Sign-Changing Solutions for Discrete Second-Order Three-Point Boundary Value Problems

Abstract

We consider the second-order three-point discrete boundary value problem. By using the topological degree theory and the fixed point index theory, we provide sufficient conditions for the existence of sign-changing solutions, positive solutions, and negative solutions. As an application, an example is given to demonstrate our main results.

1. Introduction

In this paper, we consider the following second-order three-point discrete boundary value problem (BVP):

Δ²𝑢(𝑡−1)+𝑓(𝑡,𝑢(𝑡))=0,𝑡∈[1,𝑛],𝑢(0)=0,𝑢(𝑛+1)=𝛼𝑢(𝑚), (1.1) where 𝑛∈{2,3,…}, [1,𝑛] is the discrete interval {1,2,…,𝑛}, 𝑚∈[1,𝑛], 0≤𝛼≤1, Δ𝑢(𝑡)=𝑢(𝑡+1)−𝑢(𝑡), Δ²𝑢(𝑡)=Δ(Δ𝑢(𝑡)), and 𝑓∶[1,𝑛]×𝐑→𝐑 is a continuous function.

Boundary value problems for difference equations arise in different areas of applied mathematics and physics. Existence and multiplicity of positive solutions or nontrivial solutions for boundary value problems of difference equations have been extensively studied in the literature; see [1–9] and the references therein.

On the other hand, in the existing literature, there are some papers studying the sign-changing solutions for boundary value problems of differential equations; for example, see [10–12]. But the problems of the existence of sign-changing solutions to discrete multipoint boundary value problems have received very little attention in the literature to the best knowledge of the authors. In this paper, motivated by [12, 13], we aim to study the existence of multiple sign-changing solutions to the second-order three-point discrete boundary value problem (1.1). Under some suitable conditions, we prove that the three-point discrete boundary value problem (1.1) has at least two sign-changing solutions, two positive solutions, and two negative solutions. The main approach is the topological degree theory and the fixed point index theory.

The organization of this paper is as follows. In Section 2, we present some preliminary knowledge about the topological degree theory and the fixed point index theory and use the knowledge to obtain some lemmas which are very crucial in our main results. In Section 3, by computing the topological degree and the fixed point index, we discuss the existence of multiple sign-changing solutions to BVP (1.1), and a simple example is given.

2. Preliminaries

As we have mentioned, we will use the theory of the Leray-Schauder degree and the fixed point index in a cone to prove our main existence results. Let us collect some results that will be used below. One can refer to [13–16] for more details.

Lemma 2.1 (see [13, 14]). Let 𝐸 be a Banach space and, 𝑋⊂𝐸 be a cone in 𝐸. Assume that Ω is a bounded open subset of 𝐸. Suppose that 𝐴∶𝑋∩Ω→𝑋 is a completely continuous operator. If there exists 𝑥₀∈𝑋⧵{𝜃} such that 𝑥−𝐴𝑥≠𝜇𝑥₀,∀𝑥∈𝑋∩𝜕Ω,𝜇≥0, (2.1) then the fixed point index 𝑖(𝐴,𝑋∩Ω,𝑋)=0.

Lemma 2.2 (see [13, 14]). Let 𝐸 be a Banach space and let 𝑋⊂𝐸 be a cone in 𝐸. Assume that Ω is a bounded open subset of 𝐸, 𝜃∈Ω. Suppose that 𝐴∶𝑋∩Ω→𝑋 is a completely continuous operator. If 𝐴𝑥≠𝜇𝑥,∀𝑥∈𝑋∩𝜕Ω,𝜇≥1. (2.2) then the fixed point index 𝑖(𝐴,𝑋∩Ω,𝑋)=1.

Lemma 2.3 (see [15]). Let 𝐸 be a Banach space, let Ω be a bounded open subset of 𝐸, 𝜃∈Ω, and 𝐴∶Ω→𝐸 be completely continuous. Suppose that ‖𝐴𝑥‖≤‖𝑥‖,𝐴𝑥≠𝑥,∀𝑥∈𝜕Ω, (2.3) then deg(𝐼−𝐴,Ω,𝜃)=1.

Lemma 2.4 (see [16]). Let 𝐴 be a completely continuous operator which is defined on a Banach space 𝐸. Let 𝑥₀∈𝐸 be a fixed point of 𝐴 and assume that 𝐴 is defined in a neighborhood of 𝑥₀ and Fréchet differentiable at 𝑥₀. If 1 is not an eigenvalue of the linear operator 𝐴^′(𝑥₀), then 𝑥₀ is an isolated singular point of the completely continuous vector field 𝐼−𝐴 and for small enough 𝑟>0, deg(𝐼−𝐴,𝐵(𝑥₀,𝑟),𝜃)=(−1)^𝑘, (2.4) where 𝑘 is the sum of the algebraic multiplicities of the real eigenvalues of 𝐴^′(𝑥₀) in (1,+∞).

Lemma 2.5 (see [16]). Let 𝐴 be a completely continuous operator which is defined on a Banach space 𝐸. Assume that 1 is not an eigenvalue of the asymptotic derivative. Then the completely continuous vector field 𝐼−𝐴 is nonsingular on spheres 𝑆_𝜌={𝑥∈𝐸∶‖𝑥‖=𝜌} of sufficiently large radius 𝜌 and deg(𝐼−𝐴,𝐵(𝜃,𝜌),𝜃)=(−1)^𝑘, (2.5) where 𝑘 is the sum of the algebraic multiplicities of the real eigenvalues of 𝐴^′(∞) in (1,+∞).

From [12, Lemma 2.4], we have the following lemma.

Lemma 2.6. Let 𝑋 be a solid cone of a Banach space 𝐸 (𝑋^∘ is nonempty), let Ω be a relatively bounded open subset of 𝑋, and let 𝐴∶𝑋→𝑋 be a completely continuous operator. If any fixed point of 𝐴 in Ω is an interior point of 𝑋, there exists an open subset 𝑂 of 𝐸 (𝑂⊂Ω) such that deg(𝐼−𝐴,𝑂,𝜃)=𝑖(𝐴,Ω,𝑋). (2.6)

Now we shall consider the space

𝐸={𝑢∶[0,𝑛+1]→𝐑∣𝑢(0)=0,𝑢(𝑛+1)=𝛼𝑢(𝑚)} (2.7)

equipped with the norm ‖𝑢‖=max_{𝑡∈[0,𝑛+1]}|𝑢(𝑡)|. Clearly 𝐸 is a 𝑛−dimensional Banach space. Choose the cone 𝑃⊂𝐸 defined by

𝑃={𝑢∈𝐸∣𝑢(𝑡)≥0,𝑡∈[1,𝑛]}. (2.8)

Obviously, the interior of 𝑃 is 𝑃^∘={𝑢∈𝐸∣𝑢(𝑡)>0,𝑡∈[1,𝑛]}. For each 𝑢,𝑣∈𝐸, we write 𝑢≥𝑣 if 𝑢(𝑡)≥𝑣(𝑡) for 𝑡∈[1,𝑛]. A solution 𝑢 of BVP (1.1) is said to be a positive solution (a negative solution, resp.) if 𝑢∈𝑃⧵{𝜃} (𝑢∈(−𝑃)⧵{𝜃}, resp.). A solution 𝑢 of BVP (1.1) is said to be a sign-changing solution if 𝑢∉𝑃∪(−𝑃).

Lemma 2.7. Let 𝑣∶[1,𝑛]→R be fixed. Then the problem Δ²𝑢(𝑡−1)+𝑣(𝑡)=0,𝑡∈[1,𝑛],𝑢(0)=0,𝑢(𝑛+1)=𝛼𝑢(𝑚) (2.9) has a unique solution 𝑢(𝑡)=^𝑛_𝑘=1𝐺(𝑡,𝑘)𝑣(𝑘), (2.10) where 𝐺(𝑡,𝑘) is given by 𝐺(𝑡,𝑘)=𝑘(𝑛+1−𝛼𝑚−𝑡+𝛼𝑡)𝑛+1−𝛼𝑚,𝑘∈[1,𝑡−1]∩[1,𝑚−1];𝑡(𝑛+1−𝛼𝑚−𝑘+𝛼𝑘)𝑛+1−𝛼𝑚,𝑘∈[𝑡,𝑚−1];𝑘(𝑛+1−𝛼𝑚−𝑡)+𝛼𝑚𝑡𝑛+1−𝛼𝑚,𝑘∈[𝑚,𝑡−1];𝑡(𝑛+1−𝑘)𝑛+1−𝛼𝑚,𝑘∈[𝑡,𝑛]∩[𝑚,𝑛]. (2.11)

Proof. We use a similar approach to that in [17, Lemmas 3.1, 3.3]. From Δ²𝑢(𝑡−1)+𝑣(𝑡)=0, we have Δ𝑢(1)−Δ𝑢(0)+𝑣(1)=0,Δ𝑢(2)−Δ𝑢(1)+𝑣(2)=0,…,Δ𝑢(𝑡)−Δ𝑢(𝑡−1)+𝑣(𝑡)=0. (2.12) Suming the above equations, one gets Δ𝑢(𝑡)=Δ𝑢(0)−^𝑡_𝑖=1𝑣(𝑖), (2.13) where, and in what follows, we denote ∑^𝑙_𝑘=𝑠𝑥(𝑘)=0 when 𝑙<𝑠. Again summing (2.13) from 0 to 𝑡−1, it follows that 𝑢(𝑡)=𝑢(0)+𝑡Δ𝑢(0)−^𝑡−1_𝑘=1(𝑡−𝑘)𝑣(𝑘) (2.14) where 𝑡∈[0,𝑛+1]. Since 𝑢(0)=0 and 𝑢(𝑛+1)=𝛼𝑢(𝑚), one gets 1Δ𝑢(0)=𝑛+1−𝛼𝑚^𝑛_𝑘=1𝛼(𝑛+1−𝑘)𝑣(𝑘)−𝑛+1−𝛼𝑚^𝑚−1_𝑘=1(𝑚−𝑘)𝑣(𝑘). (2.15) By (2.14) and (2.15), we have 𝑡𝑢(𝑡)=𝑛+1−𝛼𝑚^𝑛_𝑘=1(𝑛+1−𝑘)𝑣(𝑘)−𝛼𝑡𝑛+1−𝛼𝑚^𝑚−1_𝑘=1−(𝑚−𝑘)𝑣(𝑘)_{𝑡−1𝑘=1}(𝑡−𝑘)𝑣(𝑘),𝑡∈[0,𝑛+1]. (2.16) When 𝑡>𝑚, it follows from (2.16) that 𝑡𝑢(𝑡)=(𝑛+1−𝛼𝑚^𝑚−1_𝑘=1(𝑛+1−𝑘)𝑣(𝑘)+^𝑡−1_𝑘=𝑚(𝑛+1−𝑘)𝑣(𝑘)+^𝑛_𝑘=𝑡−(𝑛+1−𝑘)𝑣(𝑘))𝛼𝑡𝑛+1−𝛼𝑚^𝑚−1_𝑘=1(𝑚−𝑘)𝑣(𝑘)−^𝑚−1_𝑘=1(𝑡−𝑘)𝑣(𝑘)−^𝑡−1_𝑘=𝑚(𝑡−𝑘)𝑣(𝑘)=^𝑛_𝑘=1𝐺(𝑡,𝑘)𝑣(𝑘). (2.17) When 𝑡≤𝑚, it follows from (2.16) that 𝑡𝑢(𝑡)=(𝑛+1−𝛼𝑚^𝑡−1_𝑘=1(𝑛+1−𝑘)𝑣(𝑘)+^𝑚−1_𝑘=𝑡(𝑛+1−𝑘)𝑣(𝑘)+^𝑛_𝑘=𝑚−(𝑛+1−𝑘)𝑣(𝑘))𝛼𝑡(𝑛+1−𝛼𝑚^𝑡−1_𝑘=1(𝑚−𝑘)𝑣(𝑘)+^𝑚−1_𝑘=𝑡(𝑚−𝑘)𝑣(𝑘))−^𝑡−1_𝑘=1=(𝑡−𝑘)𝑣(𝑘)_𝑛𝑘=1𝐺(𝑡,𝑘)𝑣(𝑘). (2.18) Then, the unique solution of (2.9) can be written as 𝑢(𝑡)=∑^𝑛_𝑘=1𝐺(𝑡,𝑘)𝑣(𝑘).

Remark 2.8. Green’s function 𝐺(𝑡,𝑘) defined by Lemma 2.7 is positive on [1,𝑛]×[1,𝑛].
Define operators 𝐾,f,𝐴∶𝐸→𝐸, respectively, by
(𝐾𝑢)(𝑡)=^𝑛_𝑘=1𝐺(𝑡,𝑘)𝑢(𝑘),𝑢∈𝐸,𝑡∈[1,𝑛];(f𝑢)(𝑡)=𝑓(𝑡,𝑢(𝑡)),𝑢∈𝐸,𝑡∈[1,𝑛];𝐴=𝐾f. (2.19)(2.20) Now from Lemma 2.7, it is easy to see that BVP (1.1) has a solution 𝑢=𝑢(𝑡) if and only if 𝑢 is a fixed point of the operator 𝐴. It follows from the continuity of 𝑓 that 𝐴∶𝐸→𝐸 is completely continuous.
We shall use the following assumptions.
(𝐻₁)We have 0≤𝛼<1, or 𝛼=1,Λ=𝜙, (2.21) where Λ=𝐴∩𝐵, and 𝐴={(2𝑘−1)𝜋𝑛+1+𝑚∶𝑘=1,2,…,[𝑛+1+𝑚2{]},𝐵=2𝑡𝜋𝑛+1−𝑚∶𝑡=1,2,…,[𝑛−𝑚2]},𝑛−𝑚≥2,𝜙,𝑛−𝑚<2. (2.22)[𝑥] denotes the integer part of the real number 𝑥. (𝐻₂)For any 𝑡∈[1,𝑛], 𝑓(𝑡,0)=0; for any 𝑡∈[1,𝑛] and 𝑥∈𝐑, 𝑥𝑓(𝑡,𝑥)≥0.(𝐻₃)There exists an even number 𝑘₀∈[1,𝑛] such that1𝜆_𝑘0<𝛽₀<1𝜆_𝑘0+1, (2.23) where lim_𝑥→0(𝑓(𝑡,𝑥)/𝑥)=𝛽₀ uniformly for 𝑡∈[1,𝑛], and 𝜆_𝑘⁻¹=4sin²(𝜉_𝑘/2),𝑘∈[1,𝑛], 𝜉₁,…,𝜉_𝑛 are given in Lemma 2.9, 𝜆⁻¹_𝑛+1≜+∞.(𝐻₄)There exists an even number 𝑘₁∈[1,𝑛] such that 1𝜆_𝑘1<𝛽_∞<1𝜆_𝑘1+1, (2.24) where lim_|𝑥|→∞(𝑓(𝑡,𝑥)/𝑥)=𝛽_∞ uniformly for 𝑡∈[1,𝑛], and 𝜆₁⁻¹,…,𝜆_𝑛⁻¹,𝜆⁻¹_𝑛+1 are given in condition (𝐻₃).(𝐻₅)There exists a constant 𝑇>0 such that for any (𝑡,𝑥)∈[1,𝑛]×[−𝑇,𝑇], |𝑓(𝑡,𝑥)|<𝜌⁻¹𝑇, (2.25) where 𝜌=[(𝑛+1)²−𝛼𝑚²]²/8(𝑛+1−𝛼𝑚)².

Lemma 2.9. Suppose that (𝐻₁) holds; then there exist 𝜉₁,𝜉₂,…,𝜉_𝑛 with 0<𝜉₁<𝜉₂<⋯<𝜉_𝑛<𝜋 such that sin(𝑛+1)𝜉_𝑖=𝛼sin𝑚𝜉_𝑖, 𝑖=1,2,…,𝑛.

Proof. First, suppose that 0≤𝛼<1. Let ℎ(𝑥)=sin(𝑛+1)𝑥−𝛼sin𝑚𝑥, then we have 𝜋ℎ(𝜋2(𝑛+1))>0,ℎ(𝑛+1)≤0,ℎ(3𝜋2(𝑛+1))<0,ℎ(5𝜋2(𝑛+1))>0,ℎ(7𝜋2(𝑛+1))<0,…,(−1)^𝑛ℎ((2𝑛+1)𝜋2(𝑛+1))>0. (2.26) It follows from the intermediate value theorem that there exist 𝜉₁∈(𝜋/2(𝑛+1),𝜋/𝑛+1] and 𝜉_𝑖∈((2𝑖−1)𝜋/2(𝑛+1),(2𝑖+1)𝜋/2(𝑛+1)),𝑖=2,…,𝑛, such that sin(𝑛+1)𝜉_𝑖=𝛼sin𝑚𝜉_𝑖, 𝑖=1,2,…,𝑛.
Now suppose that 𝛼=1 and Λ=𝜙. Let Λ₁=𝐴∪𝐵. It is easy to see that there exist 𝜁_𝑖∈Λ₁,𝑖=1,2,…,𝑛 with 0<𝜁₁=(𝜋/𝑛+1+𝑚)<𝜁₂<⋯<𝜁_𝑛<𝜋 such that sin(𝑛+1)𝜁_𝑖=sin𝑚𝜁_𝑖. Then Lemma 2.9 is proved.

Remark 2.10. Condition Λ=𝜙 is reasonable. For example, let 𝑛=5, 𝑚=3, then Λ=𝜙. Let 𝑛=18, 𝑚=3, then 𝜉=𝜋/2∈Λ≠𝜙.

Lemma 2.11. Suppose that (𝐻₁) holds; then the set of eigenvalues of the linear operator 𝐾 consists of the strictly decreasing finite sequence of 𝜆_𝑘,𝑘=1,2,…,𝑛, with corresponding eigenfunctions 𝜑_𝑘(𝑡)=sin(𝑡𝜉_𝑘), where 𝜆_𝑘=(4sin²(𝜉_𝑘/2))⁻¹, 𝑘=1,2,…,𝑛, and 𝜉₁,…,𝜉_𝑛 are given in Lemma 2.9. In addition, the algebraic multiplicity of each eigenvalue 𝜆_𝑘 of the linear operator 𝐾 is equal to 1.

Proof. It is easy to see that 𝐾𝑢(𝑡)=𝜆𝑢(𝑡),𝑡∈[1,𝑛],𝑢∈𝐸 (2.27) is equivalent to the following equation: 𝜆Δ²𝑢(𝑡−1)+𝑢(𝑡)=0,𝑡∈[1,𝑛],𝑢(0)=0,𝑢(𝑛+1)=𝛼𝑢(𝑚). (2.28) By Lemma 2.9, we suppose that 𝜑_𝑘(𝑡)=sin(𝑡𝜉_𝑘) is a nontrivial solution of (2.28). Then, 𝜆(sin(𝑡+1)𝜉_𝑘−2sin𝑡𝜉_𝑘+sin(𝑡−1)𝜉_𝑘)+sin𝑡𝜉_𝑘=0. (2.29) Hence, for any 𝑘∈[1,𝑛], 𝜆=𝜆_𝑘=(4sin²(𝜉_𝑘/2))⁻¹is an eigenvalue of the linear operator 𝐾 with the corresponding eigenfunction 𝜑_𝑘(𝑡)=sin(𝑡𝜉_𝑘). Since the linear operator 𝐾 is identified with a linear transformation from 𝐑^𝑛 to 𝐑^𝑛, the set of eigenvalues of the linear operator 𝐾 consists of the strictly decreasing finite sequence of 𝜆_𝑘,𝑘=1,2,…,𝑛. Obviously, the algebraic multiplicity of each eigenvalue 𝜆_𝑘 of 𝐾 is equal to 1. This completes the proof.

Remark 2.12. When 𝛼=0, we see that BVP (1.1) is reduced to Dirichlet boundary value problem and 𝜆_𝑘=(4sin²(𝑘𝜋/2(𝑛+1)))⁻¹, 𝑘=1,2,…,𝑛. When 𝛼=1 and 𝑚=𝑛, BVP (1.1) is reduced to the focal boundary value problem and 𝜆_𝑘=(4sin²(2𝑘−1)𝜋/4𝑛+2)⁻¹, 𝑘=1,2,…,𝑛.

Lemma 2.13. Suppose that (𝐻₂) holds, and u=(𝑢₁,𝑢₂,…,𝑢_𝑛)^𝑇∈𝑃⧵{𝜃} is a solution of BVP (1.1). Then 𝑢∈𝑃^∘.

Proof. If 𝑢(𝑡)=0 for some 𝑡∈[1,𝑛], then 𝑢(𝑡+1)+𝑢(𝑡−1)=Δ²𝑢(𝑡−1)=−𝑓(𝑡,𝑢(𝑡))=0. (2.30) So 𝑢(𝑡±1)=0, and it follows that if 𝑢 is zero somewhere in [1,𝑛], then it vanishes identically in [1,𝑛].

Remark 2.14. Similarly to Lemma 2.13, we know also that if (𝐻₂) holds and 𝑢∈−𝑃⧵{𝜃} is a solution of BVP (1.1), then 𝑢∈−𝑃^∘.

Lemma 2.15. Suppose that (𝐻₂)–(𝐻₄) hold. Then the operator 𝐴 is Fréchet differentiable at 𝜃 and ∞, where operator 𝐴 is defined by (2.20). Moreover, 𝐴^′(𝜃)=𝛽₀𝐾 and 𝐴^′(∞)=𝛽_∞𝐾.

Proof. By (𝐻₃), for any 𝜀>0, there exist 𝛿>0 such that |𝑓(𝑡,𝑥)−𝛽₀𝑥|<𝜀|𝑥| for any 0<|𝑥|<𝛿, 𝑡∈[1,𝑛]. Hence, noticing that 𝑓(𝑡,0)=0 for any 𝑡∈[1,𝑛], we have ‖𝐴𝑢−𝐴𝜃−𝛽₀𝐾𝑢‖=‖𝐾(f𝑢−𝛽₀𝑢)‖≤‖𝐾‖⋅max_{𝑡∈[1,𝑛]}|𝑓(𝑡,𝑢(𝑡))−𝛽₀𝑢(𝑡)|<𝜀‖𝐾‖⋅‖𝑢‖ (2.31) for any 𝑢∈𝐸 with 0<‖𝑢‖<𝛿, where ‖𝐾‖=max_{𝑡∈[1,𝑛]}∑^𝑛_𝑘=1|𝐺(𝑡,𝑘)|. Consequently, lim_{‖𝑢‖→0}‖𝐴𝑢−𝐴𝜃−𝛽₀𝐾𝑢‖‖𝑢‖=0. (2.32) This means that the nonlinear operator 𝐴 is Fréchet differentiable at 𝜃, and 𝐴^′(𝜃)=𝛽₀𝐾.
By (𝐻₄), for any 𝜀>0, there exist 𝑀>0 such that |𝑓(𝑡,𝑥)−𝛽_∞𝑥|<𝜀|𝑥| for any |𝑥|>𝑀, 𝑡∈[1,𝑛]. Let 𝑐=max_{(𝑡,𝑥)∈[1,𝑛]×[−𝑀,𝑀]}|𝑓(𝑡,𝑥)−𝛽_∞𝑥|. By the continuity of 𝑓(𝑡,𝑥) with respect to 𝑥, we have 𝑐<+∞. Then, for any (𝑡,𝑥)∈[1,𝑛]×𝐑, |𝑓(𝑡,𝑥)−𝛽_∞𝑥|<𝜀|𝑥|+𝑐. Thus
‖𝐴𝑢−𝛽_∞𝐾𝑢‖≤‖𝐾‖⋅max_{𝑡∈[1,𝑛]}|𝑓(𝑡,𝑢(𝑡))−𝛽_∞𝑢(𝑡)|<‖𝐾‖(𝜀‖𝑢‖+𝑐) (2.33) for any 𝑢∈𝐸. Consequently, lim_{‖𝑢‖→∞}‖𝐴𝑢−𝛽_∞𝐾𝑢‖‖𝑢‖=0, (2.34) which implies that operator 𝐴 is Fréchet differentiable at ∞, and 𝐴^′(∞)=𝛽_∞𝐾. The proof is completed.

Lemma 2.16. Let 𝑇 be given in condition (𝐻₅). Suppose that (𝐻₁)–(𝐻₄) hold. Then, 𝐴(𝑃)⊂𝑃, 𝐴(−𝑃)⊂−𝑃. Moreover, one has the following.(i)There exists an 𝑟₀∈(0,𝑇) such that for any 0<𝑟≤𝑟₀, 𝑖(𝐴,𝑃∩𝐵(𝜃,𝑟),𝑃)=0,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑟),−𝑃)=0. (2.35)(ii)There exists an 𝑅₀>𝑇 such that for any 𝑅≥𝑅₀,𝑖(𝐴,𝑃∩𝐵(𝜃,𝑅),𝑃)=0,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑅),−𝑃)=0. (2.36)

Proof. By (𝐻₂) and the fact that 𝐺(𝑡,𝑘) is positive on [1,𝑛]×[1,𝑛], we get that for any 𝑡∈[1,𝑛], 𝑓(𝑡,𝑃)⊂𝑃, 𝑓(𝑡,−𝑃)⊂−𝑃, and 𝐾(𝑃)⊂𝑃, 𝐾(−𝑃)⊂−𝑃. Then 𝐴(𝑃)⊂𝑃 and 𝐴(−𝑃)⊂−𝑃.
We only need to prove conclusion (i). The proof of conclusion (ii) is similar and will be omitted here. Let 𝛾₀=inf_‖𝑢‖=1‖𝑢−𝛽₀𝐾𝑢‖. Condition (𝐻₃) yields 𝛾₀>0. It follows from (2.32) that there exists 𝑟₀∈(0,𝑇) such that
‖𝐴𝑢−𝛽₀1𝐾𝑢‖<2𝛾₀‖𝑢‖, (2.37) where 0<‖𝑢‖≤𝑟₀. Setting 𝐻(𝑠,𝑢)=𝑠𝐴𝑢+(1−𝑠)𝛽₀𝐾𝑢, then 𝐻∶[0,1]×𝐸→𝐸 is completely continuous. For any 𝑠∈[0,1] and 0<‖𝑢‖≤𝑟₀, we obtain that ‖𝑢−𝐻(𝑠,𝑢)‖≥‖𝑢−𝛽₀𝐾𝑢‖−𝑠‖𝐴𝑢−𝛽₀𝐾𝑢‖≥𝛾₀1‖𝑢‖−2𝛾₀‖𝑢‖>0. (2.38) According to the homotopy invariance of the fixed point index, for any 0<𝑟≤𝑟₀, we have 𝑖(𝐴,𝑃∩𝐵(𝜃,𝑟),𝑃)=𝑖(𝛽₀𝐾,𝑃∩𝐵(𝜃,𝑟),𝑃),𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑟),−𝑃)=𝑖(𝛽₀𝐾,−𝑃∩𝐵(𝜃,𝑟),−𝑃). (2.39)(2.40) Let 𝜑₁(𝑡)=sin(𝑡𝜉₁). Then 𝐾𝜑₁=𝜆₁𝜑₁ and 𝜑₁∈𝑃 (see Lemma 2.11 and the proof of Lemma 2.9). We claim 𝑢−𝛽₀𝐾𝑢≠𝜎𝜑₁,∀𝑢∈𝑃∩𝜕𝐵(𝜃,𝑟),𝜎≥0. (2.41) Indeed, we assume that there exist 𝑢₁∈𝑃∩𝜕𝐵(𝜃,𝑟) and 𝜎₁≥0 such that 𝑢₁−𝛽₀𝐾𝑢₁=𝜎₁𝜑₁. Obviously, 𝑢₁=𝛽₀𝐾𝑢₁+𝜎₁𝜑₁≥𝜎₁𝜑₁. Since 𝛽₀≠𝜆_𝑘⁻¹,𝑘=1,2,…,𝑛, then 𝜎₁>0. Set 𝜎_max=sup{𝜎∶𝑢₁≥𝜎𝜑₁}. It is clear that 𝜎₁≤𝜎_max<+∞ and 𝑢₁≥𝜎_max𝜑₁. Then 𝑢₁=𝛽₀𝐾𝑢₁+𝜎₁𝜑₁≥𝛽₀𝐾𝜎_max𝜑₁+𝜎₁𝜑₁=(𝛽₀𝜆₁𝜎_max+𝜎₁)𝜑₁. (2.42) Since 𝛽₀𝜆₁>1, then 𝛽₀𝜆₁𝜎_max+𝜎₁>𝜎_max, which contradicts with the definition of 𝜎_max. This proves (2.41).
It follows from Lemma 2.1 and (2.41) that
𝑖(𝛽₀𝐾,𝑃∩𝐵(𝜃,𝑟),𝑃)=0. (2.43) Similarly to (2.43), we know also that 𝑖(𝛽₀𝐾,−𝑃∩𝐵(𝜃,𝑟),−𝑃)=0. (2.44) By (2.39), (2.43), (2.40), and (2.44), we conclude 𝑖(𝐴,𝑃∩𝐵(𝜃,𝑟),𝑃)=0,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑟),−𝑃)=0. (2.45)

3. Main Results

Now with the aid of the lemmas in Section 2, we are in position to state and prove our main results.

Theorem 3.1. Assume that the conditions (𝐻₁)–(𝐻₅) hold. Then BVP (1.1) has at least two sign-changing solutions. Moreover, BVP (1.1) has at least two positive solutions and two negative solutions.

Proof. From the proof of Lemma 2.7, we have ^𝑛_𝑘=1𝑡𝐺(𝑡,𝑘)=𝑛+1−𝛼𝑚^𝑛_𝑘=1(𝑛+1−𝑘)−𝛼𝑡𝑛+1−𝛼𝑚^𝑚−1_𝑘=1(𝑚−𝑘)−^𝑡−1_𝑘=1=𝑡(𝑡−𝑘)⋅𝑛+1−𝛼𝑚𝑛(𝑛+1)2−𝛼𝑡⋅𝑛+1−𝛼𝑚𝑚(𝑚−1)2−𝑡(𝑡−1)2≤[(𝑛+1)²−𝛼𝑚²]²8(𝑛+1−𝛼𝑚)²=𝜌. (3.1) Since 𝐺(𝑡,𝑘) is positive on [1,𝑛]×[1,𝑛], by (𝐻₅), we have for any 𝑢∈𝐸 with ‖𝑢‖=𝑇, |(𝐴𝑢)(𝑡)|=|^𝑛_𝑘=1𝐺(𝑡,𝑘)𝑓(𝑘,𝑢(𝑘))|≤^𝑛_𝑘=1𝐺(𝑡,𝑘)|𝑓(𝑘,𝑢(𝑘))|<𝜌⁻¹𝑇^𝑛_𝑘=1𝐺(𝑡,𝑘)≤𝑇,∀𝑡∈[1,𝑛]. (3.2) This gives ‖𝐴𝑢‖<𝑇=‖𝑢‖. (3.3) By (3.3) and Lemmas 2.3 and 2.2, we have deg(𝐼−𝐴,𝐵(𝜃,𝑇),𝜃)=1,𝑖(𝐴,𝑃∩𝐵(𝜃,𝑇),𝑃)=1,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑇),−𝑃)=1. (3.4)(3.5)(3.6) From (𝐻₃) and Lemma 2.11, one has that the eigenvalues of the operator 𝐴^′(𝜃)=𝛽₀𝐾 which are larger than 1 are 𝛽₀𝜆₁,𝛽₀𝜆₂,…,𝛽₀𝜆_𝑘0, (3.7) From (𝐻₄) and Lemma 2.11, one has that the eigenvalues of the operator 𝐴^′(∞)=𝛽_∞𝐾 which are larger than 1 are 𝛽_∞𝜆₁,𝛽_∞𝜆₂,…,𝛽_∞𝜆_𝑘1. (3.8) It follows from Lemmas 2.4 and 2.5 that there exist 0<𝑟₁<𝑟₀ and 𝑅₁>𝑅₀ such that deg(𝐼−𝐴,𝐵(𝜃,𝑟₁),𝜃)=(−1)^𝑘0=1,deg(𝐼−𝐴,𝐵(𝜃,𝑅₁),𝜃)=(−1)^𝑘1=1, (3.9)(3.10) where 𝑟₀ and 𝑅₀ are given in Lemma 2.16. Owing to Lemma 2.16, one has 𝑖(𝐴,𝑃∩𝐵(𝜃,𝑟₁),𝑃)=0,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑟₁),−𝑃)=0,𝑖(𝐴,𝑃∩𝐵(𝜃,𝑅₁),𝑃)=0,𝑖(𝐴,−𝑃∩𝐵(𝜃,𝑅₁),−𝑃)=0. (3.11)(3.12)(3.13)(3.14) According to the additivity of the fixed point index, by (3.5), (3.11), and (3.13), we have 𝑖(𝐴,𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)),𝑃)=𝑖(𝐴,𝑃∩𝐵(𝜃,𝑇),𝑃)−𝑖(𝐴,𝑃∩𝐵(𝜃,𝑟₁),𝑃)=1−0=1,𝑖(𝐴,𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇)),𝑃)=𝑖(𝐴,𝑃∩𝐵(𝜃,𝑅₁),𝑃)−𝑖(𝐴,𝑃∩𝐵(𝜃,𝑇),𝑃)=0−1=−1. (3.15)(3.16) Hence, the nonlinear operator 𝐴 has at least two fixed points 𝑢₁∈𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)) and 𝑢₂∈𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇)), respectively. Then, 𝑢₁ and 𝑢₂ are positive solutions of BVP (1.1).
Using again the additivity of the fixed point index, by (3.6), (3.12), and (3.14), we get
𝑖(𝐴,−𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)),−𝑃)=1−0=1,𝑖(𝐴,−𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇)),−𝑃)=0−1=−1. (3.17)(3.18) Hence, the nonlinear operator 𝐴 has at least two fixed points 𝑢₃∈−𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)) and 𝑢₄∈−𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇)), respectively. Then, 𝑢₃ and 𝑢₄ are negative solutions of BVP (1.1). Let Γ₁={𝑢∈𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁Γ))∶𝐴𝑢=𝑢},₂={𝑢∈𝑃∩(𝐵(𝜃,𝑅₁)⧵Γ𝐵(𝜃,𝑇))∶𝐴𝑢=𝑢},₃={𝑢∈−𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁Γ))∶𝐴𝑢=𝑢},₄={𝑢∈−𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇))∶𝐴𝑢=𝑢}. (3.19) It follows from Lemmas 2.6, 2.13, Remark 2.14, and (3.15)–(3.18) that there exist open subsets 𝑂₁,𝑂₂,𝑂₃, and 𝑂₄ of 𝐸 such that Γ₁⊂𝑂₁⊂𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)),Γ₂⊂𝑂₂⊂𝑃∩(𝐵(𝜃,𝑅₁)⧵Γ𝐵(𝜃,𝑇)),₃⊂𝑂₃∈−𝑃∩(𝐵(𝜃,𝑇)⧵𝐵(𝜃,𝑟₁)),Γ₄⊂𝑂₄⊂−𝑃∩(𝐵(𝜃,𝑅₁)⧵𝐵(𝜃,𝑇)),deg(𝐼−𝐴,𝑂₁,𝜃)=1,deg(𝐼−𝐴,𝑂₂,𝜃)=−1,deg(𝐼−𝐴,𝑂₃,𝜃)=1,deg(𝐼−𝐴,𝑂₄,𝜃)=−1. (3.20)(3.21)(3.22)(3.23)(3.24) By (3.4), (3.21), (3.23), (3.9), and the additivity of Leray-Schauder degree, we get deg(𝐼−𝐴,𝐵(𝜃,𝑇)⧵(𝑂₁∪𝑂₃∪𝐵(𝜃,𝑟₁)),𝜃)=1−1−1−1=−2, (3.25) which implies that the nonlinear operator 𝐴 has at least one fixed point 𝑢₅∈𝐵(𝜃,𝑇)⧵(𝑂₁∪𝑂₃∪𝐵(𝜃,𝑟₁)).
Similarly, by (3.10), (3.22), (3.24), and (3.4), we get
deg(𝐼−𝐴,𝐵(𝜃,𝑅₁)⧵(𝑂₂∪𝑂₄∪𝐵(𝜃,𝑇)),𝜃)=1+1+1−1=2, (3.26) which implies that the nonlinear operator 𝐴 has at least one fixed point 𝑢₆∈𝐵(𝜃,𝑅₁)⧵(𝑂₂∪𝑂₄∪𝐵(𝜃,𝑇)). Then, 𝑢₅ and 𝑢₆ are two distinct sign-changing solutions of BVP (1.1). Thus, the proof of Theorem 3.1 is finished.

Theorem 3.2. Assume that the conditions (𝐻₁)–(𝐻₅) hold, and that 𝑓(𝑡,𝑥)=−𝑓(𝑡,−𝑥) for 𝑡∈[1,𝑛] and 𝑥∈𝐑. Then BVP (1.1) has at least four sign-changing solutions. Moreover, BVP (1.1) has at least two positive solutions and two negative solutions.

Proof. It follows from the proof of Theorem 3.1 that BVP (1.1) has at least six different nontrivial solutions 𝑢_𝑖(𝑖=1,2,…,6) satisfying that 𝑢₁,𝑢₂∈𝑃^∘,𝑢₃,𝑢₄∈−𝑃^∘,𝑢₅,𝑢₆∉𝑃∪(−𝑃),𝑟₁<‖𝑢₅‖<‖𝑇‖<‖𝑢₆‖<𝑅₁. (3.27) By the condition that 𝑓(𝑡,𝑥)=−𝑓(𝑡,−𝑥) for 𝑡∈[1,𝑛] and 𝑥∈𝐑, we know that −𝑢₅ and −𝑢₆ are also solutions of BVP (1.1). Let 𝑢₇=−𝑢₅, 𝑢₈=−𝑢₆, then 𝑢_𝑖(𝑖=1,2,…,8) are different nontrivial solutions of BVP (1.1). The proof is completed.

By the method used in the proof of Theorems 3.1 and 3.2, we can prove the following corollaries.

Corollary 3.3. Assume that the conditions (𝐻₁)–(𝐻₃) and (𝐻₅) or (𝐻₁), (𝐻₂), (𝐻₄), and (𝐻₅) hold. Then BVP (1.1) has at least one sign-changing solution. Moreover, BVP (1.1) has at least one positive solution and one negative solution.

Corollary 3.4. Assume that the conditions (𝐻₁)–(𝐻₃) and (𝐻₅) or (𝐻₁), (𝐻₂), (𝐻₄), and (𝐻₅) hold, and that 𝑓(𝑡,𝑥)=−𝑓(𝑡,−𝑥) for 𝑥∈𝐑 and 𝑡∈[1,𝑛]. Then BVP (1.1) has at least two sign-changing solutions. Moreover, BVP (1.1) has at least one positive solution and one negative solution.

Next, we present a simple example to which Theorem 3.2 can be applied.

Example 3.5. Consider the second-order three-point discrete boundary value problem Δ²𝑢(𝑡−1)+𝑓(𝑡,𝑢(𝑡))=0,𝑡∈[1,4],𝑢(0)=0,𝑢(5)=𝑢(3), (3.28) where 𝑛=4,  𝑚=3,  𝛼=1, and 𝑓(𝑡,𝑥)=𝑏𝑥1+𝑥²,|𝑥|≤20,(18−2𝑏401)𝑥+60𝑏401−360,20≤𝑥≤30,(18−2𝑏401)𝑥+360−60𝑏401,−30≤𝑥≤−20,𝑥(5+1000100+𝑥²),|𝑥|≥30, (3.29)𝑏∈(4sin²(3𝜋/16),4sin²(5𝜋/16)). Obviously, 𝛽₀=𝑏,𝛽_∞=5, and 𝑓(𝑡,𝑥)=−𝑓(𝑡,−𝑥) for all (𝑡,𝑥)∈[1,4]×𝐑.
From Lemma 2.11 and the proof of Lemma 2.9, we know that the set of eigenvalues of the linear operator 𝐾 (see (2.19)) consists of the strictly decreasing finite sequence of 𝜆_𝑘, 𝑘=1,2,3,4, where 𝜆_𝑘=(4sin²((2𝑘−1)𝜋/16))⁻¹. Then the conditions (𝐻₁)–(𝐻₄) hold. Since |𝑓(𝑡,𝑥)|≤(𝑏/2)<(3/2) for all (𝑡,𝑥)∈[1,4]×[−12,12], then (𝐻₅) holds with 𝑇=12 and 𝜌=8. Therefore, by Theorem 3.2, BVP (3.28) has at least four sign-changing solutions. Moreover, BVP (3.28) has at least two positive solutions and two negative solutions.

Acknowledgment

This project is supported by the Education Department of Fujian Province of China (no. JA06035).