Green's function on unit ball

Question

In Evans' book on PDE it is stated (p.50 in my edition) that a Green's function for solving Laplace's equation on $B_1(0)$ is given by $$G(x,y)=\Phi(y-x)-\Phi(|x|\cdot (y-\tilde x)),$$ where $x,y\in B_1(0)$, $x\not= y$, $\tilde x:=\frac{x}{|x|^2}, x\not=0$ and $\Phi$ denotes the fundamental solution of Laplace's equation on $\mathbb{R}^n\setminus\{0\}$. But why can one define $G$ for $x=0$ like this? $\Phi$ has a singularity at $x=0$, it is not even defined there.

Answer 1

The singularity at $x=0$ is removable, provided $y\ne 0$. Indeed, as $x\to 0$, we have $|x|y\to 0$. The vector $|x|\tilde x$ does not have a limit as $x\to 0$, but its magnitude stays at $1$, and $\Phi$ is radially symmetric. So, $\Phi(|x|(y-\tilde x)) $ has a limit as $x\to 0$ (it's whatever value $\Phi$ has on the unit sphere), and this is used to extend the definition of $G$ to the case $x=0$.

The only singularity $G$ has is when $y=x$, but this is to be expected from Green's function.