On a Class of Solvable Difference Equations

Abstract

A class of difference equations which can be solved in closed form is presented in this paper. Obtained formulas for the solutions are used in the study of the asymptotic behavior of their solutions in a particular case.

1. Introduction

One of the interesting problems in the theory of difference equations is to find the equations which can be solved in closed form. Such kind of equations are among the first ones which attracted attention of scientists. The reason for this is a natural expectation that obtained formulas for solutions can considerably facilitate investigation of their long-term behavior. For some classical results in the area, see, for example, book [1]. In [2], Stević solved, in an elegant way, the difference equation: 𝑥_𝑛+1=𝑏𝑥_𝑛−1𝑏+𝑐𝑥_𝑛𝑥_𝑛−1,𝑛∈ℕ₀, (1) explaining the formula in [3], which was given without any theoretical explanation. Several authors later considered some generalizations of (1) mostly by using Stević’s ideas and methods in [2] (see, e.g., [4–9] and the references therein). The idea for solving (1) was to reduce it to a nonhomogeneous linear first-order difference equation, which is solvable. Many equations and systems can be solved in a similar way (see, e.g., [4, 6–12]). For some related results and applications of linear first-order difference inequalities, see, for example, [13–15].

An interesting difference equation is the following: 𝑥_𝑛=𝑥_𝑛−𝑘𝑥_𝑛−𝑙𝑎𝑥_𝑛−𝑚+𝑏𝑥_𝑛−𝑠,𝑛∈ℕ₀, (2) where 𝑘,𝑙,𝑚, and 𝑠 are fixed natural numbers, 𝑎,𝑏∈ℝ⧵{0}, and the initial values 𝑥_−𝑖, 𝑖=1,𝜏, 𝜏∶=max{𝑘,𝑙,𝑚,𝑠} are real numbers.

Special cases of (2) frequently appear in the literature.

Example 1. The next equation 𝑥_𝑛=𝑥_𝑛−1𝑥_𝑛−22𝑥_𝑛−1+𝑥_𝑛−2,𝑛∈ℕ₀, (3) is one of them.
Equation (3) is a special case of (2) with 𝑘=𝑚,𝑙=𝑠, (4) or 𝑘=𝑠,𝑙=𝑚, (5) that is, of the equations 𝑥_𝑛=𝑥_𝑛−𝑘𝑥_𝑛−𝑙𝑎𝑥_𝑛−𝑘+𝑏𝑥_𝑛−𝑙,𝑛∈ℕ₀,𝑥_𝑛=𝑥_𝑛−𝑘𝑥_𝑛−𝑙𝑎𝑥_𝑛−𝑙+𝑏𝑥_𝑛−𝑘,𝑛∈ℕ₀. (6)(7)

Now note that (6) and (7) share a common important feature. Namely, assume that (𝑥_𝑛)_{𝑛≥−𝜏} is a well-defined solution of (6) or (7) (i.e., all its terms exist and are finite numbers). Then, if 𝑥_𝑛≠0, for every 𝑛≥−𝜏, by using the change of variables 𝑦_𝑛=1𝑥_𝑛, (8) (6) and (7) are, respectively, transformed into the following linear difference equations with constant coefficients: 𝑦_𝑛=𝑎𝑦_𝑛−𝑙+𝑏𝑦_𝑛−𝑘,𝑛∈ℕ₀,𝑦_𝑛=𝑏𝑦_𝑛−𝑙+𝑎𝑦_𝑛−𝑘,𝑛∈ℕ₀, (9)(10) which can be solved in closed form. As a consequence, (6) and (7), can be also solved in closed form. In fact, due to the known fact that polynomials of degree five and more need not be explicitly solved, we know that (9) and (10) can be theoretically solved. Nevertheless, we know some methods of how theoretically solutions of (9) and (10) can be found, as well as their form.

Note that for (3) the corresponding equation (9) is 𝑦_𝑛=𝑦_𝑛−1+2𝑦_𝑛−2. (11) The general solution of (11) is 𝑦_𝑛=𝑐₁2^𝑛+𝑐₂(−1)^𝑛, so that 𝑥_𝑛=1𝑐₁2^𝑛+𝑐₂(−1)^𝑛. (12)

A natural question and a corresponding problem are as follows.

Question 2. Are there any other values of 𝑘,𝑙,𝑚, and 𝑠 except those ones in (4) and (5), such that (2) is solvable in closed form?

Problem 3. In the affirmative case find their solutions in closed form.

The answer to the question is positive, and in recent paper [12], as a byproduct of our results therein, we have essentially shown that, in the case 𝑚=𝑘,𝑠=𝑘+𝑙, (13) (2) is solvable in closed form, and by using the described method therein, in some subcases of the case in (13) we have presented formulas for their solutions and use them in the study of their long-term behavior.

Here we continue this line of research by showing that there are some cases different from those ones in (4), (5), and (13), for which (2) is solvable in closed form.

2. Case 𝑙=𝑚=𝑘+𝑠

In this section we consider (2) in the case 𝑙=𝑚=𝑘+𝑠. (14) That is, we consider the next difference equation: 𝑥_𝑛=𝑥_𝑛−𝑘𝑥_{𝑛−𝑘−𝑠}𝑎𝑥_{𝑛−𝑘−𝑠}+𝑏𝑥_𝑛−𝑠,𝑛∈ℕ₀. (15)

First note that we may assume gcd(𝑘,𝑠)=1 (16) (the greatest common divisor of natural numbers 𝑘 and 𝑠). Otherwise, if gcd(𝑘,𝑠)=𝑟>1, then 𝑘=𝑟𝑘₁ and 𝑠=𝑟𝑠₁ for some 𝑘₁,𝑠₁∈ℕ such that gcd(𝑘₁,𝑠₁)=1. Since every 𝑛∈ℕ₀ has the form 𝑛=𝑚𝑟+𝑖, for some 𝑚∈ℕ₀ and 𝑖=0,𝑟−1, (15) can be written in the form 𝑥_{𝑚𝑟+𝑖}=𝑥_{𝑟(𝑚−𝑘1)+𝑖}𝑥_{𝑟(𝑚−𝑘1−𝑠1)+𝑖}𝑎𝑥_{𝑟(𝑚−𝑘1−𝑠1)+𝑖}+𝑏𝑥_{𝑟(𝑚−𝑠1)+𝑖}. (17)

Using the next change of variables 𝑥_𝑡^(𝑖)=𝑥_{𝑡𝑟+𝑖}, (18)𝑡≥−(𝑘₁+𝑠₁),𝑖=0,𝑟−1, in (17) we have that (𝑥_𝑡^(𝑖))_{𝑡≥−(𝑘1+𝑠1)}, 𝑖=0,𝑟−1, are 𝑟 independent solutions of the system 𝑥_𝑡=𝑥_𝑡−𝑘1𝑥_{𝑡−𝑘1−𝑠1}𝑎𝑥_{𝑡−𝑘1−𝑠1}+𝑏𝑥_𝑡−𝑠1,𝑡∈ℕ₀, (19) which is (15) with 𝑘₁ and 𝑠₁ instead of 𝑘 and 𝑠.

Hence, from now on we will assume that condition (16) holds.

2.1. A Case When Some of Initial Values Are Equal to Zero

Now assume that one of the initial values 𝑥_−𝑖, 𝑖=1,𝑘+𝑠, is equal to zero. If 𝑥_−𝑖=0 for some 1≤𝑖≤𝑘, then 𝑥_𝑘−𝑖 is not defined or 𝑥_𝑘−𝑖=0. If the later holds, then 𝑎𝑥_−𝑖+𝑏𝑥_𝑘−𝑖=0, which along with (15) implies that 𝑥_{𝑘+𝑠−𝑖} is not defined. Further, if 𝑥_−𝑖=0 for some 𝑘+1≤𝑖≤𝑘+𝑠, then 𝑥_{𝑘+𝑠−𝑖} is not defined or 𝑥_{𝑘+𝑠−𝑖}=0. If the later holds, then 𝑥_{2𝑘+𝑠−𝑖} is not defined or 𝑥_{2𝑘+𝑠−𝑖}=0. If 𝑥_{2𝑘+𝑠−𝑖}=0, then 𝑎𝑥_{𝑘+𝑠−𝑖}+𝑏𝑥_{2𝑘+𝑠−𝑖}=0, which along with (15) implies that 𝑥_{2𝑘+2𝑠−𝑖} is not defined.

On the other hand, if 𝑥_𝑛0=0 for some 𝑛₀∈ℕ₀, then from (15) we see that 𝑥_𝑛0−𝑘=0 or 𝑥_{𝑛0−𝑘−𝑠}=0. If 𝑥_𝑛0−𝑘=0, then using again (15) we see that 𝑥_{𝑛0−2𝑘}=0 or 𝑥_{𝑛0−2𝑘−𝑠}=0, while if 𝑥_{𝑛0−𝑘−𝑠}=0, then 𝑥_{𝑛0−2𝑘−𝑠}=0 or 𝑥_{𝑛0−2𝑘−2𝑠}=0. Repeating the procedure we see that there is an 𝑖₀∈{1,…,𝑘+𝑠} such that 𝑥_−𝑖0=0.

From all these considerations it follows that all solutions of (15) such that 𝑥_−𝑖=0 for some 𝑖∈{1,…,𝑘+𝑠} are not well defined and that if a solution of (15) has a zero term then it is not defined.

Hence, from now on we will consider only those solutions of (15) such that 𝑥_𝑛≠0, for every 𝑛≥−(𝑘+𝑠).

2.2. A Case When None of Initial Values Is Equal to Zero

Assume that (𝑥_𝑛)_{𝑛≥−(𝑘+𝑠)} is a solution of (15) such that 𝑥_−𝑖≠0 for every 𝑖∈{1,…,𝑘+𝑠}. Since in this case 𝑥_𝑛≠0 for every 𝑛≥−(𝑘+𝑠), then (15) can be written in the following form: 𝑥_𝑛𝑥_𝑛−𝑘=1𝑎+𝑏𝑥_𝑛−𝑠/𝑥_{𝑛−𝑘−𝑠},𝑛∈ℕ₀. (20)

The form suggests the next, natural, change of variables: 𝑦_𝑛=𝑥_𝑛𝑥_𝑛−𝑘,𝑛≥−𝑠, (21) which transforms (20) into the next difference equation of order 𝑠: 𝑦_𝑛=1𝑎+𝑏𝑦_𝑛−𝑠,𝑛∈ℕ₀. (22)

In fact (22) is essentially an equation of the first order. Indeed, if we use the following decomposition of indices 𝑛→𝑛𝑠+𝑗,  𝑗=0,𝑠−1, (22) can be written as 𝑦_𝑛^(𝑗)=1𝑎+𝑏𝑦^(𝑗)_𝑛−1,𝑛∈ℕ₀, (23) where 𝑦_𝑛^(𝑗)=𝑦_{𝑛𝑠+𝑗},𝑛≥−1,𝑗=0,𝑠−1. (24)

This means that the subsequences (𝑦_{𝑛𝑠+𝑗})_𝑛≥−1, 𝑗=0,𝑠−1, are 𝑠 independent solutions of the difference equation 𝑧_𝑛=1𝑎+𝑏𝑧_𝑛−1,𝑛∈ℕ₀, (25) of the first order.

Equation (25) is solvable, which along with (21) shows that (15) is solvable, giving another positive answer to Question 2. Indeed, (25) is a special case of the next difference equation: 𝑧_𝑛=𝛼𝑧_𝑛−1+𝛽𝛾𝑧_𝑛−1+𝛿,𝑛∈ℕ₀. (26)

Equation (26) is classical and can be solved in several ways. Since in the case 𝛾=0 (26) is a linear first-order difference equation, which is solvable, from now on we will assume that 𝛾≠0. By using the change of variables 𝑧_𝑛=𝑢_𝑛+1𝑢_𝑛+𝑐,𝑛∈ℕ₀, (27) into (26), after some calculations, we get 𝑢𝛾𝑐−𝛼^𝑛𝑢_𝑛−1𝑢+𝛾^𝑛+1𝑢_𝑛−1+𝑢𝛾𝑐+𝛿^𝑛+1𝑢_𝑛+𝑐𝛾𝑐+𝛿−𝛼𝑐−𝛽=0. (28)

By choosing 𝛿𝑐=−𝛾 (29) (here we use the assumption 𝛾≠0) we get 𝛾𝑢_𝑛+1−(𝛼+𝛿)𝑢_𝑛+𝛼𝛿−𝛽𝛾𝛾𝑢_𝑛−1=0,𝑛∈ℕ₀, (30) a second-order difference equation with constant coefficients.

It is interesting that, by using the change 𝑢_𝑛=𝑣_𝑛𝛾^𝑛, (31) (30) is transformed into the equation 𝑣_𝑛+1−(𝛼+𝛿)𝑣_𝑛+𝛼𝛿−𝛽𝛾𝑣_𝑛−1=0,𝑛∈ℕ₀. (32)

Let 𝜆₁ and 𝜆₂ be the characteristic roots of the characteristic polynomial: 𝑃(𝜆)=𝜆²−(𝛼+𝛿)𝜆+𝛼𝛿−𝛽𝛾 (33) associated with (32). Then by a well-known result on linear difference equations with constant coefficients and above transformations we get 𝑧_𝑛𝛿=−𝛾+1𝛾⋅𝑐₁𝜆₁^𝑛+1+𝑐₂𝜆₂^𝑛+1𝑐₁𝜆^𝑛₁+𝑐₂𝜆^𝑛₂, (34) if (𝛼+𝛿)²≠4(𝛼𝛿−𝛽𝛾), and 𝑧_𝑛𝛿=−𝛾+𝜆₁𝛾⋅𝑐₁̂+𝑐₂̂(𝑛+1)𝑐₁̂+𝑐₂̂𝑛, (35) if (𝛼+𝛿)²=4(𝛼𝛿−𝛽𝛾).

In our case we have that 𝛼=0, 𝛽=1, 𝛾=𝑏, and 𝛿=𝑎, so that (30) becomes 𝑏𝑢_𝑛+1−𝑎𝑢_𝑛−𝑢_𝑛−1=0,𝑛∈ℕ₀, (36) while the constant in (29) is equal to −𝑎/𝑏. The characteristic roots of the characteristic polynomial associated with (36) are 𝜆₁=𝑎+√𝑎²+4𝑏2𝑏,𝜆₂=𝑎−√𝑎²+4𝑏.2𝑏 (37) Hence, the general solution of (36) is 𝑢_𝑛=𝑐₁𝑎+√𝑎²+4𝑏2𝑏^𝑛+𝑐₂𝑎−√𝑎²+4𝑏2𝑏^𝑛, (38) if 𝑎²+4𝑏≠0, and 𝑢_𝑛=𝑐₁+𝑐₂𝑛𝑎2𝑏^𝑛=𝑐₁+𝑐₂𝑛−2𝑎^𝑛, (39) if 𝑎²+4𝑏=0.

If we choose initial values 𝑢₋₁ and 𝑢₀ such that 𝑢₋₁=1,𝑢₀=𝑧₋₁+𝑎𝑏, (40) then from (40) and (38) we get 𝑢_𝑛=1+𝜆₁𝑏𝑧₋₁+𝑎𝜆^𝑛₁−1+𝜆₂𝑏𝑧₋₁+𝑎𝜆^𝑛₂√𝑎²,+4𝑏𝑛≥−1, (41) if 𝑎²+4𝑏≠0, while from (40) and (39) we get 𝑢_𝑛=𝑧₋₁+𝑎𝑏+𝑛𝑧₋₁+𝑎𝑏+2𝑎−2𝑎^𝑛,𝑛≥−1, (42) if 𝑎²+4𝑏=0.

Using (41) and (42) in (27) we get 𝑧_𝑛𝑎=−𝑏+1+𝜆₁𝑏𝑧₋₁+𝑎𝜆₁^𝑛+1−1+𝜆₂𝑏𝑧₋₁+𝑎𝜆₂^𝑛+11+𝜆₁𝑏𝑧₋₁+𝑎𝜆^𝑛₁−1+𝜆₂𝑏𝑧₋₁+𝑎𝜆^𝑛₂,𝑛≥−1, (43) if 𝑎²+4𝑏≠0, and 𝑧_𝑛=4𝑎−2𝑎𝑧₋₁−4/𝑎+(𝑛+1)𝑧₋₁−2/𝑎𝑧₋₁−4/𝑎+𝑛𝑧₋₁,−2/𝑎𝑛≥−1, (44) if 𝑎²+4𝑏=0.

Using the fact that the sequences 𝑦_𝑛^(𝑗)=𝑦_{𝑛𝑠+𝑗}, 𝑛≥−1, 𝑗=0,𝑠−1, are solutions of (25), from formulas (43) and (44) it follows that𝑦_𝑛^(𝑗)𝑎=−𝑏+1+𝜆₁𝑏𝑦_𝑗−𝑠+𝑎𝜆₁^𝑛+1−1+𝜆₂𝑏𝑦_𝑗−𝑠+𝑎𝜆₂^𝑛+11+𝜆₁𝑏𝑦_𝑗−𝑠+𝑎𝜆^𝑛₁−1+𝜆₂𝑏𝑦_𝑗−𝑠+𝑎𝜆^𝑛₂,𝑛≥−1, (45)for 𝑗=0,𝑠−1, if 𝑎²+4𝑏≠0, and 𝑦_𝑛^(𝑗)=4𝑎−2𝑎𝑦_𝑗−𝑠−4/𝑎+(𝑛+1)𝑦_𝑗−𝑠−2/𝑎𝑦_𝑗−𝑠−4/𝑎+𝑛𝑦_𝑗−𝑠,−2/𝑎𝑛≥−1, (46) for 𝑗=0,𝑠−1, if 𝑎²+4𝑏=0.

From (21) we have that 𝑥_𝑛=𝑦_𝑛𝑥_𝑛−𝑘,𝑛≥−𝑠, (47) which implies that 𝑥_{𝑘𝑚1+𝑗1}=𝑦_{𝑘𝑚1+𝑗1}𝑥_{𝑘(𝑚1−1)+𝑗1}=𝑥_𝑗1^𝑚_−𝑘¹_𝑖=0𝑦_{𝑘𝑖+𝑗1}, (48) for 𝑚₁∈ℕ₀ and 𝑗₁=−𝑠,−𝑠+1,…,−𝑠+𝑘−1.

Although formulas (45)–(48) give solutions to (15) they are not easily applicable due to the assumption gcd(𝑘,𝑠)=1. Namely, subsequences (𝑦_𝑛^(𝑗)), 𝑗=0,𝑠−1, appearing in (45) and (46), and the subsequences (𝑦_{𝑘𝑚1+𝑗1}), 𝑗₁=−𝑠,−𝑠+1,…,−𝑠+𝑘−1, appearing in (48), are different decompositions of the sequence (𝑦_𝑛)_{𝑛≥−𝑠}, so they are not easily connected to each other.

However, if 𝑠=1 in (15), then they give complete solution of the equation. Namely, in this case (45) and (46) become 𝑦_𝑛𝑎=−𝑏+1+𝜆₁𝑏𝑦₋₁+𝑎𝜆₁^𝑛+1−1+𝜆₂𝑏𝑦₋₁+𝑎𝜆₂^𝑛+11+𝜆₁𝑏𝑦₋₁+𝑎𝜆^𝑛₁−1+𝜆₂𝑏𝑦₋₁+𝑎𝜆^𝑛₂,𝑛≥−1, (49) if 𝑎²+4𝑏≠0, and 𝑦_𝑛=4𝑎−2𝑦₋₁−4/𝑎+(𝑛+1)𝑦₋₁−2/𝑎𝑎𝑦₋₁−4/𝑎+𝑛𝑦₋₁−2/𝑎,𝑛≥−1, (50) if 𝑎²+4𝑏=0.

Using (49) and (50) in (48), we get𝑥_{𝑘𝑚1+𝑗1}=𝑥_𝑗1^𝑚_−𝑘¹_𝑖=0−𝑎𝑏+1+𝜆₁𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑖+𝑗1}₁⁺¹−1+𝜆₂𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑖+𝑗1}₂⁺¹1+𝜆₁𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑖+𝑗1}₁−1+𝜆₂𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑖+𝑗1}₂, (51)for each 𝑗₁∈{−1,0,…,𝑘−2}, and every 𝑚₁∈ℕ₀, if 𝑎²+4𝑏≠0, and𝑥_{𝑘𝑚1+𝑗1}=𝑥_𝑗1^𝑚_−𝑘¹_𝑖=04𝑎−2𝑥₋₁/𝑥_−1−𝑘−4/𝑎+𝑘𝑖+𝑗₁+1𝑥₋₁/𝑥_−1−𝑘−2/𝑎𝑎𝑥₋₁/𝑥_−1−𝑘−4/𝑎+𝑘𝑖+𝑗₁𝑥₋₁/𝑥_−1−𝑘,−2/𝑎 (52)for each 𝑗₁∈{−1,0,…,𝑘−2}, and every 𝑚₁∈ℕ₀, if 𝑎²+4𝑏=0.

3. Asymptotic Behavior of Solutions of (15) When 𝑠=1

In this section we will use formulas (51) and (52) in studying of the asymptotic behavior of solutions of (15) for the case when 𝑠=1.

The results are incorporated into two theorems. Before we formulate and prove the results we will present two auxiliary results.

Lemma 4. Assume that 𝑠=1, 𝑎,𝑏∈ℝ⧵{0}, and 𝑎+𝑏=1. Then difference equation (15) has a (not necessarily prime) 𝑘-periodic solution.

Proof. Since 𝑎+𝑏=1 it follows that 𝑎²+4𝑏≥0 (this is equivalent to (1+𝑏)²≥0). Further, in this case (25) has one or two stationary solutions. Really, if 𝑧_𝑛=𝑧, 𝑧≥−1, then it follows that 𝑏𝑧²+𝑎𝑧−1=0. That is, 𝑏𝑧²+(1−𝑏)𝑧−1=(𝑏𝑧+1)(𝑧−1)=0, from which it follows that 𝑧₁=1,𝑧₂1=−𝑏, (53) if 𝑎²+4𝑏>0, that is, if (1+𝑏)²>0, and 𝑧₁=𝑧₂=1 (54) if 𝑎²+4𝑏=0, that is, if 𝑏=−1. Hence, in both cases 𝑧_𝑛=1,𝑛≥−1, (55) is a stationary solution of (25).
By using (21), we see that, for the stationary solution of (25) given in (55), we have a solution of (15) such that 𝑥_𝑛=𝑥_𝑛−𝑘,𝑛≥−1, (56) from which the lemma follows.

Lemma 5. Assume that 𝑠=1, 𝑎,𝑏∈ℝ⧵{0}, and 𝑏−𝑎=1. Then difference equation (15) has a (not necessarily prime) 2𝑘-periodic solution.

Proof. First note that from the condition 𝑏−𝑎=1 we have also that 𝑎²+4𝑏≥0 (it is also equivalent to (1+𝑏)²≥0). On the other hand, in this case (25) also has one or two stationary solutions. Namely, if 𝑧_𝑛=𝑧, 𝑧≥−1, then it follows that 𝑏𝑧²+𝑎𝑧−1=0. That is, 𝑏𝑧²+(𝑏−1)𝑧−1=(𝑏𝑧−1)(𝑧+1)=0, from which we obtain 𝑧₁=−1,𝑧₂=1𝑏, (57) if 𝑎²+4𝑏>0, that is, if (1+𝑏)²>0, and 𝑧₁=𝑧₂=−1 (58) if 𝑎²+4𝑏=0, that is, if 𝑏=−1. Hence, in both cases 𝑧_𝑛=−1,𝑛≥−1, (59) is a stationary solution of (25).
By using (21), we see that the stationary solution of (25) given in (59) produces a solution of (15) such that 𝑥_𝑛=−𝑥_𝑛−𝑘, 𝑛≥−1, which implies that 𝑥_𝑛=𝑥_𝑛−2𝑘,𝑛≥𝑘−1, (60) from which the lemma follows.

Remark 6. By some calculation it is easy to see that −𝑎/𝑏+𝜆₁=1 if and only if 𝑎+𝑏=1 and 𝑏≥−1, while −𝑎/𝑏+𝜆₂=1 if and only if 𝑎+𝑏=1 and 𝑏≤−1.
Also, −𝑎/𝑏+𝜆₁=−1 if and only if 𝑏−𝑎=1 and 𝑏≤−1, while −𝑎/𝑏+𝜆₂=−1 if and only if 𝑏−𝑎=1 and 𝑏≥−1.

Theorem 7. Consider (15) with 𝑠=1. Assume that 𝑎,𝑏∈ℝ⧵{0}, 𝑎²+4𝑏>0, and 𝑥_−𝑖≠0,𝑖=1,𝑘+1. Let 𝜆∶=𝜆_𝑖, where |𝜆_𝑖|=max{|𝜆₁|,|𝜆₂|},𝑖∈{1,2}. Then the following statements are true.(a)If  |−𝑎/𝑏+𝜆|<1 and 1+𝜆_𝑖(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, for 𝑖=1,2, then 𝑥_𝑛→0, as 𝑛→∞.(b)If  |−𝑎/𝑏+𝜆₁|<1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, then 𝑥_𝑛→0, as 𝑛→∞.(c)If  |−𝑎/𝑏+𝜆₂|<1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, then 𝑥_𝑛→0, as 𝑛→∞.(d)If  |−𝑎/𝑏+𝜆|>1 and 1+𝜆_𝑖(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, for 𝑖=1,2, then |𝑥_𝑛|→∞, as 𝑛→∞.(e)If  |−𝑎/𝑏+𝜆₁|>1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, then |𝑥_𝑛|→∞, as 𝑛→∞.(f)If  |−𝑎/𝑏+𝜆₂|>1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, |𝑥_𝑛|→∞, as 𝑛→∞.(g)If  −𝑎/𝑏+𝜆=1 and 1+𝜆_𝑖(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, for 𝑖=1,2, then 𝑥_𝑛 converges to a (not necessarily prime) 𝑘-periodic solution of (15), as 𝑛→∞.(h)If  −𝑎/𝑏+𝜆₁=1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, then 𝑥_𝑛 is a (not necessarily prime) 𝑘-periodic solution of (15).(i)If  −𝑎/𝑏+𝜆₂=1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, then 𝑥_𝑛 is a (not necessarily prime) 𝑘-periodic solution of (15).(j)If  −𝑎/𝑏+𝜆=−1 and 1+𝜆_𝑖(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, for 𝑖=1,2, then 𝑥_𝑛 converges to a (not necessarily prime) 2𝑘-periodic solution of (15), as 𝑛→∞.(k)If  −𝑎/𝑏+𝜆₁=−1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, then 𝑥_𝑛 is a (not necessarily prime) 2𝑘-periodic solution of (15).(l)If  −𝑎/𝑏+𝜆₂=−1, 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, then 𝑥_𝑛 is a (not necessarily prime) 2𝑘-periodic solution of (15).

Proof. First note that since 𝑥_−𝑖≠0, 𝑖=1,𝑘+1, we have that 𝑥_𝑛≠0 for every 𝑛≥−(𝑘+1). Further note that the condition 𝑎²+4𝑏>0 implies that 𝜆₁,𝜆₂∈ℝ, 𝜆₁≠𝜆₂. If it were 𝜆₁=−𝜆₂, then it would be 𝑎=0, the case which is excluded. Hence we have that |𝜆₁|≠|𝜆₂|, from which it follows that min𝜆₁,𝜆₂<max𝜆₁,𝜆₂. (61)
Let𝑎^(𝑗1)_𝑚1𝑎∶=−𝑏+1+𝜆₁𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑚1+𝑗1}₁⁺¹−1+𝜆₂𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑚1+𝑗1}₂⁺¹1+𝜆₁𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑚1+𝑗1}₁−1+𝜆₂𝑏𝑥₋₁/𝑥_−1−𝑘+𝑎𝜆^{𝑘𝑚1+𝑗1}₂, (62)for 𝑚₁∈ℕ₀ and 𝑗₁=−1,0,…,𝑘−2.
Then if 1+𝜆_𝑖(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, for 𝑖=1,2, we have that for each 𝑗₁∈{−1,0,…,𝑘−2}lim_𝑚1→∞𝑎^(𝑗1)_𝑚1𝑎=−𝑏+𝜆. (63) From (51) and (63) the statements in (a) and (d) easily follow.
If 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0 and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0, then we have that for each 𝑗₁∈{−1,0,…,𝑘−2}𝑎^(𝑗1)_𝑚1𝑎=−𝑏+𝜆₁, (64) from which along with (51) the statements in (b) and (e) easily follow.
If 1+𝜆₁(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)=0 and 1+𝜆₂(𝑏(𝑥₋₁/𝑥_−1−𝑘)+𝑎)≠0, then we have that for each 𝑗₁∈{−1,0,…,𝑘−2}𝑎^(𝑗1)_𝑚1𝑎=−𝑏+𝜆₂, (65) from which along with (51) the statements in (c) and (f) follow.
(g), (j) in this case we have that for sufficiently large 𝑚₁𝑎^(𝑗1)_𝑚1𝑎=−𝑏𝜆+𝜆+𝑂min{₁,𝜆₂}max{𝜆₁,𝜆₂}^𝑘𝑚1 (66) for each 𝑗₁∈{−1,0,…,𝑘−2}, from which along with (51), Lemmas 4 and 5, and Remark 6 these two statements follow.
(h), (k) in this case we have that (64) holds, from which along with (51) these two statements follow.
(i), (l) in this case we have that (65) holds, from which along with (51) these two statements follow.

Theorem 8. Consider (15) with 𝑠=1. Assume that 𝑎,𝑏∈ℝ⧵{0}, 𝑎²+4𝑏=0, and 𝑥_−𝑖≠0, 𝑖=1,𝑘+1. Then the following statements are true.(a)If  |2/𝑎|<1, then 𝑥_𝑛→0, as 𝑛→∞.(b)If  |2/𝑎|>1, then |𝑥_𝑛|→∞, as 𝑛→∞.(c)If  |𝑎|=2, then 𝑥_𝑛→0, as 𝑛→∞.

Proof. In this case we have that 𝜆₁=𝜆₂=−2/𝑎. Let𝑏^(𝑗1)_𝑚14∶=𝑎−2𝑥₋₁/𝑥_−1−𝑘−4/𝑎+𝑘𝑚₁+𝑗₁+1𝑥₋₁/𝑥_−1−𝑘−2/𝑎𝑎𝑥₋₁/𝑥_−1−𝑘−4/𝑎+𝑘𝑚₁+𝑗₁𝑥₋₁/𝑥_−1−𝑘−2/𝑎, (67)for 𝑚₁∈ℕ₀ and 𝑗₁=−1,0,…,𝑘−2.
Then for each 𝑗₁∈{−1,0,…,𝑘−2} we have that lim_𝑚1→∞𝑏^(𝑗1)_𝑚1=2𝑎, (68) if 𝑥₋₁/𝑥_−1−𝑘≠2/𝑎, while 𝑏^(𝑗1)_𝑚1=2𝑎, (69) for 𝑗₁∈{−1,0,…,𝑘−2} and 𝑚₁∈ℕ₀, if 𝑥₋₁/𝑥_−1−𝑘=2/𝑎.
From (52), (68), and (69) the statements in (a) and (b) easily follow.
(c) in this case we have that for sufficiently large 𝑚₁𝑎^(𝑗1)_𝑚1=2𝑎−2𝑎𝑘𝑚₁1+𝑂𝑚²₁1=±1−𝑘𝑚₁1+𝑂𝑚²₁=±𝑒^{−(1/(𝑘𝑚1))+𝑂(1/𝑚21)}, (70) for each 𝑗₁∈{−1,0,…,𝑘−2}, from which along with (52) and the fact that ∑^𝑚1_𝑗=1(1/𝑗)→+∞ as 𝑚₁→∞, the statement easily follows.

Acknowledgments

This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, under Grant no. (11-130/1433 HiCi). The authors, therefore, acknowledge technical and financial support of KAU.