Interesting counterexample of Strong Maximum Principle Suppose $U=\Omega \cup (0,T)$ where $\Omega$ is a bounded domain. Let $u \in C_1^2 (U) \cap C(\overline U)$ satisfy $$u_t \le \Delta u+cu$$ in $U$ where $c \le 0$ is a constant. 
If $u \ge 0$, then $u$ attains maximum on the parabolic boundary of $U$.
But when $u<0$, it doesn't hold. 
In the $u<0$ case, can anyone give a counterexample? Thanks so much! 
 A: WLOG assume $0\in \Omega$. Consider $u(x,t)=-|x|^2 + at -b$, where $a,b \geq 0$ to be chosen. We calculate $\partial_tu -\Delta u = 2+ a$ so that the differential inequality becomes 
$$
2+ a \leq c(-|x|^2 +at -b), \qquad x\in \Omega,\ t\in (0,T).
$$
Since $c\leq 0$ we have that the RHS of our inequality is bounded below by $caT -b$, so that we need
$$
2+a \leq caT +b,
$$
but this clearly can be achieved, for any fixed $a>0$, taking $b$ large enough. Assume then that $a,b$ have been so chosen, then, by possibly making $b$ larger, we can guarantee that $u<0$ since $\Omega$ is bounded. For any fixed $t$ the maximum of $u$ is attained at $0$ and $u(0,t)=at-b$ which is an increasing function of $t$, and so the maximum is attained in $\Omega\times \{ T\}$.
Edit: As intuition think of the elliptic case: The differential inequality becomes $-\Delta u\leq cu $ and any function like $-|x|^2 - b$ satisfies this inequality for $b$ big enough. This furnishes a counterexample to the strong maximum principle for the parabolic case, since $\partial_t=0$. To treat the weak maximum principle we just add an increasing function of $t$ and tweak it to be compatible with the conditions given.
