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Density wave equation

  • Thread starter Dustinsfl
  • Start date
  • #1
699
5
Traffic is moving with a uniform density of [itex]\rho_0[/itex].
$$
\frac{\partial\rho}{\partial t} + c(\rho)\frac{\partial\rho}{\partial x} = \beta_0
$$
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 [itex]x = x_0[/itex] described by
$$
x = x_0 + u_{\text{max}}\left(1 - \frac{2\rho(x_0,0)}{\rho_{\text{max}}}\right)t - \beta_0\frac{u_{\text{max}}}{\rho_{\text{max}}}t.
$$

So we have [itex]\frac{dt}{ds} = 1[/itex], [itex]\frac{dx}{ds}=c(\rho)[/itex] and [itex]\frac{d\rho}{ds} = \beta_0[/itex].
Then [itex]t(s) = s + c[/itex] where [itex]t=s[/itex] when [itex]t(0) = 0[/itex].
Not sure how to handle the other two though.
 

Answers and Replies

  • #2
699
5
What I have done so far is:
[itex]\frac{dt}{dr} = 1\Rightarrow t = r + c[/itex] but when [itex]t = 0[/itex], we have [itex]t = r[/itex].

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

[itex]\frac{d\rho}{dr} = \beta_0\Rightarrow \rho = t\beta_0 + c[/itex]

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

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