green's second identity application

Question

I need to use the green's second identity in order to prove the following equality:

$$ \int_{\mathbb{R}^2} \ln (\sqrt{x^2+y^2})\Delta f = -2\pi f(0)$$

where $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is a smooth function with compact suuport. (And $\Delta$ denotes the laplacian operator)

So, applying the identity I have

$$\int_{\mathbb{R}^2} \ln (\sqrt{x^2+y^2}) \Delta f + f \Delta \ln (\sqrt{x^2+y^2}) dxdy = \int_{\partial \mathbb{R}^2} \ln (\sqrt{x^2+y^2}) (grad(f) \cdot n) -f( grad(\ln (\sqrt{x^2+y^2}) )\cdot n) dl$$

it's easy to show that $\Delta \ln (\sqrt{x^2+y^2}) = 0$. However I do not realize how to involve the normal vector $n$ and the compact support assumption

Answer 1

HINT 1:

The Laplcian of $\log r$ is not defined at $r=0$ and thus you need to exclude the origin in applying Green's Identity.

HINT 2:

$$\oint_{\partial R_{\epsilon}}\frac{\partial \log r}{\partial n}d\ell=2\pi$$

where $R_{\epsilon}$ is a sphere of radius $\epsilon$, centered at the origin.

SPOILER ALERT: SCROLL OVER THE SHADED AREAD TO REVEAL THE SOLUTION

Let's apply Green's Identity to the region $R>r\ge\epsilon>0$. Then, we have\begin{align}\int_{R>r\ge \epsilon} \log(r) \nabla^2f(\vec r)\,dS&=\int_{R>r\ge \epsilon} f(\vec r) \nabla^2\log(r) \,dS\\\\&-\int_0^{2\pi}\left.\left(\log(r)\frac{\partial f(\vec r)}{\partial r}-f(\vec r)\frac{\partial \log(r)}{\partial r}\right)\right|_{r=\epsilon}\epsilon\,d\phi\\\\&+\int_0^{2\pi}\left.\left(\log(r)\frac{\partial f(\vec r)}{\partial r}-f(\vec r)\frac{\partial \log(r)}{\partial r}\right)\right|_{r=R}R\,d\phi\end{align}
First, we note that $\nabla^2 \log(r)=0$ in the region $\epsilon\le r\le R$. Next, assuming that $f$ has compact support, then as $R\to \infty$, then integral over the boundary at $r=R$ vanishes. Finally, as $\epsilon \to 0$, the integration over the boundary at $r=\epsilon$ becomes $2\pi f(0)$. Putting that all together reveals that $$\int_{\mathscr{R}^2} \log(r) \nabla^2f(\vec r)\,dS=2\pi f(0)$$which was to be shown.