Given a continuous function $\rho (x,y,z)$, which is zero for $x^2+y^2+z^2 > a^2 > 0$ find $\phi$ such that:
$\nabla^2 \phi = \rho$ with $r \phi$ bounded and $r \displaystyle\frac{\partial \phi}{\partial r} \rightarrow 0$ as $r \rightarrow \infty$
I literally have no idea where to even begin with this question any help would be greatly appreciated. Thanks.
In case you have some physics background, let me reformulate your problem as follows: find the electric potential $\phi$ produced by a charge density $\rho$ which vanishes outside the ball $B$ of radius $a$ (such that $\phi$ decays as at most $r^{-1}$ at infinity, and its radial derivative decays faster that $r^{-1}$). The charge $dq$ placed at the point $\mathbf{r'}=(x',y',z')$ creates the potential $\displaystyle \frac{1}{4\pi}\frac{dq}{|\mathbf{r}-\mathbf{r'}|}$, where $$|\mathbf{r}-\mathbf{r'}|=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}.$$ Therefore the potential you are looking for will be given by $$\phi(x,y,z)=\frac{1}{4\pi}\int\!\!\!\int\!\!\!\int_B\frac{\rho(\mathbf{r'})\,dx'dy'dz'}{|\mathbf{r}-\mathbf{r'}|}.\tag{1}$$ At large distances, it will coincide with the potential of a point-like charge placed at the origin and equal to the total charge of $B$. This has necessary decay properties ($\sim r^{-1}$) so the problem is solved.
If you don't have physics background, take the formula (1) and apply to it the operator $\nabla^2$. What one has to show is that $$ \frac{1}{4\pi}\nabla^2\left(\frac{1}{|\mathbf{r}-\mathbf{r'}|}\right)=\delta(\mathbf{r}-\mathbf{r'}),$$ i.e. that $\frac{1}{4\pi|\mathbf{r}-\mathbf{r'}|}$ is the Green function of three-dimensional Laplace operator. This can be done using divergence (Gauss-Ostrogradsky) theorem.
