I'm reading the chapter on Green's functions in Strauss' PDE book. He gives the Green's function for the sphere as:
$$ G(\textbf{x}, \textbf{x}_o) = - \frac{1}{4 \pi |\textbf{x}-\textbf{x}_o|}+ \frac{a}{4 \pi |\textbf{x}_o| | \textbf{x}-\textbf{x}^{*}_o | } $$
where $\textbf{x}$ is the vector representing a point on the surface of the sphere, $\textbf{x}_o$ is the vector representing any point strictly inside the sphere, and $ \textbf{x}^*_o$ is defined as follows:
$$ \textbf{x}^*_{o} = \frac{a^2 \textbf{x}_o}{ |\textbf{x}_o |^2} $$
Strauss then asserts that when $\textbf{x}_o = \textbf{0}$ the formula reduces to
$$ G(\textbf{x}, \textbf{x}_o) = - \frac{1}{4 \pi |\textbf{x}|}+ \frac{1}{4 \pi a } $$
How do I prove this? Is there any way to do this without using spherical coordinates?
$\newcommand{\x}{\mathbf{x}}$ $\newcommand{\c}[1]{\left\langle #1 \right\rangle}$ We see that: \begin{align*} |\x_0|^2|\x - \x_0^*|^2 &= |\x_0|^2\c{\x - \x_0^*,\x - \x_0^*} \\ &= |\x_0|^2(\c{\x,\x} - 2 \c{\x,\x_0^*} + \c{\x_0^*,\x_0^*}) \\ &= |\x_0|^2\left(\c{\x,\x} - 2 \c{\x,\frac{a^2\x_0}{|\x_0|^2}} + \c{\frac{a^2\x_0}{|\x_0|^2},\frac{a^2\x_0}{|\x_0|^2}}\right) \\ &= |\x_0|^2|\x|^2 - 2a^2\c{\x,\x_0} + \frac{a^4}{|\x_0|^2}\c{\x_0,\x_0} \\ &= |\x_0|^2|\x|^2 - 2a^2\c{\x,\x_0} + a^4 \\ \end{align*} If $\x_0 = 0$, then $|\x_0|^2|\x|^2 = 0$ and $\c{\x,\x_0} = 0$, so we're left with: $$ |\x_0||\x - \x_0^*| = a^2 $$ Substitute this back into the original Green's function identity and you get the identity you want.