A week ago I became really surprised to realize that there are planar rings (annuli) for which there is an eigenvalue of the Dirichlet Laplacian with the multiplicity at least 3. I was quite sure that the situation for rings is the same as for the disk, where the multiplicity of any eigenvalue is either 1 or 2. The latter result is a consequence of Bourget’s hypothesis which states that no Bessel functions $J_\nu$ and $J_{\nu+m}$ with natural $\nu, m$ have common positive zeros. Moreover, in the well-known survey article of Grebenkov and Nguen ¹, Section 3.2 (in between (3.7) and (3.8)), it is written that eigenvalues (in rings) are either simple or twice degenerate. This assertion appears to be not true.

So, let us consider the ring $A_r := \{x \in \mathbb{R}^2: r<|x|<1\}$ and the eigenvalue problem

\[\left\{ \begin{aligned} -\Delta u &= \lambda u &&{\rm in}\ A_r, \\ u&=0 &&{\rm on }\ \partial A_r, \end{aligned} \right.\]

It is known that there is a basis of eigenfunctions of the form (up to rotation)

\[\psi_{\nu,k}(\varrho,\vartheta) = \left(J_\nu(a_{\nu,k}(r) \varrho) Y_\nu(a_{\nu,k}(r)) - J_\nu(a_{\nu,k}(r)) Y_\nu(a_{\nu,k}(r) \varrho) \right)\cos(\nu \vartheta),\]

where $\nu \in \mathbb{N} \cup \{0\}$ and $k \in \mathbb{N}$. Moreover, $\lambda_{\nu,k} = a_{\nu,k}^2(r)$ is the eigenvalue associated with $\psi_{\nu,k}$. Here $a_{\nu,k}(r)$ is the $k$-th positive zero of the following cross-product of Bessel functions of the first and second kind:

\[z \mapsto J_\nu(r z) Y_\nu(z) - J_\nu(z) Y_\nu(r z).\]

It is easy to see that any eigenvalue $\lambda_{0,k}$ has the multiplicity at least 1 (the associated eigefunction is radial) and any other eigenvalue has the multiplicity at least 2 (two associated eigenfunctions are identical up to rotation).

It is known that $a_{\nu,k}(r)$ is continuous with respect to $r \in (0,1)$, see Cochran ². Then, performing some numerical computations, we see that $a_{0,2}(0.01) \approx 6.0109$ and $a_{3,1}(0.01) \approx 6.3801$, while $a_{0,2}(0.1) \approx 6.8575$ and $a_{3,1}(0.1) \approx 6.3804$. By continuity, there exists $r_0 \in (0.01,0.1)$ such that $a_{3,1}(r_0) = a_{0,2}(r_0)$, which implies that the eigenvalue $\lambda = \lambda_{3,1} = \lambda_{0,2}$ has the multiplicity at least 3. Q.E.D.

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According to the numerical computations, $r_0 \approx 0.044951$ and $\lambda \approx 40.7064$; see the figure below.

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Actually, this fact was observed already by Kline ³ in 1948, see Fig. 1 in ³. I copypasted this figure below. (I admit that this fact can be very well known for everybody for already a couple of hundred years.)

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P.S. It would be quite interesting to know whether are there some $r \in (0,1)$ such that the multiplicity of any eigenvalue is at most 2. Intuitively, the answer have to be affimative, and it seems that higher multiplicity can occur for at most a countable set of $r$’s.

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