Partial differential equation with laplacian. Let $\Omega \subset \mathbb{R}^n$ be a bounded domain and $u$ a continuous function on $\overline{\Omega}$ such that $u \in C^2(\Omega)$. $u$ is the solution of following differential equation, 
$$ \Delta u + \sum_{i=1}^{n} a_k(x)u_{x_k} + c(x)u = 0, $$
where $c(x) < 0$ on $\Omega.$ Show that if $u = 0$ on $\partial \Omega$ then $u = 0$ on $\Omega$.
I would appreciate any tips or hints.
 A: Ok, I made a small mistake in the comments, but I've got the main idea anyway, here's the answer.
There is a global minima which is attained as $\bar \Omega$ is compact. That point is $x_0$. Suppose it lies in $\Omega$. Then, we know that at $x_0$, the quantities $\frac{\partial f}{\partial x_i} = 0$ for all $i$, and the Laplacian $\Delta u(x_0)$ is non-negative. Substituting this point $x_0$ in the equation, to get that $\Delta u(x_0) = -c(x_0)u(x_0)$. It follows that $u(x_0)$ is non-negative, so $u \geq 0$ on $\Omega$. 
Now, there is a global maximum. Suppose it's on the boundary, then the function is identically zero! So that interior minimum is also in the interior : here, the derivatives are zero but the Laplacian is non-positive, so again by substitution one sees that the local maximum must be non-positive i.e. $u \leq 0$ on $\Omega$. The upshot of all this is that $u \equiv 0 $ in $\Omega$, hence on $\bar \Omega$.
Now, if the global minima lies in $\partial \Omega$, then one gets a contradiction (using the above argument) if one assumes that the global maximum is in $\Omega$, so the global maximum must also lie in $\partial \Omega$. Since the boundary is set to $0$, these global minimum and maximum are also zero, and we are done anyway.

This is an example of the Weak Maximum principle. There is a Strong one as well, which asserts constancy of the solution under certain boundary conditions.
