Curvature of the boundary vs. normal derivative of the first eigenfunction

Question

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!

Answer 1

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.

Answer 2

You use a local system of coordinates with $x$-axis tangent to the boundary and $y$ along the normal vector; let $y=y(x)$ the equation of the curve, so $u(x,y(x))=0$ ($u=\phi_1$). Differentiating this equation twice at $x=0$ will give $u_{xx}(0) + y''(0)u_y(0)=0$, but $y''(0)=-H$, $u_y=D_nu$, $u_{xx}=-\lambda_1 u -D_{nn}u$; so you get what you wanted.

Answer 3

If $\Omega$ is piecewise smooth and the interior is arc-connected, then it is locally Lipschitz, i.e., for every $x_0 \in \partial\Omega$, there exists $r_0$ and a rotation such that $$\Omega\cap B_{r_0}(x_0) = \{x_n > g(x')\}\cap B_{r_0}(x_0),$$ with $g$ a Lipschitz function.

Then, you can use the boundary Harnack inequality for equations with right-hand side (see https://arxiv.org/pdf/1811.12908 or https://arxiv.org/pdf/2010.01064) to deduce that the first eigenfunction satisfies (locally near the boundary) $$c \leq \frac{\varphi_1}{v} \leq C,$$ where $v$ is a positive harmonic function vanishing on $\partial\Omega$.

Therefore, your question is equivalent to understanding how positive harmonic functions go to zero with homogeneous Dirichlet boundary data. For that, we have four cases:

• If the boundary is smooth ($C^2$ is more than enough), then by the Hopf-Oleinik lemma and the Lipschitz regularity of solutions, $v$ is proportional to the distance.

• If there is an inward corner, that is, $\Omega$ contains a cone with vertex at $x_0$ with an angle $\pi/\beta$, $\beta < 1$ (this looks weird but will be useful later), then the normal derivative is infinite.

To see that, consider a subsolution $w \leq v$, that will be zero on the boundary of the cone, harmonic and positive inside. Using complex analysis (the real part of a holomorphic function is harmonic), you can choose (after a rotation) $$w = c\operatorname{Re}(z^\beta)_+ = r^\beta\cos(\beta\theta),$$ and hence $v \geq cd^\beta$, that is, $v$ behaves like a power of the distance with exponent less than one, and hence the normal derivative blows up.

• If $\Omega$ has an outward corner, doing the same construction but with a cone that contains $\Omega$ gives a supersolution, and then you get that $v \leq Cd^\beta$, with $\beta > 1$, which means the normal derivative is zero.

• When the boundary is not $C^2$ but there are also no acute or obtuse cones containing it, there are some recent research results. I will mention the works of Li and Wang on boundary differentiability of harmonic functions on convex domains.