86

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

where

n

x = ( a

A

n 1

a

A

n 2

a A a

I

)x = a

y

a

y

a y

a y

,

A

n 1

n 2

n

n 1

n

n 2

n 1

2

1

0

1

0 1

A	n 1	B = 1.	From (1.5) – (1.7) we get (1.4). The lemma is proved.
	Lemma 2. Let the conditions of Lemma 1 be satisfied, and let, in addition, the
rank of the matrix
			R =	*	*	*	, , A	*n 1	*	(1.8)
					, A

of

n × n order be equal to

n,

where (*) is a transpose sign. Then:

1) there exists a row vector

= (

,

2

, ,

n

)

such as

1

= 1 y1

2 y2

n yn ;

(1.9)

2) If

y1

= x = 0, y2 = Ax = 0, , yn

= A

n 1

x = 0, then

x = 0.

Proof.

Note

that

the

rank

of

the

matrix

R = n

if

and

only

if

vectors

*

*

*

*n 1

*

are linearly independent. Since vectors

*

*

*

*n 1

*

form a

, A

, , A

, A

, , A

basis in

R

n

,

then the vector

S

*

R

n

can be represented unambiguously in the form

S

*

= 1

*

*

*

*n 1

*

.

2 A

n A

Then

= Sx = x Ax A

n 1

x = y

y

y

.

2

2

n

n

1

2

n

1 1

Now the second equation from (1.1) can

be

written as

(1.9). Let

the

vector

y = ( y , y

2

, , y

n

)* Rn . Then

1

y

x

1

y =

y2

=

Ax

=

A

*

x,

x = R

n 1

x

n 1

yn

A

A

by virtue of (1.8). Since the matrix

R

*

is non-singular then

x = R

* 1

y . Therefore, for

y = 0, x = 0.

The lemma is proved.

It follows from Lemmas 1 and 2 that if equalities (1.3) are satisfied and the rank

R = n,

then differential equation (1.1)

by a

nonsingular transformation

x = R* 1 y

reduces to the form (1.4), where = Sx can be written as (1.9).

Introducing the notation

0

1

0

0

0

0

1

0

0

0

A =

, B =

, S = ( , , ),

1

n

a0

a1

a2

an 1

1

11

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

equations of motion (1.4), (1.9) can be represented in the form

y = Ay B ( ), = Sy, ( )	.
0

(1.10)

	From
2)	A = R
	(A) =
j

*

j

(1.1),

1	*	,
	AR
(A),		j

(1.10)

B= R* 1

=1, n;

it

follows

that: 1)

A = R* AR* 1, B = R* B, S = SR* 1;

B,

*

; 3)

matrices

A,

A

are similar, therefore,

S = SR

*

* 1

B = S B.

4) SB = SR R

Lection 2.

Properties of the solutions in the main case

It can be shown that solutions of systems (1.1), (1.2), and systems (1.4), (1.9), (1.2) are limited. These properties of solutions can be used in the estimation of improper integrals.

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

function ( ) 0 and let, in addition, the equalities (1.3) be satisfied and the rank R = n. Then the estimates are correct:

| x(t) | c		0	, | x(t) | c		, t I,
				1
| y(t) | m	, |			y(t) | m , t I,
0				1
| (t) | c				, | (t) | c		, t I,
			2		3

(1.11)

(1.12)

(1.13)

where m0 = const < ,

m1 = const

< ,

ci

= const

< ,

i = 0,1,2,3.

	In addition, the functions x(t), y(t), (t), t I are uniformly continuous.
	Proof. From inclusion ( )					0	it follows that | ( (t)) | ,	0 < < ,	t I.
							*	*
Since the matrix				A	is a Hurwitz matrix, i.e. Re j ( A) < 0, a = max Re j (A) < 0				, then
							1 j n
	e At		ce(a )t ,	t, t I , c = c( ) > 0, > 0 is an arbitrarily small number [13; § 13,

inequality (7)].

The solution of the differential equation (1.1) has a form:

x(t) = eAt x0 t eA(t ) B ( ( ))d , t I.

0

12

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

Then

t

| x(t) |

e

At

| x0

| e

A(t )

B

| ( ( )) | d

0

t

c | x0

| e

(a )t

ce

(a )t

B * e

(a )

d =

0

= c | x

| e

(a )t

ce

(a )t

B

1

e

(a )t

1

=

0

*

a

a

= c | x

| e

(a )t

1

c B

1 e

(a )t

c

, t, t I ,

0

a

*

0

where e

(a )t

1, t, t I, a

< 0

. This implies the boundedness of the solution

of the system (1.1), (1.2). It follows from (1.1) that

| x(t) | A

| x(t) | B

| ( (t)) | A c

B = c

,

t,

t I,

0

*

1

| (t) | S | x(t) | c2 , | (t) | S | x(t) | c3 , t, t I.

So,

the

estimates

(1.11),

(1.13) are

proved. As

*

x(t),

t,

t I ,

then

y1 = R

| y(t) |

R

*

|

x(t) |

R

*

c

= c

,

| y

(t) | R

*

|

x(t) |

R

*

c

= c

,

t,

t I

. Therefore,

0

2

1

3

x(t),

y(t), (t), t I

estimates from (1.12) hold. From the limitations of derivatives

we get the uniform continuity of functions

x(t),

y(t), (t),

t I , respectively. The

theorem is proved.

It should be noted that: 1) From

				1
	lim	| x(t) |=| x( ) | c | x0	|		c B	* = c0 ,
	t			a

evaluation | x(t) | c0 , t, t I		we get
c0 < , a < 0;	2) From evaluation

| y(t) | m	,
0

t,

t I

we get

lim	| y(t) |=| y( ) |	R	*	lim	| x(t) | R	*	c0
t				t

= m0 ,

m0

< .

Lemma 3. Let the conditions of the lemmas 1, 2 be satisfied. Then solution of the system (1.10) the following identities hold

( (t)) = (t) a0 y1 (t) a1 y2 (t) an 1 yn (t), t I,

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

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

where (t) = yn (t), t I .

along the

(1.14)

(1.15)

(1.16)

Proof. Identities (1.4), (1.9) follow from lemmas 1, 2,. Then the last identity (1.4) implies the equality (1.14). Identities (1.15), (1.16) follow from (1.9), where(t) = yn (t), t I. The lemma is proved.

13

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

The identities (1.14)-(1.16) allow us to use the properties of the function ( )

from inclusion ( ) 0

for calculating improper integrals.

Lemma 4. Let the conditions of Lemmas 1-3 be satisfied, the matrix A is a Hur-

witz matrix, function ( )	0	.	Then:
1) for any constant matrix			Q of (n 1) (n 1) order quadratic form z* (t)Qz(t),
z(t) = ( (t), y1 (t) , yn (t)), t I			can be represented as

where 2)

where

z

*

(t)Qz(t) = q

2

(t) q y

2

q

y

2

(t)

d

*

(t)Fy(t)],

n

n

[ y

0

1

1

dt

F

is a constant matrix of

n n order;

Improper integral

z* (t)Qz(t)dt = [q 2 (t) q y2

q y2

(t)]dt

0

1 1

n n

0

0

d

l0

=

[ y

*

(t)Fy(t)]dt = y

*

( )Fy( ) y

*

(0)Fy(0),

| l

dt

0

l

*

0

= y

(t)Fy(t) | .

0

t I ,

l0 ,

0	|< ,

(1.17)

(1.18)

(1.19)

Proof. coefficients values yk ,

a) if	n,

Since the quadratic form z						*	(t)Qz(t), t I	contains terms with constant
of the product of components of the vector									z(t),	we can calculate the
y	k	y	s	,	k, s = 1, n. It is easy to verify that:
k	are odd numbers k < n , then

b) if n
y	k

c) if n

yk

d

1

n k

2

y

=

y

y

y

y

( 1) 2

y

;

k

dt

k

n

k 1

n 1

2

n k

2

is odd, k is an even number

k < n

, then

d

n k 1

n k 1

=

y

y

y

y

( 1)

2

y

y

( 1)

2

y

2

;

k

n

k 1

n 1

n

k

1

n k 1

n k 1

dt

2

2

2

is even,

k

is an odd number k < n , then

d

n k 1

n k 1

y

y2

=

y

y

y

( 1)

2 y

y

( 1) 2

;

dt

k

n

k 1

n 1

n k

1

n k 1

n k 1

2

2

2

14

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

d) if n is odd, k is an odd number k n , then

d

1

n k

y

=

y

y

y

y

( 1)

2

y

2

k

k

n

k 1

n 1

n k

;

dt

2

2

e) if k

is odd,

s

is an odd number s > k

, then

d

k s

1

k s

1

2

2

y

y

=

y

y

y

y

( 1)

2

y

y

( 1)

2

y

;

k

s

dt

k

s 1

k 1

s 2

k s

1

k s

k

s

2

2

2

f) if k

is odd,

s

is an even number

s > k , then

d

1

s

1

y

y

=

y

y

y

y

( 1)

2

y

2

k

s

k

s 1

k 1

s 2

s

;

dt

2

2

g) if k

is even,

s

is an odd number

s > k , then

d

1

k s

1

y

y

=

y

y

y

y

( 1)

2

y

2

k

s

k

s 1

k 1

s 2

;

dt

2

k s 1

2

h) if k is even,

s

is an even number

s > k

, then

d

k s

1

y

y

=

y

y

y

y

( 1)

2

y

y

k

s

k

s 1

k

1

s 2

s k

k

s

.

dt

1

2

2

In particular, for a value

n = 2,

we get:

y

=

d

( y y

) y

2

,

y

=

d

1

y

2

y y

=

d

1

y

2

2

2

2

2

,

2

1

dt

1

dt

2

1

dt 2

1

where (t) = y2 , t I;

For values n = 3, z(t) = ( (t), y (t), y

2

(t), y (t)),

(t) = y

(t)

:

1

3

3

d

1

d

2

d

1

2

y1 =

y1 y3

,

y2 =

( y2 y3 ) y3

,

y3

=

y3

;

dt

dt

2

dt 2

y y

=

d

1

y2

, y y

=

d

( y y

) y

2

, y

y

=

d

1

y2

.

2

3

2

2

3

1

2

1

1

dt

1

2

2

dt

dt 2

15

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

Substituting values

y	,
k

y	k	y	s

in the quadratic form

z	*	(t)Qz(t),

we get represen-

tation (1.17). Integrating identity (1.17), taking into account estimate (1.12), we obtain the relations (1.18), (1.19). The lemma is proved.

Lemma 5. Let the conditions of the lemmas 1, 2 be satisfied, the matrix A be a Hurwitz matrix, function ( ) 0 and let, moreover:

1) scalar continuous function V ( y) > 0	for all	y R	n	,	y 0,	V (0) = 0;

2) improper integral

V ( y(t))dt < .

0

witz

Then lim y(t) = 0. t

Proof. Note that under the matrix, function ( )

conditions of the lemmas 1, 2, the matrix	A	is a Hur-
0 , the statement of the theorem 1 is true. Consequently,

| y

as

(t) | m , | y(t) | m, t, t I.
0
Let all the conditions of the lemma be satisfied. We can show that lim			y(t) = 0.
		t
Assume the opposite i.e. lim y(t) 0.	Then there is a sequence	{tk } [0, ) such
t
| y(tk ) | > 0, k = 1,2, . Choose tk 1 tk	= m > 0. Since the function	y(t),	t I is con-

	tinuously				differentiable and					| y(t) |< m ,
										1
	t [t			m	, t			m	], k = 1,2, . Then
		k		2		k		2

t,

t I ,

then

| y(t) y(t	k	) | m	| t t	k	|,
		1

where | y(t) |=| y(tk ) of quantity m > 0. As

y(t)

| y(t)

V ( y(t))dt

0

y(tk ) | | y(tk ) |

| m	, t, t I
0

	t
	k	m
		2
		V ( y(t))dt,
k =1	t	m
	k	2
| y(t) y(t			) | m	m	=		> 0 , by choice
		k	1	2		0
, then

t
	k	m
		2

		V ( y(t))dt V	m,	V	=	min	V ( y).
		min		min		|y| m
		m
t					0	0
k		2

Then

	tk m2
V ( y(t))dt V ( y(t))dt Vmin m = .
0	k =1 t	m	k =1
	k	2

This contradicts the 2-nd condition of the lemma. The lemma is proved.

16

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

Lection 3.

Improper integrals in the basic case

Based on identities (1.14)-(1.16), estimates (1.11)-(1.13), taking into account (1.17)-(1.19), we can obtain estimates of improper integrals along the solution of system (1.10).

Theorem 2. Let the conditions of the lemmas 1 – 3 be satisfied, the matrix

A be

a Hurwitz matrix, function

( ) 0 .

Then, along the solution of system (1.10), the

improper integral

2 (t) N y2 (t) N

y2

I

1

= [ ( (t)) (t)]dt =

[N

n

(t)]dt

0

1

1

n

(1.20)

0

0

( )

l1 =

( )d = c1, | c1

|< ,

(0)

l = y

*

= y

*

( )F y( )

y

*

(0)F y(0),

| l

|< ,

(t)F y(t) |

(1.21)

1

1

0

1

1

1

where

F

is a constant matrix of

n n,

order, N

i

= N

(a

, , a

1

,

, ,

), i =

0, n.

1

i

0

n

1

n

Proof. The product

( (t)) (t) = [ (t) a y (t) a	n 1	y	(t)]
0 1		n

[ y

(t)

y

(t)

n 1

y

(t) (t)] = z* (t)Nz(t), t I ,

where

z(t) = ( (t),

1 2

2

3

n

n

Then the improper integral

( )

I1 = z

*

(t)Nz(t)dt = (

( )d = c1, | c1 |<

(t)) (t)dt =

0

0

(0)

due to finiteness

(t), t I ,

where

N is a constant matrix of (n 1)

As follows from Lemma 3.4

(Q = N ), we get the equality

z

*

(t)Nz(t) = N

2

(t) N y

2

(t) N

y

2

(t)

d

*

(t)F y(t)],

n

n

[ y

0

1 1

dt

1

Now the relations (1.20), (1.21) follow from (1.18), (1.19), where

y1(t), , yn (t)).

,
(n 1)	order.
t I.

	l1	=		d	[ y* (t)F1 y(t)]dt = y* (t)F1 y(t) |0 , | l1 |< ,
			dt
			0
	due to finiteness of	y(0)		and y( ). The theorem is proved.

Theorem 3. Let the conditions of the lemmas 1 – 3 be satisfied, matrix A be a Hurwitz matrix, function ( ) 0 . Then for any quantity 1 > 0, along the solution

of the system (1.10) the improper integral

17

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

		I		=																								( (t))]dt =																						(t)
					[ ( (t)) (t)																																						[M
																						1					2																						2															(1.22)
			2										1						1			0																								0
					0																																					0
							M				y		2		(t) M												y		2	(t)]dt							l					0,
																										n			n											2
										1		1
	l2			*														*															*	(0)F2 y(0),													| l2			|<
		= y			(t)F2 y(t) |0								= y						( )F2 y( ) y																																													(1.23)
where	F	is a constant matrix of n n order,																															M			i		= M					( , a				, , a								,		, ,					n	), i = 0, n.
	2																																									i				0							n 1			1
Proof. From inclusion														( )							0			it follows that
											( )									,													1		, ,												1
																																	0									R .
																			0		( )
Then for any quantity																	> 0 we get inequality																						( )															1		2	( ),			,				1	.
															1																																							0										R
																																												1							1
It follows that along the solution of the system (1.10) the inequality holds
									( ) (t)																		1				2	( (t)) 0,														t, t I ,
																							1				0																																					(1.24)
																	1
where
	( )				(t)					1			2			( (t)) = [ a y																		a											y			]			[ y								y
										0																																		n 1						1								2			2
			1				y	1										[ a y												0		1						y		]							n						1		1
								]							1										a																		= z				(t)Mz(t),										t I ,
																																									2					*
						n	n				1			0						0		1												n 1					n
where M is a constant matrix of (n 1) (n 1)																																					order. Further, applying Lemma 3.4,
where	Q = M ,				we get
	*	(t)Mz(t) = M										2		(t) M y							2		(t)				M												y		2	(t)					d				*		F y(t)],							t	I.
	z																																					n			n								[ y
										0									1	1																										dt								2

Now the relations (1.22), (1.23) follow from (1.17), (1.19), (1.24). The theorem is proved.

Theorem 4. Let the conditions of the lemmas 1 – 3 be satisfied matrix Hurwitz matrix, function ( ) 0 . Then for any quantities 0 , 1, , n

solution of the system (1.10) the improper integral

I3 = [ 0 (t) 1 y1 (t) n yn (t)]2 dt =

0

= [ 0 2 (t) 1 y12 (t) n yn2 (t)]dt l3 0,

0

,

A

is a along the

(1.25)

l	= y* (t)F y(t) | = y* ( )F y( ) y* (0)F y(0), | l				3		|<		(1.26)
3	3	0	3	3
where F3 is a constant matrix of n n order, i
				= i ( 0 , , n ), i = 0, n.

18

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

The proof of

[		y
0	1	1

order.

the

	y		]	2
n		n

theorem is similar

*	(t) z(t),	where
= z

to the proof of Theorem is a constant matrix of (n

3, where

1) (n 1)

Lection 4.

Absolute stability and the Iserman’s problem in the main case

Absolute stability. Based on the results presented above, in lectures 1–3, the conditions for the absolute stability of the equilibrium position of system (1.1), (1.2) can be formulated.

Theorem 5. Let the following conditions be satisfied:

1.			Matrices					A	are Hurwitz matrices, function ( ) 0														;
2.			There is a row vector = ( 1												, , n ) such as
								B = 0, AB = 0, , An 2 B = 0, = An 1B = 1;
3.			Matrix rank R =															is equal to n;
												* , A* * , , A*n 1 *
4.			The conditions of the theorems 2–4 are satisfied.
Then
		N			M		)				(t) (			N M ) y			(t) (			N		M			) y		(t)]dt < .
[(
										2						2										2
	2			0		0		0					2	1	1	1 1			2		n			n	n	n
0
																											(1.27)

where

	2

is a number.

Proof. Since the characteristic polynomial of the matrix

A

is equal to

( ) =| I

A |= a

a a ,

n

n 1

n

n 1

1

0

then with

an 1

1

an 1

1

0

1 = an 1 > 0, 2 =

> 0, 3

= an 3

an 2

an 1 > 0, , n = a0 n 1 > 0

an 3

an 2

an 5

an 4

an 3

matrix A will be a Hurwitz matrix, where yn

= a0 y1 a1 y2 an 1 yn ( ). Conse-

quently, the transformation

y = R

x is nonsingular and matrix

A is Hurwitz matrix.

*

From inclusion ( ) 0 we get a limitation on system resources. Under con-

ditions 2), 3) of the theorem, the transformation is nonsingular and identities (1.14)- (1.16) and estimates (1.11)-(1.13) are true. Since the conditions of Theorems 2-4 are satisfied, improper integrals (see (1.20)-(1.23), (1.25), (1.26))

19

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

			I									N				N y										N										y						]dt l						=				c < ,							,
					=		[								2										2															2
		2		1							2	0						2				1	1										2		n					n						2 1				2								2
						0
					I			=				[ M						2		M y								2	M										y			2	]dt l					0,				| l			|< ,
							2																													n						n					2						2
																0									1		1
												0
						I															y								y												]dt l					0,			| l			|< .
												=	[						2								2											2
										3							0								1	1								n				n							3					3
												0
Then
					I	I						I		=							N				M						)									(					N M ) y
																	[(																			2																				2
				2	1				2				3							2			0							0			0											2	1				1				1		1
						(								N		0							) y									]dt								c l						l l						< ,
																	M														2
													2			n						n				n					n						2								2 1			2		3
where	| c |< ,					| l1 |< , 2											is any number. This implies the estimate (1.27). The theo-
rem is proved.
Theorem 6. Let the conditions of the theorem 5 be satisfied, the matrix																																																											A B S,
0 < 0 ,			0			<					0	is a Hurwitz matrix, and let, moreover:
								2 N0 M0										0						0, 2 Ni Mi i																																			(1.28)
																																													> 0, i = 1, n.
Then the equilibrium position of the system (1.1), (1.2) is absolutely stable. If, in
addition, the quantity 0																= 0					,					where							> 0								is an arbitrarily small number, then
in the sector				[0, 0							] Iserman's problem has a solution.
Proof. Since the conditions of Theorem 5 are satisfied, the inequality (1.27) is
true. From (1.27) given the first inequality from (1.28), we get
																			n
																	[ ( 2 Ni Mi i ) yi2 ]dt < ,
																	0	i=1
where 2 Ni			Mi					i > 0, i = 1, n. Further, applying Lemma 5, we obtain																																																		lim y(t) = 0.
																																																										t
Note that the function														y(t),						t I						along the solution of the system (1.10) is conti-
nuously differentiable													and				satisfies											the conditions | y(t) | m0 , | y(t) | m1, t, t I.

n

Scalar function V ( y) ( 2 Ni M i i ) yi2 > 0 for any

n	,
y R

y 0, V (0) = 0.

i=1

Improper integral V ( y(t))dt < .

0

statement of Lemma 2, the limit

Consequently, lim y(t) = 0. Then, according to the

t

lim x(t;0, x0 , ) = 0, x0 , | x0 |< , , 0 .

t

20