**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 $(x,y)\in\mathbb{R}^2?$



"Combining" two differential equations into one
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 $\{\Phi = s\}$).


