On the Third Order Rational Difference Equation 1 Introduction

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 Poincaré 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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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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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

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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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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T. F. Ibrahim

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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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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T. F. Ibrahim

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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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