# Existence of Periodic Solutions for a Class of Difference Systems withp-Laplacian

Authors: Kai Chen; Qiongfen Zhang
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By applying the least action principle and minimax methods in critical point theory, we prove the existence of periodic solutions for a class of difference systems withp-Laplacian and obtain some existence theorems.


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## Body

## 1. Introduction

Consider the followingp-Laplacian difference system:
(1.1)Δ(|Δu(t-1)|p-2Δu(t-1))=∇F(t,u(t)),t∈Z,
where Δ is the forward difference operator defined by Δu(t)=u(t+1)-u(t), Δ2u(t)=Δ(Δu(t)), p∈(1,+∞) such that 1/p+1/q=1, t∈ℤ, u∈ℝN, F:ℤ×ℝN→ℝ, and F(t,x) is continuously differentiable in x for every t∈ℤ and T-periodic in t for all x∈ℝN.Whenp=2, (1.1) reduces to the following second-order discrete Hamiltonian system:
(1.2)Δ2u(t-1)=∇F(t,u(t)),t∈Z.Difference equations provide a natural description of many discrete models in real world. Since discrete models exist in various fields of science and technology such as statistics, computer science, electrical circuit analysis, biology, neural network, and optimal control, it is of practical importance to investigate the solutions of difference equations. For more details about difference equations, we refer the readers to the books [1–3].In some recent papers [4–18], the authors studied the existence of periodic solutions and subharmonic solutions of difference equations by applying critical point theory. These papers show that the critical point theory is an effective method to the study of periodic solutions for difference equations. Motivated by the above papers, we consider the existence of periodic solutions for problem (1.1) by using the least action principle and minimax methods in critical point theory.


## 2. Preliminaries

Now, we first present our main results.Theorem 2.1.
Suppose thatF satisfies the following conditions: (F1)
there exists an integerT>1 such that F(t+T,x)=F(t,x) for all (t,x)∈ℤ×ℝN;(F2)
there existf, g∈l1([1,T],ℝ+) and α∈[0,p-1) such that
(2.1)|∇F(t,x)|≤f(t)|x|α+g(t),∀(t,x)∈Z[1,T]×RN,whereℤ[a,b]:=ℤ∩[a,b] for every a,b∈ℤ with a≤b,(F3)
(2.2)liminf⁡|x|→+∞⁡|x|-qα∑t=1T‍F(t,x)>2qα(T-1)q(2p-1)/pqT∑t=1T‍fq(t),∀t∈Z[1,T].Then problem (1.1) has at least one periodic solution with period T.Theorem 2.2.
Suppose thatF satisfies (F1)  and  the following conditions:
(2.3)∑t=1T‍f(t)<Tp2p-1(T-1)p(1+q)/q;(F2)'
there existf, g∈l1([1,T],ℝ+) such that
(2.4)|∇F(t,x)|≤f(t)|x|p-1+g(t),∀(t,x)∈Z[1,T]×RN,whereℤ[a,b]:=ℤ∩[a,b] for every a,b∈ℤ with a≤b;(F4)
(2.5)liminf⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)>2pTq/p(T-1)q(2p-1)/p[Tp-2p-1(T-1)p(1+q)/q∑t=1T‍f(t)]q/p∑t=1T‍fq(t),∀t∈Z[1,T].Then problem (1.1) has at least one periodic solution with period T.Theorem 2.3.
Suppose thatF satisfies (F1), (F2),  and  the following condition:
(F5)(2.6)limsup⁡|x|→+∞⁡|x|-qα∑t=1T‍F(t,x)<-[2qα(T-1)q(2p-1)/ppT+2qα(T-1)(q-1)2(2p-1)/pqT(q-1)2/q+2qα(T-1)2p-1+(2p-1)/ppT(p+1)/q]∑t=1T‍fq(t)∀t∈Z[1,T].Then problem (1.1) has at least one periodic solution with period T.Theorem 2.4.
Suppose thatF satisfies (F1), (2.3), (F2)',  and  the following condition:
(F6)(2.7)limsup⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)<-[2p(pT)q/p(T-1)q(2p-1)/p(Tp+2p-1(T-1)p(1+q)/q∑t=1Tf(t))[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]q+2p(pT)1/pT(T-1)2p-1+(2p-1)/p[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]1+1/p+2p(pT)(q-1)2/p(T-1)(q-1)2(2p-1)/pq[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)](q-1)2/p]∑t=1T‍fq(t),∀t∈Z[1,T].Then problem (1.1) has at least one periodic solution with period T.Remark 2.5.
The lower bounds and the upper bounds of our theorems are more accurate than the existing results in the literature. Moreover, there are functions satisfying our results but not satisfying the existing results in the literature.
Let the Sobolev spaceET be defined by
(2.8)ET={u:Z⟶RN∣u(t+T)=u(t),t∈Z}.
Foru∈ET, let u¯=(1/T)∑t=1Tu(t), u=u¯+u~, and E~T={u∈ET∣u¯=0}, then ET=ℝN⊕E~T. Let
(2.9)‖u‖=(|u¯|p+∑t=1T‍|Δu~(t)|p)1/p,u∈ET.
As usual, let
(2.10)‖u‖∞=sup⁡{|u(t)|:t∈Z[1,T]},∀u∈l∞(Z[1,T],RN).
For anyu∈ET, let
(2.11)φ(u)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍F(t,u(t))=1p∑t=1T‍|Δu~(t)|p+∑t=1T‍F(t,u(t)).
To prove our results, we need the following lemma.Lemma 2.6 (see [18]).
Letu∈ET. If ∑t=1T‍u(t)=0, then
(2.12)‖u‖∞≤(T-1)(1+q)/qT‖u~‖,(2.13)‖u‖pp=∑t=1T‍|u(t)|p≤(T-1)2p-1Tp-1‖u~‖p.


## 3. Proofs

For the sake of convenience, we denote(3.1)M1=(∑t=1T‍fq(t))1/q,M2=∑t=1T‍f(t),M3=∑t=1T‍g(t).Proof of Theorem2.1.
From (F3), we can choosea1>(T-1)(2p-1)/p/T(p-1)/p such that
(3.2)liminf⁡|x|→+∞|x|-qα∑t=1T‍F(t,x)>a1q2qαqM1q.
It follows from (F2), (2.12), and (2.13) that
(3.3)|∑t=1T‍[F(t,u(t))-F(t,u¯)]|=|∑t=1T‍∫01‍(∇F(t,u¯+su~(t)),u~(t))ds|≤∑t=1T‍∫01‍f(t)|u¯+su~(t)|α|u~(t)|ds+∑t=1T‍∫01‍g(t)|u~(t)|ds≤2α∑t=1T‍f(t)(|u¯|α+|u~(t)|α)|u~(t)|+∑t=1T‍g(t)|u~(t)|≤2α|u¯|α(∑t=1T‍fq(t))1/q(∑t=1T‍|u~(t)|p)1/p+2α‖u~‖∞1+α∑t=1T‍f(t)+‖u~‖∞∑t=1T‍g(t)=2α|u¯|αM1‖u~‖p+2αM2‖u~‖∞1+α+M3‖u~‖∞≤1pa1p‖u~‖pp+a1q2qαq|u¯|qαM1q+2αM2‖u~‖∞1+α+M3‖u~‖∞≤(T-1)2p-1pa1pTp-1‖u~‖p+2αM2(T-1)(1+q)(1+α)/qT1+α‖u~‖1+α+a1q2qαq|u¯|qαM1q+M3(T-1)(1+q)/qT‖u~‖.
Hence, we have(3.4)φ(u)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍F(t,u(t))=1p∑t=1T‍|Δu~(t)|p+∑t=1T‍[F(t,u(t))-F(t,u¯)]+∑t=1T‍F(t,u¯)≥1p∑t=1T‍|Δu~(t)|p-2αM2(T-1)(1+q)(1+α)/qT1+α‖u~‖1+α+∑t=1T‍F(t,u¯)-(T-1)2p-1pa1pTp-1‖u~‖p-a1q2qαq|u¯|qαM1q-M3(T-1)(1+q)/qT‖u~‖=(1p-(T-1)2p-1pa1pTp-1)‖u~‖p-2αM2(T-1)(1+q)(1+α)/qT1+α‖u~‖1+α+|u¯|qα(|u¯|-qα∑t=1T‍F(t,u¯)-a1q2qαqM1q)-M3(T-1)(1+q)/qT‖u~‖.
The above inequality and (3.2) imply that φ(u)→+∞ as ∥u∥→∞. Hence, by the least action principle, problem (1.1) has at least one periodic solution with period T.Proof of Theorem2.2.
From (2.3) and (F4), we can choose a constant a3∈ℝ such that(3.5)a3>T1/p(T-1)(2p-1)/p[Tp-2p-1M2(T-1)p(1+q)/q]1/p>0,liminf⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)>a3q2pqM1q.
It follows from (F2)' and Lemma 2.6 that
(3.6)|∑t=1T‍[F(t,u(t))-F(t,u¯)]|=|∑t=1T‍∫01‍(∇F(t,u¯+su~(t)),u~(t))ds|≤∑t=1T‍∫01‍f(t)|u¯+su~(t)|p-1|u~(t)|ds+∑t=1T‍g(t)|u~(t)|≤∑t=1T‍∫01‍2p-1f(t)(|u¯|p-1+sp-1|u~(t)|p-1)|u~(t)|ds+∑t=1T‍g(t)|u~(t)|=∑t=1T‍2p-1f(t)(|u¯|p-1+1p|u~(t)|p-1)|u~(t)|+∑t=1T‍g(t)|u~(t)|≤2p-1|u¯|p-1(∑t=1T‍fq(t))1/q(∑t=1T‍|u~(t)|p)1/p+2p-1pM2‖u~‖∞p+M3‖u~‖∞=2p-1M1|u¯|p-1‖u~‖p+2p-1pM2‖u~‖∞p+M3‖u~‖∞≤1pa3p‖u~‖pp+a3qM1q2pq|u¯|p+2p-1pM2‖u~‖∞p+M3‖u~‖∞≤((T-1)2p-1pa3pTp-1+2p-1M2(T-1)p(1+q)/qpTp)‖u~‖p+a3qM1q2pq|u¯|p+M3(T-1)(1+q)/qT‖u~‖,
which implies that
(3.7)φ(u)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍[F(t,u(t))-F(t,u¯)]+∑t=1T‍F(t,u¯)≥(1p-(T-1)2p-1pa3pTp-1-2p-1M2(T-1)p(1+q)/qpTp)‖u~‖p-M3(T-1)(1+q)/qT‖u~‖+|u¯|p(|u¯|-p∑t=1T‍F(t,u¯)-a3qM1q2pq).
The above inequality and (3.5) imply that φ(u)→+∞ as ∥u∥→∞. Hence, by the least action principle, problem (1.1) has at least one periodic solution with period T.Proof of Theorem2.3.
First we prove thatφ satisfies the (PS) condition. Assume that {un} is a (PS) sequence of φ; that is, φ′(un)→0 as n→∞ and {φ(un)} is bounded. By (F5), we can choose a2>(T-1)(2p-1)/p/T(p-1)/p such that
(3.8)limsup⁡|x|→+∞|x|-qα∑t=1T‍F(t,x)<-(2qαa2qp+2qαa2(q-1)2q+2qαa2(T-1)2p-1pTp-1)M1q.
In a similar way to the proof of Theorem 2.1, we have
(3.9)|∑t=1T‍(∇F(t,un(t)),u~n(t))|≤(T-1)2p-1pa2pTp-1‖u~n‖p+2αM2(T-1)(1+q)(1+α)/qT1+α‖u~n‖1+α+a2q2qαq|u¯n|qαM1q+M3(T-1)(1+q)/qT‖u~n‖.
Hence, we have
(3.10)‖u~n‖p≥〈φ′(un),u~n〉=∑t=1T‍|Δun(t)|p+∑t=1T‍(∇F(t,un(t)),u~n(t))≥(1-(T-1)2p-1pa2pTp-1)‖u~n‖p-M3(T-1)(1+q)/qT‖u~n‖-2αM2(T-1)(1+q)(1+α)/qT1+α‖u~n‖1+α-a2q2qαq|u¯n|qαM1q.
From (2.13), we have
(3.11)‖u~n‖p=(∑t=1T‍|u~n(t)|p)1/p≤(T-1)(2p-1)/pT(p-1)/p‖u~n‖.
From (3.10) and (3.11), we obtain
(3.12)a2q2qαM1qq|u¯n|qα≥(1-(T-1)2p-1pa2pTp-1)‖u~n‖p-(T-1)(2p-1)/pT(p-1)/p‖u~n‖-2αM2(T-1)(1+q)(1+α)/qT1+α‖u~n‖1+α-M3(T-1)(1+q)/qT‖u~n‖≥p-1p‖u~n‖p+C1=1q‖u~n‖p+C1,
where C1=mins∈[0,+∞){(1/p-(T-1)2p-1/pa2pTp-1)sp-(2αM2(T-1)(1+q)(1+α)/q/T1+α)s1+α -[(T-1)(2p-1)/p/T(p-1)/p+M3(T-1)(1+q)/q/T]s}. Notice that a2>(T-1)(2p-1)/p/T(p-1)/p implies -∞<C1<0. Hence, it follows from (3.12) that
(3.13)‖u~n‖p≤2qαa2qM1q|u¯n|qα-qC1,(3.14)‖u~n‖≤2qα/pa2q/pM1q/p|u¯n|qα/p+C2,
where C2>0. By the proof of Theorem 2.1, we have
(3.15)|∑t=1T‍[F(t,un(t))-F(t,u¯n)]|≤2αM1|u¯n|α‖u~n‖p+2αM2‖u~n‖∞1+α+M3‖u~n‖∞≤(T-1)2p-1pa2q-1Tp-1‖u~n‖p+2αM2(T-1)(1+q)(1+α)/qT1+α‖u~n‖1+α+a2(q-1)22qαq|u¯n|qαM1q+M3(T-1)(1+q)/qT‖u~n‖.
It follows from the boundedness of φ(un), (3.13)–(3.15) that
(3.16)C3≤φ(un)=1p∑t=1T‍|Δun(t)|p+∑t=1T‍[F(t,un(t))-F(t,u¯n)]+∑t=1T‍F(t,u¯n)≤(1p+(T-1)2p-1pa2q-1Tp-1)‖u~n‖p+2αM2(T-1)(1+q)(1+α)/qT1+α‖u~n‖1+α+a2(q-1)22qαq|u¯n|qαM1q+M3(T-1)(1+q)/qT‖u~n‖+∑t=1T‍F(t,u¯n)≤(1p+(T-1)2p-1pa2q-1Tp-1)(2qαa2qM1q|u¯n|qα-qC1)+a2(q-1)22qαq|u¯n|qαM1q+∑t=1T‍F(t,u¯n)+2αM2(T-1)(1+q)(1+α)/qT1+α(2qα/pa2q/pM1q/p|u¯n|qα/p+C2)1+α×M3(T-1)(1+q)/qT(2qα/pa2q/pM1q/p|u¯n|qα/p+C2)≤(2qαa2qp+a2(q-1)22qαq+a22qα(T-1)2p-1pTp-1)M1q|u¯n|qα-(1p+(T-1)2p-1pa2q-1Tp-1)qC1+22αM2(T-1)(1+q)(1+α)/qT1+α(2qα(1+α)/pa2q(1+α)/pM1q(1+α)/p|u¯n|qα(1+α)/p+C21+α)+∑t=1T‍F(t,u¯n)+M3(T-1)(1+q)/qT(2qα/pa2q/pM1q/p|u¯n|qα/p+C2)=|u¯n|qα[|u¯n|-qα∑t=1T‍F(t,u¯n)+(2qαa2qp+a2(q-1)22qαq+a22qα(T-1)2p-1pTp-1)M1q+22α+qα(1+α)/pa2q(1+α)/pM1q(1+α)/pM2(T-1)(1+q)(1+α)/qT1+α|u¯n|α(1+α-p)(q-1)+2qα/pa2q/pM1q/pM3(T-1)(1+q)/qT|u¯n|-α]+C4,
where C3 is a positive constant and C4 is a constant. The above inequality and (3.8) imply that {u¯n} is bounded. Hence {un} is bounded by (2.13) and (3.13). Since ET is finite dimensional, we conclude that φ satisfies (PS) condition.
In order to use the saddle point theorem ([19], Theorem 4.6), we only need to verify the following conditions:(I1)
φ(u)→-∞as|u|→∞inℝN,(I2)
φ(u)→+∞as∥u∥→∞inE~T.
In fact, from (F5), we have(3.17)∑t=1T‍F(t,u)⟶-∞as|u|⟶∞inRN,
which together with (2.11) implies that
(3.18)φ(u)=∑t=1T‍F(t,u)⟶-∞as|u|⟶∞inRN.
Hence, (I1) holds.
Next, for allu∈E~T, by (F2) and (2.12), we have
(3.19)|∑t=1T‍[F(t,u(t))-F(t,0)]|=|∑t=1T‍∫01‍(∇F(t,su(t)),u(t))ds|≤∑t=1T‍f(t)|u(t)|1+α+∑t=1T‍g(t)|u(t)|≤M2‖u‖∞1+α+M3‖u‖∞≤M2(T-1)(1+q)(1+α)/qT(1+α)‖u~‖1+α+M3(T-1)(1+q)/qT‖u~‖,
which implies that
(3.20)φ(u)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍[F(t,u(t))-F(t,0)]+∑t=1T‍F(t,0)≥1p‖u~‖p-M2(T-1)(1+q)(1+α)/qT(1+α)‖u~‖1+α-M3(T-1)(1+q)/qT‖u~‖+∑t=1T‍F(t,0),
for all u∈E~T. By Lemma 2.6, ∥u∥→∞ in E~T if and only if ∥u~∥→∞, so from (3.20), we obtain φ(u)→+∞ as ∥u∥→∞ in E~T; that is, (I2) is verified. Hence, the proof of Theorem 2.3 is complete.Proof of Theorem2.4.
First we prove thatφ satisfies the (PS) condition. Assume that {un} is a (PS) sequence of φ; that is, φ′(un)→0 as n→∞ and {φ(un)} is bounded. By (2.3) and (F6), we can choose a4∈ℝ such that
(3.21)a4>p1/pT1/p(T-1)(2p-1)/p[pTp-2p-1M2(T-1)p(1+q)/q]1/p,(3.22)limsup⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)<-[2pa4q(Tp+2p-1M2(T-1)p(1+q)/q)+2pTa4(T-1)2p-1pTp-2p-1M2(T-1)p(1+q)/q+2pa4(q-1)2q]M1q.
In a similar way to the proof of Theorem 2.2, we obtain
(3.23)|∑t=1T‍(∇F(t,un(t)),u~n(t))|≤((T-1)2p-1pa4pTp-1+2p-1M2(T-1)p(1+q)/qpTp)‖u~n‖p+a4qM1q2pq|u¯n|p+M3(T-1)(1+q)/qT‖u~n‖.
Hence, we have
(3.24)‖u~n‖p≥〈φ′(un),u~n〉=1p∑t=1T‍|Δun(t)|p+∑t=1T‍(∇F(t,un(t)),u~n(t))≥(1-(T-1)2p-1pa4pTp-1-2p-1M2(T-1)p(1+q)/qpTp)‖u~n‖p-a4qM1q2pq|u¯n|p-M3(T-1)(1+q)/qT‖u~n‖,
which together with (3.11) implies that
(3.25)a4qM1q2pq|u¯n|p≥(1-(T-1)2p-1pa4pTp-1-2p-1M2(T-1)p(1+q)/qpTp)‖u~n‖p-M3(T-1)(1+q)/qT‖u~n‖-(T-1)(2p-1)/pT1/q‖u~n‖≥1q(1-2p-1M2(T-1)p(1+q)/qpTp)‖u~n‖p+C5,
where C5=mins∈[0,+∞){(1/p-(T-1)2p-1/pa4pTp-1-2p-1M2(T-1)p(1+q)/q/p2Tp)sp-[M3(T-1)(1+q)/q/T +(T-1)(2p-1)/p/T1/q]s}. It follows from (3.21) that -∞<C5<0, so, we obtain
(3.26)‖u~n‖p≤pTpa4qM1q2ppTp-2p-1M2(T-1)p(1+q)/q|u¯n|p-pTpqC5pTp-2p-1M2(T-1)p(1+q)/q,(3.27)‖u~n‖≤2p1/pTa4q/pM1q/p[pTp-2p-1M2(T-1)p(1+q)/q]1/p|u¯n|+C6,
where C6 is a positive constant. By the proof of Theorem 2.2, we have
(3.28)|∑t=1T‍(F(t,un(t))-F(t,u¯n))|≤2p-1M1|u¯|p-1‖u~‖p+2p-1pM2‖u~‖∞p+M3‖u~‖∞≤((T-1)2p-1pa4q-1Tp-1+2p-1M2(T-1)p(1+q)/qpTp)‖u~n‖p+a4(q-1)2M1q2pq|u¯n|p+M3(T-1)(1+q)/qT‖u~n‖.
It follows from the boundedness ofφ(un), (3.26), (3.27), and the above inequality that(3.29)C7≤φ(un)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍[F(t,u(t))-F(t,u¯)]+∑t=1T‍F(t,u¯)≤[1p+(T-1)2p-1pa4q-1Tp-1+2p-1M2(T-1)p(1+q)/qpTp]‖u~n‖p+∑t=1T‍F(t,u¯n)+M3(T-1)(1+q)/qT‖u~n‖+a4(q-1)2M1q2pq|u¯n|p≤[1p+(T-1)2p-1pa4q-1Tp-1+2p-1M2(T-1)p(1+q)/qpTp]×(pTpa4qM1q2ppTp-2p-1M2(T-1)p(1+q)/q|u¯n|p-pTpqC5pTp-2p-1M2(T-1)p(1+q)/q)+∑t=1T‍F(t,u¯)+a4(q-1)2M1q2pq|u¯n|p+M3(T-1)(1+q)/qT(2p1/pTa4q/pM1q/p[pTp-2p-1M2(T-1)p(1+q)/q]1/p|u¯n|+C6)=|u¯n|p{[2pa4q(Tp+2p-1M2(T-1)p(1+q)/q)+2pTa4(T-1)2p-1pTp-2p-1M2(T-1)p(1+q)/q+2pa4(q-1)2q]M1q+|u¯n|-p∑t=1T‍F(t,u¯n)+2p1/pTa4q/pM1q/pM3(T-1)(1+q)/qT[pTp-2p-1M2(T-1)p(1+q)/q]1/p|u¯n|-p+1}+C8,
where C7 is a positive constant and C8 is a constant. The above inequality and (3.22) imply that {u¯n} is bounded. Hence, {un} is bounded by (2.13) and (3.26).
Similar to the proof of Theorem2.3, we only need to verify (I1) and (I2). It is easy to verify (I1) by (F6). Now, we verify that (I2) holds. For u∈E~T, by (F2)' and (2.12), we have
(3.30)|∑t=1T‍(F(t,u(t))-F(t,0))|=|∑t=1T‍∫01‍(∇F(t,su(t)),u(t))ds|≤∑t=1T‍∫01‍f(t)sp-1|u(t)|pds+∑t=1T‍g(t)|u(t)|≤M2p‖u‖∞p+M3‖u‖∞≤M2(T-1)p(1+q)/qpTp‖u~‖p+M3(T-1)(1+q)/qpT‖u~‖.
Thus, we have
(3.31)φ(u)=1p∑t=1T‍|Δu(t)|p+∑t=1T‍(F(t,u(t))-F(t,0))+∑t=1T‍F(t,0)≥(1p-M2(T-1)p(1+q)/qpTp)‖u~‖p-M3(T-1)(1+q)/qpT‖u~‖+∑t=1T‍F(t,0),
for all u∈E~T. By Lemma 2.6, ∥u∥→∞ in E~T if and only ∥u~∥→∞. So from the above inequality, we have φ(u)→+∞ as ∥u∥→∞, that is (I2) is verified. Hence, the proof of Theorem 2.4 is complete.


## 4. Example

In this section, we give four examples to illustrate our results.Example 4.1.
Letp=5/2 and
(4.1)F(t,x)=sin⁡(2πtT)|x|5/3+(sin⁡2πtT+1)|x|4/3+(h(t),x),
where h∈l1(ℤ[1,T],ℝN) and h(t+T)=h(t). It is easy to see that F(t,x) satisfies (F1) and
(4.2)|∇F(t,x)|≤53|sin⁡2πtT||x|2/3+43|sin⁡2πtT+1||x|1/3+|h(t)|≤53(|sin⁡2πtT|+ɛ)|x|2/3+a(ɛ)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0, and a(ɛ) is a positive constant and is dependent on ɛ. The above shows that (F2) holds with α=2/3 and
(4.3)f(t)=53(|sin⁡2πtT|+ɛ),g(t)=a(ɛ)+|h(t)|.
Moreover, we have
(4.4)liminf⁡|x|→+∞⁡|x|-2α∑t=1T‍F(t,x)=T,2qα(T-1)q(2p-1)/pqT∑t=1T‍fq(t)=3×210/9(T-1)8/35(53ɛ)5/3.
We can choose ɛ suitable such that
(4.5)liminf⁡|x|→+∞⁡|x|-2α∑t=1T‍F(t,x)=T>3×210/9(T-1)8/35(53ɛ)5/3=2qα(T-1)q(2p-1)/pqT∑t=1T‍fq(t),
which shows that (F3) holds. Then from Theorem 2.1, problem (1.1) has at least one periodic solution with period T.Example 4.2.
Letp=2, then q=2. Let
(4.6)F(t,x)=16(12+sin⁡2πtT)|x|2+|x|3/2+(h(t),x),
where h∈l1(ℤ[1,T],ℝN) and h(t+T)=h(t). It is easy to see that F(t,x) satisfies (F1) and
(4.7)|∇F(t,x)|≤13|12+sin⁡2πtT||x|+32|x|1/2+|h(t)|≤13(|12+sin⁡2πtT|+ɛ)|x|+b(ɛ)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0, and b(ɛ) is a positive constant and is dependent on ɛ. The above shows that (F2)' holds with
(4.8)f(t)=13(|12+sin⁡2πtT|+ɛ),g(t)=b(ɛ)+|h(t)|.
Observe that
(4.9)|x|-p∑t=1T‍F(t,x)=|x|-2∑t=1T‍[16(12+sin⁡2πtT)|x|2+|x|3/2+(h(t),x)]=T12+T|x|-1/2+(∑t=1T‍h(t),|x|-2x).
On the other hand, if we let T=2, then we have
(4.10)∑t=1T‍f(t)=23(12+ɛ),∑t=1T‍f2(t)=19∑t=1T‍(|12+sin⁡2πtT|+ɛ)2=29(12+ɛ)2,2pTq/p(T-1)q(2p-1)/p[Tp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]q/p∑t=1T‍fq(t)=2+8ɛ+8ɛ215-6ɛ.
We can choose ɛ sufficiently small such that
(4.11)∑t=1T‍f(t)=23(12+ɛ)<2=Tp2p-1(T-1)p-1,liminf⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)=16>2+8ɛ+8ɛ215-6ɛ=2pTq/p(T-1)q(2p-1)/p[Tp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]q/p∑t=1T‍fq(t),
which shows that (2.3) and (F4) hold. Then from Theorem 2.2, problem (1.1) has at least one periodic solution with period T.Example 4.3.
Letp=2, then q=2. Let
(4.12)F(t,x)=sin⁡(2πtT)|x|7/4+(sin⁡2πtT-1)|x|3/2+(h(t),x),
where h∈l1(ℤ[1,T],ℝN) and h(t+T)=h(t). It is easy to see that F(t,x) satisfies (F1) and
(4.13)|∇F(t,x)|≤74|sin⁡2πtT||x|3/4+32|sin⁡2πtT-1||x|1/2+|h(t)|≤74(|sin⁡2πtT|+ɛ)|x|3/4+c(ɛ)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0 and c(ɛ) is a positive constant and is dependent on ɛ. The above shows that (F2) holds with α=3/4 and
(4.14)f(t)=74(|sin⁡2πtT|+ɛ),g(t)=c(ɛ)+|h(t)|.
Observe that
(4.15)|x|-qα∑t=1T‍F(t,x)=|x|-3/2∑t=1T‍[sin⁡(2πtT)|x|7/4+(sin⁡2πtT-1)|x|3/2+(h(t),x)]=-T+(∑t=1T‍h(t),|x|-3/2x).
On the other hand, we have
(4.16)[2qα(T-1)q(2p-1)/ppT+2qα(T-1)(q-1)2(2p-1)/pqT(q-1)2/q+2qα(T-1)2p-1+(2p-1)/ppT(p+1)/q]∑t=1T‍fq(t)=[2(T-1)3T+2(T-1)3/2T1/2+2(T-1)9/2T3/2]∑t=1T‍4916(|sin⁡2πtT|+ɛ)2=492ɛ2(T-1)3/2[T1/2(T-1)3/2+T+(T-1)3]16T1/2.
We can choose ɛ suitable such that
(4.17)limsup⁡|x|→+∞⁡|x|-qα∑t=1T‍F(t,x)=-T<-492ɛ2(T-1)3/2[T1/2(T-1)3/2+T+(T-1)3]16T1/2=-[2qα(T-1)q(2p-1)/ppT+2qα(T-1)(q-1)2(2p-1)/pqT(q-1)2/q+2qα(T-1)2p-1+(2p-1)/ppT(p+1)/q]∑t=1T‍fq(t),
which shows that (F5) holds. Then from Theorem 2.3, problem (1.1) has at least one periodic solution with period T.Example 4.4.
Letp=2, then q=2. Let
(4.18)F(t,x)=13(sin⁡2πtT-18)|x|2+|x|3/2+(h(t),x),
where h∈l1(ℤ[1,T],ℝN) and h(t+T)=h(t). It is easy to see that F(t,x) satisfies (F1) and
(4.19)|∇F(t,x)|≤23|sin⁡2πtT-18||x|+32|x|1/2+|h(t)|≤23(|sin⁡2πtT-18|+ɛ)|x|+d(ɛ)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0, d(ɛ) is a positive constant and is dependent on ɛ. The above shows that (F2)' holds with
(4.20)f(t)=23(|sin⁡2πtT-18|+ɛ),g(t)=d(ɛ)+|h(t)|.
Observe that
(4.21)|x|-p∑t=1T‍F(t,x)=|x|-2∑t=1T‍[13(sin⁡2πtT-18)|x|2+|x|3/2+(h(t),x)]=-T24+T|x|-1/2+(∑t=1T‍h(t),|x|-2x).
On the other hand, if we let T=2, then we have
(4.22)∑t=1T‍f(t)=43(18+ɛ),∑t=1T‍f2(t)=49∑t=1T‍(|18+sin⁡2πtT|+ɛ)2=89(18+ɛ)2,(4.23)-[2p(pT)q/p(T-1)q(2p-1)/p(Tp+2p-1(T-1)p(1+q)/q∑t=1Tf(t))[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]q+2p(pT)1/pT(T-1)2p-1+(2p-1)/p[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]1+1/p+2p(pT)(q-1)2/p(T-1)(q-1)2(2p-1)/pq[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)](q-1)2/p]∑t=1T‍fq(t)=[192+128×(1/8+ɛ)3×(8-(8/3)(1/8+ɛ))2+16(8-(8/3)(1/8+ɛ))3/2+8(8-(8/3)(1/8+ɛ))1/2]×89(18+ɛ)2.
We can choose ɛ sufficiently small such that
(4.24)∑t=1T‍f(t)=43(18+ɛ)<2=Tp2p-1(T-1)p-1,(4.25)limsup⁡|x|→+∞⁡|x|-p∑t=1T‍F(t,x)=-112<[192+128×(1/8+ɛ)3×(8-(8/3)(1/8+ɛ))2+16(8-(8/3)(1/8+ɛ))3/2+8(8-(8/3)(1/8+ɛ))1/2]×89(18+ɛ)2=-[2p(pT)q/p(T-1)q(2p-1)/p(Tp+2p-1(T-1)p(1+q)/q∑t=1Tf(t))[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]q+2p(pT)1/pT(T-1)2p-1+(2p-1)/p[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)]1+1/p+2p(pT)(q-1)2/p(T-1)(q-1)2(2p-1)/pq[pTp-2p-1(T-1)p(1+q)/q∑t=1Tf(t)](q-1)2/p]∑t=1T‍fq(t),
which shows that (F6) holds. Then from Theorem 2.4, problem (1.1) has at least one periodic solution with period T.



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*Source: 135903-2012-07-12.xml*