On the Third Order Rational Difference Equation 1 Introduction

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On the Third Order Rational Difference Equation xn+1 = xn xn−2 xn−1(a+bxn xn−2) ... n = 0, 1, ..where initial values x0 , x−1 , and x−2 are nonnegative real num-.
Int. J. Contemp. Math. Sciences, Vol. 4, 2009, no. 27, 1321 - 1334

On the Third Order Rational Difference Equation xn+1 =

xn xn−2 xn−1 (a+b xn xn−2 )

T. F. Ibrahim Department of Mathematics, Faculty of Science Mansoura University, Mansoura 35516, Egypt tfi[email protected] Abstract In this paper we will give the solutions of the rational difference equaxn xn−2 tion xn+1 = xn−1 (a+b xn xn−2 ) , n = 0, 1, ..where initial values x0 , x−1 , and x−2 are nonnegative real numbers with bx0 x−2 = −a and x−1 = 0 . Moreover we investigate some properties for this difference equation such as the local stability and the boundedness for the solutions .

Mathematics Subject Classification: 39A10 Keywords: stability, boundedness, solvability, periodic

1

Introduction

Recently there has been a lot of interest in studying the boundedness character and the periodic nature of nonlinear difference equations. For some results in this area, see for example [1-11]. Difference equations have been studied in various branches of mathematics for a long time. First results in qualitative theory of such systems were obtained by Poincar´e and Perron in the end of nineteenth and the beginning of twentieth centuries. The systematic description of the theory of difference equations one can find in books [1,6,10]. Cinar [5]gave that the solution of the difference equation xn+1 =

xn−1 . −1 + xn xn−1

Karatas et al.[8] gave that the solution of the difference equation xn+1 =

xn−5 1 + xn−2 xn−5

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T. F. Ibrahim

Aloqeili [2] has obtained the solutions of the difference equation xn+1 =

xn−1 , a − xn xn−1

In this paper we will give the solutions of the rational difference equation xn+1 =

xn xn−2 , xn−1 (a + b xn xn−2 )

(1)

n = 0, 1, ..where initial values x0 , x−1 , and x−2 are nonnegative real numbers with bx0 x−2 = −a and x−1 = 0 . Moreover we investigate some properties for this difference equation such as the local stability and the boundedness for the solutions .

2

Solvability of the difference eq(1) when a = 1, b = 1:

The following theorem give the solution of the difference equation xn+1 =

xn xn−2 xn−1 (1 + xn xn−2 )

(*)

Theorem 1 Suppose that {xn } be a solution of equation (*) where the initial values x0 , x−1 , and x−2 are nonnegative real numbers . Then the solutions of equation (*) have the form

x4n−3

 n   (1 + (4i − 5) x0 x−2 ) x0 x−2 = x−1 (1 − x0 x−2 ) i=1 (1 + (4i − 3) x0 x−2 )

(2)

x4n−2 = x−2

 n   (1 + (4i − 4) x0 x−2 ) (1 + (4i − 2) x0 x−2 ) i=1

(3)

x4n−1 = x−1

 n   (1 + (4i − 3) x0 x−2 ) (1 + (4i − 1) x0 x−2 ) i=1

(4)

x4n

 n   (1 + (4i − 2) x0 x−2 ) = x0 (1 + (4i) x0 x−2 ) i=1

(5)

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Rational difference equation

where n = 1, 2, .... . Proof. For n = 1 the result holds. Now suppose that n > 0 and that our assumption holds for n − 1. That is;

x4n−7

n−1   (1 + (4i − 5) x0 x−2 )  x0 x−2 = x−1 (1 − x0 x−2 ) i=1 (1 + (4i − 3) x0 x−2 ) n−1 

x4n−6 = x−2

i=1 n−1 

x4n−5 = x−1

i=1

x4n−4 = x0

n−1  i=1

(1 + (4i − 4) x0 x−2 ) (1 + (4i − 2) x0 x−2 ) (1 + (4i − 3) x0 x−2 ) (1 + (4i − 1) x0 x−2 )

(1 + (4i − 2) x0 x−2 ) (1 + (4i) x0 x−2 )

(6)

 (7)

 (8)

 (9)

Now we will try to prove that the eqs(2-5) hold at n . Using Eq(7) , Eq(8) and Eq(9) we have x4n−3 =  x0 =

×

n−1  i=1

x4n−4 x4n−6 x4n−5 (1 + x4n−4 x4n−6 )   n−1   (1+(4i−4)x0 x−2 (1+(4i−2)x0 i=1 n−1   (1+(4i−3)x0 x−2 ) 

(1+(4i−2)x0 x−2 ) (1+(4i)x0 x−2 )

 x−1

i=1

x−2 ) x−2 )

(1+(4i−1)x0 x−2 )

1  n−1    n−1  (1+(4i−2)x0 x−2 )   (1+(4i−4)x0 1 + x0 x −2 (1+(4i)x0 x−2 ) (1+(4i−2)x0 i=1

i=1

x0 x−2 / (1 + (4n − 4) x0 x−2 ) =  n−1   (1+(4i−3)x0 x−2 )  x−1 (1+(4i−1)x0 x−2 ) i=1



x−2 ) x−2 )



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T. F. Ibrahim

=

x0 x−2 x−1



1 (1 + (4n − 4) x0 x−2 ) (1 + x0 x−2 / (1 + (4n − 4) x0 x−2 )) n−1 

×

i=1

=

x0 x−2 x−1

=





n−1   (1 + (4i − 1) x0 x−2 )  1 1 + (4n − 4) x0 x−2 + x0 x−2 i=1 (1 + (4i − 3) x0 x−2 )

x0 x−2 x−1

=

(1 + (4i − 1) x0 x−2 ) (1 + (4i − 3) x0 x−2 )



n−1   (1 + (4i − 1) x0 x−2 )  1 1 + (4n − 3) x0 x−2 i=1 (1 + (4i − 3) x0 x−2 )

x0 x−2 x−1



 n   1 (1 + (4i − 5) x0 x−2 ) 1 − x0 x−2 i=1 (1 + (4i − 3) x0 x−2 )

Using Eq(2) , Eq(8) and Eq(9) we have x4n−2 = 

x0 x−2 x−1 (1−x0 x−2 )

=

n   (1+(4i−5)x0 i=1

(1+(4i−3)x0



x0

× 

 1+

x0 x−2 x−1 (1−x0 x−2 )



x4n−3 x4n−5 x4n−4 (1 + x4n−3 x4n−5 )

n  i=1



n−1  i=1

x−2 ) x−2 )

 

x−1

n−1 

(1+(4i−2)x0 x−2 ) (1+(4i)x0 x−2 )

1 (1+(4i−5)x0 x−2 ) (1+(4i−3)x0 x−2 )

 

x0 x−2 (1+(4n−5)x0 x−2 ) ( 1−x0 x−2 )(1+(4n−3)x0 x−2 )

i=1



x−1

(1+(4i−3)x0 x−2 ) (1+(4i−1)x0 x−2 )

n−1  i=1

(1+(4i−3)x0 x−2 ) (1+(4i−1)x0 x−2 )

(1−x0 x−2 ) (1+(4n−5)x0 x−2 )

=  n−1   (1+(4i−2)x0 x−2 )   x0 x−2 (1+(4n−5)x0 x−2 ) 1 + ( 1−x x0 (1+(4i)x0 x−2 ) 0 x−2 )(1+(4n−3)x0 x−2 ) i=1



 

(1−x0 x−2 ) (1+(4n−5)x0 x−2 )



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Rational difference equation



x−2 (1+(4n−3)x0 x−2 )

= x0 x−2 1 + (1+(4n−3)x 0

n−1 



i=1

x−2 )

n−1 

x−2 = [1 + (4n − 3) x0 x−2 + x0 x−2

= x−2

(1 + (4i) x0 x−2 ) (1 + (4i − 2) x0 x−2 )

]

i=1



(1 + (4i) x0 x−2 ) (1 + (4i − 2) x0 x−2 )



 n   (1 + (4i − 4) x0 x−2 ) (1 + (4i − 2) x0 x−2 ) i=1

Using Eq(2) , Eq(3) and Eq(9) we have x4n−1 =

x4n−2 x4n−4 x4n−3 (1 + x4n−2 x4n−4 )

 n   (1+(4i−4)x0 x−2 (1+(4i−2)x0 i=1  =

x0 x−2 x−1 (1−x0 x−2 )

×

 1 + x−2

n  i=1



  n−1   (1+(4i−2)x0 x−2 )  x0 (1+(4i)x0 x−2 ) i=1  n   (1+(4i−5)x0 x−2 )

x−2 ) x−2 )

i=1

(1+(4i−4)x0 x−2 ) (1+(4i−2)x0 x−2 )



(1+(4i−3)x0 x−2 )

1  

x0

(1+(4n−4)x0 x−2 ) (1+(4n−2)x0 x−2 )

n−1  i=1

(1+(4i−2)x0 x−2 ) (1+(4i)x0 x−2 )

  n−1 

 

(1+(4i−4)x0 x−2 ) (1+(4i)x0 x−2 )



x−1 (1 − x0 x−2 ) i=1   =n     n−1  (1+(4i−5)x0 x−2 )    (1+(4i−4)x0 x−2 )  (1+(4n−4)x0 x−2 ) 1 + x0 x−2 (1+(4n−2)x0 x−2 ) (1+(4i−3)x0 x−2 ) (1+(4i)x0 x−2 ) i=1

i=1



=

n   i=1

(1+(4n−4)x0 x−2 ) (1+(4n−2)x0 x−2 )



1 1+(4n−4)x0 x−2

x−1 (1 − x0 x−2 )    (1+(4i−5)x0 x−2 ) 0 1 + x0 x−2 (1+(4n−4)x (1+(4i−3)x0 x−2 ) (1+(4n−2)x0

x−2 ) x−2 )



1 1+(4n−4)x0 x−2

  1  n  x−1 (1 − x0 x−2 ) (1+(4n−2)x  (1 + (4i − 3) x0 x−2 ) 0 x−2 )   =  1 (1 + (4i − 5) x0 x−2 ) 1 + x0 x−2 (1+(4n−2)x i=1 0 x−2 )



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T. F. Ibrahim

 n   (1 + (4i − 3) x0 x−2 ) x−1 (1 − x0 x−2 ) = [(1 + (4n − 1) x0 x−2 )] i=1 (1 + (4i − 5) x0 x−2 )  n  x−1 (1 − x0 x−2 )  (1 + (4i − 3) x0 x−2 ) = (1 − x0 x−2 ) i=1 (1 + (4i − 1) x0 x−2 )

= x−1

 n   (1 + (4i − 3) x0 x−2 ) (1 + (4i − 1) x0 x−2 ) i=1

Using Eq(2) , Eq(3) and Eq(4) we have x4n =

x4n−1 x4n−3 x4n−2 (1 + x4n−1 x4n−3 ) 

x0 ( 1−x0 x−2 )

=

n   i=1

(1+(4i−4)x0 x−2 ) (1+(4i−2)x0 x−2 )

n   (1+(4i−5)x0 (1+(4i−1)x0

i=1

 (1 +

x0 x−2 ( 1−x0 x−2 )



x−2 ) x−2 )



n   (1+(4i−5)x0 i=1

(1+(4i−1)x0

x−2 ) x−2 )

 )

x0 (1+(4n−1)x0 x−2 )

=  n  (1+(4i−4)x0 i=1

(1+(4i−2)x0

x−2 ) x−2 )



(1 +

x0 x−2 ) (1+(4n−1)x0 x−2 )

 n   x0 (1 + (4i − 2) x0 x−2 ) = ((1 + (4n − 1) x0 x−2 ) + x0 x−2 ) i=1 (1 + (4i − 4) x0 x−2 )  n   (1 + (4i − 2) x0 x−2 ) x0 = (1 + (4n) x0 x−2 ) i=1 (1 + (4i − 4) x0 x−2 )  n   (1 + (4i − 2) x0 x−2 ) = x0 (1 + (4i) x0 x−2 ) i=1 Hence the proof is complete .

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Rational difference equation

Lemma 2 Let ρ = x0 x−2 . We have the following relations between the solutions in eqs(2-5) ρ i) x4n−3 x4n−1 = 1+ (4n−1) ρ ii) x4n x4n−2 = 1 + ρ4n ρ iii) x4n 1x4n−2 − x4n−3 1x4n−1 = 1 Proof. i) Form Eq(2) and Eq(4) we have •

 x4n−3 x4n−1 =

 n   (1 + (4i − 5) x0 x−2 ) x0 x−2 x−1 (1 − x0 x−2 ) i=1 (1 + (4i − 3) x0 x−2 ) 

×  =

=

x−1

 n   (1 + (4i − 3) x0 x−2 ) (1 + (4i − 1) x0 x−2 ) i=1

 n   (1 + (4i − 5) x0 x−2 ) x0 x−2 (1 − x0 x−2 ) i=1 (1 + (4i − 1) x0 x−2 )

(1 − ρ ) (1 + 3ρ ) ...... (1 + (4n − 5) ρ ) ρ (1 − ρ ) (1 + 3ρ ) (1 + 7ρ ) ....... (1 + (4n − 1) ρ )

=

(1 − ρ ) ρ ρ = (1 − ρ ) (1 + (4n − 1) ρ ) (1 + (4n − 1) ρ )

ii) From Eq(3) and Eq(5) we have x4n x4n−2 =    n  n    (1 + (4i − 2) x0 x−2 ) (1 + (4i − 4) x0 x−2 ) x−2 x0 (1 + (4i) x0 x−2 ) (1 + (4i − 2) x0 x−2 ) i=1 i=1

 =

 = x0 x−2

 n   (1 + (4i − 4) x0 x−2 ) (1 + (4i) x0 x−2 ) i=1



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T. F. Ibrahim



ρ (1) (1 + 4ρ) ........... (1 + (4n − 4) ρ) = (1 + 4ρ) (1 + 8ρ) .................. (1 + (4n) ρ) 1 + (4n) ρ

iii) By easy calculations from i) and ii) . x4n x4n−2 x4n−3 x4n−1

Remark 1 We note that

=

1+ (4n−1) ρ . 1 + 4n ρ

Remark 2 We note that |x4n−3 x4n−1 | → 0 as n → ∞ . Remark 3 We note that |x4n x4n−2 | → 0 as n → ∞ .

3

Solvability of the difference eq(1) when a = 1, b = −1 :

The following theorem give the solution of the difference equation xn+1 =

xn xn−2 , xn−1 (1 − xn xn−2 )

Theorem 3 Suppose that {xn } be a solution of Equation (1) and x0 = h, x−1 = k and x−2 = r.Let ρ = x0 x−2 . Then the solutions of equation (1) have the form

x4n−3

 n   (1 − (4i − 5) ρ ) hr = k (1 + ρ ) i=1 (1 − (4i − 3) ρ)

x4n−2 = r

 n   (1 − (4i − 4) ρ ) i=1

x4n−1 = k

 n   (1 − (4i − 3) ρ ) i=1

x4n = h

(1 − (4i − 2) ρ)

(1 − (4i − 1) ρ )

 n   (1 − (4i − 2) ρ ) i=1

(1 − (4i) ρ)

where n = 1, 2, .... . Proof. By using the mathematical induction as in theorem 1.

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Rational difference equation

4

Solvability of the difference eq(1) when a = 1 :

The following theorem give the solution of the difference equation(1) where a = 1 and for any value of b. Theorem 4 Suppose that {xn } be a solution of Equation (1) and x0 = h, x−1 = k , x−2 = r and ρ = x0 x−2 . Then the solutions of equation (1) have the form x1 =

x3 =

x4n−3

ρ k (a + b ρ)

k (a + b ρ) 3 a + (2a2 + 1) bρ

,

x2 =

x4 =

a2

r + 2abρ

h (a2 + 2 a b ρ) a4 + (2a3 + a + 1) bρ

4i−5 + b ρ 2a4i−6 + a ρ ⎣  = k (a + b ρ) i=2 a4i−3 + b ρ 2a4i−4 + n 

n−1 



a4i + b ρ 2a4i−1 +

i=1

x4n−1 = k

x4n

⎤ ⎦

1−a4i−4 1−a



a4i−3 + b ρ 2a4i−4 +

1−a4i−5 1−a

 a4i−5 + b ρ 2a4i−6 +

1−a4i−7 1−a

i=2 n+1  i=2

1−a4i−5 1−a

1−a4i−2 1−a

i=1 x4n−2 = r  n  a4i−2 + b ρ 2a4i−3 +

n 

1−a4i−7 1−a



⎤ 4i−4 a4i−2 + b ρ 2a4i−3 + 1−a 1−a ⎣  ⎦  =h 4i + b ρ 2a4i−1 + 1−a4i−2 a 1−a i=1 n 



where n = 2, 3, .... . Proof. By using the mathematical induction as in theorem 1. Lemma 5 The solutions {xn } of the difference eqs (1) and (*) have no prime period two solution

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Now we recall some notations and results which will be useful in our study. Definition 1 The difference equation xn+1 = F (xn , xn−1 , ..., xn−k ), n = 0, 1, ...

(10)

is said to be persistence if there exist numbers m and M with 0 < m ≤ M < ∞ such that for any initial conditions x−k , x−k+1 , ..., x−1 , x0 ∈ (0, ∞) there exists a positive integer N which depends on the initial conditions such that m ≤ xn ≤ M

for all

n ≥ N.

Definition 2 (i) The equilibrium point of Eq.(10) is locally stable if for every  > 0, there exists δ > 0 such that for all x−k , x−k+1 , ..., x−1 , x0 ∈ I with |x−k − x| + |x−k+1 − x| + ... + |x0 − x| < δ, we have |xn − x| < 

for all n ≥ −k.

(ii) The equilibrium point x of Eq.(10) is locally asymptotically stable if x is locally stable solution of Eq.(10) and there exists γ > 0, such that for all x−k , x−k+1 , ..., x−1 , x0 ∈ I with |x−k − x| + |x−k+1 − x| + ... + |x0 − x| < γ, we have lim xn = x.

n→∞

The linearized equation of Eq.(10) about the equilibrium x is the linear difference equation yn+1 =

k  ∂F (x, x, ..., x) i=0

∂xn−i

yn−i .

Theorem A [9] Assume that p, q ∈ R . Then |p| + |q| < 1 is a sufficient condition for the asymptotic stability of the difference equation xn+1 − pxn − qxn−1 = 0, n = 0, 1, ... .

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Rational difference equation

5

Local Stability of the Equilibrium Points

In this section we study the local stability of the solutions of Eq.(1). We first give the equilibrium points of Eq(1).  Lemma 6 The equilibrium points of the difference eq (1) are 0 and ± a−1 b Proof. x=

x2 x (a + bx2 )

  x2 a + bx2 = x2   x2 −1 + a + bx2 = 0  thus the equilibrium points of the difference eq (1) are 0 and ± a−1 . b Remark 4 When a = 1 , then the only equilibrium point of the difference eq (*) is 0.  Theorem 7 The equilibrium points x = ± a−1 are locally asymptotically stab ble if a  1 and in this case |a − 2a2 | ≺ 4a2 − 6a + 1 . Proof.  We will prove the theorem at the equilibrium point x = + a−1 and the b  by the same way. proof at the equilibrium point x = − a−1 b 3 let f : (0, ∞) −→ (0, ∞) be a continuous function defined by f (u, v, w) =

uw v (a + buw)

Therefore it follows that v (a + buw) w − uw (vbw) aw ∂f (u, v) = = 2 2 ∂u v (a + buw) v (a + buw)2 −uw ∂f (u, v) = 2 ∂v v (a + buw)

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au ∂f (u, v) = ∂w v (a + buw)2  At the equilibrium point x = a−1 we have b a ∂f ( x, x, x) a = p1 =  a−1 2 = ∂u (2a − 1)2 a+b b ∂f ( x, x, x) −a = = p2 ∂v (2a − 1) a ∂f ( x, x, x) = = p3 ∂w (2a − 1)2 Then the linearized equation of Eq.(1) about x =



a−1 b

is

yn+1 − p1 yn − p2 yn−1 − p3 yn−2 = 0. i. e. yn+1 −

a a a yn−1 − yn−2 = 0 2 yn + (2a − 1) (2a − 1) (2a − 1)2

Whose characteristic equation is a a a 2 λ− =0 λ3 − 2λ + (2a − 1) (2a − 1) (2a − 1)2 By the generalization of theorem A we have |p1 | + |p2 | + |p3 | ≺ 1          −a    a a        (2a − 1)2  +  (2a − 1)  +  (2a − 1)2  ≺ 1 |a| + |−a (2a − 1)| + |a| ≺ (2a − 1)2   a − 2a2  ≺ 4a2 − 6a + 1

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Rational difference equation

6

Boundedness of solutions

Here we study the boundedness of Eq.(*) . Theorem 8 Every solution of Eq.(*) is bounded from above . Proof: Let {xn }∞ n=−k be a solution of Eq.(*). It follows from Eq.(*) that xn+1 =

xn xn−2 xn xn−2 1 = ≤ xn−1 (1 + xn xn−2 ) xn−1 + xn−1 xn xn−2 xn−1

Then xn ≤

1 xn−2

for all

n ≥ 0.

This means that every solution of eq(*) is bouneded from above by M = 1 1 max( x10 , x−1 , x−2 ).

References [1] R.P. Agarwal, ” Difference Equations and Inequalities”, Dekker, New York, 1992 [2] M. Aloqeili, Dynamics of a rational difference equation, Appl. Math. Comp., 176(2) , (2006), 768-774. [3 ] C. Cinar, On the positive solutions of the difference equation axn−1 xn+1 = , Appl. Math. Comp.(in press), 2003. 1 + bxn xn−1 [4] C. Cinar, On the positive solutions of the difference equation xn−1 , Appl. Math. Comp., 150, (2004), 21-24. xn+1 = 1 + xn xn−1 xn−1 , [5] C. Cinar, On the difference equation xn+1 = −1 + xn xn−1 Appl. Math. Comp., 158, (2004) , 813-816. [6] S. Elaydi, ”An Introduction to Difference Equations”, Springer, New York, 1996. [7 ] C. H. Gibbons,M. R.S. Kulenovvic,G. Ladas,H. D. Voulov, ”On the trichotomy character of xn+1 = (α + βxn + γxn−1 ) / (A + xn ) , J. of Difference Eqs and App. 8(1) , (2002) , 75-92.

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[8] R. Karatas, C. Cinar and D. Simsek, On positive solutions of xn−5 , Int. J. Contemp. the difference equation xn+1 = 1 + xn−2 xn−5 Math. Sci., Vol. 1, no. 10, (2006), 495-500. [9] V.L. Kocic and G. Ladas, Global Behavior of Nonlinear Difference Equations of Higher Order with Applications, Kluwer Academic [10] V. Lakshmikantham, D. Trigiante, ” Theory of Difference Equations: Numerical Methods and Applications”, Academic Press, New York, 1998. [11] D. Simsek, C. Cinar and I. Yalcinkaya, On the Recursive Sexn−3 , Int. J. Contemp. Math. Sci., Vol. 1, no. quence xn+1 = 1 + xn−1 10,( 2006) , 475-480. Received: March, 2009