Maximum principle question for heat equation problem

Let $u(t, x) \in C_{t}^{1} C_{x}^{2}\left(\Omega_{T}\right) \cap C(\overline{\Omega_{T}})$ satisfies:

$$ \begin{cases}\partial_{t} u-\Delta u+c(x) u \leq 0, & (t, x) \in \Omega_{T} \\ u(t, x) \leq 0, & (t, x) \in \Gamma_{T}\end{cases} $$

where $c(x) \geq-c_{0}$ has downside bound, $c_{0}>0$. Prove:

$$ u(t, x) \leq 0, \quad(t, x) \in \Omega_{T} $$

And I found the usual method used to prove weak maximum principle does not work here ($\partial_t u \ge 0 , \Delta u(t,x)\le 0$ does not make sure $\partial_{t} u-\Delta u+c(x) u \ge 0$).

Besides, I found two answers relative to this question. One is this, but I could not find the proof in Evans' book.

And another is that, but obivously that cannot solve when $c$ is not constant.

So can you please give me a hint?

Solution 1:

I have figured it out a similar method to this answer.

Let us define $v(t,x):=e^{c_0t}u(t,x)$. Then we have:

$$\begin{aligned} &\partial_t v - \Delta v \\ =& c_0e^{c_0 t}u + e^{c_0 t}\partial_t u - e^{c_0 t}\Delta u \\ =& e^{c_0 t}(c_0 u + \partial_t u - \Delta u) \le 0 \text{ (by condition) } \end{aligned}$$

Besides, it's easy to check that $v(t,x)\le 0 , (t,x) \in \Gamma_T$. Thus, by weak maximum principle, $v(t,x)$ cannot be positive in $\Omega_T$, which is same as $u(t,x)$.

:) Wonderful.