# On a Class of Nonautonomous Max-Type Difference Equations

Authors: Wanping Liu; Xiaofan Yang; Stevo Stević
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

This paper addresses the max-type difference equationxn=max{fn/xn−kα,B/xn−mβ}, n∈ℕ0, where k,m∈ℕ, B>0, and (fn)n∈ℕ0 is a positive sequence with a finite limit. We prove that every positive
solution to the equation converges to max{(limn→∞fn)1/(α+1),B1/(β+1)} under some conditions. Explicit positive solutions to two particular cases are
also presented.


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## Body

## 1. Introduction

The study of difference equations, which usually depicts the evolution of certain phenomena over the course of time, has a long history. Many experts recently pay some attention to so-calledmax-type difference equations which stem from certain models in control theory, see, for example, [1–23] and the references therein.The study of the following family of max-type difference equations(1.1)xn=max⁡{Bn(0),Bn(1)xn-p1r1xn-q1s1,Bn(2)xn-p2r2xn-q2s2,…,Bn(k)xn-pkrkxn-qksk},n∈N0,
where pi,qi∈ℕ such that 1≤p1<⋯<pk,1≤q1<⋯<qk,ri,si∈ℝ+,k∈ℕ and Bn(i)(i=0,1,…,k) are real sequences, was proposed by S. Stević at numerous conferences, for example, [10, 11]. For some results in this direction, see [1, 2, 4, 12–23].In the beginning of the investigation the following equation was studied:(1.2)xn=max⁡{An(1)xn-1,An(2)xn-2,…,An(k)xn-k},n∈N0,
where k∈ℕ,(An(i))n∈ℕ0,i=1,…,k are real sequences and the initial values are nonzero (see, e.g., [3, 5, 6, 9] and the related references therein).In [22], Sun studied the second-order difference equation (1.3)xn=max⁡{Axn-1α,Bxn-2β},n∈N0,
with α,β∈(0,1),A,B>0, and proved that each positive solution to (1.3) converges to the equilibrium point max{A1/(α+1),B1/(β+1)}, by considering several subcases. However, the method used there is a bit complicated and difficult for extending. Hence in [14] Stević extended this, as well as the main result in [13], by presenting a more concise and elegant proof of the next theorem.Theorem 1.1 (see [14, Theorem  1]).
Every positive solution to the difference equation(1.4)xn=max⁡{A1xn-p1α1,A2xn-p2α2,…,Akxn-pkαk},n∈N0,
where pi,i=1,…,k are natural numbers such that 1≤p1<⋯<pk,k∈ℕ and Ai>0,αi∈(-1,1),i=1,…,k, converges to max1≤i≤k{Ai1/(αi+1)}.Definition 1.2.
LetF:ℕ0×ℝk→ℝ be a function of k+1 variables, then the difference equation
(1.5)xn=F(n,xn-1,…,xn-k),n∈N0,
is called nonautonomous or time variant.Note that the following nonautonomous difference equation(1.6)xn=max⁡{An(1)xn-1α1,An(2)xn-2α2,…,An(l)xn-lαl},n∈N0,
where l∈ℕ,αi∈ℝ, and (An(i))n∈ℕ0,i=1,…,l are real sequences (not all constant), is a natural generalization of (1.2), (1.3), and (1.4). It is a special case of (1.1) of particular interest.The aforementioned works are mainly devoted to the study of (1.6) with constant or periodic numerators.This paper is devoted to the study of the following nonautonomous max-type difference equation with two delays:(1.7)xn=max⁡{fnxn-kα,Bxn-mβ},n∈N0,
where k,m∈ℕ,α,β∈ℝ are fixed and (fn)n∈ℕ0 is a positive sequence with a finite limit. Inspired by the methods and proofs of the above-mentioned papers, here we try to find some sufficient conditions such that every positive solution to (1.7) converges to max{(limn→∞fn)1/(α+1),B1/(β+1)}.This paper proceeds as follows. Several useful lemmas are given in Section2. In Section 3 we establish three main results about the global attractivity of (1.7) under some conditions. Finally motivated by a recent theorem in [21], explicit solutions to two particular cases of (1.7) are presented in Section 4.


## 2. Auxiliary Results

To establish the main results in Section3, here we present several lemmas. First we extend Lemma  2.4 in [21] by proving the following result.Lemma 2.1.
Consider the nonautonomous difference equation(2.1)zn=min⁡⁡{C1(n)-α1(n)zn-1,…,Ck(n)-αk(n)zn-k},n∈N0,
where k∈ℕ and αi(n),Ci(n),i=1,2,…,k are sequences. If Ci(n)s are nonnegative sequences and there always exists i0∈{1,2,…,k} such that Ci0(n)=0 for each fixed n∈ℕ0, then
(2.2)|zn|≤max⁡⁡{|α1(n)||zn-1|-C1(n),…,|αk(n)||zn-k|-Ck(n)},n∈N0.Proof.
Suppose thatn∈ℕ0 is fixed, and denote by S⊆{1,…,k} the set of all indices for which the terms in (2.1) are negative.
IfS=∅, which means all terms in the right-hand side of (2.1) are nonnegative, then apparently
(2.3)0≤zn≤-αi0(n)zn-i0
which implies
(2.4)|zn|≤|αi0(n)||zn-i0|-Ci0(n).
Otherwise,S≠∅, which means that there exist indices such that the corresponding terms in (2.1) are negative, then we derive
(2.5)zn=min⁡j∈S{Cj(n)-αj(n)zn-j}<0.
Sinceαj(n)zn-j must be positive for j∈S, it follows from (2.5) that
(2.6)|zn|=max⁡j∈S{αj(n)zn-j-Cj(n)}=max⁡j∈S{|αj(n)||zn-j|-Cj(n)}.
Inequality (2.2) follows easily from (2.4) and (2.6).The following lemma is widely used in the literature.Lemma 2.2 (see [24]).
Let(an)n∈ℕ be a sequence of nonnegative numbers which satisfies the inequality
(2.7)an+k≤qmax⁡{an+k-1,an+k-2,…,an},forn∈N,
where q>0 and k∈ℕ are fixed. Then there exists an M≥0 such that
(2.8)an≤M(qk)n,n∈N,
which implies an→0 as n→∞ if 0<q<1.Lemma 2.3.
Assume that(xn)n≥-k is a sequence of nonnegative numbers satisfying the difference inequality
(2.9)xn≤max⁡{γ1xn-1-d1(n),…,γkxn-k-dk(n)},n∈N0,
where k∈ℕ,γi∈[0,1), and di(n),i=1,…,k are nonnegative sequences. If there exists at least one positive γi, then the sequence xn converges to zero as n→∞.Proof.
This lemma follows directly from Lemma2.2 since
(2.10)xn≤max⁡{γ1xn-1-d1(n),…,γkxn-k-dk(n)}≤max⁡{γ1xn-1,…,γkxn-k}≤γmax⁡{xn-1,…,xn-k},
where 0<γ=max{γ1,γ2,…,γk}<1.Remark 2.4.
If in Lemma2.3, we assume γi=0, i=1,…,k, then the statement also holds, since in this case, if such a sequence exists, then the solution must be trivial, that is, xn=0, n∈ℕ0 (for some results on the existence of nontrivial solutions, see, e.g., [25–27] and the references therein).
Through some simple calculations, we have the following result.Lemma 2.5.
Every positive solution(xn)n≥-1 to the first-order difference equation
(2.11)xn=A1-1/ω(xn-1)1/ω,n∈N0,
with A>0,ω>0,x-1>0, has the form
(2.12)xn=A1-1/ωn+1(x-1)1/ωn+1,n∈N0.Note that Lemma2.5 leads to the following corollary.Corollary 2.6.
Each positive solution(Zn)n≥-k to the k th-order difference equation
(2.13)Zn=A1-1/ω(Zn-k)1/ω,n∈N0,
where A>0,ω>0,k∈ℕ and the initial values Z-1,…,Z-k are positive, has the following form:
(2.14)Zn=A1-(1/ω)[n/k]+1(Zρ(n,k)-k)(1/ω)[n/k]+1,n≥0,
where [·] represents the integer part function and ρ(n,k)=n-k·[n/k].Remark 2.7.
By Corollary2.6 we have that for any positive solution (Zn)n≥-k to (2.13) the following three statements hold true if ω>1: (1)
limn→∞Zn=A;(2)
ifZi<A for every i∈{-k,…,-1}, then the subsequences (Zjk+i)j≥0 are all strictly increasing;(3)
ifZi>A for every i∈{-k,…,-1}, then the subsequences (Zjk+i)j≥0 are all strictly decreasing.


## 3. Main Results

In this section, we prove the main results of this paper, which concern the global attractivity of positive solutions to (1.7) under some conditions. In the sequel, we assume that there is a finite limit of the positive sequence (fn)n∈ℕ0 in (1.7).Theorem 3.1.
Consider (1.7), where (fn)n∈ℕ0 is a positive monotone sequence with finite limit A>0. If |α|<1,|β|<1,B>A(β+1)/(α+1), then every positive solution to (1.7) converges to B1/(β+1).Proof.
By the changexn=ynB1/(β+1), (1.7) is transformed into
(3.1)yn=max⁡⁡{Cnyn-kα,1yn-mβ},n∈N0,
with Cn=fn/B(α+1)/(β+1),n∈ℕ0. Note that the sequence (Cn)n∈ℕ0 is also monotone and limn→∞Cn=A/B(α+1)/(β+1)<1.
According to the assumption the sequence(fn)n∈ℕ0 is nondecreasing or nonincreasing. If (fn)n∈ℕ0 is nonincreasing, then for some fixed ε∈(0,B(α+1)/(β+1)-A), there exists a natural number N such that for every n≥N we have fn-A<ε, which implies
(3.2)0<Cn<1,n≥N.
On the other hand, if(fn)n∈ℕ0 is nondecreasing then obviously Cn<1 for each n∈ℕ0, hence (3.2) also holds for this case.
LetD∈(0,1) be fixed. Employing the transformation yn=Dzn, (3.1) becomes
(3.3)Dzn=max⁡⁡{CnDαzn-k,1Dβzn-m},n∈N0,
which implies
(3.4)zn=min⁡⁡{log⁡DCn-αzn-k,-βzn-m},n∈N0.
Note thatlogDCn>0 for all n≥N. From this and by Lemma 2.1 we get
(3.5)|zn|≤max⁡⁡{|α||zn-k|-log⁡DCn,|β||zn-m|},n≥N.
When both α and β are zero, it is clear that zn is always zero for n≥N. Otherwise, it follows from Lemma 2.3 that limn→∞|zn|=0, which implies
(3.6)lim⁡n→∞zn=0.
Finally, from the above two transformations we get
(3.7)lim⁡n→∞xn=B1/(β+1)lim⁡n→∞yn=B1/(β+1)Dlim⁡n→∞⁡zn=B1/(β+1).
The proof is complete.Theorem 3.2.
Consider (1.7). Let (fn)n≥-k be a positive solution to (2.13) such that fi<A (or fi>A), i=-k,…,-1, and denote
(3.8)Γ=sup⁡n≥m{ln⁡fn-m-ln⁡Aln⁡fn-ln⁡A}.
If ω>1,0<αω<1,|β|Γ<1,B<A(β+1)/(α+1), then every positive solution to (1.7) converges to A1/(α+1).Proof.
Employing the transformationxn=ynA1/(α+1), (1.7) becomes
(3.9)yn=max⁡⁡{Cnyn-kα,λyn-mβ},n∈N0,
where Cn=fn/A,n∈ℕ0 and λ=B/A(β+1)/(α+1).
Then by the changeyn=Cnzn, (3.9) is transformed into
(3.10)Cnzn=max⁡⁡{CnCn-kαzn-k,λCn-mβzn-m},n≥max⁡⁡{k,m}.
In the sequel, we proceed by considering two cases.
Case 1.
Letfi<A,i=-k,…,-1.
By Remark2.7, we have 0<Cn<1,n∈ℕ0. From (3.10) we get
(3.11)zn=min⁡⁡{1-α(log⁡CnCn-k)zn-k,log⁡Cnλ-β(log⁡CnCn-m)zn-m}=min⁡⁡{1-αωzn-k,log⁡Cnλ-β(log⁡CnCn-m)zn-m},
for n≥max{k,m}. By the change zn=gn+1/(αω+1), (3.11) becomes
(3.12)gn=min⁡⁡{-αωgn-k,Tn-β(log⁡CnCn-m)gn-m},n≥max⁡⁡{k,m},
where
(3.13)Tn=log⁡Cnλ-βlog⁡CnCn-m+1αω+1.Claim 1.
There exists an integerM>0 such that Tn>0 for every n≥M.Proof.
Sincefn=ACn, we easily have that
(3.14)|β|log⁡CnCn-m≤|β|sup⁡j≥m{log⁡CjCj-m}=|β|Γ<1,n≥m.
Hence
(3.15)0<βlog⁡CnCn-m+1αω+1<2,n≥m.
On the other hand, for ε=A(1-λ), there exists an M>0 such that for each n≥M we have fn>A-ε=Aλ, which along with the fact Cn∈(0,1),n∈ℕ0, implies that
(3.16)log⁡Cnλ>2,n≥M.
The claim follows directly from (3.15) and (3.16), as desired.
Next, from Lemma2.1 and (3.12) it follows that
(3.17)|gn|≤max⁡⁡{αω|gn-k|,β(log⁡CnCn-m)|gn-m|-Tn}≤max⁡⁡{αω|gn-k|,βΓ|gn-m|-Tn},n≥max⁡⁡{k,m,M}.
From (3.17) and by Lemma 2.3, we derive limn→∞|gn|=0. Hence
(3.18)lim⁡n→∞ln⁡yn=lim⁡n→∞(znln⁡Cn)=lim⁡n→∞zn⋅ln⁡(lim⁡n→∞Cn)=1αω+1⋅ln⁡⁡1=0,
and consequently
(3.19)lim⁡n→∞xn=A1/α+1lim⁡n→∞yn=A1/α+1.
Case 2.
Letfi>A,i=-k,…,-1.
By Remark2.7, we have Cn>1,n∈ℕ0, and (3.10) is transformed into
(3.20)zn=max⁡⁡{1-α(log⁡CnCn-k)zn-k,log⁡Cnλ-β(log⁡CnCn-m)zn-m}=max⁡⁡{1-αωzn-k,log⁡Cnλ-β(log⁡CnCn-m)zn-m},
for all n≥max{k,m}. Then employing the following change
(3.21)zn=-gn+1αω+1,
(3.20) is transformed into
(3.22)gn=min⁡{-αωgn-k,Tn-β(log⁡CnCn-m)gn-m},n≥max⁡{k,m},
where Tn=-logCnλ+(βlogCnCn-m+1)/(αω+1). In this case, Tn>0 obviously holds. The rest of the proof is similar to that of Case1 so is omitted.To illustrate Theorem3.2, we present the following example.Example 3.3.
Consider the difference equation(3.23)xn=max⁡{fnxn-1α,Bxn-mβ},n∈N0,
where B>0,m≥2, and fn=Aq1/ωn+1,ω>1,A,q>0,q≠1,n∈ℕ0.
By Theorem3.2 and through some calculations, we obtain
(3.24)Γ=sup⁡n≥m{ln⁡fn-m-ln⁡Aln⁡fn-ln⁡A}=sup⁡n≥m{ωn+1ωn-m+1}=ωm.
Hence if 0<α<1/ω,|β|<1/ωm,B<A(β+1)/(α+1), then every positive solution to (3.23) converges to A1/(α+1).Theorem 3.4.
Consider (1.7). If (fn)n∈ℕ0 is an increasing sequence converging to A,|α|,|β|<1 and B=A(β+1)/(α+1), then every positive solution to (1.7) converges to A1/(α+1).Proof.
By the changexn=ynA1/(α+1), (1.7) becomes
(3.25)yn=max⁡{Cnyn-kα,1yn-mβ},n∈N0,
where Cn=fn/A<1,n∈ℕ0. The rest of the proof is analogous to that of Theorem 3.1 and thus is omitted.


## 4. Explicit Solutions

Recently, Stević and Iričanin in [21] proved the following theorem.Theorem 4.1 (see [21, Theorem  2.8]).
Consider(4.1)xn=max⁡{xn-1a1,…,xn-kak},n∈N0,
where k∈ℕ,ai∈ℝ,i=1,…,k. Then every well-defined solution of the equation has the following form:
(4.2)xn=dn∏j=1kajin(j),
where [(n+k)/k]≤in(1)+⋯+in(k)≤n+1,n∈ℕ0,in(j)≥0,j=1,…,k, and where dn is equal to one of the initial values x-k,…,x-1.The result is interesting since (4.2) holds for all real aj’s and for all nonzero initial values if one of these exponents is negative. However, (4.2) does not give explicit solutions to (4.1) since dn’s and in(j)’s in (4.2) are uncertain. Thus the problem of finding more explicit expressions of solutions to (4.1) is of interest.In this section we find explicit solutions to the next particular cases of (4.1) (4.3)xn=max⁡{1xn-1p,1xn-2},n∈N0,(4.4)xn=max⁡{1xn-1,1xn-2p},n∈N0,
with p>1 and positive initial values x-2,x-1. First we prove a useful lemma.Lemma 4.2.
Let(xn)n∈ℕ0 be a positive solution to (4.3) or (4.4). If there exists an N∈ℕ0 such that
(4.5)xN+3=xNp,xN+4=xN+1p,xN+5=xN+2p,
then for each k∈ℕ the following equalities hold:
(4.6)xN+3k=xNpk,xN+3k+1=xN+1pk,xN+3k+2=xN+2pk.Proof.
We will only consider (4.3), because similar proof can be given to (4.4). The case k=1 obviously holds due to (4.5). Next assume that (4.6) holds for 1≤k≤m for some m∈ℕ. Then by (4.3) we derive
(4.7)xN+3(m+1)=max⁡{1xN+3m+2p,1xN+3m+1}=max⁡{1xN+2pm+1,1xN+1pm}=max⁡{1xN+2p,1xN+1}pm=xN+3pm=xNpm+1,xN+3(m+1)+1=max⁡{1xN+3(m+1)p,1xN+3m+2}=max⁡{1xN+3pm+1,1xN+2pm}=max⁡{1xN+3p,1xN+2}pm=xN+4pm=xN+1pm+1,xN+3(m+1)+2=max⁡{1xN+3(m+1)+1p,1xN+3(m+1)}=max⁡{1xN+4pm+1,1xN+3pm}=max⁡{1xN+4p,1xN+3}pm=xN+5pm=xN+2pm+1.
Thus (4.6) holds for k=m+1, finishing the inductive proof of the lemma.Proposition 4.3.
Let(xn)n∈ℕ0 be a solution to (4.3) with p>1 and positive initial values x-2,x-1, then for each k∈ℕ0 the following statements hold true.(1)
Ifx-2>x-1p,x-1≥1, then x3k=1/x-1pk+1,x3k+1=x-1pk+2,x3k+2=x-1pk+1.(2)
Ifx-2>x-1p,x-1<1, then x3k=1/x-1pk+1,x3k+1=1/x-1pk,x3k+2=x-1pk+1.(3)
Ifx-2≤x-1p,x-2px-1≥1 then x3k=1/x-2pk,x3k+1=x-2pk+1 and
(4.8)x3k+2=x-2pkifx-2≥1or=1x-2pk+2ifx-2<1.(4)
Ifx-2≤x-1p,x-2px-1<1 then x3k+1=1/x-1pk,x3k+2=x-1pk+1 and
(4.9)x3k+3=x-1pkifx-1≥1or=1x-1pk+2ifx-1<1.Proof.
(1) By the assumptionx-2>x-1p and (4.3) it follows that
(4.10)x0=max⁡{1x-1p,1x-2}=1x-1p.
Then by x-1>1 and (4.3), we have the following equalities:
(4.11)x1=max⁡{1x0p,1x-1}=max⁡{x-1p2,1x-1}=x-1p2,x2=max⁡{1x1p,1x0}=max⁡{1x-1p3,x-1p}=x-1p,x3=max⁡{1x2p,1x1}=max⁡{1x-1p2,1x-1p2}=1x-1p2=x0p,x4=max⁡{1x3p,1x2}=max⁡{x-1p3,1x-1p}=x-1p3=x1p,x5=max⁡{1x4p,1x3}=max⁡{1x-1p4,x-1p2}=x-1p2=x2p.
Hence (4.5) is satisfied for N=0. Then by Lemma 4.2 we have that
(4.12)x3k=1/x-1pk+1,x3k+1=x-1pk+2,x3k+2=x-1pk+1,k∈N0,
as desired.(2)
By similar calculations as in (1), the following equalities hold:(4.13)x0=1x-1p,x1=1x-1,x2=x-1p,x3=1x-1p2=x0p,x4=1x-1p=x1p,x5=x-1p2=x2p.
Thus (4.5) holds for N=0, and the result follows again by Lemma 4.2.(3)
By the assumptionx-2≤x-1p and (4.3) it follows that(4.14)x0=max⁡{1x-1p,1x-2}=1x-2.
Then from x-2px-1≥1 and (4.3), we get
(4.15)x1=max⁡{1x0p,1x-1}=max⁡{x-2p,1x-1}=x-2p.
Now we will consider two cases x-2≥1 and x-2<1. When x-2≥1, the following equalities hold:
(4.16)x2=max⁡{1x1p,1x0}=max⁡{1x-2p2,x-2}=x-2,x3=max⁡{1x2p,1x1}=max⁡{1x-2p,1x-2p}=1x-2p=x0p,x4=max⁡{1x3p,1x2}=max⁡{x-2p2,1x-2}=x-2p2=x1p,x5=max⁡{1x4p,1x3}=max⁡{1x-2p3,x-2p}=x-2p=x2p.
On the other hand, the case x-2<1 leads to
(4.17)x2=1x-2p2,x3=1x-2p=x0p,x4=x-2p2=x1p,x5=1x-2p3=x2p.
Hence (4.5) holds for N=0 no matter the value of x-2 is bigger or less than one. From this the result follows by Lemma 4.2.(4)
Through analogous calculations to (3), ifx-1≥1 then(4.18)x0=1x-2,x1=1x-1,x2=x-1p,x3=x-1,x4=1x-1p=x1p,x5=x-1p2=x2p,x6=x-1p=x3p.
If x-1<1 then
(4.19)x0=1x-2,x1=1x-1,x2=x-1p,x3=1x-1p2,x4=1x-1p=x1p,x5=x-1p2=x2p,x6=1x-1p3=x3p.
Hence (4.5) holds for N=1 and any x-1>0. Hence, the results also follow from Lemma 4.2, finishing the proof of the proposition.The next proposition can be similarly proved as the proof of Proposition4.3, hence the proof is omitted here.Proposition 4.4.
Let(xn)n∈ℕ0 be a solution to (4.4) with p>1 and positive initial values x-2,x-1, then for each k∈ℕ0 the following statements hold true.(1)
Ifx-2p≥x-1,x-1≥1, then x3k=1/x-1pk,x3k+1=x-1pk,x3k+2=-1pk+1.(2)
Ifx-2p≥x-1,x-1<1, then x3k=1/x-1pk,x3k+1=1/x-1pk+1,x3k+2=x-1pk+1.(3)
Ifx-2p<x-1,x-2x-1≥1 then x3k=1/x-2pk+1,x3k+1=x-2pk+1, and
(4.20)x3k+2={x-2pk+2ifx-2≥1,1x-2pk+1ifx-2<1.(4)
Ifx-2p<x-1,x-2x-1<1 then x3k+1=1/x-1pk+1,x3k+2=x-1pk+1, and
(4.21)x3k+3={x-1pk+2ifx-1≥1,1x-1pk+1ifx-1<1.Remark 4.5.
From the above propositions, we know that any positive solution(xn)n∈ℕ0 to (4.3) or (4.4) can be divided into three subsequences which have explicit expressions. If we regard the sequence (…,0,∞,∞,0,∞,∞,…) as a general periodic solution to (4.3), then the solution (xn)n∈ℕ0 eventually converges to the general period-three solution (0,∞,∞).



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