1

Indeed, occasionally (remnant of the previous version) misleading "use separation of variables" appeared. Because everybody followed and none got a correct solution, this problem was removed and FE was normalized 90->40.

There is a correct solution:

Solution is spherically symmetric because the problem is. Then

\begin{equation}

u_{tt}- \bigl(u_{rr}+\frac{2}{r}u_r\bigr)=0\qquad r>0, t>0.

\label{6-4A}

\end{equation}

Multiplying by $r$ and using (\ref{6-3}) we arrive to the first equation below:

\begin{align}

&v_{tt}-v_{rr}=0\qquad r>0,\label{6-5}\\

&v(0,t)=0,

\label{6-6}\\

&v(r,0)=g(r)=\left\{\begin{aligned} &r

\quad &&r<1,\\

&0 &&r\ge 1,\end{aligned}\right. && v_t(r,0)=0.

\label{6-7}

\end{align}

Continuing $g(r)$ as and odd function $\tilde{g}(r)=\left\{\begin{aligned} &r

\quad &&|r|<1,\\

&0 &&|r|\ge 1,\end{aligned}\right.$ and solving Cauchy problem we get

\begin{equation}

v(r,t)=\frac{1}{2}\bigl( \tilde{g}(r+t)+\tilde{g}(r-t)\bigr)=\left\{\begin{aligned}

&0 &&r>t+1,\\

&\frac{1}{2}(r-t) \qquad&&1-t<r<t+1,\\

&r &&0<r<1-t,\\

&0 && 0< r<t-1

\end{aligned}\right.

\end{equation}

and finally

\begin{equation}

u(r,t)=r^{-1}v(r,t)=\left\{\begin{aligned}

&0 &&r>t+1,\\

&\frac{1}{2r}(r-t) \qquad&&1-t<r<t+1,\\

&1 &&0<r<1-t,\\

&0 && 0< r<t-1

\end{aligned}\right.

\end{equation}

There is a correct solution:

Solution is spherically symmetric because the problem is. Then

\begin{equation}

u_{tt}- \bigl(u_{rr}+\frac{2}{r}u_r\bigr)=0\qquad r>0, t>0.

\label{6-4A}

\end{equation}

Multiplying by $r$ and using (\ref{6-3}) we arrive to the first equation below:

\begin{align}

&v_{tt}-v_{rr}=0\qquad r>0,\label{6-5}\\

&v(0,t)=0,

\label{6-6}\\

&v(r,0)=g(r)=\left\{\begin{aligned} &r

\quad &&r<1,\\

&0 &&r\ge 1,\end{aligned}\right. && v_t(r,0)=0.

\label{6-7}

\end{align}

Continuing $g(r)$ as and odd function $\tilde{g}(r)=\left\{\begin{aligned} &r

\quad &&|r|<1,\\

&0 &&|r|\ge 1,\end{aligned}\right.$ and solving Cauchy problem we get

\begin{equation}

v(r,t)=\frac{1}{2}\bigl( \tilde{g}(r+t)+\tilde{g}(r-t)\bigr)=\left\{\begin{aligned}

&0 &&r>t+1,\\

&\frac{1}{2}(r-t) \qquad&&1-t<r<t+1,\\

&r &&0<r<1-t,\\

&0 && 0< r<t-1

\end{aligned}\right.

\end{equation}

and finally

\begin{equation}

u(r,t)=r^{-1}v(r,t)=\left\{\begin{aligned}

&0 &&r>t+1,\\

&\frac{1}{2r}(r-t) \qquad&&1-t<r<t+1,\\

&1 &&0<r<1-t,\\

&0 && 0< r<t-1

\end{aligned}\right.

\end{equation}