
(5) u(r, 0=5) AjJ o(X ? r)e~ fcx »%
i= i
will formally satisfy the heat equation (1) and the condition (2);
it will also satisfy the initial condition (3) provided the coefficients
Aj can be determined so that
f(r) = j? AjJ o (\r) (0 < r < c).
j = 1
This is true, according to the Fourier-Bcssel expansion, if
(6) A * = cWXjC)V Jo r/(r)/o(v) dr (i = 1, 2 , • • • ).
The formal solution of the boundary value problem is there-
fore represented by series (5) with coefficients (6), where X 3 - are
the positive roots of equation (4). That is, our solution can
167
Sec. 68] BESSEL FUNCTIONS AND APPLICATIONS
be written
u( r> 0 = | 2
j = l
[JiO*)]'
i
e~ k ^ H r f f(r')Jo(\jr') dr*.
This result can be fully established as a solution of the boundary
value problem stated here, by following the method used in
Sec. 46. For it can be shown that the numbers l/[X 3 *JJ(X 3 c)]
are bounded for all the roots X 3 -.* Consequently the numbers
Aj/\j are bounded for all j (j = 1, 2, • • ■ ), because f(r) and
Jo(\r) are bounded. Hence for each positive number to, the
absolute values of the terms of series (5) are less than the constant
terms
M\j exp ( — kXfto)
for all r (0 ^ r ^ c) and all t (t ^ to), where M is a constant.
The series of these constant terms converges, since X/4-1 — X 3
approaches t as j increases.
Series (5) therefore converges uniformly when t > 0, and so
the function u(r, t) represented by it is continuous with respect
to r when r = c. But u(c , t ) is clearly zero; hence condition (2)
is satisfied.
Since the derivatives of /o(X 3 r) arc also bounded, it follows
in just the same way that the differentiated series converge
uniformly when t > 0, and hence that result (5) satisfies the heat
equation (1).
Finally, owing to the convergence of series (5) when t = 0,
Abel's test applies to show that u(r, +0) = u(r, 0), when
0 < r < c; hence the condition (3) is satisfied.
To determine conditions under which our solution is unique,
we should need information about the uniform convergence
of the Fourier-Bessel expansion. This matter is beyond the
scope of our introductory treatment.
PROBLEMS
1. Write the solution of the above problem when the initial tempera-
ture f(r) is a constant A, and c = 1. Give the approximate numerical
values of the first few coefficients in the series.
Ans. u = 2vl[O.SO/ 0 (2 4r)e- 6 - % ‘ - .53 / 0 (5.5r)e- 30 **
+ .43*/o(8.6r)e _7r,fci -•••].
* This can he seen, for instance, from the asymptotic formulas for \j and
J\(x) developed in Ref. 1.
168 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 69
2. Over a long solid cylinder of radius 1 at temperature A throughout
is tightly fitted a long hollow cylinder of the same material, with thick-
ness 1 and temperature B throughout. The outer surface of the latter
is then kept at temperature 1 5. • Find the temperatures in the composite
cylinder of radius 2 so formed. This is a heat problem in shrunk fittings.
(Note that it becomes a case of the problem in this section when B is
subtracted from all temperatures.)
00
Am. u(r, t) = B + — 2 - ^ where X,,
X 2 , * • • are the positive roots of J r 0 (2X) = 0.
3. Derive the formula for the potential in a cylindrical space bounded
by the surfaces r = c, z = 0, and z = b, when the first two surfaces are
kept at potential zero and the third at potential V = f(r).
00
Ans. V(r, z) = ^ A,J 0 (X J r)(sinh Xyz/sinh X,6), where X/ are the
3=1
positive roots of equation (4), and the coefficients A f are given by
equation (6).
4. Derive the formula for the steady temperatures u(r, z) in the solid
cylinder bounded by the surfaces r = 1, z = 0, and 2=1, when the
first surface is kept at temperature u = 0, the last at u = 1, and the
surface z = 0 is insulated.
69. Radiation at the Surface of the Cylinder. Let the surface
of the infinite cylinder of the last section, instead of being kept at
temperature zero, undergo heat transfer into surroundings at
temperature zero, according to Newton’s law. The flux of heat
through the surface r = c is then proportional to the temperature
of the surface; that is,
—K = Eu when r = c.
Hr J
where K is the conductivity of the material in the cylinder and
E is the external conductivity, or emissivity. Let us write
h = cE/K.
The boundary value problem for the temperature u(r, t) can
be written as follows:
( 1 )
( 2 )
( 3 )
= k(— co <■
dt \dr 2 r dr )
du(c — 0, t) , . N
c — ^ ; = -hu(c - 0, i)
u(r, +0) = f(r)
r < c, t > 0),
(t > 0 ),
(0 < r < c ).
169
Sec. 69] BESSEL FUNCTIONS AND APPLICATIONS
The particular solution of equation (1) found before,
u = Jo(Xr)e~ AXa *,
will satisfy condition (2) provided X is any root X,* of the equation
d
c^/°(Xr) = — hJo(\r ) when r = c ,
that is, of the equation
(4) Xc/' 0 (Xc) = —hJ Q (\c).
Hence the solution of the problem (l)-(3) can be written
(5) u(r, t) = V A 3 M\r)e~W
where X, are the positive roots of equation (4), and where, accord-
ing to Theorem 5,
ox? C c
Aj ' = (X?c 2 + /i 2 )[Jo(X,-c)p Jo r/o(X,T ^ (r) dr Cj - i, 2, * * ‘ )•
If h = 0, then Xi = 0, and the first term of the series in formula
(5) is the constant 4i, where
^ J^ r/(r) dr.
This is the case if the surface r = c is thermally insulated.
PROBLEMS
1. Find the steady temperatures u(r, z) in a solid cylinder bounded
by the surfaces r — 1, z — 0, and z = L if the first surface is insulated,
the second kept at temperature zero, and the last at temperature
u = f(r).
Ans. u = ~ f r'/(r') dr'
Jo
>/p (X jV ) s inh (Xyg)
[</ o(X,)] 2 sinh (X y L)
r'/ o(X,*r')/(r') dr',
where X 2 , X 3 , * * • are the positive roots of Ji(X) - 0.
2. Find the steady temperatures in a semi-infinite cylinder bounded
by the surfaces r = 1 (z ^ 0) and z = 0, if there is surface transfer of
heat at r = 1 into surroundings at temperature zero, and the base z = 0
is kept at temperature u — 1.
170 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70
3. Show that the answer to Prob. 1 reduces to u = Az/L when f(r)
is a constant A.
70. The Vibration of a Circular Membrane. A membrane,
stretched over a fixed circular frame r = c in the plane z = 0,
is given an initial displacement z = f(r, p) and released from
rest in that position. Its displacement z(r, p, t), where (r, p, z)
are cylindrical coordinates, will be found as the continuous
solution of the following boundary value problem:
m * _ JdH ,1^,1^
( ' i ' a \<9r 2 + r + r 2 d<p *)’
(2) z(c, <p, t ) = 0,
(3) --- gf - - ' = 0, *(r, ft 0) = /(r, *).
The function z = R(r)$(p)T(t ) satisfies equation (1) if
where —X 2 is any constant, according to the usual argument.
Hence T — cos (a\t) if the first of conditions (3) is to be satis-
fied. Also R and <£ must satisfy the equations
I (r'R" + rR ') + XV 2 = ~ =
where m 2 is any constant, since the member on the left cannot
vary with either p or r.
Hence
$ = A cos fji<p + B sin up.
But z must be a periodic function of p with period 2t; hence
H = n (n = 0, 1, 2, • • • ). The equation in B then becomes
Bessel's equation with the parameter X,
r 2 R" + rR' + (X 2 r 2 - n 2 )R = 0,
and so R = J*(Xr). The solution 2 = R$T will satisfy condi-
tion (2) if X is any of the roots X n; * of the equation
(4) J n (\c) =0 (n = 0, 1, 2, - • * ).
Therefore if A n ; and B n3 - are constants, the functions
J n (^njr)(A n j cos up + B n / sin np) cos ( a\ nj t )
Sec. 70] BESSEL FUNCTIONS AND APPLICATIONS 171
are solutions of (1) which satisfy all but the last of conditions (3).
The function
(5) z{r, <p,t)
00 00
= X X cos rap + B n j sin rap) cos (a\ nj t)
n= 0 j = 1
satisfies this last condition also, provided the coefficients are
such that
(6) /(r, <p)
= j £ J COS ritp -f - ^ BnjJ n (\ n jV) J Sin
when — 7r < <p ^ t, O^r^c.
For each fixed r, the right-hand member of equation (6) is the
Fourier series for /(r, <p), in the interval — tt < <p < tt, provided
the coefficients of cos n<p and sin rap are the Fourier coefficients;
that is, if
1 f 71 "
- f( r > <p) cos rap dtp
(n = 1, 2, •
• * ),
VJ-ir
d(p
(n
= 0),
i r ,r
- I f(r, (p) sin n<p dtp
n J-i r
(n = 1, 2, •
• • )•
But the left-hand member of equation (7) is a series of Bessel
functions which must represent the function of r on the right when
0 S r ^ c. It is the Fourier-Besscl expansion of that function,
provided
(9) A n j = — ----- *-- 7 — .
7T C“[tf «+j(X n yC
( 10 ) Aoj =
S' — via I rJ n (\ nj r) dr j f(r, <p) cos rup dip
AnjC)\ z Jo J-r
(n = 1,2, ■■■ ),
L-^ f\j,M dr <,) d r .
X J 7rc 2 [./i(\(i;c)]' 2 Jo .
Similarly, according to equation (8),
2 C c
(11) B n j — -7— v I tJ rt(X n y/’) ch
7rc 2 [j n+1 (x n yc)j- Jo
J fir, ip) si
sin nip dtp.
172 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70
The displacement is therefore given by formula (5) when the
coefficients have the values given by equations (9), (10), and
(11), and \nj are the positive roots of equation (4), provided
series (5) has the necessary properties of convergence, differenti-
ability, and continuity.
PROBLEMS
1. Derive the formula for the displacement of the above membrane
if the initial displacement is f(r), a function of r only. Also show that
when/(r) = AJ 0 (X*r), where X* is a root of J Q (\c) = 0, the displacement
of the membrane is periodic in time, so that the membrane gives a
musical note.
Ans.
z(r - t] = 1 2
«/o(X,-r) cos ( a\jt )
lUhc) ]*
s:
r'J 0 (\jr')f(r') dr', where X,-
are the positive roots of J 0 (Xc) = 0.
2. Find the displacements in the above membrane if at t = 0 every
point within the boundary of the membrane has the velocity dz/dt = 1
in the position 2 = 0. This is the case if the membrane and its frame
are moving as a rigid body with unit velocity and the frame is suddenly
brought to rest.
Ans.
z(r, t) =
00
2 sin dk,i
XjJifoc)
Jo (X 3 r), where X,- are the positive
roots of / 0 (Xc) = 0.
3. Derive the following formula for the temperatures in a solid
cylinder with insulated bases, if the initial temperature is u = f(r , <p), and
the surface r = c is kept at temperature zero:
u(r, <p,t) = 2 j X /»(Xnir)(A n ,* cos rup + B ni sin n<p)e- k ^ n i t ,
n—Q j=l
where A ni and B nj have the values given by equations (9), (10), and
(11), and X nj - are the positive roots of equation (4).
4. Derive the formula for the temperature u(r , 2 , t) in a solid cylinder
of radius c and altitude L whose entire surface is kept at temperature
zero and whose initial temperature is A, a constant. Show that the
formula can be written
u(r, 2 , t) = Aui{z, t)u 2 (r , t),
where U\{z , t) is the temperature in a slab whose faces 2 = 0 and 2 = L
are kept at temperature ui = 0, and whose initial temperature is Ui = 1;
while u 2 (r, t ) is the temperature in an infinite cylinder whose surface
Sue. 70] BESSEL FUNCTIONS AND APPLICATIONS
173
r = c is kept at «2 = 0, and whose initial temperature is w 2 = 1. That
is,
e -k\,k
r /-v \ ° •
5. Derive the following formula for the temperatures in an infinitely
long right-angled cylindrical wedge bounded by the surface r = c and
the planes <p = 0 and <p = tt/2, when these three surfaces are kept at
temperature zero and the initial temperature is u = /(r, <p) :
u(r, cp, t) = XX A n jJ 2 n(X TlJ r) sin (: 2n<p)e -
n = 1 j= 1
where X«, are the positive roots of J 2 n(Xc) = 0, and A n f are given by the
formula
irc 2 [J r 2 »+i(X n /c)] 2 A»/ = 8 jT sin 2mp dtp Jj r/ 2n (X n yr)/(r, <p) dr.
6. If the planes of the wedge in Prob. 5 are = 0 and <p — <p 0 , show
that the formula for the temperature will in general involve Bessel
functions of nonintegral orders. Derive the formula for u(r , <p , t) in
this case.
7. Solve Prob. 5 if the planes <p = 0 and <p = 7t/2 are insulated,
instead of being kept at temperature zero. When f(r, ip) — 1, show
that your formula reduces to
= 2 ^ /o(Xjr
c X y Ji(X y «
T ^ 0 »
where Xy are the positive roots of ,/ 0 (Xc) = 0; thus w is independent of
the angle (p .
8. Solve Prob. 5 if all three surfaces r = c, = 0, and y? = 7 t/ 2, are
insulated instead of being kept at temperature zero.
9. Let w(r, 0 be the temperature in a thin circular plate whose edge,
r = 1, is kept at temperature u — 0, and whose initial temperature is
u = 1, when there is surface heat transfer from the circular faces to
surroundings at temperature zero. The heat equation can then be
written
du d*u 1 du ,
Tt = d? + ~rd7~ hu ’
174 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 70
where h is a positive constant. Derive the following formula for
u(r, t):
u = ^e~ ht
00
2
■i = 1
/ o(X,t)
X^xCX,)
e~ x * 2< ,
vhere X 3 - are the positive roots of J o(\) = 0.
10 . Solve Prob. 9 if there is also surface heat transfer at the edge
r = 1, instead of a fixed temperature there, so that
— —hou when r = 1.
11 . Derive the following formula for the potential 7(r, z) in the semi-
infinite cylindrical space, r ^ 1, 2 ^ 0, if the surface r = 1 is kept at
potential V = 0, and the base 2 = 0 at 7 = 1 :
V =
2
2 /0 M
X,/ x(X/)
j = i
where \ 3 are the positive roots of / 0 (X) = 0.
12 . Let 7(r, 2 ) be the potential in the space inside the cylinder r = c,
when the surface r = c is kept at the potential V — f(z), where the given
function f(z) is defined for all real 2 . Derive the following formula for
V(r, 2 ):
r ■ * J. " ^ I". £§ m ^ *'
where i = *\/— T.
13 . Let 7(r, 2 ) be the potential in the semi-infinite cylindrical space
r ^ 1, 2 ^ 0, if dV/dz = 0 on the surface 2 = 0 and if, on the surface
r = 1, V = 1 when 0 < 2 < 1, while 7 = 0 when z > 1. Show that
V{r, z)
Joiiar) .
— t < cos az sin a da .
aJ oyia)
REFERENCES
1. Watson, G. N.: “ A Treatise on the Theory of Bessel Functions,” 1922.
2. Bowman, F.: “Introduction to Bessel Functions,” 1938.
3. Gray, A., G. B. Mathews, and T. M. MacRobert: “A Treatise on Bessel
Functions and Their Applications to Physics,” 1931.
4. Jahnke, E., and F. Emde: “Tables of Functions,” 1933.
CHAPTER IX
LEGENDRE POLYNOMIALS AND APPLICATIONS
71. Derivation of the Legendre Polynomials. Any solution of
the differential equation
(1) (1 - z 2 ) -2x^ + n(n + 1 )y = 0,
known as Legendre’s equation , is called a Legendre function .
Later on we shall see how this equation arises in the process of
obtaining particular solutions of Laplace’s equation in spherical
coordinates, when x is written for cos 6. We shall consider
here only the cases in which the parameter ?i is zero or a positive
integer.
To find a solution which can be represented by a power series,
if any such exist, we substitute
00
(2) V = X
into equation (1) and determine the coefficients a,-. The substi-
tution gives
00
X Ll’O' — l)a,x»' -2 (l - x 2 ) — 2ja i x i + n(n + 1 )a j x i ] = 0
3 =0
or
(3) X + 1) ~ j(j + 1 + j(j ~ l)fl,-x 3 '- 2 } = 0.
Since this must be an identity in x if our scries is to be a
solution, the coefficient of each power of x in series (3) must
vanish. Setting the total coefficient of x 3 in this scries equal to
zero gives
(j + 2)(j + + [n(n + 1) — j(j + 1 )]dj = 0
(j = 0, 1, 2, • • * ),
175
176 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 71
which is a recursion formula giving each coefficient in terms
of the second one preceding it, except for do and ai. It can be
written
(4)
2 —
(n - j)(n + j ± 1)
O' + 1 )(j + 2)
a,-
U = 0 , 1 , 2 ,
)■
The power series (2) is therefore a solution of Legendre’s
equation within its interval of convergence, provided its coeffi-
cients satisfy relation (4); this leaves a Q and cu as arbitrary
constants. But since n is an integer, it follows from relation (4)
that dn +2 = 0, and consequently
dn-j-4 — &n4-6 — * — 0.
Also, when a 0 = 0, then a 2 = a 4 = * * * =0; and when a 1 = 0,
then a 3 = a 5 = • • • =0.
Hence if n is odd and a Q is taken as zero, the series reduces
to a polynomial of degree n containing only odd powers of x.
If n is even and a\ is set equal to zero, the series reduces to a
polynomial of degree n containing only even powers of x. So
there is always a polynomial solution of equation (1), and for it
no question of convergence arises.
These polynomials can be written explicitly in descending
powers of x whether n is even or odd. All the nonvanishing
coefficients can be written in terms of a n by means of recursion
formula (4); thus
d>n — -2
n(n — 1)
'2(2 n - 1)
a n >
(n — 2 )(n — 3)
a "- 4 “ 4(2 n- 3) an_!!
_ n(n — 1 ){n — 2 ){n — 3)
2 • 4(2n - 1)(2 n - 3)
and so on. Hence the polynomial
(5)
y =
& n
X n —
^ ~ 0 a-n-2
2(2 n - 1)
. n(n — l)(n — 2 )(n — 3)
+ 2 • 4(2« — l)(2n — 3)
x n-i _
1
is a particular solution of Legendre’s equation.
Sho. 72] LEGENDRE POLYNOMIALS AND APPLICATIONS 177
Here the coefficient a n is arbitrary. It turns out to be con-
venient to give it the value
„ _ (2 n- 1)(2 n - 3)
~
do = 1 .
3 • 1
if n = 1, 2,
With this choice of a n functions (5) are known as the LeQcndTs
; polynomials :
(6) P n (x) = - 1 ) (2n - 3) • • • 1 r _ n(n - 1) _ 2
n\ L 2(2n - 1) x
. ~ 1)(« ~ 2 )(n - 3) _ 4
2 • 4(2 n - 1)(2 n - 3)
The function P n (x) is a polynomial in x of degree n, containing
only even powers of x if n is even and only odd powers if n is odd.
It is therefore an even or odd function according as n is even or
odd; that is,
Pn(-x) = (-1)-P n (*).
The first few polynomials are as follows:
Po(x) = 1, P l ( x ) = a:, P 2 (x) = - 1),
p *( x ) = i(&* 9 ~ 3x), P 4 ( x ) = %(35x 4 - 30x 2 + 3),
Pb(x) = 1 ( 63^ 6 - 70a; 3 + 15x).
PROBLEMS
1. Show that formula ((>) for P n (x) can be written in the following
compact form:
(~l)*(2n ~ 2 j)l
2 n jKn - j)\(n - 2j ) !
where m = n/2 if n is even, and m = (n — 1 ) /2 if n is odd.
2. With the aid of the formula in Prob. 1, chow that
A„(0) = (-l) n ^L=(— ,)n
P 2 n-l(0) = 0,
3 ■ 5 • • • (2n - 1)
2-A (2n) ’
(» = 1, 2, • • • ).
72. Other Legendre Functions. When n is a positive integer
or zero, wc obtained the solution y = P n (x) of Legendre’s equa-
tion by setting one of the two arbitrary constants a 0 or a, in the
178 FOURIER SERIES AND BOUNDARY PROBLEMS [Sac. 72
series solution equal to zero. If these constants are left arbitrary,
it is easily seen that the general solution of Legendre’s equation
can be written
(1) y = AP n (*) + BQ n (x),
where A and B are arbitrary constants. The functions Q n (x)
here, called Legendre’s functions of the second kind, are defined
by the following series when \x\ <1:
Q n (x) = ai x- ^ ^ - "V
if n is even; and
(» - 1 )(» + 2 )
3!
(re - 1)(» - 3)(» + 2)(» + 4) .
i — ■* - •
Q »(*)
= ao [ 1
n ( n + 1) _« , n(n - 2) (to + l)(n + 3) „
2! _l 4! x
if n is odd; and where
]■
«i = (-1)
ao = ( — 1)
1 2-4 • • • n
1 ■ 3 • 5 • • • (re - 1)’
2 - 4 • • • (re - 1)
1 • 3 • 5 • • • re
Of course P»(x) is a solution for all x. But when |x| > 1,
the above series for Q n (x) do not converge. To obtain a second
fundamental solution in that case, a series of descending powers
of x is used. The following solution so obtained is taken as the
definition of Q n (x ) when \x\ > 1 :
Qn(x) =
re!
1-3-5
(2re + 1)
2 — »— 1 + ( W + ] )^ n + 2 ) x -n-i
^ 2(2re + 3) *
I (n + l)(n + 2)(re + 3)(re + 4)
' r 2 -4(2re +3)(2re + 5)
+
Both P n (x) and Q„(x) are special cases of the function known
as the hypergeometric function.
When re is not an integer, the two fundamental solutions of
Legendre’s equation can be written as infinite series. These
Sec. 73] LEGENDRE POLYNOMIALS AND APPLICATIONS 179
are both power series when \x\ < 1; but when |x| > 1, they are
series in descending powers of x .
Of these various Legendre functions the polynomials P n (x)
are by far the most important. Let us now continue with the
study and application of those polynomials.
73. Generating Functions for P n (x ). If — 1 g x ^ 1, the
function
(1 - 2xz + z :
and its derivatives of all orders with respect to z exist when
\z\ < 1. For these functions are infinite only when
1 - 2xz + z 2 = 0,
that is, if
z — x ± '\fx 1 — 1 = cos 6 + i sin 6,
where we have written cos 6 for x. But this shows that |z| = 1.
It is shown in the theory of functions of complex variables that
such regular functions of z arc always represented by their
Maclaurin series within the region of regularity (|z| < 1, in this
case).
It will now be shown that the coefficients of the powers of z
in that series representation of the above function are the
Legendre polynomials; that is, when — 1 ^ x ^ 1 and \z\ < 1,
(1) (1 - 2xz + z 2 )~*
= Pq(x) + Pi(x)z + Pi(x)z ' 2 + • • ■ + P n {x)z n + • • ■ .
To find the coefficients, it is best to write the expansion by
means of the binomial series:
[1 - z(2x - «)]-* = 1 + 2x - z) + z*{2x - zY
+
+ -
(2 m - 1)
2"n\
z'*( 2x — z) n +
The terms in z n come from (.ho form containing z"(2x — z) n and
preceding terms, so that the total coefficient of z n is
1 -3
(2m - 1
2’*n !
+ 1 -
1 • *3
-- (2.r)'‘ -
(2?i - 3) (n - 1)
3
2“ -1 (n - 1)!
(2m - 5) (n - 2 )(n - 3)
2’- 2 (m - 2)
2!
1 !
(2x)“- 4 -
(2x) n ~
180 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 73
This can be written
1-3
■ ( 2 n - 1 )
n\
’ „ _ "(» ~ 1) „«-2
* 2(2 n - 1)
, n(n - l)(n - 2) (ft - 3) 4
" r 2 • 4(2» - 1)(2» - 3)
But this is P n (x) ; hence relation (1) is established. Incidentally,
this shows the reason for the value assigned to a n in our defi-
nition of P n (x) (Sec. 71).
For 2 = 1, expansion (1) becomes
i>n(l)s* = (1 ~ z ] )“ 1 =
or 0
Consequently
Pn(l) = 1 (n = o, 1, 2,
Likewise, putting 2 = 0 gives
• * )•
= (i+ * 2 )-* = i - 1 * 2 + H ~
+ (-i) n
1-3
(2 n — 1)
2-4
and therefore
(2n)
+
P 2 „(0) = (-1)
P 2n— 1(0) ' = 0
(2 n — 1)
(n - 1, 2, ■ • •
By differentiating equation (1) with respect to z and multiply-
ing the resulting equation by (1 — 2 xz + z 2 ), the following
identity in z is found:
00
(x - z)( 1 - 2 xz + z 2 ) -5 = (x — z) Vp„(r)z»
0
00
= (1 — 222 + 2 2 ) V nP n (x)z n ~\
0
Equating the coefficients of 2 n in the last two expressions, it
follows that
(n + l)P n+ i( 2 ) - (2 n + l) 2 P n ( 2 ) + nP n _ 1 ( 2 ) = 0
(n = 1, 2, • ■ •
this is a recursion formula for P n ( 2 ). It is valid for all values of 2 .
Sec. 74] LEGENDRE POLYNOMIALS AND APPLICATIONS 181
The result of integrating polynomial (6), Sec. 71, n times from 0
to x is
(2 n - l)(2n - 3)
( 2n)\
— x 2n —
nx
■2n-2
. »(» ~ 1 ) r 2n — 1 .
2!
and the expression in brackets differs from ( x 1 2 — l) n only by a
polynomial of degree less than n. By differentiating n times,
then, it follows that
Pn(x) = JLJ1 {X 2- x)»
2 n n ! dx n x J
This is Rodrigues 1 formula for the Legendre polynomials.
PROBLEMS
1. Show that the derivatives of Legendre polynomials have the
properties
PL( 0 ) = 0 ; PL+M = ( — '• ( 2 ” 2 w ) 1) -
The latter can be found by differentiating equation (1) with respect to x
and setting x = 0.
2. Carry out the details of the derivation of Rodrigues’ formula.
3 . Using Rodrigues’ formula, show that
K + i(a) - PL iW = (2 n -1- 1 )P n (x) (n = 1, 2, • ■ • ).
4 . Using the formula in Prob. 3, obtain the integration formula
^ P n (x) dx = 2n \ ^»+iWl (ft = J j 2, * • * ).
74. The Legendre Coefficients. When —1 ^ x ^ 1, we have
just shown that P n (:r) is the coefficient of z n in the expansion of
the generating function (1 — 2 xz + z - 2 )“* in powers of z. When
x = cos 6 = %(c i0 + c"' 16 ))
this generating function can be written
[1 — z(e i0 + e~ i0 ) + 2 2 ]~i = (1 — ze ie )~*( 1 —
and therefore as the product of the scries
1 + i«j« + 1
2-4
2«2 id
z z e
1 • 3 • • • (2n - 1)
2 - 4 • • • (2 n)
+ • • •
182 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 74
and the series
1-8 • ■ • ( 2 .- 1 )
^ 2 • 4 • • ■ (2 n) 2 6
The coefficient of z n in this product is
1 • 3 ■ • • (2n - 1)
2 • 4 • - • (2 n)
(e in> + e~ in6 )
+ \ 2^1 (e<(B_2>9 + e ~ i{n -™) +
hence this is P n (cos 0 ). Thus we have the following formula
for this function:
(1) P n ( cos 0)
= ^!2^ L C ° S + 1 • (2ra — 1) C0S (n “ W
+ i-2(2n-lKL-3) C0S (n~W + ■ ■ ■ +
where the final term T n is the term containing cos 0 if n is odd;
but it is half the constant term indicated if n is even.
These functions are called the Legendre coefficients. Tables
of their numerical values will be found in some of the more
extensive books of mathematical tables, or in Ref. 3 at the end
of this chapter.
According to formula (1), the first few functions are
P 0 (cos 0) = 1,
Pi (cos 0) = cos 6,
P 2 ( cos 0) = i(S cos 20 + 1),
P 3 (cos 6) = |(5 cos 30 + 3 cos 0),
p 4 (cos 0) = -g^(35 cos 40 + 20 cos 20 + 9).
The coefficients of the cosines in formula (1) are all positive.
Consequently P n (cos 0) has its greatest value when 0 = 0.
Since P n (l) = 1, it follows that P n ( cos 0) ^ 1. Also, each
cosine is greater than or equal to -1, so that P n ( cos 0)^-1.
That is, the Legendre coefficients are uniformly bounded as follows:
|Pn(cos 0)| ^ 1 (n = 0, 1, 2, • • • ),
for all real values of 0.
Sec. 75] LEGENDRE POLYNOMIALS AND APPLICATIONS 1S3
75. The Orthogonality of P n (x). Norms. Legendre’s equa-
tion can be written in the form
( 1 )
d_
dx
(1
+ 1 ) 2 / = 0 .
It is clearly a special case of the Sturnx-Liouville equation
(Sec. 24), in which the parameter X has been assigned the values
(2) X = n(n + 1) (n = 0, 1, 2, • • • ).
In this case r(x) = 1 — x 2 , q(x) = 0, and the weight function
p(x) = 1.
Since r(x) = 0 when x = ±1, no boundary conditions need
accompany the differential equation to form the Sturm-Liouville
problem on the interval ( — 1,1). It is only required that the
characteristic functions and their first ordered derivatives be
continuous when — 1 ^ x ^ 1. But the polynomials P n (x) are
solutions of equation (1), and, of course, they have these required
continuity properties.
The Legendre polynomials P n (x) are therefore the characteristic
functions of the Sturm-Liouville problem here, corresponding
to the characteristic numbers (2). According to Sec. 25, then,
the functions P n (x) form an orthogonal set in the interval ( — 1, 1),
with respect to the weight f unction p(x) — 1 ; that is,
(.3) Tj Pm(r)Pn(r) dx = 0 if m ^ n ( m , n = 0, 1, 2, • • ■
Furthermore, there can be no characteristic functions of the
Sturm-Liouville problem here which correspond to complex
values of the parameter X, because p(x) does not change sign.
We shall soon see that the functions P n (x) and the numbers (2)
are the only possible characteristic functions and numbers of
the problem.
To find the norm of P n (x), that is, the value of the integral in (3)
when m = n, a simple method consists first of squaring both
members of equation (1), Sec. 73, to obtain the formula
(1 — 2 xz + s 2 )" 1 = [X •
We now integrate both members here with respect to x over the
interval (—1, 1) and observe that the product terms on the right
184 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 75
vanish in view of the orthogonality property (3). Thus,
dx
1 — 2 xz + z 2
P [Pn{x)YdX
0 j — l
( 1*1 < !)•
The integral omthe left has the value
- i log (1 - 2 xz + z 2 )
-l
1 . 1+2
"I 10 *— *
- 2 ( 1+ I + I + ’ " + 5^TT + " ') <W<1) -
By equating the coefficients of z 2n in the last two series, we
have the following formula for the norm of P n (%) :
(4) JP [Pn(x)] 2 dx = (» - o, l, 2, • • • ).
The orthonormal set of functions here in the interval ( — 1, 1) is
therefore {<p n (x)} ) where
<Pn(x) = Vn + i Pn(x) (n - 0, 1, 2, • • - ).
Since [P^Oc)] 2 and the product P m (x)P n (x ), in which m and
n are both even or both odd, are even functions of .r, it follows
from formulas (4) and (3) that the polynomials of even degree,
(5)
\Z2rT+lP n (x) (ft = 0,2,4, ■ ■ • \
form an orthonormal set of functions in the interval (0, 1) ; and the
same is true for the polynomials of odd degree , represented by (5)
when ft = 1, 3, 5, • * * .
PROBLEMS
1. Establish the orthogonality property (3) by using Rodrigues’
formula for P n (x) and successive integration by parts.
2. State why it is true that
dx = 0
(ft - 1, 2, 3, ■ ■ •
3. Use the method of Prob. 1 to obtain formula (4) for the norm of
F»(s),
Sec. 76] LEGENDRE POLYNOMIALS AND APPLICATIONS 185
76. The Functions P n (x) as a Complete Orthogonal Set. Let.
us now prove the following theorem:
Theorem 1. In the interval (-1, 1) the orthogonal set of func-
tions consisting of all the Legendre polynomials
P n (x) (n = 0, 1, 2, • ■ • )
is complete with respect to the class of all functions which , together
with their derivatives of the first order , are sectionally continuous in
(-1, 1).
We are to prove that if \p(x) is a function of this class which is
orthogonal to each of the functions P n (x), then \p(x) = 0 except
at a finite number of points in the interval.
Let us suppose, then, that
(1) f\Pn(x)\ls(x) dx = 0 (n = 0, 1, 2, • ■ • ).
According to our recursion formula (Sec. 73),
(2n + 1 )xP n (x) = (n + l)P n+ i(x) + nP n ^(x)
(n = 1 , 2 , ■ • * );
and this formula can be replaced by the formula xP 0 (x) = P Y (x)
when n = 0. When vve multiply its terms by and integrate
from —1 to 1, the integrals in the right-hand member vanish, so
that
(2) P n (x)xt(x) dx = 0 (n = 0, 1, 2, • • • ).
If we suppose that the orthogonality property (1) is true
when ^(.r) then* is replaced by x k \p(x), the method just employed
clearly shows that, property (1) is true when xf^(x) is replaced by
x k+ '\f/(x). In view of equation (2), then, we conclude by
induction that, for every integer j,
f'j l > n(-r).rty(x) dx = 0 (n, j = 0, 1 , 2, ■ ■ ■ ).
As a consequence, wo have
Pntx)t(x)
( mirx )- (nnrx ) 4
2! H 4!
dx = 0;
because the power series in the brackets, representing cos mwx ,
is uniformly convergent. Moreover, the series obtained by multi-
186 FOURIER SERIES AND BOUNDARY PROBLEMS [Sue. 76
plying all terms of this series by the sectionally continuous func-
tion P n (x)\p(x) is also uniformly convergent, so it can be integrated
termwise. Thus
J l 1 P n (x)\p(x) cos mTX dx = 0 (m = 0, 1, 2, • * * );
and in just the same manner it follows that
j* 1 l Pn(z)ft(z) sin mTxdx = 0 (m = 1 , 2 , • ■ * ).
All the coefficients in the Fourier series corresponding to the
function P n (x)\p(x ) in the interval ( — 1, 1) therefore vanish.
But this function and its first derivative are sectionally con-
tinuous; hence it is represented by its Fourier series except at
the points of discontinuity of the function. Except possibly
at a finite number of points, then,
i(x) =0 (-1 S s S 1),
and the theorem is proved.
There is an interesting consequence of the above theorem.
Suppose that for some real value of A other than n(n + 1),
Legendre’s equation
( 3 )
d
dx
( 1 -*’>g] + x 1 ,-0
has a solution y = y<,(x), where y' 0 ( x) is continuous in the interval
- 1 fk x ^ 1 . Then, according to Sec. 25, y 0 (x) is orthogonal to
all the characteristic functions P n (x) already found and cor-
responding to X = n(n + 1). But this is impossible according
to Theorem 1, unless j/ 0 = 0.
Since we have already shown that X must bo real if equation (3)
is to have such a regular solution, we have the following result:
Theorem 2. The only values of X for which the Legendre equa-
tion (3) can have a non-zero solution with a continuous derivative of
the first order, in the interval — 1 ^ x g 1, are
\ = n(n + 1) (n = 0, 1, 2, • • ■ ).
It can be shown that the Legendre functions of the second
kind, Qn(x), which also satisfy equation (3) when X = n(n + 1),
become infinite at x = ±1. Consequently, the polynomials P n (x)
Sec. 77] LEGENDRE POLYNOMIALS AND APPLICATIONS 187
are , except for constant factors, the only solutions of equation (3)
which have continuous first derivatives in the interval — l^x<l.
77. The Expansion of x m . Without the use of a general expan-
sion theorem, we can easily show how every integral power of x
and therefore every polynomial, can be expanded in a finite
series of the polynomials P n (x). It will be clear that these
important expansions are valid for all x, not just for the values
of x in the interval (—1, 1).
According to its definition, the polynomial P m (x) has the form
(1) Pm(x) = ax m + bx 7> '- 2 + cx m - 4 + * • * ,
where a, 6, • • • are constants depending on the integer m.
Therefore
X m = ]-P m (x) - - T'"~“ - -*»-•* - • • -
a a a
That is, every integral power x m of x can be written as a constant
times P m (x ) plus a polynomial in x of degree m — 2. Applying
this rule to x m ~ - in the last equation, we see that x m is a linear
combination of P m (x), P n ^(x), and a polynomial of degree
m - 4. Continuing in this way, and noting that only the
alternate exponents m, m - 2, m - 4, • • ■ appear in the poly-
nomials here, it is clear that there is a finite series for x m of the
following form :
(2) x m = A„P m (x) + A„ 1 _ 2 P„,_ 2 (a:) + • • • -f T,
where the final term T is a constant A „ if m is oven, and
T = A,P,(a 0
if m is odd.
To find the value of any coefficient A wo multiply all terms
of equation (2) by P,„_ 2 ,(t) and integrate over the interval ( — 1,1),
In view of the orthogonality of the functions P„(x), this give's
f i v( x ) dec = A m -»j [Pra-s/fr)]* dx.
But the integrand on the left is an even function of x for every
integer m; and the integral on the right, the norm of P m _ 2j (x),
has the value 2 /(2m - 4j + I). Therefore,
(3) A„,- 2 j = (2m — 4j + 1) x m P m - 2 j(x) dx.
188 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 77
We shall develop here the following integration formula ,
valid for every real number r greater than — 1 :
(4)
(x) dx =
r(r — 1)
(r — n + 2)
(r + n + 1) (r + n — 1) • ■ • (r — n + 3)
(n = 2, 3, ■ • •
In view of this formula, the integral in equation (3) has the
value
m(m - 1) • ■ • (2j + 2)
(2m — 2j + l)(2m — 2j — 1)
(2j + 3)
ml
1-3-5
• (2m - 2j + 1)2 ijl
The values of the coefficients A m - % are therefore determined,
and we can write expansion (2) for any integral power of x as
follows :
xm = 1 • 3 • 5 • • • (2m + 1 ) } 2m +
+ ( 2 m - 3 ) Pm _ 2 ( x )
+ (2m - 7) gg- + ‘><f> ~ » P_, (l ) +...].
For the first few values of m, we have
1 = Po(*), X = P l(x) 7 X 2 = •§p 2 (#) + 'S'Pn(^),
* 3 = | -Pz(x) + fPi(z), Z 4 = + fP 2 (z) + *Pn(s).
Derivation of Formula (4). To obtain the integration formula
(4), let us first observe that, in view of expression (1),
(5) x'P n
( x ) dx
-£
(ax r+n + bx r+n ~ 2 +
+
) dx
+
r+w + 1 1 r n — 1 ' r + n — 3
as long as r > — 1. The last member can be written
Sir) ,
(r + n + 1 ) (r + n — 1) (r + n — 3) ■ ■ •
+
( 6 )
Sec. 78] LEGENDRE POLYNOMIALS AND APPLICATIONS 189
where
fir) = a(r + n — l)(r + n — 3) • ■ ■
+ b(r f n + l)(r + n — 3) •••+•■■ .
We can see that/(r) is a polynomial in r of degree nf 2 or (n — 1) / 2 ,
according as n is even or odd.
Now the product x n ~ 2j P n (x ) is an even function of x for every n,
when j = 1, 2, • • ■ ; so it is evident from equation (2) that
jf 1 x n-up n (x) dx = 0 (j = 1, 2, • • • ; n ^ 2j).
Therefore our integral (5) vanishes when r is replaced by n — 2,
n — 4, ft — 6, etc., down to zero or unity, and so does the poly-
nomial f(r). Also, the coefficient of the highest power of r in
fir) is a + b + c + * * * , which is P n (l) or unity. Hence,
when n = 2, 3, • ■ ■ , the factors of fir) can be shown as follows:
f(r) = (r — n + 2)(r — n + 4) • • • r,
if n is even; and
f(r) = (r - n + 2)(r - n + 4) • • • (r - 1),
if n is odd. In either case the fraction (6) can be written as
r(r — l)(r — 2) * • - (r — n + 2)
(r + + l)(r + n — 1) • ~ (r — n + 3)
(n = 2, 3, ■ • • ; r > -1).
This is the value of our integral; hence formula (4) is established.
78. Derivatives of the Polynomials. The derivative P r n {x) is a
polynomial of degree n — 1 containing alternate powers of x ,
namely, a*'* -1 , x n ~ [ \ • • • . It can therefore be written as a finite
series of Legendre polynomials:
= A n_./Vl(>) + An-lPn-zix) + ’ # •
To find the coefficient, Aj (j — n — 1, w — 3, • • • ), we
multi})ly all terms by Pj(x) and integrate; thus
A i = 2j -z 1 f x FMKW dx.
When integrated by parts, the integral here becomes
\pMPJx) 1 - f 1 P.^P^x) dx,
L J i %) l
190 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 78
and this last integral vanishes because Pj(x) is a linear combina-
tion of the polynomials P,_ i(x), P/_ 3 (rr), etc., each of which is of
lower degree than P n (x). Therefore
A l = %±±[ i -P 3 .(-l)P n (-l)]
= ^i[l - (-1)*"] = 2j + l,
since j + n = 2n — 1, 2n — 3, • * * .
Consequently we have the following expansion, valid for all x:
(1) P' n (x) = (2n - ^Pn-tCa:) + (2ft - 5)P n _ 3 (z)
+ (2n — 9)P n— c(^r) + • • • ,
ending with 3Pi(z) if n is even, and with P 0 (x) if n is odd.
. When — 1 g x ^ 1, we have seen that \P n (x)\ ^ 1; hence for
these values of x it follows from expansion (1) that
\Pin( x )\ = (4ft ” 1) + (4ft — 5) + * * * + 3 = n(2n + 1);
and similarly,
l^ 2 n+l(^)| (ft + l)(2n + 1).
Therefore,
\P 2n(. x ) | = (2ft) 2 , iP^n+iWI = (2ft + l) 2 ;
that is, for all x in the interval — 1 g x g 1,
( 2 ) l^nWI ^ ft 2 ( n = 0, 1, 2, • • • ).
Differentiating both members of expansion (1) and noting that
^ ft 2 , |P^_ 3 (3)| ^ ft 2 , etc., we see by the method used
above that
(3) \P"(x)\ ^ n 4 (-1 g x ^ 1, n = 0, 1, 2, • • • ).
Similarly, for derivatives of higher order, | P^(x)\ ^ n~ k .
Let us collect our properties on the order of magnitude of the
Legendre coefficients and their derivatives as follows:
Theorem 3. For all x in the interval — 1 g x g 1, and for
71 = 1, 2, 3, • ■ ■ , the values of the functions
i p "Wi> ^ «(*)!, ^ \p’:(x)\, • • •
can never exceed unity .
Sec. 79] LEGENDRE POLYNOMIALS AND APPLICATIONS 191
79. An Expansion Theorem. The normalized Legendre poly-
nomials were found in Sec. 75 to be
<Pn{x) = y/n + ^Pn(x) (n = 0, 1, 2, • • * ).
The Fourier constants, corresponding to the orthonormal set
here, for a function f(x) defined in the interval ( — 1, 1), are
Cn = dx = \/n + % ^ f(x)P n (x) dx .
The generalized Fourier series corresponding to f{x) is therefore
oo oo
2 C ^ X) = 2 Pn ^ dx'.
This can be written
W XAJ> n (x),
0
where
(2) An = 1 j_J(x)P n (x) dx (n = 0, 1, 2, • • • ).
The series (1) with the coefficients (2) is called Legendre's
series corresponding to the function /(z). It was shown above
that if f(x) is any polynomial, this series contains only a finite
number of terms and represents f(x) for all values of x.
It can be shown that, when -1 < x < 1, Legendre's series
converges to f(x ) under any of the conditions given earlier for
the representation of this function by its Fourier series. We now
state explicitly a fairly general theorem on such expansions, and
accept it without proof for the purposes of the present volume.*
Theorem 4. Let f(x) be bounded and mtcgrable in the interval
( 1, 1). I hen at each point , x ( — 1 < x <! 1) which is interior
to an interval in which f(x) is of bounded variation , the Legendre
series corresponding to f(x) converges to \[f(x + 0) + f(x — 0)];
that TrSy
(3) i[f(x + 0) +S(x - 0)] = V AJP n (x) ( - 1 < a < 1)
0
where the coefficients A n are given by formula (2).
* The theorem stated here is a special ease of a theorem proved in Chap.
VII of Kef. 1. The proof is lengthy and involves more advanced concepts
than wc employ in this book.
192 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 79
In particular, if f(x) is sectionally continuous in the interval
( — 1, 1), and if its derivative f'(x) is sectionally continuous in
every interval interior to ( — 1, 1), then expansion (3) is valid
whenever — 1 < x < 1. For it can be shown that the condi-
tions in the theorem are then satisfied at all points.
If /(re) is an even function, the product f(x)P n (x) is even or odd
according as n is an even or odd integer. Hence A n = 0 if n is
odd, and
(4) A n = (2 to + 1) f f(x)PJx) dx (to = 0, 2, 4, • • • );
so that expansion (3) becomes
(5) Mrta + 0) +/(*-<>)]
= ^4.o + AJPzix) + AJP±(x) + * * ■ ,
where the coefficients are defined by formula (4).
Similarly, if f(x) is an odd function , the expansion becomes
(6) Mf(a + 0) + f(x - 0)]
= AiPi(x) + AzP%(x) + A’tJPzix) + * ’ * j
where
(7) An = (2 TO + 1) jT 1 f(x)P n (x) dx (to = 1, 3, 5, ■ • • ).
In the interval (0, 1), either one of the expansions (5) or (6)
can be used , provided of course that f{x) satisfies the conditions
of the theorem in that interval. For if f(x) is defined only in
(0, 1), it can be defined in ( — 1, 0) so as to make it either even
or odd in ( — 1, 1). It was pointed out earlier (Sec. 75) that the
polynomials Pz n {x), and the polynomials P 2 n-\(x),. appearing in
expansions (5) and (6), respectively, form two sets of orthogonal
functions on the interval (0, 1).
When x = cos 6 , expansion (3) can of course be written
oo
F(6 ) = 2) AnPn(C0S e) (0 < 9 < r)
0
at points where F($) is continuous, where
A n = — F(9)P n ( cos 9) sin 9 dB (n = 0, 1, 2, - • ■
PROBLEMS
1. If f(x) = 0 when —1 < x <0, f(x) = 1 when 0 < x <1, and
./(0) = i, obtain the following expansion for f(x) when — 1 < x < 1 :
Sbc. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 193
/(*) 2 2 [Ps»(0) — P 2 » + 2 ( 0 )]P 2n+1 (a:)
0
1 3
~2 + 4 x + ^i ( _1 )”
4ra + 3 1 • 3 • • • (2n - 1) _
in + 4 2 • 4 • • • (2 n) p ^+i(x).
Suggestion: See Prob. 4, Sec. 73.
2. If /(*) = 0 when -1 < x g 0, and f(x) = x when 0
show that, when —1 < x < 1,
x < 1,
fix) = gi + ip,(*) + | P 2 (x) -
, 13 - 4 ! „ , ,
+ 2 7 ■ 4121 ^ W - • • • .
3. Expand the function /(a) = *, when 0 g x < 1, in series of
Legendre polynomials of even order, in the interval (0, 1).
80. The Potential about a Spherical Surface. Let a spherical
surface be kept at a fixed distribution of electric potential
V - F(6), where r, <p, d are spherical coordinates with the origin
at the center of the sphere. The potential at all points in the
space, assumed to be free of charges, interior to and exterior to
the surface is to be determined. It will clearly be independent
of <p; hence it must satisfy the following case of Laplace’s equation
in spherical coordinates:
( 1 )
(rV) + —
J sm g so
= o.
The potential V{r , 6) will also be required to be continuous,
together with its second-order derivatives, in every region not
containing a point ol the surface, and to vanish at points infinitely
far from the {surface. The boundary conditions are therefore
( 2 )
Iim V(r, 6) = b\6) (O<0<tt),
T — >C * '
where c is the radius of the spherical surface, and
(3) lini V(r, 0) — 0.
r — > oo
Particular solutions of equation (1) can be found by the usual
method. Setting V = R(r)0(6), equation (1) becomes
194 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 80
- — (rJt\ = 1 d f de\
Rdr 2 ^ 6 sin 9 d9 \ Sln 9 dd )
Both members here must be equal to a constant, say X; hence we
have the equations
d 2
r~(rR)=\R,
1 d / l — cos 2 9 de \ _
sin 9dd\ sin 6 d9 ) “ °‘
The first of these is Cauchy’s linear equation,
r 2 R" + 2 rR' - \R = 0 ,
which can be reduced to one with constant coefficients by substi-
tuting r = e‘. Its general solution is
R = + Br -i~VY+i'
Writing — + v/X + -J- = n, so that X = n(n + 1), we have
R(r) = Ar» + A,
where n is any constant.
Writing x for cos 0, the equation in 9 becomes, in terms of the
new parameter n,
!r] +n(n + !)e = 0,
which is Legendre’s equation. We have seen that the solution
of this equation can be continuous, together with its first ordered
derivative, in the interval — 1 g x ^ 1, or 0 ^ 0 ^ tt, only if
n is an integer. The solutions are then the Legendre poly-
nomials, which have continuous derivatives of all orders. Hence
and
n = 0, 1, 2, •
9 = P n (x) = P n { cos 0).
Thus two sets of particular solutions RQ of Laplace's equation
(1) have been determined:
(4) r n P n ( cos0);
r- n “lP„(cos 0)
in = 0 , 1 , 2 , ■ ■ • ).
Sec. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 195
In the first set the functions and their derivatives of all orders
with respect to r or 0 are continuous in every finite region; and
in the second set they are continuous in every region, finite or
infinite, not containing the origin.
Then at points inside the sphere the function
00
6) = cos 0) (r < c)
(T
satisfies (1) and (2) formally, provided J3„ can be determined so
that
oo
/(cos 0) = Yj BnC n Pn( COS 0) (0 < 0 < t),
0
where /(cos 0) = F(0). This is the expansion of the last section,
provided B n c 11 are the coefficients A n given there; that is, if
B n = ^ /(cos 0)P«( cos 0) sin 0 dd.
Hence for points inside the sphere, the solution of the problem
can be written
(5) V(r, 0) = ^ V f 1 dx
0 J -1
[r g c; F(0) = /(cos 0)].
For points exterior to th(% sphere, the functions of the second
set in (4) satisfy condition (3), and the solution can be written
OO
(6) V(r, 0) = ^ ^ / J «(cos 0) (r 1 c),
0
where
(7) An = — J" ^ f(x)P n (z 0 da;,
since the series in (6) then reduc.es to /(cos 0) when r = c.
The Solution Established. To show that our formal solution
does satisfy all the conditions of the problem, we use the same
method hero as in earlier problems (for example, Sec. 46). We
shall suppose that the given function F(0) and its derivative
F'(0) are sectionally continuous in the interval (0, r). Then
196 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 80
f{x) is sectionally continuous in the interval (“1, 1), and so is
fix), in every interval interior to ( — 1, 1).
Now consider the function V(r, 0) represented by formula (5).
When r = c, the series there converges to f(x) if —1 < x < 1.
But the sequence of functions (r/c) n (n = 0, 1, 2, * - • ) is
bounded, and monotone with respect to n; hence according to
Abel's test the series is uniformly convergent with respect to
r (0 ^ r ^ c) for each fixed x ( — 1 < x < 1). Therefore
V(c — 0, 0) = F(c, 0), and so condition (2) is satisfied.
The terms of the series in equation (5) can be written as
the product of the three factors AJn ) P n ( cos 0), and n(r/c) n .
Since the first two factors are bounded for all r, 0, and n
(» = 1 , 2 , • • ■ ),
and since the series whose general term is the last factor con-
verges when r < c, the series in equation (5) is uniformly con-
vergent when r < c. But the series of terms w*(r/c) n , for each
fixed positive k , also converges when r < c; and since n~ 2 P^(x)
and vr 4 P"(x) are uniformly bounded (Theorem 3), it follows
easily that the series in equation (5) can be differentiated term-
wise twice with respect to r and with respect to 0, when r < c.
The individual terms of that series satisfy Laplace's equation (1);
hence our function V(r, 0) satisfies that equation. Also, V(r, 0)
and its derivatives are continuous when r < c.
This establishes our solution when r < c. When r > c,
solution (6) can be proved valid in the same manner. If, as a
periodic function of the angle 0, F(9) is supposed continuous and
F'(6) sectionally continuous, it is also possible to show that the
above solutions are the only possible solutions satisfying certain
regularity conditions, essentially that V(r, 0) be continuous at
r = c (see Sec. 58).
PROBLEMS
1. If the potential is a constant Vo on the spherical surface of radius c,
show that V — Vo at all interior points, and V = 7oc/r at each exterior
point.
2. Find the steady temperatures at points within a solid sphere of
unit radius if one hemisphere of its surface is kept at temperature zero
and the other at temperature unity; that is, /(cos 0) = 0 when
?r/2 < 0 < 7t, and /(cos 0) = 1 when 0 < 0 < tt/2.
Ans. u(r, 0) = \ + f r cos 0 — | \r z P z (cos 0)
+ Hi i r 6 P 6 (cos 0) - • • • .
Sec. 80] LEGENDRE POLYNOMIALS AND APPLICATIONS 197
3. Find the steady temperatures u(r, 0) in a solid sphere of unit radius
if u = C cos 0 on the surface. Arts, u = Cr cos 0.
4. Find the potential V in the infinite region r > c, 0 ^ 0 ^ 7 t/ 2,
if F = 0 on the plane portion of the boundary (0 — tt/2, r > c) and at
r = oo, and F = /(cos 0) on the hemispherical portion of the boundary
(r = c, 0 ^ 0 ^ tt/2).
00
4ns. 7(r, 0) - 2J (4n + 3) (c/r)^P 2n+l ( C0S 5) dx.
5. Find the steady temperatures u(r, 0) in a solid hemisphere of
radius c whose convex surface is kept at temperature u = /(cos 0), if
the base is insulated; that is,
1 du n A tt
rdQ =0 when 6 = 2
Also write the result when /(cos 0) = 1.
00
Ans. u(r, 0) = X ( 4n + l)(r/c) 27 *P 2 n(cos 0) f }(x)P 2n (x) dx.
o *' u
6. Find the steady temperatures in a solid hemisphere of unit radius
if its convex surface is kept at temperature unity and its base at tem-
perature zero.
7. Show that the steady temperature u(r, 0) in a hollow sphere with
its inner surface r = a kept at temperature n = /(cos 0), and its outer
surface r — b at u = 0, is
where
, u ~
H — |*2n+l
fr2n \-l _ a 2n |-1
P»(cos 0),
An
2 n +
2
f(x)P n (x) dx.
8. If u(x, l) represents the temperature in a nonhomogeneous bar
witli ends at x = —1 and x = 1, in which the thermal conductivity is
proportional to 1 — x\ and if the lateral surface of the bar is insulated,
the heat equation has the form
du
c U = b Ox
<'->£
where b is a constant, provided the thermal coefficient c5 is constant
(Sec. 9). The ends x — ±1 arc? also insulated because the conductivity
vanishes there. If u = f(x) when t = 0, derive the following formula
198 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 81
for u(x, t) :
00 r
n(n+l)t I = —
9* When the initial temperature function in Prob. 8 is (a) f(x) — x 2 ,
(b) /(®) = £ 3 , show that the solution reduces to the following formulas,
respectively:
(a) u = | + K&c 2 — l)e _6b *,
( b ) u = + iPz(x)e~ 1%t .
81. The Gravitational Potential Due to a Circular Plate.
Another type of application of Legendre polynomials to the
solution of boundary value problems will be illustrated by the
following problem:
Find the gravitational potential due to a thin homogeneous
circular plate, or disk, of mass 5 per unit area and radius c.
Let the center of the disk be taken as the origin and the axis
as the 2 -axis, 0 = 0, where r, <p } 6 are spherical coordinates.
The potential is a function F(r, 6) independent of <p; hence it
satisfies the following form of Laplace’s equation:
(1) r ^ {rV)+ ^ele( sined I) = 0 ’
except at points in the disk. Its value at points on the positive
axis 0=0 can be found from the definition of potential by a
simple integration; thus
C c 27 rxd .
V(r, 0) = k J o ^ j= = = dx = 2 wkS(V? + c 2 - r),
where k is the gravitational constant in the definition of poten-
tial. Then F(r, 0) must be symmetric with respect to the origin
and satisfy the following boundary condition in the space
0 ^ Q < tt/ 2, r > 0:
(2) V(r, 0)=^ ( - r),
where M is the mass of the disk.
Two solutions of equation (1) were found in the last section,
namely,
Sue. 81] LEGENDRE POLYNOMIALS AND APPLICATIONS 199
( 3 )
( 4 )
V = V a„r n P n ( cos 6),
T
b n
1
P„(cos 0).
The coefficients a n , 6 n are now to be determined, if possible, so
that boundary condition (2) is satisfied. But when 0 = 0,
Pn( cos 0) = 1 and the series in (3) and (4) become power series
in r and in reciprocals of r, respectively.
Now the binomial expansions
Vr 2 + c 2 = c (l +
Vr 2 + c' 1 = r ^1 +
1 r 4 1-3 r«
2 . 4 a 4 1 2 • 4 • 6 c 6
(0 g r < c)
1 c 4 1-3 c«
2 • 4 r 4 r 2 • 4 • 6
-)
(r > c).
are absolutely convergent in the indicated intervals, and con-
vergent when r = c. Hence boundary condition (2) can be written
( 5 )
7(r, 0) =
2 Mk / r lr 2 1 r 4
c \ c 2 c 2 2 • 4 c 4
1 * 3 \
+ 2 " • 4 • 6 c” ~ ■ ■ ■ ) when 0 < r g c;
2 Mk flc 1 c* 1-3 c 5 \
c \2? 2-4r* + 2-4-6r» )
when r jjg c.
The series in (3) will then satisfy (5) for r < c if its coefficients
are identified with those of the first series in (5) ; thus a 0 = 2Mk/c f
ai = —2 Mk/c 2 , etc. Similarly, for the case r > c the series in
(4) can be used if its coefficients are taken as those in the second
series in (5), namely, b 0 — Mk , b i = 0, etc.
Hence the solution of the boundary value problem (l)-(2)
can be written as follows, when 0 ^ 0 < 7 t/ 2:
V(r, 0) =
2 Mk
- 1
1
r - Pi ( cos 6) + \ ^ P»( cos e )
2 • 4 c
7* P 4 ( COR (9) +
2 C 2
1-3 r«
2 • 4 • 6 c'
Pe( cos 0)
200 FOURIER SERIES AND BOUNDARY PROBLEMS [Sec. 81
if 0 < r < c; and
V(r, 8) =
2 Mk
c
1 c
2 r
2T4 ^ p 2( cos 9 )
+
1-3 c
2-4-6 r
-g P 4 (cos 6)
if r > c.
When r c, the convergence of the series here follows from
the absolute convergence of the series in (5) and the fact that
fPn(eos.fl)l ^ 1.
PROBLEMS
1. Derive the following formula for the gravitational potential due to
a mass M distributed uniformly over the circumference of a circle of
radius 1, when 0 ^ 8 S tt:
V(r, 8) = hM [l — ^ r 2 P 2 (cos 8) + r 4 P 4 ( cos 8) — • • • J,
if 0 ^ r < 1; and
V(r, 8) = kM [i
lP 2 (cos 8) 1 ■ 3P 4 (cos 8)
2 r 3 ' 2 • 4 7 '
if r > 1.
2. Find the gravitational potential, at external points, due to a solid
sphere, taking the unit of mass as the mass of the sphere, and the unit of
length as the radius, if the density of the sphere is numerically equal to
the distance from the diametral plane 8 = tt/2.
An,.
+
1 p4(cos 8 )
6-8 r b
1 • 3
P fi (cos 8)
6 - 8-10
3. Find the gravitational potential, at external points, due to a
hollow sphere of mass M and radii a and b, if the density is proportional
to the distance from the diametral plane 6 = t/2.
4. The points along the 2-axis, 8 = 0 or 8 = t, in an infinite solid
are kept at temperature u = Ce~ r \ Find the steady temperature
u(r, 8) at all points. Ans. u(r, 8) = <7^ (-l)»(r 2 »/w!)7 > 2 „(cos 8).
0
6. The surface 8 = t/3, r ^ 0, of an infinitely long solid cone is kept
at temperature u = Ce~'. Find the steady temperatures u(r, 8) in the
«
Ans. u(r, 6) = C g (- l)»[r»P„(cos 8)]/[n\P n (i)].
0
cone.
Sec. 81] LEGENDRE POLYNOMIALS AND APPLICATIONS 201
6. Solve Prob. 5 if the surface temperature is u(r f tt/3 ) = C/r m ,
where m is a fixed positive integer.
REFERENCES
1. Hobson, E. W.: “The Theory of Spherical and Ellipsoidal Harmonics,”
1931.
2. Whittaker, E. T., and G. N. Watson: “Modern Analysis,” 1927.
3. Byerly, W. E.: “Fourier’s Series and Spherical Harmonics,” 1893.
Appendix.
INDEX
A
Abel's test, 127
Absolute convergence, 83
Approximation in mean, 40
B
Beam, displacements in, 24
simply supported, 125
Bessel functions, of first kind, 48,
143-174
applications of, 105-174
boundedness of, 152
differentiation of, 148
expansions in, 162
integral forms of, 149
of integral orders, 145
norms of, 161
orthogonality of, 157
recursion formulas for, 148
tables of, 156n.
zeros of, 1 53
of second kind, 147
Bessel’s equation, 143, 157, 166, 170
Bessel's inequality, 41, 84
Bessel's integral form, 151
Beta function, 149
Boundary value problem, 6, 94-142,
165-174, 193-201
complete statement of, 107, 133,
138
linear, 24
units in, 100
(See also Solutions of boundary
problems)
C
Characteristic functions, 47
Characteristic numbers, 47
Characteristic numbers, of Leg-
endre's equation, 186
Closed systems, 42
of trigonometric functions, 74, 85,
88
Complete systems, 40, 42
of Legendre polynomials, 185
of trigonometric functions, 74
Conduction, equation of, 19
Conductivity, 15, 168
Convergence in mean, 42
Cylindrical coordinates, 13
Cylindrical functions ( see Bessel
functions)
D
Derivative from right, 65
Differential equation, linear, 2
linear homogeneous, 2
(See also Partial differential
equation)
Differentiation, of Fourier series, 78
of series, 5, 106, 140, 196
Diffusion, equation of, 19
Diffusivity, 19
Dini's expansion, 164
Dirichlct’s integrals, 67
E
Electrical potential, 13
(See also Potential)
Even function, 57
Expansion, Dini’s, 164
Fourier-Bessel, 162
in Fourier integrals, 88-93
in Fourier series, 53-88
in Legendre's series, 187, 190, 191
in series of characteristic func-
tions, 48, 49
203
204 FOURIER SERIES AND BOUNDARY PROBLEMS
Exponential form, of Fourier inte-
gral, 92
of Fourier series, 63
Exponential functions, 31, 45, 46
F
Flux of heat, 15
Formal solutions, 94
Fourier-Bessel expansions, 162
Fourier coefficients, 53, 70
properties of, 76, 77, 80, 85, 86
Fourier constants, 40-42
Fourier cosine integral, 91
Fourier cosine series, 57, 62, 74
Fourier integral, 88
applications of, 120-123
convergence of, 89
forms of, 91
Fourier integral theorem, 89
Fourier series, 53-88
convergence of, 70, 86
absolute, 83
uniform, 82, 86
differentiation of, 78
forms of, 61
generalized, 39
integration of, 80, 86
in two variables, 116
Fourier sine integral, 91
application of, 122
Fourier sine series, 28, 29, 57, 62, 73
in two variables, 118
Fourier theorem, 70, 86
Function space, 38, 74
Fundamental interval, 38
G
Gamma function, 145, 149
Generating function, for Bessel
functions, 147
for Legendre polynomials, 179
Gibbs phenomenon, 86
Gravitational force, 10-12
Gravitational potential, 10
(See also Potential)
Green's theorem, 18, 131, 134
H
Heat equation, 17, 19
Heat transfer, surface, 110
I
Infinite bar, temperatures in, 120
Infinite series of solutions, 5
Inner product, 34, 38
Integration, of Bessel functions,
149
of Fourier series, 80, 86
of Legendre polynomials, 181,
188
L
Lagrange's identity, 32, 71
Laplace’s equation, 12, 20
Laplacian operator, 12, 14, 15
Least squares, 41
Lebesgue integral, 43, 86
Left-hand derivative, 65-67
Legendre coefficients, 181
boundedness of, 182
Legendre functions, 175
of second kind, 178
Legendre polynomials, 175-193
applications of, 193-201
bounds of, 190
complete set of, 185
derivation of, 175
derivatives of, 189
expansions in, 187, 190, 191
generating functions for, 179
integration of, 181, 188
norms of, 183
orthogonality of, 183
recursion formula for, 180
Legendre’s equation, 175, 183, 194
characteristic numbers of, 186
general solution of, 178
Legendre’s series, 191, 192
Limit in mean, 42
Linear differential equation, 2
Lommel’s integral form, 150
INDEX
205
M
Membrane ( see Vibrating, mem-
brane)
Monotone sequence, 128, 139
N
Newton’s law, 110, 168
Norm, 34, 38
O
Odd function, 57
One-sided derivative, 65-67
Orthogonal functions, 29, 38
generated by differential equa-
tions, 46
Orthogonal sets, 34-52
Orthogonality, 34, 37, 44, 49
of Bessel functions, 157
of characteristic functions, 49
of exponential functions, 45, 46, 62
Hermitian, 45
of Legendre polynomials, 183
of trigonometric functions, 29,
39, 40, 53
with weight function, 44
Orthonormal sets, 35, 38
of Bessel functions, 161
of Legendre functions, 184
of trigonometric functions, 54, 74
P
Parseval relation, 85, 86
Parseval’s theorem, 43, 86
Partial differential equation, 2
for beam, 24
of conduction, 19
general solution of, 3
Laplace’s, 12
linear homogeneous, 2
for membrane, 23
nonliomogcneous, 100
for string, 21
types of, 24
Periodic boundary conditions, 48
Periodic extension, 96
Periodicity of function, 55
Piecewise continuous function, 65
Plucked string, 28, 98
Poisson’s equation, 13
Polar coordinates, 14
Potential, electric, in cylindrical
region, 126, 141, 174
equation of, 13
between parallel planes, 115,
116, 123
in quadrant, 124
about spherical surface, 193
in square, 137
gravitational, definition of, 10
due to disk, 198
due to hollow sphere, 200
due to ring, 200
due to sphere, 200
(See also Temperature,
steady)
R
Radiation, 110, 168, 173, 174
Right-hand derivative, 65-67
Rodrigues’ formula, 181, 184
S
Schwarz inequality, 83
Sectionally continuous functions, 64
Semi-infinite bar, temperatures in,
122
Shaft, twist in, 124
Solutions of boundary problems, 94
approximate, 97
closed form of, 96, 116, 126
established, 96, 105, 133, 141, 167,
195
superposition of, 99
uniqueness of, 105, 127—142
Sources, heat, 20, 111
Spherical coordinates, 13
String (see Vibrating string)
Sturm-Liouvillc equation, 47
Sturm-Liouvillc problem, 47-52
in Bessel’s equation, 160
in Legendre’s equation, 183
Superposition of solutions, 3, 99
206 FOURIER SERIES AND BOUNDARY PROBLEMS
T
Tchebichef polynomials, 44
Telegraph equation, 23
Temperature, steady, in cone, 200,
201
in cylinder, 168, 169
in cylindrical wedge, 126
in hemisphere, 197
in hollow sphere, 197
in infinite solid, 200
in sphere, 196, 197
variable, in bar, 104, 120-123,
197, 198
in circular plate, 173, 174
in cube, 119
in cylinder, 165, 168, 172
in cylindrical wedge, 173
in hollow sphere, 113
in infinite solid, 120-123, 125
in radiating wire, 110-112
in slab, 102-112
in sphere, 112
in square plate, 119
(See also Potential)
Termwise differentiable series, 5
U
Uniform convergence, of Fourier
series, 82, 86
of series, 105, 127
Uniqueness of solutions, 127-142
for potential, 134, 137
for temperature, 105, 130
Units, selection of, 100
V
Vectors, 34^-37
Vibrating membrane, 23
circular, 170, 172
rectangular, 116
Vibrating string, 21
with air resistance, 125
approximating problem of, 98
forced vibrations of, 100-102
problem of, 24, 28, 95-102
W
Weierstrass test, 105, 133
Weight function, 44
