# A Problem in Evans' PDE

Problem 7 in §6.6 states as follows:

Let $u\in H^1(\mathbb{R}^n)$ have compact support and be a weak solution of the semilinear PDE $$-\Delta u+c(u)=f\,\,\text{ in } \mathbb{R}^n,$$ where $f\in L^2(\mathbb{R}^n)$ and $c:\mathbb{R}\to\mathbb{R}$ is smooth, with $c(0)=0$ and $c'\geqslant 0.$ Prove $u\in H^2(\mathbb{R}^n)$.

I am trying to prove $c(u(x))\in L^2(\mathbb{R}^n)$ or $c'(u(x))\in L^{\infty}(\mathbb{R}^n)$ but both failed. I know in fact $u$ can be in $H_0^1(\mathbb{R}^n)$, but I don't know whether this is useful here. And $c(u(x))$ has compact support since $u$ has compact support. However, again, I can't figure out that $u\in L^{\infty}(\mathbb{R}^n)$. And I really don't know how to use the condition $c'\geqslant 0$.

Anyone could help me? Any advice will be appreciated.

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Have you tried a Fourier transform approach? In Fourier space the problem boils down to prove $\mathscr{F}[c(u)] \in L^2(\mathbb{R}^n)$, which looks more manageable. What do you think? – Giuseppe Negro Dec 26 '11 at 11:01
@GiuseppeNegro: Thanks. I haven't tried this approach before. Acturally I'm not familiar with Fourier transform, I mean I know some basic facts about it but I've never done any exercises. Here I still do not know how to deal with the $c(u(x))$ under the integral. I think if I can deal with this integral I can also deal with the former approach. Do you have some more details? – Y.Z Dec 26 '11 at 13:24
@GiuseppeNegro Could you please add more details about the Fourier transform approach? Thanks! – Sherry Apr 10 '15 at 6:13
@Sherry: I am afraid I cannot answer right now. I don't remember what was my idea back in '11, and frankly, there is a big possibility that it was just plain wrong. – Giuseppe Negro Apr 10 '15 at 7:20

Here's my try.

The hint in Evans suggests to mimic the proof of the $H^2$ interior regularity theorem. This seems a very good hint, so let's try to follow it.

1. (Weak form of the problem + choice of a test function) Take a test function $v\in H_0^1(\mathbb{R}^N)$, multiply both sides of equation $-\Delta u+c(u)=f$ by $v$ and integrate by parts to get: $$\tag{1} \int_{\mathbb{R}^N} \langle \nabla u,\nabla v\rangle = \int_{\mathbb{R}^N} \Big( f-c(u)\Big)\ v$$ (recall that $\text{supt }u$ is compact, hence the integrals are actually over a suitably large ball). Now, choose $|h|>0$ small, an index $n\in \{1,\ldots ,N\}$ and set $v=-D_n^{-h}[D_n^h u]$ in (1), where: $$D_k^hu(x):= \frac{u(x+h\mathbf{e}^n)-u(x)}{h}\;$$ (we use the "double" difference quotient for we want estimetes for the second derivatives of $u$); for sake of simplicity, we rewrite the resulting expression as: $$I=J\; .$$
2. (On rewriting $I$) Recalling that: $$\int_{\mathbb{R}^N} \phi\ D_n^{-h}\psi = \int_{\mathbb{R}^N} \psi\ D_n^{h}\phi\quad \text{and}\quad D_n^h\phi_{x_i}=(D_n^h\phi)_{x_i}\; ,$$ we get: $$\begin{split} I &= -\int_{\mathbb{R}^N} \langle \nabla u, \nabla (D_n^{-h} D_n^h u)\rangle \\ &= -\sum_{i=1}^N \int_{\mathbb{R}^N} u_{x_i}\ (D_n^{-h}D_n^h u)_{x_i}\\ &= -\sum_{i=1}^N \int_{\mathbb{R}^N} u_{x_i}\ D_n^{-h}(D_n^h u)_{x_i}\\ &= \sum_{i=1}^N \int_{\mathbb{R}^N} D_n^hu_{x_i}\ (D_n^h u)_{x_i}\\ &= \int_{\mathbb{R}^N} \langle D_n^h \nabla u, D_n^h \nabla u\rangle \; ,\end{split}$$ i.e.: $$\tag{I} I=\lVert D_n^h \nabla u\rVert_2^2\; .$$
3. (Estimate for $J$) We have: $$|J|=\left| \int_{\mathbb{R}^N} -f\ D_n^{-h}D_n^h u +c(u)\ D_n^{-h}D_n^h u \right| \leq \int_{\mathbb{R}^N} \Big( |f|+|c(u)|\Big) |D_n^{-h}D_n^h u|\; ;$$ using FTIC, $c(0)=0$ and $c^\prime \in L^\infty$ we find: $$|c(u(x))|=\left| \int_0^{u(x)} c^\prime (t)\ \text{d} t\right| \leq \lVert c^\prime\rVert_\infty\ |u(x)|$$ hence plugging the latter inequality in the former we obtain: $$\tag{2} |J|\leq \int_{\mathbb{R}^N} \Big( |f|+\lVert c^\prime\rVert_\infty\ |u|\Big) |D_n^{-h}D_n^h u|\; .$$ Now, from Thm 3(i), §5.8.2 we get: $$\tag{3} \int_{\mathbb{R}^N} |D_n^{-h}D_n^h u|^2\leq C_1\int_{\mathbb{R}^N} |\nabla D_n^h u|^2 \leq C_2\int_{\mathbb{R}^N} |\nabla u|^2\; ,$$ thus from (2)-(3), Cauchy inequality and an elementary inequality we infer: $$\begin{split} |J| &\leq \int_{\mathbb{R}^N} \Big( |f|+\lVert c^\prime\rVert_\infty\ |u|\Big) |D_n^{-h}D_n^h u| &\qquad \text{[by (2)]}\\ &\leq \frac{1}{2} \int_{\mathbb{R}^N} \Big( |f|+\lVert c^\prime\rVert_\infty\ |u|\Big)^2 + \frac{1}{2} \int_{\mathbb{R}^N} |D_n^{-h}D_n^h u|^2 &\qquad \text{[by Cauchy's]}\\ &\leq \int_{\mathbb{R}^N}\Big(|f|^2+\lVert c^\prime \rVert_\infty^2\ |u|^2\Big) +\frac{C_2}{2} \int_{\mathbb{R}^N} |\nabla u|^2 &\qquad \text{[by (3)]}\; ,\end{split}$$ hence: $$\tag{J} |J|\leq C_3\ \left( \lVert f\rVert_2^2+\lVert u\rVert_2^2 +\lVert \nabla u\rVert_2^2\right)$$ for some suitable constant $C_3\geq 0$.
4. (Conclusion) Therefore from $I=J$, (I) and (J): $$\lVert D_n^h u_{x_i}\rVert_2^2 \leq \lVert D_n^h \nabla u\rVert_2^2\leq C_3\ \left( \lVert f\rVert_2^2+\lVert u\rVert_2^2 +\lVert \nabla u\rVert_2^2\right)\; ,$$ for $|h|>0$ small and each $i\in \{1,\ldots, N\}$. Thm 3(ii), §5.8.2 yields $u_{x_i}\in H^1(\mathbb{R}^N)$ and $\lVert \nabla u_{x_i}\rVert_2^2 \leq C_3\ \left( \lVert f\rVert_2^2+\lVert u\rVert_2^2 +\lVert \nabla u\rVert_2^2\right)$, therefore $u\in H^2(\mathbb{R}^N)$.

What do you think?

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Thanks for your details. But I still do not know why the inequality above (2) holds. Why $c'\in L^{\infty}$(this is my question, may seems obvious for you). I know $c(u(x))$ has compact support, say, $B_R(0)$, but for $c(t),t=u(x)$, I'm not sure $t$ is in a ball. – Y.Z Dec 27 '11 at 1:39
Sorry, I misread your post so $c^\prime \in L^\infty$ appeared as an assumption to my eyes... There should be a way to overcome this difficulty, but I have to think about it. – Pacciu Dec 27 '11 at 1:50

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 $$|\xi|^2\hat u(\xi)+\widehat{c(u(x))}=\hat f(\xi)$$

It follows from the first step that $\widehat {c(u(x))} \in L^2$. 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 $u \in H^2$.

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