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.