For $x\in\mathbb R^2$, $t>0$, let $K=K(x,t)$ be the Gaussian
$$ K(x,t) = \frac1{4\pi t} \text{e}^{-|x|^2/4t}. $$
As $K$ is radial in $x$, and hiding the dependence on $t$ unless we need it, we will employ the abuse of notation $K(|x|) = K(x,t).$
Consider then the Poisson equation on $\mathbb R^2$ with right hand side $K$:
$$ \Delta \phi = K.$$
(I believe Wikipedia calls the right hand side a "charge density", but I am not fluent enough in the physics to say anything on the subject.)
I found out recently that this has an "explicit" solution in terms of the "Exponential Integral function" $E_1(s) = \int_{s}^\infty \frac{e^{-z}}z dz $. This solution is
$$ \phi(r) = \frac1{2\pi} \int_{\mathbb R^2} K(x-y) \log |y| \, dy = \frac1{2\pi} \left( \log r + \frac12 \int_{r^2/4t}^\infty \frac{e^{-z}}z dz\right).$$
Question:
How does someone find this solution without it being given to them first?
Secondary question which I suppose will be answered by the first.
Given this solution, I can verify that it satisfies $\Delta \phi = K$ at every point except $r=0$. Clearly the logarithm is meant to take care of the singularity at $r=0$. The wikipedia page on $E_1$ states that $E_1(r) + \log r$ is analytic (actually entire), and I feel like knowing that this is true, I should be able to prove it.
But I'm not seeing exactly how to use this to verify that $\Delta \phi(0,0) = \frac1{4\pi t} $?
PS for reference the Laplacian of a radial function is $\Delta f(r) = (\partial_r^2 + \frac{\partial_r }r )f(r)$.
Best Answer
I've got the solution, its actually really simple. You just treat it both as a Poisson equation and a solution to the homogeneous heat equation with initial data $\log|x|$. (This is legal following some general theory of allowable initial data which I don't remember the specifics of). Then you just need to change variables in time; $z = \frac{|x|^2}{4t}$.