2 lyapunov direct method

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


8/39Note 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).

2 lyapunov direct method

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