Consider system of : $\begin{cases} u_{tt} = \vartriangle{u} \\ u|_{t = 0} = \frac{1}{x^{2}+y^2+z^2} \\ u_t|_{t = 0} = 0 \end{cases}$
I've thought about $v = x^{2} + y^2+z^2$, then $v_{x} = 2x, v_{y} = 2y, v_{z} = 2z$. Hence we have : $\vartriangle{u} = 6u_{v} + 4v u_{v} = u_{tt}$. And addititional property is : $u|_{t=0} = \frac{1}{v}$.
But how should I finish it?
Edit :
After spherical substitution and deducing that $U(x,y,z,t) = U(r,t)$
We have : $\begin{cases} u_{tt} = u_{rr} + \frac{2}{r}u_{r} \\ u|_{t = 0} = \frac{1}{r^2} \\ u_t|_{t = 0} = 0 \end{cases}$
After substitution: $\xi = t-r$ and $\eta = t + r$ we have $u_{\xi \eta} = \frac{u_{\eta} - u_{\xi}}{\eta - \xi}$. The solution for this part is : $U(\xi,\eta) = \frac{f(\eta)+ g(\xi)}{\eta - \xi}$. Hence we have :
$U(t,r) = \frac{f(t+r) +g(t-r)}{2r}$ with initial data : $U(0,r) = \frac{1}{r^2}$ and $U_t(0,r) = 0$.
But now how can we find exact form of $f$ and $g$?