First eigenvalue of laplacian on a cylinder I am looking for a reference for the value of the first eigenvalue of the laplacian on the right cylinder
$$C(r,h) = \{ (r \cos \theta, r \sin \theta, z) : \theta \in [0, 2\pi], z \in [0,h] \}$$ 
with Neumann condition. How does it depend on $r$ and $h$?
 A: I misread with my first answer. The first eigenvalue is still 0, but to get the others, this requires some differential geometry to derive the equations, then they may be solvable with separation of variables. The Laplacian operator on the surface is derived from the metric tensor, which (I believe) is $$g = \begin{pmatrix}
r^2 & 0 \\
0 & 1
\end{pmatrix},$$
where the coordinates on the cylinder are $(\theta,z)$.
The Laplace operator is then given by
$$\Delta = \frac{1}{r^2}\partial_\theta^2+\partial_z^2$$
with periodic conditions in $\theta$ and Neumann conditions on $z$. 
Taking $\Delta u=\lambda u$ and separating variables, we get $$Z''=-\mu^2 Z, \ \ \ \ \frac{1}{r^2} \Theta''=-(\lambda-\mu^2)\Theta,$$
where $\lambda$ is the eigenvalue of the operator and $-\mu^2$ is the separation constant.
Solving the first equation with Neumann conditions, you will find that we need $\mu^2=\pi(n+1/2), \ n=0,1,2,\dots$
For the second equation with periodic conditions, you will see that we need $\lambda-\mu^2=\frac{m\pi}{r^2},\ m=1,2,\dots$
Combining, you get $\lambda_{mn} = \frac{m\pi}{r^2}+\mu^2=\pi\left(\frac{m}{r^2}+(n+1/2)\right)$
