Radially symmetric solution to the wave equation in a 3+1 dimension using D'Alembert formula

Question

I need help proving this equation which is a radially symmetric solution to the wave equation in a 3+1 dimension using D'Alembert formula. I am totally clueless as to how prove it. Thank you.

Answer 1

The standard wave equation is: $$ \Delta u = c^2 \partial_{tt} u $$ Since we are dealing with radial symmetry, let's go into spherical coordinates, where: $$ \Delta u = u_{rr} + \frac{2}{r}u_r + \frac{1}{r^2\sin(\theta}\partial_\theta \left[ \sin(\theta)u_\theta \right] + \frac{1}{r^2\sin^2(\theta)}u_{\phi\phi} $$ with $\partial_k u = u_k$. Let's assume $u$ is radially symmetric. Then $u$ will not change as $\theta$ and $\phi$ change; hence, $u_\theta=u_\phi=0$. Thus, the new wave equation can be written: \begin{align} u_{rr} + \frac{2}{r}u_r &= c^2u_{tt} \\ ru_{rr} + 2u_r &= c^2r u_{tt} \\ \partial_r(ru_r + u) &= c^2 \partial_{tt}(ru) \\ \partial_{rr}(ru) &= c^2\partial_{tt}(ru) \end{align} Let $\tilde{u}=u\,r$. Then the last equation above is just the standard 1D wave equation in $\tilde{u}$.

Thus, the solution (by D'Alembert) is: $$ \tilde{u}(r,t) = \frac{1}{2}[\tilde{f}(r-ct) +\tilde{f}(r+ct)]+\frac{1}{2c}\int\limits_{r-ct}^{r+ct} \tilde{g}(s)\,ds $$ Then: \begin{align} \tilde{u}(r,0)&=\tilde{f}(r) \implies u(r,0)=\frac{1}{r}\tilde{f}(r)=f(r)\implies f(r\pm ct) = (r\pm ct)^{-1}\tilde{f}(r\pm ct)\\ \tilde{u}_t(r,0)&=\tilde{g}(r)\implies u_t(r,0)=\frac{1}{r}\tilde{g}(r) = g(r)\implies g(s) = (s)^{-1}\tilde{g}(s) \end{align} So let's substitute these in: $$ ru(r,t) = \frac{1}{2}\left[ (r-ct)f(r-ct) + (r+ct)f(r+ct) \right] + \frac{1}{2c}\int\limits_{r-ct}^{r+ct} s{g}(s)\,ds $$ and divide by $r$ to get the result: $$ u(r,t) = \frac{1}{2r}\left[ (r-ct)f(r-ct) + (r+ct)f(r+ct) \right] + \frac{1}{2cr}\int\limits_{r-ct}^{r+ct} s{g}(s)\,ds $$