Hoi, consider $\displaystyle u= \frac{1}{|x|}e^{-|x|}$ for $x\in \mathbb{R}^3$, then one can see that $\Delta u = u$ for $|x|>0$ ( which one can see by transferring $u$ to spherical coordinates).

So can we then conclude: (*) $(1-\Delta)u$ is a distribution with support $= \left\{0\right\}$ and order $N=0$?

I want to either show that $((1-\Delta)u, \phi) = \phi(0)$ or show that (*) holds.

One of the 2 is enough to show that $(1-\Delta)u = \delta$. Thanks for any help.

  • 1
    $\begingroup$ How do you see $u$ as a distritubtion, I mean, how do you define $(u,\phi)$? I am asking it, because $u$ is not even locally integrable. $\endgroup$ – Tomás Apr 9 '13 at 18:13
  • $\begingroup$ I use it as $\int_{\mathbb{R}^3}u\phi$. Apparantly $cu$ with $c = 1/(4\pi)$ is the unique solution to $(1-\Delta)u= \delta$. $\endgroup$ – DinkyDoe Apr 9 '13 at 18:55
  • $\begingroup$ Your integral is not well defined. It can be $\infty$, so it not defines a distribution. You have to define in another way. $\endgroup$ – Tomás Apr 9 '13 at 19:12
  • 1
    $\begingroup$ @DinkyDoe: You are right for every $\phi$ with $ 0\notin\operatorname{supp }\phi$ and for such $\phi$ the integral is well-defined. But with the general definition of $(u,\phi)$, the Dirac delta enters the formula ... $\endgroup$ – Vobo Apr 9 '13 at 20:52
  • $\begingroup$ Thanks, however $(\delta, \phi) = \phi(0)$ and i would have to show the same: $((1-\Delta),\phi) = \phi(0)$ which is not so easy to see...So clearly $((1-\Delta),\phi) = \phi(0) =0$ holds true if $0\notin \text{supp} \ \phi$. So now what if $\phi(0)\neq 0$... why does then $((1-\Delta)u,\phi) = \phi(0)$ also hold true. $\endgroup$ – DinkyDoe Apr 10 '13 at 11:41

(Side note for Tomás: $u$ is locally integrable, and therefore is a distribution).

Recall that $\Delta (|x|^{-1})=-4\pi\delta_0$ (if you don't know it already, see Laplacian of the potential function). Thus, if you can show that $\Delta (u -|x|^{-1}) = u$, the conclusion $(1-\Delta) u=-4\pi \delta_0$ will follow.

The function $u(x)-|x|^{-1} = (e^{-|x|}-1)|x|^{-1} = -1 + |x|/2+\dots$ is in $W^{2,1}$ locally, which implies that its distributional Laplacian is represented by its pointwise Laplacian. See Calculation of the Laplacian of a function in $\mathbb R^3$.

  • $\begingroup$ You are right, i am doing wrong calculation. $\endgroup$ – Tomás Apr 10 '13 at 22:43
  • $\begingroup$ OK, i know now why $\Delta(|x|^{-1}) = -4\pi\delta_0$ holds... but why what you say implies the desired conclusion I dont follow $\endgroup$ – DinkyDoe Apr 11 '13 at 12:15
  • $\begingroup$ I can prove both the statements you make...but I dont see yet why it implies what we want. Thnx if you can explain that. $\endgroup$ – DinkyDoe Apr 11 '13 at 12:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.