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 6 THE SECOND ALTERNATIVE

We now discuss the problems where the Second Alternative holds. The supposition is that there is a nontrivial solution for L(y) = 0, B₁(y) = B₂(y) = 0. The Fredholm Theorems assure us that, if f is continuous, then there is a solution for L(y) = f, with B₁(y) = B₂(y) = 0 provided

for all solutions w of the equation L*(w) = 0, B₁*(w) = B₂*(w) = 0. As before, we will construct Green functions G such that, in case f satisfies the above requirement, then

provides a solution for L(y) = f.

In this second alternative, there may be many solutions for the equation L(y) = f. Consequently, we expect there may be many Green functions. In the technique developed below, G( ,t) is always in

   M= {y: B₀(y) = B₁(y) = 0 }.

This is not necessarily true for Green functions constructed by other methods: see for example the construction found by Don Jones while a graduate research assistant at GEORGIA TECH and given in an appendix.

We again divide the problems into three cases according to the nature of the boundary conditions. We shall illustrate methods of construction.

The first case to consider is where the boundary conditions arise as initial conditions. This case is not pertinent for the initial value problem has a unique solution. Thus, case one is always in the first alternative.

EXAMPLE:(Second Alternative,unmixed, two point boundary conditions)

Suppose that L(y) = y'' + y' - 2y, B₁(y) = y(0) - y'(0), and B₂(y) = y(1) - y'(1). It is the purpose of this example to show that there is no function G such that L(G(.,t))(x) = d(x,t). Note that L*(z) = z'' - z' -2z and M* = {y: 2z(1) = z'(1), 2z(0) = z'(0)}. Nontrivial functions in the nullspace of {L*, B₁*, B₂*} are multiples of e^2t. Hence, we are in the second alternative. The Fredholm Alternative theorem suggests that there will be no function G such that, if t is in (0,1), then the distribution equation L(G( ,t)) (x) = d(x,t) holds unless

Of course, the value of this integral is not zero.

For this situation, we must modify the construction of the Green function.

CONSTRUCTION OF G IN THE SECOND ALTERNATIVE, n^th ORDER

Step (1) Find the nullspace of { L*, M*}

Step (2) Find an orthonormal basis for this nullspace. Call this basis v₁,v₂,...v_m, m < n.

Step (3) Construct up such that L(u_p) = v_p, p = 1, 2, ...n.

Step (4) Construct G such that

THEOREM If 0 < t < 1, then there is G(.,t) such that

INDICATION OF PROOF. By the Fredholm Alternative Theorems, there will be such a function G provided

for all w in the nullspace of L*. This can be verified by writing w in terms of this orthonormal basis ,

and evaluating the dot product.

HOW TO CONSTRUCT G SUCH THAT

First, find linearly independent solutions y_p, p=1,...,n, of the homogeneous equation L(y) = 0. Then, find solutions u_p, p=1,..., m < n, for the equations L(u_p, p=1,...,)(x) = v_p, p=1,...,(x). It is not required that these solutions should satisfy any special boundary conditions.

The problem of finding G is now a problem of finding constants C_p and D_p such that

The constants C_p and D_p are determined by these 2n equations:

CONTINUATION OF THE PREVIOUS EXAMPLE

Recall that L(y) = y'' +y' -2y, B₁(y) = y(0) - y'(0), and B₂(y) = y(1) - y'(1). Linearly independent solutions for L(y) = 0 are e^-2x and e^x. A normalized basis for the one-dimensional nullspace of {L*,B₁*,B₂*} is \alpha e^2x where \alpha is the positive number given by

A solution u for the equation y'' +y' -2y = \alpha e^2x is u(x) = \alpha e^2x/4.

Now, G is given by:

The four constants - A,B,C, and D - can be solved by these four equations:

(1) 0 = B₁(G( ,t)) = G(0,t) - G_x(0,t) = 3A + [[alpha]]²e^t/4

(2) 0 = B₂(G( ,t)) = G(1,t) - G_x(1,t) = 3Ce^-2 + a²e^2(1+t)/4,

(3) 0 = G(t⁺,t) - G(t^-,t) = (C-A)e^-2t + (D-B)e^t, and

(4) 1 = G_x(t⁺,t) - G_x(t^-,t)= -2(C-A)e^-2t + (D-B)e^t.

Upon solving this system of four equations and four unknowns, an infinity of solutions will be found determined by these three equations:

A = - \alpha²e^2t/12, C = Ae⁴, D-B = e^-t/3.

                                                                                     QED

EXERCISE: FOR EACH OF THE FOLLOWING, GIVE L*,B₁*, B₂*, and G.

(a) L(y) = y'' +y' -2y, B₁(y) = y(0)-y'(0),B₂(y) = y(1) - y'(1).

(b) L(y) =4y'' - y, B₁(y) = y(0) - 2y'(0), B₂(y) = y(1) - 2y'(1).

(c) L(y) = y'' - 2y' - 3y, B₁(y) = 3y(0) - y'(0), B₂(y) = 3y(1) - y'(1).

EXAMPLE(Second Alternative, mixed, two point boundary conditions.)

Suppose that L(y) = y'', B₁(y) = y(0) + y(1), B₂(y) = y'(0) - y'(1). Then L*(z) = z'', B₁*(z) = z(0)- z(1), B₂*(z) = z'(0) + z'(1). All solutions of {L,B₁, B₂} are multiples of 2x - 1 . A nontrivial solution of [L*,B₁*, B₂*} is the constant function 1. Also, the function V(x) = 1 forms a basis for the null space of 0 = L*(z) in M*. The function u(x) = x²/2 satisfies L(u) = 1. Thus

G(x,t) = BLC{(A(A + Bx - x²/2 if x < t,C + Dx - x²/2 if t < x.))

We have four unknowns; we have the following four equations:

(1) 0 = G(0,t) +G(1,t) = A + C + D - 1/2

(2) 0 = dG(x,t)/dx|x=0 - dG(x,t)/dx|x=1 = B - (D-1)

(3) 0 = G(t⁺,t) - G(t^-,t) = C-A + (D-B)t

(4) 1 = dG(x,t)/dx|x=t+ - dG(x,t)/dx|x=t- = D-B.

As expected, there is an infinity of solutions to these equations which may be found by choosing D and then

B = D - 1

2C = -(t+D) + 1/2

A = C + t.

                                                                                     QED

EXERCISE: I Verify that each of these problems is second alternative and find L*,B₁*, B₂*,and G.

(1) L(y) = y'', B₁(y) = y(0) - y(1), B₂(y) = y'(0) - y'(1),

(2) L(y) = y'' + 9[[pi]]²y, B₁(y) = y(0) - y(1), B₂(y) = y'(0) + y'(1),

(3) L(y) = y'' + y' - 2y, B₁(y) = e y(0) - y(1), B₂(y) = e y'(0) - y'(1).

II. Construct L*, B* and G for each of the following L's and with periodic boundary conditions y(0) = y(1), y'(0) = y'(1):

(a) L(y) = y'',

(b) L(y) = y'' + \pi²y

(c) L(y) = 2y'' + y' - y,

NONHOMOGENEOUS BOUNDARY CONDITIONS

To solve the equations L(y) = f, B₁(y) = \alpha, B₂(y) = \beta, first construct G for the problem L(y) = f, B₁(y)= 0, B₂(y) = 0. Then construct functions z₁ and z₂ such that B₁(z₁) = 0, B₂(z₁) != 0, and B₁(z₂) != 0, B₂(z₂) = 0. The solution for the original problem is

                                                                                     QED

1. Find a formula for u if u'' = f and

(a) u(0) = u(1) = 0.

(b) u(0) = u'(0) = 0.

(c) u(0) = 3, u(1) = 5,

(d) u'(0) = 3u(0), u'(1) = 5 u(1),

(e) u(0) = u(1), u'(0) = u'(1),

(f) u(0) = 3, u'(0) = 5,

(g) u'(0) = 3, u(0) = 5,

(h) u(0) = 0, I(0,1, ) u(x) dx = 0.

(answers)

2. Find a formula for u if u'' + 9u = f and

(a) u(0) = 3, u'(1) = 5.

(b) u(0) - u'(0) = 3, u(1) = 5.

3. Find a formula for u if (x u'(x))' = f and u(1) = 0, u(2) = 5.

4. (a) Find conditions on f in order that u'' + 4 \pi² u = f, u(0)=u(1), u'(0) = u'(1) should have a solution.

(b) Give the Green function for this problem.

(c) By finding the Green function for the problem L(y) = y'', y(0) = y(1), y'(0) = y'(1), re-write this equation as an integral equation such as was studied in the previous chapter.

5. Here is a linear differential operator with boundary conditions:

L(y)(x) = (e^x y' )' and B₁(y) = y(0), B₂(y) = y'(0).

(a) Show that (e^x y')' z - y (e^xz')' = [ e^x (z y' - z' y)]'.

(b) Give L* and B*.

(c) Give the Green function for the problem L(y) = f with B₁(y) = B₂(y) = 0.

(d) Rewrite the problem (e^x y')' + sin(x) y(x) = f(x) , y(0) = y'(0) = 0 as an integral equation in the form

y = K(y) + F.

Be sure to identify K and F carefully.

6. Consider the differential equation: f is continuous on [0, \pi ] and

(sin(x) y'(x))' + 2 sin(x) y(x) = f(x)

y(0) = 0 = y( \pi ).

(a) In the context of this course, what is the appropriate space and linear operator L?

(b) What is the adjoint of L in this space? Explain your answer.

(c) Is this problem 1^st or 2^nd alternative?

(d) If possible, solve this problem with f(x) = x. If it is not possible, explain why not.

Onward to Section 3.1

Back to Section 2.5

Return to Table of Contents (Green Track)

Return to Evans Harrell's

• 4348 page
• PDE page
• home page
• home page mirror