Hello,


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.



Maximum principle corner
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. 


