homoclinic orbits of a hamiltonian system

Z. angew. Math. Phys. 50 (1999) 759–7780044-2275/99/050759-20 $ 1.50+0.20/0c© 1999 Birkhauser Verlag, Basel

Zeitschrift fur angewandteMathematik und Physik ZAMP

Homoclinic orbits of a Hamiltonian system

Yanheng Ding and Michel Willem

Abstract. We establish existence results of homoclinic orbits of the first order time-dependentHamiltonian system z = JHz(t, z), where H(t, z) depends periodically on t, H(t, z) = 1

2 zL(t)z+W (t, z), L(t) is a symmetric matrix valued function and W (t, z) satisfies certain global su-perquadratic condition. We relax partly the assumption often used before: L is independent oft and sp(JL) ∩ iR = φ.

Mathematics Subject Classification (1991). 58F05, 58E05, 70H05.

Keywords. Homoclinic orbits, Hamiltonian systems, linking theorem, concentation-compactness.

1. Introduction

In this paper we are interested in the existence of homoclinic orbit of the Hamil-tonian system:

z = JHz(t, z), (HS)

where z = (p, q) ∈ RN × RN = R2N ,J is the standard symplectic matrix in

R2N ,J =(

0 −11 0

), and H ∈ C(R×R2N ,R) is 1-periodic in t ∈ R and is of the

formH(t, z) =

12z · L(t)z +W (t, z) (1.1)

with L ∈ C(R,R4N2) being a 2N × 2N symmetric matrix valued function, and

W ∈ C1(R×R2N ,R) being globally superquadratic in z ∈ R2N . Here a homoclinicorbit of (HS) is a solution of the equation satisfying

z(t) 6≡ 0 and z(t)→ 0 as |t| → ∞.

In recent years there have been many papers devoted to the existence of homo-clinic orbits for Hamiltonian systems via critical point theory. For example, onemay refer to [AC], [CR], [D], [DG], [OW], [R] and [RT] for the second order sys-tems, and [CES], [DL], [HWy], [S] and [T] for the first order systems. We observe

760 Y. Ding and M. Willem ZAMP

that in all these results on the first order system (HS) the following condition isalways required:

(∗)L(t) ≡ L is independent of t such that sp(J L) ∩ iR = φ,where sp(J L) denotes the set of all eigenvalues of J L. Let A = −(J d

dt +L(t)) be the selfadjoint operator acting in L2(R,R2N ) with the domain D(A) =H1(R,R2N ). Then, letting σ(A) denote the spectrum of A, the assumption (∗)implies the following one

(∗)′L(t) ≡ L is independent of t and there is α > 0 such that(−α, α)∩σ(A) = φ.Consequently, the operator A : W 1,p(R,R2N )→ Lp(R,R2N ) is a homeomorphismfor all p > 1, which is important for the variational arguments.

The main purpose of the present paper is to relax the condition (∗). Naturally,we can not expect to cancel completely the restriction. For example, supposingH(t, z) = H(z) independent of t, and letting z(t) be a homoclinic orbit of (HS),one has H(z(t)) ≡ 0 in view of the conservation of energy, and so the set Ω = z ∈R2N ; H(z) = 0 % 0. If furthermore, H(z) = 1

2z · Lz + W (z),W (z) = o(|z|2)at z = 0 and W (z) ≥ a|z|µ for all z with a > 0 and µ > 2, then L must have atleast a negative eigenvalue and a nonnegative eigenvalue.

More precisely we will deal with the following case:(H1) L(t) depends periodically on t with period 1, and there is α > 0 such

that(0, α) ∩ σ(A) = φ.

We underline that, in the present case (H1), 0 may belong to the continuousspectrum of A (handling such a case via critical point theory is also of interest initself). In order to describe the difference between (∗) and (H1), one may take thefollowing example

L1 = c

(1 11 1

), c < 0.

We will show in the next section of the paper that L1 satisfies (H1). However,clearly, sp(JL1) = 0, i.e., L1 does not satisfy (∗). The matrix

L2 = diag(λ1, · · · , λ2N )

can also show the difference. Note that, λ = ±(−λiλi+N )1/2, i = 1, · · · , N , are allthe eigenvalues of JL2. L2 satisfies (∗) if and only if, without loss of generality,λ1 ≤ · · · ≤ λN < 0 < λN+1 ≤ · · · ≤ λ2N . On the other hand, L2 satisfies(H1) if λ1 ≤ · · · ≤ λN < 0 ≤ λN+1 ≤ · · · · · ·λ2N (particularly, it is allowed thatλN+1 = · · · = λ2N = 0, see the next section).

We now state the main result. Assume further on W (t, z) that the followingconditions are satisfied:

(H2) W (t, z) ∈ C1(R × R2N ,R) is 1-periodic in t, and there are a1 > 0, µ ≥γ > 2 such that for all (t, z),

a1|z|µ ≤ γW (t, z) ≤ z ·Wz(t, z);

Vol. 50 (1999) Homoclinic orbits of a Hamiltonian system 761

(H3) there are κ ∈ (1, 2) and a2 > 0 such that for all (t, z) with |z| ≥ 1

|Wz(t, z)|κ ≤ a2Wz(t, z)z;

(H4) there is a3 > 0 such that for all (t, z) with |z| ≤ 1

|Wz(t, z)| ≤ a3|z|µ−1.

We remark that (H2)− (H3) imply, letting p := 1 + 1κ−1 ,

µ ≤ p and |Wz(t, z)| ≤ a′2|z|p−1 whenever |z| ≥ 1. (1.2)

It is easy to check that, under (H2), if there is 1 + µ2 < s < 1 + µ such that

|Wz(t, z)| ≤ a2|z|s−1, ∀ (t, z) with |z| ≥ 1,

then (H3) is verified with κ = µ/(s− 1)(> 1 + 1/µ). However, the contrary is nottrue in general since in (H3), κ may arbitrarily close to 1. We shall establish thefollowing result.

Theorem 1.1. Let (H1)− (H4) be satisfied. Then (HS) has at least one homo-clinic orbit.

Remark 1.1. If there exists α > 0 such that (−α, 0) ∩ σ(A) = φ and W (t, z) :=−W (t, z) satisfies the assumptions (H2)− (H4), then the same conclusion of The-orem 1.1 remains valid.

Remark 1.2. Very recently, the paper [BD] has considered the Schrodinger equa-tion with periodic potential and nonlinearity

Su ≡ (−∆ + V (x))u = g(x, u)u(x)→ 0 as |x| → ∞,

(S)

under the assumptions that 0 ∈ R is an end point of a gap in σ(S), and g satisfiessome superlinear conditions, and obtained certain existence and multiplicity resultsfor solutions of (S).

The relations to our main result are as follows. It was known that the spectrumσ(S) of the Schrodinger operator S with periodic potentials is continuous. It isnatural to show that such a description is also valid for the selfadjoint operatorA (= −J d/dt − L(t)). Remark that, under (H1), if 0 ∈ σ(A) then there is asequence un ∈ H1 with |un|L2 = 1 and |Aun|L2 → 0. Therefore the operatorA can not lead the behavior at 0 of the equation, which brings difficulties in theusual variational arguments. As a compensation, we will show an “embedding”result of Sobolev’s type which concludes that the space of mappings u : R→ R2N

equipped with the norm


||u|| = (||A|1/2u|2L2 + |u|2Lµ)1/2, (µ > 2)

is embedded continuously in Lν for all ν ≥ µ. In addition, some concrete examplesand a class of matrix valued functions satisfying (H1) will be provided to sensethe assumption. Finally, an extension of the generalized linking theorem (dueto [KS] and [TroW]) to Banach spaces and a proper application of the so-calledconcentration compactness (see, [Li]) play also an important role for establishingTheorem 1.1.

2. Some preliminary results

Below, let | · |ν denote the Lν = Lν(R,R2N ) norm for ν ≥ 1, and (·, ·)L2 theinner product of H := L2. Let M(t) ∈ C(R,R4N2

) be a symmetric matrix valuedfunction, and let F (t) be the fundamental matrix with F (0) = I for the equation

x = JM(t)x.

M is said to have an exponential dichotomy if there is a projector P and positiveconstants K, β such that

|F (t)PF−1(s)| ≤ Ke−β(t−s) if s ≤ t|F (t)(I − P )F−1(s)| ≤ Ke−β(s−t) if s ≥ t,

(2.1)

(see [Co]).Set W 1,ν := W 1,ν(R,R2N ) for ν ≥ 1,H1 := W 1,2 and H1/2 := H1/2(R,R2N ).

For a selfadjoint operator S in H, we denote by |S| and |S|1/2 its absolute valueand square root.

Proposition 2.1. Suppose that M(t) has an exponential dichotomy and let ν ≥ 1.Then the following conclusions hold:(i) The operator

Bν : Lν ⊃W 1,ν → Lν , z 7→ −(J d

dt+M)z

has a bounded inverse B−1ν satisfying with some b = b(ν, σ) > 0

|B−1ν u|σ ≤ b|u|ν , ∀u ∈ Lν

for all σ ≥ ν;(ii) B := B2 is selfadjoint, and there are a > 0, b1 > 0, b2 > 0 such that σ(B) ∩[−a, a] = φ and

b1||u||H1 ≤ |Bu|2 ≤ b2||u||H1 for all u ∈ H1;


(iii) D(|B|1/2) = H1/2, and there are b′1, b′2 > 0 such that

b′1||u||H1/2 ≤ ||B|1/2u|2 ≤ b′2||u||H1/2 for all u ∈ H1/2.

Proof. For any u ∈ Lν, ν ≥ 1, there is a unique z ∈W 1,ν satisfying

−(J d

dt+M)z = u

given by

z(t) =∫ t

−∞F (t)PF−1(s)J uds−

∫ ∞t

F (t)(I − P )F−1(s)J uds.

Set

χ+(s) = χ−(−s) =

1 if s ≥ 00 if s < 0.

Then

z(t) =∫RF (t)PF−1(s)χ+(t− s)J u

−∫RF (t)(I − P )F−1(s)χ−(t− s)J u := z1(t) + z2(t),

and by (2.1)

|z1(t)| ≤ K∫Re−β(t−s)χ+(t− s)|u|ds,

|z2(t)| ≤ K∫Re−β(s−t)χ−(t− s)|u|ds.

Setting g+(τ) = e−βτχ+(τ) and g−(τ) = eβτχ−(τ), one has then

|z1(t)| ≤ K(g+ ∗ |u|)(t),

|z2(t)| ≤ K(g− ∗ |u|)(t)

where ∗ denotes the convolution. Observe the following∫R|g+|σ =

∫R|g−|σ =

1βσ∀σ ≥ 1,

|g±|∞ = 1.

By the convolution inequality, for any θ ≥ 1 satisfying 1/θ = 1/ν + 1/σ − 1,


|zj |θ ≤ K(βσ)−1/σ |u|ν, j = 1, 2

and for 1/ν + 1/ν′ = 1, ν > 1

|zj|∞ ≤ K(βν′)−1/ν′ |u|ν , j = 1, 2,

and also|zj|∞ ≤ K|u|1 if ν = 1, j = 1, 2.

Therefore,

|z|θ ≤ K(βσ)−1/σ |u|ν , θ, ν, σ ≥ 1 and1θ

=1ν

+1σ− 1. (2.2)

Now the conclusion (i) follows from (2.2).It is easy to verify that B is selfadjoint. Note that if there is a sequence of

positive numbers an → 0 such that σ(B)∩ [−an, an] 6= φ, then there is a sequenceun ⊂ D(A) with |un|2 = 1 and |Bun|2 → 0, contradicting (2.2). The inequalityof (ii) is clear by again (2.2).

We now verify (iii). Let Λ := − d2

dt2. Then D(Λ) = H2 and by an interpolation

theory (see [Tr, §2.5.2])

(D(Λ0),D(Λ))θ,2 = (L2,H2)θ,2 = H2θ, 0 ≤ θ ≤ 1.

On the other hand (see [Tr, Theorem 1.18.10])

(D(Λ0),D(Λ))θ,2 = D(Λθ)

and, consequentlyH2θ = D(Λθ)

equipped with the norm

||u||2D(Λθ) =∫ ∞

0(1 + λ2θ)d|Fλu|22 = |u|22 + |Λθu|22

where Fλ; −∞ < λ <∞ is the spectral family of Λ. In particular, one sees withθ = 1/4,

H1/2 = D(Λ1/4), ||u||2H1/2 ≤ |u|22 + |Λ1/4u|22.

Since |Λ1/2u|2 = |u|2 ≤ d1|Bu|2 for u ∈ H1 by (ii), one has (Λ1/2u, u)L2 ≤d2(|B|u, u)L2 (see, [Ka, Theorem 4.12]), and so |Λ1/4u|2 ≤ d2||B|1/2u|2 which,together with (2.2), implies the first inequality of (iii). In a similar way andconsidering the operator Λ := − d2

dt2+ 1, one can check the second of (iii).


The proof is complete.

Proposition 2.2. Let M(t) ∈ C(R,R2N ) be a 1-periodic symmetric matrix valuedfunction, and let B = −(J d

dt + M(t)) : L2 ⊃ H1 → L2 the selfadjoint operator.Then the spectrum σ(B) consists of continuous spectra.

Proof. Note that, since B is selfadjoint, it has no residual spectrum and theisolated points of σ(B) are eigenvalues. Hence the proof is completed if we showthat B has no eigenvalue.

Arguing indirectly, assume that B has an eigenvalue λ. Let u ∈ H1 be aneigenfunction associated with λ, i.e., u satisfies the equation

du

dt= J (M(t) + λ)u. (2.3)

Let F (t) be the fundamental matrix of (2.3) with F (0) = I. By the Floquettheory, the monodromy operator F (t) = Q(t)etΓ where Γ = lnF (1) and Q(t) isa 1-periodic continuous differentiable matrix valued function having a boundedinverse Q−1(t) (see, e.g. [RoM]). Set

y(t) = Q−1(t)u(t).

Then y(t)→ 0 as t→ ±∞ (since u ∈ H1) and y satisfies the reduction equation

dy

dt= Γy(t),

and soy(t) = etΓy(0).

Let M−,M0 and M+ be the invariant subspaces of Γ corresponding to its eigen-values with negative, 0 and positive real parts, respectively. Then R2N = M− ⊕M0⊕M+. Let P ∗, ∗ ∈ −, 0,+, be the associated projections, and set y∗ = P ∗y.Then one has

y∗(t) = etΓy∗(0), ∗ ∈ −, 0,+.Since y∗(t) → 0 as t → ±∞, there must hold y∗(t) ≡ 0 for ∗ ∈ −, 0,+, and,consequently,

y(t) = 0

for all t ∈ R, which implies that u = 0, a contradiction, ending the proof.

Now consider the matrix Ma := L(t) + aL, where a > 0, L(t) satisfies (H1)

and L =(

0 11 0

). Clearly aJ L has the eigenvalues λ1 = · · · = λN = a and

λN+1 = · · · = λ2N = −a, and its fundamental matrix is Fa = exp(at

(−1 00 1

))


Therefore aL has an expontential dichotomy. By the roughness of the expontentialdichotomy, for any

a > 4 supt∈R|L(t)|,

Ma also has an expontential dichotomy (see [Co]). We fix arbitrarily such an a.Consider the selfadjoint operator

Aa = −(J d

dt+Ma) = A− aL.

Since for u ∈ D(A)

|Aau|2 = |(A− aL)u|2 ≤ |Au|2 + a|u|2

we have, by Proposition 2.1,

b1||u||2H1/2≤ (|Aa|u, u)L2 ≤ (|A|u, u)L2 + a|u|22≤ b2||u||2H1/2 + a|u|22.

Therefore by continuity

b1||u||H1/2 ≤ ||A|1/2u|2 + a|u|2 ≤ b2||u||H1/2 + 2a|u|2 (2.4)

for all u ∈ H1/2(= D(|A|1/2)), where (and below) the symbols bi stand for genericpositive constants.

Let E(λ); λ ∈ R be the spectral family of A. We have A = U |A|, called thepolar decomposition, where U = I − E(0) − E(−0). Noting that, by (H1) andProposition 2.2, 0 is at most a continuous spectrum of A, the Hilbert space H hasan orthogonal decomposition

H = H− ⊕H+

where H± = u ∈ H; Uu = ±u. We will write in the following for u ∈ H

u = u− + u+

where u± ∈ H±. Let E be the space of the completion of D(|A|1/2) under thenorm

||u||0 = ||A|1/2u|2.

E becomes a Hilbert space under the inner product

(u, v)0 = (|A|1/2u, |A|1/2v)L2 .


By (2.4) we have for all u ∈ D(|A|1/2),

b1||u||H1/2 ≤ ||u||0 + a|u|2 ≤ b2||u||H1/2 + 2a|u|2. (2.5)

E has an orthogonal decomposition

E = E− ⊕E+

whereE± k H± ∩ D(|A|1/2). (2.6)

By assumption (H1) for u ∈ H+ ∩D(A)

||u||20 = (Au, u)L2 =∫ ∞α

λd(E(λ)u, u)L2 ≥ α|u|22

which, together with (2.5) and (2.6), shows

E+ = H+ ∩ D(|A|1/2), and || · ||0 ∼ || · ||H1/2 onE+ (2.7)

where ” ∼ ” means “equivalence”.For any ε > 0, set

H−ε = E(−ε)Hand E−ε = H−ε ∩D(|A|1/2) = H−ε ∩E−.One has similarly to (2.7)

|| · ||0 ∼ || · ||H1/2 on E−ε . (2.8)

Let H−ε := H− H−ε = H− ∩ (clH(∪λ<−εE(λ)H))⊥, where clH(F ) denotes theclosure of the set F in H, and let, for µ > 2, E−ε,µ be the completion of H−ε underthe norm

||u||µ = (||A|1/2u|22 + |u|2µ)1/2. (2.9)

Lemma 2.3. E−ε,µ ⊂ H1loc and is embedded compactly in L∞loc, and continuously

in Lν for all ν ≥ µ.

Proof. By the spectral theory of selfadjoint operators, H−ε ⊂ D(A) = H1. Letun ⊂ H−ε be a Cauchy sequence with respect to || · ||µ. Then

|A(un − um)|22 =∫ 0

−ελ2d|E(λ)(un − um)|22

≤ −ε∫ 0

−ελd|E(λ)(un − um)|22

= ε||A|1/2(un − um)|22 → 0

(2.10)


as n,m→∞. For any finite interval ω ⊂ R, one has∫ω

|un − um|2 ≤ |ω|1−2/µ|un − um|2µ → 0

which, together with (2.10), shows∫ω

|un − um|2 =∫ω

|A(un − um) + L(t)(un − um)|2

≤ 2|A(un − um)|22 + 2∫ω

|L(t)(un − um)|2 → 0

as n,m → ∞. Therefore the limit u of un with respect to || · ||µ belongs toH1(ω). Moreover, since H1(ω) is compactly embedded in L∞(ω) for any finiteinterval ω, one sees that E−ε,µ is compactly embedded in L∞loc.

By (2.10), Aun is a Cauchy sequence in L2. Hence Aun → w in L2. SinceAun → Au in L2

loc, w = Au, i.e, Au ∈ L2. Note that for any finite interval ω∫ω

|u|2 =∫ω

|Au+ Lu|2 ≤ 2∫ω

(|Au|2 + |Lu|2

)≤ d1

(∫ω

|Au|2 + |ω|1−2/µ(∫

ω

|u|µ)2/µ)

.

(2.11)

Letting τ ∈ R and integrating from τ − 1/2 to τ + 1/2 the following equality

u(τ) = u(t) +∫ τ

t

u(s)ds

yields, by Holder inequality,

|u(τ)| ≤(∫ τ+1/2

τ−1/2|u|µ

)1/µ+(∫ τ+1/2

τ−1/2|u|2

)1/2. (2.12)

Since u ∈ Lµ and Au ∈ L2, (2.11) and (2.12) show that

|u(τ)| → 0 as |τ | → ∞,

that is, u ∈ L∞. Therefore u ∈ Lµ ∩ L∞ and so u ∈ Lν for any ν ≥ µ. Replacingu by un − u in (2.11) and (2.12) one sees that E−ε,µ is continuously embedded inL∞ and so is in Lν for any ν ≥ µ.

The proof is complete.

Now let E−µ denote the completion of D(A) ∩ H− with respect to the norm|| · ||µ. Since H1/2 is continuously embedded in Lν for any ν ∈ [2,∞), by (2.8),


E−ε is a closed subspace of E−µ . Noting that E−ε,µ ⊂ E− and is orthogonal to E−εwith respect to (·, ·)0, we have

E−µ = E−ε ⊕E−ε,µ. (2.13)

We now come to the following

Proposition 2.4. Suppose (H1) is satisfied, and let Eµ denote the completion ofthe set D(A) under the norm || · ||µ given by (2.9). Then Eµ has the direct sumdecomposition

Eµ = E−µ ⊕E+, (2.14)

and Eµ is embedded continuously in Lν for any ν ∈ [µ,∞) and compactly in Lνlocfor any ν ∈ [2,∞).

Proof. In view of (2.7), (2.8), (2.13) and Lemma 2.3, E−µ and E+ are closed, and,using the decomposition of E, it is easy to check that E−µ ∩ E+ = 0, and so(2.14) holds. Using the same facts above and Lemma 2.3, one can get easily thedesired conclusion on embeding.

In the following we consider some examples of matrices satisfying (H1).

Definition 2.1. A continuous symmetric matrix valued function L(t) will becalled right (resp. left) dichotomic if there is ε > 0 such that Lε(t) := L(t) + ε hasan exponential dichotomy for each ε ∈ (0, ε] (resp. ε ∈ [−ε, 0)).

Proposition 2.5. If L(t) is right dichotomic, then it satisfies (H1).

Proof. Remark thatx = JL(t)x ⇐⇒ Aεx+ εx = 0

where

Aε := −(J d

dt+ Lε(t)

)= A− ε.

By Proposition 2.1, for any ε ∈ (0, ε], there are aε < 0 < bε, both aε and bεbeing inσ(Aε), such that (aε, bε) ⊂ ρ(Aε) := C\σ(Aε). Let λ := minε, bε. Thensince ε ∈ σ(A) if and only if 0 ∈ σ(Aε), we see that (0, λ) ⊂ ρ(A). The desiredconclusion follows.

Remark 2.1. In the same way above one can check that if L(t) is left dichotomicthen there is α > 0 such that (−α, 0) ⊂ ρ(A).

Example 1. Let

L1(t) := c

(1 11 1

), c 6= 0.


Then sp(JL1) = 0. Note that

L1ε := L1 + ε =(c+ ε cc c+ ε,

)and

sp(JL1ε) = λ±1 = . . . = λ±N = ±(−2cε)1/2.

Therefore, L1 is right (left) dichotomic if c < 0 (c > 0).

Example 2. Let L(t) be 1-perodic continuous symmetric matrix valued functionand let

L2 :=∫ 1

0L(t)dt

be its mean value. Set L2ε := L2 + ε and let

λj(ε) = αj(ε) + iβj(ε), j = 1, . . . , 2N

denote the eigenvalues of JL2ε. Assume that there is ε > 0 such that

α1(ε) ≤ . . . ≤ αN (ε) < 0 < αN+1(ε)

≤ . . . ≤ α2N (ε)∀ ε ∈ (0, ε](ε ∈ [−ε, 0)).

Then L(t) is right (left) dichotomic. In particular, if

L2 = diag(λ1, . . . , λ2N

)with λ1 ≤ . . . ≤ λN < 0 ≤ λN+1 ≤ . . . ≤ λ2N , then L(t) is right dichotomic.

3. The generalized linking theorem on Banach spaces

Let us state in reflexive Banach spaces the generalized linking theorem due toKryszewski-Szulkin [KS] (see also [TroW]) which was originally established in theframe of Hilbert spaces.

Throughout the section, let E denote a reflexive Banach space with the directsum decomposition E = X ⊕ Y, u = x + y for u ∈ E, and suppose that X has aSchauder basis e1, e2, · · · .

Define ||| · ||| : E → [0,∞) by setting

|||u||| = max||y||,∑j≥1

2−j |cj(x)|


for u =∑j≥1 cj(x)ej + y ∈ E. Then ||| · ||| is a norm on E. Below the topology

generated by ||| · ||| will be denoted by τ . Clearly, for u = x+ y ∈ E

||y|| ≤ |||u||| ≤ d1||u||

where (and in the following) di or simply d denotes a generic positive constant.Moreover if um is bounded then

um → u in τ ⇐⇒ xm x and ym → y. (3.1)

Let F j E be closed. A map h : F → E will be called τ -locally finite di-mensional if each point u ∈ E has a τ -neighborhood Wu such that h(Wu ∩ F ) iscontained in a finite-dimensional subspace of E. A map g : F → E is said to be ad-missible if it is τ -continuous and the map h = I−g, where I stands for the identitymap, is τ -locally finite-dimensional. Moreover, a map G : F × [0, 1] → E is saidto be an admissible homotopy if it is τ -continuous (i.e., G(um, tm)→ G(u, t) in τprovided um → u in τ in F and tm → t in [0, 1]), and for each (u, t) ∈ F × [0, 1],there is a neighborhood W(u,t) (in the product topology of (E, τ) and [0, 1]) suchthat the set v − G(v, s); (v, s) ∈ W(u,t) ∩ (F × [0, 1]) is contained in a finite-dimensional subspace of E. The following lemma can be proved along the sameway as [KS, Proposition 2.2].

Lemma 3.1. Let V : O → E be a vector field with O τ- open. Assume thatV is τ-locally τ-Lipschitzian and locally Lipschitzian, and each u ∈ O has a τ-neighborhood Wu which is mapped by V into a finite-dimensional subspace of E.Let F ⊂ O be closed, and assume that the solution of the Cauchy problem

d

dtη = V (η), η(u, 0) = u ∈ F,

η(u, t) exists on [0, 1] for each u ∈ F . Then the map η : F × [0, 1] → E is anadmissible homotopy.

Let Y0 be a finite dimensional subspace of Y and U an open subset of the spaceE0 = X ⊕ Y0. Suppose

(a) g : U → E0 is an admissible map;(b) g−1(0) ∩ ∂U = φ;(c) g−1(0) is τ -compact.

We remark that if (a) holds and U is bounded and convex, then (c) holds. Clearly,g−1(0) ⊂ ∪u∈g−1(0)Wu, where Wu is a τ -neighborhood of u ∈ g−1(0) which ismapped by h = I − g into a finite-dimensional subspace of E0. Thus there arepoints u1, u2, · · · , um ∈ g−1(0) such that g−1(0) ⊂ W := ∪mi=1Wui ∩ U . The setW is open and there is a finite-dimensional subspace L ⊂ E0 such that h(W ) ⊂ L.


Set wL := W ∩ L. Let gL := g|WL : WL → L. It is clear that g−1L (0) = g−1(0);

hence g−1L (0) is compact (in L). Now the degree of Kryszewski-Szulkin is defined

bydeg(g, U, 0) := degB(gL,WL, 0)

where degB stands for the usual Brouwer degree. This degree is well defined andhas the following properties.

Lemma 3.2. (i) If deg(g, U, 0) 6= 0, then g−1(0) 6= φ.(ii) If g(u) = u− u0, where u0 ∈ U , then deg(g, U, 0) = 1.(iii) Suppose that G : U × [0, 1] → E0 is an admisible homotopy such thatG−1(0) ∩ (∂U × [0, 1]) = φ and G−1(0) is τ-compact (in the product topology).Then deg(G(·, t), U, 0) is independent of t ∈ [0, 1].(iv) Suppose that U is a symmetric neighborhood of the origin (i.e., U = −U)and let g : U → E0 be an admissible odd map such that g−1(0) is τ-compact. Iffor each u ∈ U , g(u) ∈ E1, where E1 = X ⊕ Y1 and Y1 is a proper subspace of Y0,then g−1(0) ∩ ∂U 6= ∅.

For the proof, we refer to [KS, Theorem 2.4], and the details are omitted.Now we are in a position to state the generalized linking theorem. In the

following, as usual, for f ∈ C1(E,R), let fa = u ∈ E; f(u) ≥ a, f b = u ∈E; f(u) ≤ b and f ba = fa ∩ f b. f is said to be τ -upper semicontinuous if fa isτ -closed for any a ∈ R. f ′ is said to be τ -weak sequentially continuous in f ba iff ′(um) f ′(u) when um → u in τ in f ba (i.e., in the metric space (f ba, τ)).

Theorem 3.1. Assume that f ∈ C1(E,R) satisfies(f1) f is τ-upper semicontinuous and f ′ is τ-weak sequentially continuous in fd0for any d > 0;(f2) there are b, ρ > 0 such that f |Sρ ∩ Y ≥ b;(f3) there is y0 ∈ S1 ∩ Y and R > ρ such that f |∂M ≤ 0,where M := u = x+ λy0;x ∈ X, ||u|| < R, λ > 0,∂M = u = x+ λy0; x ∈ X, ||u|| = R andλ ≥ 0 or ||u|| ≤ R and λ = 0.Then there is a sequence (um) such that f ′(um)→ 0 and f(um)→ c ∈ [b, supM f ].

Proof. The proof follows the lines of [KS] and is left to the reader.

4. Proof of Theorem 1.1

In this section we assume that the conditions (H1) − (H4) are satisfied andestablish Theorem 1.1.

From now on, we consider the Banach space Eµ defined in Proposition 2.4,Section 2. Clearly Eµ is reflexive and has the direct sum decomposition

Eµ = X ⊕ Y withX := E−µ and Y := E+.


Therefore Eµ is naturally equipped with the τ -topology described in Section 3.Set

ϕ(u) =∫RW (t, u).

By assumptions and Proposition 2.4, ϕ ∈ C1(Eµ,R) and

ϕ′(u) · v =∫RWz(t, u)v ∀u, v ∈ Eµ.

Consider the functional

f(u) =12

(||u+||20 − ||u−||20)− ϕ(u)

for u = u− + u+ ∈ Eµ. Then f ∈ C1(Eµ,R). We are going to apply Theorem 3.1to f .

Lemma 4.1. f satisfies (f1), i.e., f is τ-upper semicontinuous, and f ′ is τ-weaksequentially continuous in fd0 for any d > 0.

Proof. Let c ∈ R. Assume that um ∈ fc with um → u in τ . Then c ≤ 12 ||u+

m||20 −(1

2 ||u−m||20 + ϕ(um)). Since u+m → u+, ||u+

m||0 is bounded. Since ||u−m||20 ≤||u+

m||20 − 2c, ||u−m||0 is also bounded. By (H2) one sees further that |um|µis bounded and so is ||u||µ. Therefore um u which implies um → u in Lµlocand along a subsequence um(t) → u(t) a.e. t ∈ R. Consequently, by the weaklysemicontinuity of norm and the Fatou lemma we get c ≤ f(u). This proves thefirst conclusion.

Now let um → u in τ (in fd0 ). The same argument as above shows that ||um||µis bounded, and so um u in Eµ. Then um → u in Lploc and Wz(t, um)→Wz(t, u)

in Lp/(p−1)loc . Hence f ′(um) · v → f ′(u) · v for each v ∈ Eµ proving the τ -weak

sequentially continuity of f ′.

Lemma 4.2. f satisfies (f2), i.e., there are b, ρ > 0 such that f |Sρ∩E+ ≥ b.

Proof. By (1.2) and (H4) we have

|Wz(t, z)| ≤ d(|z|µ−1 + |z|p−1) andW (t, z) ≤ d

(|z|µ + |z|p

)∀ (t, z). (4.1)

Now by (4.1), for u ∈ E+

f(u) =12||u||20 − ϕ(u) ≥ 1

2||u||20 − d(|u|µµ + |u|pp),

and the desired result follows.

Lemma 4.3. Letting y0 ∈ E+ ∩ S1, f satisfies (f3).


Proof. For u = u− + sy0, by (H2)

f(u) ≤ s2

2− 1

2||u−||20 − a1|u|µµ

≤ s2

2− 1

2||u−||20 − a′1sµ

and the lemma follows then easily.

Now by Theorem 3.1 we obtain the following

Lemma 4.4. There is a sequence un such that

f ′(un)→ 0 and f(un)→ c ∈ [b, b] (4.2)

where b := supM f .

Lemma 4.5. ||un||µ is bounded.

Proof. We have by (H2)

f(un)− 12f ′(un) · un =

∫R

(12Wz(t, un)un −W (t, un)

)≥ (

12− 1γ

)∫RWz(t, un)un (4.3)

≥ γ − 22

∫RW (t, un) ≥ a1(γ − 2)

2γ|un|µµ.

Let Γn = t; |un(t)| ≤ 1 and Γcn = R\Γn. Consider

f ′(un) · u+n = ||u+

n ||20 −∫RWz(t, un)u+

n .

(H4), the Holder inequality, (4.2) and (4.3) imply∫Γn|Wz(t, un)u+

n | ≤ d(∫

Γn|Wz(t, un)|µ/(µ−1)

)(µ−1)/µ||u+

n ||0

≤ d(

1 + ||un||(µ−1)/µµ

)||u+

n ||0

and, by the Holder inequality, (H3) and (4.3)∫Γcn|Wz(t, un)u+

n | ≤ d(∫

ΓcnWz(t, un)un

)1/κ||u+

n ||0

≤ d(

1 + ||un||1/κµ

)||u+

n ||0.


Hence||u+

n ||20 ≤ d(

1 + ||un||(µ−1)/µµ + ||un||1/κµ

)||u+

n ||0 (4.4)

In addition, we have||u−n ||20 ≤ ||u+

n ||20 − 2f(un). (4.5)

Combining (4.3), (4.4) and (4.5) yields the desired conclusion.

We need for the further arguments the following lemma which is a special caseof a more general result due to P. L. Lions [Li].

Lemma 4.6. Let a > 0 and 2 ≤ t <∞. If wn ⊂ H1 is bounded and if

supy∈R

∫B(y,a)

|wn|t → 0, n→∞

where B(y, a) is the interval (y − a, y + a), then wn → 0 in Lτ for 2 < τ <∞.

In the following, let A+ be the part of A in D(A) ∩ H+ = H1 ∩ H+ := H1+,

i.e.,

A+ =∫ ∞α

λdE(λ)

where α > 0 is the constant of (H1). Clearly, A+ : H1+ ⊂ H+ → H+ has a

bounded inverse A−1+ . Since

|A+u|22 =∫ ∞α

λd|E(λ)u|22 ≥ α|u|22 ∀u ∈ H1+

and|u|2 = |Au+ Lu|2 ≤ |A+u|2 + |Lu|2 ∀u ∈ H1

+,

there holds||u||H1 ≤ d1|A+u|2 ∀u ∈ H1

+. (4.6)

Set wn = A−1+ u+

n . Then by (4.6)

||wn||H1 ≤ d1|u+n |2. (4.7)

Lemma 4.7. There exist a sequence yn ⊂ R and a, η > 0 such that

lim infn→∞

∫B(yn,a)

|wn|2 ≥ η.

Proof. By Lemma 4.5 and (4.7), wn is bounded in H1. If the lemma is not true,then by Lemma 4.6,

wn → 0 inLτ for any τ > 2.


Then(u+n , wn)0 = −f ′(un) · wn −

∫RWz(t, un)wn → 0,

that is,|u+n |22 = (u+

n , A+wn)L2 = (u+n , wn)0 → 0.

Therefore, for any τ > 2,

|u+n |τ ≤ |u+

n |1/τ2 |u+

n |1−1/τ2(τ−1)

≤ d||u+n ||

1−1/τ0 |u+

n |1/τ2 → 0,

and then||u+

n ||20 = −f ′n(un) · u+n −

∫RWz(t, un)u+

n → 0.

We get

f(un) ≤ 12||u+

n ||20 → 0,

a contradiction.

Now we choose kn ∈ Z such that |kn − yn| = min|g − yn|; g ∈ Z, and let

zn := kn ∗ un = un(·+ kn) := z−n + z+n .

In view of the invariance of E+ under the action ∗, z+n = kn ∗ u+

n ∈ E+. Notethat A−1

+ commutes with ∗ since A does. Hence wn := kn ∗ wn = A−1+ z+

n and byLemma 4.7

|wn|L2(B(0,a+1)) ≥η

2. (4.8)

We are now in a position to prove Theorem 1.1.

Proof of Theorem 1.1. Remark that ||zn||µ = ||un||µ, and so by Lemma 4.5,||zn||µ is bounded. Hence, by Proposition 2.4, along a subsequence as n→∞,

zn → zweakly inEµ and strongly in Lνloc ∀ ν ≥ 2.

We claim thatz 6= 0. (4.9)

In fact, if not, then z+n 0 in H. So (wn, v)0 = (z+

n , v)L2 → 0 for all v ∈ H1/2,which implies that

wn → 0 weakly inH1/2 and strongly in Lνloc ∀ ν ≥ 2,

contradicting (4.8). Thus (4.9) holds.


Since ||f ′(zn)|| = ||f ′(un)|| → 0 and for any u ∈ C∞0 there holds

f ′(zn) · u = (z+n − z−n , u)0 −

∫RWz(t, zn)u,

we see as n→∞ that

(z+ − z−, u)0 =∫RWz(t, z)u.

Thus z is a nontrivial weak solution of (HS). By (4.1) and Proposition 2.4,Wz(t, z) ∈Lν for all ν ≥ 2. Hence, a standard regularity theory of solutions of ordinary differ-ential equations shows that z ∈ H1

loc∩Lν for all ν ≥ µ. By (HS) and (H3)− (H4),

|z|2 ≤ d1(|z|2 + |z|ν

)where ν = 2(p− 1), and so for all τ ∈ R

∫ τ+1/2

τ−1/2|z|2 ≤ d1

(∫ τ+1/2

τ−1/2|z|ν)2/ν

+∫ τ+1/2

τ−1/2|z|ν. (4.10)

Integrating from τ − 1/2 to τ + 1/2 the following equality

z(τ) = z(t) +∫ τ

t

z(s)ds

yields, by Holder inequality,

|z(τ)| ≤(∫ τ+1/2

τ−1/2|z|ν)1/ν

+(∫ τ+1/2

τ−1/2|z|2)1/2

. (4.11)

Since z ∈ Lν , (4.10) and (4.11) show that

|z(τ)| → 0 as |τ | → ∞,

that is, z is a homoclinic orbit of (HS).The proof is hereby completed.

Corollary 4.1. Let H(t, z) be of the form (1.1). Assume that L(t) is right di-chotomic and W (t, z) satisfies (H2) − (H3) − (H4). Then (HS) has at least onehomoclinic orbit.

Proof. It follows from Proposition 2.5 and Theorem 1.1.


Acknowledgements

The first named author was supported by the Alexander von Humboldt-Stiftungof Germany. The research was done during a visit of the first named author atthe Department of Mathematics of Louvain University, Belgium. He thanks themembers of the Department for their kind invitation and hospitality.

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Yanheng DingInst. of Math., Academia Sinica100080 Beijing, P.R. China

Willem MichelDep. Math. U.C.L.2 ch. du CyclotronB1348 Louvain-la-Neuve, Belgium

(Received: June 20, 1997)

homoclinic orbits of a hamiltonian system

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