TITLE: An interesting semi-linear PDE problem
QUESTION [6 upvotes]: Assume $u\in H^1(\mathbb{R}^n)$ has compact support, and assume that it is a weak solution of the semi linear equation
  $$
-\Delta u+c(u)=f\;\;\text{in}\;\mathbb{R^n}
$$
  where $f\in L^2(\mathbb{R^n})$ and $c:\mathbb{R}\to\mathbb{R}$ is a smooth function with $c(0)=0$ and $c'\ge 0$. Prove that $u\in H^2(\mathbb{R^n})$.

I know exactly how to prove this following the hint in textbook by "difference quotient" method. However, my friend told me it can be proved by Fourier transformation and don't assume $c' \ge 0$. 
Any hint? Thanks!

REPLY [2 votes]: Thanks for Jose27's help, I finally figured this out.
First step:
Since $u$ has compact support and $c(0)=0$, and $c(x)$ is continuous, $c(u(x))$ is continuous and has compact support and so it is in $L^2(R^n)$.
Second step:
Take Fourier transform of $$-\Delta u+c(u(x))=f(x)$$
Then you can get $$|\xi|^2\hat u(\xi)+\widehat{c(u)}(\xi)=\hat f(\xi)$$
It follows from the first step that $\widehat{c(u)}(\xi) \in L^2$. Thus $|\xi|^2\hat u(\xi) \in L^2$
We also know $\hat u \in L^2$
So $(1+|\xi|^2)\hat u(\xi) \in L^2$
It follows that $u \in H^2$.