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 chapter.

Consider a point in the interior of the domain in which a function $u$ is harmonic. Let the largest sphere around the point that stays inside the domain have radius $R$. It is to be shown that $u$ has a Taylor series around the considered point with a finite radius of convergence.

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:

$$u = (1-r^2) \int_{S} \frac{g}{\vert\vec\xi - \vec x\vert^n} \frac{{\rm d}S}{S}$$

where $r=\vert\vec{x}$, $S$ is the surface of the sphere, $n$ is the number of dimensions, and $g$ is the value of $u$ 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 $x$-axis. That gives

$$u = (1-x^2) \int_{S} \frac{g}{\vert 1 - 2 x\xi + x^2\vert^{n/2}} \frac{{\rm d}S}{S}$$

Take a factor $\vert 1+x^2\vert^{n/2}$ out of the denominator of the integrand. For what is left, define a new variable

$$\alpha \equiv \frac{2x}{1+x^2}$$

(Note that $-1<x<1$ corresponds to $-1<\alpha<1$.) That gives

$$u = \frac{1-x^2}{(1+x^2)^{n/2}} \int_{S} \frac{g}{\vert 1 - \xi\alpha\vert^{n/2}} \frac{{\rm d}S}{S}$$

Now the first factor is an analytical function of $x$ in the range $-1<x<1$ 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 $\alpha$. Note that the convergence of this power series is no worse than that of $(1-\alpha)^{-n/2}$ since

$$\int_{S} g \xi^n \frac{{\rm d}S}{S} \le \int_{S} g \frac{{\rm d}S}{S}$$

So the integral is an analytical function of $\alpha$ with a radius of convergence of 1. And $\alpha$ is in turn an analytical function of $x$. Allow $\alpha$ and $x$ to have complex values. Within a finite distance from $x=0$, $\alpha$ will be less than one in magnitude. In that range then, the integral will be an analytical function of $x$. And then so will $u$, because the factor in front of the integral is analytical too for $\vert x\vert<1$. That means that the Taylor series in terms of $x$ 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.