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

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20.2

FORMAL

CONSIOERATIONS

563

The most general form

the conjnnct for a SOLDO is

Q[u,

v*](x)

ql1

(x)u(x)v*(x)

+q12(X)U(X)v'*(x)

q21

(x)u'

(x)v*(x)

+q22(X)U'

(x)v'*(x),

which

can

he written in matrix form as

Q[u,

v*](x)

u~axv;

where

(ql1

(x)

qJ2(X)),

Q21(X)

Q22(X)

(20.15)

and

and v; have similar definitions as U

and

ahove. The vanishing of the

surface

termbecomes

(20.16)

Weneed to

translate

this equation intoa conditionon v* alone/' This is accom-

plished hy solving for two

the four quantities

u(a),

u'(a),

u(b),

and

u'(b)

terms of the other two, suhstituting the result in Equation (20.16), and setting the

coefficients of the other two equal to zero. Let us assume, as before, that the suh-

matrix B is invertible, i.e.,

u(b)

and

u'(b)

are expressible in terms of u(a) and

u'(a).

Then

-B-1Au

or ub

_u~At(Bt)-I,

and we obtain

-u~At(Bt)-labVb

u~aav~

[At

(Bt)-labVb

+aav~]

=0,

andtheconditionon v* becomes

(20.17)

We see that all factors

u have disappeared, as they should. The expandedversion

of the BCs on

v* are written as

BI[v*]

ul1v*(a)

+u12v'*(a) +

ql1v*(b)

+q12v'*(b) = 0,

B2[V*]

u2Iv*(a)

U22

v'* (a) +Q2IV*(b) +Q22V'*(b) = O.

(20.18)

adjoint

boundary

These homogeneous BCs are said to be adjoint to those

(20.12). Because of the

conditions

difference between BCs and their adjoints, the domain of a differential operator

need not be the sarne as that of its adjoint.

20.2.6.

Example.

Let Lx = d

withthehomogeneous BC,

RI[U] = au(a) - u'(a) = 0 and

R2[U]

f3u(b)

- u'(b) =

(20.19)

Wewantto calculate

Q[u, v*] andthe adjointBC, for v. By repeatedintegration byparts

[or by using

Equation

(13.23)], we obtainQ[u, u"] = u'v* - uv'*. For the

surface

termto

vanish,

we musthave

u'(a)v*(a) - u(a)v'*(a) = u'(b)v*(b) - u(b)v'*(b).

2The

boundary

conditions

on v*

should

not

depend

onthechoiceof u.

564 20.

GREEN'S

FUNCTIONS

ONE

DIMENSION

B2[V*] ={lv*(b) - v'*(b) =O.

Substitutingfrom (20.19)in thisequation,we get

u(a)[av*(a)

- v'*(a)] =u(b)[{lv*(b) - v'*(b)],

whichholdsfor arbitraryu

andonly if

Bl[V*] =

av*(a)

- v'*(a) = 0 and

(20.20)

Thisisa

special

case,in whichthe

adjoint

Bes

arethesameasthe

original

Bes

(substitute

ufor v* to see this).

Tosee

that

the

original

Bes

and

their

adjoints

neednotbethe

same,

consider

Rdu]

u'(a)

(b) =0 and

R2[U]

={lu(a) -

u'(b)

=0, (20.21)

from whichwe obtain

u(a)[{lv*(b) +v'*(a)] =

u(b)[av*(a)

+v'*(b)]. Thus,

Bl[V*] =

av*(a)

+v'*(b) =0 and BZ[v*] ={lv*(b) +v'*(a) =0,

(20.22)

mixed

and

unmixed

whichisnotthe sameas(20.21).Boundaryconditionssuchasthosein (20.19)and(20.20),

BCs

inwhicheach

equation

contains

the

function

andits

derivative

evaluated:

atthesame

point,

arecalled unmixed BCs. Onthe otherhand, (20.21) and (20.22)aremixedBCs.

III

20.2.2 Self-AdjointSOLDOs

ill

Chapter 13, we showed that a

SOLDO

satisfies the generalized Green's identity

with

w(x)

= I.

ill

fact, since u and v are real, Eqnation (13.24) is identical to

(20.11) if we set

w = 1 and

Q[u, v] =

pzvu'

- (P2V)'U +

PIUV.

(20.23)

Also, we have seen that any

SOLDO

can be

made

(formally) self-adjoint. Thus,

let us considerthe formally self-adjoint

SOLDO

Lx =

:x)

where both

p(x)

and

q(x)

are real functions and the inner productis defined with

weight

w = I.

we are interested in formally self-adjoint operators with respect

to a general weight

w > 0, we can construct

them

as follows. We first note that

Lx is formally self-adjoint with respect to a weight

unity, then

(l/w)L

is self-

adjoint withrespectto weight

w. Next, we note that Lx is formally self-adjointfor

all functions

q, in particular, for wq.

Now

we define

L(w) =

(p~)

+qw

and note that

)

is formally self-adjoint

with

respect to a weight

unity, and

therefore

ld(d)

Lx=-Li)=--

P-

wdx

(20.24)

20.3

GREEN'S

FUNCTIONS

FOR

SOlDOS 565

is formally self-adjoint with respect to weight

w(x)

For SOLDOs that are formally self-adjoint with respect to weight w, the con-

jnnct given in (20.23) reduces to

Q[u, v] =

p(x)w(x)(vu'

- uv'). (20.25)

Ri[U]

=Vi, i =

1,2.

(20.27)

common

types

boundary

conditions

fora

SOlDE

Thus, the surface term in the generalized Green's identity vanishes

and only if

p(b)w(b)[v(b)u'(b) -

u(b)v'

(b)] =

p(a)w(a)[v(a)u'

(a) -

u(a)v'

(a)].

(20.26)

The DO becomes self-adjoint if

u and v satisfy Equation (20.26) as well as the

same BCs.

can easily be shown that the following four types

BCs on

u(x)

assure the validity of Equation (20.26) and therefore define a self-adjoint operator

Lx given by (20.24):

I. The

Dirichlet

BCs: u(a) = u(b) =0

2. The

Neumann

BCs: u'(a) =u'(b) =0

General

unmixed

BCs:

au(a)

- u'(a) =fJu(b) - u'(b) =0

Periodic

BCs: u(a) =u(b) and u'(a) =u'(b)

20.3 Green's Functions for SOLDOs

We are now in a position to find the Green's function for a SOLDO. First, note

that a complete specification

Lx requires not only knowledge

po(x), Pt (x),

and

p2(x)-its

coefficient

functions-but

also knowledge

the BCs iroposed

on the solutions. The most general BCs for a SOLDO are of the type given in

Equation(20.10) with

n = 2. Thus, to specify Lxuniqnely, we considerthe system

(L; RI, R2) with data

(f;

VI,

)12).

This system defioes a unique BVP:

Lx[u]

=P2(X)

+PI(x)

po(x)u

f(x),

A necessary condition for Lx to be invertible is that the homogeneous DE

Lx[u] = 0 have only the trivial solution u =

For

u = 0 to be the only solntion,

it must be

a solntion. This means that it must meet all the conditions in Equation

(20.27).

In particular, since Ri are linearfunctionals of u, we musthave Ri[0] =

This can be stated as follows:

20.3.1.

Lemma.

Anecessary condition

for

a second-order linearDO to be invert-

ible is

for

its associated BCs to be homogeneous?

3The lemma applies to all linear DOs, not

just

second order ones.

566 20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

Thus, to study Green's functions we must restrict ourselves to problems with

homogeneous BCs. This at first may seem restrictive, since not all problems have

homogeneons BCs. Can we solve the others by the Green's function method? The

answer is yes, as will be shown later in this chapter.

The above discussionclearly indicates that the Green'sfunction of

Lx>

being its

"inverse," is defined only if we considerthe system

(L;

Rj,

Rz) withdata

(I;

0, 0).

the Green's function exists, it must satisfy the DE

Theorem 20.2.4, in which

Lx acts on G(x, y).

But

part of the definition of Lx are the BCs imposed on the

solutions. Thus, if the LHS of the DE is to make any sense,

G(x,

must also

satisfy those sarae BCs. We therefore make the following definition:

formal

definition

of 20.3.2. Definition. The Green's

function

a DO Lx is a function

G(x,

y) that

Green's

function

satisfies both the DE

8(x

- y)

LxG(x, y) =

w(x)

and, as afunction

x, the homogeneous

Bes

Ri[G] = Ofor i =

1,2

where the

Ri are defined as in Equation (20.12).

is convenient to study the Green's fumction for the adjoint of Lx simultane-

ously. Denoting this by

g(x,

y), we have

8(x

Lxg(x, y) =

w(x)

Bi[g]

= 0, for i =

1,2,

(20.28)

where

Bi are the boundary fumctionalsadjointto Ri and given in Equation(20.18).

adjoint

Green's

The function

g(x,

is known as the

adjoint

Green's

function

associated with

function

the DE

(20.27).

can

now use (20.27) and (20.28) to find the solutions to

Lx[u] =

f(x),

L1[v] = hex),

Ri[U]

= 0

Bi[V*]

= 0

for i =

1,2,

for i =

1,2.

(20.29)

With

v(x) =

g(x,

y) in Equation(20.11

)-whose

RHS is assumedto he

zero-we

getJ:

wg*(x, y)Lx[u]

wu(x)(d[gj)*dx.

Using (20.28) on the

RHSand

(20.29) on the LHS, we obtain

u(y)

= l

g*(x,

y)w(x)f(x)

dx.

Similarly, with

u(x)

G(x,

y), Equation (20.11) gives

v*(y) = l

G(x,

y)w(x)h*(x)

;

20.3

GREEN'S

FUNCTIONS

FOR

SOLO

567

or, since w

(x)

is a (positive) real function,

v(y)

= 1

G*(x,

y)w(x)h(x)dx.

These equations for

u(y)

and

v(y)

are not what we expect [see, for instance,

Equatiou (20.1)]. However,

we take into account certain properties of Green's

functions that we will discuss next, these equations become plausible.

20.3.1 Properties of Green's functions

Let us rewrite the generalized Green's identity [Equation (20.11)], with the RHS

equaltozero,as

dtw(t){v*(t)(Lt[u]))

= 1

dtw(t){u(t)(L;[vj)*j.

(20.30)

Green's

identity

This is sometimes called

Green's

identity. Substituting

G(t,

y) for

u(t)

and

g(t,

for vet) gives

8(t

y) 1

8(t

dtw(t)g*(t,

- =

dtw(t)G(t,

y) - ,

a wet) a wet)

or g*(y,

G(x,

y). A consequence

this identity is that

20.3.3. Box.

G(x,

y) mustsatisfy the adjoint

Bes

with respect to its second

argument.

for the time being we assume that the Green's function associated with a

system

(L; RI, Rz) is unique, then, since for a self-adjoint differential operator,

and Ll are identical and u and v both satisfy the same BCs, we must have

G(x,

y) =

g(x,

y) or, using

g*(y,

G(x,

y),

we get

G(x,

y) =

G*(y,

x).

particular,

the coefficient functions of Lx are all real, G(x, y) will be real, and

we have G(x,

y) = G(y,

x).

This means that G is a symmetricfunction

its two

arguments.

The last property is related to the continuity

G(x,

y) and its derivative at

x = y. For a SOLDO, we have

aZG

8(x - y)

LxG(x,

y)=

pz(x)-z

PI(x)-a

po(x)G

()'

x W x

where

PO,

PI,

and PZare assumed to be real and continuous in the interval [a, b],

and

w(x)

and

pz(x)

are assumed to be positive for all x E [a, b]. We multiply

both sides of the DE by

Jk(x)

(t)

]

hex)

--)'

where Jk(x)

exp

-(-)

pz(x

a PZ t

568 20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

notingthatdfL/dx

= (PI/P2)fL. This transforms the DE into

a [ a ] PO(X)fL(x) fL(Y)

ax fL(X) ax

G(x,

y) + P2(X)

G(x,

y) = P2(Y)W(Y)

8(x

- y).

Integrating this equation gives

a 1

Po(t)fL(t) , fL(Y)

fL(X)ijG(x,y)

()

G(t,y)dt=

()

()O(x-y)+a(y)

x a P2 t P2 y w Y (20.31)

because the primitive of 8(x - y) is O(x - y). Here

a(y)

is the "constant"

integration. First consider the case where Po = 0, for which the Green's function

will be denoted by

Go(x,

y). Then Eqnation (20.31) becomes

fL(X)aa

Go(x,y)

(;y)(

)O(x - y)

+aj(y),

P2YWY

which(since

P2,and w are continuouson [a, b],

andO(x-y)

has adiscontinuity

only at

x = y) indicates that

aGo/ax

is continuous everywhere on [a, b] except

x = y. Now divide the last equation by fL and integrate the result to get

Every term on the RHS is continuous except possibly the integral involving the

O-function. However, that integral can be written as

O(t -

/,X

-'-...,....:--'-dt = O(x - y)

a fL(t) y fL(t)

(20.32)

The O-function in front of the integral is needed to ensure that a

x as

demanded by the LHS of Equation (20.32). The RHS of Equation (20.32) is con-

tinuous at

x = y with limit being zero as x

Next, we write

G(x,

y) =

Go(x,

y) +

H(x,

y),

and apply

to both sides.

This gives

8(x

- y)

w(x)

= P2dx2 +PI

Go +poGo +

LxH(x,

8(x

- y)

w(x)

+poGo +

LxH(x,

y),

P2H"

P1H'

PoH

-PoGo·

The continuity

Go, po,

PI,

and P2 on

[a, b] implies the continuity of

because a discontinuity in H wonld entail a

delta function discontinuity in dH /

dx;

which is impossible because there are no

delta functions in the equation for

H. Since both Go and H are continuous, G

must also be continuous on [a, b].

(20.33)

existence

and

uniqueness

tor

second

order

linear

differential

operator

20.3

GREEN'S

FUNCTIONS

FOR

SOLOOS

569

We can now calcnlate the

jnmp

aG/ax

at x = y. We denote the jnmp as

(y)

and define it as follows:

, .

[aG

I aG I ]

AG (y)

Inn

-(x,

y) -

-(x,

y) .

10--+0

x=y+€

X=y-E

Dividing (20.31) by JL(x) and taking the above limit for all terms, we obtain

AG'(y)

+lim [ I 1

+' pO(t)

(t)G(t, y) dt

e....0

JL(Y+f)

pz(t)

1 1

- '

Po(t)JL(t)

G(t,

dt]

JL(Y

- f) a

pz(t)

,-"-,

JL(Y)

[O(+f)

O(-f)]

pz(y)w(y)!~

JL(Y

The second term on the LHS is zero becanse all functions are continnons at y. The

limit on the RHS is simply

JL(Y).

We therefore obtain

AG'

(y)

= 1

pz(y)w(y)

20.3.2 Construction and Uniqueness of Green's Fnnctions

We are now

ina

position to calcnlate the Green's function for a general SOLDO

and show that it is unique.

20.3.4.

Theorem.

Consider the system (L; RI, Rz) with data

(f;

0, 0), in which

Lx is a somo.

the homogeneousDE Lx[u] = 0 has nonontrivialsolution, then

the GF associated with the given system exists

and

is unique. The solution

the

system is

u(x)

= 1

dyw(y)G(x,

y)f(y)

and is also unique.

Proof

The GF satisfies the DE Lx

G(x,

y) = 0 for all x E [a, b] except x = y.

We thus divide [a, b] into two intervals, II = [a, y) and Iz =

(y,

b], and note

that a general solution to the above homogeneous DE

can

be written as a linear

combination

a basis

solutions,

and uz. Thus, we can write the solution of

the DE as

Gii», y) = CjUj(x)

+czuz(x)

Gr(x,y)

=dlul(x)+dzuz(x)

for x E II

for

x E h

570 20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

and define the GF as

G(x,

y) =

{Gl(X,

o,i»,

ifxEII,

ifxEIz,

(20.34)

where CI,cz,

di ; and dz are, in general, functions

y. To determine

G(x,

y) we

must determine four unknowns. We also have fuur relations: the continuity

the

jump

aGjax

atx

= y, and the two

BCs

RI[G] = Rz[G] =

The

continuity

ofG

gives

The

jump

aGjax

atx

= y yields

Pz(y)w(y)

Introducing bl =

- dl and bz =

- dz changes the two preceding equations

blu,

+bzuz = 0,

f f I

b,uI

+bzu

---.

pzw

These equations have a unique solution

iff

UZ)

det f f

I U

But

the determinantis simply the Wronskian

the two independentsolutions and

therefore cannot be zero. Thus, bl

(y)

and bz(y) are determined in terms

ofu"

u\'

U2.

u;.

pz.andw.

We now define

hex, y) sa

{bOI(y)U

(x)+

bz(y)uz(x)

x E

II,

ifxE/z.

so that

G(x,

y) = hex,

y)+dl(y)u,

(x)+dz(y)uz(x).

Wehave reducedthe number

unknowns to two, d, and dz. Imposing the

BCs

gives two more relations:

R,[Gj

RI[h]

+dIR,[uiJ

+dzRI[uz]

=0,

Rz[G] = Rz[h] +

dIRz[uI]

+dzRz[uz] =

Can we solve these equations and determine

and dz uniquely?We can, if

20.3

GREEN'S

FUNCTIONS

FOR

SOlDOS

571

can

shown

that this

determinant

nonzero

(see

Problem

20.5).

Having

found

the unique {bi,

dil~=l'

can

calculate ci uniquely, substitute

all

them

Equation

(20.34),

and

obtain

the

unique

G(x,

y).

That

u(x)

is also

unique

can

be shown similarly. D

20.3.5. Example. Letus calculatethe

GFfor

Lx = d

2/dx

with BCs u(a) = u(b) =

Wenote that Lx[u] = 0 with the given BCs has no nonnivial solution (verify this). Thns,

the

GF exists.TheDEfor

G(x,

y) is G" = 0 forx

-::f.

y, whose solutions are

G(x,

y) =

{CIX

if a

::::::

x < y,

dtx+d2

ify

<.x

s,b.

(20.35)

Continuityatx = y

givesqy

+c2

= dlY

+dz

or bj y

+bz

=Owithb

= Ci

-di-

The

discontinuity

of dG/ dx atx = y gives

dt - Cj =

= I

=-1

P2w

assuming

thatw = 1.

From

the

equations

abovewe also get

= y. G(x, y) mustalso

satisfy the given BCs. Thus,

G(a, y) = 0 = G(b, y). Since a S, y and b ::: y, we obtain

CIa +

=0 anddlb +dz = 0, or,

after

substituting

Ci = hi +

di,

+d2

= a - y,

bd,

+d2

The solution to these equations is dl = (y -

a)/(b

- a) and d2 =

-b(y

a)/(b

- a).

With

bl.

hz.

di.

anddzasgivenabove,we find

b-y

=b2+

=a-

b-·

-a

and

b-y

Cj=bl+dl=---

b-a

Writiog Equation (20.35)as

G(x,

y) = (Cjx +

C2)O(y

- x) +

(djx

+d2)O(X- y)

and using theidentity O(y - x) = 1 - O(x - y), we get

G(x,

y) = Cjx +

(blx

~)O(x

- y).

Usingthevaluesfoundfortheb's andc's, we

obtain

G(x,

y) = (a - x)

+ (x - y)O(x - y),

b-a

which is the sameas the GF obtained in Example 20.1.4.

III

20.3.6. Example. Letus findtheGFfor

= d

/dx

2+1

withtheBCsu(O) =

u(,,/2)

The

general

solution ofLx[u] = 0 is

u(x)

= A sin x +

Bcosx.

!fthe

BCsareimposed, we get u =

Thus,

G(x,

y) exists.The generalfonn

ofG(x,

y) is

G(x,

y) =

{CI

sinx

cosx

0::::;

x < y,

sinx

+d2COSX

Y <

x::::;

rr/2.

(20.36)

572

20.

GREEN'S

FUNCTIONS

ONE

OIMENSION

Continuity

G at x = y gives hI sin y + b: cos y = awithhi = Ci - dj. The

discontinuity

the derivative

G at x = y gives hI cos y - b: sin Y =

-I,

where we

haveset

w(x)

= 1.Solvingthese

equations

yields hI = - cosy and

= siny. The

Bes

give

G(O,y)=O

c2=0

dz=-b2=-siny,

G(rr/2,y)=0

d,=O

q=-b,=-cosy.

Substituting in Equatioo(20.36)gives

{

- COSY S

X if x -c y,

x,y)

= .

-smycosx

if y .c x,

or,

usingthetheta

function,

G(x,

y) =

-O(y

cosy sinx - O(y -

siny cosx

-[I

- O(x -

y)]

cosYsinx - O(x - y) siny cosx

-cosy

sinx +O(x -

sin(x - y).

is instructiveto verifydirectlythat

G(x,

y) satisfies

Lx[G]

~(x

- y):

Lx[G]

-cosy

(~+,)

sinx +

(::z

[O(x - y) sin(x -

y)]

[O(x - y)

sin(x -

y)]

+O(x - y) sin(x - y)

~[~(x

- y) sin(x - y)

+O(x

- y) cos(x -

y)]

+O(x - y) sin(x - y).

dx •

The

first

term

vanishes

because

the

sinevanishes attheonlypointwherethedelta

function

nonzero.

Thus,we have

Lx[G]

[~(x

- y) cos(x - y) - O(x - y) sin(x -

y)]

+O(x - y) sin(x - y)

=8(x - y)

becausethedeltafunctiondemandsthat x = y, for whichcos(x - y) = I.

The

existence and nniqueness

the Green's function

G(x,

y) in conjnnction

withits propertiesand its adjoint, implythe existenceand uniqueness

the adjoint

Green's function

g(x,

y). Using this fact, we

can

show that the condition for the

absence

a nontrivial solution for Lxlu] = 0 is also a necessary condition for

the existence

G(x,

y).

That

is,

G(x,

y) exists, then Lxlu] = 0 implies that

u = O. Suppose

G(x,

y) exists; then

g(x,

y) also exists. In Green's identity let

v =

g(x,

y). This gives an identity:

w(x)g*(x,

y)(Lxlu]) dx = l

w(x)u(x)(Lllg])*

8(x - y)

w(x)u(x)

dx =

u(y).

w(x)