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Jul 17, 2021 at 11:50 comment added student @CarloBeenakker, I don't think so. The multiplicity can be very large. For example, for rectangular domain, $\lambda_k$ has multiplicity as $\sqrt{k}$. Actually Nadirashvili proved that for general planar domain in $\mathbb{R}^n$, the multiplicity of $\lambda_k$ has upper bound $2k-1$.
Jul 17, 2021 at 6:53 comment added Carlo Beenakker the growth rate of $N_d(\lambda)$ remains the same, whether or not you count the multiplicity; identical eigenvalues have measure zero, unless they are due to some discrete symmetry of $\Omega$, but that only changes $N$ by some $\lambda$ independent factor.
Jul 17, 2021 at 2:52 history asked student CC BY-SA 4.0