2-D Poisson's equation - Green's function

[plasmoid]

In the x-y plane, we have the equation


$$\nabla^{2} \Psi = - 4\pi \delta(x- x_{0}) \delta (y- y_{0})$$

with $$\Psi = 0$$ at the rectangular boundaries, of size L.

A paper I'm looking at says that for

$$R^{2}$$ = $$(x-x_{0})^{2}$$ + $$(y-y_{0})^{2}$$ << $$L^{2}$$ ,

that is, for points very close to the source, the solution must behave as if the boundaries were at infinity, and

$$\Psi \approx -2 ln R.$$

I see that -2 ln R will satisfy the equation, but why should it be the only solution valid at R<<L? And how does it satisfy the boundary condition?

- 2 ln R goes to infinity as R goes to zero, does that have anything to do with it?

Morse and Feshbach have the same thing on Pg 798 of vol 1 of their Methods ... , but I can't see an explanation there either.

 

[gtoguy97]

When R is small compared to L, it is valid to assume that the boundaries are at infinity. This means that the solution must not depend on the size of the boundaries, so it will be of the form -2 ln R. This is because this form satisfies Laplace's equation and also satisfies the boundary condition of 0 at infinity. Furthermore, since this form goes to infinity as R goes to zero, it also satisfies the Dirac delta function in the equation.