Global Behavior of the Max-Type Difference Equation 𝑥_𝑛+1=max{1/𝑥_𝑛,𝐴_𝑛/𝑥_𝑛−1}

Abstract

We study global behavior of the following max-type difference equation 𝑥_𝑛+1=max{1/𝑥_𝑛,𝐴_𝑛/𝑥_𝑛−1}, 𝑛=0,1,…, where {𝐴_𝑛}^∞_𝑛=0 is a sequence of positive real numbers with 0≤inf𝐴_𝑛≤sup𝐴_𝑛<1. The special case when {𝐴_𝑛}^∞_𝑛=0 is a periodic sequence with period 𝑘 and 𝐴_𝑛∈(0,1) for every 𝑛≥0 has been completely investigated by Y. Chen. Here we extend his results to the general case.

1. Introduction

In the recent years, there has been a lot of interest in studying the global behavior of, the so-called, max-type difference equations; see, for example, [1–17] (see also references therein). In [1, 3–5, 7, 8], the second order max-type difference equation 𝑥_𝑛+11=max{𝑥_𝑛,𝐴_𝑛𝑥_𝑛−1},𝑛=0,1,… (1.1) has been studied for positive coefficients 𝐴_𝑛, which are periodic with period 𝑘. The case 𝑘=1 was studied in [1], the case 𝑘=2 was studied in [3], the case 𝑘=3 was studied in [4, 8], and the more difficult case 𝑘=4 was studied in [7]. Chen [5] found that every positive solution of (1.1) is eventually periodic with period 2 when {𝐴_𝑛}^∞_𝑛=0 is a periodic sequence of positive real numbers with period 𝑘≥2 and 𝐴_𝑛∈(0,1) for all 𝑛≥0. These results were also included in the recent monograph [9] along with other related references. In this paper, we study global behavior of (1.1) when {𝐴_𝑛}^∞_𝑛=0 is a sequence of positive real numbers with 0≤inf𝐴_𝑛≤sup𝐴_𝑛<1.

2. Main Results

The main results of this paper are established through the following lemmas.

Lemma 2.1. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1), then(1)𝑥_𝑛+1𝑥_𝑛≥1 for all 𝑛≥0;(2)if 𝑥_𝑘+1𝑥_𝑘>1 for some 𝑘≥1, then 𝑥_𝑘+2𝑥_𝑘+1=1.

Proof. (1) is obvious since 𝑥_𝑛+1≥1/𝑥_𝑛 for all 𝑛≥0.(2)If 𝑥_𝑘+1𝑥_𝑘>1 for some 𝑘≥1, then 𝑥_𝑘+1𝑥_𝑘−1=𝐴_𝑘. Suppose for the sake of contradiction that 𝑥_𝑘+2𝑥_𝑘+1>1, then similarly we get 𝑥_𝑘+2𝑥_𝑘=𝐴_𝑘+1 and 𝐴_𝑘+1𝐴_𝑘=𝑥_𝑘+1𝑥_𝑘−1𝑥_𝑘+2𝑥_𝑘≥1. (2.1)This is a contradiction since 𝐴_𝑘+1<1 and 𝐴_𝑘<1. The proof is complete.

Lemma 2.2. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1) and 𝑃_𝑛=max{𝑥_𝑛,𝑥_𝑛−1} for all 𝑛≥1. Then(1)𝑥_𝑛+1≤𝑃_𝑛 and 𝑃_𝑛 is nonincreasing;(2)𝑥_𝑛 is bounded, and moreover 1/𝑃₁≤𝑥_𝑛≤𝑃₁ for any 𝑛≥1.

Proof. By Lemma 2.1(1) and the assumption 𝐴_𝑛<1, we obtain that for any 𝑛≥1, 𝑥_𝑛+1𝑥=max{^𝑛−1𝑥_𝑛𝑥_𝑛−1,𝐴_𝑛𝑥_𝑛𝑥_𝑛𝑥_𝑛−1}≤max{𝑥_𝑛−1,𝑥_𝑛}=𝑃_𝑛. (2.2) Hence 𝑃_𝑛+1=max{𝑥_𝑛+1,𝑥_𝑛}≤𝑃_𝑛, (2.3) which implies that for all 𝑛≥1,𝑥_𝑛≤𝑃₁. (2.4) Furthermore, it follows that for all 𝑛≥1,𝑥_𝑛+11=max{𝑥_𝑛,𝐴_𝑛𝑥_𝑛−11}≥𝑥_𝑛≥1𝑃₁. (2.5) The proof is complete.

Remark 2.3. Note that from the proof of Lemma 2.2 we have that 𝑃₁≥1.

Remark 2.4. Various sequences which satisfy inequality in Lemma 2.2(1), that is, 𝑥_𝑛+1≤𝑃_𝑛 have been studied, for example, in [18–24].

Lemma 2.5. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1) and lim_𝑛→∞𝑃_𝑛=𝑆. Then 𝑆=limsup_𝑛→∞𝑥_𝑛.

Proof. Since 𝑃_𝑛 is a subsequence of 𝑥_𝑛, it follows that 𝑆≤limsup_𝑛→∞𝑥_𝑛. (2.6) On the other hand, by 𝑥_𝑛+1≤𝑃_𝑛 for all 𝑛≥1, we obtain limsup_𝑛→∞𝑥_𝑛≤limsup_𝑛→∞𝑃_𝑛=𝑆. (2.7) The proof is complete.

Remark 2.6. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1). By Lemma 2.2, we see that if 𝑆=limsup_𝑛→∞𝑥_𝑛 and 𝑥_𝑁<𝑆 for some 𝑁>0, then 𝑥_𝑁−1,𝑥_𝑁+1∈[𝑆,+∞). For example, if it were 𝑥_𝑁−1<𝑆, then it would be 𝑃_𝑁<𝑆, which would imply limsup_𝑛→∞𝑥_𝑛<𝑆.

Lemma 2.7. Suppose that {𝑥_𝑛}^∞_𝑛=−1 is a positive solution of (1.1) and 𝑆=limsup_𝑛→∞𝑥_𝑛. Write 𝜔(𝑥_𝑛)={𝑥∶thereexist−1≤𝑘₁<𝑘₂<⋯<𝑘_𝑛<⋯suchthatlim_𝑛→∞𝑥_𝑘𝑛=𝑥}. (2.8) Then 𝜔(𝑥_𝑛)={𝑆,1/𝑆}.

Proof. If 𝜔(𝑥_𝑛) contains only one point, we may assume by taking a subsequence that 𝐴_𝑛𝑘→𝜇(<1). By taking the limit in the following relationship: 𝑥_𝑛𝑘+11=max{𝑥_𝑛𝑘,𝐴_𝑛𝑘𝑥_𝑛𝑘−1}, (2.9) as 𝑘→∞, we obtain 1𝑆=max{𝑆,𝜇𝑆1}=𝑆, (2.10) which implies that 𝑆=1. If 𝜔(𝑥_𝑛) contains at least two points, let 𝐿∈𝜔(𝑥_𝑛)−{𝑆}, then there exists a subsequence 𝑥_𝑛𝑘 of 𝑥_𝑛 such that 𝑥_𝑛𝑘→𝐿<𝑆. (2.11) By Remark 2.6, we see that there exists 𝑁>0 such that for every 𝑛_𝑘>𝑁, 𝑥_𝑛𝑘<𝑆,𝑥_𝑛𝑘+1,𝑥_𝑛𝑘−1∈[𝑆,+∞), (2.12) from which it follows that 𝑥_𝑛𝑘+1→𝑆,𝑥_𝑛𝑘−1→𝑆. (2.13) By taking a subsequence we may assume that 𝐴_𝑛𝑘→𝜇(<1). By taking the limit in the following relationship: 𝑥_𝑛𝑘+11=max{𝑥_𝑛𝑘,𝐴_𝑛𝑘𝑥_𝑛𝑘−1}, (2.14) as 𝑘→∞, we obtain 1𝑆=max{𝐿,𝜇𝑆1}=𝐿, (2.15) which implies 1𝐿=𝑆. (2.16) The proof is complete.

Theorem 2.8. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1) and 𝑆=limsup_𝑛→∞𝑥_𝑛. Then one of the following two statements is true.(1)If there exist infinitely many 𝑛 such that 𝑥_𝑛≥𝑆 and 𝑥_𝑛+1≥𝑆, then {𝑥_𝑛}^∞_𝑛=−1 is eventually equal to 1.(2)If there exists 𝑁 such that 𝑥_𝑁+2𝑘<𝑆 and 𝑥_{𝑁+2𝑘−1}≥𝑆 for all 𝑘≥0, then 𝑥_𝑁+2𝑘→1/𝑆 and 𝑥_{𝑁+2𝑘−1}→𝑆.

Proof. (1) We assume that there exists an infinite sequence 𝑛₁<𝑛₂<𝑛₃<⋯<𝑛_𝑘<⋯ such that 𝑥_𝑛𝑘≥𝑆,𝑥_𝑛𝑘+1≥𝑆. (2.17) By taking a subsequence we may assume from Lemma 2.7 that 𝐴_𝑛𝑘→𝜇<1,𝑥_𝑛𝑘−11→𝑙∈{𝑆,𝑆}. (2.18) By taking the limit in the following relationship: 𝑥_𝑛𝑘+1𝑥_𝑛𝑘𝐴=max{1,^𝑛𝑘𝑥_𝑛𝑘𝑥_𝑛𝑘−1}, (2.19) as 𝑘→∞, we get 𝑆²=max{1,𝑆𝜇𝑙}. (2.20) Since 𝑆𝜇/𝑙∈{𝜇,𝜇𝑆²} and 𝜇<1, it follows that 𝑆²=1 and 𝜔(𝑥_𝑛)={1}.In the following, we show that {𝑥_𝑛}^∞_𝑛=−1 is eventually equal to 1. It only needs to prove that there exists 𝑁≥0 such that for all 𝑛≥𝑁,1𝑥_𝑛>𝐴_𝑛𝑥_𝑛−1. (2.21) Indeed, if there exist infinitely many 𝑛_𝑘 such that 𝑥_𝑛𝑘+1=𝐴_𝑛𝑘𝑥_𝑛𝑘−1, (2.22) by taking a subsequence we may assume that 𝐴_𝑛𝑘→𝜇<1, then it follows that 𝜇1=1,𝜇=1, (2.23) which is a contradiction. Therefore there exists 𝑁 such that for all 𝑛≥𝑁,𝑥_𝑛+1=1𝑥_𝑛. (2.24) Thus 𝑥_𝑛=𝑥_𝑁,for𝑥𝑛=𝑁+2𝑘,_𝑛=𝑥_𝑁+1,for𝑛=𝑁+2𝑘+1. (2.25) Since 𝑥_𝑛→1, we have 𝑥_𝑁+1=𝑥_𝑁=1. (2) If 𝑆=1, then the result follows from Lemma 2.7. In the following, we assume 𝑆≠1. Suppose for the sake of contradiction that there exists a subsequence 𝑥_{𝑁+2𝑘𝑖} of 𝑥_𝑁+2𝑘 such that 𝑥_{𝑁+2𝑘𝑖}→𝑆. (2.26) By taking a subsequence we may assume that 𝐴_{𝑁+2𝑘𝑖}→𝜇. (2.27) By taking the limit in the following relationship: 𝑥_{𝑁+2𝑘𝑖+1}1=max{𝑥_{𝑁+2𝑘𝑖},𝐴_{𝑁+2𝑘𝑖}𝑥_{𝑁+2𝑘𝑖−1}}, (2.28) as 𝑘_𝑖→∞, we get 1𝑆=max{𝑆,𝜇𝑆}, (2.29) which implies 𝑆=1. (2.30) This is a contradiction. The proof is complete.

Corollary 2.9. Let {𝐴_𝑛}^∞_𝑛=0 be a periodic sequence of positive real numbers, then every positive solution of (1.1) is eventually periodic with period 2.

Proof. Let {𝑥_𝑛}^∞_𝑛=−1 be a positive solution of (1.1) and 𝑆=limsup_𝑛→∞𝑥_𝑛. By Remark 2.6 and Theorem 2.8, we may assume without loss of generality that 𝑥_2𝑘<𝑆, 𝑥_2𝑘−1≥𝑆≥1 for all 𝑘≥0. Suppose for the sake of contradiction that there exists a sequence 𝑚₁<𝑚₂<⋯<𝑚_𝑘<⋯ such that(1)𝑥_𝑚𝑘+1𝑥_𝑚𝑘−1=𝐴_𝑚𝑘, and 𝑥_𝑚𝑘+1𝑥_𝑚𝑘>1;(2)𝑥_𝑛+1𝑥_𝑛=1, for 𝑛≠𝑚_𝑘.Then 𝑚_𝑘 is odd for every 𝑘≥1. Let 𝑚_𝑘=2𝑛_𝑘+1, then it follows from Lemma 2.1 that 𝑥_2𝑛𝑘+2𝑥_2𝑛𝑘=𝐴_2𝑛𝑘+1<1=𝑥_2𝑛𝑘+1𝑥_2𝑛𝑘<𝑥_2𝑛𝑘+1𝑥_2𝑛𝑘+2. (2.31) From this and by (2) it follows that 𝐴_2𝑛𝑘+1𝑥_2𝑛𝑘+2=𝑥_2𝑛𝑘<𝑥_2𝑛𝑘+2=𝑥_2𝑛𝑘+4=⋯=𝑥_2𝑛𝑘+1<𝑥_{2𝑛𝑘+1+2}=𝐴_{2𝑛𝑘+1+1}𝑥_2𝑛𝑘+1. (2.32) Therefore for every 𝑘≥1, 𝐴_2𝑛𝑘+1<𝑥²_2𝑛𝑘+2=𝑥²_2𝑛𝑘+1<𝐴_{2𝑛𝑘+1+1}, (2.33) which is a contradiction since {𝐴_𝑛}^∞_𝑛=0 is a periodic sequence. The proof is complete.

Remark 2.10. Corollary 2.9 is the main result of [5].

3. Example

In this section, we give an example for {𝐴_𝑛}^∞_𝑛=0 to be no periodic sequence.

Example 3.1. Consider 𝑥_𝑛+11=max{𝑥_𝑛,𝐴_𝑛𝑥_𝑛−1},𝑛=0,1,…, (3.1) where 𝐴_2𝑛=𝐴_2𝑛+1=(2−1/2^𝑛)(2−1/2^𝑛+1)/16 for any 𝑛≥0. Then solution {𝑥_𝑛}^∞_𝑛=−1 of (3.1) with the initial values 𝑥₋₁=1/4 and 𝑥₀=4 satisfies the following.(1)𝑥_2𝑝−1𝑥_2𝑝=1, for any 𝑝≥0.(2)𝑥_2𝑝−1<𝑥_2𝑝+1=𝐴_2𝑝𝑥_2𝑝−1<12<2<𝑥_2𝑝+2<𝑥_2𝑝,forany𝑝≥0.

Proof. By simple computation, we have 𝐴_2𝑝=(2−1/2^𝑝)(2−1/2^𝑝+1)>𝑥16²⁻¹,if(𝐴𝑝=0,₀𝑥₋₁)²,if(𝐴𝑝=1,_2𝑝−2𝐴_2𝑝−6⋯𝐴₂𝐴_2𝑝−4𝐴_2𝑝−8⋯𝐴₀𝑥₋₁)²,if𝑝≥2iseven,(𝐴_2𝑝−2𝐴_2𝑝−6⋯𝐴₄𝐴₀𝐴_2𝑝−4𝐴_2𝑝−8⋯𝐴₂𝑥₋₁)²,if𝑝≥2isodd. (3.2) It follows from (3.1) and (3.2) that 𝑥₁𝑥₋₁𝑥=max{⁻¹𝑥₀,𝐴₀}=max{𝑥²₋₁,𝐴₀}=𝐴₀,𝑥₂𝑥₁𝑥=max{1,¹𝐴₁𝑥₀𝐴}=max{1,⁰𝐴₁𝑥₋₁𝑥₀𝑥}=1,₃𝑥₁𝑥=max{¹𝑥₂,𝐴₂𝑥}=max{²¹𝑥₂𝑥₁,𝐴₂𝐴}=max{(⁰𝑥₋₁)²,𝐴₂}=𝐴₂,𝑥₄𝑥₃𝑥=max{1,³𝐴₃𝑥₂𝐴}=max{1,²𝐴₃𝑥₂𝑥₁𝑥}=1,₅𝑥₃𝑥=max{³𝑥₄,𝐴₄𝑥}=max{²³𝑥₄𝑥₃,𝐴₄𝑥}=max{(³𝑥₁𝑥₁𝑥₋₁𝑥₋₁)²,𝐴₄}𝐴=max{(²𝐴₀𝑥₋₁)²,𝐴₄}=𝐴₄,𝑥₆𝑥₅𝑥=max{1,⁵𝐴₅𝑥₄𝐴}=max{1,⁴𝐴₅𝑥₄𝑥₃𝑥}=1,₇𝑥₅𝑥=max{⁵𝑥₆,𝐴₆𝑥}=max{²⁵𝑥₆𝑥₅,𝐴₆𝑥}=max{(⁵𝑥₃𝑥₁𝑥₋₁𝑥₃𝑥₁𝑥₋₁)²,𝐴₆}𝐴=max{(⁴𝐴₀𝐴₂𝑥₋₁)²,𝐴₆}=𝐴₆,𝑥₈𝑥₇𝑥=max{1,⁷𝐴₇𝑥₆𝐴}=max{1,⁶𝐴₇𝑥₆𝑥₅}=1. (3.3) By induction, we have from (3.1) and (3.2) that for any 𝑝≥1,𝑥_4𝑝+1𝑥_4𝑝−1𝑥=max{^4𝑝−1𝑥_4𝑝,𝐴_4𝑝𝑥}=max{^24𝑝−1𝑥_4𝑝𝑥_4𝑝−1,𝐴_4𝑝}=max{𝑥²_4𝑝−1,𝐴_4𝑝}𝑥=max{(^4𝑝−1𝑥_4𝑝−3𝑥_4𝑝−5⋯𝑥₁𝑥_4𝑝−3𝑥_4𝑝−5𝑥_4𝑝−7⋯𝑥₋₁𝑥₋₁)²,𝐴_4𝑝}𝐴=max{(^4𝑝−2𝐴_4𝑝−6⋯𝐴₂𝐴_4𝑝−4𝐴_4𝑝−8⋯𝐴₀𝑥₋₁)²,𝐴_4𝑝}=𝐴_4𝑝,𝑥_4𝑝+2𝑥_4𝑝+1𝑥=max{1,^4𝑝+1𝐴_4𝑝+1𝑥_4𝑝𝐴}=max{1,^4𝑝𝐴_4𝑝+1𝑥_4𝑝𝑥_4𝑝−1𝑥}=1,_4𝑝+3𝑥_4𝑝+1𝑥=max{^4𝑝+1𝑥_4𝑝+2,𝐴_4𝑝+2𝑥}=max{^24𝑝+1𝑥_4𝑝+2𝑥_4𝑝+1,𝐴_4𝑝+2}=max{𝑥²_4𝑝+1,𝐴_4𝑝+2}𝑥=max{(^4𝑝+1𝑥_4𝑝−1𝑥_4𝑝−3𝑥_4𝑝−5⋯𝑥₁𝑥₋₁𝑥_4𝑝−1𝑥_4𝑝−3𝑥_4𝑝−5𝑥_4𝑝−7⋯𝑥₁𝑥₋₁)²,𝐴_4𝑝+2}𝐴=max{(^4𝑝𝐴_4𝑝−4⋯𝐴₄𝐴₀𝐴_4𝑝−2𝐴_4𝑝−6⋯𝐴₂𝑥₋₁)²,𝐴_4𝑝+2}=𝐴_4𝑝+2,𝑥_4𝑝+4𝑥_4𝑝+3𝑥=max{1,^4𝑝+3𝐴_4𝑝+3𝑥_4𝑝+2𝐴}=max{1,^4𝑝+2𝐴_4𝑝+3𝑥_4𝑝+2𝑥_4𝑝+1}=1. (3.4) from which the result follows. The proof is complete.

Acknowledgments

The project is supported by NNSF of China(10861002), NSF of Guangxi (0640205,0728002), and Innovation Project of Guangxi Graduate Education(2008105930701M43).