A NNALES DE L ’I. H. P., SECTION C
D. T ERMAN
Infinitely many radial solutions of an elliptic system
Annales de l’I. H. P., section C, tome 4, n
o6 (1987), p. 549-604
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Infinitely many radial solutions of
anelliptic system
D. TERMAN
Department of Mathematics Ohio State University Columbus, Ohio 43210
Dedicated to the Memory of Pro- fessor Charles Conley
Vol. 4, n° 6, 1987, p. 549-604. Analyse non linéaire
ABSTRACT. - We consider a system of
equations
of the form Au + V F(u) = 0.
In this and twosubsequent
papers we find conditions onF
( u)
to guarantee that this system hasinfinitely
many radial solutions.We also define a notion of
winding
number for each radial solution and prove that for eachpositive integer
K there exists a radial solution withwinding
number K.RESUME. - L’on considère un
systeme d’equations
de la formeu + F
( u) = ~.
Dans cet article et dans deux articles aparaitre,
l’on trouvedes conditions sur F
(u) qui garantissent
que lesysteme
a une infinite desolutions radiales. L’on definit
egalement
un nombre d’enlacements pourchaque
solution radiale et l’on demontre que pour tout entier K non nul il existe une solution radiale dont le nombre d’enlacements est K.Annales de l’lnstitut Henri Poincaré - Analyse non linéaire - 0294-1449 Vol. 4/87/06/549/56/$7,60/(6) Gauthier-Villars
1
A. Introduction
In recent years a number of authors have considered the
question
ofthe existence of radial solutions of the scalar
equation
Here A is the usual
Laplace operator
and u is a function of x E n > 1.By
a radial solution of( lA.1)
we mean a nonconstant, bounded solution of the form u(x)
= U(r),
r= II
x We also assume that each radial solution satisfies lim Ur)
= 0. Twoquestions
related to thisproblem
are( 1)
findconditions-on g (u)
to guarantee that( lA.1)
has apositive
radialsolution;
(2)
find conditions on g(u)
to guarantee that( lA.1)
hasinfinitely
many radial solutions.There have been two
general approaches
toanswering
thesequestions.
The first
approach
has been to use variational methods. For references to thisapproach
see[2], [3], [8].
The secondapproach
forstudying
theexistence of radial solutions of
(1.A1)
has beentopological
orphase
spacetechniques. Using
thesemethods,
Atkinson and Peletier[1]
have proven the existence of apositive
solution of(lA.1)
under very weak conditionson g
(u).
Jones andKupper [7]
.usetopological
methods to find conditionson g
(u)
which guarantee the existence ofinfinitely
many radial solutions of( lA.1). Moreover, they
were able to characterize the radial solutionsby
their nodalproperties;
thatis,
underappropriate
conditions on g(u),
for each
nonnegative integer
K there exists a radial solution of( lA.1)
with
precisely
K zeroes.Recently,
Brezis and Lieb[4]
have shown how the variationalapproach
can be
generalized
to systems ofelliptic equations. They
consider systems of the formwhere the m functions
f i :
[Rm -~ R aregradients
of some function FE C1 thatis,
foreach i,
They
prove that there is a function u(x)
which minimizes the action associated with( 1 A . 2) .
In this paper we show how
topological
methods can be used to prove the existence ofinfinitely
many radial solutions for a system ofequations.
We consider a system of the form
(lA.2), ( 1A.3)
and find conditions of F(U)
which guarantee the existence ofinfinitely
many radial solutions.We are also able to define a notion of
winding
number for each radialsolution,
and prove that for eachpositive integer
K there exists a radial solution withwinding
number K.B. Precise statement
of the
problem
and the first main resultThe system we consider is
where ul
and u2 are functions of x E n > 1. We assume that there existsa function
F E C2 ( (~2)
such that for i =1, 2,
for
u2) ~ By
a radial solution of(1B.1)
we mean a nonconstant bounded solution of the formWe
always
assume that a radial solutionsatisfies,
fori =1, 2,
If
(U1 (r), U2 (r))
is a radial solution of( 1 B.1 ),
andVol. 4, n° 6-1987.
then
(U 1, U2, Vi, V2)
satisfies the first order systems ofordinary
differen-tial
equations,
forr > o,
together
with theboundary
conditionsfor some real numbers
U~
andU2.
Theequalities
in( 1 B.4)
are meant tobe taken
componentwise.
We prove the existence of radial solutions of( 1 B.1 ) by proving
the existence of solutions of( 1 B. 3)
and( 1 B. 4) .
NOTATION. - For convenience we set
and
ASSUMPTIONS ON F. - We wish to assume that F looks
someting
likewhat is shown in
Figure
1. Theprecise assumptions
on Fare :(F1) FEC2 (2).
(F2)
F has at least threenondegenerate
local maxima. These are atA = (A 1, A 2) _ (o, 0), B = (B 1, B2),
andC = (C 1, C2).
F also has twosaddles. These are at
D = ( D 1, D~)
andE = ( E 1, E2) .
(F3)
F(A) F (B) F (C)
andB1 D1 A1 E1 C1.
Moreover, thereexists ao such that if a is any critical
point
of F with{A,
B,C},
thenF
(a) F (A) - ao.
Forconvenience,
we assume that F(A)=0.
(F4)
There exists W such that if KW,
then the level set{U:
F(U) >_ K~
(F6)
Letand
Suppose
that(U (r),
V(r))
is a bounded solution of( 1B.3)
with n =1which
satisfies,
fori =1, 2,
or3,
and
Then U
(r)
isidentically equal
to one of the criticalpoints A, B,
orC,
and V
(r) _ (0,0)
for all r > 0.Remarks
concerning
theseassumptions
arecertainly
in order. These remarks will begiven shortly.
We first state our first main result.Vol. 4, n° 6-1987.
THEOREM 1. - Assume that
fl
andf2 satisfy (lB.2)
where F(U2, U2) satisfies (F1)-(F6).
Then there existsinfinitely
many radial solutionsof (1B.1).
We also prove another theorem which is stated
explicitly
in Section 1 E.For that Theorem we define a notion of
winding
number for each radial solution. Our second Theorem states that for eachpositive integer
K thereexists a radial solution with
winding
number K.The
proof
of these two results issplit
into three papers. In this paperwe introduce a
family
ofequations
whichdepend
on a smallparameter
E.As
E -~ 0,
thefamily
ofequations approach (lB.1).
We prove, in this paper, that for each E there existsinfinitely
many solutions of the newequations.
In[10]
we reduce theproblem
ofproving
that for each ~ > 0 andpositive integer
K there exists a solution of the newequation
withwinding
number K to analgebraic problem.
In[9]
we solve thealgebraic problem.
Hence,[9] together
with[10]
prove that for each ~ > 0 andpositive integer
K there does indeed exist a solution of the newequations
withwinding
number K. In Section 6 of this paper we prove that asE - 0,
some sequence of the set of solutions with
winding
number K for the new’
equations
converge to a solution of(lB.1)
withwinding
number K.C. Remarks on the
assumptions
on FRemark 1. -
( F4)
will be used to prove that the set of bounded solutions of( 1 B. 3)
is compact.Remark 2. -
(F6) guarantees
that the set of bounded solutions of( 1B.3)
is not too bizarre. One may think of( lB. 3)
with n =1 asdescribing
the motion of a ball
rolling along
thelandscape
definedby
thegraph
ofF without friction. There may exist bounded solutions of
( lB.3)
becausethe ball may roll back and forth between the mountain
peaks given by
F
( A),
F(B),
and F(C). Assumption (F6) implies
that these are theonly
bounded solutions of
( lB.3)
withn =1,
besides the criticalpoints,
whichlie above F
(A) - ao
for some r. We are not interested in bounded solutions which lie below F(A) - ao
for all r, because if U(r)=(Ui (r), U 2 (r))
is aradial
solution,
then lim U(r) = A implies
that F(U (r))
> A - oco for rRemark 3. -
(F5)
is the most unreasonableassumption.
It can beweakened
slightly
as follows. Letand
I
Then
(F5) implies
that ifU E ID n N 1
orthen VF(U)
istangent to lD or lE, respectively.
Our result remains valid if this property holds for some linelD and lE through
D andE,
notnecessarily
the onesgiven by ( 1C.1).
Wechoose lD and lE
as in(lC.1) only
for convenience.We do feel that our method of
proof
should carry over to moregeneral assumptions
than(F5).
The mainplace
where(F5)
is used is to define the notion ofwinding
number. InAppendix
B we comment how one shouldbe able to weaken this
assumption.
D. The
winding
numberRecall that for the scalar
equation ( lA.1)
it has been proven that underappropriate
conditions on g(u),
for eachnonnegative integer
K thereexists a radial solution with K zeroes. We wish to prove an
analogous
result for the system
(lB.1). However,
instead ofcounting
the number ofzeroes we introduce a notion of
winding
number which measures how many times atrajectory (U 1 (r), U2 (r), V1 (r), V2 (r))
winds around inphase
space.For the scalar
equation,
the notion ofwinding
number issimple
becauseit makes sense to count the number of times a
trajectory
winds around apoint (the origin,
forexample).
For( lB.3)
thephase
space is four dimen- sional(or
five dimensional if one includes r as adependent
variable as weshall do
shortly),
and it does not make sense to count how many times atrajectory
winds around apoint. Instead,
we define two, two dimensionalplanes, PD
andPE,
and count how many times the solutions wind around theseobjects. PD
andPE
are defined as follows:and
Vol. 4, n° 6-1987.
Clearly, PD
andPE
are two dimensional.Perhaps
the mostimportant
property ofPD
andPE
isPROPOSITION 1.D.1. -
PD
andPE
are invariant with respect to theflow defined by ( 1 B. 3) . That is, if
for
some ro, thenfor
all r.Proof -
From( lB. 3)
and(F5)
we conclude that onPD
orPE,
and
These two
equalities
prove theproposition.
It now makes sense to count the number of times a radial solution winds around
PD
andP~.
This is done as follows. Letand
DEFINITION. -
Suppose
that U(r)
is a radial solution of( 1 B.1 ) .
Thenthe
winding
number of U is definedby
Remark. - The notion of
winding
number may seemcomplicated
because it involves
trajectories
in four dimensionalphase
space which is difficult topicture. However,
one caneasily
compute thewinding
numberby just considering
U(r) _ (U 1 (r), U~ (r))
in the two dimensional state space. Recall that h(U) equals
to the number of times U(z)
intersectsQo
or
Q.
Now U(zo) E QD if
andonly
if U andV 1 (zo) o.
Sincethis
implies
that U(zo)
EQo
if andonly
if U(zo)
ElD,
and atZ = Z 0’
U (z) crosses 10
fromright
to left.Similarily, U (zo) E QE
if andonly
if U(zo) E lE
and at Z=Zo, U(z)
crosseslE
from left toright.
Ofcourse, this remark
depends
on ourassumption (F5).
InAppendix B,
we will discuss how one should be able to weaken(F5).
Under the weakerassumptions
we will still be able to define twosurfaces, PD
andPE,
whichdo not intersect the radial solutions.
Hence,
we will still be able to define the notion ofwinding
number.However,
we will not be able to compute thewinding
numberby just considering
thetrajectories
in the state space.E. The second main result
THEOREM 2. - Let K be any
positive integer.
Then there exists a radialsolution,
U(r), of ( 1 B.1)
such that either h(U)
= K or h(U)
= K + 1.Remark 1. - The fact that we have either h
(U)
= K or h(U)
= K + 1may be
disturbing
because one would expect there to exist a radial solution with h(U) = K.
The reason that we obtain the weaker result is that we arecounting
the number of times solutions wind around twoobjects, namely PD
andP~.
Remark 2. - We
actually
prove that for eachinteger
K there exists atleast two radial solutions of
(lB.1),
each withwinding
number K or K + 1.We
explain why
this is true inAppendix
C.F. Reduction to a connection
problem
The
boundary
conditions(lB.4)
state that we must find atrajectory
inphase
space whichbegins (at r = o)
on theU-plane
and ends(at
r =oo)
atthe
origin.
Theseboundary
conditions are awkward to work with becausewe don’t
really
know where tobegin
on theU-plane.
It will be moreVol. 4, n° 6-1987.
convenient to transform
( 1 B. 3), ( 1 B. 4)
to aproblem
where we look for atrajectory
which connects two criticalpoints.
This is done as follows.We first consider r as a
dependent
variableby introducing
z = r as thenew
independent
variable. Wecompactify by letting
Then
( 1 B. 3)
becomeswhere
A radial solution must
satisfy
and
for some
Uo.
Now for E > 0 letwhere Pt satisfies
Consider the
system
together
with theboundary
conditionsand
It will be convenient to make the
change
in variablesThen
( 1 F. 2), (IF. 3) become,
afterdropping
thehats,
and
In this paper we prove
THEOREM 3. - There exists go such that
if
0 s go, then there existsinfinitely
many solutionsof ( lF. 5), (IF.6).
Vol. 4, n° 6-1987.
Of course, if we make the
change
in variables( lF.4),
then thisimplies
that there exists
infinitely
many solutions of(IF.2), (IF.3).
In two
subsequent
papers,[9]
and[10],
we proveTHEOREM 4. - There exists Eo such that
if 0 E Eo
and K is anypositive integer,
then there exists a solutionof( lF.2), (IF.3)
such that either h(UE)
= K or h(UE)
= K + 1. In Section 6of this paper we prove,
assuming
Theorem4,
THEOREM 5. - Let K be a
positive integer.
Then there exists a sequence such that as k - 00,{Uek (z)}
converges to a radialsolution,
U(z), of
( 1 B.1 ) . Moreover,
either h( U) = K
or h( U) = K + 1.
G.
Description
of theproof
The
proof
of Theorem 3 isquite geometrical.
The purpose of this subsection is to introduce the basicgeometrical
features of theproof.
Eachsolution of
(IF. 5) corresponds
to atrajectory
in five dimensionalphase
space. The
boundary
conditions(IF. 6) imply
that we arelooking
for atrajectory
whichapproaches (A, U, -1)
as z - - oo and(B,
(9,+ 1)
asz - + oo.
Hence,
we arelooking
for atrajectory
which lies in bothW~,
the unstable manifold at
(A, C~, -1),
andWB,
the stable manifold at(B, U, + 1).
The first step in the
proof
of Theorem 3 is to obtain apriori
boundson the bounded solutions of
(IF. 5).
This is done in Section 2. We constructa five dimensional
box, N,
which contains all of the bounded solutions of(IF. 5).
We then construct, in Section4B,
a subset ~ of theboundary
ofN with the property that each nontrivial
trajectory
inWA
canonly
leaveN
through
~. The mostinteresting
feature of ~ is that it has fourtopological
holes.We then
analyze
the unstable manifold at(A, (~, -1 ) .
Because the dimension ofWA
isthree,
we show that it ispossible
toparametrize
thenontrivial
trajectories
inWA
be thepoints
in the disctrajectory
inWA
there is awinding number,
as described in Section ID.Hence,
to each d~D we canassign
theinteger h (d)
which isequal
to thewinding
number of y(d, z).
Now the solutions of
(IF. 5), (IF. 6) correspond
to a certain subset of~,
which we denoteby
X. Thatis,
Let Y
= D~X.
We shall prove that for eachy E Y,
y(y, z)
must leave N(see Proposition 2C.1).
Because y(y, z)
canonly
leave Nthrough ~,
we have a(continuous) mapping
A : Y --~ ~ definedby,
A(y)
isequal
to theplace
where y(y, z)
leaves N.Suppose that g (s)
is a continuous function fromI = [o,1]
to Y such thatg (o) = g ( 1 ).
The( A . g) (I)
defines acontinuous,
closed curve in G. InSection 4B we define an
algebraic object,
r(g),
which describes how(A. g) (I)
winds around the four holes in G. r(g)
will be an element ofF~,
the free group on four elements. It will have thefollowing important
property:
PROPOSITION A
(see Proposition 4 B). - If gl
ishomotopic
to g2 relativeto
Y,
then r(gl)=r
The next step in the
proof
of Theorem 3 is toassign
to each element0393~F4
apositive integer
We provePROPOSITION B
(see Proposition
5A.1). -
Let M be apositive integer.
There exists a continuous
function
g : I -~ Y such that g(o) = g ( 1 )
andTheorem 3 will then follow from
Propositions
A and B.The
key
steps in theproof
ofProposition
B arePropositions
3B.1 and4C.1. In
Proposition
4C.1 we derive arelationship between I~‘ (g) f
andthe
winding
number of the varioustrajectories y (g (s), . ), 0 - s __ 1.
InProposition
3B.1 we show that there must existtrajectories
inWA
witharbitrarily high winding
number.Vol. 4, n° 6-1987.
2. THE ISOLATING NEIGHBORHOOD
A. Basic definitions
Until stated otherwise we fixE>O. Our immediate
goal
is to define aset N in
phase
space which contains all of the bounded solutions of(IF. 5).
Recall the set
N 1
defined in( lB. 5).
Letwhere V is a
large
number to bedetermined,
andRemark. - N is
topologically
a five dimensional box with two"tubes",
PD
andPE,
removed. This istopologically equivalent
to a two dimensional disc with twopoints
removed.We wish to prove
PROPOSITION 2A. I. -
IfV-, appearing
in(2A.1),
issufficiently large,
thenall solutions
of (IF.5), ( 1 F. 6)
lie in N.This result is
proved
in the next section. Theproof
is broken up .into a number of lemmas.B. Proof of
Proposition
2A.1LEMMA 2B.1. - The
projection
ontoU-space of
every bounded solutionof ( lF. 5)
lies inN l.
Moreover, there cannot exist a solutionof ( lF. 5)
whoseprojection
ontoU-space
isinternally
tangent to aN1,
theboundary of N1.
Proof -
Theproof
followsConley [5].
Choose K _ W and suppose that(U (z),
V(z),
p(z))
is a solution of(IF.5)
which satisfies for somezo~
and
Then
since the
assumption
that the level set{F (U)=K}
is conveximplies
thatThis
implies
that there cannot be any internaltangencies
on the level set{F (U) = K}
for all K _ W.On any solution which leaves the set where F > W there is a
point
whereF W and either
dFjdzO
ordFjdz>O. Suppose
thatThen
(2B.1) implies
that F(U (z))
isstrictly decreasing
for There-fore,
ifthe
solution werebounded,
it would haveto
go to a restpoint
~.where F .W.Since there aren’t any such rest
points,
the solution must be unbounded in forward time. A similarargument
shows that ifthen the solution is unbounded in backward time.
Remark. - The
proof
of this last result shows that if U(z)
leavesN 1
in forwards or backwards
time,
it can never return toN~.
LEMMA 2B.2. -
V,
as in(2A.1),
can be chosen so thatif
then U
(z)
leavesN1
in backwards time.Vol. 4, n° 6-1987.
Proof. - Suppose
thatII V (zo) II> V,
where V is to be determined.Then either
We assume that
V~ (zo) > 1 /2 V, and,
forconvenience, Zo=0.
ChooseM~
so
that ( ~J) ~ ~ M 1
inN 1. Then,
from( 1 F. 5),
as
long
asV 1 >__ o. Therefore,
ifV 1 >_ o,
thento obtain
or
which we assume to be true.
Therefore,
f or -1 -
z ~0,
whichimplies,
uponintegration,
thatLet
M2 =diameter
ofN1
and choose V so thatThen
(2B . 3)
and(2B . 4) imply
thatU ( -1) ~ N1.
Similar arguments hold for the other cases in(2B. 2).
We assume
throughout
the rest of the paper that V is chosen so that Lemma 2B. 2 is valid. We can now present theCompletion of
theproof of Proposition
2A. l. -Suppose
that(U (z), V (z), p (z))
is a solution of(IF. 5), (IF. 6).
From Lemma 2B. 1 and 2B. 2 we conclude thatfor all z. From
Corollary
1D . 2 it follows thatfor all z. From the definitions this
implies
thatfor all z.
Together
with(2B. 5)
and the definition ofN,
thisimplies
thedesired result.
C. The energy H and the critical
point
CConsider the function .
where
V, V)
is the usual innerproduct
in1R2.
If(U (z), V (z), p (z))
is asolution
of(IF. 5)
we writeH (z) = H (U (z), V (z)).
Note that on a solutionof ( l F . 5),
Therefore,
and H
(z)
isincreasing
on solutions. An immediate consequence of this is PROPOSITION 2C . 1. - Theonly
bounded solutionsof ( 1 F . 5)
which liesin the
set ~ p ~ -1 ~
are the criticalpoints
andtrajectories
which connect the criticalpoints.
Vol. 4, n° 6-19$7.
We conclude from
(2C. 3)
and(2C. 4)
that on a solution of(IF. 5), (IF. 6),
This
immediately implies
LEMMA 2C. 2. - There exists 6 such that
if (U (z), V (z), p (z))
is asolution
of ( l F . 5), (IF.6),
z.This lemma demonstrates
that,
since we areonly
interested in solutions of(IF. 5), (IF. 6),
the values of F inCs = { U : ~ ~ U-C ~) I S ~
do not matter.In
particular,
F(U)
may be chosen to bearbitrarily large
inCs.
Wechange F (U)
inCs
so that if(U (z), V (z), p(Z))EWA,
the unstable manifold at(A, (~, -1),
thenU (z) ~ C
for all z. This ispossible
for thefollowing
reason.
Suppose
that(U (z), V (z),
andU (zo) = C
for some zo.If
F ( C)
is verylarge,
then we must haveis
verylarge
for some z 1 zo.However,
as Lemma 2B. 2shows, if II
is toolarge,
thenU (z)
will leaveN~
in backwards time. The remarkfollowing
Lemma 2B. 1
implies
thatU (z)
can then never return toNi,
in backwardstime,
afterleaving N~.
This contradicts theassumption
that(U (z), V (z),
p(z))
EWA.
To make this allprecise
would be very tediousso we do not
give
the details.3. THE LOCAL UNSTABLE MANIFOLD AT A
A. A
parametrization
of thetrajectories
in the unstable manifold at
(A, C~, -1 )
Let
WA
be the unstable manifold atA = ( A, (~, -1) .
We assumethroughout
that if(U, V, p) E W A’
thenp> -1.
The behavior ofWA
near A is determinedby linearizing (IF. 5)
at A. If we setthen the linear system at A is
To compute the
eigenvalues
andeigenvectors
of this system letbe the Hessian matrix of F at A. Since F has a local maximum at
A,
itfollows that M has
negative eigenvalues,
which we denote and- ~,2,
withcorresponding eigenvectors wi
E R2 and0)2
E1R2.
Theeigenvalues
of
(3A. 1)
are thenwhere,
forf=l,2,
Eigenvectors corresponding
toi =1, 2,
areAn
eigenvector corresponding
to ~3 isWe conclude from the Stable Manifold Theorem
(see [6]),
THEOREM 3A , .1. - Near
A, WA
is aC2 injectively immersed,
three dimensionalmanifold. Moreover,
the tangent space toWA
at A is the linearsubs pace spanned by pi ,
and p3.Let
and let
loc WA equal
to the local unstable manifold at A. Animportant
consequence of Theorem 3A. 1 is
PROPOSITION 3A. 2. - 6 can be chosen so that
Vol. 4, n° 6-1987.
(a) for
each(Uo, po) E As
there exists aunique V0~R2
such that(Uo, Vo, po)
~loc03A9A, and(b) if (U (z),
V(z),
p(z))
is any nontrivialtrajectory
inWA
whichsatisfies
P
(z) ~
-1for
some z, then there exists aunique
Zo such that(U(zo),
V(zo),
P(zo))
E locWA
and(U (zo),
P(zo))
EAg.
Proof. -
This result followsimmediately
from Theorem 3A. 1 and the fact that not each of the U or p components ofp2 ,
andp3
are zero.We assume
throughout
that 8 is chosen so that theproposition
is true.The result
implies
that we mayparametrize
the nontrivialtrajectories
inWA by
thepoints
on thehemisphere As.
Letand
the
bijection given by
theProposition.
Letand let
be the
projection
map. That isLet
L3 : qfi 1
- G be definedby
We have now
parametrized
the nontrivialtrajectories
inWA by
the disc~ 1.
Instead it will be more convenient to work with the unit disc~ defined
by
Let
L4 :
~ --~ be the mapand
Lo :
~ - G the mapThen to each
point
therecorresponds
apoint Lo (d)
E locW A.
Letbe the
trajectory
inWA
which passesthrough
thepoint
For(x, y)
define thepolar
coordinatesSuppose
that FromProposition
2G.1,
either(a) y (d, z)
is a solution of(IF. 6)
or
(b)
y(d, z) eventually
leaves N.We write
where
and
For d we define
where h
(U ( d, z))
is defined as in(ID. 1).
Note that Theorem 3 is
equivalent
toproving
that X is an infinite set.close to 1
The
following
result is crucial to theproof
of the theorems.PROPOSITION 3B . 1. - Given M > 0 there exists
1)
such thatif
for
some z, then h(d) >_
M.Moreover,
iM does notdepend
on ~.Because the
proof
of this result isextremely long
and technical we savethe
proof
forAppendix
A.Vol. 4, n° 6-1987.
4. AN ALGEBRAIC INVARIANT
A. A
preliminary
resultPROPOSITION 4A. 1. - There exists Eo> 0 such that
if
0 E Eo, then thefollowing
is true. Fix i E(0, 1)
and q E Then there exists 8 = 8(q, i)
suchthat
U((r, 8), zo)=qfor
some zo andU((T, 9), z)~X2 for
z zo.Moreover,
8(q, r)
can be chosen todepend continuously
on q and i.Remark. -
X 2
was defined in(lB. 5).
Proof. -
Let Iequal
to the set ofTe(0,1)
for which the firstpart
of the lemma is true. Thatis,
for each 03C4~I and there exists8 (q, r)
such that if
d=(T, t)),
thenU (d, zo) = q
for some zo andU (d,
for z zo. We prove that I is open,
closed,
andnonempty.
We first demonstrate
why
I is open. Note that ifTe(0,l),
thenU ((i, e), z)
must leaveX2.
This is because of(2C. 2). Moreover,
if we setQ(r, e) equal
to theplace
whereU((r, 8), z)
leavesX 2,
thenQ(r, 8)
iscontinuous. This is because
assumption (F5)
and Lemma 2B. 1imply
thatU((T, e), z)
cannot betangent
toaX2.
It noweasily
follows that I is open.This also shows that
r)
can be chosen to be continuous.It is trivial to show that I is
closed,
so it remains to prove that I isnonempty.
We prove that rel if r and E aresufficiently
small. We willprove that there exists
Eo > 0
andio E (0,1)
such that if0 E Eo, 0 i io,
and
0 9 2 ~, I
isincreasing
aslong
asU((r, 8), Z)EX2.
BecauseU((r, 8), z)
musteventually
leaveX 2
thisimplies
the desired result.
Recall that A
= (0,0).
Because A is anondegenerate
local maximum ofF (U),
there existsbo > 0
such thatWe shall now show that if
re(0,1)
and8), z) ‘) ~o
for z zo, then)) U (r, e), z) ~)
isincreasing
for z zoo Fix(r, o) E ~,
setU(z)==U((r, e), z),
and assume for z zo. For z zo,
Integrating
thisequation
from - oo to z we find thatIt then follows that
for z zo.
Hence, II U (z) ~
isincreasing
aslong as ~
U(z) (I 03B40.
Since we are
trying
to provesomething
about t verysmall,
it is natural to consider whathappens
when T==0. Ifi = o, 0 9 2 ~,
andd = (i, 9), then,
from(IF. 5),
and
p (d, z)
isgoverned by
theequation
In
particular,
Note that if r=0 then all values of e determine the same
trajectory y ((o, 6), z).
From continuousdependence
of solutions ofordinary
differen-tial
equations
on initial data we concludeLEMMA 1. - Let any
neighborhood of
A’. Then there exists to such that and0 - 8 2 ~t,
theny ((i, e), Zo)E% for
some zo.This lemma demonstrates that if t is
small,
theny ((i, 9), z)
will pass close to the criticalpoint
A’. Afterpassing
close toA’, y((r, e), z)
willleave a
given neighborhood
of A’ close to the unstable manifold of A’.This will be made
precise shortly.
We first discuss the behavior ofWA,,
the unstable manifold at A’. We first prove
LEMMA 2. - Given
Ao,
there exists Eo such thatif
0 c Eo,Remark. - Note that if
(U (z), V(z), P (z)) E W A’,
thenp (z) =1
for all z.Proof. -
Letand
Vol. 4, n° 6-1987.
We prove that there exists Eo such that if
0 E Eo,
theny(z)eS’
for eachz. This
certainly implies
the desired result.We first prove that there exists Eo such that if
0 E Eo,
then S ispositively invariant;
thatis,
if for some zl, theny(z)eS
for allThis is proven
by showing
that on theboundary
ofS,
the vector fieldgiven by
theright
side of( 1 F . 5) points
into S.There are many cases to consider.
Suppose,
forexample,
thatV 1= ~, U 1, U1 > o,
and Letn = (~,, -1)
be a vectoroutwardly
normal toif Eo is
sufficiently
small. A similarproof
works in the other cases.To
complete
theproof
of Lemma2,
we show that Eo can be chosen sothat if
0 E Eo
andY(Z)EWA,
then there exists zl such thatY(Z)EWA,
for This is
proved by linearizing (IF. 5)
at A’ andshowing
that theeigenvectors corresponding
to thepositive eigenvalues point
into S. Theseeigenvectors
aregiven
as follows. Letand
be the Hessian matrix of F at A. Because
F (U)
has anondegenerate
localmaximum at A it follows that M has
negative eigenvalues,
which wedenote
by - ~,l
and -~,Z.
Thepositive eigenvalues
of( lF . 5)
linearized at A’ are a 1 and a2 where ai, i = 1,2, is thepositive
root of thepolynomial
Let w;,
i =1, 2,
be theeigenvector
of Mcorresponding
Then theeigenvector
of(IF. 5) corresponding
to (Ji isBecause each CJi -~ oo as s -
0,
it follows that each p~points
into S if E isProof. -
LetBy
Lemma2,
there exists Eo such that if 0 E Eo,II
for z zo, Asbefore,
because
U, O F (U) ~ 0
it follows isincreasing
for We assume that Eo 1. Then for E Eo,as
long
andU(Z)EX2. However,
at and(4A. 1) implies that ~V~
is thenincreasing.
for z > zo,U (z) E X2.
Anothercomputation
shows that if z >__ zo, E Eo, and U(z)
EX 2,
then
Since at zo,
we
conclude, by integrating (4A. 2), that ~ U, V)
isincreasing
for z > zo,U(Z)EX2. Theref ore, ~ U, V)
> 0 for z > zo,U(Z)EX2. Finally,
if z > zo,U(Z)EX2,
thenwhich
completes
theproof
of the lemma.Vol. 4, n° 6-1987.
We now return to the
proof
ofProposition
4A .1. Introduce newvariables
(xl,
x2, yi, y2,P)
so that nearA’,
where
Ws
is the stable manifold at A’. In these newcoordinates,
theequations
become nearA’,
where
x2), y2),
and -03B31, 03B32, -03B33,
and y4 are theeigenvalues
of(IF. 5)
at A’.Choose 81
so that then in the oldcoordinates,
By choosing bl smaller,
if necessary, andchoosing A small,
we may assume, from(4A . 3),
thattrajectories
canonly
leave.;V). through N
+.Now
points
in .;V+ lie close topoints
inWA,.
It then follows from Lemma 3 and the continuousdependence
of solutions on initial data.LEMMA 4. - Assume that 0 E Eo where Eo is
given
in Lemma 3. Then A can be chosen so thatif
y(z) = (U (z),
V(z),
p(z))
is a solutionof ( 1 F . 5)
and then there exists such that
U(z)eX2 for z2), U (z2) E aX 2, increasing for
z E(z 1, z2).
We are now
ready
tocomplete
theproof
ofProposition
4A. 1. Choose À so small thattrajectories
canonly
leave.~V’~, through .N’ +,
and thatNow fix
i E (o, to), 9 E [o, 2 ~],
and letd = (r, e).
We prove that there exists zo such that U(zo)
e U(z)
eX 2
for z zo,and ) )
U(z) ~ ~
isincreasing
forz zo. Choose z 1 so that
y ( d,
We havealready
proven that(jU(z)jj is increasing
aslong as I I U (z) I I b. By assumption,
Hence, y (d, z)
must leave~V’~.
Because~y {d, z)
canonly
leave~V’~ through
~V’ +,
there exists such thaty (d, Z2)E%+,
andis increasing
for z z2. The result now follows from Lemma 4.
Let q 1
be apoint
on the top side ofX 2,
and q2 apoint
on the bottomside of
X2.
Lete1 (r)
ande2 (t)
be continuous functions such that for eachi E (o,1),
there exists zi,i =1, 2,
such thatand
We assume, without loss of
generality,
that for each r,81 (i) 82 (i).
Let
B. r and r*
Recall that
if y e Y,
then~y (y, z)
leaves N. LetIt follows from Lemma 2B . 2 that if
yeY, z)
must leave Nthrough
8.Hence,
we have amapping
defined
by
A wherey (y, z)
leaves N. From Lemma 2B . 1 it follows that A is continuous.Let I be the unit interval and % the set of functions g : I -
Yo
such that(a) g
iscontinuous;
Vol. 4, n° 6-1987.
Note that 8 is
topologically equivalent
to an annulus with four holes removed. Forg E,
we shall define twoalgebraic objects, r* (g)
andr (g),
which indicate how the curve
( A . g) ( I)
winds around the four holes.They
will be elements of
F4,
the set of words on the four letters a,P,
y, and 8.We
begin
with the same notation. For convenience we assume thatN1
is the square:
Let
Assume that Choose E
[0, 1],
k =1, 2,
...,K,
such thatWe now define r*
(g, r~ *).
First we defineand
These
quantities
are determinedby
thefollowing
Table. In theTable,
welet,
for se[o,1],
crosses
03A6(~k)~ 03A6(~k+1)~ for
~ E ~~k~ ~k+ 1)
1
la
a 1~i ~2 C a - i
’~ 2 ~3 3
lj
P 1tf 2
~
P - 1~3 ~4
ly
Y - I~4 t~3
~
Y 1~4 ~1
1/
8 1~1 ~4 l+03B4 8 - 1
For each case not shown in the chart we let For this case we do not define
~,k = ~, (~k) because,
as we shall see, sinceek = 0
the choiceof 03BBk
doesn’t matter. Then defineHere is an
example.
In
Figure 2,
For thisexample
Note that there may be cancellations in
r* (g, r~ *). By
we meanthe element of
F~
obtainedby making
all cancellations inr* (g, r~ *).
Weshall see later that does not
depend
onr~ *.
In the aboveexample
Vol. 4, n° 6-1987.