Traffic Flow Problems - PDF Free Download

Traffic Flow Problems Nicodemus Banagaaya Supervisor : Dr. J.H.M. ten Thije Boonkkamp October 15, 2009

Outline Introduction Mathematical model derivation Godunov Scheme for the Greenberg Traffic model. Numerical experiments Higher Order Effects. Shock structure Conclusion

Introduction Definition Traffic flow Figure: copyright British library Board

Mathematical model derivation Consider the traffic flow of the cars on a highway with only one lane. Let, ρ (x, t) be the density of the cars (in vehicles per kilometre), x denote any point along the highway, t denote the time. v (x, t) denote the velocity of cars. Q (x, t) = ρ (x, t) v (x, t) denote the number of cars pass through x at time t.( flux of the vehicles given by vehicles per unit time). The number of cars which are the interval (x 1, x 2 ) at time t is x2 x 1 ρ (x, t) dx. (1) Assumption: ρ and Q are continuous functions.

Mathematical model derivation Figure: Derivation of the conservative law,where a = Q(x 1, t),b = Q(x 2, t)

Mathematical model derivation d x2 ρ (x, t) dx = Q (x 1, t) Q (x 2, t). dt x 1 Then we have, x2 x 1 [ ρ t + ] Q(x, t) dx = 0 x (2) ρ t + (ρv) = 0. x (3)

Mathematical model derivation Since x 1, x 2 R, t 1, t 2 > 0 are arbitrary, we conclude that with Initial ρ t + (ρv ) = 0, (4) x ρ (x, 0) = ρ 0 (x), x R Let flux Q = Q(ρ), then Q(ρ) = ρv (ρ). Thus equation ( 4) can be written as ρ t + c(ρ) ρ x = 0, where c(ρ) = Q (ρ).

Mathematical model derivation Figure: Flow density curve in the traffic flow Traffic models Lighthill-Whitham-Richards model ( v(ρ) = v max 1 ρ ), 0 ρ ρ max. (5) ρ max

Mathematical model derivation Greenberg model v(ρ) = a log ρmax ρ, 0 < ρ ρ max. where a is in kilometres per hour. we are going to solve this model numerically using Godunov scheme.

Godunov Scheme for the Greenberg model Godunov Scheme for Nonlinear Conservation laws consider the initial value problem u t f (u) + = 0, x R, t > 0, (6a) x u (x, 0) = u 0 (x), x R. (6b) Let us introduce a control volumes or cells V j as follows: [ V j = x j 1 2, x j+ 1 2 ), x j+ 1 2 = ( x j + x j+1 ), j = 0, ±1, ±2,..., (7)

Godunov Scheme for the Greenberg model Associated with uj n is the function ū(x, t), defined as the solution of the following initial value problem with piecewise constant initial at t = t n : ū t f (ū) + = 0, x R, t > t n, (8a) x ū ( x, t n) = uj n, x V j (j = 0, ±1, ±2,... ). (8b) To compute the numerical flux; u ū(x, t n j n, if x < x j+ 1, ) = 2 uj+1 n, if x > x j+ 1. 2

Godunov Scheme for the Greenberg model The solution of this Riemann problem is a similarity of the form ū(x, t) = u R (η; u n j, u n j+1 ), η = x x j+ 1 2 t t n. (9) Since η = 0 for x = x j+ 1, the computation of the numerical flux 2 we simply find F(uj n, uj+1 n ) = f (u R(0; uj n, uj+1 n )). (10) Thus the Godunov scheme is given by u n+1 j = uj n t ( ) F(uj n, uj+1 n x ) F(un j 1, un j ), (11) with the numerical flux as defined in (10).

Godunov Scheme for the Greenberg model And the stability condition of the method is given by t x max f (u n j ) 1, (j = 0, ±1, ±2,... ). Implementation of the scheme to the Greenberg model x = x L, t = t L/a = at L, q(x, t ) = ρ(x, t ) ρ max, ρ > 0. (12)

Godunov Scheme for the Greenberg model Where L the characteristic distance along the highway. Thus q satisfies the conservation law q f (q) + t x = 0 (13) with the flux function f (q) = q log q, q > 0. The corresponding Riemann problem has the following piecewise constant initial condition { q l, if x < 0, q(x, 0) = q r, if x > 0.

Godunov Scheme for the Greenberg model The Riemann Problem has two kinds of solutions; b(q) = f (q) = log q 1, Case 1: b(q l ) > b(q r ) = q l < q r. { q l if x/t < s, q(x, t) = q r if x/t > s. where s = f (q l ) f (q r ) q l q r. Case 2: b(q l ) < b(q r ) = q l > q r. q l if x/t < b(q l ), q(x, t) = e (1+ x t ) ifb(q l ) < x/t < b(q r ), q r if x/t > b(q r ).

Godunov Scheme for the Greenberg model In order to implement the Godunov scheme lets us consider the Riemann problem given below q t + x ( q log q) = 0, x R, t > t n, (14) q q(x, t n j n if x < x j+ 1, ) = 2 qj+1 n if x > x j+ 1. For the similarity solution of this problem we also distinguish two cases. 2

Godunov Scheme for the Greenberg model If q n j < q n j+1, q R (η; q n j, q n j+1 ) = { q n j if η < s n j, q n j+1 if η > s n j. where sj n = q j+1 log q j+1 q j log q j q j q j+1. if qj n > qj+1 n, qj n if η < b(qj n ), q R (η; qj n, qj+1 n ) = η if b(qj n ) < η < b(qj+1 n ), qj+1 n if η > b(qj+1 n ).

Godunov Scheme for the Greenberg model For the numerical flux F (uj n, uj+1 n ) = q R(0; qj n, qj+1 n ) log q R(0; qj n, qj+1 n ); If qj n < qj+1 n, then If q n j F(q n j, q n j+1 ) = { q n j log q n j if s n j > 0, q n j+1 log qn j+1 if s n j < 0. > q n j+1, then qj n log qj n if qj n < e 1, F (qj n, qj+1 n ) = e 1 if qj+1 n < e 1 < qj n, qj+1 n log qn j+1 if qj+1 n < e 1.

Numerical experiments In this case the initial condition for q is given by { 0.1 if x < 0.2, q(x, 0) = 1 if x > 0.2.

Numerical experiments In this case the initial condition for q is given by { 0.5 if x < 0.8, q(x, 0) = 0.1 if x > 0.8.

Higher Order Effects q = q(ρ x, ρ). Assumptions: q = Q(ρ) νρ x, v = V (ρ) ν ρ ρ x. There are two additional effects one may wish to include the theory: Diffusion of waves, and Response Time. To incorporate these effects, ρ t + c(ρ)ρ x = νρ xx. Dv Dt = v t + vv x = 1 τ [ v V (ρ) + ν ] ρ ρ x. (15) where τ - measure of the response time. The equation (15) is to be solved together with the conservation equation.

Higher Order Effects If equation (15) and the conservation equation are linearized for small perturbations about ρ = ρ 0, v = v 0 = V (ρ 0 ), by substituting ρ = ρ 0 + r, v = v 0 + w, and retaining only the first powers of r and w, we have ] τ (w t + v 0 w x ) = [w V (ρ 0 )r + νρ0 r x, r t + v 0 r x + ρ 0 w x = 0.

Higher Order Effects The kinematic wave speed is c 0 = ρ 0 V (ρ 0 ) + V (ρ 0 ); hence V (ρ 0 ) = (v 0 c 0 )/ρ 0. Introducing this expression and then eliminating w, we have ( ) r t + c r 0 x = ν 2 r x 2 τ t + v 2 0 r. (16) x

Higher Order Effects The effect of the finite response time τ is more complicated but can be approximated as follows. t c 0 x. (17) If this approximation is used in the right hand side of (16), the equation reduces to r t + c r [ ] 0 x = ν (v 0 c 0 ) 2 2 r τ x 2. (18)

Shock structure We need a steady profile solution of (15) and the conservation equation with ρ = ρ(x), v = v(x), X = x Ut, where U is the constant translational velocity. Then, and may be integrated to Uρ x + (vρ) x = 0 (19) ρ(u v) = A, where A is a constant. (20)

Shock structure Equation (15) becomes Since v = U A/ρ, τρ(v U)v x + νρ x + ρv Q(ρ) = 0. (21) ) (ν A2 ρ 2 τ ρ x = Q(ρ) ρu + A. (22) For τ 0, the possibility that ν A 2 τ/ρ 2 may vanish introduces the new effects. We are interested in the solution curves between ρ 1 at X = + and ρ 2 at X =. For traffic flow c (ρ) = Q (ρ) < 0, so ρ 2 < ρ 1 and the right hand side of (22) is positive for ρ 2 < ρ < ρ 1.

Shock structure If ν A 2 τ/ρ 2 remains positive in this range, then ρ x > 0 and we have a smooth profile as in the figure below. In view of (20), the condition for ν A 2 τ/ρ 2 to remain positive may be written ν > (v U) 2 τ, that is v ν/τ < U < v ν/τ. (23) Figure: Continuous wave structure

Conclusions The traffic model is based on first order approximation, and hence the original assumptions are not good approximation. Include higher order effects and shock structure in order to have a better solutions.

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