On periodic solutions of superquadratic Hamiltonian systems ∗

ftp ejde.math.swt.edu (login: ftp)

On periodic solutions of superquadratic Hamiltonian systems ^∗

Guihua Fei

Abstract

We study the existence of periodic solutions for some Hamiltonian systems ˙z =J Hz(t, z) under new superquadratic conditions which cover the caseH(t, z) =|z|²(ln(1 +|z|^p))^q withp, q >1. By using the linking theorem, we obtain some new results.

1 Introduction

We consider the superquadratic Hamiltonian system

˙

z=J Hz(t, z) (1.1)

whereH ∈C¹([0,1]×R^2N,R) is a 1-periodic function int,J =

0 −IN

IN 0

is the standard 2N×2N symplectic matrix, and

H(t, z)

|z|² →+∞as|z| →+∞uniformly int. (1.2) We assume H satisfies the following conditions.

(H1) H(t, z)≥0, for all (t, z)∈[0,1]×R^2N. (H2) H(t, z) =o(|z|²) as|z| →0 uniformly int.

In [12], Rabinowitz established the existence of periodic solutions for (1.1) under the following superquadratic condition: there existµ >0 andr₁>0 such that for all|z| ≥r₁andt∈[0,1]

0< µH(t, z)≤z·Hz(t, z). (1.3) Since then, the condition (1.3) has been used extensively in the literature; see [1-14] and the references therein.

∗Mathematics Subject Classifications: 58E05, 58F05, 34C25.

Key words: periodic solution, Hamiltonian system, linking theorem.

2002 Southwest Texas State University.c

Submitted September 20, 2001. Published Janaury 15, 2002.

1

It is easy to see that (1.3) does not include some superquadratic nonlinearity like

H(t, z) =|z|²(ln(1 +|z|^p))^q, p, q >1. (1.4) In this paper, we shall study the periodic solutions of (1.1) under some superquadratic conditons which cover the cases like (1.4). We assumeH satisfies the following condition.

(H3) There exist constantsβ >1, 1< λ <1 +^β−_β¹,c1, c2 >0 andL >0 such that

z·H_z(t, z)−2H(t, z)≥c₁|z|^β, ∀|z| ≥L, ∀t∈[0,1];

|H_z(t, z)| ≤c₂|z|^λ, ∀|z| ≥L, ∀t∈[0,1].

Theorem 1.1 SupposeH ∈C¹([0,1]×R^2N,R)is 1-periodic in t and satisfies (1.2), (H1)–(H3). Then (1.1) possesses a nonconstant 1-periodic solution.

A straightforward computation shows that ifHsatisfies (1.4), for anyT >0, the system (1.1) has a nonconstant T-periodic solution with minimal periodT.

One can see Remark 2.2 and Corollary 2.3 for more examples.

For the second order Hamiltonian system

¨

u(t) +V⁰(t, u(t)) = 0,

u(0)−u(1) = ˙u(0)−u(1) = 0˙ (1.5) we have a similar result.

Theorem 1.2 SupposeV ∈C¹([0,1]×R^N,R)is 1-periodic int and satisfies (V1) V(t, x)≥0, for all(t, x)∈[0,1]×R^N

(V2) V(t, x) =o(|x|²)as|x| →0 uniformly in t (V3) V(t, x)/|x|²→+∞ as|x| →+∞uniformly int

(V4) There exist constants1< λ≤β,d₁, d₂>0andL >0 such that x·V⁰(t, x)−2V(t, x)≥d₁|x|^β, ∀|x| ≥L, ∀t∈[0,1];

|V⁰(t, x)| ≤d₂|x|^λ, ∀|x| ≥L, ∀t∈[0,1]. (1.6) (orV(t, x)≤d2|x|^λ+1, ∀|x| ≥L, ∀t∈[0,1]). (1.7) Then (1.5) possesses a nonconstant 1-periodic solution.

We shall use the linking theorem [13, Theorem 5.29] to prove our results.

The idea comes from [11, 12, 13]. Theorem 1.1 is proved in Section 2 while the proof of Theorem 1.2 is carried out in Section 3.

2 First order Hamiltonian system

LetS¹=R/(2πZ) and E=W^1/2,2(S¹,R^2N). ThenE is a Hilbert space with normk · kand inner producth·,·i. We define

hAx, yi= Z 1

0

(−Jx, y)˙ dt, ∀x, y∈E; (2.1) f(z) = 1

2hAz, zi − Z 1

0

H(t, z)dt, ∀z∈E. (2.2) Then A is a bounded selfadjoint operator and kerA=R^2N. (H1)–(H3) imply that

|H(t, z)| ≤a1+a2|z|^λ+1, ∀z∈R^2N.

This implies thatf ∈C¹(E,R) and looking for the solutions of (1.1) is equivalent to looking for the critical points off [12, 13]. Let E⁰= ker(A),E⁺= positive definite subspace of A, andE⁻ = negative definite subspace ofA. ThenE = E⁰⊕E⁻⊕E⁺.

Lemma 2.1 Under the conditions of Theorem 1.1, f satisfies the (PS) condi- tion.

Proof. Let{zm} be a (PS)-sequence, i.e.,

|f(zm)| ≤M; f⁰(zm)→0 as m→ ∞.

We want to show that{zm} is bounded. Then by a standard argument, {zm} has a convergent subsequence [13]. Suppose{zm}is not bounded, then passing to a subsequence if necessary,kzmk →+∞as m→+∞. By (H3), there exists C3>0 such that for allz∈R^2N,t∈[0,1]

z·Hz(t, z)−2H(t, z)≥C1|z|^β−C3. Therefore, we have

2f(zm)− hf⁰(zm), zmi= Z 1

0

[zm·Hz(t, zm)−2H(t, zm)]dt

≥ Z 1

0

[C1|zm|^β−C3]dt=C1

Z 1

0

|zm|^βdt−C3. This implies

R1

0 |zm|^βdt

kz_mk →0 as m→ ∞. (2.3)

Note that from (H3), 1< λ <1 +^β−_β¹. Letα=_β(λ^β−1

−1). Then α >1, αλ−1 =α− 1

β. (2.4)

By (H3), there existsC4>0 such that

|H_z(t, z)|^α≤C₂^α|z|^λα+C₄, ∀(t, z)∈[0,1]×R^2N. (2.5) Denotezm=z_m⁺+z_m⁻+z⁰_m∈E⁺⊕E⁻⊕E⁰. We have

hf⁰(zm), z_m⁺i=hAz_m⁺, z_m⁺i − Z 1

0

[Hz(t, zm)·z⁺_m]dt

≥ hAz_m⁺, z_m⁺i − Z 1

0

|Hz(t, zm)||z⁺_m|dt

≥ hAz_m⁺, z_m⁺i −( Z 1

0

|Hz(t, zm)|^α)^α1 ·Cαkz_m⁺k,

(2.6)

whereCα>0 is a constant independent ofm. By (2.5), Z 1

0

|H_z(t, z_m)|^αdt≤ Z 1

0

(C₂^α|z_m|^λα+C₄)dt

≤C₅( Z 1

0

|z_m|^βdt)^1/β( Z 1

0

|z_m|^{(αλ−1)·β−1β} dt)^1−1β +C₄

≤C₆( Z 1

0

|z_m|^β)^1/βkz_mk^(αλ−1)+C₄. Combining this inequality with (2.3) and (2.4) yields that

(R1

0 |Hz(t, zm)|^αdt)^α1

kz_mk ≤[C6(R1

0 |zm|^βdt)^1/β

kzmk^1/β ·kzmk^(αλ−1)

kzmk^α−β1 + C4

kz_mk^α]^α1 →0 asm→ ∞. By (2.6) we have

hAz⁺_m, z⁺_mi

kz_mk kz⁺mk ≤ kf⁰(zm)k kz_m⁺k kz_mk kzm⁺k +(R1

0 |Hz(t, zm)|^αdt)^α1

kz_mk · Cαkz_m⁺k kz⁺mk →0 asm→ ∞. This implies

kz_m⁺k

kzmk →0 asm→ ∞. (2.7)

Similary, we have

kz_m⁻k

kzmk →0 asm→ ∞. (2.8)

By (H3) there existC₇, C₈>0 such that

z·Hz(t, zm)−2H(t, z)≥C7|z| −C8, ∀(t, z)∈[0,1]×R^2N.

This implies

2f(zm)− hf⁰(zm), zmi= Z 1

0

[zm·Hz(t, zm)−2H(t, zm)]dt≥ Z 1

0

[C7|zm| −C8]dt

≥ Z 1

0

[C7|z⁰_m| −C7|z_m⁺| −C7|z_m⁻| −C8]dt

≥C9kz_m⁰k −C10(kz_m⁺k+kz_m⁻k+ 1).

Therefore, by (2.7) and (2.8) kz⁰_mk

kzmk →0 asm→ ∞. Combine this with (2.7) and (2.8), we get

1 = kzmk

kzmk ≤ kz_m⁺k+kz⁻_mk+kz_m⁰k

kzmk →0 asm→ ∞,

a contradiction. Therefore,{z_m} must be bounded.

Proof of Theorem 1.1 We prove thatf satisfies the conditions of Theorem 5.29 in [13].

Step 1: By (H1)–(H3), we have

H(t, z)≤a₁+a₂|z|^λ+1, ∀(t, z)∈[0,1]×R^2N. By (H2), for anyε >0, there existsδ >0 such that

H(t, z)≤ε|z|², ∀t∈[0,1], |z| ≤δ.

Therefore, there exists M =M(ε)>0 such that

H(t, z)≤ε|z|²+M|z|^λ+1, ∀(t, z)∈[0,1]×R^2N.

Note that λ+ 1 >2. By the same arguements as in [13, Lemma 6.16], there exist ρ >0 and ˜a >0, such that forz∈E1=E⁺

f(z)≥˜a ifkzk=ρ,

i.e., f satisfies (I7)(i) in [13, Theorem 5.29] with S=∂Bρ∩E1.

Step 2: Lete∈E⁺ withkek= 1 and ˜E=E⁻⊕E⁰⊕span{e}. We denote K=

z∈E˜:kzk= 1 , λ⁻= inf

z∈E⁻,kz⁻k=1|hAz⁻, z⁻i|, γ= (kAk λ⁻ )^1/2. Forz∈K, we write z=z⁻+z⁰+z⁺∈E.˜

i) Ifkz⁻k> γkz⁺+z⁰k, by (H1) we have, for anyr >0, f(rz) = 1

2 < Arz⁻, rz⁻i+1

2hArz⁺, rz⁺i − Z 1

0

H(t, z)dt

≤ −1

2λ⁻r²kz⁻k²+1

2kAkr²kz⁺k²≤0.

ii) Ifkz⁻k ≤γkz⁺+z⁰k, we have

1 =kzk²=kz⁻k²+kz⁺+z⁰k²≤(1 +γ²)kz⁺+z⁰k², i.e.,

kz⁺+z⁰k²≥ 1

1 +γ² >0. (2.9)

Denote ˜K={z∈K:kz⁻k ≤γkz⁺+z⁰k}. Claim: There existsε1>0 such that,∀u∈K,˜

meas{t∈[0,1] :|u(t)| ≥ε1} ≥ε1. (2.10) For otherwise,∀k >0,∃uk ∈K˜ such that

meas{t∈[0,1] :|uk(t)| ≥ 1 k}< 1

k. (2.11)

Write uk = u⁻_k +u⁰_k+u⁺_k ∈ E. Notice that dim(E˜ ⁰⊕span{e}) < +∞ and ku⁰_k+u⁺_kk ≤1. In the sense of subsequence, we have

u⁰_k+u⁺_k →u⁰₀+u⁺₀ ∈E⁰⊕span{e} ask→+∞. Then (2.9) implies that

ku⁰₀+u⁺₀k²≥ 1

γ²+ 1 >0. (2.12)

Note thatku⁻_kk ≤1, in the sense of subsequenceu⁻_k * u⁻₀ ∈E⁻ ask→+∞. Thus in the sense of subsequences,

u_k* u₀=u⁻₀ +u⁰₀+u⁺₀ ask→+∞. This means thatu_k→u₀ inL², i.e.,

Z 1

0

|uk−u0|²dt→0 as k→+∞. (2.13) By (2.12) we know that ku₀k >0. Therefore,R1

0 |u₀|²dt >0. Then there exist δ₁>0,δ₂>0 such that

meas{t∈[0,1] :|u₀(t)| ≥δ₁} ≥δ₂. (2.14) Otherwise, for alln >0, we must have

meas{t∈[0,1] :|u0(t)| ≥ 1

n}= 0, i.e., meas{t∈[0,1] :|u0(t)|< 1 n}= 1;

0<

Z 1

0

|u0|²dt < 1

n² ·1→0 as n→+∞.

We get a contradiction. Thus (2.14) holds. Let Ω0={t∈[0,1] :|u0(t)| ≥δ1}, Ωk ={t∈[0,1] :|uk(t)|<1/k}, and Ω^⊥_k = [0,1]\Ωk. By (2.11), we have meas(Ω_k∩Ω₀) = meas(Ω₀−Ω₀∩Ω^⊥_k)≥meas(Ω₀)−meas(Ω₀∩Ω^⊥_k)≥δ₂−1

k. (2.15) Letkbe large enough such thatδ2−_k¹ ≥¹₂δ2andδ1−¹_k ≥ ¹₂δ1. Then we have

|uk(t)−u0(t)|²≥(δ1−1 k)²≥(1

2δ1)², ∀t∈Ωk∩Ω0. This implies that

Z 1

0

|uk−u0|²dt≥ Z

Ω_k∩Ω₀

|uu−u0|²dt≥(1

2δ1)²·meas(Ωk∩Ω0)

≥(1

2δ1)²·(δ2− 1 k)≥(1

2δ1)²(1

2δ2)>0.

This is a contradiction to (2.13). Therefore the claim is true and (2.10) holds.

Forz=z⁻+z⁰+z⁺ ∈K, let Ω˜ z ={t∈[0,1] :|z(t)| ≥ε1}. By (1.2), for M =^k_ε^A₃^k

1

>0, there existsL₁>0 such that

H(t, x)≥M|x|², ∀|x| ≥L1, uniformly in t.

Chooser1≥L1/ε1. Forr≥r1,

H(t, rz(t))≥M|rz(t)|²≥M r²ε²₁, ∀t∈Ωz. By (H1), forr≥r1

f(rz) =1

2r²hAz⁺, z⁺i+1

2r²hAz⁻, z⁻i − Z 1

0

H(t, rz)dt

≤1

2kAkr²− Z

Ω_z

H(t, rz)dt≤1

2kAkr²−M r²ε²₁·meas(Ω_z)

≤1

2kAkr²−M ε³₁r²=−1

2kAkr²<0.

Therefore, we have proved that

f(rz)≤0, for anyz∈K andr≥r₁. (2.16) Let E2=E⁻⊕E⁰, Q={re: 0≤r≤2r1} ⊕ {z∈E2 :kzk ≤2r1}. By (H1) and (2.16) we havef|∂Q≤0, i.e.,f satisfies (I₇)(ii) in [13, Theorem 5.29].

Step 3: By Lemma 2.1, f satisfies the (PS) condition. Similar to the proof of [13, Theorem 6.10], by the linking theorem [13, Theorem 5.29], there exists a critical point z^∗∈E of f such thatf(z^∗)≥˜a >0. Moreover, z^∗ is a classical

solution of (1.1) and z^∗ is nonconstant by (H1).

Remark 2.2 i) Suppose H(t, z) = ¹₂hB(t)z, zi+ ˜H(t, z) with B(t) being a 2N ×2N matrix, continuous and 1-periodic intand ˜H(t, z) satisfies (1.2) and (H1)-(H3). We have the same conclusion as Theorem 1.1. The proof is similar and we omit it.

ii) SupposeH(t, z) =H(z) is independent ont, i.e., (1.1) is an autonomous Hamiltonian system. Then under similar conditions as (1.2) and (H1)-(H3), for anyT >0, the system (1.1) has a nonconstantT-periodic solution. Moreover, if H(z)∈C²(R^2N,R) and satisfies some strictly convex conditions such asH⁰⁰(x) is positive defininte for x 6= 0, then for any T > 0, (1.1) has a nonconstant T-periodic solution with minimal periodT. We omit the proof which is similar to the one in [4, 5].

iii) Suppose (1.4) holds, i.e.,

H(t, z) =H(z) =|z|²(ln(1 +|z|^p))^q, ∀(t, z)∈[0,1]×R^2N, wherep >1 andq >1. Obviously, (1.2), (H1) and (H2) hold. Note that z·Hz(z)−2H(z) =|z|²q(ln(1 +|z|^q))^q−1 p|z|^p

1 +|z|^p ≥ |z|²pq(ln 2)^q−1

2 , ∀|z| ≥1.

|H_z(z)| ≤2(ln(1 +|z|^p))^q|z|+ p|z|^p

1 +|z|^pq(ln(1 +|z|^p))^q−1|z| ≤2|z|⁵⁴, ∀|z| ≥L, forLbeing large enough. This implies (H3). By directly computation,H⁰⁰(z) is positive definite forz6= 0. Therefore, for anyT >0, (1.1) possesses aT-periodic solution with minimal periodT.

iv) There are many examples which satisfy (H1)-(H3) and (1.2) but do not satisfy (1.3). For example

H(t, z) =|z|²ln(1 +|z|²) ln(1 + 2|z|³).

Corollary 2.3 SupposeH(t, z) =|z|²h(t, z)withh∈C¹([0,1]×R^2N,R)being 1-periodic int and satisfies

(H1⁰) h(t, z)≥0, for all(t, z)∈[0,1]×R^2N.

(H2⁰) h(t, z)→0 as|z| →0;h(t, z)→+∞as|z| →+∞. (H3⁰) There exist0≤δ <1,L >0,ε0>0 andM >0 such that

|z|^δhz(t, z)·z≥ε0, |z||hz(t, z)| ≤M h, ∀|z| ≥L;

h(t, z)

|z|^γ →0 as|z| → ∞ for anyγ >0.

Then system (1.1) possesses a nonconstant 1-periodic solution.

Proof Obviously, (H1⁰)−(H3⁰) imply (1.2), (H1) and (H2).

z·Hz(t, z)−2H(t, z) =|z|²|hz(t, z)·z≥ε0|z|^2−δ, ∀|z| ≥L;

|Hz(t, z)| ≤ |2h(t, z)||z|+|z|²|hz(t, z)|

≤(2 +M)|z|h(t, z)≤(2 +M)|z|^1+γ, ∀|z| ≥L⁰.

Let β = 2−δ andλ= 1 +γ with 0< γ <(1−δ)/(2−δ). Then (H3) holds.

By Theorem 1.1 we get the conclusion.

3 Second order Hamiltonian System

Let E = W^1,2(S¹,R^N) with the norm k · k and inner product h·,·i. Then E⊂C(S¹,R^N) andkuk²=R1

0(|u˙|²+|u|²)dt. Define hKx, yi=

Z 1

0

x·ydt, ∀x, y∈E;

f(z) = 1

2h(id−K)z, zi − Z 1

0

V(t, z)dt, ∀z∈E.

Then K is compact, ker(id−K) =R^N, and the negative definite subspace of id−K,M⁻(id−K) ={0}, i.e.,E=E⁰⊕E⁺where E⁰= ker(id−K) andE⁺ is the positive definite subspace ofid−K. Note that (V1)–(V4) imply

V(t, x)≤d₂|x|^λ+1+d₃. (3.1) This implies thatf ∈C¹(E,R) and critical points off are 1-periodic solutions of (1.5) [11].

Lemma 3.1 Suppose (V1)–(V4) hold. Thenf satisfies the (PS) condition.

Proof Let{zm} be a (PS) sequence. Suppose{zm} is not bounded. Passing to a subsequence if necessary,kzmk →+∞asm→ ∞. Then by (V4)

2f(z_m)− hf⁰(z_m), z_mi= Z 1

0

[z_m·V⁰(t, z_m)−2V(t, z_m)]dt≥d₁ Z 1

0

|z_m|^βdt−d₄. This implies

R1 0 |zm|^βdt

kz_mk →0 asm→+∞. If (1.6) holds, we have

hf⁰(zm), z_m⁺i=h(id−K)z_m⁺, z_m⁺i − Z 1

0

V⁰(t, zm)·z⁺_mdt

≥ h(id−K)z_m⁺, z_m⁺i − kz_m⁺k_∞ Z 1

0

|V⁰(t, zm)|dt

≥ h(id−K)z_m⁺, z_m⁺i −d5kz⁺_mk( Z 1

0

|zm|^λdt+d6).

Sinceλ≤β, we have

kz_m⁺k

kz_mk →0 as m→+∞. (3.2)

If (1.7) holds, we have f(zm) = 1

2h(id−K)z_m⁺, z_m⁺i − Z 1

0

V(t, zm)dt

≥ 1

2h(id−K)z_m⁺, z_m⁺i −d5

Z 1

0

|zm|^1+λdt−d7

≥ h(id−K)z_m⁺, z_m⁺i −d8kzmk Z 1

0

|zm|^λdt−d7.

Sinceλ≤β, we obtain (3.2). On the other hand, (V1)–(V4) imply x·V⁰(t, x)−2V(t, x)≥d9|x| −d10, ∀t∈S¹×R^N. 2f(z_m)− hf⁰(z_m), z_mi=

Z 1

0

[z_m·V⁰(t, z_m)−2V(t, z_m)]dt

≥d₉ Z 1

0

|z_m|dt−d₁₀

≥d₉ Z 1

0

|z⁰_m|dt−d₉ Z 1

0

|z⁺_m|dt−d₁₀

≥d9kz_m⁰k −d11kz_m⁺k −d10. This implies

kz⁰_mk

kz_mk →0 asm→+∞. (3.3)

By (3.2) and (3.3), we get a contradiction. Therefore {zm} is bounded. By a standard argument,{zm} has a convergent subsequence [11].

Proof of Theorem 1.2 As in Step 1 of the proof of Theorem 1.1, by (V2) and (3.1), there exist ˜a >0,ρ >0 such that

f(z)≥a,˜ ∀z∈E⁺ withkzk=ρ.

Choosee∈E⁺ withkek= 1. Let ˜E= span{e} ⊕E⁰ andK={u∈E˜ : kuk= 1}. Note that dim ˜E < +∞. By using similar arguments as in the proof of (2.10), there existsε1>0 such that

meas{t∈[0,1] :|u(t)| ≥ε1} ≥ε1, ∀u∈K. (3.4) By (V1), (V3) and similar arguments as in the proof of Theorem 1.1, there existsr1>0 such that

f|∂Q ≤0, where Q={re: 0≤r≤2r₁} ⊕ {z∈E⁰:kzk ≤2r₁}. Now by Lemma 3.1, [13, Theorem 5.29], and (V1),f has a nonconstant critical pointz^∗such that f(z^∗)≥˜a >0. z^∗ is 1-periodic solution of (1.5).

Remark 3.2 (i) SupposeV(t, x) =V(x) is independent ontandV(x) satisfies (V1)–(V4). Then for any T > 0, (1.5) possesses a nonconstant T-periodic solution.

(ii) There are many examples which satisfy (V1)–(V4) but do not satisfy a condition similar to (1.3). For example,

V(t, x) = [1 + (sin 2πt)²]· |x|²ln(1 + 2|x|²); or V(t, x) =|x|²ln(1 +|x|²) ln(1 + 2|x|⁴).

By using similar arguments as in the proof of Theorem 1.2, we can prove the following corollary. Details are omited.

Corollary 3.3 SupposeV(t, x) =|x|²h(t, x)withh∈C¹(S¹×R^N,R)satisfies (V1⁰) h(t, x)≥0, ∀(t, x)∈S¹×R^N.

(V2⁰) h(t, x)→0 as|x| →0; h(t, x)→+∞as|x| →+∞. (V3⁰) There existL >0,λ >0,C₁, C₂>0 such that for t∈S¹

C₁|x|(h⁰(t, x)·x)≥h(t, x), h(t, x)≤C₂|x|^λ, ∀|x| ≥L.

Then (1.5) possesses a nonconstant 1-periodic solution.

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Guihua Fei

Department of Mathematics and statistics University of Minnesota

Duluth, MN 55812, USA e-mail : [email protected]