Evans M. Harrell II and James V. Herod*

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

L(u) := x^{2} 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(x^{2}) =
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(d^{2},dx^{2}) x^{2} u(x) - 2 u(x) =**
x ^{2} 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) ** = (x ^{2}/t - t^{2}/x) / 3t^{2} **=

Why is this? for both t < x and t > x, G(x) has to be a linear
combination of x^{2} and 1/x. For x < t, the BC force the linear
combination 0 x^{2} + 0/x = 0. For x > t, we have a linear
combination a(t) x^{2} + b(t)/x, and since G(x,t) is continuous, we
need a(t) t^{2} + b(t)/t = 0, so b(t) = - a(t) t^{3}.
Finally, the jump condition means that the x derivative

G_{x}(x,t) = a(t) 2 x + (-a(t) t^{3}) (-1/x^{2})

must equal 1/t^{2} when we put x=t. This forces a(t) =
1/3t^{3}.

3. The Green function for L* is

G^{#}(x,t) ** = t ^{2}/ 3x^{3 }- 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) = x^{3} , u(1) = u'(1) = 0, is the integral of
G(x,t) t^{3} 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

**x ^{3}/4 - x^{2}/3 +
1/12x.**NAME____________________

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

L(u) := x^{2} 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 x^{2}G_{xx}(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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