Existence of Periodic Solutions for a Class of Discrete Hamiltonian Systems

Abstract

By applying minimax methods in critical point theory, we prove the existence of periodic solutions for the following discrete Hamiltonian systems Δ²𝑢(𝑡−1)+∇𝐹(𝑡,𝑢(𝑡))=0, where 𝑡∈ℤ, 𝑢∈ℝ^𝑁, 𝐹∶ℤ×ℝ^𝑁→ℝ, 𝐹(𝑡,𝑥) is continuously differentiable in 𝑥 for every 𝑡∈ℤ and is 𝑇-periodic in 𝑡;𝑇 is a positive integer.

1. Introduction

Consider the following discrete Hamiltonian system:Δ²𝑢(𝑡−1)+∇𝐹(𝑡,𝑢(𝑡))=0,𝑡∈ℤ, (1.1) where Δ is the forward difference operator defined by Δ𝑢(𝑡)=𝑢(𝑡+1)−𝑢(𝑡),Δ²𝑢(𝑡)=Δ(Δ𝑢(𝑡)),𝑡∈ℤ,𝑢∈ℝ^𝑁,𝐹∶ℤ×ℝ^𝑁→ℝ,and 𝐹(𝑡,𝑥) is continuously differentiable in 𝑥 for every 𝑡∈ℤ and is 𝑇-periodic in 𝑡; 𝑇 is a positive integer.

Difference equations usually describe evolution of certain phenomena over the course of time. For example, if a certain population has discrete generations, the size of the (𝑡+1) th generation 𝑥(𝑡+1) is a function of the 𝑡th generation 𝑥(𝑡). In fact, difference equations provide a natural description of many discrete models in real world. Since discrete models exist in various fields of science and technology such as statistics, computer science, electrical circuit analysis, biology, neural network, optimal control, and so on, it is of practical importance to investigate the solutions of difference equations. For more details about difference equations, we refer the readers to the books [1–3].

In some recent papers [4–15], the authors studied the existence of periodic solutions and subharmonic solutions of difference equations by applying critical point theory. These papers show that the critical point theory is an effective method to the study of periodic solutions for difference equations. In 2007, Xue and Tang [11] investigated the existence of periodic solutions for (1.1) and obtained the main result.

Theorem A (see [11]). Suppose that 𝐹 satisfies the following conditions: (F1)there exists a positive constant 𝑇 such that 𝐹(𝑡+𝑇,𝑥)=𝐹(𝑡,𝑥) for all (𝑡,𝑥)∈ℤ×ℝ^𝑁;(F2)there are constants 𝐿₁>0,𝐿₂>0,and0≤𝛼<1 such that |∇𝐹(𝑡,𝑥)|≤𝐿₁|𝑥|^𝛼+𝐿₂,∀(𝑡,𝑥)∈ℤ[1,𝑇]×ℝ^𝑁, (1.2) where ℤ[𝑎,𝑏]∶=∩[𝑎,𝑏]   for every 𝑎,𝑏∈ℤ with 𝑎≤𝑏;(F3)|𝑥|^−2𝛼∑^𝑇_𝑡=1𝐹(𝑡,𝑥)→+∞ as |𝑥|→+∞ for all 𝑡∈ℤ[1,𝑇].
Then problem (1.1) has at least one periodic solution with period 𝑇.

Let𝐹(𝑡,𝑥)=𝑓(𝑡)|𝑥|^7/4+sin2𝜋𝑡𝑇+1|𝑥|^3/2+(ℎ(𝑡),𝑥), (1.3) where 𝑓∶ℤ[1,𝑇]→ℝ, 𝑓(𝑡+𝑇)=𝑓(𝑡), ℎ∶ℤ[1,𝑇]→ℝ^𝑁, and ℎ(𝑡+𝑇)=ℎ(𝑡). It is easy to see that|(≤7∇𝐹𝑡,𝑥)|4𝑓(𝑡)|𝑥|^3/4+32sin2𝜋𝑡𝑇+1|𝑥|^1/2+|ℎ(𝑡≤7)|4𝑓(𝑡)+𝜀|𝑥|^3/4+𝑎(𝜀)+|ℎ(𝑡)|,∀(𝑡,𝑥)∈ℤ[1,𝑇]×ℝ^𝑁, (1.4) where 𝜀>0 and 𝑎(𝜀) is a positive constant and is dependent on 𝜀. The above inequality shows that there are functions not satisfying condition (F2). If we let ∑^𝑇_𝑡=1𝑓(𝑡)=0, 𝛼=3/4, 𝑇=2, then we have |𝑥|^𝑇−2𝛼_𝑡=1𝐹(𝑡,𝑥)=2+ℎ(1)+ℎ(2),|𝑥|^−3/2𝑥. (1.5) But the above equality does not satisfy (F3). This example shows that it is valuable to further improve conditions (F2) and (F3).

Before stating our main results, we first introduce some preliminaries.

2. Preliminaries

Let 𝐻_𝑇=𝑢∶ℤ⟶ℝ^𝑁∣𝑢(𝑡+𝑇)=𝑢(𝑡),𝑡∈ℤ. (2.1)𝐻_𝑇 can be equipped with the inner product ⟨𝑢,𝑣⟩=^𝑇_𝑡=1(𝑢(𝑡),𝑣(𝑡)),∀𝑢,𝑣∈𝐻_𝑇, (2.2) by which the norm on 𝐻_𝑇 can be reduced by ‖𝑢‖=^𝑇_𝑡=1|𝑢(𝑡)|^21/2,∀𝑢∈𝐻_𝑇, (2.3) where (⋅,⋅) and |⋅| denote the usual inner product and the usual norm in ℝ^𝑁. It is easy to see that (𝐻_𝑇,⟨⋅,⋅⟩) is a finite-dimensional Hilbert space and linear homeomorphic to ℝ^𝑁𝑇. For any 𝑟>1, define ‖𝑢‖_𝑟=^𝑇_𝑡=1|𝑢(𝑡)|^𝑟1/𝑟,∀𝑢∈𝐻_𝑇. (2.4) Obviously, ‖𝑢‖=‖𝑢‖₂ and ‖𝑢‖ is equivalent to ‖𝑢‖_𝑟. Hence, there exist two positive constants 𝐶₁, 𝐶₂, which are independent on 𝑟, such that 𝐶₁‖𝑢‖_𝑟≤‖𝑢‖≤𝐶₂‖𝑢‖_𝑟,∀𝑢∈𝐻_𝑇. (2.5) If we define ‖𝑢‖_∞=sup_{𝑡∈ℤ[1,𝑇]}|𝑢(𝑡)|, it is easy to see that for any 𝑟>1,‖𝑢‖_∞≤‖𝑢‖_𝑟,∀𝑢∈𝐻_𝑇. (2.6)

For any 𝑢∈𝐻_𝑇, let𝜑(𝑢)1=−2^𝑇_𝑡=1|Δ𝑢(𝑡)|²+^𝑇_𝑡=1[𝐹(𝑡,𝑢(𝑡))−𝐹(𝑡,0)]. (2.7) We can compute the Fréchet derivative of (2.7) as 𝜕𝜑(𝑢)𝜕𝑢(𝑡)=Δ²𝑢(𝑡−1)+∇𝐹(𝑡,𝑢(𝑡)),𝑡∈ℤ[1,𝑇]. (2.8) Hence, 𝑢 is a critical point of 𝜑 on 𝐻_𝑇 if and only if Δ²𝑢(𝑡−1)+∇𝐹(𝑡,𝑢(𝑡))=0,𝑡∈ℤ[1,𝑇],𝑢∈ℝ^𝑁. (2.9) So, the critical points of 𝜑 are classical solutions of (1.1). The following lemmas are useful in our proof.

Lemma 2.1 (see [11]). As a subspace of 𝐻_𝑇, 𝑁_𝑘 is defined by 𝑁_𝑘∶=𝑢∈𝐻_𝑇∣−Δ²𝑢(𝑡−1)=𝜆_𝑘𝑢(𝑡), (2.10) where 𝜆_𝑘=2−2cos𝑘𝜔,𝜔=2𝜋/𝑇,𝑘∈ℤ[0,[𝑇/2]],[⋅] and denote the Gauss Function. Then there hold (i)𝑁_𝑘⟂𝑁_𝑗,𝑘≠𝑗,𝑗∈ℤ[0,[𝑇/2]]; (ii)𝐻_𝑇=⊕^[𝑇/2]_𝑘=0𝑁_𝑘.

Lemma 2.2 (see [11]). Define 𝐻_𝑘∶=⊕^𝑘_𝑗=0𝑁_𝑗, 𝐻^⟂_𝑘∶=⊕^[𝑇/2]_𝑗=𝑘+1𝑁_𝑗, 𝑘∈ℤ[0,[𝑇/2]−1]; then one has ^𝑇∑_𝑡=1|Δ𝑢(𝑡)|²≤𝜆_𝑘‖𝑢‖₂,∀𝑢∈𝐻_𝑘;_𝑇𝑡∑₌₁|Δ𝑢(𝑡)|²≥𝜆_𝑘+1‖𝑢‖₂,∀𝑢∈𝐻^⟂_𝑘. (2.11)(2.12)

3. Main Results and Proofs

Theorem 3.1. Suppose that 𝐹 satisfies (F1) and the following conditions (F2)′there are 𝑝,𝑞∶ℤ[1,𝑇]→ℝ⁺, 𝛼∈[0,1) such that |∇𝐹(𝑡,𝑥)|≤𝑝(𝑡)|𝑥|^𝛼+𝑞(𝑡),∀(𝑡,𝑥)∈ℤ[1,𝑇]×ℝ^𝑁, (3.1) where ℤ[𝑎,𝑏]∶=ℤ∩[𝑎,𝑏] for every 𝑎,𝑏∈ℤ with 𝑎≤𝑏;(F3)′liminf_|𝑥|→∞|𝑥|^−2𝛼∑^𝑇_𝑡=1𝐹(𝑡,𝑥)>((𝜆_[𝑇/2]+2𝜆₁)/2𝜆²₁)∑^𝑇_𝑡=1𝑝²(𝑡) for all 𝑡∈ℤ[1,𝑇].
Then problem (1.1) has at least one periodic solution with period 𝑇.

Theorem 3.2. Suppose that 𝐹 satisfies (F1) and (F2) with 𝛼=1. Moreover, assume the following conditions hold: ^𝑇_𝑡=1𝑝(𝑡)<𝜆₁, (3.2) and (F4)liminf_|𝑥|→∞|𝑥|⁻²∑^𝑇_𝑡=1𝐹(𝑡,𝑥)>((𝜆_[𝑇/2]+𝜆₁^1/2+∑^𝑇_𝑡=1𝑝(𝑡)+𝜆₁^1/2(𝜆₁−∑^𝑇_𝑡=1𝑝(𝑡)))/2𝜆₁(𝜆₁−∑^𝑇_𝑡=1𝑝(𝑡)))∑^𝑇_𝑡=1𝑝²(𝑡) for all 𝑡∈ℤ[1,𝑇].
Then problem (1.1) has at least one periodic solution with period 𝑇.

Remark 3.3. It is easy to see that (F2)′ is more general than (F2) and (F3)′ is weaker than (F3). Theorem 3.2 is a new result, which completes Theorem A when 𝛼=1.
For the sake of convenience, we denote 𝑀₁=^𝑇_𝑡=1𝑝²(𝑡)^1/2,𝑀₂=^𝑇_𝑡=1𝑝(𝑡),𝑀₃=^𝑇_𝑡=1𝑞(𝑡). (3.3)

Proof of Theorem 3.1. First we prove that 𝜑 satisfies the (PS) condition. Suppose that a consequence {𝑢_𝑛}⊂𝐻_𝑇 is such that −𝐶₃≤𝜑(𝑢_𝑛)≤𝐶₃, where 𝐶₃>0 and 𝜑^′(𝑢_𝑛)→0 as 𝑛→∞. Then for sufficiently large 𝑛, −‖𝑢‖≤𝜑^′𝑢_𝑛,𝑢≤‖𝑢‖. (3.4) From Lemma 2.1, we can write 𝑢 as 𝑢=𝑢+𝑢∈𝐻₀⨁𝐻^⟂₀, where 𝐻₀=𝑁₀,and𝐻^⟂₀=⨁^[𝑇/2]_𝑘=1𝑁_𝑘. From (F3)′, we can choose 𝑎₁>1/𝜆²₁>0 such that liminf_|𝑥|→∞|𝑥|^𝑇−2𝛼_𝑡=1𝐹()>𝜆𝑡,𝑥^[𝑇/2]𝑎₁2+√𝑎₁𝑀²₁. (3.5) From (F2)′, (2.6), Hölder inequality, and Young inequality, we have ^𝑇_𝑡=1(𝐹(𝑡,𝑢(𝑡))−𝐹(𝑡,𝑢))=^𝑇_𝑡=1¹₀(∇𝐹(𝑡,𝑢(𝑡,𝑢(𝑡≤𝑢+𝑠))))𝑑𝑠_𝑇𝑡=110|(∇𝐹(𝑡,𝑢(𝑡,𝑢(𝑡≤𝑢+𝑠))))|𝑑𝑠_𝑇𝑡=110𝑝(𝑡)|𝑢+𝑠𝑢(𝑡)|^𝛼(𝑡)|𝑢(𝑡≤+𝑞)|𝑑𝑠_𝑇𝑡=1𝑝(𝑡)|𝑢|^𝛼+|𝑢(𝑡)|^𝛼|𝑢(𝑡)|+^𝑇_𝑡=1𝑞(𝑡𝑢(𝑡≤|)|)|𝑢|^𝛼𝑇_𝑡=1𝑝²(𝑡)^1/2𝑇_𝑡=1|𝑢(𝑡)|^21/2+𝑀₂‖𝑢‖_∞^𝛼+1+𝑀₃‖𝑢‖_∞=𝑀₁|𝑢|^𝛼‖𝑢‖+𝑀₂‖𝑢‖_∞^𝛼+1+𝑀₃‖𝑢‖_∞≤12√𝑎₁‖𝑢‖²+√𝑎₁2𝑀²₁|𝑢|^2𝛼+𝑀₂‖𝑢‖^𝛼+1+𝑀₃‖𝑢‖. (3.6) In a similar way, we have ^𝑇_𝑡=1∇𝐹𝑡,𝑢_𝑛(𝑡),𝑢_𝑛(𝑡≤1)2𝑎₁𝜆₁𝑢_𝑛²+𝑎₁𝜆₁2𝑀²₁𝑢_𝑛^2𝛼+𝑀₂𝑢_𝑛^𝛼+1+𝑀₃𝑢_𝑛. (3.7) Let 𝑢_𝑛=𝑢_𝑛+𝑢_𝑛∈𝐻₀⨁𝐻^⟂₀. From (2.12) and (3.7), we have ^𝑇_𝑡=1Δ𝑢_𝑛(𝑡),Δ𝑢_𝑛(𝑡)=^𝑇_𝑡=1Δ𝑢_𝑛(𝑡),Δ𝑢_𝑛(𝑡)≥𝜆₁𝑢_𝑛²,_𝑇𝑡=1Δ𝑢_𝑛(𝑡),Δ𝑢_𝑛(𝑡)=−𝜑^′𝑢_𝑛,𝑢_𝑛+^𝑇_𝑡=1∇𝐹𝑡,𝑢_𝑛(𝑡),𝑢_𝑛(𝑡)≤𝑢_𝑛+12𝑎₁𝜆₁𝑢_𝑛²+𝑎₁𝜆₁2𝑀²₁𝑢_𝑛^2𝛼+𝑀₂𝑢_𝑛^𝛼+1+𝑀₃𝑢_𝑛. (3.8)(3.9) It follows from (3.8) and (3.9) that 𝜆₁2𝑢_𝑛²+𝐶₄≤𝑎₁𝜆₁2𝑀²₁𝑢_𝑛^2𝛼, (3.10) where 𝐶₄=min_{𝑠∈[0,+∞)}{((𝑎₁𝜆²₁−1)/2𝑎₁𝜆₁)𝑠²−𝑀₂𝑠^𝛼+1−(1+𝑀₃)𝑠}. The fact that 𝑎₁>1/𝜆²₁>0 implies that −∞<𝐶₄<0. So it follows from (3.10) that 𝑢_𝑛²≤𝑎₁𝑀²₁𝑢_𝑛^2𝛼−2𝐶₄𝜆₁, (3.11) and so 𝑢_𝑛≤√𝑎₁𝑀₁𝑢_𝑛^𝛼+𝐶₅, (3.12) where 𝐶₅>0. It follows from the boundedness of 𝜑(𝑢_𝑛), (2.11), (3.6), (3.11), and (3.12) that 𝐶₃(𝑢)1≥𝜑=−2^𝑇_𝑡=1Δ𝑢_𝑛(𝑡)²+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(𝑡(1)−𝐹𝑡,0)=−2^𝑇_𝑡=1Δ𝑢_𝑛(𝑡)²+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(𝑡)−𝐹𝑡,𝑢_𝑛+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(1−𝐹𝑡,0)≥−2𝜆_[𝑇/2]𝑢_𝑛−12√𝑎₁𝑢_𝑛²−√𝑎₁2𝑀²₁𝑢_𝑛^2𝛼−𝑀₂𝑢_𝑛^𝛼+1−𝑀₃𝑢_𝑛+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(1−𝐹𝑡,0)≥−2𝜆_[𝑇/2]+12√𝑎₁𝑎₁𝑀²₁𝑢_𝑛^2𝛼−2𝐶₄𝜆₁+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(−√−𝐹𝑡,0)𝑎₁2𝑀²₁𝑢_𝑛^2𝛼−𝑀₂√𝑎₁𝑀₁𝑢_𝑛^𝛼+𝐶₅^𝛼+1−𝑀₃√𝑎₁𝑀₁𝑢_𝑛^𝛼+𝐶₅≥−12𝜆_[𝑇/2]𝑎₁𝑀²₁−√𝑎₁𝑀²₁𝑢_𝑛^2𝛼+𝜆_[𝑇/2]𝐶₄𝜆₁+𝐶₄𝜆₁√𝑎₁−𝑀₂2^𝛼√𝑎₁𝑀₁^𝛼+1𝑢_𝑛^𝛼(𝛼+1)+2^𝛼𝐶₅^𝛼+1−𝑀₁𝑀₃√𝑎₁𝑢_𝑛^𝛼−𝑀₃𝐶₅+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(=−𝐹𝑡,0)𝑢_𝑛^2𝛼𝑢_𝑛^𝑇−2𝛼_𝑡=1𝐹𝑡,𝑢_𝑛−12𝜆_[𝑇/2]𝑎₁+√𝑎₁𝑀²₁−𝑀₁𝑀₃√𝑎₁𝑢_𝑛^−𝛼−𝑀₂2^𝛼√𝑎₁𝑀₁^𝛼+1𝑢_𝑛^{𝛼(𝛼−1)}+𝜆_[𝑇/2]𝐶₄𝜆₁+𝐶₄𝜆₁√𝑎₁−𝑀₃𝐶₅−𝑀₂2^𝛼𝐶₅^𝛼+1−^𝑇_𝑡=1𝐹(𝑡,0). (3.13) Inequalities (3.5) and (3.13) imply that {𝑢_𝑛} is bounded. Hence, {𝑢_𝑛} is bounded by (3.12), and then {𝑢_𝑛} is bounded. Since 𝐻_𝑇 is finite dimensional, there exists a subsequence of {𝑢_𝑛} convergent in 𝐻_𝑇. Thus, we conclude that (PS) condition is satisfied.
In order to use the saddle point theorem [16, Theorem  4.6], we only need to verify the following conditions: (I1)𝜑(𝑥)→+∞as|𝑥|→∞in𝐻₀; (I2)𝜑(𝑢)→−∞as‖𝑢‖→∞in𝐻^⟂₀.
In fact, from (F3)′, we have ^𝑇_𝑡=1𝐹(𝑡,𝑥)⟶+∞as|𝑥|⟶∞in𝐻₀. (3.14) For any 𝑥∈𝐻₀, since ∑^𝑇_𝑡=1|Δ𝑥|²=0, we have 𝜑(𝑥)=^𝑇_𝑡=1[𝐹(𝑡,𝑥)−𝐹(𝑡,0)]. (3.15) It follows from (3.14) and the above inequality that 𝜑(𝑥)⟶+∞as|𝑥|⟶∞in𝐻₀. (3.16) Thus (I1) is easy to verify.
Next, for all 𝑢∈𝐻^⟂₀, from (F2)′ and (2.6), we have ^𝑇_𝑡=1(𝐹(𝑡,𝑢(𝑡))−𝐹(𝑡,0))=^𝑇_𝑡=1¹₀(((𝑡(𝑡≤∇𝐹𝑡,𝑠𝑢)),𝑢))𝑑𝑠_𝑇𝑡=1𝑝(𝑡)|𝑢(𝑡)|^𝛼+1+^𝑇_𝑡=1𝑞(𝑡)|𝑢(𝑡)|≤𝑀₂‖𝑢‖_∞^𝛼+1+𝑀₂‖𝑢‖_∞≤𝑀₂‖𝑢‖^𝛼+1+𝑀₂‖𝑢‖. (3.17) By (2.7), (2.12), and (3.17), we obtain 𝜑(𝑢)1=−2^𝑇_𝑡=1|Δ𝑢(𝑡)|²+^𝑇_𝑡=1[𝐹((𝑡(1𝑡,𝑢))−𝐹𝑡,0)]≤−2𝜆₁‖𝑢‖²+𝑀₂‖𝑢‖^𝛼+1+𝑀₂‖𝑢‖. (3.18) Since 𝜆₁>0 and 𝛼∈[0,1), we have 𝜑(𝑢)→−∞ as ‖𝑢‖→∞ in 𝐻^⟂₀. The proof of Theorem 3.1 is complete.

Proof of Theorem 3.2. By (3.2) and (F4), we can choose an 𝑎₂∈ℝ such that 𝑎₂>1𝜆₁>0,liminf_|𝑥|→∞|𝑥|^𝑇−2_𝑡=1𝐹()>1𝑡,𝑥2𝜆_[𝑇/2]+12√𝑎₂+12𝑀₂𝑎₂𝜆₁−𝑀₂+√𝑎₂2𝑀²₁. (3.19)(3.20) It follows from (F2)′ with 𝛼=1, (2.6), Hölder inequality, and Young inequality that ^𝑇_𝑡=1(𝐹(𝑡,𝑢(𝑡))−𝐹(𝑡,𝑢))=^𝑇_𝑡=1¹₀(∇𝐹(𝑡,𝑢(𝑡,𝑢(𝑡≤𝑢+𝑠))))𝑑𝑠_𝑇𝑡=110|(∇𝐹(𝑡,𝑢(𝑡,𝑢(𝑡≤𝑢+𝑠))))|𝑑𝑠_𝑇𝑡=110𝑝(𝑡)(|𝑢|+𝑠|𝑢(𝑡)|)|𝑢(𝑡)|𝑑𝑠+^𝑇_𝑡=1¹₀𝑞(𝑡𝑢(𝑡=)|)|𝑑𝑠_𝑇𝑡=1𝑝(𝑡)|𝑢|+12|𝑢(𝑡)||𝑢(𝑡)|+^𝑇_𝑡=1𝑞(𝑡𝑢(𝑡≤|)|)|𝑢|^𝑇_𝑡=1𝑝²(𝑡)^1/2𝑇_𝑡=1|𝑢(𝑡)|^21/2+12‖𝑢‖²_∞^𝑇_𝑡=1𝑝(𝑡)+‖𝑢‖_∞^𝑇_𝑡=1𝑞(𝑡)=𝑀₁|𝑢𝑢‖+𝑀|‖²2‖𝑢‖²_∞+𝑀₃‖𝑢‖_∞≤12√𝑎₂+𝑀₂2‖𝑢‖²+√𝑎₂2𝑀²₁|𝑢|²+𝑀₃‖𝑢‖. (3.21) In a similar way, we have ^𝑇_𝑡=1∇𝐹𝑡,𝑢_𝑛(𝑡),𝑢_𝑛(𝑡≤1)2𝑎₂+𝑀₂2𝑢_𝑛²+𝑎₂2𝑀²₁𝑢_𝑛²+𝑀₃𝑢_𝑛. (3.22) From (3.8) and (3.22), we have 𝜆₁𝑢_𝑛²≤^𝑇_𝑡=1Δ𝑢_𝑛(𝑡),Δ𝑢_𝑛(𝑡𝜑)=−^′𝑢_𝑛,𝑢_𝑛+^𝑇_𝑡=1∇𝐹𝑡,𝑢_𝑛(𝑡),𝑢_𝑛(𝑡)≤𝑢_𝑛+12𝑎₂+𝑀₂2𝑢_𝑛²+𝑎₂2𝑀²₁𝑢_𝑛²+𝑀₃𝑢_𝑛. (3.23) It follow from (3.23) that 12𝜆₁−𝑀₂𝑢_𝑛²+𝐶₆≤𝑎₂2𝑀²₁𝑢_𝑛², (3.24) where 𝐶₆=min_{𝑠∈[0,+∞)}{((𝜆₁𝑎₂−1)/2𝑎₂)𝑠²−(1+𝑀₃)𝑠}. The fact that 𝑎₂>1/𝜆₁>0 implies that −∞<𝐶₆<0. So it follows from (3.24) that 𝑢_𝑛²≤𝑎₂𝜆₁−𝑀₂𝑀²₁𝑢_𝑛²−2𝐶₆𝜆₁−𝑀₂, (3.25) and so 𝑢_𝑛≤𝑎₂𝜆₁−𝑀₂𝑀₁𝑢_𝑛+𝐶₇, (3.26) where 𝐶₇>0. It follows from the boundedness of 𝜑(𝑢_𝑛), (2.11), (3.21), (3.25), and (3.26) that 𝐶₃(𝑢)1≥𝜑=−2^𝑇_𝑡=1Δ𝑢_𝑛(𝑡)²+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(𝑡(1)−𝐹𝑡,0)=−2^𝑇_𝑡=1Δ𝑢_𝑛(𝑡)²+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(𝑡)−𝐹𝑡,𝑢_𝑛+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(1−𝐹𝑡,0)≥−2𝜆_[𝑇/2]𝑢_𝑛²−12√𝑎₂+𝑀₂2𝑢_𝑛²−√𝑎₂2𝑀²₁𝑢_𝑛²−𝑀₃𝑢_𝑛+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(1−𝐹𝑡,0)≥−2𝜆_[𝑇/2]+12√𝑎₂+𝑀₂2𝑎₂𝜆₁−𝑀₂𝑀²₁𝑢_𝑛²−2𝐶₆𝜆₁−𝑀₂−√𝑎₂2𝑀²₁𝑢_𝑛²−𝑀₃𝑎₂𝜆₁−𝑀₂𝑀₁𝑢_𝑛+𝐶₇+^𝑇_𝑡=1𝐹𝑡,𝑢_𝑛(=−𝐹𝑡,0)𝑢_𝑛²𝑢_𝑛^𝑇−2𝛼_𝑡=1𝐹𝑡,𝑢_𝑛−12𝜆_[𝑇/2]+12√𝑎₂+12𝑀₂𝑎₂𝜆₁−𝑀₂𝑀²₁−√𝑎₂2𝑀²₁−𝑎₂𝜆₁−𝑀₂𝑀₁𝑀₃𝑢_𝑛⁻¹+𝜆_[𝑇/2]+1√𝑎₂+𝑀₂𝐶₆𝜆₁−𝑀₂−𝑀₃𝐶₇−^𝑇_𝑡=1𝐹(𝑡,0). (3.27) Inequalities (3.20) and (3.27) imply that {𝑢_𝑛} is bounded. Hence, {𝑢_𝑛} is bounded by (3.26), and then {𝑢_𝑛} is bounded. Since 𝐻_𝑇 is finite dimensional, there exists a subsequence of {𝑢_𝑛} convergent in 𝐻_𝑇. Thus, we conclude that (PS) condition is satisfied.
In the following, we prove that 𝜑 satisfies (I1) and (I2). In fact, from (F4), we have ^𝑇_𝑡=1𝐹(𝑡,𝑥)⟶+∞as|𝑥|⟶∞in𝐻₀. (3.28) It follows from (3.27) and ∑^𝑇_𝑡=1|Δ𝑥|²=0 that 𝜑(𝑥)=^𝑇_𝑡=1[𝐹(𝑡,𝑥)−𝐹(𝑡,0)]⟶+∞as|𝑥|⟶∞in𝐻₀. (3.29) Thus (I1) is easy to verify.
Next, for all 𝑢∈𝐻^⟂₀, from (F2)′ with 𝛼=1 and (2.6), we have ^𝑇_𝑡=1(𝐹(𝑡,𝑢(𝑡))−𝐹(𝑡,0))=^𝑇_𝑡=1¹₀(((𝑡(𝑡≤1∇𝐹𝑡,𝑠𝑢)),𝑢))𝑑𝑠2^𝑇_𝑡=1𝑝(𝑡)|𝑢(𝑡)|²+^𝑇_𝑡=1𝑞(𝑡𝑢(𝑡≤1)|)|2𝑀₂‖𝑢‖²_∞+𝑀₃‖𝑢‖_∞≤12𝑀₂‖𝑢‖²+𝑀₃‖𝑢‖. (3.30) By (2.7), (2.12), and (3.30), we obtain 𝜑(𝑢)1=−2^𝑇_𝑡=1|Δ𝑢(𝑡)|²+^𝑇_𝑡=1[𝐹((𝑡(1𝑡,𝑢))−𝐹𝑡,0)]≤−2𝜆₁‖𝑢‖²+𝑀₂2‖𝑢‖²+𝑀₂‖𝑢‖. (3.31) Since 𝜆₁>𝑀₂, we have 𝜑(𝑢)→−∞ as ‖𝑢‖→∞ in 𝐻^⟂₀. The proof of Theorem 3.2 is complete.

4. Examples

In this section, we give two examples to illustrate our results.

Example 4.1. Let 𝐹(𝑡,𝑥)=sin2𝜋𝑡𝑇|𝑥|^7/4+sin2𝜋𝑡𝑇+1|𝑥|^3/2+(ℎ(𝑡),𝑥), (4.1) where ℎ∶ℤ[1,𝑇]→ℝ^𝑁 and ℎ(𝑡+𝑇)=ℎ(𝑡). It is easy to see that |(≤7∇𝐹𝑡,𝑥)|4sin2𝜋𝑡𝑇|𝑥|^3/4+32sin2𝜋𝑡𝑇+1|𝑥|^1/2+|ℎ(𝑡≤7)|4sin2𝜋𝑡𝑇+𝜀|𝑥|^3/4+𝑎(𝜀)+|ℎ(𝑡)|,∀(𝑡,𝑥)∈ℤ[1,𝑇]×ℝ^𝑁, (4.2) where 𝜀>0 and 𝑎(𝜀) is a positive constant and is dependent on 𝜀. It is easy to see that 𝐹(𝑡,𝑥) satisfies (F1). From (4.2), we can let 𝑝, 𝑞, and 𝛼 be 𝑝(𝑡)=74sin2𝜋𝑡𝑇(𝑡)(𝜀)+|ℎ(𝑡3+𝜀,𝑞=𝑎)|,𝛼=4, (4.3) which shows that (F2)′ is satisfied. Moreover, if we let 𝑇=2, then we have liminf_{|𝑥|→+∞}|𝑥|^𝑇−2𝛼_𝑡=1𝐹()𝜆𝑡,𝑥=2,₁=𝜆_[𝑇/2]𝜆=4,^[𝑇/2]+2𝜆₁2𝜆²₁^𝑇_𝑡=1𝑝²(𝑡)=147128^𝑇_𝑡=1sin2𝜋𝑡𝑇+𝜀²=147𝜀128². (4.4) If we let 𝜀²<256/147, then we obtain liminf_{|𝑥|→+∞}|𝑥|^𝑇−2𝛼_𝑡=1𝐹(𝑡,𝑥)=2>147𝜀128²=𝜆_[𝑇/2]+2𝜆₁2𝜆²₁^𝑇_𝑡=1𝑝²(𝑡), (4.5) which shows that (F3)' holds. Then from Theorem 3.1, problem (1.1) has at least one periodic solution with period 𝑇.

Example 4.2. Let 𝐹()=1𝑡,𝑥4sin2𝜋𝑡𝑇+12|𝑥|²+sin2𝜋𝑡𝑇|𝑥|^3/2+(ℎ(𝑡),𝑥), (4.6) where ℎ∶ℤ[1,𝑇]→ℝ^𝑁 and ℎ(𝑡+𝑇)=ℎ(𝑡). It is easy to see that 𝐹(𝑡,𝑥) satisfies (F1) and |(≤1∇𝐹𝑡,𝑥)|2sin2𝜋𝑡𝑇+12𝑥|+3|2sin2𝜋𝑡𝑇|𝑥|^1/2+|ℎ(𝑡≤1)|2sin2𝜋𝑡𝑇+12+𝜀|𝑥|+𝑏(𝜀)+|ℎ(𝑡)|,∀(𝑡,𝑥)∈ℤ[1,𝑇]×ℝ^𝑁, (4.7) where 𝜀>0 and 𝑏(𝜀) is a positive constant and is dependent on 𝜀. The above shows that (F2)′ holds with 𝛼=1 and 𝑝(𝑡)=12sin2𝜋𝑡𝑇+12+𝜀,𝑞(𝑡)=𝑏(𝜀)+|ℎ(𝑡)|. (4.8) Let 𝑇=2, then 𝜆₀=0,𝜆₁=𝜆_[𝑇/2]=4. Observe that |𝑥|^𝑇−2_𝑡=1𝐹(𝑡,𝑥)=|𝑥|^𝑇−2_𝑡=114sin2𝜋𝑡𝑇+12|𝑥|²+sin2𝜋𝑡𝑇|𝑥|^3/2+(ℎ(𝑡)=1,𝑥)4+^𝑇_𝑡=1ℎ(𝑡),|𝑥|⁻²𝑥. (4.9) On the other hand, we have ^𝑇_𝑡=1𝑝(𝑡)=^𝑇_𝑡=112sin2𝜋𝑡𝑇+12=1+𝜀2+𝜀,_𝑇𝑡=1𝑝²(𝑡)=^𝑇_𝑡=114sin2𝜋𝑡𝑇+12+𝜀²=1212+𝜀². (4.10) We can choose 𝜀 sufficiently small such that ∑^𝑇_𝑡=1𝑝(𝑡)<4 and liminf_{|𝑥|→+∞}|𝑥|^𝑇−2_𝑡=1𝐹()=1𝑡,𝑥4>27−2𝜀()1167−𝜀2+𝜀²=𝜆_[𝑇/2]+𝜆₁^1/2+∑^𝑇_𝑡=1𝑝(𝑡)+𝜆₁^1/2𝜆₁−∑^𝑇_𝑡=1𝑝(𝑡)2𝜆₁𝜆₁−∑^𝑇_𝑡=1𝑝(𝑡)^𝑇_𝑡=1𝑝²(𝑡), (4.11) which shows that (F4) holds. Then from Theorem 3.2, problem (1.1) has at least one periodic solution with period 𝑇.

Acknowledgment

X. H. Tang is supported by the NNSF (no. 10771215) of China.