Global Attractivity of a Family of Max-Type Difference Equations

Abstract

We propose to study a generalized family of max-type difference equations and then prove the global attractivity of a particular case of it under some parameter conditions. Through some numerical results of other cases, we finally pose a generic conjecture.

1. Introduction

The study of max-type difference equations is a hotspot in the area of discrete dynamics because such equations are often closely related to automatic control theory and competitive dynamics. For recent advances in this direction see [1–8] and the references therein.

Motivated by [9], Liu et al. [10] studied the following nonautonomous max-type difference equation:𝑦_𝑛=𝑝+𝑟𝑦_𝑛−𝑠𝑞+𝜙_𝑛𝑦_𝑛−1,…,𝑦_𝑛−𝑚+𝑦_𝑛−𝑠,𝑛∈ℕ₀, (1.1) where 𝑝≥0, 𝑟,𝑞>0, 𝑠,𝑚∈ℕ, and 𝜙_𝑛∶(ℝ⁺)^𝑚→ℝ⁺, 𝑛∈ℕ₀ are mappings satisfying the condition 𝛽min{𝑥₁,…,𝑥_𝑚}≤𝜙_𝑛(𝑥₁,𝑥₂,…,𝑥_𝑚)≤𝛽max{𝑥₁,…,𝑥_𝑚}, for some fixed 𝛽∈(0,+∞). When 𝑝=0, 𝛽∈(0,1), they proved that every positive solution to (1.1) converges to zero if 𝑟≤𝑞, while (𝑟−𝑞)/(1+𝛽) if 𝑟>𝑞. If 𝑝>0 and 𝑟𝑞≥𝑝, then each positive solution to (1.1) converges to ((𝑞−𝑟)²+4𝑝(1+𝛽)−(𝑞−𝑟))/(2(1+𝛽)), for some 𝛽∈(0,+∞), except for the case 𝑞<𝑟, 𝛽∈(𝛽₀,+∞), where 𝛽₀=4𝑝/(𝑞−𝑟)²+1. Note that the behavior of positive solutions to (1.1) for the case 𝑞<𝑟, 𝛽∈(𝛽₀,+∞), is still an unsolved open problem as was mentioned in [10].

Here, we propose to investigate the asymptotic behavior of positive solutions to the generalized family of max-type difference equations𝑥_𝑛=max_{1≤𝑖≤𝑘}𝑝_𝑖+𝑟_𝑖𝑥_𝑛−𝑠𝑞_𝑖+𝑥_𝑛−𝑠+𝑓_𝑖𝑥_𝑛−1,…,𝑥_𝑛−𝑚,𝑛∈ℕ₀, (1.2) where 𝑝_𝑖≥0, 𝑟_𝑖,𝑞_𝑖>0, 𝑠,𝑚,𝑘∈ℕ, 𝑘≥2 and the functions 𝑓_𝑖∶[0,+∞)^𝑚→[0,+∞), 𝑖=1,2,…,𝑘 satisfy the condition𝛽min𝑢₁,…,𝑢_𝑚≤𝑓_𝑖𝑢₁,𝑢₂,…,𝑢_𝑚≤𝛽max𝑢₁,…,𝑢_𝑚, (1.3) for some fixed 𝛽∈(0,1).

In this paper, we mainly consider the particular case that all 𝑝_𝑖 are zero, and then obviously (1.2) reduces to the following form:𝑥_𝑛=𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠+𝑓_𝑖𝑥_𝑛−1,…,𝑥_𝑛−𝑚,𝑛∈ℕ₀. (1.4) Let 𝑥^∗ be a nonnegative equilibrium point of (1.4), then we have𝑥^∗=𝑥^∗×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+1+𝛽𝑥^∗. (1.5) It follows directly from (1.5) that if 0<𝑟_𝑖≤𝑞_𝑖 for all 𝑖=1,2,…,𝑘, then (1.4) has the unique nonnegative equilibrium 𝑥^∗=0, while if there exists at least one 𝑗∈{1,2,…,𝑘} such that 𝑟_𝑗>𝑞_𝑗, then (1.4) has a zero equilibrium 𝑥^∗=0 and a unique positive equilibrium 𝑥^∗=max_{1≤𝑖≤𝑘}{𝑟_𝑖−𝑞_𝑖}/(1+𝛽).

Finally, the following two beautiful theorems are derived.

Theorem 1.1. Consider (1.4) with condition (1.3). If 0<𝑟_𝑖≤𝑞_𝑖 for all 𝑖=1,2,…,𝑘, then every positive solution to (1.4) converges to the unique nonnegative equilibrium zero.

Theorem 1.2. Consider (1.4) with positive initial values and positive 𝑟_𝑖 and 𝑞_𝑖. Let 𝑓_𝑖∶[0,+∞)^𝑚→[0,+∞) be functions such that for some fixed 𝛽∈(0,1), there hold 𝛽min𝑢₁,…,𝑢_𝑚≤𝑓_𝑖𝑢₁,…,𝑢_𝑚≤𝛽max𝑢₁,…,𝑢_𝑚,𝑖=1,2,…,𝑘. (1.6) If there exists at least one 𝑗∈{1,2,…,𝑘} such that 𝑟_𝑗>𝑞_𝑗, then the unique positive equilibrium of (1.4) is a global attractor.

2. Preliminary Lemmas

For the purpose of establishing the main results, some auxiliary lemmas are essential.

Lemma 2.1. Consider the first-order difference equation 𝑥_𝑛=𝑥_𝑛−1×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−1,𝑛∈ℕ₀, (2.1) with positive initial value 𝑥₋₁ and positive 𝑟_𝑖 and 𝑞_𝑖. If there exists at least one 𝑗∈{1,2,…,𝑘} such that 𝑟_𝑗>𝑞_𝑗, then lim_𝑛→∞𝑥_𝑛=max𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘. (2.2)

Proof. Suppose that max{𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘}=𝑟_𝜏−𝑞_𝜏, which is positive, for some 𝜏∈{1,2,…,𝑘}. By making the variable change 𝑥_𝑛=(𝑟_𝜏−𝑞_𝜏)𝑦_𝑛 into (2.1) and then canceling the positive term 𝑟_𝜏−𝑞_𝜏 from the resulting equation, we can derive 𝑦_𝑛=𝑦_𝑛−1×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝑦_𝑛−1,𝑛∈ℕ₀. (2.3) Note that min{𝑎₁/𝑏₁,𝑎₂/𝑏₂}≤(𝑎₁+𝑎₂)/(𝑏₁+𝑏₂)≤max{𝑎₁/𝑏₁,𝑎₂/𝑏₂} for 𝑎_𝑖,𝑏_𝑖>0, 𝑖=1,2. Then it follows from (2.3) that 𝑦_𝑛+1=max_{1≤𝑖≤𝑘}𝑞_𝑖𝑦_𝑛+𝑟_𝑖−𝑞_𝑖𝑦_𝑛𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝑦_𝑛≤max_{1≤𝑖≤𝑘}𝑞_𝑖𝑦_𝑛+𝑟_𝜏−𝑞_𝜏𝑦_𝑛𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝑦_𝑛≤max𝑦_𝑛,1. (2.4) In addition, the following two inequalities hold: 𝑦_𝑛+1−1=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑦_𝑛𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝑦_𝑛≥𝑟−1^𝜏𝑦_𝑛𝑞_𝜏+𝑟_𝜏−𝑞_𝜏𝑦_𝑛𝑞−1=^𝜏𝑦_𝑛−1𝑞_𝜏+𝑟_𝜏−𝑞_𝜏𝑦_𝑛𝑦_𝑛+1,−𝑦_𝑛=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑦_𝑛𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝑦_𝑛−𝑦_𝑛≥𝑟_𝜏𝑦_𝑛𝑞_𝜏+𝑟_𝜏−𝑞_𝜏𝑦_𝑛−𝑦_𝑛=𝑟_𝜏−𝑞_𝜏𝑦_𝑛1−𝑦_𝑛𝑞_𝜏+𝑟_𝜏−𝑞_𝜏𝑦_𝑛. (2.5)(2.6) In the following, we are confronted with three possibilities.Case 1. If there exists 𝑛₀≥−1 such that 𝑦_𝑛0=1, then it follows from (2.4) and (2.5) that 𝑦_𝑛=1 holds for all 𝑛≥𝑛₀.Case 2. If there exists 𝑛₀≥−1 such that 𝑦_𝑛0>1, then it follows from (2.5) and (2.6) that 𝑦_𝑛0≥𝑦_𝑛0+1≥𝑦_𝑛0+2≥⋯>1. (2.7) Thus there is a finite limit 𝛾=lim_𝑛→∞𝑦_𝑛≥1. By taking the limits on both sides of (2.3) and canceling the positive factor 𝛾 from the resulting equation, we obtain 1=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝛾, (2.8) which implies 𝛾=1. Because if 𝛾>1, then 1=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝛾<max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏=1, (2.9) leading to a contradiction.Case 3. If 𝑦_𝑛<1 for all 𝑛≥−1, then it follows from (2.5) and (2.6) that 𝑦₋₁<𝑦₀<𝑦₁<⋯<𝑦_𝑛<⋯<1. (2.10) Therefore, the limit of 𝑦_𝑛 exists, denoted by 0<𝛾=lim_𝑛→∞𝑦_𝑛≤1. By taking the limits on both sides of (2.3) and canceling the nonzero factor 𝛾 from the resulting equation, there hold 1=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝛾, (2.11) which implies 𝛾=1. Because if 0<𝛾<1, then 1=max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏𝛾>max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑟_𝜏−𝑞_𝜏=1, (2.12) which is a contradiction.In either of the above three cases, we get lim_𝑛→∞𝑦_𝑛=1, implying lim_𝑛→∞𝑥_𝑛=𝑟_𝜏−𝑞_𝜏.

From Lemma 2.1, we have the following result.

Lemma 2.2. Consider the 𝑠-order difference equation 𝑥_𝑛=𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠,𝑛∈ℕ₀, (2.13) with positive initial values and 𝑟_𝑖,𝑞_𝑖>0. If there exists at least one 𝑗∈{1,2,…,𝑘} such that 𝑟_𝑗>𝑞_𝑗, then lim_𝑛→∞𝑥_𝑛=max𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘. (2.14)

Proof. Let {𝑥_𝑛}_{𝑛≥−𝑠} be an arbitrary positive solution to (2.13). Apparently we know that the sequence {𝑥_𝑛}_{𝑛≥−𝑠} can be divided into 𝑠 subsequences {𝑥_{𝑗+𝑠𝑘}}_𝑘≥0, 𝑗=−𝑠,−𝑠+1,…,−1, which are, respectively, positive solutions to the first-order equation (2.1) with positive initial values 𝑥_−𝑠,𝑥_−𝑠+1,…,𝑥₋₁. According to Lemma 2.1, we derive lim_𝑘→∞𝑥_{𝑗+𝑠𝑘}=max{𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘} for all 𝑗=−𝑠,−𝑠+1,…,−1, which directly lead to lim_𝑛→∞𝑥_𝑛=max{𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘}.

Lemma 2.3. Let 𝑎>𝑏>0, 0<𝛽<1, and 0<𝜖<((1−𝛽)/(1+𝛽))(𝑎−𝑏). Define two sequences {𝑚_𝑘} and {𝑀_𝑘} in the following way: 𝑀₁𝑚=𝑎−𝑏,_𝑘=𝑀₁−𝛽𝑀_𝑘+𝜖𝑘𝑀,𝑘=1,2,…,_𝑘=𝑀₁−𝛽𝑚_𝑘−1−𝜖(𝑘−1),𝑘=2,3,…. (2.15) Then lim_𝑘→∞𝑚_𝑘=lim_𝑘→∞𝑀_𝑘.

Proof. Observe that 𝑀₂−𝑀₁()−𝜖𝑚=−𝛽1−𝛽𝑎−𝑏𝛽+1<0,_𝑘+1−𝑚_𝑘=𝛽𝑀_𝑘−𝑀_𝑘+1+𝜖𝑘()𝑀𝑘+1,𝑘=1,2,…,_𝑘+1−𝑀_𝑘=−𝛽𝑚_𝑘−𝑚_𝑘−1+𝜖𝑘(𝑘−1),𝑘=2,3,…. (2.16) It follows by induction that {𝑚_𝑘} is increasing and {𝑀_𝑘} is decreasing. Again by induction we derive 𝑚_𝑘<𝑎−𝑏 and 𝑀_𝑘>0, 𝑘=1,2,…. Hence there are two finite limits 𝜉=lim_𝑘→∞𝑚_𝑘 and 𝜂=lim_𝑘→∞𝑀_𝑘. By taking limits on both sides of (2.15), we derive 𝜉=𝑎−𝑏−𝛽𝜂,𝜂=𝑎−𝑏−𝛽𝜉, (2.17) which imply (1−𝛽)(𝜉−𝜂)=0. Therefore 𝜉=𝜂=(𝑎−𝑏)/(1+𝛽).

3. Proofs of Main Theorems

In this section, we are in a position to prove the main theorems presented in Section 1.

Proof of Theorem 1.1. Note that for the case 𝑟_𝑖<𝑞_𝑖, 𝑖=1,2,…,𝑘, the behavior of positive solutions to (1.4) is quite simple. In this case, we have that 𝑥_𝑛≤𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖=𝜇𝑥_𝑛−𝑠, (3.1) where 𝜇=max_{1≤𝑖≤𝑘}{𝑟_𝑖/𝑞_𝑖}<1. Easily the subsequences {𝑥_{𝑙𝑠+𝑗}}_𝑙∈ℕ0, 𝑗∈{0,1,…,𝑠−1} converge to zero, hence the sequence {𝑥_𝑛} also converges to zero.
For the case 𝑟_𝑖≤𝑞_𝑖, 𝑖=1,2,…,𝑘 with at least one 𝑗∈{1,2,…,𝑘} such that 𝑟_𝑗=𝑞_𝑗, we can obtain that𝑥_𝑛≤𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖=𝑥_𝑛−𝑠. (3.2) In this case, the subsequences {𝑥_{𝑙𝑠+𝑗}}_𝑙∈ℕ0, 𝑗=0,1,…,𝑠−1 are all positive and nonincreasing, thus they converge, respectively, to some nonnegative limits 𝜓_𝑗∶=lim_𝑙→∞𝑥_{𝑙𝑠+𝑗}, 𝑗=0,1,…,𝑠−1.
If we replace 𝑛 in (1.4) by 𝑠𝑙+𝑗, 𝑙∈ℕ₀ for an arbitrary fixed 𝑗∈{0,1,…,𝑠−1} and let 𝑙→∞, we can get𝜓_𝑗=𝜓_𝑗×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝜓_𝑗+𝑓_𝑖𝜓_𝑣1,…,𝜓_𝑣𝑚, (3.3) where 𝑣_𝑖∈{0,1,…,𝑠−1}, 𝑖=1,…,𝑚. Without loss of generality, assume that 𝜓_𝑗≠0, then we obtain that 𝑟1=^𝜏𝑞_𝜏+𝜓_𝑗+𝑓_𝜏𝜓_𝑣1,…,𝜓_𝑣𝑚, (3.4) with some fixed number 𝜏∈{1,2,…,𝑘}. Because 𝑟_𝜏≤𝑞_𝜏, then it follows from (3.4) that 𝑞_𝜏+𝜓_𝑗+𝑓_𝜏𝜓_𝑣1,…,𝜓_𝑣𝑚=𝑟_𝜏≤𝑞_𝜏. (3.5) Therefore we have 𝜓_𝑗+𝑓_𝜏𝜓_𝑣1,…,𝜓_𝑣𝑚=0, (3.6) leading to 𝜓_𝑗=0, which is a contradiction. Hence we have that 𝜓_𝑗=0, 𝑗=0,1,…,𝑠−1, and every positive solution to (1.4) converges to zero, if 𝑟_𝑖≤𝑞_𝑖 for all 𝑖=1,2,…,𝑘.

Proof of Theorem 1.2. Suppose that max{𝑟_𝑖−𝑞_𝑖∶𝑖=1,2,…,𝑘}=𝑟_𝜏−𝑞_𝜏>0 for some 𝜏∈{1,2,…,𝑘}. Let 𝜖 be an arbitrary fixed real number with 0<𝜖<((1−𝛽)/(1+𝛽))(𝑟_𝜏−𝑞_𝜏). Define two sequences {𝑀_𝑘} and {𝑚_𝑘} in the way shown in (2.15) with 𝑎=𝑟_𝜏,𝑏=𝑞_𝜏.

Let {𝑥_𝑛} be an arbitrary positive solution to (1.4). Next, we proceed by proving two claims.

Claim 1. There exists 𝑁₁∈ℕ such that 𝑚₁−𝜖≤𝑥_𝑛≤𝑀₁+𝜖 for all 𝑛≥𝑁₁.

Proof. Note that 𝑥_𝑛≤𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠,𝑛=0,1,2,…. (3.7) Consider the following difference equation: 𝑧_𝑛⁽¹⁾=𝑧⁽¹⁾_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑧⁽¹⁾_𝑛−𝑠,𝑛=0,1,2,…. (3.8) Let {𝑧_𝑛⁽¹⁾} be a positive solution to (3.7) with the initial values 𝑧⁽¹⁾₋₁=𝑥₋₁, 𝑧⁽¹⁾₋₂=𝑥₋₂,…,𝑧⁽¹⁾_−𝑠=𝑥_−𝑠.
Note that the mapping ℎ(𝑥)=𝑟𝑥/(𝑞+𝑥) is strictly increasing on the interval (0,+∞). It follows by induction that 𝑥_𝑛≤𝑧_𝑛⁽¹⁾ for all 𝑛≥−𝑠. By Lemma 2.2, we have lim_𝑛→∞𝑧_𝑛⁽¹⁾=𝑟_𝜏−𝑞_𝜏=𝑀₁. Hence there is an integer 𝑁^′₁∈ℕ such that 𝑥_𝑛≤𝑀₁+𝜖 for 𝑛≥𝑁^′₁.
Let 𝑡=max{𝑠,𝑚}. Note that𝑥_𝑛≥𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠+𝛽𝑀₁+𝜖,𝑛≥𝑁^′₁+𝑡. (3.9) Consider the difference equation 𝑦_𝑛⁽¹⁾=𝑦⁽¹⁾_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑦⁽¹⁾_𝑛−𝑠+𝛽𝑀₁+𝜖,𝑛≥𝑁^′₁+𝑡, (3.10) with 𝑦_𝑁⁽¹⁾_{′1+𝑡−1}=𝑥_{𝑁′1+𝑡−1}, 𝑦_𝑁⁽¹⁾_{′1+𝑡−2}=𝑥_{𝑁′1+𝑡−2},…,𝑦_𝑁⁽¹⁾_′1=𝑥_𝑁′1. Note the monotonicity of ℎ(𝑥), it follows by induction that 𝑥_𝑛≥𝑦_𝑛⁽¹⁾ for all 𝑛≥𝑁^′₁. By Lemma 2.2, we get that lim_𝑛→∞𝑦_𝑛⁽¹⁾=𝑚₁. Thus there exists an integer 𝑁₁≥𝑁^′₁ such that 𝑥_𝑛≥𝑚₁−𝜖 for all 𝑛≥𝑁₁.

Working inductively, we will reach the following claim.

Claim 2. For every 𝑘∈ℕ, there exists 𝑁_𝑘∈ℕ such that 𝑚_𝑘−𝜖𝑘≤𝑥_𝑛≤𝑀_𝑘+𝜖𝑘, (3.11) for all 𝑛≥𝑁_𝑘.

Proof. Obviously, the case 𝑘=1 follows directly from Claim 1. In the following, we proceed by induction. Assume that the assertion is true for 𝑘=𝜔(𝜔≥1). Then it suffices to prove the assertion is also true for 𝑘=𝜔+1.
Note that𝑥_𝑛≤𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠+𝛽𝑚_𝜔−𝜖/𝜔,𝑛≥𝑁_𝜔+𝑡. (3.12) Consider the difference equation 𝑧_𝑛^(𝜔+1)=𝑧^(𝜔+1)_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑧^(𝜔+1)_𝑛−𝑠+𝛽𝑚_𝜔−𝜖/𝜔,𝑛≥𝑁_𝜔+𝑡, (3.13) with 𝑧_𝑁^(𝜔+1)_{𝜔+𝑡−1}=𝑥_{𝑁𝜔+𝑡−1}, 𝑧_𝑁^(𝜔+1)_{𝜔+𝑡−2}=𝑥_{𝑁𝜔+𝑡−2},…,𝑧_𝑁^(𝜔+1)_𝜔=𝑥_𝑁𝜔. Note the monotonicity of ℎ(𝑥), it follows by induction that 𝑥_𝑛≤𝑧_𝑛^(𝜔+1) for all 𝑛≥𝑁_𝜔. By Lemma 2.2, we have that lim_𝑛→∞𝑧_𝑛^(𝜔+1)=𝑀_𝜔+1. So there is an integer 𝑁^′_𝜔+1∈ℕ such that 𝑥_𝑛≤𝑀_𝜔+1+𝜖/(𝜔+1) for all 𝑛≥𝑁^′_𝜔+1. Then note that 𝑥_𝑛≥𝑥_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑥_𝑛−𝑠+𝛽𝑀_𝜔+1+𝜖/(𝜔+1),𝑛≥𝑁^′_𝜔+1+𝑡. (3.14) Consider the following difference equation 𝑦_𝑛^(𝜔+1)=𝑦^(𝜔+1)_𝑛−𝑠×max_{1≤𝑖≤𝑘}𝑟_𝑖𝑞_𝑖+𝑦^(𝜔+1)_𝑛−𝑠+𝛽𝑀_𝜔+1+𝜖/(𝜔+1),𝑛≥𝑁^′_𝜔+1+𝑡, (3.15) with 𝑦_𝑁^(𝜔+1)_{′𝜔+1+𝑡−1}=𝑥_{𝑁′𝜔+1+𝑡−1}, 𝑧_𝑁^(𝜔+1)_{′𝜔+1+𝑡−2}=𝑥_{𝑁′𝜔+1+𝑡−2},…,𝑧_𝑁^(𝜔+1)_′𝜔+1=𝑥_{𝑁′𝜔+1}. By the monotonicity of ℎ(𝑥), it follows by induction that 𝑥_𝑛≥𝑦_𝑛^(𝜔+1) for all 𝑛≥𝑁^′_𝜔+1. By Lemma 2.2, we have that lim_𝑛→∞𝑦_𝑛^(𝜔+1)=𝑚_𝜔+1. So there is an integer 𝑁_𝜔+1≥𝑁^′_𝜔+1 such that 𝑥_𝑛≥𝑚_𝜔+1−𝜖/(𝜔+1) for all 𝑛≥𝑁_𝜔+1.

From Claim 2, we derivelim_𝑘→∞𝑚_𝑘=lim_𝑘→∞𝑚_𝑘−𝜖𝑘≤lim_𝑛→∞𝑥_𝑛≤lim_𝑛→∞𝑥_𝑛≤lim_𝑘→∞𝑀_𝑘+𝜖𝑘=lim_𝑘→∞𝑀_𝑘. (3.16) This plus Lemma 2.3 leads to thatlim_𝑛→∞𝑥_𝑛=lim_𝑘→∞𝑚_𝑘=lim_𝑘→∞𝑀_𝑘=𝑟_𝜏−𝑞_𝜏.1+𝛽 (3.17)

4. Simulations and Future Work

In the previous section, we proved the global attractivity of (1.2) when all 𝑝_𝑖 are zero. In this section, we investigate the dynamic behavior of (1.2) provided that all 𝑝_𝑖 are not zero. First, it is trivial to confirm that when all 𝑝_𝑖 are not zero, (1.2) has the following unique positive equilibrium point 𝑥^∗=max_{1≤𝑖≤𝑘}{(𝑞_𝑖−𝑟_𝑖)²+4𝑝_𝑖(1+𝛽)+𝑟_𝑖−𝑞_𝑖}/(2(1+𝛽)). In the following, some numerical results are presented.

Experiment 1. Consider the first-order difference equation 𝑥_𝑛=max0.2+0.6𝑥_𝑛−10.6+𝑥_𝑛−1+0.3𝑥_𝑛−1,𝑟𝑥_𝑛−1𝑞+𝑥_𝑛−1+0.3𝑥_𝑛−1,𝑛∈ℕ, (4.1) where 𝑟,𝑞>0 and the initial value 𝑥₀>0. (See Figures 1 and 2).

(a) Initial value 𝑥0=0.2

(b) Initial value 𝑥0=0.7

Figure 1 

𝑟=1, 𝑞=2; 𝑥∗=√26/13≈0.3922.

(a) Initial value 𝑥0=0.5

(b) Initial value 𝑥0=0.9

Figure 2 

𝑟=2, 𝑞=1; 𝑥∗=10/13≈0.7692.

Experiment 2. Consider the second-order difference equation 𝑥_𝑛=max0.5+𝑥_𝑛−21+𝑥_𝑛−2+0.5𝑥_𝑛−1,0.8+𝑟𝑥_𝑛−2𝑞+𝑥_𝑛−2+0.5𝑥_𝑛−1,𝑛≥2, (4.2) where 𝑟,𝑞>0 and the initial values 𝑥₀,𝑥₁>0. (See Figures 3 and 4).

(a) Initial values 𝑥0=0.2, 𝑥1=0.4

(b) Initial values 𝑥0=0.8, 𝑥1=0.3

Figure 3 

𝑟=1, 𝑞=2; 𝑥∗=√3/3≈0.5774.

(a) Initial values 𝑥0=1.3, 𝑥1=1.0

(b) Initial values 𝑥0=0.5, 𝑥1=0.9

Figure 4 

𝑟=2, 𝑞=1; 𝑥∗=(√5.8+1)/3≈1.1361.

Experiment 3. Consider the third-order difference equation 𝑥_𝑛=max0.5+𝑥_𝑛−31+𝑥_𝑛−3+0.9𝑥²_𝑛−1+𝑥²_𝑛−2,/23𝑥_𝑛−32+𝑥_𝑛−3+0.9𝑥²_𝑛−1+𝑥²_𝑛−2/2,𝑛≥3, (4.3) where the initial values 𝑥₀,𝑥₁,𝑥₂>0. (See Figure 5).

(a) Initial values 𝑥0=0.1, 𝑥1=0.3, 𝑥2=0.5

(b) Initial values 𝑥0=0.3, 𝑥1=0.7, 𝑥2=1.0

(c) Initial values 𝑥0=0.8, 𝑥1=0.3, 𝑥2=0.1

(d) Initial values 𝑥0=1.1, 𝑥1=0.8, 𝑥2=0.6

Figure 5 

𝑥∗=10/19≈0.5263.

Inspired by this work and the results of [10], here we pose the following conjecture.

Conjecture 4.1. Consider (1.2) with nonnegative 𝑝_𝑖 and positive 𝑟_𝑖 and 𝑞_𝑖. Let 𝑓_𝑖∶[0,+∞)^𝑚→[0,+∞), 𝑖=1,2,…,𝑘 be 𝑘 functions such that for some fixed 𝛽∈(0,1), there hold 𝛽min𝑢₁,…,𝑢_𝑘≤𝑓_𝑖𝑢₁,…,𝑢_𝑘≤𝛽max𝑢₁,…,𝑢_𝑘. (4.4) If 𝑟_𝑖𝑞_𝑖≥𝑝_𝑖 for all 𝑖=1,2,…,𝑘, then every positive solution to (1.2) converges to the equilibrium point 𝑥^∗=121+𝛽max_{1≤𝑖≤𝑘}𝑞_𝑖−𝑟_𝑖²+4𝑝_𝑖1+𝛽+𝑟_𝑖−𝑞_𝑖. (4.5)

Acknowledgments

This work was financially supported by the Fundamental Research Funds for the Central Universities (no. CDJXS10181130), the National Natural Science Foundation of China (no. 10771227), and the New Century Excellent Talent Project of China (no. NCET-05-0759).