On the Braess Paradox with Nonlinear Dynamics and Control Theory

Abstract

We show the existence of the Braess paradox for a traffic network with nonlinear dynamics described by the Lighthill–Whitham–Richards model for traffic flow. Furthermore, we show how one can employ control theory to avoid the paradox. The paper offers a general framework applicable to time-independent, uncongested flow on networks. These ideas are illustrated through examples.

Appendix: Technical Details

Lemma 6.1

Let (q) hold. Then, the speed $v = v (\rho )$, defined by

$$\begin{aligned} v (\rho ) = \left\{ \begin{array}{ll} q' (0), &{}\quad \mathrm {if}\, \rho = 0,\\ q (\rho ) / \rho , &{}\quad \mathrm {if }\rho > 0,\\ \end{array} \right. \end{aligned}$$

is well defined, continuous in $[0, \rho _m]$, strictly positive and weakly decreasing.

Proof

Continuity follows from l’Hôpital’s rule. Moreover,

$$\begin{aligned} v' (\rho ) = \left\{ \begin{array}{ll} \frac{\rho \, q' (\rho ) - q (\rho )}{\rho ^2}, &{} \text{ if }\,\rho > 0,\\ \frac{1}{2} \, q'' (0), &{} \text{ if } \rho = 0, \end{array} \right. \quad v'' (\rho ) = \left\{ \begin{array}{ll@{}} \frac{q'' (\rho )}{\rho } - 2 \frac{q' (\rho )}{\rho ^2} + 2 \frac{q (\rho )}{\rho ^3}, &{} \text{ if } \rho > 0,\\ \frac{1}{3} q''' (0), &{} \text{ if } \rho = 0. \end{array} \right. \end{aligned}$$

The concavity of $q$ implies that $q' (0) \ge q (\rho )/\rho \ge q' (\rho )$. Hence, $v' \le 0$. $\square$

Lemma 6.2

If (q) holds, the map $\rho :\vartheta \mapsto \rho (\vartheta )$ with $q\left( \rho (\vartheta )\right) = \vartheta \varphi$ satisfies:

• 1. $\rho \in \mathbf {{C}^{2}} ([0,1]; [0,1])$ and $\rho (0) = 0$;

• 2. $\rho ' (\vartheta ) > 0$ and $\rho '' (\vartheta ) > 0$ for all $\vartheta \in [0,1]$;

• 3. if $q$ is strictly convex, then $\rho '' (\vartheta ) > 0$ for all $\vartheta \in [0,1]$.

Proof

Existence and regularity of $\rho$ are immediate. By (q) and $q (\rho (\vartheta )) = \vartheta \, \varphi$, it follows that $\rho (0) = 0, \,\rho ' (\vartheta ) = \frac{\varphi }{q'\left( \rho (\vartheta )\right) } > 0$ and $\rho '' (\vartheta ) = - \frac{\varphi ^2 \, q''\left( \rho (\vartheta )\right) }{\left( q'\left( \rho (\vartheta )\right) \right) ^3} \ge 0$, and the latter inequality is strict as soon as $q$ is strictly convex. $\square$

Lemma 6.3

Let (q) hold. Then, the map $\vartheta \mapsto 1/v\left( \rho (\vartheta )\right)$ is weakly increasing. If, moreover, $q''' (\rho ) \le 0$ for $\rho \in [0,1]$, then the map $\vartheta \mapsto 1/v\left( \rho (\vartheta )\right)$ is convex.

Proof

We find $\frac{\mathrm {d}}{\mathrm {d}\vartheta } \left( \frac{1}{v\left( \rho (\vartheta )\right) }\right) = - \frac{v'\left( \rho (\vartheta )\right) \, \rho ' (\vartheta )}{\left( v\left( \rho (\vartheta )\right) \right) ^2} \ge 0$. Moreover,

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}\vartheta } \left( \frac{1}{v\left( \rho (\vartheta )\right) }\right)= & {} - \frac{v'\left( \rho (\vartheta )\right) \, \rho ' (\vartheta )}{\left( v\left( \rho (\vartheta )\right) \right) ^2} = \left( \frac{1}{q\left( \rho (\vartheta )\right) \, q'\left( \rho (\vartheta )\right) } - \frac{\rho (\vartheta )}{\left( q\left( \rho (\vartheta )\right) \right) ^2} \right) \varphi ,\\ \frac{\mathrm {d}^2}{\mathrm {d}\vartheta ^2} \left( \frac{1}{v\left( \rho (\vartheta )\right) }\right)= & {} - 2 \, \frac{\rho ' (\vartheta ) \, \varphi }{\left( q\left( \rho (\vartheta )\right) \right) ^3}\\&\times \left[ \frac{1}{2} \left( \frac{q\left( \rho (\vartheta )\right) }{q'\left( \rho (\vartheta )\right) }\right) ^2 q''\left( \rho (\vartheta )\right) + q\left( \rho (\vartheta )\right) - \rho (\vartheta ) \, q'\left( \rho (\vartheta )\right) \right] . \end{aligned}$$

Call $f (\rho ) = \frac{1}{2} \left( \frac{q(\rho )}{q'(\rho )}\right) ^2 q''(\rho ) + q(\rho ) - \rho \, q'(\rho )$. Observe that $f (0) = 0$ and

$$\begin{aligned} f' (\rho ) = \frac{1}{2} \left( \frac{q(\rho )}{q'(\rho )}\right) ^2 q'''(\rho ) + \frac{\left( q (\rho ) - \rho \, q' (\rho )\right) q'' (\rho )}{q' (\rho )} - \frac{q (\rho ) \, \left( q'' (\rho )\right) ^2}{\left( q' (\rho )\right) ^3} \le 0, \end{aligned}$$

thereby completing the proof. $\square$

Asking $q''' (\rho ) \le 0$ is sufficient, but not necessary, for the travel time convexity.

Proof of Proposition 3.1

If $f \in \mathbf {{C}^{2}} ({{\mathbb {R}}}_+; {{\mathbb {R}}})$ is convex and increasing, then also $x \mapsto x\,f (x)$ is convex and increasing. By Lemma 6.2, for $i=1, \ldots ,m$, the map $\xi \mapsto \tau _{r_i} (\xi ) \, \xi$ is convex on $[0,1]$. Hence, also the map $\vartheta \mapsto \sum _i \tau _{r_i} (\vartheta _i) \, \vartheta _i$ is convex on $[0,1]^n$. Since $\varGamma _{ij} \in \{0,1\}$, also the map $\vartheta \mapsto T (\vartheta )$ is convex. $\square$

Proof of Theorem 3.1

By Definition 3.3, the configuration $\vartheta ^N_1 = \vartheta ^N_2 = 0$ is clearly an equilibrium, the only relevant time being the equilibrium

$$\begin{aligned} \bar{\tau } = \tau _\gamma (0,0) = 2 \tau _a (1) + \tau _e (1) = 2 \, \frac{\ell }{v\left( \rho (1)\right) } + \frac{\tilde{\ell }}{\tilde{v}\left( \tilde{\rho } (1)\right) }. \end{aligned}$$

By (7), it is also a Nash point, since $\tau _a (0,0) = \tau _\beta (0,0) > \bar{\tau }$ and, by continuity, the same inequality holds in a neighborhood of $\vartheta ^N$.

Assume there exists an other equilibrium point $\bar{\vartheta }$ in the interior of $S^2$. Then, by symmetry, $\bar{\vartheta }_1 = \bar{\vartheta }_2$ and, by Definition 3.3,

$$\begin{aligned} \tau _b (\bar{\vartheta }_1) - \tau _a (1-\bar{\vartheta }_1) = \tau _e (1-2\bar{\vartheta }_1). \end{aligned}$$

(9)

By assumption, the left-hand side above is a strictly increasing function of $\vartheta _1$, while the right-hand side is weakly decreasing, so that

$$\begin{aligned} \tau _e (1-2 \bar{\vartheta }_1)\le & {} \tau _e (1) < \tau _b (0) + \tau _a (0) - 2 \tau _a (1) \qquad \quad \left[ \text{ by } (7)\right] \\\le & {} \tau _b (0) + \tau _a (0) - 2 \tau _a (0) \le \tau _b (0) - \tau _a (0) \le \tau _b (\bar{\vartheta }_1) - \tau _a (1-\bar{\vartheta }_1), \end{aligned}$$

which contradicts (9). To prove the uniqueness of the Nash points, consider the configuration $(0,1)$. In this case, the only relevant time is $\tau _\alpha (0,1)$ and

$$\begin{aligned} \tau _\alpha (1,0) = \tau _a (1) + \tau _b (1) > \tau _a (0) + \tau _b (0) = \tau _\beta (1,1), \end{aligned}$$

proving that $(1,0)$ is not a Nash point. The case of $(0,1)$ is entirely analogous.

Finally, $\tau _\alpha (1/2,1/2) = \tau _b (1/2, 1/2)$ is the globally optimal time for the case of four roads, and the leftmost bound in (7) allows to complete the proof. $\square$

Lemma 6.4

Let the travel time $\tau _a,\tau _b \in \mathbf {{C}^{0}} ([0,1]; {{\mathbb {R}}}_+)$ be non-decreasing and convex, at least one of the two being strictly convex. Then, there exists a map $\varTheta \in \mathbf {{C}^{0}} ({{\mathbb {R}}}_+;[0, 1/2])$ such that the partition $\left( \varTheta (\vartheta ), \varTheta (\vartheta )\right)$ is the point of global minimum of the mean travel time $T$ defined in (6), (5), (8) over $S^n$.

Proof

The travel time $T$ is convex by Proposition 3.1. By symmetry, its point of minimum is of the type $(\vartheta ,\vartheta )$ and if $\vartheta \in ]0, 1/2[$, then $\frac{\mathrm {d}}{\mathrm {d}\vartheta } T (\vartheta ,\vartheta )=0$. Hence,

$$\begin{aligned} T (\vartheta , \vartheta )= & {} 2(1-\vartheta ) \, \tau _a (1-\vartheta ) + 2\vartheta \, \tau _b (\vartheta ) + (1-2\vartheta ) \tilde{\tau }_e , \\ \frac{\mathrm {d}}{\mathrm {d}\vartheta } T (\vartheta ,\vartheta )= & {} 2 \left( - \tau _a (1-\vartheta ) - (1-\vartheta ) \tau _a' (1-\vartheta ) + \tau _b (\vartheta ) + \vartheta \, \tau _b' (\vartheta ) + \tilde{\tau } \right) , \\ \frac{\mathrm {d}^2}{\mathrm {d}\vartheta ^2} T (\vartheta ,\vartheta )= & {} 2\left( 2 \tau _a' (1-\vartheta ) + (1-\vartheta ) \tau _a'' (1-\vartheta ) + 2 \tau _b' (\vartheta ) + \vartheta \tau _b'' (\vartheta ) \right) > 0, \end{aligned}$$

so that $\vartheta \mapsto T (\vartheta ,\vartheta )$ is strictly convex and it admits a unique point of minimum $\varTheta (\tilde{\tau })$ in $]0, 1/2[$. By the implicit function theorem, $\varTheta$ is continuous. $\square$

Lemma 6.5

Let the travel times $\tau _a,\tau _b \in \mathbf {{C}^{0}} ([0,1]; {{\mathbb {R}}}_+)$ be non-decreasing and convex, at least one of them being strictly convex. Then, there exists a $\tilde{T} \in \mathbf {{C}^{0}} ([0, 1/2]; {{\mathbb {R}}}_+)$ such that assigning the travel time $\tilde{T} (\vartheta )$ on road $e$ makes the configuration $(\vartheta ,\vartheta )$ the unique local Nash point in the sense of Definition 3.3.

Proof

Given $\vartheta \in [0,1/2]$, we seek $\tilde{\tau }$ such that $(\vartheta ,\vartheta )$ is an equilibrium point. To this aim, we solve $\tau _a (\vartheta ,\vartheta ) = \tau _b (\vartheta ,\vartheta )$ together with $\tau _a (\vartheta ,\vartheta ) = \tau _\gamma (\vartheta ,\vartheta )$. By symmetry considerations, the former equality is satisfied for $\vartheta \in [0, 1/2]$. The latter is equivalent to: $\tau _a (1-\vartheta ) + \tau _b (\vartheta ) = 2 \tau _a (1-\vartheta ) + \tilde{\tau }$. Therefore, we set

$$\begin{aligned} \tilde{T} (\vartheta ) = \left\{ \begin{array}{l@{\quad }l} \tau _b (\vartheta ) - \tau _a (1-\vartheta ), &{}\text{ if } \tau _b (\vartheta ) \ge \tau _a (1-\vartheta ),\\ 0, &{} \text{ if } \tau _b (\vartheta ) < \tau _a (1-\vartheta ). \end{array} \right. \end{aligned}$$

By construction, $(\vartheta ,\vartheta )$ is an equilibrium configuration in the sense of Definition 3.1, once the travel time $\tilde{\tau }$ along the road $e$ is set equal end $\tilde{T} (\vartheta )$.

When $\vartheta \in ]0, 1/2[$, to prove that $(\vartheta ,\vartheta )$ is a local Nash point, thanks to the present symmetries, it is sufficient to check that for all small $\varepsilon >0$ we have $\tau _\alpha (\vartheta +\varepsilon , \vartheta ) > \tau _\gamma (\vartheta ,\vartheta ), \,\tau _\alpha (\vartheta +\varepsilon , \vartheta -\varepsilon ) > \tau _\beta (\vartheta ,\vartheta ), \,\tau _\gamma (\vartheta -\varepsilon , \vartheta ) > \tau _\alpha (\vartheta ,\vartheta )$, or,

$$\begin{aligned} \tau _b (\vartheta +\varepsilon ) - \tau _b (\vartheta ) + \tau _a (1-\vartheta ) - \tau _a(1-\vartheta -\varepsilon )> & {} 0,\\ \tau _a (1-\vartheta +\varepsilon ) - \tau _a (1-\vartheta -\varepsilon ) + \tau _b (\vartheta +\varepsilon ) - \tau _b (\vartheta -\varepsilon )> & {} 0,\\ \tau _a (1-\eta +\varepsilon ) - \tau _a (1-\vartheta )> & {} 0, \end{aligned}$$

and all these inequalities hold by the monotonicity of the travel times. $\square$

Proof of Theorem 4.1

Let $\varTheta$ and $\tilde{T}$ be as defined in Lemmas 6.4 and 6.5. Set $\varUpsilon :[0,1/2] \rightarrow [0,1/2], \,\varUpsilon = \varTheta \circ \tilde{T}$, and call $\vartheta _*$ a fixed point for $\varUpsilon$. Then, $(\vartheta _*, \vartheta _*)$ is a local Nash point, if $\tilde{\tau }_* = \tilde{T} (\vartheta _*)$ is the travel time along road $e$. $\square$