Download LIAPUNOV FUNCTIONS METHOD IN

Liapunov Functions Method In Stability Study Of
Systems
Ahmed Sabah Ahmed Al-Jilawi
University Of Babylon / College of Education For Pure Science
Abstract
The main objective of this work , is to present some aspects for construction Liapunov function for
testing stability of linear and with constant coefficients .
New approaches and modification of finding Liapunov function based on new theories with illustrative
problem are presented .
Keywords. Large scale systems; overlapping decomposition; Liapunov’s matrix function; stability;
asymptotic stability
‫الخالصة‬
‫الهدف الرئيسي لهذا البحث هو عرض بعض الطرق إلنشاء دوال ليبانوف الختبار استق اررية األنظمة الخطية ذات المعامالت الثابتة‬
. ‫كما اختص هذا البحث بتطوير بعض الطرق إلنشاء دوال ليبانوف والتي اعتمدت على أسس نظرية جديدة مع بعض المسائل التوضيحية‬.
1- Introduction
The study of equilibria plays a center role in ordinary differential equations and their
applications .An equilibrium point , however ,must satisfy a certain stability criterion in
order to be every significant physically . [S. Leela, Large-scale systems ]
Liapunov stability , is concern with every stringent restrictions on the motion .It is
required that motions (solutions to be once near together keep all future time as function
of the time .specifically .We say that a solution x (t ) of the problem under investigation
may be
dx
 f ( x , t ) is stable in the sense of liapunov , If for   0 there exists a   0
dt
Such that any solution
x  y  
 is satisfying
y
x  y  
for t=0 and satisfies
for all t  0
A second method of liapunov treats the stability problem directly in a normal way ,but
some what depend on a non constructive procedure . applications of the second method is
not limited by an artificial circumstance compared with the first method . furthermore the
second method relating to the construction or discovery of suitable testing function . In
this paper we discuss some approaches for construction Liapunov function for the both
linear and non-linear systems of differential equations For linear systems of differential
equations , we present three methods that constant a Liapunov function for a linear
system with constant coefficients depend on Hurwitz and Sylvester theories .For non
linear system of differential equations , we present three approaches based on first ,
second liapunov methods ,and the indeterminate coefficients . [M. Ikeda, D.D. ] A new
theories with their proofs are build for our approaches an illustrative problems are given
for each approach
Liapunov functions
Liapunov functions V (x) can be used to estimated basins of attraction of fixed points,
show that a system has no periodic orbits or show a system has a globally attracting fixed
point.
Definition If U be an open subset of IR n containing a fixed point x of
, x  IR n
x  f ( x)
Where f  C 1 ( IR) A function V : U  IR is a liapunov function U if
i) V  C 1 (U )
)  0
ii) V ( x
, x U , x  x
iii) V ( x)  0
dv
0
, x  U , x  x .
iv) V 
dt
note that (i-iii) can simultaneously be turn only if the system (1) has a unique fixed point
x U .
When V satisfies property iii),the function V is said to be (strictly)positive definite on U
. if this condition is weakened to
V ( x)  0 , x  U , V is said to be semi positive definite on U . Also if
condition iv ) is relaxed to
dv
V 
0
dt
,
x  U
Then V is said to be Liapunov function
2- Methods for Constructing Liapunov Functions
Stability and Liapunov functions in this section we discuss the stability of the
equilibrium points of the linear system
x  f ( x )
…(1)
The stability of any hyperbolic equilibrium point x of ( 1 ) is determined by the signs of
the real parts of the eigenvalues  j of the matrix Df ( x ). A hyperbolic equilibrium
point x is asymptotically stable if Re(  j ) < 0 for j = 1, ..., n; i.e., if
x
is a sink. And a
hyperbolic equilibrium point x is unstable if it is either a source or a saddle. The
stability of no hyperbolic equilibrium points is typically more difficult to determine. A
method, due to Liapunov, that is very useful for deciding the stability of non
hyperbolic equilibrium points is presented in this section. [ Yoshizawa ]
Example. consider the system
 x1    1 0  x1 
   
 
 x2   0  2  x2 
The only critical point is (0, 0) and it
 c1e t 
is V (c)   2t  a stable node. The flow is given by
c2 e 
2
For all  > 0, let    > 0. Then for all c  N (0)  x  R : x   and t > 0 we have

V (c)  c12 e 2t  c22 e 4t    

…(2)
and hence (0, 0) is stable. V (c)  N  (0) Moreover
V (c )  0
Lim
t 
…(3)
So we that (0, 0) is asymptotically stable.
Remark .
2
1. It can be seen from the phase portraits in that a stable node or focus of a linear system in R is
an asymptotically stable equilibrium point; an unstable node or focus or a saddle of a linear
system in R 2 is an unstable equilibrium point; and a center of a linear system in R 2 is a stable
equilibrium point which is not asymptotically stable.
2. It follows from the Stable Manifold Theorem and the Hartman-Grobman Theorem that any
sink of (1) is asymptotically stable and any source or saddle of (2) is unstable. Hence, any
hyperbolic equilibrium point of (3) is either asymptotically stable or unstable.
Theorem . [A.M. Liapunov ] If x is a stable equilibrium point of (1), then no eigenvalue of Df
( x ) has positive real part. We see that stable equilibrium points which are not asymptotically
stable can only occur at non hyperbolic equilibrium points. But the question as to whether a
nonhyperbolic equilibrium point is stable, asymptotically stable or unstable is a delicate question.
The following method, due to Liapunov (in his 1892 doctoral thesis), is very useful in answering
this question.
Theorem . [A.M. Liapunov ] ( Liapunov’s direct method )
containing x , let f  C 1 (U ) and let f ( x )  0 . Suppose further
that there exists a real valued function V  C 1 (U ) satisfying V ( x )  0 and V ( x)  0 if x  x
Then
(a) if V ( x)  0 for all x  U then x is stable;
Let E be an open subset of R
n
(b) if V ( x)  0 for all x U  X   then x is asymptotically stable;
(c)
if V ( x)  0 for all x  U  x  then x is unstable.
Remark .
1. If V ( x)  0 for all, x U then the trajectories of (1) lie on the surfaces in R n (or curves in
R 2 ) defined by V (x ) = c.
n
2. A function V : R → R satisfying the hypotheses of the above theorem is called a Liapunov
function.
3. Liapunov direct method is indeed direct in the sense that it does not require any previous
knowledge about the solutions of the system or the type of its critical point x . Instead, if one can
construct a Liapunov function for the system then one can directly determine the stability of the
critical point x . However, there is no general method for constructing a Liapunov function.
Example. Consider the system
x1   x1  x 22
x 2  x12  x 2
…(4)
The origin is an equilibrium point for this system
 1 0 
Df (0)  

 0  1
…(5)
Thus Df(0) has eigen values 1  2  1
Thus x = 0 is a sink and hence an asymptotically stable hyperbolic equilibrium point.
Let us show that V ( x)  x12  x22 is a Liapunov function for this system.
We have V  C 1 ( R 2 ) ,V (0)  0 ,V ( x)  0 for x  0 thusV ( x)  x12  x22 is a Liapunov function
for this system.
Next let us show that V ( x)  0 . Computing V ( x)  DV ( x) f ( x) we find
V ( x)  2 x12  2 x22  2 x1 x2 ( x1  x1 ) .


V ( x)  0 , we can take U  ( x1 , x2 )  R 2 : x1  1 , x2  1 :
Assume ( x1 , x2 ) U  0 then
x1x22  x22 and x12 x2  2x12
Thus x1 x2 ( x1  x2 )  x1  x2
2
V ( x)  0 in U  0
2
…(6)
Example. consider the system
x1   x23
x 2  x13
The origin is the only eigenvalues
.
…(7)
1  2  0 . thus x  0 is a nonhyperbolic equilibrium point
4
4
The function V ( x)  x1  x 2 is a liapunov function for this system V ( x)  0
Hence the solution curves lie on the closed curves
x14  x24  c
…(8)
Which encircle the origin . the origin is thus a stable equilibrium point of this system which is not
asymptotically stable .
Example. Consider the system
x1  2 x2  x2 x3
x2  x1  x1x3
…(9)
x3  x1x2
The origin is an equilibrium point for this system and
Df(0) has eigenvalues
equilibrium point .
1  0 , 2,3  2 i ; i, e , x  0 is a nonhyperbolic
so
V ( x )  c x  c 2 x  c3 x
2
1 1
 0  2 0


Df (0)   1 0 0  thus
 0 0 0


2
2
we
2
3
use
liapunov
's
method
.the
function
with positive constants c1 , c 2 and c3
usually worth a try . computing V ( x)  DV ( x) f ( x) , we find
V ( x)  2(c1  c 2  c3 ) x1 x 2 x3  (2c1  c 2 ) x1 x 2  .
… ( 10 )
c2  2c1 and c3  c1  0
Hens
if
we
is
have
V ( x)  0 for x  0 for all x  R 3 and therefore by Theorem x=0 is stable
Furthermore, choosing c1  c2  1 and c2  2 we see that the trajectories of
this system lie on the ellipsoids
x12  2 x 22  x32  c 2
… ( 11 )
Example . Consider the system
x1  2 x 2  x 2 x3  x13
x 2  x1  x1 x3  x 23
x 3  x1 x 2  x
… ( 12 )
3
3
The origin is an equilibrium point for this system and
0  2 0
Df  1 0 0
0 0 0
… ( 13 )
Thus Df(0) has eigenvalues 1  0 2,3  2i ; i, e , x  0
is a nonhyperbolic equilibrium point. So we use Liapunov’s method. The function
V ( x)  x12  2 x 22  x32 a Liapunov function. Computing V ( x)  DV ( x) f ( x)
we find
V ( x)  2( x14  2 x 24  x34 )  0
… ( 14 )
For x  0 .therefore x=0 is asymptotically stable .
3- Linear system
This approach depends on the analysis of Liapunov’s method as discussed in this
paper It can be presented as follows :
Step 1- apply Liapunov’s criterion to find the principle diagonal minors
i (i  1,2,....., n) from the characteristic polynomial
p( )  a n  a1n 1  a 2 n  2  ......a n 1  a n ,
… ( 15 )
Where the coefficients a ,s are real quantities
 
Step 2- we use following formula  i  det ckj here ckj  c jk where
(i ,j, k=1,2,….,n)
Step 3- from (1), one can solve (n-1) equations iteratively to get the values of (k ,
j=1,2,…..,n)
Step 4- construct the liapunov function in quadratic from with constant coefficients as
follows
1 n n
V ( x)   ckj xk x j
2 k 1 j 1
.
… ( 16 )
In order to check our new approach Varity of problems to demonstrate the stability of
the systems
Example 6: Consider the liner system
x1  2 x1  2 x2
x 2  x1  3x2
… ( 17 )
Which could be rewritten in matrix form
 x1    2 2  x1 
   
 

x
1

3
 x2 
 1 
where
 2 2 

A  
 1  3
The characteristic equation is given by p( )  2  5  4
… ( 18 )
Where
a  1, a1  5 and a2  4
Hence by applying Hurwitz's criterion ,we can find the principle diagonal minors
i (i  1,2) as follows
1  a1  5  0
2 
a1
a3
a
a2

5 0
1 4
 20  0,
Therefore using the formula  i  ckj (i, k , j  1,2)
  c11  5,
c11  5
Since ckj  c jk we assume c12  c12  R, and
2 
C11 C12
C21 C22

5
R
R C22
 5C22  R 2  20
… ( 19 )
20  R 2
5C22  R  20, C22 
,
5
2
Since the Liapunov function is given by
… ( 20 )
v( x) 
1 n n
 ckj xk x j.
2 k 1 j 1
Therefore
v( x) 


1
2
2
c11 x1  2c21 x2 x1  c22 x2 ,
2
 20  R 2  2 
1 2

 X 2 
And so V ( x)   5 x1  2 Rx 2 x1  

2
5



… ( 21 )
… ( 22 )
And it is clear that V is positive definite for all R ,hence the system is asymptotically
stable
Example . Consider the characteristic equation
3  32  7  18  0 ,
… ( 23 )
We have
a  1 a1  3
a2  7
a3  18
by apply Hurwitz's criterion to find the principal diagonal minors
i (i  1,2,3) ,
We have
1  a1  3  0 ,
2 
a1
a3
a  a2

 3 18
 12  18  3  0
7
1
 3 18
 3 18
a1
a3
a5
 3  a
a2
a4  1
7
0
1
 7  54  0
0
a1
a3
 3 18
0
3
We use the formula
0
 i  ckj
0
… ( 24 )
(i, j , k  1,2,3)
1  c11  3
Since ckj  c jk we can assume c12  c21  R ,
2 
Then c22 
Also we assume
c11
c12
c21 c22

… ( 25 )
3
R
R
c22
 2c22  R 2  3
3  R2
.
3
… ( 26 )
c13  c31  P and c23  c32  Q
… ( 27 )
c11
c12
c13
 3  c21 c22
3
c23  R
c31 c32
c33
P
R
3  R2
3
Q
P
Q
c33
3  R2 2
 3  c33 (3  R )  2 RPQ  c33 R  3Q  (
) P  54
3
2
2
2
Then
 3  R2  2
 P
54  2 RPQ  3Q 2  
3


c33 
3
… ( 28 )
As in previous example ,the Liapunov function is
1 n n
V x    ckj xk x j ,
2 k 1 j 1
… ( 29 )
Hence
1
2
2
2
(c11x1  2c21x2 x1  c22 x2  2c31x3 x1  2c32 x3 x2  c33 x3 )
2



 3  R2  2 


 54  2 RPQ  3Q 2  


P
3   2 
 3  R2  2
1

2

 x2  2 px3 x1  2Qx3 x2  
V x     3x1  2 Rx 2 x1  
X3 
2
3
 3 











V ( x) 
Therefore V(x) is negative definite for all R,P and Q ,and hence the system
corresponding the characteristic the characteristic is unstable ,
Example. Consider the following system
x1  ax1  bx 2
x 2  cx1  ex 2
… ( 30 )
Where a ,b, c ,and e are any real numbers .then the system is asymptotically stable if
a  e and ae  bc
Solution
First we hope to determine the conditions that the parameters of the system (30)
,which are a,b,c, and e need to be satisfied in order to get the unperturbed solution
to be asymptotically stable in accordance with equation (30) ,we
x1  x2  0
consider a Liapunov function of the form
V ( x) 
1
2
2
(x1  2x1 x2  x2 ) ,
2
… ( 31 )
Where  and  are to be determined Sylvester's criterion for the coefficient matrix




has the following form
1 
1    0 ,  2     2  0
… ( 32 )
Provided that these inequalities are satisfied , now we use Hurwitz's criterion to
determine the parameters of the system a ,b, c, and e, from the system (32) we have
the constant matrix
a b

A  
c
e


… ( 33 )
The characteristic polynomial is given by
p( )  2  (a  e)  ae  cb
… ( 34 )
Now ,using the condition of Hurwitz's criterion , we get
1  (a  e) since 1    0 from Sylvester's criterion then –(a+e)>0 ,(a+e)<0 ,
a<-e
2 
 ( a  e)
0
1
(ae  bc)
 (a  e)( ae  bc),
… ( 35 )
Since  2     2  0 (from Sylvester's criterion )
Then –(a+e)(ae-bc)>0 since –(a+e)>0
If parameters a,b, c, and e of system (35) satisfy the condition a<-e and ae>bc then
the function v given in equation (34)is positive definite , and so that the system is
asymptotically stable .
Example. Recalling the system of ordinary differential equation given in Example
x  2 x1  2 x2
x 2  x1  3x2
… ( 36 )
We have a=-2 ,b=2 ,c=1 ,and e=-3
Since a<-e -2<-(-3) then -2<3 (the first condition )
since ae> bc (-2)(-3)>(2)(1) then 6>2 (the second condition ) then the system is
asymptotically stable .
Example . Consider the liner system
x1  2 x1  2 x 2
x 2  5 x1  3 x 2
… ( 37 )
We have a=2 , b=2 , c=5 , and e=-3
Since a<-e ,(2)<-(-3) then 2<3 (the first condition )
Since ae>bc ,(2)(-3)>(2)(5) then -6>10(false) ,
Then the system is unstable
This system is very useful when studying the stability of perturbed liner system
.consider the differential system
x  Ax
… ( 38 )
When x is an n-vector and A is a real nxn constant nonsingular matrix .let the
characteristic roots of A be distinct we shall denote to te real characteristic root of A
1 , 2 ,............., r
And the complex characteristic roots by
ui  ui 1
Where
u1 , u2 ,....., um such that
i=2,4,…,m
… ( 39 )
r  m  n .define the real numbers  k and  k such that
u k 1   k 1  i k 1
, u k   k 1  i k 1
,
k=2,4,…,m
… ( 40 )
Now in order find a nonsingular constant matrix T "eigenfunction" such that
T 1 AT have the form
 1

0
 

T 1 AT   
 



0
0
... 00
...
0
2
.
K
 1 1
 1 1
22
  2 2
To this objective , consider the transformation
Now ( 41 )reduce to the form :
0










… ( 41 )
x  ty .
… ( 42 )
y  Dy
… ( 43 )
Where D  T 1 AT .for the system (43) to be asymptotically stable ,we required that
all the diagonal elements of D to be negative
Select a liapunov function
V ( y)  ( y , By )
… ( 44 )
where (44)is the usual inner product and B a rael n x n constant matrix then
… ( 45 )
v( y)  ( y, By )  ( y , By )  (Dy , By )  ( y, BDy )  ( y , ( DT B  BD ) y) ,
T
Where D is the transpose of D. in order to ensure that v is negative definite ,we
require
n
2
v( y)  ( y, y)   y j ,
j 1
… ( 46 )
Where y j are the components of y .to see the compatibility of relation (46) and (45)
we assume that the condition
D T B  BD   I
… ( 47 )
Where I is the identity matrix ,holds .After some computation , the matrix equation
(47) yields
1
 21

 0
 .
B
 .

 0

 0

0
1
...
2 2
0
0

1
2 r
1
2 1
1
2 1
0 










… ( 48 )
Assume that the matrix A is stable all i and  i are negative , and hence all the
diagonal elements of B will be positive therefore from (48), it follows that V is
positive definite .in fact V(y) takes the form
V ( y)  
1 2
1 2
1 2
1
y1 
y 2  ....... 
yr 
( y r21  y r2 2 )
2 y1
22
2r
2 1
… ( 49 )
now ,we consider the following illustrative problem
4- Conclusions
Considering the work done in this paper , we could recommended ,that the theorems
built in this work can be extended for constructing general approaches applicable to
varieties structure problems
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