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Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

where matrices A1, B1, S1 are matrices of orders

m (n 2m)

respectively,

(n 2m) (n 2m),

(n 2m) m,

x			A BES			BET				BET
								1			2
								1			2
z =	,	A =		L ES		L ET				L ET		= A (E),
		1		1			1	1		1	2	1
				L ES	I		L T			L ET
				L ES	I	m	L T			L ET
				2		m		2	1	2	2

= S

T .

B =

The equilibrium states of the system (2.126), (2.127) are denoted from solutions

of algebraic equations

A z

B (

) = 0,

= S z .

If the matrix

A = A (E)

is a

Hurwitz matrix, the function

( )

turn into zero only when

= 0,

then the

system (2.126),

(2.127)

has

the

only

equilibrium

state

z = (x

, ) = 0.

Consequently,

= 0,

= 0.

Notice, that the equilibrium state matches trivial solution

x(t) 0, (t) 0,

(t) 0,

t I

of the system (2.126), (2.127). The article studies asymptotic stability

of generally unperturbed

movement

x(t) 0,

(t) 0,

(t) 0 ,

t I

in the

system (2.126), (2.127) for all

( )

(or

z(t) 0,

t I ,

1 ).

We assume with a sufficiently small neighborhood of the point

= 0,

function

( )

can

be approximated

linear

function

( ) = ,

where

= diag ( 1,..., m ) . Consequently,

when

| |<

> 0

a sufficiently small

number, the equation of perturbed motion (2.128) can be written as

z = A z B S z = A ( )z, z(0) = z					,\| z	0	\|< ,t I
1	1	1	1	0		0

(2.129)

where A1( ) = A1 B1 S1,

E < 0

diagonal matrix determined from Hurwitz

matrix	A1( ) = A1(E) B1 S1, E
number	1 > 0 such as \| x(t) \|< 1 when

,		0	= diag(	01	,...,	0m	)	is a limit
		0		01		0m		is a limit

condition			for the matrix	A ( )	. If the
				1
0	<	0 is a Hurwitz matrix, then there is a

\|	z0 \|< 1,		moreover lim z(t) = 0. Thus,
			t

when matrix A1( ), where E 0 < 0 is a Hurwitz matrix, trivial solution z(t) 0, t I (x(t) 0, (t) 0, (t) 0,t I ) of the system (2.129) is asymptotically stable by Lyapunov when t .

141

Lectures on the stability of the solution of an equation with differential inclusions


Definition 7. Trivial solution x(t) 0, (t) 0, (t) 0,				t I , of the system
(2.126), (2.127) is		called absolutely stable, if: 1) matrices		A1 = A1(E),		A1( ),

E 0	< 0	are Hurwitz matrices	(asymptotic stability by		Lyapunov when
t ); 2)	for all	( ) 1 solution of	the differential equation		(2.128)	has the

property			lim z(t;0, z0 , ) = 0,						\| z0 \|<
			t
	(t;0, x	,	,	, ) = 0,
lim	(t;0, x	,	,	, ) = 0,	lim	(t;0, x	,	0	,	0	, ) = 0,
	0	0	0		lim	0		0		0
t					t

(or

, 0 ,

lim x t

| x0 |<

(t;0,, |

x
	0
	0

,	,
0
\|< ,

0	, ) = 0,

| |< . )

Definition 8. Conditions of absolute stability of the system (2.126), (2.127) are

called

relations,

binding

constructive

parameters

the

system

(A,B,L1

,L2 ,S,T1,T2 , 0 ),

= diag (

,...,

)

under which

the

equilibrium state

= 0,

is absolutely stable.

Problem 8. Find the condition of absolute stability of equilibrium state of the

system (2.126), (2.127).

Function

(t) = S z(t),t I

is a control formed by principle of feedback, and

matrix

S ,

of order

m (n 2m)

is a feedback matrix. Iserman’s problem consists in

choosing a feedback matrix

such that from asymptotic stability of trivial solution

z(t) 0,t I

of the

linear

system

(2.129)

for

any

, E

there followed absolute stability of trivial solution

x(t) 0, (t) 0, (t) 0,t I

of the system (2.126), (2.127),

where

= diag(

,...,

)

is a

limit diagonal

Hurwitz matrix of the matrix

A ( ),

diagonal matrix of order

m m

, with

arbitrarily small positive elements.

Definition 9. We assume that in the sector

[E,

]

Iserman’s problem has a

solution, if: 1) there exists matrix

, of order

m (n 2m)

such as

;

2) for

any

( ) = ,

solution

the system

(2.128)

asymptotically stable; 3) for any

( )

trivial solution of the system (2.126),

(2.127) is absolutely stable.

Problem 9. Find sector [E, 0

2 ], where Iserman’s problem has a solution.

Note that as follows from the work [12] Iserman’s problem not always has a

solution. The

problem

consists

of finding such a feedback matrix

order

m (2m n),

for which in the

sector

[E, 0

where

, Iserman’s problem

had a solution. Sector [E, 0 ],

means that [ , 0i ],

0i = 0i 2i , where

0 2

= ,

E = diag ( 1,..., m ),

= diag ( 01,..., 0m ),

= diag( 01,..., 0m ),

i = 1, m,

= diag( 21,..., 2m ),

E > 0,

2 > 0.

142

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Nonsingular transformation. Information about the properties of nonlinearities is contained in inclusion ( ) 1 . It is necessary to convert equation (2.128) so that explicitly present the function ( (t)), t I through phase coordinates of the

system, taking into account Hurwitz of the matrix A1 = A1(E) .

Moreover, nonsingular transformation should be such that part of the transformed system does not contain nonlinearity. With such a transformation you can get improper integrals independent of functions ( ) 1 , allowing you to expand the area of absolute stability in space of parameters of the system and get a solution to the

Iserman’s problem.
It should be noted that the existing conversion of the linear system from [13:			§5	,
				,
Theorem 2] based on the controllability of a pair	( A, B)	does not satisfy the above
requirements.

	i

Lemma 13. Let the matrices

R	n 2m	,i = 1, m such that:

b =
*
i	i

B =\|\|
1
*
1,	b
i

b1,...,bm ||,

where

j	= 0,i = 1, m,i


b Rn 2m ,i = 1.m,			vectors
i
j ,			(2.130)

where	b	b	,i	j,(*)	is
where	i	j			is

system (2.128) the following

	*	z(t) =				*
	i					i

If in addition rank			B	*	=

			1

a transpose sign. Then alongwith the identities hold.

A z(t)		(	(t)),t I = [0, ),i = 1, m
1	i	i

m	and Gram determinant

solution of the

. (2.131)

				<		,		>	<		,		2		> ...
					1		1			1			2
(	,...,	m	) =		...					...					...
1		m
				<	m		,	>	<	m		,		2	> ...
					m		1			m				2

<		,		m		>
	1
	...
<	m		,		m	>

(2.132)

then vectors		i

scalar product

Proof. As

the left

,i = 1, m

exist and they are linearly independent, where

is a

i , j ,i, j = 1, m .

= (

,...,

i = 1, m

, then by multiplying the identity from

i,n 2m

z(t) = A z(t) B ( (t)),

t I

get



z(t) = A z(t) B ( (t)),			t I ,	i = 1, m, where	*B	= ( b ,..., b ).
i	i 1	i 1			i 1	i 1	i m

From here taking into account (2.130), we get (2.131). Notice, that ratio (2.130) can be written as

143

Lectures on the stability of the solution of an equation with differential inclusions

i1b11 i 2b12 ... i,n 2mb1,n 2m = 0,

i1bi1 i 2bi 2 ... i,n 2mbi,n 2m = 1,

i1bm1 i 2bm2 ... i,n 2mbm,n 2m = 0,i = 1, m,

	b
	k1
	bk 2

bk =		, k = 1, m.


b
	k ,n 2m


If rank	B	*	= m	, then this system of equations has a solution					,i = 1, m.	From

	1							i
the condition (2.132) it follows that vectors						,i = 1, m	are linearly independent. The
					i

lemma is proved.

Lemma

14.

Let

the

vectors

,...,

such

that

b =

0,n m

k = 1, n m

b b

j , i j . Then

along

the solution

of the

system

following identities hold:

z(t) =

A z(t), k = 1, n m,t I

If in addition rank

= m

and Gram determinant

0	,	i = 1, m	,

(2.128) the

(2.133)

(	01	,...,	n m

<		01	,		01		> ...

) =		...						...
<	0n m			,		01		> ...

<		01	,		0n m		>

			...
<	0n m			,		0n m		>

(2.134)

then vectors 01,..., 0n m exist and they are linearly independent.

0,k =

(1)

(n 2m)

), k = 1, n m, then by multiplying the identity

Proof. As

( 0k ,..., 0k

from the left (2.128) taking into account that

0(1)k bj1 0(2)k bj 2

... 0(kn 2m)bj,n 2m

= 0, k = 1, n m, j = 1, m

we get

* z(t) = * A z(t), t I. From here we get identity (2.133). By the hypo-

thesis

of the

lemma

rank

B* = m

and the

following inequality holds (2.134).

144

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Consequently, vectors 0k Rn 2m ,
	k = 1, n m exist and are linearly independent.

The lemma is proved.

Lemma 15. Let the conditions of the Lemmas 2.13, 2.14 be satisfied, and let, moreover, rank of the matrix

R =\|\|	,...,	m	,	01	,...,	0,n m	\|\|
1		m		01		0,n m

(2.135)

of order (n 2m) (n 2m) be equal to	n 2m . Then equation (2.128) is equivalent
to the following system of equations

= a11 y ... a1,n 2m y

( ),...,

n 2m

= am1 y ... am,n 2m y

(

n 2m

= am 1,1 y ... am 1,n 2m y

,...,

= an 2m,1 y ... an 2m,n 2m y

m 1

n 2m

y ... s

,...,

= s

y ... s

m,n 2m

(2.136)

1,n 2m

n 2m

n 2m ,

z,i = 1, m

k = 1, n m

where

Proof.

rank

R = n 2m

then

from

(2.135) it

follows that

vectors

n 2m

i = 1, m,

n 2m

k = 1, n m form the basis in

n 2m

. Then

A z = a11

z ... a1m

a1,m 1

z ... a1,n 2m

0,n m

...

A z = am1

z ...

am,m

am,m 1

... am,n 2m

0,n m

...

A z = am 1,1

z ... am 1,m 1

... am 1,n 2m

0,n m

01 1

...

A z = an 2m,1

z ... an 2m,m 1

... an 2m,n 2m

0,n m

where

ai , j ,

i, j = 1, n 2m

are decomposition coefficients

A		,
*
1	i

i = 1, m,

A		,
*
1	0k

k = 1, n m by bases i Rn 2m ,

i = 1, m,

0k Rn 2m ,

k = 1, n m

taking into account (2.131), (2.133) we get a system of equations (2.136).

	*	R	n 2m	,	*	= (
Similarly, by decomposing vectors	S1 j	R		,	where S1	= (

. From here

S11* ,..., S1*m ),

j = 1, m by bases i , i = 1, m, 0k , k = 1, n m, we get

1 = s11 y1 ... s1,n 2m yn 2m ,..., m = sm1 y1 ... sm,n 2m yn 2m.

145

Lectures on the stability of the solution of an equation with differential inclusions

The lemma is proved.
From	Lemmas 2.13–2.15			it follows that variables	y = ( y1,..., yn 2m )	and
z = (z ,..., z		n 2m	) are related by	y = R* z . Consequently, z = (R* ) 1 y. If we denote
1		n 2m
K = R* ,	then		y = Kz, z = K 1 y. Then differential		equation (2.128)	with

nonsingular transformation is reduced to

where

y(t) = A1 y B1 ( ),

= S1( y), ( ) 1, y(0) = y0

= Kz(0),| y0 |< ,t I ,

(2.137)

= KA K

, B1

= KB

= S K

= R

A (R )

= R

= S (R )

Lecture 23.

Solution properties and improper integrals in a simple critical case

Solution properties. It can be shown that the solution of the system (2.128), and also (2.137) are limited.

Theorem			15. Let the	matrix	A = A (E)			be a Hurwitz matrix, i.e.
					1	1
Re	j	( A ) < 0,	j = 1, n 2m,	function	( )		,	and conditions of the lemmas
		1				1
1-3 are satisfied. Then following estimates are true:

| z(t) | c

,| z(t) | c

,t I

(2.138)

y(t) | c

y(t) | c ,t I

(2.139)

| (t) | c

,| (t) | c

,t I

(2.140)

where

ci = const > 0,

c < ,

i = 0,5

. Moreover, the functions

z(t), y(t), (t),t I

are uniformly continuous.

Proof. From inclusion ( ) 1

it follows that | ( (t)) | * , 0 < * < ,

t I.

matrix

Hurwitz

matrix,

i.e. Re j ( Ai ) < 0,

a =

( A ) < 0, then

A t

|| ce

(a )t

, t, t I ,

c = c( ) > 0, > 0

max

|| e 1

1 j n 2m

is an arbitrarily small number [14:§ 13, inequality (7)]. Solution of the differential equation (2.128) has the form:

z(t) = eA1t z0 t eA1(t )B1 ( ( ))d ,t I.

146

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

Then

A t

(t )

| z(t) | || e 1

||| z

|| e 1

||||

B ||| ( ( )) | d

e | z

| e(a )t ce(a )t || B ||

e (a ) d =

= c | z

| e

(a )t

[

(a )t

] =

= c | z

| e

(a )t

c ||

B ||

( (x))( 1 e

(a )t

)

c | z

|| B

(

(x)) = c

where e(a )t 1,t I , a < 0,

> 0,| ( ( )) |

, t,t I. From here we

get limitation of the solution of the system (2.128). From the equation (2.128) it follows that

| z(t) | || A

||| z(t) | || B

||| ( (t)) | || A

|| c || B ||

= c , t,t I.

Then | (t) | || S

||| z(t) | || S

|| c

= c ,

| (t) | || S

||| z(t) | || S

|| c

= c ,

t I.

function

y(t) = Kz(t),

t I ,

then

y(t) | || K |||

z(t) | || K || c0

= c2 ,

| y(t) | || K ||| z(t) | || K || c1 = c3 ,

t, t I.

So, estimates (2.138)–(2.140) are proved.

From the limitations of functions

y(t), z(t), (t),t I

we get uniform continuity

of functions

y(t), z(t), (t),t I.

The theorem is proved.

should be

noted that:

From evaluation

| z(t) |,

t I

have

lim

| z(t) |=| z( ) | c | z

c || B

= c

due to continuity

z(t),

t I ,

a < 0. Consequently, lim | y(t) |=| y( ) | || K || c0 = c2 ,

lim | (t) | || S1 || c0

where

= c4 .

Lemma 16. Let the conditions of the lemmas 13–15 be satisfied. Then along the solution of the system (2.137) the following identities hold


		(2.141)
( (t)) = H0 y(t) A11 y(t),t I ,		(2.141)

		(2.142)
	H1 y(t) = A12 y(t),t I ,	(2.142)

(t) = S11H0 y(t) S12H1 y(t),t I ,		(2.143)

147

Lectures on the stability of the solution of an equation with differential inclusions

where

H	0	= (I
	0

A11 =

	m	, O
	m	m
	a11

	...
		am1
		am1

(t) = S11H0 y(t) S12 A12 y(t),t I ,

(2.144)

A11

), H

= (O

, I

= I

, A1 =

,n 2m

n m

n 2m

n m,m

A12

...

a1,n 2m

am 1,1

...

am 1,n 2m

...

A12

...

am,n 2m

an 2m,1

...

an 2m,n 2m

S11 ...

S1,m

S1,1 m

...

S1,n 2m

S1 = S11

S12 , S11 =

... ...

...

, S12

S m1 ...

S m,m

S m,m 1

...

S m,n 2m

Proof. As conditions of the lemmas 13–15 are satisfied, then we get ratios

(2.136). Notice, that

m 1

y =

...

y =

...

, H

n 2m

Then from (2.136) it follows that

( ) =

y a11 y ... a1,n 2m y

,...,

n 2m

m ( m ) = ym am1 y1

... am,n 2m yn 2m ,

(t) = (S11, S12 ) y = (S11, S12 )

= S11H y S12H y, H y = A12 ,

H y

t I

where y = y(t), y = y(t),

. Consequently, identities (2.141)–(2.144 hold. The

lemma is proved.

Lemma 17. Let the conditions of the lemmas 13–16 be satisfied. Then for any matrices M1, M2 , N1, N2 of orders (n 2m) (n 2m), (n 2m) (n 2m),

(n 2m) (n m), (n 2m) (n m) respectively, along the solution of the system (2.137) the following identities hold

148

Chapter II. Absolute stability and Aizerman’s problem of multidimensional regulated systems

(t)M

(t) =

(t)M

y(t)], t I ïðè M

(t) y

[ y

	*	(t)N		y	*	(t)N ][H
[ y		(t)N	2	y		(t)N ][H	1	y(t) A12 y(t)] 0,t
			2			1	1
Proof. Identity (2.145) directly follows from equality

(t)M

y(t)] = y

(t)M

(t)

[ y

y(t) y

y(t) = y

(t)M

(t),t I ïðè M

= M

= y

y(t) y

	= M	*	,
1		2

I .

*	(t)M
y

(2.145)

(2.146)

2	y	(t) =
2

As follows	from identities (2.142) for any matrices	N	,	N	2	of order
		1
(n 2m) (n m) ,	(n 2m) (n m) , respectively, identity (2.146) holds. The lemma

is proved.

Improper integrals. Based on nonsingular transformation y = R* z and properties

of solutions (2.138)–(2.140) and identities (2.141)–(2.144) we can obtain evaluations of improper integrals along the solution of the system (2.137).

Theorem 16. Let the conditions of the lemmas 13–15 be satisfied, matrix

1 be a

Hurwitz

matrix,

the

function

( ) 1.

Then

for

any

diagonal

matrix

1 = diag ( 11,..., 1m )

order

m m

along

the

solution of the

system (2.137) the

improper integral

I =

( (t))

(t)dt =

[ y

(t) y

(t)

y(t)]dt =

y(t) y

( )

(

)

(2.147)

i=1

(0)

where

1 = H0* 1 S11H0 , 2 = H0* 1 S12 A12 H0* S11 1 A11, 3

= A11r1 S12 A12 .

(2.148)

= KA K 1,

Proof. Notice

that

matrix A1

where

is a

nonsingular

matrix.

Consequently,

j ( A1 ) = j ( A1 ), j = 1, n 2m , matrix

is a Hurwitz matrix if the

matrix A

a Hurwitz matrix, where

( A ) =

( A1 ), j = 1, n 2m are eigenva-

lues of the matrix A1 , Re j ( A1 ) = Re j ( A1 ) < 0, j = 1, n 2m.

| y(0) | || K ||| z0 |< .

As follows from Theorem 15, | y( ) |<|| K || c0

= c2 ,

Impro-

per integral

149

Lectures on the stability of the solution of an equation with differential inclusions

I =

( (t)) (t)dt = [H

y(t) A11 y(t)]

[S11H

y(t) S12 A12 y(t)]dt =

( )

[ y

(t) 1 y(t) y

(t) 3 y(t)]dt =

i ( i ) 1i d i

(t) 2 y(t) y

< ,

i=1

(0)

by virtue of identities (2.141)–(2.144), where

| (0) | c4

< ,

| ( ) | c4

< , matri-

ces

of orders (n 2m) (n 2m) are denoted by relations from (2.148).

The theorem is proved.

Theorem 17. Let the conditions of the lemmas 13–15 be satisfied, matrix

1 be a

Hurwitz

matrix,

the

function

( )

Then for

any diagonal

matrix

2 = diag (

21,..., 2m ) 0

of order

m m,

along the solution of the system (2.137) the

improper integral

[

( (t))

( (t)) (t)]dt =

( (t))

			1			2
		*	1	*		2
=	[ y		(t) y(t) y		(t)		y(t)
	0

y	*	(t)		y(t)]dt
			3

(2.149)

where

	*				1	H			,													*				1										*				S11H							*		S12H ,
	= H				0	H		0	,		2		= 2H													0			A11 H										2	S11H		0		H				2	S12H ,
1	0	2			0			0			2											0		2		0										0			2			0				0		2	1
										*						1				A							*					S			H					*			S			H
						3	= A							2		1				A					A						2	S	11		H	0		A			2		S		12	H	1 .		(2.150)
						3				11				2		0						11				11					2		11			0				11	2				12		1 .		(2.150)
Proof. From inclusion																	1				it follows that
Proof. From inclusion																	1				it follows that
			i ( i )												,			i										1		,					,					1
													0i		,				(				)				0i			,				i	,	i		R ,i = 1, m.
													0i														0i							i		i
						i											i					i
						i											i					i
																	1					2															Then for any value 2i 0 an
																						2
																				i			( i ),i = 1, m.
Consequently,				i i ( i ) 0i																i			( i ),i = 1, m.
																1								2
inequality holds i ( i ) 2i i																1								2				)		0,i = 1, m. From here it follows that
inequality holds i ( i ) 2i i												0i						2i i ( i										)		0,i = 1, m. From here it follows that

m			2
[ 1			2
[ 1		2i	i
	0i	2i	i
i=1
Let the 2 = diag ( 21,..., 2m ) 0,

(2.151) can be written as

(	)	(	)
i	i	i		2i

0i1 = diag ( 011

i	] 0	.	(2.151)

,..., 0m1 ). Then inequality

150

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