On a Max-Type Difference Inequality and Its Applications

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 equation 𝑥_𝑛=max{𝑥^𝑎1_𝑛−1,…,𝑥^𝑎𝑘_𝑛−𝑘}.

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: 𝑥_𝑛𝐴=max{^𝑛(1)𝑥_𝑛−1,𝐴_𝑛⁽²⁾𝑥_𝑛−2𝐴,…,^𝑛(𝑘)𝑥_𝑛−𝑘},𝑛∈ℕ₀, (1.1) where 𝑘∈ℕ,  𝐴_𝑛^(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 𝑥_𝑛=max{𝐵_𝑛⁽⁰⁾,𝐵_𝑛⁽¹⁾𝑥^𝑟1_𝑛−𝑝1𝑥^𝑠1_𝑛−𝑞1,𝐵_𝑛⁽²⁾𝑥^𝑟2_𝑛−𝑝2𝑥^𝑠2_𝑛−𝑞2,…,𝐵_𝑛^(𝑘)𝑥^𝑟𝑘_{𝑛−𝑝𝑘}𝑥^𝑠𝑘_{𝑛−𝑞𝑘}},𝑛∈ℕ₀, (1.2) where 𝑝_𝑖,𝑞_𝑖 are natural numbers such that 𝑝₁<𝑝₂<⋯<𝑝_𝑘, 𝑞₁<𝑞₂<⋯<𝑞_𝑘, 𝑟_𝑖,𝑠_𝑖∈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 𝑦_𝑛𝐴=max{𝑦_𝑛−1⋯𝑦_{𝑛−𝑚+1},1𝑦_{𝑛−𝑚−1}⋯𝑦_{𝑛−2𝑚+1}},𝑛∈ℕ₀, (1.3) arises naturally in certain models in automatic control (see [9]). By the change 𝑥_𝑛=𝑦_𝑛𝑦_𝑛−1⋯𝑦_{𝑛−𝑚+1} the equation is transformed into the equation 𝑥_𝑛𝑥=max{𝐴,^𝑛−1𝑥_𝑛−𝑚},𝑛∈ℕ₀, (1.4) 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 𝑥_𝑛𝐴=max{¹𝑥^𝛼1_𝑛−1,𝐴₂𝑥^𝛼2_𝑛−2𝐴,…,^𝑘𝑥^𝛼𝑘_𝑛−𝑘},𝑛∈ℕ₀, (1.5) where −1<𝛼_𝑖<1,𝐴_𝑖≥0,𝑖=1,…,𝑘, converges to max_{1≤𝑖≤𝑘}{𝐴^1/(𝛼𝑖_𝑖⁺¹⁾}.

Here we continue to study (1.5) by considering the cases when some of 𝛼_𝑖's are equal to one. We also give a representation of well-defined solutions of the difference equation 𝑥_𝑛=max{𝑥^𝑎1_𝑛−1,…,𝑥^𝑎𝑘_𝑛−𝑘}, where 𝑎_𝑖∈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 (𝑎_𝑛)_𝑛∈ℕ be a sequence of positive numbers which satisfy the inequality 𝑎_𝑛+𝑘≤𝑞max{𝑎_{𝑛+𝑘−1},𝑎_{𝑛+𝑘−2},…,𝑎_𝑛},for𝑛∈ℕ, (2.1) where 𝑞>0 and 𝑘∈ℕ are fixed. Then there exists an 𝑀>0 such that 𝑎_𝑛≤𝑀(√^𝑘𝑞)^𝑛,𝑛∈ℕ. (2.2)

Definition 2.1. For a sequence(𝑥_𝑛)^∞_𝑛=−𝑠, 𝑠∈ℕ₀, we say that it converges to zero geometrically if there is a𝑞∈[0,1) and𝑀>0 such that |𝑥_𝑛|≤𝑀𝑞^𝑛, (2.3) 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(𝑎_𝑛)^∞_𝑛=−𝑘 is a sequence of nonnegative numbers satisfying the difference inequality 𝑎_𝑛≤max{𝛼₁𝑎_𝑛−1−𝑑₁,…,𝛼_𝑘𝑎_𝑛−𝑘−𝑑_𝑘},𝑛∈ℕ₀, (2.4) where 𝑘∈ℕ, 𝛼_𝑖∈[0,1], 𝑑_𝑖∈ℝ₊, 𝑖∈{1,…,𝑘}, and if, for some 𝑖, 𝛼_𝑖=1, then 𝑑_i>0. Then the sequence 𝑎_𝑛 converges geometrically to zero as 𝑛→∞.

Proof. Let 𝛽∈(0,1) be chosen such that max{𝑎_−𝑘,…,𝑎₋₁𝑐}≤^𝑚,1−𝛽 (2.5) where 𝑐_𝑚=min{1,min_{𝑖∶𝛼𝑖=1}{𝑑_𝑖}}. (2.6) Then from (2.4) and using the fact that 𝑎_𝑛 are nonnegative numbers, we have that 𝑎_𝑛≤max{𝐴max_{𝑖∶𝛼𝑖∈[0,1)}{𝑎_𝑛−𝑖},max_{𝑗∶𝛼𝑗=1}{𝑎_𝑛−𝑗−𝑐_𝑚}}, (2.7) where 𝐴=max_{𝑖∶𝛼𝑖∈[0,1)}{𝛼_𝑖}.
From (2.7), (2.5) and since 0<max{𝐴,𝛽}<1, we have that 𝑎₀≤max{𝐴max_{𝑖∶𝛼𝑖∈[0,1)}{𝑎_−𝑖},max_{𝑗∶𝛼𝑗=1}{𝑎_−𝑗−𝑐_𝑚}}≤max{𝐴𝑐_𝑚,𝑐1−𝛽1−^𝑚𝛽−𝑐_𝑚,}=max{𝐴𝑐_𝑚,1−𝛽𝛽𝑐_𝑚𝑐1−𝛽}<^𝑚.1−𝛽 (2.8)
Now assume that 𝑎_𝑛≤𝑐_𝑚/(1−𝛽), for 0≤𝑛≤𝑛₀−1. Then from (2.4) we get 𝑎_𝑛0≤max{𝐴max_{𝑖∶𝛼𝑖∈[0,1)}{𝑎_𝑛0−𝑖},max_{𝑗∶𝛼𝑗=1}{𝑎_𝑛0−𝑗−𝑐_𝑚}}≤max{𝐴𝑐_𝑚,𝑐1−𝛽1−^𝑚𝛽−𝑐_𝑚,}=max{𝐴𝑐_𝑚,1−𝛽𝛽𝑐_𝑚𝑐1−𝛽}<^𝑚.1−𝛽 (2.9) Inequalities (2.8) and (2.9) along with the method of induction show that 0≤𝑎_𝑛≤𝑐_𝑚1−𝛽,for𝑛∈{−𝑠,…,−1}∪ℕ₀. (2.10)
Now note that from (2.10) we have that 𝑎_𝑛−𝑐_𝑚≤𝛽𝑎_𝑛,for𝑛∈{−𝑠,…,−1}∪ℕ₀. (2.11)
From (2.7), (2.11) and the choice of 𝑐_𝑚, it follows that for 𝑛∈ℕ₀𝑎_𝑛≤max{𝐴max_{𝑖∶𝛼𝑖∈[0,1)}{𝑎_𝑛−𝑖},𝛽max_{𝑗∶𝛼𝑗=1}{𝑎_𝑛−𝑗}}≤max{𝐴,𝛽}max_{1≤𝑖≤𝑘}{𝑎_𝑛−𝑖}. (2.12)
Applying Lemma A in inequality (2.12) with 𝑞=max{𝐴,𝛽}, the result follows.

Remark 2.3. Note that the constant 𝛽 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 𝑧_𝑛=min{𝐶₁−𝛼₁𝑧_𝑛−1,𝐶₂−𝛼₂𝑧_𝑛−2,…,𝐶_𝑘−𝛼_𝑘𝑧_𝑛−𝑘},𝑛∈ℕ₀, (2.13) where 𝑘∈ℕ,  𝐶_𝑖∈ℝ₊,𝛼_𝑗∈ℝ,  𝑖=1,…,𝑘, and there is 𝑖₀∈{1,…,𝑘} such that 𝐶_𝑖0=0. Then |𝑧_𝑛|≤max{|𝛼₁||𝑧_𝑛−1|−C₁,|𝛼₂||𝑧_𝑛−2|−𝐶₂,…,|𝛼_𝑘||𝑧_𝑛−𝑘|−𝐶_𝑘},𝑛∈ℕ₀. (2.14)

Proof. If all terms in the right-hand side of (2.13) are nonnegative then clearly 0≤𝑧_𝑛≤−𝛼_𝑖0𝑧_𝑛−𝑖0, so that |𝑧_𝑛|≤|𝛼_𝑖0||𝑧_𝑛−𝑖0|=|𝛼_𝑖0||𝑧_𝑛−𝑖0|−𝐶_𝑖0. (2.15) Otherwise, the set 𝑆⊆{1,…,𝑘} of all indices for which the terms in (2.13) are negative is nonempty, so that 𝑧_𝑛=min_𝑖∈𝑆{𝐶_𝑖−𝛼_𝑖𝑧_𝑛−𝑖}<0.
From this and since for such 𝑖∈𝑆, 𝛼_𝑖𝑧_𝑛−𝑖 must be positive, it follows that |𝑧_𝑛|=max_𝑖∈𝑆{𝛼_𝑖𝑧_𝑛−𝑖−𝐶_𝑖}=max_𝑖∈S{|𝛼_𝑖||𝑧_𝑛−𝑖|−𝐶_𝑖}. (2.16) From (2.15) and (2.16) inequality (2.14) easily follows.

By Proposition 2.2 and Lemma 2.4 we obtain the following theorem.

Theorem 2.5. Consider the difference equation 𝑥_𝑛𝐴=max{¹𝑥^𝛼1_𝑛−1,𝐴₂𝑥^𝛼2_𝑛−2𝐴,…,^𝑘𝑥^𝛼𝑘_𝑛−𝑘},𝑛∈ℕ₀, (2.17) where 𝑘∈ℕ, 0≤𝐴_𝑖≤1,−1≤𝛼_𝑖≤1,−1<𝛼_𝑖𝐴_𝑖<1 for each 𝑖∈{1,…,𝑘}, and 𝐴_𝑖=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 𝑦_𝑛=−ln𝑥_𝑛, we obtain that 𝑦_𝑛=min_{𝑖∶𝐴𝑖≠0}1{ln𝐴_𝑖−𝛼_𝑖𝑦_𝑛−𝑖},𝑛∈ℕ₀. (2.18) Now note that ln(1/𝐴_𝑖)≥0 for those 𝑖 such that 𝐴_𝑖≠0, since 𝐴_𝑖∈(0,1], and there is an 𝑆₁⊂{1,…,𝑘} such that ln(1/𝐴_𝑖)=0 when 𝑖∈𝑆₁. By Lemma 2.4 we have that for every 𝑛∈ℕ₀|𝑦_𝑛|≤max_{𝑖∶𝐴𝑖≠0}{|𝛼_𝑖||𝑦_𝑛−𝑖1|−ln𝐴_𝑖}. (2.19) From (2.19), noticing that if |𝛼_𝑖|=1 and 𝐴_𝑖≠0, then 𝐴_𝑖∈(0,1) so that ln(1/𝐴_𝑖)>0 and by applying Proposition 2.2 we obtain that |𝑦_𝑛|→0 as 𝑛→∞, from which it follows that 𝑥_𝑛=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 𝑥_𝑛𝐴=max{𝑥_𝑛−1,1𝑥^𝛼_𝑛−3},𝑛∈ℕ₀, (2.20) 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. Let 𝑦_𝑛 be a solution to the difference equation 𝑦_𝑛=max{1−𝑦_𝑛−1,−𝛼𝑦_𝑛−3},𝑛∈ℕ₀. (2.21) Then for all 𝑛∈ℕ₀, the following inequality holds: |𝑦_𝑛|≤max{|𝑦_𝑛−1|−1,𝛼|𝑦_𝑛−3|}. (2.22)Then they tried to show that 𝑦_𝑛→0 as 𝑛→∞. Note that (2.21) is obtained by the change 𝑥_𝑛=𝐴^𝑦𝑛 from (2.20), so that if it is proved that 𝑦_𝑛→0 as 𝑛→∞ then 𝑥_𝑛→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<|𝑦_𝑛−1|−1≤𝛽|𝑦_𝑛|, 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 𝑥_𝑛=max{𝑥^𝑎1_𝑛−1,…,𝑥^𝑎𝑘_𝑛−𝑘},𝑛∈ℕ₀, (2.23) where 𝑘∈ℕ, 𝑎_𝑖∈ℝ, 𝑖=1,…,𝑘. Then every well-defined solution of equation (2.23) has the following form: 𝑥_𝑛=𝑑^{∏𝑘𝑗=1𝑎𝑖𝑛𝑗(𝑗)}_𝑛, (2.24) where [𝑛+𝑘𝑘]≤𝑖_𝑛⁽¹⁾+⋯+𝑖_𝑛^(𝑘)≤𝑛+1,𝑛∈ℕ₀, (2.25)𝑖_𝑛^(𝑗)≥0, 𝑗=1,…,𝑘, and where 𝑑_𝑛 is equal to one of the initial values 𝑥_−𝑘,…,𝑥₋₁.
Moreover, if −1<𝑎_𝑖<1, 𝑖=1,…,𝑘, then 𝑥_𝑛→1 as 𝑛→∞.

Proof. The case 𝑘=1 is well known and simple. Just note that 𝑥_𝑛=𝑥^𝑎1𝑛+1₋₁. Hence assume that 𝑘≥2. We prove the result by induction. For 𝑛=0 we have 𝑥₀=max{𝑥^𝑎1₋₁,…,𝑥^𝑎𝑘_−𝑘}. (2.26) Note that 𝑥₀ can be equal to one of the numbers 𝑥^𝑎1₋₁,…,𝑥^𝑎𝑘_−𝑘 and that 𝑥^𝑎𝑖_−𝑖=𝑥^{𝑎𝑖∏𝑖≠𝑗𝑎0𝑗}_−𝑖,𝑖=1,…,𝑘, (2.27) which is nothing but formula (2.24) in this case. From this we also have that [0+𝑘𝑘]=1=𝑖₁⁽¹⁾+⋯+𝑖₁^(𝑘)=1=1+0, (2.28) which is (2.25) in this case.
Now assume that we have proved (2.24) and (2.25) for 𝑙≤𝑛−1. Then 𝑥_𝑛=max{𝑥^𝑎1_𝑛−1,…,𝑥^𝑎𝑘_𝑛−𝑘}=max{𝑑^{∏𝑘𝑗=1𝑎𝑖𝑗1𝑗(𝑗)𝑛−1+𝛿}_𝑛−1,…,𝑑^{∏𝑘𝑗=1𝑎𝑖𝑗𝑘𝑗(𝑗)𝑛−𝑘+𝛿}_𝑛−𝑘}, (2.29) where 𝛿^𝑗_𝑖 is the Kronecker symbol and [(𝑙+𝑘)/𝑘]≤𝑖_𝑙⁽¹⁾+⋯+𝑖_𝑙^(𝑘)≤𝑙, for 𝑙≤𝑛−1. Thus 𝑖⁽¹⁾_𝑛−𝑠+⋯+𝑖^(𝑘)_𝑛−𝑠+𝛿^𝑗_𝑠≤𝑛−𝑠+1+1≤𝑛+1, (2.30) for 𝑠=1,…,𝑘 and 𝑖⁽¹⁾_𝑛−𝑠+⋯+𝑖^(𝑘)_𝑛−𝑠+𝛿^𝑗_𝑠≥[𝑛−𝑠+𝑘𝑘]+1=[𝑛+𝑘−𝑠+𝑘𝑘]≥[𝑛+𝑘𝑘], (2.31)𝑠=1,…,𝑘. Hence the first statement follows by induction.
Now assume that max_{1≤𝑗≤𝑘}{|𝑎_𝑗|}<1. From this and (2.25) we have |^𝑘_𝑗=1𝑎^{𝑖𝑛(𝑗)}_𝑗|≤(max_{1≤𝑗≤𝑘}{|𝑎_𝑗|})^{[(𝑛+𝑘)/𝑘]}. (2.32) Inequality (2.32), the assumption max_{1≤𝑗≤𝑘}{|𝑎_𝑗|}<1, and (2.24) imply that 𝑥_𝑛 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 𝑎_𝑗,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 Theorem 2.8 follows easily also from Lemma A.