Show that the following problem has at most one solution:
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 $\rho\phi$ bounded and $r\dfrac{\partial\phi}{\partial r}\to0$ as $r\to\infty$.
In my understanding, the question isn't asking me to solve the equation but just prove that the solution is unique if it exists. Typically, this involves supposing that two solutions exist, considering their difference and then showing that this is identically zero by the use of some integral and then implementing boundary conditions. However I can only seem to show that solutions differ by a factor of $A/r$ where $A$ is a constant and $r$ is the distance from the origin, since the Laplacian of $1/r$ is $0$.
A useful observation: if $\phi$ is harmonic, then $r\frac{\partial\phi}{\partial r}$ is also harmonic. This follows, for example, from $$r\frac{\partial\phi}{\partial r} = \left(\frac{d}{dt}u(tx)\right)\bigg|_{t=1}$$ where $u(tx)$ is harmonic for every $t$. Or you could just write $r\frac{\partial\phi}{\partial r}$ as $xu_x+yu_y+zu_z$ and compute the Laplacian.
If $\phi$ and $\psi$ are two functions as above, their difference $\chi$ is harmonic. Since $r\frac{\partial\chi}{\partial r}$ is a harmonic function vanishing at infinity, it is identically zero. It follows that $\chi$ is constant. It can't be a nonzero constant because then $r\chi$ wouldn't be bounded.
It seems that the assumption about $\rho$ vanishing outside of some sphere is redundant for this result.