EQUILIBRIUM RESULTS FOR DYNAMIC CONGESTION GAMES ´ ERIC ´ FRED MEUNIER AND NICOLAS WAGNER Abstract. Consider the following game. Given a network with a continuum of players at some origins, suppose that players wish to reach specific destinations, but that they are not indifferent to the cost to reach their destination. They may have several possibilities (of routes), but their choices modify the travel costs on the network. Hence, each player faces the following problem: given a pattern of travel costs for the different possible routes that reach the destination, find a path of minimal cost. This kind of game belongs to the class of congestion games. In the traditional approach, each player occupies permanently the whole route he has chosen. In the present paper, the time needed to travel along the routes is taken into account: for instance, two players whose routes use the same arc may not interact if their start their trips at different times. In transportation science, the question whether there is an equilibrium and how to compute it for such a model is called the dynamic assignment problem. Up to now, there was a lack of a general model for this problem. The present paper tries to settle this point. Along the lines, we define a new class of games: dynamic congestion games, which capture this time-dependency aspect. Moreover, we prove that, under some natural assumptions, there is a Nash equilibrium. When we apply our result to the dynamic assignment problem, we get many of the previous known results.

1. Introduction 1.1. Dynamic equilibrium assignment. Consider over a time interval, say a day, a network where a set of players travel along directed paths, called routes, connecting origins to destinations. At the beginning of the day players are at origins and wish to reach a specific destination by the end of the day. In order to do so, they take a travel decision on the network, i.e. choose a route and a departure time. Yet players decisions depends on route travel time over the network, itself depending on the flow of players following each routes and thus on the decisions of the other players. Finding an equilibrium (in the Nash sense) of such a problem is, roughly speaking, the dynamic equilibrium assignment problem. Contrary to the static assignment (studied by Wardrop [22], Beckmann [2] Patricksson [16], Roughgarden and Tardos [18], Milchtaich [13] and many others), where players occupy the whole route permanently, the dynamic assignment problem represents explicitly the time needed to reach any point in the network, thus giving an accurate description of the traffic propagation on the network. Several authors have studied the dynamic assignment problem (Vickrey [21], Merchant and Nemhauser [12], Friesz and al. [4] Lindsey [9], ...) – all in the context of transport, which contrasts with the static assignment which is well studied for communication and computers networks as well. Yet, few theoretical results have been established (with the notable exceptions of Mounce [14] and Zhu et al [23]), and no general existence have been established. The purpose of the paper is to propose the right framework to study this problem and to give a general equilibrium result that cover many of the previous ones. The underlying game in the static assignment problem belongs to the class of congestion games. In our quest of a framework for the dynamic assignment problem we naturally define a new class of games: dynamic congestion games. 1.2. Dynamic congestion games. We define a dynamic congestion game to be a nonatomic (in the sense of Mas-Collel [11]) game in which each player chooses a particular sequence of operations (we will call such a sequence a job), among a finite ground set A of possible operations, together 1

with a starting time. No repetition is allowed in operation sequences, so there is a finite set R of sequences and, denoting I the set of admissible starting times, the player possible strategies are in S = I × R. Without loss of generality, let us set I = [0; hmax ]. Each operation is endowed with a processing time, which depends on the number of players having already started this operation and which, roughly speaking (a more detailed description is proposed below), is the time needed to complete this operation. Note that the terminology operation, job, completion, processing time stem from scheduling theory. Each player performs his operations in the order of the sequence: he starts operation ai+1 immediately after having finished operation ai . The player pay off associated to a specific operation depends on the players undertaking the same operation as well as the instant he has started the operation (i.e. the level of congestion varies with time). A dynamic congestion game can be seen as a temporal extension of the congestion games introduced by [17]. In the context of the dynamic assignment problem described above, the operations are the arcs of the road network, the jobs are the possible routes. Player choose strategies that maximize their payoff, that is, they choose routes that connect the origin-destination pairs they are concerned with, and starting times, in order to minimize their travel times and to leave at convenient moments. In addition to the dynamic assignment problem, which remains our main object, we may imagine other applications, for instance: Example. A set of firms are producing goods and each of them have alternative processes available, each of them requiring a different operations sequences. Both the prices (through competition effects) and the time (through queuing) required for each operation depends on the firms undertaking the operation. 1.3. Organization of the paper. Section 2 gives the main tools and notations of the paper. Since continuity results will be the main technical aspects of our work, we will carefully define in this section the topologies of our different sets. Moreover, Section 2 gives a theorem of Khan (Theorem 1), a powerful existence result on games with a continuum of agents. In Section 3, we describe precisely the model we are working with. The following section – Section 4 – exposes our two main results: the consistency of our model (Proposition 1) and the existence theorem (Theorem 2). The proofs are presented in Section 5. The proof of Theorem 2 consists mainly in establishing that Khan’s theorem can be applied. In the last section (Section 6), we will see how it covers previous results in the area of dynamic traffic assignment. 2. Mathematical tools and notations 2.1. Sets and topologies. 2.1.1. Measures. We use M(E) to denote the set of measures on a set E 1. We will systematically use the weak convergence topology on any set of measures encountered in the paper. A sequence of measure Mn defined on a set E is said to weakly converge toward a measure M on E if (i) lim supn→+∞ Mn (F ) ≤ M (F ) for any closed subset F of E, and (ii) lim supn→+∞ Mn (E) = M (E). 1The measures will be denoted by capital letters, in order to be consistent with the traditional notation for the

traffic assignment problem where cumulated flows are denoted by capital letters. A cumulated flows is a quantity of users on a time interval – in particular, it can be seen as a measure. 2

There exists a metric ρ (the Prohorov metric for instance), such that convergence for this metric is equivalent to the weak convergence. It will be used in the proof of Lemma 5. For more informations about the weak convergence, see [20]. 2.1.2. Restriction and marginal of measures. Let M be a measure on a cartesian product A × B. Then MA – also called the marginal of M on A – denotes the measure on B such that MA (I) = M (A × I) for each measurable subset I ⊆ B. Let M be a measure on R. For any h ∈ R, we denote by M |h the restriction of M on [0, h], defined such that M |h (J) := M (J∩] − ∞, h]) for all measurable subsets J of R. We extend this notation to the measure on R × R. If M is such a measure, M |h (J × R0 ) = M ((J∩] − ∞, h]) × R0 ) for all measurable subsets J of R and all subsets R0 of R. Claim 1. If h2 > h1 , then for any measure M , we have M |h1 = M |h2 |h1 . The proof is straightforward. 2.1.3. Continuous mappings. Let E and F be two topological spaces, and let C(E, F ) denote the set of all continuous maps between E and F . Given a compact subset K of E and an open subset O of F , let V (K, O) denote the set of all functions f in C(E, F ) such that f (K) is contained in O. Then the collection of all such V (K, O) is a subbase for the compact-open topology, that is the compact-open topology is the smallest topology containing all such V (K, O). The set of continuous mappings from a bounded interval I 0 to R is denoted by C(I 0 ). In this particular case, we consider to topology induced by the || · ||∞ norm. The following lemma will be useful: Lemma 1. Let I 0 be a closed interval of R and f : M(I 0 ) → C(I 0 ) and g : M(I 0 ) → C(I 0 ) be two continuous functions. Then Y 7→ f (Y ) ◦ g(Y ) is continuous. Proof. Indeed, let  > 0 and Y ∈ M(R). According to the continuity of f , there is an η1 > 0 such that ρ(Y, Y 0 ) ≤ η1 implies ||f (Y ) − f (Y 0 )||∞ ≤ /2. According to the uniform continuity of f (Y ) on the image of g(Y ), which is compact, there is an η2 > 0 such that for all h, h0 ∈ Im g(Y ), when |h − h0 | ≤ η2 , we have |f (Y )(h) − f (Y )(h0 )| ≤ /2. According to the continuity of g, there is an η3 > 0 such that ρ(Y, Y 0 ) ≤ η3 implies ||g(Y ) − g(Y 0 )||∞ ≤ η2 . Now define η := min(η1 , η3 ). For all Y 0 ∈ M(R) such that ρ(Y, Y 0 ) ≤ η, we have ||f (Y ) ◦ g(Y ) − f (Y 0 ) ◦ g(Y 0 )||∞ ≤ ||f (Y ) ◦ g(Y ) − f (Y ) ◦ g(Y 0 )||∞ +||f (Y ) ◦ g(Y 0 ) − f (Y 0 ) ◦ g(Y 0 )||∞ ≤ /2 + /2 ≤ .  2.1.4. Upper semicontinuous mappings. Assume that E is a nonempty and compact Hausdorff ¯ = R∪{−∞, +∞} is said to be upper semicontinuous if its hypograph is space. A function u : E → R ¯ is given by the set {(x, y) ∈ E × R ¯ : f (x) ≥ y}. closed. Recall the hypograph of a function f : E → R ¯ We denote by SE the space of upper semicontinuity functions E → R. The space SE is endowed with the hypotopology where to maps are “close” if their hypographs are “close” for the Hausdorff distance. 3

2.2. Games with a continuum of players and Khan theorem. One of the main approach of games with a continuum of players was introduced by Mass Collel [11] as a reformulated version of Schmeidler [19]. On the basis of Hart-Hildenbrand-Kohlberg [5], Mass Collel represents a game as a probability U on the space of utility functions U, where a utility function is a continuous function defined on the product space of M(S) and the strategy S. Given the strategy s taken by a player u and the strategy distribution of all players X ∈ M(S), u(X, s) is the utility enjoyed by the player u. A Nash equilibrium is then defined as: Definition 1. For a game U , a (Borel) probability measure D on S × U is a Nash equilibrium if (1) DU = U . (2) D (s, u) ∈ S × U : u(X, s) ≥ u(X, s0 ) for all s0 ∈ S = 1, with X := DS . Essentially, the formulation of the equilibrium states that the volume of players with an optimal decision regarding the global strategy distribution is the total volume of players. Mas-Colell [11] proved the existence of an equilibrium under the assumption of S being a compact metric space. In an attempt to generalize this approach to upper semicontinuous utility functions, Khan proposes a slightly different model. In a game “`a la Mas-Collel”, each player is characterized by a utility function u : S × M(S) → R. Khan uses an alternative view: a utility function is seen as a family of functions from S to R parameterized by elements of M(S). In simpler terms, one can rewrite u(X, s) = u ˆ(X)(s) in the previous definition, hence seeing a game as a distribution on the space of continuous mappings M(S) → SS . For the sake of readability, we will denote the set of such functions C(SS ) instead of C(M(S), SS ). In this context, a Nash equilibrium is defined exactly as above, once we have substituted u(X, s) and u(X, s0 ) respectively by u ˆ(X)(s) and u ˆ(X)(s0 ). In this framework, Khan showed the following theorem [7]: Theorem 1 (Khan). Assume that the strategy set S is a compact metric space and let U ∈ M(C(SS )) be a game. Then there exists a Nash equilibrium. 3. The model In a dynamic congestion game, we have a continuum of players, a finite set A of operations and a bounded time interval I ⊆ R. Denoting by R the set of ordered sequences of distinct operations (the set of jobs), the strategy set is S := I × R. Each player chooses strategy from S, that is, a starting time and a sequence (a1 , . . . , an ) of operations (a job) and executes it in order, starting operation ai+1 as soon as he has completed ai . The presentation of the model will be divided in two parts. First we introduce the scheduling model: given a measure on the set of strategies representing the choices of the players, how do we compute the starting times of a player for each operation of his chosen job ? Then, we define dynamic congestion games using the formalism of Definition 1. 3.1. Scheduling model. First, let us precise how we model congestion in our context. Each operation a is endowed with a processing time ta : M(R) → C(R). If Y is a measure such that for each measurable subset J, the quantity Y (J) is the number of players having started operation a at an instant in J, the value ta (Y )(h) is the time needed to process operation a when it is started at h. Such a measure Y is called the schedule of operation a. Consequently, the following question arises: once all players have made a choice of strategy, how do we deduce the starting instant of each operation for each player ? To do so, we introduce two functions that are entirely determined by the processing times ta . These two functions are the user scheduling function φa and the global scheduling function Φa . Given X ∈ M(S) (the distribution of strategies played by the players) and s = (h, a1 , . . . , an ) ∈ S (a strategy played by a particular player), we define φai (X, s) to be the instant when ai is started 4

by the particular player when the choices made by the other players are taken into account. In particular φa1 (X, s) = h. Given X ∈ M(S) (the distribution of players’ strategies), we define Φa (X) to be the schedule of operation a, defined a few lines above. It remains to explicit the scheduling functions φa and Φa . A useful notion is that of completion time, Ha : M(R) → C(R) which is defined by (1)

for Y ∈ M(R) and h ∈ R

Ha (Y )(h) := h + ta (Y )(h)

Given a schedule Y of operation a, the number Ha (Y )(h) is the instant at which the operation a is finished when it has been started at h. If we fix a subset J ∈ R, the subset Ha (Y )−1 (J) is all the instants at which operation a can be started in order to finish at some instant in J. The operations schedule (Ya )A∈A of a strategy distribution can now be formally defined. Definition 2. The operations schedule (Ya )a∈A of a strategy distribution X ∈ M(S) is a collection of (Borel) measures on R such that there exists (Yar ) for all r ∈ R and a ∈ A satisfying the system: X (2) Ya = Yar r∈R: a∈r

and for all r = (a1 , . . . , an ) ∈ R (3)

Yar1 = X r (i) Yari = Yari−1 ◦ (Hai−1 (Yai−1 ))−1 for i = 2, . . . , n (ii) Yar = 0 if a ∈ / r. (iii)

With X r the measure define by X r (J) := X(J × {r}) for all measurable subset J of R. We say that Yar is the schedule of operation a with respect to job r if, for each J ⊆ R. Indeed the quantity Yar (J) is the number of players whose chosen job is r and who start operation a for some instant in J. Similarly X r is denoted as the schedule of job r, which counts the number of players starting job r on any interval: for each J ⊆ R, the quantity X r (J) is the number of players who start job r for some instant in J. Equation (2) follows from the definition of Yar . Equality (i) of Equation (3) expresses that the number of players starting the first operation of a given job r during an interval J is the number of players starting the job r during J (the order of the operations in a job is fixed). Equality (ii) expresses that the number of players having chosen job r and starting operation ai in J is equal to the number of players having chosen job r and finishing operation ai−1 in J (in our model, there is no delay between the operations). Equality (iii) expresses that if an operation a does not belong to a job r, nobody having chosen r will process a. Until now we have no guarantee that the operations schedules and operations schedules with respect to the jobs are unambiguously defined in Definition 2. A proposition below (Proposition 1) will ensure that under five Assumptions on ta , existence and uniqueness of the solutions of system (2-3) are guaranteed for all X ∈ S. In this case, the scheduling functions Φa and φa can be properly defined through: for all a ∈ A,

Φa (X) := Ya and


φai (X, (h, a1 , . . . , an )) := Hai (Yai ) ◦ Hai−1 (Yai−1 ) ◦ . . . ◦ Ha1 (Ya1 ) (h) for all (a1 , . . . , an ) ∈ R and i ∈ {1, . . . , n}. 5

3.2. The set of utility functions. Assume given a set (ta )a∈A of completion times and the corresponding user and global scheduling functions. Each player is then identified by a collection of continuous functions u ˆr : M(Rn ) → SRn , one for each route r in R (where n denotes the number of arcs of route r). Denote by (Ur )r∈R the set of admissible functions u ˆr , assumed to be compact. (Ur )r∈R can be interpreted as the space of the player characteristics. The utility function of a player characterized by (ˆ ur )r∈R in the sense of Mass-Colell then comes from the following expression: (4)

u ˆ(X)(h, r = a1 , . . . , an ) := u ˆr (Y1 , . . . , Yn )(h1 , . . . , hn )

with hi = φai (X)(h, r) and Yi = Φai (X). u ˆ(X)(s) is the pay off of the player represented by (ˆ ur )r∈R when he plays s against the distribution X of players’ strategies. Equation (4) expresses that it simply depends on the schedule of each operations of its job and on their starting times. That is basically saying that the congestion on an operation only depends on the players actually choosing this   operations and is time-varying. The set of utility functions u ˆ is denoted U (Ur )r∈R , (ta )a∈A We are now ready to set the dynamic congestion game definition:   Definition 3. A dynamic congestion game is a probability measure U over U (Ur )r∈R , (ta )a∈A . 4. Main results 4.1. Existence and uniqueness of a schedule. To establish the existence and uniqueness of the schedule as defined in Subsection 3.1 and hence show that the scheduling function is well defined, we need to introduce five (natural) assumptions of the nature of ta , stated above. Assumption 1. [Continuity] ta : M(R) → C(R) is continuous. Assumption 2. [No infinite speed] There exists tmin > 0 such that for all Y ∈ M(R) and all h ∈ R, we have ta (Y )(h) > tmin . Assumption 3. [Finiteness] There exists a continuous map tmax : R+ → R+ such that ta (Y )(h) ≤ tmax (Y (R)) for all h ∈ R. Assumption 4. [Strict Fifoness] Let Y ∈ M(R). The map h 7→ h + ta (Y )(h) is non decreasing. Moreover, for h1 < h2 in R such that Y [h1 , h2 ] 6= 0, we have h1 + ta (Y )(h1 ) < h2 + ta (Y )(h2 ). Assumption 5. [Causality] For all h ∈ R and Y ∈ M(R), we have ta (Y |h )(h) = ta (Y )(h). Assumption 1 simply states that a small variation in the schedule of an operation leads to a small variation of the operation processing time. Assumption 2 amounts to say that the time needed to treat an operation is bounded from below. The finiteness condition (Assumption 3) assumes that if we wait for a sufficient long time, all players will finish their operations. The fifo condition (4) states that if two players starts a operation in a given order, they finish it in the same order. Finally, Assumption 5 simply implies that the processing time depends on the users that already overtook this operation, but not on the ones that will. We then have: Proposition 1. Given a strategy distribution X ∈ M(S), system (3) has a unique solution (Ya )a∈A . Moreover X 7→ Ya is continuous for each a ∈ A. Hence Φa : X ∈ M(S) 7→ Ya ∈ M(R) is welldefined and continuous for all a ∈ A. Corollary 1. φa is well-defined and continuous for all a ∈ A. 6

4.2. Existence result. Theorem 2. Given a set of operations with processing time functions (t  a )a∈A satisfying Assumptions 1-5 and a job utility function set (Ur )r∈R , every measure U on U (Ur )r∈R , (ta )a∈A ) admits a Nash equilibrium distribution. 5. Proofs We will use the following notations - Ya is the schedule of an operation a, that is the distribution over time of the players starting operation a. - Yar is the schedule of an operation a with respect to the job r, that is the distribution over time of the players having chosen job r and starting operation a. - for an operation a, Ya is the collection (Yar )r∈R . It can alternatively be seen as an element of M(R × R). 5.1. Proof of Proposition 1. We will first introduce the function ψa from M(R × R) to itself, defined as follows for each measurable subset J of R. For any r let: (5)

ψar (Ya )(J)

 :=

Yar (Ha (Ya )−1 (J)) if a ∈ r 0 if not,

P with Ya = r∈R: a∈r Yar (Equation (2)). Then ψa (Ya ) := (ψar (Ya ))r∈R , which we also see as an element of M(R × R). The fact that r ψa (Ya ) is a measure is a consequence of the continuity of Ha (Ya ) (Assumption 1). It can interpreted as a kind of “transfer function”, which, given a schedule of operation a – that is a distribution of players starting operation a – returns a distribution of players finishing operation a, and this, decomposed for each job r containing operation a. Let us first state two lemmas regarding ψa properties, used in the proof of Proposition 1. Lemma 2. Consider a bounded interval I 0 ⊆ R and an operation a. Then ψa is continuous on the set of measures having their support in I 0 . 0 Proof. Take a converging Yn → Y of Pmeasures ron R × R having their support in I and P sequence r define as usual Yn := r∈R: a∈r Yn and Y := r∈R: a∈r Y . Take an interval J = [h1 , h2 ] in I 0 and consider fn := Ha (Yn ) and f := Ha (Y ). Note that the sequence ||fn − f ||∞ converges to 0 by continuity of Ha . Choose r ∈ R. We want to prove that lim supn Ynr (fn−1 (J)) ≤ Y r (f −1 (J)). We can assume that Ynr (fn−1 (J)) 6= 0 for an infinite sequence of n, otherwise there is nothing to prove. Define now h1,n := inf fn−1 (J), h2,n := sup fn−1 (J) −1 h˜1 := inf f (J), h˜2 := sup f −1 (J) h∗1 := lim inf n h1,n , h∗2 := lim supn h2,n r −1 Under Assumption 4 and the fact that Yn (fn (J)) 6= 0 for an infinite sequence of n, we can assume that fn (h1,n ) = h1 . We have |fn (h1,n ) − f (h∗1 )| ≤ |fn (h1,n ) − f (h1,n )| + |f (h1,n ) − f (h∗1 )| hence f (h∗1 ) = h1 . Similarly, we get f (h∗2 ) = h2 . 7

Recall that by assumption, [h∗1 , h∗2 ] is a bounded interval. Let us now prove the following result: Let  > 0. There exists h01 < h∗1 and h02 > h∗2 such that Y r ([h01 , h02 ]) ≤ Y r ([h∗1 , h∗2 ]) + . Take a sequence of closed intervals (In ) converging to [h∗1 , h∗2 ] such that [h∗1 , h∗2 ] is strictly included in In for any n. According to the sequential continuity of measures (see [6], page 43) limn Y r (In ) = Y r (limn In ) = Y r ([h01 , h02 ]), so there exists n0 such that Y r (In0 ) ≤ Y r ([h∗1 , h∗2 ]) + . Take [h∗1 , h∗2 ] = In0 . Now, for n big enough, we have fn−1 (J) ⊆ [h01 , h02 ]. Hence, for n big enough Ynr (fn−1 (J)) ≤ ≤ ≤ ≤

Ynr [h01 , h02 ] (by monotonicity of a measure) Y r [h01 , h02 ] +  (Ynr converges to Y r ) Y r [h∗1 , h∗2 ] + 2 Y r [h˜1 , h˜2 ] + 2 (according to Assumption 4). 

Lemma 3. For all h ∈ R and Ya ∈ M(R), we have (6)

ψa (Ya )|Ha (Ya )(h) = ψa (Ya |h )|Ha (Ya |h)(h)

and (7)

ψa (Ya )|h+tmin = ψa (Ya |h )|h+tmin .

Let us interpret Equation (7) – Equation (6) is only a step in the proof of Equation (7). It means that the distribution of players finishing operation a before h + tmin depends only on the distribution of players starting operation a before h. Recall that tmin is a lower bound on the time needed to process an operation. Proof of Lemma 3. As soon as the first equality is true, the second one is also true, as a consequence of Claim 1 and of Assumption 2. Let us prove the first equality. Fix h ∈ R, Y, Y 0 ∈ M(R) and E a measurable subset of R. We first prove two properties. Property 1: Ha (Y |h )−1 (E) ∩ ] − ∞, h] = Ha (Y )−1 (E) ∩ ] − ∞, h]. Indeed, for h0 ≤ h, we have Ha (Y |h )(h0 ) = Ha (Y |h |h0 )(h0 ) = Ha (Y |h0 )(h0 ) = Ha (Y )(h0 ) with the help of Claim 1 for the second equality and of Assumption 5 for the first and third equalities.
Property 2: If E ⊆] − ∞, Ha (Y )(h)], and if Y 0 ≤ Y , then Y 0 Ha (Y )−1 (E) ∩ ]h, +∞[ = 0. Indeed, let h0 ∈ Ha (Y )−1 (E) ∩ ]h, +∞[. We have h0 > h and Ha (Y )(h0 ) ≤ Ha (Y )(h). According to Assumption 4, we have then Y [h, h0 ] = 0, and hence Y 0 [h, h0 ] = 0. Take now h ∈ R, Ya ∈ M(R × R), r ∈ R and J a measurable subset of R. Define E := J∩]−∞, Ha (Ya )(h)]. The set E is a measurable subset of R and it is such that E ⊆]−∞, Ha (Ya )(h)]. Note that Yar ≤ Ya when a ∈ r. ψar (Ya )|Ha (Ya )(h) (J) = Yar (Ha (Ya )−1 (E)) = Yar (Ha (Ya )−1 (E) ∩ ] − ∞, h]) +Yar (Ha (Ya )−1 (E) ∩ ]h, +∞[) r = Ya (Ha (Ya )−1 (E) ∩ ] − ∞, h]) = Yar (Ha (Ya |h )−1 (E) ∩ ] − ∞, h]) = Yar |h (Ha (Ya |h )−1 (E) ∩ ] − ∞, h]) = Yar |h (Ha (Ya |h )−1 (E) ∩ ] − ∞, h]) +Yar |h (Ha (Ya |h )−1 (E) ∩ ]h, +∞[) r = Ya |h (Ha (Ya |h )−1 (E)) = ψar (Ya |h )|Ha (Ya |h )(h) (J) 8

(by definition) (since Yar is a measure) (according to Property 2) (according to Property 1) (by definition of |h ) (by definition of |h ) (since Yar |h is a measure) (by definition).

 The following lemma states how the players strategies X induce the schedules (Yar )a∈A,r∈R on each operation a with respect to each job r. Lemma 4. Fix k ∈ N. Given a measure X ∈ M(S) (a strategy distribution), there exists a unique collection (Yar )a∈A,r∈R of elements of M(R) such that for all job r = (a1 , a2 , . . . , an )  r r   Ya1 = X |ktmin Yari = ψari−1 (Yai−1 ) for i = 2, . . . , n (Ek ) ktmin   r Ya = 0 if a ∈ /r X Yar , where (Yar )a∈A,r∈R is the solution of Moreover, for any a, the map Φka : X 7→ Ya := r: a∈r

(Ek ), is continuous. Informally, Lemma 4 says that it is possible to construct a sequence of measures on S, with each of its element representing the players progress over their jobs, with a time step of tmin . The proof of the lemma relies on Assumption 2 (an operation can not be processed at an infinite speed), which highlights the crucial importance of this assumption in our approach. Proof of Lemma 4. The proof works by induction on k. For k = 0, define Yar := 0 for all r and a. And there is no other solution. Suppose now that k ≥ 0 and that we have proved the lemma till k. Existence and continuity: Let Ya = (Ya0r )a∈A,r∈R be the solution of (Ek ). We want to prove that (Ek+1 ) has a solution. Define (Yar )a∈A,r∈R for all jobs r = (a1 , a2 , . . . , an ) by Yar1 Yari

:= X r |(k+1)tmin := ψari−1 (Ya0i−1 )

(k+1)tmin

Yar := 0

for i = 2, . . . , n if a ∈ /r

According to this definition and Lemma 2, Y depends continuously on X. Note that, according to Claim 1, we have then for all a ∈ A, r ∈ R Ya0r = Yar |ktmin

(8)

We check that Y is solution of (Ek+1 ). The first and the last equalities of (Ek+1 ) are straightforward. Let us check the second one. Let r = (a1 , a2 , . . . , an ) be a job in R. Yari

  = ψari−1 Ya0i−1   min (k+1)t r = ψai−1 Yai−1 kt min

=

ψar0

(Ya0 )|(k+1)tmin

(k+1)tmin

(by definition of Y ) (according to Equation (8)) (according to Equation (7) of Lemma 3)

Uniqueness: Assume that we have two collections (Ya )a∈A and (Z)a∈A solutions of (Ek+1 ). Yet, (Y |ktmin )a∈A and (Z|ktmin )a∈A are solutions of (Ek ). Hence, by induction,
(9) (Ya |ktmin )a∈A = Za |ktmin a∈A 9

We can write the chain of equalities for any
Yar = ψar0 Yar0 (k+1)t
min = ψar0 Ya0 |ktmin (k+1)t min
= ψar0 Za0 |ktmin (k+1)t min = Zar

a∈A (since Y is solution of (Ek+1 )) (according to Equation (7) of Lemma 3) (according to Equation (9)) (since Z is solution of (Ek+1 )). 

We are now in position to prove Proposition 1. P r Proof of Proposition 1. Recall that X(I × R) = r∈R X (I) = 1. Let τ := maxx∈[0,1] tmax (x). According to Assumption 3, for any job r = (a1 , a2 , . . . , an ), a direct induction on i leads to Yai = Yai |iτ (no one finishes operation ai after iτ ). Hence, any operations schedule (Ya )a∈A solution of (3) is solution of Equation (Ek ) for a large enough k It means that for a k large enough, we have Φka = Φa . Existence, continuity, and uniqueness are consequences of Lemma 4.  It remains to prove Corollary 1. Proof of Corollary 1. Since we have φa (X)(h, a1 , . . . , an ) := (Hai (Φai (X)) ◦ . . . ◦ Ha1 (Φa1 (X))) (h) for i such that a = ai (recall that φa (X) is only defined for strategies using operation a), the corollary is a straightforward consequence of Proposition 1.  5.2. Proof of the main theorem. Proof of Theorem 2. Theorem 2 is a direct consequence of the following lemma (Lemma 5) and of Theorem 1.    Lemma 5. The set of utility functions U (Ur )r∈R , (ta )a∈A considered in a dynamic congestion game is a measurable subset of C(SS ). Proof. For any r = a1 , . . . , an denote Φr := (Φa1 , . . . , Φan ) and φr (h) := (φa1 (h, r), . . . , φan (h, r)). To prove that X 7→ u ˆ(X) is continuous, it is enough to prove that X 7→ ur (Φr (X)) ◦ φr (X)(h) is continuous for any r (see Equation (4)). According to Proposition 1 and Corollary 1, Φr and φr are continuous. Lemma 1 then implies that S 7→ ur (Φr (X)) ◦ φr (X) is continuous. Then, the measurability comes from the fact that u 7→ (X 7→ u(Φr (X)) ◦ φr (X)(h)) is continuous (since Φr and φr are continuous) and U is compact.  6. Applications to the dynamic Wardrop assignment Our result is fairly general and notably apply to most of the problems of dynamic assignment at equilibrium found in the transportation literature. Those models, although commonly used in practice for transportation planning, lack of theoretical foundations and results of existence have been established only in very restrictive cases. In this section, we show that the existence of a solution to the most common dynamic assignment problem, the so-called dynamic Wardrop assignment problem, has a solution under the general processing time assumptions we stated earlier. Then, two common travel time models in the transportation literature are reviewed and it is shown that they are natural processing times functions in the sense stated above. 10

6.1. Dynamic Wardrop assignment with predetermined departure times. The simplest assignment model can be formulated as such. Consider a travel demand, described by flows between each origin destination pair, and assume each of them is allowed to choose his travel route, but not his departure time. We study the possible assignments of the traffic flows on the routes connecting each of the origin-destination pair. The question is the following: is there an assignment such that no route is assigned at a time h with a non zero flow of vehicles if there are routes with smaller travel times ? Such an assignment is said to verify the Dynamic Wardrop Principle. Note that the terminology in the transportation literature is variable from one authors to another and that what we call dynamic Wardrop assignment, is also termed as user equilibrium assignment ([4] or [23]). A formal statement of the dynamic assignment problem is presented below. Before doing so let us raise a few comments on the mathematical nature of traffic flows in transportation model compared to ours. Existing models represent vehicle flows by measurable functions, whereas our formulation is based on measures on I so it is useful to associate each element of L(I, R+ ), the set of positive measurable functions on I, with an element of M(I). Thus to a flow y ∈ L(I, R+ ), Rh we associate the measure Y defined by Y (] − ∞, h]) = −∞ y(h0 )dh0 . Conversely, given a measure R Y on I, there exists y ∈ L(I, R+ ) such that Y (E) = E y(h0 )dh for all measurable subsets of I if and only if Y is absolutely continuous2. To guarantee a unique mapping between an absolutely continuous measure and a measurable function, the functions equal almost everywhere in L(I, R+ ) are quotiented out. The dynamic Wardrop assignment problem can now be formulated. Consider a directed graph G = (V, A), with arc travel time functions (ta )a∈A and an origin-destination matrix (qod )o∈V,d∈V , each element of the matrix being a function in L(I, R+ ). The arc travel time functions are defined from L(I, R+ ) to C(R+ ), or using the identification exposed in the paragraph above, from the set of absolutely continuous measure on R to C(R+ ). The arc travel time functions are thus restrictions of processing time functions on L(R, R+ ). P An assignment of the traffic is an element x = (xr )r∈R of L(I, R+ )R such that r∈Ro,d xr (h) = qod (h) for all (o, d) ∈ V × V and h ∈ I, with Ro,d denoting the set of routes connecting o to d. For each route P r = a1 , . . . , an , let us define a route travel time functions tr : L(I, R+ ) → C(R) by tr (X)(h) := i=1,...,n tai (Φai (X))(φai (X)(h, r)), with X the measure associated to x. Note tr is well defined as Φai (X) returns an absolutely continuous measure if X is one. The same notations are purposely used for the players’ strategy distributions and the traffic assignments, as it is natural to interpret X r as a cumulated flow of vehicles ; X r ([0; h]) indeed counts the number of vehicles that have already entered route r. Definition 4 (Dynamic Wardrop Assignment Problem). Find an assignment x ∈ L(I, R+ )R such that whenever r, r0 ∈ Ro,d xr (h) > 0 ⇒ tr (X)(h) ≤ tr0 (X)(h), for a.e. h ∈ I The terminology predetermined departure time refers to the fact for each instant the flow of vehicles leaving an origin is fixed i.e. vehicles can their departure time. Without loss of R P not adjust generality, it is assumed in the following that od∈V ×V I qod (h) = 1. 6.2. An existence result of the dynamic Wardrop assignment problem. We claim that if the arc travel time satisfies Assumptions 1-5, in a sense to be specified below, as well as an additional one, the so-called bounded variations assumption, there exists an equilibrium assignment. 2ν is said to be absolutely continuous with respect to µ if ν(A) = 0 for every set A for which ν(A) = 0. In finite

dimensional spaces, the absolutely continuous measures with respect to the Lebesgue measure are exactly the ones that have a density. 11

Assumption 6. [Bounded variations] There is a real number K such that for any absolutely continuous measure Ya with derivative ya , the map h 7→ t(Ya )(h) is differentiable a.e. on R and 1 dta (Ya ) h 7→ (h) is smaller than or equal to K a.e. . . ya (h) dh Assumption 6 is slightly less intuitive than the ones in Section 4, but for traffic propagation it has physical sense. Intuitively,  if a flow  of vehicles x enters an arc a then the flow outcoming flow dt(x) would be something like x/ 1 + dh . Consequently, Assumption 6 implies that if the incoming flow on an arc is bounded by a constant K, then the outcoming is bounded by a constant K 0 that depends only on K and ta . In other terms, Assumption 6 insures that for a given assignment problem, there is a bound on the flows on each arc of the network (i.e. Φa (X)) so that for any traffic assignment unreasonnably hight flows of traffic won’t be observed. Let us construct a game from the assignment problem. We have already started by our choices of notations, but a few issues remain to be addressed. First, we have to extend the definition of the arc travel time functions, as they are still only defined on L(R, R+ ) – this the purpose of Subsection 6.2.1. Then, using an adequate set of utility functions (Subsection 6.2.2), Theorem 2 will tell us that there is an equilibrium, but this equilibrium is a measure D leading to an assignment X r := DS (I × {r}) that might not have a derivative in L(I, R+ ). Our equilibrium might not be an equilibrium is the sense above. This last issue is the object of Lemma 6 in Subsection 6.2.3. 6.2.1. Processing time functions. Consider K a constant and denote M≤K (R) the set of positive measurable functions essentially bounded by K, that is measures M such that for any interval J, one has M (J) ≤ Kµ(J) (where µ is the usual Lebesgue measure on R). It is easy to see that these measures are absolutely continuous and that their set is closed. As ta is continuous on M≤K (R), there exists a continuous extension t0a of ta on M(R) by Tietze-Dugundji extension theorem [3]. Since we can require that the extension stay in the convex hull of the arc travel times (defined for absolutely continuous measures) whose derivative are a.e. bounded by K, we will have continuous extensions that will satisfy Assumptions 1-3. To enforce the satisfaction of the last two assumptions, we make the following redefinition: Z h
0 ρ X|h0 , M≤K (R) dh0 , ta (X)(h) := ta (X|h )(h) + 0

where ρ is the Prohorov metric. Hence, for any constant K, we can construct well defined processing time functions for a given assignment problem that extend the arc travel time funtions. By abuse of notation, they will also be denoted ta . Note that although the extension depends on K, we have omitted any explicit reference. In this context Assumptions 1-5 can be restated for arc travel time functions. The last four assumptions have straightforward interpretations for arc travel time functions. For the first one, the previous paragraph shows that its is sufficient to have a continuous arc travel time functions for the corresponding processing time functions to be continuous. Finally, it remains to choose a constant K. What would be an appropriate value for K ? It should be high enough for so that no equilibrium assignment would induce flows Xa = Φa (X) such that xa ≥ K on a non null measurable set of R. Assumption 6 guarantees the existence of such a constant. 6.2.2. Utility functions. We can now build the game associated to the dynamic Wardrop assignment problem. Consider a distribution of users U on the set RC (stands for “route choice”) of continuous 12

utility functions of the following type  (10)

u ˆh∗ ,od (X)(h, r) =

−tr (X)(h) if h = h∗ and r ∈ Ro,d , −∞ otherwise.

The interpretation is straightforward: each player is characterized by a departure time h∗ he will always prefer, and an origin-destination pair od on which he will always travel. The utility of a travel decision is limited to the travel time on the route. Denote t˜r the map defined by X (11) t˜r (Y1 , . . . , Yn )(h1 , . . . , hn ) := tai (Φai (X))(φai (X)(h, r)) i=1,...,n

˜ for any  r = a1 , . . . , an ∈ R. Then taking Ur := {−tr } ∪ {+∞}, RC is obviously a measurable subset of U (Ur )r∈R , (ta )a∈A . Hence, we are in the framework defined in Subsection 3.2. 6.2.3. Properties of the equilibrium distribution. The set RC can be identified with V × V × I and according to the context a measure on RC is seen either as a measure on C(SS ), or as a collection of measures (Uod )o∈V,d∈V on I. The latter point of view is of particular interest because of the following proposition: Proposition 2. If U is a measure on RC seen as measure on C(SS ), the equilibrium assignment X verifies: X (12) Uod = Xr r∈Ro,d

Proof. Consider a measure U on RC such that U (C(SS )) = U {ˆ uh,od : h ∈ I, od ∈ V × V }. Let D be an associated Nash equilibrium. Recall that X := DS and U = DC(SS ) . Then for all measurable subsets E of I: X(Ro,d × E) = = = = = = =

DS (Ro,d × E) D(Ro,d × E × C(SS )) D(Ro,d × E × RC) D(Ro,d × E × {ˆ uh∗ ,od such that h∗ ∈ E}) D(S × {ˆ uh∗ ,od such that h∗ ∈ E}) U ({ˆ uh∗ ,od such that h∗ ∈ E}) Uod (E)

(by definition of X ) (by definition of a margin) (U is a measure on RC) (D is an equilibrium measure) (idem) (idem) (identifying RC with V × V × I) 

Proposition 2 simply restates in measures terms that an assignment is a decomposition of this flows over the set of the routes. An important consequence is the following lemma: Lemma 6. Let U be a measure on RC, seen as measure on C(SS ). If U is absolutely continuous, every equilibrium assignment X is also absolutely continuous. Proof. Consider an absolutely continuous measure U on C(SS ) such that U (C(SS )) = U ({ˆ uh,od : h ∈ I, od ∈ V × V }) and let X be an associated Nash equilibrium. According to Proposition 2: X Uod = Xr r∈Ro,d

Then if we have E a measurable subset of I such that Uod (E) = 0, for all r ∈ Ro,d we have X r (E) = 0. Thus, absolute continuity of U implies absolute continuity of X.  13

6.2.4. Theorem. We can now state the main result of the section: Theorem 3. Given a directed graph G = (V, A) with arc travel time functions (ta )a∈A satisfying Assumptions 1-6 and given an origin-destination matrix, there is a Wardrop assignment. Proof. Assume we are given an origin-destination matrix (qod ). Define U a measure on C(SS ) such that - U (C(SS )) = U (RC) = 1 and - Rfor a given pair od ∈ V ×V and any measurable subset J ⊆ I, we have U ({ˆ uh,od : h ∈ J}) = q (h)dh. h∈J od We have just encoded our origin-destination matrix as a measure on the set of users. Note that U is absolutely continuous. According to Theorem 2, there exists a Nash equilibrium D, and according to Lemma 6 the equilibrium assignment X := DS is absolutely continuous regarding the Lebesgue measure. Hence X admits a Radon Nikodym derivative, which we will denote x. Let h ∈ I and take any route r such that xr (h) > 0. Let od the origin-destination pair connected by r. The proof proceeds in two steps. First, we show that whenever xr is continuous in h, xr (h) > 0 ⇒ tr (X)(h) ≤ tr0 (X)(h) for all r0 ∈ Ro,d . Then, we show that this inequality holds almost everywhere. First step. Let h ∈ I be such that s is continuous in h. Now, take any route r such that xr (h) > r 0. Let od the origin-destination pair connected by r. For all  > 0,
we have X ([h − ; h + ]) > 0, 0 0 which can be rewritten D {r} × {h } × u ˆh0 ,od : h ∈ [h − , h + ] > 0. Therefore, we know that 0 for all  > 0, there is h ∈ [h − , h + ] such that u ˆh0 ,od (X)(r0 , h00 ) ≤ u ˆh0 ,od ((X))(h0 , r) for all h00 ∈ I 0 and r ∈ R, or, directly in terms of route travel times: for all  > 0, there is h0 ∈ [h − , h + ] such that tr0 (X))(h0 ) ≥ tr (X)(h0 ) for all r0 ∈ Ro,d . By continuity of h 7→ tr (X)(h), we get the required inequality. Second step. For a given r consider E the set of point such that xr (h) > 0 and tr (X)(h) > tr0 (X)(h) for a r0 on the same origin-destination pair as r. From the previous paragraph xr is discontinuous in every h ∈ E. E is measurable since tr , tr0 and xr also are. Now assume µ(E) =  6= 0, denoting µ the Lebesgue measure on R. Then, xr being measurable, there exists a set F such that the measure of its complementary µ(F c ) < /2 and xr is continuous in every h ∈ K (Lusin Theorem [10]). So E ⊆ F c , a contradiction. Hence µ(E) = 0. Thus, the required inequality is valid almost everywhere.  6.3. Two arc travel times models. The present subsection exposes two arc travel times models under which the existence of a Wardrop assignment has been shown, respectively by Zhu et al [23] (although with slightly more specific assumptions) and Mounce (on specific networks in [14] and later on general networks in [15]). We show here that both models verifies the Assumptions (1-6), and thus that those two existence results are direct consequences of Theorem 3. 6.3.1. Arc performance travel times. In an arc performance model (see for instance Friez et al [4]), travel time on an arc is assumed to depend on the volume of traffic on that same arc. More precisely a delay function Da : R+ → R∗+ taking in input a volume of traffic and returning a travel time such that Da is associated to each arc. It is assumed to be continuous, strictly increasing, positive and leading to a fifo property. Formally, the travel time model is defined by the following system: 
Va (Ya )(h) = Ya (] − ∞, h]) − Ya Ha (Ya )−1 (] − ∞, h]) (13) Ha (Ya )(h) = h + Da (Va (Ya )(h)) 14

We will refer to such a travel time function hence defined as arc performance travel time functions, in line with the classical transportation terminology. Remark that only Assumptions 1 and 5 (continuity and causality) are not straightforward. But it is possible to prove them. Proposition 3. Assuming that all users enter the network during the time interval I, the functions Ha are well-defined by Equations (13) and satisfy Assumptions 1-6. Proof. The proof works very similarly to the one of Lemma 4. Define dmin := Da (0). Then necessarily, for h ∈] − ∞, dmin ], we have Va (Ya )(h) = Ya (] − ∞, h]). Hence, we have defined Va (Ya )(h) and Ha (Ya )(h) for all h ∈] − ∞, dmin ]. Since we have Ha (Ya )−1 (] − ∞, 2dmin ]) ⊆]−∞, dmin ], we can now define Va (Ya )(h) and Ha (Ya )(h) for all h ∈] − ∞, 2dmin ]. And so on: for all k ∈ N, we have Ha (Ya )−1 (] − ∞, kdmin ]) ⊆] − ∞, (k − 1)dmin ], we can define Va (Ya )(h) and Ha (Ya )(h) for all h ∈] − ∞, (k − 1)dmin ]. Since we work by necessity and sufficient conditions, we have existence and uniqueness. Moreover, the map Va is clearly continuous. It remains to prove causality, that is, Ha (Ya |h )(h) = Ha (Ya )(h). Without loss of generality, we can assume that there is a integer kh such that h = kh dmin . Actually, a direct induction on the index k above, for k = 1 to k = kh , proves that whenever h0 ≤ h, we have Ha (Ya |h )(h0 ) = Ha (Ya )(h0 ).  6.3.2. Bottleneck travel times. Mounce uses the punctual bottleneck model to represent travel time on an arc. The arc travel time for an entrance time h is the sum of a constant travel time and a bottleneck delay. The delay arises from a limit K on the arc outcoming flow, refer to as the capacity of the arc. When traffic flow exceeds this capacity, a punctual queue starts to form at the exit of the arc. In [14], Mounce exposes how to express the travel time on an arc as a function of the cumulated volume at entrance and shows it is continuous and respect the fifo condition (Assumptions 1 and 4). The other Assumptions are also satisfied (easy verification). Hence, our model contains this model. 6.4. Further results. It’s quite clear that Theorem 2 can be applied to far more general models than the dynamic Wardrop assignment. Allowing users to choose their departure times as well as their routes, in a similar manner as done by Arnott et al or Friez et al [1, 4], is the obvious and natural next step. But much more complex models actually fit in our framework. On the demand side, the set of utility functions allows an incredibly large set of variations. For instance, utilities that varies non linearly with travel time can be considered. This is of particular importance, as studies show it is empirically relevant (see for instance [8]). Second, road pricing strategies can be embedded in the utility functions, by adding maluses on specific routes. On the supply side, the assumptions we considered are very weak and also include a wide range of particular models. The two specific cases we consider are very simple, but it is likely that more advanced traffic models, such as the ones inspired from fluid mechanics, would also fit in our framework. References [1] R. Arnott, A. de Palma, and L. Lindsey. Departure time and route choice for the morning commute. Transportation Research Part B: Methodological, 24:209–228, 1989. [2] M. Beckmann, C. B. McGuire, and C. B. Winsten. Studies in Economics of Transportation. Yale University Press, New Haven, CT, 1956. [3] J Dugundji. An extension of Tietze’s theorem. Pacific J. Math, 1(3):353–367, 1951. [4] T.L. Friesz, D. Bernstein, T.E. Smith, R.L. Tobin, and R.L. Wie. A variational inequality formulation of the dynamic network user equilibrium problem. Operations Research, 41:179–191, 1993. [5] S. Hart, W. Hildenbrand, and Kohlberg E. On equilibrium allocation as distributions on the commodity space. Journal of Mathematics economics, 1:159–167, 1974. [6] W. Hildenbrand. Core and equilibria of large economy. Princeton University Press, 1974. 15

[7] M. Ali Khan. On Cournot-Nash equilibrium distributions for games with a nonmetrizable action space and upper semicontinuous payoffs. Transactions of the American Mathematical Society, 315(1):127–146, 1989. [8] F.S. Koppelman. Non-linear utility functions in models of travel choice behavior. Transportation, 10:127–146, 1981. [9] L. Lindsey. Existence, uniqueness and trip cost function properties of user equilibrium in the bottleneck model with multiple user classes. Transportation science, 38:293–314, 2004. [10] N. Lusin. Sur les propri´et´es des fonctions mesurables. Comptes Rendus Acad. Sci. Paris, 1912. [11] A. Mas-Colell. On a theorem of Schmeidler. Journal of Mathematics economics, 13:201–206, 1984. [12] D. K. Merchant and G. L. Nemhauser. A model and an algorithm for the dynamic traffic assignment problems. Transportation Science, 12:183–199, 1978. [13] I. Milchtaich. Topological conditions for uniqueness of equilibrium in networks. Math. of Op. Res., 30:226–244, 2005. [14] R. Mounce. Convergence in a continuous dynamic queuing model for traffic networks. Transportation Research Part B: Methodological, 40:779–791, 2006. [15] R. Mounce. Existence of equilibrium in a continuous dynamic queueing model for traffic networks. Proceedings of the 4th IMA Conference on Mathematics in Transport, 2007. [16] M. Patricksson. The Traffic Assignment Problem: Models and Methods. Utrecht, 1994. [17] R.W. Rosenthal. A class of games possessing pure-strategy nash equilibria. International Journal of Game Theory, 2:65–67, 1973. [18] T Roughgarden and E. Tardos. How bad is the selfish routing ? J. of the ACM, 49:236–259, 2002. [19] D. Schmeidler. Equilibrium points on non-atomic games. J. Satistic. Phys, 7:295–300, 1973. [20] F. Topsoe. Topology and measure, volume 133 of Lecture Notes in Math. Springer-Verlag, New York, 1970. [21] W. Vickrey. Congestion theory and transport investment. American Economic Review, 59:251–261, 1969. [22] J. G. Wardrop. Some theoretical aspects of road traffic research. Proc. Inst. Civil Engrg, 2:325–378, 1952. [23] Daoli Zhu and Patrice Marcotte. On the existence of solutions to the dynamic user equilibrium problem. Transportation Science, 34(4):402–414, 2000. ´ Paris Est, LVMT, ENPC, 6-8 avenue Blaise Pascal, Cite ´ Descartes Champs-sur-Marne, Universite ´e cedex 2, France. 77455 Marne-la-Valle E-mail address: [email protected] ´ Paris Est, LVMT, ENPC, 6-8 avenue Blaise Pascal, Cite ´ Descartes Champs-sur-Marne, Universite ´e cedex 2, France. 77455 Marne-la-Valle E-mail address: [email protected]

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