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

Proof. As follows from the condition of the Theorem the following ratios are true (3.21), (3.22), (3.24), (3.25), (3.26), (3.27), (3.29), (3.30). The values

[( M

) y

(t)]dt l

n 1

[N1 yn 1

(t) ...

where R1 l1

l3 l4 .

Notice,

that

condition 2)

the

theorem

the

equality

) (M
l ,			\|
		3
N	n 1		y
	n 1
R
		1
I	1	I
	1

	H				) y					2	(t)
										2
1				1			1			n 1
l		\| , \| l						2	\| , \| l
1								2			3
2	(t)]dt l								R ;
n	(t)]dt l								R ;
n								4			1
, \| l4				\|			. When
2	I		3	I		4		R
2			3			4				1

... | ,

we have the holds, where

0, I

0, | R

| .

Then

. Consequently

[h (t) h (t) h

( (t))]

dt l

R ,

where

| l

| ,

the

function

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

t I is uniformly

continuous. From here it follows that

lim

f (t) 0

, i.e. an equality (3.43) holds.

Let

the

(t),

t I be

a solution

of the differential equation (3.44). Then

(t) (t)

when

t . By

the

hypothesis of

the

theorem,

the solution

of the

equation (3.44) is globally asymptotically stable. Consequently,

(t) when

t ,

where

( ) 0

. Then

(t)

, when

t ,

( ) (

) 0 . Consider an

autonomous

system

Ax B ( (t))

, where

( (t))

when

. As

matrix

Hurwitz

matrix,

then

x(t) x 0

when

t . So, the pair

) . It means that the solution of the system (3.2), (3.3) is globally asympto-

tically stable. The theorem is proved.

Now consider the case, when

( )d 0.

Theorem 12. Let the following conditions be satisfied:


1) matrix	А	is a Hurwitz matrix, ( ) 0 ;
1) matrix		is a Hurwitz matrix, ( ) 0 ;
2) there exists a vector			*	R	n	such that
2) there exists a vector				R		such that

B 0, AB 0,..., An 2B 0, An 1B 0;

3)rank P n;

4)( )d 0;

181

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

5) conditions of theorems 7,4 are satisfied, and let besides:

P	0,	i 1,n 1,	0
i	i		0

Then the stationary set					of the system (3.2),
stable.
Proof. From inequalities (3.24), (3.25) when								0
I			[ y		(t) y		(t) ...
I			[ y	2	(t) y	2	(t) ...
				2		2
	2		1	n 1	2 1			n 1
		0

From (3.32), (3.33) it follows that

(3.3) is globally asymptotically

0		we have
y	2	(t)]dt l	0.
	n
		2

[P y

(t)

P y

(t)

... P

(t)]dt l

2 1

n 1

Then

[( P ) y

(t) ... (P

) y

(t)]dt l

, | l

| ,

| l

| .

n 1

Further, applying Lemma

get

lim y (t) 0,

i 1,n 1 . The

theorem is

proved.

Theorem 13. Let the conditions 1) – 4)

of the Theorem 12 be satisfied, and let

besides:

1) conditions of Theorems 3,4,7 be satisfied;

2) M

0, P M

i 1,n 1.

Then the stationary set

of the system (3.2), (3.3) is globally asymptotically

stable.

Proof. As

the

proof

Theorem

10,

the

following

inequalities are true

I1 0,

0 and the value

. Then,

I2 I

. From here it follows that

) y2

) y2 (t)]dt .

[( M

(t) (P M

(t) ... (P

n 1

n 2

n 1

Further, applying Lemma 7, we get lim

The Theorem is pro-

y (t) 0,

i 1,n 2.

ved.

Theorem 14. Let the conditions 1) – 4)

of Theorem 12 be satisfied, and let

besides:

1) conditions of the Theorems 3,4,8 be satisfied;

2) L0 M0 0 0, Li

i Ni , i 1,n 1.

182

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

Then the stationary set

of the system (3.2), (3.3) is globally asymptotically

stable.

Proof. As I

, then

[( L M

) y

(t)

(L M

) y

(t) ... (L

) y

(t)]dt .

n 2

n 1

Then the statement of the theorem follows from Lemma 7. The theorem is proved.

Theorem 15. Let the conditions 1) – 4) of Theorem 12 be satisfied, and let besi-

des:

1) conditions of the Theorems 3,4,5,7 be satisfied;

Then

H	0	0, M	i	H	i	P, i 1,n 1.
	0	0	i		i	i	i
		h (t) h (t) h ( (t)) 0
		0			1		2

when

t .

If in addition the solution of the system of the second order

			t I
h h h ( ) 0,			t I
0	1	2
is globally asymptotically stable, then a stationary set				of the system (3.2), (3.3) is

globally asymptotically stable.

Proof. The proof of the Theorem is similar to the proof of Theorem 11. It is easy

to verify that

R	l	l	l
2	1	2	3

the

equality , where

I1 I2 I3 I5

I	5	R	, I	1
	5	2		1

I	2
	2

holds, where I1 0, I2		0, I3	0,
0, I3 0 . Then,	I3 I	5 R2	.

Further, by repeating the proof of Theorem 11, we get a statement of the theorem. The theorem is proved.

Theorem 13. Let the conditions 1) – 4) of Theorem 12 be satisfied, and let besides:

1) conditions of the Theorems 3,4,5,8 be satisfied;

Then

H	0	0,	M	H	L , i 1,n 1.
	0	0	i	i	i	i

h0 (t) h1 (t) h2 ( (t)) 0 when

t .

If in addition the solution of the system of the second order

h0 h1 h2 ( ) 0, t I

is globally asymptotically stable, then a stationary set of the system (3.2), (3.3) is globally asymptotically stable.

The proof of the Theorem follows from equality I1 I2 I3 I6 R3 , R3 l1 l2 l3 l6 , where I1 0, I2 0, I3 0, I6 R3 .

183

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

Lecture 29.

Problem of phase synchronization

Equation of motion of phase adjustment systems with proportionally integrated filter in an autonomous case has the form[19]

		x	1	x ( 1) ( ), x T ( ),							t I [0,	),
		x	T	x ( 1) ( ), x T ( ),							t I [0,	),
			T
where
( )				1	1	1	) / 1	d ( )	1,	( ) ( ),		1	}.
( )	0	{ ( ) C			(R	, R	) / 1	d	1,	( ) ( ),		R	}.
	0

(3.45)

(3.46)

For this example

n 1,	m 1,	x y .
		1

A	1	,	B 1, C 1,
	T

R T

, where

T 0, (0,1),

1-case. For definiteness, we choose																															the				function												( ) sin . In this																	case
			1			B T					0			. Stationary set																	{( x				,					) R			2		/			x			0,					k , k 1,2,...}
2 , R CA						B T					0			. Stationary set																	{( x				,				*	) R					/			x			0,				*	k , k 1,2,...}
																																	*						*										*						*
is a countable set. As d / d cos , d 2 / d 2																																						sin ,										\| d 2 / d 2								\| 1,						1,		2	1,
																																																														1		2
matrix A				1				0			is a Hurwitz matrix,																				y1 x.
matrix A				T				0			is a Hurwitz matrix,																				y1 x.
				T
As follows from Theorem 2, the identities hold
								( (t))										1			[ y				1			y ],				(t)									T				y						1				y			,
								( (t))													[ y		2					y ],				(t)													y			2							y			,
																1							2		T				1												1							2			1					1
																1									T																1										1
								(t)							1				[ y				1	y			],		(t)												T			y							1			y			,
								(t)											[ y	3				y		2	],		(t)															y		3								y			,
																	1			3			T			2														1						3						1				2
																	1						T																	1												1
where y		x			y					,	y			y				,			(t)				d ( (t))										.
where y		x			y				2	,	y		2	y				,			(t)														.
1									2				2			3													dt
																													dt
Improper integrals: а) we calculate

0 I							(				)					( )dt														[M					y	2		(t) M									y		2	(t) M							y		2	(t)]dt l			,
0 I		1					(				)					( )dt														[M			0		y	3		(t) M							1		y		2	(t) M						2	y			(t)]dt l			,
		1						1							1							2											0			3									1				2							2		1				1
				0																									0
where M								1				, M							2T 4				1					,	M								1					,
where M							1)2					, M						T 2 ( 1)2										,	M												1)2	,
	0			(			1)2								1			T 2 ( 1)2								1					2			(							1)2
								d			*													*
	l								[ (t)F (t)]dt (t)F (t) \|																												,			(t) ( y (t), y													2	(t), y							3	(t)),
								dt

		1													1														1						0														1
				0

184

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

y2	, \| l \|
1	1
),	t I
t I, 0

, 1 0.

is limited:

a , 0 c

| (t,

) |

due

a , and to the

																						d	2
fact that ( ) C									2				1			1					\|	d		( )				\|		1.
fact that ( ) C										(R , R ),											\|		d			2		\|		1.
																							d

b) The improper integral

0 I				[			y	(t)								y				(t)						y (t)]					2		dt					[ y		2		(t) y			2		(t) y			2	(t)]dt l		,
0 I	2			[		0	y	(t)							1	y		2		(t)						y (t)]							dt					[ y		3		(t) y			2		(t) y				(t)]dt l	2	,
	2					0	3								1			2							2 1													0		3			1		2			2	1			2
			0																																		0
																								d
																l								d	[		*		(t)F (t)]dt,										\| l				\| ,
																											*
																	2																								2
																	2						dt										2								2
																						0	dt
																						0
		0			2		1						2	2					0 2 ,				2					2			*		(t)F2 (t)							0			2	1					2
where		0			0 ,		1					1		2					0 2 ,				2					2 ,					(t)F2 (t)							0			1 y2 (	1		2		0 2 ) y1 .
c) The improper integral
																																T
																																					2						2
							I4 ( (t))																	2 (t)dt										[N1 y2 t N									2 y1 t ]dt					l4
													0																			0
								( )																									T
																																	dt
											( )										d ,								l						d	{z		* (t)F z(t)]dt, \| l										\| ,
											( )								2		d ,								l	4						{z		* (t)F z(t)]dt, \| l								4		\| ,
																			2											4										4						4
								(0)																									0

where

N		T			,	N				2		,	z* (t)F z(t)	(1 )		y2	(t),
N	1		2	2	,	N	2			2	2	,	z* (t)F z(t)		2	y2	(t),
	1	( 1)	2	2			2	T	( 1)		2		4	2( 1)	2	1
		( 1)						T	( 1)					2( 1)
								0

z(t) ( y	(t), y	(t)),	t I,	2	0
1	2			2

is any number.

M		0 :		1			2	0;
	0					2
		0	(		1)		0

								T				T					1
N M		0 :												2		4					2			0;
N M		0 :																		2	2			0;
																				2
1	1			1				( 1)	2	2		T	2	(	1)			2	1	1		0	2
								( 1)				T		(	1)
		N		M		0 :					2						1				2	0.
		N	2	M	2	0 :															2	0.
			2		2		2		T ( 1)2							T ( 1)2					2
									T ( 1)2							T ( 1)2

From (3.47) it follows that 1 02 ( 1)2 0 . Substituting the value of (3.48), we get

(3.47)

(3.48)

(3.49)

1 from

185

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

T			(	2	T	4	1)
T			(		T		1)	2	2	2		0,		0
	2	2				2		0	1	2	2	0,	2	0
( 1)	2	2			T	2		0	1	0	2		2
( 1)					T

From (3.49) we have
		2		2	2	0,		0.
						0,		0.

	T ( 1)		2	0	2		2
	T ( 1)

(3.50)

(3.51)

From (3.51) it follows that 2 ( 1)2 ( 02 22 ) T 0 . Then from (3.50) it follows

										T		1
T		(						)	2		4					2				0.
T	2	(	2				2	)					2	2		2				0.
	2		2				2						2	2
			0				2			T	2		0	1			0		2
										T
Consequently,
(1		T		4		(1 )
(1		T				(1 )			2			2	2	2	2					0.
			T		2				0	T			2	1	2		0		2	0.
					2				0				2	1			0		2

This inequality is performed for any											0 0, 1			0,		2		0,			(0,1)	. Thus, the statio-
nary set of the system (3.45), (3.46) when													( ) sin								is globally asymptotical-
ly stable for any T 0, (0,1).
2-case. Let the sin ,								(0,1).				In this case, the values
2										2													1		],
( )d 2 ,											( ) d 4[ arcsin												1	2	],
																								2

												0,5							.
																	2		.
								arcsin						1			2
								arcsin						1

The improper integral

I5 [ ( ( )) 3 (t)

4	5	T	1 2		2 1		5
4	5		1 2		2 1		5
2	( (t)) 2 (t)]dt		[P y2	t	P y2	t ]dt l		,
		0

								l T		d	{z* (t)F z(t)]dt, \| l \| ,
								l T			{z* (t)F z(t)]dt, \| l \| ,
								5	dt		5				5
								5			5				5
									0
where
P1				T			3		4			2T 2		5 , P2			3							4
P1		( 1)2					3	( 1)2				( 1)2		5 , P2		T ( 1)2					T 2 ( 1)2
	5				, z	*	(t)F z(t)				(1 ) 3				4			T	5			y	2	(t), t
				2	, z		(t)F z(t)						2			2				2		y		(t), t
	(		1)	2			5				2( 1)		2		T ( 1)	2	(	1)		2		1
	(		1)								2( 1)				T ( 1)		(	1)

z(t) ( y1(t), y2 (t)), t I.

Below are the results of applying Theorem13:

I ,

186

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

			M0 0 0 :													1								2	0;
			M0 0 0 :										( 1)2											0	0;
													( 1)2
				T																							T
P M		0 :																										2	2
P M		0 :																	4
																			4
1	1	1	1	(	1)		2				3				( 1)						2				( 1)							2		5
				(	1)										( 1)										( 1)
					(				2	T		4		1)
					(					T				1)										2	2								0,
																								2
					T		2		(						1)		2			1				1				0			2
					T				(						1)
P M			0 :			3																4
P M			0 :

2	2	2		T ( 1)								2				T		2		( 1)							2
				T ( 1)												T				( 1)
											5												1								2		0
																																	0

				(								1)			2				( 1)							2				2
				(				1				1)							( 1)
								1

From (3.52)-(3.54), we have

	( 1)					; (1 T							)			T		(1 )					2				T			(1 )				T
				2	2					2		4			2		2					2							2				2			3
1	T				0		1		0;		T				1				T		( 1)					0	3					T				0.	3
	T		T						0;		T								T		( 1)				T							T				0.
	2				2									2				2		2				2									2
		4				4		5					1					2										3			4				5

For limit values, we get:

(3.52)

(3.53)

(3.54)

					( 1)				2			2	, \|			\| ( 1)					2	T		2				2	T ( 1)										2			4		T							,
									2			2									2			2				2											2			4
				1								0			3									0				2														T								5
																																										T
											[1 T					4	(1 )]
							(1 ){				[1 T						(1 )]		2					2	2						T						2		2	}	T				2				.
						4	(1 ){								T		2		0					1	2			0		2	T								2	}	T							5	.
						4											2		0					1				0		2									2									5

										2		2																2															4				0, 5
Notice that										2		2	0	. Consequently,										\|		\|						4			5	, where
Notice that							4 4 5					3	0	. Consequently,										\|		\|										, where

																														\|		3	\|
																																3
the given inequalities, it follows that

								[1 T 4 (1 )]										2			2											2			2							2 2
			2 (1 ){															0		1		2					T							2		}		5	T					5		.
			2 (1 ){										T 2									2					T									}			T							.
\| \|													T 2												0	2																											.
\| \|	2	[	1		T (1 )2 (1 )] T (1 )															2			1					2	2								1					T (1 )											.
			1																	2			1					2									1
			T 3
0			T 3																2					T			1						0			2		T														5
In particular when 0												1		2 0 from (3.55), we get

. From

(3.55)


\| \| =	−0.5	≤	2√	, , > 0.

			1 +
	+ √1 − 2

187

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

In work [19] for the systems same result was obtained.

In general, the values

0 , 1, 2

(3.45), (3.46) based on the frequency method the

, 5 0

must be selected from the solution of the

optimization problem.

												[1 T							4	(1		)]
								2	(1 ){			[1 T								(1		)]					2			2		2												T			2			2	}				2	2	.
								2	(1 ){								T			2							0			1		2					0				2			T						2	}		5	T		5	.
																				2							0			1							0				2									2			5			5
max																																																										.
max							1																											1																		1						.
	,	,	,		2	[	1		T (1 )					2	(1 )] T (1											)				2				1										2	2							1			T (1 )
0	1	2		5		[			T (1 )						(1 )] T (1											)																			2										T (1 )
0	1	2		5
					0		T	3																						2					T									1					0		2			T				5
							T																												T																			T
	From (3.55) when									0 , we have
																			2			(		1		2			2	2										)
																										2			2
																							T		2	0			1						0			2						5
									\|		\|												T																											.
									\|		\|			1 T					4									1							2															.
														1 T									T					1							2										T

																					2					2					2								0				2
															T	3					0					2		T			1						T											5
															T													T									T
	From here when									1	0,						2			0				, we have
										1							2

																				0.5																2√							1		2
																																				2√						2			2
									\| \| =																					≤												2			0		5					.
									\| \| =																											1 +									4
													+ √1 − 2																					2		1 +									4
																																	0		(					3						) + 5
	The results of numerical calculations: when = 1 ; \| \| ≤ 0.8, = 0.6; when
=		10 ; \| \| = 0.1, = 0.1 ;																when = 0.5 ; \| \| = 0.95, = 0.8.
	The same results are close to the results from [20], obtained by qualitative nume-

rical methods. Thus, it shows the effectiveness of the proposed method in comparison with the known methods.

As an example, we consider equations of motion of the mathematical pendulum

(3.1). We have to find			кр				кр	( )	. The equation (3.1) can be represented as
(3.1). We have to find			кр				кр		. The equation (3.1) can be represented as
					x,				x x ( ),												(3.56)

where ( ) sin ,					(0, 1)					. By denoting							x y1 , we get the identities
	( ( )) y y									2	, ( (t) y					2	y	,		(t) y (t), t I,
								1		2						2	3			1
where y1 y2 ,	y2 y3 . As a result of applying Theorem 3, we have
				M		0			0 :			1	2 0;								(3.57)
						0		0				1			0
	P M	1		0 :							2			4	2		2		2	0,	(3.58)
	1	1			1			1			1			4	1			0	2

188

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

P M		0 :
					2			2
2	2	2	1	3		4	5	2

From (3.57)-(3.59) it follows that

(3.59)

	−0.5				2√(		2 2		− 2		+ 2		)
\| \| =		≤						0		1		0	2	5	.
\| \| =		≤													.
	+ √1 − 2	1		3	2	+	1	5 +		1	2		2	+ 2 0 2
			(	+	) 0	+		5 +			2	− 1		+ 2 0 2

The values					0		,			,		2	,	5		are determined													from				the solution								of		the optimization
The values					0				1			2		5		are determined													from				the solution								of		the optimization
problem															2				(									2					)
															2				(	2			2			2		2					)
	max																						0			1				0		2			5					,			0.
	max								1												1					1														,		5	0.
	,	,		,																																						5
	,	,		,				(						3	)			2											2					2	2
0	1		2			5		(							)																				2
0	1		2			5
																		0						5					2					1				0	2

In particular, when 1 0,																	2				0 , we have
			\|					\|			2											2					2				2						,			0.
																															2
													\|	4			5								0						)					5
													\|		\|					(2 2							2			4	)		2						5
																											2			4			2						5
														3																			0				5

The results of numerical calculations: = 0.1 ; | | = 0.09, кр

| | = 0.775, кр = 0.58; = 10 ; | | = 0.987, кр = 0.94.

This results are in agreement with the known estimates of values

= 0.06 ; = 1 ;

	кр	( ).
	кр

Lecture 30.

Study of the synchronization theory of complex dynamic systems

We consider the problems of global asymptotic stability of dynamic systems with cylindrical phase space, arising in synchronization tasks of various control systems. New effective criteria of global asymptotic stability of countable equilibrium state of the system based on the assessment of improper integrals along the solution of the system were proposed.

Problem statement. Consider equations of motion of dynamic systems of the following kind

d	,	d	g( , ) z*Cf ( ) ( ),	t I [0, ),
					(3.60)
dt		dt

	z Az Bf ( ) , (0) 0 , (0)	0 , z(0) z0 ,		(3.61)
where	A, B, C are constant matrices of orders	n n, n 1,	n 1,	respectively,
f ( ),	g( , ), ( ) – are continuously differentiable functions, satisfying conditions

189

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

f ( ) f ( ), ( ) (

	g( , )		,	,
			,	,
1		2

is a period, derivatives					f
is a period, derivatives

),		g( , ) g( , ), , R1,
		1			1	,	0,	g(0,		) 0,
R			, R			,	0,	g(0,		) 0,
							1
,	g	,	g	,	0 , R				1		are limited.
									1

(3.62)

(3.63)

We assume that the right parts of the system (3.60), (3.61) satisfy the local conditions of the theorem of existence of a solution and the solutions are extendable.

In particular, equations (3.60)–(3.63) describing the processes in the theory of synchronization, have numerous applications.

The system equilibrium states (3.60), (3.61) form a countable set

{ 0,		*	, z 0 / (		*	) 0}.
*			*
We say that the system (3.60), (3.61) is globally asymptotically stable, if when
t vector-functions ( (t), (t), z(t))			tend to the same point from the set				.
Problem 1. Find conditions of global asymptotic stability of the system (3.60),
(3.61) for any nonlinearities	f ( ), ( ), g( , ) ,			satisfying conditions			(3.62),

(3.63), when

	.
( )d 0, , R	.
1

(3.64)

Problem 2. Find conditions of global asymptotic stability of the system (3.60), (3.61) for any nonlinearities f ( ), ( ), g( , ) , satisfying conditions (3.62),

(3.63), when

	.
( )d , 0, , R	.
1

(3.65)

The class of ordinary differential equations (3.60), (3.61) describes dynamics of synchronous machines [4], and dynamics of control of Boissy-Sard [5]. Frequency conditions of global asymptotic stability of the system (3.60), (3.61) and condition of existence of limit cycles, circular solutions in such systems are studied in [19].

A review of the scientific literature on the research of dynamics of phase systems is given in [14]. This work is a continuation of research from [13–19, 21, 22].

	Note that checking of frequency criteria must be made for all values of parameter
,	varying in limits from to	, which is complicated. Therefore the develop-

ment of new methods of study of global asymptotic stability such systems is relevant. Nonsingular transformation. Consider equations of motion (3.61):

z Az Bf ( ) ,

z(0) z0 ,

t I [0, ).

(3.66)

The characteristic polynomial of the matrix A has the form

190

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