Staggered Conservative Scheme for Simulating the Emergence of a Jamiton in a Phantom Traffic Jam

Abstract

Traffic jams that appear without distinguishable reason is called phantom traffic jam. To study this phenomenon, macroscopic modelling using the second-order Payne-Whitham equation was adopted. In this article, the staggered conservative scheme applied on a staggered grid was implemented to solve the equation. Using this scheme, different behavior of a perturbed equilibrium solution was simulated; it might either decay or grow, depending on the critical threshold parameter. When unstable, a small perturbation was amplified into a local peak of high traffic density. This type of traveling wave is called a jamiton. On a circular road of a certain length and with a fixed number of vehicles, the growing process of these traveling jamiton waves was simulated. The shape and propagation speed of these numerical jamitons are shown to confirm the analytical formulas. A good understanding of this phenomenon may support decision-makers and engineers to determine the judicious selection of speed limits of a certain road section.

Appendix

In this Appendix, the proof of two theorems will be described.

Theorem 4

If the traveling wave solution is in a form of a left shock, then m > 0, du/dη > 0, and dρ/dη < 0 for $\eta \in \mathcal {D}$.

Proof

Assuming the shock as a left shock means that we assume (12) to be true. From Eq. 12 we have (u⁻− s) > c⁻, which then using Eq. 21 yield

$$ m=\rho^{-}(u^{-}-s)>\rho^{-}c^{-}>0. $$

(62)

The last inequality of Eq. 62 comes from the fact that both ρ⁻, and c⁻ are positive. Further, since p(ρ) is an increasing convex function, we have $p^{\prime }(\rho )>0$ and $p^{\prime \prime }(\rho )>0$, resulting

$$ \frac{d(\rho c)}{d(\rho)}=\sqrt{p^{\prime}(\rho)}+\frac{\rho p^{\prime\prime}(\rho)}{2\sqrt{p^{\prime}(\rho)}}>0, $$

(63)

which mean that ρc is an increasing function of ρ. Across the shock (9) holds which gives us ρ⁺u⁺ − ρ⁻u⁻ = s(ρ⁺ − ρ⁻) or

$$ \frac{\rho^{+}}{\rho^{-}}=\frac{u^{-}-s}{u^{+}-s}>\frac{c^{-}}{c^{+}}, $$

(64)

in which the last inequality comes from Eq. 12. Hence, we obtain ρ⁺c⁺ > ρ⁻c⁻, and since ρc is increasing, we conclude that ρ⁺ > ρ⁻, i.e. the downstream state is higher than the upstream state of the shock,

Further, along the continuity interval $\mathcal {D}$, the function ρ(η) is decreasing, and so a typical profile of a left shock (jamiton) is illustrated in Fig. 12. Furthermore, from relation (6), we can conclude u⁺ < u⁻, so du/dη > 0 on $\mathcal {D}$, and this completes our proof. □

Fig. 12

Typical profile for ρ(η) of a left shock

Theorem 5

traveling wave in a form of a right shock is mathematically inconsistent.

Proof

Assuming a right shock occur, and arguing in a similar way as in the previous theorem, Eq. 13 gives us

• (i) m < 0,

• (ii) u⁻ > u⁺, ρ⁻ > ρ⁺,

• (iii) du/dη > 0, dρ/dη > 0.

In case 2 (u⁺ + c⁺ < s < u⁻ + c⁻), we have (13) which can be re-written as

$$ \rho(u-s+c)^{-}>0>\rho(u-s+c)^{+}. $$

(65)

Rewriting (7) using (6) and its sign follows from (iii) yields

$$ \frac{du}{d\eta}=\frac{m\rho(\tilde{u}-u)}{\rho(u-s+c)(m-\rho c)}>0. $$

(66)

Since m < 0 from (i), we get mρ < 0 and also m − ρc < 0, than we may conclude that Eq. 66 is satisfied if

$$ (\tilde{u}-u)^{+}<0<(\tilde{u}-u)^{-}. $$

(67)

Denoting

$$ g(\rho)\equiv\tilde{u}-u=\tilde{u}(\rho)-s-m/\rho, $$

we conclude that g(ρ) is decreasing, since m < 0 and $\tilde {u}(\rho )$ is decreasing. Thus, g(ρ⁻) > g(ρ⁺) from (67), and the decreasing property of g gives us the conclusion that ρ⁻ < ρ⁺, which contradicts (ii). Therefore, a right shock solution must be mathematically inconsistent. □