A NNALES DE L ’I. H. P., SECTION C
P. L. F ELMER
Heteroclinic orbits for spatially periodic hamiltonian systems
Annales de l’I. H. P., section C, tome 8, n
o5 (1991), p. 477-497
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Heteroclinic orbits for spatially periodic Hamiltonian
systems
P. L. FELMER
Departamento de Matematicas, F.C.F.M., Universidad de Chile, Casilla 170 Correo 3, Santiago, Chile
Vol. 8, n° 5, 1991, p. 477-497. Analyse non linéaire
ABSTRACT. - We
study
the existence of heteroclinic orbits for a Hamil-tonian
system
-where the Hamiltonian is
periodic
in the spacevariable q
andsuperlinear
in p. We use the Saddle Point Theorem to obtain existence of solutions for a finite time
interval,
and then we obtain heteroclinic orbits as limit of them. Ourhypothesis
on H are motivatedby
the second orderLagrangean systems
on the torus.Key words : Critical Point Theory, Hamiltonian System, Heteroclinic Orbits.
RESUME. - On etudie 1’existence des orbites
heterocliniques
pour lesysteme
hamiltonien1 ~/ B1 ~ -tx /
quand
l’Hamiltonien estperiodique
parrapport
a lavariable q
etsuperli-
néaire en p. On utilise le theoreme de Point Selle pour obtenir des solutions dans un intervalle de
temps
fini et on obtient donc les orbites heterocli-niques
comme limites. Leshypotheses qu’on
utilise sur H sont motivéespar les
systemes Lagrangiens
sur le torus.Classification A.M.S. : 35560, 58 E 05, 58 F 05.
Annales de l’Institut Henri Poincaré - Analyse linéaire - 0294-1449
478 P. L. FELMER
0. INTRODUCTION
In this paper we
study
the existence of heteroclinic orbits of someautonomous Hamiltonian
systems.
Wegeneralize
results obtainedby
Rabinowitz
[8]
for theequation
with
periodic potential.
Rabinowitz studied theproblem using
a variationalapproach through
a minimizationprocedure.
In this work we consider ageneral
Hamiltoniansystem
with Hamiltonian H
periodic
inthe q
variables. In the case ofequation (HS)
a minimizationargument
can not beapplied.
Our method consists in
studying approximate problems by letting
thetime interval
being
finite. This idea has been usedby
Tanaka[10]
andRabinowitz
[9]
in thestudy
of homoclinic orbits for some second order Hamiltoniansystems
withsingular potential.
In order tostudy
theapproxi-
mate
problem
we use a version of the Saddle Point Theorem of Rabinow- itz. We obtain estimates for the criticalvalues, independent
of thelength
of the time interval. We use then the estimates in
passing
to the limit. We note that theproblem
of heteroclinicorbits,
due to the infinite timeinterval,
lacks of thecompactness
oneusually
needs to use criticalpoint theory.
At this
point
we mention the work of Coti Zelati and Ekeland[5]
wherethe
study
of homoclinic orbits for Hamiltoniansystems
is undertaken.Their
approach
is based onconvexity assumptions
on theHamiltonian,
that allows to use a dual
formulation,
and the concentratedcompactness
of P. L. Lions. Hofer andWysocki [6] generalized
the results of[5], drop- ping
theconvexity assumption; they study
theproblem considering
certainfirst order
elliptic systems.
We describe our results now. We consider a Hamiltonian H :
R,
and we denote
H(0, q) = V (q).
We note that in the case ofsystem (0 .1)
the Hamiltonian is
given by
H(p, q) 2 1 ~ p ~ 2 + V , (q)
so that H (0, q)
corre-
sponds exactly
to thepotential.
We make the
following hypotheses
on the HamiltonianAnnales de l’Institut Henri Poincaré - Analyse non linéaire
(H4)
There are constantsEo > 0
and such that everyqEM
(H5)
For constantss _ ~.,
a2 >0,
a3 > 0Here and in the future we denote
by .
the usual innerproduct
in tR" andby ] . its
norm. We normalize H so to have thatV = 0
andthat v (0) = 0.
We will prove the
following
theorem.THEOREM 0. 1. -
If (HO)-(H5)
hold then(HS)
possesses at least 2 heteroclinicorbits,
oneemanating from
0 and oneterminating
at 0.Remark 0. 1. - In Theorem 0 .1 we can
replace
0by
any otherpoint
in M.
The method used to prove Theorem 0 .1 can also be
applied
to aproblem
in which H is notperiodic
in q. ConsiderThe function
Hp (p, q)
is bounded on sets of the form[Rn,
whereBs = ~ p E ~n~~ p ~ _ ~ ~ .
(H2n)
and the set M as defined in
(H2)
is discrete and it contains at least twopoints.
Then we have
THEOREM 0 . 2. -
If (HO), (H 1 n), (H2n), (H3)-(H5)
hold thenpossesses at least 2 heteroclinic
orbits,
oneemanating from
0 and oneterminating
at 0.Coming
back to aperiodic Hamiltonian,
let us consider(H2’)
then we have
THEOREM 0 . 3. -
If
Hsatisfies (HO), (H 1 ), (H2’), (H3)-(H5)
then the(HS)
possesses at least 2 n heteroclinicorbits,
nemanating from
0 and nterminating
at 0.If
wefurther
assume(H6)
H(p, q)
= H(
- p,q),
V(p, q)
E~2n,
then there are 2 n additional het- eroclinicorbits,
nemanating from
0 and nterminating
at 0.Remark 0. 2. - If in Theorem 0. 2 we assume
(H6)
then(HS)
possesses at least 4 heteroclinic orbits.As we mention above our results
generalize
the case480 P. L. FELMER
corresponding
toequation (0.1).
Other case ofinterest,
not covered in[8],
is theLagrangean system
withLagrangean
The Hamiltonian
corresponds
to(0 . 3)
and it satisfies ourhypothesis
ifQ
and V areperiodic
in q, and
Q
ispositive
definite. TheLagrangean
of then-pendulum
hasthe form
(0.3),
and then the Hamiltonian isgiven by (0.4)
that alsosatisfies
(H2’)
and(H6).
Thus Theorem 0. 3guarantees
the existence of at least 4 n heteroclinic orbits for then-pendulum.
This resultcomplements
recent works on the forced
n-pendulum,
see Fournier and Willem[4], Chang, Long
and Zehnder[I] and Felmer [2].
This paper is divided in four sections. In Section 1 we
present
a version of the Saddle Point Theorem that we use tostudy
anapproximate problem,
for finite time intervals. In section 2 we consider the
approximate problems
and prove existence of solutions for every time interval. In Section 3 we obtain estimates on the critical values of the solutions to the
approximate problems,
that areindependent
of thelength
of the time interval. In Section 4 we let thelength
of the time interval to go toinfinity
and weprove Theorems
0.1,
0 . 2 and 0 . 3 .The author wants to express his
gratitude
to Professor Paul Rabinowitz for hishelp
and forencouraging
thewritting
of this article.1. SADDLE POINT THEOREM
The Saddle Point Theorem of Rabinowitz
[7] provides
a tool forfinding
critical
points
of functionals. Here wegive
a variation of that result thatwe will use in our
application.
We consider a Hilbert space E with inner
product ( . , . )
and normI I . ~ ( .
We assume that E has asplitting E = X Q+ Y,
where thesubspaces
Xand Y are not
necessarily orthogonal
and both of them can be infinite dimensional.Let I : E -~ ~ be a functional
having
thefollowing
structurewhere
(11)
L : E ~ E isselfadjoint, (12) b’
iscompact.
Let us consider a
family
ofbounded,
linearoperators
B(s) :
E -~ E wheres E (o, 1].
Assume that for someso E (o, 1],
B(so)
=idE,
and that Bdepends continuously
on s. We will assumeAnnales de l’Institut Henri Poincaré - Analyse non linéaire
(13)
If s E(0, 1]
and v E then the linearoperator
is
invertible,
and its inversedepends continuously
on sand v.Here
Px
denotes theprojection
of E onto X inducedby
thesplit-
ting E=XOY.
Let R>O 0 and define and=
R ~.
We define the class of functionswhere
(rl)
h isgiven by
where s : E x
[0,1] ] -~ (0,1] ]
and v : E x[0,1]
arecontinuous,
v transforms bounded sets into bounded sets,s (E x [0, 1]) stays
away from0,
K : E -~ E iscompact
and K(u, 0) = 0.
Taking v=0,
and K - 0 we obtain thath = idE belongs
to r so itis not
empty.
It can also beproved
that the class satisfies thefollowing composition property: if 11
E r has the formthen 11 (h (x, t), t) E r,
V h E r. Now we state the theoremTHEOREM 1.1
(Saddle
PointTheorem). -
Let I : E --~ I~of
classC1 satisfying
the Palais-Smale condition and(I 1 ), (12)
and(13), further
assume(14)
There are constants a > w such that, , , , - -
Then I possesses at least one critical
point
with critical valuec >
a, character- izedby
Remark 1.1. - The
only
difference with Rabinowitz Saddle Point Theorem is that here the class r isbigger.
Eventhough
this version does notproduce
ingeneral
a smaller criticalvalue,
it makes the estimates in Section 3 somewhat easier.Proof -
Theproof
goes in the standard way so that here weonly
mention the differences. Since
Q
is bounded we find that c 00. Now we show thatc > a. By (14) (i )
weonly
need to show that482 P. L. FELMER
Displaying
the form of h andusing (12), ( 1 . 3)
isequivalent
to findz~Q
such that
We
define W (x)
as the left hand side of(1. 4).
Letfor XEX and
t E [0,1].
Then we have that H is an admissible deformation andH (x, 0) = x,
Thenby
thehomotopy
invariance of theLeray-
Schauder
degree
we have1 =
deg (H (x, 0), 0, Q) = deg (H (x,1 ), 0, Q) = deg (w (x), 0, Q).
°Thus, ( 1. 4)
has at least one solution. This proves( 1. 3).
To show that cis a critical value we
proceed
in the standard way, see[7].
Weonly
notethat the deformation
provided by
the Deformation Lemma has the form(1.2)
so that for every h E r. 02. THE APPROXIMATE PROBLEM Given and T > 0 we consider the Hamiltonian
system
Let e = (2 x k)/T, e (t) = et and ç (t)
=(0, e (t))
E1R2n.
If we define the constantforcing term f= (e, 0),
we caneasily
prove that if z(t)
satisfiesthen
z (t) = z (t) + ~ (t)
satisfies(HS)T.
We devote this section to find solutions to(KS)T.
We consider the space E of functions z :[0, T] --~ 1R2n, z = (p, q)
with Fourier serieswhere al,
bj
E (~n and«
If ~ _ (cp,
with Fourier seriesAnnales de l’lnstitut Henri Poincaré - Analyse non linéaire
then we define the inner
product
that induces the norm
The space E with the inner
product ( .,. )
is a Hilbert space, that can beisometrically
embedded intoWl2°2 (S1, ~2n).
This fact allows to prove that for every 1s
oo there is aconstant as
such thathere,
and in thefuture ~ ~ . ~ ~ S
denotes the usual norm inLS (0, T; [R2n).
Wedefine the
following subspaces
ofEand
here {e1, ..., e2n}
denotes the usual basis of1R2n.
It is easy to see thatE=XCY
where andY = Eq.
Let us define now some
operators
we need later. Letz = (p, q)
and~ _ (cp,
be smooth functions inE,
then we defineand
The
symmetric
form ~ can becontinuously
extended toE X E,
and it induces alinear, bounded, selfadjoint operator
L : E -~ E definedby
thus we have
Remark 2. 1. - We observe that ~ is
negative
onE -,
and it vanisheson
Ep
andEq.
484 P. L. FELMER
We define now the
operator
Bappearing
in the Saddle Point Theoremproved
in Section 1. Given s E(0,1] we
define B(s) :
E -~ Eby
Certainly
B(s)
is abounded,
linear and itdepends continuously
on s. Wenote that .
LEMMA 2 .1. - For every v E
R+0
and s E(0,1]
the linear operatoris invertible and its inverse
depends continuously
on s and v.Proof. - By
calculationsusing
Fourier series we obtainexplicit
formulafor
B . If z = z - + z°,
z -E E - ,
thenwhere
The number
m (s, v)
iscertainly positive ~e(0,1] then B
is inverti-ble,
and its inverse isgiven by
and it
depends continuously
on sand v. DAfter this
preliminaries
we consider the variational formulation of(KS)T.
Let us assume for the moment that the Hamiltonian H satisfies the
following growth
condition(G).
There are constantsa > 0,
b>O and s> 1so that
Then we can define the functional
on E. This functional is well defined and it has the form
(1.1).
If wedefine
Annales de l’Institut Henri Poincaré - Analyse non linéaire
then
~ (z)
is of classC~
and its derivative iscompact.
See[7].
Thenby
Lemma 2 .1 we have that
IT
satisfies(Il)-(13).
Thefollowing proposition
relates the critical
points
ofIT
with the solutionsPROPOSITION 2 . 1. -
If z e E
is a criticalpoint of IT
then z(t)
isof
classC1
and itsatisfies
Proof -
See[7].
Now we
study
the existence of solutions of(KS)T.
Since theHamiltonian H does not
satisfy
thegrowth
condition(G) necessarily,
thefunctional ~f may be not well defined in E. We will make a modification
following
a trick usedby
Rabinowitz.Let K > Eo, where Eo is defined in
(H4)
and x E C°°((~ +, (~ +)
such thatx(y)=1
LetM be a constant so that
We define
The
following
lemma can beeasily proved.
LEMMA 2 . 2. - For every K > Eo, the Hamiltonian
Hx satisfies
theanalog-
ues to
(HO), (H 1 ), (H3)-(H5),
withexactly
the same constants.The
following inequalities
follow from(H3)
and the fact that K > ~o.There are constants a3 and a4
independent
of K so thatand
On the other
hand,
the definition ofHx (p, q) implies
that there areconstants as and a6
so thathere as and a6
maydepend
on K. From(2.18)
we can define the functionalWe note that
if z = (p, q)
with thenLEMMA 2 . 3. - The
functional IT
possesses at least one criticalpoint zT.
486 P. L. FELMER
Proof. - By (2.18)
we can use the sameargument given
in Lemma 3 . 2 in[3]
to show thatIT
satisfies the Palais-Smale condition.By
the structureof
IT
discussedabove,
and Lemma 2.2 weonly
need to prove(14).
Forwe have
From
(2.16),
the definitionof II liE
and the Schwarzinequality
we haveUsing (2 . 6)
and theprojection
from E - ontoE~
we findThen,
forR >__ Ro,
withRo big enough
From
(2 . 20)
and(2 . 21 )
we obtain(14),
then weapply
the Saddle Point Theorem to obtain the result. DRemark 2 . 2. -
Any R >-- Ro
will make thehypothesis (14)
of the SaddlePoint Theorem to be satisfied. In the next Lemma we will
precise
how tochoose R. This lemma will be used in the limit process and we
postpone
itsproof
to the next section.LEMMA 2.4. - For every T there is R
depending only
on T so thatfor
aconstant c
independent of
T and KIT
C.We use Lemma 2.4 to prove the
following proposition
PROPOSITION 2. 2. - For every T > 0 there is a solution zT
of
the system(KS)T
andProof. -
Theproof
consists inshowing
thatgiven
R defined inLemma 2.4 there is K
large enough
so that forz~ = qT)
we have~ pT (t) ~ K, Thus, by
the definition ofH~
we see thatAnnales de l’lnstitut Henri Poincaré - Analyse non linéaire
and
z~
is a solution of(KS)T.
Now we choose K. Since
zT - z = (p, q)
is criticalpoint
ofIT,
andusing (H 3)
forH~
and(2.16)
we haveBut this
implies
that there is a constant aioindependent
of K such thatFollowing
from here a standardargument,
andnoting
thatH~
satisfies(H 5)
with constantindependent
of K we obtainthat ]
is boundedindependent
of K. See forexample [3].
DRemark 2.3. - The constant K may
depend
onT,
however it isindependent
of R. Thus we are free to chooseR ~ Ro
withoutchanging
our conclusions.
3. ESTIMATES ON THE CRITICAL VALUE cT =
IT (zT)
In this section we
provide
aproof
of Lemma 2.4. Since our estimatesonly depend
on(H 0)-(H 5)
and their consequences(2.16)
and(2.17)
thatare
independent
of the value of K, wedrop
it from the indices. For every T we findR >_ Ro
and we construct hEr such thatfor a constant c
independent
of K and T. Let us consider a C~ functione : [0, T] -->
(~n such thatand
488 P. L. FELMER
and ~ (t) _ (o, e (t))
E E. Let us also consider y, s : R + -~ f~+, functions,
such that
with and
0 _ y (i) _ 1.
The constants R > 0 and ~>0 will be determined later. We defineClearly
the function hbelongs
to r.Proof of
Lemma 2.4. - Indoing
our estimates we consider two cases(i ) z E Q
and and(ii ) zeQ
and We will useb;
to denote several constantsindependent
of K and T.Case
(i ).
we haveh (z,
thenWe
analyse
first thequadratic
term.Since ç
hasp-component equal
to 0by (2. 8), (2.12)
and definitionof ç
we haveWe
analyse
now the last term in(3.1). By
the definitionof è
By (2.16)
we havewe recall that a3
and a4
areindependent
of K. Now we estimate the firstintegral
in(3 . 3). By (2 . 6)
we see that Choose E 1 so that0~1~0 and I V (q) I _-1 /(T -1 ) if I q ( _ E 1.
GivenE > o,
we defineAnnales de l’Institut Henri Poincaré - Analyse non linéaire
and
Since TR we have
here m
represents
theLebesgue
measure.By (2 .17) and (H 4)
if wehave
and
by (2.16)
if we haveThen, by (3 . 6), (3. 7)
andby
the choice of Si we obtainBy (3 . 3), (3 .4)
and(3. 8)
we haveBy
Holderinequality
and for T > 1 we obtainBy
Holderinequality again
we obtainThus,
from(3 . 9), (3.10)
and(3 .11 ),
for a constantb 1 o
we haveCase
(ii).
We consider now that As before we obtain490 P. L. FELMER we recall that
y _
1.By (2 .16)
we obtainAs we did for
(3 .11 )
and(3 .12)
we obtainand
And since
s _ 1
Then,
from(3.13)-(3. 17)
we obtainIf R is
large enough
then for allz~Q
with we see from(3. 18)
thatNow that we have chosen R we choose E in such a way that
1,
then from(3 .12)
and(3 .19)
we obtainwith c
independent
of T.4. THE LIMIT PROCESS AND PROOF OF THE THEOREMS In this section we
study
thesequence {zT}
as T goes toinfinity,
andwe
give
aproof
to the theoremspresented
in the introduction.By Propositions
2 .1 and2. 2,
for every T > 0 there is a solution zT of(KS)T
andwith
boundary
conditionAnnales de l’Institut Henri Poincaré - Analyse non linéaire
and
by (4 . 1 )
We assume from now on that In the
arguments
that follow we willonly
usezT,
so that no confusion will riseby dropping
the tilde.Since the
system (HS)T
isautonomous,
the energy is conservedalong
the
solutions,
then there is a constantET
such thatIn what follows we will denote
by ci
various constants that areindependent
of T.
LEMMA 4.1: l
Proof. -
Since zT satisfies(4. 2), by (4 . 4)
andusing (H 3)
we obtainFrom
(4. 5)
witht = o, (4. 3)
andhypothesis (H 3)
we findthen o.
Recalling
that V(q)
0 we obtain from(4 . 6)
thatfrom where statement
(i)
follows. Also from(4.6) (ii)
follows. To show(iii)
we note thatby (2.17)
if we havethen
Then from
(4. 7)
492 P. L. FELMER
COROLLARY 4. 1:
Proof. -
SinceqT (o) = 0
we have from(H 4)
and(2 .17)
then since oo the conclusion follows. For pT
(T)
we cangive
a similarargument.
DLet us define
The
following
lemma will be usedrepeteadly
later.LEMMA 4 . 2. - Let 0 a
d/2,
then there exist constants c(a)
> 0 andE1 so that
ifqEM
and thenProof. -
Since z-p satisfies(4. 2),
from Lemma 4.1 and theperiodicity
of H in q, we have
Then,
from(4. 8),
we haveChoose 81
so thatc4 E 1= a/2,
thenNow choose c
(a)
so thatthen
We note that since V is
periodic in q
and the set M is discrete the constantc
(a)
can be chosenindependent
ofq.
0Annales de l’Institut Henri Poincaré - Analyse non linéaire
LEMMA 4. 3. - qT is
uniformly
boundedindependent of T,
i. e. there is aconstant cs so that
Proof. -
The idea is that ifqT is
notuniformly
bounded then qT(t) spends
too much time outside Mcontradicting
Lemma 4. 1(ii).
We forma-lize this.
By
Lemma 4 .1(iii),
pT isuniformly bounded,
then since zT satisfies(4.2)
we haveAssume there is and so
that q (to) ~ = 2 N d
and~ q (t) I
2N d,
V tto.
Since qT(0)
= 0 there istl to
such thatand
I = 2 (N -1 ) d. By (4 .10) ~ t 1- to >_ 2 d/c6. By continuity
of qT and the definition of d there ist i
E(t 1, to)
so thatif
t2tl
is such that andIqT(t)I2(N-2)d,
then, by
Lemma 4 . 2 there is a constant c~ andE1 > 0
such that(ti -
El,tl
+E1)
c(t2, to)
andWe can
repeat
thisargument N/4
times to obtainfinally
By
Lemma 4 . 2(ii )
follows that N has to bebounded, completing
theproof.
DWe now
begin
the limit process. Let us consider asequence {Tm}m~N
such that oo . Let us denote Assume we have two sequences
{ tm ~ m E ~
such thatLet us now define a sequence of functions
Since
system (4 . 2)
is autonomous the functionSm (t)
is a solution for(4 . 2)
for
t E [ - tm ,
Given we consider the sequence{ ~m }
restric-ted to the interval
[ - N, N]. By
Lemmas 4 . 1 and4. 3,
and the definition494 P. L. FELMER
of Sm
we see that~m
isuniformly
bounded.By equation (4.2)
we havethat also
~m
isuniformly
bounded. Since for mlarge enough [ - N, N] c [ - tm , by
the Arzela-Ascoli theorem we find a subse- quenceuniformly convergent
to a function[ - N, N] -~ 1R2n
andthis function satisfies
(4 . 2)
in[ - N, N].
Proceeding by induction,
for every we can do the anterior proce- dure in such a way that{ ~mN + 1 ~
is asubsequence
Thenby taking
the"diagonal" subsequence
we obtain asubsequence of ~ ~m ~
wecall
{z~}
and a functionzl : f~ -~ (~2n
so thatz~
converges toz~ locally uniformly,
andz~ is a
solution of(HS).
We note thatby (s 3) I q1 (0) ==
Eoso that
z~ is
not trivial.LEMMA 4.4:
Proof -
Let For everyby
Lemma 4 .1(ii )
taking
the limit when mN 00, and thentaking
limit when N -~ oo wefind
where
z~
=(p1, q1). By (s 3)
and the definitionof Sm
we see thatand since lim
tm - tm
= oo we obtain thatI ql (t) I
Eo, ‘d t E(
- oo,0).
Letm -> o0
us assume that lim
q1 (t)
~0,
then there exist a sequencetn
~ - oo such thatTaking a d/2,
andc (a),
E1 as in Lemma4 . 2,
weobtain that
Assuming
without lost ofgenerality that
1 we see that(4 . 12)
contradicts(4 .11 ).
This proves the secondpart
of(i ).
We notethat Lemma 4. 2 was
proved only
for qT, but the sameargument
allowsto prove it for
q 1.
By
conservation of energy and Lemma4 .1,
aftertaking
the limit wefind
Annales de l’lnstitut Henri Poincaré - Analyse non linéaire
then since
lim ql (t) = o, by (H 4)
we conclude that for tlarge enough
from where lim
pl (t) = 0
follows.By
a similarargument
weT - - o0
show
(ii).
DPROPOSITION 4. l. -
Equation (HS)
possesses a heteroclinic orbitstarting
at 0 and
terminating
inM~{ 0 ~ .
Proof. -
ConsiderTm
= m, and definetm
E[0,
so thatand
define t-m
= 0. We claim that thesequences ( and ( tm }
so definedsatisfy (s 1 ), (s 2)
and(s 3).
Weonly
need to prove(s 2):
Consider theinitial value
problem
By
continuousdependence
on initialdata,
andnoting
that p --_0, q
= 0 isthe solution for
a = o,
for every K there is E > 0 sothat implies
q
(t)
EBEo (0),
V 0 t__
K.By Corollary
4.l,
for every E > 0 there is T sothat pT (o) ~ E,
from where we conclude the firstpart
of(s 2).
A similarargument
can begiven
to prove the secondpart, again
we useCorollary
4.1.Using
the limitprocedure
described before Lemma4 . 4,
and Lemma 4. 4we find a solution
zl = (p1, ql)
of(HS)
that is a heteroclinic orbit of(HS)
if lim
ql (t)
EMB~ 0 ~.
If this is not the case we end with a homoclinicorbit. Let us assume we are in this adverse situation.
Let such that
ql (t)
E(0),
V t>__ tl.
We note thattl
> 0 becauseI ql (o) ~ = Eo.
Let us consider the sequenceqm)
defined in the limitprocedure.
Sinceqm (t)
reaches 2 x keventually
there numbersim im
soand
I (t) (
Eo, V t E(’Lm , ’Lm ).
We define =tm
+im
and =tm +’Lm .
We claim that the
sequences {t1+m} and ( t1-m} satisfy (s 1), (s 2)
and(s 3).
(s 1 )
and(s 3)
areclearly
true. Let us show that(s 2)
is also satisfied.Taking
thesubsequence corresponding
to show thatlim
Tm -
= oo weproceed
as we did before.496 P. L. FELMER
We
only
need to show that goes toinfinity,
and since0 im tl,
it isenough
to show thatim
goes toinfinity. Suppose
it isbounded,
thenim
and convergethrough
asubsequence
to i - andi + ,
with
0 i - _ t 1 i + ,
butthen I q1 1 (i + ) ~ = Eo
that contradicts the definition Since{ tm+ ~ satisfy (s 1 )-(s 3)
we canrepeat
theprocedure
to obtain a solution
z2
=( p2, q2)
of(HS),
that is a heteroclinic orbit limq2 (t) E M~{ 0 ~ .
On the
contrary,
iflim q2 (t)
= 0 we obtain a second homoclinic orbit.Repeating
thisprocedure
andassuming
in each case we find a homoclinicorbit,
we obtain a sequence of homoclinic orbits. We claim that this isimpossible.
In fact we will
generate
sequences~, ~ ~, ... , ~ tm },
... whereand this fact
together
with Lemma 4 . 2 contradicts assertion(ii )
ofLemma4.1. D
Proof of
Theorem 0 . 1. -By Proposition
4. 1 there is at least oneheteroclinic orbit
emanating
from 0 andterminating
inMB{ 0 ~ .
If in
problem (HS)T
wechange
theboundary
conditionby
and we
modify
thearguments accordingly
we obtain a heteroclinic orbitemanating
inMB~ 0 ~
andterminating
in 0. DRemark 4.1. - Theorem 0.2 can be
proved
in the same way Theorem 0.1 was, with minor modifications.Proof of
Theorem 0 . 3. - Here we base theargument
in the idea usedby
Rabinowitz inproving Proposition
3.33 in[8].
Thistogether
with ananalysis
similar to the onegiven
inProposition
4.1 will build theproof.
We will be
scketchy.
We are
assuming (H 2’),
thus M consist ofinteger
translations of asingle point.
We can assume that M=Z". Let B denote the set sothat 0
and q
are connectedby
a heteroclinic orbit.By
Theorem 0.1 B is notempty.
Let A be the set of linear combinations of elements in B with coefficients in Z. We claim that A = M. If this is not the case then S =
QS.
Take 2 x k E S and consider the
problem (HS)T
with theboundary
conditionAnnales de l’lnstitut Henri Poincaré - Analyse non linéaire
As in
Proposition
4.1 we find solutions of(HS)
inR, zl, z2,
... Letqi)
the first one so thatq‘ =
limBy
definition ofT - o0
and then
q‘ ~ 2 ~c k.
Thenby going
a further translation as inProposi-
tion 4.1 we find a solution of
(HS) z‘ + 1
so that limq‘ + 1 (t) = q‘.
Iflim
q‘ + 1 (t) = qt + 1 ~ A,
thenq~ + 1- q~ ~ A,
but this isimpossible
sincezi + 1 (t)
-(0, if)
is anorbit joining
0 withq‘ + 1- if. Consequently ~~ e A.
Now we continue
generating
orbits of(HS)
whose endpoints
willalways
be in A
by
theargument given
above. Since we can not continue thisprocedure indefinitely
and forsome j qj ft A contradicting
thehypothesis.
Thus A =M,
and then we can find at least n orbitsemanating
from 0 and
terminating
inMB~ 0 ~ .
The n heteroclinic orbitsterminating
at 0 are obtained
similarly.
This proves the first assertion of Theorem 0 . 3.For the second we assume also
(H 6),
and we note that ifz
(t)
=(p (t), q (t))
is a heteroclinicorbit joining
0to q
thenz(t) _ ( - p ( - t),
q
( - t) - q) joins - q to
O. 0REFERENCES
[1] K. C. CHANG, Y. LONG and E. ZEHNDER, Forced Oscillations for the Triple Pendulum,
E.T.H. Zürich Report, August 1988.
[2] P. FELMER, Multiple Solutions for Lagrangean Systems in Tn, Nonlinear Analysis T.M.A.
(to appear).
[3] P. FELMER, Periodic Solutions of Spatially Periodic Hamiltonian Systems, Journal of Differential Equations (to appear).
[4] G. FOURNIER and M. WILLEM, Multiple Solutions of the Forced Double Pendulum Equation, Preprint.
[5] V. COTI-ZELATI and I. EKELAND, A Variational Approach to Homoclinic Orbits in Hamiltonian Systems, Preprint, S.I.S.S.A., 1988.
[6] H. HOFER and K. WYSOCKI, First Order Elliptic Systems and the Existence of Homoclinic
Orbits in Hamiltonian System, Preprint.
[7] P. RABINOWITZ, "Minimax Methods in Critical Point Theory with Applications to
Differential Equations", C.B.M.S. Regional Conference Series in Mathematics, 65, A.M.S., Providence, 1986.
[8] P. RABINOWITZ, Periodic and Heteroclinic Orbits for a Periodic Hamiltonian System, Analyse Nonlineare (to appear).
[9] P. RABINOWITZ, Homoclinic Orbits for a Class of Hamiltonian Systems, Preprint.
[10] K. TANAKA, Homoclinic Orbits for a Singular Second Order Hamiltonian System, Pre- print, 1989.
( Manuscript received December 1 3 th, 1989.)