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

Authors: Zaihong Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the existence of periodic solutions of Liénard equation with a deviating argumentx′′+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 limx→±∞g(x)=g(±∞) and limx→±∞F(x)=F(±∞) exist and are finite, where F(x)=∫0x‍f(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(-∞)]≠∫02π‍p(t)sin(θ+nt)dt, for all θ∈[0,2π],2ncos(nτ)[F(+∞)-F(-∞)]≠∫02π‍p(t)sin(θ+nt)dt, for all θ∈[0,2π],2[g(+∞)-g(-∞)]-2nsin(nτ)[F(+∞)-F(-∞)]≠∫02π‍p(t)sin(θ+nt)dt, for all θ∈[0,2π],2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]≠∫02π‍p(t)sin(θ+nt)dt, for all θ∈[0,2π].


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## Body

## 1. Introduction

We are concerned with the existence of periodic solutions of Liénard equation with a deviating argument as follows:(1)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.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 whenf(x)≡0,  for all x∈R and τ=0, (1) becomes
(2)x′′+n2x+g(x)=p(t).
Assume that limits (h1)
limx→±∞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:
(3)2[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π].
Assume that, besides (h1), the limits (h2)
limx→±∞F(x)=F(±∞)exist and are finite, where F(x)=∫0xf(u)du. It was proved in [10] that the following equation:
(4)x′′+f(x)x′+n2x+g(x(t-τ))=p(t)
has at least 2π-periodic solution provided that one of the following conditions holds:
(5)2[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π],2n[F(+∞)-F(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π].In the case whenf(x)≡0,  for all  x∈R and τ≠0, (1) becomes as follows:
(6)x′′+n2x+g(x(t-τ))=p(t).
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:
(7)2cos(nτ)[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π],2ncos(nτ)[F(+∞)-F(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π],2[g(+∞)-g(-∞)]-2nsin(nτ)[F(+∞)-F(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π],2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π].Remark 2.
Let us denote byΦ the function on the right-hand side of four inequalities above, namely,
(8)Φ(θ)=∫02π‍p(t)sin(θ+nt)dt,00θ∈[0,2π].
Then Φ can be expressed in the following form:
(9)Φ(θ)=Asinθ+Bcosθ,00θ∈[0,2π],
where
(10)A=∫02π‍p(t)cosntdt,0000B=∫02π‍p(t)sinntdt.
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:
(11)2cos(nτ)[g(+∞)-g(-∞)]∉[-ϱ,ϱ],2ncos(nτ)[F(+∞)-F(-∞)]∉[-ϱ,ϱ],2[g(+∞)-g(-∞)]0000-2nsin(nτ)[F(+∞)-F(-∞)]∉[-ϱ,ϱ],2n[F(+∞)-F(-∞)]0000-2sin(nτ)[g(+∞)-g(-∞)]∉[-ϱ,ϱ].Remark 3.
In the case whenτ=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 useR 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∥∞=max0≤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).LetX 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:
(12)ImP=KerL,0000KerQ=ImL.
Then we have the  following:
(13)X=KerL⊕KerP,0000Y=ImL⊕ImQ.
Clearly, LP=L|D(L)∩KerP→ImL is invertible. Denote by KP the inverse of LP. 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 KP(I-Q)N:Ω-→X are compact.Lemma 4 (see [11]).
Let X and Y be two real Banach spaces. Suppose thatL: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 degreedeg{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 Section2 to prove the existence of periodic solutions of (1). To this end, we first quote some notations and definitions.LetX and Y be two Banach spaces defined by the  following:
(14)X={x∈C1(R,R):x(t+2π)=x(t),∀t∈R},Y={y∈C(R,R):y(t+2π)=y(t),∀t∈R}
with the following norms
(15)∥x∥X=max{∥x∥∞,∥x′∥∞},0000∥y∥Y=∥y∥∞.
Define a linear operator
(16)L:D(L)⊂X⟶Y,0000Lx=x′′+n2x,
where D(L)={x∈X:x′′∈C(R,R)}, and a nonlinear operator
(17)N:X⟶Y,(Nx)(t)=-f(x(t))x′(t)-g(x(t-τ))+p(t).
It is easy to see that
(18)KerL=Span{sinnt,cosnt},ImL={y∈Y:∫02π‍y(t)sinntdt=0,00∫02π‍y(t)cosntdt=0}.
On the other hand, for any y∈Y, we can write the  following:
(19)y(t)=y-(t)+y~(t),00∀t∈[0,2π],
where y-(t) is defined by the  following:
(20)y-(t)=asinnt+bcosnt
with
(21)a=1π∫02π‍y(t)sinntdt,b=1π∫02π‍y(t)cosntdt,
whereas y~(t) satisfies the  following:
(22)∫02π‍y~(t)sinntdt=0,∫02π‍y~(t)cosntdt=0.
Therefore,
(23)Y=KerL⊕ImL.
It follows that L is a Fredholm map of index zero.Let us define two continuous projectorsP:X→KerL and Q:Y→Y by setting the  following:
(24)(Px)(t)=a1sinnt+b1cosnt,(Qy)(t)=y-(t)=asinnt+bcosnt,00∀t∈[0,2π]
for any x∈X and y∈Y, where constants a1 and b1 are defined as constants a and b. Obviously, ImQ=KerL.SetLP=L|D(L)∩KerP→ImL. Then LP is an algebraic isomorphism and we define KP:ImL→D(L)∩KerP by the  following:
(25)KP=LP-1.
Clearly, we have that, for any y∈ImL,
(26)(KPy)(t)=1n∫0t‍y(s)sinn(t-s)ds+12nπ∫02π‍sy(s)sinn(t-s)ds.
For any open bounded set Ω⊂X, we can prove by standard arguments that KP(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
(27)Lx=Nx.
To use Lemma 4, we embed this operator equation into an equation family with a parameter λ∈(0,1),
(28)Lx=λNx,
which is equivalent to the equation as follows:
(29)x′′+n2x+λf(x(t))x′(t)+λg(x(t-τ))0000=λp(t),00λ∈(0,1).
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):
(30)x′=y-λF(x),0000y′=-n2x-λg(x(t-τ))+λp(t).
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 transformationΦ:(r,θ)∈R+×S1→(x,y)∈R2∖{0} with S1=R/2πZ,
(31)x=1nr1/2sinθ,0000y=r1/2cosθ.
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:
(32)dθdt=n+λr-1/2g(1nr1/2(t-τ)sinθ(t-τ))sinθ-λnr-1/2F(1nr1/2sinθ)cosθ-λr-1/2p(t)sinθ,drdt=-2λr1/2g(1nr1/2(t-τ)sinθ(t-τ))cosθ-2λnr1/2F(1nr1/2sinθ)sinθ+2λr1/2p(t)cosθ.
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 r1/2, we get the  following:
(33)dr1/2dt=-λg(1nr1/2(t-τ)sinθ(t-τ))cosθ-λnF(1nr1/2sinθ)sinθ+λp(t)cosθ.
Integrating (33) and applying conditions (h1) and (h2), we get the  following:
(34)r(t)1/2=r01/2+O(1),00∀t∈[0,2π].
Furthermore,
(35)r(t)-1/2=r0-1/2+O(r0-1),00t∈[0,2π].
On the other hand, it follows from the first equation of (32) and (35) that
(36)dθdt=n+O(r0-1/2),00t∈[0,2π].
Therefore, we get the  following:
(37)θ(t)=θ0+nt+O(r0-1/2),00t∈[0,2π].
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:
(38)∫02π‍f(x(t))x′(t)sin(θ0+nt)dt000000+∫02π‍g(x(t-τ))sin(θ0+nt)dt0000=∫02π‍p(t)sin(θ0+nt)dt,∫02π‍f(x(t))x′(t)cos(θ0+nt)dt000000+∫02π‍g(x(t-τ))cos(θ0+nt)dt0000=∫02π‍p(t)cos(θ0+nt)dt.
Hence,
(39)-n∫02π‍F(x(t))cos(θ0+nt)dt000000+∫02π‍g(x(t-τ))sin(θ0+nt)dt0000=∫02π‍p(t)sin(θ0+nt)dt(40)n∫02π‍F(x(t))sin(θ0+nt)dt000000+∫02π‍g(x(t-τ))cos(θ0+nt)dt0000=∫02π‍p(t)cos(θ0+nt)dt.
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:
(41)∫02π‍f(x(t))x′(t)sin(θ0+n(t-τ))dt000000+∫02π‍g(x(t-τ))sin(θ0+n(t-τ))dt0000=∫02π‍p(t)sin(θ0+n(t-τ))dt,∫02π‍f(x(t))x′(t)cos(θ0+n(t-τ))dt000000+∫02π‍g(x(t-τ))cos(θ0+n(t-τ))dt0000=∫02π‍p(t)cos(θ0+n(t-τ))dt.
Hence,
(42)-n∫02π‍F(x(t))cos(θ0+n(t-τ))dt000000+∫02π‍g(x(t-τ))sin(θ0+n(t-τ))dt0000=∫02π‍p(t)sin(θ0+n(t-τ))dt,n∫02π‍F(x(t))sin(θ0+n(t-τ))dt000000+∫02π‍g(x(t-τ))cos(θ0+n(t-τ))dt0000=∫02π‍p(t)cos(θ0+n(t-τ))dt.Proof of Theorem1.
We shall prove the existence of periodic solutions of (1) provided that either
(43)2cosnτ[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π],
or
(44)2n[F(+∞)-F(-∞)]-2sinnτ[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,00∀θ∈[0,2π]
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 constantsM1 and M2 such that, for any 2π-periodic solution x(t) of (29),
(45)∥x∥∞<M1,0000∥x′∥∞<M2.
Assume by contradiction that (45) does not hold. Then there exists a sequence of 2π-periodic solutions {xk(t)}k=1∞ of (29) with λ=λk∈(0,1) such that
(46)∥xk∥∞+∥xk′∥∞⟶+∞,fork→∞.
Write yk(t)=xk′(t)+λkF(xk(t)). Since F is bounded on the interval (-∞,+∞), we have the  following:
(47)∥xk∥∞+∥yk∥∞⟶+∞,fork→∞.
Let (rk(t),θk(t)) be the 2π-periodic solution of (32) related to (xk(t),yk(t)). Obviously, rk(t)=n2xk2(t)+yk2(t). Then we have the  following:
(48)∥rk∥∞⟶+∞,fork⟶∞.
Without loss of generality, we also assume θk(0)∈[0,2π]. It follows from (39) that
(49)-n∫02π‍F(xk(t))cos(θk(0)+nt)dt000000+∫02π‍g(xk(t-τ))sin(θk(0)+nt)dt0000=∫02π‍p(t)sin(θk(0)+nt)dt.
From (34) and (37) we get that, for t∈[0,2π],
(50)g(xk(t-τ))0000=g(1nrk(t-τ)1/2sinθk(t-τ))0000=g(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1)).
Therefore,
(51)∫02π‍g(xk(t-τ))sin(θk(0)+nt)dt0000=∫02π‍g(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1))00000000×sin(θk(0)+nt)dt.
Obviously, we have the  following:
(52)∫02π‍g(1nrk(0)1/2sin(θk(0)+n(t-τ))+O(1))00×sin(θk(0)+nt)dt0000=1n∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))0000000000×sin(s+nτ)ds0000=cosnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))sinsds000000+sinnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))cossds.
Since θk(0)∈[0,2π], there exists a subsequence {θkj(0)} such that θkj→θ*, j→∞. By using Lebesgue dominated convergent theorem and the condition (h1), we get the  following:
(53)limj→+∞∫θkj(0)-nτ2nπ+θkj(0)-nτ‍g(1nrkj(0)1/2sins+O(1))sinsds0000=2n(g(+∞)-g(-∞)),limj→+∞∫θkj(0)-nτ2nπ+θkj(0)-nτ‍g(1nrkj(0)1/2sins+O(1))cossds=0.
Therefore,
(54)limj→+∞∫02π‍g(xkj(t-τ))sin(θkj(0)+nt)dt0000=2cosnτ[g(+∞)-g(-∞)].
Similarly, we can get the  following:
(55)limj→+∞∫02π‍F(xkj(t))cos(θkj(0)+nt)dt=0.
Hence, we obtain the  following:
(56)2cosnτ[g(+∞)-g(-∞)]=∫02π‍p(t)sin(θ*+nt)dt,
which contradicts with (43).
On the other hand, it follows from (40) that
(57)n∫02π‍F(xk(t))sin(θk(0)+nt)dt000000+∫02π‍g(xk(t-τ))cos(θk(0)+nt)dt0000=∫02π‍p(t)cos(θk(0)+nt)dt.
According to (50), we have the  following:
(58)∫02π‍g(xk(t-τ))cos(θk(0)+nt)dt0000=∫02π‍g(1nrk(0)1/2cos(θk(0)+n(t-τ))+O(1))000000000×cos(θk(0)+nt)dt0000=1n∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))0000000000×cos(s+nτ)ds0000=cosnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))cossds000000-sinnτn000000×∫θk(0)-nτ2nπ+θk(0)-nτ‍g(1nrk(0)1/2sins+O(1))sinsds.
From (53) we obtain the  following:
(59)limj→+∞∫02π‍g(xkj(t-τ))cos(θkj(0)+nt)dt0000=-2sinnτ[g(+∞)-g(-∞)].
Similarly, we have the  following:
(60)limj→+∞∫02π‍F(xkj(t))sin(θkj(0)+nt)dt0000=2[F(+∞)-F(-∞)].
It follows from (57)–(60) that
(61)2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=∫02π‍p(t)cos(θ*+nt)dt,
which contradicts with (44). Therefore, there exist positive constants M1 and M2 such that (45) holds.
(2) Letx(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 {ϱk} satisfying limk→∞ϱk=+∞ such that Nxk∈ImL with xk(t)=ϱksin(nt+α). We will prove the  following:
(62)2cosnτ[g(+∞)-g(-∞)]=∫02π‍p(t)sin(nt+α)dt,2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=∫02π‍p(t)sin(nt+α)dt.
In fact, sinceNxk∈ImL, we have the  following:
(63)∫02π‍[f(xk(t))xk′(t)+g(xk(t-τ))]sin(nt+α)dt0000=∫02π‍p(t)sin(nt+α)dt,∫02π‍[f(xk(t))xk′(t)+g(xk(t-τ))]cos(nt+α)dt0000=∫02π‍p(t)cos(nt+α)dt.
Using the same method as in step 1, we have the  following:(64)limk→∞∫02π‍[f(xk(t))xk′(t)+g(xk(t-τ))]sin(nt+α)dt0000=2cosnτ[g(+∞)-g(-∞)],limk→∞∫02π‍[f(xk(t))xk′(t)+g(xk(t-τ))]cos(nt+α)dt0000=2n[F(+∞)-F(-∞)]000000-2sin(nτ)[g(+∞)-g(-∞)].
As a consequence, (62) holds. Thus, we get a contradiction.
(3) LetM>max{nM0,M1,M2} be a sufficiently large constant (if it is necessary, M can be enlarged). Set
(65)Ω={x∈X:∥x∥∞<M,∥x′∥∞<M}.
From the conclusion in step 1 we know that
(66)Lx≠λNx,00∀x∈∂Ω∩D(L),λ∈(0,1).
From the conclusion in step 2 we know that
(67)Nx∉ImL,00∀x∈∂Ω∩KerL,
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:
(68)deg{JQN,Ω∩KerL,0}≠0.
To this end, let us define ϕ:KerL→R2, x=asinnt+bcosnt→(a,b), namely,
(69)ϕx=(a,b).
Obviously, ϕ is a linear isomorphism. For any x=asinnt+bcosnt, set
(70)(JQNx)(t)=h1(a,b)sinnt+h2(a,b)cosnt,
where
(71)h1(a,b)=1π∫02π‍[-f(x(t))x′(t)-g(x(t-τ))+p(t)]0000×sinntdth2(a,b)=1π∫02π‍[-f(x(t))x′(t)-g(x(t-τ))+p(t)]0000×cosntdt.
Define h:R2→R2 as follows:
(72)h(a,b)=ϕ∘Q∘N∘ϕ-1(a,b)=(h1(a,b),h2(a,b)).
Then we have the  following:
(73)deg{JQN,Ω∩KerL,0}=deg{h,ϕ(Ω∩KerL),0}.
To calculate deg{h,ϕ(Ω∩KerL),0}, we first estimate l1 and l2 as follows:
(74)l1(a,b)=1π∫02π‍[-f(x(t))x′(t)-g(x(t-τ))]sinntdt,l2(a,b)=1π∫02π‍[-f(x(t))x′(t)-g(x(t-τ))]cosntdt.
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→∞,
(75)∫02π‍f(x(t))x′(t)sinntdt0000=-n∫02π‍F(x(t))cosntdt0000=-n∫02π‍F(ρsin(nt+ϑ))cosntdt0000=-∫ϑ2nπ+ϑ‍F(ρsins)(cosscosϑ+sinssinϑ)ds0000=-cosϑ∫ϑ2nπ+ϑ‍F(ρsins)cossds000000-sinϑ∫ϑ2nπ+ϑ‍F(ρsins)sinsds0000=-2nsinϑ[F(+∞)-F(-∞)]+o(1).
On the other hand, we have that, for ρ→∞,
(76)∫02π‍g(x(t-τ))sinntdt0000=∫02π‍g(ρsin(n(t-τ)+ϑ))sinntdt0000=1n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)sin(s+nτ-ϑ)ds0000=cos(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)sinsds000000+sin(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)cossds0000=2cos(nτ-ϑ)[g(+∞)-g(-∞)]+o(1).
Therefore, we get the  following:
(77)l1(a,b)=2π{cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]000-cos(nτ-ϑ)[g(+∞)-g(-∞)]}+o(1).
To estimate l2, we have that, for ρ→∞,
(78)∫02π‍f(x(t))x′(t)cosntdt0000=n∫02π‍F(x(t))sinntdt0000=n∫02π‍F(ρsin(nt+ϑ))sinntdt0000=cosϑ∫ϑ2nπ+ϑ‍F(ρsins)sinsds000000-sinϑ∫ϑ2nπ+ϑ‍F(ρsins)cossds0000=2ncosϑ[F(+∞)-F(-∞)]+o(1).
Meanwhile, we get that, for ρ→∞,
(79)∫02π‍g(x(t-τ))cosntdt0000=∫02π‍g(ρsin(n(t-τ)+ϑ))cosntdt0000=1n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)cos(s+nτ-ϑ)ds0000=cos(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)cossds000000-sin(nτ-ϑ)n∫ϑ-nτ2nπ+ϑ-nτ‍g(ρsins)sinsds0000=-2sin(nτ-ϑ)[g(+∞)-g(-∞)]+o(1).
Hence, we obtain the  following:
(80)l2(a,b)=2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]000+sin(nτ-ϑ)[g(+∞)-g(-∞)]}+o(1).
Set
(81)h^1(a,b)=2π{-cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]000-cos(nτ-ϑ)[g(+∞)-g(-∞)]},h^2(a,b)=2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]000+sin(nτ-ϑ)[g(+∞)-g(-∞)]},h^(a,b)=(h^1(a,b),h^2(a,b)).
Replacing ϑ in x=ρsin(nt+ϑ) with π+ϑ, we get the  following:
(82)h^1(-a,-b)=-2π{-cos(nτ-ϑ)[g(+∞)-g(-∞)]nsinϑ[F(+∞)-F(-∞)]00000-cos(nτ-ϑ)[g(+∞)-g(-∞)]},h^2(-a,-b)=-2π{+sin(nτ-ϑ)[g(+∞)-g(-∞)]-ncosϑ[F(+∞)-F(-∞)]0000+sin(nτ-ϑ)[g(+∞)-g(-∞)]}.
As a consequence,
(83)h^1(-a,-b)=-h^1(a,b),h^2(-a,-b)=-h^2(a,b),h^(-a,-b)=-h^(a,b).
We note that, for a2+b2→+∞,
(84)h1(a,b)=h^1(a,b)+c1+o(1),h2(a,b)=h^2(a,b)+c2+o(1),
where
(85)c1=1π∫02π‍p(t)sinntdt,c2=1π∫02π‍p(t)cosntdt.
Let us consider the map H:ϕ(Ω∩KerL)×[0,1]→R2:(a,b,μ)→(h1(a,b,μ),h2(a,b,μ)) with
(86)h1(a,b,μ)=h^1(a,b)+μc1,h2(a,b,μ)=h^2(a,b)+μc2.
Obviously, H is continuous. Next, we shall prove that, for any (a,b,μ)∈∂ϕ(Ω∩KerL)×[0,1],
(87)(h1(a,b,μ),h2(a,b,μ))≠(0,0).
Otherwise, there exists some (a,b,μ)∈∂ϕ(Ω∩KerL)×[0,1] such that
(88)h1(a,b,μ)=0,h2(a,b,μ)=0.
Then we have the  following:
(89)2nsinϑ[F(+∞)-F(-∞)]000000-2cos(nτ-ϑ)[g(+∞)-g(-∞)]0000=-μπc1,-2ncosϑ[F(+∞)-F(-∞)]000000+2sin(nτ-ϑ)[g(+∞)-g(-∞)]0000=-μπc2.
Therefore, we get the  following:
(90)2cosnτ[g(+∞)-g(-∞)]=μ∫02π‍p(t)sin(nt+ϑ)dt,2n[F(+∞)-F(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000=μ∫02π‍p(t)cos(nt+ϑ)dt.
Since μ∈[0,1], we know from (90) that
(91)2cosnτ[g(+∞)-g(-∞)]∈[-ϱ,ϱ],(92)2n[F(+∞)-F(-∞)]0000-2sin(nτ)[g(+∞)-g(-∞)]∈[-ϱ,ϱ],
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),
(93)(h^1(a,b),h^2(a,b))≠(0,0),(94)(h^1(a,b)+c1,h^2(a,b)+c2)≠(0,0).
Since h^:R2→R2 is odd and (93) holds, we know from Borsuk Theorem [12] that
(95)deg(h^,ϕ(Ω∩KerL),0)=2m+1≠0,
where m is an integer.
On the other hand, we know from (94) and the expressions of h^1 and h^2 that there exists a positive constant ν, which is independent of a and b, such that, for (a,b)∈R2(96)|h^1(a,b)+c1|+|h^2(a,b)+c2|≥ν.
Consequently, we infer from the homotopy invariance of degree that, if M>max{M0,M1,M2} is large enough; then
(97)deg(JQN,Ω∩KerL,0)0000=deg(h,ϕ(Ω∩KerL),0)0000=deg(H(·,1),ϕ(Ω∩KerL),0)0000=deg(H(·,0),ϕ(Ω∩KerL),0)0000=deg(h^,ϕ(Ω∩KerL),0)0000=2m+1≠0.
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 Section3 to deal with the existence of 2π-periodic solutions of the  following equation:
(98)x′′+f(x′)+n2x+g(x(t-τ))=p(t).
Assume that the limits (h3)
limx→±∞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:
(99)2cos(nτ)[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π],2cos(nτ)[f(+∞)-f(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π],2[g(+∞)-g(-∞)]-2sin(nτ)[f(+∞)-f(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π],2[f(+∞)-f(-∞)]-2sin(nτ)[g(+∞)-g(-∞)]0000≠∫02π‍p(t)sin(θ+nt)dt,∀θ∈[0,2π].Remark 6.
In the case whenn=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].



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*Source: 906972-2013-05-08.xml*