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For a fixed $a\in \mathbb{R^n}$, we set
$$\Gamma(a,x)=\frac{1}{2\pi}log|a-x|\ \ \ \ \text{for $n=2$}$$ $$\Gamma(a,x)=\frac{1}{\omega_n(2-n)}|a-x|^{2-n}\ \ \ \ \text{for $n\ge 3$}$$ where $\omega_n$ denotes the surface are of the unit sphere in $\mathbb{R^n}$. Suppose $\Omega$ is a bounded open set in $\mathbb{R^n}$. Let $u\in C^1(\bar{\Omega})\cap C^2(\Omega)$.
Could any one give me some hints to prove that $f(x)=\Gamma(a,x)\Delta u(x)$ is Lebesgue integrable in $\Omega$. Thank you !

Omega
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  • Well if $\bar\Omega$ doesn't contain $a$ both functions are continuous over it, thus bounded, and so is the laplacian $\Delta u$, that is by hypothesis. So no problem. Otherwise I wouldn't be too sure the integrability holds. But you may try to view the integral as a limit of integrals over $\Omega$ minus a ball around $a$ and try estimating those and their limit by consequence. I think I've seen this done in class many times, this has to do with PDEs right? Fundamental solution of the Laplace equation I should say. Of course, I'm assuming you integrate in $dx$ and not $da$… – MickG Jun 02 '15 at 13:53
  • Yes. I am reading the book of Fanghua Lin. They do like you said, but I do not understand when they past to limit. If this function is not integrable, the Lebesgue dominated convergence theorem may not be applied. – Omega Jun 02 '15 at 14:00
  • What is that book's title? If the problem is in that book, maybe you could post more details about what passage is particularly problematic. Or I could find the book and the passage. – MickG Jun 02 '15 at 14:05
  • Elliptic Partial Differential Equations by Quin Hang and Fanghua Lin. I am reading theorem 1.17, section 1.3 Fundamental solutions. – Omega Jun 02 '15 at 14:07
  • Google book link. Screenshot of problematic part. – MickG Jun 02 '15 at 14:22

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The only issue to check is the origin singularity at $a \in \Omega$. So take the ball centred at $a$ with radius $\epsilon$, call it $B$. Use Holder's inequality to notice $$ \int_B |f(x)| dx \leq C(B,u) \int_B \Gamma(a,x) dx$$ where $C$ is a constant depending on $\Omega$ and $u$. Swapping to spherical coordinates around $a$ we see $dx = r^{n-1} dr d\Theta$, where $d \Theta$ is the spherical contribution. Thus $$ \int_B \Gamma(a,x) dx = \int_{S^{n-1}} \int_0^\epsilon \frac{r}{\omega_n (2-n)} dr d \Theta < \infty $$ for $n \geq 3$. You can now check the $n=2$ explicitly if you like.

Jeb
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  • If $C(B,u)$ is the maximum of $|u|$ over $B$, I'd say there is no need for Hölder's inequality. That equality says $\int|fg|\leq\sqrt{\int|f|^2}\sqrt{\int|g|^2}$, IIRC. Here we don't have that right side. – MickG Jun 02 '15 at 14:25
  • And Wikipedia agrees with my memory :). – MickG Jun 02 '15 at 14:26
  • Yeah... I'm thinking $||fg||1 \leq ||f||\infty ||g||_1$, I'm assuming that is something that Han Duy has seen. You're thinking Cauchy Schwarz – Jeb Jun 02 '15 at 14:27
  • Oh. That is a generalised version of the Hölder inequality with $p=\infty,q=1$. I usually do not think of Hölder with $p=\infty$ :). My bad. – MickG Jun 02 '15 at 15:07
  • $\Delta u$ is in $C(\Omega)$, so it is bounded in the ball $B\subset\subset \Omega$. Therefore, we do not need to use the Holder inequality in this case. – Omega Jun 03 '15 at 01:12
  • How about the domain $\Omega\setminus B$. There is nothing to ensure that $\Delta u$ is bounded in this domain. So, how do we prove that $\Gamma\Delta u$ is integrable in this domain ? – Omega Jun 03 '15 at 01:19
  • By assumption, $\Omega$ is bounded.... thus $$\int _{\Omega \setminus B} |f| dx \leq C(B,\Gamma,u) \text{Vol}(\Omega \setminus B)$$ (have a look at integration by parts if you want to explicitly compute the constant ) – Jeb Jun 03 '15 at 13:28
  • I think your answer is not true. The function $f$ is only bounded when it is continuous up to $\partial\Omega$. In our case, we do not have that because $\Delta u$ is only in $C(\Omega)$. – Omega Jun 03 '15 at 14:41
  • We only need $u \in H^1( \Omega)$ for the integration by parts to go through. You have more than enough regularity to deduce my claim. – Jeb Jun 03 '15 at 15:05
  • Could you please explain more detail about the integration by part that you are mentioning ? Thank you – Omega Jun 03 '15 at 15:38