S uc u al s abili y o he splash singula i ies
o he wa e wa es p oblem
Angel Cas o, Diego C´o doba, Cha les Fe e man,
F ancisco Gancedo, Ja ie G´omez-Se ano
Janua y 27, 2014
Abs ac
In his pape we show a s uc u al s abili y esul o wa e wa es. The main mo i-
a ion o his esul is ha we would like o exhibi a wa e wa e whose in e ace s a s
as a g aph and ends in a splash. Nume ical simula ions lead o an app oxima e solu ion
wi h he desi ed beha iou . The s abili y esul will conclude ha nea he app oxima e
solu ion o wa e wa es he e is an exac solu ion.
1 In oduc ion
The wa e wa es p oblem models he mo ion o an incomp essible luid wi h cons an
densi y ρin a domain Ω( ) wi h a ee bounda y ∂Ω( ), which sa is ies he Eule equa ion
wi h he p esence o g a i y and whose low in po en ial. The sys em, in R2, can be w i en,
a e some compu a ions, as an equa ion o he ee bounda y,
∂Ω( ) = {z(α, ) = (z1(α, ), z2(α, )) : α∈R},(1)
and an equa ion o he ampli ude o he o ici y, ω(α, ), in he ollowing way
z (α, ) = BR(z, ω)(α, ) + c(α, )zα(α, ),(2)
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, ),
(3)
whe e BR(z, ω) is he classical Bi kho -Ro in eg al
BR(z, ω)(α, ) = 1
2πPV ZR
(z(α, )−z(β, ))⊥
|z(α, )−z(β, )|2ω(β, )dβ. (4)
The unc ion c(α, ) is a bi a y since he bounda y is con ec ed by he no mal componen
o he eloci y o he luid. Also, we no ice ha , in o de o ge an explici equa ion o ∂ ω,
we need o in e he ope a o
I+T=I+ 2hBR(z, ·), zαi
1
a Xi :1401.6419 1 [ma h.AP] 24 Jan 2014
and we ha e aken he accele a ion due o g a i y and he densi y ρequal o one.
Once one has sol ed his sys em o (z, ω) he eloci y o he luid and he p essu e in he
domain Ω( ) can be eco e ed by using Bio -Sa a and Be noulli laws. Fo de ails see [3].
In he las wo decades hese equa ions ha e been in ensi ely s udied. Fo an ex ensi e
su ey abou analy ical esul s on wa e wa es see he monog aph [9].
In his pape we a e conce ned wi h he p oblem o he exis ence o wa e wa es which
s a as a g aph and become a splash cu e in ini e ime. Roughly speaking, a splash
cu e is a smoo h cu e ha collapses wi h i sel in a single poin such as he cu e o ig.
1. A igo ous de ini ion can be ound in [3] whe e he exis ence o splash singula i ies has
been shown. Cou and and Shkolle [5] ha e p o en he exis ence o splash singula i ies in
p esence o o ici y. Fe e man, Ionescu and Lie [6] ha e p o en he non exis ence o splash
singula i ies o in e nal wa es, i.e. o an in e ace be ween wo incomp essible luids.
Figu e 1: Splash singula i y. A smoo h in e ace ha collapses in a poin .
We a e in e es ed in he ollowing s a emen :
Conjec u e 1.1 The e exis ini ial da a z0(α), ω0(α)o solu ions o he wa e wa e equa ions
such ha a ime 0 he cu e z0(α)can be pa ame e ized as a g aph, he in e ace hen u ns
o e a a ini e ime T1>0, and inally p oduces a splash a a ini e ime T2> T1.
We should ema k ha his conjec u e is a combina ion o he scena ios in heo ems [3,
Theo em I.1] and [4, Theo em 7.1] and is suppo ed by nume ical e idence ha we can see
in Fig. 2. This nume ical simula ion was ca ied ou using he me hod o Beale, Hou and
Loweng ub [1].
The p oo o his conjec u e could ollow along hese lines. Fi s o all, we will mo e
backwa ds in ime, 0 being he ime o he splash, T2−T1 he ime o he u ning and T2 he
ime in which he solu ion can be pa ame e ized as a g aph. Also we w i e he wa e wa es
2
−0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4
−0.1
−0.05
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
x
y
Figu e 2: E olu ion om a g aph o a splash.
equa ion in a new domain gi en by he p ojec ion o Ω( ) by he 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 by aking he b anch o he squa e
oo abo e passing h ough his c ucial poin . The equa ion in his new domain can be
w i en as ollows:
˜z (α, ) = Q2(α, )BR(˜z, ˜ω)(α, ) + ˜c(α, )˜zα(α, ),(5)
˜ω (α, ) = −2BR (˜z, ˜ω)(α, )·˜zα(α, )−(Q2)α(α, )|BR(˜z, ˜ω)|2(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
whe e
˜z(α, ) = P(z(α, )), Q2(α, ) = 
dP
dw (P−1(˜z(α, )))
2
and α∈T.
(F om now on we will omi he supe sc ip ilde in he no a ion).
We s a compu ing a nume ical app oxima ion o a solu ion o he wa e wa es equa ion
5 ha s a s as a splash, u ns o e and inally is a g aph. Such a candida e is depic ed
3
in Fig. 2. Wi h his ap oxima ion we can cons uc explici unc ions (x, γ) ha sol e he
sys em 




x =Q2(x)BR(x, γ) + bxα+
γ =−2BR (x, γ)·xα−(Q2(x))α|BR(x, γ)|2−Q2(x)γ2
4|xα|2α
+2bBRα(x, γ)·xα+ (bγ)α−2(P−1
2(x))α+g
(6)
whe e and ga e e o s ha we hope a e small. By using he compu e we a e able o
gi e igo ous bounds o hese e o s. The ques ion we wan o answe is i he e exis s an
exac solu ion (z, ω) o he wa e wa es equa ion close o hese unc ions (x, γ). Tha means
we need o p o e he ollowing heo em:
Theo em 1.2 Le
D(α, )≡z(α, )−x(α, ), d(α, )≡ω(α, )−γ(α, ),D(α, )≡ϕ(α, )−ψ(α, )
whe e (x, γ, ψ)a e he solu ions o





































x =Q2(x)BR(x, γ) + bxα+
b=α+π
2πZπ
−π
(Q2BR(x, γ))α
xα
|xα|2dα −Zα
−π
(Q2BR(x, γ))β
xα
|xα|2dβ
| {z }
bs
+α+π
2πZπ
−π
α
xα
|xα|2dα −Zα
−π
β
xβ
|xβ|2dβ
| {z }
be
γ +2BR (x, γ)·xα=−(Q2(x))α|BR(x, γ)|2+ 2bBRα(x, γ)·xα+ (bγ)α
−Q2(x)γ2
4|xα|2α−2(P−1
2(x))α+g
ψ(α, ) = Q2
x(α, )γ(α, )
2|xα(α, )|−bs(α, )|xα(α, )|,
(7)
whe e (z, ω)a e he solu ions o (7) wi h ≡g≡0,ϕis he unc ion
ϕ=Q2
z(α, )ω(α, )
2|zα(α, )|−b(α, )|zα(α, )|,
and Eis he ollowing no m o he di e ence
E( )≡kDk2
H3+Zπ
−π
Q2σz
|zα|2|∂4
αD|2+kdk2
H2+kDk2
H3+ 1
2.
Then we ha e ha 
d
d E( )≤ C( )(E( ) + Ek( )) + cδ( )
whe e
C( ) = C(E( ),kxkH5+ 1
2( ),kγkH3+ 1
2( ),kζkH4+ 1
2( ),kF(x)kL∞( ))
and
δ( ) = (k kH5+ 1
2( ) + kgkH3+ 1
2( ))k+ (k kH5+ 1
2( ) + kgkH3+ 1
2( ))2, k big enough
4
depends on he no ms o and g, and E( )is gi en by
E( ) =kzk2
H3( ) + ZT
Q2σz
|zα|2|∂4
αz|2dα +kF(z)k2
L∞( )
+kωk2
H2( ) + kϕk2
H3+ 1
2( ) + |zα|2
m(Q2σz)( )+
4
X
l=0
1
m(ql)( )
whe e he L∞no m o he unc ion
F(z)≡|β|
|z(α, )−z(α−β, )|, α, β ∈T
measu es he a c-cho d condi ion,
σz≡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|2zα
2
(∇Q)(z)·z⊥
α+ (∇P−1
2)(z)·z⊥
α
(8)
is he Rayleigh-Taylo unc ion,
m(Q2σz)( )≡min
α∈TQ2(α, )σz(α, ),
and inally
m(ql)( )≡min
α∈T|z(α, )−ql|
o l= 0, ..., 4, wi h
q0= (0,0) , q1=1
√2,1
√2, q2=−1
√2,1
√2, q3=−1
√2,−1
√2, q4=1
√2,−1
√2,
(9)
which a e he singula poin s o he ans o ma ion P.
Rema k 1.3 We can abso b he e ms in E( )by E( ) aised o an app op ia e powe and
e ms in (x, γ)by pe o ming he spli ing kzk=kz−xk+kxk(o he analogous one o a
di e en a iable) o any no m o any quan i y ha appea s in E( ).
Theo em 1.2 was announced in [2].
I we knew C( ), ( ), g( ), k o bounds on hem, a p io i, hen we could p o ide bounds on
E( ) a any ime T. We poin ou he e ha E( ) con ols he no m k∂αz1(α)−∂αx1(α)kL∞.
Le Tgbe a ime in which he app oxima e solu ion is a g aph, i.e. ∂αx1(α, Tg)>0∀α.
Now, i E(Tg)< ∂αx1(α, Tg) hen
∂αz1(α, Tg)>−k∂αz1(α)−∂αx1(α)kL∞+∂αx1(α, Tg)>0,
and his shows ha zis a g aph. In o he wo ds, he possible se o solu ions o he wa e
wa es equa ion is a ball cen e ed a (x, γ, ζ) wi h he opology gi en by E. All o he elemen s
5

o his ball a e g aphs, he e o e he solu ion is necessa ily a g aph. Thus, he p oblem is
educed o s udy and ind bounds o C( ), ( ), g( ), k.
The ecen de elopmen s o compu e a chi ec u e ha e boos ed hei use in ma hema ics,
gi ing bi h o a ull se o new esul s only achie able by his eno mous powe . Howe e , i
has he d awback ha loa ing-poin ope a ions can no be pe o med exac ly, esul ing in
nume ical e o s. In o de o o e come his di icul y and be able o p o e igo ous esul s, we
use he so-called in e al a i hme ics, in which ins ead o wo king wi h a bi a y eal numbe s,
we pe o m compu a ions o e in e als which ha e ep esen able numbe s as endpoin s. On
hese objec s, an a i hme ic is de ined in such a way ha we a e gua an eed ha o e e y
x∈X, y ∈Y
x ? y ∈X ? Y,
o any ope a ion ?. Fo example,
[x, x]+[y,y]=[x+y, x +y]
[x, x]×[y,y] = [min{xy, xy, xy, xy},max{xy, xy, xy, xy}]
We can also de ine he in e al e sion o a unc ion (X) as an in e al I ha sa is ies ha
o e e y x∈Xwe ha e (x)∈I.
The a icle is o ganized as ollows: in sec ions 2 and 3 we gi e some de ails abou how o
con ol he e o s ,gand he cons an s ha a ise in Theo em 1.2 by using he compu e .
Finally, in sec ion 4 we gi e a comple e p oo o Theo em 1.2.
2 Bounds o ( )and g( )
2.1 Rep esen a ion o he unc ions and In e pola ion
The i s hing one has o decide is how o ep esen he da a and how o pass om
he cloud o poin s in space- ime ob ained by non- igo ous simula ion o a unc ion de ined
e e ywhe e in [−π, π]×[0, T]. We need o in e pola e in some way.
In ou case, we chose o ep esen he unc ions xand γby piecewise polynomials (splines)
o high deg ee (10) in space, and low deg ee (3) in ime. To do so, we i s in e pola e in
space o e e y node in he ime mesh. The in e pola ion is made ia B-Splines. Since he
in e pola ion is educed o sol e a linea (in e al) sys em Ac =y, whe e Ais cons an in
ime and space and ydepends on he alues o he unc ion a ime since he mesh in space
is cons an , we p econdi ion by mul iplying by he non- igo ous in e se o he midpoin s o
he en ies o A. We ema k ha he sys em is in e al-based because we need o p oduce
a cu e ha is a splash (i.e. he e ha e o be wo poin s α1, α2such ha we can gua an ee
x0(α1) = x0(α2). Finally, he sys em is sol ed using a igo ous Gauss-Seidel i e a i e me hod.
We also ema k ha he need o in e al-based calcula ions is only s ic ly necessa y a ime
= 0 since i is he only poin in which we ha e o gua an ee some equali y. By wo king
wi h mul ip ecision (1024 bi s) we can ge wid hs in he coe icien s o he o de o 10−300.
In o de o pe o m in e pola ion in ime, we ix he alues o he unc ion and i s ime
de i a i e a he mesh poin s. This gi es us lo s o sys ems o 4 equa ions ( he alues o
6
he unc ion and i s de i a i e a bo h endpoin s) and 4 unknowns ( he 4 coe icien s o he
deg ee 3 polynomial) bu wi h an explici o mula o each o hem. Wi h his me hod, ou
spline will be C1in ime bu i migh no be C2.
2.2 Rigo ous bounds o Singula in eg als
In his sec ion we will discuss he compu a ional de ails o he igo ous calcula ion o
some singula in eg als. In pa icula we will ocus on he Hilbe ans o m, bu he me hods
apply o any in eg al ke nel whose main singula i y is homogeneous o deg ee -1. Pa s o he
compu a ion ( he Npa ) a e sligh ly ela ed o he Taylo models wi h ela i e emainde
p esen ed in M. Jolde¸s’ hesis [8].
Le us suppose ha we ha e a unc ion gi en explici ly by a spline (piecewise polyno-
mial) which is Ck−1e e ywhe e and Ckexcep a ini ely many poin s ( he poin s in which
he di e en pieces o he spline a e glued oge he ). We need o calcula e igo ously he
Hilbe T ans o m o , ha is
H (x) = PV
πZT
(x)− (y)
2 an x−y
2dy,
and we wan o app oxima e i by a piecewise polynomial unc ion wi h less egula i y,
plus an e o ha can be bounded in Hq,0≤q≤c<kand in L∞. Le us assume ha he
kno s o he spline a e αi,i= 0, . . . , N −1 and ha we ix x∈[αi, αi+1] whe e he indices
a e aken modulo Nand he dis ance be ween he indices is aken o e ZN. We can spli ou
in eg al in
H (x) = PV
πZT
(x)− (y)
2 an x−y
2dy =PV
πX
jZαj+1
αj
(x)− (y)
2 an x−y
2dy
=PV
πX
|j−i|>K Zαj+1
αj
(x)− (y)
2 an x−y
2dy +PV
πX
|j−i|≤KZαj+1
αj
(x)− (y)
2 an x−y
2dy
≡H F(x) + H N(x).
Now, i we wan o exp ess H F(x) as a polynomial, i is easy since he in eg and does
no ha e a singula i y. Hence
H F(x) = PV
πX
|j−i|>K Zαj+1
αj
(x)− (y)
2 an x−y
2dy =PV
πX
|j−i|>K Zαj+1
αj
Fj(x, y)dy
=X
|j−i|>K Zαj+1
αjX
n,m
cnm(x−x∗(i))m(y−y∗(j))n+E(x, y)dy ≡P(x) + E(x),
whe e Eaccoun s o he e o and is a polynomial wi h in e al coe icien s. Typically,
we will use as he poin s o he Taylo expansions x∗(i) = αisince we will compa e he
esul ing polynomial wi h ano he one o he o m Pjbj(x−xi)jand we will also choose
7
y∗(j) = αj+αj+1
2. This choice is use ul o wo easons: i s , we will only ha e o in eg a e
hal o he e ms since he es will in eg a e o ze o; and second, he e o es ima es will
be be e o his choice o y∗(j) in he sense ha he coe icien s will be smalle . All he
compu a ions will be ca ied ou using au oma ic di e en ia ion. We should ema k ha we
can ge es ima es o he e o Ein any o he abo e men ioned no ms wi hou ha ing o
ecompu e i since he ela ion
∂q
xH F(x)−∂q
xP(x) = ∂q
xE(x)
holds o e e y q < k.
Now, we mo e on o he e m H N(x). In his case, we pe o m a Taylo expansion in
bo h he denomina o
2 an x−y
2= (x−y) + c(x−y)3, c = small (in e al) cons an
and he nume a o
(x) = (y)+(x−y) 0(y) + 1
2(x−y)2 00(y) + . . . 1
n!(x−y)k−1 k−1(η),
whe e ηbelongs o an in e media e poin be ween xand y, which we can enclose in he
con ex hull o [αi, αi+1] and [αj, αj+1] whe e he con ex hull is unde s ood in he o us.
Since ypically Kwill be e y small (compa ed o N) he e is no ambigui y in he de ini ion.
Finally, we can ac o ou (x−y) and di ide bo h in he nume a o and he denomina o .
Since we know (y) explici ly, we can pe o m he explici in eg a ion and ge a piecewise
polynomial as a esul .
2.3 Es ima es o he no m o he Ope a o I+T
In his subsec ion we will ou line how o compu e he no m o he ope a o I+T=
I+ 2hBR(z, ·), zαi. Since he ope a o Tbeha es like a Hilbe T ans o m plus smoo hing
e ms, we will desc ibe how o calcula e igo ously wi h he help o a compu e an es ima e
o he no m o i s in e se. The p ocedu e is mo e gene al and can be applied o a bigge
amily o ke nels. Le T=R/2πZ, and le A(x), B(x) be eal- alued unc ions on T. Also,
le E(x, y) be a eal- alued unc ion on T×T. We assume A, B and Ea e gi en by explici
o mulas such as as pe haps piecewise igonome ic polynomials o splines, and E(x, y) is a
igonome ic polynomial on each ec angle I×Jo some pa i ion o T×T. We suppose
A, B, E a e smoo h enough.
Le Hbe he Hilbe ans o m ac ing on unc ions on T, i.e.
H (x) = PV
2πZT
co y
2 (x−y)dy.
Assume ha Aand Bha e no common ze os on T.
Le
S (x) = A(x) (x) + B(x)H (x) + ZT
E(x, y) (y)dy, ∈L2(T).
8
Thus, Sis a singula in eg al ope a o .
We hope ha S−1exis s and has a no -so-big no m on L2, bu we don’ know his ye .
Ou goal he e is o ind app oxima e solu ions Fo he equa ion SF = o sui able
gi en ∈L2(T), and o check ha kSF − kL2(T)< δ o sui able δ. Ou compu a ion o F
will be based on heu is ic ideas, bu he compu a ion o an uppe bound o kSF − kL2(T)
will be igo ous. In ou case, A(x) = 1, B(x) = 1.
To ca y his ou , le H0⊂H1⊂L2(T) be ini e-dimensional subspaces, e.g. wi h Hi
consis ing o he span o wa ele s ( om a wa ele bases) ha ing leng hscale ≥2−Ni. He e
N1≥N0+ 3 (say). Le πibe he o hogonal p ojec ion om L2(T) o Hi, and le us sol e
he equa ion
π1Sπ1F=π0 . (10)
I is gi en explici ly in a wa ele bases, hen (10) is a linea algeb a p oblem, since
π1Sπ1is o ini e ank, and i s ma ix (in e ms o some gi en basis o H1) can be compu ed
explici ly.
•I π0 6∈ Range(π1Sπ1), hen ou heu is ic p ocedu e ails.
•I π0 ∈Range(π1Sπ1), hen we ind F∈H1such ha π1Sπ1F=π0 , i.e. π1SF =
π0 .
We hen ha e
kSF − kL2(T)≤ k(I−π1)SFkL2(T)+k(I−π0) kL2(T),
and bo h no ms on he igh -hand side may be es ima ed explici ly.
Now, ou goal is o make a heu is ic compu a ion o an ope a o o he o m
˜
S (x) = ˜
A(x) ( ) + ˜
B(x)H (x) + ZT
˜
E(x, y) (y)dy
such ha S˜
S−Ihas small no m on L2(T).
He e, we will make a heu is ic compu a ion o ˜
S; la e we will gi e a igo ous uppe
bound o he no m o S˜
S−Ion L2(T). By a heu is ic compu a ion o ˜
Swe mean a heu is ic
compu a ion o ˜
A, ˜
Band ˜
E.
We i s ind ˜
Aand ˜
Bby se ing
(A+iB)( ˜
A+i˜
B)=1⇒A˜
A−B˜
B= 1
A˜
B+B˜
A= 0
Then, his means ha
S˜
S= (A˜
A−B˜
B)+(A˜
B+B˜
A)H+ Smoo hing e ms = I+ Smoo hing e ms
So, om now on, we suppose ha ˜
Aand ˜
Ba e known. Fo he ope a o I+T, his means
˜
A= 1/2,˜
B=−1/2. We wan o compu e ˜
E. Now, le {φν}be some o hono mal basis o
9
Fi s o all, we will wo k wi h Q= 1 and la e mo e on o he case Q6= 1. We will adop
he ollowing con en ion o deno e he di e en Ke nels (in eg al ope a o s) ha appea :
Θa1,a2,a3,a4
b1,b2(α, β) = 1
(x(α)−x(β))b1(∂αx(α)−∂αx(β))a1(∂2
αx(α)−∂2
αx(β))a2
×(∂3
αx(α)−∂3
αx(β))a3(∂4
αx(α)−∂4
αx(β))a4∂b2
αγ(β)
Θa1,a2,a3,a4
b1,−1(α, β) = 1
(x(α)−x(β))b1(∂αx(α)−∂αx(β))a1(∂2
αx(α)−∂2
αx(β))a2
×(∂3
αx(α)−∂3
αx(β))a3(∂4
αx(α)−∂4
αx(β))a4.
The ope a o s o which b26=−1 will ac on Do i s de i a i es whe eas he ope a o s
o which b2=−1 will ac on do i s de i a i es. We now desc ibe how o spli he Ke nels
in such a way ha hey can be compu ed. Fo he case whe e b26=−1 we illus a e his by
spli ing Θ0,0,0,0
2,0, bu he echnique can be applied o any Ke nel.
1
2πZΘ0,0,0,0
2,0(D(α)−D(β))dβ =1
2πD(α)ZK(α, β)γ(β)dβ
| {z }
T1
−1
2πZK(α, β)γ(β)D(β)dβ
| {z }
T2
+1
2πc1(α)ZD(α)−D(β)
4 sin2α−β
2γ(β)dβ
| {z }
T3
+1
2πc2(α)ZD(α)−D(β)
2 an α−β
2γ(β)dβ
| {z }
T4
,(15)
whe e
K(α, β) = 1
(x(α)−x(β))2−c1(α)
4 sin2α−β
2−c2(α)
2 an α−β
2
c1(α) = 1
x2
α(α)
c2(α) = xαα(α)
x3
α(α).
We can hink o c1(α) and c2(α) as he Taylo coe icien s o Θ(α, β) a ound β=α. We
can bound he e ms in (15) in he ollowing way:
T4(α) = c2(α)[H(Dγ)(α)−DH(γ)(α)]
T3(α) = c1(α)[Λ(Dγ)(α)−DΛ(γ)(α)]
We ha e hen he es ima es
16

kT4kL2≤ kc2kL∞(kDkL2kγkL∞+kDkL2kHγkL∞)
kT3kL2≤ kc1kL∞(kDkL2kγαkL∞+kDαkL2kγkL∞+kDkL2kΛ(γ)kL∞).
We now mo e on o T1. We will es ima e i in he ollowing way:
ZT1D(α)dα =1
2πZ|D(α)|2ZK(α, β)γ(β)dβdα ≤1
2πkDk2
L2ZK(·, β)γ(β)dβL∞
.
To es ima e he ke nel T2we will use he Gene alized Young’s inequali y [7]:
kT2(D)k2
L2=1
4π2ZZZK(α, β)γ(β)D(β)K(α, σ)γ(σ)D(σ)dβdσdα.
De ining
˜
K(β, σ) = ZK(α, β)γ(β)K(α, σ)γ(σ)dα,
we ha e ha
kT2(D)k2
L2=1
4π2Z Z ˜
K(β, σ)D(β)D(σ)dβdσ
=1
4π2ZD(β)Z˜
K(β, σ)D(σ)dσdβ
≤1
4π2kDkL2Z˜
K(˙,σ)dσL2
≤1
4π2CkDk2
L2, C = max max
βZ|˜
K(β, σ)|dσ, max
σZ|˜
K(β, σ)|dβ
We inally show how o es ima e he Ke nels wi h b2=−1. We will do his by showing
how o es ima e Θ0,0,0,0
1,−1bu he echnique can be applied o any Ke nel.
1
2πZΘ0,0,0,0
1,−1(d(β))dβ =1
2πZK(α, β)d(β)dβ
| {z }
T1
+1
2πc1(α)Z1
2 an α−β
2d(β)dβ
| {z }
T2
,
whe e
K(α, β) = 1
(x(α)−x(β)) −c1(α)
2 an α−β
2
c1(α) = 1
xα(α).
We can easily es ima e hese wo e ms applying o T1 he same es ima es (Young’s inequali y)
as o T2in he p e ious case and by no ing ha T2is 1
2c1(α)H(d).
17
3.3 Es ima es o he linea e ms wi h Q6= 1
To pe o m he eal es ima es, whe e Q6= 1 we will use he es ima es om he p e ious
sec ions. We will explain how o pass om he o me ones o he la e ones. We will
illus a e his by compu ing he linea e ms o he Bi kho -Ro ope a o .
Fi s o all, he o al numbe o e ms will inc ease by a ac o 2, since we will ha e
Q2(z)BR(z, ω)−Q2(x)BR(x, γ)) = (Q2(z)−Q2(x))(BR(z, ω)−BR(x, γ))
| {z }
nonlinea
+Q2(x)(BR(z, ω)−BR(x, γ))
| {z }
calcula ed be o e
+ (Q2(z)−Q2(x))BR(x, γ)
| {z }
new e ms
In o de o calcula e he old e ms wi h Q6= 1, he only hing we ha e o do is o
inco po a e a ac o o ∂k
αQ2(x)(α) in he es ima es. The new e ms can easily be calcula ed
using ha , up o linea o de
(Q2(z)−Q2(x)) = 1
81 + x4
x,3x2−1
x2D+O(D2).
4 P oo o Theo em 1.2
In his sec ion, we will p o e he s abili y Theo em 1.2.
The equa ions a e:
SPLASH















z =Q2
zBR +czα
c=α+π
2πZπ
−π
(Q2BR)α
zα
|zα|2dα −Zα
−π
(Q2BR)β
zβ
|zβ|2dβ
ω +2BR ·zα=−(Q2)α|BR|2+ 2cBRα·zα+ (c$)α
−Q2$2
4|zα|2α−2(P−1
2(z))α
APPROX

































x =Q2(x)BR(x, γ) + bxα+
b=α+π
2πZπ
−π
(Q2BR)α
xα
|xα|2dα −Zα
−π
(Q2BR)β
xα
|xα|2dβ
| {z }
bs
+α+π
2πZπ
−π
α
xα
|xα|2dα −Zα
−π
β
xβ
|xβ|2dβ
| {z }
be
γ +2BR (x, γ)·xα=−(Q2(x))α|BR(x, γ)|2+ 2bBRα(x, γ)·xα+ (bγ)α
−Q2(x)γ2
4|xα|2α−2(P−1
2(x))α+g
18
whe e
BR(z, $)(α) = 1
2πPV Zπ
−π
(z(α)−z(α−β))⊥
|z(α)−z(α−β)|2$(α−β)dβ,
will be he e o o zand gwill be he e o o ω.
4.1 Compu ing he di e ence z−xand ω−γ
We de ine now:
D≡z−x, d ≡ω−γ, D ≡ ϕ−ψ
The ene gy
E( )≡1
2kDk2
L2+Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2+kdk2
H2+kDk2
H3+ 1
2
and he Rayleigh-Taylo condi ion
σz≡BR +ϕ
|zα|BRα·z⊥
α+ω
2|zα|2zα +ϕ
|zα|zαα·z⊥
α
+QBR +ω
2|zα|2zα
2
∇Q·z⊥
α−(∇P−1
2)(z)·z⊥
α
No e ha σz>0. We shall show ha

d
d E( )≤ C( )(E( ) + Ek( )) + cδ( )
whe e
C( ) = C(kxkH5+ 1
2( ),kγkH3+ 1
2( ),kψkH4+ 1
2( ),kF(x)kL∞( ))
and
δ( )=(k kH5+ 1
2( ) + kgkH3+ 1
2( ))k+ (k kH5+ 1
2( ) + kgkH3+ 1
2( ))2, k big enough
depend on he no ms o and g.
Rema k 4.1 F om now on, we will deno e E( ) + E( )kby P(E( )).
1
2
d
d kDk2
L2≤CP (E( )) + δ( ) is le o he eade . We compu e
1
2
d
d Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2=1
2Zπ
−π
(Q2
zσz)
|zα|2|∂4
αD|2+Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
αD
The i s in eg al is easy o bound by CP (E( )), we p oceed as in he local exis ence
Theo em I.7 in [3]. We spli
I=Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
αD =I1+I2+I3
19
whe e
I1=Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
α(Q2
zBR(z, ω)−Q2
xBR(x, γ))dα
I2=Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
α(czα−bxα)dα
I3=Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
α dα
We ha e:
I3≤1
2Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2dα +1
2Zπ
−π
Q2
z
|zα|2σz|∂4
α |2dα ≤CP (E( )) + kQ2
zσzkL∞
2δ( )
Thus, we a e done wi h I3. We now spli
I1= l.o. + I1,1+I1,2+I1,3+I1,4
I1,1=Zπ
−π
Q2
z
|zα|2σz∂4
αD(∂4
α(Q2
z)BR(z, ω)−∂4
α(Q2
x)BR(x, γ))dα
I1,2=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
2πZπ
−π
(∂4
αz(α)−∂4
αz(α−β))⊥
|z(α)−z(α−β)|2ω(α−β)dβ
−Q2
x
1
2πZπ
−π
(∂4
αx(α)−∂4
αx(α−β))⊥
|x(α)−x(α−β)|2γ(α−β)dβdα
I1,3=Zπ
−π
Q2
z
|zα|2σz∂4
αD
×Q2
z−1
πZπ
−π
(z(α)−z(α−β))⊥
|z(α)−z(α−β)|4(z(α)−z(α−β)) ·(∂4
αz(α)−∂4
αz(α−β))ω(α−β)dβ
+Q2
x
1
πZπ
−π
(x(α)−x(α−β))⊥
|x(α)−x(α−β)|4(x(α)−x(α−β)) ·(∂4
αx(α)−∂4
αx(α−β))γ(α−β)dβ
I1,4=Zπ
−π
Q2
z
|zα|2σz∂4
αD∂4
α(Q2
zBR(z, ∂4
αω)−Q2
xBR(x, ∂4
αγ))dα
whe e l.o. s ands o low o de e ms, nice e ms easie o deal wi h.
I1,1= l.o. + I1,1,1whe e
I1,1,1= 2 Zπ
−π
Q2
z
|zα|2σz∂4
αD(∇Q(z)·∂4
αzBR(z, ω)−∇Q(x)·∂4
αxBR(x, γ))dα
= 2 Zπ
−π
Q2
z
|zα|2σz∂4
αD∇Q(z)·∂4
αDBR(z, ω)dα
+ 2 Zπ
−π
Q2
z
|zα|2σz∂4
αD(∇Q(z)·∂4
αxBR(z, ω)−∇Q(x)·∂4
αxBR(x, γ))dα
20
≤2Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2dα k∇Q(z)BR(z, ω)kL∞
| {z }
bounded as o local exis ence
+Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2+Zπ
−π
Q2
z
|zα|2σz|∇Q(z)·∂4
αxBR(z, ω)−∇Q(x)·∂4
αxBR(x, γ)|2dα
| {z }
l.o. in Dand d
≤CP (E( ))
which means I1,1is done.
F om now on we will deno e
∆βz(α) = z(α)−z(α−β)
I1,2=I1,2,1+I1,2,2+I1,2,3+I1,2,4whe e
I1,2,1=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
2πZπ
−π
∆β∂4
αD⊥(α)
|∆βz(α)|2ω(α−β)dβdα
I1,2,2=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
2πZπ
−π
∆β∂4
αx⊥(α)1
|∆βz(α)|2−1
|∆βx(α)|2ω(α−β)dβdα
I1,2,3=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
2πZπ
−π
∆β∂4
αx⊥(α)
|∆βx(α)|2d(α−β)dβdα
I1,2,4=Zπ
−π
Q2
z
|zα|2σz∂4
αD(Q2
z−Q2
x)1
2πZπ
−π
∆β∂4
αx⊥(α)
|∆βx(α)|2γ(α−β)dβdα
I1,2,1=Zπ
−π
Q4
z
|zα|2σz∂4
αD(α)1
2πZπ
−π
∆α−β∂4
αD⊥(α)
|∆α−βz(α)|2ω(α−β)dβdα
=1
|zα|2
1
2πZπ
−πZπ
−π
∂4
αD∆α−β∂4
αD⊥(α)
|∆α−βz(α)|2Q4
z(α)σz(α)ω(β)−Q4
z(β)σz(β)ω(α)
2
+Q4
z(α)σz(α)ω(β) + Q4
z(β)σz(β)ω(α)
2
| {z }
his is ze o as in local exis ence (∂4
αD·∂4
αD⊥= 0)



dαdβ
⇒I1,2,1=1
2πZπ
−π
∂4
αD
|zα|2Zπ
−π
∆α−β∂4
αD⊥(α)
|∆α−βz(α)|2
| {z }
Hilbe ans o m
applied o ∂4
αD⊥(α)
Q4
z(α)σz(α)ω(β)−Q4
z(β)σz(β)ω(α)
2
⇒I1,2,1≤CP (E( ))
21

Fo I1,2,2we can make a ick o ge less de i a i es in x.
I1,2,2=I1
1,2,2+I2
1,2,2+I3
1,2,2
I3
1,2,2=1
2Zπ
−π
Q4
z
|zα|2σz∂4
αDω(α)1
|zα|2−1
|xα|2
Λ∂4
αx
z }| {
1
πZπ
−π
∆β∂4
αx⊥(α)
β2dβ dα
I2
1,2,2=1
2πZπ
−π
Q4
z
|zα|2σz∂4
αDZπ
−π
∆β∂4
αx⊥(α)

1
|∆βz(α)|2−1
|zα(α)|2β2+
=0
z }| {
zα·zαα
|zα|4β
−

1
|∆βx(α)|2−1
|xα(α)|2β2+
=0
z }| {
xα·xαα
|xα|4β



ω(α)dβdα
I1
1,2,2=1
2πZπ
−π
Q4
z
|zα|2σz∂4
αDZπ
−π
∆β∂4
αx⊥(α)1
|∆βz(α)|2−1
|∆βx(α)|2(ω(α−β)−ω(α)) dβdα
We use ha 
1
|zα|2−1
|xα|2≤|xα|+|zα|
|zα|2|xα|2|Dα| o ind ha
I3
1,2,2≤1
4Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2+kQzk6
L∞kσzkL∞kωk2
L∞|xα|+|zα|
|zα|2|xα|22
Sobole
inequali ies
z }| {
kDαk2
L∞
Con ol o kxkH5
z }| {
kΛ∂4
αxk2
L2
≤CP (E( ))
We can use ha
1
|∆βz(α)|2−1
|zα(α)|2β2+zα·zαα
|zα|4β≤ kzkk
C2
1
β1/2kzkC2+ 1
2kF(z)kk
L∞
and ha

1
|∆βz(α)|2−1
|zα(α)|2β2+zα·zαα
|zα|4β−1
|∆βx(α)|2−1
|xα(α)|2β2+xα·xαα
|xα|4β
≤ kzkk
C2kxkk
C2
1
β1/2kDkC2+ 1
2kF(z)kk
L∞kF(x)kk
L∞
o ind
I2
1,2,2≤1
8π2Zπ
−π
Q2
z
|zα|2σz|∂4
αD|2
+CkQzk6
L∞kσzkL∞kzkk
C2kxkk
C2kDkC2+ 1
2k∂4
αxk2
L2kF(z)kk
L∞kF(x)kk
L∞
We’ e used ha
Zπ
−π
dα Zπ
−π
∂4
αx(α−β)
|β|1/2dβ2!1/2
≤Ck∂4
αxkL2.
22
We spli u he in I1
1,2,2=I1,1
1,2,2+I1,2
1,2,2:
I1,1
1,2,2=1
2πZπ
−π
Q4
z
|zα|2σz∂4
αDZπ
−π
∆β∂4
αx⊥(α)1
|∆βz(α)|2−1
|∆βx(α)|2
×(ω(α−β)−ω(α) + ωα(α)β)dβdα
I1,2
1,2,2=1
2πZπ
−π
Q4
z
|zα|2σz∂4
αDωα(α)Zπ
−π
∆β∂4
αx⊥(α)β
|∆βz(α)|2−β
|∆βx(α)|2dβdα
Inside o he βin eg al in I1,1
1,2,2 he e is no p incipal alue, so he app op ia e es ima e
ollows:
I1,1
1,2,2≤CP (E( ))
Fo I1,2
1,2,2we p oceed as o I2
1,2,2. We decompose adding and sub ac ing 1
|zα|2β−1
|xα|2β.
Thus, we a e done wi h I1,2,2. We decompose I1,2,3=I1
1,2,3+I2
1,2,3+I3
1,2,3.
I1
1,2,3=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
2πZπ
−π
∆β∂4
αx⊥(α)
×1
|∆βx(α)|2−1
|xα|2β2+xα·xαα
|xα|4βd(α−β)dβdα
I2
1,2,3=−Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
∂4
αx⊥(α)
|xα|2
1
2πZπ
−π
∆βd(α)
β2dβdα
I3
1,2,3=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z
1
|xα|2
1
2πZπ
−π
∆β(d∂4
αx⊥)(α)
β2dβdα
I ’s easy o ob ain:
I1
1,2,3≤1
4πZπ
−π
Q2
z
|zα|2σz|∂4
αD|2dα +CkQzk6
L∞kσzkL∞kdkL∞kxkk
C2kF(x)kk
L∞kxkC2,δ k∂4
αxk2
L2
≤CP (E( ))
I2
1,2,3≤CP (E( )) analogously since kΛdkL∞≤CkdkH2
I3
1,2,3≤CP (E( )) using kΛ(d∂4
αx⊥)kL2≤CkdkH2kxkH5.
We a e done wi h I1,2,3. To deal wi h I1,2,4se use ha
Q2
z−Q2
x= 2Q((1 − )z+ x)∇Q((1 − )z+ x)·D(α) o ∈(0,1).
Then i is easy o ind
I1,2,4≤CP (E( )),
23
and we a e done wi h I1,2. We decompose I1,3as
I1,3=I1,3,1+I1,3,2+I1,3,3+I1,3,4+I1,3,5+I1,3,6
I1,3,1=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π
∆βz⊥(α)
|∆βz(α)|4∆βz(α)·∆β∂4
αD(α)ω(α−β)dβdα
I1,3,2=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π
∆βz⊥(α)
|∆βz(α)|4∆βz(α)·∆β∂4
αx(α)d(α−β)dβdα
I1,3,3=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π
∆βz⊥(α)
|∆βz(α)|4∆βD·∆β∂4
αx(α)γ(α−β)dβdα
I1,3,4=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π
∆βD⊥(α)
|∆βz(α)|4∆βx(α)·∆β∂4
αx(α)γ(α−β)dβdα
I1,3,5=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π
∆βx⊥(α)∆βx(α)·∆β∂4
αx(α)γ(α−β)
×1
|∆βz(α)|4−1
|∆βx(α)|4dβdα
I1,3,6=Zπ
−π
Q2
z
|zα|2σz∂4
αD(Q2
z−Q2
x)−1
πZπ
−π
∆βx⊥(α)
|∆βx(α)|4∆βx(α)·∆β∂4
αx(α)γ(α−β)dβdα
I1,3,j, j = 2,3,4,5,6 a e easie o deal wi h (I can be done as be o e). The e o e we
ocus on I1,3,1.
I1,3,1=I1
1,3,1+I2
1,3,1+I3
1,3,1
I1
1,3,1=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
πZπ
−π∆βz⊥(α)
|∆βz(α)|4∆βz(α)ω(α−β)
−∂αz⊥(α)
|∂αz(α)|4∂αz(α−β)ω(α)1
β2·∆β∂4
αD(α)dβdα
I2
1,3,1=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
π
∂αz⊥(α)
|∂αz(α)|4ω(α)∂4
αD(α)·Zπ
−π
∂αz(α−β)−∂αz(α)
β2dβdα
I3
1,3,1=Zπ
−π
Q2
z
|zα|2σz∂4
αDQ2
z−1
π
∂αz⊥(α)
|∂αz(α)|4ω(α)Zπ
−π
∆β(∂αz·∂4
αD)(α)
β2dβdα
In I1
1,3,1we ind a commu a o , which can be handled as be o e. I is also easy o es ima e
I2
1,3,1.
To deal wi h I3
1,3,1we emembe ha
∂αz(α)·∂4
αD(α) = ∂αz(α)·∂4
αz(α)−∂αx(α)·∂4
αx(α)−∂αD(α)·∂4
αx(α)
=−3∂2
αz(α)·∂3
αz(α)+3∂2
αx(α)∂3
αx(α)−∂αD(α)·∂4
αx(α)
24
Tha allows us o decompose u he
∂αz(α)·∂4
αD(α) = −3∂2
αz(α)·∂3
αD(α)−3∂2
αD(α)∂3
αx(α)−∂αD(α)·∂4
αx(α)
which yields
I3
1,3,1=I3,1
1,3,1+I3,2
1,3,1+I3,3
1,3,1
I3,1
1,3,1=3
πZπ
−π
Q4
z
|zα|2σz∂4
αD·∂αz⊥(α)
|∂αz(α)|4ω(α)Zπ
−π
∆β(∂2
αz·∂3
αD)(α)
β2dβdα
I3,2
1,3,1=3
πZπ
−π
Q4
z
|zα|2σz∂4
αD·∂αz⊥(α)
|∂αz(α)|4ω(α)Zπ
−π
∆β(∂2
αD·∂3
αx)(α)
β2dβdα
I3,3
1,3,1=3
πZπ
−π
Q4
z
|zα|2σz∂4
αD·∂αz⊥(α)
|∂αz(α)|4ω(α)Zπ
−π
∆β(∂αD·∂4
αx)(α)
β2dβdα
We use ha
Zπ
−π
∆β(∂2
αz·∂3
αD)(α)
β2dβ
2
L2≤C∂α(∂2
αz·∂3
αD)2
L2≤CP (E( ))
o con ol I3,1
1,3,1.I3,2
1,3,1 ollows simila ly. We con ol I3,3
1,3,1using ha
Zπ
−π
∆β(∂αD·∂4
αx)(α)
β2dβ
2
L2≤∂α(∂αD·∂4
αx)2
L2
≤ k∂αDk2
L∞k∂5
αxk2
L2+k∂2
αDk2
L∞k∂4
αxk2
L2≤CP (E( ))
This allows us o inish he es ima es o I3,3
1,3,1and I3
1,3,1. We a e done wi h I1,3,1and I1,3.
We now decompose I1,4.
I1,4=I1,4,1+I1,4,2+I1,4,3+I1,4,4
I1,4,1=Zπ
−π
Q2
z
|zα|2σz∂4
αD·Q2
zBR(z, ∂4
αd)dβdα
I1,4,2=Zπ
−π
Q2
z
|zα|2σz∂4
αD·Q2
z
1
2πZπ
−π
∆βD⊥(α)
|∆βz(α)|2∂4
αγ(α−β)dβdα
I1,4,3=Zπ
−π
Q2
z
|zα|2σz∂4
αD·Q2
z
1
2πZπ
−π
∆βx⊥(α)∂4
αγ(α−β)1
|∆βz(α)|2−1
|∆βx(α)|2dβdα
I1,4,4=Zπ
−π
Q2
z
|zα|2σz∂4
αD·(Q2
z−Q2
x)BR(x, ∂4
αγ)dα
25
hen
ψ =−Bx( )ψ−Q2
x
2|xα|∂αψ2
Q2
x−Q2
xBR
xα
|xα|+(P−1
2(z))α
|xα|
+Qx(Qx)
γ
|xα|−2bsBR ·xα
|xα|Qx(Qx)α−Qx
α
Qx
b2
s|xα|− Q3
x
|xα||BR|2Qx
α
−(bs|xα|) +E1
Wi h his o mula i is easy o ind ha
1
2
d
d Z|D|2dx ≤CP(E( )) + cδ( )
In o de o deal wi h II
II =Zπ
−π
Λ∂3
αD∂3
αD dα
we ake a de i a i e in αin he equa ion o ωand ψ o eo ganize he mos dange ous
e ms. I we ind a e m o low o de , we will deno e i by NICE. Since he equa ions o
ϕ and ψ a e analogous excep o he E1 e m, he NICE e ms a e going o be easie o
es ima e in e ms o CP(E( )) + cδ( ).
ψα =−Bx( )ψα−∂αQ2
x
2|xα|∂αψ2
Q2
x−



Q2
x



BR
xα
|xα|
| {z }
(3)
+(P−1
2(z))α
|xα|







α
+Qx(Qx)
γ
|xα|α−2bsBR ·xα
|xα|Qx(Qx)αα−Qx
α
Qx
b2
s|xα|α−Q3
x
|xα||BR|2Qx
αα
−(bs|xα|)α
|{z }
(3)
+E1
α
Expanding (3):
(3) = −Q2
xBR
xα
|xα|α−(bs|xα|)α
=−Q2
xBR α
xα
|xα|−Q2
xBR xα
|xα|α−|xα|Bx( )−(Q2
xBR)α·xα
|xα|
=−(|xα|Bx( )) + (Q2
xBR)α·xα
|xα|
+ 2(Qx(Qx) BR)α·xα
|xα|−Q2
xBR ·xα
|xα|α
We use ha xα
|xα|α
=xαα ·x⊥
α
|xα|2
x⊥
α
|xα|;xα
|xα|
=xα ·x⊥
α
|xα|2
x⊥
α
|xα|
32

o ind
ψα =−Bx( )ψα
| {z }
(4)
−∂2
α(ψ2)
2|xα|
| {z }
(5)
+∂α(Qx)α
|xα|Qx
ψ2
| {z }
(6)
−Q2
xBR ·x⊥
α
xαα ·x⊥
α
|xα|3−(|xα|Bx( ))
+ (Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3
| {z }
(13)
+ 2(Qx(Qx) BR)α
xα
|xα|
| {z }
(7)
−Q2
x
(P−1
2(z))α
|xα|α
| {z }
(8)
+Qx(Qx)
γ
|xα|α
| {z }
(9)
−2bsBR ·xα
|xα|Qx(Qx)αα
| {z }
(10)
−(Qx)α
Qx
b2
s|xα|α
|{z }
(11)
−Q3
x
|xα||BR|2(Qx)αα
| {z }
(12)
+E1
α
The e m (|xα|Bx( )) depends only on so i is no going o appea in compu ing II.
(4) = −Bx( )ψαis NICE (a he le el o ψα)
(5) = −∂2
α(ψ2)
2|xα|is a anspa en e m which is NICE (e en i we ha e o deal wi h Λ1/2)
(6) = ∂α(Qx)α
|xα|Qx
ψ2=−(Qx)2
α
|xα|(Qx)2+2(Qx)αψψα
|xα|Qx
+ψ2
Qx(Qx)α
|xα|α
The i s e m is a he le el o ∂αxso i is NICE. The second e m is a he le el o ∂αx
o ψαso i is NICE. We w i e he las one as
ψ2
Qx(Qx)α
|xα|α
=ψ2
Qx
xα·∇2Q(x)·xα
|xα|+ψ2
Qx
xα∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
The i s e m is a he le el o xαo ψso i is NICE. Fo he second e m we ha e used
ha
xα
|xα|α
=xαα ·x⊥
α
|xα|2
x⊥
α
|xα|
Finally:
(6) = NICE + ψ2
Qx
xα∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
33
(7) = 2(Qx(Qx) BR)α
xα
|xα|= 2(Qx)α(Qx) BR ·xα
|xα|+ 2Qx(Qx)
|xαα
BR ·xα
+ 2Qx(Qx) BRα·xα
|xα|
The i s e m is a he le el o xα, x , BR ∼xαso i is NICE. We use ha
(Qx) α
|xα|=(Qx)α
|xα|=(∇Q(x)·xα)
|xα|=∇Q(x)·xα
|xα| −∇Q(x)·xα1
|xα|
Using ha
xα·xα
|xα|2=Bx( ) + 1
2πZπ
−π
α·xα
|xα|2dα
and xα
|xα|
=xα ·x⊥
α
|xα|2·x⊥
α
|xα|
we ind ha
(Qx) α
|xα|=x ·∇2Q(x)·xα
|xα|+∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
+∇Q(x)·xα
|xα|Bx( ) + ∇Q(x)·xα
|xα|
1
2πZπ
−π
α·xα
|xα|2dα (19)
Tha yields
(7) = 2(Qx(Qx) BR)α
xα
|xα|= NICE + 2QxBR ·xαx ·∇2Q(x)·xα
|xα|
| {z }
NICE (a he le el o xα,x ,BR)
+ 2QxBR ·xα∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3+ 2QxBR ·xα∇Q(x)·xα
|xα|Bx( )
| {z }
NICE (a he le el o xα,x ,BR)
+ 2QxBR ·xα∇Q(x)·xα
|xα|
1
2πZπ
−π
α·xα
|xα|2dα
| {z }
pa o e o e ms
+2Qx(Qx) BRα·xα
|xα|
Finally:
(7) = 2(Qx(Qx) BR)α
xα
|xα|= NICE + 2QxBR ·xα∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3+ 2Qx(Qx) BRα·xα
|xα|
34
(8) = −Q2
x
(P−1
2(z))α
|xα|α
=−Q2
x∇P−1
2(x)·xα
|xα|α
=−2Qx∇Qx
x·xα∇P−1
2(x)·xα
|xα|
| {z }
NICE (a he le el o xα)
−Q2
xxα·∇2P−1
2(x)·xα
|xα|
| {z }
NICE (a he le el o xα)
−Q2
x∇P−1
2(x)·x⊥
α
xαα ·x⊥
α
|xα|3
which means
(8) = −Q2
x
(P−1
2(z))α
|xα|α
= NICE −Q2
x∇P−1
2(x)·x⊥
α
xαα ·x⊥
α
|xα|3
(9) = Qx(Qx)
γ
|xα|α
= (Qx)α(Qx)
γ
|xα|
| {z }
NICE (a he le el o xα,x )
+Qx
(Qx)α
|xα|γ+Qx(Qx) γ
|xα|α
We use (19) o deal wi h (Qx)α
|xα|. We ind ha
(9) = Qx(Qx)
γ
|xα|α
= NICE + Qxγ∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3+Qx(Qx) γ
|xα|α
(10) = −2bsBR ·xα
|xα|Qx(Qx)αα
=−2bsBR ·xα
|xα|(Qx)2
α
| {z }
NICE as be o e
−2bsBR ·xα
|xα|α
Qx(Qx)α
−2bsBR ·xαQx∇Qx(x)·x⊥
α
xαα ·x⊥
α
|xα|3−2bsBR ·xα
|xα|Qxxα(∇2Qx(x)) ·xα
| {z }
NICE as be o e
The e o e
(10) = −2bsBR ·xα
|xα|Qx(Qx)αα
= NICE −2bsBR ·xα
|xα|α
Qx(Qx)α
−2bsBR ·xαQx∇Qx(x)·x⊥
α
xαα ·x⊥
α
|xα|3
(11) = −(Qx)α
Qx
b2
s|xα|α
=−b2
s|xα|α
(Qx)α
Qx−b2
s|xα|2
Qx∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
−xα(∇2Q(x)·xα)
Qx
b2
s|xα|+(Qx)2
α
(Qx)2b2
s|xα|
35
The ac ha he las wo e ms a e NICE, allows us o ind ha
(11) = −(Qx)α
Qx
b2
s|xα|α
= NICE −b2
s|xα|α
(Qx)α
Qx−b2
s|xα|2
Qx∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
Finally:
(12) = −Q3
x
|xα||BR|2(Qx)αα
=−3(Qx)2(Qx)2
α|BR|2
| {z }
NICE
−Q3
x
|xα|(|BR|2)α(Qx)α
−Q3
x
|xα||BR|2xα·(∇2Q(x)·xα)
| {z }
NICE
−Q3
x|BR|2∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
which implies ha
(12) = −Q3
x
|xα||BR|2(Qx)αα
= NICE −Q3
x
|xα|(|BR|2)α(Qx)α−Q3
x|BR|2∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
We ga he all he o mulas om (4) o (12) abso bing he e o e ms by ˜
E1
αwhene e we
encoun e hem.
I yields:
ψα = NICE + ψ2
Qx∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(16)
−Q2
xBR ·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(15)
−Q2
x∇P−1
2(x)·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(15)
+Qxγ∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
| {z }
(18)
+Qx(Qx) γ
|xα|α
| {z }
(14)
+ 2QxBR ·xα∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
| {z }
(18)
+Qx(Qx) 2BRα·xα
|xα|
| {z }
(14)
−2bxBR ·xα
|xα|α
Qx(Qx)α
| {z }
(17)
−2bsBR ·xαQx∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(16)
−(b2
s|xα|)α
(Qx)α
Qx
| {z }
(17)
−b2
s|xα|2
Qx∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(16)
−Q3
x
|xα|(|BR|2)α(Qx)α
| {z }
(17)
−Q3
x|BR|2∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
| {z }
(16)
+(Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3+˜
E1
α
36
We compu e
(14) = Qx(Qx) γ
|xα|α
+Qx(Qx) 2BRα·xα
|xα|
= 2(Qx)
Qx
(Qx)2γ
2|xα|α
+ 2(Qx)
Qx
(Qx)2BRα·xα
|xα|
= 2(Qx)
Qx
ψα−2(Qx)
Qx
(Q2
x)α
γ
2|xα|−2(Qx)
Qx
(Q2
x)αBRα
xα
|xα|−2(Qx)
Qx
(|xα|Bx( ))
The las o mula allows us o conclude ha (14)=NICE. We eo ganize using (15), (16),
(17) and (18).
ψα = NICE −Q2
x(BR ·x⊥
α+∇P−1
2(x)·x⊥
α)xαα ·x⊥
α
|xα|3
−Q3|BR|2+b2
s|xα|2
Q4
x
+ 2bs
BR ·xα
Q2
x−ψ2
Q4
x∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
+ (Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3+ (Qxγ+ 2QxBR ·xα)∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
−Q3
x(|BR|2)α
|xα|+(b2
s|xα|)α
Qx
+2bsBR ·xα
|xα|α
Qx(Qx)α+˜
E1
α
We add and sub ac e ms in o de o ind he R-T condi ion. We emembe he e ha
σz=BR +ϕ
|zα|BRα·z⊥
α+ω
2|zα|2zα +ϕ
|zα|zαα·z⊥
α
+QzBR +ω
2|zα|2zα
2
∇Q(z)·z⊥
α+∇P−1
2(z)·z⊥
α
σx=BR +ψ
|xα|BRα·x⊥
α+γ
2|xα|2xα +ψ
|xα|xαα·x⊥
α
+QxBR +γ
2|xα|2xα
2
∇Q(x)·x⊥
α+∇P−1
2(x)·x⊥
α(20)
In σx he e a e e o e ms bu hey a e no dange ous. Then, we ind
ψα = NICE
−Q2
xBR +ψ
|xα|BRα·x⊥
α+γ
2|xα|2xα +ψ
|xα|xαα·x⊥
α+∇P−1
2(x)·x⊥
αxαα ·x⊥
α
|xα|3
+(Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3+Q2
xψ
|xα|BRα·x⊥
α+γ
2|xα|2xα +ψ
|xα|xαα·x⊥
αxαα ·x⊥
α
|xα|3
| {z }
(19)
−Q3|BR|2+b2
s|xα|2
Q4
x
+ 2bs
BR ·xα
Q2
x−ψ2
Q4
x∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
37

+ (Qxγ+ 2QxBR ·xα)∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
−Q3
x(|BR|2)α
|xα|+(b2
s|xα|)α
Qx
+2bsBR ·xα
|xα|α
Qx(Qx)α+˜
E1
α
Line (19) can be w i en as
(19) = (Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3+Q2
xBRα·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
+Q2
xγ
2|xα|2xα +ψ
|xα|xαα·x⊥
α
xαα ·x⊥
α
|xα|3
= (Q2
xBR)α·x⊥
α
xα ·x⊥
α
|xα|3+ (Q2
xBR)α·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
+Q2
xγ
2|xα|2xα ·x⊥
α+ψ
|xα|xαα ·x⊥
αxαα ·x⊥
α
|xα|3−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
= (Q2
xBR)α·x⊥
α
1
|xα|3xα ·x⊥
α+ψ
|xα|xαα ·x⊥
α
+Q2
xγ
2|xα|2
1
|xα|3xα ·x⊥
α+ψ
|xα|xαα ·x⊥
αxαα ·x⊥
α−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
=1
|xα|3xα ·x⊥
α+ψ
|xα|xαα ·x⊥
α(Q2
xBR)α·x⊥
α+Q2
xγ
2|xα|2xαα ·x⊥
α
−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
We expand xα o ind
(19) = 1
|xα|3(Q2
xBR)α·x⊥
α+Q2
xγ
2|xα|2xαα ·x⊥
α2
+xαα ·x⊥
α
|xα|3(Q2
xBR)α·x⊥
α+Q2
xγ
2|xα|2xαα ·x⊥
αbe
| {z }
e o e m: we inco po a e i as ˜
E2
α
−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
We deno e
Gx(α)=(Q2
xBR)α·x⊥
α+Q2
xγ
2|xα|2xαα ·x⊥
α(21)
We claim ha
Gx(α) = NICE + |xα|H(∂αψ)
ha becomes
(Gx(α))2= NICE
38
Then
(19) = NICE −2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3+˜
E2
α
We w i e
Gx(α) = 2Qx(Qx)αBR ·x⊥
α
| {z }
NICE, a he le el o xα
+Q2
x
1
2πZ(xα(α)−xα(α−β)) ·xα(α)
|x(α)−x(α−β)|2γ(α−β)dβ
| {z }
NICE, we use ha |xα|2=Ax( )
−Q2
x
1
πZ(xα(α)−xα(α−β)) ·xα(α)
|x(α)−x(α−β)|4(x(α)−x(α−β))(xα(α)−xα(α−β))γ(α−β)dβ
| {z }
NICE, we use ha |xα|2only depends on ime
+Q2
xBR(x, γα)·x⊥
α
| {z }
Hilbe ans o m applied o γα
+Q2
xγ
2|xα|2xαα ·x⊥
α
The e o e
Gx(α) = NICE + |xα|Q2
xH γ
2|xα|α+Q2
xγ
2|xα|2xαα ·x⊥
α
= NICE + |xα|HQ2
xγ
2|xα|α+Q2
xγ
2|xα|2xαα ·x⊥
α
= NICE + |xα|H(∂αψ) + H(bs|xα|2)α+Q2
xγ
2|xα|2xαα ·x⊥
α
= NICE + |xα|H(ψα)−H(Q2
xBR)α·xα+Q2
xγ
2|xα|2xαα ·x⊥
α
(Q2
xBR)α·xα= 2Qx(Qx)αBR ·xα
| {z }
NICE
+Q2
x
1
2πZ(xα(α)−xα(α−β))⊥·xα(α)
|x(α)−x(α−β)|2γ(α−β)dβ
=−Q2
x
1
πZ(x(α)−x(α−β))⊥·xα(α)
|x(α)−x(α−β)|4(x(α)−x(α−β))(xα(α)−xα(α−β))γ(α−β)dβ
| {z }
NICE, ex a cancella ion in (x(α)−x(α−β))⊥·xα(α)
+Q2
x
1
2πZ(x(α)−x(α−β))⊥·xα(α)
|x(α)−x(α−β)|2γ(α−β)dβ
| {z }
NICE, ex a cancella ion in (x(α)−x(α−β))⊥·xα(α)
This means ha
(Q2
xBR)α·xα= NICE + 1
2HQ2
x
∂2
αx⊥·xα
|xα|2γ
39
Taking Hilbe ans o ms:
−H(Q2
xBR)α·xα= NICE −1
2H2Q2
x
∂2
αx⊥·xα
|xα|2γ= NICE + 1
2Q2
x
∂2
αx⊥·xα
|xα|2γ
Using ha ∂2
αx⊥·xα=−∂2
αx·x⊥
αwe a e done. Thus (19) yields
ψα = NICE −Q2
xBR +ψ
|xα|BRα·x⊥
α
+γ
2|xα|2xα +ψ
|xα|xαα·x⊥
α+∇P−1
2(x)·x⊥
αxαα ·x⊥
α
|xα|3
−Q3
x|BR|2+b2
s|xα|2
Q4
x
+ 2bs
BR ·xα
Q2
x−ψ2
Q4
x∇Q(x)·x⊥
α
xαα ·x⊥
α
|xα|3
+ (Qxγ+ 2QxBR ·xα)∇Q(x)·x⊥
α
xα ·x⊥
α
|xα|3
−Q3
x(|BR|2)α
|xα|+(b2
s|xα|)α
Qx
+2bsBR ·xα
|xα|α
Qx(Qx)α
| {z }
(20)
−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3
| {z }
(21)
+E2
α,whe e E2
α=˜
E1
α+˜
E2
α
Fo (20) we w i e
|x |2=Q4
x|BR|2+b2
s|xα|2+ 2Q2
xbsBR ·xα
+b2
e|xα|2+ 2+ 2Q2
xBR ·xαbe+ 2bsbe|xα|2+ 2Q2
xBR · + 2bsxα· + 2bexα·
|{z }
e o e ms ˜
E3
α
⇒|x |2
Qx|xα|=Q3
x|BR|2
|xα|+b2
s|xα|
Qx
+ 2QxbsBR ·xα
|xα|+˜
E3
α
Qx|xα|
Now
(20) = NICE −(|x |2)α
Qx|xα|(Qx)α+˜
E3
α
Qx|xα|(Qx)α
which means
(20) + (21) = NICE −(|x |2)α
Qx|xα|(Qx)α−2Qx(Qx)αBR ·x⊥
α
ψ
|xα|
xαα ·x⊥
α
|xα|3+˜
E3
α
Qx|xα|(Qx)α
40
We w i e
xα = (xα ·xα)xα
|xα|2
| {z }
only depends on
+(xα ·x⊥
α)x⊥
α
|xα|2
=Bx( ) + 1
2πZπ
−π
β·xβ
|xβ|2dβxα+(Q2
xBR)α·x⊥
α+bxαα ·x⊥
α+ α·x⊥
αx⊥
α
|xα|2
=Bx( ) + 1
2πZπ
−π
β·xβ
|xβ|2dβxα+(Q2
xBR)α·x⊥
α+bsxαα ·x⊥
αx⊥
α
|xα|2
+bexαα ·x⊥
α+ α·x⊥
αx⊥
α
|xα|2
=Bx( ) + 1
2πZπ
−π
β·xβ
|xβ|2dβxα+(Q2
xBR)α·x⊥
α+Q2
xγ
2|xα|2xαα ·x⊥
α
| {z }
Gx(α) as in (21)
x⊥
α
|xα|2
−ψ
|xα|xαα ·x⊥
α
x⊥
α
|xα|2bexαα ·x⊥
α+ α·x⊥
αx⊥
α
|xα|2
W i ing x = (Q2
xBR) + bsxα+bexα+ αwe compu e
xα ·xα=Q2
xBR ·xα
| {z }
NICE



Bx( ) + 1
2πZπ
−π
β·xβ
|xβ|2dβ
| {z }
e o



+Gx(α)Q2
xBR ·x⊥
α
|xα|2
| {z }
NICE because Gxis nice
−ψ
|xα|xαα ·x⊥
αQ2
xBR ·x⊥
α
|xα|2+Q2
xBR ·x⊥
α
|xα|2
bexαα ·x⊥
α+ α·x⊥
α
| {z }
e o

x⊥
α
|xα|2
+bsBx( ) + 1
2πZπ
−π
β·xβ
|xβ|2dβ|xα|2
| {z }
NICE
+be
Bx( )
|{z}
e o
+1
2πZπ
−π
β·xβ
|xβ|2dβ
|xα|2+ˆ
E
whe e ˆ
Eis an e o e m. To simpli y we w i e
xα ·xα= NICE −ψ
|xα|xαα ·x⊥
αQ2
xBR ·x⊥
α
|xα|2+ e o s
Se ing he abo e o mula in he exp ession o (20)+(21) allows us o ind
(20) + (21) = NICE + e o s
41
∂2
α∂ (−(Q2
xBR)α·xα) = −Q2
xγ
2|xα|2Λ(∂3
αx⊥
·xα) + l.o. . + NICE
Tha gi es
∂2
α∂ (−(Q2
xBR)α·xα) = −ΛQ2
xγ
2|xα|2∂3
αx⊥
·xα+ l.o. . + NICE
which implies
H(∂2
α∂ (−(Q2
xBR)α·xα)) = ∂αQ2
xγ
2|xα|2∂3
αx⊥
·xα+ l.o. . + NICE
=−Q2
xγ
2|xα|2∂α∂3
αx ·x⊥
α+ NICE
Plugging he abo e o mula in (24) we ind ha
Q2
x
2H(∂3
αγ ) = |xα|H(∂3
αψ )−Q2
xγ
2|xα|2∂α∂3
αx ·x⊥
α+ NICE
=|xα|H(∂3
αψ )−Q2
xγ
2|xα|2∂α∂3
α(Q2
xBR)·x⊥
α−Q2
xγ
2|xα|2∂αbs∂4
αx·x⊥
α+ l.o.
+ NICE + e o s
As we did be o e, in ∂α(∂3
α(Q2
xBR)·x⊥
α), he mos dange ous e m is gi en by Q2
x1
2H(∂4
αγ),
he angen ial e ms appea , which implies
∂α(∂3
α(Q2
xBR)·x⊥
α) = Q2
x
1
2H(∂4
αγ) + NICE
and he e o e
Q2
x
2H(∂3
αγ ) = |xα|H(∂3
αψ )−Q2
xγ
2|xα|2
Q2
x
2H(∂4
αγ)−Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α+ NICE + e o s
We use (23) o ind
Q2
x
2H(∂3
αγ ) + Q2
xψ
2|xα|H(∂4
αγ)
=|xα|H(∂3
αψ )−Q2
x
2bsH(∂4
αγ)−Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α+ NICE + e o s
=|xα|H(∂3
αψ )−bs|xα|H∂4
αQ2
xγ
2|xα|−Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α+ NICE + e o s
=|xα|H(∂3
αψ )−bs|xα|H∂4
αψ−bs|xα|H(∂4
α(bs|xα|)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α
+ NICE + e o s
48

We will show ha
−bs|xα|H(∂4
α(bs|xα|)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α
is NICE and hen we a e done.
−bs|xα|H(∂4
α(bs|xα|)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α=−bsH(∂4
α(bs|xα|2)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α
=bsH(∂3
α((Q2
xBR)α·xα)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α
We epea he calcula ion o dealing wi h he mos dange ous e ms in
∂3
α((Q2
xBR)α·xα)=Λ∂4
αx⊥·xα
γQ2
x
2|xα|2+ l.o.
In he l.o. we use ha ∆βx⊥(α)·x(α) gi es an ex a cancella ion. We ind ha
bsH(∂3
α((Q2
xBR)α·xα)) −Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α
=bsH(Λ ∂4
αx⊥·xα
γQ2
x
2|xα|2)−Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α+ NICE
=−bs∂α∂4
αx⊥·xα
γQ2
x
2|xα|2−Q2
xγ
2|xα|2bs∂α∂4
αx·x⊥
α+ l.o. + NICE
Using ha ∂4
αx⊥·xα=−∂4
αx·x⊥
αwe a e done.
Acknowledgemen s
All he au ho s we e pa ially suppo ed by he g an MTM2011-26696 (Spain) and IC-
MAT Se e o Ochoa p ojec SEV-2011-0087. AC was pa ially suppo ed by he ERC g an
307179-GFTIPFD. CF was suppo ed by NSF g an DMS-09-0104.
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Angel Cas o
Depa amen o de Ma em´a icas
Uni e sidad Au ´onoma de Mad id
Ins i u o de Ciencias Ma em´a icas-CSIC
Campus de Can oblanco
Email: angel [email p o ec ed]
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: [email p o ec ed] Email: [email p o ec ed]
F ancisco Gancedo Ja ie G´omez-Se ano
Depa amen o de An´alisis Ma em´a ico Depa men o Ma hema ics
Uni e sidad de Se illa P ince on Uni e si y
C/ Ta ia, s/n 1102 Fine Hall, Washing on Rd,
Campus Reina Me cedes, 41012 Se illa P ince on, NJ 08544, USA
Email: [email p o ec ed] Email: [email p o ec ed]
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