TITLE: Why is the solution single-valued?
QUESTION [6 upvotes]: I have shown that a smooth solution of the problem $u_t+uu_x=0$ with $u(x,0)=\cos{(\pi x)}$ must satisfy the equation $u=\cos{[\pi (x-ut)]}$. Now I want to show that $u$ ceases to exist (as a single-valued continuous function) when $t=\frac{1}{\pi}$.
When $t=\frac{1}{\pi}$, then we get that $u=\cos{(\pi x-u)}$.
With single-valued function is it meant that the function is 1-1 ?
If so, then we have that $\cos{(2 \pi-u)}=\cos{(4 \pi -u)}$, i.e. for two different values of $x$, we get the same $u$, and so for $t=\frac{1}{\pi}$, $u$ is not 1-1.
But if this is meant, how are we sure that for $t \neq \frac{1}{\pi}$ the function is single-valued?

REPLY [2 votes]: Based on the proof in this post, we have the expression for the breaking time
$$
t_b = \frac{-1}{\min \partial_x u(x,0)} = \frac{1}{\pi} \, .
$$
At $t=t_b$, characteristics intersect and a shock wave occurs. This is illustrated in the figure below, where the characteristic lines are displayed in the $x$-$t$ plane:

To see that the solution deduced from the method of characteristics becomes multi-valued for $t>t_b$, one may be interested by the discussion here.