Electrostatic Potential and Compactly Supported Functions

Question

I found the following expression in IMRN International Mathematics Research Notices 2005, No. 34, page# 2064.

Let $\phi$ be a $C_c^{\infty}(\mathbb{C})$ then $$\int_{\mathbb{C}}\frac{1}{2\pi}\log|z|\Delta\phi(z)dxdy = \phi(0).$$ Here $\Delta \phi$ is the Laplacian of $\phi$.

I am not entirely sure how to get this integral. Any hint or explanation will be a great help.

See How to prove this relation between the laplacian of the logarithm and the dirac delta function? and the various linked answers. It's pretty much the same calculation as what's already written there, but another way of saying it is to consider for each $\epsilon>0$, the integral $\int_{\Bbb{R}^2}\frac{1}{2\pi}\log(|z|+\epsilon^2)\Delta \phi(z),dx,dy$. We can now shift the Laplacian over to the $\log$ term (no boundary terms arise now), and then eventually let $\epsilon\to 0^+$ (all justified due to dominated convergence theorem). — peek-a-boo, Jul 19 '21 at 10:29

score 0 · Answer 1 · answered Jul 19 '21 at 12:42

0

So, it essentially says that the $\textit{distribution}~$ Laplacian of $\displaystyle -\frac{1}{2\pi}\log\left( \frac{1}{|z|}\right)$ with respect to $z$ is the point mass at $0$. For more details, one can see Theodore W. Gamelin's book on "Complex Analysis, Page # 393-394"[1]. It's a beautiful application of Green's second identity.

. [1]: https://www.springer.com/gp/book/9780387950938

answered Jul 19 '21 at 12:42

Tapan

101

To my good fortune Prof. Rudnick the author of IMRN International Mathematics Research Notices 2005, No. 34, also provides one reference: page 95 of G. B. Folland's book "Introduction to Partial Differential Equations". Thanks to @peek-o-boo and Prof. Rudnick. – Tapan Jul 19 '21 at 13:29

Electrostatic Potential and Compactly Supported Functions

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