Subharmonics with Minimal Periods for Convex Discrete Hamiltonian Systems

Abstract

We consider the subharmonics with minimal periods for convex discrete Hamiltonian systems. By using variational methods and dual functional, we obtain that the system has a 𝑝𝑇-periodic solution for each positive integer 𝑝, and solution of system has minimal period 𝑝𝑇 as 𝐻 subquadratic growth both at 0 and infinity.

1. Introduction

Consider Hamiltonian systems 𝐽𝑢̇(𝑡)+∇𝐻(𝑡,𝑢(𝑡))=0,𝑢(0)=𝑢𝑝𝑇, (1) where 𝑢(𝑡)∈ℝ^2𝑁, 𝑡∈ℝ, ∇𝐻 stands for the gradient of 𝐻 with respect to the second variable, and 𝐽=^0−𝐼𝑁_𝐼𝑁0 is the symplectic matrix with 𝐼_𝑁 the identity in ℝ^𝑁. Moreover, 𝐻 is 𝑇-periodic in the variable 𝑡, 𝑝∈ℕ⧵{0}.

Classically, solutions for systems (1) are called subharmonics. The first result concerning the subharmonics problem traced back to Birkhoff and Lewis in 1933 (refer to [1]), in which there exists a sequence of subharmonics with arbitrarily large minimal period, using perturbation techniques. More results can be found in [1–5], where 𝐻 is convex with subquadratic growth both at 0 and infinity. Using 𝑍_𝑝 index theory and Clarke duality, Xu and Guo [1, 5] proved that the number of solutions for systems (1) with minimal period 𝑝𝑇 tends towards infinity as 𝑝→∞.

For periodic and subharmonic solutions for discrete Hamiltonian systems, Guo and Yu [6, 7] obtained some existence results by means of critical point theory, where they introduced the action functional 𝐹(𝑢)1=−2^𝑝𝑇_𝑛=1(𝐽Δ𝐿𝑢(𝑛−1),𝑢(𝑛))−^𝑝𝑇_𝑛=1𝐻(𝑛,𝐿𝑢(𝑛)). (2) Using Clarke duality, periodic solution of convex discrete Hamiltonian systems with forcing terms has been investigated in [8]. Clarke duality was introduced in 1978 by Clarke [9], and developed by Clarke, Ekeland, and others, see [10–12]. This approach is different from the direct method of variations; some scholars applied it to consider the periodic solutions, subharmonic solutions with prescribed minimal period of Hamiltonian systems; one can refer to [3, 5, 12–14] and references therein. The dynamical behavior of differential and difference equations was studied by using various methods; see [15–19]. We refer the reader to Agarwal [20] for a broad introduction to difference equations.

Motivated by the works of Mawhin and Willem [12] and Xu and Guo [21], we use variational methods and Clarke duality to consider the subharmonics with minimal periods for discrete Hamiltonian systems 𝐽Δ𝑢(𝑛)+∇𝐻(𝑛,𝐿𝑢(𝑛))=0,𝑢(𝑛)=𝑢𝑛+𝑝𝑇, (3) where 𝑢(𝑛)=^𝑢1_𝑢^(𝑛)_2(𝑛), 𝐿𝑢(𝑛)=^𝑢1_𝑢^(𝑛+1)_2(𝑛), 𝑢_𝑖(𝑛)∈ℝ^𝑁(𝑖=1,2) with 𝑁 a given positive integer, and Δ𝑢(𝑛)=𝑢(𝑛+1)−𝑢(𝑛) is the forward difference operator. 𝑝,𝑇∈ℕ⧵{0}. Moreover, hamiltonian function 𝐻 satisfies the following assumption:(A1)𝐻∶ℤ×ℝ^2𝑁→ℝ is continuous differentiable on ℝ^2𝑁, 𝐻(𝑛,⋅) convex for each 𝑛∈ℤ and 𝐻(𝑛+𝑇,𝑢)=𝐻(𝑛,𝑢) for each 𝑢∈ℝ^2𝑁;(A2) there exist constants 𝑎₁>0, 𝑎₂>0, 1<𝜃<2, such that 𝑎₁𝜃|𝑢|^𝜃()≤𝑎≤𝐻𝑛,𝑢²𝜃|𝑢|^𝜃,𝑢∈ℝ^2𝑁, (4) which implies 𝐻 subquadratic growth both at 0 and infinity.

Theorem 1. Assume (A1) holds. 𝐻(𝑛,𝑢)→+∞, 𝐻(𝑛,𝑢)/|𝑢|²→0, as |𝑢|→∞ uniformly in 𝑛∈ℤ. Then there exists a 𝑝𝑇-periodic solution 𝑢_𝑝 of (3), such that ‖𝑢_𝑝‖_∞≜max_{𝑛∈𝑍[1,𝑝𝑇]}{|𝑢_𝑝(𝑛)|}→∞, and the minimal period 𝑇_𝑝 of 𝑢_𝑝 tends to +∞ as 𝑝→∞.

Theorem 2. Under the assumptions (A1) and (A2), if 𝑎₂𝑎₁≤14𝜋sin𝑝𝑇^𝜃/21,𝑤ℎ𝑒𝑛𝑝𝑇𝑖𝑠𝑒𝑣𝑒𝑛,2𝜋sin2𝑝𝑇^𝜃/2,𝑤ℎ𝑒𝑛𝑝𝑇𝑖𝑠𝑜𝑑𝑑 (5) for given integer 𝑝>1, then the solution of (3) has minimal period 𝑝𝑇.

2. Clarke Duality and Eigenvalue Problem

First we introduce a space 𝐸_𝑝𝑇 with dimension 2𝑁𝑝𝑇 as follows: 𝐸_𝑝𝑇={𝑢(𝑛(𝑛){0},𝑢=)}∈𝑆∣𝑢𝑛+𝑝𝑇=𝑢,𝑝∈ℕ⧵,𝑛∈ℤ (6) where {𝑢(𝑛(𝑛)=𝑢𝑆=𝑢=)}∣𝑢¹(𝑛)𝑢₂(𝑛)∈ℝ^2𝑁,𝑢_𝑗(𝑛)∈ℝ^𝑁,𝑗=1,2,𝑛∈ℤ. (7) Equipped with inner product ⟨⋅,⋅⟩ and norm ‖⋅‖ in 𝐸_𝑝𝑇 as ⟨𝑢,𝑣⟩=^𝑝𝑇_𝑛=1(𝑢(𝑛)(𝑛,‖𝑢‖=,𝑣))_{𝑝𝑇𝑛=1}|𝑢(𝑛)|^21/2,∀𝑢,𝑣∈𝐸_𝑝𝑇, (8) where (⋅,⋅) and |⋅| denote the usual scalar product and corresponding norm in ℝ^2𝑁, respectively. It is easy to see that (𝐸_𝑝𝑇,⟨⋅,⋅⟩) is a Hilbert space with 2𝑁𝑝𝑇 dimension, which can be identified with ℝ^{2𝑁𝑝𝑇}. Then for any 𝑢∈𝐸_𝑝𝑇, it can be written as 𝑢=(𝑢^𝑇(1),𝑢^𝑇(2),…,𝑢^𝑇(𝑝𝑇))^𝑇, where 𝑢(𝑗)=^𝑢1_𝑢^(𝑗)_2(𝑗)∈ℝ^2𝑁, 𝑗∈𝑍[1,𝑝𝑇], the discrete interval {1,2,…,𝑝𝑇} is denoted by 𝑍[1,𝑝𝑇], and ⋅^𝑇 denotes the transpose of a vector or a matrix.

Denote the subspace 𝑌={𝑢∈𝐸_𝑝𝑇∣𝑢(1)=𝑢(2)=⋯=𝑢(𝑝𝑇)∈ℝ^2𝑁}. Let 𝑌 be the direct orthogonal complement of 𝐸_𝑝𝑇 to 𝑌. Thus 𝐸_𝑝𝑇 can be split as 𝐸_𝑝𝑇=𝑌⊕𝑌, and for any 𝑢∈𝐸_𝑝𝑇, it can be expressed in the form 𝑢=𝑢^∼+𝑢, where 𝑢^∼∈𝑌,  𝑢∈𝑌.

Next we recall Clarke duality and some lemmas.

The Legendre transform (see [12]) 𝐻^∗(𝑡,⋅) of 𝐻(𝑡,⋅) with respect to the second variable is defined by 𝐻^∗(𝑡,𝑣)=sup_{𝑢∈𝑅2𝑁}{(𝑣,𝑢)−𝐻(𝑡,𝑢)}, (9) where (⋅,⋅) denotes the inner product in ℝ^2𝑁. It follows from (A1) and (A2) that(B1)𝐻^∗(𝑛,⋅) is continuous differentiable on ℝ^2𝑁,(B2) for 𝜏=𝜃/(𝜃−1),  𝑣∈ℝ^2𝑁,  𝑛∈ℤ, one has 1𝜏1𝑎₂^𝜏−1|𝑣|^𝜏≤𝐻^∗()≤1𝑛,𝑣𝜏1𝑎₁^𝜏−1|𝑣|^𝜏. (10)

Associated with variational functional (2), the dual action functional is defined by 𝜒(𝑣)=^𝑝𝑇_𝑛=112(𝐿(((𝑛+𝐽Δ𝑣𝑛−1)),𝑣))_{𝑝𝑇𝑛=1}𝐻^∗(𝑛,Δ𝑣(𝑛)),𝑣∈𝐸_𝑝𝑇. (11) Indeed, by (11), we have 𝜒(𝑣+𝑢)=𝜒(𝑣) for any 𝑢∈𝑌 and 𝑣∈𝑌. Therefore, 𝜒 can be restricted in subspace 𝑌 of 𝐸_𝑝𝑇. Moreover, in terms of Lemma  2.6 in [8] and the following lemma, the critical points of (11) correspond to the subharmonic solutions of (3).

Lemma 3 (see [8, Theorem 1]). Assume that(H1)𝐻(𝑛,⋅)∈𝐶¹(ℝ^2𝑁,ℝ), 𝐻(𝑛,⋅) is convex in the second variable for 𝑛∈ℤ, (H2) there exists 𝛽∶ℤ→ℝ^2𝑁 such that for all (𝑛,𝑢)∈ℤ×ℝ^2𝑁, 𝐻(𝑛,𝑢)≥(𝛽(𝑛),𝑢), and 𝛽(𝑛+𝑇)=𝛽(𝑛), (H3) there exist 𝛼∈(0,2sin(𝜋/𝑝𝑇)) and 𝛾∶ℤ→ℝ⁺, such that for any (𝑛,𝑢)∈ℤ×ℝ^2𝑁, 𝐻(𝑛,𝑢)≤(𝛼/2)|𝑢|²+𝛾(𝑛), and 𝛾(𝑛+𝑇)=𝛾(𝑛), (H4) for each 𝑢∈ℝ^2𝑁, ∑^𝑝𝑇_𝑛=1𝐻(𝑛,𝑢)→+∞ as |𝑢|→∞.

Then system (3) has at least one periodic solution 𝑢, such that 𝑣=−𝐽[𝑢−(1/𝑝𝑇)∑^𝑝𝑇_𝑛=1𝑢(𝑛)] minimizes the dual action 𝜒(𝑣)=∑^𝑝𝑇_𝑛=1(1/2)(𝐿𝐽Δ𝑣(𝑛−1),𝑣(𝑛))+∑^𝑝𝑇_𝑛=1𝐻^∗(𝑛,Δ𝑣(𝑛)).

The following lemmas will be useful in our proofs, where Lemma 4 can be proved by means of Euler formula 𝑒^𝑖𝜃=cos𝜃+𝑖sin𝜃, and Lemma 5 is a Hölder inequality.

Lemma 4. For any 𝑘∈ℤ, ∑^𝑝𝑇_𝑛=1sin((2𝑘𝜋/𝑝𝑇)𝑛)=∑^𝑝𝑇_𝑛=1cos((2𝑘𝜋/𝑝𝑇)𝑛)=0.

Lemma 5. For any 𝑢_𝑗>0, 𝑣_𝑗>0, 𝑘∈ℤ, one has ∑^𝑘_𝑗=1𝑢_𝑗𝑣_𝑗≤(∑^𝑘_𝑗=1𝑢^𝑝_𝑗)^1/𝑝(∑^𝑘_𝑗=1𝑣^𝑞_𝑗)^1/𝑞, where 𝑝>1, 𝑞>1 and 1/𝑝+1/𝑞=1.

Lemma 6 (see [12, proposition 2.2]). Let 𝐻∶ℝ^𝑚→ℝ be of 𝐶¹ and convex functional, −𝛽≤𝐻(𝑢)≤𝛼𝑞⁻¹|𝑢|^𝑞+𝛾, where 𝑢∈ℝ^𝑚, 𝛼>0, 𝑞>1, 𝛽≥0, 𝛾≥0. Then 𝛼^−𝑝/𝑞𝑝⁻¹|∇𝐻(𝑢)|^𝑝≤(∇𝐻(𝑢),𝑢)+𝛽+𝛾, where 1/𝑝+1/𝑞=1.

In order to know the form of 𝑢∈𝐸_𝑝𝑇, we consider eigenvalue problem 𝐿𝐽Δ𝑢(𝑛−1)=𝜆𝑢(𝑛),𝑢𝑛+𝑝𝑇=𝑢(𝑛), (12) where 𝑢(𝑛)=^𝑢1_𝑢^(𝑛)_2(𝑛), 𝐿𝑢(𝑛−1)=^𝑢1_𝑢^(𝑛)_2(𝑛−1)∈ℝ^2𝑁, 𝑛∈ℤ, 𝜆∈ℝ. We can rewrite (12) as the following form: 𝑢₁(𝑛+1)=1−𝜆²𝑢₁(𝑛)+𝜆𝑢₂(𝑛),𝑢₂(𝑛+1)=−𝜆𝑢₁(𝑛)+𝑢₂(𝑛),𝑢₁𝑛+𝑝𝑇=𝑢₁(𝑛),𝑢₂𝑛+𝑝𝑇=𝑢₂(𝑛). (13) Denoting 𝑀(𝜆)=1−𝜆²𝐼_𝑁𝜆𝐼_𝑁−𝜆𝐼_𝑁𝐼_𝑁, (14) then problem (12) is equivalent to 𝑢(𝑛+1)=𝑀(𝜆)𝑢(𝑛),𝑢𝑛+𝑝𝑇=𝑢(𝑛). (15) Letting 𝑢(𝑛)=𝜇^𝑛𝑐 be the solution of (15), for some 𝑐∈ℂ^2𝑁, we have 𝜇𝑐=𝑀(𝜆)𝑐 and 𝜇^𝑝𝑇=1. Consider the polynomial |𝑀(𝜆)−𝜇𝐼_2𝑁|=0 and condition 𝜇^𝑝𝑇=1; it follows that 𝜇=𝑒^{2𝑘𝜋𝑖/𝑝𝑇},𝜆=2sin𝑘𝜋,.𝑝𝑇𝑘∈𝑍−𝑝𝑇+1,𝑝𝑇−1 (16)

In the following we denote by 𝜇_𝑘=𝑒^{2𝑘𝜋𝑖/𝑝𝑇}, 𝜆_𝑘=2sin(𝑘𝜋/𝑝𝑇), 𝑘∈𝑍[−𝑝𝑇+1,𝑝𝑇−1], and 𝜌∈ℝ^𝑁. By (𝑀(𝜆_𝑘)−𝜇_𝑘𝐼_2𝑁)𝑐=0, it follows that 𝑐_𝑘=𝜌𝑖𝑒^{(−𝑘𝜋𝑖/𝑝𝑇)}𝜌. (17) Thus 𝑢_𝑘(𝑛)=𝜇^𝑛_𝑘𝑐_𝑘=𝑒^{2𝑘𝜋𝑛𝑖/𝑝𝑇}𝜌𝑖𝑒^{(−𝑘𝜋𝑖/𝑝𝑇)}𝜌=cos2𝑘𝜋𝑛𝜌𝑝𝑇−sin2𝑘𝜋1𝑝𝑇𝑛−2𝜌+𝑖sin2𝑘𝜋𝑛𝜌𝑝𝑇cos2𝑘𝜋1𝑝𝑇𝑛−2𝜌. (18)

Let 𝜉_𝑘(𝑛)=cos2𝑘𝜋𝑛𝜌𝑝𝑇−sin2𝑘𝜋1𝑝𝑇𝑛−2𝜌,𝜂_𝑘=sin2𝑘𝜋𝑛𝜌𝑝𝑇cos2𝑘𝜋1𝑝𝑇𝑛−2𝜌. (19) Obviously, 𝜉_𝑘(𝑛) and 𝜂_𝑘(𝑛) satisfy (15). Moreover 𝐿𝐽Δ𝜉_𝑘(𝑛−1)=𝜆_𝑘𝜉_𝑘(𝑛), 𝐿𝐽Δ𝜂_𝑘(𝑛−1)=𝜆_𝑘𝜂_𝑘(𝑛), 𝜉_{2𝑝𝑇+𝑘}(𝑛)=𝜉_𝑘(𝑛), 𝜂_{2𝑝𝑇+𝑘}(𝑛)=𝜂_𝑘(𝑛), 𝜉_{𝑝𝑇−𝑘}(𝑛)=𝜉_𝑘(𝑛), 𝜂_{𝑝𝑇−𝑘}(𝑛)=−𝜂_𝑘(𝑛).

For 𝑘≠𝑝𝑇/2, subspace 𝑌_𝑘 is defined by𝑌_𝑘=span𝜉_𝑘(𝑛),𝜂_{𝑘+(𝑝𝑇/2)}(𝑛),𝑘∈𝑍−𝑝𝑇2+1,𝑝𝑇2⧵{0},𝜉−1𝑛∈ℤ,if𝑝𝑇iseven,span_𝑘(𝑛),𝜂_{𝑘+((𝑝𝑇+1)/2)}(𝑛),𝑘∈𝑍−𝑝𝑇2,𝑝𝑇2⧵{0},𝑛∈ℤ,if𝑝𝑇isodd, (20)where [⋅] denotes the greatest-integer function and 𝑌_𝑝𝑇/2=span𝜉_𝑝𝑇/2(𝑛),𝑌,𝑛∈ℤ_{−𝑝𝑇/2}=span𝜉_{−𝑝𝑇/2}(𝑛),𝑛∈ℤ. (21) Therefore, 𝑌=⊕𝑌_𝑘,𝑘∈𝑍−𝑝𝑇2,𝑝𝑇2⧵{0},if𝑝𝑇iseven,𝑌=⊕𝑌_𝑘,𝑘∈𝑍−𝑝𝑇2,𝑝𝑇2⧵{0},if𝑝𝑇isodd. (22) Moreover, for any 𝑢={𝑢(𝑛)}∈𝐸_𝑝𝑇, we may express 𝑢(𝑛) as 𝑢(𝑛)=^𝑝𝑇−1_{𝑘=−𝑝𝑇+1}cos2𝑘𝜋𝑛𝑎𝑝𝑇^𝑘−sin2𝑘𝜋1𝑝𝑇𝑛−2𝑎_𝑘+sin2𝑘𝜋𝑛𝑏𝑝𝑇^𝑘cos2𝑘𝜋1𝑝𝑇𝑛−2𝑏_𝑘, (23) where 𝑎_𝑘,𝑏_𝑘∈ℝ^𝑁.

Since (Δ𝑢(𝑛),Δ𝑢(𝑛))=−(Δ²𝑢(𝑛−1),𝑢(𝑛)), we consider eigenvalue problem −Δ²𝑢(𝑛−1)=𝜆𝑢(𝑛),𝑢𝑛+𝑝𝑇=𝑢(𝑛),𝑢(𝑛)∈ℝ^𝑁, (24) where Δ²𝑢(𝑛−1)=Δ𝑢(𝑛)−Δ𝑢(𝑛−1)=𝑢(𝑛+1)−2𝑢(𝑛)+𝑢(𝑛−1). The second order difference equation (24) has complexity solution 𝑢(𝑛)=𝑒^𝑖𝑛𝜃𝑐 for 𝑐∈ℂ^𝑁, where 𝜃=2𝑘𝜋/𝑝𝑇. Moreover, 𝜆=2−𝑒^−𝑖𝜃−𝑒^𝑖𝜃=2(1−cos𝜃)=4sin²(𝜃/2); that is, 𝜆=4sin²(𝑘𝜋/𝑝𝑇),  𝑘∈𝑍[0,𝑝𝑇−1].

By the previous, it follows Lemma 7.

Lemma 7. For any 𝑢∈𝐸_𝑝𝑇, one has −𝜆_max‖𝑢‖²≤∑^𝑝𝑇_𝑛=1(𝐿𝐽Δ𝑢(𝑛−1),𝑢(𝑛))≤𝜆_max‖𝑢‖², and 0≤∑^𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²≤𝜆²_max‖𝑢‖², where 𝜆_max=max_{𝑘∈[0,𝑝𝑇−1]}2sin𝑘𝜋=𝜋𝑝𝑇2,𝑖𝑓𝑝𝑇𝑖𝑠𝑒𝑣𝑒𝑛,2cos2𝑝𝑇,𝑖𝑓𝑝𝑇𝑖𝑠𝑜𝑑𝑑. (25) Moreover, if 𝑢∈𝑌, then 4sin²(𝜋/𝑝𝑇)‖𝑢‖²≤∑^𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²≤𝜆²_max‖𝑢‖².

3. Proofs of Main Results

Lemma 8. Consider ^𝑝𝑇_𝑛=1(()(𝑛𝜋𝐿𝐽Δ𝑢𝑛−1,𝑢))≥−2sin𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²,∀𝑢∈𝐸_𝑝𝑇. (26)

Proof. Letting 𝑢^∼(𝑛)=𝑢(𝑛)−(1/𝑝𝑇)∑^𝑝𝑇_𝑛=1𝑢(𝑛), then 𝑢^∼∈𝑌. By Lemmas 5 and 7, we have ^𝑝𝑇_𝑛=1(()(𝑛=𝐿𝐽Δ𝑢𝑛−1,𝑢))_{𝑝𝑇𝑛=1}𝐿𝐽Δ𝑢(𝑛−1),𝑢^∼(𝑛)≥−^𝑝𝑇_𝑛=1|𝐿𝐽Δ𝑢(𝑛−1)|^21/2⋅𝑢^𝑝𝑇_𝑛=1^∼(𝑛)^21/2≥−^𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|^21/2⋅𝜋2sin𝑝𝑇^−1𝑝𝑇_𝑛=1Δ𝑢^∼(𝑛)^21/2𝜋=−2sin𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|². (27)

Lemma 9. If there exist 𝛼∈(0,sin(𝜋/𝑝𝑇)), 𝛽≥0 and 𝛿>0, such that 𝛿|𝑢|()≤𝛼−𝛽≤𝐻𝑛,𝑢2|𝑢|²+𝛾 (28) for all 𝑛∈[1,𝑝𝑇] and 𝑢∈ℝ^2𝑁, then each solution of (3) satisfies the inequalities ^𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²≤2𝛼𝛽+𝛾𝑝𝑇sin𝜋/𝑝𝑇,sin𝜋/𝑝𝑇−𝛼_{𝑝𝑇𝑛=1}|𝐿𝑢(𝑛)|≤𝛽+𝛾𝑝𝑇sin𝜋/𝑝𝑇𝛿.sin𝜋/𝑝𝑇−𝛼 (29)

Proof. Let 𝑢 be the solution of (3). By Lemma 6, we have 1|((𝑛2𝛼∇𝐻𝑛,𝐿𝑢))|²≤(((𝑛(𝑛((𝑛)(𝑛∇𝐻𝑛,𝐿𝑢)),𝐿𝑢))+𝛽+𝛾=−𝐽Δ𝑢,𝐿𝑢))+𝛽+𝛾. (30) Obviously, |𝐽Δ𝑢(𝑛)|²=(−∇𝐻(𝑛,𝐿𝑢(𝑛)),𝐽Δ𝑢(𝑛))=|∇𝐻(𝑛,𝐿𝑢(𝑛))|² by (3), and it follows that (1/2𝛼)∑^𝑝𝑇_𝑛=1|𝐽Δ𝑢(𝑛)|²+∑^𝑝𝑇_𝑛=1(𝐽Δ𝑢(𝑛),𝐿𝑢(𝑛))≤(𝛽+𝛾)𝑝𝑇; that is, 12𝛼^𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²+^𝑝𝑇_𝑛=1(()(𝑛≤𝐿𝐽Δ𝑢𝑛−1,𝑢))𝛽+𝛾𝑝𝑇. (31) By means of Lemma 8, we have 1−𝜋2𝛼2sin𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²≤𝛽+𝛾𝑝𝑇, (32) which gives first conclusion.
Now, 𝐻(𝑛,0)≤𝛾 in view of (28); therefore by convex and Lemma 8, we have 𝛿^𝑝𝑇_𝑛=1|(𝑛≤𝐿𝑢)|−𝛽𝑝𝑇_{𝑝𝑇𝑛=1}𝐻((𝑛≤𝑛,𝐿𝑢))_{𝑝𝑇𝑛=1}[𝐻(𝑛,0)+(∇𝐻(𝑛,𝐿𝑢(𝑛)),𝐿𝑢(𝑛))]≤𝛾𝑝𝑇−^𝑝𝑇_𝑛=1(𝐽Δ𝑢(𝑛),𝐿𝑢(𝑛))=𝛾𝑝𝑇−^𝑝𝑇_𝑛=1(()(𝑛𝜋𝐽𝐿Δ𝑢𝑛−1,𝑢))≤𝛾𝑝𝑇+2sin𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑢(𝑛)|²𝛼≤𝛾𝑝𝑇+𝛽+𝛾𝑝𝑇,sin𝜋/𝑝𝑇−𝛼 (33) which gives the second conclusion. The proof is completed.

Proof of Theorem 1. Let 𝑐₁=max_𝑛∈ℤ|𝐻(𝑛,0)|. By assumption in Theorem 1, there exists 𝑅>0, such that 𝐻(𝑛,𝑢)≥1+𝑐₁, for 𝑛∈ℤ and |𝑢|≥𝑅. Moreover, there exist 𝛼∈(0,2sin(𝜋/𝑝𝑇)), 𝛾>0 such that 𝐻()≤𝛼𝑛,𝑢2|𝑢|²+𝛾,∀(𝑛,𝑢)∈ℤ×ℝ^2𝑁. (34) Thus, by convex of 𝐻, for all (𝑛,𝑢)∈ℤ×ℝ^2𝑁 with |𝑢|≥𝑅, we have 1+𝑐₁𝑅≤𝐻𝑛,|𝑢|𝑢()+𝑅≤𝐻𝑛,0|𝑢|(𝐻()(≤𝑅𝑛,𝑢−𝐻𝑛,0))|𝑢|𝐻(𝑛,𝑢)+𝑐₁. (35) Therefore there exist 𝛽>0 and 𝛿>0, such that 𝐻(𝑛,𝑢)≥𝛿|𝑢|−𝛽,∀(𝑛,𝑢)∈ℤ×ℝ^2𝑁. (36) Combining the previous argument, by Lemma 3, the system (3) has a 𝑝𝑇-periodic solution 𝑢_𝑝 such that 𝑣_𝑝=−𝐽[𝑢_𝑝−(1/𝑝𝑇)∑^𝑝𝑇_𝑛=1𝑢_𝑝(𝑛)]∈𝑌 minimizes the dual action 𝜒_𝑝𝑣_𝑝=^𝑝𝑇_𝑛=112𝐿𝐽Δ𝑣_𝑝(𝑛−1),𝑣_𝑝(𝑛)+^𝑝𝑇_𝑛=1𝐻^∗𝑛,Δ𝑣_𝑝(𝑛)on𝐸_𝑝𝑇. (37) It follows that Δ𝑢_𝑝(𝑛)=𝐽Δ𝑣_𝑝(𝑛) and 𝐽𝑣_𝑝(𝑛)=𝑢_𝑝(𝑛)−(1/𝑝𝑇)∑^𝑝𝑇_𝑛=1𝑢_𝑝(𝑛).
We next prove that ‖𝑢_𝑝‖_∞→∞ as 𝑝→∞.
Suppose not, there exist 𝑐₂>0 and a subsequence {𝑝_𝑘} such that 𝑝_𝑘→∞,𝑢_𝑝𝑘∞≤𝑐₂as𝑘⟶∞. (38) In terms of (3), it follows that ‖Δ𝑢_𝑝𝑘‖_∞≤𝑐₃ for some 𝑐₃>0, and ‖𝑣_𝑝𝑘‖_∞≤2𝑐₂, ‖Δ𝑣_𝑝𝑘‖_∞≤𝑐₃. Consequently, by 𝐻^∗(𝑛,𝑣)≥−𝐻(𝑛,0)≥−𝑐₁, we have 𝑐_𝑝𝑘=𝜒_𝑝𝑘𝑣_𝑝𝑘=^𝑝𝑘𝑇_𝑛=112𝐿𝐽Δ𝑣_𝑝𝑘(𝑛−1),𝑣_𝑝𝑘(𝑛)+^𝑝𝑘𝑇_𝑛=1𝐻^∗𝑛,Δ𝑣_𝑝𝑘(𝑛)1≥−2^𝑝𝑘𝑇_𝑛=1𝐿𝐽Δ𝑣_𝑝𝑘(𝑛−1)𝑣_𝑝𝑘(𝑛)−𝑐₁𝑝_𝑘𝑇≥−√2𝑐₂𝑐₃+𝑐₁𝑝_𝑘𝑇, (39) where 𝑛∈𝑍[1,𝑝_𝑘𝑇] and 𝐿𝐽Δ𝑣_𝑝𝑘(𝑛−1)=Δ𝑣_2,𝑝𝑘(𝑛)²+Δ𝑣_1,𝑝𝑘(𝑛−1)^21/2≤√2Δ𝑣_𝑝𝑘∞≤√2𝑐₃. (40)
By (36), if |𝑣|≤𝛿, we have (𝑣,𝑢)−𝐻(𝑛,𝑢)≤(𝑣,𝑢)−𝛿|𝑢|+𝛽≤𝛽, and 𝐻^∗(𝑛,𝑣)≤𝛽. Letting 𝜌∈ℝ^𝑁 and |𝜌|=1, in terms of (12), ℎ_𝑝 associated with 𝜆₋₁=−2sin(𝜋/𝑝𝑇) is given by ℎ_𝑝(𝑛)=𝛿⋅4sin𝜋/𝑝𝑇cos2𝜋𝑝𝑇𝑛−sin2𝜋𝑛𝜌𝑝𝑇sin2𝜋1𝑝𝑇𝑛−2+cos2𝜋1𝑝𝑇𝑛−2𝜌 (41) which belongs to 𝐸_𝑝𝑇, andΔℎ_𝑝(𝑛)²=𝛿4sin(𝜋/𝑝𝑇)²⋅𝜋2sin𝑝𝑇−sin2𝜋1𝑝𝑇𝑛+2−cos2𝜋1𝑝𝑇𝑛+2𝜌cos2𝜋𝑝𝑇𝑛−sin2𝜋𝑛𝜌𝑝𝑇²=142+sin2𝜋()𝑝𝑇2𝑛+1−sin2𝜋(⋅𝜌𝑝𝑇2𝑛)²𝛿²≤𝛿². (42)Moreover, by Lemma 4 we have ^𝑝𝑇_𝑛=1ℎ_𝑝(𝑛)²=^𝑝𝑇_𝑛=1𝛿4sin(𝜋/𝑝𝑇)²⋅2+sin2𝜋()𝑝𝑇2𝑛−1−sin2𝜋()𝜌𝑝𝑇2𝑛²=𝛿4sin(𝜋/𝑝𝑇)²2𝜌²𝛿𝑝𝑇=²𝑝𝑇8sin².𝜋/𝑝𝑇 (43) Thus 𝑐_𝑝=𝜒_𝑝(ℎ_𝑝)≤∑^𝑝𝑇_𝑛=1(1/2)(𝐿𝐽Δℎ_𝑝(𝑛−1),ℎ_𝑝(𝑛))+𝛽𝑝𝑇=∑^𝑝𝑇_𝑛=1(1/2)(−2sin(𝜋/𝑝𝑇))|ℎ_𝑝(𝑛)|²+𝛽𝑝𝑇=−𝛿²𝑝𝑇/8sin(𝜋/𝑝𝑇)+𝛽𝑝𝑇. Combining (39), we have 8(√2𝑐₂𝑐₃+𝑐₁+𝛽₁)sin(𝜋/𝑝_𝑘𝑇)≥𝛿², which is impossible as 𝑘 large. So the claim lim_𝑝→∞‖𝑢_𝑝‖_∞=∞ is valid.
It remains only to prove that the minimal period 𝑇_𝑝 of 𝑢_𝑝 tends to +∞ as 𝑝→∞.
If not, there exists 𝑇>0 and a sequence {𝑝_𝑘} such that the minimal period 𝑇_𝑝𝑘 of 𝑢_𝑝𝑘 satisfies 1≤𝑇_𝑝𝑘≤𝑇. By assumption in Theorem 1, there exists 𝛼∈(0,sin(𝜋/𝑇)) and 𝛾>0 such that 𝐻()≤𝛼𝑛,𝑢2|𝑢|²+𝛾,∀(𝑛,𝑢)∈ℤ×ℝ^2𝑁. (44) By (36) and Lemma 9 with 𝑝𝑇 replaced by 𝑇_𝑝𝑘, we get ^𝑇𝑝𝑘_𝑛=1Δ𝑢_𝑝𝑘(𝑛)²≤2𝛼𝛽+𝛾𝑇_𝑝𝑘sin𝜋/𝑇_𝑝𝑘sin𝜋/𝑇_𝑝𝑘≤()−𝛼2𝛼𝛽+𝛾𝑇sin𝜋/𝑇(),sin𝜋/𝑇−𝛼_{𝑇𝑝𝑘𝑛=1}𝐿𝑢_𝑝𝑘(𝑛≤𝑇)𝛽+𝛾_𝑝𝑘sin𝜋/𝑇_𝑝𝑘𝛿sin𝜋/𝑇_𝑝𝑘≤𝑇−𝛼𝛽+𝛾_𝑝𝑘sin(𝜋/𝑇)𝛿(()).sin𝜋/𝑇−𝛼 (45)(46)
Write 𝑢_𝑝𝑘=𝑢^∼_𝑝𝑘+𝑢_𝑝𝑘, where 𝑢_𝑝𝑘=(1/𝑇_𝑝𝑘)∑^𝑇𝑝𝑘_𝑛=1𝑢_𝑝𝑘(𝑛)=(1/𝑇_𝑝𝑘)∑^𝑇𝑝𝑘_𝑛=1𝐿𝑢_𝑝𝑘(𝑛)∈𝑌. Inequality (46) implies that 𝑢_𝑝𝑘∞≜max_{𝑛∈𝑍1,𝑇𝑝𝑘}𝑢_𝑝𝑘≤1𝑇_𝑝𝑘^𝑇𝑝𝑘_𝑛=1𝐿𝑢_𝑝𝑘(𝑛≤())𝛽+𝛾sin𝜋/𝑇𝛿(()).sin𝜋/𝑇−𝛼 (47) By Lemma 7 and (45), it follows that 𝑢^∼_𝑝𝑘²=𝑢^𝑇𝑝𝑘_𝑛=1^∼_𝑝𝑘(𝑛)²≤𝜋2sin𝑇_𝑝𝑘^{𝑇−1𝑝𝑘}_𝑛=1Δ𝑢_𝑝𝑘(𝑛)²≤(2sin(𝜋/𝑇))⁻¹2𝛼𝛽+𝛾𝑇sin(𝜋/𝑇)()≤𝛼𝑇sin𝜋/𝑇−𝛼𝛽+𝛾(),sin𝜋/𝑇−𝛼 (48) which implies that {‖𝑢^∼_𝑝𝑘‖_∞} is bounded, therefore {‖𝑢_𝑝𝑘‖_∞} is bounded; a contradiction with the second claim lim_𝑝→∞‖𝑢_𝑝‖_∞=∞. This completes the proof.

Proof of Theorem 2. Under the assumptions (A1) and (A2), all conditions in Theorem 1 are satisfied. Then, for each integer 𝑝>1, there exists a 𝑝𝑇-periodic solution 𝑢 of (3) such that 𝑣=−𝐽[𝑢−(1/𝑝𝑇)∑^𝑝𝑇_𝑛=1𝑢(𝑛)]∈𝑌 minimizes the dual action 𝜒(𝑣)=^𝑝𝑇_𝑛=112(()(𝑛+𝐿𝐽Δ𝑣𝑛−1,𝑣))_{𝑝𝑇𝑛=1}𝐻^∗(𝑛,Δ𝑣(𝑛))on𝐸_𝑝𝑇. (49)
If the critical point 𝑣 of dual action functional 𝜒 has minimal period 𝑝𝑇/𝑙∈ℕ⧵{0}, where 𝑙∈ℕ⧵{0}, then by Lemma 7 with 𝑝𝑇 replaced by 𝑝𝑇/𝑙, we have the following estimate: 4sin²𝑙𝜋𝑝𝑇^𝑝𝑇_𝑛=1|𝑣(𝑛)|²≤^𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|². (50) By Lemma 5 and the previous inequality, we have ^𝑝𝑇_𝑛=1(𝐿𝐽Δ𝑣(𝑛−1),𝑣(𝑛))≥−^𝑝𝑇_𝑛=1|𝐿𝐽Δ𝑣(𝑛−1)|^21/2⋅^𝑝𝑇_𝑛=1|𝑣(𝑛)|^21/2≥−^𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|^21/2⋅2sin𝑙𝜋𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|^21/2=−2sin𝑙𝜋𝑝𝑇^−1𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|²≥−2sin𝑙𝜋𝑝𝑇⁻¹𝑝𝑇^{(1−2/𝜏)𝑝𝑇}_𝑛=1|Δ𝑣(𝑛)|^𝜏2/𝜏, (51) where 𝜏=𝜃/(𝜃−1)>2 for 1<𝜃<2. It follows from assumption (B2) that 𝐻^∗((𝑛≥1𝑛,Δ𝑣))𝜏1𝑎₂^𝜏−1|Δ𝑣(𝑛)|^𝜏, (52) thus 𝜒(𝑣)≥−2sin𝑙𝜋𝑝𝑇⁻¹𝑝𝑇^{(𝜏−2)/𝜏𝑝𝑇}_𝑛=1|Δ𝑣(𝑛)|^𝜏2/𝜏+1𝜏1𝑎₂^{𝜏−1𝑝𝑇}_𝑛=1|Δ𝑣(𝑛)|^𝜏≥(1/𝜏−1/2)𝑝𝑇𝑎²₂^{(𝜏−1)/(𝜏−2)}sin𝑙𝜋/𝑝𝑇^{𝜏/(𝜏−2)}. (53)(54) One can obtain the previous inequality by minimizing in (53) with respect to (∑^𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|^𝜏)^1/𝜏, and the minimum is attained at (𝑝𝑇)^1/𝜏(𝑎₂)^{(𝜏−1)/(𝜏−2)}/(sin(𝑙𝜋/𝑝𝑇))^1/(𝜏−2).
On the other hand, let 𝑣(𝑛)=1𝑝𝑇cos2𝑘𝜋𝑝𝑇𝑛⋅𝑎_𝑘−sin2𝑘𝜋1𝑝𝑇𝑛−2⋅𝑎_𝑘, (55) where 𝑎_𝑘∈ℝ^𝑁, 𝑘∈𝑍[[−𝑝𝑇/2],[𝑝𝑇/2]]⧵{0}. Then 𝑣∈𝑌_𝑘, and Δ𝑣(𝑛)=−2sin𝑘𝜋1𝑝𝑇𝑝𝑇sin2𝑘𝜋1𝑝𝑇𝑛+2⋅𝑎_𝑘cos2𝑘𝜋𝑝𝑇𝑛⋅𝑎_𝑘. (56) Taking 𝑎_𝑘=(𝑑,0,…,0)^𝑇∈ℝ^𝑁, where 𝑑∈ℝ, by Lemma 4, it follows that ^𝑝𝑇_𝑛=1(()(𝑛=𝐿𝐽Δ𝑣𝑛−1,𝑣))_{𝑝𝑇𝑛=1}−Δ𝑣₂(𝑛)𝑣₁(𝑣)+Δ𝑣₁(𝑛−1)𝑣₂(𝑛=_{𝑝𝑇𝑛=1})1𝑝𝑇⋅2sin𝑘𝜋⋅𝑝𝑇cos²2𝑘𝜋|𝑑|𝑝𝑇𝑛⋅²+sin²2𝑘𝜋1𝑝𝑇𝑛−2⋅|𝑑|²=𝜆_𝑘⋅|𝑑|², (57) where 𝜆_𝑘=2sin(𝑘𝜋/𝑝𝑇) and ^𝑝𝑇_𝑛=1|Δ𝑣(𝑛)|^𝜏=^𝑝𝑇_𝑛=1𝜆_𝑘^𝜏𝑝𝑇^−𝜏/2⋅sin²2𝑘𝜋1𝑝𝑇𝑛+2+cos²2𝑘𝜋𝑛𝑝𝑇^𝜏/2|𝑑|^𝜏≤𝜆^𝜏_max⋅𝑝𝑇^1−(𝜏/2)⋅2^𝜏/2|𝑑|^𝜏. (58) Therefore, taking 𝑘=−[𝑝𝑇/2], by eigenvalue problem (24) and (B2), it follows that 𝜒(𝑣)=12^𝑝𝑇_𝑛=1(()(𝑛+𝐿𝐽Δ𝑣𝑛−1,𝑣))_{𝑝𝑇𝑛=1}𝐻^∗((𝑛1𝑛,Δ𝑣))≤−2𝜆_max⋅|𝑑|²+1𝜏1𝑎₁^{𝜏−1𝑝𝑇}_𝑛=1|Δ𝑣(𝑛)|^𝜏1≤−2𝜆_max⋅|𝑑|²+1𝜏1𝑎₁^𝜏−1𝜆^𝜏_max⋅𝑝𝑇^1−(𝜏/2)⋅2^𝜏/2|𝑑|^𝜏. (59)
Let 𝑓(𝜌) equal the right-hand side of (59) where 𝜌=|𝑑|. It is easy to see that the absolute minimum of 𝑓 is attained at 𝜌_min=(𝑎₁)^{(𝜏−1)/(𝜏−2)}(𝑝𝑇)^1/2/[𝜆^{(𝜏−1)/(𝜏−2)}_max⋅2^{𝜏/2(𝜏−2)}] and given by 𝑓_min=(1/𝜏−1/2)𝑝𝑇(𝑎²₁)^{(𝜏−1)/(𝜏−2)}/(2𝜆_max)^{𝜏/(𝜏−2)}. Hence, by (19), let 𝜉(𝑛)=𝜉_{−[𝑝𝑇/2]}(𝑛)=cos2𝑘𝜋𝑝𝑇𝑛⋅𝜌−sin2𝑘𝜋1𝑝𝑇𝑛−2,⋅𝜌 (60) where 𝜌∈ℝ^𝑁,𝑘=−[𝑝𝑇/2].
If 𝑝𝑇 is even, then 𝜉(𝑛)=(1,1)^𝑇⋅(−1)^𝑛𝜌. Set 𝑌_𝜌min=𝑣∈𝑌_{−[𝑝𝑇/2]}(𝑛)(𝑛),()∶𝑣=𝜉𝜌=𝑑,0,…,0^𝑇∈ℝ^𝑁,𝑑∈ℝ. (61) For 𝑣∈𝑌_𝜌min, we have 𝜒(𝑣)≤𝑓_min. (62) Combining (54), (59), and (62), we have (1/𝜏−1/2)𝑝𝑇𝑎²₂^{(𝜏−1)/(𝜏−2)}sin(𝑙𝜋/𝑝𝑇)^{𝜏/(𝜏−2)}≤(1/𝜏−1/2)𝑝𝑇𝑎²₁^{(𝜏−1)/(𝜏−2)}2𝜆_max^{𝜏/(𝜏−2)}. (63) By 𝜏>2, and 𝜃=𝜏/(𝜏−1), it follows that sin𝑙𝜋/𝑝𝑇2𝜆_max≤𝑎₂/𝑎₁^2/𝜃. (64) For integer 𝑝>1, 𝑇≥1, 𝑙∈ℕ⧵{0}, 𝑝𝑇/𝑙∈ℕ⧵{0}, we have 0<𝑙𝜋/𝑝𝑇≤𝜋, 0<𝜋/𝑝𝑇≤𝜋/2.
If 𝑝𝑇 is even, then 𝜆_max=2. By assumption 𝑎₂/𝑎₁≤((1/4)sin(𝜋/𝑝𝑇))^𝜃/2 we have sin(𝑙𝜋/𝑝𝑇)≤sin(𝜋/𝑝𝑇), which implies that 𝑙=1 or 𝑙=𝑝𝑇−1. If 𝑝𝑇>2, then 𝑝𝑇/𝑙=𝑝𝑇/(𝑝𝑇−1)∉ℕ. So we have 𝑙=1.
If 𝑝𝑇 is odd, then 𝜆_max=2cos(𝜋/2𝑝𝑇). By assumption 𝑎₂/𝑎₁≤((1/2)sin(𝜋/2𝑝𝑇))^𝜃/2, we have sin(𝑙𝜋/𝑝𝑇)≤sin(𝜋/𝑝𝑇), so 𝑙=1. This completes the proof.

Acknowledgments

This research is supported by the National Natural Science Foundation of China under Grants (11101187), NCETFJ (JA11144), the Excellent Youth Foundation of Fujian Province (2012J06001), and the Foundation of Education of Fujian Province (JA09152).