86

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

c

,

*

*

i = 1, n m

where

i, j = 1, n

i = 1, m,

ij

are decomposition coefficients A

i ,

A 0i ,

by bases i

Rn , i = 1, m,

0i R

n

,

i = 1, n

m. Hence taking into account (2.4), (2.6)

we get a system of equations (2.9) with respect to variables

y1

, , yn .

Similarly, by decomposing vectors

*

n

by bases i

,

0i ,

i = 1, n m

R

i = 1, m,

Si

we get

= S x = d

x d

x d

x d

x,

*

*

*

*

1

1

11 1

1m

m

1m 1

01

1n

0n m

x,

= S

x = d

*

x d

*

x d

*

x d

*

m

m

m1

1

m,m

m

m,m 1

01

m,n

0n m

where S

* = (S * , , S

* ). The Lemma is proved.

1

m

System of equations (2.9) in vector form has the form

where	A,
AA==||||cc |||,|,
	ijij

B,
i,	j

	y = Ay B ( ), = Sy,		( )			,
					0
S are constant matrices of orders			n n,		n m,
= 1, n,	SS==||||dd	|||,|, i = 1, m, j = 1, n,			I		m
				B =				,
	ijij				O
					n m,m

(2.10)

m n respectively,

I	m	is a unit matrix

of order m m,

On m,m

is a matrix of order

(n m) m

with zero

matrix

*

, then

y

*

x,

x = K

1

*

1

y,

matrices

A,

K = R

= Kx = R

y = (R )

A = KAK

1

*

)

1

AR

1

,

S = SK

1

*

)

1

,

*

= (R

= S (R

B = KB = R B. Thus,

tion (2.1) with nonlinearities (2.2) with nonsingular transformation	x
is reduced to (2.10).

elements. If the

S ,	B	are equal

differential equa-

= K	1	*	1	y
		y = (R )

Lecture 15.

Solution properties and improper integrals in general case

Solution properties. Limitation on solutions of the system (2.1), (2.2), and also equations (2.10) are considered. Identities along the solution of the equation (2.10) are obtained and its asymptotic property is examined.

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

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

| x(t) | c0 ,	| x(t) | c1, t I,	(2.11)
| y(t) | c2 ,	| y(t) | c3 , t I,	(2.12)
| (t) | c4 ,	| (t) | c5 , t I,	(2.13)

81

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

where ci = const > 0, ci < , i = 0,5. Moreover, the functions

x(t), y(t), (t), t I

are

uniformly continuous.

Proof. As matrix

matrices	A,	i.e.	Re j

(2.1) has the form:

A = KAK

(A) = Re

1

j	(

, then A) < 0,

from Hurwitz of matrix j = 1, n. Solution of the

A	we get Hurwitz of

differential equation

t, a =

x(t) = e	At	x
		0

Notice, that from Hurwitz of

t I ,		follows,		where
max	Re		( A) < 0.	Then
1 j n		j
			| x(t) | || e		At	|| | x
								0
			c | x		| e		(a )t
				0

t
eA(t ) B ( ( ))d , t I.
0
matrix				A		evaluation		|| e		At	||
> 0		is		an		arbitrarily			small
t
|	|| e		A(t )			|| || B || | ( ( )) | d
0
						t
		(a )t				*	(a )
ce				||	B || e				d =
						0

ce	(a )t	,	c = c(
number, the

) > 0,

value

where

e	(a )t

=

1,

c

t

| x	|
0

=c

I ,

e	(a )

| x	0	|

a

t	ce	(a )t	|| B ||
				*

	e	(a )t				1		c ||
					a
	c |		x		|		1		c
			0		a
	< 0,				1		> 0,
					a

		1		e	(a )t
		a
B ||			( 1 e				(a
		*
||	B ||		*	= c		,
					0

| ( (t)) |	,
*

1		=
a

)t	)

t, t I.

From here the

limitation of the solution of the system (2.1), (2.2) follows. From the equation (2.1) it follows that

		| x(t) | ||	A || | x(t) | || B || | ( (t)) | || A || c ||			B || = c , t, t I,
					0	*	1
	Then
	| (t) | || S || | x(t) | || S || c0 = c4 , | (t) | || S || | x(t) | || S || c1 = c5 , t, t I.
	As	function	y(t) = Kx(t),	t I ,	then	| y(t) | || K ||| x(t) | || K || c0 = c2 ,
| y | || K ||| x(t) | || K || c1 = c3 , t, t I.
(t),	So, estimates (2.11)–(2.13) are proven. From the limitations of functions x(t),							y(t),
	t I	we get uniform continuity of functions x(t),				y(t),	(t), t I. The theorem is

proved.

82

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

It should be noted that: 1. From evaluation

| x(t) |,

t I

we have

lim	| x(t) |=| x( ) | c | x	|	1	c || B ||	= c	,
	0		a	*	0
t

c0	c
	0

due to continuity

x(t),

t I ,

where

a

2. If

< 0.		Consequently,
x	S			= {x	R	n	||
0				0

lim	|
t

x	|
0

y(t) |=|, >

y( ) | || 0}, then

K || c0 , | | || S

lim	| (t) |=| ( ) | || s || c0 .
t
|| (c		c	|| B || ) = c4 .
		a	*

Lemma 4. Let the conditions of the lemmas 1–3 be satisfied. Then along the solution of the system (2.10)the following identities hold

( (t)) = H

y(t) A11 y(t),

t

0

H y(t) =

A12 y(t),

t I ,

1

(t) = S1H

y(t) S 2 H y(t),

t

0

1

(t) = S1H0 y(t) S 2 H1 A12 y(t),

where

H

0

H

= (I

, O

),

H

= (o

, I

),

0

m

1

n m,m

n m

m,n m

H1

I,

I,

t I,

						A11
	= I		,	A =
							,
		n
					A12

c

c

c

c

11

1n

m 1,1

m 1,n

A11

=

,

A12

=

,

S = (S1, S 2 ),

c

c

c

c

m1

mn

n1

nn

d

d

d

d

11

1m

1m 1

1n

S1 =

,

S 2

=

.

dm1

dmm

dm,m 1

dm,n

(2.14)

(2.15)

(2.16)

(2.17)

Proof. As conditions of the lemmas 1–3 are satisfied, then we get (2.9). Notice,

that

y

y

m 1

1

H

0

y

=

,

H

y =

.

1

yn

ym

Then from (2.9) it follows that

(

) = y

c

y c

y

, ,

m

(

m

) = y

m

c

y c

y

,

1

1

1

11

1

1n

n

m1

1

mn

n

H0 y

= (S1, S 2 ) y = (S1

= S1H

y S 2 H

y,

H

y = A12 y.

, S 2 )

0

1

1

H1 y

Consequently, identities (2.14)–(2.17) hold. The Lemma is proved.

83

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

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

matrices

,

M ,

N1, N2 of orders

n n,

n n,

n (n m),

n (n m)

respectively,

along the solution of the equation (2.10) the following identities hold

	( )˙( ) + ( )˙( ) =		[ ( )( )],         = , (2.18)

*	(t)N		*	(t)N ][H			(t) y(t)] 0,
[ y		2	y			y(t) A12
				1	1

Proof. Identity (2.18) directly follows from the equality

d	*	(t) y(t)] = y		*		*	(t) y(t) = y		*	*	(t)
dt	[ y				(t) y(t) y					(t) y
		*		*	*	*				*	*
		= y	(t=)Myy		(t)My(t()t) yy((tt), ty(t),I			ïðèt whereI ïðè= M.

tI.

y* (t) y(t) =

=M.

(2.19)

As follows from (2.15) for any matrices

N	,
1

N2 of orders

n (n m),

n (n m)

an identity (2.19) holds. The Lemma is proved.

Improper integrals. Based on nonsingular transformation and using properties of the solution of the system (2.1) we can obtain evaluations of improper integrals along the solution of the system (2.10).

Theorem 2. Let the conditions of the lemmas 1–3 be satisfied, the matrix A Hurwitz matrix, the function ( ) 0. Then for any diagonal matrix 1 = diag( 11, , of order m m along the solution of the system (2.10) the improper integral

be a

	1m	)

1

1

1

2

*

*

*

I

=

( (t)) (t)dt =

[ y

(t) y(t) y

(t)

y(t)

0	0
	m			( )
			i
*	(t) 3 y(t)]dt =				i ( i ) 1i d i
y						< ,
	i=1			(0)
			i

(2.20)

where

	*		S1H	,			*		S 2 A12 H	*	*				*
											S1	A11),			= A11	S 2 A12.
	= H					2	= H			0				3
1	0	1	0				0	1			1				1

(2.21)

Notice that matrix			A = KAK	1	,	where
j ( A) = j ( A), j =		. Consequently, the matrix
	1, n
matrix A is a Hurwitz matrix, where						j ( A) = j

K		is a nonsingular matrix, then
A	is a Hurwitz matrix then, when the
( A),
		j = 1, n eigenvalues of the matrix

A,

Re	( A) < 0,
j

j = 1, n.

Proof. As follows from Theorem 1, | y( ) |< c2 < ,	| y(0) | c2 < . Improper
integral (v. (2.20)).

84

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

I = * ( (t)) (t)dt = [H

0

y(t) A11 y(t)]* [S1H

0

y(t) S 2 A12 y(t)]dt =

1

1

1

0

0

= [ y

*

(t) y

(t) y

*

(t)

*

(t)

y(t)]dt =

2

y(t) y

3

1

0

m

( )

i

=

i ( i ) 1i d i < ,

i=1

(0)

i

by virtue of identities (2.14)–(2.17),

| (0) | c4

< , | ( ) | c4 < , where matrices

1, 2 , 3

of orders n n are defined by formulas (2.21), respectively. The theorem is

proved.

Theorem 3. Let the conditions of the lemmas 1–3 be satisfied, the matrix

A

be a Hur-

witz matrix, the function ( ) 0.

Then for any diagonal matrix 2 = diag( 21, , 2m ) 0

of order m m,

along the solution of the equation (2.10) the improper integral

I

=

( (t))

( (t))

( (t)) (t)]dt =

[

*

1

*

2

2

0

2

0

=

[ y

*

(t) y

(t) y

*

(t)

y(t) y

*

(t)

y(t)]dt 0,

2

3

1

0

(2.22)

where									H	,				= 2H															A11 H										S1H					H		S 2 H	,
= H						1																					1
	*																				*													*										*				(2.23)
1	0		2			0			0			2									0		2			0								0			2						0	0	2	1
										*						1							*													*
									= A11									A11 A11											S1H				A11								S	2 H .
							3				2				0													2			0									2
																																												1
Proof. From inclusion ( ) 0																					it follows that
			i ( i )											,				i									1		,					,							1	,		i = 1, m.
											0i																0i									i		R
																			(		)												i
									i
																		i		i
																1			2															Then for any value 2i 0 an ine-
Consequently, i i ( i ) 0i																		i		( i ),							i = 1, m.
quality holds		(			)							1					2			(		) 0,
																													i = 1, m. From here it follows that
	i			i				2i i			0i					2i			i		i
										m
																1					2				) i ( i								) 2i i ] 0.
										[ 0i							2i i ( i
										i=1
Let the 2 = diag( 21, , 2m ) 0,																						1												1							1			Then the given inequality
																					0				= diag ( 01 , , 0m ).
can be written as
								* ( )			2				1 ( ) * 0, , Rm .
														0														2

85

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

From here it follows that improper integral (v. (2.22))

													1 ( (t)) * ( (t)) (t)]dt 0.
						I	2	= [ * ( (t))				2
													0								2
									0
Then taking into account identities (2.14), (2.16), we have
		I2 = {[ H								*			1	[H									*
						0 y(t) A11 y(t)] 2					0					0 y(t) A11 y(t)] [H0 y(t)						A11 y(t)]
			0
		[S1H		y(t) S 2 H y(t)]}dt =											(t) y(t) y				(t)		y(t) y	(t) y(t)]dt 0,
												[ y
	2		0						1					*		1		*		2	*	3
											0
where matrices					,			2	,	3 of orders			n n are determined by the formula (2.23). The
						1
theorem is proved.
Lemma 6. Let the conditions of the lemmas 1–3 be satisfied, the matrix																								A	be a

Hurwitz matrix,

the

function

( ) 0

. Then

for

any matrices

n (n m),

n (n m)

respectively, improper integral

I

=

[ y

*

(t)P y(t) y

*

(t)P y(t)

*

(t)P y(t)]dt =

3

y

1

2

3

0

P = N H ,

P = N

A

H

*

N

,

P

1

where 1

2 1

2

2 12

1

3 = N1 A12.

Lemma’s proof directly follows from the identity (2.19).

N	,
1

0.

N	.
2

of orders

(2.24)

Lecture 16.

Absolute stability and Iserman’s problem in the general case

Absolute stability. Based on the above results on the estimation of improper integrals, and also Lemms 1–3 we can formulate conditions of absolute stability of equilibrium state of the system (2.1), (2.2).

We introduce the following notation

R = P = H *																														*				1H
																	S1H				0		H																	0		N H					,
0	1					1				1		0				1														0		2			0							1				1
										P = H *																				H *								1*
=				2				2											S			2 A12															S					A11
	0											2					0	1																	0					1
2H0* 2 0 1 A11 H0* 2 S1H0																	H0* 2 S 2 H1																		H1* N1* N2 A12 ,														(2.25)
																	*																			*
																																									1 A11
W =			3				3		P = A11											S 2 A12											A11
0												3						1																	2						0
											*
									A11 2 S1H0 A11 2 S 2 H1.

86

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

In particular when matrix							*		*
			N1 = A11 1 S 2						A11 2T
				*			1		*
		W = A11						A11	A12 K
					2		0
		0							1
*	*			*					*
	A11	2 S1H		A11		2 S 2 H1 = A11 2 S
where K1 = K1 > 0,			0
T are matrices of orders		n n,		m n			, respectively.

Then improper integral

,	then the matrix
A		0,
12
		*		S = T A12.
= A11 T A12			,
		2

Here

K	,
1

I

= I

I

I

=

[ y

*

(t)R y

(t) y

*

(t)W y(t)]dt

4

1

2

3

(t) y(t) y

0

0

0

0

m

( )

i

i ( i ) 1i d i

< .

i=1

(0)

i

(2.26)

Theorem 4. Let the conditions of the lemmas 1–6 be satisfied, the matrix		A	be a
Hurwitz matrix, the function ( ) 0 , and let, moreover, matrices R0 ,	0 of orders

n n,

where

n n such that: 1)
I	40	=

matrices R0 , 0 ,W0

R0		*	0,	2)	0		=		*
		= R0						0 .
						m		( )
		0					i
	*
	y	(t)W y(t)dt
0						i=1		(0)
							i

d 1

0 dt [ 2 y* (t) 0 y(t)]dt

are determined by the

Then the improper integral

i	(	)	1i	d	i
	i
	< ,
	formula				(2.25),		1,	2

(2.27)

0

any

diagonal matrices of orders

m m,

N1,

N

2

are any matrices of orders

n (n m),

n (n m)

, respectively.

Proof. As conditions of the lemmas 1–6 are satisfied,

then

improper

integral

I4 <

(v. (2.26)),

by virtue of estimates

(2.20), (2.22),

(2.24), where

matrices

R0 , 0 ,W0

are determined by the formula (2.25). Matrices

,

i = 1,2,3,

i

,

i = 1,2,3,

i

W =

3

3

P

are calculated by the formulas (2.21), (2.23). As matrix = * , then

0

3

0

0

y

*

(t) y(t) =

d

[

1

y

*

(t) y(t)],

t I ,

consequently, the improper integral

0

dt

2

0

d

[

1

y

*

(t) y(t)]dt =

1

y

*

(t) y(t) |

=

1

y

*

( ) y( )

1

y

*

(0) y(0) < ,

dt

0

2

0

2

0

2

0

2

0

0

by virtue of the assessment | y( ) | c2

< ,

| y(0) | c2 < .

87

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

Then

						0			0
			*					*
		[ y		(t)R y				(t) y	(t)W y(t)]dt
		0
	m i	( )					1			1
		i ( i ) 1i d i						y* ( ) 0 y( )			y* (0) 0 y(0) < .
						2				2
	i=1	(0)
	i

By the hypothesis of the theorem

y	*	(t)R y	*	(t)[	1	R	1	*	]y(t)
			(t) = y					R
		0			2	0	2	0

holds (v. (2.27))

matrix

0,	t,

R	=
0
t I

R	*	0.	Then quadratic form
0
.	Consequently, an inequality

I		=		*	(t)W y(t)dt	[ y	*	(t)R y	*	(t)W y(t)]dt < .
	40			y					y
					0			0		0
			0			0

The theorem is proved.

Theorem 5. Let the conditions of

Hurwitz matrix, the function	( ) 0
and let, moreover, matrix W	= W	*	> 0.
0	0

the lemmas 1–6 be satisfied, the matrix	A	be a

and conditions 1), 2) of theorem 4 be satisfied, Then equilibrium state of the system (2.1), (2.2)

is absolutely stable.

Proof. As all conditions of theorem 4 are satisfied, then we get inequality (2.27). Consequently, the improper integral

I		=		*	(t)T	y(t)dt < ,
	40			y
					0
			0
						*

where the scalar function V ( y) = y T0 y > 0,

T0 =

y,

1	W
2	0
		n	,
y R

1	W	*	>
2	0
y 0,

0,

V (0) = 0.

As all condi-

tions of the theorem are satisfied1, then | y(t) | c	,	| y(t) | c ,	t,	t I.	We show that
2		3

lim

y(t) = 0.

Suppose the contrary, i.e. lim

y(t) 0.

Then there exists a sequence

t

t

{tk },

tk > 0,

tk

when

k such as

| y(tk ) | > 0,

k = 1,2, .

We choose

tk 1 tk

1 > 0,

k = 1,2, .

As y(t),

t I

is

continuously differentiable, | y(t) | c2 ,

| y(t) | c3 , t I ,

then | y(t) y(tk ) | c3 | t tk

|,

t, t [tk

1

, tk

1

],

k = 1,2, . As

2

2

tk

1

> tk 1

,

tk

1

> tk 1

,

V ( y) > 0,

y R

n

,

then

2

2

tk

21

V ( y(t))dt

V ( y(t))dt,

0

k =1

tk

1

2

88

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

where

t [t	k
So, |	y

| (t)

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

		1	, tk					1	].	We can always choose the value 1 >
	2						2
	|				,	| y(t) | c					,	t [t				1	, t				1	].	As the function
				0
										2			k		2			k		2

| c			1	=		> 0,	t,
	3	2			0
0	so that the value 0						> 0.

V ( y) is continuous on the

compact set

	0	| y | c	,
		2

then there is a number

m > 0

such as

	min	V ( y) = m.
	|y| c
0	2

Then

the value of the integral

t		1
k	2
	V ( y(t))dt 1m,		k = 1,2, .
t		1
k	2

t

I

Consequently,

							0
						*
		V ( y(t))dt =			y		(t)W y(t)dt
		0		0
This contradicts the condition I								40 < ,
,	K	is a	nonsingular				matrix,	then

	t			1
	k		2
		V ( y(t))dt lim k 1m
k =1	t		1		k
	k		2
consequently, lim						y(t) = 0.
					t
lim		x(t) = 0,			,		.
t						0

= .

As x

The

(t) = K	1	y(t

theorem

),

is

proved.
From the theorems 4, 5 it follows that in cases when the matrix A		is a Hurwitz
matrix, the function ( ) 0 and the following conditions are satisfied:		*
		1) R0 = R0 0;
2) 0 = 0 ;	3) W0 = W0 > 0 equilibrium state of multidimensional controlled system
*	*

(2.1), (2.2) is absolutely stable.

Remark 1. As follows from the proof of the theorem 5, the equilibrium state of the

system (2.1), (2.2) is absolutely stable and in case, when matrix			W = W * 0, if the
			0	0
	*
surface	V ( y) = y T y = 0	does not contain whole trajectories. In	this case, integral
	0

curves "stitch" surface V ( y) = 0. Notice that surface y*T0 y = 0 does not contain whole

trajectories, if

0,

t [0, ).

y(t)Ty(t)

It should be noted that as a result of nonsingular transformation identity (2.15) is

obtained. This identity is used to determine an improper integral

I3. The improper

integral I

3

depends on arbitrary matrices N ,

N

2

of orders n (n m), n (n m) , res-

1

pectively. Matrices

R0 , 0 , W0 depend on matrices

N1, N2 . As follows from the con-

dition of theorems 4, 5 matrices 1, 2 0, N1, N2

ensure the fulfillment of conditions

R = R* 0,

= * ,

W = W * > 0. Consequently, matrices N ,

N

2

allow us to signifi-

0

0

0

0

0

0

1

cantly expand the area of absolute stability in space of constructive parameters of the

89

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

system. In the method of A.I. Lurie and in the method of V.M. Popov we have no

improper integral

matrices N1,

N2

I	3	,	consequently, we have no matrices	N1,

allowed us to solve the Iserman’s problem.

N	.
2

The presence of

Iserman’s problem. The question arises: is it possible to distinguish a class of multidimensional regulated systems, by selecting a feedback matrix S , for which Iser-

man’s problem has a solution? For this class of multidimensional regulated systems, the obtained results are necessary and sufficient conditions of absolute stability. No-

tice, that matrices

S

and

S

are related by

S = SK ,

S = SK	1	,

matrices A B 0S and

A B 0 S are similar. Indeed, if= S y for the system (2.10), then

( ) = 0 , y = Ay B (

where

) = (A

=

B	0

Sx S)y

for the system (2.1), (2.2),

,	x = Ax B ( ) = (A B	S)x.
	0

As	A = KAK	1	,	B =

follows that matrix

KB,

	0	=

S = SK	1	,

diag(	01	,

then

,

A

0m

B			S = K(A B S)K 1. From here it
0			0
) found from the condition of Hurwitz of

the matrix A B 0S	matches the value	0	is denoted from the condition of
Hurwitz of the matrix	A B 0 S.

As follows from Theorem 2 and Lemma 4 the improper integral

I

= I

I

=

[ y

*

(t)R y(t) y

*

*

(t)W y(t)]dt =

5

1

3

(t) y(t) y

1

1

1

0

m

( )

i

=

i ( i ) 1i d i

< ,

i=1

i

(0)

(2.28)

where

*

S1H

N H

,

R = P = H

0

1

1

1

0

1

1

1

*

*

*

=

P = H

S 2 H

A12 H

S1

A11 N

A12

2

0

1

0

1

2

2

2

*

W =

3

P = A11

1

S 2 A12

N A12.

1

3

1

*

*

*

In particular, when

N

= A11

S 2 A12K

,

then W = A12K

A12

1

1

2

1

2

is a matrix of order

n n.

Theorem 6. Let

the

conditions of

the lemmas 1–3 be

A, A B S, 0 0

<

0 be

Hurwitz

matrices,

the function

besides:

1) diagonal matrix

1 = diag( 11, , 1m ),

matrices N1,

N2 ,

S

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

m n respectively, such that

*	*	,
H	N
1	1

0, where

satisfied,

( ) 0

=(S1, S 2 )

K2 = K2* > 0

	matrices
,	and let

of orders

			R = R* 0,			= *;	(2.29)
			1	1	1	1
2) matrix T =	1	(W W * ) 0,		surface	V ( y) = y*T y = 0		does not contain whole
1	2	1	1		1	1

trajectories.

90