# Some Discussions on the Difference Equationxn+1=α+(xn-1m/xnk)

Authors: Awad A. Bakery
Journal: Journal of Function Spaces
 (2015)

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## Abstract

We give in this work the sufficient conditions on the positive solutions of the difference equationxn+1=α+(xn-1m/xnk),  n=0,1,…, where α, k, and m∈(0,∞) under positive initial conditions x-1,  x0 to be bounded, α-convergent, the equilibrium point to be globally asymptotically stable and that every positive solution converges to a prime two-periodic solution. Our results coincide with that known for the cases m=k=1 of Amleh et al. (1999) and m=1 of Hamza and Morsy (2009). We offer improving conditions in the case of m=1 of Gümüs and Öcalan (2012) and explain our results by some numerical examples with figures.


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## Body

## 1. Introduction

Throughout this paperl∞ and Cα denote, respectively, the class of bounded sequences of real numbers and its subclass of sequences convergent to α. Our aim in this paper is to investigate the positive solutions of the difference equation(1)xn+1=fxn,xn-1=α+xn-1mxnk,n=0,1,…,where f is a continuously differentiable function, α∈[0,∞), m, k∈(0,∞), and the initial conditions x-1, x0 are arbitrary positive numbers.Remark 1.
A pointx¯∈R is an equilibrium point of (1), if and only if it is a root for the function(2)gx=x-xm-k-α.If we replacexn and xn-1 in (1) by the variables u and v, respectively, then we have (3)p=∂f∂ux¯,x¯=-kx¯m-k-1,q=∂f∂vx¯,x¯=mx¯m-k-1.The linearized equation related with (1) about the equilibrium point x¯ is (4)yn+1=pyn+qyn-1,n=0,1,2….Its characteristic equation is(5)λ2-pλ-q=0.Definition 2.
An equilibrium pointx¯ of difference equation (1) is called locally stable if for every ɛ>0 there exists δ>0 such that if x-1,x0∈(0,∞) with |x-1-x¯|+|x0-x¯|<δ, then |xn-x¯|<ɛ for all n≥-1.Definition 3.
An equilibrium pointx¯ of (1) is called globally asymptotically stable if it is locally stable and, for every x-1,x0∈(0,∞), one has limn→∞⁡xn=x¯.Definition 4.
An equilibrium pointx¯ of difference equation (1) is called unstable if it is not locally stable.Definition 5.
A sequence(xn)n=-1∞ is said to be periodic with prime period p if p is the smallest positive integer with xn+p=xn for all n≥-1.Theorem 6 (see [1, 2]).
(i) A necessary and sufficient condition for both roots of (5) to lie in the open disk |λ|<1 is(6)p<1-q<2.
In this casex¯ is locally asymptotically stable.
(ii) A necessary and sufficient condition for one of the roots of (5) to lie outside the open disk |λ|<1 and the other root inside it is(7)p2+4q>0,p>1-q.
In this casex¯ is unstable and is called a saddle point.
(iii) A necessary and sufficient condition that a root of (5) has absolute value equal to 1 is (8)p=1-qor(9)q=-1,p≤2.In this case x¯ is called a nonhyperbolic point.By applying Theorem6 in special difference equation (1), we obtain the following results.Remark 7.
(i)x¯ is locally asymptotically stable, if and only if(10)kx¯m-k-1<1-mx¯m-k-1<2.(ii) x¯ is unstable, if and only if(11)kx¯m-k-12+4mx¯m-k-1>0,kx¯m-k-1>1-mx¯m-k-1.(iii) x¯ is called a nonhyperbolic point, if and only if(12)kx¯m-k-1=1-mx¯m-k-1or(13)mx¯m-k-1=-1,kx¯m-k-1≤2.Amleh et al. [3], for m=k=1, obtained important results for the following difference equation:(14)xn+1=α+xn-1xn.They showed in [3] that when α>1, the equilibrium point x¯=α+1 of (14) is globally asymptotically stable. When α=1, every positive solution of (14) converges to a period-two solution. Every positive solution of (14) is bounded if and only if α≥1. Finally, when 0<α<1, the equilibrium point is an unstable saddle point. Hamza and Morsy [4], for m=1, gave some results for the difference equation(15)xn+1=α+xn-1xnk∀k>0;they investigated the behavior of positive solutions of (15); they proved that when α>1, every positive solution of (15) is bounded and when α>k1/k≥1, the equilibrium point x¯ of (15) is globally asymptotically stable and they showed that (15) has periodic solutions without conditions on α and k. These results are improved in [5] by Gümüs and Öcalan; they investigated the boundedness character of positive solutions of (15); they proved that if 0≤α<1, then there exist unbounded solutions of (15) in the cases when α∈(0,∞) and k→0+ or α>1 and k→∞; then every positive solution of (15) is unbounded. When k↛0+, k↛∞, and α>1, then every positive solution of (15) is bounded. Also they show that if k↛0+, k↛∞, α>1, and α>k1+k1-k/k, then the equilibrium point x¯ of (15) is globally asymptotically stable. Finally they explained the sufficient condition for the fact that every positive solution of (15) converges to a prime two-periodic solution. In our work we prove in Sections 3 and 4 that these results are incorrect and we give the results under finer conditions and we illustrate them in some figures. Closely related equations to (1) are investigated by many authors, for example, [6–11].In this work in Section2 we study the global behavior of (1). In Section 3 we investigate the boundedness character of positive solutions of (1). Section 4 is dedicated to the periodic character of the positive solutions of (1). Finally we explain the sufficient condition that every positive solution of (1) converges to a prime two-periodic solution. Throughout the paper we denote by S the class (16)S≔xnn=-1∞:xn∈R+,xnis a solution of1and the subclass Si of the class S by (17)Si≔xnn=-1∞∈S:x0,x-1∈0,2α.The well-known inequality in [12] will be used: if 0<p<∞ and D=max⁡⁡(1,2p-1), then (18)a+bp≤Dap+bp∀a,b∈R.


## 2. The Difference Equationxn+1=α+xn-1m/xnk

In this section we give the sufficient conditions to determine the classification of the equilibrium points for (1) and its uniqueness.Lemma 8.
Letx¯ be the equilibrium point of (1) and m∈(0,k+1] for all k∈(0,∞).(i)
Ifα>(m+k-1)m+km-k/k+1-m, then the equilibrium point x¯ of (1) is locally asymptotically stable.(ii)
Ifα<(m+k-1)m+km-k/k+1-m, then the equilibrium point x¯ of (1) is unstable.(iii)
Ifα=(m+k-1)(m+k)m-k/k+1-m, then x¯ is a nonhyperbolic point.Proof.
Give an equilibrium pointx¯ of (1) and m∈(0,k+1] for all k∈(0,∞).(i)
Supposeα>(m+k-1)(m+k)m-k/k+1-m; then from (2) we have(19)gm+k1/k+1-m=m+k-1m+km-k/k+1-m-α<0,limx→∞⁡gx=∞.Hencex¯>(m+k)1/k+1-m and using Remark 7 we get that x¯ is locally asymptotically stable.(ii)
Letα<(m+k-1)(m+k)m-k/k+1-m; then from (2) we have(20)gm+k1/k+1-m=m+k-1m+km-k/k+1-m-α>0,g0<0;hencex¯<(m+k)1/k+1-m and using Remark 7 we get that x¯ is unstable.(iii)
Ifα=(m+k-1)(m+k)m-k/k+1-m, then we obtain g((m+k)1/k+1-m)=0.Sox¯=(m+k)1/k+1-m; by using Remark 7 we get that x¯ is a nonhyperbolic point.Lemma 9.
One has the following:(i)
Ifk=m=1, then (1) has a unique equilibrium point x¯=α+1.(ii)
Ifk=m, then (1) has a unique equilibrium point x¯=α+1.(iii)
Ifk<m≤k+1, then (1) has a unique equilibrium point x¯>1.(iv)
If0<m<k, then (1) has a unique equilibrium point x¯>1.Proof.
(i) Ifk=m=1, (1) gives x¯=1+α. See [3].
(ii) Ifk=m, (1) gives x¯=1+α. See [13].
(iii) Ifk<m≤k+1, the function g given by (2) is decreasing on [0,(m-k)1/m-k+1] and increasing on [(m-k)1/m-k+1,∞[. Since g(1)=-α<0 and limx→∞g(x)=∞, then g has a unique root x¯>1.
(iv) If0<m<k, the function g is increasing on [0,∞[. Since g(1)=-α<0 and limx→∞g(x)=∞, then g has a unique root x¯>1.


## 3. Global Behavior of Solutions and Boundedness

In this section, we present the conditions for the boundedness of the positive solutions and its global stability for the equilibrium point of (1).Theorem 10.
If1≤m≤k+1 and k→0+, then S∩l∞=∅.Proof.
Suppose on the contrary that{xn}n=-1∞∈S∩l∞. Then, we have(21)ψ≔lim⁡inf⁡⁡xn,Ψ≔lim⁡sup⁡⁡xn<∞.Therefore, from (1) we get (22)ψ≥α+ψmΨk.Let k→0+; this implies m→1; we obtain (23)ψ≥α+ψ⟹α≤0,which contradicts α>0, so the proof is complete.In Figures1 and 2 we illustrate the result of Theorem 10 for m=1 and small values of k=10-1, 10-2, 10-4, 10-8, and 10-13 for, respectively, α=1.5 and α=0.5, and explain that the solutions of (1) are unbounded for every α>0 and k→0+.Figure 1
Whenm=1 and k→0 the positive solution is unbounded.Figure 2
Whenm=1 and k→0 the positive solution is unbounded.Theorem 11.
Ifm>1, then S∖l∞≠∅.Proof.
For fixedm>1 let α∈(0,1). Choose δ∈(0,1-α) and let {xn}n=-1∞∈S with the initial conditions such that (24)x-1m>α+δδ1/k-αα+δk,0<α<x0<α+δ<1,hence,(25)x1=α+x-1mx0k>α+x-1mα+δk,x2=α+x0mx1k<α+α+δmα+x-1m/α+δkk<α+δ.Moreover we have(26)x3=α+x1mx2k>α+1α+δkα+x-1mα+δkm>α+x-1mα+δk,x4=α+x2mx3k<α+δmα+x-1m/α+δkk<α+δ.Therefore, by induction we can show that (27)x2n+1>x-1mα+δkn+1+α1+1α+δk+⋯+1α+δnk,α<x2n<α+δ.Hence,(28)1α+δk>1,which gives that (29)limn→∞⁡x2n+1=∞.Thus {xn}n=-1∞ is unbounded. Now, we assume that α=0 and choose the initial conditions such that(30)0<x-1≤1,x0>11-ɛk,for some0<ɛ<1.So, we obtain(31)0<x1=x-1mx0k≤1x0k<1-ɛ,x2=x0mx1k>x0m1-ɛk.Moreover we have(32)x3=x1mx2k<1x0mk<1-ɛm,x4=x2mx3k>x0m21-ɛ2mk.By induction we have (33)x2n+1∈0,1-ɛmn,x2n>x0mn1-ɛnmn-1k∀n=0,1,….Therefore since m>1, we get(34)limn→∞⁡x2n=∞,limn→∞⁡x2n+1=0.For fixed m>1 let α≥1; we deduce that (35)sup⁡n⁡xn:xn+1=α+xn-1mxnk>sup⁡n⁡xn:xn+1=α+xn-1xnk>sup⁡n⁡xn:xn+1=12+xn-1xnk.Since there exists unbounded solution of the difference equation xn+1=1/2+xn-1m/xnk, n=0,1,2,…, there exists unbounded solution of difference equation (1). This completes the proof.In Figure3 we illustrate the result of Theorem 11 for k=3, m=2, and α=0.5, with the initial conditions x-1=0.7, x0=0.54 and choosing δ=0.2; in this case the subsequence (x2n) of the solution of (1) is bounded with 0.5<x2n<0.7 for all n≥0, while the subsequence (x2n+1) is unbounded, so the terms of (x2n) appear approximately zero in comparison with the terms of (x2n+1). Also Figure 4 explains the same result for k=0.3, m=2, and α=1.5; Gümüs and Öcalan in [5] showed that when α>1 and k→∞, every solution of (1) is unbounded. Figure 5 gives a counter example of this result for k=1010, m=1, and α=1.1. We correct this result in Theorem 12 and Corollary 13.Figure 3
Whenm>1 there exists an unbounded positive solution.Figure 4
Whenm>1 there exists an unbounded positive solution.Figure 5
Whenα>1 and k→∞ the positive solution converges to α, which gives a counter example of Theorem 2.1-(b) and the results that followed in [5]. Here α=1.1, m=1, and k=1010.Theorem 12.
(a)Si⊆l∞, if 1<m<k+1 and α≥2m/k+1-m.
(b)S⊆l∞, if 0<m≤1 and α>1.Proof.
(a) From (1), xn≥α>1 for n≥0. We get for n≥0 and 1<m<k+1 that(36)x2n+1=α+x2n-1mx2nk≤α+x2n-1mαk=α+1αkα+x2n-3mx2n-2km≤α+1αkα+x2n-3mαkm≤α+2m-1αmαk+2m-1x2n-3m2αkm+1≤α+2m-1αmαk+2m-11αkm+1α+x2n-5mαkm2⋮≤α+∑n=1∞2mn+1-m/m-1-nαmnαkmn-1/m-1+2mn+1-m/m-1-nx-1mn+1αkmn+1-1/m-1≤α+∑n=1∞2mn+1-m/m-1-nαmnαkmn-1/m-1+∑n=1∞2mn+1-m/m-1-nx-1mn+1αkmn+1-1/m-1.Since α≥2m/k+1-m, we have the series(37)∑n=1∞2mn+1-m/m-1-nαmnαkmn-1/m-1<∞.To study the convergence of the series with general term an=2mn+1-m/m-1-nx-1mn+1/αk(m(n+1)-1/m-1), since (38)an+1an=2mn+1-1x-1mn+1m-1α-kmn+1for x-1∈(0,α] and since α≥2m/k+1-m, we obtain (39)an+1an≤2mn+1-1α-mn+1k-m+1≤2mn+21/m-1-1.Since m>1, then by using the ratio test we obtain (40)limn→∞⁡an+1an=0<1;then the series is convergent. And for x-1∈(α,2α] and since α≥2m/k+1-m, we obtain (41)an+1an≤2mn+1-12αmn+2-mn+1α-kmn+1≤2mn+2-1α-mn+1k+1-m≤12.So by using the ratio test we obtain (42)limn→∞⁡an+1an=12<1;then the series is convergent and hence (x2n+1)-1∞ is bounded. Also, we get(43)x2n=α+x2n-2mx2n-1k≤α+x2n-2mαk=α+1αkα+x2n-4mx2n-3km≤α+1αkα+x2n-4mαkm≤α+2m-1αmαk+2m-1x2n-4m2αkm+1≤α+2m-1αmαk+2m-11αkm+1α+x2n-6mαkm2⋮≤α+∑n=1∞2mn+1-m/m-1-nαmnαkmn-1/m-1+2mn-m/m-1-nx0mnαkmn-1/m-1≤α+∑n=1∞2mn+1-m/m-1-nαmnαkmn-1/m-1+∑n=1∞2mn-m/m-1-nx0mnαkmn-1/m-1.To study the convergence of the series with general term bn=2mn-m/m-1-nx0mn/αk(m(n)-1/m-1), since (44)bn+1bn=2mn-1x0mnm-1α-kmnfor x0∈(0,α] and since α≥2m/k+1-m, we obtain (45)bn+1bn≤2mn-1α-mnk-m+1≤2mn+11/m-1-1.Since m>1, then by using the ratio test we obtain (46)limn→∞⁡an+1an=0<1;then the series is convergent. And for x0∈(α,2α] and since α≥2m/k+1-m, we obtain (47)bn+1bn≤2mn-12αmn+1-mnα-kmn≤2mn+1-1α-mnk+1-m≤12.So by using the ratio test we obtain (48)limn→∞⁡bn+1bn=12<1;then the series is convergent and hence (x2n)0∞ is bounded.
(b) In addition, forn≥0, 0<m≤1, and α>1, we have(49)x2n+1=α+x2n-1mx2nk≤α+x2n-1mαk=α+1αkα+x2n-3mx2n-2km≤α+1αkα+x2n-3mαkm≤α+αmαk+x2n-3m2αkm+1≤α+αmαk+1αkm+1α+x2n-5mαkm2⋮≤αkαk-1+x-1αkn+1.Again, we obtain(50)x2n=α+x2n-2mx2n-1k≤α+x2n-2mαk=α+1αkα+x2n-4mx2n-3km≤α+1αkα+x2n-4mαkm≤α+αmαk+x2n-4m2αkm+1≤α+αmαk+1αkm+1α+x2n-6mαkm2⋮≤αkαk-1+x0αnk<∞,from which the proof follows.Corollary 13.
For fixedm, if α>1 and k→∞, then S⊆Cα.Proof.
Sinceα>1 and k→∞, we have {xn}n=-1∞∈S∩l∞. Then, we obtain(51)α<ψ=lim⁡inf⁡⁡xn,Ψ=lim⁡sup⁡⁡xn<∞.Hence, from (1), we get(52)Ψ≤α+Ψmψk.Let k→∞; then we obtain(53)Ψ≤α,which implies Ψ=ψ=α. So the proof is complete.Theorem 14.
The unique positive equilibriumx¯ of (1) is globally asymptotically stable, if the following conditions are satisfied:(i)
0<m≤1,(ii)
α>1 and α>(m+k-1)(m+k)m-k/k+1-m.Proof.
By Lemma8, x¯ is locally asymptotically stable. Thus, we have to prove that, for all {xn}n=-1∞∈S, there exists a unique positive equilibrium x¯ with limn→∞xn=x¯. Let {xn}n=-1∞∈S. By Theorem 12-(b), {xn}n=-1∞∈l∞. Thus, we have(54)α≤ψ=lim⁡inf⁡⁡xn,Ψ=lim⁡sup⁡⁡xn<∞.Hence from (1), we get(55)Ψ≤α+Ψmψk,ψ≤α+ψmΨk.We maintain that ψ=Ψ; otherwise Ψ>ψ. From (55), we obtain(56)Ψψk≤αψk+Ψm,Ψkψ≥αΨk+ψm.And from (56),(57)αΨkψk-1+ψk+m-1≤αψkΨk-1+Ψk+m-1.Hence, we have(58)αΨk-1ψk-1≤Ψk+m-1-ψk+m-1Ψ-ψ.For k+m≥2 and 0<m≤1, we have k≥1. Consider f(x)=xk+m-1 with x∈(0,∞); then there exists c∈(ψ,Ψ) such that(59)Ψk+m-1-ψk+m-1Ψ-ψ=k+m-1ck+m-2≤k+m-1Ψk+m-2.From (58) and (59), we obtain(60)αΨk-1ψk-1≤k+m-1Ψk+m-2,which is equivalent to(61)αk+m-1ψk-1≤Ψm-1.Using Ψ≥ψ≥α and α>1 in (61), we get(62)αk+1-m≤k+m-1.Equation (62) with the values m=0.6, k=1.5, and α=2 gives a contradiction. Thus, we find Ψ=ψ. Now, assume that 0<k+m<2, 0<m≤1, and k<1. From (58), we get (63)αΨk-1ψk-1≤k+m-1ψk+m-2.This gives(64)αΨk-1≤k+m-1ψm-1.Moreover, since xn≥1 for all n≥0, we have(65)x2n+1<α+x2n-1mαk<α+x2n-1αk.Then from (65), we get(66)α≤ψ<Ψ<αk+1αk-1.By using (64) and (66), we obtain(67)αk2-mαk-1k-1≤k+m-1α.Since α>1 and α>(m+k-1)(m+k)m-k/k+1-m, for α=2, m=1, and k=0.5, (67) is not satisfied. This is a contradiction. Thus, we find Ψ=ψ, which implies that {xn}n=-1∞ tends to the unique positive equilibrium, from which the proof follows.


## 4. Periodicity of the Solutions ofxn+1=α+(xn-1m/xnk)

In this section we study the convergence of the positive solution of (1) to a prime two-periodic solution when α>1, 0<m≤1<k.Lemma 15.
Suppose{xn}n=-1∞∈S and β>α. Then, the following conditions are contented: (i)
limn→∞x2n=β, if and only if limn→∞x2n+1=βm/k/(β-α)1/k.(ii)
limn→∞x2n+1=β, if and only if limn→∞x2n=βm/k/(β-α)1/k.Proof.
(i) Replacen by 2n+1 in (1); we deduce(68)x2n+1=x2nmx2n+2-α1/k∀n≥0.If limn→∞x2n=β, then limn→∞x2n+1=βm/k/β-α1/k and vice versa.
(ii) Replacen by 2n in (1); we have(69)x2n=x2n-1mx2n+1-α1/k∀n≥0.Let limn→∞x2n+1=β; then limn→∞x2n=βm/k/(β-α)1/k and vice versa.Theorem 16.
If these conditions are satisfied(70)α>1,0<m≤1<kand there exists a sufficient small positive number ɛ1 such that(71)α+ɛ1m/kɛ1m/k-α+ɛ1m2-k2/kɛ11/k<αɛ1m+1/k,(72)ɛ1α+ɛ1k-m<1,then, (1) has a periodic solution of prime period two.Proof.
Let{xn}n=-1∞ be a periodic solution of period two; we have(73)x-1=α+x-1mx0k=x1,x0=α+x0mx-1k=x2;then (x-1,x0) is a solution of the system(74)x=α+xmyk,y=α+ymxk.Let {xn}n=-1∞∈S. It is obvious that if (73) holds, then {xn}n=-1∞ is a periodic solution of period two. System (74) is correspondent to(75)y-α-ymxk=0,y=xmx-α1/k,so we get the equation(76)Hx=1x-α1/kxm/k-x-α1-m/kxm2-k2/k-α.Thus for 0<m≤1, we have (77)limx→α+⁡Hx=∞.Moreover, from (71) and (76) we have (78)Hα+ɛ1<0.Therefore, the equation H(x)=0 has a root x¯=α+ɛ0, where 0<ɛ0<ɛ1, in the interval (α,α+ɛ1). So, we have (79)y¯=x¯mx¯-α1/k.Let us consider the function(80)Tɛ=α+ɛm-k-ɛ.Since 0<m<k, we get(81)T′ɛ=m-kα+ɛm-k-1-1<0,Tɛ0>Tɛ1.From inequality (72), we have T(ɛ1)>0, and hence T(ɛ0)>0, which implies that (82)x¯=α+ɛ0<α+ɛ0mɛ01/k=y¯.By taking x-1=x¯, x0=y¯, then the solution {xn}n=-1∞ with initial values x-1, x0 is a prime 2-periodic solution. This completes the proof.Lemma 17.
If{xn}n=-1∞∈S which consists of at least two semicycles, then {xn}n=-1∞ is oscillatory and, except possibly for the first semicycle, every semicycle is of length 1.Proof.
The proof is directly obtained from (1).Theorem 18.
For{xn}∈S, the sequences {x2n} and {x2n+1} are ultimately monotone.Proof.
We have(83)x2n+1-x2n-1=x2n-2kx2n-1m-x2n-3m+x2n-3mx2n-2k-x2nkx2n+1x2n-2k,x2n+2-x2n=x2n-1kx2nm-x2n-2m+x2n-2mx2n-1k-x2n+1kx2n+1x2n-1k.If x1≥x-1 and x2≤x0, we obtain from (83) x3≥x1 and therefore x4≤x2. By induction we find that {x2n}n=0∞ is decreasing and {x2n+1}n=-1∞ is increasing. Correspondingly if x1≤x-1 and x2≥x0, using induction we obtain from (83) that {x2n}n=0∞ is increasing and {x2n+1}n=-1∞ is decreasing. If x1≥x-1 and x2≥x0 and x1≥x3, we have from (83) that {x2n}n=0∞ is increasing and {x2n+1}n=0∞ is decreasing. Hence, we can assume that x3≥x1≥x-1 and x0≤x2. If further x2≥x4, then {x2n}n=1∞ is decreasing and {x2n+1}n=-1∞ is increasing. So we may assume that x3≥x1≥x-1 and x0≤x2≤x4. By induction we get the result in this case. The cases x1≤x-1 and x2≤x0 can be treated in the same way.Theorem 19.
If (70), (71), and (72) are conformed, then every positive solution of (1) converges to a prime two-periodic solution.Proof.
The sequences{x2n} and {x2n+1} are eventually monotone and bounded by Theorems 18 and 12 for every positive solution of (1). So, the sequences {x2n} and {x2n+1} are convergent. Using Lemmas 15 and 17 the proof follows.Figure6 illustrates the result of Theorem 19 for α=1.1, m=0.99, and k=2.Figure 6
Existence of two-periodic positive solution,ɛ0=2.5281, x-1=x¯=3.6201, and x0=y¯=1.1903.



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*Source: 737420-2015-05-10.xml*