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Let $ M \subset \mathbb{R}^2 $ be parallelogram constructed by putting together two equilateral triangles (so that all sides of the parallelogram have length 1, and the internal angles are 60 and 120). What is the spectrum of the laplacian $ \Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} $ with dirichlet boundary conditions on $ M $?

The spectrum of the laplacian on the equilateral triangle is known, so some of the eigenfunctions - those that vanish on the diagonal - are known. But what about the whole spectrum?

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This does not appear to be known. Clearly the remaining eigenfunctions are the same as those for an equilateral triangle with Dirichlet conditions on two sides and Neumann conditions on one side. It is known that these eigenfunctions are not trigonometric, see the discussion given here: http://www.m-hikari.com/ams/ams-password-2008/ams-password57-60-2008/mccartinAMS57-60-2008.pdf. This makes a closed form seem rather unlikely.

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This may not be what you wish for, but here is a list of the first $10$ eigenvalues calculated numerically for the rhombus with sidelength $1$:

$$\begin{array}{c}24.8982\\ 52.6379\\ 71.7085\\ 83.8282\\ 122.8217\\ 140.4414\\ 169.2286\\ 169.7662\\ 210.5516\\ 228.0975\end{array}$$

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In theory, the spectrum is discrete because the Laplacian is elliptic and the domain is compact (bounded and closed), so the resolvant is compact hence the discreteness of the spectrum.

Now for computing these eigenvalues, we can use the separation of variables method to reduce the PDE equation to a pair of ordinary equations. The eigenvalues are computed using the boundary conditions.‌

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  • $\begingroup$ I think this answer is correct.. Can someone explain to me where is the error, please ? $\endgroup$ – user36539 Sep 28 '13 at 17:48
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    $\begingroup$ I did not vote here, but I think what you write seems to be known for the OP. The exact eigenvalues and eigenfunctions were the question, maybe this was the problem. $\endgroup$ – András Bátkai Sep 28 '13 at 18:23
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    $\begingroup$ May be the guy who voted doesn't know the practical details of computing these eigenvalues $\endgroup$ – user36539 Sep 29 '13 at 20:54
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    $\begingroup$ As you see from the other answer, he is not alone... $\endgroup$ – András Bátkai Sep 30 '13 at 5:12
  • $\begingroup$ I wrote down an example here of why the eigenvalues and eigenfunctions of the Laplacian are difficult to compute analytically. (I would also note that eigenvalues of finite-dimensional operators are roots of polynomials, which are also nontrivial to compute.) $\endgroup$ – Neal Nov 1 '16 at 17:42

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