## Eigenvalues and multi-dimensional Eigenfunctions

Consider the problem $u_{xx} + u_{yy} + \lambda u = 0$ on the domain $[0,L] \times [0, H]$

with $u_y(x, 0) = 0$, $u_y(x, H) = 0, u(0,y) = 0$ and $u_x(L,y) = 0$.

(a) Derive the eigenvalues $\lambda$ and multi-dimensional eigenfunctions $\phi_{\lambda}(x,y)$. Assume that $\lambda \geq 0$ and that any eigenvalues of sub-problems are non-negative.

(b) For the case $L = H = \pi$ find simple, explicit expressions for the smallest three eigenvalues $\lambda$

Here is what I have for (a)

Put $u(x,y) = X(x) Y(y)$

$\Rightarrow \ X''Y + Y''X + \lambda XY = 0$

$\Rightarrow \ \frac{X''}{X} = - \frac{Y''}{Y} - \lambda = - \mu$

So we have

$X'' + \mu X = 0, X(0) = 0, X'(L) = 0$

$Y'' +(\lambda - \mu)Y = 0, Y'(0) = 0, Y'(H) = 0$

$\Rightarrow \ X = A \cos(\sqrt{\mu}x) + B \sin(\sqrt{\mu}x)$

$X(0) = 0 \ \Rightarrow \ A = 0 \ \Rightarrow X'(L) = \sqrt{\mu} B \cos(\sqrt{\mu}L) (n - \frac{1}{2})^2$

$\Rightarrow \ X'(L) = 0 \ \Rightarrow \ \sqrt{\mu} L = (n - \frac{1}{2}) \pi \Rightarrow \ \mu = \frac{(n - \frac{1}{2})^2 \pi^2}{L^2}$

Also

$Y = C \cos (\sqrt{\lambda - \mu}y) + D \sin(\sqrt{\lambda - \mu}y)$

$Y'(0) = \sqrt{\lambda - \mu} D \cos (\sqrt{\lambda - \mu}y) \ \Rightarrow \ D = 0$

$\Rightarrow \ Y'(H) = - \sqrt{\lambda - \mu} C \sin(\sqrt{\lambda - \mu}H)$

$\Rightarrow \ \sqrt{\lambda - \mu} H = m \pi$

$\Rightarrow \ \lambda - \mu = \frac{m^2 \pi^2}{H^2}$

So

$\lambda = \frac{m^2 \pi^2}{H^2} + \mu$

$= \frac{m^2 \pi^2}{H^2} + \frac{(n - \frac{1}{2})^2 \pi^2}{L ^2}$ are the eigenvalues

Eigenfunctions:

$\phi(x,y) = X(x) Y(y)$

$= \sin \big( \frac{(n - \frac{1}{2}) \pi x}{L}\big) \sin \big(\frac{m \pi y}{H} \big)$

Really not sure if that is correct though.

For (b) I'm a little lost. If I evaluate the eigenvectors at $L = H = \pi$ I get

$m^2 + (n - \frac{1}{2})^2$

I'm really confused about the 3 smallest eigenvalues, which makes me think my answer for (a) is incorrect.