= Solution
{c}
The all-time conclusion in the original PDF is false. The <expanding-flow transformation of a wave equation> makes both the correct forward statement and a counterexample transparent. Put
$$
X=\rho e^{-t},\qquad S=e^{-t}>0,\qquad u(\rho,t)=w(X,S).
$$
For $D=\partial_t+\rho\partial_\rho$, the expanded operator from part (i) is $D^2+D-\partial_\rho^2$. In the new variables $D=-S\partial_S$ and $\partial_\rho=S\partial_X$, hence
$$
(D^2+D-\partial_\rho^2)u=S^2(w_{SS}-w_{XX}).
$$
Thus it is the ordinary <wave equation> on the half-plane $S>0$. By the <D'Alembert formula>, any $C^2$ solution has the form
$$
u(\rho,t)=F(e^{-t}(\rho+1))+G(e^{-t}(\rho-1)).
$$
At $t=0$, differentiating the zero displacement with respect to $\rho$ and using the zero velocity gives, for $-1<\rho<1$,
$$
F'(\rho+1)+G'(\rho-1)=0,\qquad
(\rho+1)F'(\rho+1)+(\rho-1)G'(\rho-1)=0.
$$
Subtracting $(\rho-1)$ times the first equation from the second gives $2F'(\rho+1)=0$, and then $G'(\rho-1)=0$. Thus $F=c$ on $(0,2)$ and $G=-c$ on $(-2,0)$. If $t\geq0$ and $-1<\rho<1$, both arguments remain in those intervals. Consequently the valid forward result is
$$
\boxed{u(\rho,t)=0\quad(-1<\rho<1,\ t\geq0).}
$$
For negative time the arguments need not stay in the initial intervals. Take $F(s)=(s-2)_+^3$ and $G=0$, where $q_+=\max\{q,0\}$. Then
$$
\boxed{u(\rho,t)=\bigl(e^{-t}(\rho+1)-2\bigr)_+^3}
$$
is a global $C^2$ solution, has the prescribed zero initial displacement and velocity on $(-1,1)$, but $u(0,-\log3)=1$. This directly disproves the assertion for $t\in\mathbb R$.
The connection with part (i) is directional <finite propagation speed>. At a future point in the strip, the two characteristics traced back to $t=0$ meet it at $\rho_0=e^{-t}(\rho+1)-1$ and $\rho_0=e^{-t}(\rho-1)+1$, both in $(-1,1)$. For negative time these intersections can lie outside that interval. The lines $\rho=\pm1$ are characteristic barriers for this forward domain of dependence, not barriers in both time directions.
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