PDEs We now consider the non-divergence form of an Elliptic PDE with . Note that a function that obtains it’s maximum at a point has Missing \begin{rcases} or extra \end{rcases}\nabla u(x_0) = 0 \\ Ju(x_0) \preceq 0 \end{rcases} \space\text{pointwise condition} $$ so that we will require the solution to be $C^2$ to make any deductions from the above observation. Given $u \in C^2(U) \cap C(\overline{U})$ then we have 1. **Weak maximum principle**: - $c \equiv 0$ $$\begin{aligned} Lu \leq 0 \space \text{in} \space U &\implies \max_{\overline{U}} u = \max_{\partial U} u \\ Lu \geq 0 \space \text{in} \space U &\implies \min_{\overline{U}} u = \min_{\partial U} u \end{aligned}$$ - $c \geq 0$ $$\begin{aligned} Lu \leq 0 \space \text{in} \space U &\implies \max_{\overline{U}} u \leq \max_{\partial U} u^+ \\ Lu \geq 0 \space \text{in} \space U &\implies \min_{\overline{U}} u \geq -\max_{\partial U} u^- \\ Lu = 0 \space \text{in} \space U &\implies \max_{\overline{U}} |u| = \max_{\partial U} |u| \end{aligned}$$ 2. **Strong maximum principle**: Let $U$ be connected, open, and bounded - $c \equiv 0$ $$\begin{aligned} Lu \leq 0 \space \text{in} \space U \space \text{and} \space u \space \text{attains its max in the interior of} \space U &\implies u \space \text{is constant in} \space U\\ Lu \geq 0 \space \text{in} \space U \space \text{and} \space u \space \text{attains its min in the interior of} \space U &\implies u \space \text{is constant in} \space U\\ \end{aligned}$$ - $c \geq 0$ $$\begin{aligned} Lu \leq 0 \space \text{in} \space U \space \text{and} \space u \space \text{attains a non-negative max in the interior of} \space U &\implies u \space \text{is constant in} \space U\\ Lu \geq 0 \space \text{in} \space U \space \text{and} \space u \space \text{attains a non-positive min in the interior of} \space U &\implies u \space \text{is constant in} \space U\\ \end{aligned}$$ 3. **Harnack's inequality**: Given $u \geq 0 \in C^2(U)$ a weak solution of $Lu = 0$ $$\sup_V u \leq C_{V, L} \inf_V u$$ for $V \subset \overline{V} \subset U$.