# Finding the Green's function for a 2D domain

To find the Green's function of the Laplacian for the full plane, I solved the problem: $-\nabla^2G(r,0)=\delta(r)$, with $r\neq0$. Thus: $$-\nabla^2G(r,0)=0\Rightarrow-G_{rr}-\frac1rG_r=0\Rightarrow G(r,0)=A\ln{r}$$

From there, I don't get how (or even why) we determine the arbitrary constant $A$ using the geometry of the problem?

The way to determine the constant $A$ is as follows. Let $D$ denote a disk of radius $R$, centred at the origin, and let $\partial D$ denote the circular boundary of this disk. By the two-dimensional divergence theorem, $$\int_{\partial D} (\nabla G ). \hat n dl = \int_{D} (\nabla^2 G) \ d^2S,$$ where $\hat n$ denotes the unit normal at each point of the circular boundary.
Now $\nabla G = (A/r )\hat{\vec r}$ (where $\hat{\vec r}$ denotes the radial unit vector), and $\hat n = \hat r$ at each point of the circular boundary, so $(\nabla G) . \hat n = (A / R) (\hat {\vec r} . \hat {\vec r}) = A / R$ at every point of the circular boundary. Therefore, $$\int_{\partial D} (\nabla G ). \hat n dl = \frac A R \int_{\partial D} dl = \frac A R \times (2\pi R) = 2\pi A.$$
Meanwhile, $$\int_{D} (\nabla^2 G )\ d^2S = \int_D (-\delta^2 (\vec r)) \ d^2 S= - 1 ,$$ and equating the two results gives $A = -1/2\pi.$
I'm not sure how to interpret the hint about "using the geometry of the problem". I suppose the fact that we chose to consider integrals on a circular disk was motivated by the geometry. But you could also argue that the geometry motivated you to seek a solution independent of $\phi$ in the first place...
• Great answer, thank you! Can you please further explain why the second integral equals $-1$ and what exactly is $d^{2}S$? – Andrew May 27 '18 at 8:26
• $d^2 S$ is the two-dimensional area element (so people write it as $d^2 A$ or $dx dy$). The second integral equals $-1$ because the defining property of the delta function $\delta^2 (\vec r)$ is that $\int_D \delta^2 (\vec r) = 1$ for any domain $D$ that contains $\vec 0$. – Kenny Wong May 27 '18 at 8:34
• The notation confused me. I've seen it simply as $dA$. Thanks! – Andrew May 27 '18 at 8:54