Existence of Periodic Solutions for a Class of Fourth-Order Difference Equation

We apply the continuation theorem of Mawhin to ensure that a fourth-order nonlinear di ﬀ erence equation of the form Δ 4 u ð k − 2 Þ − a ð k Þ u α ð k Þ + b ð k Þ u β ð k Þ = 0 with periodic boundary conditions possesses at least one nontrivial positive solution, where Δ u ð k Þ = u ð k + 1 Þ − u ð k Þ is the forward di ﬀ erence operator, α > 0, β > 0 and α ≠ β , a ð k Þ , b ð k Þ are T -periodic functions and a ð k Þ b ð k Þ > 0 . As applications, we give some examples to illustrate the application of these theorems.


Introduction
In recent years, the theory of nonlinear difference equations has been widely used in the study of discrete models in the fields of economics, neural networks, ecology, etc. For the general background of difference equations, in particular, there are many authors who have discussed the existence and multiplicity of periodic solutions for discrete boundary value problems by exploiting various methods, including the method of upper and lower solutions, Leray-Schauder degree, fixed point theory, critical theory, and variational methods; see Bereanu and Mawhin [1], Cabada and Dimitrov [2], Graef et al. [3,4], and Cai et al. [5][6][7][8][9][10] and the references therein.
Let ℕ + , Z, and ℝ denote the sets of all positive integers, integers, and real numbers, respectively. This paper considers the following fourth-order nonlinear difference equation: where α > 0, β > 0 and α ≠ β, aðkÞ, bðkÞ are T-periodic functions and ΔuðkÞ = uðk + 1Þ − uðkÞ is the forward difference operator. Equation (1) can be considered as a discrete analogue of a special case of the following fourth-order nonlinear differ-ential equation: which has been studied in [11,12] when α = 1, β = 3. In [13], Yang and Han proved the existence of a periodic solution to equation (2) when α = n, β = n + 2, where n is a positive integer.
When aðkÞ ≡ 0, β = 1, Peterson and Ridenhour [14] considered the disconjugacy of the following equation: In 2005, Cai et al. [5] studied the fourth-order nonlinear difference equation By applying the linking theorem, they obtained some criteria for the existence and multiplicity of periodic solutions of equation (4).
In fact, there is a big difference between the continuous case and the discrete case. For example, the basic ideas of calculus are not always applicable when studying difference equations, such as the embedding theorem. Therefore, we need to consider other methods to deal with the difference problem. The main tool used is the continuation theorem of Mawhin (see [15]).
Motivated by the above works, the main aim of this paper is to investigate the existence of at least one positive T-periodic solution of (1). In order to obtain the main results of (1), we assume that the coefficient functions aðkÞ and bðkÞ satisfy the following condition: F 1 : Suppose aðkÞ, bðkÞ are T − periodic functions and a ðkÞbðkÞ > 0 for all k ∈ Z. Furthermore, we assume that there exist positive constants a, A, b, B such that Let X be all real T-periodic sequences of the form u = fuðkÞg k∈Z . Then, X is a Banach space under the norm kuk The main results in this paper are stated next: Theorems 1 and 2. Theorem 1. Let F 1 hold, if α < β and the period T satisfies where R 1 = ðA/bÞ 1/ðβ−αÞ + ρ and ρ > 0 small enough such that ða/BÞ 1/ðβ−αÞ − ρ > 0; then, equation (1) admits at least one positive T-periodic solution.
Theorem 2. Let F 1 hold, if α > β and the period T satisfies where Q 1 = ðB/aÞ 1/ðα−βÞ + τ and τ > 0 small enough such that ðb/AÞ 1/ðα−βÞ − τ > 0; then, equation (1) admits at least one positive T-periodic solution. This paper is organized as follows: in Section 2, we give some lemmas needed to prove the main results. Section 3 contains the proof of Theorem 1. Section 4 contains the proof of Theorem 2. Section 5 contains the proof of Theorem 3.

Preliminary Results
In this section, we introduce some notations and wellknown results which will be used in the subsequent sections.
Definition 4 (see [7], p. 12, B.1). Let X, Y be real Banach spaces, L : DomL ⊂ X ⟶ Y be a linear mapping. The mapping L is said to be a Fredholm mapping of index zero if If L is a Fredholm mapping of index zero, then there exist continuous projectors P : X ⟶ X and Q : Y ⟶ Y such that Im P = Ker L, It follows that the restriction has an inverse which is denoted by K P .
Lemma 6 (Mawhin's continuation theorem, see [7], Theorem IV.1). Let L be a Fredholm mapping of index zero, Ω ⊂ X is an open bounded set, and let N be L − compact on Ω. Suppose Then, the equation Lu = Nu has at least one solution in Ω ∩ Dom L.
Define the operator L : X ⟶ X by setting Direct calculation shows that Thus, Note that fuðkÞg k∈Z ∈ X, it follows that Furthermore, direct calculation shows that By virtue of the above facts, we have Since dim X = T and L is a linear mapping, by the knowledge of linear algebra, we know that dim Ker L ⊕ dim Im L = dim X. It is easy to see that dim Ker L = codim Im L = 1 and dim Im L = T − 1. It follows that Im L is closed in X. Therefore, the operator L is a Fredholm operator with index zero.
Let us define N : X ⟶ X by We define P : X ⟶ Ker L and Q : X ⟶ X as follows: The operators P and Q are projections. Hence, It follows that Lj Dom L∩Ker P : ðI − PÞX ⟶ Im L has an inverse which is denoted by K P .
In view of (17) and (18), for any u ∈ X, we can see that Since the Banach space X is finite dimensional, K P is linear. By virtue of the relations (20) and (21), we see that QN and K P ðI − QÞN are continuous on X. Hence, we know that if Ω is an open and bounded subset of X, then QNð ΩÞ is bounded. It follows that is compact. Therefore, the mapping N is L-compact on Ω with any open and bounded subset Ω ⊂ X.
Lemma 7 (see [16], Lemma 2.3). Let fuðkÞg k∈Z be a real T -periodic sequence; then, 3. Proof of Theorem 1 Proof. The content of Theorem 1 is as follows: let F 1 hold, if α < β and the period T satisfies where R 1 = ðA/bÞ 1/ðβ−αÞ + ρ and ρ > 0 small enough such that ða/BÞ 1/ðβ−αÞ − ρ > 0; then, equation (1) admits at least one positive T-periodic solution. Now, we prove that the conclusion holds. We assume that α < β. From condition F 1 , we know that aðkÞbðkÞ > 0, which include both positive and negative cases. So we need to classify the cases where both aðkÞ and bðkÞ are positive and both negative. Let which is an open set in X, where where ρ > 0 small enough such that ða/BÞ 1/ðβ−αÞ − ρ > 0.
Obviously, H 1 and R 1 are well defined.

Journal of Function Spaces
By α < β, 0 < a ≤ a k ð Þ ≤ A, we obtain uniformly for k ∈ Z. Furthermore, it follows from (26), (27), and (29) that Therefore, We prove that condition (1) of Lemma 6 holds. Let 0 < λ < 1 and u be such that Summing from 2 to T + 1, we can see that Firstly, we claim that for each λ ∈ ð0, 1Þ and u ∈ ∂Ω 1 ∩ Dom L, Lu ≠ λNu. In fact, in view of (25), if u ∈ ∂Ω 1 , then kuk = H 1 or kuk = R 1 . We only prove the case of kuk = H 1 , similar to the proof of kuk = R 1 . When This is a contradiction.
If u ∈ ∂Ω 1 ∩ Ker L, then u = fH 1 g k∈Z or u = fR 1 g k∈Z . By virtue of (31), we conclude that Hence, QNu ≠ 0 for each u ∈ ∂Ω 1 ∩ Ker L. Next let us consider ðH 1 + R 1 Þ/2, the arithmetic mean of H 1 and R 1 . We define G : X × ℝ ⟶ X as follows, for all μ ∈ ½0, 1, Clearly, we find that By using the homotopy invariance theorem, it is easy to see that Therefore, conditions (1)-(3) of Lemma 6 hold for Ω 1 . Furthermore, according to the above reasoning, we deduce that (1) has at least one positive solution in Ω 1 : If the coefficient functions aðkÞ, bðkÞ are negative T -periodic functions, in view of F 1 , we have that −A ≤ aðkÞ ≤ −a < 0 and −B ≤ bðkÞ ≤ −b < 0. LetãðkÞ = −aðkÞ,bðkÞ = −bðkÞ; then, we see that It is obvious that (1) is equivalent to the equation Let 0 < λ < 1 and u be such that Summing from 2 to T + 1, we can see that Firstly, we claim that for each λ ∈ ð0, 1Þ and u ∈ ∂Ω 1 ∩ Dom L, Lu ≠ λNu. In fact, in view of (25), if u ∈ ∂Ω 1 , then kuk = H 1 or kuk = R 1 . We only prove the case of kuk = H 1 , similar to the proof of kuk = R 1 .
If max 2≤l,j≤T+1 Further, we have This is a contradiction. If max 2≤l,j≤T+1 juðlÞ − uðjÞj ≥ H 1 /2, we see from Lemma 7 that Journal of Function Spaces but this is a contradiction. Hence, for all u ∈ ∂Ω 1 and λ ∈ ð0, 1Þ, we have Therefore, we verify that condition (1) of Lemma 6 holds for Ω 1 .
The remaining proof is similar to the proof of Case 1, and so we omit it. Furthermore, according to the above reasoning, we deduce that (44) has at least one positive solution in Ω 1 :
The remaining proof is similar to the proof of Theorem 1, and so we omit it. Furthermore, we deduce that (1) has at least one positive T-periodic solution in Ω 2 : Case 2. If the coefficient functions aðkÞ, bðkÞ are negative T -periodic functions, we have that −A ≤ aðkÞ ≤ −a < 0 and − B ≤ bðkÞ ≤ −b < 0. LetãðkÞ = −aðkÞ,bðkÞ = −bðkÞ. Then, we can see that It is obvious that uniformly for k ∈ Z. The remaining proof is similar to the proof of Theorem 1, and so we omit it. Furthermore, we conclude that (1) has at least one positive T-periodic solution in Ω 2 :

Proof of Theorem 3
Proof. The content of Theorem 3 is as follows: suppose aðk ÞbðkÞ ≤ 0 and aðkÞ, bðkÞ are not identical to zero for all k ∈ Z; then, equation (1) has no positive solution. Summing equation (1) from 2 to T + 1, we obtain that In view of Hence, If aðkÞ > 0 and bðkÞ ≤ 0, it follows from (64) that (1) does not have any positive solution. Other cases are similar. 6 Journal of Function Spaces

Example
Therefore, we can prove that (65) has at least one positive T-periodic solution in Ω 1 , where Example 2. The difference equation

Data Availability
Data sharing does not apply to this article as no data set was generated or analyzed during the current study.

Conflicts of Interest
The authors declare that they have no competing interests.