MHB Characteristics and Initial Density Distribution in Traffic Flow Wave Equation

[Dustinsfl]

Suppose that along a stretch of highway the net flow of cars entering (per unit length) can be taken as a constant $\beta_0$.
The governing equation of motion is then
$$
\frac{\partial\rho}{\partial t} + c(\rho)\frac{\partial\rho}{\partial x} = \beta_0
$$
where
$$
c(\rho) = u_{\text{max}}\left(1 - \frac{2\rho}{\rho_{\text{max}}}\right).
$$
Show that the variation of the initial density distribution is given by
$$
\rho = \beta_0t + \rho(x_0,0)
$$
along a characteristic emanating from $x = x_0$ described by
$$
x = x_0 + u_{\text{max}}\left(1 - \frac{2\rho(x_0,0)}{\rho_{\text{max}}}\right)t - \beta_0\frac{u_{\text{max}}}{\rho_{\text{max}}}t.
$$

What I have done so far is:
$\frac{dt}{dr} = 1\Rightarrow t = r + c$ but when $t = 0$, we have $t = r$.

$\frac{dx}{dr} = c(\rho)\Rightarrow x = tu_{\text{max}}\left(1-\frac{2\rho}{\rho_{\text{max}}}\right)+c$ but when $t=0$, we have
$$
x = tu_{\text{max}}\left(1-\frac{2\rho}{\rho_{\text{max}}}\right) + x_0.
$$

$\frac{d\rho}{dr} = \beta_0\Rightarrow \rho = t\beta_0 + c$

How do I get to
$$
\rho(x,t) = t\beta_0 +\rho(x_0,0)
$$
and their characteristic?

 

[Sudharaka]

Hi dwsmith, :)

$\frac{d\rho}{dr} = \beta_0\Rightarrow \rho = t\beta_0 + c$

How do I get to
$$
\rho(x,t) = t\beta_0 +\rho(x_0,0)
$$
and their characteristic?

\[\rho = t\beta_0 + c\]

When, \(t=0\) we have, \(\rho(x=x_{0},t=0)=\rho(x_{0},0)\). Therefore,

\[\rho=t\beta_{0}+\rho(x_{0},0)\]

$\frac{dx}{dr} = c(\rho)\Rightarrow x = tu_{\text{max}}\left(1-\frac{2\rho}{\rho_{\text{max}}}\right)+c$ but when $t=0$, we have
$$
x = tu_{\text{max}}\left(1-\frac{2\rho}{\rho_{\text{max}}}\right) + x_0.
$$

This is incorrect. Note that \(\rho\) depends on \(t\) but you have considered it as a constant when integrating. The correct method is to substitute for \(\rho\) first so as to obtain,

\[\frac{dx}{dr}=\frac{dx}{dt}=c(\rho)=u_{\text{max}}\left(1 - \frac{2\rho}{\rho_{\text{max}}}\right)\]

\[\Rightarrow \frac{dx}{dt}=u_{\text{max}}\left(1 - \frac{2\rho(x_{0},0)}{\rho_{\text{max}}}\right)-\frac{2\beta_{0}u_{\text{max}}t}{\rho_{\text{max}}}\]

Integration gives,

\[x=x_{0}+u_{\text{max}}\left(1 - \frac{2\rho(x_{0},0)}{\rho_{\text{max}}}\right)t-\frac{\beta_{0}u_{\text{max}}t^{\color{red}2}}{ \rho_{\text{max}}}\]

Kind Regards,
Sudharaka.

 

[Dustinsfl]

\[\Rightarrow \frac{dx}{dt}=u_{\text{max}}\left(1 - \frac{2\rho(x_{0},0)}{\rho_{\text{max}}}\right)-\frac{2\beta_{0}u_{\text{max}}t}{\rho_{\text{max}}}\]

Where did this piece come from?

$$\frac{2\beta_{0}u_{\text{max}}t}{\rho_{\text{max}}}$$

 

[Sudharaka]

Where did this piece come from?

$$\frac{2\beta_{0}u_{\text{max}}t}{\rho_{\text{max}}}$$

Substitute \(\rho=t\beta_{0}+\rho(x_{0},0)\) for \(\rho\) in \(\frac{dx}{dt}=c(\rho)=u_{\text{max}} \left(1 - \frac{2\rho}{\rho_{\text{max}}}\right)\)

 

[Dustinsfl]

Substitute \(\rho=t\beta_{0}+\rho(x_{0},0)\) for \(\rho\) in \(\frac{dx}{dt}=c(\rho)=u_{\text{max}} \left(1 - \frac{2\rho}{\rho_{\text{max}}}\right)\)

How can I use Mathematica to sketch the space time diagram?

 

[Sudharaka]

How can I use Mathematica to sketch the space time diagram?

I have never used Mathematica so I won't be able to answer your question. By space-time diagram do you mean the graph between \(x\) and \(t\) ?

 

[Dustinsfl]

I have never used Mathematica so I won't be able to answer your question. By space-time diagram do you mean the graph between \(x\) and \(t\) ?

Yes

 

[Sudharaka]

Yes

In that case it is a parabola. But do you have any specific values for the constants, \(x_{0},\,u_{\text{max}},\,\rho(x_{0},0),\,\rho_{ \text{max}}\mbox{ and }\beta_{0}\) ?

 

[Dustinsfl]

In that case it is a parabola. But do you have any specific values for the constants, \(x_{0},\,u_{\text{max}},\,\rho(x_{0},0),\,\rho_{ \text{max}}\mbox{ and }\beta_{0}\) ?

No.

 

[Sudharaka]

No.

Then it seems problematic as to how you would graph this equation in any mathematical software. What you can only say is that the graph between \(x\) and \(t\) is parabolic.