Author Topic: Problem 6  (Read 15006 times)

Tharshi Srikannathasan

  • Newbie
  • *
  • Posts: 2
  • Karma: 0
    • View Profile
Problem 6
« on: September 24, 2012, 11:43:59 AM »
The answer for question b will change depending on whether we consider the highway to be one way or two way traffic. If one way then cars can only enter at a and leave at b. If we assume two way traffic, then cars can enter and leav both at a and b. The questions does not address the issue so I was wondering which one we should assume.

Thanks

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 6
« Reply #1 on: September 24, 2012, 11:46:56 AM »
One-way (no way to reverse the direction)

Aida Razi

  • Sr. Member
  • ****
  • Posts: 62
  • Karma: 15
    • View Profile
Re: Problem 6
« Reply #2 on: September 24, 2012, 09:00:05 PM »
Full solution to question 6 is attached,

Rouhollah Ramezani

  • Jr. Member
  • **
  • Posts: 13
  • Karma: 5
    • View Profile
Re: Problem 6
« Reply #3 on: September 25, 2012, 09:58:02 PM »
a) Directly implied by definition of $\rho(x,t)$ is $N(t,a,b)=\int_{a}^{b}\rho(t,x) dx$.

b) by definition of $q$ and conservation of cars we have:
$$ \frac{\partial N}{\partial t}(t,a,b)=\lim_{h \rightarrow 0} \frac{N(t+h,a,b)-N(t,a,b)}{h} \\
=\lim_{h \rightarrow 0} \frac{h(q(t,a)-q(t,b))}{h}  \\
=q(t,a)-q(t,b) $$

c) Differentiating integral form of $N(t,a,b)$ with respect to $t$
$$
\frac{\partial N}{\partial t}=\int_{a}^{b}\rho_t(t,x) dx$$
making it equal to result of part (b) we get the integral form of "conservation of cars":
$$
\int_{a}^{b}\rho_t(t,x) dx=q(t,a)-q(t,b) $$

d) RHS of above equation can be expressed as $\int_{a}^{b}-q_x(t,x) dx$. Therefore
\begin{equation} \int_{a}^{b}\rho_t(t,x) dx=\int_{a}^{b}-q_x(t,x) dx
\end{equation}
Since $a$ and $b$ are arbitrary, $(1)$ implies $\rho_t=-q_x$. The PDE $\rho_t+q_x=0$ is  conservation of cars equation.

e) Letting $q=c\rho$, we will get a first order linear PDE with constant coeffiecient:
$$\rho_t+c\rho_x=0 \\
\rightarrow \rho(t,x)=\phi(x-ct)
$$
The last equation means density of cars is moving at some constant pace $c$ in time, i.e. all cars are driving at speed $c$.
Intuitively we know that driver's speed has negative correlation with traffic density. A more realistic choice for $c$ is to let it be a monotone decreasing function of $\rho$. In this case however, conservation of cars equation is not linear anymore.  This is discussed in detail by Prof. Ivrii in last year's forum.
« Last Edit: September 16, 2013, 05:29:06 AM by Victor Ivrii »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2607
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Problem 6
« Reply #4 on: September 26, 2012, 02:20:39 AM »

Intuitively we know that driver's speed has negative correlation with traffic density. A more realistic choice for $c$ is to let it be a monotone decreasing function of $\rho$. In this case however, conservation of cars equation is not linear anymore.  This is discussed in detail by Prof. Ivrii in last year's forum.

The really interesting thing is a distinction between the speed of the individual cars $c(\rho)$ and the group speed $v(\rho)= (c(\rho)\rho)'= c'(\rho)\rho + c(\rho)$ of the group of cars. In fact the group does not have a constant "crew": if there is a place with higher density of cars than an average, it moves but the leading cars in the group have a lesser density in front, accelerate and leave the group while cars behind the group catch with it, and join it.

Such distinction is common in wave motion