A NGÉLICA M ANSILLA M ARIANA S AAVEDRA
The period function near a polycycle with two semi-hyperbolic vertices
Annales mathématiques Blaise Pascal, tome 8, n
o1 (2001), p. 93-104
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THE PERIOD FUNCTION NEAR A POLYCYCLE WITH TWO SEMI-HYPERBOLIC VERTICES
Angélica
Mansilla and Mariana Saavedra*ABSTRACT. Let P be a polycycle of an analytic vector field on an open subset of the plane Suppose that P is the union of two semi-hyperbolic singular points
(vertices
ofP)
connected by two trajectories(sides
ofP).
Assume thatone side is part of the center manifold of each vertex. Denote by L the other side.
Assume also that P is a boundary component of an annulus of periodic orbits. Let E be a Poincare section at the polycyle that intersects L. We show that the period
function defined on E has a principal part of the form
kx-n, k
>0,
n E N.1. INTRODUCTION
Consider an
analytic ordinary
differentialequation E
on an open subset of theplane IR,2. Suppose
that E has an annulus ofperiodic orbits,
notnecessarily
bounded. It is knownthat a
boundary component
union of such an annulus is apolycycle
P(cf. [P]);
that is afinite connected union of
singularities (vertices
ofP)
andintegral
curves(sides
ofP)
ofE. A
unique singular point
may be considered as apolycycle.
Let E be a small Poincaré section at thepolycycle
with a local coordinate s whoseorigin
lies at thepolycycle.
Theintegral
curve of E that passesthrough
apoint
of E is aperiodic
orbit. Theperiod function assigns
to s the(minimum) period T(s~
of thecorresponding periodic
solution.We are interested in the
qualitative
behavior ofT, mainly
in theasymptotic expansion
of T and of its
derivative,
for smallargument.
The fact that apolycycle
has aperiod
function with an
asymptotic expansion,
and also theexpansion itself,
areproblems
ofgeneral
interest.Also,
theoscillatory
character of T is of interest to us. We say thata function is
oscillatory
if the set of its criticalpoints
has accumulationpoints.
Thederivative of an
oscillatory
function either does not have anasymptotic expansion
or it hasan
asymptotic expansion identically
zero.* The research was partially supported by Dirección de Investigacion, Universidad de Concepcion, proyecto P.I.No 96.015.009.1-1. AMS subject classification: 34C, 53F
The behavior of T and of its
derivative,
as itsargument approach
zero,depends
onwhether the
polycycle
is bounded or not, and also on the type of its vertices. We would like to know how theanalytic
local invariants of thesesingular points
intervene in the behavior of such a function.The
period
function T isanalytic
at everystrictly positive
coordinate s and it isanalytic
at theorigin only
if thepolycycle
is anondegenerate
center.Generally,
both Tand its derivative grow without bound when their
arguments approach
zero. Boundedpolycycles
havenonoscillatory period functions, [C-D].
On the otherhand,
in[S, Sa]
itis
proved
that if the vertices of apolycycle (bounded
ornot)
areformally
linearizable afterdesingularization,
then T and its derivative haveasymptotic expansions
in{s }
andE IR. A consequence if such a
polycycle
is unbounded and has a finite vertex then it has anonoscillatory period
function.In this work we are interested in a class of
polycycles
withsemi-hyperbolic
vertices.We determine the
principal
part of theperiod
function for boundedpolycycles
in this class.More
precisely,
we considerpolycycles
with twosemi-hyperbolic singular points
asvertices with a side that lies in the center manifold of both vertices. We call such a side the center side of the
polycycle.
The other side is calledhyperbolic
side. We prove that theperiod
function defined on a transversal sectionthrough
thehyperbolic
side is of theform
T(s)
= +o(s)), k
>0,
n E IN.To prove this we
decompose
theperiod
function in local time functionsthrough
thepoly- cycle
sides and the saddle sectors. Theperiod
function is the sum of each of these local time functionscomposed
on theright
with anappropriate
transition map.2. PRINCIPAL PART OF T
Consider the
analytic
differentialequation E : dt
=A(x, y)
,d
=B( x, y)
on anopen subset of the
plane.
Apolycycle
P of E is a connected unionconsisting
of a finitenumber of
singularities
of E(vertices
ofP)
and theintegral
curves of E(sides
ofP)
suchthat a unilateral return map R
exists,
thatis,
there is ananalytic
curve~(~)EP
transverse to P such that the
integral
curvethrough ,(s)
intersects Eagain
for the firsttime at for each
sufficiently
small s.The
period function
T:~0, E~--~ R+,
is defined when R is theidentity
map,given by
s ~
T(s)
as theperiod
of theperiodic
orbitthrough 03B3(s).
The function T is
analytic
on]0, f[. ’But
T is notnecessarily
defined oranalytic
ats =
0,
andT (s)
may converge toinfinity
as s ~0+.
Next,
consider apolycycle
P with two vertices.Suppose
that such vertices are semi-hyperbolic singular points
of E(that is,
if a is a vertex then the Jacobian matrix of(A, B)
at a has one
eigenvalue equal
to zero and the other one diflerent tozero).
We suppose, moreover, that one side of P(center side)
is contained in the center manifold of each vertex(see Fig. 1).
ThereforeFig.1
Theorem. Let E be a Poincaré section at P such that E intersects the
hyperbolic
side.The
period function
Tdefined
on Esatisfies
T(s)
= +~~s)~~
where k >
0,
n 6 IN withe(s)
~ 0if s
~0+.
Proof. The orientation of the
trajectories
of E defines a sense of direction of P. Let a1, a2 be the vertices of P and letL1
andL2
be thehyperbolic
side and the center siderespectively,
such that the end of L1 and the start of L2 is al(see Fig. 2).
We assumethat a~ is crossed
by running
firstthrough Li
and thenthrough L2.
On aneighborhood
of each vertex
a; (a =1, 2)
of P, we choose twoanalytic
semi-transversalswhere pi, p2 E
L 1
and qi , q2 EL2.
. We choose E =E1.
..Define the functions
91:~~, aI [-~ R+ given by
s egi(s)
andS
:~0, ~1 [--~ R+ given by
s ~S(s),
such that the
positive
semi-orbit of Ethrough
03B31(s) (s ~ 0)
intersects03A01 - {q1}
at 03C01(g1(s))
and
03A32 - {p2}
at03B32(S(s)).
Define also the functions
7i
:~0, ~’I [--~ R+ given by
s Hol (s) ,
,Ti
:]0, ~1 [~ IR+ given by
It H1(03BA)
and
T2
:~o, ~2 [--~ ~.+ given by ~ H T2~) ,
, .where,
01(s)
is the timerequired
for theintegral
curvestarting
at 03B31(s)
to intersect thetransversal
II1
for the first time at ~rl~gl ~s)),
the number rl(r~)
is the timerequired
for theintegral
curvestarting
at 03C01(03BA)
E03A01
tointersect,
for the first time the transversal03A02,
andT2
(~)
is the timerequired
for theintegral
curvestarting
at y2~~)
EEZ
tointersect,
for thefirst
time,
the transversalE~
atNext,
consider theequation
E* :
dx dt
= -A( x,y);dy dt
= -B( x,y)
and define the function 03C32
:]0, ~2[~ IR+ 03BE
e03C32(03BE) ,where 03C32(03BE)
is the timerequired
forthe
integral
curve of E*starting
aty2 (~)
E E2 to intersect the transversalII~
for the first time at~r2 (gz (~)) .
The functions o~(i = 1, 2)
are called the corner passage timefunctions
relative to 03A3i and 03A0i.
Therefore,
the function T isgiven by
T(s)
= + +oz (s(S))
+r2(S(S)) (*)
Fig. 2
The functions Tl and T2 are
analytic
at zero with 0 andT2 (o) ~
0. Ineffect,
consider the function ri. From the Flow Box
Theorem, analytic
coordinates(u, v)
existon a
neighborhood
of the sideL2
of P such that the axis v = 0 is the sideL2,
whilethe semitransversals
Ili
and112
aregraphs
of theanalytic functions,
k and1,
on v > 0(the
functions k and I are defined andanalytic
on aneighborhood
of zero, and thepoints (k(o), o)
and(1(0), 0) correspond
to qi and q2respectively).
In thesecoordinates,
X(the
vector field associated with the differential
equation E)
becomesX =
A1(u,v)~ ~u,
where
~4i
is astrictly positive analytic
function. We obtain that the timerequired
for theintegral
curvestarting
at(k(v), v)
to intersectII2
at(I(v), v)
is theintegral
(v)
=l(v)k(v) 1 A1(u,v)
du .Thus T is an
analytic
function with 0. Since the coordinatechange x
~ v isanalytic
and fixes zero, we obtain that Ti is an
analytic
function on aneighborhood
of zero with0. In the same way T2 is an
analytic
function withTZ (o) ~
0.Since
g1(0)
= 0 andS(0)
= 0(with
gi and Sdefined,
at zero,by continuity),
theprincipal
parts of andr2(S(s))
are1(0)
andT2(o), respectively.
Now,
from(*),
it remains to evaluate theprincipal
part of vi(s)
+v~(S(s)).
For thatpurpose we shall need the
following propositions (to
beproved
in theappendix).
Proposition
1. There exist apositive integer
n, such that03C31(s) = k1s-n
+o(s-n)
and
03C32(03BE)
=k203BE-n
+o(03BE-n),
where
k1
andk2
arestrictly positive
numbers.Proposition
2. Thefunction
S has anasymptotic expansion
inss
and s~‘(log s)m,
where~ and ~ are
strictly positive
rational numbers and m is astrictly positive integer.
Moreprecisely,
there exists a C°°function Sl
on aneighborhood of (0, 0)
such that51 (o, o)
= 0and
S(s)
= c .s ( 1
+S1(s,
snlogs»,
where c > 0 with n is
given by Proposition
1.Hence,
from(*)
and thePropositions
1 and2,
it follows thatT(s) = k1s-n
+1(0)
+ k2c-ns-n +2(0)
+o(s-n).
Since n > 0 we have
T(s)
= ks-n +o(s-n),
with k > 0. The last
equality
constitutes the main result of thepresent
work..Remark. If the transversal section E intersects the center
side,
the calculations are morecomplicated;
in this case, we do not know how to calculate the function S defined on E.APPENDIX
I. Proof of
Propositions
I and 2On a
neighborhood
of ak(1~ =1, 2~,
there exist C°° coordinates(xk, Yk)
such that theorigin
is ak, and the semi-axes xk =0,
yk > 0 and xk >0,
yk = 0 are the sides of P. In the coordinates(x1, y1),
theequation
E isgiven by
~ ’
~-=~~/l(~l~l) ; ~=-~(~+~~)/l(~l~l) (~)
where
/i
is astrictly positive
C~function,
with ~i N andAt
]R(cf. [M]).
We canconsider that
03A31
and03A01
are thesegments
0 x11,
y1 = 1and x1
=1, 0 ~
y11, respectively.
In the coordinates
(a’2)2/2)
theequation
E*(recall
that E* =- E)
isgiven by
~ : "
~=~+~(~~,) ;
"~=-I/2(~+A2~)/2(~2~2)
where ~
is astrictly positive
C~function,
with ~2 IN and~2
IR. We supposeagain
that
03A32
and112
are thesegments
0 x21,
y2 = 1 and x2 ===1,
0 1/2I, respectively.
1
Proof c/ Proposition
Jf. The definitions of 01 and 03C32 aresimilar,
therefore we shallonly
consider the
equation
2? and omit the subindex 1 of the coordinates andAi.
From the
equation (**),
we deduce that the comer passage timefunction,
deRned onthe transversal
0~1,
the lineintegral
~’’~?-~(~
where 03B3x1 is the orbit arc of E that
joins (x1, 1)
to thepoint (1, y1 (x1)) 03A01.
A first
integral
ofequation (**)
ishence 03B3x1 is done
by
theequation y
=1 xn1),
x1 ~ x 1. We obtain theequality
~i)
=/’
~~20142014iF(~(~-)~exp(-~ - -L))~ ,
~ ~ a:~ 0 1, where jF= -
We know that two C~ functions
~
and jF2 exist onR~
such thatF(x, y)
= + +yF2(x, y).
Therefore
03C3(x1)
= F(0,0) nx-n1 - F(0,0) n
+ H1(x1)
+ H2(x1),
where
H1(x1) = 1x, x-nF1(x, x1 x)03BB exp(1 xn - 1 xn1))dx
zi x x i
and
H2(x1)
= x03BB1exp(-1 xn1) 1x1
xl ~1 x-n -03BB-1exp( 1 xn)F2(x,(x1 x)03BB exp(1 xn -
x x x1 xn1))dx.
iSince the set
{(x,(x1 x)03BB exp(1 xn - 1 xn1)) |
x1 ~ x ~1,
x1~]0,1[}
is contained in
~0,1~
x~0, l~,
a constant .k > 0 exists such thatxi a
1 _
II Fl(x~ ("")
x xxl for all x1
~]0,1[
and x E[x1,1]. Consequently
| xn1 H1(x1) |=| xn1 1x, x-nF1(x, (x1 x)03BB exp(1 xn - 1 xn1))dx |~ Kxn1 1x1
x-ndx.Xl x x xi xl
VVe conclude that =
o(x-n1). By
a similarcalculation,
we haveH2(x1)
=o(x-n1).
Thus,
Q
( x1)
= F(0,0) n x-n1 + 0(x-n1)
.Since the C°° coordinate
change
s H xl fixes zero, we obtain that Q1(s)
satisfies theequality
03C31
(s) = k1s-n
+o(s
-n),
with
k1
> 0.In the same way,
03C32(03BE)
=k203BE-n2
+o(03BE-n2),
with
k2
> 0. This provesProposition
1 ..2
Proof of Pmposition
~. Leth(K)
bethe
number such that theintegral
curvethrough
E
lI~
intersects the transversalII2
for the first time at~r~(h(~)).
The function h isa
strictly increasing analytic
function withh(0)
= 0. Recall that theintegral
curve of E*through ~y2 (~)
EE2
intersects the transversalII~
at~rZ (gz (~)).
This defines the function~
~g2 (f). Thus,
the function S satisfies theequality
S = g-12 o h o g1.
Next,
consider theexpression
of the functions g1, g2 and h in the coordinates x1, x2, y1 and y2. Thatis,
gl : xl ~ yl, g2 : x2 ~ y2 and h : y1 e y2.
Hence,
the function S isgiven by
S = g-12 o h o g1 :
To find the
expression
a?2 =S(xl ),
consider theequality
g2(x2)
=Recall
that,
on aneighborhood
of ai, a firstintegral
ofequation
E isI(x1,y1)
= y1.x03BB11 exp(-1 xn11).
So that y1 =g1(x1)
is the solution of theequation
I(1,y1) = I(x1,1).
Thus,
g1(x1)
isgiven by
g1(x1) = x03BB11 exp(1 - 1 xn11).
Similarly,
forg2(x2) :
g2(x2) = x03BB22 exp(1 - 1 xn22).
Moreover, since h : yl ~ y2 is a
strictly increasing
C°° function withh(o)
= 0(y~
and yz are C°° coordinates of II1 andIIz respectively,
and the coordinateschange
arestrictly increasing),
there exists a C°° functionh~
such that = 0 andh(y1)
=03B2y1(1
+h1(y1)),
with
p
> o.Hence,
theequation g2(x2)
=( h
og1)(x1)
isequivalent
tox03BB22 exp(1 - 1 xn22)
=03B2x03BB11 exp(1 - 1 xn11 )(1
+h1(x03BB11 exp(1- 1 xn11 )).
Applying
thelogarithmic function,
we obtain03BB2 log x2 - 1 n2
=log 03B2
+03BB1 log x1 - 1 n1
+~(x1) (1)
x2 x~ I
where
~(x1)
=10g(1
+h1(x03BB11 exp(l - 1 xn11)))
is a C~function,
which is flat at zero(that
is,
itsTaylor
series at zero isequal
tozero).
Consider the variable z
through
nz^~..
.
x2
2
1 + z
~
From
( ~ ),
we obtain03BB2n1xn11 log x1 - 03BB2xn11 log(1
+z)
- n2z =n2xn11 log 03B2
+03BB1n2xn11 log x1
+Next, put tv =
xn11 log
xl and consider the function G defined asG(x,
cu,z)
=03BB2n1w - 03BB2xn11 log(1
+z) -
n2z -n2xn11 log 03B2 - 03BB1n2w
-n2xn11 ~(x1).
The function G is C°° on a
neighborhood
of theorigin (xl, w, z) - (o, o, o),
, w ith .G(0, o, o)
= 0 andaG (0, 0, 0)
=-n2 ~
0.Thus,
fromthe Implicit
Function Theorem,there exists a C°° function z =
z(xl, w),
defined on aneighborhood
of(xl, w)
=(o, o)
such that
z(0, o)
= 0 andC(xl, w, x(xl, w))
= 0.Therefore,
~~)=l+~~log~)’ xn11
.From
this,
it follows that the function S isgiven by
S(x1) =
xn21 ( 1
+ vi(x1, xn11 log x1)) (3)
where vi is a Coo function on a
neighborhood
of(0,0)
such that = 0.Now,
the coordinatecha,nge s ~
a*i is a Coodiffeomorphism
on aneighborhood
ofzero and fixes zero, that
is,
x1 =c1s(1
+b(s)),
where c1 > 0 and b is a C~ function withb(0)
= 0.Hence, substituting
thisexpression
in(3),
we obtain that the function S satisfies theequality
5(s)
= c.sn1 n2 (1 + v(s,sn1 log s)),
where c > 0 and v is a C~ function with
v(0,0)
= 0.Note that the return function R is the
composition
of thefunction
Splus
a C°°diffeomorphism
fromE2
toE1.
Since R is theidentity
map it follows that ni = n2 and thereforewhere c > 0 and v is as above. This proves the
Proposition
2 and theproof
of the main result is nowcomplete..
Remark. If the
polycycle
have anarbitrary
number ofvertices,
it seemspossible
to find aprincipal
part of theperiod
function. Thecomposition
of the corner passage functions(gl
and g2 in the present
work),
involves thecomposition
ofexponential
functions.Therefore,
the
asymptotic expansion
of the return function(function
Shere)
is not anymore in{s }
and
~s~~ logs.
This case will be theobject
of another paper.REFERENCES
[A]
V. I. ARNOLD."Ordinary
differentialequations",
R. A.Silverman, translator,
MITPress, 1978.
[C-D]
C.CHICONE and F. DUMORTIER. "Finiteness for criticalperiods
ofplanar analytic
vector
fields",
Nonlinear Anal.Theory,
MethodsAppl. 20,
n°4, (1993),
315-335.[M]
R. MOUSSU ."Développement asymptotique
del’application
retour d’unpolycycle".
Lectures Notes in
Math.,
n° 1331(1989).
[P]
L. M. PERKO. "On the accumulation of limitcycles",
Proc. Amer. Math.Soc.,
99(1987)
515-526.[S]
M.SAAVEDRA."Développement asymptotique
de la fonctionpériode".
Thesis Doc-tor’s
degree
of theBourgogne’s University, 1995.
[Sa]
M. SAAVEDRA."Développement asymptotique
de la fonctionpériode".
C. R. Acad.Sci.
Paris,
t.319,
SérieI,
p. 563-566(1994).
Departamento
de MatematicaFacultad de Ciencias Fisicas y Matematicas Universidad de
Concepción,
ChileDepartamento
de Matematica y Estadistica Facultad deIngeniería
Universidad de la
Frontera,
ChileE-mail address:
masaaved@gauss.cfm.udec.cl
Fax number 56-41-251529.