Burgers' equation with piecewise linear initial data 
Solve
  $$ u_t + u u_x = 0 $$
  with initial data
  $$
u(x,0) = g(x) = \left\lbrace
\begin{aligned}
&0 &&\text{for}\; x < -1\\
&1-|x| &&\text{for}\; {-1}\leqslant x \leqslant 1 \\
&0 &&\text{for}\; x > 1\, .
\end{aligned}
\right.
$$
  In what region is the solution single-valued? Confirm this observation by sketching or plotting the base characteristics.

$$\frac{dt}{d\tau }=1$$
$$t=\tau $$
$$t(\tau=0)=0$$
$$\frac{dx}{d\tau}=u$$
$$x=g(\xi)\tau+\xi$$
$$x(\tau=0)=\xi$$
$$\frac{du}{d\tau}=0$$
$$u=g{\xi}$$
$$u(\tau=0)=g(\xi)$$
How should I take my "solutions" further? Any help is appreciated
 A: This problem is closely related to this one. The solution $u = g(x-ut)$ deduced from the method of characteristics is valid as long as it is single-valued, and we have
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 &&\text{for}\; x < -1\\
&\tfrac{x+1}{t+1} &&\text{for}\; {-1}\leqslant x \leqslant t\\
&\tfrac{x-1}{t-1} &&\text{for}\; t\leqslant x \leqslant 1\\
&0 &&\text{for}\; x > 1
\end{aligned}
\right.
$$
Here, the solution is single-valued for times $t<t_b$, where
$$
t_b = \frac{-1}{\inf g'} = 1
$$
is the breaking time. This time corresponds to the intersection of characteristic curves in the $x$-$t$ plane:

A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=0$ , letting $u(0)=u_0$ , we have $u=u_0$
$\dfrac{dx}{ds}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=f(u_0)+u_0s=f(u)+ut$ , i.e. $u=F(x-ut)$
$u(x,0)=\begin{cases}0&\text{for}~x<-1\\1-|x|&\text{for}~-1\leq x\leq1\\0&\text{for}~x>1\end{cases}$ :
$\therefore u=\begin{cases}0&\text{for}~x-ut<-1\\1-|x-ut|&\text{for}~-1\leq x-ut\leq1\\0&\text{for}~x-ut>1\end{cases}$
$u(x,t)=\begin{cases}0&\text{for}~x<-1~\text{or}~x>1\\\dfrac{x+1}{t+1}&\text{for}~-1\leq\dfrac{x-t}{t+1}\leq1\\\dfrac{x-1}{t-1}&\text{for}~-1\leq\dfrac{t-x}{t-1}\leq1\end{cases}$
