86

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

		*	( )	1					*	, R	m
				2 0	( ) 2 0,							.	(2.152)

From (2.152) it follows that improper integral (v. (2.149))

I =	[		*	( (t)) 1				* ( (t)) ( (t))]dt 0.
2				2	0	2
	0

From here taking into account inequalities (2.141), (2.143) we have

I2 = [H					*			1	[H		A11 y(t)]
			0 y(t)	A11 y(t)]		2 0				0 y(t)
	0
[H				*	[S11H			y(t)		S12H y(t)] dt =
	0	y(t) A11 y(t)]					0
				2							1

=

[ y

*

(t) y

(t) y

*

(t)

*

(t)

y(t)]dt 0,

2

y(t) y

3

1

0

where matrices

,

2

,

3 of

orders

(n 2m) (n 2m)

are determined

by

1

formula (2.150). The theorem is proved.

Lemma 18. Let the conditions of the lemmas 13–15 be satisfied, matrix

A

1

the

be

Hurwitz matrix, the function	( )	1	. Then for any matrices	N	, N	2
				1

(n 2m) (n m),

(n 2m) (n m)

respectively, the improper integral

*

*

*

I

3

=

[ y

(t)P y

(t)

y

(t)P y(t) y

(t)P y(t)]dt = 0

,

1

2

3

0

where

P = N H

P

= H

*

N

*

N

A

P = N

A

1

1

1

2 1 ,

2

2 12

, 3

1

12 .

Lemma’s proof directly follows from the identity (2.146).

of orders

(2.153)

Lecture 24.

Absolute stability and Iserman’s problem of multidimensional systems in a critical case

Absolute stability. Based on the above results on the estimation of improper integrals, and also Lemmas 13–15 we can formulate conditions of absolute stability of equilibrium state of the system (2.126), (2.127).

We introduce the following notation

R = P = H *									H *			1H		N H ,
						S11H		0					0
0	1	1	1	0	1				0	2	0			2	1

151

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

0 = 2 2 P2 = H0* 1 S12 A12 H0* S11 1 A11 2H0* 2 0 1 A11

H0* 2 S11H0 H0* 2 S12H1 H1N1* N2 A12 ,

						*							*		1
W =	3		3	P = A11			S11 A12				A11			2		A11
0				3	1										0

*

*

A11 2 S11H0 A11 2 S12 H1 N1 A12 .

*

*

In particular when matrix

N1

= A11 1 S12 A11 2T

A12 K1,

*

1

*

W = A11

A11

A12 K A12 0

,

2

0

0

1

*

*

*

*

where

K

= K

> 0, A

S

11

H

A

S

12

H

= A

S , S

1

= T

1

2

0

2

1

2

1

11

11

11

1

are matrices of orders

(n m) (n m), m (n m)

, respectively.

Then improper integral

(2.154)

then the matrix

(2.155)

A12.	Here	K	, T
		1

I

= I

I

I

=

[ y

*

(t)R y(t) y

*

(t) y(t)

4

2

3

1

0

0

0

m

( )

i

( i ) 1i d i

<

i=1

(0)

i

*	(t)W y(t)]dt
y
	0

. (2.156)

Theorem 18. Let the conditions of the lemmas 13–17 be satisfied, matrix	A	be a
	1

										R	,
	Hurwitz matrix, the function ( ) 1, and let, moreover, matrices												of orders
										0		0
	(n 2m) (n 2m), (n 2m) (n 2m)			such that: 1)	R R		*	0	; 2)			=		*	. Then
					0	0					0		0

improper integral

I

m i ( )

i

i=1 i (0)

	=					*	(t)W y(t)dt
40					y
										0
			0
										d			1	*
(		)				d						[		y	(t) y(t)]dt < ,	(2.157)
				1i				i		dt
	i												2		0
										0

where matrices R0 , 0 ,	W0 are determined by the formula (2.154), 1,							2 0 are
any diagonal matrices	of orders m m, N1, N2				are		any matrices	of orders
(n 2m) (n m), (n 2m) (n m) ,		respectively.
Proof. As matrix 0 = 0* , then		y* (t) y(t) =	d	[	1	y	* (t) y(t)], t I , conse-
		0	dt	2			0

quently, improper integral (v. Lemma 2.17)

152

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

=

1 2

	y	* (t) y(t)dt =						d	[	1	y* (t) y(t)]dt =
		0					dt			2		0
	0						0
*			=	1	*	( ) y( )						1	*	(0) y(0)
y	(t) y(t) |				y								y
	0	0		2			0					2		0

<

,

(2.158)

by virtue of the assessment

| y( ) | c

,

| y(0) | c2 ,

0 < c2

< .

From (2.156) taking

2

into account (2.158) we get

0

0

*

*

[ y

(t)R y(t)

y

(t)W y(t)]dt

0

m

( )

i

1

1

( )

d

y

*

( ) y( )

y

*

(0) y(0)

<

. (2.159)

i

i

1i

i

2

0

2

0

i=1

(0)

i

By the

hypothesis

of

the

theorem

matrix

R0

R0 0. Then

quadratic

form

*

y

*

(t)R y(t) = y

*

(t)[

1

(R

R

*

)]y(t)

0,

t,t I.

Consequently, from (2.159) we

0

2

0

0

get inequality

I

=

y

*

(t)W y(t)dt

[ y

*

(t)R y(t) y

*

(t)W

y(t)]dt < .

40

0

0

0

0

0

The theorem is proved.

Theorem 19. Let the conditions of

lemmas

13–17 be satisfied, matrices

A ,

1

A

( ),

0

be Hurwitz matrices, the function

( )

1

and conditions 1),

1

2)

of Theorem 18 be satisfied, and let, moreover, matrix

W W

*

> 0.

Then the

0

0

equilibrium state of the system (2.126), (2.127) is absolutely stable.

If in addition diagonal matrix

0

=

0

2

,

where

2

> 0

is a diagonal matrix

with arbitrarily small elements, then Iserman’s problem has a solution in the sector

[E, 0 ] .

Proof. As all conditions of Theorem 18 are satisfied, then we get inequality (2.157). Consequently, improper integral

				I	40	=		y* (t)W y(t)dt =
								0
							0
			1
=		y* (t)[		(W		W * )]y(t)dt =				y* (t)T y(t)dt < ,	(2.160)
			2		0			0		0
	0								0

153

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

where T0 = 12 (W0 W0* ) > 0.

y Rn 2m ,

y 0,

V (0) = 0,

Let the function	V ( y) = y*T y.	Then V ( y) > 0,			y,
	0
	where | y(t) | c2 ,			,	t,
V ( y(t))dt < ,			| y(t) | c3
0

t I = [0; ). We show that lim y(t) = 0.
	t
Suppose the contrary,	i.e.	lim y(t) 0.		Then there exists	a sequence
		t
tk ,tk > 0,tk when k such that			| y(tk ) | > 0, k = 1,2,... We choose
tk 1 tk 1 > 0, k = 1,2,...	As	y(t),t I		is continuously	differentiable,
| y(t) | c2 ,| y(t) | c3 ,t I,		then		| y(t) y(tk ) | c3 | t tk |, t,t I

[t				1
	k		2

As

	,t				1	], k
		k		2

	t				1	> t
		k		2			k

=

1	,

1,2,...

	t				1	< t		,
		k		2			k 1
				V ( y(t))dt
				0

V ( y) > 0, y R					n 2m	,
	t		1
	k	2
				V ( y(t))dt,
k =1	t		1
	k	2

then

where

| y(t) |=| y(t

) y(t) y(t

) | | y(t

) | | y(t) y(t

) | c

1

=

> 0,

k

k

k

k

0

3

2

t, t [t

1

, t

1

].

We can always choose the value

> 0

such that the

k

2

k

2

1

value

> 0.

So

| y(t) |

,

| y(t) | c

,

t [t

1

, t

1

].

As function

0

0

2

k

2

k

2

*

| y | c

,

V ( y) = y T y

is continuous on compact set

0

then there is a number

0

2

m > 0

such as

min

V ( y) = m . Then the value of the integral

0 |y| c2

t

1

k

2

V ( y(t))dt 1m,

k = 1,2,...

t

1

k

2

Consequently,

t

1

k

2

V ( y(t))dt = y* (t)W0 y(t)dt

V ( y(t))dt lim k 1m = .

0

0

k =1

1

k

t

k

2

154

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

This contradicts the condition (2.160). Consequently,

lim y(t) = 0. t

From

z(t) = K	1	y(t),

t I ,
	K is a nonsingular matrix, we have lim z(t) = 0, ,
	t

											A1( ), 0 are
1. Hence taking into account that matrices A1,
Hurwitz matrices, lim z(t,0, z0 , ) = 0,					,			1 , according to definition 7 we
		t

get that the equilibrium state of the system (2.126), (2.127) is absolutely stable. Notice

that if		=				2	,	then Iserman’s problem has a solution in the sector										[E,	0	].
			0	0
The theorem is proved.
	From theorems 18, 19 it follows that in cases when matrices A1, A1( ), E 0
are	Hurwitz			matrices,				( )				,	and			the following conditions are satisfied:				1)
											1
R R		*	0;					*	;		W W			*	> 0
				2)	=					3)							the equilibrium state of multidimensional
0	1				0			0			0		0

controlled system (2.126), (2.127) in a critical case is absolutely stable.

Remark 3. As follows from the proof of theorem 19, the equilibrium state of the

system (2.126), (2.127) is absolutely stable and in case, when			T	=	1	(W	W	*	) 0,
			0		2	0	0
	*
surface	V ( y) = y T y	does not contain whole trajectories. In		this		case, integral
	0

curves "stitch"

surface

V ( y) = 0.

Notice that surface

V ( y) = 0

does not contain

whole trajectories, if

y

*

(t)T y(t) 0,

t I = [0, ).

This condition for regulated

0

systems with nonlinearities

( )

1 is performed.

It should

be

noted

that as

a result

of nonsingular

transformation

the

identity

(2.142) is obtained. This identity is used to determine the improper integral

I

3 . The

improper

integral

I

3

depends

on

arbitrary

matrices

N

, N

2

of

orders

1

(n 2m) (n m), (n 2m) (n m)

respectively. Matrices

R

, ,W

depend on

0

0

0

matrices

N , N

2

. As

follows

from the condition

of

theorems

18,

19

matrices

1

1, 2 0,

N1,

N2

*

0

*

W0

*

ensure the fulfillment of conditions

R0 R0 0,

= 0

,

W0 > 0.

Consequently, matrices

N

, N

2

allow us to significantly expand the area of absolute

1

stability of system parameters in space. In the method of A.I. Lurie and in the method of V.M. Popov we have no improper integral I3 , consequently, we have no matrices

N1, N2 . The presence of matrices N1, N2 allowed us to solve the Iserman’s problem. Iserman’s problem. The question arises: Is it possible to distinguish a class of

multidimensional regulated systems, by selecting a feedback matrix	S	,	for which
	1

Iserman’s problem has a solution? For this class of multidimensional regulated systems, the obtained results are necessary and sufficient conditions of absolute stability.

155

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

As follows from Theorem 16 and Lemma 2.17 he improper integral

I

= I

I

=

[ y

*

(t)R y(t)

y

*

(t) y(t)

5

3

1

1

1

0

m

( )

i

=

i ( i ) 1i d i

i=1

(0)

i

y	*
<

(t)W1 y(t)]dt =,

(2.161)

where
								R = P = H													*												N
																							S11H								0								H ,
										1			1		1						0	1															2		1
										P = H *												A H*N* N
				=					2							1	S12																									2	A12 ,
				1								2			0						1		11										1					1
								W1 = 3 P3 = A11* 1 S12 A12																												N1 A12 .												(2.162)
															*									*																					*
In		particular		when						N1 = A11 1 S12											A12 K,												then						W1			= A12 K A12				0,		where
K = K	*	> 0 matrix of order (n m) (n m).
Theorem 20. Let the conditions of the lemmas 13–17 be satisfied, matrices																																																	A ,
																																																	1
A ( ),		E			0	=			0		2		be Hurwitz matrices, the function																																( )		,	and let
1																																														1
besides:
1)		diagonal		matrix						1 = diag ( 11,..., 1m ),															N					,				N			2	,	S1			= (S11, S12 ) of						orders
																											1
(n 2m) (n m),				(n 2m) (n m),												m (n 2m)													such that
														R R						*	0,								=						*	,												(2.163)
															1			1									1						1
where matrices			R ,				are determined by the formula (2.162);
			1			1
						1						*																											*
2)		Matrix	T	=			(W W						> 0		(or				surface										V ( y) = y Ty = 0 does																	not		contain
			1			2		1			1																				1
whole trajectories), where matrix														W
															1 is determined by formula (2.162).
Then in the sector								[E, ],							=			0					2	,			2			> 0						is a diagonal matrix of order
												0		0

m m with arbitrarily small positive elements, Iserman’s problem has a solution,

where 0

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

limitation

value of

the matrix

0

= diag ( 01,..., 0m ) , found from the Hurwitz condition

for the matrix

A1 ( 0 ).

Proof. From (2.161) with account for (2.163) we have

[ y* (t)R y(t) y* (t)W y(t)]dt =

I 5 =

0

1

1

m i

( )

d

[

1

y* (t) 1 y(t)]dt < ,

= i ( i ) 1i d i

(2.164)

i=1

(0)

0 dt

2

i

156

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

due to the fact that | y( ) | c

< ,

| y(0) | c

< ,

As T =

1

(W

2

2

1

2

1

T1 0 ), then from (2.164) taking into account (2.163) we get

I

=

*

(t)W y(t)dt =

*

(t)T y(t)dt

I

<

50

y

y

5

,

1

1

0

0

*

*

1

*

t,

y

(t)[

(R1

t

due to the fact that

(t)R1 y(t) = y

2

R1 )]y(t) 0,

W1 )*

I.

> 0 (or

(2.165)

Let the

function

V	(0) =
1

V1

0,

	*
( y) = y T y,
	1
	1
	(V y(t))dt <
0

y ,

R	n 2m	.

where

Then

| y(t)

V ( y)
1
| c	,
2

> 0, | y(t)

y,
| c	,
3

yt,

R	n 2m	,
t I		=

y 0, [0, ).

Further, repeating the proofs of Theorem 5, we get

lim y(t) = 0. t

Matrix

	0	=
there
( )

	0
is
=

2 is determined from the Hurwitz condition

for the matrix

A

(

1

matrix

S

1 of order

m (n 2m)

such

that

0

=

0

2

,

E

=

0

2

solution of

linear

system

z

0

	0
,
=

).	Thus,

for any A1( )z,

E	0

is asymptotically stable, for any

( )	,
1

lim	y(t) = lim Kz(t) = 0.
t	t

In this case, as follows from definition 9 Iserman’s problem has a solution. The theorem is proved.

As follows from Lemma 16, equation (2.137) can be represented as

H y = A11 y(t) ( ) = N H y N H y(t) ( ),
0		11	0	12	1

where equality

A11	= (N	, N	12	),
	11

H y	=
1
A12 =

A12 y(t) =
(N	22	, N	23

N ) .

22	H	y
	0

The

N	23	H
		1

equation

y ,

(2.166)

(2.166)

follows from

	H y				N
	y =	0			=	11
			y
	H				N	22
		1

N		H		y		I	m
12			0
N	23	H		y	O
			1			n m,m

( ), ( ) ,

1

where

N11,

N12 ,

N22 ,

N23

are

matrices

of orders

m m,

m (n m),

(n m) m,

(n m) (n m) respectively.

Theorem

21.

Let

the

conditions

of Theorem 20

be satisfied,

where

T =

1

(W W * ) = H *T H

0,

T

= T *

> 0

is a matrix of order m m,

surface

0

1

2

1

1

0

11

11

11

157

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

V ( y) = y* H *T H

0

y = 0 does not

contain whole

trajectories,

and matrix

N

23

of

1

0

11

order

(n m) (n m) is a Hurwitz matrix.

Then in the sector [E, 0 ], 0

=

2 , 2 > 0 is a diagonal matrix of order

0

m m with arbitrarily small positive elements, Iserman’s problem has a solution,

where

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

limit value of

the matrix

0

= diag (

01

,...,

0m

)

found from the Hurwitz condition for matrix A1 ( 0 ).

Proof. As follows from the condition of the Theorem, equality (2.165) holds,

where

*

H

*

T

= T

*

> 0.

Function

*

V1 ( y) = y

0T11H0 y,

V ( y) = y T y > 0,

11

11

1

11

y, y = H0 y 0,V1(0) = 0 trajectories. Then, as

lim	y(t) = lim H	0 y(t) = 0.
t	t

order		(n m) (n m)			is
lim	H y(t) = 0.		As	y(t) = (
		1
t

is proved.

when	y = 0,	surface		V ( y) = 0		does not contain whole
				1
in	the proof		of		Theorem 19, we have

Consider the second equation from (2.166). If										N	23 of
a		Hurwitz matrix, then when			lim	H	0	y(t) = 0,			limit
					t
H	0	y(t), H y(t)),	t I ,	then lim	y(t) = 0.				The theorem
		1
				t

Lecture 25.

The solution to the model problem in a critical case

The equation of a regulated system has the form

x	= x ( ), x						= x , x = ( ), = x x										x
1			1	2			3	3			1			1	1	2			1	3
( )			= { ( ) C(R					, R ) | ( ) = ( ), ( ) }
							1	1
		0									, R ) | 0 ( )								,	1
	( ) = { ( ) C(R
								1			1							2
				1												0
				(0)			= 0,| ( ) | * , R1},													(2.167)

where > 0 is an arbitrarily small			number,			= x ,			= x2 , L1 = 1, L2 = 0,
							3
S = 1, T1 = 1, T2 = 1.
For this example matrices		A1 = A1( ),		B1,		S1 , are equal to:
	1 1		1	1				1
A1 = A1 ( ) =										= ( 1	, 1,
		0	0	1		, B1	=	0	, S1			1 ).
					1			1
		1	1

The equation (2.167) in the vector form has the form

158

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

			z = A1z B1 ( ),
= S1z, z(0) = z0 ,| z0 |< , ( ) 1,t I = [0, ) ,														(2.168)
where z = (x1, x2 , x3 ). If the matrix						A1 = A1( )					is a Hurwitz matrix, the function
( )	, (0) = 0	only when				= 0 ,		then			the system			(2.168) has the only
1
equilibrium state.
Characteristic polynomial of matrix							A = A ( )					is equal to
							1	1
		( ) =| I		A |= a a a										,
							3				2
	1		3			1		1				2	3

where	a	= 1 (		),	a	= 1
	1	1	1		2

matrix	A	( )	is a Hurwitz matrix, if:
	1

А. Nonsingular transformation.

(	1	),		a	= .
		1		3	1
	< 0,		1	< 0	for any
1

Define the vectors

1

As follows from

> 0 .

R	,		R	,
3			3
		2

( )
	1
	R
	3
3

from the condition

	*	B = 1,
1		1

*B = 0,

2 1

		*	B = 0.
		3
			1

As a result, we get

	*	= (1,1,0),
1

* = (1,0, 1),

2

		*	= (1,1, 1).
		3

Matrices of transformation

	1	1	0			1		1	1
R* =					K 1 =
	1	0	1	= K,			0	1	1	,	| R |= 1.
		1					1	0
	1		1						1

Then

matrices

A1,

B1,

S

1 are equal

to:

A1 ( ) = KA K

1

,

1

S

= S K

1

,

y = Kz,

1

1

where

z = K y .

Then

equation

(2.168)

by

1

ransformation is reduced to

y

= A1 ( ) y B ( ),

1

= S

y, ( )

, y(0) = y

,| y |< ,t I = [0, )

.

1

1

0

0

The equation (2.169) can be written as

B1	= KB	,
	1

nonsingular

(2.169)

y	= (		) y [ 1 ( )]y			2	[ (			)]y	( ),
1	1	1	1	1	1		1	1	1	3

		y	2	= y y		2	y		, y		= y	,
				1			3			3	2
	= (			) y ( ) y				2		(				) y	,
= ( 1	1	1		1	1		1				1	1	1	3
	1) y1 ( 1 1)( y1				y2 y3 ) ( 1							1	1 )( y2 ) . (2.170)

159

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

Б. Solution properties. Characteristic polynomial

				( ) =| I					A1( ) |= a ( , ) a														( , ) a ( , ),
														3					2
			2					3							1						2							3
where		a ( , ) =1 ( )(											),	a ( , ) = ( )(							1	),		a ( , ) = ( ) .
		1								1		1		2									1		3					1
	Notice that as value > 0 is an arbitrarily small number, then from (2.170) we
have
		y	= y		2	( ), y				= y y					y		, y	= y , ( (t)) = y (t) y (t),
		1							2				1	2	3		3			2								1	2
	(t) = (						) y	( )( y							y	2	y		) (									)( y	(t))	, (2.171)
				1		1	1			1		1		1				3					1	1		1		2
where characteristic polynomial
											( ) = a ( )							a		( ) a					( )
														3			2											,		(2.172)
										2					1				2					3
where		a ( ) = 1 (										),	a ( ) = (					1	),				a ( ) = .						Limit value
			1						1		1			2						1			3					1
	0 is determined from the condition of Hurwitz of polynomial																									2	( ).

В. Improper integrals. As follows from Theorem 16 the improper integral

where

*	(t) 1 y(t)
= [ y
0
( (t)), (t),t I

		1			1
		I	=	( (t)) (t)dt =
				0
						( )
y	*			*	(t) 3 y(t)]dt =	( (t)) 1 d < ,
		(t) 2 y(t) y
						(0)

are denoted by identities from (2.171), matrices

	1 ( 1 1 )		0	0			1 ( 1 1 )
1						2
	=	0	0	0	,		=	0
		0	0	0				0

						0					1		1	( 1	1 )
											2
				1
		=				(		)
				2
	3				1		1	1						1 1
						0				1		1 ( 1 1 )
									2

1	1	( 1 1 )
0		0	,
0		0

		0
1
		(		) .
2	1		1	1
		0

Improper integral

160