I would like to show that the following problem has only the trivial solution :
Let be $\Omega\subset \mathbb{R}^n$ open an Lipschitz bounded, and $c$ continous and positive on $\bar{\Omega}$, s.t $\exists x_0\in\Omega,\, c(x_0)>0$. The problem then states $$ \begin{cases}\Delta u=cu & \textrm{in }\Omega\\\frac{\partial u}{\partial \nu}=0 &\textrm{on } \partial \Omega\end{cases} $$ with $\partial\nu$ the normal derivative.
I already shown that if $u>0$ for some $x\in\Omega$ then, $u>0$ on all $\Omega$ by subharmonicity, but I cannot figure out how to hanlde de $c(x_0)>0$ which should gives the conclusion.
All help welcome !
I finally found it. In fact it is just an application of the divergence Theorem :
\begin{align*} \Delta u=cu\implies\int_{\Omega}\Delta u=\int_\Omega cu \end{align*}
But $\Delta u=\textrm{div}(\nabla u)$ and then $\int_{\Omega}\Delta u=\oint_{\partial \Omega}\nabla u\cdot\nu=\oint_{\partial \Omega}\frac{\partial u}{\partial \nu}u=0$.
Then \begin{align*} \int_\Omega cu =0 \end{align*} But by continuity of $c$, there is some open neighbourhood $V$ of $x_0$ such that $c>0$ on $V$, and $u>0$ on $\Omega$ by subharmonicity. Combined with $c\geq 0$ on $\Omega$, this gives $$ 0<\int_V cu \leq \int_\Omega cu $$
which is an obvious contradiction.