12 1

C= x + (5.53)

2 3

which makes G symmetric:

± 2

2

1

’ x + x +x , 0 < x < x

3 2

G(x, x ) = . (5.54)

x2 +x 2

1

’x+ , x < x < 1,

3 2

1

It also makes G(x, x ) dx = 0.

0

x™

The modi¬ed Green function.

126 CHAPTER 5. GREEN FUNCTIONS

The solution to Ly = f is

1

y(x) = G(x, x )f (x) d + A, (5.55)

0

where A is arbitrary.

5.3 Applications of Lagrange™s Identity

5.3.1 Hermiticity of Green function

Earlier we noted the symmetry of the Green function for the Sturm-Liouville

equation. We will now establish this formally.

Let G(x, x ) obey Lx G(x, x ) = δ(x ’ x ) with homogeneous boundary

conditions B, and let G† (x, x ) obey L† G† (x, x ) = δ(x ’ x ) with adjoint

x

†

boundary conditions B . Then, from Lagrange™s identity, we have

—

Q(G, G† ) = L† G† (x, x ) G(x, x ) ’ (G† (x, x ))— LG(x, x )

dx x

—

dx δ(x ’ x )G(x, x ) ’ G† (x, x ) δ(x ’ x )

=

—

= G(x , x ) ’ G† (x , x ) . (5.56)

Thus, provided Q(G, G† ) = 0, which is indeed the case because the boundary

conditions for L, L† are mutually adjoint, we have

—

G† (x , x) = G(x, x ) , (5.57)

and the Green functions, regarded as matrices with continuous rows and

columns, are Hermitian conjugates of one another.

Example: Let

d

D(L) = {y, Ly ∈ L2 [0, 1] : y(0) = 0}.

L= , (5.58)

dx

In this case G(x, x ) = θ(x ’ x ).

Now, we have

d

L† = ’ D(L) = {y, Ly ∈ L2 [0, 1] : y(1) = 0}

, (5.59)

dx

5.3. APPLICATIONS OF LAGRANGE™S IDENTITY 127

and G† (x, x ) = θ(x ’ x).

1 1

0 x™ 1 0 x™ 1

G† (x, x )

G( x, x )

5.3.2 Inhomogeneous Boundary Conditions

Our di¬erential operators have been de¬ned with linear homogeneous bound-

ary conditions. We can, however, use them, and their Green-function in-

verses, to solve di¬erential equations with inhomogeneous boundary condi-

tions.

Suppose, for example, we wish to solve

2

’‚x y = f (x), y(0) = a, y(1) = b. (5.60)

We already know the Green function for the homogeneous boundary-condition

problem with operator

2

D(L) = {y, Ly ∈ L2 [0, 1] : y(0) = 0, y(1) = 0}.

L = ’‚x , (5.61)

It is

x(1 ’ x ), x < x ,

G(x, x ) = (5.62)

x (1 ’ x), x > x .

Now we apply Lagrange™s identity to χ(x) = G(x, x ) and y(x) to get

1

2 2

= [G (x, x )y(x)’G(x, x )y (x)]1 .

dx G(x, x ) ’‚x y(x) ’ y(x) ’‚x G(x, x ) 0

0

(5.63)

Here, as usual, G (x, y) = ‚x G(x, y). The integral is equal to

dx {G(x, x )f (x) ’ y(x)δ(x ’ x )} = G(x, x )f (x) dx ’ y(x ), (5.64)

whilst the integrated-out bit is

’(1 ’ x )y(0) ’ 0y (0) ’ x y(1) + 0y (1). (5.65)

128 CHAPTER 5. GREEN FUNCTIONS

Therefore, we have

G(x, x )f (x) dx + (1 ’ x )y(0) + x y(1).

y(x ) = (5.66)

Here the term with f (x) is the particular integral, whilst the remaining terms

constitute the complementary function (obeying the di¬erential equation

without the source term) which serves to satisfy the boundary conditions.

Observe that the arguments in G(x, x ) are not in the usual order, but, in

the present example, this does not matter because G is symmetric.

When the operator L is not self-adjoint, we need to distinguish between

L and L† , and G and G† . We then apply Lagrange™s identity to the unknown

function u(x) and χ(x) = G† (x, y).

Example: We will use the Green-function method to solve the di¬erential

equation

du

= f (x), x ∈ [0, 1], u(0) = a. (5.67)

dx

You can, we hope, write down the answer to this problem directly, but it

is interesting to see how the general strategy produces the answer. We ¬rst

¬nd the Green function G(x, y) for the operator with the corresponding ho-

mogeneous boundary conditions. In the present case, this operator is

D(L) = {u, Lu ∈ L2 [0, 1] : u(0) = 0},

L = ‚x , (5.68)

and the appropriate Green function is G(x, y) = θ(x ’ y). From G we then

—

read o¬ the adjoint Green function as G† (x, y) = G(y, x) . In the present

example, we have G† (x, y) = θ(y ’ x). We now use Lagrange™s identity in