Exercise :
Show that a smooth solution of the initial value problem $$\begin{cases} u_t + uu_x = 0 \\ u(x,0) = \cos \pi x \end{cases}$$ must satisfy $u = \cos[\pi(x-ut)]$. Also, show that when $t=1/u$, the function $u$ stops to exist (as a single-value function).
Attempt :
$$\frac{\mathrm{d}x}{u} = \frac{\mathrm{d}t}{1} = \frac{\mathrm{d}u}{0} $$
$$\implies u_1 = u, \; \; u_2 = x - ut$$
Which means that the solutions are given by $u_1 = F(u_2) \Rightarrow u = F(x-ut)$. Thus, taking into account the initial value : $F(x) = \cos \pi x.$ Now, for $x:= x-ut$ we get $F(x-ut) = \cos[\pi(x-ut)]$ which $\in C^\infty$ and thus a smooth solution.
Question : How would one proceed to show the second part of the exercise ? It gives as a hint to check the graphs of $\arccos(u)$ and $\pi(x-ut)$ as functions of $u$, which makes sense, since :
$$\arccos(u)=\pi(x-ut)$$
But I can't seem how to grasp out a graph or figure out a solution to it.
