D.1 Harmonic functions are analytic
This note shows that harmonic functions have converging Taylor series.
The proof uses the Poisson integral formula derived in a later
Consider a point in the interior of the domain in which a function
is harmonic. Let the largest sphere around the point that stays
inside the domain have radius . It is to be shown that has a
Taylor series around the considered point with a finite radius of
To do so, scale the coordinates so that the radius of the sphere
becomes 1. Move the origin of your coordinate system to the center of
the sphere. The Poisson integral formula then says:
where , is the surface of the sphere, is the
number of dimensions, and is the value of on the surface of
the sphere. Rotate the coordinate system so that the point at which
the solution is to be found is on the -axis. That gives
Take a factor out of the denominator of the integrand.
For what is left, define a new variable
(Note that corresponds to .) That gives
Now the first factor is an analytical function of in the range
and is of no concern for now. In the integral, do a Taylor
series expansion of the denominator. That gives the integral as a
power series in . Note that the convergence of this power series
is no worse than that of
So the integral is an analytical function of with a radius of
convergence of 1. And is in turn an analytical function of
. Allow and to have complex values. Within a finite
distance from , will be less than one in magnitude. In
that range then, the integral will be an analytical function of .
And then so will , because the factor in front of the integral
is analytical too for . That means that the Taylor series in
terms of converges within the indicated range.
Presumably, the radius of convergence is 1, like it is in 2 and 3
dimensions. However, the above proof shows only that it is greater
than zero. Apparently, you will need to do a separation of variables
solution to show the unit radius of convergence.