# On a Max-Type Difference Inequality and Its Applications

Authors: Stevo Stević; Bratislav D. Iričanin
Journal: Discrete Dynamics in Nature and Society
 (2010)

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## Abstract

We prove a useful max-type difference inequality which can be applied
in studying of some max-type difference equations and give an application
of it in a recent problem from the research area. We also give a representation
of solutions of the difference equationxn=max{xn−1a1,…,xn−kak}.


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## Body

## 1. Introduction

The investigation of max-type difference equations attracted some attention recently; see, for example, [1–20] and the references therein. In the beginning of the study of these equations the following difference equation was investigated:  (1.1)xn=max{An(1)xn-1,An(2)xn-2,…,An(k)xn-k},n∈ℕ0,
where k∈ℕ,  An(i),i=1,…,k, are real sequences (mostly constant or periodic), and the initial values x-1,…,x-k are different from zero (see, e.g., monograph  [6] or paper  [19] and the references therein).The study of the next difference equation(1.2)xn=max{Bn(0),Bn(1)xn-p1r1xn-q1s1,Bn(2)xn-p2r2xn-q2s2,…,Bn(k)xn-pkrkxn-qksk},n∈ℕ0,
where pi,qi are natural numbers such that p1<p2<⋯<pk, q1<q2<⋯<qk, ri,si∈ℝ+=[0,∞),i=1,…,k, and k∈ℕ, was proposed by the first author in numerous talks; see, for example, [11, 13]. For some results in this direction see [1, 4, 5, 7, 8, 12, 14–18, 20].A particular case of the difference equation(1.3)yn=max{Ayn-1⋯yn-m+1,1yn-m-1⋯yn-2m+1},n∈ℕ0,
arises naturally in certain models in automatic control (see [9]). By the change xn=ynyn-1⋯yn-m+1 the equation is transformed into the equation (1.4)xn=max{A,xn-1xn-m},n∈ℕ0,
which is a special case of (1.2) and which is a natural prototype for the equation.The following result, which extends the main result from the study in [18] was proved by the first author in [17] (see also [16]).Theorem A.
Every positive solution to the difference equation(1.5)xn=max{A1xn-1α1,A2xn-2α2,…,Akxn-kαk},n∈ℕ0,
where -1<αi<1,Ai≥0,i=1,…,k, converges to max1≤i≤k{Ai1/(αi+1)}.Here we continue to study (1.5) by considering the cases when some of αi's are equal to one. We also give a representation of well-defined solutions of the difference equation xn=max{xn-1a1,…,xn-kak}, where ai∈ℝ, i=1,…,k.


## 2. Main Results

In this section we prove the main results of this note. Before this we formulate the following very useful auxiliary result which can be found in [10] and give a definition.Lemma A.
Let(an)n∈ℕ be a sequence of positive numbers which satisfy the inequality
(2.1)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
(2.2)an≤M(qk)n,n∈ℕ.Definition 2.1.
For a sequence(xn)n=-s∞,s∈ℕ0, we say that it converges to zero geometrically if there is aq∈[0,1) andM>0 such that
(2.3)|xn|≤Mqn,
for n=-s,…,-1,0,1,….Now we are in a position to formulate and prove the main results of this note.Proposition 2.2.
Assume that(an)n=-k∞ is a sequence of nonnegative numbers satisfying the difference inequality
(2.4)an≤max{α1an-1-d1,…,αkan-k-dk},n∈ℕ0,
where k∈ℕ, αi∈[0,1], di∈ℝ+, i∈{1,…,k}, and if, for some i, αi=1, then di>0. Then the sequence an converges geometrically to zero as n→∞.Proof.
Letβ∈(0,1) be chosen such that
(2.5)max{a-k,…,a-1}≤cm1-β,
where
(2.6)cm=min{1,mini:αi=1{di}}.
Then from (2.4) and using the fact that an are nonnegative numbers, we have that
(2.7)an≤max{Amaxi:αi∈[0,1){an-i},maxj:αj=1{an-j-cm}},
where A=maxi:αi∈[0,1){αi}.
From (2.7), (2.5) and since 0<max{A,β}<1, we have that
(2.8)a0≤max{Amaxi:αi∈[0,1){a-i},maxj:αj=1{a-j-cm}}≤max{Acm1-β,cm1-β-cm,}=max{Acm1-β,βcm1-β}<cm1-β.
Now assume thatan≤cm/(1-β), for 0≤n≤n0-1. Then from (2.4) we get
(2.9)an0≤max{Amaxi:αi∈[0,1){an0-i},maxj:αj=1{an0-j-cm}}≤max{Acm1-β,cm1-β-cm,}=max{Acm1-β,βcm1-β}<cm1-β.
Inequalities (2.8) and (2.9) along with the method of induction show that
(2.10)0≤an≤cm1-β,forn∈{-s,…,-1}∪ℕ0.
Now note that from (2.10) we have that
(2.11)an-cm≤βan,forn∈{-s,…,-1}∪ℕ0.
From (2.7), (2.11) and the choice of cm, it follows that for n∈ℕ0(2.12)an≤max{Amaxi:αi∈[0,1){an-i},βmaxj:αj=1{an-j}}≤max{A,β}max1≤i≤k{an-i}.
Applying Lemma A in inequality (2.12) with q=max{A,β}, the result follows.Remark 2.3.
Note that the constantβ in the proof of Proposition 2.2 depends on initial conditions of solutions to difference equation (2.4), so that this is not a uniform constant.Lemma 2.4.
Consider the difference equation(2.13)zn=min{C1-α1zn-1,C2-α2zn-2,…,Ck-αkzn-k},n∈ℕ0,
where k∈ℕ,  Ci∈ℝ+,αj∈ℝ,  i=1,…,k, and there is i0∈{1,…,k} such that Ci0=0. Then
(2.14)|zn|≤max{|α1||zn-1|-C1,|α2||zn-2|-C2,…,|αk||zn-k|-Ck},n∈ℕ0.Proof.
If all terms in the right-hand side of (2.13) are nonnegative then clearly 0≤zn≤-αi0zn-i0, so that
(2.15)|zn|≤|αi0||zn-i0|=|αi0||zn-i0|-Ci0.
Otherwise, the set S⊆{1,…,k} of all indices for which the terms in (2.13) are negative is nonempty, so that zn=mini∈S{Ci-αizn-i}<0.
From this and since for suchi∈S, αizn-i must be positive, it follows that
(2.16)|zn|=maxi∈S{αizn-i-Ci}=maxi∈S{|αi||zn-i|-Ci}.
From (2.15) and (2.16) inequality (2.14) easily follows.By Proposition2.2 and Lemma 2.4 we obtain the following theorem.Theorem 2.5.
Consider the difference equation(2.17)xn=max{A1xn-1α1,A2xn-2α2,…,Akxn-kαk},n∈ℕ0,
where k∈ℕ, 0≤Ai≤1,-1≤αi≤1,-1<αiAi<1 for each i∈{1,…,k}, and Ai=1 for at least one i∈{1,…,k}. Then every positive solution of (2.17) converges to one.Proof.
Taking the logarithm of (2.17) and using the change yn=-lnxn, we obtain that
(2.18)yn=mini:Ai≠0{ln1Ai-αiyn-i},n∈ℕ0.
Now note that ln(1/Ai)≥0 for those i such that Ai≠0, since Ai∈(0,1], and there is an S1⊂{1,…,k} such that ln(1/Ai)=0 when i∈S1. By Lemma 2.4 we have that for every n∈ℕ0(2.19)|yn|≤maxi:Ai≠0{|αi||yn-i|-ln1Ai}.
From (2.19), noticing that if |αi|=1 and Ai≠0, then Ai∈(0,1) so that ln(1/Ai)>0 and by applying Proposition 2.2 we obtain that |yn|→0 as n→∞, from which it follows that xn=e-yn→1 as n→∞, as desired.Remark 2.6.
Recently Gelişken and Çinar in the paper:  “On the global attractivity of a max-type difference equation,’’Discrete Dynamics in Nature and Society,vol. 2009, Article ID 812674, 5 pages, 2009, have studied the asymptotic behavior to positive solutions of the difference equation
(2.20)xn=max{Axn-1,1xn-3α},n∈ℕ0,
where α∈(0,1) and A>0. They claim that if A∈(0,1), then every positive solution to (2.20) converges to one. However the proof given there cannot be regarded as complete one. Namely, they first formulated the following lemma.
Lemma 2.7.
Letyn be a solution to the difference equation
(2.21)yn=max{1-yn-1,-αyn-3},n∈ℕ0.
Then for all n∈ℕ0, the following inequality holds:
(2.22)|yn|≤max{|yn-1|-1,α|yn-3|}.Then they tried to show that yn→0 as n→∞. Note that (2.21) is obtained by the change xn=Ayn from (2.20), so that if it is proved that yn→0 as n→∞ then xn→1 as n→∞, from which the claim follows. In the beginning of the proof of the theorem they choose a number β such that 0<|yn-1|-1≤β|yn|, but do not say if these inequalities hold for all n or not, which is a bit confusing. Note that for different n the chosen number β can be different, which means that in this case β might be a function of n. Hence it is important that these inequalities hold for every n∈ℕ0∪{-2,-1}, which was not proved. This motivated us to prove Proposition 2.2 which, among others, removes the gap.Now we present a representation of solutions of a particular case of (1.5). The first author would like to express his sincere thanks to Professor L. Berg for a nice communication regarding this [2].Theorem 2.8.
Consider the equation(2.23)xn=max{xn-1a1,…,xn-kak},n∈ℕ0,
where k∈ℕ, ai∈ℝ, i=1,…,k. Then every well-defined solution of equation (2.23) has the following form:
(2.24)xn=dn∏j=1kajin(j),
where
(2.25)[n+kk]≤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.
Moreover, if-1<ai<1, i=1,…,k, then xn→1 as n→∞.Proof.
The casek=1 is well known and simple. Just note that xn=x-1a1n+1. Hence assume that k≥2. We prove the result by induction. For n=0 we have
(2.26)x0=max{x-1a1,…,x-kak}.
Note that x0 can be equal to one of the numbers x-1a1,…,x-kak and that
(2.27)x-iai=x-iai∏i≠jaj0,i=1,…,k,
which is nothing but formula (2.24) in this case. From this we also have that
(2.28)[0+kk]=1=i1(1)+⋯+i1(k)=1=1+0,
which is (2.25) in this case.
Now assume that we have proved (2.24) and (2.25) for l≤n-1. Then
(2.29)xn=max{xn-1a1,…,xn-kak}=max{dn-1∏j=1kajin-1(j)+δ1j,…,dn-k∏j=1kajin-k(j)+δkj},
where δij is the Kronecker symbol and [(l+k)/k]≤il(1)+⋯+il(k)≤l, for l≤n-1. Thus
(2.30)in-s(1)+⋯+in-s(k)+δsj≤n-s+1+1≤n+1,
for s=1,…,k and
(2.31)in-s(1)+⋯+in-s(k)+δsj≥[n-s+kk]+1=[n+k-s+kk]≥[n+kk],s=1,…,k. Hence the first statement follows by induction.
Now assume thatmax1≤j≤k{|aj|}<1. From this and (2.25) we have
(2.32)|∏j=1kajin(j)|≤(max1≤j≤k{|aj|})[(n+k)/k].
Inequality (2.32), the assumption max1≤j≤k{|aj|}<1, and (2.24) imply that xn tends to 1 as n→∞, finishing the proof of the theorem.Remark 2.9.
Note that formula (2.24) holds for each value of parameters aj,j=1,…,k, and for all solutions whose initial values are different from zero if one of these exponents is negative.Remark 2.10.
The second statement in Theorem2.8 follows easily also from Lemma A.



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