a)
\begin{equation} u_r = \frac{v_rr-v}{r^2} \end{equation}
\begin{equation} u_{rr} = \frac{v_{rr}r^2-2v_rr+2v}{r^3}\end{equation}
\begin{equation} u_{tt} = \frac{v_{tt}}{r} \end{equation}
Thus, the original equation can be written as \begin{equation} \frac{v_{tt}}{r} = c^2 [\frac{v_{rr}r^2-2v_rr+2v}{r^3} + \frac{2}{r}\frac{v_rr-v}{r^2}] \end{equation}
Then we get \begin{equation} v_{tt} = c^2v_{rr}\end{equation}
b) \begin{equation} u = \frac{1}{r}[f(r+ct)+g(r-ct)] \end{equation}
c)\begin{equation}v_{t=0} = ru_{t=0} = \phi(r) \end{equation}, thus \begin{equation} \phi(r) = r\Phi(r)\end{equation}
similarly, \begin{equation} \psi(r) = r\Psi(r)\end{equation}
Substitute, we get\begin{equation} u = \frac{1}{2r}[(r+ct)\Phi(r+ct)+(r-ct)\Phi(r-ct)]+\frac{1}{2cr}\int^{r+ct}_{r-ct}s\Psi(s)ds \end{equation}
d) not sure I understand the question. I tried to use L'Hopital rule to determine the limit of u, this seems to be well defined as long as \begin{equation} \Phi'(x)\end{equation} is well defined. However u(0,t) is infinite. That means \begin{equation} \Phi'(x)\end{equation} should be infinite? Not sure if I'm doing the right thing.
