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Finite time singularities for water waves with surface tension

Abstract

Here we consider the 2D free boundary incompressible Euler equation with surface tension. We prove that the surface tension does not prevent a finite time splash or splat singularity, i.e. that the curve touches itself either in a point or along an arc. To do so, the main ingredients of the proof are a transformation to desingularize the curve and a priori energy estimates.

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Finite time singularities for water waves with surface tension

Author: Castro Martínez, Ángel; Córdoba Gazolaz, Diego; Fefferman, Charles L.; Gancedo García, Francisco; Gómez Serrano, Javier
Publisher: AIP Publishing
Year: 2012
DOI: 10.1063/1.4765339
Source: https://idus.us.es/bitstreams/4086587a-aa20-407a-9a17-f5fbc56aa039/download
a Xi :1204.6633 2 [ma h.AP] 19 Oc 2012
Fini e ime singula i ies o wa e wa es wi h su ace ension
Angel Cas o, Diego C´o doba, Cha les Fe e man,
F ancisco Gancedo and Ja ie G´omez-Se ano
Dedica ed o Pe e Cons an in on his 60 h Bi hday
Oc obe 22, 2012
Abs ac
He e we conside he 2D ee bounda y incomp essible Eule equa ion wi h su ace
ension. We p o e ha he su ace ension does no p e en a ini e ime splash o spla
singula i y, i.e. ha he cu e ouches i sel ei he in a poin o along an a c. To do so,
he main ing edien s o he p oo a e a ans o ma ion o desingula ize he cu e and a
p io i ene gy es ima es.
Keywo ds: Eule , incomp essible, blow-up, wa e wa es, splash, spla , su ace ension.
I In oduc ion
In his pape we con inue he wo k in [8] and [9] whe e we show he o ma ion o singula i ies
o he ee bounda y incomp essible Eule equa ions. He e we p o e ha in wo space
dimensions he ee bounda y p oblem de elops ini e ime “splash” and “spla ” singula i ies
when su ace ension is aken in o accoun (see below, in Sec ion III, he p ecise de ini ion o
he splash and spla cu es).
In o de o desc ibe he e olu ion o a luid wi h a mo ing domain Ω( )⊂R2, he 2D
incomp essible Eule equa ions a e used:
( + ·∇ )(x, y, ) = −∇p(x, y, )−(0,1),(x, y)∈Ω( ) (I.1)
wi h he luid eloci y (x, y, )∈R2and he p essu e p(x, y, )∈R. The ec o −(0,1)
ep esen s he ex e nal g a i a ional o ce ( he accele a ion due o g a i y is aken equal o
one o he sake o simplici y). The ee bounda y
∂Ω( ) = {z(α, ) = (z1(α, ), z2(α, )) : α∈R}(I.2)
is smoo h and con ec ed by he eloci y ield
z (α, )·z⊥
α(α, ) = (z(α, ), )·z⊥
α(α, ),(I.3)
which is assumed o be incomp essible and i o a ional
∇· (x, y, ) = 0,∇⊥· (x, y, ) = 0,(x, y)∈Ω( ).(I.4)
1
He e we s udy he ele ance o conside ing he Laplace-Young condi ion o which he p essu e
on he in e ace ∂Ω( ) is p opo ional o i s cu a u e, meaning ha he su ace ension e ec
is conside ed:
−p(z(α, ), ) = τ
2
zαα(α, )·z⊥
α(α, )
|zα(α, )|3≡τ
2K. (I.5)
Abo e τ > 0 is he su ace ension coe icien .
The esul s in his pape can be shown o h ee di e en scena ios:
1. Ω( ) a compac domain: z(α, ) is a 2π-pe iodic unc ion in α.
2. Asymp o ically la case: z(α, )−(α, 0) →0 as α→ ∞.
3. Ω( ) pe iodic in he ho izon al a iable: z(α, )−(α, 0) is a 2π-pe iodic unc ion in α.
The p oblem o s udy he e is he po en ial o ma ion o singula i ies o he sys em (I.1-
I.5) wi h smoo h in e ace and smoo h eloci y ield wi h ini e ene gy as ini ial da a:
Ω(0) = Ω0, ∂Ω0={z0(α) : α∈R},
(x, y, 0) = 0(x, y),ZΩ0| 0(x, y)|2dxdy < +∞.(I.6)
The smoo h ini ial cu e z0(α) mus sa is y he a c-cho d condi ion:
|z0(α)−z0(β)| ≥ cAC|α−β|, o all α, β ∈R,(I.7)
whe e cAC >0 is he a c-cho d cons an . The s udy o his quan i y has been employed by
o he au ho s o p o e local exis ence (see o example [23], [24]). We will quan i y how ou
cu e z(α) sa is ies he a c-cho d condi ion h ough he ollowing quan i y
F(z) = |β|
|z(α)−z(α−β)|, α, β ∈[−π, π].
Th oughou he pape we will only ocus on scena io 3 o he sake o simplici y. F om
now on, we will deno e Ω0∩[−π, π]×Rby Ω0by abuse o no a ion (a undamen al domain
in he pe iod).
We es ablish he main esul in he pape o he sys em (I.1-I.5).
Theo em I.1 Conside z0(α)−(α, 0) ∈Hk(T) o k≥5. Then he e exis a amily o
ini ial da a sa is ying (I.6) and he a c-cho d condi ion (I.7) and a ime Ts>0such ha he
in e ace z(α, )∈Hk(T) om he unique smoo h solu ion o he sys em (I.1-I.7) on he ime
in e al [0, Ts] ouches i sel a a single poin (“splash” singula i y) o along an a c (“spla ”
singula i y) a ime =Ts.
These solu ions can be ex ended o he pe iodic 3Dse ing conside ing scena ios in a ian
unde ansla ions in one coo dina e di ec ion. In [14], Cou and-Shkolle conside addi ional
3Dsplash and spla singula i ies. The case wi h small ini ial da a was ea ed by Wu in
he wo dimensional case [25] and he h ee dimensional case was s udied by Wu [26] and
Ge main e al. [16].
2
Fo o he long ime beha iou esul s see Al a ez-Lannes [3], Cas o e al. [10] and he
e e ences he ein.
In o de o p o e his heo em we p oceed as in [8] and [9]. Using (I.4) i is easy o
decla e ha is ha monic in Ω( ). This ac allows us o in oduce he momen ω(α, ) by
elemen a y po en ial heo y as ollows:
(x, y, ) = PV
2πZR
(x−z1(β, ), y −z2(β, )))⊥
|(x, y)−z(β, )|2ω(β, )dβ, (I.8)
whe e PV deno es p incipal alue a in ini y. This momen is also known in he li e a u e
as he o ici y ampli ude. Then he sys em (I.1-I.5) is equi alen o he ollowing e olu ion
equa ions which a e only w i en in e ms o he ee bounda y z(α, ) and he ampli ude
ω(α, ):
z (α, ) = BR(z, ω)(α, ) + c(α, )zα(α, ),(I.9)
ω (α, ) = −2BR (z, ω)(α, )·zα(α, )−ω2
4|∂αz|2α(α, ) + (cω)α(α, )
+ 2c(α, )BRα(z, ω)(α, )·zα(α, )−2(z2)α(α, ) + τzαα(α, )·z⊥
α(α, )
|zα(α, )|3α
(I.10)
( o de ails see o example [12, Sec ion 2]). Abo e BR(z, ω) is he Bi kho -Ro in eg al
de ined by
BR(z, ω) = 1
2πPV ZR
(z(α, )−z(β, ))⊥
|z(α, )−z(β, )|2ω(β, )dβ, (I.11)
and c(α, ) is a bi a y since he bounda y is con ec ed by he no mal eloci y (I.3).
Local exis ence in Sobole spaces was i s achie ed by Wu [23] assuming ini ially he
a c-cho d condi ion. Fo o he a ia ions and esul s see [15, 21, 7, 27, 24, 11, 20, 13, 22, 28,
18, 6, 4, 19, 1, 2, 12].
The s a egy o he p oo o he main esul is o es ablish a local exis ence heo em om
he ini ial da a ha has a splash o a spla singula i y (no ice ha he equa ions a e ime
e e sible in a ian ). Since he cu e sel -in e sec s ( ailu e o he a c-cho d condi ion), i
is no clea i he ampli ude o he o ici y emains smoo h and he meaning o equa ions
(I.9-I.10). In o de o deal wi h hese obs acles we use a con o mal map
P(w) =  an w
21/2, w ∈C,
whose in en ion is o keep apa he sel -in e sec ing poin s aking he b anch o he squa e
oo abo e passing h ough hose c ucial poin s. He e P(z) will e e o a 2 dimensional
ec o whose componen s a e he eal and imagina y pa s o P(z1+iz2). We also make su e
ha Ω( )∪∂Ω( ) do no con ain any singula poin o he ans o ma ion P. Then po en ial
heo y helps us o ge he ollowing analogous e olu ion equa ions o he new cu e
˜z(α, ) = P(z(α, ))
3
and he new ampli ude ˜ω:
˜z (α, ) = Q2(α, )BR(˜z, ˜ω)(α, ) + ˜c(α, )˜zα(α, ),(I.12)
˜ω (α, ) = −2BR (˜z, ˜ω)(α, )·˜zα(α, )−|BR(˜z, ˜ω)|2(Q2)α(α, )−Q2(α, )˜ω(α, )2
4|˜zα(α, )|2α
+ 2˜c(α, )BRα(˜z, ˜ω)·˜zα(α, ) + (˜c(α, )˜ω(α, ))α−2P−1
2(˜z(α, ))α
+τQ3
|˜zα(α, )|3(˜zT
αHP −1
2˜zα∇P−1
1·˜zα−˜zT
αHP −1
1˜zα∇P−1
2·˜zα)α
+τQ˜zαα(α, )·˜z⊥
α(α, )
|˜zα(α, )|3α
(I.13)
whe e
Q2(α, ) = 
dP
dw (P−1(˜z(α, )))
2
,
and HP −1
ideno es he Hessian ma ix o P−1
i, which is he i- h (i={1,2}) componen
o he ans o ma ion P−1.
He e, we choose ˜c(α, ) in such a way ha |˜zα(α, )|=A( ). This pa icula choice o ˜c
was i s in oduced by Hou e al. in [17] and was la e used by Amb ose [4] and Amb ose-
Masmoudi [5]. The choice o ˜cimplies
˜c(α, ) = α+π
2πZπ
−π
(Q2BR(˜z, ˜ω))β(β, )·˜zβ(β, )
|˜zβ(β, )|2dβ
−Zα
−π
(Q2BR(˜z, ˜ω))β(β, )·˜zβ(β, )
|˜zβ(β, )|2dβ
I is easy o check ha i we ake Q≡1 in (I.12-I.13) we eco e (I.9-I.10).
We also de ine he unc ion
˜ϕ(α, ) = Q2(α, )˜ω(α, )
2|˜zα(α, )|−˜c(α, )|˜zα(α, )|(I.14)
in oduced by Beale e al. o he linea case [7] and by Amb ose-Masmoudi o he nonlinea
one [5]. This unc ion will be used o p o e local exis ence in Sobole spaces.
In he sec ions below, we show a local exis ence heo em based on ene gy es ima es. Sec-
ion III is de o ed o p o ide he app op ia e ini ial da a o he splash and spla singula i ies.
In Sec ion IV we choose an ene gy which does no need a p ecise sign on he Rayleigh-Taylo
unc ion. In Sec ion V we choose a di e en ene gy ha in ol es he sign o he Rayleigh-
Taylo unc ion and he es ima es a e uni o m wi h espec o he su ace ension coe icien .
These wo ene gies a e based on he ones ob ained in he non- ilde domain by Amb ose ([4])
and Amb ose-Masmoudi ([6]).
4
The Rayleigh-Taylo unc ion is gi en by he ollowing o mula
σ≡BR (˜z, ˜ω) + ˜ϕ
|˜zα|BRα(˜z, ˜ω)·˜z⊥
α+˜ω
2|˜zα|2˜zα +˜ϕ
|˜zα|˜zαα·˜z⊥
α
+QBR(˜z, ˜ω) + ˜ω
2|˜zα|2˜zα
2
(∇Q)(˜z)·˜z⊥
α+ (∇P−1
2)(˜z)·˜z⊥
α.
(I.15)
All solu ions ha we will conside h oughou he pape will ha e ini e ene gy, as dis-
cussed in [8]. The sys em sa is ies he conse a ion o he mechanical ene gy. We de ine i
his way: (no o be con used wi h he subsequen de ini ions o some o he ene gies, see
sec ions IV and V).
ES( ) = 1
2ZΩ ( )| (x, y, )|2dxdy +1
2Zπ
−π
(z2(α, ))2∂αz1(α, )dα +τ
2Zπ
−π|∂αz(α, )|dα
≡ Ek( ) + Ep( ) + Eτ( ),
whe e z(α, ) = (z1(α, ), z2(α, )), u(α, ) = (z(α, ), ), and Ω ( ) = Ω( )∩[−π, π]×R
is a undamen al domain in he wa e egion in a pe iod, hen i ollows ha he ene gy is
conse ed.
dEk( )
d =ZΩ ( )
(x, y, )( (x, y, ) + (x, y, )·∇ (x, y, ))dxdy
=ZΩ ( )
(x, y, )(−∇p(x, y, )−(0,1))dxdy
=−ZΩ ( )
(x, y, )(∇(p(x, y, ) + y))dxdy
=−Z∂(Ω ( ))
(x, y, )·−→
n yds +Z∂(Ω ( ))
(x, y, )·−→
nτ
2Kds
=−Zπ
−π
z2(α, )u(α, )·∂αz⊥(α, )dα +τ
2Zπ
−π
u(α, )·∂αz⊥(α, )∂2
αz(α, )·∂αz⊥(α, )
|∂αz(α, )|3dα
(I.16)
whe e we ha e used he incomp essibili y o he luid (∇· = 0) and Laplace-Young’s condi ion
o he p essu e on he in e ace. Nex
dEp( )
d =Zπ
−π
z2(α, )∂ z2(α, )∂αz1(α, )dα +1
2Zπ
−π
(z2(α, ))2∂ ∂αz1(α, )dα
=Zπ
−π
z2(α, )∂ z2(α, )∂αz1(α, )dα −Zπ
−π
z2(α, )∂αz2(α, )∂ z1(α, )dα
=Zπ
−π
z2(α, )u(α, )·∂αz⊥(α, )dα. (I.17)
5

dEτ( )
d =τ
2Zπ
−π
∂αz(α, )·∂α∂ z(α, )
|∂αz(α, )|dα =−τ
2Zπ
−π
∂2
αz(α, )·∂ z(α, )
|∂αz(α, )|dα
=−τ
2Zπ
−π
∂2
αz(α, )·u(α, )
|∂αz(α, )|dα =−τ
2Zπ
−π
∂2
αz(α, )·∂αz⊥(α, )
|∂αz(α, )|3u(α, )·∂⊥
αz(α, )dα
(I.18)
Adding all he de i a i es we ge he desi ed esul .
II P ope ies o he cu a u e in he ilde domain
In his sec ion we will ew i e he e m co esponding o he cu a u e K(z(α, )) in he new
ilde a iables ˜z(α, ).
We will p oceed s ep by s ep. Le us ecall ha he cu a u e is de ined by
K(α, ) = zαα(α, )·z⊥
α(α, )
|zα(α, )|3
We begin wi h he e m |zα(α, )|3. We ha e ha
|˜zα(α, )|2=h∂αP(z(α, )), ∂αP(z(α, ))i=h∇P(z(α, )) ·zα(α, ),∇P(z(α, )) ·zα(α, )i
Since Pand P−1a e con o mal, by he Cauchy-Riemann equa ions
∇P(z(α, ))T∇P(z(α, )) = Q2(α, )Id2,
ha implies ha
|˜zα(α, )|3=Q3(α, )|zα(α, )|3
We mo e o he o he e m
hzαα(α, ), z⊥
α(α, )i=h∂α∇P−1(˜z(α, )) ·˜zα(α, ),(∇P−1(˜z(α, )) ·˜zα(α, ))⊥i
=h∇P−1(˜z(α, )) ·˜zαα(α, ),(∇P−1(˜z(α, )) ·˜zα(α, ))⊥i
+h∂α∇P−1(˜z(α, ))·˜zα(α, ),(∇P−1(˜z(α, )) ·˜zα(α, ))⊥i ≡ W+X
Again, by he Cauchy-Riemann equa ions
W=1
Q2(α, )h˜zαα(α, ),˜zα(α, )⊥i
De eloping he e ms in X, we ge
∇P−1(˜z(α, ))·˜zα(α, ) = ˜zT
α(α, )·HP −1
1(˜z(α, )) ·˜zα(α, )
˜zT
α(α, )·HP −1
2(˜z(α, )) ·˜zα(α, ),
6
whe e HP−1
ideno es he Hessian o he i- h componen o P−1(i= 1,2). Hence, we can
w i e Xas
X=−˜zT
α(α, )·HP −1
1(˜z(α, )) ·˜zα(α, )∇P−1
2(˜z(α, )) ·˜z(α, )
+ ˜zT
α(α, )·HP −1
2(˜z(α, )) ·˜zα(α, )∇P−1
1(˜z(α, )) ·˜z(α, ).
This means ha
K(α, ) = Q(α, )˜zαα(α, )·˜z⊥
α(α, )
|˜z(α, )|3+X(α, )Q(α, )3
|˜z(α, )|3≡Q(α, )˜
K(α, ) + M(α, )
We will now y o simpli y u he by exploi ing he Cauchy-Riemann equa ions. We can
calcula e he Hessian and he g adien e ms as:
P−1
1,x (˜z) = ℜ4˜z
1 + ˜z4≡ ℜ(a)
P−1
1,y (˜z) = ℜ4i˜z
1 + ˜z4≡ −ℑ(a)
P−1
2,x (˜z) = ℑ4˜z
1 + ˜z4≡ ℑ(a)
P−1
2,y (˜z) = ℑ4i˜z
1 + ˜z4≡ ℜ(a)
P−1
1,x,x(˜z) = ℜ4(1 −3˜z4)
(1 + ˜z4)2≡ ℜ(b)
P−1
1,x,y(˜z) = ℜ4i(1 −3˜z4)
(1 + ˜z4)2≡ −ℑ(b)
P−1
2,x,x(˜z) = ℑ4(1 −3˜z4)
(1 + ˜z4)2≡ ℑ(b)
P−1
2,x,y(˜z) = ℑ4i(1 −3˜z4)
(1 + ˜z4)2≡ ℜ(b)
The e o e he Hessians a e
HP −1
1=ℜ(b)−ℑ(b)
−ℑ(b)−ℜ(b), HP−1
2=ℑ(b)ℜ(b)
ℜ(b)−ℑ(b),
Calcula ing u he :
˜zT
αHP −1
2˜zα=ℜ(b)(2˜z1
α˜z2
α) + ℑ(b)((˜z1
α)2−(˜z2
α)2)
˜zT
αHP −1
1˜zα=ℜ(b)((˜z1
α)2−(˜z2
α)2)−ℑ(b)(2˜z1
α˜z2
α)
7
X1=ℜ(a)ℜ(b)(2(˜z1
α)2˜z2
α) + ℜ(a)ℑ(b)((˜z1
α)2˜z1
α−(˜z2
α)2˜z1
α)
+ℑ(a)ℜ(b)(−2˜z1
α(˜z2
α)2) + ℑ(b)ℑ(b)((˜z1
α)2˜z2
α−(˜z2
α)2˜z2
α)
X2=ℜ(b)ℜ(b)((˜z1
α)2˜z2
α−(˜z2
α)2˜z2
α) + ℜ(a)ℑ(b)(−2˜z1
α(˜z2
α)2)
+ℑ(a)ℑ(b)(2(˜z1
α)2˜z2
α) + ℑ(a)ℜ(b)((˜z1
α)2˜z1
α−(˜z2
α)2˜z1
α)
This means
X=X1−X2= ((˜z1
α)2+ (˜z2
α)2)(˜z2
α(ℜ(a)ℜ(b) + ℑ(a)ℑ(b)) + ˜z1
α(ℜ(a)ℑ(b)−ℑ(a)ℜ(b)))
≡((˜z1
α)2+ (˜z2
α)2)hG(z),˜zαi.
We can see ha
−Qα
Q3=1
2∂α1
Q2=∂α(ℜ(a)2+ℑ(a)2)
=ℜ(a)ℜ(b)˜z1
α−ℜ(a)ℑ(b)˜z2
α+ℑ(a)ℑ(b)˜z1
α+ℑ(a)ℜ(b)˜z2
α
=hG(z),˜z⊥
αi
by he Cauchy-Riemann equa ions.
I we ake one de i a i e in space o X, we ob ain
∂αX= ((˜z1
α)2+ (˜z2
α)2)h∇G(˜z)·˜zα,˜zαi+ ((˜z1
α)2+ (˜z2
α)2)hG(˜z),˜zααi
= ((˜z1
α)2+ (˜z2
α)2)h∇G(˜z)·˜zα,˜zαi+|˜zα|3˜
KhG(˜z),˜z⊥
αi
= ((˜z1
α)2+ (˜z2
α)2)h∇G(˜z)·˜zα,˜zαi−|˜zα|3˜
KQα
Q3,
This implies
K=Q˜
K−Q3X
|˜z|3⇒Kα= (Q˜
K)α+Q3
|˜zα|h∇G(˜z)·˜zα,˜zαi− ˜
KQα= (Q˜
K)α+M1+M2
La e , we will see ha he M1is a low o de e m and can be abso bed by he ene gy.
III Ini ial da a
Fo ini ial da a we a e in e es ed in conside ing a sel -in e sec ing cu e in one poin . Mo e
p ecisely, we will use as ini ial da a splash cu es which a e de ined his way:
De ini ion III.1 We say ha z(α) = (z1(α), z2(α)) is a splash cu e i
8
1. z1(α)−α, z2(α)a e smoo h unc ions and 2π-pe iodic.
2. z(α)sa is ies he a c-cho d condi ion a e e y poin excep a α1and α2, wi h α1< α2
whe e z(α1) = z(α2)and |zα(α1)|,|zα(α2)|>0. This means z(α1) = z(α2), bu i we
emo e ei he a neighbo hood o α1o a neighbo hood o α2in pa ame e space, hen
he a c-cho d condi ion holds.
3. The cu e z(α)sepa a es he complex plane in o wo egions; a connec ed wa e egion
and a acuum egion (no necessa ily connec ed). The wa e egion con ains each poin
x+iy o which y is la ge nega i e. We choose he pa ame iza ion such ha he no mal
ec o n=(−∂αz2(α),∂αz1(α))
|∂αz(α)|poin s o he acuum egion. We ega d he in e ace o be
pa o he wa e egion.
4. We can choose a b anch o he unc ion Pon he wa e egion such ha he cu e
˜z(α) = (˜z1(α),˜z2(α)) = P(z(α)) sa is ies:
(a) ˜z1(α)and ˜z2(α)a e smoo h and 2π-pe iodic.
(b) ˜zis a closed con ou .
(c) ˜zsa is ies he a c-cho d condi ion.
We will choose he b anch o he oo ha p oduces ha
lim
y→−∞ P(x+iy) = −e−iπ/4
independen ly o x.
5. P(w)is analy ic a wand dP
dw (w)6= 0 i wbelongs o he in e io o he wa e egion.
Fu he mo e, (±π, 0) and (0,0) belong o he acuum egion.
6. ˜z(α)6=ql o l= 0, ..., 4, whe e
q0= (0,0) , q1=1
√2,1
√2, q2=−1
√2,1
√2, q3=−1
√2,−1
√2, q4=1
√2,−1
√2.
(III.1)
Mo eo e , we will de ine a spla cu e as a splash cu e bu eplacing condi ion (2) by
he ac ha he cu e ouches i sel along an a c, ins ead o a poin .
Le us no e ha in o de o measu e when he ans o ma ion Pis egula , we need o
con ol he dis ance o he poin s ql. In o de o do so, we in oduce he unc ion
m(ql)(α, )≡ |˜z(α, )−ql|
o l= 0,...,4.
We ha e pe o med nume ical simula ions, as explained in [9] wi h he ollowing ini ial
da a on he non- ilde domain:
z0
1(α) = α+1
4−3π
2−1.9sin(α) + 1
2sin(2α) + 1
4π
2−1.9sin(3α)
9
dC
d = OK + 1
|˜zα|τZQ2k+4 ˜ω2∂k
α(˜ω)∂k+1
α(˜
K) = OK + C1
IV.D De elopmen o he de i a i e in B
We s a om he de elopmen o B1,B2,B3and B4. We i ially ha e:
B1=1
τZQ2k+2Λ(∂k
α(˜ω))Q2˜ω2
|˜zα|H(∂k
α(˜
K))
B3=−2ZQ2k+2QαΛ(∂k
α(˜ω))∂k
α(˜
K)
B4= OK −Z(2k+ 2)Q2k+2QαH(∂k+1
α(˜ω))∂k
α(˜
K)
We now look a B2. We can decompose i in he ollowing way
B2= 2 ZQ2k+2Λ(∂k
α(˜ω))∂k
α(Qα˜
K+Q˜
Kα)
= OK + 2 ZQ2k+2Λ(∂k
α(˜ω))(Qα∂k
α(˜
K) + Q∂k+1
α(˜
K) + kQα∂k
α(˜
K))
= OK + B2,1+B2,2+B2,3
We can w i e down he e ms B2,1and B2,3in he o m
B2,1= 2 ZQ2k+2H(∂k+1
α(˜ω))Qα∂k
α(˜
K)
B2,3= 2kZQ2k+2H(∂k+1
α(˜ω))Qα∂k
α(˜
K)
In eg a ing by pa s in B2,2we es ablish
B2,2=−2ZQ2k+3Λ(∂k+1
α(˜ω))∂k
α(˜
K)
−2(2k+ 3) ZQ2k+2QαΛ(∂k
α(˜ω))∂k
α(˜
K)
=B2,2,1+B2,2,2
Again, B2,2,2can easily be educed o he canonical o m
B2,2,2=−2(2k+ 3) ZQ2k+2QαH(∂k+1
α(˜ω))∂k
α(˜
K)
IV.E Collec ion o he e ms
We will spli all he uncon olled e ms in o h ee ca ego ies: high o de and low o de ypes
I and II and we will see ha he sum o he e ms in each ca ego y adds up o low enough
o de e ms, deno ed by OK.
16

IV.E.1 High O de
F om A:
2ZQ2k+3 ∂k
α(˜
K)∂k
α(H(˜ωαα)) (A2)
F om B:
−2ZQ2k+3Λ(∂k+1
α(˜ω))∂k
α(˜
K) (B2,2,1)
F om C:
No e ms om C.
IV.E.2 Low O de Type I
F om A:
2Z2kQ2k+2Qα∂k
α(˜
K)∂k−1
α(H(˜ωαα)) (A1)
2Z4Q2k+2Qα∂k
α(˜
K)∂k
α(H(˜ωα)) (A3)
F om B:
−2ZQ2k+2QαΛ(∂k
α(˜ω))∂k
α(˜
K) (B3)
−Z(2k+ 2)Q2k+2QαH(∂k+1
α(˜ω))∂k
α(˜
K) (B4)
2ZQ2k+2H(∂k+1
α(˜ω))Qα∂k
α(˜
K)) (B2,1)
2kZQ2k+2H(∂k+1
α(˜ω))Qα∂k
α(˜
K) (B2,3)
−2(2k+ 3) ZQ2k+2QαH(∂k+1
α(˜ω))∂k
α(˜
K) (B2,2,2)
F om C:
No e ms om C.
IV.E.3 Low O de Type II
F om A:
No e ms om A.
F om B:
1
τZQ2k+2Λ(∂k
α(˜ω))Q2˜ω2
|˜zα|H(∂k
α(˜
K)) (B1)
F om C:
1
|˜zα|τZQ2k+4 ˜ω2∂k
α(˜ω)∂k+1
α(˜
K) (C1)
17
IV.F Regula ized sys em
Now, le ˜zε,δ,µ(α, ) be a solu ion o he ollowing sys em (compa e wi h (I.12 - I.13)):
˜zε,δ,µ
(α, ) = φδ∗φδ∗Q2(˜zε,δ,µ)BR(˜zε,δ,µ,˜ωε,δ,µ)(α, ) + φµ∗˜cε,δ,µ φµ∗∂α˜zε,δ,µ(α, ),
(IV.1)
˜ωε,δ,µ
=φδ∗φδ∗−2BR (˜zε,δ,µ,˜ωε,δ,µ)·˜zε,δ,µ
α−|BR(˜zε,δ,µ,˜ωε,δ,µ)|2(Q2(˜zεδ,µ))α
−Q2˜
(ωε,δ,µ)2
4|˜zε,δ,µ
α|2α+ 2cε,δ,µBRα(˜zε,δ,µ, ωε,δ,µ)·˜zε,δ,µ
α+cε,δ,µ ˜ωε,δ,µα−2P−1
2(˜zε,δ,µ(α, ))α
+τ Q3(˜zε,δ,µ)
|˜zε,δ,µ
α(α, )|3(˜zε,δ,µ
α)THP−1
2˜zε,δ,µ
α∇P−1
1·˜zε,δ,µ
α−(˜zε,δ,µ
α)THP−1
1˜zε,δ,µ
α∇P−1
2·˜zε,δ,µ
α)!α
+τ Q˜zε,δ,µ
αα ·(˜zε,δ,µ
α)⊥
|˜zε,δ,µ
α|3!α!−εφµ∗φµ∗Λ(˜ωε,δ,µ)1
Q2k+3 (IV.2)
˜zε,δ,µ(α, 0) = ˜z0(α) and ˜ωε,δ,µ(α, 0) = ˜ω0(α) o ε > 0, δ > 0, µ > 0. The unc ions φδand φµ
a e e en molli ie s,
˜cε,δ,µ(α) =α+π
2πZπ
−π
∂β˜zε,δ,µ(β))
|∂β˜zε,δ,µ(β)|2·φδ∗φδ∗(∂β(Q2(˜zε,δ,µ)(β)BR(˜zε,δ,µ,˜ωε,δ,µ))(β))dβ
−Zα
−π
∂β˜zε,δ,µ(β)
|∂β˜zε,δ,µ(β)|2·φδ∗φδ∗(∂β(Q2(˜zε,δ,µ)(β)BR(˜zε,δ,µ,˜ωε,δ,µ))(β))dβ,
and
cε,δ,µ(α) =α+π
2πZπ
−π
∂β˜zε,δ,µ(β))
|∂β˜zε,δ,µ(β)|2·(∂β(Q2(˜zε,δ,µ)(β)BR(˜zε,δ,µ,˜ωε,δ,µ))(β))dβ
−Zα
−π
∂β˜zε,δ,µ(β)
|∂β˜zε,δ,µ(β)|2·(∂β(Q2(˜zε,δ,µ)(β)BR(˜zε,δ,µ,˜ωε,δ,µ))(β))dβ,
The RHS o he e olu ion equa ions o ˜zε,δ,µ and ˜ωε,δ,µ a e Lipschi z in he spaces Hk+2(T)
and Hk+1
2(T) since hey a e molli ied. The e o e we can sol e (IV.1-IV.2) o sho ime,
hanks o Pica d’s heo em.
Now, we can pe o m ene gy es ima es o ge uni o m bounds in µ(we jus deal wi h a
anspo e m and a dissipa i e) and we can le µgo o ze o. The ene gy es ima es ha we
can ge a e he ollowing:
d
d k˜zε,δ,µk2
H5+kF(˜zε,δ,µ)k2
L∞+k˜ωε,δ,µk2
H3+ 1
2+
4
X
l=0
1
mε,δ,µ(ql)!( )
≤C(δ) k˜zε,δ,µk2
H5+kF(˜zε,δ,µ)k2
L∞+k˜ωε,δ,µk2
H3+ 1
2+
4
X
l=0
1
mε,δ,µ(ql)!j
( ).
18
We should no e ha o he new sys em wi hou he φµmolli ie , he leng h o he angen
ec o |∂α˜zδ|is now cons an in space and depends only on ime. Nex we will pe o m ene gy
es ima es as in he p e ious case by using he cu a u e ˜
Kδ om he cu e ˜zδ.
Simila ly, we ge (le us omi he supe sc ip δ, ε in ˜zδ,ε and ˜ωδ,ε)
•˜
K = NICE3 + Q2
2|˜zα|3φδ∗φδ∗H(˜ωαα) + 1
|˜zα|3(Q2)αφδ∗φδ∗H(˜ωα),
•∂k
α(cα˜ω) = NICE35 + Q2˜ω2
2|˜zα|H(∂k
α(˜
K)),
•∂k
α(˜c˜ωα) = NICE35 ,
and he ollowing collec ion o e ms:
IV.F.1 High O de
F om A:
2ZQ2k+3 ∂k
α(˜
K)∂k
αφδ∗φδ∗(H(˜ωαα)) (A2)
F om B:
−2ZQ2k+3Λ(∂k+1
α(˜ω))φδ∗φδ∗∂k
α(˜
K) (B2,2,1)
−2ε
τk∂k+1
α˜ωk2
L2(D)
F om C:
No e ms om C.
IV.F.2 Low O de Type I
F om A:
2Z2kQ2k+2Qα∂k
α(˜
K)∂k−1
αφδ∗φδ∗(H(˜ωαα)) (A1)
2Z4Q2k+2Qα∂k
α(˜
K)φδ∗φδ∗∂k
α(H(˜ωα)) (A3)
F om B:
−2ZQ2k+2QαΛ(∂k
α(˜ω))φδ∗φδ∗∂k
α(˜
K) (B3)
−Z(2k+ 2)Q2k+2QαH(∂k+1
α(˜ω))φδ∗φδ∗∂k
α(˜
K) (B4)
2ZQ2k+2H(∂k+1
α(˜ω))Qαφδ∗φδ∗∂k
α(˜
K)) (B2,1)
2kZQ2k+2H(∂k+1
α(˜ω))Qαφδ∗φδ∗∂k
α(˜
K) (B2,3)
−2(2k+ 3) ZQ2k+2QαH(∂k+1
α(˜ω))φδ∗φδ∗∂k
α(˜
K) (B2,2,2)
19
F om C:
No e ms om C.
IV.F.3 Low O de Type II
F om A:
No e ms om A.
F om B:
1
τZQ2k+2Λ(∂k
α(˜ω))Q2˜ω2
|˜zα|φδ∗φδ∗H(∂k
α(˜
K)) (B1)
F om C:
1
|˜zα|τZQ2k+4 ˜ω2∂k
α(˜ω)φδ∗φδ∗∂k+1
α(˜
K) (C1)
We no e ha h oughou his sec ion we ha e epea edly used he ollowing commu a o
es ima e o con olu ions:
kφδ∗(∂α g)−gφδ∗(∂α )kL2≤Ck∂αgkL∞k kL2,(IV.3)
whe e he cons an Cis independen o δ, and g.
Also using his commu a o es ima e we can ind all he cancela ions we need in he
p e ious collec ion o e ms o low o de ype I and II o ob ain a sui able ene gy es ima e.
Rega ding he high o de e ms, we will do he es ima es in de ail. We will see he need
o he dissipa i e e m since he e a e e ms ha escape o hal o a de i a i e.
A2+B2,2,1+D= 2 ZQ2k+3 ∂k
α(˜
K)φδ∗φδ∗H(∂k+2
α˜ω)
−2ZQ2k+3H(∂k+2
α(˜ω))φδ∗φδ∗∂k
α(˜
K)−2εk∂k+1
α˜ωk2
L2
= 2 Z∂k
α(˜
K)Q2k+3φδ∗φδ∗H(∂k+2
α˜ω)−φδ∗φδ∗Q2k+3H(∂k+2
α˜ω)−2εk∂k+1
α˜ωk2
L2
≤ k∂k
α˜
KkL2k∂αQ2k+3kL∞k∂k+1
α˜ωkL2−2εk∂k+1
α˜ωk2
L2≤C(ε)Ep( ),
which is uni o m in δ. This p o es ha we can pass o he limi δ→0.
Finally, by applying he a p io i ene gy es ima es o he new sys em (which only depend
on ε) we can pass o he limi ε→0 since now we don’ ha e he p e ious p oblems and
A2+B2,2,1= 0.
V Ene gy wi h he Rayleigh-Taylo condi ion
In his sec ion, we p o e local exis ence in he ilde domain, whe e he ime o exis ence does
no depend on he su ace ension coe icien . In his heo em, we need ini ial da a o sa is y
he Rayleigh-Taylo condi ion as we explain in Sec ion III. This Rayleigh-Taylo condi ion
will hold in pa icula i he su ace ension coe icien is small enough.
20
Theo em V.1 Le k≥3. Le ˜z0(α)be he image o a splash cu e by he map Ppa ame ized
in such a way ha |∂α˜z0(α)|=L
2π, whe e Lis he leng h o he cu e in a undamen-
al pe iod, and such ha ˜z0
1(α),˜z0
2(α)∈Hk+2(T). Le ˜ϕ(α, 0) ∈Hk+1
2(T)be as in (I.14)
and le ˜ω(α, 0) ∈Hk−1(T). Then he e exis a ini e ime T > 0, a ime- a ying cu e
˜z(α, )∈C([0, T]; Hk+2), and unc ions ˜ω(α, )∈C([0, T]; Hk−1)and ˜ϕ∈C([0, T ]; Hk+1
2)
p o iding a solu ion o he wa e wa e equa ions (I.12 - I.13). Assume ha ini ially, he
Rayleigh-Taylo condi ion is s ic ly posi i e.
In o de o p o e his heo em we will use he solu ions we ha e ob ained in heo em IV.1
o τ > 0. We will pe o m ene gy es ima es on hese solu ions.
V.A The ene gy
We will de ine he ene gy o k≥3 as
E2
k( ) = EE2( ) + τ|˜zα|
2ZQ2k+1 ∂k
α(˜
K)2
|{z }
A
+ZQ2k−2∂k
α( ˜ϕ)Λ(∂k
α( ˜ϕ))
| {z }
B
+|˜zα|2τZ(Ck ˜
K( )kH1+˜
K)Q2k+1∂k−1
α(˜
K)Λ(∂k−1
α(˜
K))
|{z }
C
+ 2|˜zα|ZCk ˜
K( )kH1Q2k−2∂k
α( ˜ϕ)2
|{z }
D
+|˜zα|2ZσQ2k∂k−1
α(˜
K)2
|{z }
E
+|˜zα|2
m(Q2kσ)( ),
whe e m(Q2kσ) = minα∈TQ2k(˜z(α, ))σ(α, ) and Cis a su icien ly la ge cons an such ha
Cis s ic ly posi i e. Remembe ha ˜ϕwas in oduced in Equa ion I.14.
A his poin is impo an o no ice he ollowing.
Lemma V.2 The ollowing sen ences hold.
1. Le ˜ϕ∈H3+ 1
2,˜ω∈H2and z∈Hkwi h k≥4. Then ˜ω∈H3.
2. Le ˜ϕ∈H3+ 1
2,˜ω∈H3and z∈Hkwi h k≥5. Then ˜ω∈H3.5.
3. Le ˜ω∈H3+ 1
2, and ˜z∈Hkwi h k≥5. Then ˜ϕ∈H3.5.
This lemma shows ha o a ixed τ > 0 he ene gy o his sec ion is equi alen o his one in
sec ion IV.A. This allows us o use his ene gy o ex end he solu ions o he heo em IV.1
up o a ime Twhich does no depend on τ( o a small enough τ).
V.B The ene gy es ima es
Again, we will only ocus on he new e ms (A−E) since he es ima es o he o he ones
we e p o ed in [12] and in [8].
21

V.B.1 ˜
K
P oposi ion V.3
˜
K =NICE3B +Q2
2|˜zα|3H(˜ωαα) + 1
|˜zα|3(Q2)αH(˜ωα)
=NICE3B +1
|˜zα|2H( ˜ϕαα)−1
|˜zα|(˜
K˜ϕ)α,
whe e NICE3B means ZQj∂k
α(˜
K)∂k
α(NICE3B)≤CEp
k( )
o some posi i e cons an s C, p and any j.
P oo : The i s equali y ollows om he p oo om he las sec ion since he ene gies a e
equi alen (see Lemma V.2). We now p o e he second one. We begin by using he ela ion
(I.14) o ge
˜
K = NICE3B + Q2
|˜zα|2H ˜ϕ
Q2αα+Q2
|˜zα|H ˜c
Q2αα
+2(Q2)α
|˜zα|2H ˜ϕ
Q2α+2(Q2)α
|˜zα|H ˜c
Q2α=I+J
We can easily see ha
˜cα=−˜zα
|˜zα|2·(Q2BR)α= NICE3B
since i is a he le el o ˜ωα,˜zαα bu we gain one de i a i e by mul iplying by he angen ial
di ec ion. This p o es ha
J= NICE3B + 2(Q2)α
|˜zα|2H˜ϕα
Q2.
Looking now o ˜cαα we can see ha
˜cαα =−˜zαα
|˜zα|2·(Q2BR)α−˜zα
|˜zα|2·(Q2BR)αα =I1+I2.
Using he s anda d es ima es, he only hing ha causes ouble in I1is when all he de i a-
i es hi ˜ωand he e o e
I1= NICE3B −KQ2
2|˜zα|H(˜ωα).
Rega ding I2, again, we need all he de i a i es o hi BR o ge he mos singula e ms,
which a e
22
I2= NICE3B −Q2
|˜zα|2˜zα·2
2πZπ
−π
(˜zα(α)−˜zα(β))⊥
|˜z(α)−˜z(β)|2˜ωα(α−β)dβ
−Q2
|˜zα|2˜zα·1
2πZπ
−π
(˜z(α)−˜z(β))⊥
|˜z(α)−˜z(β)|2˜ωαα(α−β)dβ
−Q2
|˜zα|2˜zα·1
2πZπ
−π
(˜zαα(α)−˜zαα(β))⊥
|˜z(α)−˜z(β)|2˜ωα(α−β)dβ
= NICE3B + 2Q2
|˜zα|2
1
2
˜zα·˜z⊥
αα
|˜zα|2H(˜ωα)−Q2
|˜zα|2
˜zα·˜z⊥
αα
|˜zα|2H(˜ωα)−Q2
|˜zα|2
1
2
˜ω
|˜zα|2˜zα·H(˜z⊥
ααα)
Collec ing all he e ms om I1and I2, we ob ain
˜cαα
Q2= NICE3B + 1
2|˜zα|H(( ˜
K˜ω)α)
= NICE3B + 1
Q2H(( ˜
K˜ϕ)α).
We can inally w i e he o al con ibu ion as
˜
K = NICE3B + Q2
|˜zα|2H˜ϕαα
Q2−Q2
|˜zα|2H4Qα˜ϕα
Q3
−1
|˜zα|(˜
K˜ϕ)α+2(Q2)α
|˜zα|2H˜ϕα
Q2
= NICE3B + Q2
|˜zα|2H˜ϕαα
Q2−1
|˜zα|(˜
K˜ϕ)α
= NICE3B + 1
|˜zα|2H( ˜ϕαα)−1
|˜zα|(˜
K˜ϕ)α
as we wan ed o p o e.

V.B.2 ˜ϕ
Th oughou his sec ion, we will use he ollowing es ima e which was p o ed in [8] o he
case wi hou su ace ension. The p oo is exac ly he same o he case wi h i .
ϕα = NICE2B + ˜ϕ˜ϕαα
|˜zα|−Q2σ˜
K+τQ2
2|˜zα|(˜
KQ)α+Mα,
whe e NICE2B means
ZQjΛ(∂k
α( ˜ϕ))∂k−1
α(NICE2B)≤CEp
k( )
o some posi i e cons an s C, p and any j.
23
V.C Calcula ions o he ime de i a i e o he ene gy
Using he p e ious lemmas and p oposi ions, we can ge he ollowing es ima es o he
de i a i e o he ene gy:
dA
d = OK + τ
|˜zα|ZQ2k+1∂k
α(˜
K)∂k
α(H( ˜ϕαα)) −τZQ2k+1∂k
α(˜
K)∂k
α(( ˜
K˜ϕ)α)
= OK + τ
|˜zα|ZQ2k+1∂k
α(˜
K)∂k
α(H( ˜ϕαα)) −τZQ2k+1∂k
α(˜
K)∂k+1
α( ˜ϕ)˜
K= OK + A1+A2
Again, we need o be ca e ul while compu ing he de i a i e o Bas in Sec ion IV. We ob ain
dB
d = 2 ZQ2k−2Λ(∂k
α( ˜ϕ))∂k−1
α( ˜ϕα ) + Z(Q2k−2)αH(∂k
α( ˜ϕ))∂k−1
α( ˜ϕα )
= OK −2ZQ2k−2Λ(∂k
α( ˜ϕ))∂k−1
α˜ϕ˜ϕαα
|˜zα|
−2ZQ2k−2Λ(∂k
α( ˜ϕ))∂k−1
α(Q2σ˜
K)
+τ
|˜zα|ZQ2k−2Λ(∂k
α( ˜ϕ))∂k
α(Q2(Q˜
K)α)
−τ
|˜zα|ZQ2kQαΛ(∂k
α( ˜ϕ))∂k
α(˜
K)
+Zτ
|˜zα|(k−1)Q2k+2QαH(∂k
α( ˜ϕ))∂k+1
α(˜
K)
= OK + B1+B2+B3+B4+B5
V.D De elopmen o he de i a i e o he B e m
We begin no icing ha B1= OK, as i was p o ed in [12]. In eg a ing by pa s in B5, we
ha e ha
B5=−Zτ
|˜zα|(k−1)Q2k+2QαH(∂k+1
α( ˜ϕ))∂k
α(˜
K)
Fu he mo e, he only singula e ms a ising om B2a e when all de i a i es hi ei he
˜
Ko σ, his gi es us
B2= OK −2ZQ2kΛ(∂k
α( ˜ϕ))∂k−1
α(σ)˜
K−2ZQ2kΛ(∂k
α( ˜ϕ))∂k−1
α(˜
K)σ= OK + B2,1+B2,2.
Howe e , he only singula e m o he Rayleigh-Taylo condi ion ha is no in Hk−1is
he one belonging o BR (˜z, ˜ω)·˜zαwhen he ime de i a i e hi s ω, his means
B2,1= OK −τZQ2kΛ(∂k
α( ˜ϕ)) ˜
KH(∂k
α(˜
KQ))
=−τZQ2k+1Λ(∂k
α( ˜ϕ)) ˜
KH(∂k
α(˜
K))
24
Finally, de eloping B3we ob ain
B3=τ
|˜zα|ZQ2k−2Λ(∂k
α( ˜ϕ))∂k
α(Q3˜
Kα)
+τ
|˜zα|ZQ2k−2Λ(∂k
α( ˜ϕ))∂k
α(Q2Qα˜
K)
=B3,1+B3,2
Modulo lowe o de e ms we can see ha
B3,2= OK + τ
|˜zα|ZQ2kQαΛ(∂k
α( ˜ϕ))∂k
α(˜
K)
We can con inue spli ing B3,1in o
B3,1= OK + τ
|˜zα|ZQ2k+1Λ(∂k
α( ˜ϕ))∂k+1
α(˜
K) + τ
|˜zα|Z3kQ2kQαΛ(∂k
α( ˜ϕ))∂k
α(˜
K)
= OK −τ
|˜zα|ZQ2k+1Λ(∂k+1
α( ˜ϕ))∂k
α(˜
K) + τ
|˜zα|Z(k−1)Q2kQαΛ(∂k
α( ˜ϕ))∂k
α(˜
K)
= OK + B3,1,1+B3,1,2
whe e in he las equali y we ha e pe o med an in eg a ion by pa s. We can obse e ha
B3,2+B4=B3,1,2+B5= 0, B3,1,1+A1= 0
We will now see ha B2,2cancels wi h he e m a ising om he de i a i e o E. Taking
in o accoun he p e ious lemmas
dE
d = 2 ZσQ2k∂k−1
α(˜
K)H(∂k+1
α( ˜ϕ)) = OK −B2,2
Finally, we will see ha he con ibu ions om he ime de i a i es o Cand Dcancel
B2,1and A2. We s a by no icing ha , modulo lowe o de e ms A2=B2,1. Fu he mo e
dC
d = OK + 2τZ(Ck ˜
K( )kH1+˜
K)Q2k+1H(∂k+1
α( ˜ϕ))Λ(∂k−1
α(˜
K))
dD
d = OK + 2τZCk ˜
K( )kH1Q2k+1 ∂k
α( ˜ϕ)∂k+1
α(˜
K),
which, by in eg a ion by pa s esul s in
dC
d +dD
d +A2+B2,1= OK.
Adding all he con ibu ions, we can bound he de i a i e in ime o he ene gy by a
powe o he ene gy.
25
[26] S. Wu. Global wellposedness o he 3-D ull wa e wa e p oblem. In en . Ma h.,
184(1):125-220, 2011.
[27] H. Yosiha a. G a i y wa es on he ee su ace o an incomp essible pe ec luid o ini e
dep h. Publ. Res. Ins . Ma h. Sci., 18(1):49-96, 1982.
[28] P. Zhang and Z. Zhang. On he ee bounda y p oblem o h ee-dimensional incomp ess-
ible Eule equa ions. Comm. Pu e Appl. Ma h., 61(7):877-940, 2008.
Angel Cas o
D´epa emen de Ma h´ema iques e Applica ions
´
Ecole No male Sup´e ieu e
45, Rue d’Ulm, 75005 Pa is
Email: cas [email protected].
Diego C´o doba Cha les Fe e man
Ins i u o de Ciencias Ma em´a icas Depa men o Ma hema ics
Consejo Supe io de In es igaciones Cien ´ı icas P ince on Uni e si y
C/ Nicol´as Cab e a, 13-15 1102 Fine Hall, Washing on Rd,
Campus Can oblanco UAM, 28049 Mad id P ince on, NJ 08544, USA
Email: dcg@icma .es Email: c @ma h.p ince on.edu
F ancisco Gancedo Ja ie G´omez-Se ano
Depa amen o de An´alisis Ma em´a ico Ins i u o de Ciencias Ma em´a icas
Uni e sidad de Se illa Consejo Supe io de In es igaciones Cien ´ı icas
C/ Ta ia, s/n C/ Nicol´as Cab e a, 13-15
Campus Reina Me cedes, 41012 Se illa Campus Can oblanco UAM, 28049 Mad id
Email: ga[email p o ec ed] Email: ja ie .gomez@icma .es
32