Behavior of solutions of a second order rational difference equation

In this paper, we solve the difference equation xn+1 = α xnxn−1 − 1 , n = 0, 1, . . . , where α > 0 and the initial values x−1, x0 are real numbers. We find some invariant sets and discuss the global behavior of the solutions of that equation. We show that when α > 2 3 √ 3 , under certain conditions there exist solutions, that are either periodic or converging to periodic solutions. We show also the existence of dense solutions in the real line. Finally, we show that when α < 2 3 √ 3 , one of the negative equilibrium points attracts all orbits with initials outside a set of Lebesgue measure zero.


Introduction
In [9], Amleh et al. studied the difference equation (1) x n+1 = α x n x n−1 + 1 , n = 0, 1, . . . , where the α is positive and the initial conditions are nonnegative real numbers. They conjectured that every solution has a finite limit but confirmed it only when α ≤ 2.
In [8], The authors studied the difference equation where the α is positive and the initial conditions are nonnegative real numbers. They conjectured that the unique positive equilibrium point is globally asymptotically stable and confirmed it only when (α − c) 2 ≤ 4. Kulenović et al. [24], studied equation (1) and gave a unified proof for all values of α that the unique equilibrium is globally asymptotically stable. 12 Behavior of solutions of a second order rational difference equation For more on difference equations with quadratic terms, see [1]- [14], [16], [17], [19], [21]- [25], [27]- [29].
In this paper, we study the difference equation (2) x n+1 = α x n x n−1 − 1 , n = 0, 1, . . . , where α > 0 and the initial conditions are real numbers. The transformation x n = y n−1 y n , with y −2 = 1, reduces the difference equation (2) into the linear third order difference equation (3) is

The characteristic equation of equation
Clear that equation (4) has a positive real root λ 0 for all values of α. Equation (4) can be written as Therefore, the roots of equation (4) are We have the following result: (4), we have the following: 3 , then equation (4) has one positive real root and two complex conjugate roots.
3 , then equation (4) has three real different roots, one of them is positive and two negative roots.
Proof. It is sufficient to see that, the discriminant of the polynomial where (8) γ 1n = c 11 λ n 0 + c 21 are such that c ij , i, j = 1, 2, 3 are given in (7).
3 , the roots of equation (4) are Then the solution of equation (2) is Using the initials y −2 , y −1 and y 0 , the values of c 1 , c 2 and c 3 in this case are: By simple calculations, we can write the solution of equation (3) in this case as y n = γ 1n y 0 + γ 2n y −1 + γ 3n y −2 , where (10) ) n n are such that c ij , i, j = 1, 2, 3 are given in (9).
3 , the roots of equation (4) are Then the solution of equation (2) is Using the initials y −2 , y −1 and y 0 , the values of c 1 , c 2 and c 3 in this case are: By simple calculations, we can write the solution of equation (3) in this case as are such that c ij , i, j = 1, 2, 3 are given in (11).
Using the previous arguments we can give the form of the forbidden set in the following result.

This equation has the characteristic equation
The solution of the characteristic equation depends on the value of its discriminant according to Lemma (1.1). When α > 2 3 √ 3 , the roots of the characteristic equation are where λ 0 is a positive real root. Then where Using the initials y −2 , y −1 and y 0 , we can find the values of c 1 , c 2 and c 3 . These values are given in formulas (6). By simple calculations, we can write the solution y n = γ 1n y 0 + γ 2n y −1 + γ 3n y −2 , where γ 1n , γ 2n and γ 3n are given in formulas (8). But as we can write y n = γ 1n Using the transformation it is clear that x n is well-defined whenever y n = 0, n ≥ −1. Therefore, the forbidden set in this case is where γ 1n , γ 2n and γ 3n are given in formulas (8). When α = 2 3 √ 3 and α < 2 3 √ 3 , the proof is similar and will be omitted. The proof is complete. (2) In this section, we shall give invariant sets for equation (2). When α > 2 3 √ 3 , we can write the constant c 1 = y 0 c 11 + y −1 c 12 + y −2 c 13 in terms of x 0 and x −1 as

Invariant sets for equation
By simple calculations, we can show that if (x 0 , x −1 ) is such that c 1 (x 0 , x −1 ) = 0, then (x 0 , x −1 ) lies on the rectangular hyperbola Consider the set Theorem 3.1. The set D 1 is an invariant for equation (2).
Proof. Let (x 0 , x −1 ) ∈ D 1 . We show that (x n , x n−1 ) ∈ D 1 for each n ∈ N . The proof is by induction on n. The point (x 0 , x −1 ) ∈ D 1 , implies But as λ 0 is a solution of equation (4), we have This implies that (x 1 , x 0 ) ∈ D 1 . Suppose now that (x n , x n−1 ) ∈ D 1 . That is Therefore, (x n+1 , x n ) ∈ D 1 and the proof is complete.
Theorem 3.2. The set D 1 is of Lebesgue measure zero.
Proof. It is required to prove that the set x is of Lebesgue measure zero. We can write We show that S + is of Lebesgue measure zero. For, let Now consider the part S n 1 of the set S 1 in the interval [n, n + 1]. For > 0, there exists δ > 0 such that |g(x) − g(y)| < 2 n+2 whenever |x − y| < δ for all x, y ∈ [n, n + 1].
Divide the interval [n, n That is

20
Behavior of solutions of a second order rational difference equation Similarly we can show that µ(S 2 ) < 2 . Therefore, µ(S + ) < and S + is of Lebesgue measure zero. The proof that S − is of Lebesgue measure zero is similar and will be omitted. Thus D 1 = S + ∪ S − is a union of two Lebesgue measure zero subsets is also of Lebesgue measure zero.
This completes the proof.
In case α = 2 3 √ 3 , consider the following subsets: Note that D 1 , D 2 and D 3 are equivalent to c 1 (x, y) = 0, c 2 (x, y) = 0 and c 3 (x, y) = 0 respectively. The two subsets D 1 and D 3 are invariant subsets for equation (2). Finally when α < 2 3 √ 3 . We shall consider the three sets By simple calculations, we can see that: Proof. The proof is similar to that of theorem (3.1) and will be omitted. (2) In this section, we shall investigate the behavior of equation (2) for all values of α.

Global behavior of equation
In the following results, we show that when α > 2 3 √ 3 , under certain conditions there exist solutions, that are either periodic or converging to periodic solutions for equation (2).
Suppose that θ = p q π is a rational multiple of π, where p and q are positive relatively prime integers such that q 2 < p < q.
(2) Assume that (x 0 , x −1 ) ∈ D 1 . Then for each i = 1, 2, . . . , q, we get Suppose that θ ∈ R − πQ is not a rational multiple of π.  Proof. We have that Clear that x 2n and x 2n+1 converge to − 1 The set of Lebesgue measure zero is the set F ∪ (∪ 3 i=1 D i ). In fact it is a union of Lebesgue measure zero subsets of R 2 .