Optima and Equilibria for a Model of Traffic Flow

Transcription

1 Optima and Equilibria for a Model of raffic Flow Alberto Bressan and Ke Han Department of Mathematics, Penn State University University Park, Pa 16802, USA s: bressan@mathpsuedu, kxh323@psuedu July 9, 2011 Abstract he paper is concerned with the Lighthill-Whitham model of traffic flow, where the density of cars is described by a scalar conservation law A cost functional is introduced, depending on the departure time and on the arrival time of each driver Under natural assumptions, we prove the existence of a unique globally optimal solution, minimizing the total cost to all drivers his solution contains no shocks and can be explicitly described We also prove the existence of a Nash equilibrium solution, where no driver can lower his individual cost by changing his own departure time A characterization of the Nash solution is provided, establishing its uniqueness Some explicit examples are worked out, comparing the costs of the optimal and the equilibrium solution he analysis also yields a strategy for optimal toll pricing 1 Introduction Aim of this paper is to analyze globally optimal solutions and Nash equilibrium solutions for a specific problem of traffic flow Car drivers starting from a location A a residential neighborhood) need to reach a destination B a working place) all at a given time here is a cost for starting early and a cost for arriving late Clearly, these costs can also account for the total time spent in the car Denoting by τ d and τ a respectively the departure and the arrival time, the total cost to an individual driver is Ψ = ϕτ d ) + ψτ a ) 11) cost An appropriate choice of the penalty functions here is 0 if s 0, ϕs) = s, ψs) = α s 2 if s 0 12) psidef 1

2 If L is the length of the highway connecting A with B, and v is the speed of cars, then τ a = τ d + L v It is now easy to compute the optimal departure time for each driver: τ opt d { = argmin ϕs) + ψ s + L ) s v he problem is that, if everyone adopts the same optimal strategy and departs exactly at the same time, a huge traffic jam is created and this strategy is not optimal anymore o resolve this issue, one needs to look at a better model, taking into account the fact that the speed of cars depends on the traffic density Call ρ = ρt, x) the density of cars at time t at the point x along the highway he Lighthill- LW, Richards Witham-Richards model [15, 16] describes the evolution of ρ in terms of the conservation law ρ t + [ρ vρ)] x = 0 13) claw1 Here the decreasing function v = vρ) describes the velocity of cars depending on the density A common choice is vρ) = a 1 ln a 2 0 ρ a 2 14) vd1 ρ he choice vρ) = 1 ρ ρ 0 ) v 0 0 ρ ρ 0 15) vd2 is also meaningful It yields the same qualitative properties and explicit solution formulas An optimization problem can now be formulated as follows A planner needs to schedule departures, in such a way that the combined total cost is as small as possible Let ρt, A) vρt, A)) = ūt) 16) dep be the departure rate from A, ie, how many drivers enter the highway per unit time Notice that the boundary condition 16) dep at x = A is meaningful provided that the characteristic speed ρ ρvρ)) is positive As in the examples 14)- vd1 15), vd2 we shall assume that the map ρ ρ vρ) is concave down, attaining a positive maximum at a point ρ > 0 We regard t ūt) as a control function In other words, ū ) is a measurable function that can be assigned at will, subject only to the constraint ūt) [0, M], M = max ρ 0 ρ vρ) 17) const1 Clearly, the incoming flux cannot be larger than the maximum flux allowed by the conservation law 13) claw1 he condition that all drivers eventually have to depart can be written as a constraint ūt) dt = κ, 18) const2 where κ is the total number of drivers Let ρ = ρt, x) be the solution of conservation law 13), claw1 defined for t, x) R [A, B], with boundary data 16) dep assigned at x = A, and let ut, x) = ρt, x) vρt, x)) t 0, x [A, B] 2

3 u ρ M ρ v ρ) ρ* fu) f *p) 0 ρ * ρ 0 M u 0 f p 0) Figure 1: Left: the function ρ ρ vρ) describing the flux of cars Middle: the function f, implicitly defined by fρvρ)) = ρ and extended according to fext 29) Right: the Legendre transform f f:t44 be the corresponding flux he total cost is then measured by Ju) = ϕt) ut, A) dt + ψt ) ut, B) dt 19) tcost It is convenient to switch the roles of the variables t, x, replacing the boundary value problem 13)- claw1 16) dep with a Cauchy problem for the conservation law describing the flux u = ρvρ), namely u x + fu) t = 0, 110) CP ut, 0) = ūt) 111) inc he function u fu) = ρ is defined as a partial inverse of the function ρ ρ vρ) = u, see Figure f:t44 1), assuming that 0 u M = max ρvρ), 0 ρ ρ 0 ρ = argmax ρvρ) ρ Remark 1 We are here making an important modeling assumption Namely, the car density never exceeds ρ, hence the characteristic speed is positive: ρ ρ vρ)) 0 As time increases, this means that characteristics move from the boundary where x = A) toward the interior of the domain Notice that this assumption is consistent with a causality principle fig 2) f:t73 By assigning the values of the incoming flux of cars at time τ, if ρ < ρ we influence the values of the solution at future times t > τ However, if ρ > ρ, this boundary data would influence the values of the solution in the past, at times t < τ A solution to 110)- CP 111) inc can be explicitly determined using the Lax formula [13] Lax Adjusting the variables so that = 0 and [A, B] = [0, L], the optimization problem can be written as minimize: Ju) = ϕt) ut, 0) dt + ψt) ut, L) dt 112) tco over all solutions whose initial data ū satisfy the constraints 17)- const1 18) const2 Under natural assumptions on the cost functions ϕ, ψ, we will show that, for every given κ > 0, this optimization problem has a unique solution he corresponding density ρ = ρt, x) 3

4 t t τ τ A B x A B x Figure 2: Left: when ρ < ρ, the characteristic speed is positive and characteristics starting from the boundary, where x = A, flow inside the domain Right: when ρ > ρ, as time increases, characteristics exit from the domain f:t73 in claw1 13) admits a simple mathematical description It has compact support and contains no shocks In absence of a higher authority who can force each driver to depart at a specific time, however, this solution is not likely to occur in practice Indeed, it yields different costs to different drivers, depending on their starting time In other words, the globally optimal solution is not a Nash equilibrium: there are drivers who can strictly reduce their individual cost, by changing the departure time o achieve a Nash equilibrium solution, one needs to modify the model, allowing for arbitrarily large departure rates We thus assume that, if drivers arrive at the beginning of the highway at a rate ūt) > M larger than the maximum flux defined at 17), const1 they join a queue he length of this queue qt) 0 and the flux of cars entering the highway are determined as follows Either qt) = 0, or qt) = d dtqt) = ūt) M Instead of 111), inc the conservation law 110) CP is solved with initial condition { M if qt) > 0, ut, 0+) = ρt, 0+)vρt, 0+)) = ūt) if qt) = 0 Remarkably, it turns out that the Lax formula provides the correct solution also for this more general model, without requiring any modification Indeed, let Ut, x) = t uτ, x) dτ be the total number of drivers that have crossed the point x along the highway at some time t Let Qt) denote the number of drivers that have started their journey at a time t joining the queue at the entrance of the highway, if there is any) hen the function U provides a solution to the Cauchy problem { U x + fu t ) = 0, Ut, 0) = Ut) = inf Qτ) + Mt τ) ; τ t 113) HJ Interpreting U = Ut, x) as the value function for an auxiliary optimization problem, for every x > 0 the solution of 113) HJ is provided by { Ut, x) = inf x f t τ ) + Qτ), 114) Lx4 τ x 4

5 where f is the Legendre transform of f Our analysis shows that, for every κ > 0, there exists a unique right-continuous, nondecreasing function t Qt), with Q ) = 0, Q+ ) = κ, 115) const3 such that the solution of 113) HJ yields a Nash equilibrium hat means: a solution where none of the drivers can lower his own cost by changing his departure time his implies that all drivers face exactly the same cost, regardless of the time at which they decide to join the queue he remainder of the paper is organized as follows Section 2 deals with an optimization problem for a scalar conservation law, where the cost depends only on the initial and terminal profile of the solution Rather than 110), CP we write the conservation law in the more familiar form u t + fu) x = 0 x R, t [0, ] 116) claw2 Of course, switching the variables t and x is purely for notational convenience Under suitable assumptions on the cost functional, we prove the existence of a unique globally optimal solution u = ut, x) By deriving a set of necessary conditions for optimality, the optimal solution can be accurately described In Section 3 we state a precise concept of Nash equilibrium solution, writing the conservation law claw2 116) in the integrated form U t + fu x ) = 0 U0, x) = Qx) 117) HJ2 For each κ > 0 we show that there exists a unique initial condition Q satisfying 115) const3 which yields a Nash equilibrium Using the Lax formula, this solution can also be described in detail In Section 4 we show how the previous results apply to the original traffic problem As an example, the globally optimal solution and the Nash equilibrium solution are explicitly computed, in the case where the cost functions are given by 12) psidef and the flux function is determined by 15) vd2 We observe that, by introducing an additional time-dependent cost φt) at the entrance of the highway at a toll booth), one can easily transform the globally optimal solution into a Nash equilibrium his provides a natural strategy for optimal toll pricing For a basic introduction to scalar conservation laws and the Lax formula we refer to the books of Evans [6] Evans or Smoller [17] Smoller An extension of the Lax formula to initial-boundary value problems was derived in [14] LF Optimality conditions for weak solutions of hyperbolic BM2, BS, U1, U2 conservation laws, also in the presence of shocks, were obtained in [3, 4, 18, 19] However, since the present problem requires only an optimal choice of the initial data for a scalar conservation law, our derivation of necessary conditions will be self-contained, relying on a direct application of the Lax formula Various optimization problems for traffic flow, based on the Lighthill-Whitham conservation FHM, GP, GHKL, HKK law model, have been considered in [9, 10, 11, 12] Global optima and equilibria for a different kind of transport problem were recently studied in [5], CCP also providing results on the asymptotic stability of Nash equilibria For an introduction to differential games and for recent applications to traffic flow on networks we refer to [2, Bgames, 7] and F [8], FKKR respectively 5

6 2 Optimal solutions to a conservation law Consider the scalar conservation law u t + fu) x = 0 21) claw Given a time > 0 and a constant κ > 0, consider the following optimization problem Among all the initial data u0, x) = ūx) 22) id4 satisfying the constraints 0 ūx) M, ūx) dx = κ, 23) inconst find one which minimizes the cost functional Jū) = ϕx)ūx) dx + ψx)u, x) dx 24) costf Here u, ) is the value at time t = of the unique entropy admissible solution to the Cauchy problem 21)- claw 22) id4 In order to prove the existence of optimal solution and derive a set of necessary conditions for optimality, the following assumptions will be used A1) he flux function f : [0, M] R is continuous, increasing, and strictly convex It is twice continuously differentiable on the open interval ]0, M[ and satisfies f0) = 0, f u) b > 0 for 0 < u < M 25) fprop A2) he cost functionals ϕ, ψ satisfy ϕ < 0, ψ, ψ 0, lim ϕx) = +, lim x ) ϕx) + ψx) = + x + 26) costp By the assumption A1) the map u f u) is strictly increasing on the open interval ]0, M[ Hence it admits one-sided limits f 0+), f M ), possibly with f M ) = + he inverse function g = f ) 1 is thus well defined on the closed possibly unbounded) interval I = In other words, for each λ I we define [ ] f 0+), f M ) 27) Idef gλ) = u if and only if f u) = λ 28) gdef Notice that I is the set of all possible characteristic speeds, as u ranges in [0, M] convenient to extend f to a function f : R R {+, by setting { f fu) = 0+)u if u < 0, fm) + f M )u M) if u > M It is 29) fext 6

7 In the case where f M ) = +, it is understood that fu) = + whenever u > M Let f p) = max {pu fu) 210) Lt u be the Legendre transform of f Notice that f p) = + if p < f 0+) or p > f M ) Calling u = up) the point where the maximum in 210) Lt is attained, we have d dp f p) = up) [0, M[ 211) df* he unique entropy-admissible solution to the Cauchy problem 21)- claw 22) id4 can now be obtained by the Lax formulas { ) x y y yt, x) = argmin t f x y + ūs)ds ; I, t t 212) ydef ) x yt, x) ut, x) = g t 213) Laxf One can show that, for all except countably many points x R, the expression on the right hand side of 212) ydef attains its global minimum at a single point yt, x) In case the minimum is attained at several points, we denote by y t, x) and y + t, x) respectively the smallest and the largest of such points One has y + t, x 1 ) y t, x 2 ) whenever x 1 < x 2 Moreover, under the assumption f u) b, the Oleinik estimates hold: ut, x 2 ) ut, x 2 ) x 2 x 1 b t for all t > 0, x 1 < x 2 214) olin Lemma 1 Let the flux function f satisfy the assumptions A1) Let u = ut, x) be a solution whose initial condition u0, x) = ūx) satisfies 23) inconst hen for any given > 0 the following holds i) here exists R large enough such that x R y, x) x for all x R 215) yxdist ii) Given any ε > 0, there exists δ > 0 such that, for every solution ũ whose initial data satisfy u0, ) ũ0, ) L 1 δ, one has [ ] ỹ, x), ỹ +, x) [ ] y, x ε) ε, y +, x + ε) + ε for all x R 216) usc Here ỹ ± is defined in the same way as y ±, replacing u with ũ Proof i) By construction, x y,x) I [0, [ 7

8 If the set of characteristic speeds I is bounded, it suffices to take R = f M) Notice that in this case yxdist 215) remains valid even without the assumption u0, x) dx = κ Next, assume that f u) + as u M Observing that f p) pm fm), we have the estimate ) x y y ) x y y Φy) = f ūs) ds M fm) ūs) ds Hence Φy) Φx) x y)m fm) x y ūs) ds > 0 as soon as x y > R = fm) + κ M Hence the minimum in 212) ydef can be attained at the point y only if 215) yxdist holds ii) Assuming that the conclusion does not hold, we shall derive a contradiction Let ū n : R [0, M] be a sequence of initial data with ū n ū L 1 0 Assume that there exists a sequence of points x n such that y + n, x n ) > y +, x n + ε) + ε 217) zin for every n 1 Here y + n refers to the initial data ū n, while y + refers to the initial data ū he case where y n, x n ) < y +, x n + ε) ε is entirely similar) By the optimality of y +, x n + ε) and y + n, x n ), we obtain f xn + ε y + ), x n + ε) + y +,x n+ε) ūs) ds f xn + ε y n + ), x n ) + y + n,x n) ūs) ds, f xn y n + ), x n ) + y + n,x n) ū n s) ds f xn y + ), x n + ε) + y +,x n+ε) ū n s) ds For notational convenience, from now on we shall write y n,ε + = y +, x n + ε), y n + = y + n, x n ) he above inequalities imply ) xn f + ε y n,ε + y + n f xn y + n,ε ) + f xn y + n ) f xn + ε y n + ) ūs) ū n s)) ds y n,ε 218) in1 By the assumption A1), the flux function f is increasing and strictly convex on [0, M], C 2 on ]0, M[ Hence f is differentiable and its derivative f ) = f ) 1 is also strictly increasing Introducing the notations z n = x n y n + )/, z n,ε = x n y n,ε)/ +, the inequality 218) in1 can be written as zn,ε+ε/ z n,ε f ) s) ds zn+ε/ z n f ) s) ds 8 y + n y n,ε ūs) ū n s)) ds 219) prop1

9 he fact that f ) is strictly increasing implies that a+ε/ a f ) s) ds b+ε/ b f ) s) ds δ, for some constant δ 0 and all a, b [f 0+), R/ ] with b a + ε By 217), zin for every n 1 we have z n + ε/ z n,ε Hence the left hand side of 219) prop1 remains δ for every n, while the right hand side goes to zero his achieves a contradiction 21 Existence of an optimal solution he goal of this section is to prove the existence of an optimal solution to the minimization problem described at 21) claw 24) costf As a preliminary, we show that under the assumptions A1)-A2) the initial data ū which provides the optimum should have bounded support Lemma 2 Let the assumptions A1)-A2) hold hen there exists a constant R 2 > 0 with the following property If ūx) is any initial data satisfying 23), inconst then there exists a second initial data ū satisfying 23), inconst supported inside the interval [ R 2, R 2 ], and such that the corresponding costs in 24) costf satisfy Jū ) Jū) 220) Jin he above inequality is strict, unless ū already vanishes outside [ R 2, R 2 ] Proof Choose a radius R 1 > 0 such that 2MR 1 = κ Let R be the constant introduced at 215) yxdist and choose a radius R 2 > R 1 + 2R large enough so that max ϕx) + max ψx) < x R 1 x [ R 1, R 1 +R] min ϕx) + x R 2 min x R 2 R ψx) 221) mm Such a radius certainly exists, because the left hand side of 221) mm is a fixed number, while by 26) costp the right hand side becomes arbitrarily large as R 2 + Now let ū ) be an initial data satisfying 23) inconst By the choice of R 1, there exists a function ū which also satisfies 23), inconst and such that ū x) ūx) if x R 1, ū x) = ūx) if R 1 < x < R 2, ū x) = 0 if x R 2 222) udprop We claim that 220) Jin holds Indeed, observe that, by finite propagation speed, for all t [0, ] we have { u t, x) ut, x) if x R 1 + R, u 223) udc t, x) ut, x) if x R 2 R, x R 1 +R ) u t, x) ut, x) dx = x R 2 R ) ut, x) u t, x) dx = x R 2 ūx) dx 224) udc2 9

10 Recalling 221), mm from 223)- udc 224) udc2 we thus obtain Jū ) Jū) = ϕx)ū x) ūx)) dx + x R 1 x R 1 +R ) ψx) u, x) u, x) dx ) ϕx)ūx) ū x)) dx ψx) u, x) u, x) dx x R 2 x R 2 R ) max ϕx) + max ψx) x R 1 x [ R 1, R 1 +R] min ϕx) + x R 2 0 ) min ψx) x R 2 R ūx) dx x R 2 ūx) dx x R 2 Observe that the above inequality is strict, except in the case where ū = ū heorem 1 Let the assumptions A1)-A2) hold hen, for any given, κ > 0, there exists an initial data ū satisfying the constraint 23) inconst and such that the corresponding entropy weak solution u = ut, x) to the conservation law 21) claw minimizes the cost functional 24) costf Proof Let ū n ) n 1 be a minimizing sequence Because of Lemma 2, it is not restrictive to assume that all functions ū n vanish outside the interval [ R 2, R 2 ] By taking a subsequence we can assume the weak convergence ū n ū for some u L 1 Since all functions ū n satisfy the constraints 23), inconst we clearly have ūx) [0, M], ūs) ds = κ, ūx) = 0 for x > R 2, Because of the Oleinik type estimates 214), olin for every t > 0 the functions u n t, ) have uniformly bounded variation In turn, this implies that the maps t u n t, ) are uniformly Lipschitz continuous from [δ, ] into L 1 R), for any fixed δ > 0 Using a version of Helly s compacness theorem see for example heorem 24 in [1]), Bbook by taking a further subsequence we can achieve the convergence u n t, ) ut, ) L 1 R) 0 for all 0 < t, 225) L1conv for some limit function u = ut, x) Here the map t ut, ) L 1 R) is Lipschitz continuous restricted to any subinterval [δ, ], with δ > 0 he fact that u is the unique entropy weak solution to the Cauchy problem 21)- claw 22) id4 can be proved by checking that the Lax identity 212)- ydef 213) Laxf is satisfied his is clear, because the weak convergence of the initial data ū n ū implies the uniform convergence of the integral functions U n x) = x ū n s) ds Ux) = x ūs) ds Moreover, L1conv 225) yields the convergence U n t, x) Ut, x), uniformly in x, for every fixed t > 0 he weak convergence ū n ū, together with the strong convergence u n, ) u, ) L 1 R) 0 and the uniform boundedness of the supports, yield ) ϕx) ū n x) dx + ψx)u n, x) dx = ϕx) ūx) dx + ψx)u, x) dx lim n 10

11 Hence u is an optimal solution 22 Necessary conditions for optimality Let u = ut, x) be an optimal solution of the conservation law 21), claw providing a minimum to the functional 24), costf subject to the constraints 23) inconst on the initial data Aim of this section is to provide a detailed description of u, deriving a set of necessary conditions for optimality For each y R, consider the maximal and minimal backward characteristics through, y) Recall that, by A1), the characteristic speed f u) is a strictly increasing function of u Hence u, ) has locally bounded variation, satisfies a one-sided Lipschitz condition, and has at most countably many downward jumps Calling u, x+) u, x ) the right and left limits of u, ) at the point x, we then define y x) y + x) = x f u, x )), = x f u, x+)), Ix) = [y x), y + x)] 226) char2 Notice that y, y + are the initial points of the minimal and maximal backward characteristics through the point, x) We recall that y is called a Lebesgue point for the function ū if 1 lim ρ 0+ ρ y+ρ y ρ Since ū L 1 R), the above limit holds at ae y R ūs) ūy) ds = 0 Remark 2 If y is a Lebesgue point for ū, then y Ix) for a unique point x In other words, y cannot be the center of a rarefaction wave Indeed, assume, on the contrary, that y Ix 1 ) Ix 2 ) for some x 1 < x 2 hen { ) x y y y = argmin f + ūs) ds for all x [x 1, x 2 ] For every ρ > 0 small, this implies: hen 1 ρ y+ρ y y y+ρ y ρ y ρ ūs) ds f x2 y ) ) f x2 y + ρ) ) ) ūs) ds f x2 y ρ) f x2 y for some ξ 1 [ x 1 y ūs) ūy) ds 1 ρ, x 1 y ρ) y+ρ y ], ξ 2 [ x 2 y + ρ) ūs) ūy) ds 1 ρ y y ρ = f ) ξ 2 ) ρ, = f ) ξ 1 ) ρ,, x 2 y ] ūs) ūy) ds ) ) f ) ξ 2 ) f ) ξ 1 ) f ) x1 y f ) x2 y, 11

12 which is impossible if y is a Lebesgue point We now derive a necessary condition valid at all Lebesgue points of ū Lemma 3 Let u = ut, x) be an optimal solution of 21), claw with initial data u0, ) = ū satisfying the constraints 23), inconst and providing a minimum to the functional 24) costf Assume that the flux function f satisfies the assumptions A1) and that the cost functions ϕ, ψ are continuous If y 1, y 2 are Lebesgue points of ū = u0, ), with { y1 Ix 1 ), y 2 Ix 2 ), { ūy1 ) > 0, ūy 2 ) < M, 227) a3 then ϕy 1 ) + ψx 1 ) ϕy 2 ) + ψx 2 ) 228) nc1 Proof Since y 1, y 2 are Lebesgue points of ū, by Remark 2, the points x 1 < x 2 in 227) a3 are uniquely determined Assuming that 228) nc1 fails, we will derive a contradiction Indeed, we will construct a new initial data ū which is slightly smaller than ū in a neighborhood of y 1 and slightly larger that ū in a neighborhood of y 2, thus yielding a lower total cost 1 Choose δ > 0 such that max ϕy) + max ψx) < y y 1 δ x x 1 δ min ϕy) + y y 2 δ min x x 2 δ ψx) 229) mm2 2 Choose points x ± 1, x± 2 where ū is continuous and such that x 1 δ < x 1 < x 1 < x + 1 < x 1 +δ, x 2 δ < x 2 < x 2 < x + 2 < x 2 +δ 230) xpm Notice that the continuity of ū at x ± i implies that the corresponding points yx 1 ) < y 1 < yx + 1 ), yx 2 ) < y 2 < yx + 2 ) are uniquely defined Consider the points { y 1 = max{yx 1 ), y 1 δ < y 1, = min{yx + 1 ), y 1 + δ > y 1, y + 1 { y 2 = max{yx 2 ), y 2 δ < y 2, = min{yx + 2 ), y 2 + δ > y 2 y By the assumption a3 227), for every ε > 0 sufficiently small we can construct a second initial condition ū such that ū y) = ūy) if y / [y 1, y+ 1 ] [y 2, y+ 2 ], ū y) ūy) if y [y 1, y+ 1 ], ū y) ūy) if y [y 2, y+ 2 ], 12

13 y + 1 y 1 [ ] ūs) ū s) ds = y + 2 y 2 [ ] ū s) ūs) ds = ε 4 By 230), xpm using part ii) of Lemma 1 we can choose ε > 0 sufficiently small such that the corresponding solutions u, u of 21) claw satisfy u, x) = u, x) for all x / [x 1 δ, x 1 + δ] [x 2 δ, x 2 + δ], u, x) u, x) for all x [x 1 δ, x 1 + δ], u, x) u, x) for all x [x 2 δ, x 2 + δ] x1 +δ x 1 δ [ ] u, x) u, x) dx = x2 +δ x 2 δ [ ] u, x) u, x) dx = ε By the strict inequality mm2 229), the above relations imply that Jū ) < Jū), contradicting the optimality of ū he following theorem yields a precise description of the optimal solution heorem 2 In addition to A1)-A2), assume that f u) + as u M hen, for every κ > 0, the optimization problem described at 21) claw 24) costf has a unique solution u = ut, x) In addition, one has: I) No shocks are present, hence u is continuous for t > 0 Moreover, sup t [0, ], x R ut, x) < M 231) <M II) For some constant c = cκ), this optimal solution admits the following characterization For every x R, let y c x) be the unique point such that ϕy c x)) + ψx) = c 232) vpp hen, along the segment with endpoints 0, y c x)),, x), the function u is defined according to: i) If x ycx) ii) If x ycx) = f v) for some v > 0, then u t, f 0+), then u t, t x + t ) y c x) t x + t ) y c x) = v for all t ]0, ], 233) ca1 = 0 for all t ]0, ] 234) ca2 13

14 Proof 1 Assume that the optimal solution contains a shock Call x the position of this shock at time t =, and let u = u, x ) > u + = u, x+) be the left and right limits of u, ) at this point Notice that these limits exists, because u, ) has bounded variation Computing the minimal and maximal backward characteristics through the point, x) we find y = y 0, x) = x f u ), y + = y + 0, x) = x f u + ) Consider any two points y 1, y 2 [y, y + ], say with y 1 < y 2 Since ϕy 1 ) > ϕy 2 ), we must have either ūy 1 ) = 0 or ūy 2 ) = M In the opposite case Lemma 3 would yield a contradiction We thus conclude that there exists a point ξ [y, y + ] such that { ūy) = 0 if y < y < ξ, ūy) = M if ξ < y < y + 235) alt We claim that alt 235) also leads to a contradiction Indeed, if ξ > y, then ) x ξ ξ ) x y f + ūs) ds < f y + ūs) ds his contradicts the optimality of y in the Lax representation 212) ydef On the other hand, if ξ = y, since both y and y + yield the minimum in 212), ydef we have 0 = f x y = x y + )/ x y )/ ) y ) x y ūs) ds f + y + + ūs) ds f ) p) dp + My + y ) 236) ii Since f ) p) = u if and only if f u) = p, the assumption f u) + as u M implies that f ) p) < M for all p f 0+) Hence the right hand side of 236) ii is strictly negative and equality cannot hold his contradiction shows that the optimal solution cannot contain shocks 2 o prove 231), <M observe that by Lemma 2 the solution u has compact support o fix the ideas, let ū = u0, ) be supported inside [ R 1, R 1 ] and let u, ) be supported inside [ R 2, R 2 ] Since lim u M f u) = +, we can choose a value h < M such that f h) > R 1 + R 2 )/ We claim that ūx) < h for all x Otherwise, choose a point x 0 such that ūx 0 ) h hen, since no shocks are present, computing the solution by the method of characteristics we find ) u t, x 0 + tf ūx 0 )) = ūx 0 ) Consider the point x = x0 + f ūx 0 )) Since x 0 R 1, the choice of h implies x > R 2 We thus have u, x ) = ūx 0 ) h > 0, contradicting the fact that the support of u, ) should be contained in [ R 2, R 2 ] his proves our claim, and hence 231) <M 14

15 3 Since no shocks are present, the solution u is continuous for t > 0 As a consequence, backward characteristics are uniquely defined In particular, at t =, for each x R the minimizer y = y, x) in the Lax representation 212) ydef is unique We recall that y, x) is the initial point on the characteristic line which ends at, x) 4 By Lemma 3 and <M 231), there exists a unique constant c such that the following holds If y = y, x) is Lebesgue point for the initial data ū, with ūy) > 0, then ϕy) + ψx) = c 237) =const Consider the map x y c x) implicitly defined by vpp 41) Since the function ϕ is strictly decreasing, the value y c x) is well defined provided that ϕ+ ) + ψx) < c For notational convenience, we set y c x) = + if ϕ+ ) + ψx) c From the assumption ψ 0 it follows that the map x y c x) is non-decreasing 5 Assume that y c x) is a Lebesgue point of ū and ūy c x)) > 0 hen 41) vpp and 237) =const together yield ϕy c x)) + ψx) = c = ϕy, x)) + ψx) Since ϕ is strictly decreasing, this yields y c x) = y, x) = x f u, x)) 238) yc herefore, along the characteristic) segment with endpoints 0, y c x)),, x) and slope f v), v = ūy c ) = u, x), the function u = ut, x) is determined by 233) ca1 6 Observe that the second identity in 238) yc always holds, because the solution u = ut, x) is continuous for t > 0 and backward characteristics are unique In this step we prove that the first equality in 238) yc remains valid as long as x ycx) > f 0+), even if y c x) is not a Lebesgue point of ū Indeed, if y, x) < y c x), let ỹ ]y, x), y c x)[ be a Lebesgue point of ū hen Since ỹ = y, x) for some x = ỹ + f ūỹ)) > x x ỹ > x y cx) > f 0+), it is clear that ūỹ) > 0 By step 5, we have ỹ = y, x) = y c x) Since x x but ỹ < y c x), recalling that ϕ is strictly decreasing we obtain ϕy, x)) + ψ x) = ϕy c x)) + ϕy, x)) < ϕy c x)) + ψx), contradicting the fact that all terms in the above expression are equal to c On the other hand, if y, x) > y c x), let ỹ ]y c x), y, x) [ be a Lebesgue point of ū hen ỹ = y, x) for some ˆx = ỹ + f u, ỹ)) < x 15

16 Since u0, ỹ) < M, using Lemma 3 and the assumptions ϕ < 0, ψ 0 we obtain c ϕỹ) + ψ x) < ϕy c x)) + ψx), again reaching a contradiction We thus conclude that, if x ycx) = f v) > f 0+), then ca1 233) holds 7 Next, consider the case where x ycx) f 0+) We claim that for all x [x, y c x) + f 0+)], x y, x) = f u, x)) = f 0+) 239) u0 Indeed, if f u, x)) > f 0+), then y, x) = x f u, x)) < y c x) ϕy, x)) + ψ x) > c As in the previous steps, we can then find a Lebesgue point ỹ ]y, x), y c x)[ such that ϕỹ) + ψ x) > c, hen ỹ = y, x), for some point x satisfying x > x, Hence ūỹ) > 0 We thus have x ỹ x ỹ > x ỹ > f 0+) 240) >cii) > f 0+) c = ϕỹ) + ψ x) > ϕy c x)) + ψx), reaching contradiction to his proves our claim Since f is strictly increasing, from 239) u0 it follows that u, x) = 0 for all x [x, y c x) + f 0+)], hence 234) ca2 holds 8 In this final step, we prove the uniqueness of the optimal solution Observe that, for each value of the constant c R, the properties 233)- ca1 234) ca2 yield a unique solution u = ut, x) of the conservation law 21) claw his solution is continuous for t > 0, while the initial condition u0, ) = ū c) ) contains at most countably many upward jumps producing centered rarefaction waves) Assume that, for some κ > 0, there exists two optimal solutions, say u 1, u 2 If u 1 u 2, these solutions will satisfy 233)- ca1 234) ca2 with different constants, say c 1 < c 2 We show that this leads to a contradiction Since ϕ is strictly decreasing, for every x R there exist unique values y c2 x) < y c1 x) such that ϕy c1 x)) + ψx) = c 1, ϕy c2 x)) + ψx) = c 2 For every x such that x yc 1 x) u 2, x) are implicitly defined f u 1, x)) = x y c 1 x) > f 0+), by the previous analysis the values of u 1, x) and < x y c 2 x) 16 = f u 2, x))

17 Since f is strictly increasing, for every x R this provides the implication u 1, x) > 0 = u 1, x) < u 2, x) herefore, by conservation of the total mass we conclude u 1 0, x) dx = u 1, x) dx < u 2, x) dx = u 2 0, x) dx his contradicts the assumption that all integrals are = κ, thus proving uniqueness Remark 3 he optimality conditions derived in heorem 2 apply to traffic flow as well as to other models, such as supply-chains All the analysis, based on the Lax formula, is valid for scalar conservation laws as long as the flux function is convex and depends only on the density of the conserved quantity On the other hand, this technique cannot be used when the flux depends also on the variables t, x, or in connection with second-order traffic models which are described by system of conservation laws o analyze these more complex situations, necessary conditions for optimality BM2, BS can still be obtained in the form of a Pontryagin maximum principle [3, 4], but only within a class of piecewise regular solutions s:nash 3 he Nash equilibrium Aim of this section is to give a precise definition of Nash equilibrium solution for the above problem of traffic flow, and prove its existence and uniqueness Introducing the integral function Ut, x) = x ut, s) ds, the conservation law claw 21) can be equivalently written as a Hamilton-Jacobi equation U t + fu x ) = 0 31) hj hroughout the following, f denotes the flux function extended to the entire real line as in 29), fext while f is the corresponding Legendre transform, defined at 210) Lt For our application to traffic flow, one should keep in mind that the x variable denotes time, while t [0, ] denotes a point along the highway Hence Ut, x) measures the total number of cars that have crossed the point t along the highway during the time interval ], x] As initial data we shall consider any bounded non-decreasing function Q : R R +, with Q ) = 0, Q+ ) = κ 32) Qinfty Here Qx) denotes the total number of cars that have entered the queue at the entrance of the highway up to time x Notice that Q is continuous except for countably many times x o fix the ideas, we shall consider the right-continuous version where Qx) = Qx+) coincides with its right limit at every x When needed, we shall denote by Qx ) = lim y x Qy) the left limit of Q at x 17

18 For a given Q ), consider the Lipschitz continuous function { Ux) = inf Qy) + Mx y) ; y x Qx) 33) indata2 Notice that Qx) Ux) measures the length of the queue at time x, while Ux) denotes the total number of drivers that have actually departed after clearing the queue) up to time x For t > 0, the entropy-admissible solution to the Cauchy problem 31), hj 33) indata2 is provided by the Lax formula: Ut, x) = min { ) x y t f + Uy) ; y R t = min { ) x y t f + Qy ) ; y R t 34) U Observe that the last two expressions in U 34) are equal because f ) p) M for all p, and f ) p) M as p + Moreover, U0+, x) = Ux) o visualize the profile of this solution at time t =, it is convenient to introduce the function ) s hs) = f 35) hdef Observe that h is a concave function Setting µ = /f 0), one has hs) = for s > µ, h µ) = 0, lim s h s) = M 36) hprop From 34) U it now follows see fig 3) f:t43 U, x) = { min Uy) hy x) y { = min Qy ) hy x) 37) U2 y In other words, U, x) is the amount by which we can shift upward the graph of h x), before hitting the graph of Q ) or, equivalently, the graph of U) From this construction it is clear that the map x U, x) is nondecreasing For solutions with different initial data one has the comparison property Qy) Qy) for all y R = U, x) Ũ, x) for all x R 38) compare Given an initial data Q ) as in 32), Qinfty for β [0, κ[ we define the points x q β), x d β), and x a β) by setting x q β) = sup{x R ; Qx) β, x d β) = sup{x R ; U0+, x) β, 39) xqdadef x a β) = sup{x R ; U, x) β In the application to traffic flow, β is a Lagrangian variable labeling a particular driver In this case, x q β) accounts for the time where this driver joins the queue, x d β) is the actual departure time and x a β) is the arrival time Remark 4 For all except countably many β, the points x q β) and x d β) are uniquely determined by the identities Qx q β) ) β Qx q β)), Ux d β)) = β 310) xqchar 18

19 Q U U,x) yx) µ x y hy x) Figure 3: Constructing the profile x U, x) using the Lax formula Since h M, the supremum in U2 37) does not change if Q ) is replaced by U ) f:t43 Moreover see fig 4), f:t45 for ae β the arrival time x a is determined as { x a β) = inf x ; Qy) β + hy x) for all y x 311) xachar More generally, for a driver that departs at time x, we define the arrival time as { Ax) = max x + µ, sup x a β) 312) Adef β<qx) In other words, if there is no traffic at all, then the total time needed for the trip is µ = [length of the highway] [maximum speed] = f 0) On the other hand, if the driver starting at time x encounters traffic, his arrival time will simply be the supremum among the arrival times of all cars departed earlier Definition We say that a bounded, nondecreasing initial data Q ) satisfying 32) Qinfty yields a Nash solution of the Cauchy problem 31)- hj 33) indata2 with initial and terminal cost functions ϕ, ψ if there exists a constant c such that: i) For almost every β [0, κ] one has ϕx q β)) + ψx a β)) = c 313) ecost ii) For all x R, there holds ϕx) + ψax)) c 314) nobetter 19

20 Qx) β+hx) β P P µ 0 q x β) a x β) x hx) Figure 4: Given the function Q ), let β [0, Q+ )] label a particular driver he starting time x q β) is then determined by the intersection of the graph of Q with the line Q = β he arrival time x a β) is determined as follows: shift the graph of the function x β + hx) horizontally, until it lies entirely below the graph of Q In the figure, the size P P of this shift yields precisely the arrival time x a β) f:t45 In connection with the traffic model, condition i) states that all drivers bear the same cost c Condition ii) says that, regardless of the starting time x, no one can achieve a cost < c heorem 3 Let the flux function f and the initial and terminal cost functions ϕ, ψ satisfy the assumptions A1)-A2) Moreover, assume that f u) + as u M hen, for every κ > 0, the Cauchy problem 31), hj 33) indata2 admits a unique Nash equilibrium solution, with initial data Q ) satisfying 32) Qinfty Proof he result will be proved in several steps he overall strategy is to show that i) o each c R there exists a unique Nash equilibrium having cost c his is determined by some initial data Q ) having total mass κc) = Q+ ) ii) For some minimum cost c 0, the map c κc) is a strictly increasing, continuous map from [c 0, + [ onto [0, + [ 1 For a fixed constant c, let Q c be the family of all bounded, non-decreasing initial data Q ) such that Q ) = 0, and such that the corresponding solution U = Ut, x) in U 34) satisfies ϕx q β)) + ψx a β)) c for ae β [0, Q+ )] 315) eless We claim that the maximum number of drivers κc) = sup {Q+ ) ; Q Q c 316) kcdef is finite Indeed, by the assumptions A2), there exists a bounded interval [x, x + ] such that ϕx) > c for all x < x, ϕx) + ψx) > c for all x > x + 20

21 By 315), eless this implies that, for every Q Q c, the corresponding function U in 33) indata2 must be constant on ], x ] and on [x +, + [ We thus obtain the bound κc) = U+ ) = + U x x)dx x + x U x x) dx Mx + x ) 2 In the remainder of the proof we show the initial data Q x) = sup {Qx) ; Q Q c 317) Q*def yields a Nash equilibrium solution U = U t, x) o begin with, we check that, if Q 1, Q 2 Q c, then the function Q 3 x) = max{q 1 x), Q 2 x) also lies in Q c Indeed, let β [0, κc)] and, to fix the ideas, assume { x q 3 β) = min x q 1 β), xq 2 β) = x q 1 β), with obvious meaning of notations hen { x a 3β) min x a 1β), x a 2β) x a 1β) Since ϕ < 0 and ψ 0, this yields herefore, Q 3 Q c as well ϕx q 3 β)) + ψxa 3β)) ϕx q 1 β)) + ψxa 1β)) c Using this property of the set Q c, we can construct a sequence functions Q n ) Q c such that, for every x R, the sequence Q n x) increases monotonically to Q x) Since all functions Q n are constant outside a bounded interval, we clearly have Q ) = 0, Q + ) = κc) We claim that Q Q c Indeed, by monotonicity and pointwise convergence we have x q nβ) x q β), for ae β [0, κc)] herefore, x a nβ) x a β) for ae β he continuity of ϕ and ψ now yields ϕx q β)) + ψx a 1β)) c for ae β [0, κc)] herefore, Q Q c 3 In this step we show that the solution U of 31) hj with initial data Q is a Nash equilibrium o prove 313) ecost we argue by contradiction Suppose there exists some β such that the points x q β), x a β) are well defined as in 310)- xqchar 311), xachar with Q, U replaced by Q, U ), but ϕx q β)) + ψx a β)) < c By the continuity of ϕ, we can choose ε > 0 small enough so that ϕx q β) ε) + ψx a β)) < c 318) epch Consider the perturbed initial data β if x [x q β) ε, x q β)], Qx) = Q x) otherwise 21

22 We claim that Q Q c Indeed, since Q Q, the corresponding solutions satisfy Ũ, x) U, x) for every x For any β [0, κc)], two cases can arise: CASE 1: x q β ) / [x q β) ε, x q β)] In this case one has x q β ) = x q β ), x a β ) x a β ) herefore, recalling that ϕ < 0 while ψ 0, we obtain ϕ x q β )) + ψ x a β )) ϕx q β )) + ψx a β )) c CASE 2: x q β ) [x q β) ε, x q β)] In this case, by epch 318) one has x q β ) x q β) ε, x a β ) x a β ) ϕ x q β )) + ψ x a β )) ϕx q β) ε) + ψx a β )) < c We conclude that Q Q c Since Qx) > Q x) for x q ε < x < x q, this contradicts the definition 317) Q*def Hence Q satisfies the condition 313) ecost It remains to prove that 314) nobetter also holds Let x R be given Again we consider two cases CASE 1: here exists a sequence of values β n Qx) satisfying 313), ecost such that x q β) x In this case we have x a β n ) Ax) By the continuity of ϕ, ψ, the inequality 314) nobetter is then an immediate consequence of 313) ecost CASE 2: Qy) = β 0 is a constant for all y in a neighborhood of x In this case, if we can still conclude ϕx) + ψax)) Ax) = inf β β 0 + xa β) 319) lima ) lim ϕx q β)) + ψx a β)) β β 0 + = c It thus remains to examine the case where 319) lima does not hold Assume that 314) nobetter fails hen there exists δ > 0 such that Ax) + δ inf β β 0 + xa β), ϕx) + ψy) c for all y [Ax), Ax) + δ] Introduce the constant η = h µ δ) and consider the function Q y) if y < x, Qy) = max { Q y), β 0 + η We claim that Q Q c Indeed, the choice of η yields if y x, β < β 0 = x q β) = x q β), x a β) = x a β), β 0 < β < β 0 + η = x q β) = x, x a β) Ax) + δ, β 0 < β = x q β) = x q β), x a β) = x a β) 22

23 he above construction shows that, if nobetter 314) fails, then Q is not maximal his completes the proof that Q provides a Nash equilibrium solution 4 In this step we prove that, for any given c R, the Nash equilibrium solution corresponding to the cost c is unique For each x R, denote by zx) the point such that ϕzx)) + ψx) = c 320) zdef Notice that zx) is uniquely defined, because ϕ is strictly decreasing If now Q is an initial data yielding a Nash solution U = Ut, x) with cost c, recalling the definitions 39) xqdadef one has x q β) = zx a β)) for ae β [0, Q+ )] 321) xbz Remark 5 Since the map β x a β) is strictly [ increasing, if Q has a jump ] at a point x 0 then ψ must be constant on the nontrivial interval x a Qx 0 )), x a Qx 0 +)) If we assume that the cost ψ is strictly increasing, then the distribution Q ) which yields a Nash equilibrium must be continuous Consider two initial data Q 1 x) and Q 2 x), yielding two Nash equilibrium solutions corresponding to the same cost c Since ϕs) + as s, there exists x R such that Q 1 x) = Q 2 x) = 0 for all x x + 1 he uniqueness property will be proved by showing that, for every ε > 0, Q 2 x) Q 1 x) < ε x x ) for all x > x 322) epsilon Indeed, if epsilon 322) fails, there exists a first point x 0 such that and moreover: either or else Q 2 has a jump at x 0 and Q 2 x ) < Q 1 x ) + ε x x ) for all x < x 0, 323) xp1 Q 2 x 0 ) = Q 1 x 0 ) + ε x 0 x ), 324) xp2 Q 2 x 0 ) < Q 1 x 0 ) + ε x 0 x ) Q 2 x 0 +) Q 1 x 0 +) + ε x 0 x ) 325) xp3 A contradiction is derived as follows Let β = Q 1 x 0 ) Choose a point ξ, β ) where the graph of the function y β +hy x a 1 β)) touches the closure of the graph of Q 1 his means β = β + hξ x q 1 β)) = Q 1ξ ) We consider three cases CASE 1: 323)- xp1 324) xp2 hold, with ξ < x 0 see fig 5, f:t46 left) Consider the higher level β = β +εξ x ) and observe that x a 2 β ) x a 1 β) Since the map x zx) defined at 320) zdef is nondecreasing, this implies zx a 2 β )) zx a 1 β)) = ξ herefore Q 2 x 0 ) Q 2 zx a 2β )) ) = β + εξ x ) < β + εx 0 x ) = Q 1 x 0 ) + εx 0 x ), 23

24 Q 2 Q 2 Q 1 β+εξ x ) β β Q 1 β 2 β 1 γ ξ x 0 ξ 2 ξ= x 0 x Figure 5: he construction used to prove uniqueness Left: CASE 1 Right: CASE 2 f:t46 in contradiction with xp2 324) CASE 2: xp1 323)- xp2 324) hold, with ξ = x 0 see fig f:t46 5, right) If Q 2 x 0 ) = β + εx 0 x ), consider the line through the point x 0, β + εx 0 x )) with slope ε, namely γx) = β + εx x ) By xp1 323) it follows Q 2 x) < Q 1 x) + εx x ) < γx) for all x [x, x 0 [ Observing that the function h ) is continuously differentiable with the above inequality implies h x) = 0 if x µ, h x) > 0 if x < µ, x a 2β + εx 0 x )) > A 1 x 0 ) = x 0 + µ Indeed, setting β 2 = Q2 x 0 ), the contact point ξ 2 defined by Q 2 ξ 2 ) = β 2 + hξ 2 x a 2β 2 )), must satisfy ξ 2 < x 0 By continuity, we can find β < β+εx 0 x ) such that x a 2 β ) > A 1 x 0 ) his implies Q 2 x 0 ) Q 2 zx a 2β )) ) = β < β + εx 0 x ), providing a contradiction CASE 3: the inequalities xp1 323) and xp3 325) hold In this case, define β 1 = Q1 x 0 ), η = { sup Q 2 x) Q 1 x) x<x 0 < εx 0 x ) 24

25 Q 2 Q 2 β β β 2 Q 1 β 2 Q 1 β 1 β 1 x ξ x 0 ξ Figure 6: he construction used to prove uniqueness, in CASE 3 Here P 1 = β 1, x a 1β 1 )), P 2 = β, x a 2β )) Since β > β 1 + η, one has x a 1β 1 ) < x a 2β ) f:t48 Since we have Q 2 x) Q 1 x) + η for all x < x 0, x a 2β + η) x a 1β) for all β < Q 1 x 0 +) Hence, for the distribution Q 1, the arrival time of a driver starting at a time x > 0 satisfies lim A 1x) x a x x 0 + 2β 1 + η) Choosing β such that we achieve the strict inequality β 1 + η < β < Q 2 x 0 +), he assumption that Q 2 is a Nash solution implies lim x x 0 + A 1x) < x a 2β 1 + η) 326) A1x zx a 2β )) = x 0 hanks to A1x 326), we can find x > x 0 such that ϕx) + ψa 1 x)) < ϕx 0 ) + ψx a 2β )) = c Hence nobetter 314) fails, contradicting the assumption that Q 1 is a Nash equilibrium Putting together all cases, we thus obtain the uniqueness of the Nash equilibrium solution, for every fixed cost c 5 By the previous steps, for each value c, there exists a unique Nash equilibrium solution with total mass Q+ ) = κc), for some non-decreasing function κ ) It is clear that κc) = 0 for 25

26 all c sufficiently large and negative, while κc) + as c + Calling c 0 = inf{c ; κc) > 0, by the characterization of Nash solution 317) Q*def and by the continuity of the cost functions ϕ, ψ it follows that κ ) is strictly increasing on [c 0, + [ o complete the proof, it remains to show that κ ) is continuous, hence it maps [c 0, + [ onto [0, + [ Observing that, for every c Q c = c >c Q c = n 1 Q c+1/n, it is clear that the map c κc) is right continuous o show that κc) = sup κc ), 327) krc c <c we proceed as follows Let ε > 0 be given Choose an initial data Q Q c such that Q+ ) = κc) Consider the Lipschitz continuous function U as in 33) indata2 and, for some fixed ε > 0, define U ε x) = 1 ε)ux) Let U, U ε be the be the corresponding solutions of 31), hj given by the formula 34) U Observe that, for a e β [0, κc)] we have x d β) = inf{x ; Ux) β = x d ε1 ε)β) We claim that, for ae β [0, κc)] one has the strict inequality x a β) = inf{x ; U, x) β > x a ε1 ε)β) = inf{x ; U ε x) 1 ε)β 328) xde = inf{x ; U ε, x) 1 ε)β 329) xae β β 1 ε)β U U ε a β+ h x β)) ξ x d β) a x β) Figure 7: Proving the strict inequality x a ε1 ε)β) < x a β) For y x d β), the graph of the function y 1 ε)β +hy x a β)) lies strictly below the graph of U ε Here the thick dotted polygonal denotes the graph of y Ux) εux d β)) f:t47 Indeed, for ae β [0, κc)] there exist a unique point x d β) such that Ux d β)) = β, U x x d β)) > 0 330) bprop 26

27 For any β as above, consider the arrival time { x a β) = inf x ; β + hy x) Uy) for all y x By the second relation in bprop 330), any point ξ, β ) where the graph of β + h x a β)) touches the graph of U must satisfy Moreover, for y x d β), one has ξ < x d β), β < β 1 ε)β + hy x a β)) Ux) εux d β)) herefore see fig 7), f:t47 for y x d β), the graph of y 1 ε)β + hy x a β)) has strictly positive distance from the graph of y U ε y) By the definition of x a ε, this yields the strict inequality 329) xae For notational convenience, in the following we write β ε new initial data U ε, we claim that = 1 ε)β In connection with the ϕx q εβ ε )) + ψx a εβ ε )) < c for ae β ε [0, 1 ε)κc)] 331) <c Indeed, since two cases can arise x d εβ ε ) = x d β), x a β) < x a β), 332) xade CASE 1: ψx a εβ ε )) < ψx a β)) hen we immediately conclude ϕx q εβ ε )) + ψx a εβ ε )) < ϕx q β)) + ψx a β)) c CASE 2: ψx a εβ ε )) = ψx a β) In this case we observe that, since ψ is continuous and nondecreasing, there can be at most countably many disjoint open intervals ]a l, b l [ such that ψ is constant on each closed interval J l = [al, b l ], l 1 Recalling that ϕ is strictly decreasing, for each l there can be at most one point y l such that ϕy l ) + ψa l ) = c Since the map β x d β) is strictly increasing, there can be at most one value β l such that x d β l ) = y l If now ψx a εβ ε )) = ψx a β)), then the two points x a β), x a εβ ε ) must lie in the same interval J l, for some l 1 Hence, either β = β l, or else ϕx d β)) + ψx a β)) < c he first possibility can occur only for countably many values of β he second alternative leads to his proves our claim <c 331) ϕx q εβ ε )) + ψx a εβ ε )) ϕx q β)) + ψx a β)) < c Based on the previous analysis, we can now choose δ > 0 such that the set B c δ = {β [0, 1 ε)κc)] ; ϕx q εβ)) + ψx a εβ)) > c δ 27

28 has measure he new initial data satisfies U δ x) = Since ε > 0 was arbitrary, this proves measb c δ ) < ε {y x ; Uy)/ B c δ U x y) dy, U δ Q c δ, U δ + ) 1 ε)u+ ) ε κc) = sup κc ) c <c herefore the map c κc) is continuous, hence surjective his completes the proof 4 Application to traffic flow In order to apply the previous results to our specific problem of traffic flow, it suffices to switch the variables t, x, and check that the flux function f, defined as a partial inverse of the map ρ ρ vρ), satisfies the assumptions A1) Under natural hypotheses on the velocity v = vρ), this is straightforward see fig 1) f:t44 Indeed, define ρ = argmax ρ ρvρ), M = ρ vρ ) = max ρ ρ vρ) Assume that the second derivative of the flux of cars satisfies 2v ρ) + ρ v ρ) γ < 0 for all ρ [0, ρ ] hen the map ρ ρ vρ) is strictly increasing and concave on [0, ρ ] he inverse function f : [0, M] [0, ρ ] is strictly increasing, uniformly convex, and satisfies f u) + as u M Hence all the assumptions used in heorems 1 3 hold In the next sections we explicitly compute the globally optimal solution and the Nash equilibrium solution for the traffic flow problem, in the case where the velocity function v and the cost functions ϕ, ψ are given by vρ) = 1 ρρ0 ) v 0, ϕt) = t, ψt) = { 0 if t < 0, t 2 if t 0 41) vpp 41 he globally optimal solution Consider the Cauchy problem { u x + fu) t = 0 t R, x [0, L], ut, 0) = ūt), 42) Cauchy 28

29 +L/v 0 τ 0 x L s τ 0 d τ s) 0 τ 1 t M _ ut) τ τ t Figure 8: he globally optimal solution f:t35 where u = ρ vρ) and fu) = ρ is the inverse of the flux function in 13) claw1 We thus have u = ρ v 0 1 ρ ), ρ = ρ 0 ρ 0 2, M = ρ 0v 0 4, fu) = ρ = ρ 0 2 f u) = ρ 0 4M 1 1 u M ) ρ [0, M], 1 1 u/m), f ) 1 s) = M ρ2 0 16Ms 2 Observe that the speed of cars is v 0 Hence the time needed to get to destination is L/v 0 Setting { c 0 = min ϕt) + ψ t + L ) { = min t + t + L ) 2 = L 1, 43) c0def t R v 0 t R v 0 v 0 4 it is clear that the total amount of cars that can incur in a cost c 0 is zero Using heorem 2, we now describe the globally optimal solution 233)- ca1 234), ca2 corresponding to a cost c > L/v 0 > c 0 see fig 8) f:t35 Define the times τ 0 = c, τ1 = sup {t ; t + t + L ) 2 c = 1 v 0 2 L 1 + v c L 44) t01def v 0 hen the flux function u = ut, x) for the optimal solution is as follows see fig 8) f:t35 ] o each terminal time s [τ 0 + L, τ v0 1 + L we can associate a unique initial time v0 τ q s) = { c if s < 0, s 2 c if s 0, 45) tqdef so that ϕτ q s)) + ψs) = c 46) costc 29