Discrete Dynamics in Nature and Society
Volume 2010 (2010), Article ID 675413, 13 pages
doi:10.1155/2010/675413
Research Article

On the Max-Type Difference Equation 𝑥 𝑛 + 1 = m a x { 𝐴 / 𝑥 𝑛 , 𝑥 𝑛 − 3 }

Bratislav D. Iričanin1 and E. M. Elsayed2

1Faculty of Electrical Engineering, University of Belgrade, Bulevar Kralja Aleksandra 73, Belgrade 11120, Serbia
2Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt

Received 29 October 2009; Revised 24 December 2009; Accepted 25 January 2010

Academic Editor: Leonid Berezansky

Copyright © 2010 Bratislav D. Iričanin and E. M. Elsayed. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We show that every well-defined solution of the fourth-order difference equation 𝑥 𝑛 + 1 = m a x { 𝐴 / 𝑥 𝑛 , 𝑥 𝑛 − 3 } ,     𝑛 ∈ ℕ 0 , where parameter 𝐴 ≥ 0 , is eventually periodic with period four.

1. Introduction

The study of max-type difference equations attracted recently a considerable attention, see, for example, [1–27], and the references listed therein. This type of difference equations stems from, for example, certain models in automatic control theory (see [28]). In the beginning of the study of these equations experts have been focused on the investigation of the behavior of some particular cases of the following general difference equation of order 𝑘 ∈ ℕ :

𝑥 𝑛 𝐴 = m a x 𝑛 ( 1 ) 𝑥 𝑛 − 1 , 𝐴 𝑛 ( 2 ) 𝑥 𝑛 − 2 𝐴 , … , 𝑛 ( 𝑘 ) 𝑥 𝑛 − 𝑘 , 𝑛 ∈ ℕ 0 , ( 1 . 1 ) where 𝑘 ∈ ℕ , 𝐴 𝑛 ( 𝑖 ) , 𝑖 = 1 , … , 𝑘 , are real sequences (mostly constant or periodic ones) and where the initial values 𝑥 − 1 , … , 𝑥 − 𝑘 are different from zero (see, e.g., [2, 3, 6, 7, 9–12, 22–25] and the references cited therein).

The study of max-type equations of the following general form

𝑥 𝑛 𝐵 = m a x 𝑛 ( 0 ) , 𝐵 𝑛 ( 1 ) 𝑥 𝑟 1 𝑛 − 𝑝 1 𝑥 𝑠 1 𝑛 − 𝑞 1 , 𝐵 𝑛 ( 2 ) 𝑥 𝑟 2 𝑛 − 𝑝 2 𝑥 𝑠 2 𝑛 − 𝑞 2 , … , 𝐵 𝑛 ( 𝑘 ) 𝑥 𝑟 𝑘 𝑛 − 𝑝 𝑘 𝑥 𝑠 𝑘 𝑛 − 𝑞 𝑘 , 𝑛 ∈ ℕ 0 , ( 1 . 2 ) where 𝑘 ∈ ℕ , 𝑝 𝑖 , 𝑞 𝑖 are natural numbers such that 𝑝 1 < 𝑝 2 < ⋯ < 𝑝 𝑘 , 𝑞 1 < 𝑞 2 < ⋯ < 𝑞 𝑘 , 𝑟 𝑖 , 𝑠 𝑖 ∈ ℝ + and 𝐵 𝑛 ( 𝑗 ) , 𝑗 = 0 , 1 , … , 𝑘 , are sequences of real numbers, was proposed by Stević in numerous talks, for example, in [13, 14]. For some results in this direction see [1, 4, 15–17, 19–21, 26, 27]. For some nonlinear difference equations related to (1.2) see, for example, [7, 15, 17, 18, 29–38].

Definition 1.1. A sequence ( 𝑥 𝑛 ) ∞ 𝑛 = − 𝑘 is said to be eventually periodic with period 𝑝 if there is an index 𝑛 0 ∈ { − 𝑘 , … , − 1 , 0 , 1 , … } such that 𝑥 𝑛 + 𝑝 = 𝑥 𝑛 for all 𝑛 ≥ 𝑛 0 . Specially, if 𝑛 0 = − 𝑘 , then the sequence ( 𝑥 𝑛 ) ∞ 𝑛 = − 𝑘 is periodic with period 𝑝 .

Motivated by some ideas due to Stević (e.g., the main lemmas there, Lemmas 3 . 1 and 3 . 2 are suggested by him), the authors of [26] considered the following second-order max-type difference equation:

𝑥 𝑛 + 1 1 = m a x 𝑥 𝑛 , 𝐴 𝑥 𝑛 − 1 , 𝑛 ∈ ℕ 0 . ( 1 . 3 ) Equation(1.3) is not difficult for handling since, by the change 𝑦 𝑛 = 𝑥 𝑛 𝑥 𝑛 − 1 , it is transformed into one of the following first-order difference equations

𝑦 𝑛 + 1 = m a x 1 , 𝐴 𝑦 𝑛 o r 𝑦 𝑛 + 1 = m i n 1 , 𝐴 𝑦 𝑛 . ( 1 . 4 ) Using these equations, it is easy to see that for the case 𝐴 = 1 every solution of (1.3) is eventually periodic with period two.

Recently, in the paper [5] it was showed that every solution of the third-order max-type difference equation

𝑥 𝑛 + 1 𝐴 = m a x 𝑥 𝑛 , 𝑥 𝑛 − 2 , 𝑛 ∈ ℕ 0 , ( 1 . 5 ) where the initial conditions 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 are arbitrary nonzero real numbers and 𝐴 ∈ ℝ , is eventually periodic with period three. The fact that all solutions of (1.5) are periodic is not a surprising fact (for an explanation see [4]).

For some recent papers on difference equations all the solutions of which are periodic see, for example, [7, 39–45] and the references cited therein.

Here we show that every well-defined solution of the following fourth-order max-type difference equation

𝑥 𝑛 + 1 𝐴 = m a x 𝑥 𝑛 , 𝑥 𝑛 − 3 , 𝑛 ∈ ℕ 0 , ( 1 . 6 ) where the parameter 𝐴 ∈ ℝ + ∪ { 0 } , is eventually periodic with period four.

Remark 1.2. Note that if 𝐴 = 0 , then (1.6) becomes 𝑥 𝑛 + 1 = 𝑥 𝑛 − 3 , from which it follows that every solution is periodic with period four. Hence, in the sequel we will consider the case 𝐴 ≠ 0 .

In the sequel we will frequently use the following simple lemma, given without a proof (for related results see [4, 5]).

Lemma 1.3. Assume that ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is a solution of (1.6) and there is 𝑘 0 ∈ ℕ 0 ∪ { − 3 , − 2 , − 1 } such that 𝑥 𝑘 0 = 𝑥 𝑘 0 + 4 , 𝑥 𝑘 0 + 1 = 𝑥 𝑘 0 + 5 , 𝑥 𝑘 0 + 2 = 𝑥 𝑘 0 + 6 , 𝑥 𝑘 0 + 3 = 𝑥 𝑘 0 + 7 . ( 1 . 7 ) Then this solution is eventually periodic with period four.

2. Main Results

In this subsection we give a specific form of the solutions of the difference equation (1.6) when the parameter 𝐴 > 0 and in each case we can deduce that every solution of this equation is periodic with period four.

Depending on the positivity of four initial values of (1.6), there are the following 16 cases to be considered:

( i ) 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 > 0 , ( i i ) 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 ( < 0 , i i i ) 𝑥 0 < 0 , 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 ( > 0 , i v ) 𝑥 − 1 < 0 , 𝑥 − 3 , 𝑥 − 2 , 𝑥 0 ( > 0 , v ) 𝑥 − 2 < 0 , 𝑥 − 3 , 𝑥 − 1 , 𝑥 0 ( > 0 , v i ) 𝑥 − 3 < 0 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 ( > 0 , v i i ) 𝑥 0 , 𝑥 − 1 < 0 , 𝑥 − 3 , 𝑥 − 2 > 0 , ( v i i i ) 𝑥 0 , 𝑥 − 2 < 0 , 𝑥 − 3 , 𝑥 − 1 ( > 0 , i x ) 𝑥 0 , 𝑥 − 3 < 0 , 𝑥 − 1 , 𝑥 − 2 > 0 , ( x ) 𝑥 − 1 , 𝑥 − 2 < 0 , 𝑥 0 , 𝑥 − 3 ( > 0 , x i ) 𝑥 − 1 , 𝑥 − 3 < 0 , 𝑥 0 , 𝑥 − 2 ( > 0 , x i i ) 𝑥 − 2 , 𝑥 − 3 < 0 , 𝑥 0 , 𝑥 − 1 ( > 0 , x i i i ) 𝑥 0 , 𝑥 − 1 , 𝑥 − 2 < 0 , 𝑥 − 3 ( > 0 , x i v ) 𝑥 0 , 𝑥 − 1 , 𝑥 − 3 < 0 , 𝑥 − 2 ( > 0 , x v ) 𝑥 0 , 𝑥 − 2 , 𝑥 − 3 < 0 , 𝑥 − 1 > 0 , ( x v i ) 𝑥 − 1 , 𝑥 − 2 , 𝑥 − 3 < 0 , 𝑥 0 > 0 . ( 2 . 1 )

First, we prove another auxiliary result.

Lemma 2.1. Assume that the parameter 𝐴 > 0 . Then every solution of (1.6) is eventually positive if initial values satisfy one of conditions (i), (iii)–(xvi).

Proof. If 𝑥 0 > 0 or 𝑥 − 3 > 0 , then 𝑥 1 𝐴 = m a x 𝑥 0 , 𝑥 − 3 > 0 . ( 2 . 2 ) From this, (1.6), and by induction it follows that 𝑥 𝑛 > 0 for every 𝑛 ∈ ℕ 0 .
If 𝑥 − 2 > 0 , then
𝑥 2 𝐴 = m a x 𝑥 1 , 𝑥 − 2 > 0 . ( 2 . 3 ) From this, (1.6), and by induction it follows that 𝑥 𝑛 > 0 for every 𝑛 ≥ 2 .
If 𝑥 − 1 > 0 , then
𝑥 3 𝐴 = m a x 𝑥 1 , 𝑥 − 1 > 0 . ( 2 . 4 ) Similar to the previous case, by induction it follows that 𝑥 𝑛 > 0 for every 𝑛 ≥ 3 .

Now, we can formulate and prove our main results.

Theorem 2.2. Assume that the parameter 𝐴 > 0 . Then every solution of (1.6) with positive initial values is eventually periodic with period four.

Proof. From (1.6), we see that 𝑥 1 𝐴 = m a x 𝑥 0 , 𝑥 − 3 . ( 2 . 5 ) We consider the following two cases. ( 𝑎 1 ) 𝑥 1 = 𝐴 / 𝑥 0 . In this case 𝐴 / 𝑥 0 ≥ 𝑥 − 3 , and we see that 𝑥 2 𝐴 = m a x 𝑥 1 , 𝑥 − 2 𝑥 = m a x 0 , 𝑥 − 2 . ( 2 . 6 ) Now, there exists two subcases. ( 𝑎 1 1 ) 𝑥 2 = 𝑥 0 , which occurs when 𝑥 0 ≥ 𝑥 − 2 . We have 𝑥 3 𝐴 = m a x 𝑥 2 , 𝑥 − 1 𝐴 = m a x 𝑥 0 , 𝑥 − 1 . ( 2 . 7 ) ( 𝑎 1 1 1 ) If 𝑥 − 1 ≥ 𝐴 / 𝑥 0 , then 𝑥 3 = 𝑥 − 1 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝐴 = m a x 𝑥 − 1 , 𝑥 0 = 𝑥 0 , 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝐴 𝑥 0 = 𝐴 𝑥 0 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝑥 = m a x 0 , 𝑥 0 = 𝑥 0 . ( 2 . 8 ) Hence, 𝑥 3 = 𝑥 − 1 , 𝑥 4 = 𝑥 0 , 𝑥 5 = 𝑥 1 , and 𝑥 6 = 𝑥 2 , which implies that ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 − 1 ) periodic solution with period four. In this case we see that the solution has the following form: 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 0 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 0 . , … ( 2 . 9 ) ( 𝑎 1 1 2 ) If 𝐴 / 𝑥 0 ≥ 𝑥 − 1 , then 𝑥 3 = 𝐴 / 𝑥 0 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝑥 = m a x 0 , 𝑥 0 = 𝑥 0 , 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝐴 𝑥 0 = 𝐴 𝑥 0 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝑥 = m a x 0 , 𝑥 0 = 𝑥 0 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 0 , 𝐴 𝑥 0 = 𝐴 𝑥 0 . ( 2 . 1 0 ) Hence, 𝑥 4 = 𝑥 0 , 𝑥 5 = 𝑥 1 , 𝑥 6 = 𝑥 2 , and 𝑥 7 = 𝑥 3 , which implies that ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 0 ) periodic solution with period four (in this case minimal period is two). This solution takes the form 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 0 , 𝐴 𝑥 0 . , … ( 2 . 1 1 ) ( 𝑎 1 2 ) 𝑥 2 = 𝑥 − 2 , which occurs when 𝑥 − 2 ≥ 𝑥 0 , and 𝑥 3 𝐴 = m a x 𝑥 2 , 𝑥 − 1 𝐴 = m a x 𝑥 − 2 , 𝑥 − 1 . ( 2 . 1 2 ) ( 𝑎 1 2 1 ) If 𝑥 − 1 ≥ 𝐴 / 𝑥 − 2 , then 𝑥 3 = 𝑥 − 1 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝐴 = m a x 𝑥 − 1 , 𝑥 0 . ( 2 . 1 3 ) ( 𝑎 1 2 1 1 ) If 𝑥 0 ≥ 𝐴 / 𝑥 − 1 , then 𝑥 4 = 𝑥 0 , and 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝐴 𝑥 0 = 𝐴 𝑥 0 . ( 2 . 1 4 ) Hence, 𝑥 2 = 𝑥 − 2 , 𝑥 3 = 𝑥 − 1 , 𝑥 4 = 𝑥 0 , and 𝑥 5 = 𝑥 1 , which implies that ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 − 2 ) periodic solution with period four. It can be written in the form 𝑥 𝑛 ∞ 𝑛 = − 3 = 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 . , … ( 2 . 1 5 ) ( 𝑎 1 2 1 2 ) If 𝐴 / 𝑥 − 1 ≥ 𝑥 0 , then 𝑥 4 = 𝐴 / 𝑥 − 1 , and 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝑥 = m a x − 1 , 𝐴 𝑥 0 = 𝐴 𝑥 0 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝑥 = m a x 0 , 𝑥 − 2 = 𝑥 − 2 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 − 2 , 𝑥 − 1 = 𝑥 − 1 , 𝑥 8 𝐴 = m a x 𝑥 7 , 𝑥 4 𝐴 = m a x 𝑥 − 1 , 𝐴 𝑥 − 1 = 𝐴 𝑥 − 1 . ( 2 . 1 6 ) Hence, 𝑥 5 = 𝑥 1 , 𝑥 6 = 𝑥 2 , 𝑥 7 = 𝑥 3 , and 𝑥 8 = 𝑥 4 , which implies that ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 1 ) periodic solution with period four. Moreover, it can be written as follows: 𝑥 𝑛 ∞ 𝑛 = − 3 = 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 − 2 , 𝑥 − 1 , 𝐴 𝑥 − 1 , 𝐴 𝑥 0 , 𝑥 − 2 , 𝑥 − 1 , 𝐴 𝑥 − 1 . , … ( 2 . 1 7 ) ( 𝑎 1 2 2 ) If 𝐴 / 𝑥 − 2 ≥ 𝑥 − 1 , then 𝑥 3 = 𝐴 / 𝑥 − 2 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝑥 = m a x − 2 , 𝑥 0 = 𝑥 − 2 , 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 − 2 , 𝐴 𝑥 0 = 𝐴 𝑥 0 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝑥 = m a x 0 , 𝑥 − 2 = 𝑥 − 2 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 − 2 , 𝐴 𝑥 − 2 = 𝐴 𝑥 − 2 , 𝑥 8 𝐴 = m a x 𝑥 7 , 𝑥 4 𝑥 = m a x − 2 , 𝑥 − 2 = 𝑥 − 2 . ( 2 . 1 8 ) As above, the solution is eventually (from 𝑥 1 ) periodic with period four and it has the form 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝐴 𝑥 0 , 𝑥 − 2 , 𝐴 𝑥 − 2 , 𝑥 − 2 , 𝐴 𝑥 0 , 𝑥 − 2 , 𝐴 𝑥 − 2 , 𝑥 − 2 , 𝐴 𝑥 0 . , … ( 2 . 1 9 ) ( 𝑎 2 ) 𝑥 1 = 𝑥 − 3 . In this case 𝑥 − 3 ≥ 𝐴 / 𝑥 0 , and we see that 𝑥 2 𝐴 = m a x 𝑥 1 , 𝑥 − 2 𝐴 = m a x 𝑥 − 3 , 𝑥 − 2 . ( 2 . 2 0 ) There again exist two subcases. ( 𝑎 2 1 ) 𝑥 2 = 𝐴 / 𝑥 − 3 , which occurs when 𝐴 / 𝑥 − 3 ≥ 𝑥 − 2 . So, 𝑥 3 𝐴 = m a x 𝑥 2 , 𝑥 − 1 𝑥 = m a x − 3 , 𝑥 − 1 . ( 2 . 2 1 ) ( 𝑎 2 1 1 ) If 𝑥 − 3 ≥ 𝑥 − 1 , then 𝑥 3 = 𝑥 − 3 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝐴 = m a x 𝑥 − 3 , 𝑥 0 = 𝑥 0 , 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝑥 − 3 = 𝑥 − 3 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝐴 = m a x 𝑥 − 3 , 𝐴 𝑥 − 3 = 𝐴 𝑥 − 3 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝑥 = m a x − 3 , 𝑥 − 3 = 𝑥 − 3 . ( 2 . 2 2 ) Then we see that the solution is 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝐴 𝑥 − 3 , 𝑥 − 3 , 𝑥 0 , 𝑥 − 3 , 𝐴 𝑥 − 3 , 𝑥 − 3 , 𝑥 0 , 𝑥 − 3 , 𝐴 𝑥 − 3 , , … ( 2 . 2 3 ) and ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 0 ) periodic solution with period four. ( 𝑎 2 1 2 ) If 𝑥 − 1 ≥ 𝑥 − 3 , then 𝑥 3 = 𝑥 − 1 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝐴 = m a x 𝑥 − 1 , 𝑥 0 = 𝑥 0 , 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝑥 − 3 = 𝑥 − 3 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝐴 = m a x 𝑥 − 3 , 𝐴 𝑥 − 3 = 𝐴 𝑥 − 3 . ( 2 . 2 4 ) So, the solution takes the following form which is an eventually (from 𝑥 − 1 ) periodic solution with period four: 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝐴 𝑥 − 3 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝐴 𝑥 − 3 . , … ( 2 . 2 5 ) ( 𝑎 2 2 ) 𝑥 2 = 𝑥 − 2 , which occurs when 𝑥 − 2 ≥ 𝐴 / 𝑥 − 3 . So, 𝑥 3 𝐴 = m a x 𝑥 2 , 𝑥 − 1 𝐴 = m a x 𝑥 − 2 , 𝑥 − 1 . ( 2 . 2 6 ) ( 𝑎 2 2 1 ) If 𝑥 − 1 ≥ 𝐴 / 𝑥 − 2 , then 𝑥 3 = 𝑥 − 1 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝐴 = m a x 𝑥 − 1 , 𝑥 0 . ( 2 . 2 7 ) ( 𝑎 2 2 1 1 ) If 𝑥 0 ≥ 𝐴 / 𝑥 − 1 , then 𝑥 4 = 𝑥 0 .
Therefore ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is a periodic solution with period four and the solution takes the form
𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 . , … ( 2 . 2 8 ) ( 𝑎 2 2 1 2 ) If 𝐴 / 𝑥 − 1 ≥ 𝑥 0 , then 𝑥 4 = 𝐴 / 𝑥 − 1 , and 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝑥 = m a x − 1 , 𝑥 − 3 = 𝑥 − 3 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝐴 = m a x 𝑥 − 3 , 𝑥 − 2 = 𝑥 − 2 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 − 2 , 𝑥 − 1 = 𝑥 − 1 , 𝑥 8 𝐴 = m a x 𝑥 7 , 𝑥 4 𝐴 = m a x 𝑥 − 1 , 𝐴 𝑥 − 1 = 𝐴 𝑥 − 1 . ( 2 . 2 9 ) Therefore, again ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is an eventually (from 𝑥 1 ) periodic solution with period four and the solution takes the form 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝐴 𝑥 − 1 , 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝐴 𝑥 − 1 . , … ( 2 . 3 0 ) ( 𝑎 2 2 2 ) If 𝐴 / 𝑥 − 2 ≥ 𝑥 − 1 , then 𝑥 3 = 𝐴 / 𝑥 − 2 , and 𝑥 4 𝐴 = m a x 𝑥 3 , 𝑥 0 𝑥 = m a x − 2 , 𝑥 0 . ( 2 . 3 1 ) ( 𝑎 2 2 2 1 ) If 𝑥 − 2 ≥ 𝑥 0 , then 𝑥 4 = 𝑥 − 2 , and 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 − 2 , 𝑥 − 3 = 𝑥 − 3 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝐴 = m a x 𝑥 − 3 , 𝑥 − 2 = 𝑥 − 2 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 − 2 , 𝐴 𝑥 − 2 = 𝐴 𝑥 − 2 , 𝑥 8 𝐴 = m a x 𝑥 7 , 𝑥 4 𝑥 = m a x − 2 , 𝑥 − 2 = 𝑥 − 2 . ( 2 . 3 2 ) Thus, the solution is in the following form which is eventually (from 𝑥 1 ) periodic with period four: 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝑥 − 2 , 𝐴 𝑥 − 2 , 𝑥 − 2 , 𝑥 − 3 , 𝑥 − 2 , 𝐴 𝑥 − 2 , 𝑥 − 2 . , … ( 2 . 3 3 ) ( 𝑎 2 2 2 2 ) If 𝑥 0 ≥ 𝑥 − 2 , then 𝑥 4 = 𝑥 0 , and 𝑥 5 𝐴 = m a x 𝑥 4 , 𝑥 1 𝐴 = m a x 𝑥 0 , 𝑥 − 3 = 𝑥 − 3 , 𝑥 6 𝐴 = m a x 𝑥 5 , 𝑥 2 𝐴 = m a x 𝑥 − 3 , 𝑥 − 2 = 𝑥 − 2 , 𝑥 7 𝐴 = m a x 𝑥 6 , 𝑥 3 𝐴 = m a x 𝑥 − 2 , 𝐴 𝑥 − 2 = 𝐴 𝑥 − 2 . ( 2 . 3 4 ) Thus, the solution is of the following form 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 , 𝑥 − 3 , 𝑥 − 2 , 𝐴 𝑥 − 2 , 𝑥 0 , 𝑥 − 3 , 𝑥 − 2 , 𝐴 𝑥 − 2 . , … ( 2 . 3 5 ) Therefore ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is eventually (from 𝑥 0 ) periodic with period four. The proof is completed.

From Lemma 1.3 and Theorem 2.2 we obtain the following result.

Theorem 2.3. Assume the parameter 𝐴 > 0 and that initial values of (1.6) satisfy one of conditions (i), (iii)–(xvi) in Lemma 2.1. Then every such solution of (1.6) is eventually periodic with period four.

Proof. If initial values of (1.6) satisfy one of conditions (i), (iii)–(xvi) in Lemma 2.1, then by the same lemma it follows that the corresponding solution is eventually positive. This means that there is 𝑘 1 ∈ ℕ 0 ∪ { − 3 , − 2 , − 1 } such that 𝑥 𝑛 > 0 for every 𝑛 ≥ 𝑘 1 . In particular, we have that 𝑥 𝑘 1 , 𝑥 𝑘 1 + 1 , 𝑥 𝑘 1 + 2 , 𝑥 𝑘 1 + 3 > 0 . Since equation (1.6) is autonomous if ( 𝑥 𝑛 ) ∞ 𝑛 = − 3 is a solution of (1.6), then 𝑦 𝑛 = 𝑥 𝑛 + 𝑘 1 + 3 is also a solution of (1.6) but such that 𝑦 − 3 , 𝑦 − 2 , 𝑦 − 1 , 𝑦 0 > 0 . Hence, the problem is reduced to the case when all the initial values are positive. Applying Theorem 2.2, the result follows.

In the next theorem we study those solutions of (1.6) such that 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 < 0 . We would like to thank Professor Stević for giving us the elegant proof below which drastically reduced our original proof.

Theorem 2.4. Assume that the parameter 𝐴 > 0 and all the initial values are negative 𝑥 − 3 , 𝑥 − 2 , 𝑥 − 1 , 𝑥 0 < 0 . Then every solution of (