FOR THOSE WHO WANT TO USE ANALYTICAL SOLUTION OF ORDINARY
DIFFERENTIAL EQUATIONS GOVERNING SYSTEM “CART-PENDULUM”
u  M ( a  b)
(1)
where u is control action, and its parameters are to satisfy:
b  0, a  (   1) g
(2)
Behavior of angle is given by (in this equation control u is already taken into account)
 
3b  3(a  g (   1))

  0 (3)
(   4 )l
(   4 )l
with initial conditions
 (0)  0, (0)  0
(4)
Behavior of the cart is defined by
4
y   l  g
(5)
3
with initial conditions
y (0)  y (0)  0
(6)
This system of equations may be solved analytically, because here we have linear
differential equations with constant coefficients, but in general case, only numerical
discrete approach may be used.
So, we seek solution of (3) in the form
  Ce pt
(7)
Substituting (7) into (3), we get
3b
3( a  g (   1))
Ce pt ( p 2 
p
)  0 (8)
(   4 )l
(   4 )l
So, we have got algebraic quadratic equation from which, in general, 2 roots may be
obtained, depending on value of discriminant
9b 2
3(a  g (   1))
(9)
D
4
2 2
(   4)l
(   4) l
Case 1. D>0 (2 distinct real roots).
3b

 D
(   4 )l
(10)
p1, 2 
2
Substituting roots (10) into (7), we get
(11)
  C1e p1t  C2 e p2t
Expression (11) has 2 not defined yet constants, values of which can be determined using
initial conditions (4):
C1  C2   (0)
(12)
p C  p C  (0)
1
1
2
2
Solving (12), we get
C 2   (0)  C1
p1C1  p 2 ( (0)  C1 )  (0)
C ( p  p )  (0)  p  (0)
1
C1 
1
2
2
(0)  p 2 (0)
p1  p 2
(0)  p 2 (0)
C 2   ( 0) 

p1  p 2
 (0)( p  p )  ((0)  p  (0))
(13)
2

p1  p 2
 (0) p1  (0) (0)  p1 (0)

p1  p 2
p 2  p1
So, this case corresponds to the solution represented by 2 exponents which grow in the
case of positive roots and vanish in the case of negative roots.
Case 2. D=0 (2 equal real roots). In the case of equality of roots (13) is not applicable
(denominators are equal to 0). In this case we have
3b

(   4 )l
p1, 2 
 p (14)
2
In such a case solution of (3) we represent as
(15)
  C1e pt  C2 te pt
with value of p defined by (14).
It may be directly checked that 2nd term in (15) also satisfies (3). From (15) and initial
conditions (4) we get:
C1   (0)
C p  C  (0)
(16)
1
1
2
2
C 2  (0)  p (0)
Case 3. D<0 (2 distinct complex roots)

p1, 2 
3b
i D
(   4 )l
,
2
where i   1 .
By introducing
3b
, (17)
2(   4)l
we can write solution in this case:
  C1e ( i )t  C2 e ( i )t
(18)
By definition,
e i  cos  i sin 
(19)
 2   D / 4,  
Substituting (19) in (18), we get
  et (C1e it  C2 e it )  et (C1 (cost  i sin t )  C2 (cost  i sin t )) 
et (cost (C1  C2 )  i sin t (C1  C2 ))  C11et cost  iC 21 et sin t
(20)
Using initial conditions (4) and (20), we can define constants C11 ,C 21 :
 (0)  C11
(0)  C 1  iC 1   (0)  iC 1 (21)
1
C 21 
2
2
(0)   (0)
i
Substituting (21) in (20), we get
(0)   (0) t
   (0)et cost 
e sin t

(22)
So, we have got solutions for equation (3) with initial conditions (4) in the Case 1
(expressions (10), (11), (13)), Case 2 (expressions (14)-(16)), Case 3 (expressions (9),
(17), (22)).
Using  , we can get u from (1), substituting  and its derivative in the right hand side of
(1).
Using  , we can get y from (1), substituting  and its 2nd derivative in the right hand side
of (5) and integrating twice right hand side. Other way, more simple, is to use discrete
representation of 2nd derivative of y as
d 2 y y (t  2t )  2 y (t  t )  y (t )

(23)
dt 2
t 2
Using (23), we can find next y, if we have 2 previous. These first 2 we can find from
initial conditions (6):
y (0)  0,
(24)
y ( t )  y (0)  ty (0)  0
Due to not zero values of the right hand side of (5), next values of y may be not zeroes.