Prove the estimate $|u(x,t)|\le Ce^{-\gamma t}$ Assume that $\Omega \subset \Bbb R^n$ is an open bounded set with smooth boundary, and $u$ is a smooth solution of
\begin{cases}
u_t - \Delta u +cu = 0  & \text{in } \Omega \times (0, \infty), \\
u|_{\partial \Omega} = 0, \\
u|_{t=0} = g
\end{cases}
and the function $C$ satisfies $c \ge \gamma \ge 0$ for some constant $\gamma$. Prove the estimate $|u(x,t)|\le Ce^{-\gamma t}$ for all $x \in \Omega, t \in [0,T]$ for any fixed $T > 0$.
Could someone please explain me how can I get the absolute value in the estimate? If I use energy estimate then I'll have the $L_2$ norm...
Any help will be much appreciated!
 A: First assume that $c \ge 0$ and show $\max u \le \max (g,0)$ and $\min u \ge \min (g,0)$.
Then apply this maximum principle to $\tilde{u} = ue^{-\gamma t}$ for $c \ge \gamma$.  
For strictly positive $c$, any interior or final-time max would give $u_{t} \ge 0$, $\Delta u \le 0$ which implies $cu \le 0$ and, thus, $u \le 0$.  Using $u=0$ at $\partial \Omega$ gives that either the $\max u$ is achieved at $t=0$ or else $\max u \le 0 \le \max (g,0)$.  A similar argument works for the $\min u$.
Now, $\tilde{u} = ue^{-\gamma t}$ solves $\tilde{u}_{t} - \Delta \tilde{u} + (c-\gamma)\tilde{u} = 0$.  So applying the maximum principle:
$\min (g,0) \le \tilde{u} \le \max (g,0)$.  
Therefore,
$\left| u(x,t) \right| \le Ce^{-\gamma t}$
with $C = \max \{\left| \max (g,0) \right|, \left| \min (g,0) \right|\}$.  
A: Let me elaborate on my comment that it is indeed possible to do this via energy methods. Let $p \geq 2$ be an even integer and differentiate under the integral and apply the chain rule to get
$$\partial_t \|u(t)\|_p^p = \int_\Omega \! p|u|^{p-1}u_t \, dx = p\int_\Omega \!|u|^{p-1}(\Delta u - cu) \, dx$$
Now integrate by parts:
$$= -p\int_\Omega \! |u|^{p-2} (\nabla u \cdot \nabla u) - p\int_\Omega \! c|u|^{p-1}u \, dx $$
The first term is less than or equal to $0$ and $c \geq \gamma \geq 0$. So
$$ \le 0 -p\gamma \int_\Omega \! |u|^p \, dx$$
Thus we have shown
$$\partial_t \|u(t)\|_p^p \le - p\gamma\|u(t)\|^p.$$
By Gronwall's inequality this implies
$$\|u(t)\|_p^p \le e^{-p\gamma t}\|u(0)\|_p^p$$
and taking $p$-th roots and inserting the initial condition
$$\|u(t)\|_p \le e^{-\gamma t}\|g\|_p.$$
Since $u$ is a priori bounded, we can take the limit as $p \to \infty$ to get
$$\|u(t)\|_\infty \le \|g\|_\infty e^{-\gamma t}.$$
This is what you want to prove with $C = \|g\|_\infty$.
