| Abstract: |
| Let $\Omega \subset \mathbb{R}^2$ be a $C^{1,\alpha}$ domain whose boundary is unbounded and connected. Suppose that $f:[0,+\infty) \to \mathbb{R}$ is $C^1$ and there exists a nonpositive prime $F$ of $f$ such that $F(0)=\sqrt{2}/2-1$. If there exists a positive bounded solution $u\in C^3$ with bounded $\nabla u$ to the overdetermined problem
$\begin{equation*}
\left\{\begin{array} {ll}
\mathrm{div} \left(\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}\right) + f(u) = 0 & \mbox{in }\; \Omega,\
u= 0 & \mbox{on }\; \partial \Omega, \
\frac{\partial u}{\partial \vec{\nu}}=1 &\mbox{on }\; \partial \Omega,
\end{array}\right.
\end{equation*}$
we prove that $\Omega$ is a half-plane. It means that a positive capillary graph whose mean curvature depends only on the height of the graph is a half-plane. |
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