On the Difference Equation 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−𝑘/(𝑥_{𝑛−𝑘+1}(𝑎+𝑏𝑥_𝑛𝑥_𝑛−𝑘))

Abstract

We show that the difference equation 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−𝑘/𝑥_{𝑛−𝑘+1}(𝑎+𝑏𝑥_𝑛𝑥_𝑛−𝑘),𝑛∈ℕ₀, where 𝑘∈ℕ, the parameters 𝑎, 𝑏 and initial values 𝑥_−𝑖, 𝑖=0,𝑘 are real numbers, can be solved in closed form considerably extending the results in the literature. By using obtained formulae, we investigate asymptotic behavior of well-defined solutions of the equation.

1. Introduction

Recently, there has been some reestablished interest in difference equations which can be solved, as well as in their applications, see, for example, [1–16]. For some old results in the topic see, for example, the classical book [17].

In recently accepted paper [18] are given formulae for the solutions of the following four difference equations: 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−3𝑥_𝑛−2±1±𝑥_𝑛𝑥_𝑛−3,𝑛∈ℕ₀, (1.1) and some of these formulae are proved by induction.

Here, we show that the formulae obtained in [18] follow from known results in a natural way. Related idea was exploited in paper [7].

Moreover, we will consider here the following more general equation: 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−𝑘𝑥_{𝑛−𝑘+1}𝑎+𝑏𝑥_𝑛𝑥_𝑛−𝑘,𝑛∈ℕ₀, (1.2) where 𝑘∈ℕ and the parameters 𝑎,𝑏 as well as initial values 𝑥_−𝑖,𝑖=0,𝑘 are real numbers, and describe the behaviour of all well-defined solutions of the equation.

For a solution (𝑥_𝑛)_{𝑛≥−𝑘}, 𝑘∈ℕ of the difference equation 𝑥_𝑛=𝑓𝑥_𝑛−1,…,𝑥_𝑛−𝑘,𝑛∈ℕ₀, (1.3) is said that it is eventually periodic with period 𝑝, if there is an 𝑛₁≥−𝑘 such that 𝑥_𝑛+𝑝=𝑥_𝑛,for𝑛≥𝑛₁. (1.4) If 𝑛₁=−𝑘, then it is said that the solution is periodic with period 𝑝. For some results in this area see, for example, [19–26] and the references therein.

2. Solutions of Equation (1.2)

By using the change of variables 𝑦_𝑛=1𝑥_𝑛𝑥_𝑛−𝑘,𝑛∈ℕ₀, (2.1) equation (1.2) is transformed into the following linear first-order difference equation: 𝑦_𝑛+1=𝑎𝑦_𝑛+𝑏,𝑛∈ℕ₀, (2.2) for which it is known (and easy to see) that 𝑦_𝑛=𝑦₀𝑎^𝑛+𝑏1−𝑎^𝑛=1−𝑎𝑏+𝑎^𝑛𝑦₀(1−𝑎)−𝑏1−𝑎,𝑛∈ℕ₀, (2.3) if 𝑎≠1, and 𝑦_𝑛=𝑦₀+𝑏𝑛,𝑛∈ℕ₀, (2.4) if 𝑎=1.

From (2.1), we have 𝑥_𝑛=1𝑦_𝑛𝑥_𝑛−𝑘=𝑦_𝑛−𝑘𝑦_𝑛𝑥_𝑛−2𝑘, (2.5) for 𝑛≥𝑘, from which it follows that 𝑥_{2𝑘𝑚+𝑖}=𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑦_{(2𝑗−1)𝑘+𝑖}𝑦_{2𝑗𝑘+𝑖}, (2.6) for every 𝑚∈ℕ₀ and 𝑖∈{𝑘,𝑘+1,…,3𝑘−1}.

Using (2.3) and (2.4) in (2.6), we get 𝑥_{2𝑘𝑚+𝑖}=𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘, (2.7) if 𝑎≠1, and 𝑥_{2𝑘𝑚+𝑖}=𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑥₀𝑥_−𝑘𝑏2𝑗−1𝑘+𝑖+1𝑥₀𝑥_−𝑘𝑏,2𝑗𝑘+𝑖+1 (2.8) if 𝑎=1, for every 𝑚∈ℕ₀ and 𝑖∈{𝑘,𝑘+1,…,3𝑘−1}.

By using formulae (2.7) and (2.8), the behavior of well-defined solutions of equation (1.2) can be obtained. This is done in the following section.

Remark 2.1. It is easy to check that the formulae in Theorems 2.1 and 4.1 from [18] are direct consequences of formula (2.8), whereas formulae in Theorem 3.1 and Theorem 5.1 from [18] are direct consequences of formula (2.7).

Remark 2.2. Note that from formula (2.8), it follows that in the case 𝑎=1, 𝑏=0, all well-defined solutions of equation (1.2) are periodic with period 2𝑘. This can be also obtained from (1.2), without knowing explicit formulae for its solutions. Namely, in this case, (1.2) can be written as follows: 𝑥_𝑛+1𝑥_{𝑛−𝑘+1}=𝑥_𝑛𝑥_𝑛−𝑘, (2.9) since we assume 𝑥_𝑛≠0, for all 𝑛≥−𝑘, from which it follows that the sequence 𝑥_𝑛𝑥_𝑛−𝑘 is constant, that is, 𝑥_𝑛𝑥_𝑛−𝑘=𝑐,𝑛∈ℕ₀ for some 𝑐∈ℝ⧵{0}. Hence, 𝑥_𝑛=𝑐𝑥_𝑛−𝑘=𝑥_𝑛−2𝑘,𝑛≥𝑘, (2.10) as claimed.

Remark 2.3. Note also that in the case 𝑎∉{0,1}, 𝑏=0, from (2.7), we have 𝑥_{2𝑘𝑚+𝑖}=𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑎^{(2𝑗−1)𝑘+𝑖}𝑎^{2𝑗𝑘+𝑖}=𝑥_𝑖−2𝑘𝑎^𝑘(𝑚+1), (2.11) for every 𝑚∈ℕ₀ and 𝑖∈{𝑘,𝑘+1,…,3𝑘−1}, from which the behaviour of the solutions in the case easily follows.

3. Asymptotic Behavior of Well-Defined Solutions of Equation (1.2)

In this section, we derive some results on asymptotic behavior of well-defined solutions of equation (1.2). We will use well-known asymptotic formulae as follows: ()𝑥ln1+𝑥=𝑥−²2+𝑂𝑥³,(1+𝑥)⁻¹=1−𝑥+𝑂𝑥², (3.1) for 𝑥→0, where 𝑂 is the Landau “big-oh” symbol.

Theorem 3.1. Let 𝑎=1 and 𝑏≠0 in (1.2). Then, every well-defined solution (𝑥_𝑛)_{𝑛≥−𝑘} of equation (1.2) converges to zero.

Proof. By formula (2.8), we have lim_𝑚→∞𝑥_{2𝑘𝑚+𝑖}=lim_𝑚→∞𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑥₀𝑥_−𝑘𝑏2𝑗−1𝑘+𝑖+1𝑥₀𝑥_−𝑘𝑏2𝑗𝑘+𝑖+1=𝑥_𝑖−2𝑘lim^𝑚_{𝑚→∞𝑗=0}𝑥1−⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏2𝑗𝑘+𝑖=𝑥_𝑖−2𝑘lim_𝑚→∞𝐶𝑚₀^𝑚_𝑗=𝑚0+1𝑥1−⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏,2𝑗𝑘+𝑖 (3.2) where 𝑚₀ is sufficiently large so that (3.1) can be applied below, and 𝐶𝑚₀=^𝑚0_𝑗=0𝑥1−⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏.2𝑗𝑘+𝑖 (3.3) Since _{∞𝑗=𝑚0+1}𝑘2𝑗𝑘+𝑖=+∞, (3.4) we conclude, using (3.1), that lim^𝑚_{𝑚→∞𝑗=𝑚0+1}𝑥1−⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏2𝑗𝑘+𝑖=lim^𝑚_{𝑚→∞𝑗=𝑚0+1}𝑥expln1−⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏2𝑗𝑘+𝑖=lim^𝑚_{𝑚→∞𝑗=𝑚0+1}−𝑥exp⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏−12𝑗𝑘+𝑖2𝑥₀𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏(2𝑗𝑘+𝑖)²𝑥+𝑂⁰𝑥_−𝑘𝑏𝑘1+𝑥₀𝑥_−𝑘𝑏(2𝑗𝑘+𝑖)³=lim^𝑚_{𝑚→∞𝑗=𝑚0+1}−𝑘exp12𝑗𝑘+𝑖+𝑂𝑗²=lim_𝑚→∞exp−^𝑚_𝑗=𝑚0+1𝑘2𝑗𝑘+𝑖^𝑚_𝑗=𝑚0+1𝑂1exp𝑗²=0. (3.5) Therefore, lim_𝑚→∞𝑥_{2𝑘𝑚+𝑖}=0, (3.6) for every 𝑖∈{𝑘,𝑘+1,…,3𝑘−1}, and consequently, lim_𝑛→∞𝑥_𝑛=0, as claimed.

Before we formulate and prove our next result, we will prove an auxiliary result which is incorporated in the lemma that follows.

Lemma 3.2. If 𝑎≠1, then equation (1.2) has 2𝑘-periodic solutions.

Proof. Let 𝑧_𝑛=𝑥_𝑛𝑥_𝑛−𝑘,𝑛∈ℕ₀. (3.7) Then 𝑧_𝑛+1=𝑧_𝑛𝑎+𝑏𝑧_𝑛,𝑛∈ℕ₀. (3.8) Since 𝑎≠1, we see that equation (3.8) has equilibrium solution as follows: 𝑧_𝑛=𝑧=1−𝑎𝑏,𝑛∈ℕ₀. (3.9) For this solution of equation (3.8), we have 𝑥_𝑛=𝑧𝑥_𝑛−𝑘=𝑥_𝑛−2𝑘,𝑛≥𝑘, (3.10) from which the lemma follows.

Remark 3.3. Note that 2𝑘-periodic solutions in the previous lemma could be prime 2𝑘-periodic. For this it is enough to choose initial conditions 𝑥_−𝑖,𝑖=0,𝑘, such that the string 𝑥_−𝑘,𝑥_−𝑘+1,…,𝑥₋₁,𝑧𝑥_−𝑘,𝑧𝑥_−𝑘+1,…,𝑧𝑥₋₁ (3.11) is not periodic with a period less then 2𝑘.

Theorem 3.4. Let |𝑎|<1 and 𝑏≠0 in (1.2). Then, every well-defined solution (𝑥_𝑛)_{𝑛≥−𝑘} of equation (1.2) converges to a, not necessarily prime, 2𝑘-periodic solution of the equation.

Proof. First note that by Lemma 3.2, in this case, there are 2𝑘-periodic solutions of the equation. We know that in this case well-defined solutions of the equation are given by formula (2.7). From this and by using asymptotic formulae (3.1), we obtain that for sufficiently large 𝑚₁𝑥_{2𝑘𝑚+𝑖}=𝑥^𝑚_{𝑖−2𝑘𝑗=0}𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘=𝑥_𝑖−2𝑘𝐶𝑚₁^𝑚_𝑗=𝑚1+11+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘/𝑏𝑥₀𝑥_−𝑘1+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘/𝑏𝑥₀𝑥_−𝑘=𝑥_𝑖−2𝑘𝐶𝑚₁^𝑚_𝑗=𝑚1+11+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎^𝑘1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑂𝑎^4𝑗𝑘=𝑥_𝑖−2𝑘𝐶𝑚₁exp1−𝑎^𝑘1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘^{𝑚𝑗=𝑚1+1}𝑎^{(2𝑗−1)𝑘+𝑖}+𝑂𝑎^4𝑗𝑘, (3.12) where 𝐶𝑚₁=^𝑚1_𝑗=0𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘. (3.13)
From (3.12) and since |𝑎|<1, it easily follows that the sequences (𝑥_{2𝑘𝑚+𝑖})_𝑚∈ℕ0 are convergent for each 𝑖∈{𝑘,𝑘+1,…,3𝑘−1}, from which the theorem follows.

Theorem 3.5. Let |𝑎|>1 and 𝑏≠0 in (1.2). Then, every well-defined solution (𝑥_𝑛)_{𝑛≥−𝑘} of equation (1.2) converges to zero.

Proof. In this case, well-defined solutions of equation (1.2) are also given by formula (2.7). Further note that for each 𝑖∈{𝑘,𝑘+1,…,3𝑘−1} holds lim_𝑗→∞𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘=1𝑎^𝑘. (3.14) Now note that 1/|𝑎|^𝑘<1, due to the assumption |𝑎|>1. Using this fact and (3.14), it follows that for sufficiently large 𝑗, say 𝑗≥𝑚₂ we have 𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘≤1211+|𝑎|^𝑘. (3.15)
From this, we have 𝑥_{2𝑘𝑚+𝑖}=𝑥_𝑖−2𝑘𝐶𝑚₂^𝑚_𝑗=𝑚2+1𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘≤𝑥_𝑖−2𝑘𝐶𝑚₂^𝑚_𝑗=𝑚2+11211+|𝑎|^𝑘=𝑥_𝑖−2𝑘𝐶𝑚₂1211+|𝑎|^{𝑘𝑚−𝑚2}⟶0, (3.16) as 𝑚→∞, where 𝐶𝑚₂=^𝑚2_𝑗=0𝑏𝑥₀𝑥_−𝑘+𝑎^{(2𝑗−1)𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘𝑏𝑥₀𝑥_−𝑘+𝑎^{2𝑗𝑘+𝑖}1−𝑎−𝑏𝑥₀𝑥_−𝑘, (3.17) from which the theorem follows.

Theorem 3.6. Let 𝑎=−1, 𝑏≠0, and 𝑘 be even in (1.2). Then, every well-defined solution (𝑥_𝑛)_{𝑛≥−𝑘} of equation (1.2) is eventually periodic with, not necessarily prime, period 4𝑘.

Proof. From (2.2), in this case, we have 𝑦_𝑛=−𝑦_𝑛−1+𝑏=𝑦_𝑛−2,𝑛≥2, (3.18) which means that the sequence 𝑦_𝑛 is two-periodic, and consequently the sequence 𝑥_𝑛𝑥_𝑛−𝑘 is two-periodic. Hence 𝑥_2𝑛𝑥_2𝑛−𝑘=𝑥₀𝑥_−𝑘,𝑥_2𝑛+1𝑥_{2𝑛−𝑘+1}=𝑥₁𝑥_−𝑘+1, (3.19) from which it follows that 𝑥_2𝑛+𝑖=𝑥_𝑖𝑥_−𝑘+𝑖𝑥_{2𝑛−𝑘+𝑖}=𝑥_{2𝑛−2𝑘+𝑖},𝑛≥𝑘−𝑖+12,𝑖∈{0,1}, (3.20) that is, the sequences 𝑥_2𝑛 and 𝑥_2𝑛+1 are 2𝑘-periodic from which the result easily follows.

Theorem 3.7. Let 𝑎=−1, 𝑏≠0, 𝑘 be odd in (1.2), and (𝑥_𝑛)_{𝑛≥−𝑘} be a well-defined solution of equation (1.2). Then the following statements are true.(a)If 𝑥₀𝑥_−𝑘=2/𝑏, then the solution is 4𝑘-periodic.(b)If |𝑏𝑥₀𝑥_−𝑘−1|<1, then 𝑥_2𝑛→0 and |𝑥_2𝑛+1|→∞, as 𝑛→∞.(c)If |𝑏𝑥₀𝑥_−𝑘−1|>1, then 𝑥_2𝑛+1→0 and |𝑥_2𝑛|→∞, as 𝑛→∞.

Proof. As in Theorem 3.6, we obtain (3.18) and consequently (3.19) holds. If 𝑘=2𝑙+1 for some 𝑙∈ℕ₀, then from (1.2) and (3.19) we have 𝑥_2𝑛𝑥_{2𝑛−2𝑙−1}=𝑥₀𝑥_−(2𝑙+1),𝑥_2𝑛+1𝑥_{2𝑛−2𝑙}=𝑥₁𝑥_−2𝑙=𝑥₀𝑥_−(2𝑙+1)𝑏𝑥₀𝑥_−(2𝑙+1).−1 (3.21)
From (3.21) we obtain 𝑥_2𝑛=𝑥₀𝑥_−(2𝑙+1)𝑥_{2𝑛−(2𝑙+1)}=𝑏𝑥₀𝑥_−(2𝑙+1)−1𝑥_{2𝑛−(4𝑙+2)},𝑥_2𝑛+1=𝑥₁𝑥_−2𝑙𝑥_{2𝑛−2𝑙}=𝑥_{2𝑛+1−(4𝑙+2)}𝑏𝑥₀𝑥_−(2𝑙+1).−1 (3.22) From relation (3.22), the statements in this theorem easily follow.

Remark 3.8. Note that the case 𝑏𝑥₀𝑥_−𝑘−1=−1 is not possible in Theorem 3.7. Namely, in this case 𝑥₀𝑥_−𝑘=0, due to the assumption 𝑏≠0 and, as we can see from (1.2), the solution is not well-defined.

Acknowledgments

The second author is supported by Grant no. P201/10/1032 of the Czech Grant Agency (Prague) and by the Council of Czech Government Grant no. MSM 00 216 30519. The fourth author is supported by Grant no. FEKT-S-11-2-921 of the Faculty of Electrical Engineering and Communication, Brno University of Technology. This paper is partially also supported by the Serbian Ministry of Science Projects III 41025, III 44006, and OI 171007.