Integral Equations and the Method of Green Functions

James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, All rights reserved.

Page maintained with additions by Evans M. Harrell, II,



This section will continue to use the basic idea that a collection of functions defined on an interval might form a vector space and that the vector space of functions might have an inner product defined on it.


Recall that L2[0,1] with the "usual" inner product is such an inner product space of functions on [0,1]. (See Chapter B2 for more discussion of inner products.) One should be aware that there are many choices that can be made for an inner product for the functions on [0,1]. One might have a weighted inner product such as we had in Rn and as was described in Chapter 1. That is, if w(x) is a continuous, non-negative function, we might have

< f , g >  = I(0,1, ) f(x) g(x) w(x) dx.

The choices for w(x) are usually suggested by the context in which the space arises.

Some inner product spaces are called HILBERT SPACES. A Hilbert space is simply a vector space on which there is an inner product and on which there is one more bit of structure. A Hilbert space is a complete inner product space. This means that if fp is an infinite sequence of vectors in the space which is Cauchy convergent - meaning that

limn,m < fn-fm , fn-fm > = limn,m | fn-fm |2 = 0

- then there is a vector g, also in the space, such that limn fn = g.

To illustrate these ideas, two examples follow. In the first, there is a sequence {fp} and a function g in the space with

limn fn = g. In the second, there is no such g.

EXAMPLE: Let E be the vector space of continuous functions on [0,1] with the usual inner product. Let

fn(x)  =   BLC{(A(1/n  if 0 <= x <= 1/n, x   if 1/n <= x <=

and let g(x) = x on [0,1]. Then limn |fn - g|2

= lim<sub>n</sub>  I(0,1, ) [ fn(x)-g(x) ]<sup>2</sup> dx                  = lim n  I(0,1/n, )  (1/n  - x)<sup>2</sup> dx = 0.

EXAMPLE. This space E of continuous functions on [0,1] with the "usual" inner product is not complete. To establish this, we provide a sequence fp for which there is no continuous function g such that limn fn = g.

fn(x) =    BLC{(A(0, n(x- 1/2) + 1/2, 1))A(if 0 <= x <= (n-1)/2n,     if
(n-1)/2n <= x <= (n+1)/2n,if (n+1)/2n <= x <= 1)

Sketch the graphs of f1, f2, and f3 to see that the limit of this sequence of functions is not continuous.

The Riesz Representation Theorem is an important idea in a Hilbert space. You will recall that in Rn, this result declared that if L is a linear function from Rn to R then there is a vector v in Rn such that L(x) = < x, v > for each x in Rn. (More description of the analogue of Riesz representation for Rn.)

In general, where the vector space is not Rn, more is required.

THEOREM If {E, < , >} is a Hilbert space, then these are equivalent:

(a) L is a continuous, linear function from E to R (or the complex numbers C), and

(b) there is a member v of E such that L(x) = < x, v > for each x in E.

It is not hard to show that (b) implies (a). To show that (a) implies (b) is more interesting. The argument uses the fact that E is complete and can be found in any introduction to Hilbert space or functional analysis.

EXAMPLE. Since the space of continuous functions on [0,1], denoted C[O,1], with the usual inner product is not a Hilbert space, one should expect that there might be a linear function L from C[0,1] to R for which there is no v in C[0,1] such that

L(f) = I(0,1, ) f(x) <b>v</b>(x) dx

for each f in C[0,1]. In fact, here is such an L:


L(f) = I(3/4,1, ) f(x) dx.

The candidate for v is v(x) =1 on [3/4, 1] and 0 on [0, 3/4). But this v is not continuous! It is only piecewise continuous.

DEFINITION. The Heaviside function H is defined as follows:

H(x)  =  BLC{(A(1 if x > 0,0 if x < 0)).

Note that

I(0,1, ) f(x) H(x- 3/4) dx = I(3/4,1, ) f(x) dx.

Thus, the Heaviside function provides an element v for which the linear function

L(f)  = I(3/4,1, )  f(x) dx

has a Riesz representation. As noted, v is not in C[0,1].

It is not always possible to have a piecewise continuous v which will rectify the situation. Consider the following linear function: L(f) = f(1/2). It is not so hard to see that there is no piecewise continuous function v on [0,1] having the property that for every continuous f,

I(0,1, ) f(x) <b>v</b>(x) dx = f(1/2).

DEFINITION. The symbol \delta is used to denote the "generalized" function which has the property that

I(0,1, ) f(x) d(x,a) dx = f(a)


for some suitably large class of functions f, when 0 < a < 1. It is no surprise that some effort has been made to develop a theory of generalized functions in which the delta function can be found. Generalized functions are also called "distributions". While the delta function is not a well-defined function of the familiar type, it can be manipulated like a function in most cases, provided that in the end it will be integrated, and that the other quantities in the integral with it will be continuous.

A suggestive analogy is that of complex numbers. Like complex numbers, generalized functions are idealizations which do not describe real, physical things, but they have been found tremendously useful in applied mathematics. Most famously they were used by Dirac in his quantum mechanics; it is less well known that mathematical physicist Kirchhoff used them decades earlier in his work on optics. The theory of distributions is attractive and establishes a precise basis for the ideas which these notes will use. It is the choice of this course and these notes, however, to use the delta function without exploring the mathematical framework in which it should be studied.

We shall return to the delta function when its properties are needed to understand how to construct Green functions.


1. Calculate the following integrals:

(a)  i(-[[infinity]],[[infinity]], d(x) exp(-x<sup>2</sup>) dx);    
(b)  i(-[[infinity]],[[infinity]], d(x-1) exp(-x<sup>2</sup>) dx);    
(c)  i(-[[infinity]],[[infinity]], d(x-1) exp(-x<sup>2</sup>) H(x-t) dx), for
all t != 1;

2. Let <f,g> denote the standard inner product for 0 <= x <= 1.

Find a generalized function g(x) such that

<f,g> =  f(f(1,2)) + i(-f(1,3),f(1,3), x f(x) dx)

Onward to Section 2.2

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