Let $u:R^2 \rightarrow R$ be a $C^2$ function with a compact support. I want to show that $u(x) = \frac{1}{2\pi} \int_{R^2} log(|y-x|)\Delta u(y)dy$, when $x, y \in R^2$.
This look like it should be solved with harmonic functions and with Green's formulas.
I declare $v(x,y) = log(|y-x|)$ which is harmonic in $R^2$ \ $\{ x \}$.
Let $G_{R,\epsilon}$ be $B(x,R)$ \ $\bar B(x,\epsilon)$.
From Green's 3rd formula I get: $$ \int_{\partial G_{R,\epsilon}} u<\nabla v, N> - v<\nabla u,N> dS = \int_{\partial G_{R,\epsilon}} u\Delta v - v\Delta u = -\int_{\partial G_{R,\epsilon}} v\Delta u$$ when $N$ is the outer unit normal.
Now: $$ \int_{\partial G_{R,\epsilon}} u<\nabla v, N> - v<\nabla u,N> dS = \int_{|y-x| = R} (u<\nabla v, N> - v<\nabla u,N>)dS - \int_{|y-x| = \epsilon} (u<\nabla v, N> - v<\nabla u,N>)dS$$
I need to somehow evaluate those integrals so that when I take $R \rightarrow \infty$ and $\epsilon \rightarrow 0$ I will get $-2\pi u(x) $.
However, the only integral I managed to evaluate is $\int_{|y-x| = \epsilon} v<\nabla u,N>)$ which I showed that goes to $0$. But with the others I didn't have much luck.
Because I got stuck with $log(R)$ who doesn't go to zero and because $<\nabla, N>$ ends being not so nice of an expression.
I fit is needed: $\nabla v = (\frac{y_1-x_1}{(y_1-x_1)^2 + (y_2-x_2)^2}, \frac{y_2-x_2}{(y_1-x_1)^2 + (y_2-x_2)^2}$ and the outer Normals are $\frac{y}{R}$ and $\frac{y}{\epsilon}$
Help would be appreciated.
Let $g(x,y)=\frac1{2\pi}\log(|x-y|)$. Then, note that we have
$$\int_{\mathbb{R}^2\setminus{B(x,\epsilon)}}\left ( u(y)\nabla^2g(x,y)-g((x,y)\nabla^2u(y) \right)\,dy=-\oint_{\partial B(x,\epsilon)}\left(u(y)\nabla g(x,y)-g(x,y)\nabla u(y)\right)\cdot \frac{x-y}{|x-y|}\,d\ell\tag1$$
On $\mathbb{R}^2\setminus{B(x,\epsilon)}$, $\nabla^2 (x,y)=0$ and $|x-y|=\epsilon$. Furthermore, $\nabla g(x,y)\cdot \frac{x-y}{|x-y|}=\frac1{2\pi\epsilon}$. Hence, as $\epsilon\to0$,
$$\lim_{\epsilon\to0}\int_{\mathbb{R}^2\setminus{B(x,\epsilon)}}\left ( u(y)\nabla^2g(x,y)-g((x,y)\nabla^2u(y) \right)\,dy=-\frac1{2\pi}\int_{\mathbb{R}^2}\log(|x-y|)\nabla^2u(y)\,dy\tag2$$
and
$$\lim_{\epsilon\to0}\oint_{\partial B(x,\epsilon)}\left(u(y)\nabla g(x,y)-g(x,y)\nabla u(y)\right)\cdot \frac{x-y}{|x-y|}\,d\ell=-u(x)\tag3$$
Using $(2)$ and $(3)$ in $(1)$ reveals
$$u(x)=\frac1{2\pi}\int_{\mathbb{R}^2}\log(|x-y|)\nabla^2u(y)\,dy$$
as was to be shown!