# Some Notes on the Difference Equationxn+1=α+(xn−1/xnk)

Authors: Mehmet Gümüş; Özkan Öcalan
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## Abstract

We investigate the behavior of the solutions of the recursive sequencexn+1=α+xn-1/xnk,n=0,1,…, where α∈[0,∞), k∈(0,∞), and the initial conditions x-1,x0 are arbitrary positive numbers. Included are results that considerably improve those in the recently published paper by Hamza and Morsy (2009).


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## Body

## 1. Introduction

Our aim in this paper is to give some remarks for the positive solutions of the difference equation(1.1)xn+1=α+xn-1xnk,n=0,1,…,
where α∈[0,∞), k∈(0,∞), and the initial conditions x-1,x0 are arbitrary positive numbers. Amleh et al. in [1] obtained important results for the difference equation
(1.2)xn+1=α+xn-1xn,
which guide many authors. It was proved in [1] that, when α>1, the equilibrium x¯=α+1 of  (1.2) is globally asymptotically stable. When α=1, every positive solution of (1.2) converges to a period-two solution. Every positive solution of (1.2) is bounded if and only if α≥1. Finally, when 0<α<1, the equilibrium is an unstable saddle point. Closely related equations to (1.1) are investigated by many authors, for example, [2–10].In [4] the authors investigated the behavior of positive solutions of (1.1). It was proved in [4] that, when α≠1, every positive solution of (1.1) is bounded and when α>k1/k≥1, the equilibrium x¯ of (1.1) is globally asymptotically stable. But in [4] the authors obtain some incorrect results for the boundedness character and the global stability of solutions of (1.1), and it is not shown that (1.1) has periodic solutions with conditions of α and k.Our aim here is to improve and correct these results and extend some of the results in [4].We say that the equilibrium pointx¯ of the equation
(1.3)xn+1=F(xn,xn-1,…,xn-k),n=0,1,…,
is the point that satisfies the condition
(1.4)x¯=F(x¯,x¯,…,x¯).
A positive semicycle of a solution {xn}n=-1∞ of (1.1) consists of a “string” of terms {xl,xl+1,…,xm} all greater than or equal to x¯, with l≥-1 and m≤∞, such that
(1.5)eitherl=-1orl>-1,xl-1<x¯,eitherm=∞orm<∞,xm+1<x¯.
A negative semicycle of a solution {xn}n=-1∞ of (1.1) consists of a “string” of terms {xl,xl+1,…,xm} all less than x¯, with l≥-1 and m≤∞, such that
(1.6)eitherl=-1orl>-1,xl-1≥x¯,eitherm=∞orm<∞,xm+1≥x¯.
A solution {xn}n=-k∞ of (1.1) is called nonoscillatory if there exists N≥-1 such that either
(1.7)xn>x¯∀n≥N
or
(1.8)xn<x¯∀n≥N.{xn}n=-1∞ of (1.1) is called oscillatory if it is not nonoscillatory. We say that a solution {xn}n=-1∞ of (1.1) is bounded and persists if there exist positive constants P and Q such that P≤xn≤Q for n=-1,0,….The linearized equation for (1.1) about the positive equilibrium x¯ is
(1.9)yn+1+kx¯-kyn-x¯-kyn-1=0,n=0,1,….
We need the following lemmas, which were given in [4].Lemma 1.1.
Letx¯ be the equilibrium point of (1.1).(i)
Ifk(1+k)(1-k)/k<α, then the equilibrium point x¯ of (1.1), is locally asymptotically stable.(ii)
Ifk(1+k)(1-k)/k>α, then the equilibrium point x¯ of  (1.1) is unstable.Lemma 1.2.
The following statements are true.(i)
Ifk=1, then (1.1) has a unique equilibrium point x¯=α+1.(ii)
Ifk≠1, then (1.1) has a unique equilibrium point x¯>1.Lemma 1.3.
Let{xn}n=-1∞ be a solution of (1.1), 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 one.The paper is organized as follows. In Section2 we investigate the boundedness character of positive solutions of (1.1). We prove that if 0≤α<1, then there exist unbounded solutions of (1.1) and when the cases either α∈(0,∞) and k→0+ or α>1 and k→∞, then every positive solution of (1.1) is unbounded. We show that when α>1 and k↛0+,k↛∞, then every positive solution of (1.1) is bounded. Also we show that if α>1, α>k(1+k)(1-k)/k and k↛0+,k↛∞, then the equilibrium point x¯ of (1.1) is globally asymptotically stable. Section 3 is devoted to the periodic character of the positive solutions of (1.1). Finally we show that a sufficient condition that every positive solution of (1.1) converges to a prime two periodic solution.


## 2. Boundedness and Global Stability of (1.1)

In this section, we present some results for the boundedness character of positive solutions and global stability of the equilibrium point of (1.1).Theorem 2.1.
Consider (1.1). Then, the following statements are true.(a)
Ifα∈(0,∞) and k→0+, then every positive solution of (1.1) is unbounded.(b)
Ifα>1 and k→∞, then every positive solution of (1.1) is unbounded.Proof.
(
a
) On the contrary, we assume that {xn}n=-1∞ is a positive bounded solution of (1.1). Then, we have
(2.1)s=lim⁡inf⁡xn,S=lim⁡sup⁡xn<∞.
Thus, from (1.1) we get
(2.2)s≥α+sSk.
Let k→0+; then we obtain
(2.3)s≥α+s,α≤0,
which contradicts α>0, so the proof is complete.
(
b
) Again we assume that {xn}n=-1∞ is a positive bounded solution of (1.1). Then, we have
(2.4)α<s=lim⁡inf⁡xn,S=lim⁡sup⁡xn<∞.
Thus, from (1.1), we have
(2.5)S≤α+Ssk.
Let k→∞; then
(2.6)S≤α,
which contradicts α<s, so the proof is complete.Now, we show that if0≤α<1, then there exist positive solutions of (1.1) that are unbounded.Theorem 2.2.
One has(2.7)0≤α<1.
Then there exist positive solutions of (1.1) that are unbounded.Proof.
Assume thatα∈(0,1). Choose δ∈(0,1-α), and let {xn}n=-1∞ be a solution of (1.1) with the initial conditions such that
(2.8)x-1>((α+δδ)1/k-α)(α+δ),α<x0<α+δ.
Then,
(2.9)x1=α+x-1x0k>α+x-1(α+δ)k,x2=α+x0x1k<α+α+δ(α+x-1/(α+δ)k)k<α+δ.
Further we have
(2.10)x3=α+x1x2k>α+x1(α+δ)k>α+1(α+δ)k(α+x-1(α+δ)k)>α+x-1(α+δ)k,x4=α+x2x3k<α+α+δ(α+x-1/(α+δ)k)k<α+δ.
Therefore, working inductively we can prove that for n=0,1,…(2.11)x2n+1>α+x2n-1(α+δ)k>α+x-1(α+δ)k,α<x2n<α+δ.
Hence,
(2.12)x2n+1>x-1((α+δ)k)n+1+α(1+1(α+δ)k+⋯+1((α+δ)k)n).
Since
(2.13)1(α+δ)k>1,
which implies that
(2.14)lim⁡n→∞x2n+1=∞{xn}n=-1∞ is unbounded. For α∈(0,1), the proof is complete.
Now, we assume thatα=0 and choose the initial conditions such that
(2.15)0<x-1≤1,x0>1(1-ε)1/kforsome0<ε<1.
So, we have
(2.16)0<x1=x-1x0k≤1x0k<1-ε,x2=x0x1k>x0(1-ε)k.
Further we have
(2.17)x3=x1x2k<1x0k<1-ε,x4=x2x3k>x0/(1-ε)k(1-ε)k=x0(1-ε)2k.
By induction we have
(2.18)x2n+1∈(0,1-ε),x2n>x0((1-ε)k)n∀n=0,1,….
Thus,
(2.19)lim⁡n→∞x2n=∞,lim⁡n→∞x2n+1=0.
This completes the proof.The following theorem is given in [4].Theorem 2.3.
Suppose thatα≠1; then; every positive solution of (1.1) is bounded.In [4] this result is not correct. So, we give the following theorem for the boundedness of (1.1).Theorem 2.4.
Suppose thatα>1, k↛0+, and k↛∞; then every positive solution of (1.1) is bounded.Proof .
From (1.1), xn≥α for n≥0. Thus, from (1.1), without loss of generality, we obtain for n≥0(2.20)x2n+1=α+x2n-1x2nk≤α+x2n-1αk,x2n=α+x2n-2x2n-1k≤α+x2n-2αk.
From (2.20) using induction, we obtain
(2.21)x2n+1≤α(1+1αk+1(αk)2+⋯+1(αk)n)+x-1(αk)n+1≤αkαk-1+x-1(αk)n+1,x2n≤α(1+1αk+1(αk)2+⋯+1(αk)n-1)+x0(αk)n≤αkαk-1+x0(αk)n.
From which the proof follows.Actually, Hamza and Morsy in [4] obtained global stability of the equilibrium point of (1.1). But the result does not include the case k∈(0,1) and some parts of its proof are incorrect. So, here we will obtain global stability of the equilibrium point of (1.1) when k∈(0,1).Theorem 2.5.
Consider (1.1). Let k↛0+ and k↛∞. Suppose that
(2.22)α>1,α>k(1+k)(1-k)/k
hold. Then, the unique positive equilibrium x¯ of (1.1) is globally asymptotically stable.Proof.
By Lemma1.1, x¯ is locally asymptotically stable. Thus, it is enough for the proof that every positive solution of (1.1) tends to the unique positive equilibrium x¯. Let {xn}n=-1∞ be a solution of (1.1). By Theorem 2.4, {xn}n=-1∞ is bounded. Thus, we have
(2.23)1<s=lim⁡inf⁡xn,S=lim⁡sup⁡xn<∞.
Then, from (2.23), we get
(2.24)S≤α+Ssk,s≥α+sSk.
We claim that s=S, otherwise S>s. From (2.24), we obtain
(2.25)Ssk≤αsk+S,Sks≥αSk+s,
And, from (2.25),
(2.26)αSksk-1+sk≤αskSk-1+Sk.
Thus,
(2.27)αSk-1sk-1(S-s)≤Sk-sk.
Assume that k≥1. We consider f(x)=xk with x∈(0,∞); then there exists a,c∈(s,S) such that
(2.28)Sk-skS-s=kck-1≤kSk-1.
From (2.27) and (2.28), we obtain
(2.29)αSk-1sk-1≤kSk-1,
which is equivalent to
(2.30)αsk-1≤k.
Since we have S,s≥α, we get
(2.31)αk≤k.
Since α>1 and α>k(1+k)(1-k)/k, for some values α and k, (2.31) is not satisfied. This is a contradiction. Thus, we find S=s.
Now, assume thatk<1. Then, from (2.27) and arguing as above, we get
(2.32)αSk-1sk-1≤ksk-1.
Furthermore, we have
(2.33)x2n+1<α+x2n-1αk.
We consider the following difference equation:
(2.34)ym+1=α+ymαk.
Every positive solution of the previous equation converges to αk+1/(αk-1). It follows that S=limsupxn≤αk+1/(αk-1). Then, we obtain that
(2.35)(αk+1αk-1)k-1≤Sk-1≤kα.
Thus,
(2.36)α(αk+1αk-1)k-1≤k.
Since α>1 and α>k(1+k)(1-k)/k, for some values α and k, (2.36) is not satisfied. So s=S, which implies that {xn}n=-1∞ tends to the unique positive equilibrium. From which the proof follows.Remark 2.6.
Consider (1.1), where 0≤α<1. Let {xn}n=-1∞ be a solution of (1.1). If {xn}n=-1∞ is bounded, then it is stable too.


## 3. Periodicity of the Solutions of (1.1)

In this section we investigate the periodicity of (1.1) when α>1 and k>1,k↛∞.We need the following lemma whose proof follows by simple computation and thus it will be omitted.Lemma 3.1.
Let{xn}n=-k∞ be a solution of (1.1), and let L>α. Then, the following statements are satisfied.(i)
limn→∞x2n=L if and only if limn→∞x2n+1=L1/k/(L-α)1/k,(ii)
limn→∞x2n+1=L if and only if limn→∞x2n=L1/k/(L-α)1/k.Theorem 3.2.
Consider (1.1), where
(3.1)α>1,k>1,k↛∞.
Assume that there exists a sufficient small positive number ε1 such that
(3.2)(α+ε1)1/k-(α+ε1)(1-k2)/k<αε11/k,(3.3)ε1<(α+ε1)1-k.
Then, (1.1) has a periodic solution of prime period two.Proof.
Let{xn}n=-1∞ be a solution of (1.1). It is obvious that if
(3.4)x-1=α+x-1x0k,x0=α+x0x-1k
hold, then {xn}n=-1∞ is a periodic solution of period two. Consider the system(3.5)x=α+xyk,y=α+yxk.
Then this system is equivalent to
(3.6)y-α-yxk=0,y=(xx-α)1/k,
and so we get the equation
(3.7)F(x)=(xx-α)1/k-α-x(1-k2)/k(x-α)1/k
and obtain
(3.8)F(x)=1(x-α)1/k(x1/k-x(1-k2)/k)-α.
Thus,
(3.9)lim⁡x→α+F(x)=∞.
Moreover, from (3.2) we can show that
(3.10)F(α+ε1)<0.
Therefore, the equation F(x)=0 has a solution x¯=α+ε0, where 0<ε0<ε1, in the interval (α,α+ε1). So, we have
(3.11)y¯=(x¯x¯-α)1/k.
We now consider the function
(3.12)K(ε)=(α+ε)1-k-ε.
Since from (3.1)
(3.13)K′(ε)=(1-k)(α+ε)-k-1<0,
we have
(3.14)K(ε0)>K(ε1).
From (3.3), we have K(ε1)>0, and thus K(ε0)>0, which implies that
(3.15)x¯=α+ε0<(α+ε0ε0)1/k=y¯.
Hence, if 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.In the following theorem we will generalize the result due to Stević [6, Theorem 2].Theorem 3.3.
For every positive solution{xn} of (1.1), the sequences {x2n} and {x2n+1} are eventually monotone.Proof.
We have(3.16)x2n+1-x2n-1=x2n-2k(x2n-1-x2n-3)+x2n-3(x2n-2k-x2nk)(x2nx2n-2)k,x2n+2-x2n=x2n-1k(x2n-x2n-2)+x2n-2(x2n-1k-x2n+1k)(x2n+1x2n-1)k.
If x1≥x-1 and x2≤x0, we obtain from (3.16) x3≥x1 and consequently x4≤x2. By induction we obtain
(3.17)x0≥x2≥⋯≥x2n≥⋯,⋯≥x2n+1≥x2n-1≥⋯≥x1≥x-1.
Similarly if x1≤x-1 and x2≥x0, using induction we obtain from (3.16)
(3.18)x0≤x2≤⋯≤x2n≤⋯,⋯≤x2n+1≤x2n-1≤⋯≤x1≤x-1.
If x1≥x-1, x2≥x0 and x1≥x3, we can obtain from (3.16)
(3.19)x0≤x2≤⋯≤x2n≤⋯,⋯≤x2n+1≤x2n-1≤⋯≤x1.
Hence, we may assume that x3≥x1≥x-1 and x0≤x2. If further x2≥x4, then
(3.20)x2≥⋯≥x2n≥⋯,⋯≥x2n+1≥x2n-1≥⋯≥x1≥x-1.
So we may assume that x3≥x1≥x-1 and x0≤x2≤x4. By induction we obtain the result in this case.
The casesx1≤x-1 and x2≤x0 can be treated similarly.Theorem 3.4.
Consider (1.1) where (3.1), (3.2), and (3.3) hold. Then, every positive solution of (1.1) converges to a prime two periodic solution.Proof.
By Theorem3.3, for every positive solution of (1.1) the sequences {x2n} and {x2n+1} are eventually monotone. By Theorem 2.4, the sequences {x2n} and {x2n+1} are bounded. Hence, the sequences {x2n} and {x2n+1} are convergent. Using Lemmas 1.3 and 3.1 the result follows.



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*Source: 258502-2012-08-08.xml*