Today we consider Poissson equation, for example, solving for the gravity potential function when there exists a source distribution $$ \nabla u(x,y,z) = f(x,y,z), \quad u(x,y,z) \to 0, \z{ as } |(x,y,z)| \to \infty. $$

The solution is given by the Green's function $$ G(\vec x; \vec x_0)$$ which solves the equation $$ \nabla u (\vec x) = \delta(\vec x- \vec x_0) $$ with the same decay condition as we go to infinity.

Question: what if we do not impose the decay condition? Is the solution still unique?

The solution is $$ G(\vec x; \vec x_0) = \frac{-(4\pi)^{-1} }{|\vec x - \vec x_0}}. $$

Recall that the idea of Green's function, is to decompose the 'source term', $f(\vec x)$, as 'sum' (more precisely integral) of the $\delta$-function, with the obvious coefficient, $f$ itself $$ f(\vec x) = \int f(\vec x') \delta(\vec x - \vec x') d \vec x'.$$ Then by the superposition principle, the solution $u$ is 'sum' (more precisely integral) of the Green function, with the same coefficient, $f$ itself $$ u(\vec x) = \int f(\vec x') G(\vec x, \vec x') d \vec x' = \int f(\vec x') \frac{-(4\pi)^{-1} }{|\vec x - \vec x'}} d \vec x'.$$

Suppose the source function $f(\vec x)$ is given as $f(r, \theta, \phi)$. We now write $G(\vec x, \vec x')$ using spherical coordinate. Assume $\vec x$ has spherical coordinate $R=|\vec x|$, $\theta=0$ ($\phi$ doesn't matter). And assume $|\vec x| \gg |\vec x'|$ when $f(\vec x')\neq 0$, ie., we are sufficiently far away from the source. Then, we have $$ \frac{1}{|\vec x - \vec x'|} = \frac{1}{\sqrt{ |\vec x|^2 + |\vec x'|^2 - 2\vec x \cdot \vec x' } } = \frac{1}{\sqrt{ R^2 + r^2 - 2R r \cos \theta } } = \sum_{l=0}^\infty \frac{r^l P_l(\cos \theta)}{R^{l+1}}.$$