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May 8, 2019 at 9:34 comment added Liviu Nicolaescu Think of the special situation in my answer when $\Omega$ is the unit disk in the plane and f$u$ is radially symmetric. Look at the Taylor expansion in $r$ near $r=1$, $$u=u(1)+\partial_r(1)(r-1)+\partial^2_ru(1)(r-1)^2/2+O(\;(r-1)^3\;),$$ where $|r-1|={\rm dist}\;(x,\partial \Omega)$, $\partial_r=\partial_\nu$.
May 8, 2019 at 9:27 comment added Liviu Nicolaescu Think Taylor expansion in normal direction.
May 8, 2019 at 7:01 comment added user139845 Why do you need that?
May 8, 2019 at 0:35 comment added Liviu Nicolaescu You also need $\frac{\partial^2 u}{\partial \nu^2}\neq 0$.
May 8, 2019 at 0:33 history edited Liviu Nicolaescu CC BY-SA 4.0
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May 7, 2019 at 22:23 comment added user139845 For example, what if we require also $\nabla u = 0$ on $\partial \Omega$?
May 7, 2019 at 20:16 comment added user139845 Is there any way to recover the square under additional assumptions?
May 7, 2019 at 20:16 history edited Liviu Nicolaescu CC BY-SA 4.0
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May 7, 2019 at 20:07 history answered Liviu Nicolaescu CC BY-SA 4.0