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I finally decided to post the following naive question but will if consensus is that it is out of the scope of this site , it will be immediately deleted.

Suppose $\Omega\subset\mathbb R^2$ is a bounded simply connected domain with smooth boundary and let $0<\lambda_1^D(\Omega)\leq\lambda_2^D(\Omega)\leq\cdots$ and $0=\lambda_1^N(\Omega)\leq\lambda_2^N(\Omega)\leq\cdots$ denote the eigenvalues of Laplacian with Dirichlet and Neumann boundary conditions, respectively. From Szego inequality we know that $\lambda_{k+1}^N(\Omega)\leq\lambda_k^D(\Omega)$ for $k=1,2,\cdots$.

$\textbf{Q.}$ Does there exists a universal constant $\alpha$(i.e. the one the doesnt depend on $\Omega$) so that $$\lambda_2^N(\Omega)\geq\alpha\lambda_1^D(\Omega)?$$

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No. Consider a thin rectangle, with side lengths $a\ll 1$ and $1$. Then $\lambda_1^D=\pi^2(1+1/a^2)$ (eigenfunction $\sin \pi x\sin \pi y/a$), $\lambda_2^N = \pi^2$ (eigenfunction $\cos \pi x$), and now you're in trouble when $a\to 0+$.

This domain does not have a smooth boundary, but of course you can iron out the corners and still get almost the same eigenvalues.

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