Eventually Periodic Solutions for Difference Equations with Periodic Coefficients and Nonlinear Control Functions

Abstract

For nonlinear difference equations of the form 𝑥_𝑛=𝐹(𝑛,𝑥_𝑛−1,…,𝑥_𝑛−𝑚), it is usually difficult to find periodic solutions. In this paper, we consider a class of difference equations of the form 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛𝑓(𝑥_𝑛−𝑘), where {𝑎_𝑛},{𝑏_𝑛} are periodic sequences and 𝑓 is a nonlinear filtering function, and show how periodic solutions can be constructed. Several examples are also included to illustrate our results.

1. Introduction

There are good reasons to find “eventually periodic solutions” of difference equations of the form 𝑥_𝑛=𝐹(𝑛,𝑥_𝑛−1,𝑥_𝑛−2,…,𝑥_𝑛−𝑚),𝑛∈{0,1,2,…}. (1.1) For instance, the well-known logistic population model 𝑥_𝑛=𝜆𝑥_𝑛−1(𝐿−𝑥_𝑛−1),𝑛∈{0,1,2,…} (1.2) is of the above form, and the study of the existence of its periodic solutions leads to chaotic solutions. As another example in [1], Chen considers the equation 𝑥_𝑛=𝑥_𝑛−1+𝑔(𝑥_{𝑛−𝑘−1}),𝑛∈{0,1,2,…}, (1.3) where 𝑘 is a nonnegative integer, and 𝑔∶𝐑→𝐑 is a McCulloch-Pitts type function 𝑔(𝜉)=−1,𝜉∈(𝜎,∞),1,𝜉∈(−∞,𝜎], (1.4) in which 𝜎∈𝐑 is a constant which acts as a threshold. Chen showed that all solutions of (1.3) are eventually periodic and pointed out that such a result may lead to more complicated dynamical behavior of a more general neural network. Recently, Zhu and Huang [2] discussed the periodic solutions of the following difference equation: 𝑥_𝑛=𝑎𝑥_𝑛−1+(1−𝑎)𝑓(𝑥_𝑛−𝑘),𝑛∈{0,1,2,…}, (1.5) where 𝑎∈(0,1),𝑘 is a positive integer, and 𝑓∶𝐑→𝐑 is a signal transmission function of the form (1.9). In particular, they obtained the following theorem.

Theorem A. Let 𝑝,𝑞∈{0,1,2,…}. If 𝜅∈(𝑎^𝑝+1,𝑎^𝑝(1−𝑎^𝑘−1)(1−𝑎^{𝑘+𝑝−1}))∩(1−𝑎^𝑞+𝑎^{𝑝+𝑞+𝑘}𝑎,1−^𝑞+1(1−𝑎^𝑘+𝑝)1−𝑎^{2𝑘+𝑝+𝑞}), (1.6) then (1.5) has an eventually (2𝑘+𝑝+𝑞)-periodic solution -periodic solution {𝑥_𝑛}^∞_𝑛=−𝑘.

In this paper, we consider the following delay difference equation: 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛𝑓(𝑥_𝑛−𝑘),𝑛∈{0,1,2,…}, (1.7) where {𝑎_𝑛}^∞_𝑛=0 and {𝑏_𝑛}^∞_𝑛=0 are positive 𝜔-periodic sequences such that 𝑎_𝑛+𝑏_𝑛≤1 for 𝑛≥0.

The integer 𝑘 is assumed to satisfy 𝑘=𝑙𝜔+1, (1.8) for some nonnegative integer 𝑙. The function 𝑓 can be chosen in a number of ways. Here, 𝑓 is a filtering function of the form 𝑓(𝑥)=1,𝑥∈(0,𝜅],0,𝑥∈(−∞,0]∪(𝜅,∞), (1.9) where the positive number 𝜅 can be regarded as a threshold term. Therefore, if 𝜔=1, then 𝑎_𝑛=𝑎,𝑏_𝑛=𝑏, and 𝑘=𝑙+1 so that (1.7) reduces to 𝑥_𝑛=𝑎𝑥_𝑛−1+𝑏𝑓(𝑥_{𝑛−𝑙−1}), (1.10) which includes (1.5) as a special case.

When 𝑙=0, we have 𝑥_𝑛=𝑎𝑥_𝑛−1+𝑏𝑓(𝑥_𝑛−1), (1.11) which will also be included in the following discussions.

Let Ω denote the set of real finite sequences of the form {𝜙_−𝑘,𝜙_−𝑘+1,…,𝜙₋₁}. Given 𝜙={𝜙_−𝑘,…,𝜙₋₁}∈Ω, if we let 𝑥_−𝑘=𝜙_−𝑘,…,𝑥₋₁=𝜙₋₁, then we may compute 𝑥₀,𝑥₁,… successively from (1.7) in a unique manner. Such a sequence 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 is called a solution of (1.7) determined by 𝜙∈Ω. Recall that a positive integer 𝜂 is a period of the sequence {𝑥_𝑛}^∞_𝑛=−𝑘 if 𝑥_𝜂+𝑛=𝑥_𝑛 for all 𝑛≥−𝑘 and that 𝜏 is the least period of {𝑥_𝑛}^∞_𝑛=−𝑘 if 𝜏 is the least among all periods of {𝑥_𝑛}^∞_𝑛=−𝑘. The sequence {𝑥_𝑛}^∞_𝑛=−𝑘 is said to be 𝜏-periodic if 𝜏 is the least period of {𝑥_𝑛}^∞_𝑛=−𝑘. In case {𝑥_𝑛}^∞_𝑛=−𝑘 is not periodic, it may happen that for some 𝑁≥−𝑘, the subsequence {𝑥_𝑛}^∞_𝑛=𝑁 is 𝜏-periodic. Such a sequence is said to be eventually 𝜏-periodic. In other words, let us call {𝑦_𝑗}^∞_𝑗=−𝑘 a translate of {𝑥_𝑛}^∞_𝑛=−𝑘 if 𝑦_𝑗=𝑥_{𝑗+𝑁+𝑘} for 𝑗∈{−𝑘,−𝑘+1,…}, where 𝑁 is some integer greater than or equal to −𝑘. Then, {𝑥_𝑛}^∞_𝑛=−𝑘 is eventually 𝜏-periodic if one of its translates is 𝜏-periodic.

We will seek eventually periodic solutions of (1.7). This is a rather difficult question since the existence question depends on the sequences {𝑎_𝑛},{𝑏_𝑛}, the “delay” 𝑘, and the control term 𝜅.

Throughout this paper, empty sums are taken to be 0 and empty products to be 1. We will also need the following elementary facts. If the real sequence {𝑥_𝑛}^∞_𝑛=−1 satisfies the recurrence relation 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛,𝑛∈{0,1,2,…}, (1.12) then 𝑥₀=𝑎₀𝑥₋₁+𝑏₀,𝑥₁=𝑎₁𝑥₀+𝑏₁=𝑎₁(𝑎₀𝑥₋₁+𝑏₀)+𝑏₁=𝑎₁𝑎₀𝑥₋₁+𝑎₁𝑏₀+𝑏₁,𝑥₂=𝑎₂𝑥₁+𝑏₂=𝑎₂(𝑎₁𝑎₀𝑥₋₁+𝑎₁𝑏₀+𝑏₁)+𝑏₂=𝑎₂𝑎₁𝑎₀𝑥₋₁+𝑎₂𝑎₁𝑏₀+𝑎₂𝑏₁+𝑏₂, (1.13) and by induction, 𝑥_𝑛=𝛼_0,𝑛𝑥₋₁+𝛼_0,𝑛𝛼_0,0𝑏₀+𝛼_0,𝑛𝛼_0,1𝑏₁𝛼+⋯+^0,𝑛𝛼_0,𝑛𝑏_𝑛=𝛼_0,𝑛(𝑥₋₁+𝑏₀𝛼_0,0+𝑏₁𝛼_0,1𝑏+⋯+^𝑛𝛼_0,𝑛), (1.14) where 𝛼_0,𝑗=^𝑗_𝑛=0𝑎_𝑛,𝑗∈{0,1,2,…}. (1.15) Since {𝑎_𝑛} and {𝑏_𝑛} are positive 𝜔-periodic sequences, we see further that 𝛼_{0,𝑚𝜔+𝑖}=(𝛼_0,𝜔−1)^𝑚𝛼_0,𝑖,𝑖∈{0,…,𝜔−1};𝑚∈{0,1,2,…}, (1.16) that ^{𝑚𝜔+𝑖}_𝑗=0𝑏_𝑗𝛼_0,𝑗𝑏=(⁰𝛼_0,0𝑏+⋯+^𝜔−1𝛼_0,𝜔−1𝑏)+(^𝜔𝛼_0,𝜔𝑏+⋯+^2𝜔−1𝛼_0,2𝜔−1)𝑏+⋯+(^{(𝑚−1)𝜔}𝛼_{0,(𝑚−1)𝜔}𝑏+⋯+^𝑚𝜔−1𝛼_{0,𝑚𝜔−1}𝑏)+(^𝑚𝜔𝛼_0,𝑚𝜔𝑏+⋯+^{𝑚𝜔+𝑖}𝛼_{0,𝑚𝜔+𝑖})𝑏=(⁰𝛼_0,0𝑏+⋯+^𝜔−1𝛼_0,𝜔−11){1+𝛼_0,𝜔−11+⋯+(𝛼_0,𝜔−1)^𝑚−1}+1(𝛼_0,𝜔−1)^𝑚{𝑏₀𝛼_0,0𝑏+⋯+^𝑖𝛼_0,𝑖} (1.17) for 𝑖∈{0,…,𝜔−1} and 𝑚∈{0,1,2,…}, and that 𝑥_{𝑚𝜔+𝑖}=𝛼_{0,𝑚𝜔+𝑖}(𝑥₋₁+^{𝑚𝜔+𝑖}_𝑗=0𝑏_𝑗𝛼_0,𝑗)=(𝛼_0,𝜔−1)^𝑚𝛼_0,𝑖𝑥₋₁+𝛼_0,𝜔−11−𝛼^𝑚_0,𝜔−11−𝛼_0,𝜔−1𝛼_0,𝑖𝛽_0,𝜔−1+𝛼_0,𝑖𝛽_0,𝑖 (1.18) for 𝑖∈{0,…,𝜔−1} and 𝑚∈{0,1,2,…}, where 𝛽_0,𝑗=^𝑗_𝑘=0𝑏_𝑘𝛼_0,𝑘,𝑗∈{0,1,…,𝜔−1}. (1.19)

2. Convergence of Solutions

The filtering function 𝑓 will return 0 for inputs that fall below 0 or above the threshold constant 𝜅. For this reason, we will single out some subsets of Ω as follows: Ω₋={{𝜙_−𝑘,…,𝜙₋₁}∈Ω∣𝜙_𝑖Ω≤0,−𝑘≤𝑖≤−1},_∗={{𝜙_−𝑘,…,𝜙₋₁}∈Ω∣0<𝜙_𝑖Ω≤𝜅,−𝑘≤𝑖≤−1},₊={{𝜙_−𝑘,…,𝜙₋₁}∈Ω∣𝜙_𝑖>𝜅,−𝑘≤𝑖≤−1}. (2.1)

Let 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 be the solution of (1.7) determined by 𝜙∈Ω₋. By (1.7), 𝑥₀=𝑎₀𝑥₋₁+𝑏₀𝑓(𝑥_−𝑘)=𝑎₀𝑥₋₁𝑥≤0,₁=𝑎₁𝑥₀+𝑏₁𝑓(𝑥_−𝑘+1)=𝑎₁𝑥₀=𝑎₁𝑎₀𝑥₋₁≤0. (2.2) By induction, we may see that 𝑥_𝑛=𝑎_𝑛𝑎_𝑛−1⋯𝑎₁𝑎₀𝑥₋₁≤0,𝑛∈{0,1,2,…}. (2.3) Since 0≤lim_𝑛→∞𝑎₀𝑎₁⋯𝑎_𝑛−1𝑎_𝑛≤lim_𝑛→∞(max{𝑎₀,𝑎₁,…,𝑎_𝜔−1})^𝑛+1=0, (2.4) we see that lim_𝑛→∞𝑥_𝑛=0.

Next, let 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 be the solution of (1.7) determined by 𝜙∈Ω_∗. If 𝜅≥1, then by (1.7), 0<𝑥₀=𝑎₀𝑥₋₁+𝑏₀≤𝑎₀𝜅+𝑏₀=𝑎₀𝜅−𝑎₀+𝑎₀+𝑏₀≤𝑎₀(𝜅−1)+1≤𝜅,0<𝑥₁=𝑎₁𝑥₀+𝑏₁≤𝑎₁(𝑎₀𝜅+𝑏₀)+𝑏₁=𝑎₁𝜅+𝑏₁≤𝜅. (2.5) By induction, we see that 0<𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛≤𝑎_𝑛𝜅+𝑏_𝑛≤𝜅,𝑛∈{0,1,2,…}. (2.6) By (1.7), we see that 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛𝑓(𝑥_𝑛−𝑘)=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛,𝑛∈{0,1,2,…}. (2.7) In view of (1.18), we see further that lim_𝑚→∞𝑥_{𝑚𝜔+𝑖}=𝐴_𝑖,𝑖∈{0,1,…,𝜔−1}, (2.8) where 𝐴_𝑖=𝛼_0,𝑖(𝛼_0,𝜔−1𝛽_0,𝜔−11−𝛼_0,𝜔−1+𝛽_0,𝑖),𝑖∈{0,1,…,𝜔−1}. (2.9)

Next, let 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 be the solution of (1.7) determined by 𝜙∈Ω₊. Then, by (1.7), 𝑥₀=𝑎₀𝑥₋₁+𝑏₀𝑓(𝑥_−𝑘)=𝑎₀𝑥₋₁, (2.10) and by induction, 𝑥_𝑛=𝑎_𝑛𝑎_𝑛−1⋯𝑎₀𝑥₋₁,𝑛∈{0,1,2,…}. (2.11) Although 𝑥₋₁>𝜅, since (2.4) holds, we see that {𝑥_𝑛} is a strictly decreasing sequence tending to 0. Hence, there is a nonnegative integer 𝑗 such that 𝑥_𝑗−1>𝜅 but 𝑥_𝑗≤𝜅. Then, 𝜅≥𝑥_𝑗>𝑥_𝑗+1>𝑥_𝑗+2>⋯>𝑥_{𝑗+𝑘−1}. If we let 𝜙={𝑥_𝑗,𝑥_𝑗+1,…,𝑥_{𝑗+𝑘−1}}, then 𝜙∈Ω_∗. If 𝜅≥1, then by what we have shown above, the solution {𝑥^{0𝑥002𝑑𝑐}_𝑛} of (1.7) determined by 𝜙 satisfies lim𝑥_𝑚→∞^{0𝑥002𝑑𝑐}_{𝑚𝜔+𝑖}=𝐴_𝑖 for 𝑖∈{0,1,…,𝜔−1}. By uniqueness, ^0𝑥0𝑥^02𝑑𝑐_𝑛=𝑥_{𝑛+𝑗+𝑘} for 𝑛≥0. In other words, the translate {𝑥^{0𝑥002𝑑𝑐}_𝑛} of the solution {𝑥_𝑛}^∞_𝑛=−𝑘 satisfies lim𝑥_𝑚→∞^{0𝑥002𝑑𝑐}_{𝑚𝜔+𝑖}=𝐴_𝑖 for 𝑖∈{0,1,…,𝜔−1}.

We summarize the above discussions by means of the following result.

Lemma 2.1. A solution 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 determined by 𝜙∈Ω₋ will tend to 0; and if 𝜅≥1, then a solution 𝑥={𝑥_𝑛}^∞_𝑛=−𝑘 determined by 𝜙∈Ω_∗∪Ω₊ will satisfy (2.8) or one of its translates will satisfy it.

Lemma 2.2. If 0<𝜅<min{1,max{𝐴₀,𝐴₁,…,𝐴_𝜔−1}}, then for any solution {𝑥_𝑛} of (1.7) determined by a 𝜙∈Ω_∗∪Ω₊, there exists an integer 𝑚∈{0,1,…} such that {𝑥_𝑚−𝑘,…,𝑥_𝑚−1}∈Ω_∗ and 𝑥_𝑚∈(𝜅,1).

Proof. First let {𝑥_𝑛}^∞_𝑛=−𝑘 be the solution of (1.7) determined by a 𝜙∈Ω_∗. If 𝑥_𝑛∈(0,𝜅] for all 𝑛∈{−𝑘,−𝑘+1,…}, then 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛𝑓(𝑥_𝑛−𝑘)=𝑎_𝑛𝑥_𝑛+𝑏_𝑛,𝑛∈{0,1,2,…}, (2.12) so that by (1.18), we see that lim_𝑚→∞𝑥_{𝑚𝜔+𝑖}=𝐴_𝑖,𝑖∈{0,1,…,𝜔−1}. (2.13) But, this is contrary to our assumption that 0<𝜅<min{1,max{𝐴₀,𝐴₁,…,𝐴_𝜔−1}}. Hence, there is some nonnegative integer 𝑚 such that 𝑥_𝑛∈(0,𝜅] for 𝑛∈{−𝑘,−𝑘+1,…,𝑚−1} but 𝑥_𝑚∈(−∞,0]∪(𝜅,∞). Note that 𝑥_𝑚=𝑎_𝑚𝑥_𝑚−1+𝑏_𝑚𝑓(𝑥_𝑚−𝑘)>0, (2.14) which implies that 𝑥_𝑚∈(𝜅,∞). Moreover, since 𝑥_𝑚−1∈(0,𝜅]⊂(0,1), we then have 𝑥_𝑚=𝑎_𝑚𝑥_𝑚−1+𝑏_𝑚<𝑎_𝑚+𝑏_𝑚≤1, (2.15) so that 𝑥_𝑚∈(𝜅,1).
Next, let {𝑥_𝑛}^∞_𝑛=−𝑘 be the solution of (1.7) determined by a 𝜙∈Ω₊. As seen in the discussions immediately preceding Lemma 2.1, there is a nonnegative integer 𝑗 such that {𝑥_𝑗,𝑥_𝑗+1,…,𝑥_{𝑗+𝑘−1}}∈Ω_∗. If 𝑥_𝑛∈(0,𝜅] for all 𝑛∈{𝑗,𝑗+1,…}, then as we have just explained, a translate {𝑥^{0𝑥002𝑑𝑐}_𝑛} of {𝑥_𝑛} will satisfy lim𝑥_𝑚→∞^{0𝑥002𝑑𝑐}_{𝑚𝜔+𝑖}=𝐴_𝑖,𝑖∈{0,1,…,𝜔−1}. (2.16) This is again a contradiction. Hence, we may conclude our proof in a manner similar to the above discussions. The proof is complete.

From the proof of Lemma 2.2, we see that if 𝜅∈(0,min{1,max{𝐴₀,𝐴₁,…,𝐴_𝜔−1}}), then to study the limiting behavior of a solution {𝑥_𝑛}^∞_𝑛=−𝑘 determined by 𝜙 in Ω_∗∪Ω₊, we may assume without loss of generality that 𝜙∈Ω_∗ and 𝑥₀∈(𝜅,1). As an example, let us consider (1.11), where we recall that 𝑎,𝑏>0 and 𝑎+𝑏≤1.

Example 2.3. Let 𝑎𝑏/(1−𝑎²)≤𝜅<𝑏/(1−𝑎²). Then, (1.11) has a 2-periodic solution {𝑥_𝑛}^∞_𝑘=−1 with 𝑥₋₁∈(0,𝜅] and 𝑥₀∈(𝜅,1). Indeed, let us choose 𝑥₋₁=𝑎𝑏/(1−𝑎²) (and hence, 𝑥₀=𝑏/(1−𝑎²)). Then, 0<𝑥₋₁=𝑎𝑏1−𝑎²≤𝜅,𝜅<𝑥₀=𝑎𝑥₋₁𝑏+𝑏=1−𝑎²<1. (2.17) Furthermore, 𝑥₁=𝑎𝑥₀=𝑎𝑏1−𝑎²𝑥∈(0,𝜅],₂=𝑎𝑥₁+𝑏=𝑎⋅𝑎𝑏1−𝑎²𝑏+𝑏=1−𝑎²=𝑥₀, (2.18) so that 𝑥₁=𝑥₃=𝑥₅=⋯ and 𝑥₂=𝑥₄=𝑥₆=⋯ and 𝑥₁≠𝑥₂.

3. Existence of Eventually Periodic Solutions

Recall that 𝐺^[0](𝑢)=𝑢,𝐺^[1](𝑢)=𝐺(𝑢),𝐺^[2](𝑢)=(𝐺∘𝐺)(𝑢)=𝐺(𝐺(𝑢)),…,𝐺^[𝑗](𝑢)=𝐺(𝐺^[𝑗−1](𝑢)) are the zeroth, first, second, and so forth and the 𝑗th iterate of the function 𝐺(𝑢). Also, recall the fact that if {𝑢_𝑛}^∞_𝑛=0 is a sequence that satisfies 𝑢_𝑛+1=𝐺(𝑢_𝑛),𝑛∈{0,1,2,…}, (3.1) then {𝑢_𝑛} is a 𝜏-periodic sequence if and only if 𝑢₀=𝐺^[𝜏](𝑢₀𝑢),₀≠𝐺^[𝑗](𝑢₀),𝑗=1,2,…,𝜏−1. (3.2) For convenience, denote 𝛼_𝑛=^𝑛_𝑗=1𝑎_𝑗,𝛽_𝑛=^𝑛_𝑗=1𝑏_𝑗𝛼_𝑗,𝑛∈{1,2,…}. (3.3) Since 𝛼_𝑛𝛽_𝑛+𝛼_𝑛=𝑎₁⋯𝑎_𝑛+𝑎₂⋯𝑎_𝑛𝑏₁+𝑎₃⋯𝑎_𝑛𝑏₂+⋯+𝑏_𝑛≤1, (3.4) we see that 𝛼_𝑛𝛽_𝑛1−𝛼_𝑛≤1,𝑛∈{1,2,…}. (3.5)

Theorem 3.1. Let 𝑘=𝑙𝜔+1,𝑝=𝜏𝜔−1, and 𝑞=𝜎𝜔−1, where 𝑙,𝜏,𝜎∈{1,2,…,𝑘−1}. Let 𝐼₁(𝑝)=[𝛼^𝜏_𝜔(𝛼^𝑙_𝜔+(1−𝛼^𝑙_𝜔)1−𝛼_𝜔𝛼_𝜔𝛽_𝜔𝛼),^𝑝(1−𝛼^𝑙_𝜔)𝛼_𝜔𝛽_𝜔(1−𝛼_𝑝𝛼^𝑙_𝜔)(1−𝛼_𝜔)𝐼),₂𝛼(𝑝,𝑞)=[𝑀,^{𝜔𝜏+𝜎+𝑙}(1−𝛼^𝑙_𝜔)+(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔), (3.6) where 𝑀=max{𝛼_𝑛𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑛𝛽_𝑛∶𝑛∈{0,1,…,𝑞}}. (3.7) If 𝜅∈𝐼₁(𝑝)∩𝐼₂(𝑝,𝑞) and 𝛼0<𝜅<min{^𝑛𝛽_𝑛1−𝛼_𝑛∶𝑛∈{1,2,…,𝑘−1}}, (3.8) then (1.7) has an eventually (2𝑘+𝑝+𝑞)-periodic solution {𝑥_𝑛}^∞_𝑛=−𝑘 (which can be explicitly generated).

Proof. From the condition that 𝑙,𝜏,𝜎∈{1,2,…,𝑘−1}, we have 𝑘−1≥𝜔. By (3.5), we see that 1≥𝐴₀=𝛼_0,0(𝛼_0,𝜔−1𝛽_0,𝜔−11−𝛼_0,𝜔−1+𝛽_0,0)=11−𝛼_𝜔(𝑎₀𝛼_𝜔𝛽_0,𝜔−1+𝑏₀−𝑏₀𝛼_𝜔)=11−𝛼_𝜔(𝑎₀𝛼_𝜔(𝑏₀𝑎₀+𝑏₁𝑎₀𝑎₁𝑏+⋯+^𝜔−1𝑎₀⋯𝑎_𝜔−1)+𝑏₀−𝑏₀𝛼_𝜔)=𝛼_𝜔𝛽_𝜔1−𝛼_𝜔>𝜅. (3.9) Hence, 𝜅<max{𝐴₀,𝐴₁,…,𝐴_𝜔−1}. Thus, 0<𝜅<min{1,max{𝐴₀,𝐴₁,…,𝐴_𝜔−1}}. By Lemmas 2.1 and 2.2, we may look for our desired eventually periodic solution {𝑥_𝑛}^∞_𝑛=−𝑘 determined by 𝜙∈Ω_∗ such that 𝑥₀∈(𝜅,1).
Define 𝑔_𝑛(𝑢)=𝛼_𝑛𝑢+𝛼_𝑛𝛽_𝑛ℎfor𝑛∈{0,1,2,…},_𝑛(𝑢)=𝑎_𝑛𝑢for𝑛∈{0,1,2,…}, (3.10) and the mapping 𝑔 by 𝑔(𝑥)=(𝑔_𝑞+1∘(ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙]∘𝑔_𝑘−1)(𝑥). (3.11) We will show that 𝑔(𝑥)=𝛼_𝜔^{𝜏+𝜎+𝑙}(𝛼^𝑙_𝜔𝛼𝑥+^𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔𝛼)+^𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔, (3.12) and that 𝑔 maps 𝐷₀=(𝜅,1) into 𝐷₀ with a fixed point 𝑥^∗∈𝐷₀, where 𝑥^∗=𝛽_𝜔𝛼_𝜔^{𝜏+𝜎+𝑙+1}(1−𝛼^𝑙_𝜔)+𝛽_𝜔𝛼_𝜔(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔). (3.13) The first assertion is easy to show. Indeed, since 𝑔_𝑘−1(𝑥)=𝛼_𝑘−1𝑥+𝛼_𝑘−1𝛽_𝑘−1,(ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙](𝑥)=(𝑎_𝜔⋯𝑎₁)^𝜏+𝑙𝑥=𝛼_𝜔^𝜏+𝑙𝑔𝑥,_𝑞+1(𝑥)=𝛼_𝑞+1𝑥+𝛼_𝑞+1𝛽_𝑞+1, (3.14) we see that ((ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙]∘𝑔_𝑘−1)(𝑥)=𝛼_𝜔^𝜏+𝑙(𝛼_𝑘−1𝑥+𝛼_𝑘−1𝛽_𝑘−1),𝑔(𝑥)=𝛼_𝑞+1𝛼_𝜔^𝜏+𝑙(𝛼_𝑘−1𝑥+𝛼_𝑘−1𝛽_𝑘−1)+𝛼_𝑞+1𝛽_𝑞+1=𝛼_𝜔^{𝜏+𝜎+𝑙}(𝛼^𝑙_𝜔𝛼𝑥+^𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔𝛼)+^𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔. (3.15)
We now show the second assertion. Note that the linear maps 𝑔_𝑛 and ℎ_𝑛 satisfy 𝑔_𝑚𝜔(𝑢)=𝛼_𝑚𝜔𝑢+𝛼_𝑚𝜔𝛽_𝑚𝜔=𝛼^𝑚_𝜔𝑢+(1−𝛼^𝑚_𝜔)1−𝛼_𝜔𝛼_𝜔𝛽_𝜔ℎ,𝑚∈{0,1,2,…},_𝑚𝜔∘ℎ_𝑚𝜔−1∘⋯∘ℎ₁(𝑢)=𝛼^𝑚_𝜔𝑢,𝑚∈{0,1,2,…}. (3.16) Let 𝑔_𝑛(𝐷₀)=𝐷_𝑛 for 𝑛∈{1,…,𝑘−1}. Since 𝜙∈Ω_∗ and 𝑥₀∈𝐷₀, it is clear that the solution {𝑥_𝑛} of (1.7) satisfies 𝑥_𝑛=𝑔_𝑛(𝑥₀),𝑛∈{1,…,𝑘−1}. (3.17) Moreover, it is easy to prove that 𝐷_𝑛=(𝑔_𝑛(𝜅),𝑔_𝑛(1)),𝑛∈{1,…,𝑘−1}. (3.18) Indeed, we have 𝜅<𝛼_𝑛𝜅+𝛼_𝑛𝛽_𝑛=𝑔_𝑛(𝜅)<𝛼_𝑛+𝛼_𝑛𝛽_𝑛=𝑔_𝑛(1)<𝛼_𝑛+𝛽_𝑛≤1,𝑛∈{1,2,…,𝑘−1}. (3.19) That is, 𝐷_𝑛⊂𝐷₀ holds for all 𝑛∈{0,…,𝑘−1}. Let 𝑛₁ be the largest integer such that 𝑥_𝑛>𝜅 for 𝑛∈{0,1,…,𝑛₁+𝑘−1}. Then, from (1.7), we can obtain 𝑥_{𝑛+𝑘−1}=𝑎_{𝑛+𝑘−1}⋯𝑎_𝑘(𝛼^𝑙_𝜔𝑥₀+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔),𝑛∈{1,2,…,𝑛₁+𝑘}, (3.20) which implies that 𝑥_{𝑛+𝑘−1}∈𝐷_{𝑛+𝑘−1} for 𝑛∈{1,2,…,𝑛₁+𝑘}, where 𝐷_{𝑛+𝑘−1}=𝑎_𝑛⋯𝑎₁𝑔_𝑘−1(𝐷₀)=(𝛼_𝑛(𝛼^𝑙_𝜔𝛼𝜅+^𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔),𝛼_𝑛(𝛼^𝑙_𝜔+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔)). (3.21) Since 𝜅∈𝐼₁(𝑝), we have 𝛼𝜅<^𝑝(1−𝛼^𝑙_𝜔)𝛼_𝜔𝛽_𝜔(1−𝛼_𝑝𝛼^𝑙_𝜔)(1−𝛼_𝜔), (3.22) that is, 𝜅<𝛼_𝑝(𝛼^𝑙_𝜔𝜅+(1−𝛼^𝑙_𝜔)𝛼_𝜔𝛽_𝜔(1−𝛼_𝜔))<𝛼_𝑝−1(𝛼^𝑙_𝜔𝜅+(1−𝛼^𝑙_𝜔)𝛼_𝜔𝛽_𝜔(1−𝛼_𝜔))<⋯<𝛼^𝑙_𝜔𝜅+(1−𝛼^𝑙_𝜔)(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔,𝛼_𝑝(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)<𝛼_𝑝−1(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)<⋯<𝛼^𝑙_𝜔+(1−𝛼^𝑙_𝜔)(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔≤1, (3.23) which shows that 𝐷_{𝑛+𝑘−1}⊂𝐷₀ for 𝑛∈{0,1,…,𝑝}. Thus, 𝑛₁≥𝑝 and 𝑥_{𝑛+𝑘−1}∈𝐷_{𝑛+𝑘−1}⊂(0,𝜅]for𝑛∈{𝑝+1,…,𝑝+𝑘}. (3.24) In fact, from 𝜅∈𝐼₁(𝑝), we have 𝑥_𝑝+𝑘=𝑎_𝑝+𝑘𝑥_{𝑝+𝑘−1}+𝑏_𝑝+𝑘𝑓(𝑥_𝑝)=𝑎_𝑝+𝑘𝑥_{𝑝+𝑘−1}=𝑎_𝑝+𝑘𝑎_{𝑝+𝑘−1}𝑥_{𝑝+𝑘−2}=⋯=𝑎_𝑝+𝑘⋯𝑎_𝑘(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)=𝛼_𝑝+1(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)=𝛼^𝜏_𝜔(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)𝑥≤𝜅,_𝑝+𝑘+1=𝑎_𝑝+𝑘+1𝑥_𝑝+𝑘+𝑏_𝑝+𝑘+1𝑓(𝑥_𝑝+1)=𝑎_𝑝+𝑘+1𝑥_𝑝+𝑘<𝜅, (3.25) and, by induction, 𝑥_{𝑝+2𝑘−1}=𝑎_{𝑝+2𝑘−1}𝑥_{𝑝+2𝑘−2}+𝑏_{𝑝+2𝑘−1}𝑓(𝑥_{𝑝+𝑘−1})=𝑎_𝑝+2𝑘+1𝑥_{𝑝+2𝑘−2}<𝜅. (3.26) Then, it is easy to see that 𝑛₁=𝑝.
Taking 𝑛=𝑝+𝑘 in (3.20), we have 𝑥_{2𝑘+𝑝−1}=𝑎_{2𝑘+𝑝−1}⋯𝑎_𝑘𝑔_𝑘−1(𝑥₀)=𝑎_𝑘+𝑝⋯𝑎₁𝑔_𝑘−1(𝑥₀)=𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔𝑥₀+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔). (3.27) Let 𝑛₂ be the largest integer such that 𝑥_{𝑛+2𝑘+𝑝−1}∈(0,𝜅] for 𝑛∈{0,1,…,𝑛₂}. Then, it follows from (1.7) that 𝑥_{𝑛+2𝑘+𝑝−1}=^{𝑛+2𝑘+𝑝−1}_{𝑗=2𝑘+𝑝}𝑎_𝑗𝑥_{2𝑘+𝑝−1}+^{𝑛+2𝑘+𝑝−1}_{𝑗=2𝑘+𝑝}𝑎_𝑗^{𝑛+2𝑘+𝑝−1}_{𝑗=2𝑘+𝑝}𝑏_𝑗𝑎_2𝑘+𝑝⋯𝑎_𝑗=𝛼_𝑛𝑥_{2𝑘+𝑝−1}+𝛼_𝑛𝛽_𝑛=𝛼_𝑛𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑛𝛽_𝑛=𝑔_𝑛(𝑥_{2𝑘+𝑝−1}) (3.28) for 𝑛∈{1,2,…,𝑛₂+𝑘}. This implies that 𝑥_{𝑛+2𝑘+𝑝−1}∈𝐷_{𝑛+2𝑘+𝑝−1} for 𝑛∈{1,2,…,𝑛₂+𝑘}, where 𝐷_{𝑛+2𝑘+𝑝−1}=(𝑔_𝑛(ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙]𝑔_𝑘−1)(𝐷₀).
Substituting (3.21) with 𝑛₁=𝑝 into (3.28), we have 𝐷_{𝑛+2𝑘+𝑝−1}=(𝑔_𝑛∘(ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙]𝑔_𝑘−1(𝜅),𝑔_𝑛(ℎ_𝜔∘⋯∘ℎ₁)^[𝜏+𝑙]𝑔_𝑘−1(1))=(𝛼_𝑛𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔𝜅+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑛𝛽_𝑛,𝛼_𝑛𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑛𝛽_𝑛) (3.29) for 𝑛∈{1,2,…,𝑛₂+𝑘}. Since 𝜅∈𝐼₂(𝑝,𝑞), we have 𝛼_𝑛𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑛𝛽_𝑛≤𝜅for𝑛∈{0,1,…,𝑞}. (3.30) From (3.29), we further have 𝑥_{𝑛+2𝑘+𝑝−1}∈𝐷_{𝑛+2𝑘+𝑝−1}⊂(0,𝜅]for𝑛∈{0,1,…,𝑞}. (3.31) By (3.8), (3.24), (3.28), and (3.31) as well as 𝜅∈𝐼₂(𝑝,𝑞), we have 𝑥_{2𝑘+𝑝+𝑞}=𝑎_{2𝑘+𝑝+𝑞}𝑥_{2𝑘+𝑝+𝑞−1}+𝑏_{2𝑘+𝑝+𝑞}=𝑎_{2𝑘+𝑝+𝑞}(𝛼_𝑞𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝑞𝛽_𝑞)+𝑏_{2𝑘+𝑝+𝑞}=𝑎_𝑞+1(𝛼_𝑞𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝑎_𝑞+1𝛼_𝑞𝛽_𝑞)+𝑏_𝑞+1=𝛼_𝜔^{𝜏+𝜎+𝑙}(𝛼^𝑙_𝜔𝑥₀+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+1−𝛼^𝜎_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔>𝛼_𝜔^{𝜏+𝜎+𝑙}(𝛼^𝑙_𝜔𝜅+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+1−𝛼^𝜎_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔𝑥>𝜅,_{2𝑘+𝑝+𝑞+1}=𝑎_{2𝑘+𝑝+𝑞+1}𝑥_{2𝑘+𝑝+𝑞}+𝑏_{2𝑘+𝑝+𝑞+1}>𝑎₁𝜅+𝑏₁𝑥>𝜅,_{2𝑘+𝑝+𝑞+2}=𝑎_{2𝑘+𝑝+𝑞+2}𝑥_{2𝑘+𝑝+𝑞+1}+𝑏_{2𝑘+𝑝+𝑞+2}>𝑎₂(𝑎₁𝜅+𝑏₁)+𝑏₂⋮𝑥>𝜅,_{2𝑘+𝑝+𝑞+𝑘−1}=𝑎_{2𝑘+𝑝+𝑞+𝑘−1}𝑥_{2𝑘+𝑝+𝑞+𝑘−2}+𝑏_{2𝑘+𝑝+𝑞+𝑘−1}=^{2𝑘+𝑝+𝑞+𝑘−1}_{𝑗=2𝑘+𝑝+𝑞+1}𝑎_𝑗𝑥_{2𝑘+𝑝+𝑞}+^{2𝑘+𝑝+𝑞+𝑘−1}_{𝑗=2𝑘+𝑝+𝑞+1}𝑎_𝑗^{2𝑘+𝑝+𝑞+𝑘−1}_{𝑗=2𝑘+𝑝+𝑞+1}𝑏_𝑗𝑎_{2𝑘+𝑝+𝑞+1}⋯𝑎_𝑗=𝛼_𝑘−1𝑥_{2𝑘+𝑝+𝑞}+𝛼_𝑘−1𝛽_𝑘−1>𝛼_𝑘−1𝜅+𝛼_𝑘−1𝛽_𝑘−1>𝜅. (3.32)
Hence,𝑥_{𝑛+2𝑘+𝑝−1}∈𝐷_{𝑛+2𝑘+𝑝−1}⊂𝐷₀for𝑛∈{𝑞+1,…,𝑞+𝑘}, (3.33) which implies that 𝑛₂=𝑞. In particular, taking 𝑛=𝑞+1 in (3.33) and (3.28), we have, respectively, 𝑥_{2𝑘+𝑝+𝑞}∈𝐷_{2𝑘+𝑝+𝑞}⊂𝐷₀,𝑥_{2𝑘+𝑝+𝑞}=𝑔(𝑥₀)=𝛼_𝜔^{𝜏+𝜎+𝑙}(𝛼^𝑙_𝜔𝑥₀+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔𝛼)+^𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔. (3.34)
Since 𝑔 is a linear map sending 𝐷₀ into 𝐷₀, then it is easy to see that it has a unique fixed point 𝑥^∗ in 𝐷₀ which satisfies (3.13).
Next, we assert that there is a 𝜙^∗∈Ω_∗ such that the solution {𝑥_𝑛} determined by 𝜙^∗ satisfies 𝑥₀=𝑥^∗, and that {𝑥_𝑛} is a periodic solution of (1.7) with minimal period 2𝑘+𝑝+𝑞. To see this, we choose 𝜙₋₁=(𝑥^∗−𝑏₀)/𝑎₀ and arbitrary 𝜙₋₂,…,𝜙_−𝑘∈(0,𝜅]. Then, clearly, the solution {𝑥_𝑛} of (1.7) determined by 𝜙_−𝑘,…,𝜙₋₁ will satisfy 𝑥₀=𝑥^∗. Furthermore, we may show that 𝑥₋₁=𝜙₋₁∈(0,𝜅]. Indeed, from 𝛼_𝜔^{𝜏+𝜎+𝑙}+𝛼_𝜔>𝛼^𝜎_𝜔+𝛼_𝜔^{𝜏+𝜎+2𝑙+1}, (3.35) we have 𝛼_𝜔^{𝜏+𝜎+𝑙}(1−𝛼^𝑙_𝜔)+(1−𝛼^𝜎_𝜔)>1−𝛼_𝜔−𝛼_𝜔^{𝜏+𝜎+2𝑙}+𝛼_𝜔^{𝜏+𝜎+2𝑙+1},𝛼_𝜔^{𝜏+𝜎+𝑙}(1−𝛼^𝑙_𝜔)+(1−𝛼^𝜎_𝜔)1−𝛼_𝜔^{𝜏+𝜎+2𝑙}>(1−𝛼_𝜔𝛼)=^𝜔𝛽_𝜔(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔>𝑏₀(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔, (3.36) hence, 𝑥^∗=𝛽_𝜔𝛼_𝜔^{𝜏+𝜎+𝑙+1}(1−𝛼^𝑙_𝜔)+𝛽_𝜔𝛼_𝜔(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔)>𝑏₀. (3.37) Thus, 𝜙₋₁=(𝑥^∗−𝑏₀)/𝑎₀>0. Next, from 0≤𝛼_𝜔^{𝜏+𝜎+2𝑙}−𝛼_𝜔^{𝜏+𝜎+3𝑙}, (3.38) we get 𝛼_𝜔^𝜏+𝑙−1≤𝛼_𝜔^𝜏+2𝑙+𝛼_𝜔^𝜏+𝑙(1−𝛼^𝑙_𝜔)−1−𝛼_𝜔^{2𝜏+𝜎+4𝑙}−𝛼_𝜔^{2𝜏+𝜎+3𝑙}(1−𝛼^𝑙_𝜔)+𝛼_𝜔^{𝜏+𝜎+2𝑙}, (3.39) so that 𝛼_𝜔^𝜏+𝑙−11−𝛼_𝜔^{𝜏+𝜎+2𝑙}≤𝛼_𝜔^𝜏+2𝑙+𝛼_𝜔^𝜏+𝑙(1−𝛼^𝑙_𝜔𝛼)−1,1+^{𝜔𝜏+𝜎+𝑙}−𝛼^𝜎_𝜔1−𝛼_𝜔^{𝜏+𝜎+2𝑙}≤𝛼_𝜔^{𝜎+𝜏+2𝑙}+𝛼_𝜔^{𝜎+𝜏+𝑙}(1−𝛼^𝑙_𝜔)+1−𝛼^𝜎_𝜔,𝛼_𝜔^{𝜏+𝜎+𝑙+1}(1−𝛼^𝑙_𝜔)+𝛼_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔^{𝜏+𝜎+2𝑙}≤𝛼_𝜔^{𝜎+𝜏+2𝑙+1}+𝛼_𝜔^{𝜎+𝜏+𝑙+1}(1−𝛼^𝑙_𝜔)+𝛼_𝜔(1−𝛼^𝜎_𝜔𝛼),_{𝜔𝜏+𝜎+𝑙+1}(1−𝛼^𝑙_𝜔)+𝛼_𝜔(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔)≤𝛼_𝜔^{𝜎+𝜏+2𝑙+1}1−𝛼_𝜔+𝛼_𝜔^{𝜎+𝜏+𝑙+1}(1−𝛼^𝑙_𝜔)1−𝛼_𝜔+𝛼_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔. (3.40) On the other hand, by (3.5), we have 𝛼_𝜔1−𝛼_𝜔≤1𝛽_𝜔, (3.41) so that 𝛼_𝜔^{𝜏+𝜎+𝑙+1}𝛽_𝜔(1−𝛼^𝑙_𝜔)+𝛼_𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔)≤𝛼_𝜔^{𝜎+𝜏+2𝑙}+𝛼_𝜔^{𝜎+𝜏+𝑙+1}𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔+𝛼_𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔=𝛼_𝜔^{𝜎+𝜏+𝑙}(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔𝛼)+^𝜔𝛽_𝜔(1−𝛼^𝜎_𝜔)1−𝛼_𝜔=𝛼_𝜔^{𝜎+𝜏+𝑙}(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝛼_𝜎𝜔𝛽_𝜎𝜔=𝑎₀𝛼_𝑞𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔1−𝛼_𝜔𝛼_𝜔𝛽_𝜔)+𝑎₀𝛼_𝑞𝛽_𝑞+𝑏₀. (3.42) In view of our assumption that 𝜅∈𝐼₂(𝑝,𝑞), we may now see that 𝑥₋₁≤𝜅.
In view of the above discussions, we see that 0<𝑥_𝑛≤𝜅 for 𝑛∈{−𝑘,…,−1},𝑥_𝑛>𝜅 for 𝑛∈{0,…,𝑝+𝑘−1}, and 0<𝑥_𝑛≤𝜅 for 𝑛∈{𝑝+𝑘,…,𝑝+2𝑘+𝑞−1}. Since 𝑥^∗ is the unique fixed point of 𝑔(𝑥) in 𝐷₀, we have 𝑔(𝑥^∗)=𝑥^∗,𝑔^[2](𝑥^∗)=𝑥^∗,…,𝑔^[𝑛](𝑥^∗)=𝑥^∗, and so forth, and hence, 𝑥_{2𝑘+𝑝+𝑞}=𝑔(𝑥^∗)=𝑥^∗,𝑥_{2𝑘+𝑝+𝑞+1}=𝑎_{2𝑘+𝑝+𝑞+1}𝑥_{2𝑘+𝑝+𝑞}+𝑏_{2𝑘+𝑝+𝑞+1}𝑓(𝑥_{𝑘+𝑝+𝑞+1})=𝑎₁𝑥^∗+𝑏₁=𝑥₁⋮𝑥>𝜅,_{𝑝+2𝑘+𝑞+𝑘−1}=𝑎_{𝑝+2𝑘+𝑞+𝑘−1}𝑥_{𝑝+2𝑘+𝑞+𝑘−2}+𝑏_{𝑝+2𝑘+𝑞+𝑘−1}𝑓(𝑥_{𝑝+𝑞+2𝑘−1})=𝑎_𝑘−1𝑥_{𝑝+2𝑘+𝑞+𝑘−2}+𝑏_𝑘−1=𝑎_𝑘−1𝑥_𝑘−2+𝑏_𝑘−1=𝑥_𝑘−1𝑥>𝜅,_{𝑝+2𝑘+𝑞+𝑘}=𝑎_{𝑝+2𝑘+𝑞+𝑘}𝑥_{𝑝+2𝑘+𝑞+𝑘−1}+𝑏_{𝑝+2𝑘+𝑞+𝑘}𝑓(𝑥_{𝑝+𝑞+2𝑘})=𝑎_𝑘𝑥_𝑘−1=𝑥_𝑘𝑥>𝜅,_{𝑝+2𝑘+𝑞+𝑘+1}=𝑎_{𝑝+2𝑘+𝑞+𝑘+1}𝑥_{𝑝+2𝑘+𝑞+𝑘}+𝑏_{𝑝+2𝑘+𝑞+𝑘+1}𝑓(𝑥_{𝑝+𝑞+2𝑘+1})=𝑎_{𝑝+2𝑘+𝑞+𝑘+1}𝑥_{𝑝+2𝑘+𝑞+𝑘}=𝑎_𝑘+1𝑥_𝑘=𝑥_𝑘+1⋮𝑥>𝜅,_{𝑝+2𝑘+𝑞+𝑘+𝑝−1}=𝑎_{𝑝+2𝑘+𝑞+𝑘+𝑝−1}𝑥_{𝑝+2𝑘+𝑞+𝑘+𝑝−2}+𝑏_{𝑝+2𝑘+𝑞+𝑘+𝑝−1}𝑓(𝑥_{𝑝+𝑞+2𝑘+𝑝−1})=𝑎_{𝑝+2𝑘+𝑞+𝑘+𝑝−1}𝑥_{𝑝+2𝑘+𝑞+𝑘+𝑝−2}=𝑎_{𝑘+𝑝−1}𝑥_{𝑘+𝑝−2}=𝑥_{𝑘+𝑝−1}𝑥>𝜅,_{𝑝+2𝑘+𝑞+𝑘+𝑝}=𝑎_{𝑧𝑝+2𝑘+𝑞+𝑘+𝑝}𝑥_{𝑝+2𝑘+𝑞+𝑘+𝑝−1}+𝑏_{𝑝+2𝑘+𝑞+𝑘+𝑝}𝑓(𝑥_{𝑝+𝑞+2𝑘+𝑝})=𝑎_𝑝+𝑘𝑥_{𝑝+𝑘−1}+𝑏_𝑝+𝑘𝑓(𝑥_𝑝)=𝑎_𝑘+𝑝𝑥_{𝑘+𝑝−1}=𝑥_𝑘+𝑝⋮𝑥≤𝜅,_{2𝑘+𝑝+𝑞+2𝑘+𝑝+𝑞−1}=𝑎_{2𝑘+𝑝+𝑞−1}𝑥_{2𝑘+𝑝+𝑞+2𝑘+𝑝+𝑞−2}+𝑏_{𝑝+2𝑘+𝑞−1}𝑓(𝑥_{2𝑘+𝑝+𝑞+𝑘+𝑝+𝑞−1})=𝑎_{2𝑘+𝑝+𝑞−1}𝑥_{2𝑘+𝑝+𝑞−2}+𝑏_{2𝑘+𝑝+𝑞−1}=𝑥_{2𝑘+𝑝+𝑞−1}𝑥≤𝜅,_{2(2𝑘+𝑝+𝑞)}=𝑔^[2](𝑥^∗)=𝑔(𝑥_{2𝑘+𝑝+𝑞})=𝑔(𝑥^∗)=𝑥^∗, (3.43) and so forth. Thus, 𝑥_𝑛>𝜅for𝑛∈{0,…,𝑝+𝑘−1},0<𝑥_𝑛𝑥≤𝜅for𝑛∈{𝑝+𝑘,…,𝑝+2𝑘+𝑞−1},_𝑛>𝜅for𝑛∈{𝑝+2𝑘+𝑞,…,𝑝+2𝑘+𝑞+𝑝+𝑘−1},0<𝑥_𝑛≤𝜅for𝑛∈{𝑝+2𝑘+𝑞+𝑝+𝑘,…,2(𝑝+2𝑘+𝑞)−1}, (3.44) and so forth.
By induction, we may see that 𝑥_𝑛>𝜅 for 𝑛∈{𝑚(𝑝+2𝑘+𝑞),…,𝑚(𝑝+2𝑘+𝑞)+𝑝+𝑘−1},0<𝑥_𝑛≤𝜅 for 𝑛∈{𝑚(𝑝+2𝑘+𝑞)+𝑝+𝑘,…,(𝑚+1)(𝑝+2𝑘+𝑞)−1}, where 𝑚∈{0,1,2,…}, and 𝑥_{𝑛(2𝑘+𝑝+𝑞)}=𝑥^∗,𝑥_{𝑛(2𝑘+𝑝+𝑞)+1}=𝑥₁,…,𝑥_{𝑛(2𝑘+𝑝+𝑞)+2𝑘+𝑝+𝑞−1}=𝑥_{2𝑘+𝑝+𝑞−1}. This shows that {𝑥_𝑛} is an eventually periodic solution of (1.7), whose minimal period is 2𝑘+𝑝+𝑞. The proof is complete.

We remark that in the above result, 𝑙 cannot be 0. We may, however, show the following by similar considerations.

Theorem 3.2. Let 𝑘=1,𝐼₁𝛼=[^3𝜔𝛽_𝜔(1−𝛼³_𝜔),𝛼_2𝜔−1𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔𝐼),₂𝛼=[𝑀,^2𝜔−1𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔), (3.45) where 𝑀=max{𝛼_𝑛𝛼³_𝜔𝛽_𝜔1−𝛼³_𝜔+𝛼_𝑛𝛽_𝑛∶𝑛∈{0,1,…,𝜔−1}}. (3.46) If 𝜅∈𝐼₁∩𝐼₂ and 𝛼0<𝜅<min{^𝑛𝛽_𝑛1−𝛼_𝑛∶𝑛∈{1,2,…,𝜔}}, (3.47) then (1.7) has an eventually 3𝜔-periodic solution {𝑥_𝑛}^∞_𝑛=−𝑘 (which can be generated explicitly).

Proof. Similar to the proof of the Theorem 3.1, set (3.10) and define the mapping 𝑔 by 𝑔(𝑥)=𝑔_𝜔∘(ℎ_𝜔∘ℎ_𝜔−1∘⋯∘ℎ₁)^[2](𝑥). (3.48) We may show that 𝑔(𝑥)=𝛼_𝜔(𝛼²_𝜔𝑥)+𝛼_𝜔𝛽_𝜔, (3.49) and that 𝑔 maps 𝐷₀=(𝜅,1) into 𝐷₀ with a unique fixed point 𝑥^∗∈𝐷₀, where 𝑥^∗=𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔. (3.50) Let us choose 𝑥₋₁=1𝑎₀(𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔−𝑏₀). (3.51) By 𝛼_𝜔>𝛼_𝜔(1−𝛼³_𝜔𝑏)>⁰𝛼_𝜔𝛽_𝜔𝛼_𝜔(1−𝛼³_𝜔𝑏)=⁰𝛽_𝜔(1−𝛼³_𝜔), (3.52) we have 𝛼_𝜔𝛽_𝜔(1−𝛼³_𝜔)>𝑏₀, (3.53) and hence, 𝑥₋₁=1𝑎₀(𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔−𝑏₀)>0. (3.54) Since 𝛼³_𝜔(𝑏₀+𝛼_𝜔𝛽_𝜔−1)=𝛼⁴_𝜔𝛽_𝜔, (3.55) then 𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔−𝑏₀=𝛼⁴_𝜔𝛽_𝜔1−𝛼³_𝜔+𝛼_𝜔𝛽_𝜔−1, (3.56) and hence, 𝑥₋₁=1𝑎₀(𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔−𝑏₀)=𝛼_𝜔−1(𝛼³_𝜔𝛽_𝜔1−𝛼³_𝜔)+𝛼_𝜔−1𝛽_𝜔−1𝑥≤𝜅,₀=𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔𝑥>𝜅,₁=𝑎₁𝑥₀=𝑎₁𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔⋮𝑥>𝜅,_2𝜔−1=𝑎_2𝜔−1⋯𝑎₁𝛼_𝜔𝛽_𝜔1−𝛼³_𝜔𝑥>𝜅,_2𝜔=𝛼³_𝜔𝛽_𝜔1−𝛼³_𝜔⋮𝑥≤𝜅,_3𝜔−1=𝛼_𝜔−1𝛼³_𝜔𝛽_𝜔1−𝛼³_𝜔+𝛼_𝜔−1𝛽_𝜔−1𝑥≤𝜅,_3𝜔=𝛼⁴_𝜔𝛽_𝜔1−𝛼³_𝜔+𝛼_𝜔𝛽_𝜔=𝑥₀,⋮ (3.57) so that {𝑥_𝑛} is an eventually 3𝜔-periodic solution of the system (1.7).

4. Examples and Remarks

Let {𝑎_𝑛},{𝑏_𝑛} be 2-periodic sequences, 𝑘=3,𝑝=1,𝑞=1, and 𝐼₁=[𝛼₂(𝛼₂+𝛼₂𝛽₂𝑎),¹𝛼₂𝛽₂1−𝑎₁𝛼₂𝐼),₂=[𝛼²₂(𝛼₂+𝛼₂𝛽₂𝛼),³²+11−𝛼⁴₂𝛼₂𝛽₂)∩[𝛼₁𝛼²₂(𝛼₂+𝛼₂𝛽₂)+𝛼₁𝛽₁,𝛼³₂+11−𝛼⁴₂𝛼₂𝛽₂). (4.1) Suppose 𝜅∈𝐼₁∩𝐼₂ and 𝛼0<𝜅<min{¹1−𝛼₁𝛽₁,𝛼₂1−𝛼₂𝛽₂}. (4.2) Consider the following “delay” difference equation: 𝑥_𝑛=𝑎_𝑛𝑥_𝑛−1+𝑏_𝑛𝑓(𝑥_𝑛−3),𝑛∈{0,1,2,…}. (4.3) We can check that (4.3) has an eventually 8-periodic solution {𝑥_𝑛}^∞_𝑛=−3 with 𝑥₀∈(𝜅,1).

In fact, as in the proof of Theorem 3.1, let 𝑥^∗=𝛼⁴₂𝛽₂+𝛼₂𝛽₂1−𝛼⁴₂,𝜙₋₁=𝑥^∗−𝑏₀𝑎₀, (4.4) and 𝜙₋₂,𝜙₋₃ be arbitrary numbers in (0,𝜅]. Then, as shown in the proof of Theorem 3.1, the solution of (4.3) determined by 𝜙₋₃,𝜙₋₂,𝜙₋₁ satisfies 𝑥₋₁=𝜙₋₁∈(0,𝜅] and 𝑥₀=𝑥^∗.

Since 𝜅∈𝐼₂, we have 𝑥₀>𝜅. On the other hand, by (3.3), and (𝛼³₂+1)/(1+𝛼²₂)(1+𝛼₂)<1, hence 𝜅<𝑥₀𝑥<1,₁=𝑎₁𝑥₀+𝑏₁𝑓(𝑥₋₂)=𝑎₁𝑥₀+𝑏₁>𝑎₁𝜅+𝑏₁𝑥>𝜅,₂=𝑎₂𝑥₁+𝑏₂𝑓(𝑥₋₁)=𝑎₂𝑥₁+𝑏₂=𝑎₂(𝑎₁𝑥₀+𝑏₁)+𝑏₂=𝑎₁𝑎₂𝑥₀+𝑎₂𝑏₁+𝑏₂=𝛼₂𝑥₀+𝛼₂𝛽₂>𝛼₂𝜅+𝛼₂𝛽₂𝑥>𝜅,₃=𝑎₃𝑥₂+𝑏₃𝑓(𝑥₀)=𝑎₃𝑥₂>𝑎₁(𝛼₂𝜅+𝛼₂𝛽₂𝑥)>𝜅,₄=𝑎₄𝑥₃+𝑏₄𝑓(𝑥₁)=𝑎₄𝑥₃=𝑎₂𝑎₁(𝛼₂𝑥₀+𝛼₂𝛽₂)=𝛼₂(𝛼₂𝑥₀+𝛼₂𝛽₂)<𝛼₂(𝛼₂+𝛼₂𝛽₂𝑥)≤𝜅,₅=𝑎₅𝑥₄+𝑏₅𝑓(𝑥₂)=𝑎₅𝛼₂(𝛼₂𝑥₀+𝛼₂𝛽₂)=𝑎₁𝛼₂(𝛼₂𝑥₀+𝛼₂𝛽₂)<𝛼₂(𝛼₂+𝛼₂𝛽₂𝑥)≤𝜅,₆=𝑎₆𝑥₅+𝑏₅𝑓(𝑥₃)=𝑎₆𝑥₅=𝑎₂𝑎₁𝛼₂(𝛼₂𝑥₀+𝛼₂𝛽₂)=𝛼²₂(𝛼₂𝑥₀+𝛼₂𝛽₂)<𝛼²₂(𝛼₂+𝛼₂𝛽₂𝑥)≤𝜅,₇=𝑎₇𝑥₆+𝑏₇𝑓(𝑥₄)=𝑎₇𝛼²₂(𝛼₂𝑥₀+𝛼₂𝛽₂)+𝑏₇=𝛼₁𝛼²₂(𝛼₂𝑥₀+𝛼₂𝛽₂)+𝑏₁=𝛼₁𝛼²₂(𝛼₂𝑥₀+𝛼₂𝛽₂)+𝛼₁𝛽₁<𝛼₁𝛼²₂(𝛼₂+𝛼₂𝛽₂)+𝛼₁𝛽₁𝑥≤𝜅,₈=𝑎₈𝑥₇+𝑏₈𝑓(𝑥₅)=𝑎₈𝑥₇+𝑏₈=𝑎₈(𝛼₁𝛼²₂(𝛼₂𝑥₀+𝛼₂𝛽₂)+𝛼₁𝛽₁)+𝑏₈=𝛼³₂(𝛼₂𝑥₀+𝛼₂𝛽₂)+𝛼₂𝛽₂=𝛼³₂(𝛼₂𝛼⁴₂𝛽₂+𝛼₂𝛽₂1−𝛼⁴₂+𝛼₂𝛽₂)+𝛼₂𝛽₂=𝛼⁴₂𝛽₂+𝛼₂𝛽₂1−𝛼⁴₂=𝑥₀,𝑥₉=𝑎₉𝑥₈+𝑏₉𝑓(𝑥₆)=𝑎₉𝑥₈+𝑏₉=𝑎₉𝑥₀+𝑏₉=𝑎₁𝑥₀+𝑏₁=𝑥₁,𝑥₁₀=𝑎₁₀𝑥₉+𝑏₁₀𝑓(𝑥₇)=𝑎₁₀𝑥₉+𝑏₁₀=𝑎₂𝑥₉+𝑏₂=𝑎₂𝑥₁+𝑏₂=𝑥₂,𝑥₁₁=𝑎₁₁𝑥₁₀+𝑏₁₁𝑓(𝑥₈)=𝑎₁₁𝑥₁₀=𝑎₉𝑥₂=𝑥₃,⋮ (4.5) so that {𝑥_𝑛} is an eventually 8-periodic solution of the system (4.3).

Next, let 𝑎_𝑛≡𝑎 and 𝑏_𝑛≡1−𝑎 in (1.7). We have 𝛼_𝑛1−𝛼_𝑛𝛽_𝑛=𝑎^𝑛1−𝑎^𝑛(𝑏𝑎+𝑏𝑎²𝑏+⋯+𝑎^𝑛𝛼)=1,_𝜏𝜔(𝛼^𝑙_𝜔+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔)=𝑎^𝑝+1(𝛼^𝑙_𝜔+1−𝛼^𝑙_𝜔)=𝑎^𝑝+1,𝛼_𝑝𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)(1−𝛼_𝑝𝛼^𝑙_𝜔)(1−𝛼_𝜔)=𝑎^𝑝(1−𝑎^𝑘−1)1−𝑎^{𝑝+𝑘−1},𝛼_𝑖𝛼_𝜔^𝜏+𝑙(𝛼^𝑙_𝜔+𝛼_𝜔𝛽_𝜔(1−𝛼^𝑙_𝜔)1−𝛼_𝜔)+𝛼_𝑖𝛽_𝑖=𝑎^{𝑝+𝑘+𝑖}+1−𝑎^𝑖≤𝑎^{𝑝+𝑞+𝑘}+1−𝑎^𝑞𝛼,𝑖∈{0,…,𝑞−1},_{𝜔𝜏+𝜎+𝑙}(1−𝛼^𝑙_𝜔)+(1−𝛼^𝜎_𝜔)(1−𝛼_𝜔^{𝜏+𝜎+2𝑙})(1−𝛼_𝜔)𝛼_𝜔𝛽_𝜔𝑎=1−^𝑞+1(1−𝑎^𝑘+𝑝)1−𝑎^{𝑝+𝑞+2𝑘}. (4.6) Hence, 𝐼₁(𝑝)=(𝑎^𝑝+1,𝑎^𝑝(1−𝑎^𝑘−1)(1−𝑎^{𝑘+𝑝−1})),𝐼₂(𝑝,𝑞)⊃(1−𝑎^𝑞+𝑎^{𝑝+𝑞+𝑘}𝑎,1−^𝑞+1(1−𝑎^𝑘+𝑝)1−𝑎^{2𝑘+𝑝+𝑞}). (4.7) Form the above, we can see that Theorem A is just a special case of Theorem 3.1, hence Theorem 3.1 is an extension of Theorem A.

Further, if 𝑘=1 in (1.7), then the intervals 𝐼₁ and 𝐼₂ in Theorem 3.2 are, respectively, 𝐼₁=𝐼₂𝑎=[²1+𝑎+𝑎²,𝑎1+𝑎+𝑎²). (4.8)

Corollary 4.1. Let 𝑎_𝑛≡𝑎,𝑏_𝑛≡1−𝑎, and 𝑘=1. If 𝑎𝜅∈(0,1)∩[²1+𝑎+𝑎²,𝑎1+𝑎+𝑎²), (4.9) then (1.7) has an eventually 3-periodic solution {𝑥_𝑛}^∞_𝑛=−𝑘 (which can be generated explicitly).

As our final remark, note that under the conditions of Theorems 3.1 or 3.2 if {𝑥^′_𝑛} is an arbitrary solution of (1.7) with 𝑥^′_−𝑘,…,𝑥^′₋₂,𝑥^′₋₁∈(0,𝜅] such that 𝑥^′₀∈(𝜅,1), then in view of the proofs of Theorems 3.1 or 3.2, lim_𝑗→∞𝑔^[𝑗](𝑥^′₀)=𝑥^∗=𝑥₀. (4.10) This shows, by means of the continuity properties of the maps 𝑔_𝑛 and ℎ_𝑛, that lim_𝑛→∞|𝑥^′_𝑛−𝑥_𝑛|=0. Note that the requirement 𝑥^′_−𝑘,…,𝑥^′₋₂,𝑥^′₋₁∈(0,𝜅] with 𝑥^′₀∈(𝜅,1) is the same as requiring 𝑥^′₋₁=1𝑎₀(𝑥^′₀−𝑏₀1)∈(𝑎₀(𝜅−𝑏₀1),𝑎₀(1−𝑏₀))∩(0,𝜅]. (4.11) In other words, let {𝑥^′_𝑛} be a solution determined by 𝜙_−𝑘,…,𝜙₋₁∈(0,𝜅] such that 𝜙₋₁1∈(𝑎₀(𝜅−𝑏₀1),𝑎₀(1−𝑏₀))∩(0,𝜅], (4.12) then {𝑥^′_𝑛} will be “attracted” to the periodic solution {𝑥_𝑛} in the proofs of Theorems 3.1 or 3.2. We remark that (1−𝑏₀)/𝑎₀>0. Thus, if 1𝑎₀(𝜅−𝑏₀)≤𝜅, (4.13)then the above intersection is nonempty. And, if 𝜅−𝑏₀1≤0,𝑎₀(1−𝑏₀)>𝜅, (4.14) then (1𝑎₀(𝜅−𝑏₀1),𝑎₀(1−𝑏₀))∩(0,𝜅]=(0,𝜅]. (4.15) Since 𝑎₀ and 𝑏₀ can be chosen in arbitrary manners in Theorems 3.1 and 3.2, such additional conditions can easily be achieved once 𝜅 is determined.

We may illustrate the above discussions by the following example. Let 𝑘=1 and 𝑎_𝑛=1/2=𝑏_𝑛 for all 𝑛∈{0,1,2,…}. According to Corollary 4.1, if 1𝜅∈(0,1)∩[7,271)=[7,27), (4.16)then the solution {𝑥_𝑛} of (1.7) determined by 𝑥₀=𝑥^∗ in (3.50), that is, 𝑥₋₁=1/7, is eventually 3-periodic. Furthermore, let {𝑥^′_𝑛} be the solution determined by 𝑥^′₋₁=𝜙₋₁. If 𝜙₋₁≤0, then by Lemma 2.1, lim_𝑛→∞𝑥^′_𝑛=0. If 𝜙₋₁∈(2𝜅−1,1)∩(0,𝜅]=(0,𝜅], (4.17) then the solution {𝑥^′_𝑛} will satisfy lim_𝑛→∞|𝑥^′_𝑛−𝑥_𝑛|=0. If 𝜙₋₁>𝜅, then by Lemma 2.2, a translate {𝑦_𝑛} of {𝑥^′_𝑛} will satisfy lim_𝑛→∞|𝑦_𝑛−𝑥_𝑛|=0.

Acknowledgment

Project supported by the National Natural Science Foundation of China (10661011).