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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: enter image description here

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!

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    $\begingroup$ The first paper that leaps to mind is this by Mazzeo-Rowlett on how the heat trace for smooth domains can limit to the heat trace of polygons. Maybe you'll get mileage out of approximating a domain by a piecewise-linear domain. $\endgroup$
    – Neal
    Nov 20 '17 at 17:30
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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.

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    $\begingroup$ So, we get $D_n u(0) = -e^{-CH}$, and if $C>0$ is assumed to be fixed, then the desired dynamics w.r.t. to $H$ can be indeed traced. But why $C$ is fixed along the boundary? It seems that $C$ depends on $H$ by itself, which looks like an obstacle to establish a rigorous proof. Could you please provide some additional details in this direction? $\endgroup$ Mar 31 '19 at 16:16

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