Autar Kaw
Humberto Isaza
http://nm.MathForCollege.com
Transforming Numerical Methods Education for STEM Undergraduates
http://nm.MathForCollege.com
An Example to Show How to Reduce Coupled Differential Equations to a
Set of First Order Differential Equations
3
Figure 1: A school bus.
Figure 2: A model of 1/4th of suspension system of bus.
Problem Statement:
A suspension system of a bus can be modeled as above. Only
1/4th of the bus is modeled. The differential equations that
govern the above system can be derived (this is something
you will do in your vibrations course) as
d 2 x1
 dx dx 
M 1 2  B1  1  2   K1  x1  x2   0
dt
dt 
 dt
(1)
 dx 2 dx 1 
d 2 x2
 dx2 dw 




M2

B


K
x

x

B

  K 2  x 2  w  0
1
1
2
1
2
dt 
dt 
dt 2
 dt
 dt
(2)
x1 (0)  0, v1 (0)  0, x2 (0)  0, v2 (0)  0
Where
M1  body
M 2  suspension mass
K1  spring constant of suspension system
K 2  spring constant of wheel and tire
B1  damping constant of suspension system
B2  damping constant of wheel and tire
x1  displacement of the body mass as a function of time
x2  displacement of the suspension mass as a function of time
w  input profile of the road as a function of time
The constants are given as
m1  2500 kg
m2  320 kg
K1  80,000 N/m
K 2  500,000 N/m
B1  350 N - s/m
B2  15,020 N - s/m
Reduce the simultaneous differential equations (1) and (2) to simultaneous first order
differential equations and put them in the state variable form complete with
corresponding initial conditions.
Solution
Substituting the values of the constants in the two differential equations (1) and (2)
gives the differential equations (3) and (4), respectively.
d 2 x1
 dx dx 
2500 2  350 1  2   80000 x1  x2   0
dt
dt 
 dt
320
d 2 x2
 dx2 dx1 
 dx2 dw 



350


80000
x

x

15020




  500000( x2  w)  0
2
1
dt 2
dt
dt
dt
dt




(3)
(4)
,
,
Since w is an input, we take it to the right hand side to show it as a forcing function and
rewrite Equation (4) as
d 2 x2
dx 
dw
 dx
 dx 
320 2  350 2  1   80000 x2  x1   15020 2   500000 x2  15020
 500000 w
dt
dt 
dt
 dt
 dt 
(5)
Now let us start the process of reducing the 2 simultaneous differential equations
{Equations (3) and (5)} to 4 simultaneous first order differential equations.
Choose
dx1
 v1
dt
(6)
dx2
 v2
dt
(7)
then Equation (3)
d 2 x1
 dx dx 
2500 2  350 1  2   80000x1  x 2   0
dt 
dt
 dt
can be written as
2500
dv1
 350v1  v2   80000x1  x2   0
dt
2500
dv1
 350(v1  v2 )  80000( x1  x2 )
dt
dv1
 0.14v1  v2   32x1  x2 
dt
(8)
and Equation (5)
d 2 x2
dx 
dw
 dx
 dx 
320 2  350 2  1   80000 x2  x1   15020 2   500000 x2  15020
 500000 w
dt
dt 
dt
 dt
 dt 
can be written as
320
dv2
dw
 350v2  v1   80000x2  x1   15020v2  500000 x2  15020
 500000w
dt
dt
320
dv2
dw
 350(v2  v1 )  80000( x2  x1 )  15020v2  500000 x2  15020
 500000 w
dt
dt
dv2
dw
 1.09375v2  v1   250 x2  x1   46.9375v2  1562.5 x2  46.9375
 1562.5w
dt
dt
(9)
The 4 simultaneous first order differential equations given by Equations 6 thru 9
complete with the corresponding initial condition then are
dx1
 v1  f1 t , x1 , x2 , v1 , v2 , x1 0   0
dt
dx2
 v2  f 2 t , x1 , x2 , v1 , v2 , x2 0  0
dt
dv1
 0.14v1  v2   32 x1  x2   f 3 t , x1 , x2 , v1 , v2 , v1 0  0
dt
dv2
dw
 1.09375v2  v1   250 x2  x1   46.9375v2  1562.5 x2  46.9375
 1562.5w
dt
dt
 f 4 t , x1 , x2 , v1 , v2 , v2 0   0
(10)
(11)
(12)
(13)
Assuming that the bus is going at 60 mph, that is, approximately 27 m/s, it takes
6m
 0.22 s
27 m / s
to go through one period. So the frequency
1
0.22
 4.545 Hz
f 
The angular frequency then is
  2    4.545
 28.6 rad / s
Giving
w  0.01sin t 
 0.01sin 28.6t 
and
dw
 0.286 cos28.6t 
dt
To put the differential equations given by Equations (10)-(13) in matrix form, we
rewrite them as
dx1
 v1  1v1  0v2  0 x1  0 x2 , x1 0   0
dt
dx2
 v2  0v1  1v2  0 x1  0 x2 , x2 0   0
dt
(10)
(11)
dv1
 0.14v1  0.14v2  32 x1  32 x2 , v1 0   0
dt
(12)
dv2
dw
 250 x1  1812.5 x2  1.09375v1  48.03125v2  1562.5w  46.9375
dt
dt
 f 4 t , x1 , x2 , v1 , v2 , v2 0   0
(13)
In state variable matrix form, the differential equations are given by
 dx1 
 dt 
0

0
1
0
  x1  
 dx   0


2
 x 

  0
0
0
0
1


2
  
 dt   


0
 dv1   32
32
 0.14
0.14   v1 

dw
   1562.5w  46.9375 
 dt  
250

1812
.
5
1
.
09375

48
.
03125
  v2  
 dv  
dt 
2


 dt 
where w  0.01sin( 28.6t )
and the corresponding initial conditions are
 x1 (0)  0
 x (0) 0
 2  
 v1 (0)  0

  
 v2 (0)  0
See how MATLAB ode45 is used to solve such problems
http://www.math.purdue.edu/academic/files/courses/past//2004fall/MA266/ode45.pdf