86

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

Function

( )		= { ( ) C(R		, R ) | ( ) = ( ), 0 ( )								,
			1		1						2
	0		, R		, (0) = 0,	| ( ) |	,	0 <		0
0,										< }
				1
						*			*

(1.42)

where

> 0

is an arbitrarily small number. Notice that

( ) = { ( ) C(R			, R ) | 0 ( )					, 0,
		1	1				2
1	R	, (0) = 0, | ( ) |		, 0 <	0	< }
,
	1
			*		*

(1.43)

The automatic control systems encountered in practice are systems with limited

resources, and for such systems the function	( )	satisfies conditions (1.42), (1.43).

As the value		*	,	0 <	*	< , > 0	is an arbitrarily	small number, then
inclusions (1.42), (1.43) contain all nonlinearities from the sector								[0,	0	].

As follows from the inclusions (1.42), (1.43) equations of motion (1.41) can be represented as

z = A z B ( ), = Sz,

z(0) = z

,

| z

|< , t I,

1

1

0

0

where

x

A B D

B E

B

=

B

S = D

E ,

z =

,

A =

,

,

1

D

E

1

1

(1.44)

> 0 is an arbitrarily small number, function

( )

,

A = A ( ).

1

1

1

The equilibrium state of the system (1.41), (1.42) is determined by the solution of

algebraic equations A z

B (

*

) = 0,

*

= Sz .

If the matrix

A

is a Hurwitz

1 *

1

*

1

matrix, function ( ) 1 turns to zero only when

= 0,

then the system (1.41),

(1.42) has a single equilibrium state = ( , ) = 0. Consequently, x* = 0, * = 0, where * = 0.

Notice that the equilibrium position corresponds to the trivial solutions x(t) 0,

(t) 0,		t		I	of the system (1.41), (1.42). In this work, we study the asymptotic
stability of				the	whole unperturbed motion x(t) 0, (t) 0, t I of					the system
(1.41), (1.42) for any ( ) 0 (or z(t) 0, t I ,								( ) 1 ).
We assume that for a sufficiently small neighborhood of the point										= 0, the
function ( )					can be approximated by a linear function ( ) = . Consequently,

31

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

when | |< , > 0 is an arbitrarily small number, and the perturbed motion equation (1.44) can be written in the form

z = A z B Sz = A ( )z,			z(0) = z	, | z	0	|< , t I ,
1	1	1	0

(1.45)

where A ( ) = A B S,					<	0	,		0
1	1	1
Hurwitz matrix	A1	( ).	If the matrix A1( ),
then there is a number			1	> 0	such that | x(t) |<

is a limit value					determined from the
0	< 0 ,				is a Hurwitz matrix,
1 when | x0	|< 1		, moreover lim z(t) = 0.
					t

So when the matrix

A ( ),

where

0

<

0

is a Hurwitz matrix, the trivial

1

solution

z

(t)

0,

t

I

(

x(t) 0, (t) 0,

t I

) of the system (1.45) is asymp-

totically stable by Lyapunov for t .

Definition 4. Trivial

solution

x(t) 0 ,

(t) 0,

t I

of the system (1.41),

(1.42) is called absolutely stable if: 1) matrices

A

,

A

( ),

0

,

0

>

0

1

1

are

Hurwitz matricies

(asymptotic Lyapunov

stability

for

t );

2)

for

all

( ) 1

solution

of

the

differential equation

(1.44) possesses the

property

lim z(t;0, z0 , ) = 0,

| z0

|<

(or lim x(t;0, x0 , 0 , ) = 0,

lim (t;0, x0 , 0 , ) = 0,

t

t

t

,

0

,

|

x

|< ,

|

0

|<

).

0

Definition 5. The conditions for the absolute stability of the system (1.41), (1.42)

are

the

relations

connecting

the constructive

parameters

of

the

system

( A, B, D, E,

0

),

at which the equilibrium state

(x*

= 0, * = 0) is absolutely stable.

Problem 3. Find the condition for the absolute stability of the equilibrium state of

the system (1.41), (1.42).

Function

(t) = Sz(t),

t I

is a control formed by the principle of feedback,

and the vector string S = (S1, , Sn 1 ) R

n 1

is called a feedback vector. Iserman's

problem is that how to choose feedback vectors

S R

n 1

,

so that from the asymptotic

stability of the trivial solution

z(t) 0,

t I

of the linear system (1.45) for any

,

0

,

=

0

we

get

absolute stability

of

the trivial

solution

x(t) 0,

0

t I

( ) 0,

of the system (1.41), (1.42), where

0 is a limiting value of the Hur-

witz matrix

A1( ),

> 0 is an arbitrarily small number.

Definition

6. We assume that in the sector

[ ,

0

]

Iserman's problem has a

solution if: 1) there is a feedback vector S Rn 1

such that

2) for any

0 = 0 ;

( ) = , 0 0 the solution of system (1.41) is asymptotically stable;

3) for any ( ) 1 the trivial solution of system (1.41), (1.42), (1.43) is absolutely stable.

Problem 4. Find the sector [ , 0	= 0 ], where Iserman's problem has a
solution.

32

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

Note that, as follows from [11], the Iserman problem does not always have a solution. The problem consists in finding such a feedback vector S Rn 1, for which

in sector [ , 0 ],

Iserman's problem has a solution, where > 0 is an

0 = 0

A1( ).

arbitrarily small number, 0 is the limiting value of the Hurwitz matrix

Nonsingular

transformation. The

characteristic

polynomial of

the matrix

A1 = A1

( )

is equal to

( ) =| I

n 1

A |= n 1 a

n a

n 1 a a ,

1

n

n 1

1

0

where

I

n 1

is an identity matrix of order

(n 1) (n 1),

a = a ( ),

i = 1, n.

As

i

i

follows from the Hamilton-Cayley theorem, the matrix

(A1) = 0

. Then

A

n 1

=

a

A

n

a

A

n 1

a A a

I

.

n

n 1

n 1

1

1

1

0

Lemma 6. Let the row vector

= ( ,

, ,

,

) R

n 1

such that

2

n

n 1

1

n 1

n

0.

B = 0, A B = 0, , A

B = 0, A B

(1.46)

1

1

1

1

1

1

1

Then equation (1.44) can be represented as

		y	= y		,		y	2	= y	, , y		n	= y		n 1	,
y		1		2					3							B ( ),
		= a y		a y					a y					A
															n
	n 1	0	1			1		2		n	n 1				1	1

(1.47)

where

y1 =

Proof.

get

z, y		= A z, , y		n	z, z = z(t),	y	= y (t), i = 1, n 1.
	2		n 1	= A
		1		1		i	i

Consider the first equation from (1.44). Multiplying it from left by

,

we

z = A z B ( ) = A z, z(0) = z				, t I,
1	1	1	0

(1.48)

by virtue of equality B1		= 0,		where z = y1			, A1z = y2. Consequently,
t I.
Differentiating identity (1.48) by t , we obtain
y	= z = A z = A [A z B ( )] = A2 z = y ,								y (0) = A z ,
2	1	1	1	1			1	3	2	1	0

where A1B1 = 0. In a similar way we obtain the following system of equations:

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

t I,

differential

y	= z = A2 z = A3 z = y			,	y	(0) = A2 z		, ,
3	1	1	4		3	1	0

33

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

			y	= y		n 1	, y	= a y a y An B ( ),
			n				n 1	0 1 1 2	1 1
where	y	n 1	(0) = An z		.	The lemma is proved.
			1	0

Lemma 7. Let the conditions of Lemma 6 be satisfied, and let, in addition, the rank of the matrix

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

(1.49)

of (n 1) (n 1)

1. there exists

order be equal to a row vector =

n

(	0	,

1,

1

where

, ,	n

(*)

) R

is a transpose sign. Then:

n 1	such that

= 0 y1 1 y2

n yn 1;

(1.50)

2. if

y1

= z = 0, y2 = A1z = 0, , yn 1

n

z = 0.

= A1 z = 0, then

Proof.

Notice

that

rank

R = n 1

then

and

only

then

the

vectors

*

, A

, ,

A

*

* *

*n *

are linearly independent. Since vectors

form a

, A1

, , A1

1

1

basis in

R

n 1

,

then vector

S

*

R

n 1

can be represented unambiguously in the form

S

*

= 0

*

*

*

*n

*

.

Then

1 A1

n A1

= Sz = z A* * z

A*n * z =

y y

2

n

y

n 1

.

0

1 1

n 1

0 1

1

Now the second equation from (1.44) can be written as (1.50).

is controllable. Out

On the other hand, it follows from (1.49) that the pair

(

, A )

*

*

1

of control of pairs (

* , A* )

n

z = 0,

it follows that the equalities z = 0, A z = 0, , A

1

1

1

which entail

z = 0.

Consequently, from

y	i	= 0,

i = 1, n 1

it follows that

z = 0.

The lemma is proved.

From lemmas 6, 7 it follows that if equalities (1.46) are satisfied and the rank

R = n 1,

then the system (1.41) is equivalent to the system (1.47), (1.50). Moreover,

from lim yi (t) = 0,

i = 1, n 1

it follows that lim z(t) = 0.

t

t

Introducing the notation

0

1

0

0

0

0

0

1

0

0

0

A =

, B =

, S = ( , , , ),

0 1

n

a0

a1

a2

an

n

A1

B1

equations of motion (1.47), (1.50) can be represented as:

34

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

y = Ay B ( ), = S y, ( )	.
1

(1.51)

Consider the differential equations (1.44), (1.51). As follows from the lemmas 6,

7, if z(t) 0, t I is a trivial solution to the equation (1.44), then				y(t) 0,	t I
is a trivial solution to the equation (1.51). If | |< ,					<
	( ) = ,					0	,
1				0
then the equation of perturbed motion can be written as

y = Ay B S y = A( ) y,	y(0) = y	, | y	|<	, t I,	(1.52)
	0	0	2

where A( ) = A B S, 0 < 0 , 1 > 0 is an arbitrarily small number. If

the matrix

| y(t) |<	2

A( ) when

is a

| y	0

Hurwitz			matrix,	then
|<	2	,	moreover	lim
				t

there is

y(t) = 0.

a number	2 > 0	such that
In this	case, the	solution

y(t) 0, t .

t I

of the system (1.52) is asymptotically stable by Lyapunov when

From (1.44), we get (1.51): 1)

*	A R	* 1	,
A = R
	1

*	B	,
B = R
	1

S = SR	* 1	;

2)	A1 = R	* 1	*	,
			AR

B1 = R

* 1

B,

*

; 3) matrices

A1,

A

are similar, therefore

j ( A1 ) = j ( A),

S = SR

j = 1, n 1;

4) SB1 = S B.

Notice that the nonsingular transformation above is different from the transfor-

mation from [12;

§

5, theorem 2].

Lecture 7.

Properties of solutions in a simple critical case

We can show that solutions of the system (1.41), (1.42), as well as (1.47), (1.50), are limited. These properties can be used when estimating improper integrals.

Theorem 8. Let the matrix

A1

be a Hurwitz matrix, i.e.

Re	( A ) < 0,
j	1

j = 1, n 1,

function ( ) 1 , and let, in addition, equalities (1.46)take place and the rank

R = n 1. Then true estimates
| z(t) | c0 , | z(t) | c1,	t,	t I = [0, ),	(1.53)

			| yi (t) | mi1, | yi (t) | mi2 , i = 1, n 1, t I,							(1.54)
			| (t) | c2 , | (t) | c3 , t,					t I ,		(1.55)
where mi1, mi 2 = const < ,								i = 0,1,2,3, functions		z(t),
				i = 1, n 1, ci = const < ,
			(t), t I
yi (t), i = 1, n 1,				are uniformly continuous.

35

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

Proof.

From inclusion

( ) 1

it follows that

| ( (t)) |

t,

t I.

As matrix

A = A ( )

is a Hurwitz matrix, i.e.

a = max

1

1

1 j n 1

then

e A1t

ce (a )t

t, t I , c = c( ) > 0, > 0 is

an

arbitrarily

[13; § 13, inequality (7)].

The solution of differential equation (1.44) can be written as:

A t

t

(t )

0

A

1

z(t) = e 1 z

e 1

B ( ( ))d , t I.

0

*	, 0 <	*	< ,

Re	(A ) < 0,
j	1

small number

Then

A t

t

(t )

| z(t) |

| z

|

A

| B

|| ( ( )) | d c | z

| e

(a )t

e

1

0

e 1

0

1

0

t

1

1

ce

(a )t

B

e

(a )

d

= c | z

| e

(a )t

ce

(a )t

B

e

(a )t

=

*

0

*

1

1

a

0

a

= c | z

| e

(a )t

1

c

B

( 1 e

(a )t

) c

,

t,

t I ,

0

*

a

1

0

(a )t

where

e

1,

t, t I , a < 0.

Hence, we get a limited solution of the sys-

tem (1.44). Consequently, we get a limited solution of the system (1.41), (1.42). It follows from (1.44) that

|

z(t) |

A

|

z(t) | | B

|| ( (t)) |

A

c | B

|

*

= c

,

t, t I ,

1

1

1

0

1

1

| (t) |

S

| z(t) | c

,

| (t) |

S

| z(t) | c

,

t,

t I

2

3

From limitation

z(t), (t),

t I

it follows that they are uniformly continuous.

y (t) = z(t),

y

(t) = A z(t), , y

n

z(t),

t I ,

As

2

n 1

(t) = A

then

1

1

1

| y1 (t) | | || z(t) | | | c0 = m11, | y2 (t) | | | A1

| z(t) | m21, ,

| y

n 1

(t) | | | An

| z(t) | m

n 1,1

, t,

t I.

1

| yi (t) | |

yi 1 (t) | mi 2 ,

From (1.47) it follows that

i = 1, n, t,

t I ,

| y

(t) | | a

|| y (t) | | a

|| y

(t) |

m

, t, t I.

n 1

0

n

n 1

An B

*

n 1,2

1

1

From the limitations of derivatives

(t), i = 1, n 1 we get uniform continuity of

yi

function

yi (t), i = 1, n 1, t I. The theorem is proved.

36

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

It should be noted that: 1) from estimation | z(t) | c0 ,

t,

t I we get

1

c

< ,

lim | z(t) |=| z( ) | c | z0

|

c B1 * = c0 ,

0

a

t

t, t I

we get

lim |

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

*

lim | z

t

t

Theorem

9.

Let

the

conditions

of

the

n

0.

Then along the solution of the

= A1 B1

identities are true

a < 0; 2) from estimation | y(t) | m0 ,

		*
	(t) | R		c0	= m0 ,	m0 < .

lemmas 6, 7 be satisfied, value system (1.47), (1.50) the following

where

=

( (t) = 1 (t)

1a y (t)

1a y

(t), t I,

0

1

n n 1

(t) =

y (t) y

(t)

n

y

n 1

(t),

t I ,

0 1

1 2

(t) =

0

y

(t) y

(t)

n 1

y

n 1

(t) (t),

t I ,

2

1 3

n

(t) = y

(t),

t I.

n 1

(1.56)

(1.57)

(1.58)

Proof. Along the solution of the system (1.51) (v. (1.47), (1.50)) the following identity is true

y

n 1

(t) = (t)

= a

y (t) a y

(t) a

y

(t) ( (t)),

t I,

0

1

1 2

n

n 1

n

0.

where

= A B

From here we get the identity (1.56). Identity (1.57) follows

1

1

from (1.50). As

(t) =

0

y

(t) y

2

(t)

n

y

n 1

(t), t I ,

then and identity

1

1

(1.58) is true. The theorem is proved.

Lemma 8. Let the conditions of the lemmas 6, 7 be satisfied, matrix

A1 be a Hur-

witz matrix, function ( ) 1. Then for any constant matrix

Q

of (n 2) (n 2)

order quadratic form

*

(t)Q (t), (t) = ( (t), y1 (t), , yn 1 (t))

can be represented

as

*

(t)Q (t) = q

2

(t) q y

2

(t) q

y

2

(t)

d

*

(t)Fz(t)], t

I = [0, ), (1.59)

n 1

n 1

[z

0

1

1

dt

where F is a constant matrix of order

(n 1) (n 1).

Proof. Let n 1 = n1. It is easy to verify that:

a) if n1, k are odd numbers and k n1, then

n1 k

d

1

y2

n k

2

y

=

y

y

y

y

( 1)

, k m <

1

;

n1 1

k

dt

k

n1

k 1

2

n1 k

2

2

37

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

b) if n1

is odd, k

is an even number and k < n1, then

n k 1

d

1

y

=

y

y

y

y

( 1)

2

y

y

k

k

1

k 1

n

1

k 1

n k 1

n

dt

n

1

1

1

2

2

n k 1

1

n k 1

( 1)

2

y

2

,

k m <

;

n k

1

1

2

1

2

c) if n1

is

even, k

is an odd number, and k < n1, then

n k 1

d

1

y

=

y

y

y

y

( 1)

2

y

y

1

n

1

k

k

k 1

n k 1

n k 1

dt

n

1

1

1

2

2

n k 1

1

n k 1

( 1)

2

y

2

,

k m <

;

n k 1

1

2

1

2

d) if

n1

is even,

k

is an even number, and

k n1,

then

n k

d

1

1

n k

y

=

y

y

y

y

( 1)

2

y

2

,

k m <

;

1

n k

1

k

k

k

1

n 1

dt

n

1

2

1

2

2

e) if

k

is odd,

s

is an odd number and

s > k,

then

d

k s

1

k s

1

k s

2

y

y

=

y

y

y

y

( 1) 2

y

y

( 1)

2

y

, k m <

;

k

s

dt

k

s 1

k 1

s 2

k s

1

k s

k

s

2

2

2

2

f) if k is odd, s

is an even number, and

s > k,

then

d

1

k s 1

1

k s 1

y2

y

y

=

y

y

y

y

( 1)

2

, k m <

;

k

s

dt

k

s 1

k 1

s 2

2

k s 1

2

2

g) if k is

even, s

is an odd number and s > k,

then

d

1

k

s 1

2

k s 1

y

y

=

y

y

y

y

( 1)

2

y

, k m <

;

k

s

dt

k

s 1

k 1

s 2

2

k

s 1

2

2

38

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

h) if k is even, s

is an even number and s > k, then

d

k s

2

y

,

y

y

=

y

y

y

( 1)

2 y

k

s

dt

k

s 1

k 1

s 2

k

s

2

In particular, when

n1

= 3,

we get:

y

=

d

( y y

1

1

dt

1

3

2

y

=

1 d

( y

2

),

y y

=

1 d

( y

2

),

y y

=

d

( y y

) y

2

,

3

2

3

2

2

3

2 dt

1

2 dt

1

1

dt

1

when n1 = 4 we get:

k m

y	2	),
	2
y		2	y	3

	<	k s				.
				2
	y			=	d		( y
			2		dt				2
	=	1 d			( y		2	),
		2 dt					2

	y	) y	2	,
			3
	3

= y3 ;

y

=

d

( y y

y

y

) y

2

,

y

=

d

( y

y

1

y

2

),

y

=

d

( y

y

) y

2

,

4

2

3

2

2

4

3

3

3

4

4

1

dt

1

3

dt

2

dt

y

=

1 d

( y

2

),

y y

=

1 d

( y

2

),

y y

=

d

( y y

) y

2

,

y y

=

d

( y y

1

y

2

),

4

4

2

3

2

2

4

3

2

2 dt

1

2 dt

1

1

dt

1

1

dt

1

2

y

y

=

d

( y

y

) y

2

,

y

y

=

1 d

( y

2

).

2

4

2

3

3

4

3

dt

3

2 dt

As the quadratic form

*

(t)Q (t)

contains terms with constant coefficients of the

product of components of the vector

(t),

then true representation of the form (1.59).

The lemma is proved.

Lemma 9. Let the conditions of lemma 8 be satisfied. Then the equality holds

*

(t)Q (t)dt

=

[q

2

(t) q y

2

(t) q

y

2

]dt l

,

(1.60)

n 1

0

1

1

n 1

0

0

0

d

l

=

[ y

*

(t)Fy(t)]dt = y

*

( )Fy( ) y

*

(0)Fy(0) = y

*

(t)Fy(t) |

,

| l

|< .

(1.61)

0

dt

0

0

0

Proof. Integrating identity (1.59), taking into account the estimate (1.54), where

| yi

(0) | mi,1 ,

|

yi ( ) | mi,1 ,

i = 1, n 1,

we

obtain

the

relations

(1.60),

(1.61). The

lemma is proved.

Lemma 10. Let the following conditions hold:

1. vector function

y(t) = ( y1

(t), , yn 1 (t)),

t I

= [0, )

is

bounded

| y(t) | a,

t I

and is continuously differentiable, moreover

|

t I ,

a < ,

c < ;

y(t) | c,

2. scalar continuous function V (x) > 0 for any

x R

n 1

,

x

0,

V (0) = 0;

3. improper integral

Then lim y(t) = 0.

t

V ( y(t))dt < .

0

39

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

Proof. Let the conditions of Lemmas 1)-3) be fulfilled. Let us show that

lim

y(t) = 0.

t

Suppose the contrary i.e.

lim y(t) 0.

Then there is a sequence {tk } [0, ) such

t

that | y

k

(t) | > 0,

k = 1,2, . We choose

tk 1 tk

m > 0.

As y(t), t I

is conti-

nuously

differentiable and

| y(t) |< c,

t, t I ,

then

| y(t) y(t

k

) | c | t t

k

|,

t [t

m

, t

m

],

k = 1,2, .

Then

k

2

k

2

t

m

k

2

V ( y(t))dt

V ( y(t))dt,

0

k =1

t

m

k

2

where			| y(t) |=| y(tk ) y(t) y(tk ) | | y(tk ) | |
t	m
k		2
		V ( y(t))dt V		m, V =		min V (x), then
			min	min	0	|x| a
t		m
k		2

y(t) y(tk ) | c	m	= 0 > 0. As
2
V ( y(t))dt = .
0

This contradicts the third condition of the lemma. The proof of Lemma 10 for the

case when

y(t),

t I

is the scalar function is given in [2].

Lecture 8.

Improper integrals and absolute stability in a simple critical case

Based on identities (1.56)–(1.58), assessments (1.53)–(1.55) and taking into account (1.59)–(1.61) estimates of improper integrals can be obtained along the solution of the system (1.51).

Theorem 10. Let the conditions of lemmas 6, 7 be satisfied, matrices	A,	A1	be

Hurwitz matricies, function

( ) 1.

Then along the solution of the system (1.51)

improper integral

I

=

( (t)) (t)dt =

[N

2

(t) N y

2

(t)

N

y

2

(t)]dt

l

=

1

n 1

n 1

0

1 1

1

0

0

( )

c

2

=

( )d

( )d = c1 , | c1 |< ,

(0)

c

2

l

*

*

*

(0)F y(0),

| l |< ,

= y

(t)F y(t) |

= y

( )F y( ) y

1

1

0

1

1

1

where F1

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

Proof. Products

( (t)) (t) = 1[ (t) a y (t) a y

n 1

(t)]

0

1

n

(1.62)

(1.63)

40