Atlanta University CenterDigitalCommons@Robert W. Woodruff Library, AtlantaUniversity Center

ETD Collection for AUC Robert W. Woodruff Library

9-1-1969

Fourier series and boundary value problemsJamalendin Sadr-Ghadar-GhadrAtlanta University

Follow this and additional works at: http://digitalcommons.auctr.edu/dissertations

Part of the Mathematics Commons

This Thesis is brought to you for free and open access by DigitalCommons@Robert W. Woodruff Library, Atlanta University Center. It has beenaccepted for inclusion in ETD Collection for AUC Robert W. Woodruff Library by an authorized administrator of DigitalCommons@Robert W.Woodruff Library, Atlanta University Center. For more information, please contact cwiseman@auctr.edu.

Recommended CitationSadr-Ghadar-Ghadr, Jamalendin, "Fourier series and boundary value problems" (1969). ETD Collection for AUC Robert W. WoodruffLibrary. Paper 2182.

FOURIER 3EHIE8 AND BOUNDARY VALUE PROBLEMS

SUBMITTED TO THE FACULTY OF ATLANTA UNIVERSITY

IN PARTIAL PULPILLMEMT OF THE REQUIREMENTS FOR

THE DEGflEE OF MASTER OF SCIENCE

BI

JAMALEDIN SADfl-GHADAR-GHADR

DEPARTMENT OF MATHEMATICS

ATLANTA, GEO SGIA

SEPTEMBER 1969

V- e>

TABLE OF CONTENTS

ACKNOWLEDGEMENTS 1

Chapter

I. FOURIER SERIES 1

Definition of Fourier series

Odd function

Even function

Properties of Fourier series

II. PARTIAL DIFFERENTIAL EQUATIONS ............... 13

Definition of Partial Differential Equations

String Equation

Heat Equation

BIBLIOGRAPHY 30

ACKNO WLEDGEWENTS

1 wish to express my sincere thanks to Dr. Warsi, of

Atlanta University , my thesis advisor, for his assistance

and expert direction in the preparation of this thesis. In

addition, I also thank Malya Momaya for her assistance in

typing.

CHAPTER I

FOURIER SERIES

1. • Definition of Fourier Series. The trigonometric series

ij-a+Ca cos x+b sin x)+(slcos 2x+b, sin 2x)+---+(a cos nx+

bn sin nx)

is called Fourier deiies provided its coefficients are given

by the foumulas

a—L- ff(x) cos nxdx (n=0ti,2,...

-^r[tM sin nxdx (n=l,2,...)

where f(x) is some function defined in the interval (-f^

2. Odd Function. Suppose a function f is defined on a

set D. If for any x element of D there exists (-x) an

element of D such that f(-x)=-f(x) the function f is called

an Odd Function.

Example.

f(x)=X is an odd function because

if f(x)=X then f(-x)=-x=wf(x)

{ft)

Example.

f(x)-sin x is an odd function also, because

f(-x)=sin (-x)=-sinx=-f(x),

ftt)

If a function is an odd function, then its integral

from -1 to 1 is equal to zero.

Example?

Show the two odd functions f(x)=x and f(x)=sin x

their integral from -1 to 1 is equal to zero.

■1 rl

f(x)dx=-1

If f is an odd function deiined in (-TF,W) * then

Fourier coefficients a and b become

a=0n

(n=0,l,2,•

-=- ) f(x) sin nxdx

and its Fourier series, a-

s (a cos nx+bsin nx), becomes

K n n

2L ^ sin nx,n=i

n=l

b sin nx is called "Fourier sine series"»

Example s

Find the Fourier sine series of the odd function

f(x)=x which is defined in (-fT,V).

f(-x)=(-x)=-x=-f(x)

a =0n

-TT

b =-^- \ x sin nxdxn TT }0

n=i.,2,3i

-2

n

2 / -x cos nx , iif/—n "+

.cos

TT

TT

- cos n

n.+ 0

4-(-d n+1

Fourier sine series = \ (-i)

n=i

n+i

sin nx.

3« Even Function. Suppose a function f is defined on

a set D and if every x element of D there exist (-x) element

of D such that f(-x)=f(x), then the function f is called an

"Even Function"

Example %

f (x)=3T is an "even function" because f(-x) = (~xf=^=f(x)

f fC*)z

Example 5

f(x)=jx/ Is an "even function" also

f(-x)=f(x)

Example s

f(x)=cos x is another "even function"

f(-x)=cos (-x)= cos x* f(x)

f(-x)=f(x)

\

6

If a function f is an even function then its integral

from -1 to 1 is equal to twice the integral from zero to lj

'1 flf(x)dx=2 V f(x)dxt

Exam-lei

Show the integral of f(x)=cos x from -1 to 1 is equal

to twice the integral from zero to 1;

11cos xdx=sin x = sin l-(sin (-1))=2 sin 1 (l)

/-I J-l

i1 r i12 1 cos xdx=2 sin x =2(sin l-0)=2 sin 1 (2)

Jo l Jo

From <$i) and (2) it follows that

ri ri

cos xdx=2 \ cos xdx.

The Fourier coefficients for function f which is even a

and defined in the interval (-TT,TT) are

9 rtra =— V f(x) cos nxdx (n=0, l ,2, ... )

B ~n Jo

b^O (1,2,3,...)

Also, the Fourier oeries,-^—+ ") (&n cos nx+^sin nx),

n=i

becomesj-s-+ S a cos nx.

n=l

S a

7

Is called n the Fourier Cosine series.

Example s

Find Fourier series for f(x)=cos x defined in interval

(HfT.TT)

Since f(x)=cos x is an even function in interval (— Tr,n)

the full Fourier series becomes half range fourier cosine

series:

b=Q

2\ cos x cos nxdx

0 if n=i ( 0 if n=i

0 n=1 H if 1-1

Thus the Fourier aeries "becomes s

V— u K\ (a cos nx+b sin nx) =«_+ ^__ a. cos nx

n=i n=i

-s—+ cos x = cos x.

^' Properties of Fourier Merles. In establishing the

uniqueness of solutions of boundary value problems in prrtial

differential equations of boundary it is sometimes helpful

to know conditions in a function under which its Fourier

8

series will converge uniformly over the fundamental intervals.

Such conditions are presented in the following section.

Differentiations and integration of Fourier series are not

involved in our later applications j nevertiieless we shell

extend our theory here to cover those 'basic operations on

the series.

If A and JB (n=lf2,..»m) represent real numbers, the

quadratic equation,

m m m m

]Tn=l n=l n=l n=i

—Bx=x. then A x+B=0 for each n. Thus the ratio__n_ must be

can not have distinct real roots. Far if it has real root

—BA x+B0 for each n. Thus the ration

independent of n and equal to that number xq , n for each

nonzero A. In case A=0 for some n. &=0. The discriminant of

the quadratic equation is, therefore, negative or zero?

that is,

m ~ m m

][I iI in=l n=l n=l

The above condition is known as cauchy's inequality.

When m=3, it simply states that the square of the inner

product of two vectors does not exceed the product of the

square of their lengths. The corresponding property for

inner products of function is the dchwarz inequality. We

shall use Cauchy's inequality in proving the following theorem.

Theorem.

Let f be a continuous function on the interval -fl^x^n

such that f(-lT)=f(lT)f and let its derivative f be sectionally

continuous on that interval. Then the series ?

21 V^L\ n

converges, where ^ and b^are the Fourier coefficients given

f f=_l_{ f(x) cos nxdXf ^=_L_U(X) sln nydx#

'-IT

From the comparison test we note that each of the series

fn=i n=l

convergence as a consequence of the convergence of series

7 ■-„n=l

The Fourier coefficients of f are given by

Also, when n=l,2f..., we see by integration by parts that

(x) sin ri3CCiX +JLJfr(x) cos nx 1n iff ^i

n \ 11 <rT£= - fl. \ f(x) cos nxdx + -JL-lf(x) sin nx

10

We note that the condition f(-fl)=f(n), under which the

periodic extension of £ is also continuous, is necessary

if equations

sln nxdx +~ fCx) cos nx-J-ii

is to reduce to the form =nbn.n

Now let om denote the partial sum of the infinite series

^ • In view of relation^ -~-( f(x) sin nxdxn=i Jn

-_^_ f(x) cos nxtt J-n

=n n

-r. f11 1 ff l"Hand ^=-~l f(x) cos nxdx + ip ^(x) sin (nx)J = -n

-t-n

we obtain

m

n n /___ n f n

n=l

dlnce Sm^Gf ifc folloW3 fro» the Cauchy's inequality, that

^L n=l n=i

The first 3urn on the right Is bounded for all m because the

infinite series of positive terms -L. converges.rr

From Bessels inequality for the seconally continuous

11

function f, with respect to the orthonormal set

11-1-=—. —^— cos Ix, —— sin nx (K,n=i,2,...) on the

m

interval (-H,TT), we find that for every m, \ f 2+F ZJL.

n=l

' 9\f(x) dx, since^sO. Therefore, the tight-hand member of

i A

conditions sm

,n=l

2 2 1s7\ c^ +ir I , is bounded for all ms

n=l

and so is s.3D

Since s is a 'bounded and nondecreasing sequence, its

o$.

limit exists; that lss series \ \/ a +ly converges, &.s

n=l

stated in the theorem.

Theorems Under the conditions stated in the previous

theorem the convergence of the Fourier Series

Q0_

la,+y (a^cos nxib sin nx) to f(x) on the interval

n=l

is absolute and uniform with respect to x on that

Interval,

The conditions on f and f ensure the continuity, and

existance of one-sided derivatives, of the periodic extension

of f for all x. It follows from the Fourier Theorem thet

12

the Fourier Series converges to f(x) every where on the inter-

al-TT^x^fl, Now|^cos nx H-T^aln (nx)j^jaj+ jhj and the series

of the constants |aj+ |bi converges. Therefore, from the

comparision test and Weierst»as M-test, it is clerr that the

convergence of the Fourier series is absolute and uniform

as stated in the theorem.

The tests appling as well to show the convergence is

absolute and uniform. Therefore, the Fourier series is given

f (x)=-2-* + Ya cos nx + Y bn sin(nx) (--^^x^ri).

n=1.n=l

Both the series converge absolutely and uniformly.

*

CHAPTEH II

PARTIAL DIFFERENTIAL EQUATIONS

1» Definition of Partial Differential Equations. An

equation in a function of two or more variable and partial

derivative is called a ijajj^ial differential equation. The

order of a partial differential equation, as in the case of

an ordinary differential equation, is that of the highest

ordered derivative appearing in it. Thus the equation 1^=0

is one of the second order. A partial differential equation

is linear, if it is of the first degree In the unknown fun

ction and its derivitives, the equation u+u +u=0 is ajgr jjy isx.

linear partial differential equation. The equation (u)> u=0-ft. -A.

is not linear. The general linear independent partial diff

erential equation of the second ordert in two independent

variables x and y, is

^ EV Fu = Gf (I)where the letter A,B,C,D,E,F,Gt represent functions of x and

A V BV CV

We attempt to find the general solution of form (I) in

cases, A=0 and B=C=D=E=F=G=0, B=l and A=C=D=E=F=G=0, and in

case C=l; A=B=D=E=F=G=0.

13

u =0 Is a differential equation of form (1) where A=i

and B=C=D=E=F=G=0. Thus,

^=0 Imply u=f(y)

u.= f^y) Imply u=x^(y )+£p(y) where T and fg are two ©rb-

itary functions.

u =0 is a differential equation of form (I) in case whe

re B=l and A=C=D=E=F=G=0. Hence,

u =0 imply u=^(y)

imply u= h^j)dy+Y^x) where,

:) and h and h^ are arbitary functions such

that

u =0 is a differential equation of form (I) in case C=i

and A=B=D=E=F=G=0. Thus,

imply ^(x)

) imply u=y^(x)+^(x), where g and & are arbitary

functions of x.

Knowing the general solution of the above cases of form

(I) will help us in finding the general solution of form (I)

where D=E=F=G=0, Au^+Bu +Cu =0. in finding the general

solution of the equation Au +Bu +Cu =0, we may let the

unknown function, u(x,y), equal to v(f,g) where f=ax+by and

g=cx+dy for constant a,b,c,d such that ac=bd.

Now,

u(x,y)=v(f,g),

15

If we substitute in equation Au+Bu+Cu =0 for u , u and uWWW 3K W w»

we get the following equation.

Again,

(Aa>Bab+cb2)vff+ (2Aac+Bad+Bt>c+2Cbd)v_ + (Ac^-Bcd+Cd^v =0,

We may choose constants g,b,c and d such that two of the

three coefficients vanish and the above differential equatio

ns comes in the form of one of the following.

where ft, IL and EL are not zero. Now,

Hvff=0t then vff=0

Hv_.=0 implies v(f 9g)=fLj (gj+L,(g) for arbitary functi

ons L and L, Also,

ii^=0 implies %=0, and

v«=0 implies v=tL(f)+L(g) where Land L are arbitary

functions. Consequently,

Furthermore,

mr =0 implies V=o» %° and V=o implies v(f fg)=

+Lif) where Land L-are arbitary functions.

16

Hence,

u(x,y)=v(f,g)=(cx+dy)JLf

Au+Bu+Cu is called the principle p_art of the differe

ntial equations.of the form (1). We use only the principle

part in finding the type of differential equation. If B^AC 0

then the differential equation is of Hyperbolic type. If

&4AC=0, then the differential equation is of Parabolic type.

If 1^-^Ac 0, then the differential equation is of Elliptic

type.

Example:

Find the type of the differential equations

In this case A=i, B=x+y~ and C=i8 jJ-ij-AC o or x^-y2-4 0

the partial differential equation is of Hyperbolic type. For

xiy-^=0 it is Parabolic type, and is elliptic when

17

The partial differential equation of form (1) can be

solved by a method called "oeperation of Variables81, especia

lly when B=0 in the equation of form (I). Let us show that

for any equation of form ai^+2bu +c|r+dwou+fu=0, we can

find a function u(x,y)=X(x)-Y(y) where X is a function of in

dependent varible y. Thus,

u(x,y)=X(x)Y(y),

U|=X(x)Y(y),

u=X(x)Y'(y)

U

aX(X)Y9y)+2bX/(x)Y'(y)+GX(x)Y'(y)+dx1x)Y(y)+eX(x)Yfy)+fXCx)Y(y)=0

for a fixed y we can get Y(y), Y(y) equal to constants ai , a-

©^respectively. Consider,

X(x)+2baJ((x)+caJ((x)+daX(x)+ea^-(x)+faX(x)=0aa

The above equation Is an ordinary differential equation of

second order and X(x) can be found by method of solution of

ordinary differential equations. Also, for a fixed x we are

able to find constants b^ 1^ tu for X(x), X(x), and X(x)

respectively.

Now consider,

)+(bY()+db/()+^Y()fbi()=0 or

The above equation also is an ordinary differential equation

and Y(y) can be found by the method of solution of an ordina

ry differential equation.

18

2» String EguatjLgn. Now we show how "otring Equation"

with "boundary conditions can be solved by usage of "separati

on of variables" and Fourier aeries.

A tightly stretched string, whose position of equilibri

um in some interval on the x- axis, is vibrating in the xy

plane, Each point of the string, with coordinates (x,$) in the

equilibrium position, has a transverse displacement J(x,t) at

time t. We assume that the displacement y are small relative

to the length of the string, that slopes are small, and that

other conditions are such that the movement of each point is

essentially in the direction of the y axis. Then at time t

the point has coordinates (x,y).

Let the tension V of the string be sufficient enough for

the string to be have as if it were perfectly flexible} that

is, at each point the part of the string on the left of that

point exerts the force of magnitude P in the tangential dire

ction upon the part on the right? the effect of bending mome

nts at the point can be neglected, the magnitude of the x

component of the tensile force is denoted by the following

figure

19

Our final assumption is that H is constant, that is, that

the variation of H with x and t aan be neglected.

Those ideal assumptions are severe; but they are justif

ied in many applications. They are adequately satisfied, for

instance, by strings of musical instruments under ordinary

conditions of operation. Mathematically, the assumptions

lead to a partial differential equation in y(x,t) which is

linear. Now let V(x,-fc ) denote the y component of the tensile

force exerted by the left-hand portion of the string on the

right-hand portion at the point (x,y). We take the positive

sence of V as that of the y axis. If o{ is the i&lope angle of

the string at the point (x,y) at time t, then -V/H=tana|=y

as indicated in the figure. Thus the y component V or the

force exerted by the part of the string on the left of a poi

nt (x,y) upon the part on right, at time t, is given by the

formula. V(x,t)=-Eyx(x, t) (H>0).

This is the basic formula for deriving the equation of

root ion of the string. It is also used in setting up certain

types of boundary conditions.

Suppose that all external forces such as the weight of

the string and resistance forces, which act on the string,

other than forces as the end points, can be neglected.

Consider a segment of the string not containing an end point,

whose projection on the x axis has length ^x. dince x

components of displacements are negligible, the mess oj? the

segment is d ^x, where the constant d is the mass of the

20

string per unit length. At time t the y component of the

force exerted by the string on the segment at the left-hand

end (x,y) is V(x,t)» v(x, t)=-Hyx(x, t). The y component of the

force exerted by the string on the other end of the segment

is -V(x+ x,£)» where the negative sign signifies that the

force is exerted by the right-hand part on the left-hand

part at that point. The acceleration of the end (x,y) in y

direction is y..(x,t)« According to Newton's second law of

motion (mass times acceleration equals force). Then

d.Axy..(x, t)=-Hy(x, t)+HyJx+ x,t)IX. X A

Happroximately, when x is small. Hence yLL.(x,t)=d

+ x,t)-yx(x,t)= hv (Xft)

x 4*

at each point where the partial derivatives exist.

Thus the function y(x,t), representing the transverse

displacements in a stretched string under the conditions sta

ted above, satisfies the wave equation,

V8^ inhere a2=*[ > 0

at points where no external forces act on the string. The

constant a has the physical dimension of velocity.

Let us find a formula for the tranverse displacements

y(x, ) of a string stretched between the points(0,0) and

(c,0) if the string is initially displaced into s position

y=f(x) and released, at rest from that position. We assume

no external forces act along the string. Then the function

y satisfies the wave equation

(i) ytt(x,t)=a?y3K(x,-l) where 0<x^c and t>0.

It must also satisfy boundary conditions:

21

(ii) y(0,i)=0=y(o,t)i yt(x,O)=O and

(iii) y(x,o)=f(x) for O^x^c where the perscribed dis

placement function f is continuous on the Interval O^x ^c

and f(O)=f(c)=O

In order to find an extensive set odC particular sol

utions of all homogeneous conditions (i) and (il) in the abo

ve boundary value problem, using ordinary differential equat

ion, we first determine those functions of f(x,t)=X(x)T( )

which satisfy those conditions. We also note that X Is a

function of x alone and T a function of t alone.

r, then X(x)T(t)=aX(x)T(t).

o

We divide both lide above equation baXT and vie get

4 , Xx_(x)x(x)

since the member on the left Is a function of x alone, It

can not vary with t. However, it is equal to a function of

t alone and so It can not vary with x. Hence both members

must be contants, valve which we write as -IC, in common j

that is, X(x)=-I&(x), T(t)=-l42L'(t). That is,

If i is to satisfy the first of conditions (Ii) then,

X(0)=T(t) must vanish for all t Is t>0. The T(t)=O for

all t is trivial since the function i=0 always satisfies

linear homogeneous equations. Hence X(0)=0. Likewise, the

last two of conditions (ii) are satisfied by Y if X(c)=0

/

and T(0)=0

Thus 1 satisfies conditions (i) and |.ii) are so when

X and T satisfy these two homogeneous problems:

22

X(x)+tfk(aO=0, X(0)=0, X(c)=O

T(t)+IraT(t)=ft, T(O)=0> where the parameter IT has the same

value in both problems. To find nontrlvial solutions of this

pair of problems we first note that problem T(t)+iftQ'(t)=0,

T(0)=0 has only one boundary condition and therefore msny

solutions for each value of Ir. Since problem X+ia.=0, X(0)=0,

X(c)=0 has two boundary conditions, it may have nontrivial s

solutions for exceptional values of IT.

If K=0, the differential equation in problem, X+KX=Oj

X(0)=0, then X(x)=0, becomes X(x)=O and its general solution

is X=Ax+B. dince B=0 if X(0)=0, then X(c)=A(c)+0=Ac=0 and

since c is not equal to zero, A must be equal to zero.

Therefore this problem has just the trivial solution X(x)=0,

when X=0.

4 9If K^O the general solution of X+KX=0 is X(x)=c. .

^ cosKx, then X(0)=0 if the constant c_ equal solution of 0

X.(c) = O if sin Xc=Oj that is, Kc=nn(n=i, 2, . . . ).

Then, except for a constant factor,

X=sin ^H (n=i,2,...).

The number & /g* for which problem, X+l3c=0; X(O)=Os X(c)=0,

has nontrivial solutions are called eigenvalues of that prob

lem, and functions.

X=sin -£- are the corresponding eigenfunctions.

When IT<^Of let us write i&-I% where ¥ is real. Then X=a

sin npx is the solution of X-iaC=0 Which satisfies the condit

ion X(0)=0. dince sin nuc=0, o,=0 if X(c)=O. Thus problem

* -z 23X+lfeQ, X(0)=0, X(c)=O has no negative eigen values.

9 9 2 '' ?When K=ri c, problem X+KX=0 Is a distinct problem for

each different positive Integern. For a fixed Integer n, it

has the solution X=sin ~~, and problem T+irafT=0, T(0)-0

becomes T+ £'2=0, T(0)=0; except for a constant factor,

T=cos ^? ,

Therefore, each function of the Infinite set

satisfies all the homogeneous conditions (I) and (II).

The function y(x,tj= V" bn sin nTrI. cos -£5^£—~ c c

also satisfies conditions (1) and (II), provided the coeffic

ient bn can be restricted adequately. That function will

satisfy the remaining condition (III) If can be represented.

In the form

« / \ V " . .MYf(x)=2_ ^ sin c (O^x^p).

Now by Half fiange Fourier dine deries

bn= c f(x) sin ncx dx (n=i,2,...)

If f(x)= V" bn sin ~i is to be valid.

n=i

The formal solution of our boundary value problem In the

displacement of string is given "by

y(x,t)= JJ ^sin ££F cos SH&t where the soeffic-n=l " " c

ients have the values

bn= -• \ f(x) sin ^ dx (n=i,2,...)c

The function y is required to satisfy all conditions

of the boundary value problem

(A) ytt(x,f;)=a2y3K(x,i) (0<x<cst>0)8

(B) y(0,t)=0, y(cti)=O

(C) y(x,O) =f(x), yt(x,0) = 0

We have used separation of variables, superposition, and

the orthogonality of the functions sin (HlL^ to arrive at aG

formal solution -

(D) y(x,i)=n=i

bn sin ^ cos JS&Pc c

(E) b= | f(x) sinn c jo

Where,

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

The given function f is to be continuous on the Interval

[c). Also, f{O)=f(c)=O. We assume that f is at least sec-

tionally continuous on the Intervals Under those conditions

ere now know that f Is represented by its Fourier dine Series

on that interval. The coefficients In that series ere the

lumbers b^ given by equation (js). Hence when t=0, the series

'25

in formula (D) does converge to f(x). Hence y(x,0)=f(x)

when

The nature of problem calls for salutlon y(x,t) which

is continuous in t at t=0. Then the prescribed boundary

values are limiting valves at the boundr&es y(0,t)=y(+0,t),

y(c,f)=y(c-O,t), etc.

To show that formula D represents a solution we should

prove that the series converges to a continuous functions

y(x, ) which, together with its partial derivatives satisfies

the wave equation A and the boundary conditions But the

series may not be twice differentiable with respect to x

and t even though the series has a sum y(x,"t) that may

satisfy the wave equation, in the case of the plucked, string

for instance, the coefficient b are propertional to

n^sin (—•) so that, after differentiating the series for

y(x, ) twice with respect to either x or t, the resulting se

ries fails to converge.

it is possible to sum the series in formula (D) that

is, to represent its sum without using infinite series.

This will simplify the verification of the solution,

dince,

2 sin iill^- cos ^^ = sin >s "'"'+ sinc c c

equation (D) can be written in the form

OO

(F) y45~ bn sin—„ (x-a*) +J Y bn sin ^~ (x+at)c/ c ic

n=t n=l

26

Let f(x) be defined for all real x by the sine sftries

for f or f

(G) F(x) y bn sin (IH5E) (-oo</x<^).n=l c

Then F(x) is the odd periodic extension^ with period 2c,

of f(x)t

F(x)=f(x) when O^x^p,

(H) F(-x)=-F(x)f F(x-2o)=F(x), for all x.

In view of equation (G), formula (F) can be written

(I) y(xf4)=| F(x-at)+F(x+at) .

Thus, series (D) is summed with the aid of the function

F defined by equation (G). The convergence of series (G),

Our conditions on f are such that its extension F is

continuous for all x. Hence F(x-at) and F(x+at) and therefore

the function y given by formula (1), are continuous functions

of x and t. Form either formula (D) or (I) we find that y(0,

t)=0, y(c, )=0, and y(x,0)=f(x). Note that when x=c in

formula (I), we can write F(c-at)=-F(at-c)=-F(et+c).

Since F(-x)=-F(x)f -F(-x)=-P(x) holds, whenever F(x) ex

ists where the prime denotes the derivative with respect to

the argument of F. F is even function. Likewise F is an

odd function*

In case, f and f are continuous when 0 x c and f(O)=f(

c)=0. the functions F(x) and F(x) are continuous for ell

as indicated in ,the figure below.

27

Thus,

yt(x,t)=|--F(x-at) +F(x+at) ,

y is continuous for all x and t, and y.(x, 0)=0.

P(x-at) and F(x+at) satisfy the wave equation (A).

Therefore y satisfies that equation, as well as all boundary

conditions. The function y given by formula (I) is then

established as a solution of our boundary value problem.

If the condition on f and f are relaxed by merely

requirng those two functions to be sectionally continuous,

we find that at each instant t there may be at most a

finite number of points x(0/x^c) where the partial derivativ

es of y fail to exist. Except at those points, our function

satisfies the wave equation and the conditions are ;fc(x,Q)=0.

The other boundary conditions are satisfied as before. In

this case we have a solution of our boundary value problem

in a broader sense.

3' The Heal Equation. The lateral surface of a solid

right cylinder of length is insulated. The initial tempera

ture distribution within the bar is a prescribed function f

of the distance x from the base x=0. At the instant t=0

the temperature of both bases x=0 and x= is brought to

zero and kept at that valve. If no heat is generated in the

solid, the temperature should be given by a function uCx^t)

that satisfies the heat equation

u^(x,t)=Ku<jc(x,t) (o£t<rr, fc>o)

and the boundary conditions

28

,t) = O, u(lT-O,t) = O

u(x,+O)=f(x)

The constant K is the diffusivity of the material,

we have written the boundary conditions as limits in order to

indicate continuity properties that should be satisfied by u.

The problem is also that of determining temperatures u(x,f;)

in a slab bounded by the planes x=0 and x=TT, initially at

temper?ture f(x), with its faces kept at temperature zero.

By separation of variables we find that X(x)T(t) satisf

ies the homogeneous conditions if

X(x)+KX(x)=0, X(O)=X(TI) = O, T(t)=-KT(t).

The otorm-Liouville problem has the eigen values K=rr

and eigen functions sin nx (n=i. ,2, ...). Corresponding

functions T, form equation (5), are

exp(-nTCt). Formally then, the function

u(x,t)= J~ bn sin nx (Ox).

We assume"that f and f are sectionally continueus. Then

f is represented !:>y its Fourier dine Series, where

2 C11bn= _ \f(x) sin (nx)dx

Formula u(x,t)=^[ bn e sin nx, with bn defined as above.

We have seen earlSer that bn->0 as n-> c^ 5 hence those

coefficients are bounded, for all n, lbnl M, where M is some

constant. Whenever t^to where to is a positive constant,

4 ^Since the infinite series with constant terms e""n^to

averges, according to the ratio test, the Weierstrass test

co_

29

ensures the uniform convergence of series with respect to x

and t when O^x^, t^toj> 0. The terms of that series are cont

inuous functions! hence the series converges to a continuous

function u(xf^) when t to, that is, whenever t 5 since to

is an arbitarary positive number, in particular, u(+0,t)

-u(0,t) when t>0, and since k(o,t)=0, the first conditions,

i(+0,t)=0, is satisfied by the function u9 .Similarly, the

isecond of those conditions is satisfied.

2

The series with terms ne -""■"", Or rte Zo, also conver

ges. Hence the series can be differentiated twice with res pe

st to x and once with respect to t, when t 0, because the

series of derivatives converge uniformly when t to. But the

;erms of the series satisfies that homogeneous differential

equation.

It remains to show that u satisfies the initial conditi

on, u(x, + 0)=f (x) . For each fixed x (0<(x<Tl) the series with t

term bn sin nx converges to f(x). At a point of discontinuity

we define f(x) as the mean of the values f(x+0) and f(x-0).

According to Abless test, the new series formed by

nultipling the terms of a convergent series by corresponding

members of a bounded sequence of functions of t, such as e

, whose values never increase with n, is uniformly

onvergent with respect to t. Our series therefore converges

liniformly.

BIBLIOGRAPHY-

BOOKS

Churchill, Huel V., Fourier 3erles and Boundary Value^Problems, New Yorks McGraw-Hill Book~c£7y T^ffT

Churchill, Ruel V., Fourier Series, and Boundary V&lueProblems. New Yorki McGraw-Hill BookTo"., 191*1.

Hogosinke, Merner, Fourier Series. New iork: Chelseatubllshing Co., 1950.