I'm working on a paper about General Relativity and I am stuck in proving this result. I have this PDE:
$$\Box\phi = \frac{4m}{r^3}$$
Where $m$ is a positive parameter (basically, the mass), and $r$ is a generic spatial coordinate. Treat it as if it were $x$ or $y$ or whatever.
By consequence, the D'Alembertian operator is $\Box = \Box(r, t)$, and what I have is basically a non homogeneous wave equation.
I have the solution of this, which is
$$\phi = at + \log\left(1 - \frac{2m}{r}\right) + A\left(r + 2m\log(r - 2m)\right) + B$$
Where $a, A, B$ are constants.
Any hint? Or any reference in which I might learn a bit about solving D'Alembertian non homogeneous equations?
Hint: For convenience, we will set $c=1$. Let us begin by rewriting the inhomogeneous wave equation in spherical coordinates to get \begin{align} \Box \phi = \left(\partial_{tt}-\partial_{rr}-\frac{2}{r}\partial_r+\text{ terms involving angular derivatives}\right)\phi = f(r). \end{align}
To simplify matter, we will only look at radial solutions. Thus, our equation becomes \begin{align} \partial_{tt}\phi - \frac{2}{r}\partial_{r}\phi -\partial_{rr}\phi = f(r) \ \ \Rightarrow& \ \ \partial_{tt}(r\phi) - \partial_{rr}(r\phi) = rf(r) \end{align} where the last implication comes from multiplying both side by $r$. Set $\psi = r\phi$, then we see that $\psi$ satisfies the 1D wave equation \begin{align} \partial_{tt}\psi-\partial_{rr}\psi = rf(r). \end{align} Let us prescribe the problem with some generic conditions say \begin{align} &\partial_{tt}\psi-\partial_{rr}\psi = rf(r) \ \ \text{ for } r>0\\ &\psi(r, 0) = g(r) \ \ \text{ and } \ \ \psi_t(r, 0)= h(r) \end{align} which is just solving the 1D wave equation on the half-line. Moreover, since we have a half-line problem, we need to also specify the boundary condition at $r=0$, but for us $\psi(0, t) = 0$ since $\psi = r\phi$.
Reference: I think Partial Differential Equations: An Introduction, Second edition by Walter A. Strauss is an excellent text to learn basics about the linear wave equations.