Existence of One-Signed Solutions of Discrete Second-Order Periodic Boundary Value Problems

Abstract

We prove the existence of one-signed periodic solutions of second-order nonlinear difference equation on a finite discrete segment with periodic boundary conditions by combining some properties of Green's function with the fixed-point theorem in cones.

1. Introduction

Let ℝ be the set of real numbers, ℤ be the integers set, 𝑇,𝑎,𝑏∈ℤ with 𝑇>2,𝑎>𝑏, and [𝑎,𝑏]_ℤ={𝑎,𝑎+1,…,𝑏}.

In recent years, the existence and multiplicity of positive solutions of periodic boundary value problems for difference equations have been studied extensively, see [1–5] and the references therein. In 2003, Atici and Cabada [2] studied the existence of solutions of second-order difference equation boundary value problemΔ²𝑦(𝑛−1)+𝑎(𝑛)𝑦(𝑛)+𝑓𝑛,𝑦(𝑛)=0,𝑛∈[1,𝑇]_ℤ,𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇), (1.1) where 𝑎, 𝑓 satisfy

(H1) 𝑎∶[1,𝑇]_ℤ→(−∞,0]and𝑎(⋅)≢0;

(H2) 𝑓∶[1,𝑇]_ℤ×ℝ→ℝ is continuous with respect to 𝑦∈ℝ.

The authors obtained the existence results of solutions of (1.1) under conditions (H1), (H2), and the used tool is upper and lower solutions techniques.

Naturally, whether there exists the Green function 𝐺(𝑡,𝑠) of the homogeneous linear boundary value problem corresponding to (1.1) if 𝑎(𝑛)≥0? Moreover, if the answer is positive, whether 𝐺(𝑡,𝑠) keeps its sign? To the knowledge of the authors, there are very few works on the case 𝑎(𝑛)≥0.

Recently, in 2003, Torres [6] investigated the existence of one-signed periodic solutions for second-order differential equation boundary value problem𝑥^′′(𝑡)((𝑡[],𝑥(0)(𝑇)=𝑓𝑡,𝑥)),𝑡∈1,𝑇=𝑥,𝑥^′(0)=𝑥^′(𝑇), (1.2) by applying the fixed-point theorem in cones, and constructed Green’s function of𝑥^′′(𝑡)(𝑡)𝑥(𝑡)[],𝑥(0)(𝑇)+𝑎=0,𝑡∈1,𝑇=𝑥,𝑥^′(0)=𝑥^′(𝑇), (1.3) where 𝑎∈𝐿^𝑝(0,𝑇) satisfies either

(H3) 𝑎≤0, 𝑎(⋅)≢0 on [0,𝑇];

(H4) 𝑎≥0, 𝑎(⋅)≢0 on [0,𝑇] and ‖𝑎‖_𝑝≤𝐾(2𝑝^∗) for some 1≤𝑝≤+∞.

Motivated by Torres [6], in Section 2, the paper gives the new expression of Green’s function of the linear boundary value problemΔ²𝑦(𝑡−1)+𝑎(𝑡)𝑦(𝑡)=0,𝑡∈[1,𝑇]_ℤ,𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇), (1.4)(1.5) where 𝑎∈Λ⁺∪Λ⁻ andΛ⁻=𝑎∣𝑎∶[1,𝑇]_ℤ⟶(](⋅),Λ−∞,0,𝑎≢0₊=𝑎∣𝑎∶[1,𝑇]_ℤ⟶[0,∞),𝑎(⋅)≢0,max_{𝑡∈[1,𝑇]ℤ}|𝑎(𝑡)|<4sin²𝜋,2𝑇 (1.6) and obtains the sign properties of Green’s function of (1.4), (1.5).

In Section 3, we obtain the existence of one-signed periodic solutions of the discrete second-order nonlinear periodic boundary value problem Δ²𝑦(𝑡−1)=𝑓𝑡,𝑦(𝑡),𝑡∈[1,𝑇]_ℤ,𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇), (1.7) where 𝑓∶[1,𝑇]_ℤ×ℝ→ℝ is continuous. For related results on the associated differential equations, see Torres [6].

2. Preliminaries

Let𝐸=𝑦∣𝑦∶[0,𝑇+1]_ℤ⟶ℝ,𝑦(0)=𝑦(𝑇),𝑦(1)=𝑦(𝑇+1) (2.1) be a Banach space endowed with the norm ‖𝑦‖=max_{𝑡∈[0,𝑇+1]ℤ}|𝑦(𝑡)|.

We say that the linear boundary value problem (1.4), (1.5) is nonresonant when its unique solution is the trivial one. If (1.4), (1.5) is nonresonant, and let ℎ∶[1,𝑇]_ℤ→ℝ, by the virtue of the Fredholm’s alternative theorem, we can get that the discrete second-order periodic boundary value problemΔ²𝑦(𝑡−1)+𝑎(𝑡)𝑦(𝑡)=ℎ(𝑡),𝑡∈[1,𝑇]_ℤ𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇) (2.2)(2.3) has a unique solution 𝑦,𝑦(𝑡)=^𝑇_𝑠=1𝐺(𝑡,𝑠)ℎ(𝑠),𝑡∈[0,𝑇+1]_ℤ, (2.4) where 𝐺(𝑡,𝑠) is Green’s function related to (1.4), (1.5).

Definition 2.1 (see [7]). We say that a solution 𝑦 of (1.4) has a generalized zero at 𝑡₀ provided that 𝑦(𝑡₀)=0 if 𝑡₀=0 and if 𝑡₀>0 either 𝑦(𝑡₀)=0 or 𝑦(𝑡₀−1)𝑦(𝑡₀)<0.

Theorem 2.2. Assume that the distance between two consecutive generalized zeros of a nontrivial solution of (1.4) is greater than 𝑇. Then Green’s function 𝐺(𝑡,𝑠) has constant sign.

Proof. Obviously, 𝐺 is well defined on [0,𝑇+1]_ℤ×[1,𝑇]_ℤ. We only need to prove that 𝐺 has no generalized zero in any point. Suppose on the contrary that there exists (𝑡₀,𝑠₀)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ such that (𝑡₀,𝑠₀) is a generalized zero of 𝐺(𝑡,𝑠). It is well known that for a given 𝑠₀∈[1,𝑇]_ℤ, 𝐺(𝑡,𝑠₀) as a function of 𝑡 is a solution of (1.4) in the intervals [0,𝑠₀−1]_ℤ and [𝑠₀+1,𝑇+1]_ℤ such that 𝐺0,𝑠₀=𝐺𝑇,𝑠₀,𝐺1,𝑠₀=𝐺𝑇+1,𝑠₀. (2.5)
Case 1 (𝐺(𝑡₀,𝑠₀)=0,(𝑡₀,𝑠₀)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ). If 𝑡₀∈[𝑠₀+1,𝑇+1]_ℤ, we can construct 𝑦(𝑡)=𝐺𝑡,𝑠₀,𝑡∈𝑠₀,𝑇+1_ℤ,𝐺𝑡−𝑇,𝑠₀,𝑡∈𝑇+1,𝑠₀+𝑇_ℤ. (2.6) Consequently, 𝑦 is a solution of (1.4) in the whole interval [𝑠₀,𝑠₀+𝑇]_ℤ. Since 𝑦(𝑡₀)=0, we have Δ²𝑦(𝑡₀−1)=−𝑎(𝑡)𝑦(𝑡₀)=0, that is, 𝑦(𝑡₀−1)𝑦(𝑡₀+1)<0. Moreover, 𝑦(𝑠₀)=𝑦(𝑠₀+𝑇), so there at least exists another generalized zero 𝑡₁∈[𝑠₀+1,𝑠₀+𝑇]_ℤ of 𝑦. Note that the distance between 𝑡₀ and 𝑡₁ is smaller than 𝑇, which is a contradiction.
Analogously, if 𝑡₀∈[0,𝑠₀−1]_ℤ, we get a contradiction by the same reasoning with 𝑦(𝑡)=𝐺𝑡+𝑇,𝑠₀,𝑡∈𝑠₀−𝑇,0_ℤ,𝐺𝑡,𝑠₀,𝑡∈0,𝑠_0ℤ. (2.7)
If 𝑡₀=𝑠₀, we can apply 𝑦 as defined (2.7). Since 𝑦(𝑡₀)=𝑦(𝑠₀)=0 and 𝑦(𝑠₀−𝑇)=𝑦(𝑠₀), which contradicts with the hypothesis.
Case 2 (𝐺(𝑡₀−1,𝑠₀)𝐺(𝑡₀,𝑠₀)<0,(𝑡₀,𝑠₀)∈[1,𝑇+1]_ℤ×[1,𝑇]_ℤ). If 𝑡₀∈[𝑠₀+1,𝑇+1]_ℤ, we can construct 𝑦 defined as (2.6). It is not difficult to verify that 𝑦 is a solution of (1.4) in the whole interval [𝑠₀,𝑠₀+𝑇]_ℤ. Also, we have that 𝑦(𝑡₀−1)𝑦(𝑡₀)<0, that is, 𝑡₀ is a generalized zero of 𝑦. Moreover, 𝑦(𝑠₀)=𝑦(𝑠₀+𝑇), so there at least exists another generalized zero 𝑡₁∈[𝑠₀+1,𝑠₀+𝑇]_ℤ of 𝑦. Note that the distance between 𝑡₀ and 𝑡₁ is smaller than 𝑇, which is a contradiction.
Similarly, if 𝑡₀∈[1,𝑠₀−1]_ℤ, we can get a contradiction by the same reasoning as 𝑦 defined (2.7).
If 𝑡₀=𝑠₀, we can construct 𝑦 defined by (2.7). Since 𝑦(𝑡₀−1)𝑦(𝑡₀)=𝑦(𝑠₀−1)𝑦(𝑠₀)<0 and 𝑦(𝑠₀−𝑇)=𝑦(𝑠₀), it is clear that there exists another generalized zero 𝑡₁∈[𝑠₀−𝑇,𝑠₀]_ℤ of 𝑦. Note that the distance between 𝑡₀ and 𝑡₁ is smaller than 𝑇, this contradicts with the hypothesis.

To apply the above result, we are going to study the two following cases.

Corollary 2.3. If 𝑎∈Λ⁻, then 𝐺(𝑡,𝑠)<0  for all (𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.

Proof. If 𝑎∈Λ⁻, by [7, Corollary 6.7], it is easy to verify that (1.4) is disconjugate on [0,𝑇+1]_ℤ, and any nontrivial solution of (1.4) has at most one generalized zero on [0,𝑇+1]_ℤ. Hence, by Theorem 2.2, Green’s function 𝐺(𝑡,𝑠) has constant sign. We claim that the sign is negative. In fact, 𝑦(𝑡)=∑^𝑇_𝑠=1𝐺(𝑡,𝑠) is the unique 𝑇-periodic solution of the equation Δ²𝑦(𝑡−1)+𝑎(𝑡)𝑦(𝑡)=1, (2.8) and summing both sides of (2.8) from 𝑡=1 to 𝑡=𝑇, we can get ^𝑇_𝑡=1𝑎(𝑡)𝑦(𝑡)=𝑇>0. (2.9) Since 𝑎(𝑡)<0,𝑦(𝑡)<0 for some 𝑡∈[1,𝑇]_ℤ, and as a consequence 𝐺(𝑡,𝑠)<0 for all (𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.

Remark 2.4. If 𝑎(⋅)≡𝑎₀ (𝑎₀ is a negative constant), then by computing we can obtain 𝐺()=−𝜆𝑡,𝑠^1𝑡−𝑠+𝜆₁^{𝑇−𝑡+𝑠}𝜆₁−𝜆₁⁻¹𝜆^𝑇₁−𝜆−1,1≤𝑠≤𝑡≤𝑇+1,_1𝑠−𝑡+𝜆₁^{𝑇−𝑠+𝑡}𝜆₁−𝜆₁⁻¹𝜆^𝑇₁−1,0≤𝑡≤𝑠≤𝑇, (2.10) where 𝜆₁=(2−𝑎₀+𝑎²₀−4𝑎₀)/2>1. Obviously, 𝐺(𝑡,𝑠)<0,(𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.

If 𝑎≥0, then the solutions of (1.4) are oscillating, that is, there are infinite zeros, and to get the required distance between generalized zeros, 𝑎 should satisfy Λ⁺.

Corollary 2.5. If 𝑎∈Λ⁺, then 𝐺(𝑡,𝑠)>0  for all (𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.

Proof. We claim that the distance between two consecutive generalized zeros of a nontrivial solution 𝑦 of (1.4) is strictly greater than 𝑇. In fact, it is not hard to verify that Δ²𝑦(𝑡−1)+‖𝑎‖𝑦(𝑡)=0 is disconjugate on [0,𝑇+1]_ℤ under assumption ‖𝑎‖<4sin²(𝜋/2𝑇). Since 𝑎(𝑡)≤‖𝑎‖,𝑡∈[1,𝑇]_ℤ, by Sturm comparison theorem [7, Theorem 6.19], (1.4) is disconjugate on [0,𝑇+1]_ℤ, that is, any nontrivial solution of (1.4) has at most one generalized zero on [0,𝑇+1].
Hence, by Theorem 2.2, 𝐺(𝑡,𝑠) has constant sign on [0,𝑇+1]_ℤ×[1,𝑇]_ℤ, and the positive sign of 𝐺 is determined as the proof process of Corollary 2.3.

Remark 2.6. If 𝑎(⋅)≡𝑎(𝑎isapositiveconstant), and 0<𝑎<4sin²(𝜋/2𝑇), then by computing we can obtain 𝐺()=[𝜃([𝜃(𝑡,𝑠sin𝑡−𝑠)]+sin𝑇−𝑡+𝑠)](([𝜃([𝜃(2sin𝜃1−cos𝜃𝑇)),1≤𝑠≤𝑡≤𝑇+1,sin𝑠−𝑡)]+sin𝑇−𝑠+𝑡)]2sin𝜃(1−cos(𝜃𝑇)),0≤𝑡≤𝑠≤𝑇, (2.11) where 𝜃=arccos((2−𝑎)/2) and 0<𝜃<𝜋/𝑇. Clearly, 𝐺(𝑡,𝑠)>0,(𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.
If 𝑎(⋅)≡𝑎 and 𝑎=4sin²(𝜋/2𝑇), then 𝜃=𝜋/𝑇, and by computing we get 𝐺()=1𝑡,𝑠()𝜋2sin𝜋/𝑇sin𝑇(1𝑡−𝑠),1≤𝑠≤𝑡≤𝑇+1,()𝜋2sin𝜋/𝑇sin𝑇(𝑠−𝑡),0≤𝑡≤𝑠≤𝑇. (2.12) Obviously, Green’s function 𝐺(𝑡,𝑠)=0 for 𝑡=𝑠 and 𝐺(𝑡,𝑠)>0 for 𝑡≠𝑠.
If 𝑎(⋅)≡𝑎 and 𝑎=4sin²(𝜋/𝑇), then 𝜃=2𝜋/𝑇, and it is not difficult to verify that 𝜑(𝑡)=sin2𝜋𝑇𝑡,𝜓(𝑡)=cos2𝜋𝑇𝑡,𝑡∈[0,𝑇+1]_ℤ (2.13) are nontrivial solutions of (1.4), (1.5). That is, the problem (1.4), (1.5) has no Green’s function.
If 𝑎(⋅)≡𝑎and4sin²(𝜋/2𝑇)<𝑎<4sin²(𝜋/𝑇), then Green’s function may change its sign. For example, let 𝑇=6,𝑎=4sin²(𝜋/8)=2−√2, it is easy to verify that 2−√3=4sin²(𝜋/12)<𝑎<4sin²(𝜋/6)=1 and 𝜃=𝜋/4, thus 𝐺()=𝜋𝑡,𝑠sin4(𝜋𝑡−𝑠−1),1≤𝑠≤𝑡≤𝑇+1,sin4(𝑠−𝑡−1),0≤𝑡≤𝑠≤𝑇. (2.14) Clearly, 𝐺(𝑡,𝑠)=−sin(𝜋/4)<0 for 𝑡=𝑠, 𝐺(𝑡,𝑠)=0 for |𝑡−𝑠|=1, and 𝐺(𝑡,𝑠)=sin(𝜋/4)>0 for |𝑡−𝑠|=2.
Consequently, 𝑎∈Λ⁺ is the optimal condition of 𝐺(𝑡,𝑠)>0,(𝑡,𝑠)∈[0,𝑇+1]_ℤ×[1,𝑇]_ℤ.

Next, we provide a way to get the expression of 𝐺(𝑡,𝑠). Let 𝑢 be the unique solution of the initial value problemΔ²𝑢(𝑡−1)+𝑎(𝑡)𝑢(𝑡)=0,𝑡∈[1,𝑇]_ℤ,𝑢(0)=0,Δ𝑢(0)=1, (2.15) and 𝑣 be the unique solution of the initial value problemΔ²𝑣(𝑡−1)+𝑎(𝑡)𝑣(𝑡)=0,𝑡∈[1,𝑇]_ℤ,𝑣(𝑇)=0,Δ𝑣(𝑇)=−1. (2.16)

Lemma 2.7. Let 𝑎∈Λ⁻∪Λ⁺. Then Green’s function 𝐺(𝑡,𝑠) of (1.4), (1.5) is explicitly given by 𝐺()=[𝑢(𝑠)(𝑠𝑢(𝑡)(𝑡𝑡,𝑠+𝑣)][+𝑣)]𝑣(0(1)(−1)[2+𝑣−𝑢𝑇+1)]𝑣(0)𝑢(𝑡)𝑣(𝑠)𝑢(𝑠)𝑣(𝑡),0≤𝑡≤𝑠≤𝑇,,1≤𝑠≤𝑡≤𝑇+1. (2.17)

Proof. Suppose that Green’s function of (1.4), (1.5) is of the form 𝐺()=𝛼(𝑠)𝑢(𝑡)(𝑠)𝑣(𝑡−1𝑡,𝑠+𝛽)𝑣(0)𝑢(𝑡)𝑣(𝑠)𝑢(𝑠)𝑣(𝑡),0≤𝑡≤𝑠≤𝑇,,1≤𝑠≤𝑡≤𝑇+1, (2.18) where 𝛼(𝑠),𝛽(𝑠) can be determined by imposing the boundary conditions.
From the basis theory of Green’s function, we know that 𝐺(0,𝑠)=𝐺(𝑇,𝑠),𝐺(1,𝑠)=𝐺(𝑇+1,𝑠),∀𝑠∈[1,𝑇]_ℤ,𝑣(0)=𝑢(𝑇). (2.19) Hence, 𝛽(𝑠)𝑣(0)=𝐺(0,𝑠)=𝐺(𝑇,𝑠)=𝛼(𝑠)𝑢(𝑇),𝑠∈[1,𝑇]_ℤ, combining with 𝑣(0)=𝑢(𝑇), we can get 𝛼(𝑠)=𝛽(𝑠),𝑠∈[1,𝑇]_ℤ. (2.20) Moreover, since 𝐺(1,𝑠)=𝐺(𝑇+1,𝑠), it follows that 𝛼(𝑠)=𝑢(𝑠)+𝑣(𝑠)𝑣(0(1)(.)[2+𝑣−𝑢𝑇+1)] (2.21)

Note that 𝛼(⋅) has the same sign with 𝑎(⋅). In fact, by the comparison theorem [7, Theorem 6.6], it is easy to prove that 𝑢,𝑣≥0 on [0,𝑇]_ℤ. If 𝑎(𝑡)≥0, thenΔ²𝑢(𝑡−1)=−𝑎(𝑡)𝑢(𝑡)≤0,Δ𝑢(𝑡)≤Δ𝑢(𝑡−1),𝑡∈[1,𝑇]_ℤ. (2.22) Thus Δ𝑢(𝑇)<Δ𝑢(0)=1. Similarly, we can get that Δ𝑣(0)>Δ𝑣(𝑇)=−1. Since 𝑣(0)=𝑢(𝑇), we have2+𝑣(1)−𝑢(𝑇+1)=2+Δ𝑣(0)−Δ𝑢(𝑇)>0. (2.23) That is 𝛼(𝑡)=(𝑢(𝑡)+𝑣(𝑡))/(𝑣(0)[2+𝑣(1)−𝑢(𝑇+1)])>0.

If 𝑎(⋅)≤0, by the similar method, we can prove 𝛼(⋅)<0.

Lemma 2.8. Let 𝑎∈Λ⁻∪Λ⁺. Then the periodic boundary value problem (2.2), (2.3) has the unique solution 𝑦(𝑡)=^𝑇_𝑠=1𝐺(𝑡,𝑠)ℎ(𝑠),𝑡∈[0,𝑇+1]_ℤ, (2.24) where 𝐺(𝑡,𝑠) is defined by (2.17).

Proof. We check that 𝑦 satisfies (2.2). In fact, 𝑦(𝑡)=^𝑇_𝑠=1(𝑢(𝑡)(𝑡𝛼(𝑠)ℎ(𝑠)−1+𝑣))𝑣(0)^𝑡−1_𝑠=1𝑢(𝑡)𝑣(𝑠)ℎ(𝑠)−1𝑣(0)^𝑇_𝑠=𝑡𝑢(𝑠)𝑣(𝑡)ℎ(𝑠)=(𝑢(𝑡)+𝑣(𝑡))^𝑇_𝑠=1𝛼(𝑠)ℎ(𝑠)−𝑢(𝑡)𝑣(0)^𝑡−1_𝑠=1𝑣(𝑠)ℎ(𝑠)−𝑣(𝑡)𝑣(0)^𝑇_𝑠=𝑡𝑢(𝑠)ℎ(𝑠),𝑦(𝑡+1)=(𝑢(𝑡+1)+𝑣(𝑡+1))^𝑇_𝑠=1𝛼(𝑠)ℎ(𝑠)−𝑢(𝑡+1)𝑣(0)^𝑡_𝑠=1𝑣(𝑠)ℎ(𝑠)−𝑣(𝑡+1)𝑣(0)^𝑇_𝑠=𝑡+1𝑢(𝑠)ℎ(𝑠),𝑦(𝑡−1)=(𝑢(𝑡−1)+𝑣(𝑡−1))^𝑇_𝑠=1𝛼(𝑠)ℎ(𝑠)−𝑢(𝑡−1)𝑣(0)^𝑡−2_𝑠=1𝑣(𝑠)ℎ(𝑠)−𝑣(𝑡−1)𝑣(0)^𝑇_{𝑠=𝑡−1}𝑢(𝑠)ℎ(𝑠),Δ²𝑦()(𝑡)𝑦(𝑡)()−((𝑡𝑦(𝑡)()=Δ𝑡−1+𝑎=𝑦𝑡+12−𝑎))+𝑦𝑡−1₂𝑢(𝑡−1)+𝑎(𝑡)𝑢(𝑡)+Δ²𝑣(𝑡−1)+𝑎(𝑡)𝑣(𝑡)^𝑇_𝑠=1𝛼(𝑠)ℎ(𝑠)−Δ²𝑢(𝑡−1)+𝑎(𝑡)𝑢(𝑡)𝑣(0)^𝑡−2_𝑠=1𝑣(𝑠)ℎ(𝑠)−Δ²𝑣(𝑡−1)+𝑎(𝑡)𝑣(𝑡)𝑣(0)^𝑇_𝑠=𝑡+1𝑢(𝑠)ℎ(𝑠)−𝑢(𝑡−1)ℎ(𝑡−1)𝑣(0)Δ²𝑣()(𝑡)𝑣(𝑡−𝑢(𝑡)ℎ(𝑡)𝑡−1+𝑎)𝑣(0)Δ²𝑣()(𝑡)𝑣(𝑡+𝑢(𝑡)𝑣()ℎ(𝑡)()𝑣(𝑡)ℎ(𝑡)𝑡−1+𝑎)𝑡−1−𝑢𝑡−1𝑣(0)=ℎ(𝑡)𝑣(0)𝑢(𝑡)𝑣(𝑡)𝑢()𝑣()=ℎ(𝑡)𝑡−1𝑡−1𝑣(0)𝑢(1)𝑣(1)𝑢(0)𝑣(0)=ℎ(𝑡). (2.25) On the other hand, it is easy to verify that 𝑦(0)=𝑦(𝑇),𝑦(1)=𝑦(𝑇+1).

Denote that𝑚=min_{[]𝑡,𝑠∈1,𝑇ℤ}𝐺(𝑡,𝑠),𝑀=max_{[]𝑡,𝑠∈1,𝑇ℤ}𝐺(𝑡,𝑠). (2.26) As a direct application, we can compute the maximum and the minimum of the Green’s function when 𝑎(⋅)≡𝑎₀, it follows that0>𝑚≥−2𝜆₁^𝑇/2𝜆₁−𝜆₁⁻¹𝜆^𝑇₁𝜆−1,𝑀=−^𝑇1+1𝜆₁−𝜆₁⁻¹𝜆^𝑇₁,−1 (2.27) where 𝜆₁ is defined in Remark 2.4. Similarly, when 0<𝑎(⋅)≡𝑎<4sin²(𝜋/2𝑇), we can get1𝑚=2sin𝜃cot𝜃𝑇21>0,𝑀≤(),2sin𝜃sin𝜃𝑇/2 (2.28) where 𝜃 is defined in Remark 2.6.

3. Main Results

In this section, we consider the existence of one-signed solutions of (1.7). The following well-known fixed-point theorem in cones is crucial to our arguments.

Theorem 3.1 (see [8]). Let 𝐸 be a Banach space and 𝐾⊂𝐸 be a cone. Suppose Ω₁ and Ω₂ are bounded open subsets of 𝐸 with 𝜃∈Ω₁,Ω₁⊂Ω₂. Assume that 𝐴∶𝐾∩(Ω₂⧵Ω₁)→𝐾 is a completely continuous operator such that either(i)‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁ and ‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂ or(ii)‖𝐴𝑢‖≥‖𝑢‖,𝑢∈𝐾∩𝜕Ω₁ and ‖𝐴𝑢‖≤‖𝑢‖,𝑢∈𝐾∩𝜕Ω₂.
Then 𝐴 has a fixed point in 𝐾∩(Ω₂⧵Ω₁).

Theorem 3.2. Assume that there exist 𝑎∈Λ⁺ and 0<𝑟<𝑅 such that 𝑓(𝑡)𝑚𝑡,𝑦+𝑎𝑦≥0,∀𝑦∈𝑀𝑀𝑟,𝑚𝑅,𝑡∈[1,𝑇]_ℤ. (3.1) If one of the following conditions holds:
(i)𝑓(𝑡)𝑀𝑡,𝑦+𝑎𝑦≥𝑇𝑚²𝑚𝑦,∀𝑦∈𝑀𝑟,𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≤𝑀𝑇𝑀𝑦,∀𝑦∈𝑅,𝑚𝑅,𝑡∈[1,𝑇]_ℤ, (3.2)(ii)𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≤𝑚𝑇𝑀𝑦,∀𝑦∈𝑀𝑟,𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)𝑀𝑡,𝑦+𝑎𝑦≥𝑇𝑚²𝑀𝑦,∀𝑦∈𝑅,𝑚𝑅,𝑡∈[1,𝑇]_ℤ, (3.3)then problem (1.7) has a positive solution.

Proof. From Corollary 2.5, we get that 𝑀>𝑚>0. It is easy to see that the equation Δ²𝑦(𝑡−1)=𝑓(𝑡,𝑦(𝑡)) is equivalent to Δ²𝑦(𝑡−1)+𝑎(𝑡)𝑦(𝑡)=𝑓𝑡,𝑦(𝑡)+𝑎(𝑡)𝑦(𝑡). (3.4) Define the open sets Ω₁=𝑦∈𝐸∶𝑦<𝑟,Ω₂=𝑦<𝑀𝑦∈𝐸∶𝑚𝑅, (3.5) and the cone 𝑃 in 𝐸, 𝑃=𝑦∈𝐸∶min_{[]𝑡∈0,𝑇+1ℤ}𝑦(𝑡)>𝑚𝑀𝑦. (3.6) Clearly, if 𝑦∈𝑃∩(Ω₂⧵Ω₁), then (𝑚/𝑀)𝑟≤𝑦(𝑡)≤(𝑀/𝑚)𝑅,forall𝑡∈[0,𝑇+1]_ℤ.
From Lemma 2.8, we define the operator 𝐴∶𝐸→𝐸 by 𝐴𝑦(𝑡)=^𝑇_𝑠=1𝐺(𝑡,𝑠)𝑓𝑠,𝑦(𝑠)+𝑎(𝑠)𝑦(𝑠),𝑡∈[0,𝑇+1]_ℤ. (3.7) From (3.1), if 𝑦∈𝑃∩(Ω₂⧵Ω₁), then (𝑡)≥𝑚𝐴𝑦𝑀𝑀^𝑇_𝑠=1𝑓(𝑠(𝑠)𝑦(𝑠>𝑚𝑠,𝑦)+𝑎)𝑀max_{[]𝑡∈0,𝑇+1ℤ}^𝑇_𝑠=1𝐺(𝑓(𝑠)(𝑠)𝑦(𝑠=𝑚𝑡,𝑠)𝑠,𝑦+𝑎)𝑀𝐴𝑦. (3.8) Thus 𝐴(𝑃∩(Ω₂⧵Ω₁))⊂𝑃. Moreover, 𝐸 is a finite space, it is easy to prove that 𝐴∶𝑃∩(Ω₂⧵Ω₁)→𝑃 is a completely continuous operator. Clearly, 𝑦 is the solution of problem (1.7) if and only if 𝑦 is the fixed point of the operator 𝐴.
We only prove (i). (ii) can be obtained by the similar method. If 𝑦∈𝜕Ω₁∩𝑃, then ‖𝑦‖=𝑟 and (𝑚/𝑀)𝑟≤𝑦(𝑡)≤𝑟 for all 𝑡∈[0,𝑇+1]_ℤ. Therefore, from (i), 𝐴𝑦(𝑡)≥𝑚^𝑇_𝑠=1𝑓(𝑠(𝑠)𝑦(𝑠≥𝑀𝑠,𝑦)+𝑎)𝑇𝑚^𝑇_𝑠=1𝑦(𝑠)≥𝑟=𝑦. (3.9) If 𝑦∈𝜕Ω₂∩𝑃, then ‖𝑦‖=(𝑀/𝑚)𝑅 and 𝑅≤𝑦(𝑡)≤(𝑀/𝑚)𝑅 for all 𝑡∈[0,𝑇+1]_ℤ. As a consequence, 𝐴𝑦(𝑡)≤𝑀^𝑇_𝑠=1𝑓(𝑠(𝑠)𝑦(𝑠≤1𝑠,𝑦)+𝑎)𝑇^𝑇_𝑠=1𝑦(𝑠)≤𝑦. (3.10) From Theorem 3.1, 𝐴 has a fixed point 𝑦∈𝑃∩(Ω₂⧵Ω₁) and satisfies 𝑚𝑀(𝑡)≤𝑀𝑟≤𝑦𝑚𝑅. (3.11) Therefore, 𝑦 is a positive solution of (1.7).

Similar to the proof of Theorem 3.2, we can prove the following.

Corollary 3.3. Assume that there exist 𝑎∈Λ⁺ and 0<𝑟<𝑅 such that 𝑓(𝑡)−𝑀𝑡,𝑦+𝑎𝑦≤0,∀𝑦∈𝑚𝑚𝑅,−𝑀𝑟,𝑡∈[1,𝑇]_ℤ. (3.12) If one of the following conditions holds
(i)𝑓(𝑡)𝑀𝑡,𝑦+𝑎𝑦≤𝑇𝑚²𝑚𝑦,∀𝑦∈−𝑟,−𝑀𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≥−𝑀𝑇𝑀𝑦,∀𝑦∈𝑚𝑅,−𝑅,𝑡∈[1,𝑇]_ℤ, (3.13)(ii)𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≥𝑚𝑇𝑀𝑦,∀𝑦∈−𝑟,−𝑀𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)𝑀𝑡,𝑦+𝑎𝑦≤𝑇𝑚²−𝑀𝑦,∀𝑦∈𝑚𝑅,−𝑅,𝑡∈[1,𝑇]_ℤ, (3.14)then (1.7) has a negative solution.

Applying the sign properties of 𝐺(𝑡,𝑠) when 𝑎∈Λ⁻ and the similar argument to prove Theorem 3.2 with obvious changes, we can prove the following.

Theorem 3.4. Assume that there exist 𝑎∈Λ⁻ and 0<𝑟<𝑅 such that 𝑓(𝑡)𝑀𝑡,𝑦+𝑎𝑦≤0,∀𝑦∈𝑚𝑚𝑟,𝑀𝑅,𝑡∈[1,𝑇]_ℤ. (3.15) If one of the following conditions holds
(i)𝑓(𝑡)𝑚𝑡,𝑦+𝑎𝑦≤𝑇𝑀²𝑀𝑦,∀𝑦∈𝑚𝑟,𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≥𝑚𝑇𝑚𝑦,∀𝑦∈𝑅,𝑀𝑅,𝑡∈[1,𝑇]_ℤ, (3.16)(ii)𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≥𝑀𝑇𝑚𝑦,∀𝑦∈𝑚𝑟,𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)𝑚𝑡,𝑦+𝑎𝑦≤𝑇𝑀²𝑚𝑦,∀𝑦∈𝑅,𝑀𝑅,𝑡∈[1,𝑇]_ℤ, (3.17)then (1.7) has a positive solution.

Proof. Since 𝑚<𝑀<0, define the open sets Ω₁=𝑦∈𝐸∶𝑦<𝑟,Ω₂=𝑦<𝑚𝑦∈𝐸∶𝑀𝑅, (3.18) and define the cone 𝑃 in 𝐸, 𝑃=𝑦∈𝐸∶min_{[]𝑡∈0,𝑇+1ℤ}𝑦(𝑡)>𝑀𝑚𝑦. (3.19) If 𝑥∈𝑃∩(Ω₂⧵Ω₁), then 𝑀𝑚(𝑡)≤𝑚𝑟≤𝑦𝑀𝑅,∀𝑡∈[0,𝑇+1]_ℤ. (3.20) Define the operator 𝐴 as (3.7), and the proof is analogous to that of Theorem 3.2 and is omitted.

Corollary 3.5. Assume that there exist 𝑎∈Λ⁻ and 0<𝑟<𝑅 such that 𝑓(𝑡)−𝑚𝑡,𝑦+𝑎𝑦≥0,∀𝑦∈𝑀𝑀𝑅,−𝑚𝑟,𝑡∈[1,𝑇]_ℤ. (3.21) If one of the following conditions holds
(i)𝑓(𝑡)𝑚𝑡,𝑦+𝑎𝑦≥𝑇𝑀²𝑀𝑦,∀𝑦∈−𝑟,−𝑚𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≤−𝑚𝑇𝑚𝑦,∀𝑦∈𝑀𝑅,−𝑅,𝑡∈[1,𝑇]_ℤ, (3.22)(ii)𝑓(𝑡)1𝑡,𝑦+𝑎𝑦≤𝑀𝑇𝑚𝑦,∀𝑦∈−𝑟,−𝑚𝑟,𝑡∈[1,𝑇]_ℤ,𝑓(𝑡)𝑚𝑡,𝑦+𝑎𝑦≥𝑇𝑀²−𝑚𝑦,∀𝑦∈𝑀𝑅,−𝑅,𝑡∈[1,𝑇]_ℤ, (3.23)then (1.7) has a negative solution.

Example 3.6. Let us consider the periodic boundary value problem Δ²𝑦(𝑛−1)=𝑓𝑛,𝑦(𝑛),𝑛∈[1,𝑇]_ℤ,𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇), (3.24) where 𝑓𝑛,𝑦=2𝑛²√+3𝑦,𝑦∈22𝑟,𝑟,𝑛∈[1,𝑇]_ℤ,2𝑛²+3𝑦𝑅−𝑦𝑅−𝑟−0.25𝑦𝑦−𝑟[][]𝑅−𝑟,𝑦∈𝑟,𝑅,𝑛∈1,𝑇_ℤ,−0.25𝑦,𝑦∈𝑅,√2𝑅,𝑛∈[1,𝑇]_ℤ. (3.25)
Consider the auxiliary problem Δ²𝑦(𝑛−1)+𝑎(𝑛)𝑦(𝑛)=𝑓𝑛,𝑦(𝑛)+𝑎(𝑛)𝑦(𝑛),𝑛∈[1,𝑇]_ℤ,𝑦(0)=𝑦(𝑇),Δ𝑦(0)=Δ𝑦(𝑇), (3.26) take 𝑟=2√2,𝑅=8√2,𝑇=3,𝑎(𝑛)≡𝑎=2−√3<4sin²(𝜋/2𝑇),𝜃=𝜋/6,cos𝜃=(2−𝑎)/2=√3/2,sin𝜃=1/2,𝑚=(1/2sin𝜃)cot(𝜃𝑇/2)=1,𝑀=1/(2sin𝜃sin(𝜃𝑇/2))=√2. By computing, 𝑓(𝑛,𝑦)+𝑎(𝑛)𝑦≥0,𝑦∈[(√2/2)𝑟,√2𝑅],𝑛∈[1,𝑇]_ℤ;𝑓(𝑛,𝑦)+𝑎(𝑛)𝑦≥(√2/3)𝑦,𝑦∈[(√2/2)𝑟,𝑟],𝑛∈[1,𝑇]_ℤ; 𝑓(𝑛,𝑦)+𝑎(𝑛)𝑦≤(1/3√2)𝑦,𝑦∈[𝑅,√2𝑅],𝑛∈[1,𝑇]_ℤ. Consequently, from Theorem 3.2, the problem (3.24) has a positive solution.

Acknowledgment

This work was supported by the NSFC (no. 11061030) and the Fundamental Research Funds for the Gansu Universities.