If this condition is satis¬ed then

x

y(x) = f (x) dx (5.3)

0

satis¬es both the di¬erential equation and the boundary conditions at x =

0, 1. If this condition is not satis¬ed, y(x) is not a solution, because y(1) = 0.

Initially we will discuss only solutions of Ly = f with homogeneous

boundary conditions. After we have understood how to do this, we will

extend our methods to deal with di¬erential equations with inhomogeneous

boundary conditions.

5.2 Constructing Green Functions

We wish to solve Ly = f , a di¬erential equation with homogeneous boundary

conditions, by ¬nding an inverse operator L’1 , so that y = L’1 f . This inverse

5.2. CONSTRUCTING GREEN FUNCTIONS 117

operator L’1 will be represented by an integral kernel

(L’1 )x,y = G(x, y), (5.4)

with the property

Lx G(x, y) = δ(x ’ y). (5.5)

Here, the subscript x on L indicates that L acts on the ¬rst argument of G.

Then

y(x) = G(x, y)f (y) dy (5.6)

will obey

δ(x ’ y)f (y) dy = f (x).

Lx y = Lx G(x, y)f (y) dy = (5.7)

The problem is how to construct G(x, y). There are three necessary ingredi-

ents:

• the function χ(x) ≡ G(x, y) must have some discontinuous behaviour

at x = y in order to generate the delta function;

• away from x = y, the function χ(x) must obey Lχ = 0;

• the function χ(x) must obey the homogeneous boundary conditions

required of y at the ends of the interval.

The last ingredient ensures that the resulting solution, y(x), obeys the bound-

ary conditions. It also ensures that the range of the integral operator, G,

coincides with the domain of L, a prerequisite if the product LG = I is

to make sense. The manner in which these ingredients are assembled to

construct G(x, y) is best explained through examples.

5.2.1 Sturm-Liouville equation

We want to ¬nd a function G(x, x ) such that χ(x) = G(x, x ) obeys

Lχ = (pχ ) + qχ = δ(x ’ x ), (5.8)

The function χ(x) must also obey the homogeneous boundary conditions that

are to be imposed on the solutions of Ly = f .

Now (5.8) tells us that χ(x) must be continuous at x = x . For if not, the

two di¬erentiations applied to a jump function would give us the derivative

of a delta function, and we want only a plain δ(x ’ x ). If we write

AyL (x)yR (x ), x < x ,

G(x, x ) = (5.9)

AyL (x )yR (x), x > x ,

118 CHAPTER 5. GREEN FUNCTIONS

then χ(x) = G(x, x ) is automatically continuous at x = x . We take yL (x)

to be a solution of Ly = 0, chosen to satisfy the boundary condition at the

left hand end of the interval. Similarly yR should solve Ly = 0 and satisfy

the boundary condition at the right hand end. With these choices we satisfy

(5.8) at all points away from x = x .

To ¬gure out how to satisfy the equation exactly at the location of the

delta-function, we integrate (5.8) from x ’ µ to x + µ and ¬nd that

x +µ

[pχ ]x ’µ = 1 (5.10)

This determines the constant A via

Ap(x ) yL (x )yR (x ) ’ yL (x )yR (x ) = 1. (5.11)

We recognize the Wronskian W (yL , yR ; x ) on the left hand side of this equa-

tion. We therefore have

1

y (x)yR (x ), x<x,

Wp L

G(x, x ) = (5.12)

1

y (x )yR (x), x>x.

Wp L

Now, for the Sturm-Liouville equation, the product pW is constant. This

follows from Liouville™s formula,

p1

x

W (x) = W (0) exp ’ dx , (5.13)

p0

0

and from p1 = p0 = p in the Sturm-Liouville equation. Thus

p(0)

W (x) = W (0) exp ’ ln(p(x)/p(0) = W (0) . (5.14)

p(x)

The constancy of pW means that G(x, x ) is symmetric:

G(x, x ) = G(x , x). (5.15)

This is as it should be. The inverse of a symmetric matrix (and the real,

self-adjoint, Sturm-Liouville operator is the function-space analogue of a real

symmetric matrix) is itself symmetric.

The solution to

Ly = (p0 y ) + qy = f (x) (5.16)

5.2. CONSTRUCTING GREEN FUNCTIONS 119

is therefore

1 b x

y(x) = yL (x) yR (x )f (x ) dx + yR (x) yL (x )f (x ) dx . (5.17)

Wp x a

Take care to understand the ranges of integration in this formula. In the ¬rst

integral x > x and we use G(x, x ) ∝ yL (x)yR (x ). In the second integral

x < x and we use G(x, x ) ∝ yL (x )yR (x). It is easy to get these the wrong

way round.

It is necessary that the Wronskian W (yL , yR ) not be zero. This is reason-

able. If W were zero then yL ∝ yR , and a single function satis¬es both Ly = 0

and the boundary conditions. This means that the di¬erential operator L

has a zero-mode, and there can be no unique solution to Ly = f .

Example: Solve

2

’‚x y = f (x), y(0) = y(1) = 0. (5.18)

We have

yL = x

’ yL yR ’ yL yR ≡ 1. (5.19)

yR = 1 ’ x

We ¬nd that

x(1 ’ x ), x < x ,

G(x, x ) = (5.20)

x (1 ’ x), x > x ,

x™

0 1

The function χ(x) = G(x, x ) .

and

x 1

y(x) = (1 ’ x) (1 ’ x )f (x ) dx .

x f (x ) dx + x (5.21)

0 x

5.2.2 Initial Value Problems