How to compute the first eigenvalue of Laplace operator in an ellipse? Let $\mathcal{E}$ be an ellipse in the $\mathbb{R}^2$ plane with center in $o=(0,0)$, given focal distance $c\geq 0$ and given area $A>0$.
It is a fact that the eigenvalue problem for the Laplace operator with homogeneous Dirichlet boundary condition, i.e.:
$$\begin{cases} u_{xx}+u_{yy}=-\lambda u, &\text{in } \mathcal{E}, \\ u=0, &\text{on } \partial \mathcal{E}, \end{cases} $$
has solutions for infinite positive values of $\lambda$: these values are called eigenvalues of the Laplace operator in $\mathcal{E}$; in particular, there exists an eigenvalue $\lambda_1(c)$ which is the smallest one: $\lambda_1(c)$ is called the first eigenvalue of the Laplace operator and it has some nice properties (e.g. it is simple, for the eigenspace associated to $\lambda_1(c)$ is one dimensional).
Now, if $\mathcal{E}$ is a circle (this can happen iff $c=0$) it is well known that $\lambda_1(0)=\frac{\pi}{A}\ j_{0,1}^2$, where $j_{0,1}\approx 2.40483$ is the first zero of the Bessel function $\text{J}_0(x)$.
The questions I'm interested in are the following:


*

*What happens to $\lambda_1(c)$ if $\mathcal{E}$ is a "true" ellipse (i.e. if $c>
0$)? Can it be evaluated explicitly (in terms of some special functions)?

*And how different values of $c$ bias the value of $\lambda_1(c)$ around $0$?


It is known that $\lambda_1(c)\geq \lambda_1(0)$, with equality iff $\mathcal{E}$ is a circle (i.e., iff $c=0$; this is the famous Faber-Krahn inequality), but it also seems quite obvious that $\lambda_1(c)$ has to exhibit a sort of continuity in $0$: in fact one expects that $\lim \limits_ {c\to 0^+} \lambda_1(c) = \lambda_1(0)$...
Now, I did some researches on the net.
In the case $c>0$, one can introduce the elliptic coordinates $(\mu ,\nu)$:
$$\begin{cases} x=c\cosh \mu \cos \nu, \\ y=c\sinh \mu \sin \nu ,\end{cases} $$
so that equation $u_{xx}+u_{yy}=-\lambda u$ transforms into:
$$u_{\mu\mu} +u_{\nu \nu} =-c^2 \lambda (\sinh^2 \mu +\sin^2 \nu) u $$
which is harder to solve with separation of variables than the equation for the circle; neverthless separation of variables applies and yields a couple of so-called Mathieu's differential equations, which are a sort of ugly counterpart of Bessel's differential equation...
But then I cannot figure out how to compute $\lambda_1(c)$ (neither for fixed $c$ nor for varying $c$)!
Do I have to use some tables (like the ones in Abramowitz & Stegun, §20)?
And, in the positive case, how they can be used?
If you have any reference it could be worth reading, please feel free to suggest.
Thanks in advance for your help.
 A: I like this question! I didn't have the stamina to look into the issue too precisely, but here's what I got so far from Abramowitz and Stegun. As you said, the eigenfunction will be described by Mathieu functions. More precisely, on page 722 they study the "wave equation for the elliptic cylinder" (I would call that the Helmholtz equation...) using separation of variables. For your case you want to take $\varphi = $ constant because you have only two variables. This means that your eigenfunction is given by $f(u)g(v)$ where $f$ and $g$ solve the modified Mathieu and Mathieu equation respectively as in 20.1.6. In the equations you have a parameter $q$ which depends on the ellipse (I think for you $q = \lambda_1(c) \cdot c^2/4$) and you have a parameter $a$ which comes from separation of variables and which we must determine along with $\lambda_1(c)$. You can also find these equations on the wikipedia page for Mathieu functions.
Here $u$ is the "radial" variable and $v$ is the "angular" variable (the picture in the wikipedia page for elliptical coordinates which you linked to above  depicts these nicely). This means that $g$ must be $2\pi$ periodic to define a function on the ellipse, and $f$ must vanish at $1$ to satisfy the Dirichlet boundary condition (to simplify this part I've renormalized your area so the ellipse passes through $(x,y) = (c\cosh 1,0)$). Also, both must be positive inside the ellipse because the ground state is always positive. Over the next few pages you have a study telling you which $a$ lead to periodic solutions. The only one which is never vanishing is $a_0(q)$ and this gives a solution $g(v) = \textrm{ce}_0(v)$ which is graphed in figure 20.2. 
Now we've identified the angular part of the eigenfunction, namely $g(v) = \textrm{ce}_0(v)$, and the parameter $a = a_0(q) \approx - q^2/2$ (for $q$ small -- this is equivalent to $c$ being small) which has an expansion given in 20.2.25. To fix the eigenvalue $\lambda_1(c)$ we plug this into the equation for $f(u)$, and we see what is the smallest positive value which gives us a solution which has $f'(0) = 0$ (so the solution is smooth) and has $f(1) = 0$.
If my calculation is right, $f(\pm iv)$ solves the same equation as $g(v)$, so we may take (this part is not precise! -- you will need to make some kind of approximation argument here) $f(u) = \textrm{ce}_0(iu)$. Then the expansion 20.2.27 says that if $q$ is small then $f(u) = \textrm{ce}_0(iu) \approx 1 - \frac q 2 \cosh(2u)$ which has $f'(0) = 0$. If $f(1) = 0$ then $q = 2/\cosh2$, and we get $\lambda_1(c) = \frac 8 {c^2 \cosh(2)}$. To compare with your answer for a circle, we must renormalize the area. This ellipse has area $A =\pi c^2 \sinh 1 \cosh 1$, giving
$$ \lambda_1(c) \approx \frac {8 \pi \sinh 1 \cosh 1}{A \cosh 2} \approx 3.9 \frac \pi A$$
This is not the right answer (should have gotten 5.8 rather than 3.9), and I think the issue here comes from the fact that the approximation $\textrm{ce}_0(iu) \approx 1 - \frac q 2 \cosh(2u)$ is not good near $u=1$, even if $q$ is very small. What you really need is info on the zeros of Mathieu functions, which I'm too tired to look up... do post again if you find anything!
A: Every decade for the last fifty years, one to several people have worked on finding an easy answer to this problem. Is there a simple formula that would enable one to compute any eigenvalue of the ellipse (along with its associated eigenfunction)? The Mathieu functions can be used, but the algorithm is not "simple" in that Mathieu functions are considered by some (like me) to be the most difficult special functions to work with. 
By simple, I mean: Specify only the two semi-axes (or ellipse area and eccentricity) and to which circle mode it would correspond, i.e., either J_m(j_mn r/R)*cos(m th) or J_m(j_mn r/R)*sin(m th), and from that compute the eigenvalue. Dan Henry (about 1985) came closest to solving that, but 2xTroesch (1973) was also close. Being now my turn, this is my first stab at it: https://arxiv.org/abs/1802.07768 (which has relevant references).
