Global asymptotic stability in a rational dynamic equation on discrete time scales by Engineering Journal- IJOER

International Journal of Engineering Research & Science (IJOER)

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

Global asymptotic stability in a rational dynamic equation on discrete time scales x( (t )) 

ax(t )  bx(  (t )) , t T c  ( x(  (t )))m

Sh. R. Elzeiny Department of Mathematics, Faculty of Science, Al-Baha University, Kingdom of Saudi Arabia E-mail: shrelzeiny@yahoo.com

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 )) 

where a, b, c

ax(t )  bx(  (t )) , t  T, c  ( x(  (t )))m

0 and m  1.

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

INTRODUCTION

The theory of time scales, which has recently received a lot of attention, was introduced by Stefan Hilger in his PhD thesis [15] in order to unify continuous and discrete analysis. The theory of dynamic equations not only unifies the theories of differential equations and difference equations, but also it extends these classical cases to cases in between, e.g., to so-called q-difference equations. Since then several authors have expounded on various aspects of this new theory, see the survey paper by Agarwal et al. [2] and the references cited therein. Many other interesting time scales exist, and they give rise to many applications, among them the study of population dynamic models (see [8]). A book on the subject of time scales by Bohner and Petreson [5] summarizes and organizes much of the time scales calculus (see also [4]). The study of rational dynamic equations on time scales goes to back to Elzeiny [13]. For the notions used below we refer the reader to [5] and to the following a short introduction to the time scale calculus. Definition 1.1: A time scale is an arbitrary nonempty closed subset of the real numbers

. 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 theCantor 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. 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

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 (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

sup T and  (t )  t , thent is called right-dense, and if t

inf T and  (t )  t , then t is called left-dense.

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

f :T 

and if

is a function, then the function f



:T 

is defined by

f  (t )  f ( (t )) for all t T , i. e., f   f  is the composition function of f with  . Example 1.1: Let us briefly consider the two examples (i) If

T

,  (t )  inf{s 

, then we have for any t 

 (t )  t.

Hence every point t 

The graininess operator

  0 for all t  .

and similarly

(ii) If

T

and T  . :s

t}  inf(t, )  t,

is left-dense and right-dense.

T  , then we have for any t 

 (t )  inf{s  : s t}  inf{t  1, t  2,....}  t  1, and similarly  (t )  t 1. scattered and right-scattered. The graininess operator   1 for all t  . Example 1.2: Consider the time scale T 

2 0

 {n2 : n 

Hence, every point t 

is left-

}. Then, we have

 (n2 )  (n  1)2 ,  (n2 )  (n  1) 2 , and  (n2 )  2n  1 for n 

. Thus,

 (t )  ( t  1)2 ,  (t )  ( t  1) 2 , and  (t )  2 t  1. Example 1.3: Let T 

/ 4  {k / 4 : k  }. Then, we have for t  T ,

 (t )  inf{s T : s t}  inf{t  n / 4 : n  }  t  1/ 4, and  (t )  t  h / 4. Hence, every point so that

t  T is isolated and  (t )   (t )  t  t  1/ 4  t  1/ 4 for t T ,

 in this example is constant.

Example 1.4: Consider the time scale

t T  2 , we get  (t )  2t ,  (t )  , and  (t )  t for t  T . 2

Continuity on time scales is defined in the following manner. 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

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and we say

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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 continuous at

f :[a, b]T 

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

is continuous provided

a , and f is right continuous at b .

In the time scale calculus, the functions are right-dense continuous which we now define. Definition 1.4: A function

f :T 

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

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

right-dense points in

f :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 

x( (t )) 

, the rational dynamic equation on discrete time scales

ax(t )  bx(  (t )) , t  T, c  ( x(  (t )))m

where a, b, c

0 and m  1. becomes the recursive sequence xn1 

axn  bxn1 , n  1, 2,..., c  xnm1

(1.1)

where a, b, c  0 and m  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.[20]. 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, 3, 6, 7, 9- 12, 14, 16, 18, 19, 21-26]. This paper addresses, the global stability, periodicity character and boundedness of the solutions of the rational dynamic equation on discrete time scales

x( (t )) 

ax(t )  bx(  (t )) , t T , c  ( x(  (t )))m

(1.2) Page | 3

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T

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0 and m  1.

where a, b, c When

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and m  1 , our equation reduces to equation which examined by Yang et al. [24].

Here, we recall some notations and results which will be useful in our investigation. Let I be some interval of real numbers and let

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 , has a unique solution

(1.3)

{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



z (t )  uzt  vz (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 [17]):

(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  if x(t   )  x(t ) for all t T .

(1.6)

{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. (b) A solution

Definition 1.8: An interval J  I is called invariant for equation (1.3) if every solution with initial conditions For a real number

( x(  (t0 )), x(t0 ))  J  J satisfies

{x(t ) : t T } of equation (1.3)

x(t )  J for all t T .

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 [17].

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II. 2.1

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MAIN RESULTS

Local asymptotic stability of the equilibrium points

Consider the rational dynamic equation on discrete time scales

x( (t )) 

ax(t )  bx(  (t )) , t T , c  ( x(  (t )))m

(2.1)

where a, b, c  0 and m  1. . Let a 

a c 1 , c  , and d   . Then equation (2.1) can be rewritten as b b b

ax(t )  x(  (t )) , t T . c  d ( x(  (t )))m

x( (t ))  The change of variables y(t )  (d )

1/ m

x(t ), followed by the change x(t )  y(t ) reduces the above equation to

x( (t ))  where p  a  Now, when

(2.2)

px(t )  x(  (t )) , t T , q  ( x(  (t )))m

(2.3)

a c , and q  c  . Hereafter, we focus our attention on equation (2.3) instead of equation (2.1). b b

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

  0, and   m p  q  1, and when

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

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

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

q  1. Furthermore, suppose that, m 

Then, in this case, we consider

x(t )  0 for all t T .

When

2k  1 , k, n  2n



p  q  1, equation (2.3) has unique equilibrium point   0.

When

q  1, however, equation (2.3) has the following two equilibrium points:

  0, and   m p  q  1. The local asymptotic behavior of

 0

is characterized by the following result.

THEOREM 2.1:

max{1, p  1} , then  is locally asymptotically stable.

(1) If

(2) If

p  1 q 1, then  is a repeller.

(3) If

p  1 , then  is a saddle point. Page | 5

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

z ( (t ))  Now, when T 

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 0

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

(2.5)

p 1    0, where z (t )   t , q q

(2.6)

zn1  The characteristic equation associated with equation (2.5) is

2  when

T  r , r 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  when

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

(2.8)

T  h , then equation (2.4) becomes zh ( n1) 

p 1 zhn  zh ( n1)  0, n  . q q

(2.9)

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

(2.11)

p 1    0, where z (t )   t . q q

(2.12)

The characteristic equation associated with equation (2.9) is

2 

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

2 

p 1    0, for all t  T . q q

(2.13)

p 1 Let u   , and v  . q q (1) The result follows from Theorem 1.2 (1) and the following relations Page | 6

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| u | (1  v) 

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p 1 ( p  1)  q  (1  )  q q q

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0, and v 

1 q

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

| u |  |1  v | (3) Note that

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

Now, the local asymptotic behavior of Theorem 2.2 : Assume that

then both

m( p  q 1)

 and 

(iii) If either



and

p 4 p 2  4q 0, and u 2  4v  ( ) 2   q q q2

 are characterized by the following result.

p  1. p  2 or 3 p 1 q 1, where m  1,

are locally asymptotically stable.

(ii) If either m( p  q  1) then both

1 q

p  1  p 2 / 4. The result follows from Theorem 1.2 (3) and the following relations

| u |  |1  v |

(i) If either

0, and | v |

p  2 or q

max{1, 3 p 1}, where m  1,

are repeller.

p 1 q 3 p 1, where m  1, then  is a saddle point.

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

z ( (t )) 



p m( p  q  1)  1 z (t )  z (  (t ))  0. p 1 p 1

(2.14)

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

2  Let u 

p m( p  q  1)  1   0, t  T . p 1 p 1

p m( p  q  1)  1 and v  . p 1 p 1

(i) Assume

m( p  q 1)

| u | (v  1) |

p  2, then, p 1. Then, from Theorem 1.2 (1) and the following relations

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

and

v we conclude that



m( p  q  1)  1 1, p 1

is locally asymptotically stable. Assume

3 p 1 q 1, where m  1, then, p 1. Then, from

Theorem 1.2 (1) and the following relations

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| u | (v  1) |

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p ( p  q) p ( p  q)  p  1 3 p  1  q | (  1)  ( ) p 1 p 1 1 p p 1 1 p

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and

pq q p  p 1 1  p

v we conclude that (ii) Assume



is locally asymptotically stable. Similarly, we can prove that

m( p  q 1)

| u | (v  1) |

 is locally asymptotically stable.

p  2, then, p 1. Then, from Theorem 1.2 (2) and the following relations

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

and

| v | we conclude that



is a repeller. Assume that q

m( p  q  1)  1 1, p 1

max{1,3 p 1}, where m  1. Then, from Theorem 1.2 (2) and the

following relations

| u | (v  1) | and when

when

p ( p  q) p 2 p  q  1 (3 p  1)  q ||  1|  | | p 1 p 1 | p  1| p 1 | p  1|

p 1, we have q  p

p 1, we have q 1

we conclude that



2 p 1  0, so | v ||

pq q p | p 1 p 1

| v ||

pq q p | p 1 1  p

p, so

is a repeller. Similarly, we can prove that

 is a repeller.

p 1. Note that the given condition is equivalent to p | 2 p  q 1| . Besides, we have p(3 p  4) p 1  . Then, from Theorem 1.2 (3) and the following relations 4( p  1)

(iii) If

| u |  | v  1||

p pq p | 2 p  q  1| ||  1|| | p 1 p 1 p 1 p 1 u 2  4v  (

we conclude that



0, and

p 2 4( p  q) 1 )   [4( p  1)q  p(3 p  4)] 0, p 1 p 1 ( p  1)2

is a saddle point.

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p 1. Note that the given condition is equivalent to p | 2 p  q 1| . Besides, we have p(4  3 p) q 3 p 1  . Then, from Theorem 1.2 (3) and the following relations 4(1  p) Assume that

| u |  | v  1||

p pq p | 2 p  q  1| ||  1|   p 1 p 1 1 p 1 p u 2  4v  (

we conclude that 2.2



0, and

p 2 4( p  q) 1 )   [ p(4  3 p)  4(1  p)q] 0, p 1 p 1 (1  p)2

is a saddle point.

Boundedness of solutions of equation (2.3)

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

p  1  q and m is positive rational number and numerator's even positive integers, i  1, 2 . Then the solution of equation (2.3) is bounded for all t  T .

Theorem 2.3: Suppose that

Proof: We argue that Case (1): If T 

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

. Given any initial conditions

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

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, induction on

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

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

| pxn1  xn2 | p | xn1 |  | xn2 | ( p  1) A   | q  xnm2 | | q  xnm2 | | q  xnm2 |

(2.15)

Since,

| q  xnm2 | q  xnm2  q

1 1  . m | q  xn2 | q

(2.16)

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

| xn | Case(2): If

( p  1) A q

A for all n.

T  h  {hk : h 0 and k  }. 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 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), Page | 9

International Journal of Engineering Research & Science (IJOER)

| xhn |

| pxh ( n1)  xh ( n 2) | |qx

m h ( n  2)



ISSN: [2395-6992]

p | xh ( n1) |  | xh ( n 2) | |qx

m h ( n  2)

| q  xhm( n2) | q  xhm( n2)  q

Since,

[Vol-2, Issue-12, December- 2016]

( p  1) A . | q  xhm( n 2) |

(2.17)

1 1  . m | q  xh ( n2) | q

(2.18)

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

( p  1) A q

| xhn | Case (3): If T 

2 0

 {n2 : n 

| xk | A for all k  n2 , n  for

A for all n.

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

by induction on

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 and | x( n1)2 | A. (n  1) in equation (2.3),

Now, we consider xn2 , where we put n instead of

| xn2 |

| px( n 1)2  x( n 2)2 | |qx

m ( n  2)2

p | x( n 1)2 |  | x( n 2)2 |



|qx

m ( n  2)2

| q  x(mn2)2 | q  x(mn2)2  q

Since,

( p  1) A . | q  x(mn 2)2 |

(2.19)

1 1  . m | q  x( n2)2 | q

(2.20)

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

( p  1) A q

| xk | Case(4):

T  r , r 1.

| xk | A for all k  r n , n  for

Given

any

initial

by induction on

A for all k.

conditions

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

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 |

p | xr n1 |  | xr n2 |

| q  xrmn2 | q  xrmn2  q,

Since,

|qx

m r n 2

( p  1) A . | q  xrmn2 |

p 1 1  . m | q  xr n2 | q

(2.21)

(2.22)

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

| xk |

( p  1) A q

A for all k.This completes the proof. Page | 10

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2.3

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

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

q 3 p 1, where q  p 2  p and m  1. Proof: First, suppose that, there exists a prime period two solution

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

3 p  1. We see from equation (2.3) that

 Then,

p   p  and   . q  q 

 q   2  p   , and  q  2  p  , hence, q(  )  (  )(  )  p(   )  (   ).

Since,

   0, q  (  )  1  p. Therefore,    ( p  q  1),

(2.23)

and so,

  

pq(   )  q(   )  p(   )(  ) . Since,  [q 2  q(  )   ]  pq  q  p(  ). (q   )(q  )

Hence,

  q( p  q  1)  ( p  q)(  ).

(2.24)

From (2.23) and (2.24), we obtain

  q( p  q  1)  ( p  q)( p  q  1). Then,   p( p  q  1). (2.25) It is clear now, from equation (2.23) and (2.25) that

 and 

are the two distinct roots of the quadratic equation

t 2  ( p  q  1)t  p( p  q  1)  0, So,

( p  q  1)(q  1  3 p)

Second, suppose that

 Therefore

0. Then, q

3 p  1.

q 3 p  1. We will show that equation (2.3) has a prime period two solutions. Assume that

( p  q  1)  ( p  q  1)(q  1  3 p) ( p  q  1)  ( p  q  1)(q  1  3 p) , and   . 2 2

 and 

are distinct real numbers. Set,

x(  (t0 ))   and x(t0 )   , t0 T . We wish to show that

x( (t0 ))  x(  (t0 ))  . Page | 11

International Journal of Engineering Research & Science (IJOER)

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[Vol-2, Issue-12, December- 2016]

It follows from equation (2.3) that

x( (t0 ))    

px(t0 )  x(  (t0 )) q  x(  (t0 )) p[( p  q  1)  ( p  q  1)(q  1  3 p) (q  p  1)  ( p  q  1)(q  1  3 p)



[( p  q  1)  ( p  q  1)(q  1  3 p)] (q  p  1)  ( p  q  1)(q  1  3 p)

(q  p  1)(1  p)  ( p  1) ( p  q  1)(q  1  3 p ) (q  p  1)  ( p  q  1)(q  1  3 p) (q  p  1)(1  p )  ( p  1) ( p  q  1)(q  1  3 p )  [(q  p  1)  ( p  q  1)(q  1  3 p)] (q  p  1) 2  ( p  q  1)(q  1  3 p)



2( p 2  p  q )[( p  q  1)  ( p  q  1)(q  1  3 p)] 4( p 2  p  q)



( p  q  1)  ( p  q  1)(q  1  3 p)  x(  (t0 ))   , q  p 2  p. 2

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

  0.

p  1. Then for any positive real number

A  (q  ( p  1))1/ m , The interval O(0, A)  ( A, A) is invariant for eq. (2.3). Proof: We consider Case (1): If T 

T  , h  {hk : h

0 and k  },

. Given any initial conditions

2 0

 {n2 : n 

}, and r , r 1.

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

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, induction on

| 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2

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

Page | 12

International Journal of Engineering Research & Science (IJOER)

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

p 1  1. | q  xnm2 |

(2.26)

Thus,

| pxn 1  xn 2 | p | xn 1 |  | xn 2 | p 1  ( )  max{| xn 1 |,| xn 2 |} m m | q  xn 2 | | q  xn 2 | | q  xnm2 |

| xn |

A for all n  .

max{| xn 1 |,| xn 2 |} Case (2): If

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

| xhn | A for all n 

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

by induction on

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 .

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

Now, we consider

q  xhm( n2)

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

p 1  1. | q  xhm( n2) |

(2.27)

Thus,

| xhn |

| pxh ( n 1)  xh ( n 2) | |qx

m h ( n  2)



p | xh ( n 1) |  | xh ( n 2) | |qx

m h ( n  2)

2 0

 {n2 : n 

| xk | A for all k  n2 , n  for

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

A for all n  .

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

(

0 0

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

by induction on

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

q  x(mn2)2

(n  1) in equation (2.3). Since,

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

p 1  1. | q  x(mn2)2 |

(2.28)

Thus,

| xn2 |

| px( n 1)2  x( n 2)2 | |qx

m ( n  2)2



max{| x( n 1)2 |,| x( n  2)2 |}

p | x( n 1)2 |  | x( n 2)2 | |qx

m ( n  2)2

(

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

A for all k  n 2 . Page | 13

International Journal of Engineering Research & Science (IJOER)

Case (4): If

[Vol-2, Issue-12, December- 2016]

T  r , r 1 . Given any initial conditions | x1/ r | A and | x1 | A , we argue that

| xk | A for all k  r n , n  for

ISSN: [2395-6992]

by induction on

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

q  xrmn2

(n  1) in equation (2.3). Since,

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

p 1  1. | q  xrmn2 |

(2.29)

Thus,

| xr n |

| pxr n1  xr n2 | |qx

m r n2



p | xr n1 |  | xr n2 | |qx

m r n2

max{| xr n1 |,| xr n2 |}

(

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

Hence, Page | 14

International Journal of Engineering Research & Science (IJOER)

ISSN: [2395-6992]

1 p q  (max{| x1 |,| x0 |})m

[Vol-2, Issue-12, December- 2016]

1. This means that R1  (0,1).

n . From equation (2.3), we have 1 p 1 | xn | ( p | xn 1 |  | xn 2 |)  ( ) max{| xn 1 |,| xn 2 |} m | q  xn 2 | | q  xnm2 | max{| xn 1 |,| xn 2 |}, from (2.26).

Now, we prove that (2.32), by induction on

(2.33) (2.34)

Then, from (2.33), we obtain

| x1 | (

p 1 ) max{| x0 |,| x1 |}. | q  xm1 |

(2.35)

But

| q  xm1 | q  (max{| x0 |,| x1 |}) m

q  Rm  1  p

0. Then,

| q  xm1 | q  (max{| x0 |,| x1 |}) m

From (2.35), we have

| x1 | (

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

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

Then, from (2.33), we have

| xn | (

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

 R1  max{R1( n 1) / 2 , R1( n 2) / 2 } max{| x0 |,| x1 |}  R1n / 2 max{| x0 |,| x1 |}. Page | 15

International Journal of Engineering Research & Science (IJOER)

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

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

Then, R1  (0,1), and | xhn | R1

hn / 2

(2.36)

 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 q  (max{| x h |,| x0 |})m

1 p q  (max{| x h |,| x0 |})m

Now, we prove that (2.37), by induction on

| xhn |

q  Rm  1  p

Rm . Thus,

0. Hence,

1. This means that R1  (0,1).

n . From equation (2.3), we have

p | xh ( n 1) |  | xh ( n 2) | |qx

m h ( n  2)

(

p 1 ) max{| xh ( n 1) |,| xh ( n 2) |} | q  xhm( 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  xmh |

(2.40)

But

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

q  Rm  1  p

0. Then, | q  xmh | q  (max{| x0 |,| x h |}) m

From (2.40), we have

| xh | (

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

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International Journal of Engineering Research & Science (IJOER)

| x2 h | (

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

p 1 p 1 ) max{| xh |,| x0 |}  max{| xh |,| x0 |} m | q  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,

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

Then, from (2.38), we have

| xhn | (

p 1 ) max{| xh ( n 1) |,| xh ( n 2) |}  R1 max{| xh ( n 1) |,| xh ( n 2) |} | q  xhm( n 2) |

 R1h max{| xh ( n 1) |,| xh ( n  2) |}  R1h  max{R1h ( n 1) / 2 , R1h ( n 2) / 2 }  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 

}, 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  n and n  2

} be a solution of this equation and

R1 

1 p . q  (max{| x1 |,| x0 |})m

(2.41)

Then,

| 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

Hence, Page | 17

International Journal of Engineering Research & Science (IJOER)

ISSN: [2395-6992]

1 p q  (max{| x1 |,| x0 |})m

[Vol-2, Issue-12, December- 2016]

1. This means that R1  (0,1).

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

| xn2 |

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

(2.43)

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

(2.44)

(

Then, from (2.43), we obtain

| x1 | (

p 1 ) max{| x0 |,| x1 |}. | q  x1m |

(2.45)

But

| q  x1m | q  | x1m |  q  (max{| x0 |,| x1 |})m

q  Rm  1  p

| q  x1m | q  (max{| x0 |,| x1 |}) m

0.Then,

From (2.45), we have

| x1 | (

p 1 p 1 ) max{| x0 |,| x1 |}  max{| x0 |,| x1 |} m | q  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 |} m | q  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 

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International Journal of Engineering Research & Science (IJOER)

So,

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[Vol-2, Issue-12, December- 2016]

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

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(mn 2)2 |

 max{R1( n 1) / 2 , R1( n  2) / 2 }  max{| x0 |,| x1 |}}  R1( n 2) 2

 R1n

max{| x0 |,| x1 |}, n  1, 2,...

max{| x0 |,| x1 |}, n  3, 4,...

This completes the inductive proof of (2.42). Lemma 2.4: Assume that

T  r , 1 r  2, 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  r , n  n

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

R1 

1 p . q  (max{| x1/ r |,| x1 |})m

Then, R1  (0,1), and | xk | R1

k /2

(2.46)

 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

q  (max{| x1/ r |,| x1 |})m

q  Rm  1  p

0. Hence, 0

Rm. Thus,

1 p q  (max{| x1/ r |,| x1 |})m

1. This means that

R1  (0,1). Now, we prove that (2.47), by induction on

n . From equation (2.3), we have

1 ( p | xr n1 |  | xr n2 |) | q  xrmn2 |

| xr n |

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

(2.48)

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

(2.49)

(

Then, from (2.48), we obtain

| xr | (

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

(2.50)

But Page | 19

International Journal of Engineering Research & Science (IJOER)

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

[Vol-2, Issue-12, December- 2016]

q  Rm  1  p

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

0.Then,

From (2.50), we have

| xr | (

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

p 1  max{R1r / 2 max{| x1/ r |,| x1 |},| x1 |} q  (max{max{| x1/ r |,| x1 |},| x1 |}) m p 1  R1r / 2  max{| x1/ r |,| x1 |} q  (max{| x1/ r |,| x1 |}) m R12 r / 2  max{| x1/ r |,| x1 |}  R1r Thus, the inequality (2.47) holds for

R1  R1r / 2  max{| x1/ r |,| x1 |}

 max{| x1/ r |,| x1 |}.

k  r , r 2 . Suppose that the inequality (2.47) holds for

k  r n2 and k  r n1 , n  , respectively. By (2.49), we have | xr n | max{| xr n1 |,| xr n2 |} for all n. So,

p 1 p 1 p 1    R1. m m | q  xr n2 | q  | 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  xrmn2 |

 max{R1r  R1r

n2

 R1r

n1

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

n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

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

of the equation (2.3) is a global attractor. Page | 20

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Proof: From Lemmas

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[Vol-2, Issue-12, December- 2016]

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  q  1, where p

q  1.

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  }, the family of invariant intervals centered at

 {n2 : n 

2 0

}, and r , r

1, m  1 . This means that, we determine

  p  q  1. To this end, we establish the following relation.

px(t )  x(  (t )) p    px(t )  q p  q  (  1)  (  1) q  x(  (t )) q q  x(  (t )) q px(t )  q p  q px(t )  q p  q p  q p  q   (  )(  ) q  x(  (t )) q   q  x(  (t )) q  x(  (t )) q  x (  (t )) q   p p  q  ( x(t )   )  ( x(  (t ))   ). q  x(  (t )) (q  x(  (t )))(q   )

x( (t ))   

p  q  (1   )(q   )  ( p  q)(q   ), the above equality is reduced to

In view of

x( (t ))   

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

| x( (t ))   |

Now, we are ready to describe the family of nested invariant intervals centered at

(2.51)

.

p 1/ 2, and 3 p 1 q 1  p. Then for every positive number

Theorem 2.7: Assume that

A  min{q  (3 p 1),(1  p)  q,1  2 p},

(2.52)

the interval

O( , A)  (  A,   A) is invariant for equation (2.3). Proof: Case (1): If T 

, given any initial conditions

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

| xn   | A for all n by inductionon n. The proof is similar to the proof of Theorem 4.2 in [24 ] and will be omitted. Case 2: If

T  h , given any initial conditions

| x h   | A and | x0   | A, Page | 21

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[Vol-2, Issue-12, December- 2016]

we argue that

| xhn   | A for all n by inductionon n. It follows from the given initial assertion is true for

n  1,0. Suppose the asseration is true for

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

q  xh ( n 2)

xhn where we put hn instead of h(n  1) in equation (2.51). Since,

q  (   A)  q  ( p  q  1)  A

A  (1  p)  A  (1  p)  0,

we have | q  xh ( n 2) | (q  xh ( n 2) ). Now, the condition A  q  (1  2 p ) is equivalent to  q  (   A)  p  (1  2 p)  A  0,

the condition A  q  (3 p  1) is equivalent to p  q  (q  (   A)  p ), and the condition A  (1  p)  q is equivalent to p  q  (q  (   A)  p ). So,

| p  q | (q  (   A)  p). Thus,

| q  xh ( n  2) | ( p  | p  q |)  (q  xh ( n 2) )  p  | p  q | q  (   A)  p  | p  q |  q  (   A)  p  q  (   A)  p  0. From this we deduce that

p | p  q | 1. | q  xh ( n 2) | By (2.51), we have

| xhn   |

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

 max{| xh ( n 1)   |,| xh ( n 2)   |}

This completes the inductive proof. When

T

2 0

and r , r 1, the proof will be omitted.

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

.

T  , p 1/ 2, and 3 p 1 q 1  p. Consider equation (2.3) with the restriction that f : O( , R)  O( , R)  O( , R), where

Let

R  min{q  (3 p 1),(1  p)  q,1  2 p}.

(2.53)

p | p  q | . 1  p  max{| x1   |,| x0   |}

(2.54)

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

Page | 22

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Then,

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[Vol-2, Issue-12, December- 2016]

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 [24 ] and will be omitted.

T  h  {hk : h  (0,1) and k  }, p 1/ 2, and 3 p 1 q 1  p, and (2.53)

Lemma 2.9: Assume that

holds. 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 M

Then,

p | p  q | . 1  p  max{| x h   |,| x0   |}

(2.56)

M  (0,1), and | xhn   | M hn / 2  max{| x h   |,| x0   |}, n  1, 2,...(2.57)

Proof: Note that

1  p  | x h   | ( p  | p  q |) 1  p  R  ( p  | p  q |) 1  2 p  | p  q |  R. When, p  q, 1  p  | x h   | ( p  | p  q |)

(q  (3 p  1))  R  0.

When, p

q, 1  p | x h   | ( p | p  q |)

So,

p | p  q | 1. 1  p  | x h   |

Similarly,

((1  p)  q)  R  0.

p | p  q | 1. Hence, M  (0,1). 1  p  | x h   |

Next, we prove equation (2.57) by induction on n. By (2.51), where we put have | xh   |

hn instead of h(n  1) , we

p | p  q |  max{| x0   |,| x h   |}. Since, q  x h

q  x h  (q   _  ( x h   )  1  p  ( x h   )  1  p  max{| x0   |,| x h   |} 1 p  R M

(1  2 p )  R  0, we derive

p | p  q | p | p  q |  1  p  max{| x0   |,| x h   |} q  x h

Similarly, we have M

p | p  q | q  x0

0.Then,

| xh   | M  max{| x0   |,| x h   |}  M h / 2  max{| x0   |,| x h   |}. And we have, Page | 23

International Journal of Engineering Research & Science (IJOER)

| x2 h   |

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[Vol-2, Issue-12, December- 2016]

p | p  q |  max{| x0   |,| xh   |} q  x0

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

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 (2.51) and the inductive hypothesis, we have p | q  p | | xhn   |  max{| xh ( n 1)   |,| xh ( n 2)   |} q  xh ( n 2) 

p | q  p | max{M h ( n 1) / 2 , M h ( n 2) / 2 } max{| x0   |,| x h   |}}. q  xh ( n 2)

Now, fromTheorem (2.7), we have

| xhn   | max{| xh ( n 1)   |,| xh ( n 2)   |}. By induction on n, we can prove | xhn   | max{| x0   |,| x h   |}, hence,

q  xh ( n 2)  (q   )  ( xh ( n 2)   )  1  p  ( xh ( n 2)   )  1  p  max{| x0   |,| x h   |} 1  p  R we conclude 0

(1  2 p)  R  0,

p | q  p |  M , then q  xh ( n  2)

| xhn   | M  max{M h ( n 1) / 2 , M h ( n 2) / 2 } max{| x0   |,| x h   |}}  M h / 2  max{M h ( n 1) / 2 , M h ( n 2) / 2 } max{| x0   |,| x h   |}}  M nh / 2  max{M h / 2 ,1} max{| x0   |,| x h   |}}  M nh / 2  max{| x0   |,| x h   |}}. This completes the inductive proof of equation (2.57).

T

Lemma 2.10: Assume that

2 0

 {n2 : n 

}, p 1/ 2, and 3 p 1 q 1  p, and (2.53) holds. Consider

equation (2.3) with the restriction that

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

{xk : k  n2 and n 

} be a solution of this equation and M

Then,

p | p  q | . 1  p  max{| x1   |,| x0   |}

(2.58)

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

(2.59) Page | 24

International Journal of Engineering Research & Science (IJOER)

Lemma 2.11: Assume that

ISSN: [2395-6992]

[Vol-2, Issue-12, December- 2016]

T  r , 1 r  2, p 1/ 2, and 3 p 1 q 1  p, and (2.53) holds. 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 M

Then,

p | p  q | . 1  p  max{| x1/ r   |,| x1   |}

(2.60)

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

(2.61)

p 1/ 2, and 3 p 1 q 1  p, and (2.53) holds. Consider equation (2.3) with the

Theorem 2.9: Assume that 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.

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

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