Shock-wave solution for PDE $u_t+(u-1)u_x=2$ I want to solve the following PDE initial value problem
$u_t+(u-1)u_x=2$
and
$u (x,0)=\begin{cases}
1 & \text{for } x <0,\\
1-x & \text{for } 0<x <1\\
0 & \text{for } 1 <x
\end{cases}$
However, I find that I have intersecting characteristics between $x=t^2$ and $x=t^2-t+1$. 
How would I apply the shockwave method in this case since the PDE is given? Is it possible to solve this PDE as is?
 A: Following the comments, we transform this problem by setting $v=u-1$ as
$$
v_t + v v_x = 2, \qquad v(x,0) = \left\lbrace
\begin{aligned}
&0 &&\text{for } x<0\\
&{-x} &&\text{for } 0<x<1\\
&{-1} &&\text{for } 1<x\\
\end{aligned}\right.
$$
The breaking time for the Burgers equation above is $t_b = -1/\inf v_x(x,0) = 1$. Hence, there is indeed a shock formation. Before the shock, the solution is given by the method of characteristics. Here, the characteristic curves are $x = v(x_0,0) t + t^2 + x_0$ along which $v = v(x_0,0)+2t$. Hence, for $t<1$, the solution reads
$$
v(x,t) = \left\lbrace
\begin{aligned}
&2t &&\text{for } x<t^2\\
&\tfrac{t^2-x}{1-t} + 2t &&\text{for } t^2<x<t^2-t+1\\
&{-1}+2t &&\text{for } t^2-t+1<x\\
\end{aligned}\right.
$$
When characteristic curves intersect, the Rankine-Hugoniot condition gives the shock speed $\dot x_s(t) = \frac{1}{2}(2t - 1 + 2t)$ with $x_s(1) = 1$. Therefore, for $t\geq 1$, the solution reads
$$
v(x,t) = \left\lbrace
\begin{aligned}
&2t &&\text{for } x< \tfrac{1}{2}(4t-1)\\
&{-1}+2t &&\text{for } \tfrac{1}{2}(4t-1)<x\\
\end{aligned}\right.
$$
To recover $u$, add $1$ to the values of $v$ above.
A: @user393349 : Sorry, something has gone wrong during re-typing my answer . I first closed it and later I re-open it to continue the typing. The main part was done when abruptly all disappeared. Probably it was closed by someone else. 
All my latex is lost. It took me a so long time that I will not write it again. 
The only think remaining is a graph where the results are summarized. This shows where and when a blow-up point appears. I don't know if this can help you (I hope so).
 
If this figure is of no interest for you, I suppose that the answer will be definitively deleted. Good continuation.
A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=2$ , letting $u(0)=u_0$ , we have $u=u_0+2s=u_0+2t$
$\dfrac{dx}{ds}=u-1=u_0+2s-1$ , letting $x(0)=f(u_0)$ , we have $x=s^2+(u_0-1)s+f(u_0)=t^2+(u-2t-1)t+f(u-2t)=(u-1)t-t^2+f(u-2t)$ , i.e. $u=2t+F(x+t^2-(u-1)t)$
$u(x,0)=\begin{cases}1&\text{when}~x<0\\1-x&\text{when}~0<x<1\\0&\text{when}~1<x\end{cases}$ :
$\therefore u=\begin{cases}1&\text{when}~x+t^2-(u-1)t<0\\2t+1-x-t^2+(u-1)t&\text{when}~0<x+t^2-(u-1)t<1\\0&\text{when}~1<x+t^2-(u-1)t\end{cases}$
Hence $u=\begin{cases}1&\text{when}~x<-t^2\\\dfrac{x+t^2-t-1}{t-1}&\text{when}~0<x+t^2-\dfrac{(x+t^2-t-1)t}{t-1}+t<1\\0&\text{when}~x>1-t^2-t\end{cases}$
