Kinematic waves and fluid-dynamics

The Lighthill-Whitham-Richards equation

The intuition for kinematic waves is easy to understand. Start with five vehicles of velocity zero in five adjoining cells. In the first time step, only the first vehicle can move. In the second time step, the second vehicle can start, etc. However, in the meantime it can happen that another vehicle joins the queue at the tail.

Given the right conditions (more vehicles joining at the tail than leaving at the head), this results in a cluster of vehicles of velocity zero and that cluster will move against the traffic direction. Note that the vehicle composition of this cluster is constantly changing - from the perspective of a driver, you join the jam from the end, the jam ``moves through you'', and then you can start again (look at the two trajectories in the lower part of Fig. 27.4 for an illustration). This is a standard wave phenomenon.

A detailed introduction into such waves can for example be found by Haberman (52). Here, we will just [[word?]] give an overview for people who have some prior knowledge about partial differential wave equations.

One way to see all the connections [[word?]] is to start from the standard equation of continuity, which needs to be fulfilled as long as our traffic obeys mass conservation (no vehicles leaving or joining). This equation is (equation of continuity). This equation can be easily understood when it is discretized (with discretization constants $\Delta t = 1$ and $\Delta x = 1$):

$$N_{t+1}(x) = N_t(x) - \Big( q_t(x+\frac{1}{2}) - q_t(x-\fra... ... \Big) \\ = N_t(x) + q_t(x-\frac{1}{2}) - q_t(x+\frac{1}{2})$$ (27.25)

where $N_t(x)$ is the number of vehicles in a spatial interval of size $\Delta x = 1$. The notation mirrors the computational implementation, where the spatial index would be represented by an array index, while the temporal index would typically not show up at all. The equation states that the number of vehicles at time $t+1$ is equal to the number of vehicles at time $t$, plus what flows in from the left, and minus what flows out to the right.

We now need a relation between $q$ and $\rho$. Let us assume that $q$ is a function of $\rho$ only, i.e. the total differential is $dq = \frac{dq}{d\rho} \, d\rho$. The meaning of this (instantaneous velocity adaptation) will be discussed below. The resulting theory is also called the Lighthill-Whitham-Richards (LWR) theory (70). [[Richards ref]] The equation of continuity can immediately re-written as (LWR equation), where $q(\rho )$ is some externally given but as of yet unspecified function.

Figure 27.9: Illustration of Eq. (27.26).

Linearization

Since we now have a fully defined partial differential equation, we can try to understand some of it. A typical first step is ``linearization''. For this, $\rho$ is replaced by $\overline \rho + \rho'$, with $\partial_t \overline{\rho}=0$ (stationary) and $\partial_x \overline{\rho}=0$ (homogeneous); this is always possible. One now assumes that $\rho'$ is small. Functions in $\rho$ are Taylor-expanded:

$$F(\rho) = F(\overline{\rho}) + \rho' {dF \over d\rho}(\overline \rho) + ... \ ;$$ (27.26)

in our case, we need $F = dq/d\rho$. This results in

$$\partial_t \rho' + \Big( {dq \over d\rho}(\overline\rho) +... ...} (\overline\rho) + \ldots \Big) \, \partial_x \rho' = 0 \ .$$ (27.27)

Finally, higher-order terms (i.e. which contain products of $\rho'$) are dropped, resulting in

$$\partial_t \rho' + {dq \over d\rho}(\overline\rho) \, \partial_x \rho' = 0 \ .$$ (27.28)

This is now a linear equation in $\rho'$, since in each term $\rho'$ occurs at most once. In such cases, one knows that one can make the ansatz

$$\rho' = A \, e^{i (\omega t - kx)} \ .$$ (27.29)

If one has never seen this before, it is probably impossible to explain this in two minutes.^27.6 Inserting Eq. (27.31) into Eq. (27.30) leads to

$$\omega - {dq \over d\rho}(\overline\rho) \, k = 0$$ (27.30)

and therefore to This is the phase velocity of the travelling wave. That is, this wave will travel in traffic direction when $q(\overline\rho)$ is increasing ( $\frac{dq}{d\rho}(\overline\rho)$ positive), and against the traffic direction when $q(\overline\rho)$ is decreasing (Fig. 27.10).

Figure 27.10: Phase speeds of kinematic waves

Macroscopic shocks

Linearization is not very useful for traffic, since it assumes small $\rho'$, which is often not fulfilled in traffic. Let us thus look at a macroscopic front with speed $c$. Let us go to the same reference system as the front. In that reference system, the flow to the left of the front needs to be the same as the flow to the right of the front, because otherwise there would either be an excess or a lack of ``material'' at the front. Let us denote variables in the reference system of the front with a tilde. In equations, the statement means

$$\tilde q_l = \tilde q_r \ .$$

Now $\tilde q = \rho \, \tilde v$, where the density $\rho$ does not need a tilde because it is independent from the speed of the reference system. That is, one has

$$\rho_l \, \tilde v_l = \rho_r \, \tilde v_r \ .$$

For the translation into a non-moving coordinate system, one has $\tilde v = v + c$, and therefore

$$\rho_l \, (v_l + c) = \rho_r \, (v_r + c)$$ (27.31)

Rearranging yields One can see geometrically that this is just the slope of the line connecting the corresponding points on the fundamental diagram (Fig. 27.11).

Figure 27.11: Speed of discontinuous fronts

The deterministic CA in terms of kinematic waves

We can now analyse our deterministic CA (Sec. 27.4.2) in terms of kinematic waves (see also Fig. 27.5):

• In the laminar regime, we have $dq/d\rho = v_{max}$. This means that our waves have the same speed as the traffic -- that is, they are the ``clusters'' or ``platoons'' of cars.

• In the congested regime, $dq/d\rho = -1$. This can be seen in the space-time diagram via the fact that the ``patterns'' move backwards one cell in each time step (Fig. 27.4 bottom).

• With respect to our introductory problem with the five cars: The jam has density $\rho=1$ and speed $v=0$, thus also $q=0$. Outflow from the jam is eventually at $v=v_{max}$ and $\rho = 1/(v_{max}+1)$ (this can be seen by following the dynamics). In consquence,
  

  $${\Delta q \over \Delta \rho} = {v_{max}/(v_{max}+1) - 0 \over 1/(v_{max}+1) - 1} = -1 \ .$$ (27.32)

  
  
  Thus, the downstream front of the jam moves backwards with speed $c = -1$. -- One could also have seen that by noticing that the outflow is equal to the maximum flow in this model, and then do the geometric solution similar to Fig. 27.11.

  [[might be good to do xfig here too]]

  The inflow is somewhere on the ``laminar'' branch of the fundamental diagram. That means that the slope of the line connecting to $(\rho=0,q=0)$ is either $-1$ or less steep. The inflow front thus moves backwards with speed $1$ or less -- that is, the jam will eventually vanish except when inflow is exactly equal to maximum flow.

One can treat queues at traffic lights similarly. While the traffic light is red, $q_{out}=0$ and thus the outflow front does not move (which we know since the first car is waiting at the red light). The inflow front moves backwards with $c_{in} = q_{in}/(\rho_{in}-1)$.

Once the traffic light turns green, the outflow front now moves backwards with $-1$, while the inflow front keeps moving backwards with $c_{in}$. The situation remains like that until the outflow front catches up with the inflow front. And if the traffic light turns red before that, one needs to include that effect (Fig. 27.12).

Figure 27.12: Traffic light in terms of kinematic waves

More advanced fluid-dynamical models

The kinematic theory is entirely sufficient to understand the most important theoretical aspects of traffic flow. This section goes a little bit beyond that, by providing an outlook what else could be done.

The STCA and in particular the slow-to-start model are not entirely described by the kinematic theory. This is in part due to the stochastic elements, which are not captured in the equation. It is also due to the hysteresis which is displayed by the slow-to-start model (Fig. 27.7) but not by kinematic theory. This motivates to look for fluid-dynamical equations for traffic that capture effects beyond the kinematic theory. Two extensions of the kinematic theory will be discussed.

Addition of diffusive terms

Diffusive terms can be justified for many reasons. The result is an equation like

$$\partial_t \rho + \partial_x q = D \partial_x^2 \rho \ .$$ (27.33)

The wave solution after linearization now is [[check]]

$$\rho' = A \, e^{-k^2 D t} \, e^{i(\omega t - kx)}$$ (27.34)

which means that it has the same phase velocity $c=dq/d\rho$ as before but in addition a decreasing amplitude -- waves slowly die out.

Addition of inertia

Above, we have assumed that flow $q$ is a function of the density $\rho$ only. This is in general not true -- if a driver suddenly comes into denser traffic, she/he will need some time to adjust; the same is true if density suddenly decreases. That means that velocity will be delayed in its adaptation to density.

A way to capture this is to add an equation for the velocity. One can for example use the car following equation (27.15)

$$a = {Dv \over Dt} = {1 \over \tau} \Big( V(\Delta x) - v \Big) \ .$$ (27.35)

The translation of the particle-oriented $Dv/Dt$ into the fluid-dynamical $\partial_t v + v \, \partial_x v$ yields

$$\partial_t v + v \, \partial_x v = {1 \over \tau} \Big( V(\Delta x) - v \Big) \ .$$ (27.36)

We need however $V(\rho)$ instead of $V(\Delta x)$, and we also need $\rho$ measured at the location of the vehicle and not in the middle between two vehicles, where $\Delta x$ is measured.^27.7This is the mathematical reason for what is usually called the anticipation term

$$- {c_0^2 \over \rho} \, \partial_x \rho \ .$$ (27.39)

If density goes up in the driving direction, then $\partial_x \rho$ is positive, thus the term causes negative acceleration, which is plausible.

In addition, we will again add a diffusion term, $\nu \, \partial_x^2 v$. Overall, one obtains the momentum equation Note that we still need to specify $V(\rho)$, which is the same information as $q(\rho )$ introduced after Eq. (27.25). The only difference is that we now allow that it can take some time until velocities have adjusted accordingly. Indeed, the relaxation time is $\tau$. If we let $\tau$ go to zero, then the momentum equations becomes $v = V(\rho)$, which means instantaneous adaptation.

There is quite a lot of theory about this equation and its meaning for traffic (e.g. 61,53). Much of the behavior of the micro-simulation models can be explained using these equations; in fact, much of it was first observed in the fluid-dynamical equations. This, however, would be a full class in traffic flow theory and would thus go beyond the scope of this text.

[[breakdown and recovery. do I really want that for this text?]]