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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

where

a1 = 1 ( 1 2 1 2 ),

a2 = ( 1 2 ) 2[ 1( 1 2 ) 2 ( 2 1 2 1) 1 2 2 1 ],

a3 = 2 1 ( 1 2 ) 3[( 1 2 )( 1 2 2 1 ) ( 1 2 )( 1 2 2 1 ) ( 1 2 )( 2 1 2 1 )],

> 0 is an arbitrarily small number. As follows from	( ),	matrix	A ( )	is a
	1		1

Hurwitz matrix, if	( 1 2 ) > 0,	1( 1 2 ) > 0, for any arbitrarily small number
> 0.										1
						1				1
						1
1. Nonsingular transformation. Matrix B1 = B11			B12 ,	B	=	1		B	=
1. Nonsingular transformation. Matrix B1 = B11			B12 ,	B	=	1	,	B	=		0 .
				11				12
						0
						0					1

										R				,				R		,
Define the vectors													3						3
Define the vectors										1							2
	*	B	= 0;		б)			*	B		= 0,				*	B			= 1;
		B	= 0;					2	B		= 0,				2	B			= 1;
1		12						2	11						2		12
			1				1									1

		=	0	,	2	=			1	,		3	=			1 .
1					2							3
			1						0							1
			1						0							1

Matrix of transformation

в)

3	,
R
B
11


from the conditions: a)						*	B	= 1,
							B	= 1,
					1		11
= 0,	*	B	= 0.	As a result, we get
= 0,	3	B	= 0.
	3	12

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

		1	0	1		1	1	1			0	1	1

Then matrices

A1,

B1,

are equal to:

						1 ( )							1 (							)		1 (									)
										1		1				1			1					1			1			1
										1		1				1			1					1			1			1
A1	= KA K 1				=		1 (		2			)	1 (					2		2	)	1 (			2				2		2		) ,
			1						2		2							2		2					2				2		2
								1								1									1
								1								1									1

					1		0
										= S Ê 1

B1	= KB =				0 1 , S1								=		1			1			1		1			1				1		1 ,
			1								1					2			2 2					2 2
					0								2			2			2 2					2 2									2
					0		0
			1 ( )									1 (					)			1 (								)
		A11	1 ( )									1 (					)			1 (								)
		A11	=				1			1				1		1							1	1			1			,
				1 ( 2 2 )							1 ( 2 2 )
				1 ( 2 2 )							1 ( 2 2 )								1 ( 2 2 2 )

131

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

													1		0	0
	= ( 1,1,1), H		1		0	0					H0
A12			1		0	0		H		= (0,0,1),	H0	=		0	1	0
A12		0	=				,	H	1	= (0,0,1),		=		0	1	0	,
		0							1
				0	1	0					H1			0	0	1
														0	0	1

												y
												y
S11	=	1	1			1	1 , S12	=	1 1	1 , H	0	y = 1	, H		y = y ,
				2					2 2		0			1	3
	2 2			2			2		2 2	2		y2
where	y = Kx,		x = K		1	y.
	y = Kx,		x = K			y.

Then equation (2.112) by nonsingular transformation reduces to

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

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

2 = ( 2 2 ) y1 ( 2 2 ) y2 ( 2 2 2 ) y3.

2.Solution properties. Matrix

							1 ( )( )
								1		1			1
								1		1			1
A1 ( ) = A1 B1 S1 = 1 ( 2 )( 2 2 )
								1
								1

1 ( )(					)		1 ( )(								)
1		1		1			1		1		1		1
1		1		1			1		1		1		1
1 (	2	)(	1		2	)	1 (	)(	2			2		2		) .
	2		1		2		2		2			2		2
1								1
1								1

Characteristic polynomial

(2.113)

2 ( ) =| I3 A1( ) |=| I3 A1( ) |=

= 3 a1( , 1, 2 ) 2 a2 ( , 1, 2) a2 ( , 1, 2 ).

Notice, that as > 0 is an arbitrarily small number, then from (2.113) we have

y1 = y1 y2 y3 1( 1),

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

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

132

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

2 = ( 2 2 ) y1 ( 2 2 ) y2 ( 2 2 2 ) y3.					(2.114)
where the characteristic polynomial

2 ( ) =\| I3 A1( ) \|=\| I3 A1( ) \|=
= 3 a1 ( 1, 2 ) 2 a2 ( 1, 2 ) a3 ( 1, 2 ).					(2.115)

The limit diagonal matrix		0

of polynomial		2	( )	from (2.115),

= diag (	01	,	02	)
	01		02

where matrix

is determined from the Hurwitz


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

As follows from (2.114) the following identities hold:

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

	2	(t) = (		2			2

y			= y (t),		y	2
	1		1

y2 y3 ,

=( 1 1

) y (		2
1		2
= y	(t),
2

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

) y (			) y	2	(			) y	,
1	1	1		2	1	1	1	3

	2	) y	2	(		2	2	2	) y	,
									3
y3 = y3 (t),					t I = [0, ],					(2.116)

where		0,
where
A	( )).
1

(	2	) > 0,		(		) > 0
1	2		1	1	2

(Hurwitz conditions for matrix

3. Improper integrals. As follows from Theorem 9, improper integral

where I11,	I12 is equal to:

		(t)dt =		*					*		*	(t) 13
I11 = 1 ( 1 ( )) 11 1		(t)dt =	y		(t) 11 y(t) y						(t) 12 y(t) y	(t) 13
0			0
				( )
			1
		=				(	)	d	1	< ,
					1	1		11	1
				(0)
			1
		t I are denoted by identities (2.116), matrices

where 1 ( (t)), 1(t),

I	= I	I	,
1	11	12

y(t) dt =

133

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

																											1
															1 1														( 1 1 )					0
															1 1												2		( 1 1 )					0
																											2
															1
															1
												=				(							)							0				0					,
												=			2	(							)							0				0					,
								11							2				1			1																11
																		0												0				0


					2 1 2 1 1															2 1 2 1 1													2 1 1 1

		=																																														,
				12								1		1										1				1								1				1						11
												0													0												0


									1				1				1						1				1				1		1				1					1
									1				1				1						1				1				1		1				1					1
					=													1												1											1				11	;
				13							1			1				1				1				1				1			1			1					1				11

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

I	=			( ( ))								(t)dt =					[ y				*	(t)				y	(t) y					*	(t)										*	(t)					y(t)]dt =
I	=		2	( ( ))						2		(t)dt =					[ y					(t)		21		y	(t) y						(t)	22	y(t) y									(t)			23		y(t)]dt =
12			2			12				2														21										22													23
		0															0
							2		( )
							2
					=							2	( 2 ) 12d 2										< ,						1 = diag ( 11, 12 ),
							2		(0)
							2

where

										0			1	(
										0			2	(	2		2
								1					2
						=		1	(			)
						=		2	(			)
					21			2		2	2				2	2
					21					2	2				2	2
										0					0


	(			2			2 )				2 2
		2	2 2 2 2							2 2 2 2 2
22 =		2	2 2 2 2							2 2 2 2 2
22 =					0								0
					0								0



				2			2 2				2 2 2
			2		2		2				2 2 2
23	=		2		2		2				2 2 2
23	=		2		2		2				2 2 2
			2		2		2				2 2 2

)
)	0

	0		,
	0		,
		12
		12
	0

2		2
2			2
2			2
	2		2	2	12	,
		0			12
		0

2 2 2
2	2	2
2	2	2		.
				.
2	2	2	12
2	2	2

The improper integral

I3 = [ y* (t)N2 y* (t)N1 ][H1 y(t) A12 y(t)]dt =

134

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


			=	[ y		*	(t)P y			(t) y				*	(t)P y(t) y						*	(t)P y(t)]dt = 0,
			=	[ y			(t)P y			(t) y					(t)P y(t) y							(t)P y(t)]dt = 0,
								1								2								3
				0
where
		0			0			1	N
		0			0			2	N		21						N							N							N
								2											21								21							21
								1									N22							N22							N22
		0			0			1	N
P =		0			0				N
											22																											,
								2			22	, P =																										,
1	1			1				2						2		N		N					N			N				N			N
																N	23	N		11			N		23	N				N		23	N			13
		N21			N22			N23									23			11					23				12			23				13
	2	N21		2	N22			N23
	2			2


		N						N					N				N			N
		N								11			12				11				13
										11			12				11				13
			11						2				2				2			2								N11						N21
									2				2				2			2								N11						N21
		N		N													N			N
		N		N					N								N			N								N12						N22
P =		11			12				N								12				13							N12						N22
		11			12												12				13

		2		2								12					2			2			, N			=				, N	2	=						.
3		2		2													2			2					1				N		2				N
		N		N				N					N																N						N		23
		N		N	13			N					N	13															13								23
		11			13					12				13			N13
		2		2					2				2				N13
		2		2					2				2

Then the improper integral


I		= I	I		=		[ y		*	(t)R y			(t)		y	*	(t) y(t) y							*	(t)W y(t)]dt =
I	5	= I	I	3	=		[ y			(t)R y			(t)		y		(t) y(t) y								(t)W y(t)]dt =
	5	1		3								1						1									1
							0
						( )										2	( )
					1											2
			=					(			)	d		1				2	(	2	)	12	d			2	< ,
							1			1		11		1				2		2		12				2
						(0)										2	(0)
					1											2

(2.117)

where

														R =								21		P =
														1				11				21			1
				(				)												1	(					)							1		(					)			1	N
				(			1	)				11									(	2				)				12					(		1			)		11		N	21
							1			1		11								2		2		2						12			2				1		1			11	2		21
	1									1										2													2										2
	1	(				)				1	(				)									(										)									1	N
=	2	(				)				2	(				)									(										)									2	N
=	2		2		2			12		2			1	1		11													2				2			12							2		22
			2		2			12					1	1		11													2				2			12									22
							1		N21																				1			N22											1	N23
							2		N21																				2			N22											2	N23
							2																						2														2


															N		21		b					N			21				b						N		21		b
																	21			1							21					1							21			1
															N22 b2									N22 b2													N22 b2
		1 = 12 22 P2 = N														23		N					N		23			N 2								N		23		N			,
																23				11					23							1						23				13

   ,

b1 = ( 2 2 ) 12 (2 1 2 1 1) 11,

b2 = ( 2 2 2 2 2 ) 12 ( 1 1) 11,

135

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

					N11 a1						N11			N12		a1	N11		a1		N13

											2			2			2			2
											2			2			2			2
W =			P =		N11			N12		a	N a						N12		a		N13	,
W =			P =	2						a	N a						2		a			,
1	13	23	3	2		2			1			12			1		2	1		2
					N					N	N					N
					11		a1			13	12		a1			13	N13 a1
				2			a1		2		2		a1		2		N13 a1
				2					2		2				2

a1 = ( 2 2 2 ) 12 ( 1 1 1) 11.

		Here	11	,	12		,		N		,	N			,		N	13	,		N	21	,	N	22	,	N	23		are arbitrary parameters,
			11		12				11				12					13				21			22			23
a	feedback			matrix.								If the						equalities							are			satisfied					N	22	b = N				21
a	feedback			matrix.								If the						equalities							are		,	satisfied						22			2		21
N		N = N						b ,				N				N					= N				b		,										=		.
																												then			the		matrix					*
	23	11				21			1					23				12						22		2		then			the		matrix				1	1
from (2.117) it follows
																													( )
																				(t)W y(t) dt =								1
				I 5 =					*										*													(	)	d
				I 5 =					y	(t)R y(t)							y															(	)	d		1
									1			1										1									1	1	11			1
								0																					(0)
																												1

S1 isb1, Then

			2	( )																		d				1
			2
							(						)		d												*	(t) y(t)						dt < ,
						2	(				2		)	12	d		2										y	(t) y(t)						dt < ,
						2					2			12			2									2				1
			2	(0)																	0	dt				2
				(0)																	0

due to the fact that	\|	y( ) \| c									< ,				\| y(0) \| c									< . Matrix							R = R					*	0,
due to the fact that	\|	y( ) \| c									< ,				\| y(0) \| c									< . Matrix							R = R						0,
								2															2									1			1
	( )									11			0,					11		(			)(						2		2	)	12
				1		1				11								11			1					1			2		2		12
	1														1												2
	1	(							)						1	(							)				0,			\|	R \| 0.
		(			2			2	)			12				(			1				)		11		0,			\|	R \| 0.
	2				2			2				12			2				1			1			11							1
	2														2

(2.118)

when

(2.119)

Consequently, when performing inequalities (2.102) from (2.101) it follows that the improper integral

					y				(t)W y(t) dt < .
I		=	*	(t)W y(t)dt		*	(t)R y	*
I	50	=	y	(t)W y(t)dt			(t)R y	(t) y
	50			1			1		1

(2.120)

If arbitrary parameters and elements of feedback matrix S1 are selected so that: a)

*	*	*		3
		y W y = 0,
either W1 = W1	> 0; b) either W1 = W1 0 (surface		y R		does not contain
		1

whole trajectories), then all conditions of theorem 6 are satisfied. Consequently, in the

sector [ , 0 ],

0 = 0 2 ,

2 > 0

is a

diagonal matrix with arbitrarily small

positive elements, Iserman’s problem has a

solution. Limit values 01,

02 are

determined from the Hurwitz

condition

for

polynomial 2 ( ) from (2.115) and

136

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

condition

(		2	) > 0,
1		2

	(		) > 0.
1	1	2

For positive definiteness of the matrix

W	= W	*	> 0,

1	1

it is necessary and sufficient that the following inequalities are fulfilled:

N a > 0,		(N a )( N a ) (				N11	N12	a )2	> 0, \| W \|> 0.
N a > 0,		(N a )( N a ) (						a )2	> 0, \| W \|> 0.
11	1	11	1	12	1	2	2	1	1
						2	2

Absolute stability. Below we present the results of study of absolute stability of equilibrium state of the system (2.111) with non fixed elements of feedback matrix

S1.

The improper integral

																																																										*																			(t) dt
			I			=											(					(t))												1								( (t))																*	( (t))
																																		1
				2												1																		01																								1
				2												1				1										21				01							1																	1												21					1
										0
																																																																														(t) dt =
												(						(t))													1								(					(t))														(					(t))
										2		(					2	(t))							22						02						2		(				2	(t))												2		(				2	(t))												2
										2							2								22						02						2						2													2						2								22					2
					0
										y																																																									y(t) dt 0,
					=									*		(t) y(t) y																				*		(t)																				*		(t)
					=											(t) y(t) y																						(t)						2	y(t) y															(t)						3
																					1																							2																						3
								0
where
															1											0									0
															01											0									0																														1
									21						01																																										2								1												(					)
																													1																												2								01								21				(				1	)
											0																		1						0																										21				01								21								1						1
		=									0												22						02						0														=						2									1															(											)
		=																					22						02											,						2			=						2							22		1										22					(			2							2	)
			1								0															0									0											2																22				02								22								2							2
											0															0									0																																					0
																																																																								0


		2									1											(																)					2											1											(																						)
		2									01									21		(							1						1			)					2						21					01									21		(						1														1		)
						21					01									21									1						1														21					01									21								1									1					1
			2							1																																											1
										1											(													)										2									1											(																					)			,
					22					02									22		(					2							2	)										2				22				02										22		(				2									2						2		)			,
					22					02									22							2							2															22				02										22						2									2						2
																		0																																														0
																		0																																														0

																																					c11									c12										c13
																												3
																												3					=				c21									c22										c23					,
																																					c									c										c
																																									31							32									33
						c =															1				(							1	)											21							22							1				(					2									2			)		22		,
							11										21			01												1								1				21							22							02									2									2					22
c = c			=										1							1							(2																						)													1					1			( 2																						))			,
c = c			=																								(2																						)																					( 2																						))			,
12		21						21					01							2				21									1								1						1									22					02						2													2						2					2			22
c = c =													1							1							(2							2																)													1						1			(2									2												)			,
c = c =																											(2							2																)																						(2									2												)			,
31	13							21					01							2				21									1										1						1									22					02						2											2								2				2		22
					c		22			= 1																				21		(								1	)												22							1			(						2									2		)		22		,
							22									21				01										21										1					1								22						02										2									2				22

137

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

c = c		= 1						1			( 2				2					)				1		1	(2					2				))		,
c = c		= 1									( 2				2					)				1			(2					2				))		,
32	23			21	01			2		21				1	1				1				22		02	2			2		2				2		22
	c =					1				21		(	1				1	)			22	1			(	2		2		2	)		22	.
		33			21	01				21			1		1		1				22		02			2		2		2			22
The improper integral


	I	4	= I I			2	I		3	= [ y* (t)R y						(t) y* (t) y(t) y* (t)W y(t)]dt
		4		1		2			3						0										0					0

2		( )
2	i
			i ( i ) 1i d i	< ,	1 = diag ( 11		, 12 ),		2	= diag ( 21, 22 ),
i=1		(0)
	i
where
					R = R			=
					0	1	1

(2.121)

			(			)							1				1	(				)					1	(					)		1	N
			(	1		)			11			21	01					(	2		2	)	12					(		1		1	)	11		N	21
				1		1			11			21	01				2		2		2		12				2			1		1		11	2		21
	1									1							2										2								2
=	1	(				)				1	(				)			(						)							1				1	N
	2	(				)				2	(				)			(						)											2	N
	2		2		2		12			2		1		1		11				2			2			12			22		02				2		22
			2		2		12					1		1		11				2			2			12			22		02						22	,
						1		N21														1			N22										N23
						2		N21														2			N22										N23
						2																2

															=				2	=
														0			1		2
	N			b 2					1			( )						N				b 2					1					(				)
	N			b 2								( )						N				b 2										(				)
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
= N				b	2				1			(					)	N				b	2					1					(			)
= N		22		b	2	22			02		22	(	2			2	)	N			22	b	2		22			02			22		(	2	2	)
		22		2		22			02		22		2			2					22	2			22			02			22			2	2
					N			N															N				N
					N		23	N		11													N			23	N
							23			11																23			12

				N			b 2						1							(								)
						21	1		21				01					21						1		1	1
						21	1		21				01					21						1		1	1
				N			b		2					1							(							) ,
				N		22	b		2	22				02					22		(		2			2	2	) ,
						22	2			22				02					22				2			2	2
									N							N
									N				23			N
													23						13
									W0 = W1 3 =
											N		11				N													N
	N11 a1 c11												11						12			a1			c12					11		a1
	N11 a1 c11											2						2				a1			c12				2			a1
												2						2											2
	N11		N12																											N12
=	N11		N12		a c							N a c																		N12	a
=					a c			21				N a c													22						a
	2		2			1		21								12						1			22				2			1
	2		2																										2
		N11		a1 c31											N12			a1 c32									N13 a1
		N11													N12

	2															2

c13

c23 .

c33

138

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

If the equalities are satisfied

			N						b			2											1											(															) = N												b 2																	1									(									),
			N			22			b			2						22					02								22			(				2								2			) = N									21			b 2																	01						21			(				1				1	),
						22				2								22					02								22							2								2												21								1							21					01						21							1				1
												N							N									= N												b 2																		1											(																	),
												N			23				N							11		= N									21			b 2																	01									21			(					1											1	),
															23											11											21								1								21				01									21								1					1						1
												N								N									= N													b							2											1											(																	),						(2.122)
												N				23				N									= N									22				b							2					22						02									22		(				2						2					2		),
																23										12												22								2								22						02									22						2						2					2
then matrix									= *.										Then from (2.121) we have
									0				0
																																						2						i	( )																											d					1
																																						2						i
										*	R y										*																																																													*
		I		=											y W y]dt																																				(				)						d															[			y y] < .
		I	4	=					[ y						y W y]dt																																				(			i	)				1i		d					i										[			y y] < .															(2.123)
			4									0														0																								i				i					1i							i						dt					2							0										(2.123)
								0																														i=1							(0)																										0	dt					2
								0																														i=1							(0)																										0
																																												i
		Matrix							R = R*							0 when inequalities are satisfied
									0				0
			( )																										0, ( )																																				1			(														)											1
																																																																	1																												1
							1			1			11								12						01														1								1			11							12					01											2					2				12						22			02
																	1																										1																						2
																	1			(													)										1			(										)												0,								\|	R				\| 0.													(2.124)
																				(					2							2	)				12									(				1					1	)					11							0,								\|	R				\| 0.
																	2								2							2					12						2							1					1						11																		0
																	2																										2
		When equalities (2.122) and inequalities (2.124) are valid, as follows from
(2.123) the improper integral:
																																								(t)W y(t) dt I 4
																									I				=									*																													< .
																									I		40		=						y																																< .
																											40																				0
																																	0
		Matrix						W		= W					*				> 0							then and only then, when inequalities are satisfied:
								W		= W									> 0
									0				0
												N													c							> 0,											(N							c													)( N												a c										)
													11									1						11																			11							1							11										12							1					22
																														N11										N12							a c													2				> 0,									\| W \|> 0.																					(2.125)
																																															a c									21								> 0,									\| W \|> 0.																					(2.125)
																																2										2									1					21																					0
																																2										2
		If arbitrary parameters																										11			,					12		,						21				>		0,						22					>		0,						N					,			N					,		N			13		,		N	21		,			N	22	,
		If arbitrary parameters																										11								12								21												22															11								12								13					21						22
N	23	,			,					,				1								2		,							2	,							2					are						selected													so,							that						equalities													(2.122)								and
	23		1						1					1								2									2								2					are						selected													so,							that						equalities													(2.122)								and
inequalities (2.124), (2.125) are satisfied and Hurwitz condition																																																														for the matrix																				A ( )
																																																														for the matrix																						1
( ( 1 2 ) > 0,												1( 1								2 ) > 0),																	then the equilibrium state of the system (2.111) is

absolutely stable. If when these conditions are met matrix																																																															0			= 0								2 ,									2					> 0,

then Iserman’s problem has a solution. All of the above statements remain true in case,

139

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

when matrix W = W * 0, surface		y*W y = 0, y R3	does not contain whole trajec-
0	0	0

tories.

A new method of study of absolute stability of equilibrium state of multidimensional regulated systems with limited nonlinearities in a simple critical case based on nonsingular transformation equations of motion was developed, properties of solutions and evaluations of improper integrals along the solution of the system were studied.

The class of multidimensional nonlinear regulated systems in a simple critical case, for which Iserman’s problem has a solution was determined. For this class of regulated systems necessary and sufficient conditions of absolute stability were obtained.

Key results, obtained in the work and efficiency of the proposed method are shown on examples.

Lecture 22.

Absolute stability and Iserman’s problem of multidimensional systems in a critical case. Problem statement. Nonsingular transformation

Problem statement. Equations of motion of multidimensional regulated systems in a critical case have a form:

x = Ax B ( ), = L1 ( ),

			),
= L ( ), = Sx T T ,t I = [0,			),
2	1	2

(2.126)

where

A, B,

L ,

S ,

T ,

are constant matrices of orders

n n,

n m,

m m,

m m, m n,

m m,

m m

respectively, matrix

is a Hurwitz matrix,

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

j = 1, m,

j ( A)

are eigenvalues of the matrix A.

Function

( )

= ( ) = ( ( ),...,

( )) C(R

) | ( ) = E ( ),

( ) = (

1( 1),...,

m ( m )) 1,

(0) = 0, , Rm ,

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

(

)

Rm ,i = 1, m,

,0 <

}

(0) = 0,| ( ) |

< ; , = (

,...,

) R

(2.127)

where,

E = diag ( 1

,..., m ), i

= ,

> 0 is an arbitrarily small number.

i = 1, m,

Practical systems of automatic control relate to the systems with limited resources

and for such systems function

( )

satisfies the conditions (2.127).

As follows from inclusions (2.127) equations of motion (2.126) can be repre-

sented as

z = A1z B1 ( ), = S1z, z(0) = z0 ,| z0 |< ,t I, ( ) 1 ,

(2.128)

140

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			(			)							1				1	(				)					1	(					)		1	N
			(	1		)			11			21	01					(	2		2	)	12					(		1		1	)	11		N	21
				1		1			11			21	01				2		2		2		12				2			1		1		11	2		21
	1									1							2										2								2
=	1	(				)				1	(				)			(						)							1				1	N
	2	(				)				2	(				)			(						)											2	N
	2		2		2		12			2		1		1		11				2			2			12			22		02				2		22
			2		2		12					1		1		11				2			2			12			22		02						22	,
						1		N21														1			N22										N23
						2		N21														2			N22										N23
						2																2

															=				2	=
														0			1		2
	N			b 2					1			( )						N				b 2					1					(				)
	N			b 2								( )						N				b 2										(				)
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
= N				b	2				1			(					)	N				b	2					1					(			)
= N		22		b	2	22			02		22	(	2			2	)	N			22	b	2		22			02			22		(	2	2	)
		22		2		22			02		22		2			2					22	2			22			02			22			2	2
					N			N															N				N
					N		23	N		11													N			23	N
							23			11																23			12

			(			)							1				1	(				)					1	(					)		1	N
			(	1		)			11			21	01					(	2		2	)	12					(		1		1	)	11		N	21
				1		1			11			21	01				2		2		2		12				2			1		1		11	2		21
	1									1							2										2								2
=	1	(				)				1	(				)			(						)							1				1	N
	2	(				)				2	(				)			(						)											2	N
	2		2		2		12			2		1		1		11				2			2			12			22		02				2		22
			2		2		12					1		1		11				2			2			12			22		02						22	,
						1		N21														1			N22										N23
						2		N21														2			N22										N23
						2																2

															=				2	=
														0			1		2
	N			b 2					1			( )						N				b 2					1					(				)
	N			b 2								( )						N				b 2										(				)
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
= N				b	2				1			(					)	N				b	2					1					(			)
= N		22		b	2	22			02		22	(	2			2	)	N			22	b	2		22			02			22		(	2	2	)
		22		2		22			02		22		2			2					22	2			22			02			22			2	2
					N			N															N				N
					N		23	N		11													N			23	N
							23			11																23			12

			(			)							1				1	(				)					1	(					)		1	N
			(	1		)			11			21	01					(	2		2	)	12					(		1		1	)	11		N	21
				1		1			11			21	01				2		2		2		12				2			1		1		11	2		21
	1									1							2										2								2
=	1	(				)				1	(				)			(						)							1				1	N
	2	(				)				2	(				)			(						)											2	N
	2		2		2		12			2		1		1		11				2			2			12			22		02				2		22
			2		2		12					1		1		11				2			2			12			22		02						22	,
						1		N21														1			N22										N23
						2		N21														2			N22										N23
						2																2

															=				2	=
														0			1		2
	N			b 2					1			( )						N				b 2					1					(				)
	N			b 2								( )						N				b 2										(				)
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
			21	1		21			01		21		1		1						21	1		21			01			21				1	1
= N				b	2				1			(					)	N				b	2					1					(			)
= N		22		b	2	22			02		22	(	2			2	)	N			22	b	2		22			02			22		(	2	2	)
		22		2		22			02		22		2			2					22	2			22			02			22			2	2
					N			N															N				N
					N		23	N		11													N			23	N
							23			11																23			12