A Note on Periodic Solutions of Second Order Nonautonomous Singular Coupled Systems

Abstract

We establish the existence of periodic solutions of the second order nonautonomous singular coupled systems 𝑥^′′+𝑎₁(𝑡)𝑥=𝑓₁(𝑡,𝑦(𝑡))+𝑒₁(𝑡) for a.e. 𝑡∈[0,𝑇], 𝑦^′′+𝑎₂(𝑡)𝑦=𝑓₂(𝑡,𝑥(𝑡))+𝑒₂(𝑡) for a.e. 𝑡∈[0,𝑇]. The proof relies on Schauder's fixed point theorem.

1. Introduction

Some classical tools have been used in the literature to study the positive solutions for two-point boundary value problems of a coupled system of differential equations. These classical tools include some fixed point theorems in cones for completely continuous operators and Leray-Schauder fixed point theorem; for examples, see [1–3] and literatures therein.

Recently, Schauder's fixed point theorem has been used to study the existence of positive solutions of periodic boundary value problems in several papers; see, for example, Torres [4], Chu et al. [5, 6], Cao and Jiang [7], and the references contained therein. However, there are few works on periodic solutions of second-order nonautonomous singular coupled systems. In these papers above, there are the major assumption that their associated Green's functions are positive. Since Green's functions are positive, in the paper, we continue to study the existence of periodic solutions to second-order nonautonomous singular coupled systems in the following form: 𝑥^′′+𝑎₁(𝑡)𝑥=𝑓₁(𝑡,𝑦(𝑡))+𝑒₁𝑦(𝑡)fora.e.𝑡∈[0,𝑇],_′′+𝑎₂(𝑡)𝑦=𝑓₂(𝑡,𝑥(𝑡))+𝑒₂(𝑡)fora.e.𝑡∈[0,𝑇], (1.1) with 𝑎₁,𝑎₂,𝑒₁,𝑒₂∈𝐿¹[0,𝑇], 𝑓₁,𝑓₂∈Car([0,𝑇]×(0,+∞),(0,+∞)). Here we write 𝑓∈Car([0,𝑇]×(0,+∞),(0,+∞)) if 𝑓∶[0,𝑇]×(0,+∞)→(0,+∞) is an 𝐿¹-caratheodory function, that is, the map 𝑥↦𝑓(𝑡,𝑥) is continuous for a.e. 𝑡∈(0,1) and the map 𝑡↦𝑓(𝑡,𝑥) is measurable for all 𝑥∈(0,+∞), and for every 0<𝑟<𝑠 there exists ℎ_𝑟,𝑠∈𝐿¹(0,𝑇) such that |𝑓(𝑡,𝑥)|≤ℎ_𝑟,𝑠(𝑡) for all 𝑥∈[𝑟,𝑠] and a.e. 𝑡∈[0,𝑇]; here “for a.e." means “for almost every".

This paper is mainly motivated by the recent papers [4–6, 8, 9], in which the periodic singular problems have been studied. Some results in [4–6, 9] prove that in some situations weak singularities may help create periodic solutions. In [6], the authors consider the periodic solutions of second-order nonautonomous singular dynamical systems, in which the scalar periodic singular problems have been studied by Leray-Schauder alternative principle, a well-known fixed point theorem in cones, and Schauder's fixed point theorem, respectively.

The remaining part of the paper is organized as follows. In Section 2, some preliminary results will be given. In Sections 3–5, by employing a basic application of Schauder's fixed point theorem, we state and prove the existence results for (1.1) under the nonnegative of the Green's function associated with (2.1)-(2.2). Our view point sheds some new light on problems with weak force potentials and proves that in some situations weak singularities may stimulate the existence of periodic solutions, just as pointed out in [9] for the scalar case.

To illustrate our results, for example, we can select the system 𝑥^′′+𝑎₁(𝑡)𝑥=𝑦^−𝛼1+𝑒₁𝑦(𝑡),_′′+𝑎₂(𝑡)𝑦=𝑥^−𝛼2+𝑒₂(𝑡), (1.2) with 𝑎₁,𝑎₂,𝑒₁,𝑒₂∈ℂ[0,𝑇], 0<𝛼_𝑖<1, 𝑖=1,2. Here we emphasize that in the new results 𝑒₁,𝑒₂ do not need to be positive.

Let us fix some notation to be used in the following: given 𝑎∈𝐿¹(0,1), we write 𝑎≻0 if 𝑎≥0 for a.e. 𝑡∈[0,1] and it is positive in a set of positive measures. For a given function 𝑝∈𝐿¹[0,𝑇], we denote the essential supremum and infimum by 𝑝^∗ and 𝑝_∗, if they exist. The usual 𝐿^𝑝-norm is denoted by ‖⋅‖_𝑝. The conjugate exponent of 𝑝 is denoted by 𝑝∶1/𝑝+1/𝑝=1.

2. Preliminaries

We consider the scalar equation 𝑥^′′+𝑎_𝑖(𝑡)𝑥=𝑒_𝑖(𝑡),𝑖=1,2, (2.1) with periodic boundary conditions 𝑥(0)=𝑥(𝑇),𝑥^′(0)=𝑥^′(𝑇). (2.2) In this paper, we assume that the following standing hypothesis is satisfied. (𝐻₁) The Green function 𝐺_𝑖(𝑡,𝑠), associated with (2.1)-(2.2), is nonnegative for all (𝑡,𝑠)∈[0,𝑇]×[0,𝑇], 𝑖=1,2.

In other words, the (strict) antimaximum principle holds for (2.1)-(2.2). Under the conditions (𝐻₁), the solution of (2.1)-(2.2) is given by𝑥(𝑡)=^𝑇₀𝐺_𝑖(𝑡,𝑠)𝑒_𝑖(𝑠)𝑑𝑠. (2.3)

For a nonconstant function 𝑎(𝑡), there is an 𝐿^𝑝-criterion proved in [9], which is given in the following lemma for the sake of completeness. Let 𝐊(𝑞) denote the best Sobolev constant in the following inequality:𝐶‖𝑢‖²_𝑞≤‖𝑢^′‖²₂,∀𝑢∈𝐻¹₀(0,𝑇). (2.4) The explicit formula for 𝐊(𝑞) is𝐊(𝑞)=2𝜋𝑞𝑇^1+2/𝑞(2)2+𝑞^1−2/𝑞(Γ(1/𝑞))Γ(1/2+1/𝑞)²4if1≤𝑞<∞,𝑇if𝑞=∞, (2.5) where Γ is the Gamma function. See [10].

Lemma 2.1. For each 𝑖=1,2, assume that 𝑎_𝑖(𝑡)≻0 and 𝑎_𝑖∈𝐿^𝑝[0,𝑇] for some 1≤𝑝≤∞. If ‖𝑎_𝑖‖_𝑝≤𝐊(2𝑝), (2.6) then the standing hypothesis (𝐻₁) holds.

We define the function 𝛾_𝑖∶ℝ→ℝ by𝛾_𝑖(𝑡)=^𝑇₀𝐺_𝑖(𝑡,𝑠)𝑒_𝑖(𝑠)𝑑𝑠,𝑖=1,2, (2.7) which is the unique 𝑇-periodic solution of𝑥^′′+𝑎_𝑖(𝑡)𝑥=𝑒_𝑖(𝑡). (2.8) Throughout this paper, we use the following notations:𝛾_𝑖∗=min_𝑖,𝑡𝛾_𝑖(𝑡),𝛾^∗_𝑖=max_𝑖,𝑡𝛾_𝑖(𝑡). (2.9)

3. The Case 𝛾_1∗≥0,𝛾_2∗≥0

Theorem 3.1. Assume that (𝐻₁) is satisfied; furthermore, we assume that there exist 𝑏_𝑖≻0, 𝑏̂_𝑖≻0, and 0<𝛼_𝑖<1 such that (𝐻2)𝑏̂0≤^𝑖(𝑡)𝑥^𝛼𝑖≤𝑓_𝑖𝑏(𝑡,𝑥)≤^𝑖(𝑡)𝑥^𝛼𝑖,∀𝑥>0,a.e.𝑡∈(0,𝑇),𝑖=1,2. (3.1)If 𝛾_1∗≥0, 𝛾_2∗≥0, then there exists a positive 𝑇-periodic solution of (1.1).

Proof. A 𝑇-periodic solution of (1.1) is just a fixed point of the completely continuous map 𝐴(𝑥,𝑦)=(𝐴𝑥,𝐴𝑦)∶𝐶_𝑇×𝐶_𝑇→𝐶_𝑇×𝐶_𝑇 defined as (𝐴𝑥)(𝑡)∶=^𝑇₀𝐺₁(𝑡,𝑠)[𝑓₁(𝑠,𝑦(𝑠))+𝑒₁=(𝑠)]𝑑𝑠_𝑇0𝐺₁(𝑡,𝑠)𝑓₁(𝑠,𝑦(𝑠))𝑑𝑠+𝛾₁(𝑡),(𝐴𝑦)(𝑡)∶=^𝑇₀𝐺₂(𝑡,𝑠)[𝑓₂(𝑠,𝑥(𝑠))+𝑒₂=(𝑠)]𝑑𝑠_𝑇0𝐺₂(𝑡,𝑠)𝑓₂(𝑠,𝑥(𝑠))𝑑𝑠+𝛾₂(𝑡). (3.2)
By a direct application of Schauder's fixed point theorem, the proof is finished if we prove that 𝐴 maps the closed convex set defined as 𝐾={(𝑥,𝑦)∈𝐶_𝑇×𝐶_𝑇∶𝑟₁≤𝑥(𝑡)≤𝑅₁,𝑟₂≤𝑦(𝑡)≤𝑅₂,∀𝑡∈[0,𝑇]}, (3.3) into itself, where 𝑅₁>𝑟₁>0, 𝑅₂>𝑟₂>0 are positive constants to be fixed properly. For convenience, we introduce the following notations: 𝛽_𝑖(𝑡)=^𝑇₀𝐺_𝑖(𝑡,𝑠)𝑏_𝑖(𝑠)𝑑𝑠,𝛽_𝑖(𝑡)=^𝑇₀𝐺_𝑖(𝑡,𝑠)𝑏̂_𝑖(𝑠)𝑑𝑠,𝑖=1,2. (3.4) Given (𝑥,𝑦)∈𝐾, by the nonnegative sign of 𝐺_𝑖 and 𝑓_𝑖, 𝑖=1,2, we have (𝐴𝑥)(𝑡)=^𝑇₀𝐺₁(𝑡,𝑠)𝑓₁(𝑠,𝑦(𝑠))𝑑𝑠+𝛾₁(𝑡)≥^𝑇₀𝐺₁𝑏̂(𝑡,𝑠)¹(𝑠)𝑦^𝛼1(𝑠)𝑑𝑠≥^𝑇₀𝐺₁𝑏̂(𝑡,𝑠)¹(𝑠)𝑅^𝛼1₂𝑑𝑠≥𝛽_1∗⋅1𝑅^𝛼1₂, (3.5) and note for every (𝑥,𝑦)∈𝐾 that (𝐴𝑥)(𝑡)=^𝑇₀𝐺₁(𝑡,𝑠)𝑓₁(𝑠,𝑦(𝑠))𝑑𝑠+𝛾₁≤(𝑡)_𝑇0𝐺₁𝑏(𝑡,𝑠)¹(𝑠)𝑦^𝛼1(𝑠)𝑑𝑠+𝛾^∗₁≤^𝑇₀𝐺₁𝑏(𝑡,𝑠)¹(𝑠)𝑟^𝛼1₂𝑑𝑠+𝛾^∗₁≤𝛽^∗₁⋅1𝑟^𝛼1₂+𝛾^∗₁. (3.6)
Also, follow the same strategy, (𝐴𝑦)(𝑡)=^𝑇₀𝐺₂(𝑡,𝑠)𝑓₂(𝑠,𝑥(𝑠))𝑑𝑠+𝛾₂(𝑡)≥^𝑇₀𝐺₂𝑏̂(𝑡,𝑠)²(𝑠)𝑥^𝛼2(𝑠)𝑑𝑠≥^𝑇₀𝐺₂𝑏̂(𝑡,𝑠)²(𝑠)𝑅^𝛼2₁𝑑𝑠≥𝛽_2∗⋅1𝑅^𝛼2₁,(𝐴𝑦)(𝑡)=^𝑇₀𝐺₂(𝑡,𝑠)𝑓₂(𝑠,𝑥(𝑠))𝑑𝑠+𝛾₂≤(𝑡)_𝑇0𝐺₂𝑏(𝑡,𝑠)²(𝑠)𝑥^𝛼2(𝑠)𝑑𝑠+𝛾^∗₂≤^𝑇₀𝐺₂𝑏(𝑡,𝑠)²(𝑠)𝑟^𝛼2₁𝑑𝑠+𝛾^∗₂≤𝛽^∗₂⋅1𝑟^𝛼2₁+𝛾^∗₂. (3.7) Thus (𝐴𝑥,𝐴𝑦)∈𝐾 if 𝑟₁,𝑟₂,𝑅₁, and 𝑅₂ are chosen so that 𝛽_1∗⋅1𝑅^𝛼1₂≥𝑟₁,𝛽^∗₁⋅1𝑟^𝛼1₂+𝛾^∗₁≤𝑅₁,𝛽_2∗⋅1𝑅^𝛼2₁≥𝑟₂,𝛽^∗₂⋅1𝑟^𝛼2₁+𝛾^∗₂≤𝑅₂. (3.8) Note that 𝛽_𝑖∗,𝛽_𝑖∗>0 and taking 𝑅=𝑅₁=𝑅₂, 𝑟=𝑟₁=𝑟₂, 𝑟=1/𝑅, it is sufficient to find 𝑅>1 such that 𝛽_1∗⋅𝑅^1−𝛼1≥1,𝛽^∗₁⋅𝑅^𝛼1+𝛾^∗₁𝛽≤𝑅,_2∗⋅𝑅^1−𝛼2≥1,𝛽^∗₂⋅𝑅^𝛼2+𝛾^∗₂≤𝑅, (3.9) and these inequalities hold for 𝑅 being big enough because 𝛼_𝑖<1.

4. The Case 𝛾_1∗<0<𝛾^∗₁,𝛾_2∗<0<𝛾^∗₂

Theorem 4.1. Assume (𝐻₁) and (𝐻₂) are satisfied. If 𝛾_1∗<0<𝛾^∗₁, 𝛾_2∗<0<𝛾^∗₂, and 𝛾_1∗≥𝑟₁₀−𝛽_1∗⋅𝑟^𝛼1𝛼2₁₀(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁₀)^𝛼1,𝛾_2∗≥𝑟₂₀−𝛽_2∗⋅𝑟^𝛼1𝛼2₂₀(𝛽^∗₁+𝛾^∗₁𝑟^𝛼1₂₀)^𝛼2, (4.1) where 0<𝑟₁₀<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^1+𝛼1=𝛼₁𝛼₂𝛽^∗₂𝛽_1∗, (4.2) and 0<𝑟₂₀<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^1+𝛼2=𝛼₁𝛼₂𝛽^∗₁𝛽_2∗, (4.3) then there exists a positive 𝑇-periodic solution of (1.1).

Proof. We follow the same strategy and notation as in the proof of ahead theorem. In this case, to prove that 𝐴∶𝐾→𝐾, it is sufficient to find 𝑟₁<𝑅₁, 𝑟₂<𝑅₂ such that 𝛽_1∗𝑅^𝛼1₂+𝛾_1∗≥𝑟₁,𝛽^∗₁𝑟^𝛼1₂+𝛾^∗₁≤𝑅₁,𝛽_2∗𝑅^𝛼2₁+𝛾_2∗≥𝑟₂,𝛽^∗₂𝑟^𝛼2₁+𝛾^∗₂≤𝑅₂. (4.4)(4.5)
If we fix 𝑅₁=𝛽^∗₁/𝑟^𝛼1₂+𝛾^∗₁, 𝑅₂=𝛽^∗₂/𝑟^𝛼2₁+𝛾^∗₂, then the first inequality of (4.5) holds if 𝑟₂ satisfies 𝛾_2∗≥𝑔(𝑟₂)∶=𝑟₂−𝛽_2∗⋅𝑟^𝛼1𝛼2₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^𝛼2. (4.6) According to 𝑔^′(𝑟₂)=1−𝛽_2∗⋅1(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^2𝛼2⋅[𝛼₁𝛼₂𝑟^𝛼1𝛼2₂⁻¹(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^𝛼2−𝑟^𝛼1𝛼2₂𝛼₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^𝛼2−1𝛼₁𝛾^∗₁𝑟^𝛼1₂⁻¹]𝛽=1−^2∗𝛼₁𝛼₂𝑟^𝛼1𝛼2₂⁻¹(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^𝛼2𝑟[1−^𝛼12𝛾^∗₁𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂]=1−𝛼₁𝛼₂𝛽^∗₁𝛽_2∗𝑟^𝛼1𝛼2₂⁻¹(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^−1−𝛼2, (4.7) we have 𝑔^′(0)=−∞, 𝑔^′(+∞)=1; then there exists 𝑟₂₀ such that 𝑔^′(𝑟₂₀)=0, and 𝑔^′′(𝑟₂)=−[𝛼₁𝛼₂𝛽^∗₁𝛽_2∗(𝛼₁𝛼₂−1)𝑟^𝛼1𝛼2₂⁻²(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^−1−𝛼2+𝛼₁𝛼₂𝛽^∗₁𝛽_2∗𝑟^𝛼1𝛼2₂⁻¹(−1−𝛼₂)(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^−2−𝛼2𝛾^∗₁𝛼₁𝑟^𝛼1₂⁻¹]>0. (4.8) Then the function 𝑔(𝑟₂) possesses a minimum at 𝑟₂₀, that is, 𝑔(𝑟₂₀)=min_{𝑟2𝜖(0,+∞)}𝑔(𝑟₂). Note 𝑔^′(𝑟₂₀)=0; then we have 1−𝛼₁𝛼₂𝛽^∗₁𝛽_2∗𝑟^{𝛼1𝛼2−1}₂₁(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂₁)^−1−𝛼2=0, (4.9) or equivalently, 𝑟^{1−𝛼1𝛼2}₂₀(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂₀)^1+𝛼2=𝛼₁𝛼₂𝛽^∗₁𝛽_2∗. (4.10) Similarly, 𝛾_1∗≥𝑔(𝑟₁)∶=𝑟₁−𝛽_1∗⋅𝑟^𝛼1𝛼2₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^𝛼1. (4.11)𝑔(𝑟₁₀)=min_{𝑟1𝜖(0,+∞)}𝑔(𝑟₁), and 𝑟^{1−𝛼1𝛼2}₁₀(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁₀)^1+𝛼1=𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (4.12)
Taking 𝑟₁=𝑟₁₀ and 𝑟₂=𝑟₂₀, then the first inequality in (4.4) and (4.5) holds if 𝛾_1∗≥𝑔(𝑟₁₀), 𝛾_2∗≥𝑔(𝑟₂₀), which are just condition (4.1). The second inequalities hold directly by the choice of 𝑅₁ and 𝑅₂, and it would remain to prove that 𝑟₁₀<𝑅₁ and 𝑟₂₀<𝑅₂. This is easily verified through elementary computations 𝑅₁=𝛽^∗₁𝑟^𝛼1₂₀+𝛾^∗₁=𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂₀𝑟^𝛼1₂₀=(𝛼₁𝛼₂𝛽^∗₁𝛽_2∗)^1/(1+𝛼2)⋅𝑟^{(𝛼1𝛼2−1)/(1+𝛼2)}₂₀𝑟^𝛼1₂₀=(𝛼₁𝛼₂𝛽^∗₁𝛽_2∗)^1/(1+𝛼2)⋅𝑟^{−(1+𝛼1)/(1+𝛼2)}₂₀. (4.13) The proof is the same as that in 𝑅₁,𝑅₂=(𝛼₁𝛼₂𝛽^∗₂𝛽_1∗)^1/(1+𝛼1)⋅𝑟^{−(1+𝛼2)/(1+𝛼1)}₁₀.
Next,we will prove 𝑟₁₀<𝑅₁, 𝑟₂₀<𝑅₂, or equivalently, 𝑟₁₀𝑟^{(1+𝛼1)/(1+𝛼2)}₂₀<(𝛼₁𝛼₂𝛽^∗₁𝛽_2∗)^1/(1+𝛼2),𝑟₂₀𝑟^{(1+𝛼2)/(1+𝛼1)}₁₀<(𝛼₁𝛼₂𝛽^∗₂𝛽_1∗)^1/(1+𝛼1). (4.14) Namely, 𝑟^1+𝛼2₁₀𝑟^1+𝛼1₂₀<𝛼₁𝛼₂𝛽^∗₁𝛽_2∗,𝑟^1+𝛼1₂₀𝑟^1+𝛼2₁₀<𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (4.15) On the other hand, 𝑟^{1−𝛼1𝛼2}₂₀(𝛽^∗₁)^1+𝛼2≤𝛼₁𝛼₂𝛽^∗₁𝛽_2∗. (4.16) Then 𝑟₂₀≤(𝛼₁𝛼₂(𝛽^∗₁)^−𝛼2𝛽_2∗)^{1/(1−𝛼1𝛼2)}. (4.17) Similarly, 𝑟₁₀≤(𝛼₁𝛼₂(𝛽^∗₂)^−𝛼1𝛽_1∗)^{1/(1−𝛼1𝛼2)}. (4.18) By (4.17) and (4.18), 𝑟^1+𝛼2₁₀𝑟^1+𝛼1₂₀≤(𝛼₁𝛼₂(𝛽^∗₂)^−𝛼1𝛽_1∗)^{(1+𝛼2)/(1−𝛼1𝛼2)}×(𝛼₁𝛼₂(𝛽^∗₁)^−𝛼2𝛽_2∗)^{(1+𝛼1)/(1−𝛼1𝛼2)}. (4.19) Now if we can prove (𝛼₁𝛼₂(𝛽^∗₂)^−𝛼1𝛽_1∗)^{(1+𝛼2)/(1−𝛼1𝛼2)}(𝛼₁𝛼₂(𝛽^∗₁)^−𝛼2𝛽_2∗)^{(1+𝛼1)/(1−𝛼1𝛼2)}<𝛼₁𝛼₂𝛽^∗₁𝛽_2∗, (4.20) then 𝑟^1+𝛼2₁₀𝑟^1+𝛼1₂₀<𝛼₁𝛼₂𝛽^∗₁𝛽_2∗. (4.21) In fact, (𝛼₁𝛼₂)^{(2+𝛼2+𝛼1−1)/(1−𝛼1𝛼2)}𝛽⋅(^1∗𝛽^∗₁)^{(1+𝛼2)/(1−𝛼1𝛼2)}𝛽⋅(^2∗𝛽^∗₂)^{𝛼1(1+𝛼2)/(1−𝛼1𝛼2)}<1, (4.22) since 𝛽𝑖∗≤𝛽^∗_𝑖, 𝑖=1,2. Similarly, we have 𝑟^1+𝛼1₂₀𝑟^1+𝛼2₁₀<𝛼₁𝛼₂𝛽^∗₂𝛽_1∗; we omit the details. Now we can obtain 𝑟₁₀<𝑅₁, 𝑟₂₀<𝑅₂. The proof is complete.

5. The Case 𝛾^∗₁≤0,𝛾_2∗<0<𝛾^∗₂  (𝛾^∗₂≤0,𝛾_1∗<0<𝛾^∗₁)

Theorem 5.1. Assume (𝐻₁) and (𝐻₂) are satisfied. If 𝛾^∗₁≤0, 𝛾_2∗<0<𝛾^∗₂, and 𝛾_2∗1≥(1−𝛼₁𝛼₂)[𝛼₁𝛼₂𝛽_2∗(𝛽^∗₁)^𝛼2]^{1/(1−𝛼1𝛼2)},𝛾_1∗≥𝑟₁₁−𝛽_1∗⋅𝑟^𝛼1𝛼2₁₁(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁₁)^𝛼1, (5.1) where 0<𝑟₁₁<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^1+𝛼1=𝛼₁𝛼₂𝛽^∗₂𝛽_1∗, (5.2) then there exists a positive 𝑇-periodic solution of (1.1).

Proof. In this case, to prove that 𝐴∶𝐾→𝐾, it is sufficient to find 𝑟₁<𝑅₁,𝑟₂<𝑅₂ such that 𝛽_1∗𝑅^𝛼1₂+𝛾_1∗≥𝑟₁,𝛽^∗₁𝑟^𝛼1₂≤𝑅₁,𝛽_2∗𝑅^𝛼2₁+𝛾_2∗≥𝑟₂,𝛽^∗₂𝑟^𝛼2₁+𝛾^∗₂≤𝑅₂. (5.3)
If we fix 𝑅₁=𝛽^∗₁/𝑟^𝛼1₂, 𝑅₂=𝛽^∗₂/𝑟^𝛼2₁+𝛾^∗₂, then the first inequality of (6.4) holds if 𝑟₂ satisfies 𝛾_2∗≥𝑟₂−𝛽_2∗𝑅^𝛼2₁=𝑟₂−𝛽_2∗(𝛽^∗₁)^𝛼2⋅𝑟^𝛼1𝛼2₂, (5.4) or equivalently 𝛾_2∗≥𝑓(𝑟₂)∶=𝑟₂−𝛽_2∗(𝛽^∗₁)^𝛼2⋅𝑟^𝛼1𝛼2₂. (5.5) Then the function 𝑓(𝑟₂) possesses a minimum at 𝑟₂₁=[𝛼₁𝛼₂⋅𝛽_2∗(𝛽^∗₁)^𝛼2]^{1/(1−𝛼1𝛼2)}, (5.6) that is, 𝑓(𝑟₂₁)=min_{𝑟2𝜖(0,+∞)}𝑓(𝑟₂).
On the analogy of (5.4), we obtain 𝛾_1∗≥𝑟₁−𝛽_1∗⋅𝑟^𝛼1𝛼2₁(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁)^𝛼1, (5.7) or equivalently, 𝛾_1∗≥ℎ(𝑟₁)∶=𝑟₁−𝛽_1∗⋅𝑟^𝛼1𝛼2₁(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁)^𝛼1. (5.8) According to ℎ^′(𝑟₁)∶=1−𝛼₁𝛼₂𝛽^∗₂𝛽_1∗𝑟^𝛼1𝛼2₁⁻¹(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁)^−1−𝛼1, (5.9) we have ℎ^′(0)=−∞, ℎ^′(+∞)=1; then there exists 𝑟₁₁ such that ℎ^′(𝑟₁₁)=0, and ℎ^′′(𝑟₁)=−[𝛼₁𝛼₂𝛽^∗₂𝛽_1∗(𝛼₁𝛼₂−1)𝑟^𝛼1𝛼2₁⁻²(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^−1−𝛼1+𝛼₁𝛼₂𝛽^∗₂𝛽_1∗𝑟^𝛼1𝛼2₁⁻¹(−1−𝛼₁)(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^−2−𝛼1𝛾^∗₂𝛼₂𝑟^𝛼2₁⁻¹]>0. (5.10) Then the function ℎ(𝑟₁) possesses a minimum at 𝑟₁₁, that is, ℎ(𝑟₁₁)=min_{𝑟1𝜖(0,+∞)}𝑓(𝑟₁).
Note ℎ^′(𝑟₁₁)=0; then we have 1−𝛼₁𝛼₂𝛽^∗₂𝛽_1∗𝑟^{𝛼1𝛼2−1}₁₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁₁)^−1−𝛼1=0. (5.11) Namely, 𝑟^{1−𝛼1𝛼2}₁₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁₁)^1+𝛼1=𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (5.12) Taking 𝑟₂=𝑟₂₁ and 𝑟₁=𝑟₁₁, then the first inequality in (5.3) hold if 𝛾_2∗≥ℎ(𝑟₂₁) and 𝛾_1∗≥ℎ(𝑟₁₁) which are just condition (5.1). The second inequalities hold directly by the choice of 𝑅₂ and 𝑅₁, so it would remain to prove that 𝑅₁=𝛽^∗₁/𝑟^𝛼1₂₁>𝑟₁₁, 𝑅₂=𝛽^∗₂/𝑟^𝛼2₁₁+𝛾^∗₂>𝑟₂₁. Now we turn to prove that 𝑅₁>𝑟₁₁, 𝑅₂>𝑟₂₁.
First, 𝑅₁=𝛽^∗₁𝑟^𝛼1₂₁=𝛽^∗₁{[𝛼₁𝛼₂⋅𝛽_2∗/(𝛽^∗₁)^𝛼2]^{1/(1−𝛼1𝛼2)}}^𝛼1=𝛽^∗₁[𝛼₁𝛼₂⋅𝛽_2∗/(𝛽^∗₁)^𝛼2]^{𝛼1/(1−𝛼1𝛼2)}=(𝛽^∗₁)^{1+(𝛼1𝛼2)/(1−𝛼1𝛼2)}(𝛼₁𝛼₂⋅𝛽_2∗)^{𝛼1/(1−𝛼1𝛼2)}=(𝛽^∗₁)^{1/(1−𝛼1𝛼2)}[(𝛼₁𝛼₂⋅𝛽_2∗)^𝛼1]^{1/(1−𝛼1𝛼2)}𝛽=[^∗1(𝛼₁𝛼₂⋅𝛽_2∗)^𝛼1]^{1/(1−𝛼1𝛼2)}1=[(𝛼₁𝛼₂)^𝛼1⋅𝛽^∗₁(𝛽_2∗)^𝛼1]^{1/(1−𝛼1𝛼2)}>[𝛼₁𝛼₂⋅𝛽_1∗(𝛽^∗₂)^𝛼1]^{1/(1−𝛼1𝛼2)}=𝑟₁₁, (5.13) since 𝛽_𝑖∗≤𝛽^∗_𝑖, 𝑖=1,2.
On the other hand, 𝑅₂=𝛽^∗₂𝑟^𝛼2₁₁+𝛾^∗₂=𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁₁𝑟^𝛼2₁₁. (5.14) By (5.2), we have 𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁₁=(𝛼₁𝛼₂𝛽^∗₂𝛽_1∗)^1/(1+𝛼1)𝑟^{(𝛼1𝛼2−1)/(1+𝛼1)}₁₁. (5.15) Combing (5.14) and (5.15), 𝑅₂=(𝛼₁𝛼₂𝛽^∗₂𝛽_1∗)^1/(1+𝛼1)𝑟^{−(1+𝛼2)/(1+𝛼1)}₁₁. (5.16) In what follows, we will verify that 𝑅₂>𝑟₂₁. In fact, (𝛼₁𝛼₂)^{(2+𝛼2+𝛼1)/(1−𝛼1𝛼2)−1}𝛽⋅(^2∗𝛽^∗₂)^{(1+𝛼1)/(1−𝛼1𝛼2)}𝛽⋅(^1∗𝛽^∗₁)^{𝛼2(1+𝛼1)/(1−𝛼1𝛼2)}<1, (5.17) since 𝛽_𝑖∗≤𝛽^∗_𝑖,𝑖=1,2. Thus (𝛼₁𝛼₂𝛽^{∗(−𝛼2)}₁𝛽_2∗)^{(1+𝛼1)/(1−𝛼1𝛼2)}⋅(𝛼₁𝛼₂𝛽^{∗(−𝛼1)}₂𝛽_1∗)^{(1+𝛼2)/(1−𝛼1𝛼2)}<𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (5.18) On the other hand, 𝑟^{1−𝛼1𝛼2}₂₁𝛽^∗(1+𝛼2)₁≤𝛼₁𝛼₂𝛽^∗₁𝛽_2∗,𝑟^{1−𝛼1𝛼2}₁₁𝛽^∗(1+𝛼1)₂≤𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (5.19) Thus one can see easily that 𝑟₂₁≤(𝛼₁𝛼₂𝛽^{∗(−𝛼2)}₁𝛽_2∗)^{1/(1−𝛼1𝛼2)},𝑟₁₁≤(𝛼₁𝛼₂𝛽^{∗(−𝛼1)}₂𝛽_1∗)^{1/(1−𝛼1𝛼2)}. (5.20) From (5.20), 𝑟^1+𝛼2₁₁𝑟^1+𝛼1₂₁≤(𝛼₁𝛼₂𝛽^{∗(−𝛼1)}₂𝛽_1∗)^{(1+𝛼2)/(1−𝛼1𝛼2)}(𝛼₁𝛼₂𝛽^{∗(−𝛼2)}₁𝛽_2∗)^{(1+𝛼1)/(1−𝛼1𝛼2)}. (5.21) Combing (5.18) and (5.21), 𝑟^1+𝛼2₁₁𝑟^1+𝛼1₂₁<𝛼₁𝛼₂𝛽^∗₂𝛽_1∗. (5.22) Therefore, 𝑟₂₁𝑟^{(1+𝛼2)/(1+𝛼1)}₁₁<(𝛼₁𝛼₂𝛽^∗₂𝛽_1∗)^1/(1+𝛼1). (5.23) Recall (5.16), we obtain 𝑟₂₁<𝑅₂ immediately. The proof is complete.

Similarly, we have the following theorem.

Theorem 5.2. Assume (𝐻₁) and (𝐻₂) are satisfied. If 𝛾^∗₂≤0, 𝛾_1∗<0<𝛾^∗₁, and 𝛾_1∗1≥(1−𝛼₁𝛼₂)⋅[𝛼₁𝛼₂𝛽_1∗(𝛽^∗₂)^𝛼1]^{1/(1−𝛼1𝛼2)},𝛾_2∗≥𝑟₂₁−𝛽_2∗⋅𝑟^𝛼1𝛼2₂₁(𝛽^∗₁+𝛾^∗₁𝑟^𝛼1₂₁)^𝛼2, (5.24) where 0<𝑟₂₁<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^1+𝛼2=𝛼₁𝛼₂𝛽^∗₁𝛽_2∗, (5.25) then there exists a positive 𝑇-periodic solution of (1.1).

6. The Case 𝛾_1∗≥0,𝛾_2∗<0<𝛾^∗₂(𝛾_2∗≥0,𝛾_1∗<0<𝛾^∗₁)

Theorem 6.1. Assume (𝐻₁) and (𝐻₂) are satisfied. If 𝛾_1∗≥0, 𝛾_2∗<0<𝛾^∗₂, and 𝛾_2∗≥𝑟₂₂−𝛽_2∗⋅𝑟^𝛼1𝛼2₂₂(𝛽^∗₁+𝛾^∗₁𝑟^𝛼1₂₂)^𝛼2, (6.1) where 0<𝑟₂₂<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂)^1+𝛼2=𝛼₁𝛼₂𝛽^∗₁𝛽_2∗, (6.2) then there exists a positive 𝑇-periodic solution of (1.1).

Proof. The following proof is the same as the proof of ahead theorem. In this case, to prove that 𝐴∶𝐾→𝐾, it is sufficient to find 𝑟₁<𝑅₁, 𝑟₂<𝑅₂ such that 𝛽_1∗𝑅^𝛼1₂≥𝑟₁,𝛽^∗₁𝑟^𝛼1₂+𝛾^∗₁≤𝑅₁,𝛽_2∗𝑅^𝛼2₁+𝛾_2∗≥𝑟₂,𝛽^∗₂𝑟^𝛼2₁+𝛾^∗₂≤𝑅₂. (6.3)(6.4)
If we fix 𝑅₁=𝛽^∗₁/𝑟^𝛼1₂+𝛾^∗₁, 𝑅₂=𝛽^∗₂/𝑟^𝛼2₁+𝛾^∗₂, then the first inequality of (6.4) satisfies 𝛽_2∗𝛽⋅(^∗1𝑟^𝛼1₂+𝛾^∗₁)^−𝛼2+𝛾_2∗≥𝑟₂, (6.5) or equivalently 𝛾_2∗≥𝑙(𝑟₂)∶=𝑟₂−𝛽_2∗(𝛽^∗₁+𝛾^∗₁𝑟^𝛼1₂)^𝛼2⋅𝑟^𝛼1𝛼2₂. (6.6) Then the function 𝑙(𝑟₂) possesses a minimum at 𝑟₂₂, that is, 𝑙(𝑟₂₂)=min_{𝑟2𝜖(0,+∞)}𝑙(𝑟₂).
Note 𝑙^′(𝑟₂₂)=0; then we have 1−𝛼₁𝛼₂𝛽^∗₁𝛽_2∗𝑟^{𝛼1𝛼2−1}₂₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂₂)^−1−𝛼2=0. (6.7) Therefore, 𝑟^{1−𝛼1𝛼2}₂₂(𝛽^∗₁+𝛾^∗₁⋅𝑟^𝛼1₂₂)^1+𝛼2=𝛼₁𝛼₂𝛽^∗₁𝛽_2∗. (6.8) Note that 𝛽𝑖∗,𝛽_𝑖∗>0, 𝑖=1,2. And taking 𝑟₂=𝑟₂₂, 𝑅₁=𝛽^∗₁/𝑟^𝛼1₂₂+𝛾^∗₁, 𝑟₁=1/𝑅₂, it is sufficient to find 𝑟₁<𝑅₁, 𝑟₂<𝑅₂ such that 𝑅^𝛼1₂⁻¹≤𝛽^∗₁,𝑅^𝛼2₂𝛽^∗₂+𝛾^∗₂≤𝑅₂, (6.9) and these inequalities hold for 𝑅₂ being big enough because 𝛼_𝑖<1. The proof is completed.

Similarly, we have the following theorem.

Theorem 6.2. Assume (𝐻₁) and (𝐻₂) are satisfied. If 𝛾_2∗≥0, 𝛾_1∗<0<𝛾^∗₁, and 𝛾_1∗≥𝑟₁₂−𝛽_1∗⋅𝑟^𝛼1𝛼2₁₂(𝛽^∗₂+𝛾^∗₂𝑟^𝛼2₁₂)^𝛼1, (6.10) where 0<𝑟₁₂<+∞ is a unique positive solution of the equation 𝑟^{1−𝛼1𝛼2}₁(𝛽^∗₂+𝛾^∗₂⋅𝑟^𝛼2₁)^1+𝛼1=𝛼₁𝛼₂𝛽^∗₂𝛽_1∗, (6.11) then there exists a positive 𝑇-periodic solution of (1.1).

Acknowledgments

The work was supported by Scientific Research Fund of Heilongjiang Provincial Education Department (no. 11544032), a grant from the Ph.D. Programs Foundation of Ministry of Education of China (no. 200918), Key Subject of Chinese Ministry of Education (no. 109051), and NNSF of P. R. China (no. 10971021).