86

Since matrices

A,

A ( ), 0

0

<

0

are Hurwitz matrices, then, by definition 1,

1

the trivial solution

x(t) 0,

t I of the system (1.1), (1.2) is absolutely stable.

When 0

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

of definition 3 are satisfied, therefore, in the sector [0,

0

] Iserman's problem has a

1.

Matrices A are Hurwitz matrices, function

( )

0

;

2.

There is a row vector

= ( 1, , n )

such as

n 2

B = 0,

n 1

B = 1;

B = 0, AB = 0, , A

= A

3.

Matrix rank R =

*

* *

*n 1

*

is equal to

n;

, A

, , A

4.

The conditions of Theorems 2-4 are satisfied, where N0 0, N1 0, N2 > 0, , Nn > 0.

Then in the sector [0, 0

] Iserman's problem has a solution, where 0 = 0 ,

> 0 is an arbitrarily small number, the value 0

is the limiting value of the Hurwitz

matrix

A ( ) = A B S,

0

0

0

.

1

Proof. Let the conditions of the theorem be satisfied. Then the improper integral

2

N y2

y2

I =

( (t)) (t)dt =

[N

N

n

]dt l = c, | l |< , | c |< ,

1

0

1 1

n

1

1

0

0

By the hypothesis of the theorem N0

0, N1 0, , Nn

> 0.

Then from (1.29)

for

N

> 0 it follows that

1

[N0 2

N1 y12 Nn yn2 ]dt <

.

(1.30)

0

Further, repeating the proofs of Theorem 6, from (1.30) we have lim y(t) = 0. As

t

(t) = 1 y1 (t) n yn (t), t I ,

then

lim (t) = 0. Then

lim

( (t)) = 0 due to

t

t

continuity of the function

( ).

In this case, the value

0

is determined from the

Hurwitz condition for the matrix

A B S, consequently,

0

= 0 . In the sector

Consider the case when

N = 0.

In this case, inequality (1.30) can be written as

1

It follows from (1.31) that

lim

y (t) = 0, , lim

y

(t) = 0.

From equation (1.30)

t

2

t

n

for

t ,

we get

lim

y (t) = lim

y (t) =

0,

lim

y

(t) = lim

y

(t) = lim

y (t) = 0, ,

t

1

t

2

t

2

t

1

t

3

lim

y

(t) = lim

y

(n 1)

(t) = lim

y

(t) = 0.

Let lim

y (t) = y

.

Then from the last

n 1

n

t

t

1

t

t

1

1*

equation (1.4) for

t ,

we get

0 = a

y

(

*

),

where

*

= lim

(t) = y

,

0

1*

t

1 1*

(

*

)

. This is only possible when

*

= 0, y

= 0. Consequently, lim y(t) = 0

0

1*

t

and according to the proof above in the sector

[0,

0

]

Iserman's problem has a solu-

tion. The theorem is proved.

It should be noted that the quantities

Ni

= Ni (a0 , , an 1

, 1, , n ), i = 0, n ,

therefore, the fulfillment of inequalities N0

0, N1

0, N2

> 0, , Nn

> 0 depends

̇ =

+

,  ̇ = −2

− 1.03

− 0.03

− 0.75 ( ),

1

1

2

2

1

2

3

(1.32)

̇ = −0.01

− 1.01

− 0.25 ( ),   =

+

,   ( ) .

3

2

3

2

3

0

1

1

0

0

= (−2

−1.03

−0.03) ,    = (−0.75) ,    = (0

1 1),    = 1,   = 3.

0

−0.01

−1.01

−0.25

where 0 = 0.98, 1 = 1, 2 = 1.04. Matrix

A

is a Hurwitz matrix.

The characteristic equation of the matrix

A B S

is equal to:

1( ) = | 3 − − | = 3 + (1.04 + ) 2 + + (0.98 − 0.5 ) = 0.

Matrix

A B S

is a Hurwitz matrix if 0 ≤ < 1.96 = 0.

As

0;

2

0,

2

B = 1.

4

;

2

;2

=

3

; 2 , then B = 0, AB =

A

Vectors A =

3

3

,

A

2

0;

2

; 2

=

3

. Matrix

0

4/3

0

R =

2/3

2/3

2/3

| R |=

32

0.

,

9

2

2

2

( ( )) = ( ) + 0.98 1( ) + 2( ) + 1.04 3( ),    ,

(1.34)

∞

∞

= ∫ ( ( )) ̇( ) = ∫

[− 2( ) + 0.5 2( ) + 0.5 2( )] +

1

2

3

0

0

d

( )

1

( )d = c,

| c |< .

dt

[F (t)]dt =

0

(0)

2

2

d

2 y2

(t)

3 y3

(t)]dt

F3 (t) dt,

0

dt

where

=

2

,

=

2

,

=

2

2

,

=

2

2

,

F

*

(t)F y(t),

t I.

0

1

2

3

3

0

2

= y

0

1

2

1

3

3

3

N

0

M

0

0 :

2

1 2 0;

(1.35)

2

0

1

0

0

−

− Γ

> 0: 0.9604

−1 − 0.49

− 2

> 0;

(1.36)

2

1

1

1

1

0

1

1

− − Γ

> 0: 0.5

− 1.0384

−1

− 0.46

− 2

+ 2

> 0;

(1.37)

2

2

2

2

2

1

0

1

2

1

3

−

− Γ

> 0: 0.5

2

− 0.9184

−1 + 1.04

− 2 + 2

> 0.

(1.38)

2

3

3

3

1

0

1

3

0

2

Solving the system of algebraic equations (1.35)-(1.38) we find the limiting value

= 1.24 when

0

=

2

= 0,

= 3.514739229024956

2,

= 1.571428571428574 ,

1

1

3

1

= −1

= 2.834467120181413 2.Thus, for example

(1.32), the value

0

= 1.24 − ,

2

1

0

1

> 0

is an arbitrarily small number. Define the value

0

from the Hurwitz condition

for the matrix

A1( ) = A B S,

0 <

0 .

The characteristic equation of the matrix

A B S

is equal to

1

( ) = |

− − | = 2 + (1.04 + ) 2 + + (0.98 − 0.5 ) = 0.

3

It follows that the matrix

A B S

is a Hurwitz matrix if 0 ≤ < 1.96 = 0.

Notice that 0

= 1.24 − < 0 = 1.96. Consequently, in the sector

[0,

0

]

Aizer-

man's problem has no solution.

Consider the absolute stability of the equilibrium system (1.32), when the func-

tion

( ) 0

1

1

2

1

= { ( ) C(R

, R )/0 ( )

, (0) = 0, , R }.

0

Unlike inclusion ( ) 0

, function

( )

0

is not limited. The ratio of the

~

~

( )

input transfer function

to the output

is equal to

( ) = ( − )−1 =

2 − 0.5

= −

̃

,

3

3 + 1.04 2 + + 0.98

̃( ̃)

where

p

~

~

~

are function images

( ), ( (t)), t I

is a complex variable, ,

( )

respectively.

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

0

is de-

termined from the condition

1

Re(1 i q)W (i ) > 0, < < ,

0

Where

( ) = ( ) + ( ) =

1.04 2 − 0.46 − 0.49

+

3

2

− 1.0384 + 0.9604

− 2

− 0.9184

+

− 1.5 + 0.5

, = 2

,

3

2

− 1.0384 + 0.9604

− 0.9184

q

is a real number.

Modified

frequency

response

W* (i ) = X iY , X = ReW (i ), Y = ImW (i ).

=

1.04 2 − 0.46 − 0.49

, =

− 2 − 1.5 + 0.5

,