Positive Periodic Solution of Second-Order Coupled Systems with Singularities

Abstract

This paper establishes the existence of periodic solution for a kind of second-order singular nonautonomous coupled systems. Our approach is based on fixed point theorem in cones. Examples are given to illustrate the main result.

1. Introduction

We are concerned with the existence of positive 𝑇-periodic solution for the second-order nonautonomous singular coupled systems: 𝑢^′′(𝑡)=𝑓₁(𝑡,𝑣(𝑡))+𝑐₁(𝑡)(),𝑣,𝑡∈0,𝑇_′′(𝑡)=𝑓₂(𝑡,𝑢(𝑡))+𝑐₂(𝑡)(),𝑢(0)(𝑇),𝑡∈0,𝑇=𝑢,𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇), (1) where 𝑓_𝑖∈𝐶((0,𝑇)×[0,+∞),[0,+∞)), 𝑐_𝑖∶(0,𝑇)→[0,+∞), 𝑖=1,2, are Lebesgue integrable, 𝑓_𝑖 may be singular at 𝑡=0, 𝑇 and 𝑐_𝑖 can have finitely many singularities.

Singular differential equations or systems arise from many branches of applied mathematics and physics such as gas dynamics, Newtonian fluid mechanics, and nuclear physics, which have been widely studied by many authors (see [1–7] and references therein). Some classical-tools have been used to study the positive solutions for two point nonperiodic boundary value problems of coupled systems [8, 9]. However, there are few works on periodic solutions of second order nonautonomous singular coupled systems of type (1).

In the recent papers [10, 11], the periodic solutions of singular coupled systems 𝑥^′′+𝑎₁(𝑡)𝑥=𝑓₁𝑡,𝑦+𝑐₁(𝑡),𝑦^′′+𝑎₂(𝑡)𝑦=𝑓₂(𝑡,𝑥)+𝑐₂(𝑡), (2) were proved by using some fixed point theorems in cones for completely continuous operators, where 𝑎₁,𝑎₂,𝑐₁,𝑐₂∈𝐿¹(0,𝑇), 𝑓₁,𝑓₂∈Car([0,𝑇]×(0,+∞),(0,+∞)). When the Green's function 𝐺_𝑖(𝑡,𝑠) (𝑖=1,2), associated with the periodic boundary problem 𝑥^′′+𝑎_𝑖(𝑡)𝑥=𝑐_𝑖(𝑡)(0)(𝑇),𝑥,𝑥=𝑥_′(0)=𝑥^′(𝑇), (3) is nonnegative for all (𝑡,𝑠)∈[0,𝑇]×[0,𝑇] and 𝑓_𝑖 (𝑖=1,2) satisfies weak singularities 𝑏̂0≤^𝑖(𝑡)𝑥^𝛼𝑖≤𝑓_𝑖()≤𝑏𝑡,𝑥_𝑖(𝑡)𝑥^𝛼𝑖,∀𝑥>0,a.e.𝑡∈(0,𝑇), (4) where 0<𝛼_𝑖<1, 𝑏̂_𝑖,𝑏_𝑖∈𝐿¹(0,𝑇), 𝑏̂_𝑖≥0, 𝑏_𝑖≥0, and 𝑏̂_𝑖, 𝑏_𝑖 are strictly positive on some positive measure subsets of (0,𝑇), some sufficient conditions for the existence of periodic solutions of (2) were obtained in [10, 11].

Motivated by the papers [9–11], we consider the existence of positive 𝑇-periodic solution of (1). Owing to the disappearing of the terms 𝑎_𝑖(𝑡) (𝑖=1,2) in (2), the methods in [9, 10] are no longer valid. In present paper, we will deal with the periodic solutions of (1) under new conditions. Let 𝑘 be a constant satisfying 0<𝑘<𝜋/𝑇. Denote by 𝐺(𝑡,𝑠) the Green function of 𝑥^′′(𝑡)+𝑘²𝑥(𝑡)𝑥(0)(𝑇)=0,=𝑥,𝑥^′(0)=𝑥^′(𝑇), (5) which can be expressed by 𝐺()=1𝑡,𝑠()()(),2𝑘1−cos𝑘𝑇()(),sin𝑘𝑡−𝑠+sin𝑘𝑇−𝑡+𝑠0≤𝑠≤𝑡≤𝑇,sin𝑘𝑠−𝑡+sin𝑘𝑇−𝑠+𝑡0≤𝑡≤𝑠≤𝑇. (6) By a direct computation, we can get 𝑚∶=min_{0≤𝑡,𝑠≤𝑇}𝐺()=1𝑡,𝑠2𝑘cot𝑘𝑇2,𝑀∶=max_{0≤𝑡,𝑠≤𝑇}𝐺()=1𝑡,𝑠(),𝑚2𝑘sin𝑘𝑇/2𝜎∶=𝑀=cos𝑘𝑇2∈(0,1). (7) Assume 𝑓_𝑖 (𝑖=1,2) satisfies the following conditions.(H₁) For 𝑡∈(0,𝑇),𝑓_𝑖(𝑡,1)>0 and there exist constants 𝜇₁≥𝜇₂>1, 𝜆₁≥𝜆₂>1 such that, for any constants 0≤𝜌_𝑖≤1, 𝑖=1,2, 𝜌^𝜇1₁𝑓₁(𝑡,𝑢)≤𝑓₁𝑡,𝜌₁𝑢≤𝜌^𝜇2₁𝑓₁(),()∈()×[),𝜌𝑡,𝑢𝑡,𝑢0,𝑇0,+∞_𝜆12𝑓₂(𝑡,𝑢)≤𝑓₂𝑡,𝜌₂𝑢≤𝜌^𝜆2₂𝑓₂(𝑡,𝑢),(𝑡,𝑢)∈(0,𝑇)×[0,+∞). (8)(9)(H₂)∫^𝑇₀𝑐⁻₁(𝑡)𝑑𝑡=𝑟₁>0,∫^𝑇₀𝑐⁻₂(𝑡)𝑑𝑡=𝑟₂>0 and ^𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡)𝑘𝑑𝑡<²𝑇𝑚𝑟₁2𝑀²𝑟₂/𝑚+1^𝜇1_𝑇0,𝑓₂(𝑡,1)+𝑐⁺₂(𝑡)𝑘𝑑𝑡<²𝑇𝑚𝑟₂2𝑀²𝑟₁/𝑚+1^𝜆1, (10)where 𝑐⁺_𝑖(𝑡)=max{𝑐_𝑖(𝑡),0}, 𝑐⁻_𝑖(𝑡)=max{−𝑐_𝑖(𝑡),0}, 𝑖=1,2.

Remark 1. For any 𝜌₁≥1, we get from (8) that 𝜌^𝜇2₁𝑓₁(𝑡,𝑢)≤𝑓₁𝑡,𝜌₁𝑢≤𝜌^𝜇1₁𝑓₁(𝑡,𝑢),(𝑡,𝑢)∈(0,𝑇)×[0,+∞). (11)
For any 𝜌₂≥1, we get from (9) that 𝜌^𝜆2₂𝑓₂(𝑡,𝑢)≤𝑓₂𝑡,𝜌₂𝑢≤𝜌^𝜆1₂𝑓₂(𝑡,𝑢),(𝑡,𝑢)∈(0,𝑇)×[0,+∞). (12)

Typical functions that satisfy (8) or (9) are those taking the form 𝑓_𝑖(𝑡,𝑢)=^𝑛_𝑘=1𝑎_𝑖𝑘(𝑡)𝑢^𝑙𝑖𝑘,𝑡∈(0,𝑇), (13) where 𝑎_𝑖𝑘∈𝐶(0,𝑇), 𝑎_𝑖𝑘(𝑡)>0, 𝑙_𝑖𝑘>1, 𝑖=1,2; 𝑘=1,2,…,𝑛.

Definition 2. Supposing that (𝑢,𝑣)∈𝐶¹[0,𝑇]∩𝐶²(0,𝑇)×𝐶¹[0,𝑇]∩𝐶²(0,𝑇) satisfies (1) and 𝑢(𝑡)>0, 𝑣(𝑡)>0 for any 𝑡∈[0,𝑇], then one says that (𝑢,𝑣) is a 𝐶¹[0,𝑇]×𝐶¹[0,𝑇] positive solution of system (1).
By using fixed point theorem in cones, we are able to prove the following result.

Theorem 3. Assume that (H₁), (H₂) hold. Then (1) has at least one positive 𝑇-periodic solution.

The proof of Theorem 3 will be given in Section 3 of this paper.

2. Preliminaries

Lemma 4 (see [12]). Let 𝑋 be a Banach space, 𝐾 a cone in 𝑋, Ω₁, Ω₂ two nonempty bounded open sets in 𝐾, 𝜃∈Ω₁⊂Ω₁⊂Ω₂. 𝑇∶Ω₂/Ω₁→𝐾 is a completely continuous operator. If(i)𝑇(𝑥)≠𝜆𝑥, 𝑥∈𝜕Ω₁, 𝜆>1,(ii)𝑇(𝑥)≠𝜆𝑥, 𝑥∈𝜕Ω₂, 0<𝜆<1, inf_{𝑥∈𝜕Ω2}‖𝑇𝑥‖>0,then 𝑇 has a fixed point in Ω₂/Ω₁.

Lemma 5. If 𝑓_𝑖(𝑡,𝑢) satisfy (H₁), then, for 𝑡∈(0,𝑇), 𝑓_𝑖(𝑡,𝑢) are increasing on 𝑢 and, for any [𝛼,𝛽]⊂(0,𝑇), lim_𝑢→+∞𝑓_𝑖(𝑡,𝑢)𝑢=+∞,(𝑖=1,2) (14) uniformly with respect to 𝑡∈[𝛼,𝛽].

Proof. We only deal with 𝑓₁. Without loss of generality, let 0≤𝑥≤𝑦. If 𝑦=0, we get 𝑓₁(𝑡,𝑥)≤𝑓₁(𝑡,𝑦). If 𝑦≠0, let 𝑐₀=𝑥/𝑦, then 0≤𝑐₀≤1. From (8), we get 𝑓₁(𝑡,𝑥)=𝑓₁𝑡,𝑐₀𝑦≤𝑐^𝜇2₀𝑓₁𝑡,𝑦≤𝑓₁𝑡,𝑦, (15) which means 𝑓₁(𝑡,𝑢) is increasing on 𝑢.
Assume 𝑢>1. It follows from (11) that 𝑓₁(𝑡,𝑢)≥𝑢^𝜇2𝑓₁(𝑡,1). Thus 𝑓₁(𝑡,𝑢)𝑢≥𝑢^𝜇2−1𝑓₁(𝑡,1),𝑡∈(0,𝑇). (16) From (H₁), for any [𝛼,𝛽]⊂(0,𝑇), we obtain 𝑓₁(𝑡,𝑢)𝑢≥𝑢^𝜇2−1𝑓₁(𝑡,1)≥max_{𝑡∈𝛼,𝛽}𝑓₁(𝑡,1)𝑢^𝜇2−1>0. (17) Therefore lim_𝑢→+∞𝑓₁(𝑡,𝑢)𝑢=+∞ (18) uniformly with respect to 𝑡∈[𝛼,𝛽].

Let 𝑋=𝐶[0,𝑇]. We know 𝑋 is a Banach space with the norm ‖𝑢‖=max_{𝑡∈[0,𝑇]}|𝑢(𝑡)|. Define the sets 𝑃={𝑢∈𝑋∶𝑢(𝑡)≥0},𝑄=𝑢∈𝑃∶min_{0≤𝑡≤𝑇}𝑢(𝑡)≥𝜎‖𝑢‖. (19) It is easy to check that 𝑃, 𝑄 are cones in 𝑋 and 𝑄⊂𝑃. Throughout this paper, we consider the space 𝑋×𝑋. It is easy to see 𝑋×𝑋 is a Banach space with the norm, ‖(𝑢,𝑣)‖₀=‖𝑢‖+‖𝑣‖,(𝑢,𝑣)∈𝑋×𝑋. (20)

We can get the conclusion that 𝑃×𝑃, 𝑄×𝑄 are cones in 𝑋×𝑋 and 𝑄×𝑄⊂𝑃×𝑃.

For any 𝑢∈𝑋, define the function [𝑢(𝑡)]^∗=𝑢(𝑡)(𝑡)(𝑡),𝑢≥0,0,𝑢<0. (21) Then the solution of periodic boundary value problem 𝑥^′′(𝑡)+𝑘²𝑥(𝑡)=−𝑐⁻₁(𝑡),𝑥(0)=𝑥(𝑇),𝑥^′(0)=𝑥^′(𝑇) (22) can be expressed by 𝑥₁(𝑡)=−∫^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₁(𝑠)𝑑𝑠, and the solution of periodic boundary value problem 𝑥^′′(𝑡)+𝑘²𝑥(𝑡)=−𝑐⁻₂(𝑡),𝑥(0)=𝑥(𝑇),𝑥^′(0)=𝑥^′(𝑇) (23) can be expressed by 𝑥₂(𝑡)=−∫^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₂(𝑠)𝑑𝑠. Obviously, 𝑥_𝑖(𝑡)<0, 𝑖=1,2. Then 𝑢(𝑡)+𝑥₁(𝑡)^∗(𝑡)≤‖𝑢‖,𝑣(𝑡)+𝑥≤𝑢₂(𝑡)^∗≤𝑣(𝑡)≤‖𝑣‖. (24) From (11), (12), and Lemma 5, we have 𝑓₁𝑡,𝑣(𝑡)+𝑥₂(𝑡)^∗≤𝑓₁(𝑡,‖𝑣‖)≤𝑓₁(‖𝑣‖)≤𝑣‖)𝑡,+1(‖+1^𝜇1𝑓₁(),𝑓𝑡,1₂𝑡,𝑢(𝑡)+𝑥₁(𝑡)^∗≤𝑓₂(𝑡,‖𝑢‖)≤𝑓₂(‖𝑢‖)≤𝑢‖)𝑡,+1(‖+1^𝜆1𝑓₂(𝑡,1). (25)(26) Then, for any fixed (𝑢,𝑣)∈𝑃×𝑃, it follows from (H₂) that 0≤^𝑇₀𝐺()×𝑘𝑡,𝑠₂𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠)𝑑𝑠≤𝑀^𝑇₀𝑘²‖𝑢‖+(‖𝑣‖+1)^𝜇1𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)𝑑𝑠≤𝑘²𝑀𝑇‖𝑢‖+𝑀(‖𝑣‖+1)^𝜇1𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)𝑑𝑠<+∞,0≤^𝑇₀𝐺()×𝑘𝑡,𝑠₂𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑐⁺₂(𝑠)𝑑𝑠≤𝑀^𝑇₀𝑘²‖𝑣‖+(‖𝑢‖+1)^𝜆1𝑓₂(𝑠,1)+𝑐⁺₂(𝑠)𝑑𝑠≤𝑘²𝑀𝑇‖𝑣‖+𝑀(‖𝑢‖+1)^𝜆1𝑇₀𝑓₂(𝑠,1)+𝑐⁺₂(𝑠)𝑑𝑠<+∞. (27)

Thus, we can define the operator 𝑇∶𝑃×𝑃→𝑃×𝑃, 𝑇(𝑢,𝑣)=(𝑇₁𝑢,𝑇₂𝑣) by 𝑇₁𝑢(𝑡)=^𝑇₀𝐺()×𝑘𝑡,𝑠₂𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠𝑇)𝑑𝑠,₂𝑣(𝑡)=^𝑇₀𝐺()×𝑘𝑡,𝑠₂𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑐⁺₂(𝑠)𝑑𝑠, (28) for (𝑢,𝑣)∈𝑃×𝑃. Then, we have the following lemma.

Lemma 6. Assuming that (H₁), (H₂) hold, then 𝑇 has a fixed point if and only if 𝑢^′′(𝑡)+𝑘²𝑢(𝑡)=𝑘²𝑢(𝑡)+𝑥₁(𝑡)^∗+𝑓₁𝑡,𝑣(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡)(),𝑣,𝑡∈0,𝑇_′′(𝑡)+𝑘²𝑣(𝑡)=𝑘²𝑣(𝑡)+𝑥₂(𝑡)^∗+𝑓₂𝑡,𝑢(𝑡)+𝑥₁(𝑡)^∗+𝑐⁺₂(𝑡)(),𝑢(0)(𝑇),𝑡∈0,𝑇=𝑢,𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇) (29) has one positive 𝑇-periodic solution.

Lemma 7. Assuming that (H₁), (H₂) hold, then 𝑇(𝑄×𝑄)⊂𝑄×𝑄 and 𝑇∶𝑄×𝑄→𝑄×𝑄 is completely continuous.

Proof. For (𝑢,𝑣)∈𝑄×𝑄, we have 𝑇₁𝑢(𝑡)≥𝑚^𝑇₀𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁×𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠)𝑑𝑠=𝜎^𝑇₀max_{0≤𝑡,𝑠≤𝑇}𝐺(𝑡,𝑠)𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠)𝑑𝑠=𝜎𝑇₁𝑢,𝑇₂𝑣(𝑡)≥𝑚^𝑇₀𝑘²𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑐⁺₂(𝑠)𝑑𝑠=𝜎^𝑇₀max_{0≤𝑡,𝑠≤𝑇}𝐺()×𝑘𝑡,𝑠₂𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑐⁺₂(𝑠𝑇)𝑑𝑠=𝜎₂𝑣. (30) Then, we can get ‖𝑇(𝑢,𝑣)‖₀=𝑇₁𝑢,𝑇₂𝑣₀=𝑇₁𝑢+𝑇₂𝑣≥𝜎𝑇₁𝑢+𝑇₂𝑣=𝜎‖𝑇(𝑢,𝑣)‖₀, (31) which means 𝑇(𝑄×𝑄)⊂𝑄×𝑄.
Let 𝐵⊂𝑄×𝑄 be any bounded set. Then there exists a constant 𝑁 such that, for any (𝑢,𝑣)∈𝐵, ‖(𝑢,𝑣)‖₀=‖𝑢‖+‖𝑣‖≤𝑁. (32) From (27), we have 𝑇₁𝑢(𝑡)≤𝑘²𝑀𝑇‖𝑢‖+𝑀(‖𝑣‖+1)^𝜇1×^𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)𝑑𝑠≤𝑘²𝑀𝑇𝑁+𝑀(𝑁+1)^𝜇1×^𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠𝑇)𝑑𝑠,₂𝑣(𝑡)≤𝑘²𝑀𝑇‖𝑣‖+𝑀(‖𝑢‖+1)^𝜆1×^𝑇₀𝑓₂(𝑠,1)+𝑐⁺₂(𝑠)𝑑𝑠≤𝑘²𝑀𝑇𝑁+𝑀(𝑁+1)^𝜆1×^𝑇₀𝑓₂(𝑠,1)+𝑐⁺₂(𝑠)𝑑𝑠. (33) Let 𝑁=𝑘²𝑀𝑇𝑁+𝑀(𝑁+1)^𝜆1+𝜇1×^𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)𝑑𝑠+^𝑇₀𝑓₂(𝑠,1)+𝑐⁺₂(𝑠)𝑑𝑠<+∞. (34) Thus ‖𝑇(𝑢,𝑣)‖₀=𝑇₁𝑢+𝑇₂𝑣≤𝑁, (35) which implies that 𝑇(𝐵) is bounded.
Next we prove that 𝑇(𝐵) is equicontinuous. For any (𝑢,𝑣)∈𝐵, 𝑡∈[0,𝑇], we know 𝑇₁𝑢(𝑡)=^𝑇₀𝐺()×𝑘𝑡,𝑠₂𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠=1)𝑑𝑠2𝑘(1−cos𝑘𝑇)×^𝑡₀(()(×𝑘sin𝑘𝑡−𝑠+sin𝑘𝑇−𝑡+𝑠))₂𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠+1)𝑑𝑠2𝑘(1−cos𝑘𝑇)×^𝑇_𝑡(()(×𝑘sin𝑘𝑠−𝑡+sin𝑘𝑇−𝑠+𝑡))₂𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠𝑇)𝑑𝑠,₁𝑢^′(𝑡)1=−()×2sin𝑘𝑇/2_𝑡0()sin2𝑘𝑡−𝑠−𝑘𝑇2×𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠+1)𝑑𝑠()×2sin𝑘𝑇/2_𝑇𝑡()sin2𝑘𝑠−𝑡−𝑘𝑇2×𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠)𝑑𝑠. (36) From (25) and (H₂), we have 𝑇₁𝑢^′(𝑡≤1)2sin(𝑘𝑇/2)×^𝑡₀𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠+1)𝑑𝑠()×2sin𝑘𝑇/2_𝑇𝑡𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠=1)𝑑𝑠()×2sin𝑘𝑇/2_𝑇0𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑐⁺₁(𝑠≤1)𝑑𝑠()×2sin𝑘𝑇/2_𝑇0𝑘²‖𝑢‖+(‖𝑣‖+1)^𝜇1×𝑓₁(𝑠,1)+𝑐⁺₁(𝑠≤𝑘)𝑑𝑠₂𝑇()‖𝑢‖+12sin𝑘𝑇/2()𝑣‖)2sin𝑘𝑇/2(‖+1^𝜇1×^𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠≤𝑘)𝑑𝑠₂𝑇()12sin𝑘𝑇/2𝑁+()()2sin𝑘𝑇/2𝑁+1^𝜇1×^𝑇₀𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)𝑑𝑠<+∞. (37)
Using the same method, we can obtain |(𝑇₂𝑣)^′(𝑡)|<+∞. Therefore, 𝑇(𝐵) is equicontinuous. According to Ascoli-Arzela theorem, 𝑇(𝐵) is a relatively compact set.
Next, we prove that 𝑇∶𝑄×𝑄→𝑄×𝑄 is continuous. Suppose (𝑢_𝑛,𝑣_𝑛),(𝑢₀,𝑣₀)∈𝑄×𝑄, (𝑢_𝑛,𝑣_𝑛)→(𝑢₀,𝑣₀), 𝑛→+∞, that is, 𝑢_𝑛→𝑢₀, 𝑣_𝑛→𝑣₀, 𝑛→+∞. We know that there exists a constant 𝐿>0 such that 𝑢₀≤𝐿,𝑢_𝑛≤𝐿,𝑣₀𝑣≤𝐿,_𝑛≤𝐿,𝑛=1,2…. (38)
We shall prove 𝑇(𝑢_𝑛,𝑣_𝑛)→𝑇(𝑢₀,𝑣₀), 𝑛→+∞, that is, 𝑇₁𝑢_𝑛⟶𝑇₁𝑢₀,𝑇₂𝑣_𝑛⟶𝑇₂𝑣₀,𝑛⟶+∞. (39)
We first deal with 𝑇₁𝑢_𝑛→𝑇₁𝑢₀, 𝑛→+∞. Otherwise, there exist 𝜀₀>0, {𝑡_𝑛}∈[0,𝑇] such that |𝑇₁𝑢_𝑛(𝑡_𝑛)−𝑇₁𝑢₀(𝑡_𝑛)|≥𝜀₀. Without loss of generality, we can assume 𝑡_𝑛→𝑡₀∈[0,𝑇]. We know 𝑇₁𝑢_𝑛𝑡_𝑛−𝑇₁𝑢_𝑛𝑡₀𝑇⟶0,₁𝑢₀𝑡_𝑛−𝑇₁𝑢₀𝑡₀⟶0,𝑛⟶∞. (40)
Next, we show 𝑇₁𝑢_𝑛(𝑡₀)−𝑇₁𝑢₀(𝑡₀)→0, 𝑛→∞. In fact, 𝑇₁𝑢_𝑛𝑡₀−𝑇₁𝑢₀𝑡₀=^𝑇₀𝐺𝑡₀×𝑘,𝑠₂𝑢_𝑛(𝑠)+𝑥₁(𝑠)^∗−𝑘²𝑢₀(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣_𝑛(𝑠)+𝑥₂(𝑠)^∗−𝑓₁𝑠,𝑣₀(𝑠)+𝑥₂(𝑠)^∗𝑑𝑠. (41) Let 𝑟_𝑛(𝑠)=𝐺𝑡₀×𝑘,𝑠₂𝑢_𝑛(𝑠)+𝑥₁(𝑠)^∗−𝑘²𝑢₀(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣_𝑛(𝑠)+𝑥₂(𝑠)^∗−𝑓₁𝑠,𝑣₀(𝑠)+𝑥₂(𝑠)^∗. (42) Since 𝑣_𝑛(𝑠)+𝑥₂(𝑠)^∗−𝑣₀(𝑠)+𝑥₂(𝑠)^∗=𝑣_𝑛(𝑠)+𝑥₂(𝑠)+𝑣_𝑛(𝑠)+𝑥₂(𝑠)2−𝑣₀(𝑠)+𝑥₂(𝑠)+𝑣₀(𝑠)+𝑥₂(𝑠)2=𝑣_𝑛(𝑠)+𝑥₂(𝑠)−𝑣₀(𝑠)+𝑥₂(𝑠)2+𝑣_𝑛(𝑠)−𝑣₀(𝑠)2≤𝑣_𝑛(𝑠)−𝑣₀(𝑠)⟶0,𝑛⟶+∞, (43) and 𝑓₁ is continuous, we know 𝑟_𝑛(𝑠)→0, 𝑛→∞. From (24), we know 𝑢_𝑛(𝑠)+𝑥₁(𝑠)^∗≤𝐿,𝑢₀(𝑠)+𝑥₁(𝑠)^∗𝑣≤𝐿,_𝑛(𝑠)+𝑥₂(𝑠)^∗𝑣≤𝐿<𝐿+1,₀(𝑠)+𝑥₂(𝑠)^∗≤𝐿<𝐿+1. (44) It can be inferred from (8) that 𝑟_𝑛(𝑠)≤𝑀𝑘²𝑢_𝑛(𝑠)+𝑥₁(𝑠)^∗−𝑘²𝑢₀(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣_𝑛(𝑠)+𝑥₂(𝑠)^∗−𝑓₁𝑠,𝑣₀(𝑠)+𝑥₂(𝑠)^∗≤2𝑀𝑘²𝐿+2𝑀(𝐿+1)^𝜇1𝑓₁(𝑠,1). (45)
Set 𝐹(𝑠)=2𝑀𝑘²𝐿+2𝑀(𝐿+1)^𝜇1(𝑓₁(𝑠,1)+𝑐⁺₁(𝑠)). Thus, we get 𝑟_𝑛(𝑠)≤𝐹(𝑠),𝑠∈(0,𝑇),𝑛=1,2,…. (46) From (H₂), we know ∫^𝑇₀𝐹(𝑠)𝑑𝑠<+∞. Using Lebesgue-dominated convergence theorem, we get 𝑇₁𝑢_𝑛𝑡₀−𝑇₁𝑢₀𝑡₀=^𝑇₀𝑟_𝑛(𝑠)𝑑𝑠⟶0,𝑛⟶+∞. (47)
From (40) and (47), we obtain 𝜀₀≤𝑇₁𝑢_𝑛𝑡_𝑛−𝑇₁𝑢₀𝑡_𝑛=𝑇₁𝑢_𝑛𝑡_𝑛−𝑇₁𝑢_𝑛𝑡₀+𝑇₁𝑢_𝑛𝑡₀−𝑇₁𝑢₀𝑡₀+𝑇₁𝑢₀𝑡₀−𝑇₁𝑢₀𝑡_𝑛≤𝑇₁𝑢_𝑛𝑡_𝑛−𝑇₁𝑢_𝑛𝑡₀+𝑇₁𝑢_𝑛𝑡₀−𝑇₁𝑢₀𝑡₀+𝑇₁𝑢₀𝑡₀−𝑇₁𝑢₀𝑡_𝑛⟶0,𝑛⟶∞, (48) which is a contradiction. Thus, we know 𝑇₁𝑢_𝑛→𝑇₁𝑢₀, 𝑛→+∞. Using the same method, we can obtain 𝑇₂𝑣_𝑛→𝑇₂𝑣₀, 𝑛→+∞. Then 𝑇𝑢_𝑛,𝑣_𝑛=𝑇₁𝑢_𝑛,𝑇₂𝑣_𝑛⟶𝑇₁𝑢₀,𝑇₂𝑣₀=𝑇𝑢₀,𝑣₀,𝑛⟶+∞, (49) which means 𝑇∶𝑄×𝑄→𝑄×𝑄 is continuous. Therefore 𝑇∶𝑄×𝑄→𝑄×𝑄 is a completely continuous operator.

3. Proof of Theorem 3

We proceed to prove Theorem 3 in two steps.

(1) Let Ω₁={(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖<2𝑀²𝑟₁/𝑚,‖𝑣‖<2𝑀²𝑟₂/𝑚}. We can get 𝑇(𝑢,𝑣)≠𝜆(𝑢,𝑣),(𝑢,𝑣)∈𝜕Ω₁,𝜆>1. (50) For (𝑢,𝑣)∈𝜕Ω₁, we have the following two cases.

Case I. One has {(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖=2𝑀²𝑟₁/𝑚,‖𝑣‖≤2𝑀²𝑟₂/𝑚}. Under this condition, we can get 𝑇₁𝑢≠𝜆𝑢,(𝑢,𝑣)∈𝜕Ω₁,𝜆>1. (51) Otherwise, there exist (𝑢₀,𝑣₀)∈𝜕Ω₁, 𝜆₀>1 such that 𝑇₁𝑢₀=𝜆₀𝑢₀. As 𝑢₀(𝑡)+𝑥₁(𝑡)=𝑢₀(𝑡)−^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₁(𝑠)𝑢𝑑𝑠≥𝜎₀−𝑀𝑟₁>0, (52) then [𝑢₀(𝑡)+𝑥₁(𝑡)]^∗=𝑢₀(𝑡)+𝑥₁(𝑡). By a direct computation, we know 𝑢₀=(1/𝜆₀)𝑇₁𝑢₀ satisfies 𝑢₀^′′(𝑡)+𝑘²𝑢₀(𝑡)=1𝜆₀𝑘²𝑢₀(𝑡)+𝑥₁(𝑡)+𝑓₁𝑡,𝑣₀(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡,𝑢₀)(0)=𝑢₀(𝑇),𝑢^′₀(0)=𝑢^′₀(𝑇). (53) Then, we get 𝑢₀^′′(𝑡)=1𝜆₀𝑘²𝑢₀(𝑡)+𝑥₁(𝑡)+𝑓₁𝑡,𝑣₀(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡)−𝑘²𝑢₀(𝑡)≤𝑘²𝑢₀(𝑡)+𝑘²𝑥₁(𝑡)+𝑓₁𝑡,𝑣₀(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡)−𝑘²𝑢₀(𝑡)=𝑘²𝑥₁(𝑡)+𝑓₁𝑡,𝑣₀(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡). (54)
Since [𝑣₀(𝑡)+𝑥₂(𝑡)]^∗≤𝑣₀(𝑡)<2𝑀²𝑟₂/𝑚+1, integrating both sides of (54) on [0,𝑇], we get 0≤𝑘^2𝑇₀𝑥₁(𝑡)+𝑑𝑡_𝑇0𝑓₁𝑡,𝑣₀(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡)𝑑𝑡≤𝑘^2𝑇₀𝑥₁(𝑡)𝑑𝑡+2𝑀²𝑟₂𝑚+1^𝜇1×^𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡)𝑑𝑡. (55) Then 2𝑀²𝑟₂𝑚+1^𝜇1𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡)𝑑𝑡≥−𝑘^2𝑇₀𝑥₁(𝑡)𝑑𝑡=𝑘^2𝑇₀^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₁(𝑠)𝑑𝑠𝑑𝑡≥𝑘²𝑇𝑚𝑟₁. (56) That is, ^𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡𝑘)𝑑𝑡≥²𝑇𝑚𝑟₁2𝑀²𝑟₂/𝑚+1^𝜇1, (57) which contradicts with (H₂).

Case II. One has {(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖≤2𝑀²𝑟₁/𝑚,‖𝑣‖=2𝑀²𝑟₂/𝑚}. Under this condition, we can get 𝑇₂𝑣≠𝜆𝑣,(𝑢,𝑣)∈𝜕Ω₁,𝜆>1. (58) Otherwise, there exist (𝑢₀,𝑣₀)∈𝜕Ω₁, 𝜆₀>1 such that 𝑇₂𝑣₀=𝜆₀𝑣₀. As 𝑣₀(𝑡)+𝑥₂(𝑡)=𝑣₀(𝑡)−^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₂(𝑠)𝑣𝑑𝑠≥𝜎₀−𝑀𝑟₂>0, (59) then [𝑣₀(𝑡)+𝑥₂(𝑡)]^∗=𝑣₀(𝑡)+𝑥₂(𝑡). By a direct computation, we know 𝑣₀=(1/𝜆₀)𝑇₂𝑣₀ satisfies 𝑣₀^′′(𝑡)+𝑘²𝑣₀(𝑡)=1𝜆₀𝑘²𝑣₀(𝑡)+𝑥₂(𝑡)+𝑓₂𝑡,𝑢₀(𝑡)+𝑥₁(𝑡)^∗+𝑐⁺₂(𝑡,𝑣₀)(0)=𝑣₀(𝑇),𝑣^′₀(0)=𝑣^′₀(𝑇). (60)
For [𝑢₀(𝑡)+𝑥₁(𝑡)]^∗≤𝑢₀(𝑡)<2𝑀²𝑟₁/𝑚+1, using the same method as condition I, we obtain ^𝑇₀𝑓₂(𝑡,1)+𝑐⁺₂(𝑡𝑘)𝑑𝑡≥²𝑇𝑚𝑟₂2𝑀²𝑟₁/𝑚+1^𝜆1, (61) which is also a contradiction.

(2) Choose an interval [𝛼,𝛽]⊂(0,𝑇) satisfying 𝛽−𝛼=𝑇/2. Set 𝑀>4𝑀/𝑇𝑚². From Lemma 5, there exists 𝑅₁>2𝑀²𝑟/𝑚, 𝑟=max{𝑟₁,𝑟₂} such that 𝑓_𝑖(𝑡,𝑢)≥𝑀𝑢,𝑢≥𝑅₁,𝑡∈𝛼,𝛽,𝑖=1,2. (62)

Let 𝑅≥(2𝑀/𝑚)𝑅₁>𝑅₁>2𝑀²𝑟/𝑚. Define Ω₂={(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖<𝑅,‖𝑣‖<𝑅}. We can get 𝑇(𝑢,𝑣)≠𝜆(𝑢,𝑣),(𝑢,𝑣)∈𝜕Ω₂,0<𝜆<1. (63)

For (𝑢,𝑣)∈𝜕Ω₂, we have the following two cases.

Case I. One has {(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖=𝑅,‖𝑣‖≤𝑅}. Under this condition, we know 𝑢(𝑡)+𝑥₁(𝑡)≥𝜎‖𝑢‖−𝑀𝑟₁𝑚≥𝜎𝑅−𝑀𝑟>𝑅2𝑀>𝑅₁>0. (64) Thus [𝑢(𝑡)+𝑥₁(𝑡)]^∗=𝑢(𝑡)+𝑥₁(𝑡). Furthermore, we obtain from (62) that 𝑇₂𝑣(𝑡)=^𝑇₀𝐺(𝑡,𝑠)𝑘²𝑣(𝑠)+𝑥₂(𝑠)^∗+𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)+𝑐⁺₂(𝑠≥)𝑑𝑠_𝛽𝛼𝐺(𝑡,𝑠)𝑓₂𝑠,𝑢(𝑠)+𝑥₁(𝑠)≥𝑑𝑠𝑀𝑇𝑅𝑚².4𝑀 (65) On the other hand, for 0<𝜆<1, we know 𝜆𝑣(𝑡)<𝑣(𝑡)≤𝑅. From the choice of 𝑀, we get 𝑇₂𝑣≠𝜆𝑣.

Case II. One has {(𝑢,𝑣)∈𝑄×𝑄∶‖𝑢‖≤𝑅,‖𝑣‖=𝑅}. Under this condition, we get 𝑣(𝑡)+𝑥₂(𝑡)≥𝜎‖𝑣‖−𝑀𝑟₂𝑚≥𝜎𝑅−𝑀𝑟>𝑅2𝑀>𝑅₁>0. (66) Thus [𝑣(𝑡)+𝑥₂(𝑡)]^∗=𝑣(𝑡)+𝑥₂(𝑡). From (62), we get 𝑇₁𝑢(𝑡)=^𝑇₀𝐺(𝑡,𝑠)𝑘²𝑢(𝑠)+𝑥₁(𝑠)^∗+𝑓₁𝑠,𝑣(𝑠)+𝑥₂(𝑠)+𝑐⁺₁(𝑠≥)𝑑𝑠_𝛽𝛼𝐺(𝑡,𝑠)𝑓₁𝑠,𝑢(𝑠)+𝑥₁(𝑠≥)𝑑𝑠𝑀𝑇𝑅𝑚².4𝑀 (67) For 0<𝜆<1, 𝜆𝑢(𝑡)<𝑢(𝑡)≤𝑅, from the choice of 𝑀, we know 𝑇₁𝑢≠𝜆𝑢.
Furthermore, we can obtain ‖𝑇(𝑢,𝑣)‖₀=𝑇₁𝑢+𝑇₂𝑣≥𝑀𝑇𝑅𝑚²,()4𝑀𝑢,𝑣∈𝜕Ω₂. (68) It implies inf_{(𝑢,𝑣)∈𝜕Ω2}‖𝑇(𝑢,𝑣)‖₀>0.
From Lemma 4, we know 𝑇 has a fixed point (𝑢,𝑣)̃ in Ω₂/Ω₁. For (𝑢,𝑣)̃∈Ω₂/Ω₁, we have the following three cases.

Case 1. One has (𝑢,𝑣)̃∈𝑄×𝑄∶2𝑀²𝑟𝑚≤‖𝑢‖≤𝑅,‖𝑣‖̃<2𝑀²𝑟𝑚∶=𝑆₁. (69)

Case 2. One has (𝑢,𝑣)̃∈𝑄×𝑄∶‖𝑢‖<2𝑀²𝑟𝑚,2𝑀²𝑟𝑚≤‖𝑣‖̃≤𝑅∶=𝑆₂. (70)

Case 3. One has (𝑢,𝑣)̃∈𝑄×𝑄∶2𝑀²𝑟𝑚≤‖𝑢‖≤𝑅,2𝑀²𝑟𝑚≤‖𝑣‖̃≤𝑅∶=𝑆₃. (71)
Next, we show Cases 1 and 2 are impossible. In Case 1, we have 𝑢(𝑡)+𝑥₁(𝑡)=𝑢(𝑡)−^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₁(𝑠)‖𝑢‖𝑑𝑠≥𝜎−𝑀𝑟₁>0. (72) It follows that [𝑢(𝑡)+𝑥₁(𝑡)]^∗=𝑢(𝑡)+𝑥₁(𝑡). By a direct computation, we know 𝑢=𝑇₁𝑢 satisfies 𝑢^′′(𝑡)+𝑘²𝑢(𝑡)=𝑘²𝑢(𝑡)+𝑥₁(𝑡)+𝑓₁𝑡,𝑣̃(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡),𝑢(0)=𝑢(𝑇),𝑢^′(0)=𝑢^′(𝑇). (73) Then, we get 𝑢^′′(𝑡)=𝑘²𝑥₁(𝑡)+𝑓₁𝑡,𝑣̃(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡). (74) Since [𝑣(̃𝑡)+𝑥₂(𝑡)]^∗≤𝑣(̃𝑡)<2𝑀²𝑟₂/𝑚+1, integrating both sides of (74) on [0,𝑇], we get 0=𝑘^2𝑇₀𝑥₁(𝑡)+𝑑𝑡_𝑇0𝑓₁𝑡,𝑣̃(𝑡)+𝑥₂(𝑡)^∗+𝑐⁺₁(𝑡)𝑑𝑡≤𝑘^2𝑇₀𝑥₁(𝑡)𝑑𝑡+2𝑀²𝑟₂𝑚+1^𝜇1×^𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡)𝑑𝑡. (75) Then 2𝑀²𝑟₂𝑚+1^𝜇1𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡)𝑑𝑡≥−𝑘^2𝑇₀𝑥₁(𝑡)𝑑𝑡=𝑘^2𝑇₀^𝑇₀𝐺(𝑡,𝑠)𝑐⁻₁(𝑠)𝑑𝑠𝑑𝑡≥𝑘²𝑇𝑚𝑟₁. (76) That is, ^𝑇₀𝑓₁(𝑡,1)+𝑐⁺₁(𝑡𝑘)𝑑𝑡≥²𝑇𝑚𝑟₁2𝑀²𝑟₂/𝑚+1^𝜇1, (77) which contradicts with (H₂). Using the same method, we can prove that Case 2 is also impossible. Therefore, Case 3 is satisfied and 𝑇 has a fixed point (𝑢,𝑣)̃ in Ω₂/Ω₁ satisfying 2𝑀²𝑟𝑚≤‖𝑢‖≤𝑅,2𝑀²𝑟𝑚≤‖𝑣‖̃≤𝑅. (78) Since 𝑢(𝑡)+𝑥₁(𝑡)≥𝜎‖𝑢‖−𝑀𝑟₁𝑣̃(𝑡)+𝑥>0,₂(𝑡)≥𝜎‖𝑣‖̃−𝑀𝑟₂>0, (79) from Lemma 6, we know 𝑢(𝑡),𝑣(̃𝑡) satisfy 𝑢^′′(𝑡)+𝑘²𝑢(𝑡)=𝑘²𝑢(𝑡)+𝑥₁(𝑡)+𝑓₁𝑡,𝑣̃(𝑡)+𝑥₂(𝑡)+𝑐⁺₁(𝑡),𝑣̃^′′(𝑡)+𝑘²𝑣̃(𝑡)=𝑘²𝑣̃(𝑡)+𝑥₂(𝑡)+𝑓₂𝑡,𝑢(𝑡)+𝑥₁(𝑡)+𝑐⁺₂(𝑡),𝑢(0)=𝑢(𝑇),𝑢^′(0)=𝑢^′(𝑇),𝑣̃(0)=𝑣̃(𝑇),𝑣^′̃(0)=𝑣^′̃(𝑇). (80)
Let (𝑢^∗(𝑡),𝑣^∗(𝑡))=(𝑢(𝑡)+𝑥₁(𝑡),𝑣(̃𝑡)+𝑥₂(𝑡)). For 𝑐₁(𝑡)=𝑐⁺₁(𝑡)−𝑐⁻₁(𝑡),𝑐₂(𝑡)=𝑐⁺₂(𝑡)−𝑐⁻₂(𝑡), (81) we obtain 𝑢^∗′′(𝑡)=𝑓𝑡,𝑣^∗(𝑡)+𝑐₁(𝑡)(),𝑣,𝑡∈0,𝑇_∗′′(𝑡)=𝑓𝑡,𝑢^∗(𝑡)+𝑐₂(𝑡)(),𝑢,𝑡∈0,𝑇_∗(0)=𝑢^∗(𝑇),𝑢^∗′(0)=𝑢^∗′(𝑇),𝑣^∗(0)=𝑣^∗(𝑇),𝑣^∗′(0)=𝑣^∗′(𝑇). (82)
This means (𝑢^∗,𝑣^∗) is one positive 𝑇-periodic solution of (1).

4. Applications of Theorem 3

Finally, we give some examples as the applications of Theorem 3: ()𝑢𝑒1^′′(𝑡)=𝑘²𝑇𝑚𝑣^3/22𝑀²√𝑇/𝑚+1²√𝑡−2√𝑡,𝑣^′′(𝑡)=𝑘²𝑇𝑚𝑢^3/22𝑀²√𝑇/𝑚+1²√𝑡−2√𝑡,𝑢(0)=𝑢(𝑇),𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇).()𝑢𝑒2^′′(𝑡)=𝑘²𝑇𝑚𝑣^3/22𝑀²√𝑇/𝑚+1²√−2𝑇−𝑡√𝑡,𝑣^′′(𝑡)=𝑘²𝑇𝑚𝑢^3/22𝑀²√𝑇/𝑚+1²√−2𝑇−𝑡√𝑡,𝑢(0)=𝑢(𝑇),𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇). (83)

Choosing 𝑓₁(𝑡,𝑣)=𝑘²𝑇𝑚𝑣^3/2/(2𝑀²√𝑇/𝑚+1)²√𝑡, 𝑓₂(𝑡,𝑢)=𝑘²𝑇𝑚𝑢^3/2/(2𝑀²√𝑇/𝑚+1)²√𝑇−𝑡, 𝑐⁻_𝑖(𝑡)=2/√𝑡, 𝑐⁺_𝑖(𝑡)≡0, 𝑟_𝑖=∫^𝑇₀𝑐⁻_𝑖(𝑡)𝑑𝑡4√𝑇, 𝑖=1,2, and 𝜇₁=2>𝜇₂=5/4>1, 𝜆₁=2>𝜆₂=5/4>1, then (H₁) is satisfied. Notice (H₂) also holds, since ^𝑇₀𝑓_𝑖(𝑡,1)+𝑐⁺_𝑖(𝑡)𝑑𝑡=2𝑘²𝑇√𝑇𝑚2𝑀²√𝑇/𝑚+1²,𝑖=1,2. (84)

Existence of the positive 𝑇-periodic solutions is guaranteed from Theorem 3. We can also consider the following examples and the same result can be obtained: ()𝑢𝑒3^′′(𝑡)=𝑘²𝑇𝑚𝑣^3/22𝑀²√𝑇/𝑚+1²√−2𝑇−𝑡√𝑡,𝑣^′′(𝑡)=𝑘²𝑇𝑚𝑢^3/22𝑀²√𝑇/𝑚+1²√𝑡−2√,𝑢(0)(𝑇)𝑇−𝑡=𝑢,𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇),()𝑢𝑒4^′′(𝑡)=𝑘²𝑇𝑚𝑣^3/22𝑀²√𝑇/𝑚+1²√−2𝑇−𝑡√,𝑣𝑇−𝑡_′′(𝑡)=𝑘²𝑇𝑚𝑢^3/22𝑀²√𝑇/𝑚+1²√−2𝑇−𝑡√,𝑢(0)(𝑇)𝑇−𝑡=𝑢,𝑢^′(0)=𝑢^′(𝑇),𝑣(0)=𝑣(𝑇),𝑣^′(0)=𝑣^′(𝑇). (85)(86)

Acknowledgments

The author is grateful to the referees for valuable comments and useful remarks on the paper. Research supported by China Postdoctoral Science Foundation (2012M510341), Beijing Natural Science Foundation (1112006) and the Grant of Beijing Education Committee Key Project (KZ20130028031).