subharmonic oscillations of a class of hamiltonian systems

Nonlinear Analysis 68 (2008) 2697–2708www.elsevier.com/locate/na

Subharmonic oscillations of a class of Hamiltonian systems

Mohsen Timoumi∗

Department of Mathematics, Faculty of Sciences, 5000 Monastir, Tunisia

Received 23 September 2006; accepted 15 February 2007

Abstract

In this work, we prove the existence of subharmonics of the Hamiltonian system x = J H ′(t, x) when the Hamiltonian His coercive or resonant in a part of the variables uniformly in the other part. We derive the existence of such solutions by usingminimax theory.c© 2007 Elsevier Ltd. All rights reserved.

Keywords: Subharmonics; Hamiltonian systems; Minimax theory

1. Introduction

Let H : R × R2N−→ R, (t, x) 7−→ H(t, x) be a continuous function T -periodic (T > 0) in the first variable,

differentiable in the second variable, and H ′(t, x) =∂H∂x (t, x) is continuous. Consider the Hamiltonian system of

ordinary differential equations:

x = J H ′(t, x) (H)

where x ∈ R2N , J is the standard symplectic matrix:

J =

(0 −ININ 0

),

IN being the identity matrix of order N .There have been different works devoted to subharmonic solutions, i.e. of distinct kT -periodic solutions, of

Hamiltonian systems. They have been obtained by using many different techniques, for example, Morse theory andminimax theory. However, most of the results proving the existence of subharmonics have made use of a coercivity ora resonance assumption in all the variables on H , see [1–3,6–8]. The main goal of this work is to derive the existenceof such solutions for (H) when the Hamiltonian H is coercive or resonant only in a part of the variables.

In Section 3, we study an unbounded case and, in Section 4, a bounded case. For the resolution, we will apply aminimax theorem to a sequence of strongly indefinite functionals.

∗ Tel.: +216 96509963; fax: +216 73500278.E-mail address: m [email protected].

0362-546X/$ - see front matter c© 2007 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2007.02.016

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2698 M. Timoumi / Nonlinear Analysis 68 (2008) 2697–2708

2. Preliminaries

We will recall a minimax theorem: “generalized saddle point theorem [4]”, which will be useful in the proof of ourresults.

Given a Banach space E and a complete connected Finsler manifold V of class C2, we consider the spaceX = E × V . Let E = W ⊕ Z (topological direct sum) and En ⊕ Zn be a sequence of closed subspaces withZn ⊂ Z , Wn ⊂ W , 1 ≤ dim Wn < ∞. Define

Xn = En × V .

For f ∈ C1(X,R), we denote by fn = f|Xn . Then we have fn ∈ C1(Xn,R), for all n ≥ 1.

Definition 2.1. Let f ∈ C1(X,R). The function f satisfies the Palais–Smale condition with respect to (Xn) at a levelc ∈ R if every sequence (xn) satisfying

xn ∈ Xn, f (xn) −→ c, f ′n(xn) −→ 0

has a subsequence which converges in X to a critical point of f . The above property will be referred as the (P S)∗ccondition with respect to (Xn).

Theorem 2.1 (Generalized Saddle Point Theorem). Assume that there exist r > 0 and α < β ≤ γ such that

(a) f satisfies the (P S)∗c condition with respect to (Xn) for every c ∈ [β, γ ],(b) f (w, v) ≤ α for every (w, v) ∈ W × V such that ‖w‖ = r ,(c) f (z, v) ≥ β for every (z, v) ∈ Z × V ,(d) f (w, v) ≤ γ for every (w, v) ∈ W × V such that ‖w‖ ≤ r .

Then f −1([β, γ ]) contains at least cuplength (V )+ 1 critical points of f .

Remark 2.1. In (c) we may replace Z by ϕ + Z , ϕ ∈ W . Likewise in (b) we may replace W by W + ψ,ψ ∈ Z .

Now, for giving a variational formulation of (H), some preliminary materials on functional spaces and norms areneeded.

Consider the Hilbert space E = W12 ,2(S1,R2N ), S1

= R/T Z and the quadratic form Q defined in E by

Q(x) =12

∫ T

0(J x/x)dt

where (./.) inside the sign integral is the inner product in R2N . If x ∈ E , then x has a Fourier expansion

x(t) ≈

∑m∈Z

exp(

2πT

mt J

)xm,

where xm ∈ R2N and∑

m∈Z(1 + |m|)|xm |2 < ∞. By an easy calculation, we obtain

Q(x) = −π∑m∈Z

m|xm |2. (2.1)

Therefore Q is a continuous quadratic form on E .Consider the subspaces of E :

E0= R2N

E−=

{x ∈ E/x(t) =

∑m≥1

exp(

2πT

mt J

)xm a.e.

}

E+=

{x ∈ E/x(t) =

∑m≤−1

exp(

2πT

mt J

)xm a.e.

}.

M. Timoumi / Nonlinear Analysis 68 (2008) 2697–2708 2699

Then E = E0⊕ E+

⊕ E−. It is not difficult to verify that E0, E−, E+ are respectively the subspaces of E on whichQ is null, negative definite and positive definite, and these subspaces are orthogonal with respect to the bilinear form:

B(x, y) =12

∫ T

0(J x/y)dt, x, y ∈ E

associated with Q. If x ∈ E+ and y ∈ E−, then B(x, y) = 0 and Q(x + y) = Q(x)+ Q(y). It is also easy to checkthat E0, E− and E+ are mutually orthogonal in L2(S1,R2N ). It follows that if x = x+

+ x−+ x0

∈ E , then theexpression

‖x‖2

= Q(x+)− Q(x−)+ |x0|2

is an equivalent norm in E .

Proposition 2.2 ([5]). E is compactly embedded in L2(S1,R2N ). In particular there is a constant α > 0 such that

‖x‖L2 ≤ α‖x‖

for all x ∈ E.

3. Unbounded Hamiltonian

Consider a decomposition R2N= A ⊕ B of R2N with

A = space ≺ ei1 , . . . , ei p �, B = space ≺ ei p+1 , . . . , ei2N �

where 0 ≤ p ≤ 2N − 2, such that there exists an integer i0 ∈ {1, . . . , N } satisfying the condition ei0 , ei0+N ∈ B,where (ei )1≤i≤2N is the standard basis of R2N .

Let H : R × R2N−→ R, (t, x) 7−→ H(t, x) be a continuous function T -periodic (T > 0) in the first variable,

differentiable in the second variable, and H ′(t, x) =∂H∂x (t, x) is continuous. Consider the following assumptions:

(H0) H is periodic in the variables xi1 , . . . , xi p .(H1) There exists g ∈ L2([0, T ],R+) such that

∀x ∈ R2N ,∥∥H ′(t, x)

∥∥ ≤ g(t) a.e. t ∈ [0, T ].

(H2) Either

(i) lim|PB (x)|→+∞

H(t, x) = +∞, uniformly in (t, PA(x)) ∈ R × A,

or

(ii) lim|PB (x)|→+∞

H(t, x) = −∞, uniformly in (t, PA(x)) ∈ R × A,

where PA (resp. PB) is the projection of X into A(resp. B).

Our first main result in this section is:

Theorem 3.1. Assume (H0), (H1), (H2) hold. Then for all integer k ≥ 1, the Hamiltonian system

x = J H ′(t, x) (H)

possesses at least (p + 1) kT -periodic solutions geometrically distinct x1k , . . . , x p+1

k satisfying∥∥x i

k

∥∥∞

−→ +∞ ask −→ ∞, for all i = 1, . . . , p + 1, where ‖x‖∞ = supt∈R |x(t)|.

Now, consider the following assumption:

(H′

2) Either

(i′) lim|PB (x)|→+∞

≺ H ′(t, x), PB(x) �= +∞, uniformly in (t, PA(x)) ∈ R × A,

or

(ii′) lim|PB (x)|→+∞

≺ H ′(t, x), PB(x) �= −∞, uniformly in (t, PA(x)) ∈ R × A.


Remark 3.1. If the assumptions (H0), (H1) and (H′

2) hold, then the assumption (H2) holds. In fact, assume(H0), (H1) and (H

′

2)(i′

) hold and let R > 0 be such that

≺ H ′(t, x), PB(x) �≥ 1, for |PB(x)| ≥ R.

Then, for |PB(x)| ≥ R, we have

H(t, x) = H(t, PA(x))+

∫ R|PB (x)|

0≺ H ′(t, PA(x)+ s PB(x)), PB(x) � ds

+

∫ 1

R|PB (x)|

≺ H ′(t, PA(x)+ s PB(x)), PB(x) � ds

≥ H(t, PA(x))− g(t)R + Log(

|PB(x)|

R

),

and since H is bounded in R × A, (H2)(i) follows.

Our second main result in this section is:

Theorem 3.2. If the assumptions (H0), (H1) and (H′

2) hold, then the results of Theorem 3.1 are true. Moreover,for any sufficiently large prime number k and for any i = 1, . . . , p + 1, kT is the minimal period of x i

k , wherex i

k, i = 1, . . . , p + 1 are the solutions of the system (H) obtained in Theorem 3.1.

In the case of convexity, we have

Corollary 3.3. If H is convex or concave in the second vatiable and satisfies assumptions (H0), (H1), (H2), thenthe conclusion of Theorem 3.2 is true.

Example 3.1. Let A : R × Rn−→ R, (t, r) 7−→ A(t, r) be a continuous function periodic in the variables t, r ,

differentiable in the variable r and ∂A∂r is continuous; for example A(t, r) = cos(t) cos(k.r)v, where k, v are two fixed

vectors in Rn . Consider the Hamiltonian

H(t, r, p) = [1 + |p − A(t, r)|2]12 .

It is easy to see that H is convex and satisfies (H0), (H1) and (H2) with A = Rn× {0} and B = {0} × Rn .

Proof of Theorem 3.1. By making the change of variables t 7−→tk , the system (H) transforms to

u = k J H ′(kt, u). (Hk)

Hence, to find kT -periodic solutions of (H), it suffices to find T -periodic solutions of (Hk).

Remark 3.2. If u(t) is a periodic solution of (Hk), then u(−t) is a periodic solution of the system

u = −k J H ′(−kt, u).

Moreover, −H(−t, x) satisfies (H2)(i) whenever H(t, x) satisfies (H2)(ii). Hence, in the following, we will assumethat H satisfies (H2)(i).

Consider the family of functionals ( fk)k∈N defined on the space E = W12 ,2(S1,R2N ) by:

fk(u) =12

∫ T

0(J u/u)dt + k

∫ T

0H(kt, u)dt.

It is well known that fk ∈ C1(X,R) and critical points of fk are T -periodic solutions of (Hk).We will show that, for every positive integer k ≥ 1, there exist at least (p + 1) T -periodic solutions u1

k, . . . , u p+1k

of (Hk) geometrically distinct such that for all i = 1, . . . , p + 1, the sequence (uik) has the following property

limk→∞

1k

fk(uik) = +∞. (3.1)


This will be obtained by using some estimates on the critical levels of fk given by Theorem 2.1, and implies that forall i = 1, . . . , p + 1, the sequence (‖ui

k‖∞)k∈N goes to infinity as k goes to infinity.Now, let W = E−, Z = E+

⊕ B and V be the quotient space

V = A/{x ∼ x + ei , i = i1, . . . , i p}

which is nothing but the torus T p, and let us fix k ≥ 1. We regard the functional fk as defined on the spaceX = (W ⊕ Z)× V as follows

fk(u + v) =12

∫ T

0(J u/u)dt + k

∫ T

0H(kt, u + v)dt

and we apply Theorem 2.1 to this functional, with respect to the sequence of subspaces Xn = En × V , where

En =

{x ∈ E/x(t) =

∑|m|≤n

exp(

2πT

mt J

)um a.e.

}, n ≥ 0.

First, we will check the Palais–Smale condition.

Lemma 3.1. For all level c ∈ R, the functional fk satisfies the (P S)∗c condition with respect to the sequence (Xn)n∈N.

Proof. Let (un, vn)n∈N be a sequence such that for all n ∈ N, (un, vn) ∈ Xn and

fk(un + vn) −→ c and f ′

k,n(un + vn) −→ 0 as n −→ ∞,

where fk,n is the functional fk restricted to Xn . Let

un(t) =

∑|m|≤n

exp(

2πT

mt J

)un,m

be an element of En . Applying f ′

k,n(un + vn) to u+n − u−

n , we obtain

f ′

k,n(un + vn).(u+n − u−

n ) = π∑

1≤|m|≤n

|m| |un,m |2+ k

∫ T

0H ′(kt, un + vn).(u

+n − u−

n )dt.

It follows from (H1) that the sequence (un), where un = un − u0n , with u0

n is the mean value of un , is bounded in X .By applying mean value theorem to H(kt, .), we obtain

fk(un + un) = −π∑

1≤|m|≤n

m|un,m |2+ k

∫ T

0H(kt, u0

n + vn)dt + k∫ T

0H ′(kt, yn).undt

and since ( fk(un + vn)) is bounded, we deduce from assumptions (H1), (H2) and Proposition 2.2 that (u0n) has to be

bounded. So (un) is bounded in X . Going, if necessary, to a subsequence, we can assume that un ⇀ u, u0n −→ u0

and vn −→ v in X . Notice that

Q(u+n − u+) = ( f ′

k,n(un + vn)− f ′

k,n(u + v)).(u+n − u+)

− k∫ T

0(H ′(kt, un + vn)− H ′(kt, u + v)).(u+

n − u+)dt

which implies that u+n −→ u+ in X . Similarly, u−

n −→ u− in X . It follows that un −→ u in X and f ′

k(u + v) = 0.So fk satisfies the (P S)∗c condition for all c ∈ R. The Lemma 3.1 is proved. �

Now, by mean value theorem and assumption (H1), it is easy to check that there exist constants a, b, c ∈ R suchthat

∀(u, v) ∈ W × V, fk(u + v) ≤ −π

2‖u‖

2+ a ‖u‖ + b

and

∀u = u++ u0

∈ Z , v ∈ V, fk(u + v) ≥ +π

2‖u+

‖2− c‖u+

‖ + k∫ T

0H(kt, u0

+ v)dt,


where c is a constant. Then it is clear from assumption (H2)(i), that the functional fk satisfies assumptions (a), (b),(c) of Theorem 2.1. It follows, from Lemma 3.1, that the functional fk satisfies all assumptions of Theorem 2.1, so fk

has at least (p + 1) critical points u1k, . . . , u p+1

k ∈ X satisfying for all i = 1, . . . , p + 1,

fk(uik) ≥ inf

u∈Z×Vfk(u).

Therefore the system (Hk) possesses at least (p + 1) T -periodic solutions geometrically distinct.It remains to prove that for all i = 1, . . . , p + 1, the sequence (ui

k), defined above, satisfies property (3.1). For this,we will need the following two lemmas.

Lemma 3.2. Let i0 ∈ {1, . . . , N } be such that ei0 , ei0+N ∈ B and given

u(t) = e2πT t J ei0 + u+(t)+ u0

+ v

with u+∈ X+, u0

∈ B, v ∈ V . Then we have

PB(u(t)) 6= 0, for a.e. t ∈ [0, T ].

Proof. Assume PB(u(t)) = 0 for a.e. t ∈ [0, T ]. We have

u+(t) ≈

∑m≤−1

exp(

2πT

mt J

)um a.e. t ∈ [0, T ]

≈

∑m≤−1

[cos

(2πT

mt

)um + sin

(2πT

mt

)J um

]a.e. t ∈ [0, T ],

where um ∈ R2N . Denote um =∑N

j=1 αm, j e j +∑N

j=1 βm, j e j+N , with αm, j , βm, j ∈ R, then

u+(t) ≈

N∑j=1

∑m≤−1

[αm, j cos

(2πT

mt

)− βm, j sin

(2πT

mt

)]e j

+

N∑j=1

∑m≤−1

[βm, j cos

(2πT

mt

)+ αm, j sin

(2πT

mt

)]e j+N a.e.

But PB(u(t)) = 0 implies Pei0(u(t)) = 0 and Pei0+N (u(t)) = 0, which gives us

cos(

2πT

t

)+

∑m≤−1

[αm,i0 cos

(2πT

mt

)− βm,i0 sin

(2πT

mt

)]+ Pei0

(u0) = 0

and

sin(

2πT

t

)+

∑m≤−1

[βm,i0 cos

(2πT

mt

)+ αm,i0 sin

(2πT

mt

)]+ Pei0+N (u

0) = 0

then we obtain 1 + α−1,i0 = 0, 1 − α−1,i0 = 0 which is impossible. The proof of Lemma 3.2 is complete. �

Lemma 3.3. Suppose that H ∈ C1(R × R2N ,R) satisfies (H0) and (H2), then

infu∈Z×V

fk(√

kϕ + u)

k−→ +∞, as k −→ +∞, (3.2)

where ϕ(t) =1

√π

e2πT t J ei0 , with i0 ∈ {1, . . . , N } is such that ei0 , ei0+N ∈ B.

Proof. Arguing by contradiction and assuming that there exist sequences k j −→ ∞, (u j ) ⊂ Z × V , and a constantC1 ∈ R such that

fk j (√

k jϕ + u j ) ≤ k j C1, ∀ j ∈ N. (3.3)


Taking u j =√

k j (u+

j + u0j + v j ) with u+

j ∈ E+, u0j ∈ B, v j ∈ V , we obtain, by an easy calculation

fk j (√

k jϕ + u j ) = k j

[‖u+

j ‖2− 1 +

∫ T

0H(k j t,

√k j (ϕ + u+

j + u0j + v j ))dt

]. (3.4)

By assumptions (H0) and (H2)(i), the Hamiltonian H is bounded from below, so we deduce from (3.4) that thereexists a constant C2 > 0 such that

fk j (√

k jϕ + u j ) ≥ k j [‖u+

j ‖2− C2]. (3.5)

Combining this with (3.3), we conclude that (u+

j ) is bounded in X . Taking a subsequence if necessary we can findu+

∈ E+ such that

u+

j (t) −→ u+(t) as j −→ ∞ for a.e. t ∈ [0, T ]. (3.6)

We claim that (u0j ) is also bounded in X . Indeed, if we suppose otherwise, (3.6) implies that√

k j |PB(ϕ(t)+ u+

j (t)+ u0j + v j )| −→ +∞ as j −→ ∞ for a.e. t ∈ [0, T ]. (3.7)

Consequently, by (H2) and Fatou’s lemma, we obtain∫ T

0H(k j t,

√k j (ϕ(t)+ u+

j (t)+ u0j + v j ))dt −→ ∞ as j −→ ∞ (3.8)

and we deduce from equality (3.4) that

1k j

fk j (√

k jϕ + u j ) −→ ∞ as j −→ ∞ (3.9)

which contradicts (3.3) and proves our claim. Going if necessary to a subsequence, we can assume that there existsu0

∈ B and v ∈ V such that for a.e t ∈ [0, T ]

u+

j (t)+ u0j + v j −→ u(t) = u+(t)+ u0

+ v, as j −→ ∞. (3.10)

By Lemma 3.2, we know that PB(ϕ(t)+ u(t)) 6= 0 for almost every t ∈ [0, T ]. Therefore√k j |PB(ϕ(t)+ u+

j (t)+ u0j + v j )| −→ +∞ as j −→ ∞, for a.e. t ∈ [0, T ]. (3.11)

Using (3.11), (H2) and Fatou’s lemma, we obtain (3.8), which contradicts (3.3). Therefore (3.2) must hold andLemma 3.3 is proved. �

Now, Theorem 2.1 and Remark 2.1 imply that for all integer k ≥ 1 and all i = 1, . . . , p + 1, we have

fk(uik) = bi

k ≥ infu∈Z×V

fk(√

kϕ + u), (3.12)

with x ik(t) = ui

k(tk ) a kT -periodic solution of (H). Therefore, we obtain from Lemma 3.3

bik

k−→ +∞, as k −→ ∞. (3.13)

We claim that∥∥x i

k

∥∥∞

=∥∥ui

k

∥∥∞

−→ ∞ as k −→ ∞. Indeed, if we suppose otherwise, (uk) possesses a boundedsubsequence (ukn ). Since

fkn (uikn)

kn= −

12

∫ T

0H ′(kn t, ui

kn).ui

kndt +

∫ T

0H(kn t, ui

kn)dt

the sequencebi

knkn

is bounded, contrary to (3.12). Consequently, we have

limk→∞

‖x ik‖∞ = lim

k→∞‖ui

k‖∞ = +∞, (3.14)

that concludes the proof of Theorem 3.1. �


Proof of Theorem 3.2. Consider the functional

gk(x) =12

∫ kT

0(J x/x)dt +

∫ kT

0H(t, x)dt

defined in the space Xk defined as the space X introduced above, with E = W12 ,2(S1,R2N ) and S1

= R/kT Z. Letx i

k(t) = uik(

tk ), i = 1, . . . , p + 1, the sequences obtained above. It is easy to see that for all i = 1, . . . , p + 1 and

k ≥ 1, x ik is a critical point of gk and

limk→∞

1k

gk(xik) = +∞. (3.15)

In a first step, we will show that the set ST of T −periodic solutions of (H) is bounded in X . Assume by contradictionthat there exists a sequence (xk) in ST such that ‖xk‖ −→ ∞ in X . Let us write xk = x+

k + x−

k + x0k + vk where

x+

k ∈ E+, x−

k ∈ E−, x0k ∈ B and vk ∈ V . Multiplying both sides of the identity

J xk + H ′(t, xk) = 0 (3.16)

by x+

k − x−

k and integrating, we obtain

‖xk‖2+

∫ T

0≺ H ′(t, xk), x+

k − x−

k � dt = 0.

Using assumption (H1), we easily deduce that (xk) is bounded in X . Therefore (xk) is bounded in L∞(0, T ) and theidentity

‖xk‖2

= ‖xk‖2+ |x0

k |2+ |vk |

2

shows that∣∣x0

k

∣∣ −→ ∞ as k −→ ∞ and so

min0≤t≤T

|xk(t)| −→ ∞, as k −→ ∞. (3.17)

Multiplying (3.16) by PB(xk) and integrating, we obtain∫ T

0≺ J xk, PB(xk) � dt +

∫ T

0≺ H ′(t, xk), PB(xk) � dt = 0. (3.18)

Since (xk) is bounded in X , we deduce from (3.18) that the sequence (∫ T

0 ≺ H ′(t, xk), PB(xk) � dt) is also bounded.

But this is in contradiction with (3.17) and assumption (H′

2)(i). Consequently ST is bounded. As a consequence,g1(ST ) is bounded, and since for any x ∈ ST one has gk(x) = kg1(x), we have

∃c > 0 : ∀x ∈ ST , ∀k ≥ 1,1k

|gk(x)| ≤ c. (3.19)

Consequently, (3.15) and (3.19) show that for k sufficiently large and i = 1, . . . , p + 1, we have x ik 6∈ ST . So if k is

chosen to be a prime number, the minimal period of x ik has to be kT and the proof is complete. �

Proof of Corollary 3.3. If H is convex in the second variable x , then, for every x ∈ R2n , we have

H(t, PA(x)) ≥ H(t, x)+ ≺ H ′(t, x),−PB(x) �

which shows that assumptions (H0), (H2)(i) imply (H′

2)(i′

).Similarly, If H is concave in the second variable x , then, for every x ∈ R2n , we have

−H(t, PA(x)) ≥ −H(t, x)+ ≺ −H ′(t, x),−PB(x) �

which shows that assumptions (H0), (H2)(ii) imply (H′

2)(ii′

).We conclude that if H is convex or concave and satisfies assumptions (H0), (H1) and (H2) then it satisfies

assumptions (H0), (H1) and (H′

2) and the conclusion of Theorem 3.2 follows. �


4. Bounded Hamiltonian

Let H : R × R2N−→ R, (t, x) 7−→ H(t, x) be a continuous function, T -periodic (T > 0) in the first variable,

differentiable with respect to the second variable, and H ′(t, x) =∂H∂x (t, x) is continuous. Consider a decomposition

R2N= A ⊕ B of R2N as in Section 3 and consider the following assumptions:

(H0) H is periodic in the variables xi1 , . . . , xi p .

(H1) H(t, x) 6= 0,∀t ∈ R,∀x ∈ R2N .(H2) lim|PB (x)|→∞ H(t, x) = 0, uniformly in t ∈ R and PA(x) ∈ A.(H3) lim|PB (x)|→∞ H ′(t, x) = 0, uniformly in t ∈ R and PA(x) ∈ A.

Our main result in this section is:

Theorem 4.1. Assume (H0), (H1), (H2), (H3) hold. Then for all integer k ≥ 1, the Hamiltonian system

x = J H ′(t, x) (H)

possesses at least (p + 1) kT -periodic solutions geometrically distinct x1k , . . . , x p+1

k . Furthermore, (H) possesses asequence (x j ) of distinct k j T -periodic solutions, with k j −→ ∞, as j −→ ∞.

Proof of Theorem 4.1. By assumption (H1), we can assume that

∀t ∈ R, ∀x ∈ R2N , H(t, x) > 0. (4.1)

The other case is the same.As in Section 3, to find kT -periodic solutions of (H), it suffices to find T -periodic solutions of the transformed

system

u = k J H ′(kt, u). (Hk)

For this, we will apply Theorem 2.1 to the family of continuously differentiable functionals

fk(u) =12

∫ T

0(J u/u)dt + k

∫ T

0H(kt, u)dt

defined on the space X = (W ⊕ Z)× V introduced above, with W = E−⊕ B and Z = E+. We have

∀(u, v) ∈ Z × V, fk(u + v) =12

‖u‖2+ k

∫ T

0H(kt, u + v)dt.

Let (un, vn) be a minimizing sequence: fk(un + vn) −→ infZ×V fk . Since, by assumptions (H0) and (H2), theHamiltonian H is bounded, then the sequence (un) is bounded in X . Therefore, up to a subsequence, there existsu ∈ Z such that (un) is weakly convergent to u. Moreover, the embedding map X −→ L2, u 7−→ u is compact, soun −→ u in L2. Then, by taking a subsequence if it is necessary, we can assume un(t) → u(t) a.e., vn −→ v and byLebesgue’s theorem, we have∫ T

0H(kt, un + vn)dt −→

∫ T

0H(kt, u + v)dt.

Now, it is not difficult to see that fk(un + vn) −→ fk(u + v) and then fk achieves its minimum on Z × V at u + v.We then have

βk = infZ×V

fk ≥

∫ T

0H(kt, u + v)dt > 0. (4.2)

Let 0 < αk < βk , we have for (u = u−+ u0, v) ∈ W × V

fk(u) = −12

∥∥u−∥∥2

+ k∫ T

0H(kt, u−

+ u0+ v)dt. (4.3)


By assumptions (H0), (H2) and (4.3), it is easy to see that

lim‖u−‖→∞

fk(u + v) = −∞, lim|u0|→∞

fk(u + v) ≤ 0.

So there exists a constant rk > 0 such that

∀(u, v) ∈ W × V, ‖u‖ = rk ⇒ fk(u + v) ≤ αk . (4.4)

The functional fk is also bounded from above on Brk × V by a constant γk ≥ βk , where Brk is the closed disc in Wcentered in zero with radius rk . Therefore fk verifies assumptions (b), (c) and (d) of Theorem 2.1.

The next result establishes the Palais–Smale condition.

Lemma 4.1. For all c 6= 0, fk satisfies the Palais–Smale condition at level c with respect to the sequence of subspaces

Xn =

{u ∈ E / u(t) =

∑1≤|m|≤n

exp(

2πmt

TJ

)um

}× V .

Proof. Let c 6= 0 and (un) be a sequence such that

∀n ∈ N, un ∈ Xn, fk(un) −→ c, f ′

k,n(un) −→ 0,

where fk,n is the restriction of fk to the subspace Xn . Set un = un + u0n + vn , with u0

n ∈ B, vn ∈ V . By formula (2.1),we have

f ′

k,n(un)(u+n − u−

n ) = π∑

1≤|m|≤n

|m|∣∣um

∣∣2+ k

∫ T

0H ′(kt, un).(u

+n − u−

n )dt (4.5)

and we deduce, from assumptions (H0) and (H3), the inequality

‖un‖2

≤ const ‖un‖ .

So (un) is bounded in X and since X is compactly embedded in L2 we can assume that un(t) −→ u(t) a.e. We claimthat (uo

n) is bounded. Otherwise, by taking a subsequence if necessary and using Lebesgue’s theorem, we obtain fromassumptions (H2), (H3)

limn→∞

∫ T

0H(kt, un(t))dt = 0, (4.6)

and

limn→∞

∫ T

0

∣∣H ′(kt, un(t))∣∣2 dt = 0. (4.7)

Since (un) is bounded, then by Holder’s inequality and Proposition 2.2 there exists a constant C > 0 such that∣∣∣∣∫ T

0H ′(kt, un).(u

+n − u−

n )dt

∣∣∣∣ ≤ C∥∥H ′(kt, un)

∥∥L2 .

We deduce from (4.7) that∫ T

0H ′(kt, un).(u

+n − u−

n )dt −→ 0, as n −→ ∞. (4.8)

Elsewhere, we have∣∣ f ′

k,n(un).(u+n − u−

n )∣∣ ≤

∥∥ f ′

k,n(un)∥∥

X∗‖un‖ ≤ M

∥∥ f ′

k,n(un)∥∥

and so

f ′

k,n(un).(u+n − u−

n ) −→ 0 as n −→ ∞. (4.9)


Consequently, we deduce from (4.5), (4.8) and (4.9) that (un) goes to zero in X as n −→ ∞. Therefore by (4.6) weconclude that

fk(un) = π∑

1≤|m|≤n

m∣∣um

∣∣2+ k

∫ T

0H(kt, un)dt −→ 0 as n −→ ∞

which contradicts the fact that fk(un) −→ c 6= 0. Then (un) is bounded in X . Going, if necessary, to a subsequence,we can assume that un ⇀ u weakly, u0

n −→ u0 and vn −→ v in X . Notice that

Q(u+n − u+) = ( f ′

k(un)− f ′

k(u)).(u+n − u+)− k

∫ T

0(H ′(kt, un)− H ′(kt, u)).(u+

n − u+)dt

which implies that u+n −→ u+ in X . Similarly u−

n −→ u− in X . It follows that un −→ u in X and f ′

k(u) = 0. So fksatisfies the (P S)∗c condition for all level c 6= 0. �

Consequently the functional fk verifies all the assumptions of Theorem 2.1, so fk has at least cuplength (V ) + 1critical points satisfying:

fk(u) ≥ infZ×V

fk = βk, (4.10)

and since V is the torus T p, then cuplength(V ) = p and the first part of Theorem 4.1 is proved.Finally, we will prove the existence of subharmonics for the system (H). For this, we will need:

Lemma 4.2. Assume H satisfies (4.1) and assumptions (H0), (H2). Then, for all ε > 0, there exists an integer k0 ∈ Nsuch that

supu∈W×V

fk

(√kεπϕ + u

)k

< 2ε, ∀k ≥ k0, (4.11)

where ϕ = exp( 2πT t J )ei0 , with i0 ∈ {1, . . . , N } is such that ei0 , ei0+N ∈ B.

Proof. Arguing by contradiction, we suppose there exist sequences (k j ) ⊂ N, k j −→ ∞, as j −→ ∞ and(u j ) ⊂ W × V such that for every large j ∈ N

fk j

(√k j

√ε

πϕ + u j

)≥ 2εk j . (4.12)

By taking u j =√

k j (u−

j + u0j + v j ), with u−

j ∈ E−, u0j ∈ B and v j ∈ V and using formula (2.1), we obtain

fk j

(√k j

√ε

πϕ + u j

)= k j

[ε −

∥∥∥u−

j

∥∥∥2+

∫ T

0H

(k j t,

√k j

√ε

πϕ + u j

)dt

]. (4.13)

Since H is bounded, we deduce, from (4.12) and (4.13), that the sequence (u−

j ) is bounded in X . Without loss ofgenerality, we may suppose that there exists u−

∈ X− such that

u−

j (t) −→ u−(t) as j −→ ∞ for a.e. t ∈ [0, T ].

We claim that (u0j ) is also bounded in X . Otherwise, we may assume that

√k j

∣∣∣∣√ ε

πϕ(t)+ u−

j (t)+ u0j + v j

∣∣∣∣ −→ +∞ as j −→ ∞ for a.e. t ∈ [0, T ].

Consequently, by assumption (H2) and Lebesgue’s theorem, we obtain∫ T

0H

(k j t,

√k j

(√ε

πϕ + u−

j + u0j + v j

))dt −→ 0 as j −→ ∞.


It follows from (4.13) that, for j sufficiently large, we have

fk j

(√k j

√ε

πϕ + u j

)< 2εk j . (4.14)

But (4.14) contradicts (4.12), thus (u0j ) must be bounded in X and we may suppose, without loss of generality, that

there exists u0∈ B such that, for a.e. t ∈ [0, T ]√

ε

πϕ(t)+ u−

j (t)+ u0j + v j −→ u(t) =

√ε

πϕ(t)+ u−(t)+ u0

+ v as j −→ ∞.

Since by Lemma 3.2, PB(u(t)) 6= 0 for a.e. t ∈ [0, T ], we get, by the same argument used above, that (4.14) holdsfor j sufficiently large, which contradicts (4.12) and concludes Lemma 4.2. �

Now, we apply induction to obtain a family (x j ) of distinct k j T -periodic solutions of (H) with k j −→ ∞ asj −→ ∞. We will use property (4.10)

For k = 1, f1 possesses a critical point u1 ∈ X satisfying

f1(u1) ≥ β1 > 0.

Furthermore, x1 = u1 is a T -periodic solution of (H). Next, we consider m ≥ 1 and suppose that (x j ), 1 ≤ j ≤ m, isa family of distinct k j T -periodic solutions of (H), with k j ≥ j, 1 ≤ j ≤ m, k1 = 1, such that

fk j (u j ) > 0 for 1 ≤ j ≤ m, (4.15)

where u j (t) = x j (k j t). Taking 0 < ε < min{

fk j (u j )

k j/ 1 ≤ j ≤ m

}, by Lemma 4.2, there exists km+1 ≥ m + 1 such

that for every u ∈ W

fkm+1

(√km+1

√ε

πϕ + u

)< 2εkm+1.

By Lemma 4.2 and Theorem 2.1 (Remark 2.1), fkm+1 possesses a critical point um+1 satisfying

0 < βkm+1 ≤ fkm+1(um+1) < 2εkm+1, (4.16)

and xm+1(t) = um+1(t

km+1) is a km+1T -periodic solution of (H). If, for some j ∈ {1, . . . ,m} , xm+1(t) =

um+1(t

km+1) = u j (

tk j) = x j (t), we would have

fkm+1(um+1) =km+1

k jfk j (u j ) ≥ 2εkm+1.

But this contradicts (4.16). Thus, (x j ), 1 ≤ j ≤ m + 1, is a family of distinct k j T -periodic solutions of (H), withk j ≥ j, 1 ≤ j ≤ m + 1. The proof of Theorem 4.1 is complete. �

Acknowledgment

I wish to thank the referee for suggestions and corrections.

References

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subharmonic oscillations of a class of hamiltonian systems

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