On the Rational Dynamic Equation on Discrete Time Scales Ijoer sep 2015 3

International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

On the Rational Dynamic Equation on Discrete Time Scales ax  t   bx    t   x( (t ))  , t T c  d ( x(t )) m1 ( x(  (t ))) m2

Sh. R. Elzeiny Department of Mathematics, Faculty of Science, Al-Baha University, Kingdom of Saudi Arabia

Abstract— In this paper, we study the global stability, periodicity character and some other properties of solutions of the rational dynamic equation on discrete time scales

x( (t )) 

ax  t   bx    t  

c  d ( x(t ))m1 ( x(  (t )))m2

where a  0, b, c, d

, t  T,

0 and m1  1, m2  1.

Keywords— Rational dynamic equation, Time scales, Equilibrium point, Global attractor, Periodicity, Boundedness, Invariant interval.

I.

INTRODUCTION

The study of dynamic equations on time scales, which goes back to its founder Stefan Hilger [16], is an area of mathematics that has recently received a lot of attention. It has been created in order to unify the study of differential and difference equations. Many results concerning differential equations carry over quite easily to corresponding results for difference equations, while other results seem to be completely different from their continuous counterparts. The study of dynamic equations on time scales reveals such discrepancies, and helps avoid proving results twice-once for differential equations and once again for difference equations.

The general idea is to prove a result for a dynamic equation where the domain of the unknown function is a so-called time scale, which may be an arbitrary closed subset of the reals. This way results not only related to the set of real numbers or set of integers but those pertaining to more general time scales are obtained. The three most popular examples of calculus on time scales are differential calculus, difference calculus, and quantum calculus. Dynamic equations on a time scale have enormous potential for applications such as in population dynamics. For example, it can model insect population that are continuous while in season, die out in say winter, while their eggs are incubating or dormant, and then hatch in a new season, giving rise to a nonoverlapping population (see Dosly et al. [10]).

Several authors have expounded on various aspects of this new theory, see the survey paper by Agarwal et al. [3] and the references cited therein. The first course on dynamic equations on time scales as in Bohner et al. [7]. For advance of dynamic equations on time scales as in Bohner et al. [6]. For completeness, we give a short introduction to the time scale calculus. the introduction of the paper should explain the nature of the problem, previous work, purpose, and the contribution of the paper. The contents of each section may be provided to understand easily about the paper.

Definition 1.1: A time scale is an arbitrary nonempty closed subset of the real numbers

, , ,

0

. Thus

,

i. e., the real numbers, the integers, the natural numbers, and the nonnegative integers are examples of time scales, as are [0,1]  [2,3], [0,1]  , and the Cantor set , while

,

\ ,

, (0,1),

i. e., the rational numbers, the irrational numbers, the complex numbers, and the open interval between 0 and 1, are not time scales.

Throughout this paper, a time scale is denoted by the symbol T and has the topology that it inherits from the real numbers with the standard topology.

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To reference points in the set T, the forward and backward jump operators are defined.

Definition 1.2: For t  T , , the forward operator  :T  T is defined by

 (t )  inf{s  T : s

t},

and backward operator  :T  T is defined by

 (t )  sup{s  T : s

t}.

If T has a maximum t  , and a minimum t  , then  (t  )  t  , and  (t  )  t  . When  (t )  t then t is called right scattered. When  (t )  t then t is called left scattered.

Points t such that

 (t ) t

 (t ),  (t ) t  sup T , or  (t )

t  inf T ,

are called isolated points. If a time scale consists of only isolated points, then it is an isolated (discrete) time scale.

Also, if t sup T and  (t )  t , then t is called right-dense, and if t inf T and  (t )  t , then t is called leftdense. Points t that are either left-dense or right-dense are called dense.

Finally, the graininess operator  : T  [0, ) is defined by  (t )   (t )  t and if f : T  

then the function f : T 

is a function,

is defined by

f  (t )  f ( (t )) for all t  T ,  i.e f  f  is the composition function of f with  .

Definition 1.3: Assume f : T 

is a function and let t T . If t is an isolated point, then we define

lim f (s)  f (t ) s t

and we say f is continuous at t . When t is not isolated point, then when we write

lim f (s)  L, s t

it is understood that s approaches t in the time scale ( s  T , s  t ) . We say f is continuous on T . provided

lim f ( s)  f (t ) for all t  T . s t

In particular, we have that any function defined on an isolated time scale (since all of its points are isolated points) is continuous. We say f :[a, b]T 

is continuous provided f is continuous at each point in (a, b)T ,

f is left continuous at a , and f is right continuous at b .

In the time scale calculus, the functions are right-dense continuous which we now define.

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Definition 1.4: A function f : T 

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is called right-dense continuous or briefly rd-continuous provided it is

continuous at right-dense points in T . and its left-sided limits exist (finite) at left-dense points in T . . The set of rd-continuous functions f : T  will be denoted in this paper by

Crd  Crd (T )  Crd (T , ) In the next theorem we see that the jump operator is rd-continuous.

Theorem 1.1 (Bohner et al. [5]): The forward operator  :T  T is increasing, rd-continuous, and

 (t )  t for all t  T , and the jump operator is discontinuous at points which are left-dense and right-scattered.

Remark 1.1: The graininess function  : T  [0,  ) is rd-continuous and  is discontinuous at points in T that are both left-dense and right-scattered.

When T 

, the rational dynamic equation on discrete time scales

x( (t )) 

ax(t )  bx(  (t )) , t  T , where a  0, b, c, d c  d ( x(t )) m1 ( x(  (t ))) m2

0 and m1 , m2  1,

becomes the recursive sequence

xn1 

where a  0, b, c, d

axn  bxn1 , n  1, 2,..., c  dxnm1 xnm21

(1.1)

0 and m1 , m2  1 .

Now, the difference equations (as well as differential equations and delay differential equations) model various diverse phenomena in biology, ecology, physiology, physics, engineering and economics, etc.[21]. The study of nonlinear difference equations is of paramount importance not only in their own right but in understanding the behavior of their differential counterparts.

There is a class of nonlinear difference equations, known as the rational difference equations, each of which consists of the ratio of two polynomials in the sequence terms. There has been a lot of work concerning the global asymptotic behavior of solutions of rational difference equations [1, 2, 4, 8, 9, 11--15, 17-27].

This paper addresses, the global stability, periodicity character and boundedness of the solutions of the rational dynamic equation on discrete time scales

x( (t )) 

where a  0, b, c, d When T 

ax(t )  bx(  (t )) , t T , c  d ( x(t )) m1 ( x(  (t ))) m2

(1.2)

0 and m1 , m2  1 .

and m1  m2  1 , our equation reduces to equation which examined by Yang et al. [27].

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International Journal of Engineering Research & Science (IJOER)

Also, when T 

[Vol-1, Issue-6, September- 2015]

and m1  m2  b  c  d  1 and a  0 , our equation reduces to equation which examined

by Cinar [8].

Here, we recall some notations and results which will be useful in our investigation.

Let I be some interval of real numbers and let f

be a continuous function defined on I  I . Then, for initial

conditions x(  (t0 )), x(t0 )  I , it is easy to see that the dynamic equation on discrete time scales

x( (t ))  f ( x(t ), x(  (t ))), t  T ,

(1.3)

has a unique solution {x(t ) : t  T } , which is called a recursive sequence on time scales.

Definition 1.5: A point  is called an equilibrium point of equation (1.3) if

  f ( ,  ).

That is, x(t )   for t  T , is a solution of equation (1.3), or equivalently,  is a fixed point of f .

Assume  is an equilibrium point of equation (1.3) and u   f x (t ) (, ), and v    f x (  (t )) (, ) . Then the linearized equation associated with equation (1.3) about the equilibrium point  is

zx ( (t ))  uzx (t )  vzx (  (t ))  0.

(1.4)

The characteristic equation associated with equation (1.4) is

 2  u  v  0.

(1.5)

Theorem 1.2 (Linearized stability theorem [18]):

(1) If | u | 1  v and v 1, then  is locally asymptotically stable.

(2) If | u | |1  v | and | v | 1, then  is a repeller.

(3) If | u | |1  v | and u 2 4v, then  is a saddle po int .

(4) If | u ||1  v |, then  is a non  hyperbolic po int .

Definition 1.6: We say that a solution {x(t ) : t  T } of equation (1.3) is bounded if

| x(t ) | A for all t  T .

Definition 1.7: (a) A solution {x(t ) : t  T } of equation (1.2) is said to be periodic with period υ if

x(t   )  x(t ) for all t  T .

(1.6)

(b) A solution {x(t ) : t  T } of equation (1.3) is said to be periodic with prime period  , or  -cycle if it is periodic with period  and  is the least positive integer for which (1.6) holds.

Definition 1.8: An interval J  I

is called invariant for equation (1.3) if every solution {x(t ) : t  T } of

equation (1.3) with initial conditions ( x(  (t0 )), x(t0 ))  J  J satisfies x(t )  J for all t  T .

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For a real number x(t0 ) and a positive number R , let O( x(t0 ), R)  {x(t ) :| x(t )  x(t0 ) | R}.

For other basic terminologies and results of difference equations the reader is referred to [18].

II MAIN RESULTS 2.1 Local asymptotic stability of the equilibrium points Consider the rational dynamic equation on discrete time scales

x( (t )) 

where a  0, b, c, d Let a 

ax(t )  bx(  (t )) , t T , c  d ( x(t )) m1 ( x(  (t ))) m2

(2.1)

0 and m1 , m2  1 .

a c d , c  , and d   . Then equation (2.1) can be rewritten as b b b

x( (t )) 

ax(t )  x(  (t )) , t T . c  d ( x(t )) m1 ( x(  (t ))) m2

(2.2)

The change of variables y(t )  (d )1/ m x(t ), m  m1  m2 , followed by the change y (t )  x (t ) reduces the above equation to

x( (t ))  where p  a 

px(t )  x(  (t )) , t T , q  ( x(t )) m1 ( x(  (t ))) m2

(2.3)

a c , and q  c  . Hereafter, we focus our attention on equation (2.3) instead of equation b b

(2.1). Assume that,

m1 

k1 k2 , and m2  , k1 , k2  2n1  1 2n2  1



, and n1 , n2 



 {0}.

When m is the ratio of odd positive integers, equation (2.3) has only two equilibrium points

  0, and   m p  1  q .

When m is positive rational number and numerator's even positive integers, equation (2.3) has unique equilibrium point

  0 if q  p  1, and three equilibrium points:

  0,   m p  1  q , and    m p  1  q if p  1 q.

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Furthermore, suppose that, at least one of m1 and m2 

2k3  1 , k3 , n3  2n3



.

Then, in this case, we consider x(t )  0 for all t  T .

When q  p  1, equation (2.3) has unique equilibrium point

  0.

When p  1

q, however, equation (2.3) has the following two equilibrium points:

  0, and   m p  1  q .

The local asymptotic behavior of

 0

is characterized by the following result.

THEOREM 2.1:

p  1 , then  is locally asymptotically stable.

(2) If q 1  p , then  is a repeller.

(3) If | q  1| p , then  is a saddle point.

(1) If q

Proof: The Linearized equation associated with equation (2.3) about the equilibrium point   0 is

z ( (t ))  Now, when T 

p 1 z (t )  z (  (t ))  0, t  T . q q

(2.4)

, then equation (2.4) becomes

p 1 zn  zn 1  0, n  . q q

zn 1 

(2.5)

The characteristic equation associated with equation (2.5) is

2  when T  r , r

p 1    0, where z (t )   t , q q

(2.6)

1, then equation (2.4) becomes zr n1 

p 1 zr n  zr n1  0, n  . q q

(2.7)

The characteristic equation associated with equation (2.7) is

2 

p 1    0, where z (t )   logr (t ) , q q

(2.8)

when T  h , then equation (2.4) becomes

zh ( n 1) 

p 1 zhn  zh ( n 1)  0, n  . q q

(2.9) Page | 6

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The characteristic equation associated with equation (2.9) is

2  and when T 

2 0

p 1    0, where z (t )   t / h , q q

(2.10)

, then equation (2.4) becomes

z( n 1)2 

p 1 zn2  z( n 1)2  0, n  q q

0

.

(2.11)

The characteristic equation associated with equation (2.9) is

2 

p 1    0, where z (t )   t . q q

(2.12)

Hence, the characteristic equation associated with equation (2.4) is

2  Let u  

p 1    0, for all t  T . q q

(2.13)

p 1 , and v   . q q

(1) The result follows from Theorem 1.2 (1) and the following relations

| u | (1  v) 

p 1 ( p  1)  q  (1  )  q q q

0, and v  

1 q

0 1.

(2) The result follows from Theorem 1.2(2) and the following relations

| u |  |1  v |

p 1 p q  1 p 1  q q  (1  p)  |1  |  | |   q q q q q q q

0,

and

| v |

1 q

1.

)3( The result follows from Theorem 1.2 (3) and the following relations

| u |  |1  v |

p 1 p q  1 p  | q  1|  |1  |  | | q q q q q

0,

and|

p 4 u 2  4v  ( ) 2  q q

0.

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Now, the local asymptotic behavior of β and γ are characterized by the following result. that heorem 2.2: Assume that

p  1 q, m1

p , and m2 p 1 q

p2 , p 1 q

then both  and  are locally asymptotically stable.

Proof : The Linearized equation associated with equation (2.3) about the equilibrium  is

z ( (t )) 

m1 ( p  1  q)  p m ( p  1  q)  1 z (t )  2 z (  (t ))  0, t  T . p 1 p 1

(2.14)

Hence, the characteristic equation associated with equation (2.14) is

2  Let u 

m1 ( p  1  q)  p m ( p  1  q)  1  2  0, for all t  T . p 1 p 1

m1 ( p  1  q)  p m ( p  1  q)  1 and v  2 p 1 p 1

Then, from Theorem 1.2 (1) and the following relations

m1 ( p  1  q )  p m ( p  1  q)  1 p  m1 ( p  1  q ) m2 ( p  1  q )  p | ( 2  1)  ( ) p 1 p 1 p 1 p 1  m( p  1  q )  0, p 1

| u | (v  1) |

and

v

m2 ( p  1  q)  1 1, p 1

we conclude that  is locally asymptotically stable. Similarly, we can prove that  is locally asymptotically stable.

2.2

Boundedness of solutions of equation (2.3) In this section, we study the boundedness of solution of equation (2.3).

Theorem 2.3: Suppose that

p  1  q and mi is positive rational number and numeraor's even positive integers,

i  1, 2 . Then the solution of equation (2.3) is bounded for all t T .

Proof: We argue that

| x(t ) | A for all t  T by induction on t .

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Case(1): If T 

. Given any initial conditions

[Vol-1, Issue-6, September- 2015]

| x1 | A and | x0 | A , we argue that | xn | A for all n 

induction on n . It follows from the given initial conditions that this assertion is true for true for

by

n  1, 0. Suppose the assertion is

n  2 and n  1(n  1). That is,

| xn 2 | A and | xn 1 | A. Now, we consider

xn , where we put n instead of (n  1) in equation (2.3), | xn |

1 ( p  1) A ( p | xn1 |  | xn2 |)  m1 m2 | q  xn1 xn2 | | q  xnm11 xnm22 |

(2.15)

Since,

| q  xnm11 xnm22 | q  xnm11 xnm22  q

0,

1 1  . m1 m2 | q  xn1 xn2 | q

(2.16)

From (2.15) and (2.16), we obtain,

| xn | Case(2): If

T  h  {hk : h

| xhn | A for all n 

( p  1) A q

A for all n.

0 and k  }. Given any initial conditions | x h | A and | x0 | A , we argue that

by induction on n . It follows from the given initial conditions that this assertion is true for

n  1, 0. Suppose the assertion is true for n  2 and n  1(n  1). That is,

| xh( n2) | A and | xh( n1) | A. Now, we consider

xhn , where we put h n instead of h(n  1) in equation (2.3),

| xhn |

1 |qx

m1 m2 h ( n 1) h ( n  2)

x

|

( p | xh ( n 1) |  | xh ( n 2) |)

( p  1) A . | q  xhm(1n 1) xhm(2n 2) |

(2.17)

Since,

| q  xhm(1n1) xhm(2n2) | q  xhm(1n1) xhm(2n2)  q

1 |qx

m1 m2 h ( n 1) h ( n  2)

x

1  . | q

0,

(2.18)

From (2.17) and (2.18), we obtain,

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( p  1) A q

| xhn |

T

Case (3): If

2 0

 {n2 : n 

| xk | A for all k  n2 , n  for

0

0

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A for all n.

}. Given any initial conditions | x1 | A and | x0 | A , we argue that

k . It follows from the given initial conditions that this assertion is true

by induction on

k  0,1. Suppose the assertion is true for (n  2) 2 and (n  1) 2 . That is,

| x( n  2)2 | A and | x( n 1)2 | A. Now, we consider

| xn2 |

xn2 , where we put n instead of (n  1) in equation (2.3),

1 |qx

m1 ( n 1)2

m2 ( n  2)2

x

|

( p | x( n 1)2 |  | x( n 2)2 |)

( p  1) A . | q  x(mn11)2 x(mn2 2)2 |

(2.19)

Since,

| q  x(mn11)2 x(mn22)2 | q  x(mn11)2 x(mn22)2  q

1 |qx

m1 ( n 1)2

m2 ( n  2)2

x

0,

1  . | q

(2.20)

From (2.19) and (2.20), we obtain,

| xk | Case(4):

If

T r , r

1.

| xk | A for all k  r n , n  for

Given

any

by induction on

( p  1) A q initial

A for all k .

conditions

| x1/ r | A and | x1 | A ,

we

argue

that

k . It follows from the given initial conditions that this assertion is true

k  1/ r ,1. Suppose the assertion is true for k  r n 2 and k  r n 1 , where n  . That is,

| xr n2 | A and | xr n1 | A. Now, we consider

xr 2 , where we put n instead of (n  1) in equation (2.3), | xr n |

1 |qx x

m1 m2 r n1 r n2

|

( p | xr n1 |  | xr n2 |)

( p  1) A . | q  xrmn11 xrmn22 |

(2.21)

Since,

| q  xrmn11 xrmn22 | q  xrmn11 xrmn22  q,

p 1 1  . m1 m2 | q  xr n1 xr n2 | q

(2.22) Page | 10

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From (2.21) and (2.22), we obtain,

| xk |

( p  1) A q

A for all k .

This completes the proof.

(2.3) Periodic solutions of equation In this section, we study the existence of periodic solutions of equation (2.3).

The following theorem states the necessary and sufficient conditions that this equation has a periodic solutions.

Theorem 2.4: Equation (2.3) has prime period two solutions if and only if

p  q 1 and 0

p  1, where m1  m2  the ratio of odd positive integers.

Proof: First, suppose that, there exists a prime period two solution

...,  , ,  , ,..., of equation (2.3). We will prove that



p  q 1. We see from equation (2.3) that p   p  and   , k  m1  m2 . k q  ( ) q  ( ) k

Then,

  

Since,

   0, 1 

p (   )  (  ) . q  ( ) k

1 p . q  ( ) k

Therefore,

  k 1  p  q ,

(2.23)

and so,

  

( p  1)(  ) . q  ( ) k

Hence,

(  )[q  p  1  ( ) k ]  0.

(2.24)

From (2.23) and (2.24), we obtain

(  )[q  p  1  1  p  q ]  0. Then, (   )(2 p)  0.

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Since,

p

[Vol-1, Issue-6, September- 2015]

0, (  )  0.

(2.25)

 and 

It is clear now, from equation (2.23) and (2.25) that

are the two distinct roots of the quadratic equation

t 2  k 1  p  q  0 or t 2  k p  q  1  0, So,

p  q  1 0. Then, p  q 1. p  q 1 . We will show that equation (2.3) has a prime period two solutions. Assume that

Second, suppose that

  k p  q  1 and    k p  q  1.

Therefore

 and 

are distinct real numbers. Set,

x(  (t0 ))   and x(t0 )   , t0  T .

We wish to show that

x( (t0 ))  x(  (t0 ))  .

It follows from equation (2.3) that

x( (t0 ))  Since,

(1  p) k p  q  1 (1  p) k p  q  1 px(t0 )  x(  (t0 ))   . q  ( x(t0 ) x(  (t0 )) k q  ( p  q  1) 1 p

p  1 , x( (t0 ))  k p  q  1  .

Similarly as before one can easily show that

x( (t ))   , where x(  (t ))   and x(t )   .

Thus equation (2.3) has the prime period two solution

...,  , ,  , ,..., where

 and 

are the distinct roots of the quadratic equation (2.3). The proof is completed.

To examine the global attractivity of the equilibrium points of equation (2.3), we first need to determine the invariant intervals for equation (2.3).

2.4

Invariant intervals and global attractivity of the zero equilibria

In this subsection, we determine the family of invariant intervals centered at Theorem 2.5: Assume

q

  0.

p  1 . Then for any positive real number

A  (q  ( p  1))1/ m , Page | 12

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The interval

[Vol-1, Issue-6, September- 2015]

O (0, A)  (  A, A) is invariant for equation (2.3).

Proof: We consider

T  , h  {hk : h

0 and k  },

Case (1): If T 

. Given any initial conditions

2 0

 {n2 : n 

0

}, and r , r 1.

| x1 | A and | x0 | A , we argue that | xn | A for all n 

induction on n . It follows from the given initial conditions that this assertion is true for true for

by

n  1, 0. Suppose the assertion is

n  2 and n  1(n  1). That is,

| xn2 | A  (q  ( p  1))1/ m and | xn1 | A  (q  ( p  1))1/ m .

Now, we consider

xn , where we put n instead of (n  1) in equation (2.3). Since, q  xnm11 xnm22

q  Am  q  (q  ( p  1))  p  1 0,

p 1  1. | q  xnm11 xnm22 |

(2.26)

Thus,

| xn |

1 | q  xnm11 xnm22 |

(

( p | xn 1 |  | xn  2 |)

p 1 ) max{| xn 1 |,| xn 2 |} | q  xnm11 xnm22 |

max{| xn 1 |,| xn  2 |} Case (2): If

T  h  {hk : h

| xhn | A for all n 

A for all n  .

0 and k  }. Given any initial conditions | x h | A and | x0 | A , we argue that

by induction on n . It follows from the given initial conditions that this assertion is true for

n   h, 0. Suppose the assertion is true for n  2 and n  1(n  1). That is,

| xh ( n2) | A  (q  ( p  1))1/ m and | xh ( n1) | A  (q  ( p  1))1/ m .

Now, we consider

xhn , where we put h n instead of h(n  1) in equation (2.3). Since, q  xhm(1n1) xhm(2n 2)

p 1 |qx x

m1 m2 h ( n 1) h ( n  2)

|

 1.

q  Am  q  (q  ( p  1))  p  1 0,

(2.27)

Thus,

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| xhn | (

1 |qx

m1 m2 h ( n 1) h ( n  2)

x

|

p 1 |qx x

|

m1 m2 h ( n 1) h ( n  2)

( p | xh ( n 1) |  | xh ( n 2) |)

) max{| xh ( n 1) |,| xh ( n  2) |}

max{| xh ( n 1) |,| xh ( n  2) |} Case (3): If

T

2 0

 {n2 : n 

| xk | A for all k  n2 , n  for

0

0

[Vol-1, Issue-6, September- 2015]

A for all n  .

}. Given any initial conditions | x1 | A and | x0 | A ,

by induction on

we argue that

k . It follows from the given initial conditions that this assertion is true

k  0,1. Suppose the assertion is true for (n  2) 2 and (n  1) 2 . That is,

| x( n2)2 | A  (q  ( p  1))1/ m and | x( n1)2 | A  (q  ( p  1))1/ m .

Now, we consider

xn2 , where we put n instead of (n  1) in equation (2.3). Since, q  x(mn11)2 x(mn22)2

q  Am  q  (q  ( p  1))  p  1 0,

p 1  1. |qx x(mn22)2 |

(2.28)

m1 ( n 1)2

Thus,

| xn2 | (

1 |qx

m1 ( n 1)2

x(mn22)2 |

( p | x( n 1)2 |  | x( n 2)2 |)

p 1 ) max{| x( n 1)2 |,| x( n 2)2 |} |qx x(mn22)2 | m1 ( n 1)2

max{| x( n 1)2 |,| x( n 2)2 |} Case

(4):

If

T r , r

| xk | A for all k  r n , n  for

A for all k  n 2 .

1 . Given any initial conditions by induction on

| x1/ r | A and | x1 | A ,

we

argue

that

k . It follows from the given initial conditions that this assertion is true

k  1/ r ,1. Suppose the assertion is true for k  r n 2 and k  r n 1 , where n  . That is,

| xr n2 | A  (q  ( p  1))1/ m and | xr n1 | A  (q  ( p  1))1/ m .

Now, we consider

xr 2 , where we put n instead of (n  1) in equation (2.3). Since, q  xrmn11 xrmn22

p 1  1. | q  xrmn11 xrmn22 |

q  Am  q  (q  ( p  1))  p  1 0,

(2.29)

Thus, Page | 14

International Journal of Engineering Research & Science (IJOER)

| xr n | (

1 |qx

m1 m2 r n1 r n2

x

|

[Vol-1, Issue-6, September- 2015]

( p | xr n1 |  | xr n2 |)

p 1 ) max{| xr n1 |,| xr n2 |} | q  xrmn11 xrmn22 |

max{| xr n1 |,| xr n2 |}

A for all n.

This completes the proof.

Now, we investigate the global attractivity of the equilibrium point Lemma 2.1: Assume that

T ,q

  0.

p  1, and m is positive rational number and numerator's even positive integers.

Furthermore, suppose that

R  (q  ( p  1))1lm ,

(2.30)

and consider equation (2.3) with the restriction that

f : O(0, R)  O(0, R)  O(0, R).

Let

{xn } be a solution of this equation and R1 

1 p . q  (max{| x1 |,| x0 |})m

(2.31)

Then,

R1  (0,1) and | xn | R1n / 2  max{| x1 |,| x0 |}, n  1, 2,...

.

(2.32)

Proof: From Theorem 2.5, we have

| xn | A  R, n  1, 0,1, 2,... where we put n instead of

,

( n  1) in equation (2.3). Then,

(max{| x1 |,| x0 |})m

Rm .

Thus,

q  (max{| x1 |,| x0 |})m

q  Rm  1  p

0.

Hence,

0

1 p q  (max{| x1 |,| x0 |})m

1. This means that R1  (0,1).

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[Vol-1, Issue-6, September- 2015]

Now, we prove that (2.32), by induction on n . From equation (2.3), we have

1

| xn |

| q  xnm11 xnm22 |

( p | xn 1 |  | xn  2 |)

p 1 ) max{| xn 1 |,| xn  2 |} | q  xnm11 xnm22 |

(2.33)

max{| xn 1 |,| xn 2 |}, from (2.26).

(2.34)

(

Then, from (2.33), we obtain

| x1 | (

p 1 ) max{| x0 |,| x1 |}. | q  x0m1 xm12 |

(2.35)

But

| q  x0m1 xm12 | q  | x0m1 || xm12 | q  (max{| x0 |,| x1 |}) m

q  Rm  1  p

0.

Then,

| q  x0m1 xm12 | q  (max{| x0 |,| x1 |})m

0.

From (2.35), we have

| x1 | (

p 1 p 1 ) max{| x0 |,| x1 |}  max{| x0 |,| x1 |} m1 m2 | q  x0 x1 | q  (max{| x0 |,| x1 |}) m

 R1  max{| x0 |,| x1 |} R11/ 2  max{| x0 |,| x1 |}, And from (2.33) and (2.34), we obtain

| x2 | (

p 1 p 1 ) max{| x1 |,| x0 |}  max{| x1 |,| x0 |} m1 m2 | q  x1 x0 | q  (max{| x1 |,| x0 |}) m

p 1  max{max{| x0 |,| x1 |},| x0 |} q  (max{max{| x0 |,| x1 |},| x0 |}) m p 1  max{| x1 |,| x0 |}  R1  max{| x1 |,| x0 |}. q  (max{| x1 |,| x0 |}) m Thus, the inequality (2.32) holds for

n  1, 2. Suppose that the inequality (2.32) holds for n  1 and n  2 ( n  3) ,

respectively. By (2.34), we have

| xn | max{| x0 |,| x1 |} for all n.

So,

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[Vol-1, Issue-6, September- 2015]

p 1 p 1 p 1    R1. m1 m2 m1 m2 | q  xn1 xn2 | q | xn1 | | xn2 | q  (max{| x0 |,| x1 |}) m

0

Then, from (2.33), we have

| xn | (

p 1 ) max{| xn 1 |,| xn  2 |}  R1 max{| xn 1 |,| xn  2 |} | q  xnm11 xnm22 |

 R1 max{R1( n 1) / 2 max{| x0 |,| x1 |}, R1( n  2) / 2 max{| x0 |,| x1 |}}  R1n / 2 max{R11/ 2 max{| x0 |,| x1 |}, max{| x0 |,| x1 |}}  R1n / 2 max{| x0 |,| x1 |}.

This completes the inductive proof of (2.32).

Lemma 2.2: Assume that

T  h  {hk : h  (0,1) and k  }, q

p  1, and m is positive rational number and

numerator's even positive integers. Furthermore, suppose that (2.30) holds and consider equation (2.3) with the restriction that

f : O(0, R)  O(0, R)  O(0, R).

Let

{xkn : h  (0,1) and n  } be a solution of this equation and R1 

1 p . q  (max{| x h |,| x0 |})m

(2.36)

Then,

R1  (0,1) and | xhn | R1hn / 2  max{| xh |,| x0 |}, n  1, 2,...

.

(2.37)

Proof: From Theorem 2.5, we have

| xhn | A  R, n  1, 0,1,...

Where we put hn instead of

,

h( n  1) in equation (2.3). Then,

(max{| x h |,| x0 |})m

Rm .

Thus,

q  (max{| x h |,| x0 |})m

q  Rm  1  p

0.

Hence,

0

1 p q  (max{| x h |,| x0 |})m

1. This means that R1  (0,1).

Page | 17

International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

Now, we prove that (2.37), by induction on n . From equation (2.3), we have

1

| xhn | (

|qx

m1 m2 h ( n 1) h ( n  2)

x

|

p 1 |qx x

|

m1 m2 h ( n 1) h ( n  2)

( p | xh ( n 1) |  | xh ( n  2) |)

) max{| xh ( n 1) |,| xh ( n  2) |}

(2.38)

max{| xh ( n 1) |,| xh ( n  2) |}, from (2.27).

(2.39)

Then, from (2.38), we obtain

| xh | (

p 1 ) max{| x0 |,| x h |}. | q  x0m1 xmh2 |

(2.40)

But

| q  x0m1 xmh2 | q  | x0m1 || xmh2 | q  (max{| x0 |,| x h |})m

q  Rm  1  p

0.

Then,

| q  x0m1 xmh2 | q  (max{| x0 |,| x h |})m

0.

From (2.40), we have

| xh | (

p 1 p 1 ) max{| x0 |,| x h |}  max{| x0 |,| x h |} m1 m2 | q  x0 x h | q  (max{| x0 |,| x h |}) m

 R1  max{| x0 |,| x h |} R1h / 2  max{| x0 |,| x h |}, And from (2.38) and (2.39), we obtain

| x2 h | (

p 1 p 1 ) max{| xh |,| x0 |}  max{| xh |,| x0 |} m1 m2 | q  xh x0 | q  (max{| xh |,| x0 |}) m

p 1  max{max{| x0 |,| x h |},| x0 |} q  (max{max{| x0 |,| x h |},| x0 |}) m p 1  max{| x h |,| x0 |}  R1  max{| x h |,| x0 |} q  (max{| x h |,| x0 |}) m R12 h / 2  max{| x h |,| x0 |}.

Thus, the inequality (2.37) holds for

n  1, 2. Suppose that the inequality (2.37) holds for h( n  1) and h( n  2) ( n  3) ,

respectively. By (2.39), we have

| xn | max{| x0 |,| x h |} for all n.

So, Page | 18

International Journal of Engineering Research & Science (IJOER)

0

p 1 |qx x

m1 m2 h ( n 1) h ( n  2)

|

[Vol-1, Issue-6, September- 2015]

p 1 p 1   R1. m1 m2 q  | xh ( n 1) | | xh ( n 2) | q  (max{| x0 |,| x h |}) m



Then, from (2.38), we have

| xhn | (

p 1 |qx x

m1 m2 h ( n 1) h ( n  2)

|

) max{| xh ( n 1) |,| xh ( n  2) |}

 R1 max{| xh ( n 1) |,| xh ( n 2) |}  R1h max{| xh ( n 1) |,| xh ( n 2) |}  R1h max{R1h ( n 1) / 2 max{| x0 |,| x h |}, R1h ( n 2) / 2 max{| x0 |,| x h |}}  R1hn / 2 max{R1h / 2 max{| x0 |,| x h |}, max{| x0 |,| x h |}}  R1hn / 2 max{| x0 |,| x h |}.

This completes the inductive proof of (2.37).

Lemma 2.3: Assume that

T

2 0

 {n2 : n 

0

}, q

p  1, and m is positive rational number and numerator's even

positive integers. Furthermore, suppose that (2.30) holds and consider equation (2.3) with the restriction that

f : O(0, R)  O(0, R)  O(0, R).

Let

{xk : k  n2 and n 

0

R1 

} be a solution of this equation and

1 p . q  (max{| x1 |,| x0 |})m

(2.41)

Then,

R1  (0,1) and | xk | R1k / 2  max{| x1 |,| x0 |}, k  1, 4,...

.

(2.42)

Proof: From Theorem 2.5, we have

| xk | A  R, k  1, 2,... where we put n instead of

,

( n  1) in equation (2.3). Then,

(max{| x1 |,| x0 |})m

Rm .

Thus,

q  (max{| x1 |,| x0 |})m

q  Rm  1  p

0.

Hence,

0

1 p q  (max{| x1 |,| x0 |})m

1. This means that R1  (0,1).

Page | 19

International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

Now, we prove that (2.42), by induction on k . From equation (2.3), we have

| xn2 |

1 |qx

m1 ( n 1)2

x(mn22)2 |

( p | x( n 1)2 |  | x( n 2)2 |)

p 1 ) max{| x( n 1)2 |,| x( n 2)2 |} |qx x(mn22)2 |

(2.43)

max{| x( n 1)2 |,| x( n 2)2 |}, from (2.28).

(2.44)

(

m1 ( n 1)2

Then, from (2.43), we obtain

| x1 | (

p 1 ) max{| x0 |,| x1 |}. | q  x0m1 x1m2 |

(2.45)

But

| q  x0m1 x1m2 | q  | x0m1 || x1m2 | q  (max{| x0 |,| x1 |}) m

q  Rm  1  p

0.

Then,

| q  x0m1 x1m2 | q  (max{| x0 |,| x1 |}) m

0.

From (2.45), we have

| x1 | (

p 1 p 1 ) max{| x0 |,| x1 |}  max{| x0 |,| x1 |} m1 m2 | q  x0 x1 | q  (max{| x0 |,| x1 |}) m

 R1  max{| x0 |,| x1 |} R11/ 2  max{| x0 |,| x1 |}, And from (2.43) and (2.44), we obtain

| x4 | (

p 1 p 1 ) max{| x1 |,| x0 |}  max{| x1 |,| x0 |} m1 m2 | q  x1 x0 | q  (max{| x1 |,| x0 |}) m

p 1  max{R1 max{| x0 |,| x1 |},| x0 |} q  (max{max{| x0 |,| x1 |},| x0 |}) m p 1  R1  max{| x1 |,| x0 |} q  (max{| x1 |,| x0 |}) m

R12  max{| x1 |,| x0 |}

R14 / 2  max{| x1 |,| x0 |}.

Thus, the inequality (2.42) holds for

n  1, 2. Suppose that the inequality (2.42) holds for (n  1)2 and ( n  2) 2 ,

respectively. By (2.44), we have

| xn2 | max{| x( n 1)2 |,| x( n  2)2 |} for all n 

0

.

So, Page | 20

International Journal of Engineering Research & Science (IJOER)

0

[Vol-1, Issue-6, September- 2015]

p 1 p 1 p 1    R1. m2 m1 m2 |qx x( n2)2 | q  | x( n1)2 | | x( n2)2 | q  (max{| x0 |,| x1 |}) m m1 ( n 1)2

Then, from (2.43), we have

| xn2 | (

p 1 ) max{| x( n 1)2 |,| x( n 2)2 |}  R1 max{| x( n 1)2 |,| x( n 2)2 |} |qx x(mn2 2)2 | m1 ( n 1)2

 max{R1( n 1)  R1( n  2)  R1n

2

/2

2

/2

2

/2

max{| x0 |,| x1 |}, R1( n 2)

2

/2

max{| x0 |,| x1 |}}

max{| x0 |,| x1 |}, n  3, 4,...

max{| x0 |,| x1 |}, n  1, 2,...

.

This completes the inductive proof of (2.42).

T  r , 1 r  2, q

Lemma 2.4: Assume that

p  1, and m is positive rational number and numerator's even

positive integers. Furthermore, suppose that (2.30) holds and consider equation (2.3) with the restriction that

f : O(0, R)  O(0, R)  O(0, R).

Let

{xk : k  r n , n 

and 1 r  2} be a solution of this equation and R1 

1 p . q  (max{| x1/ r |,| x1 |})m

(2.46)

Then,

R1  (0,1) and | xk | R1k / 2  max{| x1/ r |,| x1 |}.

(2.47)

Proof: From Theorem 2.5, we have

| xk | A  R, k  r n , n  , and 1 r  2, where we put n instead of

( n  1) in equation (2.3). Then,

(max{| x1/ r |,| x1 |})m

Rm .

Thus,

q  (max{| x1/ r |,| x1 |})m

q  Rm  1  p

0.

Hence,

0

1 p q  (max{| x1/ r |,| x1 |})m

1. This means that R1  (0,1).

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[Vol-1, Issue-6, September- 2015]

Now, we prove that (2.47), by induction on n . From equation (2.3), we have

| xr n |

1 |qx

m1 m2 r n1 r n2

x

|

( p | xr n1 |  | xr n2 |)

p 1 ) max{| xr n1 |,| xr n2 |} | q  xrmn11 xrmn22 |

(2.48)

max{| xr n1 |,| xr n2 |}, from (2.29).

(2.49)

p 1 ) max{| x1/ r |,| x1 |}. | q  x1m1 x1/m2r |

(2.50)

(

Then, from (2.48), we obtain

| xr | (

But

| q  x1m1 x1/m2r | q | x1m1 || x1/m2r | q  (max{| x1/ r |,| x1 |}) m

q  Rm  1  p

0.

Then,

| q  x1/m1r x1m2 | q  (max{| x1/ r |,| x1 |})m

0.

From (2.50), we have

| xr | (

p 1 p 1 ) max{| x1/ r |,| x1 |}  max{| x1/ r |,| x1 |} m1 m2 | q  x1/ r x1 | q  (max{| x1/ r |,| x1 |}) m

 R1  max{| x1/ r |,| x1 |} R1r / 2  max{| x1/ r |,| x1 |}, And from (2.48) and (2.49), we obtain

| xr 2 | (

p 1 p 1 ) max{| xr |,| x1 |}  max{| xr |,| x1 |} m1 m2 | q  xr x1 | q  (max{| xr |,| x1 |}) m

p 1  max{R1r / 2 max{| x1/ r |,| x1 |},| x1 |} m q  (max{max{| x1/ r |,| x1 |},| x1 |}) p 1  R1r / 2  max{| x1/ r |,| x1 |} m q  (max{| x1/ r |,| x1 |}) R1  R1r / 2  max{| x1/ r |,| x1 |} R12 r / 2  max{| x1/ r |,| x1 |}  R1r Thus,

k r

the n2

inequality

and k  r

n 1

(2.47)

holds

for

R1r / 2  R1r / 2  max{| x1/ r |,| x1 |} 2

/2

 max{| x1/ r |,| x1 |}.

k  r, r 2.

Suppose

that

the

inequality

(2.47)

holds

for

, n  , respectively. By (2.49), we have

| xr n | max{| xr n1 |,| xr n2 |} for all n.

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[Vol-1, Issue-6, September- 2015]

So,

0

p 1 p 1 p 1    R1. m1 m2 m1 m2 | q  xr n1 xr n2 | q  | xr n1 | | xr n2 | q  (max{| x1/ r |,| x1 |}) m

Then, from (2.48), we have

| xr n | (

p 1 ) max{| xr n1 |,| xr n2 |}  R1 max{| xr n1 |,| xr n2 |} | q  xrmn11 xrmn22 |

 max{R1r  R1r

n2

 R1r

n

/2

/2

n1

/2

max{| x1/ r |,| x1 |}, R1r

n2

/2

max{| x1/ r |,| x1 |}}

 max{| x1/ r |,| x1 |}}, n  3, 4,...

 max{| x1/ r |,| x1 |}, n  1, 2,3,... .

This completes the inductive proof of (2.47).

Theorem 2.6: Assume that

q

p  1, and m is positive rational number and numerator's even positive integers.

Furthermore, suppose that (2.30) holds, and consider equation (2.3) with the restriction that

f : O(0, R)  O(0, R)  O(0, R).

Then the equilibrium point

 0

Proof: From Lemmas

of the equation (2.3) is a global attractor.

2.i, i  1, 2,3, 4, x(t )    0 when t  , then the equilibrium point   0 of the

equation (2.3) is a global attractor.

Next, we determine the family of invariant intervals centered at 2.5

  m ( p  1)  q when p  1 q.

Invariant intervals and global attractivity of the nonzero equilibria

In this subsection, we consider the discrete time scales

T  , h  {hk : h

0 and k  },

2 0

 {n2 : n 

0

},

and r , r 1, m1  m2  1, and ( p  1)

This means that, we determine the family of invariant intervals centered at

  ( p  1)  q when p  1 q.

q.

To this

end, we establish the following relation.

px(t )  x(  (t )) p     q  x(t ) x(  (t )) q   2 px(t )  x(  (t )) p   x(  (t )) p   x(  (t )) p    (  )(  ) q  x(t ) x(  (t )) q   x(  (t )) q   x(  (t )) q   2

x( (t ))   



pq  x 2 (  (t )) ( x(t )   ) (q  x(t ) x(  (t )))(q   x(  (t ))) 

In view of

q  p 2 ( x(  (t ))   ). (q   x(  (t )))(q   2 )

 2  p  1  q,

the above equality is reduced to Page | 23

International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

pq  x 2 (  (t )) x( (t ))    ( x(t )   ) (q  x(t ) x(  (t )))(q   x(  (t ))) q p  ( x(  (t ))   ). q   x(  (t )) Thus,

pq  x 2 (  (t )) || x(t )   | (q  x(t ) x(  (t )))(q   x(  (t ))) q p | || x(  (t ))   | q   x(  (t ))

| x( (t ))   ||

pq  x 2 (  (t )) q p  (| || |) (q  x(t ) x(  (t )))(q   x(  (t ))) q   x(  (t ))  max{| x(  (t ))   |,| x(t )   |}.

(2.51)

Consider the function

g ( x, y ) 

pq  y 2 | pq|  ,x (q  xy )(q   y ) (q   y )

0, y  (0, pq ).

(2.52)

The following result is obvious.

Lemma 2.5: The function g is strictly decreasing in x and decreasing in y .

Consider the function

pq  x 2 | pq| h ( x )  g ( x, x )   , x  (0, pq ).

2 (q  x )(q   x) (q   x) We have the following result.

Lemma 2.6: Assume that Proof: Clearly,

h

q  p. Then, h( x) 1 for x  (0, pq ).

is strictly decreasing in x . Since

h( x )  The result follows from

q  p, then

pq  x 2 pq  .

2 (q  x )(q   x) (q   x)

h(0)  1.

Now, we are ready to describe the family of nested invariant intervals centered at Theorem 2.7: Assume that

max{1, p} q



.

p  1.

Page | 24

International Journal of Engineering Research & Science (IJOER)

i- If

q

[Vol-1, Issue-6, September- 2015]

4( p  1) , then for every positive number A   , the interval 4 p O(  , A)  (   A,   A)

is invariant for equation (2.3). ii- If

q

4( p  1) , then for every positive number A  4 p

pq   , the interval

O(  , A)  (   A,   A) is invariant for equation (2.3).

Proof: First, we note that

pq   2  pq  [( p  1)  q]  (1  p)(q  1) That is,



between

0 and

pq . Now assume 

A  min{ , pq   }  { Case (1): If T 

0.

4( p 1) , 4 p 4( p 1) if q .

4 p

if q  pq  

, given any initial conditions

| x1   | A and | x0   | A, we argue that

| xn   | A for all n by induction on n.

The proof is similar to the proof of Theorem 4.4 in [27 ] and will be omitted.

Case 2: If

T h

, given any initial conditions

| x h   | A and | x0   | A, we argue that

| xhn   | A for all n by induction on n.

It follows from the given initial assertion is true for Suppose the asseration is true for

n  1, 0.

h(n  2) and h(n  1). . That is,

| xh( n2)   | A and | xh( n1)   | A. Now, we consider

xhn where we put hn instead of h( n  1) in equation (2.51), Lemma 2.5 and Lemma 2.6, we have Page | 25

International Journal of Engineering Research & Science (IJOER)

| xhn   | (| (

pq  xh2( n  2)

||

(q  xh ( n 1) xh ( n 2) )(q   xh ( n 2) ) pq  xh2( n  2) (q  xh ( n 1) xh ( n 2) )(q   xh ( n 2) )



[Vol-1, Issue-6, September- 2015]

q p |)  max{| xh ( n 1)   |,| xh ( n 2)   |} q   xh ( n 2)

q p )  max{| xh ( n 1)   |,| xh ( n 2)   |} q   xh ( n 2)

 g ( xh ( n 1) , xh ( n  2) )  max{| xh ( n 1)   |,| xh ( n 2)   |}  g (min{xh ( n 1) , xh ( n 2) }, min{xh ( n 1) , xh ( n 2) })  max{| xh ( n 1)   |,| xh ( n 2)   |}  h(min{xh ( n 1) , xh ( n 2) })  max{| xh ( n 1)   |,| xh ( n 2)   |}  max{| xh ( n 1)   |,| xh ( n 2)   |}

A.

This completes the inductive proof.

When

T

2 0

and r , r 1,

the proof will be omitted.

By an analogous argument to that for Theorem 2.7, we can obtain the following result about the family of nested invariant intervals centered at  .

Theorem 2.8: Assume that

i- If

q

max{1, p} q

p  1.

4( p  1) , then for every positive number A |  |, the interval 4 p

O( , A)  (  A,   A) is invariant for equation (2.3). ii- If

q

4( p  1) , then for every positive number A  4 p

pq  |  |, the interval

O( , A)  (  A,   A) is invariant for equation (2.3).

Now, we investigate the global attractivity of the equilibrium point Lemma 2.8: Assume that

T  , and max{1, p} q

 and  .

p  1. Suppose that

R  min{ , pq   },

(2.53)

and consider equation (2.3) with the restriction that

f : O (  , R )  O (  , R )  O (  , R ).

Let

{xn } be a solution of this equation and

Page | 26

International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0 M {

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  (2   x0 ) )(q   (2   x0 )) q   (2   x0 ) pq  x21 q p  if | x1   | | x0   |& x1   , 2 (q  x1 )(q   x1 ) q   x1 pq  x21 q p  2 (q  (2   x1 ) )(q   (2  x1 )) q   (2   x1 )

if | x1   | | x0   |& x1   .

(2.54)

Then,

M  (0,1) and | xn   | M n / 2  max{| x1   |,| x0   |}, n  1, 2,...

.

(2.55)

The proof is similar to the proof of Lemma 6.1 in [27 ] and will be omitted.

Lemma 2.9: Assume that

T  h  {hk : h  (0,1) and k  }, and max{1, p} q

p  1. Suppose that (2.53)

holds and consider equation (2.3) with the restriction that

f : O (  , R )  O (  , R )  O (  , R ).

Let

{xhn : h  (0,1) and n  } be a solution of this equation and pq  x02 q p  if | x h   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0

M {

pq  x02 q p  if | x h   | | x0   |& x0   , 2 (q  (2   x0 ) )(q   (2   x0 )) q   (2   x0 ) pq  x2h q p  if | x h   | | x0   |& x h   , 2 (q  x h )(q   x h ) q   x h pq  x2h q p  2 (q  (2  x h ) )(q   (2  x h )) q   (2   x h )

if | x h   | | x0   |& x h   .

(2.56)

Then,

M  (0,1) and | xhn   | M hn / 2  max{| xh   |,| x0   |}, n  1, 2,...

Proof: The result that

.

(2.57)

M  (0,1) follows from Lemma 2.6. Next, we prove equation (2.57) by induction on n. First, set

  x0 x  A { 0   x h x h  

if | x h   | | x0   |& x0   , if | x h   | | x0   |& x0   , if | x h   | | x0   |& x h   , if | x h   | | x0   |& x h   .

Page | 27

International Journal of Engineering Research & Science (IJOER)

Then,

[Vol-1, Issue-6, September- 2015]

A   . By Lemma 2.5 and Theorem 2.7, we derive that for all n , pq  xh2( n  2) (q  xh ( n 1) xh ( n  2) )(q   xh ( n  2) )

where we put

hn

instead of



q p  M, q   xh ( n 2)

h( n  1) . By equation (2.51), we have

| xh   | (|

pq  x2h q p || |)  max{| x0   |,| x h   |} (q  x0 xh ( n2) )(q   x h ) q   x h

 M  max{| x0   |,| x h   |}  M h / 2  max{| x0   |,| x h   |}, and | x2 h   | (|

pq  x02 q p || |)  max{| x0   |,| xh   |} (q  x0 xh )(q   x0 ) q   x0

 M  max{| x0   |, M  max{| x0   |,| xh   |}}  M  max{| x0   |,| xh   |}  M 2 h / 2  max{| x0   |,| xh   |}.

So equation (2.57) holds for n  1, 2. Suppose that equation (2.57) holds for h(n  1) and h(n-2), respectively.

By the inductive hypothesis, we have | xhn   | (

pq  xh2( n  2) (q  xh ( n 1) xh ( n 2) )(q   xh ( n 2) )



q p )  max{| xh ( n 1)   |,| xh ( n  2)   |} q   xh ( n 2)

 M max{M h ( n 1) / 2  max{| x0   |,| x h   |}, M h ( n 2) / 2  max{| x0   |,| x h   |}}  M h max{M h ( n 1) / 2  max{| x0   |,| x h   |}, M h ( n  2) / 2  max{| x0   |,| x h   |}}  M nh / 2 max{M h / 2  max{| x0   |,| x h   |}, max{| x0   |,| x h   |}}  M nh / 2 max{| x0   |,| x h   |}.

This completes the inductive proof of equation (2.57).

Lemma 2.10: Assume that

T

2 0

 {n2 : n 

0

}, and max{1, p} q

p  1. Suppose that (2.53) holds and

consider equation (2.3) with the restriction that

f : O (  , R )  O (  , R )  O (  , R ).

Let

{xk : k  n2 and n 

0

} be a solution of this equation and

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International Journal of Engineering Research & Science (IJOER)

[Vol-1, Issue-6, September- 2015]

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0 M {

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  (2  x0 ) )(q   (2   x0 )) q   (2   x0 ) pq  x12 q p  if | x1   | | x0   |& x1   , 2 (q  x1 )(q   x1 ) q   x1 pq  x12 q p  2 (q  (2  x1 ) )(q   (2   x1 )) q   (2  x1 )

if | x1   | | x0   |& x1   .

(2.58)

Then,

M  (0,1) and | xk   | M k / 2  max{| x1   |,| x0   |}, k  1, 4,...

Lemma 2.11: Assume that

T  r , 1 r  2, and max{1, p} q

.

(2.59)

p  1. Suppose that (2.53) holds and consider

equation (2.3) with the restriction that

f : O (  , R )  O (  , R )  O (  , R ).

Let

{xk : k  r n , 1 r  2, and n  } be a solution of this equation and pq  x12 q p  if | x1/ r   | | x1   |& x1   , 2 (q  x1 )(q   x1 ) q   x1

M {

pq  x12 q p  if | x1/ r   | | x1   |& x1   , 2 (q  (2   x1 ) )(q   (2   x1 )) q   (2   x1 ) pq  x1/2 r q p  if | x1/ r   | | x1   |& x1/ r   , 2 (q  x1/ r )(q   x1/ r ) q   x1/ r pq  x1/2 r q p  2 (q  (2   x1/ r ) )(q   (2   x1/ r )) q   (2   x1/ r )

if | x1/ r   | | x1   |& x1/ r   .

(2.60)

Then,

M  (0,1) and | xk   | M k / 2  max{| x1/ r   |,| x1   |}.

Theorem 2.9: Assume that

max{1, p} q

(2.61)

p  1, and (2.53) holds. Consider equation (2.3) with the restriction that

f : O (  , R )  O (  , R )  O (  , R ).

Then the equilibrium point Proof: From Lemmas



of the equation (2.3) is a global attractor.

2.i, i  8,9,10,11, we obtain x(t )   when t  , then the equilibrium point β of the

equation (2.3) is a global attractor.

At the end of this section , we can establish the following results related to the global attractivity of



.

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International Journal of Engineering Research & Science (IJOER)

Lemma 2.12: Assume that

T  , and max{1, p} q

[Vol-1, Issue-6, September- 2015]

p  1. Suppose that

R  min{|  |, pq  |  |},

(2.62)

and consider equation (2.3) with the restriction that

f : O( , R)  O( , R)  O( , R).

Let

{xn } be a solution of this equation and pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0

N {

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  (2  x0 ) )(q   (2  x0 )) q   (2  x0 ) pq  x21 q p  if | x1   | | x0   |& x1   , 2 (q  x1 )(q   x1 ) q   x1 pq  x21 q p  2 (q  (2  x1 ) )(q   (2  x1 )) q   (2  x1 )

if | x1   | | x0   |& x1   .

(2.63)

Then,

N  (0,1) and | xn   | N n / 2  max{| x1   |,| x0   |}, n  1, 2,...

Lemma 2.13: Assume that

.

(2.64)

T  h  {hk : h  (0,1) and k  }, and max{1, p} q

p  1. Suppose that (2.62)

holds and consider equation (2.3) with the restriction that

f : O( , R)  O( , R)  O( , R).

Let

{xhn : h  (0,1) and n  } be a solution of this equation and pq  x02 q p  if | x h   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0

N {

pq  x02 q p  if | x h   | | x0   |& x0   , 2 (q  (2  x0 ) )(q   (2  x0 )) q   (2  x0 ) pq  x2h q p  if | x h   | | x0   |& x h   , 2 (q  x h )(q   x h ) q   x h pq  x2h q p  2 (q  (2  x h ) )(q   (2  x h )) q   (2  x h )

if | x h   | | x0   |& x h   .

(2.65)

Then,

N  (0,1) and | xhn   | N hn / 2  max{| xh   |,| x0   |}, n  1, 2,...

.

(2.66)

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International Journal of Engineering Research & Science (IJOER)

T

Lemma 2.14: Assume that

2 0

 {n2 : n 

0

[Vol-1, Issue-6, September- 2015]

}, and max{1, p} q

p  1. Suppose

that (2.62) holds and

consider equation (2.3) with the restriction that

f : O( , R)  O( , R)  O( , R).

Let

{xk : k  n2 and n 

0

} be a solution of this equation and pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  x0 )(q   x0 ) q   x0

N {

pq  x02 q p  if | x1   | | x0   |& x0   , 2 (q  (2  x0 ) )(q   (2  x0 )) q   (2  x0 ) pq  x12 q p  if | x1   | | x0   |& x1   , 2 (q  x1 )(q   x1 ) q   x1 pq  x12 q p  2 (q  (2  x1 ) )(q   (2  x1 )) q   (2  x1 )

if | x1   | | x0   |& x1   .

(2.67)

Then,

N  (0,1) and | xk   | N k / 2  max{| x1   |,| x0   |}, k  1, 4,...

Lemma 2.15: Assume that

T  r , 1 r  2, and max{1, p} q

.

(2.68)

p  1. Suppose that (2.62) holds and consider

equation (2.3) with the restriction that

f : O( , R)  O( , R)  O( , R).

Let

{xk : k  r n , 1 r  2, and n  } be a solution of this equation and pq  x12 q p  if | x1/ r   | | x1   |& x1   , 2 (q  x1 )(q   x1 ) q   x1

N {

pq  x12 q p  if | x1/ r   | | x1   |& x1   , 2 (q  (2  x1 ) )(q   (2  x1 )) q   (2  x1 ) pq  x1/2 r q p  if | x1/ r   | | x1   |& x1/ r   , 2 (q  x1/ r )(q   x1/ r ) q   x1/ r pq  x1/2 r q p  2 (q  (2  x1/ r ) )(q   (2  x1/ r )) q   (2  x1/ r )

if | x1/ r   | | x1   |& x1/ r   .

(2.69)

Then,

N  (0,1) and | xk   | N k / 2  max{| x1/ r   |,| x1   |}.

Theorem 2.10: Assume that

max{1, p} q

(2.70)

p  1, and (2.62) holds. Consider equation (2.3) with the restriction that

f : O( , R)  O( , R)  O( , R).

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International Journal of Engineering Research & Science (IJOER)

Then the equilibrium point Proof: From Lemmas

[Vol-1, Issue-6, September- 2015]

 of the equation (2.3) is a global attractor.

2.i, i  12,13,14,15, we obtain x(t )   when t  , then the equilibrium point  of the

equation (2.3) is a global attractor.

ACKNOWLEDGEMENTS I thank my teacher Prof. E. M. Elabbasy.

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