introduction - roma tre university

MULTIBUMP SOLUTIONS HOMOCLINIC TO PERIODICORBITS OF LARGE ENERGY IN A CENTER MANIFOLD

VITTORIO COTI ZELATI AND MARTA MACRI

1. Introduction

Variational methods have been successfully used to prove existence of solutionshomoclinic to stationary, hyperbolic rest points of periodically forced Hamiltoniansystems starting with the papers [7, 13] and [24, 23], see also [15]. A very nat-ural and interesting generalization of such a problem is that of finding solutionshomoclinic to invariant tori. Such a problem has been extensively studied usingperturbative methods, also in connection with Arnold’s diffusion (see the papers[12, 4]). Variational methods have been used to tackle such a problem in the papers[19, 6, 22, 8, 9, 10, 11]. In all these papers the invariant tori to which the solutionsare asymptotic have some minimizing properties.

In a couple of recent papers (see [1, 2]), Patrick Bernard has proved, usingvariational methods, the existence of solutions homoclinic to periodic orbits ina center manifold for a class of autonomous Hamiltonian systems. One of theinteresting features of this result is the fact that in this case the periodic solutionshave no minimizing property.

Such a problem has also been investigated using perturbative methods, see thepapers [4, 5, 16, 17, 18, 20] and the recent [25] (where also infinite-dimensionalHamiltonian systems are considered). The results in these papers show that oneshould expect to find a lot of homoclinic solutions in this setting and even chaoticbehavior.

In this paper we show existence of multibump solutions, homoclinic to periodicorbits in a center manifold, in a non-perturbative setting. This is a first step inproving that such a class of systems has a chaotic behavior. The systems we considerare in the class of Hamiltonians studied by Bernard in [1, 2] (see also [14]).

To be more specific, we consider the Hamiltonian system

H(x, x, q, q) =12(x2 + ω2x2) +

12q2 − (1 + δ(x))V (q),

where we assume that V ∈ C2(R) satisfies(V1) V (q + 2π) = V (q) for all q ∈ R;(V2) 0 = V (0) < V (q) for all q ∈ R \ 2πZ;(V3) V ′′(0) = µ > 0;

while on δ ∈ C2(R) we assume(δ1) 0 ≤ δ ≤ δ(s) ≤ δ for all s ∈ R;(δ2) δ = lims→−∞ δ(s) < lims→+∞ δ(s) = δ;(δ3) δ − δ ≤ 1 + δ;(δ4) δ′(s) → 0 as s→ ±∞;

Remark 1.1. Such assumptions are satisfied, for example, if V (q) = (1− cos q) andδ(s) = (π + 2 arctan s)/2π.

Supported by MIUR, project “Variational Methods and Nonlinear Differential Equations”.

1

2 VITTORIO COTI ZELATI AND MARTA MACRI

To such an Hamiltonian correspond the following Lagrangian

L(x, x, q, q) =12(x2 − ω2x2) +

12q2 + (1 + δ(x))V (q),

and the corresponding Euler-Lagrange equations

(EL)

{q(t) = (1 + δ(x(t)))V ′(q(t))x(t) + ω2x(t) = δ′(x(t))V (q(t))

Let us remark that the point P0 = (x = 0, x = 0, q = 0, q = 0) is a saddle-centerstationary point for the associated Hamiltonian system.

Such a stationary point has a one dimensional stable and unstable manifoldand a center manifold, which in this particular situation is simply the manifoldq = 0, q = 0, which is foliated by the periodic orbits xR(t) = R cos(ωt + φ),x = −R sin(ωt+φ), orbits which are hyperbolic with respect to their energy surface.

Then, just by dimension arguments, one expects to find solutions homoclinic toperiodic orbits in the center manifold, while in general there could be no homoclinicto P0.

An interesting result contained in the paper [3] shows that there are Hamiltoniansystems which have solutions homoclinic to xR for all R > 0 and small (even if theremight be no homoclinic to P0).

In this paper we look for multibump solutions which are homoclinic to periodicorbits in the central manifold of the saddle-center critical point (x, x, q = 0, q = 0),that is we look for solutions (x(t), q(t)) of (EL) such that, for given k ∈ N andτ > 0, satisfy

(1.2)

limt→−∞ q(t) = 0 and limt→+∞ q(t) = 2kπ,q(t) ≈ 2jπ for all t ∈ [αj , αj + τ ], j = 0, . . . , klimt→−∞ |x(t)−R cos(ωt+ ϕ1)| = 0,limt→+∞ |x(t)−R cos(ωt+ ϕ2)| = 0,

for some R > 0, αj ∈ R, ϕ1, ϕ2 ∈ [0, 2π]. Remark that the energy of such a solutionis given by

E =12ω2R2.

Our result is the following:

Theorem 1.3. There is ω1 such that, for all 0 < ω < ω1, k ∈ N and τ > 0 thereis R and T such that for all R ≥ R there is a solution (x(t), q(t)) of system (EL)satisfying

limt→−∞

q(t) = 0 and limt→+∞

q(t) = 2kπ;

|q(t)− 2jπ| < 1 for all t ∈ [jT , jT + τ ], j = 0, . . . , k12(x(t)2 + ω2x(t)2) +

12q(t)2 − (1 + δ(x(t))V (q(t))) =

12R2ω2 + εR

where εR → 0 as R→ +∞.

Remark 1.4. We esplicitly remark that:• using the above theorem, we can find, for fixed k ∈ N and for ω sufficiently

small, infinitely many R’s for which there is a k-bump solution of system(EL) which satisfy the boundary conditions (1.2);

• for fixed ω small, we can find a k bump solution only if R ≥ R. SinceR depends on k, we cannot prove existence of solutions having infinitelymany bumps. In particular we cannot show that the system is chaotic. We

HOMOCLINIC MULTIBUMP SOLUTIONS OF LARGE ENERGY 3

recall that this has been shown to hold true by Sere in [24] for Hamiltoniansystems with an hyperbolic rest point;

• the q part of our multibump solutions always goes from 0 to 2kπ withk jumps of +2π. We believe that our approach can be used also to findmultibump solutions which goes from 0 to 2kπ after an arbitrary largenumber of jumps of ±2π.

In order to prove this result, we set

x(t) = R cosωt+ y(t)

and look for solutions of the boundary value problem

(BVP)

q(t) = (1 + δ(R cosωt+ y))V ′(q(t))y + ω2y = δ′(R cosωt+ y)V (q(t))q(0) = q(T )− 2kπ = 0y(0)− y(T ) = y(0)− y(T ) = 0

when T = ω−1π(2N + 1) for N ∈ N, N large enough, and then pass to the limit asT → +∞.

Existence of a solution for this boundary value problem will be proved using crit-ical point theory, in particular a min-max procedure similar to the one introducedby Bernard in [2] and subsequently used in [14] to find one-bump solutions. In thiscase, however, a particular care has to be put in such a procedure in order to proveexistence of a sequence of critical points along which it is possible to pass to thelimit as T →∞.

Remark 1.5. Let us point out, for future reference, that (V1)–(V3) implies that(V4) There exists γ such that |V ′(q)|2 ≤ 2γ2(1 + δ)V (q), for all q;

and that there is a η0 ∈]0, 1[ such that

(1.6)µ

2≤ V ′′(q) ≤ 2µ for all min

k∈Z|q − 2kπ| ≤ η0;

let us also remark that, choosing η0 eventually smaller, it results

(1.7) V (q) ≥ µ

4η20 for all q such that min

k∈Z|q − 2πk| ≥ η0.

2. Some estimates

In this section we give some preliminary estimates. We start by analyzing thesolutions of the boundary value problem

(BVP)

q(t) = (1 + δ(R cosωt+ y))V ′(q(t))y + ω2y = δ′(R cosωt+ y)V (q(t))q(0) = q(T )− 2kπ = 0y(0)− y(T ) = y(0)− y(T ) = 0.

We recall that for all T = 2πN+πω the solution of{

y + ω2y = f(t)y(0)− y(T ) = y(0)− y(T ) = 0

is given by

y(t) =12ω

∫ t

0

f(s) sinω(t− s) ds− 12ω

∫ T

t

f(s) sinω(t− s) ds

=∫ T

0

f(s)G(t− s) ds,


where

G(s) =

{sin ωs

2ω s ≥ 0− sin ωs

2ω s ≤ 0.

Since |G(s)| ≤ 12ω for all s, we have

Lemma 2.1. Assume T = ω−1π(2N + 1) and y is a solution of

(2.2)

{y + ω2y = δ′(R cosωt+ y)V (q(t))y(0)− y(T ) = y(0)− y(T ) = 0.

Then

‖y‖∞ ≤ 12ω‖δ′‖∞

∫ T

0

V (q(t)) dt

‖y‖∞ ≤ 12‖δ′‖∞

∫ T

0

V (q(t)) dt.

Proof. For all t ∈ [0, T ], we have

|y(t)| ≤∫ T

0

|δ′(R cosωs+ y(s))G(t− s)|V (q(s)) ds

≤ 12ω‖δ′‖∞

∫ T

0

V (q(s)).

Similarly one deduces the estimate on y(t). �

Let us give some definitions that will be useful later.

Γk,T = {q ∈ H1(0, T ) | q(0) = 0, q(T ) = 2kπ},Γ∞ = {q ∈ H1

loc | limt→−∞

q(t) = 0, limt→+∞

q(t) = 2π},

ET = {y ∈ H1(0, T ) | y(0) = y(T ), y(0) = y(T )}.

Given C1, ν, D > 0 and T of the form T = ω−1π(2N + 1), we let

B(k,C1,ν,D, T ) ={q ∈ Γk,T

∣∣∣ ∫ T

0

[12q2 + (1 + δ)V (q)] ≤ kC1,

and ∃0 = a0 ≤ b0 ≤ a1 ≤ b1 ≤ . . . ≤ bk = T,

such that aj − bj−1 ≤ D and∫

[aj ,bj ]

[12q2 + (1 + δ)V (q)] ≤ ν,

}.

Lemma 2.3. Given ω > 0, k ∈ N, ν > 0, D > 0 and C1, there exists R0 > 0 suchthat, for all T of the form T = ω−1π(2N + 1), R ≥ R0, and q ∈ B(k,C1, ν,D, T )and for all y solutions of (2.2) we have that

‖y‖∞ <32ω‖δ′‖∞(k + 1)ν,∣∣∣∣∣

∫ T

0

(y2 − ω2y2)

∣∣∣∣∣ < 92ω‖δ′‖2∞(k + 1)2ν2,

∫ T

0

(y2 + y2) <9

4ω2T (1 + ω2)‖δ′‖2∞(k + 1)2ν2 +

92ω‖δ′‖2∞(k + 1)2ν2.

Proof. Fix q ∈ B(k,C1, ν,D, T ). By lemma 2.1 we know that

‖y‖∞ ≤ 12ω‖δ′‖∞kC1.


Take 0 < σ < 1/4 such that

σ <Dω

4and

4σω

(1 +Dω

π) <

ν

‖V ‖∞

and let

Aσ = {t ∈ [0, T ] | | cosωt| ≤ σ}.

Since σ < 1/4, then Aσ is the union of disjoint intervals centered on t =ω−1π(m + 1/2) of length less than 4σ/ω. In each interval ∆i = [bi−1, ai] thereare, at the most, [π−1Dω] + 1 such intervals ([α] denotes the integer part of α).Hence

|Aσ ∩∆i| ≤4σω

(1 +

[Dω

π

])≤ 4σ

ω

(1 +

Dω

π

)<

ν

‖V ‖∞.

We deduce∣∣∣∫ T

0

δ′(R cosωs+ y(s))V (q(s)) ds∣∣∣ ≤

≤k∑

i=1

∫∆i∩Aσ

|δ′(R cosωs+ y(s))|V (q(s)) ds+

+k∑

i=1

∫∆i\Aσ

|δ′(R cosωs+ y(s))|V (q(s)) ds+

+k∑

i=0

∫ bi

ai

|δ′(R cosωs+ y(s))|V (q(s)) ds

< ‖δ′‖∞‖V ‖∞k∑

i=1

|∆i ∩Aσ|+ kC1 sup{|δ′(s)| | |s| ≥ Rσ − ‖y‖∞}+

+ ‖δ′‖∞(k + 1)ν.

By (δ4), there exists R0 such that

|δ′(s)| ≤ ‖δ′‖∞νC1

for all |s| ≥ R0σ −kC1‖δ′‖∞

2ω

hence for all R ≥ R0 and ‖y‖∞ ≤ 12ω‖δ

′‖∞kC1 ≡ C2k

C1 sup{|δ′(s)| | |s| ≥ Rσ − ‖y‖∞} ≤ C1 sup{|δ′(s)| | |s| ≥ Rσ − kC2} ≤ ‖δ′‖∞ν;

so that ∣∣∣∣∣∫ T

0

δ′(R cosωs+ y(s))V (q(s)) ds

∣∣∣∣∣ < 3‖δ′‖∞(k + 1)ν.

Therefore for all t ∈ [0, T ],

|y(t)| =

∣∣∣∣∣∫ T

0

δ′(R cosωs+ y(s))V (q(s))G(t− s) ds

∣∣∣∣∣≤ 1

2ω

∣∣∣∣∣∫ T

0

δ′(R cosωs+ y(s))V (q(s)) ds

∣∣∣∣∣ < 32ω‖δ′‖∞(k + 1)ν,


and, integrating by parts,∣∣∣∣∣∫ T

0

(y2 − ω2y2) dt

∣∣∣∣∣ =∣∣∣∣∣∫ T

0

(−y − ω2y)y dt

∣∣∣∣∣=

∣∣∣∣∣∫ T

0

−δ′(R cosωt+ y(t))V (q(t))y(t) dt

∣∣∣∣∣≤ ‖y‖∞

∫ T

0

|δ′(R cosωt+ y(t))|V (q(t)) dt

<92ω‖δ′‖2∞(k + 1)2ν2.

The bound on the H1 norm of y is then a simple consequence of∫ T

0

(y2 + y2) =∫ T

0

(y2 − ω2y2) +∫ T

0

(1 + ω2)y2.

�

We conclude this section giving some properties of the functions q ∈ Γk,T .

Lemma 2.4. If q ∈ Γk,T and I is an interval of length |I| ≥ 1 such that∫I

[q2

2+ (1 + δ)V (q)] <

η20

4min{µ

4,

√µ

2}

thenmink∈Z

|q(t)− 2πk| ≤ η0 for all t ∈ I.

Proof. By contradiction, let us assume that there exists t0 ∈ I such that

mink|q(t0)− 2πk| > η0.

Let us first assume that mink |q(t) − 2πk| > η02 for all t ∈ I. If t ∈ I is such that

mink |q(t)− 2πk| ≥ η0, using (1.7) it follows that

V (q(t)) ≥ µ

4η20 ;

if t ∈ I is such that η02 < mink |q(t)− 2πk| ≤ η0, using (1.6), we have that

V (q(t)) ≥ µ

4η20

4.

Then, it follows that ∫I

[q2

2+ (1 + δ)V (q)] ≥ µ

4η20

4,

contradiction.If there exists t1 ∈ I such that q(t1) = η0

2 , then we can assume that q(t) ∈ [η02 , η0]

in some interval [t1, t2] ⊂ I; then, by Holder inequality, follows that

η02

= |q(t2)− q(t1)| = |∫ t2

t1

q(t)dt| ≤(∫ t2

t1

q2(t)dt)1/2√

t2 − t1,

so that ∫ t2

t1

q2(t)2

dt ≥ η20

8(t2 − t1);

moreover it results ∫ t2

t1

V (q) ≥ µ

4η20

4(t2 − t1);


since the real function g(x) = ax + bx assumes, for x > 0, the minimum value 2

√ab,

we have that∫ t2

t1

[q2

2+ (1 + δ)V (q)] ≥ η2

0

8(t2 − t1)+µ

4η20

4(t2 − t1) ≥

η20

4

√µ

2,

contradiction. �

Remark 2.5. Let us point out that the proof of the previous lemma implies thatfor all q such that q(c) = η0/2 and q(d) = η0 we have∫ d

c

[12q2 + (1 + δ)V (q)] ≥ η2

0

4

√µ

2.

Definition 2.6. Let η ≤ η0, we say that q ∈ Γk,T jumps from 2`π+η to 2(`+1)π−ηin the interval [α, β] ⊂ [0, T ], if q(α) = 2`π + η, q(β) = 2(` + 1)π − η and q(t) ∈[2`π + η, 2(`+ 1)π − η] for all t ∈ [α, β].

Lemma 2.7. Let η ≤ η0. If q ∈ Γk,T jumps from 2`π + η to 2(` + 1)π − η in theinterval [α, β], then∫ β

α

[q2

2+ (1 + δ)V (q)] ≥ 1

γ(2‖V ‖∞ − V (η)− V (2π − η))

with γ as in (V4) and

β − α ≤ (min{V (s) | s ∈ [η, 2π − η]})−1

∫ β

α

V (q).

Proof. Let s ∈ [η, 2π − η] be such that V (s) = ‖V ‖∞ and t ∈ [α, β] be such thatq(t) = s+ 2π`. Then, using (V4) and the periodicity of V , we have

‖V ‖∞ − V (η) =∫ t

α

d

dsV (q(s)) ds =

∫ t

α

V ′(q(s))q(s) ds

≤ γ

2

∫ t

α

|q(t)|2 ds+12γ

∫ t

α

|V ′(q(s))|2 ds

≤ γ

∫ t

α

[|q|2

2+ (1 + δ)V (q)] ds,

so that

2‖V ‖∞ − V (η)− V (2π − η) ≤ γ

∫ β

α

[|q|2

2+ (1 + δ)V (q)].

Finally ∫ β

α

V (q) ≥ (β − α) min{V (s) | s ∈ [η, 2π − η]}.

�

3. Variational setting

For every T = ω−1π(2N + 1) and for all q ∈ Γk,T and y ∈ ET we let

f(y, q) =12

∫ T

0

(y2 − ω2y2)dt+∫ T

0

[12q2 + (1 + δ(R cosωt+ y))V (q)]dt.


We introduce the following notations that will be used later

Q(y) =∫ T

0

(y2 − ω2y2)dt ∀y ∈ ET ,

J(y, q) =∫ T

0

[12q2 + (1 + δ(R cosωt+ y))V (q)]dt ∀(y, q) ∈ ET × Γk,T ,

J(R, q) =∫ T

0

[12q2 + (1 + δ(R cosωt))V (q)]dt ∀q ∈ Γk,T ,

J(q) =∫ T

0

[12q2 + (1 + δ)V (q)]dt ∀q ∈ Γk,T ,

J(q) =∫ T

0

[12q2 + (1 + δ)V (q)]dt ∀q ∈ Γk,T .

With these notations,

f(y, q) =12Q(y) + J(y, q) ∀(y, q) ∈ ET × Γk,T .

Moreover, let us remark that by assumption (δ3) follows that

(3.1) J(q) = J(q) +∫ T

0

(δ − δ)V (q) ≤ J(q) +∫ T

0

(1 + δ)V (q) ≤ 2J(q).

Lemma 3.2. f ∈ C1(ET × Γk,T ). If ∇f(y, q) = 0, then (y(t), q(t)) solves (BVP)and (R cosωt+ y(t), q(t)) solves (EL).

Proposition 3.3. Assume (yn, qn) ∈ ET × Γk,T is such that

f(yn, qn) → c,∂f

∂y(yn, qn) → 0.

Then (yn, qn) is bounded in ET × Γk,T , yn → y in ET , qn → q in L∞, qn ⇀ q inH1. Moreover (y, q) is solution of (2.2).

Proof. Recall (see for example [14]) that associated to quadratic form Q(y) thereis the splitting ET = E+

T ⊕ E−T and that there exist λ−(T ), λ+(T ) > 0 such that

λ−(T )‖y−‖2H1 ≤ −Q(y−) ≤ max{1, ω2}‖y−‖2H1 ,

λ+(T )‖y+‖2H1 ≤ Q(y+) ≤ max{1, ω2}‖y+‖2H1 .

Then

εn‖y−n ‖H1 ≥∣∣∣∣〈∂f∂y (yn, qn), y−n 〉

∣∣∣∣=

∣∣∣∣∣∫ T

0

(y−2

n − ω2y−2

n ) +∫ T

0

δ′(R cosωt+ yn)y−n V (qn)

∣∣∣∣∣≥ |Q(y−n )| − ‖δ′‖∞‖y−n ‖L2

(∫ T

0

V (qn)2)1/2

≥ λ−(T )‖y−n ‖2H1 − ‖δ′‖∞‖V ‖∞√T‖y−n ‖H1

and the boundedness of ‖y−n ‖H1 follows. Similarly one finds that ‖y+n ‖H1 is bounded

and hence yn is bounded in ET . Since∫ T

0

q2n = 2f(yn, qn)−Q(yn)− 2∫ T

0

(1 + δ(R cosωt+ yn))V (qn)

≤ 2(c+ 1) + max{1, ω2}‖yn‖2H1 ≤ cost,


and since qn(0) = 0, we have that qn is bounded in H1(0, T ). We can then deducethat, up to a subsequence, qn → q in L2, uniformly and weakly in H1, yn → y inL2, uniformly and yn ⇀ y weakly in H1. Since∫ T

0

|yn − y|2 +∫ T

0

|yn − y|2 =∫ T

0

yn(yn − y)−∫ T

0

y(yn − y) +∫ T

0

|yn − y|2,

recalling that∫ T

0|yn − y|2 → 0 as well as (by weak convergence)

∫ T

0y(yn − y) → 0,

to prove that yn → y in H1 it is enough to prove that∫ T

0

yn(yn − y) → 0.

Since∫ T

0

yn(yn − y) = 〈∂f∂y

(yn, qn), yn − y〉+ ω2

∫ T

0

yn(yn − y) +

−∫ T

0

δ′(R cosωt+ yn)V (qn)(yn − y),

the result follows because yn− y is bounded, ∂f∂y (yn, qn) → 0, yn → y in L2 and the

sequences yn and δ′(R cosωt+ yn)V (qn) are bounded in L∞. Then yn → y in H1.Finally, let ϕ ∈ H = {ϕ ∈ H1(0, T ) | ϕ(0) = ϕ(T )} be a test function; it results

〈∂f∂y

(yn, qn), ϕ〉 =∫ T

0

(ynϕ− ω2ynϕ) +∫ T

0

δ′(R cosωt+ yn)V (qn)ϕ

→∫ T

0

(yϕ− ω2yϕ) +∫ T

0

δ′(R cosωt+ y)V (q)ϕ;

hence y is a weak solution of (2.2) and, by standard regularity argument, a classicalsolution of (2.2). �

Corollary 3.4. Let ω > 0, k ∈ N, T = ω−1π(2N + 1) and C1, ν, D > 0 be given.Then for all R ≥ R0 (R0 given by lemma 2.3) there is a χ > 0 such that∥∥∥∥∂f∂y (y, q)

∥∥∥∥ ≥ χ

for all (y, q) ∈ ET ×B(k,C1, ν,D, T ) such that |f(y, q)| ≤ kC1 and

‖y‖∞ ≥ 32ω‖δ′‖∞(k + 1)ν or |Q(y)| ≥ 9

2ω‖δ′‖2∞(k + 1)2ν2.

Proof. If not, it exists (yn, qn) ∈ ET ×B(k,C1, ν,D, T ) such that

∂f

∂y(yn, qn) → 0, f(yn, qn) → c,

‖yn‖∞ ≥ 32ω‖δ′‖∞(k + 1)ν or |Q(yn)| ≥ 9

2ω‖δ′‖2∞(k + 1)2ν2.

By proposition 3.3, yn → y in H1, qn → q uniformly and weakly in H1 and (y, q)is a solution of (2.2). It results

|Q(y)| = limn→+∞

|Q(yn)| ≥ 92ω‖δ′‖2∞(k + 1)2ν2,

or

‖y‖∞ = limn→+∞

‖yn‖∞ ≥ 32ω‖δ′‖∞(k + 1)ν.

Now we show that q ∈ B(k,C1, ν,D, T ) so that we get a contradiction withlemma 2.3.


Since qn → q in L∞(0, T ), it follows that (1 + δ)V (qn(t)) → (1 + δ)V (q(t)) foralmost every t ∈ (0, T ) and then, by Fatou lemma, we have that∫ T

0

(1 + δ)V (q) ≤ lim infn→+∞

∫ T

0

(1 + δ)V (qn).

Therefore, since qn ⇀ q in H1, by semicontinuity of the norm with respect to theweak convergence, we have that∫ T

0

[12q2 + (1 + δ)V (q)] ≤ lim inf

n→+∞

∫ T

0

12q2n + lim inf

n→+∞

∫ T

0

(1 + δ)V (qn)

≤ lim infn→+∞

∫ T

0

[12q2n + (1 + δ)V (qn)] ≤ kC1.

Since qn ∈ B(k,C1, ν,D, T ) there are points 0 = an,0 ≤ bn,0 ≤ an,1 ≤ . . . ≤ bn,k =T , such that an,i − bn,i−1 ≤ D and

∫ bn,i

an,i[ 12 q

2 + (1 + δ)V (q)] ≤ ν for all n ∈ N andi = 1, . . . , k. We can assume, up to subsequence, that an,i → ai, bn,i−1 → bi−1 andai − bi−1 ≤ D. We claim that

(3.5)∫ bi

ai

[12q2 + (1 + δ)V (q)] ≤ ν.

Indeed for all θ > 0 there exists m(θ) such that for all n ≥ m(θ) and i = 1, . . . , kwe have [ai + θ, bi − θ] ⊂ [an,i, bn,i] and then∫ bi−θ

ai+θ

[12q2n + (1 + δ)V (qn)] ≤

∫ bn,i

an,i

[12q2n + (1 + δ)V (qn)] ∀n ≥ m(θ).

Then, for all θ > 0 we have that∫ bi−θ

ai+θ

[12q2 + (1 + δ)V (q)] ≤ lim inf

n→+∞

∫ bi−θ

ai+θ

[12q2n + (1 + δ)V (qn)]

≤ lim infn→+∞

∫ bn,i

an,i

[12q2n + (1 + δ)V (qn)] ≤ ν.

Since θ > 0 is arbitrary the claim follows.�

Corollary 3.6. (PS) holds.

Proof. If (yn, qn) is such that ∇f(yn, qn) → 0 and f(yn, qn) → c, then by proposi-tion 3.3, yn → y in ET , qn → q in H1 weak and L∞ strong. To show that qn → qin H1, it is enough to show (as in proposition 3.3) that

(3.7)∫ T

0

qn(qn − q) → 0.

Since∫ T

0

qn(qn − q) = 〈∂f∂q

(yn, qn), qn − q〉 −∫ T

0

(1 + δ(R cosωt+ yn))V ′(qn)(qn − q),

it is immediate to deduce that (3.7) holds. �


4. The min-max procedure

We will follow a min-max procedure similar to that we have used in [14], whichwas closely related to the one introduced by P. Bernard in [1, 2].

Let

Hk = {h ∈ C(E−T , ET × Γk,T ) | ∃L > 0, ∃q ∈ Γk,T s. t. h(y) = (y, q), ∀‖y‖ ≥ L},and

ck(T ) = infh∈Hk

maxy∈E−

T

f(h(y)).

We also setck(R, T ) = inf

q∈Γk,T

J(R, q),

ck(T ) = infq∈Γk,T

J(q), ck(T ) = infq∈Γk,T

J(q).

We observe that, since πE− ◦ h : E−T → E−T is the identity on the boundary of theball of radius L, there is y0 ∈ E−T such that πEh(y0) ∈ E+

T . Then, from

J(q) ≤ J(y, q) ≤ J(q) ∀q ∈ Γk,T , ∀y ∈ E,we have that

ck(T ) = infq∈Γk,T

J(q) = infh∈Hk

maxy∈E−

T

{12Q(πEh(y)) + J(πΓh(y))}

≤ infh∈Hk

maxy∈E−

T

{12Q(πEh(y)) + J(h(y))} = ck(T ) ≤ ck(T ).

Proposition 4.1. For all T = ω−1π(2N + 1) it exists (yT , qT ) such that

f(yT , qT ) = ck(T ), ∇f(yT , qT ) = 0

and hence (yT , qT ) is a solution of (BVP).

Proof. We first observe that ck(T ) ≥ ck(T ) > 0. The proposition then follows from(PS) and the fact that the class Hk is invariant under deformations which leave thesublevels f0 = {(y, q) | f(y, q) ≤ 0} fixed. Indeed, for every h ∈ Hk, it exists L ≥ Lsuch that f(y, q) ≤ 0 for all ‖y‖ ≥ L. �

Lemma 4.2. Let

c = infq∈Γ∞

∫R

(q2

2+ (1 + δ)V (q)

), c = inf

q∈Γ∞

∫R

(q2

2+ (1 + δ)V (q)

)c(R,∞) = inf

q∈Γ∞

∫R

(q2

2+ (1 + δ(R cosωt))V (q)

).

Then c < c, and such minima are achieved on q, q and qR respectively. Moreoverthere is L > 0 (which does not depend on R) such that for q = q, q and qR thereare τ1 and τ2 ∈ R, τ2 − τ1 ≤ L such that q(τ1) = η0 and q(τ2) = 2π − η0, and

|q(t)| ≤ η0ea(t−τ1), for all t ≤ τ1,

|q(t)− 2π| ≤ η0e−a(t−τ2), for all t ≥ τ2,

where a =√µ(1 + δ)/2.

Proof. The existence of the minimizers is well known, see for example [21] (onecould also use proposition 4.1 to prove it). The fact that L does not depend on Ris a consequence of lemma 2.7. Then

c− c ≥∫

R[˙q2

2+ (1 + δ)V (q)]−

∫R[˙q2

2+ (1 + δ)V (q)] = (δ − δ)

∫RV (q) > 0.

The exponential estimates are a simple consequence of maximum’s principle. �


Remark 4.3. Let qR be the minimizer in lemma 4.2. Then 0 ≤ qR(t) ≤ η0 for allt ≤ τ1 and from the equation we immediately deduce that qR is convex in (−∞, τ1).Moreover if 0 < qR(τ) = η ≤ η0 for some τ < τ1, then, taking as a test function

q0(t) =

0 t ≤ τ − 1η(t− τ + 1) τ − 1 ≤ t ≤ τ

qR(t) t ≥ τ

we have that∫ τ

−∞

(12|qR(t)|2 + (1 + δ(R cosωt))V (qR(t))

)dt

≤∫ τ

−∞

(12|q0(t)|2 + (1 + δ(R cosωt))V (q0(t))

)dt

≤ 12η2 + (1 + δ)V (η) ≤

(12

+ (1 + δ)µ)η2

and hence ∫ τ

−∞

(12|qR(t)|2 + (1 + δ)V (qR(t))

)dt ≤ 2

(12

+ (1 + δ)µ)η2.

Remark 4.4. For all η ≤ η0, if q ∈ Γk,T jumps from 2(j − 1)π + η to 2jπ − η in aninterval [αj , βj ] then∫ βj

αj

12q2 + (1 + δ(R cosωt)V (q)) > c(R,∞)− η2(1 + (1 + δ)µ).

Indeed, let us define the following function of Γ∞

p(t) =

0 t ∈ (−∞, αj − 1]η(t− αj + 1) t ∈ [αj − 1, αj ]q(t)− 2(j − 1)π t ∈ [αj , βj ]2π − η(βj + 1− t) t ∈ [βj , βj + 1]2π t ∈ [βj + 1,+∞)

it results∫ βj

αj

12q2 + (1 + δ(R cosωt)V (q)) > J(R, p)

−∫ αj

αj−1

[12η2 + (1 + δ)V (η(t− αj + 1))]

−∫ βj+1

βj

[12η2 + (1 + δ)V (2π − η(βj + 1− t))]

≥ c(R,∞)− η2 − (1 + δ)µ∫ αj

αj−1

η2(t− αj + 1)2

− (1 + δ)µ∫ βj+1

βj

η2(βj + 1− t)2

≥ c(R,∞)− η2 − (1 + δ)µη2 = c(R,∞)− η2(1 + (1 + δ)µ).

Let us also remark that there exists η1 ≤ η0 such that

c− η2(1 + 2(1 + δ)µ) > 0

for all η ≤ η1.


Proposition 4.1 gives us a solution of (BVP) on the interval [0, T ]. In order tofind a homoclinic solution of (EL) satisfying (1.2), we have to pass to the limit asT = ω−1π(2N + 1) → +∞. Unfortunately, we do not known if such a sequenceof solutions converges to what we are looking for. For this reason we have to find,for every T , a “suitable” critical point. This will be possible provided ω is smallenough and R is large enough.

We now let

(4.5) C1 = 2(1 + c− c+ δ)(1 + c(R,∞)) + 4

and show

Lemma 4.6. For all ρ ∈ (0, 1), ω > 0, k ∈ N, ν > 0 and η ∈ (0, η1], there are D,`0 ∈ N and R1 (which we assume to be ≥ R0) such that, for all ` ≥ `0, R ≥ R1 andT ≥ T` = 4πk`/ω we can find q ∈ B(k,C1 − 2, ν,D, T ) such that

(4.7) J(y, q) ≤ kc(R,∞) +ρ

2for all ‖y‖∞ ≤ 1,

and

(4.8) q(t) =

0 0 ≤ t ≤ 2Nπ/ω2jπ t = 2(N + j`)π/ω, j = 0, 1, . . . , k2kπ 2(N + k`)π/ω ≤ t ≤ T

.

Such a q jumps from 2jπ + η to 2(j + 1)π − η in the interval [αj+1, βj+1] ⊂[2π(N + j`+ 2)/ω, 2π(N + j`+ `0 − 2)/ω].

Proof. Let us fix ρ, ω, k, ν and η. Let T = ω−1π(4N + 1) and qR, τ1, τ2 be givenby lemma 4.2. Choose τ ′1 < τ1 < τ2 < τ ′2 so that qR jumps from η to 2π − η in aninterval contained in [τ ′1, τ

′2] and

12qR(τ ′1)

2 + (1 + δ)V (qR(τ ′1)) < min{ ρ

32k,ν

4

}12|qR(τ ′2)− 2π|2 + (1 + δ)V (qR(τ ′2)) < min

{ ρ

32k,ν

4

},

(4.9)

It is clear from lemma 4.2, that D = τ ′2 − τ ′1 does not depend on R. Eventuallytranslating qR by an integer multiple of the period 2π/ω, we can assume that1 + 4π/ω ≤ τ ′1 ≤ 1 + 6π/ω.

Let `0 ∈ N be the smallest ` ∈ N such that τ ′2 ≤ 2π(`0 − 2)/ω − 1. Then2π`0/ω ≥ τ ′2 − τ ′1 + 2 + 8π/ω. For all ` ≥ `0 and T = (4N + 1)π/ω ≥ T` = 4π`k/ω,we define q ∈ Γk,T as follows:

q(t) =

0 0 ≤ t ≤ 2Nπ/ωlinear 2Nπ/ω ≤ t ≤ 2Nπ/ω + 1qR(t− 2Nπ/ω) 2Nπ/ω + 1 ≤ t ≤ 2(N + `)π/ω − 1linear 2(N + `)π/ω − 1 ≤ t ≤ 2(N + `)π/ω2π t = 2(N + `)π/ωlinear 2(N + `)π/ω ≤ t ≤ 2(N + `)π/ω + 12π + qR(t− 2(N + `)π/ω) 2(N + `)π/ω + 1 ≤ t ≤ 2(N + 2`)π/ω − 1. . .

2kπ 2(N + k`)π/ω ≤ t ≤ T

.

We now show that we can find R1 ≥ R0 such that for all ` ≥ `0, T ≥ T`, R ≥ R1,q satisfies (4.7) and q ∈ B(k,C1 − 2, ν,D, T ) provided D = τ ′2 − τ ′1.


To prove (4.7) let us estimate∫ 2(N+`)π/ω

2Nπ/ω

q2

2+ (1 + δ(R cosωt+ y))V (q)

=∫ τ ′1+2Nπ/ω

2Nπ/ω

q2

2+ (1 + δ(R cosωt+ y))V (q)

+∫ τ ′2+2Nπ/ω

τ ′1+2Nπ/ω

qR(t− 2πN/ω)2

2+ (1 + δ(R cosωt))V (qR(t− 2πN/ω))

+∫ τ ′2+2Nπ/ω

τ ′1+2Nπ/ω

(δ(R cosωt+ y)− δ(R cosωt))V (qR(t− 2πN/ω))

+∫ 2(N+`)π/ω

τ ′2+2Nπ/ω

q2

2+ (1 + δ(R cosωt+ y))V (q)

The first and the last term can be estimated as follows∫ τ ′1+2Nπ/ω

2Nπ/ω

q2

2+ (1 + δ(R cosωt+ y))V (q)

≤ 12qR(1)2 + (1 + δ)V (qR(1)) +

12qR(τ ′1)

2 + (1 + δ)V (qR(τ ′1)) <ρ

8k.

Then we can estimate the second term as follows:∫ τ ′2+2Nπ/ω

τ ′1+2Nπ/ω

qR(t− 2πN/ω)2

2+ (1 + δ(R cosωt))V (qR(t− 2πN/ω)) ≤ c(R,∞)

To estimate the third term, we let Aσ = {t ∈ R | cosωt = 0} + (−σ, σ). Then|Aσ ∩ [τ ′1, τ

′2]| ≤ 4`0σ. We then choose σ so small that

|Aσ ∩ [τ ′1, τ′2]| ‖δ′‖∞‖V ‖∞ <

ρ

16k,

and R1 so large that, for all t ∈ R \Aσ,

|δ′(R cosωt+ y)|C1 <ρ

16kfor all ‖y‖∞ ≤ 1 and R ≥ R1.

We then have∫ τ ′2+2Nπ/ω

τ ′1+2Nπ/ω

(δ(R cosωt+ y)− δ(R cosωt))V (qR(t− 2πN/ω))

≤∫

Aσ∩[τ ′1,τ ′2]

|δ′(R cosωt+ θy)| ‖y‖∞V (qR)

+∫

[τ ′1,τ ′2]\Aσ

|δ′(R cosωt+ θy)| ‖y‖∞V (qR)

≤ |Aσ ∩ [τ ′1, τ′2]| ‖δ′‖∞‖V ‖∞ +

ρ

16k<

ρ

8kHence we have, after repeating the argument in the intervals [2(N + j`)π/ω, 2(N +(j + 1)`)π/ω], j = 0, . . . , k − 1, that equation (4.7) holds.

From the above estimates we also have that∫ 2(N+(j+1)`)π/ω

2(N+j`)π/ω

q2

2+ V (q) ≤ c(R,∞) +

ρ

2k

and hence, setting c0 = 0, cj = 2(N + j`)π/ω, ck = T we have that∫ cj

cj−1

q2

2+ (1 + δ)V (q) ≤ (1 + δ)(c(R,∞) + 1) ≤ C1 − 2.


To show that q ∈ B(k,C1 − 2, ν,D, T ), arguing as above, we find∫[ 2Nπ

ω ,2(N+`)π

ω ]\[τ ′1,τ ′2]+2Nπ

ω

q2

2+ (1 + δ)V (q) < ν.

�

Lemma 4.10. For all ρ ∈ (0, 1), ω > 0, k ∈ N, ν > 0, η ∈ (0, η1], ` > `0,T ≥ T` and R ≥ R1, take q ∈ B(k,C1 − 2, ν,D, T ) given by lemma 4.6. We recallthat such a q jumps from 2jπ + η to 2(j + 1)π − η in the interval [αj+1, βj+1] ⊂[2π(N + j`+ 2)/ω, 2π(N + j`+ `0 − 2)/ω] and is such that

J(y, q) ≤ kc(R,∞) +ρ

2for all ‖y‖∞ ≤ 1,

and

q(t) =

0 0 ≤ t ≤ 2Nπ/ω2jπ t = 2(N + j`)π/ω2kπ 2(N + k`)π/ω ≤ t ≤ T

.

Then there is a function h1 : E−T → ET such that(a) if h : E−T → ET × Γk,T is defined by h(y) = (h1(y), q), then h ∈ Hk;(b) maxy∈E−

Tf(h(y)) ≤ kc(R,∞) + ρ.

In particularck(T ) ≤ kc(R,∞) + ρ.

Proof. To show the existence of such a function h1 we will use the function q givenby lemma 4.6 and then deform, using a suitable pseudo-gradient, the identity mapId: E−T → E−T .

First of all we fix ν satisfying

(4.11)32ω‖δ′‖∞(k + 1)ν < ρ,

92ω‖δ′‖2∞(k + 1)2ν2 <

ρ

16, ν <

ρ

4and then let q be the function defined in lemma 4.6. Since q ∈ B(k,C1− 2, ν,D, T )and R ≥ R0(ω), we know from corollary 3.4 that for all y such that 0 ≤ f(y, q) ≤kC1 and

‖y‖∞ ≥ 32ω‖δ′‖∞(k + 1)ν or |Q(y)| ≥ 9

2ω‖δ′‖2∞(k + 1)2ν2,

it exists χ > 0 such that ∥∥∥∥∂f∂y (y, q)∥∥∥∥ ≥ χ > 0.

Let ϕ, ψ : [0,+∞) → [0, 1] be

ϕ(s) =

0 s ≤ 02s/c 0 ≤ s ≤ c/21 c/2 ≤ s ≤ kC1

kC1 + 1− s kC1 ≤ s ≤ kC1 + 10 s ≥ kC1 + 1

, ψ(s) =

0 0 ≤ s ≤ 1s− 1 1 ≤ s ≤ 21 s ≥ 2

and define the vector field v : ET → ET

v(y) = −[ψ

(4ω‖y‖∞

3‖δ′‖∞(k + 1)ν

)+ ψ

(ω|Q(y)|

‖δ′‖2∞(k + 1)2ν2

)]ϕ(f(y, q))∂f

∂y (y, q)

1 +∥∥∥∂f

∂y (y, q)∥∥∥ .

Since v is a bounded locally lipschitz function of y, the Cauchy problem{dηds (s, y) = v(η(s, y))η(0, y) = y


has a unique solution for every y ∈ ET , defined on [0,+∞).We claim that, setting τ3 = k(C1 − c(R,∞))(1 + χ)/χ2, then

f(η(τ3, y), q) ≤ kc(R,∞) + ρ

for all y such that f(y, q) ≤ kC1.First of all, we have that

df

ds(η(s, y), q) = 〈∂f

∂y(η, q),

dη

ds(s, y)〉 = 〈∂f

∂y(η, q), v(η(s, y))〉

= −[ψ + ψ]ϕ

∥∥∥∂f∂y

∥∥∥2

1 +∥∥∥∂f

∂y

∥∥∥ ≤ 0

and hence f(η(s, y), q) is a non increasing function of s and the claim follows forall y such that f(y, q) ≤ kc(R,∞) + ρ.

Take now any y ∈ ET such that

kc(R,∞) + ρ < f(y, q) ≤ kC1.

Assume, by contradiction, that

f(η(s, y), q) > kc(R,∞) + ρ, ∀s ∈ [0, τ3].

Let us remark that if ‖η(s, y)‖∞ ≥ 32ω‖δ

′‖∞(k + 1)ν, then∥∥∥∥∂f∂y (η(s, y), q)∥∥∥∥ ≥ χ and ψ

(4ω‖η(s, y)‖∞

3‖δ′‖∞(k + 1)ν

)= 1.

On the other hand, if ‖η(s, y)‖∞ < 32ω‖δ

′‖∞(k + 1)ν (< ρ < 1), then, fromlemma 4.6

12Q(η(s, y)) = f(η(s, y), q)− J(η(s, y), q)

> kc(R,∞) + ρ− kc(R,∞)− ρ

2=ρ

2

>36ω‖δ′‖2∞(k + 1)2ν2,

so that ∥∥∥∥∂f∂y (η(s, y), q)∥∥∥∥ ≥ χ and ψ

(ω|Q(η(s, y))|

‖δ′‖2∞(k + 1)2ν2

)= 1.

Therefore, in any case, for all s ∈ [0, τ3]∥∥∥∥∂f∂y (η(s, y), q)∥∥∥∥ ≥ χ

and

ψ

(ω|Q(η(s, y))|

‖δ′‖2∞(k + 1)2ν2

)+ ψ

(4ω‖η(s, y)‖∞

3‖δ′‖∞(k + 1)ν

)≥ 1.

We also have that, for all s ∈ [0, τ3]

ϕ(f(η(s, y), q)) = 1.

Then

kC1 − kc(R,∞)− ρ > f(y, q)− f(η(τ3, y), q) = −∫ τ3

0

df

ds(η(s, y), q)ds

≥∫ τ3

0

‖∂f∂y (η(s, y), q)‖2

1 + ‖∂f∂y (η(s, y), q)‖

ds ≥ χ2

1 + χτ3 = kC1 − kc(R,∞),

a contradiction which proves the claim.


We now define

(4.12) h : y ∈ E−T 7→ h(y) = (η(τ3, y), q) ∈ ET × Γk,T .

Since f(y, q) ≤ kC1 for all y ∈ E−T , we have that

(4.13) f(h(y)) ≤ kc(R,∞) + ρ ∀y ∈ E−T .

Also, there exists L large such that

f(y, q) < 0 ∀‖y‖ ≥ L

and hence, by definition of η,

h(y) = (y, q) ∀‖y‖ ≥ L,

which shows that h ∈ Hk and the lemma follows. �

5. Some additional estimates

We now have to fix some constants. First of all, given η1 as in remark 4.4 we fixρ0 in such a way that

ρ0 < c− η21(1 + 2(1 + δ)µ).

Then let

(5.1) ρ = min{ρ0,c− c

6,η20

20µ

4,η20

24

√µ

2,16c,

13} > 0.

We now take η2 ≤ η1 such that

(5.2) η22(1 + (1 + δ)µ) <

ρ

4;

then we know from remark 4.4 that every q ∈ Γk,T which jumps from 2jπ + η2 to2(j + 1)π − η2 in the interval [α, β], is such that

(5.3)∫ β

α

[q2

2+ (1 + δ(R cosωt))V (q)] ≥ c(R,∞)− ρ

4.

From now on, given q ∈ Γk,T jumping from 2π(j − 1) + η2 to 2jπ − η2 for somej = 1, . . . , k, we will use the notation [αj , βj ] to denote the (largest) intervals whereq jumps, and cj to denote (one of the) points where q(cj) = 2jπ.

We remark here that such an interval [αj , βj ] is unique and is a subset of [cj−1, cj ](for all choices of cj−1 and cj) whenever∫ cj

cj−1

[q2

2+ (1 + δ(R cosωt))V (q)] ≤ c(R,∞) + 3ρ.

Indeed we know from (5.3) that a jump of 2π− 2η2 costs at least c(R,∞)− ρ/4, sothe claim follows since 2c(R,∞)− ρ/2 > c(R,∞) + 3ρ.

We also recall that, if ∫ cj

cj−1

[q2

2+ (1 + δ)V (q)] ≤ C1

where C1 is given by (4.5), from lemma 2.7 follows that

(5.4) βj − αj ≤ L := (min{V (s) | s ∈ [η2, 2π − η2]})−1C1.

We also define

(5.5) ω1 =π

6LWe now give an estimate of the critical level c(R,∞) for ω < ω1.


Lemma 5.6. Let ρ be as in (5.1). Then we can find R2 (which we assume to belarger than R1) such that for all 0 < ω < ω1 and R ≥ R2, we have

c(R,∞) ≤ c+ ρ.

Proof. Let q be a minimizer for∫R

12q2 + (1 + δ)V (q),

(see lemma 4.2). We can assume that q(0) = π. If q jumps from η2 to 2π − η2 inthe interval [α, β], we have that −L ≤ α < 0 < β ≤ L; then the definition of η2and calculations similar to those in remark 4.3 show that∫

R\[α,β]

12q2 + (1 + δ)V (q) <

ρ

2.

Then

c(R,∞) ≤ J(R, q(· − π/ω)) =∫

R

12q2 + (1 + δ(−R cosωt))V (q)

=∫

R\[α,β]

12q2 + (1 + δ(−R cosωt))V (q)

+∫ β

α

12q2 + (1 + δ)V (q) +

∫ β

α

(δ(−R cosωt)− δ)V (q)

≤ ρ

2+ c+

∫ β

α

(δ(−R cosωt)− δ)V (q).

Since ω1 is such that cosωt > 1/2 for all 0 < ω < ω1, t ∈ [−L,L], the lemmafollows by taking R large enough. �

Lemma 5.7. Let ρ as in (5.1), and assume q ∈ Γ1,∞ jumps from η2 to 2π − η2 inthe interval [α, β] ⊂ [− π

4ω + 2mπω , π

4ω + 2mπω ] for some m ∈ Z.

Then, provided R ≥ R3 (which we assume ≥ R2), we have that∫ β

α

[12q2 + (1 + δ(R cosωt)V (q))] > c− ρ.

Proof. Let q be a function equal to q in the interval [α, β] which decays exponentiallyto 0 and 2π outside such an interval. Then, since cosωt ≥ 1/2 for all t ∈ [α, β], wecan choose R3 such that, for all R ≥ R3∫ β

α

∣∣δ(R cosωt)− δ∣∣V (q) <

ρ

2,

and ∫ β

α

12q2 + (1 + δ(R cosωt))V (q) =

∫ β

α

12q2 + (1 + δ)V (q)

+∫ β

α

(δ(R cosωt)− δ)V (q)

>

∫R

12

˙q2 + (1 + δ)V (q)− ρ

2− ρ

2≥ c− ρ

and the lemma follows. �

We now fix ω > 0 and k ∈ N and take ν satisfying32ω‖δ′‖∞(k + 1)ν < ρ,

92ω‖δ′‖2∞(k + 1)2ν2 <

ρ

16, ν <

ρ

4,

32ω

(c+ 1)‖δ′‖2∞(k + 1)2ν <ρ

16.

(5.8)


Then the following holds

Lemma 5.9. For q ∈ Γk,T , let 0 = c0 ≤ c1 ≤ . . . ≤ ck = T be such that q(cj) = 2πjand assume that q jumps from 2(j − 1)π + η2 to 2jπ − η2 in the interval [αj , βj ] ⊂[cj−1, cj ]. If

(5.10)∫ cj

cj−1

[q2

2+ (1 + δ(R cosωt))V (q)] < c(R,∞) + 3ρ

then

(5.11)∫

[cj−1,cj ]\[αj ,βj ]

[q2

2+ (1 + δ(R cosωt))V (q)] ≤ 4ρ,

and |q(t)− 2(j − 1)π| < η0 for all t ∈ [cj−1, αj) and |q(t)− 2jπ| < η0 for allt ∈ (βj , cj ].

Let (y, q) ∈ ET × Γk,T be such that

f(y, q) ≤ kc(R,∞) + 2ρ

|Q(y)| ≤ 92ω‖δ′‖2∞4(k + 1)2ν2,

‖y‖∞ <3ω‖δ′‖∞(k + 1)ν.

Then for any choice of 0 = c0 ≤ c1 ≤ . . . ≤ ck = T such that q(cj) = 2πj (5.10)holds.

Proof. Using (5.3) we have

c(R,∞) + 3ρ >∫ βj

αj

[12q2 + (1 + δ(R cosωt))V (q)]

+∫


[12q2 + (1 + δ(R cosωt))V (q)]

> c(R,∞)− ρ

4+∫


[12q2 + (1 + δ(R cosωt))V (q)],

and (5.11) follows.Then, arguing by contradiction, let us assume that there exists t0 ∈ [cj−1, αj)

such that q(t0) = 2(j − 1)π + η0. Then there is an interval [a, b] ⊂ [cj−1, t0] suchthat

q(a) = 2(j − 1)π +η02, q(b) = 2(j − 1)π + η0,

q(t)− 2(j − 1)π ∈ [η02, η0] for all t ∈ [a, b].

Then, using the estimate (5.3), the remark 2.5 and by the choice of ρ it follows that

c(R,∞) + 3ρ >∫ βj

αj

[12q2 + (1 + δ(R cosωt))V (q)]

+∫ b

a

[12q2 + (1 + δ(R cosωt))V (q)]

≥ c(R,∞)− ρ

4+η20

4

√µ

2> c(R,∞)− ρ

4+ 6ρ = c(R,∞) +

234ρ,

contradiction.To prove the second part of the lemma, we first remark that, using (5.8) and by

the choice of ρ we have∫ T

0

V (q) ≤ J(y, q) = f(y, q)− 12Q(y) ≤ kc(R,∞) + 2ρ+

ρ

8≤ k(c+ 1).


Then, for any choice of 0 = c0 ≤ c1 ≤ . . . ≤ ck = T such that q(cj) = 2πj

kc(R,∞) + 2ρ ≥∫ T

0

[q2

2+ (1 + δ(R cosωt))V (q)]

+∫ T

0

(δ(R cosωt+ y)− δ(R cosωt))V (q) +12Q(y)

≥∫ cj+1

cj

[q2

2+ (1 + δ(R cosωt))V (q)] + (k − 1)c(R,∞)

− ‖δ′‖∞‖y‖∞∫ T

0

V (q)− 94ω‖δ′‖2∞4(k + 1)2ν2,

so that

∫ cj+1

cj

[q2

2+ (1 + δ)V (q)] ≤

∫ cj+1

cj

[q2

2+ (1 + δ(R cosωt))V (q)]

≤ c(R,∞) + 2ρ+3ω‖δ′‖2∞(k + 1)νk(c+ 1) +

94ω‖δ′‖2∞4(k + 1)2ν2

< c(R,∞) + 3ρ.

(5.12)

�

6. Pseudo-gradient vector field

Take ρ as (5.1), η2 as in (5.2), L as in (5.4), ω < ω1 (ω1 as in (5.5)), k ∈ N, ν asin (5.8). Let R1, `0 and D be given by lemma 4.6 (corresponding to such ω, k, η2,ρ, and ν), and define

(6.1) D = max{D,L+ 2 + 200ρK/ν},where K = max{1, 2µ−1(1+δ)−1}. For notational convenience, we will denote sucha D simply D from now on.

Throughout all this section, ρ, η2, L, ω, k, ν, and D will be fixed as above. Wealso fix C1 as in (4.5) and ` ≥ `0 and T ≥ T`. Let us point out that ρ does notdepend on k, while ν, R and T do. Since in the end we will be interested in T →∞,this will mean that we can fix ω and carry on our construction for a given k onlyif R is large enough.

For all T > T` we now define a function w : ET ×Γk,T → ET ×Γk,T that will beused to define a pseudo-gradient vector field v.

Case 1. (y, q) is such that

• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ,

• ∃0 = a0 ≤ b0 ≤ . . . ≤ ak ≤ bk = T ∈ [0, T ] s.t.

|aj − bj−1| ≤ D and∫

[aj ,bj ]

[q2

2+ (1 + δ)V (q)] ≤ ν,

• ‖y‖∞ <32ω‖δ′‖∞2(k + 1)ν and |Q(y)| < 9

2ω‖δ′‖2∞4(k + 1)2ν2.

In this case we letw(y, q) = ∇f(y, q).

Remark that from (5.8) and (4.5) we have that∫ T

0

[q2

2+ (1 + δ)V (q)] ≤ kc(R,∞) + 2ρ− 1

2Q(y)

< kc(R,∞) + 2ρ+ρ

8<k

2(C1 − 4)

(6.2)


and hence J(q) < k(C1 − 4) and q ∈ B(k,C1 − 4, ν,D, T ) in this case.


• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ,

• q ∈ B(k,C1, 2ν,D, T )

• ‖y‖∞ ≥ 32ω‖δ′‖∞2(k + 1)ν or |Q(y)| ≥ 9

2ω‖δ′‖2∞4(k + 1)2ν2.

We let

w(y, q) =(∂f

∂y(y, q), 0

).


• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ,

• ∀0 = a0 ≤ b0 ≤ . . . ≤ ak ≤ bk = T ∈ [0, T ] such that |aj − bj−1| ≤ D

there is j such that∫

[aj ,bj ]

[q2

2+ (1 + δ)V (q)] > ν,

• ∃0 = a0 ≤ b0 ≤ . . . ≤ ak ≤ bk = T ∈ [0, T ] such that

|aj − bj−1| ≤ D and∫

[aj ,bj ]

[q2

2+ (1 + δ)V (q)] ≤ 2ν,

• ‖y‖∞ <32ω‖δ′‖∞2(k + 1)ν and |Q(y)| < 9

2ω‖δ′‖2∞4(k + 1)2ν2.

Also in this case, arguing as in case 1, we have that q ∈ B(k,C1 − 4, 2ν,D, T ).Moreover from lemma 5.9 we deduce that, for any choice of 0 = c0 ≤ c1 ≤ . . . ≤ck = T such that q(cj) = 2πj,∫ cj+1

cj

[q2

2+ (1 + δ)V (q)] ≤

∫ cj+1

cj

[q2

2+ (1 + δ(R cosωt))V (q)] < c(R,∞) + 3ρ,

that|q(t)− 2jπ| ≤ η0 for all t ∈ [βj , αj+1]

and that ∫[cj−1,cj ]\[αj ,βj ]

[q2

2+ (1 + δ(R cosωt))V (q)] ≤ 4ρ.

Let γj = (αj + βj)/2 for j = 1, . . . , k, and define

(6.3)

a0 = 0, b0 = max{0, γ1 −D

2},

aj = γj +D

2, bj = γj+1 −

D

2if γj+1 − γj ≥ D

aj = bj =γj + γj+1

2if γj+1 − γj < D

ak = min{T, γk +D

2}, bk = T

Since aj − bj−1 ≤ D, we know that there is (at least one) j such that

(6.4)∫ bj

aj

[q2

2+ (1 + δ)V (q)] > ν.

Obviously aj 6= bj so that using the definitions we have

|[βj , aj ]| = aj − βj =D

2− βj − αj

2≥ D

2− L

2≥ 1 +

100ρkν

.


Then there is a subinterval [βj , βj + 1] ⊂ [βj , aj ] such that∫ βj+1

βj

[q2

2+ (1 + δ)V (q)] <

ν

8K.

Indeed, otherwise

8ρ >∫ aj

βj

[q2

2+ (1 + δ)V (q)] ≥ ν

8K[aj − βj ] ≥

ν

8K100ρKν

> 8ρ.

Similarly, there exists [αj+1, αj+1 + 1] ⊂ [bj , αj+1] such that∫ αj+1+1

αj+1

[q2

2+ (1 + δ)V (q)] <

ν

8K.

We set:wj(y, q) = (0, (q − 2jπ)χj)

where

χj =

0 0 ≤ t ≤ βj

t− βj βj ≤ t ≤ βj + 11 βj + 1 ≤ t ≤ αj+1,

1− t+ αj+1 αj+1 ≤ t ≤ αj+1 + 10 αj+1 + 1 ≤ T

Finally, if (6.4) holds for j ∈ A ⊂ {0, 1, . . . , k} we set

w(y, q) =∑j∈A

wj(y, q).

Other cases. That is, at least one of the following hold:

• f(y, q) ≤ ck(T )− 2ρ or f(y, q) ≥ kc(R,∞) + 2ρ

• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ and∫ T

0

[q2

2+ (1 + δ)V (q)] > kC1

• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ and∫ T

0

[q2

2+ (1 + δ)V (q)] ≤ kC1

and for all choiches of 0 = a0 ≤ b0 ≤ . . . ≤ ak ≤ bk = T ∈ [0, T ] such that

|aj − bj−1| ≤ D one can find j such that∫

[aj ,bj ]

[q2

2+ (1 + δ)V (q)] > 2ν,

we letw(y, q) = ∇f(y, q).

For all (y, q) ∈ ET × Γk,T we let

W(y, q) =w(y, q)

1 + ‖w(y, q)‖.

In the following lemma we show some properties of the function W in the set

(ET × Γk,T )∗ = {(y, q) ∈ ET × Γk,T | ∇f(y, q) 6= 0}.

Lemma 6.5. Let C ∈ (0, 2] be such that C < ν(16ρK + ν)−1/2/4 where K =max{1, 2µ−1(1 + δ)−1}. Then the following properties hold for all (y, q) ∈ (ET ×Γk,T )∗

(a) ‖W(y, q)‖ < 2C min{‖∇f(y, q)‖, C};


(b) with χ = χ2ν given by corollary 3.4 we have

〈W,∇f〉 > 13

min{‖∇f‖2, ‖∇f‖, χ2

1 + χ,

ν

4[1 + (16ρK + ν)1/2]};

(c) πΓW(y, q) = 0 for all (y, q) such that

• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ,

• q ∈ B(k,C1, 2ν,D, T ),

• ‖y‖∞ ≥ 32ω‖δ′‖∞2(k + 1)ν or |Q(y)| ≥ 9

2ω‖δ′‖2∞4(k + 1)2ν2.

(d) if (y, q) are such that

• ck(T )− 2ρ < f(y, q) < kc(R,∞) + 2ρ,

• q ∈ B(k,C1, 2ν,D, T ),



[aj ,bj ]

[q2

2+ (1 + δ)V (q)] > ν,

• ‖y‖∞ <32ω‖δ′‖∞2(k + 1)ν and |Q(y)| < 9

2ω‖δ′‖2∞4(k + 1)2ν2,

then, letting

(6.6) Ha,b(q) =∫ b

a

[q2

2+ (1 + δ)V (q)],

we have that〈H ′

a,b(q), πΓW(y, q)〉 > ν

8whenever [aj , bj ] ⊂ [a, b] for some j ∈ A (where aj, bj are defined in (6.3) andsatisfy (6.4)).

Proof. Let us first prove that (a) and (b) hold. It is trivial whenever

W(y, q) =∇f(y, q)

1 + ‖∇f(y, q)‖.

In case W(y, q) = (∂f∂y /(1 + ‖∂f

∂y ‖), 0), we have

‖W(y, q)‖ =‖∂f

∂y ‖1 + ‖∂f

∂y ‖≤ 2C

min{‖∂f∂y‖, C} ≤ 2

Cmin{‖∇f‖, C},

which proves (a).To prove (b) in this case, we observe that whenever w(y, q) = (∂f

∂y , 0), we haveby corollary 3.4, that ‖∂f

∂y ‖ ≥ χ > 0 and hence (b) follows.In case w(y, q) =

∑j∈A(0, (q − 2jπ)χj), we are in case 3, so that∫

[cj ,cj+1]\[αj+1,βj+1]

[q2

2+ (1 + δ)V (q)] ≤ 4ρ, j = 0, . . . , k − 1,

and |q(t)− 2jπ| < η0 for all t ∈ [βj , αj+1]. Hence, for all such t

V (q(t)) ≥ µ

4(q(t)− 2jπ)2 V ′(q(t))(q(t)− 2jπ) ≥ µ

2(q(t)− 2jπ)2 ≥ V (q(t))

2.

We have also defined, always in case 3, whenever j is such that (6.4) holds, βj andαj+1 such that∫ βj+1

βj

[q2

2+ (1 + δ)V (q)] <

ν

8K

∫ αj+1+1

αj+1

[q2

2+ (1 + δ)V (q)] <

ν

8K.


Hence, letting Aj = [βj , βj + 1] ∪ [αj+1, αj+1 + 1] and v = q − 2jπ

‖vχj‖2 ≤∫ αj+1

βj+1

[q2 + v2] +∫

Aj

[q2 + 2v2 + 2|qv|]

≤∫ αj+1+1

βj

[q2 + v2] +∫

Aj

[q2 + 2v2]

≤ 2K∫ αj+1+1

βj

[q2

2+ (1 + δ)V (q)] + 4K

∫Aj

[q2

2+ (1 + δ)V (q)]

≤ 16Kρ+ 4Kν

4K= 16ρK + ν.

so that‖w(y, q)‖ = ‖

∑j∈A

(0, (q − 2jπ)χj)‖ ≤ |A| (16ρK + ν)1/2;

(|A| denotes the cardinality of A). On the other hand, recalling that, for all j ∈ A∫ αj+1

βj+1

[q2

2+ (1 + δ)V (q)] > ν,

we have

〈∂f∂q,(q − 2jπ)χj〉 =

∫ αj+1+1

βj

[qd

dt(vχj) + (1 + δ(R cosωt+ y))V ′(q)vχj ]

≥∫ αj+1

βj+1

[q2 + (1 + δ(R cosωt+ y))V ′(q)v]

+∫

Aj

[q2χj − |vq|+ (1 + δ(R cosωt+ y))V ′(q)vχj ]

≥∫ αj+1

βj+1

[q2 + (1 + δ)V (q)

2]− 1

2

∫Aj

q2 + v2

≥ 12

∫ αj+1

βj+1

[q2

2+ (1 + δ)V (q)]−K

∫Aj

[q2

2+ (1 + δ)V (q)]

>ν

2−K

ν

4K=ν

4

(6.7)

and hence‖W(y, q)‖‖∇f(y, q)‖ ≥ 〈∇f(y, q),W(y, q)〉

=∑j∈A

11 + ‖w(y, q)‖

〈∂f∂q, (q − 2jπ)χj〉

> |A| 11 + ‖w(y, q)‖

ν

4≥ |A|

1 + |A| (16ρK + ν)1/2

ν

4;

this proves (b) and implies

‖∇f(y, q)‖ > 1‖W(y, q)‖

|A|1 + ‖w(y, q)‖

ν

4

=|A|

‖w(y, q)‖ν

4≥ |A||A| (16ρK + ν)1/2

ν

4.

Since

(6.8) ‖W(y, q)‖ =‖w(y, q)‖

1 + ‖w(y, q)‖≤ 1,

we have also

‖∇f(y, q)‖ > ‖W(y, q)‖(16ρK + ν)1/2

ν

4


and therefore

‖W(y, q)‖ < 4(16ρK + ν)1/2

ν‖∇f(y, q)‖ < 2

C‖∇f(y, q)‖;

then (a) is proved by (6.8) and the last inequality.It is immediate to check that also (c) holds.To prove that (d) holds, let us remark that case 3 holds, so that w(y, q) =∑j∈A(q− 2jπ)χj . A calculation similar to that of (6.7) yields, whenever [aj , bj ] ⊂

[a, b] for some j ∈ A,

〈H ′a,b(q), vχj〉 =

∑j∈A

∫[a,b]∩[βj ,αj+1+1]

[qd

dt(vχj) + (1 + δ)V ′(q)vχj ] >

ν

8.

�

In the following lemma we construct the pseudo-gradient vector field by smooth-ing the function W defined above.

Lemma 6.9. There exists a locally Lipschitz continuos vector field v : (ET ×Γk,T )∗ → ET × Γk,T such that for all (y, q) ∈ (ET × Γk,T )∗ we have(a) ‖v(y, q)‖ < 2

C min{‖∇f(y, q)‖, C}, C as in lemma 6.5;(b) 〈v(y, q),∇f(y, q)〉 > 1

3 min{‖∇f(y, q)‖2, ‖∇f(y, q)‖, χ2

1+χ ,ν

4[1+(16ρK+ν)1/2]};

(c) πΓ(v(y, q)) = 0 for all (y, q) such that

• ck(T )− 32ρ < f(y, q) < kc(R,∞) +

32ρ,

• q ∈ B(k,C1 − 1,74ν,D, T )

• ‖y‖∞ ≥ 232ω‖δ′‖∞2(k + 1)ν or |Q(y)| ≥ 2

92ω‖δ′‖2∞4(k + 1)2ν2.

(d) if (y, q) are such that

• ck(T )− 32ρ < f(y, q) < kc(R,∞) +

32ρ,

• q ∈ B(k,C1 − 1,74ν,D, T )



[aj ,bj ]

[q2

2+ (1 + δ)V (q)] >

54ν,

then we have that〈H ′

a,b(q), πΓv(y, q)〉 ≥ 0whenever [aj , bj ] ⊂ [a, b] for some j ∈ A (where aj, bj are defined as in (6.3)—remark that they are well defined in this situation—and Ha,b(q) has been defined in(6.6)). Moreover

〈H ′a,b(q), πΓv(y, q)〉 >

ν

16if aj, bj satisfy the previous condition and

•‖y‖∞ <12

32ω‖δ′‖∞2(k + 1)ν and |Q(y)| < 1

292ω‖δ′‖2∞4(k + 1)2ν2.

Proof. By continuity for all (y, q) ∈ (ET × Γk,T )∗, there exists a neighborhood of(y, q), Uy,q, such that for all (y, q) ∈ Uy,q,(i) ‖W(y, q)‖ < 2

C min{‖∇f(y, q)‖, C}, C as in lemma 6.5,(ii) 〈W(y, q),∇f(y, q)〉 > 1

3 min{‖∇f(y, q)‖2, ‖∇f(y, q)‖, χ2

1+χ ,ν

4[1+(16ρK+ν)1/2]};


moreover if (y1, q1), (y2, q2) ∈ Uy,q, then we can assume that∫ T

0

12

∣∣q21 − q22∣∣+ (1 + δ) |V (q1)− V (q2)| <

ν

4,

|J(y1, q1)− J(y2, q2)| <ν

4,

‖y1 − y2‖∞ <32ω‖δ′‖∞(k + 1)ν,

|Q(y1)−Q(y2)| < min{ν4,9ω‖δ′‖2∞(k + 1)2ν2},

‖H ′a,b(q)−H ′

a,b(qi)‖ ≤ν

16.

(6.10)

Since (ET × Γk,T )∗ is paracompact, there is a locally finite refinement {Ui}i∈I

of the open cover {Uy,q}(y,q)∈(ET×Γk,T )∗ , that is

• for all i ∈ I, there exists (yi, qi) such that Ui ⊂ Uyi,qi;

• for all (y, q) ∈ (ET ×Γk,T )∗, there exists U neighborhood of (y, q) such thatU ∩ Ui 6= ∅ for finitely many i ∈ I only.

Then take a Lipschitz continuos partition of unity {ϕi}i∈I subordinate to {Ui},that is, a family of Lipschitz continuos functions 0 ≤ ϕi ≤ 1 such that

• suppϕi ⊂ Ui;•∑

i∈I ϕi(y, q) = 1 for all (y, q) ∈ (E × Γ)∗.

Let, for all (y, q) ∈ (E × Γ)∗,

v(y, q) =∑i∈I

ϕi(y, q)W(yi, qi).

Since this is locally a finite sum, then v(y, q) is locally Lipschitz.Moreover, for all (y, q) ∈ (E × Γ)∗

‖v(y, q)‖ = ‖∑

i

ϕi(y, q)W(yi, qi)‖ ≤∑

i

ϕi(y, q)‖W(yi, qi)‖

<2C

min{‖∇f(y, q)‖, C}∑

i

ϕi(y, q) =2C

min{‖∇f(y, q)‖, C}

(remark that (y, q) ∈ Uyi,qi whenever ϕi(y, q) 6= 0) and this proves (a).Similarly

〈v(y, q),∇f(y, q)〉 =∑

i

ϕi(y, q)〈W(yi, qi),∇f(y, q)〉

>13

min{‖∇f(y, q)‖2, ‖∇f(y, q)‖, χ2

1 + χ,

ν

4[1 + (16ρK + ν)1/2]}

(6.11)

proves (b).Now let us show that (c) holds. Let us take (y, q) satisfying the conditions in

(c), and let us show that (yi, qi) satisfies the assumptions of point (c) of lemma6.5 if (y, q) ∈ Ui ⊂ Uyi,qi

. From

|f(yi, qi)− f(y, q)| ≤ 12|Q(yi)−Q(y)|+ |J(yi, qi)− J(y, q)| < ν

8+ν

4<ρ

2

we deduce that

(6.12) ck(T )− 2ρ < f(yi, qi) < kc(R,∞) + 2ρ.


We also have that∫ T

0

[q2i2

+ (1 + δ)V (qi)] =∫ T

0

[q2

2+ (1 + δ)V (q)]

+∫ T

0

[q2i − q2

2+ (1 + δ)(V (qi)− V (q))]

≤ k(C1 − 1) +ν

4< kC1.

(6.13)

Since (y, q) satisfies the assumptions of point (c), we can find 0 = a0 ≤ b0 ≤ . . . ≤ak ≤ bk = T ∈ [0, T ] such that |aj − bj−1| ≤ D and∫

[aj ,bj ]

[q2

2+ (1 + δ)V (q)] ≤ 7

4ν.

Then ∫ bj

aj

[q2i2

+ (1 + δ)V (qi)] ≤∫ bj

aj

[q2

2+ (1 + δ)V (q)]+

+∫ bj

aj

[|q2i − q2|

2+ (1 + δ) |V (qi)− V (q)|]

<74ν +

ν

4= 2ν;

(6.14)

moreover either ‖y‖∞ ≥ 2 32ω‖δ

′‖∞2(k + 1)ν and

‖yi‖∞ ≥ ‖y‖∞ − ‖y − yi‖∞ >32ω‖δ′‖∞2(k + 1)ν,

or |Q(y)| ≥ 2 92ω‖δ

′‖2∞4(k + 1)2ν2 and

|Q(yi)| ≥ |Q(y)| − |Q(yi)−Q(y)| > 92ω‖δ′‖2∞4(k + 1)2ν2.

Hence πΓW(yi, qi) = 0 and

πΓv(y, q) =∑

i

ϕi(y, q)πΓW(yi, qi) = 0.

Now let us prove that (d) holds. Let (y, q) ∈ Ui ⊂ Uyi,qifor some i ∈ I satisfy

assumptions of point (d). Then, arguing as we did to prove point (c) we have that

ck(T )− 2ρ < f(yi, qi) < kc(R,∞) + 2ρ,

qi ∈ B(k,C1, 2ν,D, T ),

for all 0 = a0 ≤ b0 ≤ · · · ≤ ak ≤ bk = T ∈ [0, T ] such that

|aj − bj−1| ≤ D there is j such that∫ bj

aj

[q2i2

+ (1 + δ)V (qi)] > ν.

Then (yi, qi) satisfies either assumption (c) of lemma 6.5, which implies

πΓW(yi, qi) = 0 and 〈H ′a,b(q), πΓW(yi, qi)〉 = 0

or assumption (d) of lemma 6.5, so that

〈H ′a,b(qi), πΓW(yi, qi)〉 >

ν

8whenever [aj , bj ] ⊂ [a, b] for some j ∈ A. Moreover, since

|〈H ′a,b(q)−H ′

a,b(qi), πΓW(yi, qi)〉| ≤ ‖H ′a,b(q)−H ′

a,b(qi)‖ ≤ν

16,


we have that in such a case

〈H ′a,b(q), πΓW(yi, qi)〉 = 〈H ′

a,b(qi), πΓW(yi, qi)〉+ 〈H ′a,b(q)−H ′

a,b(qi), πΓW(yi, qi)〉

>ν

8− ν

16=

ν

16.

We can repeat the same arguments for all i ∈ I such that (y, q) ∈ Ui ⊂ Uyi,qi

and we obtain for all y ∈ E

〈H ′a,b(q), πΓv(y, q)〉 =

∑i∈I

ϕi(y, q)〈H ′a,b(q), πΓW(yi, qi)〉 ≥ 0.

If

•‖y‖∞ <12

32ω‖δ′‖∞2(k + 1)ν and |Q(y)| < 1

292ω‖δ′‖2∞4(k + 1)2ν2,

holds, then (yi, qi) satisfies (d) of lemma 6.5, and also the last part of (d) follows.�

Using the pseudo-gradient we have defined, we show that the following deforma-tion lemma holds:

Proposition 6.15. For all N neighborhood of the critical set

K = {(y, q) | ∇f(y, q) = 0 and ck(T )− 32ρ ≤ f(y, q) ≤ kc(R,∞) +

32ρ}

there exists θ > 0 and a continuos one-parameter family of homeomorphism φ(·, t)of ET × Γk,T , 0 ≤ t < +∞ with the properties

(a) φ(y, q, t) = (y, q) if t = 0 or ∇f(y, q) = 0 or f(y, q) ≤ ck(T )− 32ρ or f(y, q) ≥

kc(R,∞) + 32ρ;

(b) f(φ(y, q, t)) is non increasing in t for all (y, q) ∈ ET × Γk,T ;(c) for all t ≥ 0 such that ck(T )−ρ < f(φ(y, q, t)) < kc(R,∞)+ρ and φ(y, q, t) /∈ Nwe have that

d

dtf(φ(y, q, t)) < −1

3min{θ2, χ2

1 + χ,

ν

4[1 + (16ρK + ν)1/2]};

(d) φ : ET × Γk,T × [0,+∞[→ ET × Γk,T has the semi-group property, that isφ(·, t) ◦ φ(·, s) = φ(·, s+ t) for all s, t ≥ 0;(e) if q ∈ B(k,C1 − 2, ν,D, T ) and y ∈ ET is such that f(y, q) < kc(R,∞) + ρ,then

πΓφ(y, q, s) ∈ B(k,C1 − 1,32ν,D, T ) for all s ≥ 0;

(f) let (y, q) be such that q ∈ B(k,C1−2, ν,D, T ), ck(T )−ρ < f(y, q) < kc(R,∞)+ρ. Set (ys, qs) = φ(y, q, s). Then, for all s ≥ 0, if qs(cs,j) = 2jπ, we have∫ cs,j+1

cs,j

12q2s + (1 + δ(R cosωt))V (qs) < c(R,∞) + 3ρ,

and exists ∆ > 0 such that, if [αs,j , βs,j ] denotes the interval where qs jumps from2(j − 1)π + η2 to 2jπ − η2, then

(6.16) [αs,j , βs,j ] ∩ [αs+t,j , βs+t,j ] 6= ∅ for all 0 ≤ t ≤ ∆,

provided the same holds for s = 0.As a consequence, if R ≥ R3 we have that 2mπ

ω + π12ω ≤ αs,j < βs,j ≤ 2(m+1)π

ω −π

12ω for some m ∈ Z and for all s ≥ 0.


Proof. Let Nκ ⊂ N be a κ-neighborhood of the set of critical points K. Then, since(PS) holds, we can find θ ∈ (0, 1) such that ‖∇f(y, q)‖ > θ for all (y, q) /∈ Nκ andsuch that ck(T )− 3

2ρ < f(y, q) < kc(R,∞) + 32ρ.

Let ξ be a Lipschitz continuos function on ET ×Γk,T such that 0 ≤ ξ ≤ 1, ξ ≡ 1outside Nκ, ξ ≡ 0 in Nκ/2. Also let ϕ be a Lipschitz continuos function on R suchthat 0 ≤ ϕ ≤ 1, ϕ(s) ≡ 0 if s ≤ ck(T ) − 3

2ρ or s ≥ kc(R,∞) + 32ρ, ϕ(s) ≡ 1 if

ck(T ) − ρ ≤ s ≤ kc(R,∞) + ρ. Finally, let v : (ET × Γk,T )∗ → ET × Γk,T be thepseudo-gradient vector field already defined and

e(y, q) =

{−ξ(y, q)ϕ(f(y, q))v(y, q) if (y, q) ∈ (ET × Γk,T )∗

0 otherwise.

By the choice of ϕ and ξ, the vector field e vanishes identically (and thereforeis Lipschitz continuos) near critical points. Hence e is locally Lipschitz-continuosthroughout ET ×Γk,T . Moreover, since ‖v‖ < 2, also ‖e‖ ≤ 2 is uniformly bounded.Hence there exists a global solution φ : ET × Γk,T × R → ET × Γk,T of the initialvalue problem {

∂∂tφ(y, q, t) = e(φ(y, q, t))φ(y, q, 0) = (y, q).

The function φ is continuos in (y, q), differentiable in t and has the semi-groupproperty φ(·, t) ◦ φ(·, s) = φ(·, s+ t) for all s, t ≥ 0. In particular for any t ∈ R themap φ(·, t) is a homeomorphism and (d) follows.

Property (a) is trivially satisfied.Property (b) is satisfied because

d

dtf(φ(y, q, t)) = 〈∇f(φ(y, q, t)),

∂φ

∂t(y, q, t)〉

= 〈∇f(φ(y, q, t)), e(φ(y, q, t))〉 ≤ 0.

Similarly, using point (b) of lemma 6.9 we deduce that point (c) holds.Now let us show that property (e) holds. We use the notation ys = πEφ(y, q, s)

and qs = πΓφ(y, q, s). By contradiction, assume that there exists s such that

qs = πΓφ(y, q, s) /∈ B(k,C1 − 1,32ν,D, T ).

Then

(6.17) J(qs) > k(C1 − 1)

or, for all choices of 0 = a0 ≤ b0 ≤ . . . ≤ bk = T such that aj − bj−1 ≤ D there is jsuch that

(6.18)∫ bj

aj

[q2s2

+ (1 + δ)V (qs)] >32ν.

Let us consider the case (6.17) first. Then we can assume that there is s0 < s suchthat J(qs0) = k(C1 − 1) − 1 and for all s ≤ s0 there are 0 ≤ as,0 ≤ bs,0 ≤ . . . ≤bs,k = T such that as,j − bs,j−1 ≤ D and∫ bs,j

as,j

[q2s2

+ (1 + δ)V (qs)] ≤32ν;

then, using (3.1), (4.5), (5.1), (5.8), we have

−Q(ys0) = 2[J(ys0 , qs0)− f(ys0 , qs0)] ≥ 2J(qs0)− 2f(ys0 , qs0)

≥ J(qs0)− 2(kc(R,∞) + ρ) = k(C1 − 1)− 1− 2kc(R,∞)− 2ρ

≥ 4ρ > 6492ω‖δ′‖2∞(k + 1)2ν2,


so that

|Q(ys0)| ≥ 6492ω‖δ′‖2∞(k + 1)2ν2.

Then, by point (c) of lemma 6.9, it follows that πΓv(ys0 , qs0) = 0 and then J(qs0)can not increase. Then this case can never happen.

Suppose now that (6.18) holds. There exists s0 < s such that qs ∈ B(k,C1 −1, 3

2ν,D, T ) for all s ∈ [0, s0) and

• ∀0 = a0 ≤ b0 ≤ . . . ≤ bk = T such that aj − bj−1 ≤ D∫ bj

aj

[q2s0

2+ (1 + δ)V (qs0)] ≤

32ν for all j

and there is κ such that∫ bκ

aκ

[q2s0

2+ (1 + δ)V (qs0)] =

32ν

• ∀s ∈ (s0, s], ∀0 = a0 ≤ b0 ≤ . . . ≤ bk = T such that aj − bj−1 ≤ D

there is j such that∫ bj

aj

[q2s2

+ (1 + δ)V (qs)] >32ν

• qs ∈ B(k,C1 − 1,74ν,D, T ) ∀s ≤ s.

In this case (ys0 , qs0) is such that either point (c) of lemma 6.9 holds (in whichcase πΓv(ys0 , qs0) = 0 and qs = qs0 for all s in a right neighborhood of s0) or point(d) of lemma 6.9 holds. In the latter case, let us take aj , bj be defined as in (6.3).Then 〈H ′

aκ,bκ(qs0), πΓv(ys0 , qs0)〉 ≥ 0, showing that qs ∈ B(k,C1 − 1, 3

2ν,D, T ) fors in a right neighborhood of s0. So also (6.18) cannot hold and point (e) follows.

Let us prove point (f).We observe that for all s ≥ 0 we have that ck(T )− 3

2ρ < f(ys, qs) < kc(R,∞)+ 32ρ

(by point (a)) and qs ∈ B(k,C1 − 1, 32ν,D, T ) (by point (e)).

Then, by the way the pseudo-gradient vector field has been constructed, we havethat dqs/ds 6= 0 if and only if

‖ys‖∞ < 232ω‖δ′‖∞2(k + 1)ν and |Q(ys)| < 2

92ω‖δ′‖2∞4(k + 1)2ν2,

and the first statement of point (f) follows arguing as in lemma 5.9. Equation(6.16), which gives some sort of continuity of the jump intervals, follows from thefact that qs is Lipschitz in s (e is bounded) and the uniqueness of the jump intervals[αs,j , βs,j ].

Then, using lemma 5.7 one deduce that [αs,j , βs,j ] 6⊂ [ 2mπω − π

4ω ,2mπ

ω + π4ω ]. Since

βs,j − αs,j ≤ L ≤ π/6ω we have that [αs,j , βs,j ] ∩ [ 2mπω − π

12ω ,2mπ

ω + π12ω ] = ∅ and

(6.16) proves the last statement of point (f). �

7. Existence of critical points

Let the constants ρ, η2, L, ω, k, ν, `0 and D be fixed as at the beginning ofsection 6.

We want to show that for R ≥ R3 (given by lemma 5.7) we can find a criticalpoint (yN , qN ) ∈ ETN

×Γk,TN, with TN = ω−1π(4N +1), such that qN jumps from

η2 to 2kπ − η2 in an interval of bounded length and containing the point T/2.More precisely, we prove that

Theorem 7.1. Given ω < ω1, k ∈ N and ` ≥ `0, for all N large enough we canfind a critical point (yN , qN ) of the functional f(y, q) such that(a) qN ∈ B(k,C1, 2ν,D, T );(b) |qN (t)| ≤ η0 for all t ∈ [0, 2Nπ/ω];


(c) |qN (t)− 2jπ| ≤ η0 for all t ∈ [2(N + j`)π/ω − π/ω, 2(N + j`)π/ω + π/ω];(d) |qN (t)− 2kπ| ≤ η0 for all t ∈ [2(N + k`)π/ω, 4(N + 1)π/ω];(e) f(yN , qN ) ≤ kc(R,∞) + ρ.

Proof. We will prove the theorem by contradiction. We assume that there is nocritical point of f satisfying point (a) through (e). Then, since (PS) holds, thereis a N neighborhood of the critical set K introduced in proposition 6.15 such that

(y, q) /∈ N for all (y, q) satisfying (a)–(e).

Consider the path y− ∈ E− 7→ h(y−) = (h1(y−), q) ∈ Hk which we have built inlemma 4.10 (for the current choice of ω, k, `, T and ρ given by (5.1), η = η2 given by(5.2), ν given by (5.8)), and let for s ≥ 0 and y− ∈ E−, (yy−,s, qy−,s) = φ(h(y−), s),where φ is the flow defined in proposition 6.15.

We claim that, for all s ≥ 0 and y− ∈ E− such that ck(T ) − ρ < f(h(y−)),(yy−,s, qy−,s) satisfies (a)–(e).

Since qy−,0 = q ∈ B(k,C1 − 2, ν,D, T ) and f(yy−,0, qy−,0) = f(h(y−)) <kc(R,∞) + ρ for all y− ∈ E− (see lemma 4.10), we know from point (e) of propo-sition 6.15 that qy−,s ∈ B(k,C1 − 1, 3

2ν,D, T ) for all s ≥ 0 and for all y− ∈ E−,and (a) follows.

Point (e) follows trivially from point (b) of proposition 6.15.Since q jumps from 2jπ + η2 to 2(j + 1)π − η2 in the interval [αj+1, βj+1] ⊂

[2π(N+j`+2)/ω, 2π(N+j`+`0−2)/ω], from point (f) of proposition 6.15, we havethat qy−,s jumps from 2jπ+η2 to 2(j+1)π−η2 in the interval [αy−,s,j+1, βy−,s,j+1] ⊂[2π(N + j`+ 1)/ω, 2π(N + j`+ `0 − 1)/ω].

In order to prove that qy−,s satisfies point (b), (c) and (d) we remark that,arguing as in lemma 4.6 we have∫ cj+1

cj

12q2 + (1 + δ(R cosωt))V (q) < c(R,∞) +

ρ

2k,

and using again point (f) of proposition 6.15, we can apply lemma 5.9.From the claim follows that (yy−,s, qy−,s) /∈ N for all s ≥ 0. Point (c) of

proposition 6.15 shows then that

d

dtf(φ(y, q, t)) < −θ ≡ −1

3min{θ2, χ2

1 + χ,

ν

4[1 + (16ρK + ν)1/2]}

provided f(yy−,s, qy−,s) ≥ ck(T )− ρ. This implies that

f(yy−,s0 , qy−,s0) < ck(T )− ρ for all y− ∈ E−

if s0θ > kc(R,∞)− ck(T ) + 2ρ, contradiction which proves the theorem. �

Corollary 7.2. Under the assumption of theorem 7.1, the critical point (yN , qN )whose existence follows from theorem 7.1 is such that, for all N large

(a)∫ T

012 q

2N + (1 + δ)V (qN ) ≤ kc(R,∞) + 1;

(b) Q(yN ) ≤ ε1,R;(c) ‖yN‖∞ ≤ ε2,R;(d) ‖yN‖∞ ≤ ε3,R;

where ε1,R, ε2,R and ε3,R → 0 as R→ +∞.


Proof. Since (yN , qN ) is a critical point and hence yN a solution of (2.2), we havefrom lemma 2.3 that since qN ∈ B(k,C1, 2ν,D, T )

‖yN‖∞ <3ω‖δ′‖∞(k + 1)ν,∣∣∣∣∣

∫ T

0

( ˙yN2 − ω2y2

N )

∣∣∣∣∣ < 18ω‖δ′‖2∞(k + 1)2ν2 <

ρ

4.

We have then that∫ T

0

12q2N + (1 + δ)V (qN ) ≤

∫ T

0

12q2N + (1 + δ(R cosωt+ yN ))V (qN )

= f(yN , qN )− 12Q(yN ) ≤ kc(R,∞) + ρ+

ρ

8and point (a) follows.

In order to prove points (b), (c), and (d), we fix ε > 0. As in lemma 2.1, wehave that

‖yN‖∞ ≤ 12ω

∫ T

0

|δ′(R cosωt+ yN )|V (qN ),

‖yN‖∞ ≤ 12

∫ T

0

|δ′(R cosωt+ yN )|V (qN ),

|Q(yN )| ≤ ‖yN‖∞∫ T

0

|δ′(R cosωt+ yN )|V (qN ).

Letting yN (t) = yN (t + 2πN/ω) and qN (t) = qN (t + 2πN/ω) (recall that T =(4πN + π)/ω), we have that∫ T

0

|δ′(R cosωt+ yN )|V (qN ) =∫ (2N+1)π/ω

−2πN/ω

|δ′(R cosωt+ yN )|V (qN ).

We also remark that, as in lemma 4.2, the following exponential estimates holdsfor some α > 0 and T large:

|qN (t)| ≤ η0eαt t ≤ −T

|qN (t)− 2kπ| ≤ η0e−αt t ≥ T .

Let us take T > log(µ‖δ′‖∞η20/εα)1/2α and N such that 2πN > T ω, we have that∫

[−2πN/ω,(2N+1)π/ω]\[−T ,T ]

|δ′(R cosωt+ yN )|V (qN ) < ε.

We now let Aσ = {t ∈ [−T , T ] such that |cosωt| ≤ σ}. Then, for σ small enough(see also the proof of lemma 2.3)∫

[−T ,T ]∩Aσ

|δ′(R cosωt+ yN )|V (qN ) < ‖δ′‖∞‖V ‖∞4σω

(1 +2T ωπ

) < ε.

Let us remark that T ≈ − log ε and σ ≈ −ε/ log ε.Finally, always as in lemma 2.3, we have that∫[−T ,T ]\Aσ

|δ′(R cosωt+ yN )|V (qN ) < sup{|δ′(s)| | s ≥ Rσ − ‖yN‖∞}kC1 < ε

for R large enough, (if δ′(s) ≈ 1/s, then R ≈ −ε−2 log ε) and ‖yN‖∞ < 3ε/2ω,‖yN‖∞ < 3ε/2 and |Q(yN )| ≤ 9ε2/2ω. �


Theorem 7.3. For ω < ω1, k ∈ N, ` ≥ `0 and R ≥ R3 there exists (x, q) solutionof the problem {

q(t) = (1 + δ(x))V ′(q(t))x+ ω2x = δ′(x)V (q(t))

satisfying the conditions

(a) limt→−∞ q(t) = 0, limt→+∞ q(t) = 2kπ;(b) |q(t)| ≤ η0 for all t ≤ 0;(c) |q(t)− 2jπ| ≤ η0 for all t ∈ [2j`π/ω − π/ω, 2j`π/ω + π/ω];(d) |q(t)− 2kπ| ≤ η0 for all t ≥ 2k`π/ω;(e) there is R such that limk→±∞

∣∣x(t+ k)− R cos(t+ θ±)∣∣ = 0, where

∣∣R− R∣∣ ≤

εR → 0 as R→ +∞.

Proof. We known that for all TN = π(4N + 1)/ω, there exists (yN , qN ) solution ofq(t) = (1 + δ(R cosωt+ y))V ′(q(t))y + ω2y = δ′(R cosωt+ y)V (q(t))q(0) = q(TN )− 2kπ = 0y(0)− y(TN ) = y(0)− y(TN ) = 0.

satisfying properties (a)–(e) of theorem 7.1 and (a)–(d) of corollary 7.2.For all N let τN = TN/2− π/2ω = 2πN/ω and set

qN (t) = qN (t+ τN ) ∀t ∈ [−τN , TN − τN ].

Then, from∫ TN−τN

−τN

˙q2N ≤ 2kC1 and points (b)–(d) of theorem 7.1 we have that,since obviously |q(t)| ≤ 2kπ+ 1 (one could also show that 0 ≤ q(t) ≤ 2kπ for all t),

‖qN‖2H1(a,b) =∫ b

a

[q2N + q2N ]dt ≤ 2kC1 + (2kπ + 1)2(b− a) ∀a < b ∈ R

and hence, for all a < b ∈ R, ‖qN‖H1(a,b) is uniformly bounded. This implies that,up to a subsequence, qN → q weakly in H1(a, b) and uniformly on compact sets.

We also have that, since qN jumps from η0 to 2kπ − η0 in a subinterval of thecompact set [0, 2k`π/ω], q(t) 6≡ 0 and satisfies point (b)–(d).

Similarly, we deduce from point (c)–(d) of corollary 7.2 that yN converges weaklyin H1(a, b) and uniformly on compact sets to y. It is then easy to prove that (y, q)is a solution of {

q(t) = (1 + δ(R cosωt+ y))V ′(q(t))y + ω2y = δ′(R cosωt+ y)V (q(t))

Using also the fact that qN (t) is exponentially decreasing to 0 (increasing to 2kπ)outside the set [0, 2k`π/ω], one then deduce that q(t) → 0 as t→ −∞, q(t) → 2kπas t→ +∞, and that y(t− n) tends, as n→ ±∞, to a solution of

y + ω2y = 0,

say ρ± cos(ωt + ϕ±) as n → ±∞. We also know that (x(t), q(t)) = (R cosωt +y(t), q(t)) is a solution of (EL). Since energy is conserved for such an equation,we immediately have that

∣∣x(t− n)− R cos(ωt+ θ±)∣∣ → 0 as n → ±∞ for some

θ± ∈ [0, 2π].Finally, we remark that, since yN (t) is a periodic solution of (BVP), xN (0) =

R cosω0+yN (0) = R+yN (0), xN (TN ) = −R+yN (0), xN (0) = yN (0) = yN (TN ) =


xN (TN ). Using energy conservation for (EL), we then have that

EN =12(xN (0)2 + ω2xN (0)2) +

12q2N (0)− (1 + δ(xN (0)))V (qN (0))

=12(yN (0)2 + ω2(R+ yN (0))2) +

12q2N (0)

=12(xN (T )2 + ω2xN (T )2) +

12q2N (T ) + (1 + δ(xN (T )))V (qN (T ))

=12(yN (0)2 + ω2(−R+ yN (0))2) +

12q2N (T ).

From which we deduce that

Rω2yN (0) +12q2N (0) = −Rω2yN (0) +

12q2N (TN ).

As a consequence

EN =12R2ω2 +

12(yN (0)2 + ω2yN (0)2) +

14q2N (0) +

14q2N (TN ).

Since EN → ω2R2/2 as N → ∞, qN (t) decays exponentially to 0 and using theestimates (c) and (d) of corollary 7.2, we immediate have that the last statementof the theorem follows. �

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E-mail address, Coti Zelati: [email protected]

E-mail address, Macrı: [email protected]

Dipartimento di Matematica e Applicazioni “R. Caccioppoli”, Universita di Napoli“Federico II”, via Cintia, 80126 Napoli, Italy

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