First Dirichlet eigenvalue on regular polygons

Question

Assume $\lambda(P)$ is the first Dirichlet eigenvalue of a regular polygon $P$. Let $u$ be the corresponding eigenfunction, normalized by $\|u\|_{L^2(P)}=1$, and $\partial_{\nu}u$ be its normal derivative on the boundary. Is the following estimate correct: \begin{eqnarray*} \lambda(P)\geq \|\partial_{\nu} u\|^2_{L^{\infty}(\partial P)}\vert P\vert\quad? \end{eqnarray*} Here $\vert P\vert$ denotes the area of $P$.

Answer 1

Ignoring the normalization, your inequality writes $$\lambda(P)\|u\|_{L^2(P)}^2\ge\|\partial_\nu u\|_{L^\infty(\partial P)}^2|P|\qquad?$$ Good news: this is scaling invariant.

Bad news: this is false for a square $P=(-\pi/2,\pi/2)^2$. Then $u(x,y)=\cos x\cos y$,$\lambda(P)=2$, $|P|=\pi^2$ and $$\|u\|_{L^2}^2=\left(\int_{-\pi/2}^{\pi/2}\cos^2 x\,dx\right)^2=\frac{\pi^2}4\,,\qquad\|\partial_\nu u\|_{L^\infty}^2=1.$$

Perhaps your inequality is correct with an extra constant factor.Perhaps also you should think to an inequality of the form $$\lambda(P)\|u\|_{L^2(P)}\ge c\|\partial_\nu u\|_{L^\infty(\partial P)},$$ which is scaling invariant too. For instance the limit case of the unit disk gives $u(x,y)=a(r)$ with $$a''+\frac1r\,a'=-\lambda a \quad a'(0)=0,\quad a(1)=0.$$ Then \begin{eqnarray*} \|\partial_\nu u\|_{L^\infty(\partial P)} & = & |a'(1)|=\left|\int_0^1(ra'(r)'dr\right| \\ & = & \lambda\left|\int_0^1ra(r)dr\right|\le\frac\lambda{2\sqrt\pi}\,\|u\|_{L^2(P)}. \end{eqnarray*}

Answer 2

There is a more general reason why any such statement will fail: If you consider a $P$ consisting of $N$ separate copies of the same basic region $P_0$, then $|P|=N|P_0|$, while everything else in your inequality is independent of $N$, so the inequality will fail for large $N$. (This problem has a degenerate ground state, but of course you could address this by slightly changing the shapes.)

If you connect the components by thin tubes, then you now have a connected $P$ and are still approximately in the situation described above.

Answer 3

The quantity $\lambda(P)|P|$ is scale invariant since $\lambda(tP)=\frac{1}{t^2}\lambda(P)$. Therefore, its derivative with perturbations preserving the regularity of the polygon equals zero.

The derivative of the eigenvalue can be found using the shape derivative formula:

$$\lambda'(P)(\theta) = -\int_{\partial P} (\partial_n u)^2 (\theta \cdot n)$$

It is immediate to notice that a vector field which corresponds to dilations verifies $\theta \cdot n = 1$.

In the same case the shape derivative of the area gives

$$ |P|'(\theta) = \int_{\partial P} \theta \cdot n.$$

This gives $\lambda(P)|P|'=-|P|\lambda'(P)(\theta) = |P|\int_{\partial P}(\partial_n u)^2$.

In the end: $\lambda(P)=|P| \frac{1}{|\partial P|}\int_{\partial P}(\partial_n u)^2$.

If my reasoning is correct, this is an equality (always better than an inequality :) )

Therefore the inequality the OP wants goes in the other sense (the average is smaller than the $L^\infty$ norm).