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Burgers' equation:

$u_t+u u_x=0$

This solution $u(t,x)=c(1-\frac{2 \rho(t,x)}{\rho_0})$ should be verified. $c$ is the maximal velocity and $\rho_0$ is the maximal vehicle density, by which the traffic comes to a standstill.

Now calculating the partial derivitives:

$u_t=-\frac{2c}{\rho_0}\rho_t(t,x)$, $u_x=-\frac{2c}{\rho_0}\rho_x(t,x)$

Plugging the solution in the PDE:

$-\frac{2c}{\rho_0} \rho_t(t,x)-\frac{2c^2}{\rho_0} \rho_x(t,x)+\frac{4c^2}{\rho_0^2}\rho_x(t,x) \rho(t,x)=0$

Multiplying by $-\frac{\rho_0}{2c}$ yields:

$\rho_t(t,x)+\rho_x(t,x)(c-\frac{2c}{\rho_0} \rho(t,x))=0$

Now, I don't have an idea of what to do next.

Isaac Jérome
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    (The guy's name was Burgers, not Burger.) – Hans Lundmark May 28 '23 at 14:23
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    I edited the question, thanks for the remark :) – Isaac Jérome May 28 '23 at 14:37
  • You made a mistake in putting $u_t$ and $u_x$ into the PDE. The resulting equation isn't $-\frac{2c}{\rho_0} \rho_t(t,x)-\frac{2c^2}{\rho_0} \rho_x(t,x)+\frac{4c^2}{\rho_0^2}\rho_x(t,x)=0$ but is : $$-\frac{2c}{\rho_0} \rho_t(t,x)-\frac{2c^2}{\rho_0} \rho_x(t,x)+\frac{4c^2}{\rho_0^2}\rho(t,x)\rho_x(t,x)=0$$ which isn't linear. As a consequence what you finally found is of no use. – JJacquelin May 29 '23 at 07:49
  • True, thanks for mentioning that, I will edit the question accordingly – Isaac Jérome May 29 '23 at 12:43
  • OK. Now the PDE $\rho_t(t,x)+\rho_x(t,x)(c-\frac{2c}{\rho_0} \rho(t,x))=0$ is correct. Nevertheless what you are looking for is still as obscure as before because without initial and/or boundary conditions they are infinity many solutions $\rho(t,x)$. For example $\rho(t,x)=$constant satisfies the PDE (among infinity many other solutions). – JJacquelin May 29 '23 at 14:18

2 Answers2

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By setting $u = c \left( 1-2\rho/\rho_0\right)$, we find $$ u_t = -2c \rho_t/\rho_0, \quad u_x = -2c \rho_x/\rho_0 . $$ Hence, substitution in the Burgers equation $u_t + uu_x = 0$ produces $$ \rho_t + c \left( 1-2\rho/\rho_0\right) \rho_x = 0 \, . $$ where multiplication by $-\rho_0/(2c)$ was performed. This equation takes the form of the Lighthill-Whitham-Richards (LWR) traffic flow model, where $\rho$ is the car density. In fact, it can be rewritten as $$ \rho_t + (\rho v)_x = 0, \qquad v = c \left( 1-\rho/\rho_0\right) $$ where $v$ is the car velocity. It equals $c$ on empty roads ($\rho=0$), and it vanishes on saturated roads ($\rho = \rho_0$). The quantity $Q = \rho v$ is the flux of cars.

EditPiAf
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$$u_t-uu_x=0$$ The solution on implicit form is : $$t\,u(t,x)-x=F\big(u(t,x)\big)$$ In which $F$ is an arbitrary function.

With the specified constraint $u(t,x)=c(1-\frac{2 \rho(t,x)}{\rho_0})$ $$t\,c(1-\frac{2 \rho(t,x)}{\rho_0})-x=F\left(c(1-\frac{2 \rho(t,x)}{\rho_0})\right)$$ Since they are infinity many functions $F$ they are infinity many solutions $\rho(t,x)$.

For example one of the simplest is with $F=C=$constant , leading to $\rho(t,x)=\frac{\rho_0}{2}\left(1-\frac{x+C}{c\,t} \right)$

Obviously this is non-sens. Probably there is something missing in the wording of the question to overcome the indeterminancy.

JJacquelin
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  • Thank you. Our task is to show that the given solution $u(t,x)=c(1-\frac{2 \rho(t,x)}{\rho_0})$ is valid for the Burger's equation – Isaac Jérome May 29 '23 at 12:49
  • The velocity: $v(t,x)=c(1-\frac{ \rho(t,x)}{\rho_0})$ and $u(t,x)=v(t,x)-c \frac{\rho(t,x)}{\rho_0}=c(1-\frac{2 \rho(t,x)}{\rho_0})$ – Isaac Jérome May 29 '23 at 13:00
  • On mathematical viewpoint $u = c \left( 1-2\rho/\rho_0\right)$ isn't a solution of the PDE. This is a change of variable which transforms the original PDE into another PDE. – JJacquelin May 29 '23 at 14:34