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Казахский национальный университет им. аль-Фараби

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is any number.

Then the equilibrium state of the system (1.76), (1.77) is absolutely stable. If in

addition the value

0 = 0 , where > 0 is an arbitrarily small number, then in

the sector

Iserman’s problem has a solution.

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Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

Proof. Along the solution of the system (1.86) the following identity is true

(t) = a0 y (t) a1 y

(t) an 1 y

(t) ( (t)),

t I,

n 2

where

n 1

B1.

From here the identity (1.91) follows. Identity (1.92) follows from

= A1

(1.84). As (t) = y (t)

(t)

n 1

(t)

n 2

(t),

where

n 2

(t)

= (t),

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

then we get identity (1.93). The lemma is

proved.

Theorem 16. Let the conditions of the lemmas 1, 2 be satisfied, matrix

( )

be a Hurwitz matrix, the function

( )

Then for any constant matrix

order

(n 3) (n 3)

the quadratic form

(t)Q (t),

(t) = ( (t), y1 (t), , yn 2 (t)),

t I

can be represented as

(t)Q (t) = q

(t)

q y

(t) q

(t)

[

(t)F (t)],

t I ,

n 1

n 2

1 1

(1.94)

where

is a constant matrix of order

2) (n 2).

Improper integral

(t)Q (t)dt =

(t) q y

(t) q

(t)]dt l

| l

|< ,

(1.95)

n 2

1 1

[ y

(t)Fy(t)]dt = y

(t)Fy(t) |

= y

( )Fy( ) y

(0)Fy(0).

(1.96)

Proof. Let

n 2 = n .

1 It is easy to verify that:

a) if n2 , k

are odd numbers, k n2 ,

then

n1 k

yk yn yk 1 yn 1

...

yn2

;

b) if

is an odd,

is an even number,

k < n

then

n1 k 1

yn2

yk yn

yk 1 yn 1 ... 1

k 1 yn k 1

k 1 ;

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

c) if n1

is an even, k

is an odd number, k < n1, then

k 1

y y

... 1

;

k 1

n 1

n k 1

d) if

is an odd,

is an odd number,

k n ,

then

y y

...

;

n k

e) if

is an odd,

is an odd number,

s > k,

then

k s

( 1)

;

s 1

k 1

s 2

k s

f) if

is an odd,

is an even number,

s > k,

then

( 1) 2

;

s 1

k 1

s 2

g) if

is an even,

is an odd number,

s > k,

then

k s

( 1)

k 1

;

k s 1

h) if

is an even,

is an even number,

s > k,

then

s k

( 1) 2

k 1

s 2

s k

In particular, when n1 = 3 we get:

y =

( y y

y2 ); y

( y

) y

2 ; y

1 d

( y

2 );

2 dt

y y

( y

2 ); y y =

( y y

) y2

; y

( y

2 );

2 dt

Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

when n1 = 4 we get:

y	=	d	( y y		y		y	) y	2	;	y		=	d	( y		y		1	y	2	);

				4		2			3			2				2		4			3
1		dt	1				3							dt					2

( y

) y

; y

1 d

( y

);

y y

1 d

( y

);

2 dt

y y

( y y

) y

; y y

( y y

);

( y

) y

;

y y

1 d

( y

2 dt

As the quadratic form

(t)Q (t)

contains terms with constant coefficients of the

product of components of vector

(t),

then the true form is (1.94).

By integrating identity

(1.94)

taking

into

account

evaluation (1.89), where

| y(0) | m , | y( ) | m ,

obtain

the

relations (1.95),

(1.96). The theorem is

proved.
Lemma 14. Let the following conditions hold:
1) vector function y(t) = ( y1 (t), , yn 2 (t)), t I			= [0, )
t I and is continuously differentiable, moreover		\| y	(t) \| c,

2) scalar continuous function	V (x) > 0, x,	x R		n 2	, x

is bounded

t I ,
	a < ,
0,	V (0) =

| y(t) | a, c < ; 0;

3) improper integral

V ( y(t))dt < .

Then

lim y(t) = 0. t

Proof. Let the conditions of Lemmas

1) – 3) be satisfied. Let us show that

lim y(t) = 0.

Suppose

the

contrary

i.e.

lim y(t) 0.

Then there is

sequence

} I = [0, )

such that

y(t) |

> 0,

k = 1,2, .

We choose

k 1

m > 0.

y(t), t I

is continuously

differentiable and

| y(t) |< c,

t I ,

then

| y(t) y(tk ) | c | t tk

|, t [tk

, tk

k = 1,2, . Then

V ( y(t))dt V ( y(t))dt,

k =1 tk

where | ||y|y(y(t|t)t()yt|=|)(ty)=||(y|=|ty()y(t|=|t(ty)t()ty)()yty(y)(tt)y(t)t()ty)(yty()y(tyt(t|(t)yt)|()||)tt|)|||)y|y=|(y(t|t(t(yt)t)(|)t|||)|||y|y(y(|tt()tyt)()t)yy(yt(yt(tt()t)|)||)|| | y(cctc)

|===| y(>t>>)00.0...y(t

) | c

kkk k k

kkkkk

kkkk k

kkkkk

k222

00000

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

t		m
t	k	2
	k

		V ( y(t))dt V
		min
t		m
t
k		2
		2

V	=	min	V (x),
min		\|x\| a
		\|x\| a
	0

then V ( y(t))dt = . This

contradicts the third condition of the Lemma. The Lemma is proved.

Lecture 12.

Improper integrals and absolute stability in a critical case

Based on identities (1.91)–(1.93), assessments (1.88)–(1.90) and taking into account (1.94)–(1.96) we can obtain evaluations of improper integrals along the solution of the system (1.86).

Theorem 17. Let the conditions of the lemmas 11, 12 be satisfied, matrices

A ( )

be Hurwitz matrices, the function ( )

Then along the solution of the

system (1.86) improper integral

= ( (t)) (t)dt = [N

(t) N y

(t)

(t)]dt l

n 2

(1.97)

( )

( )d = c1,

| c1 |< ,

(0)

l = y* (t)F y(t) | = y* ( )F y( ) y* (0)F y(0),

| l |< ,

(1.98)

where

is a constant matrix of order

(n 2)

(n 2).

t I ,

(t),

Proof. Products ( (t)) (t) =

(t)N (t),

where

( (t)),

t I

are

defined by formulas (1.91), (1.93), respectively. Then improper integral

( )

I1 =

( )d = c1,

| c1 |< ,

(t)N (t)dt = ( (t)) (t)dt =

(0)

due to limited (t), t I ,

where N is a constant matrix of order

(n 3) (n 3).

As it follows from Theorem 2 (Q = N ), the following equality holds

* (t)N (t) = N 2 (t) N y2 (t) N

(t)

[ y* (t)F y(t)],

t I.

n 2

Then the ratios (1.97), (1.98) follow from (1.94)–(1.96). The theorem is proved.

Theorem 18. Let the conditions of the lemmas 11, 12 be satisfied, matrices

A1( )

be Hurwitz matrices, the function ( ) 1. Then for any value 1 > 0 along-

with the solution of the system (1.86) the improper integral

Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

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

=		[M		2	(t) M y	2	(t) M					y	2	(t)]dt l		0,
=		[M			(t) M y		(t) M				n 2	y	n 2	(t)]dt l	2	0,
			0		1 1						n 2		n 2		2
		0

(1.99)

(0)F2 y(0),

| l2 |< ,

(1.100)

= y

(t)F2 y(t) |0

= y

( )F2 y( ) y

where M

= M

( ),

> 0, i =

0, n 2,

is a constant matrix of order

2) (n 2).

Proof. As ( )

, then

, ,

R . Then ( )

( ),

1 2

( )

, R

(t)M (t) =

Along

the

solution

the

system

(1.86),

= ( (t)) 1

( (t)),

t I ,

where

( )(t),

(t),

t I

are

defined

(t) 1 0

formulas (1.91), (1.92), respectively. Further, in a similar way, as in the proof Theorem 3, we obtain the relations (1.99), (1.100). The theorem is proved.

Theorem 19. Let the conditions of the lemmas 11, 12 be satisfied, matrices

A	( )	be Hurwitz matrices, the function ( )	.	Then for any values
1		1

along the solution of the system (1.86) the improper integral

	0	,	1	, ,	n 2

[ (t)

y (t)

(t)]

dt =

n 2

[

(t) y

(t)

(t)]dt l

n 2

(1.101)

l3 = y	*			*	(0)F3 y(0),			\| l3 \|< ,		(1.102)
		(t)F3 y(t) \|0 = y *( )F3 y( ) y
where i ( 0 , 1, , n 2 ), i = 0, n 2, F3			is a constant matrix of order						(n 2) (n 2).
Proof. Product [ 0 1 y1 n 2 yn 2 ] =							(t) (t),	where		is a constant
			2			*
matrix of order (n 3) (n 3). . Next			by analogy, as in the proof of Theorem 4, we

obtain the relations (1.101), (1.102). The theorem is proved.

Absolute stability. Based on the results outlined above, we can formulate conditions of absolute stability of equilibrium states of the system (1.76), (1.77).

Theorem 20. Let the following conditions hold e:

1. matrices

A	( ),
1

0 <	0	<	0

are Hurwitz matrices, the function

( ) 1, > 0 is an arbitrarily small number;

2.there is a vector * Rn 2 such that

B1 = 0, A1B1 = 0, , A1n B1 = 0, A1n 1B 0;

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

							R =		, A				, , A											n 2;
3. rank of the matrix								*			*	*			*n 1		*	is equal to
3. rank of the matrix											1				1			is equal to
4. conditions of Theorems 17-19 are satisfied, and let besides:
	2	N	0	M	0	0,				2	N M			1	0,				2	N	2	M	2	> 0,
	2		0		0		0			2	1			1		1			2		2		2	2
	2	N	i	M	i	> 0,		i = 3, n 2,
	2		i		i		i

(1.103)

Proof. Under conditions 1) - 4) of the Theorem we get

	I		I		I					=					N		M			)							(t) ( N M										) y						(t)
	I		I		I					=		[(			N		M			)						2	(t) ( N M										) y					2	(t)
2		1		2					3					2		0			0			0										1	1			1			1		1
			(					N				0			) y				(t) (						N				M			) y				(t)
			(					N			M				) y			2	(t) (						N				M			) y			2	(t)
																		2																	2
(							2			2	M		2				2	2	) y					3				3			3		3	l	3				l		< ,
(						N					M								) y		2		(t)]dt								c1			l	l				l		< ,
																					2
					2			n 2					n 2				n 2				n 2									2				2 1				2		3

(1.104)

where

\| c \|	<
1

\| l	\|< ,
i

= 1,2,3,

	2

is any number.

From (1.104) taking into account first inequality from (1.103), we get

n 2

(

) y2

(t)

dt < ,

i=1

where

N M

> 0, i = 3, n 2.

Hence

2 1

in case when

2 Ni

Mi i

> 0,

i = 1, n 2,

we get

[

2 (t)dt < , i = 1, n 2.

i i

Then as follows from lemma 4 the limit lim

yi (t) = 0, i = 1, n 2. Consequently,

according to

Lemma

the

limit

lim z(t) = 0.

matrices

A ( ),

0 < 0

0 are

Hurwitz matrices, then

according to definition 7 trivial

solution

z(t) 0, t I

the

system

(1.76),

(1.77)

absolutely

stable,

where

z(t) = (x(t), (t), (t)),

t I.

Consider the case when 2 N1

M1 1

= 0, 2 N2

M2 2 = 0. In this case as proved

				y1	(t) = 0, lim y2 (t) = 0.
above lim	yi	(t) = 0, i = 3, n 2. We only have to prove lim		y1	(t) = 0, lim y2 (t) = 0.
t			t		t

Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems


As \| ( ) \| * , \| (t) \| c2 ,	t, t I is uniformly continuous by t, t I. Then
	n 1
function yn 2 (t) = a0 y1(t) a0 y2 (t) A1 ( (t)), t I			is uniformly continuous
by t due to uniform continuity of	yi (t), i =		t I. Then according to
		1, n 2, ( (t)),


the Barbalat Lemma [14; § 21, Lemma 1] equality lim yn 2 (t) = 0						entails lim yn 2 (t) = 0.
			t			t

Then when t we get	0 = a0 lim y (t) a0 lim y				(t) An 1B lim ( (t)),
		t 1		t 2		1 1 t

lim (t) =	1	lim y		(t)
t		t	1


2	lim y2 (t),	where a0 = 1 , a1 = 2 , > 0 is an
	t

arbitrarily small number, lim y	(t) = lim y		(t), lim y		(t) = lim y		(t) = 0.	Then
t 1	t	2	t	2	t	3

lim

then

y1(t

So,

y2 (t) = y2 = const, lim y1 (t) = y2 .

As the system (1.76), (1.77) is autonomous,

its

properties

solutions

do not

depend

on the

start time. Consequently,

) = y2t y1, y1

= const,

due to limited

, t, t I ,

value

= 0.

| y (t) | m

lim y

(t) = 0,

lim

y (t) = y

0 = a0

n 1

B lim ( (t)), lim (t) =

a0	=
	1

is an arbitrarily small number, then when

lim a0	= 0
t

and the

following equality holds 0 = lim ( (t)) = (	1	y	). Hence it follows that	y	1	= 0,	due
t	1	1			1
t
to the fact that (0) = 0 only when = 0, 1	is any number.
Notice that equality to zero of the sum 2 N1			M1 1 = 0, 2 N2 M2 2 = 0					is

one of the features of the critical case in comparison with the main case from [6–9].

In case

= 0 , > 0 is an arbitrarily small number, meets all conditions of

the definition 3, consequently in the sector	[ ,	0	]	Iserman’s problem has a solu-
		0

tion. The theorem is proved.

Lecture 13.

Aizerman’s problem. The solution to the model problem in a critical case

The question arises: is it possible to distinguish a class of regulated systems, for which Iserman’s problem has a solution, without resorting to checking conditions of absolute stability from Theorem 6.

Theorem 21. Let the following conditions hold:

	matrices A, A ( ), 0 <
1.	matrices A, A ( ), 0 <		0	<	0	are Hurwitz matrices, the function
	1		0		0
1	, > 0 is an arbitrarily small number;
2. there is a vector * Rn 2		such that

B1 = 0, A1B1 = 0, , A1n B1 = 0, A1n 1B1 0;

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

3. rank of the matrix

R =	*	*	*	, , A	*n 1	*
		, A				1
		1		1

is equal to n 2;

4. the conditions of the

theorem

are

satisfied. There is

feedback vector

S R

n 2

and number

such as

N 0,

2 N2 0,

2 Ni

> 0, i = 3, n 2.

Then in the sector

[ ,

> 0

is an arbitrarily small number, Iserman’s

problem

has a solution,

the

value

limiting value

Hurwitz matrices

A1( ) =

A B S			,
1	1	1

	0	.
	0

Proof. Let the conditions of the theorem hold. Then improper integral

		I =										N			(t) N y								(t) N						y		(t)
		I =	( (t)) (t)dt =							[		N		2	(t) N y							2	(t) N						y	2	(t)
	2			2							2		0						2	1	1						2	2		2
			0		N		y			0				=			c1,			\| l	\|< , \| c \|< .
					N		y	2	(t)]dt l					=			c1,			\| l	\|< , \| c \|< .
								2
					2	n 2		n 2					2 1			2				1
Hence taking into account that											2	N	0		0,			2	N 0,					2	N	2	0,			2	N	i	> 0,
											2		0					2	1					2		2				2		i

we get

i = 3, n 2,


		N	2	(t)dt <	,						N y			2		(t)dt <			,			N		y	2	(t)dt < ,
	2	N		(t)dt <	,						N y					(t)dt <			,		2	N	2	y	2	(t)dt < ,
	2	0									2	1	1								2		2		2
0									0											0

							N y			2	(t)dt < , i = 3, n 2.
						2	N y				(t)dt < , i = 3, n 2.
						2		i	i
				0
Consider two cases: 1) case when													2		N		i	> 0,	i = 0, n 2;					2) case when
Consider two cases: 1) case when													2				i		i = 0, n 2;					2) case when

	N = 0,
2	1

2	= 0,
2

In the

2	N	i

first

> 0, i

case

= 3, n 2.

the implemented inequalities


	2
	2 Ni yi	(t)dt < ,
0

i = 1, n 2.

Consequently, lim yi (t) = 0, i = 1, n 2,

the value

0 is determined from the condi-

tions of Hurwitz matrices

A ( ). Then

, > 0 and Iserman’s problem

has a solution.

In the second case implemented inequalities

N y2 (t)dt < , i = 3, n 2. Then

i i

lim y		(t) =
t	i
t

condition

0,
	i = 3, n 2.

lim yn 2 (t) = 0,

Next repeating the proofs of the Theorem 6 from the

we get lim	y1	(t) = 0, lim y2 (t) = 0. Then in the sector
t		t

[ , 0 ] Iserman’s problem has a solution. The theorem is proved.

Solution of model problems. The effectiveness of the proposed method of defining the condition of absolute stability and solution of Aizerman’s problem will be shown on examples.

Chapter 1. Absolute stability and Aizerman’s problem of one-dimensional regulated systems

Example 1. Differential equations of regulated system have a form

x	= x	( ),	x	= x		,	x	= ( ), = x			x	x	,	t I = [0, ),	(1.105)
1	1		2		3		3			1	2	3			(1.105)
where ( ) =				( ),	( )				.	For this example, the source data is as follows:
where ( ) =				( ),				1		For this example, the source data is as follows:

A = 1, B = 1, 1 = 0, 2 = 1, S = , 1 = , 2 = .

1. Equations (1.105) in vector form have a form where

z = A1z B1 ( ),

= S1z,

		1		0	0			1					x
														1
														1
A	=		0	0	1	,	B =	0	,	S	= ( , , ),	z =		x .
1							1			1				2
			0	0	0			1						x
			0	0	0			1						x
														3

Characteristic polynomial of the matrix

is equal to

( ) =\| I		2	( 1).
	3	A \|=
		1

2. Vectors

= (1;1; 1),

A1 = ( 1;0;1),

A	2	= (1;0;0),

1

2	= = 1.
A B	= = 1.
1

Matrices

	1		1	1			1	1	1			0		0	1
						R* =	1				R* 1 =						\| R \|= 1 0,
R =		1	0	0	,	R* =	1	0	1	,	R* 1 =		1	1	0	,	\| R \|= 1 0,

		1	1	0			1	0	0				0	1	1
		1	1	0			1	0	0				0	1	1

3. Nonsingular transformation. Vector where

*	z,
y = R

y =

A1 y B ( ),

= S1 y,

			0		1	0				0
	= R* A R* 1								= R*B =				= S R* 1
A1	= R* A R* 1	=		0	0	1	,	B1	= R*B =	0	,	S1	= S R* 1	= ( , , ),
	1								1				1
				0	0					1
				0	0	1				1

= ,	= ,	= , ( )	.
		0

4. For this example equation (1.86) has a form

where	( ) = ( ),

y = A1 ( ) y B1 ( ),

= S1 y,

	0		1	0				0
A1 ( ) =							B =			S1 = ( , , ).
A1 ( ) =		0	0	1		,	B =	0	,	S1 = ( , , ).
							1
			( )	1 ( )				1
			( )	1 ( )				1


Characteristic polynomial of the matrix							is equal to 2 ( ) =\| I3 A1( ) \|=
Characteristic polynomial of the matrix					A1 ( )		is equal to 2 ( ) =\| I3 A1( ) \|=

3 (1 ) 2 . Necessary and sufficient conditions for Hurwitz of matrix

A1 ( ) : < 0, < 0, ( ) > 0, , , > 0 is an arbitrarily small number.

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

Limit value

0 is determined

from

inequalities:

1 > 0,

< 0, < 0,

(1 )( ) > 0.

From these

inequalities

get: 1) if

< 0, < 0,

> 0, 0,

then

0 = ;

< 0,

< 0, > 0, > 0, then

10 = min , .

5. Along the solution of the system y = A1 ( ) y B1 (

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

the following identities are true:

), = S1 y (or

= y (t) y		(t)
1	2

= y2 ,

y3 (t))

where

(

(t)) = (t) y (t) ( ) y				(t)
		1	2
) = y (t) y		(t) y	(t), = y
1	2	3		2
, = , = .
			Notice that

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

( (t)) y	y	(1 ) y .
1	2	3

6. Improper integrals:

	1																				3									1						1
I	1	=																			3		(t)]dt							1		= c1,				1		\|< ,		\| c1 \|< ,
I		=		( ( )) (t)dt =							[	2		(t) ( ) y								2	(t)]dt				l					= c1,			\| l			\|< ,		\| c1 \|< ,
												2										2
			0							0
															2														2											2
0 I				=	[ ( (t))				(t)				1		2	( (t))]dt =							[						2	y	2		(t)		(					2	) y	2	(t)
					1								1										1								2											2
			2		1						1		0										1							1												2
					0																		0
								2																		2												2
					(			2	) y	2	(t)]dt						1	[		2	(t)			2		2	y		2	(t) (					2			2	2
										2							1			2				2					2						2
										3					1		0											1
																		0
					2	2	) y	2	(t) (1 2									2	2	2 ) y					2	(t)]dt l								,		\| l			\|< ,
					2		) y		(t) (1 2											2 ) y						(t)]dt l								,		\| l			\|< ,

								2																	3								2				2

(1.106)

(1.107)

0 I							y		y ]								y			y					y			]dt l , \| l			\|< ,
0 I			= [				y		y ]		dt = [						y			y					y			]dt l , \| l			\|< ,
									2							2			2					2			2
		3		0		1	1		3 3					0			1	1				2		2	3	3			3	3
			0									0
																															(1.108)
where			= 2	, = 2				,	= 2			2	3	,			= 2	2				0	2	.
		0	0		1		1	2	2			1	3			3	3					0	2
7. Aizerman’s problem. Solution of Aizerman’s problem was obtained from
boundedness of					the		improper					integral			I1		along				the				solution of				the		system

	( ) y B ( (t)), (t) = S									1		y. From (1.106) it follows that
y = A1	( ) y B ( (t)), (t) = S											y. From (1.106) it follows that


			[ 2 (t) ( ) y2								(t)]dt = c1				l , \| c1				l		\| \| c1				\| \| l	\|< ,					(1.109)
									3								1			1					1
			0

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