Structural stability for the splash singularities of the water waves problem

Structural stability for the splash singularities of the water waves problem Angel Castro, Diego C´ordoba, Charles Fefferman, Francisco Gancedo, Javier G´omez-Serrano January 27, 2014 Abstract In this paper we show a structural stability result for water waves. The main motivation for this result is that we would like to exhibit a water wave whose interface starts as a graph and ends in a splash. Numerical simulations lead to an approximate solution with the desired behaviour. The stability result will conclude that near the approximate solution to water waves there is an exact solution. 1 Introduction The water waves problem models the motion of an incompressible fluid with constant density ρin a domain Ω(t) with a free boundary ∂Ω(t), which satisfies the Euler equation with the presence of gravity and whose flow in potential. The system, in R2, can be written, after some computations, as an equation for the free boundary, ∂Ω(t) = {z(α, t) = (z1(α, t), z2(α, t)) : α∈R},(1) and an equation for the amplitude of the vorticity, ω(α, t), in the following way zt(α, t) = BR(z, ω)(α, t) + c(α, t)zα(α, t),(2) ωt(α, t) = −2BRt(z, ω)(α, t)·zα(α, t)−ω2 4|∂αz|2α(α, t)+(cω)α(α, t) + 2c(α, t)BRα(z, ω)(α, t)·zα(α, t)−2(z2)α(α, t), (3) where BR(z, ω) is the classical Birkhoff-Rott integral BR(z, ω)(α, t) = 1 2πPV ZR (z(α, t)−z(β, t))⊥ |z(α, t)−z(β, t)|2ω(β, t)dβ. (4) The function c(α, t) is arbitrary since the boundary is convected by the normal component of the velocity of the fluid. Also, we notice that, in order to get an explicit equation for ∂tω, we need to invert the operator I+T=I+ 2hBR(z, ·), zαi 1 arXiv:1401.6419v1 [math.AP] 24 Jan 2014 and we have taken the acceleration due to gravity and the density ρequal to one. Once one has solved this system for (z, ω) the velocity of the fluid and the pressure in the domain Ω(t) can be recovered by using Biot-Savart and Bernoulli laws. For details see [3]. In the last two decades these equations have been intensively studied. For an extensive survey about analytical results on water waves see the monograph [9]. In this paper we are concerned with the problem of the existence of water waves which start as a graph and become a splash curve in finite time. Roughly speaking, a splash curve is a smooth curve that collapses with itself in a single point such as the curve of fig. 1. A rigorous definition can be found in [3] where the existence of splash singularities has been shown. Coutand and Shkoller [5] have proven the existence of splash singularities in presence of vorticity. Fefferman, Ionescu and Lie [6] have proven the non existence of splash singularities for internal waves, i.e. for an interface between two incompressible fluids. Figure 1: Splash singularity. A smooth interface that collapses in a point. We are interested in the following statement: Conjecture 1.1 There exist initial data z0(α), ω0(α)of solutions of the water wave equations such that at time 0the curve z0(α)can be parameterized as a graph, the interface then turns over at a finite time T1>0, and finally produces a splash at a finite time T2> T1. We should remark that this conjecture is a combination of the scenarios in theorems [3, Theorem I.1] and [4, Theorem 7.1] and is supported by numerical evidence that we can see in Fig. 2. This numerical simulation was carried out using the method of Beale, Hou and Lowengrub [1]. The proof of this conjecture could follow along these lines. First of all, we will move backwards in time, 0 being the time of the splash, T2−T1the time of the turning and T2the time in which the solution can be parameterized as a graph. Also we write the water waves 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 Figure 2: Evolution from a graph to a splash. equation in a new domain given by the projection of Ω(t) by the conformal map P(w) = tan w 21/2, w ∈C, whose intention is to keep apart the self-intersecting point by taking the branch of the square root above passing through this crucial point. The equation in this new domain can be written as follows: ˜zt(α, t) = Q2(α, t)BR(˜z, ˜ω)(α, t) + ˜c(α, t)˜zα(α, t),(5) ˜ωt(α, t) = −2BRt(˜z, ˜ω)(α, t)·˜zα(α, t)−(Q2)α(α, t)|BR(˜z, ˜ω)|2(α, t)−Q2(α, t)˜ω(α, t)2 4|˜zα(α, t)|2α + 2˜c(α, t)BRα(˜z, ˜ω)·˜zα(α, t) + (˜c(α, t)˜ω(α, t))α−2P−1 2(˜z(α, t))α where ˜z(α, t) = P(z(α, t)), Q2(α, t) =  dP dw (P−1(˜z(α, t))) 2 and α∈T. (From now on we will omit the superscript tilde in the notation). We start computing a numerical approximation of a solution to the water waves equation 5 that starts as a splash, turns over and finally is a graph. Such a candidate is depicted 3 in Fig. 2. With this aproximation we can construct explicit functions (x, γ) that solve the system      xt=Q2(x)BR(x, γ) + bxα+f γt=−2BRt(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) where fand gare errors that we hope are small. By using the computer we are able to give rigorous bounds for these errors. The question we want to answer is if there exists an exact solution (z, ω) of the water waves equation close to these functions (x, γ). That means we need to prove the following theorem: Theorem 1.2 Let D(α, t)≡z(α, t)−x(α, t), d(α, t)≡ω(α, t)−γ(α, t),D(α, t)≡ϕ(α, t)−ψ(α, t) where (x, γ, ψ)are the solutions of                                      xt=Q2(x)BR(x, γ) + bxα+f 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 fα xα |xα|2dα −Zα −π fβ xβ |xβ|2dβ | {z } be γt+2BRt(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, t) = Q2 x(α,t)γ(α,t) 2|xα(α,t)|−bs(α, t)|xα(α, t)|, (7) where (z, ω)are the solutions of (7) with f≡g≡0,ϕis the function ϕ=Q2 z(α, t)ω(α, t) 2|zα(α, t)|−b(α, t)|zα(α, t)|, and Eis the following norm for the difference E(t)≡kDk2 H3+Zπ −π Q2σz |zα|2|∂4 αD|2+kdk2 H2+kDk2 H3+ 1 2. Then we have that  d dtE(t)≤ C(t)(E(t) + Ek(t)) + cδ(t) where C(t) = C(E(t),kxkH5+ 1 2(t),kγkH3+ 1 2(t),kζkH4+ 1 2(t),kF(x)kL∞(t)) and δ(t) = (kfkH5+ 1 2(t) + kgkH3+ 1 2(t))k+ (kfkH5+ 1 2(t) + kgkH3+ 1 2(t))2, k big enough 4 depends on the norms of fand g, and E(t)is given by E(t) =kzk2 H3(t) + ZT Q2σz |zα|2|∂4 αz|2dα +kF(z)k2 L∞(t) +kωk2 H2(t) + kϕk2 H3+ 1 2(t) + |zα|2 m(Q2σz)(t)+ 4 X l=0 1 m(ql)(t) where the L∞norm of the function F(z)≡|β| |z(α, t)−z(α−β, t)|, α, β ∈T measures the arc-chord condition, σz≡BRt(z, ω) + ϕ |zα|BRα(z, ω)·z⊥ α+ω 2|zα|2zαt +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 the Rayleigh-Taylor function, m(Q2σz)(t)≡min α∈TQ2(α, t)σz(α, t), and finally m(ql)(t)≡min α∈T|z(α, t)−ql| for l= 0, ..., 4, with 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 are the singular points of the transformation P. Remark 1.3 We can absorb the terms in E(t)by E(t)raised to an appropriate power and terms in (x, γ)by performing the splitting kzk=kz−xk+kxk(or the analogous one for a different variable) for any norm or any quantity that appears in E(t). Theorem 1.2 was announced in [2]. If we knew C(t), f(t), g(t), k or bounds on them, a priori, then we could provide bounds on E(t) at any time T. We point out here that E(t) controls the norm k∂αz1(α)−∂αx1(α)kL∞. Let Tgbe a time in which the approximate solution is a graph, i.e. ∂αx1(α, Tg)>0∀α. Now, if E(Tg)< ∂αx1(α, Tg) then ∂αz1(α, Tg)>−k∂αz1(α)−∂αx1(α)kL∞+∂αx1(α, Tg)>0, and this shows that zis a graph. In other words, the possible set of solutions of the water waves equation is a ball centered at (x, γ, ζ) with the topology given by E. All of the elements 5 of this ball are graphs, therefore the solution is necessarily a graph. Thus, the problem is reduced to study and find bounds for C(t), f(t), g(t), k. The recent developments of computer architecture have boosted their use in mathematics, giving birth to a full set of new results only achievable by this enormous power. However, it has the drawback that floating-point operations can not be performed exactly, resulting in numerical errors. In order to overcome this difficulty and be able to prove rigorous results, we use the so-called interval arithmetics, in which instead of working with arbitrary real numbers, we perform computations over intervals which have representable numbers as endpoints. On these objects, an arithmetic is defined in such a way that we are guaranteed that for every x∈X, y ∈Y x ? y ∈X ? Y, for any operation ?. For 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 define the interval version of a function f(X) as an interval Ithat satisfies that for every x∈Xwe have f(x)∈I. The article is organized as follows: in sections 2 and 3 we give some details about how to control the errors f,gand the constants that arise in Theorem 1.2 by using the computer. Finally, in section 4 we give a complete proof of Theorem 1.2. 2 Bounds for f(t)and g(t) 2.1 Representation of the functions and Interpolation The first thing one has to decide is how to represent the data and how to pass from the cloud of points in space-time obtained by non-rigorous simulation to a function defined everywhere in [−π, π]×[0, T]. We need to interpolate in some way. In our case, we chose to represent the functions xand γby piecewise polynomials (splines) of high degree (10) in space, and low degree (3) in time. To do so, we first interpolate in space for every node in the time mesh. The interpolation is made via B-Splines. Since the interpolation is reduced to solve a linear (interval) system Ac =y, where Ais constant in time and space and ydepends on the values of the function at time tsince the mesh in space is constant, we precondition by multiplying by the non-rigorous inverse of the midpoints of the entries of A. We remark that the system is interval-based because we need to produce a curve that is a splash (i.e. there have to be two points α1, α2such that we can guarantee x0(α1) = x0(α2). Finally, the system is solved using a rigorous Gauss-Seidel iterative method. We also remark that the need for interval-based calculations is only strictly necessary at time t= 0 since it is the only point in which we have to guarantee some equality. By working with multiprecision (1024 bits) we can get widths in the coefficients of the order of 10−300. In order to perform interpolation in time, we fix the values of the function and its time derivative at the mesh points. This gives us lots of systems of 4 equations (the values of 6 the function and its derivative at both endpoints) and 4 unknowns (the 4 coefficients of the degree 3 polynomial) but with an explicit formula for each of them. With this method, our spline will be C1in time but it might not be C2. 2.2 Rigorous bounds for Singular integrals In this section we will discuss the computational details of the rigorous calculation of some singular integrals. In particular we will focus on the Hilbert transform, but the methods apply to any integral kernel whose main singularity is homogeneous of degree -1. Parts of the computation (the Npart) are slightly related to the Taylor models with relative remainder presented in M. Jolde¸s’ thesis [8]. Let us suppose that we have a function fgiven explicitly by a spline (piecewise polynomial) which is Ck−1everywhere and Ckexcept at finitely many points (the points in which the different pieces of the spline are glued together). We need to calculate rigorously the Hilbert Transform of f, that is Hf(x) = PV πZT f(x)−f(y) 2 tan x−y 2dy, and we want to approximate it by a piecewise polynomial function with less regularity, plus an error that can be bounded in Hq,0≤q≤c<kand in L∞. Let us assume that the knots of the spline are αi,i= 0, . . . , N −1 and that we fix x∈[αi, αi+1] where the indices are taken modulo Nand the distance between the indices is taken over ZN. We can split our integral in Hf(x) = PV πZT f(x)−f(y) 2 tan x−y 2dy =PV πX jZαj+1 αj f(x)−f(y) 2 tan x−y 2dy =PV πX |j−i|>K Zαj+1 αj f(x)−f(y) 2 tan x−y 2dy +PV πX |j−i|≤KZαj+1 αj f(x)−f(y) 2 tan x−y 2dy ≡HfF(x) + HfN(x). Now, if we want to express HfF(x) as a polynomial, it is easy since the integrand does not have a singularity. Hence HfF(x) = PV πX |j−i|>K Zαj+1 αj f(x)−f(y) 2 tan 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), where Eaccounts for the error and is a polynomial with interval coefficients. Typically, we will use as the points for the Taylor expansions x∗(i) = αisince we will compare the resulting polynomial with another one of the form Pjbj(x−xi)jand we will also choose 7 y∗(j) = αj+αj+1 2. This choice is useful for two reasons: first, we will only have to integrate half of the terms since the rest will integrate to zero; and second, the error estimates will be better for this choice of y∗(j) in the sense that the coefficients will be smaller. All the computations will be carried out using automatic differentiation. We should remark that we can get estimates for the error Ein any of the above mentioned norms without having to recompute it since the relation ∂q xHfF(x)−∂q xP(x) = ∂q xE(x) holds for every q < k. Now, we move on to the term HfN(x). In this case, we perform a Taylor expansion in both the denominator 2 tan x−y 2= (x−y) + c(x−y)3, c = small (interval) constant and the numerator f(x) = f(y)+(x−y)f0(y) + 1 2(x−y)2f00(y) + . . . 1 n!(x−y)k−1fk−1(η), where ηbelongs to an intermediate point between xand y, which we can enclose in the convex hull of [αi, αi+1] and [αj, αj+1] where the convex hull is understood in the torus. Since typically Kwill be very small (compared to N) there is no ambiguity in the definition. Finally, we can factor out (x−y) and divide both in the numerator and the denominator. Since we know f(y) explicitly, we can perform the explicit integration and get a piecewise polynomial as a result. 2.3 Estimates of the norm of the Operator I+T In this subsection we will outline how to compute the norm of the operator I+T= I+ 2hBR(z, ·), zαi. Since the operator Tbehaves like a Hilbert Transform plus smoothing terms, we will describe how to calculate rigorously with the help of a computer an estimate for the norm of its inverse. The procedure is more general and can be applied to a bigger family of kernels. Let T=R/2πZ, and let A(x), B(x) be real-valued functions on T. Also, let E(x, y) be a real-valued function on T×T. We assume A, B and Eare given by explicit formulas such as as perhaps piecewise trigonometric polynomials or splines, and E(x, y) is a trigonometric polynomial on each rectangle I×Jof some partition of T×T. We suppose A, B, E are smooth enough. Let Hbe the Hilbert transform acting on functions on T, i.e. Hf(x) = PV 2πZT cot y 2f(x−y)dy. Assume that Aand Bhave no common zeros on T. Let Sf(x) = A(x)f(x) + B(x)Hf(x) + ZT E(x, y)f(y)dy, f ∈L2(T). 8 Thus, Sis a singular integral operator. We hope that S−1exists and has a not-so-big norm on L2, but we don’t know this yet. Our goal here is to find approximate solutions Fof the equation SF =ffor suitable given f∈L2(T), and to check that kSF −fkL2(T)< δ for suitable δ. Our computation of F will be based on heuristic ideas, but the computation of an upper bound for kSF −fkL2(T) will be rigorous. In our case, A(x) = 1, B(x) = 1. To carry this out, let H0⊂H1⊂L2(T) be finite-dimensional subspaces, e.g. with Hi consisting of the span of wavelets (from a wavelet bases) having lengthscale ≥2−Ni. Here N1≥N0+ 3 (say). Let πibe the orthogonal projection from L2(T) to Hi, and let us solve the equation π1Sπ1F=π0f. (10) If fis given explicitly in a wavelet bases, then (10) is a linear algebra problem, since π1Sπ1is of finite rank, and its matrix (in terms of some given basis for H1) can be computed explicitly. •If π0f6∈ Range(π1Sπ1), then our heuristic procedure fails. •If π0f∈Range(π1Sπ1), then we find F∈H1such that π1Sπ1F=π0f, i.e. π1SF = π0f. We then have kSF −fkL2(T)≤ k(I−π1)SFkL2(T)+k(I−π0)fkL2(T), and both norms on the right-hand side may be estimated explicitly. Now, our goal is to make a heuristic computation of an operator of the form ˜ Sf(x) = ˜ A(x)f(f) + ˜ B(x)Hf(x) + ZT ˜ E(x, y)f(y)dy such that S˜ S−Ihas small norm on L2(T). Here, we will make a heuristic computation of ˜ S; later we will give a rigorous upper bound for the norm of S˜ S−Ion L2(T). By a heuristic computation of ˜ Swe mean a heuristic computation of ˜ A, ˜ Band ˜ E. We first find ˜ Aand ˜ Bby setting (A+iB)( ˜ A+i˜ B)=1⇒A˜ A−B˜ B= 1 A˜ B+B˜ A= 0 Then, this means that S˜ S= (A˜ A−B˜ B)+(A˜ B+B˜ A)H+ Smoothing terms = I+ Smoothing terms So, from now on, we suppose that ˜ Aand ˜ Bare known. For the operator I+T, this means ˜ A= 1/2,˜ B=−1/2. We want to compute ˜ E. Now, let {φν}be some orthonormal basis for 9 First of all, we will work with Q= 1 and later move on to the case Q6= 1. We will adopt the following convention to denote the different Kernels (integral operators) that appear: 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 operators for which b26=−1 will act on Dor its derivatives whereas the operators for which b2=−1 will act on dor its derivatives. We now describe how to split the Kernels in such a way that they can be computed. For the case where b26=−1 we illustrate this by splitting Θ0,0,0,0 2,0, but the technique can be applied to any Kernel. 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 tan α−β 2γ(β)dβ | {z } T4 ,(15) where K(α, β) = 1 (x(α)−x(β))2−c1(α) 4 sin2α−β 2−c2(α) 2 tan α−β 2 c1(α) = 1 x2 α(α) c2(α) = xαα(α) x3 α(α). We can think of c1(α) and c2(α) as the Taylor coefficients of Θ(α, β) around β=α. We can bound the terms in (15) in the following way: T4(α) = c2(α)[H(Dγ)(α)−DH(γ)(α)] T3(α) = c1(α)[Λ(Dγ)(α)−DΛ(γ)(α)] We have then the estimates 16 kT4kL2≤ kc2kL∞(kDkL2kγkL∞+kDkL2kHγkL∞) kT3kL2≤ kc1kL∞(kDkL2kγαkL∞+kDαkL2kγkL∞+kDkL2kΛ(γ)kL∞). We now move on to T1. We will estimate it in the following way: ZT1D(α)dα =1 2πZ|D(α)|2ZK(α, β)γ(β)dβdα ≤1 2πkDk2 L2ZK(·, β)γ(β)dβL∞ . To estimate the kernel T2we will use the Generalized Young’s inequality [7]: kT2(D)k2 L2=1 4π2ZZZK(α, β)γ(β)D(β)K(α, σ)γ(σ)D(σ)dβdσdα. Defining ˜ K(β, σ) = ZK(α, β)γ(β)K(α, σ)γ(σ)dα, we have that 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 finally show how to estimate the Kernels with b2=−1. We will do this by showing how to estimate Θ0,0,0,0 1,−1but the technique can be applied to any Kernel. 1 2πZΘ0,0,0,0 1,−1(d(β))dβ =1 2πZK(α, β)d(β)dβ | {z } T1 +1 2πc1(α)Z1 2 tan α−β 2d(β)dβ | {z } T2 , where K(α, β) = 1 (x(α)−x(β)) −c1(α) 2 tan α−β 2 c1(α) = 1 xα(α). We can easily estimate these two terms applying to T1the same estimates (Young’s inequality) as for T2in the previous case and by noting that T2is 1 2c1(α)H(d). 17 3.3 Estimates for the linear terms with Q6= 1 To perform the real estimates, where Q6= 1 we will use the estimates from the previous sections. We will explain how to pass from the former ones to the latter ones. We will illustrate this by computing the linear terms of the Birkhoff-Rott operator. First of all, the total number of terms will increase by a factor 2, since we will have Q2(z)BR(z, ω)−Q2(x)BR(x, γ)) = (Q2(z)−Q2(x))(BR(z, ω)−BR(x, γ)) | {z } nonlinear +Q2(x)(BR(z, ω)−BR(x, γ)) | {z } calculated before + (Q2(z)−Q2(x))BR(x, γ) | {z } new terms In order to calculate the old terms with Q6= 1, the only thing we have to do is to incorporate a factor of ∂k αQ2(x)(α) in the estimates. The new terms can easily be calculated using that, up to linear order (Q2(z)−Q2(x)) = 1 81 + x4 x,3x2−1 x2D+O(D2). 4 Proof of Theorem 1.2 In this section, we will prove the stability Theorem 1.2. The equations are: SPLASH                zt=Q2 zBR +czα c=α+π 2πZπ −π (Q2BR)α zα |zα|2dα −Zα −π (Q2BR)β zβ |zβ|2dβ ωt+2BRt·zα=−(Q2)α|BR|2+ 2cBRα·zα+ (c$)α −Q2$2 4|zα|2α−2(P−1 2(z))α APPROX                                  xt=Q2(x)BR(x, γ) + bxα+f b=α+π 2πZπ −π (Q2BR)α xα |xα|2dα −Zα −π (Q2BR)β xα |xα|2dβ | {z } bs +α+π 2πZπ −π fα xα |xα|2dα −Zα −π fβ xβ |xβ|2dβ | {z } be γt+2BRt(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 where BR(z, $)(α) = 1 2πPV Zπ −π (z(α)−z(α−β))⊥ |z(α)−z(α−β)|2$(α−β)dβ, fwill be the error for zand gwill be the error for ω. 4.1 Computing the difference z−xand ω−γ We define now: D≡z−x, d ≡ω−γ, D ≡ ϕ−ψ The energy E(t)≡1 2kDk2 L2+Zπ −π Q2 z |zα|2σz|∂4 αD|2+kdk2 H2+kDk2 H3+ 1 2 and the Rayleigh-Taylor condition σz≡BRt+ϕ |zα|BRα·z⊥ α+ω 2|zα|2zαt +ϕ |zα|zαα·z⊥ α +QBR +ω 2|zα|2zα 2 ∇Q·z⊥ α−(∇P−1 2)(z)·z⊥ α Note that σz>0. We shall show that  d dtE(t)≤ C(t)(E(t) + Ek(t)) + cδ(t) where C(t) = C(kxkH5+ 1 2(t),kγkH3+ 1 2(t),kψkH4+ 1 2(t),kF(x)kL∞(t)) and δ(t)=(kfkH5+ 1 2(t) + kgkH3+ 1 2(t))k+ (kfkH5+ 1 2(t) + kgkH3+ 1 2(t))2, k big enough depend on the norms of fand g. Remark 4.1 From now on, we will denote E(t) + E(t)kby P(E(t)). 1 2 d dt kDk2 L2≤CP (E(t)) + δ(t) is left to the reader. We compute 1 2 d dt Zπ −π Q2 z |zα|2σz|∂4 αD|2=1 2Zπ −π (Q2 zσz)t |zα|2|∂4 αD|2+Zπ −π Q2 z |zα|2σz∂4 αD∂4 αDt The first integral is easy to bound by CP (E(t)), we proceed as in the local existence Theorem I.7 in [3]. We split I=Zπ −π Q2 z |zα|2σz∂4 αD∂4 αDt=I1+I2+I3 19 where 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 αfdα We have: I3≤1 2Zπ −π Q2 z |zα|2σz|∂4 αD|2dα +1 2Zπ −π Q2 z |zα|2σz|∂4 αf|2dα ≤CP (E(t)) + kQ2 zσzkL∞ 2δ(t) Thus, we are done with I3. We now split I1= l.o.t + 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 where l.o.t stands for low order terms, nice terms easier to deal with. I1,1= l.o.t + I1,1,1where 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 for local existence +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.t in Dand d ≤CP (E(t)) which means I1,1is done. From now on we will denote ∆βz(α) = z(α)−z(α−β) I1,2=I1,2,1+I1,2,2+I1,2,3+I1,2,4where 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 } this is zero as in local existence (∂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 } Hilbert transform applied to ∂4 αD⊥(α) Q4 z(α)σz(α)ω(β)−Q4 z(β)σz(β)ω(α) 2 ⇒I1,2,1≤CP (E(t)) 21 For I1,2,2we can make a trick to get less derivatives 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 that  1 |zα|2−1 |xα|2≤|xα|+|zα| |zα|2|xα|2|Dα|to find that 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 Sobolev inequalities z }| { kDαk2 L∞ Control of kxkH5 z }| { kΛ∂4 αxk2 L2 ≤CP (E(t)) We can use that 1 |∆βz(α)|2−1 |zα(α)|2β2+zα·zαα |zα|4β≤ kzkk C2 1 β1/2kzkC2+ 1 2kF(z)kk L∞ and that  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∞ to find 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’ve used that Zπ −π dα Zπ −π ∂4 αx(α−β) |β|1/2dβ2!1/2 ≤Ck∂4 αxkL2. 22 We split further 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 of the βintegral in I1,1 1,2,2there is no principal value, so the appropriate estimate follows: I1,1 1,2,2≤CP (E(t)) For I1,2 1,2,2we proceed as for I2 1,2,2. We decompose adding and subtracting 1 |zα|2β−1 |xα|2β. Thus, we are done with 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 It’s easy to obtain: 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(t)) I2 1,2,3≤CP (E(t)) analogously since kΛdkL∞≤CkdkH2 I3 1,2,3≤CP (E(t)) using kΛ(d∂4 αx⊥)kL2≤CkdkH2kxkH5. We are done with I1,2,3. To deal with I1,2,4se use that Q2 z−Q2 x= 2Q((1 −t)z+tx)∇Q((1 −t)z+tx)·D(α) for t∈(0,1). Then it is easy to find I1,2,4≤CP (E(t)), 23 and we are done with 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 are easier to deal with (It can be done as before). Therefore we focus 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 find a commutator, which can be handled as before. It is also easy to estimate I2 1,3,1. To deal with I3 1,3,1we remember that ∂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 That allows us to decompose further ∂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 that Zπ −π ∆β(∂2 αz·∂3 αD)(α) β2dβ 2 L2≤C∂α(∂2 αz·∂3 αD)2 L2≤CP (E(t)) to control I3,1 1,3,1.I3,2 1,3,1follows similarly. We control I3,3 1,3,1using that 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(t)) This allows us to finish the estimates for I3,3 1,3,1and I3 1,3,1. We are done with 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 then ψt=−Bx(t)ψ−Q2 x 2|xα|∂αψ2 Q2 x−Q2 xBRt xα |xα|+(P−1 2(z))α |xα| +Qx(Qx)t γ |xα|−2bsBR ·xα |xα|Qx(Qx)α−Qx α Qx b2 s|xα|− Q3 x |xα||BR|2Qx α −(bs|xα|)t+E1 With this formula it is easy to find that 1 2 d dt Z|D|2dx ≤CP(E(t)) + cδ(t) In order to deal with II II =Zπ −π Λ∂3 αD∂3 αDtdα we take a derivative in αin the equation for ωand ψto reorganize the most dangerous terms. If we find a term of low order, we will denote it by NICE. Since the equations for ϕtand ψtare analogous except for the E1term, the NICE terms are going to be easier to estimate in terms of CP(E(t)) + cδ(t). ψαt =−Bx(t)ψα−∂αQ2 x 2|xα|∂αψ2 Q2 x−    Q2 x    BRt xα |xα| | {z } (3) +(P−1 2(z))α |xα|        α +Qx(Qx)t 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|)αt |{z } (3) +E1 α Expanding (3): (3) = −Q2 xBRt xα |xα|α−(bs|xα|)αt =−Q2 xBRtα xα |xα|−Q2 xBRtxα |xα|α−|xα|Bx(t)−(Q2 xBR)α·xα |xα|t =−(|xα|Bx(t))t+ (Q2 xBR)α·xα |xα|t + 2(Qx(Qx)tBR)α·xα |xα|−Q2 xBRt·xα |xα|α We use that xα |xα|α =xαα ·x⊥ α |xα|2 x⊥ α |xα|;xα |xα|t =xαt ·x⊥ α |xα|2 x⊥ α |xα| 32 to find ψαt =−Bx(t)ψα | {z } (4) −∂2 α(ψ2) 2|xα| | {z } (5) +∂α(Qx)α |xα|Qx ψ2 | {z } (6) −Q2 xBRt·x⊥ α xαα ·x⊥ α |xα|3−(|xα|Bx(t))t + (Q2 xBR)α·x⊥ α xαt ·x⊥ α |xα|3 | {z } (13) + 2(Qx(Qx)tBR)α xα |xα| | {z } (7) −Q2 x (P−1 2(z))α |xα|α | {z } (8) +Qx(Qx)t 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 term (|xα|Bx(t))tdepends only on tso it is not going to appear in computing II. (4) = −Bx(t)ψαis NICE (at the level of ψα) (5) = −∂2 α(ψ2) 2|xα|is a transparent term which is NICE (even if we have to deal with 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 first term is at the level of ∂αxso it is NICE. The second term is at the level of ∂αx or ψαso it is NICE. We write the last 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 first term is at the level of xαor ψso it is NICE. For the second term we have used that 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)tBR)α xα |xα|= 2(Qx)α(Qx)tBR ·xα |xα|+ 2Qx(Qx)t |xαα BR ·xα + 2Qx(Qx)tBRα·xα |xα| The first term is at the level of xα, xt, BR ∼xαso it is NICE. We use that (Qx)tα |xα|=(Qx)αt |xα|=(∇Q(x)·xα)t |xα|=∇Q(x)·xα |xα|t−∇Q(x)·xα1 |xα|t Using that xα·xαt |xα|2=Bx(t) + 1 2πZπ −π fα·xα |xα|2dα and xα |xα|t =xαt ·x⊥ α |xα|2·x⊥ α |xα| we find that (Qx)tα |xα|=xt·∇2Q(x)·xα |xα|+∇Q(x)·x⊥ α xαt ·x⊥ α |xα|3 +∇Q(x)·xα |xα|Bx(t) + ∇Q(x)·xα |xα| 1 2πZπ −π fα·xα |xα|2dα (19) That yields (7) = 2(Qx(Qx)tBR)α xα |xα|= NICE + 2QxBR ·xαxt·∇2Q(x)·xα |xα| | {z } NICE (at the level of xα,xt,BR) + 2QxBR ·xα∇Q(x)·x⊥ α xαt ·x⊥ α |xα|3+ 2QxBR ·xα∇Q(x)·xα |xα|Bx(t) | {z } NICE (at the level of xα,xt,BR) + 2QxBR ·xα∇Q(x)·xα |xα| 1 2πZπ −π fα·xα |xα|2dα | {z } part of error terms +2Qx(Qx)tBRα·xα |xα| Finally: (7) = 2(Qx(Qx)tBR)α xα |xα|= NICE + 2QxBR ·xα∇Q(x)·x⊥ α xαt ·x⊥ α |xα|3+ 2Qx(Qx)tBRα·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 (at the level of xα) −Q2 xxα·∇2P−1 2(x)·xα |xα| | {z } NICE (at the level of 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)t γ |xα|α = (Qx)α(Qx)t γ |xα| | {z } NICE (at the level of xα,xt) +Qx (Qx)αt |xα|γ+Qx(Qx)tγ |xα|α We use (19) to deal with (Qx)αt |xα|. We find that (9) = Qx(Qx)t γ |xα|α = NICE + Qxγ∇Q(x)·x⊥ α xαt ·x⊥ α |xα|3+Qx(Qx)tγ |xα|α (10) = −2bsBR ·xα |xα|Qx(Qx)αα =−2bsBR ·xα |xα|(Qx)2 α | {z } NICE as before −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 before Therefore (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 fact that the last two terms are NICE, allows us to find that (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 that (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 gather all the formulas from (4) to (12) absorbing the error terms by ˜ E1 αwhenever we encounter them. It yields: ψαt = NICE + ψ2 Qx∇Q(x)·x⊥ α xαα ·x⊥ α |xα|3 | {z } (16) −Q2 xBRt·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 xαt ·x⊥ α |xα|3 | {z } (18) +Qx(Qx)tγ |xα|α | {z } (14) + 2QxBR ·xα∇Q(x)·x⊥ α xαt ·x⊥ α |xα|3 | {z } (18) +Qx(Qx)t2BRα·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 xαt ·x⊥ α |xα|3+˜ E1 α 36 We compute (14) = Qx(Qx)tγ |xα|α +Qx(Qx)t2BRα·xα |xα| = 2(Qx)t Qx (Qx)2γ 2|xα|α + 2(Qx)t Qx (Qx)2BRα·xα |xα| = 2(Qx)t Qx ψα−2(Qx)t Qx (Q2 x)α γ 2|xα|−2(Qx)t Qx (Q2 x)αBRα xα |xα|−2(Qx)t Qx (|xα|Bx(t)) The last formula allows us to conclude that (14)=NICE. We reorganize using (15), (16), (17) and (18). ψαt = NICE −Q2 x(BRt·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 xαt ·x⊥ α |xα|3+ (Qxγ+ 2QxBR ·xα)∇Q(x)·x⊥ α xαt ·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 subtract terms in order to find the R-T condition. We remember here that σz=BRt+ϕ |zα|BRα·z⊥ α+ω 2|zα|2zαt +ϕ |zα|zαα·z⊥ α +QzBR +ω 2|zα|2zα 2 ∇Q(z)·z⊥ α+∇P−1 2(z)·z⊥ α σx=BRt+ψ |xα|BRα·x⊥ α+γ 2|xα|2xαt +ψ |xα|xαα·x⊥ α +QxBR +γ 2|xα|2xα 2 ∇Q(x)·x⊥ α+∇P−1 2(x)·x⊥ α(20) In σxthere are error terms but they are not dangerous. Then, we find ψαt = NICE −Q2 xBRt+ψ |xα|BRα·x⊥ α+γ 2|xα|2xαt +ψ |xα|xαα·x⊥ α+∇P−1 2(x)·x⊥ αxαα ·x⊥ α |xα|3 +(Q2 xBR)α·x⊥ α xαt ·x⊥ α |xα|3+Q2 xψ |xα|BRα·x⊥ α+γ 2|xα|2xαt +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 xαt ·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 written as (19) = (Q2 xBR)α·x⊥ α xαt ·x⊥ α |xα|3+Q2 xBRα·x⊥ α ψ |xα| xαα ·x⊥ α |xα|3 +Q2 xγ 2|xα|2xαt +ψ |xα|xαα·x⊥ α xαα ·x⊥ α |xα|3 = (Q2 xBR)α·x⊥ α xαt ·x⊥ α |xα|3+ (Q2 xBR)α·x⊥ α ψ |xα| xαα ·x⊥ α |xα|3 +Q2 xγ 2|xα|2xαt ·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xαt ·x⊥ α+ψ |xα|xαα ·x⊥ α +Q2 xγ 2|xα|2 1 |xα|3xαt ·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xαt ·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 xαt to find (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 } error term: we incorporate it as ˜ E2 α −2Qx(Qx)αBR ·x⊥ α ψ |xα| xαα ·x⊥ α |xα|3 We denote Gx(α)=(Q2 xBR)α·x⊥ α+Q2 xγ 2|xα|2xαα ·x⊥ α(21) We claim that Gx(α) = NICE + |xα|H(∂αψ) that becomes (Gx(α))2= NICE 38 Then (19) = NICE −2Qx(Qx)αBR ·x⊥ α ψ |xα| xαα ·x⊥ α |xα|3+˜ E2 α We write Gx(α) = 2Qx(Qx)αBR ·x⊥ α | {z } NICE, at the level of 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 that |xα|2=Ax(t) −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 that |xα|2only depends on time +Q2 xBR(x, γα)·x⊥ α | {z } Hilbert transform applied to γα +Q2 xγ 2|xα|2xαα ·x⊥ α Therefore 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, extra cancellation 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, extra cancellation in (x(α)−x(α−β))⊥·xα(α) This means that (Q2 xBR)α·xα= NICE + 1 2HQ2 x ∂2 αx⊥·xα |xα|2γ 39 Taking Hilbert transforms: −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 that ∂2 αx⊥·xα=−∂2 αx·x⊥ αwe are done. Thus (19) yields ψαt = NICE −Q2 xBRt+ψ |xα|BRα·x⊥ α +γ 2|xα|2xαt +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 xαt ·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,where E2 α=˜ E1 α+˜ E2 α For (20) we write |xt|2=Q4 x|BR|2+b2 s|xα|2+ 2Q2 xbsBR ·xα +b2 e|xα|2+f2+ 2Q2 xBR ·xαbe+ 2bsbe|xα|2+ 2Q2 xBR ·f+ 2bsxα·f+ 2bexα·f |{z } error terms ˜ E3 α ⇒|xt|2 Qx|xα|=Q3 x|BR|2 |xα|+b2 s|xα| Qx + 2QxbsBR ·xα |xα|+˜ E3 α Qx|xα| Now (20) = NICE −(|xt|2)α Qx|xα|(Qx)α+˜ E3 α Qx|xα|(Qx)α which means (20) + (21) = NICE −(|xt|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 write xαt = (xαt ·xα)xα |xα|2 | {z } only depends on t +(xαt ·x⊥ α)x⊥ α |xα|2 =Bx(t) + 1 2πZπ −π fβ·xβ |xβ|2dβxα+(Q2 xBR)α·x⊥ α+bxαα ·x⊥ α+fα·x⊥ αx⊥ α |xα|2 =Bx(t) + 1 2πZπ −π fβ·xβ |xβ|2dβxα+(Q2 xBR)α·x⊥ α+bsxαα ·x⊥ αx⊥ α |xα|2 +bexαα ·x⊥ α+fα·x⊥ αx⊥ α |xα|2 =Bx(t) + 1 2πZπ −π fβ·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+fα·x⊥ αx⊥ α |xα|2 Writing xt= (Q2 xBR) + bsxα+bexα+fαwe compute xαt ·xα=Q2 xBR ·xα | {z } NICE    Bx(t) + 1 2πZπ −π fβ·xβ |xβ|2dβ | {z } error    +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+fα·x⊥ α | {z } error  x⊥ α |xα|2 +bsBx(t) + 1 2πZπ −π fβ·xβ |xβ|2dβ|xα|2 | {z } NICE +be Bx(t) |{z} error +1 2πZπ −π fβ·xβ |xβ|2dβ |xα|2+ˆ E where ˆ Eis an error term. To simplify we write xαt ·xα= NICE −ψ |xα|xαα ·x⊥ αQ2 xBR ·x⊥ α |xα|2+ errors Setting the above formula in the expression of (20)+(21) allows us to find (20) + (21) = NICE + errors 41 ∂2 α∂t(−(Q2 xBR)α·xα) = −Q2 xγ 2|xα|2Λ(∂3 αx⊥ t·xα) + l.o.t. + NICE That gives ∂2 α∂t(−(Q2 xBR)α·xα) = −ΛQ2 xγ 2|xα|2∂3 αx⊥ t·xα+ l.o.t. + NICE which implies H(∂2 α∂t(−(Q2 xBR)α·xα)) = ∂αQ2 xγ 2|xα|2∂3 αx⊥ t·xα+ l.o.t. + NICE =−Q2 xγ 2|xα|2∂α∂3 αxt·x⊥ α+ NICE Plugging the above formula in (24) we find that Q2 x 2H(∂3 αγt) = |xα|H(∂3 αψt)−Q2 xγ 2|xα|2∂α∂3 αxt·x⊥ α+ NICE =|xα|H(∂3 αψt)−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.t + NICE + errors As we did before, in ∂α(∂3 α(Q2 xBR)·x⊥ α), the most dangerous term is given by Q2 x1 2H(∂4 αγ), the tangential terms appear, which implies ∂α(∂3 α(Q2 xBR)·x⊥ α) = Q2 x 1 2H(∂4 αγ) + NICE and therefore Q2 x 2H(∂3 αγt) = |xα|H(∂3 αψt)−Q2 xγ 2|xα|2 Q2 x 2H(∂4 αγ)−Q2 xγ 2|xα|2bs∂α∂4 αx·x⊥ α+ NICE + errors We use (23) to find Q2 x 2H(∂3 αγt) + Q2 xψ 2|xα|H(∂4 αγ) =|xα|H(∂3 αψt)−Q2 x 2bsH(∂4 αγ)−Q2 xγ 2|xα|2bs∂α∂4 αx·x⊥ α+ NICE + errors =|xα|H(∂3 αψt)−bs|xα|H∂4 αQ2 xγ 2|xα|−Q2 xγ 2|xα|2bs∂α∂4 αx·x⊥ α+ NICE + errors =|xα|H(∂3 αψt)−bs|xα|H∂4 αψ−bs|xα|H(∂4 α(bs|xα|)) −Q2 xγ 2|xα|2bs∂α∂4 αx·x⊥ α + NICE + errors 48 We will show that −bs|xα|H(∂4 α(bs|xα|)) −Q2 xγ 2|xα|2bs∂α∂4 αx·x⊥ α is NICE and then we are 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 repeat the calculation for dealing with the most dangerous terms in ∂3 α((Q2 xBR)α·xα)=Λ∂4 αx⊥·xα γQ2 x 2|xα|2+ l.o.t In the l.o.t we use that ∆βx⊥(α)·x(α) gives an extra cancellation. We find that 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.t + NICE Using that ∂4 αx⊥·xα=−∂4 αx·x⊥ αwe are done. Acknowledgements All the authors were partially supported by the grant MTM2011-26696 (Spain) and ICMAT Severo Ochoa project SEV-2011-0087. AC was partially supported by the ERC grant 307179-GFTIPFD. CF was supported by NSF grant DMS-09-0104. References [1] J. T. Beale, T. Y. Hou, and J. Lowengrub. Convergence of a boundary integral method for water waves. SIAM J. Numer. Anal., 33(5):1797–1843, 1996. [2] A. Castro, D. C´ordoba, C. Fefferman, F. Gancedo, and J. G´omez-Serrano. Splash singularity for water waves. Proceedings of the National Academy of Sciences, 109(3):733–738, 2012. [3] A. Castro, D. C´ordoba, C. Fefferman, F. Gancedo, and J. G´omez-Serrano. Finite time singularities for the free boundary incompressible Euler equations. Ann. of Math. (2), 178(3):1061–1134, 2013. 49 [4] ´ A. Castro, D. C´ordoba, C. Fefferman, F. Gancedo, and M. L´opez-Fern´andez. Rayleightaylor breakdown for the Muskat problem with applications to water waves. Ann. of Math. (2), 175:909–948, 2012. [5] D. Coutand and S. Shkoller. On the Finite-Time Splash and Splat Singularities for the 3-D Free-Surface Euler Equations. Comm. Math. Phys., 325(1):143–183, 2014. [6] C. Fefferman, A. D. Ionescu, and V. Lie. On the absence of “splash” singularities in the case of two-fluid interfaces. arXiv preprint arXiv:1312.2917, 2013. [7] G. B. Folland. Introduction to partial differential equations. Princeton University Press, Princeton, NJ, second edition, 1995. [8] M. Joldes. Rigorous polynomial approximations and applications. PhD thesis, ´ Ecole normale sup´erieure de Lyon, 2011. [9] D. Lannes. The Water Waves Problem: Mathematical Analysis and Asymptotics. Mathematical Surveys and Monographs. Amer Mathematical Society, 2013. Angel Castro Departamento de Matem´aticas Universidad Aut´onoma de Madrid Instituto de Ciencias Matem´aticas-CSIC Campus de Cantoblanco Email: angel [email protected] Diego C´ordoba Charles Fefferman Instituto de Ciencias Matem´aticas Department of Mathematics Consejo Superior de Investigaciones Cient´ıficas Princeton University C/ Nicol´as Cabrera, 13-15 1102 Fine Hall, Washington Rd, Campus Cantoblanco UAM, 28049 Madrid Princeton, NJ 08544, USA Email: [email protected] Email: [email protected] Francisco Gancedo Javier G´omez-Serrano Departamento de An´alisis Matem´atico Department of Mathematics Universidad de Sevilla Princeton University C/ Tarfia, s/n 1102 Fine Hall, Washington Rd, Campus Reina Mercedes, 41012 Sevilla Princeton, NJ 08544, USA Email: [email protected] Email: [email protected] 50