Consider a standard 3D cylinder $C_{R,a} := B_R \times [0,a]$, where $B_R \subset \mathbb{R}^2$ is a disk of radius $R>0$, and $a>0$ is the height of the cylinder. It is not hard to see (we will discuss it below) that the spectrum of the Laplace operator on $C_{R,a}$ under zero Dirichlet conditions consists of eigenvalues

$\lambda_{m,n,k}(R,a) = \frac{j_{m,n}^2}{R^2} + \frac{k^2 \pi^2}{a^2}, \quad m \in \mathbb{N} \cup \{0\}, ~ n,k \in \mathbb{N},$

where $j_{m,n}$ is the $n$-th positive zero of the Bessel function $J_{m}$ of order $m$. That is, the first term corresponds to the spectrum of $B_R$, and the second term - to the spectrum of $[0,a]$.

Recalling the properties of the spectrum of $B_R$, we know that the multiplicity of any $\lambda_{m,n,k}(R,a)$ can be, at least, one or two. Let us give the following more detailed result whose proof follows directly from the expression of $\lambda_{m,n,k}(R,a)$.

Lemma. Let $R,a>0$. If there exist $m_i,n_i,k_i$, $i=1,2$, such that $\lambda_{m_1,n_1,k_1}(R,a) = \lambda_{m_2,n_2,k_2}(R,a)$, then

$a = \pi R \left(\frac{k_1^2-k_2^2}{j_{m_2,n_2}^2-j_{m_1,n_1}^2}\right)^{1/2}.$

In particular, if $R>0$ is fixed, then the set of $a$’s for which the multiplicity of some $\lambda_{m,n,k}(R,a)$ exceeds two is at most countable. The same holds true if we fix $a>0$ and vary $R>0$. Moreover, clearly, analogous assertions can be stated for higher-dimensional cylinders $B_R \times [0,a]$, where a ball $B_R \subset \mathbb{R}^N$.

Let us draw the dependence of several first eigenvalues $\lambda_{m,n,k}(1,a)$ with respect to $a \in (0,4)$.

Let us draw the dependence of several first eigenvalues $\lambda_{m,n,k}(R,1)$ with respect to $R \in (0,4)$. Predictably, the plot looks similarly.

Let us show that eigenvalues $\lambda_{m,n,k}(R,a)$ on $C_{R,a}$ are given by the formula above. Since $C_{R,a}$ has a separable geometry, let us look for eigenfunctions in the form $u(x,y,z) = v(x,y) w(z)$. Then, we get

$-\frac{1}{v}\Delta_{x,y} v - \frac{w''}{w} = \lambda.$

Because of the boundary conditions, we see that $w(z) = \sin\left(\frac{\pi k z}{a}\right)$ and $-\frac{w’’}{w} = \frac{\pi^2 k^2}{a^2}$. Thus,

$-\Delta_{x,y} v = \left(\lambda - \frac{\pi^2 k^2}{a^2}\right) v \quad \text{in } B_R.$

This is the classical eigenvalue problem in the disk $B_R$, and we very well know that

$\lambda - \frac{\pi^2 k^2}{a^2} = \frac{j_{m,n}^2}{R^2}, \quad m \in \mathbb{N} \cup \{0\}, ~ n \in \mathbb{N}.$

This is, the result follows.