Differential Operators

Integral Equations and the Method of Green Functions

James V. Herod*

*(c) Copyright 1993,1994,1995 by James V. Herod, herod@math.gatech.edu. All rights reserved.

Page maintained with additions by Evans M. Harrell, II, harrell@math.gatech.edu.



As suggested in the first two chapters the role of the adjoint of a linear function will be critical. If L is a linear function that is defined on some subspace of E, then the task of finding the adjoint L* will involve finding not only how the adjoint is defined, but also what is the subspace composing the domain of L*.

Consider the differential operator L given by

L(y) = ISU(p=0,n, ) ap(x) y<sup>(p)</sup>(x)  = an(x)y<sup>(n)</sup>(x) +
... +a1(x)y'(x) + a0(x)y(x).

One often defines the formal adjoint L* of L by

L*(y) =  ISU(p=0,n, ) (-1)<sup>p</sup>(ap(x) y(x))<sup>(p)</sup> =(-1)<sup>n</sup>(an(x) y(x))<sup>(n)</sup> +...+a0(x) y(x).


The second order operator L(y) = a2(x) y''(x) + a1(x) y'(x) + a0(x) y(x), according to this formula, will have formal adjoint

L*(y) = (a2(x) y(x))'' - (a1(x) y(x))' + a0(x) y(x).

If L is not defined on all of E, but just some subspace, or manifold M, then one must find where L* is defined. We denote the domain of L* by M*.

DEFINITION. Suppose that L is defined on a manifold, or subspace, M, and that L* is defined on a manifold M*. Then L* is THE adjoint of L if

I(0,1, ) [v L(u) - L*(v) u] = 0


for all u in M and v in M*.

EXAMPLE Let L(u) = u''(x) + 3x u'(x) + x2 u(x) be defined on the manifold consisting of all functions u on [0,1] which satisfy u(0) = u'(1) and u(1) = u'(0),

M = {u:u(0) = u'(1), u(1) = u'(0)}.

We indicated how to find L* and M*.

L*(v) = v'' - (3x v(x))' + x2v(x).

The manifold M* is chosen to make the equation

< L(u), v> = < u, L*(v) >


I(0,1, )[L(u)(x) v(x) - u(x) L*(v)(x) ] dx = I(0,1, )B([vu'' - uv''] + [3xu' v + (3x v)' u]) dx

= I(0,1, )(u'v - uv')' dx  + I(0,1, )[u(x) 3x v(x)]'  dx  = [u'(1) v(1) - u(1) v'(1)] - [u'(0) v(0) - u(0)
v'(0)] + u(1) 3 v(1)  - u(0) 3.0.v(0)

Since u is in M, u'(1) = u(0) and u(1) = u'(0), so that

< L(u), v > - < u, L*(v) > = [v(1) + v'(0)]u(0) - v'(1) + v(0) - 3v(1)] u'(0).

In order for this last line to be zero for all u in M, the manifold M* should consist of functions f such that

v(1) + v'(0) = 0 and v'(1) + v(0) = 3v(1).


These previous example has introduced the type problems that we shall consider. We will provide a method to get solutions for ordinary differential equations with boundary conditions. The problem, together with the boundary conditions, defines the operator L and a manifold M.


The problem y''+ 3y' + 2y = f , y(0) = y'(0) = 0 leads to an operator L and a manifold M given by

L(y) = y'' + 3y'+2y and M = {y: y(0) = y'(0) = 0 }.

The problem is to solve the following equation: given a continuous function f, find y in M such that L(y) = f. The technique is to construct a function G such that u is given by

u(x) = I(0,1, )  G(x,t) f(t) dt.

For such problems, we will have techniques to construct G. In this case

G(x,t) =    BLC{(A(0,e<sup>t-x</sup> - e<sup>2(t-x)</sup>))A( if 0 < x
< t < 1, if 0 < t < x < 1).

Most frequently, we will consider three types of boundary conditions illustrated below for a second order problem:

Initial conditions

BLC{(A( 0 = B1(u) = u(0), 0 = B2(u) = u'(0)))

unmixed, two point boundary conditions

BLC{(A(0 = B1(u) = au(0)+bu'(0),0 = B2(u) = cu(1)+du'(1)))

mixed, two point boundary conditions

BLC{(A(0 = B1(u) = ,0 = B2(u) =))A(a11u(0)+a12u'(0)+b11u(1)+b12u'(1),a21u(0)+a22u'(0+b21u(1)+b22u'(1))


(1) Compute the formal adjoint for each of the following:

(a) L(y) = x2 y'' + x y' + y (b) L(y) = y'' + 9 \pi2 y

(c) L(y) = (ex y'(x))' + 7 y(x) (d) L(y) = y'' + 3y' + 2y

(2) Argue that L is formally self adjoint if it has constant coefficients and derivatives of even order only.

(3) Suppose that L(y) = y'' + 3y' + 2y and y(0)= y'(0) = 0.

Find conditions on v which assure that

I(0,1, )[vL(y) - L*(v) y] = 0.

(4) Let L(u) = u'' + u. The formal adjoint of L is given by L*(v) = v''+ v. For each manifold M given below, find M* such that L* on M* is the adjoint of L on M.

(a) M = {u: u(0)=u(1)=0},

(b) M = {u: u(0)=u'(0)=0}

(c) M = {u: u(0)+3u'(0)=0, u(1)-5u'(1)=0},

(d) M = {u: u(0)=u(1), u'(0)=u'(1) }.


(5) Let L and M be as given below; find L* and M*.

(a) L(u)(x) = u''(x) + b(x) u'(x) + c(x) u(x),

M = {u: u(0)=u'(1), u(1) = u'(0) }.

(b) L(u)(x) = -(p(x) u'(x))' + q(x) u(x);

M = {u: u(0) = u(1), u'(0) =u'(1) }.

       (Hint) (c) L(u)(x) = u''(x);

M = {u: u(0) + u(1) = 0, u'(0) - u'(1) = 0 }


(6) Verify that for L, M, and u as given in the TYPICAL PROBLEM above, u is in M and L(u) = f. (Recall Exercise 3 in the section AN INTRODUCTION TO THE PROBLEMS OF GREEN'S FUNCTIONS of these notes.)

Suppose G is as in TYPICAL PROBLEM and

z(x) = I(0,1, )G*(x,t) h(t) dt.

Show z solves L* on M*.

Onward to Section 2.3

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