$How does the traffic flow model arrive at the scaled equation?$

Question

$How does the traffic flow model arrive at the scaled equation?$

Consider the model for traffic flow with density $ρ(x, t)$ that satisfies
$ρ_t + (ρV_*(1 − ρ/ρ_*))_x = 0$
where $V_*$ was a given maximum speed and $ρ_*$ a given maximum density. Show that by scaling $ρ$ and a combination of $x$ (space) and $t$ (time) we can arrive at the scaled equation
$u_t + (u − u^2)_x = 0$

Partial Differential Equations Differential Equations Ordinary Differential Equations

Chdogordon

153

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Since $\rho / \rho_*$ appears in the equation we look for some $u$ of the form $$ u(x,t):= {1 \over \rho _*} \rho (ax,bt) $$ for some constants $a,b$. Now we have $u_t =b\rho _t / \rho _* $. So $\rho _t = \rho _* u_t /b$, and the equation for $u$ becomes $$ \rho _* u_t/b + (V_* \rho _* u(1-u))_x = \rho _t + (V_* \rho (1-\rho /\rho _* ))_x = 0 $$ Now if we choose $b$ such that $\rho _* /b = V_* \rho_*$ then we can take $ V_* \rho_*$ outside the derivative (since it is constant) and factor it out of the equation to get $$ u_t + ( u(1-u))_x=0 $$ as desired. Hence we can set $b=1/V_*$. And since $a$ did not appear in our calculations we set it equal to 1 (i.e. we do not scale w.r.t. $x$). So the final answer becomes $$ u(x,t):= {1 \over \rho _*} \rho (x,t/V_*)$$

Martin

1.7K

$How does the traffic flow model arrive at the scaled equation?$

Answer

Related Questions

Search