Interesting problem appeared recently. It’s equivalent to solving the electrostatic potential above a plane given the surface potential on that plane.
Say the plane is at and we wish to find the electrostatic potential
for
given
. We start with the divergence theorem
where is a well behaved vector field defined in the volume
bounded by the closed surface
. Let
where
and
are arbitrary scalar fields. Then
and
where is the normal derivative at the surface
(directed outwards from inside the volume
). Substituting the above equations back into the divergence equation, we find Green’s first identity:
Interchanging and
in the above and subtracting the two equations (the
term cancels) we obtain Green’s theorem (or Green’s second identity):
We now choose where
is the observation point and
is the integration variable and let
, the electrostatic potential. Now use the facts that
and
and we find
If lies within the volume
, we obtain:
If lies outside of
then the left hand side of the above equation is 0.
This equation is not a solution of the boundary value problem but is simply an integral statement since the right hand side is an overspecification of the problem. We can define a Green’s function
where
and the function satisfies Laplace equation inside the volume
:
We see if we use and the freedom that
allows us, we can now use Green’s theorem to eliminate one or the other surface terms. If we take
and plug this back into Green’s theorem we find
Since we are interested in the Dirichlet boundary condition (that is we are given the potential on ), we demand
for on
. Then we have
If one is interested in the Neuman boundary condition (that is we have the normal component of the field on ), then we can define the corresponding Green’s function:
for on
. Similarly as before, we can use Green’s theorem to find
where is the average value of the potential over the surface- typically an irrelevant constant.
For the Dirichlet case, it can be shown via Green’s theorem, that while for the Neuman case,
can be forced to have the same property.
In our problem, consists of the upper half infinite volume
. If we assume that
falls to zero at large
the surface integral only survives for
. Since we are specifying the potential on the
plane we need to find
. By the method of images, we can immediately write it down
We see that when
as required. We also note that the second term on the right hand side does satisfy Laplace’s equation in
. Therefore by uniqueness we have the correct Green’s function. We next need to compute
Plugging this back into the equation for and assuming there are no charges in the volume (
) we find
This is the solution to our problem. We see that given we have an integral equation to find
for
. Furthermore, we see the Green’s function only depends on the differences of
and
, or equivalently,
is simply a 2D convolution of
and the kernel (
being viewed as simply a parameter for the surface integral to determine where in
the potential should be evaluated at).
If one is more interested in the electrostatic field (vs. potential), we use and we can find
where again we can view these as 2D convolutions. We also note that can be shown for
, as required.
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