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

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20.3

GREEN'S

FUNCTIONS

FOR

SOlDOS

573

In particular, if

Lx[u]

= 0, theu

u(y)

=0 for all y. We have proved the following

result.

20.3.7.

Proposition.

The

Lx[u]

= 0

implies

that

u sa 0 if

and

only

ifthe

corresponding to

and

the homogeneous

Bes

exist.

is sometimes stated that the Green's function

a SOLDO with constant

coefficients depends on the difference

x - y. This statement is motivated by the

observation that

(x)

is a solution

Lx[u]

=a2

+aou =f

ix),

then

u(x

is the solution

a2u"

alu'

+aou =

I(x

ao,

ai,

and a2

are constant. Thus,

G(x)

is a solution

Lx[G] =

8(x)

[again assuming that

w(x)

=I], thenit seemsthatthe solution

Lx[G] =

8(x

is simply

G(x

y).

This is clearly wrong, as Examples 20.3.5 and 20.3.6 showed. The reason is,

course, the BCs. The fact that G(x - y) satisfies the right DE does not guarantee

that it also satisfies the right BCs. The following example, however, shows thatthe

conjecture is true for a homogeneous

initial

value

problem.

20.3.8.

Example.

Themoslgeneralfonn for the OF is

{

CI UI(X) +C2U2(x)

x < Y,

x,y)

dlul(x)

+d2U2(x)

Y < x

TheIVP conditionG(a, y) = 0 =

G'(a,

y) implies

CIUI(a)

+c2u2(a)

= 0

and

Linear

independence

of Ul andU2

implies

(

UI (a) U2(a»)

det

u;(a)

ui(a)

= W(a;

uI,

U2) ""

Hence,

= C2 = 0 istheonly

solution.

Thisgives

(20.37)

{

ffa5x<y,

G(x,

y) =

dIUI(X)

+d2U2(x)

y < x

Continuity of G at x = y yields

dlUI

(y) + d2U2(y) = 0, while the discontinuity jump

conditionin thederivative gives

diU;

(y) +d2ui(Y) = I. Solvingthese twoequations, we

get

(20.38)

U;(Y)U2(Y) -

ui(y)uI

(y)'

d2=

= U2(Y) ,

U;(Y)U2(Y) -

ui(y)uI(Y)

Substituting thisin (20.37)gives

G(x,

y) = [U2(Y)UI(x) - UI(Y)U2(X)]

e(x

_ y).

u; (Y)U2(Y) -

ui(y)uI

(y)

574 20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

Equation

(20.38)

holdsfor any

SOLDO

withthe givenBes. Wenowusethe factthatthe

SOLDO

hasconstantcoefficients. In thatcase,weknowtheexactformof

zej

and"2.

There

are

twocasesto

consider:

the characteristic polynomial

Lxhastwodistinctroots Al and

}..2,

thenUl (x) =

AtX

and

U2(X)

= e

A2X

Writing

Al = a +b

and)..2

= a - b and

substituting

the

exponential

functions

andtheir

derivatives

Equation

(20.38)yields

[

eCa- b)Ye ca+b)X _

e(a+b)ye(a-b

G(x,

y) = 2be

2ay

e(x

- y)

[e(a+b)(X-y) -

e(a-b)(x-Y)j

e(x

- y),

whichis a

function

of x - y alone.

Al = ).2 =

then

Ul(X)

= e

u2(x)

= xe

and

substitution

of these

functions

Equation

(20.38)

gives

G(x,

y) =(x -

y)e'(x-Y)e(x

- y).

20.3.3 Inhomogeneous

Bes

III

solves

inhomogeneous

BCs

well

So far we have concentrated on problems with homogeneous BCs, Ri [u] = 0,

for i =

1,2.

What

the BCs are inhomogeneous?

turns out that the Green's

function method, even though it was derived for homogeneous BCs, solves this

kind

problem as well! The secret

this success is the generalized Green's

identity.

Suppose we are interested in solving the DE

Lx[u] =

f(x)

with

Ri[U] = Yi

for i =

1,2,

and we have the GF for

(with homogeneous BCs,

course). We can substitute

v =

g(x,

y) =

G*(y,

x) in the generalized Green's identity and use the DE to

obtain

w(x)G(y,

x)f(x)

dx -l

w(X)u(x)(Lr[gj)*

dx =

Q[u,

g*(x,

Y)]I;~~,

or, using

d[g(x,

y)]

= 8(x -

y)/w(y),

u(y)

= t

w(x)G(y,

x)f(x)

- Q[u,

g*(x,

Y)]I;~~.

To evaluate the surface term, let us write the BCs iu matrix form [see Equation

(20.17)]:

+BUb = Y '* Ub =

B-Iy

B-IAu

+BGb = 0 '* A'(B')-IOb9); +

Oa9;

= 0,

(

20.3

GREEN'S

FUNCTIONS

FOR

SOlOOS 575

where Y is a column vector composed

and

Y2,

and we have assumed that

G(x,

y) and

g*(x,

y) satisfy, respectively, the homogeneons BCs (with Y = 0)

and their adjoints. We have also assumed that the 2 x 4 matrix of coefficients has

rank 2, and withontloss of generality,let Bbe the invertible 2 x 2 submatrix. Then,

assuming the general form

the surface term as in Equation (20.15), we obtain

)][x~b

to * to *

u,g

x,Y

x=a=ub

b9b-ua

aga

(

-I -I

A)'

a * t *

= B Y - B U

b9b - u

y'(B,)-I

Ob9);

U~,[At(B')-IOb9);

+Oa9~]

= 0

because

satisfies

homogeneous

adjoint

(20.39) .

where

* (

g*(b,

y) ) (

G(y,

b) )

sa a * = a .

(x, y)[x=b

axG(y,X)[x~b

It follows that

Q[u,

g*(x,

Y)]I;'~~

is given entirely in terms of G, its derivative, the

coefficient functions

the DE (hidden in the matrix

0),

the homogeneous BCs

(hidden in B), and the constants

and

Y2.

The fact that g* and

agO

lax

appear

to be evaluated at

x = b is due to the simplifying (but harmless) assumption that

Bis invertible, i.e., that

u(b)

and

u'(b)

can be written in terms of

ural

and

u'(a).

Of course, this may not be possible; then we have to find another pair of the four

quantities in terms of the other two, in which case the matrices and the vectors

will change but the argument, as well as the conclusion, will remain valid. We can

now write

u(y)

= l

w(x)G(y,x)j(x)dx

ytM9*,

(20.40)

where ageneralmatrix Mhas beenintroduced, and the subscript

bhas beenremoved

to encompasscaseswhere submatricesotherthan Bare invertible. Equation (20.40)

shows that

u can be determined completely once we know

G(x,

y),

even though

the BCs are inhomogeneous.

In practice, there is no need to caicolate

We can (

use the expression for

Q[u, gO] obtained from the Lagrange identity

Chapter

13 and evaluate it at

b and a. This, in general, involves evaluating u and G and

their derivatives at

a and b. We know how to handle the evaluation

G because

we can actually constroct it

(if

it exists). We next find two of the four' quantities

corresponding to

u in terms of the other two and insert the result in the expression

for

Q[u,

gO].Equation (20.39) then guaranteesthat the coefficientsof the othertwo

terms will be zero. Thus, we

can

simply drop all the terms in

Q[u,

gO]containing

a factor of the other two terms.

(20.41)

576

20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

Specifically, we

use

the

conjunct

for

a formally self-adjoint

SOLDO

[see Equa-

tion (20.26)]

and

g*(x,

y) =

G(y,

obtain

u(y)

= l

w(x)G(y,x)f(x)dx

{

]}X~b

p(x)w(x)

[G(y,

u(x)h(y,

x~a

Intercbanging x

and

y gives

u(x)

= l

w(y)G(x,

y)f(y)

{

[

du]}Y=b

p(y)w(y)

u(y)-a

(x,

y) -

G(x,

y)-d

y y y=a

This

equation is valid only

for

a self-adjoint

SOLDO.

That

is,

using

it requires

casting

the

SOLDO

into a self-adjoint

form

(a process

that

is always possible, in

light

Theorem

13.5.4).

setting

f(x)

= 0, we

can

also

obtain

the

solution to a homogeneous

Lx[u] = 0

that

satisfies

the

inhomogeneous

Bes.

20.3.9. Example. Let us findthe solution of the simpleDE d

2u/dx

f(x)

subjectto

the simple inhomogeneousBCs

u(a) =

and u(b) =

Y2.

The GF for this problem has

been calculatedin Examples20.1.5 and 20.3.5.Let us begin bycalculating the surfaceterm

in Equation (20.41).Wehave

p(y)

= I, and we set w(y) = I, then

surfaceterm

= u(b) oG I -

G(x,

b)u'(b) - u(a) oG I + G(x, a)u'(a)

y=b

y~a

oG I - Yl oG I + G(x, a)u'(a) - G(x, b)u'(b).

y~b

y~a

That

the

unwanted

(and

unspecified)

terms

are

zerocanbeseenby

observing

that

G(x,

a) =

g*(a, x) =(g(a, x))*, and that g(x, y) satisfiesthe BCsadjointto the homogeneous BCs

(obtained when

Yt = 0). In this particular and simple case, the BCs happen to be self-

adjoint (DirichletBCs). Thus,

u(a) =u(b) =0 implies that g(a, x) = g(b, x) =0 for all

x E [a, b]. (In amore

general

casethecoefficient

ofu'(a)

wouldbemore

complicated,

but

still zero.) Thus, we fioallyhave \

surfaceterm

oG I - Yl oG I .

oy y=b oy y=a

Now,using the expressionfor G(x, y) obtained in Examples 20.1.5 and 20.3.5,we get

a-x

x-a

- =

---

- 9(x - y) - (x - y)8(x - y) =

- 8(x -

y).

b-a

Thus,

soI

x-a

y=b =

b-a'

OGI

-a

x - b

----1---

oy y=a -

b-a

b-a'

20.4

EIGENFUNCTION

EXPANSION

GREEN'S

FUNCTIONS

577

Substituting in Equation

(20.41),

we get

u(x) = f

G(x,y)f(y)dy

~=~IX+

bY~=:Y2.

(Compare

thiswiththeresultobtainedin Example

20.1.5.)

III

Green's functions have a very simple and enlighteningphysical interpretation.

AninhomogeneousDE such as

Lx[u]

f(x)

can be interpretedas ablackbox

(L,)

that determines a physical quantity (u) when there is a source

(f)

of that physical

quantity. For instance, electrostatic potential is a physical quantity whose source

is charge; a magnetic field has an electric current as its source; displacements and

velocities have forces as their sources; and so forth. Applying this interpretation

and assuming that

w(x)

= 1, we have

G(x,

as the physical quantity, evaluated

x when its source 8(x - y) is located at y. To be more precise, let us say that

the strength of the source is

and it is located at Yl; then the source becomes

818(x -

Yl).

The physical quantity, the Green's function, is then 81G(X,

Yl),

because of the linearity of Lx:

G(x,

y) is a solution of

[u] = 8(x - y), then

G(x,

Yl)

is a solution of

Lx[u]

= 818(x -

y).

there are many sources located

{yil[:,1

with corresponding strengths {8il[:,1' then the overall source f as a

function of

x becomes

f(x)

L,[:,t

8i8(X - Yi), and the corresponding physical

quantity

u(x)

becomes

u(x)

L,[:,1

8iG(X,

Yi).

Sincethe source S; is located at Yi, it is more natural to define a function 8

(x)

and write S; = 8(Yi). When the number

point sources goes to infinity and Yi

becomes a smoothcontinuous variable, the sumsbecome integrals,

and-

we have

f(x)

= l

8(y)8(x

- y)

dy,

u(x)

= l

8(y)G(x,

dy.

The first integral shows that 8

(x)

= f

(x).

Thus, the second integral becomes

u(x)

f(y)G(x,

y) dy which is precisely what we obtained formally.

20.4 Eigenfunction Expansion

Green'sFunctions

Green'sfunctions areinversesof differential

operators.

Inverses

operators

in a

Hilbert space are best studied in terms of resolvents. This is becauseif an operator

Ahasaninverse, zerois initsresolventset, and

Thus,

it is

instructive

to discuss

Green's

functions

in thecontext of theresolvent

of a differential operator. We will cousider only the case where the eigenvalues are

discrete, for example, when Lxis a Sturm-Liouville operator.

Formally, we have

(L - A

l)R)JL)

= 1, wbich leads to the DE

8(x - y)

(Lx - A)R).(x, y) = ,

w(x)

(20.42)

578 20.

GREEN'S

FUNCTIONS

ONE

DIMENSION

where R).(x, y) = (xl R,,(L)ly). The DE simply says that R,,(x, y) is the

Green's function for the operator

So we can rewrite the equation as

a, - A)G,,(X, y) =

8(x

y)jw(x)

where

- A is a DO having some ho-

mogeneous BCs. The GF G"

(x,

y) exists

and only

(Lx - A)[u] = 0 has no

nontrivial solution, which is true only if

Ais not an eigenvalue

Lx.We choose

the BCs in such a way that

becomes self-adjoint.

Let

{Anl~l

be the eigenvalues of the system

Lx[u]

= AU, {Ri[U] =

OIT=l'

and let the

u~k)

(x)

be the corresponding eigenfunctions. The index k distinguishes

among the linearly independent vectors corresponding to the same eigenvalue

An.

Assuming that Lhas compact resolvent (e.g., a Sturm-Liouville operator), these

eigenfunctions form a complete set for the subspace of the Hilbert space that

consists of those functions that satisfy the same BCs as the

u~k)

(x). In particnlar,

G,,(x,

y) can be expanded in terms of

u~k)(x).

The expansion coefficients are,

course,functions of y. Thus,we can write

G,,(x,

= L

~>~k)(y)u~k)(x)

n=l

where

a~k)(y)

w(x)u~(k)(x)G,,(x,

dx.

Using Green's identity, Equation

(20.30), and the fact that

Anis real, we have

b .

Ana~k)(y)

= 1

W(X)[AnU~k\X)]'G,,(x,

y)dx

= l

w(x)G).(x,

y){Lx[u~k)(x)]}·

= l

w(x)[u~k)(x)]'Lx[G,,(x,

y)]

= l

w(x)u~(k)(x)

[8(X

+AG,,(X,

y)]

w(x)

u~(k)(y)

+Al

w(x)u~(k)(x)G).(x,

u~(k)(y)

Aa~k)(y).

Thus,

a~k)(y)

u~(k)(y)j(An

- A), and the expansionfor the Green's function is

G ( )

u~(k)(y)u~k\x)

" x, y - L... L... A _ A .

n=l

This expansion is valid as long as An

A for any n = 0, I, 2,

....

But this is

precisely the condition that ensures the existence of an inverse for L-

A1.

An interesting result is obtained from Equation (20.42)

A is considered a

complex variable.

In that case, G" (x,

has (infinitely many) simple poles at

20.4

EIGENFUNCTION

EXPANSION

GREEN'S

FUNCTIONS

579

{A"}~l'

The residue at the pole A" is -

u:(k)(y)u~k)(x).

c; is a contour

having the poles

{A"};:'=l in its interior, then, by the residue theorem, we have

In particular, if we let

00,

we obtain

S(x -

w(x)

(20.43)

where Coo is

any contour that encircles all the eigenvalues, and in the last step

we used the completeness

the eigenfunctions. Equation (20.43) is the infinite-

dimensional analogue

Equation (16.12) with f (A) = 1 whenthe latterequation

is sandwiched between

(x Iand I

y).

20.4.1.

Example.

Consider the DO Lx = d

2/dx

with BCs u(O) =

ural

This is

an S-L operator with eigenvalues and normalized eigenfunctions

(mr)2

An-

and

(2 .

(mr

)

Un(X)=v~sm

for n = 1,2,

....

Equation (20.42) becomes

sin(mrx/a)

sin(mry/a)

)J ,y) -

/)2

n=l

I\,

- n1f a ,

which leads to

1 i

()d

1 i [2

sin(mrx/a)sin(mry/a)]

x Y A=-

2rri

Coo

2:rri Coo a

n=l

A-

(mrja)2

(~)

I:sin

(mr

sin

(""

2:Jr:i

n=l

a a ie

A-

(nJrja)2

= _

(~)

f:sin

(n"

sin

(n"

Res [ 1

n=l

a a ').-

(mf/a)

l=A

= -

(~)

sinC"

SinC"

y).

a n=l a a

eigenfunclion

expansion

The RHS is recognized as

-8(x

- y).

zero is not an eigenvalue

Lx, Equation (20.42) yields

u:(k)(y

)u~k)

(x)

G(x,y)=Go(x,y)=LL

n=l

(20.44)

which is an expression for the Green's function

Lx in terms

its eigenvalues

and eigenfunctions.

580 20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

20.5 Problems

20.1. Using the GF method, solve the DE

Lxu(x)

duf

f(x)

subject to

the BC

u(O) =a. Hint: Consider the function

vex)

u(x)

- a.

20.2. Solvethe problem of Example 20.1.4 subjectto the BCs

u(a)

u'(a)

Show that the corresponding GF also satisfies these BCs.

20.3. Show that the IVP with data 10;0, 0,

...

has only u

0 as a solution.

Hint: Assume otherwise, add

u to the solution

the inhomogeneous equation,

and invoke uniqueness.

20.4.

In this problem, we generalize the concepts

exactness and integrating

factor to a NOLDE. The DO

Lin)

I:~=o

pkCx)d

/dx

is said to be exact if

there exists a DO Min-I) sa

I:~;;6

ak(x)d

/dx

such that

VUE

en[a, b].

(a) Show that

Lin)

is exact iff

I:~~o(

_I)md

Pm/dxm

(b) Show that there exists an integrating factor for

Lin)-that

is, a function

/L(x)

such that /L(x)Li

exact-if

and only

/L(x)

satisfies the DE

The DO Ni

is the formal adjoint

L¥').

20.5. Assuming that

Lx[u]

= 0 has no nonttivial solution, show that the matrix

where UI and

U2 are independent solutions

Lx[u]

= 0 and Ri are the bound-

ary functionals, has a nonzero determinant. Hint: Assume otherwise and show

that the system of homogeneous linear equations

aRI[uI]

,BRI[U2]

= 0 and

aR2[u!l

+,BR2[U2] = 0 has a nontrivial solution for

(a,,8).

Reacha contradiction

by considering

u =

aUI

+,Bu2as a solution of

Lx[u]

20.6. Determine the formal adjoint of each

the operators in (a) through (d)

below (i) as a differential operator, and (ii) as an operator, that is, including the

BCs. Whichoperatorsare formally self-adjoint?Whichoperators are self-adjoint?

(a)

Lx = d

2/dx

+I in [0, 1] with

BC,

u(O) =

u(l)

(b) Lx = d

2/dx

in [0, 1]

withBCs

u(O) = u'(O) =

in [0, 00]

withBCs

u(O) =

(d) Lx = d

3/dx

sinxd/dx

+3 in [0, IT] with BCs u(O) =

u'(O)

= 0,

u"(O)

4u(IT)

with R[v]

20.5

PROBLEMS

581

20.7. Show that the Dirichlet, Neumann, generalumnixed, and periodicBCs make

the following formally self-adjoint SOLDO self-adjoint:

=.!..~

(p~)

+q.

wdx

20.8. Using a procedure similar to that described in the text for SOLDOs, show

that for the FOLDO

= PI dj dx +

(a) the indefinite GF is

G(x,

y) sa

{'(y)

[e(x

y)]

C(y),

PI(Y)w(y)

{'(x)

po(t)

]

where

{'(x)

= exp

PI(t)

(b) and the GF itselfis discontinuous at x = y with

. 1

Inn

[G(y

+E, y) -

G(y

- E,

y)]

()

()'

,-+0

PI Y w Y

(c)

For

the homogeneous BC

R[u] sa

O<lu(a)

0<2u'(a)

+f3tu(b) +

fhu'(b)

= 0

construct

G(x,

y) and show that

G(x,

y) =

()

( ) v

(x)e

(x - y) +

C(y)v(x),

PI Y w Y v Y

where

v(x)

is any solution to the homogeneous DE Lx[u] = 0 and

C(y)

f!(v(b)

{hv'(b)

R[v]PI

(y)w(y)v(y)

(d) Show directly that Lx[G] =

~(x

y)jw(x).

20.9. Let Lx be a NOLDO with constant coefficients. Show that

u(x)

satisfies

Lx[u] =

f(x),

then

u(x

- y) satisfies Lx[u] =

f(x

y).

(Note that no BCs are

specified.)

20.10. Find the GF for

jdx

+1 with BCs u(O) =u'(O) =

Show that

it can be written as a function of

x - y only.

20.11. Find the GF for Lx =d

jdx

withBCs u(O) =

u(a)

20.12. Find the GF for Lx =d

jdx

- k

with BCs u(oo) =

u(-oo)

20.13. Find the GF for

(djdx)(xdjdx)

given the condition that

G(x,

y) is

finite at

x = 0 and vanishes at x = I.

582 20.

GREEN'S

FUNCTIONS

ONE

DIMENSION

20.14. Evaluate the GF and the solutions for each of the following DEs in the

interval [0, I].

(a)

u" - k

u =

u(O) - u'(O) = a,

u(l)

= b.

(b) u" =

u(O) = u'(O) = O.

(d)

u" +!J)2u =

f(x),

for x > 0; u(O) = a, u'(O) = b.

(e) u(4) =

u(O) = 0, u'(O) =

2u'(I),

u(l)

= a, u"(O) =

20.15. Use eigenfunctionexpansion

the GF to solve the BVP u" =

u(0) =0,

u(l)

2u'(I)

Additional Reading

I. Dennery, P.and Krzywicki,A. Mathematicsfor Physicists, Harperand Row,

1967. Contains an exposition

Green's functions for second-order linear

differential equations.

2. Roach, G.

Green's Functions, Van Nostrand, 1970. A readable introduc-

tion to Green's functions, especially those we have called "indefinite" (no

bonndary conditions specified).