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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