Periodic Solutions for a Class of 𝑛-th Order Functional Differential Equations

Abstract

We study the existence of periodic solutions for n-th order functional differential equations 𝑥^(𝑛)(𝑡)=∑^{𝑛−1𝑖=0}𝑏_𝑖[𝑥^(𝑖)(𝑡)]^𝑘+𝑓(𝑥(𝑡−𝜏(𝑡)))+𝑝(𝑡). Some new results on the existence of periodic solutions of the equations are obtained. Our approach is based on the coincidence degree theory of Mawhin.

1. Introduction

In this paper, we are concerned with the existence of periodic solutions of the following 𝑛-th order functional differential equations: 𝑥^(𝑛)(𝑡)=^𝑛−1_𝑖=0𝑏_𝑖𝑥^(𝑖)(𝑡)^𝑘+𝑓(𝑥(𝑡−𝜏(𝑡)))+𝑝(𝑡),(1.1) where 𝑏_𝑖,𝑖=0,1,…,𝑛−1 are constants, 𝑘 is a positive odd, 𝑓∈𝐶¹(𝑅,𝑅) for ∀𝑥∈𝑅, 𝑝∈𝐶(𝑅,𝑅) with 𝑝(𝑡+𝑇)=𝑝(𝑡).

In recent years, there are many papers studying the existence of periodic solutions of first-, second- or third-order differential equations [1–12]. For example, in [5], Zhang and Wang studied the following differential equations: 𝑥^(𝑡)+𝑎𝑥^{2𝑘−1}(𝑡)+𝑏𝑥^2𝑘−1(𝑡)+𝑐𝑥^2𝑘−1(𝑡)+𝑔𝑡,𝑥𝑡−𝜏₁,𝑥^𝑡−𝜏₂=𝑝(𝑡).(1.2) The authors established the existence of periodic solutions of (1.2) under some conditions on 𝑎,𝑏,𝑐, and 2𝑘−1.

In [13–24], periodic solutions for 𝑛, 2𝑛, and 2𝑛+1 th order differential equations were discussed. For example, in [22, 24], Pan et al. studied the existence of periodic solutions of higher order differential equations of the form 𝑥^(𝑛)(𝑡)=^𝑛−1_𝑖=1𝑏_𝑖𝑥^(𝑖)(𝑡)+𝑓𝑡,𝑥(𝑡),𝑥𝑡−𝜏₁(𝑡),…,𝑥𝑡−𝜏_𝑚(𝑡)+𝑝(𝑡).(1.3) The authors obtained the results based on the damping terms 𝑥^(𝑖)(𝑡) and the delay 𝜏_𝑖(𝑡).

In present paper, by using Mawhin’s continuation theorem, we will establish some theorems on the existence of periodic solutions of (1.1). The results are related to not only 𝑏_𝑖 and 𝑓(𝑡,𝑥) but also the positive odd 𝑘. In addition, we give an example to illustrate our new results.

2. Some Lemmas

We investigate the theorems based on the following lemmas.

Lemma 2.1 (see [17]). Let 𝑛₁>1,𝛼∈[0,+∞) be constants, 𝑠∈𝐶(𝑅,𝑅) with 𝑠(𝑡+𝑇)=𝑠(𝑡), and 𝑠(𝑡)∈[−𝛼,𝛼],forall𝑡∈[0,𝑇]. Then for ∀𝑥∈𝐶¹(𝑅,𝑅) with 𝑥(𝑡+𝑇)=𝑥(𝑡), one has ^𝑇0||𝑥(𝑡)−𝑥(𝑡−𝑠(𝑡))||^𝑛1𝑑𝑡≤2𝛼^𝑛1^𝑇0||𝑥^(𝑡)||^𝑛1𝑑𝑡.(2.1)

Lemma 2.2. Let 𝑘≥1,𝛼∈[0,+∞)be constants, 𝑠∈𝐶(𝑅,𝑅) with 𝑠(𝑡+𝑇)=𝑠(𝑡), and 𝑠(𝑡)∈[−𝛼,𝛼],forall𝑡∈[0,𝑇]. Then for ∀𝑥∈𝐶¹(𝑅,𝑅) with 𝑥(𝑡+𝑇)=𝑥(𝑡), one has ^𝑇0||𝑥^𝑘(𝑡)−𝑥^𝑘(𝑡−𝑠(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡≤2𝛼^{(𝑘+1)/𝑘}𝑘^1/𝑘(𝑘−1)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡+^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡.(2.2)

Proof. Let 𝐹(𝑡)=𝑥^𝑘(𝑡). By Lemma 2.2, one has ^𝑇0||𝑥^𝑘(𝑡)−𝑥^𝑘(𝑡−𝑠(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡=≤2^𝑇0𝛼||𝐹(𝑡)_{(𝑘+1)/𝑘}−𝐹(𝑡−𝑠(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^𝑇0=2𝛼_{(𝑘+1)/𝑘}||𝐹^(𝑡)||^{(𝑘+1)/𝑘}𝑑𝑡^𝑇0=2𝛼_{(𝑘+1)/𝑘}||𝑘𝑥^𝑘−1(𝑡)𝑥^(𝑡)||^{(𝑘+1)/𝑘}𝑑𝑡𝑘^{(𝑘+1)/𝑘}^𝑇0||𝑥(𝑡)||^{((𝑘−1)(𝑘+1))/𝑘}||𝑥^(𝑡)||^{(𝑘+1)/𝑘}𝑑𝑡.(2.3) By inequality 𝑥𝑦≤𝑥^𝑝𝑝+𝑦𝑞^𝑞,𝑥≥0,𝑦≥0,1𝑝+1𝑞=1,(2.4) one has ||𝑥(𝑡)||^{((𝑘−1)(𝑘+1))/𝑘}||𝑥^(𝑡)||^{(𝑘+1)/𝑘}≤(𝑘−1)||𝑥(𝑡)||^𝑘+1𝑘+||𝑥^(𝑡)||^𝑘+1𝑘.(2.5) Thus we obtain ^𝑇0||𝑥^𝑘(𝑡)−𝑥^𝑘(𝑡−𝑠(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡≤2𝛼^{(𝑘+1)/𝑘}𝑘^1/𝑘(𝑘−1)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡+^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡.(2.6)

Lemma 2.3. If 𝑘≥1 is an integer, 𝑥∈𝐶^𝑛(𝑅,𝑅), and 𝑥(𝑡+𝑇)=𝑥(𝑡), then ^𝑇0||𝑥^(𝑡)||^𝑘𝑑𝑡^1/𝑘≤𝑇^𝑇0||𝑥^(𝑡)||^𝑘𝑑𝑡^1/𝑘≤⋯≤𝑇^𝑛−1^𝑇0||𝑥^(𝑛)(𝑡)||^𝑘𝑑𝑡^1/𝑘.(2.7)

The proof of Lemma 2.3 is easy, here we omit it.

We first introduce Mawhin’s continuation theorem.

Let 𝑋 and 𝑌 be Banach spaces, 𝐿∶𝐷(𝐿)⊂𝑋→𝑌 are a Fredholm operator of index zero, here 𝐷(𝐿) denotes the domain of 𝐿. 𝑃∶𝑋→𝑋,𝑄∶𝑌→𝑌 be projectors such that Im𝑃=Ker𝐿,Ker𝑄=Im𝐿,𝑋=Ker𝐿⊕Ker𝑃,𝑌=Im𝐿⊕Im𝑄.(2.8) It follows that 𝐿∣_{𝐷(𝐿)∩Ker𝑃}∶𝐷(𝐿)∩Ker𝑃⟶Im𝐿(2.9) is invertible, we denote the inverse of that map by 𝐾_𝑝. Let Ω be an open bounded subset of 𝑋, 𝐷(𝐿)∩Ω≠∅, the map 𝑁∶𝑋→𝑌 will be called 𝐿-compact in Ω, if 𝑄𝑁(Ω) is bounded and 𝐾_𝑝(𝐼−𝑄)𝑁∶Ω→𝑋 is compact.

Lemma 2.4 (see [25]). Let 𝐿 be a Fredholm operator of index zero and let 𝑁 be 𝐿-compact on Ω. Assume that the following conditions are satisfied: (i)𝐿𝑥≠𝜆𝑁𝑥,forall𝑥∈𝜕Ω∩𝐷(𝐿),𝜆∈(0,1);(ii)𝑄𝑁𝑥≠0,forall𝑥∈𝜕Ω∩Ker𝐿;(iii)deg{𝑄𝑁𝑥,Ω∩Ker𝐿,0}≠0,then the equation 𝐿𝑥=𝑁𝑥 has at least one solution in Ω∩𝐷(𝐿).

Now, we define 𝑌={𝑥∈𝐶(𝑅,𝑅)∣𝑥(𝑡+𝑇)=𝑥(𝑡)} with the norm |𝑥|_∞=max_{𝑡∈[0,𝑇]}{|𝑥(𝑡)|} and 𝑋={𝑥∈𝐶^𝑛−1(𝑅,𝑅)∣𝑥(𝑡+𝑇)=𝑥(𝑡)} with norm ‖𝑥‖=max{|𝑥|_∞,|𝑥^|_∞,…,|𝑥^(𝑛−1)|_∞}. It is easy to see that 𝑋,𝑌 are two Banach spaces. We also define the operators 𝐿 and 𝑁 as follows: 𝐿∶𝐷(𝐿𝑁)⊂∶𝑋𝑋⟶⟶𝑌𝑌,,𝐿𝑁𝑥𝑥==𝑥−^(𝑛),_𝑛−1𝐷(𝐿)={𝑥∣𝑥∈𝐶^𝑛(𝑅,𝑅),𝑥(𝑡+𝑇)=𝑥(𝑡)},_𝑖=1𝑏_𝑖𝑥^(𝑖)(𝑡)^𝑘−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))+𝑝(𝑡).(2.10) It is easy to see that (1.1) can be converted to the abstract equation 𝐿𝑥=𝑁𝑥. Moreover, from the definition of 𝐿, we see that ker𝐿=𝑅, dim(ker𝐿)=1, Im𝐿={𝑦∣𝑦∈𝑌,∫^𝑇0𝑦(𝑠)𝑑𝑠=0} is closed, and dim(𝑌⧵Im𝐿)=1, one has codim(Im𝐿)=dim(ker𝐿). So 𝐿 is a Fredholm operator with index zero. Let 𝑃∶𝑋⟶ker𝐿,𝑃𝑥=𝑥(0),𝑄∶𝑌⟶𝑌⧵Im𝐿,𝑄𝑦=1𝑇^𝑇0𝑦(𝑡)𝑑𝑡,(2.11) and let 𝐿|_{𝐷(𝐿)∩Ker𝑃}∶𝐷(𝐿)∩Ker𝑃⟶Im𝐿.(2.12) Then 𝐿|_{𝐷(𝐿)∩Ker𝑃} has a unique continuous inverse 𝐾_𝑝. One can easily find that 𝑁 is 𝐿-compact in Ω, where Ω is an open bounded subset of 𝑋.

3. Main Result

Theorem 3.1. Suppose 𝑛=2𝑚+1, 𝑚>0 an integer and the following conditions hold:(𝐻₁)The function 𝑓 satisfies lim_𝑥→∞|||𝑓(𝑡,𝑥)𝑥^𝑘|||≤𝛾,(3.1)||𝑓(𝑡,𝑥)−f(𝑡,𝑦)||≤𝛽||𝑥^𝑘−𝑦^𝑘||,(3.2) where 𝛾≥0.(𝐻₂)||𝑏₀||>𝛾+𝜃₂.(3.3)(𝐻₃)There is a positive integer 0<𝑠≤𝑚 such that 𝑏𝑏_2𝑠≠0,if𝑠=𝑚,_2𝑠≠0,𝑏_2𝑠+𝑖=0,𝑖=1,2,…,2𝑚−2𝑠,if0<𝑠<𝑚.(3.4)(𝐻₄)𝐴₂(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}+𝛾+𝜃²𝐴₁(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇^2𝑠𝐴¹(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_2𝑠||,if1<𝑠≤𝑚,₁𝑇^𝑘+𝛾+𝜃²𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏₂||,if𝑠=1,(3.5) where 𝐴₁(𝑠,𝑘)=∑^𝑠𝑖=1|𝑏_𝑖|𝑇^{(𝑠−𝑖)𝑘}, 𝐴₂(𝑠,𝑘)=∑^{𝑠−2𝑖=1}|𝑏_𝑖|𝑇^{(𝑠−𝑖)𝑘}, 𝜃₁=2^{𝑘/(𝑘+1)}𝛽|𝜏(𝑡)|_∞𝑘^1/(𝑘+1), 𝜃₂=2^{𝑘/(𝑘+1)}𝛽|𝜏(𝑡)|_∞𝑘^1/(𝑘+1)(𝑘−1)^{𝑘/(𝑘+1)}. Then (1.1) has at least one 𝑇-periodic solution.

Proof. Consider the equation 𝐿𝑥=𝜆𝑁𝑥,𝜆∈(0,1),(3.6) where 𝐿 and 𝑁 are defined by (2.10). Let Ω₁=𝑥∈𝐷(𝐿)Ker𝐿,𝐿𝑥=𝜆𝑁𝑥forsome𝜆∈(0,1).(3.7) For 𝑥∈Ω₁, one has 𝑥^(𝑛)(𝑡)=𝜆^2𝑠_𝑖=0𝑏_𝑖𝑥^(𝑖)(𝑡)^𝑘+𝜆𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))+𝜆𝑝(𝑡),𝜆∈(0,1).(3.8) Multiplying both sides of (3.8) by 𝑥(𝑡), and integrating them on [0,𝑇], one has for 𝜆∈(0,1)^𝑇0𝑥^(𝑛)(𝑡)𝑥(𝑡)𝑑𝑡=𝜆^2𝑠_𝑖=0𝑏_𝑖^𝑇0𝑥^(𝑖)+𝜆_𝑇0(𝑡)^𝑘𝑥(𝑡)𝑑𝑡𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))𝑥(𝑡)𝑑𝑡+𝜆^𝑇0𝑝(𝑡)𝑥(𝑡)𝑑𝑡.(3.9) Since for any positive integer 𝑖, ^𝑇0𝑥^(2𝑖−1)(𝑡)𝑥(𝑡)𝑑𝑡=0,(3.10) and in view of 𝑛=2𝑚+1 and 𝑘 is odd, it follows from (3.3) and (3.9) that ||𝑏₀||≤^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡_2𝑠_𝑖=1≤||𝑏_𝑖||^𝑇0||𝑥^(𝑖)(𝑡)||^𝑘||𝑥(𝑡)||𝑑𝑡+^𝑇0||𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||||𝑥(𝑡)||𝑑𝑡+^𝑇0||𝑝(𝑡)||||𝑥(𝑡)||𝑑𝑡_2𝑠_𝑖=1+||𝑏_𝑖||^𝑇0||𝑥^(𝑖)(𝑡)||^𝑘||𝑥(𝑡)||𝑑𝑡+^𝑇0||𝑓(𝑡,𝑥(𝑡))||||𝑥(𝑡)||𝑑𝑡_𝑇0||𝑓(𝑡,𝑥)−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||||𝑥(𝑡)||𝑑𝑡+^𝑇0||𝑝(𝑡)||||𝑥(𝑡)||𝑑𝑡.(3.11) By using Hölder inequality and Lemma 2.1, from (3.11), we obtain ||𝑏₀||≤^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡_𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)^2𝑠_𝑖=1||𝑏_𝑖||^𝑇0||𝑥^(𝑖)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥)−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑝(𝑡)||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}≤_𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)^2𝑠_𝑖=1||𝑏_𝑖||𝑇^{(2𝑠−𝑖)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥)−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+||𝑝(𝑡)||_∞𝑇^{𝑘/(𝑘+1)}.(3.12) So ||𝑏₀||^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝐴₁(2𝑠,𝑘)^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+𝑢₁,(3.13) where 𝑢₁ is a positive constant. Choosing a constant 𝜀>0 such that 𝛾+𝜀+𝜃₂<||𝑏₀||,(3.14)𝐴₂(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}+𝛾+𝜀+𝜃²𝐴₁(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂+𝑘||𝑏₀||𝑇^2𝑠𝐴¹(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_2𝑠||,if1<𝑠≤𝑚,₁𝑇^𝑘+𝛾+𝜀+𝜃²𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−(𝛾+𝜀)−𝜃₂+𝑘||𝑏₀||𝑇²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−(𝛾+𝜀)−𝜃₂^{(𝑘−1)/𝑘}<||𝑏₂||,if𝑠=1,(3.15) for the above constant 𝜀>0, we see from (3.1) that there is a constant 𝛿>0 such that ||𝑓(𝑡,𝑥(𝑡))||<(𝛾+𝜀)||𝑥(𝑡)||^𝑘,for||𝑥(𝑡)||>𝛿,𝑡∈[0,𝑇].(3.16) Denote Δ₁=𝑡∈[0,𝑇]∶||𝑥(𝑡)||≤𝛿,Δ₂=𝑡∈[0,𝑇]∶||𝑥(𝑡)||>𝛿.(3.17) Since ^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡≤≤𝑓_Δ1||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡+_Δ2||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡_𝛿=𝑓^{(𝑘+1)/𝑘}𝑇+(𝛾+𝜀)^{(𝑘+1)/𝑘}^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡_𝛿^{(𝑘+1)/𝑘}𝑇+(𝛾+𝜀)^{(𝑘+1)/𝑘}^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡,(3.18) using inequality (𝑎+𝑏)^𝑙≤𝑎^𝑙+𝑏^𝑙for𝑎≥0,𝑏≥0,0≤𝑙≤1,(3.19) it follows from (3.18) that ^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}≤𝑓_𝛿𝑇^{𝑘/(𝑘+1)}+(𝛾+𝜀)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}.(3.20) From (3.2) and by Lemma 2.2, one has ^𝑇0||𝑓(𝑡,𝑥(𝑡))−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}≤𝛽^𝑇0||𝑥^𝑘(𝑡)−𝑥^𝑘(𝑡−𝜏(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}≤2^{𝑘/(𝑘+1)}𝛽||𝜏(𝑡)||_∞𝑘^1/(𝑘+1)(𝑘−1)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡+^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤2^{𝑘/(𝑘+1)}𝛽||𝜏(𝑡)||_∞𝑘^1/(𝑘+1)(𝑘−1)^{𝑘/(𝑘+1)}^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤2_{𝑘/(𝑘+1)}𝛽||𝜏(𝑡)||_∞𝑘^1/(𝑘+1)(𝑘−1)^{𝑘/(𝑘+1)}^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+2^{𝑘/(𝑘+1)}𝛽||𝜏(𝑡)||_∞𝑘^1/(𝑘+1)𝑇^{(2𝑠−1)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}=𝜃₂^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+𝜃₁𝑇^{(2𝑠−1)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}.(3.21) Substituting the above formula into (3.13), one has ||𝑏₀||−(𝛾+𝜀)−𝜃₂^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝐴₁(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+𝑢₂,(3.22) where 𝑢₂ is a positive constant. That is ^𝑇0||𝑥(𝑡)||^k+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝐴¹(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+𝑢₃,(3.23) where 𝑢₃ is a positive constant.

On the other hand, multiplying both sides of (3.8) by 𝑥^(2𝑠)(𝑡), and integrating on [0,𝑇], one has ^𝑇0𝑥=^(𝑛)(𝑡)𝑥^(2𝑠)(𝑡)𝑑𝑡_2𝑠_𝑖=0𝑏_𝑖^𝑇0𝑥^(𝑖)(𝑡)^𝑘𝑥^(2𝑠)(𝑡)𝑑𝑡+^𝑇0𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))𝑥^(2𝑠)(𝑡)𝑑𝑡+^𝑇0𝑝(𝑡)𝑥^(2𝑠)(𝑡)𝑑𝑡.(3.24)

If 1<𝑠≤𝑚, since ^𝑇0𝑥^(2𝑚+1)(𝑡)𝑥^(2𝑠)(𝑡)𝑑𝑡=0,^𝑇0𝑥^(2𝑠−1)(𝑡)^𝑘𝑥^(2𝑠)(𝑡)𝑑𝑡=0,(3.25)^𝑇0[𝑥(𝑡)]^𝑘𝑥^(2𝑠)(𝑡)𝑑𝑡=−𝑘^𝑇0[𝑥(𝑡)]^𝑘−1𝑥^(2𝑠−1)(𝑡)𝑥^(𝑡)𝑑𝑡,(3.26) by using Hölder inequality and Lemma 2.1, from (3.23), one has ||𝑏_2𝑠||≤^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡_𝑇0||𝑥^(2𝑠)(𝑡)||^2𝑠−2_𝑖=1+𝑘||𝑏₀||𝑏_𝑖||||𝑥^(𝑖)(𝑡)||^𝑘+||𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||+||𝑝(𝑡)||𝑑𝑡≤||^𝑇0||𝑥(𝑡)||^𝑘−1||𝑥^(2𝑠−1)(𝑡)||||𝑥^(𝑡)||𝑑𝑡_𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)^2𝑠−2_𝑖=1||𝑏_𝑖||𝑇^{(2𝑠−𝑖)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+^𝑇0||𝑓(𝑡,𝑥(𝑡))−𝑓(𝑡,𝑥(𝑡−𝜏))||^{(𝑘+1)/𝑘}𝑑𝑡^{𝑘/(𝑘+1)}+||𝑝(𝑡)||_∞𝑇^{𝑘/(𝑘+1)}+𝑘||𝑏₀||||𝑥^(𝑡)||_∞^𝑇0||𝑥(𝑡)||^𝑘−1||𝑥^(2𝑠−1)(𝑡)||𝑑𝑡.(3.27) Since 𝑥(0)=𝑥(𝑇), there exists 𝜉∈[0,𝑇] such that 𝑥^(𝜉)=0. So for 𝑡∈[0,𝑇]𝑥^(𝑡)=𝑥^(𝜉)+^𝑡𝜉𝑥^(𝜎)𝑑𝜎.(3.28) Using Hölder inequality and Lemma 2.1, one has ||𝑥^(𝑡)||_∞≤^𝑇0||𝑥^(𝑡)||𝑑𝑡≤𝑇^{𝑘/(𝑘+1)}^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)≤𝑇^{2𝑠−1−(1/(𝑘+1))}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1).(3.29) Using inequality 𝑇1^𝑇0||||𝑥(𝑡)||^𝑟||^1/𝑟≤1𝑇^𝑇0|||||𝑥(𝑡)||_𝑙|||^1/𝑙for0≤𝑟≤𝑙,∀𝑥∈𝑅.(3.30) and applying Hölder inequality and by Lemma 2.1, we obtain ^𝑇0||𝑥(𝑡)||^𝑘−1||𝑥^(2𝑠−1)(𝑡)||𝑑𝑡≤^𝑇0||𝑥(𝑡)||^𝑘𝑑𝑡^{(𝑘−1)/𝑘}^𝑇0||𝑥^(2𝑠−1)(𝑡)||^𝑘𝑑𝑡^1/𝑘≤𝑇^1/(𝑘+1)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{(𝑘−1)/(𝑘+1)}^𝑇0||𝑥^(2𝑠−1)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)≤𝑇^1+1/(𝑘+1)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{(𝑘−1)/(𝑘+1)}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1).(3.31) Substituting the above formula, (3.20), (3.27), and (3.30) into (3.26), one has ||𝑏_2𝑠||≤^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡_𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)𝐴²(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+(𝛾+𝜀)+𝜃₂^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+||𝑝(𝑡)||_∞+𝑓_𝛿𝑇^{𝑘/(𝑘+1)}+𝑘||𝑏₀||𝑇^2𝑠^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^2/(𝑘+1)^𝑇0|||||𝑥(𝑡)||_𝑘+1|||𝑑𝑡^{(𝑘−1)/(𝑘+1)}.(3.32) Then, one has ||𝑏_2𝑠||−𝐴₂(2𝑠,𝑘)−𝜃₁𝑇^{(2𝑠−1)𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝑘||𝑏₀||𝑇^2𝑠^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)^𝑇0|||||𝑥(𝑡)||_𝑘+1|||𝑑𝑡^{(𝑘−1)/(𝑘+1)}+(𝛾+𝜀)+𝜃₂^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+𝑢₄,(3.33) where 𝑢₄ is a positive constant. Using inequality (𝑎+𝑏)^𝑙≤𝑎^𝑙+𝑏^𝑙for𝑎≥0,𝑏≥0,0≤𝑙≤1,(3.34) it follows from (3.23) that ^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{(𝑘−1)/(𝑘+1)}≤𝐴¹(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂^{(𝑘−1)/𝑘}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{(𝑘−1)/(𝑘+1)}+𝑢₅,(3.35) where 𝑢₅ is a positive constant. Substituting the above formula and (3.23) into (3.33), one has ⎧⎪⎨⎪⎩||𝑏_2𝑠||−𝐴₂(2𝑠,𝑘)−𝜃₁𝑇^{(2𝑠−1)𝑘}−𝛾+𝜀+𝜃²𝐴₁(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂−𝑘||𝑏₀||𝑇^2𝑠𝐴¹(2𝑠,𝑘)+𝜃₁𝑇^{(2𝑠−1)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂^{(𝑘−1)/𝑘}⎫⎪⎬⎪⎭_𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝑢₅𝑘||𝑏₀||𝑇^2𝑠^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)+𝑢₆,(3.36) where 𝑢₆ is a positive constant.

If 𝑠=1, since ∫^𝑇0[𝑥^(𝑡)]^𝑘𝑥^(𝑡)𝑑𝑡=0,∫^𝑇0[𝑥(𝑡)]^𝑘𝑥^(𝑡)𝑑𝑡=−𝑘∫^𝑇0[𝑥(𝑡)]^𝑘−1[𝑥^(𝑡)]²𝑑𝑡, from (3.24), one has 𝑏₂^𝑇0𝑥^=−𝑘(𝑏𝑡₀)^𝑘+1𝑑𝑡^𝑇0[𝑥(𝑡)]^𝑘−1𝑥^(𝑡)²𝑑𝑡−^𝑇0𝑓(𝑡,𝑥(𝑡−𝜏))𝑥^(𝑡)𝑑𝑡+^𝑇0𝑝(𝑡)𝑥^(𝑡)𝑑𝑡.(3.37) Applying the above method, one has ||𝑏²||−𝜃₁𝑇^𝑘−𝛾+𝜀+𝜃²𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−(𝛾+𝜀)−𝜃₂−𝑘||𝑏₀||𝑇²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘×_𝑇0||𝑏₀||−(𝛾+𝜀)−𝜃₂^{(𝑘−1)/𝑘}||𝑥^(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝑢₇𝑘||𝑏₀||𝑇²^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)+𝑢₈,(3.38) where 𝑢₇,𝑢₈ is a positive constant. Hence there is a constant 𝑀₁,𝑀₂>0 such that ^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡≤𝑀₁,(3.39)^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡≤𝑀₂.(3.40) From (3.5), using Hölder inequality and Lemma 2.1, one has ^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡≤^2𝑠_𝑖=0+||𝑏_𝑖||^𝑇0||𝑥^(𝑖)(𝑡)||^𝑘𝑑𝑡+^𝑇0||𝑓(𝑡,𝑥(𝑡))||𝑑𝑡_𝑇0≤||𝑓(𝑡,𝑥(𝑡))−𝑓(𝑡,𝑥(𝑡−𝜏(𝑡)))||𝑑𝑡+^𝑇0||𝑝(𝑡)||𝑑𝑡_2𝑠_𝑖=1||𝑏_𝑖||𝑇^{(2𝑠−𝑖)𝑘+1/(𝑘+1)}+𝜃₁𝑇^{(2𝑠−1)𝑘+1/(𝑘+1)}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+||𝑏₀||+(𝛾+𝜀)+𝜃₂𝑇^1/(𝑘+1)^𝑇0≤_2𝑠||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}+||𝑝(𝑡)||_∞+𝑓_𝛿𝑇_𝑖=1||𝑏_𝑖||𝑇^{(2𝑠−𝑖)𝑘+1/(𝑘+1)}+𝜃₁𝑇^{(2𝑠−1)𝑘+1/(𝑘+1)}𝑀¹^{𝑘/(𝑘+1)}+||𝑏₀||+(𝛾+𝜀)+𝜃₂𝑀₂^{𝑘/(𝑘+1)}+||𝑝(𝑡)||_∞+𝑓_𝛿𝑇=𝑀,(3.41) where 𝑀 is a positive constant. We claim that ||𝑥^(𝑖)(𝑡)||≤𝑇^{𝑛−𝑖−1}^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡,𝑖=1,2,…,𝑛−1.(3.42) In fact, noting that 𝑥^(𝑛−2)(0)=𝑥^(𝑛−2)(𝑇), there must be a constant 𝜉₁∈[0,𝑇] such that 𝑥^(𝑛−1)(𝜉₁)=0, we obtain ||𝑥^(𝑛−1)(𝑡)||=||||𝑥^(𝑛−1)𝜉₁+^𝑡𝜉₁𝑥^(𝑛)(𝑠)𝑑𝑠||||≤||𝑥^(𝑛−1)𝜉₁||+^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡=^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡.(3.43) Similarly, since 𝑥^(𝑛−3)(0)=𝑥^(𝑛−3)(𝑇), there must be a constant 𝜉₂∈[0,𝑇] such that 𝑥^(𝑛−2)(𝜉₂)=0, from (3.43) we get ||𝑥^(𝑛−2)(𝑡)||=||||𝑥^(𝑛−2)𝜉₂+^𝑡𝜉₂𝑥^(𝑛−1)(𝑠)𝑑𝑠||||≤^𝑇0||𝑥^(𝑛−1)(𝑡)||𝑑𝑡≤𝑇^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡.(3.44) By induction, we conclude that (3.42) holds. Furthermore, one has ||𝑥^(𝑖)(𝑡)||_∞≤𝑇^{𝑛−𝑖−1}^𝑇0||𝑥^(𝑛)(𝑡)||𝑑𝑡≤𝑇^{𝑛−𝑖−1}𝑀,𝑖=1,2,…,𝑛−1.(3.45) It follows from (3.39) that there exists a 𝜉∈[0,𝑇] such that |𝑥(𝜉)|≤𝑀^21/(𝑘+1). Applying Lemma 2.1, we get ||𝑥(𝑡)||_∞≤𝑥(𝜉)+^𝑡𝜉𝑥^(𝑡)𝑑𝑡≤𝑀^21/(𝑘+1)+𝑇^{𝑘/(𝑘+1)}^𝑇0||𝑥^(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)≤𝑀^21/(𝑘+1)+𝑇^{2𝑠−1+(𝑘/(𝑘+1))}^𝑇0||𝑥^(2𝑠)(𝑡)||^𝑘+1𝑑𝑡^1/(𝑘+1)=𝑀^21/(𝑘+1)+𝑇^{2𝑠−1+(𝑘/(𝑘+1))}𝑀^11/(𝑘+1).(3.46) It follows that there is a constant 𝐴>0 such that ‖𝑥‖≤𝐴. Thus Ω₁ is bounded.

Let Ω₂={𝑥∈Ker𝐿,𝑄𝑁𝑥=0}. Suppose 𝑥∈Ω₂, then 𝑥(𝑡)=𝑑∈𝑅 and satisfies 𝑄𝑁𝑥=1𝑇^𝑇0−𝑏₀𝑑^𝑘−𝑓(𝑡,𝑑)+𝑝(𝑡)𝑑𝑡=0.(3.47) We will prove that there exists a constant 𝐵>0 such that |𝑑|≤𝐵. If |𝑑|≤𝛿, taking 𝛿=𝐵, we get |𝑑|≤𝐵. If |𝑑|>𝛿, from (3.47), one has ||𝑏₀||||𝑑||^𝑘=||||1𝑇^𝑇0≤1[−𝑓(𝑡,𝑑)+𝑝(𝑡)]𝑑𝑡||||𝑇^𝑇0||𝑓(𝑡,𝑑)||𝑑𝑡+||𝑝(𝑡)||_∞≤(𝛾+𝜀)||𝑑||^𝑘+||𝑝(𝑡)||_∞.(3.48) Thus ||𝑑||≤||𝑝(𝑡)||^∞||𝑏₀||−(𝛾+𝜀)^1/𝑘.(3.49) Taking [|𝑝(𝑡)|_∞/(|𝑏₀|−(𝛾+𝜀))]^1/𝑘=𝐵, one has |𝑑|≤𝐵, which implies Ω₂ is bounded. Let Ω be a nonempty open bounded subset of 𝑋 such that Ω⊃Ω₁∪Ω₂. We can easily see that 𝐿 is a Fredholm operator of index zero and 𝑁 is 𝐿-compact on Ω. Then by the above argument, we have(i)𝐿𝑥≠𝜆𝑁𝑥,forall𝑥∈𝜕Ω∩𝐷(𝐿),𝜆∈(0,1), (ii)𝑄𝑁𝑥≠0,forall𝑥∈𝜕Ω∩Ker𝐿.

At last we will prove that condition (iii) of Lemma 2.4 is satisfied. We take 𝐻(𝑑,𝜇)𝐻=∶𝜇𝑑(Ω+∩1K−e𝜇r𝐿)×[0,1]⟶Ker𝐿,𝑇^𝑇0−𝑏₀𝑑^𝑘−𝑓(𝑡,𝑑)+𝑝(𝑡)𝑑𝑡.(3.50) From assumptions (𝐻₁) and (𝐻₂), we can easily obtain 𝐻(𝑑,𝜇)≠0,forall(𝑑,𝜇)∈𝜕Ω∩Ker𝐿×[0,1], which results in deg{𝑄𝑁,Ω∩Ker𝐿,0}=deg{𝐻(⋅,0),Ω∩Ker𝐿,0}=deg{𝐻(⋅,1),Ω∩Ker𝐿,0}≠0.(3.51)

Hence, by using Lemma 2.2, we know that (1.1) has at least one 𝑇-periodic solution.

Theorem 3.2. Suppose 𝑛=4𝑚+1, 𝑚>0 an integer and conditions (𝐻₁),(𝐻₂) hold. If(𝐻₅)there is a positive integer 0<𝑠≤𝑚 such that 𝑏_4𝑠−3≠0,𝑏_{4𝑠−3+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+3,(3.52)(𝐻₆)𝐴₂(4𝑠−3,𝑘)+𝜃₁𝑇^{(4𝑠−4)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−3,𝑘)+𝜃₁𝑇^{(4𝑠−4)𝑘}||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇^4𝑠−3𝐴¹(4𝑠−3,𝑘)+𝜃₁𝑇^4𝑠−4||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_4𝑠−3||,if1<𝑠≤𝑚,₁+𝛾+𝜃²𝐴₁(1,𝑘)+𝜃₁||𝑏₀||−𝛾−𝜃₂<||𝑏₁||,if𝑠=1,(3.53) then (1.1) has at least one 𝑇-periodic solution.

Proof. From the proof of Theorem 3.1, one has ^𝑇0||𝑥(𝑡)||^𝑘+1𝑑𝑡^{𝑘/(𝑘+1)}≤𝐴¹(4𝑠−3,𝑘)+𝜃₁𝑇^{(4𝑠−4)𝑘}||𝑏₀||−(𝛾+𝜀)−𝜃₂^𝑇0||𝑥^(4𝑠−3)(𝑡)||^𝑘+1𝑑𝑡^𝑘/(k+1)+𝑢₉,(3.54) where 𝑢₉ is a positive constant. Multiplying both sides of (3.8) by 𝑥^(4𝑠−3)(𝑡), and integrating on [0,𝑇], one has ^𝑇0𝑥^(𝑛)(𝑡)𝑥^(4𝑠−3)(𝑡)𝑑𝑡=−𝜆^4𝑠−3_𝑖=0𝑏_𝑖^𝑇0𝑥^(𝑖)−𝜆_𝑇0(𝑡)^𝑘𝑥^(4𝑠−3)(𝑡)𝑑𝑡𝑓(𝑡,𝑥(𝑡−𝜏))𝑥^(4𝑠−3)(𝑡)𝑑𝑡+𝜆^𝑇0𝑝(𝑡)𝑥^(4𝑠−3)(𝑡)𝑑𝑡.(3.55) Since ^𝑇0𝑥^(4𝑚+1)(𝑡)𝑥^(4𝑠−3)(𝑡)𝑑𝑡=(−1)^{2𝑚−2𝑠+2}^𝑇0𝑥^{(2𝑚+2𝑠−1)}(𝑡)²𝑑𝑡,(3.56) then it follows from (3.55) and (3.56) that 𝑏_4𝑠−3^𝑇0||𝑥^(4𝑠−3)(𝑡)||^𝑘+1𝑑𝑡≤−^4𝑠−4_𝑖=0𝑏_𝑖^𝑇0𝑥^(𝑖)−(𝑡)^𝑘𝑥^(4𝑠−3)(𝑡)𝑑𝑡_𝑇0𝑓(𝑡,𝑥(𝑡−𝜏))𝑥^(4𝑠−3)(𝑡)𝑑𝑡+^𝑇0𝑝(𝑡)𝑥^(4𝑠−3)(𝑡)𝑑𝑡.(3.57)

By using the same way as in the proof of Theorem 3.1, the following theorems can be proved in case 1<𝑠≤𝑚 or 𝑠=1.

Theorem 3.3. Suppose 𝑛=4𝑚+1, 𝑚>0 for a positive integer and conditions (𝐻₁),(𝐻₂) hold. If(𝐻₇)there is a positive integer 0<𝑠≤𝑚 such that 𝑏_4𝑠−1≠0,𝑏_{4𝑠−1+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+1,(3.58)(𝐻₈)𝐴₂(4𝑠−1,𝑘)+𝜃₁𝑇^{(4𝑠−2)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−1,𝑘)+𝜃₁𝑇^{(4𝑠−2)𝑘}||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇^4𝑠−1𝐴¹(4𝑠−1,𝑘)+𝜃₁𝑇^{(4𝑠−2)𝑘}||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏_4𝑠−1||,(3.59) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.4. Suppose 𝑛=4𝑚+3,  𝑚≥0 an integer and conditions (𝐻₁),(𝐻₂) hold. If(𝐻₉)there is a positive integer 0≤𝑠≤𝑚 such that 𝑏_4𝑠+1≠0,𝑏_4𝑠+1+𝑖=0,𝑖=1,2,…,4𝑚−4𝑠+1,(3.60)(𝐻₁₀)𝐴₂(4𝑠+1,𝑘)+𝜃₁𝑇^4𝑠𝑘+𝛾+𝜃²𝐴₁(4𝑠+1,𝑘)+𝜃₁𝑇^4𝑠𝑘||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇^4𝑠+1𝐴¹(4𝑠+1,𝑘)+𝜃₁𝑇^4𝑠𝑘||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_4𝑠+1||,if0<𝑠≤𝑚,₁+𝛾+𝜃²𝐴₁(1,𝑘)+𝜃₁||𝑏₀||−𝛾−𝜃₂<||𝑏₁||,if𝑠=0,(3.61) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.5. Suppose 𝑛=4𝑚+3,𝑚>0 an integer and conditions (𝐻₁),(𝐻₂) hold. If(𝐻₁₁) there is a positive integer 0<𝑠≤𝑚 such that 𝑏_4𝑠−1≠0,𝑏_{4𝑠−1+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+3,(3.62)(𝐻₁₂)𝐴₂(4−1,k)+𝜃₁𝑇^{(4𝑠−2)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−1,𝑘)+𝜃₁𝑇^{(4𝑠−2)𝑘}||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||𝑇^4𝑠−1𝐴¹(4𝑠−1,𝑘)+𝜃₁𝑇^{(4𝑠−2)𝑘}||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏_4𝑠−1||,(3.63) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.6. Suppose 𝑛=4𝑚, 𝑚>0 an integer and conditions (𝐻₁) hold. If (𝐻₁₃)𝑏₀>𝛾+𝜃₂,(3.64)(𝐻₁₄)there is a positive integer 0<𝑠≤2𝑚 such that 𝑏𝑏_2𝑠−1≠0,if𝑠=2𝑚,_2𝑠−1≠0,𝑏_{2𝑠−1+𝑖}=0,𝑖=1,2,…,4𝑚−2𝑠,if0<𝑠<2𝑚,(3.65)(𝐻₁₅)𝐴₂(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}+𝛾+𝜃²𝐴₁(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}𝑏₀−𝛾−𝜃₂+𝑘𝑏₀𝑇^2𝑠−1𝐴¹(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_2𝑠−1||,if1<𝑠≤2𝑚,₁+𝛾+𝜃²𝐴₁(1,𝑘)+𝜃₁𝑏₀−𝛾−𝜃₂<||𝑏₁||,if𝑠=1,(3.66) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.7. Suppose 𝑛=4𝑚+2,𝑚>0 an integer and conditions (𝐻₁) hold. If (𝐻₁₆)−𝑏₀>𝛾+𝜃₂,(3.67)(𝐻₁₇)there is a positive integer 0<𝑠≤2𝑚+1 such that 𝑏𝑏_2𝑠−1≠0,if𝑠=2𝑚+1,_2𝑠−1≠0,𝑏_{2𝑠−1+𝑖}=0,𝑖=1,2,…,4𝑚−2𝑠,if0<𝑠<2𝑚+1,(3.68)(𝐻₁₈)𝐴₂(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}+𝛾+𝜃²𝐴₁(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}−𝑏₀−𝛾−𝜃₂−𝑘𝑏₀𝑇^2𝑠−1𝐴¹(2𝑠−1,𝑘)+𝜃₁𝑇^{(2𝑠−2)𝑘}−𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_2𝑠−1||,if1<𝑠≤2𝑚+1,₁+𝛾+𝜃²𝐴₁(1,𝑘)+𝜃₁−𝑏₀−𝛾−𝜃₂<||𝑏₁||,if𝑠=1,(3.69) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.8. Suppose 𝑛=4𝑚, 𝑚>0 is an integer, and conditions (𝐻₁),(𝐻₁₃) hold. If(𝐻₁₉)there is a positive integer 0<𝑠≤𝑚 such that 𝑏_4𝑠−2≠0,𝑏_{4𝑠−2+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+1,(3.70)(𝐻₂₀)𝐴₂(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}𝑏₀−𝛾−𝜃₂+𝑘𝑏₀𝑇^4𝑠−2𝐴¹(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝛾+𝜃₂<||𝑏_4𝑠−2||,if1<𝑠≤𝑚,𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘𝑏₀−𝛾−𝜃₂+𝑘𝑏₀𝑇²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏₂||,if𝑠=1,(3.71) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.9. Suppose 𝑛=4𝑚, 𝑚>1 an integer and conditions (𝐻₁),(𝐻₁₃) hold. If(𝐻₂₁)there is a positive integer 1<𝑠≤𝑚 such that 𝑏_4𝑠−4≠0,𝑏_{4𝑠−4+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+3,(3.72)(𝐻₂₂)𝐴₂(4𝑠−4,𝑘)+𝜃₁𝑇^{(4𝑠−5)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−4,𝑘)+𝜃₁𝑇^{(4𝑠−5)𝑘}𝑏₀−𝛾−𝜃₂+𝑘𝑏₀𝑇^4𝑠−4𝐴¹(4𝑠−4,𝑘)+𝜃₁𝑇^{(4𝑠−5)𝑘}𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏_4𝑠−4||,(3.73) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.10. Suppose 𝑛=4𝑚+2, 𝑚≥1 an integer and conditions (𝐻₁),(𝐻₁₆) hold. If(𝐻₂₃)there is a positive integer 1≤𝑠≤𝑚 such that 𝑏_4𝑠≠0,𝑏_4𝑠+𝑖=0,𝑖=1,2,…,4𝑚−4𝑠+1,(3.74)(𝐻₂₄)𝐴₂(4𝑠,𝑘)+𝜃₁𝑇^{(4𝑠−1)𝑘}+𝛾+𝜃²𝐴₁(4𝑠,𝑘)+𝜃₁𝑇^{(4𝑠−1)𝑘}−𝑏₀−𝛾−𝜃₂−𝑘𝑏₀𝑇^4𝑠𝐴¹(4𝑠,𝑘)+𝜃₁𝑇^{(4𝑠−1)𝑘}−𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏_4𝑠||,(3.75) then (1.1) has at least one 𝑇-periodic solution.

Theorem 3.11. Suppose 𝑛=4𝑚+2, 𝑚≥1 is an integer, and conditions (𝐻₁),(𝐻₁₆) hold. If(𝐻₂₅)there is a positive integer 1≤𝑠≤𝑚 such that 𝑏_4𝑠−2≠0,𝑏_{4𝑠−2+𝑖}=0,𝑖=1,2,…,4𝑚−4𝑠+3,(3.76)(𝐻₂₆)𝐴₂(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}+𝛾+𝜃²𝐴₁(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}−𝑏₀−𝛾−𝜃₂−𝑘𝑏₀𝑇^4𝑠−2𝐴¹(4𝑠−2,𝑘)+𝜃₁𝑇^{(4𝑠−3)𝑘}−𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}𝜃<||𝑏_4𝑠−2||,if1<𝑠≤𝑚,₁𝑇^𝑘+𝛾+𝜃²𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘−𝑏₀−𝛾−𝜃₂−𝑘𝑏₀𝑇²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘−𝑏₀−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏₂||,ifs=1,(3.77) then (1.1) has at least one 𝑇-periodic solution.

The proofs of Theorem 3.3–3.11 are similar to that of Theorem 3.1.

Example 3.12. Consider the following equation: 𝑥⁽⁵⁾(𝑡)+300𝑥^(𝑡)³+150𝑥^(𝑡)³+1100[𝑥(𝑡)]³+1300(sin𝑡)𝑥𝑡−1𝜋0³=cos𝑡,(3.78) where 𝑛𝜋/=105,𝑘=3,𝑏₄=𝑏₃=0,𝑏₂=300,𝑏₁=1/50,𝑏₀=1/100,𝑓(𝑡,𝑥)=1/300(sin𝑡)𝑥³,𝑝(𝑡)=cos𝑡,𝜏(𝑡)=. Thus, 𝑇=2𝜋,𝛾=1/300,𝐴₁(2,𝑘)=|𝑏₁|(2𝜋)³+|𝑏₂|=1/50×(2𝜋)³+200. Obviously assumptions (H₁)–(H₃) hold and 𝜃₁𝑇^𝑘+𝛾+𝜃²𝐴₁(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−𝛾−𝜃₂+𝑘||𝑏₀||(2𝜋)²𝐴¹(2,𝑘)+𝜃₁𝑇^𝑘||𝑏₀||−𝛾−𝜃₂^{(𝑘−1)/𝑘}<||𝑏₂||.(3.79) By Theorem 3.1, we know that (3.78) has at least one 2𝜋-periodic solution.