Hassani S. Mathematical Physics: A Modern Introduction to Its Foundations

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17.2

FREDHOLM

INTEGRAL

EQUATIONS

503

The reader may feel uneasy that the functions

j (x) and

j (t) appearing in

a degenerate kemel are arbitrary to within a multiplicative function. After

all, we

can multiply

<Pj(x) by a nonzero function, and divide

Vrj(t)

by the same function,

and get the same kemel. Such a change clearly alters the matrices

A and B and

therefore seems likely to change the solution,

u(x).

That this is not the case is

demonstrated in Problem 17.2.

In fact, it can be shown quite generally that the

transformations described above do not change the solution.

As the alert reader may have noticed, we have been avoiding the problem

of solving the eigenvalue (characteristic) problem for integral operators. Such

a problem is nontrivial, and the analogue

the finite-dimensional case, where

one works with determinants and characteristic polynomials, does not exist.

exception is a degenerate hermitian'' kemel, i.e., a kemel

the form K (x, t) =

L:?~l

hi(X)h~(t).

Substituting this in the characteristic-value equation

u(x)

Alb

K(x,

t)u(t)

dt,

we obtain

u(x)

= A

L:?=l

hi (x)

h~(t)u(t)

dt.

Defining

I-'i

h~(t)u(t)

and substituting it backin the equation gives

u(x)

= A

I)i(x)l-'i.

i=l

Multiplying this equation by A

-lh;;(x)

and integrating over x yields

(17.26)

This is an eigenvalueequation for the hermitian n x n

matrix

Mwith elements

mi],

which, by spectral theorem for hermitian operators, can be solved.

fact, the

matrix need not be hetmitian; as long as it is

normal, the eigenvalue problem can

be solved. Once the eigenvectors and the eigenvalues are found, we can substitute

them in Equation (17.26) and obtain

(x).

We expect to find a finite number of

eigenfunctions and eigenvalues. Our analysis of compactoperatorsincludedsuch a

case. That analysis also showed that the entire(infinite-dimensional) Hilbert space

could be written as the direct sum of eigenspaces that are finite-dimensional for

nonzeroeigenvalues. Therefore, we expect the eigenspace corresponding to the

zero eigenvalue (or infinite characteristic value) to be infinite-dimensional. The

following example illustrates these points.

17.2.6. Example. Let us

find

the

nonzero

characteristic

values

and

corresponding

eigen-

functions of the kernel

K(x,

t) = 1 +sin(x + t) for

-Jl'

::0

Actually,

the

problem

of a

degenerate

kernel

that

leadsto anormal

matrix,

described

below,canalsobesolved.

504 17.

INTEGRAL

EQUATIONS

We are seeking functions u and scalars Asatisfying

u(x)

= AK[u](x), or

u(x)

= A

[I +sin(x +

t)lu(t)

dt.

Expanding sin(x +

t),

we obtain

u(x)

=)"

[1 +sinx cos t +cos x

sint]u(t)

dt,

-lu(x)

ILl

+IL2sinx +

JL3

COSX,

(17.27)

(17.28)

where

1'1 = f':.n

u(t)

dt,

!'2

= f':.n

u(t)

cost

dt,

and

!'3

= f':.n

u(t)

sin t dt. Integrate

both sides

Equation (17.28) withrespect to x from

-rr

to obtain

A-I

1 =

23l'

Similarly,

multiplying

bysin x andcosx and

integrating

yields

and

(17.29)

1'1

0, we get A

-I

= 2,.., which, whensubstituted in (17.29),

yields!'2

= 1'3 =

thus

have,

asa first

solution,

All

2Jr

and

lUI)

a(6),

where

a is an

arbitrary

constant.

Equation (17.28) now gives

Ai1uI(X)

= 1'10 or UI(X) = cj , where

is an arbitrary

constant

tobe

determined.

On the other hand, 1'1 =0

-I

2,... ThenEquation (17.29) yields A

-I

±,..

and

1'2 = ±1'3·For A

-I

1.+

=,..,Equation (17.28) gives

u(x)

u+(x)

c+(sinx

cosx),

andforA-I

).,=1

-Jr,

it yields

u(x)

u_(x)

= c_{sinx - COS

x),

where

c± are

arbitrary

constants

to bedetermined by

normalization

of eigenfunctions. The

normalized

eigenfunctions are

U1=

u±(x)

~(sinx

± cos

x).

v 2,..

Direct

substitution

in the

original

integral

equation

easily

verifies

that

UI,'

U+,

and

u.:

are

eigenfunctions

ofthe

integral

equation

withtheeigenvalues

calculated

above.

Letusnow

consider

thezero

eigenvalue

(or

infinite

characteristic

value).

Divideboth

sides

Equation (17.27) by Aand take the limit

00.

Then the integral equation

becomes

[1+sinx

cost

+cos x

sint]u(t)

Thesolutions

u(t)

to this

equation

wouldspantheeigenspace

corresponding

to thezero

eigenvalue,

infinite

characteristic

value.

pointed

out above

that

thiseigenspace is

expected

to be

infinite-dimensional.

This.

expectation

borne

out once we note

that

all

functions

of the

form

sin

orcosnt withn

2 makethe

above

integral

zero;

and

there

are

infinitely many such

functions.

III

17.3

PROBLEMS

505

17.3 Problems

17.1. Use mathematical induction to derive Equation (17.5).

17.2. Repeat part (a) of Example 17.2.5 using

¢2(x)

'h(t)

= t

so that we still have

K(x,

t) =

4>1

(X)1/tl(t) +¢2(X)1/r2(t).

17.3. Use the spectral theoremfor compacthermitian operators to show that

the

kernel of a Hilbert-Schmidt operator has a finite number of nonzero eigenvalues,

then the kernel is separable. Hint: See the discussion at the beginning of Section

17.2.1.

17.4. Use the method of successive approximations to solve the Volterra equation

u(x)

= Ag

u(t)dt.

Then derive a DE equivalent to the Volterra equation (make

sure to include the initial condition), and solve it.

17.5. Regard the Fourier transform,

I 1

F[f](x)

ixy

fey)

,J2ir

-00

as anintegral

operator.

(a) Show that F

2[f](x)

fe-x).

(b) Deduce, therefore. that the only eigenvalues of this operator are A= ±I,

±i.

(c)

Let

f(x)

be any even function of

Show that an appropriate choice of a can

make

u = f +

aF[f]

an eigenfunction of F. (This shows that the eigenvalues of

Fhave infinite multiplicity.)

17.6. For what values of Adoes the following integral equation have a solution?

u(x)

= A

10"

sin(x +

t)u(t)

What

is that solution? Redo the problemusing a Neumann series expansion. Under

whatcondition is theseries

convergent?

17.7.

is possible to mnltiply the functions 4>j(x) by

rj(x)

and 1/rj(t) by

I!yj(t)

and still get the same degenerate kernel,

K(x,

t) =

I:j=1

4>j

(x)1/rj(t). Show that

such arbitrariness, although affecting the matrices A and S, does not change the

solution of the Fredholm problem

u(x)

Alb

K(x,

t)u(t)dt

f(x).

17.8. Show, by direct substitution, that the solutionfound in Example 17.2.4 does

satisfy its integral equation.

506 17. INTEGRAL

EQUATIONS

17.9. Solve

u(x)

= !

f~1

(x +

t)u(t)

+x.

17.10. Solve

u(x)

),.

xtu(t)

+x using the Neumann series method.

For

. what values

of)"

is the series convergent? Now find the eigenvalues and eigen-

functions

the kernel and solve the problem using these eigenvalues and eigen-

functions.

17.11. Solve

u(x)

),.

K(x,

t)u(t)dt+x

where", is any realnumberexcept

a negative integer, and

K(x,

t) =

e-(x+I).

For

what

values

of)"

does the integral

equation have a solution?

17.12. Solve the integral equations

(a)

u(x)

+),.

fol

xtu(t)dt.

(c)

u(x)

= x

fol

xtu(t)

dt.

(b)

u(x)

=),.

fo"

sin(x -

t)u(t)

dt.

(d)

u(x)

= x +foX

u(t)

dt.

17.13. Solve the integral equation

u(x)

= x

+),.

f~(x

+t)tu(t)

dt,

keeping terms

upto),.2.

17.14. Solve the integralequation

u(x)

ixi

+),.

f.""oo

e-1x-11u(t)

dt,

assuming

that

f remains finite as x

-->

±oo.

17.15. Solve the integral equation

u(x)

ixi

+),.

u(t)

cosxt

dt,

assuming

that

f remains finite as x -->

±oo.

Additional Reading

1. DeVito, C. Functional Analysis and Linear Operator Theory, Addison-

Wesley, 1990.

2. Jiirgen, K.

Linear Integral Operators, Pitman, 1982. Translated from its

original German, this is a thorough (but formal) introduction to integral

operators and equations.

3. Zeidler, E.

AppliedFunctional Analysis, Springer-Verlag, 1995.

18,

Sturm-Liouville Systems: Formalism

The linear operators discussed in the last two chapters were exclusively integral

operators. Most applications

physical interest, however, involve differential

operators (DO). Unfortunately,differentialoperatorsare unbounded. We noted that

complications arise when one abandons the compactness property of the operator,

e.g., sums tum into integrals and one loses one's grip over the eigenvalues of

noncompact operators. The transition to unbounded operators further complicates

malters. Fortunately, the formalism of one type of DOs that occurmost frequently

in physics can be stndied in the context

compactoperators. Such a stndy is our

aim for this chapter.

18.1 Unbounded Operators with Compact Resol-

vent

domain

ofa

linear

operator

As was pointed out in Example 16.2.7, the derivative operator cannot be defined

for

all functions in .(,2(a, b). This motivates the following:

18.1.1. Definition.

Let

D be a linear manifold1 in the Hilbert space

11:.

A linear

map

T :

-->

11:

will be calleda linear operator

11:.

is called the domain

ofT

and

often denoted by

:D(T).

18.1.2.

Example.

The domainof the derivative operatorD, as anoperatoron£,2(a, b),

cannot be the entire space. On the other hand, Dis defined on the linear manifold M in

£,2(a. b) spannedby (ei2nrrx/Lj wiih L = b - a. As we saw in Cbapter8, :M is dense

1A linear manifoldofan infinite-dimensional normedvectorspace

'\7

is a propersubsetthat is a vectorspacein its ownright,

butls notnecessarilyclosed.

As opposed to on X.

508

18. STURM-LIOUVILLE SYSTEMS: FORMALISM

(see Definition 16.4.5 and the discussion following it) in

~2(a,

b). This is the essence

Fourier

series:

That

every

function

int.}(a, b) canbe

expanded

in (i.e.,

approximated

by)

Fourier

series. It

turns

out thatmanyunbounded

operators

on a Hilbertspace sharethe

same

property,

namely

that

their

domains

aredensein the

Hilbert

space.

Another important property

Fourier expansion is the fact that if the function is

differentiable,

thenone can

differentiate

both sides, i.e., one can

differentiate

Fourier

expansion

term

such

operation

makes

senseforthe

original

function.

Define

the sequence

m) by

fm(x) = L uni2rrnx/L,

n=-m

1 l

a'l =

.JL

a !(x)e-l271:nxjLdx.

Then wecan statethepropertyaboveasfollows:Soppose

(fm)

is in

M.lflim

[m = I and

lim

I:"

=g, thenI' = g andI E M. Manyunboundedoperatorssharethisproperty. iii

18.1.3. Definition. Let

be a linear manifold in the Hilbert space Jf. Let T :

--'>

be a linear operator in Jf. Suppose that for any sequence {Iu

n))

in D,

both {Iu

))

and {Tlu

)}

converge in

Jf,

i.e.,

lim lu

)

= lu)

and

lim

T lu

= Iv) .

difference

between

hermitian

and

self-adjoint

operators

closed

operator

We say

thatT

is closed

iflv)

:DandTlu)

= [u).

Notice that we cannot

demand

that Iv) be in

for a general operator. This, as

we saw in the precedingexample, will

not

be appropriatefor unboundedoperators.

The

restriction

the domain

unbounded

operator is necessitated by

the

fact that the action

the

operatoron a vectorin

the

Hilbert space in general takes

that vector

out

the space.

The

following theorem (see [DeVi 90, pp.

251-252]

for a

proot)

shows

why

this is necessary:

18.1.4.

Theorem.

A closed linear operator in

that is defined at every point

(so that :D=

Jf)

is bounded.

Thus,

we are interested in unbounded operators (for instance, differential

operators), we

have to restrict

their

domains. In particular, we have to accept

the

possibility

an operatorwhose adjointhas a different domain.f

18.1.5. Definition.

Let Tbea linear operatorin

Jf.

Weshall say thatTishermitian

ifTt

isan extensionofT,i.e., :D(T) C :D(Tt)

andTt

lu) = T

lu)foralllu)

E :D(T).

is called self-adjoint if:D(T) = :D(Tt).

operators

with

compact

resolvent

As we shall see shortly, certain types

Sturm-Liouville

operators, although

unbounded, lend themselves to a study within

the

context

compact

operators.

18.1.6. Definition.

A hermitian linear operator T in a Hilbert space

is said to

have a compactresolvent

there is a I-' E P(T) for which the resolvent R" (T) is

compact.

3Thissubtle

difference

between

hermitian

and

self-adjoint

is statedheremerelytowarnthe

reader

andwill be

confined

to the

present

discussion. Thetwo

qualifiers

will be

(ab)used

interchangeably

intherestof thebook.

18.1

UNBOUNOEO

OPERATORS

WITH

COMPACT

RESOLVENT

509

An immediate consequence

this definition is that R, (T) is compact for all

A E

pen.

To see this, note that

R,(T)

is bounded by Definition 16.3.1.

Now

use

Equation (16.9) and write

The RHS is a product

a bounded"

and

a compactoperator, and therefore

must

compact.

The

compactness

the resolventcharacterizesits spectrumby Theorem

16.7.8. As the following theorem shows, this in

tnrn

characterizes the spectmm

the operators with compact resolvent.

18.1.7.

Theorem.

Let T be an operator with compact resolvent R,(T) where A E

pm.

Then 0

I" E

p(R,m)

and only

(A + 1/1") E peT). Similarly,

'lOis

an eigenvalue

ofR,(T)

ifand

only

(A+ 1/1") is an eigenvalue

ofT.

Furthermore, the eigenvectors

ofR,(T)

corresponding to I" coincide with those

T corresponding to (A+1/1").

Proof

The

proof

consists

a series

two-sided implications involving defini-

tions. We give the

proof

the first part, the second

part

being very similar:

I" E

(R,

(T))

iff

R,(T)

-1"1

is invertible.

R,(T)

-1"1

is invertible iff

(T - A

1)-1

-1"1

is invertible.

(T -

Al)-1

-1"1

is invertible iff

l-I"(T-Al)

invertible.

l-I"(T-Al)

is invertible iff

-l-T+Al

is invertible.

(±H)l-T

is invertible

iff

(±

E peT).

Comparing the LHS

the first line

with

the

RHS

the last line, we obtain the

first part

the theorem. D

A consequence

this theorem is

that

the eigenspaces

an (unbounded)

operator with compact resolvent are finite-dimensional, i.e., such an operator has

only finitely many eigenvectors correspondingto each

its eigenvalues.Moreover,

arranging the eigenvalues I"n

the resolvent in decreasing order (as done in

Theorem 16.7.8), we conclude that the eigenvalues

T can be arranged in a

sequencein increasing order

theirabsolutevalues and the limit

this sequence

is infinity.

18.1.8.

Example.

Considerthe operatorT in .(,2(0,1) definedbyS

= - r having

the domain

2J(T)

E .(,2(0,I) Ir E .(,2(0, I), frO) =

f(l)

OJ.

The readermay

4rhe sumoftwo

bounded

operators

bounded.

5Weshall

depart

from

our

convention

here

and

shallnotusethe

Dirac

bar-ket

notation

although

theuseof

abstract

operators

encourages

theiruse.The

reason

is thatin this

example,

we aredealingwithfunctions, andit is more

convenient

undress

the

functions

fromtheirDirac

clothing.

510 18. STURM-LIOUVILLE

SYSTEMS:

FORMALISM

check

that

zerois not an

eigenvalue

Therefore,

we maychoose

Ro(n

= T-

shallstudya

systematic

way of

finding

inverses

of some specific

differential

operators

the

upcoming

chapters

Green's

functions.

At thispoint,sufficeit tosay

that

T-

canbe

written

as aHilbert-Schmidt

integral

operator

withkernel

K(x,t)

{X(I-t)

if 0

t s I,

(I -

x)t

if 0

Thus,ifT

g, i.e.,if

-g,

then

rig

K[g],

i.e.,

I(x)

K[g](x)

K(x,

t)g(t)

fox

x)tg(t)

{(I

x)tg(t)

dt.

isreadilyverified thatK[g](O) =

K[g](l)

=0 and

!,,(x)

K[g]"(X)

-g.

We can now use Theorem 18.1.7

with)"

= 0 to find all the eigenvalues ofT:

/Ln

is an

eigenvalue

of T if andonly

Jkn

is aneigenvalue of

T-

These

eigenvalues shouldhave

finite-dimensional

eigenspaces, andwe shouldbe ableto

arrange

themin

increasing

order

magnitude

without

bound.

Toverifythis,we solve

-J.L!.

whose

solutions

are

/Ln

7f2

andIn(x) = sinmfx. Note

that

there

is onlyoneeigenfunction

corresponding

toeach

eigenvalue.

Therefore,

theeigenspaces are

finite-

(one-)dimensional.

l1li

The example above is a specialcase of a large class of DOs occurring in math-

ematical physics. Recall from Theorem 13.5.4 that all linear second-order differ-

ential equations can be made self-adjoint. Moreover, Example 13.4.12 showed

that any SOLDE can be transformed into a form in which the first-derivative term

is absent. By dividing the DE by the coefficient

the second-derivative term

necessary, the study of the mostgeneral second-orderlinear differential operators

boils down to that of the so-called

Sturm-Liouville

(S-L)

operators

which are assumed to be self-adjoint. Differential operators are necessarily ac-

companiedby boundary conditions that specify their domains. So, to be complete,

let us assume that the DO in Equation (18.1) acts on the subset

£,z(a, b) con-

sisting of functions

u that satisfy the following so-called

separated

boundary

conditions:

Sturm-Liouville

operators

separated

boundary

conditions

q(x),

alu(a)

fhu'(a)

= 0,

azu(b) +fJzu'(b) = 0,

(18.1)

(18.2)

regular

Sturm-Liouville

systems

where

aI,

az,

fh,

and

fJz

are real constants with the property that the matrix of

coefficients has no zero rows. The collection

the DO and the boundaryconditions

above is called a

regular

Sturm-Liouville

system.

We now show that the DO of a regular Sturm-Liouville system has compact

resolvent. First observe that by adding

au-with

a an arbitrary number different

from all eigenvalues of the

DO-to

both sides

the eigenvalue equation u" -

18.1

UNBOUNDED

OPERATORS

WITH

COMPACT

RESOLVENT

511

qu = AU,we canassume" thatzerois notaneigenvalue of

Lx.

Next,suppose that

Ut(x)

and

uz(x)

are ihe two linearly independent solutions

ofihe

homogeneous

DE satisfying ihe first and ihe second boundary conditions of Equation (18.2),

respectively. The operator whose kernel is

K(x,

t) =

{-U1(X)UZ(t)/w(a)

a::o x

::0

t::o b,

-UI(t)UZ(x)/W(a)

a::o t::o x

::0

in which W is ihe Wronskian

the solutions, is a Hilbert-Schmidt operator and

iherefore compact. We now show ihat

K(x,

t) is ihe resolvent Ro(L

) =

L.;-t

sa K

of our DO. To see this, write Lxu = v, and

uz(x)

Ut(x)

u(x)

K[v](x)

---

Ut(t)v(t)dt

-(-)

uz(t)v(t)dt.

W(a)

Wax

Differentiating this once gives

(x)

u';

(x)

(t)v(t)

_t_

uz(t)v(t)

dt,

W(a)

and a second differentiation yields

u"(x)

uI(t)v(t)dt

uz(t)v(t)dt

v(x).

W(a)

The last equationfollows from ihe fact ihatthe Wronskian

Uz-

u;u

I is constant

for a DE

the form u"

-qu

By substituting

qu;

andu~

qu:

in ihe

last equation, we verify ihat

u =

K[v]

is indeed a solution

ofihe

Sturm-Liouvi1le

system Lxu = v.

Next, we show that ihe eigensolutions of ihe S-L system are nondegenerate,

i.e., ihe eigenspaces are one-dimensional. Suppose

and

are any two eigen-

functions corresponding to ihe same eigenvalue. Then boihmust satisfy ihe same

DE and the Same boundary conditions; in particular, we must have

at!t(a)

fhI{(a)

(!t(a)

I{(a))

(at)

(0)

at!z(a)

fhI~(a)

= 0

jz(a)

I~(a)

fJt

- 0 .

(18.3)

and

fJI

are not boih zero, ihe

Wronskian-ihe

determinant

ihe matrix

above-must

vanish. Therefore, ihe two functions must be linearly dependent.

Finally, recall ihat a Hilbert space on which a compact operator K is defined

can be written as a direct sum

ihe latter's eigenspaces. More specifically,

'£-7=0

EIlJV(

where each

JV(

j is finite-dimensional for j = 1, 2,

...

, and

Although

this will change

q-and

the originaloperator-no

information

will be lost becausetheeigenvectors will be the

sameandall eigenvalues will be

changed

bya.

(18.4)

512

lB.

STURM-LIOUVILLE

SYSTEMS:

FORMALISM

N can befinite or infinite.

N is finite, then

Mo,

which can be considered as

the eigenspace of zero eigenvalne,7 will be infinite-dimensional.

is finite-

dimensional (or absent), then N mnst be infinite, and the eigenvectors of K will

span the entire space, i.e., they will form a complete orthogonal system. We now

show that this holds for the regular Sturm-Liouville operator.

Jacques Charles FraucoisSturm

(1803-1855)madethefirst

accurate

determination

of the velocity of soundin

water

1826,

working

with the Swiss engineer

Daniel

Colladon.

became

French

citizenin 1833 and

worked

Paris

at the

EcolePolytechnique

where

became

professor

in 1838.In

1840he succeeded

Poisson

in the

chair

of mechanics in the

Pacuue

des Sciences,

Paris.

The

problems

of determining theeigenvalues andeigen-

functions

of an

ordinary

differential

equation

with

boundary

conditions andof

expanding

a given

function

in tenus of an

infinite

seriesof theeigenfunctions, which

date

from

about

1750,becamemore

prominent

asnewcoordinate systemswere

introduced

andnewclasses

functions

aroseas theeigenfunctions of

ordinary

differential

equations.

Sturm

andhis

friend

Joseph

Liouville

decided

tackle

the

general

problem

for

any

second-order

linear

differential

equation.

Sturm

hadbeen

working

since1833on

problems

partial

differential

equations,

pri-

marily

ontheflowof heatin a

bar

variable

density,

andhencewasfully

aware

of the

eigenvalue

and

eigenfunction

problem.

The

mathematical

ideas

applied

tothis

problem

areclosely

tohis

investigations

the

reality

and

distribution

of the

roots

algebraic

equations.

His

ideas

differential

equations,

he says,came

from

the

study

difference

equations

anda

passage

tothe

limit.

Liouville,

informed

Sturm

ofthe

problems

hewas

working

on, tookup thesame

subject.

The

results

their

jointworkwas

published

several

papers

which

are

quite

detailed.

Suppose that the above Hilbert-Schmidt operator K has a zero eigenvalue.

Then, there must exists a nonzero function

v such that

K[v](x)

= 0, i.e.,

U2(X)1

Ut(x)

(t)v(t)

U2(t)V(t)

= 0

W(a)

for all x. Differentiate this twice to get

U"(x)

u"(x)

- .Ji(a)

(t)v(t)

~(a)

x U2(t)V(t)

v(x)

Now substitute

= qUj and U

= qU2 in this equation and use Equation (18.4)

to conclude that v =

This is impossible because no eigenvector can be zero.

7The

reader

recalls

that

whenKactsonMo, it yieldszero.