Question

Problem 2. (15 points) Solve the following Laplace's equation in a cube as outlined below. au au au 2,2 + a2 + a2 = 0, on 0<x<1, 0<y<1, 0<?<1, (0, y, z) = (1, y, z) = 0, (x, 0, 2) = u(x, 1, ) = 0, (x, y,0) = 0, u(x,y, 1) = x. (a) Seek a solution of the form u(x, y, z) = F(x) G(v) H(-). Show that with the appropriate choice of separation constants, you can obtain the following ODES: F"(x) = -AFX). G"(y) = -1G(y), H") (+) H ) (b) Find the ODE boundary conditions for F(x), G(y) and H() that are consistent with the homogeneous (i.e., the zero) boundary conditions on u(x, y, *). Hint: F and G will have two boundary conditions, H(2) will only have one. The non-homogeneous boundary condition will be used later. (c) Solve the eigenvalue problems for F(x) and Gy). Clearly state the eigenvalues and in terms of indices n and m, as well as the associated eigenvectors. Make sure to consider possible solutions for > 0, 1 = 0 and x > 0, = 0 (you do not need to consider <0 or <0).

Answer 1

$a) \\ Let \, u(x,y,z)=F(x)G(y)H(z) \\ \\ Substitute \, this\, in\, Laplace \,equation \,to\, get \\ \\ F''(x)G(y)H(z) + F(x)G''(y)H(z) +F(x)G(y)H''(z) =0 \\ \\ Divided \, this \, by \, F(x)G(y)H(z)\, to\, get \\ \\ \frac{ F''(x) }{F(x)} + \frac{ G''(y) }{G(y)} + \frac{ H''(z) }{H(z)} = 0 \\ \\$In above equation each fraction has to be constant. This is due to the fact that the first fraction is independent of y,z and the second fraction is independent of x,z and the third fraction is independent of x,y.

$Let \\ \\ \frac{ F''(x) }{F(x)} = -\lambda \\ \frac{ G''(y) }{G(y)} = -\mu \\ \\ Then \\ \\ \frac{ H''(z) }{H(z)} = \lambda +\mu\\ \\$

Here the contents $\lambda , \mu$

are assumed to be nonnegative. Now the above equations will take the following form.

$F''(x)= -\lambda F(x) \\ \\ G''(y) = -\mu G(y) \\ \\ H''(z) = (\lambda +\mu ) H(z) \\ \\$

b) ODE boundary conditions

We have u(x,y,z) = F(x)G(y)H(z)

Substitute this in the boundary conditions to obtain the ODE boundary conditions as follows.

$\frac{\partial }{\partial x}u(0,y,z) =\frac{\partial }{\partial x}u(1,y,z) =0 \\ \\ F'(0)G(y)H(z) = F'(1)G(y)H(z) =0$this has to be true for every y,z in the given range. Hence,

Similarly

$\frac{\partial }{\partial y}u(x,0,z) =u(x,1,z) =0 \\ \\ F(x)G'(0)H(z) = F(x)G(1)H(z) =0$

This has to be true for every x,z hence we get

The boundary conditions for H(z) follows from the last relationship.

ur, y,0) = 0 F(x)G(y) H'(0) = 0 or HO) = 0 ull, y, 1) = 1 F(x)G(y)H(1) = x H(1) = F(c) Gy)

c) Solution for F, and G

$Let \, \lambda = 0\, then F''(x) = 0 \\ F(x) = a+ bx\\ Applying \,F'(1)=0 \, give \, b=0\, hence \\ F(x)= a$

$Let \, \lambda = k^2\, then F''(x) = -k^2 F(x)\\ F(x) = a_k\cos(kx)+ b_k\sin(kx)\\ Applying \,F'(0)=0 \, give \, b_k=0\, hence \\ F(x)= a_k\cos(kx) \\ \\ Applying \,F'(1)=0 \, give \, -a_k\sin(k)=0\, hence \\ k = m\pi \\ eigenvalues \, are \\ \lambda = k^2 = m^2\pi^2$

Similarly we can solve for G also.

$Let \, \mu = k^2\, then G''(y) = -k^2 G(y)\\ G(y) = a_k\cos(ky)+ b_k\sin(ky)\\ \,G'(0)=0 \, give \, b_k=0\, hence \\ G(y)= a_k\cos(ky) \\ \\ Applying \,G(1)=0 \, give \, a_k\cos(k)=0\, hence \\ k = \frac{n\pi}{2}\\ eigenvalues \, are \\ \mu = k^2 = \frac{n^2\pi ^2}{4}$