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I am trying to solve a problem, but am really confused on what I'm supposed to prove. The question is:

Prove in $2$D that $\Delta \log(|x|)=2\pi\delta_0$ in the distributional sense.

Now, let $x=\langle\alpha,\beta\rangle$. Then I figured out: $$\langle\Delta\log(|x|),\phi\rangle = \iint_{\mathbb R^2}\log(\sqrt{\alpha^2+\beta^2})(\phi_{\alpha\alpha}+\phi_{\beta\beta})d\alpha d\beta$$ But where am I supposed to go from here? Do I use integration by parts?

Lê Thành Đạt
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Felicio Grande
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  • In general if $d$ is a distribution then $\langle \Delta d,f \rangle = \langle d,\Delta f \rangle$; this is formally justified by integration by parts but is really just a definition. So you need to show that if $f$ is smooth with compact support then $\int_{\mathbb{R}^2} \log(|x|) \Delta f(x) dx = 2\pi f(0)$. – Ian Nov 13 '17 at 01:35
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    One way is to write it as an integral over a disk centered at the origin plus an integral over the complement of this disk, then apply the divergence theorem, then shrink the disk to be arbitrarily small. (Note by the way that this is done in detail in textbooks e.g. Evans PDE.) – Ian Nov 13 '17 at 01:37
  • I think I understand how to show that the integral over the disk goes to 0, but I dont understand stil how to apply the DT to the double integral of $log(|x|)\cdot\Delta\phi$ to get $2\pi\phi(0)$? – Felicio Grande Nov 13 '17 at 02:31
  • The integral over the disk does not go to zero, this is part of the point. The integral over the complement of the disk is given only by a certain integral over the boundary of the disk (because integration by parts moves the Laplacian over to the logarithm again, and the Laplacian of that logarithm is zero away from the origin). – Ian Nov 13 '17 at 02:33
  • @Felicio Grande Just a tip for the future: It is rarely advisable to use abbreviations because you might be the only one who understands what is meant by it. I mean... It is you who wants the problem to be solved, right? So, you should do everything possible to be understood. – Friedrich Philipp Nov 13 '17 at 02:36
  • which abbreviation have I used? @FriedrichPhilipp – Felicio Grande Nov 13 '17 at 02:37
  • You have not written much. So,it is actually pretty easy to spot it, but ok. It's DT. – Friedrich Philipp Nov 13 '17 at 02:38
  • @Ian I'm very confused, I have read up on this proof, and the way it is constructed in one of the textbooks I am looking at separates the double integral to domains $D(0,\epsilon)$ and $\mathbb R^2/D(0,\epsilon)$, and then it is to show that the integral over $D(0,\epsilon)$ goes to $0$ as $\epsilon$ goes to $0$, and that you get $2\pi\delta_0$ over the complement of the disk... – Felicio Grande Nov 13 '17 at 02:40
  • @FriedrichPhilipp ah I see, sorry about any confusion, I was just directly answering Ian and thought that DT would be understandable – Felicio Grande Nov 13 '17 at 02:41
  • @FelicioGrande Oh, sorry, yes, I misspoke a bit. Yes, the singularity at the origin is integrable, so shrinking the disk shrinks the integral. Then on the complement of the disk you can integrate by parts to recover only a boundary term, which you can evaluate. – Ian Nov 13 '17 at 02:42
  • @FelicioGrande "...thought that DT would be understandable" Of course, you did. That's what I meant. You thought so because you know what it means. Well, but others might not. For example, I still don't know what it is. – Friedrich Philipp Nov 13 '17 at 02:47
  • @FriedrichPhilipp it stands for Divergence Theorem... the Theorem Ian mentioned in the comment right above the comment I made that I was answering to. Once again, sorry if it caused any confusion. – Felicio Grande Nov 13 '17 at 02:48
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    @FelicioGrande No need to say sorry. As I wrote, it was just a tip for further conversations on MSE (Math Stack Exchange ;o)). – Friedrich Philipp Nov 13 '17 at 02:50
  • @Ian I'm having trouble solving the integrals, how does this lead to $2\pi\phi(0)$? – Felicio Grande Nov 13 '17 at 03:16
  • You do the integral over the complement of the disk. You convert to polar coordinates and then integrate by parts. The "bulk" integral vanishes because $\log(r)$ has zero Laplacian away from $r=0$. Then you approximate the integral over the circle in a way that becomes exact as $\epsilon \to 0$ (using the fact that $\phi$ is smooth). – Ian Nov 13 '17 at 03:23
  • Is the correct formula:$$\iint_{\Omega}\log(|\vec x|)\cdot\Delta\phi d\vec x = \oint_{\partial\Omega}\log(|\vec x|)\nabla\phi\cdot\vec n ds - \iint_{\Omega}\nabla(\log(|\vec x|))\nabla\phi d\vec x$$ – Felicio Grande Nov 13 '17 at 03:26
  • That's one integration by parts. You need to integrate by parts in the bulk integral again, so that the Laplacian makes its way all the way back to the logarithm. – Ian Nov 13 '17 at 03:28
  • so the second integral would become:$$\iint_{\Omega}\nabla(\log(\vec x)\nabla\phi d\vec x = \oint_{\partial\Omega}\nabla\log(|x|)\phi\cdot \vec n ds +\iint_{\omega}\Delta(\log|\vec x|)\phi d\vec x$$ – Felicio Grande Nov 13 '17 at 03:30
  • That's more or less right, except for some sign errors and a bit of sloppiness as to what the dot product acts on. Now you need to show that the integral involving $\nabla \phi$ vanishes in the limit, and that the integral involving $\nabla \log$ goes to $2\pi \phi(0)$ in the limit. – Ian Nov 13 '17 at 04:21

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