Positive Solutions to a Second-Order Discrete Boundary Value Problem

Abstract

We are concerned with second-order discrete boundary value problems and obtain some sufficient conditions for the existence of at least one positive solution by using the fixed point theorem due to Krasnosel'skii on a cone.

1. Introduction

Boundary value problems for difference equations have been studied extensively by many authors, for example, [1–10] to name a few. Many techniques arose in the studies of this kind of problem. For example, Agarwal et al. [1] employed the critical point theory to establish the existence of multiple solutions of some regular as well as singular discrete boundary value problems. Cai and Yu [2] applied the Linking Theorem and the Mountain Pass Lemma in the critical point theory to study second-order discrete boundary value problems and obtained some new results for the existence of solutions. Li and Sun [3, 4] were concerned with discrete system boundary value problems and gave some sufficient conditions for the existence of one or two positive solutions by using a nonlinear alternative of Leray-Schauder type and Krasnosel'skii's fixed point theorem in a cone. Pang et al. [5] provided sufficient conditions for the existence of at least three positive solutions for quasilinear boundary value problems for finite difference equations by using a generalization of the Leggett-Williams fixed point theorem due to Avery and Peterson. Du [6], Lin and Liu [7] discussed triple positive solutions of some second-order discrete boundary value problems by making use of the Leggett-Williams fixed-point theorem, respectively.

This paper deals with the following three-point boundary value problem for second-order difference equation of the form Δ²𝑦()(𝑘)𝑓𝑦(𝑘{},𝑦(0)(0)()(𝑛),𝑘−1+ℎ)=0,𝑘∈1,…,𝑇−𝛼Δ𝑦=0,𝑦𝑇+1=𝛽𝑦 (1.1) where Δ𝑦(𝑘−1)=𝑦(𝑘)−𝑦(𝑘−1), Δ²𝑦(𝑘−1)=𝑦(𝑘+1)−2𝑦(𝑘)+𝑦(𝑘−1),𝑘∈{1,…,𝑇}.

Throughout this paper, we will assume that the following conditions are satisfied:(A1)𝑇≥3 is a fixed positive integer, 𝑛∈{2,…,𝑇−1}, constant 𝛼,𝛽>0 such that 𝐻∶=𝑇+1−𝛽𝑛+𝛼(1−𝛽)>0 and 𝑇+1−𝛽𝑛>0;(A2)𝑓∈𝐶([0,+∞),[0,+∞)), 𝑓 is either superlinear or sublinear, that is, either 𝑓₀=0, 𝑓_∞=∞ or 𝑓₀=∞, 𝑓_∞=0, where 𝑓₀=lim_𝑢→0+𝑓(𝑢)𝑢,𝑓_∞=lim_𝑢→∞𝑓(𝑢)𝑢; (1.2)(A3)ℎ(𝑘) is nonnegative on {1,…,𝑇} and ℎ(𝑘)≡0 does not hold on {𝑛,…,𝑇}.

In the paper, we show the existence of positive solutions of (1.1) under some assumptions. We also establish the associate Green's function. Readers may find that it is useful to define a cone on which a positive operator was defined, and a fixed point theorem due to Krasnosel'skii [11] will be applied to yield the existence of at least one positive solution.

2. Preliminary and Green's Function

Let 𝐍 be the nonnegative integers; we let 𝐍_𝑖,𝑗={𝑘∈𝐍∶𝑖≤𝑘≤𝑗} and 𝐍_𝑝=𝐍_0,𝑝.

By a positive solution 𝑦 of problem (1.1), we mean 𝑦∶𝐍_𝑇+1→𝑅, 𝑦 satisfies the first equation of (1.1) on 𝐍_1,𝑇, 𝑦 fulfills 𝑦(0)−𝛼Δ𝑦(0)=0, 𝑦(𝑇+1)=𝛽𝑦(𝑛), and 𝑦 is nonnegative on 𝐍_𝑇+1 and positive on 𝐍_1,𝑇.

We shall need the following fixed point theorem due to Krasnosel'skii [8, 11].

Theorem A. Let 𝐸 be a Banach space, and let 𝐾⊂𝐸 be a cone in 𝐸. Assume that Ω₁ and Ω₂ are open subsets of 𝐸 with 0∈Ω₁, Ω₁⊂Ω₂, and let 𝐴∶𝐾⋂(Ω₂⧵Ω₁)→𝐾 be a completely continuous operator such that either(1)‖𝐴𝑢‖≤‖𝑢‖, 𝑢∈𝐾⋂𝜕Ω₁ and ‖𝐴𝑢‖≥‖𝑢‖, 𝑢∈𝐾⋂𝜕Ω₂ or(2)‖𝐴𝑢‖≥‖𝑢‖, 𝑢∈𝐾⋂𝜕Ω₁ and ‖𝐴𝑢‖≤‖𝑢‖, 𝑢∈𝐾⋂𝜕Ω₂. Then 𝐴 has a fixed point in 𝐾⋂(Ω₂⧵Ω₁).

Lemma 2.1 (see [7]). The function 𝐺()=1𝑘,𝑙𝐻(𝑙+𝛼)𝑇+1−𝑘−𝛽(𝑛−𝑘),𝑙∈𝐍_1,𝑘−1∩𝐍_1,𝑛−1,(𝑙+𝛼)(𝑇+1−𝑘)+𝛽(𝑛+𝛼)(𝑘−𝑙),𝑙∈𝐍_{𝑛,𝑘−1},(𝑘+𝛼)𝑇+1−𝑙−𝛽(𝑛−𝑙),𝑙∈𝐍_{𝑘,𝑛−1},(𝑘+𝛼)(𝑇+1−𝑙),𝑙∈𝐍_𝑘,𝑇∩𝐍_𝑛,𝑇, (2.1) is the Green's function of the problem −Δ²𝑦(𝑘−1)=0,𝑘∈𝐍_1,𝑇,𝑦(0)−𝛼Δ𝑦(0)=0,𝑦(𝑇+1)=𝛽𝑦(𝑛). (2.2)

Remark 2.2. We observe that the condition 𝐻>0 and 𝑇+1−𝛽𝑛>0 implies 𝐺(𝑘,𝑙) is positive on 𝐍_𝑇+1×𝐍_1,𝑇, which means that the finite set 𝐺(𝑘,𝑙)𝐺(𝑘,𝑘)∶𝑘∈𝐍_𝑇+1,𝑙∈𝐍_1,𝑇 (2.3) takes positive values. Then we let 𝑀₁𝐺()=min𝑘,𝑙𝐺(𝑘,𝑘)∶𝑘∈𝐍_𝑇+1,𝑙∈𝐍_1,𝑇, (2.4)𝑀₂𝐺()=max𝑘,𝑙𝐺(𝑘,𝑘)∶𝑘∈𝐍_𝑇+1,𝑙∈𝐍_1,𝑇. (2.5)

3. Main Results

Theorem 3.1. Assume that (A1)–(A3) hold, then problem (1.1) has at least one positive solution.

Proof. In the following, we denote 𝑚=min_{𝑘∈𝐍𝑛,𝑇}𝐺(𝑘,𝑘),𝑀=max_{𝑘∈𝐍𝑇+1}𝐺(𝑘,𝑘). (3.1) Then 0<𝑚<𝑀.
Let 𝐸 be the Banach space defined by 𝐸={𝑦∶𝐍_𝑇+1→𝑅}. Define𝐾=𝑦∈𝐸∶𝑦(𝑘)≥0,for𝑘∈𝐍_𝑇+1andmin_{𝑘∈𝐍𝑛,𝑇}𝑦(𝑘)≥𝜎𝑦, (3.2) where 𝜎=𝑀₁𝑚/𝑀₂𝑀∈(0,1), ‖𝑦‖=max_{𝑘∈𝐍𝑇+1}|𝑦(𝑘)|. It is clear that 𝐾 is a cone in 𝐸.
We define the operator 𝑆∶𝐾→𝐸 by 𝑆𝑦(𝑘)=^𝑇_𝑙=1𝐺(𝑘,𝑙)ℎ(𝑙)𝑓𝑦(𝑙),𝑘∈𝐍_𝑇+1. (3.3) It is clear that problem (1.1) has a solution 𝑦 if and only if 𝑦∈𝐸 is a solution of the operator equation 𝑦(𝑘)=(𝑆𝑦)(𝑘). We shall now show that the operator 𝑆 maps 𝐾 into itself. For this, let 𝑦∈𝐾; from (A2),(A3), we find 𝑆𝑦(𝑘)=^𝑇_𝑙=1𝐺(𝑘,𝑙)ℎ(𝑙)𝑓𝑦(𝑙)≥0,for𝑘∈𝐍_𝑇+1. (3.4) From (2.5), we obtain 𝑆𝑦(𝑘)=^𝑇_𝑙=1𝐺(𝑘,𝑙)ℎ(𝑙)𝑓𝑦(𝑙)≤𝑀₂^𝑇_𝑙=1𝐺(𝑘,𝑘)ℎ(𝑙)𝑓𝑦(𝑙)≤𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝑓𝑦(𝑙),for𝑘∈𝐍_𝑇+1. (3.5) Therefore 𝑆𝑦≤𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝑓𝑦(𝑙). (3.6) Now from (A2), (A3), (2.4), and (3.6), for 𝑘∈𝐍_𝑛,𝑇, we have 𝑆𝑦(𝑘)≥𝑀₁^𝑇_𝑙=1G(𝑘,𝑘)ℎ(𝑙)𝑓𝑦(𝑙)≥𝑀₁𝑚^𝑇_𝑙=1ℎ(𝑙)𝑓𝑦(𝑙≥𝑀₁)𝑚𝑀₂𝑀𝑆𝑦=𝜎𝑦. (3.7) Then min_{𝑘∈𝐍𝑛,𝑇}𝑆𝑦(𝑘)≥𝜎𝑆𝑦. (3.8) From (3.4) and (3.6), we obtain 𝑆𝑦∈𝐾. Hence 𝑆(𝐾)⊆𝐾. Also standard arguments yield that 𝑆∶𝐾→𝐾 is completely continuous.

Case 1. Suppose 𝑓 is superlinear. Now since 𝑓₀=0, we may choose 𝐶₁>0 such that 𝑓(𝑢)≤𝛿₁𝑢, for 0<𝑢≤𝐶₁, where 𝛿₁ satisfies 𝛿₁𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)≤1. (3.9) Let 𝑦∈𝐾 be such that ‖𝑦‖=𝐶₁; by using (2.5) and (3.9), we have 𝑆𝑦(𝑘)≤𝑀₂^𝑇_𝑙=1𝐺(𝑘,𝑘)ℎ(𝑙)𝑓𝑦(𝑙)≤𝛿₁𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝑦(𝑙)≤𝛿₁𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝑦≤𝑦. (3.10) Now if we let Ω₁=𝑦∈𝐸∶𝑦<𝐶₁, (3.11) then 𝑆𝑦≤𝑦,for𝑦∈𝐾∩𝜕Ω₁. (3.12) Next since 𝑓_∞=∞, there exists 𝐶₂>0, such that 𝑓(𝑢)≥𝛿₂𝑢, for 𝑢≥𝐶₂, where 𝛿₂>0 satisfying 𝛿₂𝑀₁𝜎^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)≥1. (3.13) Let 𝐶₂=max{2𝐶₁,𝐶₂/𝜎} and Ω₂={𝑦∈𝐸∶‖𝑦‖<𝐶₂}, and let 𝑦∈𝐾 and ‖𝑦‖=𝐶₂, then min_{𝑘∈𝐍𝑛,𝑇}𝑦(𝑘)≥𝜎𝑦≥𝐶₂. (3.14) Applying (2.4) and (3.13), one has 𝑆𝑦(𝑛)=𝑀₁^𝑇_𝑙=1𝐺(𝑛,𝑙)ℎ(𝑙)𝑓𝑦(𝑙)≥𝑀₁^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)𝑓𝑦(𝑙)≥𝛿₂𝑀₁^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)𝑦(𝑙)≥𝛿₂𝑀₁𝜎^𝑇_𝑙=𝑛𝐺()ℎ(𝑙𝑦≥𝑦.𝑛,𝑛) (3.15) Thus 𝑆𝑦≥𝑦,𝑦∈𝐾∩𝜕Ω₂. (3.16) In view of (3.12) and (3.16), it follows from Theorem A that 𝑆 has a fixed point 𝑦∈𝐾⋂(Ω₂⧵Ω₁) such that 𝐶₁≤‖𝑦‖≤𝐶₂.

Case 2. Suppose 𝑓 is sublinear case. Since 𝑓₀=∞, we may choose 𝐶₃>0 such that 𝑓(𝑢)≥𝛿₃𝑢 for 0<𝑢≤𝐶₃, where 𝛿₃>0 satisfying 𝛿₃𝑀₁𝜎^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)≥1, (3.17)Ω₃={𝑦∈𝐸∶‖𝑦‖<𝐶₃}; let 𝑦∈𝐾 and ‖𝑦‖=𝐶₃. Using (2.4) and (3.17), one has 𝑆𝑦(𝑛)≥𝑀₁^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)𝑓𝑦(𝑙)≥𝛿₃𝑀₁^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)y(𝑙)≥𝛿₃𝑀₁𝜎^𝑇_𝑙=𝑛𝐺(𝑛,𝑛)ℎ(𝑙)𝑦≥𝑦. (3.18) Then ‖𝑆𝑦‖≥‖𝑦‖, 𝑦∈𝐾⋂𝜕Ω₃.
In view of 𝑓_∞=0, there exists 𝐶₄>0 such that 𝑓(𝑢)≤𝛿₄𝑢 for 𝑢≥𝐶₄, where 𝛿₄>0 satisfying 𝛿₄𝑀₂𝑀^𝑇_𝑙=𝑛ℎ(𝑙)≤1. (3.19) There are two subcases to consider, that is, 𝑓 is bounded and 𝑓 is unbounded.
Subcase 2.1. Suppose 𝑓 is bounded, that is, 𝑓(𝑦)≤𝐿 for all 𝑦∈[0,∞) for some 𝐿>0. Let 𝐶₄=max2𝐶₃,𝐿𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙). (3.20) Then, for 𝑦∈𝐾 and ‖𝑦‖=𝐶₄, one has 𝑆𝑦(𝑘)≤𝑀₂^𝑇_𝑙=1𝐺(𝑘,𝑘)ℎ(𝑙)𝑓𝑦(𝑙)≤𝐿𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)≤𝐶₄=𝑦. (3.21) Hence ‖𝑆𝑦‖≤‖𝑦‖.
Subcase 2.2. Suppose 𝑓 is unbounded, that is, there exists 𝐶₄>max{2𝐶₃,𝐶₄/𝜎} such that 𝑓(𝑢)≤𝑓(𝐶₄) for all 0<𝑢≤𝐶₄. Then, for 𝑦∈𝐾 with ‖𝑦‖=𝐶₄, from (2.5) and (3.19), we have 𝑆𝑦(𝑘)≤𝑀₂^𝑇_𝑙=1𝐺(𝑘,𝑘)ℎ(𝑙)𝑓𝑦(𝑙)≤𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝑓𝐶₄≤𝛿₄𝑀₂𝑀^𝑇_𝑙=1ℎ(𝑙)𝐶₄≤𝐶₄=𝑦. (3.22) Thus in both Subcases 2.1 and 2.2, we may put Ω₄={𝑦∈𝐸∶‖𝑦‖<𝐶₄}. Then 𝑆𝑦≤𝑦,𝑦∈𝐾∩𝜕Ω₄. (3.23) By using the fixed point Theorem A, it follows that problem (1.1) has at least one positive solution, such that 𝐶₃≤‖𝑦‖≤𝐶₄. The proof is finished.
Finally, we give an example to demonstrate our main result.

Example 3.2. Consider the following three-point boundary value problem: Δ²𝑦()+2𝑘−1−𝑘²+10𝑘+33^1.5𝑦+20^1.5=0,𝑘∈𝐍_1,8,𝑦(0)−139Δ𝑦(0)=0,𝑦(9)=22𝑦(4),37 (3.24) where 𝑇=8, 𝑛=4, 𝛼=13/9, 𝛽=22/37, 𝑇+1−𝛽𝑛+𝛼(1−𝛽)=800/111>0, 𝑇+1−𝛽𝑛=245/37>0, ℎ(𝑘)=2/(−𝑘²+10𝑘+33)^1.5, 𝑘∈𝐍_1,8, 𝑓(𝑦) = (𝑦+20)^1.5, then 𝑓 is superlinear. Conditions of Theorem 3.1 are all satisfied. Then problem (3.24) has at least one positive solution 𝑦. Indeed 𝑦=−𝑘²+10𝑘+13 is one such positive solution.

Acknowledgments

This project is supported by the National Natural Science Foundation of China (11101349, 11071205), the NSF of Jiangsu Province (BK2008119gundong), and the NSF of the Education Department of Jiangsu Province (11KJB110013).