1
$\begingroup$

If we have the heat equation on a simply connected (and smooth) domain $\Omega$ with boundary $\Gamma$ with Robin boundary conditions, so that: $$ \epsilon u_t - \Delta u = -1 \quad on \quad \Omega \\ \nabla u \cdot n + u = Q \quad on \quad \Gamma \\ u(0) = u_0 \quad on \quad \Omega \cup \Gamma. $$ Then can we get an $L^p (0, T; L^2(\Omega))$ bound on $u$ that does not depend on $1/ \epsilon$? (Here $p > 0$, and maybe $\infty$). An $L^p (0, T; L^2(\Gamma))$ bound would be just as good.

If I follow usual energy techniques, then I can use Gronwall's inequality to give a bound, but it depends on $1 / \epsilon$. I don't want it to depend on $\epsilon$, as I will later be taking $\epsilon \rightarrow 0$.

Thanks!

$\endgroup$
  • $\begingroup$ I don't think you will get anything better than $p=\infty$. Note that you can remove the parameter $\varepsilon$ by rescaling time in your equation. So any estimate you will get independent of $\varepsilon$ will also need to hold in a rescaled version. Using this and the scaling behaviour of $L^p$ for $p< \infty$ you can probably derive a contradiction. $\endgroup$ – mlk Jul 10 '17 at 14:10
  • $\begingroup$ $p = \infty$ would be fine. Re-scaling time is a good idea, however I guess that it would only hold for $[0, T_\epsilon]$, with $T_\epsilon \rightarrow 0$ as $\epsilon \rightarrow 0$? $\endgroup$ – Josiki Jul 10 '17 at 14:48
  • $\begingroup$ You are right, I kind of only had $T=\infty$ in my head, for some reason. But I think the limit is the other way round, assuming $u$ is a solution until time $T$, then $v(t,x) = u(\varepsilon t,x)$ is a solution for $\varepsilon = 1$ until the time $T_\varepsilon = T/\varepsilon \to \infty$ for $\varepsilon \to 0$. Effectively the derivative $u_t$ is of order $1/\varepsilon$, so rescaling kind of slows the evolution down. $\endgroup$ – mlk Jul 10 '17 at 18:31
  • $\begingroup$ Yes, I think I agree! Although I think I would then have $v_t - \Delta v = -1$ on $[0, T_\epsilon ]$. If I then use the usual energy techniques, and Gronwall's inequality, then I would get something like: $$\sup_{[0, T]} ||u||_{L^2}^2 \leq \sup_{[0, T_\epsilon ]} || v ||_{L^2}^2 \leq C T_\epsilon , $$ and the right hand side goes to $\infty$ as $\epsilon \rightarrow 0$. Am I missing something? $\endgroup$ – Josiki Jul 11 '17 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.