**(a)**: $v = ur$, then $u = \frac{v}{r}$ and

\begin{align}

&u_t = \frac{ v_t }{r},\notag \\

&u_{tt} = \frac{1}{r}{v_{tt}}\label{A}\\

&u_r = - \frac{v}{{{r^2}}} + \frac{{{v_r}}}{r},\label{B}\\

&u_{rr} = \frac{{2v}}{{{r^3}}} - \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r}.\label{C}

\end{align}

Plugging (\ref{A})--(\ref{C}) into the equation we have

$$

\frac{{{v_{tt}}}}{r} = {c^2}(\frac{{2v}}{{{r^3}}} - \frac{{2{v_r}}}{{{r^2}}} + \frac{{{v_{rr}}}}{r} + \frac{{2{v_r}}}{{{r^2}}} - \frac{{2v}}{{{r^3}}})

\frac{{{v_{tt}}}}{r} = {c^2}(\frac{{{v_{rr}}}}{r})$$

Hence, ${v_{tt}} = {c^2}{v_{rr}}$.

**(b)** From (a), known that ${v_{tt}} = {c^2}{v_{rr}}$:

$$v(x,t) = f(x + ct) + g(x - ct)$, $u =r^{-1} \left[ f(r + ct) + g(r - ct) \right]$$

where $f$ and $g$ are arbitrary functions.

**(c)** $\phi (r) = v(r,0) = ru(r,0) = r\Phi ( r ) $, $\varphi (r) = {v_t}(r,0) = r{u_t}(r,0) = r\Psi (r )$,

$$

u(r,t) = \frac{1}{2}\left[ {(r + ct)\Phi (r + ct) + (r - ct)\Psi (r - ct)} \right] + \frac{1}{{2cr}}\int\limits_{r - ct}^{r + ct} {s\Phi (s)ds} .$$

**(d)** If this solution is continuous at $r=0$, $\lim_{r \to 0} u(r,t)$ should exist; so, $\lim_{r \to 0} \left[ {f(r + ct) + g(r - ct)} \right] = 0$. If $\lim_{r \to 0} \left[ {f(r + ct) + g(r - ct)} \right] \ne 0$, then $\lim_{r \to 0} u(r,t)=\infty$.

Therefore, $f(ct) + g( - ct) = 0$, that is, $f(ct) = - g( - ct)$. So $g(r - ct) = - f(ct - r)$,

$$u(r,t) = r^{-1}\left[ f(r + ct) - f(ct - r)\right].$$

I fixed LaTeX. Please do not use the crapware you used to produce that code! It was produced by some aliens and is not for human use! V.I.