Elliptic Operator , a priori bounds

Question

suppose we have the following operator : \begin{equation*} L [u] = \sum_{i,j=1}^{n} a_{ij}(x)u_{x_ix_j}(x) + \sum_{i=1}^{n} b_i(x) u_{x_i}(x) + c(x) u(x) = f(x) , x \in \Omega \end{equation*} where $c(x)\leq -c_0<0$ in addition \begin{equation} u\Big|_{\partial \Omega} = 0 , \end{equation} I want to prove that \begin{equation} u(x) \leq \max \Big\{0,\max_{\Omega} \frac{f(x)}{-c_0} \Big\}, \forall x \in \overline\Omega \end{equation} I've tried many things which follow the idea that i want a function $S$ such that $S\Big|_{\partial \Omega}=\max_{\Omega} \frac{f(x)}{-c_0} $ and $L[S]\geq 0$ and $S\geq u$. From the maximum principle we have that there is no positive maximum of S in $\Omega$, so positive maximum doesn't exist ($S\geq 0$) or it exists at the boundary. In other words : \begin{equation} u \leq S \leq \max\Big\{ 0 , \max_{\partial \Omega}S \Big\}= \max\Big\{ 0, \max_{\Omega} \frac{f}{-c_0}\Big\} \end{equation}

Answer 1

There is three cases : 1) positive maximum doesn't exist ( so $u \leq 0$ ) 2) positive maximum is at boundary ( there is no way to be at the boundary because $u|_{\partial \Omega}=0$)

3) positive maximum exist at the internal point So , let's denote with $x_0$ the point at which the positive maximum is located so we then have :

\begin{equation*} \sum_{i,j=1}^{n} a_{ij}u_{x_ix_j}(x_0) + \sum_{i=1}^{n} \beta_{i}u_{x_i}(x_0) + c(x_0)u(x_0) = f(x_0) \end{equation*} but obviously $\sum_{i=1}^{n} \beta_{i}u_{x_i}(x_0)=0$ because $x_0$ is located the maximum , and for the same reason $\sum_{i,j=1}^{n} a_{ij}u_{x_ix_j}(x_0)\leq 0$. Also we have from our hypothesis that $u(x_0) \leq 0 $ and we can conclude that $f(x_0) \leq 0$ . So \begin{equation} c(x_0)u(x_0) \geq f(x_0) \end{equation} \begin{equation} u(x_0) \leq \frac{f(x_0)}{c(x_0)} \leq\frac{f(x_0)}{-c_0} \leq \frac{\max_{\Omega}f}{-c_0} \end{equation} The last holds because $c(x)\leq -c_0 \Rightarrow \frac{1}{c(x)}\geq \frac{1}{-c_0} \Rightarrow \frac{f(x_0)}{c(x)}\leq \frac{f(x_0)}{-c_0}$ .