# 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 $$u(x,t)=\begin{cases} \frac{1}{4} & xt \end{cases}$$.

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, $$u(x,t)=\begin{cases} \frac{1}{4} & xx>t\\ 0 & x>0\end{cases}$$ 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, $$u(x,t)=\begin{cases}u_l & f'(u_l)t\\ v(z) & f'(u_l)tf'(u_r)t\end{cases}\ =\ \begin{cases}1 & x<-t\\ \frac{x+t}{2t} & -tt \end{cases}$$

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.

• Could you give more details how "the light is green/red" is reflected in your equation? It seems that you are computing shock/dispersion waves, does the Rankine-Hugoniot equation appear in your calculus? – LutzL Jan 8 at 12:19
• @LutzL this is part of where my confusion lies - I am unsure how the red\green light part works in relation to the equations. Yes I originally thought of using the Rankine-Hugoniot Equation to get $s=y'=\frac{u_r+u_l}{2}$ but I was unsure on where to apply it. – KieranSQ Jan 8 at 12:45
• You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar? – LutzL Jan 8 at 12:53
• I’m working from something similar - notes directly from my lecture. – KieranSQ Jan 8 at 12:55
• The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf. – LutzL Jan 8 at 13:03

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 $$t_{init}=1$$, $$x_{init} giving $$x-t/2=x_{init}-t_{init}/2$$, $$x=x_{init}+(t-1)/2$$,

• $$z=1$$ for $$a_0, leading to $$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 $$v(\frac43)=-\frac13$$ $$C=-\frac{\sqrt3}{2},~~ \boxed{v(t)=-\frac{\sqrt{3(t-1})}2+\frac{t-1}2}.$$

• how did you get that $x-\frac{t}{2}=x_i-\frac{t_i}{2}$? I followed in the first part but now I am stuck. Thanks in advance! – KieranSQ Jan 9 at 18:16
• We have $z=\frac14$ and $x+(2z-1)t=const.$ so that $x-\frac12t$ is constant and thus also the same where the characteristic curve crosses the boundary. – LutzL Jan 9 at 19:40
• My last question, how do you get $u_r=\frac{u}{1-t}$? I cannot see why - apologies for all of the questions. – KieranSQ Jan 9 at 21:57
• The line $x+(2z-1)t=c$ that goes through the rarefaction center $(x,t)=(0,1)$ requires $c=(2z-1)$, so that $x=-(2z-1)(t-1)$, the slope of that line is thus $\dot x=1-2z=\frac{x}{t-1}$. I see that re-using $u$ is a bad idea, I'll change that to $v$. – LutzL Jan 9 at 22:06