TRAVELING SOLUTIONS FOR A MULTI-ANTICIPATIVE CAR-FOLLOWING TRAFFIC MODEL

. In this paper, we consider a steady state multi-anticipative traﬃc model and we provide necessarily and suﬃcient conditions for the existence of traveling solutions. In our work, the word “traveling” means that the distance between two consecutive vehicles travels continuously between two diﬀerent states. As application to our result, we show that taking a strictly concave optimal velocity, we can construct a traveling solution such that the distance between two vehicles decreases. The existence, uniqueness and the study of the asymptotic behavior of such solutions is done at the level of the Hamilton-Jacobi equation

described by the optimal velocity function, which depends on the headway distance, and each driver controls the velocity using this function. The OVM is found to accurately characterize the jamming transition.
Lenz et al. noted in [12] that correlations between cars have generally been ignored in the majority of traffic models, despite the fact that it's known from experience that drivers frequently detect two or more nearby vehicles ahead. Multi-vehicle interactions result from this, and they have an impact on the phase separation and the fundamental diagram. For these reasons, authors in [12] introduced the multi-anticipative car-following model in which multivehicle interactions are added to the Bando microscopic car-following model. They demonstrated that responding to more than one vehicle ahead causes the dynamical behavior to stabilize and to increase the stable region.
In this paper, we construct traveling solutions for the steady state of the multi-anticipative car-following model which can be seen as a phase transition between two traffic states. The multi-anticipative car-following model (see [12]) is a second order microscopic model given by where 1) U i denotes the position of the i-th vehicle, U i its velocity, U i its acceleration.
2) n ≥ 1 is a positive integer number.
3) V represents the optimal velocity function. 4) s k is a positive parameter such that n k=1 s k is the overall sensitivity and s k s 1 ≤ 1 for k ∈ [2, ..., n].
In this model, the driver do not only react on the dynamics of the leading vehicle U i+1 but also takes into consideration up to n cars ahead.

Main results
In this paper, we consider the steady state multi-anticipative car-following model with s k = s for all k ∈ [1, ..., n]. The model is given by For model (2.1), we construct traveling solutions U i of (2.1) satisfying By traveling solutions we mean that U i has the following form U i (t) = u i + t T + ct with u(y + 1) − u(y) → b as y → −∞, u(y + 1) − u(y) → a as y → +∞.
The interpretation of (2.2) is that a shock occurred and the interdistance is b far before the shock, and is a far after it. Moreover, the speed of the shock is c. The reader can also notice that we have U i (t + T ) = U i+1 (t) + ct and this means that T is the time period for which each vehicle will replace the one preceding him, up to a shift of a distance cT . We look for particular shock solutions of (2.1) of the form We obtain our results in the framework of viscosity solutions and we refer the reader to reference [3,5,6] for a full presentation of this theory. To prove our results, the following assumptions on the function F (p) = sV (p) are imposed.
with equality if and only if p ∈ [a, b], Remark 2.1 (Comments on assumptions (A)). The regularity assumption (A1) provides regular viscosity solutions. Assumption (A2) allows us to get strong comparison results for equation (2.3) which is crucial to prove many parts of our main results. We will show (see Thm. 2.4) that our traveling solutions exist if and only if assumption (A3) is satisfied. Finally, we will use assumption (A4) to get exponential asymptotics of the solution at infinity. Remark 2.2. As example of general functions F satisfying (A3) and (A4), we can consider F ∈ C 1 (R) and strictly concave. We recall that Using the mean value theorem, there exists e ∈ (a, b) such that 1 T = nF (e). Using that F is strictly concave, we have and strictly decreasing in (e, b]. In particular, taking the optimal velocity used in [12] and given by with h = constant, we remark that V (and so F ) is strictly concave for p > h.
The main result of this paper is given in the following theorem.
Assume that (A) holds for some a, b, c ∈ R and T > 0. There exists a concave solution u ∈ C 2 (R) of (2.3) satisfying for some constant C > 0, Moreover, we have ii) (Uniqueness.) The solution u is unique (up to translation and addition of constants) among the solutions v with v ∈ C 1 (R) such that |v −ū| ≤ C for some constant C > 0.
In the following theorem, we provide necessarily conditions to construct traveling solutions for (2.1) and it can be seen as a justification of our choice to impose assumptions (A3) and (A4).
As a consequence of Theorem 2.4, we have the following non-existence result: Corollary 2.5. Assume (A1) and (A2). We assume that there existsā,b,d ∈ R, and T > 0 such that with equality at least for p =ā,b,d withd ∈ (ā,b) .
Then there is no solution u of (2.3) satisfying Remark 2.6 (Application to traffic modeling). Planning and managing the infrastructure for directing traffic on roads require the use of traffic models. To model the traffic at large scale (like a city), traffic engineers use macroscopic models. These models use aggregate quantities such as density, average speed and flow of vehicles by making the analogy between the flow of traffic and the flow of a continuous fluid. However, a disadvantage of such models is that their assumptions are not easy to justify because, in their formulation, the dynamics of vehicles are not described individually. A way to justify the assumptions of macroscopic models is to derive them from microscopic ones so that the behavior of every single vehicle can be considered with high precision.
As a contribution of our study, the choice of strictly concave flux function in the macroscopic LWR model [18] given by where ρ is the unknown representing the cars' density and f is the flux function is justified. Shock waves are byproducts of traffic congestion and queuing. They are transition zones between two traffic states that move through a traffic environment like, as their name states, a propagating wave. That is, they form both when a queue is forming and when it is dissipating. In 2008, a team of Japanese researchers [21] demonstrated, through a life-size experiment (see the video of the experiment by hitting the link Shockwave traffic), the reality of a particularly frustrating phenomenon for drivers: at high density, traffic is unstable and traffic waves can arise even in the absence of bottleneck, such as a reduction in the number of lanes, for example following a car accident. These are, therefore, so-called "ghost" traffic waves, with no visible external cause.
It is known in traffic modeling that only upward jumps in the density are admissible which means a < b. Actually, one can imagine cars lined up in front of a red light (discontinuous initial data with downward jump). At t = 0, the red light turns to green but we do not observe this jump if the cars' density moves forward. Actually, we observe that the cars spread out.
Using the homogenization results of [9] (or [7]), the scalar conservation law (2.8) with f (p) = npF 1 p could be derived from the microscopic model (2.1). Hence, shocks that are traveling from a to b are expected to exist at the microscopic level and we prove this fact for F satisfying assumptions (A). Our justification of the strict concavity of the flux function f could be formulated in the following way: for a strictly concave function F satisfying (A1) and (A2), we can construct discrete shocks traveling from a to b with a < b. In particular, F being strictly concave implies that Organization of the paper. In Section 3, we define the viscosity solutions of (2.3) and state the crucial proposition: the strong comparison principle. In Section 4, we prove the exponential behavior of the solution at ±∞. In Section 5, we first show that the interdistance is strictly monotone and then we prove Theorem 2.4. Finally, in Section 6, we prove Theorem 2.3 by constructing a suitable viscosity subsolution.
Normalization. Up to consider a new force functioñ and replacing F byF , we can assume that T = 1 and c = 0.
This allows us to present our results in the rest of the paper considering the following equation

Viscosity solution
In this section, we first give the definition of viscosity solutions of (2.9). We then state different comparison principles.
1) We say that u is a viscosity subsolution (resp. supersolution) of (2.9) if u is a upper-semi continuous (resp. lower-semi continuous) and if for all test function ϕ ∈ C 1 (R) such that u − ϕ attains a local maximum (resp. local minimum) at some point x 0 , we have 2) We say that u is a viscosity solution of (2.9) if u ∈ L ∞ loc (R) and u is a viscosity subsolution and viscosity supersolution of (2.9).
Proof. We will first show that u is locally lipschitz. Using (2.9) and that u is locally bounded, for any R > 0, there exists L R > 0 such that This implies that u satisfies in the sense of viscosity solutions the following: This implies that Let us now define the function Since The comparison principle (see [3]) implies that v is constant. This implies u ∈ C 1 (R). Moreover, using F ∈ C 1 (R) and equation (2.9), we deduce that u ∈ C 2 (R).
Since we will use the strong comparison principle for two equations ((2.9) and (4.4)), we will state the following proposition for a general function L.

Asymptotic profile
In this section, we study the asymptotic behavior of the solution of (2.9). We have the following proposition.
Proposition 4.1. Assume (A) (for T = 1). Let u be a solution of (2.9) and let G be the function defined by G(y) = u(y + 1) − u(y). We assume that and that for y ∈ R, Then there exists K, γ > 0 and c 1 , c 2 ∈ R such that The proof of this proposition can be derived from the following lemma.
Lemma 4.2. Assume (A) (for T = 1 and c = 0). Let u be a solution of (2.9) satisfying (4.1) and G satisfying (4.2). We remark that G satisfies Recalling that nF (a) > 1 > nF (b), let ε > 0 small enough such that We have the following: (4.5) Then there exists a constant C > 0 such that for all y ≥ 0, Then there exists a constant C > 0 such that for all y ≤ 0, Proof of Lemma 4.2. We will only prove part 2) since the proof of part 1) can be done in the same way (even simpler). Using (4.1), let y 0 < 0 small enough be such that for all y ≤ y 0 , We will prove that for y ≤ y 0 , If (4.10) is true, we obtain (4.8) for all y ∈ R because we can easily check that for y > y 0 , We define the following function We then define M = sup y≤y0 ϕ(y).
We will prove that M ≤ 0. Assume by contradiction that M > 0. Using that G(y) → b as y → −∞, we deduce that M is reached at some point x. If x = y 0 , we get We deduce that x = y 0 and writing the viscosity inequality, we get We claim that x + k < y 0 for all k ∈ [1, ..., n]. If ∃k ∈ [1, ..., n] such that x + k ≥ y 0 , we get for C 1 big enough. This implies that which contradicts the fact that ϕ(x) > 0. We deduce that x + k < y 0 for all k ∈ [1, ..., n] and using that we obtain that Using assumption (A2) and the above inequality, we get (4.12) Using (4.11), we remark that we will get a contradiction if we prove We have (4.14) To get (4.13), we have to prove that n k=1 F (p k )(e γk − 1) γk < 1.
This implies that Using that G(x + l) ≥ b − ε, we deduce that Finally, we have that where we use the condition on γ in (4.7). We deduce that M ≤ 0 and in particular, we get (4.8).

Proof of Theorem 2.4
In this section, we prove our first main result, Theorem 2.4. We will first prove that the bounded interdistance is monotone. Let u be a solution of (2.9). We define the function G(y) = u(y + 1) − u(y). (5.1) The function G satisfies We have the following proposition.
Then we have that Proof. Let y 0 ∈ R. We have We will prove the first inequality of (5.4), the second one can be done similarly. Let x 0 , x 1 ∈ R such that x 1 > x 0 and Let η > 0 and m = G(x 0 ). We define By contradiction, we assume that M > 0.
Case 1: M is reached for some pointx > x 0 . Writing the viscosity inequality, we get Using that G(x) ≥ G(x + k) for all k ∈ [1, ..., n], we get a contradiction in the above inequality if G(x) > G(x + 1). If G(x) = G(x + 1), we deduce that M = G(x + 1) − m − η and so we can write the viscosity inequality using the function G at the pointx + 1. Continuing in the same way, we construct a sequence x n =x + n such that M = G(x n ) − m − η. We then define the following function Using the fact that G is a bounded Lipshitz continuous function, we have (up to passing to the limit on a subsequence) The stability of viscosity solutions implies (see [3]) that G ∞ solves (5.2). In addition, using the definition of M , we also have for x ∈ R, above properties of G , we remark that G is not strictly increasing and G is not constant implies that G has a global maximum at some point x 0 with Using the strong comparison principle (Prop. 3.3) and that G(x 0 ) is a solution of (5.2), we get G(x) = G(x 0 ) for all x ∈ R which gives a contradiction. Similarly, we can prove that if G is non-increasing, then G is strictly decreasing or constant.
We are now ready to prove Theorem 2.4 and since we are working with T = 1 and c = 0, we recall the Theorem 2.4 for T = 1 and c = 0.
Theorem 5.2. Assume (A1) and (A2). Let u ∈ C 1 (R) be a solution of (2.9) such that G(y) = u(y + 1) − u(y) is a bounded function. There exitsā,b ∈ R such that Moreover, ifā <b, then we have u is strictly decreasing on R and p ≤ nF (p) for p ∈ [ā,b] with equality if and only if p =ā,b.
Ifā >b, then we have u is strictly increasing on R and p ≥ nF (p) for p ∈ [ā,b] with equality if and only if p =ā,b.
Ifā =b, then we have u is constant on R.
Proof. Using Proposition 5.1, we deduce that the limit of G at ±∞ exist, Case 1: if G is strictly decreasing. In this caseā <b. Using that we deduce that u is strictly decreasing. Therefore u (±∞) exist and Taking y to +∞ in (5.8), we obtainā = nF (ā).
Similarly, we can prove thatb = nF (b). It remains to show that p < nF (p) if p ∈ (ā,b). We define the function u(t, y) = u(t + y).
We remark thatû is a viscosity solution of We rescaleû in the following wayû ε (t, y) = εû t ε , y ε .
As ε goes to zero, we have thatû ε → u 0 with As in Theorem 1.3 in [9], we can prove that u 0 is a viscosity solution of Testing u 0 from above with any test function of the form we deduce that p ≤ nF (p) for all p ∈ [ā,b] with equality for p =ā,b.
Using Proposition 3.4, we obtainũ ≤ v on R.
Using (5.9), we remark that m = min R (v −ũ) exists. We have for all y ∈ R, with equality at some point x 0 . Using Proposition 3.3, we get that Taking y → −∞, we get a contradiction and this implies that nF (d) >d.
Case 2: if G is strictly increasing. Similar to the above case.
Case 3: if G is constant. We obtain that u is constant.

Proof of Theorem 2.3
In this section, we prove Theorem 2.3. We construct the solution via Perron's method (see [3]). First, the following lemma provides a viscosity supersolution. Then,ū is a viscosity supersolution of (2.9).
Proof. Using (A3), we can easily check that y → ay and y → by are two solutions of (2.9). Then, using the stability of viscosity solutions (see [3]), we deduce thatū is a viscosity supersolution of (2.9).
We turn now to the construction of the subsolution. In order to do this, we need the following lemma.
The operator Φ is defined on the set which is a closed subset of the Banach space C([−δ, n]). We can easily check that we have which shows that Φ is a contraction on X for small δ. Using the fixed point theorem, there exists h ∈ X such that h = Φ(h). This gives a solution of (6.1) on [−δ, 0). To construct a solution on (−∞, 0), we use an iteration argument by proceeding as above on the intervals −(k + 2) Proposition 6.3. Assume (A). For y ≥ 0, we define g(y) = ay − δe −γy for y ≥ 0. Using Lemma 6.2, we extend g by continuity on y < 0 as the solution of Then, g is a viscosity subsolution of (2.9).
Proof. By construction (see Lem. 6.2), we need to show that g is a viscosity subsolution for y ≥ 0.
On the one hand, we have g (y) = a + δγe −γy .
On the other hand, we have Using (A4), we obtain (6.2) for δ, γ small enough.
Case 2: y = 0: Let ϕ ∈ C 1 (R) be a test function such that Then, we have which is the desired result.
Proposition 6.4. LetF be a truncation of F defined as follows: Let g be the viscosity subsolution of (2.9) provided in Proposition 6.3 replacing F byF . We define for y ∈ R the function , H(y) = g(y + 1) − g(y).
Then, the function H is non-increasing on R. Moreover, we have b ≥ H(y) ≥ a for all y < −1.
Proof. If y < 0, we have that k .
Using that F (b) ≥F ≥ F (a), we get Let us now prove that H is non-increasing on (−∞, 0). First, let us recall that the following holds: Let x 0 < 0. We want to prove that By contradiction, assume that M > 0. We recall that we have the following: From (6.4), we remark that H is non-increasing for x ≥ 0. Hence, M is reached at some pointx ≤ 0. Moreover, ifx = x 0 , we get M = 0. We deduce thatx > x 0 .
Case 1:x = 0. Using that H reaches its maximum on (x 0 , +∞) at the point 0, we deduce that This implies From the proof of Proposition 6.3 (case 1), we have which gives a contradiction.
Case 3:x = −1. Using that H reaches its maximum on (x 0 , +∞) at the point −1, we deduce that This implies which gives (using the first line in (6.3)), Using that H(−1) ≥ H(−1 + k), and thatF is non-decreasing, we deduce that But we have H(−1) ≥ H(0) which implies that H reaches its maximum at 0. This gives a contradiction thanks to case 1.
If H(x) > H(x + 1), we obtain a contradiction using that F (H(x)) >F (H(x + 1)). (6.6) In fact, we know that H(x) ∈ [a, b] for all x < −1 which implies If H(x + 1) < a then (6.6) is true since H(x) ≥ a. If H(x + 1) ≥ a, we use the strict monotony of the function F on [a, b].
If H(x) = H(x + 1), we deduce that M = H(x + 1) − H(x 0 ) and so we can write the viscosity inequality using the function H at the pointx + 1. Continuing in the same way, we obtain a contradiction once we reach x n =x + n with n ∈ N such that x n ≥ −1.
As a consequence of the preceding proposition, we have the following corollary. Corollary 6.5. Assume (A). Let g be the viscosity subsolution provided by Proposition 6.4. Then, we have g is a viscosity subsolution of (2.9) g is concave on R g (+∞) = H(+∞) = a g (−∞) = H(−∞) = b.

(6.7)
Proof. The proof of this corollary is divided into the following steps: Step 1: H (−∞) and g (−∞) exist. Using that H is non-increasing and a ≤ H(y) ≤ b on (−∞, −1), we deduce that H (−∞) exists. Moreover, for x < 0, we have g (y) = Step 2: d = b. Passing to the limit, we have that Using that H is non-increasing and its definition for x ≥ 0, we deduce that d = b.
Step 3: g is a subsolution of (2.9). Since H is non-increasing, H(−∞) = b and H(+∞) = a, we deduce that H(y) ∈ [a, b] for all y ∈ R. Using thatF (p) = F (p) in [a, b], we get the desired result.
We can now otain the proof of the main result.
Using Proposition 3.2, we have that u ∈ C 2 (R