Global Behavior of a Rational Difference Equation with Quadratic Term

In this paper, we determine the forbidden set, introduce an explicit formula for the solutions and discuss the global behavior of all solutions of the difference equation xn+1 = axnxn−1 bxn − cxn−2 , n = 0, 1, . . . where a, b, c are positive real numbers and the initial conditions x−2, x−1, x0 are real numbers.


Introduction
Li and Zhu [10] discussed the global asymptotic stability of the difference equation x n x n−1 + a x n + x n−1 , n = 0, 1, . . .
In this paper, we derive the forbidden set, introduce an explicit formula for the solutions and discuss the global behavior of all solutions of the difference equation (1.1) x n+1 = ax n x n−1 bx n − cx n−2 , n = 0, 1, . . .
where a, b, c are positive real numbers and the initial conditions x −2 , x −1 , x 0 are real numbers.

Forbidden Set and Solutions of Qquation (1.1)
In this section we derive the forbidden set and introduce an explicit formula for the solutions of the difference equation (1.1).
Suppose that x 0 x −1 = 0. Then we have the following: Therefore, we can start with the nonzero initial conditions x −1 , x 0 , x 1 , which we shall discuss. Now suppose that x −i = 0 for all i = 0, 1, 2. Using the substitution r n = xn x n−1 , equation (1.1) becomes Now using the substitution l n = 1 rnr n−1 , we can obtain the linear nonhomogeneous difference equation

The solution of equation (2.2) is
But l n = 1 rnr n−1 = x n−2 xn . Therefore, Therefore, y n 0 +1 is undefined. On the other hand, from equation (1.1) we have that For a fixed n 0 ∈ N, suppose that we have y n 0 +1 is undefined. This implies that bx n 0 − cx n 0 −2 = 0.
That is Hence using equation (2.4), we have the following: Then These observations lead us to conclude the following result.

Global Behavior of Equation (1.1)
In this section, we investigate the global behavior of equation (1.1) with a = c, using the explicit formula of its solution.
We can write the solution of equation (1.1) in the form: Then the following statements are true.
(1) If a < c, then {x n } ∞ n=−2 converges to zero. (2) If a > c, then we have the following: Proof.

Case a + c = b
In order to discuss the case when a + c = b with a > c, we need to remember the behavior of equation (2.1) with a + c = b with a > c.
Many authors [1][2][3][4][5][6][7][8][9] discussed the behavior of the solutions of some special cases of the equation where A, B, C are real numbers, but with reducing the numbers of parameters to one or two. We shall derive only some results concern the behavior of the solutions of equation (2.1), that we shall use.
The following theorem gives the solution of equation (2.1).
where θ = a+c−bα α(a+c) . The solution of equation (2.1) can be written as: Now assume that a + c = b and α = 0. Then we have is not a period-2 solution. We claim that, there exists j 0 ∈ N such that γ i (j) > 0 for all j ≥ j 0 . For, let a i (j) = θ(− c a ) 2j+i−1 + 1 and b i (j) = θ(− c a ) 2j+i + 1. Then we can write We have the following situations: • If θ < 0, then, we have the following: But as a 1 (j) converges to 1, there exists j 1 ∈ N such that a 1 (j) > 0 for all j ≥ j 1 . Therefore, γ 1 (j) = a 1 (j) b 1 (j) > 0 for all j ≥ j 1 . -If i = 2, then a 2 (j) > 0 for all j ∈ N. But as b 2 (j) converges to 1, there exists j 2 ∈ N such that b 2 (j) > 0 for all j ≥ j 2 . Therefore, γ 2 (j) = a 2 (j) b 2 (j) > 0 for all j ≥ j 2 . In all cases, there exists a natural number j 0 = max{j 1 , j 2 } such that γ i (j) = a i (j) b i (j) > 0, i = 1, 2 for all j ≥ j 0 . • If θ > 0, the situation is similar and will be omitted. Claim is complete. Now for each i ∈ {1, 2}, we have for large m We shall test the convergence of the series ∞ j=j 0 |ln γ i (j)|. Since lim j→∞ ln γ i (j+1) ln γ i (j) = 0 0 , using L'Hospital's rule we obtain Then from d'Alembert's test that the series ∞ j=j 0 |ln γ i (j)| is convergent, it follows that there exist 2 real numbers ρ i ∈ R such that Now we are ready to introduce the main results in this section. Proof. Assume that a + c = b. If α = 1, then θ = 0. Therefore,  (1) If a < c, then {x n } ∞ n=−2 converges to zero. (2) If a > c, then {x n } ∞ n=−2 converges to a period-2 solution {µ 0 , µ 1 } such that µ 1 = µ 0 ρ 1 , where ρ 1 is as in Theorem (4.2). Proof.
(1) The proof is similar to that in theorem (3.1).
By an argument similar to that in theorem (4.2), there exists j 0 ∈ N such that, β i (j) > 0, for all j ≥ j 0 . Hence We shall test the convergence of the series ∞ j=j 0 |ln β i (j)|.
It follows from D' Alemberts' test that the series ∞ j=j 0 |ln β i (j)| is convergent.