Consider the PDE initial value problem $$ \begin{cases}u_t+(x^2-1)u_x=0, & x,t\in\mathbb{R}\\u(x,0)=u_0(x)\in C^1(\mathbb{R}), & t=0\end{cases}\tag{1}\label{eq1} $$ I use the method of characteristics, i.e., start by making the ansatz $u(s)=u(x(s),t(s))$ and use the chain rule to get $$ \frac{du}{ds}=\frac{dx}{ds}\frac{\partial u}{\partial x} + \frac{dt}{ds}\frac{\partial u}{\partial t}. $$ This gives me the three equations $$ \begin{align*} &\frac{dx}{ds}=x^2-1,\\ &\frac{dt}{ds}=1,\\ &\frac{du}{ds}=0 \end{align*} $$ together with three initial conditions $$ x(0)=\xi,\quad t(0)=0,\quad u(0)=\xi. $$
The second equation immediately gives $ds=dt$, hence I am left with the two equations $$ \begin{align*} &\frac{dx}{dt}=x^2-1,\\ &\frac{du}{dt}=0\implies u=C_1, \end{align*} $$ where $C_1$ is a constant.
Solving the first equation gives me $$ \frac{1}{2}\ln(x-1)-\frac{1}{2}\ln(x+1)=\frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)=t+C\tag{2}\label{eq2} $$ which yields $$ \frac{x-1}{x+1}=C_2e^{2t}.$$ Hence the general solution is given by $$ u(x,t)=u_0\left(e^{-2t}\frac{x-1}{x+1}\right).\tag{3}\label{eq3} $$
Let $X(t;\xi)$ be the associated flow map, i.e. $\frac{d}{dt}u(X(t),t)=0$ and $X(0,\cdot)=\textrm{Id}$.
Prove that for all $t\neq 0$, $X(t,\cdot)$ is injective but not surjective. Deduce that there is no smooth solution of (\ref{eq1}) defined for all $(x,t)\in\mathbb{R}^2$.
I am not sure what exactly is meant by "associated flow". I guess it is the flow which is associated to the ODE on the characteristic curves.
Probably the proof is related to the fact that, using (\ref{eq2}), the constant $$ C=\frac{1}{2}\ln\left(\frac{x-1}{x+1}\right)-t $$ is defined for $\lvert x\rvert>1$ only. This should mean that there are no characteristic curves going through $x\in [-1,1]$ and hence the flow is not surjective for $t\neq 0$ (for $t=0$ it is the identity and hence bijective). Thus the maximal domain connected to $\{t=0\}$ through the flow map should be $\mathbb{R}\setminus [-1,1]$. On this domain, the solution should be given by (\ref{eq3}).