Maximum principle for a “modified” laplacian

Let $\Omega \in \mathbb{R}^{n}$ be bounded dmain, given $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be a solution of $$\Delta u+c(x)u=0$$ where $c(x)\leq 0 \in \Omega.$ Shown that $u=0$ on $\partial \Omega$ implies $u=0$ in $\Omega$

My approach: Prove by contradiction: if $u(x_0)<0$ for some $x_0 \in \Omega$, then $u$ must obtain its minimum in $\Omega$, Let the minimum be $x_1.$ Second derivative test implies $\Delta u(x_1)\geq 0.$ But, $\Delta u(x_1)=-c(x_1)u(x_1)\leq0$. So if $\Delta u(x_1)>0$ then we obtain a contradiction. The next step is to overcome the possibility that $\Delta u(x_1)=0$, how may I proceed?

(Note: the case $x_1$ with $u(x_1)>0$ is a maximum can be proved similarly)

• It is easier to use an energy approach. Multiply the equation by $u$, integrate by parts and obtain $\int (\nabla u)^2 \; dx = \int c(x) u^2 \; dx \leq 0$... – uvs Nov 4 '14 at 11:16
• – Michael Medvinsky Sep 29 '15 at 23:53