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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!

Tony
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    If the derivative $c'$ is bounded then we get that $|c(u)|\leq |u|$ so that $c(u) \in L^2$ and so it's Fourier transform is in $L^2$. If $c'$ is not bounded however I don't see how to conclude: For example $c(t)=t^p$ for $p>n/(n-2)$ will have $c(v)\notin L^2$ for many $v\in H^1$ (this is not a counterexample though). – Jose27 Apr 10 '15 at 04:32
  • Could you ask your friend how he/she bypasses this problem? It would be interesting to know if such an argument can be made to work in general. – Jose27 Apr 12 '15 at 00:31
  • The Fourier transform can be defined as a bijective isometry in $L^2$, or more generally on tempered distributions, so the only problem is to conclude that $\mathcal{F}(c(u))\in L^2$. As I mentioned, I don't see how to do this without a bounded derivative of $c$. – Jose27 Apr 13 '15 at 03:12
  • @Jose27 I just don't understand why we can take Fourier transform of $-\Delta u+c(u)=f;;\text{in};\mathbb{R^n}$. Since $u$ is a weak solution, it doesn't satisfy this equation, it satisfies $B[u,v]=<f,v>$. – Sherry Apr 13 '15 at 12:25
  • It satisfies the equation in the sense of (tempered) distributions, which is enough to apply the Fourier transform. – Jose27 Apr 13 '15 at 17:05
  • Possible duplicate of this question – Winther Apr 17 '15 at 00:12

1 Answers1

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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 $u \in H^2$.

Sherry
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  • How do you get the term $c(\widehat{u(x)})$? $c$ is not a linear function. – Hans Apr 17 '15 at 00:20
  • @It is the Fourier transform of $c(u(x))$. I think it has nothing to do with linear. – Sherry Apr 17 '15 at 01:10
  • Sorry. I misread it as function $c$ evaluated at the Fourier transform of $u$. I suggest you write it as $\widehat{c(u)}(\xi)$. – Hans Apr 17 '15 at 01:21
  • @Hans Thanks! I have edited it. – Sherry Apr 17 '15 at 01:24
  • $c(u)$ is not continuous, since $u$ is a priori not so. I still think the argument for $\hat{c(u)}\in L^2$ requires something delicate. – Jose27 Apr 18 '15 at 04:47
  • @Jose27 What does $u$ is a priori not so mean? – Sherry Apr 18 '15 at 05:27
  • Sorry that was poorly written. What I meant was that we only know that $u$ is an $H^1$ function so that, if the dimension is $\geq 2$, we can't guarantee the boundedness or continuity of $u$. – Jose27 Apr 18 '15 at 06:58