# Existence of Subharmonic Periodic Solutions to a Class of

Authors: Xiao-Bao Shu; Yongzeng Lai; Fei Xu
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By introducing subdifferentiability of lower semicontinuous
convex functionφ(x(t),x(t−τ)) and its conjugate function, as well
as critical point theory and operator equation theory, we obtain the existence of
multiple subharmonic periodic solutions to the following second-order nonlinear
nonautonomous neutral nonlinear functional differential equation x″(t)+x″(t−2τ)+f(t,x(t),x(t−τ),x(t−2τ))=0, x(0)=0.


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## Body

## 1. Introduction

The existence of periodic solutions for differential system has received a great deal of attention in the last few decades. Different from ordinary differential equations and partial differential equations that do not contain delay variate, it is very difficult to study the existence of periodic solutions for functional differential equations. For this reason, many mathematicians developed different approaches such as the averaging method [1], the Massera-Yoshizawa theory [2, 3], the Kaplan-York [4] method of coupled systems, the Grafton cone mapping method [5], the Nussbaum method of fixed point theory [6] and Mawhin [7] coincidence degree theory.However, the critical point theory was rarely used in the literature, and most of the existing results were established for autonomous functional differential equations while little was done for nonautonomous equations via critical point theory.In this paper, by using critical point and operator equation theories, we study the existence of the following second-order nonlinear and nonautonomous mixed-type functional differential equation:(1.1)x′′(t)+x′′(t-2τ)+f(t,x(t),x(t-τ),x(t-2τ))=0,x(0)=0.Our basic assumptions are the following(A1)~(A3) and (A4)~(A6) to be given in Section 4:(A1)
f(t,x1,x2,x3)∈C(ℝ4,ℝ), and ∂f(t,x1,x2,x3)/∂t≠0;(A2)
there exists a continuously differentiable functionF(t,x1,x2)∈C1(ℝ3,ℝ) such that
(1.2)F2′(t,x1,x2)+F1′(t,x2,x3)=f(t,x1,x2,x3),
where F2′(t,x1,x2) and F1′(t,x2,x3) denote ∂F(t,x1,x2)/∂x2 and ∂F(t,x2,x3)/∂x2, respectively;(A3)
F(t+τ,x1,x2)=F(t,x1,x2) for all x1,x2∈ℝ.


## 2. Variational Structure

Fixγ>1,τ>0, where γ is a positive integer and (2.1)H01[0,2γτ]={x(t)∈L2[0,2γτ]∣x′∈L2[0,2γτ],x(t)is2γτ-periodicfunctionintx(0)=0,andx(t)hascompactsupporton⁡[0,2γτ]x′∈L2[0,2γτ]}.It is obvious thatH01[0,2γτ] is a Sobolev space by defining the inner product (·,·) and the norm ∥·∥ as follows: (2.2)〈x,y〉H01[0,2γτ]=∫02γτx′(t)y′(t)dt,‖x‖H01[0,2γτ]=(∫02γτ|x′(t)|2dt)1/2,∀x,y∈H01[0,2γτ].Moreover,x(t)∈H01[0,2γτ] can be expressed as (2.3)x(t)=a0+∑k=1∞(akcos⁡kπγτt+bksin⁡kπγτt).Let us consider the functionalI(x) defined on H01[0,2γτ] by (2.4)I(x)=∫02γτ[x′(t)x′(t-τ)-F(t,x(t),x(t-τ))]dt.For allx,y∈H01[0,2γτ] and ε>0, we know that (2.5)I(x+εy)=I(x)+ε(∫02γτ[(F(t,x(t)+εy(t),x(t-τ)+εy(t-τ))-F(t,x(t),x(t-τ)))x′(t)y′(t-τ)+x′(t-τ)y′(t)-(F(t,x(t)+εy(t),x(t-τ)+εy(t-τ))-F(t,x(t),x(t-τ)))x′(t)y′(t-τ)+x′(t-τ)]dt∫02γτy′(t)y′(t-τ)dt.)+ε2∫02γτy′(t)y′(t-τ)dt.It is then easy to see that(2.6)〈I′(x),y〉=∫02γτ[x′(t)y′(t-τ)+x′(t-τ)y′(t)-F1′(t,x(t),x(t-τ))y(t)-F2′(t,x(t),x(t-τ))y(t-τ)]dt,
where I′(x) denotes the Frechet differential of the function I(x). By the periodicity of F(t,u1,u2), x(t), and y(t), we have (2.7)∫02γτx′(t)y′(t-τ)dt=∫02γτx′(t)dy(t-τ)=x′(t)y(t-τ)[]|02γτ,-∫02γτx′′(t)y(t-τ)dt=-∫02γτx′′(t+τ)y(t)dt,∫02γτx′(t-τ)y′(t)dt=-∫02γτx′′(t-τ)y(t)dt.
Similarly, we have (2.8)∫02γτF2′(t,x(t),x(t-τ))y(t-τ)dt=∫-τ(2γ-1)τF2′(t+τ,x(t+τ),x(t))y(t)dt=∫02γτF2′(t,x(t+τ),x(t))y(t)dt.
Hence, (2.9)〈I′(x),y〉=∫02γτ[-x′′(t+τ)-x′′(t-τ)-Fu1′(t,x(t)x(t-τ))-Fu2′(t,x(t+τ),x(t))]y(t)dt.
Therefore, the Euler equation corresponding to the functional I(x) is (2.10)x′′(t+τ)+x′′(t-τ)+[Fu1′(t,x(t),x(t-τ))+Fu2′(t,x(t+τ),x(t))]=0.It is not difficult to see that (2.10) is equivalent to (1.1). Thus, system (1.1) is the Euler equation of the functional I(x). It follows that it is possible to obtain 2γτ-periodic solutions of system (1.1) by seeking critical points of the functional I(x).SinceI(x) has neither a supremum nor an infimum, we do not seek critical points of the functional I(x) by the extremum method. But we may use operator equation theory. First via the dual variational principle, we obtain new operator equations (see (4.16)) related to (1.1). Then solutions to system (1.1) are obtained by seeking critical points of operator equation (4.16).In this paper, our main tool is the following.Lemma 2.1 (Maintain Pass Theorem).
LetH be a real Banach space. If I(·)∈C1(H,ℝ) satisfies the Palais-Smale condition as well as the following additional conditions:(1)
there exist constantsρ>0 and a>0 such that I(x)≥a,forallx∈∂Bρ, where Bρ={x∈H:∥x∥H<ρ},(2)
I(θ)≤0 and there exists x0∈¯Bρ such that I(x0)≤0, then c=infh∈Γsups∈[0,1]I(h(s)) is a positive critical value of I, where(2.11)Γ={h∈C([0,1],H)∣h(0)=θ,h(1)=x0}.The rest of this paper is organized as follows. Subdifferentiability of lower semicontinuous convex functionφ(x(t),x(t-τ)) and its conjugate function are introduced in Section 3. In Section 4, we first give the definition of the weak solution to (1.1), then we establish the new operator equation (4.16) related to (1.1) by the conjugate function of F(t,x(t),x(t-τ)) and show that we can obtain solution to (1.1) from the solution to operator equation (4.16). In Section 5, by seeking critical points of operator equation (4.16), we obtain the result that there exist multiple subharmonic periodic solutions to system (1.1). Finally in Section 6, an example and a remark are given to illustrate our result.


## 3. The Subdifferentiability and the Conjugate Function of the Lower Semicontinuous Convex Functionφ(x(t),x(t-τ))

LetX be a space of all given n×τ-periodic functions in t and a Banach space, where n∈N is a positive integer. Denote ℝ¯=ℝ∪{+∞}. Let φ:X2→ℝ¯ be a lower semicontinuous convex function. Generally, φ is not always differentiable in conventional sense, but we may generalize the definition of “derivative” as follows.Definition 3.1.
Let(x1*,x2*)∈X*×X*. We say that (x1*,x2*) is a subgradient of φ at point (x0(t),x0(t-τ))∈X×X if
(3.1)φ(x0(t),x0(t-τ))+〈x1*,x(t)-x0(t)〉+〈x2*,x(t-τ)-x0(t-τ)〉≤φ(x(t),x(t-τ)).
For all x0(t)∈X, the set of all subgradients of φ at point (x0(t),x0(t-τ)) will be called the sub-differential of φ at point (x0(t),x0(t-τ)) and will be denoted by ∂φ(x0(t),x0(t-τ)).
By the definition of Subdifferentiability of functionφ, we may define its conjugate function φ* by
(3.2)φ*(x1*,x2*)=sup⁡{〈x1*,x(t)〉+〈x2*,x(t-τ)〉-φ(x(t),x(t-τ))},
where 〈·〉 denotes the duality relation of X* and X. So it is not difficult to obtain the following propositions.Proposition 3.2.
φ* is a lower semicontinuous convex function (φ* may have functional value +∞, but not functional value -∞).Proposition 3.3.
Ifφ≤ψ, then φ*≥ψ*.Proposition 3.4 (Yang inequality).
One has(3.3)φ(x(t),x(t-τ))+φ*(x1*,x2*)≥〈x1*,x(t)〉+〈x2*,x(t-τ)〉.Proposition 3.5.
One has(3.4)φ(x(t),x(t-τ))+φ*(x1*,x2*)=〈x1*,x(t)〉+〈x2*,x(t-τ)〉⟺(x1*,x2*)∈∂φ(x(t),x(t-τ)).Proposition 3.6.
φ* does not always equal +∞.Proof.
Letx0(t)∈X and β∈ℝ such that φ(x0(t),x0(t-τ))<+∞ and β0<φ(x0(t),x0(t-τ)). We consider the two convex sets in X2×ℝ defined by
(3.5)A=epi⁡φ≜{(x(t),x(t-τ),β)∈X2×R∣:φ(x(t),x(t-τ))<+∞,β>φ(x(t),x(t-τ))},B={(x0(t),x0(t-τ),β0)}.
By Hahn-Banach Theorem, we know that there exists (f1,f2,k)∈X*×X*×ℝ and α∈ℝ such that
(3.6)〈f1,x(t)〉+〈f2,x(t-τ)〉+kβ>α,∀(x(t),x(t-τ),β)∈eipφ,〈f1,x0(t)〉+〈f2,x0(t-τ)〉+kβ0<α.
So, we have
(3.7)〈f1,x0(t)〉+〈f2,x0(t-τ)〉+kφ(x0(t),x0(t-τ))>α>〈f1,x0(t)〉+〈f2,x0(t-τ)〉+kβ0.
Thus k>0, and
(3.8)〈-1kf1,x(t)〉+〈-1kf2,x(t-τ)〉-φ(x(t),x(t-τ))<-αk.Sinceφ* is a lower semicontinuous convex function that does not always equal +∞, we may define its conjugate function φ** by
(3.9)φ**(x1**,x2**)=sup⁡(x1*,x2*)∈X*×X*{〈x1**,x1*〉+〈x2**,x2*〉-φ(x1*,x2*)},
where 〈·〉 denotes the duality relation of X** and X*.Theorem 3.7.
Letφ be a lower semicontinuous convex function that does not always equal +∞, then φ**=φ.Proof.
We divide our proof into two parts. First we show thatφ**=φ holds when φ>0 and then φ**=φ holds for all lower semicontinuous convex functions φ that do not always equal +∞.
(i) The case whenφ>0.
From the definition ofφ** and Yang inequality, it is obvious that φ**≤φ holds.
Next, to proveφ**≥φ holds, suppose to the contrary that there exist a point (x0(t),x0(t-τ))∈X2, such that φ**(x0(t),x0(t-τ))<φ(x0(t),x0(t-τ)) holds.
Consider the two convex sets(3.10)A=eipφ≜{(x(t),x(t-τ),β)∈X2×R∣φ(x(t),x(t-τ))<+∞,β≥φ(x(t),x(t-τ))},B0={(x0(t),x0(t-τ),φ**(x0(t),x0(t-τ)))}.
By the Hahn-Banach Theorem, we know that there exist (g1,g2,k*)∈X*×X*×ℝ and α1∈ℝ such that
(3.11)〈g1,x(t)〉+〈g2,x(t-τ)〉+k*β>α1,∀(x(t),x(t-τ),β)∈eipφ,(3.12)〈g1,x0(t)〉+〈g2,x0(t-τ)〉+k*φ**(x0(t),x0(t-τ))<α1.
So, it follows that k*≥0. Let ε>0. Using φ>0 and (3.11), one obtains that
(3.13)〈g1,x(t)〉+〈g2,x(t-τ)〉+(k*+ε)φ(x(t),x(t-τ))≥α1,∀(x(t),x(t-τ))∈D(φ).
Thus, we have
(3.14)φ*(-g1k*+ε,-g2k*+ε)≤-α1k*+ε.
Then, by the definition of φ**, we obtain that
(3.15)φ**(x0(t),x0(t-τ))≥〈-g1k*+ε,x0(t)〉+〈-g2k*+ε,x0(t-τ)〉-φ*(-g1k*+ε,-g2k*+ε)≥〈-g1k*+ε,x0(t)〉+〈-g2k*+ε,x0(t-τ)〉+α1k*+ε.
That is to say,
(3.16)〈g1,x0(t)〉+〈g2,x0(t-τ)〉+(k*+ε)φ**(x0(t),x0(t-τ))≥α1,∀ε>0,
which is a contradiction to (3.12).
(ii) For allφ, by Proposition 3.6, we know that D(φ*)≠∅. Choose (x10*,x20*)∈D(φ*), and define function φ¯ by
(3.17)φ¯(x(t),x(t-τ))=φ(x(t),x(t-τ))-〈x10*,x(t)〉-〈x20*,x(t-τ)〉+φ*(x10*,x20*).
Then φ¯ is a lower semicontinuous convex function that is not always equal to +∞ and satisfies φ¯≥0. By the result of (i), we have that φ¯**=φ¯. On the other hand, we have
(3.18)φ¯*(x1*,x2*)=φ*(x1*+x10*,x2*+x20*)-φ*(x10*,x20*),φ¯**(x(t),x(t-τ))=φ¯(x(t),x(t-τ))-〈x10*,x(t)〉-〈x20*,x(t-τ)〉+φ*(x10*,x20*).
That is, φ¯**=φ.Corollary 3.8.
Letφ be a lower semicontinuous convex function that is not always equal to +∞. Then (x1*,x2*)∈∂φ(x(t),x(t-τ)) if and only if
(3.19)(x(t),x(t-τ))∈∂φ*(x1*,x2*).Proof.
One has(3.20)(x1*,x2*)∈∂φ(x(t),x(t-τ))⟺〈x1*,x(t)〉+〈x2*,x(t-τ)〉=φ(x(t),x(t-τ))+φ*(x1*,x2*)=φ**(x(t),x(t-τ))+φ*(x1*,x2*)⟺(x(t),x(t-τ))∈∂φ*(x1*,x2*).


## 4. Weak Solution to (1.1)

Define an operatorA=d2/dt2. By (2.6) and (4.1)〈u(t),A(ω(t))〉=∫02γτu(t)ω′′(t)dt=u(t)(ω′(t))[]|02γτ-∫02γτω′(t)u′(t)dt=-∫02γτu′(t)dω(t)=-u′(t)ω(t)[]|02γτ+∫02γτω(t)du′(t)=〈Au(t),ω(t)〉
as well as (4.2)〈u(t),A(ω(t-τ))〉=〈Au(t+τ),ω(t)〉,〈u(t-τ),A(ω(t))〉=〈A(u(t-τ)),ω(t)〉
we may define a weak solution to (1.1) as follows.Definition 4.1.
Foru∈Lp[0,2γτ], we say that u is a weak solution to (1.1), if
(4.3)〈u(t),A(ω(t-τ))〉+〈u(t-τ),A(ω(t))〉+〈ω(t),F1′(t,u(t),u(t-τ))〉+〈ω(t-τ),F2′(t,u(t),u(t-τ))〉=0,for all ω(t)∈D(A)∩Lp[0,2γτ], where
(4.4)〈u(t),υ(t)〉=∫02γτu(t)υ(t)dt,
when u(t)∈Lp[0,2γτ],υ(t)∈Lq[0,2γτ], where 2<p<+∞,(1/p)+(1/q)=1.
Our objective is to define the conjugate function ofF(t,x(t),x(t-τ)) by using the definition of Subdifferentiability of lower semicontinuous convex function and by making use of the dual variational structure. So we add three more conditions on function F(t,x(t),x(t-τ)) as follows: (A4)
u=(u1,u2)→F(t,u1,u2) is a continuously differentiable and strictly convex function and satisfies(4.5)F(t,0,0)=0,F1′(t,0,0)=F2′(t,0,0)=0∀t∈[0,2γτ];(A5)
forα2=1/p, there exist constants M,C>0, such that when |u|=u12+u22≥C we have
(4.6)F(t,u1,u2)≤α2[F1′(t,u1,u2)u1+F2′(t,u1,u2)u2],F(t,u1,u2)≤M|u|(1/α2);(A6)
(4.7)lim⁡|u|→0F(t,u1,u2)|u|2=0.
The conjugate function of functionF(t,x(t),x(t-τ)) is defined by
(4.8)H(t,ω(t),ω(t-τ))=sup⁡x(t)∈Lp[0,2γτ]{〈ω(t),x(t)〉+〈ω(t-τ),x(t-τ)〉-F(t,x(t),x(t-τ))},
for t∈[0,2γτ].
ThenH is a continuously differentiable and strictly convex function. By the duality principle (Corollary 3.8), we have that
(4.9)(ω(t),ω(t-τ))=(F1′(t,x(t),x(t-τ)),F2′(t,x(t),x(t-τ)))⟺(H1′(t,ω(t),ω(t-τ)),H2′(t,ω(t),ω(t-τ)))=(x(t),x(t-τ)),
where H1′(t,ω(t),ω(t-τ)) and H2′(t,ω(t),ω(t-τ)) denote ∂H(t,ω(t),ω(t-τ))/∂ω(t) and ∂H(t,ω(t),ω(t-τ))/∂ω(t-τ), respectively.Example 4.2.
LetF(x(t),x(t-τ))=(1/p)(x2(t)+x2(t-τ))p. Then
(4.10)H(ω(t),ω(t-τ))=1q(ω2(t)+ω2(t-τ))q.Proof.
The above expression holds since(4.11)H(ω(t),ω(t-τ))=sup⁡x(t)∈Lp[0,2γτ]{〈ω(t),x(t)〉+〈ω(t-τ),x(t-τ)〉-1p(x2(t)+x2(t-τ))p}=sup⁡λ>0{ω2(t)+ω2(t-τ)λ-1pλp}=1q(ω2(t)+ω2(t-τ))q.LetR(A) denote the value field of operator A. Then R(A) is a closed set. Let P be the orthogonal projection operator of R(A) and K̂=A-1P. Then it is not difficult to see that K̂ maps continuous continuation into a compact operator of Lq[0,2γτ]→Lq[0,2γτ].Let(4.12)E={(υ(t),υ(t-τ))∈(Lq[0,2γτ])2υ(0)=0∣〈ϕ(t),υ(t)〉=〈ϕ(t),υ(t-τ)〉=〈ϕ(t-τ),υ(t)〉=0,∀ϕ(t)∈R(A)∩Lp[0,2γτ],ϕ(0)=0(υ(t),υ(t-τ))∈(Lq[0,2γτ])2υ(0)=0},
where ℜ(A)={u∈D(A)|A(u(t)+u(t-2τ))=0}.Remark 4.3.
In fact, for allx(0)=0,x(t)∈Lp[0,2γτ], or x(t)∈Lq[0,2γτ], x(t) can be expressed as
(4.13)x(t)=a0+∑k=1∞(akcos⁡kπγτt+bksin⁡kπγτt).
So, it follows that
(4.14)〈υ(t),ϕ(t)〉=0⟺〈ϕ(t-τ),υ(t)〉=〈ϕ(t),υ(t-τ)〉=〈ϕ(t),υ(t)〉=0.
Thus E can also be expressed as
(4.15)E={(υ(t),υ(t-2τ))∈Lq[0,2γτ]×Lq[0,2γτ],v(0)=0∣〈ϕ(t),υ(t)〉=0,∀ϕ(t)∈R(A)∩Lp[0,2γτ],ϕ(0)=0}.
We want {(υ(t),υ(t-τ)),(χ(t),χ(t-τ))} to satisfy
(4.16)χ(t)=K̂(υ(t-τ))+H1′(t,υ(t),υ(t-τ)),χ(t-τ)=K̂(υ(t))+H2′(t,υ(t),υ(t-τ)),
where (υ(t),υ(t-τ))∈E,χ(t)∈ℜ(A)∩Lp[0,2γτ], that is, (χ(t),χ(t-τ))∈E⊥.
If{(υ(t),υ(t-τ)),(χ(t),χ(t-τ))} is a solution to (4.16) and if u(t)=H1′(t,υ(t),υ(t-τ)), u(t-τ)=H2′(t,υ(t),υ(t-τ)), then by the duality principle and (4.16), we have
(4.17)〈u(t),A(z(t-τ))〉+〈u(t-τ),A(z(t))〉+〈z(t),F1′(t,u(t),u(t-τ))〉+〈z(t-τ),F2′(t,u(t),u(t-τ))〉=〈H1′(t,υ(t),υ′(t-τ)),A(z(t-τ))〉+〈H2′(t,υ(t),υ(t-τ)),A(z(t))〉+〈z(t),υ(t)〉+〈z(t-τ),υ(t-τ)〉=〈χ(t)-K̂(υ(t-τ)),A(z(t-τ))〉+〈χ(t-τ)-K̂(υ(t)),A(z(t))〉+〈z(t),υ(t)〉+〈z(t),υ(t)〉=-〈υ(t),z(t)〉-〈υ(t),z(t)〉+〈z(t),υ(t)〉+〈z(t),υ(t)〉=0∀z(t)∈D(A)∩Lp[0,2γτ].
Thus, u(t) is a weak solution to (1.1).


## 5. Existence of Solutions to Operator Equation (4.16)

In this section, we discuss the existence of solutions to operator equation (4.16) by using the critical point theory. Our main result is the following.Theorem 5.1.
Under assumptions(A1)~(A6), problem (1.1) has at least one nontrivial weak 2γτ-periodic solution.
To prove this theorem, we state and prove the following lemmas first.
Letυ=(υ(t),υ(t-τ)), and
(5.1)K(υ(t)υ(t-τ))=(0K̂K̂0)(υ(t)υ(t-τ))=(K̂υ(t-τ)K̂υ(t)).It is not difficult to verify that〈K(υ),ψ〉=〈υ,K(ψ)〉=〈K̅υ(t-τ),ψ(t)〉+〈K̅υ(t),ψ(t-τ)〉, where ψ=(ψ(t),ψ(t-τ)), that is, the operator K is a symmetric operator.We may obtain solutions to (4.16) by seeking critical points of functional J(υ) defined by
(5.2)J(υ)=12〈K(υ),υ〉+∫02γτH(t,υ)dt=12〈K̂υ(t-τ),υ(t)〉+12〈K̂υ(t),υ(t-τ)〉+∫02γτH(t,υ(t),υ(t-τ))dt∀(υ(t),υ(t-τ))∈E.J may be regarded as the restriction to E of function Ĵ defined on Lq[0,2γτ]×Lq[0,2γτ] since both functions have identical component on E. Moreover
(5.3)Ĵ′(υ)=K(υ)+H′(υ).
Since
(5.4)〈Ĵ′(υ)-J′(υ),z〉=0,∀υ∈E,z=(z(t),z(t-τ))∈E,
there exist χ(t)∈ℜ(A) and χυ=(χυ(t),χυ(t-τ))∈E⊥ such that
(5.5)Ĵ′(υ)-J′(υ)=χυ.So ifυ* is a critical point of J′(υ*)=0 on E, then there exists χυ**=(χυ**(t),χυ**(t-τ))∈E⊥ such that
(5.6)K(υ*)+H′*=χυ**.
Hence, {υ*,χυ**} is a solution to (4.16). That is, {(υ*(t),υ*(t-τ)),(χυ**(t),χυ**(t-τ))} is a solution to (4.16).Lemma 5.2.
The following two conditions are equivalent:(1)
F(t,u1,u2)≤α2[F1′(t,u1,u2)u1+F2′(t,u1,u2)u2],forallt∈[0,2γτ], when |u|=u12+u22≥C;(2)
F(t,βu1,βu2)≥β1/α2F(t,u1,u2)>0,forallβ≥1,t∈[0,2γτ],|u|≥C.Proof.
For allu=(u1,u2),|u|≥C, let Φ(β)=F(t,βu1,βu2),Ψ(β)=β1/α2F(t,u1,u2).
(2)⇒(1). By Φ(β)≥Ψ(β),forallβ≥1 and Φ(1)=Ψ(1), it is easy to see that Φ′(1)≥Ψ′(1). That is,
(5.7)F1′(t,u1,u2)u1+F2′(t,u1,u2)u2≥1α2F(t,u1,u2).(1)⇒(2). By
(5.8)Φ′(β)=F1′(t,βu1,βu2)u1+F2′(t,βu1,βu2)u2=1β[F1′(t,βu1,βu2)βu1+F2′(t,βu1,βu2)βu2]≥1α2βΦ(β),
it follows that
(5.9)F(t,βu1,βu2)≥β1/α2F(t,u1,u2)>0,∀β≥1,t∈[0,2γτ].Lemma 5.3.
LetF(t,u1,u2) satisfy assumptions (A4) and (A5). Then there exist constants m>0 and M>0 such that
(5.10)F(t,u1,u2)≥m(u12+u22)1/α2∀t∈[0,2γτ],when|u|≥C,|F′(t,u1,u2)|≤(21/α2M-m)(u12+u22)(1/α2)-1,when|u|≥C,
where |F′(t,u1,u2)|=|F1′(t,u1,u2)|2+|F2′(t,u1,u2)|2.Proof.
Let(5.11)m=min⁡(u1,u2)∈∂BCF(t,u1,u2)C1/α2,
where BC denotes the ball of radius C centered at the origin. By (A4), one knows m>0. On the other hand, by Lemma 5.2 and (A5), we have
(5.12)F(t,u1,u2)≥F(t,Cu1u12+u22,Cu2u12+u22)(u12+u22C)1/α2≥m(u12+u22)1/α2.
By the convexity of functionF, one has
(5.13)F(t,u1,u2)+F1′(t,u1,u2)(z1-u1)+F2′(t,u1,u2)(z2-u2)≤F(t,z1,z2).
Letz=(z1,z2) run all over the ball B|u|(z) of radius |u| centered at u=(u1,u2), and choose the maximum of F1′(t,u1,u2)(z1-u1)+F2′(t,u1,u2)(z2-u2). Then it is not difficult to see that
(5.14)|F′(t,u1,u2)|u12+u22≤M(z12+z22)1/α2-m(u12+u22)1/α2.
By z≤2|u| we have that
(5.15)|F′(t,u1,u2)|≤(21/α2M-m)(u12+u22)(1/α2)-1.Lemma 5.4.
H∈C1(ℝ3,ℝ) is a strictly convex function and satisfies (1)
(5.16)H1′(t,0,0)=H2′(t,0,0)=0,H(t,0,0)=0,∀t∈[0,2γτ];(2)
(5.17)Cα2M|ω|1/(1-α2)-C1≤H(t,ω(t),ω(t-τ))≤Cα2m|ω|1/(1-α2)+C2;(3)
(5.18)Cα2′|ω|α2/(1-α2)-C4≤|H′(t,ω(t),ω(t-τ))|≤Cα2(21/(1-α2)m-1M)×|ω|α2/(1-α2)+C3;(4)
(5.19)lim⁡|ω|→0H(t,ω(t),ω(t-τ))|ω|2=∞,
where C1,…,C4 are constants, Cα2 and Cα2′ are constants, depending on α2, |ω|=ω2(t)+ω2(t-τ) and |H′(t,ω(t),ω(t-τ))|=|H1′(t,ω(t),ω(t-τ))|2+|H2′(t,ω(t),ω(t-τ))|2.Proof.
By Corollary3.8 and F1′(t,0,0)=F2′(t,0,0)=0, we have H1′(t,0,0)=H2′(t,0,0)=0, for all t∈[0,2γτ]. Moreover, by the definition of H, we know that H(t,0,0)=0.
Now we show that (5.17) holds. By (A5) one obtains F(t,u1,u2)≤M|u|1/α2+C1,forallu=(u1,u2)∈ℝ2. So, by Proposition 3.3 and Example 4.2, it is easy to see that
(5.20)H(t,ω(t),ω(t-τ))≥Cα2M|ω|1/(1-α2)-C1,
where Cα2=21/(1-α2)/Mα2/(1-α2)-1(α2α2/(1-α2)-α21/(1-α2)).
Similar arguments to the proof of Lemma5.3 show that there exists a constant C2 such that
(5.21)F(t,u1,u2)≥m|u|1/α2-C2.
Therefore, it follows that
(5.22)H(t,ω(t),ω(t-τ))≤Cα2m|ω|1/(1-α2)+C2.
Next we show that (5.18) holds. Again as in the proof of Lemma 5.3, we can estimate H′ by
(5.23)|H′(t,ω(t),ω(t-τ))|≤Cα2(21/(1-α2)m-1M)|ω|α2/(1-α2)+C3,
where C3=max{C1+C2,sup|ω|<1|H′(t,ω(t),ω(t-τ))||}. By Lemma 5.3 again and the duality principle
(5.24)(u1,u2)=(H1′(t,ω(t),ω(t-τ)),H2′(t,ω(t),ω(t-τ)))⟺(ω(t),ω(t-τ))=(F1′(t,u1,u2),F2′(t,u1,u2))
when |H′(t,ω(t),ω(t-τ))|≥C, we have
(5.25)|ω|≤(21/α2M-m)∣H′(1/α2)-1.
Since there exists a constant MC such that when |u|=u12+u22=|H′(t,ω(t),ω(t-τ))|≤C, we have
(5.26)|ω|=|F′(t,u1,u2)|≤MC.
Choose
(5.27)Cα2′=(21/α2M-m)α2/(α2-1),C4=Cα2′MCα2/(1-α2).
Then it is not difficult to see
(5.28)|H′(t,ω(t),ω(t-τ))|≥Cα2′|ω|α2/(1-α2)-C4.
Finally, we show that (5.19) holds. By (A6), for all ε>0, there exists δ>0 so that when |u|=u12+u22<δ, we have
(5.29)F(t,u1,u2)≤εu12+u22.
Now, for all K>0, choose ε=1/4K and let η=2εδ(ε). Then whenever ω2(t)+ω2(t-τ)<η, we have
(5.30)H(t,ω(t),ω(t-τ))≥14ε(ω2(t)+ω2(t-τ))=K|ω|2.
That is,
(5.31)lim⁡|ω|→0H(t,ω(t),ω(t-τ))|ω|2=∞.Lemma 5.5.
There exist constantsCδ and Cδ′ depending on δ, such that
(5.32)H(t,ω(t),ω(t-τ))≥{Cδ|ω|2,when|ω|≤δ,Cδ′|ω|q,when|ω|>δ,
and when δ→+0,Cδ→+∞.Proof.
By (5.19), we have
(5.33)lim⁡|ω|→0H(t,ω(t),ω(t-τ))|ω|2=∞.
So, as δ→+0, we obtain Cδ≜inf{H(t,ω(t),ω(t-τ))/|ω|2:|ω|≤δ}→+∞. That is,
(5.34)H(t,ω(t),ω(t-τ))≥Cδ|ω|2,
when |ω|≤δ.
We next show the validity of the second part of the inequality.
For allω0=(ω0(t),ω0(t-τ)),|ω0|=1, let ϕω0(β)=H(t,βω0(t),βω0(t-τ)). Then
(5.35)ϕω0′(β)=H1′(t,βω0(t),βω0(t-τ))ω0(t)+H2′(t,βω0(t),βω0(t-τ))ω0(t-τ).
Since ϕω0 is a convex function for all β>0, we have
(5.36)H1′(t,βω0(t),βω0(t-τ))ω0(t)+H2′(t,βω0(t),βω0(t-τ))ω0(t-τ)≥1βϕω0(β).
So, by (5.34) we obtain
(5.37)H1′(t,δω0(t),δω0(t-τ))ω0(t)+H2′(t,δω0(t),δω0(t-τ))ω0(t-τ)≥Cδ⋅δ.
By the convexity of H again, it is easy to see that
(5.38)H(t,sω0(t),sω0(t-τ))≥H1′(t,δω0(t),δω0(t-τ))(s-δ)ω0(t)+H2′(t,δω0(t),δω0(t-τ))(s-δ)ω0(t-τ)+H(t,δω0(t),δω0(t-τ))≥Cδ⋅δ(s-δ)+Cδδ2=Cδδs,∀s>0.
So, we further obtain that
(5.39)H(t,ω(t),ω(t-τ))≥Cδ⋅δ|ω|.
By Lemma 5.4, there exists T>0 such that
(5.40)H(t,ω(t),ω(t-τ))≥Cα22M|ω|q.
Let Cδ′=min{Cα2/2M,T1-qδCδ}. Then by (5.34), (5.39), and (5.40), we have
(5.41)H(t,ω(t),ω(t-τ))≥{Cδ|ω|2,|ω|≤δ,Cδ′|ω|q,|ω|>δ.Lemma 5.6.
Letυm=(υm(t),υm(t-τ))⇀υ=(υ(t),υ(t-τ)) (weakly convergent sequence on Lq([0,2γτ])2) and satisfy
(5.42)∫02γτH(t,υm)dt⟶∫02γτH(t,υ)dtasm⟶∞.
Then
(5.43)∫02γτH(t,υm-υ)dt⟶0asm⟶∞.Proof.
(I) First, we show that the terms in{H(t,υm)} have equicontinuous integrals, that is, for all ε>0, there exists δ>0 such that
(5.44)∀m,∫ΩH(t,υm)dt<εwhenμ(Ω)<δ.
SinceH is convex, we have
(5.45)H1′(t,υ)(υ(t)-υm(t))+H2′(t,υ)(υ(t-τ)-υm(t-τ))≤H(t,υm)-H(t,υ).
So, from υm⇀υ and the above equality, we have
(5.46)∫02γτH(t,υ)dt≤lim⁡m→∞¯∫02γτH(t,υm)dt.
By H≥0 and the assumption that ∫02γτH(t,υ)dt→∫02γτH(t,υm)dt, it is not difficult to see
(5.47)lim⁡m→∞∫ΩH(t,υm)dt=∫ΩH(t,υ)dtfor all  measurable  setsΩ.
Now suppose to the contrary that{H(t,υm)} does not have equicontinuous integrals, that is, there exists ε0>0 and functions υmk=(υmk(t),υmk(t-τ)), as well as measurable sets Ωk, such that the following inequalities
(5.48)∫ΩH(t,±υ)dt<ε0,forallmeasurablesetsΩ,μ(Ω)<δ,∫ΩkH(t,υmk)dt≥ε0,μ(Ωk)<δ2k
hold. Then choose Ω0=⋃k=1∞Ωk. It is not difficult to obtain μ(Ω0)<δ and
(5.49)∫Ω0H(t,υmk)dt≥∫ΩkH(t,υmk)dt≥ε0,
which is contradictory to (5.46) and (5.47).
(II) For allb>0, we divide [0,2γτ] into the following three subsets:
(5.50)Q1={t∈[0,2γτ]∣|υ|=υ2(t)+υ2(t-τ)>b},Q2m={t∈[0,2γτ]∣|υ|≤b,|υm-υ|≥δ},Q3m={t∈[0,2γτ]∣|υ|≤b,|υm-υ|<δ},
where |υm-υ|=(υm(t)-υ(t))2+(υm(t-τ)-υ(t-τ))2. By inequality (5.17), we know that there exist constants K and L such that
(5.51)H(t,2z(t),2z(t-τ))≤KH(t,z(t),z(t-τ))+L,∀(z(t),z(t-τ))∈R2.
By the convexity of the function H on Q1, we obtain
(5.52)H(t,υm-υ)≤12[H(t,2υm)+H(t,-2υ)]≤K2(H(t,υm)+H(t,-υ))+L.
So by (I), we may choose a constant b large enough and fixed such that μ(Q1) is small enough so that
(5.53)∫Q1H(t,υm-υ)dt≤K2∫Q1(H(t,υm)+H(t,-υ))dt+Lμ(Q1)<ε3.
For the above chosen constant b, now choose δ small enough so that
(5.54)∫Q3mH(t,υm-υ)dt<ε3.
For the chosen constantsb and δ, let
(5.55)κ=inf⁡|ω-z|≥δ,|z|≤b[H(t,ω)-H(t,z)-H1′(t,ω)(ω(t)-z(t))-H2′(t,ω)(ω(t-τ)-z(t-τ))].
Then κ>0. Now
(5.56)κμ(Q2m)≤∫Q2m[H(t,υm)-H(t,υ)-H1′(t,υ)(υm(t)-υ(t))-H2′(t,υ)(υm(t-τ)-υ(t-τ))]dt≤∫02γτ[H(t,υm)-H(t,υ)-H1′(t,υ)(υm(t)-υ(t))-H2′(t,υ)(υm(t-τ)-υ(t-τ))]dt.
Thus we have μ(Q2m)→0 as m→∞. Hence, it is easy to see that ∫Q2mH(t,υm)→0 as m→∞.
By repeating the above argument onQ1, we conclude that there exists an n0 such that when m>n0, we have
(5.57)∫Q2mH(t,υm-υ)dt<ε3.
From (5.53), (5.54), and (5.57), we obtain that
(5.58)lim⁡m→∞∫02γτH(t,υm-υ)dt=0.Corollary 5.7.
υm=(υm(t),υm(t-τ))→υ=(υ(t),υ(t-τ))(Lq([0,2γτ])×Lq([0,2γτ])) if and only if
(5.59)∫02γτH(t,υm-υ)dt=0.Proof.
(⇒). υm→υ implies that υm⇀υ (weakly). By inequality (5.17) and the continuity of the composition operator, one gets H(t,υm)→H(t,υ)(L1([0,2γτ])). That is, ∫02γτH(t,υm)dt→∫02γτH(t,υ)dt. So, by Lemma 5.6, one obtains the conclusion.
(⇐). By Lemma 5.5, there exist constants B1 and B2>0 such that
(5.60)∫02γτH(t,υ)dt≥B1∫|υ|≥δ|υ|qdt+B2∫|υ|<δ|υ|2dt≥B1∫|υ|≥δ|υ|qdt+B2(2γτ)2/(q-2)(∫|υ|<δ|υ|qdt)2/q≥Cδmin⁡{∫02γτ|υ|qdt,(∫02γτ|υ|qdt)2/q}.
Choose δ small enough. Then Cδ>0 is a constant and it is not difficult to see that the conclusion is correct.We next use the Maintain Pass Theorem to prove Theorem5.1 in three steps.(i) We show thatJ satisfies the P. S. condition in E. Let {υn=(υn(t),υn(t-τ))}⊂E, and let the constants C1,C2 satisfy
(5.61)C1≤J(υn)≤C2,J′(υn)⟶θ.
That is, we want to show that {υn} has a convergent subsequence in E.First, we show that{υn} is bounded. In fact, by
(5.62)zm=Kυm+H′(t,υm)-χm⟶θ,C1≤12〈Kυm,υm〉+∫02γτH(t,υm)dt≤C2,
where zm=(zm(t),zm(t-τ)),υm=(υm(t),υm(t-τ)),χm=(χm(t),χm(t-τ)), we know that there exists n(ε)>0 for all ε>0 such that the following inequality holds when m≥m(ε):
(5.63)∫02γτH(t,υm)dt-12[H1′(t,υm)υm(t)+H2′(t,υm)υm(t-τ)]≤C2+ε2(‖υm(t)‖Lq+‖υm(t-τ)‖Lq)=C2+ε‖υm(t)‖Lq.On the other hand, by Lemmas5.2 and 5.3, there exist constants α2,C3,C4 and C5 such that
(5.64)H(t,ω)-12H1′(t,ω)ω(t)-12H2′(t,ω)ω(t-τ)=12z(t)F1′(t,z(t),z(t-τ))+12z(t-τ)F2′(t,z(t),z(t-τ))-F(t,z(t),z(t-τ))≥(12α2-1)F(t,z(t),z(t-τ))-C3≥m|z|1/α2(12α2-1)-C4≥|ω|q-C5,
where ω(t)=F1′(t,z(t),z(t-τ)), ω(t-τ)=F2′(t,z(t),z(t-τ)); z(t)=H1′(t,ω(t),ω(t-τ)), z(t-τ)=H2′(t,ω(t),ω(t-τ)); |ω|=ω2(t)+ω2(t-τ); |z|=z2(t)+z2(t-τ).So by (5.63) and (5.64), it is easy to see that
(5.65)‖υm(t)‖Lq[0,2γτ]=‖υm(t-τ)‖Lq[0,2γτ]≤C6(constant)
That is, {υn} is bounded. We next show that {υn} has a convergent subsequence. Since Lq[0,2γτ] is a reflexive Banach space, there exists a subsequence of {υn} which is weakly convergent in Lq[0,2γτ]. We denote it by {υmk}, that is to say, υmk(t)⇀υ*(t),υmk(t-τ)⇀υ*(t-τ). On one hand, by the convexity of the function H, we have
(5.66)H(t,υ*(t),υ*(t-τ))+H1′(t,υ*(t),υ*(t-τ))(υmk(t)-υ*(t))+H2′(t,υ*(t),υ*(t-τ))(υmk(t-τ)-υ*(t-τ))≤H(t,υmk(t),υmk(t-τ)). So we have
(5.67)∫02γτH(t,υ*(t),υ*(t-τ))dt≤lim⁡k→∞¯∫02γτH(t,υmk(t),υmk(t-τ))dt.
On the other hand, by convexity of the function H again, we obtain
(5.68)H(t,υ*(t),υ*(t-τ))≥H(t,υmk(t),υmk(t-τ))+H1′(t,υmk(t),υmk(t-τ))(υ*(t)-υmk(t))+H2′(t,υmk(t),υmk(t-τ))(υ*(t-τ)-υmk(t-τ))=H(t,υmk(t),υmk(t-τ))+(-Kυmk+zmk+χmk)⋅(υ*-υmk).
Since operators A and K are compact and (zmk(t),zmk(t-τ))→θ, we have
(5.69)lim⁡k→∞¯∫02γτH(t,υmk(t),υmk(t-τ))dt≤∫02γτH(t,υ*(t),υ*(t-τ))dt.By (5.67), (5.69), and applying Lemma 5.6 and Corollary 5.7, it is not difficult to see that
(5.70)(υmk(t),υmk(t-2τ))⟶(υ*(t),υ*(t-τ)).(ii) We show that there exist constants ρ,r>0, such that
(5.71)J|∂Ωr≥ρ>0,
where ∂Ωr={(υ(t),υ(t-τ))∈Lq[0,2γτ]×Lq[0,2γτ]∣∥υ(t)∥Lq[0,2γτ]=∥υ(t-τ)∥Lq[0,2γτ]=r}.Letβ=∥K̂∥£(Lp,Lq). Choose δ>0 such that the constant Cδ is large enough and choose r small enough so that when ∥υ(t)∥Lq=r, by Lemma 5.5, there exists constant C7>0 satisfying
(5.72)Cδ∫|υ|<δ|υ(t)|2dt-4β(∫|υ|<δ|υ(t)|qdt)2/q≥C7(∫|υ|<δ|υ(t)|qdt)2/q,Cδ′∫|υ|≥δ|υ(t)|qdt-4β(∫|υ|≥δ|υ(t)|qdt)2/q≥C7(∫|υ|≥δ|υ(t)|qdt)2/q,
where |υ|=υ2(t)+υ2(t-τ).By (5.72) and inequalities
(5.73)ac+bc≤(a+b)c≤2c(ac+bc),
where a,b>0, and c>1, we have
(5.74)J(υ)=12〈K̂(υ(t)),υ(t-τ)〉+12〈K̂(υ(t-τ)),υ(t)〉+∫02γτH(t,υ(t),υ(t-τ))≥-β2‖υ(t)‖Lq[0,2γτ]2-β2‖υ(t)‖Lq[0,2γτ]2+Cδ∫|υ|<δ|υ(t)|2dt+Cδ∫|υ|<δ|υ(t-τ)|2dt+Cδ′∫|υ|≥δ(υ2(t)+υ2(t-τ))qdt≥-β2‖υ(t)‖Lq[0,2γτ]2-β2‖υ(t)‖Lq[0,2γτ]2+Cδ∫|υ|<δ|υ(t)|2dt+Cδ∫|υ|<δ|υ(t-τ)|2dt+Cδ′∫|υ|≥δ|υ(t)|qdt≥C7[(∫|υ|<δ|υ(t)|qdt)2/q+(∫|υ|≥δ|υ(t)|qdt)2/q]+Cδ∫|υ|<δ|υ(t-τ)|2dt≥C722/q‖υ(t)‖Lq[0,2γτ]2=C722/qr2.
Thus (5.71) is true with ρ=(C7/22/q)r2.(iii)  It is obvious thatJ(θ)=0, and J(υ) is an even function in υ.Remark 5.8.
Letυj(t)=sin(jπ/γτ)t,j=1,2,…. Then by
(5.75)Aυj(t)=-(jπγτ)2sin⁡jπγτt,
we have
(5.76)K̂υj(t)=-(γτjπ)2sin⁡jπγτt.
Hence, we obtain
(5.77)K̂υj(t)=-(γτjπ)2[sin⁡2jπγcos⁡jπγτ(t-2τ)+cos⁡2jπγsin⁡jπγτ(t-2τ)],K̂υj(t-2τ)=-(γτjπ)2[cos⁡2jπγsin⁡jπγτt-sin⁡2jπγcos⁡jπγτt].
Therefore, it is easy to see
(5.78)J(υj)=12〈K̂υj(t-2τ),υj(t)〉+12〈K̂υj(t),υj(t-2τ)〉+∫02γτH(t,υj)dt=-12(γτjπ)2cos⁡2jπγ[∫02γτ|υj(t)|2dt+∫02γτ|υj(t-2τ)|2dt]+∫02γτH(t,υj)dt.
We may choose jk, such that cos2jkπ/γ>0. Let υjk(t)=sin(jkπ/γτ)t. By (5.78), we have
(5.79)J(λυjk)≤-λ12(γτjπ)2cos⁡2jπγ[∫02γτ|ϕ(t)|2dt+∫02γτ|ϕ(t-2τ)|2dt]+Cα2m∫02γτ(ϕ2(t)+ϕ2(t-2τ))qdt+2γτC2⟶-∞,(λ⟶+∞).
From (i), (ii), (iii), and the Maintain Pass Theorem, we conclude that problem (4.16) has at least one nontrivial 2γτ-periodic solution. Thus, Theorem 5.1 holds.


## 6. Example

In this section, we present a remark and an example to illustrate our main result.Remark 6.1.
For assumptions(A1)~(A3), function F is a solution to the following partial differential equation:
(6.1)F2′(t,x1,x2)+F1′(t,x2,x3)=f(t,x1,x2,x3).
In some special cases, F can be easily determined. For example, if there exist continuously differential functions g,h∈C1(ℝ2,ℝ) such that (∂g/∂u)(u,v)=(∂h/∂v)(u,v) and f(t,x1,x2,x3)=α(t)[g(x1,x2)+h(x2,x3)], where α(t) is continuously differentiable and τ-periodic in t, then
(6.2)F(t,x1,x2)=α(t)[∫0x1h(s,x2)ds+∫0x2g(0,s)ds]
or
(6.3)F(t,x1,x2)=α(t)[∫0x2g(x1,s)ds+∫0x1h(s,0)ds].
For example,
(1) if g(u,v)=u2-2uv-v2,h(u,v)=u2+2uv-v2 and f(t,x1,x2,x3)=cos(2τt/π)[x12-2x1x2+2x22x3-x32], then
(6.4)F(t,x1,x2)=cos⁡2τtπ[∫0x2g(x1,s)ds+∫0x1h(s,0)ds]=cos⁡2τtπ[∫0x2(x12-2x1s-s2)ds+∫0x1s2ds]=cos⁡2τtπ(x12x2-x1x22-13x23+13x13);
(2)  if g(u,v)=2vcosu-u2sinv,h(u,v)=2ucosv-v2sinu and f(t,x1,x2,x3)=cos(2τt/π)[2x2cosx1-x12sinx2+2x2cosx3-x32sinx2], then
(6.5)F(t,x1,x2)=cos⁡2τtπ[∫0x2g(x1,s)ds+∫0x1h(s,0)ds]=cos⁡2τtπ[∫0x2(2scos⁡x1-x12sin⁡s)ds+∫0x12sds]=cos⁡2τtπ(x22cos⁡x1+x12cos⁡x2).
One can check easily that F satisfies assumption (A2).Finally, as an application, we consider the following example.Example 6.2.
Consider the equation(6.6)x′′(t)+x′′2(2τtπ)x(t-τ)[(x2(t)+x2(t-τ))(p/2)-1+(x2(t-τ)+x2(t-2τ))(p/2)-1]=0,
with x(0)=0, where p>2. Then g(u,v)=pv(u2+v2)(p/2)-1 and h(u,v)=pu(u2+v2)(p/2)-1.
Moreover,g and h satisfy the condition (∂g/∂u)(u,v)=(∂h/∂v)(u,v). So we can choose
(6.7)F(t,x(t),x(t-τ))=p(1+sin⁡22τtπ)[∫0x(t-τ)s(x(t)2+s2)(p/2)-1ds+∫0x(t)sp/2ds]=(1+sin⁡22τtπ)(x(t)2+x(t-τ)2)p/2.
It is obvious that assumptions (A1)~(A3) and the following (A4′)~(A6′) hold:(A4′)
x=(x(t),x(t-τ))→F(t,x(t),x(t-τ)) is a continuously differentiable and strictly convex function and also satisfies
(6.8)F(t,0,0)=0,F1′(t,0,0)=F2′(t,0,0)=0∀t∈[0,2γτ];(A5′)
forα2=1/p<1/2, there exist constants M,C>0, such that when |x|=x2(t)+x2(t-τ)≥C we have
(6.9)F(t,x(t),x(t-τ))≤α2[F1′(t,x(t),x(t-τ))x(t)+F2′(t,x(t),x(t-τ))x(t-τ)],F(t,x(t),x(t-τ))≤M|x|1/α2;(A6′)
(6.10)lim⁡|x|→0F(t,x(t),x(t-τ))|x|2=0.
Thus, (6.6) has at least one nontrivial weak 2γτ-periodic solution by Theorem 5.1.



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*Source: 404928-2012-05-07.xml*