86

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

[

0

y

2

(t)

n 1

y

n 1

(t) (t)] = * (t)N (t), t I ,

n

where (t) = ( (t), y1 (t), , yn 1 (t)). Then improper integral

( )

c

2

I1 =

*

( )d

( )d = c1,

| c1 |< ,

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

0

0

(0)

c

2

due to the fact that

c2

(t) c2 ,

t I ,

where

N

is a constant matrix of order

(n 2) (n 2).

As follows from lemma 9

(Q = N ),

the following equality holds

*

(t)N (t) = N

2

(t) N y

2

(t)

N

y

2

(t)

d

[ y

*

(t)F y(t)],

t I.

n 1

n 1

0

1 1

dt

1

Then the ratios (1.62), (1.63) follow from (1.60), (1.61), where

d

l

=

*

*

;

| l

|<

[ y

(t)F y(t)]dt = y

(t)F y(t) |

1

dt

1

1

0

1

0

due to limited

y(0)

and

y

( ).

The theorem is proved.

Notice that

( )

(0)

c

2

( )d = c1.

I1 = ( ) dt =

( )d =

( )d

0

(0)

( )

c

2

Then for

2

0,

2

I

1

c ,

if

2

< 0,

then

2 I1

=| 2

| ( I1 ) | 2 | c1.

2

1

Consequently, for any value 2

value

2

I

1

=

2

c .

1

Theorem 11. Let the conditions of lemmas 6, 7 be satisfied, matrices

A,

A1 be

Hurwitz matricies, function ( ) 1.

Then for any value

1

> 0

along the solution

of the system (1.51) improper integral

1 2 ( (t))]dt =

I

2

=

[ ( (t)) (t)

1

1

0

0

(1.64)

= [M 2

(t) M

1

y 2 (t) M

n 1

y 2

(t)]dt l

2

0,

0

1

n 1

0

l

2

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

| l

2

|< ,

(1.65)

2

0

2

2

41

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

where Mi	= Mi ( 1 ), 1 > 0, i = 0, n 1,						F2	is a constant matrix of order (n 1) (
N0 0.
Proof. From inclusion ( ) 1 it follows that
	( )				,				1	,	1
									0		, R .
				0		( )
Then for any value 1 > 0											1		2	( ),	,
				an equality holds ( ) 1 > 1 0

Hence it follows that along the solution of the system (1.51) the inequality holds

n 1)

R	.
1

where

and M

			( (t)) (t)										1			2	(
													0
				1							1
( (t)) (t)							1			2	( (t)) = (
							0
			1	1
	n	y	n 1	) 1		1 1 ( a y
					1			0								0	1
is a constant matrix of order															(n

(t)) 0,

(t) 0, t, t I ,

(1.66)

a y

a

y

)

1

(

y

n

n 1

0

0 1

1

1

a

n

y

n 1

)2

= *

(t)M (t), t I ,

2) (n 2).

Next applying lemma 8, where

Q = M , t I.

we get

*	2		2	(t) M		y	2	(t)	d	*	(t)F y(t)],
	(t)M (t) = M	(t) M y			n 1		n 1			[ y
	0	1	1						dt		2

Then the ratios (1.64), (1.65) follow from (1.60), (1.61), (1.66). The theorem is proved.

Theorem 12. Let the conditions of the lemmas be satisfied 6, 7, matrices

																		,		, ,
	be Hurwitz matrices, function ( ) 1. Then for any values																0		1		n 1
	the solution of the system (1.51) improper integral
								y (t)					(t)]2 dt =
	I	3	= [ (t)						n 1	y	n 1
				0			1 1
			0
			= [ 2		(t) y2 (t)							y2		(t)]dt l	3	0,
				0			1 1		n 1			n 1
			0

A, A1

along

(1.67)

l = y*

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

3

|< ,

(1.68)

3

3

0

3

3

where (

0

,

, ,

n 1

), i = 0, n 1,

F is a constant matrix of order (n 1) (n 1)

i

1

3

The

proof of

the

theorem is

similar

to

the

proof of

theorem

4,

where

[

y

n 1

y

n 1

]2 = * (t) (t),

is

a constant matrix

of

order

0

1 1

(n 2) (n 2).

42

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

Absolute stability. Based on the results described above, conditions for the absolute stability of the equilibrium position can be formulated for the system (1.41), (1.42).

Theorem 13. Let the following conditions hold:

1. matrices A1, A1( ), 0 < 0 < 0 are Hurwitz matrices, the function

( ) 1 ,							is an arbitrarily small number;
		2. there is a vector							*		R	n 1		such that
							B = 0, A B = 0, , An 1B = 0, An B 0;
							1					1	1			1		1					1
		3. rank of the matrix								R =						, , A*n *					is equal to n 1;
												* , A* *
															1	1
		4. estimates (1.62), (1.64), (1.67) are true, and let besides:
	2	N	0	M		0	0,			2	N M			1	0,		2	N	i		M	i	> 0,	i = 2, n 1.	(1.69)
							0				1				1								i

Then the equilibrium state of the system (1.41), (1.42) is absolutely stable. If in

addition the value

the sector	[ ,	0	]

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

Iserman’s problem has a solution.

Proof. Let the conditions 1) - 4) of the theorem hold. Then

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

(1.70)

where	| c	|< ,	| l1 |< , | l2 |< ,	|
	1

inequalities (1.63), (1.65), (1.68). From (1.70) taking into account

l	3	|< ,	2	is any number,		2	I	1	|	2	| c	1	,

the first inequality from (1.69), we get

due to

where		N
	2	1

2 Ni Mi i

n 1

( 2 Ni Mi i ) yi2

(t) dt < ,

0

i=1

M

1

0,

2

N

i

M

i

> 0,

i = 2, n 1.

Hence in case when

1

i

> 0,

i = 1, n 1,

we get

[(

2

N

i

M

i

) y2 (t)]dt < , i = 1, n 1.

i

i

0

43

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

Then as follows from lemma 10 the limit

by Lemma 7 the limit	lim z(t) = 0.
	t

	lim y	i
	t
	As

(t) = 0,

i =

matrices

1, n 1.

A	=
1

Consequently,

A ( ),	A	( ),
1	1

0 <

0

<

0

are Hurwitz matrices, then according to Definition 4, a trivial

solution

z(t) 0,

t

I

of the

system (1.41), (1.42)

is absolutely

stable,

where

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

t I.

Consider

the case

when

2 N1 M1

1 = 0. In

this

case as

proved

above

lim

y (t) = 0,

i = 2, n 1.

It

remains to

prove that

lim y1

(t) = 0.

As

| |

*

,

i

t

t

| (t) | c	,
2

t,

t I ,

then function

( (t)),

t I

is uniformly continuous in

t,

t I.

Then function

y	= a y (t) a y		(t) A B ( (t)),
			n
n 1	0 1	1 2	1	1

t I

is

uniformly continuous by t t I. Then according to

lim y	n 1	(t) = 0	entails
t

due the

lim t

to uniform continuity Barbalat lemma [14;

y	n 1	(t) = 0.	Then

y	(t),
i

§	21,

when

i = 1, n

Lemma

t

1,

( (t)),

1] equality we get

0 = a		lim y (t) A			B lim ( (t)),
				n
	0	t	1	1	1	t

lim (t) =	0	lim y (t).
t		t	1

Hence, taking into

account that	a	= a	( ) =| A ( ) |	is an arbitrarily small number,	y1 (t),
	0	0	1

limited function, we get	0 = lim (	0	y (t)).	As	( ) = 0	only when
	t		1

lim y		(t) = 0.	Notice that equality to zero of the sum		2	N M	1	= 0
t	1					1		1

t I	is a
= 0,	then

is one of

the features of a simple critical case compared to the main case from [6–9].

In case

0

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

definition 6 are satisfied, consequently in the sector	[ ,	0	]	Iserman’s problem

has a solution. The theorem is proved.

Lecture 9.

Aizerman’s problem. The solution of the model problem for a simple critical case

The question arises: is it possible to single out a class of regulated systems for which Iserman’s problem has a solution, without resorting to checking the conditions of absolute stability from Theorem 13.

Theorem 14. Let the following conditions hold:

1)matrices A1

( ) 1, > 0

1				0
		0 < 0 <
, A	( ),

is an arbitrarily small number,

are Hurwitz matrices, the function

A1 = A1( );

2) there is a vector * Rn 1 such that

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

44

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

3) rank of the matrix R =

*

*

*

*n

*

is equal to

n 1;

, A1

, , A1

4) the condition (1.62) is satisfied. There

is a

feedback vector S Rn 1 and

number 2 such that 2 N0

0, 2 N1 0, 2 N2

> 0, , 2 Nn 1 > 0.

Then in the sector

[ ,

0

],

0 = 0

, > 0

is an arbitrarily small number,

Iserman’s problem has a

solution,

the

value

0

is a limiting value of Hurwitz

matrices

A1 ( ) =

A B S,
1	1

	0	.

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

		I	=					N		(t) N y			(t)		N		y	(t)
				( (t)) (t)dt =		[
									2			2					2
	2	1			2		2	0		2	1	1		2		2	2
				0		0

Hence taking we get

2	N	n

into

	y	2	(t)]dt l				|
1		n 1		2 1
account that					2	N		0

	2

| c1, 0,

	| l
	2
2	N
	1

|<

, 0,

| c1 |< .
2	N	2	0, ,	N	n 1	> 0,
			2

		2	(t)dt < ,				2	(t)dt < ,				2	(t)dt < , i = 2, n 1.
		N				N y					N y
	2	0			2	1	1			2	i	i
0				0					0

Consider two cases: 1) case when 2 Ni > 0, i = 1, n 1;

	2	N	i	> 0, i = 2, n 1.

In the first case we have the inequality

2) case when

	N = 0,
2	1

			N	y		2	(t)dt < ,
		2
			i		i
	0
Consequently,	lim yi (t) = 0,			i		= 1, n 1,
	t
conditions of Hurwitz matrices			A1		( ). Then

blem has a solution.

In the second case we have the inequality

2 Ni yi2 (t)dt < ,

i = 1, n 1.

the value		0	is

	0	=	0	, >

i = 2, n 1.

determined from the

0	and Iserman’s pro-

0

Then lim yi (t) = 0, i = 2, n 1. Next repeating the proofs of theorem 6 from the
t
condition lim y
	n 1	(t) = 0,	we get lim y (t) = 0.			Then in the sector [ ,		0	]
t					t 1

Iserman’s problem has a solution. The fulfillment of condition 4) is enough to check for the values 2 = 1, or 2 = 1. The theorem is proved.

45

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

Solution of model problem. The effectiveness of the proposed method for determining the conditions of absolute stability and solving the Aizerman problem will be shown on examples.

Example 1. The differential equation of the regulated system has a form

x1 = x1 x2 , x2 = 3x1 2x2 ( ), = ( ), = 3x1 2x2 2x3 , (1.71)

where ( ) = ( ), ( ) 1. For this example, the source data is as follows

1		1			0		D = ( 3, 2),	E = 2, n = 2.
A =			,	B =		,
	3	2			1

1.The characteristic

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

Matrix

polynomial of the matrix It is easy to check that matrix

	A	has a form
A	is Hurwitz matrix.

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

Function		z(t),	t I	is a solution of the differential equation
						z =
= Sz.	Characteristic			polynomial of the matrix	A1 ( )

A ( )z B ( ),
1	1
is	equal to

((())=|)=| III AAA((())|=)|= (1(1 ))) 222...
					333	222
1	11	3	33	1	11

For Hurwitz of the matrix

A	( )
1

it is

necessary	and	sufficient, that 1	> 0, 2 > 0, (1 ) > 0. Hence, it
matrix A1 ( )		is a Hurwitz matrix	with an arbitrarily small	> 0.
a = 2 ,	a1 = 1 , a2 = 1, > 0 is an arbitrarily small number.
0

follows that Coefficients

String vector

= (	,	,	) = ( 2,1, 1),
1	2	3

A1 = (1,0,0),

A	2	= (1, 1,0),
1

2		0.
A B = 1
1	1

Matrix

									2		1	1					2		1	1					0		1	0
	*	*		*	*2		*								*								* 1
R = (								) =		1	0	1 ,		R		=		1	0	0	,	R		=		0	1	1 ,
		, A			, A
		1			1
										1	0	0						1	1	0						1	1
																												1

where

| R |= 1,

rangR = 3.

Vectors

		, A		, A
	*	*	*	*2	*
		1		1

form a basis in R3. Based on

ratios	A = R* A R* 1 , B = R* B ,					S = SR* 1 from (1.71), we get
	1		1
y1	= y2 , y2		= y3 , y3 = a0 y1
						a1 y2 a3 y3 ( ), = S y = 2y1 y2 , (1.72)

46

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

where

		0	1	0				0
A =							B =				S = (2, 1, 0).
		0	0	1		,			0	,
		a	a	a
					2				1
		0	1

2. From (1.72) it follows that identities are true:

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

where (t) = y3 (t), t I. As > 0 is an arbitrarily small number, then we accept a0 = 0, a1 = 1, a2 = 1. As rangR = 3, then the systems (1.71), (1.72) are equivalent

and from

y(t) = ( y1 (t), y2 (t), y3 (t)) 0

when

z(t) = (x1(t), x2 (t), (t)) 0

when t .

3. Let us calculate improper integrals

I

, I

2

,

I

.

1

3

a) Improper integral

1

0

2

2

I

=

( (t)) (t)dt =

[N

2

(t) N

y

2

(t)

0

0

d

where

N

= 0,

N2

= 2,

N

= 1,

1

1

0

3

l

=

dt

[F (t)]dt,

0

t 0 it follows

N y		2	(t)]dt l			= c1	,
3	3				1
				( )
c1 =					( )d ,		| c
							1
				(0)

that

|< ,

F (t) =	3	y	2	(t)	1	y	2	(t),
			2				3
1	2				2

t I.

Notice that for any number

	2

the product

	2	I	1	=	2	c1,

|	c1 |< .
2

b) Improper integral

where

I

M

2

0

										2
=	[ ( (t)) (t)						1				( (t))]dt =				[M				2	(t) M y					2	(t)
		1			1		0									0							1	1
	0														0
		M		2	y2 (t) M			3	y	2	(t)]dt l		2	0, | l			2	|< ,
					2				3
																										d
	1	, M			= 0, M 2						1	1,	M3					1		1	, l		=				[F	(t)]dt.
= 1 0			1			= 1 0									= 1 0							2
																										dt	2
																									0

c) Improper integral

I	3	= [			1	y		2	y	2		3	y ]2 dt = [ 2			y2		y2		y2 ]dt l 0, | l |< ,
			0			1							3		0	1	1	2	2		3	3		3	3
		0												0
where 0				= 02 , 1			= 12 , 2					= 22		2 1 3 , 3		= 32	2 0 2 , l				=	d		[F (t)]dt.
																						dt
																				3				3
																						0

47

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

4. Let us denote limit value 0 from Hurwitz matrices A1( ).									Characteristic
equation for matrices		A1( ) has a form
1 ( ) =| I3 A1 ( ) |= 3 2 (1 ) 2 = 0.
Limit value	0	is determined from condition						(1 ) 2 > 0.	Hence, it fol-
lows that limit value			0	= 1. So, matrix			A ( ) is a Hurwitz matrix, if			0	,
							1
> 0 is an arbitrarily small number.

5. Absolute stability. Conditions of theorem 6.1 (v. (1.69)) will be written as:

			N		M			0 :					1					2	0;
		2		0			0						0					0
								0		1
			N M					0 :		2	= 0;
		2		1		1		1		1
			N		M			> 0 :											1			2	2				> 0;
		2		2			2					2				1			0			2				3
								2																1
			N		M			> 0 :						2								1		2	2					> 0.
		2		3			3					2						1				0		3					2
								3												1							0

(1.73)

Solving a system of algebraic equations (1.73) we find the limit value

0max

= 1

when 3 = 0, 2 = 2 0 ,

1 =

2

2

lim y2

(t) = 0,

lim y3 (t) = 0.

0 > 0, 2 =

2

0 . Consequently,

t

t

It remains to prove that

lim y

(t) = 0. As

y

= y

2

,

lim y

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

= const.

1

2

1

t

1

t

t

1

Let us show that

y

1

= 0.

Notice that

y

= a

y (t) a y

(t) a y

(t) ( (t)), t I

is a uniformly

3

0

1

1

2

3

3

continuous function. Then, according to the Barbalat lemma [14;

§

21, Lemma 1]

condition

lim

y

(t) = 0,

entails

lim y

3

(t) = 0.

Passing

t

3

t

lim y

3

(t) = a

0

lim y (t) a

lim y

2

(t) a lim y

3

(t) lim ( (t))

t

t

1

1

t

3

t

t

to the limit

we get

0 = a0 y lim ( (t)),

t

lim (t) = 2lim y			(t) lim y	2	(t) = 2 y	.
t	t	1	t		1

Hence, taking into

account that

a = 2 , > 0 is

an arbitrarily small number, y (t),

t I is a limited

0

1

lim y (t) = y

= 0.

function, we get lim (2 y

) = 0.

As ( ) = 0, only when = 0, we get

1

t

1

t

1

Thus, equilibrium states x* = 0, * = 0 of the system (1.71) are absolutely stable

in the sector [ ,

],

= 1 , > 0, > 0 is an arbitrarily small number.

0

=

0

0

(0;1)

Then in the

sector

Iserman’s problem has a solution,

0max = 0 = 1.

Consequently, the necessary and sufficient condition for absolute stability is obtained.

	6. We study absolute stability of equilibrium states of the system (1.71)		by
	~	~
	applying the theorem of V.M. Popov [15]. The ransfer function from input		to
		( )

the output ~ is equal to

48

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

				p 2					~
W ( p) = S( A pI	)	1	B =				=			,
				2				~	~
3			p( p		p	1)
									( )

where

~

~

are function images (t),

p is a complex variable.

( p), ( p)

respectively. It is easy to verify that

~

1

~

~

( p) =

1

~

~

( p) =

1

y ( p) =

2

,

y

2

2

,

y

2

1

p( p

p 1)

p

p 1

3

p( p

p

According to the frequency condition of V.M. Popov [15] the value mined from the condition

( (t)), t I
		~
		.
1)
	0	is deter-

						1		Re(1 i q)W (i ) 0, < < ,
						0
where																						3
																								2
			W (i ) = ReW (i ) iImW (i ),													ReW (i ) =									,
																						3	2
								ImW (i ) =					(2 3 )				,				2	.
																				=
													3	2
Modified frequency response										W* (i ) = X iY ,									X = ReW (i ),						Y =			ImW (i ).
For variables			X ,Y				the		frequency condition of									V.M. Popov						can be				written	as
	1	> 0 for all						,	0 < . As
X = qY 0
								3						(2 3 )
									2
			X =								, Y =								,	0 < ,
								3	2					3		2
then the value =							2	0 for which Y = 0,							is equal					1 = 2/3.				For these values					1
value	X ( 1) =		33/7,		ImW = 0. Notice that value													1	33/7.				Then				≤ 0.212 < 1,
																		0								0
								*

where the value 0 = 1 is obtained by the method proposed above.

To obtain a modified frequency response we calculate the values X ( ), Y ( ) :

1.	= 0,	X (0) = 3,	Y (0) = 2;

2.= 0.1, (0.1) = −3.406, (0.1) = 1.868;

3.= 0.4, (0.4) = −4.474, (0.4) = 1.053;

4.= 0.6, (0.6) = −4.737, (0.6) = 0.263;

5.= 2/3, (2/3) = −4.714, (2/3) = 0;

6.= 0.7, (0.7) = −4.683, (0.7) = −0.126;

7.= 0.8, (0.8) = −4.523, (0.8) = −1.476;

8.= 1, X (1) = 4, Y (1) = 1;

9.= , X ( ) = 0, Y ( ) = 0;

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

Based on these data, we built modified frequency response and found values

1 = 33/7,

0

q = 0. Consequently, the value 0 according to the frequency condition

of V.M. Popov is equal to 0 = 337 = 0.212 and the value 0 found by the proposed method is equal to 0 = 1 , > 0 is an arbitrarily small number. This means that

the proposed method allows us to outline, in the space of system parameters, the area of absolute stability wider than that given by the known methods (method of A.I. Lurie, method of V.M. Popov). It is shown, that the frequency condition of V.M. Popov is not effective for the study of absolute stability of regulated systems in a simple critical case with limited nonlinearity.

Aizerman’s problem. The solution to Aizerman’s problem was obtained from

the boundedness of the improper integral								I1	along the solution of the system. Notice
that improper integral
I			( (t)) (t)dt		[N	2	(t) N y			2	(t) ... N		y	2	(t)]dt l
												n 1		n 1
1					0				1 1						1
		0		0

c2

( )d = c1, | c1 |< , | l1 |< .

								c			2
Consequently,
[N				2	(t)		N y			2			(t) N						y		2		(t)]dt = c1		l	, | c l					|< .
																	n 1				n 1
	0							1	1																1			1		1
0
Then
I			( (t)) (t)dt												[N			2	(t)				2	(t)	... N				y	2		(t)]dt
																						N y						n 1		n 1
1															0								1 1
		0												0
								c
						1							2										1
					l								( )d = c1,									| l |<		, |	c1 |< ,
								c				2
							[N		2			(t) N						y		2		(t)]dt = c1			l		< ,
																n 1				n 1
								0																	1
						0

where | c1 l1 | | c1 | | l1 |< .

Example 2. We consider the equations of the following kind

̇ =		−	, ̇ = 3		− 2 + ( ), ̇= ( ), = −0.2				+ 0.4		− 0.5 ,
1	1	2	2	1	2			1		2
				, ( ) = + ( ),			( ) Φ1.				(1.74)
This	example			differs	from the previous example in			that		= S1x, where
1 = (−0.2		0.4	−0.5). Matrix

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