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!
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 $a=1/2$