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Казахский национальный университет им. аль-Фараби

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Chapter IV. Global asymptotic stability of multidimensional dynamic systems with cylindrical …

Theorem 1. Let the matrix	A	be a Hurwitz matrix, i.e. Re j	(A) < 0,	j = 1, n and

conditions of the lemmas 1–3 be satisfied. Then following estimates are true:

\| x(t) \| c		,	\| x(t) \| c	,	t I,	(4.14)
0			1			(4.14)
\| y(t) \| c2	,		\| y(t) \| c3 ,		t I,	(4.15)

	,	t I,	(4.16)
\| (t) \| c4

where	ci = const > 0,	c		< ,		i = 0,5.		Moreover, the functions		x(t),	y(t ), (t),
where	ci = const > 0,	i				i = 0,5.		Moreover, the functions		x(t),	y(t ), (t),
t I	are uniformly continuous.
Proof. Notice that				periodic			continuously		differentiable	function	( ) 0 is
limited, i.e. \| ( ) \| ,			0 < < ,				, Rn .		Solution of the differential equation
(4.1) with respect to x(t),					t I		has the form
								t

		x(t) = eAt x						eA(t ) B ( ( ))d , t I.
							0
								0

From here

e At c( )e(a )t

taking into

, t I ,

account 0 is an

that matrix	A	is a

arbitrarily small number,

Hurwitz		matrix and
a =	max	Re	(A) < 0,
	max	j
	1 j n

| ( (t

where

)) | , t I , we get

x(t) \| c( ) \| x		\| e	(a )t
	0

c	= const,	0 < c	< ,
0		0

c( ) \| x	0

0 < c( )

\| e	(a

< .

)t	B
)t

So,

	1
	a


x(t) \| c		,
	0

e	(a

t,

)t		1


		a
	t I ,

x = Ax B ( ),

then

\| x(t) \|	A c	0	B = c .
		0	1

Consequently,

\| x(t) \| c	,
1

t I , , 0. From | x(t) | c0 , | x(t) | c1, t, t I there follows evaluation (4.14) and uniform continuity of the function x(t), t I. By the hypothesis of the theorem

conditions of the lemmas 1–3 are satisfied, consequently, matrices A, A are similar

i.e.

	(A) =	(A),
j	j


j =	1, n	, yi = i* x, i = 1, m,			ym i = 0*i x,

i = 1, n m,

y = Kx = P* x. Then

\| y(t) \| P*	\| x(t) \| P* c	0	= c	,	t,	t I ,	\| y(t) \| P*	\| x(t) \| P* c = c		,
		0	2					1	3

t, t I. Thus, estimates (4.15) are proved. Finally from (4.13) it follows that

| (t) | C | y(t) | R | ( (t)) | Cc2 R = c4 , t, t I , , 0.

211

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

Consequently, we

x(t),

y(t ),

have an evaluation (4.10). From the limitation of derivatives

I ,

we get uniform inequalities

x(t),

y(t ), (t),

t I.

The theorem is proved.

It should

noted

that:

From evaluation

x(t) | c

t I

we have

lim | x(t) |=| x( ) | c( ) |

B = c0

virtue

injustice | x(t) |,

t I , a < 0;

lim

| y(t) |=| y( ) |=

K c

lim

| (t) |=| ( ) | c .

Lemma 4. Let the conditions of the lemmas 1–3 be satisfied. Then along the

solution of the system (4.13) the following identities hold

( (t)) = H0 y(t) An y(t),

t I,

(4.17)

(t) = A

y(t),

t I,

(4.18)

(t) = (C RAH )y(t) RH0 y(t),

t I,

(4.19)

where

A11

= (I

, O

= (O

, I

= I

A =

n m

m,n m

n m,m

A12

C C

m 1,1

m 1,n

A11 =

C =

A12 =

Cn1 Cnn

Cm1 Cmn

dm1 dmn

Proof. As conditions of the lemmas 1–3 are satisfied, then we get ratio (4.11).

Notice that

m 1

y =

Then from (4.1) it follows that

1( 1) = y1 c11 y1 c1n yn , , m ( m ) = ym cm1 y1 cmn yn ,

= cyR ( ),

H y = A12 y,

where i ( i ) = i ( i (t)),

yi = yi (t),

= yi (t),

t I. From here

i =

1, m

i =

1, n

i = 1, n,

we get identities (4.17) – (4.19). The lemma is proved.

212

Chapter IV. Global asymptotic stability of multidimensional dynamic systems with cylindrical …

Lemma 5. Let the conditions of the lemmas 1–4 be satisfied. Then

* ,

M M * , 1 , N1, N2 , of orders

n n,

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

along the solution of the system (4.13)

the identities hold.

1 d

[ y

y],

1 d

( y

My),

t I ,

2 dt

[ y

]

y y(t),

y y(t

for matrices n (n m)

	(4.20)
),	(4.21)

[ y

][H1 y(t) A12

y(t)] 0,

(4.22)

(t ) N2 y

(t ) N1

Proof. Identity (4.20) directly follows from equality

[ y

y] y y y y 2 y

My]

and

identity

(4.21)

from

equality

[ y

My y

2 y

My,

[ y

t I.

follows

from

formula (4.18)

for

any

1 y] y

1 y y

1 y,

matrices

, N

of orders

n (n m),

n (n m)

the identity (4.22) holds. The

lemma is proved.

Lemma 6. Let Hurwitz matrix, the

the conditions of the lemmas 1–3 be satisfied, the matrix

function	( ) 0.	Then function	z(t) = ( y(t), y(t)),

be a

limited,

i.e.

z(t) | a,

t I ,

continuously

differentiable,

moreover

| z(t) |=| ( y(t), y(t)) | c,

t I , 0 < a < , 0 < c < .

Proof. Let the conditions of the Lemma be satisfied. We show that the derivative

( (t)) = ( (

), ,

(

)),

t I

is limited. Indeed,

(

(t))

(

(t))

(t)

(

(t)) =

t I ,

i = 1, m.

d i

|), |

= max (|

|,|

(t) | c

, i = 1, m,

t I ,

by virtue of ( ) 0 ,

| (t) |

c , t

(see theorem 1 from [16]), then | ( (t)) | m ,

t I ,

0 < m1 < . From the first equation for identities (4.3), we have

y(t) = Ay(t) B ( (t)),

t I.

From here it follows that | y(t) |

A | y |

B | ( (t)) |

A c3

B m1 = m2 ,

0 < m2 < , t,

t I. From evaluation

| y(t) |

c2 , |

y(t) | c3 ,

| y(t) | = m1 , t I

213

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

we get | z(t) | = | ( y(t), y(t)) | a,

| z(t) | = | ( y(t), y(t)) | c,

t I. the lemma

is proved.

Lemma 7. Let the conditions of Lemma 1 be satisfied, and let besides:

1) scalar continuous function W (z) > 0, z,	z R2n , W (0) = 0;
2) \| z(t) \| a, \| z(t) \| c, t I ;


3) improper integral
3) improper integral				W (z(t))dt < .
				0
Then	lim	z(t) = 0,	where
Then	lim		where		z(t) = ( y(t), y(t)),
	t

t I.

Proof. Let the conditions of Lemmas 1) – 3) be satisfied. We show that

lim

t
k

z(t) = 0,	(lim y(t) = 0,	lim	y(t)

	t	t

Suppose the contrary, i.e.	lim	z(t)

	t
when k such as	\| z(tk ) \|

0).


0.	Then
	Then
> 0,		k =

there exists a sequence

1,2, . We choose	t	k

{t		k	},

1	t			k

t	k	> 0,
	k

	1	> 0,
	1

k = 1,2, .

z(t),

t I is continuously differentiable

z(t) | a, | z(t) |

t I ,

then

| z(t) z(t

) | c | t t

t [tk

, tk

k = 1,2, .

> tk 1

< tk 1, W (z) > 0,

z R

then

																			t		1
																			t		1
																			k	2
																				2
													W (z(t))dt								W (z(t))dt,
													0					k =1
																			t	1
																			k
																				2
where			\|	y(t) \|=\| y(tk ) y(t) y(tk ) \| \| y(tk ) \| \| y(t) y(tk ) \| c																							1	=		> 0,
																											1
																										2			0
																										2
t,	t [tk					1	, tk				1	].	We can always choose the value											1	> 0 so, that the value
						1					1
					2					2
					2					2
0 >	0.	So	\| z(t) \|					,	\| z(t) \| c,							t [tk		1	, tk			1	].
			\| z(t) \|					,
							0
																	2				2
	As the function W ( z)												is continuous on the compact set 0												\| z \| c, then there is a
number m > 0 such as												min				W (z) = m. Then the value of the integral
										01 \|z\| c
												t		1
												t
												k	2
													2
															W[z(t)]dt 1m, k = 1, 2, .
															W[z(t)]dt 1m, k = 1, 2, .
												t	1
												k
													2

214

Chapter IV. Global asymptotic stability of multidimensional dynamic systems with cylindrical …

Consequently,

	t
	t	1
		1
	k	2
		2
W[z(t)]dt			W[z(t)]dt	k( m) = .
				lim	1
0	k =1			k
0	t
	t	1
	k
		2

This contradicts condition 3) of the Lemma. The lemma is proved.

Lecture 34.

Global asymptotic stability of multidimensional dynamic systems I. Auxiliary lemmas

Based on nonsingular transformation and using properties of the solution of the system (4.1), (4.2) we can obtain estimates of improper integrals along the solution of the system (4.13), separately for two cases:

( i )d i

= 0, i1, m;

á)

( i )d i

0, i1, m;

where i	1	,		is a period of the function i ( i ).
	R		i
Theorem 2. Let the conditions of lemmas 1–3 be satisfied, the matrix
Hurwitz	matrix,			the function	( ) 0.	Then for any diagonal

A be a matrix

	1	= diag(	11	, ,	1m		) > 0
	1		11		1m
improper integral

		I1		= [ y		*	(t) 1
		I1		= [ y			(t) 1
				0

where

of order	m m,	along the solution of

y(t) y

(t)

y y

(t)

(t) y

(t)

y(t) y

(t)

y(t)]dt 0,

the system (4.13) the

4	y	(t)
4

(4.23)

																																*
						= (C R A11 )* (																										*								H * R*
				1		= (C R A11 )* (									1		1			2	)H			0	,					2	= A11					H		0		H * R*								H			0
				1										1	1		1			2				0						2				1				0				0		1			1				0
	H			* R*				H	0		,		3	= (C R A11 )*															A 2(C R A11 )*																				RH				0
				0		1	2		0				3												1		1				11												1	1			2						0
																																	*
(C R A11 )*																					* R*												*				A11 H * R*
(C R A11 )*								1		2		A11 ,			4	= H					* R*					1		A11 A11							1		A11 H * R*											1				RH		0
								1		2					4					0					1	1									1							0				1		1			2			0
					R8									= H *
H			0		R8		2	A11 ,					5	= H *					H			0	,			6		= (C R A11 )																(C R A11 ),
			0			1	2						5			0		1				0				6															1	1	2
		1			= diag ( 11 , , 1m ),											2		=			diag ( 21 , , 2m ),																		1		< ,				2					< .

(4.24)

215

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

Proof. From inclusion ( ) 0 it follows that

	i	i
	i	i	0,
1i			2i
	i	i
	i	i

where

d ( (t))

= (t) =	d ( (t))	,	= (t),	t I ,
	dt

d ( (t))

		(t)
	dt	(t)
=	dt	= (t) ,	t I.
=	d	= (t) ,	t I.
	dt

(4.25)

Multiplying identity (4.25) by																			2				(t),				we get
Multiplying identity (4.25) by																							(t),				we get
						(								̇						̇			−					̇) ≥ 0, , .
						(				̇− )(													−				2	̇) ≥ 0, , .
								1																			2
Then for any value										1i	> 0		the inequality holds
Then for any value										1i			the inequality holds
				(					̇− ̇)										(̇−												̇) ≥ 0, , .																										(4.26)
						1									1													2
From (4.26) it follows that
				(								̇											̇					̇) ≥ 0, , ,																													(4.27)
				(			̇− )											( −										̇) ≥ 0, , ,																													(4.27)
						1										1											2
where 1 = (11, … , 1 ), 1 = (11, … , 1 ) > 0, 2 = (21, … , 2 ).
Then the improper integral
		I	=		( )											( )dt =
		I	=		( )											( )dt =																								[
													*																															*	*								*
		1						1							1											2																			1		1								1
				0																																		0
																																						̇] ≥ 0,
										− ̇													̇+ ̇															̇] ≥ 0,
													1			1					2												1			2
															= (t) = (																			R								11)y(t) RH0 y(t), t I
where = (t) H0 y(t) A11 y,															= (t) = (															C				R						A		11)y(t) RH0 y(t), t I															by virtue
of identities (4.17), (4.18), respectively,																												*	=							,				*			=			.		From here due to the fact
																											1						1								2				2
that:
			*
			*							= y* (C RA11)*																			1	H						y	yHR*														1	H		y
					1	1																					1		1				0												0				1		1		0
								*											*																	*					*				*
					y					(C RA11)															1	A11y							y						H				R					1	A11 y,
																								1	1																0				1			1
		*				= y* H							*									y 2 y*													11*										y y*							11*
		*				= y* H							*		1	H					0	y 2 y*										A			11*				1		H			0	y y*					A		11*		1		A11 y,
				1									0		1						0																		1					0										1

*		= y* (C RA11)*															1							(C RA11)y y*H																						R										(C RA11)y
1	1	2													1		1					2																							0						1	1		2

			y* (C RA11)*																						RH			0	y				y				HR*														RH			0		y,
																1					1			2				0													0				1			1	2					0

216

Chapter IV. Global asymptotic stability of multidimensional dynamic systems with cylindrical …

= y						(C RA11) H y y										H	R H y
*	1	2			*				*	1	2	0		*		*		*		1	2	0
	1	2								1	2	0				0				1	2	0
	y		*				*						*	H	*	*
	y			(C RA11)					2	A11 y y				H	0	R				2	A11 y,
							1		2						0		1			2
we have an assessment (4.23) where i,											___	are denoted by formula (4.24).
we have an assessment (4.23) where i,								i		1,6		are denoted by formula (4.24).
Theorem 3. Let the conditions of lemmas 1-5 be satisfied, matrix be a Hurwitz
matrix, the function	( ) 0.						Then for any matrices N1												, N 2 of orders				n (n m),

n (n m)

respectively, the improper integral

I 2

(t)P2 y(t) y

(t)P3 y(t)]dt 0,

(4.28)

(t)P1 y(t) y

P1 N2 H1 ,

A12

P N

A .

where

Proof.

follows

from

identity

(4.22)

the

equality

holds

N2 H1 y(t) y

(t)(N

(t)N1

A12 y(t) 0, t, t I. From

A12 H1

N1 ) y(t) y

here it follows that

(t)P2 y(t) y

(t)P3 y(t) 0,

t, t I.

Consequently, the

P2 y(t)

improper integral

The lemma is proved.

Consider the case when

= 0,

i = 1, m.

(4.29)

( )d

Theorem 4. Let the conditions of Theorem 2 be satisfied, the function

( ) 0

satisfy

the

condition

(4.29), where

( ) = ( (

), ,

(

)). Then

for

any

diagonal

matrix

2 = diag(

21, ,

2m )

along

the

solution

the

system (4.13)

the improper

integral

																	( )

I	3	= [ y* (t)T			* y	(t) y* (t)T y(t) y* (t)T y(t)]dt =												* ( )	2	d
	3			1					2						3				2
			0														(0)
			*
	T = A				RH				(C R A	*		H		,		*			T = H
where	T = A			2	RH		0		(C R A	)	2	H	0	,	T2	= A11 2	(C R A11),		T = H
where		1	11	2			0		11		2		0		T2	= A11 2	(C R A11),			3
						( )						m		( )
								* ( ) 2d =							i ( i ) 2i d i .
							(0)					i=1 (0)

(4.30)

*		RH		,
	2		0
0

217

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

Proof. Notice that

		i	( )	(0)	i

(

)

(t)dt =

(

)

(

)

2i i

(0)

(k 1)

( )

(

(0)

(0) k

(0) (k 1)

( )

(

< ,

(0) (k 1)

where values

(0)	i	(0) (k 1)	i

(

)

= 0,

(

= 0,

(0)

(0) k

(4.31)

for any any i

k = ( )

0, 1, 2, ,	i	is a period of the function i ( i ). In other words, for
there is such	a	natural number k,	where the length of a segment

[ i (0) (k 1) i , i ( )]

is less than

The improper integral

* ( (t))

(t)dt =

(

(t))

(t)dt =

i=1

= [H

[(C

R A11 ) y(t) RH

0 y(t) A11 y(t)]

0 y(t)]dt =

[ y

(t)T y(t) y

(t)T

y(t) y

(t)T y(t)]dt =

( )

=* ( ) 2 d <

(0)


by virtue of inequalities (4.31), where ( (t)) = H0 y(t) A11y(t),			y = (C RA11)y(t)

t I. So, it is inequality (4.30) (see (4.32)) is proved. The theorem is proved. Consider the case when

(4.32)

RH	0	y	(t),
	0

i	i

		i	( i )d i	= i 0,	i , i R1, i = 1, m.	(4.33)

218

Chapter IV. Global asymptotic stability of multidimensional dynamic systems with cylindrical …

Define the values i ,

so that

| i ( i ) | d i , i =

i = 1, m.

i =

Theorem 5. Let the conditions of Theorem 2 be satisfied, the function ( ) 0

satisfy the condition (4.33), values i , i

such that i

i i = 0, i = 1, m. Then for any

diagonal

matrices

= (31, … , 3 ),

4 = (41, … , 4 ) > 0,

5 = (51, … , 5 ) > 0

such

that

44 5 − (5)(5) > 0,

= (

, … , ), along the solution of the system (4.13) the improper integral

( )

[ y

(t) y

(t) y(t) y

(t) y

(t)]dt

(

)

i=1

(0)

(4.34)

where

	(	) =	(	)	i	\|	(	) \|,
i	i	i	i		i	i	i

i = 1, m,

	___	Q	*	Q	M	,
	= N 1	Q		Q	M	,

1			2	3	1

	___
	= N 2	Q	M	2	,
2		1		2

___
3 = N 3	Q4	M 0 .
			__
Here matrices			N i ,

i = 1,3,

Qi ,

i = 1,4,

M	,
i

i = 0,2 are equal to:

___		*											*					__					*							___		*
		*																					*
N 1	= A11		RH												H		,	N 2		= A11					(C R A11 ),										RH			,
N 1	= A11		RH		0			(C R A11 )							H		,	N 2		= A11				3	(C R A11 ),					N 3 = H					RH		0	,
		3			0									3		0								3								0		3			0

Q = (C RA11)*								(C RA11),						Q = HR							(C RA11),					Q = (C RA11)*							5	RH		,
1							5								2			0		5							3						5		0
			*	*											*										*					*
	Q	= H	*	*			RH			,	M				*			H	,	M			= 2A11			H	,	M		= A11		A11.
	Q	= H		R		5	RH		0	,	M	0	= H					H	,	M		1	= 2A11			H	,	M	2	= A11	4	A11.
	4	0				5			0			0			0	4		0				1			4	0			2		4

Proof. Notice, that function							(	),
Proof. Notice, that function						i	i
	( )				( )
i				i
		i ( i )d i	=		i ( i )d
	(0)				(0)
i				i

From here it follows that

i= 1, m

has the property
( )
	\| i ( i ) \| d i = i i i	= 0.
(0)

	m	i		( )		m
					i ( i ) 3i d i	= i ( i (t)) 3i i (t)dt <	(4.35)
i=1			i	(0)		i=1 0
i=1						i=1 0
for any numbers 3i ,
for any numbers 3i ,		i = 1, m. This inequality follows from (4.14) after replacement of

i ( i ) by i ( i ), i = 1, m.

219

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

Consider the function

S (t) = (			(t))		(t)	2	(
						2
i	i	i		3i i	4i	i

=	2	( )			2		\| ( ) \|
	2				2
4i	i	i	5i		i	i	i		i			3i	i
			i 3i					2
			i 3i					2
			2
									(		) 0,
									(		) 0,
			4	5i		4i		i		i
				5i

	4					> 0,
				2	2
due to the fact that		4i	5i	i	3i		i = 1, m0.

)

(t)

( )

( ) |

t I , i = 1, m,

From here it follows that

	(	(t))	3i		(t)
i	i		3i	i

Summarizing by			i
Summarizing by
	*	( (t)) (t)		*
	*			*
		3

	2	( )		2	(t
	2			2
4i	i	i	5i	i

in limits from 1 to	m
in limits from 1 to
( (t)) ( (t))	*	(t
	*
4

)	(	i	)	3i		,
i		i		3i	i
in (4.36), we get
			m
) (t) =					(		i
5			i				i
			i=1

i = 1, m.

)		,
	3i i

(4.36)

i	=	(t),	t I.
i	i

Integrating by

I	4	=
	4

t	in limits from 0 to
	in limits from 0 to
				3
				3	(t)
[		( (t))			(t)
	*
0
	m
	m
			(	(t))		(t)
		i	i		3i i
0	i=1
0


	in (4.37), we get
*	( (t)) ( (t))							*	(t) (t)]dt
*								*
			4							5
		m
		m
dt =					(	(t))				(t)dt < ,
				i	i		3i		i
		i=1	0
			0

(4.37)

(4.38)

	4 4i 5i	*
by virtue of inequalities (4.35). From inequalities	4 4i 5i	i

that

(

)(

) > 0.

As identities hold

(t) = y

(t) N

( (t))

1 y(t)

(t) N 2 y(t) y

(t) (t) = y

(t)Q y(t) y

(t)Q

y(t) y

(t)Q y

(t)

(t)M

( (t)) ( (t)) = y

y(t) y

(t)M y(t) y

then we have an evaluation (4.34). The theorem is proved.

2					i = 1, m it follows
3i > 0,					i = 1, m it follows
__			y(t), t I ,
) N 3			y(t), t I ,

y	*	Q			y(t),	t I,
y		Q			y(t),	t I,
				4
(t)M			2	y(t),		t I,
			2

220

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\| y(t) \| P*	\| x(t) \| P* c	0	= c	,	t,	t I ,	\| y(t) \| P*	\| x(t) \| P* c = c		,
		0	2					1	3