GEOMETRY AND A PRIORI ESTIMATES FOR FREE BOUNDARY PROBLEMS OF THE EULER S EQUATION

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1 GEOMETRY AND A PRIORI ESTIMATES FOR FREE BOUNDARY PROBLEMS OF THE EULER S EQUATION JALAL SHATAH AND CHONGCHUN ZENG Abstract. In this paper we derive estimates to the free boundary problem for the Euler equation with surface tension, and without surface tension provided the Rayleigh-Taylor sign condition holds. We prove that as the surface tension tends to zero, when the Rayleigh-Taylor condition is satisfied, solutions converge to the Euler flow with zero surface tension. 1. Introduction In this paper we study free boundary problems of the Euler s equation in vacuum: { v t + v v = p, x Ω t R n (E) v = 0, x Ω t. where for every t R, v(t, ) is the velocity field of an incompressible inviscid fluid in a moving domain (bounded and connected) Ω t R n, n, and p(t, ) is the pressure. The boundary of the domain Ω t moves with the fluid velocity and the pressure at the boundary is given by the surface tension, that is (BC) D t = t + v is tangent to Ω t R n+1 p(t, x) = ɛ κ(t, x), x Ω t, 0 ɛ 1 where κ(t, x) is the mean curvature of the boundary Ω t at x Ω t, and D t is the material derivative. This is equivalent to saying the velocity of Ω t is given by v N where N is the unit normal to Ω t. The case ɛ = 0 corresponds to the zero surface tension problem. In the presence of surface tension we will derive energy estimates that bound the Sobolev norms of the velocity and the boundary. In addition we will show that if the Rayleigh-Taylor sign condition is verified, then some of these bounds are independent of ɛ. In this case we conclude that as ɛ 0 solutions of the problem with surface tension converge to solutions of the zero surface tension problem. We do not include the effects of gravity in (E) as it will only contribute lower order terms to our estimates. The free boundary value problem for (E) has been studied intensively by many authors. In the absence of surface tension the earliest mathematical results on the well posedness of the water waves problem were given by V. I. Nalimov [NA74] where he considered the irrotational problem in dimensions with small data in some Sobolev space (see also H. Yoshihara [YO8]). The first break through in solving the well posedness for the irrotational problem, no surface tension, for general data came in the work of S. J. Wu [WU97, WU99] who solved the problem in all dimensions. For the general problem with no surface tension D. Christodoulou and H. Lindblad [CL00] were the first to obtain energy estimates based on the geometry of the moving domain, assuming the Rayleigh-Taylor sign condition for rotational flows. H. Lindblad [LI05] proved existence of solutions for the general problem. In the absence of this condition D. Ebin [EB87] proved that the problem The first author is funded in part by NSF DMS The second author is funded in part by NSF DMS and the Sloan Fellowship. 1 t

2 SHATAH AND ZENG is ill-posed. There is also the work of K. Beyer and M. Günther [BG98, BG00] on well posedness which we will comment more on at the end of the introduction. For the problem with surface tension H. Yoshihara [YO83], T. Iguchi [IG01] and D. Ambrose [AM03] solved the well posedness irrotational problem in dimensions under varying assumptions on the initial data. B. Schweizer [SC05] proved existence for the general 3 dimensional problem. Recently D. Ambrose and N. Masmoudi [AM05a] proved that as ɛ 0 solutions of the dimensional irrotational problem converges to solutions of the zero surface problem by writing the equation in terms of the arc length of the fluid boundary. There are many other works on this problem we mention the work of D. Lannes [LA05], T. J. Beal, T. Hou, and J. Lowengrub [BHL93], W. Craig [CR85], G. Shnider and E. C. Wayne [SW0], M. Ogawa and A. Tani [OT0]. During the writing of this manuscript we were informed and received reprints of several related work by D. Coutand and S. Shkoller [SC05], D. Ambrose and N. Mamoudi [AM06], and P. Zhang and Z. Zhang [ZZ06]. In D. Coutand and S. Shkoller work they proved local well posedness of the general problem using Lagrangian coordinates. Our results differs from those mentioned above in that we can obtain the ɛ 0 limit for the general problem. Our approach to the problem is based on the well known fact that the free boundary problem (E, BC) has a Lagrangian formulation given by u t I(u) = Ω0 dydt ɛ S(u)dt, where u(t, ) Γ = {Φ : Ω 0 R n, volume preserving homeomorphisms}, and S(u) is the surface area of u( Ω t ). Critical points of I satisfy (E-L) t u t + q + ɛ S (u) = 0, where q is the Lagrange multiplier due to the constraint u Γ. Writing v = u t u 1 and changing to Eulerian coordinates we obtain t u t D t v, q p v,v = 1 tr(dvdv), S (u) J κ H where 1 is the inverse Laplacian with zero Dirichlet data, κ the mean curvature, and κ H is the harmonic extension of κ into Ω t. Thus (E-L) is the Euler equation with the pressure p given by p = p v,v + ɛ κ H. It is important to note that this derivation splits the pressure into two terms, the first p v,v is the Lagrange multiplier, and the second κ H is due to surface tension. Thus these two terms will be treated differently in the energy estimates. Using this variational derivation, one can interpret the Lagrange multiplier as the second fundamental form of the manifold Γ L (Ω 0, R n ) and rewrite (E-L) as D t u t + ɛ S (u) = 0. where D is the Riemannian connection induced on Γ by the embedding in L. The above form of the equation makes it relatively easy to identify the correct linearized problem D t w + R(u t, w)u t + ɛ D S(u)( w) = 0, w(t, ) T u(t, ) Γ, where R is the curvature tensor of the infinite dimensional manifold Γ L. Keeping the highest order terms in the above equation we obtain (LN) D t w + R 0 (v) w + ɛ A w = lower order terms, Symbols in the Lagrangian description have a bar, e.g. D, while their Eulerian counterparts do not, D.

3 WATER WAVES 3 where R 0 (v) is a first order differential operator and A is a third order differential operator. In Eulerian coordinates these terms are given by R 0 (v)(w, w) = N p v,v w N ds, A (u)(w, w) = u(ω 0 ) u(ω 0 ) w N ds where N is the unit normal and is the tangential gradient on the boundary of u(ω 0 ). Here once again we are led in a natural way to distinguish the two problems in the following manner. 1) For ɛ > 0 two time derivatives are associated with A,which is a positive semi-definite operator similar to three spatial differentiation, thus roughly speaking, t ( x ) 3. Therefore one may be led to believe that the regularity of the Lagrangian coordinates given by t u = v is 3 order better than v, which reflects the regularizing effect of the surface tension. However this is not true for the Lagrangian coordinates since A is degenerate, and the regularity improvement of the Ω t is geometric and is not reflected in the Lagrangian coordinates system. See Section 5 for examples. Thus Eularian coordinates are more suitable to use than Lagrangian coordinates for our estimates. ) For ɛ = 0 the leading term involves R 0 (v) and thus the Rayleigh-Taylor instability may occur unless we impose the condition (RT) N p v,v (t, x) > a > 0 x Ω t. In this case two time derivatives are associated with R which is a positive semi-definite operator similar to one spatial differentiation. Thus, t ( x ) 1 and comments similar to above hold on the regularity of Ω t. 3) For ɛ > 0 one can directly obtain nonlinear estimates that depend on ɛ by multiplying (E-L) by (D S) k S. 4)The control that any power R 0 (v), with (RT) condition, can give over vector fields is limited by the smoothness of the boundary Ω t. This fact makes the velocity field v inappropriate vector field to estimate because it is smoother than what these operators allow. 5) Since A and R 0 (v) are degenerate for fields which are tangential to the boundary Ω t one needs to add the vorticity ω which controls the rotational part of the velocity which is tangential to the boundary. These facts imply that a natural energy to control is 1 E = A k 1 D t J + ɛ A k 1 J dx + 1 R 0(v)(A k 1 J, A k 1 J) + ω H 3k 1 (Ω) Ω where J = κ H is less smooth than v and satisfies D t J + R 0 (v) J + ɛ A J = lower order terms, In addition to the geometry of Γ the geometry of Ω t plays a crucial role in the estimates. The appearance of κ in the surface tension and κ H in the energy make the study of the geometry of the boundary as well as the study of the harmonic extension and Dirichlet-Neumann operators on Ω t central to the estimate. In using Lagrangian coordinates these operators may be hidden but can not be avoided. Based on these estimates one can construct an existence proof using the following iteration method. Since the acceleration of the boundary is given by the surface tension plus lower order terms, in the first step of the iteration we evolve the boundary using this evolution. In the second step of the iteration we establish the evolution of the velocity in the interior. This will appear in a forthcoming paper. Finally after this work was completed A. Mielke pointed out to the second author that a similar geometric approach had been used by K. Beyer and M. Günther to study the irrotational problem

4 4 SHATAH AND ZENG by reducing it to the boundary[bg98, BG00]. Indeed they proved local well posedness for star shaped domains with surface tension and studied the linearized flow for any irrotational flow. In fact they derived the principle part of the curvature of Γ for irrotational problem which of course coincides with our R 0 (v) acting on gradient vector fields. This paper is organized as follows. In section of the paper we give the intuition behind the energy estimates by computing the geometry of Γ. In section 3 we present the geometric computation of the moving boundary. In section 4 we present our energy estimates. In sections 5 and 6 we present some examples and basic analytic and geometric calculations. Notations. All notations will be defined as they are introduced. In addition a list of symbols will be given at the end of the paper for a quick reference. Here we ll present some standard notations and conventions used throughout the paper. All constants will be denoted by C which is a generic bound depending only on the quantities specified in the context. We follow the Einstein convention where we sum upon repeated indices. For a domain Ω t and x Ω t we denote by N(t, x) the outward unit normal, Π the second fundamental form where Π(w) = w N T x Ω t for w T x Ω t, and κ the mean curvature given by the trace of Π, i.e., κ = trπ. The regularity of the domains Ω t is characterized by the local regularity of Ω as graphs. In general, an m-dimensional manifold M R n is said to be of class C k or H s, s > n, if, locally in linear frames, M can be represented by graphs of Ck or H s mappings, respectively. For Ω, throughout this paper we will only use these local graph coordinates in orthonormal frames.. The geometry behind the Energy In this section, we heuristically outline our geometric point of view on the free boundary problems of the Euler s equation and the intuition leading to the energy estimates in the following two sections. Though the discussion in this section are mostly in Lagrangian coordinates, the estimates are actually done in Eulerian coordinates in the next two sections..1. Lagrangian formulation of the problem. One of the fundamental properties of the inviscid fluid motion is the law of energy conservation. Multiplying the Euler s equation (E) by v, integrating on Ω t, and using (BC), we obtain the conserved energy E 0 : v (.1) E 0 = E 0 (Ω t, v(t, )) = Ωt v dx + ɛ ds Ω t Ωt dx + ɛ S( Ω t ). The main difficulty of these problems is handling the free boundary. A traditional way to avoid this difficulty is to consider the Lagrangian coordinates. Let u(t, y), y Ω 0, be the Lagrangian coordinate map solving dx (.) = v(t, x), x(0) = y, dt then we have v = u t u 1 and for any vector field w(t, x), x Ω t, it is clear that (.3) D t w t w + v w = (w u) t u 1 Therefore, the Euler s equation can be rewritten as (.4) u tt = ( p) u, u(0) = id Ωt, p = tr((dv) ), p Ωt = κ, where κ is the mean curvature of Ω t. Since v(t, ) is divergence free, then u(t, ) is volume preserving. Let Γ {Φ : Ω 0 R n Φ is a volume preserving homeomorphism}. As a manifold, the tangent space of Γ is given by divergence free vector fields: T Φ Γ = { w : Ω 0 R n w = 0, where w = ( w Φ 1 )}.

5 WATER WAVES 5 For the remainder of this section we follow the following convention: for any vector field X : Φ(Ω 0 ) R n its description in Lagrangian coordinates is given by X = X Φ. With slight abuse of notation, we also let S(Φ) = Φ(Ω 0 ) ds, i.e. the surface area of Φ(Ω 0). Thus, the energy E 0 takes the following form in the Lagrangian coordinates: (.5) E 0 = E 0 (u, u t ) = 1 u t dy + ɛ S(u), (u, u t ) T Γ Ω 0 where the volume preserving property of u is used. This conservation of energy suggests: 1) T Γ be endowed with the L metric; and ) the free boundary problem of the Euler s equation has a Lagrangian action u t I(u) = Ω0 dxdt ɛ S(u)dt, u(t, ) Γ. Let D denote the covariant derivative associated with the metric on Γ, then a critical path u(t, ) of I satisfies (.6) Dt u t + ɛ S (u) = 0. In order to verify that the Lagrangian coordinate map u(t, ) of a solution of (E) and (BC) is indeed a critical path of I, it is convenient to calculate D and S by viewing Γ as a submanifold of the Hilbert space L (Ω 0, R n ). Computing (T Φ Γ). For any vector field X : Φ(Ω) R n we form the Hodge decomposition X = w ψ, ψ = 1 X, w = 0, where 1 is the inverse Laplacian on Ω t with zero Dirichlet data. Therefore if Φ Γ, then w = w Φ T Φ Γ. This implies that normal space of T Φ Γ at Φ is (T Φ Γ) = { ( ψ) Φ ψ (Φ(Ω0 )) 0}. since the Hodge decomposition is orthogonal in L and Φ is volume preserving. Computing D t. Given a path u(t, ) Γ and v = u t. Suppose w(t, ) T u(t) Γ, then the covariant derivative D t w and the second fundamental form II u(t) ( w, v) satisfy w t = D t w + II u(t) ( w, v), Dt w T u(t) Γ, II u(t) ( w, v) (T u(t) Γ). Let v = u t u 1 = v u 1 and w = w u 1 which are in the Eulerian coordinates. Then from the Hodge decomposition we have (.7) Dt w = w t II( w, v), II( w, v) = ( p w,v ) u, p w,v = 1 tr(dwdv). As we do the estimates in the Eulerian coordinates, sometimes it is more convenient to use (.8) D t w = ( D t w) u 1 = D t w + p w,v. Computing S (u). By the variation of surface area formula and for any w T u Γ we have (.9) < S (u), w > L (Ω 0 )= κ(w) ds = κ H w dx Ω t Ω t where κ is the mean curvature, and κ H is its harmonic extension. Since ( κ H ) u T u Γ, we obtain (.10) S (u) = ( κ H ) u J u. This vector field J, divergence free on Ω t, is very important for it connects the free boundary Euler s flow with the geometry of Ω t and even of Γ as we will see later in section 3.. Combining (.10) and (.7) with w = u t, we obtain that the equation (.6) for critical paths of I becomes (.11) u tt = D t v u = ( p v,v ɛ J) u, v = u t u 1,

6 6 SHATAH AND ZENG which is equivalent to (.4). Therefore, the free boundary problem (E) and (BC) is a lagrangian system on Γ given by (.6). If ɛ = 0 equation (.6) becomes the geodesic equation on Γ, which is a well-known fact... Linearization. In order to analyze the free boundary problems of the Euler s equation, it is natural to start with the linearization. The Lagrangian formulation provides a convenient frame work for this purpose. From (.6), the linearized equation is (.1) D t w + R(u t, w)u t + ɛ D S(u)( w) = 0, w(t, ) T u(t, ) Γ, where R is the curvature tensor of the infinite dimensional manifold Γ. Below we calculate D S(u), which is viewed as a linear operator on T u Γ, and R. Computing D S(u). Let g(s, ) be a geodesic on Γ, g(0) = u. Let w = g s and Ω s = g(s, )(Ω 0 ). From (.8) we have D s w = (D s w + p w,w ) g = 0. Differentiating (.9), D S(u)( w, w) = d κw N ds. ds Ω s and substitute the expressions for D s N, D s S, and D s κ from (3.1), (3.), and (3.7) we obtain D S(u)( w, w) = κw (κw + D w ) + κd s w N + κw D s N + w D s κ ds Ω s = κw (κw + D w ) κ N p w,w κ w w N Ω s ( ) + w Ωs w w Π + (D Π)(w ) ds, where Π is the second fundamental form of Ω s. Needless to say that this is a very complicated expression for D S(u)( w, w). We will show that D S(u) is a differential operator and will single out its leading order part. Let us assume that Ω s is a sufficiently smooth domain, then from the trace theorem, D S(u)( w, w) Ω s w + κ N p w,w ds C w H 1 (Ω. s) The third term on the left side is estimated by applying the Divergence Theorem twice, κ N p w,w ds = Ω s κ H p w,w κ H tr(dw) dx = Ω s w w κ H dx Ω s κ w w NdS Ω s Therefore, we obtain D S(u)( w, w) Ω s w ds C w L ( Ω s) w H 1 ( Ω s) C w H 1 (Ω. s) Much as in the derivation of (.9), for a general u Γ we derive an self-adjoint operator T u Γ ( ) A (u)( w) = H( u(ω0 )(w u(ω0 )) ) u which satisfies A (u)( w, w) = u(ω) w ds for any w T u Γ and w = w u 1. In the Eulerian coordinates, A takes the form A (u)(w) = H( u(ω0 )(w u(ω0 )) ), w : u(ω 0 ) R n satisfying w = 0. Since D S(u) is self-adjoint, then (.13) D S(u) = A + at most nd order diff. operators A (u) on

7 WATER WAVES 7 Computing R. In the linearized equation (.1), we need to calculate R(u)(u t, w)u t for a linearized solution w(t, ). Therefore, we may again assume that u(t, ) is a sufficiently smooth critical path of the action I, thus v v u is smooth as well, and study the operator R(u)(u t, )u t on w. Here we apply a well-known formula in Riemannain geometry formally. For any v, w T u Γ, let v = v u 1 and w = w u 1 R(u)( v, w) v w = II u ( v, v) II u ( w, w) II u ( v, w) = p v,v p w,w p v,w dx. For smooth v and w L (u(ω 0 )), clearly p v,w L (u(ω 0 )) C w L (u(ω t)). As for the term, p v,v p w,w dx = p v,v tr(dw) dx = w w p v,v ; dx u(ω 0 ) = u(ω 0 ) u(ω 0 ) u(ω 0 ) u(ω 0 ) ( N p v,v )(w ) ds + u(ω 0 ) D p v,v (w, w) dx. Much as in the derivation of (.9), we derive an self-adjoint operator R 0 (v) on T u Γ, depending on u and v, ( ) R 0 (v)( w) = H( N p v,v (w u(ω0 )) ) u which satisfies R 0 (v)( w, w) = u(ω) In the Eulerian coordinates, R0 (v) takes the form N p v,v w ds. R 0 (v)(w) = H( N p v,v (w u(ω0 )) ), w : u(ω 0 ) R n satisfying w = 0. Therefore, in a very rough sense, (.14) R(u)( v, w) v = R0 ( v) + bounded operators where we used the fact that < R(u)( v, ) v, > is self-adjoint. 3. The geometry of evolving domains Suppose Ω t R n is a family of smooth domains with the parameter t, moving with a smooth velocity vector field v(t, x), x Ω t. We calculate various quantities related to the evolution of the geometry of the domain, which are essential in the energy estimate of the free boundary problem of the Euler s equations Material derivative D t. For any x 0 Ω t0, the particle path x(t) is the solution of the ODE: x t = v(t, x) x(t 0 ) = x 0 and the material derivative D t = t + v is differentiation along the direction of x(t) in the space time domain in R n R. Clearly, x(t) Ω t if x 0 Ω t0. Calculations of D t N and D t S. At any x 0 Ω t0, D t N(t 0, x 0 ) N(t 0, x 0 ) since N(t, x) 1. To derive D t N(t 0, x 0 ), let τ(t) T x(t) Ω t be a solution to the linearized particle path ODE: D t τ = τ v τ(t 0 ) = τ 0 T x0 Ω t0. At (t 0, x 0 ), D t N τ 0 = D t (N τ) N D t τ = (Dv) (N) τ 0. Therefore, we have (3.1) D t N = ((Dv) (N)). From standard calculations for hypersurfaces (3.) D t ds = (v κ + D v )ds.

8 8 SHATAH AND ZENG Covariant differentiation D t. For the family of hypersurfaces Ω t with the velocity field v we define parallel transport along the material line x(t) as follows. Given a tangent vector τ 0 T x0 Ω t0, let τ(t) be the solution of the following ODE: (3.3) D t τ T x(t) Ω t D t τ = ( τ v N)N, τ(t 0 ) = τ 0. It is easy to verify that τ(t) T x(t) Ω t and that this transport preserves the inner product. A natural connection between T Ω t R n for different t along the materials lines is provided by the above parallel transport which induces the covariant differentiation D t, the projection of D t in R n acting on w T x(t) Ω t R n. This covariant differentiation induces the covariant differentiations of linear (multilinear) operators on tensor products of T Ω t and T Ω, which will also denoted by D t. Calculation of D t Π and D t κ. Given τ T x0 Ω t0, let τ(t) be its parallel transport along the material line x(t) which enable us to compute (D t Π)(τ) = D t (Π(τ)) = (D t τ N) = ( τ D t N + [Dt,τ]N) = D τ D t N + Π(([D t, τ]) ) From (3.3) and (3.1), we have (3.4) (3.5) [D t, τ] = D t τ τ ( t + v) = ( τ v N)N τ v = ( τ v), (D t Π)(τ) = D τ (((Dv) (N)) ) Π(( τ v) ). To calculate D t κ at (t 0, x 0 ), we take an orthonormal frame {τ 1,..., τ n 1 } of T x0 Ω t0 and parallel transport it into an orthonormal frame along x(t). Thus D t κ = D t (Π(τ i ) τ i ) = (D t Π)(τ i ) τ i and (3.5) give slightly different but useful forms for D t κ (3.6) (3.7) D t κ = D ((Dv) (N)) Π(τ i ) τi v = Ω v N Π ((D T Ωt )v) D t κ = Ω v v Π + (D Π)(v ). Calculations of commutators involving D t. In the following, we will calculate the commutators of D t with operators H, N, and Ω, to show that they are of lower orders. [D t, H]f = 1 (Dv D f H + f H v). To start, we write the basic formula for any function f(t, x), x Ω t, (3.8) D t f = D t f (Dv) ( f). For the tangential gradient, using f = f ( N f)n, it is straight forward to obtain (3.9) D t f = D t f ((Dv) ( f)) Let f(t, x), x Ω t, be a smooth function. Recall f H = H(f) represents the harmonic extension of f into Ω t. We have (3.10) D t f H = D t f H + v D f H + v f H = Dv D f H + f H v which implies D t f H = H(D t f) + 1 D t f H = H(D t f) + 1 (Dv D f H + f H v). Therefore we can write (3.11) D t H(f) = H(D t f) + 1 (Dv D f H + f H v). [D t, 1 ]g = 1 (Dv D 1 g + v 1 g). Next, we calculate [D t, 1 ]. Let g(t, x), x Ω t be a smooth function and φ = 1 g. From the first half of (3.10) where f = 0 was not used, D t g = D t φ = D t φ Dv D φ v φ.

9 Since D t φ Ωt = 0, we obtain WATER WAVES 9 (3.1) D t 1 g = 1 D t g + 1 (Dv D 1 g + v 1 g) [D t, N ]f = N 1 (Dv D f H + f H v) f H N v f v N. To calculate the commutator of [D t, N ], from (3.1), (3.8) and (3.11), we have Thus, D t ( f H N) = N D t f H f H N v + f H D t N = N [H(D t f) + 1 (Dv D f H + f H v)] f H N v f v N. (3.13) D t N (f) = N (D t f) + N 1 (Dv D f H + f H v) f H N v f v N. [ Ωt, D t ]f = D f ((D T Ωt )v) + f Ωt v κ f v N. In order to calculate the commutator [ Ωt, D t ] at x 0 Ω t0, take an orthonormal frame {τ 1,..., τ n 1 } of T x0 Ω t. We first extend this to an orthonormal frame to T x Ω t0 for all x Ω t0 close to x 0 by parallel transporting {τ 1,..., τ n 1 } along geodesics on Ω t0 starting from x 0. Parallel transporting them again along the material lines x(t), we obtain an orthonormal frame {τ 1,..., τ n 1 } of T x Ω t for all (t, x) near (t 0, x 0 ). From the standard Riemannian geometry, this orthonormal frame satisfies the property that, at (t 0, x 0 ), Dτ j = 0 and [τ i, τ j ] = D τi τ j D τj τ i = 0, which will be used repeatedly. For any smooth function f(t, x) defined on Ω t, at (t 0, x 0 ), D t Ωt f =D t ( τj τj f Dτj τ j f) = τj D t τj f + [Dt,τ j ] τj f [Dt,D τj τ j ]f = τj τj D t f + τj [Dt,τ j ]f + [Dt,τ j ] τj f [Dt,D τj τ j ]f = Ωt D t f + D f(τ j, [D t, τ j ]) + Dτj [D t,τ j ] [D t,d τj τ j ]f. For any vector field τ(t, x) T x Ω t, it is easy to see that [D t, τ] T x Ω t since (a) τ, t + v T ( t Ω t ) [D t, τ] T ( t Ω t ) and (b) [D t, τ] = D t τ τ v does not have t component. Thus, D τj [D t, τ j ] [D t, D τj τ j ] T Ω t and we can drop all the normal components in its calculation. Using D τj τ j = τj τ j + κn and Dτ j = 0 at (t 0, x 0 ), we obtain at (t 0, x 0 ), D τj [D t, τ j ] [D t, D τj τ j ] = ( τj [D t, τ j ] D t ( τj τ j + κn) ) =( τj D t τ j τj τj v D t τj τ j κd t N) = ( Ωt v) + κ((dv) (N))). Therefore, from (3.4), (3.14) D t Ωt f = Ωt D t f D f ((D T Ωt )v) f Ωt v + κ f v N. Calculation of D t κ. This calculation starts with formula (3.6). Since Π : T Ω t T Ω t then Π D T Ωt v = Π T Ωt v. Let {τ 1,..., τ n 1 } be an orthonormal frame which is the parallel transport of an orthonormal frame T x0 Ω t0 along the material line x(t) Ω t. From (3.1), (3.5), (3.8), and (3.6), we have at (t 0, x 0 ), D t κ = D t Ωt v N Ωt v D t N (D t (Π(τ i ))) τi v Π(τ i ) D t ( τi v) = D t Ωt v N + Ωt v (Dv) (N) + D τi (((Dv) (N)) ) τi v (3.15) + Π(( τi v) ) τi v Π(τ i ) Dtτi v Π(τ i ) τi D t v + Π(τ i ) (Dv) (τ i ) = D t Ωt v N Π (D T Ωt0 D t v) + Ωt v (Dv) (N) + [D (((Dv) (N)) ) + Π((D T Ωt0 v) )] (D T Ωt0 v) + Π ((Dv) T Ωt ). To compute D t Ωt v N from (3.14) we need the general formula D f(τ, τ ) = D f(τ, τ ) (Π(τ) τ ) N f.

10 10 SHATAH AND ZENG for any τ, τ T x0 Ω t0. Therefore, (3.16) D t Ωt v N = N Ωt D t v + N D v(τ i, ( τi v) ) ( N v N)(Π (D T Ωt0 v)) + N v(( Ωt v) ) κ ( v) (N). When v and Ω t satisfy the Euler s equation, the expression for D t κ can be written as (3.17) D t κ = N Ωt D t v + ɛ Π (D T Ω J) + r where we signaled out the important terms in the above equation 3.. An important vector field for the water wave problem. Since J = κ H appears in the Euler s equation as a force generated by the surface tension and its regularity is closely related to that of Ω t, we will study the temporal evolution of J for the rest of this section. Computing D t J and D t J. From (3.6), (3.8), (3.11) and the definition of J, (3.18) D t J = D t κ H (Dv) J = H(D t κ) + 1 [Dv DJ + J v] (Dv) J = H( Ωt v N) H(Π ((D T Ωt )v)) + 1 [Dv DJ + J v] (Dv) J. From (3.7), a slightly different way to write D t J is (3.19) D t J = H( Ωt v ) + H[ v Π + (D Π)(v )] + 1 [Dv DJ + J v] (Dv) J. Generally, when the surface tension is of order O(1), it is sufficient to consider D t J. However, when there is no surface tension or the surface tension converges to 0, we have to calculate D t J. Differentiating (3.18), we obtain (3.0) D t J =D t ( D t κ H (Dv) J) = D t κ H (Dv) D t κ H (Dv) D t J (D t Dv) J = D t κ H (Dv) D t J ((Dv) ) J (DD t v (Dv) ) J. Since κ H is harmonic, from (3.11) and (3.18), D t κ H =D t [ H(Dt κ) + 1 (Dv DJ + J v) ] (3.1) =H(D t κ) + 1 (Dv D + v )H(D t κ) + D t 1 (Dv DJ + J v) =H(D t κ) + 1 (Dv D + v )[D t J 1 (Dv DJ + J v) (Dv) J] + D t 1 (Dv DJ + J v) A more explicit expression of H(D t κ) can be derived from (3.17). The D t 1 is another term in the above equation that can be explicitly calculated from (3.1) and (3.10) to write (3.) D t 1 (Dv DJ + J v) = 1 D t (Dv DJ + J v) + 1 [(Dv D + v ) 1 (Dv DJ + J v)] = 1 [(DD t v (Dv) ) DJ + Dv (DD t J DJDv) + D t J v + J ( D t v v v i v j ij v)] + 1 [(Dv D + v ) 1 (Dv DJ + J v)]

11 WATER WAVES 11 Computing D t J and D tt J with divergence free v. In the rest of this section, assume v = 0. Given any vector field w defined on Ω t with w = 0, let D t w denote the divergence free part of D t w. It is easy to calculate that (3.3) D t w = D t w + p v,w, p v,w = tr(dvdw), p v,w Ωt = 0. As J is divergence free, we will decompose material derivatives of J into the divergence parts and gradient parts, i.e. we consider D t J and D tt J, the covariant derivatives defined in (3.3). Then we have (3.4) D t J = D t J + p v,j. For the second order derivative, (3.5) D t J = D t D t J + p v,dtj = D t J + D t p v,j + p v,dtj. where p v,j and p v,dtj are defined as in (3.3). Using (3.8), (3.1) (3.6) D t p v,j = D t p v,j (Dv) p v,j = D t 1 (Dv DJ) (Dv) p v,j = 1 D t (Dv DJ) + 1 (Dv D + v D)p v,j (Dv) p v,j = 1 [(DD t v (Dv) ) DJ + Dv (DD t J DJDv)] + 1 (Dv D + v D)p v,j (Dv) p v,j. Using the above calculations, we will show in Lemma 4.4 that J satisfies the linearized Euler s equation with lower order terms. The estimates on J and D t J from the linearized Euler s Equation will imply the estimates on the geometry of the moving domain and the velocity fields. 4. Main Results In this section, we will derive local energy estimates and prove convergence theorems. We show that solutions of (E) with boundary condition (BC) are locally bounded (4.1) v(t, ) H 3k (Ω t ) and Ω t H s 0, s 0 = 3k or 3k + 1 for ɛ = 0 or > 0 where k is an integer satisfying 3k > n +1 (equivalently 3k n + 3 ). When ɛ > 0, this estimate is obtained without any additional assumption and it may depend on ɛ. To derive a priori estimates independent of ɛ, we assume the Rayleigh-Taylor sign condition (RT): N p v,v (t, x) > a > 0 x Ω t for some constant a. Definition of the energies and statements of the theorems. The conserved energy of the Euler s equation is given by 1 E 0 = E 0 (Ω, v) = v dx + ɛ S(Ω), Ω where S(Ω) = Ω ds is the surface area. Higher order energies are based on the linearized Euler flow and thus involve the differential operators D t, A, and R 0 (v). Recall that, for any vector field w H s (Ω) with w = 0, A is given by (4.) A (w) = H Ω w. A is a semi-positive definite self-adjoint third order differential operator if w Ω is not smoother than N H s 0 1 ( Ω). In fact, it is positive definite acting on the irrotational part w ir of w, i.e. w ir = HN 1 w, see (6.0) for details. Also recall that R 0 (v), which depends on Ω as well as on a vector field v H 3k (Ω) with v = 0, is given by (4.3) R 0 (v)(w) = H(( N p v,v )w ) = H(( N 1 tr(dvdv))w ).

12 1 SHATAH AND ZENG By Lemma 6.4, N p v,v H 3k 3 ( Ω) and therefore R 0 (v) is a first order self-adjoint differential operator if w is not smoother than p v,v. Under the sign assumption (RT), R 0 (v) is semi-positive definite, and like A, it is positive definite on the irrotational part w ir of w. Let ω v : R n R n, often simply written as ω for short, represent the curl or vorticity of a vector field v defined on Ω, i.e. ω(x) Y = X v Y Y v X for any vector X, Y R n. Viewing ω as a matrix, its entries are ω j i = ω( x i ) x j = i v j j v i. Definition 4.1. For any domain Ω in H s 0, s 0 = 3k if ɛ = 0 or s 0 = 3k + 1 if ɛ > 0, and any vector field v H 3k (Ω) with v = 0, define the energies E(Ω, v) and E RT (Ω, v), often written as E and E RT for short, E = Ω 1 A k 1 D t J + ɛ A k 1 J dx + ω H 3k 1 (Ω), E RT = where D t is the divergence free part of D t J defined in (3.3), Set E = E + E RT. D t J = D t J + p v,j = D t J 1 tr(dvdj) Ω 1 R 0(v)A k 1 J A k 1 Jdx To estimate terms in the energy we need to consider the following type of H s 0 neighborhoods of Ω, a bounded connected domain in R n, which are bounded in H s for some s s 0. Definition 4.. Let Λ = Λ(Ω, s 0, s, L, δ) be the collection of all domains Ω satisfying (A1) there exists a diffeomorphism F : Ω Ω R n, so that F id Ω H s 0 ( Ω ) < δ; (A) the mean curvature κ of Ω satisfies κ H s ( Ω) < L. Fix 0 < δ 1 and let L 0 = 1 + κ(0, ) H 3k 5 ( Ω) and Λ 0 Λ(Ω 0, 3k 1, 3k 1, L 0, δ). By Lemma 6.4, and equations (6.10) and (6.19) we have (4.4) (4.5) A L(H s (Ω),H s 3 (Ω)) C, s [3, 3k 1] R 0 (v) L(H s (Ω),H s 1 (Ω)) C N p v,v H 3k 3 ( Ω) C v H 3k (Ω), s [1, 3k 1] (4.6) p v,v H 3k (Ω) C (Dv) H 3k (Ω) C v H 3k 5 8 (Ω) by Sobolev inequalities where C is uniform in Ω Λ 0. Here used the fact 3k n + 3. The norm H3k 5 8 in (4.6) is chosen for convenience; any norm H 3k 1+α would work with α > 1 4. The next proposition gives bounds on the velocity and mean curvature in terms of these energies. Proposition 4.1. For Ω Λ 0 with Ω H s 0, we have and, if we also assume (RT), ɛ κ H 3k 1 ( Ω) 3E + C 0ɛ, v H 3k (Ω) C 0(E + E 0 ) κ H 3k ( Ω) C E RT + C 0, for some constant C, C 0 > 0. C depends only on a in assumption (RT) and C 0 depends only on the set Λ 0. The proof of this proposition will be given below. Using this result we will prove the following three theorems. The first theorem holds when there is surface tension which makes the regularity of Ω t better (in H 3k+1 ) but the bound on κ H 3k 1 ( Ω t) depends on ɛ, the strength of of the surface tension.

13 WATER WAVES 13 Theorem 4.1. Assume ɛ > 0 and fix δ > 0 sufficiently small. Then there exists L ɛ > 0 such that, if a solution of (E) and (BC) is given by Ω t with Ω t H 3k+1 and v(t, ) C(H 3k (Ω t )), then there exists t > 0, depending only on v(0, ) H 3k (Ω t), L ɛ, and the set Λ 0, such that, for all t [0, t ], (4.7) Ω t Λ 0 and κ H 3k 1 ( Ω t) L ɛ, E(Ω t, v(t, )) E(Ω 0, v(0, )) + C 1 + t 0 P ɛ (E 0, E(Ω t, v(t, ))) dt where P ɛ ( ) is a polynomial of positive coefficients determined only by ɛ and the set Λ 0 and C ɛ is an constant determined only by ɛ, v(0, ) H 3k 3 (Ω 0 ), and the set Λ 0. Since the domain is evolving, the above continuity assumption of v in t means that there exists an extension of v to [0, T ] R n which is continuous in H 3k (R n ). The second theorem holds under the assumption (RT) and the estimates are uniform in ɛ [0, 1]. As it does not take the advantage of the surface tension even if it is present, the bound on the regularity of Ω t is only in H 3k. Theorem 4.. Assume ɛ [0, 1] and (RT) holds. Fix sufficiently small δ > 0. There exists L > 0 such that, if a solution of (E) and (BC) is given by Ω t with Ω t H 3k and v(t, ) C(H 3k (Ω t )), then there exists t > 0, depending only on v(0, ) H 3k (Ω t), L, and the set Λ 0, such that, for all t [0, t ], (4.8) Ω t Λ 0 and κ H 3k ( Ω t) L, E(Ω t, v(t, )) E(Ω 0, v(0, )) + t 0 P (E(Ω t, v(t, ))) dt where P ( ) is a polynomial of positive coefficients uniform in ɛ, determined by the set Λ 0. An immediate consequence of the above theorem is convergence of solution as the surface tension approaches 0. Theorem 4.3. Assume (RT) holds. Fix the initial data Ω 0 H 3k+1 and v(0, ) H 3k (Ω 0 ). As ɛ 0, subject to a subsequence, the solution of (E) and (BC) with vanishing surface tension converges to a solution of (E) and (BC) for ɛ = 0 weakly in the space of Ω t H 3k and v(t, ) H 3k. The above convergence of Ω t is in the sense of local coordinates and the convergence of v can be obtained by using the Lagrangian coordinates u(t, y) which is also in H 3k. We also observe that the neighborhood Λ 0 of the domains Ω does not have to be centered at Ω 0. Thus, since the constants involved in the energy estimates only depend on the neighborhoods and the norm of the initial velocity, these estimates provide a basis for a continuation argument local in time. Proof of Proposition 4.1. From the definition of A and R 0 (v), it is easy to obtain ɛ N ( Ω N ) k 1 κ L ( Ω) = ɛ A k 1 J L (Ω) E, N p v,v N ( Ω N ) k 1 κ ds =< R 0 (v)a k 1 J, A k 1 J > L (Ω) = E RT. Ω To estimate κ in either H 3k ( Ω) or H 3k 1 ( Ω), it is sufficient to use the estimates on functions and operators defined on Ω considered as in H 3k 1 only. Therefore, the inequality for κ in Proposition 4.1 follows from (RT) and the fact that N behaves as a first order derivative (Theorem 6.1, (6.16), and (6.15)). To estimate v, it is easy to calculate that (4.9) v i = j ω i j which is part of the energy. Therefore, we only need to show that some boundary data of v is controlled by E and the conserved energy E 0. This boundary data of v turns out to be N v. Since

14 14 SHATAH AND ZENG N v = (Dv) (N) + ω(n) and ω is controlled by E, it suffices to estimate ν = (Dv) (N). Step 1. Tangential curl ων T x Ω be defined as of ν. Let ν be the tangential component of ν and ω ν (x) : T x Ω ω ν (x)(x) Y = D X ν Y D Y ν X = X ν Y Y ν X for any x Ω and X, Y T x Ω. To obtain a more explicit form of ω ν, let X and Y be extended to tangent vector fields on a neighborhood of x on Ω by parallel transport along geodesics on Ω emitting from x. From the definition of ν, we have (4.10) ω ν (X) Y = X ( Y v N) D X Y ν Y ( X v N) + D Y X ν =Π(X) Y v Π(Y ) X v. Therefore, by Sobolev inequalities, there exists C > 0 uniform in Ω Λ 0 so that ω ν H 3k 5 ( Ω) C Π D v H 3k 5 ( Ω) C v H 3k 1 8 (Ω) since 3k n + 3. Again here the norm H3k 1 8 is chosen to illustrate that the term is lower order. In fact any H 3k α with 0 < α < 1 4 works. Step. Divergence D ν. At any x 0 Ω, let {X 1,..., X n 1 } be an orthonormal frame of T x0 Ω. We extend them to orthonormal frames of T x Ω at each x in a neighborhood of x 0 in Ω by parallel transport along geodesics on Ω emitting from x 0. At x 0, (4.11) D ν = D Xi ν X i = D Xi ( Xi v N) = Ω v N + (D T Ωt )v Π. To control the first term on the right side, we use (3.4) and (3.18) to obtain D t J + H( Ω v N) H 3k 3 (Ω) C v H 3k 1 8 (Ω) where we also used Lemma 6.4 for the estimates. To get boundary estimates, note that D t J N and N H( Ω v N) are well defined in H 3k 7 ( Ω) even if k = 1 since they are divergence free. and thus combining the above inequality with the identity for D ν (4.11), we obtain N (D ν ) + (D t J) H 3k 7 ( Ω) C v H 3k 1 8 (Ω) Step 3. Control of ν. Using the same frame as in Step, at x 0, D ν H 3k 5 ( Ω) C(E + v H 3k 1 8 (Ω) ). Ω ν X i =D Xj D Xj ν X i = Xj (D Xj ν X i ) = Xj (D Xi ν X j + ω ν (X j ) X i ) One can also write = Xi (D ν ) + R(X i, X j )ν X j + (D Xj ω ν )(X j ) X i. Ω ν = (D ν ) + Ric(( v) (N) ) + (D Xj ω ν )(X j ) where Ric is the Ricci curvature of Ω. From the estimate on ω ν and D ν, with a uniform constant C > 0. ν H 3k 3 ( Ω) C(E + v H 3k 1 8 (Ω) ) Step 4. Normal component ν = N v N of ν. This will be estimated by calculating the divergence of ν in two ways. Recall N H = (NH 1,..., N H n ) denotes the harmonic extension of N into Ω. Let ν also be extended to (Dv) (N H ). Near any x 0 Ω, let {X 1,..., X n 1 } be the orthonormal frame of T x Ω constructed above and let X n = N H. On one hand, at x 0, ν = Xi ((Dv) (N H )) X i = Xi ( Xi v N H ) N Dv( Xi X i ) =( Xi ω)(x i ) N + Dv DN H

15 where the first term in the last line follows from (4.9). Therefore WATER WAVES 15 ν ( Xi ω)(x i ) N H 3k 5 ( Ω) C v H 3k 1 8 (Ω). On the other hand, by decomposing ν into the tangential and normal parts, one may calculate ν alternatively ν = D ν + κν + N ν N = D ν + κν + N ( NH v N H ) N Dv(N (N)) which along with the previous identity implies D ν + N ( NH v N H ) ( Xi ω)(x i ) N H 3k 5 ( Ω) C v H 3k 1 8 (Ω). Since ω is controlled by E and D ν has been estimated in Step, we have N ( NH v N H ) Ḣ 3k 5 ( Ω) C(E + v H 3k 1 8 (Ω) ) The following decomposition trick on Ω has been used many times in the basic estimates in Sections 6 and 3, N ( NH v N H ) =N ( N v N) + N 1 ( NH v N H + tr((dn H ) DvDN H ) + D v(n H ) DN H + D v( N H x i, x i ) N H). Using (4.9) again, we have with a uniform C > 0, N (ν ) H 3k 5 ( Ω) = N ( N v N) Ḣ 3k 5 ( Ω) C(E + v H 3k 1 8 (Ω) ) From Step 3 and Step 4 above, we have ν Ḣ 3k 3 ( Ω) C(E + v H 3k 1 8 (Ω) ) which implies the same estimate of the boundary data N v on Ω. Combining it with the Poisson equation (4.9), we obtain v H 3k (Ω) C(E + v H 3k 1 8 (Ω) ). Since v H 3k 1 8 (Ω) β v H 3k (Ω) + C β v L (Ω), proposition 4.1 follows immediately. In the proof of Theorem 4.1 and Theorem 4., we will need the following lemmas. Lemma 4.. For any Ω Λ 0 with κ H s ( Ω), s [3k 5, 3k 3 ], we have Π H s ( Ω t) + N H s+1 ( Ω t) C(1 + κ H s ( Ω t)) for some C > 0 uniform in Ω Λ 0. Proof. We only need to prove the estimates for Π. This is obvious if n =. For n 3, we use identity (6.5): Ω Π = D κ + ( Π I κπ)π. Since Ω Λ 0, we have and it implies Π H 3k 5 ( Ω t) + κ H 3k 5 ( Ω t) C Π I κπ H s 1 ( Ωt) C, where s 1 = min{6k 5 n 1, 3k 5 } as long as 3k 5 n 1. Therefore, for s [3k 5, 3k 3 n 1 ]\{ }, ( Π I κπ)π H s 3 ( Ωt) C Π H s ( Ωt) s 3 = min{s 1, s 1 + s n 1 }.

16 16 SHATAH AND ZENG Since s 3 + s min{1, 3k 1 n } > 0 and s 1 + > 3k 3, the estimate of Π can be improved to H s ( Ω t ) by bootstrap on 1 Ω t. The exceptional cases of the indices can be handled similarly. Although this regularity of Π in terms of κ follows directly from Proposition 6.3, the point of this lemma is that the constant C > 0 depends only on Λ 0, i.e., Ω H 3k 1. This is also the point of the following lemma. Lemma 4.3. For Ω Λ 0 with Ω H 3k and v H 3k (Ω), we have NH p v,v H 3k 1 (Ω) + D p v,v H 3k 3 (Ω) C(1 + κ H 3k ( Ω)) v H 3k (Ω). for some C > 0 uniform in Ω Λ 0. Proof. The idea of the proof is to use p v,v Ω0 = 0 and the identity f = H(f Ω ) + 1 f for any f : Ω R. On the one hand, notice on Ω, N NH p v,v = N (N) p v,v + D p v,v (N, N) = N (N) p v,v + p v,v κ N p v,v, which, along Lemma 4. and (4.6), implies N NH p v,v H 3k ( Ω) C(1 + κ H 3k ( Ω)) v H 3k (Ω). On the other hand, using Lemma 4. and (4.6) as well, NH p v,v H 3k 5 (Ω) = N H tr(dv) + D p v,v DN H H 3k 5 (Ω) C(1 + κ H 3k ( Ω)) v H 3k (Ω). Therefore. we obtain the estimates on NH p v,v The estimate of D p v,v is also achieved similarly. Firstly, D p v,v H 3k 7 (Ω) = D tr(dv) H 3k 7 (Ω) C v H 3k (Ω). For the boundary value of D p v,v, we first consider D p v,v (X, X) at x Ω with X T x Ω. As usual, extend X to a vector fields in a neighborhood of x on Ω by parallel transporting X along geodesics emitting from x. Thus, Also, we have D p v,v (X, X) = X X p v,v X Xp v,v = N p v,v Π(X, X). D p v,v (N, N) = p v,v κ N p v,v = tr(dv) κ N p v,v D p v,v (N, X) = X N p v,v X Np v,v = X N p v,v. Therefore, from the above estimate on NH p v,v, we obtain the estimate on D p v,v. The following lemma is the most important observation of this paper, which states that J is a solution of the linearized Euler s equation up to lower order terms. For the rest of this section, let Q denote a generic positive polynomial in v H 3k (Ω t), κ H 3k ( Ω t), and ɛ κ H 3k 1 ( Ω t) with coefficients depending only on the set Λ 0. Lemma 4.4. Suppose a solution of the Euler s equation is given by Ω t Λ 0 with Ω t H s 0, s 0 = 3k if ɛ = 0 or s 0 = 3k + 1 if ɛ > 0, and v(t, ) H 3k (Ω t ), then we have D t J + R 0 (v)j + ɛ A J H 3k 3 (Ω t) (1 + D t J H 3k 3 (Ω t))q. Proof. We start our proof by two simple observations. First we recall that for any vector field w defined on Ω t satisfying w = 0, we write p v,w = 1 tr(dvdw) (4.1) p v,w H s+1 (Ω t) C v H 3k (Ω t) w H s (Ω t), s [0, 3k 1],

17 where C > 0 is uniform in Ω Λ 0. by (3.4) we have WATER WAVES 17 Second the first order derivative D t J is lower order since (4.13) D t J D t J H s (Ω t) Q s [0, 3k 1], and by (3.18) and (3.6) (4.14) D t J H 3k 3 ( Ω t) C v H 3k (Ω t). To verify the lemma we consider the expression for D t J given in (3.5) and keep the least regular terms. Thus D t J D t J H 3k 3 (Ω t) Q + D t p v,j H 3k 3 (Ω t) D t p v,j is given by (3.6) and can be estimated using (4.1), lemma 4.3 and Euler s equation D t p v,j H 3k 3 (Ω t) Q + C DD t v DJ H 3k 4 (Ω t) Q. To estimate D t J given in (3.0), we use (4.6) and Euler s equation to obtain To estimate D t κ H we use (3.1) and (3.) Combine these inequalities, we obtain D t J D t κ H H 3k 3 (Ω t) Q. D t κ H H(D t κ) H 3k 3 (Ω t) Q. D t J H(D t κ) H 3k 3 (Ω t) Q. The term D t κ has been calculated explicitly in (3.17) which yields, (4.15) D t J + H( Ωt D t v N ɛ Π (D T Ωt J)) H 3k 3 (Ω t) Q. To deal with Ωt D t v N we use Euler s equation to obtain that on the boundary Ωt D t v N ɛ Π (D T Ωt J)) = N Ωt ( p v,v ) ɛ Ω N (κ) + ɛ J Ωt N. The last term can be bounded by the identity Ωt N = Π N + κ, and Lemma 4. ɛ J Ωt N H 3k 5 ( Ω t) = ɛ κ ɛ N (κ) Π H 3k 5 ( Ω t) < Q Substituting the above into (4.15) and Lemma 4. (4.16) D t J H(N Ωt ( p v,v )) + ɛ A J H 3k 3 (Ω t) Q. As we are very close to the final desired form, the second term on the above left side has to be related to R 0. Using formula (6.13), on Ω t, we have N Ωt p v,v = N p v,v + κn N p v,v + N D ( p v,v )(N, N) =N (tr(dv) ) + N (κ H NH p v,v + D p v,v (N H, N H )) N p v,v J κ p v,v N N H D p v,v (N, N N H ) Keeping the least regular terms and using lemma 4.3, implies (4.17) N Ωt p v,v + N p v,v J H 3k 5 ( Ω t) Q + N ( κh NH p v,v + D p v,v (N H, N H ) ) H 3k 5 ( Ω t) Let f = κ H NH p v,v + D p v,v (N H, N H ) defined on Ω t, since p v,v Ωt = 0 then f Ωt H 3k 3 ( Ω t) = p v,v Ωt p v,v H 3k 3 ( Ω t) = tr(dv) H 3k 3 ( Ω t) C v H 3k (Ω t). Moreover it is easy to check from Lemma 4.3, f H 3k 3 (Ω t) Q.

18 18 SHATAH AND ZENG which implies N f H 3k 5 ( Ω t) Q. Therefore (4.17) together with the definition of R 0 and the half derivative behavior of H implies R 0 J + H(N Ωt ( p v,v )) H 3k 3 (Ω t) Q which together with (4.16) concludes the estimate in the statement of the lemma. Proof of Theorems 4.1 and 4.. To prove Theorem 4.1, in addition to Proposition 4.1, we need the following: a) the estimates on the Lagrangian coordinates map and consequently κ H 3k 5 ( Ω t ), b) estimates on ω = Dv (Dv), and c) commutators involving D t, mostly have been done in Section 3. In the following all constant C > 0 will be determined only by the set Λ 0. Estimate of the Lagrangian coordinate map u(t, y). From our assumption on v, the ODE u t (t, y) = v(t, u(t, y)) solving u is well-posed. Since u(t, ) : Ω 0 Ω t is volume preserving and 3k > n + 1, it is easy to derive, for any s [0, 3k], and f H s (Ω t ) f u(t, ) H s (Ω 0 ) C f H s (Ω t) u(t, ) s H 3k (Ω 0 ), where C > 0 depends only on s. The proof follows simply from induction and interpolation. By duality, for s [0, 3k], Therefore, (4.18) u(t, ) I H 3k (Ω 0 ) C f u(t, ) H s (Ω 0 ) C f H s (Ω t) u(t, ) 1 s H 3k (Ω 0 ). t 0 v(t, ) H 3k (Ω t) u(t, ) 3k H 3k (Ω 0 ) dt, where C > 0 depends only on n and k. Let µ > 0 be a positive large number to be specified later, (4.19) t 0 = sup{t v(t, ) H 3k (Ω t) < µ, t [0, t]}, We have t 0 > 0 due to the continuity of v(t, ) in H 3k (Ω t ). Then, for all t [0, t 0 ], u(t, ) I H 3k (Ω 0 ) µ t 0 u(t, ) 3k H 3k (Ω 0 ) dt. Therefore, from ODE estimates, there exists t 1 > 0 and C > 0 which depend only on v(0, ) H 3k (Ω t) such that, for all 0 t min{t 0, t 1 }, (4.0) u(t, ) I H 3k (Ω 0 ) C t. It implies the mean curvature estimate, for all 0 t min{t 0, t 1 }, (4.1) κ(t, ) H 3k 5 ( Ω t) κ(0, ) H 3k 5 ( Ω 0 ) + C 3t. Here it is easy to see from local coordinates constructed in Section 6 that C 3 is determined only by v(0, ) H 3k (Ω t) and the set Λ 0. Therefore, there exists t > 0 determined only by v(0, ) H 3k (Ω t) and the set Λ 0 such that Ω t Λ 0 for 0 t min{t 0, t }. Evolution of the curl ω = Dv (Dv). From equations (E) and (3.8), we have D t ω = DD t v (DD t v) + ((Dv) ) (Dv) = ((Dv) ) (Dv) = (Dv) ω ωdv. It is clear how to obtain the estimate of ω in terms of v: differentiating the above equation 3k 1 times, multiplying it by D 3k 1 ω and integrating it on Ω t, we have d (4.) ω dt H 3k 1 (Ω dx C v t) H 3k (Ω t) ω H 3k 1 (Ω. t) Ω t

19 WATER WAVES 19 The commutator involving D t. function f defined on Ω t, First, from (3.14) and (3.13), it is easy to verify that, for any (4.3) [D t, Ωt ] L(H s 1 ( Ωt),H s 1 ( Ω t)) C v H 3k (Ω t) s 1 ( 7 3k, 3k 1 ], [D t, N ] L(H s ( Ωt),H s 1 ( Ω t)) C v H 3k (Ω t) s (1, 3k 1 ]. To extend the range of s, we use the weak form of [D t, N ]: g[d t, N ]fds = g((dv) (Dv) )( f) N + gn (f) N v N ds Ω t Ω t + g H f H v Dv( f H ) g H + g H 1 D (Dv D f H + f H v)dx. Ω t To conclude that the above estimate for [D t, N ] holds for s = 1. By interpolation, (4.4) [D t, N ] L(H s ( Ω t),h s 1 ( Ω t)) C v H 3k (Ω t), s [ 1, 3k 1 ]. Evolution of E: first look. Recall the expression of E RT and E written as E = I 1 + I : I 1 = ɛ A k 1 J L (Ω = ɛ t) κ N ( Ωt N ) k 1 κ ds I = 1 Ω t A k 1 D t J L (Ω, t) E RT = 1 < (R 0(v))A k 1 J, A k 1 J > 1 L (Ω t)= N p v,v ( N Ωt ) k 1 N (κ) ds. Ω t Also recall that d dt ds = (D v + κv )ds. Since v Ωt = D v + κv + N v N = 0 then κv + D v H 3k 3 ( Ω t) = Nv N H 3k 3 ( Ω t) C v H 3k (Ω t) and thus d dt ds would not complicate the estimates since 3k n + 3. I: d dt I 1 ɛ < A k 1 J, A k 1 D t J > L (Ω t) Q. To prove the inequality I, we use (4.3) and (4.4) to obtain and from (4.1) and (4.13), we have d dt I 1 ɛ < ( Ωt N ) k 1 κ, N D t κ > L ( Ω t) Q, (4.5) D t J H(D t κ) H 3k 3 (Ω t) C v H 3k (Ω t) J H 3k 3 (Ω t). It implies the estimate I for d dt I 1. d II: dt I < A k 1 D t J, A k 1 Dt J > L (Ω t) Q. If k = 1, which may happen when n =, 3, d dt I =< D t J, Dt J > L (Ω t), where we used the fact < D t J, (Dt D t D t )J > L (Ω t)= 0. If k > 1, I = 1 (D t J) ( Ωt )( N Ωt ) k 3 (D t J) ds. Ω t

20 0 SHATAH AND ZENG From (4.3) and (4.4), we obtain d dt I D t ((D t J) ) ( Ωt )( N Ωt ) k 3 (D t J) ds Q. Ω t On Ω t, D t (D t J N) = (D t D t J) N (DtJ) v N = (D t J) N + N p v,dtj (DtJ) v N, which implies, along with (4.1), the estimate II for I. III: d dt E RT < A k 1 R 0 (v)j, A k 1 D t J > L (Ω t) Q. In general, for any function f(t, ) defined on Ω t with Ω t Λ 0 satisfying Ω t fds = 0, we have d N p v,v f ds = N p v,v f (κv + D v ) N p v,v fd t f f D t ( N p v,v )ds dt Ω t Ω t Therefore, we obtain from (4.6), d N p v,v f ds N p v,v fd t f f D t ( N p v,v ) ds Q. dt Ω t Ω t Commuting D t with and 1 by (3.8) and (3.1), D t (N p v,v ) = N ( D t p v,v (Dv) p v,v ) = N p v,v N v N + N 1 (Dv D p v,v + v p v,v ) N 1 (D t tr(dv) ), and using Euler s equation (.11) to get 1 D ttr(dv) = tr(dv) 3 D p v,v Dv ɛ D κ H Dv. Therefore, d N p v,v f ds N p v,v fd t f ɛ f N 1 (D κ H Dv)dS Q dt Ω t Ω t Substituting f = ( N Ωt ) k 1 N (κ), we obtain d dt E RT N p v,v ( N Ωt ) k 1 N (κ) D t ( N Ωt ) k 1 N (κ) ds Q. Ω t From (4.3), (4.4), and (4.5), d dt E RT < R 0 (v)a k 1 J, A k 1 D t J > L (Ω t) Q. In order to apply lemma 4.4 to the estimate III, we need to estimate I 3 < A k 1 D t J, A k 1 R 0 (v)j > L (Ω t) < R 0 (v)a k 1 J, A k 1 D t J > L (Ω t) = (D t J) ( Ωt N ) k ( N p v,v N κ) Ω t ( ) (D t J) ( Ωt N ) k 1 ( N p v,v )N ( Ωt N ) k 1 κ ds. Our strategy will be to move N and the multiplication operator by N p v,v in the first integrand by commuting them with N and Ωt. Thus, we need to estimate [ Ωt, N p v,v ], [N, N p v,v ],

21 and [ Ωt, N ]. Using Lemma 4.3, (6.1), and (6.16), we have, WATER WAVES 1 n 1 [ Ωt, N p v,v ] L(H s ( Ω t),h s 3 Q s (3 3k, ) ( Ω t)) [N, N p v,v ] L(H s ( Ω t),h s 1 Q s [1 ( Ω t)), n 1 ) [ Ωt, N ] L(H s ( Ω t),h s 5 C s (7 3k, 3k 1). ( Ω t)) Therefore we obtain I 3 Q, which implies the estimate III on d dt E RT. Evolution of E. Combining the estimates I, II, III, and Lemma 4.4, we obtain (4.6) d dt E Q, where Q = Q( v H 3k (Ω t), κ H 3k ( Ω t), ɛ κ H 3k 1 ( Ω t)) is a polynomial with positive coefficients that depend only on Λ 0. This inequality on [0, min{t 0, t }] where t 0 is defined in (4.19) and t is determined only by v(0, ) H 3k (Ω t) and the set Λ 0. Proof of Theorem 4.. Assume (RT) holds. From Proposition 4.1, (4.6) implies (4.8). In addition, by choosing µ large enough compared to the initial data, t 0 is bounded below by a constant t > 0 depending only on v(0, ) H 3k (Ω 0 ) and the set Λ 0. Theorem 4. follows immediately. Proof of Theorem 4.1. Assume ɛ > 0. From inequality (4.6) and Proposition 4.1, we obtain (4.7) E(t) E(0) + E RT (t) E RT (0) t 0 Q ɛ ( v H 3k (Ω t ), κ H 3k 1 ( Ω t )) dt, where we use Q ɛ to represent the dependence Q on ɛ. From proposition 6.6 and (4.6) we have E RT C N p v,v L ( Ω t) κ H 3k ( Ω t) C v H 3k 5 8 (Ω t) κ H 3k ( Ω t). Interpolating v between H 3k (Ω t ) and H 3k 3 (Ω t ) and κ between H 3k 1 ( Ω t ) and H 3k 5 ( Ω t ), we obtain from Proposition 4.1, E RT 1 E + C 1(1 + v m H 3k 3 (Ω t) ) for some integer m > 0 where the constant C 1, which include κ H 3k 5 ( Ω t), is determined only by E 0 and the set Λ 0. Since D t v = p v,v ɛ J is controlled by E in H 3k 3 (Ω t ) due to Proposition 4.1, we can use the Lagrangian coordinate map u(t, ) to estimate v(t, ) H 3k 3 (Ω t) v(0, ) H 3k 3 (Ω 0 ). Through a similar procedure of the derivation of (4.0) and using Proposition 4.1, there exists t 3 > 0, depending only on v(0, ) H 3k (Ω t) and the set Λ 0 so that for 0 t min{t 0, t 3 }, v(t, ) m v(0, t H 3k 3 (Ω ) m t) H 3k 3 (Ω 0 ) Q ɛ dt for some polynomial Q ɛ with positive coefficients. Therefore, E RT 1 t E + C 1(1 + v(0, ) m ) + Q H 3k 3 ɛ dt 1 (Ω 0 ) E + C 1 + where C 1 is determined only by v(0, ) H 3 k 3 (Ω 0 ) and the set Λ 0. Thus E(Ω t, v(t, )) E(Ω 0, v(0, )) + C ɛ + 0 t 0 0 Q ɛ dt. t 0 Q ɛ dt,