TRAFFIC FLOW MODELS USING SECOND-ORDER ORDINARY DIFFERENTIAL EQUATIONS

Vane PetrosyanMariam Balabanyan

Overview

• What is traffic flow?• Mathematical model of traffic flow• Application of the model to investigate stopping

distances– Solving an initial value problem using a linear, non-

homogenous, constant-coefficient, second-order differential equation

What is traffic flow and why is it important?

Exploration of interactions between vehicles, drivers, and infrastructure in a general environment

Goal of researchers: Generate a useful model Promote a road network with efficient

movement of traffic and nominal traffic congestion interference

Three variables: speed, flow and concentration

A Mathematical Approach

Microscopic models consider the behavior of individual vehicles and their motion in relation to each other

Car-following model that describes how driver behavior is influenced

011

21

2

xxdt

dx

dt

xd

Achieving a simplified version

= sensitivity coefficient T = reaction time d = preferred separation between the two

vehicles = position of the lead car = position of the following car

0x

1x

))()(()(

1021

2

dtxtxdt

Ttxd

Inclusion of the reaction time T makes it difficult to solve the equation (although it should be included in more realistic models)

We will consider the quick thinking driver version, where T=0

))()(()(

1021

2

dtxtxdt

txd

Our preferred separation tends to depend on our speed

Short distance in slow moving traffic

Greater distance in faster moving traffic

Assume that the following driver’s preferred separation is dependent on her velocity

dt

tdxd

)(1

Substituting d into the original model, yields the simplified version

))()(()(

1021

2

dtxtxdt

txd

))(

)()(()( 1

1021

2

dt

tdxtxtx

dt

txd

dt

tdxtxtx

dt

txd )()()(

)( 1102

12

)()(

)()(

01

121

2

txdt

tdxtx

dt

txd

011

21

2

xxdt

dx

dt

xd = sensitivity coefficient

(’s value is directly proportional to the reaction of the following driver to the relative velocity between vehicles)

= preferred temporal separation (seconds)

= position of the lead vehicle

0x

So the model we will be using is…

Application to investigate stopping distances

“Consider two cars driving along the road at constant velocity U m/s and separated by a distance D meters. At time t = 0 we assume the following car is at the origin t 0, D = . How does the following car respond?” (McCartney 591)

0x

Becomes the following initial value problem

U

dt

dx

01 0)0( x

Dxdt

dx

dt

xd 1

121

2

01

121

2

xdt

dx

dt

xd

stetx )(1

11

21

2

xdt

dx

dt

xd )()()(

2

2st

stst

edt

ed

dt

ed ststst esees 2

stess )( 2 steSince is never zero, we must have

)(0 2 ss

First we must find the general solution of the corresponding homogenous equation.

From the characteristic polynomial, we can deduce the roots of the equation

So we are given 3 cases:

2

42

1

r

2

42

2

r

=4, two repeated roots

2>4, two real and distinct roots

2<4, two complex roots

Case 1: 2=4

DAxp

DAtececx rtrt 21

rtrth tececx 21

rtrtrt rtececercx 221 rtrtrt tercercecrx 2

2212

Dxdt

dx

dt

xd 1

121

2

DDAtececrtececectercercecr rtrtrtrtrtrtrtrt )()()( 212212

2212

DDA 2

D

D

D

DA

2

1A

1

A

1

21 Dtececx rtrt

Dtececx rtrt 21

Plug in the initial condition, to solve for the value of c1:

0)0( x

00)0( 02

01 Dececx rr

0)0( 1 Dcx

Dc 1

Now use the initial condition given for x’(0)=U to solve for c2

U

dt

dx

01

rtrtrt rtececercx 221

UercecDrex rrr 02

02

0 00

UcDrx 20DrUc 2

DteDrUDex rtrt )(

So substituting c1 and c2 into the equation yields:

this equation can be rewritten as

UteDe

DeDx tt

t

/2

/2/2 2

which can then be simplified to

tUD

DeDtxt

)2(

)(/2

1

Case 2:

DtDtDUe

txt

)4(2

1cosh)4()4(

2

1sinh2

)4()( 222

2

)2/1(

1

42

• Has two real and distinct roots

• Solution includes hyperbolic sine and cosine function

Case 3: 42

DtDtDUe

txt

)4(2

1cos)4()4(

2

1sin2

)4()( 222

2

)2/1(

1

•2<4, two complex roots

What this means physically? By making case 3 equal to D we ultimately see

the following:

0= asin(x) - bcos(x)

0= atan(x) – b

tan(x)= b/a

So there are infinite times where tan(x) = b/a

DtDtDUe

txt

)4(2

1cos)4()4(

2

1sin2

)4()( 222

2

)2/1(

1

Conclusion

The mathematical model will only be physically meaningful if we restrict parameter values

The driver may be quick thinking but the vehicle will infinitely collide

the stopping distance is proportional to the initial velocity of the driver