86

Chapter IІI. Study of stability of solutions of dynamic systems with cylindrical phase space

( ) det( I

n

A) n a

n 1

n 1

... a a

.

1

0

As follows from Theorem of Hamilton-Cayley,

( A) 0.

Then

A

n

a

A

n 1

a

A

n 2

... a A a

I

,

n 1

n 2

n

1

0

where In

is a unit matrix of order n n.

Lemma 1. Let there exists a constant vector

*

R

n

such that

n 2

B 0,

n 1

B 0,

B 0, AB 0, ... , A

A

(3.67)

where (*) is a transpose sign,

(

,...,

n

)

is a row vector. Then equation (3.66)

1

can be represented as

y y

2

,

y

2

y

, ... ,

y

n 1

y

n

,

3

(3.68)

y

a

y

a y

... a

y

f ( ) .

n

2

n 1

n

0

1

1

where z y1, Az y2

n 2

z yn ,

z z(t), yi

yi (t),

i 1, n.

, ... , A

Proof. Multiplying

from

the

left

by

of the system (3.66) we get

z Az Bf ( ) Az,

where

B 0

by

a

condition

(3.67) of

the Lemma.

From here taking into account that

y

z,

y

Az, we get

y

y

.

2

2

As z Az,

1

1

then

z Az y		A[Az Bf ( ) ] A	z ABf ( ) A z,
		2	2
	2

where AB 0 by a

y

y

,

2

z , etc. As a result, we have

condition (3.67). Consequently,

2

3

where

y3 A

y

y

,

where

y

A

n 2

z,

y

A

n 1

z.

Then

n 1

n 1

n

n

y

A

n

1

z A

z A

Bf ( ) A

z f ( ) ,

n

n 1

n

n

where An 1B 0. As

An z ( a

n 1

An 1

a

n 2

An 2

... a A a

0

I

n

)z

1

a0 y1 a1 y2 ... an 1 yn ,

then

yn a0 y1 a1 y2 ... an 1 yn f ( ) .

Thus, when conditions are satisfied (3.67) the system (3.66) can be reduced to (3.68). the lemma is proved.

Lemma 2. Let the conditions of Lemma 1 be satisfied, and let, moreover, the rank of the matrix

191

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

of order

n n

R ||	*	,	*	*	*n 1		*	||
			A		, ... , A

be equal to n. Then

(3.69)

1)

*

z,

z R

* 1

y,

where y y(t)

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

t I,

z z(t) (z1

(t), ... , zn

(t))

y R

solutions of differential equations (3.68) and (3.66), respectively,

2) if

lim y(t) 0,

then

lim z(t) 0.

t

t

Proof. As

y

z, y

2

Az, ... ,

y

n 1

An 2 z, y

n

An 1z, then

1

y

z

1

y

y2

Az

A

z R

*

z.

n 1

n 1

yn

A

z

A

By the hypothesis of the Lemma, the matrix

R

is nonsingular. Consequently,

there exist inverse matrices R 1, R* 1. Then z R* 1 y.						On the other hand, from (3.69) it
follows that the pair		*	*				*	*
	(		, A ) is controllable. From the controllability of the pair (					, A ) it
follows that equalities		z 0, Az 0, ... , A		n 1	z 0	entail z 0. Consequently, from
lim y(t) 0 we get	lim z(t) 0. The lemma is proved.
t	t

	From Lemmas 1, 2 it follows that if the equalities are satisfied (3.67) and
R n, then the system (3.68) is equivalent to the system (3.66). Moreover,
lim	y(t) 0	follows	lim z(t) 0.
t			t

rank from

Solution properties. Consider the differential equation (3.68).

Lemma 3. Let the conditions of the lemmas 1, 2 be satisfied, the value

0.

Then along the solution of the system (3.68) the following equality holds

f ( (t)) (t)

1

[ (t) a

y

(t) a y

(t) ... a

y

(t)], t I,

(3.70)

0

2

n 1

n

1

1

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

Proof. Along the solution of the system (3.68) the following identities hold

yi (t) yi 1 (t), i 1, n 1, yn (t) (t) a0 y1 (t) ... an 1 yn (t) f ( (t)) (t),

t I,

where An 1 B 0. From the last identity, we get (3.70). The lemma is proved. Improper integrals. Let the conditions of the lemmas 1, 2 be satisfied. Now

equations of motion of the system (3.60), (3.61) can be represented in a form

	d	, g( , ) y* R 1Cf ( ) ( ),	t I [0, ),	(3.71)
	dt

192

Chapter IІI. Study of stability of solutions of dynamic systems with cylindrical phase space

where

y

y

2

,

y

2

y

, ... , y

n 1

y

n

,

1

3

y

n

a

0

y

a y

2

... a

n 1

y

n

f ( ) , t I ,

1

1

z R

* 1

y,

z

*

*

R

1

.

y

(3.72)

Theorem 1. Let the conditions of the lemmas 1, 2 be satisfied. Then along the solution of the system (3.71), (3.72) the equality is satisfied

T

I1

lim [g( , ) ( ) y

*

R

1

Cf ( ) ]dt

T

(3.73)

0

lim

1

2

(T )

1

2

(0)

,

2

2

T

where (t), (t), y y(t),

(t) (t), t I.

Proof. Multiplying the second equation from (3.71) by (t),

we get

g( , ) y*R 1Cf ( ) ( ) , .

Integrating by

t the given identity in limits from 0 to T, T we get the

equality (3.73). The theorem is proved.

Theorem 2. Let the conditions of the lemmas 1-3 be satisfied. Then along the solution of the system (3.71), (3.72) the improper integral

2

T

T

1

1

*

1

I

lim

y

(t)R

Cf ( (t)) (t)dt lim

[M

y

(t) ...

T

0

T

0

M

y

2

(t)]dt lim

y

*

(T )Ny(T ) y

*

(0)Ny(0),

n

n

T

(3.74)

where values

M	,..., M	,
1	n

are constant matrices

N

of order

n n

denoted through

parameters

R, C, a0 ,..., an 1,

Proof. Let the vectors

Then

of the system (3.71), (3.72).
		a	0				y
		a						1
a			1		Rn ,	y		.
							yn
	an 1

y*R 1Cf ( ) y*R 1C[ 1( a* y)],

where y y(t),

f ( ) f ( (t)), (t),

(t), t I , the product

f ( (t)) (t),

t I

is determined by formula (3.70). Consequently, the improper integral

193

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

T

I2

lim

y* R 1Cf ( (t)) (t)dt I21

I22 ,

T

0

where

T

T

I21

lim

y

*

(t)R

1

1

I22

lim

y

*

(t)R

1

1

a

*

y(t)dt.

C

(t)dt,

C

T

0

T

0

(3.75)

(3.76)

C

1

We calculate

the

integral

I

.

Denote

by

C R

1

1

R

n

.

21

C

Cn

T

n

I

21

lim

y

j

(t)C

j

y

n

(t)dt.

Notice, that:

T

0

j 1

T

T

d

1

2

1

1

j n :

yn yn dt

yn

dt

yn (T )

yn

(0);

0

0

dt

2

2

2

T

T

d

j n 1 :

yn 1 yn dt

( yn yn 1 ) yn 1 yn dt

0

0

dt

T

d

T

2

yn 1 yn ;

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

0

dt

0

T

T

d

1

2

j n 2 :

yn 2 yn (t)dt

yn 2 yn

yn 1

dt;

0

0

dt

2

T

T

d

T

j n 3 :

yn 3 yndt

2

(t)dt

dt

( yn 3 yn yn 2 yn 1 )dt yn 1

0

0

0

and so on.

As a result, we get (v. (3.76))

Then

T

T

I

lim

y

*

(t)R

1

1

[ y

2

... y

2

]dt

21

C

(t)dt lim

n

T

T

1 1

n

0

0

lim

y

*

(T )P y(T )

y

*

(0)P y(0),

T

0

0

where 1

,..., n , the matrix P0

of order n n is denoted by vector

C R

n

.

(3.77)

We calculate the integral I 22 . Denote by Then

D R	1	1	*	|| D ||, i, j 1, n.
	C	a
				ij

T

I22 lim y* (t)Dy(t)dt

T

0

T	n	n	ij j
T 1
lim		y (t)D y		(t)dt.
0	i 1	j 1

194

Chapter IІI. Study of stability of solutions of dynamic systems with cylindrical phase space

It is easy to verify that:

	T							T									T		d				1												1						1
		1	2	dt						1			1	dt												2				dt								2		(T )			2		(0);
																									1												1					1
		y y								y y									dt				2	y											2	y					2	y
	0							0									0
										T			1		3					T			1		2								T			2
												y y dt										( y y					)dt								y	2	dt;
										0										0													0
										T									T			d										1
												1		4	dt										1				3						y	2			dt
																																				2
											y y										dt			y y								2
										0									0
and so on.
Now the improper integral I															22		can be represented as
						T																										T
I		lim					y	*	(t)R				1			1		a	*	y(t)dt					lim									[L y				2		... L y				2		]dt
	22													C																														n
			T																									T							1			1				n
						0																										0
lim			y	*	(T )P y(T ) y											*	(0)P y(0),
	T							1													1

(3.78)

where have

L ,..., L		,
1	n

are numbers, matrix

P 1

is denoted by matrix

D.

From (3.75)-(3.78) we

2

T

T

1

lim

*

1

I

y

(t)R

Cf ( (t)) (t)dt lim

[(

T

0

T

0

... (

L ) y

2

(t)]dt lim

y

*

(T )( P P ) y(T )

n

n

1

T

0

1

Theorem 3. Let the conditions of the lemmas 1-3 solution of the system (3.71), (3.72) the improper integral

L ) y			2	(t) ...
1		1
y	*	(0)( P			P ) y(0).
				0	1

be satisfied. Then along the

T

T

I

3

lim

g( (t), (t)) (t)dt lim [M

1

y2

(t) ...

T

T

1

0

0

... M n yn2 (t)]dt

1

lim

2 (T ) lim y*

(T )Ny(T )

2 T

T

(T )

1

lim

( )d

2 (0) y* (0)Ny(0).

T

2

(0)

Proof. From equalities (3.73) taking into account (3.74), and also

we get

lim T

T

0

T			(T )
lim	( (t)) (t)dt	lim			( )d ,
T		T
0			(0)
	T
g( (t), (t)) (t)dt lim		[M y	2	(t) ... M		n	y2	(t)]dt
	T	1 1					n
	0

(3.79)

195

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

						(T )		1					1
lim	y	*	(T )Ny(T ) y	*	(0)Ny(0) lim	( )d lim				2					2	(0).
								2			(T )		2
T					T	T
						(0)

From here we get the ratio (3.79). The theorem is proved.

Lecture 31.

Global asymmetric stability of complex dynamic systems

Consider the solution of Problem 1.

Theorem 4. Let the following conditions be satisfied:

1)there exists a vector

n 1	B 0;
A

2) the value	0;
1

3) the rank of the matrix

4)		( )d 0;

	*	R	n	such that B 0, AB 0, ... ,	n 2	B 0,
					A
R || *,				A* *, ... , A*n 1 * || is equal to n;

		5) Values	Mi 0, i 1, n;
		Then a stationary set
stable.
		Proof. As follows from
	2	g( , )		, R ,
			2		1
1			2

matrix N 0.

of the system (3.60), (3.61) is globally asymptotically

restrictions	(3.63)	the following inequalities				are true
1	As		0,	then	the	product
R .
		1

g( , ) 0, , R	, , R .
1	1

As

( )d 0,

then

	(T )
| lim	( )d | .
T	(0)

From (3.79) it follows that

T

T

n

lim

g( (t), (t)) (t)dt lim

M y2 (t)dt

T

T

i

i

0

0 i 2

1

(T )

1

lim 2 (T ) lim y*

(T )Ny(T ) lim

( )d

2

(0) y* (0)Ny(0) 0, (3.80)

2

2 T

T

T

(0)

g( (t), (t)) (t) 0,

t, t I.

due to

the fact that

As

Mi 0, i 1, n, matrix

N 0,

then inequality (3.80)

is possible

then

and

only

then, when functions

yi (t), t I, (t), t I are limited. Indeed, if yi (t), t I, (t), t I

not limited, then

T

0 lim

g( (t), (t)) (t)dt .

T

0

It is impossible. So, it is proved, that || y(t) || ,

| (t) | , t, t I. Then

from (3.80) it follows that

196

Chapter IІI. Study of stability of solutions of dynamic systems with cylindrical phase space

T

0 lim

g( (t), (t)) (t)dt .

(3.81)

T

0

According

to the statement

of the

problem, partial

derivatives

g( (t), (t))

,

g( (t), (t))

, t I

are limited. Consequently, the function

g( (t), (t))

is uniform-

mly continuous

by

, . From

limitation

(t) (t), t I

it

follows that the

function

(t), t I

is a uniformly continuous function. As | (t)

| ,

t, t I ,

g( , ) g( , ),

then the function

g( (t), (t)), t I

is limited. Then as

follows from (3.71), the inequality holds

| (t) | |

*

(t)R

1

g( (t), (t)) | | y

Cf ( (t)) | | ( (t)) | ,

where periodic functions

f ( ), ( ), R

1

are limited.

From here it follows that the function (t), t I is uniformly continuous. So, it

is proved, that

functions (t), (t), t I

are uniformly continuous. Then from

(3.81) it follows that

lim g( (t), (t)) 0,

lim (t) 0.

T

T

As g( , )

is a periodic function by ,

where

g( ,

*

) 0, (

*

)

from (3.82) we have

lim (t)

.

t

*

On the other hand from (3.79) we get the inequality

T

T

n

I3

lim g( (t), (t)) (t)dt lim

Mi

2

(t)dt ,

yi

T

T

i 2

0

0

where M

i

0, i 1,n,

g( (t), (t)) (t) 0,

t I. Then

T

n

i

i

lim

M y

2

(t)dt

.

T

0

i 2

Consider equation (3.72), where functions

yi (t), t I, i 1, n are limited,

is limited. Then

| yn (t) | a0 || y1(t) | ... | an 1 || yn (t) | | || f ( (t)) || (t) | , t, t I.

(3.82)

0,

then

(3.83)

(t),

t I

197

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

From here it follows that the function

y	(t),	t I
n

is uniformly continuous. As

y

y

2

, ...

, y

n 1

y

n

,

then functions

yi (t), i 1, n 1 are also uniformly continuous.

1

Then from (3.83) it follows that

lim

y

(t) 0, i

1, n.

t

i

So it is proved, that

lim

y(t) 0,

lim (t) 0, lim (t) * , where ( * ) 0.

t

t

t

This means that stationary set

of the system (3.60), (3.61) is globally asympto-

tically stable. The theorem is proved.

Consider the solution of Problem 2, when (v. (3.65))

( )d ,

0.

Improper integrals. Based on Lemmas 1-3 and Theorems 1-3, we can obtain

evaluations of improper integrals.

Theorem 5. Let the nonlinearity

g( , )

satisfy the condition (3.63). Then for

any numbers

1

0,

2

0

along the solution of the system (3.60) the improper

integral

I

lim

T

(

)

(t) (

)g( (t), (t)) (t) dt 0.

(3.83)

2

4

T

1

1

2

2

2

1

0

Proof. As follows from the condition (3.63) along the solution of the system (3.60) the following inequalities are true

	2	(t) g( (t), (t)) (t) 0,		2	(t) g( (t), (t)) (t) 0,	t,	t I [0, ).
1			2

(3.85)

		Multiply the first inequality from (3.85) by 1 0, and the second inequality by
	2	0 and summarize them, we get inequality

			( )				2	(t) (				)g( (t), (t)) (t) 0, t,									t I.		(3.86)
			1	1	2	2					2		1
Integrating by				in limits from 0 to										T inequality (3.86) from the transition to the
limit when T we get an assessment (3.83). The theorem is proved.
Theorem 3. Let the numbers
																		2		2
( )d ,					( )d ,												, 3 , 4			3	0, 5	0.
		0					0											4	5
Then along the solution of the system (3.60) the improper integral
I5 lim	T		3 ( (t)) (t) 4							( (t)) 5							(t) dt C1	,			С1 const,
T									2							2							(3.87)
	0
where (t), (t) (t), t I,												C1		.

198

Chapter IІI. Study of stability of solutions of dynamic systems with cylindrical phase space

Proof. Notice, that for function ( ) ( )

d

0

( ), R10,

the integral

(3.88)

due to the fact that . Along the solution of the system (3.60) we consider the

function

S(t) ( )		2	( )	2	( ) ,	t 0,
3	4		5		3

(3.89)

where (t), (t) (t), ( ) ( (t)),	( ) ( (t)), t I.
We show that, S(t) 0, t, t I. Really, as

( )		( ) ( ) ,
3	3	3

then S(t) 3		( )					2	( )				2		,		t I. From here taking into account that
						4					5
																					2		2			2
									2																			2
																		3					3						,
			3												5
								5									2	5					4
																									5

we get

												2			2 2
S t	5					3					( )				3			2
																	4
			2			5									4 5
					2			2	4 4 5			0.
due to the fact that					3										Consequently,
Then from (3.88) it follows that
					( )								2	( )			2
									3			4				5			3

						3
	5				2	5
S(t) 0,				t I.
( ) ,	t I.

2

( ) , t I

(3.90)

By integrating inequality (3.89) by

t

in limits from 0 to

T

taking into account

(3.88), we get an assessment (3.87). The theorem is proved.

Theorem 7. Let the conditions of the lemmas 1-3 and Theorem 5 be satisfied.

Then for any numbers									1	0,				2	0 along the solution of the system (3.71), (3.72)
the improper integral
I6 lim				T	2 ( 1 1 2 2 ) 2 (t) 1													2 ( 2			1 ) g( (t), (t)) (t)
		T
				0
1 ( (t)) (t) 1 M1 y12 (t) ... M n yn2 (t) dt																														(3.91)
					1		2									*						1		2			*
		lim						(T )				lim			y		(T )Ny(T )				1				(0)	y		(0)Ny(0)		,
	T				2								T						1
																					2
where (t) (t),						t	I .

199

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

Proof. As conditions of the lemmas 1-3 are satisfied, then we get the ratio (3.79). From statements of Theorem 5 we get the evaluation (3.83). Then improper integral

I6 1 I3			2 I4	lim	T	1 g( (t), (t)) (t) 1 ( (t)) (t) 1 M	1 y12	(t) ... M n yn2 (t) dt
				T
					0
lim	T		2 ( 1 1	2 2 ) 2 (t) 2 ( 2 1 )g( (t), (t)) (t) dt
T
	0

				1		2				*				1		2		*
1	lim						(T )		lim y		(T )Ny(T )	1					(0) y		(0)Ny(0) .
				2										2
									T
	T

From here evaluation (3.91) follows. The theorem is proved.

Theorem 8. Let the conditions of Theorem 7 be satisfied, values

	1	0,

	2		0,							1		0,							2		0, 0,												0,			2						0.				Then along the solution of the
																											1					2						2			1	1
system (3.71), (3.72) the improper integral
											T
I			lim											[										(					)]g( (t), (t)) (t)																[M		y	2	(t) ... M									y	2	(t)]
		7															1						2			2		1																		1										n			n
						T																																						1			1
											0
																																									1
												2				2											( (t))						dt lim											(T ) y						(T )Ny(T )										(3.92)
												1														2																	2						*
			4					(												)															T			1			2
						2				2					2
																			1			1
							1				2		(0)				y					*	(0)Ny(0)									,			0			,
				1			2
																								T
where								C						lim									1		( )d ,										( )d 0,										C		I				5	,	I	5		is				defined by
										1						T																														1
																								0										0

formula (3.87).

Proof. As conditions of Theorem 7 are satisfied, then we have evaluation (3.91). From (3.91) when

						,									(								),				32 2										12 2			0
			3							5				2			2			2		1			4
					1																		1							4 5			4 2 ( 2 2 1 1 )
we have
I	6	lim T [						1				2	(	2					1	)]g( (t), (t)) (t)												[M				y 2	(t) ... M	n		y 2	(t)]
		T																													1				1 1					n
					0
				( (t)) dt											T
		4	2							lim						[											2			(t)		4		2		( (t))]dt
																			3 ( ) (t) 5
												T
															0
lim						1	2			(T ) y* (T )Ny(T )																	1		2 (0) y* (0)Ny(0) .
	T				1																					1
					2																						2

200