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consider a parabolic boundary value problem, for instance

$-\partial_tu+\Delta u=0$, in $\Omega$,

$\partial_\nu u=0$ on $\partial\Omega$,

in a domain $Q=(0,T)\times\Omega$, where $\Omega\subset R^n$ is bounded. Now suppose $u$ is smooth and attains its maximum in $(T,x_0)$ for some $x_0\in\partial\Omega$. Is it possible to say anything about the normal derivative $\partial_\nu u(T,x_0)$ similarly as in Hopf's boundary point lemma?

(Hopf's boundary point lemma is not applicable, since there is no circle tangent to $\partial Q$ at $(T,x_0)$ entirely contained in $Q$!)

I came across this problem by the following thought:

By classical maximum principles the maximum can't be attained in $\Omega$, so it has to be located on the boundary. What if the maximum is at the (in time) top end? Although the initial distribution is strictly negative, it could be positive without violating the boundary condition $\partial_\nu u=0$ on $\partial\Omega$, so there would be a change in signs. Of course one can exclude this by applying positive operator theory, but is there an elementary proof for this?

I would be grateful for valuable comments on this.

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3 Answers 3

up vote 1 down vote accepted

Pick 0< $\tau$ < T, and let $$\phi(\tau)=\max_{x\in\bar\Omega,0\le t\le\tau} u(x,t).$$ If this maximum is assumed at any point where $x\in\Omega$ and $t>0$, then $u$ must be constant. Suppose the maximum is at a point $(x_0,\tau)$, where $x_0\in\partial\Omega$, and $u$ is not constant. Then the function $\phi(\tau)$ must be increasing. This is possible only if $u_t(x_0,\tau)\ge 0$. But then the elliptic Hopf lemma gives a contradiction to the boundary condition.

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Nice idea, haven't thought of using the elliptic Hopf lemma! In the meantime I found out that the parabolic Hopf lemma still applies in my situation. Instead if the ball K in the proof of Thm 3 on page 170 of WP one can also choose a cylinder $(0,\tau)\times\Omega)$ and consider the body enclosed by the wall of that cylinder, the ball around the maximum on the boundary from the proof and the plain $\{\tau\}\times\Omega$. Using a modified auxiliary function, one still can prove that $\partial_\nu>0$ (or <0 depending on whether you want positivity or negativity). Thank you very much! –  Marc May 6 '11 at 7:32

Yes, there is an analogue to Hopf lemma. In most general form (degenerate operators...) is it done in the works of Kamynin and Khimchenko. Boundary required to be smooth enough, e.g. from parabolic class $C^{1,\alpha}$. That's an analogue for Lapunov class for elliptic problems. For $C^1$ boundaries Hopf lemma doesn't hold.

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I think what you are looking for is the parabolic Hopf lemma, given for instance in the book of Protter and Weinberger.

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You are right. However the parabolic Hopf lemma in PW requires a circle tangent to $\partial Q$ through $(T,x_0)$ which lies entirely in $Q$. So I am looking for a suitable extension of this result for corners. –  Marc May 4 '11 at 19:10
There seems to be no exactly such result for domains with corners. For example, function $f(x,y)=xy$ in the positive angle $\{x>0,y>0\}$ satisfy Laplace equation (stationary heat) equation. It is smooth and has a local minimum at the origin, but its derivatie in any direction is zero. –  Andrew May 5 '11 at 9:39
For domains of $C^1$ class there is a result of Nadirashvili, however. It states that if $x_0$ is a point of solution's local minimum at the boundary then in any neibourhood there is a point in which normal derivative is positive. For parabolic equations it was generalized by Kamynin. Since domains with angles is sort of good Lipschitz, may be something analogous holds there. For the above example it does :) –  Andrew May 5 '11 at 9:42

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