# On the Difference Equationxn=anxn-k/(bn+cnxn-1⋯xn-k)

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

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## Abstract

The behavior of well-defined solutions of the difference equationxn=anxn-k/(bn+cnxn-1⋯xn-k),n∈ℕ0, where k∈ℕ is fixed, the sequences an, bn and cn are real, (bn,cn)≠(0,0), n∈ℕ0, and the initial values x-k,…,x-1 are real numbers, is described.


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## Body

## 1. Introduction

Recently there has been a huge interest in studying nonlinear difference equations and systems (see, e.g., [1–33] and the references therein). Here we study the difference equation
(1.1)xn=xn-kbn+cnxn-1⋯xn-k,n∈N0,
where k∈ℕ is fixed, the sequences bn and  cn are real, (bn,cn)≠(0,0), n∈ℕ0, and the initial values x-k,…,x-1 are real numbers. Equation (1.1) is a particular case of the equation
(1.2)xn=a^nxn-kb^n+c^nxn-1⋯xn-k,n∈N0,
with real sequences a^n, b^n and c^n. For a^n=0, n∈ℕ0, the equation is trivial, and, for a^n≠0,n∈ℕ0, it is reduced to equation (1.1) with bn=b^n/a^n and cn=c^n/a^n.Equation(1.3)xn=xn-kb+cxn-1⋯xn-k,n∈N0,
where b,c∈ℝ, which was treated in [32], is a particular case of equation (1.1).As in [32], here, we employ our idea of using a change of variables in equation (1.1) which extends the one in our paper [21] and is later also used, for example, in [4]. For similar methods see also [22, 25]. Equation (1.3) in the case k=2 was also studied in [1, 2], in a different way. The case when the sequences bn and  cn are two-periodic was studied in [31] (some related results are also announced in talk [3]). For related symmetric systems of difference equations, see [27, 29]. For some other recent results on difference equations and systems which can be solved, see, for example, [6, 7, 20–22, 30, 31, 33]. Some classical results can be found, for example, in [11].Equation (1.1) is a particular case of the equation
(1.4)yn=f(yn-1,…,yn-k,n)yn-k,n∈N0,
where f:ℝk+1→ℝ is a continuous function. Numerous particular cases of (1.4) have been investigated, for example, in [9, 21, 23]. In this paper we adopt the customary notation ∏i=k+1k‍gi=1 and ∑i=k+1k‍gi=0.


## 2. Casecn=0,n∈ℕ0

Here we consider the casecn=0,n∈ℕ0. In this case equation (1.1) becomes
(2.1)xn=xn-kbn,n∈N0,bn≠0,n∈ℕ0, from which it follows that for each i∈{1,…,k}(2.2)xkm-i=x-i∏j=1mbkj-i,m∈N0.
Using formula (2.2) the following theorem can be easily proved.Theorem 2.1.
Consider equation (1.1) with cn=0, bn≠0,n∈ℕ0. Then the following statements are true:  (a)
if(2.3)liminf⁡m→∞|bkm-i|=pi>1,for somei∈{1,…,k}, then xkm-i→0 as m→∞;(b)
if for eachi∈{1,…,k}, the limits pi in (2.3) are greater than 1, then xn→0 as n→∞;(c)
ifbkm-i=1, for every m∈ℕ and for some i∈{1,…,k}, then xkm-i=x-i, m∈ℕ0;(d)
ifbkm-i=-1, for every m∈ℕ and for some i∈{1,…,k}, then xkm-i=(-1)mx-i, m∈ℕ0;(e)
if(2.4)limsup⁡m→∞|bkm-i|=qi∈[0,1),andx-i≠0, for some i∈{1,…,k}, then |xkm-i|→∞, as m→∞;(f)
if for eachi∈{1,…,k}, the limits qi in (2.4) belong to the interval [0,1) and x-i≠0, then |xn|→∞ as n→∞.


## 3. Casebn=0,n∈ℕ0

In this section we consider the casebn=0,n∈ℕ0. Note that in this case equation (1.1) becomes
(3.1)xn=xn-kcnxn-1⋯xn-k+1xn-k,n∈N0,
where cn≠0,n∈ℕ0. If xn is a well-defined solution of equation (3.1) (i.e., a solution with initial values x-i≠0, i=1,…,k, which implies xn≠0, n∈ℕ0), then
(3.2)xn=1cnxn-1(xn-2⋯xn-k+1)=cn-1xn-2⋯xn-kcnxn-2⋯xn-k+1=cn-1cnxn-k,n∈N.Hence for eachi∈{0,1,…,k-1}(3.3)xkm-i=x-i∏j=1m‍ckj-i-1ckj-i,m∈N0.Using formula (3.3) we easily prove the next theorem.Theorem 3.1.
Consider equation (1.1) with bn=0, cn≠0, n∈ℕ0. Then the following statements are true:  (a)
if(3.4)liminf⁡m→∞|ckm-ickm-i-1|=p^i>1,for somei∈{0,1,…,k-1}, then xkm-i→0 as m→∞;(b)
if, for eachi∈{0,1,…,k-1}, the limits p^i in (3.4) are greater than 1, then xn→0 as n→∞;(c)
ifckm-i-1=ckm-i, for every m∈ℕ and for some i∈{0,1,…,k-1}, then xkm-i=x-i, m∈ℕ0;(d)
ifckm-i-1=-ckm-i, for every m∈ℕ and for some i∈{0,1,…,k-1}, then xkm-i=(-1)mx-i, m∈ℕ0;(e)
if(3.5)limsup⁡m→∞|ckm-ickm-i-1|=q^i∈[0,1),andx-i≠0 for some i∈{0,1,…,k-1}, then |xkm-i|→∞, as m→∞;(f)
if, for eachi∈{0,1,…,k-1}, x-i≠0 and the limits q^i in (3.5) belong to the interval [0,1), then |xn|→∞ as n→∞.


## 4. Casebn≠0 and cn≠0

The case whenbn≠0 and cn≠0 for every n∈ℕ0 is considered in this section.Ifx-i0=0 for some i0∈{1,…,k}, then from (1.1) we have that
(4.1)xkm-i0=0,form∈N0.
From (4.1) and (1.1) we have that for each i∈{1,…,k}∖{i0}(4.2)xkm-i=xk(m-1)-ibkm-i=x-i∏j=1mbkj-i,m∈N0.
From (4.1) we see that, for i=i0, (4.2) also holds. Hence Theorem 2.1 can be applied in this case. Note that if xn=0 for some n∈ℕ0, then (1.1) implies that there is an i0∈{1,…,k} such that x-i0=0, and by the previous consideration we have that (4.2) also holds.Ifx-i≠0, for each i∈{1,…,k}, then for every well-defined solution we have xn≠0 for n≥-k (note that there are solutions which are not well defined, that is, those for which xn-1⋯xn-k=-bn/cn, for some n∈ℕ0).Multiplying equation (1.1) by xn-1⋯xn-k+1 and using the transformation
(4.3)yn=1xnxn-1⋯xn-k+1,n≥-1,
we obtain equation
(4.4)yn=bnyn-1+cn,n∈N0.
Note that from (4.3), for every well-defined solution (xn)n≥-k of equation (1.1) such that x-i≠0, for each i∈{1,…,k}, it follows that yn≠0, n≥-1.Sincebn≠0, n∈ℕ0, we have that
(4.5)yn=(∏i=0n‍bi)(y-1+∑j=0n‍cj∏i=0jbi),n∈N0.From (4.3) and (4.5) we have that
(4.6)xn=1ynxn-1⋯xn-k+1=yn-1ynxn-k=y-1+∑j=0n-1(cj/∏i=0jbi)bn(y-1+∑j=0n(cj/∏i=0jbi))xn-k,
for every n∈ℕ0.Hence, from (4.6), we obtain that
(4.7)xmk-i=x-i∏l=1m‍1/α+∑j=0kl-i-1(cj/∏i=0jbi)bkl-i(1/α+∑j=0kl-i(cj/∏i=0jbi)),
for every m∈ℕ0 and each i=1,2,…,k, where
(4.8)α=∏l=1k‍x-l.


## 5. Casebn=1,n∈ℕ0

Here we consider the casebn=1, n∈ℕ0. In this case, from (4.7) we have that for each i∈{1,…,k}(5.1)xmk-i=x-i∏l=1m‍1+α∑j=0kl-i-1cj1+α∑j=0kl-icj,m∈N0.
Note that this formula includes also the case when x-i0=0 for some i0∈{1,…,k}.Now we formulate and prove a result in this case by using formula (5.1).Theorem 5.1.
Consider equation (1.1) with bn=1, n∈ℕ0, signcn=signc0, n∈ℕ, α≠0, and
(5.2)α∑j=0n‍cj≠-1,n∈N0.
Then the following statements hold:  (a)
if for somei∈{1,…,k}(5.3)∑l=1∞‍αckl-i1+α∑j=0kl-icj=+∞,(5.4)lim⁡l→∞αckl-i1+α∑j=0kl-icj=0,thenxmk-i→0 as m→∞;(b)
if (5.3) and (5.4) hold for every i∈{1,…,k}, then xn→0 as n→∞;(c)
if for somei∈{1,…,k} the sum
(5.5)∑l=1∞‍αckl-i1+α∑j=0kl-icjconverges, then the sequencexmk-i is also convergent;(d)
if the sum in (5.5) is finite for every i∈{1,…,k}, then the sequences xkm-i are convergent.Proof.
Let(xn)n≥-k be a solution of equation (1.1). Using condition signcn=signc0, n∈ℕ, it is easy to see that if (5.4) holds for some i∈{1,…,k}, there is an m0∈ℕ such that for j≥m0+1 the terms in the product in (5.1) are positive and that the following asymptotic formula
(5.6)ln⁡(1+x)=x+O(x2)
can be used with x being the fraction in the limit (5.4). From (5.1) and (5.6) we have that
(5.7)|xkm-i|=|x-i|∏l=1m‍|1+α∑j=0kl-i-1cj1+α∑j=0kl-icj|=|x-i|c(m0)exp⁡(∑l=m0+1m‍ln⁡1+α∑j=0kl-i-1cj1+α∑j=0kl-icj)=|x-i|c(m0)exp⁡(∑l=m0+1m‍ln⁡(1-αckl-i1+α∑j=0kl-icj))=|x-i|c(m0)exp⁡(-∑l=m0+1m‍αckl-i(1+o(1))1+α∑j=0kl-icj),
where
(5.8)c(m0)=∏l=1m0‍|1+α∑j=0kl-i-1cj1+α∑j=0kl-icj|.
Using formula (5.7), the assumptions regarding the sum ∑j=m0+1∞‍(αckl-i/(1+α∑j=0kl-i‍cj)) and the comparison test for the series whose terms are of eventually the same sign, the results in the theorem easily follow.


## 6. Casebn=-1,n∈ℕ0

Here we consider the casebn=-1, n∈ℕ0. In this case from (4.7) we have
(6.1)xmk-i=(-1)mx-i∏l=1m‍1+α∑j=0kl-i-1(-1)j+1cj1+α∑j=0kl-i(-1)j+1cj,
for every m∈ℕ0 and each i=1,2,…,k, where α is defined by (4.8).Theorem 6.1.
Consider equation (1.1) with α≠0, bn=-1, n∈ℕ0, and
(6.2)α∑j=0n‍(-1)j+1cj≠-1,n∈N0.
Then the following statements hold:  (a)
if for somei∈{1,…,k}(6.3)∑l=1∞‍α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cj=+∞,(6.4)lim⁡l→∞α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cj=0,(6.5)∑l=1∞‍ckl-i2(1+α∑j=0kl-i(-1)j+1cj)2<+∞,thenxmk-i→0 as m→∞;(b)
if for everyi∈{1,…,k}, (6.3), (6.4), and (6.5) hold, then xn→0 as n→∞;(c)
if for somei∈{1,…,k}(6.6)∑l=1∞‍α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cj=-∞,conditions (6.4) and (6.5) hold, and x-i≠0, then |xmk-i|→∞ as m→∞;(d)
if, for everyi∈{1,…,k}, conditions (6.4), (6.5), and (6.6) hold, and x-i≠0, i∈{1,…,k}, then |xn|→∞ as n→∞;(e)
if for somei∈{1,…,k} the sum
(6.7)∑l=1∞‍α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cjconverges and condition (6.5) holds, then the sequences x2mk-i and x(2m+1)k-i are also convergent;(f)
if, for everyi∈{1,…,k} the sum in (6.7) converges and condition (6.5) holds, then the sequences x2km-j,j=1,…,2k are convergent.Proof.
Let(xn)n≥-k be a solution of equation (1.1). By (6.4) we see that irrespectively on i∈{1,…,k}, there is an m1∈ℕ such that for j≥m1+1 the terms in the product in (6.1) belong to the interval (1/2,3/2) and that asymptotic formulae
(6.8)ln⁡(1+x)=x-x22+O(x3)
can be used with -x being the fraction in (6.4). From this and (6.1) we have that
(6.9)|xkm-i|=|x-i|∏l=1m‍|1+α∑j=0kl-i-1(-1)j+1cj1+α∑j=0kl-i(-1)j+1cj|=|x-i|c1(m1)exp⁡(∑l=m1+1m‍ln⁡1+α∑j=0kl-i-1(-1)j+1cj1+α∑j=0kl-i(-1)j+1cj)=|x-i|c1(m1)exp⁡(∑l=m1+1m‍ln⁡(1-α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cj))=|x-i|c1(m1)exp⁡(-∑l=m1+1m‍(α(-1)kl-i+1ckl-i1+α∑j=0kl-i(-1)j+1cj+α2ckl-i2(1+o(1))2(1+α∑j=0kl-i(-1)j+1cj)2)),
where
(6.10)c1(m1)=∏l=1m1‍|1+α∑j=0kl-i-1(-1)j+1cj1+α∑j=0kl-i(-1)j+1cj|.
Using formula (6.9), the assumptions of the theorem and some well-known convergence tests for series, the results in (a)–(f) easily follow.


## 7. Casebn=bn+k,cn=cn+k,n∈ℕ0

In this section we consider equation (1.1) for the case bn=bn+k, cn=cn+k, n∈ℕ0, that is, when the sequences bn and cn are k-periodic.First we show the existence ofk-periodic solutions of equation (4.4). If
(7.1)(y¯0,y¯1,…,y¯k-1)
is such a solution, then we have that
(7.2)y¯1=b1y¯0+c1,y¯2=b2y¯1+c2,…,y¯0=bky¯k-1+ck.
By successive elimination, or by Kronecker theorem (note that system (7.2) is linear), we get
(7.3)y¯i=∑j=0k-1cσ[j](i)∏s=0j-1bσ[s](i)1-∏j=1kbj,i=1,k¯,
if ∏j=1k‍bj≠1, where σ is the permutation defined by
(7.4)σ(i)=i-1,i=2,k¯,σ(1)=k,
and σ[i]=σ∘σ[i-1],σ[0]=Id, where Id denotes the identity.It is easy to see that (4.4) along with k periodicity of sequences bn and cn implies
(7.5)ykm+i=(∏j=1k‍bj)yk(m-1)+i+∑j=0k-1‍cσ[j](i)∏s=0j-1‍bσ[s](i),
for every m∈ℕ0 and i∈{1,2,…,k}, such that k(m-1)+i≥-1.Since (7.5) is a linear first-order difference equation, we have that when ∏j=1k‍bj≠1, its general solution is
(7.6)ykm+i=(∏j=1k‍bj)myi+1-(∏j=1kbj)m1-∏j=1kbj∑j=0k-1‍cσ[j](i)∏s=0j-1‍bσ[s](i).By lettingm→∞ in (7.6) we obtain the following corollary.Corollary 7.1.
Consider equation (4.4) with bn=bn+k, cn=cn+k, n∈ℕ0. Assume that
(7.7)|∏j=1k‍bj|<1.
Then for every solution yn of the equation we have that
(7.8)lim⁡m→∞ykm+i=y¯i,
for every i∈{1,2,…,k}, that is, yn converges to the k-periodic solution in formula (7.3).Let(7.9)Li:=∑j=0k-1‍cσ[j](i)∏s=0j-1‍bσ[s](i),i=1,k¯,q:=∏j=1k‍bj.From now on we will use the following convention: ifi,j∈ℕ0, then we regard that Lj=Li, if i≡j(modk). Also if a sequence (mj)j∈ℕ0 is defined by the relation mj=f(Lj), where f is a real function, then we will assume that mj=mi, if i≡j(modk).Using (7.6) and notation (7.9) in the relation xn=(yn-1/yn)xn-k (see (4.6)), for the case q≠1, we have that
(7.10)xkm+i=xi-k∏j=0m‍(yi-1-Li-1/(1-q))qj+Li-1/(1-q)(yi-Li/(1-q))qj+Li/(1-q)=xi-k∏j=0m‍Li-1Li⋅1+((1-q)yi-1/Li-1-1)qj1+((1-q)yi/Li-1)qj,
for every m∈ℕ0 and each i∈{2,…,k}, and
(7.11)xkm+1=x1-k∏j=0m‍(yk-Lk/(1-q))qj-1+Lk/(1-q)(y1-L1/(1-q))qj+L1/(1-q)=x1-k∏j=0m‍LkL1⋅1+((1-q)yk/Lk-1)qj-11+((1-q)y1/L1-1)qj.Now we present some results, which are applications of formulae (7.10) and (7.11).


### 7.1. Caseq=-1

Ifq=-1, then by (7.5) we get
(7.12)ykm+i=-yk(m-1)+i+Li=Li-(Li-yk(m-2)+i)=yk(m-2)+i,m∈N,
for k(m-2)+i≥-1; that is, ykm+i is two-periodic for each i∈{1,…,k}. Hence yn is a 2k-periodic solution of equation (4.4), in this case.Hence from the relationxn=(yn-1/yn)xn-k (see (4.6)), for each i∈{1,2,…,k}, we have
(7.13)xkm-i=ykm-i-1ykm-ixkm-i-k=ykm-i-1ykm-iyk(m-1)-i-1yk(m-1)-ixkm-i-2k,
for k(m-1)≥i.From (7.13) and by 2k periodicity of yn, we get
(7.14)x2kl+j=(yj-1yj+k-1yjyj+k)lxj,l∈N0,
for each j∈{-k+1,…,-1,0,1,…,k}.From (7.14), the behavior of solutions of equation (1.1), in this case, easily follows. For example, if
(7.15)pj:=yj-1yj+k-1yjyj+k=1,
for each j∈{-k+1,…,-1,0,1,…,k}, then the solution (xn)n≥k of (1.1) is 2k-periodic.


### 7.2. Caseq=1

Ifq=1 and α≠0, then from (7.5) we obtain
(7.16)ykm+i=yk(m-1)+i+Li,m∈N0,i=1,k¯,
when k(m-1)+i≥-1, from which along with (4.6), it follows that
(7.17)xkm+i=xi∏j=1m‍yi-1+jLi-1yi+jLi,m∈N,i=2,k¯,xkm+1=x1∏j=1m‍yk+(j-1)Lky1+jL1,m∈N.Corollary 7.2.
Consider equation (1.1). Let q=1, α≠0, and p^i:=Li-1/Li, i∈{1,…,k}. Then the following statements hold true.  (a)
If|p^i|<1, for some i∈{1,…,k}, then xkm+i→0 as m→∞.(b)
If|p^i|>1, or Li=0 and Li-1≠0, for some i∈{1,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(c)
Ifp^i=1, for some i∈{2,…,k}, and (yi-1-yi)/Li>0, then |xkm+i|→∞ as m→∞, if xi≠0.(d)
Ifp^i=1, for some i∈{2,…,k}, and (yi-1-yi)/Li<0, then xkm+i→0 as m→∞.(e)
Ifp^i=1, for some i∈{2,…,k}, and yi-1=yi, then the sequence (xkm+i)m∈ℕ0 is convergent.(f)
Ifp^1=1, and (yk-L1-y1)/L1>0, then |xkm+1|→∞ as m→∞, if x1≠0.(g)
Ifp^1=1, and (yk-L1-y1)/L1<0, then xkm+1→0 as m→∞.(h)
Ifp^1=1, and yk=L1+y1, then the sequence (xkm+1)m∈ℕ0 is convergent.Proof.
The statements in (a) and (b) follow from the facts that(7.18)lim⁡j→∞|yi-1+jLi-1yi+jLi|=|p^i|,i∈{2,…,k}
if Li≠0,
(7.19)lim⁡j→∞|yi-1+jLi-1yi+jLi|=+∞,i∈{2,…,k}
if Li=0 and Li-1≠0,
(7.20)lim⁡j→∞|yk+(j-1)Lky1+jL1|=|p^1|,
if L1≠0, and
(7.21)lim⁡j→∞|yk+(j-1)Lky1+jL1|=+∞,
if L1=0 and Lk≠0.
Now assume thatp^i=1 and let (xn)n≥-k be a solution of equation (1.1). It is easy to see that there is an m2∈ℕ such that for j≥m2+1 the terms in the products in (7.17) are positive and that the following asymptotic formulae
(7.22)(1+x)-1=1-x+O(x2),ln⁡(1+x)=x+O(x2)
can be applied with x=(yi-1-yi)/(jLi), wheni∈{2,…,k}or with x=(yk-L1-y1)/(jL1). Using these formulae, for the case i∈{2,…,k}, we have that
(7.23)xkm+i=xi∏j=1m‍yi-1+jLi-1yi+jLi=xic(m2)exp⁡(∑j=m2+1m‍ln⁡yi-1+jLi-1yi+jLi)=xic(m2)exp⁡(∑j=m2+1m‍ln⁡(1+yi-1-yijLi+O(1j2)))=xic(m2)exp⁡(∑j=m2+1m‍(yi-1-yijLi+O(1j2))),
where
(7.24)c(m2)=∏j=1m2‍yi-1+jLi-1yi+jLi.
Lettingm→∞ in (7.23), using the facts that
(7.25)∑j=m2+1m‍1j⟶+∞asm⟶∞
and that the series ∑j=m2+1∞‍O(1/j2) converges, we get statements (c)–(e).
Ifp^1=1, that is L1=Lk≠0, then by using (7.22) we get
(7.26)xkm+1=x1∏j=1m‍yk+(j-1)Lky1+jL1=x1d(m2)exp⁡(∑j=m2+1m‍ln⁡yk+(j-1)Lky1+jL1)=x1d(m2)exp⁡(∑j=m2+1m‍ln⁡(1+yk-L1-y1jL1+O(1j2)))=x1d(m2)exp⁡(∑j=m2+1m‍(yk-L1-y1jL1+O(1j2))),
where
(7.27)d(m2)=∏j=1m2‍yk+(j-1)Lky1+jL1.
Letting m→∞ in (7.26), using (7.25) and the fact that the series ∑j=m2+1∞‍O(1/j2) converges, we get statements (f)–(h), as desired.


### 7.3. Caseq≠±1

Ifq≠±1, then from (7.6) we get
(7.28)ykm+i=qmsi+ti,m∈N0,
where
(7.29)si=yi+Liq-1,ti=Li1-q,i=1,k¯,
from (4.6) it follows that
(7.30)xkm+i=xi∏j=1m‍qjsi-1+ti-1qjsi+ti,m∈N0,
for i∈{2,…,k}, and
(7.31)xkm+1=x1ykqs1+t1∏j=2m‍qj-1sk+tkqjs1+t1,m∈N0.
Note that ti=tj, if i≡j(modk).Corollary 7.3.
If0<|q|<1, α≠0, and qjsi+ti≠0, for every j∈ℕ0 and i∈{1,…,k}, then the following statements hold true.(a)
If|ti-1|<|ti|, for some i∈{1,…,k}, we have that xkm+i→0 as m→∞.(b)
If|ti-1|>|ti|, and si≠0 if ti=0 for some i∈{1,…,k}, we have that |xkm+i|→∞ as m→∞, if xi≠0.(c)
Ifti-1=ti≠0, for some i∈{1,…,k}, then xkm+i is convergent.(d)
Ifti-1=ti=0, and |si-1|<|si| for some i∈{2,…,k}, then |xkm+i|→0 as m→∞.(e)
Ift1=tk=0, and |sk|<|qs1|, then xkm+1→0 as m→∞.(f)
Ifti-1=ti=0, and |si-1|>|si| for some i∈{2,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(g)
Ift1=tk=0, and |sk|>|qs1|, then |xkm+1|→∞ as m→∞, if x1≠0.(h)
Ifti-1=ti=0, and si-1=si≠0  for some i∈{2,…,k}, then xkm+i is constant.(i)
Ift1=tk=0, and sk=qs1≠0, then xkm+1 is constant.(j)
Ifti-1=ti=0, and si-1=-si≠0  for some i∈{2,…,k}, then xkm+i=(-1)mxi.(k)
Ift1=tk=0, and sk=-qs1≠0, then xkm+1=x1yk(-1)m-1/(qs1).(l)
Ifti-1=-ti≠0, for some i∈{1,…,k}, then the subsequences x2km+i and x2km+k+i are convergent.Proof.
Since we have that(7.32)lim⁡j→∞qjsi-1+ti-1qjsi+ti=ti-1ti,i∈{2,…,k}
when ti≠0,
(7.33)lim⁡j→∞|qjsi-1+ti-1qjsi+ti|=+∞,i∈{2,…,k}
when |ti-1|>ti=0 and si≠0,
(7.34)lim⁡j→∞qj-1sk+tkqjs1+t1=tkt1,
when t1≠0, and
(7.35)lim⁡j→∞|qj-1sk+tkqjs1+t1|=+∞,
when |tk|>t1=0 and s1≠0, the statements in (a) and (b) easily follow from (7.30)–(7.35).
(c) Ifti-1=ti≠0, then
(7.36)xkm+i=xi∏j=1m‍(1+qj(si-1-siti)+o(qj)),
for i∈{2,…,k}, and if t1=tk≠0, then
(7.37)xkm+1=x1ykqs1+t1∏j=2m‍(1+qj-1(sk-qs1t1)+o(qj))
from which the statement in (c) easily follows.
(d)–(k) Ifti-1=ti=0, then
(7.38)xkm+i=xi∏j=1m‍si-1si,
for i∈{2,…,k}, and if t1=tk=0, then
(7.39)xkm+1=x1ykqs1∏j=2m‍skqs1
from which the statements in (d)–(k) easily follow.
(l) Ifti-1=-ti≠0, then we have that
(7.40)xkm+i=xi∏j=1m‍[-(1-qj(si-1+siti)+o(qj))]
for i∈{2,…,k}, and if t1=-tk≠0, then
(7.41)xkm+1=x1ykqs1+t1∏j=2m‍[-(1-qj-1(sk+qs1t1)+o(qj))]
from which the statement in (l) easily follows.Corollary 7.4.
If|q|>1 and α≠0, and qjsi+ti≠0, for every j∈ℕ0 and i∈{1,…,k}, then the following statements hold true.  (a)
If|si-1|<|si|, for some i∈{2,…,k}, then xkm+i→0 as m→∞.(b)
If|sk|<|qs1|, then xkm+1→0 as m→∞.(c)
If|si-1|>|si|, or si=0, si-1≠0 and ti≠0, for some i∈{2,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(d)
If|sk|>|qs1|, or if s1=0, sk≠0 and t1≠0, then |xkm+1|→∞ as m→∞, if x1≠0.(e)
Ifsi-1=si≠0, for some i∈{2,…,k}, then the sequence (xkm+i)m∈ℕ0 is convergent.(f)
Ifsi-1=si=0 and |ti-1|<|ti|  for some i∈{2,…,k}, then xkm+i→0 as m→∞.(g)
Ifs1=sk=0 and |tk|<|t1|, then xkm+1→0 as m→∞.(h)
Ifsi-1=si=0 and |ti-1|>|ti| for some i∈{2,…,k}, then |xkm+i|→+∞ as m→∞, if xi≠0.(i)
Ifs1=sk=0 and |tk|>|t1|, then |xkm+1|→+∞ as m→∞, if x1≠0.(j)
Ifsi-1=si=0 and ti-1=ti for some i∈{2,…,k}, then the sequence xkm+i is constant.(k)
Ifs1=sk=0 and t1=tk, then the sequence xkm+1 is constant.(l)
Ifsi-1=si=0 and ti-1=-ti for some i∈{2,…,k}, then the sequence xkm+i is two-periodic.(m)
Ifs1=sk=0 and t1=-tk, then the sequence xkm+1 is two periodic.(n)
Ifsk=qs1≠0, then the sequence (xkm+1)m∈ℕ0 is convergent.(o)
Ifsi-1=-si≠0, for some i∈{2,…,k}, then the sequences (x2km+i)m∈ℕ0 and (x2km+k+i)m∈ℕ0, are convergent.(p)
Ifsk=-qs1≠0, then the sequences (x2km+1)m∈ℕ0 and (x2km+k+1)m∈ℕ0, are convergent.Proof.
(a)–(d) These statements follow correspondingly from the next relations (which are derived using formulae (7.30) and (7.31)):
(7.42)lim⁡j→∞qjsi-1+ti-1qjsi+ti=si-1si
for i∈{2,…,k} if si≠0, and
(7.43)lim⁡j→∞|qjsi-1+ti-1qjsi+ti|=+∞
for i∈{2,…,k} if si=0, si-1≠0 and ti≠0;
(7.44)lim⁡j→∞qj-1sk+tkqjs1+t1=skqs1,
if s1≠0, and
(7.45)lim⁡j→∞|qj-1sk+tkqjs1+t1|=+∞
if s1=0, sk≠0 and t1≠0.
(e) Ifsi-1=si≠0, then from (7.30) we get
(7.46)xkm+i=xi∏j=1m‍(1+q-j(ti-1-tisi-1)+o(q-j)),
for i∈{2,…,k}, from which (e) follows.
(f)–(m) Ifsi-1=si=0 for some i∈{2,…,k}, then for i∈{2,…,k} we have
(7.47)qjsi-1+ti-1qjsi+ti=ti-1ti
while when s1=sk=0, we have
(7.48)qj-1sk+tkqjs1+t1=tkt1
from which the statements (f)–(i) easily follow.
(n) Ifsk=qs1≠0, then we have
(7.49)xkm+1=x1ykqs1+t1∏j=2m‍(1+q-j(tk-t1s1)+o(q-j)),
from which along with the assumption |q|>1 the statement follows.
(o) and (p) Ifsi-1=-si≠0, then
(7.50)xkm+i=xi∏j=1m‍[-(1-q-j(ti-1+tisi)+o(q-j))]
for i∈{2,…,k}, and
(7.51)xkm+1=x1ykqs1+t1∏j=2m‍[-(1-q-j(tk+t1s1)+o(q-j))].
From (7.50) and (7.51) the statements in (o) and (p) correspondingly follow.


## 7.1. Caseq=-1

Ifq=-1, then by (7.5) we get
(7.12)ykm+i=-yk(m-1)+i+Li=Li-(Li-yk(m-2)+i)=yk(m-2)+i,m∈N,
for k(m-2)+i≥-1; that is, ykm+i is two-periodic for each i∈{1,…,k}. Hence yn is a 2k-periodic solution of equation (4.4), in this case.Hence from the relationxn=(yn-1/yn)xn-k (see (4.6)), for each i∈{1,2,…,k}, we have
(7.13)xkm-i=ykm-i-1ykm-ixkm-i-k=ykm-i-1ykm-iyk(m-1)-i-1yk(m-1)-ixkm-i-2k,
for k(m-1)≥i.From (7.13) and by 2k periodicity of yn, we get
(7.14)x2kl+j=(yj-1yj+k-1yjyj+k)lxj,l∈N0,
for each j∈{-k+1,…,-1,0,1,…,k}.From (7.14), the behavior of solutions of equation (1.1), in this case, easily follows. For example, if
(7.15)pj:=yj-1yj+k-1yjyj+k=1,
for each j∈{-k+1,…,-1,0,1,…,k}, then the solution (xn)n≥k of (1.1) is 2k-periodic.


## 7.2. Caseq=1

Ifq=1 and α≠0, then from (7.5) we obtain
(7.16)ykm+i=yk(m-1)+i+Li,m∈N0,i=1,k¯,
when k(m-1)+i≥-1, from which along with (4.6), it follows that
(7.17)xkm+i=xi∏j=1m‍yi-1+jLi-1yi+jLi,m∈N,i=2,k¯,xkm+1=x1∏j=1m‍yk+(j-1)Lky1+jL1,m∈N.Corollary 7.2.
Consider equation (1.1). Let q=1, α≠0, and p^i:=Li-1/Li, i∈{1,…,k}. Then the following statements hold true.  (a)
If|p^i|<1, for some i∈{1,…,k}, then xkm+i→0 as m→∞.(b)
If|p^i|>1, or Li=0 and Li-1≠0, for some i∈{1,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(c)
Ifp^i=1, for some i∈{2,…,k}, and (yi-1-yi)/Li>0, then |xkm+i|→∞ as m→∞, if xi≠0.(d)
Ifp^i=1, for some i∈{2,…,k}, and (yi-1-yi)/Li<0, then xkm+i→0 as m→∞.(e)
Ifp^i=1, for some i∈{2,…,k}, and yi-1=yi, then the sequence (xkm+i)m∈ℕ0 is convergent.(f)
Ifp^1=1, and (yk-L1-y1)/L1>0, then |xkm+1|→∞ as m→∞, if x1≠0.(g)
Ifp^1=1, and (yk-L1-y1)/L1<0, then xkm+1→0 as m→∞.(h)
Ifp^1=1, and yk=L1+y1, then the sequence (xkm+1)m∈ℕ0 is convergent.Proof.
The statements in (a) and (b) follow from the facts that(7.18)lim⁡j→∞|yi-1+jLi-1yi+jLi|=|p^i|,i∈{2,…,k}
if Li≠0,
(7.19)lim⁡j→∞|yi-1+jLi-1yi+jLi|=+∞,i∈{2,…,k}
if Li=0 and Li-1≠0,
(7.20)lim⁡j→∞|yk+(j-1)Lky1+jL1|=|p^1|,
if L1≠0, and
(7.21)lim⁡j→∞|yk+(j-1)Lky1+jL1|=+∞,
if L1=0 and Lk≠0.
Now assume thatp^i=1 and let (xn)n≥-k be a solution of equation (1.1). It is easy to see that there is an m2∈ℕ such that for j≥m2+1 the terms in the products in (7.17) are positive and that the following asymptotic formulae
(7.22)(1+x)-1=1-x+O(x2),ln⁡(1+x)=x+O(x2)
can be applied with x=(yi-1-yi)/(jLi), wheni∈{2,…,k}or with x=(yk-L1-y1)/(jL1). Using these formulae, for the case i∈{2,…,k}, we have that
(7.23)xkm+i=xi∏j=1m‍yi-1+jLi-1yi+jLi=xic(m2)exp⁡(∑j=m2+1m‍ln⁡yi-1+jLi-1yi+jLi)=xic(m2)exp⁡(∑j=m2+1m‍ln⁡(1+yi-1-yijLi+O(1j2)))=xic(m2)exp⁡(∑j=m2+1m‍(yi-1-yijLi+O(1j2))),
where
(7.24)c(m2)=∏j=1m2‍yi-1+jLi-1yi+jLi.
Lettingm→∞ in (7.23), using the facts that
(7.25)∑j=m2+1m‍1j⟶+∞asm⟶∞
and that the series ∑j=m2+1∞‍O(1/j2) converges, we get statements (c)–(e).
Ifp^1=1, that is L1=Lk≠0, then by using (7.22) we get
(7.26)xkm+1=x1∏j=1m‍yk+(j-1)Lky1+jL1=x1d(m2)exp⁡(∑j=m2+1m‍ln⁡yk+(j-1)Lky1+jL1)=x1d(m2)exp⁡(∑j=m2+1m‍ln⁡(1+yk-L1-y1jL1+O(1j2)))=x1d(m2)exp⁡(∑j=m2+1m‍(yk-L1-y1jL1+O(1j2))),
where
(7.27)d(m2)=∏j=1m2‍yk+(j-1)Lky1+jL1.
Letting m→∞ in (7.26), using (7.25) and the fact that the series ∑j=m2+1∞‍O(1/j2) converges, we get statements (f)–(h), as desired.


## 7.3. Caseq≠±1

Ifq≠±1, then from (7.6) we get
(7.28)ykm+i=qmsi+ti,m∈N0,
where
(7.29)si=yi+Liq-1,ti=Li1-q,i=1,k¯,
from (4.6) it follows that
(7.30)xkm+i=xi∏j=1m‍qjsi-1+ti-1qjsi+ti,m∈N0,
for i∈{2,…,k}, and
(7.31)xkm+1=x1ykqs1+t1∏j=2m‍qj-1sk+tkqjs1+t1,m∈N0.
Note that ti=tj, if i≡j(modk).Corollary 7.3.
If0<|q|<1, α≠0, and qjsi+ti≠0, for every j∈ℕ0 and i∈{1,…,k}, then the following statements hold true.(a)
If|ti-1|<|ti|, for some i∈{1,…,k}, we have that xkm+i→0 as m→∞.(b)
If|ti-1|>|ti|, and si≠0 if ti=0 for some i∈{1,…,k}, we have that |xkm+i|→∞ as m→∞, if xi≠0.(c)
Ifti-1=ti≠0, for some i∈{1,…,k}, then xkm+i is convergent.(d)
Ifti-1=ti=0, and |si-1|<|si| for some i∈{2,…,k}, then |xkm+i|→0 as m→∞.(e)
Ift1=tk=0, and |sk|<|qs1|, then xkm+1→0 as m→∞.(f)
Ifti-1=ti=0, and |si-1|>|si| for some i∈{2,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(g)
Ift1=tk=0, and |sk|>|qs1|, then |xkm+1|→∞ as m→∞, if x1≠0.(h)
Ifti-1=ti=0, and si-1=si≠0  for some i∈{2,…,k}, then xkm+i is constant.(i)
Ift1=tk=0, and sk=qs1≠0, then xkm+1 is constant.(j)
Ifti-1=ti=0, and si-1=-si≠0  for some i∈{2,…,k}, then xkm+i=(-1)mxi.(k)
Ift1=tk=0, and sk=-qs1≠0, then xkm+1=x1yk(-1)m-1/(qs1).(l)
Ifti-1=-ti≠0, for some i∈{1,…,k}, then the subsequences x2km+i and x2km+k+i are convergent.Proof.
Since we have that(7.32)lim⁡j→∞qjsi-1+ti-1qjsi+ti=ti-1ti,i∈{2,…,k}
when ti≠0,
(7.33)lim⁡j→∞|qjsi-1+ti-1qjsi+ti|=+∞,i∈{2,…,k}
when |ti-1|>ti=0 and si≠0,
(7.34)lim⁡j→∞qj-1sk+tkqjs1+t1=tkt1,
when t1≠0, and
(7.35)lim⁡j→∞|qj-1sk+tkqjs1+t1|=+∞,
when |tk|>t1=0 and s1≠0, the statements in (a) and (b) easily follow from (7.30)–(7.35).
(c) Ifti-1=ti≠0, then
(7.36)xkm+i=xi∏j=1m‍(1+qj(si-1-siti)+o(qj)),
for i∈{2,…,k}, and if t1=tk≠0, then
(7.37)xkm+1=x1ykqs1+t1∏j=2m‍(1+qj-1(sk-qs1t1)+o(qj))
from which the statement in (c) easily follows.
(d)–(k) Ifti-1=ti=0, then
(7.38)xkm+i=xi∏j=1m‍si-1si,
for i∈{2,…,k}, and if t1=tk=0, then
(7.39)xkm+1=x1ykqs1∏j=2m‍skqs1
from which the statements in (d)–(k) easily follow.
(l) Ifti-1=-ti≠0, then we have that
(7.40)xkm+i=xi∏j=1m‍[-(1-qj(si-1+siti)+o(qj))]
for i∈{2,…,k}, and if t1=-tk≠0, then
(7.41)xkm+1=x1ykqs1+t1∏j=2m‍[-(1-qj-1(sk+qs1t1)+o(qj))]
from which the statement in (l) easily follows.Corollary 7.4.
If|q|>1 and α≠0, and qjsi+ti≠0, for every j∈ℕ0 and i∈{1,…,k}, then the following statements hold true.  (a)
If|si-1|<|si|, for some i∈{2,…,k}, then xkm+i→0 as m→∞.(b)
If|sk|<|qs1|, then xkm+1→0 as m→∞.(c)
If|si-1|>|si|, or si=0, si-1≠0 and ti≠0, for some i∈{2,…,k}, then |xkm+i|→∞ as m→∞, if xi≠0.(d)
If|sk|>|qs1|, or if s1=0, sk≠0 and t1≠0, then |xkm+1|→∞ as m→∞, if x1≠0.(e)
Ifsi-1=si≠0, for some i∈{2,…,k}, then the sequence (xkm+i)m∈ℕ0 is convergent.(f)
Ifsi-1=si=0 and |ti-1|<|ti|  for some i∈{2,…,k}, then xkm+i→0 as m→∞.(g)
Ifs1=sk=0 and |tk|<|t1|, then xkm+1→0 as m→∞.(h)
Ifsi-1=si=0 and |ti-1|>|ti| for some i∈{2,…,k}, then |xkm+i|→+∞ as m→∞, if xi≠0.(i)
Ifs1=sk=0 and |tk|>|t1|, then |xkm+1|→+∞ as m→∞, if x1≠0.(j)
Ifsi-1=si=0 and ti-1=ti for some i∈{2,…,k}, then the sequence xkm+i is constant.(k)
Ifs1=sk=0 and t1=tk, then the sequence xkm+1 is constant.(l)
Ifsi-1=si=0 and ti-1=-ti for some i∈{2,…,k}, then the sequence xkm+i is two-periodic.(m)
Ifs1=sk=0 and t1=-tk, then the sequence xkm+1 is two periodic.(n)
Ifsk=qs1≠0, then the sequence (xkm+1)m∈ℕ0 is convergent.(o)
Ifsi-1=-si≠0, for some i∈{2,…,k}, then the sequences (x2km+i)m∈ℕ0 and (x2km+k+i)m∈ℕ0, are convergent.(p)
Ifsk=-qs1≠0, then the sequences (x2km+1)m∈ℕ0 and (x2km+k+1)m∈ℕ0, are convergent.Proof.
(a)–(d) These statements follow correspondingly from the next relations (which are derived using formulae (7.30) and (7.31)):
(7.42)lim⁡j→∞qjsi-1+ti-1qjsi+ti=si-1si
for i∈{2,…,k} if si≠0, and
(7.43)lim⁡j→∞|qjsi-1+ti-1qjsi+ti|=+∞
for i∈{2,…,k} if si=0, si-1≠0 and ti≠0;
(7.44)lim⁡j→∞qj-1sk+tkqjs1+t1=skqs1,
if s1≠0, and
(7.45)lim⁡j→∞|qj-1sk+tkqjs1+t1|=+∞
if s1=0, sk≠0 and t1≠0.
(e) Ifsi-1=si≠0, then from (7.30) we get
(7.46)xkm+i=xi∏j=1m‍(1+q-j(ti-1-tisi-1)+o(q-j)),
for i∈{2,…,k}, from which (e) follows.
(f)–(m) Ifsi-1=si=0 for some i∈{2,…,k}, then for i∈{2,…,k} we have
(7.47)qjsi-1+ti-1qjsi+ti=ti-1ti
while when s1=sk=0, we have
(7.48)qj-1sk+tkqjs1+t1=tkt1
from which the statements (f)–(i) easily follow.
(n) Ifsk=qs1≠0, then we have
(7.49)xkm+1=x1ykqs1+t1∏j=2m‍(1+q-j(tk-t1s1)+o(q-j)),
from which along with the assumption |q|>1 the statement follows.
(o) and (p) Ifsi-1=-si≠0, then
(7.50)xkm+i=xi∏j=1m‍[-(1-q-j(ti-1+tisi)+o(q-j))]
for i∈{2,…,k}, and
(7.51)xkm+1=x1ykqs1+t1∏j=2m‍[-(1-q-j(tk+t1s1)+o(q-j))].
From (7.50) and (7.51) the statements in (o) and (p) correspondingly follow.



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*Source: 409237-2012-08-16.xml*