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# Periodic and subharmonic solutions for a class of second-order p-Laplacian Hamiltonian systems

## Abstract

In this paper, the periodic and subharmonic solutions are investigated for a class of second-order non-autonomous ordinary differential equations with a p-Laplacian. With the perturbation technique and the dual least action principle, some existence results are given of solutions to the convex p-Laplacian systems.

## 1 Introduction

In this paper, we consider the second-order ordinary differential equations with a p-Laplacian

(1)

where $p>1$, $x=( x 1 , x 2 ,…, x N )$, $Φ p (x)= | x | p − 2 x$. Here $|⋅|$ stands the Euclidean norm in $R N$. In what follows we always suppose that $F:R× R N →R$, $(t,x)→F(t,x)$ is measurable and T-periodic in t for every $x∈ R N$, continuously differentiable and convex in x for a.e. $t∈R$. When $p=2$, (1) reduces to the second-order Hamiltonian system

(2)

Since F is T-periodic in t, it is natural to seek T-periodic solutions of (1) and (2). Also, F is kT-periodic for $k∈N$, one can search the kT-periodic solutions, which is called subharmonic solutions. By a subharmonic solution, it means a kT periodic solution with $k⩾2$ an integer, that is, the minimal period is strictly greater than T. When $k=1$, it is a periodic solution or harmonic. Clearly, a solution x of (1) over $[0,kT]$ verifying $x(0)=x(kT)$ and $x ˙ (0)= x ˙ (kT)$ can be extended by kT-periodicity over to give a kT-periodic solution. So, it is fine to study the periodic boundary value of problem of (1) over $[0,kT]$ to present conditions for the existence of the periodic and subharmonic solutions of (1). For the study of the subharmonic solutions, we refer to – for a few examples.

The dual action principle was firstly introduced by Clarke  and developed by Clarke and Ekeland –, which is from the spirit of optimal control theory and convex analysis; see . Following this use and the direct variational method, the periodic solutions are obtained for the Hamiltonian system of (2) in , , ,  and the references therein. Mawhin and Willem  presented the existence results of solutions to the more general systems by using such a principle and the perturbation technique argument. In 2007, Tian and Ge  generalized the existence results to the p-Laplacian system (1). By transforming the variable, they found a first-order Hamiltonian system equivalent to the second-order p-Laplacian system (1). Then by applying the Clarke duality, the perturbation technique, and the least action principle, they obtained the existence result of the periodic solution. The authors  also discussed the multi-point boundary value problem of a second-order differential equation with a p-Laplacian.

Motivated by the work listed above, we aim to discuss the periodic and subharmonic solutions to the p-Laplacian systems of (1). The proofs are depending on the dual least action principle and the perturbation arguments. To the best of our knowledge, there is no paper discussing the subharmonics solutions of p-Laplacian system by using the dual least action principle. This paper is a first try. The existence of the periodic solutions obtained in this paper slightly improves the result in . New arguments are considered for the posterior estimates of the periodic solutions and the existence of subharmonic solutions are presented, which extend those in , .

Other existence results for periodic and subharmonic solutions of the p-Laplacian differential equation using other variational methods can be found in , –.

The paper is organized as follows. In Section 2, we establish the variational structure of (1) and transfer the existence of the solution into the existence of a critical point of the corresponding functional. The dual action is mainly discussed here. In Section 3, sufficient conditions are presented to guarantee the existence of the periodic solution of (1). We also estimate the prior bounds of all the periodic solutions of (1). The existence of the subharmonic solutions are given in the last section.

## 2 Preliminary and dual action

Let $k⩾1$ be any integer and $p>1$ a constant. The Sobolev space $W k T 1 , p (0,kT; R N )$ is the space of functions $x:[0,kT]→ R N$ with $x∈ L p (0,kT; R N )$ having a weak derivative $x ˙ ∈ L p (0,kT; R N )$ and $x(0)=x(kT)$. The norm over $W k T 1 , p$ is defined by

$∥ x ∥ W k T 1 , p = ( ∫ 0 k T | x ( t ) | p d t + ∫ 0 k T | x ˙ ( t ) | p d t ) 1 p .$

Here, we recall that

$|x|= ∑ i = 1 N | x i | 2 , ∥ x ∥ L p = ( ∫ 0 k T | x ( t ) | p d t ) 1 p , ∥ x ∥ ∞ = max t ∈ [ 0 , k T ] |x(t)|.$

Consider the spaces X and Y defined by

$X = { u = ( u 1 , u 2 ) : [ 0 , k T ] → R 2 N , u 1 ∈ W k T 1 , p ( 0 , k T ; R N ) , u 2 ∈ W k T 1 , q ( 0 , k T ; R N ) } , Y = { v = ( v 1 , v 2 ) : [ 0 , k T ] → R 2 N , v 1 ∈ W k T 1 , q ( 0 , k T ; R N ) , v 2 ∈ W k T 1 , p ( 0 , k T ; R N ) } ,$

with the norm $∥ u ∥ X = ∥ u 1 ∥ W k T 1 , p + ∥ u 2 ∥ W k T 1 , q$ and $∥ v ∥ Y = ∥ v 1 ∥ W k T 1 , q + ∥ v 2 ∥ W k T 1 , p$, respectively. Here q is a constant such that $1 p + 1 q =1$. It is easy to verify that X and Y are reflexive Banach spaces and $X ∗ =Y$.

For any $h=( h 1 , h 2 )∈ L 1 (0,kT; R N × R N )$, the mean value is defined by

$h ¯ = 1 k T ∫ 0 k T h(t)dt.$

Similarly, we can define $h ¯ 1$ and $h ¯ 2$. We denote by $Y ˜$ the subspace of Y by

$Y ˜ ={v∈Y: v ¯ =0}.$

We easily find the following inequality.

### Lemma 2.1

For every$u 1 ∈ W k T 1 , p$and$u ¯ 1 =0$, we have

$∥ u 1 ∥ L p ⩽kT ∥ u ˙ 1 ∥ L p .$

For every$u 2 ∈ W k T 1 , q$and$u ¯ 2 =0$, we have

$∥ u 2 ∥ L q ⩽kT ∥ u ˙ 2 ∥ L q .$

Let $u 1 =x$, $α u 2 = Φ p ( x ˙ )$, then the second-order p-Laplacian system (1) can be changed to the first-order ones,

${ u ˙ 1 = Φ q ( α u 2 ( t ) ) , u ˙ 2 = − 1 α ∇ F ( t , u 1 ( t ) ) ,$
(3)

where $α>0$ is a parameter. Define $H i :[0,kT]× R N →R$, $i=1,2$, and $H:[0,kT]× R 2 N →R$ by

$H 1 ( t , u 1 ) = 1 α F ( t , u 1 ) , H 2 ( t , u 2 ) = α q − 1 q | u 2 | q , H ( t , u ) = H 1 ( t , u 1 ) + H 2 ( t , u 2 )$

for $u=( u 1 , u 2 )∈ R 2 N$. Then the system (3) can be written by

$J u ˙ (t)+∇H ( t , u ( t ) ) =0,$
(4)

where $J= ( 0 N I N − I N 0 N )$ is the symplectic matrix. Obviously, $J 2 =− I 2 N$ and $(Ju,v)=−(u,Jv)$ for all $u,v∈ R 2 N$.

By the kT-periodic property, we have the Hamiltonian action of (4) on X given by

$ψ(u)= ∫ 0 k T ( − 1 2 ( J u ˙ ( t ) , u ( t ) ) − H ( t , u ( t ) ) ) dt,$
(5)

where $(⋅,⋅)$ is the inner product in $R 2 N$. Because the first part of ψ is indefinite, the dual least process is applied to discuss the existence of the periodic solutions of (4).

Let $Γ 0$ the set of all convex lower semi-continuous functions from $R 2 N$ to $(−∞,∞]$ whose effective domain (not equal to ∞) is nonempty. Then the Fenchel transform $H ∗ (t,⋅)$ of $H(t,⋅)∈ Γ 0$ is defined by

$H ∗ (t,v)= sup u ∈ R 2 N ( ( v , u ) − H ( t , u ) ) .$
(6)

Similarly, we also can define

$F ∗ (t, v 1 )= sup x ∈ R N ( ( v 1 , x ) − F ( t , x ) ) .$

For more properties of the Fenchel transform, we refer to , . By Theorem 2.2 and Proposition 2.4 in , we have the following conclusion.

### Lemma 2.2

Suppose the following condition holds.

(A0):There exist positive constants α, δ, and positive functions$β,γ∈ L q (0,kT;[0,+∞))$such that

$δ p | x | p −β(t)⩽F(t,x)⩽ α p | x | p +γ(t)$

holds for a.e. $t∈[0,kT]$and$x∈ R N$.

Then

Furthermore, we have$F ∗ (t,⋅)∈ C 1 ( R N )$and

From the definition of $H 2$ and Lemma 2.2, we have the following result.

### Corollary 2.3

$H 2 ∗ (t, v 2 )= 1 p α | v 2 | p$.

Easily we find $H ∗ (t,v)= 1 α F ∗ (t, v 1 )+ H 2 ∗ (t, v 2 )$. So when F satisfies the condition (A0), $H ∗ (t,v)$ is continuously differentiable in v for a.e. $t∈[0,kT]$ and

$H ∗ (t,v)=(v,u)−H(t,u)⇔v=∇H(t,u)⇔u=∇ H ∗ (t,v).$
(7)

Let $v=−Ju$ and by duality, we have

$ψ ( u ) = ∫ 0 k T ( − 1 2 ( J u ˙ ( t ) , u ( t ) ) − H ( t , u ( t ) ) ) d t = ∫ 0 k T ( 1 2 ( v ˙ ( t ) , u ( t ) ) − H ( t , u ( t ) ) ) d t = ∫ 0 k T ( − 1 2 ( v ˙ ( t ) , u ( t ) ) + ( v ˙ ( t ) , u ( t ) ) − H ( t , u ( t ) ) ) d t = ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , v ( t ) ) + H ∗ ( t , v ˙ ( t ) ) ) d t .$

So the dual action can be defined on Y by

$X(v)= ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , v ( t ) ) + H ∗ ( t , v ˙ ( t ) ) ) dt.$
(8)

$X$ is the functional we needed since the critical points of $X$ on Y coincide with the solutions of (4). Because $X(v+c)=X(v)$, it suffices to find a critical point in $Y ˜$. Similarly to the discussion of the related lemmas in reference , we have the following results.

### Lemma 2.4

Suppose the condition (A0) holds. Then$X$is continuously differential on$Y ˜$. For any$h∈ Y ˜$, we have

$〈 X ′ ( v ) , h 〉 = ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , h ( t ) ) + ( ∇ H ∗ ( t , v ˙ ( t ) ) − 1 2 J v ( t ) , h ˙ ( t ) ) ) dt.$
(9)

### Lemma 2.5

Suppose the condition (A0) holds. If$v∈ Y ˜$is a critical point of$X$, then the function$u(t)=∇ H ∗ (t, v ˙ (t))$is the kT-periodic solution of (4) in X.

### Lemma 2.6

For every$u∈X$, we have

$∫ 0 k T ( J u ˙ ( t ) , u ( t ) ) dt⩾− 2 k T p ∥ u ˙ 1 ∥ L p p − 2 k T q ∥ u ˙ 2 ∥ L q q .$

For every$v∈Y$, we have

$∫ 0 k T ( J v ˙ ( t ) , v ( t ) ) dt⩾− 2 k T q ∥ v ˙ 1 ∥ L q q − 2 k T p ∥ v ˙ 2 ∥ L p p .$

### Remark 2.1

There are some extended versions of the inequalities in Lemmas 2.1 and 2.6; see , .

## 3 Periodic solutions

In this section, we discuss the kT-periodic solution of (4). Here we note that the $H(t,u)$ is T-periodic in t for each $u∈ R 2 N$.

### Theorem 3.1

Suppose the following conditions are satisfied.

(A1):There exists$l∈ L p q (0,kT; R N )$such that for all$x∈ R N$and a.e. $t∈[0,kT]$, one has

$F(t,x)⩾ ( l ( t ) , Φ p ( x ) ) .$

(A2):There exists$α∈(0, ( k T ) − max { p , q } / q )$, $γ∈ L max { p , q } (0,kT; R N )$such that for all$x∈ R N$and a.e. $t∈[0,kT]$, one has

$F(t,x)≤ α 2 p | x | p +γ(t).$

(A3):$∫ 0 k T F(t,x)dt→+∞$as$|x|→+∞$, $x∈ R N$.

Then the problem (4) has at least one solution$u=( u 1 , u 2 )∈X$such that$u 1$is the kT-periodic solutions of (1) and

$v(t)=−J [ u ( t ) − 1 k T ∫ 0 k T u ( s ) d s ]$

minimizes the dual action

$X: Y ˜ →(−∞,∞],v↦ ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , v ( t ) ) + H ∗ ( t , v ˙ ( t ) ) ) dt.$

### Proof

From the discussion in Section 2, we can see that if the problem (4) has one solution $u=( u 1 , u 2 )∈X$, then $u 1 ∈ W k T 1 , p$ is for the kT-periodic solutions of (1). Now we need to prove the problem (4) has at least one kT-periodic solution in X. The proof is divided into three parts.

Step 1: Existence of a solution for the perturbed problem.

Choose $ϵ 0 >0$ such that

$0<α+ ϵ 0

and for any $0<ϵ< ϵ 0$, let

$H 1 ϵ (t, u 1 )= H 1 (t, u 1 )+ ϵ p | u 1 | p .$

Clearly, $H 1 ϵ (t, u 1 )$ is strictly convex and continuously differentiable in $u 1$ for a.e. $t∈[0,kT]$. From (A1) and (A2), we have

$− 1 α |l(t)| | u 1 | p − 1 + ϵ p | u 1 | p ⩽ H 1 ϵ (t, u 1 )⩽ α + ϵ p | u 1 | p + 1 α γ(t).$

Because the function $g(s)= ϵ 2 p s p − 1 α |l(t)| s p − 1$, $s>0$, attains its minimum at $s=2(p−1)|l(t)|/αϵ$, we have

$− 1 α p ( 2 ( p − 1 ) α ϵ ) p − 1 |l(t) | p + ϵ 2 p | u 1 | p ⩽ H 1 ϵ (t, u 1 )⩽ α + ϵ p | u 1 | p + 1 α γ(t).$

Let $H ϵ (t,u)= H 1 ϵ (t, u 1 )+ H 2 (t, u 2 )$. Easily we find $H ϵ ∗ (t,v)= H 1 ϵ ∗ (t, v 1 )+ H 2 ∗ (t, v 2 )$, where $v=( v 1 , v 2 )∈Y$. From Lemma 2.4 and Lemma 2.5, we find that the perturbed dual action

$X ϵ (v)= ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , v ( t ) ) + H ϵ ∗ ( t , v ˙ ( t ) ) ) dt$
(10)

is continuously differentiable on $Y ˜$ and if $v ϵ ∈ Y ˜$ is a critical point of $X ϵ$, the function $u ϵ$ defined by

$u ϵ (t)=∇ H ϵ ∗ ( t , v ˙ ϵ ( t ) )$

is a solution of

${ J u ˙ ( t ) + ∇ H ϵ ( t , u ( t ) ) = 0 , u ( 0 ) = u ( k T ) ,$
(11)

i.e.,

${ u ˙ 1 ϵ ( t ) − Φ q ( α u 2 ϵ ( t ) ) = 0 , u ˙ 2 ϵ ( t ) + ϵ Φ p ( u 1 ϵ ( t ) ) + 1 α ∇ F ( t , u 1 ϵ ( t ) ) = 0 , u ϵ ( 0 ) = u ϵ ( k T ) ,$
(12)

where we rewrite $u ϵ =( u 1 ϵ , u 2 ϵ )∈X$. Meanwhile, from Lemma 2.2, we have

$H ϵ ∗ ( t , v ˙ ( t ) ) ⩾ 1 q ( α + ϵ ) − q / p | v ˙ 1 (t) | q + 1 p α − 1 | v ˙ 2 (t) | p − 1 α γ(t),$

which together with (10) and Lemma 2.6 implies that

$X ϵ ( v ) ⩾ 1 q ( ( α + ϵ ) − q / p − k T ) ∫ 0 k T | v ˙ 1 ( t ) | q d t + 1 p ( α − 1 − k T ) ∫ 0 k T | v ˙ 2 ( t ) | p d t − ∫ 0 k T 1 α γ ( t ) d t ⩾ δ 10 ∥ v ˙ 1 ∥ L q q + δ 20 ∥ v ˙ 2 ∥ L p p − γ 0$
(13)

with $δ 10 >0$, $δ 20 >0$. Thus every minimizing sequence of $X ϵ$ on $Y ˜$ is bounded by (13) and Lemma 2.1. From the continuity of $H ϵ$ and the definition of $H ϵ ∗$, we can see that the second term of $X ϵ$ is weakly lower semi-continuous on $Y ˜$. Meanwhile the first part of $X ϵ$ is weakly continuous. So $X ϵ$ is weakly lower semi-continuous. This implies that $X ϵ$ has a minimum at some $v ϵ ∈ Y ˜$. So (11) has a solution $u ϵ =∇ H ϵ ∗ (t, v ˙ ϵ (t))$.

Step 2: Estimation of $u ϵ$.

It follows from a similar discussion to Lemma 2.2 that

$|∇ H 1 (t, u 1 )|⩽ ( p α − q p ( | u 1 | + | l ( t ) | | u 1 | p − 1 + γ ( t ) ) ) p − 1 +1.$

So for any $u 1 ∈ R N$, $∇ H 1 (t, u 1 )∈ L q [0,kT]$. Then the function

$H ¯ 1 : R N →R, u 1 → ∫ 0 k T H 1 (t, u 1 )dt$

is continuously differentiable. By (A3), $H ¯ 1$ has a minimum at some point $u ¯ 1 ∈ R N$ such that

$∫ 0 k T ∇ H 1 (t, u ¯ 1 )dt=0.$

Easily, $∇ H 2 (t, u 2 )= Φ p (α u 2 )$ and

$∫ 0 k T ∇ H 2 (t,0)dt=0.$

So, let $u ¯ =( u ¯ 1 ,0)∈ R 2 N$, and then $v ˙ (t)=∇H(t, u ¯ )$ has a unique solution $ω∈ Y ˜$. By duality, we also have

$H ∗ (t, ω ˙ )=( ω ˙ , u ¯ )−H(t, u ¯ ),$

and $H ∗ (⋅, ω ˙ (⋅))∈ L 1 [0,kT]$. From the inequality $H(t, u ¯ )⩽ H ϵ (t, u ¯ )$, we have $H ϵ ∗ (t, ω ˙ )⩽ H ∗ (t, ω ˙ )$, and from (13), we obtain

$δ 10 ∥ v ˙ 1 ϵ ∥ L q q + δ 20 ∥ v ˙ 2 ϵ ∥ L p p − γ 0 ⩽ X ϵ ( v ϵ )⩽ X ϵ (ω)⩽X(ω)= c 1 <+∞.$

By Lemma 2.1, we have $∥ v ϵ ∥ Y ⩽ c 2$, where $c 1$ and $c 2$ are constants independent of ϵ, as well as the following constants $c i$, $i=3,4,5$. Furthermore, from $u ˙ ϵ =J v ˙ ϵ$, we have

$∥ u ˙ 1 ϵ ∥ L p + ∥ u ˙ 2 ϵ ∥ L q = ∥ v ˙ 1 ϵ ∥ L q + ∥ v ˙ 2 ϵ ∥ L p ⩽ ∥ v ϵ ∥ Y ⩽ c 2 ,$

and from $∫ 0 k T u ˜ ϵ (t)dt=0$, i.e.

$| u ˜ ϵ (t)|⩽| ∫ 0 k T u ˜ ˙ ϵ (t)dt|⩽ ( k T ) 1 / q ∥ u ˜ ˙ 1 ϵ ∥ L p + ( k T ) 1 / p ∥ u ˜ ˙ 2 ϵ ∥ L q ,$

we have $∥ u ˜ ϵ ∥ X ⩽ c 3$.

Meanwhile, by the convexity of $F(t,⋅)$, we have

$H ( t , u ¯ ϵ / 2 ) ⩽ 1 2 H ( t , u ϵ ( t ) ) + 1 2 H ( t , − u ˜ ϵ ( t ) ) ⩽ 1 2 H ( t , u ϵ ( t ) ) − 1 2 H ( t , 0 ) + 1 2 H ( t , 0 ) + α 2 p | u ˜ 1 ϵ ( t ) | p + 1 2 α γ ( t ) + α q − 1 2 q | u ˜ 2 ϵ ( t ) | q ⩽ 1 2 ( ∇ H ( t , u ϵ ( t ) ) , u ϵ ( t ) ) + α 2 p | u ˜ 1 ϵ ( t ) | p + α q − 1 2 q | u ˜ 2 ϵ ( t ) | q + 1 α γ ( t ) = 1 2 ( − J u ˙ ϵ ( t ) , u ϵ ( t ) ) − ϵ ( Φ p ( u 1 ϵ ( t ) ) , u 1 ϵ ( t ) ) + α 2 p | u ˜ 1 ϵ ( t ) | p + α q − 1 2 q | u ˜ 2 ϵ ( t ) | q + 1 α γ ( t ) .$

So,

$∫ 0 k T F ( t , u ¯ ϵ / 2 ) d t ⩽ ∫ 0 k T 1 2 ( − J u ˙ ϵ ( t ) , u ϵ ( t ) ) d t + α 2 p ∥ u ˜ ϵ ∥ L p p + α q − 1 2 q ∥ u ˜ 2 ϵ ∥ L q q + γ 0 ⩽ 1 2 ∥ u ˜ 1 ϵ ∥ L p ∥ u ˜ ˙ 2 ϵ ∥ L q + 1 2 ∥ u ˜ ˙ 1 ϵ ∥ L p ∥ u ˜ 2 ϵ ∥ L q + α 2 p ∥ u ˜ ϵ ∥ L p p + α q − 1 2 q ∥ u ˜ 2 ϵ ∥ L q q + γ 0 ⩽ c 4 .$

This together with condition (A3) implies that $| u ¯ ϵ |$ is bounded. Consequently,

$∥ u ϵ ∥ X ⩽ ∥ u ¯ ϵ ∥ X + ∥ u ˜ ϵ ∥ X ⩽ c 5 .$

Step 3: Existence of a solution for the problem (4).

Since $∥ u ϵ ∥ X$ is bounded, there is a sequence $ϵ n →0$ ($n→∞$) with $ϵ n ∈(0, ϵ 0 )$ and a function $u∈X$ such that

Moreover, from $v ˙ ϵ =−J u ˙ ϵ$, we have

$v ϵ (t)=−J ( u ϵ ( t ) − u ¯ ϵ ) .$

So ${ v ϵ n }$ converges weakly to $v(t)=−J(u(t)− u ¯ )$. From (12), we have the integrated form

${ u 1 ϵ n ( t ) − u 1 ϵ n ( 0 ) − ∫ 0 t Φ q ( α u 2 ϵ n ( s ) ) d s = 0 , u 2 ϵ n ( t ) − u 2 ϵ n ( 0 ) + ϵ n ∫ 0 t Φ p ( α u 1 ϵ n ( s ) ) d s + ∫ 0 t 1 α ∇ F ( s , u 1 ϵ n ( s ) ) d s = 0 .$

Because $u ϵ n$ converges weakly to u in X, $u ϵ n$ converges uniformly to u in $C k T ∞$ (see Proposition 1.2 in ). So, let $n→∞$ and we have

${ u 1 ( t ) − u 1 ( 0 ) − ∫ 0 t Φ q ( α u 2 ( s ) ) d s = 0 , u 2 ( t ) − u 2 ϵ n ( 0 ) + 1 α ∫ 0 t ∇ F ( s , u 1 ( s ) ) d s = 0 ,$

that is, u is a solution of (4) in X.

Finally, we show $v=−J(u(t)− u ¯ )$ minimizes the dual action $X$ on $Y ˜$. Because $v ϵ n$ is a minimum of $X ϵ n$ and $H ϵ n ∗ (t,v)⩽ H ∗ (t,v)$, we have

$X ϵ n ( v ϵ n )⩽ X ϵ n (h)⩽X(h),∀h∈ Y ˜ .$

By the duality between $u ϵ n$ and $v ˙ ϵ n$ and the definition of $H ϵ n (t,v)$, we have

$X ϵ n ( v ϵ n ) = ∫ 0 k T ( 1 2 ( J v ˙ ϵ n ( s ) , v ϵ n ( s ) ) + ( v ˙ ϵ n ( s ) , u ϵ n ( s ) ) − H ϵ n ( s , u ϵ n ( s ) ) ) d s = ∫ 0 k T ( 1 2 ( J v ˙ ϵ n ( s ) , v ϵ n ( s ) ) + ( u ϵ n ( s ) , v ˙ ϵ n ( s ) ) − H ( s , u ϵ n ( s ) ) − ϵ n p | u 1 ϵ n ( s ) | p ) d s .$

Moreover, as $v ϵ n$ converges weakly to v in $Y ˜$, $J v ˙ ϵ n$ converges weakly to Jv. Letting $n→∞$ we obtain, by $v ˙ (t)=∇H(t,u(t))$,

$lim n → ∞ X ϵ n ( v ϵ n ) = lim n → ∞ ∫ 0 T ( 1 2 ( J v ˙ ϵ n ( s ) , v ϵ n ( s ) ) + ( u ϵ n ( s ) , v ˙ ϵ n ( s ) ) − H ( s , u ϵ n ( s ) ) − ϵ n p | u 1 ϵ n ( s ) | p ) d s = ∫ 0 T ( 1 2 ( J v ˙ ( s ) , v ( s ) ) + H ∗ ( s , v ˙ ( s ) ) ) d s = X ( v ) .$

So $X(v)⩽X(h)$ for all $h∈ Y ˜$. The proof is complete. □

### Remark 3.1

Theorem 3.1 still holds if condition (A1) is changed to:

(A4): There exists $l∈ L p q / ( p − m + 1 ) (0,kT; R N )$ such that for all $x∈ R N$ and a.e. $t∈[0,kT]$, one has

$F(t,x)⩾ ( l ( t ) , Φ m ( x ) ) ,$

where m is an integer such that $2⩽m⩽p$.

When the parameter α is smaller, we can obtain the prior bound for all the solutions of the p-Laplacian system (1).

### Theorem 3.2

If there exist$α∈(0,min{ ( 2 k T ) − p / q ,p ( 2 k T ) − 1 })$, $β⩾0$, $γ⩾0$, and$δ>0$such that

$δ|x|−β⩽F(t,x)⩽ α 2 p | x | p +γ$

for a.e. $t∈[0,kT]$and$x∈ R N$, then each solution of (1) satisfies the inequalities

$∫ 0 k T | x ˙ (t) | p dt⩽ p k T ( β + γ ) p − 2 k T α ,$
(14)
$∫ 0 k T |x(t)|dt⩽ k T ( β + γ ) δ ( 1 − 2 k T α q / p ) .$
(15)

### Proof

We still set $u 1 =x$, $α u 2 = Φ p ( x ˙ )$, and the equalities (3) hold. Easily we find

$1 q α − 2 q p |∇F(t, u 1 ) | q ⩽ ( ∇ F ( t , u 1 ) , u 1 ) +β+γ.$

It follows from

$( ∇ H ( t , u ) , u ) = 1 α ( ∇ F ( t , u 1 ) , u 1 ) + ( Φ q ( α u 2 ) , u 2 )$

that

$1 q α − 1 − 2 q p |∇F(t, u 1 ) | q + ( Φ q ( α u 2 ) , u 2 ) ⩽ ( ∇ H ( t , u ) , u ) + β α + γ α .$

That is,

$1 q α − q p | u ˙ 2 (t) | q + α q − 1 | u 2 | q ⩽ ( − J u ˙ ( t ) , u ( t ) ) + β α + γ α .$

Integrating over $[0,kT]$ and using Lemma 2.6, we have

$1 q α − q p ∥ u ˙ 2 ∥ L q q + α q − 1 ∥ u 2 ∥ L q q ⩽ − ∫ 0 k T ( J u ˙ ( t ) , u ( t ) ) d t + k T ( β + γ ) α ⩽ 2 k T p ∥ u ˙ 1 ∥ L p p + 2 k T q ∥ u ˙ 2 ∥ L q q + k T ( β + γ ) α = 2 k T p α q ∥ u 2 ∥ L q q + 2 k T q ∥ u ˙ 2 ∥ L q q + k T ( β + γ ) α .$

So,

$( 1 q α − q p − 2 k T q ) ∥ u ˙ 2 ∥ L q q + ( α q − 1 − 2 k T p α q ) ∥ u 2 ∥ L q q ⩽ k T ( β + γ ) α .$

Hence

$∥ u 2 ∥ L q q ⩽ p k T ( β + γ ) α q ( p − 2 k T α ) , ∥ u ˙ 2 ∥ L q q ⩽ q k T ( β + γ ) α 1 − q / p − 2 k T α$

and

$∥ u ˙ 1 ∥ L p p = ∥ Φ q ( α u 2 ) ∥ L p p = α q ∥ u 2 ∥ L q q ⩽ p k T ( β + γ ) p − 2 k T α .$

Meanwhile, by the convexity of F, we have

$δ ∫ 0 k T | u 1 ( t ) | d t − β k T ⩽ ∫ 0 k T F ( t , u 1 ( t ) ) d t ⩽ ∫ 0 k T ( F ( t , 0 ) + ( ∇ F ( t , u 1 ( t ) ) , u 1 ( t ) ) ) d t ⩽ γ k T + α ∫ 0 k T ( ∇ H ( t , u ( t ) ) , u ( t ) ) d t − α ∫ 0 k T ( Φ q ( α u 2 ) , u 2 ) d t = γ k T − α ∫ 0 k T ( J u ˙ ( t ) , u ( t ) ) d t − α q ∥ u 2 ∥ L q q ⩽ γ k T + α 2 k T q ∥ u ˙ 2 ∥ L q q ⩽ γ k T + 2 k 2 T 2 ( β + γ ) α − q / p − 2 k T ,$

which completes the proof. □

## 4 Subharmonic solutions

### Theorem 4.1

Assume that$F:R× R N →R$is continuous. Suppose further that

$F(t,x)→+∞$
(16)

and

$F ( t , x ) | x | p →0$
(17)

as$|x|→+∞$uniformly in$t∈R$. Then for each$k∈R∖{0}$, the system (1) has a kT-periodic solution$x k$such that

$∥ x k ∥ ∞ →+∞$

and such that the minimal period$T k$of$x k$tends towhen$k→+∞$.

### Proof

Let $c 1 = max t ∈ R |F(t,0)|$. Then condition (16) implies that there exists a $r>0$ such that

$F(t,x)⩾1+ c 1$

for all $t∈R$ and $x∈ R N$ with $|x|>r$. By the convexity of F, we have

$1 + c 1 ⩽ F ( t , r | x | x ) ⩽ r | x | F ( t , x ) + ( 1 − r | x | ) F ( t , 0 ) ⩽ r | x | F ( t , x ) + ( 1 − r | x | ) c 1 ,$

i.e.

$F(t,x)⩾ 1 r |x|+ c 1$

for all $t∈R$ and $x∈ R N$ with $|x|>r$. Furthermore, the continuity of F implies that there exist positive constants δ, β such that

$F(t,x)⩾δ|x|−β$
(18)

for all $t∈R$ and $x∈ R N$.

By the condition (17), there exists $α∈(0, ( 2 k T ) − max { p , q } / q )$ such that

$F(t,x)⩽ α 2 p | x | p +γ$
(19)

for all $t∈R$ and $x∈ R N$. Then from Theorem 3.1, we can see that the problem (4) has at least one solution $u k =( u 1 k , u 2 k )∈X$ such that $u 1 k$ is the kT-periodic solution of (1) and

$v k (t)=−J [ u k ( t ) − 1 k T ∫ 0 k T u k ( s ) d s ]$

minimizes the dual action

$X: Y ˜ →(−∞,∞],v↦ ∫ 0 k T ( 1 2 ( J v ˙ ( t ) , v ( t ) ) + H ∗ ( t , v ˙ ( t ) ) ) dt.$

Next we estimate the upper bound of $c k =X( v k )$. For any $h=( h 1 , h 2 )∈ Y ˜$, we have

$c k = X ( v k ) ⩽ X ( h ) = ∫ 0 k T ( 1 2 ( J h ˙ ( t ) , h ( t ) ) + H ∗ ( t , h ˙ ( t ) ) ) d t = ∫ 0 k T ( 1 2 ( J h ˙ ( t ) , h ( t ) ) + 1 α F ∗ ( t , h ˙ 1 ( t ) ) + 1 p α | h ˙ 2 | p ) d t .$
(20)

Let $ρ=( ρ 1 , ρ 2 )∈ R N × R N$ with $| ρ i |=1$, $i=1,2$, and $ϵ⩽2πδ/kT$. Define the function

$h(t)= ϵ k T 2 π ( ρ cos 2 π k T t + J ρ sin 2 π k T t ) .$

Obviously, $h∈ Y ˜$ and $J h ˙ (t)=− 2 π k T h(t)$. Meanwhile, from (18), we can see that when $z∈ R N$ with $|z|⩽δ$, we have

$F ∗ (t,z)⩽ sup x ∈ R n ( ( z , x ) − F ( t , x ) ) ⩽ sup x ∈ R n ( ( z , x ) − δ | x | + β ) ⩽β.$

So it follows from (20) that

$c k ⩽ X ( h ) ⩽ ∫ 0 k T ( 1 2 ( J h ˙ ( t ) , h ( t ) ) + β α + 1 p α | h ˙ 2 | p ) d t = − ϵ 2 T 2 4 π k 2 + ( β α + ϵ p p α ) T k .$
(21)

If $∥ x k ∥ ∞ →+∞$ does not hold when $k→+∞$. Then there exist a subsequence ${ k n }$ and a constant $c 1 >0$ such that

$∥ x k n ∥ ∞ ⩽ c 1 .$

From (1), we have

$∥ d d t Φ p ( x ˙ k n ) ∥ ∞ ⩽ c 2 ,$

and so $∥ x ˙ k n ∥ ∞ ⩽ c 3$ for some constants $c 2 , c 3 >0$. This implies that

$∥ v k n ∥⩽ c 4 ,∥ v ˙ k n ∥⩽ c 5 .$

On the other hand, the inequality

$c k n = X ( v k n ) = ∫ 0 k n T ( 1 2 ( J v ˙ k n ( t ) , v k n ( t ) ) + H ∗ ( t , v ˙ k n ( t ) ) ) d t ⩾ ∫ 0 k n T ( − 1 2 ∥ v ˙ k n ∥ ∞ ∥ v k n ∥ ∞ − H ( t , 0 ) ) d t ⩾ − ( 1 2 c 4 c 5 + γ α ) T k n$
(22)

is incompatible with (21) when n is sufficiently large. Thus $∥ x k ∥ ∞ →+∞$ when $k→+∞$.

It remains to prove that the minimal period $T k$ of $x k$ tends to +∞ as $k→+∞$. If not, there exist $R>0$ and a subsequence ${ k n }$ such that the minimal period $T k n$ of $x k n$ is less than R. Meanwhile, by (18), (19), and Theorem 3.2, we have

$∫ 0 T k n | x ˙ (t) | p dt⩽ α q ∥ u 2 ∥ L q q ⩽ p T k n ( β + γ ) p − 2 T k n α ,$
(23)
$∫ 0 T k n |x(t)|dt⩽ T k n ( β + γ ) δ ( 1 − 2 T k n α q / p ) .$
(24)

We notice that

$∥ x k n ∥ ∞ = ∥ x ¯ k n + x ˜ k n ∥ ∞ ⩽ | x ¯ k n | + ∥ x ˜ k n ∥ ∞ ⩽ 1 T k n ∫ 0 T k n | x ( t ) | d t + T 1 / q ( ∫ 0 T k n | x ˙ ( t ) | p d t ) 1 / p .$

Inequalities (23) and (24) imply that ${ ∥ x k n ∥ ∞ }$ is bounded, which is a contraction. The proof is complete. □

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## Acknowledgements

This research is supported by the National Natural Science Foundation of China (No. 11101385) and by the Beijing Higher Education Young Elite Teacher Project.

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