Infinitely many homoclinic orbits of a Hamiltonian system with symmetry

Nonlinear Analysis 38 (1999) 391 – 415

In nitely many homoclinic orbits of a Hamiltonian system with symmetry Yanheng Ding a;∗ , Mario Girardi b a

Institute of Mathematics, Academia Sinica, 100080 Beijing, People’s Republic of China b Dip. di Mat., Univ. degli Studi Roma Tre, 00146 Rome, Italy Received 29 December 1997; accepted 14 March 1998

Keywords: Hamiltonian system; Symmetry; Homoclinic orbits

1. Introduction We are interested in the multiplicity of homoclinic orbits of the Hamiltonian system: z˙ = JHz (t; z);


0 −I

(HS)

where z = (p; q) ∈ RN ×RN = R2N , J = I 0 is the standard symplectic structure on R2N , H ∈ C(R×R2N ; R) is 1-periodic in t ∈ R and H (t; z) = 12 zL(t)z + W (t; z)

(1.1)

2

with L ∈ C(R; R4N ) being a 2N ×2N symmetric matrix valued function, and W ∈ C 1 (R×R2N ; R) being globally superquadratic in z ∈ R2N . Here by a homoclinic orbit of (HS) we mean a solution of the equation satisfying z(t) 6≡ 0

and

z(t) → 0 as |t| → ∞:

There have been many papers devoted to the study on the existence and multiplicity of homoclinic orbits for Hamiltonian systems via critical point theory, see, e.g., [1, 4– 6, 13–15] for the second-order systems, and [3, 7, 9, 17, 18], for the rst-order systems. We note that in all these results on the rst-order system (HS) the following condition is always required:  ∩ iR = ∅, (∗) L(t) ≡ L is independent of t such that sp(JL) ∗

Corresponding author. E-mail:[email protected]

0362-546X/99/$ - see front matter ? 1999 Elsevier Science Ltd. All rights reserved. PII: S 0 3 6 2 - 5 4 6 X ( 9 8 ) 0 0 2 0 4 - 1

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Y. Ding, M. Girardi / Nonlinear Analysis 38 (1999) 391 – 415

 denotes the set of all eigenvalues of JL.  Let A =−(J(d=dt)+ L(t)) be where sp(JL) 2 2N the self-adjoint operator acting in L (R; R ) with the domain D(A)= H 1 (R; R2N ). Then, letting (A) denote the spectrum of A, the assumption (∗) implies that (∗)0 L(t) ≡ L is independent of t and there is ¿0 such that (−; ) ∩ (A) = ∅. Consequently, the operator A : W 1; p (R; R2N ) → Lp (R; R2N ) is a homeomorphism for all p¿1, which is important for the variational arguments in the papers. Very recently, the paper [8] has considered relaxing condition (∗) and obtained the existence of one homoclinic orbit under the following condition: (H1) L(t) depends periodically on t with period 1, and there is ¿0 such that (0; ) ∩ (A) = ∅; and some superquadraic assumptions on W (t; z). Note that, in the case (H1), 0 may be a (continuous) spectrum of A. Let us consider the example   I I ; c¡0: L1 = c I I We will show in the next section that L1 satis es (H1). However, clearly, sp(JL1 ) = {0}, i.e., L1 does not satisfy (∗). Another example is the matrix L2 = diag(1 ; : : : ; 2N ) with 1 ≤ · · · ≤ 2N . Note that,  = ±(−i i+N )1=2 , i = 1; : : : ; N , are all the eigenvalues of JL2 . L2 satis es (∗) if and only if 1 ≤ · · · ≤ N ¡0¡N +1 ≤ · · · ≤ 2N . On the other hand, L2 satis es (H1) if 1 ≤ · · · ≤ N ¡0 ≤ N +1 ≤ · · · · · · 2N (particularly, in (H1) we allow N +1 = · · · = 2N = 0, see the next section). The main purpose of the present paper is to establish the existence of in nitely many homoclinic orbits of (HS) provided the conditions of [8] and an additional evenness assumption on W (t; z). Recall that, based on the periodic condition, if z is a homoclinic orbit then for any  ∈ Z,  ∗ z := z(· + ) is also a homoclinic orbit. Letting O(z) = { ∗ z;  ∈ Z}, two homoclinic orbits z1 and z2 of (HS) are said to be geometrically distinct if O(z1 ) 6= O(z2 ). We shall establish the following result. Theorem 1.1. Let H (t; z) be of the form (1.1) such that (H1) holds. Suppose moreover that W (t; z) satisÿes: (H2) W (t; z) ∈ C 1 (R×R2N ; R) is 1-periodic in t; and there are a1 ¿0;  ≥ ¿2 such that for all (t; z); a1 |z| ≤ W (t; z) ≤ Wz (t; z)z;

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(H3) there are  ∈ (1; 2) and a2 ¿0 such that for all (t; z) with |z| ≥ 1 |Wz (t; z)| ≤ a2 Wz (t; z)z; (H4) there is a3 ¿0 and ¿0 such that for all (t; z) and v ∈ R2N with |v| ≤  |Wz (t; z + v) − Wz (t; z)| ≤ a3 |v|(|v|−2 + |z|−2 + |z|1=(−1) ); (H5) W (t; z) = W (t; −z) for all (t; z). Then (HS) has inÿnitely many geometrically distinct homoclinic orbits. We note that (H2) and (H3) imply, letting p := 1 + 1=( − 1), ≤p

and

|Wz (t; z)| ≤ a02 |z|p−1 whenever |z| ≥ 1

(1.2)

and, letting z = −v, (H4) implies |Wz (t; z)| ≤ a03 |z|−1

whenever |z| ≤ 1:

(1.3)

A typical example satisfying (H2)–(H5) is the following: W (t; z) = a(t)(|z| + |z|p ); where 2¡ ≤ p¡∞ and 0¡a(t) ∈ C(R; R) is 1-periodic. Remark 1.1. If there exists ¿0 such that (−; 0) ∩ (A) = ∅ and W (t; z) := −W (t; z) satisÿes the assumptions (H2)–(H5), then the same conclusion of Theorem 1.1 remains valid. Remark 1.2. In the paper [17], SÃerÃe obtained inÿnitely many geometrically distinct homoclinic orbits for (HS) under the assumptions (∗) and that W is C 2 ; W (t; ·) is strictly convex, a1 |z| ≤ W (t; z) ≤ a2 |z| and Wz (t; z)z ≥ W (t; z); but without evenness condition on W (t; ·). 2. Some preliminary results In this section we refer mainly to the paper [8] and, for the reader’s convenience, some of the results, together with the proofs, will be provided here. 2 Let M (t) ∈ C(R; R4N ) be a symmetric matrix valued function, 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 positive constants K, such that |F(t)PF −1 (s)| ≤ Ke− (t−s) |F(t)(I − P)F −1 (s)| ≤ Ke− (s−t) (see [2]).

if s ≤ t; if s ≥ t

(2.1)

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Below, let | · | denote the L = L (R; R2N ) norm for  ≥ 1, and (·; ·)L2 the inner product of H := L2 . Set W 1;  := W 1;  (R; R2N ) for  ≥ 1, H 1 := W 1; 2 and H 1=2 := H 1=2 (R; R2N ). For a self-adjoint operator S in H, we denote by |S| and |S|1=2 its absolute value and square root. Proposition 2.1. Suppose that M (t) has an exponential dichotomy and let  ≥ 1. Then the following conclusions hold: (i) The operator B : L ⊃ W 1;  → L , z 7→ −(J(d=dt) + M )z has a bounded inverse B−1 such that for all  ≥  there is b = b(; )¿0 satisfying |B−1 u| ≤ b|u|

∀u ∈ L ;

(ii) B := B2 is self-adjoint, and there are a¿0; b1 ¿0; b2 ¿0 such that (B) ∩ [−a; a] = ∅

and

b1 kukH 1 ≤ |Bu|2 ≤ b2 kukH 1 ∀u ∈ H 1 ;

(iii) D(|B|1=2 ) = H 1=2 ; and there are b01 ; b02 ¿0 such that b01 kukH 1=2 ≤ kB|1=2 u|2 ≤ b02 kukH 1=2

∀u ∈ H 1=2 :

Proof. For any u ∈ L ,  ≥ 1, there is a unique z ∈ W 1;  satisfying   d − J +M z=u dt given by Z z(t) =

t

−∞

Z F(t)PF −1 (s)Ju ds −

Set

t

( +

−

 (s) =  (−s) = Then

Z

z(t) =

R

∞

1

if s ≥ 0;

0

if s¡0:

F(t)(I − P)F −1 (s)Ju ds:

Z F(t)PF −1 (s)+ (t − s)Ju − F(t)(I − P)F −1 (s)− (t − s)Ju R

:= z1 (t) + z2 (t); and by Eq. (2.1) Z |z1 (t)| ≤ K e− (t−s) + (t − s)|u| ds R

Z |z2 (t)| ≤ K e− (s−t) − (t − s)|u| ds: R

and

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Setting g+ () = e−  + () and g− () = e  − (), one then has |z1 (t)| ≤ K(g+ ∗ |u|)(t)

and

|z2 (t)| ≤ K(g− ∗ |u|)(t);

where ∗ denotes the convolution. Observe the following: Z Z 1 +  |g | = |g− | = ∀ ≥ 1 and |g± |∞ = 1:  R R By the convolution inequality, for any  ≥ 1 satisfying 1= = 1= + 1= − 1, |zj | ≤ K( )−1= |u| ;

j = 1; 2

and for 1= + 1=0 = 1; ¿1 0

|zj |∞ ≤ K( 0 )−1= |u| ;

j = 1; 2

and also |zj |∞ ≤ K|u|1

if  = 1; j = 1; 2:

Therefore, |z| ≤ K( )−1= |u| ;

; ;  ≥ 1 and

1 1 1 = + − 1:   

(2.2)

Now conclusion (i) follows from Eq. (2.2). It is easy to verify that B = B2 is self-adjoint. Clearly, 0 ∈= (B) by (i). Moreover, the inequality of (ii) follows again from Eq. (2.2). We now verify (iii). Let  := −d2 =dt 2 . Then D() = H 2 and by an interpolation theory (see [19, Section 2.5.2]) (D(0 ); D()); 2 = (L2 ; H 2 ); 2 = H 2 ;

0 ≤  ≤ 1:

On the other hand (see [19, Theorem 1.18.10]) (D(0 ); D()); 2 = D( ) and, consequently, H 2 = D( ) equipped with the norm Z ∞ 2 (1 + 2 ) d|F u|22 = |u|22 + | u|22 ; kukD( ) = 0

where {F ; −∞¡¡∞} is the spectral family of . In particular, one sees with  = 14 , H 1=2 = D(1=4 );

2

kukH 1=2 ≤ |u|22 + |1=4 u|22 :

˙ 2 ≤ d1 |Bu|2 for u ∈ H 1 by (ii), one has (1=2 u; u)L2 ≤ d2 (|B|u; u)L2 Since |1=2 u|2 = |u| (see, [10, Theorem 4.12]), and so |1=4 u|2 ≤ d2 ||B|1=2 u|2 which, together with Eq. (2.2),

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implies the rst inequality of (iii). In a similar way and considering the operator ˜ := − d2 =dt 2 + 1; one can check the second inequality of (iii).  Proposition 2.2. Let M (t) ∈ C(R; R2N ) be a 1-periodic symmetric matrix valued function; and let B = − (J(d=dt) + M (t)) : L2 ⊃ H 1 → L2 be the self-adjoint operator. Then the spectrum (B) consists of continuous spectra. Proof. Note that, since B is selfadjoint, it has no residual spectrum and the isolated points of (B) are eigenvalues. Hence the proof is completed if we show that B has no eigenvalue. Arguing indirectly, assume that B has an eigenvalue . Let u ∈ H 1 be an eigenfunction associated with , i.e., u satis es the equation du = J(M (t) + )u: dt

(2.3)

Let F(t) be the fundamental matrix of Eq. (2.3) with F(0) = I: By the Floquet theory, the monodromy operator F(1) = Q(t)et where = ln F(1) and Q(t) is a 1-periodic continuous dierentiable matrix valued function having a bounded inverse Q−1 (t) (see, e.g. [16]). Set y(t) = Q−1 (t)u(t): Then y(t) → 0 as t → ±∞ (since u ∈ H 1 ) and y satis es the reduction equation dy = y(t); dt correand so y(t) = et y(0): Let M − ; M 0 and M + be the invariant subspaces of sponding to its eigenvalues with negative, 0 and positive real parts, respectively. Then R2N = M − ⊕ M 0 ⊕ M + . Let P ∗ ; ∗ ∈ {−; 0; +}; be the associated projections, and set y∗ = P ∗ y: Then one has y∗ (t) = et y∗ (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) + aJ;˜ where a ¿ 0; L(t) satis es (H1) and
J˜ = 01 10 . Clearly aJJ˜ has the eigenvalues 1 = · · · =  N = a and N +1 = · · · =
−1 0 2N = −a; and its fundamental matrix is Fa = exp at 0 1 : Therefore aJ˜ has an exponential dichotomy. By the roughness of the exponential dichotomy, for any a ¿ 4 sup |L(t)|; t∈R

Ma also has an exponential dichotomy (see [2]). We x arbitrarily such an a. Consider the self-adjoint operator   d Aa = − J + Ma = A − aJ:˜ dt

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˜ 2 ≤ |Au|2 + a|u|2 for u ∈ D(A) we have, by Proposition 2.1, Since |Aa u|2 = |(A − aJ)u| b1 kuk2H 1=2 ≤ (|Aa |u; u)L2 ≤ (|A|u; u)L2 + a|u|22 ≤ b2 kukH 1=2 + a|u|22 : Therefore by continuity b1 kukH 1=2 ≤ ||A|1=2 u|2 + a|u|2 ≤ b2 kukH 1=2 + 2a|u|2

(2.4)

for all u ∈ H 1=2 (= D(|A|1=2 )); where (and below) the symbols bi stand for generic positive constants. Let {E();  ∈ R} be the spectral family of A. We have A = U |A|; called the polar decomposition, where U = I − E(0) − E(− 0): Noting that, by (H1) and Proposition 2.2, 0 is at most a continuous spectrum of A, the Hilbert space H has an orthogonal decomposition H = H− ⊕ H+ ; where H± = {u ∈ H; Uu = ± u}: We will write in the following u = u− + u+ for u ∈ H; where u± ∈ H± : Let E be the space of the completion of D(|A|1=2 ) under the norm kuk0 = ||A|1=2 u|2 : E becomes a Hilbert space under the inner product (u; v)0 = (|A|1=2 u; |A|1=2 v)L2 : By Eq. (2.4) we have for all u ∈ D(|A|1=2 ); b1 kukH 1=2 ≤ kuk0 + a|u|2 ≤ b2 kukH 1=2 + 2a|u|2 :

(2.5)

E has an orthogonal decomposition E = E− ⊕ E+; where E ± ⊇ H± ∩ D(|A|1=2 ):

(2.6)

By assumption (H1) for u ∈ H+ ∩ D(A) Z ∞  d(E()u; u)L2 ≥ |u|22 kuk20 = (Au; u)L2 = 

which, together with Eqs. (2.5) and (2.6), shows E + = H+ ∩ D(|A|1=2 )

and

k · k0 ∼ k · kH 1=2 on E + ;

where “∼” means “equivalent to”.

(2.7)

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Proposition 2.3. Suppose (H1) is satisÿed; and let E denote the completion of the set D(A) under the norm kuk = (||A|1=2 u|22 + |u|2 )1=2 :

(2.8)

Then E has the direct sum decomposition E = E− ⊕ E + ;

(2.9)

where E− denotes the completion of D(A) ∩ H− with respect to the norm k · k ; and E is embedded continuously in L for any  ∈ [; ∞) and compactly in Lloc for any  ∈ [2; ∞): Proof. Since E ⊂ E; it is easy to see that we have Eq. (2.9). Taking into account Eq. (2.7), we only need to show that E− possesses the desired embedding property. For any  ¿ 0; set H− = E(−)H and E− = H− ∩ D(|A|1=2 ) = H− ∩ E − . One has similarly to Eq. (2.7) on E− :

k · k0 ∼ k · kH 1=2

(2.10)

− − Let H := H− H− and let E;− be the completion of H under norm (2.8). Clearly,

E− = E− ⊕ E;− : 1 and is embedded compactly in Hence it suces by Eq. (2.10) to show that E;− ⊂ Hloc ∞  Lloc ; and continuously in L for all  ≥ : − − By the spectral theory of self-adjoint operators, H ⊂ D(A) = H 1 : Let (un ) ⊂ H be a Cauchy sequence with respect to k · k : Then Z 0 2 d|E()(un − um )|22 |A(un − um )|22 = −

Z ≤ −

0

−

 d|E()(un − um )|22

= ||A|1=2 (un − um )|22 → 0 as n; m → ∞. For any nite interval ⊂ R; one has Z |un − um |2 ≤ | |1−2= |un − um |2 → 0

which, together with Eq. (2.11), shows Z Z |u˙n − u˙m |2 = |A(un − um ) + L(t)(un − um )|2



≤ 2|A(un −

um )|22

Z +2



|L(t)(un − um )|2 → 0

(2.11)

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as n; m → ∞. Therefore, the limit u of (un ) with respect to k · k belongs to H 1 ( ). Moreover, since H 1 ( ) is compactly embedded in L∞ ( ) for any nite interval , one sees that E;− is compactly embedded in L∞ loc . By Eq. (2.11), (Aun ) is a Cauchy sequence in L2 . Hence Aun → w in L2 . Since Aun → Au in L2loc ; w = Au; i.e, Au ∈ L2 . Note that for any nite interval

Z Z Z |u| ˙ 2 = |Au + Lu|2 ≤ 2 (|Au|2 + |Lu|2 )





Z ≤ d1



2

1−2=

Z

|Au| + | |





Letting  ∈ R and integrating from  − Z  u(s) ˙ ds u() = u(t) +

1 2

2= !

|u|

to  +

: 1 2

(2.12)

the following equality:

t

yields, by Holder’s inequality, !1= Z |u()| ≤

r+1=2

r−1=2



|u|

Z +

r+1=2

r−1=2

!1=2 2

|u| ˙

:

(2.13)

Since u ∈ L and Au ∈ L2 ; Eqs. (2.12) and (2.13) show that |u()| → 0

as || → ∞;

that is, u ∈ L∞ . Therefore u ∈ L ∩ L∞ and so u ∈ L for any  ≥ . Replacing u by un − u in Eqs. (2.12) and (2.13) one sees that E;− is continuously embedded in L∞ and so is in L for any  ≥ . Next, we consider some examples of matrices satisfying (H1). We recall the following notion introduced in [8]. Deÿnition 2.1. A continuous symmetric matrix valued function L(t) is said to be right (resp. left) dichotomic if there is  ¿ 0 such that L (t) := L(t) +  has an exponential dichotomy for each  ∈ (0;  ] (resp.  ∈ [−; 0)): Proposition 2.4. If L(t) is right dichotomic; then it satisÿes (H1). Proof. Note that x˙ = JL(t)x ⇔ A x + x = 0; where

  d A := − J + L (t) = A − : dt

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By Proposition 2.1, for any  ∈ (0;  ], there are a ¡ 0 ¡ b ; both a and b being in (A ), such that (a ; b ) ⊂ (A ) := C\(A ): Let  := min{; b}. Then since  ∈ (A)  ⊂ (A). The desired conclusion follows. if and only if 0 ∈ (A ); we see that (0; ) Remark 2.1. In the same way as above one can check that if L(t) is left dichotomic then there is  ¿ 0 such that (−; 0) ⊂ (A): Example 1. Let  I L1 (t) := c I

I I

 ;

c 6= 0:

Then sp(JL1 ) = {0}. Note that   c+ c L1 := L1 +  = c c+ and sp(JL1 ) = {±1 = · · · = ±N = ±(−2c)1=2 }: Therefore, L1 is right (left) dichotomic if c ¡ 0 (c ¿ 0): Example 2. Let L(t) be 1-periodic continuous symmetric matrix valued function and let Z 1 L(t) dt L2 := 0

be its mean value. Set L2 := L2 +  and let j () = j () + i j ();

j = 1; : : : ; 2N

denote the eigenvalues of JL2 . Assume that there is  ¿ 0 such that 1 () ≤ · · · ≤ N () ¡ 0 ¡ N +1 () ≤ · · · ≤ 2N ()

∀  ∈ (0;  ] ( ∈ [−; 0)):

Then L(t) is right (left) dichotomic. In particular, if L2 = diag (1 ; : : : ; 2N ) with 1 ≤ · · · ≤ N ¡ 0 ≤ N +1 ≤ · · · ≤ 2N ; then L(t) is right dichotomic. 3. Critical values of a symmetric functional To establish our main result we use variational arguments. The homoclinic orbits of (HS) will be obtained as critical points of a functional de ned on a Banach space. So we discuss here such kind of functionals. Our consideration was motivated by [11]. Throughout the section, let E denote a re exive Banach space with the direct sum decomposition E = X ⊕ Y; u = x + y for u ∈ E; and suppose that dim Y = ∞ and X has

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a Schauder basis {e1 ; e2 ; : : :}. Let PX and PY denote the projections onto X and Y , respectively. In the following, as usual, for f ∈ C 1 (E; R), let fa = {u ∈ E; f(u) ≥ a}; fb = {u ∈ E; f(u) ≤ b}; fab = fa ∩ fb ; K = {u ∈ E; f0 (u) = 0}; Ka = K ∩ fa and Kab = K ∩ fab . f0 is said to be weakly sequentially continuous if f0 (um ) * f0 (u) when um * u. For c ∈ R, by a (PS)c sequence we mean a sequence (um ) ⊂ E satisfying f(um ) → c and f0 (um ) → 0. Deÿnition 3.1. Given ¿0, a set F ⊂ E will be called a (PS) neighborhood absorber upto if, for any neighborhood O := {u ∈ E; ||u − F||¡}, any (PS)c sequence with c ≤ has a subsequence lying in O . A set F is said to be isolated if inf {ku − vk; u; v ∈ F; u 6= v}¿0. It will be seen that the functional f associated to (HS) satis es the following conditions (with E = E ; X = E− and Y = E + ): (f1) for any (um ) ⊂ fa , if (PY um ) is bounded then (um ) is bounded, if um * u and PY um → PY u then u ∈ fa ; (f 2) f0 is weakly sequentially continuous; (f 3) for each nite-dimensional subspace Y0 ⊂ Y; f(u) → −∞ as u ∈ X ⊕ Y0 and ||u|| → ∞; (f 4) there exist b; ¿0 such that f|S ∩Y ≥ b, where S = @B(0; ). These conditions are almost sucient to ensure the existence of an unbounded critical value sequence of even functionals satisfying the usual (PS) condition. Unfortunately, the functional we will encounter does not satisfy the (PS). However, as a substitution, the following assumption works. (f 5) for each ¿0 there is a symmetric (PS) neighborhood absorber F upto such that PY F is isolated. Theorem 3.1. Let f ∈ C 1 (E; R) be even with f(0) = 0. Assume f statisÿes (f 1)– (f 5). Then f possesses a sequence (cn ) of critical values satisfying cn → ∞ as n → ∞. In fact, we shall show a slight more general result. For this purpose, some notations are in order. P Recall that x = j≥1 cj (x)ej for all x ∈ X . De ne     X 2−j |cj (x)| |||u||| = max ||y||;   j≥1

P

for u = j≥1 cj (x)ej +y ∈ E. Then |||·||| is a norm on E. Below the topology generated by ||| · ||| will be denoted by . Clearly, for u = x + y ∈ E ||y|| ≤ |||u||| ≤ d||u||; where d is a positive constant. Moreover, if (um ) is bounded then 

um → u ⇔ xm * x

and ym → y:

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f is said to be -upper semicontinuous if fa is -closed for any a ∈ R: f0 is said to be  -weak sequentially continuous in fab if f0 (um ) * f0 (u) when um → u (in the metric b space (fa ; )). Also, a set F ⊂ E will be called a (PS) -neighborhood absorber up to if, for any -neighborhood N := {u ∈ E; |||u − F|||¡}, any (PS)c sequence with c ≤ has a subsequence lying in N . Clearly K ∩ f ⊂ F. We remark that assumptions (f1) and (f 2) imply that (f 1)0 f is -upper semicontinous, and f0 is -weak sequentially continuous in f0d for any d¿0. Also clearly (f1) and (f5) imply that (f 5)0 any (PS)c sequence (un ) with (PY un ) being bounded is bounded, and for any ¿0 there is a symmetric (PS) -neighborhood absorber F up to such that PY F is isolated. Therefore, Theorem 3.1 is a consequence of the following one. Theorem 3.2. Let f ∈ C 1 (E; R) be even with f(0) = 0. Assume that (f 1)0 ; (f 3), (f 4) and (f 5)0 are satisÿed. Then f has a sequence (cn ) of critical values satisfying cn → ∞ as n → ∞. The remainder of this section is devoted to the proof of Theorem 3.2. First we need other notions referring to the paper [11]. Let F ⊂ E be closed. A map h : F → E is called -locally nite-dimensional if each point u ∈ F has a -neighborhood Wu such that h(Wu ∩ F) is contained in a nitedimensional subspace of E. A map g : F → E is said to be admissible if it is -continuous and the map h = id − g is -locally nite dimensional. Moreover, a map G : F × [0; 1]→ E is said to be an admissible homotopy if it is -continuous (i.e.,   G(um ; tm ) → G(u; t) provided um → u in F and tm → t in [0,1]), and for each (u; t) ∈ F × [0; 1], there is a neighborhood W(u; t) (in (E; ) × [0,1]) such that the set {v − G(v; s); (v; s) ∈ W(u; t) ∩ (F × [0; 1])} is contained in a nite-dimensional subspace of E. The following lemma can be proved in the same way of Kryszewski and Szulkin [11, Proposition 2.2]. Lemma 3.1. Let O ⊂ E be a -open set and V : O → E a vector ÿeld. Assume that V is -locally -Lipschitzian and locally Lipschitzian; and each u ∈ O has a -neighborhood Wu such that V (Wu ) is contained in a ÿnite-dimensional subspace of E. Let F ⊂ O be closed; and assume that the solution of the Cauchy problem d = V (); dt

(u; 0) = u ∈ F;

(u; t) exists on F × [0; T ] for some T ¿0. Then the map  : F × [0; 1] → E; (u; t) := (u; Tt) is an admissible homotopy. Let Y0 be a nite-dimensional subspace of Y and U an open subset of the space E0 = X ⊕ Y0 . Let G(U ) denote the set of all maps g : U → E0 satisfying (a) g is admissible; (b) g−1 (0) ∩ @U = ∅; (c) g−1 (0) is -compact.

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For each g ∈ G(U ), since g−1 (0) is -compact, there is a -open set N ⊂ E such that g− (0) ⊂ W := N ∩ U and (id−g)(W ) is contained in a nite-dimensional subspace L of E0 . Set WL := W ∩ L. Let gL := g|WL : WL → L. It is clear that gL−1 (0) = g−1 (0); hence, gL−1 (0) is compact (in L). Now a degree of the type of Kryszewski–Szulkin is de ned by deg(g; U; 0) := degB (gL ; WL ; 0); where degB stands for the usual Brouwer degree. This degree is well de ned and it is not dicult to check the following lemma using the properties of Brouwer degree (cf. [11] or [20]). Lemma 3.2. The degree possesses the following properties: (i) If deg(g; U; 0) 6= 0; then g−1 (0) 6= ∅. (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 admissible homotopy such that G −1 (0) ∩ (@U × [0; 1]) = ∅ and G −1 (0) is -compact (in the product topology). Then deg(G(·; t); U; 0) is independent of t ∈ [0; 1]. (iv) Suppose that U is a symmetric (i.e. U = −U ) neighborhood of the origin and let g : U → E0 be an admissible odd map such that g−1 (0) is -compact. If for each u ∈ U ; g(u) ∈ E1 ; where E1 = X ⊕ Y1 and Y1 is a proper subspace of Y0 ; then g−1 (0) ∩ @U 6= ∅. Now we turn to consider the functional f ∈ C 1 (E; R) given in Theorem 3.2. Let  := {B ⊂ E; B is closed and B = −B}; and

(B) := min{k ∈ N; there is an odd continuous g : B → Rk \{0}} the genus of B ∈ . Moreover, for each B ∈ , set H(B) := {g : B → E; g satis es (g1) − (g3)}; where (g1) g(B) is closed and g is a homeomorphism of B onto g(B); (g2) g is an odd admissible map; (g3) for any u ∈ B; f(g(u)) ≤ f(u). Note that if B1 ⊂ B2 then H(B2 ) ⊂ H(B1 ) via the restriction. We will use the following pseudoindex for B ∈ :

∗ (B) := min (g(B) ∩ S ∩ Y ); g∈H(B)

where S is the sphere of (f4). In addition, letting B ∈  be xed and setting B := {B ∈ ; B ⊂ B}; we will also use the related pseudoindex for B ∈ B

∗B (B) := min (g(B) ∩ S ∩ Y ): g∈H(B)

The following properties follow directly from the de nitions (cf. [11]).

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Lemma 3.3. Let B; C ∈ . (i) If ∗ (B) 6= 0; then B 6= ∅; (ii) If B ⊂ C then ∗ (B) ≤ ∗ (C); (iii) If g ∈ H(B); then ∗ (g(B)) ≥ ∗ (B). Lemma 3.4. Let B; C ∈ B . (i) ∗B (B) ≥ ∗ (B); (ii) If B ⊂ C; then ∗B (B) ≤ ∗B (C); (iii) If g ∈ H(B) and g(B) ⊂ B; then ∗B (g(B)) ≥ ∗B (B); (iv) ∗B (B ∪ C) ≤ ∗B (B) + (C). Note that, by (f3) and (f4), for any k-dimensional subspace Yk of Y , letting Ek := X ⊕ Yk , there is Rk ¿ such that sup

Ek \B(0; Rk )

f¡ inf f: B(0; )

Set Bk := B(0; Rk ) ∩ Ek = {u ∈ Ek ; ||u|| ≤ Rk }. The following result is a generalized version of [11, Lemma 4.8] for our setting. It can easily be checked by Lemma 3.2(iv) and so its proof is omitted. Lemma 3.5. ∗ (Bk ) ≥ k. The proof of Theorem 3.2. will be proceeded indirectly, i.e., we will have a contradiction if K ⊂ fc for some c ∈ R. The following lemma will play an important role. Lemma 3.6. Let 0¡c¡ − 2. (a) If Kc = ∅ then there is a map g ∈ H(f ) such that g(f ) ⊂ fc . (b) If Kc 6= ∅ then there exist ¿0; a symmetric -open set N with (N) ≤ 1 and a g ∈ H(f ) such that g(fa+ \N) ⊂ fa− for any  ∈ (0; ) and a ∈ [c + ; − 1]. Proof. We adopt an argument similar to [11, Lemma 4.6]. Note that there is r¿0 such that |||u||| ≥ r for all u ∈ fc (otherwise there were un ∈ fc with |||un ||| → 0, then 0 = f(0) ≥ c by (f1)0 , a contradiction). Recall that there is a (PS) -neighborhood absorber F up to the level + 1 such that FY := PY F is isolated. If FY 6= {0}, we also assume that 0¡r¡ inf {||y − y0 ||; y; y0 ∈ FY ; y 6= y0 }. Set, for y ∈ FY and s¿0, Ay (s) := X ⊕ BY (y; s), where BY (y; s) = B(y; s) ∩ Y , and  r  r  [ [ and V := X ⊕ : BY y; BY y; N0 := X ⊕ 8 16 y∈F y∈F Y

Y

Then V ⊂ N0 are -neighborhoods of F, and there is ¿0 such that ||f0 (u)|| ≥  for all u ∈ f +1 \V. 1 0  For any u ∈ fc +1 \K, there is w(u) ∈ E, ||w|| = 1, such that f0 (u)w(u)¿ 2 ||f (u)||. +1 0 0 0 Set w(u) := 2w(u)\||f  (u)||. By (f 1) , the function f (·)w(u) : fc → R is -conti-

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nuous. Therefore u has a -open neighborhood Uu ⊂ E such that f0 (v)w(u)¿1; v ∈ fc +1 ∩ Uu ; r |||v − u|||¡ ; v ∈ fc +1 ∩ Uu : 16

(3.1) (3.2)

We may also ask Uu ⊂ Ay if moreover u ∈ Ay . Set U0 := f−1 (−∞; c). Take a -locally -open re nement (Oj )j∈J of the -open covering (Uu )u∈f +1 \K ∪ {U0 } of the (metric) c space f +1 \Kc , and a -locally -Lipschitzian partition of unity (j )j∈J subordinated to (Oj ). Set wj := w(uj ) if Oj ⊂ Uuj for some uj ∈ fc +1 and wj := 0 if Oj ⊂ U0 . Let O S be a symmetric -open set such that f +1 \Kc ⊂ O ⊂ j∈J Oj and de ne X j (u)wj and V (u) := 12 (V (u) − V (−u)) V (u) := j∈J

for u ∈ O. Then V is odd, -locally -Lipschitzian (and thus locally Lipschitzian); f0 (u)V (u) ≥ 0 for all u ∈ O and f0 (u)V (u)¿1 when u ∈ fc +1 \K by Eq. (3.1) and the oddness of f0 ; each u ∈ f +1 \Kc has a -neighborhood Wu on which V is -Lipschitzian and such that V (Wu ) is contained in a nite-dimensional subspace of E. Next, if Kc 6= ∅, let : E → [0; 1] be an even -locally -Lipschitzian function such that (u) = 0 if |||u − Kc ||| ≤ r=10 and (u) = 1 if |||u − Kc ||| ≥ r=8. If Kc = ∅, let ≡ 1. Let t (u) = (u; t) denote the ow of the Cauchy problem d = − ()V (); dt

(u; 0) = u ∈ f +1

with the right maximal existing interval [0, !+ ). Plainly, !+ (u) = !+ (−u) and (−u; t) = (u; t) since V is odd. We rst check that !+ (u) = ∞ for any u ∈ f +1 . Assume by contradiction that tn !+ (u)¡∞ for some u ∈ f +1 P . Then there is tn % !+ (u) with vn =  (u) such that 1 (vn )¿0 and kV (vn )k = 2 k j∈J (j (vn ) − j (−vn ))wj k → ∞. Hence, without loss of generality, we can assume for each n there is a jn ∈ J such that jn (vn )¿0; vn ∈ Uun (un = ujn ) with k|vn − un k| ¡ r=16 (see Eq (3.2)), and kwjn k = 2=kf0 (un )k → ∞. Thus (un ) is a (PS)a -sequence (a ∈ [c; + 1]) and so we can assume un ∈ V. Now if un ∈ Ay (r=16) with some y ∈ FY xed for almost all n then one sees in view of (f 1)0  and (f 5)0 that un −→ u = x +y ∈ Kc and, consequently, k|vn −Kc k| ¡ r=8 which yields (vn ) = 0, a contradiction. So (un ), resp. (vn ), enters in nitely many sets of the form Ay (r=16), resp. of the form Ay (r=8). This and (f 5)0 imply that there must be a sequence of intervals (tn1 ; tn2 ) with tn1 → !+ (u) such that t (u) leaves Ayn1 (r=8) at tn1 , enters Ayn2 (r=8) 2

at tn2 and t (u) ∈= V for all t ∈ (tn1 ; tn2 ). Clearly, kt1 (u)−tn k ≥ 3r=4, and by Eq. (3.2) and the construction of V; kV (t (u))k ≤ max{kwj k = kw(uj )k; k|uj − t (u)k| ¡ r=16} ≤ 2= for all t ∈ (tn1 ; tn2 ). Thus, Z tn2 2(t 2 − tn1 ) 3r kV (t (u)k dt ≤ n ≤ ; (3.3) 4  tn1 again a contradiction since tn2 − tn1 → 0 (n → ∞).

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Next we claim that for any T ¿ 0 the map h : f → f ; h(u) := (u; T ), belongs to H(f ). In fact, since  : f +1 × [0; 1] → f +1 ; (u; t) := (u; Tt), is an admissible homotopy by Lemma 3.1, h = (·; 1) |f is admissible, odd, homeomorphic and satis es f(h(u)) ≤ f(u). It remains to check that h(f ) is closed. To this end let ’ : E → [0; 1] be an even Lipschitzian function such that ’|f +1 ≡ 0 and ’ |fB ≡ 1, and let t (u) = (u; t) be the solution of d = − ’() ()V (); dt

(u; 0) = u ∈ E:

Arguing similarly as above, one can show that t (u) exists for all t ∈ R. Assume vn := h(un ) = (un ; T ) = (un ; T )→ v, where un ∈ f . Then un = (vn ; −T ) → (v; −T ) = u ∈ f . Thus h(u) = v, i.e., v ∈ f . Therefore h(f ) is closed. We now prove (a). Assume Kc = ∅. Let u ∈ f . If f( s (u)) ¿ c for all s ∈ [0; t] then Z t f0 ( s (u))V ( s (u)) ds = f(u) − f(t (u))¡ − c: t≤ 0

Hence, setting T := − c; (f ; T ) ⊂ fc and the map g(·) := (·; T ) is as required. Finally, assume Kc 6= ∅. We show that g(·) := (·; 1)|f satis es the requirement of (b). If FY = {0} we set N = ∅. If FY \{0} 6= ∅ let r  r [ [ =X ⊕ : Ay BY y; N := 4 4 y∈FY \{0}

y∈FY \{0}

Since for y ∈ FY ; Ay (r=4) is contractible and N = satisfying   1 r 1 ; min ;  ¡ kPY k 2 32

S y∈FY \{0}

 = 1. Take  ¿ 0 Ay ; (N)

 and a ∈ [c + ; − 1] where kPY k denotes the operator norm. Fix arbitrarily  ∈ (0; ) and let u ∈ fa+ \N. Assume that there is t ¿ 0 such that kPY t (u)−yk = r=8 for some y ∈ FY and  s (u) ∈= N0 for s ∈ [0; t) (the other case is simpler). Similar to Eq. (3.3) one has ( ( s (u)) = 1) Z t 2tkPY k r t kV ( s (u))k ds ≤ ≤ kPY (u −  (u))k ≤ kPY k 8  0 and f(t (u)) = f(u) −

Z 0

t

f0 ( s (u))V ( s (u)) ds = f(u) − t ≤ a +  − t:

(3.4)

This implies f(t (u))≤a+−r=16kPY k¡ a−. Thus if t ≤ 1 then f(1 (u)) ≤ f(t (u)) ≤ a − , and if t¿1 then Eq. (3.4) implies f(1 (u)) ≤ a − (1 − )¡a − . Remark 3.1. In Lemma 3.6(b) if moreover c¡b¡d¡ −2; where b is from (f4) and d ∈ R; then we may take  so that also  ¡ b−c and; consequently; g (fa+ \N) ⊂ fa−

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for all  ∈ (0;  ] and a ∈ [b; − 1]. Furthermore; if Kd = ∅ then the argument of (a) implies that g (fa+ ) ⊂ fa− for all  ∈ (0; 1) and a ∈ [d + 1; − 1]. As a consequence of Lemma 3.6(a) we have: Corollary 3.1. There is at least one u ∈ K such that f(u) ≥ b=2 where b is from (f 4). Proof. Take e ∈ Y with kek = 1, and set E1 = X ⊕ Re. By (f 3) there is R ¿ such that sup f (E1 \B) ≤ 0, where B := B(0; R) ∩ E1 . Let c = b=2 and d ¿ 2 + sup f(B). We show Kc 6= ∅. Arguing indirectly, suppose Kc = ∅. By Lemma 3.6(a) there is an odd admissible g : fd → fd such that g(fd ) ⊂ fc . Since B is -compact, there is a nite-dimensional subspace L of E such that (id-g)(B) ⊂ L. Without loss of generality we assume Re ⊂ L. Let D := {u ∈ B ∩ L: kg(u)k¡} ⊂ L1 := L ∩ E1 = PX L ⊕ Re: Note that g(B ∩ L) ⊂ L since (id-g)(B) ⊂ L. Since g is odd, 0 ∈ D. Let O denote the component of D containing 0. Then O is a bounded open neighborhood of 0 in L1 . Hence, (@O) = dim L1 . We claim g(@O) ∩ @B(0; ) ∩ Y 6= ∅:

(3.5)

If this is proved then we see b ≤ sup f ◦ g(B) ≤ sup f ◦ g(fd ) ≤ c ¡ b; a contradiction, and so Kc 6= ∅ proving the corollary. It remains to prove Eq. (3.5). By de nition, g(@O) ⊂ @B(0; ). If g(@O) ∩ Y = ∅ then the map h := PX ◦ g : @O → PX L = PX L1 is odd and h(@O) ⊂ PX L1 \ {0}. Thus, dim L1 = (@O) ≤ (PX L1 \{0}) = dim L1 − 1; a contradiction. Now we are ready to prove the main result. Proof of Theorem 3.2. Arguing indirectly, assume that there is d ¿ b, where b is from (f 4), such that sup f(K) ¡ d. Let bk :=

inf

sup f(u);

∗ (B)≥k u∈B

which is a well-de ned real number for each k ∈ N by Lemma 3.5. Moreover, b ≤ bk ≤ bk+1

∀k ∈ N:

(3.6)

Indeed, bk ≤ bk+1 follows directly from the de nition. If ∗ (B) ≥ k then g(B) ∩ S ∩ Y 6= ∅ for any g ∈ H(B), and so there is u ∈ B such that g(u) ∈ S ∩ Y which implies f(u) ≥ f(g(u)) ≥ b by (f4). Hence bk ≥ b.

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We encounter two cases: Case 1: There is k ∈ N such that a := bk ¿ d + 1; or Case 2: For all k ∈ N; bk ≤ d + 1. Suppose Case 1 occurs. Take ¿ a + 1. Then ¿ d + 2. By de nition, for each  ¿ 0 there is B ⊂ fa+ with ∗ (B) ≥ k. In view of Lemma 3.3(ii) one has

∗ (fa+ ) ≥ ∗ (B) ≥ k: Since Ka−1 = ∅, by Lemma 3.6(a) (with c = a − 1), there is g ∈ H(f ) such that, for any  ∈ (0; 1); g(fa+ ) ⊂ fa− . Consequently, it follows from Lemma 3.3(iii) that k ≤ ∗ (fa+ ) ≤ ∗ (g(fa+ )) ≤ ∗ (fa− ) which implies a = bk ≤ a − , a contradiction. Now suppose that Case 2 occurs. By Eq. (3.6) the sequence (bk ) is nondecreasing and bounded, and so it is convergent. Let a := limk→∞ bk . Then b ≤ a ≤ d + 1. By the de nition and Lemma 3.3(ii) one has ∗ (fa+ ) ≥ k for all ¿0 and k ≥ 1. Thus,

∗B (fa+ ) ≥ ∗ (fa+ ) = ∞

∀0¡¡1:

(3.7)

Take any ¿ d + 2 and set B := fd+2 ⊂ f : By Eq. (3.7), the following sequence b0k :=

inf

sup f(u)

∗ (B)≥k u∈B B

is well de ned. It is also apparent by de nition that b ≤ b0k ≤ b0k+1 ≤ a

∀k ≥ 1:

This implies b ≤ a0 := limk→∞ b0k ≤ a¡ − 1. Since, for all  ¿ 0 small enough, 0 0 0 B = fd+2 ⊃ fa + ⊃ fbk + , one sees ∗B (fa + ) = ∞. By virtue of Lemma 3.6(b) (with c = b=2), there exist  ¿ 0; g ∈ H(B) and a -open set N with (N) ≤ 1 such that 0 0 0 0 0 g(fa + \ N) ⊂ fa − and g(B) ⊂ B. Since fa + = (fa + \ N) ∪ (N ∩ fa + ), it follows from Lemma 3.4(iv), (iii), and (ii) that 0

0

0

∞ = ∗B (fa + ) ≤ ∗B (fa + \ N) + (N ∩ fa + ) 0

0

≤ ∗B (g(fa + \ N)) + 1 ≤ ∗B (fa − ) + 1: 0

This implies that ∗B (fa − ) = ∞ and, consequently, b0k ≤ a0 −  for all k, inducing a contradiction since b0k → a0 . The proof is hereby completed.

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4. Homoclinic orbits In this section we assume that the conditions (H1) – (H5) are satis ed and establish Theorem 1.1. From now on, we consider the Banach space E de ned in Proposition 2.3, Section 2. E is re exive and has the direct sum decomposition with X := E− and Y := E + :

E = X ⊕ Y

Recall that Eqs. (1.2) and (1.3) imply that for all (t; z) |Wz (t; z)| ≤ a03 (|z|−1 + |z|p−1 ): Set

(4.1)

Z ’(u) =

R

W (t; u):

By assumptions, Eq. (4.1) and Proposition 2.3, ’ ∈ C 1 (E ; R) and Z Wz (t; u)v ∀u; v ∈ E : ’0 (u)v = R

Consider the functional f(u) = 12 (ku+ k20 − ku− k20 − ’(u) for u = u− + u+ ∈ E . Then f ∈ C 1 (E ; R). Lemma 4.1. Nontrivial critical points of f are homoclinic orbits of (HS). Proof. Let z ∈ E be a nontrivial critical point of f. Then, for all v ∈ D(A) = H 1 , there holds Z Wz (t; z)v: (4.2) (z; Av)L2 = R


˜ where J˜ = 0 1 , has an exponential diTake a¿0 large so that Ma := L(t) + a J, 1 0 chotomy as before. Recall that by Proposition 2.1 the operator Aa := − (J(d=dt) + −1 Ma ) = A − a J˜ : L ⊃ W 1; → L has a bounded inverse A−1 a so that |Aa u| ≤ d |u| for  ∞ all u ∈ L where  ≥ . Hence by Eq. (4.2) for any v ∈ C0 Z ˜ 2 (z; v)L2 + a(A−1 ( Jz); v) = A−1 L a a Wz (t; z)v; R

we see that ˜ z = A−1 a (−a Jz + Wz (t; z)):

(4.3)

Note that Proposition 2.4 and Eq. (4.1) imply Wz (t; z) ∈ L for all  ≥ 0 := =( − 1). Thus Eq. (4.3) implies z ∈ W 1; for all  ≥ . Acting Aa on the two sides of Eq. (4.3) shows that z solves (HS). Now, similarly to the veri cations of (2.12) and (2.13), one sees that z is a homoclinic orbit of (HS) ending the proof.

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Let K be the set of critical points of f and set F := K=Z, the set consisting of arbitrarily chosen representatives of the orbits of K. By (H 5) we may assume that F = − F. In view of the invariance of f under the shift ∗, O(u1 ) 6= O(u2 )

if u1 ; u2 ∈ K

with f(u1 ) 6= f(u2 ):

(4.4)

By virtue of (H2) Z f(u) = ( 12 Wz (t; u)u − W (t; u))¿0 R

for all u ∈ F\{0}. These, together with the form of f, imply that f(0) = 0, K ⊂ f0 and K ∩ X = {0}. Moreover, by [8], K 6= {0}. Theorem 1.1 will be proved by showing that #F = ∞, that is, K is an in nite set. To this end, arguing by contradiction, we suppose that (#)

#F¡∞:

Then b := inf f(K\{0})¿0. Below, we are going to apply Theorem 3.1 to f. Lemma 4.2. f satisÿes (f1). Proof. Let (um ) ⊂ fa . If (PY um ) is bounded then the form of f implies that Z kPX um k20 + 2 W (t; um ) = kPY um k20 − 2f(um ) ≤ Const: R

which, together with (H2), shows that (um ) is bounded in E . Now assume um * u and PY um → PY u. Then um → u in Lloc and along a subsequence um (t) → u(t) a:e: t ∈ R. Consequently, by the weak lower semicontinuity of norm and the Fatou lemma we get a ≤ f(u). This proves the conclusion. Lemma 4.3. f satisÿes (f2), i.e.; f0 is weakly sequentially continuous. . Hence Proof. Let um * u. Then um → u in Lploc and Wz (t; um ) → Wz (t; u) in Lp=(p−1) loc f0 (um )v → f0 (u)v for each v ∈ E proving the lemma. Lemma 4.4. f satisÿes (f3). Proof. Let Y0 be a nite-dimensional subspace of Y = E + , and let E0 := X ⊕ Y0 . By (H2), f(u) ≤ 12 (kPY uk20 − kPX uk20 ) − a1 |u| which, jointly with dim(Y0 )¡∞, yields the desired conclusion. Lemma 4.5. f satisÿes (f4), i.e.; there are b; ¿0 such that f|S ∩ E + ≥ b. Proof. By Eq. (1.2) and (H4) we have W (t; z) ≤ d(|z| + |z|p )

∀(t; z);

(4.5)

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where (and below) d stands for a generic positive constant. Thus, for u ∈ E + , f(u) = 12 kuk20 − ’(u) ≥ 12 kuk20 − d(|u| + |u|pp ); and the desired result follows. In the following, let [r] denote the integer part of r ∈ R. Lemma 4.6. Suppose that (#) holds. Let (un ) ⊂ E be a (PS)c sequence. Then either (i) un → 0 (corresponding to c = 0); or (ii) c ≥ b and there are l ≤ [c=b]; vi ∈ F\{0} (i = 1; : : : ; l); a subsequence denoted again by (un ); and l sequences (ain ) ⊂ Z (i = 1; : : : ; l) such that

l

X

ain ∗ vi → 0;

un −

i=1



|ain − ajn | → ∞ (i 6= j) and l X

f(vi ) = c:

i=1

Proof. By (H2), 1 f(un ) − f0 (un )un = 2

Z R

( 12 Wz (t; un )un − W (t; un ))

−2 ≥ 2

Z R

Wz (t; un )un ≥

a1 ( − 2) |un | : 2

(4.6)

Setting n := kf0 (un )k; we see |un | ≤ d|f(un ) + n kun k|:

(4.7)

Letting n := {t; |un (t)| ≤ 1} and cn = R\ n ; by Eq. (1.3) and (H3), the Holder inequality and Eq. (4.7), there holds (0 = =( − 1)) Z Z 1−1=  Wz (t; un )un+ ≤ a03 |un | |un+ | + R

!1=

Z

n

c n

Wz (t; un )un

|un+ |0

≤ d(|f(un ) + n kun k|1−1= + |f(un ) + n kun k|1= )kun+ k0 :

(4.8)

It is easy to see that kun k20 = kun− + un+ k20 ≤−2f(un ) + 2kun+ k20 Z 0 + ≤ −2f(un ) + 2f (un )un + 2 Wz (t; un )un+ R

(4.9)

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which, together with Eqs. (4.7) and (4.8), implies that (un ) is bounded. Now, letting n → ∞ in Eq. (4.6) yields c ≥ 0. Moreover, Eqs. (4.6) – (4.9) show that c = 0 if and only if un → 0. R∞ Let A+ :=   dE() be the part of A in H+1 := D(A) ∩ H+ = H 1 ∩ H+ where  is from (H1). Then A+ : H+1 → H+ has a bounded inverse A−1 + and kukH 1 ≤ d|A+ u|2 for all u ∈ H+1 . + + Consider c¿0. Set wn := A−1 + un . Then kwn kH 1 ≤ d|un |2 . We claim that there exist (yn ) ⊂ R and a; ¿0 such that Z |wn |2 ≥ : (4.10) lim inf n→∞

B(yn ; a)

In fact, otherwise, by the concentration compactness principle (see [12]), wn → 0 in L for all  ∈ (2; ∞). Then Z |un+ |22 = (un+ ; wn )0 = f0 (un )wn + Wz (t; un )wn → 0: R

Therefore, by interpolation, |un+ | → 0 for all ¿2. This implies Z kun+ k20 = f0 (un )un+ + Wz (t; un )un+ → 0: R

1 + 2 2 kun k0

→ 0; a contradiction. We get f(un ) ≤ Choose kn ∈ Z such that |kn − yn | = min{|g − yn |; g ∈ Z}; and set zn := kn ∗ un = zn− + zn+ . In view of the invariance of E ± under the action ∗ and noting that A−1 + commutes −1 + + with ∗, it follows that w n := kn ∗ wn = A−1 + zn = kn ∗ A+ un ; and by Eq. (4.10) |w n |L2 (B(0; a+1)) ≥ =2:

(4.11)

Now along a subsequence zn → z ∈ K weakly in E and in Lloc for all  ∈ [2; ∞). Using Eq. (4.11) and arguing as above, it is not dicult to check z + 6= 0 and so z 6= 0. Next, setting u1n := zn − z; we show f(u1n ) → c − f(z);

(4.12)

f0 (u1n ) → 0:

(4.13)

For any  ∈ (0; ); where  is from (H4), choose R¿0 such that, letting TR = [−R; R] and TRc = R\TR ; kzkL∞ (TRc ) ¡

and

kzkL (TRc ) ¡:

(4.14)

Then by Eq. (4.5) Z W (t; z) ≤ d; TRc

and by the mean value theorem and Eq. (4.1) Z Z |z|(|z|−1 + |u1n |−1 + |u1n |p−1 ) ≤ d: (W (t; u1n + z) − W (t; u1n )) ≤ d Tc Tc R

R

Y. Ding, M. Girardi / Nonlinear Analysis 38 (1999) 391 – 415

413

Since u1n → 0 in Lp (TR ) we have Z (W (t; u1n + z) − W (t; u1n ) − W (t; z) ≤  TR

for n large. Hence, Z (W (t; u1n + z) − W (t; u1n ) − W (t; z)) → 0 R

as n → ∞:

(4.15)

Since − + − u1n ; z)0 f(zn ) = f(u1n ) + f(z) + (u1n Z − (W (t; u1n + z) − W (t; u1n ) − W (t; z)); R

Eq. (4.12) follows from Eq. (4.15) and that u1n * 0. To verify Eq. (4.13) let ’ ∈ E with k’k = 1. Using Eqs. (1.3), (4.14) and (H4), one gets Z Wz (t; z)’ ≤ dkzk−1 L (TRc ) k’k ≤ d; Tc R

Z (Wz (t; u1n + z) − Wz (t; u1n ))’ Tc R Z |z|(|z|−2 + |u1n |−2 + |u1n |p−1 )|’| ≤ d; ≤d TRc

that is, Z (Wz (t; u1n ) − Wz (t; zn ) − Wz (t; z))’ ≤ d: Tc R

On the other hand, since u1n → 0 and zn → z in Lp (TR ); Z (Wz (t; u1n ) − Wz (t; zn ) − Wz (t; z))’ ≤ d TR

for n large. So Z sup (Wz (t; u1n ) − Wz (t; zn ) − Wz (t; z))’ → 0 k’k≤1

R

as n → ∞:

Therefore, Eq. (4.13) follows from that Z 0 0 f (u1n )’ = f (zn )’ − (Wz (t; u1n ) − Wz (t; zn ) − Wz (t; z))’ R

and Eq. (4.16).

(4.16)

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Y. Ding, M. Girardi / Nonlinear Analysis 38 (1999) 391 – 415

Now if f(z) = c; then proceeding, the rst paragraph above shows u1n → 0 and nishes the proof. If c¿f(z); then we can repeat the above argument. After at most [c=b ] such procedures one gets the desired conclusion. Given l ∈ N and a nite set A ⊂ E, let ) ( j X ki ∗ ai ; 1 ≤ j ≤ l; ki ∈ Z; ai ∈ A : [A; l] := i=1

Plainly one veri es that (see [4]) inf {ka − a0 k0 ; a; a0 ∈ [PY A; l]; a 6= a0 }¿0:

(4.17)

As a consequence of Lemma 4.6, we have that, assuming (#), if (un ) is a (PS)c sequence then kun − [F; l]k → 0

(4.18)

provided that l ≥ [c=b]. Lemma 4.7. Assume (#) holds. Then for any ¿0; letting l := 1 + [ =b]; the set F := [F; l] is a (PS) neighborhood absorber upto such that PY F = [PY F; l] is isolated. Proof. This is a consequence of Eqs. (4.17) and (4.18). Now we are in a position to prove Theorem 1.1. Proof of Theorem 1.1. Assume that F is nite, i.e., (#) holds. Then f satis es the assumptions of Theorem 3.1. Therefore f possesses a sequence of critical values, cn → ∞; a contradiction. The proof is completed. Corollary 4.1. Let H (t; z) be of the form (1.1). Assume that L(t) is right dichotomic and W (t; z) satisÿes (H2)–(H5). Then (HS) has inÿnitely many geometrically distinct homoclinic orbit. Proof. It follows from Proposition 2.4 and Theorem 1.1.

Acknowledgements Supported by the Alexander von Humboldt-Stiftung of Germany. He would like to thank the members of the Dipartimento di Matematica, Universita di Roma Tre for their kind invitation and hospitality. The authors thank Prof. T. Bartsch for many useful discussions.

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References [1] A. Ambrosetti, V. Coti-Zelati, Multiplicite des orbites homoclines pour des systemes conservatifs, C. R. Acad. Sci. Paris 314 (1992) 601– 604. [2] W.A. Coppel, Dichotomics in Stability Theory, Lecture Notes in Maths, vol. 629, Springer, Berlin, 1978. [3] V. Coti-Zelati, I. Ekeland, E. Sere, A variational approach to homoclinic orbits in Hamiltonian systems, Math. Ann. 228 (1990) 133–160. [4] V. Coti-Zelati, P.H. Rabinowitz, Homoclinic orbits for second order Hamiltonian systems possessing superquadratic potentials, J. Amer. Math. Soc. 4 (1991) 693–727. [5] Y.H. Ding, Existence and multiplicity results for homoclinic solutions to a class of Hamiltonian systems, Nonlinear Anal. TMA 25 (1995) 1095–1113. [6] Y.H. Ding, M. Girardi, Periodic and homoclinic solutions to a class of Hamiltonian systems with the potentials changing sign, Dynamic Systems Appl. 2 (1993) 131–145. [7] Y.H. Ding, S.J. Li, Homoclinic orbits for rst order Hamiltonian systems, J. Math. Anal. Appl. 189 (1995) 585– 601. [8] Y.H. Ding, M. Willem, Homoclinic orbits of a Hamiltonian system, ZAMP, to appear. [9] H. Hofer, K. Wysocki, First-order elliptic systems and the existence of homoclinic orbits in Hamiltonian systems, Math. Ann. 228 (1990) 483–503. [10] T. Kato, Perturbation Theory for Linear Operators, Springer, New York, 1966. [11] W. Kryszewski, A. Szulkin, Generalized linking theorem with an application to semilinear Schrodinger equation, Preprint. [12] P.L. Lions, The concentration-compactness principle in the calculus of variations, Ann. Inst. Henri Poincare, Analyse Non Lineaire 1 (1984) 109 –145 and 223– 283. [13] W. Omana, M. Willem, Homoclinic orbits for a class of Hamiltonian systems, Dierential Integral Equations 5 (1992) 1115–1120. [14] P.H. Rabinowitz, Homoclinic orbits for a class of Hamiltonian systems, Proc. Roy. Soc. Edinburgh 114 (1990) 33–38. [15] P.H. Rabinowitz, K. Tanaka, Some results on connecting orbits for a class of Hamiltonian systems, Math. Z. 206 (1991) 473– 499.  [16] N. Rouche, J. Mawhin, Equations Dierentielles Ordinaires, Masson et Cie, 1973. [17] E. Sere, Existence of in nitely many homoclinic orbits in Hamiltonian systems, Math. Z. 209 (1992) 27– 42. [18] K. Tanaka, Homoclinic orbits in a rst order superquadratic Hamiltonian system: Convergence of subharmonic orbits, J. Dierential Equations 94 (1991) 315–339. [19] H. Triebel, Interpolation Theory, Function Spaces, Dierential Operators, North-Holland, Amsterdam, 1978. [20] M. Willem, Minimax Theorems, Birkhauser, Basel, 1996.