# Mariam Balbanyan and Vane Petrosyan

```TRAFFIC FLOW MODELS USING
SECOND-ORDER ORDINARY
DIFFERENTIAL EQUATIONS
Vane Petrosyan
Mariam 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, nonhomogenous, 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
2
d x1
dx1


x

x
1
0


2 
dt
dt
Achieving a simplified version
d x1 (t  T )
  ( x0 (t )  x1 (t )  d )
2
dt
2





 = sensitivity coefficient
T = reaction time
d = preferred separation between the two vehicles
= position of the lead car
x0 = position of the following car
x1


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
2
d x1 (t )
  ( x 0 (t )  x1 (t )  d )
2
dt
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

dx1 (t )
d 
dt
Substituting d into the original model, yields
the simplified version
d 2 x1 (t )
  ( x 0 (t )  x1 (t )  d )
2
dt
d 2 x1 (t )
dx1 (t )
  ( x0 (t )  x1 (t )  
)
2
dt
dt
d 2 x1 (t )
dx1 (t )
 x0 (t )  x1 (t )  
2
dt
dt
d 2 x1 (t )
dx1 (t )
 x1 (t )  
 x0 (t )
2
dt
dt
So the model we will be using is…
2
d x1
dx1

x1  x0

2 
dt
dt

 = sensitivity coefficient



(’s value is directly proportional to the reaction of the following driver
to the relative velocity between vehicles)
 = preferred temporal separation (seconds)
x0 = position of the lead vehicle
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
x0
 the following
at the origin t 0, D = . How does
car
respond?” (McCartney 591)
Becomes the following initial value problem
2
d x1
dx1

x1  D

2 
dt
dt
dx1 0 
U

dt
x(0)  0
First we must find the general solution of the
corresponding homogenous equation.
d 2 x1
dx1



2 
dt
dt
x0
1
x1 (t )  e st
d 2 x1
dx1
d 2 (e st )
d (e st )

 (e st )

x1 


2

2 
dt
dt
dt
dt
 s e  se  e
2 st
st
st
 (s 2  s   )e st
st
Since e is never zero, we must have
0  (s  s   )
2
From the characteristic polynomial, we can deduce the
roots of the equation

     2  4
r1 
2

So we are given 3 cases:
2 =4, two repeated roots
2&gt;4, two real and distinct roots
2&lt;4, two complex roots

     2  4
r2 
2

Case 1: 2=4
xh  c1e  c2te
rt
rt
x p  DA
x  c1ert  c2tert  DA
x  rc1ert  c2ert  c2rtert
x  r 2c1ert  rc2ert  c2r 2tert
d 2 x1
dx1

x1  D

2 
dt
dt
(r 2c1ert  rc2ert  c2r 2tert )   (c1ert  c2ert  c2rtert )   (c1ert  c2tert  DA)  D
 2 DA  D
 2 DA D

D
D
A  1
A
1

x  c1e  c2te  D
rt
rt
1

x  c1ert  c2tert  D
Plug in the initial condition, to solve for the value of c1:
x(0)  0
x(0)  c1er 0  c2 0er 0  D  0
x(0)  c1  D  0
c1   D
Now use the initial condition given for x’(0)=U to solve for c2
dx1 0 
U
dt 
x  rc1ert  c2ert  c2rtert
x0  Drer 0  c2er 0  c2r 0er 0  U
x0  Dr  c2  U
c2  U  Dr
So substituting c1 and c2 into the equation yields:
x  Dert  (U  Dr)tert  D
this equation can be rewritten as
 2 / t
x  D  De

2 De2t /

 e 2t / Ut
which can then be simplified to
x1 (t )  D  e
2 t / 
(2 D  U ) 

t
D



Case 2:   4
2
x1 (t ) 
e  (1/ 2)  t

1

1

 2U  D sinh 
 (  2  4)t    (  2  4) D cosh 
 (  2  4)t    D
2

2

 (  2  4) 
• Has two real and distinct roots
• Solution includes hyperbolic sine and cosine function
Case 3:   4
2
x1 (t ) 
e  (1/ 2 )  t

1

1

 2U  D sin 
 (4   2 )t    (4   2 ) D cos 
 (4   2 )t    D
2

2

 (4   2 ) 
•2&lt;4, two complex roots
What this means physically?

By making case 3 equal to D we ultimately see the
following:
x1 (t ) 

e  (1/ 2 )  t

1

1

 2U  D sin 
 (4   2 )t    (4   2 ) D cos 
 (4   2 )t    D
2

2

 (4   2 ) 
0= asin(x) - bcos(x)
0= atan(x) – b
tan(x)= b/a
So there are infinite times where tan(x) = b/a
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
```