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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

Then in the sector [0, 0 ],

	0	=	0	,
	0		0

= diag(	, ,	m	),
1		m

	i	> 0,
	i

i = 1, m

are

arbitrarily small numbers, Iserman’s problem		has a solution,
		0 = diag( 01, , 0m ),
	0 = diag( 01, , 0m ) the limit value of the matrix	0 = diag( 01, , 0m ),

from the Hurwitz condition for the matrix A B 0S.

Proof. From (2.28) when we have the condition (2.29), we get

where found

	5				1
	5			*	1			*
I		=	[ y		(t)R y(t) y					(t
			0
						d			1
						d	[		1
						dt	[		2
					0	dt			2
					0
due to the fact that					\| y( ) \| c2 < ,
then

			m		( )
	1		m	i		i
	1					i
)W y(t)]dt							(
			i=1		(0)
				i
*	(t) y(t)]dt <			,
y	(t) y(t)]dt <			,
	1
\| y(0) \| c2		< .	As matrix

i	)	1i	d	i

T =	1		*	) 0
T =		(W W		) 0
1	2	1	1
	2

= V

( y(t))dt =

(t)T y(t)dt I

where V1( y(t)) =

(t)T1 y(t),

t I. As surface V1

does not contain whole

( y) = y T1 y = 0

trajectories, then repeating the proofs of Theorem 5, we get

lim y(t) = 0.

As follows

from [14] the sufficient condition for the absence of whole trajectories

on the set

is the fulfillment of inequality

V1( y) = y T1 y = 0

y(t) = 2 y

(t)T [ Ay(t) B ( (t))]

t I = [0, ).

It can be shown that for the system (2.10) the given inequality is performed. The theorem is proved.

As follows from Lemma 4, equation (2.10) can be represented as

H y = A1H y A2 H y ( ),			H y = A3H y A4 H y,			( )	,
0	0	1	1	0	1	0

(2.30)

where A11 = ( A1, A2 ),

m m,

m (n m),

lows from equality

A12 = ( A3 , A4 ),

m) m,

(n m) (n

A1,


	A4
A2 , A3 ,	A4	are matrices of orders

respectively. The equation (2.30) fol-

H0 y
			A1
y =		=
	H1 y
		A3

A 2 A4

H0 y			Im

H	y	O
1			n m,m

( ), ( ) .0

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

Theorem 7. Let

the

conditions

of Theorem 6

be satisfied, where

the matrix

T11

surface

V1 ( y) = y

0 y = 0 does not contain whole

T1 = H0T11 H0

= T11

H0T11H

trajectories, the matrix

is a Hurwitz matrix.

Then in the sector

[0, 0 ],

0 = 0 ,

> 0

is a diagonal matrix of order

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

Proof. As

follows

from the

condition

the

Theorem, the

inequality holds

< ) , where V1

is a matrix of order

(n m) (n m),

(I50

( y) = y H0T11H

0 y, T11 = T11

V (y) > 0, y,

y R

V (0) = 0,

except for the surface V (y) = 0, which does not con-

tain whole trajectories. Then according to the statement of the Theorem lim

y(t) = 0.

Consider

the

second

equation

from

(2.30).

If the

matrix

of order

(n m) n m is a Hurwitz matrix, then when

lim

y(t) = 0,

limit

lim H y(t) = 0.

As y(t) = (H0 y(t), H1 y(t)),

t I ,

then

lim y(t) = 0,

, 0 .

The theorem is proved.

It should be noted that matrices 1, N1,

2 , S

ensure the fulfillment of conditions

R = R* 0, = *

T =

W * ) 0.

As follows from Theorem 6, Iserman’s

problem has a solution. Consequently, Theorem 6 gives necessary and sufficient con-

ditions of absolute stability, and matrices	N	,	N	2	expand areas of absolute stability in
	1			2

space of parameters.

For multidimensional regulated systems with limited nonlinearities we get improper integral I1 I2 < without any additional requirement on values of constructive parameters of the system.

Lecture 17.

The solution of model problems in the general case

The constructiveness of the proposed method for solving the Iserman’s problem is shown in examples.

1. Iserman’s problem for a one-dimensional system.

The equation of a one-dimensional system has the form

x	= x	x		,	x	2	= 2x					1.03x		2	0.03x		0.75 ( ),
1	1		2			2				1				2		3
		x			= 0.01x					2	1.01x				0.25 ( ),
			3							2			3
	=		x		x			2	x				, ( )			,	t [0, ).
		1		1			1	2		1		3			0

(2.31)

For this example matrices A, B, S are equal:

	1	1	0	0

A =	2	1.03	0.03 , B =	0.75	, S = ( 1	, 1	,	1 ).
	0	0.01	1.01
	0	0.01	1.01	0.25

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

the equation (2.31) in vector form can be written as

x = Ax B ( ), = Sx, ( )	.
0

А. Nonsingular transformation. The characteristic polynomial of the matrix A is equal

( ) =\| I		A \|=	1.04	0.98 = a		a a		,
		3	2	3	2
	3			2		1	0

where a0		= 0.98, a1 = 1,	a2 = 1.04.		Matrix		A is a Hurwitz matrix.
	Define the vectors		3		3	, 3	R	3		*
			1 R	, 2 R					from the conditions 1
*				*					*	*
	B = 0.			1	= (0, 1, 1),
3		As a result, we get						2 = (0,1, 3),		3 = (1,1,

transformation

B = 1,		*	B = 0,
		2

3).	Matrix of

	0000	0000	1111

RRR=R=\|\|==\|\|\|\|\|\|, ,,,, ,,, \|\|=\|\|\|=\|\|=\|= 1111		1111	1111, ,\|,,R\|\|R\|R\|=R\|=\|=\|=44440.0.0.0.
1 1112 2223 333

	1111 3333 3333

Consequently, the rank of the matrix

is equal to

n = 3.

Matrices

	0		1	1			0	1	1
K = R* =				3		K 1 = R* 1 =	3/4
		0	1		,			1/4	0 .

		1	1	3			1/4	1/4	0

Matrices

A, B, S

are equal

	1.04		2	2				1
A =		0	1	2	= KAK 1
A =		0	1	2	= KAK 1	,	B = KB =		0	,

		0.75	0.25	1					0
		0.75	0.25	1					0

S = SR 1 = ( 34 1 14 1, 1 14 1 14 1 , 1 ),

A11	= ( 1.04 2	2), A12		0	1	2	H		= (1 0 0),	H		0	1	0
A11	= ( 1.04 2	2), A12	=			,	H	0	= (1 0 0),	H	1	=		,
				0.75	0.25			0			1		0
				0.75	0.25	1						0	0	1
where	y = Kx, x = K 1 y. Then equation (2.31) by nonsingular transformation is redu-
ced to

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

y	= 1.04 y			2 y			2	2 y			3	( ),						y		2	= y				2		2 y			,		y	3	= 0.75y							0.25y	2		y	,
1			1				2				3									2					2				3				3						1			2		3
			= ((						3					1			) y	(										1				1			) y				y		),
			= ((													1	) y	(						1										1	) y		2		y	3	),
									4	1				4		1	1							1				4	1			4		1			2	1		3
									4					4														4				4
= (		3		1		) y			(						1				1				)( y					2 y				)				( 0.75y					0.25y			y		),
= (					1	) y			(				1									1	)( y				2	2 y				)				( 0.75y					0.25y		2	y		),
		4	1	4	1		1						1		4		1		4			1					2				3				1				1				2	3
		4		4											4				4
							( ) = y								1.04 y						2 y					2		2 y			,	t [0, ).
														1					1							2			3

(2.32)

Б. Solution properties. Matrix

	1		1		0
A ( ) = A B S =	2 0.75		1.03 0.75
A ( ) = A B S =	2 0.75		1.03 0.75	0.03	0.75	.
1		1	1		1
	0.25		0.01 0.25	1.01	0.25
	0.25		0.01 0.25	1.01	0.25
	1		1		1

Characteristic polynomial

			3					2	(1 0.75 1 )
2 ( ) =\| I3 A1 ( ) \|= (1.04 0.75 1 0.25 1 )									(1 0.75 1 )
(0.98 0.765			0.75 0.25			3		2				,
(0.98 0.765			0.75 0.25		) = a			2	a1 a0			,
		1	1	1
where
a	2	= 1.04 0.75 0.25 ,				a1	= 1	0.75			,
			1		1					1
	a0 = 0.98 0.765			0.75 0.25						.
			1			1			1

(2.33)

The limit value

	0

is determined from the condition of Hurwitz for polynomial

2 ( ).

В. Improper integrals. As follows from Theorem 2 the improper integral

where

I	1	=
	1

( (t)),

(t)

y(t) y

(t)

y(t)]dt =

( (t)) dt =

[ y

(t) y(t) y

( )

( ) 1d < ,

(0)

t I is determined by identities from (2.32). Matrices

(t),

		(1)		(1)		(1)				(2)		(2)	(2)			(3)	(3)	(3)
		11		12		13				11		12	13			11	12	13
1	=	(1)21		(1)22		(1)23	, 2		=	(2)21		(2)22	(2)23	,	3 =	(3)21	(3)22	(3)23	,
		31		32		33				31		32	33			31	32	33
		(1)		(1)		(1)				(2)		(2)	(2)			(3)	(3)	(3)
(1)	= 1 (		3	1	1	1 ),		(1)	= 0,				i 1,		j 1;
(1)			3		1			(1)
where 11			4		4			ij			i, j = 1,3,
											i, j = 1,3,

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

11(2)

(2) 13

( 0.75 1 0.78 1 0.26 1

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

(2) 121,

=	( 0.75	1	1.25	0.75	1	),
1		1	1		1

j 1;

(3)	= 0.78		,
	= 0.78		,
11	1	1

(3)	=	( 1.14 0.13 0.13			),
				1
12	1	1	1

(3)	=	(0.75		0.26	0.26		),
			1			1
13	1			1

(3) 21

(3) 31

=	,
(3)
12

=	,
(3)
13

	=
(3)
22

	=
(3)
32

1	( 1.5	0.5
1	1	1

		,		=	(
(3)			(3)
	23		33	1	1

0.5	1	),
	1

	1	),
	1

(3)	=		(0.75		0.75	0.75		),
		1		1			1
23					1

Improper integral

													n				n																m					m
														11			12																			11			12
														11			12																			11			12
I	3	=		[( y		y		2	y	3	)			n			n		( y							y		2		y )				m				m			]
	3				1			2		3				21			22						1					2		3						21			22
				0										n			n																	m				m
														n			n																	m				m
														31			32																			31			32
														y		y			y
															2			2					3								dt =
																															dt =

										y			0.75y 0.25y													y
										y			0.75y 0.25y											2		y
											3						1							2						3

				=	[ 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
											m				m											n						n
														11			12												11				12
														11			12												11				12
							N =						m		m			,	N			2	=				n					n			,
									1					21			22					2							21				22
													m		m												n					n
													m		m												n					n
														31			32												31				32
											0						n11/2													n12		/2

					P1		=		n11/2								n21									(n22 n31)/2											,
									n			/2		(n			n			)/2										n
										12					22			31													32
					0.75n									n				0.25n																	2n			n
								12									11									12											11			12
								12									11									12											11			12
P =			0.75n				m							n			0.25n						22		m							2n					n			m		,
2					22					11				21									22							21					21			22			31
			0.75n				m							0.25n					m							n						n				m					2n
			0.75n				m							0.25n					m							n						n				m					2n
					32					12				p			32		p				22			p				31			32				32				31
														p			(3)		p		(3)					p		(3)
																11				12							13
																11				12							13
												P = p					(3)		p		(3)					p		(3)		,
												P = p					21		p		22					p		23		,
													3				21				22							23
															p		(3)		p		(3)					p		(3)
															p		31		p		32					p		33
																	31				32							33

where

p	(3)	= 0.75m	,

11		12

p	(3)	=	1	( m	0.25m	0.75m		),

							22
12			2	11	12

p13(3) = 12 (2m11 m12 0.75m32 );

p(3)	= p(3)	,	p(3)	= m	0.25m ,	p(3)	=	1	(2m	m	m	0.25m );
p(3)	= p(3)	,	p(3)	= m	0.25m ,	p(3)	=		(2m	m	m	0.25m );
21	12		22	21	22	23		2	11	31	22	32
								2
p(3)	= p(3)	,	p(3)	= 2m	m .
32	23		33	31	32

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

Then the improper integral

= I

[ y

(t)R y

(t) y

(t) y(t)

(t)W y(t)]dt =

( )

= ( ) 1d < ,

(0)

where R1

= P,

1 = 2

P2 , W1 =

P3.

Transformation of matrix

W =

P . If we select elements of the matrix

equal (v. (2.28))

to be

		m	= 1.04			(					1					1			),			m			= 0.75k 1.04								,
		m	= 1.04			(		1						1				1	),			m			= 0.75k 1.04								,
		11			1			1			4			1		4		1					12					1		1		1
											4					4
		m	= k 2				(					1					1				),			m		= 0.25k			2					,
		m	= k 2				(			1										1	),			m	22	= 0.25k			2					,
		21				1				1			4		1		4			1					22			1			1	1
													4				4
		m		= 2k 2					(						1					1				), m				= k 2			,
		m		= 2k 2				1	(			1											1	), m			22	= k 2		1	,
			31					1				1			4	1				4			1				22	1	1	1
															4					4
where k	> 0,	k > 0 are any numbers, then the matrix																							W		has the form
1																									1
					0.5625k								0.1875k														0.75k
								1													1							1
								1													1							1
		W =			0.1875k							k 0.0625k													2k 0.25k .
		1							1													1							1
					0.75k							2k				0.25k											k	4k
					0.75k							2k				0.25k											k	4k
							1															1					1

(2.34)

Notice,	that	matrix				W1 0	for	all
*		2 y	)	2	k (0.75y		0.25y
y W y = k( y	2							2
1		3			1	1

	k		> 0,
		1
y	)	2	0.
y	)		0.
3

k > 0.

Matrices

Quadratic form

R1,

are equal

			3		1					n11/2				n12/2
	1	(		1		1 )
	1	(	4	1	4	1 )
			4		4
													1
R =			n /2							n			1	(n n ) ,
R =			n /2							n		2		(n n ) ,
1				11						21		2		22	31
			n12/2						1	(n22 n31)				n32
									1
									2
									2
								11		12	13

				1	=			21		22	23	,
							31			32
							31			32	33

where 11 = 0.75 1 0.78 1 0.26 1		0.75n12,		12 = 1( 0.75 1 1.25 1 0.75 1 n11 0.25n12 ),
13	= 1( 1 1 1) 2n11 n12;	21	= 0.75n22 m11,		22 = n21 0.25n22 m21,
23	= 2n21 n22 m31; 31 = 0.75n32 m12 ,		32	= 0.25n32 n31	m22 , 33 = n32 2n31 m32 ,

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

m	,
11

m	,
21

m	,
31

m	,
22

m32 are determined by the formula (2.34). As follows from the

condition of Theorem 6 there must be fulfilled the equality equalities hold

=		.
	*
1	1

Consequently, the

0.75n22 m11 = 1 ( 0.75 1	1.25 1 0.75 1 ) n11 0.25n12 ,
0.75n32 m12 = 1 ( 1 1		1 ) 2n11 n12 ,	(2.35)
0.25n32 m22 n31 = 2n21 n22 m31.
Consider a special case, when n	= 0, n	= 0. In this case from (2.35) we have
11	12

0.75n	m	=		( 0.75					1	1.25					0.75			1	),
22	11		1						1					1				1
0.75n			m			=		(		1			1			1	),
	32			12		1				1			1			1
0.25n		m			n		= 2n					n				m .
	32		22			31					21			22			31

(2.36)

where

m11,

m12 ,

m22 ,

m31

are determined by the formula (2.34). The solution of the

system of algebraic equations (2.36) with respect to variables	n	,	n	,	n21	is:
	22		32		n21	is:

n		=	(0.38666...		1	2.0133...				0.6233...	1	),
22		1			1				1		1
n	= k			( 0.05333...			1	1.333...		1.333...			1	)
32		1	1				1			1			1

n	= 0.125k	k	(0.25	0.25	1	) n	n .
21	1	1	1		1	31	22

(2.37)

(2.38)

Hence in particular, when

n	= n	,
31	22

the matrix

		(	3		1		)	0	0
	1	(				1	)	0	0
	1		4	1	4	1
R =			4	0	4			n
R =				0				n	0 .
1								21
				0				0	n
									32

(2.39)

				*	0 when
As follows from (2.39) matrix			R1 = R1
1 (	3	1	1	1 ) 0,		n21 0,	n32 0.	(2.40)
	4		4

From ratios (2.37)–(2.40) it follows that the set of values 1, 1, 1 for which Iserman’s problem has a solution belongs to the set

= {( 1, 1, 1 ) / 1 ( 34 1 14 1 ) 0, n21 0, n32 0},

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

where values n21,			n	are defined by formulas (2.38), (2.37), respectively, when the
where values n21,			32
surface	y Wy = 0 does not contain whole trajectories. The limit value							0 correspon-
	*
ding to	the	triple	(	, ,	1	) is determined from the condition	of	Hurwitz	for
			1	1	1
polynomial		2 ( )	from (2.33). When these conditions are met in the sector [0, 0						]

Iserman’s problem has a solution. The sufficient condition of that surface y*Wy = 0

does not contain whole trajectories is

(t)Wy

(t)

0, t I = [0, ).

In this case, phase

coordinates of the system (2.31) stitch surface

y Wy = 0. For this example specified

condition is performed.

In particular triple (

0.02

= 0.02) .

Indeed,

(	3		1		) = 0.01		> 0,
	4	1	4	1		1

	1	> 0,
	1

n	= k	0.07111...	1	> 0,
32	1		1

n	= 0.125k	k 0.00333...	1	> 0.
21	1		1

As values

a2	= 1.04 0.01 ,

= 1 0.5 ,

a0	= 0.98 0.5 ,

then the limit value

	0	= 1.96.
	0

In the sector

[0, 1.96 ],

where

> 0 is an arbitrarily small number, Iserman’s problem has a solution.

2. Iserman’s problem for a two-dimensional system. The equation of a regulated system has the form

x		= x						x			(				),		x		2	= x				x		2			(	2		) (							),	x		= x		2	x		,
1					1					3		1		1					2				1			2		2		2					1			1			3			2		3
									1	= 1x						1	x	2			1		x ,				2	= 2 x							2	x	2			2	x	,
( )									1				1			1		2			1			3			2				1				2		2			2	3								,
( )							= { ( ) = ( ( ),																( )) C(R								, R			)/0 ( )															,
																													2				2															2
0	(				0								,		1			1			2				2		, (0) = 0,												1		1		1			01		1
0	(			)									,	R						,				R			, (0) = 0,							\| ( ) \|								, 0 <					< }.
												2							1							1
2			2			2					02	2			1							2																			*					*

For this example matrices	A, B, S	are equal:

(2.41)

1 0				1								1

A =	1	1		0	, B = (B1				, B2 ), B1 =			1	,
	0	1										0
	0	1		1								0
			0
							1
			1 ,							1	1
	B =		1 ,		S =						.
	2									2
			0			2				2	2
			0

It is necessary to select the values 1,								1,		1 ,	2 ,	2 ,		2 so that for the

system (2.41) Iserman’s problem had a solution. In other words, to find necessary and sufficient conditions of absolute stability of the equilibrium state of the system (2.41).

А. Nonsingular transformation. Define the vectors								R3	,	2	R3	,		3	R3
								1		2				3
from the conditions: а) *B = 1,		*B = 0;		б) B = 0,		*B = 1; в)			* B = 0,				3* B2 = 0.
1	1	1	2	2	1	2	2		3	1

As a result, we get

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

		1				1			0

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

Then matrix of transformation

Matrices

K =

R	*

						1		1	0				1		1	0
											K 1
= (		,	2	,	) =		0	1	0	,	K 1	=		0	1	0 .
	1		2	3
							0	0	1					0	0	1
							0	0	1					0	0	1
S	are equal

						1		0		1							1		0
		A = KAK 1
		A = KAK 1		=			1	1		1	,		B = KB =					0	1	,

							1	1										0	0
							1	1	1									0	0

		S = SK 1 =		1				1
		S = SK 1 =								1		1		=		1		1		1 ,
				2				2 2									2
				2				2 2				2			2		2			2
		1	0 1											1 0			0
A11		1	0 1		A12 = (1 1 1),						H			1 0			0		H		= (0 0 1),
A11	=		,		A12 = (1 1 1),						H	0	=					,	H	1	= (0 0 1),
												0			0 1		0			1
		1 1 1													0 1		0


S1	=	1	1		S 2	=
S1	=			,	S 2	=

		2	2
		2	2

1 2


	S = (S1, S 2 ).
,	S = (S1, S 2 ).

where to.

y = Kx,	x = K	1	y.	Then equation (2.41) by nonsingular transformation reduces

y	= y	y	(		),	y	2	= y		y	2	y		(	2	),	y	= y	y	2	y	,
1	1	3	1	1			2		1		2	3	2		2		3	1		2	3
	1 = 1 y1 1 y2 1 y3 ,								2 = 2 y1 2 y2 2 y3 , ( ) 0 ,

					,		=		,			2 ,			= 2		,		=		.
			= 1								=
where	1	= 1,				1		1		2				2		2		2		2
			1	1
Б. Solution properties. The characteristic polynomial of the matrix																		A is equal to

			( ) =\| I3 A \|=\| I3 A \|= 3										3 2 3 2,
where a0		= 2, a1	= 3, a2	= 3.	Matrices				A, A are				Hurwitz matrices. Consequently,

conditions of Theorem 1 are satisfied, estimates (2.11) – (2.13) are true and identities

1 ( 1(t)) = y1(t) y1 (t) y3 (t), 2 ( 2 (t)) = y2 (t) y1(t) y2 ( y) y3 (t), t I,

1(t) = 1 y1(t) 1 y2 (t) 1 y3 (t), 2 (t) = 2 y1(t) 2 y2 (t) 2 y3 (t), t I,

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

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

	2	(t) =	2	y	(t)	2	y	2	(t)	2	[ y (t) y		(t) y	(t)],	t I.
	2		2	1		2		2		2	1	2	3

В. Improper integrals. As follows from Theorem 2, improper integral

where	I		,	I	12	are equal:
	11				12

I		= ( (t))						1	(t)dt
11					1		11	1
				0

where matrices	,		,
11		12		13


=	[ y		*	(t)			y			*	(t)	*	(t)	y(t
=	[ y			(t)			y	(t) y			(t)	y(t) y	(t)	y(t
					11						12		13
	0
		( )
	1
=					(	)		d	1	< ,
				1	1			11	1
		(0)
	1

are calculated by the formula (2.21):

I	= I	I	12	,
1	11		12

)]dt =

I12 = 0

where 11 are equal to:

														/2				0
													1		1
													1		1
					=		11		/2						0			0	,					=		11
		11					11				1											12				11
											0				0			0
											0				0			0
																							/2
																				1					1
																				1					1
													13 = 11 1/2												0
																										/2
																				1						/2
																				1					1

	(			(t))								(t)dt =				[ y	*	(t)				y	(t) y				*
2	(		2	(t))							2	(t)dt =				[ y		(t)			21	y	(t) y
2			2					12			2										21
																0
										( )
									1
					=					2 ( 2 ) 12d 2 < ,														1		=
											(0)
									1
		21	=			,								22	=		,							23	=
		21			1						12			22		2			13					23

										0
1					1		1
1					1		1
			1				0			0	,
			1
										0
	1					1				0
	1					1		1
		1
		1
	/2 .
1

	1
	1
(t)				y		(t) y		*	(t)			y(t)]dt =
(t)		22		y		(t) y			(t)		23	y(t)]dt =
		22									23

diag (		11	,	12	).

3	according to the formula (2.21)

		0		2 /2	0				2		2 2	0				2		2		0

21	= 12		2 /2	2	0	,	22	= 12		2	2	0	,	23	= 12		2		2	0	.
			0	0	0					2	2 2	0					0	0
			0	0	0					2	2 2	0					0	0		2

From (2.24) it follows that improper integral


	3		3	N2i ( y3 y1 y2
I3 = yi N1i ( y3 y1 y2 y3 ) yi				N2i ( y3 y1 y2	y3 ) dt =
0	i=1		i=1

		*P y	y* (t)P y y*P y]dt = 0,
	= [ y	*P y	y* (t)P y y*P y]dt = 0,
		1	2	3
	0

100

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