Structural stability for the splash singularities of the water waves problem

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. Re e ences [1] J. T. Beale, T. Y. Hou, and J. Loweng ub. Con e gence o a bounda y in eg al me hod o wa e wa es. SIAM J. Nume . Anal., 33(5):1797–1843, 1996. [2] A. Cas o, D. C´o doba, C. Fe e man, F. Gancedo, and J. G´omez-Se ano. Splash singu- la i y o wa e wa es. P oceedings o he Na ional Academy o Sciences, 109(3):733–738, 2012. [3] A. Cas o, D. C´o doba, C. Fe e man, F. Gancedo, and J. G´omez-Se ano. Fini e ime singula i ies o he ee bounda y incomp essible Eule equa ions. Ann. o Ma h. (2), 178(3):1061–1134, 2013. 49 [4] ´ A. Cas o, D. C´o doba, C. Fe e man, F. Gancedo, and M. L´opez-Fe n´andez. Rayleigh- aylo b eakdown o he Muska p oblem wi h applica ions o wa e wa es. Ann. o Ma h. (2), 175:909–948, 2012. [5] D. Cou and and S. Shkolle . On he Fini e-Time Splash and Spla Singula i ies o he 3-D F ee-Su ace Eule Equa ions. Comm. Ma h. Phys., 325(1):143–183, 2014. [6] C. Fe e man, A. D. Ionescu, and V. Lie. On he absence o “splash” singula i ies in he case o wo- luid in e aces. a Xi p ep in a Xi :1312.2917, 2013. [7] G. B. Folland. In oduc ion o pa ial di e en ial equa ions. P ince on Uni e si y P ess, P ince on, NJ, second edi ion, 1995. [8] M. Joldes. Rigo ous polynomial app oxima ions and applica ions. PhD hesis, ´ Ecole no male sup´e ieu e de Lyon, 2011. [9] D. Lannes. The Wa e Wa es P oblem: Ma hema ical Analysis and Asymp o ics. Ma h- ema ical Su eys and Monog aphs. Ame Ma hema ical Socie y, 2013. 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] 50