Radial Limits of Capillary Surfaces at Corners

TL;DR

This paper investigates the behavior of solutions to a prescribed mean curvature equation near corners in the domain boundary, establishing the existence and nature of radial limits under various angular conditions.

Contribution

The authors prove the existence and characterize the behavior of radial limits of capillary surfaces at corners with specific angular measures, extending previous results to new angular ranges.

Findings

Radial limits exist for angles greater than π/2, regardless of boundary conditions.

Radial limits also exist for angles between π/4 and π/2, with specific behavior depending on contact angles.

The results are independent of boundary behavior on the opposite side of the corner.

Abstract

Consider a solution $f\in C^{2}(\Omega)$ of a prescribed mean curvature equation \[ {\rm div}\left(\frac{\nabla f}{\sqrt{1+|\nabla f|^{2}}}\right)=2H(x,f) \ \ \ \ {\rm in} \ \ \Omega\subset R^{2}, \] where $\Omega$ is a domain whose boundary has a corner at ${\cal O}=(0,0)\in\partial\Omega$ and the angular measure of this corner is $2\alpha,$ for some $\alpha\in (0,\pi).$ Suppose $\sup_{x\in\Omega} |f(x)|$ and $\sup_{x\in\Omega} |H(x,f(x))|$ are both finite. If $\alpha>\frac{\pi}{2},$ then the (nontangential) radial limits of $f$ at ${\cal O},$ \[ Rf(\theta) = \lim_{r\downarrow 0} f(r\cos(\theta),r\sin(\theta)), \] were recently proven by the authors to exist, independent of the boundary behavior of $f$ on $\partial\Omega,$ and to have a specific type of behavior. Suppose $\alpha\in \left(\frac{\pi}{4},\frac{\pi}{2}\right],$ the contact angle $\gamma(\cdot)$ that the graph of $f$ makes with one side of $\partial\Omega$ has a limit (denoted $\gamma_{2}$) at ${\cal O}$ and \[ \pi-2\alpha < \gamma_{2} <2\alpha. \] We prove that the (nontangential) radial limits of $f$ at ${\cal O}$ exist and the radial limits have a specific type of behavior, independent of the boundary behavior of $f$ on the other side of $\partial\Omega.$ We also discuss the case $\alpha\in \left(0,\frac{\pi}{2}\right].$