Solution of 3D wave equation $u_{tt}=\Delta u$ Find the solution $u=u(x,y,z,t)$ of the problem:
$$\frac{\partial^2u }{\partial t^2}=\Delta u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}$$
satisfy conditions 
$$\left\{\begin{matrix}u(x, y, z, 0)=\varphi(r)\\\frac{\partial u }{\partial t}(x, y, z,0)=\psi(r)\end{matrix}\right. ,\quad r=\sqrt{x^2+y^2+z^2}$$
 A: Since the initial conditions are dependent only on $r$, we transform to spherical coordinates wherein $u=u(r,t)$.  
Then, we have in spherical coordinates
$$\frac1{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u(r,t)}{\partial r}\right)-\frac{\partial^2 u(r,t)}{\partial t^2}=0 \tag 1$$
We can rewrite the first term on the left-hand side of $(1)$ in a form that facilitates analysis.  To that end, we see that $(1)$ is equivalent to  
$$\frac{\partial^2 (ru(r,t))}{\partial r^2}-\frac{\partial^2 (ru(r,t))}{\partial t^2}=0 \tag 2$$
Note that $(2)$ is simply a one-dimensional wave equation for $ru(r,t)$.  Hence, we can write a general solution to $(2)$ as 
$$u(r,t)=\frac{f(r+t)+g(r-t)}{r} \tag 3$$
Applying the initial conditions to $(3)$ yields 
$$f(r)=\frac12 r\phi(r)+\int_a^r r'\psi(r')\,dr'\tag 4$$
and
$$g(r)=\frac12 r\phi(r)-\int_a^r r'\psi(r')\,dr'\tag 5$$
Finally, using $(4)$ and $(5)$ in $(3)$ reveals
$$\bbox[5px,border:2px solid #C0A000]{u(r,t)=\frac12\,\frac{(r+t)\phi(r+t)+(r-t)\phi(r-t)+\int_{r-t}^{r+t}r'\psi(r')\,dr'}{r}}$$
And we are done!
