I am motivated by the following question:

Given a uniform heat source in a convex domain, and suppose that the outside temperature is equal to $0$, can we determine where the long-time temperature inside the domain changes fastest on the boundary?

Mathematically, we let $\Omega \subset \mathbb{R}^n$ be a smooth convex domain, and let $u$ be the function satisfying \begin{align} \begin{cases} -\Delta u=1,\quad &\mbox{in $\Omega$}\\ u=0\quad &\mbox{on $\partial \Omega$}. \end{cases} \end{align} The question is, how to determine the set $$\Big\{p \in \partial \Omega: \Big|\frac{\partial u}{\partial \nu}(p)\Big|=\max_{\sigma \in \partial \Omega}\Big|\frac{\partial u}{\partial \nu}(\sigma)\Big| \Big\}?$$ My intuition is that, the set should coincide with the points on the boundary with minimal mean curvature. However, I cannot prove this claim. Any comments or ideas are really appreciated.

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    $\begingroup$ Note that this is not a local problem, so I would not expect that the answer can be given in terms of local characteristics, such as curvature. To be specific: a tiny modification of the domain near an extremal point $p$ should not affect the answer much, but it can significantly change the curvature near $p$. $\endgroup$ May 5 '21 at 8:35
  • $\begingroup$ @MateuszKwaśnicki, Numerical results on ellipses do imply that the conjecture is true, at least on this class of domains, so I'm guessing this is valid for general convex domains. $\endgroup$
    – student
    May 5 '21 at 9:29
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    $\begingroup$ For an ellipse, $u$ is just a quadratic function, no need for numerical experiments. $\endgroup$ May 5 '21 at 9:33
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    $\begingroup$ and the symmetries of the ellipse can mislead the intuition. try experimenting with more generic domains, perhaps? $\endgroup$ May 5 '21 at 9:40
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    $\begingroup$ The fact that this holds for the ellipse is essentially due to the fact that in this case, the solution $u$ has constant Hessian. (Your ellipse can be written as the level set of some quadratic function. Up to a multiplicative scaling and a constant addition, that defining quadratic function is the solution to your PDE.) $\endgroup$ May 5 '21 at 13:18

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