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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!

Pukki
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  • You can use energy methods to get an estimate for the $L^p$ norm for all $p > 1$. Then take the limit as $p \to \infty$ to estimate the supremum. This idea is explained in detail in the (excellent) book Partial Differential Equations by J. Rauch. –  Nov 15 '13 at 01:07
  • @brom Thank you! But if I use enegry estimate, then I'll get the norm of g on the right hand side... – Pukki Nov 15 '13 at 01:44
  • There may be a typo on the definition of $C$ and what $C$ satisfies. – user37238 Nov 15 '13 at 13:24

2 Answers2

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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 $C = \max \{\left| \max (g,0) \right|, \left| \min (g,0) \right|\}$.

jpb
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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 $C = \|g\|_\infty$.