Suppose $U=\Omega \times (0,\infty),$ $\Omega$ is a bounded domain in $\mathbb{R}^n$ and $u \in C^{2,1}(\overline{U})$ satisfies $u_t=\Delta u - u^3$ in $U$ with $u(x,t)=0$ on $\partial \Omega \times (0,\infty).$ Construct an energy functional and use it to show that $u(x,t) \rightarrow 0$ as $t \rightarrow \infty.$

I haven't learnt constructing such a functional. I'm studying for my preliminaries. Any help is much appreciated.

  • $\begingroup$ What is the $\delta$ in the equation? $\endgroup$ – Harald Hanche-Olsen Sep 6 '16 at 19:56
  • $\begingroup$ My mistake. That's the Laplacian. $\endgroup$ – user358174 Sep 6 '16 at 20:42
  • $\begingroup$ Do you know a way to do this in the case of the heat equation? $\endgroup$ – Ian Sep 6 '16 at 20:46
  • $\begingroup$ @Ian I do because when the equation is given in terms of (say) $u_t-ku_{xx}=0$ for $0 \leq x \leq l,~t >0$ we do have the forumula $E[u](t)=\frac{1}{2} \int_{0}^{1} [u(x,t)]^{2}~dx.$ $\endgroup$ – user358174 Sep 6 '16 at 21:16
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    $\begingroup$ The intution that @Ian is driving at here, is this: You may think of the equation as a combination of the heat equation $u_t=\Delta u$ (which you already know how to handle) and $u_t=-u^3$, which is actually an ODE. You can solve the latter explicitly, but more importantly, since the RHS has the opposite sign of $u$, it will drive the value of $u$ towards zero. In particular the term $-u^3$ will also help to decrease the usual energy functional. For details, see the answer that Ian provided. $\endgroup$ – Harald Hanche-Olsen Sep 7 '16 at 11:29

If $E[u](t)=\int_\Omega |u(x,t)|^2 dx$ then

$$\frac{dE}{dt}=2 \int_\Omega u(x,t) u_t(x,t) dx = 2 \int_\Omega u(x,t) (\Delta u(x,t) - u(x,t)^3) dx \\ = -2 \int_\Omega |\nabla u(x,t)|^2 dx - 2 \int_\Omega u(x,t)^4 dx \leq 0$$

with equality iff $u(\cdot,t)=0$ (strictly speaking if $u(x,t)=0$ for a.e. $x$, but you have enough regularity that this just means $u(\cdot,t)=0$).

  • $\begingroup$ thanks again. If I ask one more question, the problem says $u(x,t) \rightarrow \infty$ as $t \rightarrow \infty,$ but what you've proved here is $u(x,t) \equiv 0.$ Should'nt the question have worded differently ? $\endgroup$ – user358174 Sep 7 '16 at 21:36
  • $\begingroup$ I haven't, actually. I showed that $\frac{dE}{dt} \leq 0$ and that $\frac{dE}{dt}<0$ unless $u=0$. You now still need to show that $E \to 0$ (which is not immediate from here, it requires some estimation still) so that you can conclude that $u \to 0$. $\endgroup$ – Ian Sep 8 '16 at 0:52

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