SUB-RIEMANNIAN SPHERE IN MARTINET FLAT CASE
A. AGRACHEV, B. BONNARD, M. CHYBA, AND I. KUPKA
Abstract. This article deals with the localsub-Riemannian geometry on R3 (D g) where Dis the distributionker !, !being the Martinet one-form: dz; 12y2dx andg is aRiemannian metricon D :We prove that we can takeg as a sum of squaresadx2+cdy2:Then we analyze theat casewherea=c= 1:We parametrize the set ofgeodesicsusing elliptic integrals. This allows to compute the exponential mapping,the wave front,theconjugate and cut loci,and thesub-Riemannian sphere.
A direct consequence of our computations is to show that the sphere is not sub-analytic. Some of these computations are generalized to a one parameter deformation of the at case.
1. Introduction
In this article we consider the sub-Riemannian geometry (
MDg
) whereM
is the real analytic manifold R3D
is the distribution ker! !
being Martinet one-formdz
;12y
2dx
whereq
= (xyz
) are the coordinates ofR3 andg
is an analytic Riemannian metric onD:
Since onD dz
= 12y
2dx g
can be writtena
(q
)dx
2+ 2b
(q
)dxdy
+c
(q
)dy
2:
Whena b c
are not depending onz
the problem is said isoperimetric.An admissible curve is an absolutly continuous curve
: 0T
] 7;! R3 such that _(t
) =d
(t
)dt
2D
((t
))nf0g for almost everyt:
Let () be the scalar product dened byg:
The length of an admissible curve is:L
() =Z T0
(_
(t
)_(
t
))12dt
and the energy of is:E
() =Z T
0
(_
(t
)_(
t
))dt:
Let
q
0q
12R3a minimizing curve joiningq
0 toq
1 is an admissible curve : 0T
]7;! R3 joiningq
0 toq
1 with minimal length and the length of denes the sub-Riemannian distanced
SR betweenq
0 andq
1:
Andrei Agrachev, Steklov Mathematical Institute, 42 Vavilova Street, GSP-1, Moscow 117 966 Russia. E-mail: agrachev@mi.ras.ru.
Bernard Bonnard, Universite de Bourgogne, Laboratoire de Topologie, UMR 5584 du CNRS, BP 400, 21004 Dijon Cedex France. E-mail: bbonnard@u-bourgogne.fr.
Monique Chyba, Section de Mathematiques, Universite de Geneve, 2-4 rue du Lievre, Case 240, 1211 Geneve 24, Suisse. E-mail: chyba@sunny.unige.ch.
Ivan Kupka, Universite de Paris VI, Departement de math., 4, place Jussieu, 75232 Paris Cedex 05, France. E-mail: kupka@mathp6.jussieu.fr.
Received by the journal November 22, 1996. Revised November 10, 1997. Accepted for publication November 28, 1997.
c
Societe de Mathematiques Appliquees et Industrielles. Typeset by LATEX.
In the Martinet at case, a standard existence theorem due to Filippov 26] and the maximum principle 32] allows to show the existence of such minimizer in the analytic category. Hence in the sequel we shall restrict our study to analytic admissible curves.
Let
F
1F
2 be two analytic vector elds such thatD
= SpanfF
1F
2g:
Hence an admissible (analytic) curve is solution of the control system:d
(t
)dt
=2
X
i=1
u
i(t
)F
i((t
))where
u
: 0T
]7;! (u
1(t
)u
2(t
)) is the unique analytic control associated to:
The two following remarks can save a lot of computations in sub- Riemannian geometry. First, the length of a curve is not depending on its parametrization. Hence, we can assume the curves parametrized by arc- length(_
(t
)_(
t
)) = 1and in this case, the length minimizing problem is equivalent to the time optimal control problem. Secondly, if all the curves are dened on the same interval, the length and energy minimizing problems are equivalent.
To carry out the computations of minimizers, it is convenient to use max- imum principle. According to this principle, minimizers can be selected among a restricted set of curves. Let us introduce the Hamilton function:
H
(qpu
) =hpF
(q
)u
i;(F
(q
)uF
(q
)u
)where
p
= (p
xp
yp
z) 2 R3 is a constant equals to 0 or 12, (p
) is non zero,hi is the standard scalar product andF
(q
)u
denotesP2i=1u
iF
i(q
):
A bi-extremal is an absolutly continuous curve (p
)dened on 0T
] where (pu
) is solution for almost everyt
of the following equationsd dt
=@H
@p
(pu
)dp dt
=;@H
@
(pu
) (1.1)@H
@u
(pu
) = 0:
(1.2)Its projection
on R3 is called a geodesic. From the maximum principle, a minimizer is a geodesic.When we study the bi-extremal curves, we must distinguish between two cases. If
= 0, the bi-extremal is called abnormal and if 6= 0it is called normal and the associated geodesics are respectively called abnormal and normal. A geodesic is called strictly abnormal (respectively strictly normal) if it is the projection of an abnormal (respectively normal) bi-extremal but not of a normal (respectively abnormal) one.In the normal case, equation (1.2) can be solved as follows. Since it is linear with respect to
u
, the control solution can be computed as an analytic maping ^u
: (p
)7;! R2:
We plug ^u
inH
and we setH
n(p
) =H
(p
^u
)where= 12:
Now using (1.1) and (1.2), we observe that the bi- extremal (p
) is solution of the analytic dierential Hamiltonian equationd dt
=@H
n@p
(p
)dp dt
=;@H
n@
(p
):
(1.3)Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
This computation is straightforward if
F
1F
2 are taken orthonormal. If we setP
i = hpF
i(q
)i fori
= 12the Hamilton functionH
n takes the form1
2(
P
12+P
22):
Similarly, the computations of abnormal bi-extremals is straightforward.
When
= 0, the constraint (1.2) ish
pF
i()i= 0i
= 12 (1.4) and if we dierentiate twice equation (1.4), we get that abnormal geodesics are contained in the set det(F
1F
2F
1F
2]) = 0 which is the planey
= 0 corresponding to the points where!
is not a contact form and are the linesz
=z
0:
The sety
= 0which has an important geometric meaning is called the Martinet surface.Consider now arc-length parametrized curves. We x
q
0 2 R3 and let (tq
0p
0)p
(tq
0p
0) be the solution of (1.3) starting att
= 0 from (q
0p
0):
It is contained in the level setH
n= 12:
The exponential mapping is the map:expq0 : (
p
0t
)7;!(tp
0q
0):
The point
q
1 is said to be conjugate toq
0 along if there exists (p
0t
1) such that (t
) = expq0(p
0t
)q
1 = expq0(p
0t
1)t
1>
0 and the exponential mapping is not an immersion at (p
0t
1):
The conjugate locusC
(q
0) is the set of rst conjugate points along the curves when we consider all the geodesics starting fromq
0. Observe that the exponential mapping is dened onC
RwhereC
is the cylinder parametrized byP
12+P
22= 1 atq
0ifF
iare taken orthonormal. The non compact nature of
C
is the main problem when we study the exponential mapping. Let be a geodesic corresponding to a normal or an abnormal bi-extremal and starting fromq
0:
The rst pointq
1 6=q
0 where ceases to be minimizing is called the cut point and the set of such points when varies form the cut-locusL
(q
0):
The sub- Riemannian sphere with radiusr >
0 is the setS
(q
0r
) of points which are at sub-Riemannian distancer
fromq
0:
The wave front, of lengthr
, is the setW
(q
0r
) of end-points of geodesics with lengthr
starting fromq
0:
One of the main problem in sub-Riemannian geometry is the study of the exponential mapping in order to give estimates and a geometric description of the conjugate and cut loci, of the sphere and the wave front. Related works to this problem are the following. In a pionnering contribution 11], Brockett has analyzed the Heisenberg case, where
M
is the 3-dimensional Heisenberg group,D
is a left invariant contact distribution andg
is left invariant. Very recently, the sub-Riemannian sphere with small radius was computed, whenD
is ker! !
being a contact form in R3andg
is a generic metric 2], 17]. It appears that the general case is a perturbation of the Heisenberg case. The object of this article is to pursue the analysis by considering the Martinet situation whereD
= ker! !
beingdz
;12y
2dx:
The main results of this article are threefolds.
First we clarify the geometry of the problem by computing a normal form for the pair (
Dg
)the group actionG
being induced by the following two set of transformations(i)
q
7;!Q
='
(q
) where'
is a germ of analytic dieomorphism on R3 preserving 0Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
(ii) feedback transformation of the form
u
7;!v
= (q
)u
whereq
7;! (q
) 2GL
(2R)is a germ at 0 of an analytic mapping, and is preserving the metricg:
IfD
= SpanfF
1F
2g whereF
1F
2 are orthonormal vector elds, (q
)2O
(2) the set of orthogonal matrices.We can formulate a rst main result.
Theorem 1.1. Let (
Dg
) be such thatD
= ker! !
=dz
; 12y
2dx
andg
is an analytic germ at 0 of a Riemannian metric onD
:a
(q
)dx
2 + 2b
(q
)dxdy
+c
(q
)dy
2:
Then under the action ofG
we may assumeb
0 anda
(0) =c
(0) = 1:
Technically the proof of this result is relevant for two reasons. First, we need to parametrize the set of germs of dieomorphisms preserving the distribution ker
!
and we improve a similar result obtained in 33], where only those tangent to the identity are computed. This parametrization is important to any geometric problem dealing with Martinet type distribu- tions. Secondly, we show that the problem of computing our normal form is related to a Cauchy problem for a singular partial dierential equation of the Briot-Bouquet type. This type of problem is studied in 18] and it allows to use a Cauchy-Kowaleska type theorem to get a convergent normal form in our situation.In this normal form, the distribution is entirely normalized. Let us in- troduce the two vector elds:
F
1 =@
@x
+y
2 2@
@z
andF
2 =@
@y
such thatD
= SpanfF
1F
2g:
Hence the length ofF
1F
2 is given by(
F
1F
1) =a
(F
2F
2) =c
and we get two orthonormal vector elds by settingF
e1 =F
1=
pa F
e2 =F
2=
pc
and the two vector elds (
F
e1F
e2) are a representation of (Dg
):
Using the following gradation for the variables: the weight ofx y
is one and the weight ofz
if three, the associated weights for vector elds being;1 for @x@ ,@y@ and;3 for @z@ , we get a gradated normal form using (
F
e1F
e2):
With this gradation, the at caseg
=dx
2+dy
2 corresponds precisely to collect only the terms of weight -1 in the normal form. Hence in terms of normal form, the at case in the Martinet case is the equivalent of the Heisenberg case when we analyse the contact situation as in 2], 17].Therefore, the Martinet at case has to be carefully studied. The second part of our work is to get a complete descrition of the conjugate and cut loci and of the sphere in this situation. We prove
Theorem 1.2. The intersection of the conjugate-locus
C
(0) and the cut- locusL
(0) is empty. The cut locusL
(0) is the Martinet surfacey
= 0 minus the abnormal geodesicz
= 0:
The intersection of the sphereS
(0r
)r >
0 with the Martinet surfacey
= 0 is a closed cuvek
7;!c
(k
) around 0 which is the union ofL
(0) intersected with the sphere and two pointsx
=r z
= 0 representing the trace of the abnormal geodesic on the sphere. The graph ofc
is not sub-analytic. Hence the sphere and the sub-Riemannian distance are not sub-analytic.Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
This result illustrates the main dierence between the contact and the Martinet case. In the Martinet case there exist abnormal geodesics. One object of this article is to prove that abnormal geodesics will cause a log- arithmic singularity for the exponential mapping. The main technical tool to handle this problem is to use models where the set of geodesics can be computed by quadratures. In the Martinet at case we need elliptic inte- grals. It is a precise measure of the transcendence of the problem. Using the catalogue of properties and estimates concerning elliptic integrals, we shall be able to estimate the conjugate-locus and to compute the cut-locus.
Indeed our study will show that the cut-locus has a singularity similar to the singularity of Poincare-Dulac return mapping for planar algebraic dieren- tial equations when computed for a section to a separatrix connecting two saddles 22].
The Martinet at case is not a stable model to understand in general the role of abnormal geodesics in sub-Riemannian geometry when
D
is a Martinet type distribution. Indeed in the at case an abnormal geodesic is not strictly abnormal. Hence a third contribution of our work is to build a one-parameter deformation of the at case (D
"g
") where for"
6= 0each abnormal geodesic is strictly abnormal. The deformation is constructed in order to get the set of geodesics integrable by quadratures. If we require to have a parametrization of geodesics with the lowest transcendence (this shall be precised later), the model chosen is(
D
"g
") whereD
" = SpanfF
1F
2gF
1= (1+"y
)@
@x
+y
2 2@
@z
andF
2 =@
@y
and the metricg
" is dened by takingF
1 andF
2 as orthonormal vector elds. In this model the Martinet surface is the sety
= 0when jy
j<
j2"j and the abnormal geodesics are the linesy
= 0z
=z
0. Ifis the abnormal geodesic starting from 0 and parametrized by arc-length:t
7;!(t
00) we prove the following result.Theorem 1.3. Let
M
0. Then the abnormal geodesic :t
7! (t
00) isC
0-isolated in the set of geodesics of length less thanM
.A similar result has been obtained in the unpublished preprint 23], but we use a simpler model which shortens the proof. Also we evaluate the exponential mapping in the neighborhood of the abnormal geodesic. In particular we show the persistence for
"
6= 0 of the logarithmic term which causes the fact that the sphere is not sub-analytic in the Martinet at case.Besides it will allow to evaluate the trace of the sub-Riemannian sphere with the Martinet plane
y
= 0near the abnormal geodesic.These three theorems are a rst step in order to understand the general case. Our approach is a mechanical approach to the problem founded on an analysis of the whole set of geodesics which are solutions of a dierential equation integrable by quadratures. This approach has the advantage to select geodesics which play a central role in our analysis. Of course, it has to be completed by purely analytical methods to analyze the general situation. It will be the purpose of a series of forthcoming articles.
The organisation of this article is the following. In section 2, we compute a normal form for the sub-Riemannian geometry (
Dg
)whereD
is a MartinetEsaim: Cocv, December 1997, Vol. 2, pp. 377{448
type distribution. In section 3 we present two geometric properties which are elementary but crucial in the eective computation of the conjugate locus. In section 4, we analyse the Martinet at case: we parametrize the set of geodesics and we evaluate the conjugate locus. Then we describe the cut-locus and the sphere. The section 5 is devoted to the analysis of a one-parameter deformation of the at case. Endly an appendix contains commented numerical simulations concerning the conjugate locus and the sub-Riemannian sphere using algorithms related to our work.
2. Sub-Riemannian geometry with a Martinet type distribution
The analysis of a geometry contains three steps: compute a complete set of invariants, nd a normal form and analyze the structure of the orbits set.
The section is devoted to the rst two steps in a sub-Riemannian geometry with a Martinet type distribution.
2.1. Geodesics
Definition 2.1. Consider a control system dened on an open set
U
Rn:dx
(t
)dt
=m
X
i=1
u
i(t
)F
i(x
(t
)) (2.1) wherex
= (x
1:::x
n) are the coordinates in Rn fF
1:::F
mg arem C
1 independant vector elds onU
and the set of admissible controlsU is the set of smooth mappingsu
: 0T
(u
)] ! Rm:
LetD
be the distribution SpanfF
1:::F
mg:
A smooth sub-Riemannian metricg
is aC
1positive def- inite quadratic form onD:
To compute geodesics it is convenient to assume that theF
i are taken orthonormal. The length and the energy of a smooth admissible curve are thenL
() =Z T
0 Xm
i=1
u
2i(t
)12dt E
() =Z T
0
m
X
i=1
u
2i(t
)dt:
To compute the geodesics we shall use the following notations:
T
U
is the cotangent space and (xp
) its coordinates, wherep
= (p
1:::p
n):
The Liouville formis the form onT
U
:=Pni=1p
idx
iandT
U
is endowed with its canonical structure dened by the two form!
=d:
LetH
:T
U
7;!R be a smooth function,~H
denote the Hamiltonian vector eld dened byi
~H(!
) =;dH:
IfH
1H
2 are smooth functions onT
U
the Poisson-bracket isdH
1(~H
2) =fH
1H
2g:
Having restricted
U
if necessary, letfF
m+1:::F
ngben
;m
smooth vector elds onU
such that for eachx
2U
SpanfF
i(x
)i
= 1:::n
g =T
xU
tangent space atx:
We introduce then
functions onT
U P
i =hp F
i(x
)i where hiis the standard inner product and we setH
(xpu
) =Xmi=1
u
iP
i+Xmi=1
u
2iwhere
is a scalar equal to 0 or;12 and the vector (p
) is non zero.Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
Definition 2.2. A (smooth) bi-extremal is a curve
t
7;! (x
(t
)p
(t
)) onT
U
such that (x
(:
)p
(:
)u
(:
)) is solution of the following equationsdx dt
=@H
@p dp
dt
=;@H
@x
(2.2)@H
@u
= 0:
(2.3)It is called normal if
=;12 and abnormal if= 0:
Its projection on thex
- space is called a normal (resp. abnormal) geodesic. A geodesic can be both normal and abnormal. If it is a projection of an abnormal bi-extremal but not a normal one it is called strictly abnormal. It is called strictly normal if it is not abnormal.Proposition 2.3. Let(
xp
) be a normal bi-extremal. Then the correspond- ing control can be writtenu
^ = (P
1:::P
m) and the curve (xp
) is solution of the Hamiltonian vector eld~H
n whereH
n= 12Pmi=1P
i2:
HenceH
n is a rst integral and arc-length parametrized bi-extremals are contained in the level setH
n= 12:
Proof. From @H@u = 0 =;12
we getu
i =P
ifori
= 1:::m:
Replacingu
i byP
i inH
we obtainH
= 12Pmi=1P
i2:
Since @H@u = 0 the curvet
7;!(x
(t
)p
(t
)) is solution of the Hamiltonian vector eld.Proposition 2.4. Assume
m
= 2. LetS
i =f(xp
)2T
U
ffP
1P
2gP
ig= 0gfori
= 12 and setS
=S
1\S
2:
Letz
:t
7;!(x
(t
)p
(t
)) be an abnormal bi-extremal such thatx
_(t
)6= 0 for eacht
and the curvet
!z
(t
) is contained inT
U
nS:
Then the controlu
= (u
1u
2) can be computed as the unique solution on the projective spaceP
2 of the equationu
1ffP
1P
2gP
1g(z
) +u
2ffP
1P
2gP
2g(z
) = 0:
If we set
H
a =P
1ffP
1P
2gP
2g;P
2ffP
1P
2gP
1g thent
7;!z
(t
) is a reparametrized solution of~H
a starting att
= 0 from the setP
1=P
2 =fP
1P
2g= 0:
Proof. Since
= 0H
is reduced to P2i=1u
iP
i and @H@u = 0 impliesP
1(z
(t
)) =P
2(z
(t
)) = 0:
Dierentiating with respect to
t
and usingu
(t
)6= 0we getf
P
1P
2g(z
(t
)) = 0:
If we dierentiate once more, we obtain the equation
2
X
i=1
u
i(t
)ffP
1P
2gP
ig(z
(t
)) = 0:
Since
t
7;!z
(t
) lies inT
U
nS
it can be solved as follows. One may assumeff
P
1P
2gP
1g(z
(t
))6= 0henceu
1 =;u
2ffP
1P
2gP
2gff
P
1P
2gP
1g:
Therefore
t
7;!z
(t
) is a reparametrized solution corresponding to the vector eld~H
a:
Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
2.2. Computations in the Martinet case.
Definition 2.5. We denote by
q
= (xyz
) the coordinates inR3p
= (p
xp
yp
z) being the dual variable. The Lie bracket of two smooth vector eldsF G
is computed with the conventionFG
](q
) =@F
@q
(q
)G
(q
);@G
@q
(q
)F
(q
):
In particular if
P
i=hpF
i(q
)ione has the identity fP
1P
2g=hp F
1F
2]i:
LetF
1F
2 be the germs at 0 of two smooth vector elds and setD
= SpanfF
1F
2g:
AssumeD
is of rank 2 at 0. The point 0 is called a Darboux point if the vector eldsF
1F
2F
1F
2] are independant at 0. ThenD
= ker!
, where!
is a contact form. The point 0 is called a Martinet point if(i)
F
1F
2F
1F
2] are dependant at 0,(ii) one of the two determinants
D
i= det(F
1F
2F
1F
2]F
i]) fori
= 1 2 is non zero at 0.It is well-known 33] that there exists a smooth (resp. analytic if
F
1F
2 are analytic) system of coordinates at 0 such thatD
can be written ker!
where!
is Martinet canonical form. HenceD
is generated byF
1F
2 whereF
1 =@
@x
+y
2 2@
@z F
2 =@
@y:
Consider now a germ of a smooth metric onD:
It is dened byg
=a
(q
)dx
2+2b
(q
)dxdy
+c
(q
)dy
2wherea b c
are germs at 0 of smooth functions. Ifg
=dx
2+dy
2the metric is said at.Lemma 2.6. Assume that
D
= ker!
where!
=dz
;12y
2dx:
Then the set : det(F
1F
2F
1F
2]) = 0 called the Martinet surface is the planey
= 0:
The abnormal geodesics are the linesy
= 0z
=z
0:
Proof. We use proposition 2.4. Abnormal bi-extremals are solutions of
P
1 =P
2 =fP
1P
2g= 0 X2i=1
u
iffP
1P
2gP
ig= 0:
Computing we getF
1F
2] =y @@z
F
1F
2]F
1] = 0F
1F
2]F
2] =@
Since
p
6= 0 the abnormal geodesics are contained in the Martinet surface@z:
det(
F
1F
2F
1F
2]) = 0 that isy
= 0:
The associated controls satisfyu
1D
1+u
2D
2 = 0 wherey
= 0:
Since
D
1 = 0 andD
2 6= 0 we getu
2 = 0:
Choosing a parametrization we get the curves :t
7;!(t
+x
00z
0):
UsingP
1 =P
2 = 0the adjoint vector satisesp
x=p
y = 0:
Lemma 2.7. Assume
g
=dx
2 +dy
2:
Then the abnormal geodesics are not strictly abnormal.Proof. The Hamilton function with
=;12 is:H
=p
xu
1+p
yu
2+p
zy
22
u
1;12(u
21+u
22):
Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
Solving
@H
@u
= 0we obtain:u
2 =p
yu
1=p
x+p
zy
2 2:
Normal bi-extremals are solutions of:x
_ =p
x+p
zy
22
p
_x = 0y
_ =p
yp
_y = ;(p
x+p
zy
2 2 )p
zy z
_ =y
22 (
p
x+p
zy
22 ) _
p
z = 0:
(2.4) We check easily that the lines
x
=t
+x
0y
= 0z
=z
0 are solutions.Indeed they correspond to
u
1 =1u
2 = 0:
Hence, withy
= 0we obtainp
y = 0p
x = 1 andp
z arbitrary constant. This proves that they are not strictly abnormal.Definition 2.8. Consider the control systems dened by (2.1) where
U
is a neighborhood of 0 and theF
i's are germs of smooth (resp. analytic) vector elds onU:
The set of such control systems can be identied to the set C ofm
vector elds (F
1:::F
m):
The feedback groupG
f is the transformation group induced by the following actions on C:(i)
Q
='
(q
) where'
is a smooth (resp. analytic) germ of dieomor- phism preserving 0,(ii) feedback transformations
v
= (q
)u
where (q
) 2GL
(m
R) andq
7;!(q
) is a smooth (resp. analytic) germ at 0.We can dene the action of
G
f on the set of germs at 0 of Hamiltonian vector elds as follows. The action of a feedback is trivial and a dieo- morphism acts by the canonical symplectic dieomorphism~'
induced by':
Giving a germ at 0 of a metricg
onD
= SpanfF
1F
mgwe denote byG
the subgroup ofG
f using only feedback transformations preserving the metricg:
In invariant theory, covariants are mappings which commute with the respective actions of
G:
Both normal and abnormal bi-extremals provide covariants. First we haveLemma 2.9. Assume that 0 is a Martinet point. Then the two mappings
1 :C7;! Martinet surface, 2 :C7;! unparametrized lines corresponding to abnormal geodesics, are covariants.Proof. This result can be deduced from 6]. Indeed abnormal geodesics are depending only on
D:
To the control system (2.1), wherem
= 2 we can associate two ane control systems by takingu
1 = 1 andu
2 = 1:
This x locally the parametrization of admissible trajectories. The abnormal geodesics correspond to singular trajectories and1 2are covariants for the induced action of the feedback group. Also observe that in the classication of 6], they correspond to exceptional trajectories, because the Hamiltonian is zero.Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
Lemma 2.10. The mapping
3 : (Dg
)7;!~H
n is a covariant.Proof. The proof is a straightforward computation. It is simplied if we use orthonormal vector elds. Then a feedback
v
= (q
)u
preservingg
is such that (q
)2O
(m
R)i.et =I:
2.3. Normal form: isoperimetric case.
2.3.1. Statement of the problem. When computing a normal form for a pair (
Dg
)near a Martinet point we have dierent choices using the previous covariants. We can normalize the distributionD
or the metricg:
We adopt the rst choice. Our study is localized near 0 andD
is normalized to ker!
where!
=dz
;12y
2dx:
We shall rst analyze the isoperimetric case, which is the important case in a gradated normal form and the generalization is straightforward. We use the following relations. The set G is the set of metrics onD
represented asa
(q
)dx
2+ 2b
(q
)dxdy
+c
(q
)dy
2 wherea b c
are germs at 0 ofC
1 functions. The set Gi is the subset of metrics called isoperimetric such thata b c
are not depending onz:
We denote by D the set of germs at 0 ofC
1 dieomorphisms on R3 preserving 0. The setDi is the subset of dieomorphisms 2 D : (
xyz
) 7;! (XYZ
) whereX
z =Y
z = 0:
To compute a normal form for (Dg
) we use dierent steps.Lemma 2.11. Take 2D: (
xyz
)7;!(XYZ
) such that (i)X
x(0)6= 0(ii)
dZ
;Y
22
dX
=dz
;y
2 2dx:
Then is dened by the following equations
Y
2=y
2Xx
=+y
2
yZ
=z
;y
3 4ywhere
is any germ at 0 of a function(yX
)7;!Rsuch that (0) = 0 and X(0)>
0:
Proof. By denition
dZ
;Y
22
dX
=dz
;y
2 2dx:
Hence
Y dX
^dY
=ydx
^dy
andy
= 0 is equivalent toY
= 0.Since
;
y
22
dx
=;d
(xy
22 ) +
xydy
we getd
Z
+xy
2 2=
Y
22
dX
+dz
+xydy:
(2.5) AssumeX
x(0) 6= 0 hence we can choosey z X
as coordinates on R3:
We introduce the functionS
dened byS
(yzX
) =Z
+xy
2 2:
From (2.5), we obtainS
X =Y
22
S
y =xy S
z = 1Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
hence
S
can be writtenS
=z
+y
22
(yX
)X(0)
>
0 andxY
are dened by the relations:Y
2 =y
2Xx
=+y
2
y (2.6)and
Z
is dened byZ
=S
;xy
22
:
(2.7)Definition 2.12. Without loosing any generality, we can choose the branch
Y
=y
pX:
Therefore the germ is uniquely dened by the germ and is called the generating function of:
We denote by M the set of such germs of dieomorphism satisfying (i) and (ii).Proposition 2.13. Let
g
2 Gi of the forma
(xy
)dx
2 + 2b
(xy
)dxdy
+c
(xy
)dy
2:
If : (xyz
) 7;! (XYZ
) is an element of M, whose gen- erating function is , transformingg
in a sum of squaresA
(XY
)dX
2+C
(XY
)dY
2then is solution of a partial dierential equation of the form3
y+y
y2 =F
(y
Xy
yy
XXy
Xy)where
F
is smooth ifa b c
are smooth and analytic ifa b c
are analytic, neary
= 0:
Proof. We have
x
=+y
2
yY
=y
pX thendx
= X +y
2XydX
+3y+y
y2 2dy dY
=y
XX2p
XdX
+pX +y
2 Xyp
Xdy:
We want to transform
g
into a sum of squaresA
(XY
)dX
2+C
(XY
)dY
2 using a germ 2M:
SinceXy
is a system of coordinates, usingAdX
2+C
y
2XXp
XdX
+pX +y
2 Xyp
Xdy
2=
a
h(X +y
2
Xy)dX
+ (3y+y
y22 )
dy
i2+
b
h(2X +y
Xy)dX
+ (3y+y
y2)dy
idy
+cdy
2 we obtain the relationsA
+C y
42X XX2 =a
(X +y
2Xy)2Cy
XX2
X (2X +y
Xy) =a
2(2
X +y
Xy)(3y+y
y2) +b
(2X+y
Xy)C
(2X +y
Xy)24
X =a
4(3
y+y
y2)2+b
(3y+y
y2) +c:
Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
By assumption
X(0)>
0hence fory
small enough, we haveX+y
2Xy 6= 0:
Simplifying in the second equation, we getCy
XX X =a
(3y+y
y2) + 2b:
Using this equation
A
can be computed asA
=a
X+y
2
Xy
2
;
y
4XXh
a
(3y+y
y2) + 2b
i:
(2.8) Now, from the third equationC
is given byC
= Xa
(3y+y
y2)2+ 4b
(3y+y
y2) + 4c
](2
X+y
Xy)2 (2.9)and since
X(0) 6= 0 is solution fory
small enough of the following compatibility equation(2
X+y
Xy)2a
(3y+y
y2) + 2b
]=
y
XXa
(3y+y
y2)2+ 4b
(3y+y
y2) + 4c
]:
(2.10) This equation can be solved as follows. We set = 3y+y
y2:
Hence (2.10) can be written asA
2+B+C = 0 (2.11)where A
BC are functions ofa b c
and of the jet dened byA =
ay
XXB = 4
by
XX;a
(2X +y
Xy)2C = 4
cy
XX;2b
(2X+y
Xy)2:
If (Xy
) = 0one has A= 0 andB=;4
a
(0)X2(0) C=;8b
(0)2X(0)where
X(0)>
0:
Sinceg >
0we havea
(0)>
0 andB6= 0 near 0. Therefore at 0, solution of (2.11) is given by0=;2b=a:
Now (2.11) can be written as an equationG
(Jab
) = 0where
J
is a vector in the jet of and is of the form (Xy
XXy
Xy) and we have at (Xy
) = (00):@G @
=B(0)6= 0:
Hence using the implicit function theorem, there exists a smooth (resp.
analytic) function
F
ifa b c
are smooth (resp. analytic) such that 3y+y
y2 =F
(y
Xy
yy
XXy
Xy) fory
small enough.More precisely
F
can be computed as follows. The discriminant of (2.11) is =B2;4AC:
Neary
= 0>
0 and the equation has two real roots: "= ;B+"
Bp1;4AC=
B22A
"
=1:
Esaim: Cocv, December 1997, Vol. 2, pp. 377{448
When
y
! 0A! 0 and " behaves like " 2A =:
Hence, wheny
!0the root corresponding to"
=;1 tends to the innity and the root corresponding to"
= +1 tends to;CB
=;2
b
Thereforea :
3
y+y
y2 = B(;1 +p1;4AC=
B2)2A (2.12)
where the right side can be expand in a power series of (
y
Xy
yy
XXy
Xy) converging near (00 X(0)000) when X(0)>
0:
Proposition 2.14. Consider the Cauchy problem for the singular partial dierential equation:
3
y+y
y2 =F
(y
Xy
yy
XXy
Xy) (Xy
= 0) =0(X
):
Let
P
= (0 0(0) 00(0)0 00):
IfF
is smooth atP
then there exists an unique power series(Xy
) =Pn0 ynn!n(X
) solution of the problem. If0 is analytic at 0 andF
is analytic atP
then the power series is converging at 0. Hence the Cauchy problem admits an unique analytic solution at 0.Proof. Our singular particular equation is an equation of high-order Briot- Bouquet type and we can apply results from section 5 and 6 of 18]. First we compute a formal solution. Indeed consider the equation
3
y +y
y2 =F
(y
Xy
yy
XXy
Xy)where
(Xy
) = Pn0 n(nX! )y
n and 0(X
) is given. Evaluating aty
= 0 we get3
1=F
(P
e) whereP
e= (0000000):
Dierentiating with respect to
y
and evaluating aty
= 0 we obtain 42=F
1(P
e) + (F
2(P
e) +F
4(P
e))1+ (F
3(P
e) +F
6(P
e))10+F
5(P
e)000 whereF
i is the partial derivative ofF
with respect to the ith-argument.Hence, we can compute
2:
More generally, by successive dierentiation, with respect toy
and evaluating aty
= 0we get a recurrence relation(
n
+ 2)n= n(0 1:::
n;1)n
1:
Moreover the power series
(Xy
) =Pn0 n(nX!)yn is such that each nare analytic in a common disk centered at 0 and
belongs to a Gevrey classs s
1, wheres
1 is the smallest integer such that the power seriesX
n0
n(X
)y
n (n
!)s1 is converging at (Xy
) = (00):
The integer
s
1 can be easily computed. Indeed we observe that 1 is function of 0 00 2 is function of 0 00 000 and more generally n is function of 0(k) fork
n:
This ensures thats
1 = 1 as in the Cauchy- Kowaleska theorem andis analytic at 0. For instance consider the equation3
y+y
y2 =y
XXEsaim: Cocv, December 1997, Vol. 2, pp. 377{448