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A traffic flow governed by the equations below is observed at $t = 0$ to have a dip in the density distribution at $x = 0$ given by:

$$ \rho (x,t=0) = \left\{\begin{aligned} &a &&: x \lt 0\\ &a- \frac {a}{2}(1+x)&&: -1<x< 0\\ &a- \frac {a}{2}(1-x)&&: 0<x<1\\ &a &&: 1<x\\ \end{aligned} \right.$$

where $0<a<1$

Where the equation of characteristics is given by $x-(1-2 \rho)t=x_0$

I am am trying to find the trajectory of the shock for $0<a<1$. I have worked out that the shock appears in the 3rd region in the above function so I have subbed the values $x_0=1$ and $x_0=0$ into my equation for the characteristic for the 3rd region and have worked out that they are equal when $t=1/a$ and $x= \frac {1-a}{a}$.

For the trajectory, I am trying to work out $ \frac {dS}{dt}= \frac {[q]}{[ \rho ]}$ where this represents the change in the flux given by the flux in the 4th region minus the flux in the 2nd region (left and right of the shock). The same idea for $[ \rho ]$.

Then to find the position of the shockwave, integrate with respect to $t$ and then solve for the constant using the initial conditions above.

I'm sure this is the correct method however the answer that I get is very messy and complicated and does not seem right.

Any help would be great!

juper
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1 Answers1

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It might not be as friendly a problem as it looks. Your procedure is correct, and when I went through it I got the following ODE $$ {\frac {\rm d}{{\rm d}t}}x \left( t \right) ={\frac {ax(t)-2\,{a}^{2}t +at-3\,a+2}{2(at+1) }}$$

Does that look at all familiar? It does not look very nice except for the special case $a=1/2$

Philip Roe
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  • How did you arrive at $ \frac {d}{dt} x(t)$? I thought the shock was $\frac {dS}{dt}$ I get confused between $x(t)$ and the position of the characteristics $x_0$. – juper May 05 '17 at 21:02
  • I am using dx/dt to define the shock speed which is $1-(\rho_L+\rho_R)$ – Philip Roe May 05 '17 at 21:29
  • I have been using that $ \frac {dS}{dt}$ as the velocity of the shock (trajectory) – juper May 05 '17 at 22:11
  • Then that's just notation. your $ S$ is my $x$. I was a bit lazy. I should probably have used $x_s$ – Philip Roe May 05 '17 at 22:56
  • To work out the trajectory I used $ \frac {dS}{dt}= \frac {[q]}{[ \rho ]}$ which I have got to be $\frac {a(1-a)-( \frac {a}{2}(1-x)(1- \frac {a}{2}(1-x))}{a- \frac {a}{2}(1-x)}$ as $a- \frac {a}{2}(1+x) = \frac {a}{2}(1-x)$ and then I just simplified this. Does this look right to you? Thanks for the help by the way. Note: My flux is given by $ \rho (1-\rho)$ – juper May 06 '17 at 10:50
  • @juper: This expression does not look correct, because it does not contain $t$. The general shockfree solution to that traffic equation is $\rho=F(x-(1-2\rho)t)$. where $F$ is an arbitrary function. For example, in region 2, $\rho=a/2(1-x+(1-2\rho)t$ which can be solved to get $$\rho=a/2(1-x+2t)/(1+at)$$ This should be one of your inputs to the shock trajectory equation. – Philip Roe May 06 '17 at 16:49
  • Yes that sounds right to me. I am confused on this topic to be honest. I have the characteristic equation for any region to be $x=(1-2 \rho )t +x_0$ and then for region 2, I have simply subbed in what $\rho$ is for the region and I get $x -t(1-ax+ax_0)=x_0$. Could you explain where i'm going wrong? Should the $x$ in the bracket be $x$ or $x_0$? – juper May 06 '17 at 17:16
  • Why does the denisty on the left of the shock have to contain $t$ when I have the density on the left of the shock given in the fucntion? – juper May 06 '17 at 17:22
  • and how is $\rho$ in region 2 given by what you've stated? I thought it's just how it is in the function. Sorry for the questions and thanks a lot for your time. – juper May 06 '17 at 17:26
  • What you are calling the characteristic equation only gives the current location of the characteristic line. You also need that the density does not change along this line, so that $\rho(x,t)=\rho(x_0,0)=F(x-(1-2\rho)t,0)$ which is what I gave. Clearer now? – Philip Roe May 06 '17 at 17:31
  • Oh right I see. So wheny I am working out $\frac {dS}{dt}$ and I use the density on the left of the shock, when I am subbing in to find my differences in flux, I should use that $x(t)=x-(1-2 \rho)t$ ? – juper May 06 '17 at 17:34
  • You are looking for on ODE $\tfrac{dS}{dt}= f(x,t)$, so you need to know $\rho(x,t)$ on both sides of the shock. – Philip Roe May 06 '17 at 17:43
  • I reread your points and I completely misinterpreted where you got $\rho$ from. I didnt realise that it was just the equation of the charcteristics rearranged for $\rho$. It now makes complete sense. Thanks for your time and effort! – juper May 06 '17 at 20:23