Discrete Dynamics in Nature and Society

Discrete Dynamics in Nature and Society / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 624157 | 12 pages | https://doi.org/10.1155/2011/624157

Global Structure of Positive Solutions of a Discrete Problem with Sign-Changing Weight

Academic Editor: Zengji Du
Received27 Apr 2011
Accepted28 Jun 2011
Published15 Aug 2011

Abstract

Let 𝑇>5 be an integer, 𝕋={1,2,…,𝑇}. We are concerned with the global structure of positive solutions set of the discrete second-order boundary value problems Δ2𝑢(𝑡−1)+𝑟𝑚(𝑡)𝑓(𝑢(𝑡))=0,𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0, where 𝑟∈ℝ is a parameter, 𝑚∶𝕋→ℝ changes its sign and 𝑚(𝑡)≠0for𝑡∈𝕋.

1. Introduction

Let 𝑇>5 be an integer, 𝕋={1,2,…,𝑇}. In this paper, we are concerned with the global structure of positive solutions set of the discrete second-order boundary value problems Δ2𝑢(𝑡−1)+𝑟𝑚(𝑡)𝑓(𝑢(𝑡))=0,𝑡∈𝕋,(1.1)𝑢(0)=𝑢(𝑇+1)=0,(1.2) where 𝑟∈ℝ is a parameter, 𝑚∶𝕋→ℝ changes its sign, and 𝑚(𝑡)≠0 for 𝑡∈𝕋.

The boundary value problems with sign-changing weight arise from a selection-migration model in population genetics, see Fleming [1]. 𝑚 changes sign corresponds to the fact that an allele 𝐴1 holds an advantage over a rival allele 𝐴2 at the same points and is at a disadvantage at others. The parameter 𝑟 corresponds to the reciprocal of the diffusion. So, the existence and multiplicity of positive solutions with sign-changing weight in continuous case has been studied by many authors, see, for example, [2–8] and the references therein.

For the discrete case, there are many literature dealing with difference equations similar to (1.1) subject to various boundary value conditions. We refer to [9–17] and the references therein. However, there are few papers to discuss the existence of positive solutions to (1.1) and (1.2) if 𝑚(𝑡) changes sign on 𝕋. Maybe the main reason is the spectrum of the following linear eigenvalue problems Δ2𝑢(𝑡−1)+𝜆𝑚(𝑡)𝑢(𝑡)=0,𝑡∈𝕋,𝑢(0)=𝑢(𝑇+1)=0(1.3) is not clear when 𝑚 changes its sign on 𝕋.

In 2008, Shi and Yan [18] investigated the spectrum of the second-order boundary value problems of difference equations [],−Δ𝑝(𝑡−1)Δ𝑢(𝑡−1)+ğ‘ž(𝑡)𝑢(𝑡)=𝜆𝑚(𝑡)𝑢(𝑡),𝑡∈𝕋,𝑝(0)Δ𝑢(0)−𝑝(𝑇)Δ𝑢(𝑇)=𝐾𝑢(0)𝑢(𝑇)(1.4) where 𝐾=𝑘11𝑘12𝑘21𝑘22 is a symmetric and positive definite 2×2 matrix. 𝑝∶{0,1,…,𝑇}→ℝ, ğ‘žâˆ¶ğ•‹â†’â„, and 𝑚∶𝕋→ℝ satisfy(A1)𝑚∶𝕋→ℝ changes its sign, and 𝑚(𝑡)≠0 for 𝑡∈𝕋.(A2)𝑝(𝑡−1)≥0,ğ‘ž(𝑡)≥0,𝑝(𝑡−1)+ğ‘ž(𝑡)>0,for𝑡∈𝕋, and 𝑝(0)+𝑘11≠0, 𝑝(𝑇)>0.

They proved that (1.4) has 𝑇 eigenvalues 𝜆1≤𝜆2≤⋯≤𝜆𝑇 which have 𝑇 corresponding linearly independent and orthogonal eigenfunctions.

However, it's easy to see that (1.3) is not included in (1.4) under the condition (A2). Also, Shi and Yan [18] provided no information about the sign of the eigenvalues and no information about the corresponding eigenfunctions.

In this paper, we will show that (1.3) has two principal eigenvalues 𝜆𝑚,−<0<𝜆𝑚,+, and the corresponding eigenfunctions we denote by 𝜓𝑚,− and 𝜓𝑚,+ don't change their sign on 𝕋∶={0,1,…,𝑇,𝑇+1}. Based on this result, using Rabinowitz's global bifurcation theorem [19], we will discuss the global structure of positive solutions set of (1.1) and (1.2).

The assumptions we are interested in this paper are as follows:(H1)𝑓∈𝐶(ℝ,ℝ) with 𝑠𝑓(𝑠)>0 for 𝑠≠0;(H2)𝑓0=lim|𝑠|→0(𝑓(𝑠)/𝑠)∈(0,∞);(H3)ğ‘“âˆž=lim|𝑠|→+∞(𝑓(𝑠)/𝑠)=0;(H3′)ğ‘“âˆž=∞.

Our main result is the following.

Theorem 1.1. Assume that (H1), (H2) hold. (i)If (H3) holds, then there exist 0<𝜆∗≤𝜆𝑚,+/𝑓0 and 𝜆𝑚,−/𝑓0≤𝜆∗<0 such that (1.1), (1.2) has at least one positive solution for 𝑟∈(−∞,𝜆∗)∪(𝜆∗,∞).(ii)If (H3′) holds, then there exist 𝜌∗≤𝜆𝑚,−/𝑓0<0 and 𝜌∗≥𝜆𝑚,+/𝑓0>0 such that (1.1), (1.2) has at least one positive solution for 𝑟∈(𝜌∗,0)∪(0,𝜌∗).

The rest of the paper is arranged as follows. In Section 2, the existence of two principal eigenvalues of (1.3), and some properties of these two eigenvalues will be discussed. In Section 3, we will prove our main result.

2. Existence of Two Principal Eigenvalues to (1.3)

Recall that 𝕋={1,2,…,𝑇} and 𝕋={0,1,…,𝑇+1}. Let 𝑋={𝑢∶𝕋→ℝ∣𝑢(0)=𝑢(𝑇+1)=0}. Then 𝑋 is a Banach space under the norm ‖𝑢‖𝑋=max𝑡∈𝕋|𝑢(𝑡)|. Let 𝑌={𝑢∣𝑢∶𝕋→ℝ}. Then 𝑌 is a Banach space under the norm ‖𝑢‖𝑌=max𝑡∈𝕋|𝑢(𝑡)|.

It's well known that the operator 𝜒∶𝑋→𝑌, 𝜒(0,𝑢(1),𝑢(2),…,𝑢(𝑇),0)=(𝑢(1),𝑢(2),…,𝑢(𝑇))(2.1) is a homomorphism.

Define the operator 𝐿∶𝑋→𝑌 by 𝐿𝑢(𝑡)=−Δ2𝑢(𝑡−1),𝑡∈𝕋.(2.2)

In this section, we will discuss the existence of principal eigenvalues for (1.3) with 𝑚∶𝕋→ℝ that changes its sign. The main idea we will use aries from [20, 21].

Theorem 2.1. Equation (1.3) has two principal eigenvalues 𝜆𝑚,+ and 𝜆𝑚,− such that 𝜆𝑚,−<0<𝜆𝑚,+, and the corresponding eigenfunctions, we denoted by 𝜓𝑚,+ and 𝜓𝑚,− do not change sign on 𝕋.

Proof. Consider, for fixed 𝜆, the eigenvalue problems 𝑢𝐿𝑢−𝜆𝑚(𝑡)𝑢(𝑡)=𝜇𝑢(𝑡),𝑡∈𝕋,(0)=𝑢(𝑇+1)=0.(2.3) By Kelley and Peterson [22, Theorem  7.6], for fixed 𝜆, (2.3) has 𝑇-simple eigenvalues 𝜇𝑚,1(𝜆)<𝜇𝑚,2(𝜆)<⋯<𝜇𝑚,𝑇(𝜆),(2.4) and the corresponding eigenfunction 𝜓𝑚,𝑘(𝜆,𝑡) has exactly 𝑘−1 simple generalized zeros.
Thus, 𝜆 is a principal eigenvalue of (1.3) if and only if 𝜇𝑚,1(𝜆)=0.
On the other hand, let 𝑆𝑚,𝜆=𝑇𝑡=0||||Δ𝜙(𝑡)2−𝜆𝑇𝑡=1𝑚(𝑡)𝜙(𝑡)2∶𝜙∈𝑋,𝑇𝑡=1𝜙(𝑡)2=1.(2.5) Clearly, 𝑆𝑚,𝜆 is bounded below. Then 𝜇𝑚,1(𝜆)=inf𝑆𝑚,𝜆.
For fixed 𝜙∈𝑋, ∑𝜆→𝑇𝑡=0|Δ𝜙(𝑡)|2∑−𝜆𝑇𝑡=1𝑚(𝑡)𝜙2(𝑡) is an affine and so concave function. As the infimum of any collection of concave functions is concave, it follows that 𝜆→𝜇𝑚,1(𝜆) is a concave function. Also, by considering test functions 𝜙1,𝜙2∈𝑋 such that ∑𝑇𝑡=1𝑚(𝑡)𝜙21(𝑡)<0 and ∑𝑇𝑡=1𝑚(𝑡)𝜙22(𝑡)>0, it is easy to see that 𝜇𝑚,1(𝜆)→−∞ as ğœ†â†’Â±âˆž. Thus, 𝜆→𝜇𝑚,1(𝜆) is an increasing function until it attains its maximum and is a decreasing function thereafter.
Since 𝜇1(0)>0, 𝜆→𝜇𝑚,1(𝜆) must have exactly two zeros. Thus, (1.3) has exactly two principal eigenvalues, 𝜆𝑚,+>0 and 𝜆𝑚,−<0, and the corresponding eigenfunctions we denoted by 𝜓𝑚,+ and 𝜓𝑚,− don't change sign on 𝕋.

Remark 2.2. From the proof of Theorem 2.1, it is not difficult to see that the following results hold.(i)If 𝑚(𝑡)≥0 on 𝕋 and there exist at least one point 𝑡0∈𝕋 such that 𝑚(𝑡0)>0, then (1.3) has only one principal eigenvalue 𝜆𝑚,+>0.(ii)If 𝑚(𝑡)≤0 on 𝕋 and there exist at least one point 𝑡0∈𝕋 such that 𝑚(𝑡0)<0, then (1.3) has only one principal eigenvalue 𝜆𝑚,−<0.

Now, we give some properties for the above principal eigenvalue(s).

Theorem 2.3. Let 𝑚∶𝕋→ℝ change its sign. Assume that there exists ğ‘šî…žâˆ¶ğ•‹â†’â„ such that 𝑚(𝑡)â‰¤ğ‘šî…ž(𝑡) for 𝑡∈𝕋. Then the followings hold. (i)If ğ‘šî…ž changes sign on 𝕋, then 𝜆𝑚′,−≤𝜆𝑚,−, 𝜆𝑚′,+≤𝜆𝑚,+;(ii)If ğ‘šî…žâ‰¥0, then 0<ğœ†ğ‘šî…ž,+≤𝜆𝑚,+.

Proof. For convenience, we only prove the case (i). It can be seen that for 𝜆>0, 𝑆𝑚,𝜆≥𝑆𝑚′,𝜆, which implies 𝜇𝑚,1(𝜆)≥𝜇𝑚′,1(𝜆) and consequently, 𝜆𝑚,+≥𝜆𝑚′,+.
On the other hand, for 𝜆<0, 𝑆𝑚,𝜆≤𝑆𝑚′,𝜆, which implies 𝜇𝑚,1(𝜆)≤𝜇𝑚′,1(𝜆) and consequently, 𝜆𝑚,−≥𝜆𝑚′,−.

Suppose that 𝕋0={ğ‘Ž+1,ğ‘Ž+2,…,𝑏−1} is a strict subset of 𝕋, and 𝑚𝕋0 denote the restriction of 𝑚 on 𝕋0. Consider the linear eigenvalue problems Δ2𝑢(𝑡−1)+𝜆𝑚(𝑡)𝑢(𝑡)=0,𝑡∈𝕋0,𝑢(ğ‘Ž)=𝑢(𝑏)=0.(2.6)

Then we get the following result.

Theorem 2.4. (i) If 𝑚(𝑡)>0 for 𝑡∈𝕋0, then (2.6) has only one positive principal eigenvalue 𝜆𝑚𝕋0,+ such that 0<𝜆𝑚,+<𝜆𝑚𝕋0,+.
(ii) If 𝑚(𝑡)<0 for 𝑡∈𝕋0, then (2.6) has only one principal eigenvalue 𝜆𝑚𝕋0,− such that 𝜆𝑚𝕋0,−<𝜆𝑚,−.
(iii) If 𝑚 changes its sign on 𝕋0, then (2.6) has two principal eigenvalue 𝜆𝑚𝕋0,−<0 and 𝜆𝑚𝕋0,+>0 such that 𝜆𝑚,+<𝜆𝑚𝕋0,+,𝜆𝑚𝕋0,−<𝜆𝑚,−.(2.7)

Proof. Consider the following problems: −Δ2𝑢(𝑡−1)−𝜆𝑚(𝑡)𝑢(𝑡)=𝜇𝑢(𝑡),𝑡∈𝕋0,𝑢(ğ‘Ž)=𝑢(𝑏)=0.(2.8) Let 𝜇𝑚𝕋0,𝑘 denote the 𝑘th eigenvalue of (2.8), and 𝜓𝑚𝕋0,𝑘 the corresponding eigenfunction which has exactly 𝑘−1 generalized zeros in (ğ‘Ž,𝑏). Let 𝐸={𝑢∶{ğ‘Ž,…,𝑏}→ℝ∣𝑢(ğ‘Ž)=𝑢(𝑏)=0} be a Banach space under the norm ‖𝑢‖𝐸=max{|𝑢(𝑡)|∶𝑡∈{ğ‘Ž,ğ‘Ž+1,…,𝑏−1,𝑏}}. Let 𝑆𝑚𝕋0,𝜆=𝑏−1𝑡=ğ‘Ž||||Δ𝜙(𝑡)2−𝜆𝑏−1𝑡=ğ‘Ž+1𝑚(𝑡)𝜙(𝑡)2∶𝜙∈𝐸,𝑏−1𝑡=ğ‘Ž+1𝜙(𝑡)2=1.(2.9) We get 𝜇𝑚𝕋0,1=inf𝑆𝑚𝕋0,𝜆. Similar to the proof of Theorem 2.1, we get the following assertions.(i)If 𝑚(𝑡)>0 for 𝑡∈𝕋0, then (2.6) has only one principal eigenvalue 𝜆𝑚𝕋0,+>0.(ii)If 𝑚(𝑡)<0 for 𝑡∈𝕋0, then (2.6) has only one negative principal eigenvalue 𝜆𝑚𝕋0,−<0.(iii)If 𝑚 changes its sign on 𝕋0, then (2.6) has two principal eigenvalue 𝜆𝑚𝕋0,−<0 and 𝜆𝑚𝕋0,+>0.
Now, we prove that the inequalities in (i), (ii), and (iii) hold. For convenience, suppose that ∑𝑏−1𝑡=ğ‘Ž+1𝜓2𝑚𝕋0,1(𝜆,𝑡)=1.
Let 𝜓𝑚𝕋0,1 denote the extension of 𝜓𝑚𝕋0,1 by zero on 𝕋, that is, 𝜓𝑚𝕋0,1𝜓(𝜆,𝑡)=𝑚𝕋0,1(𝜆,𝑡),𝑡∈𝕋0,0,𝑡∈𝕋⧵𝕋0.(2.10) Then, 𝜇𝑚𝕋0,1(𝜆)=𝑏−1𝑡=ğ‘Ž|||Δ𝜓𝑚𝕋0,1|||(𝜆,𝑡)2−𝜆𝑏−1𝑡=ğ‘Ž+1𝑚(𝑡)𝜓2𝑚𝕋0,1=(𝜆,𝑡)𝑇𝑡=1|||Δ𝜓𝑚𝕋0,1|||(𝜆,𝑡)2−𝜆𝑇𝑡=1𝑚(𝑡)𝜓2𝑚𝕋0,1(𝜆,𝑡)>inf𝑣∈𝑋𝑇𝑡=1||||Δ𝑣(𝜆,𝑡)2−𝜆𝑇𝑡=1𝑚(𝑡)𝑣2(𝜆,𝑡),𝑇𝑡=1𝑣2(𝜆,𝑡)=1=𝜇𝑚,1(𝜆),(2.11) which implies the desired results.

3. The Proof of the Main Result

First, we deal with the case 𝑟>0.

Recall that 𝐿∶𝑋→𝑌, 𝐿𝑢(𝑡)=−Δ2𝑢(𝑡−1).(3.1) Let 𝜁,𝜉∈𝐶(ℝ,ℝ) be such that 𝑓(𝑢)=𝑓0𝑢+𝜁(𝑢),𝑓(𝑢)=ğ‘“âˆžğ‘¢+𝜉(𝑢).(3.2) Clearly, lim|𝑢|→0𝜁(𝑢)𝑢=0,lim|𝑢|â†’âˆžğœ‰(𝑢)𝑢=0.(3.3) Let ̃𝜉(𝑢)=max0≤𝑠≤𝑢||||𝜉(𝑠).(3.4) Then, ̃𝜉 is nondecreasing and lim|𝑢|â†’âˆžÌƒğœ‰(𝑢)𝑢=0.(3.5) Let us consider 𝐿𝑢−𝜆𝑚(𝑡)𝑟𝑓0𝑢−𝜆𝑚(𝑡)𝑟𝜁(𝑢)=0,(3.6) as a bifurcation problem from the trivial solution 𝑢≡0.

Equation (3.6) can be converted to the equivalent equation 𝑢(𝑡)=𝜆𝐿−1𝑚(⋅)𝑟𝑓0𝑢(⋅)+𝑚(⋅)𝑟𝜁(𝑢(⋅))(𝑡).(3.7) Further, we note that ‖𝐿−1[𝑚(⋅)𝜁(𝑢(⋅))]‖=𝑜(‖𝑢‖) for 𝑢 near 0 in 𝑋, since ‖‖𝐿−1[]‖‖𝑚(⋅)𝜁(𝑢(⋅))=max𝑡∈𝕋|||||𝑇𝑠=1|||||𝐺(𝑡,𝑠)𝑚(𝑠)𝜁(𝑢(𝑠))≤𝐶⋅max𝑠∈𝕋||||(𝑚(𝑠)‖𝜁𝑢(⋅))‖,(3.8) where 1𝐺(𝑡,𝑠)=(𝑇+1(𝑇+1−𝑡)𝑠,0≤𝑠≤𝑡≤𝑇+1,𝑇+1−𝑠)𝑡,0≤𝑡≤𝑠≤𝑇+1.(3.9)

The results of Rabinowitz [19] for (3.6) can be stated as follows: from (𝜆𝑚,+/𝑟𝑓0,0), there emanates an unbounded continua ğ’ž+ of positive solutions in ℝ×𝑋.

It is clear that any solution of (3.6) of the form (1,𝑢) yields a solution 𝑢 of (1.1) and (1.2). So, we focus on the shape of ğ’ž+ under the conditions (H1)–(H3) or (H1)–(H3′), and we will show that ğ’ž+ crosses the hyperplane {1}×𝑋 in ℝ×𝑋.

Lemma 3.1. Suppose that (H1)–(H3) hold. Let 𝐽=[ğ‘Ž,𝑏] be a given compact interval in (0,∞). Then for all 𝑟∈𝐽, there exists 𝑀𝐽>0 such that all possible positive solution 𝑢 of (1.1) and (1.2) satisfy ‖𝑢‖𝑋≤𝑀𝐽.

Proof. Suppose on the contrary that there exists a sequence {𝑦𝑛} of positive solutions for (3.6) with {𝜇𝑛}⊂𝐽 and â€–ğ‘¦ğ‘›â€–ğ‘‹â†’âˆž. Let 𝛼∈(0,1/𝑏𝑄), where ∑𝑄=𝑇𝑠=1𝐺(𝑠,𝑠)|𝑚(𝑠)|. Then, by (H3), there exists 𝑢𝛼>0 such that 𝑢>𝑢𝛼 implies 𝑓(𝑢)<𝛼𝑢.
Let 𝐾𝛼=max𝑢∈[0,𝑢𝛼]𝑓(𝑢) and let 𝐴𝑛={𝑡∈𝕋∣𝑦𝑛(𝑡)≤𝑢𝛼} and 𝐵𝑛={𝑡∈𝕋∣𝑦𝑛(t)>𝑢𝛼}. Then we have 𝑦𝑛(𝑡)=𝜇𝑛𝑇𝑠=1𝑦𝐺(𝑡,𝑠)𝑚(𝑠)𝑓𝑛(𝑠)=𝜇𝑛𝐴𝑛𝑦𝐺(𝑡,𝑠)𝑚(𝑠)𝑓𝑛(𝑠)+𝜇𝑛𝐵𝑛𝑦𝐺(𝑡,𝑠)𝑚(𝑠)𝑓𝑛(𝑠)≤𝜇𝑛𝐾𝛼𝑄+𝜇𝑛𝐵𝑛𝑦𝐺(𝑡,𝑠)𝑚(𝑠)𝑓𝑛(𝑠)(3.10) for 𝕋𝑡∈. Thus, 1𝜇𝑛≤𝐾𝛼𝑄‖‖𝑦𝑛‖‖𝑋+𝐵𝑛𝑓𝑦𝐺(𝑡,𝑠)𝑚(𝑠)𝑛(𝑠)‖‖𝑦𝑛‖‖𝑋.(3.11) On 𝐵𝑛,𝑦𝑛(𝑠)>𝑢𝛼 implies 𝑓(𝑦𝑛(𝑠))/‖𝑦𝑛‖𝑋<𝑓(𝑦𝑛(𝑠))/𝑦𝑛(𝑠)<𝛼. Thus, 1𝜇𝑛≤𝐾𝛼𝑄‖‖𝑦𝑛‖‖𝑋+𝛼𝐵𝑛||||≤𝐾𝐺(𝑠,𝑠)𝑚(𝑠)𝛼𝑄‖‖𝑦𝑛‖‖𝑋+𝛼𝑄.(3.12) Since 0<ğ‘Žâ‰¤ğœ‡ğ‘›â‰¤ğ‘ for all 𝑛, we have 1/𝜇𝑛≥1/𝑏 for all 𝑛, and, thus, 1𝑏≤𝐾𝛼𝑄‖‖𝑦𝑛‖‖𝑋+𝛼𝑄.(3.13) By the fact â€–ğ‘¦ğ‘›â€–ğ‘‹â†’âˆž as ğ‘›â†’âˆž, we get 1𝑏1≤𝛼𝑄<𝑏.(3.14) This contradiction completes the proof.

Lemma 3.2. Suppose that (H1)–(H3) hold. Then Projâ„ğ’ž+⊃[𝜆𝑚,+/𝑟𝑓0,+∞).

Proof. Assume on the contrary that sup{𝜆∣(𝜆,𝑦)âˆˆğ’ž+}<∞, then, there exists a sequence {(𝜇𝑛,𝑦𝑛)}âˆˆğ’ž+ such that limğ‘›â†’âˆžâ€–â€–ğ‘¦ğ‘›â€–â€–ğ‘‹=∞,𝜇𝑛<𝐶0(3.15) for some positive constant 𝐶0 independent of 𝑛, since ğ’ž+ is unbounded. On the other hand, 𝜇𝑛>0 for all 𝑛∈ℕ, since (0,0) is the only solution of (3.6) for 𝜆=0 and ğ’ž+∩({0}×𝑋)=∅. Meanwhile, {(𝜇𝑛,𝑦𝑛)} satisfy Δ2𝑦𝑛(𝑡−1)+𝜇𝑛𝑓𝑦𝑚(𝑡)𝑛(𝑡)𝑦𝑛𝑦(𝑡)𝑛𝑦(𝑡)=0,𝑡∈𝕋,𝑛(0)=𝑦𝑛(𝑇+1)=0.(3.16) By (H3), there exist a positive constant 𝐿𝑓>0 such that 𝑓(𝑢)≤𝐿𝑓𝑢. Define a function 𝜒∶𝕋→[0,∞) by 𝐿𝜒(𝑡)=𝑓,𝑚(𝑡)>0,0,𝑚(𝑡)<0.(3.17) Then, 𝑚(𝑡)(𝑓(𝑦𝑛(𝑡))/𝑦𝑛(𝑡))≤𝜒(𝑡)𝑚(𝑡). Let 𝛾+ be the principal eigenvalue of linear eigenvalue problem Δ2𝑣(𝑡−1)+𝛾𝜒(𝑡)𝑚(𝑡)𝑣(𝑡)=0,𝑡∈𝕋,𝑣(0)=𝑣(𝑇+1)=0.(3.18) Then, by Remark 2.2, 𝛾+>0. Subsequently, by Theorem 2.3, we know that 𝛾+≤𝜇𝑛.(3.19) This combine with Lemma 3.1, limğ‘›â†’âˆžğœ‡ğ‘›=∞, which contradicts (3.15). Thus, Projâ„ğ’ž+⊃𝜆𝑚,+𝑟𝑓0,∞.(3.20)

Now, by Lemma 3.2, ğ’ž+ crosses the hyperplane {1}×𝑋 in ℝ×𝑋, and, then, Theorem 1.1(i) holds. To obtain Theorem 1.1(ii), we need to prove the following Lemma.

Lemma 3.3. Suppose that (H1), (H2), and (H3′) hold. Then Projâ„ğ’ž+⊃(0,𝜆𝑚,+/𝑟𝑓0).

Proof. Let {(𝜇𝑛,𝑦𝑛)}âŠ‚ğ’ž+ be such that |𝜇𝑛|+â€–ğ‘¦ğ‘›â€–ğ‘‹â†’âˆž as ğ‘›â†’âˆž. Then, Δ2𝑦𝑛(𝑡−1)+𝜇𝑛𝑦𝑚(𝑡)𝑓𝑛(𝑦𝑡)=0,𝑡∈𝕋,𝑛(0)=𝑦𝑛(𝑇+1)=0.(3.21) If {‖𝑦𝑛‖} is bounded, say, ‖𝑦𝑛‖≤𝑀1, for some 𝑀1 independent of 𝑛, then we may assume that limğ‘›â†’âˆžğœ‡ğ‘›=∞.(3.22) Note that 𝑓𝑦𝑛(𝑡)𝑦𝑛(𝑡)≥inf𝑓(𝑠)𝑠∣0<𝑠≤𝑀1>0.(3.23) Then, there exist two constants 𝑀3>0,𝑀2>0 such that 0<𝑀2<𝑓𝑦𝑛(𝑡)𝑦𝑛(𝑡)<𝑀3.(3.24) Define two functions 𝜒1,𝜒2∶𝕋→(0,∞) by 𝜒1=𝑀2𝑀,𝑚(𝑡)>0,3𝜒,𝑚(𝑡)<0.2=𝑀3𝑀,𝑚(𝑡)>0,2,𝑚(𝑡)<0.(3.25) Let 𝜂∗,𝜂∗ be the positive principal eigenvalue of the linear eigenvalue problems Δ2𝑣(𝑡−1)+𝜂𝜒2(Δ𝑡)𝑚(𝑡)𝑣(𝑡)=0,𝑡∈𝕋,(3.26)𝑣(0)=𝑣(𝑇+1)=0,(3.27)2𝑣(𝑡−1)+𝜂𝜒1(𝑡)𝑚(𝑡)𝑣(𝑡)=0,𝑡∈𝕋,(3.28)𝑣(0)=𝑣(𝑇+1)=0,(3.29) respectively.
Combining (3.22) and (3.24) with the relation Δ2𝑦𝑛(𝑡−1)+𝑟𝜇𝑛𝑓𝑦𝑚(𝑡)𝑛(𝑡)𝑦𝑛𝑦(𝑡)𝑛𝑦(𝑡)=0,𝑡∈𝕋,𝑛(0)=𝑦𝑛(𝑇+1)=0,(3.30) using Theorem 2.3, we get 𝜂∗𝑟≤𝜇𝑛≤𝜂∗𝑟.(3.31) This contradicts (3.22). So, {‖𝑦𝑛‖𝑋} is bounded uniformly for all 𝑛∈ℕ.
Now, taking {(𝜇𝑛,𝑦𝑛)}âŠ‚ğ’ž+ be such that ‖‖𝑦𝑛‖‖𝑋⟶+∞,as𝑛⟶+∞.(3.32) We show that limğ‘›â†’âˆžğœ‡ğ‘›=0.
Suppose on the contrary that, choosing a subsequence and relabeling if necessary, 𝜇𝑛≥𝑏0 for some constant 𝑏0>0. By (3.32), there exists 𝑡0∈𝕋 such that 𝑦𝑛(𝑡0)=‖𝑦𝑛‖𝑋 and 𝑦𝑛(𝑡0)→+∞ as 𝑛→+∞. Thus, 𝜇𝑛𝑓𝑦𝑛𝑡0𝑦𝑛𝑡0⟶+∞,as𝑛⟶+∞.(3.33)
Now, the proof can be divided into two cases.
Case 1 (𝑚(𝑡0)>0). Consider the following linear eigenvalue problems Δ2𝑣(𝑡−1)+𝛼𝑚(𝑡)𝑣(𝑡)=0,𝑡=𝑡0,𝑣𝑡0𝑡−1=𝑣0+1=0.(3.34) By Theorem 2.4, (3.34) has a positive principal eigenvalue 𝛼+, and 𝜇𝑛𝑓𝑦𝑛𝑡0𝑦𝑛𝑡0≤𝛼+,(3.35) which contradicts (3.33).Case 2 (𝑚(𝑡0)<0). Since (𝜇𝑛,𝑦𝑛) is a solution of (3.7), we get 0≥Δ2𝑦𝑛𝑡0−1=−𝜇𝑛𝑡𝑟𝑚0𝑓𝑦𝑛𝑡0>0.(3.36) This is a contradiction.
Thus, limğ‘›â†’âˆžğœ‡ğ‘›=0.

At last, we deal with the case 𝑟<0.
Let us consider 𝐿𝑢−𝜆𝑟𝑚(𝑡)𝑓0𝑢−𝜆𝑟𝑚(𝑡)𝜁(𝑢)=0(3.37) as a bifurcation problem from the trivial solution 𝑢≡0. The results of Rabinowitz [19] for (3.37) can be stated as follows: from (𝜆𝑚,−/−𝑟𝑓0,0), there emanates an unbounded continua ğ’žâˆ’ of positive solutions in ℝ×𝑋.
It is clear that any solution of (3.37) of the form (−1,𝑢) yields a solution 𝑢 of (1.1) and (1.2). Now, our proofs focus on the shape of ğ’žâˆ’. It will be proved that when (H1)–(H3) hold, then Projâ„ğ’žâˆ’âŠƒ(−∞,𝜆𝑚,−/−𝑟𝑓0), and when (H1)–(H3′) hold, Projâ„ğ’žâˆ’âŠƒ(𝜆𝑚,−/−𝑟𝑓0,0), that is, ğ’žâˆ’ crosses the hyperplane {−1}×𝑋 in ℝ×𝑋. Since the proof is similar to the case 𝑟>0, we omit it.

Remark 3.4. As an application of Theorem 1.1, let us consider nonlinear discrete problem with indefinite weight Δ2𝑢𝑓(𝑡−1)+𝑟𝑚(𝑡)(𝑢(𝑡))=0,𝑡∈𝕋,𝑢(0)=0,𝑢(3)=0,(3.38) where 𝕋={1,2}, 𝑚∶𝕋→ℝ with 𝑚(1)=1 and 𝑚(2)=−1, and ⎧⎪⎨⎪⎩[],𝑓(𝑠)=𝑠,𝑠∈−1,12𝑠1+𝑠4,𝑠∈(−∞,−1)∪(1,∞).(3.39) Then, 𝑓0𝑓=1,∞=0.(3.40)
To find the principal eigenvalues of linear eigenvalue problem Δ2𝑢(𝑡−1)+𝜆𝑚(𝑡)𝑢(𝑡)=0,𝑡∈𝕋,(3.41)𝑢(0)=0,𝑢(3)=0,(3.42) we rewrite (3.41) to the recursive sequence 𝑢𝑢(𝑡+1)=2−𝜆𝑚(𝑡)(𝑡)−𝑢(𝑡−1).(3.43) This together with the initial value condition 𝑢(0)=0,𝑢(1)=1(3.44) imply that 𝑢𝑢(2)=2−𝜆𝑚(1)(1)−𝑢(0)=2−𝜆,𝑢(3)=2−𝜆𝑚(2)𝑢(2)−𝑢(1)=3−𝜆2.(3.45) The last equation together with the boundary value condition 𝑢(3)=0 imply that 𝜆𝑚,−√=−3,𝜆𝑚,+=√3.(3.46) Thus by Theorem 1.1(i), (3.38), has a positive solution if √𝑟∈(−∞,−√3)∪(3,∞).

Acknowledgments

The authors are grateful to the anonymous referee for their valuable suggestions. R. Ma is supported by NSFC (11061030).

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