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This question is motivated by an inverse coefficient problem, for which it is useful to find solutions to a particular PDE so that the gradient of the solution does not vanish at all, or at least too often.

Suppose $\Omega \subset \mathbb{R}^d$ with $d \ge 2$ is a bounded open set with at least $C^1$ boundary. Suppose $f \colon \partial \Omega \to \mathbb{R}$ is at least $C^1$ smooth.

Consider the $p$-conductivity equation $$ \nabla \cdot (\sigma |\nabla u|^{p-2} \nabla u) = 0 $$ with the Dirichlet boundary conditions $f$. Suppose that $\Omega$, $f$ and $\sigma$ are smooth enough so that we have $u \in C^1(\overline{\Omega})$. Here $\sigma \in L^\infty(\Omega)$ is bounded from below by a positive constant; $\sigma$ is called the conductivity. We suppose it is continuous, at least near the boundary.

When can we say $\nabla u = 0$ only rarely on $\partial \Omega$? What conditions must $f$ satisfy? To be more precise, I would like the set where $\nabla u \neq 0$ (intersected with $\partial \Omega$) to be dense in $\partial \Omega$.

My best guess is that if I select nonconstant boundary values $f$, then $\nabla u$ should vanish only rarely. If $p = 2$ and everything is smooth, then unique continuation over a hypersurface suggests this.

Another approach would be to consider $\nabla f$ restricted to $\partial \Omega$ and argue that, for general (generic?) boundary values $f$, this gradient rarely vanishes. Is there a way to state something like this formally, and is this even true?

The weakest relevant statement would be to prove that there exist boundary values with the desired property. This seems to be true, at least.

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  • $\begingroup$ I am not sure I understand your question. The tangential derivatives along $\partial \Omega$ can be computed using only $f$, right? thus a nonconstant $f$ gives a nonzero $\nabla u$ near the boundary $\endgroup$ – Piero D'Ancona Oct 27 '17 at 4:28
  • $\begingroup$ @PieroD'Ancona Yes, unless the boundary of the set happens to exactly follow a level set of the function. Can you guarantee this does not happen? $\endgroup$ – Tommi Brander Oct 27 '17 at 5:16
  • $\begingroup$ Intuitively, it seems that this is almost always the case, but I would like to know if this intuition can be made formal. $\endgroup$ – Tommi Brander Oct 27 '17 at 5:16
  • $\begingroup$ what do you mean? if $u=f$ at the boundary and $f$ is nonconstant, then the boundary is not a level set of $u$ $\endgroup$ – Piero D'Ancona Oct 27 '17 at 12:42
  • $\begingroup$ @PieroD'Ancona Under what assumptions can you guarantee that such an $f$ exists? Is there a sense in which almost all boundary functions have that property? (Also, at minima and maxima of $f$ the gradient could still be zero even for nonconstant $f$.) $\endgroup$ – Tommi Brander Oct 27 '17 at 13:35

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