TITLE: Curvature of the boundary vs. normal derivative of the first eigenfunction
QUESTION [9 upvotes]: Disclaimer. I posted this question in Math.SE, but it haven't received enough attention.
Let $\varphi_1$ be the first eigenfunction of the zero Dirichlet Laplacian in a planar bounded domain $\Omega$. That is,
$$
\left\{
\begin{aligned}
-\Delta \varphi_1 &= \lambda_1 \varphi_1 &&\text{in } \Omega \subset \mathbb{R}^2,\\
\varphi_1 &= 0 &&\text{on } \partial \Omega.
\end{aligned}
\right.
$$
If $\Omega$ is piece-wise smooth and has corners, and we look at the plot of $\varphi_1$, then we see that its normal derivative tends to zero near exterior (outward) corners, and tends to infinity near interior (inward) corners. 
See the plot for the standard L-shape: 

This fact suggests that, in a smooth domain, there should be some connection between curvature of the boundary at a point and the normal derivative of $\varphi_1$ at this point. 
That is, if the curvature is big positive, then the normal derivative is close to zero. And if the curvature is big negative, then the normal derivative is large. 
However, I was not able to find corresponding inequalities in the literature. (Although I believe that such results should be well-known.)
I would appreciate some references to such facts and related results in this direction. 
Thanks!

REPLY [2 votes]: I do not there is a strong relation between the two notions in the general case. Curvature is obviously a local object. On the other hand, the behaviour of the first Dirichlet eigenfunction near a boundary point $p$ is highly non-local.
For example, if $\Omega$ is the union of two balls of different radii connected by a narrow channel, then the normal derivative at the boundary of the smaller ball will converge to zero as the width of the channel goes to zero, despite the fact that the curvature remains unchanged.
That said, curvature and the normal derivative should be related to each other for domains "without thin channels" — for example, for NTA domains. A possible approach would be via results similar to the boundary Harnack inequality. Unfortunately, I am not aware of any references in that direction.