Traffic Waves

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Equations

We consider a single lane of road, and we measure distance along the road with the variable [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t }[/math] is time. We define the following variables

[math]\displaystyle{ \begin{matrix} &\rho(x,t) &: &\mbox{car density (cars/km)} \\ & v(\rho) &: &\mbox{car velocity (km/hour)} \\ & q(x,t) =\rho v &: &\mbox{car flow rate (cars/hour)} \\ \end{matrix} }[/math]

If we consider a finite length of road [math]\displaystyle{ x_1\leq x \leq x_2 }[/math] then the net flow of cars in and out must be balanced by the change in density. This means that

[math]\displaystyle{ \frac{\partial}{\partial t} \int_{x_1}^{x_2} \rho(x,t) dx = -q(x_2,t) + q(x_1,t) }[/math]

We now consider continuous densities (which is obviously an approximation) and set [math]\displaystyle{ x_2 = x_1 + \Delta x }[/math] and we obtain

[math]\displaystyle{ \frac{\partial}{\partial t} \rho(x_1,t) = -\frac{q(x_2,t) + q(x_1,t)}{\Delta x} }[/math]

and if we take the limit as [math]\displaystyle{ \Delta x \to 0 }[/math] we obtain the differential equation

[math]\displaystyle{ \frac{\partial \rho}{\partial t} + \frac{\partial q}{\partial x} = 0 }[/math]

Note that this equation has been derived purely from the need to conserve cars and it currently is not possible to solve until we have derived a connection between [math]\displaystyle{ \rho }[/math] and [math]\displaystyle{ q }[/math].

Relationship between [math]\displaystyle{ \rho }[/math] and [math]\displaystyle{ q }[/math]

The simplest relationship between [math]\displaystyle{ \rho }[/math] and [math]\displaystyle{ q }[/math] is derived from the following assumptions

• When the density [math]\displaystyle{ \rho = 0 }[/math] the speed is [math]\displaystyle{ v=v_0 }[/math]
• When the density is [math]\displaystyle{ \rho = \rho_{\max} }[/math] the speed is [math]\displaystyle{ v=0 }[/math]
• The speed is a linear function of [math]\displaystyle{ \rho }[/math] between these two values.

This gives us

[math]\displaystyle{ v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}} }[/math]

Equation for [math]\displaystyle{ \rho }[/math] only

If we substitute the expression for [math]\displaystyle{ q }[/math] into our differential equation we obtain

[math]\displaystyle{ \frac{\partial \rho}{\partial t} + \frac{\partial }{\partial x} \left(v(\rho)\rho\right) = 0 }[/math]

which gives us

[math]\displaystyle{ \frac{\partial \rho}{\partial t} + \left(v^{\prime}(\rho)\rho + v(\rho)\right) \frac{\partial \rho }{\partial x} = 0 }[/math]

or

[math]\displaystyle{ \frac{\partial \rho}{\partial t} + c(\rho)\frac{\partial \rho }{\partial x} = 0 }[/math]

where [math]\displaystyle{ c(\rho) = \left(v^{\prime}(\rho)\rho + v(\rho)\right) }[/math] is the kinematic wave speed. Note that this is not the speed of the cars, but the speed at which disturbances in the density travel.

Small Amplitude Disturbances

We can linearise the model by assuming that the variation in density is small so that we can write

[math]\displaystyle{ \rho = \rho_0 + \tilde{\rho} }[/math]

where we assume that [math]\displaystyle{ \tilde{\rho} }[/math] is small. This allows us to write the equations as

[math]\displaystyle{ \frac{\partial \tilde{\rho}}{\partial t} + c(\rho_0) \frac{\partial \tilde{\rho}}{\partial x} = 0 }[/math]

where the main difference between this and the full equation is that we assume that [math]\displaystyle{ c }[/math] is a constant. This linearises the equation.

Under these assumptions the solution to the equation is

[math]\displaystyle{ \tilde{\rho} = f(x - c(\rho_0)t) }[/math]

where [math]\displaystyle{ f }[/math] is determined by the initial condition. This represents disturbances which travel with speed [math]\displaystyle{ c(\rho_0) }[/math] in the positive [math]\displaystyle{ x }[/math] direction.

We now consider the characteristic curves which are curves along which the density [math]\displaystyle{ \rho }[/math] is a constant. These are give by

[math]\displaystyle{ x = X(t) = x_0 + c(\rho_0) t. }[/math]

Nonlinear Initial Value Problem

We wish to solve

[math]\displaystyle{ \frac{\partial \rho}{\partial t} + c(\rho) \frac{\partial \rho}{\partial x} = 0 }[/math]

subject to the initial conditions

[math]\displaystyle{ \rho(x,0) = \rho_0(x) \, }[/math]

It turns out that the concept of characteristic curves is very important for this problem.

If we want [math]\displaystyle{ \rho(X(t),t) }[/math] to be a constant then we require

[math]\displaystyle{ \frac{d}{dt}\rho(X(t),t) = \frac{d X}{dt} \frac{\partial \rho}{\partial x} + \frac{\partial \rho}{\partial t} = 0. }[/math]

Comparing this to the governing partial differential equation we can see that we require

[math]\displaystyle{ \frac{d X}{dt} = c(\rho) }[/math]

This means that the characteristics are straight lines with

[math]\displaystyle{ X(t) = x_0 + c(\rho_0(x_0))t \, }[/math]

This does not allow us to write down a solution to the initial value problem, all we can do is write

[math]\displaystyle{ \rho(x_0 + c(\rho_0(x_0))t,t) = \rho(x_0)\, }[/math]

which allows us to calculate the solution stepping forward in time, but not to determine the solution given a value of [math]\displaystyle{ (x,t) }[/math].

The characteristics are a family of straight lines which will all have different slopes. If two characteristics meet, our solution method will break down because there will be two values of the density [math]\displaystyle{ \rho }[/math]. This gives rise to a shock. It turns out that this is a problem with the equations themselves and not with the solution method and requires special methods.

Case when no shocks are formed

The characteristic curves will fill the space without meeting provided that the wave speed [math]\displaystyle{ c(\rho_0) }[/math] is a monotonically increasing function of the distance [math]\displaystyle{ x }[/math]. If we work with our previous model we have

[math]\displaystyle{ v(\rho) = v_0\frac{\rho_{\max} - \rho}{\rho_{\max}} }[/math]

and

[math]\displaystyle{ c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}} }[/math]

so that [math]\displaystyle{ c }[/math] is a monotonically decreasing function of density [math]\displaystyle{ \rho }[/math]. This means that the wave speed [math]\displaystyle{ c(\rho_0) }[/math] will be a monotonically increasing function of the distance [math]\displaystyle{ x }[/math] if an only if the density is a monotonically decreasing function. In this case the solution can be calculated straightforwardly by expansion of the initial density.

Riemann problem and the expansion fan

We can consider a simple problem in which there is a jump in the initial density

[math]\displaystyle{ \rho(x,0) = \left\{ \begin{matrix} \rho_{L},& x \lt 0 \\ \rho_{R},& x \gt 0 \end{matrix} \right. }[/math]

where [math]\displaystyle{ \rho_{L} \gt \rho_{R} }[/math] so that we do not form a shock. In this case the characteristics on each side of [math]\displaystyle{ x=0 }[/math] have a different slope and the question is what happens between. It is easiest to think about the following problem

[math]\displaystyle{ \rho(x,0) = \left\{ \begin{matrix} \rho_{L},& x \lt -\epsilon \\ \frac{\rho_{R}-\rho_{L}}{2\epsilon}x + \frac{\rho_{R}+\rho_{L}}{2} & -\epsilon \leq x \leq \epsilon \\ \rho_{R},& x \gt \epsilon \end{matrix} \right. }[/math]

We can then see that we have lines of uniformly varying slope for [math]\displaystyle{ -\epsilon\lt x\lt \epsilon }[/math] with slope between [math]\displaystyle{ c(\rho_L) }[/math] and [math]\displaystyle{ c(\rho_R) }[/math]. If we then take the limit as [math]\displaystyle{ \epsilon \to 0 }[/math] we obtain an expansion fan emanating from [math]\displaystyle{ x=0 }[/math].

If we assume that

[math]\displaystyle{ c(\rho) = v_0\frac{\rho_{\max} - 2\rho}{\rho_{\max}} }[/math]

then we know that on the lines of the expansion fan (which all start at [math]\displaystyle{ x=0 }[/math] we have [math]\displaystyle{ c(\rho) = x/t }[/math]. We can rearrange this and solve for [math]\displaystyle{ x }[/math] and obtain [math]\displaystyle{ \rho = 1/2 \rho_{\max} (1-x/v_0 t) }[/math]

The solution is then given by

[math]\displaystyle{ \rho(x,t) = \left\{ \begin{matrix} \rho_{L},& x \lt c(\rho_L) t\\ \frac{ \rho_{\max}}{2} (1-x/v_0 t),& c(\rho_L) t \leq x \leq c(\rho_R) t\\ \rho_{R},& x \gt c(\rho_R) t \end{matrix} \right. }[/math]

This solution is known as an expansion fan.

Shocks

So far we have only considered the case when [math]\displaystyle{ c(x_0) }[/math] is monotonically increasing so that two characteristics never cross. We now consider the case when characteristics can meet. We can easily see that the first characteristics to meet will be neighbouring characteristics. Consider two characteristics with

[math]\displaystyle{ X_1(t) = x_0 + c(x_0)t\, }[/math] [math]\displaystyle{ X_2(t) = x_0 + \delta x + c(x_0+\delta x)t\, }[/math]

Then these curves will meet at time [math]\displaystyle{ T }[/math] where

[math]\displaystyle{ x_0 + c(x_0)T = \delta x + c(x_0+\delta x)T\, }[/math]

which implies that

[math]\displaystyle{ T = -\frac{1}{c^{\prime}(x_0)} }[/math]

Note the following

• If [math]\displaystyle{ c^{\prime}(x) \lt 0 }[/math] then no shock will form.
• The shock first forms at [math]\displaystyle{ T_{\min} = \min_{-\infty \lt t \lt \infty}\frac{1}{c^{\prime}(x_0)} }[/math] .

Shock Fitting

If we calculate the solution using our formula

[math]\displaystyle{ \rho(x_0 + c(\rho_0(x_0))t,t) = \rho(x_0)\, }[/math]

then we find that the solution becomes multivalued in the case when a shock forms. We then have to fit a shock which we do by imposing the condition that equal areas are removed and added to the solution. This corresponds to the condition that the number of cars must be conserved

Speed of the shock

If we consider the case when there is a shock at [math]\displaystyle{ s(t) }[/math] with [math]\displaystyle{ \rho = \rho^{-} }[/math] at [math]\displaystyle{ s=s^{-} }[/math] and [math]\displaystyle{ \rho = \rho^{+} }[/math] at [math]\displaystyle{ s=s^{+} }[/math]. If we substitute this into the governing integral equation we obtain

[math]\displaystyle{ \frac{\partial}{\partial t} \left( \int_{x_1}^{s(t)} + \int_{s(t)}^{x_2}\right) \rho(x,t)dx = q(x_1,t) - q(x_2,t) }[/math]

and hence

[math]\displaystyle{ \int_{x_1}^{x_2} \frac{\partial \rho(x,t)}{\partial t} dx + \frac{ds}{dt}\rho^{-} - \frac{ds}{dt}\rho^{+} = q(x_1,t) - q(x_2,t) }[/math]

If we now take the limit as [math]\displaystyle{ x_1\to x_2 }[/math] we obtain

[math]\displaystyle{ \frac{ds}{dt}\rho^{-} - \frac{ds}{dt}\rho^{+} = q(\rho^{-}) - q(\rho^{-}) }[/math]

so that

[math]\displaystyle{ \frac{ds}{dt} = \frac{q(\rho^{-}) - q(\rho^{-})} {\rho^{-} - \rho^{+}} }[/math]

Riemann problem

We now consider the Riemann problem

[math]\displaystyle{ \rho(x,0) = \left\{ \begin{matrix} \rho_{L},& x \lt 0 \\ \rho_{R},& x \gt 0 \end{matrix} \right. }[/math]

where [math]\displaystyle{ \rho_{L} \lt \rho_{R} }[/math]. In this case a shock forms immediately and the characteristics terminate at the shock. The shock moves with constant speed given by the equation for the motion of the shock.