Let $u$ solve the following elliptic equation (where $c$ is continuous, non-negative): $$ \sum_{i,j = 1, ... ,d}a_{ij}\partial_{ij}u(x) - c(x)u(x) \geq 0 $$ So that the matrix $a_{ij}$ is positive-definite on the interior of the domain $\Omega$. Prove that if $u \leq 0$ on $\partial \Omega$ we have that $u \leq 0$ on all of $\Omega$.
My attempt: If the matrix $a_{ij}$ is positive definite, and hence can be diagonalized with positive eigen-values. Hence, If we define $v$ to have laplacian $\sum_i a_{ii}\partial u_{ii}$, then we can ascertain some information about $v$, I'm not sure what however.
I will assume $u \in C^2(\Omega) \cap C^0(\overline \Omega).$ The argument I present is given in theorem 3.1 of Gilbarg and Trudinger's text.
The idea is simple; take the global maximum of $u$ and by considering the signs in the above equation, we reach a contradiction. The fact that our inequalities are not strict makes difficult however, so we need to preturb $u$ to get something strict.
Let $\Omega_+ = \{ x \in \Omega : u(x) > 0\}$ and assume it is non-empty. Then on $\Omega_+$ we have, $$ \sum_{ij} a^{ij}\partial_{ij} u \geq cu \geq 0, $$ with $u \equiv 0$ on $\partial \Omega_+.$ Now observe that since $(a^{ij})$ is positive definite in $\Omega,$ we have $a_{11} > 0$ in $\Omega$ also. Hence for all $\varepsilon > 0$ we get, $$ \sum_{ij} a^{ij}\partial_{ij}(u + \varepsilon e^{x_1}) \geq cu + \varepsilon a_{11}e^{x_1} > 0 $$ in $\Omega_+.$
Choose $\varepsilon > 0$ such that $u + \epsilon e^{x_1}$ attains its global maximum in $\Omega_+.$ We can do this as there exists $x \in \Omega_+$ so we choose $\varepsilon > 0$ such that, $$ \varepsilon \sup_{\partial \Omega_+} e^{x_1} < u(x). $$ So if we let $v(x) = u(x) + \varepsilon e^{x_1}$ we get, $$ \sup_{\partial \Omega_+} v < u(x) \leq v(x). $$ Now let $x_0 \in \Omega_+$ be a global maximum, which exists as $u$ is continuous on the compact set $\overline \Omega_+$ and does not attain its maximum on the boundary. We know that, $$ \sum_{ij} a^{ij}\partial_{ij}v(x_0) > 0. $$ But we know that $\partial_{ij}v(x_0)$ is non-positive definite (being a maximum), so the above quantity must be non-negative. This gives the desired contradiction.
The above ended up being longer that I expected, so here's a quick summary.
Define $\Omega_+ = \{ x \in \Omega_+ : u(x) > 0 \}$ and note $\sum_{ij}a^{ij}\partial_{ij}u \geq 0$ in $\Omega_+.$
For any $\varepsilon > 0,$ $\sum_{ij}a^{ij}\partial_{ij}(u+\varepsilon e^{x_1}) > 0$ in $\Omega_+.$
By choosing $\varepsilon > 0$ to be sufficiently small, we have $v = u + \varepsilon e^{x_1}$ attains its maximum in the interior of $\Omega_+.$
Then $\sum_{ij} a^{ij}\partial_{ij}v > 0$ gives a contradiction, since the Hessian of $v$ is non-positive definite at any maximum point.