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Assume $u_0\in L^1\cap L^\infty (\mathbb{R}^d)$ and consider the equation without diffusion $$\frac{\partial u}{\partial t}=-u^p,\;\;\;t\geq 0, x\in \mathbb{R}^d.$$ Show that $$\int_{\mathbb{R}^d}u(t,x)dx\rightarrow 0, \mbox{ as }t\rightarrow \infty.$$

So far I tried to use maximum principle to prove that $u\in L^\infty$, for every $t>0$. For that purpose I think is enough to consider the problem $u_t=0,\,u(x,0)=u_0(x)$. Then I want to use Gronwall using the $L^\infty$ bound, but I haven't been able to do so. Please help!

Charles
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    Hi: You need to add your work and get us to the point where you are stuck. If you don't, this is likely to attract closure votes. You will get better answers in general if you include your work. – rschwieb Dec 31 '13 at 16:05
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    Thank you. I put my work above. – Charles Dec 31 '13 at 16:12
  • Since there is no mention of $\frac\partial{\partial x}$, it is an ODE. It can be very easily explicitly solved using first semester ODE techniques. – Stephen Montgomery-Smith Dec 31 '13 at 16:48
  • I'm not sure about that, because you need a condition of integrability of the solution in the $x$ variable, and if you just solve the ode, you won't end up with a solution integrable. – Charles Dec 31 '13 at 19:29
  • And also, the solution has to be compatible with the initial condition, which depends in the variable $x$. – Charles Dec 31 '13 at 19:31
  • When you solve the ODE, the answer includes a constant. That is the quantity that will depend upon $x$. – Stephen Montgomery-Smith Dec 31 '13 at 19:35
  • Following that the solution has the form: u_0(x)f(t), where f is the solution of the ODE. But, you need u_0(x)=-u_0(x)^p in order to has a solution of the problem? – Charles Dec 31 '13 at 19:41
  • Charles - when replying to me, put in @StephenMontgomery-Smith. That way I'll get a notification of your message. Otherwise I might not see it. – Stephen Montgomery-Smith Dec 31 '13 at 19:49
  • Charles - if the solution had the form $u_0(x) f(t)$ (which it would in the case $p=1$), then you could argue $\int u(x,t) , dx = f(t) \int u_0(x) dx$. – Stephen Montgomery-Smith Dec 31 '13 at 19:55
  • @StephenMontgomery-Smith thank you! – Charles Dec 31 '13 at 21:19

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The formula $u^p$ is not meaningful without assuming that $u\ge 0$ or that $p$ is an integer. I assume $u\ge 0$, because otherwise the result isn't true (take $u_0 $ negative; the solution of $u_t=-u^2$ grows in absolute value).

The ODE $u_t=-u^p$ is autonomous with the only equilibrium at $0$. Since $u_t<0$ when $u>0$, solutions with positive initial value decrease to $0$. (I'd like to point out that it is not necessarily to solve the ODE to arrive at this conclusion.) Thus, $u$ converges to $0$ pointwise as $t\to\infty$.

Since $u(\cdot,t)\le u_0$ for all $t$ and $u_0\in L^1$, the dominated convergence theorem applies: the integral of $u$ also tends to $0$.

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