Solution of 3D wave equation $u_{tt}=\Delta u$

Question

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 $$\left\{\begin{matrix}u(x, y, z, 0)=\varphi(r)\\\frac{\partial u }{\partial t}(x, y, z,0)=\psi(r)\end{matrix}\right. ,\quad r=\sqrt{x^2+y^2+z^2}$$

Answer 1

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!