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.

  • $\begingroup$ 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? $\endgroup$ – LutzL Jan 8 at 12:19
  • $\begingroup$ @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. $\endgroup$ – KieranSQ Jan 8 at 12:45
  • $\begingroup$ You are working from the slides macs.hw.ac.uk/~lb138/slides_ch6_extra.pdf or something similar? $\endgroup$ – LutzL Jan 8 at 12:53
  • $\begingroup$ I’m working from something similar - notes directly from my lecture. $\endgroup$ – KieranSQ Jan 8 at 12:55
  • $\begingroup$ The model is explained in macs.hw.ac.uk/~lb138/slides_ch6.pdf. $\endgroup$ – 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 $t_{init}=1$, $x_{init}<a_0=-\frac14$ 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 $v(\frac43)=-\frac13$ $$ C=-\frac{\sqrt3}{2},~~ \boxed{v(t)=-\frac{\sqrt{3(t-1})}2+\frac{t-1}2}. $$

  • $\begingroup$ 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! $\endgroup$ – KieranSQ Jan 9 at 18:16
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    $\begingroup$ 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. $\endgroup$ – LutzL Jan 9 at 19:40
  • $\begingroup$ My last question, how do you get $u_r=\frac{u}{1-t}$? I cannot see why - apologies for all of the questions. $\endgroup$ – KieranSQ Jan 9 at 21:57
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    $\begingroup$ 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 $v$. $\endgroup$ – LutzL Jan 9 at 22:06

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