A Method for Constructing Green Functions

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.



There are, of course, several methods of constructing Green functions. The one we will present first, and emphasize, is the one students seem to prefer. Perhaps this is because it is easy to remember and has an inherent simplicity. Other methods will be included in these notes for comparison. There are ideas which the other methods use that are important.

As before, we assume a certain form for the differentiable operator L:

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).

We suppose that an(x) is not zero on [0,1] and that each term of the sequence ap(x), p=0,..., n, has at least n continuous derivatives. We discuss the construction of the Green function in three cases depending on the nature of the boundary conditions. Until further notice, we assume the first alternative holds and will repeat this warning for emphasis. We continue to denote M and M* as the manifolds associated with {L,B} and {L*,B*}, respectively.

In most of our examples, and in the majority of applications, the differential equations are of second order. Ultimately, this arises from Newton's force law, F = m a, which is second order, since acceleration is a second derivative.

Let's begin by describing the algorithm for constructing G for second-order problems. We'll discuss while this works below

The function G depends on two variables and has the following properties: if t is in (0,1) , then

G(x,t), F([[partialdiff]](G(x,t)),[[partialdiff]]x)<sup>, and
 </sup> exist for 0 < x < t and for t < x < 1. Further suppose that these derivatives have a continuous extension to the triangular region 0 <= x <= t and t <= x <= 1. The effect of this extension is that




for p = 1,2

At the boundary we shall insist that G(x,t) be continuous. For the partial Gx, however, we require a special jump discontinuity as follows:

Pick t in [0,1].
(a)  L(G(.,t))(x) = 0 for 0 < x < t and for t < x < 1,
(b)  G(.,t) is in M, 
(c)  G(x,t) is a continuous function.
(d)     [[partialdiff]](G(t<sup>+</sup>,t))/[[partialdiff]]x  -
[[partialdiff]](G(t<sup>-</sup>,t))/[[partialdiff]]x<sup>  </sup>= 1/a2(t).

Here is what happens if there are derivatives of higher orders:

The function G should be constructed on [0,1]x[0,1] to have the following properties: if t is in (0,1) and 0 < p < n then




At this point, all we have asked of G is that it should have n continuous partials on the closed triangles 0 < x < t and t < x < 1. The requirement along the boundary will be that for p <= n-2, we have continuity. For example at p = 0, the effect is that G(t+,t ) = G(t-,t). Indeed, G(t+,t) = G(t,t-) = G(t,t+) = G(t-,t). And, this happens for the pth partials up to p = n-2. For the (n-1)th partial, we allow a jump discontinuity as prescribed in the summary below:

Pick t in [0,1].
(a)  L(G(.,t))(x) = 0 for 0 < x < t and for t < x < 1,
(b)  G(.,t) is in M, 
(c)  for 0 <= p <= n-2,
[[partialdiff]]<sup>p</sup>(G(t<sup>+</sup>,t))/[[partialdiff]]x<sup>p</sup>  =

-   [[partialdiff]]<sup>n-1</sup>(G(t<sup>-</sup>,t))/[[partialdiff]]x<sup>n-1
</sup>= 1/an(t).

Before showing that the above recipe really does provide solutions to the nth order equation, it would be well do to some examples, beginning with the TYPICAL PROBLEM:

EXAMPLE (First alternative, initial conditions): Here is the problem: given f continuous on [0,1], construct y such that

y'' + 3y' + 2y = f with y(0) = y'(0) = 0.

Let's identify the important parts here.

L(y) = y'' + 3y' + 2y, B1(y) = y(0) , B2(y) = y'(0),

and M = {y: y(0) = y'(0) = 0}.

We are in the first alternative because for this {L,B1,B2} the system L(y) = 0, B1(y) = B2(y) = 0 has only one solution and it is zero. We construct G step-by-step from the above directions.

To follow the directions of step (a) we need the general solution of the homogeneous equation L(y) = 0, that is, we need the general solution of the homogeneous equation

y'' + 3y' +2y = 0.

It's not so hard to see that linearly independent solutions for this equation are e-2x and e-x. (In this age, if you prefer, you can find these solutions with Maple, using the command dsolve or Mathematica, using DSolve.) Thus G satisfies step (a) if

G(x,t)  =  BLC{(A( Ae<sup>-2x</sup> + Be<sup>-x</sup>  for x < t,
Ce<sup>-2x</sup>  + De<sup>-x </sup> for t < x))

Note that A, B, C, and D are constant in x, but may change with t. We chall determine the four unknowns from the continuity and jump conditions.

To follow the directions of step (b) which requires that G(.,t) be in M, we need

G(0,t) = 0 and Gx(x,t)) | x=0 = 0

The implications of this are that

A + B = 0 and -2A - B = 0.

This implies that A = 0 and B = 0.

To follow the directions of step (c) which requires that G(t+,t) = G(t-,t), we need

(C - A)e-2t +(D - B)e-t = 0.

Or, knowing that A = B = 0,

Ce-2t + De-t = 0.

To follow the directions of step (d) which requires that

    Gx| x=t+ - Gx(x,t))|x=t- = 1,

we need

-2(C - A)e-2t - (D - B)e-t = 1.

Knowing that A = B = 0,

-2Ce-2t-De-t = 1.

This gives two equations, two unknowns in C and D. The solution is

C = -e2t and D = et.

Try to get an over view of this example: after finding two linearly independent solutions of the second order equation L(y) = 0, we would know G provided we solved for A, B, C, and D. Steps b, c, and d gave four equations in these four unknowns. Written in matrix form,

B(ACO4( 1, 1, 0, 0, -2, -1, 0, 0, 0, 0, e<sup>-2t</sup>, e<sup>-t</sup>, 0,
0, -2e<sup>-2t</sup>, -e<sup>-t</sup>))B(A(A,B,C,D)) =  B(A(0,0,0,1)).

The problem has been reduced to a matrix equation! We have solved the equations to find A = 0, B = 0, c = -e2t, and d = et. The end result is:

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

We are confident that if f is continuous then the equation y = Gf provides a solution for L(y) = f because of Exercise 3 of the Introduction.


Suppose that L(y)(x) = a2(x)y''(x) + a1(x)y'(x) + a0(x) y(x). Let

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

Since G( , t) is in M then u is in M. It remains to see that L(u) = f. Note that

u(x) =  I(0,x, ) G(x,t) f(t) dt +  I(x,1, ) G(x,t) f(t) dt, u'(x) = ...

= I(0,x, ) [[partialdiff]](G(x,t))/[[partialdiff]]x f(t) dt
+<sup>   </sup>I(x,1, ) [[partialdiff]](G(x,t))/[[partialdiff]]x f(t) dt.

This last equality holds because of the assumption that G(x,x-) = G(x,x+).


 u''(x)  = .. =   f(x)/a2(x) + I(0,x, )
[[partialdiff]]<sup>2</sup>(G(x,t))/[[partialdiff]]x<sup>2</sup> f(t) dt +
I(x,1, ) [[partialdiff]]<sup>2</sup>(G(x,t))/[[partialdiff]]x<sup>2</sup> f(t)

This last equality uses the condition Gx(x,x-)) - Gx(x,x+)) = 1/a2(x). Finally, we use the fact that G(x, t), as a function of x, satisfies L(y) = 0 on [0,x] and on [x,1] to get that L(u) = f.

EXAMPLE( first alternative; unmixed, two point boundary conditions):

We will construct the Green function for the problem

y'' + 3y' + 2y = f with y(0) = 0 and y(1) = 0.

Here are the important parts:

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

and M = {y: y(0) = y(1) = 0}.

A little bit of work needs to be done to verify that we are in the first alternative. If L(y) = 0, then there are numbers a and b such that

y(x) = a e-2x + b e-x.

To require that B1(y) = 0 and B2(y) = 0 requires that

0 = a + b

and 0 = a e-2 + b e-1.

The only solution to this pair of equations is a = 0 = b, which verifies that we are in the first alternative.

The construction for G is as before:

G(x,t)  =  BLC{(A( Ae<sup>-2x</sup> + Be<sup>-x</sup>  for x < t,
Ce<sup>-2x</sup>  + De<sup>-x </sup> for t < x))

The two boundary conditions and the continuity conditions lead to the equations 0 = A + B

0 = C e-2 + D e-1

0 = (C-A) e-2t + (D-B) e-t

1 = -2(C-A) e-2t - (D-B) e-t.

Certainly, these equations can be solved, although the details are tedious. Here is a better idea. Instead of choosing e-2t and e-t as linearly independent solutions of the equation L(y) = 0, choose another pair having these properties:

   u1(0) = 0 and u1(1) >< 0

   u2(0) >< 0 and u2(1) = 0.

(For this example, u1(t) = e-2t - e-t and u2(t) = e-2(t-1) - e-(t-1).)

Now, make up G this way,

G(x,t) = BLC{(A(Au1(x) + Bu2(x) for  x < t,Cu1(x) + Du2(x)<sup>   </sup>for  x > t))

Apply the boundary conditions:

0 = Bu2(0), which implies that B = 0,

and 0 = Cu1(1), which implies that C = 0.

The continuity conditions give the two equations

0 = G(t<sup>+</sup>,t) - G(t<sup>-</sup>,t) = Du1(t) - Au0(t); and 1/a2(t) = [[partialdiff]](G(t<sup>+</sup>,t))/[[partialdiff]]x -
[[partialdiff]](G(t<sup>-</sup>,t))/[[partialdiff]]x = D u'1(t) - A u'0(t).

From these equations we get

A = u2(t) /a2(t)w(t)

and D = u1(t) / a2(t)w(t)


w(t) = det  B(ACO2( u1(t), u2(t), u1'(t), u2'(t)))

is called the Wronskian of u0 and u1.

Here is the final result:

G(x,t) =  BLC{(A(u2(t) u1(x)/a2(t)w(t) for  x < t,u1(t) u2(x)/a2(t)w(t)  for  t < x.)).

There is one more important piece of information that you will learn, or be reminded of, if we work out both parts of the formula for G. Recall that

u1(t) = e-2t - e-t and u2(t) = e-2(t-1) - e-(t-1).

And now, to compute w(t). The chore of that computation seems too tedious to be fun. Not to worry! Look up "Wronskian" is some good sophomore differential equations book and you will find a convenient formula :

W(t) = W(0)  exp( I(0,t, ) -[a_{n-1}(s)/a_n(s)] ds ).


Notice that this is particularly simple if an-1 = 0, as often happens: The Wronskian is then a constant.

Now the computation is easy:

W(0) =   detB(ACO2( 0,  e<sup>2</sup> - e<sup>1, </sup>-1,  -2e<sup>2 </sup>+e<sup>1</sup>)) = e<sup>2</sup> -e<sup>1</sup>.

and        w(t) = (e2 - e1)e-3t.


F(u1(t) u2(x),w(t))   = (e<sup>2+t</sup> - e<sup>1+2t</sup>) (e<sup>-2x</sup> - e<sup>-x</sup>) / (e<sup>2</sup> - e<sup>1</sup>).... =  e<sup>2(t-x) </sup>(e<sup>1-t</sup> - 1)(1 - e<sup>x</sup>)/(e-1)


F(u1(t) u2(x),w(t))  = (e<sup>t</sup> - e<sup>2t</sup>) (e<sup>2-2x</sup> - e<sup>1-x</sup>) /(e<sup>2</sup> - e)... =  e<sup>t-x</sup> (1-e<sup>t</sup>) (e<sup>1-x</sup> - 1)/ (e-1).

We got more from this example than the answer: we also got the following quick method that works for this type problem.


Pick u1 and u2 such that B1(u1) = 0, B2(u1) >< 0, B2(u1) = 0, and B1(u2) >< 0.


G(x,t) =  BLC{(A(u2(t) u1(x)/a2(t)w(t), for  x < t,u1(t) u2(x)/a2(t)w(t), for  t < x.)).

where w is the Wronskian of u1 and u2.

EXAMPLE (first alternative; mixed, two point boundary conditions):


L(y) = y'', B1(y) = y(0) + y(1), and B2(y) = y'(0) + y'(1).

First, we verify that we have the first alternative B2 supposing that

L(y) = 0 and B1(y) = B2(y) = 0.

Then y(x) = a + bx , for constants a and b. Since

y(0) + y(1) = 0

then 2a + b = 0.


y'(0) + y'(1) = 0,

then 2a = 0.

These two equations imply that a = b = 0. We now begin the construction of the Green function.

Pick 0 < t < 1:

G(x,t) =  BLC{(A(A + Bx,  for x < t,C + Dx  for t < x))

We have four constants to determine; here are four equations:

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

0 = Gx(0,t)) + Gx(1,t)) = B + D,

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

1/a2(t) = Gx(t+,t) - Gx(t-,t)) = D - B.

The solution for these four equations is A= (2t - 1)/4, B = -1/2,

C = -(2t + 1)/4, and D = 1/2.


(Review the general method or ad hoc method for constructing Green functions.) 1. Find a Green function such that if f is continuous, then the equation y = Gf provides a solution for L(y) = f, y(0) = y'(0) = 0, where L is as defined below. In each case, first give L* and M* and verify that the first alternative holds.

(a) L(y) = y'' (b) L(y) = y'' + 4 \pi 2y (c) L(y) = 2y'' + y' - y (d) L(y)(x) = (exy'(x))'.        (Answers)

2. (Ad hoc method) Suppose that u is a function on [0,1] which

satisfies L(u) = 0, u(0) = 0, and u'(0) = 1/a2(t). Let H(x,t) = 0 if x < t and = u(x-t) if t < x. Show that H is a Green function for the problem {L, B1(y) = y(0), B2(y) = y'(0) }.

3.. (Equivalent integral equation) Let G(x,t) be the Green function for problem I(a) above. Suppose that b and f are continuous functions. Let h be the function given by

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

and H(x,t) be the function given by H(x,t) = G(x,t) b(t) (Note: Here H is not the Heaviside function). Show these are equivalent:

(a) y''(x) - b(x)y(x) = f(x), y(0) = y'(0) = 0, and

(b) y(x) = I(0,1, ) H(x,t) y(t) dt + h(x).

4. Construct L*, B*, and G for the following :

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

(b) L(y) = y'', B1(y) = y(0) + y'(0), B2(y) = y(1) + y'(1).

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

(d) L(y) = y'' + 4 \pi 2y, B1(y) = y(0) + y'(0), B2(y) = y(1) - y'(1).

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

(f) L(y)(x) = (ex y'(x))', B1(y) = y(0) + y'(0), B2(y) = y(1) - y'(1).


5. So that you will remember why we are constructing Green functions, use the above result to provide a solution for the equation y''(x) = x2, y(0) + y(1) = 0, and y'(0) + y'(1) = 0.

Onward to Section 2.5

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