I am looking through V. Fock's 1935 paper on the symmetry of the Hydrogen atom and am stuck on the topic of Green's functions.
The setup is as follows:
The 4D Laplace equation is $\boldsymbol{\nabla}^2 u = 0$, for some function $u(\textbf{x})$. To find the Green's function, let the RHS be $\delta^{(4)}(\textbf{x} - \textbf{x}')$, the 4D delta function. This gives the Green's function as \begin{align} G(\textbf{x}, \textbf{x}') = -\frac{1}{4\pi^2}\frac{1}{(\textbf{x} - \textbf{x}')^2}. \end{align} However, as far as I understand, we may also add a function $F(\textbf{x}, \textbf{x}')$ that satisfies $\boldsymbol{\nabla}^2F = 0$, so there is some freedom in fixing $G(\textbf{x}, \textbf{x}')$. As far as I understand, we can use this freedom to satisfy boundary conditions.
The following is directly copied from the paper, with my own comments and questions in parentheses (note $r = |\textbf{x}|$):
Let us introduce the function \begin{align} G = \frac{1}{2R^2}+\frac{1}{2R_1^2}, \end{align} where $R^2$ and $R_1^2$ are ($R^2$ is $(\textbf{x} - \textbf{x}')^2$) \begin{align} R^2 = r^2 + r'^2 - 2rr'\cos\omega ; \qquad R_1^2 = 1-2rr'\cos\omega + r^2r'^2. \end{align}
This function can be called the Green's function of the third kind (I haven't been able to find this terminology explained) because it satisfies the boundary condition on the sphere surface \begin{align} \frac{\partial G}{\partial r'} + G = 0 \qquad\text{ at }\qquad r'=1. \end{align}
By the Green theorem the function, which is harmonic inside the sphere, can be expressed through the boundary value of $u + \frac{\partial u}{\partial r}$ as (note $2\pi^2$ is the surface area of the 3-sphere, and $d\Omega$ is an element of the surface of the 3-sphere) \begin{align} u(\textbf{x}) = \frac{1}{2\pi^2}\int \left(u + \frac{\partial u}{\partial r'}\right)_{r' = 1}G\,d \Omega'. \end{align}
I assume $G$ is chosen such that, $\frac{\partial G}{\partial r'} + G = 0$ at $r'=1$, but I don't understand why that condition is important. I also don't understand where the final equation for $u(\textbf{x})$ has come from.