Existence of Subharmonic Periodic Solutions to a Class of Second-Order Non-Autonomous Neutral Functional Differential Equations

Abstract

By introducing subdifferentiability of lower semicontinuous convex function 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)) and its conjugate function, as well as critical point theory and operator equation theory, we obtain the existence of multiple subharmonic periodic solutions to the following second-order nonlinear nonautonomous neutral nonlinear functional differential equation 𝑥^′′(𝑡)+𝑥^′′(𝑡−2𝜏)+𝑓(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏),𝑥(𝑡−2𝜏))=0, 𝑥(0)=0.

1. Introduction

The existence of periodic solutions for differential system has received a great deal of attention in the last few decades. Different from ordinary differential equations and partial differential equations that do not contain delay variate, it is very difficult to study the existence of periodic solutions for functional differential equations. For this reason, many mathematicians developed different approaches such as the averaging method [1], the Massera-Yoshizawa theory [2, 3], the Kaplan-York [4] method of coupled systems, the Grafton cone mapping method [5], the Nussbaum method of fixed point theory [6] and Mawhin [7] coincidence degree theory.

However, the critical point theory was rarely used in the literature, and most of the existing results were established for autonomous functional differential equations while little was done for nonautonomous equations via critical point theory.

In this paper, by using critical point and operator equation theories, we study the existence of the following second-order nonlinear and nonautonomous mixed-type functional differential equation:𝑥^′′(𝑡)+𝑥^′′()((𝑡)()(𝑥(0)𝑡−2𝜏+𝑓𝑡,𝑥,𝑥𝑡−𝜏,𝑥𝑡−2𝜏))=0,=0.(1.1)

Our basic assumptions are the following (𝐴₁)∼(𝐴₃) and (𝐴₄)∼(𝐴₆) to be given in Section 4:(𝐴₁)𝑓(𝑡,𝑥₁,𝑥₂,𝑥₃)∈𝐶(ℝ⁴,ℝ), and 𝜕𝑓(𝑡,𝑥₁,𝑥₂,𝑥₃)/𝜕𝑡≠0;(𝐴₂) there exists a continuously differentiable function 𝐹(𝑡,𝑥₁,𝑥₂)∈𝐶¹(ℝ³,ℝ) such that 𝐹^′₂𝑡,𝑥₁,𝑥₂+𝐹^′₁𝑡,𝑥₂,𝑥₃=𝑓𝑡,𝑥₁,𝑥₂,𝑥₃,(1.2) where 𝐹^′₂(𝑡,𝑥₁,𝑥₂) and 𝐹^′₁(𝑡,𝑥₂,𝑥₃) denote 𝜕𝐹(𝑡,𝑥₁,𝑥₂)/𝜕𝑥₂ and 𝜕𝐹(𝑡,𝑥₂,𝑥₃)/𝜕𝑥₂, respectively;(𝐴₃)𝐹(𝑡+𝜏,𝑥₁,𝑥₂)=𝐹(𝑡,𝑥₁,𝑥₂) for all 𝑥₁,𝑥₂∈ℝ.

2. Variational Structure

Fix 𝛾>1,𝜏>0, where 𝛾 is a positive integer and 𝐻¹₀0,2𝛾𝜏=𝑥(𝑡)∈𝐿²0,2𝛾𝜏∣𝑥^′∈𝐿²(𝑡)𝑥(0)(𝑡).0,2𝛾𝜏,𝑥is2𝛾𝜏-periodicfunctionin𝑡=0,and𝑥hascompactsupporton0,2𝛾𝜏(2.1)

It is obvious that 𝐻¹₀[0,2𝛾𝜏] is a Sobolev space by defining the inner product (⋅,⋅) and the norm ‖⋅‖ as follows: 𝑥,𝑦_{𝐻10[0,2𝛾𝜏]}=₀^2𝛾𝜏𝑥^′(𝑡)𝑦^′(𝑡)‖𝑥‖𝑑𝑡,_{𝐻10[0,2𝛾𝜏]}=₀^2𝛾𝜏𝑥^′(𝑡)²𝑑𝑡^1/2,∀𝑥,𝑦∈𝐻¹₀0,2𝛾𝜏.(2.2)

Moreover, 𝑥(𝑡)∈𝐻¹₀[0,2𝛾𝜏] can be expressed as 𝑥(𝑡)=𝑎₀+^∞_𝑘=1𝑎_𝑘cos𝑘𝜋𝛾𝜏𝑡+𝑏_𝑘sin𝑘𝜋𝑡𝛾𝜏.(2.3)

Let us consider the functional 𝐼(𝑥) defined on 𝐻¹₀[0,2𝛾𝜏] by 𝐼(𝑥)=₀^2𝛾𝜏𝑥^′(𝑡)𝑥^′(𝑡−𝜏)−𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑑𝑡.(2.4)

For all 𝑥,𝑦∈𝐻¹₀[0,2𝛾𝜏] and 𝜀>0, we know that 𝐼𝑥+𝜀𝑦=𝐼(𝑥)+𝜀₀^2𝛾𝜏𝑥^′(𝑡)𝑦^′(𝑡−𝜏)+𝑥^′(𝑡−𝜏)𝑦^′(𝑡)−𝐹𝑡,𝑥(𝑡)+𝜀𝑦(𝑡),𝑥(𝑡−𝜏)+𝜀𝑦(𝑡−𝜏)−𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑑𝑡+𝜀²₀^2𝛾𝜏𝑦^′(𝑡)𝑦^′(𝑡−𝜏)𝑑𝑡.(2.5)

It is then easy to see that 𝐼^′(𝑥),𝑦=₀^2𝛾𝜏𝑥^′(𝑡)𝑦^′(𝑡−𝜏)+𝑥^′(𝑡−𝜏)𝑦^′(𝑡)−𝐹^′₁(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑦(𝑡)−𝐹^′₂(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑦(𝑡−𝜏)𝑑𝑡,(2.6) where 𝐼^′(𝑥) denotes the Frechet differential of the function 𝐼(𝑥). By the periodicity of 𝐹(𝑡,𝑢₁,𝑢₂), 𝑥(𝑡), and 𝑦(𝑡), we have ₀^2𝛾𝜏𝑥^′(𝑡)𝑦^′(𝑡−𝜏)𝑑𝑡=₀^2𝛾𝜏𝑥^′(𝑡)𝑑𝑦(𝑡−𝜏)=𝑥^′(𝑡)𝑦(𝑡−𝜏)₀^2𝛾𝜏,−₀^2𝛾𝜏𝑥^′′(𝑡)𝑦(𝑡−𝜏)𝑑𝑡=−₀^2𝛾𝜏𝑥^′′(𝑡+𝜏)𝑦(𝑡)𝑑𝑡,_02𝛾𝜏𝑥^′(𝑡−𝜏)𝑦^′(𝑡)𝑑𝑡=−₀^2𝛾𝜏𝑥^′′(𝑡−𝜏)𝑦(𝑡)𝑑𝑡.(2.7) Similarly, we have ₀^2𝛾𝜏𝐹^′₂(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑦(𝑡−𝜏)𝑑𝑡=^{(2𝛾−1)𝜏}_−𝜏𝐹^′₂(()(𝑡𝑦(𝑡)=𝑡+𝜏,𝑥𝑡+𝜏,𝑥))𝑑𝑡_02𝛾𝜏𝐹^′₂(𝑡,𝑥(𝑡+𝜏),𝑥(𝑡))𝑦(𝑡)𝑑𝑡.(2.8) Hence, 𝐼^′(𝑥),𝑦=₀^2𝛾𝜏−𝑥^′′(𝑡+𝜏)−𝑥^′′(𝑡−𝜏)−𝐹^′_𝑢1(𝑡,𝑥(𝑡)𝑥(𝑡−𝜏))−𝐹^′_𝑢2(𝑡,𝑥(𝑡+𝜏),𝑥(𝑡))𝑦(𝑡)𝑑𝑡.(2.9) Therefore, the Euler equation corresponding to the functional 𝐼(𝑥) is 𝑥^′′(𝑡+𝜏)+𝑥^′′(𝑡−𝜏)+𝐹^′_𝑢1(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))+𝐹^′_𝑢2(𝑡,𝑥(𝑡+𝜏),𝑥(𝑡))=0.(2.10)

It is not difficult to see that (2.10) is equivalent to (1.1). Thus, system (1.1) is the Euler equation of the functional 𝐼(𝑥). It follows that it is possible to obtain 2𝛾𝜏-periodic solutions of system (1.1) by seeking critical points of the functional 𝐼(𝑥).

Since 𝐼(𝑥) has neither a supremum nor an infimum, we do not seek critical points of the functional 𝐼(𝑥) by the extremum method. But we may use operator equation theory. First via the dual variational principle, we obtain new operator equations (see (4.16)) related to (1.1). Then solutions to system (1.1) are obtained by seeking critical points of operator equation (4.16).

In this paper, our main tool is the following.

Lemma 2.1 (Maintain Pass Theorem). Let 𝐻 be a real Banach space. If 𝐼(⋅)∈𝐶¹(𝐻,ℝ) satisfies the Palais-Smale condition as well as the following additional conditions:(1)there exist constants 𝜌>0 and 𝑎>0 such that 𝐼(𝑥)≥𝑎,forall𝑥∈𝜕𝐵_𝜌, where 𝐵_𝜌={𝑥∈𝐻∶‖𝑥‖_𝐻<𝜌},(2)𝐼(𝜃)≤0 and there exists 𝑥₀∈𝐵_𝜌 such that 𝐼(𝑥₀)≤0, then 𝑐=inf_ℎ∈Γsup_𝑠∈[0,1]𝐼(ℎ(𝑠)) is a positive critical value of 𝐼, where Γ=ℎ∈𝐶([0,1],𝐻)∣ℎ(0)=𝜃,ℎ(1)=𝑥₀.(2.11)

The rest of this paper is organized as follows. Subdifferentiability of lower semicontinuous convex function 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)) and its conjugate function are introduced in Section 3. In Section 4, we first give the definition of the weak solution to (1.1), then we establish the new operator equation (4.16) related to (1.1) by the conjugate function of 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)) and show that we can obtain solution to (1.1) from the solution to operator equation (4.16). In Section 5, by seeking critical points of operator equation (4.16), we obtain the result that there exist multiple subharmonic periodic solutions to system (1.1). Finally in Section 6, an example and a remark are given to illustrate our result.

3. The Subdifferentiability and the Conjugate Function of the Lower Semicontinuous Convex Function 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))

Let 𝑋 be a space of all given 𝑛×𝜏-periodic functions in 𝑡 and a Banach space, where 𝑛∈𝑁 is a positive integer. Denote ℝ=ℝ∪{+∞}. Let 𝜑∶𝑋²→ℝ be a lower semicontinuous convex function. Generally, 𝜑 is not always differentiable in conventional sense, but we may generalize the definition of “derivative” as follows.

Definition 3.1. Let (𝑥^∗₁,𝑥^∗₂)∈𝑋^∗×𝑋^∗. We say that (𝑥^∗₁,𝑥^∗₂) is a subgradient of 𝜑 at point (𝑥₀(𝑡),𝑥₀(𝑡−𝜏))∈𝑋×𝑋 if 𝜑𝑥₀(𝑡),𝑥₀(𝑡−𝜏)+𝑥^∗₁,𝑥(𝑡)−𝑥₀(𝑡)+𝑥^∗₂,𝑥(𝑡−𝜏)−𝑥₀(𝑡−𝜏)≤𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)).(3.1) For all 𝑥₀(𝑡)∈𝑋, the set of all subgradients of 𝜑 at point (𝑥₀(𝑡),𝑥₀(𝑡−𝜏)) will be called the sub-differential of 𝜑 at point (𝑥₀(𝑡),𝑥₀(𝑡−𝜏)) and will be denoted by 𝜕𝜑(𝑥₀(𝑡),𝑥₀(𝑡−𝜏)).
By the definition of Subdifferentiability of function 𝜑, we may define its conjugate function 𝜑^∗ by 𝜑^∗𝑥^∗₁,𝑥^∗₂=sup𝑥^∗₁,𝑥(𝑡)+𝑥^∗₂,𝑥(𝑡−𝜏)−𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)),(3.2) where ⟨⋅⟩ denotes the duality relation of 𝑋^∗ and 𝑋. So it is not difficult to obtain the following propositions.

Proposition 3.2. 𝜑^∗ is a lower semicontinuous convex function (𝜑^∗ may have functional value +∞, but not functional value −∞).

Proposition 3.3. If 𝜑≤𝜓, then 𝜑^∗≥𝜓^∗.

Proposition 3.4 (Yang inequality). One has 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))+𝜑^∗𝑥^∗₁,𝑥^∗₂≥𝑥^∗₁,𝑥(𝑡)+𝑥^∗₂,𝑥(𝑡−𝜏).(3.3)

Proposition 3.5. One has 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))+𝜑^∗𝑥^∗₁,𝑥^∗₂=𝑥^∗₁,𝑥(𝑡)+𝑥^∗₂(⟺𝑥,𝑥𝑡−𝜏)_∗1,𝑥^∗₂∈𝜕𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)).(3.4)

Proposition 3.6. 𝜑^∗ does not always equal +∞.

Proof. Let 𝑥₀(𝑡)∈𝑋 and 𝛽∈ℝ such that 𝜑(𝑥₀(𝑡),𝑥₀(𝑡−𝜏))<+∞ and 𝛽₀<𝜑(𝑥₀(𝑡),𝑥₀(𝑡−𝜏)). We consider the two convex sets in 𝑋²×ℝ defined by 𝐴=epi𝜑≜𝑥(𝑡),𝑥(𝑡−𝜏),𝛽∈𝑋²(𝑥(𝑡)((𝑥(𝑡)(,𝑥×ℝ∣∶𝜑,𝑥𝑡−𝜏))<+∞,𝛽>𝜑,𝑥𝑡−𝜏))𝐵=₀(𝑡),𝑥₀(𝑡−𝜏),𝛽₀.(3.5) By Hahn-Banach Theorem, we know that there exists (𝑓₁,𝑓₂,𝑘)∈𝑋^∗×𝑋^∗×ℝ and 𝛼∈ℝ such that 𝑓₁,𝑥(𝑡)+𝑓₂(𝑥(𝑡)()𝑓,𝑥𝑡−𝜏)+𝑘𝛽>𝛼,∀,𝑥𝑡−𝜏,𝛽∈eip𝜑,₁,𝑥₀(𝑡)+𝑓₂,𝑥₀(𝑡−𝜏)+𝑘𝛽₀<𝛼.(3.6) So, we have 𝑓₁,𝑥₀(𝑡)+𝑓₂,𝑥₀(𝑡−𝜏)+𝑘𝜑𝑥₀(𝑡),𝑥₀(𝑡−𝜏)>𝛼>𝑓₁,𝑥₀(𝑡)+𝑓₂,𝑥₀(𝑡−𝜏)+𝑘𝛽₀.(3.7) Thus 𝑘>0, and −1𝑘𝑓₁(𝑡+−1,𝑥)𝑘𝑓₂((𝑥(𝑡)(𝛼,𝑥𝑡−𝜏)−𝜑,𝑥𝑡−𝜏))<−𝑘.(3.8)

Since 𝜑^∗ is a lower semicontinuous convex function that does not always equal +∞, we may define its conjugate function 𝜑^∗∗ by 𝜑^∗∗𝑥₁^∗∗,𝑥₂^∗∗=sup_{𝑥∗1,𝑥∗2∈𝑋∗×𝑋∗}𝑥₁^∗∗,𝑥^∗₁+𝑥₂^∗∗,𝑥^∗₂−𝜑𝑥^∗₁,𝑥^∗₂,(3.9) where ⟨⋅⟩ denotes the duality relation of 𝑋^∗∗ and 𝑋^∗.

Theorem 3.7. Let 𝜑 be a lower semicontinuous convex function that does not always equal +∞, then 𝜑^∗∗=𝜑.

Proof. We divide our proof into two parts. First we show that 𝜑^∗∗=𝜑 holds when 𝜑>0 and then 𝜑^∗∗=𝜑 holds for all lower semicontinuous convex functions 𝜑 that do not always equal +∞.
(i) The case when 𝜑>0.
From the definition of 𝜑^∗∗ and Yang inequality, it is obvious that 𝜑^∗∗≤𝜑 holds.
Next, to prove 𝜑^∗∗≥𝜑 holds, suppose to the contrary that there exist a point (𝑥₀(𝑡),𝑥₀(𝑡−𝜏))∈𝑋², such that 𝜑^∗∗(𝑥₀(𝑡),𝑥₀(𝑡−𝜏))<𝜑(𝑥₀(𝑡),𝑥₀(𝑡−𝜏)) holds.
Consider the two convex sets 𝐴=eip𝜑≜𝑥(𝑡),𝑥(𝑡−𝜏),𝛽∈𝑋²(𝑥(𝑡)((𝑥(𝑡)(,𝐵×ℝ∣𝜑,𝑥𝑡−𝜏))<+∞,𝛽≥𝜑,𝑥𝑡−𝜏))₀=𝑥₀(𝑡),𝑥₀(𝑡−𝜏),𝜑^∗∗𝑥₀(𝑡),𝑥₀(𝑡−𝜏).(3.10) By the Hahn-Banach Theorem, we know that there exist (𝑔₁,𝑔₂,𝑘^∗)∈𝑋^∗×𝑋^∗×ℝ and 𝛼₁∈ℝ such that 𝑔₁,𝑥(𝑡)+𝑔₂,𝑥(𝑡−𝜏)+𝑘^∗𝛽>𝛼₁𝑥(𝑡)()𝑔,∀,𝑥𝑡−𝜏,𝛽∈eip𝜑,(3.11)₁,𝑥₀(𝑡)+𝑔₂,𝑥₀(𝑡−𝜏)+𝑘^∗𝜑^∗∗𝑥₀(𝑡),𝑥₀(𝑡−𝜏)<𝛼₁.(3.12) So, it follows that 𝑘^∗≥0. Let 𝜀>0. Using 𝜑>0 and (3.11), one obtains that 𝑔₁,𝑥(𝑡)+𝑔₂,𝑥(𝑡−𝜏)+𝑘^∗+𝜀𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))≥𝛼₁,∀(𝑥(𝑡),𝑥(𝑡−𝜏))∈𝐷𝜑.(3.13) Thus, we have 𝜑^∗−𝑔₁𝑘^∗𝑔+𝜀,−²𝑘^∗𝛼+𝜀≤−¹𝑘^∗+𝜀.(3.14) Then, by the definition of 𝜑^∗∗, we obtain that 𝜑^∗∗𝑥₀(𝑡),𝑥₀()≥−𝑔𝑡−𝜏¹𝑘^∗+𝜀,𝑥₀(𝑡+−𝑔)²𝑘^∗+𝜀,𝑥₀(𝑡−𝜏)−𝜑^∗−𝑔₁𝑘^∗𝑔+𝜀,−²𝑘^∗≥−𝑔+𝜀₁𝑘^∗+𝜀,𝑥₀(𝑡+−𝑔)²𝑘^∗+𝜀,𝑥₀(+𝛼𝑡−𝜏)¹𝑘^∗.+𝜀(3.15) That is to say, 𝑔₁,𝑥₀(𝑡)+𝑔₂,𝑥₀(𝑡−𝜏)+𝑘^∗+𝜀𝜑^∗∗𝑥₀(𝑡),𝑥₀(𝑡−𝜏)≥𝛼₁,∀𝜀>0,(3.16) which is a contradiction to (3.12).
(ii) For all 𝜑, by Proposition 3.6, we know that 𝐷(𝜑^∗)≠∅. Choose (𝑥^∗₁₀,𝑥^∗₂₀)∈𝐷(𝜑^∗), and define function 𝜑 by 𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))=𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))−𝑥^∗₁₀,𝑥(𝑡)−𝑥^∗₂₀,𝑥(𝑡−𝜏)+𝜑^∗𝑥^∗₁₀,𝑥^∗₂₀.(3.17) Then 𝜑 is a lower semicontinuous convex function that is not always equal to +∞ and satisfies 𝜑≥0. By the result of (i), we have that 𝜑^∗∗=𝜑. On the other hand, we have 𝜑^∗𝑥^∗₁,𝑥^∗₂=𝜑^∗𝑥^∗₁+𝑥^∗₁₀,𝑥^∗₂+𝑥^∗₂₀−𝜑^∗𝑥^∗₁₀,𝑥^∗₂₀,𝜑^∗∗(𝑥(𝑡),𝑥(𝑡−𝜏))=𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))−𝑥^∗₁₀,𝑥(𝑡)−𝑥^∗₂₀,𝑥(𝑡−𝜏)+𝜑^∗𝑥^∗₁₀,𝑥^∗₂₀.(3.18) That is, 𝜑^∗∗=𝜑.

Corollary 3.8. Let 𝜑 be a lower semicontinuous convex function that is not always equal to +∞. Then (𝑥^∗₁,𝑥^∗₂)∈𝜕𝜑(𝑥(𝑡),𝑥(𝑡−𝜏)) if and only if (𝑥(𝑡),𝑥(𝑡−𝜏))∈𝜕𝜑^∗𝑥^∗₁,𝑥^∗₂.(3.19)

Proof. One has 𝑥^∗₁,𝑥^∗₂∈𝜕𝜑(𝑥(𝑡),𝑥(𝑡−𝜏))⟺𝑥^∗₁,𝑥(𝑡)+𝑥^∗₂((𝑥(𝑡)(,𝑥𝑡−𝜏)=𝜑,𝑥𝑡−𝜏))+𝜑^∗𝑥^∗₁,𝑥^∗₂=𝜑^∗∗(𝑥(𝑡),𝑥(𝑡−𝜏))+𝜑^∗𝑥^∗₁,𝑥^∗₂⟺(𝑥(𝑡),𝑥(𝑡−𝜏))∈𝜕𝜑^∗𝑥^∗₁,𝑥^∗₂.(3.20)

4. Weak Solution to (1.1)

Define an operator 𝐴=𝑑²/𝑑𝑡². By (2.6) and ⟨𝑢(𝑡),𝐴(𝜔(𝑡))⟩=₀^2𝛾𝜏𝑢(𝑡)𝜔^′′(𝑡)𝑑𝑡=𝑢(𝑡)𝜔^′(𝑡)₀^2𝛾𝜏−₀^2𝛾𝜏𝜔^′(𝑡)𝑢^′(𝑡)𝑑𝑡=−₀^2𝛾𝜏𝑢^′(𝑡)𝑑𝜔(𝑡)=−𝑢^′(𝑡)𝜔(𝑡)₀^2𝛾𝜏+₀^2𝛾𝜏𝜔(𝑡)𝑑𝑢^′(𝑡)=⟨𝐴𝑢(𝑡),𝜔(𝑡)⟩(4.1) as well as ⟨𝑢(𝑡),𝐴(𝜔(𝑡−𝜏))⟩=⟨𝐴𝑢(𝑡+𝜏),𝜔(𝑡)⟩,⟨𝑢(𝑡−𝜏),𝐴(𝜔(𝑡))⟩=⟨𝐴(𝑢(𝑡−𝜏)),𝜔(𝑡)⟩(4.2) we may define a weak solution to (1.1) as follows.

Definition 4.1. For 𝑢∈𝐿^𝑝[0,2𝛾𝜏], we say that 𝑢 is a weak solution to (1.1), if ⟨𝑢(𝑡),𝐴(𝜔(𝑡−𝜏))⟩+⟨𝑢(𝑡−𝜏),𝐴(𝜔(𝑡))⟩+𝜔(𝑡),𝐹^′₁((𝑡)(+𝜔()𝑡,𝑢,𝑢𝑡−𝜏))𝑡−𝜏,𝐹^′₂(𝑡,𝑢(𝑡),𝑢(𝑡−𝜏))=0,(4.3)for all 𝜔(𝑡)∈𝐷(𝐴)∩𝐿^𝑝[0,2𝛾𝜏], where ⟨𝑢(𝑡),𝜐(𝑡)⟩=₀^2𝛾𝜏𝑢(𝑡)𝜐(𝑡)𝑑𝑡,(4.4) when 𝑢(𝑡)∈𝐿^𝑝[0,2𝛾𝜏],𝜐(𝑡)∈𝐿^𝑞[0,2𝛾𝜏], where 2<𝑝<+∞,(1/𝑝)+(1/𝑞)=1.
Our objective is to define the conjugate function of 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)) by using the definition of Subdifferentiability of lower semicontinuous convex function and by making use of the dual variational structure. So we add three more conditions on function 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)) as follows: (𝐴₄)𝑢=(𝑢₁,𝑢₂)→𝐹(𝑡,𝑢₁,𝑢₂) is a continuously differentiable and strictly convex function and satisfies 𝐹(𝑡,0,0)=0,𝐹^′₁(𝑡,0,0)=𝐹^′₂(𝑡,0,0)=0∀𝑡∈0,2𝛾𝜏;(4.5)(𝐴₅) for 𝛼₂=1/𝑝, there exist constants 𝑀,𝐶>0, such that when |𝑢|=𝑢²₁+𝑢²₂≥𝐶 we have 𝐹𝑡,𝑢₁,𝑢₂≤𝛼₂𝐹^′₁𝑡,𝑢₁,𝑢₂𝑢₁+𝐹^′₂𝑡,𝑢₁,𝑢₂𝑢₂,𝐹𝑡,𝑢₁,𝑢₂≤𝑀|𝑢|^(1/𝛼2);(4.6)(𝐴₆)lim_|𝑢|→0𝐹𝑡,𝑢₁,𝑢₂|𝑢|²=0.(4.7)
The conjugate function of function 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)) is defined by 𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))=sup_{𝑥(𝑡)∈𝐿𝑝0,2𝛾𝜏}{⟨𝜔(𝑡),𝑥(𝑡)⟩+⟨𝜔(𝑡−𝜏),𝑥(𝑡−𝜏)⟩−𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))},(4.8) for 𝑡∈[0,2𝛾𝜏].
Then 𝐻 is a continuously differentiable and strictly convex function. By the duality principle (Corollary 3.8), we have that (𝜔(𝑡),𝜔(𝑡−𝜏))=𝐹^′₁(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)),𝐹^′₂((𝑡)(⟺𝐻𝑡,𝑥,𝑥𝑡−𝜏))_′1(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)),𝐻^′₂(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))=(𝑥(𝑡),𝑥(𝑡−𝜏)),(4.9) where 𝐻^′₁(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)) and 𝐻^′₂(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)) denote 𝜕𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))/𝜕𝜔(𝑡) and 𝜕𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))/𝜕𝜔(𝑡−𝜏), respectively.

Example 4.2. Let 𝐹(𝑥(𝑡),𝑥(𝑡−𝜏))=(1/𝑝)(𝑥²(𝑡)+𝑥²(𝑡−𝜏))^𝑝. Then 𝐻(𝜔(𝑡)(=1,𝜔𝑡−𝜏))𝑞𝜔²(𝑡)+𝜔²(𝑡−𝜏)^𝑞.(4.10)

Proof. The above expression holds since 𝐻(𝜔(𝑡),𝜔(𝑡−𝜏))=sup_{𝑥(𝑡)∈𝐿𝑝[0,2𝛾𝜏]}𝜔(𝑡)(𝑡+⟨𝜔()(−1⟨,𝑥)⟩𝑡−𝜏,𝑥𝑡−𝜏)⟩𝑝𝑥²(𝑡)+𝑥²(𝑡−𝜏)^𝑝=sup_𝜆>0𝜔²(𝑡)+𝜔²()1𝑡−𝜏𝜆−𝑝𝜆^𝑝=1𝑞𝜔²(𝑡)+𝜔²(𝑡−𝜏)^𝑞.(4.11)

Let 𝑅(𝐴) denote the value field of operator 𝐴. Then 𝑅(𝐴) is a closed set. Let 𝑃 be the orthogonal projection operator of 𝑅(𝐴) and 𝐾=𝐴⁻¹𝑃. Then it is not difficult to see that 𝐾 maps continuous continuation into a compact operator of 𝐿^𝑞[0,2𝛾𝜏]→𝐿^𝑞[0,2𝛾𝜏].

Let 𝐸=(𝜐(𝑡),𝜐(𝑡−𝜏))∈𝐿^𝑞0,2𝛾𝜏²𝜐(0)𝜙(𝑡)(𝑡=𝜙(𝑡)(=𝜙()(𝑡(𝑡)(𝐴)=0∣,𝜐),𝜐𝑡−𝜏)𝑡−𝜏,𝜐)=0,∀𝜙∈ℜ∩𝐿^𝑝0,2𝛾𝜏,𝜙(0)=0,(4.12) where ℜ(𝐴)={𝑢∈𝐷(𝐴)|𝐴(𝑢(𝑡)+𝑢(𝑡−2𝜏))=0}.

Remark 4.3. In fact, for all 𝑥(0)=0,𝑥(𝑡)∈𝐿^𝑝[0,2𝛾𝜏], or 𝑥(𝑡)∈𝐿^𝑞[0,2𝛾𝜏], 𝑥(𝑡) can be expressed as 𝑥(𝑡)=𝑎₀+^∞_𝑘=1𝑎_𝑘cos𝑘𝜋𝛾𝜏𝑡+𝑏_𝑘sin𝑘𝜋𝑡𝛾𝜏.(4.13) So, it follows that 𝜐(𝑡),𝜙(𝑡)=0⟺𝜙(𝑡−𝜏),𝜐(𝑡)=𝜙(𝑡),𝜐(𝑡−𝜏)=𝜙(𝑡),𝜐(𝑡)=0.(4.14) Thus 𝐸 can also be expressed as 𝐸=(𝜐(𝑡),𝜐(𝑡−2𝜏))∈𝐿^𝑞0,2𝛾𝜏×𝐿^𝑞(0)𝜙(𝑡)(𝑡(𝑡)(𝐴)0,2𝛾𝜏,𝑣=0∣,𝜐)=0,∀𝜙∈ℜ∩𝐿^𝑝0,2𝛾𝜏,𝜙(0)=0.(4.15) We want {(𝜐(𝑡),𝜐(𝑡−𝜏)),(𝜒(𝑡),𝜒(𝑡−𝜏))} to satisfy 𝜒(𝑡)=𝐾(𝜐(𝑡−𝜏))+𝐻^′₁((𝑡)(,𝜒()=𝐾(𝜐(𝑡𝑡,𝜐,𝜐𝑡−𝜏))𝑡−𝜏))+𝐻^′₂(𝑡,𝜐(𝑡),𝜐(𝑡−𝜏)),(4.16) where (𝜐(𝑡),𝜐(𝑡−𝜏))∈𝐸,𝜒(𝑡)∈ℜ(𝐴)∩𝐿^𝑝[0,2𝛾𝜏], that is, (𝜒(𝑡),𝜒(𝑡−𝜏))∈𝐸^⟂.
If {(𝜐(𝑡),𝜐(𝑡−𝜏)),(𝜒(𝑡),𝜒(𝑡−𝜏))} is a solution to (4.16) and if 𝑢(𝑡)=𝐻^′₁(𝑡,𝜐(𝑡),𝜐(𝑡−𝜏)), 𝑢(𝑡−𝜏)=𝐻^′₂(𝑡,𝜐(𝑡),𝜐(𝑡−𝜏)), then by the duality principle and (4.16), we have ⟨𝑢(𝑡),𝐴(𝑧(𝑡−𝜏))⟩+⟨𝑢(𝑡−𝜏),𝐴(𝑧(𝑡))⟩+𝑧(𝑡),𝐹^′₁(𝑡,𝑢(𝑡),𝑢(𝑡−𝜏))+𝑧(𝑡−𝜏),𝐹^′₂((𝑡)(=𝐻𝑡,𝑢,𝑢𝑡−𝜏))_′1𝑡,𝜐(𝑡),𝜐^′(𝑡−𝜏),𝐴(𝑧(𝑡−𝜏))+𝐻^′₂((𝑡)((𝑧(𝑡+⟨𝑧(𝑡)(𝑡+⟨𝑧()(=𝜒(𝑡)−𝐾(𝜐((𝑧(+𝜒()−𝐾(𝜐(𝑡(𝑧(𝑡+⟨𝑧(𝑡)(𝑡+⟨𝑧(𝑡)(𝑡⟨𝜐(𝑡)(𝑡−⟨𝜐(𝑡)(𝑡+⟨𝑧(𝑡)(𝑡+⟨𝑧(𝑡)(𝑡(𝑡)(𝐴)𝑡,𝜐,𝜐𝑡−𝜏)),𝐴)),𝜐)⟩𝑡−𝜏,𝜐𝑡−𝜏)⟩𝑡−𝜏)),𝐴𝑡−𝜏))𝑡−𝜏)),𝐴)),𝜐)⟩,𝜐)⟩=−,𝑧)⟩,𝑧)⟩,𝜐)⟩,𝜐)⟩=0∀𝑧∈𝐷∩𝐿^𝑝0,2𝛾𝜏.(4.17) Thus, 𝑢(𝑡) is a weak solution to (1.1).

5. Existence of Solutions to Operator Equation (4.16)

In this section, we discuss the existence of solutions to operator equation (4.16) by using the critical point theory. Our main result is the following.

Theorem 5.1. Under assumptions (𝐴₁)∼(𝐴₆), problem (1.1) has at least one nontrivial weak 2𝛾𝜏-periodic solution.
To prove this theorem, we state and prove the following lemmas first.
Let 𝜐=(𝜐(𝑡),𝜐(𝑡−𝜏)), and 𝐾𝜐(𝑡)𝜐()=0𝐾𝜐(𝑡)𝜐()=()𝑡−𝜏𝐾0𝑡−𝜏(𝑡)𝐾𝜐𝑡−𝜏𝐾𝜐.(5.1)

It is not difficult to verify that ⟨𝐾(𝜐),𝜓⟩=⟨𝜐,𝐾(𝜓)⟩=⟨𝐾𝜐(𝑡−𝜏),𝜓(𝑡)⟩+⟨𝐾𝜐(𝑡),𝜓(𝑡−𝜏)⟩, where 𝜓=(𝜓(𝑡),𝜓(𝑡−𝜏)), that is, the operator 𝐾 is a symmetric operator.

We may obtain solutions to (4.16) by seeking critical points of functional 𝐽(𝜐) defined by 𝐽(𝜐)=12⟨𝐾(𝜐),𝜐⟩+₀^2𝛾𝜏𝐻()=1𝑡,𝜐𝑑𝑡2()(𝑡+1𝐾𝜐𝑡−𝜏,𝜐)2(𝑡)(+𝐾𝜐,𝜐𝑡−𝜏)_02𝛾𝜏𝐻(𝑡,𝜐(𝑡),𝜐(𝑡−𝜏))𝑑𝑡∀(𝜐(𝑡),𝜐(𝑡−𝜏))∈𝐸.(5.2)

𝐽 may be regarded as the restriction to 𝐸 of function 𝐽̂ defined on 𝐿^𝑞[0,2𝛾𝜏]×𝐿^𝑞[0,2𝛾𝜏] since both functions have identical component on 𝐸. Moreover 𝐽^′̂(𝜐)=𝐾(𝜐)+𝐻^′(𝜐).(5.3) Since 𝐽^′̂(𝜐)−𝐽^′(𝜐),𝑧=0,∀𝜐∈𝐸,𝑧=(𝑧(𝑡),𝑧(𝑡−𝜏))∈𝐸,(5.4) there exist 𝜒(𝑡)∈ℜ(𝐴) and 𝜒_𝜐=(𝜒_𝜐(𝑡),𝜒_𝜐(𝑡−𝜏))∈𝐸^⟂ such that 𝐽^′̂(𝜐)−𝐽^′(𝜐)=𝜒_𝜐.(5.5)

So if 𝜐^∗ is a critical point of 𝐽^′(𝜐^∗)=0 on 𝐸, then there exists 𝜒^∗_𝜐∗=(𝜒^∗_𝜐∗(𝑡),𝜒^∗_𝜐∗(𝑡−𝜏))∈𝐸^⟂ such that 𝐾𝜐^∗+𝐻^′∗=𝜒^∗_𝜐∗.(5.6) Hence, {𝜐^∗,𝜒^∗_𝜐∗} is a solution to (4.16). That is, {(𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)),(𝜒^∗_𝜐∗(𝑡),𝜒^∗_𝜐∗(𝑡−𝜏))} is a solution to (4.16).

Lemma 5.2. The following two conditions are equivalent:(1)𝐹(𝑡,𝑢₁,𝑢₂)≤𝛼₂[𝐹^′₁(𝑡,𝑢₁,𝑢₂)𝑢₁+𝐹^′₂(𝑡,𝑢₁,𝑢₂)𝑢₂],forall𝑡∈[0,2𝛾𝜏], when |𝑢|=𝑢²₁+𝑢²₂≥𝐶;(2)𝐹(𝑡,𝛽𝑢₁,𝛽𝑢₂)≥𝛽^1/𝛼2𝐹(𝑡,𝑢₁,𝑢₂)>0,forall𝛽≥1,𝑡∈[0,2𝛾𝜏],|𝑢|≥𝐶.

Proof. For all 𝑢=(𝑢₁,𝑢₂),|𝑢|≥𝐶, let Φ(𝛽)=𝐹(𝑡,𝛽𝑢₁,𝛽𝑢₂),Ψ(𝛽)=𝛽^1/𝛼2𝐹(𝑡,𝑢₁,𝑢₂).
(2)⇒(1). By Φ(𝛽)≥Ψ(𝛽),forall𝛽≥1 and Φ(1)=Ψ(1), it is easy to see that Φ^′(1)≥Ψ^′(1). That is, 𝐹^′₁𝑡,𝑢₁,𝑢₂𝑢₁+𝐹^′₂𝑡,𝑢₁,𝑢₂𝑢₂≥1𝛼₂𝐹𝑡,𝑢₁,𝑢₂.(5.7)(1)⇒(2). By Φ^′𝛽=𝐹^′₁𝑡,𝛽𝑢₁,𝛽𝑢₂𝑢₁+𝐹^′₂𝑡,𝛽𝑢₁,𝛽𝑢₂𝑢₂=1𝛽𝐹^′₁𝑡,𝛽𝑢₁,𝛽𝑢₂𝛽𝑢₁+𝐹^′₂𝑡,𝛽𝑢₁,𝛽𝑢₂𝛽𝑢₂≥1𝛼₂𝛽Φ𝛽,(5.8) it follows that 𝐹𝑡,𝛽𝑢₁,𝛽𝑢₂≥𝛽^1/𝛼2𝐹𝑡,𝑢₁,𝑢₂>0,∀𝛽≥1,𝑡∈0,2𝛾𝜏.(5.9)

Lemma 5.3. Let 𝐹(𝑡,𝑢₁,𝑢₂) satisfy assumptions (𝐴₄) and (𝐴₅). Then there exist constants 𝑚>0 and 𝑀>0 such that 𝐹𝑡,𝑢₁,𝑢₂≥𝑚𝑢²₁+𝑢²₂^1/𝛼2|𝑢|𝐹∀𝑡∈0,2𝛾𝜏,when≥𝐶,_′𝑡,𝑢₁,𝑢₂≤2^1/𝛼2𝑀−𝑚𝑢²₁+𝑢²₂^{(1/𝛼2)−1},when|𝑢|≥𝐶,(5.10) where |𝐹^′(𝑡,𝑢₁,𝑢₂)|=|𝐹^′₁(𝑡,𝑢₁,𝑢₂)|²+|𝐹^′₂(𝑡,𝑢₁,𝑢₂)|².

Proof. Let 𝑚=min_{𝑢1,𝑢2∈𝜕𝐵𝐶}𝐹𝑡,𝑢₁,𝑢₂𝐶^1/𝛼2,(5.11) where 𝐵_𝐶 denotes the ball of radius 𝐶 centered at the origin. By (𝐴₄), one knows 𝑚>0. On the other hand, by Lemma 5.2 and (𝐴₅), we have 𝐹𝑡,𝑢₁,𝑢₂≥𝐹𝑡,𝐶𝑢₁𝑢²₁+𝑢²₂,𝐶𝑢₂𝑢²₁+𝑢²₂𝑢²₁+𝑢²₂𝐶^1/𝛼2≥𝑚𝑢²₁+𝑢²₂^1/𝛼2.(5.12)
By the convexity of function 𝐹, one has 𝐹𝑡,𝑢₁,𝑢₂+𝐹^′₁𝑡,𝑢₁,𝑢₂𝑧₁−𝑢₁+𝐹^′₂𝑡,𝑢₁,𝑢₂𝑧₂−𝑢₂≤𝐹𝑡,𝑧₁,𝑧₂.(5.13)
Let 𝑧=(𝑧₁,𝑧₂) run all over the ball 𝐵_|𝑢|(𝑧) of radius |𝑢| centered at 𝑢=(𝑢₁,𝑢₂), and choose the maximum of 𝐹^′₁(𝑡,𝑢₁,𝑢₂)(𝑧₁−𝑢₁)+𝐹^′₂(𝑡,𝑢₁,𝑢₂)(𝑧₂−𝑢₂). Then it is not difficult to see that 𝐹^′𝑡,𝑢₁,𝑢₂𝑢²₁+𝑢²₂≤𝑀𝑧²₁+𝑧²₂^1/𝛼2−𝑚𝑢²₁+𝑢²₂^1/𝛼2.(5.14) By 𝑧≤2|𝑢| we have that 𝐹^′𝑡,𝑢₁,𝑢₂≤2^1/𝛼2𝑀−𝑚𝑢²₁+𝑢²₂^{(1/𝛼2)−1}.(5.15)

Lemma 5.4. 𝐻∈𝐶¹(ℝ³,ℝ) is a strictly convex function and satisfies (1)𝐻^′₁(𝑡,0,0)=𝐻^′₂(𝑡,0,0)=0,𝐻(𝑡,0,0)=0,∀𝑡∈0,2𝛾𝜏;(5.16)(2)𝐶_𝛼2𝑀|𝜔|^{1/(1−𝛼2)}−𝐶₁((𝑡)(≤𝐶≤𝐻𝑡,𝜔,𝜔𝑡−𝜏))^𝛼2𝑚|𝜔|^{1/(1−𝛼2)}+𝐶₂;(5.17)(3)𝐶^′_𝛼2|𝜔|^{𝛼2/(1−𝛼2)}−𝐶₄≤𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))≤𝐶_𝛼22^{1/(1−𝛼2)}𝑚−1𝑀×|𝜔|^{𝛼2/(1−𝛼2)}+𝐶₃;(5.18)(4)lim_|𝜔|→0𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|𝜔|²=∞,(5.19) where 𝐶₁,…,𝐶₄ are constants, 𝐶_𝛼2 and 𝐶^′_𝛼2 are constants, depending on 𝛼₂, |𝜔|=𝜔²(𝑡)+𝜔²(𝑡−𝜏) and |𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|=|𝐻^′₁(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|²+|𝐻^′₂(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|².

Proof. By Corollary 3.8 and 𝐹^′₁(𝑡,0,0)=𝐹^′₂(𝑡,0,0)=0, we have 𝐻^′₁(𝑡,0,0)=𝐻^′₂(𝑡,0,0)=0, for all 𝑡∈[0,2𝛾𝜏]. Moreover, by the definition of 𝐻, we know that 𝐻(𝑡,0,0)=0.
Now we show that (5.17) holds. By (𝐴₅) one obtains 𝐹(𝑡,𝑢₁,𝑢₂)≤𝑀|𝑢|^1/𝛼2+𝐶₁,forall𝑢=(𝑢₁,𝑢₂)∈ℝ². So, by Proposition 3.3 and Example 4.2, it is easy to see that 𝐻((𝑡)(≥𝐶𝑡,𝜔,𝜔𝑡−𝜏))^𝛼2𝑀|𝜔|^{1/(1−𝛼2)}−𝐶₁,(5.20) where 𝐶_𝛼2=2^{1/(1−𝛼2)}/𝑀^{𝛼2/(1−𝛼2)−1}(𝛼^{𝛼2/(1−𝛼2)}₂−𝛼^{1/(1−𝛼2)}₂).
Similar arguments to the proof of Lemma 5.3 show that there exists a constant 𝐶₂ such that 𝐹𝑡,𝑢₁,𝑢₂≥𝑚|𝑢|^1/𝛼2−𝐶₂.(5.21) Therefore, it follows that 𝐻((𝑡)(≤𝐶𝑡,𝜔,𝜔𝑡−𝜏))^𝛼2𝑚|𝜔|^{1/(1−𝛼2)}+𝐶₂.(5.22)
Next we show that (5.18) holds. Again as in the proof of Lemma 5.3, we can estimate 𝐻^′ by 𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))≤𝐶_𝛼22^{1/(1−𝛼2)}𝑚−1𝑀|𝜔|^{𝛼2/(1−𝛼2)}+𝐶₃,(5.23) where 𝐶₃=max{𝐶₁+𝐶₂,sup_|𝜔|<1|𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))||}. By Lemma 5.3 again and the duality principle 𝑢₁,𝑢₂=𝐻^′₁(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)),𝐻^′₂((𝑡)(⟺(𝜔(𝑡)(=𝐹𝑡,𝜔,𝜔𝑡−𝜏)),𝜔𝑡−𝜏))^′₁𝑡,𝑢₁,𝑢₂,𝐹^′₂𝑡,𝑢₁,𝑢₂(5.24) when |𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|≥𝐶, we have |𝜔|≤2^1/𝛼2𝑀−𝑚∣𝐻^{′(1/𝛼2)−1}.(5.25) Since there exists a constant 𝑀_𝐶 such that when |𝑢|=𝑢²₁+𝑢²₂=|𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|≤𝐶, we have |𝜔|=𝐹^′𝑡,𝑢₁,𝑢₂≤𝑀_𝐶.(5.26) Choose 𝐶^′_𝛼2=2^1/𝛼2𝑀−𝑚^{𝛼2/(𝛼2−1)},𝐶₄=𝐶^′_𝛼2𝑀^{𝛼2/(1−𝛼2)}_𝐶.(5.27) Then it is not difficult to see 𝐻^′(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))≥𝐶^′_𝛼2|𝜔|^{𝛼2/(1−𝛼2)}−𝐶₄.(5.28)
Finally, we show that (5.19) holds. By (𝐴₆), for all 𝜀>0, there exists 𝛿>0 so that when |𝑢|=𝑢²₁+𝑢²₂<𝛿, we have 𝐹𝑡,𝑢₁,𝑢₂≤𝜀𝑢²₁+𝑢²₂.(5.29) Now, for all 𝐾>0, choose 𝜀=1/4𝐾 and let 𝜂=2𝜀𝛿(𝜀). Then whenever 𝜔²(𝑡)+𝜔²(𝑡−𝜏)<𝜂, we have 𝐻((𝑡)(≥1𝑡,𝜔,𝜔𝑡−𝜏))𝜔4𝜀²(𝑡)+𝜔²(𝑡−𝜏)=𝐾|𝜔|².(5.30) That is, lim_|𝜔|→0𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|𝜔|²=∞.(5.31)

Lemma 5.5. There exist constants 𝐶_𝛿 and 𝐶^′_𝛿 depending on 𝛿, such that 𝐻((𝑡)(≥𝐶𝑡,𝜔,𝜔𝑡−𝜏))^𝛿|𝜔|²|𝜔|𝐶,when≤𝛿,_′𝛿|𝜔|^𝑞,when|𝜔|>𝛿,(5.32) and when 𝛿→+0,𝐶_𝛿→+∞.

Proof. By (5.19), we have lim_|𝜔|→0𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))|𝜔|²=∞.(5.33) So, as 𝛿→+0, we obtain 𝐶_𝛿≜inf{𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))/|𝜔|²∶|𝜔|≤𝛿}→+∞. That is, 𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))≥𝐶_𝛿|𝜔|²,(5.34) when |𝜔|≤𝛿.
We next show the validity of the second part of the inequality.
For all 𝜔₀=(𝜔₀(𝑡),𝜔₀(𝑡−𝜏)),|𝜔₀|=1, let 𝜙_𝜔0(𝛽)=𝐻(𝑡,𝛽𝜔₀(𝑡),𝛽𝜔₀(𝑡−𝜏)). Then 𝜙^′_𝜔0𝛽=𝐻^′₁𝑡,𝛽𝜔₀(𝑡),𝛽𝜔₀(𝑡−𝜏)𝜔₀(𝑡)+𝐻^′₂𝑡,𝛽𝜔₀(𝑡),𝛽𝜔₀(𝑡−𝜏)𝜔₀(𝑡−𝜏).(5.35) Since 𝜙_𝜔0 is a convex function for all 𝛽>0, we have 𝐻^′₁𝑡,𝛽𝜔₀(𝑡),𝛽𝜔₀(𝑡−𝜏)𝜔₀(𝑡)+𝐻^′₂𝑡,𝛽𝜔₀(𝑡),𝛽𝜔₀(𝑡−𝜏)𝜔₀()≥1𝑡−𝜏𝛽𝜙_𝜔0𝛽.(5.36) So, by (5.34) we obtain 𝐻^′₁𝑡,𝛿𝜔₀(𝑡),𝛿𝜔₀(𝑡−𝜏)𝜔₀(𝑡)+𝐻^′₂𝑡,𝛿𝜔₀(𝑡),𝛿𝜔₀(𝑡−𝜏)𝜔₀(𝑡−𝜏)≥𝐶_𝛿⋅𝛿.(5.37) By the convexity of 𝐻 again, it is easy to see that 𝐻𝑡,𝑠𝜔₀(𝑡),𝑠𝜔₀(𝑡−𝜏)≥𝐻^′₁𝑡,𝛿𝜔₀(𝑡),𝛿𝜔₀(𝑡−𝜏)(𝑠−𝛿)𝜔₀(𝑡)+𝐻^′₂𝑡,𝛿𝜔₀(𝑡),𝛿𝜔₀(𝑡−𝜏)(𝑠−𝛿)𝜔₀(𝑡−𝜏)+𝐻𝑡,𝛿𝜔₀(𝑡),𝛿𝜔₀(𝑡−𝜏)≥𝐶_𝛿⋅𝛿(𝑠−𝛿)+𝐶_𝛿𝛿²=𝐶_𝛿𝛿𝑠,∀𝑠>0.(5.38) So, we further obtain that 𝐻(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏))≥𝐶_𝛿⋅𝛿|𝜔|.(5.39) By Lemma 5.4, there exists 𝑇>0 such that 𝐻((𝑡)(≥𝐶𝑡,𝜔,𝜔𝑡−𝜏))^𝛼2|𝜔|2𝑀^𝑞.(5.40) Let 𝐶^′_𝛿=min{𝐶_𝛼2/2𝑀,𝑇^1−𝑞𝛿𝐶_𝛿}. Then by (5.34), (5.39), and (5.40), we have 𝐻((𝑡)(≥𝐶𝑡,𝜔,𝜔𝑡−𝜏))^𝛿|𝜔|²,|𝜔|𝐶≤𝛿,_′𝛿|𝜔|^𝑞,|𝜔|>𝛿.(5.41)

Lemma 5.6. Let 𝜐_𝑚=(𝜐_𝑚(𝑡),𝜐_𝑚(𝑡−𝜏))⇀𝜐=(𝜐(𝑡),𝜐(𝑡−𝜏)) (weakly convergent sequence on 𝐿^𝑞([0,2𝛾𝜏])²) and satisfy ₀^2𝛾𝜏𝐻𝑡,𝜐_𝑚𝑑𝑡⟶₀^2𝛾𝜏𝐻(𝑡,𝜐)𝑑𝑡as𝑚⟶∞.(5.42) Then ₀^2𝛾𝜏𝐻𝑡,𝜐_𝑚−𝜐𝑑𝑡⟶0as𝑚⟶∞.(5.43)

Proof. (I) First, we show that the terms in {𝐻(𝑡,𝜐_𝑚)} have equicontinuous integrals, that is, for all 𝜀>0, there exists 𝛿>0 such that ∀𝑚,_Ω𝐻𝑡,𝜐_𝑚𝑑𝑡<𝜀when𝜇(Ω)<𝛿.(5.44)
Since 𝐻 is convex, we have 𝐻^′₁(𝑡,𝜐)𝜐(𝑡)−𝜐_𝑚(𝑡)+𝐻^′₂(𝑡,𝜐)𝜐(𝑡−𝜏)−𝜐_𝑚(𝑡−𝜏)≤𝐻𝑡,𝜐_𝑚−𝐻(𝑡,𝜐).(5.45) So, from 𝜐_𝑚⇀𝜐 and the above equality, we have ₀^2𝛾𝜏𝐻(𝑡,𝜐)𝑑𝑡≤lim_𝑚→∞0^2𝛾𝜏𝐻𝑡,𝜐_𝑚𝑑𝑡.(5.46) By 𝐻≥0 and the assumption that ∫₀^2𝛾𝜏𝐻(𝑡,𝜐)𝑑𝑡→∫₀^2𝛾𝜏𝐻(𝑡,𝜐_𝑚)𝑑𝑡, it is not difficult to see lim_𝑚→∞Ω𝐻𝑡,𝜐_𝑚𝑑𝑡=_Ω𝐻(𝑡,𝜐)𝑑𝑡forallmeasurablesetsΩ.(5.47)
Now suppose to the contrary that {𝐻(𝑡,𝜐_𝑚)} does not have equicontinuous integrals, that is, there exists 𝜀₀>0 and functions 𝜐_𝑚𝑘=(𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)), as well as measurable sets Ω_𝑘, such that the following inequalities _Ω𝐻(𝑡,±𝜐)𝑑𝑡<𝜀₀,forallmeasurablesetsΩ,𝜇(Ω)<𝛿,_Ω𝑘𝐻𝑡,𝜐_𝑚𝑘𝑑𝑡≥𝜀₀,𝜇Ω_𝑘<𝛿2^𝑘(5.48) hold. Then choose Ω₀=⋃^∞_𝑘=1Ω_𝑘. It is not difficult to obtain 𝜇(Ω₀)<𝛿 and _Ω0𝐻𝑡,𝜐_𝑚𝑘𝑑𝑡≥_Ω𝑘𝐻𝑡,𝜐_𝑚𝑘𝑑𝑡≥𝜀₀,(5.49) which is contradictory to (5.46) and (5.47).
(II) For all 𝑏>0, we divide [0,2𝛾𝜏] into the following three subsets: 𝑄₁=𝑡∈0,2𝛾𝜏∣|𝜐|=𝜐²(𝑡)+𝜐²(),𝑄𝑡−𝜏>𝑏_𝑚2=𝑡∈0,2𝛾𝜏∣|𝜐|≤𝑏,𝜐_𝑚,𝑄−𝜐≥𝛿_𝑚3=𝑡∈0,2𝛾𝜏∣|𝜐|≤𝑏,𝜐_𝑚−𝜐<𝛿,(5.50) where |𝜐_𝑚−𝜐|=(𝜐_𝑚(𝑡)−𝜐(𝑡))²+(𝜐_𝑚(𝑡−𝜏)−𝜐(𝑡−𝜏))². By inequality (5.17), we know that there exist constants 𝐾 and 𝐿 such that 𝐻(𝑡,2𝑧(𝑡),2𝑧(𝑡−𝜏))≤𝐾𝐻(𝑡,𝑧(𝑡),𝑧(𝑡−𝜏))+𝐿,∀(𝑧(𝑡),𝑧(𝑡−𝜏))∈ℝ².(5.51) By the convexity of the function 𝐻 on 𝑄₁, we obtain 𝐻𝑡,𝜐_𝑚≤1−𝜐2𝐻𝑡,2𝜐_𝑚(≤𝐾+𝐻𝑡,−2𝜐)2𝐻𝑡,𝜐_𝑚+𝐻(𝑡,−𝜐)+𝐿.(5.52) So by (I), we may choose a constant 𝑏 large enough and fixed such that 𝜇(𝑄₁) is small enough so that _𝑄1𝐻𝑡,𝜐_𝑚𝐾−𝜐𝑑𝑡≤2_𝑄1𝐻𝑡,𝜐_𝑚+𝐻(𝑡,−𝜐)𝑑𝑡+𝐿𝜇𝑄₁<𝜀3.(5.53) For the above chosen constant 𝑏, now choose 𝛿 small enough so that _𝑄𝑚3𝐻𝑡,𝜐_𝑚𝜀−𝜐𝑑𝑡<3.(5.54)
For the chosen constants 𝑏 and 𝛿, let 𝜅=inf_{|||𝑧|𝜔−𝑧≥𝛿,≤𝑏}𝐻(𝑡,𝜔)−𝐻(𝑡,𝑧)−𝐻^′₁(𝑡,𝜔)(𝜔(𝑡)−𝑧(𝑡))−𝐻^′₂(𝑡,𝜔)(𝜔(𝑡−𝜏)−𝑧(𝑡−𝜏)).(5.55) Then 𝜅>0. Now 𝜅𝜇𝑄^𝑚₂≤_𝑄𝑚2𝐻𝑡,𝜐_𝑚−𝐻(𝑡,𝜐)−𝐻^′₁(𝑡,𝜐)𝜐_𝑚(𝑡)−𝜐(𝑡)−𝐻^′₂(𝑡,𝜐)𝜐_𝑚()()≤𝑡−𝜏−𝜐𝑡−𝜏𝑑𝑡_02𝛾𝜏𝐻𝑡,𝜐_𝑚−𝐻(𝑡,𝜐)−𝐻^′₁(𝑡,𝜐)𝜐_𝑚(𝑡)−𝜐(𝑡)−𝐻^′₂(𝑡,𝜐)𝜐_𝑚(𝑡−𝜏)−𝜐(𝑡−𝜏)𝑑𝑡.(5.56) Thus we have 𝜇(𝑄^𝑚₂)→0 as 𝑚→∞. Hence, it is easy to see that ∫_𝑄𝑚2𝐻(𝑡,𝜐_𝑚)→0 as 𝑚→∞.
By repeating the above argument on 𝑄₁, we conclude that there exists an 𝑛₀ such that when 𝑚>𝑛₀, we have _𝑄𝑚2𝐻𝑡,𝜐_𝑚𝜀−𝜐𝑑𝑡<3.(5.57)
From (5.53), (5.54), and (5.57), we obtain that lim_𝑚→∞0^2𝛾𝜏𝐻𝑡,𝜐_𝑚−𝜐𝑑𝑡=0.(5.58)

Corollary 5.7. 𝜐_𝑚=(𝜐_𝑚(𝑡),𝜐_𝑚(𝑡−𝜏))→𝜐=(𝜐(𝑡),𝜐(𝑡−𝜏))(𝐿^𝑞([0,2𝛾𝜏])×𝐿^𝑞([0,2𝛾𝜏])) if and only if ₀^2𝛾𝜏𝐻𝑡,𝜐_𝑚−𝜐𝑑𝑡=0.(5.59)

Proof. (⇒). 𝜐_𝑚→𝜐 implies that 𝜐_𝑚⇀𝜐 (weakly). By inequality (5.17) and the continuity of the composition operator, one gets 𝐻(𝑡,𝜐_𝑚)→𝐻(𝑡,𝜐)(𝐿¹([0,2𝛾𝜏])). That is, ∫₀^2𝛾𝜏𝐻(𝑡,𝜐_𝑚)𝑑𝑡→∫₀^2𝛾𝜏𝐻(𝑡,𝜐)𝑑𝑡. So, by Lemma 5.6, one obtains the conclusion.
(⇐). By Lemma 5.5, there exist constants 𝐵₁ and 𝐵₂>0 such that ₀^2𝛾𝜏𝐻(𝑡,𝜐)𝑑𝑡≥𝐵_{1|𝜐|≥𝛿}|𝜐|^𝑞𝑑𝑡+𝐵_2|𝜐|<𝛿|𝜐|²𝑑𝑡≥𝐵_{1|𝜐|≥𝛿}|𝜐|^𝑞𝑑𝑡+𝐵₂2𝛾𝜏^2/(𝑞−2)_|𝜐|<𝛿|𝜐|^𝑞𝑑𝑡^2/𝑞≥𝐶_𝛿min₀^2𝛾𝜏|𝜐|^𝑞𝑑𝑡,₀^2𝛾𝜏|𝜐|^𝑞𝑑𝑡^2/𝑞.(5.60) Choose 𝛿 small enough. Then 𝐶_𝛿>0 is a constant and it is not difficult to see that the conclusion is correct.

We next use the Maintain Pass Theorem to prove Theorem 5.1 in three steps.

(i) We show that 𝐽 satisfies the P. S. condition in 𝐸. Let {𝜐_𝑛=(𝜐_𝑛(𝑡),𝜐_𝑛(𝑡−𝜏))}⊂𝐸, and let the constants 𝐶₁,𝐶₂ satisfy 𝐶₁≤𝐽𝜐_𝑛≤𝐶₂,𝐽^′𝜐_𝑛⟶𝜃.(5.61) That is, we want to show that {𝜐_𝑛} has a convergent subsequence in 𝐸.

First, we show that {𝜐_𝑛} is bounded. In fact, by 𝑧_𝑚=𝐾𝜐_𝑚+𝐻^′𝑡,𝜐_𝑚−𝜒_𝑚𝐶⟶𝜃,₁≤12𝐾𝜐_𝑚,𝜐_𝑚+₀^2𝛾𝜏𝐻𝑡,𝜐_𝑚𝑑𝑡≤𝐶₂,(5.62) where 𝑧_𝑚=(𝑧_𝑚(𝑡),𝑧_𝑚(𝑡−𝜏)),𝜐_𝑚=(𝜐_𝑚(𝑡),𝜐_𝑚(𝑡−𝜏)),𝜒_𝑚=(𝜒_𝑚(𝑡),𝜒_𝑚(𝑡−𝜏)), we know that there exists 𝑛(𝜀)>0 for all 𝜀>0 such that the following inequality holds when 𝑚≥𝑚(𝜀): ₀^2𝛾𝜏𝐻𝑡,𝜐_𝑚1𝑑𝑡−2𝐻^′₁𝑡,𝜐_𝑚𝜐_𝑚(𝑡)+𝐻^′₂𝑡,𝜐_𝑚𝜐_𝑚(𝑡−𝜏)≤𝐶₂+𝜀2𝜐_𝑚(𝑡)_𝐿𝑞+𝜐_𝑚(𝑡−𝜏)_𝐿𝑞=𝐶₂+𝜀𝜐_𝑚(𝑡)_𝐿𝑞.(5.63)

On the other hand, by Lemmas 5.2 and 5.3, there exist constants 𝛼₂,𝐶₃,𝐶₄ and 𝐶₅ such that 𝐻()−1𝑡,𝜔2𝐻^′₁()𝜔(𝑡)−1𝑡,𝜔2𝐻^′₂()𝜔()=1𝑡,𝜔𝑡−𝜏2𝑧(𝑡)𝐹^′₁((𝑡)(+1𝑡,𝑧,𝑧𝑡−𝜏))2𝑧(𝑡−𝜏)𝐹^′₂((𝑡)(((𝑡)(≥1𝑡,𝑧,𝑧𝑡−𝜏))−𝐹𝑡,𝑧,𝑧𝑡−𝜏))2𝛼₂−1𝐹(𝑡,𝑧(𝑡),𝑧(𝑡−𝜏))−𝐶₃≥𝑚|𝑧|^1/𝛼212𝛼₂−1−𝐶₄≥|𝜔|^𝑞−𝐶₅,(5.64) where 𝜔(𝑡)=𝐹^′₁(𝑡,𝑧(𝑡),𝑧(𝑡−𝜏)), 𝜔(𝑡−𝜏)=𝐹^′₂(𝑡,𝑧(𝑡),𝑧(𝑡−𝜏)); 𝑧(𝑡)=𝐻^′₁(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)), 𝑧(𝑡−𝜏)=𝐻^′₂(𝑡,𝜔(𝑡),𝜔(𝑡−𝜏)); |𝜔|=𝜔²(𝑡)+𝜔²(𝑡−𝜏); |𝑧|=𝑧²(𝑡)+𝑧²(𝑡−𝜏).

So by (5.63) and (5.64), it is easy to see that 𝜐_𝑚(𝑡)_{𝐿𝑞[0,2𝛾𝜏]}=𝜐_𝑚(𝑡−𝜏)_{𝐿𝑞[0,2𝛾𝜏]}≤𝐶₆(constant)(5.65) That is, {𝜐_𝑛} is bounded. We next show that {𝜐_𝑛} has a convergent subsequence. Since 𝐿^𝑞[0,2𝛾𝜏] is a reflexive Banach space, there exists a subsequence of {𝜐_𝑛} which is weakly convergent in 𝐿^𝑞[0,2𝛾𝜏]. We denote it by {𝜐_𝑚𝑘}, that is to say, 𝜐_𝑚𝑘(𝑡)⇀𝜐^∗(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)⇀𝜐^∗(𝑡−𝜏). On one hand, by the convexity of the function 𝐻, we have 𝐻𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)+𝐻^′₁𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)𝜐_𝑚𝑘(𝑡)−𝜐^∗(𝑡)+𝐻^′₂𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)𝜐_𝑚𝑘(𝑡−𝜏)−𝜐^∗(𝑡−𝜏)≤𝐻𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏).(5.66) So we have ₀^2𝛾𝜏𝐻𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)𝑑𝑡≤lim_𝑘→∞0^2𝛾𝜏𝐻𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)𝑑𝑡.(5.67) On the other hand, by convexity of the function 𝐻 again, we obtain 𝐻𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)≥𝐻𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)+𝐻^′₁𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)𝜐^∗(𝑡)−𝜐_𝑚𝑘(𝑡)+𝐻^′₂𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)𝜐^∗(𝑡−𝜏)−𝜐_𝑚𝑘(𝑡−𝜏)=𝐻𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)+−𝐾𝜐_𝑚𝑘+𝑧_𝑚𝑘+𝜒_𝑚𝑘⋅𝜐^∗−𝜐_𝑚𝑘.(5.68) Since operators 𝐴 and 𝐾 are compact and (𝑧_𝑚𝑘(𝑡),𝑧_𝑚𝑘(𝑡−𝜏))→𝜃, we have lim_𝑘→∞0^2𝛾𝜏𝐻𝑡,𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−𝜏)𝑑𝑡≤₀^2𝛾𝜏𝐻𝑡,𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏)𝑑𝑡.(5.69)

By (5.67), (5.69), and applying Lemma 5.6 and Corollary 5.7, it is not difficult to see that 𝜐_𝑚𝑘(𝑡),𝜐_𝑚𝑘(𝑡−2𝜏)⟶𝜐^∗(𝑡),𝜐^∗(𝑡−𝜏).(5.70)

(ii) We show that there exist constants 𝜌,𝑟>0, such that 𝐽|_𝜕Ω𝑟≥𝜌>0,(5.71) where 𝜕Ω_𝑟={(𝜐(𝑡),𝜐(𝑡−𝜏))∈𝐿^𝑞[0,2𝛾𝜏]×𝐿^𝑞[0,2𝛾𝜏]∣‖𝜐(𝑡)‖_{𝐿𝑞[0,2𝛾𝜏]}=‖𝜐(𝑡−𝜏)‖_{𝐿𝑞[0,2𝛾𝜏]}=𝑟}.

Let 𝛽=‖𝐾‖_{£(𝐿𝑝,𝐿𝑞)}. Choose 𝛿>0 such that the constant 𝐶_𝛿 is large enough and choose 𝑟 small enough so that when ‖𝜐(𝑡)‖_𝐿𝑞=𝑟, by Lemma 5.5, there exists constant 𝐶₇>0 satisfying 𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡)|²𝑑𝑡−4𝛽_|𝜐|<𝛿|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞≥𝐶_7|𝜐|<𝛿|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞,𝐶^′_{𝛿|𝜐|≥𝛿}|𝜐(𝑡)|^𝑞𝑑𝑡−4𝛽_{|𝜐|≥𝛿}|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞≥𝐶_{7|𝜐|≥𝛿}|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞,(5.72) where |𝜐|=𝜐²(𝑡)+𝜐²(𝑡−𝜏).

By (5.72) and inequalities 𝑎^𝑐+𝑏^𝑐≤(𝑎+𝑏)^𝑐≤2^𝑐𝑎^𝑐+𝑏^𝑐,(5.73) where 𝑎,𝑏>0, and 𝑐>1, we have 𝐽(𝜐)=12𝐾(𝜐(𝑡(+1)),𝜐𝑡−𝜏)2𝐾(𝜐(𝑡−𝜏)),𝜐(𝑡)+₀^2𝛾𝜏𝐻((𝑡)(𝛽𝑡,𝜐,𝜐𝑡−𝜏))≥−2‖𝜐(𝑡)‖²_{𝐿𝑞0,2𝛾𝜏}−𝛽2‖𝜐(𝑡)‖²_{𝐿𝑞0,2𝛾𝜏}+𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡)|²𝑑𝑡+𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡−𝜏)|²𝑑𝑡+𝐶^′_{𝛿|𝜐|≥𝛿}𝜐²(𝑡)+𝜐²(𝑡−𝜏)^𝑞𝛽𝑑𝑡≥−2‖𝜐(𝑡)‖²_{𝐿𝑞0,2𝛾𝜏}−𝛽2‖𝜐(𝑡)‖²_{𝐿𝑞0,2𝛾𝜏}+𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡)|²𝑑𝑡+𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡−𝜏)|²𝑑𝑡+𝐶^′_{𝛿|𝜐|≥𝛿}|𝜐(𝑡)|^𝑞𝑑𝑡≥𝐶_7|𝜐|<𝛿|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞+_{|𝜐|≥𝛿}|𝜐(𝑡)|^𝑞𝑑𝑡^2/𝑞+𝐶_{𝛿|𝜐|<𝛿}|𝜐(𝑡−𝜏)|²≥𝐶𝑑𝑡₇2^2/𝑞‖𝜐(𝑡)‖²_{𝐿𝑞0,2𝛾𝜏}=𝐶₇2^2/𝑞𝑟².(5.74) Thus (5.71) is true with 𝜌=(𝐶₇/2^2/𝑞)𝑟².

(iii) It is obvious that 𝐽(𝜃)=0, and 𝐽(𝜐) is an even function in 𝜐.

Remark 5.8. Let 𝜐_𝑗(𝑡)=sin(𝑗𝜋/𝛾𝜏)𝑡,𝑗=1,2,…. Then by 𝐴𝜐_𝑗(𝑡)=−𝑗𝜋𝛾𝜏²sin𝑗𝜋𝛾𝜏𝑡,(5.75) we have 𝐾𝜐_𝑗(𝑡)=−𝛾𝜏𝑗𝜋²sin𝑗𝜋𝛾𝜏𝑡.(5.76) Hence, we obtain 𝐾𝜐_𝑗(𝑡)=−𝛾𝜏𝑗𝜋²sin2𝑗𝜋𝛾cos𝑗𝜋()𝛾𝜏𝑡−2𝜏+cos2𝑗𝜋𝛾sin𝑗𝜋(,𝛾𝜏𝑡−2𝜏)𝐾𝜐_𝑗(𝑡−2𝜏)=−𝛾𝜏𝑗𝜋²cos2𝑗𝜋𝛾sin𝑗𝜋𝛾𝜏𝑡−sin2𝑗𝜋𝛾cos𝑗𝜋𝑡.𝛾𝜏(5.77) Therefore, it is easy to see 𝐽𝜐_𝑗=12𝐾𝜐_𝑗(𝑡−2𝜏),𝜐_𝑗(𝑡+1)2𝐾𝜐_𝑗(𝑡),𝜐_𝑗(𝑡−2𝜏)+₀^2𝛾𝜏𝐻𝑡,𝜐_𝑗1𝑑𝑡=−2𝛾𝜏𝑗𝜋²cos2𝑗𝜋𝛾₀^2𝛾𝜏𝜐_𝑗(𝑡)²𝑑𝑡+₀^2𝛾𝜏𝜐_𝑗(𝑡−2𝜏)²𝑑𝑡+₀^2𝛾𝜏𝐻𝑡,𝜐_𝑗𝑑𝑡.(5.78) We may choose 𝑗_𝑘, such that cos2𝑗_𝑘𝜋/𝛾>0. Let 𝜐_𝑗𝑘(𝑡)=sin(𝑗_𝑘𝜋/𝛾𝜏)𝑡. By (5.78), we have 𝐽𝜆𝜐_𝑗𝑘1≤−𝜆2𝛾𝜏𝑗𝜋²cos2𝑗𝜋𝛾₀^2𝛾𝜏𝜙(𝑡)²𝑑𝑡+₀^2𝛾𝜏𝜙(𝑡−2𝜏)²+𝐶𝑑𝑡_𝛼2𝑚₀^2𝛾𝜏𝜙²(𝑡)+𝜙²(𝑡−2𝜏)^𝑞𝑑𝑡+2𝛾𝜏𝐶₂⟶−∞,(𝜆⟶+∞).(5.79)
From (i), (ii), (iii), and the Maintain Pass Theorem, we conclude that problem (4.16) has at least one nontrivial 2𝛾𝜏-periodic solution. Thus, Theorem 5.1 holds.

6. Example

In this section, we present a remark and an example to illustrate our main result.

Remark 6.1. For assumptions (𝐴₁)∼(𝐴₃), function 𝐹 is a solution to the following partial differential equation: 𝐹^′₂𝑡,𝑥₁,𝑥₂+𝐹^′₁𝑡,𝑥₂,𝑥₃=𝑓𝑡,𝑥₁,𝑥₂,𝑥₃.(6.1) In some special cases, 𝐹 can be easily determined. For example, if there exist continuously differential functions 𝑔,ℎ∈𝐶¹(ℝ²,ℝ) such that (𝜕𝑔/𝜕𝑢)(𝑢,𝑣)=(𝜕ℎ/𝜕𝑣)(𝑢,𝑣) and 𝑓(𝑡,𝑥₁,𝑥₂,𝑥₃)= 𝛼(𝑡)[𝑔(𝑥₁,𝑥₂)+ℎ(𝑥₂,𝑥₃)], where 𝛼(𝑡) is continuously differentiable and 𝜏-periodic in 𝑡, then 𝐹𝑡,𝑥₁,𝑥₂=𝛼(𝑡)^𝑥1₀ℎ𝑠,𝑥₂𝑑𝑠+^𝑥2₀𝑔(0,𝑠)𝑑𝑠(6.2) or 𝐹𝑡,𝑥₁,𝑥₂=𝛼(𝑡)^𝑥2₀𝑔𝑥₁,𝑠𝑑𝑠+^𝑥1₀ℎ(𝑠,0)𝑑𝑠.(6.3) For example,
(𝟙) if 𝑔(𝑢,𝑣)=𝑢²−2𝑢𝑣−𝑣²,ℎ(𝑢,𝑣)=𝑢²+2𝑢𝑣−𝑣² and 𝑓(𝑡,𝑥₁,𝑥₂,𝑥₃)=cos(2𝜏𝑡/𝜋)[𝑥²₁−2𝑥₁𝑥₂+2𝑥₂2𝑥₃− 𝑥²₃], then 𝐹𝑡,𝑥₁,𝑥₂=cos2𝜏𝑡𝜋^𝑥2₀𝑔𝑥₁,𝑠𝑑𝑠+^𝑥1₀ℎ(𝑠,0)𝑑𝑠=cos2𝜏𝑡𝜋^𝑥2₀𝑥²₁−2𝑥₁𝑠−𝑠²𝑑𝑠+^𝑥1₀𝑠²𝑑𝑠=cos2𝜏𝑡𝜋𝑥²₁𝑥₂−𝑥₁𝑥²₂−13𝑥³₂+13𝑥³₁;(6.4)
(𝟚) if 𝑔(𝑢,𝑣)=2𝑣cos𝑢−𝑢²sin𝑣,ℎ(𝑢,𝑣)=2𝑢cos𝑣−𝑣²sin𝑢 and 𝑓(𝑡,𝑥₁,𝑥₂,𝑥₃)=cos(2𝜏𝑡/𝜋)[2𝑥₂cos𝑥₁− 𝑥²₁sin𝑥₂+2𝑥₂cos𝑥₃−𝑥²₃sin𝑥₂], then 𝐹𝑡,𝑥₁,𝑥₂=cos2𝜏𝑡𝜋^𝑥2₀𝑔𝑥₁,𝑠𝑑𝑠+^𝑥1₀ℎ(𝑠,0)𝑑𝑠=cos2𝜏𝑡𝜋^𝑥2₀2𝑠cos𝑥₁−𝑥²₁sin𝑠𝑑𝑠+^𝑥1₀2𝑠𝑑𝑠=cos2𝜏𝑡𝜋𝑥²₂cos𝑥₁+𝑥²₁cos𝑥₂.(6.5) One can check easily that 𝐹 satisfies assumption (𝐴₂).

Finally, as an application, we consider the following example.

Example 6.2. Consider the equation 𝑥^′′(𝑡)+𝑥^′′22𝜏𝑡𝜋𝑥(𝑡−𝜏)𝑥²(𝑡)+𝑥²(𝑡−𝜏)^(𝑝/2)−1+𝑥²(𝑡−𝜏)+𝑥²(𝑡−2𝜏)^(𝑝/2)−1=0,(6.6) with 𝑥(0)=0, where 𝑝>2. Then 𝑔(𝑢,𝑣)=𝑝𝑣(𝑢²+𝑣²)^(𝑝/2)−1 and ℎ(𝑢,𝑣)=𝑝𝑢(𝑢²+𝑣²)^(𝑝/2)−1.
Moreover, 𝑔 and ℎ satisfy the condition (𝜕𝑔/𝜕𝑢)(𝑢,𝑣)=(𝜕ℎ/𝜕𝑣)(𝑢,𝑣). So we can choose 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))=𝑝1+sin²2𝜏𝑡𝜋₀^{𝑥(𝑡−𝜏)}𝑠𝑥(𝑡)²+𝑠^{2(𝑝/2)−1}𝑑𝑠+₀^𝑥(𝑡)𝑠^𝑝/2=𝑑𝑠1+sin²2𝜏𝑡𝜋𝑥(𝑡)²+𝑥(𝑡−𝜏)^2𝑝/2.(6.7) It is obvious that assumptions (𝐴₁)∼(𝐴₃) and the following (𝐴^′₄)∼(𝐴^′₆) hold:(𝐴^′₄)𝑥=(𝑥(𝑡),𝑥(𝑡−𝜏))→𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏)) is a continuously differentiable and strictly convex function and also satisfies 𝐹(𝑡,0,0)=0,𝐹^′₁(𝑡,0,0)=𝐹^′₂(𝑡,0,0)=0∀𝑡∈0,2𝛾𝜏;(6.8)(𝐴^′₅) for 𝛼₂=1/𝑝<1/2, there exist constants 𝑀,𝐶>0, such that when |𝑥|=𝑥²(𝑡)+𝑥²(𝑡−𝜏)≥𝐶 we have 𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))≤𝛼₂𝐹^′₁(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))𝑥(𝑡)+𝐹^′₂((𝑡)(𝑥(,𝐹((𝑡)(|𝑥|𝑡,𝑥,𝑥𝑡−𝜏))𝑡−𝜏)𝑡,𝑥,𝑥𝑡−𝜏))≤𝑀^1/𝛼2;(6.9)(𝐴^′₆)lim_|𝑥|→0𝐹(𝑡,𝑥(𝑡),𝑥(𝑡−𝜏))|𝑥|²=0.(6.10)
Thus, (6.6) has at least one nontrivial weak 2𝛾𝜏-periodic solution by Theorem 5.1.

Acknowledgments

The project is supported by NNSF of China (10771055), Doctoral Fund of Ministry of Education of China (200805321017).