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Казахский национальный университет им. аль-Фараби

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

As follows from Theorem 9 and Lemma 12 the improper integral

= I

[ y

(t)R y

(t)

m i

( )

i=1

(0)

y	*		*	(t)W y(t)]dt =
y		(t) y(t) y		(t)W y(t)]dt =
		1		1
( i ) 1i d i			< ,	(2.88)

where

R = P =

S11H

S12 A12 H

S11

A11

A12 ,

= H

= 3 P3

= A11 1 S12 A12

A12.

(2.89)

In particular when

then

where

= K

> 0

N = A11

S12 A12K,

= A12K A12 0,

is a matrix of order

n n.

Theorem 13. Let the conditions of the lemmas 7–11 be satisfied, matrices A ( ),

A1( ),

0 = 0

2 be

Hurwitz matrices,

the function

( ) 1,

and

let

besides:

1) diagonal matrix

= diag ( 11, , 1m ), matrices

N1,

S1 = (S11, S12 )

of orders

(n m) n,

m (n m)

such that

R1 R1

= 1

(2.90)

where matrices

R ,

are determined by formula (2.89);

matrix

T =

(W W * ) > 0

(or

surface

V ( y) = y*T y = 0

does not

contain

whole trajectories), where matrix

W is determined by formula (2.89).

Then in the sector

[ , 0 ],

0 2 , 2 > 0

is a diagonal matrix of order

m m

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

= diag (

, ,

where

0 = diag ( 01, , 0m )

limitation value of the matrix

found from the condition of Hurwitz of the matrix

(

Proof. From (2.86) when we have condition (2.90), we have

( )

I 5 =

(t)W y(t)]dt

(

)

[ y

(t)R y(t) y

i=1

(0)

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

(2.91)

dt 2

due to the fact that | y( ) | c

< ,

| y(0) | c

< . As matrix

T =

(W W * ) > 0

(or T1 0 ), then from (2.91) it follows that

121

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


I		=		*	(t)W y(t)dt =		*	(t)T y(t)dt I 5	< ,
I	50	=		y	(t)W y(t)dt =		y	(t)T y(t)dt I 5	< ,
	50				1			1
			0			0

(2.92)

due to the fact that

(t)R y

(t) = y

(t)

)

(t) 0,

(R R

t I.

Let the

		*		y R	n m	. Then from (2.92) we have V ( y)
function V ( y) = y T y,				y R		. Then from (2.92) we have V ( y)
	1	1									1

y 0,	V (0) = 0,						where					,
	V (0) = 0,	V ( y(t))dt <					where	\| y(t) \| c	,	\| y(t) \| c		,
	1		1					2			3
		0

> 0,t,

y, t

y R	n m	,

I = [0; ).

Further, repeating the proofs of Theorem 12, we get	lim y(t) =
	t

is determined from the condition of Hurwitz of the matrix										A (
is determined from the condition of Hurwitz of the matrix										1
a matrix	S	1	of order m (n m)			such	as =	0	2 ,
							0	0
	=			2	solution of	linear	system
	=		0	2	solution of	linear	system		z = A
0			0							1

0.	Matrix	0	=	0	2
		0		0

	0	).	Thus, there exists

for			any ( ) = ,
( )z,			0	is

asymptotically stable, for any

( )	,
1

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

In this case, as

follows from definition 6, Iserman’s problem has a solution. The theorem is proved. As follows from Lemma 2.10, equation (2.65) can be represented as

H	y = N H		y N H y ( ),		H y	= N	22	H	y N	23	H y,
0	11	0	12	1	1		22	0		23	1

(2.93)

where

A11	= (N	, N	12	),
	11		12

A12	= (N	22	, N	23	).
		22		23

The equation (2.93) follows from equality

H		y		N			N		H		y
y =	0		=			11		12		0		I	m	( ),	( )	,


	H	y			N		N			H	y	O			1
						22		23
	1									1		n,m

where	N ,	N	,	N	22	, N	23	are matrices of orders	m m,	m n,
where	11	12			22		23	are matrices of orders
respectively.
Theorem 14.				Let		the	conditions of Theorem 13 be satisfied,

n m,

where

n n,

matrix

			1	(W W * ) = H					*T							*
T =											H	0	0,	T11 = T11			> 0		be a matrix of order				m m,		surface
T =											H		0,	T11 = T11			> 0		be a matrix of order				m m,		surface
1			2	1		1			0	11
			2
V ( y) = y* H					*T H		0	y = 0			do not contain whole trajectories, matrix												N	23	of order
1				0		11	0																	23
n n
			be a Hurwitz matrix.
		Then in the sector							[ ,			0	],	=	0		2	,		2	> 0	is a diagonal matrix of order
		Then in the sector										0	0		0		2			2		is a diagonal matrix of order
m m				with arbitrarily small elements, Iserman’s problem has a solution, where

																						0 = diag ( 01,..., 0m ) found
	0 = diag( 01,..., 0m )									is a limit value of the matrix												0 = diag ( 01,..., 0m ) found

from the condition of Hurwitz of the matrix A1( 0 ) .

122

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

Proof. As follows from the condition of the Theorem, the following inequality

holds (2.92), where

*	*	H		y,
V ( y) = y	H	H		y,
1	0	11	0

T	= T	*	> 0.

11	11

Function

*		y > 0,
V ( y) = y		y > 0,
1	11

y,	y = H	y
	0

trajectories.

0,	V1(0) = 0	when	y

Then, as in the


= 0,	surface		V ( y) = 0		does not contain whole
			1
proof		of		Theorem 12, we have

lim	y(t) = lim H	0 y(t) = 0.	Consider the second equation from (2.63). If the matrix
t	t

of order

n n

is a

Hurwitz matrix, then when

lim H0 y(t) = 0,

limit

lim H y(t) = 0.

y(t) = (H

y(t), H y(t)),

t I ,

then

lim

y(t) = 0.

The theorem is

proved.

Lecture 21.

The solution of model problems in a simple critical case

Constructiveness of the proposed method of solution of Aizerman’s problem and defining the condition of absolute stability will be shown on examples. Below we present solutions to Aizerman’s problem of one-dimensional and two-dimensional regulated systems.

Iserman’s problem for one-dimensional system. The equation of one-dimen- sional regulated system has the form

x = x x		,	x	= 3x 2x		( ),	= ( ),	= x x
1	2		2	1	2			1	1	1	2

( )		= { ( ) C(R			, R ) \| ( ) = ( ), \| ( ) \|							,
				1	1
	0	0 <		< , , R },							*
		0 <		< , , R },					t I = [0, ),
							1
			*				, R ) \| 0				,
		( ) = { ( ) C(R					, R ) \| 0				,
						1	1			2
				1					0
				(0)	= 0,	\| ( ) \|		,	1
				(0)	= 0,	\| ( ) \|		,	R },
							k
where > 0 is an arbitrarily small number.
For this example, matrices					A = A ( ), B , ,				S are equal:
					1	1	1		1

1 ,

(0) = 0,

(2.94)

					1	1	0
				3	1	2 1	1
				3	1	2 1	1
A1 = A1 ( ) =					1	1		,
					1	1	1


	0
	1
	1		= 1,			S1 = ( 1, 1, 1 ).
B1 =	1	,	= 1,			S1 = ( 1, 1, 1 ).

123

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

The equation (2.94) in vector form can be written as:

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

(2.95)

where z = (x1, x2 , ).			If the matrix			A1	= A1( )	is a Hurwitz matrix, the function
( ) 1,		(0) = 0	only when	= 0,			then the system (2.95) has the only equili-
( ) 1,		(0) = 0	only when	= 0,			then the system (2.95) has the only equili-
brium state	z* = (x1, x2, * ) = 0.				Characteristic polynomial of matrix A1 = A1( ) is
equal to			( ) =\| I		A \|= a a a ,
			( ) =\| I		A \|= a a a ,
							3	2
			1	3		1	1	2	3

where a1 = 1 ( 1 1),

a2 = 1 ( 1 1 1),

arbitrarily small number. As follows from 1( ) matrix

if		1	< 0	for any > 0.

A. Nonsingular transformation. Define the vectors

a3 = 1,	> 0 is an
A1( ) is a Hurwitz matrix,

R	3	,		R	3	,		R	3
	3				3				3
1			2				3

from the conditions

	*	B
	*
	1	1

*B 2 1

	*	B
	*
	3	1

. As a result, we get

	*	= (1,0,1),

1

* = (0, 1,1),

		*	= (1, 1,1).
		3

Matrix of transformation

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

		1	1	1		1	1				1	1
		1	1	1		1	1	1			1	1	1

Then matrices

A1,

B1,

are equal

= KA1

B1 =

		1
	1
K	1	=
K		=

		1

KB1	=	0	,
		0
		0

(			)	1 (				)	2 (									)
	1	1			1		1					1		1			1
	1	1			1		1					1		1			1
2					3							5						,

1					2							3
1					2							3

	= S K 1			= (		,				,
S1	= S K 1			= (		,		1		,	1				1	),
	1			1	1			1	1		1		1		1

where	y = Kz, z = K	1	y.

reduced to

Then equation (2.94) by nonsingular transformation is

y1 = [ 1 ( 1 1)]y1 [ 1 ( 1 1)]y2 [2 ( 1 1 1)]y3 ( ),

y2 = 2y1 3y2 5y3 ,

y3 = y1 2y2 3y3 ,

124

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

			= ( 1	1) y1 ( 1 1) y2 ( 1 1 1 ) y3.										(2.96)
B. Solution properties. Matrix

							A1 ( ) = A1 B1 S1 =
	1 ( )( 1 1 )						1 ( )( 1 1 )		2 ( )( 1				1	1 )
												5
	=		2					3				5		.
			1					2				3
			1					2				3

Characteristic polynomial
	( ) =\| I		A1( ) \|=\| I			A ( ) \|= a ( , ) a				( , ) a ( , ),
								3	2
2		3			3	1		1	2				3
where a ( , ) =1 ( )(						), a		( , ) =1 ( )(		1	),	a ( ,	) = ( )		.
	1		1		1	2		1	1	1		3		1

Notice, that as value	> 0 is an arbitrarily small number, then from (2.96) we
have

y	= y y	2	2y	( ),	y	2	= 2y 3y	2	5y	,	y	= y 2y	2	3y	,
1	1	2	3			2	1	2	3		3	1	2	3

			= (			) y (			1	) y	2	(				1	) y	,
			1		1	1		1	1		2		1	1		1	3
= (		) y	(		)(2 y 3y		2	5y	) (						)( y		2y		2	3y ),
1	1	1	1	1		1	2	3				1	1	1	1				2	3

where characteristic polynomials

(2.97)

where	a ( ) = 1
	1

The limit value

	( ) = a ( ) a		( ) a	( ),
	3	2			(2.98)
2	1	2	3		(2.98)

(			),	a	( ) = 1 (			),	a	( ) =	.
1		1		2	1	1	1		3	1
	0	is determined from the condition of Hurwitz of polynomial
	0	is determined from the condition of Hurwitz of polynomial

2 ( ). As follows from (2.97) the following identities hold:

( (t)) = y1(t) y1 y2 2y3 ,

(t) = ( 1 1 ) y1 (t) ( 1 1) y2 (t) ( 1 1 1 ) y3 (t),

(t) = ( 1 1 ) y1 (t) ( 1 1 )[2 y1 (t) 3y2 (t) 5y3 (t)]

( 1 1 1 )[ y1 (t) 2y2 (t) 3y3 (t)],

t I = [0, ).

(2.99)

C. Improper integrals. As follows from Theorem 9 the improper integral

125

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

					(t) 3 y(t) dt =
	*		*	*
I1 = ( (t)) 1	(t)dt = y	(t) 1 y(t) y		(t) 2 y(t) y
0	0

where

( (t)),

(t)

( )

=( ) 1d < ,

(0)

is determined by identities (2.99), matrices

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

										( 2									)							( 2																	)				(2 4																		)
									1					1			1								1							1							1			1								1				1					1					1
									1					1			1								1							1							1			1								1				1					1					1
		2		=										0																							0																				0
														0																							0																				0
														0																							0																				0
								(												)								1								(				4									)				1					(4				5 4							)
							1	(				1							1	)																(			1	4							1		)									(4			1	5 4					1		)
							1					1			1				1										2						1				1			1					1						2				1				1				1		1
				1																									2																								2
		=		1					( 4													)												( 2														)					1			(4 7 4
	3	=		2					( 4													)												( 2														)					2			(4 7 4													) .
				2				1			1					1				1											1							1			1					1							2				1			1					1		1
								1			1					1				1											1							1			1					1											1			1					1		1
				1				(4					5 4											)			1								(4					7 4											)						( 4					6 4						)
								(4			1		5 4										1	)											(4				1	7 4										1	)				1		( 4			1		6 4				1		)
				2		1					1					1							1				2					1							1			1								1					1					1					1	1
				2																							2
Improper integral
																																		(t) 2 y (t) 3y																	(t) 5 y								(t)
																y* (t)N											y							(t) 2 y (t) 3y																	(t) 5 y								(t)
		I	3	=				y* (t)N							2	y* (t)N															2									1										2								3					dt =
			3												2										1					y					(t) y (t) 2 y															(t) 3y								(t)
					0																									y			3		(t) y (t) 2 y													2		(t) 3y								(t)
					0																												3							1								2									3
																																									(t)P3 y(t) dt = 0,
													*											*																*
									y					(t)P1 y(t) y													(t)P2 y(t) y
									0
where
																			m11										m12															n11									n12

															N1 =				m21									m22								, N2					= n21												n22				,
																			m										m															n									n
																								31									32														31						32
																					0														1		n11											1			n12

																																			2													2
																																			2													2
																		1																									1
														P =				1				n													n								1		(n							n
																																																									)	,
																			2																								2														)	,
															1				2				11														21						2						31					22
																		1				n12					1			(n31 n22 )																				n32

																		2									2
																		2									2

126

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

		2n11 n12			3n11 2n12			5n11 3n12
		2n21 n22 m11			3n21 2n22 m21			5n21 3n22 m31
P2	=	2n21 n22 m11			3n21 2n22 m21			5n21 3n22 m31			,
		2n n m			3n 2n m			5n 3n m
		31	32	12	31	32	22	31	32	32


				2m	m
				11	12
				11	12
P =	1	( 3m		2m	2m	m			)
P =	2	( 3m		2m	2m	m			)
3	2		11	21	12		22
3			11	21	12		22
	1		(5m	2m	3m	m		)
			(5m	2m	3m	m		)
	2		11	31	12	32
	2

1	( 3m		2m		2m			m	)	1		(5m	2m			3m	m	)
	( 3m		2m		2m			m	)			(5m	2m			3m	m	)
2	11		21			12		22		2		11	31			12	32
2										2
		3m		2m						1	(5m		3m		3m		2m		)
		3m		2m						2	(5m		3m		3m		2m		)
			21			22				2		21	21			22	32
			21			22						21	21			22	32
1	(5m	3m		2m			2m		)				5m	3m
	(5m	3m		2m		22	2m		)				5m	3m
2	21		31			22		32					31			32
2

Then improper integral

(t)W y(t) dt =

= I

(t)R y

(t) y

(t) y(t) y

( )

( ) 1d < ,

(0)

(2.100)

where	R =			P	,
	1	1		1

P	equal to
3

	=	2	P ,
1		2	2

W =		P .
1	3	3

If we select elements of the matrix

				m	= 2k				(			),	m = 3k					(			),
				11				1	1		1			21			1		1	1
		m		= k			(				),	m		= 2k			(						),
		12			1	1	1		1	1		22		1	1			1	1		1
			m = 5k 2					(			),	m		= 3k 2		(						),
				31			1		1	1		32		1	1			1	1	1
where	k	> 0,	k > 0 are any numbers, then the matrix												W	has the form
where	1		k > 0 are any numbers, then the matrix												1	has the form

(2.101)

				9k k			6k 2k		10k 3k
						1	1			1
						1	1			1
		W =		6k 2k			9k 4k		15k 6k .
		1				1	1			1
			10k 3k				15k 6k		25k 9k
			10k 3k				15k 6k		25k 9k
						1		1		1
Notice, that matrix W1 > 0 for all k1 > 0, k > 0, indeed,
					4k k		6k 2k				2
					4k k		6k 2k
= 4k k		> 0,		=		1	1	= kk > 0,				.
= 4k k		> 0,	2	=				= kk > 0,		\|W \|= 163kk		.
1	1		2		6k	2k	9k 4k		1	1	1
					6k	2k	9k 4k
						1	1
Quadratic form
	y*W1 y = k( 2 y1					3y2	5 y3 )2 k1 ( y1 2 y2 3y3 )2 .					(2.102)

127

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

Matrices R1, 1 are equal

				(			)		n					n
				(			)		11					12
									11					12
			1		1	1			2					2
				n					2				1	2
				n									1
R = P =					11				n					(n	n	)
R = P =					2				n				2	(n	n	)	,
1	1	1			2				21				2	31	22
1	1	1							21					31	22
				n			1
					12			(n		n		)		n
								(n		n	22	)		n
					2		2	31			22			32
					2		2

							2n				n		( 2						)
									11		12	1				1		1
									11		12	1				1		1
	=			2	P =					2n		n		m
		1		2		2					21	22				11
										2n		n		m
										2n		n		m
											31	32				12
3n	2n		( 2						)		5n	3n			(2 4						)
11	12	1			1	1		1			11		12			1		1	1	1
11	12	1			1	1		1			11		12			1		1	1	1
	3n	2n			m							5n				3n		m			,
	21		22			21							21				22		31
	3n	2n			m							5n				3n		m
	3n	2n			m							5n				3n		m
	31		32			22							31				32		32

where

m	,
11

m	,	m	,	m	,	m	,
21		31		12		22

m32

are determined by the formula (2.101). As

follows from the condition of Theorem 13, the equality 1 = 1* must be fulfilled. Consequently, the following equalities are true

2n21 n22 m11 = 2n21 n22 2k 1( 1 1) =

=3n11 2n12 1( 1 1 2 1),

2n31 n32 m12 = 2n31 n32 k1 1( 1 1 1 ) =

=5n11 3n12 1(2 1 1 4 1),

(2.103)

(2.104)

3n	2n	m	= 3n	2n	2k		(			) =
31	32	22	31	32	1	1	1	1	1

= 5n
21

Consider a special case, when (2.105), we have

3n	5k
22

n	= 0,
11

2	(
1	1

n	= 0.
12

	1	).	(2.105)

In this case, from (2.103)–

2n21 n22 = 2k 1( 1 1),

2n31 n32 = k1 1 ( 1 2 1 3 1 ),

3n31 2n32 5n21 3n22 = 2k1 5k 1 ( 1 1 1 ). (2.106)

Matrix R1 can be written as

128

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


		(
	1		1
R =			0
1
			0
			0

As follows from (2.107) matrix

1 )

R	=
1

n21

12 (n31 n22 )

*	0	when
R1


0
0

(n n		) .	(2.107)
31	22
n32
n32

	(		) 0,	n	0,	n	0,	n	n	= 2n	,	n	n .
1	1	1		21		32		31	22	21		32	21

(2.108)

Solutions of the system of algebraic equations (2.106) are:

	n	= 3k		(				),	n	= 2k		(			),
	21		1	1	1		1		31		1		1	1
n	= 4k		(2 3			1	),	n	= 4k k				(			),
22		1	1	1		1		32		1		1		1	1

(2.109)

where	k > 0,	k	> 0	are arbitrary numbers. From ratios (2.107)– (2.109) it follows
where		1

that the set of values		1	,		,		1	, for which Iserman’s problem has a solution,
that the set of values		1		1			1	, for which Iserman’s problem has a solution,

belongs to the set

= {(	,	,	) \|	1	< 0,		(		) 0,	n	n	0}.
1	1	1		1		1	1	1		32	21

(2.110)

Here

< 0

follows from Hurwitz of the matrix

= A ( )

for any arbitrarily

small

value

> 0,

inequalities

(

) 0,

= 2n

ensure the fulfillment of the condition

= R

Values

from

(2.109) provide symmetry of the matrix

i.e.

As follows from (2.102)

matrices

W = W

> 0

when

k > 0,

k > 0.

Thus, all conditions of theorem 13 are

satisfied,

consequently,

for

all

( 1, 1, 1) Iserman’s problem has a

solution.

Notice, that for every triple

( 1, 1, 1) limit value

is determined from the

condition of Hurwitz of polynomial

( )

from (2.98).

In particular triples

( = 0.1; = 1;

= 0.5) .

Indeed,

= 0.5 < 0,

1( 1 1) = 1.5 1 > 0,

1 < 0,

n21

= 3k 1.6 1 > 0,

= 4k k

0.6

> 0

when

= 7k 2k1 < 0,

k > 0,

k1 > 0.

Since

the

values

= 1 1.5 ,

a2 = 1 0.4 ,

a3 = 0.5 ,

then

the

limit

value

0 = 0.88.

the

sector

[ , 0.88 ],

> 0 is an arbitrarily

small

number,

Iserman’s

problem

has a

solution.

129

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

For triple ( 1 = 0.2; 1 = 0.4; 1								= 0.5) . a1		= 1 0.1 ,			a2 = 1 1.1 ,
a3 = 0.5 , the limit value
a3 = 0.5 , the limit value			0 = .
Iserman’s problem for two-dimensional system. The equation of the regulated
system has the form
x = x 1( 1) 2 ( 2 ),				1 = 1( 1), 2 = 2 ( 2 ),							1 = 1x1 1 1 1 2 ,
		2 = 2 x1			2 1 2 2 , t I = [0, ],
									)) C(R2 , R2 ) \| (
( )	0	= { ( ) = ( (				),	(	2	)) C(R2 , R2 ) \| (			) =			(	),
	0		1		1	2		2		1	1	1	1	1	1

	( ) =	2			2	(	2	),	(	) = (				(	),		2	(	2	)),		(0) = 0,
2	2	2		,	2		2		= (		,		1	1		,	2		2			< },
	\| ( ) \|			,		,			= (		,		) R			,		0 <				< },
															2
			*							1		2									*

(2.111)

where > 0 is an arbitrarily small number,

For this example, matrices	A	= A ( ),
For this example, matrices	1	1

= B1,

diag (	,	2	),		1	=	2	= .
1		2			1		2

,	S	are equal to:
	1	are equal to:

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

	1		1			x
B =		1	0		z =				S
B =		1	0	,	z =			,	S	=	1	1
1							1		1
		0	1								2	2
											2	2
							2
							2

1 2

.

The equation (2.111) in the vector form can be written as:

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

If the matrix

A = A ( )

is a Hurwitz matrix, the function

( ) = { ( ) C(R2 , R2 ) | 0

(

)

, i = 1,2,

0i i

= (

(0) = 0, | ( ) |

0 <

< },

(2.112)

where (0) = 0 only when = 0, then the system (2.112) has the only equilibrium state z* = (x*, * ) = 0.

Characteristic polynomial of matrix A1 = A1( ) is equal to

1 ( ) =| I3 A1 |= 3 a1 2 a2 a3 ,

130

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