Question

Green's function for the Poisson equation in three dimensions was obtained integrating the Laplacian in spherical coordinates, assuming that there was a charge punctual at the origin.
 
a) Consider the case where there is a point charge at the origin in a plane.
  Taking now the Laplacian in polar coordinates, what would be the function of
  Green?

  b) If you now have a charge at the origin of a straight line between −∞ and + ∞, which
  Would it be Green's function?

Answer 1

a) The Poisson equation in two dimension is

Here q is the magnitude of point charge at the origin. The Green's function for this equation will satisfy

and

Equation (3) is the boundary condition, which differs from the 3d case, where the Green's function itself goes to zero. It turns out that in 2d it is impossible to find a Green's function for Poisson equation that goes to zero at infinity. Consider a disk of radius , centered at the origin, then

This equation just comes from the definition of delta function, the integral is over the surface area of the disk. Using equation (2), we can write

To obtain the last term in equation (5), we have used the Green's theorem in 2d. is the unit normal to the boundary of the disk, therefore . Also, since the point charge is at the origin, we have circular symmetry, therefore we expect the Green's function to take the form , hence equation (5) becomes

Integrating equation (6), we get

A is a constant. It can be checked that equation (7) satisfies, the boundary condition in equation (3), regardless of the value of A.

b) Now, we have to work in 1d, therefore the Poisson equation becomes

Consider an interval where , then by definition of delta function

Here, we have used the Green's theorem again, this time in 1d, the boundary of the region is just the end points of the interval and the direction of normal changes sign as we move across the origin. Since there is no difference between the positive and negative axis, we expect

Therefore equation (9) becomes

The potential due to point charge does not go to zero as , nor does its derivative. This seems strange, but the electric field found from this, does satisfy the Gauss' law. Also the potential goes to zero at the origin instead of infinity, which is what we expect therefore I think must be excluded from the domain of G.