# On the Difference Equationxn+1=xnxn-k/(xn-k+1a+bxnxn-k)

Authors: Stevo Stević; Josef Diblík; Bratislav Iričanin; Zdenĕk Šmarda
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We show that the difference equationxn+1=xnxn-k/xn-k+1(a+bxnxn-k),n∈ℕ0, where k∈ℕ, the parameters a, b and initial values x-i, i=0,k̅ are real numbers, can be solved in closed form considerably extending the results in the literature. By using obtained formulae, we investigate asymptotic behavior of well-defined solutions of the equation.


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## Body

## 1. Introduction

Recently, there has been some reestablished interest in difference equations which can be solved, as well as in their applications, see, for example, [1–16]. For some old results in the topic see, for example, the classical book [17].In recently accepted paper [18] are given formulae for the solutions of the following four difference equations:
(1.1)xn+1=xnxn-3xn-2(±1±xnxn-3),n∈N0,
and some of these formulae are proved by induction.Here, we show that the formulae obtained in [18] follow from known results in a natural way. Related idea was exploited in paper [7].Moreover, we will consider here the following more general equation:(1.2)xn+1=xnxn-kxn-k+1(a+bxnxn-k),n∈N0,
where k∈ℕ and the parameters a,b as well as initial values x-i,i=0,k¯ are real numbers, and describe the behaviour of all well-defined solutions of the equation.For a solution(xn)n≥-k, k∈ℕ of the difference equation
(1.3)xn=f(xn-1,…,xn-k),n∈N0,
is said that it is eventually periodic with period p, if there is an n1≥-k such that
(1.4)xn+p=xn,forn≥n1.
If n1=-k, then it is said that the solution is periodic with period p. For some results in this area see, for example, [19–26] and the references therein.


## 2. Solutions of Equation (1.2)

By using the change of variables(2.1)yn=1xnxn-k,n∈N0,
equation (1.2) is transformed into the following linear first-order difference equation:
(2.2)yn+1=ayn+b,n∈N0,
for which it is known (and easy to see) that
(2.3)yn=y0an+b1-an1-a=b+an(y0(1-a)-b)1-a,n∈N0,
if a≠1, and
(2.4)yn=y0+bn,n∈N0,
if a=1.From (2.1), we have
(2.5)xn=1ynxn-k=yn-kynxn-2k,
for n≥k, from which it follows that
(2.6)x2km+i=xi-2k∏j=0my(2j-1)k+iy2jk+i,
for every m∈ℕ0 and i∈{k,k+1,…,3k-1}.Using (2.3) and (2.4) in (2.6), we get
(2.7)x2km+i=xi-2k∏j=0mbx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k),
if a≠1, and
(2.8)x2km+i=xi-2k∏j=0mx0x-kb((2j-1)k+i)+1x0x-kb(2jk+i)+1,
if a=1, for every m∈ℕ0 and i∈{k,k+1,…,3k-1}.By using formulae (2.7) and (2.8), the behavior of well-defined solutions of equation (1.2) can be obtained. This is done in the following section.Remark 2.1.
It is easy to check that the formulae in Theorems 2.1 and 4.1 from [18] are direct consequences of formula (2.8), whereas formulae in Theorem 3.1 and Theorem 5.1 from [18] are direct consequences of formula (2.7).Remark 2.2.
Note that from formula (2.8), it follows that in the case a=1, b=0, all well-defined solutions of equation (1.2) are periodic with period 2k. This can be also obtained from (1.2), without knowing explicit formulae for its solutions. Namely, in this case, (1.2) can be written as follows:
(2.9)xn+1xn-k+1=xnxn-k,
since we assume xn≠0, for all n≥-k, from which it follows that the sequence xnxn-k is constant, that is, xnxn-k=c,n∈ℕ0 for some c∈ℝ∖{0}. Hence,
(2.10)xn=cxn-k=xn-2k,n≥k,
as claimed.Remark 2.3.
Note also that in the casea∉{0,1}, b=0, from (2.7), we have
(2.11)x2km+i=xi-2k∏j=0ma(2j-1)k+ia2jk+i=xi-2kak(m+1),
for every m∈ℕ0 and i∈{k,k+1,…,3k-1}, from which the behaviour of the solutions in the case easily follows.


## 3. Asymptotic Behavior of Well-Defined Solutions of Equation (1.2)

In this section, we derive some results on asymptotic behavior of well-defined solutions of equation (1.2). We will use well-known asymptotic formulae as follows:
(3.1)ln⁡(1+x)=x-x22+O(x3),(1+x)-1=1-x+O(x2),
for x→0, where O is the Landau “big-oh” symbol.Theorem 3.1.
Leta=1 and b≠0 in (1.2). Then, every well-defined solution (xn)n≥-k of equation (1.2) converges to zero.Proof.
By formula (2.8), we have
(3.2)lim⁡m→∞x2km+i=lim⁡m→∞xi-2k∏j=0mx0x-kb((2j-1)k+i)+1x0x-kb(2jk+i)+1=xi-2klim⁡m→∞∏j=0m(1-x0x-kbk1+x0x-kb(2jk+i))=xi-2klim⁡m→∞C(m0)∏j=m0+1m(1-x0x-kbk1+x0x-kb(2jk+i)),
where m0 is sufficiently large so that (3.1) can be applied below, and
(3.3)C(m0)=∏j=0m0(1-x0x-kbk1+x0x-kb(2jk+i)).
Since
(3.4)∑j=m0+1∞‍k2jk+i=+∞,
we conclude, using (3.1), that
(3.5)lim⁡m→∞∏j=m0+1m(1-x0x-kbk1+x0x-kb(2jk+i))=lim⁡m→∞∏j=m0+1mexp⁡[ln⁡(1-x0x-kbk1+x0x-kb(2jk+i))]=lim⁡m→∞∏j=m0+1mexp⁡[-x0x-kbk1+x0x-kb(2jk+i)-12(x0x-kbk1+x0x-kb(2jk+i))2lim⁡m→∞∏j=m0+1m‍expexp+O((x0x-kbk1+x0x-kb(2jk+i))3)]=lim⁡m→∞∏j=m0+1mexp⁡[-k2jk+i+O(1j2)]=lim⁡m→∞exp⁡[-∑j=m0+1m‍k2jk+i]∏j=m0+1mexp⁡[O(1j2)]=0.
Therefore,
(3.6)lim⁡m→∞x2km+i=0,
for every i∈{k,k+1,…,3k-1}, and consequently, limn→∞xn=0, as claimed.Before we formulate and prove our next result, we will prove an auxiliary result which is incorporated in the lemma that follows.Lemma 3.2.
Ifa≠1, then equation (1.2) has 2k-periodic solutions.Proof.
Let(3.7)zn=xnxn-k,n∈N0.
Then
(3.8)zn+1=zna+bzn,n∈N0.
Since a≠1, we see that equation (3.8) has equilibrium solution as follows:
(3.9)zn=z-=1-ab,n∈N0.
For this solution of equation (3.8), we have
(3.10)xn=z-xn-k=xn-2k,n≥k,
from which the lemma follows.Remark 3.3.
Note that2k-periodic solutions in the previous lemma could be prime 2k-periodic. For this it is enough to choose initial conditions x-i,i=0,k¯, such that the string
(3.11)(x-k,x-k+1,…,x-1,z-x-k,z-x-k+1,…,z-x-1)
is not periodic with a period less then 2k.Theorem 3.4.
Let|a|<1 and b≠0 in (1.2). Then, every well-defined solution (xn)n≥-k of equation (1.2) converges to a, not necessarily prime, 2k-periodic solution of the equation.Proof.
First note that by Lemma3.2, in this case, there are 2k-periodic solutions of the equation. We know that in this case well-defined solutions of the equation are given by formula (2.7). From this and by using asymptotic formulae (3.1), we obtain that for sufficiently large m1(3.12)x2km+i=xi-2k∏j=0mbx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k)=xi-2kC(m1)∏j=m1+1m1+a(2j-1)k+i((1-a-bx0x-k)/bx0x-k)1+a2jk+i((1-a-bx0x-k)/bx0x-k)=xi-2kC(m1)∏j=m1+1m(1+a(2j-1)k+i(1-ak)(1-a-bx0x-k)bx0x-k+O(a4jk))=xi-2kC(m1)exp⁡((1-ak)(1-a-bx0x-k)bx0x-k∑j=m1+1m‍(a(2j-1)k+i+O(a4jk))),
where
(3.13)C(m1)=∏j=0m1bx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k).
From (3.12) and since |a|<1, it easily follows that the sequences (x2km+i)m∈ℕ0 are convergent for each i∈{k,k+1,…,3k-1}, from which the theorem follows.Theorem 3.5.
Let|a|>1 and b≠0 in (1.2). Then, every well-defined solution (xn)n≥-k of equation (1.2) converges to zero.Proof.
In this case, well-defined solutions of equation (1.2) are also given by formula (2.7). Further note that for each i∈{k,k+1,…,3k-1} holds
(3.14)lim⁡j→∞bx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k)=1ak.
Now note that 1/|a|k<1, due to the assumption |a|>1. Using this fact and (3.14), it follows that for sufficiently large j, say j≥m2 we have
(3.15)|bx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k)|≤12(1+1|a|k).
From this, we have(3.16)|x2km+i|=|xi-2k|C(m2)∏j=m2+1m|bx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k)|≤|xi-2k|C(m2)∏j=m2+1m(12(1+1|a|k))=|xi-2k|C(m2)(12(1+1|a|k))m-m2⟶0,
as m→∞, where
(3.17)C(m2)=∏j=0m2|bx0x-k+a(2j-1)k+i(1-a-bx0x-k)bx0x-k+a2jk+i(1-a-bx0x-k)|,
from which the theorem follows.Theorem 3.6.
Leta=-1, b≠0, and k be even in (1.2). Then, every well-defined solution (xn)n≥-k of equation (1.2) is eventually periodic with, not necessarily prime, period 4k.Proof.
From (2.2), in this case, we have
(3.18)yn=-yn-1+b=yn-2,n≥2,
which means that the sequence yn is two-periodic, and consequently the sequence xnxn-k is two-periodic. Hence
(3.19)x2nx2n-k=x0x-k,x2n+1x2n-k+1=x1x-k+1,
from which it follows that
(3.20)x2n+i=xix-k+ix2n-k+i=x2n-2k+i,n≥[k-i+12],i∈{0,1},
that is, the sequences x2n and x2n+1 are 2k-periodic from which the result easily follows.Theorem 3.7.
Leta=-1, b≠0, k be odd in (1.2), and (xn)n≥-k be a well-defined solution of equation (1.2). Then the following statements are true.(a)
Ifx0x-k=2/b, then the solution is 4k-periodic.(b)
If|bx0x-k-1|<1, then x2n→0 and |x2n+1|→∞, as n→∞.(c)
If|bx0x-k-1|>1, then x2n+1→0 and |x2n|→∞, as n→∞.Proof.
As in Theorem3.6, we obtain (3.18) and consequently (3.19) holds. If k=2l+1 for some l∈ℕ0, then from (1.2) and (3.19) we have
(3.21)x2nx2n-2l-1=x0x-(2l+1),x2n+1x2n-2l=x1x-2l=x0x-(2l+1)bx0x-(2l+1)-1.
From (3.21) we obtain
(3.22)x2n=x0x-(2l+1)x2n-(2l+1)=(bx0x-(2l+1)-1)x2n-(4l+2),x2n+1=x1x-2lx2n-2l=x2n+1-(4l+2)bx0x-(2l+1)-1.
From relation (3.22), the statements in this theorem easily follow.Remark 3.8.
Note that the casebx0x-k-1=-1 is not possible in Theorem 3.7. Namely, in this case x0x-k=0, due to the assumption b≠0 and, as we can see from  (1.2), the solution is not well-defined.



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*Source: 108047-2012-08-14.xml*