## Problem

Determine the Green function of the two dimensional Laplace operator

$\Delta_2 = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}$

## Solution

Recall that the Green function $G$ has to satisfy

$\Delta G(\vec x) = \delta^{(2)}(\vec x)$

As an ansatz, we assume that $G$ depends only on the magnitude of $\vec x$, i.e. $G(\vec x) = G(|\vec x|=\rho)$.  Then let us use Gauss theorem with a “volume” $D(\rho)$ being a disk with radius $\rho$.
Then

$1 = \int_{D(\rho)}d^2x'\, \vec{\nabla}^{\,\prime}\cdot\vec{\nabla}^{\,\prime} G(\rho') = \oint_{\partial D(\rho)} d\vec{\sigma}^{\,\prime}\cdot\vec{\nabla}^{\,\prime} G(\rho')\,.$

Since the boundary of the disk is the circle and $\vec{\nabla}^{\,\prime} G(\rho') = G'(\rho')\hat{\rho}^{\,\prime}$, one obtains

$G'(\rho) = \frac{1}{2\pi\rho}$

from which follows that $G(\rho)$ is given by (up to functions that lie in the kernel of the two-dimensional Laplace operator)

$G(\rho) = \frac{\log(\rho/\lambda)}{2\pi}\,,$

where $\lambda$ is some length scale in order to make the logarithm well-defined.