# Existence of Periodic Solutions for a Class of Discrete Hamiltonian Systems

Authors: Qiongfen Zhang; X. H. Tang; Qi-Ming Zhang
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

By applying minimax methods in critical point theory, we prove
the existence of periodic solutions for the following discrete Hamiltonian systemsΔ2u(t-1)+∇F(t,u(t))=0, where t∈ℤ, u∈ℝN, F:ℤ×ℝN→ℝ, F(t,x) is continuously differentiable in x for every t∈ℤ and is T-periodic in t;T is a positive integer.


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## Body

## 1. Introduction

Consider the following discrete Hamiltonian system:(1.1)Δ2u(t-1)+∇F(t,u(t))=0,t∈Z,
where Δ is the forward difference operator defined by Δu(t)=u(t+1)-u(t),Δ2u(t)=Δ(Δu(t)),t∈ℤ,u∈ℝN,F:ℤ×ℝN→ℝ,and F(t,x) is continuously differentiable in x for every t∈ℤ and is T-periodic in t; T is a positive integer.Difference equations usually describe evolution of certain phenomena over the course of time. For example, if a certain population has discrete generations, the size of the(t+1) th generation x(t+1) is a function of the tth generation x(t). In fact, 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, optimal control, and so on, 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 [4–15], 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. In 2007, Xue and Tang [11] investigated the existence of periodic solutions for (1.1) and obtained the main result.Theorem A (see [11]).
Suppose thatF satisfies the following conditions: (F1)
there exists a positive constantT such that  F(t+T,x)=F(t,x) for all  (t,x)∈ℤ×ℝN;(F2)
there are constantsL1>0,L2>0,and0≤α<1 such that
(1.2)|∇F(t,x)|≤L1|x|α+L2,∀(t,x)∈Z[1,T]×RN,
where  ℤ[a,b]≔∩[a,b]   for every  a,b∈ℤ with  a≤b;(F3)
|x|-2α∑t=1TF(t,x)→+∞ as  |x|→+∞ for all  t∈ℤ[1,T].
Then problem (1.1) has at least one periodic solution with period  T.Let(1.3)F(t,x)=f(t)|x|7/4+(sin2πtT+1)|x|3/2+(h(t),x),
where f:ℤ[1,T]→ℝ, f(t+T)=f(t), h:ℤ[1,T]→ℝN, and h(t+T)=h(t). It is easy to see that(1.4)|∇F(t,x)|≤74|f(t)||x|3/4+32|sin2πtT+1||x|1/2+|h(t)|≤74(|f(t)|+ε)|x|3/4+a(ε)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0 and a(ε) is a positive constant and is dependent on ε. The above inequality shows that there are functions not satisfying condition (F2). If we let ∑t=1Tf(t)=0, α=3/4, T=2, then we have (1.5)|x|-2α∑t=1TF(t,x)=2+(h(1)+h(2),|x|-3/2x).
But the above equality does not satisfy (F3). This example shows that it is valuable to further improve conditions (F2) and (F3).Before stating our main results, we first introduce some preliminaries.


## 2. Preliminaries

Let(2.1)HT={u:Z⟶RN∣u(t+T)=u(t),t∈Z}.HT can be equipped with the inner product (2.2)〈u,v〉=∑t=1T(u(t),v(t)),∀u,v∈HT,
by which the norm on HT can be reduced by (2.3)‖u‖=(∑t=1T|u(t)|2)1/2,∀u∈HT,
where (·,·) and |·| denote the usual inner product and the usual norm in ℝN. It is easy to see that (HT,〈·,·〉) is a finite-dimensional Hilbert space and linear homeomorphic to ℝNT. For any r>1, define (2.4)‖u‖r=(∑t=1T|u(t)|r)1/r,∀u∈HT.
Obviously, ∥u∥=∥u∥2 and ∥u∥ is equivalent to ∥u∥r. Hence, there exist two positive constants C1, C2, which are independent on r, such that (2.5)C1‖u‖r≤‖u‖≤C2‖u‖r,∀u∈HT.
If we define ∥u∥∞=supt∈ℤ[1,T]|u(t)|, it is easy to see that for any r>1,(2.6)‖u‖∞≤‖u‖r,∀u∈HT.For anyu∈HT, let(2.7)φ(u)=-12∑t=1T|Δu(t)|2+∑t=1T[F(t,u(t))-F(t,0)].
We can compute the Fréchet derivative of (2.7) as (2.8)∂φ(u)∂u(t)=Δ2u(t-1)+∇F(t,u(t)),t∈Z[1,T].
Hence, u is a critical point of φ on HT if and only if (2.9)Δ2u(t-1)+∇F(t,u(t))=0,t∈Z[1,T],u∈RN.
So, the critical points of φ are classical solutions of (1.1). The following lemmas are useful in our proof.Lemma 2.1 (see [11]).
As a subspace ofHT,  Nk is defined by
(2.10)Nk:={u∈HT∣-Δ2u(t-1)=λku(t)},
where λk=2-2coskω,ω=2π/T,k∈ℤ[0,[T/2]],[·] and denote the Gauss Function. Then there hold (i)
Nk⊥Nj,k≠j,j∈ℤ[0,[T/2]];(ii)
HT=⊕k=0[T/2]Nk.Lemma 2.2 (see [11]).
DefineHk:=⊕j=0kNj,  Hk⊥:=⊕j=k+1[T/2]Nj,  k∈ℤ[0,[T/2]-1]; then one has
(2.11)∑t=1T|Δu(t)|2≤λk‖u‖2,∀u∈Hk;(2.12)∑t=1T|Δu(t)|2≥λk+1‖u‖2,∀u∈Hk⊥.


## 3. Main Results and Proofs

Theorem 3.1.
Suppose thatF satisfies  (F1)  and the following conditions (F2)′
there arep,q:ℤ[1,T]→ℝ+,  α∈[0,1) such that
(3.1)|∇F(t,x)|≤p(t)|x|α+q(t),∀(t,x)∈Z[1,T]×RN,
where  ℤ[a,b]:=ℤ∩[a,b] for every  a,b∈ℤ with  a≤b;(F3)′
liminf|x|→∞|x|-2α∑t=1TF(t,x)>((λ[T/2]+2λ1)/2λ12)∑t=1Tp2(t) for all  t∈ℤ[1,T].
Then problem (1.1) has at least one periodic solution with period  T.Theorem 3.2.
Suppose thatF satisfies  (F1) and  (F2) with  α=1. Moreover, assume the following conditions hold:
(3.2)∑t=1Tp(t)<λ1,
and (F4)
liminf|x|→∞|x|-2∑t=1TF(t,x)>((λ[T/2]+λ11/2+∑t=1Tp(t)+λ11/2(λ1-∑t=1Tp(t)))/2λ1(λ1-∑t=1Tp(t)))∑t=1Tp2(t) for all  t∈ℤ[1,T].
Then problem (1.1) has at least one periodic solution with period  T.Remark 3.3.
It is easy to see that (F2)′ is more general than (F2) and (F3)′ is weaker than (F3). Theorem3.2 is a new result, which completes Theorem A when α=1.
For the sake of convenience, we denote(3.3)M1=(∑t=1Tp2(t))1/2,M2=∑t=1Tp(t),M3=∑t=1Tq(t).Proof of Theorem3.1.
First we prove thatφ satisfies the (PS) condition. Suppose that a consequence {un}⊂HT is such that -C3≤φ(un)≤C3, where C3>0 and φ′(un)→0 as n→∞. Then for sufficiently large n,
(3.4)-‖u‖≤〈φ′(un),u〉≤‖u‖.
From Lemma 2.1, we can write u as u=u¯+ũ∈H0⨁H0⊥, where H0=N0,andH0⊥=⨁k=1[T/2]Nk. From (F3)′, we can choose a1>1/λ12>0 such that
(3.5)lim⁡⁡inf⁡|x|→∞⁡|x|-2α∑t=1TF(t,x)>(λ[T/2]a12+a1)M12.
From (F2)′, (2.6), Hölder inequality, and Young inequality, we have
(3.6)|∑t=1T(F(t,u(t))-F(t,u¯))|=|∑t=1T∫01(∇F(t,u¯+sũ(t)),ũ(t))ds|≤∑t=1T∫01|(∇F(t,u¯+sũ(t)),ũ(t))|ds≤∑t=1T∫01(p(t)|u¯+sũ(t)|α+q(t))|ũ(t)|ds≤∑t=1Tp(t)(|u¯|α+|ũ(t)|α)|ũ(t)|+∑t=1Tq(t)|ũ(t)|≤|u¯|α(∑t=1Tp2(t))1/2(∑t=1T|ũ(t)|2)1/2+M2‖ũ‖∞α+1+M3‖ũ‖∞=M1|u¯|α‖ũ‖+M2‖ũ‖∞α+1+M3‖ũ‖∞≤12a1‖ũ‖2+a12M12|u¯|2α+M2‖ũ‖α+1+M3‖ũ‖.
In a similar way, we have
(3.7)∑t=1T(∇F(t,un(t)),ũn(t))≤12a1λ1‖ũn‖2+a1λ12M12|u¯n|2α+M2‖ũn‖α+1+M3‖ũn‖.
Let un=u¯n+ũn∈H0⨁H0⊥. From (2.12) and (3.7), we have
(3.8)∑t=1T(Δun(t),Δũn(t))=∑t=1T(Δũn(t),Δũn(t))≥λ1‖ũn‖2,(3.9)∑t=1T(Δun(t),Δũn(t))=-〈φ′(un),ũn〉+∑t=1T(∇F(t,un(t)),ũn(t))≤‖ũn‖+12a1λ1‖ũn‖2+a1λ12M12|u¯n|2α+M2‖ũn‖α+1+M3‖ũn‖.
It follows from (3.8) and (3.9) that
(3.10)λ12‖ũn‖2+C4≤a1λ12M12|u¯n|2α,
where C4=mins∈[0,+∞){((a1λ12-1)/2a1λ1)s2-M2sα+1-(1+M3)s}. The fact that a1>1/λ12>0 implies that -∞<C4<0. So it follows from (3.10) that
(3.11)‖ũn‖2≤a1M12|u¯n|2α-2C4λ1,
and so
(3.12)‖ũn‖≤a1M1|u¯n|α+C5,
where C5>0. It follows from the boundedness of φ(un), (2.11), (3.6), (3.11), and (3.12) that
(3.13)C3≥φ(u)=-12∑t=1T|Δun(t)|2+∑t=1T[F(t,un(t))-F(t,0)]=-12∑t=1T|Δũn(t)|2+∑t=1T[F(t,un(t))-F(t,u¯n)]+∑t=1T[F(t,u¯n)-F(t,0)]≥-12λ[T/2]‖ũn‖-12a1‖ũn‖2-a12M12|u¯n|2α-M2‖ũn‖α+1-M3‖ũn‖+∑t=1T[F(t,u¯n)-F(t,0)]≥-(12λ[T/2]+12a1)(a1M12|u¯n|2α-2C4λ1)+∑t=1T[F(t,u¯n)-F(t,0)]-a12M12|u¯n|2α-M2(a1M1|u¯n|α+C5)α+1-M3(a1M1|u¯n|α+C5)≥(-12λ[T/2]a1M12-a1M12)|u¯n|2α+λ[T/2]C4λ1+C4λ1a1-M2(2α(a1M1)α+1|u¯n|α(α+1)+2αC5α+1)-M1M3a1|u¯n|α-M3C5+∑t=1T[F(t,u¯n)-F(t,0)]=|u¯n|2α[|u¯n|-2α∑t=1TF(t,u¯n)-(12λ[T/2]a1+a1)M12-M1M3a1|u¯n|-α-M22α(a1M1)α+1|u¯n|α(α-1)∑t=1TF(t,u¯n)]+λ[T/2]C4λ1+C4λ1a1-M3C5-M22αC5α+1-∑t=1TF(t,0).
Inequalities (3.5) and (3.13) imply that {u¯n} is bounded. Hence, {ũn} is bounded by (3.12), and then {un} is bounded. Since HT is finite dimensional, there exists a subsequence of {un} convergent in HT. Thus, we conclude that (PS) condition is satisfied.
In order to use the saddle point theorem [16, Theorem  4.6], we only need to verify the following conditions: (I1)
φ(x)→+∞as|x|→∞inH0;(I2)
φ(u)→-∞as∥u∥→∞inH0⊥.
In fact, from (F3)′, we have(3.14)∑t=1TF(t,x)⟶+∞as|x|⟶∞inH0.
For any x∈H0, since ∑t=1T|Δx|2=0, we have
(3.15)φ(x)=∑t=1T[F(t,x)-F(t,0)].
It follows from (3.14) and the above inequality that
(3.16)φ(x)⟶+∞as|x|⟶∞inH0.
Thus (I1) is easy to verify.
Next, for allu∈H0⊥, from (F2)′ and (2.6), we have
(3.17)|∑t=1T(F(t,u(t))-F(t,0))|=|∑t=1T∫01(∇F(t,su(t)),u(t))ds|≤∑t=1Tp(t)|u(t)|α+1+∑t=1Tq(t)|u(t)|≤M2‖u‖∞α+1+M2‖u‖∞≤M2‖u‖α+1+M2‖u‖.
By (2.7), (2.12), and (3.17), we obtain
(3.18)φ(u)=-12∑t=1T|Δu(t)|2+∑t=1T[F(t,u(t))-F(t,0)]≤-12λ1‖u‖2+M2‖u‖α+1+M2‖u‖.
Since λ1>0 and α∈[0,1), we have φ(u)→-∞ as ∥u∥→∞ in H0⊥. The proof of Theorem 3.1 is complete.Proof of Theorem3.2.
By (3.2) and (F4), we can choose an a2∈ℝ such that
(3.19)a2>1λ1>0,(3.20)lim⁡⁡inf⁡|x|→∞⁡|x|-2∑t=1TF(t,x)>[(12λ[T/2]+12a2+12M2)a2λ1-M2+a22]M12.
It follows from (F2)′ with α=1, (2.6), Hölder inequality, and Young inequality that
(3.21)|∑t=1T(F(t,u(t))-F(t,u¯))|=|∑t=1T∫01(∇F(t,u¯+sũ(t)),ũ(t))ds|≤∑t=1T∫01|(∇F(t,u¯+sũ(t)),ũ(t))|ds≤∑t=1T∫01p(t)(|u¯|+s|ũ(t)|)|ũ(t)|ds+∑t=1T∫01q(t)|ũ(t)|ds=∑t=1Tp(t)(|u¯|+12|ũ(t)|)|ũ(t)|+∑t=1Tq(t)|ũ(t)|≤|u¯|(∑t=1Tp2(t))1/2(∑t=1T|ũ(t)|2)1/2+12‖ũ‖∞2∑t=1Tp(t)+‖ũ‖∞∑t=1Tq(t)=M1|u¯|‖ũ‖+M22‖ũ‖∞2+M3‖ũ‖∞≤(12a2+M22)‖ũ‖2+a22M12|u¯|2+M3‖ũ‖.
In a similar way, we have
(3.22)∑t=1T(∇F(t,un(t)),ũn(t))≤(12a2+M22)‖ũn‖2+a22M12|u¯n|2+M3‖ũn‖.
From (3.8) and (3.22), we have
(3.23)λ1‖ũn‖2≤∑t=1T(Δun(t),Δũn(t))=-〈φ′(un),ũn〉+∑t=1T(∇F(t,un(t)),ũn(t))≤‖ũn‖+(12a2+M22)‖ũn‖2+a22M12|u¯n|2+M3‖ũn‖.
It follow from (3.23) that
(3.24)12(λ1-M2)‖ũn‖2+C6≤a22M12|u¯n|2,
where C6=mins∈[0,+∞){((λ1a2-1)/2a2)s2-(1+M3)s}. The fact that a2>1/λ1>0 implies that -∞<C6<0. So it follows from (3.24) that
(3.25)‖ũn‖2≤a2λ1-M2M12|u¯n|2-2C6λ1-M2,
and so
(3.26)‖ũn‖≤a2λ1-M2M1|u¯n|+C7,
where C7>0. It follows from the boundedness of φ(un), (2.11), (3.21), (3.25), and (3.26) that
(3.27)C3≥φ(u)=-12∑t=1T|Δun(t)|2+∑t=1T[F(t,un(t))-F(t,0)]=-12∑t=1T|Δũn(t)|2+∑t=1T[F(t,un(t))-F(t,u¯n)]+∑t=1T[F(t,u¯n)-F(t,0)]≥-12λ[T/2]‖ũn‖2-(12a2+M22)‖ũn‖2-a22M12|u¯n|2-M3‖ũn‖+∑t=1T[F(t,u¯n)-F(t,0)]≥-(12λ[T/2]+12a2+M22)(a2λ1-M2M12|u¯n|2-2C6λ1-M2)-a22M12|u¯n|2-M3(a2λ1-M2M1|u¯n|+C7)+∑t=1T[F(t,u¯n)-F(t,0)]=|u¯n|2[∑t=1TF(t,u¯n)|u¯n|-2α∑t=1TF(t,u¯n)-(12λ[T/2]+12a2+12M2)a2λ1-M2M12-a22M12-a2λ1-M2M1M3|u¯n|-1∑t=1TF(t,u¯n)]+(λ[T/2]+1a2+M2)C6λ1-M2-M3C7-∑t=1TF(t,0).
Inequalities (3.20) and (3.27) imply that {u¯n} is bounded. Hence, {ũn} is bounded by (3.26), and then {un} is bounded. Since HT is finite dimensional, there exists a subsequence of {un} convergent in HT. Thus, we conclude that (PS) condition is satisfied.
In the following, we prove thatφ satisfies (I1) and (I2). In fact, from (F4), we have
(3.28)∑t=1TF(t,x)⟶+∞as|x|⟶∞inH0.
It follows from (3.27) and ∑t=1T|Δx|2=0 that
(3.29)φ(x)=∑t=1T[F(t,x)-F(t,0)]⟶+∞as|x|⟶∞inH0.
Thus (I1) is easy to verify.
Next, for allu∈H0⊥, from (F2)′ with α=1 and (2.6), we have
(3.30)|∑t=1T(F(t,u(t))-F(t,0))|=|∑t=1T∫01(∇F(t,su(t)),u(t))ds|≤12∑t=1Tp(t)|u(t)|2+∑t=1Tq(t)|u(t)|≤12M2‖u‖∞2+M3‖u‖∞≤12M2‖u‖2+M3‖u‖.
By (2.7), (2.12), and (3.30), we obtain
(3.31)φ(u)=-12∑t=1T|Δu(t)|2+∑t=1T[F(t,u(t))-F(t,0)]≤-12λ1‖u‖2+M22‖u‖2+M2‖u‖.
Since λ1>M2, we have φ(u)→-∞ as ∥u∥→∞ in H0⊥. The proof of Theorem 3.2 is complete.


## 4. Examples

In this section, we give two examples to illustrate our results.Example 4.1.
Let(4.1)F(t,x)=sin⁡2πtT|x|7/4+(sin⁡2πtT+1)|x|3/2+(h(t),x),
where h:ℤ[1,T]→ℝN and h(t+T)=h(t). It is easy to see that
(4.2)|∇F(t,x)|≤74|sin⁡2πtT||x|3/4+32|sin2πtT+1||x|1/2+|h(t)|≤74(|sin2πtT|+ε)|x|3/4+a(ε)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ε>0 and a(ε) is a positive constant and is dependent on ε. It is easy to see that F(t,x) satisfies (F1). From (4.2), we can let p, q, and α be
(4.3)p(t)=74(|sin⁡2πtT|+ε),q(t)=a(ε)+|h(t)|,α=34,
which shows that (F2)′ is satisfied. Moreover, if we let T=2, then we have
(4.4)lim⁡⁡inf⁡|x|→+∞⁡|x|-2α∑t=1TF(t,x)=2,λ1=λ[T/2]=4,λ[T/2]+2λ12λ12∑t=1Tp2(t)=147128∑t=1T(|sin2πtT|+ε)2=147128ε2.
If we let ɛ2<256/147, then we obtain
(4.5)lim⁡⁡inf⁡|x|→+∞⁡|x|-2α∑t=1TF(t,x)=2>147128ε2=λ[T/2]+2λ12λ12∑t=1Tp2(t),
which shows that (F3)' holds. Then from Theorem 3.1, problem (1.1) has at least one periodic solution with period T.Example 4.2.
Let(4.6)F(t,x)=14(sin⁡2πtT+12)|x|2+sin⁡(2πtT)|x|3/2+(h(t),x),
where h: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)|≤12|sin2πtT+12||x|+32|sin2πtT||x|1/2+|h(t)|≤12(|sin⁡2πtT+12|+ε)|x|+b(ε)+|h(t)|,∀(t,x)∈Z[1,T]×RN,
where ɛ>0 and b(ε) is a positive constant and is dependent on ε. The above shows that (F2)′ holds with α=1 and
(4.8)p(t)=12(|sin⁡2πtT+12|+ε),q(t)=b(ε)+|h(t)|.
Let T=2, then λ0=0,λ1=λ[T/2]=4. Observe that
(4.9)|x|-2∑t=1TF(t,x)=|x|-2∑t=1T(14(sin⁡2πtT+12)|x|2+sin⁡(2πtT)|x|3/2+(h(t),x))=14+(∑t=1Th(t),|x|-2x).
On the other hand, we have
(4.10)∑t=1Tp(t)=∑t=1T12(|sin⁡2πtT+12|+ε)=12+ε,∑t=1Tp2(t)=∑t=1T14(|sin⁡2πtT+12|+ε)2=12(12+ε)2.
We can choose ε sufficiently small such that ∑t=1Tp(t)<4 and
(4.11)lim⁡⁡inf⁡|x|→+∞⁡|x|-2∑t=1TF(t,x)=14>27-2ε16(7-ε)(12+ε)2=λ[T/2]+λ11/2+∑t=1Tp(t)+λ11/2(λ1-∑t=1Tp(t))2λ1(λ1-∑t=1Tp(t))∑t=1Tp2(t),
which shows that (F4) holds. Then from Theorem 3.2, problem (1.1) has at least one periodic solution with period T.



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*Source: 463480-2011-08-03.xml*