# Global Dynamics, Boundedness, and Semicycle Analysis of a Difference Equation

Authors: Abdul Qadeer Khan; Hamdy El-Metwally
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

In this paper, we explore local stability, attractor, periodicity character, and boundedness solutions of the second-order nonlinear difference equation. Finally, obtained results are verified numerically.


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## Body

## 1. Introduction

For decades, the qualitative analysis of difference equations has been steadily increasing. This is due to the fact that difference equations appear as mathematical models in statistical problems, queuing theory, combinatorial analysis, electrical networks, genetics in biology, probability theory, economics, psychology, stochastic time series, sociology, geometry, number theory, etc. Precisely, there is an increasing interest in the qualitative analysis of difference equations. For instance, Devault et al. [1] have explored boundedness, existence of unbounded solutions, persistence, and global attractivity results for following nonautonomous difference equation:(1)xn+1=pn+xn−1xn,where pn is a positive bounded sequence and initial conditions are positive. Amleh et al. [2] have explored global stability, periodic nature, and boundedness character of the following difference equation:(2)xn+1=α+xn−1xn,where α∈0,∞ and x−1andx0 are positive constants. DeVault et al. [3] have explored boundedness, periodic character, and global stability of the following difference equation:(3)xn+1=p+xn−kxn,where k∈2,3,…,, p is positive, and initial conditions are arbitrary positive numbers. Berenhaut and Stević [4] have explored the behaviour of the following difference equation:(4)xn+1=A+xn−2xn−1p,where pandA∈0,∞,p≠1, and x−1andx0∈0,∞. Stević [5] has explored the behavior of the following difference equation:(5)xn+1=a+xn−1xn,where α is a negative number. For more results on the behavior of the difference equation, we refer the reader to recent published articles [6–10] and books [11–13]. Motivated from aforementioned studies, we explore the behavior of the following difference equation:(6)xn+1=an+xnpxn−1p,n=0,1,…,where p is a nonnegative real number. Moreover, initial conditions x−1andx0 are positive real numbers, and an is a nonnegative periodic sequence with(7)an=α,ifnis even,β,ifnis odd,where αandβ∈0,∞.


## 2. Dynamics of Solutions of Equation (6)

In this study, we consider the following three cases of the functionan.


### 2.1. Case 1:an=a∈R+

In this case, (6) becomes(8)xn+1=a+xnpxn−1p,n=0,1,….It is easy to see thatx¯=a+1 is the only positive fixed point of (8).Now, the functionf:0,∞2⟶0,∞ is defined by(9)fx,y=a+xpyp.Therefore,(10)∂fx,y∂x=pxp−1yp,∂fx,y∂y=−pxpyp+1.Now,(11)∂fx¯,x¯∂x=pa+1,∂fx¯,x¯∂y=−pa+1.So, the linearized equation of (8) about x¯=a+1 is(12)yn+1−pa+1yn+pa+1yn−1=0.Theorem 1.
(i)
Ifp<a+1, then x¯=a+1 of (8) is locally asymptotically stable, and so it is also called a sink(ii)
Ifp>a+1, then x¯=a+1 of (8) is unstable and is called a repeller(iii)
Ifp=a+1, then x¯=a+1 of (8) is unstable and is called a nonhyperbolic pointProof.
(i)
We setp1=p/a+1 and p2=−p/a+1. Now,(13)p1<1−p2⟺pa+1<1+pa+1⟺0<1,and also,(14)1−p2<2⟺pa+1<1,which is valid if(15)p<a+1.So, by Theorem 1.1.1 (a) and (c) of [13], one can obtain that x¯=a+1 is locally asymptotically stable when p<a+1.(ii)
Again,(16)p2−1=pa+1−1>0⟺pa+1>1,p1−1−p2=pa+1−1−pa+1=−1<0,and then,p1<1−p2. Thus, by Theorem 1.1.1 (d) of [13], x¯=a+1 is unstable (repeller point) when p>a+1.(iii)
Note that(17)p2=−1⟺−pa+1=−1⟺−p=−a+1⟺p=a+1,p1−2≤0⟺pa+1−2≤0⟺p≤2a+1.Thus, by Theorem 1.1.1 (e) of [13], x¯=a+1 is unstable (repeller point) when p=a+1.Theorem 2.
Positive solution of (8) is bounded and persists if 0<p<1.Proof.
We obtain from (8) that(18)xn+1>a,∀n≥0.
Hence,xn persists. Then, again, from (8), it follows that(19)x2n+1≤a+x2nap,n=0,1,…,.
Now considering(20)yn+1=a+ynap,∀n≥0.
If the solution of (20) with y0=x0 is yn, then(21)x2n+1≤yn+1,Resp.x2n+2≤yn+1.
Now, we have to show thatyn is bounded. Let(22)fx=a+xpap,and then,(23)f′x=1appxp−1>0,f″x=1appp−1xp−2<0.
Therefore,f is nondecreasing and concave. Therefore, one gets y∗ as the unique fixed point of fy=y. Moreover, f also satisfies(24)fy−yy−y∗<0,y∈0,∞.
By Theorem 2.6.2 of [14], y∗ is a global attractor for all positive solutions of (20), and hence, it is bounded. So, from (8), xn is also bounded.Theorem 3.
Letp≥4, and then, (8) has unbounded solutions.Proof.
It is noted that following holds:(25)xn+1>xnpxn−1p,n∈N,for every solution xnn=−1∞ of (8). Let yn=lnxn. Then, it follows from (25) that(26)yn+1>pyn−pyn−1.
Now, roots of(27)pλ=λ2−pλ+p,are given by(28)λ1,λ2=p±p2−4p2.
Sincep≥4, we have that λ1>1 and(29)λ2=2pp+−4p+p2.
Therefore, both roots ofpλ are positive if p≥4. Moreover, (26) can also written as(30)yn+1−λ1yn−λ2yn−λ1yn−1>0.
Then, we see that(31)xn+1xnλ1>xnxn−1λ1λ2.
It follows that(32)xnxn−1λ1>xn−1xn−2λ1λ2>⋯>x1x0λ1λ2>x0x−1λ1λ2.
Choosex−1 and x0 so that(33)x0>1,x0=x−1λ1.
It follows from this and (32) that(34)xn>x0x−1λ1λ2xn−1λ1=xn−1λ1>⋯>x0λ1n,and consequently,(35)xn>x0λ1n,n∈N.
It follows by lettingn⟶∞ in (35) that xn⟶∞ as n⟶∞, and hence, it follows from this result.Theorem 4.
Leta≥1 and 0<p<1; then, x¯=a+1 of (8) is globally asymptotically stable.Proof.
By Theorem1 (i), x¯=a+1 is a sink. Hence, it is enough to prove further that xnn=−1∞ of (8) tends to x¯=a+1. Recall that xnn=−1∞ of (8) is bounded by Theorem 2. Thus,(36)a≤s=liminfxn,S=limsupxn<∞.
Then, from (8), we get(37)S≤a+Spsp,s≥a+spSp.
Now, claiming thatS=s, otherwise, S>s. From (37), we obtain(38)spS<spa+Sp,sSp>Spa+sp.
Since0<p<1 holds, then(39)s1−p<S1−p,or equivalently(40)sSp<Ssp.
It follows from (38) and (40) that(41)Spa+sp≤spa+Sp.
Hence,(42)Spa−1≤spa−1,which is impossible for a≥1. This is contradiction, and hence, the result follows.Theorem 5.
Every positive solution of (8) oscillates about a+1=x¯ with semicycles of length two or three, and extreme of every semicycle occurs at the first or the second term.Proof.
Let the positive solution of (8) is xnn=−1∞. First, we prove that every positive semicycle except possibly the first term has two or three terms. Assuming xσ−1<x¯ and xσ≥x¯ for some σ∈ℕ, we obtain from (8) that(43)xσ+1=a+xσpxσ−1p>a+1=x¯.
Ifxσ+1>xσ, then we have(44)xσ+2=a+xσ+1pxσp>a+1=x¯.
On the contrary, sincep∈0,1, we see that(45)xσ+2=a+xσ+1pxσp≤a+xσ+1px¯p≤a+xσ+1pa+1≤xσ+1.
So,x¯<xσ+2<xσ+1. Therefore,(46)xσ+3=a+xσ+2pxσ+1p<a+1=x¯.Theorem 6.
Equation (8) has no periodic solution having prime period two.Proof.
Let(47)…,η1,η2,η1,η2,…,be a periodic solution of period two of (8). It follows that(48)η1=a+η2η1p,η2=a+η1η2p,which implies that(49)η2=a+1η1−a.
Substituting from (49) into (48) and after some calculation, we get(50)η1−ap+1η1p=aη1−a+1p.
From (50), one has(51)fη1=p+1lnη1−a+plnη1−plnaη1−a+1=0.
Obviously,η1=a+1 is a solution of (51). But one has to prove that this is the unique solution of (51). Now,(52)f′η1=η1−aaη1+paη1−a+1+p+1η1η1η1−aaη1−a+1.
Thus,f′η1>0 for η1∈a,∞. This implies that, on a,∞, f is strictly nondecreasing. Hence, x¯=a+1 is the unique solution of (51), and consequently, a+1,a+1 is the unique solution of (48) completing the theorem’s proof.


### 2.2. Case 2:an be a Function of Period Two

We will explore dynamics of equation (6) when an is a periodic sequence having period two with αandβ∈0,∞ and α≠β. Consider a2n=α and a2n+1=β. Then, we have(53)x2n+1=α+x2npx2n−1p,n=0,1,…,x2n+2=β+x2n+1px2np,n=0,1,…,.By separating the even-indexed and odd-indexed terms, equation (6) now becomes(54)un+1=α+unpvnpvn+1=β+vnpunp,n=0,1,…,where u¯,v¯=α+1,β+1 is the unique fixed point of system (54).Theorem 7.
Ifp<β+1α+1/α+1p+β+1p, then Eu¯,v¯=α+1,β+1 of (54) is a sink.Proof.
We consider the mapT on 0,∞×0,∞, which is described as follows:(55)Tu,v=T1u,vT2u,v=α+upvpα+vpup.
Then,(56)∂T1∂u=−pup−1vpup2,∂T1∂v=pvp−1up,∂T2∂u=pup−1vp,∂T2∂v=−pvp−1upvp2.
Therefore, the Jacobian matrix ofT evaluated at Eu¯,v¯=α+1,β+1 is(57)JEu¯,v¯=−pup−1vpup2pvp−1uppup−1vp−pvp−1upvp2,and the auxiliary equation associated with u¯,v¯ is(58)λ2−λpβ+1p−1α+1+α+1p−1β+1,and then, we obtain(59)λ1=0,λ2=pβ+1p−1α+1+α+1p−1β+1.
It follows by Corollary 1.3.1 of [14] that u¯,v¯=α+1,β+1 of (54) is locally stable if(60)p<β+1α+1α+1p+β+1p.
Then, the proof is completed.


### 2.3. Case 3: A Positive Bounded Sequence isan

We assume thatan is positive bounded with(61)limn⟶∞infan=a≥0,limn⟶∞supan=b<∞,for some real constants a and b.Theorem 8.
xnn=−1∞ of (6) is bounded and persists if 0<p<1.Proof.
Its proof is same as proof of Theorem2, and hence, it is omitted.Lemma 1.
Assume (61) is satisfied, and if(62)λ=limn⟶∞infxn,η=limn⟶∞supxn,then(63)ab−1b−1≤λ≤η≤ab−1a−1.Proof.
Letɛ>0 for n≥N0ɛ, and we get(64)λ−ɛ≤xn≤μ+ɛ,a−ɛ≤an≤b+ɛ.
Therefore,(65)xn+1≥a−ɛ+λ−ɛη+ɛp.
Taking thelimn⟶∞inf for (65), we obtain(66)λ≥a−ɛ+λ−ɛη+ɛp.
Sinceɛ>0 is arbitrary, it follows that(67)λ≥a+ληp.
Similarly,(68)η≤b+ηλp.
We get from inequalities (67) and (68) that(69)ληp≥aηp+λp,ηλp≤bλp+ηp.
Since0<p<1 holds, we get(70)λ1−p≤η1−p,or equivalently(71)ληp≤ηλp.
It follows from equation (69) that(72)aηp+λp≤bλp+ηp.
So,(73)ηpa−1≤λpb−1,and one has(74)ηλp≤b−1a−1,ληp≥a−1b−1.
We have from (67), for all n>N0ɛ,(75)λ≥a+ληp≥a+a−1b−1=ab−1b−1.
Similarly, we obtain from (68) that(76)η≤ab−1a−1.
This completes the proof.
Now, we will explore attractively of solutions of equation (6).
Letx¯n represent the arbitrary positive solution of (6). Now, one can find appropriate conditions such that x¯n attracts all positive solutions of (6), that is,(77)limn⟶∞xnx¯n=1.
Now, defineyn:(78)yn=xnx¯n,n=−1,0,1,…,and then, equation (6) becomes(79)yn+1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p.Lemma 2.
Letx¯n be a positive solution of (6), and then,(i)
y¯=1 is the positive fixed point of (79).(ii)
If for somen and yn−1≥yn, then yn+1<1. Moreover, if for some n and yn−1<yn, then yn+1≥1.(iii)
Every semicycle, except first one, of any oscillatory solution of (79) contains exactly one term.Proof.
(i)
The proof of (i) is trivial.(ii)
Ifyn−1≥yn, then yn/yn−1<1 and(80)yn+1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p<an+x¯n/x¯n−1pan+x¯n/x¯n−1p=1.In a similar way, the case is the same whenyn−1<yn is proven.(iii)
Letyn be an eventually oscillatory solution of (27) such that yn−1<1 and yn≥1. It follows from part (ii) that yn+1<1. So, the positive semicycle has exactly one term. In similar way, one can prove for the negative semicycle.Lemma 3.
Every nonoscillatory solution of (79) converges to 1.Proof.
Assumingyn be a nonoscillatory solution of (79). We may assume, without losing generality, that yn<1, for n≥N0. Clearly, for n≥N0, one has yn+1<yn; otherwise, there exists k>N0 such that yk−1≤yk, and it follows by Lemma 1 (ii) that yk+1≥1, that is, not possible. As yn is decreasing and yn<1, it converges. Assume(81)limn⟶∞yn=τ,where 0<τ≤1. We have to prove that τ=1. Since(82)limn⟶∞ynyn−1=1,for ɛ>0 and n, the sufficiently large one has(83)ynyn−1p−1<ɛ.
Hence,(84)yn+1−1=x¯n/x¯n−1pan+x¯n/x¯n−1pynyn−1p−1≤ynyn−1p−1<ɛ.
So, we obtainlimn⟶∞yn=1.Lemma 4.
Ifyn is a positive solution of (79) and suppose that there exists m∈1,2,…,, s.t.,(85)y2m−1<1,y2m≥1,then(86)y2n−1≥1,yn<1,n=m,m+1,…,.
Moreover, if(87)y2m−1≥1,y2m<1,then(88)y2n−1<1,yn≥1,n=m,m+1,…,.Proof.
Ifyn be a solution of (79) such that (85) holds for m∈1,2,…,, then we obtain(89)y2m−1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p≥an+x¯n/x¯n−1pan+x¯n/x¯n−1p=1,and by working inductively, one can prove that (86) is satisfied.
In similar way, one can prove that if (87) holds for m∈1,2,…,, then (88) is satisfied.Theorem 9.
Ifx¯n is a particular positive solution of (6) and yn is a positive solution of (79) and suppose that 0<p≤1/2 or 1/2<p<1, a>1, and aa−1>b−1, then(90)limn⟶∞yn=1.Proof.
Ifyn is a solution of (79), then it is enough to prove that(91)limn⟶∞yn=1.
Assuming there existsm∈1,2,…, such that (86) or (88) hold. We may also assume that (86) holds for m∈1,2,…,, and 0<p≤1/2 holds. Let(92)μ=limn⟶∞infyn,θ=limn⟶∞supyn,and also, let(93)k1=limn⟶∞infx¯n,k2=limn⟶∞supx¯n,(94)k=k2k1.
Now, considering(95)Fx,y,z=x+ypzpx+yp,for x,y,andz>0, then one has(96)∂F∂x=yp1−zpx+yp2,∂F∂y=pxyp−1zp−1x+yp2.
Thus, it can be observed that(i)
F is nonincreasing in x and nondecreasing in y if z>1(ii)
F is nonincreasing in y and nondecreasing in x if z<1
Letn≥m. Using (79), one has(97)y2n+1=Fa2n,x¯2nx¯2n−1,y2ny2n−1,y2n+2=Fa2n+1,x¯2n+1x¯2n,y2n+1y2n.
Since (85) holds, so by Lemma 4, one can obtain(98)y2n−1y2n<1,y2ny2n−1≥1,n≥m.
Using (61) and (92)–(95) and the monotone properties of F, we get(99)θ≤Fa,k,θμ=a+θ/μpkpa+kp,μ≥Fa,k,μθ=a+μ/θpkpa+kp,or(100)θμp≤aμp+θpkpa+kp,μθp≥aθp+μpkpa+kp.
Then,(101)aθpμp−1+μ2p−1kp≤θpμp≤aμpθp−1+θ2p−1kp.
Hence,(102)aθpμp−1+μ2p−1kp≤aμpθp−1+θ2p−1kp.
So,(103)θpaμp−1+μp−1μθpkp≤μpaθp−1+θp−1θμpkp,or(104)θμpaμθp−1−kp≤a−μθp−1kp.
Thus,(105)aθμ−kpθμp≤a−θμ1−pkp,aθμ−1≤kpθμp−θμ1−p.
But, from0<p≤1/2, one can obtain p≤1−p. This implies that(106)aθμ−1≤0,or(107)θ≤μ.
Thus, we get thatθ=μ. The proof is completed.
Now, suppose that1/2<p<1, a>1, and aa−1>b−1 hold. Then, using relations (61) and (92)–(95) and η/λp≤b−1/a−1, λ/ηp≥a−1/b−1 holds; we obtain(108)θ≤Fa,ηλ,θμ=a+η/λpθ/μpa+η/λp≤a+b−1/a−1θ/μpa+b−1/a−1,μ≥Fa,ηλ,μθ=a+η/λpμ/θpa+η/λp≥a+b−1/a−1μ/θpa+b−1/a−1,or(109)μpθ≤aμpa+b−1/a−1+b−1/a−1θpa+b−1/a−1,(110)μθp≥aθpa+b−1/a−1+b−1/a−1μpa+b−1/a−1.
Sinceμ≤θ, it follows that μθp≤θμp. Therefore, from equation (109), we get(111)aθpa+b−1/a−1+b−1/a−1μpa+b−1/a−1≤aμpa+b−1/a−1+b−1/a−1θpa+b−1/a−1,or(112)aθpa+b−1/a−1−b−1/a−1θpa+b−1/a−1≤aμpa+b−1/a−1−b−1/a−1μpa+b−1/a−1.
Since1/2<p<1, a>1, and aa−1>b−1 hold, then from (112), we have θ≤μ; so, θ=μ. Then, the proof is completed.Theorem 10.
Assume that0<p<1, and an is a periodic sequence, s.t., an+2=an for all n=0,1,…,and then, (6) has a periodic solution of the prime period two.Proof.
For (6) possesses a periodic solution xn having prime period two, one can find positive numbers x0andx−1, s.t.,(113)x−1=x1=a0+x0x−1p,x0=x2=a1+x1x0p.
Or, equivalently,(114)x−1=a0+x0x−1p,x0=a1+x−1x0p.
Now, one has to prove that system (114) is consistent. From (114), one gets(115)x−1−a0x0−a1=1,and from this, it follows that(116)x−1−a0p+1=a1x−1−a0+1px−1p,(117)x0−a1p+1=a0x0−a1+1px0p.
Define(118)Fx=x−a0p+1−a1x−a0+1pxp,x>a0.
Then,(119)Fa0=−1a0<0,Fa0+1=−a1+1pa0+1p+1>0,where a1<a0. So, F has a zero, say x−1∈a0,a0+1, and in view of equation (115) and (116), one can get that (6) has a two-periodic solution. Now, assuming that a1>a0, one can define(120)Hx=−a0x−a1+1pxp+x−a1p+1,x>a1.
Then,(121)Ha1=−1a1<0,Ha1+1=−a0+1pa1+1p+1>0.
Thus,H has a zero, say x0∈a1,a1+1, and in view of equations (115) and (116), one can get that (6) has a two-periodic solution.


## 2.1. Case 1:an=a∈R+

In this case, (6) becomes(8)xn+1=a+xnpxn−1p,n=0,1,….It is easy to see thatx¯=a+1 is the only positive fixed point of (8).Now, the functionf:0,∞2⟶0,∞ is defined by(9)fx,y=a+xpyp.Therefore,(10)∂fx,y∂x=pxp−1yp,∂fx,y∂y=−pxpyp+1.Now,(11)∂fx¯,x¯∂x=pa+1,∂fx¯,x¯∂y=−pa+1.So, the linearized equation of (8) about x¯=a+1 is(12)yn+1−pa+1yn+pa+1yn−1=0.Theorem 1.
(i)
Ifp<a+1, then x¯=a+1 of (8) is locally asymptotically stable, and so it is also called a sink(ii)
Ifp>a+1, then x¯=a+1 of (8) is unstable and is called a repeller(iii)
Ifp=a+1, then x¯=a+1 of (8) is unstable and is called a nonhyperbolic pointProof.
(i)
We setp1=p/a+1 and p2=−p/a+1. Now,(13)p1<1−p2⟺pa+1<1+pa+1⟺0<1,and also,(14)1−p2<2⟺pa+1<1,which is valid if(15)p<a+1.So, by Theorem 1.1.1 (a) and (c) of [13], one can obtain that x¯=a+1 is locally asymptotically stable when p<a+1.(ii)
Again,(16)p2−1=pa+1−1>0⟺pa+1>1,p1−1−p2=pa+1−1−pa+1=−1<0,and then,p1<1−p2. Thus, by Theorem 1.1.1 (d) of [13], x¯=a+1 is unstable (repeller point) when p>a+1.(iii)
Note that(17)p2=−1⟺−pa+1=−1⟺−p=−a+1⟺p=a+1,p1−2≤0⟺pa+1−2≤0⟺p≤2a+1.Thus, by Theorem 1.1.1 (e) of [13], x¯=a+1 is unstable (repeller point) when p=a+1.Theorem 2.
Positive solution of (8) is bounded and persists if 0<p<1.Proof.
We obtain from (8) that(18)xn+1>a,∀n≥0.
Hence,xn persists. Then, again, from (8), it follows that(19)x2n+1≤a+x2nap,n=0,1,…,.
Now considering(20)yn+1=a+ynap,∀n≥0.
If the solution of (20) with y0=x0 is yn, then(21)x2n+1≤yn+1,Resp.x2n+2≤yn+1.
Now, we have to show thatyn is bounded. Let(22)fx=a+xpap,and then,(23)f′x=1appxp−1>0,f″x=1appp−1xp−2<0.
Therefore,f is nondecreasing and concave. Therefore, one gets y∗ as the unique fixed point of fy=y. Moreover, f also satisfies(24)fy−yy−y∗<0,y∈0,∞.
By Theorem 2.6.2 of [14], y∗ is a global attractor for all positive solutions of (20), and hence, it is bounded. So, from (8), xn is also bounded.Theorem 3.
Letp≥4, and then, (8) has unbounded solutions.Proof.
It is noted that following holds:(25)xn+1>xnpxn−1p,n∈N,for every solution xnn=−1∞ of (8). Let yn=lnxn. Then, it follows from (25) that(26)yn+1>pyn−pyn−1.
Now, roots of(27)pλ=λ2−pλ+p,are given by(28)λ1,λ2=p±p2−4p2.
Sincep≥4, we have that λ1>1 and(29)λ2=2pp+−4p+p2.
Therefore, both roots ofpλ are positive if p≥4. Moreover, (26) can also written as(30)yn+1−λ1yn−λ2yn−λ1yn−1>0.
Then, we see that(31)xn+1xnλ1>xnxn−1λ1λ2.
It follows that(32)xnxn−1λ1>xn−1xn−2λ1λ2>⋯>x1x0λ1λ2>x0x−1λ1λ2.
Choosex−1 and x0 so that(33)x0>1,x0=x−1λ1.
It follows from this and (32) that(34)xn>x0x−1λ1λ2xn−1λ1=xn−1λ1>⋯>x0λ1n,and consequently,(35)xn>x0λ1n,n∈N.
It follows by lettingn⟶∞ in (35) that xn⟶∞ as n⟶∞, and hence, it follows from this result.Theorem 4.
Leta≥1 and 0<p<1; then, x¯=a+1 of (8) is globally asymptotically stable.Proof.
By Theorem1 (i), x¯=a+1 is a sink. Hence, it is enough to prove further that xnn=−1∞ of (8) tends to x¯=a+1. Recall that xnn=−1∞ of (8) is bounded by Theorem 2. Thus,(36)a≤s=liminfxn,S=limsupxn<∞.
Then, from (8), we get(37)S≤a+Spsp,s≥a+spSp.
Now, claiming thatS=s, otherwise, S>s. From (37), we obtain(38)spS<spa+Sp,sSp>Spa+sp.
Since0<p<1 holds, then(39)s1−p<S1−p,or equivalently(40)sSp<Ssp.
It follows from (38) and (40) that(41)Spa+sp≤spa+Sp.
Hence,(42)Spa−1≤spa−1,which is impossible for a≥1. This is contradiction, and hence, the result follows.Theorem 5.
Every positive solution of (8) oscillates about a+1=x¯ with semicycles of length two or three, and extreme of every semicycle occurs at the first or the second term.Proof.
Let the positive solution of (8) is xnn=−1∞. First, we prove that every positive semicycle except possibly the first term has two or three terms. Assuming xσ−1<x¯ and xσ≥x¯ for some σ∈ℕ, we obtain from (8) that(43)xσ+1=a+xσpxσ−1p>a+1=x¯.
Ifxσ+1>xσ, then we have(44)xσ+2=a+xσ+1pxσp>a+1=x¯.
On the contrary, sincep∈0,1, we see that(45)xσ+2=a+xσ+1pxσp≤a+xσ+1px¯p≤a+xσ+1pa+1≤xσ+1.
So,x¯<xσ+2<xσ+1. Therefore,(46)xσ+3=a+xσ+2pxσ+1p<a+1=x¯.Theorem 6.
Equation (8) has no periodic solution having prime period two.Proof.
Let(47)…,η1,η2,η1,η2,…,be a periodic solution of period two of (8). It follows that(48)η1=a+η2η1p,η2=a+η1η2p,which implies that(49)η2=a+1η1−a.
Substituting from (49) into (48) and after some calculation, we get(50)η1−ap+1η1p=aη1−a+1p.
From (50), one has(51)fη1=p+1lnη1−a+plnη1−plnaη1−a+1=0.
Obviously,η1=a+1 is a solution of (51). But one has to prove that this is the unique solution of (51). Now,(52)f′η1=η1−aaη1+paη1−a+1+p+1η1η1η1−aaη1−a+1.
Thus,f′η1>0 for η1∈a,∞. This implies that, on a,∞, f is strictly nondecreasing. Hence, x¯=a+1 is the unique solution of (51), and consequently, a+1,a+1 is the unique solution of (48) completing the theorem’s proof.


## 2.2. Case 2:an be a Function of Period Two

We will explore dynamics of equation (6) when an is a periodic sequence having period two with αandβ∈0,∞ and α≠β. Consider a2n=α and a2n+1=β. Then, we have(53)x2n+1=α+x2npx2n−1p,n=0,1,…,x2n+2=β+x2n+1px2np,n=0,1,…,.By separating the even-indexed and odd-indexed terms, equation (6) now becomes(54)un+1=α+unpvnpvn+1=β+vnpunp,n=0,1,…,where u¯,v¯=α+1,β+1 is the unique fixed point of system (54).Theorem 7.
Ifp<β+1α+1/α+1p+β+1p, then Eu¯,v¯=α+1,β+1 of (54) is a sink.Proof.
We consider the mapT on 0,∞×0,∞, which is described as follows:(55)Tu,v=T1u,vT2u,v=α+upvpα+vpup.
Then,(56)∂T1∂u=−pup−1vpup2,∂T1∂v=pvp−1up,∂T2∂u=pup−1vp,∂T2∂v=−pvp−1upvp2.
Therefore, the Jacobian matrix ofT evaluated at Eu¯,v¯=α+1,β+1 is(57)JEu¯,v¯=−pup−1vpup2pvp−1uppup−1vp−pvp−1upvp2,and the auxiliary equation associated with u¯,v¯ is(58)λ2−λpβ+1p−1α+1+α+1p−1β+1,and then, we obtain(59)λ1=0,λ2=pβ+1p−1α+1+α+1p−1β+1.
It follows by Corollary 1.3.1 of [14] that u¯,v¯=α+1,β+1 of (54) is locally stable if(60)p<β+1α+1α+1p+β+1p.
Then, the proof is completed.


## 2.3. Case 3: A Positive Bounded Sequence isan

We assume thatan is positive bounded with(61)limn⟶∞infan=a≥0,limn⟶∞supan=b<∞,for some real constants a and b.Theorem 8.
xnn=−1∞ of (6) is bounded and persists if 0<p<1.Proof.
Its proof is same as proof of Theorem2, and hence, it is omitted.Lemma 1.
Assume (61) is satisfied, and if(62)λ=limn⟶∞infxn,η=limn⟶∞supxn,then(63)ab−1b−1≤λ≤η≤ab−1a−1.Proof.
Letɛ>0 for n≥N0ɛ, and we get(64)λ−ɛ≤xn≤μ+ɛ,a−ɛ≤an≤b+ɛ.
Therefore,(65)xn+1≥a−ɛ+λ−ɛη+ɛp.
Taking thelimn⟶∞inf for (65), we obtain(66)λ≥a−ɛ+λ−ɛη+ɛp.
Sinceɛ>0 is arbitrary, it follows that(67)λ≥a+ληp.
Similarly,(68)η≤b+ηλp.
We get from inequalities (67) and (68) that(69)ληp≥aηp+λp,ηλp≤bλp+ηp.
Since0<p<1 holds, we get(70)λ1−p≤η1−p,or equivalently(71)ληp≤ηλp.
It follows from equation (69) that(72)aηp+λp≤bλp+ηp.
So,(73)ηpa−1≤λpb−1,and one has(74)ηλp≤b−1a−1,ληp≥a−1b−1.
We have from (67), for all n>N0ɛ,(75)λ≥a+ληp≥a+a−1b−1=ab−1b−1.
Similarly, we obtain from (68) that(76)η≤ab−1a−1.
This completes the proof.
Now, we will explore attractively of solutions of equation (6).
Letx¯n represent the arbitrary positive solution of (6). Now, one can find appropriate conditions such that x¯n attracts all positive solutions of (6), that is,(77)limn⟶∞xnx¯n=1.
Now, defineyn:(78)yn=xnx¯n,n=−1,0,1,…,and then, equation (6) becomes(79)yn+1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p.Lemma 2.
Letx¯n be a positive solution of (6), and then,(i)
y¯=1 is the positive fixed point of (79).(ii)
If for somen and yn−1≥yn, then yn+1<1. Moreover, if for some n and yn−1<yn, then yn+1≥1.(iii)
Every semicycle, except first one, of any oscillatory solution of (79) contains exactly one term.Proof.
(i)
The proof of (i) is trivial.(ii)
Ifyn−1≥yn, then yn/yn−1<1 and(80)yn+1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p<an+x¯n/x¯n−1pan+x¯n/x¯n−1p=1.In a similar way, the case is the same whenyn−1<yn is proven.(iii)
Letyn be an eventually oscillatory solution of (27) such that yn−1<1 and yn≥1. It follows from part (ii) that yn+1<1. So, the positive semicycle has exactly one term. In similar way, one can prove for the negative semicycle.Lemma 3.
Every nonoscillatory solution of (79) converges to 1.Proof.
Assumingyn be a nonoscillatory solution of (79). We may assume, without losing generality, that yn<1, for n≥N0. Clearly, for n≥N0, one has yn+1<yn; otherwise, there exists k>N0 such that yk−1≤yk, and it follows by Lemma 1 (ii) that yk+1≥1, that is, not possible. As yn is decreasing and yn<1, it converges. Assume(81)limn⟶∞yn=τ,where 0<τ≤1. We have to prove that τ=1. Since(82)limn⟶∞ynyn−1=1,for ɛ>0 and n, the sufficiently large one has(83)ynyn−1p−1<ɛ.
Hence,(84)yn+1−1=x¯n/x¯n−1pan+x¯n/x¯n−1pynyn−1p−1≤ynyn−1p−1<ɛ.
So, we obtainlimn⟶∞yn=1.Lemma 4.
Ifyn is a positive solution of (79) and suppose that there exists m∈1,2,…,, s.t.,(85)y2m−1<1,y2m≥1,then(86)y2n−1≥1,yn<1,n=m,m+1,…,.
Moreover, if(87)y2m−1≥1,y2m<1,then(88)y2n−1<1,yn≥1,n=m,m+1,…,.Proof.
Ifyn be a solution of (79) such that (85) holds for m∈1,2,…,, then we obtain(89)y2m−1=an+x¯n/x¯n−1pyn/yn−1pan+x¯n/x¯n−1p≥an+x¯n/x¯n−1pan+x¯n/x¯n−1p=1,and by working inductively, one can prove that (86) is satisfied.
In similar way, one can prove that if (87) holds for m∈1,2,…,, then (88) is satisfied.Theorem 9.
Ifx¯n is a particular positive solution of (6) and yn is a positive solution of (79) and suppose that 0<p≤1/2 or 1/2<p<1, a>1, and aa−1>b−1, then(90)limn⟶∞yn=1.Proof.
Ifyn is a solution of (79), then it is enough to prove that(91)limn⟶∞yn=1.
Assuming there existsm∈1,2,…, such that (86) or (88) hold. We may also assume that (86) holds for m∈1,2,…,, and 0<p≤1/2 holds. Let(92)μ=limn⟶∞infyn,θ=limn⟶∞supyn,and also, let(93)k1=limn⟶∞infx¯n,k2=limn⟶∞supx¯n,(94)k=k2k1.
Now, considering(95)Fx,y,z=x+ypzpx+yp,for x,y,andz>0, then one has(96)∂F∂x=yp1−zpx+yp2,∂F∂y=pxyp−1zp−1x+yp2.
Thus, it can be observed that(i)
F is nonincreasing in x and nondecreasing in y if z>1(ii)
F is nonincreasing in y and nondecreasing in x if z<1
Letn≥m. Using (79), one has(97)y2n+1=Fa2n,x¯2nx¯2n−1,y2ny2n−1,y2n+2=Fa2n+1,x¯2n+1x¯2n,y2n+1y2n.
Since (85) holds, so by Lemma 4, one can obtain(98)y2n−1y2n<1,y2ny2n−1≥1,n≥m.
Using (61) and (92)–(95) and the monotone properties of F, we get(99)θ≤Fa,k,θμ=a+θ/μpkpa+kp,μ≥Fa,k,μθ=a+μ/θpkpa+kp,or(100)θμp≤aμp+θpkpa+kp,μθp≥aθp+μpkpa+kp.
Then,(101)aθpμp−1+μ2p−1kp≤θpμp≤aμpθp−1+θ2p−1kp.
Hence,(102)aθpμp−1+μ2p−1kp≤aμpθp−1+θ2p−1kp.
So,(103)θpaμp−1+μp−1μθpkp≤μpaθp−1+θp−1θμpkp,or(104)θμpaμθp−1−kp≤a−μθp−1kp.
Thus,(105)aθμ−kpθμp≤a−θμ1−pkp,aθμ−1≤kpθμp−θμ1−p.
But, from0<p≤1/2, one can obtain p≤1−p. This implies that(106)aθμ−1≤0,or(107)θ≤μ.
Thus, we get thatθ=μ. The proof is completed.
Now, suppose that1/2<p<1, a>1, and aa−1>b−1 hold. Then, using relations (61) and (92)–(95) and η/λp≤b−1/a−1, λ/ηp≥a−1/b−1 holds; we obtain(108)θ≤Fa,ηλ,θμ=a+η/λpθ/μpa+η/λp≤a+b−1/a−1θ/μpa+b−1/a−1,μ≥Fa,ηλ,μθ=a+η/λpμ/θpa+η/λp≥a+b−1/a−1μ/θpa+b−1/a−1,or(109)μpθ≤aμpa+b−1/a−1+b−1/a−1θpa+b−1/a−1,(110)μθp≥aθpa+b−1/a−1+b−1/a−1μpa+b−1/a−1.
Sinceμ≤θ, it follows that μθp≤θμp. Therefore, from equation (109), we get(111)aθpa+b−1/a−1+b−1/a−1μpa+b−1/a−1≤aμpa+b−1/a−1+b−1/a−1θpa+b−1/a−1,or(112)aθpa+b−1/a−1−b−1/a−1θpa+b−1/a−1≤aμpa+b−1/a−1−b−1/a−1μpa+b−1/a−1.
Since1/2<p<1, a>1, and aa−1>b−1 hold, then from (112), we have θ≤μ; so, θ=μ. Then, the proof is completed.Theorem 10.
Assume that0<p<1, and an is a periodic sequence, s.t., an+2=an for all n=0,1,…,and then, (6) has a periodic solution of the prime period two.Proof.
For (6) possesses a periodic solution xn having prime period two, one can find positive numbers x0andx−1, s.t.,(113)x−1=x1=a0+x0x−1p,x0=x2=a1+x1x0p.
Or, equivalently,(114)x−1=a0+x0x−1p,x0=a1+x−1x0p.
Now, one has to prove that system (114) is consistent. From (114), one gets(115)x−1−a0x0−a1=1,and from this, it follows that(116)x−1−a0p+1=a1x−1−a0+1px−1p,(117)x0−a1p+1=a0x0−a1+1px0p.
Define(118)Fx=x−a0p+1−a1x−a0+1pxp,x>a0.
Then,(119)Fa0=−1a0<0,Fa0+1=−a1+1pa0+1p+1>0,where a1<a0. So, F has a zero, say x−1∈a0,a0+1, and in view of equation (115) and (116), one can get that (6) has a two-periodic solution. Now, assuming that a1>a0, one can define(120)Hx=−a0x−a1+1pxp+x−a1p+1,x>a1.
Then,(121)Ha1=−1a1<0,Ha1+1=−a0+1pa1+1p+1>0.
Thus,H has a zero, say x0∈a1,a1+1, and in view of equations (115) and (116), one can get that (6) has a two-periodic solution.


## 3. Numerical Simulation

In this section, we will provide some simulation in order to verify obtained theoretical results, for these following cases are to be considered for the completeness of this section ():Case 1: Ifa=2 and p=1.4<a+1=3, then Figure 1 implies that fixed point x¯=3 of equation (8) is a sink. This simulation agrees with the conclusion of Theorem 1.Case 2: Ifa=2 and p=2.1<a+1=3, then Figure 2 implies that fixed point x¯=3 of equation (8) is a sink. This simulation again agrees with the conclusion of Theorem 1.Case 3: Ifα=0.6,β=0.103,andp=0.4, then Figure 3 implies that fixed point u¯,v¯=1.6,1.103 of system (54) is a sink. Additionally, if α=0.6,β=0.103,andp=0.4, then p=0.4<β+1α+1/β+1p+α+1p=0.7854629670125335, and hence, this simulation agrees with the conclusion of Theorem 7.Case 4: Ifα=0.26,β=0.1,andp=0.04, then Figure 4 implies that fixed point u¯,v¯=1.26,1.1 of system (54) is a sink. Additionally, if α=0.26,β=0.1,andp=0.04, then p=0.04<β+1α+1/β+1p+α+1p=0.6884879889459452, and hence, this simulation again agrees with the conclusion of Theorem 7.Figure 1
Phase portrait of equation (8) with x−1=1.1andx0=1.7.Figure 2
Phase portrait of equation (8) with x−1=1.9andx0=2.9.Figure 3
Phase portrait of system (54) with u0,v0=1.2,1.9.
(a)(b)Figure 4
Phase portrait of system (54) with u0,v0=0.3,1.9.
(a)(b)



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