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The Setup: Suppose $\Omega$ is a bounded, open, connected, simply connected subset of $\mathbb{R}^2$ with smooth boundary. Suppose that I am given a function $\Phi:\mathbb{R}^2\to\mathbb{R}$ and two continuous functions $f,g:\mathbb{R}\to\mathbb{R}$ satisfying $$\begin{cases}\Delta\Phi=f(\Phi),& \text{ if }(x,y)\in\Omega,\\ \Delta\Phi=g(\Phi),&\text{ if }(x,y)\in\text{Int}(\mathbb{R}^2\setminus\Omega).\end{cases}$$ Most importantly, $\Phi$ is of class at least $C^1(\mathbb{R}^2)$ (as I am not disregarding weak solutions, one can take $\Phi\in H^2(\mathbb{R}^2)$ for instance). In other words, the values of $\Phi$ on $\partial\Omega$ are chosen so that $\Phi\in C^1(\mathbb{R}^2)$.

The Question: Can we find a function $h:\mathbb{R}\to\mathbb{R}$ such that $\Delta\Phi=h(\Phi)$ for all $(x,y)\in\mathbb{R}^2?$

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  • $\begingroup$ I think I am mis-understanding the question. You specify that $\Phi$ is given, and then ask if we can impose regularity conditions on $\Phi$; and you ask for a function $h$, but without regularity conditions, so that it seems that defining $h$ to be the patch-together of $f$ and $g$ should trivially do it. $\endgroup$
    – LSpice
    Commented Sep 13, 2019 at 19:17
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    $\begingroup$ @LSpice: I don't think $h$ can be dependent on the coordinate in $\mathbb{R}^2$. So if there exists $z\in \Omega$ and $w\in \Omega^C$ such that $\Phi(z) = \Phi(w) = P$, but $f(P) \neq g(P)$, you cannot just "patch together" $f$ and $g$. // Your answer of course would work if $\Phi(\Omega)$ and $\Phi(\Omega^C)$ are disjoint. $\endgroup$
    – Willie Wong
    Commented Sep 13, 2019 at 23:00
  • $\begingroup$ @WillieWong, good point; indeed I was misunderstanding. $\endgroup$
    – LSpice
    Commented Sep 14, 2019 at 1:09
  • $\begingroup$ @WillieWong Exactly. One cannot patch $f$ and $g$ as they do not depend on the coordinates of $\Bbb R^2$. But that does not imply that $h$ doesn't exist. Also, I am investigating the case where $\Phi(\Omega)$ and $\Phi(\Omega^c)$ are not disjoint. $\endgroup$
    – UserA
    Commented Sep 14, 2019 at 4:12

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When $\Phi(\Omega)$ and $\Phi(\Omega^c)$ overlap, $h$ does not necessarily exist. Take for example $\Omega = B_1$ and $\Phi = (1-|x|^2)^2$. Then we may take $f(s) = 8(1-2\sqrt{|s|})$ and $g(s) = 8(1+2\sqrt{|s|})$. Since $f$ and $g$ disagree on $\Phi(B_1) \cap \Phi\left(\overline{B_1}^c\right) = (0,\,1]$, the function $h$ cannot exist.

(In particular, for each $s \in (0,\,1)$, $\Delta \Phi$ is different on the two circles that comprise the level set $\{\Phi = s\}$).

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