Suppose that $\Omega\subset \mathbb{R}^n$ is open and bounded. Let $u\in C^2(\Omega)\cap C^0(\bar\Omega)$ is a solution for the equation $\triangle u+\sum_{k=1}^na_ku_{x_k}+c(x)u=0$ where $c(x)<0$ in $\Omega$. Prove that $u=0$ in $\partial\Omega$ implies $u=0$ in $\Omega$.
I ask for some advice to attack this problem, I do not know how to begin.
Thanks!
I'm guessing you mean $u\in C^2(\Omega)\cap C^0(\bar\Omega)$. $u$ is a continuous function on $\bar\Omega$, which is compact, hence reaches its extrema.
Let $x_0$ be a maximum point. If $x_0\in\Omega$, then $Du(x_0)=0$ and $\Delta u(x_0)\leq 0$. By plugging this into your equation, we get $$ \underbrace{\Delta u(x_0)}_{\leq 0} + \underbrace{c(x_0)}_{<0}u(x_0) = 0, $$ and necessarily, $u(x_0)\leq0$. If $x_0\notin \Omega$, then $x_0\in \delta\Omega$ and $u(x_0)=0$.
Conversely, if $x_1$ is a minimum point, if it is inside $\Omega$, then $$ \underbrace{\Delta u(x_1)}_{\geq 0} + \underbrace{c(x_1)}_{<0}u(x_1) = 0, $$ meaning that $u(x_1)\geq0$. Like before, if $x_1\in\delta\Omega$, then $u(x_1)=0$.
Finally, for all $x\in\bar\Omega$, $$ 0\leq u(x_1) \leq u(x) \leq u(x_0) \leq 0 $$ and $u\equiv0$.
