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.

SECTION 2. DIFFERENTIAL OPERATORS AND THEIR ADJOINTS

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

One often defines the formal adjoint L* of L by

                                                                                                                 (2.3)

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

                                                                                                                 (2.4)

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) >

satisfied.

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).

                                                                                   QED

TYPICAL PROBLEM:

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

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

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

Initial conditions

unmixed, two point boundary conditions

mixed, two point boundary conditions

EXERCISES 2.2:

(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) = (e^x 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

(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) }.

       (Answers)

(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 }

       (Answers)

(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

Show z solves L* on M*.

Onward to Section 2.3

Back to Section 2.1

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