da gugo82 » 23/04/2013, 18:47
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:
\[
\max_{a,b>0, ab=A} \mu_2(a,b) = \mu_2(\sqrt{A}, \sqrt{A}) = \frac{\pi^2}{A}\; .
\]
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)