[EX] First Eigenvalue of the Neumann Laplacian

[gugo82]

Exercise:

1. Let \(\mathcal{R}(a,b)\subseteq \mathbb{R}^2\) be any open rectangle with sides of length $a,b>0$.
Using the separation of variables find the first nonzero eigenvalue \(\mu_2(a,b)\) of the Neumann Laplacian in \(\mathcal{R}(a,b)\), i.e. find the smallest $\mu >0$ s.t. the problem:
\[ \tag{N}
\begin{cases} -\Delta u= \mu\ u &\text{, in } \mathcal{R}(a,b)\\
\frac{\partial u}{\partial \nu} =0 &\text{, on } \partial \mathcal{R}(a,b)
\end{cases}
\]
(where $\frac{\partial u}{\partial \nu}$ denotes the outward normal derivative) has some nonconstant solutions with separate variables.*

2. Are there values of $a,b>0$ which minimize/maximize $\mu_2(a,b)$ under measure constraint, i.e. under condition \(| \mathcal{R}(a,b) |=ab=\text{constant}\)?


__________
* I used $\mu_2$ because, actually, the first eigenvalue of the Neumann Laplacian is always $\mu_1=0$. In fact, every constant function $u(x,y)=c$ is a solution of the eigenvalue problem (N) associated with the $0$ eigenvalue and it can be easily proved that every other eigenvalue of (N) is positive.

[gugo82]

1. Without loss of generality assume \(\mathcal{R}(a,b)=]0,a[\times ]0,b[\).

Using the separation of variables technique, Neumann eigenvalue problem (N) rewrites as a couple of bundary values problems (BVPs), namely:
\[
\begin{cases}
X^{\prime \prime} (x)+\alpha X(x)=0\\
X^\prime (0)=0\\
X^\prime (a)=0
\end{cases} \quad \text{and}\quad
\begin{cases}
Y^{\prime \prime} (y)+(\mu - \alpha) Y(y)=0\\
Y^\prime (0)=0\\
Y^\prime (b)=0
\end{cases}
\]
where \(\alpha \in \mathbb{R}\) is a suitable separation constant.

The first BVP has nontrivial solutions iff \(\alpha \geq 0\): in particular, if \(\alpha=\alpha_0=0\), then any constant function satisfies the problem; on the other hand, if \(\alpha >0\), then nontrivial solutions can be found only iff:
\[
\alpha =\alpha_k= \frac{\pi^2}{a^2}\ k^2
\]
with \(k\in \mathbb{N}\setminus \{0\}\), and they are of the type:
\[
X_k(x) := A_k\ \cos \left( \frac{\pi}{a}\ k\ x\right)
\]
where \(A_k\) is any constant.

The same can be said about the second BVP. In particular, it has nontrivial solutions iff \(\mu-\alpha_k>0\): in particular, it has only constant solutions iff \(\mu -\alpha_k=0\), i.e. iff \(\mu=\mu_{k,0}=\alpha_k=\frac{\pi^2}{a^2}\ k^2\); while it possesses nonconstant solutions iff:
\[
\mu=\mu_{k,h}=\frac{\pi^2}{a^2}\ k^2 + \frac{\pi^2}{b^2}\ h^2
\]
with \(h\in \mathbb{N}\setminus \{0\}\), and the solutions are of the type:
\[
Y_h (y)=C_h\ \cos \left( \frac{\pi}{b}h\ y\right)
\]
where \(C_h\) is any constant.

Therefore, the first eigenvalue which corresponds to nonconstant eigenfunctions is:
\[
\begin{split}
\mu_2(a,b) &:= \min \left\{ \frac{\pi^2}{a^2},\ \frac{\pi^2}{b^2}\right\} \\
&= \frac{\pi^2}{\max \{a^2, b^2\}}\\
&= \frac{\pi^2}{(\max \{a,b\})^2}\; .
\end{split}
\]

2. Choose a constant \(A>0\).
It follows from previous computations that \(\inf_{a,b>0, ab=A} \mu_2(a,b)=0\) because \(\max \{a,b\}\) can tend to \(+\infty\) on the constraint.
On the other hand, it is really simple to prove that \(\mu_2(a,b)\) achieves its maximum when \(a=b\); therefore:
\[
\max_{a,b>0, ab=A} \mu_2(a,b) = \mu_2(\sqrt{A}, \sqrt{A}) = \frac{\pi^2}{A}\; .
\]