# On a Class of Solvable Difference Equations

Authors: Stevo Stević; Mohammed A. Alghamdi; Naseer Shahzad; Dalal A. Maturi
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

A class of difference equations which can be solved in closed form is presented in this paper. Obtained formulas for the solutions are used in the study of the asymptotic behavior of their solutions in a particular case.


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## Body

## 1. Introduction

One of the interesting problems in the theory of difference equations is to find the equations which can be solved in closed form. Such kind of equations are among the first ones which attracted attention of scientists. The reason for this is a natural expectation that obtained formulas for solutions can considerably facilitate investigation of their long-term behavior. For some classical results in the area, see, for example, book [1]. In [2], Stević solved, in an elegant way, the difference equation:
(1)xn+1=bxn-1b+cxnxn-1,n∈ℕ0,
explaining the formula in [3], which was given without any theoretical explanation. Several authors later considered some generalizations of (1) mostly by using Stević’s ideas and methods in [2] (see, e.g., [4–9] and the references therein). The idea for solving (1) was to reduce it to a nonhomogeneous linear first-order difference equation, which is solvable. Many equations and systems can be solved in a similar way (see, e.g., [4, 6–12]). For some related results and applications of linear first-order difference inequalities, see, for example, [13–15].An interesting difference equation is the following:(2)xn=xn-kxn-laxn-m+bxn-s,n∈ℕ0,
where k,l,m, and s are fixed natural numbers, a,b∈ℝ∖{0}, and the initial values x-i, i=1,τ¯, τ:=max{k,l,m,s} are real numbers.Special cases of (2) frequently appear in the literature.Example 1.
The next equation(3)xn=xn-1xn-22xn-1+xn-2,n∈ℕ0,
is one of them.
Equation (3) is a special case of (2) with
(4)k=m,l=s,
or
(5)k=s,l=m,
that is, of the equations
(6)xn=xn-kxn-laxn-k+bxn-l,n∈ℕ0,(7)xn=xn-kxn-laxn-l+bxn-k,n∈ℕ0.Now note that (6) and (7) share a common important feature. Namely, assume that (xn)n≥-τ is a well-defined solution of (6) or (7) (i.e., all its terms exist and are finite numbers). Then, if xn≠0, for every n≥-τ, by using the change of variables
(8)yn=1xn,
(6) and (7) are, respectively, transformed into the following linear difference equations with constant coefficients:
(9)yn=ayn-l+byn-k,n∈ℕ0,(10)yn=byn-l+ayn-k,n∈ℕ0,
which can be solved in closed form. As a consequence, (6) and (7), can be also solved in closed form. In fact, due to the known fact that polynomials of degree five and more need not be explicitly solved, we know that (9) and (10) can be theoretically solved. Nevertheless, we know some methods of how theoretically solutions of (9) and (10) can be found, as well as their form.Note that for (3) the corresponding equation (9) is
(11)yn=yn-1+2yn-2.
The general solution of (11) is yn=c12n+c2(-1)n, so that
(12)xn=1c12n+c2(-1)n.A natural question and a corresponding problem are as follows.Question 2.
Are there any other values ofk,l,m, and s except those ones in (4) and (5), such that (2) is solvable in closed form?Problem 3.
In the affirmative case find their solutions in closed form.The answer to the question is positive, and in recent paper [12], as a byproduct of our results therein, we have essentially shown that, in the case
(13)m=k,s=k+l,
(2) is solvable in closed form, and by using the described method therein, in some subcases of the case in (13) we have presented formulas for their solutions and use them in the study of their long-term behavior.Here we continue this line of research by showing that there are some cases different from those ones in (4), (5), and (13), for which (2) is solvable in closed form.


## 2. Casel=m=k+s

In this section we consider (2) in the case
(14)l=m=k+s.
That is, we consider the next difference equation:
(15)xn=xn-kxn-k-saxn-k-s+bxn-s,n∈ℕ0.First note that we may assume(16)gcd(k,s)=1
(the greatest common divisor of natural numbers k and s). Otherwise, if gcd(k,s)=r>1, then k=rk1 and s=rs1 for some k1,s1∈ℕ such that gcd(k1,s1)=1. Since every n∈ℕ0 has the form n=mr+i, for some m∈ℕ0 and i=0,r-1¯, (15) can be written in the form
(17)xmr+i=xr(m-k1)+ixr(m-k1-s1)+iaxr(m-k1-s1)+i+bxr(m-s1)+i.Using the next change of variables(18)xt(i)=xtr+i,t≥-(k1+s1),i=0,r-1¯, in (17) we have that (xt(i))t≥-(k1+s1), i=0,r-1¯, are r independent solutions of the system
(19)xt=xt-k1xt-k1-s1axt-k1-s1+bxt-s1,t∈ℕ0,
which is (15) with k1 and s1 instead of k and s.Hence, from now on we will assume that condition (16) holds.


### 2.1. A Case When Some of Initial Values Are Equal to Zero

Now assume that one of the initial valuesx-i, i=1,k+s¯, is equal to zero. If x-i=0 for some 1≤i≤k, then xk-i is not defined or xk-i=0. If the later holds, then ax-i+bxk-i=0, which along with (15) implies that xk+s-i is not defined. Further, if x-i=0 for some k+1≤i≤k+s, then xk+s-i is not defined or xk+s-i=0. If the later holds, then x2k+s-i is not defined or x2k+s-i=0. If x2k+s-i=0, then axk+s-i+bx2k+s-i=0, which along with (15) implies that x2k+2s-i is not defined.On the other hand, ifxn0=0 for some n0∈ℕ0, then from (15) we see that xn0-k=0 or xn0-k-s=0. If xn0-k=0, then using again (15) we see that xn0-2k=0 or xn0-2k-s=0, while if xn0-k-s=0, then xn0-2k-s=0 or xn0-2k-2s=0. Repeating the procedure we see that there is an i0∈{1,…,k+s} such that x-i0=0.From all these considerations it follows that all solutions of (15) such that x-i=0 for some i∈{1,…,k+s} are not well defined and that if a solution of (15) has a zero term then it is not defined.Hence, from now on we will consider only those solutions of (15) such that xn≠0, for every n≥-(k+s).


### 2.2. A Case When None of Initial Values Is Equal to Zero

Assume that(xn)n≥-(k+s) is a solution of (15) such that x-i≠0 for every i∈{1,…,k+s}. Since in this case xn≠0 for every n≥-(k+s), then (15) can be written in the following form:
(20)xnxn-k=1a+b(xn-s/xn-k-s),n∈ℕ0.The form suggests the next, natural, change of variables:(21)yn=xnxn-k,n≥-s,
which transforms (20) into the next difference equation of order s:
(22)yn=1a+byn-s,n∈ℕ0.In fact (22) is essentially an equation of the first order. Indeed, if we use the following decomposition of indices n→ns+j,  j=0,s-1¯, (22) can be written as
(23)yn(j)=1a+byn-1(j),n∈ℕ0,
where
(24)yn(j)=yns+j,n≥-1,j=0,s-1¯.This means that the subsequences(yns+j)n≥-1, j=0,s-1¯, are s independent solutions of the difference equation
(25)zn=1a+bzn-1,n∈ℕ0,
of the first order.Equation (25) is solvable, which along with (21) shows that (15) is solvable, giving another positive answer to Question 2. Indeed, (25) is a special case of the next difference equation:
(26)zn=αzn-1+βγzn-1+δ,n∈ℕ0.Equation (26) is classical and can be solved in several ways. Since in the case γ=0 (26) is a linear first-order difference equation, which is solvable, from now on we will assume that γ≠0. By using the change of variables
(27)zn=un+1un+c,n∈ℕ0,
into (26), after some calculations, we get
(28)(γc-α)unun-1+γun+1un-1+(γc+δ)un+1un+c(γc+δ)-αc-β=0.By choosing(29)c=-δγ
(here we use the assumption γ≠0) we get
(30)γun+1-(α+δ)un+αδ-βγγun-1=0,n∈ℕ0,
a second-order difference equation with constant coefficients.It is interesting that, by using the change(31)un=vnγn,
(30) is transformed into the equation
(32)vn+1-(α+δ)vn+(αδ-βγ)vn-1=0,n∈ℕ0.Letλ~1 and λ~2 be the characteristic roots of the characteristic polynomial:
(33)P(λ)=λ2-(α+δ)λ+(αδ-βγ)
associated with (32). Then by a well-known result on linear difference equations with constant coefficients and above transformations we get
(34)zn=-δγ+1γ·c1λ~1n+1+c2λ~2n+1c1λ~1n+c2λ~2n,
if (α+δ)2≠4(αδ-βγ), and
(35)zn=-δγ+λ~1γ·c^1+c^2(n+1)c^1+c^2n,
if (α+δ)2=4(αδ-βγ).In our case we have thatα=0, β=1, γ=b, and δ=a, so that (30) becomes
(36)bun+1-aun-un-1=0,n∈ℕ0,
while the constant in (29) is equal to -a/b. The characteristic roots of the characteristic polynomial associated with (36) are
(37)λ1=a+a2+4b2b,λ2=a-a2+4b2b.
Hence, the general solution of (36) is
(38)un=c1(a+a2+4b2b)n+c2(a-a2+4b2b)n,
if a2+4b≠0, and
(39)un=(c1+c2n)(a2b)n=(c1+c2n)(-2a)n,
if a2+4b=0.If we choose initial valuesu-1 and u0 such that
(40)u-1=1,u0=z-1+ab,
then from (40) and (38) we get
(41)un=(1+λ1(bz-1+a))λ1n-(1+λ2(bz-1+a))λ2na2+4b,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhn≥-1,
if a2+4b≠0, while from (40) and (39) we get
(42)un=(z-1+ab+n(z-1+ab+2a))(-2a)n,n≥-1,
if a2+4b=0.Using (41) and (42) in (27) we get
(43)zn=-ab+(1+λ1(bz-1+a))λ1n+1-(1+λ2(bz-1+a))λ2n+1(1+λ1(bz-1+a))λ1n-(1+λ2(bz-1+a))λ2n,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.hhhhhn≥-1,
if a2+4b≠0, and
(44)zn=4a-2a(z-1-4/a+(n+1)(z-1-2/a))(z-1-4/a+n(z-1-2/a)),=========================n≥-1,
if a2+4b=0.Using the fact that the sequencesyn(j)=yns+j, n≥-1, j=0,s-1¯, are solutions of (25), from formulas (43) and (44) it follows that(45)yn(j)=-ab+(1+λ1(byj-s+a))λ1n+1-(1+λ2(byj-s+a))λ2n+1(1+λ1(byj-s+a))λ1n-(1+λ2(byj-s+a))λ2n,n≥-1,for j=0,s-1¯, if a2+4b≠0, and
(46)yn(j)=4a-2a(yj-s-4/a+(n+1)(yj-s-2/a))(yj-s-4/a+n(yj-s-2/a)),hhhhhhhhhhhhhhhhhhhhhhn≥-1,
for j=0,s-1¯, if a2+4b=0.From (21) we have that
(47)xn=ynxn-k,n≥-s,
which implies that
(48)xkm1+j1=ykm1+j1xk(m1-1)+j1=xj1-k∏i=0m1‍yki+j1,
for m1∈ℕ0 and j1=-s,-s+1,…,-s+k-1.Although formulas (45)–(48) give solutions to (15) they are not easily applicable due to the assumption gcd(k,s)=1. Namely, subsequences (yn(j)), j=0,s-1¯, appearing in (45) and (46), and the subsequences (ykm1+j1), j1=-s,-s+1,…,-s+k-1, appearing in (48), are different decompositions of the sequence (yn)n≥-s, so they are not easily connected to each other.However, ifs=1 in (15), then they give complete solution of the equation. Namely, in this case (45) and (46) become
(49)yn=-ab+(1+λ1(by-1+a))λ1n+1-(1+λ2(by-1+a))λ2n+1(1+λ1(by-1+a))λ1n-(1+λ2(by-1+a))λ2n,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhln≥-1,
if a2+4b≠0, and
(50)yn=4a-2(y-1-4/a+(n+1)(y-1-2/a))a(y-1-4/a+n(y-1-2/a)),n≥-1,
if a2+4b=0.Using (49) and (50) in (48), we get(51)xkm1+j1=xj1-k∏i=0m1‍(-ab+(1+λ1(b(x-1/x-1-k)+a))λ1ki+j1+1-(1+λ2(b(x-1/x-1-k)+a))λ2ki+j1+1(1+λ1(b(x-1/x-1-k)+a))λ1ki+j1-(1+λ2(b(x-1/x-1-k)+a))λ2ki+j1),for each j1∈{-1,0,…,k-2}, and every m1∈ℕ0, if a2+4b≠0, and(52)xkm1+j1=xj1-k∏i=0m1‍(4a-2((x-1/x-1-k)-4/a+(ki+j1+1)((x-1/x-1-k)-2/a))a((x-1/x-1-k)-4/a+(ki+j1)((x-1/x-1-k)-2/a))),for each j1∈{-1,0,…,k-2}, and every m1∈ℕ0, if a2+4b=0.


## 2.1. A Case When Some of Initial Values Are Equal to Zero

Now assume that one of the initial valuesx-i, i=1,k+s¯, is equal to zero. If x-i=0 for some 1≤i≤k, then xk-i is not defined or xk-i=0. If the later holds, then ax-i+bxk-i=0, which along with (15) implies that xk+s-i is not defined. Further, if x-i=0 for some k+1≤i≤k+s, then xk+s-i is not defined or xk+s-i=0. If the later holds, then x2k+s-i is not defined or x2k+s-i=0. If x2k+s-i=0, then axk+s-i+bx2k+s-i=0, which along with (15) implies that x2k+2s-i is not defined.On the other hand, ifxn0=0 for some n0∈ℕ0, then from (15) we see that xn0-k=0 or xn0-k-s=0. If xn0-k=0, then using again (15) we see that xn0-2k=0 or xn0-2k-s=0, while if xn0-k-s=0, then xn0-2k-s=0 or xn0-2k-2s=0. Repeating the procedure we see that there is an i0∈{1,…,k+s} such that x-i0=0.From all these considerations it follows that all solutions of (15) such that x-i=0 for some i∈{1,…,k+s} are not well defined and that if a solution of (15) has a zero term then it is not defined.Hence, from now on we will consider only those solutions of (15) such that xn≠0, for every n≥-(k+s).


## 2.2. A Case When None of Initial Values Is Equal to Zero

Assume that(xn)n≥-(k+s) is a solution of (15) such that x-i≠0 for every i∈{1,…,k+s}. Since in this case xn≠0 for every n≥-(k+s), then (15) can be written in the following form:
(20)xnxn-k=1a+b(xn-s/xn-k-s),n∈ℕ0.The form suggests the next, natural, change of variables:(21)yn=xnxn-k,n≥-s,
which transforms (20) into the next difference equation of order s:
(22)yn=1a+byn-s,n∈ℕ0.In fact (22) is essentially an equation of the first order. Indeed, if we use the following decomposition of indices n→ns+j,  j=0,s-1¯, (22) can be written as
(23)yn(j)=1a+byn-1(j),n∈ℕ0,
where
(24)yn(j)=yns+j,n≥-1,j=0,s-1¯.This means that the subsequences(yns+j)n≥-1, j=0,s-1¯, are s independent solutions of the difference equation
(25)zn=1a+bzn-1,n∈ℕ0,
of the first order.Equation (25) is solvable, which along with (21) shows that (15) is solvable, giving another positive answer to Question 2. Indeed, (25) is a special case of the next difference equation:
(26)zn=αzn-1+βγzn-1+δ,n∈ℕ0.Equation (26) is classical and can be solved in several ways. Since in the case γ=0 (26) is a linear first-order difference equation, which is solvable, from now on we will assume that γ≠0. By using the change of variables
(27)zn=un+1un+c,n∈ℕ0,
into (26), after some calculations, we get
(28)(γc-α)unun-1+γun+1un-1+(γc+δ)un+1un+c(γc+δ)-αc-β=0.By choosing(29)c=-δγ
(here we use the assumption γ≠0) we get
(30)γun+1-(α+δ)un+αδ-βγγun-1=0,n∈ℕ0,
a second-order difference equation with constant coefficients.It is interesting that, by using the change(31)un=vnγn,
(30) is transformed into the equation
(32)vn+1-(α+δ)vn+(αδ-βγ)vn-1=0,n∈ℕ0.Letλ~1 and λ~2 be the characteristic roots of the characteristic polynomial:
(33)P(λ)=λ2-(α+δ)λ+(αδ-βγ)
associated with (32). Then by a well-known result on linear difference equations with constant coefficients and above transformations we get
(34)zn=-δγ+1γ·c1λ~1n+1+c2λ~2n+1c1λ~1n+c2λ~2n,
if (α+δ)2≠4(αδ-βγ), and
(35)zn=-δγ+λ~1γ·c^1+c^2(n+1)c^1+c^2n,
if (α+δ)2=4(αδ-βγ).In our case we have thatα=0, β=1, γ=b, and δ=a, so that (30) becomes
(36)bun+1-aun-un-1=0,n∈ℕ0,
while the constant in (29) is equal to -a/b. The characteristic roots of the characteristic polynomial associated with (36) are
(37)λ1=a+a2+4b2b,λ2=a-a2+4b2b.
Hence, the general solution of (36) is
(38)un=c1(a+a2+4b2b)n+c2(a-a2+4b2b)n,
if a2+4b≠0, and
(39)un=(c1+c2n)(a2b)n=(c1+c2n)(-2a)n,
if a2+4b=0.If we choose initial valuesu-1 and u0 such that
(40)u-1=1,u0=z-1+ab,
then from (40) and (38) we get
(41)un=(1+λ1(bz-1+a))λ1n-(1+λ2(bz-1+a))λ2na2+4b,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhn≥-1,
if a2+4b≠0, while from (40) and (39) we get
(42)un=(z-1+ab+n(z-1+ab+2a))(-2a)n,n≥-1,
if a2+4b=0.Using (41) and (42) in (27) we get
(43)zn=-ab+(1+λ1(bz-1+a))λ1n+1-(1+λ2(bz-1+a))λ2n+1(1+λ1(bz-1+a))λ1n-(1+λ2(bz-1+a))λ2n,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.hhhhhn≥-1,
if a2+4b≠0, and
(44)zn=4a-2a(z-1-4/a+(n+1)(z-1-2/a))(z-1-4/a+n(z-1-2/a)),=========================n≥-1,
if a2+4b=0.Using the fact that the sequencesyn(j)=yns+j, n≥-1, j=0,s-1¯, are solutions of (25), from formulas (43) and (44) it follows that(45)yn(j)=-ab+(1+λ1(byj-s+a))λ1n+1-(1+λ2(byj-s+a))λ2n+1(1+λ1(byj-s+a))λ1n-(1+λ2(byj-s+a))λ2n,n≥-1,for j=0,s-1¯, if a2+4b≠0, and
(46)yn(j)=4a-2a(yj-s-4/a+(n+1)(yj-s-2/a))(yj-s-4/a+n(yj-s-2/a)),hhhhhhhhhhhhhhhhhhhhhhn≥-1,
for j=0,s-1¯, if a2+4b=0.From (21) we have that
(47)xn=ynxn-k,n≥-s,
which implies that
(48)xkm1+j1=ykm1+j1xk(m1-1)+j1=xj1-k∏i=0m1‍yki+j1,
for m1∈ℕ0 and j1=-s,-s+1,…,-s+k-1.Although formulas (45)–(48) give solutions to (15) they are not easily applicable due to the assumption gcd(k,s)=1. Namely, subsequences (yn(j)), j=0,s-1¯, appearing in (45) and (46), and the subsequences (ykm1+j1), j1=-s,-s+1,…,-s+k-1, appearing in (48), are different decompositions of the sequence (yn)n≥-s, so they are not easily connected to each other.However, ifs=1 in (15), then they give complete solution of the equation. Namely, in this case (45) and (46) become
(49)yn=-ab+(1+λ1(by-1+a))λ1n+1-(1+λ2(by-1+a))λ2n+1(1+λ1(by-1+a))λ1n-(1+λ2(by-1+a))λ2n,hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhln≥-1,
if a2+4b≠0, and
(50)yn=4a-2(y-1-4/a+(n+1)(y-1-2/a))a(y-1-4/a+n(y-1-2/a)),n≥-1,
if a2+4b=0.Using (49) and (50) in (48), we get(51)xkm1+j1=xj1-k∏i=0m1‍(-ab+(1+λ1(b(x-1/x-1-k)+a))λ1ki+j1+1-(1+λ2(b(x-1/x-1-k)+a))λ2ki+j1+1(1+λ1(b(x-1/x-1-k)+a))λ1ki+j1-(1+λ2(b(x-1/x-1-k)+a))λ2ki+j1),for each j1∈{-1,0,…,k-2}, and every m1∈ℕ0, if a2+4b≠0, and(52)xkm1+j1=xj1-k∏i=0m1‍(4a-2((x-1/x-1-k)-4/a+(ki+j1+1)((x-1/x-1-k)-2/a))a((x-1/x-1-k)-4/a+(ki+j1)((x-1/x-1-k)-2/a))),for each j1∈{-1,0,…,k-2}, and every m1∈ℕ0, if a2+4b=0.


## 3. Asymptotic Behavior of Solutions of (15) When s=1

In this section we will use formulas (51) and (52) in studying of the asymptotic behavior of solutions of (15) for the case when s=1.The results are incorporated into two theorems. Before we formulate and prove the results we will present two auxiliary results.Lemma 4.
Assume thats=1, a,b∈ℝ∖{0}, and a+b=1. Then difference equation (15) has a (not necessarily prime) k-periodic solution.Proof.
Sincea+b=1 it follows that a2+4b≥0 (this is equivalent to (1+b)2≥0). Further, in this case (25) has one or two stationary solutions. Really, if zn=z, z≥-1, then it follows that bz2+az-1=0. That is, bz2+(1-b)z-1=(bz+1)(z-1)=0, from which it follows that
(53)z1=1,z2=-1b,
if a2+4b>0, that is, if (1+b)2>0, and
(54)z1=z2=1
if a2+4b=0, that is, if b=-1. Hence, in both cases
(55)zn=1,n≥-1,
is a stationary solution of (25).
By using (21), we see that, for the stationary solution of (25) given in (55), we have a solution of (15) such that
(56)xn=xn-k,n≥-1,
from which the lemma follows.Lemma 5.
Assume thats=1, a,b∈ℝ∖{0}, and b-a=1. Then difference equation (15) has a (not necessarily prime) 2k-periodic solution.Proof.
First note that from the conditionb-a=1 we have also that a2+4b≥0 (it is also equivalent to (1+b)2≥0). On the other hand, in this case (25) also has one or two stationary solutions. Namely, if zn=z, z≥-1, then it follows that bz2+az-1=0. That is, bz2+(b-1)z-1=(bz-1)(z+1)=0, from which we obtain
(57)z1=-1,z2=1b,
if a2+4b>0, that is, if (1+b)2>0, and
(58)z1=z2=-1
if a2+4b=0, that is, if b=-1. Hence, in both cases
(59)zn=-1,n≥-1,
is a stationary solution of (25).
By using (21), we see that the stationary solution of (25) given in (59) produces a solution of (15) such that xn=-xn-k, n≥-1, which implies that
(60)xn=xn-2k,n≥k-1,
from which the lemma follows.Remark 6.
By some calculation it is easy to see that-a/b+λ1=1 if and only if a+b=1 and b≥-1, while -a/b+λ2=1 if and only if a+b=1 and b≤-1.
Also,-a/b+λ1=-1 if and only if b-a=1 and b≤-1, while -a/b+λ2=-1 if and only if b-a=1 and b≥-1.Theorem 7.
Consider (15) with s=1. Assume that a,b∈ℝ∖{0}, a2+4b>0, and x-i≠0,i=1,k+1¯. Let λ:=λi, where |λi|=max{|λ1|,|λ2|},i∈{1,2}. Then the following statements are true.(a)
If|-a/b+λ|<1 and 1+λi(b(x-1/x-1-k)+a)≠0, for i=1,2, then xn→0, as n→∞.(b)
If|-a/b+λ1|<1, 1+λ1(b(x-1/x-1-k)+a)≠0, and 1+λ2(b(x-1/x-1-k)+a)=0, then xn→0, as n→∞.(c)
If|-a/b+λ2|<1, 1+λ1(b(x-1/x-1-k)+a)=0, and 1+λ2(b(x-1/x-1-k)+a)≠0, then xn→0, as n→∞.(d)
If|-a/b+λ|>1 and 1+λi(b(x-1/x-1-k)+a)≠0, for i=1,2, then |xn|→∞, as n→∞.(e)
If|-a/b+λ1|>1, 1+λ1(b(x-1/x-1-k)+a)≠0, and 1+λ2(b(x-1/x-1-k)+a)=0, then |xn|→∞, as n→∞.(f)
If|-a/b+λ2|>1, 1+λ1(b(x-1/x-1-k)+a)=0, and 1+λ2(b(x-1/x-1-k)+a)≠0, |xn|→∞, as n→∞.(g)
If-a/b+λ=1 and 1+λi(b(x-1/x-1-k)+a)≠0, for i=1,2, then xn converges to a (not necessarily prime) k-periodic solution of (15), as n→∞.(h)
If-a/b+λ1=1, 1+λ1(b(x-1/x-1-k)+a)≠0, and 1+λ2(b(x-1/x-1-k)+a)=0, then xn is a (not necessarily prime) k-periodic solution of (15).(i)
If-a/b+λ2=1, 1+λ1(b(x-1/x-1-k)+a)=0, and 1+λ2(b(x-1/x-1-k)+a)≠0, then xn is a (not necessarily prime) k-periodic solution of (15).(j)
If-a/b+λ=-1 and 1+λi(b(x-1/x-1-k)+a)≠0, for i=1,2, then xn converges to a (not necessarily prime) 2k-periodic solution of (15), as n→∞.(k)
If-a/b+λ1=-1, 1+λ1(b(x-1/x-1-k)+a)≠0, and 1+λ2(b(x-1/x-1-k)+a)=0, then xn is a (not necessarily prime) 2k-periodic solution of (15).(l)
If-a/b+λ2=-1, 1+λ1(b(x-1/x-1-k)+a)=0, and 1+λ2(b(x-1/x-1-k)+a)≠0, then xn is a (not necessarily prime) 2k-periodic solution of (15).Proof.
First note that sincex-i≠0, i=1,k+1¯, we have that xn≠0 for every n≥-(k+1). Further note that the condition a2+4b>0 implies that λ1,λ2∈ℝ, λ1≠λ2. If it were λ1=-λ2, then it would be a=0, the case which is excluded. Hence we have that |λ1|≠|λ2|, from which it follows that
(61)min{|λ1|,|λ2|}<max{|λ1|,|λ2|}.
Let(62)am1(j1)≔-ab+(1+λ1(b(x-1/x-1-k)+a))λ1km1+j1+1-(1+λ2(b(x-1/x-1-k)+a))λ2km1+j1+1(1+λ1(b(x-1/x-1-k)+a))λ1km1+j1-(1+λ2(b(x-1/x-1-k)+a))λ2km1+j1,for m1∈ℕ0 and j1=-1,0,…,k-2.
Then if1+λi(b(x-1/x-1-k)+a)≠0, for i=1,2, we have that for each j1∈{-1,0,…,k-2}(63)limm1→∞am1(j1)=-ab+λ.
From (51) and (63) the statements in (a) and (d) easily follow.
If1+λ1(b(x-1/x-1-k)+a)≠0 and 1+λ2(b(x-1/x-1-k)+a)=0, then we have that for each j1∈{-1,0,…,k-2}(64)am1(j1)=-ab+λ1,
from which along with (51) the statements in (b) and (e) easily follow.
If1+λ1(b(x-1/x-1-k)+a)=0 and 1+λ2(b(x-1/x-1-k)+a)≠0, then we have that for each j1∈{-1,0,…,k-2}(65)am1(j1)=-ab+λ2,
from which along with (51) the statements in (c) and (f) follow.
(g), (j) in this case we have that for sufficiently largem1(66)am1(j1)=-ab+λ+O((min{|λ1|,|λ2|}max{|λ1|,|λ2|})km1)
for each j1∈{-1,0,…,k-2}, from which along with (51), Lemmas 4 and 5, and Remark 6 these two statements follow.
(h), (k) in this case we have that (64) holds, from which along with (51) these two statements follow.
(i), (l) in this case we have that (65) holds, from which along with (51) these two statements follow.Theorem 8.
Consider (15) with s=1. Assume that a,b∈ℝ∖{0}, a2+4b=0, and x-i≠0, i=1,k+1¯. Then the following statements are true.(a)
If|2/a|<1, then xn→0, as n→∞.(b)
If|2/a|>1, then |xn|→∞, as n→∞.(c)
If|a|=2, then xn→0, as n→∞.Proof.
In this case we have thatλ1=λ2=-2/a. Let(67)bm1(j1)≔4a-2((x-1/x-1-k)-4/a+(km1+j1+1)((x-1/x-1-k)-2/a))a((x-1/x-1-k)-4/a+(km1+j1)((x-1/x-1-k)-2/a)),for m1∈ℕ0 and j1=-1,0,…,k-2.
Then for eachj1∈{-1,0,…,k-2} we have that
(68)limm1→∞bm1(j1)=2a,
if x-1/x-1-k≠2/a, while
(69)bm1(j1)=2a,
for j1∈{-1,0,…,k-2} and m1∈ℕ0, if x-1/x-1-k=2/a.
From (52), (68), and (69) the statements in (a) and (b) easily follow.
(c) in this case we have that for sufficiently largem1(70)am1(j1)=2a-2akm1+O(1m12)=±(1-1km1+O(1m12))=±e-(1/(km1))+O(1/m12),
for each j1∈{-1,0,…,k-2}, from which along with (52) and the fact that ∑j=1m1‍(1/j)→+∞ as m1→∞, the statement easily follows.



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*Source: 157943-2013-12-05.xml*