Integral Equations and the Method of Green's Functions

James V. Herod*

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

Page maintained by Evans M. Harrell, II,


The following three problems illustrate, in a simple way, the primary concerns of this course. The first is a problem about matrices and vectors, and it will be our guide to solving integral equations and differential equations.

SAMPLE PROBLEM 1: Let B = B(ACO2(  3, -2, -1,  1) ).

Suppose v is a vector in R2. If u is a vector then

(a)   B(ACO2( 1,  2, 1,  3))   u = v

if and only if

(b) v is a vector and u = Bv.

The equivalence of these two is easy to establish. Even more, given only statement (a), you should be able to construct B such that statement (b) is equivalent to statement (a).

SAMPLE PROBLEM 2: Let K(x,t) = 1 + x t. The function u is a solution for

(a)  		u(x) =   I(0,1, K(x,t)u(t) dt) + x^2   for x in [0,1]

if and only if

(b)  		u(x) = x^2 - (25+12x)/18  for  x in [0,1].

If one supposes u is as in (b), then the integral calculus should show that u satisfies (a). On the other hand, the task of deriving a formula for u from the relationship in (a) involves techniques which we will discuss in this course.


K(x,t) = BLC{(A( x(1-t) if x  t ,, t(1-x) if t  x.))

Suppose f is continuous on [0,1]. The function g is a solution for

(a) g''= -f and g(0) = g(1) = 0

if and only if

(b)  g(x) = I(0,1, K(x,t)f(t)dt).


(a)=>(b) Suppose that f is continuous on [0,1] and g'' = -f with g(0) = g(1) = 0. Suppose also that K is as given by sample problem (3). Then

 I(0,1,K(x,t)f(t)dt) =  -I(0,1,K(x,t)g''(t)dt) =  -(1-x) I(0,x, t g''(t)dt) - x I(x,1, (1-t) g''(t)dt)

Using integration by parts this last line can be rewritten as

-(1-x) ([x g'(x)- 0 g'(0)] - I(0,x,g'(t)dt))- x ([(1-1)g'(1) - (1-x)g'(x)] + I(x,1,g'(t)dt))

= -(1-x)[x g'(x) -(g(x)-g(0)}

-x[-(1-x) g'(x) + (g(1) - g(x))]

= (1-x) g(x) + x g(x) = g(x).

To get the last line we used the assumption that g(1) = g(0) = 0.

(b)=>(a) Again, suppose that f is continuous and, now, suppose that

g(0) = g(1) = 1, etc.

As you can see, it is not hard to show that these two statements are equivalent. Before the course is over then, given statement (a), you should be able to construct K such that statement (b) is equivalent to statement (a). Perhaps you can do this already.

SAMPLE PROBLEM 4: Let u(x,y) = e-y sin(x) for y >= 0 and all x. Then

u_xx + u_yy = 0><p>
				u(x,0) = sin(x).<p>
	This result can be verified by simple calculus.<p>
	It gives insight into the unifying ideas of this course to realize each of
these sample problems as being concerned with an equation of the form Lu=v.  It
is a worthwhile exercise to reformulate each of the problems in this form.<p>

Find B such that, if v is in R2, then these are equivalent:

(a) u is a vector and Au = v.

(b) v is a vector and u = Bv.

2. Let K be as in SAMPLE PROBLEM 2. Show that if u(x) =

3x2 - (25 + 12x )/6 then u solves the equation

u(x) = I(0,1, )K(x,t)f(t)dt  + 3x2

3. Let

K(x,t) = BLC{(A(    0        if 0 < x < t < 1, e^{t-x} - e^{2(t-x)} if 0 < t < x < 1.))

Suppose that f is continuous on [0,1]. Show these are equivalent:

a) y'' + 3y' + 2y = f, y(0) = y'(0) = 0   and  (b)  y(x) = I(0,1, K(x,t)f(t)dt)

4. Let u(r,\theta) = r sin(\theta). Show that

F(1,r) F(partial (rF(partialu,partial r)),r)  +  F(1,r^2)  F(partial^2u,partial theta^2) = 0with u(1,[[theta]]) = sin([[theta]]).

Onward to Chapter I

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