- #1

- 699

- 5

$$

\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.