Evans M. Harrell II and James V. Herod*
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 - 1 + 4 x
1/4 x - 1/12 ----------
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.
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):_________
Back to Compendium of Problems
Return to Table of Contents (Green Track)
Return to Evans Harrell's