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

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 min_{1≤𝑖≤𝑛}𝑎^𝑖,1𝑎_𝑖≤𝑓𝑎¹,…,𝑎_𝑛≤max_{1≤𝑖≤𝑛}𝑎^𝑖,1𝑎_𝑖,(1.1) which is closely related to the well-known part metric [1] and has important applications in the study of difference equations [2–13]. Below are three previously known PMR inequality chains: min_1≤𝑖≤4𝑎^𝑖,1𝑎_𝑖≤𝑎¹+𝑎₂+𝑎₃𝑎₄𝑎₁𝑎₂+𝑎₃+𝑎₄≤max_1≤𝑖≤4𝑎^𝑖,1min_{1≤𝑖≤𝑘}𝑎_𝑖(see[5]),(1.2)𝑎^𝑖,𝑎1_𝑖≤𝑎¹+⋯+𝑎_𝑘−2+𝑎_𝑘−1𝑎_𝑘𝑎₁𝑎₂+𝑎₃+⋯+𝑎_𝑘≤max_{1≤𝑖≤𝑘}𝑎^𝑖,1min_1≤𝑖≤5𝑎_𝑖(see[11]),(1.3)𝑎^𝑖,𝑎1_𝑖≤(1+𝑤)𝑎¹𝑎₂𝑎₃+𝑎₄+𝑎₅𝑎₁𝑎₂+𝑎₁𝑎₃+𝑎₂𝑎₃+𝑤𝑎₄𝑎₅≤max_1≤𝑖≤5𝑎^𝑖,1𝑎_𝑖,1≤𝑤≤2(see[13]).(1.4)min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖≤ℎ^𝑤(𝑎₁,…,𝑎_2𝑝−1)≤max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖,(1.5)ℎ_𝑤 In this article, we establish the following PMR inequality chain: 𝑝−2≤𝑤≤𝑝−1 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 𝑓𝑎₁,…,𝑎_2𝑝−1||_{𝑖1∼𝑖𝑟}=𝑓𝑎₁,…,𝑎_2𝑝−1||_{𝑎𝑖𝑗=𝑚,1≤𝑗≤𝑟}.(2.1).

2. Main Results

This section establishes the main results of this paper. For a function 𝑎₁,…,𝑎_𝑛,𝑏₁,…,𝑏_𝑛>0, let min_{1≤𝑖≤𝑛}{𝑎_𝑖/𝑏_𝑖}≤(𝑎₁+⋯+𝑎_𝑛)/(𝑏₁+

Lemma 2.1. Let ⋯+𝑏_𝑛)≤max_{1≤𝑖≤𝑛}{𝑎_𝑖/𝑏_𝑖}. Then 𝑎₁/𝑏₁=⋯=𝑎_𝑛/𝑏_𝑛𝑝≥3. One equality in the chain holds if and only if 𝑤>0.

For ℎ_𝑤∶(ℜ₊)^2𝑝−1→ℜ₊ and ℎ_𝑤𝑎¹,…,𝑎_2𝑝−1=(1+𝑤)∏^𝑝𝑖=1𝑎_𝑖+∏^{2𝑝−1𝑖=𝑝+1}𝑎_𝑖×∑^{2𝑝−1𝑖=𝑝+1}1/𝑎^𝑖∏^𝑝𝑖=1𝑎_𝑖×∑^𝑝𝑖=11/𝑎^𝑖+𝑤∏^{2𝑝−1𝑖=𝑝+1}𝑎_𝑖.(2.2), define a function ℎ_𝑤𝑎¹,…,𝑎₅=(1+𝑤)𝑎¹𝑎₂𝑎₃+𝑎₄+𝑎₅𝑎₁𝑎₂+𝑎₁𝑎₃+𝑎₂𝑎₃+𝑤𝑎₄𝑎₅ℎ_𝑤,𝑎¹,…,𝑎₇=(1+𝑤)𝑎¹𝑎₂𝑎₃𝑎₄+𝑎₅𝑎₆+𝑎₅𝑎₇+𝑎₆𝑎₇𝑎₁𝑎₂𝑎₃+𝑎₁𝑎₂𝑎₄+𝑎₁𝑎₃𝑎₄+𝑎₂𝑎₃𝑎₄+𝑤𝑎₅𝑎₆𝑎₇.(2.3) as follows: ℎ_𝑤=ℎ_𝑤(𝑎₁,…,𝑎_2𝑝−1) Below are two examples of this function: 𝑎_𝑟

For brevity, let ℎ_𝑤. Note that, for each 𝑎_𝑟, ℎ_𝑤 is linear fractional in 𝑎_𝑟. As a consequence, 𝑝≥3 is monotone in 𝑎₁,…,𝑎_2𝑝−1>0. Through simple calculations, we get the following two lemmas.

Lemma 2.2. Let 𝑚=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖}, 1≤𝑟≤𝑝, ℎ_𝑝−2, ℎ_𝑝−2≤(𝑝−1)/∑^{𝑝𝑖=1,𝑖≠𝑟}(1/𝑎_𝑖).
(1)If ℎ_𝑝−2 is increasing in a_r, then 𝑎_𝑟. The equality holds if and only if ℎ_𝑝−2 is constant in 𝑎_𝑟.(2)If ℎ_𝑝−2≤ℎ_𝑝−2|_{𝑎𝑟=𝑚} is strictly decreasing in 𝑎_𝑟=𝑚, then 𝑝≥3. The equality holds if and only if 𝑎₁,…,𝑎_2𝑝−1>0.

Lemma 2.3. Let 𝑚=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖}, 𝑝+1≤𝑟≤2𝑝−1, ℎ_𝑝−2, 𝑎_𝑟.
(1)If ℎ_𝑝−2≤∑^{2𝑝−1𝑖=𝑝+1,𝑖≠𝑟}(1/𝑎_𝑖)/(𝑝−2) is increasing in ℎ_𝑝−2, then 𝑎_𝑟. The equality holds if and only if ℎ_𝑝−2 is constant in 𝑎_𝑟.(2)If ℎ_𝑝−2≤ℎ_𝑝−2|_{𝑎𝑟=𝑚} is strictly decreasing in 𝑎_𝑟=𝑚, then 𝑝≥3. The equality holds if and only if 𝑎₁,…,𝑎_2𝑝−1>0.

Theorem 2.4. Let min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}≤ℎ_𝑝−2≤, max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}. Then 𝑎₁=⋯=𝑎_2𝑝−1=1𝑚=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖}. One of the two equalities holds if and only if 𝑀=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖}.

Proof. Let ℎ_𝑝−2≤max{𝑀,1/𝑚}, min{𝑀,1/𝑚}≤ℎ_𝑝−2.
We prove only 𝑖₁,…,𝑖_2𝑝−1 because 1,2,…,2𝑝−1 can be proved similarly. We proceed by distinguishing two possible cases.
Case 1. There is a permutation 1≤𝑘≤2𝑝−1 of 𝑎_𝑖𝑘=𝑚 such that for each ℎ_𝑝−2|_{𝑖1∼𝑖𝑘−1}, either 𝑎_𝑖𝑘 or ℎ_𝑝−2≤ℎ_𝑝−2||_𝑖1≤⋯≤ℎ_𝑝−2||_{𝑖1∼𝑖2𝑝−1}=12𝑚+𝑚1≤max𝑚,𝑚1≤max𝑀,𝑚1.(2.4) is strictly decreasing in 𝑖₁,…,𝑖_𝑟. Then1,2,…,2𝑝−1
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 1≤𝑘≤𝑟 (𝑎_𝑖𝑘=𝑚) such that (a) for each ℎ_𝑝−2|_{𝑖1∼𝑖𝑘−1}, either 𝑎_𝑖𝑘 or 𝑡∈{1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟} is strictly decreasing in 𝑎_𝑖𝑡≠𝑚, and (b) for each ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}, 𝑎_𝑡 and 𝑡∈{1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟} is increasing in ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}. One of the following two cases must occur.
Case 1. There is 𝑎_𝑡 such that 𝑡∈{1,…,𝑝}−{𝑖₁,…,𝑖_𝑟} is strictly increasing in ℎ_𝑝−2≤ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}<(𝑝−1)∑^{𝑝𝑖=1,𝑖≠𝑡}1/𝑎^𝑖||_{𝑖1∼𝑖𝑟}≤max_{1≤𝑖≤𝑝,𝑖≠𝑡}𝑎^𝑖|^𝑖₁^∼𝑖_𝑟≤max𝑀,1𝑚.(2.8). If 𝑡∈{𝑝+1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟}, it follows by (2.5), (2.6), and Lemma 2.2 that ℎ_𝑝−2≤ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}<∑^{2𝑝−1𝑖=𝑝+1,𝑖≠𝑡}1/𝑎^𝑖(𝑝−2)||_{𝑖1∼𝑖𝑟}≤max_{𝑝+1≤𝑖≤2𝑝−1,𝑖≠𝑡}1𝑎_𝑖|^𝑖₁^∼𝑖_𝑟≤max𝑀,1𝑚.(2.9) A contradiction occurs. Whereas if 𝑡∈{1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟}, it follows by (2.5), (2.7), and Lemma 2.3 that ℎ_𝑝−2|_{𝑖1∼𝑖𝑟} Again a contradiction occurs.
Case 2. For each 𝑎_𝑡, {1,…,𝑝}⊆{𝑖₁,…,𝑖_𝑟} is constant in 𝑡∈{1,…,𝑝}−{𝑖₁,.
First, let us show that …,𝑖_𝑟}. Otherwise, there is ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}=(𝑝−1)∑^{𝑝𝑖=1,𝑖≠𝑡}1/𝑎^𝑖||_{𝑖1∼𝑖𝑟}.(2.10)𝑠∈{1,…,𝑝}−{𝑖₁,…,𝑖_𝑟,𝑡}. By Lemma 2.2, we have ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}If there is 𝑎_𝑠, it follows from (2.10) that {1,…,𝑝}−{𝑖₁,…,𝑖_𝑟}={𝑡} is strictly increasing in max𝑀,1𝑚=ℎ^𝑝−2≤ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}=ℎ_𝑝−2(𝑎₁,…,𝑎_2𝑝−1)||_{𝑎𝑖1=⋯=𝑎𝑖𝑟=𝑚}=𝑚<𝑀,(2.11), a contradiction occurs. So, {1,…,𝑝}⊆{𝑖₁,…,𝑖_𝑟} and thus 𝑡∈{𝑝+1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟} from which a contradiction follows. So, ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}=∑^{2𝑝−1𝑖=𝑝+1,𝑖≠𝑡}1/𝑎^𝑖(𝑝−2)||_{𝑖1∼𝑖𝑟}.(2.12).

According to the previous argument, there is 𝑠∈{𝑝+1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟,𝑡}. By Lemma 2.3, we get ℎ_𝑝−2|_{𝑖1∼𝑖𝑟} If there is 𝑎_𝑠, it follows from (2.12) that {𝑝+1,…,2𝑝−1}−{𝑖₁,…,𝑖_𝑟}={𝑡} is strictly decreasing in 𝑎₁=⋯=𝑎_𝑡−1=𝑎_𝑡+1=⋯=𝑎_2𝑝−1=𝑚.(2.13), a contradiction. So, ℎ_𝑝−2=ℎ_𝑝−2||_{𝑖1∼𝑖𝑟}=(𝑝−1)𝑚³+𝑚+(𝑝−2)𝑎_𝑡𝑝𝑚²+(𝑝−2)𝑚𝑎_𝑡.(2.14) and thus ℎ_𝑝−2|_{𝑖1∼𝑖𝑟} By (2.13) and (2.2), we get 𝑎_𝑡 Since (d/d𝑎_𝑡)ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}=((𝑝−1)(𝑝−2)𝑚²(1−𝑚²))/ is constant in [𝑝𝑚²+(𝑝−2)𝑚𝑎_𝑡]², and 𝑚=1ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}=1/𝑚, we derive ℎ_𝑝−2=max{𝑀,1/𝑚}. From (2.12) and (2.13), we get 𝑚=1. Since ℎ_𝑝−2|_{𝑖1∼𝑖𝑟}=1/𝑚=1≥𝑀, all equalities in chains (2.5) and (2.7) hold. These plus 𝑀=𝑚=1 yield 𝑎_𝑡=1=𝑚, from which we derive ℎ_𝑝−2=max{𝑀,1/𝑚}. So, 𝑎₁=⋯=𝑎_2𝑝−1=𝑚. This is a contradiction. Claim 1 is proved.

By Claim 1 and ℎ_𝑝−2(𝑚,…,𝑚)=(𝑚+1/𝑚)/2=𝑚, all equalities in (2.4) must hold. This plus Lemma 2.2 yields 𝑎₁=⋯=𝑎_2𝑝−1=1 and 𝑝≥3. This implies 𝑎₁,…,𝑎_2𝑝−1>0.

Theorem 2.5. Let min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}≤ℎ_𝑝−1≤, max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}. Then, 𝑎₁=⋯=𝑎_𝑝=1/𝑎_𝑝+1=⋯=1/𝑎_2𝑝−1. One of the two equalities holds if and only if ℎ_𝑝−1≤max𝑎₁,…,𝑎_𝑝,1𝑎_𝑝+1,…,1𝑎_2𝑝−1≤max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1ℎ_𝑝−1𝑎_𝑖,≥min𝑎₁,…,𝑎_𝑝,1𝑎_𝑝+1,…,1𝑎_2𝑝−1≥min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖.(2.15)𝑝≥3.

Proof. By Lemma 2.1 and (2.2), we get 𝑝−2≤𝑤≤𝑝−1 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,…,𝑎_2𝑝−1>0, 𝑎_𝑘=ℎ_𝑤𝑎^{𝑘−2𝑝+1},…,𝑎_𝑘−1,𝑘=2𝑝,2𝑝+1,….(2.16), min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}≤𝑎_𝑘≤max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}. Let 𝑘=2𝑝,2𝑝+1,… Then 𝑘≥2𝑝+1, 𝑎₁=⋯=𝑎_2𝑝−1=1. If ℎ_𝑤, then one of the two equalities holds if and only if 𝑎_2𝑝≥minℎ_𝑝−2𝑎¹,…,𝑎_2𝑝−1,ℎ^𝑝−1𝑎¹,…,𝑎_2𝑝−1≥min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_2𝑝𝑎_𝑖,≤maxℎ_𝑝−2𝑎¹,…,𝑎_2𝑝−1,ℎ^𝑝−1𝑎¹,…,𝑎_2𝑝−1≤max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_2𝑝+1𝑎_𝑖,≥minℎ_𝑝−2𝑎²,…,𝑎_2𝑝,ℎ^𝑝−1𝑎²,…,𝑎_2𝑝≥min_{2≤𝑖≤2𝑝}𝑎^𝑖,1𝑎_𝑖≥min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_2𝑝+1𝑎_𝑖,≤maxℎ_𝑝−2𝑎²,…,𝑎_2𝑝,ℎ^𝑝−1𝑎²,…,𝑎_2𝑝≤max_{2≤𝑖≤2𝑝}𝑎^𝑖,1𝑎_𝑖≤max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖.(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 𝑎_𝑘≥minℎ_𝑝−2𝑎^{𝑘−2𝑝+1},…,𝑎_𝑘−1,ℎ^𝑝−1𝑎^{𝑘−2𝑝+1},…,𝑎_𝑘−1≥min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎𝑎_𝑖,(2.18)_𝑘≤maxℎ_𝑝−2(𝑎_{𝑘−2𝑝+1},…,𝑎_𝑘−1),ℎ_𝑝−1(𝑎_{𝑘−2𝑝+1},…,𝑎_𝑘−1)≤max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖.(2.19), 𝑎_2𝑝+1=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}, 𝑎₁=⋯=𝑎_2𝑝−1=1𝑎_2𝑝+1=maxℎ_𝑝−2𝑎²,…,𝑎_2𝑝,ℎ^𝑝−1𝑎²,…,𝑎_2𝑝=max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖.(2.20)

Claim 2. If 𝑎_2𝑝+1=ℎ_𝑝−2(𝑎₂,…,𝑎_2𝑝)=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖} , then 𝑎₂=⋯=𝑎_2𝑝=1.

Proof of Claim 2. By (2.19), we get 𝑎_2𝑝+1=1 Here, we encounter two possible cases. Case 1. 1=𝑎_2𝑝+1=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}=max{𝑎₁,. By Theorem 2.4, we get 1/𝑎₁} and, hence, 𝑎₁=1. Then 𝑎_2𝑝+1=ℎ_𝑝−1(𝑎₂,…,𝑎_2𝑝)=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}𝑎₂=⋯=𝑎_𝑝+1=1𝑎_𝑝+2=⋯=1𝑎_2𝑝,(2.21), implying 𝑎_2𝑝+1=ℎ_𝑝−1(𝑎₂,…,𝑎_2𝑝)=𝑎₂.(2.22).Case 2. max_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖=𝑎^2𝑝+1=1𝑎_2𝑝≤1≤max_{1≤𝑖≤2𝑝−1}minℎ_𝑝−2𝑎¹,…,𝑎_2𝑝−1,ℎ^𝑝−1𝑎¹,…,𝑎_2𝑝−1𝑎^𝑖,𝑎1_𝑖.(2.23). By Theorem 2.5, we get minℎ_𝑝−2𝑎¹,…,𝑎_2𝑝−1,ℎ^𝑝−1𝑎¹,…,𝑎_2𝑝−1=min_{1≤𝑖≤2𝑝−1}𝑎^𝑖,1𝑎_𝑖.(2.24) and consequently, ℎ_𝑝−2(𝑎₁,…,𝑎_2𝑝−1)=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖} Then, 𝑎₁=⋯= Hence, all equalities in this chain hold. In particular, we have 𝑎_2𝑝−1=1 If ℎ_𝑝−1(𝑎₁,…,𝑎_2𝑝−1)=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}, it follows from Theorem 2.4 that 𝑎₁=⋯=𝑎_𝑝=1𝑎_𝑝+1=⋯=1𝑎_2𝑝−1.(2.25)𝑎₁=⋯=𝑎_2𝑝−1=1. Now, assume that 𝑎_𝑘=max_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}. By Theorem 2.5, we get 𝑘≥2𝑝+1 Equations (2.21) and (2.25) imply that 𝑎₁=⋯=𝑎_2𝑝−1=1. Claim 2 is proven.

By Claim 2 and working inductively, we get that if 𝑎₁=⋯=𝑎_2𝑝−1=1 for some

𝑎_𝑘=min_{1≤𝑖≤2𝑝−1}{𝑎_𝑖,1/𝑎_𝑖}, then 𝑘≥2𝑝+1.

Similarly, we can show that 𝑝≥3 if 𝑝−2≤𝑤≤𝑝−1 holds for some 𝑥_𝑛=ℎ_𝑤𝑥^{𝑛−2𝑝+1},…,𝑥_𝑛−1,𝑛=1,2,...,(2.26).

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

Theorem 2.7. Let 𝑦_𝑛=𝑦_𝑛−𝑘+𝑦_𝑛−𝑚/1+𝑦_𝑛−𝑘𝑦_𝑛−𝑚, . 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).