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I.\342\200\231

Georgi

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FOLIRIER

SERIES))

FOURIER

SERIES)

GEORGI P. TOLSTOV) of Mathematics

Professor

Moscow State Translated

University)

from the A.

Richard

Russian by

Silverman)

DOVER PUBLICATIONS, NEW

YORK))

INC.

Copyright All rights

International

A. Silverman.

by Richard

1962 \302\251

reserved

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Published in Canada by General ComPublishing Don Mills, Toronto, pany, Ltd., 30 Lesmill Road,

Ontario.)

This Dover edition,

lirst published in 1976, is an with slight corrections, of republication, Inc., originally published by Prentice-Hall, Englewood Cliffs, New Jersey in 1962.)

unabridged the work

International Library

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10014))

PREFACE)

AUTHOR\342\200\231S

on Fourier series,which originally in appeared has already been translated into Chinese, I am very grateful to German, Polish and Rumanian. Dr. R. A. Silverman and to the Prentice-Hall Publishing for undertaking an to prepare and publish Company version of the It would second Russian edition. English be most to serve the gratifying to me if the book were needs of American readers. I would like to thank and V. Y. Kozlov, L. A. Tumarkin I) A. I. Plesner for the helpful advice they gave me while My book

Russian,

was

this book.

writing

G. P. '1\342\200\230.)

PREFACE)

TRANSLATOR\342\200\231S

The

present volume

translations

of

monographs

in

engineering,

under

Professor tion

to the

is

the

second

the

\357\254\201elds of

will constitute

English-language

literature

series of

textbooks

mathematics,

my editorship.

Tolstov\342\200\231s book

a new

in

Russian

outstanding

physics

and and

It is hoped that a valuable addi-) on

Fourier

series. V))

'rnANsLAron\342\200\231s pnancn)

vl

The

1. To

changes, made

two

following

are worth

Tolstov\342\200\231s consent,

value

the

enhance

with

Professor

mentioning:) of the English-language

been a large number of extra problemshave L. Shields of the by myself and Professor Allen of Michigan. We haye consulted a variety University of sources, in particular, A Collection of Problems in

edition,

added

Mathematical Physics by kaya,

Y. S.

and

most of the

been taken.)

2. To

keep the four

Russian

original

chapters

(8 and

N. Lebedev,

number

chapters have

Skals-

end of Chapter

9) of the presentedition.)

also added a Bibliography, for collateral and supplementary

should be noted that contain material ofa more without

I. P.

from which

of cross references to a 11) of the (8 and 9, 10 and to make two been combined

I have

omitted

1955),

at the

appearing

problems

9 have

minimum,

N.

(Moscow, U\357\254\202yand

loss

sections

containing

suggestions

it Finally, with asterisks which can be nature,

reading.

marked

advanced

of continuity.

R. A.

s.))

CONTENTS)

FOURIER SERIES Page 1. 1: 2: Harmonics, 3. 3: Trigonometric and Series, 6. 4: A More Precise Terminology. Polynomials 5: The Basic Trigo8. Integrability. Series of Functions, nometric The Orthogonality of Sines and Cosines, System. 10. 6: Fourier Series for Functions of Period 21:,12. 7: on an Interval of Series for Functions De\357\254\201ned Fourier Limits. 8: Right-hand and Left-hand Jump 21:, 15. Length and Piecewise Smooth 17. 9: Smooth Discontinuities, for the Convergence of Functions, 18. 10: A Criterion 21. 12: 11: Even and Odd Functions, 19. Fourier Series, of Expansions in Cosine and Sine Series,22. 13:Examples Form of a Fourier Fourier Series, 24. 14:The Complex of Period 21, 35. Problems, 38.) Series,32. 15:Functions TRIGONOMETRIC

1)

Periodic

1.

Functions,

ORTHOGONAL SYSTEMSPage

1:

41.

De\357\254\201nitions,41.

Orthogonal System, 42. 3: Some Simple Orthogonal Systems, 44. 4: Square Inte50. 5: The The Schwarz Inequality, grable Functions. 51. 6: Bessel\342\200\231s Error and its Minimum, Mean Square in the 7: 53. Convergence Systems. Complete Inequality, of Complete Systems, 57. Properties Mean, 54. 8: Important 9: A Criterion for the Completeness of a System, 58. \"10: 63.) 60. The Vector Analogy, Problems,

2: Fourier

Series with

OF

CONVERGENCE

SERIES Page66. l : A 66.

2: The cos

L\342\200\235f(x)

Limit

nx dx

to an

Respect

as

and

TRIGONOMETRIC Consequence

n->oo

of

FOURIER

Bessel\342\200\231s Inequality,

of the Trigonometric sin

L\"f(x) of Cosines. Auxiliary

nx

dx, 57.

Integrals)

3: Formula

for)

71. 4: The Integrals, 72. Sum of a Fourier Series, 5: Right-Hand and Left-Hand Derivatives, 73. 6: A Suf\357\254\201cient at a Con-) Series for Convergence of a Fourier Condition for Convergence Condition 7: A Su\357\254\201icient 75. Point, tinuity

the

Integral

Sum

Formula

for the Partial

vii))

viii)

CONTENTS)

of

a-

6 and 7, 78. 9: 79.

10: Absolute

and

of

Integrable Derivative,

tions,

2!, 94.

91.

IS: A

Discontinuous),

Smooth Function of the Fourier

Period

21:

Localization

Expansions

Remark Concerning

Seriesof a

an

Absolutely of the Results

with

82. 12:Generalization

Series

Fourier of Period

of the

Convergence

13: The

90. 14: Principle, of Unbounded FuncFunctions

of

Period

Problems,94.) WITH

SERIES

TRIGONOMETRIC 4)

Uniform

Convergence

Function

or

a

Series of

Fourier

(Continuous

Piecewise

21:,80. 11:Uniform of Sec. 11, 85. Examples of Fourier

77. 8: in Secs.

Discontinuity, Proved

the

of

Convergence

Seriesofa Continuous, Continuous

of

Point

Conditions Su\357\254\201icient

the

Function

Smooth

Piecewise

at a

Series

Fourier

Generalization of

COEFFICIENTS Page 97.

1:

DECREASING

Abel\342\200\231s Lemma,

97.

2:

Formula for the Sum of Sines. Auxiliary 98. Inequalities, Series with Monotonically 3: Convergence of Trigonometric Decreasing Coe\357\254\202icients,100. \"4: Some Consequences of the of Ftmctions of Sec. 3, 103. 5: Applications of a Theorems to the Evaluation of Certain TrigonoComplex Variable Form of the Results of metric Series, 105. 6: A Stronger Sec. 5, 108. Problems,112.)

OPERATIONSON

SERIES Page 115. 1: by Trigonometric Polynomials, 2: Completeness of the Trigonometric 115. System, 117. 3: Theorem. The Most Important Consequencesof Parscval\342\200\231s of the Trigonometric System, 119. \"4: the Completeness of Functions by Polynomials, 120. 5: Approximation and Subtraction of Fourier Series, Multiplication Addition of 122. \"6: Products of a Fourier Series by a Number, of Fourier Fourier Series, 125. Series, 123. 7: Integration 8: Diiferentiation of Fourier Series. The Case of a Conof tinuous Function of Period 21:, 129. *9:Differentiation De\357\254\201ned on the Fourier Series. The Case of a Function of Fourier Differentiation Interval [-1:, 1:],132. \"10: Series. De\357\254\201ned on the Interval [0, 1:], 137. The Case of a Function ll: Improving the Convergence of Fourier Series, 144. 12: A List of Trigonometric Expansions, 147. 13:Approximate 152.) Calculation of Fourier Coe\357\254\202icients, 150. Problems,

FOURIER TRIGONOMETRIC of the Problem, 155. 2: Arithmetic Means, 156. 3: The Integral)) OF

SUMMATION

SERIES Page155.

The Method

FOURIER

of Functions

Approximation

of

1: Statement

Ix)

CONTENTS

for

Formula

Mean

Arithmetic

the

Series, 157.

Fourier Method

4:

of\342\200\230 Arithmetic

of the

Means,

Sums

Partial

Series

of Fourier

Summation

158.

5:

of\342\200\230 a by

the

of

Abel\342\200\231s Method

163. 7: Application Summation, 162. 6: Poisson\342\200\231s Kernel, Method to the Summation of Fourier of Abel\342\200\231s 164. Series, Problems, 170.)

DOUBLE FOURIER INTEGRAL

Page

SERIES.

173. l:

Variables, 173. 2: The Basic Variables. Double Trigonometric

The

Formula

Integral

4: Double

178.

in Two System in Two Series, 175. 3:

Systems

Trigonometric Fourier

the Partial Sums of a Double Series. A Convergence Criterion,

for

Fourier

Trigonometric

FOURIER

THE

Orthogonal

Fourier Seriesfor a Function

Different

with

in x and y, 180. 5: The Fourier as Integral a Limiting Case of the Fourier Series, 180. 6: Improper on a Parameter, 182. 7: Two Lemmas, Integrals Depending 185. 8: Proof of the Fourier Theorem, 188. 9: Integral Different Forms of the Fourier Integral \"10: 189. Theorem, The Fourier Transform, 190. \342\200\230'11: The Function, Spectral Periods

193. Problems, 195.) AND BESSELFUNCTIONS Page 197. 1: Bessel\342\200\231s Equation,

The First

Kind of Nonnegative 201.

Function,

4: Bessel

FOURIER-BESSEL SERIES 2: Bessel Functions of 197. Order,

198.

Functions of

Negative Order, 202. 5: The General 203. 6: Bessel Functions of Equation,

7: Relations

the

3: The

Gamma of

Kind

First

of Bessel\342\200\231s Solution the Second Kind, 204.

between Bessel Functions of Different Orders, 8: Bessel Functions of the First Kind of Half-Integral 207 . 9: Asymptotic Formulas for the Bessel Functions Order, 208. 10: Zeros of the Bessel Functions and Related Func215. 213. ll: Parametric Form of Bessel\342\200\231s tions, Equation, 12: Orthogonality of the Functions J ,,().x), 216. 13: Evalua-)

205.

tion

of the

Integral Ll xJ,,2().x)dx, 218.

dx, 219. Integral J: xJ,,2(>.x) 220. for 16: Criteria Series,

\"14:Bounds

15:De\357\254\201nitionof the

Convergence

which Guarantees

Bessel Series,225.

Uniform

Convergence

the)

Fourier-Bessel)

and Bessel\342\200\231s BesselSeries,221. \"17: Inequality quences, 223. \"18:The Order of Magnitude of

cients

for

of Fourierits Consethe

of a

Coeffi-

Fourier-

of the \"19: The Order of Magnitude Fourier-Bessel Coefficients of a Twice Differentiable Function, of the Fourier-Bessel 228. \"20: The Order of Magnitude Several)) Which is Differentiable of a Function Coefficients

XCONTBNT8)

Times, 231.

\"21: Term by

BesselSeries,234. Type,

\"23: Extension

237.

Series of

Fourier-Bessel Bessel

22:

Problems,

of Fourier-

of

on De\357\254\201ned

the

Second

Secs. 17-21to

the SecondType, 239.

of Functions

Expansions

[0, I], 241.

Diiferentiation Term Series Fourier-Bessel of the Results of

24:Fourier-

the

Interval

243.)

METHOD AND ITS APPLICAEIGENFUNCTION 245. TIONS T0 MATHEMATICAL PHYSICS Page 245. 2: The 1: The Gist of the Part 1: THEORY. Method, 250. 3: of the Boundary Value Statement Usual Problem, and 250. 4: Eigenfunctions of Eigenvalues, The Existence 251. 5: Sign of the Eigenvalues, 254. Their orthogonality, 255. with Respect to the Eigenfunctions, Series 6: Fourier Method Always Lead to a Solution 7: Does the Eigenfunction of the Problem ?, 258. 8: The Generalized Solution, 261. 9: 264. 10: Supplementary The Inhomogeneous Problem, of Remarks, 266. Part II: APPLICATIONS, 268. 11: Equation of a String, Vibrations a Vibrating String, 268. 12: Free of a String, 273. 14:Equation 269. 13:Forced Vibrations 15: Free 275. Vibrations of a Rod, of the Longitudinal 16: Forced Vibrations of a Rod, Vibrations of a Rod, 277. 282. 18: 17: Vibrations of a Rectangular 280. Membrane, a Circular of Radial Vibrations Membrane, 288. 19: 291. of a Circular Membrane (GeneralCase), Vibrations in a Flow of Heat Flow in a Rod, 296. 21:Heat Equation Rod with Ends Held at Zero Temperature, 297. 22: Heat Ends Held at Constant Temperature, Flow in a Rod with 299. 23: Heat Flow in a Rod Whose Ends are at Speci\357\254\201ed 301. 24. Heat Flow in a Rod Whose Variable Temperatures, the Surrounding Heat Freely with Ends Medium, Exchange 26: Heat Flow 306. 301. 25: Heat Flow in an In\357\254\201nite Rod, 27: 310. in a Circular Cylinder WhoseSurfaceis Insulated, Whose Surface Exchanges Heat Flow in a Circular Cylinder Heat with the Surrounding Medium, 312. 28: Steady-State 316.) 313. Flow in a Circular Cylinder, Heat Problems, THE

20:

ANSWERS

TO PROBLEMS

BIBLIOGRAPHY

INDEX

Page

Page

333.))

331.)

Page319.

FOURIER

SERIES))

1)

TRIGONOMETRIC

SERIES)

FOLIRIER

I. Periodic

Functions)

A function

is calledperiodicif

f (x)

f(x

there

+ 7?

exists

a constant

T > 0 for

which)

(1-1))

=f(x).

of de\357\254\201nitionof

x both f (x). (It is understoodthat constant T is called a period of the function functions are sin x, cos x, tan x, The most familiar f (x). periodic etc. Periodic functions arise in many of mathematics to applications and It is clear that the sum, difference, problemsof physics engineering. of two functions of period T is again a function of product, or quotient T. period If we plot a periodic function y = f (x) on any closed interval a < x < a + T,-we can obtain the entire graph of f (x) by periodic repetition of the portion of the graph correspondingto a < x < a + T (seeFig.1). If T is a period of the function f(x), then the numbers 2T, 3T,4T, . . . are also periods. This follows immediately the graph of a by inspecting or from the series of equalitiesl) periodicfunction for

any

and x

x in

the domain

+ T lie in

f(x) 1 We the

suggest

following

that

this

domain.)

==f(x

+ T)

the reader

equalities: f(x)

=f(x

Such a

=f(x + 2T) =f(x

prove the

- T)=f(x

validity

not

+ 37) =)

of these equalities but only \342\200\231

- 2T)=f(x- 3T)=--I)

also of

2

TRIGONOMETRIC FOURIER

period,sois kT,

CI-IAP. I)

Thus, if T (1.1). by repeated use of the condition k is any positive integer, i.e., if a periodexists,

obtained

are

which

SERIES

where

is a it is

not unique.) I)

I-

./Y\\/\342\200\235\\) \\/

Fromuz

Next,

\\/V\342\200\235)

\\l\342\200\230\342\200\231\342\200\231'\342\200\231\\l

V\342\200\235

the

we note

(x) is integrable other interval of the

If f

on any same

function f(x)

of any

property

following

1)

of period T:)

on any interval of length T, then it is integrable and the value the is the same, integral length, of

i.e.,) =

ax.

j\"\"\342\200\231f 1, compression along the x-axis by a factor uniform expansion along the x-axisby a factor l/co if co < 1. Figure shows the harmonic y = sin 3x, of period T = consider the harmonic Now, y = sin (cox + cp), and set cox + cp = coz, so

reduces

to a

and

21:/3.

4(b)

that x graph along

= z \342\200\224 cp/co.

We already know the graph of sin coz. of y = sin (cox + cp) is obtained by shifting the graph the x-axis by the amount \342\200\224cp/co. Figure 4(c) represents = sin

y

with

(3x

21:/3 and initial phase -n:/3. the graph of the harmonic

period Finally,

y =

y =

(3x

the

harmonic)

obtained

all ordinates

multiplying

from

the

by

+)

summarized as follows:)

may be

results

These

2sin

(cox + cp) is

sin

A

harmonic y = sin (cox + cp) by A. Figure 4(d) showsthe harmonic)

number

of y = sin cox

+)

of the

that

the

Therefore,

graph of the harmonic y = A sin (cox + cp) is obtained from the graph the of familiar sine curve compression (or expansion) along by uniform the axes plus a shift the x-axis.) coordinate along The

Using a

A sin

Then,

from

formula

well-known

(cox +

=

cp)

cox sin

A(cos

ourselves

we convince

= A

that

sin

every

to

sufficient

cos cp).

\342\200\224\342\200\224\342\200\224\342\200\224 a . A.=\342\200\231\\/a2+b2,

which

From

for the

cp now

harmonic

is easily on,

in

the

form (2.2))

The result

To

prove

this,

is)

b

a

COS9=Z=

b

9)

found.

we shall shown

represented

(2.2)is a harmonic.

9

SlnQ=Z=

(2.1)

cp,

+ b sin cox.

of the form for A and B. (2.1)

solve

A cos

can be

harmonic

function

every

Conversely,

b =

cp,

a cos cox

from

sin cox

+

cp

write

setting

a

it is

we

trigonometry,

write harmonics in 4(d), this form

2sin

(3x

+

=

\\/\302\247cos

form

the

in Fig.

(2.2).

is)

3x

+ sin

3x))

For example,

6

'r1uooNom=:r1uc

and

to

also be convenient

will

It

= 21,then,

set T

If we

since

the harmonic

therefore,

CI-IAP. 1)

SERIES 1-\342\200\230ounnzn

213/(0,we

with

period

a cos

the

introduce

explicitly

T =

T

period

T =

21 can

be written

(2.3))

\"-13-C

3. Trigonometric the period

Given

Polynomialsand T=

2!,consider

with

frequencies

co,,

b,,

= -trk/I and

Series)

k

Sin,-\342\200\230T35

periods T,, = T= 21 =

number

the an

integral sum

every

T = 21is simultaneously a multiple of a period is again of the form

.s',,(x)=

A

+

harmonics)

the

.

k

a,, cos\342\200\230:-If+

kg] (ak

(2.2).

as)

b sin

+

in

have)

. . .)

1, 2, =

21:/cok

21/k.

(3.1) Since

kTk,)

of all the

period

cos

=

(k

11:5

+

harmonics (3.1),for 1). Therefore,

(see Sec.

a period

sin

bk

Elli\342\200\230).

A is a constant, is a function of period 21, since it is a sum of functions of period 21. (The addition of a constant obviously does not destroy in fact, a constant can be regarded as a function for which periodicity; any is a period.) The function number s,,(x) is called a trigonometric polynomial of order n (and period 21). Even it is a sum of various a trigonometric harmonics, though polynomial in general represents a function of a much more complicated nature than a simple harmonic. By suitably the constants A, a1, bl, choosing = s,,(x) with graphs quite form functions unlike the a2, b2, . . . we can y smooth and symmetric graph of a simple For example, harmonic. Fig. 5 where

shows the

polynomial)

trigonometric

y = sinx The

trigonometric in\357\254\201nite

+ }sin2x

series) \302\260\302\260

A +

+ isin 3x.

cos

k2-:l(a,,

rrkx

.

+

bk sin

rrkx

-1-)))

\342\200\224I\342\200\224

S5C-3

TRIGONOMETRIC

of period also represents a function 21. which are sums of such in\357\254\201nite trigonometric arises naturally: diverse. Thus, the following question

(if it

The

converges)

tions

Frourua

T =

FOURIER

nature

of func-

is even

series

Can

7)

SERIES

any

more given)

5)

21 be represented

as the sum of a trigonometric a such series? We shall see later that representationis in fact possible for a functions. class of wide very This means For the time being, suppose that f (x) belongsto this class. functions a sum of as of as a sum can be that i.e., harmonics, expanded f(x) = f (x) is obtained The graph of the function structure. with a very simple y to give a of these harmonics. Thus, of the graphs as a \342\200\234superposition\342\200\235 a complicated oscillatorymotion we can represent mechanical interpretation, Howoscillations which are particularly of individual simple. f (x) as a sum are series that applicable only to ever, one must not imagine trigonometric In fact, the case. the from far is This oscillation being phenomena. in useful is also series of a many studying very trigonometric concept function

of period

phenomena of a quite

nature.

different

If) \302\260\302\260

=

f(x\342\200\231)A

then,

setting

=

t \342\200\230ttx/I

\302\242(x>ax.

we have =

l|

zero

approach

function

the Fourier

for

formulas

theorem as follows: The

and

function,

function. integrable the function be

converge

to zero

Sum of Cosines. Auxiliary

as n

this integrable

section, we

now It should be absolutely intewe have

\342\200\224> oo.)

Integrals)

that) \302\260

( \302\247+cosu+cos2u+---+cosnu=)

To

prove

this, we

denote the

2Ssin;-=

sin%+

+

sum

on

the left by

S.

+

.

%)

Obviously we have)

Zcosusing)

2cos2usin;+-~-+

2cosnusin;-))

(3.1))

72

OF TRIGONOMETRIC

CONVERGENCE

Cl-IAP. 3)

SERIES

the formula

Applying

=

2cosatsinB to

FOURIER

on the

product

every

=

-

+

sin;

2Ssin;

\342\200\224

\342\200\224 (3))

sin(ct

+2 (sin (I! -1-)u

\342\200\224

+

sing)2

2

(sin\303\251u

+

(3)

obtain

we

right,

+

sin(at

-

\342\200\224 sin

sin\303\251u)

= sin

2

(I!

+---

2

2

(sin\303\251u

(I!

+2

l)u

\302\260)

Therefore)

_

S _

as was to Next,

sin (n

2 sin

+ })u') (u/2)

be proved. two more

we prove

(3.1) overthe

formulas.

auxiliary

[\342\200\224-it, -it] and

interval

dividing the result

1 1' sin (n + 1~)u -n 2 sin (u/2)

1 =

the Integrating we obtain by 1:,

equality

du,

(32))

1-:

for

any

n whatsoever

(since the

see

that

the integrand

in

cosines vanish). the sign of

of the

integrals

is even

(3.2)

(since

changing

and denominator and leaves their

sign of both the numerator Therefore we have) changed). 1 E

4.

\342\200\230The

Integral

Let

f (x)

have period f(x)

21:and

~

+ 229

the

ratio

un-

(33))

Fourier Series)

Partial Sum of a

for the

changes

1. 2

=

o\"sin(n+\302\247)udu=lJ\302\260sin(n+\302\253})ud sin (u/2) \"'\342\200\224\342\200\2342 1'!) -.. 2 sin (u/2)

Formula

It is easyto 14

that)

suppose

[V18) (ak I! 1) R\342\200\230

cos kx

+

sin

bk

kx).)

Writing

s,,(x) and

substituting

s,,(x) =

-517:

=

+

(a,, cos kx

%\342\200\224\302\260 =1)

I:

the expressions for the

J:f(t)

dt +

Fourier

1 \342\200\230N) [ \302\2431

+

L::f(t) +

bk

kx),

coe\357\254\202icients,

cos kt

J:;f(t)

sin

dt-cos

sin kt

dt

we

obtain)

kx)

-

sin

kn]))

oonvenomce or TRIGONOMETRIC

sec. 4 R

1

=

or

using

[5

formula

+ sin kt sin

kx

cos

kx)]

dt

\342\200\224

dt,

x)]

(3.1) ''

_

functions

sin [(n

_l \342\200\230 1: -..f(\342\200\230) 2

change variablesby

the

(coskt

cos k(t kg \342\200\230

+

=

s,,(x)

Now,

2:)1)

k

1! -1:f(t)

\"\"~(\")

We now

+

\"

1

=

73

saunas

\"

1

_nf(t)

.;

FOURIER

,1:

f (x

+

t

setting

+

sin

\342\200\224 \302\247)(t x)]

\342\200\224 x =

u.

+ u)

f(x

d\"

\342\200\224 [\302\253}(t x)]

The result

is)

du.)

and)

u)

sin (rt + \302\247)u 2 sin (u/2))

are periodic

21: [see (3.1)],and the interval 21:. Therefore,the integral over this interval [-\342\200\230It, integral over the interval 11:] (see Ch. 1, Sec. 1) and

is

same

the

we

a, with period

variable

the

in

of length

\342\200\224 \342\200\224 [\342\200\2241tx, 1: x] is

as the

obtain)

s,,(x) = This

for the partial

integral formula

which the

under

conditions

establish

+

f(x

u) sum

du.

of

(4.1)

series allows us to f (x) can

a Fourier

convergence of the

series

to be

guaranteed.)

5.

Right-Hand

and

Suppose that

the

f (x +

0) = f (x).

point x if the

Derivatives)

Left-Hand function

is continuous from that f (x) has a right-hand

f(x)

we say

Then,

the

right

at

derivative

x, i.e., at the

limit)

lim)

f(x

\"

f(x\342\200\231 =f.\302\243(x)

(5.1))

\"2,\342\200\231

u->0 u>0)

exists

f(x),

and is

If \357\254\201nite.

and if the

f(x)

is continuous

from

the

left

at x,

i.e., f (x

\342\200\224 =

0)

limit) \342\200\230\342\200\234 f\342\200\230\342\200\231\342\200\230

(5.2))

=r:

\342\200\234Z,\342\200\230'f\342\200\230\342\200\231\342\200\230\342\200\231

:33

exists and is

then \357\254\201nite,

we say

that f (x) has a

left-hand

derivative

at

x.))

or mroouommuc

oouvenoeuca

74

can.

seams

rouruan

3)

case where f_{_(x) = f1(x), the function f (x) obviously has an at x, which is equal to the common value of the rightderivative ordinary hand and left-hand derivatives at x, i.e., the curve at y = f (x) has a tangent the point with abcissa x. In the case where and both exist but f ,;(x) f1(x) are unequal, the curve f (x) has a \342\200\234corner\342\200\235 but we can still speak of righthand and left-hand tangents (as indicatedby the arrows in Fig. 31).) the

In

y=f(x))

V)

Froune

has a jump

let x be a point where f(x) of (5.1), the limit)

Now

instead

+

f(\"

lim

31)

if

Then,

discontinuity.

\" \302\260) \342\200\234\342\200\231 \"K\" =

14

u\342\200\224>0

(5.3))

f,\342\200\231,(x)

u>0)

and is \357\254\201nite, we again say that f (x) has a Similarly, if instead of (5.2),the limit)

exists

\"

0)

u

u\342\200\224>0

at x.

derivative

\342\200\230

+ \342\200\234) f(\342\200\231\342\200\230 f(\"

lim

right-hand

= fj(x)

(5.4))

u is

that f

say

right-hand

at x

equal for x 2

x0 to the curve y = f.(x), equal 0) for x = x0. (The function f..(x) an example, consider the function)

in Fig.

32.

This

to to

the existence

for

x < 1 x = 1

V3:

for

x >

a jump

f_(x)

= \\/x =

for

\342\200\224x3 for

existence

of a

tangent

to

equal

< xo.)

1,)

discontinuity

obviously

f,.(x)

(Thus, the the

f (x) for x < x0 and is de\357\254\201ned for x only

0 has

function

to the curve y = f.,.(x),

(x0 + 0) for x = x0. x0.) In just the sameway,

\342\200\224x3 for

shown

discontinuity

to f

= x0 is equivalent

f(x) =

at x. The x = x0 is

derivative

left-hand

a point of at x = x0

x 2

l,

x <

l.))

at x

= l,

and

SEC. 6

OF TRIGONOMBTRIC FOURIER

CONVERGENCE

SERIES)

we have)

Therefore,

=

rm)

=

f1(l)

The corresponding

(\342\200\2243x1),,.1

\303\251s)

=

-3.

by arrows

indicated

are

tangents

= (-1%)\342\200\235!

in

the

\357\254\201gure.)

I) 4*

/

;,

.) 3

FIGURE 32)

6.

A

Su\357\254\202icient

at a

prove the

THEOREM. Then, tives

Let f

at

Fourier Series

important)

following

integrable function of period 21:. absolutely the right-hand and left-hand derivapoint where to the value f (x). In series of f (x) converges where f (x) has a derivative.) case at every point

(x) be an

every continuity the Fourier exist,

particular, Proof. hand

of a

Convergence

Point)

Continuity

We now

for

Condition

and

this

is the

Let x be a continuity left-hand derivatives

lim

of f

point

exist.

We

=

-'\302\273'..(x)

(x), where both the to prove

have

right-

that)

f (x).

It-NI)

By

(4.1),

this is equivalent

to

+

\"1112 _} f_:f(x

Moreover, since it

follows

f(x) = we can

from

,1,

that)

proving

du

u)

=

(6.1))

f(x).

(3.2) that) an.

_\",r(x>

write)

+ [f(x \"lino 11: ff\342\200\234

u)

\342\200\224

f(x)]

du

= o,

(6.2)))

76

OF TRIGONOMETRIC

CONVERGENCE

of

instead

FOURIER

Thus, the problem has

(6.1).

CI-IAP. 3)

SERIES

been reduced to

proving

(6.2).

We begin by

proving

function)

the

that

+

_f(x

+ _f(x \342\200\230

14) -f(x) \342\200\230 \342\200\234\342\200\231(\342\200\234)

2 sin

Since f integrable. derivative at the point x, the ratio) _

+

f(x

2 sin

u

is absolutely \357\254\201xed)

(x

left-hand

14

--f(x)

14)

(u/2)

(x) has a

(u/2)

(\"3\342\200\231)

and

right-hand

a

f(x))

u?\342\200\230

as

remains bounded 8 > 0 such that

u\342\200\224> 0.

+ u)

f(x

_

u

- 8 0,3 and therefore is a is unde\357\254\201ned only for u = 0. in (6.3) is absolutely integrable, being

u)

ratios)

0), (8.2))

f(x+u)'_f(x_0)

(u x,

of

the

The

the

>

x.

,

of

,

the

interval

[a, b], derivative

x = b.

the conditions exists at x =

Therefore,

f In

the

of the a and

criteria

of

However, if the inter(x) is piecewise smooth this case, the Fourier

everywhere.)

criterion formulated pertaining to the absolute of the Fourier series in the case where f (x) is convergence will be proved in the next section.))

we \357\254\201nally,

1, Sec. uniform

at a point

derivative both at a corner and at a of the left-hand derivative is

on the whole x-axis, sincef (x) is periodic.

continuous,

+ 0). f\342\200\231(x

Similarly,

Secs. 6 and 7 cannot be applied at these points. val [a, b] isoflength 21:, then it is easily seenthat

Thus,

use

existence

imply only that the right-hand left-hand derivative exists at

seriesconverges

of f (x), we

same way.

As for the end points of theorem

E

.

a right-hand

discontinuity.

in

proved

a

< b, so that is obvious at

2::

words, f (x)has

In other

This

a

=hmf(\302\247)=f(x+0).

\342\200\234

m point

< x

a

that have

have)

. hm f(x+u)\342\200\224f(x+0)

and

and

.\342\200\230::8)

(x), we

of f

discontinuity

7.

=

um /'(a)

in

corners

the

At

\342\200\230:38

since

every x

Secs.6 and

\342\200\234

Ch.

=a

consequenceof the fact 1, Sec.9) must

for

derivative

\342\200\235 + \342\200\234\342\200\231 f(\342\200\231\342\200\230 f(x)

lim

in

at x

(see Ch.

[a, b]

only apply the theorems of where f (x) has a derivative. rule and obtain) L\342\200\231Hospital\342\200\231s

we

that

may fail

convergence

(The

discontinuity.

b.))

have

proved

10. The secondpart

the

\357\254\201rst part

of the

criterion,

of the

or nuoonommuc

oouvanonncn

80

a

Uniform

and

Absolute

I0.

Continuous,

3)

of the Fourier Seriesof of Period 21:

Convergence

Function

Smooth

Piecewise

cmr.

saunas

romuen

be a continuous, piecewisesmooth function of period 211:. f(x) Then, derivative f \342\200\231(x) exists is a and everywhere except at the corners of f bounded function (see Ch. 1, the formula for Therefore, 9). applying Let

(x)

the

Sec.

a,, =

11:

f (x)

E:

=.-

=

obtain)

.. nx

sin

E1;L:f'(;c)

nx

sin

dx,)

dx)

:1: _\342\200\230:f(x)

'73-\"

The terms Thus, denoting

we

cos nx dx

sin

[f(x)

1%

b,,

permissible becauseof Ch. 1,Sec. 4),

(which is

parts

by

integration

the

+ nx]::\357\254\202

in the

brackets

in

cos

[f(x)

Fourier

f\342\200\231 (x) cos

L;

7-31;

formulas vanish.

of both

side

right-hand

the function

coe\357\254\202icientsof

f

nx dx.)

\342\200\231(x) by a,',

and

we b,\342\200\231,,

\357\254\201nd that)

bi. \342\200\224

a,, =

b,,

-5.

Since f \342\200\231(x) is bounded theorem of Sec.1 that

and the

_

0.\342\200\231. \342\200\224

(n

7

hence square

_ \342\200\224

l, 2, . .

(10.1))

.).

integrable, it

follows

from

the

series)

3; (a;,2 +

(10.2)

b,\342\200\231,3)

n=-I)

converges.

obvious inequalities)

Next, consider the

11

,

,

2a,\342\200\231,| 1

=\"~\"\342\200\230|7r+;i>\302\260\342\200\231)

(\"'~\";)

n (|b;.|_.1.)2=b;_2_.2_'_bz'-_'+_12.;o, which

n

n)

imply la}.

b.'.

1

.

1

.

(n=1,2,...).

-;\342\200\230-1+|\342\200\224n-|].Therefore, [f\342\200\231(x) integrable may also converges in this case [when f (x) is a continuous function of (l0.3) To

Remark.

prove of the

the

21:].

period

We now given

consider

a very

a trigonometric

series)

simple

+

%

If the

1.

nx +

(a,, cos

fact.

important

very

sin

b,,

Suppose we are

(10.4))

nx)

\"\302\2431

which is not assumedin advance Then the following result holds:) THEOREM

but

seriesof any

the Fourier

to be

function.

series) Q)

2

then the

converges,

(lanl +

series (10.4)converges sum of which 1, Sec. 6).)

therefore

has

a continuous.

Theorem

1 of

Ch.

(10-5))

lbnl)

and

absolutely

is the

it

uniformly,

Fourier series

and

(see

Since

Proof.

la, cosnx +

sin

b,,

nxl

a,, cos

<

nxl +

|b,,sin

nx|



any

SERIES)

(x + u)oo(u) cos mu]u=e

=

du

gives)

[\342\200\224 f

'1\"

(1 1.2)

+

The term

[f(x + u)o)(u)]\342\200\231 f easily seen that the integral =

it is not

+ u)oo(u)]'

[f(x

L\342\200\234

for su\357\254\201icient

our

Now

we

(11.2) LEMMA

that

are

purposes. have shown

bounded, 2.

The

the

bounded

Proof.

for

-1:

<

M

+ u)| |f\342\200\231(x

<

M

=

const,)

\"

mu)!

ff\"

In

du +M((3

du +

fact,)

do

\342\200\224 at))

on)) Mas \342\200\224

is periodic. f (x), and hence I f \342\200\231(x)|, < 21:, which is not essentialbut

\342\200\224 at

that the term of

validity

(11.3)

u)a)\342\200\231(u),

bounded.

absolute values which

by (11.3))

that

brackets

in

and

is

the integral

is obvious.)

(11.1)

integral) 1*

I _

is

right

i.e., have

dul

(3

+

is also

then,

cos mu

where we have used the fact We have also assumed that

in

on the

are bounded, + u)oo\342\200\231(u) exceed some constant M. But

oo(u)]'

+ f(x

+ u)co(u) \342\200\231(x

and f(x

a)(u)

u)

Since)

bounded.

is obviously

brackets

in

[f(x +

% If

cos mu du.)

< u <

mt

sin

dt

J0 2--\342\200\224\342\200\224\342\200\224sin 0/2)

1: and any

(11.4))

m.

We have)

1=

dt

+

]:c..(:)

sin mt dt,

\342\200\230\302\260'i\342\200\234t'\"\342\200\230 where) l)

=

\342\200\234 0bearbitrary.

of F(x)

By

converges uniformly

the theorem to

F(x),

so

that)

|1,|

for all x, Now

providedthat n is su\357\254\202icientlylarge. 8. Then (x) let a + 8 < x < b \342\200\224 I2 =

If -8 <

(12.4))

oo. The same is

mu

true

the

integral

du)

of I2.

Thus, whether

or))

SEC. 14

the partial

not

depends on the

sums of

\342\200\224> oo of

+

\342\200\224 x

8,

+ 8]

values of the of the point x. This

I4. Examples of

the

at

limit

SERIES)

point

x

du,)

u)

only the

involves

which

have a the integral)

series

Fourier

the

as m

behavior

J:f(x 1-\342\200\230:

[x

OF TRIGONOMETRIC FOURIBR

CONVERGENCE

f (x) in the neighborhood the localization principle.)

function proves

Fourier Series Expansions

of

Functions)

Unbounded

This Example 1. Let f(x) = \342\200\224ln |2 sin (x/2)|.7 becomes in\357\254\201niteat x = 2k1r (k = 0, 11, :2, periodic, since)

. ..).

is even

function Moreover,

and

f(x)

is

= Zsinxgz\357\254\202

Zsingl.

|-Zsin;

=|2sin(;+-n:)|=

sothat ln

=

2sinx

ln

:2\342\200\234

The graph of f(x)

is shown

in

37.)

Fig.

Frame

37)

to prove that f (x) is integrable, it is suf\357\254\201cient [0, 1:/3] (see the graph of f (x)). Clearly,

To prove

on the interval \342\200\224

ln

2

J\342\200\230:/3

sin;

dx =

\342\200\224

ln

L\342\200\234/3 (2

[xln

= sln In x

denotes the

natural

logarithm

that it is we

\"/3 x

x=*/3

(2s1n\302\247)L_\342\200\230 +L

(2sin;-) of x.

+

integrable

have)

dx

sin

. x

_

\342\200\224 \342\200\224

7

2sin;-)

Ln/3

(Translator)))

dx,)

cos (x/2) dx

92

FOURIBR

OF TRIGONOMETRIC

CONVERGENCE

CHAP.

SERIES

3)

2 sin (x/2) > I for since where we have dropped the absolute value sign, 3 sin ln 0 < x < 1:/3. As 3 \342\200\224> the 0, (2 (s/2)) approaches zero, quantity as can easily be veri\357\254\201ed while the last integral rule, L\342\200\231Hospital\342\200\231s by using to the

converges

integral) x cos (x/2)) 2 sin (x/2) dx\342\200\231)

F/3 0 which

since the

has meaning,

obviously

In

(see

the

dx) \342\200\2315\342\200\230

[0, -n:/3]. Moreover, f(x) is it does not change sign

interval

interval [0, \302\253/31,since

on the

integrable

absolutely

there

is integrable on

f (x)

i.e.,

exists,

2 sin

J\342\200\230:/3

y-13

is bounded.\342\200\234 Thus)

integrand

Fig. 37).

is even,

Sincef(x)

have)

we

(n=l,2,...),)

b,,=0 a,, =

(n

= 0,

\342\200\224;2-:Kln(2sin-\342\200\231;)cosnxdx

of

First

all, we

calculate the

l,2,...).)

integral)

I=f:1n(2sin\302\247)dx)

'' =fo the last

denote

We

. x (ln2+lns1n-2-)dx=1tln2+_[o integral

by

\"/2

.

\"/3 2L Y =

Therefore

\"/2

ln

1: t

cosi-dt

1: In 2

+

3

implies Recall

t

2]:/zlncositdt.)

gives \"

=

u

.

2L/zln smidu

=

3

.

t

2L/zlnsinidt.

so that 21\342\200\231,

Y = This

.t

= 2t:)

x

substitution

(2s1n\302\247cos\302\247)dt)

21:/zlnsin\303\251dt u

x

.

lnsmidx.)

ln

+

+

-:rln2 t =

substitution

The

make the

lns1ntdt=2Io

Y=2Jo =

Y and

W

that I = 0, so that

= a0

-1:

ln

2.

0.)

that)

..\342\200\234\342\200\230.\342\200\230a2\342\200\230ss?\302\247\"\342\200\224co)

Since < M\302\260\342\200\230ns |\302\260'u\302\242n' we

have

lim o',,a,, =

0.

ll-Fm

Therefore

lim

.9, = .9

18-50)

exists,

the series

i.e.,

s satis\357\254\201es the

for the Sumof Sines.Auxiliary

2. Formula We now

(1.1) converges,and

prove the

sinx+

sin2x

inequality

(1.2).)

Inequalities

formula)

+---+

sinnx

=c\302\260s(x/2)

2 sin

cos-9' (x/2)

+

\357\254\201x)

(24)))

sec. 2

so, we denote the

To do

=

have)

formula

the

Using

Then, we obviously

2sin2xsin; +---+ 2sinnxsin;-)

+

2sinxsin-E

2Ssin;

99)

coemcnmrs

DBCREASING

the left by S.

on

sum

wm-1

saunas

TRIGONOMETRIC

= cos(az -

2sinazsin(3

\342\200\224

+

cos(az

B)

B),)

obtain)

we

\342\200\224

= 2Ssin\342\200\231-Z;

(cos;

-

+

(cosgx

2

+

2 l)x

(it

+

(it

%)x

\342\200\224 cos

+---)

cosgx)

\342\200\224 cos

(cos (n

= cosf

It follows

\342\200\224

+

cos-3-x)

'

that)

S

which

=

cos (x/2)

\342\200\224 cos

(n +

2 sin

\357\254\201x)

')

(x/2)

(2.1).

proves

Since obviously) cos (n |cos (x/2) \342\200\224

2 sin

I 2k-at for x a\303\251

+

(x/2)

(k =

0,

|cos \342\200\230Dxl < \342\200\230 I

|2 sin

:1, i2, . . .),we \"

.

sin kx

S

x

96

2k-tr,

now

We

This

2k1c.

x =

(For

the

recall

+

the

<

Ch.

3, Sec.

Bin (x/2)|)

inequality)

1 (2.2))

WW

for every shows that the sum of sinesis bounded vanishes and hence is also bounded.)

x. \357\254\201xed

sum

the formula)

.)

&+cosx+msh+---+cosnx=

(see

1

\302\247)x|

(x/2)|

obtain

lg] for

+ |cos (n

(x/2)|

3), from

at once

it follows

which \"

H +

cos

kxl

<

that)

l

(2.3))

,2\342\200\230

for x n

+

95

2k1:

4}, and

(k = 0, i hence is not

l, :2, . . .). (For x = bounded as n increases.)))

2k-:-:,

the

sum

is obviously

3.

saunas wmr

nuoonommuc

I00

of

Convergence

can.

conmcnmrs

DBCRBASING

Series

Trigonometric

with

4)

Monotonically

Decreasing

Coefficients)

the two

Consider

trigonometric series)

% + \"Z1 2

a,, cos

sin

b,,

(3.1)

nx,

(3.2)

nx,

n==l)

which

we do

not even assumeto 1.

THEOREM

the

any

case

of any functions.)

series

Fourier

the coe\357\254\202icientsa,, and b,, are positive and decrease as n -> oo,1 then the series (3.1) and (3.2) converge in possibly the values x = 2kn: (k = 0, i 1, :2, .

to zero

monotonically

for

If

be the

x,

except

. .)

of the series (3.1).)

Proof.

the

If

sums of the

theorem follows from

case,considerthe

coe\357\254\202icientsa,,

Theorem 1 of

and

b,,

3, Sec.

Ch.

converge,

then

the

In the general

10.

series)

(3.3))

4}+cosx+cos2x+-~-+cosnx+---,

whose partial sums a,,(x) are bounded to prove for any x 96 2l\342\202\2541't. Then, the theorem for the series(3.1),we apply Abel's lemma. By (2.2), the same argument to the series applies (3.2).)

Remark. Of course, Theorem

1 and the theorems which follow valid if the requirement that the coe\357\254\202icients be nonincreasing is satis\357\254\201ed not for all n, but from a certain value of n. In only starting is the theorem true in the case where some of the early particular,

remain

For

coe\357\254\201icientsvanish.

the series)

example,

icosnx lnn)

n=2)

converges

for x

and hence

diverges.)

We

now

make

1I.e.,a; 2 a2

2k-it. a\303\251

Theorem

2 - - -, b;

For

x = 2k1r, this

1 more

2 b2 2

- -- and

series

becomes)

precise.) =

lim an lim b,. = 0. n\342\200\224>\302\256 n\342\200\224>\302\256))

(Translator)

sec.

3

2. If

THEOREM

to

monotonically

on any

uniformly

(k =

x =2k1r

interval [a, b] which

i2,...).) of the

sums

every

same

for

the series(3.1). Let

\342\200\224

s,,(x)

series

(3.1),

=

,,.,.1

't'...(x) = where

x-axis

case, it is

since the

sums of the

and o,,.,.,,,(x)are

o,,(x)

by

series

(3.1)

prove

interval

< x < b, 1):: +

-

6n+m(x)

partial

1 of

Theorem

the theorem for The [0, 21:]. proof is the we con\357\254\201ne to the case of ourselves to su\357\254\202icient

we cos

a,,+2

consider (n +

Abel's lemma. To do this,

and apply

then

converge,

b,,

the whole

(n +

cos

and

a,,

on

For a

0 be arbitrary.

s >

s(x)

of the

coe\357\254\202icients

[a, b] contained in the both series, and therefore

interval

and decrease b,, are positive the series (3.1) and (3.2) converge does not contain points of the form

\342\200\224> oo, then

(3.1) and (3.2) convergeuniformly Ch. 3, Sec. 10. In the general and (3.2) are periodicfunctions,

IOI)

coerrrcnmrs

DBCREASING

and

coe\357\254\202icients a,,

as n

0, :1, the

If

Proof.

the

zero

wrra

seams

TRIGONOMBTRIC

sums

the remainder)

2)x + - - we

(3.4)

set)

0..(x).)

of

the

series (3.3).

Then

by

(2.3),

g 'an+m(x)|

|Tm(x)'

Since0

sin; where

p. is

the smaller

+ |6n(x)' <

of the numbers

sin

Frame

setting

Therefore,

M =

1/51.,

|r,,,(x)|

for all x

in

the

interval

[a, b].

we

0,)

(a/2)

and sin

(b/2) (seeFig.38).)

38)

\357\254\201nd

< M

= const)

Thus, Abel's lemma

is applicable

to

the))

TRIGONOMETRIC saunas wrm

I02

numbers

. . ., a,,+,,,, . . .

a,,+2,

a,,.,_,,

|S(x) for

any

for

all

and for

x in

the

interval

[a, b].

[a, b], the

x in

Theorems

\"'

2

n

n(x)| < 3

monotonically

w

=

f(x)

21: are period x = 2k1I: (k = of

+ Z g9 u=l continuous

the

If

g(x) =

nx,

2

b,,

sin

nx

n=-I

all

for

sums of the

theorem follows from

positive and decrease

w cos

a,,

0, i1, i2,...).)

Proof.

series(3.1).

imply)

3. If the coe\357\254\201icier ts a,, and b, are to zero as n \342\200\224> oo, then the fz: .ctions)

THEOREM

have

\342\200\224> co.

convergence of the

the uniform

proves 1 and

n

\342\200\224> 0 as

an

inequality)

|S(x)

holds, which

for .. su\357\254\202\342\200\230 ciently large

gives

< Mam

Man+1 < 3) In other words,

n.

su\357\254\202iciently large any

Since

AP.

series (3.4),ar

and to the

- s..(x)|

c

COEFFICIENTS

DBCRBASING

x, except

coe\357\254\202icients a,,

Theorem 1 of

possibly for

and

b,,

converge,

the

values

then

the

10. In the general we 2k-rt in an can include interval case, [a, b] which does any point xo a\303\251 two not contain the -the of x= 2k-tr. In this form interval, points series converge uniformly are their sums Theorem and hence 2) (by at continuous continuous Ch. Sec. In are 1, (see 4). particular, they x = x0. Since .1\342\200\230 a point of the form from , is any point different x = 2k1c, the tneorem is proved.) Ch.

3, Sec.

I A

\\

\\

\\\342\200\230,21r

FIGURE

As

:1!

J\302\243\\

0

consider

an illustration,

the

\\61r

39)

expressions) Q)

f (x

=

\342\200\224 ln

|2 sin;

g(x)=1t\342\200\224\303\251x=

=

Zsinnnx

\"Z1

\357\254\201_

41r

coin)\342\200\230.

(0 0, there

existsan Ia

3

I.e.,

points which

are not

4''\342\200\234, 112 +\302\260\302\260\"I-11,,

singularities.

integer

N such

\342\200\224

o',,|

<

E.

(Translator)))

that if

n

2

N)

(6.3))

SEC.

TRIGONOMETRIC

5

where

a,, is

of the

series

series

+

an-I-2

in

'

un-I-2 +

' + \302\260

=

an-I-ml

monotonically is applicable

lemma

It

follows

moreover

that

r\"+3, . .

as n to the series) '_\342\200\224.

this

a,,(r)

letting

< r <

1.

..(r)| =

that

Noting

a(l)

assert that the inequality r < 1. This proves that r S 1 and hence implies

0 < r < 1.

This

(0 < r

the partial sum

denote

we can 0 <

any r

(0 < r

< 1),Abel\342\200\231s

. . ..)

the series

s

<

er\"+1

|\302\260(r)

0 <

oo for

+

un+lrIl-I-I

|R,,(r)| < Now,

(6.4), just

to zero

Rn(r)

lknl +

<

Rn+m|

numbers)

r\"+1, decrease

\342\200\224

IRII

series

Since the

value.

absolute

l929\302\260\302\260\302\260)\302\260

have

sum of the

every partial

Thus,

+\"\302\260s

(m=

an-I-l+un-I-2+\302\260\302\260'+un+m

+

the remainder

Consider

(6.1).

=\302\260\"'\302\260'n=un+1

(6.3) we obviously |un+l

COEFFICIENTS)

sums

its partial

By

DECREASING

(6.1))

Rn and

of the

the partial sum

WITH

SERIES

< 1).

of the

|R..(r)| <

series

(6.2),

8

(6-5))

a,,(l) = a,,, and

= a and

we have

using

(6.3),

(6.5) holds everywhere on the interval on the series (6.2) convergesuniformly that the function a(r) is continuous on

the lemma.)

proves

To prove the propositionformulated that the series)

at

of

the beginning

this

section,

we assume

co+clz+c2z2+...+c\"zn+...) converges the

at

a point

co + is a

z

lying

on

the circle

C.

By the lemma just

function)

clrzz + czrzzz +

- - -

+

c,,r~z\"

for 0 < r < 1,and = lim (co + clrz + czrzzz

function of r

continuous

lim

r->1 rl

r p,,,(x)

0, there = co

of Sec.

results

1, we can obtain the

is continuous

which

on

the

interval

exists a polynomial)

+ clx

+ - - -c,,,x\"',

which

|f(x) everywhere 2 See 3

Polynomials)

rather simple consequenceof the which is often very useful:) result

[a, b].

for

zero,

identically

to

vanishes identically.)

the function

TI-IEOREM

f

to all

the trigonometric system and 4 also be can an orthogonal Theorem \342\200\234enlarged\342\200\235 system. as follows: coe\357\254\201icientsof a continuous function If all the Fourier

paraphrased

(x) is

the

zero.)

identically

which is

function

This

in

given

of

is orthogonal

which

be

its

from

the method

using

by

function

yet

chapter; however,

consequences of the completeness 1 and 2 of Ch. 2, Sec.8):)

In other words, exceptfor the trivial add an extra continuous function

we cannot still

does not mean that we from a knowledge of its Fourier

this but \342\200\234de\357\254\201ned\342\200\235

say

determine

on the

the footnote to proposition

(\357\254\201anslator)))

-

pm(x)|

<

8

(4-1))

interval [a, b].) Ch. 2,

is

Sec. 7.

usually

(Translator)

known

as the

Weierstrass approximation

theorem.

4

sec.

ON FOURIBR

OPERATIONS

By making the

Proof.

seams

transformation)

mm)

t=n;::

or

equivalently)

=

x

we take the interval t-axis. (If the length this

b\342\200\224a)

t +

the x-axis is less than We now set)

b] of of [a, b]

transformation.)

\"

theorem of

2 of the

Corollary

a, interval

the

into

[a,

we do

21:,

[0, 1:]

not have

of the

to make

= F(t).)

\342\200\234z + \342\200\230N) a) f(\342\200\235

By

|2|

Sec.1, there

a trigonometric

exists

polynomial)

o',,(t) = for

cos kt

2 (at, k=l

+

at,

(3,, sin

+

kt),

which

\342\200\224

o..(t)|

Ira)

everywhere on [0,1:]. so > 0 which is so small

it is

known

(lakl +

for

cosz

a positive

number

the condition)

that

that

(4.3))

-3

n \357\254\201xed, we choose

Holding

wk; Now,

<

lam < \302\247-

(4.4)

all z

24

z3

-1--2\342\200\224!+4\342\200\224!----,

z5)

23

s1nz=z\342\200\224\302\247\342\200\224\302\247+5\342\200\224!----,)

where

these

series converge

on 0

in particular,

< z < mt.

uniformly

But

on

then,

every

for

interval

\357\254\201xed n and

of

\357\254\201nite length,

all

su\357\254\202iciently

\342\200\231

larg

we

have)

kzz

k-:4

kzzzz

co,..,-(1-.,;.+%-...+(-.y\357\254\201);*'-.7) n=l

or \302\24331 _

if

12

12

E (_

,_,

THEOREM

1. Let

nx_

Case of a Continuous

21:)

Period

of

Function

sin

1),,\342\200\234 \342\200\230n_3)

Series. The

of Fourier

Differentiation

8.

_\342\200\230

f (x) be a

continuous an function of period 21:, with derivative not exist at certain absolutely integrable (which may points\342\200\230). Then the Fourier series of f \342\200\231(x) can be obtained from the Fourier series of the function f (x) by term by term dt\357\254\201erentiation.)

Let)

Proof}

f(x)

where we can

Sec.11.

of all,

we

Let

write

a}, and

In other

+

2 (a,, cos nx n=l

1}\342\200\231

the

equality

b}, denote

the

because Fourier

+

b,,

sin

(8.1)

nx),

of the theorem coe\357\254\202icients of

f

of Ch. 3, First \342\200\231(x).

have)

as

5

=

=1

dx \342\200\230N) f;f\342\200\231(x)

=/or)

\342\200\224/(-20

number words, 1\342\200\230 \342\200\231(x) may not exist at a \357\254\201nite

=

o.)

of points (in

each period).))

I30

rounmn

on

OPERATIONS

c1-mp.

cosnx

a;, =11:[;f'(x) =

dx

+

cos 11:

nx]::t

[f(x)

=

nx dx

sin

nb,,,

$\302\243;f(x)

(8.2))

=

b,',

5

by parts, we obtain

integrating

Then,

saunas

gt]-:nf\342\200\231(x)sinnxdx

=

:5 [f(x)

sin

\342\200\224

= -na,,.)

nx dx

cos

3-\342\200\230J-:nf(x)

Therefore) Q)

-

at

follows

an

n

b,, go

na,, =

to zero

lim

=

1/n as

than

faster

Let f (x) be a continuous where m - l derivatives absolutely integrable (the

derivatives, is

derivative

(8.3))

3

nb,, n\342\200\224rco)

THEOREM 2. m

13.) Fourier

to

zero

co-

as

Sec.2), wecan write) lim

a,, and

differentiation.

from (8.2). Moreover, since the absolutely integrable function converge

n-no i.e.,

term

once

Ch. 3,

oo (see

->

by term by

b,,=)

n

e\357\254\202icientsof

nx),)

of Theorem1

b5.) d,,=-\342\200\224:

which

a,,

(8.1)

the conditions

Under

Remark.

\342\200\224 sin

\342\200\231(x)

the series obtainedfrom

this is

and

2 n(b,, cos nx n= 1

~

f

0,

oo.) n\342\200\224>

function are

of period continuous

m\342\200\231th derivative

21:, which

has

and the m\342\200\231th may not exist at

certain points). Then, the Fourier be series can of all m derivatives obtained by term by term di\357\254\202erentiation of the Fourier series of f (x), where all the series, except possibly the last, to the corresponding converge derivatives. the Fourier the function Moreover, f (x) coe\357\254\202icients of relations

the

satisfy

lim nma,

=

ll-PG

To prove

Proofl

cations of Theorem term ing

by term

follows

\342\200\224

1)th

n\"'b,,

it is

\357\254\201rst assertion,

The

di\357\254\201'erentiation,

derivatives

to the (m

the 1.

lim ll\">@

convergence except

from the

derivative].))

possibly

differentiability

= 0.

(8.4)

su\357\254\202icient to make m appliof all the series obtained by the last, to the correspondof these derivatives [up

opemmous

sec. 8

The relations

(8.4) are obtained times.

(8.3) m

relations

the

a

_

of successively applying

a result

as

Thus)

an

b3

02

bf.

n-\342\200\224---1-1-\342\200\234\342\200\230m' n

n

0

'

n

(8.5)

n\"

b=\357\254\202__5__\357\254\202_m_\357\254\202, \"

a;, . . a,\342\200\231,,

where

f 0\(x)")

is

implies

(8.4).)

f(\"')(x), taken

the

of Theorem

series for f (x) differentiation,

the

2,

by term

term

uniformly

converge

last,

Fourier coappropriate Since the sign. proper 0 as n-> co, which (3,,\342\200\224>

with

from f (x) by

functions

the

coe\357\254\202icients of

0 and a.,,\342\200\224>

Under the conditions series obtained

the

except possibly for

(this follows

from

3, Sec. 11).)

Ch.

The next propositionis in THEOREM 3.

Given

%

if

(3,,

n\"')

the

denote

integrable,

absolutely

Remark. all

and

an

function

the

e\357\254\202icients of

and

while

the Fourier

. are

b,\342\200\231,, b;, . .

.,

. . .,

\342\200\231(x), f \"(x),

n3

n2

n

f

|3l)

seams

rounnan

ON

nx +

(a,, cos

2:)

series)

trigonometric

+

converse of Theorem

b,,

sin

(8.6)

nx),

\"\302\2431

relations

the

by satis\357\254\201ed

is a

continuous

which

can

Proof

s

M

coe\357\254\202icientsa,,

and

< M,

|n\"'a,,| are

the

sense the

a certain

the

|n'\"b,,|

be obtained

by term

Denote the

tefm

by

of the

sum

then

b,,,

21:, with

of period

function

(m 2 m

2; M = const) the sum

of

the

(8.7) series

(8.6))

\342\200\224 2 continuous

di\357\254\202erentiation

derivatives, the series (8.6).)

of

series (8.6) by f (x).

By

(8.7),

we can

write)

f(x

=

+ 329

Z (3% cos nx

+

sin n\342\200\224\" nae):)

where

|a,,| < If we formally coef\357\254\201cientshave

differentiate

absolute

M, |B,,|< this

M,

series,

values which M)

5:))

M

= const.)

then after k differentiations do not exceedthe numbers)

the

I32

on

open/moms

saunas

FOURIBR

can.

of the

that the sum for k = l, 2, .

5)

of the coe\357\254\202icients conby Theorem 1 of Ch. 3, Sec. the series obtained by term by term differentiation of the = series (8.6) are uniformly for k . m 2. But 1, 2,. convergent it follows from Theorem 2 of Ch. 4 that the function f (x) then, It

follows

verges

. .,

10,

Therefore,

.,

1,Sec.

is differentiable tives

values

absolute

\342\200\224 2.

m

\342\200\224 2 times

m

are continuous,

and

(and hence

continuous), by term

term

the

that

the

that

derivais

differentiation

legitimate.)

*9.

of Fourier

Differentiation

Interval

the

De\357\254\201nedon

Series. The Caseof a

[\342\200\224-n:, 1:])

1. Let f (x) be a continuous an absolutely integrable

THEOREM

~

where

a,, and c is

derivative

(which

the interval

may not

exist at

Then

f\342\200\231(x)

constant

on de\357\254\201ned

function

[\342\200\224\342\200\230lt, 1:] with

certain points).

Function

g

+

2

+

[(nb,,

(\342\200\224 l)\"c)

cos nx

\342\200\224sin

na,,

nx],

(9.1)

n =- I)

b,, are the Fourier given by the formula) c =

coe\357\254\202icientsof

the function

f (x) and

the

(9.2)

,1,Lr(n>-f(-\302\253)1.

Let

Proofl

f

~

cos nx (a,\342\200\231,

\342\200\231(x) + 3;\342\200\230

\"$1

+ bj, sin

nx),

so that) a{,

Obviously

=) 1

dx

=

\342\200\230I'll ff\357\254\202x)

,1,Lr

\342\200\224 r(-\302\253)1.)

we have7)

f The Fourier

-

\342\200\231(x) 956

series of the

~

+

bf,

sin

nx).

(9.3))

function)

-

I;

cos nx (a,\342\200\231,

3

(f\342\200\231(x)

dx =

f(x)

\342\200\224 -

f(0)

(9.4))

7 it is still possible to transpose terms from the rightAlthough (9.3) is not an equality, of hand side to the left-hand side, as can be veri\357\254\201ed by calculating the Fourier coe\357\254\201icients the function appearing in the left-hand side. In our case,the constant term of the function \342\200\224 and all the other coe\357\254\201icients remain the same as for f\342\200\231(x).)) f\342\200\231(x) \302\253lab vanishes,

openxnons

9

sec.

1-\342\200\230ounnzn saunas

ON

I33)

the can be obtained from (see integration series (9.3) by term by term the series (9.3) can be Theorem 3 of Sec.7). Therefore, conversely, of the series for the function obtained by term differentiation by term

But

(9.4).

we have)

f(x) = and

by

+

(a,, cos nx

2

929

+

sin

b,,

nx),

n=-I)

(7.7))

-

=

\342\200\224 \342\200\224

f(x)

1(0)

/(0)

\342\200\230-3\342\200\231

Q

+

cos

2 n=l)

nx +

+ (_ 1)::

(b,

[an

I

sin

-33))

mt]-

Therefore

\342\200\224 ~ \342\200\231(x) (\342\200\230-3

sin nx + (nb,, 2 [\342\200\224na,, n= 1)

f

which

(9.1), if we

implies

If c

COROLLARY.

f In

other

words, term

differentiated f(-nz)

=

continuous and

Fourier seriesof avoid

absolutely

Sec.2).

~

if f(1t) = f(\342\200\2241c),the But this by term.

1 is

nx],

= a6.

n(b,, cos nx

2

henceTheorem

cos (\342\200\224l)\"a6)

then i.e., iff(1:)= f(\342\200\224-rc),

\342\200\231(x) n==l

Theorem

Remark.

Fourier seriesof knowledge of the

can

= 0,

the periodic

f(--rc),

constant

set c

+

\342\200\224 sin

series

Fourier

gives)

nx).

a,,

is

(9.1)

immediately

extension of f(x) onto the 1 of Sec. 8 can be applied.) especially important

in

the

can be

of f(x)

clear, since if whole x-axis is

case

where the

is given instead of f (x) itself, for it shows that a f(x) Fourier series of f (x) is all that is needed to form the

in this case, the calculation of the However, f \342\200\231(x). if we use the formula (9.2). We turn out to be di\357\254\202icult the use of (9.2)by recalling that the Fourier coefficients of an function converge to zero as n -> oo (see Ch. 3, integrable follows from (9.1) that) it Therefore,

c can

lim

[nb,. +

(\342\200\224 l)\"c]

=

0,)

whence)

c = It

is usually

an easy

matter

lim to

[(\342\200\224l)\"'\342\200\235nb,,].)

calculate

this limit.

Moreover,

it

is not))

opammons

I34

to see

hard

Cl-IAP.

that the existence of this limit is equivalent to the quantity which for even and odd n separately, are equal in

absolute value and oppositein applications, one often series of f (x). In this

sign.

that we

1 presupposes that f (x) and [-\342\200\230It, 1:]

Theorem

on

5)

limits

having

nb,,

snnnas

rounmn

ON

has an

encounters

the

case,

f(x) is continuous In the

the function

that

know

derivative.

absolutely

integrable

situations

where we

know

only

the Fourier

becomes more

complicated,i.e., ascertain whether f (x)

problem

Fourier serieswe have to its derivative is absolutely integrable, and if the answer is a\357\254\202irmative, we have to form the Fourier series of the derivative. The following theorem often helpsto solve this problem:) we are given the series) THEOREM 2. Suppose that of the

a knowledge

from

is differentiable

whether

and

(a,, E; + \"\302\2431

If

+

cos nx

b,,

sin

(9.5)

nx).

series

the

+

E

+

[(nb,,

\342\200\224sin

nx

cos (\342\200\2241)\"c)

(9.6)

nx],

na,,

\"\302\2431

where

c =

is the

of an

series

Fourier

series (9.5) is

f(x),

continuous

is

which

and obviously

Proof.

We



(10.2)

,3,f(o>.

Proof. If ~

f\342\200\231(x)

%6

+

cos

nx,

Elm\342\200\231.

then

f We

know

\342\200\224 ~ \342\200\231(x)

a5.

3

%

nx.

cos

(10.3)

n=l

that

co

-

2k

-

co

=

=

k;,

_

[1

:13

(-0,?

sltznx

\302\2432,

(10.4)

(0 < x < 1:). [See

(13.9)

and

(13.11) of

\342\200\224dx

I:

= f(x)

Ch. 1, Sec.13.] Therefore \342\200\224 -

=

E

b,, \"\302\260'

-

\342\200\224

sinnx

a6

,\302\247f(o>E

=

[1

\342\200\224 \"Z1

which is the

1(0)

1\342\200\230;

(f\342\200\231(x)

1%

[nb,,

Fourier seriesof the x [0

{(0)

(\342\200\2241)\"'\342\200\235 gill\342\200\230 \"\302\260'

3

-

(10.5))

sin nx) (\342\200\2241>\"1

,,)

+

(a6

+

sin nx T,) 1%:/(o))(\342\200\2241)\"]

n

function \342\200\231 (L6 \342\200\224 dx.)

(f(x)

2)

the series (10.3) tenn series can also be obtained by integrating be obtained can Sec. and by) hence, ( 10.3) by conversely, (see 7), Therefore differentiating (10.5) term by term. But this term

\342\200\224 ~ f\342\200\231(x) 3: 97\"\342\200\231 n=-1))

\342\200\224

[nbn

%f(0)

+

(a6

+

cos %f(0))(\342\200\2241)\"]

nx,)

sec.

on

opmmous

to

(10.1) and (10.2)if

which implies

d = %

COROLLARY.

smuas

I39)

set)

we

f(0).)

If (n=l,2,...),

\342\200\224d+(c+d)(\342\200\224l)\"=0

then, instead

FOURIBR

of (10.1),we

obtain) Q)

~

f\342\200\231(x)

the Fourier

i.e.,

of the

entiation Proof:

already considered c = 0. But from

d by

using

di\357\254\201\342\200

to the condition) = 0)

(10.2).

Instead of

Remark.

term

In fact, for even n, we immediately = 0 or d = O, and we have \342\200\2242d n, f(0)

odd

for

then

by

1.

Theorem

in

obtain

simply by term

=f(W)

f(0)

2 f(1c)follows

nb,, cos nx,)

is obtained series of f \342\200\231(x) Fourier series off (x).)

case corresponds

This

2

n= I

(10.2),we

c and

the constants

detennine

can

formulas)

the

c =

\342\200\224 lim

where n is even,)

nb,,,

II->0)

d

To seethis,

we

=

note

tegrable function f

lim \302\247(

nb,,

\342\200\224

c),

where

n is

Fourier coefficientsof the

that the

to zero

\342\200\231(x) converge

as

n \342\200\224> 00.

(10.6))

odd.)

in-

absolutely

Therefore,

it follows

from (10.1)that) + c)

(nb,,

for even n,

which

the

implies lim

which

implies

\357\254\201rst formula

the second formula (10.6). 2 have the following

3.

Suppose

(10.6),

while for

odd

n

(nb,,\342\200\224c\342\200\2242d)=0,

Theorems1 and THEOREM

= 0

that we are

given

converses:) the

series)

G 92

2

+

a,, cos \"Z

nx.

(10.7)))

I40 opemmous on rounnan If the

CHAP. 5

saunas

series1\302\260 @

na,, sin nx

\342\200\224

2

n

series of an Fourier seriesof the

Fourier

is the

is the

This theorem is a by

on

and obviously

setting

b,,

=

4.

THEOREM

[0, 1t].11 =

f

\342\200\231(x) cp(x)

simple

consequence

= l,2,..

0 (n

i=

limits (10.6) exist and +

g is

2=

is the

the

continuity

Proof.

of

=

5

continuity

b,,

series)

the

given

nx.

sin

(10.8)

\342\200\224 d

+ (c

+ d)(-1)\"]

cosnx

(10.9)

I)

+ E

and

function,

Sec. 7 to

2 of \342\200\224 d

+ (c

[nb,,

0,n>0,) and

c,,,,,,

b,,,,,,

and

1,2,...

The Fourier seriesof

m>0,n=0orm=0,n>0,

for

1

= 0,

m=n=0,

for \302\253}

f (x, y)

n

by the

calculated

are

d,,,,,

formulas

l,2,....

=0,

can be written

more

compactly

in the

complex

form) Q)

y) ~

f(x.

2

(2.4))

c...e\302\253-x+~\302\273.

m,n=-no)

where)

c... =

y>e-Wu\302\273dx

I\342\200\230,-. fKff(x.

(2.5))

We

the proof

leave

to the

result

[it

to go

is recommended

from

Ch. 5,

it

can

be

that the system

shown

(2.1) is complete.

This

that)

LJf3(x,

y) dx dy

=

5:12-r:2A,f,o

S

m=I + m,

+ E

+

415130

+

so

reader

i2,...).)

(2.3)].

As in means

of this

n=0, :1,

i2,...;

i1,

(m=0,

(2.4)to

dy)

+ 2:333\342\200\234,

S =n

S

21c\342\200\231A\302\247,,)

21933,,)

n=l

\302\253=,

12:... +

that)

,\342\200\224f. jxfmx.

y) dx

dy

=

2 m,

x... a:..

+

12.3..

+

c.?.. +

43...).

(2.6)

n=-0

variables. for the case of two (2.6) expresses Parseval\342\200\231stheorem of the trigonoanalogs of all the results implied by the completeness remain metric system in Ch. 5 for the case of one variable) valid, (proved the way in which the results are stated.)) provided that we suitably modify

Formula

The

DOUBLE FOURIER

The

3.

CHAP.

INTEGRAL

for the Partial Fourier Series. A

Formula

Integral

Trigonometric

7)

a Double

of Sum_\302\247

Criterion)

Convergence

expansionof the form (2.3), where it is assumed 21: in x and in y. If f(x, y) is de\357\254\201ned on both only period x onto to and the whole (with respect y) xy-plane. periodically

that we have an

Suppose that

THE FOURIER

SERIES.

y) is of

f(x,

K, we extend Now let)

it

y) =

s,,,,,(x,

cosp.x 5: 7.,,,[a,,,

S

u.=0 v=0

vy +

cos

b,,, sin

+ c,,, cos ux sin vy

s...,.(x. y)

= ,-,1,

l

1

=

t)

t) cos u(s -

x)

cos

+

2 cos v(t

.55 LJf(s,

[5

\"'

+

cos

\342\200\234Z1

Recalling the formula for the E

- x

s

Setting

sum

'[(

Sm

-1 \"

\"\"\"(\"\342\200\231 3\342\200\231)

=

\342\200\235 LJf(S\342\200\231

u,

p.(s

= v t \342\200\224 y

of

\342\200\224

x)]

cosines

1

[5

px sin

d,,,, sin

+

1,

Jxjns.

ion...

in

vy)

1,2,. . .; n = 0, series. According

quantities s,,,,,(x, y) (m = 0, sums of the double Fourier partial

The

cos

y.x

are called the

2,...) to

(2.2))

vcz

- y)

as at)

\" \342\200\224ds

3, Sec. 3), we

(see Ch.

[(sS\342\200\224 x)\342\200\230/f

and using the

[E1

y )] -

dt.

y)]

9:1)

+l\342\200\230)('')]'[(+%)(t\342\200\230

Zlsin

vy].)

have)

4\342\200\230 \"\342\200\230\30

y)/2]

periodicity of the

we

integrand,

obtain)

This formula is completelyanalogous formula (4.1) of to the corresponding Ch.3, proved for the case of one variable. The similar to that used in result can be proved by a method following Ch. 3, Secs. 6 and 7:) Let f

THEOREM. with

bounded

partial

function (x, y) bea continuous 8f derivatives 0f / ex and

on de\357\254\201ned

/ By.

Then,

point f (x, y) at every interior the mixed partial derivative 32f/axay f (x, y) is ofperiod21:in x and in y and has continuous partial series of f (x,y) Bf/0x, Bf/By, 31f/ Bxay, then the Fourier f (x, y) everywhere.)

series

of f

(x, y) convergesto of

neighborhood

For

the

reader

make the following

which

who is not accustomed to dealing to avoid confusion: The remark,

with formula))

a square the

K,

Fourier

of K in a exists. If derivatives converges

double

to

series, we

DOUBLE FOURIER

SEC. 3

THE

SERIES.

INTEGRAL)

FOURIER

Q

f(x, y)

=

Z

A,,,,,

mx cos

a,,,,, cos

ny

+

sin

b,,,,,

mx cos

ny

m, n=0)

+ c,,,,,cos mx

lim

more

that

precisely,

f (x.

0, there

8 >

any

given

y).

y) =. -'\302\273'\302\273...(x.

m-no) R--bw)

or

d,,,,,sin mx sin ny]

that

means

the

ny +

sin

existsa

number

N

that

such

inequality) \342\200\224

for

holds

K

square

m 2 N,

n

All that

N.

2

S

Smn(x9

|.f(x9

(-1: < x < 1:,-1: < y

<

8)

has been said above concerning 1:) is also applicable to every

the square

Q(a oo, then \302\260\302\260 _\342\200\230= -4\302\273 0 (j\"f(:)

1/21:

_To

formulas

prove

the last

show

that

two

em du ax)

X)

we

formulas,

differentiating

the

= 3 F(x).

use integration

original

function f

by

parts.

These

to)) (x) corresponds

DOUBLE FOURIBR

'92

spondsto

F(x)

dividing

operations on the

matical

its transform

basis for the operational calculus,

a

mathematics. Next, we consider transforms

of

corre-

algebraic operations on of the

is the of applied

\357\254\201nal result)

branch

form. The

dilferent

somewhat

a

7)

mathe-

complicated

important

very

f (x)

integrating

taking the inverse transform

then

(and

while reducing

to simple

function

original

CHAP.

INTEGRAL

FOURIER

transform F(x) by x/i, by x/i. This idea of

its Fourier

multiplying

THE

SERIES.

function)

J

1-5:f(u)

is called the (Fourier) cosinetransform theorem holds for f (x), then integral

of

cosine transforms

F are

f and

from

f (x). If (9.3) that)

the

cos xx ax,

m)

cosine transform

the Fourier

is itself

functions

(10.4))

the function

of

A/E

du

Au

it follows

.\342\200\224.

f(x)

i.e.,f (x)

cos

=.-

F0)

of

Fourier

(10.5))

In F().). each other.

other words, the Similarly, the

function)

=

0(1)

=

f(x)

as in the

just

sin M du

f(u)

the (Fourier) sine transform

is called

i.e.,

J;

of f(x),

(10.6))

and (9.4)

gives)

0(1) sin x).at,

J;

case of cosinetransforms,

f and

0,

a derivative

This

function

is integrable we) parts,

by

Integrating

\357\254\201nd

=

2

2

s)

j;c

2

no

.

E

by

2.

\"\"\342\200\234\"\342\200\234\342\200\234\"\342\200\234=\302\273/;;aT+T2')) \342\200\234\342\200\231W=\302\273/.:J.,

10

sec.

Then

and (10.7)

(10.5)

saunas.

rounmn

DOUBLE

INTEGRAL

\342\200\230rm: romunn

I93)

give

2a woosxxdl

= \342\200\230=\342\200\230\342\200\230'\342\200\231

7;

.,

> \302\260>= 0\342\200\230

Tm

ooxsinxxdl e_\342\200\234=g 1!:

o

(x>0))

7.2

a3 +

Example 2. If

1 f

f(x) =

x

< a,

< x = a,

> a,

obviously)

and

cos xudu =

=

Fa)

\342\200\234\342\200\234x\"\".

by (10.5))

f(x)

In

0

for

for x

0 then

for

sin ax cosx).dA _

=)

Era

\342\200\230It o

for x

particular,

=

R)

1

a,

x > a.)

T\342\200\234

a = 1-,we obtain oosin).

'2\"'f.. The

x < a,

x =

\302\260\302\260sin2aA

1:

*II.

0 <

for

a)

1

setting

for

0 for

5-:J.. and

1 Q

T\342\200\234)

Function)

Spectral

It is easyto seethat f(x)

can be

(9.1)

equation

=

dx

in

written

f(u)emx-~>

511-:

the form)

du

(11.1)

since

emx-'0 = cosA(x = cos

since the

and

integral

\342\200\224

u)

Mu

the

containing

+ isin

A(x

\342\200\224 \342\200\224 isin

x)

)\\(x

sine vanishes

\342\200\224

u)

\342\200\224

u),)

(see Sec. 9).

Now

we

set3)

Am

=

3 Sometimesthe factor 1/21: is omitted in in (11.3).))

then reappears

f(u)e-W

du.

(11.2))

5\342\200\231;

the formula

for .40.);

of course, the

factor

I94

nounuz

FOURIBR

engineering,

density or

spectrum)

of f

(x). f(x)

(11.2)is the

analog

Fourier

complex

INTEGRAL

FOURIBR \342\200\230me

(which is in general and is called the

function

This

seams.

great importancein

=

spectral

(11.2))

(11.3))

Amenx an;

(14.7) of and

(11.1)and

electrical

the

(synonymously,

According to

of formula coe\357\254\201icients),

is of

complex)

spectralfunction

Ch. 1 (which

gives

the values

analog of formula

is the

(11.3)

1)

CI-IAP.

of

(14.6)

the)

of

Ch. 1.)

Example.

Find the spectral function 1

f(x) =

{O

in

shown

Fig.

the function)

of

for

|x| <

for

|x|

a,

> a,

42.)

_-_-_-___J) ._-____-_L vs\342\200\230 o)

.

FIGURE 42)

According

to (11.2)) \342\200\230ha)

A(;)=%

:03-iAudu=%t[e;M -\342\200\230mu-\342\200\224a

__1_eW\342\200\224e-\"~\302\260_1sina). \342\200\2302n' -1\342\200\230: 2.\342\200\231) ix and shown

therefore in

Fig.

in this case A0.) turns 43 (for a = 1).)

out

to be

real.

The graph of A0.) is

.1. 17\342\200\231)

=-31r

- 21r'\302\245-/-tr

13-12

0| Frounz

43))

-

1!\342\200\231 31r)

DOUBLE roumen

PROBLEMS

I95)

\342\200\230rm: I-\342\200\230OURIER INTEGRAL

SERIES.

PROBLEMS

1.

if f

that

Show

(x, y) = ~

g(x)

where)

g(x)h(y),

~

E

3:

1:00 ame\342\200\230\"\"\342\200\230.

b..e\342\200\231\"\342\200\235.)

Q)

y) ~

f (x.

2

a...b\302\273e\342\200\234'\"\"*\"\"\342\200\231-)

m.Il=-Q)

2. Show that

if

y) =

f (x,

g(x +

where)

y),

Q

:(x) ~

2 a...e'\"\"\342\200\230.) m=\342\200\224w)

then Q

y) ~

f(x. 3.

the following

Expand

-1: <

f(x.y)

Find

the Fourier sine

+

the

in

f(x)-{O

5. Find the

transform

<

x

0

(3.1))

0.)

now

We

verify

that

2.) =

[\342\200\224e-x]::;\302\260

1,)

I\342\200\230(p+1)=j;\302\260e-xxndx.

by parts,

we obtain

+ 1) I\342\200\230(p

on the

\357\254\201rst term

is just

integral

+ 1) I\342\200\230(p

in view

=

e-xx\302\273-I + p [\342\200\224e-xxn]\"\342\200\230\302\260\302\260 j\302\260\302\260 0) x=0

right vanishes

This I\342\200\230(p).

3) If p is a positive

or,

only for

given

I\342\200\230(l)

Integrating

The

meaning

the formula

de\357\254\201ned by

dx. e\342\200\224xx\302\273-I

I:

properties

2)

the

if x

that

take

will

values,

complex

not

< 0 and if p is not an integer, values [see (2.7)]. In order to complex we shall consider J,(x) only for x 2 0 (when p is

be noted

It should

Remark.

then in general

integer,

=p(p =p1\342\200\230(p)

proves then

(provided,of course,that

p

> 0),

and

Property 2. using Property 2 we obtain -

-1)I\342\200\230(p

1) =---=p(p

-1)-\"2-1-I\342\200\230(1).

of Property 1 + l) I\342\200\230(p

Thus, for p properties.))

dx.

> 0, the

function

= p!

de\357\254\201ned by

(3.1)

actually

has the required

BESSEL runcnons

202

the function

extend

To

seams

FOURIBR-BESSEL

AND

I\342\200\230( p) to

values of p,

all real

8)

CI-IAP.

we

the

with

start

formula)

F(p +

1) = pm\302\273)

01\342\200\230)

+ l)_ = I\342\200\230(p I\342\200\230(P)

(3.2))

P)

-1 < p < 0, the 1 < 1. Therefore, + p

since of (3.2) has meaning, for be used to de\357\254\201ne (3.2) I\342\200\230(p) -1 in the that as p \342\200\224> numerator the right-hand 0,

If

we

Incidentally,

then

0 <

< p

< 0.

can

note

1 while

(3.2) approaches

side

right-hand

denominator

the

of

side

0, and hence we have)

approaches =

co.) I\342\200\230(0)

Now let

-1.

-2 < p <

-1

Then

of (3.2) again has meaning. Therefore, -2 < p < -1. Ifp\342\200\224> 1, then it hence we have)

+ 1<

< p

0, and

(3.2) that

from

follows

the

side

right-hand

for used to de\357\254\201ne I\342\200\230(p)

can be

(3.2)

co, P(p)\342\200\224>

and

= oo.) I\342\200\230(\342\200\224 1)

we

can

p =

0, -

p >

0 and

its very

then

(3.2)

Functions

of

solution

m

note

=

+ oo

of the

First Kind

equation

(1.1),

be applicable

Replacing

(1.1).

L4\")

We

forth.

so

and

all

P(p) has the three propertieslisted

construction,

Since p2 appears in siderationsof Sec.2 will a

-3 < p < -2,

negative values of p, with In other words, (3.1) can be used to for all real p. can be used to de\357\254\201ne I\342\200\230(p)

de\357\254\201ne for I\342\200\230(p)

1,-2, . ...

Bessel

4.

the values

we consider

Next, step,

p

by

of

it is to

-p

in (2.8)

for de\357\254\201ne I\342\200\230(p)

Moreover, by

Order)

to

expect

as p

and

natural

that the will

also

+ 1)\342\200\230 I\342\200\230(mfI-\342\200\2241l))\302\243\"(i/:):+r:zm

for

to

- 1,

G\342\200\234))

= 0, 1,2, . . .,p the p + quantity p and m m zero. and value + values the + Therefore, 1) p I\342\200\230( negative terms in the series (4.1) these values of m, and the corresponding to be zero. Thus, for integral p)

that for

integral

1 takes

are taken

con-

lead

gives)

=

,2,

by

for

=

above.)

Negative

- p as well

step

Thus,

I\342\200\230(p)oo

\302\260\302\260

(-1>m(

_\"

J-P0\342\200\230)

2:0

P(m +

/2)-\302\273+=m

+ l)l\"(x\342\200\224p

m +

1)\342\200\231))

SEC. 4

AND FOURIER-BESSEL

FUNCTIONS

BESSEL

SERIES)

or,ifwesetm=p+k) \302\260\302\260

_

_ 1) ,, \342\200\230(

(-1>k(x/2>v+='=

J-9\302\260\342\200\230)

=

k +1)

+ +1)I\342\200\230(p

,;oP(k

(4,2))

(\342\200\224 l)PJ,(x).)

is not an integer, then ratio test shows that integer and for all x if p If p

of the

none

The

the

denominators

in

becomes

(4.1)

(4.1) converges for all

series

x

96

0 if

in\357\254\201nite.

p is not an

is an integer [see (4.2)]. is function again called a Bessel of the first kind, this J_,(x) that J.,(x) is direct time is veri\357\254\201ed substitution order It easily by of p. of ( 1.1). We leave this veri\357\254\201cation to the reader.1 a solution actually the formulas For what follows, it is useful to observe that (2.8) and (4.1) can be combined in one formula) function

The

M\") = where the

end

5.

The

the number p can be either of Sec. 2 appliesto the

General

+

\":0 l\"(m(-If

of fractional

J,,(x)

=

C1 and

If p

C2 are arbitrary

2 0 is an dependent,

linearly

solution. solution solution

integer,

and

C1J,,(x)

> 0 is not

function

J,(x)

Thus, if p has the

should be noted

Then,

cannot

vanishes,

is not

an

whereas the

integer,

\342\200\230I\342\200\231 C2J_,,(x),)

constants

then

hence

[cf. (l.2)]. and J_,,(x) are by (4.2) the functions J,(x) in this case, (5.1) doesnot give the general to \357\254\201nd another particular p we have

Therefore, for integral of (1.1), which is linearly independent the is so-called Besselfunction Y,,(x)

in Sec.

integer. i.e., there

form)

of J,,(x). of

the

that for some values of p (e.g., for integral since 2 ceasesto be legitimate when applied to \342\200\224p, the \357\254\201nal formula zero denominators [seee.g.(2.4)]. Nevertheless, p is changed to -p, and in fact gives a solutionof (1.1),asnoted.)) 1 It

given

an

dependent,

CJ_,(x).)

To see this, we observe that [see (2.8) and (4.l)]. J_,(x) becomesin\357\254\201nite general solution of Bessel\342\200\231s equation (1.1)

where

sign.)

Equation)

for x = 0, the

=

made at

Bessel\342\200\231s

Consider \357\254\201rst the case where the number p the functions and J_,,(x) cannot be linearly J,,(x) exist a constant C such that)

y

The remark values of p of either

or negative.

positive case

Solution of

(43)) 1)\342\200\231

This particular

second

kind, to

be)

p), part of the argument it leads to fractions with

(2.8) has

meaning

when

204

BESSEL FUNCTIONS

next section.

in the

discussed

of (1.1)has

FOURIER-BESSBL

AND

Thus,when

For

=

Y,(x)

Second

J;,(x) cot

pr:

C, Y,,(x).) Kind)

second kind is obtainedfrom and C, C2, i.e., we set)

p, the Bessel function choice of the constants

fractional

by a special

(5.1)

C1J,(x) +

of the

Functions

Bessel

general solution

an integer, the

p is

8)

form)

the

y =

6.

CHAP.

SERIES

of the

_

csc p1: =) JP(x)

\342\200\224

J_,,(x)

J-P(x)_

P\"

is indeterminate; in fact, the numerator reduces to vanishes according to (4.2), and the denominator the following question: Can the indeterminacy also vanishes. suggests the limit of the ratio (6.1) as p approaches integral be \342\200\234removed\342\200\235 by taking solution for integral As we and will this limit values, give the required p?

For

(6.1)

p,

integral

\342\200\224

(\342\200\224 1)PJ,(x)

which

J_,,(x) This

now show, the answer is to

According

\"\"0\"

in

the

a\357\254\202irmative.

rule) L\342\200\231Hospital\342\200\231s

'

-

_

\"

(3/3P)[Jp(-V)

\302\260\302\260sP7\302\247 *,-p(-\342\200\231\342\200\230)]

.\342\200\230.\342\200\230.\342\200\230f.\342\200\230. (6/ep)sinpn)

=

cos pace/ap>J.(x)

lim

,.,,.

\342\200\224

v=J.(x>

7:

'(\"

__

aP)Jp(-x)

cos

sin pn

- ca/ap)J..(x>

p1!

\342\200\230

1)\"

_

77('

1)\"

(3/aP)J\342\200\224p(-70]p-n

them series (2.8)and (4.1) in this expression, differentiate set the arbitrary index p equal to the integral respect to p, and then index n. After some manipulations, the details of which we omit (since they) involve special properties of the gamma and are rather tedious), we function

We substitute the

with

obtain)

Y,,(x) =

-

1

\"ff 11: \"\"30

C) on

/2)n+2m

(_l)m(

Ego, where

\342\200\224.

+ 1-2tJ,,(x)(lnJ-g

m!(nJ-cI-m)!

C = 0.57721566490l532- - -

particular,

when Yo(x)

n

=

(1.-...\"'__-1)\342\200\231

m!

n-I-ml

(,2,

(\342\200\231\302\247\342\200\230)\342 m

1

+ I?

,2,

I-5)\342\200\231)

is the so-called Euler\342\200\231s constant.

In

0)

= \302\247:Jo(x) (In;

+

C)) +%+%+~~+

%))) -?=,\302\247.\342\200\230-\342\200\231\303\251-\"(\342\200\231-\302\243)\342\200\231\"'(1

sec. 6

BESSEL

AND FOURIBR-BESSEL

wucnous

saunas

205)

= n of the function Y,,(x) into the equation (1.1) with Substitution p is a shows that of solution the functions }\342\200\231,,(x)actually (1.1). Moreover, since for x = O, J,,(x) has a J,,(x) and Y,,(x) cannot be linearly dependent, \357\254\201nite whereas value, Y,,(x) becomes in\357\254\201nite. Therefore, Y,,(x) is the resolution of of the end Sec. The quired particular (1.1)(cf. 5). graph of the function in Fig. 45.) y = Yo(x) is shown

45)

Fromm

7.

For

Bessel Functionsof

between

Relations

any p, we

have the

formulas

=

[x\302\273J.(x>1

g

=

[x-PJ,(x)]

formulas hold

for the

(7.1))

x\302\273J,.1 oo for

Si\342\200\234 + (\342\200\230

m

+ 3) (\342\200\230

4:.)

315\302\260\342\200\230

as b

limit

the

converges. In o:(b) and hence for

note that the resulting of a \357\254\201nite limit

\342\200\224> co, and

This implies the

integral

improper

b

= ln

at(x)

existence

as

We set)

o:(b).

lim az(b) =

A,

b\342\200\224>oo)

A

where

as we

Just the

since

otherwise

ln oz(x)

= ln A

a\303\251 0,

oo as b\342\200\224> oo.

In oc(b)->

sin (t

+ m

+

(t

Then + 8)

2) 8)tcos

proved the boundedness of the function)

of the

function

1;(x),

boundedness

=

mx CPO\342\200\230)

sin (t

(t + 8)dt. 8)t;:os

+

Then

in oc(x) = In A

+

Eggs

whence)

a(x)

= A

According to Taylor's theorem,

exp [ 00.

(9.15), it

Therefore, we substitute The result is) (9.6).

is useful the

to have

expressions

z\342\200\231=A(l+\302\247%))cos(x+co+3%2)-)

If we

transform

this

we

the function

It equation

follows

way

as

we transformed

(9.14)

= .4

cos (x

+

co)

(9.18))

+\"1(;\342\200\230\342\200\224).

oo.) s(x) is boundedas x\342\200\224>

of Bessel

Zeros

I0.

same

obtain)

2'

where

in the

expression

to (9.15)),

led

[which

Functions and Related Functions)

at once from formula that (9.15) any solution of Bessel\342\200\231s an in\357\254\201nite number of positive zeros and that zeros are these zerosof the function sin (x + co),i.e., to the numbers of the form)

has

close to the

k,,=mI:\342\200\224o),)

where

n is

an integer.

just one zero near each have the same positive

We

k,,. zeros,

now

show

Since, it is

that for

su\357\254\202iciently large

according to (9.1),the to su\357\254\202icient

prove

u, there

functions

this for the

is

y and 2 function))

BESSBL FUNCTIONS

AND

FOURIER-BESSEL

CHAP.

SERIES

8)

arbitrarily large n, there were a pair of zerosof the function 2 near that 2\342\200\231 has a zero theorem k,,, then it would follow from Rolle\342\200\231s the value near 2,. But by (9.18) this is impossible, sincenear 2,, = mt \342\200\224 co, of z\342\200\231 is near A cos mt, provided n is su\357\254\202iciently large. that for Thus, all the the numbers zeros of the function lie near n, k,,, su\357\254\202icientlylarge y and there is only one zero near It follows that the distance between each k,,. the consecutive zeros of the function 1: as the distancefrom y approaches In particular, considerations these origin increases. apply to the functions the numbers k,, have the values) J,(x) and Y,(x), where according to (9.17), 2.

If for

the

point

k,,

___

mt+2

P.L'_\342\200\231_\342\200\230,

4

_

_P_\342\200\231_\342\200\230 1\342\200\230 k,,\342\200\224mc

2+4)

for J,,(x) and Y,,(x), respectively. Below,we shall be concerned with the positive zeros of J,,(x). (It should be noted that by (4.3), the positive and negative zeros of J,(x) are located

and Rolle\342\200\231s with respect to the origin.) It follows from (7.2) there is at least one zero of the function x-PJ,,+1(x) between any two consecutive zeros of the function i.e., there is at least one zero x-PJ,,(x), of the function zeros of the function J,,+1(x) between any two consecutive If we replace p by p + 1 in (7.1), we obtain) J,,(x). symmetrically theorem

that

[x\342\200\235+lJp+1(x)] \342\200\230\302\2432

=

xp+1Jp(x)-)

that there is at From this we conclude as before between any two consecutive zerosof the

least onezeroof the

function

Thus, the zeros of J,,(x) and J,,.,_1(x) \342\200\234separateeach other.\342\200\235More precisely, one and one zero of J,,,,1(x) appears between any two consecutive positive zeros of only in zeros and cannot have any positive J,+,(x) J,(x). Moreover, J,,(x) J,(x)

common. J ,\342\200\231,(x) would

function

In fact, if J,,(x) and J,,+,(x)both vanished also vanish at xo. But this is impossible,

J,,+1(x).

> 0, then by (7.8), since by the uniqueness = 0, equation, J,(xo)

at x0

theorem for solutions of a secondorder differential = 0 would is obviously J ,\342\200\231,(xo) imply that J,,(x) 5 0, which

false.

at J ;(x). Consider next the zeros of the function theorem, By Rolle\342\200\231s of zeros consecutive two least one zero of J;(x) lies between any J,(x). set of positive zeros. like J,,(x), the function J ;(x) has an in\357\254\201nite Therefore, Finally,

we

investigate

the

of

the

\342\200\230zeros

function

\342\200\224 HJ,(x), xJ,\302\247(x)

where

which is often encountered in the applications. constant, a function simultanehave just seen,the functions J,(x) and J;(x) cannot vanish as we pass at once that J,(x) must change sign ously (for x > 0). It follows Let 1,, A2, . . . , A\", . . . denote)) through a value of x for which J,,(x) vanishes.

H is a As we

10

SEC.

the

BESSEL FUNCTIONS

zeros of J,(x) arrangedin does not change sign,

positive

function

of interest to us), J,,(x)is

of p

the

where

As

positive.]

which

\357\254\201rst term,

for p

in fact

positive

for 0

determines

the

For 0

order.

increasing and

J,,(x)

FOURIER-BESSEL

AND

of

J,,(x)

for x

2 | 5)

< 11,the

< x

values

only

(the

< 1,. [Seeformula

< x sign

> -1

SERIES

(4.3),

near zero, is

to negative values, through 7.1,J,,(x) goesfrom positive to from values, etc., so positive negative 7.2, J,,(x) goes

we pass

as we pass through that)

>

<

0, J,\342\200\231,()\\2)

J,\342\200\231,(7.,)O,

<

J,\342\200\231,().;,)0,...

we have

Clearly,

- HJ,(x)1...,=

[xJ;.(x) function

the

Therefore,

= 7.1,A2,

for x

xJ;,(x)

and

J,\342\200\231,(x)

be

\342\200\224 xJ,\342\200\231,(x)HJ,(x)

increases,

just as

prove

fact

this

Let the

the

function

the

vanishes at least once between This shows

each

\342\200\224 xJ,\342\200\231,(x)HJ,,(x)

that

pair also

zeros. between consecutive zeros of the 1: as the distance from

distance

approaches

zeros of J,,(x).

of the

case

we

However,

both

origin shall not

here.)

Parametric

H.

in

negative and positive

is alternately

HJ,(x)

213, . . ..

the

that

shown

\342\200\224

and hence

7.3, . . .,

of consecutive zeros 7.1,7.2, has an in\357\254\201nite set of positive

It can

x.J;,.)

Form of

Bessel\342\200\231s

Equation)

function y(x) be any solution = y()tx), and set Ax = y + t3-dd-3

of t.

+' (:2

t%

Bessel\342\200\231s equation

Consider

(1.1).

Since obviously) \342\200\224

p2)y

= 0,

(11.1))

substituting)

\"_J\342\200\231_liJ1 _

into

(11.1), and

Ax

using

= t,

+

x2%:%

Thus,

if the

y().x) is a

function

solution

y(x) of

the

.431- _

dt2 7. dx\342\200\231

dt

is a

we

xgg

141).\342\200\231 2&3

dx2)

obtain)

+

=

(\357\254\201x? \342\200\224p2)y

0.)

solution of Bessel\342\200\231s equation

(1.1),

the function

equation)

+ 90\" xzy\342\200\235

called the parametric form

+ (73162-

\342\200\231s equation, of Bessel

=

(11-2))

P\342\200\231)?0.

with

parameter

7..))

nasser. nmcrrons

2| 6

I2.

of the

Orthogonality A and

Let

= y

functions

two

be

p.

z =

and J,,()\302\273x)

crap.

saunas

AND_l3OURIBR-BBSSEL

8

FunctionsJ,(>.x)) Consider the

numbers.

nonnegative

> -1.

J,(y.x), where p

Sec.

to

According

functions

11, these

the equations

obey

-

=

+ Wx\342\200\231p\342\200\231)y0. + xy\342\200\231 xzy\342\200\235

x32\" +

\342\200\224

x2\342\200\231 + (pix?

p1)z

= 0

or

P:)

-

=

+ y\342\200\231 -Azxy.) xy\342\200\235 3;)\342\200\231 \342\200\224 = xz\342\200\235 + z\342\200\231 [\302\2432

the

Subtract

\357\254\201rst equation

The result

by y.

\342\200\224p.3xz.)

by z from the secondequation

multiplied

multiplied

is)

-

-

+ 02' x(yz\342\200\2352y\342\200\231)

- Mxyz.)

=

03 2y\342\200\231)

01' \342\200\224 = + 02\342\200\231 (12 2y\342\200\231) x02\342\200\231zy')\342\200\231

so

that

- zy')]'

We now integrate (12.1)from

Sinceby

[0,

=

(12.1)

p.3)xyz.

1, obtaining

-1, the integrand since y =

\342\200\224

M)

J;

xyz dx.

(12.2))

is actually integrable on from it follows

in (12.2)

J,,().x),z = J,(p.x),

In fact,

1].

-

(X2

= 02

zyoljjf,

p >

hypothesis

interval

0 to

\342\200\224

[x

-1.

instead

Therefore,

-

we can

of (12.2),

zy91....1 =

[x(yz\342\200\231

as

0

-

um

I;

write)

xyz

\302\253be.

(12.4)))

sec.

12

We

now

note

other

on the

=

,0\302\273).

2

saunas

I 7)

=

[zlx=.1

Jp(p\342\200\230)9

hand)

.(Ax) =

=

y\342\200\231

g

=

2'

so

AND FOURIBR-BESS!-3L

that)

D'L=1 while

FUNCTIONS

BESSBL

AJ,',(Ax).)

=

5-,

.r.(;\302\273x>

uJ;(;\302\273x>)

that =

=

1 [z\342\200\231].

[.v\342\200\231]x..i7J,',(7~).

becomes

(12.4)

Thus,

uJ,\342\200\231.(t\302\242)-

\342\200\224 \342\200\224 = (A2 M) xJp(>~x)Jp(v~x) A-\342\200\231p(s\302\273)J.\342\200\231.(7~) s\302\273Jp(>~)J.\342\200\231.(s\302\273) L\342\200\230,

So far,

1) A and J,(p.) = 0, A vanishes.

been

p. have

and

A

restrictions on

certain

impose

p.

are

a\303\251 p..

Noting

be

functions also

might

that

z1=

-1/2,

A and

=

(12.6))

=

12 \342\200\230/-;Jp(P-3\342\200\230)) \342\200\230/;\342\200\231Jp(7~x).

in the usual

orthogonal

we

= o.

xJ,,(Ax)J,,(p.x) dx

now

We

three cases: i.e., J,,(A) = 0, J,,(x), left-hand side of (12.5)

absent in the integrand, the functions J,(Ax) and J,,(p.x) the usual the in sense. In the case, we say that present orthogonal we x. Of with and are course, weight orthogonal J,(p.x) J,,(Ax) say that the functions)

It should be noted

where

numbers.

the following

different zeros of the function such values of A and p., the that A3 \342\200\224 75 0. we then obtain) 9.\342\200\231

31 are

nonnegative Consider

(1..

(12.5))

x were

factor

would

arbitrary and

For

J;

If the

A

ax.

p. are

obtain

sense.

according

1/%sinAx,

of the the functions) numbers

21 =

2/1:cosAx,

to (8.1),

22 =

form

22

mt.

=

for p

= 1/2,2,and

z; reduce

to)

A/msinpx,) According

to

(8.2), for

p =

x/2/\342\200\224-nzcos y.x,)

case A and y. are numbers of the form (n + 3})-n:. Thus, in these we case i.e., in the \357\254\201rst functions, special cases, we obtain the trigonometric obtain the functions are sin mcx (except for a factor), which orthogonal which are also on [0, l], and in the second case the functions cos (n + 1-)-n:x, on [0, 1].)) orthogonal where

in this

I)

2 I8

A and

2)

zeros of the

different

are

p.

=

In

are orthogonal

with

let

3) Finally,

J,(Ax)

zeros of the

two different

p. be

and

we

hence

and

and

still)

J,,(;.tx)

x.)

weight A

9\342\200\230 O\302\273 it)-)

side of (12.5)also vanishes, Thus, here again the functions

this case, the left-hand the formula (12.6).

obtain

8)

J ;(x):

function

= 0

0. J;(u)

1,30\302\273)

can.

seams

AND FOURIBR-BESSEL

FUNCTIONS

BBSSEL

function

\342\200\22

xJ,',(x)

HJ,(x):)

AJ;,(A)

\342\200\224

= 0,

-

=

HJ,(A)

s\302\273J,f.(:t)

Multiply

the by

multiplied

by J,,(p.) and

\357\254\201rst equation

The result

J,,(A).

in this

again obtain the with

orthogonal

I3.

weight

-

I

=

7Jp(P-)J,\302\247(7\\)

left-hand

the

second equation

0-)

of (12.5) also vanishes, and the functions J,(Ax) and J,(ux)

we

side,

i.e.,

(12.6),

Integral

p. are different,

A and

from

it

are

x.)

of the

Evaluation

When

case, the formula

subtract

is)

it-7.(7~)J;(P-)

Therefore,

0-)

H-\342\200\231p(:t)

xJ\302\247(Ax)

leads

(12.5)

xJp(P'x)JpO\342\200\230x) (1)

dx)

to)

\342\200\224 itJp(A)J\302\243(u)_)

= u,J;.(x)

dx

_

12) P\342\200\230;

on the right becomes since the A, the fraction indeterminate, p. \342\200\224> numerator and denominator both approach zero. To \342\200\234remove\342\200\235 this A = const and p.\342\200\224> A. we use L\342\200\231Hospital\342\200\231s rule, letting indeterminacy, When

This

gives)

dx = I; x,:(M)

7~J\302\243(:t)J$(>~)

I2

=

-

-

\357\254\201n} II

_

ma)

Jp(A)J\302\243(:t))

Mp(;\357\254\202JZ(u)

_

I J,J..(x)

W)

M,.(x>2J{J,T.(7~) in the

Since

(13.3)) x by

\357\254\201rst replace

A

in

present case J,,(A)

=

\342\200\230 \"-\342\200\231n+1(7~)-

=

pJ,.(7~)

0, we have)

-

=

Jp+1(7~),)

1,30\302\273)

that

xJ\302\247(xx)

f;

2) If A

is a

zero of the

function

dx [0 x.r;(xx) Bounds for

*l4. The

following

and

M

then 2

1

5 (1

inequality, valid for K

O

=

(13.4)

)J;+,(x).

\342\200\224

{-2)

(13.5))

13(1).

the IntegralI (: xJ,\342\200\231,(Ax) dx)

K

Here K >

=

dx

J;,(A),

1

we

(1

obtaining)

(7.8),

so

\342\200\224

then

J,,(A),

1

can

+

conclusions:

[0 xJ;(xx) formula

2

\342\200\231

5 [J,2(x)

Thus, we can draw the following 1) If A is a zero of the function

This

J30),

(1

(13.1) that

from

follows

2

_

> 0

suf\357\254\201cientlylarge

1

< [0 xJ3(>.x)

dx

<

are constants(which

A, will

be

M 7-

can

useful

later)

(14.1) depend

on p).

Obviously)

have) \342\200\230

2

[0 xJp(Ax)

According to the

asymptotic

dx =

formula

1

\"

2

\342\200\230xi [0 :J,,(z)

(9.16)

lJp(t)l <

2A) \342\200\230-/7))

dt.

(14.2)

220

BBSSBL

for

[2

hence

t, and

su\357\254\202icientlylarge

CHAP. 8

seams

AND FOURIER-BBSSBL

FUNCTIONS

M

<

tJ\302\247(t)dt

j2

dt

(M = const).)

m

=

In view

in (14.1). this implies the right-hand of (14.2), inequality formula same the other the On (9.16)) hand, by asymptotic =

tJ\357\254\201(t)

for

large

for

t, i.e.,

A

sin (t

(A

=

A3

sin\342\200\231 (t +

2

A2

sinz

>

t

(z +

[M

co) +

(0)

2Arsin(t + co)

+

t

f

:2)

(L = const)

-9;

A

dt

:J\302\247(z)

92)

But then

10.

A

>

[0 ugmdt

+ co) +

2

L

_

s1n3(t + co) \342\200\224dt)

[M (A2

7)

xx

(K=const,K>

o),)

=A3Kosin3(t+\302\242o)dt\342\200\224L(ln).\342\200\224lnAo)2

and

view

in

I5.

this implies the

of (14.2),

De\357\254\201nition

of

to Sec. 12,the

is devoted

the

arranged

J,().,x),

in increasing

/3

order.

According

..

To give the weight of the system (15.1), in

I 1}

expansions with the positive roots of

of Fourier . . . be

. . . . , J,().,,x),

system on [0, 1],with of the functions appearance

an orthogonal

idea of

1,, 1)

to the study A3, . . ., 7.,,,

functions)

J,()\\1x), form

Let

> -

of (14.1).)

Series)

Fourier-Bessel

The rest of this chapter respect to Besselfunctions. = 0 (p the equation J,,(x)

side

left-hand

t/|(xgX))

Frame 46))

x.

(15.1))

reader some Fig.

46 we)

sec. 15

22!)

seams

AND FOURIER-BESSEL

FUNCTIONS

nassar.

the graphs of the functions J1(X,x), J1(X2x), J1()\\3x) on the interval and more The functions J1(A,,x)(n = 3, 4, . . .) on [0, 1] become of \342\200\234oscillations\342\200\235 increases. more complicated, i.e., the number is absolutely For any function f (x) which integrable on [0, 1],we can form to the the Fourierthe Fourier series with system (15.1), or brie\357\254\202y, respect Bessel series) draw

1].

[0,

f(x) ~ where the

+ czwzx) + - - -. e1J.(x1x>

constants)

dx

xf(x)J,(>~..x)

ff,

xJ\302\247()\\,,x)

K

._.

dx)

dx xf(x)J,,().,,x) $120\") I;

are called the Fourier-Bessel coe\357\254\201icientsof f (x). formal obtained by using the following argument: =

f(x)

We [0,

(15.2)

sides

both

multiply

orthogonality

of (15.4)

that term

1], assuming

by

term

x) of

weight

(with

xf(x)J.(x.x>

J;

+

c1J,(A,x) by

and

(15.4))

justi\357\254\201ed.

the system (15.1),the dx =

over the

integrate

is

integration

be

coe\357\254\202icients can

of (15.2), we write)

Instead + - - -.

cz.I,,()\\2x)

xJ,,(A,,x)

These

(15.3))

result

interval

of the

Because is)

ax.) c. I; xJ:(x.x>

[See equation (l3.4).] actually holds and if the series on the right converges is known to be legitimate, and then term by term integration uniformly, hence the coe\357\254\202icientsc,, must be given by the formula However, (15.3). just as in the case of ordinary orthogonal systems (seeCh.2, Sec.2), we \357\254\201rst use (15.3) to form the series (15.2),and only later examine the convergence of the to f (x).) series which

implies

If the

(15.3).

(15.4)

equality

I6. Criteria for the Convergence

of

We now state without proof the most important vergence of a Fourier-Besselseriesto the function

Thesecriteria

are

of trigonometric

proofs are and

hence

much

function

with

which

criteria

from we are

the con-

for

which it is

formed.

familiar

the

in

case

Fourier series (seeCh. 3, Secs.9 and 12). However, the than in the case of Fourier-Besselseries, more complicated

we omit

THEOREM tinuous

to those

analogous

Series)

Fourier-Bessel

them.)

1. Let

f (x)

on [0,

1].

be a Then

piecewise

the

smooth,

Fourier-Bessel

continuous

series (p

or discon-

2

\342\200\224 \302\253}) of))

222

FUNCTIONS

BESSEL

f (x)

converges for

point

of continuity of f

discontinuity

= 0 if

note

We

values).

in\357\254\201nite for

become

be

the interval series (p 2

on

Fourier-Bessel

[a +

subinterval

3.

THEOREM

0

where

(x) be

Let f

the

every

subinterval

[a

condition

system (15.1)

Remark.

f (1)

= 0 is quite

for

1/xf(x)

(x) converges uniformly

derivative

integrable

condition

the

to formula

p+1

= 0.

f(l) uniformly

converges

on

8 > 0.)

all the

since

natural,

functions

of the

3, it

is

suf\357\254\201cient to

require

= x? (p 2

\342\200\224

only absolute

.)

J,,(A2x), .

. .,

-}) for

0 <

. .. J,,().,,x).

dx

J1 0 xP+1J,().,,x)

1,2, . . .).

(n =

But we have)

I1ox

every

[0, 1], let f (x) be on the interval

on

integrable

\342\200\224\342\200\224-2\342\200\224\342\200\224 J%+lQn)

the)

on

(15.3) c,, =

According

integrable Then,

x = 1.)

J,(7qx), formula

1.

< b s

< a

the function f (x) Expand series with respect to the system)

Example. Fourier-Bessel

By

where

In Theorems 2 and of

integrability

every

point of

8 > 0.)

absolutely

1],

+ 8,

vanish

0

1, and let f(x) satisfy series (p 2 - i) off (x)

a <

s

of f

an absolutely

Fourier-Bessel

Then

-})

an absolutely

have

and where

\342\200\224

8, b \342\200\224 8], where have

and

The

(x) at

every

piecewise

be continuous [a, b],

Let f (x)

THEOREM 2.

[a, 1],

at

piecewise

require \342\200\224 8 [8, 1 8] where

continuous

equals f 0)]

of f (x) on [0, 1], it is smoothness subinterval smoothness of f (x) on every > 0, in addition to the:requirement that f (x) [or even) [0, 1].) absolutely integrable on the whole interval

suf\357\254\201cient to

derivative

sum

its

Moreover,

J}[f(x+ 0) + f(x -

to zero to zero for x = 1,and converges converges of the system vanish for these (since all the functions of the system (15.1) that for p < 0 all the functions it is meaningless to talk of x = 0 [see (4.l)], so that series at x = 0.)

convergenceof the Remark. Instead of

1/:_cf(x)]

8)

CI-IAP.

(x).

> 0

p

1.

00))

{5)

therefore)

n, and

large

2

1imc\342\200\224'%=0

II-twin) 01')

= o. n-no

Even

more

(17.2))

7\302\273,)

said: It follows from

can be

the

formula

asymptotic

(9.16)

that)

IJ,,(x)| all

for

x, and

su\357\254\202icientlylarge

n is su\357\254\202icientlylarge.

Then,

2.4 -\342\200\230/:9

x > hence for every \357\254\201xed

IJ.I

if

<

0)

(17.3))

<

%

using (17.2), we obtain =

0

(17.4))

n1\302\247_Ig|c.Jp(A..x)|

Thus, if f (x) is square converges of f (x) always this will also be the case for x obviously for x = 0). system (15.1) vanish Finally,

for every \357\254\201xed x > 0. the then

the

Fourier-Bessel

I; or by

dx =

xf(x)J,(>~.x)

c..J;

the general

integrable,

= it

term of

(x > 0). If p 0 (since all the functions from (15.3) that) follows to zero

series

> 0, of

dx.

xJ3(A.x)

(14.1)

I for sufliciently

,1,

large n, so that

by

ff, u1_i\302\247I;



where

0 and

c are constants,

and

the series

then

czJ,(7.2x)+ - - - +

c,J,,(7.1x) +

convergesabsolutely

(13.1))

on [0,

uniformly

c,,J,(7.,,x)

+ ---

(18.2)

1].)

Proof. For p 2 0,the function J,,(x) is bounded in the neighborhood = 0. By the asymptotic formula (9.16), J,(x) is bounded for large x, i.e.,) Therefore, J,,(x) is bounded for all (positive)

of x x.

|c,.1p(7\302\273.x)|

so

that

<

=

(L

|\302\242..|L

\302\260\302\260nSt).

by (18.1))

L

Ic.J.(x.x>I-n:

But

since

n >

m (where m

is some

Sec. 9),

oo (see n\342\200\224>

n

is large,

follows

1,, 2

-5n

and)

L

(18.3))

<

7\302\273. zgto)

for

Consequently,

inequality

is the

have

large n we

<

|c,,J,,()\\,,x)|

Theorem 1 now

follows

for

that

(h=const).

).,,>7.,,,+(n-m)=n+h

Therefore, if

it

\357\254\201xed number),)

from

general term

H 2;; the fact

(H = const).) that

of a convergent

the

right-hand numerical

side of series.))

this

226

\"

Ifp 2

THEOREM 2.

\342\200\224\302\247andif

<

|c,,|

for all su\357\254\202icientlylarge

czx/3:J,,(7.2x)

+

and uniformly on (4.2) converges absolutely 1], where 8 > 0.)

interval

[8,

p 2 -15, the function (4.3)]. According to

For

Proof.

equation

1/:'cJ,,(x) is

bounded for

(positive) x, and hencefor

x.

large

It follows

and

x in

any

const))

L

<

(18.6)

7.;

that ,, L

<

L

-51;;

(18.3) \342\200\224

(H = const).

H

J,,().,,x)|

|c,,\\/x

of

0 x\342\200\224>

(9.16),)

[0, 1] we have)

|c,,\\/.7cJ,()\\,,x)| <

ical

as

sub-

J,,(x) is bounded for all

(L =

< L

on every

formula

asymptotic

series

- - - (18.5))

is bounded

1/:_cJ,,(x) the

the

This implies

[0, 1].

uniformly

Therefore,

|x/3cJ,,(A,,x)|

The

interval

the

then

constants,

c,,\\/3-:J,,(A,,x) +

- - - +

i.e.,) IX/K,J,,(A,,x)I

or by

c are

series

the

[see

0 and

absolutely

converges that

(18.4))

e >

n, where

J,,(7.,x) +

cn/3:

8 emu\302\273.

SERIES AND l-\342\200\230OURIBR-BESSEL

FUNCTIONS

EEssEL

s W

(18.7))

term of a convergent numerright-hand side of (18.7)is the general This proves the \357\254\201rst theorem. The secondpart) of the part the theorem is a consequenceof the following obtained inequality, series.

from (18.7):)

THEOREM

3.

H

<

Ic.J.(A.x>I

W, \342\200\230,3; Ifp

>

-1

H

<

(8

< x

< 1).)

\342\200\230,\302\247n,,,

and_if)

C

lcnl <

\342\200\231)

for

all

(18.2) 8 >

su\357\254\202icientlylarge

converges 0.))

n, where

absolutely and

e >

0 and

uniformly

c are on

constants, then interval [8,

every

the

series

1], where

SEC. 18

FUNCTIONS

BESSEL

Let

Proof.

8 <

1.

x <

By

the

AND FOURIER-BESSEL

formula

asymptotic

SERIES)

(9.16),)

lJp(x)l g) all su\357\254\202iciently large x, i.e., for x 2 x0. Moreover, 1,8 2 x0 holds for all su\357\254\202icientlylarge n. Therefore hence for 8 < x < 1) 7.,,x 2 x0 a fortiori, and

for

the

if x

inequality

2 8, we

have

2A

2A

< \\ I-\342\200\231p(7\302\273.x)| \342\200\230mt\342\200\231 \"7:-5)

so that

2.4

,,

lc..J,(A..x)|g 7-5

or 1

2Ac

< 'cnJPO\342\200\230nx)|

By

(18.3),

7?

this implies)

H

(H =

|c,,J,(7.,,x)|< 3;; for

all

su\357\254\202iciently large

the right-hand side numerical series.) we

x.

of this

Remark. In the conditions(18.1)and write simply n instead of 1,,[in view

I

The

2.

the

1, 2, and

3,

and

A +...

_1(\342\200\231T;x_)

on

[0, 1],

since here p

= l,

e

=

1,

series)

J_l/4(1lx)+ absolutely

of Theorems

that

of (l8.3)].)

J

A

.1_(2.22;x.).+...+

is absolutely and uniformly convergent and hence Theorem 1 is applicable.

while

(18.4)

fact

the

a convergent

The series J1()\342\200\230lx)+

Example

const))

Theorem 3 now follows from of series is the general term

can

Example 1.

is

33;\342\200\230

uniformly

)+...+ convergent

on every

+...) interval

[8,

1],

series)

+

\302\273\\/;;_]_l/4()\342\200\230lx)

+...+

II))

+...)

8 >

0,

Theorem 2

convergent on the whole

and uniformly

is absolutely

3. The

and

absolutely

Here Theorem

3 is

1].

Here

a Twice

+...)

+...+ on every

convergent

uniformly

interval

[8,

1],

8 >

0.

applicable.)

The Order of of

[0,

series

]_3/4(1lx)+

*l9.

interval

8)

is applicable.)

Example

is

CHAP.

SERIES

AND FOURIER-BESSEL

BESSEL FUNCTIONS

Coe\357\254\201icients Magnitudeof the Fourier-Bessel

Function)

Differentiable

on the F (x) be a twice Let di\357\254\202erentiable function de\357\254\201ned = 0, that F(l) = 0, and such [0, 1], such that F(0) = F\342\200\231(0) at is bounded (the secondderivative not exist certain may points). if A is a zero of the function J,,(x), where p > -1, the inequality)

LEMMA.

interval F\"(x) Then,

L1)

x/3: F(x)J,,().x) dx

<

(R

X15-Q,-5

= const)

(19.1))

holds\342\200\230.

According

Proof.

to equations

= V?

z(t) satis\357\254\201es the

J,(t))

P2 \342\200\224

t:

(1

set

function

equation) + z\342\200\235(t)

If we

(9.1) and (9.2),the

t =

xx,

%)

z(t) =

0.)

then)

,

ldz

\342\200\230='xz;\342\200\231

,,

ldzz

\342\200\230(\342\200\230>-X57372\342\200\231

sothat

ldzz

3-}

m;2+(\342\200\230

_

\302\260

ma)\342\200\231

01')

3%

+

(A2

z =

\342\200\230\"2 \342\200\224 \342\200\230Z\342\200\231

i)

o.

(19.2)))

19

sec.

the

Thus,

from V Ax

2 =

1/5

also

\\/3cJ,()\302\273x)

-1-

I =

I;

since

factor. obtaining)

_

Z

x2

A2

229)

saunas

But (19.2). it differs)

equation

satis\357\254\201es (19.2),

by a constant set p2 \342\200\224 i = m in (19.2) Z =

It follows

satis\357\254\201es the

J,,(Ax)

only

J,,0x)

now

We

2 =

function

function

the

then

AND rounmn-nassu

FUNCTIONS

BESSBL

z\342\200\235).

that)

1/3:F(x)J,,(hx)dx =

=

dx

F(x)z

Ll)

53-2

2

L1) F(x)(%

\342\200\224dx.)

2\")

Since \342\200\224 = F\"z \342\200\224 (F\342\200\231z Fz\342\200\231)\342\200\231 Fz\342\200\235,

we

have 1

1

II \342\200\224 F \357\254\201fo \342\200\234F(x)-32-; (x))z

I=

=

I

1

i5

m

+ (F

I! \342\200\224 F

0 (F(x)?

dx

172)]

+ [F

(x))zdx

I I \342\200\224

I

I

I x=l \342\200\224

F213,.

But) - F2\342\200\231= IF\342\200\231! 1526

in view of the

-

[F'(1)z(1)

2)

-

[F\342\200\231(0)z(0)

=

F(0)z\342\200\231(0)l

=

= o and

1,0) and

Pa) is \357\254\201nite;

is \357\254\201nite; z\342\200\231(l)

= 3) (0 < 0 < By F\342\200\231(x)xF\342\200\235(0x) Taylor\342\200\231s formula, = = 2 1/3': J,(Ax)

(4.3)].

[see equation

=

formula, Taylor\342\200\231s

=

series

power

xP+(3/2)

=f\"*\342\200\234\"(0)

=f\342\200\231(l)

Then the following

these

with

Coe\357\254\201icients of Magnitude of the Fourier-Bessel Which is Differentiable Several Times)

1. Let

THEOREM that

and

18.)

Sec.

of

twice

=

Remark. If we use Theorem 2 of Sec. 18, then on f (x) and with p 2 \342\200\224 we obtain absolute and \302\247, of the series (18.5) on the whole interval [0, 1].)

\"20.

and

0, f (1) = 0, and \342\200\231(0) not exist at certain points).

(0) = f

0.)

and Theorem3 of Sec.18. whole

Sec. 16is a

is continuous

which

second derivative may series of f (x) converges 1] where 0 < 8 if p

Fourier-Bessel

on every subinterval interval [0, 1] If p Proof.

be a function [0, 1], let

f (x)

be bounded (the

the

Then,

2 of

Theorem

1:)

the interval

on

dzjferentiable let f \"(x)

to F(x).) supplementing

Theorem

of

true if we impose the requirements of) = \\/:6 f (x) instead of on f (x), since

F(x)

lemma

proposition

following

consequence

function

the

applied

actually

(19.5).)

inequality

1 remains

Theorem

Remark. the

SERIES)

not

exist

at certain

points);

=

0=f\"-\342\200\234\342\200\235(1) is

by satis\357\254\201ed

the Fourier-Bessel

coe\357\254\202icients

of f (x):) C)

lC,,|

S

-Kim

= const).

(20.1)))

BESSBL FUNCTIONS

to see that the function F(x) = 1/} f (x) also of the theorem. In particular, F(x) satis\357\254\201es the the lemma of Sec.19and hence satis\357\254\201es i.e.,) (19.3),

conditions

satis\357\254\201es the

of

conditions

dx = J; x:r

1)

o(\302\247F\342\200\224 F\342\200\235)zdx,) =5\342\200\230-3

\\/3cJ,()\\,,x . then we

last integral,

the

in

1

dx)

\302\253/3:

I;

denotes

the

function

in

have)

1

1

I =

If F,

Flzdx.)

\342\200\230[0 \342\200\230E

the function

Since

F,

of the lemma, this

the conditions

satis\357\254\201es all

time

gives)

(19.3)

1 I =

1

T: Io

we have

where

F22

dx,

written

n=\302\247n\342\200\224m)

> 2, then F3 again satis\357\254\201es the conditions of the lemma, and the s repeat argument times, \357\254\201nally obtaining)

If s

we can

1

1

I=:'2-;J\342\200\230oF_,Zdx,)

where)

m)

is a

bounded

function.

follows

that)

It

];F,z

dx|

< L]; |z| dx

(L

=

const).

By (19.4)

dx < x/\342\200\224M/A, I; |z| so

that

we

have)

\"'

Lx/Tl? \342\200\230))

(M

= const),)

in

fact

8)

20

sec.

\302\253FUNCTIONS AND

BESSEL

FOURIER-BESSEL

233

SERIES

But

I 0\" l

| f;xf(x)Jp(1..x) _\342\200\224 I

L:

dx| ,)

xJ\302\247(A,,x)

dxl

since)

and

1

dx [0 xJ,2,().,,x) (14.1) we

to equation

according

2

K

(K > 0),)

E

obtain) \357\254\201nally

Lx/I11) < |\342\200\230\302\273| \342\200\230T $2.-\342\200\224 n(x). Proof.

-1,

(In

uniformly;)

1)

this

the

(20.3)

inequality

is no

there

case,

In Sec. 18 (seethe J,(x) is bounded for p

proof

holds for

the

inequality

for

p >

-1,

(18.6), we it

follows

need

x<

1)

Hence,

<

the

-

(20.1)

apply

(L

x<

1)

inequality L

since

to get

the

that

(20.2) is an satis\357\254\201es

J,(7.,,x)

(20.3).

Finally,

formula (9.16)that)

the asymptotic

from

|J,(A,,x)I

for every x (0 < (20.3).))

only

1), we saw

of Theorem

2 0. immediate consequenceof (20.1). Forp 2

function

each x (0 <

in x.))

uniformity

= const),

(20.4))

-\342\200\230%C

and

n >

n(x).

Again

using

(20.1),

we obtain

234

FUNCTIONS

BESSEL

*2|. Term by Given a

AND

Term

of

Differentiation

=

for the

conditions \357\254\201nd su\357\254\202icient

from

formula

2

(7.8)

=

> -1,

that p

equality

fl c.1.J;(2.x>.

- 71..xJ,.+1(k..x)|

|pJ,(>~.x)

< assume

of the

(21.2)

that)

|1..xJ;.(1..x)|

We

(211)

validity

(c..J.'=

=

f\342\200\231(x)

follows

Series)

c.J..

\"Z1

It

Fourier-Bessel

8

Fourier-Besselseries) \357\254\201x)

we now

cmp.

seams

rounmn-nnssm.

+

In-7p(l..x)|

so that

+

p

0. Therefore,the

l >

(21.3))

|>~.x\342\200\224\342\200\231p+1(>~..x)|-

quantity

'1/7?:-7\342\200\230 -7p+1(7\\nx)|

by the

is bounded,

where

asymptotic formula

(9.16),

+ I pJ,(7.,,x)| |).,,xJ;(7\\,,x)|\342\202\254 consider only values of x such

we

hence)

and

(H = const),

H

\\/7 that

1. If

x <

0 <

(21.4))

C |C,,|

where s

> 0 and

C are

constants, .

\342\202\254

9

then for x 012

<

>

0)

J 0~..x)

+

CH

|cn)\342\200\230nJp()\342\200\230nx)| 2)+\302\242 \342\200\230Ly.\342\200\231 x']'+\302\242x\342\200\231) 19

or

by (20.4))

+)

<

Ic.2.J;(x.x)I

(3535

CH

A

7\\,',+\302\242x')

[see (18.3)]that the series (21.2) converges for 0 < x < l on every subinterval [8, 1] where 0 < 8 < l. The uniformly converges last fact implies the validity of (21.2) for 0 < x < 1. As for the validity of (21.2) for x = 0, if p < 1, p a\303\251 0, it is easy to see for x = 0, since J,,(x) = x\342\200\231.,,x) where H. A remark 1 of Sec. 16 also applies to Theorem

there are also two

of

orthogonality

Thus,

the

simplicity,

\342\200\234additional\342\200\2

we

restrict

made after the statement of the present theorem, and moreover, theorems completely analogous to Theorems 2 and 3 of condition p > H in both cases and the supplementary condition f (1) = 0 in the second case. (The remark made Theorems 2 and 3 of Sec. 16remains in force.))

Make a seriesexpansion system)

like that

of the

function f

(x) =

x!\342\200\231 (0 <

x <

1)

to the

J,,(7.,x), J,(7.2x),. . ., J,().,,x), where

< H, the

for

1,, are the

roots of the

equation

. . .,

= 0 J ,\342\200\231,(x) (p

(22.6)

>

0).))

23

sec.

BESSEL

By formula

SERIES

239)

(22.5)) 1

21,2,

and

AND FOURIER-BESSEL

FUNCTIONS

by (16.1))

f (1)xP+1J,(A,,x) by the

Therefore,

\302\243\342\200\224nJ,+,(A,,).

theorem given above, we can write)

ask whether since for p = 0)

now

We

negative,

dx =

this

holds

expansion

=

Jp-I-l()\342\200\230n J10\302\273:

=

for p

= 0.

The

answer

is

=

0) -J(\342\200\231)0\342\200\230n)

used equation (7.8)]. Thus, the side of (22.7) right-hand side is f(x) = x0 = 1. The reason why left-hand the = = = H fails for 0 In our case that H. is the so 0, expansion p following: p must the be Then, as remarked above, the system (22.6) supplemented by = x0 = 1. If we denote the function to this x\342\200\235 coef\357\254\201cientcorresponding we have

[where

is zero, while

function

by

the

co, we

obtain)

= co

sincef(x) =

1.

Moreover,

c,,

dx) J: xf(x)-l I x-13 dx I0

= 0 if n

_.

= 1, 2, . .

..

Therefore,

instead

of

(22.7) we obtain the expansion)

l=l+0+0+---,) which

is obviously

*23.

Extension Series

valid!)

of the Results of Secs. |7\342\200\2242|to of the Second Type)

Fourier-Bessel

of Secs. 17and 18, we essentially made no use whatsoever that the numbers 1,,are the roots of the equation J,,(x) = 0. assumption series of the these are equally true for Fourier-Bessel theorems Therefore, second type. However,in the lemma of Sec. 19, we did use the condition = 0 to prove the formula we now need a new lemma, Thus, (19.3). J,,(A) which we proceed to prove.)) In the

of the

theorems

240

that

=

F(0)

di\357\254\202erentiable function \342\200\224 + \302\247)F(l) F\342\200\231(1)(H

(the second

a root

if A is

of the

derivative

p >

-1,

it

arrive 1

1:57

at the

=

HJ,(x

<

0,)

(R =

1%

dx|

(23.1))

const).

1) \342\200\224 zdx o)

(F2; x)

[F '(1)Z(1) -

+

\342\200\224 [F\342\200\231zFz\342\200\231]\"\342\200\234 x==0)

F\342\200\235)

19,

that

that

equality)

as in the lemma of Sec. the last term vanishes.

just

[0, 1],

and such

at certain points).

not exist

may

\342\200\224

F(x)J,().x)

\302\2431/.7

We

0,

that)

follows

I=)

Proof.

on de\357\254\201ned =

8)

equation)

xJ;(x) where

CHAP.

SERIES

twice

=

F\342\200\231(0)0,

is bounded F\342\200\235(x) Then,

a

F(x) be

Let

LEMMA.

such

AND FOURIER-BBSSEL

FUNCTIONS

BESSBL

whole

the

and

F(1)z'(l)l-

term We begin by calculating the \357\254\201rst

reduces to showing

problem

This last term

difference)

is the

-

[F'(0)Z(0)

(23-2))

F(0)z'(0)l.

Since 2

in brackets.

=

,

\\/:_cJ,().x

we have)

2(1) =

-\342\200\231p(7\\).

=

+

w.=1..,)

+ >J;,m =

= we

where

\357\254\201rst term

used the condition in brackets is just) have

-

term in

second

follows

proof The

by the

vanishes

which

lemma

brackets

by just

the

just proved

HJ,0)

the

But

then

leads to the following

The fact that indeed the rest of

lemma.

and vanishes, as in the method

(23.2)

same

= 0.

\342\200\224

the

%)F(l)l-7p(?\\).)

of

hypothesis in

)J;,(7t)

(H +

[F\342\200\231(1)

(H + 91.0),)

the the

lemma of Sec.19.)

result:)

on the Let f (x) be a twice differentiable function de\357\254\201ned = = = such) and such that O, f \342\200\231(1) 0,5 f \342\200\231(0) Hf(l) [0, 1], f(0)

THEOREM 1.

interval 5

The

naturally

\342\200\224 f\342\200\231(l) Hf (1) the applications.))

condition in

= 0 appears

arti\357\254\201cial at this

point, but

it

arises

quite

sec. 24 that f

is bounded \342\200\235(x)

Then

the

(the second

derivative

at certain points).

not exist

may

24!)

saunas

inequality)

IC,,| S is

AND rounmn-nessan

FUNCTIONS

BESSBL

Fourier

by the

obeyed

C

=

W

const),)

f (x)

coe\357\254\201icients of

with

respect

to the

system

=

and

(22.3).

Proof. If F(x) = 1/;f (x), bounded.

F\"(x) is

then

F(0) =

obviously

F\342\200\231(0)0

Moreover)

Fee) =

+

~/:'cf'(x>.

{}\342\200\230-\342\200\224f;

so that -

(H +

F\342\200\231(1)

~i)F(1)

= lf(1) Thus, the

lemma

dx

formula

(H +

be applied to the

can

L1)xf(x)J(A,,x)

Since by

-

+f\342\200\231(1)

I;

1 and

THEOREM

[0, 1],

such

f \"(x) is Then

the

lutely

and

and

that

const).)

2

(K

>

0),)

T\"

= f

twice

=

0, \342\200\231(0)

dt\357\254\202erentiable function \342\200\224 = f\342\200\231(1)Hf(l)

secondderivative may series of f (x) of the

not

[0, 1]if p 2 0.5) 1 and 2 of Sec.21apply

exist at certain

second

type

points). abso-

converges

1,ifp > -1

< 8<

where0

[8, 1],

on the interval 0 and such that

whole interval

Theorems second

equally

well to

Fourier-Bessel

type.)

Fourier-Bessel

Expansions

of Functions

De\357\254\201ned

on

the

[0,1])

Let f (x) be an [0, I], and set x = It need

dx

on every subinterval

uniformly

Finally

Interval

(R =

%

of Sec. 18imply)

f (x) be a

f(0)

Fourier-Bessel

series of the 24.

<

is)

account of (22.5).

the results

2. Let

bounded (the

on the

dx

= 0.

Hf(1)

the result

F(x);

K xJ\302\247(A,,x)

(23.3), taking

Theorem

function

We F(x)J(A,,x)

1

obtain

-

=f\342\200\231(1)

(14.1))

Jo we

l)f(1)

integrable

absolutely or

t =

x/I.

5 If the conditionp > H appearing not have 1\342\200\230 (x) as its sum.))

function

Then the function in the

on de\357\254\201ned

= f(l 0

10.

the function

Expand

x <

x\342\200\230? (0 <

1) in

series

Fourier-Bessel

of the

\357\254\201rst

kind.

the function

11. Expand the

x!\342\200\231 (0 <

x <

1) in

series with

Fourier-Bessel

respect to

system \302\260 - \302\260 9 Jp()\342\200\230lx)9 Jp()\342\200\2302x)9

are the roots Use the example

Hint.

12. Expand

= 3.

Use

Hint.

13. Let

function

the

with p

kind,

of the

the 1,,

where

1,,

A2,

. . . be

\342\200\235+R\302\242'+Q\302\243_1\342\200\235__

,.(x>. 0)

term

by

twice

term

with respect

to x and

8214

83a

Bu

Qu--5t\342\200\2242) P-'\342\200\224+R5;+ ax? \302\260\302\260 \342\200\224

8711,,

+

Page}?

\302\260\302\260

3a,, Rngo\357\254\201

''\302\260

''\302\260 \342\200\2248211,,

+

Qngoun

go?)

\302\260\302\260

=

0214,,

8n,

6 8x) \"575)) 07ru,,)_ Z(P\342\200\224x\342\200\2242+R\342\200\224\342\200\224+Qu,,\342\200\224) \"=0

me

248

METHOD BIGBN1-\342\200\230UNCTION

can-.

APPLICATIONS

rrs

AND

9)

term in parentheses in the last sum'vanishes [because u,, is a of (l.1)], the entire sum vanishes, which means that the function is a solution of (1.1). Moreover, each term of the series (1.12) since (1.12) conditions (1.3), the sum of the series, i.e., the function satis\357\254\201es the boundary each

Since

solution

u,

also

by

conditions.

satis\357\254\201es these

We must

now

expressions

t).

u,,(x,

This can be achieved

conditions (1.4).

values of the

choosing the for the functions

suitably

initial

the

satisfy

constants

A,, and

With this

in

mind,

B,, appearing in the we require that the

relations)

0) =

u(x.

f(x) =

i

..(x>T.(o).

n=0) (1.13) =

=

so)

3\342\200\224\"

that the functions f (x) and g(x) can to the eigenfunctions (1.10). The possimaking such expansionscan be proved under rather broad conditions

which is

hold,

be expanded of

bility

on the

\302\247o..(x>T:.(o))

in

to

equivalent

requiring

with respect

series

coe\357\254\202icients in

(1.1) and on the

equation

which

functions

are

to be

expanded. Thus let) =

f(x)

c...(x>.)

3,0

g(x) = we need

Then

c...(x>.

Z0

only set

: in

order

to

(1.14)

\357\254\201nd A,, and

B,,.

we.

(n

=

Hence, if (1.11) holds, we

0) =

f(x) =

in

\302\260\342\200\231

=

Mg;

=

we

$0

(1.15)

..),

0, 1,2,.

have

A.,..

B..~/T...(x>.

so that

A,, =

C,,, B,,=

c,, (n

_

\342\200\224

0,1,

(1.16))

...).

2,

VA.\"

Our results

are

based

and can be differentiated

fore, the

coe\357\254\202icientsA,,

on the supposition by term twice

term

and

B,, just

found

that with

must

the

series

respect

be such

(1.12) converges

and t. Thereas to guarantee that)) to x

me

sec.

1

this

is the

do

often converge

make

case. However,in this

have

not

that

know

We

order to avoid

words

functions.

boundary

concerning

By the conditions

in order.

are

the coefficients A,, and B, problems, Whether or not the series (1.12)will in Sec. 7. In the meantime, discussed we

de\357\254\201ne discontinuous

confusion,a few

conditions

initial

be

often

series

249)

actual

property.

case will remarks:

in such a the following

AND 113 APPLICATIONS

METHOD

EIGENFUNCTION

(1.3) and

Hence,

in

conditions and (1.4),

we mean

more precisely) at

lim .

Y 1

\"(x, 0

3},

of (1.3),

instead

t) +

u(x,

(3

o,)

.

8u(x, t) = 0 + 8 l \342\200\224\342\200\224-\342\200\224 .13}, 3x)

t)

=

lim

f(x),

=

g(x)

3\342\200\230i\302\247-\"*\342\200\224\342\200\230) 1 :\342\200\224>o

:\342\200\224>0

In other words, in as etc., are to be interpreted

of (1.4).

0u(a, t)/ax,

3\342\200\230$\"*\342\200\224\342\200\2303 =

and lim u(x,

instead

lim

a) X\342\200\224>d

(1.3)

and

the limits

the values

(1.4),

to

which

u(x,

of

u(a,

t),

t), au(x, t)/Bx, b, t > 0 con-

etc., convergeas the point (x, t) lying in the region a < x < It is quite clear that such verges to the correspondingboundary point. only an interpretation of the boundary and initial conditions can correspond to the physical of the problem. In the same way , when we say content that in the region the function a < J: < b, t 2 0, we u(x, t) is continuous mean that u(x, t) is continuous in the region a < x < b, t > 0 and that the limit

lim

u(x,

t)

(a <

x < b,

t >

0))

x\342\200\230Vxo (\342\200\230\"510)

a

has

the

for every point (xo, to) on the boundary of the region} to show that the boundary conditions vary continuously as to) is moved along the boundary.

\357\254\201nite value

Then, it point

is easy (xo,

similar by the solution of equation (1.1),or of some equasolution which is mean a continuous in the sense just always It is easy to see that if such a solution is to exist, then the bound-

Subsequently,

tion, we shall

indicated. ary conditions and the

and (b, 0) in continuity.

t) = 0,

such Thus,

initial

conditions

a way that the boundary for example, if the

must

at the \342\200\234agree\342\200\235 values

do

points not undergo

conditions (1.3)

have

the

(a, 0) a disform

u(0, u(l, t) 0, and the conditions (1.4) have the form u(x, 0) = x + 3u(x, 0)/at = x2, then it is clear that the boundary values a undergo discontinuity at the points (0, 0) and (1, 0), so that the problem certainly cannot have a continuous solution.) =

1,

2

a jump

The analytic expression on the boundary.))

for u(x,

t) may

not be a continuous

function,

i.e., may

undergo

THE mcmruncnou

250

Usual

The

2.

AND rrs APPLICATIONS

METHOD

of the

Statement

9) emu\302\273.

BoundaryValue

Problem)

P in equation (1.1) does not vanish. that the function nor gains eigenvalues and eigenloses equation (1.7) neither function. We now if we multiply all its terms by a nonvanishing we can transform out such a multiplication, (1.7) into by carrying

shall assume

We

Obviously, functions

show that the form)

=

+ q

> 0, p\342\200\231 is continuous

+ rQIl> =

of

of

0.

\342\200\224).r

set)

and

q = according

Then,

(2.3)

rQ.

to (2.2), =

\342\200\224w9) p\302\242\342\200\235+p\342\200\231d)\342\200\231

is just

which The

with

(2.1),

are still

3. The We

basic

(real

given

by

posed

the requirements same formulas

coe\357\254\202icientssatisfying

conditions

boundary

the desired equation (2.1). value problem is usually

boundary

the

equation of the form just stated. The boundary (1.9).) for an

Existenceof Eigenvalues)

shall

not give

value

a completeproof of the

problem

under

existence

consideration;

idea of such a proof. Thus, or complex) and \357\254\201nd a solution =

3.

in equation

satisfying

of eigenvalues

for the

instead we just describethe value (2.1) we give 1.a \357\254\201xed the = -a\

conditions)

sec. 3

rm:

i.e.,

(3.1))

boundary conditions (1.9). (The prime respect to x.) As A changes, d>(x, A) also continues to satisfy the condition (3.1). Thus, a

with

differentiation

changes, but

= 0,

A) [3\342\200\231(a,

\357\254\201rst of the

the A) satis\357\254\201es

(x,

denotes

25|)

Obviously

+

oc(x, A).

denote

We

monnruncrron

nevertheless

of x and the parameterA satis\357\254\201es the \357\254\201rst of the conditions it is shown that (In the theory of differential equations, of a power series in A, and hence is d>(x, A) can be represented in the form an analytic function of A for all values of A.) We now form the function) function

known

(1.9) for

A.

any

A) +

y(x,

(3.2)

8\342\200\231(x, A)

and set

D(A) = a

D(A) is

D(A) =

is

an

obviously

and (3.3)

met.

Thus,

variable

of one

function

known

y(b,

of our

eigenvalue

are satis\357\254\201edsimultaneously, the

problem

of the roots of D(A).

By

has an in\357\254\201nite set of sequenceof the form)

A) +

y(b,

A) +

8\342\200\231(b, A). A,

and

A

= 0

existence of eigenvalues

using

this fact,

it

can

which

of A, (3.1)

values

such

both of the

i.e.,

of the

for

(3.3))

8\342\200\231(b, A)

problem, since for

real eigenvalues

value of

every

conditions (1.9) are to a study that the problem can be written as a reduces

be shown

(see Sec.4), which

)\\o\342\200\230P\" [\342\200\230I\" \"I\342\200\231],,_.,, d>\"I\342\200\231],,.,,, 0.

The

Proof.

on and do not vanish [3, which satisfy the homogeneous system)

numbers

to (1.5),

according

the

contradicting one eigenfunction

eigenvalue.)

Ltd\302\273)

12 dx

by

what

has just

\342\200\224 a.(x\302\2731= dx)

a.(x>12

be approximated form

<

s.)

in the

mean

F(x)/V

r(x)

(6.8).

Then the

function.

dx

function

been proved, there

exists

a 0, series of the form (7.1),where)

of the boundary theorem and

x<

from (7.4)

follows

value

function

the

. .)

1, 2,. Theorem

by

(7.4))

problem. 2 of

t) can

u(x,

Sec. 6, for be expandedin a

0, 1,2,. . .),

(n =

ru(x, t),,(x)dx

=

T,,(t)

= 0,

(n

\342\200\2247.,,r,,

holds

It

APPLICATIONS

rrs

AND

(7.5)

that

=

r,.

\342\200\224 ;\342\200\230;L.

so

that

no

=

-

If

z>L ax,

u(x,

or by (4.3)) x=-b)

=

T, (z)

The

last term

Thus we

\342\200\224

vanishes becauseof

other hand,

where the differentiation of our hypotheses obtain

a <

region

x <

..(x>L(u)

-,1-I\342\200\235

b

1

\342\200\224

X30)

Sec. 4.

(7.6)

dx.

(7.7)

the

behind

2

,,(x)

twice

93-;,,(x)

fr

with respect to

t

dx,

gives

(7.3))

integral sign is legitimate Comparing 83:4/atz.

because

(7.7) and since u(x, t) is continuous in the Furthermore, 0 and since lim u(x, t) = f(x), it follows from

concerning 314/

(7.2).

:r,,(z) =

r\342\200\224>0

dx.

8214

3;

(7.5)

differentiating

b, t

r

in

at

and

\342\200\234\342\200\231\302\260)

(7.5)that lim

2 of

Lemma

-

T,';(z) =

(7.8),we

(,,g\342\200\224\302\247 ;.u)])

we obtain

(7.3),

using

7',,(z) = On the

[p

have)

T..(z) =

whence,

-\342\200\224

dx +

,,(x)L(u)

lim

ru(x,

f\"a) r\342\200\224>0

rf(x),,(x)

t),,(x) dx dx =

c,,

(n =

o, 1,2,. . .),

(7.9)))

ms eromnmcnou

7

sec.

the C,

where

are the Fourier coe\357\254\202icientsof

T,,(t) is

function

continuous,

we

show

=

c,, are the

the

completesthe

proof

Fourier

On the

be found

(n =

0, 1,2,. ..).

(n =

0, l, 2, . . .),)

f (x).

Sincethe

relations)

in

coe\357\254\202icients

of

the

function

This

g(x).

theorem.)

of the

Thus, if the problemin solution can

to the

26!)

that c,, T',\342\200\231.(0)

where

AND rrs APPLICATIONS

the function

is equivalent

this

= C,,

T,,(0) Similarly,

METHOD

we are

which

of a

the form

interested has

series(1.12)by

a solution

the

often

all, the of Sec. 1. u(x, t) which as be regarded at

method

this method leads to a function everywhere. Such a u(x,t) cannot of the problem in the exact senseof the word, since u(x, t) certainly must the differential equation! However, becauseof the theorem satisfy to look for an exact solution, since if just proved, in this case it is useless such it would have to coincidewith u(x, t). This compels us to be existed, satis\357\254\201ed with the function u(x, t) that we have already found; we shall call this function solution of the problem. that a generalized It can be shown with our hypotheses, the series (l.12) obtained of Sec. 1 by the method or de\357\254\201nes a function to which it converges either in the ordinary sense always in the mean, and thereforethe eigenfunction a method always gives generalized if it fails to give an exact solution.) solution,

other

hand, quite a derivative

does not a solution

have

8. The

Generalized Solution)

is the practical value of the generalizedsolution described above? it represent anything or engineer, of use to the physicist or is it of mathematical is in fact valuable, interest? The generalizedsolution purely as will emerge from the following theorem:) What

Does

Let)

THEOREM.

u(x,

0 ~

3

T.(:>.

\342\200\224

g...(x>1=dx

262

METHOD

\342\200\230rm: EIGENFUNCTION

and g,,,(x) converge in

if f,,,(x)

i.e.,

g(x), respectively,

m

as

is either

exact

the

the

to

boundary

t) =

We recall

Proof.

in

T...(z>..)

m \342\200\224> oo.)

Tn(0) =

0: )\342\200\230nTn

=

Cm

cn

= 0, (n +

Tfnn

( 1.1), subject

that)

=

T7: +

f(x) and

if)

or the generalized solution of the equation conditions (1.3) and the initial conditions) u.. co,\" and

u..(x.

APPLICATIONS

rrs

AND

)\\nTmn = 0:

=

Tmn(o)

T

Cmns

(n

l,2,...),

= = 0,

(8.2))

cmn

l,2,...),)

where C,,,c,,,C,,,,,, c,,,,, are the Fourier coeflicients of g(x), f,,,(x), g,,,(x),respectively. Since)

the

functions

f (x),

Ao=.)

,$:\342\200\230o(c,.

If (see

-

g...1=

in.

Sec. 6). In view

of

c,,.,.>=)

these formulas

(8.1),

imply

that)

Q)

=

\342\200\224

2 (0,, m\342\200\224>ao lim

c,,,,,)2

o,)

\"=0)

(8.5)

00 \"\"

2 (cu

=

0:

cm\302\273):

m\342\200\224>ao \"=0

whence

lim C,,,,, =

C,,,

\"\"\342\200\231\302\260\302\260

c,,,,, =

lim

Then,

(8.4)

(8.3),

by

(8.6)

c,,.

and (8.6) \"

Tmn] = 09

\"1'i_I\342\200\231n\302\260\302\260[Tn

where =

:r,,,,,]2 [11,\342\200\224

\342\200\224

cos

c,,,,,)

[(c,,

sin

\302\253/11: +

\302\253/17,

t]2

\"\"\"/'f?\"'\" \"

(8.8)

g

for

n >

N.

r[u(x.

J\342\200\235 \"

All

0

2 that

_ [(53,

4.

cm\:")

remains

- u.,.1= dx

VAT.)

is to

consider the

= E (21.n=0)

relation)

T.....>2 Q)

=

i

n=0

(Tn

\"'

Tmn)2 +

2

n==N+l))

(Tu

_

Tmn)2)

1113 momruucrron

264

Sec. 6).

3 of

Theorem

(see

AND rrs APPLICATIONS

METHOD

(8.5),

By

-

If r[u(x.2) that the

which

means

as m

\342\200\224> oo.

function

9)

and (8.8)

u..(x. 012 dx\342\200\224>.o.)

u,,,(x,

in the

t) converges

mean to u(x, t)

theorem.)

proves the

This

(8.7),

cruur.

as follows: we have just proved can be summarized If f,,,(x) is close or even closeness close to the sense is and g(x) of uniform g,,,(x) (in f (x) closeness in the mean), then the function u,,,(x, t) is close to u(x, t) in the mean. and engineering, the that in actual \302\273problems of physics We now observe not exact, but rather represent approxifunctions f (x) and g(x) are in general to mations certain exact functions. Nevertheless, by the theorem just if of to the conditions even the solution (1.1), equation subject proved, is and not solution but a an exact generalized solution, it will only (1.3) (1.4), of uniform closeness or closeness in the mean) di\357\254\201'er the sense only slightly (in value of from the true solution of the problem. Thisconstitutes the practical What

to

generalized solutions. we

Finally,

consequence of the

one further

note

theorem

proved

above:)

and g,,,(x) are chosenin such a way that the exact solutions of the problems corresponding be chosen !9), then the exact or general(such f,,,(x) and g,,,(x) can always ized solution of equation (1.1), subject to the conditions (1.3) and (1.4), is the limit of the exact solutions u,,,(x, t) as f,,,(x) \342\200\224> g(x) f (x) and g,,,(x)-\342\200\224>

If

the

either

u,,,(x,

uniformly

It follows

9.

The

f,,,(x)

functions

functions

t) are

or

in

mean.)

the

immediately that

is

solution

generalized

unique.)

Problem)

Inhomogeneous

Instead of equation

the

(1.1),

consider

the more Bzu

Bu

0714

general equation)

P5?+REc+Qu=-aT5+F(x,t)

subject to vibration

tions,

the

same

problems,

while

multiplied

and initial

boundary equation

equation (1.1) by the function)

corresponds to the \342\200\230R

__1. 9 For

example, we can choose the series of f (x) and g(x).))

Fourier

the

(9.-1) \342\200\230corresponds to

r\342\200\224Pexp{Lo

of the

conditions

functions

(1.3) and (1.4). In forced vibra-

case of

of free

case

vibrations.

When

-3

}gdx}\342\200\224P)

f.,.(x) and

g,..(x)to be the

mth partial

sums

sec. 9

EIGBNFUNCTION

THE

(see Sec.

2), (9.1) takes the

Suppose now has

that

=

of the

+

value problem correspondingto equation of a series expansionin terms for the equation) value problem

that

t) has

F(x,

boundary

L()

Then, for t >

Sec. 2).

(see

(9.2))

t).

rF(x,

rg\303\251\342\200\230

the boundary

a solution and

eigenfunctions

265

form

L(u)

(9.1)

AND rrs APPLICATIONS

METHOD

0, we

u(x, :)

=

the

\342\200\224m1>)

write

u(x,

=3 n=0

series)

t) as the

(9.3)

T..(t>,..

where

r,,(z) = which

given

the

proof

=

mo [see (7.6)]

=

ff

\342\200\224 .1. 3.3.22

3:2),,(x)

1,, .,

where we have used(9.2). di\357\254\201erentiating

if we F(x,

the argument

(9.4)

dx

\342\200\224

\" Bu/at

b

rF(x,

t),,(x)

dx,

and 8114/8:3 are

to t

(9.5))

bounded,we

that)

azu

\342\200\234P-a7(D,,(x)dX.)

set)

t) = =

1\302\253*,,(:)

(9.5)

Repeating

obtain)

with respect

twice ,,

then

6.

(9.4))

,.(x>L_)

= ES 37*u(x and is also directed very small, we can

along

the

x-axis.

E/p

p

is the

= a3

and

density dividing

Ax,) the

Regarding

81\302\242: -\"'= ES '5?\342\200\231

being)

(14.1)

Ax,)

of the material from which by Ax and s, we obtain 3:2

is the equation for free for as the equation

sameform

AB as

azu

Ex-5

the

\342\200\224 .32.\342\200\234 _ a2 _a.2_.u

which

element

write)

pS Ax

where

t) 5|.-x:Ax,

t))

vibrations

free vibrations

the of

rod. a

made.

Setting

(14.2)

Bx?\342\200\231)

of

rod is

string.))

This equation has the

SEC. 14

rod is acted upon

If the

volume,

is the

This

We now rod

the

boundary

'5?

The

rod

t)s Ax,)

F(x,

Case 2) One end of the

on the freeend of the 3) Both endsare free.)

(the force

We

(14.4)

=

u(l, t)

= 0; and the other

= -0,

t) =) 0

8u(I,

(14.4))

ex)

hence Bu/Bx

is zero and

rod

end is free:)

=

shall devote our attention to Case 2, where the boundary are met. The initial conditions have the familiar form) u(x.

i.e.,the

initial

are

0) =

= sec).

f(x).

and initial velocities

displacements

0);

conditions

(14.5))

of the crosssectionsof the

speci\357\254\201ed.)

Free

Vibrations

in the

As

t)_)

P)

is fastened,

rod

u(0, t)

Case

+ F(x,

at both ends:

is fastened

u(0, t)

solutions

of a

case of the form)

Rod)

(see Sec.

string

vibrating

and arrive at the

=

for particular

(x)T(t),

equations d)\"

+

12(1) =

T\342\200\235 + a3A3T

the

12), we look

of the

u(x, t)

with

t) per unit

equation for forced oscillationsof the rod [cf. (1 1.3)]. the problem of \357\254\201nding the displacement of the cross sections pose at any time t, with speci\357\254\201ed and boundary initial conditions. The conditions can take various forms:)

Case 1)

I5.

F(x,

that)

3314 _ Bzu \342\200\224 G2

rod

+

-a7:5Ax

-372'

of

APPLICATIONS)

have)

= Es 3214

811:

psAx -57

so

we

ITS

force of amount

an additional

by

of (14.1),

instead

then

METHOD AND

EIGENFUNCTION

THE

boundary

0,

= 0,

(15.1) (15.2)

conditions (0) =

'(I)=

0.

(15.3)))

THE EIGENFUNCTION

has the

(15.1)

Equation

= (x)

C, cos

Assuming \357\254\201nd that

=0

C1 =

=

0,

C2

=

const).

must have

= Iwe

x

and

const,

C21 \342\200\231(I)

=

cos M

0.)

=

(n

o,1,2,...),

=Q\"\342\200\2243;7l)\342\200\224\342\200\231\342\200\224\342\200\230

sin 7t,,x =

=

,,(x)

values of

negative

9)

be identically zero, we C, 96 0, since otherwise (x) would n is an integer. where We write) (2n + 1)1c/2,

that M =

A,,

(The

(C1 =

C2 sin Ax

+

Ax

=

(0)

CHAP.

solution

to (15.3), for x

According

ITS APPLICATIONS

METHOD AND

sin

Q-n-'\342\200\224\"\342\200\224l)1-x= 21)

:1give

no

For

new eigenfunctions.)

(15.4)

2, . . .).)

0,1,

(n

A

=

1,, equation

(15.2) leads to) = A,,

T,,(t) and

cos

+

aA,,t

B,, sin

(n =

aA,,t

0, 1,2, . . .),

therefore

u,,(x, t)

To

= our

solve

a).,,t)sin A,,x form the series) B,, sin

a7\\,,t +

cos

(A,,

we

problem,

(n

= 0,

l, 2, . . .).

Q)

= 14%\342\200\230)

and

Z

11-:0)

=

u,,(x, t)

a7t,,t +

B,, sin

sin

aA,,t)

7t,,x,

(15.5))

that)

require

u(x, 0)

M

(A,, cos \"go

= Z [ \":0

at

=

=

(-A,,aA,,

B,,ax,, sin

2

go

sin

A,,

)\\,,x

a)x,,t +

sin

= f

(x),)

B,,a)x,,cos aA,,t)

sin

Ax]

(=0)

= g(x).)

x,,x

n=-0)

A calculation

of the

system {sin 7\\,,x}

Fourier coefficientsof

f (x)

and g(x)

with

gives)

_

L,

f (x) sin

= J:

dx

(n=0,1,2,...),)

Ll

B,,a7t,,

)t,,x

sin?

A,,x

g(x) sin [1

dx)

)\\,,x

sin? ).,,x dx)

dx)

(n =

0,1, 2,.

. .).))

respect

to the

SEC. 15

AND ITS APPLICATIONS)

EIGENI-\342\200\230UNCTION METHOD

THE

But) . sin?

7.,,x dx

=

L\342\200\231

and

1

I

dx =

\342\200\224 cos 2A,,x) 51; (1

5.)

therefore)

B,, =

1,2,...),

(n=0,

.4,,=%fo'f(x)sinx,,xdx

.

2

257L:

g(x)s1n

(2n +

7.,,x dx

(15.6) (n=0l2...) 9

0

=-\342\200\2244j\342\200\224J\342\200\230lg(x)sin7\\\"xdx

9

9

(1 5.4)].

[see

Thus, the solution of our problem is given by and B,, are determined from (15.6),i.e.,the vibrational of separate harmonic vibrations) superposition (A,, cos

=

u,,

aA,,t + B,,sin a7.,,t)

formula motion

sin

(15.5), where A,,) of the rod is a

(15.7)

Aux,

or u,,

=

H,, sin

(aA,,t +

a,,)sin Aux,

.

A

where

H,, =

\342\200\224\342\200\224

+

x/A3

B},

sin a:,, =

cos

a.,,

=

B

F:\302\273

The amplitude

which

depends

As for

t.

time

of the

vibrational

motion

H,,|sinA,,x|

=

described

H,, sin

on the position x of the frequency, it is given

is

|.) the

only

by (15.7)

cross

section,

and not on the

by)

_(2n+l)a1r_(2n+l)1t

E)

a),,=a7\\,,\342\200\224-p\342\200\224\342\200\224\342\200\224\342\200\224\302\247I\342\200\224\342\200\224 P9)

and

hence

the period is

T\" =

2;: = a),,

In n =

case of a 0; it has amplitude) the

vibrating

41 (2n + rod,

l)a the

A0 sin

=

41

2n + l J1

fundamental

TEX) 21:))

E)

mode is

obtained for

280

\342\200\230rm: BIGENFUNCTION

METHOD

AND

APPLICATIONS

rrs

9) cum\302\273.

frequency)

and

period)

\342\200\230l'o=4lJ%\302\260)

Therefore,

to a node at

(x =

as

mode corresponds the fundamental and an at the free end (x = I), antinode 0),

II]]]]I|l|lll||Jl||||||lIl||ll

I

II

I I

IJ

x=O

the

end \357\254\201xed

in Fig. 50.)

shown

ll x=l)

l\357\254\201oUne50)

Next,

x =

0 and

of a

Vibrations

Forced

I6.

the case where the rod is suspended force is the force of gravity, i.e.,)

we consider the

Rod)

perturbing

F (x.

where F(x, t) is the

is the

force per

due to

acceleration

unit

gravity.

end

t) = 93.) this

the density

p is

volume,

In

the

from

the equation

case,

of

the

rod,

and

g

for the vibrations

takes the form azu

33a

=

0257 -372\342\200\231

[see (l4.3)], again conditions

to the

subject

(14.5).

+ g

boundary conditions

(14.4)and

t) =

u(x,

:0 \"3

T,,(t)

t) = = F(x, \342\200\224\342\200\224

sin

initial

7t,,x,

\302\260\302\260

.

F,, sin ).,,x,)

g

\"20

P)

where) 1

gsinxuxdx

F,,

the

We set)

-I\302\260\342\200\224\342\200\224\342\200\224\342\200\224-\342\200\224\342\200\224=3g-

J;

sin? 7t,,x

dx))

A\

(n=0,l,2,...).)

(16.2)

16

sec.

the series

Substituting

the

(16.2) in

and

(16.1)

AND rrs APPLICATIONS

terms to

all the

transposing

28|)

obtain)

we

left,

METHOD

EIGENFUNCTION

THE

\342\200\224

+

[V18) It) o) (T;

a2)\\,2,T,,

sin

= 0,

lax

11)

whence)

(n=0,l,2,...).)

T;+a27\\,2,T,,\342\200\224l2\342\200\224f-=0

The solution T,,

of this

= A,,

has

the form)

B,, sin

aA,,t +

equation

cos

+

aA,,t

2g

=

(n

0,1,

2,. .

.).)

la2A,3,)

To

satisfy

(14.5), we require that)

the conditions

u(x, 0) =

3

x..x =

sin

12(0)

f(x).)

n=-0)

0) =

0u(x, at)

of the

A calculation

system {sin Aux}

Fourier

coef\357\254\201cients of

= A.

+

=

T,\342\200\231,(0) B,,aA,,

so

to the

respect

sin x..x ax.)

=

% ,\342\200\224f,g7,_

=

fo'f

g(x)

I

B,,

we

with

7\\,,x dx

sin

that)

2 . A,, = 7 f(x) sin [0

Therefore,

and g(x)

f(x)

gives)

mo)

[cf. (15.6)],

= g(x).)

sin A,,x T,\342\200\231,(O) S\342\200\230 n=-0)

g)

.

2

=

2

-

)\\,,x dx

g(x) s1nA,,xdx

ml;

(n

=

0,1,2,...).)

have)

0

u(x,

1) =

Z (A,, cos a).,,t +

B,,

sin

aA,,t) sin

Aux

n=0)

0)

_ ;g_ la? The giving

reader

will recognize at

the solution

2

cosa7\\,,t

once that

sin

\302\260\302\260

A,,x

1;\342\200\231;

the

of the problem of the

\357\254\201rst sum

free

sin

E on the

vibrations

7\\,,x. A},)

Ia3n=0

right is the function

of a

rod with

the

same))

ml; nromruncnon

282 conditions

the second and

Hence,

(14.5).

due to the

AND 113 APPLICATIONS

METHOD

terms

third

CHAP.

9)

the corrections

give

of gravity.)

force

of a Rectangular Membrane)

I7. Vibrations

we mean an elastic\357\254\201lm by a closed plane curve. supported is at rest, all its points lie in one plane, which we take to be the xy-plane. If the membrane is displaced from its equilibrium it begins to vibrate. We consider released, position and then only small vibrations of the membrane, assuming that the area of the membrane does not to and that each of its points vibrates in a direction change perpendicular the xy-plane. Let u(x,y, t) denote the displacement at time t of the point from its equilibrium position. Then, by a derivation (x, y) of the membrane similar to that made in the case of the string, one \357\254\201nds that the equation for the free vibrations of the membrane has the form) By a

membrane,

membrane

the

When

87-1:

3314

=

8324

+

C2

-875

while the equation for

forced vibrations

8114=

and

F(x,

solution

the

of

the

in

area

time

any

of the u(x.y.

initial

surface density, membrane. 9 is its

membrane,

on the

membrane can be posedas follows: the (17.1)or (17.2),i.e.,\357\254\201nd displacement

boundary (\357\254\201xed)

the

y, t)

To)

acting

Find

of the

condition)

0

(17.3)

membrane, and the 0) =f(x.

initial

conditions

(17-4))

y)

of the membrane) and

displacement Bu

to the

t, subject 14 =

(specifying

is

vibrating

of equation

points of the membrane at met on the

membrane F(x,

+

+

is the forceper unit

The problem the

3311

0311

the tension

T is

T/p,

y, t)

of the

C2

W where c2 =

W):

,

, 0

=

(17.5))

g(x.y)

-\342\200\224\342\200\230\342\200\224\342\200\231g-31-\342\200\231

(specifying the We

of a

now rectangle

initial

study

of the

velocities

the case

R:

considered

0

s x

points of the

of the free vibrations

< a, 0 < y

<

b.

of

membrane). a membrane

The problem

differs

in the from

shape the

on three rather that the function u depends problem than two variables. method Nevertheless, we again apply the eigenfunction and begin by looking for particular solutions of the form) in

Part

I in

14

=

4\342\200\231(x. J\342\200\231)T(t).

(17-6)))

SEC. 17 which

are

boundary

result

EIGENFIJNCTION

THE

in

AND ITS APPLICATIONS)

METHOD

not identically zero and which of the rectangle R. Differentiating we obtain) (17.1), 0. The

_a_\342\200\234\342\200\230.~.\342\200\230*1-3. I9)

(21.5) is

for t 2 to > 0, uniformly convergent for the series obtained by term by term of times). number differentiation of (21.5) with to x and t (any respect term and the term sum of is the the series Therefore, continuous, by (21.5) is legitimate (cf. Sec. 10).) differentiation

Heat

22.

This

boundary

Flow problem

with

is true

same

in a

ConstantTemperatures)

Rod with Ends Held at in

consists

of equation

solution

the \357\254\201nding

the

initial

t) =

A =

const,

u(l,

= B

t)

= const,

the

(22.1)

condition

u(x, 0) = f (x). We look

(20.4), with

conditions\342\200\235)

u(0, and

series

the

for a solution

in

the

form

u(x, t)

=

of a

Z

(22.2)

series

T,,(t) sin

n==l

#7\342\200\230.

(22.3)

where

:r,(z) = 19

The

like

u(x,

z) sin

dx.

these.))

(22.4))

\342\200\234\342\200\224\342\200\231I\342\200\231\342\200\224\342\2

boundary conditions in this problem, and from those considered previously.

a di\357\254\202erent form cases

\302\247

also in We

the problem of Sec. 23, have show below how to deal with

THE EIGENFUNCTION

twice, we

by parts

Integrating

METHOD

obtain)

x='

I

T\"

2

=

I3 6u(x, t)

_I_l_I\302\2437_C_,_L) C0821}

1m

[_

I)

CHAP. 9)

ITS APPLICATIONS

AND

0x

[ nzznz

x=0

sin

-71tnx]x=\342\200\231 x=0)

.. .11.. 12.2-.\342\200\234 sin 1% 1r3n2 0 8x3

Since

u(x, t)

satis\357\254\201es equation

and the

(20.4)

2

I

Q.\342\200\230 sin TE

a21t2n3 o at

n=%u\342\200\224pwm-

have)

m)

I)

to t, we obtain)

with respect

(22.4)

Differentiating

conditions (22.1),we

.__g.._

3.

d

I)

2 tau . nnx , _ \342\200\224 7L a\342\200\224ts1n\342\200\224l\342\200\224dx,

Tn so

that

l

l ;'n[/1

-

57'\302\273

- (-1) ,,B]

I2

-

,

\302\2475;;5n\342\200\224zT...

or

2\"2 \342\200\234z\342\200\235 2\342\200\2342\"\342\200\235 \342\200\224 = [A 1\",, (\342\200\224 l)\"B].

+ 1\",\342\200\231,

12

This equation has the

solution)

T,,

To

satisfy

the

(22.5))

12)

= Ane\"

:52.

condition

initial

u(x,

o) =

A

+ 2

:9

\342\200\224 (\342\200\224l)\"B_)

(22.6))

\357\254\201n

we require

(22.2),

that

= f(x). 2 T,,(O)sin \342\200\231-\342\200\230ll\342\200\231-\342\200\231f

am!)

A calculation

{sin

(-nznx/1)}

of the

T,,(0) and

= A,,

coe\357\254\202icients of

respect to

f (x) with

the

system

+

= 2\302\243\342\200\230_1f:';'._1)\"\302\243

f(x)

sin

dx,

\342\200\231-5%-\342\200\2315

hence ,,

Thus,

Fourier

gives)

the solution

are determined from

I

.._2

7 [0

f(x)

. sin

-rcnx

-7-

dx

of our problemis given the

formulas

- 2A

\342\200\224(\342\200\224l)\"B

(22.7))

\342\200\224\342\200\224n\342\200\224,;-\342\200\224\342\200\22

by

(22.6) and

the series

(22.7).))

(22.3), where the T,,

sec.23

THE EIGENFUNCTION

Flow

Heat

23.

Variable

problem,

to

is required

it

\357\254\201nd the

u(0. t) = \"4\302\2731dz.)

gives)

:r,,(o)

with

312\342\200\230mo

(23.2), we requirethat)

o) =

u(x,

-

of heat

with

heat

through \357\254\202owing

a surrounding an area

gaseous

sin the time At

formula)

Q

= H(u

\342\200\224

uo)s

At,

(24.1)))

is

um

302

nrommncnou

u is

where

rounding

AND rrs APPLICATIONS

METHOD

the temperature of the and H is a constant

body,

medium

called

CHAP.

uo is the temperature the emissivity.

In the case of heat \357\254\202ow in a rod whose lateral ends exchange heat freely with the surrounding of (20.1)and (24.1) leads to the boundary conditions)

whose

H(u for

x =

0,

x =

1.

Setting

h =

medium,

but

a comparison

8

\342\200\224=

uo)

K-3-2)

\342\200\224=

uo)

(h >

H/K

-

-33 x

\342\200\224K%-\342\200\231:)

0), these

+

assume

that

=

(24.2) 0.)

uo)L-1

= 0.

uo

::=0

\342\200\224

h(u

become)

= 0,

\342\200\224

h(u

3-3\342\200\230

let us

conditions

110)]

an First,

surfaceis insulated,

sur-

and)

H(u

for

of the

9)

The

conditions

boundary

then

take the

form)

3

= 0,

\342\200\224 hu

[\303\251lc

*=\302\260

(24.3)

an

= 0,

-I-a\342\200\224 3\342\200\230 hit]

while

x=l

is

the initial condition

14(x.0) = f (x). as before.

Following our

equation (20.4)of

the

usual

method,

the

conditions

for particular solutions of

form)

= ,,(x_)}

A,,x +

1:sin ).,,x}

gives

_ ,4 ,, _

dx

f(x)..(x)

fa\342\200\231 .j____)

,

-_-

(n

1, 2,.

. .).

(24.13)

[0 d>:(x)

dx)

the

Thus,

of our problem

solution

coeflicientsare determined The integral The equation)

the

in

from

is given

the

by

series (24.12), where the

(24.13).)

can be calculated

of (24.13)

denominator

(I); +

as

follows:

=

0 )\342\200\230?n(Dn

implies that =

A\357\254\201dlf, -\342\200\224,,;.)

we have)

Therefore,

=

\342\200\224

,'. 32:6

>.,=,fo\342\200\231:dx

+

;,2dx.

Io\342\200\231

(24.14)

But

(D, =

).,,cos).,,x+ hsinxnx,

=

sin ).,,x 111,\342\200\231, -1,\342\200\231, which

means

+ In,

cos )\\,,x,

that

Ago: +

of = 13+ hzxg,

(24.15)

hence)

and

xi 20

According

dx ,,;,];:.',.

(24.16)

conditions and (24.15)

the boundary

from

follows

it

305

APPLICATIONS

implies

dx \302\247

other hand,

On the

AND us

METHOD

EIGENFUNCTION

that

=

2.: +

+ h2\302\247 ).f,\302\247

hzx\357\254\201,

or

0%

at both x

= 0 and

= I.

x

= xi,

(24.17))

writing the

Therefore,

in the

conditions

boundary

form) \"

=

0. [\342\200\230Dn\342\200\230p:'a h\342\200\230D:2Jx=-o =

\"' hq\342\200\231;2.]x-=1 0. [\342\200\230pn\342\200\230p:'a

and using

that (24.17), we \357\254\201nd =

1,.:. \342\200\230:1. -21:13.

this expression in

Substituting

gives

(24.16)

of (24.13), we

instead

Thus,

2h_)

11:):

can write) cos A,,x +

2

f(x)(A,,

=

A\"

+

(13 +

= Lz\357\254\201dx

(x3, +

h

}.,,x) dx

sin

M): + 2h (n

we

Next, heat

consider

a medium

with

heat

exchanges

reduced

with

solvedby u =

function

1, 2,

. . .).

12

=

v

+

substitution)

w,

only on

v(x) depends

the

making

x and

satis\357\254\201es the

= 0, 12'\342\200\231

with the

boundary

-

w

satis\357\254\201es the

equation

(24.19)

conditions

\342\200\224 [v\342\200\231 h(v

while

(24.13))

the case where the end of the rod at x = 0 exchanges at the temperature uo, while the end of the rod at x = I a medium at the temperature ul. This problem can be

to the problem just

where the

=

uo)],,.o =

0,

+ h(v [v\342\200\231

- u1)],,., =

equation)

8w Bt

.3\302\273 =

32w

a 2 1--, Bx\342\200\231))

0,

(24.20)

306

METHOD AND 113 APPLICATIONS

\342\200\230ml; monuruncnou

the boundary

with

9) cum\302\273.

conditions = 0,

\342\200\224

hw] [-8527\342\200\230

x-=-0

= 0

+

x

:\342\200\224w hw] x-=-I

and

condition)

the initial

W(x. 0) to (24.19),

According

the

= f(x) v =

function

v

=

110:)-

v(x) has the

form

+ B,)

Ax

where the constants A and B are determined from the This leads to the system of equations)

conditions(24.20).

A\342\200\224h(B-uo)=0,

A+h(AI+B\342\200\224ll1)--0,)

which can be solvedvery easily. is of the type discussed above.)

25. Heat Flow

in

an

case of an in\357\254\201nite there rod, to reduces a solution problem \357\254\201nding and

t

>

0, and

are no boundary conditions, of equation (20.4) which is

we look for

(-00 < x < 00).

= f(x) solutions

particular

u(x, t) = Substituting

this

w

expression

and

the

for de\357\254\201ned

condition)

satis\357\254\201es the

u(x,0) As usual,

value problemfor

In\357\254\201niteRod)

In the

all x

the boundary

Then,

in (20.4)

of the

(25.1)

form

(x)T(t).

gives)

= a3 \"T, QT\342\200\231

whence > 0 is for t 0 This follows from the 2 to convergent (where to arbitrary). the

integral

inequalities

|(A()t) cos

Ax

[(A(7t) cos Ax

B(7t) sin

+

+

sin Bo\302\273)

)tx)|e-am\342\200\230<

2Ce'0\342\200\235*\342\200\235' < 2Ce'\302\260\342\200\231*\342\200\235*o,)

).x)e\342\200\2300\342\200\235*\342\200\235'])

Hg.

g

K.

Ax +

cos

31\"\342\200\231)

sin

B(7\\)

2C)\\ne-(2212!

<

2C)\\ne-azlzto,

)tx)e\342\200\2342\342\200\2313']

<

2Ca2\"*A3\"'

e-am\342\200\230 < 2Ca2m).3me-\302\260\"7\\2'o,)

fact that the majorizing functions on the right are integrable in A 0 to co. Thus, we need only apply Theorems 4 and 3 of Ch. 7, Sec. 6. It should that although our argument be noted shows that u(x, t) is a solution of equation (20.4), it does not show that) and

the

from

\"Or. 0

133 However, Using

this

relation is true and can easilybe proved. we can rewrite the solution of our problem

(25.6),

u(x,

To

further

to change 3> 0

z) =

transform the

order

1

co

1-:f0

7t(x

su\357\254\202icientlylarge

dv

dx

co

f_wf(v)

this expression, of integration.

f(v)

we begin

To show this,

<

L00 v)e-\302\242\342\200\231*2'dA|

I (where

L\342\200\230-t L\342\200\235

cos we \342\200\224 o)e-am:

\342\200\224

Utoocos

for

= f (x)-)

t >

cos x(x

proving

by

we

note

as)

dv. that

(25.7) it is

\357\254\201rst that

e-0\342\200\235\342\200\231~\342\200\230\302\260\"dA < e

0 is \357\254\201xed).Therefore) \342\200\224

<

o)e-aw:

ail

possible

for every

|f(v)|

1-\342\200\230:

dv,))

SEC. 25

which it follows But then)

from

as

the

that

METHOD AND

EIGENFUNCTION

THE

in the

integral

APPLICATIONS)

ITS

side

left-hand

to zero

converges

I\342\200\224> oo.

dv

cos

11:

-

A(x

J:\302\260f(v)

0\302\260

lim 1

=

dx) v)e-am\342\200\230

dv

7: -co) l\342\200\224\302\273oo

dl 11:

Here the

(25.7)].

[sec the

f(v) cos

do v)e-am\342\200\230

= u(x,

is legitimate,

of integration

order

in

dl) }.(x \342\200\224 v)e-02*\342\200\235!

cos ).(x -

f(v)

J:

change

fol

t))

because

integral)

f (12)

I: is uniformly the

convergent in is dominated

integrand

u(x, z) = of (25.7). It

instead

0 <

A for

A

\342\200\224 dv) v)e-021\342\200\235!

from the fact that Ch. 7, Sec. 6).

This follows

I.

<

Theorems 4 and

(v)| (see

|f

by

)(x

2 of

write)

can

we

Thus,

cos

11:

the inner

that

out

turns

cos x(x

dv

f(v)

\342\200\224 dz, v)e\342\200\224am=

In

be evaluated.

can

integral

(25.8)

fact, set

am/t~=z, so

A(x\342\200\224v)=y.z

that)

dz J)\342\200\230-=

-

\342\200\230L

E

.

ax/?)

we have)

Then

no

L

cos 7x(x -

1

v)e

with

Differentiation

\342\200\234*1\342\200\235! dx =

co

e-=2 cos p.z dz

L a\342\200\224\342\200\230/.\342\200\224t

respect to p. behind the = \342\200\224e-=2 z I\342\200\231(p.) L\342\200\235

this

x\342\200\224v

=

,

is legitimate

differentiation

convergent We

now

in

integral

sign

=

1

Z7; I(p.). gives

sin y.z dz;

because the

resulting

integral

is uniformly

(1..)

integrate

I , (u)

=

1

by parts, . sm

5 Ir\342\200\234

M

u.z1,-..

obtaining)

\342\200\224

% If

(25.9))

e-=2 cos

uzdz =

gm\302\273).))

me

3 I0

It follows

that

=

I(p.) To

\357\254\201nd C, we

set p.

= 0.

Ce-it\342\200\235/4.

This

gives)

c =

1(0) =

which, as we know,

an integral

e-3\342\200\235 dz,

Ch.

(see \302\247\\/-T

equals

1(9)= W5 and

CHAP. 9

APPLICATIONS

rrs

METHOD AND

BIGBNFUNCTION

8, Sec.

8). Therefore)

e-'**\342\200\230\342\200\231'/\342\200\230\342\200\230.

by (25.9))

cos 7.(x \342\200\224 d). v)e-am\342\200\231

J?

this expression

Substituting

u(x, t)

in (25.8), we

=)

= 515

e-0:3?\342\200\230) \\/1-:/\342\200\224t

\357\254\201nd) \357\254\201nally

1

2a

f(v)e'(L4;'3-9

dv

\302\260) (2510)

\302\260

f\302\260\302\260 -00 Wt\342\200\230:

the one hand, formula (25.10) shows that as t increases, 0, u(x, t) \342\200\224> the heat the rod. On the other hand, (25.10) i.e., \342\200\234spreadsout\342\200\235 along \342\200\235 shows that the heat is \342\200\234transmitted In fact, instantaneously along the rod. let the initial this be for v and zero outside < < temperature positive xo x, interval. Then the subsequent distribution of temperature is given by the On

formula)

u(x, t) which it is

from

large

trarily

that

2171/; u(x,

t) >

f

0'0.) (\3-sag\342\200\230;")

\302\2433

0 for

small

arbitrarily

t >

0 and arbi-

x.)

Heat Flow

26.

clear

=

in

a

Circular

Cylinder

Whose

Surface is

Insulated)

of a circularcylinder of radius lbe directed along the z-axis, be insulated let the length of the cylinder be in\357\254\201nite). (or that the initial temperature distribution and the boundary conditions Suppose are independent of 2. Then, it can be shown that the equation for heat \357\254\202ow is) Let

and

the axis

its ends

let

au.

where

9)

,(a2u

K/cp, K is the conductivity c is its heat capacity, and

a3 =

is made,

a

azu)

of the material from which the rod 9 its density. Thus, the temperature is))

SEC. 26

EIGENFUNCTION

THE

METHOD

AND ITS APPLICATIONS

I)

of z (a fact which is of course a consequence of the assumptions and we are with a problem in the plane. If just made), essentially dealing we to polar coordinates by setting x = r cos 0, y = r sin 6, then, go over instead of equation (26.1),we obtain) independent

\342\200\224

We

now

heat

0.

\357\254\202ow equation

that the

further

assume

independent of

Then,

obviously

takes

the form Bu

2

T9?=\"

We also assume rounding

and

initial

u is

a

'-)

+

only

are and t, and the

conditions

boundary

function

of r

1

(26.2))

3\342\200\231: 7 7' (6324 8a).)

of the cylinder is insulated

surface

the

that

r3 802

r Br

Br?

at

1E .

191

E

2

_

Q

from

the

aua. o

=)

0

(26.3))

Br and (absence of heat \357\254\202ow), condition) the by

the initial

that

0) =

u(r, We

look

for particular

solutions of the u(r,

Substituting

sur-

medium, i.e.,)

this

t) =

in (16.2)

expression

= a2 RT\342\200\231

temperature

distribution

is given

(26.4)

f (r). form

R(r)T(t). gives)

+

(RT

$127).)

whence R\342\200\235 + (1/r)R\342\200\231 =

T

T\342\200\231

-12

'aTT=

_\342\200\224 COnSt,

so that +

R\"

121: =

+

\302\24312\342\200\231 T\342\200\231 + a27.3T

Equation

(26.5)

is the

= 0.

parametric form

p = 0 (seeCh.8, Sec.11). Itsgeneral =

(26.6))

of Bessel\342\200\231s equation,

solution

'|' R0\342\200\230)C1100\342\200\231)

(26.5)

o,

is)

C2 Yo(7\\\\302\253))

with

index

3 I2

\342\200\230run BIGENFUNCTION

nth positive

the

is

=

C,

Taking

l, we

= 0-

1604) M

= 0.

we r\342\200\224>0,

the boundary

Therefore, 1:.=

Cl-IAP. 9)

have to set C, condition (26.3) that)

as

co Yo().r)\342\200\224>

Since

\357\254\201nd from

AND 113 APPLICATIONS

METHOD

root of the

J

equation

=

{,(y.)

We

0.

write)

_Pm

*~-7')

R..(r> = where )1 =

p.,,

1,, in

=

the

).,,I is

nth

for

particular

of the

zero

positive

=

equation

(26.2), subject of the t) =

function

Setting

J(,(p.).

to

the

. . .).

1, 2,

condition

(26.7))

(26.3),

we have

found

form)

=

l, 2,.

(n A,,Jo(7.,,r)e-0\342\200\235?\342\200\230-\342\200\234at

. .).

(26.8)

the series)

form

now

=

(n

A,,e-02*?-I

solutions u,,(r,

We

1. 2. . - .).

(n =

-70

equation (26.6), we \357\254\201nd T,,(t)

Thus,

=

Jo(7~n\")

u(r, t) =

S 1 A,,.Io(7.,,r)e-0\342\200\235~3',

(26.9)

n= and

the

to satisfy

condition

initial

we require

(26.4),

0) =

u(r,

=

that

(26.10))

f(r).

21/1n\302\253\342\200\231o(7~n')

A calculation

of the

Fourier

coe\357\254\202icients of

respect to

f (r) with

the

system

{Jo(7\302\273.r)}gives) 2)

A,, =

(secCh.8, Sec.24).

Therefore,

series (26.9), where the

(n =

rf(r)Jo(7\\,.r)dr

Fm

the solution

coef\357\254\201cients A,,

are

of our

(26.11))

1,2,...)

problemis given

by

the

from the formula

determined

(26.11).) 27.

Heat

Flow

Heat

with

This problem

Circular Cylinder Whose Medium) Surrounding

in a the

reduces

to

solving

Surface

(26.2)

equation

Exchanges

with

the

boundary

condition) +

hu(l,

z) =

@g;\342\200\224\342\200\230)

o

(27.1)))

sec.27 the previous

with

and

Repeatingthe

of Sec.

argument

we again

and

3 I 3)

condition)

initial

u(r, o)

(26.6),

AND rrs APPLICATIONS

METHOD

\342\200\234rm; EIGENFUNCTION

= f(r).

(27.2))

26, we again obtain equations

(26.5) and

\357\254\201nd that)

= Jo(N)-

R(r)

The condition(27.1)gives = 0

+ hJo()J)

).J(,()J)

or = 0.)

+ hlJo()J) )JJ(\342\200\231,(7J)

Therefore, the

number

p.

= 11

must be

a root

of

it-16(9) + hUo(:t) = We now

the

equation)

0-

(27-3)

write 2.,

R,.

=

=

Jo(7~n\")

__

\357\254\201n, I)

(n =

1..

1. 2. . . .),)

is the For nth positive root of equation A = 1,, (27.3). p.,, = and (26.8) of equation 1,2, . . .), the solution (n by (26.7), (26.6) is given the condition We again givesparticular solutions of (26.2)satisfying (27.1). form the series(26.9), and that the relation (26.10) be satis\357\254\201ed. A require calculation of the Fourier coe\357\254\202icients of f (r) with respect to the system

where

{Jo(A,,r)}

leads

to the

A \"

formula) 2

=

+ l2[J6\342\200\231(u..)

(see Ch.

L:tf(r)Jo(A,,r)

dr

(27.4))

the of equation (26.2), subject to the solution Thus, and (27.2), is given by the series (26.9), where the coefficients from (27.4), and the numbers p.,, are the roots of equation

8, Sec. 24).

conditions

(27.1)

are detennined (27.3).) 28.

J%(w.)l)

Steady-State

We now assume of the cylinder, and

Heat Flow

in

that

a constant

that

the

a Circular

Cylinder)

temperature is maintained

distribution

of temperature

on

the surface

is independentof z.))

nroauruncnou

ma

3|-1

a

after

Then

AND rrs APPLICATIONS

METHOD

at every point of the cylinder, on t. Thus, instead of equation

or

in

alu

Byz-)

the

1 Ba 7 7,

+

372

on the

temperature

look

be

0) =

(23.1))

by speci\357\254\201ed

condition

the

=f(9).

(28-2)

gives)

= o,

+ imp\"

+ imp

MD

o.

R(r)(0).

in (28.1)

expression

=

of the form

u(r, this

8114

:5 5.72\342\200\231

boundary

for particular solutions

Substituting

1

+

140.0) and

to

ceases

coordinates

polar

Let

u

0.

@411\342\200\230-

3x3

is

de\357\254\201nite temperature

i.e., the function (26.1), we have)

established depend

a

of time,

interval

su\357\254\202iciently long

9) emu\302\273.

r

r3)

whence \342\200\224

=

%-

\342\200\224g-)5

= -13

= const,

(28.3)

so that r3R\"

lb\"

The solution of

follows

It

must

=

=

0.

+

Bsin

= A

(28.5)

cos

A0

A0.)

the physical meaning of the problem that the function 21:, and hence A must be an integer. (Incidentally, would not have been periodicif we had taken the constant

(0)

be positive.)

n, equation

we write)

Thus, A,,

cos

n0 +

B,

sinn\357\254\201

(28.4) takes the r3R\"

+ rR'

(n

= O, 1,

\342\200\224 n3R

..).

= 0,

is

direct

R,

2,.

(28.6)

form

a second order linear differential equation. r\" and r-\" satisfy substitution that the functions fore, for n > 0, the general solution of (28.7)is) which

(28.4)

period

d>,,(0)=

For A

12(1)

from

have

we note that in (28.3) to

+

(28.5)is) (9.)

113(0)

- HR = 0,

+ rR'

= C,.r\" +

D,,r'\".

(28.7)) It can this

be

equation.

veri\357\254\201ed by

There-

(28.8)))

sec. 28

run

Since r-n->

EIGBNFUNCTION

00 as r -> 0, we

D,, =

to set

have

AND rrs APPLICATIONS

METHOD

0. For n

=

0, we

3 I5

easily

\357\254\201nd

that

R0 =

C0 +

and hence we must again take D0 = 0. conditions D,, = 0 (n = 0, 1, 2, . . .), we (28.1)as)

0) =

u,,(r,

now

form the

the

satisfy

+

n0

(3,,

sin

n0)r\",)

21

(28.2), we

condition

+

g3

3

Fourier

+

(ot,,cosn0

coe\357\254\202icientsof

f(0)cosn6d0

=1}-cf\342\200\230

=

9,1\"

l, 2, . . .),

(n =

n0)r\"

(1,, cos

+ 9%\342\200\231

=

of the a,,z~

sin

series)

boundary

u(I,0) A calculation

(3,,

write

7\302\260

u(r, 0) = and to

+

(28.8), (28.9), and the the particular solutions of

(28.6),

Using

can

\342\200\230*0

=

110 We

(on,cos n0

I\342\200\230,) D0111

require that) =f(0

[3,,sinn6)I\"

f (0)

.)

gives)

= a,,

(n =

o,1,2,...),)

=

(n =

1, 2,.

b,,

:\342\200\224cf_\342\200\231;f(e)sinnede

. .),)

so that an =

u

bn

=

is\302\273 \342\200\230'77\342\200\231 F\302\260

Therefore

u(r, e)

For r < I, this with

respect

0.(x)

and normalized 0 on the interval

of

eigenfunctions

[0, 1],

the

subject to

differential

the

following

conditions:)

boundary

a)

eigenvalues

\"(x)

c, x2-> c in

of Prob.

=

subsequent Prob.

5.))

A

cot act upon the motion of the string.

sin

mcrmnmcnou

me

pnonuams

AND 113 APPLICATIONS

METHOD

3 I7)

7. Consider a rod of length I, Young\342\200\231s modulus E, cross section 3 and density fastened at the end x = 0. Suppose that the rod is stretched by a force F acting on the end x = I, and then is suddenly released at time t = 0. Write the initial conditions and \357\254\201nd the subsequent longitudinal vibrations of the rod. Hint. The amount by which the rod is initially stretched is Fl/Es.)

8. Let the free end of the rod of Prob. 7 receive a sudden impulse t = 0. Find the subsequent longitudinal vibrations of the rod. Hint. Solve the problem with the initial conditions)

P at

p,

time

u(x.0)=0. 0

_

for0