Traffic Flow Problems - Faculteit Wiskunde en … · Trafﬁc Flow Problems Nicodemus Banagaaya Supervisor : Dr. J.H.M. ten Thije Boonkkamp October 15, 2009

/centre for analysis, scientific computing and applications

Traffic Flow ProblemsNicodemus Banagaaya

Supervisor : Dr. J.H.M. ten Thije Boonkkamp

October 15, 2009


Outline

IntroductionMathematical model derivationGodunov Scheme for the Greenberg Traffic model.Numerical experimentsHigher Order Effects.Shock structureConclusion


Outline



Outline



Outline



Outline



Outline



Outline



Outline



IntroductionDefinitionTraffic flow

Figure: copyright British library Board


Mathematical model derivationConsider the traffic flow of the cars on a highway with only onelane. 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 passthrough x at time t .( flux of the vehicles given by vehiclesper unit time).

The number of cars which are the interval (x1, x2) at time t is∫ x2

x1

ρ (x , t) dx . (1)

Assumption: ρ and Q are continuous functions.





x1

ρ (x , t) dx . (1)






x1

ρ (x , t) dx . (1)






x1

ρ (x , t) dx . (1)






x1

ρ (x , t) dx . (1)






x1

ρ (x , t) dx . (1)






x1

ρ (x , t) dx . (1)



Mathematical model derivation

Figure: Derivation of the conservative law,where a = Q(x1, t),b = Q(x2, t)



ddt

∫ x2

x1

ρ (x , t) dx = Q (x1, t)−Q (x2, t) .

Then we have, ∫ x2

x1

[∂ρ

∂t+

∂

∂xQ(x , t)

]dx = 0 (2)

∂ρ

∂t+

∂

∂x(ρv) = 0. (3)



ddt

∫ x2

x1

ρ (x , t) dx = Q (x1, t)−Q (x2, t) .

Then we have, ∫ x2

x1

[∂ρ

∂t+

∂

∂xQ(x , t)

]dx = 0 (2)

∂ρ

∂t+

∂

∂x(ρv) = 0. (3)



Since x1, x2 ∈ R, t1, t2 > 0 are arbitrary, we conclude that

∂ρ

∂t+

∂

∂x(ρv ) = 0, (4)

with Initialρ (x ,0) = ρ0 (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).



Figure: Flow density curve in the traffic flow

Traffic modelsLighthill-Whitham-Richards model

v(ρ) = vmax

(1− ρ

ρmax

), 0 ≤ ρ ≤ ρmax . (5)



Greenberg modelv(ρ) = a log ρmax

ρ , 0 < ρ ≤ ρmax .

where a is in kilometres per hour.we are going to solve this model numerically using Godunovscheme .


Godunov Scheme for the Greenberg model

Godunov Scheme for Nonlinear Conservation lawsconsider the initial value problem

∂u∂t

+∂f (u)

∂x= 0, x ∈ R, t > 0, (6a)

u (x ,0) = u0 (x),∀x ∈ R. (6b)

Let us introduce a control volumes or cells Vj as follows:

Vj =[xj− 1

2, xj+ 1

2

), xj+ 1

2=(xj + xj+1

), j = 0,±1,±2, . . . ,

(7)



Associated with unj is the function u(x , t), defined as the

solution of the following initial value problem with piecewiseconstant initial at t = tn:

∂u∂t

+∂f (u)

∂x= 0, x ∈ R, t > tn, (8a)

u(x , tn) = un

j , x ∈ Vj (j = 0,±1,±2, . . . ). (8b)

To compute the numerical flux;

u(x , tn) =

unj , if x < xj+ 1

2,

unj+1, if x > xj+ 1

2.



The solution of this Riemann problem is a similarity of the form

u(x , t) = uR(η; unj ,u

nj+1), η =

x − xj+ 12

t − tn . (9)

Since η = 0 for x = xj+ 12, the computation of the numerical flux

we simply find

F (unj ,u

nj+1) = f (uR(0; un

j ,unj+1)). (10)

Thus the Godunov scheme is given by

un+1j = un

j −4t4x

(F (un

j ,unj+1)− F (un

j−1,unj )), (11)

with the numerical flux as defined in (10).



And the stability condition of the method is given by

4t4x

max |f ′(unj )| ≤ 1, (j = 0,±1,±2, . . . ).

Implementation of the scheme to the Greenberg model

x∗ =xL, t∗ =

tL/a

=atL, q(x∗, t∗) =

ρ(x∗, t∗)ρmax

, ρ > 0. (12)



Where L the characteristic distance along the highway. Thus qsatisfies the conservation law

∂q∂t∗

+∂f (q)

∂x∗= 0 (13)

with the flux function f (q) = −q log q, q > 0.The corresponding Riemann problem has the followingpiecewise constant initial condition

q(x ,0) =

{ql , if x < 0,qr , if x > 0.



The Riemann Problem has two kinds of solutions;b(q) = f ′(q) = − log q − 1,

Case 1: b(ql) > b(qr ) =⇒ ql < qr .

q(x , t) =

{ql if x/t < s,qr if x/t > s.

where s = f (ql )−f (qr )ql−qr

.

Case 2: b(ql) < b(qr ) =⇒ ql > qr .

q(x , t) =

ql if x/t < b(ql),

e−(1+ xt ) ifb(ql) < x/t < b(qr ),

qr if x/t > b(qr ).



In order to implement the Godunov scheme lets us consider theRiemann problem given below

∂q∂t

+∂

∂x(−q log q) = 0, x ∈ R, t > tn, (14)

q(x , tn) =

qnj if x < xj+ 1

2,

qnj+1 if x > xj+ 1

2.

For the similarity solution of this problem we also distinguishtwo cases.



If qnj < qn

j+1,

qR(η; qnj ,q

nj+1) =

{qn

j if η < snj ,

qnj+1 if η > sn

j .

where snj =

qj+1 log qj+1−qj log qjqj−qj+1

.if qn

j > qnj+1,

qR(η; qnj ,q

nj+1) =

qn

j if η < b(qnj ),

η if b(qnj ) < η < b(qn

j+1),

qnj+1 if η > b(qn

j+1).



For the numerical fluxF (un

j ,unj+1) = −qR(0; qn

j ,qnj+1) log qR(0; qn

j ,qnj+1);

If qnj < qn

j+1, then

F (qnj ,q

nj+1) =

{−qn

j log qnj if sn

j > 0,−qn

j+1 log qnj+1 if sn

j < 0.

If qnj > qn

j+1, then

F (qnj ,q

nj+1) =

−qn

j log qnj if qn

j < e−1,

e−1 if qnj+1 < e−1 < qn

j ,

−qnj+1 log qn

j+1 if qnj+1 < e−1.


Numerical experiments

In this case the initial condition for q is given by

q(x ,0) =

{0.1 if x < 0.2,1 if x > 0.2.


Numerical experiments

In this case the initial condition for q is given by

q(x ,0) =

{0.5 if x < 0.8,0.1 if x > 0.8.


Higher Order Effects

q = q(ρx , ρ).Assumptions:q = Q(ρ)− νρx , v = V (ρ)− ν

ρρx .There are two additional effects one may wish to include thetheory: Diffusion of waves, and Response Time. To incorporatethese effects,

ρt + c(ρ)ρx = νρxx .

DvDt

= vt + vvx = −1τ

[v − V (ρ) +

ν

ρρx

]. (15)

where τ - measure of the response time. The equation (15)is to be solved together with the conservation equation.



If equation (15) and the conservation equation are linearized forsmall perturbations about ρ = ρ0, v = v0 = V (ρ0), bysubstituting

ρ = ρ0 + r , v = v0 + w ,

and retaining only the first powers of r and w , we have

τ (wt + v0wx ) = −[w − V ′(ρ0)r +

ν

ρ0rx

],

rt + v0rx + ρ0wx = 0.



The kinematic wave speed is

c0 = ρ0V ′(ρ0) + V (ρ0);

hence V ′(ρ0) = −(v0 − c0)/ρ0. Introducing this expression andthen eliminating w , we have

∂r∂t

+ c0∂r∂x

= ν∂2r∂x2 − τ

(∂

∂t+ v0

∂

∂x

)2

r . (16)



The effect of the finite response time τ is more complicated butcan be approximated as follows.

∂

∂t≈ −c0

∂

∂x. (17)

If this approximation is used in the right hand side of (16) , theequation reduces to

∂r∂t

+ c0∂r∂x

=[ν − (v0 − c0)2τ

] ∂2r∂x2 . (18)


Shock structure

We need a steady profile solution of (15) and the conservationequation with

ρ = ρ(X ), v = v(X ), X = x − Ut ,

where U is the constant translational velocity.Then,

−Uρx + (vρ)x = 0 (19)

and may be integrated to

ρ(U − v) = A, where A is a constant. (20)


Shock structureEquation (15) becomes

τρ(v − U)vx + νρx + ρv −Q(ρ) = 0. (21)

Since v = U − A/ρ,(ν − A2

ρ2 τ

)ρx = Q(ρ)− ρU + A. (22)

For τ 6= 0, the possibility that ν − A2τ/ρ2 may vanish introducesthe new effects.We are interested in the solution curves between ρ1 at X = +∞and ρ2 at X = −∞. For traffic flow c′(ρ) = Q′′(ρ) < 0, soρ2 < ρ1 and the right hand side of (22) is positive forρ2 < ρ < ρ1.


Shock structureIf ν − A2τ/ρ2 remains positive in this range, then ρx > 0 and wehave a smooth profile as in the figure below. In view of (20), thecondition for ν − A2τ/ρ2 to remain positive may be written

ν > (v − U)2τ, that is v −√ν/τ < U < v −

√ν/τ . (23)

Figure: Continuous wave structure


Conclusions

The traffic model is based on first order approximation, andhence the original assumptions are not goodapproximation.Include higher order effects and shock structure in order tohave a better solutions.


Thank you

Documents

Traffic Flow Problems - Faculteit Wiskunde en … · Trafﬁc Flow Problems Nicodemus Banagaaya Supervisor : Dr. J.H.M. ten Thije Boonkkamp October 15, 2009