Periodic Solutions of a Nonautonomous Plant-Hare Model with Impulses

Abstract

A plant-hare model subjected by the effect of impulses is studied in this paper. Sufficient conditions are obtained for the existence of at least one positive periodic solution.

1. Introduction

Classical predator-prey model has been well studied (e.g., see [1–8] and the references cited therein). To explore the impact of plant toxicity on the dynamics of plant-hare interactions, Gao and Xia [9] consider a nonautonomous plant-herbivore dynamical system with a toxin-determined functional response: 𝑁̇(𝑡)=𝑟(𝑡)𝑁(𝑡)1−𝑁(𝑡)𝐾(𝑁(𝑡𝑃(𝑡),𝑃̇(𝑡)(𝑡)𝐶(𝑁(𝑡𝑃(𝑡)(𝑡)𝑃(𝑡),𝐶(𝑁(𝑡(𝑁(𝑡𝑓(𝑁(𝑡−𝐶))=𝐵))−𝑑))=𝑓))1−)),𝑓(𝑁(𝑡=(𝑡)4𝐺))𝑒𝛿𝑁(𝑡),1+ℎ𝑒𝛿𝑁 (1) where 𝑁(𝑡) denotes the density of plant at time 𝑡, 𝑃(𝑡) denotes the herbivore biomass at time 𝑡, 𝑟(𝑡) is the plant intrinsic growth rate at time 𝑡, 𝑑(𝑡) is the per capita rate of herbivore death unrelated to plant toxicity at time 𝑡, 𝐵(𝑡) is the conversion rate at time 𝑡, 𝑒 is the encounter rate per unit plant, 𝛿 is the fraction of food items encountered that the herbivore ingests, 𝐾 is the carrying capacity of plant, 𝐺 measures the toxicity level, and ℎ is the time for handing one unit of plant. To explore the impact of environmental factors (e.g., seasonal effects of weather, food supplies, mating habits, harvesting, etc.), the assumption of periodicity of parameters is more realistic and important. To this reason, they assumed that 𝑟(𝑡), 𝑑(𝑡), and 𝐵(𝑡) are continuously positive periodic functions with period 𝜔 and 𝑒, 𝛿, 𝐾, 𝐺, ℎ are five positive real constants.

However, birth of many species is an annual birth pulse, for having more accurate description of the system, we need to consider using the impulsive differential equations. To see how impulses affect the differential equations, for examples, one can refer to [10–17]. Motivated by the above-mentioned works, in this paper, we consider the above system with impulses: 𝑁̇(𝑡)=𝑁(𝑡)𝑟(𝑡)1−𝑁(𝑡)𝐾−4𝐺𝑒𝛿𝑃(𝑡)+(4𝐺ℎ−1)𝑒²𝛿²𝑁(𝑡)𝑃(𝑡)4𝐺(1+ℎ𝑒𝛿𝑁(𝑡))²,𝑃̇(𝑡)=𝑃(𝑡)4𝐺𝑒𝛿𝐵(𝑡)𝑁(𝑡)+(4𝐺ℎ−1)𝑒²𝛿²×𝐵(𝑡)𝑁²(𝑡)4𝐺(1+ℎ𝑒𝛿𝑁(𝑡))²⁻¹−𝑑(𝑡),𝑡≠𝑡_𝑘;Δ𝑁𝑡_𝑘=𝑁𝑡⁺_𝑘−𝑁𝑡⁻_𝑘=𝑐_1𝑘𝑁𝑡_𝑘,Δ𝑃𝑡_𝑘=𝑃𝑡⁺_𝑘−𝑃𝑡⁻_𝑘=𝑐_2𝑘𝑃𝑡_𝑘,𝑡=𝑡_𝑘,𝑘=1,2,…, (2) where the assumptions on 𝑟, 𝑑, 𝐵, 𝑒, 𝛿, 𝐾, 𝐺, and ℎ are the same as before, 𝑐_𝑗𝑘∈(−1,∞)(𝑗=1,2,𝑘∈ℕ=1,2,…), {𝑡_𝑘}∈ℕ is a strictly increasing sequence with 𝑡₁>0, and lim_𝑘→∞𝑡_𝑘=∞. We further assume that there exists a 𝑞∈ℕ such that 𝑐_{𝑗(𝑘+𝑞)}=𝑐_𝑗𝑘(𝑗=1,2) and 𝑡_𝑘+𝑞=𝑡_𝑘+𝜔 for 𝑘∈ℕ.

Without loss of generality, we will assume 𝑡_𝑘≠0 for 𝑘=1,2,…, and [0,𝜔]∩{𝑡_𝑘}={𝑡₁,𝑡₂,…,𝑡_𝑚}; hence 𝑞=𝑚.

2. Preliminaries

In this section, we cite some definitions and lemmas.

Let 𝑃𝐶_𝜔 denote the space of 𝜔-periodic functions 𝜓∶ℝ→ℝ which are continuous for 𝑡≠𝑡_𝑘, are continuous from the left for 𝑡∈ℝ, and have possible discontinuities of the first kind at points 𝑡=𝑡_𝑘; that is, the limit from the right of 𝑡_𝑘 exists but may be different from the value at 𝑡_𝑘. We also denote 𝑃𝐶¹_𝜔={𝜓∈𝑃𝐶_𝜔∶𝜓̇∈𝑃𝐶_𝜔}.

For the convenience, we list the following definitions and lemmas.

Definition 1 (see [10]). The set 𝔽⊂𝑃𝐶_𝜔 is said to be quasi-equicontinuous in [0,𝜔] if for any 𝜖>0 there exists a 𝛿>0 such that if 𝑥∈𝔽; 𝑘∈ℤ; 𝜏₁,𝜏₂∈(𝑡_𝑘−1,𝑡_𝑘)∩[0,𝜔] and |𝜏₁−𝜏₂|<𝛿, then 𝑥𝜏₁−𝑥𝜏₂<𝜖. (3)

Lemma 2 (see [10]). The set 𝔽⊂𝑃𝐶_𝜔 is relatively compact if and only if (1)𝔽 is bounded, that is, ‖𝑥‖≤𝑀, for each 𝑥∈𝔽, and some 𝑀>0;(2)𝔽 is quasi-equicontinuous in [0,𝜔].

Lemma 3 (see [11]). Assume that 𝜓∈𝑃𝐶¹_𝜔, then the following inequality holds: sup_{[]𝑠∈0,𝜔}𝜓(𝑠)−inf_{[]𝑠∈0,𝜔}𝜓(𝑠)≤12^𝜔₀𝜓̇(𝑠)𝑑𝑠+^𝑚_𝑘=1Δ𝜓𝑡_𝑘. (4) Before starting the main result, for the sake of convenience, one denotes 1𝑓=𝜔^𝜔₀𝑓(𝑡)𝑑𝑡𝑓∈𝑃𝐶_𝜔,𝑐_𝑗=^𝑚_𝑘=1ln1+𝑐_𝑗𝑘𝐶,𝑗=1,2,₁=^𝑚_𝑘=1ln1+𝑐_1𝑘+𝑐₁,𝐶₂=^𝑚_𝑘=1ln1+𝑐_2𝑘−𝑐₂. (5)

3. Existence of Positive Periodic Solutions

In order to obtain the existence of positive periodic solutions of (2), for convenience, we will summarize in the following a few concepts and results from [18] that will be basic for this section.

Let 𝑋, 𝑌 be normed vector spaces, let 𝐿∶Dom𝐿⊂𝑋→𝑌 be a linear mapping, and 𝑁∶𝑋→𝑌 a continuous mapping. The mapping 𝐿 is called a Fredholm mapping of index zero if dimKer𝐿=codimIm𝐿<+∞ and Im𝐿 is closed in 𝑌. If 𝐿 is a Fredholm mapping of index zero, there exist continuous projectors 𝑃∶𝑋→𝑋 and 𝑄∶𝑌→𝑌 such that Im𝑃=Ker𝐿, Ker𝑄=Im𝐿=Im(𝐼−𝑄). It follows that 𝐿∣dom𝐿∩Ker𝑃∶(𝐼−𝑃)𝑋→Im𝐿 is invertible. We denote the inverse of that map by 𝐾_𝑝. If Ω is an open bounded subset of 𝑋, then the mapping 𝑁 will be called 𝐿-compact on Ω if 𝑄𝑁(Ω) is bounded and 𝐾_𝑝(𝐼−𝑄)𝑁∶Ω→𝑋 is compact. Since Im𝑄 is isomorphic to Ker𝐿, there exists an isomorphism 𝐽∶Im𝑄→Ker𝐿.

Lemma 4 (see [18]). Let Ω⊂𝑋 be an open and bounded set. Let 𝐿 be a Fredholm mapping of index zero and let 𝑁 be 𝐿-compact on Ω. Assume(a) for each 𝜆∈(0,1), 𝑥∈𝜕Ω∩Dom𝐿, 𝐿𝑥≠𝜆𝑁𝑥;(b) for each 𝑥∈𝜕Ω∩Ker𝐿, 𝑄𝑁𝑥≠0;(c)deg{𝐽𝑄𝑁,Ω∩Ker𝐿,0}≠0. Then 𝐿𝑥=𝑁𝑥 has at least one solution in Ω∩Dom𝐿.

If 𝑓 is a continuous 𝜔-periodic function, then we set 1𝑓=𝜔^𝜔₀𝑓(𝑡)𝑑𝑡. (6) The following assumptions are valid throughout this paper:(𝐴₁)(4𝐺𝑑ℎ²/(4𝐺ℎ−1))<𝐵<2ℎ(𝑑−(𝑐₂/𝜔)), (𝐴₂)𝐾(1+(𝑐₁/𝑟𝜔))−𝑒^𝐻1>0, 𝐻₁=max{|ln𝐾+2𝑟𝜔|,|ln(2(𝑑−(𝑐₂/𝜔)−(𝐵/2ℎ))/𝑒𝛿𝐵)−2𝑟𝜔|}.

For convenience, we introduce two numbers as follows: 𝑢₊=4𝐺𝑒𝛿𝐵−8𝐺ℎ𝑒𝛿𝑑−𝑐₂/𝜔+√Δ24𝐺𝑑ℎ²𝑒²𝛿²−(4𝐺ℎ−1)𝑒²𝛿²𝐵,𝑢₋=4𝐺𝑒𝛿𝐵−8𝐺ℎ𝑒𝛿𝑑−𝑐₂/𝜔−√Δ24𝐺𝑑ℎ²𝑒²𝛿²−(4𝐺ℎ−1)𝑒²𝛿²𝐵, (7) where Δ=(4𝐺𝑒𝛿𝐵−8𝐺ℎ𝑒𝛿(𝑑−(𝑐₂/𝜔)))²−16𝐺𝑑[4𝐺𝑑ℎ²𝑒²𝛿²−(4𝐺ℎ−1)𝑒²𝛿²𝐵].

Theorem 5. In addition to (𝐴₁), (𝐴₂), suppose that (𝐴₃)112ℎ<𝐺<ℎ. Then system (2) has at least one positive 𝜔-periodic solution.

Remark 6. If the impulsive operators disappear, then 𝑐₁=𝑐₂=0. Then Theorem 5 reduces to the main results in Gao and Xia [9]. This implies that our result generalizes the previous one. It shows that the impulses do affect the system indeed.

Proof. Making the change of variables 𝑁(𝑡)=exp𝑢₁(𝑡),𝑃(𝑡)=exp𝑢₂(𝑡). (8) Then, system (2) can be rewritten as 𝑢̇₁(𝑡)(𝑡)−𝑟(𝑡)=𝑟𝐾exp𝑢₁(𝑡)−4𝐺𝑒𝛿exp𝑢₂(𝑡)+(4𝐺ℎ−1)𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²≡𝑓₁(𝑡),𝑢̇₂(𝑡)=−𝑑(𝑡)+4𝐺𝑒𝛿𝐵(𝑡)exp𝑢₁(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²𝐵(𝑡)exp2𝑢₁(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹≡𝑓₂(𝑡),𝑡≠𝑡_𝑘,Δ𝑢₁𝑡_𝑘=ln1+𝑐_1𝑘,Δ𝑢₂𝑡_𝑘=ln1+𝑐_2𝑘,𝑡=𝑡_𝑘. (9) Take 𝑋=𝑥=𝑢₁,𝑢₂^𝑇∶𝑢_𝑗∈𝑃𝐶_𝜔,𝑗=1,2,𝑥(𝑡+𝜔)=𝑥(𝑡),𝑌=𝑋×ℝ^2𝑞 (10) and define ‖𝑥‖₀=²_𝑗=1sup_{[]𝑡∈0,𝜔}𝑢_𝑗(𝑡),𝑥=𝑢₁,𝑢₂𝑦∈𝑋,₁=‖𝑥‖₀+^𝑞_𝑗=1𝜉_𝑗,𝑦=𝑥,𝜉₁,…,𝜉_𝑞∈𝑌. (11) Both (𝑋,‖⋅‖) and (𝑌,‖⋅‖₁) are Banach spaces.
Define {̇}𝐿𝑢Dom𝐿=𝑥∈𝑋∶𝑥∈𝑋,𝐿∶Dom𝐿⟶𝑌,₁,𝑢₂^𝑇=𝑢̇₁𝑢̇₂,Δ𝑢₁𝑡_𝑘Δ𝑢₂𝑡_𝑘^𝑞_𝑘=1; (12)𝑁∶𝑋→𝑌, 𝑁𝑢₁𝑢₂=𝑓₁(𝑡)𝑓₂(𝑡),ln1+𝑐_1𝑘ln1+𝑐_2𝑘^𝑞_𝑘=1; (13)𝑃∶𝑋→𝑋, 𝑃((𝑢₁,𝑢₂)^𝑇)=(𝑢₁,𝑢₂)^𝑇; 𝑄∶𝑌→𝑌, 𝑄𝑢₁𝑢₂,𝑚_𝑘𝑛_𝑘^𝑞_𝑘=1=𝑢₁+1𝜔^𝑞_𝑘=1𝑚_𝑘𝑢₂+1𝜔^𝑞_𝑘=1𝑛_𝑘,00^𝑞_𝑘=1. (14) It is not difficult to show that Ker𝐿={𝑥∣𝑥∈𝑋,(𝑢₁,𝑢₂)^𝑇∈𝑋∶(𝑢₁(𝑡),𝑢₂(𝑡))^𝑇=(𝑐₁,𝑐₂)^𝑇∈ℝ²,𝑡∈ℝ}.=𝑢Im𝐿₁𝑢₂,𝑚_𝑘𝑛_𝑘^𝑞_𝑘=1𝜔∈𝑌∶𝑢₁+∑^𝑞_𝑘=1𝑚_𝑘𝜔=0𝑢₂+∑^𝑞_𝑘=1𝑛_𝑘.=0 (15) Since Im𝐿 is closed in 𝑌, 𝑃 and 𝑄 are continuous projectors such that Im𝑃=Ker𝐿,Ker𝑄=Im𝐿=Im(𝐼−𝑄),dimKer𝐿=codimIm𝐿=2. (16) It follows that 𝐿 is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to 𝐿) 𝐾_𝑝∶Im𝐿→Dom𝐿∩Ker𝑃 exists, which is given by 𝐾_𝑝𝑢₁𝑢₂,𝑚_𝑘𝑛_𝑘^𝑞_𝑘=1=^𝑡0𝑢₁(𝑠)𝑑𝑠+_{0<𝑡𝑘<𝑡}𝑚_𝑘−1𝜔^𝑞_𝑘=1𝑚_𝑘−^𝑡₀𝑢₁(𝑠)𝑑𝑠_𝑡0𝑢₂(𝑠)𝑑𝑠+_{0<𝑡𝑘<𝑡}𝑛_𝑘−1𝜔^𝑞_𝑘=1𝑛_𝑘−^𝑡₀𝑢₂(𝑠).𝑑𝑠 (17) Then 𝑄𝑁∶𝑋→𝑌 and 𝐾_𝑝(𝐼−𝑄)𝑁∶𝑋→𝑋 are defined by 𝑢𝑄𝑁¹𝑢₂=𝑓₁+𝑐₁𝜔𝑓₂+𝑐₂𝜔,00^𝑞_𝑘=1,𝐾_𝑝()𝑁𝑢𝐼−𝑄¹𝑢₂=^𝑡0𝑓₁(𝑠)𝑑𝑠+_{0<𝑡𝑘<𝑡}ln1+𝑐_1𝑘𝑡0𝑓₂(𝑠)𝑑𝑠+_{0<𝑡𝑘<𝑡}ln1+𝑐_2𝑘+12−𝑡𝜔^𝜔0𝑓₁(𝑡)𝑑𝑡+𝑐_1𝜔0𝑓₂(𝑡)𝑑𝑡+𝑐₂−1𝜔^𝜔0𝑡0𝑓₁(𝑠)𝑑𝑠𝑑𝑡+𝑐_1𝜔0𝑡0𝑓₂(𝑠)𝑑𝑠𝑑𝑡+𝑐₂. (18) Clearly, 𝑄𝑁 and 𝐾_𝑝(𝐼−𝑄)𝑁 are continuous. By using the Arzela-Ascoli theorem (see [10]), it is not difficult to prove that 𝐾_𝑝(𝐼−𝑄)𝑁(Ω) is compact for any open bounded set Ω⊂𝑋. Moreover, 𝑄𝑁(Ω) is bounded. Therefore, 𝑁 is 𝐿-compact on Ω with any open bounded set Ω⊂𝑋.
Now, we reach the position to search for an appropriate open, bounded subset Ω for the application of the continuation theorem.
Corresponding to the operator equation 𝐿𝑥=𝜆𝑁𝑥, 𝜆∈(0,1), we have 𝑢̇₁(𝑡)𝑟(𝑡)−𝑟(𝑡)=𝜆𝐾exp𝑢₁(𝑡)−4𝐺𝑒𝛿exp𝑢₂(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹,𝑢̇₂(𝑡)(𝑡)+(𝑡)𝑢=𝜆−𝑑4𝐺𝑒𝛿𝐵exp₁(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²𝐵(𝑡)exp2𝑢₁(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹,𝑡≠𝑡_𝑘,Δ𝑢₁𝑡_𝑘=𝜆ln1+𝑐_1𝑘,Δ𝑢₂𝑡_𝑘=𝜆ln1+𝑐_2𝑘,𝑡=𝑡_𝑘. (19) Suppose 𝑥=(𝑢₁(𝑡),𝑢₂(𝑡))^𝑇∈𝑋 is a solution of (19) for a certain 𝜆∈(0,1). Integrating the first equation of (19) over the interval [0,𝜔], we obtain ^𝜔₀𝑟(𝑡)𝐾exp𝑢₁(𝑡)𝑑𝑡+^𝜔₀4𝐺𝑒𝛿exp𝑢₂(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹=𝑑𝑡𝑟𝜔+𝑐₁. (20) Similarly, integrating the second equation of (19) over the interval [0,𝜔], we obtain 4𝐺𝑒𝛿𝐵(^𝜔₀𝑡)exp𝑢₁(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²𝐵(𝑡)exp2𝑢₁(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹=𝑑𝑡𝑑𝜔−𝑐₂. (21) It follows from the first equation of (19) and (20) and (𝐴₂) that 𝑢̇^𝜔₀₁(𝑡)𝑑𝑡=𝜆^𝜔₀𝑟(𝑡)−𝑟(𝑡)𝐾exp𝑢₁(𝑡−𝑢)4𝐺𝑒𝛿exp₂(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹<𝑑𝑡_𝜔0𝑟(𝑡)𝑑𝑡+^𝜔₀𝑟(𝑡)𝐾exp𝑢₁(𝑡+)𝑑𝑡_𝜔04𝐺𝑒𝛿exp𝑢₂(𝑡)+(4𝐺ℎ−1)×𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)×4𝐺1+ℎ𝑒𝛿exp𝑢₁(𝑡)²⁻¹=𝑑𝑡_𝜔0𝑟(𝑡)𝑑𝑡+𝑟𝜔+𝑐₁=2𝑟𝜔+𝑐₁. (22) That is, 𝑢̇^𝜔₀₁(𝑡)𝑑𝑡<2𝑟𝜔+𝑐₁. (23) Similarly, it follows from the second equation of (19) and (21) and (𝐴₂) that 𝑢̇^𝜔₀₂(𝑡)𝑑𝑡<2𝑑𝜔−𝑐₂. (24) Since (𝑢₁(𝑡),𝑢₂(𝑡))^𝑇∈𝑋, there exists 𝜉_𝑖,𝜂_𝑖∈[0,𝜔] such that 𝑢_𝑖𝜉_𝑖=inf_{[]𝑡∈0,𝜔}𝑢_𝑖(𝑡),𝑢_𝑖𝜂_𝑖=sup_{[]𝑡∈0,𝜔}𝑢_𝑖(𝑡),𝑖=1,2. (25)
From (20), we see that 𝑟𝜔+𝑐₁≥^𝜔₀𝑟(𝑡)𝐾exp𝑢₁(𝑡)𝑑𝑡, (26) which implies 𝑐𝑟+¹𝜔≥𝑟𝐾exp𝑢₁𝜉₁. (27) So 𝑢₁𝜉₁𝑐≤ln𝐾1+¹𝑟𝜔∶=Δ₁₁. (28) This, combined with (23), gives 𝑢₁(𝑡)≤𝑢₁𝜉₁+^𝜔₀𝑢̇₁(𝑡)𝑑𝑡<Δ₁₁+2𝑟𝜔∶=𝐻₁₁. (29) Similarly, it follows from (21) that 𝑑𝜔−𝑐₂≤^𝜔₀4𝐺𝑒𝛿𝐵(𝑡)exp𝑢₁(𝑡)8𝐺ℎ𝑒𝛿exp𝑢₁(𝑡)+𝑑𝑡_𝜔04𝐺ℎ𝑒²𝛿²𝐵(𝑡)exp2𝑢₁(𝑡)8𝐺ℎ𝑒𝛿exp𝑢₁(𝑡)𝑑𝑡, (30) which implies 𝑐𝑑−²𝜔≤𝐵+2ℎ𝑒𝛿𝐵2exp𝑢₁𝜂₁. (31) It follows from (𝐴₁) that 𝑢₁𝜂₁≥ln22𝑑−𝐵/2ℎ−𝑐₂/𝜔𝑒𝛿𝐵∶=Δ₁₂. (32) This, combined with (23), gives 𝑢₁(𝑡)≥𝑢₁𝜂₁−^𝜔₀𝑢̇₁(𝑡)𝑑𝑡>Δ₁₂−2𝑟𝜔∶=𝐻₁₂. (33) It follows from (29) and (33) that max_{[]𝑡∈0,𝜔}𝑢₁(𝑡)<max𝐻₁₁,𝐻₁₂∶=𝐻₁. (34)
On the other hand, it follows from (21) and (34) that 𝑟𝜔+𝑐₁≤𝑟𝜔𝐾exp𝐻₁+^𝜔₀4𝐺𝑒𝛿exp𝑢₂(𝑡)+4𝐺𝑑𝑡_𝜔04𝐺ℎ𝑒²𝛿²exp𝑢₁(𝑡)+𝑢₂(𝑡)8𝐺ℎ𝑒𝛿exp𝑢₁(𝑡)𝑑𝑡, (35) which implies 𝑐𝑟+¹𝜔≤𝑟𝐾exp𝐻₁+32𝑒𝛿exp𝑢₂𝜂₂. (36) It follows from (𝐴₂) that 𝑢₂𝜂₂2≥ln𝑟1+𝑐₁/𝑟𝜔−𝑒^𝐻1/𝐾3𝑒𝛿∶=Δ₂₂. (37) This, combined with (25), gives 𝑢₂(𝑡)≥𝑢₂𝜂₂−^𝜔₀𝑢̇₂(𝑡)𝑑𝑡>Δ₂₂−2𝑑𝜔∶=𝐻₂₂. (38)
Similarly, it follows from (21) and (34) that 𝑟𝜔+𝑐₁≥^𝜔₀4𝐺𝑒𝛿exp𝑢₂𝜉₂4𝐺1+ℎ𝑒𝛿exp𝐻₁²𝑑𝑡, (39) which implies 𝑐𝑟+¹𝜔≥𝑒𝛿exp𝑢₂𝜉₂1+ℎ𝑒𝛿exp𝐻₁². (40) So 𝑢₂𝜉₂≤ln𝑟1+𝑐₁/𝑟𝜔1+ℎ𝑒𝛿exp𝐻₁²𝑒𝛿∶=Δ₂₁. (41) This, combined with (25), gives 𝑢₂(𝑡)≤𝑢₂𝜉₂+^𝜔₀𝑢̇₂(𝑡)𝑑𝑡<Δ₂₁+2𝑑𝜔∶=𝐻₂₁. (42) It follows from (38) and (42) that max_{[]𝑡∈0,𝜔}𝑢₂(𝑡)<max𝐻₂₁,𝐻₂₂∶=𝐻₂. (43) Now, let us consider 𝑄𝑁𝑥 with 𝑥=(𝑢₁,𝑢₂)^𝑇∈ℝ². Note that 𝑄𝑁𝑢₁,𝑢₂=𝑟−𝑟𝐾exp𝑢₁−4𝐺𝑒𝛿exp𝑢₂+(4𝐺ℎ−1)𝑒²𝛿²exp𝑢₁+𝑢₂4𝐺1+ℎ𝑒𝛿exp𝑢₁²+𝑐₁𝜔,−𝑑+4𝐺𝑒𝛿𝐵exp𝑢₁+(4𝐺ℎ−1)𝑒²𝛿²𝐵exp2𝑢₁4𝐺1+ℎ𝑒𝛿exp𝑢₁²+𝑐₂𝜔. (44) It follows from (𝐴₁), (𝐴₂), and (𝐴₃) that 𝑢₊<0, which implies that the equation 𝑄𝑁(𝑢₁,𝑢₂)=0 has only one solution 𝑢=ln𝑢₋,ln4𝐺𝑟−(𝑟/𝐾)𝑢₋1+ℎ𝑒𝛿𝑢₋²4𝐺𝑒𝛿+(4𝐺ℎ−1)𝑒²𝛿²𝑢₋. (45) Choose 𝐶>0 such that 𝐶>ln4𝐺𝑟−(𝑟/𝐾)𝑢₋1+ℎ𝑒𝛿𝑢₋²4𝐺𝑒𝛿+(4𝐺ℎ−1)𝑒²𝛿²𝑢₋. (46) Set 𝐻=‖(𝐻₁,𝐻₂)^𝑇‖+𝐶; then ‖𝑥‖<𝐻. Let Ω=𝑥(𝑡)=𝑢₁(𝑡),𝑢₂(𝑡)^𝑇∈𝑋∶‖𝑥(𝑡)‖<𝐻. (47) It is clear that Ω verifies the requirement (a) in Lemma 4. When 𝑥∈Ker𝐿∩𝜕Ω, 𝑥 is a constant with ‖𝑥‖=𝐻. Then 𝑄𝑁𝑥≠0 for 𝑥∈Ker𝐿∩𝜕Ω. Simple computation shows that deg{𝐽𝑄𝑁,Ω∩Ker𝐿,0}≠0. Here, 𝐽 is taken as the identity mapping since Im𝑄=Ker𝐿.

By now, we have proved that Ω verifies all the requirements in Lemma 4. Hence, (2) has at least one 𝜔-periodic solution in Ω.

Acknowledgment

This work is supported by NNSFC and the Natural Science Foundation of Fujiang Province (2013J01010).