Finding the time when the speed of discontinuity becomes time-dependent in traffic flow 
I am trying to use the following conservation law:
$$u_t+f(u)_x=0 \ \ \ \ \text{where} \   \ \ f(u)=u(1-u).$$
IC: $u(x,0)=\frac{1}{4}$ for   BC: $u(0,t)=1$  for $t>0$.

I found the solution as $u(x,t)=\begin{cases} \frac{1}{4} & x<t\\
1 & x>t \end{cases}$.
Where my confusion lied was how you used s to get the bounds above.
Then we are tasked with the following questions:

Suppose that at $t_0=1$ the red light turns green - where t=0 the traffic is stopped - with a solution of,
$u(x,t)=\begin{cases} \frac{1}{4} & x<t\\
1 & 0>x>t\\
0 & x>0\end{cases}$
where the boundaries come from the first part.

At the first discontinuity there is a constant speed. I need to find the time at which the speed is non-constant.
Find the curve $y(t)$ describing the movement.
Now for the first part I have done the following. Let $u(x,t)=v(x/t)=v(z)$. Then I substituted this into the first equation to get,
$$\frac{xv'}{t^2}+\frac{v'}{t}-\frac{2vv'}{t}=0,$$ noting that we have used the chain rule to elimnate $v_t,v_x$. Then we can solve this ODE to get $$v=\frac{x+t}{2t}.$$ Then, from the Riemann Problem we have,
$$u(x,t)=\begin{cases}u_l & f'(u_l)t\\
v(z) & f'(u_l)t<x<f'(u_r)t\\
u_r & x>f'(u_r)t\end{cases}\ =\ \begin{cases}1 & x<-t\\
\frac{x+t}{2t} & -t<x<t\\
0 & x>t \end{cases}$$
From here I am confused on how to find $t_1$ and then how to progress onto finding the curve. Any tips would be appreciated.
 A: The Lagrange equations are $\frac{dt}1=\frac{dx}{1-2z}=\frac{dz}0$ with $z=u(x(t),t)$ constant along the characteristic curves. So that $$z=\phi(x-(1-2z)t)~~\text{ or }~~ x+(2z-1)t=\psi(z).$$

For $x<0$, $t>0$ there are two values for $z$ giving possibly characteristic curves going through each point $(x,t)$, 


*

*$z=1$ from the vertical boundary at $x_{init}=0$, leading to $x+t=t_{init}>0$, $x=t_{init}-t$ for $t>t_{init}$, and 

*$z=\frac14$ for $t_{init}=0$ giving $x-t/2=x_{init}<0$, $x=x_{init}+t/2$ for $t<-2x_{init}$.
These two phases collide, starting at $(x,t)=(0,0)$. The phase boundary $x=v(t)$ has, by the Rankine-Hugoniot condition for a shock wave, as speed the mean of the speeds of the phases, $$\frac{dv}{dt}=\frac{-1+1/2}2=-\frac14.$$ With that the solution of the PDE is
$$
u(x,t)=\begin{cases}\frac14,&x<-\frac t4,\\ 1, & -\frac t4 < x < 0.\end{cases}
$$

Then at $t=1$ the block at $x=0$, that is the boundary condition fixing the value there, is removed. The existing solution for $x<0$ is extended to the whole real axis via $u(x,1)=0$ for $x>0$. This gives a shock from the existing phase boundary starting at $a_0=-\frac14$ and a rarefaction wave at $x=0$. The available values of $z$ are


*

*$z=\frac14$ for $t_{init}=1$, $x_{init}<a_0=-\frac14$ giving $x-t/2=x_{init}-t_{init}/2$, $x=x_{init}+(t-1)/2$,

*$z=1$ for $a_0<x_{init}<0$, leading to $x+t=x_{init}+t_{init}>0$, $x=x_{init}+1-t$, and 

*$z\in [0,1]$ along the curves $x=(t-1)(1-2z)$,

*$z=0$ for $t-1>x>0$.
The remainder of the phase $z=1$ is used up where its left boundary $x=1-t$ for $x_{init}=0$ meets the shock wave at the right boundary $x=-\frac t4$. This happens at $t=\frac43$. After that the phase $z=\frac14$ collides with the rarefaction wave, the mean of the speeds at $(x,t)=(v(t),t)$ is
$$
v'(t)=\frac{\frac12+\frac{v(t)}{t-1}}2\iff \left(\frac{v}{\sqrt{t-1}}\right)'=\frac1{4\sqrt{t-1}}
\\
\implies \frac{v(t)}{\sqrt{t-1}}=C+\frac{\sqrt{t-1}}2
$$
and thus with $v(\frac43)=-\frac13$
$$
C=-\frac{\sqrt3}{2},~~ \boxed{v(t)=-\frac{\sqrt{3(t-1})}2+\frac{t-1}2}.
$$
