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Казахский национальный университет им. аль-Фараби

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

	0k	(1)	,...,
Proof. As		= ( 0k

(2.56) taking into account that

1	b		(2)	b
1			(2)
0k		j1	0k

we get

(n n)	),
0k	),	then

	...
		(n m)
j 2		0k
	*	z(t) =	*
	0k	z(t) =	0k
	0k		0k

by multiplying the identity from the left

b	jn m	= 0, k = 1, n j = 1, m,
	jn m
A z(t), t I.
	1

From here we get identity (2.61). By the hypothesis of the lemma the inequality

(2.62) holds and rank	B	*	= m.	Consequently, vectors		R	,	k = 1, n	exist and are
					0k
	1					n m

linearly independent. The lemma is proved.

Lemma 19. Let the conditions of the Lemmas 7, 8 be satisfied, and let, moreover, the rank of the matrix

==\|\|\| ,,...,, ,, ,,...,,
11	mm 0101	00nn

|||

(2.63)

of order	(n m) (n m)	be equal to	(n m).	Then equation (2.56) is equivalent to the

following system of equations

= c

... c

(

),..., y

= c

y ... c

(

1n m

n m

mn 1

n m

m 1

= c

... c

n m

,..., y

n m

= c

n m,1

... c

n m

m 1,1

m 1,n m

m n,n m

= d

... d

n m

,...,

= d

... d

m,m n

m n

(2.64)

1n m

* z,

= *

where y

i = 1, m,

k = 1, n.

Proof. As

rank

R = n m,

then

from

(2.63)

follows

that

vectors

Rn m ,

i = 1, m,

Rn m ,

i = 1, n

form a basis in Rn m . Then

A z = c

z ... c

z c

1m 1 01

1n m

.........

A z = c

z ... c

z c

m,m

m,m 1

mn m

* A z = c

*z ... c

* z c

m 1,1

m 1,m

m 1,m 1

m 1,n m

.........

A z = c

z ... c

z c

0n 1

m n,1

m n,m

m n,m 1

m n,n m

where c

i, j = 1, n m are decomposition coefficients

i = 1, m,

i = 1, n

Rn m ,

by bases

i = 1, m,

i = 1, m. From here taking into account (2.59),

(2.61) we get a system of equations (2.64) with respect to the listed

y1, , yn m .

Similarly, by decomposing vectors

S * Rn m

by bases

i = 1, m,

i = 1, n

we get

111

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

= S z = d

z ... d

z d

z ... d

11 1

1m 1

1n m

= S

z = d

z ... d

z d

z ... d

m,m 1

m,n m

where S

= (S1

, , S1 ). the lemma is proved.

The system of equations (2.64) in the vector form is

where

y = A1,

A1 y B1,

B1 S1 ,

( ), = S1 y, ( )	, y(0) = y		, \|
1		0
are constant matrices of		orders

y	\|< , t I,
0
(n m) (n m),

(2.65)

(n m) m,

m (n m)

respectively,

AA1 1==|| |c| cij||,||,

i, j = 1, n m,

SS1

1==\|\|\|\|dd \|\|\|\|,	,
ij ij

i = 1, m,

j = 1, n m,

B1 =

zero

A1,

I m is a unit matrix of order

m m,

On,m is a matrix of order

n m with

n,m

then

z = K

y matrices

elements. If the matrix K = R

y = Kz = R

y = (R )

B1,

are equal A1 = KA1K

)

B1 = KB1

B1,

S1 = S1K

= R

A1 (R

= R

= S1 (R )

Thus, differential equation (2.56) with nonlinearities (2.55) with nonsingular transfor-

mation

z = K	1	*	1	y
z = K		y = (R )		y

lead to the form (2.65).

Lecture 19.

Solution properties in the general case and improper integrals in a simple critical case

Solution properties. Limitation on solutions of the system (2.56), (2.55), and also equations (2.65) is considered. Identities along the solution of the equation (2.65) and examined its asymptotic property are obtained.

Theorem 8. Let	the matrix A1 = A1( ) be a Hurwitz matrix,	i.e. Re j ( A1 ) < 0,
j = 1, n m and conditions of the Lemmas 7–9 are satisfied. Then following estimates
are true:
	\| z(t) \| c0 , \| z(t) \| c1, t I,		(2.66)
	\| y(t) \| c2 , \| y(t) \| c3 , t I,		(2.67)
	\| (t) \| c4 , \| (t) \| c5 , t I,		(2.68)
where ci = const > 0,
where ci = const > 0,	ci < , i = 0,5. Moreover, the functions z(t),	y(t), (t),	t I

are uniformly continuous.

112

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

Proof. As matrix

A1 = KA1K

then from Hurwitz of the matrix

we get

Hurwitz of the matrix

i.e.

(A ) = Re

(A ) <

j = 1, n.

Solution of the differential equation (2.56) has the form:

A t

A (t )

z(t) = e 1

( ( ))dx, t I.

Notice, that from Hurwitz of the matrix

A t

|| ce

(a )t

we get evaluation || e 1

c = c( ) > 0,

t I ,

where

> 0

arbitrarily

small

number,

value

a = max Re

( A ) < 0.

Then

1 j n m

A t

(t )

| z(t) | || e

1 |||

|| e

||||

B ||| ( ( )) | d

c |

| e

(a )t

|| B

(a )

d = c | z

| e

(a )t

where

ce(a )t

e	(a )t	1,

|| *[

I ,

e (a

0,
	a

a ] = c | z0 |

> 0,	\| ( (t))

*	,	t
*

B1 ||

* = c0 ,

I. From here we

get limitations of solution of the system (2.56). From the equation (2.56) follows that

\| z(t) \| \|\| A	\|\|\| z(t) \| \|\| B	\|\|\| ( (t)) \| \|\| A	\|\| c \|\| B		\|\| = c , t, t I.
1	1	1	0	1	1

Then

| (t) | || S |||

z(t) | || S || c0

= c4 ,

| (t) | || S ||| z(t) | || S || c

= c , t I.

As func-

tion

y(t) = Kz(t),

t I ,

then | y(t) | ||

K ||| z(t) | || K || c0 = c2 ,

y(t) ||

K ||| z(t) | ||

K | c1

= c3 ,

t I. So, proven estimates (2.66)– (2.68). From the limitations of functions

z(t),

t I

y(t),

(t),

t I.

y(t),

(t),

we get uniform continuity of functions

z(t),

The

theorem is proved.

should

be noted

that:

From

evaluation

| z(t) |,

t I ,

have

z(t),

t I ,

lim |

z(t) |=| z( ) | c | z

c || B ||

= c0 ,

c c0 due

continuity

where a < 0. Consequently, lim | y(t) |=| y( ) | || K || c0 ,

lim

| (t) | || S || c0 .

Lemma 10. Let the conditions of the Lemmas 7–9 be satisfied. Then along the

solution of the system (2.65) the following identities hold

(2.69)

( (t)) = H0 y(t) A11 y(t), t I,

113

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

where


H	0	= (I
	0

(t) =
m	, O
	m

H1 y(t) = A12 y(t), t I ,

(t) = S11H0 y(t) S12H1 y(t), t

S11H y(t) S12 A12 y(t), t I,
		0
		0
	),		H		= (O	, I	),	H	0
									0	= I		,
,n m				1						= I	n m	,
,n m				1	n,m	n					n m
								H1

		A11
A1	=
A1	=		,
		A12
		A12

(2.70)

(2.71)

(2.72)

...

1n m

m 11

m 1n m

A11

... ...

...

A12

...

mn m

n m1

n mn m

...

1m 1

1n m

S1 = (S11, S12 ), S11

... ...

...

, S12

...

... .

dm1 ...

dmm 1

...

dmm

dmn m

Proof. As conditions of the Lemmas 7–9 are satisfied, then we get (2.64). Notice,

that

m 1

...

y =

... .

yn m

Then from (2.64) it follows that

(

) = y

y ... c

n m

,...,

(

) = y

y ... c

n m

11 1

1n m

mn m

y S12H y, H y

= (S11, S12 ) y = (S11, S12 )

= S11H

= A12 y.

Consequently, identities (2.69)–(2.72) hold. The lemma is proved.

Lemma 11. Let the conditions of the lemmas be satisfied 2.2–2.3. Then for any

matrices , M ,

N1,

2 of

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

(n m) n,

(n m) n respectively,

along

the

solution of the equation (2.65) the

following

identities hold

y* (t)My(t) y* (t) y(t) =

[ y* (t) y(t)], t I ïðè M = * ,

(2.73)

[ y* (t)N

y* (t)N ][H

y(t) A12 y(t)] 0, t I.

(2.74)

114

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

Proof.

Identity

(2.73)

directly

follows

from

equality

d	*	(t) y(t)) =
dt	( y	(t) y(t)) =
dt


t I when	*
t I when
(n m) n,	(n

y	*	*	*	*	*	*	(t)My
y		(t) y(t) y	(t) y(t) = y	(t) y	(t) y	(t) y(t) = y	(t)My

= M .	As it follows from (2.70) for any matrices	N	,
		1
m) n	an identity (2.74) holds. The lemma is proved.

(t)

N	2

y	*	(t) y	(t),

of orders

Improper integrals. Based on nonsingular transformation (2.64) and using properties of solution (2.66)–(2.68) and identity (2.69)–(2.72) we can obtain evaluations

of improper integrals along the solution of the system (2.65).

Theorem 94. Let the conditions of the Lemmas 7–9 be satisfied, matrix

be a

Hurwitz

matrix,

the

function

( ) 1.

Then for

any

diagonal

matrix

= diag(

, ,

)

order m m along the

solution of

the

system (2.65)

the

improper integral


I =		*	( (t)) (t)dt = [ y* (t) y(t) y* (t) y(t) y* (t) y(t)]dt =
I =		*	( (t)) (t)dt = [ y* (t) y(t) y* (t) y(t) y* (t) y(t)]dt =
1			1			1	2	3
1			1			1	2	3
	0			0
				m		( )
				m	i
				=		i ( i ) 1i d i	< ,
				i=1		(0)
					i

(2.75)

where

	*			,		*			*				*
			S11H					S12 A12 H		S11	A11,		= A11	S12 A12.
	= H				2	= H			0			3
1	0	1	0			0	1			1			1

(2.76)

Proof. Notice, that matrix

A1	= KA K	1	,

	1

where K is a nonsingular matrix, then

j = 1, n m.

j (A1 ) = j (A1 ),

Consequently, matrix A1

is a Hurwitz matrix then, when

matrix

is a Hurwitz matrix, where

j ( A1 ) = j ( A1 ),

j = 1, n m eigenvalues of

A1, Re j ( A1 ) < 0,

| y( ) |<|| K || c0 < ,

matrix

j = 1, n m. As it follows from Theorem 8,

| y0 | ||

K |||

|< . Improper integral:

I =

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

11 y(t)]* [

11H y(t)

12 y(t)]dt =

( )

[ y* (t) 1 y(t) y* (t) 2 y(t) 3 y(t)]dt = i ( i ) 1i d i < ,

i=1

(0)

by virtue of identities (2.69)–(2.72), where | (0) | c4

< ,

| ( ) | c4 < ,

matrices

, 2 ,

of orders

(n m) (n m) are denoted by relations from (2.76). The

theorem is proved.

Theorem 10. Let the conditions of the Lemmas 7–9 be satisfied, matrix

A1 be a

Hurwitz

matrix, the

function ( ) 1. Then

for

any

diagonal

matrix

2 = diag ( 21,..., 2m ) 0 of order

m m,

along the solution of the equation (2.65) the

improper integral

115

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

							*											*
		I		=	[		*	( (t))			1							*	( (t)) (t)]dt =
											1
			2								0	( (t))
			2							2	0										2
					0


		= [ y* (t) y(t) y*										(t)		2	y(t) y* (t)						y(t)]dt 0,					(2.77)
								1						2						3
			0
where
	*			1	H	,						*				1	A11		*		S11H		*		S12H	,
	= H			0	H	,			2	= 2H						0	A11	H		2	S11H	0	H	2	S12H	,
1	0	2		0	0				2			0	2			0			0	2		0	0	2	1

** *

3 = A11 2 0 1 A11 A11 2 S12 H1 A11 2 S11H0 .

Proof. From inclusion			( )				1		it follows that
Proof. From inclusion							1		it follows that
i ( i		)			,	i					1	,		,		1	, i = 1, m.
				0i	,						0i	,		,	i	R	, i = 1, m.
				0i				(		)	0i		i		i
	i							(		)
	i					i			i

(2.78)

Consequently, i lity holds i ( i ) 2i i

	(		)				1
	(		)				0i
i		i					0i
		1				2		(
		1
		0i		2i		i
			m
							oi
					[			1
			i=1

2	( i		) , i = 1, m . Then for any value 2
i
i ) 0 , i = 1, m . From here it follows
		2		(	)	(	)	i] 0.
		i
21				i	i	i		2i

i	0

that

an inequa-

(2.79)

Let the 2 = can be written as

diag (	21	,...,	2m	) 0,
	21		2m

1	= diag (	1	,...,	1	).
o		01		0m

Then inequality (2.79)

From (2.80) it follows that improper integral (v. (2.77))

(2.80)

		*				*	( (t)) 2 (t) dt 0.
		*		1		*
I	2 =		( (t))	2 0	( (t))
	0

From here taking into account identities (2.69), (2.71), we have

2 = { H

0 y(t) A11 y(t)

2 0

H0 y(t) A11 y(t)

2 S11H0 y(t) S12 H1 y(t) }dt =

H0 y(t) A11 y(t)

(t) 3 y(t) dt 0,

= y

(t) 1 y(t) y

(t) 2 y(t) y

116

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

where matrices 1, 2 , 3 of orders	(n m) (n m)	are determined by the formula
(2.78). The theorem is proved.
Lemma 12. Let the conditions of the lemmas be satisfied 2.57–2.69, matrix			A1 be

a Hurwitz matrix, the function

( )

. Then for any matrices

(n m) n,

(n m) n

respectively, improper integral

[ y

(t)P y

(t) y

(t)P y(t)]dt = 0,

(t)P y(t) y

where P = N

H ,

P = H * N * N

P = N A12.

A12 ,

Lemma’s proof directly follows from the identity (2.74).

N	,
1

N	2

of orders

(2.81)

Lecture 20.

Absolute stability and Iserman’s problem for multidimensional systems in a simple critical case

Absolute stability. Based on the above results on the estimation of improper integrals, and also Lemmas 2.77–2.89 we can formulate conditions of absolute stability of the equilibrium state of the system (2.53), (2.54).

We introduce the following notation

*		1
A11
	2	0

				R = P = H S11H																														H										1			H					N H							,
																							*															*						1
				0					1						1			1					0			1							0					0		2				0						0						2		1
				=																								*															*					S11 A11
				=									2				2	P = H													S12 A12 H																	S11 A11
					0								2				2					2						0		1													0			1
				*					1										*			S11H											*			S12H							H							*	N			*		N			A12 ,
			2H			2			0		A11 H										2	S11H					0		H						2	S12H							H								N					N		2	A12 ,
				0		2			0										0		2						0						0		2						1							1					1					2
																																			*
											W					=			3				3		P = A11														S12				A12
															0				3				3						3									1
							*							1		A						*				S				H						*				S					H					N A .
				A						2				1		A				A				2		S	11			H		0	A						2	S		12			H			1		N A .												(2.82)
							11			2				0		11						11		2			11					0				11			2			12						1								1	12					(2.82)
In particular when matrix																								*										*						then the matrix																				*		1	*
																	N = A11											S			A11 T,																												W = A11			1	A11 A12 K A12 0,
																	N = A11											S			A11 T,																												W = A11			0	A11 A12 K A12 0,
																		1								1			12							2																							0	2		0	1
*				where																		*												*															*									*		S1	= T A12.
*	0,							K1			= K1 > 0,											*			S11H									*															*				S					*
A11 A12 K A12	0,							K1			= K1 > 0,									A11					S11H								A11				S12 H = A11																S		1 = A11 T A12 ,
																*
1																							2							0						2					1											2							2
Here K1,				T are matrices of orders n n,																														m n respectively. Then improper inte-
gral


	I	4	= I I				2	I				3			= [ y* (t)R y(t) y* (t) y(t) y* (t)W y(t)]dt
		4		1			2					3											0															0																		0
																0
																				m i ( )																																										(2.83)

																		i ( i ) 1i d i																				< .
																		i ( i ) 1i d i																				< .
																				i=1			(0)
																						i

Theorem 11. Let the conditions of the lemmas 7–11 be satisfied, matrix A1 be a Hurwitz matrix function ( ) 1, and let, moreover, matrices R0 , 0 of orders

117

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

(n m) (n m),

per integral

(n m) (n m)

m i ( )

i (

i=1 i (0)

be such that: 1)

(t)W y(t)dt

)

[

(t) y(t)]dt

2)0 = 0*.

Then impro-

(2.84)

where matrices			R			,		,			W						are determined by the formula (2.82),																												,	2	0 are any
					0					0				0																														1		2
diagonal matrices of orders																		m m,							N			,	N		2		are any matrices of orders (n m) n,
																		m m,									1				2
(n m) n , respectively.
Proof. As matrix																	*	,		then				y		*	(t) y(t) =										d		[	1			*	(t) y(t)], t	I ,		consequen-
Proof. As matrix												=						,		then				y			(t) y(t) =												[			y		(t) y(t)], t	I ,		consequen-
										0						0													0								dt			2				0
																																					dt			2
tly, the improper integral
																												d		1
													y		*					0								d		1			*					0
													y			(t) y(t) =													[				y		(t) y(t)]dt =
											0																0	dt		2
											0																0
	=		1		y		*	(t) y(t) \|											=		1			*	( ) y( )												1		y		*	(0) y(0) < ,					(2.85)
	=				y			(t) y(t) \|										0	=				y		( ) y( )														y			(0) y(0) < ,					(2.85)
			2							0								0			2								0								2							0
			2																		2																2
by virtue of the assessment															\|\|yy(( ))\|\| cc											<< ,,					\|		\|yy(0)\| \| cc									<< ..
																								22																22
From (2.83) taking into account (2.85) we get
																							0												0
																			*				0							*					0
																[ y				(t)R y(t) y												(t)W y(t)]dt
															0
	m			( )
	m		i																			1																1
										(	)					d						1		y	*	( ) y( )												1		y		*	(0) y(0) < .				(2.86)
																									*																	*
																			i
								i		i				11					i			2							0									2						0
	i=1			(0)																		2																2
			i
By the hypothesis of the theorem, the matrix																																				R0			R0 0. Then the quadratic form
																																											*
y* (t)R y(t) = y* (t)[								1		(R		R* )]y(t) 0,																	t,					t I. Consequently, from (2.86) we
y* (t)R y(t) = y* (t)[										(R		R* )]y(t) 0,																	t,					t I. Consequently, from (2.86) we
0								2		0							0
								2
get inequality


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

The theorem is proved.

Theorem 12. Let the conditions of the lemmas 7–11 be satisfied, matrices A1 ( ), A1( ), 0 be Hurwitz matrices, the function ( ) 1 and conditions 1, 2 of

118

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

Theorem 11 be satisfied, and let, moreover, matrix	W W * > 0. Then equilibrium
	0	0
state of the system (2.53), (2.54) is absolutely stable.
If in addition the diagonal matrix 0 = 0 2 ,	2 > 0	is a diagonal matrix with

arbitrarily small elements, then Iserman’s problem has a solution in the sector [ , 0 ].

Proof. As all conditions of theorem 11 are satisfied, we get inequalities (2.84). Consequently, the improper integral

where


										I			=		*	(t)W y(t)dt =
										I	40		=		y	(t)W y(t)dt =
											40					0
														0
								1
					*			1		0		0						*	0
			=		y	(t)[		2	(W W					)]y(t)dt =				y	(t)T y(t)dt < ,
				0				2									0
				0													0
		1																	*
T	=		(W W			*	) > 0.			Let the function							V ( y) = y T y.			Then
T	=		(W W				) > 0.			Let the function									0	Then
0		2	0		0
		2

V ( y) > 0,

(2.87)


y R	n m	, y 0, V (0) 0,
y R		, y 0, V (0) 0,		V ( y(t))dt <
				0
t I = [0, ].			We show that	lim	y(t) = 0.
			We show that	lim
				t
Suppose the contrary, i.e. lim						y(t) 0.
Suppose the contrary, i.e. lim
				t

,	where	\| \|yy(t()t)\| \| cc,		, \| \|yy((tt))\| \| cc,, t,
		2	2					33
Then there exists a sequence					t	k	,	t	k	> 0,
Then there exists a sequence						k			k

t
k

when

such as

\| y(t	k	) \| > 0,
	k

k = 1,2,...

We choose

t	k 1	t	k		1	>
	k 1		k		1

\| y(t) \| c	,
3

k = 1,2,... As

0,		k =
t I ,
t				1


	k		2

1,2,...

then

> t	k 1	,
	k 1

y(t),

t I

continuously differentiable,

y(t) | c

| y(t) y(t

) | c

| t t

t [t

, t

< tk 1,

V ( y) > 0,

y R

n m

then

V ( y(t))dt

V ( y(t))dt,

k =1

where | y(t) |=|

t [t			1	, t
t [t			1	, t
	k	2			k
		2

So, | y(t) | 0 ,

nuous on the

y(t

) y(t) y(t

) | | y(t

) | | y(t) y(t

) | c

> 0, t,

We can always choose the value 1 > 0 so that the value

> 0.

| y(t) | c ,

t [t

, t

is conti-

As function V ( y) = y T y

compact set

0 | y | c2 ,

then

there is

a number

m > 0

such as

119

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

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

Then the value of the integral

t		1

k	2

	V ( y(t))dt 1m,
t		1

k	2

k = 1,2,...

Con-

sequently,

V ( y(t))dt =

(t)W y(t)dt

V ( y(t))dt

lim

m = .

k =1

This contradicts the condition (2.87). Consequently,

lim y(t) = 0. As z(t)

= K

y(t),

t I ,

К is a nonsingular matrix, then lim z(t) = 0.

from here taking into

account that matrices

A ( ),

are Hurwitz matrices, lim z(t,0, z

, ) = 0,

according to definition 4 we get that the equilibrium state of the system (2.53), (2.54) is absolutely stable. The theorem is proved.

	From the theorems 11, 12 it follows that in cases when matrices A ( ),				A ( ),
				1	1
0		are Hurwitz matrices,	( ) 1	and the following conditions are satisfied:
1)	*	0; 2) 0 = 0 ; 3) W0	W0 > 0	the equilibrium state of the multidimensio-
1)	R0 R0	0; 2) 0 = 0 ; 3) W0	W0 > 0	the equilibrium state of the multidimensio-
		*	*

nal system (2.53), (2.54) in a simple critical case is absolutely stable.

Remark 2. As follows from the proof of the theorem 12, the equilibrium state of

the system (2.53), (2.54)	is absolutely stable and in case, when matrix T0 0, if
*	does not contain whole trajectories. In this case, the integral
surface V ( y) = y T0 y = 0

curves "stitch" surface

V ( y) = 0. Notice, that the surface V ( y) = 0 does not contain

whole trajectories, if y* (t)T y(t)

0, t [0, ).

It should be noted that as a result of nonsingular transformation the identity (2.70)

is obtained. This identity is used for defining an

improper

integral

Improper

integral I3 depends on arbitrary matrices N1

of orders

(n m) n,

(n m) n ,

respectively. Matrices

R ,

depend on matrices

N ,

. As follows from the

condition of theorems 11, 12 matrices 1, 2

N1,

ensure the fulfillment of

consequently, matrices

N1,

N2 allow

conditions R0 R0 0,

= 0 ,

W0 W0 > 0,

us to significantly expand the area of absolute stability of the system parameters in space. In the method of A. I. Lurie and the method of V. M. Popov an improper

integral	I	3	,		is absent, consequently, we have no matrices N1, N2 . The presence of

matrices	N			,	N	2	allowed to solve the Iserman’s problem.
			1

Iserman’s problem. The question arises: is it possible to distinguish a class of multidimensional regulated systems, by selecting a feedback matrix S1 , for which

Iserman’s problem has a solution? For this class of multidimensional regulated systems, the obtained results are necessary and sufficient conditions of absolute stability.

120

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\| z(t) \| \|\| A	\|\|\| z(t) \| \|\| B	\|\|\| ( (t)) \| \|\| A	\|\| c \|\| B		\|\| = c , t, t I.
1	1	1	0	1	1