Global Asymptotic Stability for a Fourth-Order Rational Difference Equation

Abstract

Our aim is to investigate the global behavior of the following fourth-order rational difference equation: 𝑥_𝑛+1=(𝑥_𝑛𝑥_𝑛−2𝑥_𝑛−3+𝑥_𝑛+𝑥_𝑛−2+𝑥_𝑛−3+𝑎)/(𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3+1+𝑎), 𝑛=0,1,2,… where 𝑎∈[0,∞) and the initial values 𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀∈(0,∞). To verify that the positive equilibrium point of the equation is globally asymptotically stable, we used the rule of the successive lengths of positive and negative semicycles of nontrivial solutions of the aforementioned equation.

1. Introduction

There has been a great interest in studying global behaviors of rational difference equations. One can easily see that it is hard to understand thoroughly the global behaviors of solutions of rational difference equations although they have simple forms. And there has not been any general method to identify the global behaviors of rational difference equations of order greater than one and so far [1–3].

Let us consider the following fourth-order difference equation: 𝑥_𝑛+1=𝑥_𝑛𝑥_𝑛−2𝑥_𝑛−3+𝑥_𝑛+𝑥_𝑛−2+𝑥_𝑛−3+𝑎𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3+1+𝑎,𝑛=0,1,2,…, (1.1) where 𝑎∈[0,∞) and the initial values 𝑥₋₃,𝑥₋₂,𝑥₋₁,𝑥₀∈(0,∞) in this paper. By determining the rule for the positive and negative semicycles, we assigned the global behavior of the positive equilibrium point. The unique positive equilibrium point 𝑥 of (1.1) is obtained 𝑥=1 by solving 𝑥=𝑥³+3𝑥+𝑎3𝑥²+1+𝑎. (1.2)

Definition 1.1. A solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is said to be eventually trivial if 𝑥_𝑛 is eventually equal to 𝑥=1; otherwise, the solution is said to be nontrivial [4–6].

Definition 1.2. A solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is said to be eventually positive if 𝑥_𝑛 is eventually greater than 𝑥=1.
A solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is said to be eventually negative if 𝑥_𝑛 is eventually less than 𝑥=1 [1, 2, 4–6].

Definition 1.3. A positive semicycle of a solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) consists of a “string” of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚}, all greater than or equal to the equilibrium point 𝑥, with 𝑙≥−3 and 𝑚<∞ such that either𝑙=−3or𝑙>−3,𝑥_𝑙−1<𝑥,either𝑚=∞or𝑚<∞,𝑥_𝑚+1<𝑥. (1.3) A negative semicycle of a solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) consists of a “string” of terms {𝑥_𝑙,𝑥_𝑙+1,…,𝑥_𝑚}, all greater than or equal to the equilibrium point 𝑥, with 𝑙≥−3 and 𝑚<∞ such that either𝑙=−3or𝑙>−3,𝑥_𝑙−1≥𝑥,either𝑚=∞or𝑚<∞,𝑥_𝑚+1≥𝑥. (1.4) And also the lengths of a semicycle is (𝑚−𝑙+1), the number of the terms contained in it. And we denote that the lengths of a positive semicycle are by (𝑚−𝑙+1)⁺ and the lengths of a negative semicycle are by (𝑚−𝑙+1)⁻ [1, 2, 4–6].

Definition 1.4. A solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is called nonoscillatory about 𝑥, or simply nonoscillatory, if there exists 𝑁≥−3 such that either 𝑥_𝑛≥𝑥,∀𝑛≥𝑁 (1.5) or 𝑥_𝑛<𝑥,∀𝑛≥𝑁. (1.6) Otherwise, the solution {𝑥_𝑛}^∞_𝑛=−3 is called oscillatory about 𝑥, or simply oscillatory [1, 2].

Lemma 1.5. A positive solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is eventually trivial if and only if (𝑥₋₃−1)(𝑥₋₂−1)(𝑥₋₁−1)(𝑥₀−1)=0. (1.7)

Proof. To prove the lemma, first we assume that (𝑥₋₃−1)(𝑥₋₂−1)(𝑥₋₁−1)(𝑥₀−1)≠0 and then we must show that 𝑥_𝑛≠1 for any 𝑛≥1.
Assume that for some 𝑁≥1, 𝑥_𝑁=1 and that 𝑥_𝑛≠1 for −3≤𝑛≤𝑁−1. So 𝑥_𝑁=𝑥_𝑁−1𝑥_𝑁−3𝑥_𝑁−4+𝑥_𝑁−1+𝑥_𝑁−3+𝑥_𝑁−4+𝑎𝑥_𝑁−1𝑥_𝑁−3+𝑥_𝑁−1𝑥_𝑁−4+𝑥_𝑁−3𝑥_𝑁−4+1+𝑎=1, (1.8) and we obtain (𝑥_𝑁−4−1)(𝑥_𝑁−3−1)(𝑥_𝑁−1−1)=0; hence 𝑥_𝑁−4=1,𝑥_𝑁−3=1,𝑥_𝑁−1=1, when we solve the equation above. It is easy to see that 𝑥_𝑁−4=1, 𝑥_𝑁−3=1, or 𝑥_𝑁−1=1 contradicts with 𝑥_𝑛≠1 for −3≤𝑛≤𝑁−1.
If (1.7) holds, it is clear that the following conclusions hold:(i)if 𝑥₋₃=1, 𝑥_𝑛=1 for 𝑛≥1,(ii)if 𝑥₋₂=1, 𝑥_𝑛=1 for 𝑛≥1,(iii)if 𝑥₋₁=1, 𝑥_𝑛=1 for 𝑛≥2,(iv)if 𝑥₀=1, 𝑥_𝑛=1 for 𝑛≥1.It is obvious that if the initial conditions do not satisfy (1.7), then the positive solution {𝑥_𝑛}^∞_𝑛=−3 of (1.1) is eventually nontrivial.

Lemma 1.6. If {𝑥_𝑛}^∞_𝑛=−3 is a nontrivial positive solution of (1.1), then the following conclusions are satisfied:(i)(𝑥_𝑛+1−1)(𝑥_𝑛−1)(𝑥_𝑛−2−1)(𝑥_𝑛−3−1)>0,(ii)(𝑥_𝑛+1−𝑥_𝑛)(𝑥_𝑛−1)<0,(iii)(𝑥_𝑛+1−𝑥_𝑛−2)(𝑥_𝑛−2−1)<0,(iv)(𝑥_𝑛+1−𝑥_𝑛−3)(𝑥_𝑛−3−1)<0.

Proof. (i) The proof of the inequality (i) is obtained by subtracting 1 from (1.1) 𝑥_𝑛+1𝑥−1=^𝑛𝑥_𝑛−2𝑥_𝑛−3+𝑥_𝑛+𝑥_𝑛−2+𝑥_𝑛−3+𝑎𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3=+1+𝑎−1(𝑥_𝑛−1)(𝑥_𝑛−2−1)(𝑥_𝑛−3−1)𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3.+1+𝑎 (1.9) The dominator of this fraction is positive so (𝑥_𝑛+1−1)(𝑥_𝑛−1)(𝑥_𝑛−2−1)(𝑥_𝑛−3−1)>0 also.
(ii) If we subtract 𝑥_𝑛 from (1.1), we obtain 𝑥_𝑛+1−𝑥_𝑛=𝑥_𝑛𝑥_𝑛−2𝑥_𝑛−3+𝑥_𝑛+𝑥_𝑛−2+𝑥_𝑛−3+𝑎𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3+1+𝑎−𝑥_𝑛=−(𝑥_𝑛−1)[𝑥_𝑛−2(𝑥_𝑛+1)+𝑥_𝑛−3(𝑥_𝑛+1)+𝑎]𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3.+1+𝑎 (1.10) The expression [𝑥_𝑛−2(𝑥_𝑛+1)+𝑥_𝑛−3(𝑥_𝑛+1)+𝑎]/(𝑥_𝑛𝑥_𝑛−2+𝑥_𝑛𝑥_𝑛−3+𝑥_𝑛−2𝑥_𝑛−3+1+𝑎) is positive, and so we get (𝑥_𝑛+1−𝑥_𝑛)(𝑥_𝑛−1)<0. (1.11) The proofs for inequalities (iii) and (iv) are similar to the one for (ii).

2. Main Results and Their Proofs

The trajectory of (1.1) and global asymptotic stability of the positive solution are considered in this part of the paper.

Theorem 2.1. Let {𝑥_𝑛}^∞_𝑛=−3 be a strictly oscillatory solution of (1.1). Then the positive and negative semicycles of (1.1) are …,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,… or …,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,… or …,2⁻,1⁺,2⁻,1⁺,2⁻,1⁺,2⁻,1⁺,… or …,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,….

Proof. Assume that {𝑥_𝑛}^∞_𝑛=−3 is a strictly oscillatory solution of (1.1), then the initial values must satisfy one of the following four cases: (i)𝑥_𝑝−3>1, 𝑥_𝑝−2<1, 𝑥_𝑝−1>1, 𝑥_𝑝>1,(ii)𝑥_𝑝−3>1, 𝑥_𝑝−2<1, 𝑥_𝑝−1>1, 𝑥_𝑝<1,(iii)𝑥_𝑝−3>1, 𝑥_𝑝−2<1, 𝑥_𝑝−1<1, 𝑥_𝑝>1,(iv)𝑥_𝑝−3>1, 𝑥_𝑝−2<1, 𝑥_𝑝−1<1, 𝑥_𝑝<1.If (i) occurs, it follows from Lemma 1.6(i) that 𝑥_𝑝−3>1,𝑥_𝑝−2<1,𝑥_𝑝−1>1,𝑥_𝑝>1,𝑥_𝑝+1<1,𝑥_𝑝+2>1,𝑥_𝑝+3𝑥>1,_𝑝+4<1,𝑥_𝑝+5>1,𝑥_𝑝+6>1,𝑥_𝑝+7<1,𝑥_𝑝+8>1,𝑥_𝑝+9𝑥>1,_𝑝+10<1,𝑥_𝑝+11>1,𝑥_𝑝+12>1,𝑥_𝑝+13<1,𝑥_𝑝+14>1,𝑥_𝑝+15𝑥>1,_𝑝+16<1,𝑥_𝑝+17>1,𝑥_𝑝+18>1,𝑥_𝑝+19<1,𝑥_𝑝+20>1,𝑥_𝑝+21>1,…. (2.1) It means that the rule for the lengths of positive and negative semicycles of the solution of (1.1) occurs successively as …,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,….
If (ii) happens, the positive and negative semicycles are 𝑥_𝑝−3>1,𝑥_𝑝−2<1,𝑥_𝑝−1>1,𝑥_𝑝<1,𝑥_𝑝+1>1,𝑥_𝑝+2<1,𝑥_𝑝+3>1,𝑥_𝑝+4𝑥<1,_𝑝+5>1,𝑥_𝑝+6<1,𝑥_𝑝+7>1,𝑥_𝑝+8<1,𝑥_𝑝+9>1,𝑥_𝑝+10<1,𝑥_𝑝+11𝑥>1,_𝑝+12<1,𝑥_𝑝+13>1,𝑥_𝑝+14<1,𝑥_𝑝+15>1,𝑥_𝑝+16<1,𝑥_𝑝+17𝑥>1,_𝑝+18<1,𝑥_𝑝+19>1,𝑥_𝑝+20<1,𝑥_𝑝+21>1,𝑥_𝑝+22<1,𝑥_𝑝+23>1,…. (2.2) The regulation for the lengths of positive and negative semicycles which occur successively is …,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,….
The other cases can be shown similarly.

Theorem 2.2. The positive equilibrium point of (1.1) is globally asymptotically stable.

Proof. Let us show that the positive equilibrium point 𝑥=1 of (1.1) is locally asymptotically stable and also globally attractive, then it is globally asymptotically stable. The linearized equation of (1.1) about the positive equilibrium 𝑥=1 is 𝑦_𝑛+1=𝑞₀𝑦_𝑛+𝑞₁𝑦_𝑛−1+𝑞₂𝑦_𝑛−2+𝑞₃𝑦_𝑛−3, (2.3) where 𝑞_𝑖=(𝜕𝐹/𝜕𝑢_𝑖)(𝑥,𝑥,𝑥,𝑥) and 𝐹(𝑢₀,𝑢₁,𝑢₂,𝑢₃)=(𝑢₀𝑢₂𝑢₃+𝑢₀+𝑢₂+𝑢₃+𝑎)/(𝑢₀𝑢₂+𝑢₀𝑢₃+𝑢₂𝑢₃+1+𝑎).
And we obtain 𝑦_𝑛+1=𝑞₀𝑦_𝑛+𝑞₁𝑦_𝑛−1+𝑞₂𝑦_𝑛−2+𝑞₃𝑦_𝑛−3=0⋅𝑦_𝑛+0⋅𝑦_𝑛−1+0⋅𝑦_𝑛−2+0⋅𝑦_𝑛−3=0, (2.4) thereby 𝑥=1 is locally asymptotically stable.
Now we must show that lim_𝑛→∞𝑥_𝑛=𝑥=1. The proof is as follows.(1)If initial values of (1.1) satisfy (1.7), then 𝑥_𝑛=1 according to Lemma 1.5, so lim_𝑛→∞𝑥_𝑛=𝑥=1. (2)If the initial values of (1.1) do not satisfy (1.7), then for any solution of (1.1), 𝑥_𝑛≠1 for 𝑛≥−3. (i)If the solution is nonoscillatory about the positive equilibrium point of (1.1), then {𝑥_𝑛} is monotonic and bounded because of Lemma 1.6. So, the limit lim_𝑛→∞𝑥_𝑛=𝐿 (2.5) exists and is finite. If we take limits on both sides of (1.1), then we obtain 𝐿=(𝐿³+3𝐿+𝑎)/(3𝐿²+1+𝑎) and thereby 𝐿=1. So lim_𝑛→∞𝑥_𝑛=𝐿=1.(ii)If the solution is strictly oscillatory, then trajectory structure of nontrivial solutions of (1.1) is …,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,… or …,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,2⁺,1⁻,… or …,2⁻,1⁺,2⁻,1⁺,2⁻,1⁺,2⁻,1⁺,… or …,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,1⁺,1⁻,….First, we investigate the case where the rule of the trajectory structure is …,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,3⁺,3⁻,… in a period. We denote positive semicycles by {𝑥_𝑝,𝑥_𝑝+1,𝑥_𝑝+2}⁺ and negative semicycles by {𝑥_𝑝+3,𝑥_𝑝+4,𝑥_𝑝+5}⁻. The rule for the positive and negative semicycles can be “periodically’’ expressed as follows: {𝑥_𝑝+6𝑛,𝑥_𝑝+6𝑛+1,𝑥_𝑝+6𝑛+2}⁺,{𝑥_𝑝+6𝑛+3,𝑥_𝑝+6𝑛+4,𝑥_𝑝+6𝑛+5}⁻,𝑛=0,1,…. (2.6) By using Lemma 1.6 we obtain(i)𝑥_𝑝+6𝑛+2<𝑥_𝑝+6𝑛+1<𝑥_𝑝+6𝑛; 𝑥_𝑝+6𝑛+3<𝑥_𝑝+6𝑛+4<𝑥_𝑝+6𝑛+5,(ii)𝑥_𝑝+6𝑛+6<𝑥_𝑝+6𝑛+2; 𝑥_𝑝+6𝑛+9>𝑥_𝑝+6𝑛+5.This relations give rise to 𝑥_𝑝+6𝑛+6<𝑥_𝑝+6𝑛+2<𝑥_𝑝+6𝑛+1<𝑥_𝑝+6𝑛,𝑥_𝑝+6𝑛+3<𝑥_𝑝+6𝑛+4<𝑥_𝑝+6𝑛+5<𝑥_𝑝+6𝑛+9. (2.7) It is easy to see that {𝑥_𝑝+6𝑛}^∞_𝑛=0 is decreasing with its lower bound 1 because of the inequality 𝑥_𝑝+6𝑛+6<𝑥_𝑝+6𝑛+2<𝑥_𝑝+6𝑛+1<𝑥_𝑝+6𝑛. Hence the limit exists and is finite. The limit is then lim_𝑛→∞𝑥_𝑝+6𝑛=lim_𝑛→∞𝑥_𝑝+6𝑛+1=lim_𝑛→∞𝑥_𝑝+6𝑛+2=𝐿. (2.8) Similarly {𝑥_𝑝+6𝑛+3}^∞_𝑛=0 is increasing and its upper bound 1 because of the inequality 𝑥_𝑝+6𝑛+3<𝑥_𝑝+6𝑛+4<𝑥_𝑝+6𝑛+5<𝑥_𝑝+6𝑛+9. So, the limit lim_𝑛→∞𝑥_𝑝+6𝑛+3=𝑀 exists and is finite. And we derive lim_𝑛→∞𝑥_𝑝+6𝑛+4=lim_𝑛→∞𝑥_𝑝+6𝑛+5=𝑀.
Now, we must show that 𝐿=𝑀=1. To do this, let us take 𝑥_𝑝+6𝑛+6=𝑥_𝑝+6𝑛+5𝑥_𝑝+6𝑛+3𝑥_𝑝+6𝑛+2+𝑥_𝑝+6𝑛+5+𝑥_𝑝+6𝑛+3+𝑥_𝑝+6𝑛+2+𝑎𝑥_𝑝+6𝑛+5𝑥_𝑝+6𝑛+3+𝑥_𝑝+6𝑛+5𝑥_𝑝+6𝑛+2+𝑥_𝑝+6𝑛+3𝑥_𝑝+6𝑛+2+1+𝑎, (2.9) and taking the limit on both sides of the above equality, we obtain 𝐿=𝑀⋅𝑀⋅𝐿+𝑀+𝑀+𝐿+𝑎𝑀⋅𝑀+𝑀⋅𝐿+𝑀⋅𝐿+1+𝑎. (2.10) By solving this equation we have (𝐿−1)[2𝑀(𝐿+1)+𝑎]=0, (2.11) and then 𝐿=1. Similarly to obtain 𝑀 by using 1≤𝑥_𝑝+6𝑛+2<𝑥_𝑝+6𝑛+1 we are taking 𝑥_𝑝+6𝑛+5=𝑥_𝑝+6𝑛+4𝑥_𝑝+6𝑛+2𝑥_𝑝+6𝑛+1+𝑥_𝑝+6𝑛+4+𝑥_𝑝+6𝑛+2+𝑥_𝑝+6𝑛+1+𝑎𝑥_𝑝+6𝑛+4𝑥_𝑝+6𝑛+2+𝑥_𝑝+6𝑛+4𝑥_𝑝+6𝑛+1+𝑥_𝑝+6𝑛+2𝑥_𝑝+6𝑛+1+1+𝑎. (2.12) And taking the limit on both sides of the above equality we get 𝑀=𝑀⋅𝐿⋅𝐿+𝑀+𝐿+𝐿+𝑎𝑀⋅𝐿+𝑀⋅𝐿+𝐿⋅𝐿+1+𝑎. (2.13) By solving this equation we have (𝑀−1)[2𝐿(𝑀+1)+𝑎]=0 and 𝑀=1. Hence, we derive 𝐿=𝑀=1, so lim_𝑛→∞𝑥_𝑛=1. (2.14) It can be shown that lim_𝑛→∞𝑥_𝑛=1 for the other rules of the trajectory structures with the same manner. Therefore, the positive equilibrium point is globally asymptotically stable.