# A proof for a version of the maximum principle

Let $U=\Omega \times (0,\infty),$ where $~\Omega$ is a bounded domain in $\mathbb{R}^n$ and $u \in C^{2,1}(\overline{U})$ satisfies $$u_t \leq \Delta u+cu~~\text{in}~U,$$ where $c \leq 0$ is a constant.

Question: If $u \geq 0,$ show that the weak maximum principle holds for $u.$

So basically, I want to prove that the $\max$ of $u$ in $\overline{U}$ is attained in $\partial U.$ I've got no clue to even how to start this problem. I think I just haven't quite got my head around some of this stuff yet so if you can please provide a detailed proof, I'm much appreciated.

Assume that the maximum is attained at an interior point. Then $u_t=0$ and $\Delta u\leq0$ at this point. So from the given equation and the fact that $c\leq0$ it follows that $u\leq0$. This is a contradiction if we assume $u$ to be not identically zero.

• Put $u_t=0$, $\Delta u\leq 0$ and $c\leq 0$ in $u_t\leq \Delta u+ cu$ to get $u\leq 0$. Since it is given that $u\geq 0$ in $U$ we get $u=0$ at this point and similarly on every point in the interior. This implies either $u$ is identically zero or maximum is attained on the boundary. – Mathew George Jan 2 '17 at 9:19

Fix $T\in(0,\infty)$ and assume $(x_*,t_*)$ is a maximum point of $u$ in the domain $\Gamma_T = [0,T)\times \Omega$. If the maximum point is not on the boundary then we have $u_t(x_*,t_*) \geq 0$ as otherwise

$$u(x_*,t_* - \delta) = u(x_*,t_*) -\delta u_t(x_*,t_*) + \mathcal{O}(\delta^2) > u(x_*,t_*)$$ for a sufficiently small $\delta > 0$ contradicting $(x_*,t_*)$ being a maximum point. Next we use that $\Delta u(x_*,t_*) \leq 0$ since the Laplacian is negative at a maximum point. Conbinding this with $u_t \leq \Delta u + cu$ shows that

$$u_t(x_*,t_*) \geq 0 \geq \Delta u(x_*,t_*) \geq u_t(x_*,t_*) - cu(x_*,t_*)$$

Now if $c < 0$ and $u\geq 0$ then $-cu(x_*,t_*) \geq0$ and we have a contradiction unless $u(x_*,t_*) = 0$ which is only the case if $u\equiv 0$.

If $c=0$ then we can take $w = u - \epsilon t$ for some $\epsilon > 0$ to get the inequality $w_t = u_t - \epsilon \leq \Delta w - \epsilon < \Delta w$. As above this leads to $w_t(x_*,t_*) \geq 0 \geq \Delta w(x_*,t_*) > w_t(x_*,t_*)$ which gives us a contradiction to $(x_*,t_*)$ being a maximum point of $w$. Taking $\epsilon\to 0^+$ gives the same result for $u$: the maximum point has to be attained on the boundary.

Finally since $T>0$ was arbitrary it follows that $\sup_{(t,x)\in \Gamma_\infty} u = \sup_{(t,x)\in\partial \Gamma_\infty} u$ so the weak maximum principle holds for $u$.

• Thank you very much. Earlier proof was much shorter but from this I get to learn some techniques. – user399481 Jan 3 '17 at 0:55