Neutral Operator and Neutral Differential Equation

Abstract

In this paper, we discuss the properties of the neutral operator (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), and by applying coincidence degree theory and fixed point index theory, we obtain sufficient conditions for the existence, multiplicity, and nonexistence of (positive) periodic solutions to two kinds of second-order differential equations with the prescribed neutral operator.

1. Introduction

In [1], Zhang discussed the properties of the neutral operator (𝐴₁𝑥)(𝑡)=𝑥(t)−𝑐𝑥(𝑡−𝛿), which became an effective tool for the research on differential equations with this prescribed neutral operator, see, for example, [2–5]. Lu and Ge [6] investigated an extension of 𝐴₁, namely, the neutral operator 𝐴₂𝑥(𝑡)=𝑥(𝑡)−∑^𝑛_𝑖=1𝑐_𝑖𝑥(𝑡−𝛿_𝑖) and obtained the existence of periodic solutions for a corresponding neutral differential equation.

In this paper, we consider the neutral operator (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), where 𝑐 is constant and |𝑐|≠1, 𝛿∈𝐶¹(ℝ,ℝ), and 𝛿 is an 𝜔-periodic function for some 𝜔>0. Although 𝐴 is a natural generalization of the operator 𝐴₁, the class of neutral differential equation with 𝐴 typically possesses a more complicated nonlinearity than neutral differential equation with 𝐴₁ or 𝐴₂. For example, the neutral operators 𝐴₁ and 𝐴₂ are homogeneous in the following sense (𝐴_𝑖𝑥)^′(𝑡)=(𝐴_𝑖𝑥^′)(𝑡) for 𝑖=1,2, whereas the neutral operator 𝐴 in general is inhomogeneous. As a consequence many of the new results for differential equations with the neutral operator 𝐴 will not be a direct extension of known theorems for neutral differential equations.

The paper is organized as follows: in Section 2, we first analyze qualitative properties of the neutral operator 𝐴 which will be helpful for further studies of differential equations with this neutral operator; in Section 3, by Mawhin's continuation theorem, we obtain the existence of periodic solutions for a second-order Rayleigh-type neutral differential equation; in Section 4, by an application of the fixed point index theorem we obtain sufficient conditions for the existence, multiplicity, and nonexistence of positive periodic solutions to second-order neutral differential equation. Several examples are also given to illustrate our results. Our results improve and extend the results in [1, 2, 4, 7].

2. Analysis of the Generalized Neutral Operator

Let 𝐶_𝜔={𝑥∈𝐶(ℝ,ℝ)∶𝑥(𝑡+𝜔)=𝑥(𝑡),𝑡∈ℝ} with norm ‖𝑥‖=max_{𝑡∈[0,𝜔]}|𝑥(𝑡)|. Then (𝐶_𝜔,‖⋅‖) is a Banach space. A cone 𝐾 in 𝐶_𝜔 is defined by 𝐾={𝑥∈𝐶_𝜔∶𝑥(𝑡)≥𝛼‖𝑥‖,forall𝑡∈ℝ}, where 𝛼 is a fixed positive number with 𝛼<1. Moreover, define operators 𝐴,𝐵∶𝐶_𝜔→𝐶_𝜔 by (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)),(𝐵𝑥)(𝑡)=𝑐𝑥(𝑡−𝛿(𝑡)). (2.1)

Lemma 2.1. If |𝑐|≠1, then the operator 𝐴 has a continuous inverse 𝐴⁻¹ on 𝐶_𝜔, satisfying
(1)𝐴⁻¹𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=1𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖,for|𝑐|<1,∀𝑓∈𝐶_𝜔,−𝑓(𝑡+𝛿(𝑡))𝑐−^∞_𝑗=11/𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1𝛿𝐷_𝑖,for|𝑐|>1,∀𝑓∈𝐶_𝜔. (2.2)(2)|(𝐴⁻¹𝑓)(𝑡)|≤‖𝑓‖/|1−|𝑐‖,forall𝑓∈𝐶_𝜔. (3)∫^𝜔₀|(𝐴⁻¹𝑓)(𝑡)|𝑑𝑡≤1/|1−|𝑐||∫^𝜔₀|𝑓(𝑡)|𝑑𝑡,forall𝑓∈𝐶_𝜔.

Proof. We have the following cases
Case 1 (|𝑐|<1). Let 𝑡−𝛿(𝑡)=𝑠 and 𝐷_𝑗=𝑠−∑^𝑗−1_𝑖=1𝛿(𝐷_𝑖), 𝑗=1,2,…. Therefore, 𝐵^𝑗𝑥(𝑡)=𝑐^𝑗𝑥𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖,_∞𝑗=0𝐵^𝑗𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=1𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖. (2.3) Since 𝐴=𝐼−𝐵, we get from ‖𝐵‖≤|𝑐|<1 that 𝐴 has a continuous inverse 𝐴⁻¹∶𝐶_𝜔→𝐶_𝜔 with 𝐴⁻¹=(𝐼−𝐵)⁻¹=𝐼+^∞_𝑗=1𝐵^𝑗=^∞_𝑗=0𝐵^𝑗, (2.4) where 𝐵⁰=𝐼. Then 𝐴⁻¹𝑓(𝑡)=^∞_𝑗=0𝐵^𝑗𝑓(𝑡)=^∞_𝑗=0𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖, (2.5) and consequently 𝐴⁻¹𝑓(𝑡)=^∞_𝑗=0𝐵^𝑗𝑓(𝑡)=^∞_𝑗=0𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖≤𝑓1−|𝑐|. (2.6) Moreover, ^𝜔₀𝐴⁻¹𝑓(𝑡)𝑑𝑡=^𝜔₀^∞_𝑗=0𝐵^𝑗𝑓𝑡)≤(𝑑𝑡_{∞𝑗=0𝜔0}𝐵^𝑗𝑓(𝑡=)𝑑𝑡_{∞𝑗=0𝜔0}𝑐^𝑗𝑓𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖≤1𝑑𝑡1−|𝑐|^𝜔₀𝑓(𝑡)𝑑𝑡. (2.7)
Case 2 (|𝑐|>1). Let 𝐸∶𝐶_𝜔⟶𝐶_𝜔,(𝑡)(𝑡)−1𝐸𝑥)(=𝑥𝑐𝑥((𝑡,𝐵𝑡+𝛿))₁∶𝐶_𝜔⟶𝐶_𝜔,𝐵₁𝑥(𝑡)=1𝑐𝑥(𝑡+𝛿(𝑡)). (2.8) By definition of the linear operator 𝐵₁, we have 𝐵^𝑗₁𝑓(𝑡)=1𝑐^𝑗𝑓𝑠+^𝑗−1_𝑖=1𝛿𝐷_𝑖, (2.9) where 𝐷_𝑖 is defined as in Case 1. Summing over 𝑗 yields ^∞_𝑗=0𝐵^𝑗₁𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=11𝑐^𝑗𝑓𝑠+^𝑗−1_𝑖=1𝛿𝐷_𝑖. (2.10) Since ‖𝐵₁‖<1, we obtain that the operator 𝐸 has a bounded inverse 𝐸⁻¹, 𝐸⁻¹∶𝐶_𝜔⟶𝐶_𝜔,𝐸⁻¹=𝐼−𝐵₁⁻¹=𝐼+_𝑗=1𝐵^𝑗₁, (2.11) and forall𝑓∈𝐶_𝜔 we get 𝐸⁻¹𝑓(𝑡)=𝑓(𝑡)+^∞_𝑗=1𝐵^𝑗₁𝑓(𝑡). (2.12) On the other hand, from (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), we have (𝑡)(𝑡)((𝑡𝑥((𝑡−1𝐴𝑥)(=𝑥−𝑐𝑥𝑡−𝛿))=−𝑐𝑡−𝛿))𝑐𝑥(𝑡), (2.13) that is, (𝐴𝑥)(𝑡)=−𝑐(𝐸𝑥)(𝑡−𝛿(𝑡)). (2.14) Let 𝑓∈𝐶_𝜔 be arbitrary. We are looking for 𝑥 such that (𝐴𝑥)(𝑡)=𝑓(𝑡). (2.15) that is, −𝑐(𝐸𝑥)(𝑡−𝛿(𝑡))=𝑓(𝑡). (2.16) Therefore, (𝑡)𝑓((𝑡𝐸𝑥)(=−𝑡+𝛿))𝑐=∶𝑓₁(𝑡), (2.17) and hence 𝑥(𝑡)=𝐸⁻¹𝑓₁(𝑡)=𝑓₁(𝑡)+^∞_𝑗=1𝐵^𝑗₁𝑓₁(𝑡)𝑓((𝑡=−𝑡+𝛿))𝑐−^∞_𝑗=1𝐵^𝑗₁𝑓(𝑡+𝛿(𝑡))𝑐, (2.18) proving that 𝐴⁻¹ exists and satisfies 𝐴⁻¹𝑓𝑡)𝑓((𝑡(=−𝑡+𝛿))𝑐−^∞_𝑗=1𝐵^𝑗₁𝑓(𝑡+𝛿(𝑡))𝑐𝑓((𝑡=−𝑡+𝛿))𝑐−^∞_𝑗=11𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1𝛿𝐷_𝑖,𝐴⁻¹𝑓𝑡=−𝑓((𝑡()𝑡+𝛿))𝑐−^∞_𝑗=11𝑐^𝑗+1𝑓𝑠+𝛿(𝑡)+^𝑗−1_𝑖=1𝛿𝐷_𝑖≤𝑓|𝑐|.−1 (2.19) Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced.

Lemma 2.2. If 𝑐<0 and |𝑐|<𝛼, one has for 𝑦∈𝐾 that 𝛼−|𝑐|1−𝑐²𝑦≤𝐴⁻¹𝑦(𝑡)≤1|𝑐|𝑦.1− (2.20)

Proof. Since 𝑐<0 and |𝑐|<𝛼<1, by Lemma 2.1, we have for 𝑦∈𝐾 that 𝐴⁻¹𝑦(𝑡)=𝑦(𝑡)+^∞_𝑗=1𝑐^𝑗𝑦𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖=𝑦(𝑡)+_𝑗≥1even𝑐^𝑗𝑦𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖−_𝑗≥1odd|𝑐|^𝑗𝑦𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖≥𝛼𝑦+𝛼_𝑗≥1even𝑐^𝑗𝑦−𝑦_𝑗≥1odd|𝑐|^𝑗=𝛼1−𝑐²𝑦−|𝑐|1−𝑐²𝑦=𝛼−|𝑐|1−𝑐²𝑦. (2.21)

Lemma 2.3. If 𝑐>0 and 𝑐<1 then for 𝑦∈𝐾 one has 𝛼𝑦≤𝐴1−𝑐⁻¹𝑦(𝑡)≤1𝑦.1−𝑐 (2.22)

Proof. Since 𝑐>0 and 𝑐<1, 𝛼<1, by Lemma 2.1, we have for 𝑦∈𝐾 that 𝐴⁻¹𝑦(𝑡)=𝑦(𝑡)+_𝑗≥1𝑐^𝑗𝑦𝑠−^𝑗−1_𝑖=1𝛿𝐷_𝑖≥𝛼𝑦+𝛼𝑦_𝑗≥1𝑐^𝑗=𝛼𝑦.1−𝑐 (2.23)

3. Periodic Solutions for Neutral Differential Equation

In this section, we consider the second-order neutral differential equation(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))^′′=𝑓𝑡,𝑥^′(𝑡)+𝑔(𝑡,𝑥(𝑡−𝜏(𝑡)))+𝑒(𝑡), (3.1) where 𝜏,𝑒∈𝐶_𝜔 and ∫^𝜔₀𝑒(𝑡)𝑑𝑡=0; 𝑓 and 𝑔 are continuous functions defined on ℝ² and periodic in 𝑡 with 𝑓(𝑡,⋅)=𝑓(𝑡+𝜔,⋅),𝑔(𝑡,⋅)=𝑔(𝑡+𝜔,⋅), 𝑓(𝑡,0)=0, 𝑓(𝑡,𝑢)≥0, or 𝑓(𝑡,𝑢)≤0 for all (𝑡,𝑢)∈ℝ².

We first recall Mawhin's continuation theorem which our study is based upon. Let 𝑋 and 𝑌 be real Banach spaces and 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 a Fredholm operator with index zero, where 𝐷(𝐿) denotes the domain of 𝐿. This means that Im𝐿 is closed in 𝑌 and dimKer𝐿=dim(𝑌/Im𝐿)<+∞. Consider supplementary subspaces 𝑋₁, 𝑌₁, of 𝑋, 𝑌 respectively, such that 𝑋=Ker𝐿⊕𝑋₁, 𝑌=Im𝐿⊕𝑌₁, and let 𝑃₁∶𝑋→Ker𝐿 and 𝑄₁∶𝑌→𝑌₁ denote the natural projections. Clearly, Ker𝐿∩(𝐷(𝐿)∩𝑋₁)={0}, thus the restriction 𝐿_𝑃1∶=𝐿|_{𝐷(𝐿)∩𝑋1} is invertible. Let 𝐿_𝑃⁻¹₁ denote the inverse of 𝐿_𝑃1.

Let Ω be an open bounded subset of 𝑋 with 𝐷(𝐿)∩Ω≠∅. A map 𝑁∶Ω→𝑌 is said to be 𝐿-compact in Ω if 𝑄₁𝑁(Ω) is bounded and the operator 𝐿_𝑃⁻¹₁(𝐼−𝑄₁)𝑁∶Ω→𝑋 is compact.

Lemma 3.1 (Gaines and Mawhin [8]). Suppose that 𝑋 and 𝑌 are two Banach spaces and 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 is a Fredholm operator with index zero. Furthermore, Ω⊂𝑋 is an open bounded set, and 𝑁∶Ω→𝑌 is 𝐿-compact on Ω. Assume that the following conditions hold:
(1)𝐿𝑥≠𝜆𝑁𝑥,forall𝑥∈𝜕Ω∩𝐷(𝐿),𝜆∈(0,1);(2)𝑁𝑥∉Im𝐿,forall𝑥∈𝜕Ω∩Ker𝐿;(3)deg{𝐽𝑄₁𝑁,Ω∩Ker𝐿,0}≠0, where 𝐽∶Im𝑄₁→Ker𝐿 is an isomorphism.
Then the equation 𝐿𝑥=𝑁𝑥 has a solution in Ω∩𝐷(𝐿).

In order to use Mawhin's continuation theorem to study the existence of 𝜔-periodic solutions for (3.1), we rewrite (3.1) in the following form:𝐴𝑥₁^′(𝑡)=𝑥₂(𝑡),𝑥^′₂(𝑡)=𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝑒(𝑡). (3.2) Clearly, if 𝑥(𝑡)=(𝑥₁(𝑡),𝑥₂(𝑡))^⊤ is an 𝜔-periodic solution to (3.2), then 𝑥₁(𝑡) must be an 𝜔-periodic solution to (3.1). Thus, the problem of finding an 𝜔-periodic solution for (3.1) reduces to finding one for (3.2).

Recall that 𝐶_𝜔={𝜙∈𝐶(ℝ,ℝ)∶𝜙(𝑡+𝜔)≡𝜙(𝑡)} with norm ‖𝜙‖=max_{𝑡∈[0,𝜔]}|𝜙(𝑡)|. Define 𝑋=𝑌=𝐶_𝜔×𝐶_𝜔={𝑥=(𝑥₁(⋅),𝑥₂(⋅))∈𝐶(ℝ,ℝ²)∶𝑥(𝑡)=𝑥(𝑡+𝜔),𝑡∈ℝ} with norm ‖𝑥‖=max{‖𝑥₁‖,‖𝑥₂‖}. Clearly, 𝑋 and 𝑌 are Banach spaces. Moreover, define 𝐿∶𝐷(𝐿)=𝑥∈𝐶¹ℝ,ℝ²∶𝑥(𝑡+𝜔)=𝑥(𝑡),𝑡∈ℝ⊂𝑋⟶𝑌 (3.3) by (𝐿𝑥)(𝑡)=𝐴𝑥₁^′(𝑡)𝑥^′₂(𝑡) (3.4) and 𝑁∶𝑋→𝑌 by(𝑡)=𝑥𝑁𝑥)(²(𝑡)𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝑒(𝑡). (3.5) Then (3.2) can be converted to the abstract equation 𝐿𝑥=𝑁𝑥. From the definition of 𝐿, one can easily see that Ker𝐿≅ℝ²,Im𝐿=𝑦∈𝑌∶^𝜔₀𝑦₁(𝑠)𝑦₂(𝑠)00𝑑𝑠=. (3.6) So 𝐿 is a Fredholm operator with index zero. Let 𝑃₁∶𝑋→Ker𝐿 and 𝑄₁∶𝑌→Im𝑄₁⊂ℝ² be defined by 𝑃₁𝑥=𝐴𝑥₁(0)𝑥₂(0);𝑄₁1𝑦=𝜔^𝜔₀𝑦₁(𝑠)𝑦₂(𝑠)𝑑𝑠, (3.7) then Im𝑃₁=Ker𝐿, Ker𝑄₁=Im𝐿. Setting 𝐿_𝑃1=𝐿|_{D(𝐿)∩Ker𝑃1} and 𝐿_𝑃⁻¹₁∶ Im𝐿→𝐷(𝐿) denotes the inverse of 𝐿_𝑃1, then 𝐿_𝑃⁻¹₁𝑦𝑡)=𝐴(⁻¹𝐹𝑦₁(𝑡)𝐹𝑦₂𝑡),(𝐹𝑦₁(𝑡)=^𝑡₀𝑦₁(𝑠)𝑑𝑠,𝐹𝑦₂(𝑡)=^𝑡₀𝑦₂(𝑠)𝑑𝑠. (3.8) From (3.5) and (3.8), it is clear that 𝑄₁𝑁 and 𝐿_𝑃⁻¹₁(𝐼−𝑄₁)𝑁 are continuous and 𝑄₁𝑁(Ω) is bounded, and then 𝐿_𝑃⁻¹₁(𝐼−𝑄₁)𝑁(Ω) is compact for any open bounded Ω⊂𝑋 which means 𝑁 is 𝐿-compact on Ω.

Now we give our main results on periodic solutions for (3.1).

Theorem 3.2. Suppose there exist positive constants 𝐾₁,𝐷,𝑀,𝑏 with 𝑀>‖𝑒‖ such that:
(H₁)|𝑓(𝑡,𝑢)|≤𝐾₁|𝑢|+𝑏, for (𝑡,𝑢)∈ℝ×ℝ;
(H₂) sgn𝑥⋅𝑔(𝑡,𝑥)>‖𝑒‖, for |𝑥|>𝐷;
(H₃)𝑔(𝑡,𝑥)≥−𝑀, for 𝑥≤−𝐷 and 𝑡∈ℝ.
Then (3.1) has at least one solution with period 𝜔 if 0<𝜔^1/2(1+|𝑐|)^1/22𝐾₁/(|1−|𝑐||−|𝑐|𝛿₁)<1, where 𝛿₁=max_{𝑡∈[0,𝜔]}|𝛿′(𝑡)|.

Proof. By construction (3.2) has an 𝜔-periodic solution if and only if the following operator equation 𝐿𝑥=𝑁𝑥 (3.9) has an 𝜔-periodic solution. From (3.8), we see that 𝑁 is 𝐿-compact on Ω, where Ω is any open, bounded subset of 𝐶_𝜔. For 𝜆∈(0,1] define Ω₁=𝑥∈𝐶_𝜔∶𝐿𝑥=𝜆𝑁𝑥. (3.10) Then 𝑥=(𝑥₁,𝑥₂)^⊤∈Ω₁ satisfies 𝐴𝑥₁^′(𝑡)=𝜆𝑥₂(𝑡),𝑥^′₂(𝑡)=𝜆𝑓𝑡,𝑥^′₁(𝑡)+𝜆𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝜆𝑒(𝑡). (3.11) We first claim that there is a constant 𝜉∈ℝ such that 𝑥₁(𝜉)≤𝐷. (3.12) In view of ∫^𝜔₀(𝐴𝑥₁)^′(𝑡)𝑑𝑡=0, we know that there exist two constants 𝑡₁,𝑡₂∈[0,𝜔] such that (𝐴𝑥₁)^′(𝑡₁)≥0,(𝐴𝑥₁)^′(𝑡₂)≤0. From the first equation of (3.11), we have 𝑥₂(𝑡)=(1/𝜆)(𝐴𝑥₁)^′(𝑡), so 𝑥₂𝑡₁=1𝜆𝐴𝑥₁^′𝑡₁𝑥≥0,₂𝑡₂=1𝜆𝐴𝑥₁^′𝑡₂≤0. (3.13) Let 𝑡₃,𝑡₄∈[0,𝜔] be, respectively, a global maximum and minimum point of 𝑥₂(𝑡). Clearly, we have 𝑥₂𝑡₃≥0,𝑥^′₂𝑡₃𝑥=0,₂𝑡₄≤0,𝑥^′₂𝑡₄=0. (3.14) Since 𝑓(𝑡,𝑥^′₁)≥0 or 𝑓(𝑡,𝑥^′₁)≤0, w.l.o.g., suppose 𝑓(𝑡,𝑥^′₁)≥0, for (𝑡,𝑥^′₁)∈[0,𝜔]×ℝ. Then −𝑔𝑡₃,𝑥₁𝑡₃−𝜏𝑡₃−𝑒𝑡₃=𝑓𝑡,𝑥^′₁𝑡₃𝑔𝑡≥0,₃,𝑥₁𝑡₃−𝜏𝑡₃≤−𝑒𝑡₃≤‖𝑒‖. (3.15) From (𝐻₂) we see that 𝑥₁𝑡₃−𝜏𝑡₃<𝐷. (3.16) Similarly, we have 𝑔𝑡₄,𝑥₁𝑡₄−𝜏𝑡₄≥−𝑒𝑡₄≥−‖𝑒‖, (3.17) and again by (𝐻₂), 𝑥₁𝑡₄−𝜏𝑡₄<−𝐷. (3.18)
Case 1. If 𝑥₁(𝑡₃−𝜏(𝑡₃))∈(−𝐷,𝐷), define 𝜉=𝑡₃−𝜏(𝑡₃), obviously |𝑥₁(𝜉)|≤𝐷.
Case 2. If 𝑥₁(𝑡₃−𝜏(𝑡₃))<−𝐷, from (3.18) and the fact that 𝑥 is a continuous function in ℝ, there exists a constant 𝜉 between 𝑥₁(𝑡₃−𝜏(𝑡₃)) and 𝑥₁(𝑡₄−𝜏(𝑡₄)) such that |𝑥₁(𝜉)|=𝐷. This proves (3.12).
Choose an integer 𝑘 and a constant 𝑡₅∈[0,𝜔] such that 𝜉=𝜔𝑘+𝑡₅, then |𝑥₁(𝜉)|=|𝑥₁(𝑡₅)|≤𝐷. Hence 𝑥₁(𝑡)≤𝐷+^𝜔₀𝑥^′₁(𝑠)𝑑𝑠. (3.19) Substituting 𝑥₂(𝑡)=(1/𝜆)(𝐴𝑥₁)^′(𝑡) into the second equation of (3.11) yields 1𝜆𝐴𝑥₁(𝑡)^′′=𝜆𝑓𝑡,𝑥^′₁(𝑡)+𝜆𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝜆𝑒(𝑡), (3.20) that is, 𝐴𝑥₁(𝑡)^′′=𝜆²𝑓𝑡,𝑥^′₁(𝑡)+𝜆²𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝜆²𝑒(𝑡). (3.21) Integrating both sides of (3.21) over [0,𝜔], we have ^𝜔₀𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡=0. (3.22) On the other hand, multiplying both sides of (3.21) by (𝐴𝑥₁)(𝑡) and integrating over [0,𝜔], we get ^𝜔₀𝐴𝑥₁(𝑡)^′′𝐴𝑥₁(𝑡)𝑑𝑡=−^𝜔₀𝐴𝑥₁^′(𝑡)²𝑑𝑡=−𝜆^2𝜔₀𝑓𝑡,𝑥^′₁(𝑡)𝐴𝑥₁(𝑡)𝑑𝑡−𝜆^2𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝐴𝑥₁(𝑡)𝑑𝑡−𝜆^2𝜔₀𝑒(𝑡)𝐴𝑥₁(𝑡)𝑑𝑡. (3.23) Using (𝐻₁), we have ^𝜔₀𝐴𝑥₁^′(𝑡)²𝑑𝑡≤^𝜔₀𝑓𝑡,𝑥^′₁(𝑡)𝑥₁(𝑡)−𝑐𝑥₁(𝑡−𝛿(𝑡))𝑑𝑡+^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑥₁(𝑡)−𝑐𝑥₁((𝑡+𝑡−𝛿))𝑑𝑡_𝜔0|𝑒(𝑡)|𝑥₁(𝑡)−𝑐𝑥₁((𝑡≤(|𝑐𝑥𝑡−𝛿))𝑑𝑡1+|)₁𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+𝑏𝜔+^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+𝜔‖𝑒‖. (3.24) Besides, we can assert that there exists some positive constant 𝑁₁ such that ^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡≤2𝜔𝑁₁+𝜔𝑏+𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡. (3.25) In fact, in view of condition (𝐻₁) and (3.22) we have ^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))−𝐾₁𝑥^′₁(𝑡)−𝑏𝑑𝑡≤^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))−𝑓𝑡,𝑥^′₁(𝑡)𝑑𝑡≤^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝑓𝑡,𝑥^′₁(𝑡)𝑑𝑡=0. (3.26) Define 𝐸₁=𝑡∈[0,𝜔]∶𝑥₁((𝑡;𝐸𝑡−𝜏))>𝐷₂=𝑡∈[0,𝜔]∶𝑥₁(𝑡−𝜏(𝑡))≤𝐷∪𝑡∈[0,𝜔]∶𝑥₁(𝑡−𝜏(𝑡))<−𝐷. (3.27) With these sets we get _𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡≤𝜔max𝑀,sup_{[],𝑥𝑡∈0,𝜔1((𝑡𝑡−𝜏))≤𝐷}𝑔𝑡,𝑥₁._𝐸1𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))−𝐾₁𝑥^′₁(𝑡=)−𝑏𝑑𝑡_𝐸1𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))−𝐾₁𝑥^′₁(𝑡)−𝑏𝑑𝑡≤−_𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))−𝐾₁𝑥^′₁(𝑡)−𝑏𝑑𝑡≤_𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝐾₁𝑥^′₁(𝑡)+𝑏𝑑𝑡, (3.28) which yields _𝐸1𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡≤_𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+_{𝐸1∪𝐸2}𝐾₁𝑥^′₁(𝑡=)+𝑏𝑑𝑡_𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+𝜔𝑏+𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡. (3.29) That is, ^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡=_𝐸1𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+_𝐸2𝑔𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡≤2_𝐸2𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+𝜔𝑏+𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡≤2𝜔max𝑀,sup_{[],𝑥𝑡∈0,𝜔1((𝑡𝑡−𝜏))<𝐷}𝑔𝑡,𝑥₁+𝜔𝑏+𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡=2𝜔𝐷₁+𝜔𝑏+𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡, (3.30) where 𝑁₁=max{𝑀,sup_{𝑡∈[0,𝜔],|𝑥1(𝑡−𝜏(𝑡))|<𝐷}|𝑔(𝑡,𝑥₁)|}, proving (3.25).
Substituting (3.25) into (3.24) and recalling (3.19), we get ^𝜔₀𝐴𝑥₁^′(𝑡)²𝑑𝑡≤(1+|𝑐|)𝑥₁₀2𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+2𝜔𝑏+2𝜔𝑁₁+𝜔max_{[]𝑡∈0,𝜔}|𝑒(𝑡=(|𝑐)|1+|)2𝐾₁𝑥₁₀^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+2𝜔𝑏𝑥₁₀+2𝜔𝑁₁𝑥₁₀+𝜔𝑥₁₀max_{[]𝑡∈0,𝜔}|𝑒(𝑡≤(|𝑐)|1+|)2𝐾₁𝐷+^𝜔₀𝑥^′₁(𝑡)𝑑𝑡^𝜔₀𝑥^′₁(𝑡+)𝑑𝑡2𝜔𝑏+2𝜔𝑁₁+𝜔max_{[]𝑡∈0,𝜔}|𝑒(𝑡)|𝐷+^𝜔₀𝑥^′₁(𝑡=(|𝑐)𝑑𝑡1+|)2𝐾₁𝐷^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+2𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡²+𝑁₂^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+𝑁₂𝐷=2𝐾₁(1+|𝑐|)^𝜔₀𝑥^′₁(𝑡)𝑑𝑡²+(1+|𝑐|)𝑁₂+2𝐾₁𝐷^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+(1+|𝑐|)𝑁₂𝐷, (3.31) where 𝑁₂=2𝜔𝑏+2𝜔𝑁₁+𝜔‖𝑒‖. Since (𝐴𝑥)(𝑡)=𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)), we have 𝐴𝑥₁^′(𝑡)=𝑥₁(𝑡)−𝑐𝑥₁(𝑡−𝛿(𝑡))^′=𝑥^′₁(𝑡)−𝑐𝑥^′₁(𝑡−𝛿(𝑡))+𝑐𝑥^′₁(𝑡−𝛿(𝑡))𝛿^′(𝑡)=𝐴𝑥^′₁(𝑡)+𝑐𝑥^′₁(𝑡−𝛿(𝑡))𝛿^′(𝑡),𝐴𝑥^′₁(𝑡)=𝐴𝑥₁^′(𝑡)−𝑐𝑥^′₁(𝑡−𝛿(𝑡))𝛿^′(𝑡). (3.32) By applying Lemma 2.1, we have ^𝜔₀𝑥^′₁(𝑡)𝑑𝑡=^𝜔₀𝐴⁻¹𝐴𝑥^′₁(𝑡≤∫)𝑑𝑡_𝜔0𝐴𝑥^′₁(𝑡)𝑑𝑡||𝑐=∫1−||_𝜔0𝐴𝑥₁^′(𝑡)−𝑐𝑥^′₁(𝑡−𝛿(𝑡))𝛿^′(𝑡)𝑑𝑡||𝑐≤∫1−||_𝜔0𝐴𝑥^′₁(𝑡)𝑑𝑡+|𝑐|𝛿₁∫^𝜔₀𝑥^′₁(𝑡)𝑑𝑡||𝑐,1−|| (3.33) where 𝛿₁=max_{𝑡∈[0,𝜔]}|𝛿′(𝑡)|. Since 0<𝜔^1/2(1+|𝑐|)^1/22𝐾₁/(|1−|𝑐||−|𝑐|𝛿₁), then |1−|𝑐||−|𝑐|𝛿₁>0, so we get ^𝜔₀𝑥^′₁(𝑡∫)𝑑𝑡≤^𝜔0𝐴𝑥₁^′(𝑡)𝑑𝑡|1−|𝑐||−|𝑐|𝛿₁≤𝜔^1/2∫^𝜔₀𝐴𝑥₁^′(𝑡)²𝑑𝑡^1/2|1−|𝑐||−|𝑐|𝛿₁. (3.34) Applying the inequality (𝑎+𝑏)^𝑘≤𝑎^𝑘+𝑏^𝑘 for 𝑎,𝑏>0, 0<𝑘<1, it follows from (3.31) and (3.34) that ^𝜔₀𝑥^′₁(𝑡≤𝜔)𝑑𝑡_1/2|1−|𝑐||−|𝑐|𝛿₁(1+|𝑐|)^1/22𝐾₁^𝜔₀𝑥^′₁(𝑡)𝑑𝑡+(1+|𝑐|)^1/2𝜔₀𝑥^′₁(𝑡)𝑑𝑡^1/2×𝑁₂+2𝐾₁𝐷^1/2+(1+|𝑐|)^1/2𝑁₂𝐷^1/2. (3.35) Since 𝜔^1/2(1+|𝑐|)^1/22𝐾₁/(|1−|𝑐||−|𝑐|𝛿₁)<1, it is easy to see that there exists a constant 𝑀₁>0 (independent of 𝜆) such that ^𝜔₀𝑥^′₁(𝑡)𝑑𝑡≤𝑀₁. (3.36) It follows from (3.19) that ‖𝑥₁‖≤𝐷+^𝜔₀𝑥^′₁(𝑡)𝑑𝑡≤𝐷+𝑀₁∶=𝑀₂. (3.37)
By the first equation of (3.11) we have ∫^𝜔₀𝑥₂(𝑡)𝑑𝑡=∫^𝜔₀(𝐴𝑥₁)′(𝑡)𝑑𝑡=0, which implies that there is a constant 𝑡₁∈[0,𝜔] such that 𝑥₂(𝑡₁)=0, hence ‖𝑥₂‖≤∫^𝜔₀|𝑥^′₂(𝑡)|𝑑𝑡. By the second equation of (3.11) we obtain 𝑥^′₂(𝑡)=𝜆𝑓𝑡,𝑥^′₁(𝑡)+𝜆𝑔𝑥₁(𝑡−𝜏(𝑡))+𝜆𝑒(𝑡). (3.38) So, from (𝐻₁) and (3.25), we have 𝑥₂₀≤^𝜔₀𝑓𝑡,𝑥^′₁(𝑡)𝑑𝑡+^𝜔₀𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))𝑑𝑡+^𝜔₀|𝑒(𝑡)|𝑑𝑡≤2𝐾₁𝑀₁+2𝜔𝑏+2𝜔𝑁₁+𝜔‖𝑒‖∶=𝑀₃. (3.39) Let 𝑀₄=𝑀²₂+𝑀²₃+1,Ω={𝑥=(𝑥₁,𝑥₂)^⊤∶‖𝑥₁‖<𝑀₄,‖𝑥₂‖<𝑀₄}, then forall𝑥∈𝜕Ω∩Ker𝐿𝑄₁1𝑁𝑥=𝜔^𝜔₀𝑥₂(𝑡)𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁(𝑡−𝜏(𝑡))+𝑒(𝑡)𝑑𝑡. (3.40) If 𝑄₁𝑁𝑥=0, then 𝑥₂(𝑡)=0,𝑥₁=𝑀₄ or −𝑀₄. But if 𝑥₁(𝑡)=𝑀₄, we know 0=^𝜔₀𝑔𝑀₄𝑑𝑡, (3.41) that is, 𝑔(𝑀₄)=0. From assumption (𝐻₂), we know 𝑀₄≤𝐷, which yields a contradiction, one can argue similarly if 𝑥₁=−𝑀₄. We also have 𝑄₁𝑁𝑥≠0, that is, forall𝑥∈𝜕Ω∩Ker𝐿,𝑥∉Im𝐿, so conditions (1) and (2) of Lemma 3.1 are both satisfied. Define the isomorphism 𝐽∶Im𝑄₁→Ker𝐿 as follows: 𝐽𝑥₁,𝑥₂^⊤=𝑥₂,𝑥₁^⊤. (3.42) Let 𝐻(𝜇,𝑥)=𝜇𝑥+(1−𝜇)𝐽𝑄₁𝑁𝑥,(𝜇,𝑥)∈[0,1]×Ω, then, forall(𝜇,𝑥)∈(0,1)×(𝜕Ω∩Ker𝐿), 𝐻𝜇,𝑥=𝜇𝑥₁(𝑡)+1−𝜇𝜔^𝜔₀𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁((𝑡(𝑡𝑥𝑡−𝜏))+𝑒)𝑑𝑡𝜇+1−𝜇₂(𝑡). (3.43) We have ∫^𝜔₀𝑒(𝑡)𝑑𝑡=0. So, we can get 𝐻𝜇,𝑥=𝜇𝑥₁(𝑡)+1−𝜇𝜔^𝜔₀𝑓𝑡,𝑥^′₁(𝑡)+𝑔𝑡,𝑥₁((𝑡𝑥𝑡−𝜏))𝑑𝑡𝜇+1−𝜇₂(𝑡),∀𝜇,𝑥∈(0,1)×(𝜕Ω∩Ker𝐿). (3.44) From (𝐻₂), it is obvious that 𝑥^⊤𝐻(𝜇,𝑥)>0, forall(𝜇,𝑥)∈(0,1)×(𝜕Ω∩Ker𝐿). Hence deg𝐽𝑄₁{𝐻()}{𝐻()}{}𝑁,Ω∩Ker𝐿,0=deg0,𝑥,Ω∩Ker𝐿,0=deg1,𝑥,Ω∩Ker𝐿,0=deg𝐼,Ω∩Ker𝐿,0≠0. (3.45) So condition (3) of Lemma 3.1 is satisfied. By applying Lemma 3.1, we conclude that equation 𝐿𝑥=𝑁𝑥 has a solution 𝑥=(𝑥₁,𝑥₂)^⊤ on Ω∩𝐷(𝐿), that is, (3.1) has an 𝜔-periodic solution 𝑥₁(𝑡).

By using a similar argument, we can obtain the following theorem.

Theorem 3.3. Suppose there exist positive constants 𝐾₁,𝐷,𝑀,𝑏 with 𝑀>‖𝑒‖ such that:
(H₁)|𝑓(𝑡,𝑢)|≤𝐾₁|𝑢|+𝑏, for (𝑡,𝑢)∈ℝ×ℝ;(H₂) sgn𝑥⋅𝑔(𝑡,𝑥)>‖𝑒‖, for |𝑥|>𝐷,(H₃)𝑔(𝑡,𝑥)≤𝑀, for 𝑥≥𝐷 and 𝑡∈ℝ,
then (3.1) has at least one solution with period 𝜔 if 0<𝜔(1+|𝑐|)^1/22𝐾₁/(|1−|𝑐||−|𝑐|𝛿₁)<1.

Remark 3.4. If ∫^𝜔₀𝑒(𝑡)𝑑𝑡≠0 and 𝑓(𝑡,0)≠0, the problem of existence of 𝜔-periodic solutions to (3.1) can be converted to the existence of 𝜔-periodic solutions to (𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))^′′=𝑓₁𝑡,𝑥^′(𝑡)+𝑔₁(𝑡,𝑥(𝑡−𝜏(𝑡)))+𝑒₁(𝑡), (3.46) where 𝑓₁(𝑡,𝑥)=𝑓(𝑡,𝑥)−𝑓(𝑡,0),𝑔₁(𝑡,𝑥)=𝑔(𝑡,𝑥)+(1/𝜔)∫^𝜔₀𝑒(𝑡)𝑑𝑡+𝑓(𝑡,0), and 𝑒₁(𝑡)=𝑒(𝑡)−(1/𝜔)∫^𝜔₀𝑒(𝑡)𝑑𝑡. Clearly, ∫^𝜔₀𝑒₁(𝑡)𝑑𝑡=0 and 𝑓₁(𝑡,0)=0, and (3.46) can be discussed by using Theorem 3.2 (or Theorem 3.3).

4. Positive Periodic Solutions for Neutral Equations

Consider the following second-order neutral functional differential equation:(𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))^′′=−𝑎(𝑡)𝑥(𝑡)+𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))), (4.1) where 𝜆 is a positive parameter; 𝑓∈𝐶(ℝ,[0,∞)), and 𝑓(𝑥)>0 for 𝑥>0; 𝑎∈𝐶(ℝ,(0,∞)) with max{𝑎(𝑡)∶𝑡∈[0,𝜔]}<(𝜋/𝜔)², 𝑏∈𝐶(ℝ,(0,∞)), 𝜏∈𝐶(ℝ,ℝ), 𝑎(𝑡), 𝑏(𝑡), and 𝜏(𝑡) are 𝜔-periodic functions.

Define the Banach space 𝑋 as in Section 2, and let 𝐶⁺_𝜔={𝑥∈𝐶(ℝ,(0,∞))∶𝑥(𝑡+𝜔)=𝑥(𝑡)}. Denote 𝑀=max{𝑎(𝑡)∶𝑡∈[0,𝜔]},𝑚=min{𝑎(𝑡)∶𝑡∈[0,𝜔]},𝛽=√1𝑀,𝐿=2𝛽sin𝛽𝜔/2,𝑙=cos𝛽𝜔/2()𝑘2𝛽sin𝛽𝜔/2,𝑘=𝑙𝑀+𝑚+𝐿𝑀,₁=𝑘−𝑘²−4𝐿𝑙𝑀𝑚𝑙[(𝑐2𝐿𝑀,𝛼=𝑚−𝑀+𝑚)||](|𝑐.𝐿𝑀1−|) (4.2) It is easy to see that 𝑀,𝑚,𝛽,𝐿,𝑙,𝑘,𝑘₁>0.

Now we consider (4.1). First let 𝑓₀=lim_𝑥→0𝑓(𝑥)𝑥,𝑓_∞=lim_𝑥→∞𝑓(𝑥)𝑥,𝑓₀=lim_𝑥→0𝑓(𝑥)𝑥,𝑓_∞=lim_𝑥→∞𝑓(𝑥)𝑥, (4.3) and denote 𝑖₀=numberof0'sin𝑓₀,𝑓_∞,𝑖₀=numberof0'sin𝑓₀,𝑓_∞;𝑖_∞=numberof∞'sin𝑓₀,𝑓_∞,𝑖_∞=numberof∞'sin𝑓₀,𝑓_∞. (4.4) It is clear that 𝑖₀,𝑖₀,𝑖_∞,𝑖_∞∈{0,1,2}. We will show that (4.1) has 𝑖₀ or 𝑖_∞ positive 𝑤-periodic solutions for sufficiently large or small 𝜆, respectively.

In the following we discuss (4.1) in two cases, namely, the case where 𝑐<0 and 𝑐>−min{𝑘₁,𝑚/(𝑀+𝑚)} (note that 𝑐>−𝑚/(𝑀+𝑚) implies 𝛼>0; 𝑐>−𝑘₁ implies |𝑐|<𝛼) and the case where 𝑐>0 and 𝑐<𝑚𝑖𝑛{𝑚/(𝑀+𝑚),(𝐿𝑀−𝑙𝑚)/((𝐿−𝑙)𝑀−𝑙𝑚)} (note that 𝑐<𝑚/(𝑀+𝑚) implies 𝛼>0; 𝑐<(𝐿𝑀−𝑙𝑚)/((𝐿−𝑙)𝑀−𝑙𝑚) implies 𝛼<1). Obviously, we have |𝑐|<1 which makes Lemma 2.1 applicable for both cases and also Lemmas 2.2 or 2.3, respectively.

Let 𝐾={𝑥∈𝑋∶𝑥(𝑡)≥𝛼‖𝑥‖} denote the cone in 𝑋 as defined in Section 2, where 𝛼 is just as defined above. We also use 𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖<𝑟} and 𝜕𝐾_𝑟={𝑥∈𝐾∶‖𝑥‖=𝑟}.

Let 𝑦(𝑡)=(𝐴𝑥)(𝑡), then from Lemma 2.1 we have 𝑥(𝑡)=(𝐴⁻¹𝑦)(𝑡). Hence (4.1) can be transformed into𝑦^′′(𝑡)+𝑎(𝑡)𝐴⁻¹𝑦(𝑡)=𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡)), (4.5) which can be further rewritten as𝑦^′′(𝑡)+𝑎(𝑡)𝑦(𝑡)−𝑎(𝑡)𝐻𝑦(𝑡)=𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡)), (4.6) where 𝐻(𝑦(𝑡))=𝑦(𝑡)−(𝐴⁻¹𝑦)(𝑡)=−𝑐(𝐴⁻¹𝑦)(𝑡−𝛿(𝑡)).

Now we discuss the two cases separately.

4.1. Case  I

Assume 𝑐<0 and 𝑐>−min{𝑘₁,𝑚/(𝑀+𝑚)}.

Lemma 4.1 (see [7]). The equation 𝑦^′′(𝑡)+𝑀𝑦(𝑡)=ℎ(𝑡),ℎ∈𝐶⁺_𝜔, (4.7) has a unique 𝜔-periodic solution 𝑦(𝑡)=_𝑡^𝑡+𝜔𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠, (4.8) where 𝐺()=))𝑡,𝑠cos𝛽((𝜔/2+𝑡−𝑠[].2𝛽sin𝛽𝜔/2,𝑠∈𝑡,𝑡+𝜔 (4.9)

Lemma 4.2 (see [7]). One has∫_𝑡^𝑡+𝜔𝐺(𝑡,𝑠)𝑑𝑠=1/𝑀. Furthermore, if max{𝑎(𝑡)∶𝑡∈[0,𝜔]}<(𝜋/𝜔)², then 0<𝑙≤𝐺(𝑡,𝑠)≤𝐿 for all 𝑡∈[0,𝜔] and 𝑠∈[𝑡,𝑡+𝜔].

Now we consider𝑦^′′(𝑡)+𝑎(𝑡)𝑦(𝑡)−𝑎(𝑡)𝐻𝑦(𝑡)=ℎ(𝑡),ℎ∈𝐶⁺_𝜔, (4.10) and define operators 𝑇,𝐻∶𝑋→𝑋 by (𝑇ℎ)(𝑡)=_𝑡^𝑡+𝜔𝐺(𝑡,𝑠)ℎ(𝑠)𝑑𝑠,𝐻𝑦(𝑡)=𝑀−𝑎(𝑡)𝑦(𝑡)+𝑎(𝑡)𝐻𝑦(𝑡). (4.11) Clearly 𝑇,𝐻 are completely continuous (𝑇ℎ)(𝑡)>0 for ℎ(𝑡)>0 and ‖𝐻‖≤(𝑀−𝑚+𝑀(|𝑐|/(1−|𝑐|))).

By Lemma 4.1, the solution of (4.10) can be written in the form𝑦(𝑡)=(𝑇ℎ)(𝑡)+𝑇𝐻𝑦(𝑡). (4.12) In view of 𝑐<0 and 𝑐>−min{𝑘₁,𝑚/(𝑀+𝑚)}, we have𝑇𝐻≤‖𝑇𝐻≤|𝑐|‖𝑀−𝑚+𝑚𝑀(1−|𝑐|)<1, (4.13) and hence 𝑦(𝑡)=𝐼−𝑇𝐻⁻¹(𝑇ℎ)(𝑡). (4.14) Define an operator 𝑃∶𝑋→𝑋 by (𝑃ℎ)(𝑡)=𝐼−𝑇𝐻⁻¹(𝑇ℎ)(𝑡). (4.15) Obviously, for any ℎ∈𝐶⁺_𝜔, if max{𝑎(𝑡)∶𝑡∈[0,𝜔]}<(𝜋/𝜔)², 𝑦(𝑡)=(𝑃ℎ)(𝑡) is the unique positive 𝜔-periodic solution of (4.10).

Lemma 4.3. 𝑃 is completely continuous and (𝑡)≤(𝑡)≤𝑀(|𝑐𝑇ℎ)(𝑃ℎ)(1−|)(𝑐|‖‖𝑚−𝑀+𝑚)|𝑇ℎ,∀ℎ∈𝐶⁺_𝜔. (4.16)

Proof. By the Neumann expansion of 𝑃, we have 𝑃=𝐼−𝑇𝐻⁻¹𝑇=𝐼+𝑇𝐻+𝑇𝐻²+⋯+𝑇𝐻^𝑛𝑇𝑇𝐻+⋯=𝑇+𝑇𝐻𝑇+²𝑇+⋯+𝑇𝐻^𝑛𝑇+⋯. (4.17) Since 𝑇 and 𝐻 are completely continuous, so is 𝑃. Moreover, by (4.17), and recalling that ‖𝑇𝐻‖≤(𝑀−𝑚+𝑚|𝑐|)/𝑀(1−|𝑐|)<1, we get (𝑡)≤(𝑡)≤𝑀(|𝑐𝑇ℎ)(𝑃ℎ)(1−|)(𝑐|‖‖.𝑚−𝑀+𝑚)|𝑇ℎ (4.18)

Define an operator 𝑄∶𝑋→𝑋 by𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡)). (4.19)

Lemma 4.4. One has 𝑄(𝐾)⊂𝐾.

Proof. From the definition of 𝑄, it is easy to verify that 𝑄𝑦(𝑡+𝜔)=𝑄𝑦(𝑡). For 𝑦∈𝐾, we have from Lemma 4.3 that 𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦((𝑡(𝑡)𝑓𝐴𝑡−𝜏))≥𝑇𝜆𝑏⁻¹𝑦(𝑡−𝜏(𝑡))=𝜆_𝑡^𝑡+𝜔𝐺(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))𝑑𝑠≥𝜆𝑙^𝜔₀𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))𝑑𝑠. (4.20) On the other hand, 𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦((𝑡≤𝑀(|𝑐𝑡−𝜏))1−|)(𝑐|𝑇(𝑡)𝑓𝐴𝑚−𝑀+𝑚)|𝜆𝑏⁻¹𝑦((𝑡𝑀(|𝑐𝑡−𝜏))=𝜆1−|)𝑚−(𝑀+𝑚)|𝑐|max_{[]𝑡∈0,𝜔𝑡}^𝑡+𝜔𝐺(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠𝑀(|𝑐(𝑠−𝜏))𝑑𝑠≤𝜆1−|)(𝑐|𝐿𝑚−𝑀+𝑚)|^𝜔₀𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))𝑑𝑠. (4.21) Therefore, (𝑡)≥𝑙[(𝑐𝑄𝑦𝑚−𝑀+𝑚)||](|𝑐,𝐿𝑀1−|)𝑄𝑦=𝛼𝑄𝑦 (4.22) that is, 𝑄(𝐾)⊂𝐾.

From the continuity of 𝑃, it is easy to verify that 𝑄 is completely continuous in 𝑋. Comparing (4.6) to (4.10), it is obvious that the existence of periodic solutions for (4.6) is equivalent to the existence of fixed points for the operator 𝑄 in 𝑋. Recalling Lemma 4.4, the existence of positive periodic solutions for (4.6) is equivalent to the existence of fixed points of 𝑄 in 𝐾. Furthermore, if 𝑄 has a fixed point 𝑦 in 𝐾, it means that (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solutions of (4.1).

Lemma 4.5. If there exists 𝜂>0 such that 𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡))≥𝐴⁻¹𝑦(𝑡−𝜏(𝑡))𝜂,for𝑡∈[0,𝜔],𝑦∈𝐾, (4.23) then |𝑐|𝑄𝑦≥𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦,𝑦∈𝐾. (4.24)

Proof. By Lemmas 2.2, 4.2, and 4.3, we have for 𝑦∈𝐾 that 𝑄𝑦(𝑡)=𝑃𝜆𝑏(𝑡)𝑓𝐴⁻¹𝑦((𝑡(𝑡)𝑓𝐴𝑡−𝜏))≥𝑇𝜆𝑏⁻¹𝑦(𝑡−𝜏(𝑡))=𝜆_𝑡^𝑡+𝜔𝐺(𝑡,𝑠)𝑏(𝑠)𝑓𝐴⁻¹𝑦(𝑠−𝜏(𝑠))𝑑𝑠≥𝜆𝑙𝜂^𝜔₀𝑏(𝑠)𝐴⁻¹𝑦((𝑠|𝑐|𝑠−𝜏))𝑑𝑠≥𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦. (4.25) Hence |𝑐|𝑄𝑦≥𝜆𝑙𝜂𝛼−1−c^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦,𝑦∈𝐾. (4.26)

Lemma 4.6. If there exists 𝜀>0 such that 𝑓𝐴⁻¹𝑦(𝑡−𝜏(𝑡))≤𝐴⁻¹𝑦(𝑡−𝜏(𝑡))𝜀,for𝑡∈[0,𝜔],𝑦∈𝐾, (4.27) then 𝑄𝑦≤𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦𝑚−𝑀+𝑚)|,𝑦∈𝐾. (4.28)

Proof. By Lemmas 2.2, 4.2, and 4.3, we have (𝑡𝑀(|𝑐𝑄𝑦)≤𝜆1−|)(𝑐|𝐿𝑚−𝑀+𝑚)|^𝜔₀𝑏(𝑠)𝑓𝐴⁻¹𝑦((𝑠𝑀(|𝑐𝑠−𝜏))𝑑𝑠≤𝜆1−|)𝑚−(𝑀+𝑚)|𝑐|𝐿𝜀^𝜔₀𝑏(𝑠)𝐴⁻¹𝑦(𝑠−𝜏(𝑠))𝑑𝑠≤𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦.𝑚−𝑀+𝑚)| (4.29)

Define 𝐹(𝑟)𝑓(𝑡)𝑟=max∶0≤𝑡≤|𝑐|,𝑓1−₁(𝑟)𝑓(𝑡)∶|𝑐|=min𝛼−1−𝑐²𝑟𝑟≤𝑡≤|𝑐|.1− (4.30)

Lemma 4.7. If 𝑦∈𝜕𝐾_𝑟, then 𝑄𝑦≥𝜆𝑙𝑓₁(𝑟)^𝜔₀𝑏(𝑠)𝑑𝑠. (4.31)

Proof. By Lemma 2.2, we obtain ((𝛼−|𝑐|)/(1−𝑐²))𝑟≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤𝑟/(1−|𝑐|) for 𝑦∈𝜕𝐾_𝑟, which yields 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥𝑓₁(𝑟). The lemma now follows analog to the proof of Lemma 4.5.

Lemma 4.8. If 𝑦∈𝜕𝐾_𝑟, then (|𝑐𝐹(𝑟)𝑄𝑦≤𝜆𝐿𝑀1−|)𝑚−(𝑀+𝑚)|𝑐|^𝜔₀𝑏(𝑠)𝑑𝑠. (4.32)

Proof. By Lemma 2.2, we can have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤𝑟/(1−|𝑐|) for 𝑦∈𝜕𝐾_𝑟, which yields 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤𝐹(𝑟). Similar to the proof of Lemma 4.6, we get the conclusion.

We quote the fixed point theorem which our results will be based on.

Lemma 4.9 (see [9]). Let 𝑋 be a Banach space and 𝐾 a cone in 𝑋. For 𝑟>0, define 𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖<𝑟}. Assume that 𝑇∶𝐾_𝑟→𝐾 is completely continuous such that 𝑇𝑥≠𝑥 for 𝑥∈𝜕𝐾_𝑟={𝑢∈𝐾∶‖𝑢‖=𝑟}.
(i)If ‖𝑇𝑥‖≥‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=0.(ii)If ‖𝑇𝑥‖≤‖𝑥‖ for 𝑥∈𝜕𝐾_𝑟, then 𝑖(𝑇,𝐾_𝑟,𝐾)=1.

Now we give our main results on positive periodic solutions for (4.1).

Theorem 4.10. (a) If 𝑖₀=1 or 2, then (4.1) has 𝑖₀ positive 𝜔-periodic solutions for 𝜆>1/𝑓₁(1)𝑙∫^𝜔₀𝑏(𝑠)𝑑𝑠>0;
(b) If 𝑖_∞=1 or 2, then (4.1) has 𝑖_∞ positive 𝜔-periodic solutions for 0<𝜆<(𝑚−(𝑀+𝑚)|𝑐|)/𝐿𝑀(1−|𝑐|)𝐹(1)∫^𝜔₀𝑏(𝑠)𝑑𝑠;
(c) If 𝑖_∞=0 or 𝑖₀=0, then (4.1) has no positive 𝜔-periodic solutions for sufficiently small or sufficiently large 𝜆>0, respectively.

Proof. (a) Choose 𝑟₁=1. Take 𝜆₀=1/𝑓₁(𝑟₁)𝑙∫^𝜔₀𝑏(𝑠)𝑑𝑠>0, then for all 𝜆>𝜆₀, we have from Lemma 4.7 that 𝑄𝑦>𝑦,for𝑦∈𝜕𝐾_𝑟1. (4.33)
Case 1. If 𝑓₀=0, we can choose 0<𝑟₂<𝑟₁, so that 𝑓(𝑢)≤𝜀𝑢 for 0≤𝑢≤𝑟₂, where the constant 𝜀>0 satisfies 𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠𝑚−(𝑀+𝑚)|𝑐|<1. (4.34) Letting 𝑟₂=(1−|𝑐|)𝑟₂, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤𝜀(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₂ for 𝑦∈𝜕𝐾_𝑟2. In view of Lemma 4.6 and (4.34), we have for 𝑦∈𝜕𝐾_𝑟2 that 𝑄𝑦≤𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦<𝑦.𝑚−𝑀+𝑚)| (4.35) It follows from Lemma 4.9 and (4.33) that 𝑖𝑄,𝐾_𝑟2,𝐾=1,𝑖𝑄,𝐾_𝑟1,𝐾=0, (4.36) thus 𝑖(𝑄,𝐾_𝑟1⧵𝐾_𝑟2,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟1⧵𝐾_𝑟2, which means (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-positive solution of (4.1) for 𝜆>𝜆₀.
Case 2. If 𝑓_∞=0, there exists a constant 𝐻>0 such that 𝑓(𝑢)≤𝜀𝑢 for 𝑢≥𝐻, where the constant 𝜀>0 satisfies 𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠𝑚−(𝑀+𝑚)|𝑐|<1. (4.37) Letting 𝑟₃=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤𝜀(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟3. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟3. Thus by Lemma 4.6 and (4.37), we have for 𝑦∈𝜕𝐾_𝑟3 that 𝑄𝑦≤𝜆𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦<𝑦.𝑚−𝑀+𝑚)| (4.38) Recalling from Lemma 4.9 and (4.33) that 𝑖𝑄,𝐾_𝑟3,𝐾=1,𝑖𝑄,𝐾_𝑟1,𝐾=0, (4.39) then 𝑖(𝑄,𝐾_𝑟3⧵𝐾_𝑟1,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟3⧵𝐾_𝑟1, which means (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-positive solution of (4.1) for 𝜆>𝜆₀.
Case 3. If 𝑓₀=𝑓_∞=0, from the above arguments, there exist 0<𝑟₂<𝑟₁<𝑟₃ such that 𝑄 has a fixed point 𝑦₁(𝑡) in 𝐾_𝑟1⧵𝐾_𝑟2 and a fixed point 𝑦₂(𝑡) in 𝐾_𝑟3⧵𝐾_𝑟1. Consequently, (𝐴⁻¹𝑦₁)(𝑡) and (𝐴⁻¹𝑦₂)(𝑡) are two positive 𝜔-periodic solutions of (4.1) for 𝜆>𝜆₀.
(b) Let 𝑟₁=1. Take 𝜆₀=(𝑚−(𝑀+𝑚)|𝑐|)/𝐿𝑀(1−|𝑐|)𝐹(𝑟₁)∫^𝜔₀𝑏(𝑠)𝑑𝑠>0; then by Lemma 4.8 we know if 𝜆<𝜆₀ then 𝑄𝑦<𝑦,𝑦∈𝜕𝐾_𝑟1. (4.40)
Case 1. If 𝑓₀=∞, we can choose 0<𝑟₂<𝑟₁ so that 𝑓(𝑢)≥𝜂𝑢 for 0≤𝑢≤𝑟₂, where the constant 𝜂>0 satisfies |𝑐|𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠>1. (4.41) Letting 𝑟₂=(1−|𝑐|)𝑟₂, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥𝜂(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₂ for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.5 and (4.41), |𝑐|𝑄𝑦≥𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦>𝑦. (4.42) It follows from Lemma 4.9 and (4.40) that 𝑖𝑄,𝐾_𝑟2,𝐾=0,𝑖𝑄,𝐾_𝑟1,𝐾=1, (4.43) which implies 𝑖(𝑄,𝐾_𝑟1⧵𝐾_𝑟2,𝐾)=1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟1⧵𝐾_𝑟2. Therefore, (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (4.1) for 0<𝜆<𝜆₀.
Case 2. If 𝑓_∞=∞, there exists a constant 𝐻>0 such that 𝑓(𝑢)≥𝜂𝑢 for 𝑢≥𝐻, where the constant 𝜂>0 satisfies |𝑐|𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠>1. (4.44) Letting 𝑟₃=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥𝜂(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟3. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟3. Thus by Lemma 4.5 and (4.44), we have for 𝑦∈𝜕𝐾_𝑟3 that |𝑐|𝑄𝑦≥𝜆𝑙𝜂𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦>𝑦. (4.45) It follows from Lemma 4.9 and (4.40) that 𝑖𝑄,𝐾_𝑟3,𝐾=0,𝑖𝑄,𝐾_𝑟1,𝐾=1. (4.46) that is, 𝑖(𝑄,𝐾_𝑟3⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟3⧵𝐾_𝑟1. That means (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (4.1) for 0<𝜆<𝜆₀.
Case 3. If 𝑓₀=𝑓_∞=∞, from the above arguments, 𝑄 has a fixed point 𝑦₁ in 𝐾_𝑟1⧵𝐾_𝑟2 and a fixed point 𝑦₂ in 𝐾_𝑟3⧵𝐾_𝑟1. Consequently, (𝐴⁻¹𝑦₁)(𝑡) and (𝐴⁻¹𝑦₂)(𝑡) are two positive 𝜔-periodic solutions of (4.1) for 0<𝜆<𝜆₀.
(c) By Lemma 2.2, if 𝑦∈𝐾, then (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖>0 for 𝑡∈[0,𝜔].
Case 1. If 𝑖₀=0, we have 𝑓₀>0 and 𝑓_∞>0. Let 𝑏₁=min{𝑓(𝑢)/𝑢;𝑢>0}>0, then we obtain 𝑓(𝑢)≥𝑏₁𝑢,𝑢∈[0,+∞). (4.47) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (4.1) for 𝜆>𝜆₀, where 𝜆₀=(1−𝑐²)/𝑙𝑏₁(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)𝑑𝑠>0. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡∈[0,𝜔], then by Lemma 4.5, if 𝜆>𝜆₀ we have 𝑦=𝑄𝑦≥𝜆𝑙𝑏₁𝛼−|𝑐|1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦>𝑦, (4.48) which is a contradiction.
Case 2. If 𝑖_∞=0, we have 𝑓₀<∞ and 𝑓_∞<∞. Let 𝑏₂=max{𝑓(𝑢)/𝑢∶𝑢>0}>0, then we obtain 𝑓(𝑢)≤𝑏₂𝑢,𝑢∈[0,∞). (4.49) Assume 𝑦(𝑡) is a positive 𝜔-periodic solution of (4.1) for 0<𝜆<𝜆₀, where 𝜆₀=(𝑚−(𝑀+𝑚)|𝑐|)/𝑏₂𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠. Since 𝑄𝑦(𝑡)=𝑦(𝑡) for 𝑡∈[0,𝜔], it follows from Lemma 4.6 that 𝑦=𝑄𝑦≤𝜆𝑏₂𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦<𝑦,𝑚−𝑀+𝑚)| (4.50) which is a contradiction.

Theorem 4.11. (a) If there exists a constant 𝑏₁>0 such that 𝑓(𝑢)≥𝑏₁𝑢 for 𝑢∈[0,+∞), then (4.1) has no positive 𝜔-periodic solution for 𝜆>(1−𝑐²)/𝑙𝑏₁(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)𝑑𝑠.
(b) If there exists a constant 𝑏₂>0 such that 𝑓(𝑢)≤𝑏₂𝑢 for 𝑢∈[0,+∞), then (4.1) has no positive 𝜔-periodic solution for 0<𝜆<(𝑚−(𝑀+𝑚)|𝑐|)/𝑏₂𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠.

Proof. From the proof of (c) in Theorem 4.10, we obtain this theorem immediately.

Theorem 4.12. Assume 𝑖₀=𝑖₀=𝑖_∞=𝑖_∞=0 and that one of the following conditions holds:
(1)𝑓₀≤𝑓_∞;(2)𝑓₀>𝑓_∞;(3)𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞;(4)𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀.
If 1−𝑐²𝑙(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)𝑑𝑠max𝑓₀,𝑓₀,𝑓_∞,𝑓_∞(𝑐|<𝜆<𝑚−𝑀+𝑚)|𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠min𝑓₀,𝑓₀,𝑓_∞,𝑓_∞, (4.51) then (4.1) has one positive 𝜔-periodic solution.

Proof. We have the following cases.
Case 1. If 𝑓₀≤𝑓_∞, then 1−𝑐²𝑓_∞𝑙(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)(𝑐|𝑑𝑠<𝜆<𝑚−𝑀+𝑚)|𝑓₀𝐿𝑀∫^𝜔₀𝑏(𝑠).𝑑𝑠 (4.52) It is easy to see that there exists an 0<𝜀<𝑓_∞ such that 1−𝑐²𝑓_∞−𝜀𝑙(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)(𝑐|𝑑𝑠<𝜆<𝑚−𝑀+𝑚)|𝑓₀+𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠).𝑑𝑠 (4.53) For the above 𝜀, we choose 𝑟₁>0 such that 𝑓(𝑢)≤(𝑓₀+𝜀)𝑢 for 0≤𝑢≤𝑟₁. Letting 𝑟₁=(1−|𝑐|)𝑟₁, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤(𝑓₀+𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟1. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₁ for 𝐾∈𝜕𝐾_𝑟1. Thus by Lemma 4.6 we have for 𝑦∈𝜕𝐾_𝑟1 that 𝑄𝑦≤𝜆𝑓₀+𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦<𝑦.𝑚−𝑀+𝑚)| (4.54)
On the other hand, there exists a constant 𝐻>0 such that 𝑓(𝑢)≥(𝑓_∞−𝜀)𝑢 for 𝑢≥𝐻. Letting 𝑟₂=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥(𝑓_∞−𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2, we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.5, for 𝑦∈𝜕𝐾_𝑟2𝑄𝑦≥𝜆𝑙𝑓_∞|𝑐|−𝜀𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦>𝑦. (4.55) It follows from Lemma 4.9 that 𝑖𝑄,𝐾_𝑟1,𝐾=1,𝑖𝑄,𝐾_𝑟2,𝐾=0, (4.56) thus 𝑖(𝑄,𝐾_𝑟2⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟2⧵𝐾_𝑟1. So (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (4.1).
Case 2. If 𝑓₀>𝑓_∞, in this case, we have 1−𝑐²𝑓₀𝑙(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)(𝑐|𝑑𝑠<𝜆<𝑚−𝑀+𝑚)|𝑓_∞𝐿𝑀∫^𝜔₀𝑏(𝑠).𝑑𝑠 (4.57) It is easy to see that there exists an 0<𝜀<𝑓₀ such that 1−𝑐²𝑓₀−𝜀𝑙(𝛼−|𝑐|)∫^𝜔₀𝑏(𝑠)(𝑐|𝑑𝑠<𝜆<𝑚−𝑀+𝑚)|𝑓_∞+𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠).𝑑𝑠 (4.58) For the above 𝜀, we choose 𝑟₁>0 such that 𝑓(𝑢)≥(𝑓₀−𝜀)𝑢 for 0≤𝑢≤𝑟₁. Letting 𝑟₁=(1−|𝑐|)𝑟₁, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≥(𝑓₀−𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟1. By Lemma 2.2, we have 0≤(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≤‖𝑦‖/(1−|𝑐|)≤𝑟₁ for 𝑦∈𝜕𝐾_𝑟1. Thus we have by Lemma 4.5 that for 𝑦∈𝜕𝐾_𝑟1𝑄𝑦≥𝜆𝑙𝑓₀|𝑐|−𝜀𝛼−1−𝑐^2𝜔₀𝑏(𝑠)𝑑𝑠𝑦>𝑦. (4.59) On the other hand, there exists a constant 𝐻>0 such that 𝑓(𝑢)≤(𝑓_∞+𝜀)𝑢 for 𝑢≥𝐻. Letting 𝑟₂=max{2𝑟₁,𝐻(1−𝑐²)/(𝛼−|𝑐|)}, we have 𝑓((𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)))≤(𝑓_∞+𝜀)(𝐴⁻¹𝑦)(𝑡−𝜏(𝑡)) for 𝑦∈𝐾_𝑟2. By Lemma 2.2 we have (𝐴⁻¹𝑦)(𝑡−𝜏(𝑡))≥((𝛼−|𝑐|)/(1−𝑐²))‖𝑦‖≥𝐻 for 𝑦∈𝜕𝐾_𝑟2. Thus by Lemma 4.6, for 𝑦∈𝜕𝐾_𝑟2𝑄𝑦≤𝜆𝑓_∞+𝜀𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠(𝑐|𝑦.𝑚−𝑀+𝑚)| (4.60) It follows from Lemma 4.9 that 𝑖𝑄,𝐾_𝑟1,𝐾=0,𝑖𝑄,𝐾_𝑟2,𝐾=1. (4.61) Thus 𝑖(𝑄,𝐾_𝑟2⧵𝐾_𝑟1,𝐾)=−1 and 𝑄 has a fixed point 𝑦 in 𝐾_𝑟2⧵𝐾_𝑟1, proving that (𝐴⁻¹𝑦)(𝑡) is a positive 𝜔-periodic solution of (4.1).
Case 3. One has 𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞. The proof is the same as in Case 1.
Case 4. One has 𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀. The proof is the same as in Case 2.

4.2. Case  II

Assume 𝑐>0 and 𝑐<𝑚𝑖𝑛{𝑚/(𝑀+𝑚),(𝐿𝑀−𝑙𝑚)/(𝐿−𝑙)𝑀−𝑙𝑚}.

Define 𝑓₂(𝑟)𝑓(𝑡)∶𝛼=min𝑟1−𝑐𝑟≤𝑡≤.1−𝑐 (4.62) Similarly as in Section 4.1, we get the following results.

Theorem 4.13. (a) If 𝑖₀=1 or 2, then (4.1) has 𝑖₀ positive 𝜔-periodic solutions for 𝜆>1/𝑓₂(1)𝑙∫^𝜔₀𝑏(𝑠)𝑑𝑠>0.
(b) If 𝑖_∞=1 or 2, then (4.1) has 𝑖_∞ positive 𝜔-periodic solutions for 0<𝜆<(𝑚−(𝑀+𝑚)𝑐)/𝐿𝑀(1−𝑐)𝐹(1)∫^𝜔₀𝑏(𝑠)𝑑𝑠.
(c) If 𝑖_∞=0 or 𝑖₀=0, then (4.1) has no positive 𝜔-periodic solution for sufficiently small or large 𝜆>0, respectively.

Theorem 4.14. (a) If there exists a constant 𝑏₁>0 such that 𝑓(𝑢)≥𝑏₁𝑢 for 𝑢∈[0,+∞), then (4.1) has no positive 𝜔-periodic solution for 𝜆>(1−𝑐)/𝑙𝛼𝑏₁∫^𝜔₀𝑏(𝑠)𝑑𝑠.
(b) If there exists a constant 𝑏₂>0 such that 𝑓(𝑢)≤𝑏₂𝑢 for 𝑢∈[0,+∞), then (4.1) has no positive 𝜔-periodic solution for 0<𝜆<(𝑚−(𝑀+𝑚)𝑐)/𝑏₂𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠.

Theorem 4.15. Assume 𝑖₀=𝑖₀=𝑖_∞=𝑖_∞=0 hold and that one of the following conditions holds:
(1)𝑓₀≤𝑓_∞;(2)𝑓₀>𝑓_∞;(3)𝑓₀≤𝑓_∞≤𝑓₀≤𝑓_∞;(4)𝑓_∞≤𝑓₀≤𝑓_∞≤𝑓₀.
If 1−𝑐𝑙𝛼∫^𝜔₀𝑏(𝑠)𝑑𝑠max𝑓₀,𝑓₀,𝑓_∞,𝑓_∞()𝑐<𝜆<𝑚−𝑀+𝑚𝐿𝑀∫^𝜔₀𝑏(𝑠)𝑑𝑠min𝑓₀,𝑓₀,𝑓_∞,𝑓_∞, (4.63) then (4.1) has one positive 𝜔-periodic solution.

Remark 4.16. In a similar way, one can consider the second-order neutral functional differential equation (𝑥(𝑡)−𝑐𝑥(𝑡−𝛿(𝑡)))^′′−𝑎(𝑡)𝑥(𝑡)=−𝜆𝑏(𝑡)𝑓(𝑥(𝑡−𝜏(𝑡))).

5. Examples

Example 5.1. Consider the following equation:𝑥(𝑡)1−15𝑥𝑡−60sin4𝑡^′′=𝑥^′(𝑡)𝑥()sin4𝑡+arctan𝑡−sin4𝑡1+cos³(4𝑡)+cos4𝑡. (5.1) Comparing (5.1) to (3.1), we have 𝜔=𝜋/2, 𝑓(𝑡,𝑥)=𝑥(𝑡)sin4𝑡, 𝑔(𝑡,𝑥)=arctan(𝑥/(1+cos³(4𝑡))), 𝑐=15, 𝛿(𝑡)=(1/60)sin4𝑡, 𝜏(𝑡)=sin4𝑡, 𝑒(𝑡)=cos4𝑡 and 𝛿₁=max_{𝑡∈[0,𝜔]}|(1/15)cos4𝑡|=1/15, and we can easily choose 𝐷>𝜋/2 and 𝑀=𝜋/2 such that (𝐻₂)and(𝐻₃) holds. Regarding assumption (𝐻₁) note that 𝑓𝑡,𝑥^′(𝑡)≤𝑥^′(𝑡), (5.2) that is, (𝐻₁) holds with 𝐾₁=1,𝑏=0, and 𝜔^1/2(1+|𝑐|)^1/22𝐾₁|1−|𝑐||−|𝑐|𝛿₁=𝜋/2(1+15)^1/2√2||−()=4√1−151/15⋅15𝜋13<1. (5.3) Hence by Theorem 3.2, (5.1) has at least one 𝜋/2-periodic solution.

Example 5.2. Consider the following neutral functional differential equation: 𝑢(𝑡)+7𝑢()30𝑡−sin𝑡^′′+1𝑢(𝑡)()𝑢16=𝜆1−sin𝑡²(𝑡−𝜏(𝑡))𝑎^{𝑢(𝑡−𝜏(𝑡))}, (5.4) where 𝜆 and 0<𝑎<1 are two positive parameters, 𝜏(𝑡+2𝜋)=𝜏(𝑡).
Comparing (5.4) to (4.1), we see that 𝛿(𝑡)=sin𝑡, 𝑐=−7/30, 𝑎(𝑡)≡1/16, 𝑏(𝑡)=1−sin𝑡, 𝜔=2𝜋, 𝑓(𝑢)=𝑢²𝑎^𝑢. Clearly, 𝑀=1/16<(𝜋/2𝜋)²=1/4, 𝑓₀=0, 𝑓_∞=0, 𝑖₀=2. By Theorem 4.10, we easily get the following conclusion: (5.4) has two positive 𝜔-periodic solutions for 𝜆>1/4𝜋𝑟₁, where 𝑟₁=min{𝑓(0.27),𝑓(30/23)}.
In fact, by simple computations, we have 1𝑀=𝑚=116,𝛽=41,𝐿=√2𝛽sin𝛽2𝜋/2=22,𝑙=cos𝛽2𝜋/2√2𝛽sin𝛽2𝜋/2=2,𝑘=2+28,𝑘₁=√2+1−√328,𝛼=√23|𝑐|=72,𝑘30<𝑚𝑖𝑛₁,𝑚=√𝑀+𝑚2+1−√32,|𝑐|=7<830√23𝑓2=𝛼,₁(1)𝑓(𝑡)()√=min∶0.27≈8/232−(7/30)1−(7/30)²≤𝑡≤30𝑓()23=min0.27,𝑓3023=𝑟₁,1𝑓₁(1)𝑙∫^𝜔₀𝑏(𝑠)=1𝑑𝑠4𝜋𝑟₁. (5.5)

Acknowledgment

This research is supported by the National Natural Science Foundation of China (no. 10971202).