0
$\begingroup$

Let $\phi\in C^{\infty}(\overline{\Omega})$ be such that $\Delta\phi=0$ on $\Omega$. I am trying to solve a question which asks me to prove that $-\frac{1}{4\pi}\int_{\Omega}\frac{\Delta\phi(y)}{|y-x|}\,dy=0$.

If we use integration by parts, then this yields

$$\begin{aligned}&-\frac{1}{4\pi}\left(\Delta\phi(y)\cdot\log|y-x|\bigg|_{y\in\Omega}-\int_{\Omega}\log|y-x|\cdot\nabla\Delta\phi(y)\,dy\right) \\ &=\frac{1}{4\pi}\int_{\Omega}\log|y-x|\cdot\nabla\Delta\phi(y)\,dy,\end{aligned}$$

but this avenue wouldn't appear to yield any fruit. Might anyone recommend a better suggestion?

  • 0
    For any $x$, or for any $x$ from somewhere? – 2017-01-09
  • 0
    @kolobokish I would *assume* $x\in\Omega$. – 2017-01-09
  • 1
    Do I understand it correctly? There is only one point, where we have something like $\frac{0}{0}$, for all other points it's explicitly $0$. – 2017-01-09
  • 0
    @kolobokish Honestly, I didn't pose the question; it's from a homework sheet of mine and I was also dumbfounded (which is why I decided to post it on here). – 2017-01-09
  • 0
    Maybe we both miss something. ) – 2017-01-09

1 Answers 1

2

Maybe I'm misinterpreting your question, but if $\Delta \varphi =0$ in $\Omega$, then $\Delta \varphi(y)/|x-y| =0$ for all $y \in \Omega$, and so $$ \int_{\Omega} \frac{\Delta \varphi(y)}{|x-y|} dy = \int_\Omega 0 dy =0 $$ for all $x$. There's no need to integrate by parts.

  • 0
    But would it than be an exercise?)) It seem to me, that there is something wrong in formulation in the book, or may be we understand something wrong. – 2017-01-09
  • 0
    @kolobokish Ah, I just realized that you pointed this out in your comment. Sorry, I didn't mean to steal your thunder. :) I'm completely confused about the statement of this problem. If this is the totality of an "exercise" then I would guess it has been given in error. – 2017-01-09
  • 0
    Yes, I also thought of this, but that cannot be what the question expects us to show. I guess there is an error there. – 2017-01-09