Using Greens function to solve homogenous wave equation with inhomogeneous boundary conditions I have derived the Green's function for the 3D wave equation as
$$G(x,y,t,\tau)=\frac{\delta\left(|x-y|-c(t-\tau)\right)}{4\pi c|x-y|}$$
and I'm trying to use this to solve
$$u_{tt}-c^2\nabla^2u=0 \hspace{10pt}u(x,0)=0\hspace{10pt} u_t(x,0)=f(x)$$
but I'm not sure how to proceed. I tried converting the problem to a forced wave equation with homogenous boundary conditions by setting $u'=u-tf(x)$ so $u'_{tt}-c^2\nabla^2u'=c^2t\nabla^2f(x)$ and then integrating to find
$$u=f(x)t+\int_0^\infty\int_{\mathbb{R}^3}G(x,y,t,\tau)c^2\tau\nabla^2f(y)$$
but when I evaluate this integral I don't seem to recover the causal structure of the wave equation. I'm meant to find that $u=\langle f\rangle t$ where $\langle f\rangle$ is the average value of $f$ on a sphere of radius $ct$ around $x$.
Firstly, is this the right approach to using Green's functions here?
And if so how can I recover the causal structure of the problem?
 A: You look for the solution of the classical Cauchy problem, which is given by Kirchhoff's formula in $R^3$
If
$$u_{tt}-c^2\nabla^2u=F(x,t), \hspace{10pt}u(x,0)=u_0(x),\hspace{10pt} u_t(x,0)=u_1(x)$$
The solution is given by
$$U(x,t)=\int\int G(|x-y|,t-\tau)\Bigl(F(y,\tau)+u_0(y)\delta'(\tau)+u_1(y)\delta(\tau)\Bigr)\,dy\,d\tau$$
Where
$$G(|x-y|,t-\tau)=\frac{\delta\left(|x-y|-c(t-\tau)\right)}{4\pi c|x-y|}$$
is Green function of the equation
$$u_{tt}-c^2\nabla^2u=\delta(x,t)$$
The proof is simple - you have to show that the function $\bar u(x,t)$ ($\bar u=u(x,t), t\geqslant0$ and $=0$ if $t<0$) - is the solution of the equation
$$\bar u_{tt}-c^2\nabla^2\bar u=\bar F(x,t)+u_0(x)\delta'(t)+u_1(x)\delta(t)$$
where $\bar F(x,t)=F(x,t)$, if $t\geqslant0$ and $=0$ if $t<0$
It means you have to solve the same equation, but with a modified RHS.
I also found a direct and nice solution (which uses FT) exactly for your case ($F(x,t)=0, u_0=0$) here:
https://www.math.usm.edu/lambers/mat417/spr14/lecture17.pdf
Hopefully this will hepl.
