Here's the solution.
First of all, let us note that the radius of \(D(l)\) is exactly \(l/\sqrt{\pi}\).
Step 1. Now we turn to problem (**), looking for its eigenvalues and eigenfunctions.
Because \(D(l)\) is radially symmetric, it is convenient to look for solutions of (**) in polar coordinates.
So, let \(u(x,y)=u(r\cos \theta,r\sin \theta)=v(r,\theta)\) be a candidate solution; applying the
chain rule we find:
\[\begin{align} v_r &= u_x \cos \theta +u_y \sin \theta \\ v_\theta &= -u_x r \sin \theta +u_y r \cos \theta \\
v_{rr} &= u_{xx} \sin^2 \theta +2u_{x,y} \sin \theta \cos \theta +u_{yy} \sin^2 \theta \\ v_{\theta \theta} &= r^2 (u_{xx} \sin^2 \theta -2u_{x,y} \sin \theta \cos \theta +u_{yy} \sin^2 \theta ) -r(u_x \cos \theta +u_y\sin \theta)\end{align}\]
therefore:
\[\Delta u(x,y) =v_{rr}(r,\theta)+\frac{1}{r}\ v_r (r,\theta)+\frac{1}{r^2}v_{\theta \theta} (r,\theta)\]
and the PDE in (**) transforms into:
\[r^2 v_{rr}(r,\theta)+r\ v_r (r,\theta)+v_{\theta \theta} (r,\theta) =\lambda r^2 v(r,\theta) \; ,\]
so that problem (**) rewrites:
\[\tag{***} \begin{cases} r^2 v_{rr}(r,\theta)+r\ v_r (r,\theta)+v_{\theta \theta} (r,\theta) =\lambda r^2 v(r,\theta) &\text{, in } ]0,l/\sqrt{\pi}[\times ]0,2\pi[ \\ v(r,0)=v(r,2\pi) &\text{, for } 0\leq r\leq l/\sqrt{\pi} \\ v(l/\sqrt{\pi},\theta) =0 &\text{, for } 0\leq \theta \leq 2\pi \; .\end{cases}\]
PDE in (***) separates variables, hence we can look for solutions in the form \(v(r,\theta)=R(r) \Theta (\theta)\); such a function solves the PDE iff there exists a constant \(\omega\) (which is usually called
separation constant) such that:
\[\frac{r^2}{R}R^{\prime \prime} +\frac{r}{R} R^\prime +\lambda r^2 =\omega=\frac{1}{\Theta} \Theta^{\prime \prime}\]
that is iff \(R\) and \(\Theta\) solve:
\[\begin{align} r^2R^{\prime \prime} +rR^\prime +(\lambda r^2-\omega)R&=0 & \Theta^{\prime \prime} +\omega\ \Theta&=0 \; .\end{align}\]
Moreover, \(R(r)\Theta (\theta)\) solves (***) iff \(R\) and \(\Theta\) satisfy the following conditions:
\[\begin{align} &\begin{cases} R \text{ is continuous in } [0,l/\sqrt{\pi}] \\ R(l/\sqrt{\pi})=0 \end{cases} &, & \Theta(0)=\Theta(2\pi)\; ;\end{align}\]
therefore, problem (**) is transformed into a couple of BV problems for two ODEs, namely:
\[\tag{I} \begin{cases} \Theta^{\prime \prime} +\omega\ \Theta =0 &\text{, in } ]0,2\pi[ \\ \Theta (0)=\Theta (2\pi)\end{cases}\]
and:
\[\tag{II} \begin{cases} r^2\ R^{\prime \prime} +r\ R^\prime +(\lambda\ r^2-\omega)R =0 &\text{, in } ]0,l/\sqrt{2\pi}[ \\ R \text{ is continuous} &\text{, in } [0,l/\sqrt{\pi}] \\ R(l/\sqrt{\pi})=0\; . \end{cases}\]
Problem (I) can be easily solved: in fact, (I) has nontrivial solution iff \(\omega =n^2\) for some \(n\in \mathbb{N}\) and its general solution corresponding to \(\omega =n^2\) equals:
\[\Theta (\theta) =A_n \cos n\theta +B_n \sin n\theta\]
where \(A_n,B_n\) are arbitrary real constants.
Now we turn to (II). If we set \(\omega =n^2\) in the ODE, we find the Bessel-type equation:
\[r^2\ R^{\prime \prime} +r\ R^\prime +(\lambda r^2-n^2)R =0\]
which can be put into canonical form:
\[\rho^2\ \ddot{R}+\rho\ \dot{R}+(\rho^2-n^2)R=0\]
with the substitution \(\rho=\sqrt{\lambda}\ r\) (here the dot denotes derivative w.r.t. \(\rho\)); thus it becomes clear that the ODE has solution:
\[R(r)=C_n\ \text{J}_n(\sqrt{\lambda}\ r) +D_n\ \text{Y}_n (\sqrt{\lambda}\ r)\]
where \(C_n,D_n\) are arbitrary constants and \(\text{J}_n,\ \text{Y}_n\) are Bessel functions of order \(n\) of the first and second kind.
Such a function fulfill the continuity constraint in (II) iff \(D_n=0\) (for \(\text{Y}_n\) diverges when the argument approaches zero), and it satisfies \(R(l/\sqrt{\pi})=0\) iff \(\lambda\) is chosen among the solutions of equation \(J_n(\sqrt{\lambda}\ l/\sqrt{\pi})=0\). From the general theory of Bessel functions it follows that \(\lambda\) has to satisfy \( \sqrt{\lambda}\ l/\sqrt{\pi}=j_{n,k}\), where \(j_{n,k}>0\) is the \(k\)-th nontrivial zero of \(\text{J}_n\), therefore we have:
\[R(r)=C_k\ \text{J}_n (\sqrt{\lambda_{n,k}}\ r)\]
with \(\lambda_{n,k}=\pi\ j_{n,k}^2/l^2\).
Finally, we can return to problem (**). Summing up what we wrote, the eigenvalues of (**) are:
\[\lambda_{n,k} = \pi\ j_{n,k}^2/l^2\]
(with \(n=0,1,2,\ldots \) and \(k=1,2,\ldots\))) and any eigenfunction \(u_{n,k} (x,y)\) corresponding to \(\lambda_{n,k}\) is given in polar coordinates by:
\[u_{n,k} (r,\theta) =(A_{n,k}\ \cos n\theta +B_{n,k}\ \sin n\theta )\ \text{J}_n \left( \frac{\sqrt{\pi}\ j_{n,k}}{l}\ r\right) \; .\]
Step 2. Now we determine the first eigenvalue \(\lambda_1^\star (l)\).
The following classical results about eigenfunctions corresponding to the first eigenvalue of the Dirichelet Laplacian will be useful:
Let \(\Omega \subset \mathbb{R}^N\) be a sufficiently smooth nonempty open set.
Then all the eigenfunctions of the homogeneous Dirichlet Laplacian in \(\Omega\) are of class \(C^\infty\).
Let \(B\subset \mathbb{R}^N\) be an open ball and \(u\) an eigenfunction corresponding to the first eigenvalue of the homogeneous Dirichlet Laplacian in \(B\).
If \(u\) is of class \(C^2\), then \(u\) is a radial function (in other words, \(u(x,y)=v(r)\) with \(r=\sqrt{x^2+y^2}\)).
[For a proof, which make use of the
moving plane method (a geometric method based on the
maximum principle), the reader may refer to § 9.5.2 in Evans,
Partial Differential Equations - Second Edition, GSM-19, AMS, Providence.]
Let \(u_{n,k}\) be an eigenfunction corresponding to \(\lambda_1^\star (l)\); the
angular part \(A_n \cos n\theta +B_n\sin n\theta\) of \(u_{n,k}\) must reduce to a constant (for \(u_{n,k}\) has to be radial), therefore \(n\) has to be zero and:
\[
u_{0,k}(r,\theta) =A_{0,k}\ \text{J}_0 (\sqrt{\pi}\ j_{0,k}/l\ r)\; .
\]
Thus the first eigenvalue we are looking for is the least element of the set \(\{\pi\ j_{0,k}^2/l^2\}_{k=1,2,\ldots}\), which is the one correspondig to \(k=1\), i.e.:
\[\lambda_1^\star (l)=\frac{\pi}{l^2}\ j_{0,1}^2 \approx 5.783\ \frac{\pi}{l^2} \; .\]
Step 3. Finally, we can compare \(\lambda_1(l,l)\) with \(\lambda_1^\star (l)\).
Remembering that \(\lambda_1(l,l)=2 \pi^2/l^2\), we easily find \(\lambda_1^\star (l)<\lambda_1(l,l)\).
