# Periodic Solutions for a Class ofn-th Order Functional Differential Equations

Authors: Bing Song; Lijun Pan; Jinde Cao
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

We study the existence of periodic solutions forn-th order functional differential
equations x(n)(t)=∑i=0n−1bi[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.


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## Body

## 1. Introduction

In this paper, we are concerned with the existence of periodic solutions of the followingn-th order functional differential equations: (1.1)x(n)(t)=∑i=0n-1bi[x(i)(t)]k+f(x(t-τ(t)))+p(t),
where bi,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: (1.2)x′′′(t)+ax′′2k-1(t)+bx′2k-1(t)+cx2k-1(t)+g(t,x(t-τ1),x′(t-τ2))=p(t).
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 (1.3)x(n)(t)=∑i=1n-1bix(i)(t)+f(t,x(t),x(t-τ1(t)),…,x(t-τm(t)))+p(t).
The authors obtained the results based on the damping terms x(i)(t) and the delay τi(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 bi 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]).
Letn1>1,α∈[0,+∞) be constants, s∈C(R,R) with s(t+T)=s(t), and s(t)∈[-α,α],for allt∈[0,T].  Then for ∀x∈C1(R,R) with x(t+T)=x(t), one has
(2.1)∫0T|x(t)-x(t-s(t))|n1dt≤2αn1∫0T|x′(t)|n1dt.Lemma 2.2.
Letk≥1,α∈[0,+∞)be constants, s∈C(R,R) with s(t+T)=s(t), and s(t)∈[-α,α],for allt∈[0,T]. Then for ∀x∈C1(R,R) with x(t+T)=x(t), one has
(2.2)∫0T|xk(t)-xk(t-s(t))|(k+1)/kdt≤2α(k+1)/kk1/k[(k-1)∫0T|x(t)|k+1dt+∫0T|x′(t)|k+1dt].Proof.
LetF(t)=xk(t). By Lemma 2.2, one has
(2.3)∫0T|xk(t)-xk(t-s(t))|(k+1)/kdt=∫0T|F(t)-F(t-s(t))|(k+1)/kdt≤2α(k+1)/k∫0T|F′(t)|(k+1)/kdt=2α(k+1)/k∫0T|kxk-1(t)x′(t)|(k+1)/kdt=2α(k+1)/kk(k+1)/k∫0T|x(t)|((k-1)(k+1))/k|x′(t)|(k+1)/kdt.
By inequality
(2.4)xy≤xpp+yqq,x≥0,y≥0,1p+1q=1,
one has
(2.5)|x(t)|((k-1)(k+1))/k|x′(t)|(k+1)/k≤(k-1)|x(t)|k+1k+|x′(t)|k+1k.
Thus we obtain
(2.6)∫0T|xk(t)-xk(t-s(t))|(k+1)/kdt≤2α(k+1)/kk1/k[(k-1)∫0T|x(t)|k+1dt+∫0T|x′(t)|k+1dt].Lemma 2.3.
Ifk≥1 is an integer, x∈Cn(R,R), and x(t+T)=x(t), then
(2.7)(∫0T|x′(t)|kdt)1/k≤T(∫0T|x′′(t)|kdt)1/k≤⋯≤Tn-1(∫0T|x(n)(t)|kdt)1/k.The proof of Lemma2.3 is easy, here we omit it.We first introduce Mawhin’s continuation theorem.LetX 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 (2.8)Im⁡P=Ker⁡L,Ker⁡Q=Im⁡L,X=Ker⁡L⊕Ker⁡P,Y=Im⁡L⊕Im⁡Q.
It follows that (2.9)L∣D(L)∩Ker⁡P:D(L)∩Ker⁡P⟶Im⁡L
is invertible, we denote the inverse of that map by Kp. Let Ω be an open bounded subset of X, D(L)∩Ω¯≠Ø, the map N:X→Y will be called L-compact in Ω¯, if QN(Ω¯) is bounded and Kp(I-Q)N:Ω¯→X is compact.Lemma 2.4 (see [25]).
LetL 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,for allx∈∂Ω∩D(L),λ∈(0,1);(ii)
QNx≠0,for allx∈∂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 defineY={x∈C(R,R)∣x(t+T)=x(t)} with the norm |x|∞=maxt∈[0,T]{|x(t)|} and X={x∈Cn-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:
(2.10)L:D(L)⊂X⟶Y,Lx=x(n),D(L)={x∣x∈Cn(R,R),x(t+T)=x(t)},N:X⟶Y,Nx=-∑i=1n-1bi[x(i)(t)]k-f(t,x(t-τ(t)))+p(t).
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,∫0Ty(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
(2.11)P:X⟶ker⁡L,Px=x(0),Q:Y⟶Y∖Im⁡L,Qy=1T∫0Ty(t)dt,
and let
(2.12)L|D(L)∩Ker⁡P:D(L)∩Ker⁡P⟶Im⁡L.
Then L|D(L)∩KerP has a unique continuous inverse Kp. 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.
Supposen=2m+1, m>0 an integer and the following conditions hold:(H1)
The functionf satisfies
(3.1)lim⁡x→∞|f(t,x)xk|≤γ,(3.2)|f(t,x)-f(t,y)|≤β|xk-yk|,
where γ≥0.(H2)
(3.3)|b0|>γ+θ2.(H3)
There is a positive integer0<s≤m such that
(3.4)b2s≠0,ifs=m,b2s≠0,b2s+i=0,i=1,2,…,2m-2s,if0<s<m.(H4)
(3.5)A2(2s,k)+θ1T(2s-1)k+(γ+θ2)(A1(2s,k)+θ1T(2s-1)k)|b0|-γ-θ2+k|b0|T2s[A1(2s,k)+θ1T(2s-1)k|b0|-γ-θ2](k-1)/k<|b2s|,if1<s≤m,θ1Tk+(γ+θ2)(A1(2,k)+θ1Tk)|b0|-γ-θ2+k|b0|T2[A1(2,k)+θ1Tk|b0|-γ-θ2](k-1)/k<|b2|,ifs=1,
where A1(s,k)=∑i=1s|bi|T(s-i)k,  A2(s,k)=∑i=1s-2|bi|T(s-i)k, θ1=2k/(k+1)β|τ(t)|∞k1/(k+1),  θ2=2k/(k+1)β|τ(t)|∞k1/(k+1)(k-1)k/(k+1). Then (1.1) has at least one T-periodic solution.Proof.
Consider the equation(3.6)Lx=λNx,λ∈(0,1),
where L and N are defined by (2.10). Let
(3.7)Ω1={x∈D(L)Ker⁡L,Lx=λNxfor  someλ∈(0,1)}.
For x∈Ω1, one has
(3.8)x(n)(t)=λ∑i=02sbi[x(i)(t)]k+λf(t,x(t-τ(t)))+λp(t),λ∈(0,1).
Multiplying both sides of (3.8) by x(t), and integrating them on [0,T], one has for λ∈(0,1)(3.9)∫0Tx(n)(t)x(t)dt=λ∑i=02sbi∫0T[x(i)(t)]kx(t)dt+λ∫0Tf(t,x(t-τ(t)))x(t)dt+λ∫0Tp(t)x(t)dt.
Since for any positive integer i,
(3.10)∫0Tx(2i-1)(t)x(t)dt=0,
and in view of n=2m+1 and k is odd, it follows from (3.3) and (3.9) that
(3.11)|b0|∫0T|x(t)|k+1dt≤∑i=12s|bi|∫0T|x(i)(t)|k|x(t)|dt+∫0T|f(t,x(t-τ(t)))||x(t)|dt+∫0T|p(t)||x(t)|dt≤∑i=12s|bi|∫0T|x(i)(t)|k|x(t)|dt+∫0T|f(t,x(t))||x(t)|dt+∫0T|f(t,x)-f(t,x(t-τ(t)))||x(t)|dt+∫0T|p(t)||x(t)|dt.
By using Hölder inequality and Lemma 2.1, from (3.11), we obtain
(3.12)|b0|∫0T|x(t)|k+1dt≤(∫0T|x(t)|k+1dt)1/(k+1)[∑i=12s|bi|(∫0T|x(i)(t)|k+1dt)k/(k+1)+(∫0T|f(t,x(t))|(k+1)/kdt)k/(k+1)+(∫0T|f(t,x)-f(t,x(t-τ(t)))|(k+1)/kdt)k/(k+1)+(∫0T|p(t)|(k+1)/kdt)k/(k+1)]≤(∫0T|x(t)|k+1dt)1/(k+1)[∑i=12s|bi|T(2s-i)k(∫0T|x(2s)(t)|k+1dt)k/(k+1)+(∫0T|f(t,x(t))|(k+1)/kdt)k/(k+1)+(∫0T|f(t,x)-f(t,x(t-τ(t)))|(k+1)/kdt)k/(k+1)+|p(t)|∞Tk/(k+1)∑i=12s].
So
(3.13)|b0|(∫0T|x(t)|k+1dt)k/(k+1)≤A1(2s,k)(∫0T|x(2s)(t)|k+1dt)k/(k+1)+(∫0T|f(t,x(t))|(k+1)/kdt)k/(k+1)+(∫0T|f(t,x(t))-f(t,x(t-τ(t)))|(k+1)/kdt)k/(k+1)+u1,
where u1 is a positive constant. Choosing a constant ε>0 such that
(3.14)γ+ε+θ2<|b0|,(3.15)A2(2s,k)+θ1T(2s-1)k+(γ+ε+θ2)(A1(2s,k)+θ1T(2s-1)k)|b0|-(γ+ε)-θ2+k|b0|T2s[A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2](k-1)/k<|b2s|,if1<s≤m,θ1Tk+(γ+ε+θ2)(A1(2,k)+θ1Tk)|b0|-(γ+ε)-θ2+k|b0|T2[A1(2,k)+θ1Tk|b0|-(γ+ε)-θ2](k-1)/k<|b2|,ifs=1,
for the above constant ε>0, we see from (3.1) that there is a constant δ>0 such that
(3.16)|f(t,x(t))|<(γ+ε)|x(t)|k,for|x(t)|>δ,t∈[0,T].
Denote
(3.17)Δ1={t∈[0,T]:|x(t)|≤δ},Δ2={t∈[0,T]:|x(t)|>δ}.
Since
(3.18)∫0T|f(t,x(t))|(k+1)/kdt≤∫Δ1|f(t,x(t))|(k+1)/kdt+∫Δ2|f(t,x(t))|(k+1)/kdt≤(fδ)(k+1)/kT+(γ+ε)(k+1)/k∫0T|x(t)|k+1dt=(fδ)(k+1)/kT+(γ+ε)(k+1)/k∫0T|x(t)|k+1dt,
using inequality
(3.19)(a+b)l≤al+blfora≥0,b≥0,0≤l≤1,
it follows from (3.18) that
(3.20)(∫0T|f(t,x(t))|(k+1)/kdt)k/(k+1)≤fδTk/(k+1)+(γ+ε)(∫0T|x(t)|k+1dt)k/(k+1).
From (3.2) and by Lemma 2.2, one has
(3.21)(∫0T|f(t,x(t))-f(t,x(t-τ(t)))|(k+1)/kdt)k/(k+1)≤β[∫0T|xk(t)-xk(t-τ(t))|(k+1)/kdt]k/(k+1)≤2k/(k+1)β|τ(t)|∞k1/(k+1)[(k-1)∫0T|x(t)|k+1dt+∫0T|x′(t)|k+1dt]k/(k+1)≤2k/(k+1)β|τ(t)|∞k1/(k+1)[(k-1)k/(k+1)(∫0T|x(t)|k+1dt)k/(k+1)+(∫0T|x′(t)|k+1dt)k/(k+1)]≤2k/(k+1)β|τ(t)|∞k1/(k+1)(k-1)k/(k+1)(∫0T|x(t)|k+1dt)k/(k+1)+2k/(k+1)β|τ(t)|∞k1/(k+1)T(2s-1)k(∫0T|x(2s)(t)|k+1dt)k/(k+1)=θ2(∫0T|x(t)|k+1dt)k/(k+1)+θ1T(2s-1)k(∫0T|x(2s)(t)|k+1dt)k/(k+1).
Substituting the above formula into (3.13), one has
(3.22)[|b0|-(γ+ε)-θ2](∫0T|x(t)|k+1dt)k/(k+1)≤[A1(2s,k)+θ1T(2s-1)k](∫0T|x(2s)(t)|k+1dt)k/(k+1)+u2,
where u2 is a positive constant. That is
(3.23)(∫0T|x(t)|k+1dt)k/(k+1)≤A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2(∫0T|x(2s)(t)|k+1dt)k/(k+1)+u3,
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
(3.24)∫0Tx(n)(t)x(2s)(t)dt=∑i=02sbi∫0T[x(i)(t)]kx(2s)(t)dt+∫0Tf(t,x(t-τ(t)))x(2s)(t)dt+∫0Tp(t)x(2s)(t)dt.If1<s≤m, since (3.25)∫0Tx(2m+1)(t)x(2s)(t)dt=0,∫0T[x(2s-1)(t)]kx(2s)(t)dt=0,(3.26)∫0T[x(t)]kx(2s)(t)dt=-k∫0T[x(t)]k-1x(2s-1)(t)x′(t)dt,
by using Hölder inequality and Lemma 2.1, from (3.23), one has (3.27)|b2s|∫0T|x(2s)(t)|k+1dt≤∫0T|x(2s)(t)|[∑i=12s-2|bi||x(i)(t)|k+|f(t,x(t-τ(t)))|+|p(t)|]dt+k|b0|∫0T|x(t)|k-1|x(2s-1)(t)||x′(t)|dt≤(∫0T|x(2s)(t)|k+1dt)1/(k+1)[∑i=12s-2|bi|T(2s-i)k(∫0T|x(2s)(t)|k+1dt)k/(k+1)+(∫0T|f(t,x(t))|(k+1)/kdt)k/(k+1)+(∫0T|f(t,x(t))-f(t,x(t-τ))|(k+1)/kdt)k/(k+1)+|p(t)|∞Tk/(k+1)∑i=12s-2]+k|b0||x′(t)|∞∫0T|x(t)|k-1|x(2s-1)(t)|dt.
Since x(0)=x(T), there exists ξ∈[0,T] such that x′(ξ)=0. So for t∈[0,T](3.28)x′(t)=x′(ξ)+∫ξtx′′(σ)dσ.
Using Hölder inequality and Lemma 2.1, one has (3.29)|x′(t)|∞≤∫0T|x′′(t)|dt≤Tk/(k+1)(∫0T|x′′(t)|k+1dt)1/(k+1)≤T2s-1-(1/(k+1))(∫0T|x(2s)(t)|k+1dt)1/(k+1).
Using inequality (3.30)(1T∫0T||x(t)|r|)1/r≤(1T∫0T||x(t)|l|)1/lfor0≤r≤l,∀x∈R.
and applying Hölder inequality and by Lemma 2.1, we obtain (3.31)∫0T|x(t)|k-1|x(2s-1)(t)|dt≤(∫0T|x(t)|kdt)(k-1)/k(∫0T|x(2s-1)(t)|kdt)1/k≤T1/(k+1)(∫0T|x(t)|k+1dt)(k-1)/(k+1)(∫0T|x(2s-1)(t)|k+1dt)1/(k+1)≤T1+1/(k+1)(∫0T|x(t)|k+1dt)(k-1)/(k+1)(∫0T|x(2s)(t)|k+1dt)1/(k+1).
Substituting the above formula, (3.20), (3.27), and (3.30) into (3.26), one has (3.32)|b2s|∫0T|x(2s)(t)|k+1dt≤(∫0T|x(2s)(t)|k+1dt)1/(k+1){(∫0T|x(2s)(t)|k+1dt)(∫0T||x(t)|k+1|dt)[A2(2s,k)+θ1T(2s-1)k](∫0T|x(2s)(t)|k+1dt)k/(k+1)+[(γ+ε)+θ2](∫0T|x(t)|k+1dt)k/(k+1)+(|p(t)|∞+fδ)Tk/(k+1)(∫0T|x(2s)(t)|k+1dt)1(∫0T|x(2s)(t)|k+1dt)}(∫0T|x(2s)(t)|k+1dt)[A2(2s,k)+θ1T(2s-1)k](∫0T|x(2s)(t)|k+1dt)kk+1(∫0T|x(2s)(t)|k+1dt)+k|b0|T2s(∫0T|x(2s)(t)|k+1dt)2/(k+1)(∫0T||x(t)|k+1|dt)(k-1)/(k+1).
Then, one has (3.33)[|b2s|-A2(2s,k)-θ1T(2s-1)k](∫0T|x(2s)(t)|k+1dt)k/(k+1)≤k|b0|T2s(∫0T|x(2s)(t)|k+1dt)1/(k+1)(∫0T||x(t)|k+1|dt)(k-1)/(k+1)+[(γ+ε)+θ2](∫0T|x(2s)(t)|k+1dt)k/(k+1)+u4,
where u4 is a positive constant. Using inequality (3.34)(a+b)l≤al+blfora≥0,b≥0,0≤l≤1,
it follows from (3.23) that (3.35)(∫0T|x(t)|k+1dt)(k-1)/(k+1)≤[A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2](k-1)/k(∫0T|x(2s)(t)|k+1dt)(k-1)/(k+1)+u5,
where u5 is a positive constant. Substituting the above formula and (3.23) into (3.33), one has (3.36){[A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2]k-1k|b2s|-A2(2s,k)-θ1T(2s-1)k-(γ+ε+θ2)(A1(2s,k)+θ1T(2s-1)k)|b0|-(γ+ε)-θ2-k|b0|T2s[A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2](k-1)/k[A1(2s,k)+θ1T(2s-1)k|b0|-(γ+ε)-θ2]k-1k}(∫0T|x(2s)(t)|k+1dt)k/(k+1)≤u5k|b0|T2s(∫0T|x(2s)(t)|k+1dt)1/(k+1)+u6,
where u6 is a positive constant.Ifs=1, since ∫0T[x′(t)]kx′′(t)dt=0,∫0T[x(t)]kx′′(t)dt=-k∫0T[x(t)]k-1[x′(t)]2dt, from (3.24), one has (3.37)b2∫0T[x′′(t)]k+1dt=-kb0∫0T[x(t)]k-1[x′(t)]2dt-∫0Tf(t,x(t-τ))x′′(t)dt+∫0Tp(t)x′′(t)dt.
Applying the above method, one has (3.38){|b2|-θ1Tk-(γ+ε+θ2)(A1(2,k)+θ1Tk)|b0|-(γ+ε)-θ2-k|b0|T2[A1(2,k)+θ1Tk|b0|-(γ+ε)-θ2](k-1)/k}×(∫0T|x′′(t)|k+1dt)k/(k+1)≤u7k|b0|T2(∫0T|x′′(t)|k+1dt)1/(k+1)+u8,
where u7,u8 is a positive constant. Hence there is a constant M1,M2>0 such that (3.39)∫0T|x(2s)(t)|k+1dt≤M1,(3.40)∫0T|x(t)|k+1dt≤M2.
From (3.5), using Hölder inequality and Lemma 2.1, one has (3.41)∫0T|x(n)(t)|dt≤∑i=02s|bi|∫0T|x(i)(t)|kdt+∫0T|f(t,x(t))|dt+∫0T|f(t,x(t))-f(t,x(t-τ(t)))|dt+∫0T|p(t)|dt≤[∑i=12s|bi|T(2s-i)k+1/(k+1)+θ1T(2s-1)k+1/(k+1)](∫0T|x(2s)(t)|k+1dt)k/(k+1)+[|b0|+(γ+ε)+θ2]T1/(k+1)(∫0T|x(t)|k+1dt)k/(k+1)+(|p(t)|∞+fδ)T≤[∑i=12s|bi|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)|∞+fδ)T=M,
where M is a positive constant. We claim that (3.42)|x(i)(t)|≤Tn-i-1∫0T|x(n)(t)|dt,i=1,2,…,n-1.
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 (3.43)|x(n-1)(t)|=|x(n-1)(ξ1)+∫ξ1tx(n)(s)ds|≤|x(n-1)(ξ1)|+∫0T|x(n)(t)|dt=∫0T|x(n)(t)|dt.
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 (3.44)|x(n-2)(t)|=|x(n-2)(ξ2)+∫ξ2tx(n-1)(s)ds|≤∫0T|x(n-1)(t)|dt≤T∫0T|x(n)(t)|dt.
By induction, we conclude that (3.42) holds. Furthermore, one has (3.45)|x(i)(t)|∞≤Tn-i-1∫0T|x(n)(t)|dt≤Tn-i-1M,i=1,2,…,n-1.
It follows from (3.39) that there exists a ξ∈[0,T] such that |x(ξ)|≤M21/(k+1). Applying Lemma 2.1, we get (3.46)|x(t)|∞≤x(ξ)+∫ξtx′(t)dt≤M21/(k+1)+Tk/(k+1)(∫0T|x′(t)|k+1dt)1/(k+1)≤M21/(k+1)+T2s-1+(k/(k+1))(∫0T|x(2s)(t)|k+1dt)1/(k+1)=M21/(k+1)+T2s-1+(k/(k+1))M11/(k+1).
It follows that there is a constant A>0 such that ∥x∥≤A. Thus Ω1 is bounded.LetΩ2={x∈KerL,QNx=0}. Suppose x∈Ω2, then x(t)=d∈R and satisfies (3.47)QNx=1T∫0T[-b0dk-f(t,d)+p(t)]dt=0.
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
(3.48)|b0||d|k=|1T∫0T[-f(t,d)+p(t)]dt|≤1T∫0T|f(t,d)|dt+|p(t)|∞≤(γ+ε)|d|k+|p(t)|∞.
Thus
(3.49)|d|≤[|p(t)|∞|b0|-(γ+ε)]1/k.
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,for allx∈∂Ω∩D(L),λ∈(0,1),(ii)
QNx≠0,for allx∈∂Ω∩KerL.At last we will prove that condition (iii) of Lemma2.4 is satisfied. We take (3.50)H:(Ω∩Ker⁡L)×[0,1]⟶Ker⁡L,H(d,μ)=μd+1-μT∫0T[-b0dk-f(t,d)+p(t)]dt.
From assumptions (H1) and (H2), we can easily obtain H(d,μ)≠0,for all(d,μ)∈∂Ω∩KerL×[0,1], which results in (3.51)deg⁡{QN,Ω∩Ker⁡L,0}=deg⁡{H(⋅,0),Ω∩Ker⁡L,0}=deg⁡{H(⋅,1),Ω∩Ker⁡L,0}≠0.Hence, by using Lemma2.2, we know that (1.1) has at least one T-periodic solution.Theorem 3.2.
Supposen=4m+1, m>0 an integer and conditions (H1),(H2) hold. If(H5)
there is a positive integer0<s≤m such that(3.52)b4s-3≠0,b4s-3+i=0,i=1,2,…,4m-4s+3,(H6)
(3.53)A2(4s-3,k)+θ1T(4s-4)k+(γ+θ2)(A1(4s-3,k)+θ1T(4s-4)k)|b0|-γ-θ2+k|b0|T4s-3[A1(4s-3,k)+θ1T4s-4|b0|-γ-θ2](k-1)/k<|b4s-3|,if1<s≤m,θ1+(γ+θ2)(A1(1,k)+θ1)|b0|-γ-θ2<|b1|,ifs=1,
then (1.1) has at least one T-periodic solution.Proof.
From the proof of Theorem3.1, one has
(3.54)(∫0T|x(t)|k+1dt)k/(k+1)≤A1(4s-3,k)+θ1T(4s-4)k|b0|-(γ+ε)-θ2(∫0T|x(4s-3)(t)|k+1dt)k/(k+1)+u9,
where u9 is a positive constant. Multiplying both sides of (3.8) by x(4s-3)(t), and integrating on [0,T], one has
(3.55)∫0Tx(n)(t)x(4s-3)(t)dt=-λ∑i=04s-3bi∫0T[x(i)(t)]kx(4s-3)(t)dt-λ∫0Tf(t,x(t-τ))x(4s-3)(t)dt+λ∫0Tp(t)x(4s-3)(t)dt.
Since
(3.56)∫0Tx(4m+1)(t)x(4s-3)(t)dt=(-1)2m-2s+2∫0T[x(2m+2s-1)(t)]2dt,
then it follows from (3.55) and (3.56) that
(3.57)b4s-3∫0T|x(4s-3)(t)|k+1dt≤-∑i=04s-4bi∫0T[x(i)(t)]kx(4s-3)(t)dt-∫0Tf(t,x(t-τ))x(4s-3)(t)dt+∫0Tp(t)x(4s-3)(t)dt.By using the same way as in the proof of Theorem3.1, the following theorems can be proved in case 1<s≤m or s=1.Theorem 3.3.
Supposen=4m+1, m>0 for a positive integer and conditions (H1),(H2) hold. If(H7)
there is a positive integer0<s≤m such that
(3.58)b4s-1≠0,b4s-1+i=0,i=1,2,…,4m-4s+1,(H8)
(3.59)A2(4s-1,k)+θ1T(4s-2)k+(γ+θ2)(A1(4s-1,k)+θ1T(4s-2)k)|b0|-γ-θ2+k|b0|T4s-1[A1(4s-1,k)+θ1T(4s-2)k|b0|-γ-θ2](k-1)/k<|b4s-1|,
then (1.1) has at least one T-periodic solution.Theorem 3.4.
Supposen=4m+3,  m≥0 an integer and conditions (H1),(H2) hold. If(H9)
there is a positive integer0≤s≤m such that
(3.60)b4s+1≠0,b4s+1+i=0,i=1,2,…,4m-4s+1,(H10)
(3.61)A2(4s+1,k)+θ1T4sk+(γ+θ2)(A1(4s+1,k)+θ1T4sk)|b0|-γ-θ2+k|b0|T4s+1[A1(4s+1,k)+θ1T4sk|b0|-γ-θ2](k-1)/k<|b4s+1|,if0<s≤m,θ1+(γ+θ2)(A1(1,k)+θ1)|b0|-γ-θ2<|b1|,ifs=0,
then (1.1) has at least one T-periodic solution.Theorem 3.5.
Supposen=4m+3,m>0 an integer and conditions (H1),(H2) hold. If(H11)
there is a positive integer0<s≤m such that
(3.62)b4s-1≠0,b4s-1+i=0,i=1,2,…,4m-4s+3,(H12)
(3.63)A2(4-1,k)+θ1T(4s-2)k+(γ+θ2)(A1(4s-1,k)+θ1T(4s-2)k)|b0|-γ-θ2+k|b0|T4s-1[A1(4s-1,k)+θ1T(4s-2)k|b0|-γ-θ2](k-1)/k<|b4s-1|,
then (1.1) has at least one T-periodic solution.Theorem 3.6.
Supposen=4m, m>0 an integer and conditions (H1) hold. If (H13)
(3.64)b0>γ+θ2,(H14)
there is a positive integer0<s≤2m such that
(3.65)b2s-1≠0,ifs=2m,b2s-1≠0,b2s-1+i=0,i=1,2,…,4m-2s,if0<s<2m,(H15)
(3.66)A2(2s-1,k)+θ1T(2s-2)k+(γ+θ2)(A1(2s-1,k)+θ1T(2s-2)k)b0-γ-θ2+kb0T2s-1[A1(2s-1,k)+θ1T(2s-2)kb0-γ-θ2](k-1)/k<|b2s-1|,if1<s≤2m,θ1+(γ+θ2)(A1(1,k)+θ1)b0-γ-θ2<|b1|,ifs=1,
then (1.1) has at least one T-periodic solution.Theorem 3.7.
Supposen=4m+2,m>0 an integer and conditions (H1) hold. If (H16)
(3.67)-b0>γ+θ2,(H17)
there is a positive integer0<s≤2m+1 such that
(3.68)b2s-1≠0,ifs=2m+1,b2s-1≠0,b2s-1+i=0,i=1,2,…,4m-2s,if0<s<2m+1,(H18)
(3.69)A2(2s-1,k)+θ1T(2s-2)k+(γ+θ2)(A1(2s-1,k)+θ1T(2s-2)k)-b0-γ-θ2-kb0T2s-1[A1(2s-1,k)+θ1T(2s-2)k-b0-γ-θ2](k-1)/k<|b2s-1|,if1<s≤2m+1,θ1+(γ+θ2)(A1(1,k)+θ1)-b0-γ-θ2<|b1|,ifs=1,
then (1.1) has at least one T-periodic solution.Theorem 3.8.
Supposen=4m, m>0 is an integer, and conditions (H1),(H13) hold. If(H19)
there is a positive integer0<s≤m such that
(3.70)b4s-2≠0,b4s-2+i=0,i=1,2,…,4m-4s+1,(H20)
(3.71)A2(4s-2,k)+θ1T(4s-3)k+(γ+θ2)(A1(4s-2,k)+θ1T(4s-3)k)b0-γ-θ2+kb0T4s-2[A1(4s-2,k)+θ1T(4s-3)kb0-γ-θ2](k-1)/k<|b4s-2|,if1<s≤m,(γ+θ2)(A1(2,k)+θ1Tk)b0-γ-θ2+kb0T2[A1(2,k)+θ1Tkb0-γ-θ2](k-1)/k<|b2|,ifs=1,
then (1.1) has at least one T-periodic solution.Theorem 3.9.
Supposen=4m, m>1 an integer and conditions (H1),(H13) hold. If(H21)
there is a positive integer1<s≤m such that
(3.72)b4s-4≠0,b4s-4+i=0,i=1,2,…,4m-4s+3,(H22)
(3.73)A2(4s-4,k)+θ1T(4s-5)k+(γ+θ2)(A1(4s-4,k)+θ1T(4s-5)k)b0-γ-θ2+kb0T4s-4[A1(4s-4,k)+θ1T(4s-5)kb0-γ-θ2](k-1)/k<|b4s-4|,
then (1.1) has at least one T-periodic solution.Theorem 3.10.
Supposen=4m+2, m≥1 an integer and conditions (H1),(H16) hold. If(H23)
there is a positive integer1≤s≤m such that
(3.74)b4s≠0,b4s+i=0,i=1,2,…,4m-4s+1,(H24)
(3.75)A2(4s,k)+θ1T(4s-1)k+(γ+θ2)(A1(4s,k)+θ1T(4s-1)k)-b0-γ-θ2-kb0T4s[A1(4s,k)+θ1T(4s-1)k-b0-γ-θ2](k-1)/k<|b4s|,
then (1.1) has at least one T-periodic solution.Theorem 3.11.
Supposen=4m+2, m≥1 is an integer, and conditions (H1),(H16) hold. If(H25)
there is a positive integer1≤s≤m such that
(3.76)b4s-2≠0,b4s-2+i=0,i=1,2,…,4m-4s+3,(H26)
(3.77)A2(4s-2,k)+θ1T(4s-3)k+(γ+θ2)(A1(4s-2,k)+θ1T(4s-3)k)-b0-γ-θ2-kb0T4s-2[A1(4s-2,k)+θ1T(4s-3)k-b0-γ-θ2](k-1)/k<|b4s-2|,if1<s≤m,θ1Tk+(γ+θ2)(A1(2,k)+θ1Tk)-b0-γ-θ2-kb0T2[A1(2,k)+θ1Tk-b0-γ-θ2](k-1)/k<|b2|,ifs=1,
then (1.1) has at least one T-periodic solution.The proofs of Theorem3.3–3.11 are similar to that of Theorem 3.1.Example 3.12.
Consider the following equation:(3.78)x(5)(t)+300[x′′(t)]3+150[x′(t)]3+1100[x(t)]3+1300(sin⁡t)[x(t-π10)]3=cos⁡t,
where n=5,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)=π/10. Thus, T=2π,γ=1/300,A1(2,k)=|b1|(2π)3+|b2|=1/50×(2π)3+200. Obviously assumptions (H1)–(H3) hold and
(3.79)θ1Tk+(γ+θ2)(A1(2,k)+θ1Tk)|b0|-γ-θ2+k|b0|(2π)2[A1(2,k)+θ1Tk|b0|-γ-θ2](k-1)/k<|b2|.
By Theorem 3.1, we know that (3.78) has at least one 2π-periodic solution.



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*Source: 916279-2011-10-11.xml*