ORDINARY DI FFERENTIAL EOUATIONS

LAPLACE TRANSFORMS

SPECIAL FUNCTIONS

FOURI ER SERI TS AND TRANSFOR|vlS

EI GENFUNCT I ON EXPANS I ONS

LrcruRE Norrs FoR A|\4n 95s (1983)

P, G, SnFFMAN

IVOE L gMYTH

- j

i .

:{

ORDT}IANI DIFFERENTIA], EQUATIONS

IAPI,ACE TRANSFORUS

SPECIAI FUNCTIONS

FOURIER SERIES A}TD TRANSFORMS

EIGENI'I]NCEION EXPAI{S]ONS

Lecture Notes for Alvla 95b (1995)

P. G. Saffnan

-1-

I . First Order Linear ODE

We begin our study of differential equations by discussing the

l inear f i rst order problem

In many cases' P(x) and r(x) are continuous, and this continuity

guarantees that (1) has sol.utions. For continuous p(x) and r(x),

the general forrn of the solution can be fouad. Define

^xP(x) = J p( ()dg G\

Xg

Then

(r)

y = e-r{x) /* "HE) , (6)d( * Ae-Hx)

x9

is a solut ion.for each value of the integrat ion constant A. I f there

is an initial (or boundary) condition y(xo) = yo, then the unique

solution is

Y = yo "-Hx)

* "-q*, dr- eq() r(od6

where does the result come from? l{ult iplying both sides of

(1) by an integrat j -ng factor .P(*) , g ives

.P (y ' * n,") = $? (" ty) =

"P,This integrates immediately into

DvDe-y=/ le"rd6+A

^0

(3)

(41

-2-

Alternatively, consider the homogeneous equation

**-pv=o

The general solut ion is y = uo "-P(-)

, where u6

constant. Now make a guess. Perhaps the solution

geneous problem can be written as

Y = u(x) u-Hx) ,

where u(x) is a function, and not just a constant.

variation , of pararyters technique.

. - f -P -Pu'e ' - pue +

Hence u' = ""P

and

is an

of the

arbi trary

inhomo-

(5)

Substituting ( 5)

.PP1re =r

This is the

into ( l ) gives

" "Pd( + A

Although (1) can be solved when p(x) and r(x) are continuous,

discontinuities in p(x) and r(x) don't always deny the existence of

solut ions. The case of piecewise cont inuous p(x) and r(x) is easy

to diagnose. (1) can be solved on intervals where p(x) and r(x) are

both continuous. On each of these intervals, there is a solution

with an arbitrary integration constant. At a point x, where p or r

has a jurnp, the integration constants can be chosea so that the so-

lutions for x ( x, and x ) *J agree at * = *J The result is a

continuous solution y(x) , which rnay have discontinuities in slope

where p and r have jurnps.

If p(xo ) = s , tJlen x6 is called a singular point of the

xu=J

Xg

3-

equation (l). Singular points cause a variety of behaviors in the so-

lution. Sometimes it is irnpossible to satisfy a boundary condition

y(xo) L yo , or when i t is possible, tJ.e solut ion may not be'nique.

J.vfex i ) y ' **=O hassolut ions y=Ae6x- I f A+0,

r(o+1 and v(o-) are both iafinite. rf y is to be finite at x = 0 ,

then A must be 0, which irnplies y = O In this case, it is

irnpossible to satisfy y(0) = yo for yo any finite, non-zero Dlrrrr-

ber.

' ex i i ) y ' - "*=0, v)0 hassolut ions y=Axv I t is

clear ly impossible to sat isfy y(0) * 0 , but y(0) = Q al lows in-

finitely many solutions, one for each value of A.

The behavior of the solution near a singularity is best strr.died

by generalizing the problern from real variables into complex variables.

suppose p(z) and r(z) are analyt ic in a sirnply connected region

D. Then

is a differential eguation in the complex plane that determines w

as an analytic function of z. Given a boundary condition w(zo ) =

wo , there is a unique solution given by

wlzl =wo e-P(z) + e-P(z) J ,G).qg)dE (71Zg

where

, .2Plz) = J p(6)dg

2g

(5)

(8)

-4-

since p(z) and r(z) are anaryt ic in the sirnpry connected

region D, the integrars frorn zs to z are path independent and de-

fine analytic functions of z for zeD. Hence w{z) is an analytic

funct ionfor zeD.

rf p(z) has por.es in D, w(z) rnay acquire singular i t ies.

exi i i )#- rYr= 0 hassolut ions w=Azv. r f y isnotaninte-

g€r, z = 0 is a branch point of w. r f v is a negat ive integer,

z = 0 is a pole, and i f v is a posit ive integer, then w is anarv-

t ic for al l z.

I t is possibre to use the exact formula (?) to study the

solutions near a singular point, U'iit a general rnethod is simpler. It is

sufficient to study the homogeneo,us eguation, since the solution to

the inhomogeneous equation folLows frorn variation of pararneters.

Specif ical ly, consider

#-p(z) w=o (e)

and has a pole at z = 0.where p(z) is analyt ic in 0 < l " l ( R ,

7ni

-?-

The solution w is analytic in any sirnply connected region D ex-

cluding 0, and is uniquely def ined by w(zo ), 26 € D. w(zo " 'o ' )

is found by analytic continuation: Given w(20 ), the solution is uni-

quely def ined in Do. In part icular, w(zr) is uniquely def ined. 21

l ies in Dr. Hence, w(21) uniquely def ines the solut ion in Dr

Using this solut ion val id in Dl, calculate w(zo "Zoi)

. In general ,

7*iv(zo e-" ' ) * w(zs) (Look at exarnple i i i for instance, when v is

aot an integer. ) . Hence, w(zo uZ[ i l = I w(zo), where ) is 4 corrr-

plex constant that is not necessarily unity. In fact, a quick check of

tte exact solution shows that ), = e+(t)df , where the contour is a

loop enclosing z = 0 Although w(z) is not single valued, we can

choose o so that zc w(z) is.

, Zt i ,a , ?z. i(ze-"-)- w(ze-"-) = .zr ia ) w(z) =

za w(zl i f I = " 'zr ia

or r = , f tos

Siace za w(zl

expansion t a-6

is single valued

mz . tslencenx

o<l" l <R, i thasa Laurent

(10 )6

-aFw(z)=zL"*-6

o< l" l <R r f a =Q when trr-(-N, then

w(z) = "- t -N

(bo * b, z +. . . ) = r-" h(r)

vhere h(z) is analyt ic. In this case, z = 0 is cal led a

singularitv. The necessary and sufficient condition for z =

be a regular singularity is that p(z) has a sirnple pole at

(11 )

regular

0to

z=0

- 5a-

thn order Iinear equation

The equation

- ,--\ dnv . dt-l.,arr(x)g*+ a-_., t")a# + " . . + aq (x)y = r(x)dx* n-r ' 'dxn

is a non-homogeneous linear ordinary differential equation of rrth-order.

our principal concern will be with n = z, but we will in sirnple cases

consider examples with n = 3 or 4.

The eqtration is equivarent to a linear system of n first order

equations for n unknowns. Define

Yl=Y, Yz=Yt, Yl=y"+.. .yn=iOlf .*dx

Then

a dY'n-F = - arr-lYn - an-zYn-l - as Y1 * r

d&rt -1

= Yn

:

dv,6=Fz

The system can be written generally as

dyi

6i = f i (Yt ,yz . . . r r r ;x) i = 1, . . . n.

I'or purposes of nurnerical analysis, the eqr:ations are more conveniently

written in this form.

F"

-o-

II. Second Order Equation with Constant Coefficients

The solution of the linear first order problern is usually spe-

cified by a single integration coastant, or by a single boundary con-

dition y(xo) = yo solutions to higher order problerns need rnore

integration constants or boundlty conditions to be specified uniquely.

New concepts are introduced'to handle the diversity of solutions. As

a basic example, consider the second order equation

y"+ay'+by=0

where a

y=kelx

and b are constants. Direct substitution shows that

is a solution if

(1)

(2)t rz*aI+b=Q

Thia quadratic has eitJrer distinct real roots, roots that are complex

conjugates of each other, or a double root. These three cases r€-

sult in three different types of solution. If the roots L1 and 12 are

real and distinct, then there are two solutions{r,.

u1 = e) ' lx and u2 = elzx (3)

then "l '

*

real valued

from the

u1 = j ls l r* + "-rzx)

= uP* "o"

qx

u2 = f,{.rt- - u-l 'zx) = ePx sin qx

If tr1 and 12 are complex conjugates of each other,

and ul t* are cof i rplex valued solut ions. To recover

solutions, write lt - p * iq , ).2 = p - iq and note

linsarify of (1) that the surns

{4)

also satisfy the equation. In the case of a doubl'e root ).1 = ).2 = )t ,

the present rnethod gives only one solution u, = utr* But another

is e:qpected. y = etr* siq qx

solves the equation in the case of com-q

plex conjugate roots t. i iq . As 9 * 0 , the roots merge into a

double root, so it is reasonabLe to propose that

Lx sin qx --^l'xU2 = Ilrn e ':-r' = :(e'- (5)

q*o q

is a second solution, Alternatively, rnultiply the old solution by a

function v(x) , and deterrnine , ":

that the product is a new

solutioa. This is a good trick to use on any second order linear

equation. In the case that L is a double root, the equation is

y, , - ZLy,+).2 y =0. Subst i tut ing y = r" t r* intothisequat ion

gives (y" + ?1.v,+ l .zv) - 2) . (v '* lv)+ 12 v = 0 or v" = Q. Hence,

y=a:x+p

Given a pair of solutions ur(x) and u2(x) , the linearity of

the equat ion guarantees that u(x) = cl u1(x) + c2 u2(x) is also a

solution for any choice of constants c, and c2 On the other

hand, when is it possible to write every solution as a linear coln-

bination of ur (x) and u2 (x)?

u1(x) and uz(x) are said to be l inearlv dependent when there

are constants a and B , not both zero, such that a ur (x) +

F uz (x) = 0 If such a and B can't be found, ur (x) and u2 (x)

are @ How does one test for l inear dependence?

suppose {r (x) , . . . 0*(x) are m tirnes differentiable. lhen*

these S's are l inearly dependent i f f

6T-t)

The result can be used to show that all soLutions of (1) can be ex-

pressed as linear combinations of linearlyindependent solutions

ur (x) and u2 (x) . Let y be any solution of (1) . Then

o'In

d'.Tn

(mo'rn

Qr

.q'

u1

ui

-8-

(6)=0

-1)

u2Y

uj v '

111

ult y'

U2

uj =0

ul ' uJt y" au l*tnr 1 auj *bu2

since the third row is a linear combination of the first two rows.

Since u1 and u2 are linearly independent, the first minor is non-

z,eto. This means that the first two colurnns are linearly indepen-

dent. But then the third colurnn rnust be a linear combination of the

first two and it follows that y = aur (x) + Bu2 (x) Notice that the

argurnent would remain intact t' if a and b were replaced by

functions of x.

A sirnple criteria for linear independence would be useful.

u1 and u2 are linearly independent if the l^trcnskian

W(x) = +0

6:rarrt, Differential and Integral Calcrrh:s Vol. 2, p 440.

ayr*by

u1

ui

u2

uj

(7)

-9-

No* ${u1

,'.l'

U2

uj'

u1

aul *bu1

rl2

aujtbu2= - aW. Hence ,

W(x) =ke-ax , (8)

or in the case that a is a function a(x)

x' I a(g)d6

lll(x) = e "'

(e)

Frorn these expressions for W, i.t fol.lows that W(x) is either

ident ical ly 0, or is non-zero for al l x. I lence, to check the

linear independence of two solution", *it

i" sufficient to cornpute the

Wrpnskian at a single point x6 . Two solutions are linearly inde-

pendent if and only if their Wronskian is non- zero at some xe.

What boundary conditions uniquely specify solutions? Two

solutions that satisfy

ur(xo) = I

u i (xo) -- 0

ua (xo )

uj(xo )

=Q

=|

(10)

are called a fundamental set. Assuming such u1 and u2 exist,

, l n,W(xs) = l ; i l

= ! . Hence, the members of a fundarnental set are

linearly independent, and it follows that the general solution is

Y(x) = k1ul(x) + k2u2(x). At x = Q' t

y(al - kt u1(a) + k, u2 (a) and

y'(a) = k, u i (a) * kz uj(a) ,Yi

But W(a) * 0. Hence,The deterrninant of this systern is W(a)

i t is possible to

that the solution

by its value and

through when a

Conside r

-10-

find kr and k, given y(a)

to the hornogeneous equation

derivative at a point. Again,

and b are functions of x.

the inhomogeneous equation

and y ' (a) This shows

is cornpletely specif ied

this argument goes

y"*al , '+by=r(x)

Any solution v(x) of (11) is called a particular integral.

be anorher solution of (11). It is easy to see tJrat y(x)

solution of tlre hornogeneous equation. Hence, the general

of (11) is

(lt I

Let y

y(x) is a

solution

y(x) = y(x) * c1u1 (x) t c2 u2(x)

where u1(x) and dx) are independent solutions of the homogeneous

problem. As in the hornogeneous case, the solut ion is determined

uniquely by its value and derivative at a point.

How does one compute part icular integrars? Recal l ing the

variat ion of pararneters apprtach for the f i rst order problern, a

natural guess would be

y(x) - a(x) u(x) + F(x) v(x)

where a(x) and F(x) are funct ions to be determined. From (12),

i t is c lear that there are many part icular integrals. This means

that e and p can't 6e deterrnined uniquely by the differential equa-

tion, so there is sorne freedom in choosing a and B. The

condition

ar(x) u(x) + F'(x) v(x) = Q

(12)

(13 )

(14)

11-

is i rnposed.

Differentiating (13) twice then gives

y'=au' f Fv' and

y" =a' ut * Ft v, + aurt + F v, ,

Upon substituting these into (Il) ,

arur* F 'v ' = r (15)

results. (14) and (15) form a system

atu i P' v ="0(161

a' ur + p ' v = r

which can be solved for et and B', because w = u v, - v u, rr o.

The result is

a'(x) =+kfg(17)

F'(x)=$ff i

Hence, a part icular integral is

' y(x) = - u(x) 1F r( l , f$E) d( x

,*o wE '5 * "(*) {o

r (O u(4) ; ,w(E) sb

(18)

.?.

-L?-

III Laplace Transforrn

Let f(t) be a piecewise continuous function on

new function defined by

qF(s) = f 4q u-" t dt

0

I o,o). The

is called the Laplace transforrn of f. Alternate notations are i(")

and f(s) There is a quest ion of convergence. For what s is

F(s) real ly def ined? I f there is an M such that f ( t ) e-Mt - 0

as t - 61 then the integral in (1) defines an analytic frrnction fc

Res)M. Afunct ion f ( t ) thatsat isf ies f ( t ) e-Mt*0 as t*€

for some M is said to be of exponential order one.

It is clear frorn the definition that f(t) uniquely defines its

Laplace transform i(") . fhe converse is less obvious, but t rue.

If i(") - O, (or vanishes on a set with a limit point), then Lerche's

Theorern* guarantees that f ( t ) = O I f F(") = E(") , then f( t) =

g(t) fol lows frorn applying Lerche's theorern to i (") - Ett l . Hence,

the Laplace transform i(")*"uniquely specif ies the funct ion f( t ) .

Given F(s), how does one compute f ( t )? This is accorn-

plished by the Mellin inversion forrnula:

(1)

f ( t ) =#

where c is chosen so that the

singular i t ies. In the case where

is suff ic ient to choose c ) M.

F(s) est d.s

contour is to the r ight of F's

f is of exponent ial order one, i t

Anonr igorous proof of the inversion

"c *io

Jc -16

(z l

* Carslaw and Jaeger, Operat ional Methods of Appl ied Mathematics,Appendix 1.

[3-

forrnula can be outlined briefly.

L ? -st "c* iof r J "-o" at 1[ . E@\ ezr d.z

U C-16

F(z) dz / " - t "

-z l t d. t., p*ioI I

ZdJc -16

(3)

assuming this change in the order of integration is valid. If s ) c,

then the t integral can be evaruated and the expression in (3) be-

colrre "

*

t ,c* ioo F(z\fr j

" az

c -16

cauchy's theorem provides the final b1ow. Applied to the contour in

the f igure, i t says

Lrf rJ.

"R

F(z)d.zs-z

If f(t) is of exponential order one, then it

provethat F(z) -0 as z+6 in Rez)c. In

,l- * O and, one finds that 1 rS+b f(r)*R fr Jc- i*# dz = F(s)

R -c.

is not \ard to

this case,

in the lirnit as

-I4-

are related to each other

F(s) e"t d" and

areca11edLap1ace@ofeachother.Bookscontain

tables of transforrns comrnonly used in applications. The little table

below gives some of the most comnnon ones.

s-ai 'a

)7w i\s1fr7

00

The restr ict ions on s define the regions in which I f l t l "-"t

dt"0

convergent. But the expressions for F(s) c lear ly def ine funct ions

al l s. In these exarnples, analyf , ic cont inuat ion gives F(s) for

9.

Here are sorne propert ies of Laplace transforrns that are es-

sent ial in appl icat ions: The Laplace transforrn of " t t

f ( t ) is c lear-

Iy F(s-a) This is the shi f t ing propertv. An integrat ion by parts

yields the Laplace transforrn of a derivat ive:

Any f(t) and F(s) that by

., ..c *ioor ( t ) =# f - i *

F(s) = f "-"t r(t) at

(4)

i )

i i )

111 ,l

f ( t ) =1 ,

. tn- t

I- : : Res-

rn!- rn+l ',

s

s)0

Res)0

F(s )

F(s)f( t )

f ( t ) ate,

f "i"\co"

oo€0o

i i t t l " - t t

dt = [e- t t r ( t ) ] + s I r t t l " -" t

dt ordo"o

L. T. ( i ( t ) ) = s F(s) - f (o)

iv) f ( t ) =

F(s) =

F(s) =

s)a

Res)0at,

at

1S

for

all

(5)

-15 -

be used to find the Laplace transforrn of aThis result can now

second der ivat ive.

L.T. ( f ( t ) )

L. T.( f ( t ) )

The proof is an

H(s)

f (0)) - f (0) or

s f (0) - f (0) (5)

= s(s

-^z-D

F(s)

F(s)

Differentiating both sides of

f -sf af" 6F(s) = I f ( t ) e-"" dt , one obtains #

= - l t f ( t ) . - " t dt-0

,?_-___ aF :- ^L^ ? -_r__^ L_^-^_r_-_^ ^r L t A

, 8F,Hence, - 5; is the Laplace transforgr of t f(t) , and - #

(- #)

a2F= ff, is the Laplace transforrn of t(t f(t)') = tz f(t)

If f(t) and g(t) are continuous for t >- 0 , and of expor€rr-

tial order one, then their q,envolulie4 is defined by

Th(t) = ( f*g) = | f ( r l g( t - r ) dr

b

(?)t

={

(8)

f . ( t - r ) g(r l dr

The convolution theorem says that

exe

)--

H(s) = F(s) G(s)

rcise in double integrals.

ootf e-" t at f qt-r) g(r) dr =b6a

,

a

a

A., t . ' r i l "- "tf(t-")

A

r Tg(rr az lrr€'0-r

u-"td" f r (") . -" t da = G(s) F(s)b

eG)dr=

-st , .e dt=

f e7t-0

t -T)

at, l

t

16-

The last step was accornplished with

As a sirnple application of the

calculate the Laplace transforrn of an

Hence,

a change of variable

convolution theorem,

indefinite integral.

t -T

one

Note

=(t

can

thatt

J' r(t) at - I ,ft f(t)5

Hence, i ts Laplace transform is F(s).

The theory of the Laplace transform gives an outline of its

basic Propert ies. An appreciat ion of i ts role and capabi l i t ies corrres

only with the applications. The study of the applicaf,ions begins with

sorne simple exarnples.

i ) y ' (x) + a y(x) - s in px

Y(0) = Q

ts

(e)

I f y(x) has Laplace transform

has Laplac-e transform s y(s) - y(O) +

sin px has Laplace transform 4-s'+[3,

sY+ay=-*P- "zl_pz

Y = 146fr(;;r ,LEJ:i

Y(s), then y ' (x) + a y(x)

aY(s) =sY(s) +aY(s)

Hence, (9) implies

or

"ir + Ffu')di= 74p

Y(x) = Vfu (" -* - cos F* +

Ug sin px)

The Laplace transform method can be applied to

tial equations. Notice that the boundary conditions

incorporated.

systems of di f feren-

are autornatically

-t7 -

ex i i )

x(0) = Y(0)=0,

=f

-0

x(0) = y(0)

\IIII

)I

=Q I)

. )xf a- y

, 'y-a-x

bsy =;

sx-0

(10)

L. T. (t i)

L. T. (x)

L. r . ( i )

-x(0)=sx,

- sx(O) - i . (o l

and L. T. (v)

=sx

=3x

-sy

and

,= s-x

=sz;

and

Similarly,

Hence, (1.0) impl ies

s2x + aZ

"ry - ^,

These algebraic eguat ions yield

F(s)

IJ-4

IG=f,'

I

b baz;(F+?| ano Y =!ei;r1;5

f ( t )

ate

.atte

-xte sin wtw

-xte cos wt

The inversion can be accomplished in var ious ways. One

method is to decornpos" i and I into partial fractions and then in-

vert each term separately. Here is a table showing terrns that typi-

cal ly ar ise in part ial f ract ion decomposit ions of Laplace transforms,

together wi th their inverses:

;'TE+*t

s-x;ffiF;r

* = rGFT

-18 -

In this exarnple,

:_ b _ b b s*=;Tf iGzI =;%'- ; r - - 'Fk,

i=;r1f , ; ; ry=# *3*ts-*=

Hence,

x=$-bcgsazt and

y =S -S sinazt

After performing a detbiled calculation, it is good to make a quick

test of the answer's plausibility. one convenient test is to check the

dimensions. Looking at the systern (10), one sees that az has di-I

mensions # , and b has dirnensions distance

Err"z- On the othe r

hand, the proposed answer d.ernands that i is a distance and that

bV rs a vetoci ty. A glance shows that this is t rue.

The preceeding example br ings up a key point. The systern

of differential equations for x and y yierds a systern of algebraic

equations for i atta |. This conversion of analysis into algebra

via the Laplace transform occurs quite general ly in l inear problerns

sith constant coefficients.

k i '

(lt I

eE. i i i) f " ,u,vi(0) =

(Lz1I

o' Yi = cr(t) )

Ii =1, . . . , r r r .

)

-19 _

The Laplace transforrns y. ,(s) sat isfy anI

t ions:

" t i *ot i ) r r (s) = cr(s)

algebraic system of equa-

a. . k. or1JJ

(13)

(sy-1) =

(sy) +y=0 or

(r+1

,YaLj

I t "J

If . (A=.,) is non-singular, i t is possible to solve for the yr(s). TheU 'J

expressions for | . , t r ) are rat ional funct ions of s and can, conse-J

quently, be decornposed into partial fractions.

The Laplace transforrn has shoon its virtue in solving problerns

with constant coefficients. What happens when it is applied to

ytt

y(0)

t t i -sy(o) -y ' (o)

dvsv=- i

GS

xy'+y

I , y ' (o) =,I ,which has

L.T. (y ' ) =

Ax- fr (sv)

a coeff icient that varies l inearly in

sy - y(0) = sy - 1 , and L. T. (xY')

Hence, (13) implies

x?

d=--cts

d.e;

This is a f i rst order equat ion. Applyrng the Laplace transforrn has

reduced the order of the problern by one. To solve (14), a boundary

iondit ion is needed. I f y is bounded by an exponent ial (h 'e. :

l r (x1 l< rn ekx) , then

-20-

l i t" l | -( rn *0

as s + oo. Hence. the reQrl i red boundary condit ion is i (**) - 0

The integrating

y(+€) = Q is

s2

factor of (14) is - 2

? (k-s)x 1J e ' dx =

;T0

FJeD,tr

s2)

Y(s) = - €-

sz-)y(s) = ,12 e-

tz"s

' tJe - do or

6

Hence, the solution with

- tZ>Af

-u2 ctu

(15 )

(15) '

(17 )

The integral in (15)' is so corrunon that it

cornplementarv error function.

has been given the narne

(r5)

The error function is defined by

t l l r2

erf x =-f- / " -E- dEt7r 6

h-ote that erf (0) = Q and erf (oo)

frnaction is defined in terrns of" erf

er fcx-1-erf

The complernentary error1

as

?Jex

2x=-r'{1t

i {ence, the solut ion for i t "1 can be wri t ten as

s2)-

i t " l - 6 u" erfc t#lv4

The inversion can be accomplished wit t r tables or contour integrat ion.

But i : r this case, a tr ick is faster. Looking at (15), i t is c lear that

the solution can be written as

zt-

F (sz - f t /Zy(s) =J el- -"-da

s

With the change of variable o = s * t, this becomes

r*

Hence, y(x) =;Z

How does erfc (x) behave for large x? An integration by

parts shows that

? -Lz "-* t t ? u '*

)e 'dE - t -2J? dgxx

The integral on the right hand side can be estirnated:

?" 'L ' ? r -72 r -xzJ

--z

di < J F e -- d6 =Z? e --xbi4

Hence,

*

i t " l = j " -st "-Z u,

W -/z o-xiJ" 'dE==-+ error,x

where the error is less than # u-* ' in absolute value. Not ice

that the correct ion dies to zero faster than the f i rst terrn + lx

as x + € . Hence, the first terrn becomes a good approxirnation

when x is large. Instead of estimating

, oo -Lzr rezr zrdl ,x: 'P

it is possible to do another integration by parts, and refine the

(18 )

-zz-

allproxinution given in (18). The result is

"€ -L2,- -x2rL I 'J e 'd6=" ' - t* '4 'Fj +error '

x

where the error is now less than #u-"t . This process of integrating

by parts can be continued indefinitely. After rn integrations by parts,

one finds that

^€ rz --zr | 1 3) e- -* .i6 =

"-^ r* - i'Vt - "'AE

x

3.5-w . 3 .5.7rzryTa

(2 0)

1),

le

( -

is

ml

ss3 '5' " ' ( 'z#+l)

' This series diverges fortnan -]@shere the error

a.ll x, but if we take only a fixed number of terms and let X - 6,

the error becornes srnall, and the resulting approxirnation is quite

god. -

(20) is an exarnple of an asvmptotic expansion.

As an exarnple of a @ equation, consider

(zL)

on the value at the

y(t-1) is

dt

i ' tt l = y(t-1)

where the rate of change at tirne t

earlier tirne t-l The LaPlace tran

€/ v( t - l ) " -" t

dt = "-" - I6- I

depends

sforrn of

- cf

y(t) e

0= .- t

{ r t t l " - t t dt + e-s }1s1

data. In order to evaluate

-steoE

initial

0I vfti

It

There is a quest ion of

, -23-

y(t) rnustbegivenfor - l<t<O Suppose y=l for

-1<t<0 Then

pit. (y(t-t)) = : - + + "- ' |1"1

Hence, taking the Laplace transform of (21) yields

r ^ -sst- l= l -" : +u-" i , or'ss

11' S 7 -9.

. . s(s -e )

{Hence

t ^c* io ^sty( t ) =1.* l E ds

61l l ! - s.

c-1ql s(E-e )

' i i

Inversion in closed form is not po".irt!. But a good approxirnation

can be obtained in the l i rni t t -6. The integrand has pole" " j

that are located at the zeros of s(s-e-s1 As Re g * *oo ,

s(s-e-") - +* . Hence al l the , j are conf ined to some lef t hal f '

plane Re s ( k A detailed study shows that one pole lurks at

s =.567, whi le the rest have Re s ( 0. The contour integrar equals

the surn of the residues at the sj. Hence

"; tY(t1 =I f I# ' 'T

J' J '

The dorninant term in the sum is :

.s67 te_ : L.LZI e.567 til56?)(1Tt67i- i

All tJle other terms correspond to "j -(

0 , and rernain bounded as

.:|'

24-

t - o. The f inal result is

y( t ) - l .Lzs e '567 t . " t *r-m.

The example illustrates the need to approximate

f(t) when the inversion canrt be done exactly. The two cases that

readily lend thernselves to analysis are the limits t * 0 and

t -o

The behavior of f(t) for small t depends on the bahavior

or F( s ) for large s. If F( s ) is analytic at os then

r . (s)=pf Gz)

ia the exterior of some circle l" | = ft . Due to the uniform con-

vergence of. (?21, the contour integral

a"* i* . , . a* , st

J (L f ) e-- dsc-1€ s

cau be done termwise, and

d'ge a tm-lr(t) = I, fui- e3lI

is irnrnediate. If F(s) has a developernent in fractional Dowers

9AaF(s) = ) ' -+? v (24)

l rns

for Re s ) I\d , then termwise inversion is stil l valid.y-1

F a-tmr ( t )=I fu_I - ' rn '

finds

(z5l

Il (241 is an asymptotic expansion of F(s) in the tirnit s + @ r

/A

;dot{

--€r

c,L

c,

A

f,r

-/-)^

-'-/

- ia

c

o

25-

(not necessari ly convergent), then (25) is an asymptot ic expansion of

f(t) in the lirnit t * 0 .

The approximation of f(t) in the limit t *oo involves the

asymptotic evaluafion of integrals. The general theory is outsid.e

the scope of this course, but the main results can be outlined. The

singular i t ies of F(s) determines the behavior of f ( t ) as ! - s ,

and the singularity with the largest real part gives the dominant be-

havior ' suppose F(s) has two singurar i t ies s ' and s., witJr

Ress)Resz

F(s) est ds =

F(s) ' . " t d"

. ' c* iof ( t ) =#J

c -1s

mtC1

by Cauchy's theorern. As d * oo r the contributions frorn OA

_26-

and 0' A' become negl ig ib le because F(s)

as kn s * * o . The contribution from the

A A' goes to 0 as b - - o because ut t

Re s +- oo Consequent ly,

usual ly goes to zero

vert ical port ion of

becornes smal l as

LfZf i J l

L2

f ( t ) =*L,

F(s) est ds * ste d.sF(s)

rfiere Lr arrd [2 are the loops shown in the figure below:

Before evaluating these loop integrals, note that the contributions from

L^aLareo(." l t )ando(" ' ' t ) respect ive1y.SinceRes1>

Bo sz, the contr ibut ion from L, c lear ly dominates as t -* o.

I f s1 is a pole, then the contr ibut ion from L, is "" t t

'

t=s F(s) To i l lustrate the case that s1 is a branch point, sup-t[=s I

1rcse that F(s) = k(s-sr) t-1 1 srnal ler terrns in a neighborhood of

-27 -

sl , w' i th a)0. Then

ostt ^ - (s-s1)tf ( t ) -*kJ dsLr (s-" t ) r -a

The loop integral was done in AMa 95a. Ornitbing the details of the

evaluation, the final result is

f ( t ) - * "" t t t - " f (a) s in zr a

Let g(x) and k(x, y) be known functions.

^b

ni

J t(*, y) f(y) dy = g(x) (26')a

is called an integral equation of the first kind" k(x, y) is called the

kernel. It is generally irnpossible to solve (26) tor the unlcnown

funct ion f(y). But in some special cases (eg: kernel s ingular, range

of integration infinite), solutions can be found. one special case

can be treated with Laplace transforrns. If. k(x, y) = Q for y > x

and k(x, y) = k(x-y) for y ( x, then k(x, y) is called a convolution

kernel. (261 can be wri t ten as

)</ k(x-y) f (y) dy = g(x) (z7l

6

Taking the Laplace transforrn of (2?) gives

7t ̂ ,k(s) f (s) = g(s) , or f (s) = :13/

k(s)

Now g(y) can be found by inversion.

-27 a-

For an advanced rigorous treatment of the Laplace

transform, see lviddler, Laplace Transform, Princeton University

Press. There are inversion formulae which involve only values

of F(s) on the real l ine. Thus

n+l

when this l imit exists. See Chapter 7 of Widd,er for further

developments. A simpler proof can be found in Chapter 13 of

Advanced Calculus by l{ idder. Apart from a few special cases

where the derivative can be found explicit ly in simple form,

the formula is useless for computing or estimating the inverse.

(I do not know if this formula can be derived from the contour

integral . )

r (r) = r im (-1) n I t" l f* t l

n + @ n! L t" 'J (3)

-28-

fV Second Order Equation 'W'ith Variable Coefficients

By generalizing the second order problem

y"*p(x)yr*q(x)y=9 ( l )

frorn real variable to cornplex variables, it is possible to probe more

deeply the nature of the solutions. Let p(z) and g(z) be analytic in a

simply connected region D. A natural generalization of (l) is

dzw .di# r p(z)# * q(z)w = 0, e)

which determines w as an analytic furrction of z. (z) can be written

as a systern. Let wr = \m and *, r#. g = f i :) satisf ies

dg

E;. = AE, (3)

where

lo r \A=l I

\ -o -P /The main existence theorem is due to picard. It says that (3) has a

unique solution in D that satisfies w(zo ) = yo . To prove this result,

consider the iterative scheme, (carled picard i teration)

*( l ) = go

a*(2)^;

at l (2\

T = Ag'- ' , gt- ' ( to) = go

::

a*(t)

# = 4*(n- l ) , g( t ) (zo ) = go .

: :

- i i

Another way of writing €# - 4*(n-1), *(t)("0 ) = :yo is

_29_

g(") = Eo + [] oyb-L)p1ar.- -v

tZO

Define lg l = { l * , l t * l * r l r .For each z e D, def ine

lel =r;l#u p, q are bounded in D, trr"r, i is possible to find. M > o such that

lal . r"r irr"iau D. I:r this case, legl . rurlgl and (4) impties

ls ' - s '-r l . * l lJ s("-1) - _(n-z) l lu, l . (s)The path from zo to z carL be chosen freely. Assume ttrat it is the

straight line from zs to z.

For n = 2, eguation (4) reads

y(r)-g(1)= I looyouu.

Hence,

fg( t ) - w(r) l . vr l r - ,o l .For n = 3, equation (5) reads

ls( t ) - y(zr l . i t r ls(r) - w(1) | la, l .

But lg( ' ) - w(1) I . r t t l " ' uol , so th is becornes

lE( ' ) - s(2) l . * , I :J" - ,o l la, l = yp lz =znlz .

Proceeding by induction, it is possible to prove that

ls(') - s('-t)I . *-il,= j-,i. ,"-t .Now let s_ = *(n) - *(n-1). Thenn

(4)

n

*( t )= L", .r?r K

z-z:T) l is r:niforrnly conver-

gent by comparison

Hence,

exponential series.

E(z) = t irr, g(t){")n+6

exists and is analytic in z e D,. Letting n + 6 in (4) gives

nz

Y=So t JroAwdz.

This equation says that g satisfies the differential equation and the

init ial condit ion g(zo) = yo. The existence is proved. lrterpreting

this reeult in terms of the second order eguation, one concludes that

dzw "dg*q(z)w=oA7' Ptz)a,

w(26) = a, w'(zo) j U

has a solution. Now recall that the solution of a second order l inear

equation is specified uniquely by its value and derivative at a point.

Hence the solut ion w = w(z rarbrzs) is r :nique. The proof shows that

w is analytic in z. But it is also possible to show that w is analytic

in a,b,zs. I f p = p(z,L) is analyt ic in a pararneter \ , then w is

analytic in I as well.

The existence theorem can easily be generalized, to non-Linear

equat ions. Let

since l .kl .c{

-30-

€

, the series ' I,dr"o

fr (yr yz . . . yrr, x)

:

frr(yr yz yrr, x)

I, l

and f (y,x1 = {*'.(,' I

\

:II

Yt\-ni

v=

l , ' ' rI 'lYzi

\

\

\It

tI . .

IThe system

- 31-

dv

# = !ft,'=lbas a r:nique solution in sorne neighborhood of xs that satisfies

lfro) =

1o if I satisfies a Lipschitz condition

l l t tG),*) - l (x,(z),*) l < KlfG) - t( ' , ) lfior I(1), f(z) in a neighborhood of 1o and x in a neighborhood of x6.

The proof is similar to the one already given, and can be for:nd in Birkhoff-

Bota, Chapt. 4 or l:rce, Chapt. 3..

z=zo isanordinarvpointof wtr*pw,*qw=0 i f p and q are

eaalytic at zo. Let zo = Q and suppose that p, g have Taylor series

p(z) =po f prz | . . . , q(z) = go * grz+ . . .

in l" l . A. Then w(z) is analyt ic in lr l .n and consequently has a

Taylor series

w(z)=ao+alz+., ,

r lso val id in lu I a n. Thes e series for p, g and w can be substituted

inb the differential equation to get

(2a2 * 6a3z + . . . ) + (po + ptz | . . . ) (ar t Za2z | . . . ) +

(qo + 9. ,2 t . . l ' ) ( .0 | a2z r . . . ) = o

collecting like powers of z and equating the coefficients of each power

of z to zero gives

2a2 = - arpo - aogo

6a,= -prar -Z?rpo-aogr -argo.

:

xn general, "r,

is a l inear combination of 4oor .. . an_l. This arlows

rb+ "',

to be found recursively, given the values of &e and 41 .

Tbe result can be expressed in the forrn

-32-

.r . = "ourr(po

. . . pn-3, qo . . . qn_Z) +

arvn(po . . . pn_z, go . . . grr_z).

Hence,

aewl (2.) * a1w2@). (6)

Notice that wr (z) and wz@) are d,etermined by p(z) and g(z) a10ne.I f 40 = 1, €Ll = 0, then w(0) = 1 and wr(g) = g. But &o = l , 3r = 0irnplies w(z) = wr (z). Hence w1 (0) = 1, w1 r(0) = g. similarly,w2 (0) = 6, w2t(0) = l . The w1 and.*w2 form a f i :ndamental set. Theirpower series have radii of convergence not less than R. (For directproof, see Birkhoff and Rota, Ghapt. 3).

Rettrrning to real variables,

y" *p(x)yr*q(x)y= g (7)has two power series sorutions about an ordinary point xe. Both serieshave radii of convergence not less (maybe more) than the distance of x6to the nearest singularity of p(z) or q(z). There is a solution thatsat isf ies y(xo)= 1, y ' (xo) = o and another that sat isf ies y(xo) = 0,y ' (xo ) = 1.

The second order equation (2) can be written in d.ifferent equivalentforms' Multiplying both sid.es of. (7) by the integrating factor

" =

'4*n(g)dggives Py,' + ppy' + qy - O. But p, = pp, hence,

(Py') ' fQy=0,

where Q = qP. (8) is cal led the self-adjoint form of (Z). The !eU-adjoint forrn w'1 play an important role in eigenvarue probrems.

€\-r .,.t oo

w = 4olorrrt + a1)vrrzn =00

(8)

-33-

Another useful forrn is obtained with the transformation

1 aX--z t^ P(GdLy(x)=u(x)e u

.

Calculation gives

| - i Iy' = (u' - ino)" and

y,,=.(u, , - pu,-+ -Lrn,o). * f .

Elence, (7) becomes

utr*J(x)u=0, (e)rhere J(x) = , - E' - ,' . The virtue of this trans.formation from y to

r is that the eqr:ation for u has p = O. !f p, q are analytic at xo,

t'nen J(x) is clearly analytic at xe. Any regular point of (?) is also a

regular point of (9)

rsolated singularities of p and q are called singular points of

tTi. suppose that z = o is a singular point. Then p,g are analytic in

Ir i . n, but rnay have poles or essential singularit ies at 0. A branch

?oint is excluded because it is not an isolated singularity. Let zs be

..y point in 0 . lrl < R and let D be a neighborhood. of zs that

cxcludes 0. rt is possible to find linearly independent solutions wr (z)

rwr wz(z) that are valid in D. w1 and w2 can be continued analytically

rhout the origin to give another pair of solutions i, (r) = wr keZri) and+ )q2uz{z) = wl (ze'" ' ) that are also val id in D. These new solut ions ;r

*' '.f wz must be linear cornbinations of the old solutions wr and w2

bacar:'se w1 and w2 are a linearly independ,ent pair. specificalry,

wr (zo " 'o i )

= i r ( ro) = awr @o) + Fwr (zo)

wz (zo.?oi) = iz bi l = y*, ko) + 6w2 (zs ) .

Tllhis can be w.ritten as

-34-

la P\I I is called the ci.rcuit matrix.lv 6,r I

It the circuit matrix has distinct eigenvalues l,r and tr2, then it can

be diagonalized. rn this case, it is possible to find. a pair of linearly

independent solrrtions

u1 =AW1 *bw2

t l2=CW1 *dw2

for which

lf the circuit matrix has a double eigenvalue l, = \r = Lz, then

diagonalization may sti1l be possible.but in general,l tur / | / \

f : ) = [") l * ) (u)\ t i \ r r / \o, /

is the best.that can be done

Here are some examples to illustrate these d.ifferent cases:

ex' i) w" - I 1P, = g has solutionsz 4zo

13,

u1(z)=sz and uz@l=12.

or1r"zoi1= - ur (z) and the same holds for u2.

Hence,

l - t o \e= | I\0 -Ll

The circuit rnatrix is diagonal even though the eigenvalues are

equal.

I i

/* t \t lt l

\* r /

/ ; ' \ la pl - | = l

\*r / \v 6

(10)

t , r ,lE\ / r ' o\ 1u, \t t=t l l l

\ t r l \o *r l \ " r / '

ex. ii)

-35-

7wrr +f t '2w= 0 has solut ions

YZ-

1zur (z) = i3 , uzkl = 27. Hence

,_ -Zt i , Tt i /3ut(ze )=e ut@) and

er" iii)

- - _ z log z __ ,_^Znt i , z !os. zlr = z, uz = -Z-ff. uz(ze -) =

-;f

+ z.

Eeuce, l t o\e= i I

\1 r l

In the case that the eigenvalues trl and \z are distinct, define

\ (z l = r- t t o, (u) and f2 @) = z 'cz u, 121,

where

" , =

#tog 1. , and n, = $; log L2.

Now (10) irnplies that r,.rqrua'i1= l.u1 (z). Hence,

e t- } t i , , Zr i . 'a Zr i .r l (ze )=\ze ) u1 (ze )=

! 'n 'H

)r1u1 (z) = u 'n ' ur (z) = \ (z) ,

eud it follows that fi is single valued and analytic 0 . Ir l . n. The

satne holds for fz@1. Expressing the result in a different way,

ut (z) = ,n' \ (z) and'

uz k) = ,n ' fz @),

rnb,ere fi and f,z are analytic in 0 . l" | < R.

. . r l r "zoiy = "4r i /3oz@).

This gives

i ezri/s o \Q= | I

\ o "anft l*" - # *# = O. There are two solutions

36-

In the case of a double eigenvalue, (11) irnplie s

'ot) =

^or (z) and

t") = ..r(z) + ).u2 (z),

As before u1 (z) = utf1r1, *h.t e a = *rlog )r. aad f(z) is analytic in

0 < l " l <R. Now def ine

s(zl = z-a luz(zl - f f iq@)).

Then)a2

g(ze'" '1 =

_Crz ' r - i los z , , ,=1- {Iu2 (z) + u1 (z) + l)u1 (z} =

z-a(u2tu1 - } f f " , @l) = g(z).

Hence, g(z) is single valued and aaalytic in 0 < lz| <R, }r surnhary,

u1 (z) = za f.121 a".d

w2@) = zag(zl + *.1i", t 't

where f and g are analytic i.a O < lzl <4.

The fi:nctions { and f2 1 ot f aad g rnay have isolated

singularities at z = Q. If the siagularitie g are no worse than poles, then

z = 0 is called a regular gingufaritv. Si:cce f1 and f2 or f and g are

analytic in a neighborhood of z = O, they have Laurent expansionsco

X"rr"t. !f, z = 0 is a regular singularity these Laurent series must-6

Fn-rt runcate L^o u" = , - ' t (bo+br z + . , . ) . I lence, f .1, f .2, f . and, .g are

-rnall of the forrn u-^ q(rl, where 9 is analytic and non-zero at

z -- 0, In the case of distinct eigenvalues, this irnplie s that

us (z) = unt f.r 1u1 = utt "-*F,

(zl = zct F1 (z) and

ur (ze

Wlze

-37 -

lz @, = ut" F, (") ,

where F1 and F2 are analytic and non-zero at z = 0. In the case of

a double eigenvalue, the correspondirg statement ig

u1 (z) = zcF1z1 and

n2 @\ = o"ooclu1 + u1 (z) log z,

shere F and G are analytic and nonzero at z=0 and n is a non-

zelo integer. Why is n + 0? $ n = 0, then it is possible to elirxinate

the constant terrn of G(z) by subtracting the proper rnultiple of u1

from u2. The resulting solution,

E, = ""+1G121

+ u, (z l tog z

has n=1. So evenwhen n=0 is possible, i t is bet ter to take n+ 0

to avoid redrmdancy.

A regular singular point can be detected from the coefficients

p(z) and q(z) in the equation. A necessary and sufficient condition for

a regular singularity at z= 0 is that zp(z) and zz 9(z) are finite at

z = 0, or equivalently, p(z) has no worse than a sirnple pole at z = 0

aad q(z) has no worse thaa a double pole. The necessity is easily

demonstrated. Let w be a sol.ution. Then

w=Aur*Buz

w'=Aui lBul

w'r=Aul lBul

and consequently,

]^ 'utu2

wr ui ul

w" ui ui

This determinant eauation. can be written as

-38-

w" + #+ iiga)w'+ figl:-$$1-)w = o.'ul ui - ui u2 . 'u1 ui _ uiu2 , '.

Henc e,

o = ur ufl - tlf uz "rru- ut ui - uiuz

o = ui'ui - uiui .- ut ui - uiu2

lf the solutions u1 and u2 are ""r8r1"1

and zc2F, (z), then sub_

stitution give s

plzy=*-*J1eg1

c,@.=*+ oet.l f u1 = zcF1z1 and uz = zc+nQ(z) - ur (z) log z, then there are two

cases to consider. l f n * 0 or G = 0, subst i tut ion gives

I-Zc - az 1p =

---

+ O(l) and c, = ;z+ o(-z).

I f n > 0,

n=t-z:-n+o(r) and

q = c(c-+ n) + otl l .

The sufficiency is rnore invorved. Briefly, the idea is to show that the

solut ions w must sat isfy lwl < Xlz l -M fo, sorne M > 0. A bormd

like this excludes the possibility of an essential singularity at z = O.

The detailed proof can be for:ld irl Birkhoff and Rota, p. 253. An

alternate method of studying the solutions near a regular singular point

is to substitute the series

*= u"("0 +arz+,. . ) , as iE 0

w'= uc- l (cao + (c * 1)a1 z+.. . , )

w" = zc-Z(c(c - l )ao + c(c f l )a l z + , . . )

-39-

P=;(po+ptz+.. . )

l

. q=VGo +qrz+ ' . . )

into the equaf,ion and deterrnine c and the a; so that w is formally a

solution. This is Frobenius' Method. With his approach, it is necessary

to show that the resulti' 'g series for w is convergent. The differential

equation can be written as

zzwtt + z (zplwt t (22 q)w = 0 or

zc1{c1c - l )as * c(c * 1)a1 + . . . } +

{po + pr z + . , . } {caq + (c + I larz + . , . } +

{qo + qrz+.. .J {ao + a1 z +. . . } ) = O

Collecting like powers of z and equati-ng the coefficients to zero gives

as{c(c-1)+psc+qo} =O

a1 {1c + l )c + po(c + 1) + qo} = ( - cpr - 9r )ao

an{(c + n) (c + n - 1) + po (c + n) + qo} =

I inear cornbinat ion of ao, al . . . an_l .

Since ao * 0,

c(c - I ) + Poc + qo = 0

This is the indicial equation, whose roots cl and c2 are the possible

ralues of c. Note that the sum of the roots is c1 * cz = 1- po and the

product is c 'c2 = qo.

lf c1 * c2 and cr - cz is not an integer, then there are two

liaearly independent solutions.

* = u", 1t + a{ l )o + . . . ) and

*=r"2p+rtz)r+. . . ) .

-40 -

Il this case, the coeflicient rnultiplying a' in (12) is never zero. Hence,

the coefficient" "G) "rrd "G)

are well defined.

lf the indicial equation has a double root cr = cz = c, then

u= zc( l + a)z+,. . ) is one solut ion, To f ind the other, the oldtr ick

w = vu avails.

wr=v'u*ulv

wtr = vuu+ Zvrur* vur.

Sub stituting into the equation gives

z2 uv' t * vt (Zzz,ar * z(ps * pt z | . . . )u) = 0 or

vft -

Zu' pd , 7., ,1F--T---(pr +pzz+ $. . )=-f f - f -Ot" l .

One integ ration gives

o, = , 'P,o "-

1+{uH'u .trz '

when c is a double root, the fo rmula for the s urn of the roots gives

Po = 1 - 2c. Recal l that u = otq@), where g is analyt ic. Hence,

, , - " : : - r . - . .o. tdz = _Lo@t,z-" oZ

where O(z) is analytic at z = O and O(0) = f. lO(rl can be wlitten

"" Lrr . r+azz+.. . Hence,

v= log z+ arz+$zz + . , .

=logz+zVlz),

and.

w= ulog z + ,c+1y1u1,

Suppose c1 and c2 are distinct, but c1 - cz = N, where N

is a posi t ive intege!. u1 (z) = 2cr (a[ I ) + a{ l )z + . . . ) i6 always a solut ion.111

The a*' are well determined becauge the coefficient of ao in (12)

_4L-

aever vanishes. The natulal guess at a second solution is

u2(z) = 7c2 @[z) + a?)r+ . . . ) .

But Dow there is a problern in determining the coeffici.rrt" "(Z).

c2 * N = cl is a root of t.he ildicial equation. Hence,

(c2 + N)(c2 + N - 1) + po(cz + N) + qo = 6,

and (IZ ) reads

a*. 0 = linear combination of ao , . , aN_ 1. (13 )

If ttre right hand side of (13) is zero, tJren a* is arbitrary. But if a

value for ao is chosen, the recursion can be continued. The resulting

solution is

1tz@) = zcz lao + arz +, . . ) + u"211*zN + . . .1.

. c ' . . N. . cr . .The second tert ' r z- ' (b*z- '+, . , )=z- ' (bN*. . ) is a l inear mult ip le

of u1 . The u1 and u2 obtained in this manner are linearly independent.

The cilcuit rnatrix has equal eigenvalues, but is diagonal. If the right

band side of (13) is not 0, then the power series solution fails, and one

lesolts to w = vul again. This tirne

vr' 2u' n". . .==-- -pzzz *vuz

Hence

- -Povt = +6(z\u-

rrhere E(z) is analytic at z = 0

su:3r of the roots,

-D^=+v@).z'

and \Il (0) = l.

t 14\

From the fo rmula for the

Po=1+N-Zc1

Fith this value of p6 , (14) become s

Iv' = 5fu1v(z).z

-+ L-

Integlation give s

--- I , 31 "N-z ar.r- l

"=F*$+' * t* fo losz+x(z)

This discussion has concentrated on the nature of solutions neara finite point. The results can be extended to include the behavior nearz = q. lI p and q are analytic ia lol > R, it is aatutal to ask ifz = o is a regular point or a singular point. ,An obvioug guess is that

o is a regular poi:rt when p and q are analytic at €. This is not

true. For a cor:ntelexarrple, note that the solution of wr * w = 0 isw=Acos z+ B sin z, which has an essent ia l s ingular i ty at z=a.

The problern is studied with a chanle of variable , = f. ff..r, quesf,ions

about z=€ become quest ions about l=Q.

ogi=*f #= - 6'€f ",,a#= - , ' * , - , '#, = s+ff +zs'#.

Hence,

t'$i*+ 26':f - nqrre$f + s(i)* = o or

:iv ,f,-?,#*f*= o, (15 )

where P(6) = n() ""a a(g) = S( i) . From (15), i t fo l tows that z = o isan ordinary point if

fi" - | ^"a $ "t. analytic at L = o, or equivalengy,

if. zzp(zl - 2z and, zaq(z) ate bounded at co. r: trris case, there are twosolutions

w=l*9, * . . . and w=b, * ! , *z __ , iz , . . .

o is a regular singularity if lE, - Z and

S are analytic at.jid = 0. Irt

the z variable, these requirements are zzq (o and zp <oo. Thesolutions are of the forrn

.+ 5- '

C, . i r ,w= z [ r+- '+. . .1 ot

c , . a ' c+n,- b '. f r - z (r*-u '* . . . ) Logz+ z (1 +- ' i , , .1.

3 is possible to find a power series solution of the inhomogeneous

wtr+pwl+qw=r

v\-l nT=Z' l , t z .

o--

i z = 0 is an ordinary point, then

* = " \+z 1^o + arz + . , . )

. toiution. If z6 is a regular singularity, with

t .p=i(po + prz+., . ) and

s = !2, (co + qrz + . . , ) ,

: : r r" * z(p6 + ptz+ . . . )w' * (qo + qrz * . . . ' )w =

6v*2s n

L/ i0

co

on of this equation suggests that w = u\+ z\a^rn

is sti l l the

power series for the solution. It is, provided that y + Z,

-- - ale not roots of the indicial equation. In the case that sorne

&cs solve the indicial equation, it is necessary to introduce

tertns. Here are sorne exarnples to il lurninate these

,. }wt 2w*" - ;+-= 2.

:@cs of the indicial equation are 1 and 2 , and y = 1. Hence,

I,

-44-

w= 23 (ao + arz+.. . ) is appropr iate. The resul t ing solut ion is

ex. ) * ' , -2{ *#=r.

c1 =1, c2=2, and. y=0. A ser ieE start ing wi th w = zz fa i ls . B

s, = oz (ao + alz + . . . ) log z gives w=zz Log z.

U the fotrn of the solution cannot be guessed in this way the

rnethod of variation of pararneters will always work.

z3

vzt,

-45 -

ln

rt

Besselr s equation,

, 'z* ' ,*?+(1 -7)w=0 ( t )

is of the chiefest irnportance. Its solutions are known as Bessel

Fuactionsl How does Besselrs equation arise in nr..rt".J*

exarnple, consider the two dimensional He lmholtz equation

Vz q + k2 q = g.

In polar cooldinates (r,0), this equation i€

{ f -+*- + { f+ kze = e.maay ptoblerns, 9 = f(r) cos n0 ol f ( r) s in nO. Thele are solut ions

( l - t r o. t n- . -

- r - ' , - -TrI(--"5r i=v.d.r- r Clr

With a change of variable x = kr, this becornes

-1-- . . -Ttr--r , I=v.o.x- x ctiK

Bessel's equation has a regular siagularity at 0 and an

irregular singularity at o.' Hence, it is natural to seek power series

solutions about z = 0. Substitutins

w = z*(ao + arz + . . .1

iato Besse| s equation gives an indicial equation

azo (cz - vzl=O

aad the recur sion formulas

a1 ((c+ l )2 -y21= 6

an-Z. _-- . ' - , . i ,_

n (c+nr- - y- '

A standard tert is G. N. Watsonr s Bessel Funct ions.

-46-

Lf ao ra 0, then c = * y aad ar, = 1s1. Hence,

?r=-#'

to - (- r)k_=_".-ff i .

The resulting solution is

Jr(z)=w=

rU

rfu€) {r - #q3) r r1rr+)rr,-2.;6,' - ...t.This solutioa exists prorrided v iJ not a negative intege!. trr(z) iscalled tlle Bessel function of, the first kind of order v , Jr(z) andJ-r(z) are teo linearly i:rdependent eolutions if v+ n. Jr.(z) behavesnlike z' nea! z = 0 and is finite, while J_, behaves like z-v and isinfiaite at 0. To illustrater

rL@ =Es2z and

J-y(4 =.pcos z .

Suppose y is an intege! n. Then

rn(z) = *,.","rt - *6,' +,;6Th6Tz, ,ir, * ...jis one solution. But it can be shown that J, _ (_ l)oJr, ." ? _ _ n.Hence, J_n does not provide a gecoad independent solution. To fiad asecond solutioa, it is possible to use the procedure developed in thegeneral theory. lf the roots of the indicial equation diffe r by an intege!,and u1 (z) is the series solution, then there is another soluti6n oftheform

3r=ao

(z)

(o-nT k

Ltakz ,0

u2(z)=st(z) logzrz

This u2 (z)

coelficients

This give s

can be substituted irlto Bessel s equation to find the

a Al ternat ivelv. one can use the old t r ick w = vtz lJ (z l .n ' ' n '

But neither of theee rnetlhods has the

Frobeniusr approach. Define

Yv@l = f r ! { ,1, (z) cos vr - t _r(z l \ .

The Y, (z) are called Neumann functions* or Bessel fimctions of the

second kind. If v + n, Y, (z) is a second linearly independent solution

of Bessels equation, ltrhat happens as v - n? Defiae the Bessel

oPerato r

B =zz$'*r9*t-r , .n o.z- oz

Bo J, = (v2 -na ) J, and .Bo J_, = (va -r2 ) J-u Henc e,

1vt =# and v=

z\J- lz tn '"z ArI i>-

JZO LJ A(L) '

convenience and sirnplicity of

vr - J . ,1

as v +11

func ti on

(3)

(4',)

Conse-

and

As , +l1r Ju -Jr l ,

quently, the right hanci

Br1 Yu *0 as v -n.

so one conclude s

Y = --L (v2 -rLz l (J cos-v s i ' r v l t \ - - , r -v

J_u - ( - r ,

s ide of (4)

But Y

B

Jn , cos rv

goes lo ze ro

is an analytic of

B Y =0nn

* Alx alte rnate notation fo!

, where

tl:e Neumann functions is Nr( z ) .

_48-

, . I= I i rn . - :g iJ cosvzr-J )nSlnvn'v-v '

v-n

Evideatly, Yo(z) ie the desired second solutioa. The limit in (5)

can be evaluated by LrHopital's rule to get

(5)

Yo = * f* ,-,," +J _.As an exarnple, consider Yo(z) .

({*J.=o = (;'*d v1p (rJ.=o

Hence,

,. =+ ipJv=0

can be written as

v. log

v

,vNotice ttrat (+)

'4

lI

. . . )

)

Diffe reutiating,

1)ffiTiiJ "

Iy+1

vle

(1og

I. . . )

I

+

rI1 l

tag

)

+

IIv+1)

is wlitten

.i={ro* e,

f .+ *''

+

I

+

z\=tz'

, z, 'u',gz

4u4

og

z.zz,z'

-49 -

,=o =

f"; - ri'rJ IsQ) +

la" ' ' i f &;- )I Iv+t) =

/ . - ' , ' u. . Hence r , ( t ) =

/e- t roe t at

is Eulerrs constant.

\ = ,5772

Frnally,

z( - \Y6(z) =

i i t " t ' + v! Jo 121 +

\J

Z"" l - , ,Zrz L 3 ]_ Z- \ L - \At TTi\a , + , . .,

, . . r . . t t_ . )

Tbe procedure for calculating the rest of the yrr'" is similiar.

dr-r'i-nant behavior near t, = O is given by

?alYrr(z) = - ;Fi)T ," -

smaller terrns

AJ

'8v'

rbat is f ' {1)?

= - y, where y

What do the Bessel functions look

traphs* giwe the behavior for real x.

large x, the Bessel functions behave

ekqrouitdr and Stegun is a good solrcefsstions.

Iike ? The suc ceeding

Notic e the oscillations. For

like cPs I "116

*P-Ivx vx

for tables and crraphs of

T - 5U-

A remarkable property of Bessel functions is that

z , . L,t i r - ; l 90.

e '=)JLn

( r ) t t (6)

2.. L, \t-;',t

e- ' is called the generatiag functiou. The right hadd side of

(6) can be regarded as t.l..e Laurent expansion of the generating func _

tion. Hence, to prove (6), it is sufficient to show

J lz ln '

=+ dant .t

z,. | ., lt'; )

-_, .1+n+lt

(71

whe re the contour is a

change of variable u =

sirnple closed culve enclosing zeto.

i " t ,

With a

*raTfl ,t

t tt-; Ie-----i-- dt

t-- -

bec orne s

_ 51_

zz_n . .^u ^-6/cv

t?) o ------...----- -a J n+l

22-4"

e

(8)I

-zZot +u

(e)I

F

in powe rs so (8) bec orne scan be expanded

n6,2, \\ , , f /

l<--u

bec orne s

-n eq

l( =u

e-n+k+l -*

By the re sidue the orem,

Hence, (9 )

u,eoul

. '_rn*k+l (n+k)!+dLltl 't

i1tri i"+kjf

zk$l = Jo (z)

iag the

iA

The generating function provides a

propert ies of the Bessel funct ions

. r nen

c onvenient tool

. l r , (z) . In (7 ) ,

for shrdy-

let

J (z\ =n

I

( ' " - t " r . i , " i r o do =

1 "Zfr

; nzn,; JO "o"

(n 0 - z sin e) de - f J sin (n

Si-nce sin(n 0 - z sin 0) i s pe riodic, the

be changed from 0 and Zr to -r and, r.

,11

J s in(n0-zsin0) d0=Q_1t

0- z s in 0) d0

l irnits of integration can

But

becau se

This is an integral representation of

jcos(nO-xsin0) l

Diffe rentiating

lr'n

z

itt-fr fn

<1 for x,

both side s of

J lz l t - - = )

e

(6)

Jn . Note that

real . Hence,

with re spect to

' (z) tn or

the integral is

-52-

odd. Hence,

| ^-4( r l =*

J cos (n 0 - z s in 0) d0-" - 0

Jn

Jn

k) l < t

give s

recurrence tela _

n _rQ) - Jrr*r(z)) tn ' (z) tn

the coef of tn on both sides yields the

\ - t , -L z\')n

f -

n

Equating

tion

?Tt-r- "n - "n_l

Diffe rentiating (6) with respect to

He nce,

" n+l

t instead of z gives

(10 )

ior$t ) r , , t '= n - lr or

zn - lt - -

z "n - 'n- l

t ru. , r rorn ( I l )

+Tn+1

give s

(r)

Subtracting equation

-5 3-

ZJ n+r?n

z - t r

nr

7'n - 'n

(rz)

In particular,

( I2) can be

( r3 )

J = (- I )nJn -n

T-

tf Z) allows the dete rmination of J n +l frorn Jn .

Jr = - Jd , which is also apparent frorn the graphs.

rewritten as

dz'-J l= 'z J

n nj-r

I -€t k+l = -n

(13) become s

Witl a Little help from the identity

Si : rc e (13) is val id for a l l integers n, (14) is val id for

ihese recurrence forrnula are also val id fot the Y (z\

:eed a little ca re.

To study the behavior of Bessel funct ions as

convenient to d.efine y = ^[x Jn (x) Then ,; *

+ t

::::lie s

-z-+v" + { I _ -- I . ) v = 0

lquat ion (15) has an i r regular s ingular i ty at co , hence, a

ser ies is inappropr iate. As x -€, i t seems reasonable

solutions should behave like sin x and cos x. This leads

-{-D,satz

(14)

all inte 8e r k.

but the pr oofs

nZtL -

_-- ' - ,J

( i5 )

power

that the

to fhe

"="o"*toFor simplicity, as

x

Srrrne n

y"+(1

-54-

1r. .1+ s in x (b"

J r-= 0. Substihrting

.1,f - - t - rv=u+* '

L

x

{16) into

) (16)

ll

r )

l

( t^ \

{F-. . . }\ - ,

z\

{+. . . . ILx- )

r4rb" b,(=- f

' - i

I atx- +)g

grve s

zsinx[*-?-. . .J-

2cosx(*-3-. . ) -

. . " ' ( -*+f f+ )-

cos x

sitr x

g1n x

y"+y

1@

Notice that ?e and b6 can be chosen arbitrarily. This is no sur-

prize, since one expects two linearly independent solutions. Equating. is

the coefficients of *i ."d *I to ze ro gives a set of recursionxx

formula:

cos x sln x

-ZA, + !

-4b2 * 221

nx

l-

+;

+ zb,

x

' r^

4^z

a9 and

**=o

Given bo , these cau be solved. In fact, the value of ao

dete rmine s the

bo determines

ao =1, bo =

-55-

sequence ao, bl , a2, b3 . . . , whi le the value

the remaining sequence b6, a1, b2, a3 . , .

0, tlxe resulti.ng candidate for a solution is

Iyr = cos "

+ A sin x + , . .

of

II

t -

rg

)a

ao = 0 , bo = I leads to a. second candidate

yz = s in "

- * cos x +. . .

In order for these series to be solutions, they must converge. If

the general terrns in these series are c ompr:.ted., and the convergence

checked, it becomes apparent that the series d.iverge for all x.

This does not mean that this effort has been a waste, Although

dive rgent, the series provide asymptotic expansi.ons of the solufions

yt and y2 in the limit x + oo .

Jo (x) is relaied to y, (x) and

ro(x) =#for, ,* , *

where A and B are constants, The

can be employed to find the value s of

Js(x) = * f "" . (x s in 0) de = 3

b1l

Y2 (x) b

L:Yz (xl ))

inte g ral

A and

7l

I COS'0

(17)

representat ion of Jo

B.

(x s in 0) d0

To resolve the behavior as x - co , a suitable change of variable is, ,2

aecessary. Let s in0=1-: Thenx

cos u2 + s in x s i .n, "{*Jo (x) = ': ,.6 (c os x

5

- f f i J (cos

u- | r l -,VX

I

' - Zx'

x cos uz f s in x s i .n u2) du as x + oo (18)

-56 -

Now

f"o" o' au =;f"i.', - d. = i .E

Hence, (18) becomes

ro(x) - *H-.4 = uF.." rx - it

as x + € . Using (19), it is pos sible to dete rmine

(17). As x+6, yr(x) -cosx, Yz(x) - s in x

P - l - .tn

of the remaining

forrnula J .. =n+l

(19 )

the constants in

Hence, A =

The asymptotic behavior

witJ the help of the recurrence

The result is

cos

latg z l ( n - 6

Consider the solutions

the farniliar sin x, cos x

(20)

One pair of solutions

ix -ixe , e . I f rese

1-n

.J

can be found

-J-

Jo (x) -

The result (20) is actually more general than the arguments leading

to it indicate. First, (20) is a valid asyrnptotic formula when n

is replaced by non-integer v. Second, it rernains true when x is

replaced by complex z, provided that l^rg " l 1 t _ 6, 6 > 0

Xfith (20) tJ:us generalized, it is possible to find asymptotic forrnula s

for the Yr, (z) from Frobenius' definition. The result is

E\j; sin (z Ly (zu

t* -V-L+l

v7l

z

IO

1n

1S

Y"+Y=0.

Anothe r pair is

-57 -

are related to each other by

1X =cosx+lsrnx

n( ' ) (*) -v

The Hankel functions

x- ix inx

= Jvkl + i Y, (z) and

=J(zl- iY(z)vv

i t " - t -x\

- i (xe

-ixe = cos

This motivates the definition of the Hankel functions

nf') t,)

*(z l p l

Notic e that

12 |H; ',(x)

,_--:-

\7t'x

as x * + co .f r l l : l

H'r ' (z l and H* ' (z) are a se-

cond l inear ly independent pair of solut ions to Bessel 's equat ion. They

sat isfy the same recurrence relat ions as J, and Yr. Their rnain

appl icat ion comes in wave propagat ion problems with cyl indr ical

syrrunetry,

The o rthogonalitv properties of the Be ssel functions can be

demonstrated via the eigenvalue problern

_ vn _ t \z4'

,, ', * )L. 1 112-x6z

- -j- J Y = U t)) \

(23)Y(a) = 0, Y(0) < o

0 -<y

(a

_58-

The general solut ion to (ZZ) is

y=AJ, ' (kx) + B yr, (kx)

Since Y(0) < o, B =0 ( I t is a general rule that boundedness

of solutions at a singular point is equi.valent to a bound.ary condition).

Hence y=AJ'" (kx) But y(a) =O impt ies J," (ka) =0 This

equat ion is sat isf ied when ka = 6r , (z . . . , where 6r are the

roots of J' There is a countable iafioity of solutions. Let

Yr(x) = J," (6rx) and y"(x) = Jr, (6"x) The self adjoint form of

(z2l is

^2x - =-) v = 0

f

*o*, +(k2x-}- ' r , ,=o (zs\

Mult ip l ied by ys, th ig becornes

v" $ t*ft, -

(kz x - #, ", y" = o

I J- ,

d(xH)+(k,

I, . s.atisfies (24) with k = k,

Similarly,

where k

r" $ t* *, - (q * o*')

", I" = o ,

= + . Subtracting (26) from (2?) gives

(24')

(26l

(z7l

d oYr A A.,yr;! (x -a;) - y" ft (xs*) = (r<r" - k?) x yr ys (n)

Intdgtating both side s

lx (v v - v- 's ' r 's

i f r+ s, then k ra k

of (28)

y-)J-0

-59 _

frorn

=(k2 -r

glve s

vdx

0toa

t1) f *v-

0

The le{t hand side is the result of an integration

Y"(a) = Yr(a) = 0 '

["(v" vi - vi v"))

Hence,

by parts. Since

tK- - K- ' In

:<v v dx = 0' r 'a

and

I xv v = uU

There are sorne special functions that are closely related to

Bessel funct ions. Def ine

' '_ I

I (x) = e-2 vnL J ( ix) =vv

I /xv I , I /x\z 1 /x\4=_i l; I r f '7'-"- '- r= | r '=:"- ':-- lv :Iv+I) \z/ )

(v+I) \a i

. t lv+L)\v+. | \ . t\

There I . . (x) are cal led modif ied Bessel funcf ion+ or

of cornplex argurnent. Notice that the Ir(x) are real

for x ) 0 They satisfy the equation

jBessel funct ions

and positive

z2 wu + z w' - (zz + uz) w = 0 (29)

(30)

provides a second solution of. (291, If v = n, Io = I_rr.. In this

case, a second solution Ko (x) is obtained by taking the tirnit of

(23t

symptotic formulas

-60-

I (x) ea "^ J ( ix)'v

I (x) - I (x)1l -v ' v '

A tx, = ? - .--v I SU3. yT

x,tx l -*$) for v)o

raKo(x) - - I t "e { + y }61=1

)\ /

The behavior as x * + 6 follows from the a

(20) and (2I) that were der ived for J and Yvv

I..(x) = .-b rr s 6xl -'VV

"-i" {= cos (ix - + il

If v is not an integer n,

as v *n ,

Near x=0,

r rxl - 1f1'v

Now

- 6L-

e-*cos

- :-v-:t7

1,/1.

z I rvi xe- e

Hence, ( 23 ) be c ome s

L-sing this result for lr,

iound:

.11

-x

- Ixv t altx

+T

.1f :t:- e

1t\

, i - t Lzrxf-- ; -+-;-

-z 1lv7 e-x

. l I -hv+--- e| ! a1tx

-x

the asyrrrptotic behavior of K can be

. .n-vv11.

s1n y7t-Ea-x

\2x

important cla s s

sider the three

-62-

of speci.al functions. To see

dimensional Laplace equation

s are another

how they ar ise, con-

in spher ical c oordinate s.

u = rv y(x) , x = cos 0 is.a solut ion i f

(1 - :*) yt t - ?ryt + v(v +1) y = 0

Legendrers equation. There are regular singularities at

and x = 6 . The indici{l equation corresponding to

ie cz = 0 Hence there is one solution that is resular

and another solution with a logarithmic singula fity a!

x = 1 The same statements hold for x = -l , In general, the

solut ion which is regular at x=l is s ingular at x=- l and

vice versa. Solutions that are regular at both x = I and x = - I

ar ise only in the special cages y = n or - (n + l ) In these

special cases, there are polynomial solutions p (x) of degree n .

They are normalized so that po(l) = 1. It is these Legend.re polv_

nomials pn(x) that warrent our chiefest concern. In most physical

A , . ^ Ar. 8 .";. n fu., . I 0.. _ .,5; (r- sln u 6;) + ffi (sur o Ub., - ;i;-- E7- - "

o (x) = I o

6 n! dx

are given by

f ,o -',i'

(1)

This is

x =+ I

x =t I

at x=

problems, the solution u = ru y (cos 0) is regular at 0 = 0 ,

Hence y(x) must be regular at x = * I , and this forces v

be a Legendre polynomial.

The Legendre polynornials Rodrigues formula

(z l

-63-

l i i s resul t can be establ ished via a t r ick: y(x) = (x. - r )

time s with the

sati sfie s

(3)

help of

( l -xz )yu + 2( n -1):<y' + Zn y

Iach term in (3) can be diffe rentiated rn

- : ahni f zre rr la

l le re sul t is

ct,- (uv) =clx

.,r.,, (n ) * ,,.,r,rr(n -1) * o$J .,,' ..{n -2) * (n)1tv

( l -xz1"(n)" - Zn x "(n) ' -

n(n- l ) y( t )*

Z(n -1) x y( t ) ' * zn (n - I ) " ( :n

) * zn "(n

) - o

c:ich simplifie s to

/- \ r r -

(n) , (n)( l -xz;

" ' " ' - 2x y*- ' + n(n-I) y"" = O

lc check the normalization,

y( ' ) (*) = i : (x+r)n (x- i )nclx

(x+l)n n ! + n (x+I)n

/- \=-e3ce y ' \ I l = L n! . A

l t t ( - t ) = 1-z)n n! Hence,

, ' 'nPn (-r ' = (-L '

ll i s could have also been surmized from

shows that

'1 n!(x- l ) +. . .

similar calculation

the evenness of the P:n's

-64-

and the oddness of the pz1g1's.

I f n)m,

tJ.l Pn(x) P,,'(x) dx =

1 I ^n

_ rrn

Ztt- ,, '! m! J-l

d*t. '- -^'

dr.t t^-

(- l) ' J'- 1r-l)n + (*-r)* ax el2o** ,,! m! J-l " u*m+n

t^

112 -1)m is a polynornial of degree $2m Therefore, its n+mth

deriv4ive ig zero since n > rn Hence,

"1J, rr, (*) prrr(x) dx = o- l

It is cleat that the same result holds i.f rn < n We conclude that

the prr(x) form a set of orthogonal pollmomials on (_1, l).

Sett ing n=m in (4) gives

If p2 (*) d*=

" I - ! l

o$ 1r- r*-u '4 ,"-r)n dx =Z" (n! )z " - l dx '

Lznlt ft (r_,*)o o*2"(n!12 --1 (5)

Now

I

l1r ' t r f

- l - - l

t_| (t-", )" dx +

-?n | +n

-65-

( I -x- I dx =

, .n- I -4n x- (l -x- ) O.x =I x(I-xz

-Zr:

AlI irnportant

expansion theorern.

f (x)

Hence,

n-r

Witlx this value for I r r , (5) becomes

I

{, ni t*r

, , , .n- l .i l - : f ) dx --

n-I

the Legendre polynornials is the

cont inuous. on -1 ( x '< 1, then

(x), where (6)

f (x) p (x) dx_n

^lznl

n atL+t

)

2ox =-

4rt+ |

prope rty of

1r r tx l rs

s=/ a D0..

Izn+I f= --'i- I

L"t

The convergence is uniform in intervals where f(x) is srnooth. The

Weierstrase approxirnation theorern* provides a plausibility argurnent

for this result. Suppose f(x) is continuous on a S x S b Then

for any € > 0, there is an integer N = N(e) and a pol lmorrr ia l

p(x) of degree N such that

=TouilnT-5na-trilbert, p. 65.

-OO-

l i (x) - p(x) | < e on a (x (b

In words, f(x) can be unifo rrnly approximated to any degree of ac-

curacy by polynomials. Since any polynomial can be written as a

sum of Legendre polynornials, it follows that f(x) can be approxi-

mated to any degree of accuracy by sums of Legendre polynornials '

The coefficients an are di ctated by the orthogonality pro-

perties of Legendre polynornials. Multiplying (5) by Prr(x) and

integrat ing {rom -1 (x( l g ives

{

l iqJ. t(*) p,r(x) dx = I^ ".,

p (x)- lJ

else O'< = .--.-zl.t+ I

p (x) dx'n

^1I pz (x)

' - l - l r

(7)

II f n * t r , f p. . (*) p-(x) dx = O

_1 t t tL

Hence, (11) becomes

(8)

have

$0

a gene rating functi on

D(x, t_n

1I r r* t o (x) dx=" 3,_l "^' ,lt--' --- 'u 21t+I

The Legendre polynornials

G(x,t) = i_+tr

This result can be established by an appealing physical atgument.

The potentiil field of a point charge is p = fR = distance from the

1charge. Hence, Vz (i) = O . Now adoPt a polar coordinate systern

(r ,0, e) whose or igin is a uni t d istance from the charge.

R -- Gz::iJ"-=;-5"

Any s olution of

and 0 inside

_O'-

Laplace's equat ion that is a

r ( 1 can be exPressed bY

ec) a r"p (cos 0)? n -n

srnooth function

a ser ies

used to f ind re currence

can be tewri t ten as

g=) a r^ 'o (cos 0)

t "

ia r ( 1. To find the 3-, set 0 = 0. Then

,Q0.-eq

; ; := i aor"P,r( l ) =t ""

Hence, the "r,

are all uni tY' so

rSn.-1- = ) r.. n_ (x\

t- ?.tt].- +!- a,fx 1.,

Hence,

The generating function in (8) can

relations for the Legendre polynomials'

( l '2x t + tz)Ge (x, t1

Differentiating with resPect to t gives

(e)

be

(8)

(r - 2x t + t') 3f * ,t - x) G = o

6

Substituting G -- )- P- (x)-T

side of this equation to be

tirag the coefficient of tn

tn into this equation allows the left hand

expressed as a power ser ies in t . Equa-

to zero gives the recurrence relat ion

_og-

(2n + l ) x p_-(x) = (n + 1)^npn+l(x) + n pll_t(x) (10 )

This equation allows Pn+l to be calculated from Pr, "td Prr-l .

If (9) is differentiated with respect to x instead of t, a

second recursion fo rrnula results!

P," = Poif ' 2x P' * Pr-, (11)

The recurgion forrnula s are useful in studying the zeros of

pn . Specifically, we want to show that Pn+l has one zero be-

tween consequtive zeros of Pn"" and that pn has n distinct zeros

in -1 ( x ( I To dernonstrate the ProPosition for n = ?, note

the graphs of pz (x) and p3 (x):

Now suppose that pu(x) has u roots in -1 (x( l for u<n

and that pn has one root between consecutive roots of pn-I. Let

Lt, Lz and (3 be three consequtive roots of pn_1 Since pn_l

has n- l roots in -1 ( x ( L, Lt , lz, h are sirnple roots. pn

has one root 6f in 6r < x ( 6z and another root L: in (z 1 * <

The succeeding graph illustrates the situation:

-69 -

Using the recursion relat ion (8),

(n + r) prr*, tEIl + " nrr-t ((i) = o

(n + 1) prr*, tgll + " nrr-1 ((I) = o

Hence, pn+l

D , change s_ n-I

changes sign

root between

accounts for

more roots.

and pn_l have oPPosite

*f

sign between ( , and (2 ,

+signs at Lt and

it f olIows thatt

/ q i i^a92

' ntr

between 6l "na gI ' Hence, Pn+l has at least one

6l ara EJ . Since pn has n roots, this argurnent

n - 1 roots of Pn+l. It remains to account for two

. the n- l roots that areaccounted for

21 the two rernaining roots

I

- t u-

By an argument sirnilar in spirit to the preceeding, it is possible to

show tiat pn+l has one root in 6-"* ( x < 1 and anothe r in

-I ( x ( ( min,where (rnax (Emin) is the largest (smallest) root

of pn N9* n + 1 roots of pn+l have been found. Since

pn+l is a polynomial of degree n * l, all the roots have been

accounted for, Hence, tlxere are no additional roots in Sl < * < EI .

This means pn+l has exactly one root between two consecutive

roots of pn .

The Legendre polynornials;r.re an exarnple of a general class of

ortho gonal polynomials.

suc h that

-oJ W(x)Q, ' (x)Q*(x)dx

- 0 i f n#m,a

where a, b, W(x) are given.

Other irrrportant exarnple s are:

Herrni te polynornials a=-ao, b=€, w = e 'xz/2

Laguerre polynomials a=0, b=o, w=e-x

Tchebycheff polynomials a=-I , b=1, w=1f - 53z1- i

The Tchebyeheff polynomials have the mini- max propelty thatn nl l

P = x" + arx" ' '+ . . . * . r , deviates least f rom zero in {_ 1, l ) , i .e.

max f f I is a rninirnum, when p is the Tchebychell polynomial of

degree n. See Courant-Hi lbert p. 89.

These are polynornials Qrr(x) of degree n

-71-

VII. Fourier Serie s

If the tlisonometric serie s

€o$+I(a cosnx+b sinnx)z #t ' n n

coavelges, it defines a periodic fi:nction f(x), with period Zzr. lf the

convergence to f(x) is u:ri.forrn, the coefficients an and b' can be

erptessed itr terrns of f(x). To do this, note that the trigonornetric

f imct ions 1, cos x, cos 2x, s in x, s in Zx, . . . are orthogonal on

0(x(ztr . That is,

) , i Zn' :dx=J cos mx cos nx cx =

J sln firr su:. nx

00

7d

,[ "o" -* sfut nx dx = 0

0

iJ rn# n and

cos2 mx dx = sinz nx dx = z'

iJ n > 0,

Multiplying both sides of the equation

. "Sf(x) =

7 + ) (a- cos nx + b. sin nx)d=l "

by cos mx and integlating frorn 0 to Z1t gives

21 2r| . . - z^ t

J f (x) cos mxdx=] J cos mxdx+00

?t@t

t (a- cos nx + b- sin nx)dx =J cos f f i , - , n no n=r

(1)

,z

2ttr

0

2r

f

0

(z l

I

0

c o s :irx d.x * f,, t.r,n=r

-72-

I cos mxcosnxdx+ b

a 1t i I n>-IN

)tr

J cos mx sia nx dx) =

I aad

2t

/ f t" la" = aoz i f m = 0,o

The iaterchaage of sr:.:a'nation a'.d integratioa is permissibre due to the

unifo rrrr convergeace of tb.e seriee (I). Heace,

2zr1r

"_ =

i J f(x) cos mx dx.

If (2) is rnultiplied by sia mr< iastead of coe mx, aad tbe preceedi,,g

calcrrlatioa rcpeated., oDe obtainBo

2trbrr, =

i J f(x) sia mx dx.0

Now suppose f(x) is aay piecewise coutinuoss fi:actioa that is

absolutely integrable oa 0-<x< Zr ' . ( i .e. / l f t " l la" <o), I t ie

possible to calculate the coefficieats a- aado b- accordiag to (3) aadl lN

(a) aud constluct the &urier series of f(x):

p1*1 =g .,- $ ,- L - '-t * 3r("o

cos n:r + bo sia nx)

The crucial questioas are: When doeg thie series converge?

Aad if it does, whea is F(x) = f(x)? Ansvrera are provided by the

Theorem : U f(x) is piecewise differeatiable aad absolutely iotegrable

on 0 -<

x < 2", then the foutier setieg coaverges uaiformly oa iate

where f(x) is differentiable, I:1 t*ris case, F(x) = g1*1. If f(x) has a

sirnple jump at x = a, then the fourier serieg converges to

F(a) = +(f(a + o) - f (a - o)) ,

_Weinberger, Chapt. fV, Churchill, Chapt. W. (Fourier Series aadBoundary Value Problerng)

(3)

l4\

- I ) -

but the convergence is not uniforrn in any neighborhood of x = a.

t3)

:edi- t

tremma

Hence

The proof rests on a key result known as the Riemann Lebesguet -

**I I t J l f (x) ldx

(o, thenaL

lirn J f(x) sin nx dx =n+co a

"bIirn J f(x) cos nx dx = 0.n+€a

A weak proof can be outlined: Let f(x) be continuously

^bdi l ferent iable in a(x- b. Choose any €>0, Sinc e J l f1x1 lax

exists, there is a 6>0 for which

a+6

J l r (x) ld*. e.

Now

nb - . inx.J !(xre dx =

b. , , l t lx,r (xre dx.

afo

fJa

" , . lnx,r(xre cx f I

a+6

lhl f - - , inx.lJ r tx le cxla

1bu1*1.t*u*1.a+6 |

(5)

€+

-{r integration by parts shows that

br " , . rna,.J- r (xre cu. =

a+o(6)

**Whittaker and Watson, Modern Analysis,Churchi l l , Four ier Ser ies, p. 86.

,. f(x) -inx.' u rb f,(x) inx,t;- ' l "+O -

JU --"- "

dx-

I

p. 172.

Both terms on the right hand side of (6) approach zero as rr + €.

Hence, there is an N ) 0 Euch that

l^u I| .[ r1x1"mxax I .. ir ,, r lr.la+6 |

Sub stitutiag this regult into (5) gives

l -b IlJ- r1"pua*l .2" i r orN.l ; l

Hence,h| . -

/ - f lx leuax - 0 as rr+oor

whicb is equivalent torb

"b ' tJ f(x) cos nx dx = .[ f1"1

"io nx dx * 0 as n * o.aa

To study the convergence of the fourier series, consider thepartial surn

Na^ FrsN(x) = 7L + L@^ cos nx + b- sin nx; =

1--

Ztt '.Tl r r I

+ J t(t){i + ) cos nt cos nx + sin nt sin nx} dt =o1

ztN| | - l e

;J f@{i + )cos n(t _ x)}dt . F)01

By taking the real part of the identity

f l . *"=t-" i (T+r)vi r t

o I - e ' '

t \ t 1.r $_ _ rs infN+7)vZ' ' Lccs " ' t= i - .I srn t

it follows tiat

Hence, (7 ) becomes

(8)

sN(x) f(t)

'|sin{ (N + i)(t - x)}---r----

srn 7 ( t - x l(e)

2t

zl t J0

I .z)v

Integlating both sides of (8) frorn a to Zit + q glve s

a sin(N +

sin j

Henc e,

Z1t+1rzr=* |z, l

ady.

, , r ;* s in6 + | )y

f(x) = :- J f(x) -6"

=-'{. slnt

Ztr 1, , .I r - , .

srn(N +2,( t_ x) .

rt J itxt ..-T:..:(it.

o slrt 7(t _ x)

Subtracting (10) frorn (t) gives

s*(x) - f(x) =

{f( t) . - f(x)} "tn6

+ }11t _ x1at.sin j ( t-x)

(10 )

(1r )

Z1r1t

ZTJ0

U 0<x<22,

f /r ' l - f /Y\g(t) =..:.I;:,:#

sint( t - x)

is absolutely integrable in inte rvals containing every point 0 -<t S Zz,

with the possible exception of t = x. If f(t) is diffe rentiable at t = x.

then g(t) is continuous at t = x provided we define

c(x) = zf ' (x) '

Hence, the right hand side of (ll) goes to zero as N * o by the

Riernann Lebesgue L ernrna. We have proved that s*(x) * f(x) as

N-co, provided 0(x(Z1t arrd f ' (x) exists. I f f ' (x) is p iecewise

smooth, the r . h. s. is O(1/N) uni formly in x, Hence convergence

is uniform.

I

- aO-

If f(t) has a jurnp at t = x, the argurnent must be

Extend f( t ) outside of 0-<t<21t by def in iag f(x * Zr) = f(x) .

be written ae

sin N + |)r

"*zgives

rnodified.

Then (9) can

s*(x) = fi_fo rtr * *y (ElA-

Integlating both sidee of (8) frorn 0 to tr

Hence

, .z s in(N + |)zJ _7'n7=; '0 su:' Z

r r t . p i ' I6* | )" .f (x+ 0) =+ J fe + 0)a-dr.

o sln Z

Sirnilarly,

r eo s in6 + |1rf (x - 0) =: ) r .@ - 0)-

_1t "bZ

(U), (I3), (14) can now be cornbi:red to yield

s*(x) - | t r tx+ ol+ r(x- o)) = (rs)

L rr f , (*+ 7) - f (x+o) , - , . i ,f rJ

-

s in (N+)irdr 'o .b i

I 10 f (x+ z)- f (x- o) . - , . l ,fr J ' -- ' ; . : sin (N + 2ir dr '

_1r "D Z

for 0 < x< Z1r. Now suppose f(t) has right hand aad left hand

derivaf,ives at t = x. That is,

lirn f(x + 6) - f(x - 0). ! t

o+u o

lirn f(x + 6) - f(x - 0)

6*0- 6

(13 )

(r4)AQ

- I I -

bohh exist. Then

c'(d=S$sln Z

is cont i : ruous on 0(7-<,r and

g'( t l=S#sln Z

is contiuuous on - 1r -< r( 0'

.Hence' the right hand side of (15) goes to

zero as N - o bY the Riemana l-'ebesgue Lemrna ' We conclude

s*tx) * | t r t* + o) + r(x - o)) (16)

as N-co, provided 0€x( 2r arrd ' f has r ight and lef t hand

derivatives at r.. (16) can be used to clea! up the cases x = 0 and

z= Zn. Frorn the exteoded definition of f' f(0-) = f(21t)' Hence

s*(o) * 1t101 = |1r to* l + f (o-)) =

I l t tot * r tzol l .

simi lar ly, f . (Tn+1= f(0), and

F(2tr1 =lc?orl + f (27r-\) =

;i,,0, * ,,'",,.Suppose f(x) is def ined on -L-<x<L' Then 9(y) = f#)

is defined on - tr < y < rr. E(y) hae the Fourier series expansion

ao

vlfl = |

+ X{"r, cos nY + b' sin nY)' (1?)

1

where cos ny dy and1t

'I /'a^ =; ) ElYl

(r8)

1t. l tb- =; J g (y) sin nY dY. (19 )

-78_

Let y = {x.

rhen (17), (18), (19) becorne

a(20 )

(z4l

the fourie r series of

l f (x) , 0<x<LI

(x)=\I

l11t-"1 ' -L<x<o

odd ( i . e. f (x) = - f ( - x)) ' t "hen

a =0

. z , l - - . . nt : . .b,,=i J r tx) s in;ax0

) is- L <x<L, (23

Sirnilarly, if f(x

f(x) = ? + Xtao "o" f + bn sinT)

' , "where % =

i _i f(x) cos offax ana (2r)

L

u = I f f(*) "io $a*. Gzlt r t - i

Suppo se f(x) is even' That is, f (x) = f(-x) . Then (21) ar l .d (?21

irnply

z rL - . . i * .""

= I J

t(x) cos -t-ax anc

b = u.n

Henc e,

coa^ \.r n?l*

r(x) =- ' + Lzn cos-- .

1

(z3l

This is called a Fourier cosine series. If f(x) is a general function on

andcoSr . t'tzx

Itx.l = L bn sln

T- .

I

?his is " $!g!39-993!g. If f(x) is a general function on

- L < x < L, (24) is the Fourier ser ies of

I f (z l , 0<x<L\i

c(*) = {I. . - f ( -x) , -L<x<0

Sorne graphs provide a good illustration of the differences betrneen the

Foul ier ser ies, Four ier cosine ser ies and Four ier s ine ser ies:

-79_

Fourier ser ies of f (x)

Fourier cosine ser ies of f (x)

t

-80-

Fourier sine series of f (x)

Notice that info rmation about f(x) in - L

sine and co sine series.

Another useful form of the Fourier

< :< ( 0 is lost i.rt' the fourier

oo ..i

f (*) =? * I ("- cosnx+ b-i 'n n

is obtained by rnaking the substitutions

s erie s

sin nx) (25\

L inxcos nx =Z(e + - lnx.e)

The result is

sin nx = *," t"" - u-h*

f (x) = sl inx

-co(26\

where

n

I t

l i@) -\It i .l t (a

- ib) i fn>0nn

+ib ) i f n<0 and-n -n

a^co =T

(25) is cal led a complex Fourier ser ies. The fulct ions

n = -€ + co ale oTthogonal on 0 -<

x ( 2r. That is,

- 81-

Multiplying (26)

2r (0. . ' ,#nf lrua llTut, I 'I e e dx= (

- | r -o f" ' t l r= n

. - irnxby . --'-- and integrating from 0 to Z1l

_Lt

) ,,r

" =.1 f . ' - i lT lx '

rn Tr J tllrle cx'

0

f(x) is real, then inspection of (Z?) reveals that c0 is

Suppo se (26) converges

2tr 2r

J ea"= "f rX.0

€

Zlt)ccL' nn

r:nifo rmlv in 0 -< x -<

2?'. Then

inx. ,F- - imx. .,( 4ctne ldx =

give s

(27 |

real and that

(29)

= ,e'+ f {arrz + b,,z)) .

This result is known as Parseva s theorern. It is actually true provided

-Z1t| -2.J fz dx ( .o, even if the Fourier series does not converge pointwise.

0Suppose we atternpt to approximate f(x) by a finite trigonornetric

sutn

N@nF

A * L"n

cos nx + F' sin nx. (28)

What is the rrbestrr way to do this? One cliterion is to minirnize the

"mean square deviatioa'l

zlnN

/ frt"t - ? - Lr"ncos kx + Fu sin kx))2dx =.o l

Ztt N

I t" * + ,l loo" + po21 + $aozol

N

Zzr) , (a,_a,_ + B'_b., . ) - l tcroao,IJ KK KI.

and b are the farniliar r.oulier coefficients. . By elernentary:phere a

-82-

methods, one can show that the right hand side of (29) is rninimurn whenok =

"k and Bk = bk. With thege choices of an and pO, the trigonometlic

surn (28) gives the best mean squale approxirnatlon to f(x). The lefthand side of (29) is always greate! than or equal to zero. Hence, settineot = .t and pO = bt h (Z!) gives Bessgl 's ineqr:al i tv:

Z7rN

| {"o. + unrl + $ -. } / r.a*.

Parseval's theorern states that the left hand

convergen to the right hand side as N * o.

lirnN*o

side of Bessel s inequality

Equivalently,Zlt Nf , " a^ Yr

! *, - Z- - L@n cos nx + bo sin nx))zdx = 0.

This is called convergeace in the rnean, By virtue of tJ:is convergencein the rnean, the trigonometric fiurctions cos n:(, sin nx,n = 0,1, , , . are said to be cornplete in L, norm.

It is now time to seei sorne Fourier series in action:

ex. l ) Let f (x) = 3 in 0<x<Z?r. The troul ier coeff ic ientsare

2tr1rao=i J xdx = ?r

0

)r

"n=iJxcosnxdx=00

Zt

u_--- , x s ln n:a ctx =0

l . * "o"

,o.2o7t - -''o---l o

2r.1 1 cosnx, Z, iJ ---ox= - ;

0

Heace, f (x) =; has the Fourier ser ies

-83-

f (x l=v-2

for 0 ( :x1Ztt . The convergence

I:spect ion reveals F(0) = P12o1 =

1F(0) = P1271 =;( f (0) + f . (azr l l= n.

and F(x):

For f (x) = x, Parseval s theorem reads) r

Since fr (x) = 1,

terrrrwise to obtaia

€

L= - a /_J cos1

This is patent ly fa lse. The ser ies on

failure is due to the discontinuities in

ex. Z) f (x) = x in - 1t <x<1t.

The Four ier ser ies of f (x) is

€-$1,/ . - t l Oft__

I xzax0

as\I tzun' 6I

=+ = l-ltznz + 4

one rnig ht be ternpted to differentiate F(x)

00\1 sin nx

nI

is uni forrn in 6<x<Z7r-6, 6>0,

r. This tallies with the general. result

The succeeding graph displays f(x)

t lx)

f lt)

IIK.

the right hand

F(x) at x= Q

side diverges. The

and. x= Ztt .

I

F(x) =2l t - t )o* l sln nx

The ser ies

!epre sents

'84-

differs frorn the one given in the first

f(x) in a diffe rent interval. A graph

exarnple because F(x)

illustrates the difference:

The convergence of the series is r:niforrn in

terrn by terrn diffe rentiation of thc series is

tinuitie s in F(x) al x. = + 1t.

ex.3) f (x) = lx l in - i r<x.<1t.

The Fourier coefficients are

ao=1l

- 7t + 6 < x < 7 - 6. Again,

not valid because of discon-

This tirne, terrn by telrn

z f1ln1lJ

x cos nxdx=

1l,a i

;frt x s1n nxl o -

,- '2 (cos nzr - 1) =frHence, f(x) has the Fourier series

alltYt J

0

6

r (*)= -a XI

The convergence is unifo rrn in -

dilferentiation of F(x) give s

cos ( 2n-1) x,

r(2n-1)-

1t < a, < 1t.

ao4sr

1

sln (2n-1) x-(zffF'(x)

r 'f ' (x) =(

L-'In this case, notice that F(x)

which is a valid representation

-85-

of the derivative

in 0<x< 7r

in -z < 3 < g

is continuous, as evidenced by the

F(ut

I

graph:

These exarnples raise a question3

differentiation of Fourier series valid?

fourier series with coefficients e,, and

0 ( x -<

Ztr, and. f.(Ol = 112v1,

2ra =: I

ta 1fr0

tQ1

- 'f I f(x) cos nx1 +

0

'When is terrn by terrn

Suppose f ' (x) has a convergent

9o, If f1x1 is contiauous in

f ' (x)

2

iJ0

cos nx dx =

f (x) s in nx dx=nb

Similarly,

Hence,

0 = - na

€d sr-

E L ("r, cos n:< + bn sin nx) =1

sr,L (n b- c o s lx'r - na_ sin nx).!n

I

terrn by terrn differentiation is valid i{We conclude that f (x) is con-

T - ;+ sin nx'r $ rr - "o" ,,*))

even iJ the four ier ser ies (30) does not converge.

The proof is simple. Sinc e F'(x) is piecewise c ontinuous,

g(x) = F(x) - "9* t" represented by a Fourier series

^-An \ag(x) = ?o + )r(An cos nx +Bn sin nx).

I

The coefficients A and B are easilv calculated:nn

Z7r

A =_: I q(x) cos nxdx=n " l

I , g(x) sin nx, zz I?L.---- j - i

0

2r

t . - , , a^. s innx,- | t t txt - =.w I -qx

=ZN

0

Ztt

B,,=*J g(x) s in nx dx =

0

ran - f . (Zt t l .aa-; ; - +; '

Z7rr . . s in nx,I g ' (x) - (1x =

II

0

b

-J i1 11>- 1.n

- 86-

tinuous in 0 -<

x < zz. f(0) = f.(zrl, and f (x) has a convergent Fouriet

ser ies.

Integlation is less dernanding. Suppose f(x) is piecewise con-

tinuous and integrable, aad has Fourier series

coI t .T

*an + ) . (a_ cos nx + b_ sin nx).! - , 'n n

x 'F(x)= 1 f(x)dx =

0

(30)

Then

Sirnilarly,

(31)

-6 t -

) 'n

l .ao=iJ f (x)dx=]rpry.

0

Hence, R = " t ; r - .s r_n

n rr u_/ r .

With these values of Ao aad Bn, (31) becornes

€,P1111 = 3S + 4o . T,

ot -^- -- . ' " r ,z f r L l ' : co8 nr. +- s in nx).'1 'n

Since F(0) = 0, setting x = 0 in i:he above equation gives€,

+ = X+ cos rx*'1

Hence,

ia- b-+ L; s in nx +; i ( t - cos nx)) .

I

Ztr N

* I ,"(x)dx = fn2 laoz + b,rz).o1

F(x) =

lfhat error ig incurred by tn:ncating a Fourier series to itsfirst N terrrrs? Suppo se !(x) is continuous and periodic, and thatf (x) has a Fourier ser ies, Under these hypothesis, f r (x) isrepre sented by

€

f ' (x) = X(D b' ibs nx - n ao sin nx),I

where a' and b' are the coefficients in the Fourier series of f(x).Applyiag Parseva s theorem to f' gives

&nX

z

(32)

N

s*(x) = ? - I

(a' cos nx + bn sin nx).i - -

r nen

_ 88_

M

ls*(x) - s*(x) | = | f . r ,

"o" nx+ bn sin nxl . (33)

N+1

At this poiat, the Cauchv Schwartz inequality is helpful, Given a andn9' , r n=Ir . . .N, i t says

NNN

lX"oool -. rf,oo, lltf po, t*.I . l 1

Hence,

la. cos ax + bn sin nxl <

I(arrz + brr2 1;1cos2 nx + sia2 nx1; = larrz * brr. )i,

and (32 ) becomes

, . . Y - - !I sM(x) - sN(x) | *< ], (an + bnz')z

N+l

Ms1 l - - - I= Lf ;X" ' (anz+bn2Dz.

N+1

Applying the cauchy schwartz inequality once rrrole, this becomes

I s*(x) - s;(x)l <

MMrl

1 1-\ L t lz(1, a2 {arrz + b,,2))t.N+l N+l

But (32) irnplies

Ztt Mla

* J t,, (x)dx > l,n, (" " + b_ z ).r N*''r

n n' '

Hence,

I s*(x) _ s*(x) l <M Ztr

1,r '1.1 . r . . , +

-(L31"(J

f ' (x)dx)2.ot N+l o

_ 89_

L etti.ag M * co gives the final result:

00 ZIt

If f(x) has a jurnp at x = a, then the Fourier

does not converge r:aiformly to f(x) in a neighbothood

exarnple is provided by Gibb's phenomenon. Let

r(x, =

x<0

Now

Hence. s- . (x)' atn- I'

corresponds to the

f r1x1 - s*(x)l * t, X f,r i t I rz(x)dx13.tr2 N+l o

.7r z. "6

c izr 2n+

"2,,,-rffi) =#J ftffa'=0

?f ' s in t Z { s:nt , .iJ

- - -Tar*;J -_ot=Zrn sini ' :- ztrL

1.179 as tn + co.

ser ies of f (x)

of a. A conclete

x)0

[ ' '

t.,The Fourier series of f(x) is

f@) =! (s inx+ " t r** " tor t* * . . . , .

On the other hand, there ia an identity

cos 7+ cos 3r+ . . . cos(Zm - lh =+ srt ] -zm? .

Integ rating both sides of this identity frorn 0 to x gives

srrrr-r(x) =rr l t " t** "- ' ** . , . " t i !1Ti t , " ,=

2 tx sin ltnr ,1f J gin 7

0

si-- , (x) = Z "!1 zrro .ZIn- L' ,? Srn x

has extrerna when sin Zmx = 0, or

tallest maximum. We calcuLate

k?Ix==_axn

(34)

- 90-

This result shows 1fie non- unifo rrn

gence were uaiform, the lirniting value of

not l . I79.

coavergence. If the. conver_.7f

"Zrrr_ t$;) would be I and

(34) can be understood on a heuristic basis. The succeeding graph shows

Notice that there is always an

7r

J s i*-1(x)0

undershoot near x = 0. Since

"77-Jf-(x)dx=2,0

the undershoot nea! x = 0 rnust be balanced by an overshoot, whichoccurs in the lirst hurnp centered at :< =

*.Proof of Parseva s Theorern.

Given f (x), piecewise continuous

obviously construct f.(x), where f.(x)

I 'o ,, ' f . )a dx < e

for any € > 0. By the

0

proof of Besselrs inequal i ty

e. Here S*"

and square integrable, we c arl

is periodic and Cl , such that

J {r - s*)'zax -. /tr - sN€)zdx

".rl a - 1)zax +.2! {tr_ s*.)zdx _< € + €

when N > N(e) such that second iategral is less thandenotes sum of N terrns of Fourier series of f.(x),

formly, because f. is C I and periodic,

s2rrr- 1(x) neat x=0,

S*a * fu, uni-

- 90a-

Order of magnitude of Fourier coeff ic ients.

Suppose f , f ' , , . . f (k- I ) are cont inuous and per iodi" , "nd

f (k)

is p,w. cont inuous and integrable. Then the Fourier coeff ic ients are

^. -k. ku(n ) , i .e. D3.-0 as n-o. For on repeated iategrat ion by

parts

ra = (- t )k 1zt , t t ) cos -- . r - .

" --f

JO r s' llx dx'

and resul t fo l lows f tom Riemann-Lebesque Lemrna.

I

- 9 l -

Let f (x) be def ined in _m<x(o, I . I : the iaterval _L<x<L,f(x) can be represented by a Fourier ser ies

€ - azral_

f (x) = ) .c e !gn

-cO

where

(1)

(z l

r(x) =+ _i_ l"r,r,.-#,,-*,u*. (3)-€

-LThe Fourier ser ies of f (x) is per iodic with per iod ZL. f (x) is ingeneral aperiodic. Hence, the Fourier series usually contains no infor_mat ion about f (x) outs id.e of _L(x<L, I t is tempt ing to let L*oo,i'' the hope of obtaining a representation of f(x) that is varid everywhere.Let k =f f anaoo=i. (3) can be wrf t ren as

,Lt (x) = h X--ou / r {61.- ik{6-*)ur, ( - - < n < _)

k=# _L

oof r . , . r .

J l r (x, ldr< exists, then_co

L

/ r1g1"-ik(t-x)dg "or*-L

.erges to a func t ion g(k,x) as L _o. Hence,

we expect

LTLI ^

_uX_+

" t r=fr J t (L l" "dg.- I

Substituting (Z) ir:to (1) gives

lsrZ7I LJ

K=:IJ

g(k, r . )Ak - f (x) as L * cq .

-9a-

But the left hand gide of the above erpression is a Riemann surn foro

1a

fr J e(k, x)dk. Sinc e Ak - 0 as L - co, we expecr-a

6

f(x\ = -- J e(k,x)dt =-€

(o6

L f t 1" r,r , ,-- ik((-x) *̂E) dk.-€ -co

at | = :<, (4) generaliies iato

| t r t " * o) + f (x - o)) =

co6

* t t / i1s1.- ik(t-")ag1ar,

(41

If f(l) has a jump

(5)-€ -oO

which slerns to be a reasonable extrapolation in view of our experience

with Fourier serie s.

The argurnent leading up to (5) is naught but heuristic speculation.

It lernains to conjure up a proof. Suppose f(x) is piecewise smooth andco

J Jf(6) ldL <o.

Then

_00

_oo

/ 11g1.-ft(6-")4,.oo

is wriformly convergent by the comparison te st. Henc e,A€

* t t / r1g1"- i t (6-")a61ar=_A _rc

oA

I uglt I "-ik(6-x)646s =

-co -A

'I

6

r

-95-

€

I I xtt$ds = rA(x). rcl-oO

r s inAx. zr\ow J .-- dx =

Z was done last quarter as easy rneat for the0

Residue theorern. Hence

€

ry =I I n*+ or${{(1- rDagx

f(x - o) x

T=* J t@ - o) s in{A(L- x)}dto_u - ' l -x

Subtracting (?) and (8) from (6) gives.'

fA(x) - , ( f (x + 0) + f (x - o)) =

€r Fl f \ dr-- ,

^ \J :ff:/ sin {A(E - x)}d6 +i 5-x

x! r l l3)_f(x_0)r J --:t_< sin tA(6 - x)l d6'

_co

If f(!) has right and left hand derivatives at ! = :<, 15..,

" r , , , - f (L l - f (x + o)rsr- . f f i

is cont inuous in !>x and

e' 1r1 = f (6) ' - f0< - o)5-^,

is continuous in | -< x. Hence, the right hand side of (!) goes to zero

as A * o by the Riernann Lebe sgue lemrna.

The Fourier integral forrnula can be tephrased. Let€

rGt) =l f " - ikr t* la*-'

- 21,

-!* s r(x'q:< ' (10 )

If f(x) is absolutely integrable, then F(k) is defined for real k, and

(7)

(8)

I1t

(e)

-94-

analytic c ontinuation defines it for cornplex k. F(k) is called the

Fourier Transform of f(x). Sub stituting the definition (10) into the

Fourier integral fo rmula givesco

f(x) = rFG)eikxdk..;

The pair (10) and (11) is equivalent to the Fourier i:etegral formula.

ex. ) Let f (x) = 1r16-ax?. Then

F(k)

lf we require f(x)dx = l , then N =G . Hence,

(r)

e2n

N r -ax2- ikx-=fr J e dx=-a

kz oo , ik._Ta f

_atx + E) .

J e ( l ) .=

-6

lN 4a-- e- {ta

6

I-€

(Lz)

The succ eeding graphs

I

l4

F - axzr(xr = J?e

. - --.l: . I +eF(k) =

77e (13)

show the rnain features of f(x) and F(k):

-95-

F (r l

zF

limiting behavior. It is

are the se:

x* 0

x= 0

for al l a.

ernbodie s this

as a - 0.

)

I

irnpoltant feature s of f(xr10,Ir(x) * iIL- '

a

/ r1*P* =-cO

Drrec 4elte fun.ilion 6 (x)

The

rne

a - ao.

of 6 (x)

6 (x)dx

defined by the prope rtie s'.I

I 0, x* 0j

o(x) = 1Il - ' x=o

From (I3), we see that F(k) - h ""'I

fi:nction 7ft is the Fourier transforrn

r r i - t*f i= -*-J"

a

. I O(x 'dx = r ,_co

Hence, the conetant

. Formally, we have

(14)

00

6(*) =* J" ikar 0s)-€

Notice that the integral on the right hand side of (15) is divergent when

considered from the viewpoint of legitimate analysis, In the forrnalisrn

-yo-

of delta firnction theory, we say that (15) gives rneaning to the divergent

integral.

Here are some properties of Fourie! tlansforms that

to the applications. Denote the Fourier transforrn of y bygrdx

€o

f-6

I21t

€., | - ikx . -lK J e y(x)dx.

-co

so the first terrn is zero. The

the Fourier traasforrn of , we perforrn a:r integration by parts:

- ikrc . . . -e y' (x)dx =

1 . - ilol €

717-LY|.xle I +-co

!r rnost applications, y(+ o ) = 0,

eecond terrn is iky, Hence,

n'F.r. (tf) = ikr.

- , dv . .If ai (+ co ) = 0, the preceeding result can be used to find the Fourier

transform of a second derivative:

Suppose f(x) has Fourie! transforrn F(k) and g(x) has Fourie!

transforrn G(k), then the cgnvolution€

f+e = f f { ()n(* - r tat- -ahas Fourier transform 2"J.(k)G(k). The proof ie a straightforward

calculation:

rft)c(t)eikat =

coi l -

^ :1-"

c(k)e'*{ J f(g)e- '^sd6} dk-o

€

r(E) { /c(k)eik(x- 6)64 69 =-€

are es sential

t '

€

IJ

-€

€

f rJ-€

a1r

--:- |21r J

-€

€co1r:br l^:1-r

;E J fG 6)e"-rdE =* | f ( ! )e- '^56E =-€ ' '

-aO

F(k), or G(k) = TiR).

Substituting g(- 6) = f(E) and c(k) = F(k) inro (12) gives€6( ' , . r ' r . L c , - , . , , .

J l rG) l ,ak = hJ l f (x) lzdx.-€ -€

The Fourier transforrn is a useful tool for solving differential

equations on - oo ( x ( co in those cases where the solution vanishes as

l* l * - .

As an exarnple, consider

dzvEit - a"Y = g(x) on -oo (x(co,

y(€o) = y(- ao) = 0

Taking the Fourier transform o{ both sides of (18) gives

- (k2+az);=; . ' r=

-+=.Inverting,

, -

Let g(x) = 6(x). Then

->t-

'T* Jr tElet*-E)dE.-€

Parsevalrs theorern for Foutier transforms

case of the preceeding result. Let x = 0 inoo oOr1

J F(k)c(k)dk=fi J t{Ltz{-_oo -ao

Set g(- 6) = T(g). t t ret t

€

eel =# /Ergt . ik6as=-€

(15)

as a spec ialcan be proved

(16 ) . lhen

g)d6. (17 )

(18 )

(1e )

co

_co

g(k) = 12It

ikx ,.

I

c

The right hand side is evaluated by contour integration. The integrand

has poles at k = + a. If x t 0, "ik

is exponentially srnall in the

uppe! half plane, so it is appropriate to close the contour about k = ia

like this:

Evaluating the residue at k = ia sives

- 98-

ao .,- 1KX

Y=-fr J 14;zox'-6

y=-|e"*, *<0.

(20) and (21) can be cornbined into a single staternent,

i -a lx lY=-Ee

a graph of the solution and its derivative appears below:

If x < 0, then we cloge the contour about k = - ia to obtain

(z0l

(z 1)

2

-99-

^l '

The most strikiag featute is the discontinuity of theThis can be understood, by a sirnple argurnent. LetPositive nurnber. Integrating both Jides of

d2v&f,-a 'y=6(x)

f rom e to -e gives

-€rgv" , r.c iJ -a 'Jydx=L.

€ -€ -€f4s €-u, J ydx-0, hence,

-€

; , . er1*gl

- l as e-0.I (

-€

This resul t says that yr(x) jumps by + I atfrom the graph. Since y,(x) is finite, .

€r '

' r Y ' (x)dx =;Y1 -0

-€ _€

der ivat ive at x=0.

€ be a small

x = 0, This is apparent

= 0. There o bservat ionsas €-0. Hence, y(x) is cont inuous at xsugge€t an alternate approach for solvins

d2 , ,

f f -azy=61*1,First, we note that for x > 0 and x

Y(+o)= 0.

< 0,

ez .-f , ; f -azy=e.

Hence, y

To satisfy

A=R

=A,eax+

Y(+ o) =

Hence

_ _axIJ. E

- 100-

i l x) ^axU and y=A_e

Continuity of y

+ B_e-ax in x < o.

at x=0 requires

"{

Ae x>0

The derivative

A-

must jurnp by+

. 0 'l -+ l. d.:c.

^ -

, x<0

-alx l

1 at x = 0, hence

ZA a=1, or

A-

+

1

The final answe!,

I -a lx Iv=-Ee

agrees with the result fo und by the transform method.

Consider a generalization of the preceeding problern:

dava+ - a 'y = o(x - E), y(* o) = 0. (zz)ot<-

solut ion is cal led the Greenrs f i rnct ion and is denoted by G(xl6).

x =x-L. (22) becornes

a{ - azy = 6(x ) , y(*o) = 0,. $r. 'x -

. , * ,which has the solut ion y = - j :e-alx l . H.rrce,

r - l - - v lc(xl6l = -7|e- ' l * -st . (23\

What is the virfue of the green's function? Let us return to the

inhomogenous problem

rne

I

d2v+-a-v=g{x).dx'

(?41

- 10i-

y(o) = y(- o) = 0.

By the convolution theolern,co

g(x)*6(x) : /s(L)o(x- 6)a6-co

tas Zr # = Gft) as its Fourier transform. Hence,

€

e$t= I s(g)6(x- E)dg.-€

Let L = *, -

", . By the definition of green, s f rmction,

r ,c(x lg)=6(x_6).Multiplying this equation by g(6) an{. integrating from IL = o give s

co

fr c(E)LG(xl6)dg =

*--

J e(6)6(x - Oag = g(x) or-a

@

l.l , (J e(t)c(x lOdO = g(x).

)

This heuristic argurnent suggests thata

y = J g(6)c(xlE)aE t" rhe solut ion

-cO

of the inhornogeneous problem (241, (ZS\, This can beaox

v= [z.G9r.x l r ; ,at* 1.x -co

(2sl

= -€ to

verified directly.

(z6l

v' = / g(E)# dL + 1. , . . .

- g(xXc(x,x+) _ c(x,x- [ .But c(x lg) is cont inuous in ! . Hence, G(x,x+) _ G(x,x-1= 6.Finally,

€f

^2 ^Y" = / g(6)o, Yar +x o:a_

:a

I€

- g(x){ff(x,x*t - ffr*,*-11

-r02-

the last term, consider the followiag graph:To evaluate

**t -#(x,x-)

= ietsr $$ ue

(27\,

- i - t l l=-r

x

J + g(*)-6

( * ,1 ' )

( t , t ' )

(27)

1G

) r

3-9.) r

Hence,Evideutln (x,

Cornbiaing (261 and

y" 'aZ c(E )

"" t f f -azc=o for L

g(x) '

-e90:(

xda+ f +

Hence,

c(x)

ytt -az y=

( r(#-. 'o)\ )

>x and((x.

6

xt -

Suppose f (x) is'' even. W'e can always write its Fourie r

(28)

t ransform as

I "6 '0

* Io tl*l "-'ffi

d* - + i__f1x1 e-ikx d*

Since f(-x) = 1(76;, the lattel' integral can be n,ritten

* ( n-", .ik d* = + Irrl*; eikx &

Hence, (28) can be wl i t ten as

,-L i u*, fut* * "-t*l a* = ! l$r(*) "o" k*. ruo t J

- nt | - . - - . dx

tn2

The latter integral is called the cosine transform of f.,

noted by Fc(k). Notice that Fc(k) is an even function

allows us to write tlre inversion forrnula

f (x1 = f€ r' /L r -ikx -,'-J tct , ( , e dk as

flxl = Z .F"(k) cos kx dk

Sirnilarly, if f(x) is odd.

6

f-0

This is called the

allows us to write

Fs&)

Fourier transform can be written as

f(x) 5in kx dx

and is de-

of k. This

its

€t-0

I

t

sine transform of f. The

the inversion formula as

oddness of Fs (k)

The Laplace transform can be

of tlre Fourier transform. Suppose f(x)

as x - + oo. Let Gc(k) be

f(x) e-cx That is3

t (x l=z/r"{ t ) " i t t

,€

Io'|

=.-

kx dk

conside red

=0 for x

the Fourie r

as a special case

< 0 and f (x)e- c l

t ransform of

cc(k) = +€o

f (x) e-cx "- ik*

d*

f(x) e -(c +ik)x dx

Gc (k) is analytic in Irn k < 0. Let s :

.F(s) = Zz ^ 's-c ' 16 "" . (T '=J 1(x)

o

c + ik.

-S:< .edx

Then

'104-

is analyt ic ia Re s ) c. We recognize F(s) as the Laplace trans-

forrn of f(x), Now by the inversion fo rrnula for Fourier transforrns.

-cx " ' rQ ^ " ' ikx "-e - t (x) =

J uc(r( , e cK

Pe.rforrning the change of

tegral g ives

Ina

6(x- B ), y(to) <

s = c * ik in tlle preceeding in-variable

c +i6fcc -16

,s-c, (s-c)x .c tT, e- ds-cx " , , I

e t (x, =-

sx.-E{s) e dx

g: q +i lc

previous example, we saw that yt ' 'az y =

oo has a r:nique solution G(* | S ) .

Let us try to solve

y" + az v = 6(x-!) , Y(+o) ( o

of the forrn

r .c+ioo11y1 =* |

al l t JC -16

l l t . t :c

I

a

ar|

,I

t ,1

ua

1

a

I

The solution is

(29')

fe." i " (*-e ) * " .u- ia(x- ! )t +- - -+= x>g

y=(

["a-" i " (*-e I *

" -u- ia(x- ! ) , x< g

Notice that boundedne s s at o is satisfied for any value s of A*, B*,A_, B_ . Continuity at x = E implie s

A++B+=A_+B_ (30)

r- . r r*The jurnp condition I dv I I

. LdJ ,_

= t g ives anothe r equat ion

( ia A* - ia B*) - ( ia A- - * "- ,

= , . ,

ia (A+ - A_) - ia (B+ - B_) = r (3Ir

(30) and (31) provide only two relat ions between A+, B+, A_, B-Appalently the solution to (29) is not unique and more cond.itions musrbe imposed to ensure uniqueness.

Suppose we impose the gausalitv condition y _ 0 asx + - oo. Then A- = B_ = 0. (30) and (31) become

Hence, A+=-B-

- . tu5-

A, + B. = o' r+

I

A. - B =+' r+Ia

, and

sir t a(x- ! ) ,

_ lZia

( t

=l 'II

x>E

x< E

- 106-

Alternately, we can irnpo se the radiation condition

(, iax.

, -J ' x*co

l- - t .*L*"

, x+-co.

Then B* = A_ = 0 so (30) and (31) now read

A+=B-

a*+a-=| ; .

Heuce, A+=B-=fr aoa

f J-" i^e-el , x> LI)

v= (. l

[ ; " - t " t*- ' ) ' * '6 '

- l0?-

I :^ l - - r IY =

Zta er4 l 'a-9 | .

l fow can these results be obtained by the Fourier trans-

forrn method? Taking the Fourier transform of y,, + az y = 6(x)

give s

(ar -kz)f =*, " , i=miTr)

Hence, the formal solution is

v=+f ! - [ dk ezl

This is aa improper integral due to poles on the axis of integration.

In hindsight, it is easy to see why difficulty arises in applying Fourier

transforms to the current problem. The solution is not unique and

does not vanish at x = + 6 . But i t is not yet t ime for despair .

Suppose we change the pat}l of integ ration in (32) so that the poles are

avoided. The succeeding f igure shows two ways of doing this:

i \

i i )a ' a.rConsider the path i). If x ) 0, we close it in the upper half plane

and the solution is the surn of the resid.ues at k = + a. Hence.

y = + f -"- t "* * u iaxl = ! s in ax)^

i f x>0. I f

th i s case, no

sut nary,

[ l " rax,x)o

"=("

L. , x(0

This is the solution that satisfies the causality condition. In a similiar

fashion, it is possible to calculate the solution that corresponds to

path i i ) , The resul t is

ia lx Ie"

This is the solution that satisfies the radiation condition.

The concept of the Fourier transforrn canbe extended to

functions of several variables. The Fourier transform of f(x, y) is

defined to be

F(k, r ) - i (kx+l v)r(x, y, e ' dx dy (33)

If l(x, y) has continuous partial derivitives and

a

JJ ltt*, y)l dx dy < o ,

then there is an inversion forrnula

-108-

x ( 0, the path is closed in the lower half plane. In

poles are enclosed. Hence, y = 0 i f x ( 0. I l l

't

V=-

€t (I

+1t -

r(x, y) = f1 re, 11 .i(k+ly) ak al (34)

Suppose f(x, y) radially symetric.

write the integral

That is, f (x,

(33) in polar

I

r t I th( V# + t' ) It is natural

coordinates, Let

angle between X

I=xf+

and 5,

_109_

yf ana b-

t=JWd+rf .

and K

Let € be the

=.f i - f i .

tThen kx+Iy=k.1,

r d0 dr Hence, the

H(K) = 4+

= l f l l f l cos 0 = K r cos

integral in (33 ) becornes

f1-"r t t . l " - i Kr cos o r do

6"0

0, and dxdy =

this substitution, (3)

(37t

aeJ a,

I"Jwith

r so. , I f , " - ,zF Jo "n(,) L- o

Similarly, (34) caa be written as

- ( .?, ,

h(") ={KH(K)tro" '

Now f"" t * r cos o de - ) - r .to

and (4) become

Kr cos 0

Kr cos x

(Kr)

r ,9oFI(X) = * / rh(r) Je (Kr) dr and

00h(r) = 2o f xntxt J6 (Kr) dK

-0

H(K) is called the Hankel Transform of h(r). It is useful in pro_

blerns with circular syrrretry. (5) and (6) rernain true whet:. Jq is re-

placed by J,

-110-

IX Eigenfunction ExPansions

and Sturm Liouville Theorv

A sequence of c ornplex valued functions

S x ( b is called aa orthonorrnal set if theona

whe re

b({,, +J = J" +;t*l T*{*) a* = 6,,,,,,

=l l l

+rn

a sum of the

d (x)'n-

{ oot"D defined

inner product

6 's'n That is,

(r)

Suppose f(x) can be

Multiplyiog (2) E(*l and illteg!ating

I t , "I-nrn I

I o ' o

expressed as

nf (x) = 7 c-

. -d(21

by QJx) and illtegrat

b .' eo.

J tl" l4](x) ax = | c

f rorn x=atos=b gives

(3)t -^

(Z) and (3) can be exPressed c ornpactly as

6(x-[) =l +-t* l 4-rel0-

Multiplyiag both side s of (a) by f(g) and

tog=b gives

integrating frorn

t4\

S,,(x) r.,4$0

f ( r ) d (r) drf (x)

-11I-

which is equivalent to (2) and (3).

Many exarnple s of orthonorrnal sets have been encountered

previously. Four ier ser ies are expansions in orthonolrnal funct ions.

For the Four ier s ine ser ies. the relevant orthonorrnal set is

6 (x) ='n- sin n x , 0 -< x

-< n

For this orthonorrnal set, ( l ) - (3) read

stnnxsrnlnxc:x=o2/ '

- l1f "o

f (x)Feq

=ri : ) , cYoTN

where

(1')

(z ' )

( ' n, s in n x d.x (3 ' )

Equivalent resul ts were obtained dur ing our study of Four ier ser ies.

Similarly, the orthonormal functions corre sponding to the Fourie r

coslne ser les

6 (x) ='n

xSr

cosine )

d (x) ='n '

(Complex Four ie r)

is not restr icted to certain se-

Il fact, we previously studied

and cornplex Four ier ser ies are

t , -l r / r n=oI vn

) "n 0s

\

I 'F"o"nx,n)o (Four ier

I v"L

I inxV; e---- on - z Sx (n

The concept of an orthonorrnal set

lluence s of trigonometric func tions,

2n+1_T

-ll2-

a non-trigonometric example in detail.

The Legendre Polynornial s satisfy

I

J_, nrrt*l Prrr(x) dx z= ZiTr 6nrn

We conjectured the expansion

Rqf(x) = | ao

2n{Ia =-

, ,a

theorern

Pn(x) ,

I

I f(x)

whe re

p (x) dxn

and

Evidently,

orr(x) = eo(x)

is an orthonormal set on (-1, 1).

In the preceeding exarnples, the orthonorrnal sets have

countably rnany members. Is it possible to extend the concept of

orthonormal set to include the case of unc ountably many members ?

The Fourier transform Eive s us a prototyPe, Let

ikx4(k, x) = h whe re

the role of the integer n-oo(k( oo, -oo(x(co. k PlaYs

encountered in the previous exarnples.

Recall the forrnulas

f (x) =,90I

J-a

F(k)

6t

ikxe ofa

-ikxrtx, e

(5)

F(k) dk (6)

- ] IJ-

Setting . C(h) = '6' r(t) , tbese can be rewrimen as

r(x1 = /s c(k) g (k, x) dk_00

€c(k) = J r(x) $- (r, x) ax

-6

(?) and (8) are the analogues. of

af(x) = I c- S- (x) , whe re

0-

-bc'-n =

J a r(x l Sa (x) dx

Suppose F(k) = 61L - 111 . Then (5) implies f(x; = uftx and (6)bec orne s

6(k-K) =+r u- i (k-K)xo*

(7)

(8)

Wri t ten as

(e)

we recognize it as the analog of

L

6'"tt* = J" 0n (x) Om(x) &<

Finally, the farniliar re sult

6(x - ! ) = J- / ' " ik(x-

O . ,u. ' Ztr J-q -

can De recast as

-tL4-

6(x-E)= rooIr -6 S(k, x) g1r, gy ar

Its counte rnart in the

There are other

rnembe rs, The

transforrng are

exarnple s of

orthonormal

+m(g)

orthonormal sets

set corresponding

(10 )

with uncountably rnany

to the siae and cosine

discreet case is

tq_ E) = i+-(x)'0

6(x

$(k, x) = sinkx 0 -<x

( o.

(sine transform)

0G, x) = cos k x on 0 €x ( o .

(cogiae traasforrn)

The Hankel trans{orrn provides a non-trigonometric example. If

F(k) = 2z vk H (k), and

f ( r )=f11r1

then equatione (3?), (38) of Chapte r 8 becorne

F(k) = f6I, .6

poI

! -6

f(r) 0(k, r) dr

F(k) Q(k, r) dk

and

_ t15_

In what conter.t do orthonormal

Conside r the eigenvalue problem

y"+Iy=9

y(0) = y(7) =a

sets of functions a rise ?

The solutions must be of the. form y = sin 16 x to satisfy the dilfe rential equatiou and the boundary conditiolr y(0) = O The secondboundary condition ]r(r) = g forces sin yl 7 = g Hence,I = )r, = n2. we say that the problem ([) has eigenvalues l = )o,with eigenfunctions y = sin n x -r We recogb.ize that these eigen_functions forrn the orthogonal set of the sine series. There isan important class of eigenvalue problems whose ergenfunctions foraorthonormal sets. These are the next objects of our stud.v.

Let y = u(x) and y = v(x) be any two functionstisfy the same homogeneous boundary conditions y(a) = y(b)Define the ope rato r

L=*rnf ; r - r

where p and q are known functi ons of x. Then

(v, L u) - (L v, u) = (r2)

1i" * tn *r - " fi rn"n:,J .- =

Int"$g-o4cr l"=ol ' dx -dx' l- Ja

(u)

that sa-

It is by virtue of (lZ) that L is said to be sel f adioint ,

- L. to-

Conside r the eigenvalue problem

L(y) + l . p y = 0, y(a) =y(b) = 0 ( r3)

I f -o(a(b( oo, and p, pr .O in a Sx Sb, then theproblem (13) is called a Regular Sturrn Liouville svstem. The charac -ter of its solutions are outlined in the

Theorem: There are an infinite numbe r of eigenvaluest - trn for which the Shrrrn_ Liouville system (13) has a solutiony = ulx (x) This sequence o{ I's has no accurnulation point e:{celrE) = o The eigenfunctions uo (x) are orthogonal with weighting

P(x) That is.

I^ p&) u,',(x) urr(x) dx = 9

whenever n * rn. Consequently,

Qo(x) = y'llil ,',,1*y / Il. ,r*, u: (x) dxt -

is an ort$.onormal set.'- [urthermore, the urr(x) are complete.f (x) is smooth in a:<x:<b, then

qqf(x)=)" , , (x)

0-

whe re

If

u /o",,

= f, p(x) f(x)

"nt*l e/ Jl pel u2(x) dx

As an example,, I ,Jr - q / q, I' ' -a; rxa; , e =;

consider a Sturm-Liouville problem with

and a=1, b=Z The problem can

be wri t ten as

- lJ- l -

*z 4,'| + x Er * )w = oqx- dx ' ' ,

y( l )=y(2) =e

(14) is an Euler equation. Its general solution is

y = A sin (vl fog. x) + B cos (vl tog xl

The boundary condition at x = I forces B = 0. Hence, y(2) = Oimplie s

sin (vl log Z)" = 0

vl log2=17

Hence, the eigenvalues are

- - zI = ) = (:g /':.)

n ' tog / , '

The eigenfunction corresponding to ) = )r, is

urr(x) = sin (ffi tos x)

From the theorern on Sturm Liouville systems, we can conclude:

,, Qn(x) = un(x) /* i, + *

is an orthonormal set on I < x < 2

i i) If f(x) is smooth on I _< x _< 2 , then there is an expansion

i(x) = | c- u- (x)n-

(14 )

l faol

0t-or o in a <x

b is

<b,

-118-

infinite, if P or p as surrre t.Le values

then the problem (13) is called a qslg

The results of the preceeding theorern hold

for the singular case provided that all the eigeafunctions are square

integrable. That is,

o(x) u2 (x) dx ( cn'

for all eigenfunctions urr(x)'

ex) The problern

bI

Jz

*t t t -* ' )**) +)v=Q,

y(-1) , Y(1) < o

is a singular Sturrn Liouville system because P =l - xz = 0 when

x = -l or 1 . The di{ferential equation is Legendre's equation' SoIu-

t ions that are bounded on -1 (x Sl occur when I = n(n + 1)

Evidently, lr, = n(n + 1)* ale the eigenvalues' The corresponding

eigenfunctions are uu = pn(x) They are clearly square integrable '

The orthogonality and completeness were studied in Chapter 6'

ex) Bessels equat ion,

f r t* f f r+)xv=Q

together with the boundary conditions y(0) < o ' y(1) = 0

' is a

Singular Stu rrn- Liouville system because p = P = x are zero at

x = 0 This problern was solved ia Chapter 5' The eigenvalues

are )r , = j i r , where j ' is the nth zero of Jo(x) ' The eigenfunct ions'

ur, = J6 (in x), are square integrable. The orthogonality Property is

II

_119_

i]

Jo * "*{*) urr(x) dx = o

If f(x) is srnooth on 0 -< x

-< I ,

Four ier Bessel se r ie s

i f mr.n

then it can be expanded in a

i f * ) = lan Jo ( jn *) in 0-<x-<t ,'0

where

, r f ,""

= {

x f(x) Jo (j' x) 4s / Ji * J6 (ir, *) a*I '

In a Shrrm-Liouville problem, the orthogonality of the eigen_functions derives frorn the self adjoinb:ess of L. Let )r, be rneigenvalue of the Sturm_Liouville system (13). The correspondingeigenfunction urr(x) sati sfie s

L t,,' * trrt

Recalling the definition

pu =0'n

b(u, v) =J u; ax

we write

Similarly,

\um' |uJ

- t n

(rs )

(L u*, uo)

- )* (P o- '

u . u) =md

(u*' P un)n'

0-P

- ) . m

(16 )

a s su.rning

The left haud

(24) reads

and the

_ 120_

is realvalued, Subtracting

(u^, L

(l.trr

un) - (L um, un) =

- rrr) (urr, , P uJ

side of (18) is zero because

(16) from (17) give s

Lis

(r8)

self adjoint, hence,

(19 )

(1,,

( r -Tn

- l*) (u*, n

b

*){c"*I , ,

ot)

dx

srnce p+0 and

This statement says

blem are real.

is proven.

in equation (19), then

b

^o'{Punundx=o

lur, le ax ;r o

un+0 in a -<x -<b.

that the eigenvalues of

Hence, ir, - )o = 0

a Sturm-LiouviLle pro_

orthogonality

I f rn =n

b, ou

(r

Now

bi l i ix=J"

TI Je(

Are the eigenfunctioas

Liouville problem is linear, the

within a multiplicative c onstant,

to the question is no. But it is

tions that are real. Suppose I

ing eigenfr:nction u(x), Then

necessari ly real? Since a Sf,urm_

eigenfunctions are determined. to

which may be complex. The answer

always po s sible to find eigenfunc -is an eigenvalue with correspond_

- l? l -

*Atof f l+6p-q1 G=0,

Y" + IY = I

Y(o) = Y(h)

y ' (0) = y ' (zz)

adjoint in the sense that

)

(20)

is sel f

$tn$;r +(rP-q) u=0,

. . r - \ - , ,4L\ - nu\4/ - s\v/ - v

fi p, p, q are real, tJ:.en

u(a)=s( l )=a

Since ) is real, f = f , and i satisfies the same equation and

boundary conditions as u. Hence i is also an eigenfunction with

eigenvalue ). Finally, note from the linearity of the diffe rential

equation and boundary conditions that j1r.r + i1 = "u

o is an eigen-

function.

In sorne problems, there rnay be two or rnore linearly in-

dependent eigenfunctions with the sarne eigenvalue. Such an eigen-

value is cal led degenerate, In th is case, i t is possible to f ind an

orthonorrnal basis of eigenfunctions. By taking lirre ar cornbinations

of the basis members, i t is possible to generate al l the eigenfunc-

t j .ons corresponding to the degenerate eigenvalue.

ex) The eigenvalue problem

A+(pd.x '-

r"*

, '

-122-

(rr, t") - (u'r, v) = Q

for any two functions u and v that satisfy the boundary condi-

tions (20). Hence, eigenfunctions with different eigenvalues are

orthogonal What are the eigenfunctions corresponding to the

single eigenvalue ) = n2? A real, orthogonal basis of eigenfunctions

is

urr(x) = cos n x , vrr(x) sln nx

Suppose u(x) and v(x) satisfy

- ,+gru=Udx

and

,.1i (pdx -_

dv, ,

- r fctx 52 " - v

respectively, with gt -<

there is at least one root

gz , and p > 0 in a -<x (b Then

of v(x) between any fwo roots of u(x).

I

In heuristic lan guage, v(x) oscillate s more rapid.ly than u(x). Thisresul t is known as Sturmrs f i rst comparison theorem. We give aproof by contradict ion. First , note that

d--_oJ< {p (o '

" - u. , ' } =

u') + pu,v,-"stn v,) -p r ,v,=

v(- gr u) - u(-ge v) = uv (cz : gr)

v+(pctx '_

- tz3-

consecut ive zeros of u. Integrat ing (Zl)

give s

(ztl

xI

x

and x, be two

=xl tox=xz

| , -xz "xzLP (u ' v - r u. ' )Jxr = J*, , . , (gz _ gr)dx (22')

I f v*0 in

suppose v ) 0

u)0inxr(

xr and x, are

u'(x, ) ( 0

?. :<x -<b, then v

, Since x1 and x2

x ( xz Assume u

simple ze ros of u,

has only one sign. Let us

are consecut ive zeros of u,

> 0 in x l <x \< xz I f

then u'(x1 ) ) 0 and

-L24-,,1 .t

a^' (z , ) (2.-) t Q

tr this case, the left haud side ot (22) is

p (x2 ) u ' (xz ) v(xz ) - p(x1 ) u ' (x1 ) w(x1 )< 0

But the right hand side of (22) is clearly greater than zero. The

contradiction forces ug to conclude that v(x) has a zero in

xt \<x \<xz, The other possibi l i t ies u(0 or v(0 are easi ly

reduced to the case just coasidered. As an example, suppose that

u ( 0 , Thea we can replace u by -u and appeal to the preceed-

. ing argument because -u has the same zeros as u.

lf, in addition to the hypotheses of the comparison theorem,

u and v satisfy u(a) =. v(31 = I , then v vanishes before u ,

This is a trivial corollary of the cornparison theorem. Between

a and the next zero of u, there is a zeto of v,

Piconers Modification'* is a more gene ral ve rsion of ttre

cornparison theorem. If u and v satisfy

d du.6; (Pr a;) + gr u = 0 and

dclx o.$| t+gzv=o

o

'I

Ince, p, 225

re spectively, with gr -<

then tfre result of the old

The diffe rential

be written as

-tz5-

Ez and 0( pz < pr i r : a Sx _<b,

comparison theorem still holds.

equation of a Sturm-Liouville problem can

where g=Ip-9 wi th p)

ing I causes the solutions to

Picone's modification allows us

successive zeros vanishes as I

the boundary coadition y(a) = 0

then between any two zeros

solutions oI (24) are sirnple

(o-.* ti) + h pmin - 9rrr.*) u = o

u(a) = 0,

d'-q.x

ddx gY=0,, dv.rp e;, +

0. We

become

to show

, and

see imrnediately

more osci.llatory.

that the distance

If y satisfie s

u sati sfie s

(23)

that inc reas -

In fact,

between

(23) wi th

(24)

.! neof u, there is one zero

sinusoids with pe riod

of y.

T =2t

Since T goes to zero as x -6, the f lyswatte r*pr inc ip le al lows

us to conclude that the distance between the successive zeros of v

goes to zero as ) -o

Since y is more oscillatory than u, the maximum se -parat ion of y,s zeros is

* I f a(b(c and a then b - c.

T7 =n

dl

.i

{

t

,II

T,ll:+'.,'

ii;t

{:l:

t,fr:

;'

t^, I l -* i -

n7 \ /-----:+s:--- < x - a < nz\ / l ^ -q nY rmax Tnln

Simila rly,

shows

Hence,

a c ornpa ri s on

A AAA

*(p +)dx ' rnln qx

tl1at the

i fx n

miairnurn

is the

- LZ6-

of y with t.Le solution of

+ ( lo - q ) u = 0- rnax 1n1n

u(a) = g

separat ion of y 'a zeros iE

nth zeto of y,

\ rrnint \ : -''

V ) P rr.r"* - 9roin

P*"*(25],

I P -i''-

%.r"*

Now suppose rr'e are dealing with a Sturrn. Liouville problern with

hornogeneous boundary eonditions y(a) = y(b) = 0 The first eigen-

func tion has no zeros ia a < x < b . The second eigenfr:nction has

oue zero, and the nth eigenJunction has n - I zeros.

ex)

X" , / (^! tO

a{(o) 2 t (2n)

3 O

Hence, *r, = b

now implie s

1t

is equal ioi f r the nth ei.senvalue ) (25)

I

The central concept of the proof involve s the Rayleigh euo-

which is defined as

b. . t , . t f+qq-)dx/ l pdrdx

ta

in the case of homogeneous bouadary conditions y(a) = y(l) : g

IJ p, p, q>0 and O+0, then R(S) > 0 rn th is case, i tseems reasonable to ask for a function $ that minimize s R(O)

subject to the constraints Q(a) = {(b) = a. This problem can be

ved rigorously by the Calculus of Variations. *

Hu"., we outline

solution via a plausibility Suppo se

n'

whe re tl:e u (x) are then

zation condition

a1q1 =/tpq,z

eigenfunctions which satisfv the normali -

s ol-

the

-LZT-

l \

' l l - \! ) rnz \- -

. ' t 1 I ,nr lz lp,''; \ i;:6' Pmir, * %'i' i -'lr,

(p*.- ( (--J P*.* * q,,r.*

i .\ ) "*" t )Given rnore general boundary conditions, it is possible to find sirnilar

bounds on the l,, by slightly more complicated. arguments. The

main point to rernember is that ln groqrs apploximately as n2

for the regular Sturm-Liouville problern.

The eigenfunctions of a Sturm-Liouville problem are corn_plete.

argur:nent.

NO= /c u

rn

u2 dx = In

bl "ra.

x Courant and Hilbert, Chapte r IV.

- 128-

An integration bY Parts shows that

"b - -b P Al" p +'2 dx = [p o+']. - J. Q *

(P Q') dx

The first terrn on tlre right hand side is zero because $(a) = Q(b) = 0'

In the second terrn,

d,?- tD O,ox_

=*(o)cclt( -' (i

!n-nn

sr.-J nn

,,.\ ,, P ot

A

dx(p u ' )-nnn'

- /-t c

Hence, the second term is

I()cu)Ocl .pu-cqu)dx=Ja'u fo rn"L n. n ' n n - nrnn-

bbsl f . f

I .., ",r,

[r,J P or' oo 6* - J" 9 u' u* dx) =IIr, n zL

bF

" c ( I o - / qu u dx) =

mfn-m -n 'm n ra ' n rn

I r . . - I " " Fo' t r u dx,r- t t ,rl* o rrr .,a - n ln

\r ,b /bThe terrn I "r, "r' J. I oo u* dx is J" e f a* in disguise.

Hence, the numerator of the Rayleigh Quotient is

1. (o o 'z + o tp) dx = f I .zra '' fiJ n n

A sirnple calculation shows that the denorrrinator is

I o rtr- (u< = / c-re ' 4 nn

_LZ9 _

Hence,

R(O) = ) ' \ cz / l cz' !nn!nnn

Assrrrne that the eigenwalues are labeled so that

' l hen

R(+) - r, =Jrn" - r.,) "i/L.X o!

R({) =r, *"flrn" - rr) cf, rJr.l

The right hand side ie clearly minimized by choosing "r,

= 0 for

n >- 2 . Hence, it see:rrs reasonable that the minirnurn value of

R(Q) is the first eigenvalue I I and tJ:at the function { which

minimizes R(+) is the first eigenf unction u, (x). Our argurnent

fails to be a proof though, It is not clear that all { with C(a) =

Q(b) = 0 can be wri t ten as I . r , o ' This is precisely the com-n

pleteness property that v/e are trying to prove!

Suppose we introducel.an additional constraint on Q :

(S, pur) = J^q gu1 dx;0

,Fu 9=rcnun, then cr =0n

Hence.

-150-

Evidently, R(4) is rninirnurn when c: = cr : . . . = 0 This

means that 4 = uz and tl1e minirnurn value of R( O) is I , In

general, we have the minimurn principle

ln = rnin R( 9)

when $ is subject to the constraints Q(a) = +(b)

(4, eup =o for k=1,2, . . . n- I The {

rninimum is achieved is the nth eigeufunction r,,.

Let f(x) be a srnooth function on d -<

x ( b with f(a) =

f(b) = O. Def ine

=0 and

for which the

p utJ =0 for

allows us to

lz6)

n-r )

-. I . "n "nl, ) a*K=I

)

one can show that the

., .I ( r1alr- J" (Pli ' -. I cn 5)2 + q (r

" L R=r.

I

qlO(x) = f (x) - ) , c,- t ,_

tKt.

where the c, are Fourier coefficientsK

b., = | o fu. dxK .,/a ' R

It is easy to see that 0(a) = O(b) = 0 and t!.at (0,

k=1,2, . . . n- l , "Hence, the rninimum pr inciple

conclude tr.n < R(0) , or written out in full,

h n - l

f- o(r -"f c, u. )z dx -<Ja k_=1

K K

After a somewhat messy cornputation,*,

hand side of (26) is bounded above by

T--ffim6;rger, p. 156.

r ight

Now recall that

- IJI-

,bM=+[{pr ' ,+qr+dx

L Jd --n

, - | f ,n, Iprrr.* f

'b-a '

as n* o.and

n- l

p\r - / c,_ \_) . ctx <= *0k=l ^

ra i'

lr

p12dx-) . i f oozd*=k-=-=n

K "a n

trPr- dx-, / cf

kl.tr *

)r t

Eranra I -

. ,)"--n

b

T^ (27 )

as n - co . This result shows that the eigenfunctions are complete

in the sense of mean square. But i t remains to show that the ser ies

1 ct u1. (x) converges Dointwise in a S x S bE-

The resul t in (Z7l can be interpreted. in another way. A

simple calculation shows

b,- P (1 - / c- u. )" dx =

k?rkK

b

T^b

L

provided that theb

Ju e ,tt dx = 1

eigenfunctions are norrnalized so that

Hence, we obtain Parsewal 's theorern, which says

s"b) c?=IoFa*

r-?r K '/a

f(x) was as sumed to be srnooth. But theIn this argument,

- !32-

result still rernains true if f is allowed to be piecewise smooth'

We now apply Parsevalrs tleorern in e stablishing the Point-

wise convergence of the eigenfunction expansion I cn un (x)k

This expansion converges uni formly on asx(b i f

M), c, ,+ (x) -O as N' M-o

k-=N K

MThe sum I c'- l+- (x) can be rewritten a 3

N

f ,ru "n, ,$,

tc5

By

M

IouMtL ri

N-

the Schwartz inequality,

ur- (x)cn) (j-)

r u-2 (x) r. i)z (f #,'

Hence, the eigenfuncti.on exPansion "

ott",if u&)"tiforrnly if it can be

shown that I fl "t ( o and that I a? converges uniforrnlv

in a Sx -<b.

The first task is accornplished by replacing f with

L(f) =! t9 tpSl -qO in Parseva|s theorem. I f f ' (x) isp P 'd.x ' ' dx

piecewise smooth, then

Iu?-=d,,#fu*=foHu*,t rK.

whe re

I

t

b.

blxr

"b=ioJa

\) dx

t-, u- dxpK

L

=-t1 J" p

b= I rT. f l

f u dx =n

\dx=

-I , c,Kl<

Hence,

dx( oo

For the second task, consider the

which is the solution of

I l . t "?ft- r( K = 1b (r-rr.'a p

I f I =0 is not an

It

L

is possible to f ind

vr =0 and

and v2 (b) = 0

l ' ' ' t.,ti, t*l

I :, ;* ,.,(" o w(x)

*,oS-qc=6(x-r)

G(a/ l ) = G(b/t l

eigenval ue of

Ly=-I py

y(a) =y(b)=0

then we can construct the Green's

vr (x) and v2 (x) such that vr (x)

v1 (a) = Q , whi le vz (x) sat isf ies

Then it can be shown that

func ti on.

satisfie s

u v2 -

x -<4

Greenrs funct ion G(x/ L t ,

G(x/ 91 =

x>l

-134-

whe re w(x) = yyaelelian of v, aod v. . In this problem w(x) =

const. . ^ -effi

t 0 for the regular Sturm-Liouville problern. The main

point of this constrrctioa is to note that the Green's function is con-

tinuous, so that its Fourie r coefficieats can be c ornputed without any

trouble. Thege coefficients are

uz (L)n'-

t2- . c onve rge 3

^bbJ. p to G dx = - q J. L(uo) G dx =

1b_ L G dx = - r- I u (x) 6(x- E) dxt

^r, " n

u (L)n ' - '

n

Hence, the expansion of G(x/ E) is

. . u-(x) u-(6)- l-f*

We can apply Parsevalrs theorem to find that

buniformly to I" , o, d:f as !t * 6

So far, the Rayleigh quotient has appeared as a tool in the

construction of forrnal proofs. It has practical aspects as wel1. Let

us choose a set of functions or rr(x) which satisfy roo(a) = oo(b) = 0.

Construct a surn S(x) = I a' or,'' (x) and calculate the Rayleigh quo-n"

tient R(S) correspondiag to the Shrrrn- Liouville problem

$tnf f l +np-q) y=o

t "b_gt

I Ja -n

xn

y(a)=y(b)=a

- tJ5-

The minimum of R(O) often provid.es a good appro:dmation to the

lowest eigenvalue I t, even when { is a quite crud.e guess at theform of the first eigenfunction ur (x)

ex). The lowest eigenvalue of y" + y = O,

y(1) = y(-r) =0 is 4zr r =?=2.48

Let us make

Q = c r (l-x2 )

Hence

a guess at the form of the lowest

'f c2 x(l-xz ). For this problem,

eigeafunction:

P=1, p=1, q=O

I

II

[.,

S'2 ax 7 5c7 + 3c4^--F-..---+Ic i + cz '

R(Q) = I

I 02 d*

z 5+3t; ,

7 1 1212el

R(+) is minimized when 9- = 0 Hence, we est imate the lowest

Notice that this e stirnate iseigenvalue to be lr :

slightly larger than the

Eigenfunctioa

geneous problem

gl

2 ) - ' .J

true value.

expansions can be used to solve the inhomo-

L(y) +

y(a)

I pv=f(x)

=y(b)=0 ( 28)

be

n

the eigenvalues of the

and the eigenfuactions

corre sponding homogeneous problern

uo(x). f(x) has an expansion

- r5b-

f(x) = p(x) un (x) (29)

which is val id in a<x<b even i f f (a) , f (b)+ 0 (But in that

case, the expansion wonrt converge unifo rrnly aear x = a and

x = b) We seek an eigenfunction expansioa of the solution:

y=) a_ u-- (x) (30)nu

Notice that the boundary conditions are automatically satisfied. Sub-

stituting (2!) and (30) into (28) gives

L(uo) + ) p unn

,, (r - lrr) uo u

Hence,

,n'- tr, = iT

n

and the s olution is

nn

sl s1) a u =o ) c

n

=pI.n

r\PLA

provided l. + ln Does this expansion converge? Since

sv=L ( 3 l)

as

F

n

D-6'

u.nn

the expansion for y converges by compalison to

then the expansion (31) fails, unlessI f I =r

- ! ) t -

b

, =

J^ uv (x) f (x) dx = 0 In th is case, a solut ion is obtained by

deleting the vth terrn frorn the sum (31). But this answer is not uni_

que. Since ) = )r, the homogeneous problern has solutions

y = A ur, (x) Evidently,

-- F "t'Y=L j -

" . (x) +Auu (x)

n+Y 'Y ' 'n

is a solution of (28) for all values of A.

We briefly sumrntrize the results. The inhornogeneous

problem (28) has a urrique solution if the corresponding hornogeneous

problem has no non-trivial solution. (i. ".

) + )n). lf the horno-

geneous problem does have a non-tr iv ia l solut ion (e. i . I = ) ; ) ,

then (28) has no solution unless "

= f u (x) f(x) dx = 0 InvJav-

the case ",

= 0, the solution is not unique, This collection of

results is of the chiefe st irnportance, and has been granted the title

ex) y"+)y=0, y(0) = y(1) =0 has eigenvalues

trr , =n2 zz and ei .genfunct ions un=sinnnx.I f )+)n, then

the inhomogeneous problem y' , + t ry = t , y(O) = y( i ) = 0 has a uni-

que solution. By elementary methods, it is found to be

y = | - | .o" vl x + ff i f t--t sinvl x

1 = (Znz)e , then there are solut ions

B sin Znzx because f ',r rr, (*) d- =

f

1._y = 7f-T2 (1

sin Znzx dx =

cos Znnx)

. Notice

- 138-

that the coefficient ffi$+ of the sin vl x term vanishes

)i* (2s7)2 tf ). = ((Zn + lhrll , then there are no solutions

^l ,J7

cause J

oZrr*f(*) U* = Jo si : r (2n+1) zx dx =fr t O. Not ice

*f-t blows up as 1 - 1(2n+t)z)2

ex) If L(y) = O , y(a) = y(b) = 0, has no non-trivial

tion, thea the re is a G reenr s function G(*/ El which satisfie s

L(G) = 6(x - 9) , c la/ l ) = c(b/6) = o .

be-

that

solu-

If y = u(x) ieb

unless J" o(*)

i ts easy to see

L(y) + p In

has no solution uale s s

by elernentary rnethod s.

0

This can be proved

be two linearly

nontrivial solution, then there is no Green's function

6(x - 6) dx = u(!) = 0 In the pecul iar case u(g) =

how a Greenrs function arises. Witness this graph:

\ ^ l\ t

t'- v

0,

c. l t ,

An irnportant part of the Fredholrn Alte rnative theorem is

the result that

y = f(x) '

b

/ ur, (x)

Let V1

Y(a) =

f(x) dx

(x) and

Y(b) =

(x)V2

-139-

independent solutioas of L(y) + pln y = 0 which satisfy v1 (a)

v, (a) * 0, v j (a) = a. The Wronskian of v1 and v2 is'I

Yv = vr vi - vz ti = Fft;

t O . The general solution to

geneous problem can be found by variation of parameters,

sul t is

= ur, v i (a) = 0,

the inhomo -

The re-

Y = - vr (x)

:(v, (x) I

-JT

v2 (t) dt +x

I" t(

f (E)

E)

where

at x=

Now suppose vr

boundary c ondition

are arbi t rary constants.

.r = a. Hence,

x.r- v1(x) Jn r( l ) uz (g) dg +

^xvz (x) Ja f(t) v1 (t) dl

eigenfunction, Then

= b force s

b

""$) I^ r (O v1 (() d(

The boundary c ondition

vr (b) : 0 and the

w(b) = vr (b) v l6)

0 Hence

a and r

force s

! -

vr( l ) d l

0 would imply

w(t) ={ 1

d( =o

is an

atx

Y(b) =

Now v2 (b) :r 0 because v2 (b) =

vz (b) vi(b) = 0 , while in truth,

;'J" f (c;

" '191

H (x)

-14 0-

Remarks on seneral ized funct ions.

The purpose of general ized funct ions is to extend the

defj-nition or concept of functj-on so that the analytical pro-

cesses for smooth, inf in i te ly di f ferent iable, absolutely inte-

grable funct ions remain true. Thus given f (x) is a general ized

funct ion, we require that f r (x) = df /dx alrr /ays exists, in some

sense, and

+ const. (1)

Ex(i) f (x) = H(x)

where H(x) is the Heaviside funct ion or step funct ion.

"x or -r At ox

In the

and i-s

x=0

ndy

ordinary seose,

not def ined for x

does not af fect the

is not sat isf ied.

U:I

(x) =g for x>0 and x<0,

= 0. The lack of def in i- t ion at

def in i l ion of the Riemann integral

Using generaLized funct ions, we

H'(x) = 61*1

rnen

H (x) = "x " , , -J --o

(x, ox -1 ,x

,x

>0

<0

l

and (1) is sat isf ied.

1) erent iated

Ex( i i ) f (x) = sgn x = +I , X > 0,

= -1, x < 0.

H(x) - H(-x) = -1 + 2 H(x)

x = 26 (x)

x=/ l -zo(x)dx-1.

Generalized. functions are constructedrules always apply.

sgnx=

f,- SqnLlr(

s9n

so that the follol':

' (but they need not be cont inuous!) see ex( i ) .2)

colunute

d& E u,, (x) = X ui (x)

-?r ( x ( Tr has the Fourier cosine ser iesEx ( i i i ) l * l ,

dx r- ' l

t* t = i - f r .o"*. T*.

-*

+. . . ) .= sgn x has the Fourier s ine ser ies (_T < x < T,,

sgn x = f t " : .n * * s i l3x * s i l5x +. . . ) .

-T < X <

(2)

(3)

Eq. (3) is the term by term di f ferent iat ion of (2\ .

Di f ferent iate again. Fron ex( i i ) ,

6(x) = f ; { .o"* + cos3x + cos5x +. . . ) (4)

The r .h.s. of (4) is d ivergent in the ordinaxy sense. Asa general ized funct ion, i t def j .nes 6(x) . The fol lowing resuJ.tcan be proved easi ly for an arbi t rary smooth funct ion f (x) :

-L42-

l i r t -

Nf (0) = ; ;1 ; /_. f (x) (xcos (2n-1) x) dx .' '1

This is the sense in which (4) has meaning.

3) !'ou!ier transforms alvrays exist.

F(k) = |, r];rt*1.-ik*a*

always defines a generalized function F(k) such that

f (x) = / l-r tr. l " ikx* , f , (x) = /]- i :<r (k) eikxax (5 )

etc. , are always true.

Ex(iv) From the definit ion of 6(x).

/l-o l*i.-ik*a* = 1,'.F.r. (6 (x) ) = +

(s)

I lence,

6(x) = l - r-2n ' - -

o' tx) = I- r-2r ' - -

Taking the complex conjugate

f olIo!'rs that the F . T . of x

Ex (v) Treated .s oibitr.ty

f (x) = g-x,

. ik*dk,

ik eikxdk .

and interchanging

is i6 ' (k) .

functions

irk,and

x> 0

x<0

r (k) = *" , r*rn, '

f ' (x)

Treated as

f (x)

f ' (x)

= -e - , x

=0 , x

generaliz ed

= e ^s(x),

= -e ^H (x)

(x)) = 1-zr'(1+ik ) I ikF (k) .>0

<0

funct ions,

-co<x<

+e-- at& H (x) = e-*t t (*)

= e ^t i (x)

e ' -6 (x)

6 (x)

.d

since e-x = 1 when d(x) I 0.

r . t . (e-xt(x ' " 1 1 ik) + c{x)) = -h-(r* i , . \ * h = #tr** l= ikr(k)

How does one establ ish r igorously that operat ions withgeneralized functions are consistent and that the correctanswer is obtained t hen the rules are followed properly? Thereare several approaches but the proper treatment is beyond thescope of AMa95. The following three books rreat the subjectin di f ferent vrays. They are r isted j -n order of increasinqabstraction and complexity.

(caawLighthi l - l Four ier analysis and general ized funct ions.

Jones General ized funct ions l . l .4rnr_p,u.

cel fandand Shi lov General ized funct ions Vo1 . t .

Probably the best theoretical approach is to define ageneral ized funct ion as a cont inuous l iaear funct ional (dis_

tr ibut ion) on the space of 'good, funct ions, which are smoothfunct ions of f in i te extent (compact support) . An elementary

versi-on of th is approach (see Lighthi l l ) def j .ned tbem as thel imit of sequences. I f

J.Im .€n*- J--9n (x) G(x)dx exists for a sequence

gn(x) of smooth algebraical ly bounded funcrrons and everygood funct ion G(x) , the sequence is said to def ine ageneral ized funct ion g(x), and the l imit is wr i t ten

.rgGdx , and g = {Sr.,}

- r43-

d(x) = U+ "-*2t,)

Thus

e+0

- f=i,', r',* )-L '" Jn

-144-

(Conpare the definition of a real number as the limit of a

sequence of rational numbers) .

- I f f (x) is ordinary, i t is obvious that there exists

a sequence gn(x) such that

lim ,ii- fsrrea* = .rg G dx = .ff G dx

for all G, so the ordinary functions are included in the class

of generalized functions. Note the important fact that no meaning

is to be given to the value of a generalized function at a point.

Generalized functions are operators, not point value/ functj.ons.

However, if

. rgcdx=.f fGi lx

for al l good G which vanish for x outside (a,b), and t

is ori l inary in (arb), then rre can say g = f for x in (a,b) .

f,n this sense, 6(x) = 0 for x I 0, but strictly no rneaning

can be given to 6 (0) .

To see that S(x) alhrays has a derivative, note that+

.rsicdx = -/gnG'dx

which has a limit as n-+.o because G, is a good function,

and 9' is defined to be tSi] .

D-- /.-.i r itII ^

6 era co At?Ex(v:-) t-co d; G dx = -/__H # ax = -/o E; dx = G(0)

= / l -d (x)c (x) dx

/ltrTlrus 6(x) =

# .

To evaluate a generalized. function, ne can always integrate

by parts until ordinary functions are obtained. fhere is a

- 145-

theorem (Jones p 75) that i f

then there exists an ordinary

g i .s a general ized

piecewise cont inuous f (x) such

/eGdx = (-I)n,rf * u* ,dxt'

so that every g is the derivative of an ordinary f.

ceneralized functj-ons can be eoormousLy useful .provide a usefuL shorthand and certain results becomeparent, but there are snags, 6f . r rg.r . . and di f f icul t ies.

Mult ip l icat ion and div is ion of general ized funct ions may notbe def ined, or make sel tse, or b6 unigue. Thus, no meaning can

f- . 1 ' )De glven to [6 (x)J ' , and t / l " l is not unique (Lighthi l l

p. 39).

As an example of the use of general ized funct ions, wederive the poisson Sunmation Formula.

The Fouri.er series expansi.on

2rf (x) = I e-rnxfT f (v) o lnY. l , , ,r_TJ-\) , rc aY , _?I < x < ?T

is equivalent to the st,atement

2n6 (x-y) = t e-1n(x-Y), -TT < x < 7r

since mult ip ly ing by f (y)

pul l ing y = o,

and integrat ing gives (7). Thus,

2r6(x) = i . - i t ' * , -1r < x < 7T. (9)

But the r ight hand ":.a.

i" periodic with period 2tr. There-fore (9) realLy says

Xe*"^=

This is the poisson formula. To put j.t in its usual form,

(7 ' )

(8)

2r X 6 (x-2nn ) (10)

-146-

let f (ax), F (k,/a) be E'ourier transforms, for arbitrary

constant (real) a. Multiply by f (ax) and integrate

f l - r ' r -- r s,( : : ) = X f(2nra) . (11)a__a

-@

Ex(vi i ) 2 | -k2 /4

Let f (1) = E-x-r F(k)=*"^

; e-4r2a2n2 =

*^ i "'n2/+"2

From (I1)

(L2)

ff a is small (1arge) the left hand side is slolrly (quickly)

convergent and conversely for the right hand side.

Extensions to functions of several variable are straiqht-

forcard. Thus, 6 (x) 6 (y) has meaning:

' I /G(x 'Y)6(x)6(Y)dxdY = G(0,0) .

But care nay be needed with coordinate transformations if these

are singular. h polar coordinates rr9:

* l6(x)6(yI = Z+? 6(r+0) .