Solvability of a Class of Generalized Neumann Boundary Value Problems for Second-Order Nonlinear Difference Equations

Abstract

This paper is motivated by Rachnkovab and Tisdell (2006) and Anderson et al. (2007). New sufficient conditions for the existence of at least one solution of the generalized Neumann boundary value problems for second order nonlinear difference equations ∇Δ𝑥(𝑘)=𝑓(𝑘,𝑥(𝑘),𝑥(𝑘+1)), 𝑘∈[1,𝑛−1], 𝑥(0)=𝑎𝑥(1), 𝑥(𝑛)=𝑏𝑥(𝑛−1), are established.

1. Introduction

Recently, there have been many papers discussed the solvability of two-point or multipoint boundary value problems for second-order or higher-order difference equations, we refer the readers to the text books [1, 2] and papers [3–8] and the references therein.

In a recent paper [3], Anderson et al. studied the following problem: (𝑘)(𝑘)(𝑘)(0)(𝑛)∇Δ𝑦=𝑓𝑘,𝑦,Δ𝑦,𝑘=1,…,𝑛−1,Δ𝑦=Δ𝑦=0, (1.1) where (𝑘)=𝑦()(𝑘)(𝑘)=𝑦()(𝑘)()Δ𝑦𝑘+1−𝑦,for𝑘=0,…,𝑛−1,0,for𝑘=𝑛,∇Δ𝑦𝑘+1−2𝑦+𝑦𝑘−1,for𝑘=1,…,𝑛−1,0,for𝑘=0or𝑛. (1.2) The following result was proved.

Theorem ART
Suppose that 𝑓 is continuous and there exist constants 𝛼≤0, 𝐾≥0 such that 𝑓𝑡,𝑝,𝑞−𝑝≤𝛼2𝑝𝑓𝑡,𝑝,𝑞+𝑞²+𝐾,𝑡,𝑝,𝑞∈{1,…,𝑛−1}×𝑅². (∗) Then BVP(1.1) has at least one solution.

The methods in [3] involved new inequalities on the right-hand side of the difference equation and Schaefer's Theorem in the finite-dimensional space setting.

In [7], the following discrete boundary value problem (BVP) involving second order difference equations and two-point boundary conditions ∇Δ𝑦_𝑘ℎ²=𝑓𝑡_𝑘,𝑦_𝑘,Δ𝑦_𝑘ℎ𝑦,𝑘=1,…,𝑛−1,₀=0,𝑦_𝑛=0, (1.3) was studied, where 𝑛≥2 an integer, 𝑓 is continuous, scalar-valued function, the step size is ℎ=𝑁/𝑛 with 𝑁 a positive constant, the grid points are 𝑡_𝑘=𝑘ℎ for 𝑘=0,…,𝑛. The differences are given by Δ𝑦_𝑘=𝑦_𝑘+1−𝑦_𝑘,𝑘=0,…,𝑛−1,0,𝑘=𝑛,∇Δ𝑦_𝑘=𝑦_𝑘+1−2𝑦_𝑘+𝑦_𝑘−1,𝑘=1,…,𝑛−1,0,𝑘=0or𝑘=𝑛. (1.4)

The following two results were proved in [7].

Theorem RT
Let 𝑓 be continuous on [0,𝑁]×𝑅² and 𝛼,𝛽, and, 𝐾 be nonnegative constants. If there exist 𝑐,𝑑∈[0,1) such that 𝑓(𝑡,𝑢,𝑣)≤𝛼|𝑢|^𝑐+𝛽|𝑣|^𝑑+𝐾,(𝑡,𝑢,𝑣)∈[0,𝑁]×𝑅², (1.5) then the discrete BVP(1.3) has at least one solution.

Theorem RT
Let 𝑓 be continuous on [0,𝑁]×𝑅² and 𝛼,𝛽, and 𝐾 nonnegative constants. If 𝑓(𝑡,𝑢,𝑣)≤𝛼|𝑢|+𝛽|𝑣|+𝐾,(𝑡,𝑢,𝑣)∈[0,𝑁]×𝑅²,𝛼𝑁²8+𝛽𝑁2<1, (1.6)(1.7) then the discrete BVP(1.3) has at least one solution.

In paper [8], Cabada and Otero-Espinar studied the existence of solutions of a class of nonlinear second-order difference problem with Neumann boundary conditions by using upper and lower solution methods. Assuming the existence of a pair of ordered lower and upper solutions 𝛾 and 𝛽, they obtained optimal existence results for the case 𝛾≤𝛽 and even for 𝛾≥𝛽.

In this paper, we study the following boundary value problem for second-order nonlinear difference equation (𝑘)((𝑘)([],𝑥(0)(1)(𝑛)(),∇Δ𝑥=𝑓𝑘,𝑥,𝑥𝑘+1)),𝑘∈1,𝑛−1=𝑎𝑥,𝑥=𝑏𝑥𝑛−1 (1.8) where 𝑎,𝑏∈𝑅, 𝑛≥2is an integer, and 𝑓 is continuous, scalar-valued function. We note that when 𝑎=𝑏=1, BVP(1.8) becomes the following BVP: (𝑘)((𝑘)([],(0)(),∇Δ𝑥=𝑓𝑘,𝑥,𝑥𝑘+1)),𝑡∈1,𝑇−1Δ𝑥=0=Δ𝑥𝑛−1 (1.9) which is called Neumann boundary value problem of difference equation and is a special case of BVP(1.1). When 𝑎=𝑏=0, BVP(1.8) is changed to (𝑘)((𝑘)([],𝑥(0)(𝑛),∇Δ𝑥=𝑓𝑘,𝑥,𝑥𝑘+1)),𝑡∈1,𝑇−1=0=𝑥 (1.10) which is the so-called Dirichlet problem for discrete difference equations and is a special case of BVP(1.3).

The purpose of this paper is to improve the assumptions (*), (1.5), and (1.6) in the results in paper [3, 5, 7–9], by using Mawhin′s continuation theorem of coincidence degree, to establish sufficient conditions for the existence of at least one solution of BVP(1.8). It is interesting that we allow 𝑓 to be sublinear, at most linear or superlinear.

This paper is organized as follows. In Section 2, we make the main results, and in Section 3, we give some examples, which cannot be solved by theorems in [5, 7, 9], to illustrate the main results presented in Section 3.

2. Main Results

To get the existence results for solutions of BVP(1.8), we need the following fixed point theorems.

Let 𝑋 and 𝑌 be Banach spaces, 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 a Fredholm operator of index zero, and 𝑃∶𝑋→𝑋, 𝑄∶𝑌→𝑌 projectors such that Im𝑃=Ker𝐿,Ker𝑄=Im𝐿,𝑋=Ker𝐿⊕Ker𝑃,𝑌=Im𝐿⊕Im𝑄. It follows that 𝐿|_{𝐷(𝐿)∩Ker𝑃}∶𝐷(𝐿)∩Ker𝑃→Im𝐿 is invertible; we denote the inverse of that map by 𝐾_𝑝.

If Ω is an open bounded subset of 𝑋, 𝐷(𝐿)∩Ω≠∅, the map 𝑁∶𝑋→𝑌 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾_𝑝(𝐼−𝑄)𝑁∶Ω→𝑋 is compact.

Lemma 2.1 . (see [9]). Let 𝐿 be a Fredholm operator of index zero, and let 𝑁 be 𝐿-compact on Ω. Assume that the following conditions are satisfied:
(i)𝐿𝑥≠𝜆𝑁𝑥 for every (𝑥,𝜆)∈[(𝐷(𝐿)⧵Ker𝐿)∩𝜕Ω]×(0,1);
(ii)𝑁𝑥∉Im𝐿 for every 𝑥∈Ker𝐿∩𝜕Ω;
(iii)deg(∧𝑄𝑁|_Ker𝐿,Ω∩Ker𝐿,0)≠0, where ∧∶Ker𝐿→𝑌/Im𝐿 is the isomorphism.
Then the equation 𝐿𝑥=𝑁𝑥 has at least one solution in 𝐷(𝐿)∩Ω.

Lemma 2.2 . (see [9]). Let 𝑋 and 𝑌 be Banach spaces. Suppose 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 is a Fredholm operator of index zero with Ker𝐿={0}, 𝑁∶𝑋→𝑌 is 𝐿-compact on any open bounded subset of 𝑋. If 0∈Ω⊂𝑋 is an open bounded subset and 𝐿𝑥≠𝜆𝑁𝑥forall𝑥∈𝐷(𝐿)∩𝜕Ωand𝜆∈[0,1], then there is at least one 𝑥∈Ω so that 𝐿𝑥=𝑁𝑥.
Let 𝑋=𝑅^𝑛+1,𝑌=𝑅^𝑛−1 be endowed with the norms ‖𝑥‖=max_{[]𝑛∈0,𝑛}|𝑥(𝑛)|,𝑦=max_{[]𝑘∈1,𝑛−1}𝑦(𝑘) (2.1) for 𝑥∈𝑋 and 𝑦∈𝑌, respectively. It is easy to see that 𝑋 and 𝑌 are Banach spaces. Choose 𝐷(𝐿)={𝑥∈𝑋∶𝑥(0)=𝑎𝑥(1),𝑥(𝑛)=𝑏𝑥(𝑛−1)}. Let 𝐿∶𝑋→𝑌,𝐿𝑥(𝑘)=∇Δ𝑥(𝑘),𝑥∈𝐷(𝐿), and 𝑁∶𝑋→𝑌 by 𝑁𝑥(𝑘)=𝑓(𝑘,𝑥(𝑘),𝑥(𝑘+1)).

Consider the following problem: ∇Δ𝑥(𝑘)=0,𝑥(0)=𝑎𝑥(1),𝑥(𝑛)=𝑏𝑥(𝑛−1). (2.2) It is easy to see that problem (2.2) has a unique solution 𝑥(𝑘)=0 if and only if (1−𝑎)[(𝑛−1)𝑏−𝑛]≠𝑎(1−𝑏). (2.3) If (2.3) holds, we call BVP(1.8) at nonresonance case. If (1−𝑎)[(𝑛−1)𝑏−𝑛]=𝑎(1−𝑏), then problem (2.2) has infinite nontrivial solutions. At this case, we call BVP(1.8) at resonance case. In this paper, we establish sufficient conditions for the existence of solutions of BVP(1.8) at nonresonance case, that is, (1−𝑎)[(𝑛−1)𝑏−𝑛]≠𝑎(1−𝑏), and at resonance case, 𝑎=𝑏=1. It is similar to get existence results for the existence of solutions at resonance case when (1−𝑎)[(𝑛−1)𝑏−𝑛]=𝑎(1−𝑏) and 𝑎≠1,𝑏≠1.

Lemma 2.3. Suppose 𝑎=𝑏=1. Then the following results are valid.
(i)Ker𝐿={𝑥=(𝑐,…,𝑐)∈𝑋∶𝑐∈𝑅}.
(ii)Im𝐿={𝑦∈𝑌∶∑^𝑛−1_𝑖=1𝑦(𝑖)=0}.
(iii)𝐿 is a Fredholm operator of index zero.
(iv) There are projectors 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌 such that Ker𝐿=Im𝑃,Ker𝑄=Im𝐿. Furthermore, let Ω⊂𝑋 be an open bounded subset with Ω∩𝐷(𝐿)≠∅; then 𝑁 is 𝐿-compact on Ω.
(v)𝑥∈𝐷(𝐿) is a solution of 𝐿(𝑥)=𝑁(𝑥) which implies that 𝑥 is a solution of BVP(1.8).

The projectors 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌, the isomorphism ∧∶Ker𝐿→𝑌/Im𝐿, and the generalized inverse 𝐾_𝑝∶Im𝐿→𝐷(𝐿)∩Im𝑃 are as follows: (𝑛)(1),(𝑛)=1𝑃𝑥=𝑥𝑄𝑦𝑛−1^𝑛−1_𝑖=1𝑦(𝑖),∧(𝑐)𝐾=𝑐,_𝑝𝑦(𝑛)=^𝑘𝑠_𝑠=1𝑖=1𝑦(𝑖). (2.4)

Lemma 2.4. Suppose (1−𝑎)[(𝑛−1)𝑏−𝑛]≠𝑎(1−𝑏). Then the following results are valid.
(i)𝑥∈𝐷(𝐿) is a solution of 𝐿(𝑥)=𝑁(𝑥) which implies that 𝑥 is a solution of BVP(1.8).
(ii) Ker𝐿={0}.
(iii) 𝐿 is a Fredholm operator of index zero, 𝑁 is 𝐿-compact on each open bounded subset of 𝑋.

Suppose(A)there exist numbers 𝛽>0, 𝜃≥1, nonnegative sequences 𝑝(𝑛),𝑞(𝑛),𝑟(𝑛), functions 𝑔(𝑛,𝑥,𝑦), ℎ(𝑛,𝑥,𝑦) such that 𝑓(𝑛,𝑥,𝑦)=𝑔(𝑛,𝑥,𝑦)+ℎ(𝑛,𝑥,𝑦) and 𝑔𝑛,𝑥,𝑦𝑥≥𝛽|𝑥|^𝜃+1,ℎ𝑛,𝑥,𝑦≤𝑝(𝑛)|𝑥|^𝜃+𝑞(𝑛)𝑦^𝜃+𝑟(𝑛), (2.5) for all 𝑛∈{1,…,𝑛−1},(𝑥,𝑦)∈𝑅²;(B)there exists a constant 𝑀>0 so that 𝑐^𝑛−1_𝑖=1𝑓(𝑛,𝑐,𝑐)>0 (2.6) for all |𝑐|>𝑀 or 𝑐^𝑛−1_𝑖=1𝑓(𝑛,𝑐,𝑐)<0 (2.7) for all |𝑐|>𝑀.

Theorem L
Suppose 𝑎²≤1,𝑏²≤1, and that (𝐴) and (𝐵) hold. Then BVP(1.8) has at least one solution if 𝑝+𝑞max|𝑏|^𝜃+1,1<𝛽. (2.8)

Proof. To apply Lemma 2.1, we consider 𝐿𝑥=𝜆𝑁𝑥 for 𝜆∈[0,1].
Step 1. Let Ω₁={𝑥∈𝑋∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈[0,1]}. For 𝑥∈Ω₁, we have 𝑥()(𝑘)()((𝑘)([],𝑥(0)(1),𝑥(𝑛)().𝑘+1−2𝑥+𝑥𝑘−1=𝜆𝑓𝑘,𝑥,𝑥𝑘+1)),𝑘∈1,𝑛−1=𝑎𝑥=𝑏𝑥𝑛−1 (2.9) So [𝑥(𝑘+1)−2𝑥(𝑘)+𝑥(𝑘−1)]𝑥(𝑘)=𝜆𝑓(𝑘,𝑥(𝑘),𝑥(𝑘+1))𝑥(𝑘),𝑘∈[1,𝑛−1]. (2.10) It is easy to see that 2^𝑛−1_𝑛=1[𝑥()(𝑘)(𝑥(𝑘)=𝑘+1−2𝑥+𝑥𝑘−1)]_{𝑛−1𝑛=1}−[𝑥(𝑘+1)]²+2𝑥(𝑘)𝑥(𝑘+1)−[𝑥(𝑘)]²−[𝑥(𝑘−1)]²+2𝑥(𝑘−1)𝑥(𝑘)−[𝑥(𝑘)]²+[𝑥(𝑘+1)]²−2[𝑥(𝑘)]²+[𝑥(𝑘−1)]²=^𝑛−1_𝑛=1−[𝑥(𝑘+1)−𝑥(𝑘)]²−[𝑥(𝑘−1)−𝑥(𝑘)]²+[𝑥(𝑘+1)]²−2[𝑥(𝑘)]²+[𝑥(𝑘−1)]²=^𝑛−1_𝑛=1−[𝑥(𝑘+1)−𝑥(𝑘)]²−[𝑥(𝑘−1)−𝑥(𝑘)]²+[𝑥(𝑛)]²−[𝑥(𝑛−1)]²−[𝑥(1)]²+[𝑥(0)]²=^𝑛−1_𝑛=1−[𝑥(𝑘+1)−𝑥(𝑘)]²−[𝑥(𝑘−1)−𝑥(𝑘)]²+𝑏²−1[𝑥(𝑛−1)]²+𝑎²−1[𝑥(1)]². (2.11) Since 𝑎²≤1,𝑏²≤1, we get ^𝑛−1_𝑛=1[𝑥(𝑘+1)−2𝑥(𝑘)+𝑥(𝑘−1)]𝑥(𝑘)≤0. (2.12) So, we get 𝜆^𝑛−1_𝑛=1𝑓(𝑘,𝑥(𝑘),𝑥(𝑘+1))𝑥(𝑘)≤0. (2.13) Then ^𝑛−1_𝑛=1𝑔(𝑘,𝑥(𝑘),𝑥(𝑘+1))+ℎ(𝑘,𝑥(𝑘),𝑥(𝑘+1))𝑥(𝑘)≤0. (2.14) It follows that 𝛽^𝑛−1_𝑘=1|𝑥(𝑘)|^𝜃+1≤−^𝑛−1_𝑛=1ℎ((𝑘)(𝑥(𝑘)≤𝑘,𝑥,𝑥𝑘+1))_{𝑛−1𝑛=1}𝑝(𝑘)|𝑥(𝑘)|^𝜃+1+𝑞(𝑘)|𝑥(𝑘+1)|^𝜃|𝑥(𝑘(𝑘𝑥(𝑘≤𝑝)|+𝑟)|)|_{𝑛−1𝑛=1}|𝑥(𝑘)|^𝜃+1+𝑞^𝑛−1_𝑘=1|𝑥(𝑘+1)|^𝜃|𝑥(𝑘)|+^𝑛−1_𝑘=1𝑟(𝑘)|𝑥(𝑘)|. (2.15) For 𝑥_𝑖≥0,𝑦_𝑖≥0, Holder′s inequality implies ^𝑠_𝑖=1𝑥_𝑖𝑦_𝑖≤^𝑠_𝑖=1𝑥^𝑝_𝑖^1/𝑝𝑠_𝑖=1𝑦^𝑞_𝑖^1/𝑞,1𝑝+1𝑞=1,𝑞>0,𝑝>0. (2.16) It follows that 𝛽^𝑛−1_𝑘=1|𝑥(𝑘)|^𝜃+1≤𝑝^𝑛−1_𝑛=1|𝑥(𝑘)|^𝜃+1+𝑞^𝑛−1_𝑘=1|𝑥(𝑘+1)|^{𝜃+1𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}+‖𝑟‖^𝑛−1_𝑘=1|𝑥(𝑘=‖𝑟))|‖(𝑛−1^{𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}+𝑝^𝑛−1_𝑛=1|𝑥(𝑘)|^𝜃+1+𝑞|𝑏|^𝜃+1|𝑥(1)|^𝜃+1+^𝑛−2_𝑘=1|𝑥(𝑘+1)|^{𝜃+1𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}≤‖𝑟‖(𝑛−1)^{𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}+𝑝^𝑛−1_𝑛=1|𝑥(𝑘)|^𝜃+1+𝑞max|𝑏|^𝜃+1,1^𝑛−1_𝑘=1|𝑥(𝑘)|^{𝜃+1𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}=‖𝑟‖(𝑛−1)^{𝜃/(𝜃+1)𝑛−1}_𝑘=1|𝑥(𝑘)|^{𝜃+11/(𝜃+1)}+𝑝^𝑛−1_𝑛=1|𝑥(𝑘)|^𝜃+1+𝑞max|𝑏|^𝜃+1,1^𝑛−1_𝑘=1|𝑥(𝑘)|^𝜃+1. (2.17) It follows from (2.8) that there exists a constant 𝑀₁>0 such that ^𝑛−1_𝑘=1|𝑥(𝑘)|^𝜃+1≤𝑀₁. (2.18) Hence |𝑥(𝑘)|≤(𝑀₁/(𝑛−1))^1/(𝜃+1) for all 𝑘∈{1,…,𝑛−1}. Hence ||𝑥||≤(𝑀₁/(𝑛−1))^1/(𝜃+1). So Ω₁ is bounded.
Step 2. Prove that the set Ω₂={𝑥∈Ker𝐿∶𝑁(𝑥)∈Im𝐿} is bounded.
For 𝑥∈Ker𝐿, we have 𝑥(𝑘)=𝑐 for 𝑘∈[0,𝑛]. Thus we have 𝑁𝑥(𝑘)=𝑓(𝑘,𝑐,𝑐). 𝑁(𝑥,𝑦)∈Im𝐿 implies that ^𝑛−1_𝑘=1𝑓(𝑛,𝑐,𝑐)=0. (2.19) It follows from condition (𝐵) that |𝑐|≤𝑀. Thus Ω₂ is bounded.
Step 3. Prove the set Ω₃={𝑥∈Ker𝐿∶±𝜆∧(𝑥)+(1−𝜆)𝑄𝑁(𝑥)=0,∃𝜆∈[0,1]} is bounded.
If the first inequality of (𝐵) holds, let Ω₃={𝑥∈Ker𝐿∶𝜆∧(𝑥)+(1−𝜆)𝑄𝑁(𝑥)=0,∃𝜆∈[0,1]}. (2.20)
We will prove that Ω₃ is bounded. For 𝑥(𝑘)=𝑐 for 𝑘∈[0,𝑛] such that 𝑥∈Ω₃, and 𝜆∈[0,1], we have −(1−𝜆)^𝑛−1_𝑘=1𝑓(𝑛,𝑐,𝑐)=𝜆𝑐(𝑛−1). (2.21)
If 𝜆=1, then 𝑐=0. If 𝜆≠1, then 0>−(1−𝜆)𝑐^𝑛−1_𝑘=1𝑓(𝑛,𝑐,𝑐)=𝜆𝑐²𝑇≥0, (2.22) a contradiction.
If the second inequality of (𝐵) holds, let Ω₃={𝑥∈Ker𝐿∶−𝜆∧(𝑥)+(1−𝜆)𝑄𝑁(𝑥)=0,∃𝜆∈[0,1]}. (2.23) Similarly, we can get a contradiction. So Ω₃ is bounded.
Step 4. Obtain open bounded set Ω such (i), (ii), and (iii) of Lemma 2.1.
In the following, we will show that all conditions of Lemma 2.1 are satisfied. Set Ω an open bounded subset of 𝑋 such that Ω⊃⋃³_𝑖=1Ω_𝑖. We know that 𝐿 is a Fredholm operator of index zero and 𝑁 is 𝐿-compact on Ω. By the definition of Ω, we have Ω⊃Ω₁ and Ω⊃Ω₂, thus 𝐿(𝑥)≠𝜆𝑁(𝑥) for 𝑥∈(𝐷(𝐿)⧵Ker𝐿)∩𝜕Ω and 𝜆∈(0,1); 𝑁(𝑥)∉Im𝐿 for 𝑥∈Ker𝐿∩𝜕Ω.
In fact, let 𝐻(𝑥,𝜆)=±𝜆∧(𝑥)+(1−𝜆)𝑄𝑁(𝑥). According the definition of Ω, we know Ω⊃Ω₃, thus 𝐻(𝑥,𝜆)≠0 for 𝑥∈𝜕Ω∩Ker𝐿, thus by homotopy property of degree, deg𝑄𝑁|_Ker𝐿(𝐻())(𝐻())(),Ω∩Ker𝐿,0=deg⋅,0,Ω∩Ker𝐿,0=deg⋅,1,Ω∩Ker𝐿,0=deg±∧,Ω∩Ker𝐿,0≠0. (2.24) Thus by Lemma 2.1, 𝐿(𝑥)=𝑁(𝑥) has at least one solution in 𝐷(𝐿)∩Ω, which is a solution of BVP(1.8). The proof is completed.

Theorem L
Suppose 𝑎²≤1,𝑏²≤1, (1−𝑎)[(𝑛−1)𝑏−𝑛]≠𝑎(1−𝑏), and that (𝐴) holds. Then BVP(1.8) has at least one solution if (2.8) holds.

Proof. To apply Lemma 2.2, we consider 𝐿𝑥=𝜆𝑁𝑥 for 𝜆∈[0,1]. Let Ω₁={𝑥∈𝑋∶𝐿𝑥=𝜆𝑁𝑥,𝜆∈[0,1]}. For 𝑥∈Ω₁, we get (2.9) and (2.10). using the methods in the proof of Theorem LX1, we get that Ω₁ is bounded. Let Ω be a nonempty open bounded subset of 𝑋 such that Ω⊃Ω₁ centered at zero. It is easy to see that 𝐿 is a Fredholm operator of index zero and 𝑁 is L-compact on Ω. One can see that 𝐿𝑥≠𝜆𝑁𝑥forall𝑥∈𝐷(𝐿)∩𝜕Ωand𝜆∈[0,1]. Thus, from Lemma 2.2, 𝐿𝑥=𝑁𝑥 has at least one solution 𝑥∈𝐷(𝐿)∩Ω, so 𝑥 is a solution of BVP(1.8). The proof is complete.

3. An Example

In this section, we present an example to illustrate the main results in Section 2.

Example 3.1. Consider the following problem: 𝑥(𝑘+1)−2𝑥(𝑘)+𝑥(𝑘−1)=𝛽[𝑥(𝑘)]^2𝑚+1+𝑝(𝑘)[𝑥(𝑘)]^2𝑚+1+𝑞(𝑘)[𝑥(𝑘+1)]^2𝑚+1(𝑘),[],𝑥(0)(1),𝑥(𝑛)(),+𝑟𝑘∈1,𝑛−1=𝑎𝑥=𝑏𝑥𝑛−1 (3.1) where 𝑛≥2,𝑚≥1 are integers and 𝛽>0, 𝑝(𝑛), 𝑞(𝑛), 𝑟(𝑛) are sequences. Corresponding to the assumptions of Theorem L1, we set 𝑓𝑘,𝑥,𝑦=𝛽𝑥^2𝑚+1+𝑝(𝑘)𝑥^2𝑚+1+𝑞(𝑘)𝑦^2𝑚+1(𝑘),𝑔+𝑟𝑘,𝑥,𝑦=𝛽𝑥^2𝑚+1,ℎ𝑘,𝑥,𝑦=𝑝(𝑘)𝑥^2𝑚+1+𝑞(𝑘)𝑦^2𝑚+1+𝑟(𝑘), (3.2) and 𝜃=2𝑚+1. It is easy to see that (𝐴) holds, and 𝑓(𝑛,𝑐,𝑐)=𝑐^2𝑚+1𝛽+𝑝(𝑘)𝑐^2𝑚+1+𝑞(𝑘)𝑐^2𝑚+1+𝑟(𝑘) (3.3) implies that there is 𝑀>0 such that 𝑐∑^𝑛−1_𝑖=1[𝑐^2𝑚+1𝛽+𝑝(𝑘)𝑐^2𝑚+1+𝑞(𝑘)𝑐^2𝑚+1+𝑟(𝑘)]>0 for all 𝑛∈[1,𝑛−1] and |𝑐|>𝑀.

It follows from Theorem L2 that (3.1) has at least one solution if 𝑎²≤1,𝑏²≤1,(1−𝑎)[(𝑛−1)𝑏−𝑛]≠𝑎(1−𝑏) and ||𝑝||+||𝑞||max{|𝑏|^𝜃+1,1}<𝛽. BVP(3.1) has at least one solution if 𝑎=𝑏=1 and ||𝑝||+||𝑞||max{|𝑏|^𝜃+1,1}<𝛽.

Remark 3.2. It is easy to see that BVP(3.1) when 𝑎=𝑏=0 cannot be solved by using theorems obtained in paper [7]. BVP(3.1) when 𝑎=𝑏=1 cannot be solved by the results obtained in paper [3].

Acknowledgments

This paper is supported by Natural Science Foundation of Hunan province, China (no. 06JJ5008) and Natural Science Foundation of Guangdong province (no. 7004569).