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8/14/2019 2 Lyapunov Direct Method
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7-4 Lyapunov Direct MethodThere are two Lyapunov methods for stability analysis.
Lyapunov direct method is the most effective method for
studying nonlinear and time-varying systems and is a basic
method for stability analysis and control law desgin.
The f i rst methodusually requires the analytical solution of
the differential equation. It is an indirect method.
In the second method, it is not necessary to solve the
differential equation. Instead, a so-called Lyapunov functionis constructed to check the motion stability. Therefore, it is
said to be direct method.
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Because v(x) is positive definite and is negative definite, x
converges to zero. Note that the analytical solution of the
differential equation is not required.
Example:Consider the stability of zero-solution of the
following system
5x x
First of all, consider a positive definite function
2( ) v x x
( ) 0 0v x x
22 10 0 0
v xx x x
v
It is clear that, and .
The derivative of v is computed as
( ) 0 0v x x
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Similar to the above example,
construct a function called v-function
1x
2xExample: Consider the small-
damping vibration system
1 2
2 1 2
0.5damping ratiox x
x x x
2 21 2 1 1 2 2( , ) 3 2 2 v x x x x x x
Study the stability at equilibrium
x1=0, x2=0.
It is easy to verify that the function is positive definite.
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Generally speaking, it is difficult or sometimes
impossible to solve a differential equation. Hence, it is
almost impossible to get the analytical solution of v.However, the derivative of v can be computed as
2 2
1 2 1 2 2 1 2 1 2 1 21 2
(6 2 ) (2 4 )( ) 2( )
= + = + + + - - = - +
v v
v x x x x x x x x x x xx x
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1x
2x
When x1and x2 are not zero at the
same time, the inequality dv/dt
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The following figure is a sketch map of v(x) = Ci > 0 where
v(x) is in a positive definite quadratic form. From the figure
we can see that it is a set of close curves.
1 2 3 4 5 6 7C C C C C C C
C1
C2
C3
C4
C5
C6
C7v(x) is usually in
quadratic form and v(x)=C>0) represents a
set of hyperspheres.
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1. Defini tion of sign functions
0(0, ) 0= " v t t t
2 2
1 2 02
1
( , ) ( ), 01= + >+v x t x x t t t
2 21 2 1 2( ) 2v x x x x x
First of all, consider a real-valued function v(x,t) defined
on , where . Assume that v(x,t) is acontinuous single-valued function with
. For example
0,< W x t t 0W>
Defini tion 7-12
(1)Suppose v is the function of x, when
and v(x)=0 has nonzero solutions , then v(x)is said to be a constantly positive (constantly negative)
function.
Example: is a constantly positive function.
) 0( 0) v x
< Wx
x 0
8/14/2019 2 Lyapunov Direct Method
8/39Note that in the example .
The constantly positive (constantly negative) function is alsosaid to be positive (negative) semidefinite function and they
are both said to be constant sign functions.
Example: is a positive definite function.
(2) v(x,t) is said to be constantly positive (constantly
negative) if holds for .
Example: is a constantly positive
function.
2 21 2( )v x x x
0 < Wt t x,2 2
1 2 02
1( , ) ( ), 0
1v x t x x t t
t
lim ( , ) 0 =t v x t
If for and x=0 if and only if v(x)=0, then v(x)
is said to be positive definite (negative definite).
) 0( 0)v x x 00 and the bound of uj
is composed of v=0 and x = .
0v
v(x)>0
0v
0v
0v
Theorem 7-20*
Theorem 7-21*
Theorem 7-22**
v(x)>0
v(x)>0 Asymptotically stable
It is not zero at all time.
asymptotically stable.
stable
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f
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Theorem 7-25The zero-solution of the time-invariant
system is asymptotically stable if and only if for
any positive definite matrix N=NTthe Lyapunovequation
Ax x
T A M MA N 7-44
3. v function in quadratic form of linear system
Why should we study this problem?
The matrix A is often not precisely known in the control law
design. Theorem 7-25 has great significance in control law
design.
has a unique and positive definite solution M=MT.
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1 11 22 11 22 12 21
11 22
det( ) 4( )( )
4( ) det( )
A
A
a a a a a a
a a
If detA10, the equations have unique solutions
2 221 22 12 22 21 11
2 21 12 22 21 11 11 22
det( ( )2
det( ) ( ) det(
+ + - +- =
- + + +
A)
M A A)
a a a a a a
a a a a a a
From Sylvester criterion, M is positive definite if
2 2 2 221 22 21 22
11
11 22
2[det( ) ] det( )0 1
det( ) 2( ) det( )
a a a am
a a
1
A A
A A
The determinant of A1is
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Therefore, the matrix A should satisfy the conditions (3)
and (4) when the system is asymptotically stable.
11 22 12 21det 0 3A a a a a
2 211 22 12 21
211 22
( ) ( )det( ) 0 2
4( ) det( )
a a a a
a a
M
A
11 22( ) 0 4a a
From (2) it follows that
From (1) and (3), we can obtain that
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Note that Theorem 7-25 does not mean
Example 7-10
1 1 1 2,
1 3 2 5
A M
It is clear that the eigenvalues of A have negative real
parts and M is positive definite. However, from Equation
(7-44)
2 2
2 26
-=
N
is not positive definite.
A is asymptotically stable and M is positive definite, then
N in Equation (7-44) must be positive definite.
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has a positive definite symmetric solution if and only if
is asymptotically stable.= Ax x
Theorem 7-26 If N of Equation (7-44) in Theorem 7-25 is
positive semidefinite symmetric matrix, and xTNx is not
identically zero along any nonzero solution of =Ax, then
the matrix equationx
ATX+XA=N 7-46
Remark: The condition the xTNx is not identically zero
along any nonzero solution of can not be omitted.
Example 1:A is asymptotically stable and N is positive
semidefinite, then M may not be positive definite.
x A x
11 0 1 0 0
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1. If xTNx is identically zero along any nonzero solutionof , it is easy to compute
10 0Tx x x N
2 20 1( 0)t
x e x becuase x
20 20 0x x x
12
1 0 1 0 0,
1 1 0 0 0 0
A N M
x A x
However,
If
is a nonzero solution,
then xTNx is identically zero along a nonzero solution of
the differential equation, which does not satisfy the
condition of the theorem.
d f d d f
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1
2 1 0 1 00 ,0 0 0 10 0
A N M
Example 2N is positive semidefinite and M is positive definite,
then A may not be asymptotically stable.
Analysis1. Let xTNx=0.
0.51 10 10 1, 0 0 ;
-= = tx e x x x
2. Let C=[1 0], then N=CTCand (A, C) is unobservable.
Because xTNx=x12, xTNx=x12is identically zero
2 20 20, 0,= x x x
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xTNx is not identically zero along any nonzero solution
of the differential equation and (A, C) is observable,
which satisfies the condition of the theorem.
1 12 4
1 14 4
1 1 1 0,
0 1 0 0
- = = = -
A N M
Example 3
( ) (0) (0)0
- -
-
= =
At t
t
t
e tex t e x x
e
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Conclusion:xTNx is not identically zero along any
nonzero solution of the differential equation can be
replaced by the condition that (A, C) is observable,where N=CTC.
Theorem 7-26*The zero-solution of time-invariant
differential equation is asymptotically stable if and
only if under the condition that (A,N) is observable, where
N is positive semidefinite, Lyapunov equation (7-44)
Ax x
A M MA NT 744
has a unique positive definite solution M.
P f
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Proof
1) Because N is positive semidefinite, it can be decomposed
as
N=CTC
0 0
A A A AM= N C C
T Tt t t T t e e dt e e dt
is a unique positive definite solution and satisfies Equation (7-
44). Here, N is an arbitrarily positive semidefinite matrix such
that (AN) is observable.
2) Necessity:The proof is similar to Theorem 7-25. Because
the zero-solution of the system is asymptotically stable, M
determined by
It is easy to prove that (A, C) is observable if (A, N) is
observable.
3) S ffi i If d h di i h N i i i
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3) Sufficiency: If under the condition that N is positive
semidefinite such that (A,N) is observable, the solution M
of Equation (7-44) is positive definite. Now, we need to
prove that the system is asymptotically stable. Considerv(x)=xTMx dv/dt = xTNx. Hence, we only need to prove
that xTNx0 x00.
00 0 0 (*)T T T t x x x x x e x AN C C C C
0 0 0| 0A
C Ct
te x x
0 0 0 00 | 0A A
CA CA CAt t
te x e x x
1 0 0CAn x
The derivatives of the two sides are computed as
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Which implies that is zero, i.e. dv/dt is not
identically zero along the nonzero solution of the
differential equation. From Theorem 7-21**, the system is
asymptotically stable. Q.E.D
0T
x xN
0 0
1
0 0
C
CA
CAn
x x
Remark: From (A,C) is observable if and only if ={0}.
4 Ab t L f ti
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4. About Lyapunov function
3. It should be emphasized that the conditions of Theorem 7-
20*7-22** are all sufficient conditions. This means that
a system may be stable even if a v function cannot be
constructed.
1. A scalar function is said to be Lyapunov function if by
employing the function, the stability of a system can be
determined without computing the analytical solution ofthe differential equation;
2. The construction of the v function is a complex problem.
Even if the v function exists in theory, it is still a hard
work to find an analytic expression. It is not practical tofind a general method to construct the v function.
However, for linear systems, there are some methods to
construct their v functions.
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A better Lyapunov function has less conservative and
yields a better result.
5. For time-varying function v(x, t), the theorems are also
different from the theorems of the systems with fixed
coefficients. When the stability theorems on time-varying
systems are used, we should pay more attention todecrescent(1) or Class-K function bound(2).