Sign of coefficient in the weak maximum principle

Question

( Weak maximum principle) Let $u \in C^{2}(\Omega)\cap C(\bar{\Omega})$ satisfies $Lu \geq 0$ with $c(x) \leq 0$ Em $\Omega$. Then $u$ attains in $\partial \Omega$ its nonnegative maximum in $\bar{\Omega}$.

Where:

• $\Omega$ is a bounded connected domain in $\mathbb{R}^{n}$ and satisfies

$a_ij(x)\xi_{i}\xi_{j} \geq \lambda |\xi|^{2}$ for al $x\in \Omega$ and $\xi \in \mathbb{R}^{n}$;

• $Lu=a_{ij}D_{ij}u +b_{i}uD_{i}u + cu$ (Einstein notation)

and $a_{ij},b_{i} and $c$ are continuous functions.

My question is wh: why it is required $c\leq 0$ in $\Omega$ in the weak maximum principle?

Answer 1

Lemma: Let $L$ be a linear, second-order operator. Let $u$ satisfy $Lu > 0$ (resp. $Lu \geq 0$) in a domain $\Omega$. Then, $u$ cannot attain a nonnegative (resp. positive) maximum in $\Omega$ if the zero-th order coefficient in $L$, call it $c\equiv c(x)$, satisfies $c \leq 0$ (resp. $c < 0$).

Here's a rough idea of the proof in the case of $Lu > 0$, and why the sign of $c$ appears in it.

Let $x$ be a point in the interior at which a nonnegative maximum is attained. At $x$, the first derivative terms are zero. Similarly, the second derivative terms are less than or equal to zero (by a linear transformation). This leaves us with $Lu(x)\leq c(x)u(x)\leq 0$, a contradiction.