On a Higher-Order Nonlinear Difference Equation

Abstract

This paper shows that all positive solutions of a higher-order nonlinear difference equation are bounded, extending some recent results in the literature.

1. Introduction

There is a considerable interest in studying nonlinear difference equations nowadays; see, for example, [1–40] and numerous references listed therein.

The investigation of the higher-order nonlinear difference equation 𝑥_𝑛𝑥=𝐴+^{𝑝𝑛−𝑚}𝑥^𝑟_𝑛−𝑘,𝑛∈ℕ₀, (1.1) where 𝐴,𝑟>0 and 𝑝≥0, and 𝑘,𝑚∈ℕ, 𝑘≠𝑚, was suggested by Stević at numerous talks and in papers (see, e.g., [20, 28, 30, 34–38] and the related references therein).

In this paper we show that under some conditions on parameters 𝐴,𝑟, and 𝑝 all positive solutions of the difference equation 𝑥_𝑛𝑥=𝐴+^𝑝𝑛−1𝑥^𝑟_𝑛−𝑘,𝑛∈ℕ₀, (1.2) where 𝑘∈ℕ⧵{1}, are bounded. To do this we modify some methods and ideas from Stević's papers [30, 35–37]. Our motivation stems from these four papers.

The reader can find results for some particular cases of (1.2), as well as on some closely related equations treated in, for example, [1, 2, 5–11, 18–20, 26, 30, 33–35, 38, 40].

2. Main Result

Here we investigate the boundedness of the positive solutions to (1.2) for the case 0<𝑝<(𝑟𝑘^𝑘/(𝑘−1)^𝑘−1)^1/𝑘. The following result completely describes the boundedness of positive solutions to (1.2) in this case. The result is an extension of one of the main results in [35].

Theorem 2.1. Assume, 𝑝,𝑟>0 and 𝑘∈ℕ⧵{1}. Then every positive solution of (1.2) is bounded if 0<𝑝<(𝑟𝑘^𝑘(𝑘−1)^𝑘−1)^1/𝑘. (2.1)

Proof. First note that from (1.2) it directly follows that 𝑥_𝑛>𝐴,for𝑛∈ℕ₀. (2.2) Using (1.2), it follows that 𝑥_𝑛𝑥=𝐴+^𝑝𝑛−1𝑥^𝑟_𝑛−𝑘𝑥=𝐴+(^𝑛−1𝑥^𝑟/𝑝_𝑛−𝑘)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘+𝑥^𝑝_𝑛−2𝑥^𝑟/𝑝_𝑛−𝑘𝑥^𝑟_{𝑛−𝑘−1})^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝑥+(^𝑛−2𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1})^𝑝)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}+𝑥^𝑝_𝑛−3𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝑥^𝑟_{𝑛−𝑘−2})^𝑝)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝑥+(^𝑛−3𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2})^𝑝)^𝑝)^𝑝. (2.3) After 𝑘 steps we obtain the following formula 𝑥_𝑛𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(𝑥^{𝑟/𝑝𝑘−1}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}+𝑥^𝑝_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−1}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}𝑥^𝑟_{𝑛−(2𝑘−1)})^𝑝⋯)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(𝑥^{𝑟/𝑝𝑘−1}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}+𝑥^{𝑝−(𝑟/𝑝𝑘−1)}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}𝑥^𝑟_{𝑛−(2𝑘−1)})^𝑝⋯)^𝑝. (2.4)
Two subcases can be considered now.
Case 1 (𝑟≥𝑝^𝑘). If 𝑟≥𝑝^𝑘, then by (2.2) equality (2.4) implies that 𝑥_𝑛𝐴<𝐴+(𝐴^𝑟/𝑝𝐴+(𝐴^{𝑟/𝑝2+𝑟/𝑝}𝐴+(𝐴^{𝑟/𝑝3+𝑟/𝑝2+𝑟/𝑝}𝐴+⋯+(𝐴^{𝑟/𝑝𝑘−1+𝑟/𝑝𝑘−2+⋯+𝑟/𝑝}+1𝐴^{𝑟/𝑝𝑘−1+𝑟/𝑝𝑘−2+⋯+𝑟/𝑝+𝑟−𝑝})^𝑝⋯)^𝑝<∞, (2.5) for 𝑛≥2𝑘−1. This means that (𝑥_𝑛) is a bounded sequence.Case 2 (𝑝^𝑘>𝑟). In this case we have 𝑟𝑝−𝑝^𝑘−1>0. (2.6)
From (2.4) and (1.2) we further obtain 𝑥_𝑛𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(𝑥^{𝑟/𝑝𝑘−1}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}+𝑥^{𝑝−𝑟/𝑝𝑘−1}_𝑛−𝑘𝑥^{𝑟/𝑝𝑘−2}_{𝑛−𝑘−1}⋯𝑥^𝑟/𝑝_{𝑛−(2𝑘−2)}𝑥^𝑟_{𝑛−(2𝑘−1)})^𝑝⋯)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(∏^𝑘−2_𝑗=0𝑥^𝑧0(𝑗)_{𝑛−𝑘−𝑗}+𝑥^{𝑝−𝑧0(0)}_𝑛−𝑘(∏^𝑘−2_𝑗=1𝑥^𝑧0(𝑗)_{𝑛−𝑘−𝑗})𝑥^𝑟_{𝑛−(2𝑘−1)})^𝑝⋯)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(∏^𝑘−2_𝑗=0𝑥^𝑧0(𝑗)_{𝑛−𝑘−𝑗}𝑥+(^𝑛−𝑘(∏^𝑘−2_𝑗=1𝑥^{𝑧0(𝑗)/𝑝−𝑧0(0)}_{𝑛−𝑘−𝑗})𝑥^{𝑟/(𝑝−𝑧0(0))}_{𝑛−(2𝑘−1)})^{𝑝(𝑝−𝑧0(0))})^𝑝⋯)^𝑝𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(∏^𝑘−2_𝑗=0𝑥^𝑧1(𝑗)_{𝑛−𝑘−1−𝑗}+𝑥^{𝑝−𝑧1(0)}_{𝑛−𝑘−1}∏^𝑘−2_𝑗=1𝑥^𝑧1(𝑗)_{𝑛−𝑘−1−𝑗}𝑥^𝑟_𝑛−2𝑘)^{𝑝(𝑝−𝑧0(0))})^𝑝⋯)^𝑝𝐴=⋯=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(∏^𝑘−2_𝑗=0𝑥^{𝑧𝑚(𝑗)}_{𝑛−𝑘−𝑚−𝑗}+𝑥^{𝑝−𝑧𝑚(0)}_{𝑛−𝑘−𝑚}(∏^𝑘−2_𝑗=1𝑥^{𝑧𝑚(𝑗)}_{𝑛−𝑘−𝑚−𝑗})𝑥^𝑟_{𝑛−2𝑘+1−𝑚})^{𝑝∏𝑚−1𝑖=0(𝑝−𝑧𝑖(0))})^𝑝⋯)^𝑝, (2.7) for each 𝑘∈ℕ⧵{1} and every 𝑛≥2𝑘+𝑚−1, where the sequences (𝑧_𝑚^(𝑗)),𝑗=0,1,…,𝑘−2, satisfy the system 𝑧⁽⁰⁾_𝑚+1=𝑧_𝑚⁽¹⁾𝑝−𝑧_𝑚⁽⁰⁾,𝑧⁽¹⁾_𝑚+1=𝑧_𝑚⁽²⁾𝑝−𝑧_𝑚⁽⁰⁾,…,𝑧^(𝑘−3)_𝑚+1=𝑧_𝑚^(𝑘−2)𝑝−𝑧_𝑚⁽⁰⁾,𝑧^(𝑘−2)_𝑚+1=𝑟𝑝−𝑧_𝑚⁽⁰⁾, (2.8) and the initial values are given by 𝑧₀^(𝑗)=𝑟𝑝^{𝑗+1−𝑘},𝑗=0,1,…,𝑘−2. (2.9)
Note that 𝑝^𝑘>𝑟 implies that 𝑧₀⁽⁰⁾<𝑝. Assume 𝑧_𝑚⁽⁰⁾<𝑝 for every 𝑚∈ℕ₀.
By a direct calculation it follows that 𝑧₀^(𝑗)<𝑧₁^(𝑗), 𝑗=0,1,…,𝑘−2, which, along with (2.8) implies that (𝑧_𝑚^(𝑗)),𝑗=0,1,…,𝑘−2, are strictly increasing sequences.
From system (2.8), we have, 𝑧⁽⁰⁾_𝑚+1=𝑟(𝑝−𝑧_𝑚⁽⁰⁾)(𝑝−𝑧⁽⁰⁾_𝑚−1)⋯(𝑝−𝑧⁽⁰⁾_{𝑚−𝑘+2}),𝑚≥𝑘−2. (2.10)
If it were 𝑧_𝑚⁽⁰⁾<𝑝, 𝑚∈ℕ₀, then there was lim_𝑚→∞𝑧_𝑚⁽⁰⁾=𝑧∈(0,𝑝]. (2.11) Clearly 𝑧 is a solution of the equation 𝑓(𝑥)=𝑥(𝑝−𝑥)^𝑘−1−𝑟=0. (2.12)
Since 𝑓(0)=𝑓(𝑝)=−𝑟, (2.13) and 𝑓^′(𝑥)=(𝑝−𝑥)^𝑘−2(𝑝−𝑘𝑥), (2.14) we see that the function 𝑓 attains its maximum at the point 𝑥=𝑝/𝑘.
Further, by assumption (2.1) we get 𝑝𝑓(𝑘)=(𝑘−1)^𝑘−1𝑘^𝑘(𝑝^𝑘𝑘−𝑟^𝑘(𝑘−1)^𝑘−1)<0, (2.15) which along with (2.13) implies that (2.12) does not have solutions on (0,𝑝], arriving at a contradiction.
This implies that there is a fixed index 𝑚₀∈ℕ such that 𝑧_𝑚⁽⁰⁾₀₋₁<𝑝,𝑧_𝑚⁽⁰⁾₀≥𝑝. (2.16) From this, inequality (2.2), and identity (2.7) with 𝑚=𝑚₀, it follows that 𝑥_𝑛𝐴=𝐴+(𝑥^𝑟/𝑝_𝑛−𝑘𝐴+(𝑥^𝑟/𝑝2_𝑛−𝑘𝑥^𝑟/𝑝_{𝑛−𝑘−1}𝐴+(𝑥^𝑟/𝑝3_𝑛−𝑘𝑥^𝑟/𝑝2_{𝑛−𝑘−1}𝑥^𝑟/𝑝_{𝑛−𝑘−2}𝐴+⋯+(∏^𝑘−2_𝑗=0𝑥^{𝑧𝑚0(𝑗)}_{𝑛−𝑘−𝑚0−𝑗}+𝑥^{𝑝−𝑧𝑚0(0)}_{𝑛−𝑘−𝑚0}(∏^𝑘−2_𝑗=1𝑥^{𝑧𝑚0(𝑗)}_{𝑛−𝑘−𝑚0−𝑗})𝑥^𝑟_{𝑛−2𝑘+1−𝑚0})^{𝑝∏𝑚0−1𝑖=0(𝑝−𝑧𝑖(0))})^𝑝⋯)^𝑝𝐴≤𝐴+(𝐴^𝑟/𝑝𝐴+(𝐴^{𝑟/𝑝2+𝑟/𝑝}𝐴+(𝐴^{𝑟/𝑝3+𝑟/𝑝2+𝑟/𝑝}𝐴+⋯+(𝐴^{∑𝑘−2𝑗=0𝑧𝑚0(𝑗)}+1𝐴^{𝑟+𝑧𝑚0(0)−𝑝+∑𝑘−2𝑗=1𝑧𝑚0(𝑗)})^{𝑝∏𝑚0−1𝑖=0(𝑝−𝑧𝑖(0))})^𝑝⋯)^𝑝<∞ (2.17) for 𝑛≥2𝑘+𝑚₀−1.
From (2.17) the boundedness of the sequence (𝑥_𝑛) directly follows, as desired.

Acknowledgment

The research was partially supported by the Serbian Ministry of Science, through The Mathematical Institute of SASA, Belgrade, Project no. 144013.