Lazer-Leach Type Conditions on Periodic Solutions of Liénard Equation with a Deviating Argument at Resonance

Abstract

We study the existence of periodic solutions of Liénard equation with a deviating argument 𝑥′′+𝑓(𝑥)𝑥^′+𝑛²𝑥+𝑔(𝑥(𝑡−𝜏))=𝑝(𝑡), where 𝑓,𝑔,𝑝∶𝑅→𝑅 are continuous and 𝑝 is 2𝜋-periodic, 0≤𝜏<2𝜋 is a constant, and 𝑛 is a positive integer. Assume that the limits lim_𝑥→±∞𝑔(𝑥)=𝑔(±∞) and lim_𝑥→±∞𝐹(𝑥)=𝐹(±∞) exist and are finite, where 𝐹(𝑥)=∫^𝑥₀𝑓(𝑢)𝑑𝑢. We prove that the given equation has at least one 2𝜋-periodic solution provided that one of the following conditions holds: 2cos(𝑛𝜏)[𝑔(+∞)−𝑔(−∞)]≠∫₀^2𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡, for all 𝜃∈[0,2𝜋],2𝑛cos(𝑛𝜏)[𝐹(+∞)−𝐹(−∞)]≠∫₀^2𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡, for all 𝜃∈[0,2𝜋],2[𝑔(+∞)−𝑔(−∞)]−2𝑛sin(𝑛𝜏)[𝐹(+∞)−𝐹(−∞)]≠∫₀^2𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡, for all 𝜃∈[0,2𝜋],2𝑛[𝐹(+∞)−𝐹(−∞)]−2sin(𝑛𝜏)[𝑔(+∞)−𝑔(−∞)]≠∫₀^2𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡, for all 𝜃∈[0,2𝜋].

1. Introduction

We are concerned with the existence of periodic solutions of Liénard equation with a deviating argument as follows: 𝑥^′′+𝑓(𝑥)𝑥^′+𝑛²𝑥+𝑔(𝑥(𝑡−𝜏))=𝑝(𝑡), (1) where 𝑓,𝑔,𝑝∶𝐑→𝐑 are continuous and 𝑝 is 2𝜋-periodic, 0≤𝜏<2𝜋 is a constant, and 𝑛 is a positive integer.

In recent years, the periodic problem of Liénard equations with a deviating argument has been widely studied because of its background in applied sciences (see [1–8] and the references cited therein).

In the case when 𝑓(𝑥)≡0, for all 𝑥∈𝐑 and 𝜏=0, (1) becomes 𝑥^′′+𝑛²𝑥+𝑔(𝑥)=𝑝(𝑡). (2) Assume that limits (ℎ₁)  lim_𝑥→±∞𝑔(𝑥)=𝑔(±∞)exist and are finite. Lazer and Leach [9] proved that (2) has at least one 2𝜋-periodic solution provided that the following condition holds: 2𝑔()(≠+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋]. (3) Assume that, besides (ℎ₁), the limits (ℎ₂)  lim_𝑥→±∞𝐹(𝑥)=𝐹(±∞)exist and are finite, where 𝐹(𝑥)=∫^𝑥₀𝑓(𝑢)𝑑𝑢. It was proved in [10] that the following equation: 𝑥^′′+𝑓(𝑥)𝑥^′+𝑛²𝑥+𝑔(𝑥(𝑡−𝜏))=𝑝(𝑡) (4) has at least 2𝜋-periodic solution provided that one of the following conditions holds: 2𝑔()(≠+∞−𝑔−∞)_02𝜋𝑝(𝑡)()[],[𝐹()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋2𝑛+∞−𝐹−∞)]_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋]. (5)

In the case when 𝑓(𝑥)≡0, for all 𝑥∈𝐑 and 𝜏≠0, (1) becomes as follows: 𝑥^′′+𝑛²𝑥+𝑔(𝑥(𝑡−𝜏))=𝑝(𝑡). (6) When the condition (ℎ₁) holds, it was proved in [5] that (6) has at least one 2𝜋-periodic solution provided that the condition (3) holds.

In the present paper, we deal with the existence of periodic solutions of (1) by assuming (ℎ₁) and (ℎ₂). By using the continuation theorem [11], we prove the following result.

Theorem 1. Assume that the conditions (ℎ₁) and (ℎ₂) hold. Then (1) has at least one 2𝜋-periodic solution provided that one of the following conditions is satisfied: (𝑔()(≠2cos𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)()[],(𝐹()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋2𝑛cos𝑛𝜏)[+∞−𝐹−∞)]_02𝜋𝑝(𝑡)()[],2𝑔()((𝐹()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋+∞−𝑔−∞)−2𝑛sin𝑛𝜏)[+∞−𝐹−∞)]_02𝜋𝑝(𝑡)()[],[𝐹()((𝑔()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋]. (7)

Remark 2. Let us denote by Φ the function on the right-hand side of four inequalities above, namely, Φ(𝜃)=₀^2𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,𝜃∈[0,2𝜋]. (8) Then Φ can be expressed in the following form: Φ(𝜃)=𝐴sin𝜃+𝐵cos𝜃,𝜃∈[0,2𝜋], (9) where 𝐴=₀^2𝜋𝑝(𝑡)cos𝑛𝑡𝑑𝑡,𝐵=₀^2𝜋𝑝(𝑡)sin𝑛𝑡𝑑𝑡. (10) Obviously, the value of Φ forms a closed interval [−𝜚,𝜚] with 𝜚=𝐴²+𝐵². Therefore, the four conditions in Theorem 1 are equivalent to the following conditions, respectively: (𝑔()(∉,(𝐹()(∉,2𝑔()((𝐹()(∉,[𝐹()((𝑔()(∉.2cos𝑛𝜏)+∞−𝑔−∞)−𝜚,𝜚2𝑛cos𝑛𝜏)[+∞−𝐹−∞)]−𝜚,𝜚+∞−𝑔−∞)−2𝑛sin𝑛𝜏)[+∞−𝐹−∞)]−𝜚,𝜚2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)−𝜚,𝜚 (11)

Remark 3. In the case when 𝜏=0, the four conditions in Theorem 1 reduce to the conditions (5). Therefore, Theorem 1 generalizes the result in [10].

Throughout this paper, we always use 𝐑 to denote the real number set. For a multivariate function 𝜁 depending on 𝑟, the notation 𝜁=𝑜(1) always means that, for 𝑟→∞, 𝜁→0 holds uniformly with respect to other variables, whereas 𝜁=𝑂(1) (or 𝜁=𝑂(𝑟⁻¹)) always means that 𝜁 (or 𝑟⋅𝜁) is bounded for 𝑟 large enough. For any continuous 2𝜋-periodic function 𝜓(𝑡), we always set ‖𝜓‖_∞=max_{0≤𝑡≤2𝜋}|𝜓(𝑡)|.

2. Basic Lemmas

It is well known that continuation theorems play an important role in studying the existence of periodic solutions of differential equations. We now introduce a continuation theorem which will be used to prove the existence of periodic solutions of (1).

Let 𝑋 and 𝑌 be two real Banach spaces and let 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 be a Fredholm operator with index zero, where 𝐷(𝐿) denotes the domain of 𝐿. This means that Im𝐿 is a closed subspace of 𝑌 and dimker𝐿=codimIm𝐿<+∞. Let 𝑃∶𝑋→Ker𝐿, 𝑄∶𝑌→𝑌 be two linear continuous projectors satisfying the following: Im𝑃=Ker𝐿,Ker𝑄=Im𝐿. (12) Then we have the following: 𝑋=Ker𝐿⊕Ker𝑃,𝑌=Im𝐿⊕Im𝑄. (13) Clearly, 𝐿_𝑃=𝐿|_{𝐷(𝐿)∩Ker𝑃}→Im𝐿 is invertible. Denote by 𝐾_𝑃 the inverse of 𝐿_𝑃. Let Ω⊂𝑋 be an open bounded set. A continuous map 𝑁∶Ω→𝑌 is said to be 𝐿-compact on Ω if both 𝑄𝑁∶Ω→𝑌 and 𝐾_𝑃(𝐼−𝑄)𝑁∶Ω→𝑋 are compact.

Lemma 4 (see [11]). Let X and Y be two real Banach spaces. Suppose that 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 is a Fredholm operator with index zero and 𝑁∶Ω→𝑌 is 𝐿-compact on Ω, where Ω is an open bounded subset of 𝑋. Moreover, assume that all the following conditions are satisfied:(1)𝐿𝑥≠𝜆𝑁𝑥, for all 𝑥∈𝜕Ω∩𝐷(𝐿), 𝜆∈(0,1);(2)𝑁𝑥∉Im𝐿, for all 𝑥∈𝜕Ω∩Ker𝐿;(3)The Brouwer degree deg{𝐽𝑄𝑁,Ω∩Ker𝐿,0}≠0, where 𝐽∶Im𝑄→Ker𝐿 is an isomorphism.Then equation 𝐿𝑥=𝑁𝑥 has at least one solution on 𝐷(𝐿)∩Ω.

3. Main Results

In this section, we will use the continuation theorem introduced in Section 2 to prove the existence of periodic solutions of (1). To this end, we first quote some notations and definitions.

Let 𝑋 and 𝑌 be two Banach spaces defined by the following: 𝑋=𝑥∈𝐶¹()()(𝑡),()()(𝑡)𝐑,𝐑∶𝑥𝑡+2𝜋=𝑥,∀𝑡∈𝐑𝑌=𝑦∈𝐶𝐑,𝐑∶𝑦𝑡+2𝜋=𝑦,∀𝑡∈𝐑 (14) with the following norms ‖𝑥‖_𝑋=max‖𝑥‖_∞,𝑥^′_∞,𝑦_𝑌=𝑦_∞. (15) Define a linear operator 𝐿∶𝐷(𝐿)⊂𝑋⟶𝑌,𝐿𝑥=𝑥^′′+𝑛²𝑥, (16) where 𝐷(𝐿)={𝑥∈𝑋∶𝑥^′′∈𝐶(𝐑,𝐑)}, and a nonlinear operator (𝑡)(𝑥(𝑡𝑥𝑁∶X⟶𝑌,𝑁𝑥)(=−𝑓))^′(𝑡)−𝑔(𝑥(𝑡−𝜏))+𝑝(𝑡). (17) It is easy to see that Ker𝐿=Span{sin𝑛𝑡,cos𝑛𝑡},Im𝐿=𝑦∈𝑌∶₀^2𝜋𝑦(𝑡)sin𝑛𝑡𝑑𝑡=0,_02𝜋𝑦(𝑡)cos𝑛𝑡𝑑𝑡=0. (18) On the other hand, for any 𝑦∈𝑌, we can write the following: 𝑦(𝑡)=𝑦(𝑡)+𝑦^∼(𝑡),∀𝑡∈[0,2𝜋], (19) where 𝑦(𝑡) is defined by the following: 𝑦(𝑡)=𝑎sin𝑛𝑡+𝑏cos𝑛𝑡 (20) with 1𝑎=𝜋₀^2𝜋𝑦(𝑡)1sin𝑛𝑡𝑑𝑡,𝑏=𝜋₀^2𝜋𝑦(𝑡)cos𝑛𝑡𝑑𝑡, (21) whereas 𝑦^∼(𝑡) satisfies the following: ₀^∼2𝑦^𝜋(𝑡)sin𝑛𝑡𝑑𝑡=0,₀^∼2𝜋𝑦(𝑡)cos𝑛𝑡𝑑𝑡=0. (22) Therefore, 𝑌=Ker𝐿⊕Im𝐿. (23) It follows that 𝐿 is a Fredholm map of index zero.

Let us define two continuous projectors 𝑃∶𝑋→Ker𝐿 and 𝑄∶𝑌→𝑌 by setting the following: (𝑃𝑥)(𝑡)=𝑎₁sin𝑛𝑡+𝑏₁(𝑡)=cos𝑛𝑡,𝑄𝑦𝑦(𝑡)=𝑎sin𝑛𝑡+𝑏cos𝑛𝑡,∀𝑡∈[0,2𝜋] (24) for any 𝑥∈𝑋 and 𝑦∈𝑌, where constants 𝑎₁ and 𝑏₁ are defined as constants 𝑎 and 𝑏. Obviously, Im𝑄=Ker𝐿.

Set 𝐿_𝑃=𝐿|_{𝐷(𝐿)∩Ker𝑃}→Im𝐿. Then 𝐿_𝑃 is an algebraic isomorphism and we define 𝐾_𝑃∶Im𝐿→𝐷(𝐿)∩Ker𝑃 by the following: 𝐾_𝑃=𝐿_𝑃⁻¹. (25) Clearly, we have that, for any 𝑦∈Im𝐿, 𝐾_𝑃𝑦𝑡)=1(𝑛^𝑡₀𝑦(𝑠)()+1sin𝑛𝑡−𝑠𝑑𝑠2𝑛𝜋₀^2𝜋𝑠𝑦(𝑠)sin𝑛(𝑡−𝑠)𝑑𝑠. (26) For any open bounded set Ω⊂𝑋, we can prove by standard arguments that 𝐾_𝑃(𝐼−𝑄)𝑁 and 𝑄𝑁 are compact on the closure Ω. Therefore, 𝑁 is 𝐿-compact on Ω.

It is noted that (1) is equivalent to the operator equation 𝐿𝑥=𝑁𝑥. (27) To use Lemma 4, we embed this operator equation into an equation family with a parameter 𝜆∈(0,1), 𝐿𝑥=𝜆𝑁𝑥, (28) which is equivalent to the equation as follows: 𝑥^′′+𝑛²𝑥+𝜆𝑓(𝑥(𝑡))𝑥^′(𝑡)(𝑥((𝑡)().+𝜆𝑔𝑡−𝜏))=𝜆𝑝,𝜆∈0,1 (29) In the following, we will prove some new results on the existence of periodic solutions of (1) by using the continuation theorem. Consider the equivalent system of (29): 𝑥^′=𝑦−𝜆𝐹(𝑥),𝑦^′=−𝑛²𝑥−𝜆𝑔(𝑥(𝑡−𝜏))+𝜆𝑝(𝑡). (30) Let 𝑥(𝑡) be any (possible) 2𝜋-periodic solution of (29). Set 𝑦(𝑡)=𝑥^′(𝑡)+𝜆𝐹(𝑥(𝑡)). Then (𝑥(𝑡),𝑦(𝑡)) is a 2𝜋-periodic solution of (30).

Now, let us introduce a transformation Φ∶(𝑟,𝜃)∈𝐑⁺×𝑆¹→(𝑥,𝑦)∈𝐑²⧵{0} with 𝑆¹=𝐑/2𝜋𝐙, 1𝑥=𝑛𝑟^1/2sin𝜃,𝑦=𝑟^1/2cos𝜃. (31) Under the transformation Φ, if |𝑥(𝑡)|+|𝑦(𝑡)|≠0, for 𝑡∈[0,2𝜋], then the 2𝜋-periodic solution (𝑥(𝑡),𝑦(𝑡)) of (30) can be expressed in the form (𝑟(𝑡),𝜃(𝑡)) satisfying the equations as follows: 𝑑𝜃𝑑𝑡=𝑛+𝜆𝑟^−1/2𝑔1𝑛𝑟^1/2(𝑡−𝜏)sin𝜃(𝑡−𝜏)sin𝜃−𝜆𝑛𝑟^−1/2𝐹1𝑛𝑟^1/2sin𝜃cos𝜃−𝜆𝑟^−1/2𝑝(𝑡)sin𝜃,𝑑𝑟𝑑𝑡=−2𝜆𝑟^1/2𝑔1𝑛𝑟^1/2(𝑡−𝜏)sin𝜃(𝑡−𝜏)cos𝜃−2𝜆𝑛𝑟^1/2𝐹1𝑛𝑟^1/2sin𝜃sin𝜃+2𝜆𝑟^1/2𝑝(𝑡)cos𝜃. (32) Let us set (𝑟₀,𝜃₀)=(𝑟(0),𝜃(0)) with 𝑟₀=𝑛²𝑥²(0)+𝑦²(0). Without loss of generality, we always assume 𝜃(0)∈[0,2𝜋]. Dividing the second equation of (32) by 𝑟^1/2, we get the following: 𝑑𝑟^1/21𝑑𝑡=−𝜆𝑔𝑛𝑟^1/2()()1𝑡−𝜏sin𝜃𝑡−𝜏cos𝜃−𝜆𝑛𝐹𝑛𝑟^1/2sin𝜃sin𝜃+𝜆𝑝(𝑡)cos𝜃. (33) Integrating (33) and applying conditions (ℎ₁) and (ℎ₂), we get the following: 𝑟(𝑡)^1/2=𝑟₀^1/2+𝑂(1),∀𝑡∈[0,2𝜋]. (34) Furthermore, 𝑟(𝑡)^−1/2=𝑟₀^−1/2+𝑂𝑟₀⁻¹,𝑡∈[0,2𝜋]. (35) On the other hand, it follows from the first equation of (32) and (35) that 𝑑𝜃𝑟𝑑𝑡=𝑛+𝑂₀^−1/2,𝑡∈[0,2𝜋]. (36) Therefore, we get the following: 𝜃(𝑡)=𝜃₀+𝑛𝑡+𝑂𝑟₀^−1/2,𝑡∈[0,2𝜋]. (37) The estimations (34)–(37) will be used to obtain apriori bounds of 2𝜋-periodic solutions of (29). Multiplying both sides of (29) by sin(𝜃₀+𝑛𝑡) and cos(𝜃₀+𝑛𝑡), respectively, and integrating over the interval [0,2𝜋], we obtain the following: ₀^2𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)sin𝜃₀++𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))sin𝜃₀=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)sin𝜃₀+𝑛𝑡𝑑𝑡,_02𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)cos𝜃₀++𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))cos𝜃₀=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)cos𝜃₀+𝑛𝑡𝑑𝑡. (38) Hence, −𝑛₀^2𝜋𝐹(𝑥(𝑡))cos𝜃₀++𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))sin𝜃₀=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)sin𝜃₀𝑛+𝑛𝑡𝑑𝑡_02𝜋𝐹(𝑥(𝑡))sin𝜃₀++𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))cos𝜃₀=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)cos𝜃₀+𝑛𝑡𝑑𝑡. (39)(40) Multiplying both sides of (29) by sin(𝜃₀+𝑛(𝑡−𝜏)) and cos(𝜃₀+𝑛(𝑡−𝜏)), respectively, and integrating over the interval [0,2𝜋], we obtain the following: ₀^2𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)sin𝜃₀()++𝑛𝑡−𝜏𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))sin𝜃₀(=+𝑛𝑡−𝜏)𝑑𝑡_02𝜋𝑝(𝑡)sin𝜃₀+𝑛(𝑡−𝜏)𝑑𝑡,_02𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)cos𝜃₀(++𝑛𝑡−𝜏)𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))cos𝜃₀()=+𝑛𝑡−𝜏𝑑𝑡_02𝜋𝑝(𝑡)cos𝜃₀+𝑛(𝑡−𝜏)𝑑𝑡. (41) Hence, −𝑛₀^2𝜋𝐹(𝑥(𝑡))cos𝜃₀()++𝑛𝑡−𝜏𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))sin𝜃₀(=+𝑛𝑡−𝜏)𝑑𝑡_02𝜋𝑝(𝑡)sin𝜃₀()𝑛+𝑛𝑡−𝜏𝑑𝑡,_02𝜋𝐹(𝑥(𝑡))sin𝜃₀(++𝑛𝑡−𝜏)𝑑𝑡_02𝜋𝑔(𝑥(𝑡−𝜏))cos𝜃₀()=+𝑛𝑡−𝜏𝑑𝑡_02𝜋𝑝(𝑡)cos𝜃₀+𝑛(𝑡−𝜏)𝑑𝑡. (42)

Proof of Theorem 1. We shall prove the existence of periodic solutions of (1) provided that either 𝑔()(≠2cos𝑛𝜏+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋], (43) or [𝐹()(𝑔()(≠2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋] (44) holds by using (39) and (40). The other cases can be handled similarly by using (43) and (44). We proceed in three steps.
(1) We prove that there exist positive constants 𝑀₁ and 𝑀₂ such that, for any 2𝜋-periodic solution 𝑥(𝑡) of (29), ‖𝑥‖_∞<𝑀₁,𝑥^′_∞<𝑀₂. (45) Assume by contradiction that (45) does not hold. Then there exists a sequence of 2𝜋-periodic solutions {𝑥_𝑘(𝑡)}^∞_𝑘=1 of (29) with 𝜆=𝜆_𝑘∈(0,1) such that 𝑥_𝑘∞+𝑥^′_𝑘∞⟶+∞,for𝑘→∞. (46) Write 𝑦_𝑘(𝑡)=𝑥^′_𝑘(𝑡)+𝜆_𝑘𝐹(𝑥_𝑘(𝑡)). Since 𝐹 is bounded on the interval (−∞,+∞), we have the following: 𝑥_𝑘∞+𝑦_𝑘∞⟶+∞,for𝑘→∞. (47) Let (𝑟_𝑘(𝑡),𝜃_𝑘(𝑡)) be the 2𝜋-periodic solution of (32) related to (𝑥_𝑘(𝑡),𝑦_𝑘(𝑡)). Obviously, 𝑟_𝑘(𝑡)=𝑛²𝑥²_𝑘(𝑡)+𝑦²_𝑘(𝑡). Then we have the following: 𝑟_𝑘∞⟶+∞,for𝑘⟶∞. (48) Without loss of generality, we also assume 𝜃_𝑘(0)∈[0,2𝜋]. It follows from (39) that −𝑛₀^2𝜋𝐹𝑥_𝑘(𝑡)cos𝜃_𝑘(0)++𝑛𝑡𝑑𝑡_02𝜋𝑔𝑥_𝑘(𝑡−𝜏)sin𝜃_𝑘(0)=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)sin𝜃_𝑘(0)+𝑛𝑡𝑑𝑡. (49)
From (34) and (37) we get that, for 𝑡∈[0,2𝜋], 𝑔𝑥_𝑘()1𝑡−𝜏=𝑔𝑛𝑟_𝑘(𝑡−𝜏)^1/2sin𝜃_𝑘()1𝑡−𝜏=𝑔𝑛𝑟_𝑘(0)^1/2sin𝜃_𝑘(0)+𝑛(𝑡−𝜏)+𝑂(1). (50) Therefore, ₀^2𝜋𝑔𝑥_𝑘(𝑡−𝜏)sin𝜃_𝑘(0)=+𝑛𝑡𝑑𝑡_02𝜋𝑔1𝑛𝑟_𝑘(0)^1/2sin𝜃_𝑘(0)((1𝜃+𝑛𝑡−𝜏)+𝑂)×sin_𝑘(0)+𝑛𝑡𝑑𝑡. (51) Obviously, we have the following: ₀^2𝜋𝑔1𝑛𝑟_𝑘(0)^1/2sin𝜃_𝑘(0)()(1)𝜃+𝑛𝑡−𝜏+𝑂×sin_𝑘(0)=1+𝑛𝑡𝑑𝑡𝑛^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2(1)()=sin𝑠+𝑂×sin𝑠+𝑛𝜏𝑑𝑠cos𝑛𝜏𝑛×^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2(1)+sin𝑠+𝑂sin𝑠𝑑𝑠sin𝑛𝜏𝑛×^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2sin𝑠+𝑂(1)cos𝑠𝑑𝑠. (52) Since 𝜃_𝑘(0)∈[0,2𝜋], there exists a subsequence {𝜃_𝑘𝑗(0)} such that 𝜃_𝑘𝑗→𝜃_∗, 𝑗→∞. By using Lebesgue dominated convergent theorem and the condition (ℎ₁), we get the following: lim_𝑗→+∞^{2𝑛𝜋+𝜃𝑘𝑗}_𝜃^{(0)−𝑛𝜏}_{𝑘𝑗(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘𝑗(0)^1/2(1)𝑔()(),sin𝑠+𝑂sin𝑠𝑑𝑠=2𝑛+∞−𝑔−∞lim_𝑗→+∞^{2𝑛𝜋+𝜃𝑘𝑗}_𝜃^{(0)−𝑛𝜏}_{𝑘𝑗(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘𝑗(0)^1/2sin𝑠+𝑂(1)cos𝑠𝑑𝑠=0. (53) Therefore, lim_𝑗→+∞0^2𝜋𝑔𝑥_𝑘𝑗(𝑡−𝜏)sin𝜃_𝑘𝑗(0)𝑔()(.+𝑛𝑡𝑑𝑡=2cos𝑛𝜏+∞−𝑔−∞) (54) Similarly, we can get the following: lim_𝑗→+∞0^2𝜋𝐹𝑥_𝑘𝑗(𝑡)cos𝜃_𝑘𝑗(0)+𝑛𝑡𝑑𝑡=0. (55) Hence, we obtain the following: 2cos𝑛𝜏𝑔(+∞)−𝑔(−∞)=₀^2𝜋𝑝(𝑡)sin𝜃_∗+𝑛𝑡𝑑𝑡, (56) which contradicts with (43).
On the other hand, it follows from (40) that 𝑛₀^2𝜋𝐹𝑥_𝑘(𝑡)sin𝜃_𝑘(0)++𝑛𝑡𝑑𝑡_02𝜋𝑔𝑥_𝑘(𝑡−𝜏)cos𝜃_𝑘(0)=+𝑛𝑡𝑑𝑡_02𝜋𝑝(𝑡)cos𝜃_𝑘(0)+𝑛𝑡𝑑𝑡. (57) According to (50), we have the following: ₀^2𝜋𝑔𝑥_𝑘(𝑡−𝜏)cos𝜃_𝑘(0)=+𝑛𝑡𝑑𝑡_02𝜋𝑔1𝑛𝑟_𝑘(0)^1/2cos𝜃_𝑘(0)((1𝜃+𝑛𝑡−𝜏)+𝑂)×cos_𝑘(0)=1+𝑛𝑡𝑑𝑡𝑛^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2(1)()=sin𝑠+𝑂×cos𝑠+𝑛𝜏𝑑𝑠cos𝑛𝜏𝑛×^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2(1)−sin𝑠+𝑂cos𝑠𝑑𝑠sin𝑛𝜏𝑛×^{2𝑛𝜋+𝜃𝑘}_𝜃^{(0)−𝑛𝜏}_{𝑘(0)−𝑛𝜏}𝑔1𝑛𝑟_𝑘(0)^1/2sin𝑠+𝑂(1)sin𝑠𝑑𝑠. (58) From (53) we obtain the following: lim_𝑗→+∞0^2𝜋𝑔𝑥_𝑘𝑗(𝑡−𝜏)cos𝜃_𝑘𝑗(0)𝑔()(.+𝑛𝑡𝑑𝑡=−2sin𝑛𝜏+∞−𝑔−∞) (59) Similarly, we have the following: lim_𝑗→+∞0^2𝜋𝐹𝑥_𝑘𝑗(𝑡)sin𝜃_𝑘𝑗(0)[𝐹()(.+𝑛𝑡𝑑𝑡=2+∞−𝐹−∞)] (60) It follows from (57)–(60) that [𝐹()((𝑔()(=2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)cos𝜃_∗+𝑛𝑡𝑑𝑡, (61) which contradicts with (44). Therefore, there exist positive constants 𝑀₁ and 𝑀₂ such that (45) holds.
(2) Let 𝑥(𝑡)=𝜚sin(𝑛𝑡+𝛼), where 𝛼 is an arbitrary constant. We will prove that there exists 𝑀₀>0 such that, for 𝜚≥𝑀₀, 𝑁𝑥∉Im𝐿. Otherwise, there exits a sequence {𝜚_𝑘} satisfying lim_𝑘→∞𝜚_𝑘=+∞ such that 𝑁𝑥_𝑘∈Im𝐿 with 𝑥_𝑘(𝑡)=𝜚_𝑘sin(𝑛𝑡+𝛼). We will prove the following: 2cos𝑛𝜏𝑔(+∞)−𝑔(−∞)=₀^2𝜋𝑝(𝑡)()[𝐹()((𝑔()(=sin𝑛𝑡+𝛼𝑑𝑡,2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝑛𝑡+𝛼)𝑑𝑡. (62)
In fact, since 𝑁𝑥_𝑘∈Im𝐿, we have the following: ₀^2𝜋𝑓𝑥_𝑘(𝑡)𝑥^′_𝑘(𝑡)+𝑔𝑥_𝑘()()=𝑡−𝜏sin𝑛𝑡+𝛼𝑑𝑡_02𝜋𝑝(𝑡)sin(𝑛𝑡+𝛼)𝑑𝑡,_02𝜋𝑓𝑥_𝑘(𝑡)𝑥^′_𝑘(𝑡)+𝑔𝑥_𝑘()()=𝑡−𝜏cos𝑛𝑡+𝛼𝑑𝑡_02𝜋𝑝(𝑡)cos(𝑛𝑡+𝛼)𝑑𝑡. (63)
Using the same method as in step 1, we have the following: lim_𝑘→∞0^2𝜋𝑓𝑥_𝑘(𝑡)𝑥^′_𝑘(𝑡)+𝑔𝑥_𝑘()()𝑔()(,𝑡−𝜏sin𝑛𝑡+𝛼𝑑𝑡=2cos𝑛𝜏+∞−𝑔−∞)lim_𝑘→∞0^2𝜋𝑓𝑥_𝑘(𝑡)𝑥^′_𝑘(𝑡)+𝑔𝑥_𝑘()()[𝐹()((𝑔()(.𝑡−𝜏cos𝑛𝑡+𝛼𝑑𝑡=2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞) (64)
As a consequence, (62) holds. Thus, we get a contradiction.
(3) Let 𝑀>max{𝑛𝑀₀,𝑀₁,𝑀₂} be a sufficiently large constant (if it is necessary, 𝑀 can be enlarged). Set Ω=𝑥∈𝑋∶‖𝑥‖_∞<𝑀,𝑥^′_∞<𝑀. (65) From the conclusion in step 1 we know that 𝐿𝑥≠𝜆𝑁𝑥,∀𝑥∈𝜕Ω∩𝐷(𝐿),𝜆∈(0,1). (66) From the conclusion in step 2 we know that 𝑁𝑥∉Im𝐿,∀𝑥∈𝜕Ω∩Ker𝐿, (67) which implies 𝑄𝑁𝑥≠0 for any 𝑥∈𝜕Ω∩Ker𝐿. Since Im𝑄=Ker𝐿, we can take an isomorphism 𝐽=identity∶Im𝑄→Ker𝐿. In what follows, we will prove the following: deg{𝐽𝑄𝑁,Ω∩Ker𝐿,0}≠0. (68) To this end, let us define 𝜙∶Ker𝐿→𝐑², 𝑥=𝑎sin𝑛𝑡+𝑏cos𝑛𝑡→(𝑎,𝑏), namely, 𝜙𝑥=(𝑎,𝑏). (69) Obviously, 𝜙 is a linear isomorphism. For any 𝑥=𝑎sin𝑛𝑡+𝑏cos𝑛𝑡, set (𝐽𝑄𝑁𝑥)(𝑡)=ℎ₁(𝑎,𝑏)sin𝑛𝑡+ℎ₂(𝑎,𝑏)cos𝑛𝑡, (70) where ℎ₁()=1𝑎,𝑏𝜋₀^2𝜋−𝑓(𝑥(𝑡))𝑥^′(𝑡)(𝑥((𝑡ℎ−𝑔𝑡−𝜏))+𝑝)×sin𝑛𝑡𝑑𝑡₂()=1𝑎,𝑏𝜋₀^2𝜋−𝑓(𝑥(𝑡))𝑥^′(𝑡)−𝑔(𝑥(𝑡−𝜏))+𝑝(𝑡)×cos𝑛𝑡𝑑𝑡. (71) Define ℎ∶𝐑²→𝐑² as follows: ℎ(𝑎,𝑏)=𝜙∘𝑄∘𝑁∘𝜙⁻¹(𝑎,𝑏)=ℎ₁(𝑎,𝑏),ℎ₂(𝑎,𝑏). (72) Then we have the following: deg{𝐽𝑄𝑁,Ω∩Ker𝐿,0}=degℎ,𝜙(Ω∩Ker𝐿),0. (73) To calculate deg{ℎ,𝜙(Ω∩Ker𝐿),0}, we first estimate 𝑙₁ and 𝑙₂ as follows: 𝑙₁()=1𝑎,𝑏𝜋₀^2𝜋−𝑓(𝑥(𝑡))𝑥^′(𝑡)(𝑥(𝑙−𝑔𝑡−𝜏))sin𝑛𝑡𝑑𝑡,₂()=1𝑎,𝑏𝜋₀^2𝜋−𝑓(𝑥(𝑡))𝑥^′(𝑡)−𝑔(𝑥(𝑡−𝜏))cos𝑛𝑡𝑑𝑡. (74) Write 𝑥=𝜌sin(𝑛𝑡+𝜗) with 𝜌=𝑎²+𝑏², 𝜗=arctan(𝑏/𝑎) or 𝜗=𝜋+arctan(𝑏/𝑎). Then we have that, for 𝜌→∞, ₀^2𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)sin𝑛𝑡𝑑𝑡=−𝑛₀^2𝜋𝐹(𝑥(𝑡))cos𝑛𝑡𝑑𝑡=−𝑛₀^2𝜋𝐹𝜌sin(𝑛𝑡+𝜗)cos𝑛𝑡𝑑𝑡=−_𝜗^{2𝑛𝜋+𝜗}𝐹𝜌sin𝑠(cos𝑠cos𝜗+sin𝑠sin𝜗)𝑑𝑠=−cos𝜗_𝜗^{2𝑛𝜋+𝜗}𝐹𝜌sin𝑠cos𝑠𝑑𝑠−sin𝜗_𝜗^{2𝑛𝜋+𝜗}𝐹[𝐹()((1).𝜌sin𝑠sin𝑠𝑑𝑠=−2𝑛sin𝜗+∞−𝐹−∞)]+𝑜 (75) On the other hand, we have that, for 𝜌→∞, ₀^2𝜋𝑔(𝑥(=𝑡−𝜏))sin𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑛()=1𝜌sin𝑡−𝜏+𝜗)sin𝑛𝑡𝑑𝑡𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔()=()𝜌sin𝑠sin𝑠+𝑛𝜏−𝜗𝑑𝑠cos𝑛𝜏−𝜗𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔+()𝜌sin𝑠sin𝑠𝑑𝑠sin𝑛𝜏−𝜗𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔(𝑔()((1).𝜌sin𝑠cos𝑠𝑑𝑠=2cos𝑛𝜏−𝜗)+∞−𝑔−∞)+𝑜 (76) Therefore, we get the following: 𝑙₁()=2𝑎,𝑏𝜋[𝐹()((𝑔()((1).𝑛sin𝜗+∞−𝐹−∞)]−cos𝑛𝜏−𝜗)+∞−𝑔−∞)+𝑜 (77) To estimate 𝑙₂, we have that, for 𝜌→∞, ₀^2𝜋𝑓(𝑥(𝑡))𝑥^′(𝑡)cos𝑛𝑡𝑑𝑡=𝑛₀^2𝜋𝐹(𝑥(𝑡))sin𝑛𝑡𝑑𝑡=𝑛₀^2𝜋𝐹𝜌sin(𝑛𝑡+𝜗)sin𝑛𝑡𝑑𝑡=cos𝜗_𝜗^{2𝑛𝜋+𝜗}𝐹𝜌sin𝑠sin𝑠𝑑𝑠−sin𝜗_𝜗^{2𝑛𝜋+𝜗}𝐹[𝐹()((1).𝜌sin𝑠cos𝑠𝑑𝑠=2𝑛cos𝜗+∞−𝐹−∞)]+𝑜 (78) Meanwhile, we get that, for 𝜌→∞, ₀^2𝜋𝑔(𝑥(=𝑡−𝜏))cos𝑛𝑡𝑑𝑡_02𝜋𝑔(𝑛()=1𝜌sin𝑡−𝜏+𝜗)cos𝑛𝑡𝑑𝑡𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔()=()𝜌sin𝑠cos𝑠+𝑛𝜏−𝜗𝑑𝑠cos𝑛𝜏−𝜗𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔−()𝜌sin𝑠cos𝑠𝑑𝑠sin𝑛𝜏−𝜗𝑛^{2𝑛𝜋+𝜗−𝑛𝜏}_{𝜗−𝑛𝜏}𝑔(𝑔()((1).𝜌sin𝑠sin𝑠𝑑𝑠=−2sin𝑛𝜏−𝜗)+∞−𝑔−∞)+𝑜 (79) Hence, we obtain the following: 𝑙₂()=2𝑎,𝑏𝜋[𝐹()((𝑔()((1).−𝑛cos𝜗+∞−𝐹−∞)]+sin𝑛𝜏−𝜗)+∞−𝑔−∞)+𝑜 (80) Set ℎ₁()=2𝑎,𝑏𝜋[𝐹()((𝑔()(,ℎ𝑛sin𝜗+∞−𝐹−∞)]−cos𝑛𝜏−𝜗)+∞−𝑔−∞)₂()=2𝑎,𝑏𝜋[𝐹()((𝑔()(,ℎ()=ℎ−𝑛cos𝜗+∞−𝐹−∞)]+sin𝑛𝜏−𝜗)+∞−𝑔−∞)𝑎,𝑏₁(𝑎,𝑏),ℎ₂(𝑎,𝑏). (81) Replacing 𝜗 in 𝑥=𝜌sin(𝑛𝑡+𝜗) with 𝜋+𝜗, we get the following: ℎ₁()2−𝑎,−𝑏=−𝜋[𝐹()((𝑔()(,ℎ𝑛sin𝜗+∞−𝐹−∞)]−cos𝑛𝜏−𝜗)+∞−𝑔−∞)₂()2−𝑎,−𝑏=−𝜋[𝐹()((𝑔()(.−𝑛cos𝜗+∞−𝐹−∞)]+sin𝑛𝜏−𝜗)+∞−𝑔−∞) (82) As a consequence, ℎ₁(−𝑎,−𝑏)=−ℎ₁(),ℎ𝑎,𝑏₂(−𝑎,−𝑏)=−ℎ₂(),ℎ()ℎ().𝑎,𝑏−𝑎,−𝑏=−𝑎,𝑏 (83) We note that, for 𝑎²+𝑏²→+∞, ℎ₁(𝑎,𝑏)=ℎ₁(𝑎,𝑏)+𝑐₁(1),ℎ+𝑜₂(𝑎,𝑏)=ℎ₂(𝑎,𝑏)+𝑐₂+𝑜(1), (84) where 𝑐₁=1𝜋₀^2𝜋𝑝(𝑡)𝑐sin𝑛𝑡𝑑𝑡,₂=1𝜋₀^2𝜋𝑝(𝑡)cos𝑛𝑡𝑑𝑡. (85) Let us consider the map 𝐻∶𝜙(Ω∩Ker𝐿)×[0,1]→𝐑²∶(𝑎,𝑏,𝜇)→(ℎ₁(𝑎,𝑏,𝜇),ℎ₂(𝑎,𝑏,𝜇)) with ℎ₁𝑎,𝑏,𝜇=ℎ₁(𝑎,𝑏)+𝜇𝑐₁,ℎ₂𝑎,𝑏,𝜇=ℎ₂(𝑎,𝑏)+𝜇𝑐₂. (86) Obviously, 𝐻 is continuous. Next, we shall prove that, for any (𝑎,𝑏,𝜇)∈𝜕𝜙(Ω∩Ker𝐿)×[0,1], ℎ₁𝑎,𝑏,𝜇,ℎ₂𝑎,𝑏,𝜇≠(0,0). (87) Otherwise, there exists some (𝑎,𝑏,𝜇)∈𝜕𝜙(Ω∩Ker𝐿)×[0,1] such that ℎ₁𝑎,𝑏,𝜇=0,ℎ₂𝑎,𝑏,𝜇=0. (88) Then we have the following: [𝐹()((𝑔()(2𝑛sin𝜗+∞−𝐹−∞)]−2cos𝑛𝜏−𝜗)+∞−𝑔−∞)=−𝜇𝜋𝑐₁,[𝐹()((𝑔()(−2𝑛cos𝜗+∞−𝐹−∞)]+2sin𝑛𝜏−𝜗)+∞−𝑔−∞)=−𝜇𝜋𝑐₂. (89) Therefore, we get the following: 2cos𝑛𝜏𝑔(+∞)−𝑔(−∞)=𝜇₀^2𝜋𝑝(𝑡)()[𝐹()((𝑔()(sin𝑛𝑡+𝜗𝑑𝑡,2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)=𝜇₀^2𝜋𝑝(𝑡)cos(𝑛𝑡+𝜗)𝑑𝑡. (90) Since 𝜇∈[0,1], we know from (90) that 𝑔()(∈,[𝐹()((𝑔()(∈,2cos𝑛𝜏+∞−𝑔−∞)−𝜚,𝜚2𝑛+∞−𝐹−∞)]−2sin𝑛𝜏)+∞−𝑔−∞)−𝜚,𝜚 (91)(92) where 𝜚 is given in Remark 2. From Remark 2 we know that (91) and (92) contradict with (43) and (44).
In particular, we have that, for (𝑎,𝑏)∈𝜕𝜙(Ω∩Ker𝐿), ℎ₁(𝑎,𝑏),ℎ₂()≠(),ℎ𝑎,𝑏0,0₁(𝑎,𝑏)+𝑐₁,ℎ₂(𝑎,𝑏)+𝑐₂≠(0,0). (93)(94) Since ℎ∶𝐑²→𝐑² is odd and (93) holds, we know from Borsuk Theorem [12] that degℎ,𝜙(Ω∩Ker𝐿),0=2𝑚+1≠0, (95) where 𝑚 is an integer.
On the other hand, we know from (94) and the expressions of ℎ₁ and ℎ₂ that there exists a positive constant 𝜈, which is independent of 𝑎 and 𝑏, such that, for (𝑎,𝑏)∈𝐑²ℎ₁(𝑎,𝑏)+𝑐₁+ℎ₂(𝑎,𝑏)+𝑐₂≥𝜈. (96) Consequently, we infer from the homotopy invariance of degree that, if 𝑀>max{𝑀₀,𝑀₁,𝑀₂} is large enough; then ()()𝐻()()𝐻()()()deg𝐽𝑄𝑁,Ω∩Ker𝐿,0=degℎ,𝜙Ω∩Ker𝐿,0=deg⋅,1,𝜙Ω∩Ker𝐿,0=deg⋅,0,𝜙Ω∩Ker𝐿,0=degℎ,𝜙Ω∩Ker𝐿,0=2𝑚+1≠0. (97) Therefore, all conditions of Lemma 4 are satisfied. Thus, (1) has at least one 2𝜋-periodic solution.

4. Remarks

We can use the method developed in Section 3 to deal with the existence of 2𝜋-periodic solutions of the following equation: 𝑥^′′+𝑓𝑥^′+𝑛²𝑥+𝑔(𝑥(𝑡−𝜏))=𝑝(𝑡). (98) Assume that the limits (ℎ₃)  lim_𝑥→±∞𝑓(𝑥)=𝑓(±∞)exist and are finite. We can prove the following theorem.

Theorem 5. Assume that the conditions (ℎ₁) and (ℎ₃) hold. Then (98) has at least one 2𝜋-periodic solution provided that one of the following conditions holds: (𝑔()(≠2cos𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)()[],(𝑓()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋2cos𝑛𝜏)+∞−𝑓−∞)_02𝜋𝑝(𝑡)()[],2𝑔()((𝑓()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋+∞−𝑔−∞)−2sin𝑛𝜏)+∞−𝑓−∞)_02𝜋𝑝(𝑡)()[],2𝑓()((𝑔()(≠sin𝜃+𝑛𝑡𝑑𝑡,∀𝜃∈0,2𝜋+∞−𝑓−∞)−2sin𝑛𝜏)+∞−𝑔−∞)_02𝜋𝑝(𝑡)sin(𝜃+𝑛𝑡)𝑑𝑡,∀𝜃∈[0,2𝜋]. (99)

Remark 6. In the case when 𝑛=1, the third and the fourth condition in Theorem 5 are identical to the related conditions in [6]. But the first and the second condition in Theorem 5 did not appear in [6].

Acknowledgments

The research was supported by Research Fund for the Doctoral Program of Higher Education of China, no. 11AA0013, Beijing Natural Science Foundation (Existence and multiplicity of periodic solutions in nonlinear oscillations), no. 1112006, and the Grant of Beijing Education Committee Key Project, no. KZ201310028031.