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
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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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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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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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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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