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Discrete Dynamics in Nature and Society
Volume 2008, Article ID 193872, 7 pages
http://dx.doi.org/10.1155/2008/193872
Research Article

A New Part-Metric-Related Inequality Chain and an Application

1School of Computer and Information, Chongqing Jiaotong University, Chongqing 400074, China
2College of Computer Science, Chongqing University, Chongqing 400044, China
3Department of Computer Science, Hong Kong Baptist University, Kowloon, Hong Kong

Received 28 September 2007; Accepted 6 November 2007

Academic Editor: Stevo Stevic

Copyright © 2008 Xiaofan Yang et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

Part-metric-related (PMR) inequality chains are elegant and are useful in the study of difference equations. In this paper, we establish a new PMR inequality chain, which is then applied to show the global asymptotic stability of a class of rational difference equations.

1. Introduction

A part-metric related (PMR) inequality chain is a chain of inequalities of the form min1𝑖𝑛𝑎𝑖,1𝑎𝑖𝑎𝑓1,,𝑎𝑛max1𝑖𝑛𝑎𝑖,1𝑎𝑖,(1.1) which is closely related to the well-known part metric [1] and has important applications in the study of difference equations [213]. Below are three previously known PMR inequality chains: min1𝑖4𝑎𝑖,1𝑎𝑖𝑎1+𝑎2+𝑎3𝑎4𝑎1𝑎2+𝑎3+𝑎4max1𝑖4𝑎𝑖,1𝑎𝑖(see[5]),(1.2)min1𝑖𝑘𝑎𝑖,1𝑎𝑖𝑎1++𝑎𝑘2+𝑎𝑘1𝑎𝑘𝑎1𝑎2+𝑎3++𝑎𝑘max1𝑖𝑘𝑎𝑖,1𝑎𝑖(see[11]),(1.3)min1𝑖5𝑎𝑖,1𝑎𝑖(1+𝑤)𝑎1𝑎2𝑎3+𝑎4+𝑎5𝑎1𝑎2+𝑎1𝑎3+𝑎2𝑎3+𝑤𝑎4𝑎5max1𝑖5𝑎𝑖,1𝑎𝑖,1𝑤2(see[13]).(1.4)min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖𝑤(𝑎1,,𝑎2𝑝1)max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖,(1.5)𝑤 In this article, we establish the following PMR inequality chain: 𝑝2𝑤𝑝1 where 𝑝=3 will be defined in the next section, . When , chain (1.5) reduces to chain (1.4). On this basis, we prove that the difference equation with positive initial conditions admits a globally asymptotically stable equilibrium 𝑓𝑎1,,𝑎2𝑝1||𝑖1𝑖𝑟𝑎=𝑓1,,𝑎2𝑝1||𝑎𝑖𝑗=𝑚,1𝑗𝑟.(2.1).

2. Main Results

This section establishes the main results of this paper. For a function 𝑎1,,𝑎𝑛,𝑏1,,𝑏𝑛>0, let min1𝑖𝑛{𝑎𝑖/𝑏𝑖}(𝑎1++𝑎𝑛)/(𝑏1+

Lemma 2.1. Let +𝑏𝑛)max1𝑖𝑛{𝑎𝑖/𝑏𝑖}. Then 𝑎1/𝑏1==𝑎𝑛/𝑏𝑛𝑝3. One equality in the chain holds if and only if 𝑤>0.

For 𝑤(+)2𝑝1+ and 𝑤𝑎1,,𝑎2𝑝1=(1+𝑤)𝑝𝑖=1𝑎𝑖+2𝑝1𝑖=𝑝+1𝑎𝑖×2𝑝1𝑖=𝑝+11/𝑎𝑖𝑝𝑖=1𝑎𝑖×𝑝𝑖=11/𝑎𝑖+𝑤2𝑝1𝑖=𝑝+1𝑎𝑖.(2.2), define a function 𝑤𝑎1,,𝑎5=(1+𝑤)𝑎1𝑎2𝑎3+𝑎4+𝑎5𝑎1𝑎2+𝑎1𝑎3+𝑎2𝑎3+𝑤𝑎4𝑎5,𝑤𝑎1,,𝑎7=(1+𝑤)𝑎1𝑎2𝑎3𝑎4+𝑎5𝑎6+𝑎5𝑎7+𝑎6𝑎7𝑎1𝑎2𝑎3+𝑎1𝑎2𝑎4+𝑎1𝑎3𝑎4+𝑎2𝑎3𝑎4+𝑤𝑎5𝑎6𝑎7.(2.3) as follows: 𝑤=𝑤(𝑎1,,𝑎2𝑝1) Below are two examples of this function: 𝑎𝑟

For brevity, let 𝑤. Note that, for each 𝑎𝑟, 𝑤 is linear fractional in 𝑎𝑟. As a consequence, 𝑝3 is monotone in 𝑎1,,𝑎2𝑝1>0. Through simple calculations, we get the following two lemmas.

Lemma 2.2. Let 𝑚=min1𝑖2𝑝1{𝑎𝑖}, 1𝑟𝑝, 𝑝2, 𝑝2(𝑝1)/𝑝𝑖=1,𝑖𝑟(1/𝑎𝑖).
(1)If 𝑝2 is increasing in ar, then 𝑎𝑟. The equality holds if and only if 𝑝2 is constant in 𝑎𝑟.(2)If 𝑝2𝑝2|𝑎𝑟=𝑚 is strictly decreasing in 𝑎𝑟=𝑚, then 𝑝3. The equality holds if and only if 𝑎1,,𝑎2𝑝1>0.

Lemma 2.3. Let 𝑚=min1𝑖2𝑝1{𝑎𝑖}, 𝑝+1𝑟2𝑝1, 𝑝2, 𝑎𝑟.
(1)If 𝑝22𝑝1𝑖=𝑝+1,𝑖𝑟(1/𝑎𝑖)/(𝑝2) is increasing in 𝑝2, then 𝑎𝑟. The equality holds if and only if 𝑝2 is constant in 𝑎𝑟.(2)If 𝑝2𝑝2|𝑎𝑟=𝑚 is strictly decreasing in 𝑎𝑟=𝑚, then 𝑝3. The equality holds if and only if 𝑎1,,𝑎2𝑝1>0.

Theorem 2.4. Let min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}𝑝2, max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}. Then 𝑎1==𝑎2𝑝1=1𝑚=min1𝑖2𝑝1{𝑎𝑖}. One of the two equalities holds if and only if 𝑀=max1𝑖2𝑝1{𝑎𝑖}.

Proof. Let 𝑝2max{𝑀,1/𝑚}, min{𝑀,1/𝑚}𝑝2.
We prove only 𝑖1,,𝑖2𝑝1 because 1,2,,2𝑝1 can be proved similarly. We proceed by distinguishing two possible cases.
Case 1. There is a permutation 1𝑘2𝑝1 of 𝑎𝑖𝑘=𝑚 such that for each 𝑝2|𝑖1𝑖𝑘1, either 𝑎𝑖𝑘 or 𝑝2𝑝2||𝑖1𝑝2||𝑖1𝑖2𝑝1=121𝑚+𝑚1max𝑚,𝑚1max𝑀,𝑚.(2.4) is strictly decreasing in 𝑖1,,𝑖𝑟. Then1,2,,2𝑝1
Case 2. There is a partial permutation of () such that (a) for each , either or is strictly decreasing in , and (b) for each , and is increasing in . Then
Since , there is . If , it follows from (2.5) and Lemma 2.2 that Whereas if , it follows from (2.5) and Lemma 2.3 that Hence, is proven.
Second, we prove that if . The claim of “ if ” can be treated similarly. To this end, we need to prove the following.

Claim 1. If , then there is a permutation of such that for each , either or is strictly decreasing in .

Proof of Claim 1. On the contrary, assume that Claim 1 is not true. Then there is a partial permutation of 1𝑘𝑟 (𝑎𝑖𝑘=𝑚) such that (a) for each 𝑝2|𝑖1𝑖𝑘1, either 𝑎𝑖𝑘 or 𝑡{1,,2𝑝1}{𝑖1,,𝑖𝑟} is strictly decreasing in 𝑎𝑖𝑡𝑚, and (b) for each 𝑝2|𝑖1𝑖𝑟, 𝑎𝑡 and 𝑡{1,,2𝑝1}{𝑖1,,𝑖𝑟} is increasing in 𝑝2|𝑖1𝑖𝑟. One of the following two cases must occur.
Case 1. There is 𝑎𝑡 such that 𝑡{1,,𝑝}{𝑖1,,𝑖𝑟} is strictly increasing in 𝑝2𝑝2||𝑖1𝑖𝑟<(𝑝1)𝑝𝑖=1,𝑖𝑡1/𝑎𝑖||𝑖1𝑖𝑟max1𝑖𝑝,𝑖𝑡𝑎𝑖|𝑖1𝑖𝑟1max𝑀,𝑚.(2.8). If 𝑡{𝑝+1,,2𝑝1}{𝑖1,,𝑖𝑟}, it follows by (2.5), (2.6), and Lemma 2.2 that 𝑝2𝑝2||𝑖1𝑖𝑟<2𝑝1𝑖=𝑝+1,𝑖𝑡1/𝑎𝑖||(𝑝2)𝑖1𝑖𝑟max𝑝+1𝑖2𝑝1,𝑖𝑡1𝑎𝑖|𝑖1𝑖𝑟1max𝑀,𝑚.(2.9) A contradiction occurs. Whereas if 𝑡{1,,2𝑝1}{𝑖1,,𝑖𝑟}, it follows by (2.5), (2.7), and Lemma 2.3 that 𝑝2|𝑖1𝑖𝑟 Again a contradiction occurs.
Case 2. For each 𝑎𝑡, {1,,𝑝}{𝑖1,,𝑖𝑟} is constant in 𝑡{1,,𝑝}{𝑖1,.
First, let us show that ,𝑖𝑟}. Otherwise, there is 𝑝2||𝑖1𝑖𝑟=(𝑝1)𝑝𝑖=1,𝑖𝑡1/𝑎𝑖||𝑖1𝑖𝑟.(2.10)𝑠{1,,𝑝}{𝑖1,,𝑖𝑟,𝑡}. By Lemma 2.2, we have 𝑝2|𝑖1𝑖𝑟If there is 𝑎𝑠, it follows from (2.10) that {1,,𝑝}{𝑖1,,𝑖𝑟}={𝑡} is strictly increasing in 1max𝑀,𝑚=𝑝2𝑝2||𝑖1𝑖𝑟=𝑝2(𝑎1,,𝑎2𝑝1)||𝑎𝑖1==𝑎𝑖𝑟=𝑚=𝑚<𝑀,(2.11), a contradiction occurs. So, {1,,𝑝}{𝑖1,,𝑖𝑟} and thus 𝑡{𝑝+1,,2𝑝1}{𝑖1,,𝑖𝑟} from which a contradiction follows. So, 𝑝2||𝑖1𝑖𝑟=2𝑝1𝑖=𝑝+1,𝑖𝑡1/𝑎𝑖||(𝑝2)𝑖1𝑖𝑟.(2.12).

According to the previous argument, there is 𝑠{𝑝+1,,2𝑝1}{𝑖1,,𝑖𝑟,𝑡}. By Lemma 2.3, we get 𝑝2|𝑖1𝑖𝑟 If there is 𝑎𝑠, it follows from (2.12) that {𝑝+1,,2𝑝1}{𝑖1,,𝑖𝑟}={𝑡} is strictly decreasing in 𝑎1==𝑎𝑡1=𝑎𝑡+1==𝑎2𝑝1=𝑚.(2.13), a contradiction. So, 𝑝2=𝑝2||𝑖1𝑖𝑟=(𝑝1)𝑚3+𝑚+(𝑝2)𝑎𝑡𝑝𝑚2+(𝑝2)𝑚𝑎𝑡.(2.14) and thus 𝑝2|𝑖1𝑖𝑟 By (2.13) and (2.2), we get 𝑎𝑡 Since (d/d𝑎𝑡)𝑝2|𝑖1𝑖𝑟=((𝑝1)(𝑝2)𝑚2(1𝑚2))/ is constant in [𝑝𝑚2+(𝑝2)𝑚𝑎𝑡]2, and 𝑚=1𝑝2|𝑖1𝑖𝑟=1/𝑚, we derive 𝑝2=max{𝑀,1/𝑚}. From (2.12) and (2.13), we get 𝑚=1. Since 𝑝2|𝑖1𝑖𝑟=1/𝑚=1𝑀, all equalities in chains (2.5) and (2.7) hold. These plus 𝑀=𝑚=1 yield 𝑎𝑡=1=𝑚, from which we derive 𝑝2=max{𝑀,1/𝑚}. So, 𝑎1==𝑎2𝑝1=𝑚. This is a contradiction. Claim 1 is proved.

By Claim 1 and 𝑝2(𝑚,,𝑚)=(𝑚+1/𝑚)/2=𝑚, all equalities in (2.4) must hold. This plus Lemma 2.2 yields 𝑎1==𝑎2𝑝1=1 and 𝑝3. This implies 𝑎1,,𝑎2𝑝1>0.

Theorem 2.5. Let min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}𝑝1, max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}. Then, 𝑎1==𝑎𝑝=1/𝑎𝑝+1==1/𝑎2𝑝1. One of the two equalities holds if and only if 𝑝1𝑎max1,,𝑎𝑝,1𝑎𝑝+11,,𝑎2𝑝1max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖,𝑝1𝑎min1,,𝑎𝑝,1𝑎𝑝+11,,𝑎2𝑝1min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.15)𝑝3.

Proof. By Lemma 2.1 and (2.2), we get 𝑝2𝑤𝑝1 The second claim follows immediately from Lemma 2.1.
We are ready to present the main result of this paper.

Theorem 2.6. Let 𝑎1,,𝑎2𝑝1>0, 𝑎𝑘=𝑤𝑎𝑘2𝑝+1,,𝑎𝑘1,𝑘=2𝑝,2𝑝+1,.(2.16), min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}𝑎𝑘max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}. Let 𝑘=2𝑝,2𝑝+1, Then 𝑘2𝑝+1, 𝑎1==𝑎2𝑝1=1. If 𝑤, then one of the two equalities holds if and only if 𝑎2𝑝min𝑝2𝑎1,,𝑎2𝑝1,𝑝1𝑎1,,𝑎2𝑝1min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖,𝑎2𝑝max𝑝2𝑎1,,𝑎2𝑝1,𝑝1𝑎1,,𝑎2𝑝1max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖,𝑎2𝑝+1min𝑝2𝑎2,,𝑎2𝑝,𝑝1𝑎2,,𝑎2𝑝min2𝑖2𝑝𝑎𝑖,1𝑎𝑖min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖,𝑎2𝑝+1max𝑝2𝑎2,,𝑎2𝑝,𝑝1𝑎2,,𝑎2𝑝max2𝑖2𝑝𝑎𝑖,1𝑎𝑖max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.17).

Proof. Regard 𝑘=2𝑝 as a linear fractional function in w, which is monotone in w. By Theorems 2.4 and 2.5, we obtain 2𝑝+1, Working inductively, we conclude that for 𝑎𝑘min𝑝2𝑎𝑘2𝑝+1,,𝑎𝑘1,𝑝1𝑎𝑘2𝑝+1,,𝑎𝑘1min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖𝑎,(2.18)𝑘max𝑝2(𝑎𝑘2𝑝+1,,𝑎𝑘1),𝑝1(𝑎𝑘2𝑝+1,,𝑎𝑘1)max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.19), 𝑎2𝑝+1=max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}, 𝑎1==𝑎2𝑝1=1𝑎2𝑝+1=max𝑝2𝑎2,,𝑎2𝑝,𝑝1𝑎2,,𝑎2𝑝=max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.20)

Claim 2. If 𝑎2𝑝+1=𝑝2(𝑎2,,𝑎2𝑝)=max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖} , then 𝑎2==𝑎2𝑝=1.

Proof of Claim 2. By (2.19), we get 𝑎2𝑝+1=1 Here, we encounter two possible cases. Case 1. 1=𝑎2𝑝+1=max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}=max{𝑎1,. By Theorem 2.4, we get 1/𝑎1} and, hence, 𝑎1=1. Then 𝑎2𝑝+1=𝑝1(𝑎2,,𝑎2𝑝)=max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}𝑎2==𝑎𝑝+1=1𝑎𝑝+21==𝑎2𝑝,(2.21), implying 𝑎2𝑝+1=𝑝1(𝑎2,,𝑎2𝑝)=𝑎2.(2.22).Case 2. max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖=𝑎2𝑝+1=1𝑎2𝑝1min𝑝2𝑎1,,𝑎2𝑝1,𝑝1𝑎1,,𝑎2𝑝1max1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.23). By Theorem 2.5, we get min𝑝2𝑎1,,𝑎2𝑝1,𝑝1𝑎1,,𝑎2𝑝1=min1𝑖2𝑝1𝑎𝑖,1𝑎𝑖.(2.24) and consequently, 𝑝2(𝑎1,,𝑎2𝑝1)=min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖} Then, 𝑎1== Hence, all equalities in this chain hold. In particular, we have 𝑎2𝑝1=1 If 𝑝1(𝑎1,,𝑎2𝑝1)=min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}, it follows from Theorem 2.4 that 𝑎1==𝑎𝑝=1𝑎𝑝+11==𝑎2𝑝1.(2.25)𝑎1==𝑎2𝑝1=1. Now, assume that 𝑎𝑘=max1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}. By Theorem 2.5, we get 𝑘2𝑝+1 Equations (2.21) and (2.25) imply that 𝑎1==𝑎2𝑝1=1. Claim 2 is proven.

By Claim 2 and working inductively, we get that if 𝑎1==𝑎2𝑝1=1 for some

𝑎𝑘=min1𝑖2𝑝1{𝑎𝑖,1/𝑎𝑖}, then 𝑘2𝑝+1.

Similarly, we can show that 𝑝3 if 𝑝2𝑤𝑝1 holds for some 𝑥𝑛=𝑤𝑥𝑛2𝑝+1,,𝑥𝑛1,𝑛=1,2,...,(2.26).

As an application of Theorem 2.6, we have the following theorem.

Theorem 2.7. Let 𝑦𝑛=𝑦𝑛𝑘+𝑦𝑛𝑚/1+𝑦𝑛𝑘𝑦𝑛𝑚, . The difference equation with positive initial conditions admits the globally asymptotically stable equilibrium c = 1.

The proof of this theorem is similar to those in [11, 13], and hence is omitted.

Acknowledgments

The authors are grateful to the anonymous referees for their valuable comments and suggestions. This work is supported by Natural Science Foundation of China (10771227), Program for New Century Excellent Talent of Educational Ministry of China (NCET-05-0759), Doctorate Foundation of Educational Ministry of China (20050611001), and Natural Science Foundation of Chongqing CSTC (2006BB2231).

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