Test 1

## Integral Equations and the Method of Green's Functions Evans M. Harrell II and James V. Herod*

SAMPLE TEST SOLUTIONS

Background. The operator to consider in this test is defined by

L(u) := x2 u''(x) - 2 u(x),

for 1 < x < 3. (It begins at 1 because the equation becomes singular when x=0.) Perhaps it is helpful to notice that L(x2) = L(1/x) = 0. You do not have to show this.

Impose boundary (initial) conditions that u(1) = u'(1) = 0 for L on this page.

Find the following:

1. L*(u) = f(d2,dx2) x2 u(x) - 2 u(x) = x2 u''(x) + 4 x u'(x)

with conditions on u: u(3) = u'(3) = 0

2. The Green function for L is

G(x,t) = 0        for x < t

G(x,t) = (x2/t - t2/x) / 3t2 = x2/ 3t3 - 1/3x        for x > t

Why is this? for both t < x and t > x, G(x) has to be a linear combination of x2 and 1/x. For x < t, the BC force the linear combination 0 x2 + 0/x = 0. For x > t, we have a linear combination a(t) x2 + b(t)/x, and since G(x,t) is continuous, we need a(t) t2 + b(t)/t = 0, so b(t) = - a(t) t3. Finally, the jump condition means that the x derivative

Gx(x,t) = a(t) 2 x + (-a(t) t3) (-1/x2)

must equal 1/t2 when we put x=t. This forces a(t) = 1/3t3.

3. The Green function for L* is

G#(x,t) = t2/ 3x3 - 1/3t for x < t

G#(x,t) = 0 for x > t

Among the many correct ways to get this answer is the method from the chapters on integral operators; just interchange x and t. The inverse of the adjoint is the adjoint of the inverse.

4. The solution of L(u) = x3 , u(1) = u'(1) = 0, is the integral of G(x,t) t3 dt for 1 < t < 3, which is

> int(t^3 * ( x^2/(3 * t^3) - 1/(3*x) ), t = 1..x);

3

3 - 1 + 4 x

1/4 x - 1/12 ----------

x

Perhaps you would prefer to write this as

x3/4 - x2/3 + 1/12x.NAME____________________

In problem 5, we consider different boundary conditions, but still

L(u) := x2 u''(x) - 2 u(x).

The boundary conditions are of the strange form:

9 u(3) = u(1)

9 u'(3) - u'(1) = 12 u(3)

You may take as given that L*(1) = 0. You do not have to show this, and you do not have to get involved in the strange boundary conditions.

5.

a) Give an example of a function f(x) for which

L(u) = f(x)

has a solution.

ANSWER: f(x) = sin( \pi x) or any other function with average 0 (to be orthogonal to 1).

b) What specific differential equation must be satisfied by a function G(x,t) such that a solution to part a) is given by

u(x) = the integral from 1 to 3 of G(x,t) f(t) dt ?

ANSWER x2Gxx(x,t) - 2 G(x,t) = \delta(x-t) - 1/2

(because the normalized version of the function v(x) = 1 for 1 < x < 3 is 1/Sqrt(2)).

It is not necessary to solve for G or for u. (But I will be suitably impressed if you do!)

FORMULA OR KEY FACT (FOR POSSIBLE PARTIAL CREDIT):_________

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