Test 1

Integral Equations and the Method of Green's Functions

Evans M. Harrell II and James V. Herod*

*(c) Copyright 1996 by Evans M. Harrell II, harrell@math.gatech.edu. All rights reserved.


1. (This problem is taken almost directly from Herod's notes.)

Suppose that L[u] = uxx - uyy - uzz -

M = {u: u(0,y,z) = 0, u(1,y,z) = 0,
   u(x,y,0) = u(x,y,1),
   uz(x,y,0) = uz(x,y,1)
   uy(x,0,z) = 2 u(x,0,z)
   uy(x,1,z) = -1 u(x,1,z) }

Give L*. Find F such that v L[u} - L*[v] u = --.F. What is M*?


a) L* v = L v (the operator is formally self-adjoint)

b) F = {v ux - u vx, u vy - v uy, u vz - v uz}

b) M* = M (the operator is actually self-adjoint)


a) Find the Green function for Poisson's equation

grad2u = f

in the three-dimensional region z > 0, x > 0, with boundary conditions

uz(x,y,0) = 0, u(0,y,z) = 0.


With the method of images, the Neumann boundary condition requires an even reflection while the Dirichlet boundary condition requires an odd reflection. There are three image charges, because when both reflections are used, another image is created. G(P,Q) = (1/(4 pi)) (-1/((x-a)2 + (y-b)2 + (z-c)2)1/2 + 1/((x+a)2 + (y-b)2 + (z-c)2)1/2 - 1/((x-a)2 + (y-b)2 + (z+c)2)1/2 + 1/((x+a)2 + (y-b)2 + (z+c)2)1/2 )

Is the solution to Poisson's equation unique? No, not without further conditions, since there are many harmonic functions, and any of them could be added to the solution using this Green function, and Poisson's equation would be unaffected. The solution given by this Green function could be uniquely specified by any of several "finite energy conditions," and would require a related condition on f.

3. Background. The operator to consider in this problem is defined by

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

Impose boundary (initial) conditions that u(0) = u(1) = 0 for L.

Find the following:

a) L*(u) = u'' + u' - 2 u

with conditions on u: u(0) = u(1) = 0

b) The Green function for L is

G(x,t) = a(t) uleft(x<) uright(x>),

where a(t) = e2/(3 et(e3-1) ) and, as usual, x< = min(x,t), x> = max(x,t)

c) The Green function for L* is

G#(x,t) = (interchange x and t in part b)

d) The solution of L(u) = ex , u(0) = u(1) = 0, is:

u(x) = ex + (e sinh(1)/(e3 - 1) e2x + (e3 - e2 - 2)/(2 (e3 - 1))

(There are other, equally good ways to write the answers to problem 3.

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