Existence for $-\Delta u + (g(x)-\Delta u)^+\varphi(u) = f(x)$

Let $\Omega$ be a smooth bounded domain. Consider the equation $$-\Delta u + (g(x)-\Delta u)^+\varphi(u) = f(x)$$ $$u|_{\partial\Omega} = 0$$ where $f,g$ are smooth functions on $\Omega$ and $\varphi$ is a real valued map which is such that $0 \leq \varphi \leq M$ for some constant $M$.

How do I show existence for this equation?

My problem here is the positive-part term there. I tried to apply a Tikhonov fixed point approach in the spaces $H^2$ to the PDE $$-\Delta u + (g(x)-\Delta w)^+\varphi(u) = f(x)$$ but this requires some weak continuity: if $w_n \rightharpoonup w$ in $H^2$, I would have to show that the corresponding solutions $u_n \to u$ in some space, where $u$ solves the limit problem. But I cannot say anything about the weak convergence of the $(\cdot)^+$ term.

Any existence in Holder or Sobolev spaces would be helpful. Thanks

• Is $\varphi$ at least continuous? – fedja Jun 2 '18 at 1:33
• @fedja Yes, it is a smooth function. – C_Al Jun 2 '18 at 11:53
• This is just a partial answer, let me come back to this problem in a short while. I assume that $f$ is non-negative. First of all, remark that from the strong maximum principle we have $u > 0$ on $\Omega$. Conversely, if $v$ denotes the solution to \begin{eqnarray} -\Delta v &=& f(x)~\text{on } \Omega\\ v &=& 0~\text{on } \partial\Omega, \end{eqnarray} we have $u \leq v$ once again from the maximum principle. – Romain Gicquaud Jun 4 '18 at 8:17
• @romaingicquaud thanks for your attention but how does that help? – C_Al Jun 5 '18 at 8:12
• @C_Al: the usual strategy in such a context is to construct barriers (aka sub and super-solutions) and try to set up an iteration method. Yet, it is not clear whether a maximum principle holds for this equation. I was unable to go further. – Romain Gicquaud Jun 7 '18 at 14:42