Consider the heat equation $\frac{\partial u}{\partial t}=\frac{1}{2}\Delta u$ in a bounded domain (say the interval [0,$\pi$]) with boundary condition $$\frac{\partial u}{\partial \vec{n}}=u^2$$ with a non-negative initial condition $f$, where $\vec{n}$ is the inward unit normal.

How fast will $u(t,x)$ decay? Is it exponential in $t$ (i.e. $u(t,x)\leq C_1e^{-C_2t}$), as in the Dirichlet or the Robin boundary condition? Or does it decay like a polynomial? Thanks!

  • $\begingroup$ Is it known that such a solution exists (and for all $t>0$)? $\endgroup$ – Andrew Feb 13 '14 at 9:26
  • $\begingroup$ Yes, at least in the following weak sense: $u(t,x)$ is the unique element in $C([0,T]\times \bar{D})$ which solves the following integral equation: $$u(t,x)= P_tf(x)- \int_0^t \int_{\partial D}p(t-s,x,y)u^2(s,y)d\sigma(y)ds.$$ This can be checked using a Gronwall type argument. $\endgroup$ – Fantastic Feb 13 '14 at 10:52
  • $\begingroup$ Have you attempted a literature search? $\endgroup$ – Michael Renardy Feb 13 '14 at 20:17

For the inward pointing unit normal, it seems we arrive at the ($L^2(\Omega)$) energy identity, $$\frac{d}{dt}\|u\|^2 + \|\nabla u\|^2 + \int_\Gamma u^2 u dx =0.$$ Assuming the "solution" $u$ is nonnegative and continuous in time, we rewrite the boundary integral using the MVT for Integrals (a suitable version of this) as follows: for each $t>0$, there is $\xi(t)\in\Gamma$ in which $$\int_\Gamma u^2(t,x) u(t,x) d\sigma=u(t,\xi(t))\int_\Gamma u^2(t,x)d\sigma.$$ Hence, $$u(t,\xi(t))=\frac{\langle u^3(t),1 \rangle_\Gamma}{\|u(t)\|^2_\Gamma}.$$ The identity now reads $$\frac{d}{dt}\|u(t)\|^2 + \|\nabla u(t)\|^2 + u(t,\xi(t))\|u(t)\|^2_\Gamma =0.$$ Let $\lambda>0$ be the best constant so that the Poincare/Sobolev-inequality holds, $$\lambda\int_\Omega u^2 dx \leq \int_\Omega |\nabla u|^2dx + \int_\Gamma u^2 d\sigma.$$ Define $m(t):=\min\{1,u(t,\xi(t))\}\geq0$. Considering the above, we arrive at the estimate $$\frac{d}{dt}\|u(t)\|^2+\lambda m(t)\|u(t)\|^2\leq 0.$$

  • $\begingroup$ He specified that n is the inward normal, so you have the opposite sign. That is, the heat flux is outward, and one should expect decay rather than blowup. $\endgroup$ – Michael Renardy Feb 13 '14 at 20:03
  • $\begingroup$ I think the calculations are correct, though It does NOT tell us the decay rate. Now we have $$\|u(t)\|^2\leq C\,\exp^{-\lambda \int_0^tm(s)ds}$$ Unless we know a lower bound for $m(t)$, it is as hard as the original question. Thanks for the elegant calculation though. $\endgroup$ – Fantastic Feb 14 '14 at 3:25

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