On the Max-Type Difference Equation 𝑥_𝑛+1=max{𝐴/𝑥_𝑛,𝑥_𝑛−3}

Abstract

We show that every well-defined solution of the fourth-order difference equation 𝑥_𝑛+1=max{𝐴/𝑥_𝑛,𝑥_𝑛−3},    𝑛∈ℕ₀, 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_𝑛𝐴=max{^𝑛(1)𝑥_𝑛−1,𝐴_𝑛⁽²⁾𝑥_𝑛−2𝐴,…,^𝑛(𝑘)𝑥_𝑛−𝑘},𝑛∈ℕ₀, (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=max{𝐵_𝑛⁽⁰⁾,𝐵_𝑛⁽¹⁾𝑥^𝑟1_𝑛−𝑝1𝑥^𝑠1_𝑛−𝑞1,𝐵_𝑛⁽²⁾𝑥^𝑟2_𝑛−𝑝2𝑥^𝑠2_𝑛−𝑞2,…,𝐵_𝑛^(𝑘)𝑥^𝑟𝑘_{𝑛−𝑝𝑘}𝑥^𝑠𝑘_{𝑛−𝑞𝑘}},𝑛∈ℕ₀, (1.2) where 𝑘∈ℕ, 𝑝_𝑖,𝑞_𝑖 are natural numbers such that 𝑝₁<𝑝₂<⋯<𝑝_𝑘, 𝑞₁<𝑞₂<⋯<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 𝑛₀∈{−𝑘,…,−1,0,1,…} such that 𝑥_𝑛+𝑝=𝑥_𝑛 for all 𝑛≥𝑛₀. Specially, if 𝑛₀=−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+11=max{𝑥_𝑛,𝐴𝑥_𝑛−1},𝑛∈ℕ₀. (1.3) Equation(1.3) is not difficult for handling since, by the change 𝑦_𝑛=𝑥_𝑛𝑥_𝑛−1, it is transformed into one of the following first-order difference equations

𝑦_𝑛+1=max{1,𝐴𝑦_𝑛}or𝑦_𝑛+1=min{1,𝐴𝑦_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+1𝐴=max{𝑥_𝑛,𝑥_𝑛−2},𝑛∈ℕ₀, (1.5) where the initial conditions 𝑥₋₂,𝑥₋₁,𝑥₀are 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+1𝐴=max{𝑥_𝑛,𝑥_𝑛−3},𝑛∈ℕ₀, (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 𝑘₀∈ℕ₀∪{−3,−2,−1} such that 𝑥_𝑘0=𝑥_𝑘0+4,𝑥_𝑘0+1=𝑥_𝑘0+5,𝑥_𝑘0+2=𝑥_𝑘0+6,𝑥_𝑘0+3=𝑥_𝑘0+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,𝑥₀>0,(ii)𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀(<0,iii)𝑥₀<0,𝑥₋₃,𝑥₋₂,𝑥₋₁>0,(iv)𝑥₋₁<0,𝑥₋₃,𝑥₋₂,𝑥₀(>0,v)𝑥₋₂<0,𝑥₋₃,𝑥₋₁,𝑥₀>0,(vi)𝑥₋₃<0,𝑥₋₂,𝑥₋₁,𝑥₀(>0,vii)𝑥₀,𝑥₋₁<0,𝑥₋₃,𝑥₋₂>0,(viii)𝑥₀,𝑥₋₂<0,𝑥₋₃,𝑥₋₁(>0,ix)𝑥₀,𝑥₋₃<0,𝑥₋₁,𝑥₋₂>0,(x)𝑥₋₁,𝑥₋₂<0,𝑥₀,𝑥₋₃(>0,xi)𝑥₋₁,𝑥₋₃<0,𝑥₀,𝑥₋₂>0,(xii)𝑥₋₂,𝑥₋₃<0,𝑥₀,𝑥₋₁(>0,xiii)𝑥₀,𝑥₋₁,𝑥₋₂<0,𝑥₋₃>0,(xiv)𝑥₀,𝑥₋₁,𝑥₋₃<0,𝑥₋₂(>0,xv)𝑥₀,𝑥₋₂,𝑥₋₃<0,𝑥₋₁>0,(xvi)𝑥₋₁,𝑥₋₂,𝑥₋₃<0,𝑥₀>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 𝑥₀>0 or 𝑥₋₃>0, then 𝑥₁𝐴=max{𝑥₀,𝑥₋₃}>0. (2.2) From this, (1.6), and by induction it follows that 𝑥_𝑛>0 for every 𝑛∈ℕ₀.
If 𝑥₋₂>0, then
𝑥₂𝐴=max{𝑥₁,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
𝑥₃𝐴=max{𝑥₁,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 𝑥₁𝐴=max{𝑥₀,𝑥₋₃}. (2.5) We consider the following two cases. (𝑎₁)𝑥₁=𝐴/𝑥₀. In this case 𝐴/𝑥₀≥𝑥₋₃, and we see that 𝑥₂𝐴=max{𝑥₁,𝑥₋₂}=max{𝑥₀,𝑥₋₂}. (2.6)Now, there exists two subcases. (𝑎₁₁)𝑥₂=𝑥₀, which occurs when 𝑥₀≥𝑥₋₂. We have 𝑥₃𝐴=max{𝑥₂,𝑥₋₁𝐴}=max{𝑥₀,𝑥₋₁}. (2.7)(𝑎₁₁₁)If 𝑥₋₁≥𝐴/𝑥₀, then 𝑥₃=𝑥₋₁,and𝑥₄𝐴=max{𝑥₃,𝑥₀𝐴}=max{𝑥₋₁,𝑥₀}=𝑥₀,𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝐴𝑥₀𝐴}=𝑥₀,𝑥₆𝐴=max{𝑥₅,𝑥₂}=max{𝑥₀,𝑥₀}=𝑥₀. (2.8) Hence, 𝑥₃=𝑥₋₁, 𝑥₄=𝑥₀, 𝑥₅=𝑥₁, and 𝑥₆=𝑥₂, 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,𝑥₀,𝐴𝑥₀,𝑥₀,𝑥₋₁,𝑥₀,𝐴𝑥₀,𝑥₀,…). (2.9)(𝑎₁₁₂)If 𝐴/𝑥₀≥𝑥₋₁, then 𝑥₃=𝐴/𝑥₀,and𝑥₄𝐴=max{𝑥₃,𝑥₀}=max{𝑥₀,𝑥₀}=𝑥₀,𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝐴𝑥₀𝐴}=𝑥₀,𝑥₆𝐴=max{𝑥₅,𝑥₂}=max{𝑥₀,𝑥₀}=𝑥₀,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₀,𝐴𝑥₀𝐴}=𝑥₀. (2.10) Hence, 𝑥₄=𝑥₀, 𝑥₅=𝑥₁, 𝑥₆=𝑥₂, and 𝑥₇=𝑥₃, which implies that (𝑥_𝑛)^∞_𝑛=−3 is an eventually (from 𝑥₀) periodic solution with period four (in this case minimal period is two). This solution takes the form (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝐴𝑥₀,𝑥₀,𝐴𝑥₀,𝑥₀,𝐴𝑥₀,𝑥₀,𝐴𝑥₀,…). (2.11)(𝑎₁₂)𝑥₂=𝑥₋₂, which occurs when 𝑥₋₂≥𝑥₀, and 𝑥₃𝐴=max{𝑥₂,𝑥₋₁𝐴}=max{𝑥₋₂,𝑥₋₁}. (2.12)(𝑎₁₂₁)If 𝑥₋₁≥𝐴/𝑥₋₂, then 𝑥₃=𝑥₋₁,and 𝑥₄𝐴=max{𝑥₃,𝑥₀𝐴}=max{𝑥₋₁,𝑥₀}. (2.13)(𝑎₁₂₁₁)If 𝑥₀≥𝐴/𝑥₋₁, then 𝑥₄=𝑥₀,and𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝐴𝑥₀𝐴}=𝑥₀. (2.14) Hence, 𝑥₂=𝑥₋₂, 𝑥₃=𝑥₋₁, 𝑥₄=𝑥₀, and 𝑥₅=𝑥₁, 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,𝑥₀,𝐴𝑥₀,𝑥₋₂,𝑥₋₁,𝑥₀,𝐴𝑥₀,…). (2.15)(𝑎₁₂₁₂)If 𝐴/𝑥₋₁≥𝑥₀, then 𝑥₄=𝐴/𝑥₋₁,and𝑥₅𝐴=max{𝑥₄,𝑥₁}=max{𝑥₋₁,𝐴𝑥₀𝐴}=𝑥₀,𝑥₆𝐴=max{𝑥₅,𝑥₂}=max{𝑥₀,𝑥₋₂}=𝑥₋₂,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₋₂,𝑥₋₁}=𝑥₋₁,𝑥₈𝐴=max{𝑥₇,𝑥₄𝐴}=max{𝑥₋₁,𝐴𝑥₋₁𝐴}=𝑥₋₁. (2.16) Hence, 𝑥₅=𝑥₁, 𝑥₆=𝑥₂, 𝑥₇=𝑥₃, and 𝑥₈=𝑥₄, which implies that (𝑥_𝑛)^∞_𝑛=−3 is an eventually (from 𝑥₁) periodic solution with period four. Moreover, it can be written as follows: (𝑥_𝑛)^∞_𝑛=−3=(𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝐴𝑥₀,𝑥₋₂,𝑥₋₁,𝐴𝑥₋₁,𝐴𝑥₀,𝑥₋₂,𝑥₋₁,𝐴𝑥₋₁,…). (2.17)(𝑎₁₂₂)If 𝐴/𝑥₋₂≥𝑥₋₁, then 𝑥₃=𝐴/𝑥₋₂,and𝑥₄𝐴=max{𝑥₃,𝑥₀}=max{𝑥₋₂,𝑥₀}=𝑥₋₂,𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₋₂,𝐴𝑥₀𝐴}=𝑥₀,𝑥₆𝐴=max{𝑥₅,𝑥₂}=max{𝑥₀,𝑥₋₂}=𝑥₋₂,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₋₂,𝐴𝑥₋₂𝐴}=𝑥₋₂,𝑥₈𝐴=max{𝑥₇,𝑥₄}=max{𝑥₋₂,𝑥₋₂}=𝑥₋₂. (2.18) As above, the solution is eventually (from 𝑥₁) periodic with period four and it has the form (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝐴𝑥₀,𝑥₋₂,𝐴𝑥₋₂,𝑥₋₂,𝐴𝑥₀,𝑥₋₂,𝐴𝑥₋₂,𝑥₋₂,𝐴𝑥₀,…). (2.19)(𝑎₂)𝑥₁=𝑥₋₃. In this case 𝑥₋₃≥𝐴/𝑥₀, and we see that 𝑥₂𝐴=max{𝑥₁,𝑥₋₂𝐴}=max{𝑥₋₃,𝑥₋₂}. (2.20)There again exist two subcases. (𝑎₂₁)𝑥₂=𝐴/𝑥₋₃, which occurs when 𝐴/𝑥₋₃≥𝑥₋₂. So, 𝑥₃𝐴=max{𝑥₂,𝑥₋₁}=max{𝑥₋₃,𝑥₋₁}. (2.21)(𝑎₂₁₁)If 𝑥₋₃≥𝑥₋₁, then 𝑥₃=𝑥₋₃,and𝑥₄𝐴=max{𝑥₃,𝑥₀𝐴}=max{𝑥₋₃,𝑥₀}=𝑥₀,𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝑥₋₃}=𝑥₋₃,𝑥₆𝐴=max{𝑥₅,𝑥₂𝐴}=max{𝑥₋₃,𝐴𝑥₋₃𝐴}=𝑥₋₃,𝑥₇𝐴=max{𝑥₆,𝑥₃}=max{𝑥₋₃,𝑥₋₃}=𝑥₋₃. (2.22) Then we see that the solution is (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑥₋₃,𝐴𝑥₋₃,𝑥₋₃,𝑥₀,𝑥₋₃,𝐴𝑥₋₃,𝑥₋₃,𝑥₀,𝑥₋₃,𝐴𝑥₋₃,…), (2.23) and (𝑥_𝑛)^∞_𝑛=−3 is an eventually (from 𝑥₀) periodic solution with period four. (𝑎₂₁₂)If 𝑥₋₁≥𝑥₋₃, then 𝑥₃=𝑥₋₁,and𝑥₄𝐴=max{𝑥₃,𝑥₀𝐴}=max{𝑥₋₁,𝑥₀}=𝑥₀,𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝑥₋₃}=𝑥₋₃,𝑥₆𝐴=max{𝑥₅,𝑥₂𝐴}=max{𝑥₋₃,𝐴𝑥₋₃𝐴}=𝑥₋₃. (2.24) 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,𝑥₀,𝑥₋₃,𝐴𝑥₋₃,𝑥₋₁,𝑥₀,𝑥₋₃,𝐴𝑥₋₃,…). (2.25)(𝑎₂₂)𝑥₂=𝑥₋₂, which occurs when 𝑥₋₂≥𝐴/𝑥₋₃. So, 𝑥₃𝐴=max{𝑥₂,𝑥₋₁𝐴}=max{𝑥₋₂,𝑥₋₁}. (2.26)(𝑎₂₂₁)If 𝑥₋₁≥𝐴/𝑥₋₂, then 𝑥₃=𝑥₋₁,and 𝑥₄𝐴=max{𝑥₃,𝑥₀𝐴}=max{𝑥₋₁,𝑥₀}. (2.27)(𝑎₂₂₁₁)If 𝑥₀≥𝐴/𝑥₋₁, then 𝑥₄=𝑥₀.
Therefore (𝑥_𝑛)^∞_𝑛=−3 is a periodic solution with period four and the solution takes the form
(𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,…). (2.28)(𝑎₂₂₁₂)If 𝐴/𝑥₋₁≥𝑥₀, then 𝑥₄=𝐴/𝑥₋₁,and𝑥₅𝐴=max{𝑥₄,𝑥₁}=max{𝑥₋₁,𝑥₋₃}=𝑥₋₃,𝑥₆𝐴=max{𝑥₅,𝑥₂𝐴}=max{𝑥₋₃,𝑥₋₂}=𝑥₋₂,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₋₂,𝑥₋₁}=𝑥₋₁,𝑥₈𝐴=max{𝑥₇,𝑥₄𝐴}=max{𝑥₋₁,𝐴𝑥₋₁𝐴}=𝑥₋₁. (2.29) Therefore, again (𝑥_𝑛)^∞_𝑛=−3 is an eventually (from 𝑥₁) periodic solution with period four and the solution takes the form (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑥₋₃,𝑥₋₂,𝑥₋₁,𝐴𝑥₋₁,𝑥₋₃,𝑥₋₂,𝑥₋₁,𝐴𝑥₋₁,…). (2.30)(𝑎₂₂₂)If 𝐴/𝑥₋₂≥𝑥₋₁, then 𝑥₃=𝐴/𝑥₋₂,and 𝑥₄𝐴=max{𝑥₃,𝑥₀}=max{𝑥₋₂,𝑥₀}. (2.31)(𝑎₂₂₂₁)If 𝑥₋₂≥𝑥₀, then 𝑥₄=𝑥₋₂,and𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₋₂,𝑥₋₃}=𝑥₋₃,𝑥₆𝐴=max{𝑥₅,𝑥₂𝐴}=max{𝑥₋₃,𝑥₋₂}=𝑥₋₂,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₋₂,𝐴𝑥₋₂𝐴}=𝑥₋₂,𝑥₈𝐴=max{𝑥₇,𝑥₄}=max{𝑥₋₂,𝑥₋₂}=𝑥₋₂. (2.32) Thus, the solution is in the following form which is eventually (from 𝑥₁) periodic with period four: (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑥₋₃,𝑥₋₂,𝐴𝑥₋₂,𝑥₋₂,𝑥₋₃,𝑥₋₂,𝐴𝑥₋₂,𝑥₋₂,…). (2.33)(𝑎₂₂₂₂)If 𝑥₀≥𝑥₋₂, then 𝑥₄=𝑥₀,and𝑥₅𝐴=max{𝑥₄,𝑥₁𝐴}=max{𝑥₀,𝑥₋₃}=𝑥₋₃,𝑥₆𝐴=max{𝑥₅,𝑥₂𝐴}=max{𝑥₋₃,𝑥₋₂}=𝑥₋₂,𝑥₇𝐴=max{𝑥₆,𝑥₃𝐴}=max{𝑥₋₂,𝐴𝑥₋₂𝐴}=𝑥₋₂. (2.34) Thus, the solution is of the following form (𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀,𝑥₋₃,𝑥₋₂,𝐴𝑥₋₂,𝑥₀,𝑥₋₃,𝑥₋₂,𝐴𝑥₋₂,…). (2.35) Therefore (𝑥_𝑛)^∞_𝑛=−3 is eventually (from 𝑥₀) 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 𝑘₁∈ℕ₀∪{−3,−2,−1} such that 𝑥_𝑛>0 for every 𝑛≥𝑘₁. In particular, we have that 𝑥_𝑘1,𝑥_𝑘1+1,𝑥_𝑘1+2,𝑥_𝑘1+3>0. Since equation (1.6) is autonomous if (𝑥_𝑛)^∞_𝑛=−3 is a solution of (1.6), then 𝑦_𝑛=𝑥_𝑛+𝑘1+3 is also a solution of (1.6) but such that 𝑦₋₃,𝑦₋₂,𝑦₋₁,𝑦₀>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,𝑥₀<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,𝑥₀<0. Then every solution of (1.6) is eventually periodic with period four.

Proof. Since 𝑥₋₃<0, 𝑥₋₂<0, 𝑥₋₁<0, 𝑥₀<0, and 𝐴>0, by induction we have 𝑥_𝑛<0 for each 𝑛∈ℕ. By the change 𝑥_𝑛=−√𝐴/𝑧_𝑛, (1.6) becomes 𝑧_𝑛+1=(min{𝑧_𝑛,1𝑧_𝑛−3})⁻¹1=max{𝑧_𝑛,𝑧_𝑛−3}, (2.36) where 𝑧_𝑛>0 for every 𝑛≥−3. Hence by Theorem 2.2 the result follows.

Since all the cases are reduced to the case when all initial values are positive, it is of interest to investigate this case in more details. The next theorem describes all eventually constant solutions of (1.6). Our idea stems from [46] (see also [47–50]).

Theorem 2.5. Assume that the parameter 𝐴>0. Then all positive solutions to (1.6) which are eventually equal to the positive equilibrium have the following form: (1,𝑐,𝑏,1,1,1,…),forsome𝑏,𝑐∈(0,1]. (2.37)

Proof. First note that by the change 𝑥_𝑛=√𝐴𝑦_𝑛 the problem is reduced to the case 𝐴=1. Assume that 𝑥_𝑛=1 for 𝑛≥𝑘 and 𝑥_𝑘−1=𝑎≠1. Then 1=𝑥_𝑘=max{1/𝑎,𝑥_𝑘−4}, 𝑘≥1, which implies 𝑥_𝑘−4=1 and 𝑎>1. On the other hand, we have 1=𝑥_𝑘+3=max{1/𝑥_𝑘+2,𝑥_𝑘−1}=max{1,𝑎}, so that 𝑎<1, which is a contradiction. Hence 𝑘≤0, that is, (𝑥_𝑛)=(𝑑,𝑐,𝑏,1,1,1,…). Since 𝑥₁=1, we get 𝑥₋₃=1 (case 𝑘=1 above). Since 1=𝑥₂=max{𝑥₁,𝑥₋₂}=max{1,𝑐} and 1=𝑥₃=max{𝑥₂,𝑥₋₁}=max{1,𝑏}, it follows that 𝑏,𝑐∈(0,1], as claimed.

3. Conclusions and Future Works

We finish this article with some comments which can motivate further works.

Case 𝐴<0
The case when the parameter 𝐴 is negative is not treated in this article. By similar calculations as above, we have managed to show that all the solutions of (1.6) are eventually periodic with period four in many subcases. However, there are too many subcases and calculations without some new ideas, so we decided not to present any result when 𝐴<0. We conjecture that every solution of (1.6), when 𝐴<0, is eventually periodic with the same period four. What is more interesting is to find a reasonably short proof of the conjecture without using tiresome calculations similar to those in Theorem 2.2.

Periodicity
We also want to mention that in [4] we proved that every solution of the equation 𝑥_𝑛+1𝐴=max{𝑥_𝑛−𝑘,𝑥_𝑛−𝑙},𝑛∈ℕ₀, (3.1) where 𝑘,𝑙∈ℕ₀, is periodic. It is of some interest to find the minimal period of the equation, as well as to get a general result concerning this problem.

Acknowledgments

The authors would like to express their sincere thanks to the anonymous referees for numerous comments which considerably improved this article. The research of the first author was partly supported by the Serbian Ministry of Science, through The Mathematical Institute of SASA, Belgrade, Project no. 144013.