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Consider the PDE initial value problem $$\frac{\partial u}{\partial t}+(1-2u)\frac{\partial u}{\partial x}=0,$$ $$u(x,0)=f(x),$$ with the initial conditions for traffic congestion:

$$f(x)=\begin{cases} \frac{1}{3} & \text{if }\left|x\right|\ge 1\\ \frac{1}{2}\left( \frac{5}{3}-\left| x \right| \right) &\text{if } \left|x\right|\le 1\\ \end{cases}$$

This equation can be solved by the method of characteristics to obtain $$u(x,t)=f(x)+f(x-t(1+2u))$$

Now to find the time $t_s$ and position $x_s$ of the initial shock formation, we can do the following:

$$u(x,t) = f(x)+f(\tau)$$ $$x=(1-2f(\tau))t+\tau$$ $$u_x(x,t) = f'(x) + f'(\tau)\tau_x$$ $$1=-2tf'(\tau)\tau_x+\tau_x=\tau_x(1-2tf'(\tau))$$ $$\tau_x=\frac{1}{1-2tf'(\tau)}$$ $$u_x = f'(x)+\frac{f'(\tau)}{1-2tf'(\tau)}$$

Now to find the time and position of the initial shock formation, we need

$$t_s=\min_\tau\left( -\frac{1}{f'(\tau)} \right)$$

But the question is: how do we find $f(\tau)$?

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  • $\begingroup$ Well if you have $f(x)$, $f(\tau)$ is the same, just replace $x$ by $\tau$ :-) $\endgroup$
    – Shawn
    Aug 7, 2016 at 4:52
  • $\begingroup$ $x$ and $\tau$ are two completely different variables here. $\tau = x-(1-2u)t$. $\endgroup$
    – sequence
    Aug 7, 2016 at 4:58
  • $\begingroup$ If you last equation is filly correct (I didn't check the above math), it does not matter. What matters is the definition of $f$. You can express $f'$ "easily" from the definition of $f$, then plug that in the last equation, find it's minimum (if any/if it exists) by minimizing over $\tau$ and that will give you $t_s$. I didn't fully understand the math above though so maybe the last equation means different to you. $\endgroup$
    – Shawn
    Aug 7, 2016 at 5:21
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    $\begingroup$ @sequence : did you check if your result $u(x,t)=f(x)+f(x-t(1+2u))$ agrees with the PDE ? You just have to put it into the PDE and see if it is satisfied. If you had done this check, you would have seen that your result is false. $\endgroup$
    – JJacquelin
    Aug 7, 2016 at 7:33
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    $\begingroup$ Now, $u(x,t)=f(x-t(1-2u))$ is OK. In the wording of your question, you should correct the equation $u(x,t)=f(x)+f(x-t(1+2u))$ which is false : why is there $f(x)$ in it ?. $\endgroup$
    – JJacquelin
    Aug 7, 2016 at 7:55

2 Answers 2

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The general solution of the PDE is : $$u(x,t)=F(x-t(1-2u))$$ where $F$ is any differentiable function.

Do not confuse $F$ with $f$ appearing in the initial condition : $$f(x)=\begin{cases} \frac{1}{3} & \text{if }\left|x\right|\ge 1\\ \frac{1}{2}\left( \frac{5}{3}-\left| x \right| \right) &\text{if } \left|x\right|\le 1\\ \end{cases}$$ The initial condition is : $u(x,0)=F(x)=f(x)$ which determines $F$ as : $$F(X)=\begin{cases} \frac{1}{3} & \text{if }\left|X\right|\ge 1\\ \frac{1}{2}\left( \frac{5}{3}-\left| X \right| \right) &\text{if } \left|X\right|\le 1\\ \end{cases}$$ In the particular solution fitting to the initial condition, $X\neq x$ but $X=x-t(1-2u)$ $$u(x,t)=\begin{cases} \frac{1}{3} & \text{if }\left|x-t(1-2u)\right|\ge 1\\ \frac{1}{2}\left( \frac{5}{3}-\left| x-t(1-2u) \right| \right) &\text{if } \left|x-t(1-2u)\right|\le 1\\ \end{cases}$$ You see that it is very different from the cases $|x|\ge 1$ and $|x|\le 1$ because $u$ is involved into them.

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This is a sketch of the characteristic curves $x(t) = (1-2f(x_0))t+x_0$ in the $x$-$t$ plane:

characteristics

Characteristics seem to intersect at $t=1$. This can be recovered via the expression of the breaking time \begin{aligned} t_s &= \frac{-1}{\min\, (1-2f)'(x)} \\ &= \frac{1/2}{\max f'(x)} \\ &= 1 \, . \end{aligned}

N.B. : this is the LWR traffic flow model.

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