$u$ is a $C^2$ solution of $u_t - \Delta u = f(u)$ and $u=0$ on $\partial\Omega \times (0,\infty)$. Show if $u(x,0) \geq 0$, then $u(x,t) \geq 0$ Let $\Omega$ be a bounded open subset of $\mathbb R^n$ with a smooth boundary $\partial \Omega$ and let $f: \mathbb R \rightarrow \mathbb R$ be a function such that its derivative is bounded, i.e. $|f'| \leq M$. $f(0)=0$. Assume $u$ is a $C^2$ solution of $u_t - \Delta u = f(u)$ in $\Omega \times (0,\infty)$ and $u=0$ on $\partial\Omega \times (0,\infty)$.
Show that if $u(x,0) \geq 0$ for all $x \in \Omega$, then $u(x,t) \geq 0$ for all $x \in \Omega$ and $t>0$.
So far I've tried using energy method, i.e. multiply by $u$ and integrate (then use $|f(u)| \leq K|u|$), which gives me some results of the $L^2$ norm of $u$. I wanted to prove $u_t >0$, but it may not work. What are the other possible directions?
 A: First suppose $v$ is a strict supersolution, i.e. $v_t - \Delta v > f(v)$,
with boundary and initial conditions $v > 0$. If $v$ ever failed to be positive,
by compactness and continuity it must do so for the first time at some interior point. At this point we would have:


*

*$v=0$ by continuity of $v$ (and thus $f(v) = 0$), 

*$\Delta v \ge 0$ because $v$ is at a spatial minimum, and

*$\partial_t v \le 0$ because this is the first time $v$ is not positive.


But putting these facts together would give $\partial_t v - \Delta v \le f(v)$, a contradiction. Thus we have shown that strict supersolutions stay positive.
Now, to make this work with $u_t - \Delta u = f(u)$,
we can study the peturbation $$v_{\epsilon}\left(x,t\right)=u\left(x,t\right) + \epsilon e^{2Mt}$$
for arbitrary $\epsilon > 0$. Since the function we've added is positive, this satisfies $v > 0$ at the initial time and on the boundary; and thanks to the assumption on $f$ we have $$|f(u) - f(v)| \le M|u - v| = \epsilon Me^{2Mt}. \tag{1}$$ Thus
$$\begin{align} \partial_t v - \Delta v &= \partial_t u - \Delta u + 2 \epsilon M e^{2 M t} \\
&= f(u) + 2 \epsilon M e^{2Mt} \\
&\ge f(v) - \epsilon M e^{2Mt} + 2\epsilon M e^{2Mt} > f(v); \tag{by 1}
\end{align}$$
so $v$ is a strict supersolution and thus $v_\epsilon > 0$ for every $\epsilon > 0$. Sending $\epsilon \to 0$, we conclude that $u \ge 0$.
