On the Cauchy problem for gravity water waves

Transcription

1 On the Cauchy problem for gravity water waves T Alazard, N Burq, C Zuily Abstract We are interested in the system of gravity water waves equations without surface tension Our purpose is to study the optimal regularity thresholds for the initial conditions In terms of Sobolev embeddings, the initial surfaces we consider turn out to be only of C 3/2+ǫ -class for some ǫ > 0 and consequently have unbounded curvature, while the initial velocities are only Lipschitz We reduce the system using a paradifferential approach 1 Introduction We are interested in this work in the study of the Cauchy problem for the water waves system in arbitrary dimension, without surface tension An important question in the theory is the possible emergence of singularities (see [15, 16, 24, 56, 19]) and as emphasized by Craig and Wayne [29], it is important to decide whether some physical or geometric quantities control the equation In terms of the velocity field, a natural criterion (in view of Cauchy-Lipschitz theorem) is given by the Lipschitz regularity threshold Indeed, this is necessary for the fluid particles motion (ie the integral curves of the velocity field) to be well-defined In terms of the free boundary, there is no such natural criterium In fact, the systematic use of the Lagrangian formulation in most previous works [8, 52, 53], and the intensive use of Riemannian geometry tools (parallel transport, vector fields,) by Shatah-Zeng [47, 48, 49], Christodoulou Lindblad [20] or Lindblad [39] seem to at least require bounded curvature assumptions (see also [23] where a logarithmic divergence is allowed) In this direction, the beautiful work by Christodoulou Lindblad [20], gives a priori bounds as long as the second fundamental form of the free surface is bounded, and the first-order derivatives of the velocity are bounded This could lead to the natural conjecture that the regularity threshold for the water waves system is indeed given by Christodoulou Lindblad s result and that the domain has to be assumed to be essentially C 2 Our main contribution in this work is that this is not the case and that the relevant threshold is actually only the Lipschitz regularity of the velocity field Indeed (see Theorem 12), our local existence result involves assumptions which, in view of Sobolev embeddings, require only (in terms of Hölder regularity) the initial free domain to be C 3/2+ǫ for some ǫ > 0 As an illustration of the relevance of the analysis of low regularity solutions in a domain with a rough boundary, let us mention that in a forthcoming paper, we shall give an application of our analysis to the local Cauchy theory of three-dimensional TA was supported by the French Agence Nationale de la Recherche, projects ANR-08-JCJC and ANR-08-JCJC

2 gravity water waves in a canal This question goes back to the work by Boussinesq at the beginning of the 20 th century (see [14]) Our analysis require the introduction of new techniques and new tools In [1, 2] we started a para-differential study of the water waves system in the presence of surface tension and were able to prove that the equations can be reduced to a simple form (11) t u+t V u+it γ u = f, where T V is a para-product and T γ is a para-differential operator of order 3/2 Here the main step in the proof is to perform the same task without surface tension, with T γ of order 1/2 It has to be noticed however that performing our reduction is considerably more difficult here than in our previous papers ([1, 2]) Indeed, in the case with non vanishing surface tension, the natural regularity threshold forces the velocity field to be Lipschitz while the domain is actually much smoother (C 5/2 ) In the present work, the velocity field is also Lipschitz, but the domain is merely C 3/2 To overcome these difficulties, we had to give a micro-local description (and contraction estimates) of the Dirichlet-Neumann operator which is non trivial in the whole range of C s domains, s > 1 (see the work by Dahlberg-Kenig [30] and Craig-Schanz-Sulem [27] for results on the Dirichlet-Neumann operator in Lipschitz domains) We think that this analysis is of independent interest Finally, let us mention that, as we proceed by energy estimates, our results are proved inl 2 -basedsobolevspacesandourinitial data(η,v)whichdescriberespectively the initial domain as the graph of the function η and the trace of the initial velocity on the free surface, are assumed to be in H s+1 2(R d ) H s (R d ),s > 1+ d 2 The gravity water waves system enjoys a scaling invariance for which the critical threshold is s c = d 2 (in other terms our well-posedness result is 1/2 above the scaling critical index) 11 Assumptions on the domain Hereafter, d 1, t denotes the time variable and x R d and y R denote the horizontal and vertical spatial variables We work in a time-dependent fluid domain Ω located underneath a free surface Σ and moving in a fixed container denoted by O This fluid domain Ω = {(t,x,y) [0,T] R d R : (x,y) Ω(t)}, is such that, for each time t, one has Ω(t) = {(x,y) O : y < η(t,x)}, where η is an unknown function and O is a given open domain which contains a fixed strip around the free surface Σ = {(t,x,y) [0,T] R d R : y = η(t,x)} This implies that there exists h > 0 such that, for all t [0,T], { } (12) Ω h (t) := (x,y) R d R : η(t,x) h < y < η(t,x) Ω(t) We also assume that the domain O (and hence the domain Ω(t)) is connected Remark 11 (i) Two classical examples are given by O = R d R (infinite depth case) or O = R d [ 1,+ ) (flat bottom) Notice that, in the following, no regularity assumption is made on the bottom Γ := O (ii) Notice that Γ does not depend on time However, our method applies in the case where the bottom is time dependent (with the additional assumption in this case that the bottom is Lipschitz) 2

3 12 The equations Below we use the following notations = ( xi ) 1 i d, x,y = (, y ), = 1 i d 2 x i, x,y = + 2 y We consider an incompressible inviscid liquid, having unit density The equations by which the motion is to be determined are well known Firstly, the Eulerian velocity field v: Ω R d+1 solves the incompressible Euler equation (13) t v +v x,y v + x,y P = ge y, div x,y v = 0 in Ω, where ge y is the acceleration of gravity (g > 0) and where the pressure term P can be recovered from the velocity by solving an elliptic equation The problem is then given by three boundary conditions They are v n = 0 on Γ, (14) t η = 1+ η 2 v ν on Σ, P = 0 on Σ, wherenandν aretheexteriorunitnormalstothebottomγandthefreesurfaceσ(t) The first condition in (14) expresses the fact that the particles in contact with the rigid bottom remain in contact with it Notice that to fully make sense, this condition requires some smoothness on Γ, but in general, it has a weak variational meaning (see Section 3) The second condition in (14) states that the free surface moves with the fluid and the last condition is a balance of forces across the free surface Notice that the pressure at the upper surface of the fluid may be indeed supposed to be zero, provided we afterwards add the atmospheric pressure to the pressure so determined The fluid motion is supposed to be irrotational The velocity field is therefore given by v = x,y φ for some potential φ: Ω R satisfying x,y φ = 0 in Ω, n φ = 0 on Γ Using the Bernoulli integral of the dynamical equations to express the pressure, the condition P = 0 on the free surface implies that t η = y φ η φ on Σ, (15) t φ+ 1 2 x,yφ 2 +gy = 0 on Σ, n φ = 0 on Γ, where recall that = x Many results have been obtained on the Cauchy theory for System (15), starting from the pioneering works of Nalimov [44], Shinbrot [50], Yoshihara [57], Craig [25] In the framework of Sobolev spaces and without smallness assumptions on the data, the well-posedness of the Cauchy problem was first proved by Wu for the case without surface tension (see [52, 53]) and by Beyer- Günther in [12] in the case with surface tension Several extensions of their results have been obtained by different methods (see [22, 31, 32, 33, 35, 41, 54, 55, 59] for recent results and the surveys [11, 29, 36] for more references) Here we shall use the Eulerian formulation Following Zakharov [58] and Craig Sulem [28], we reduce the analysis to a system on the free surface Σ(t) = {y = η(t,x)} If ψ is defined by ψ(t,x) = φ(t,x,η(t,x)), then φ is the unique variational solution of x,y φ = 0 in Ω, φ y=η = ψ, n φ = 0 on Γ 3

4 Define the Dirichlet-Neumann operator by (G(η)ψ)(t,x) = 1+ η 2 n φ y=η(t,x) = ( y φ)(t,x,η(t,x)) η(t,x) ( φ)(t,x,η(t,x)) For the case with a rough bottom, we recall the precise construction later on (see 31) Now (η,ψ) solves (see [28] or [36, chapter 1] for instance) t η G(η)ψ = 0, (16) t ψ +gη ψ 2 1 ( ) 2 η ψ +G(η)ψ 2 1+ η 2 = 0 13 The Taylor condition Introduce the so-called Taylor coefficient (17) a(t,x) = ( y P)(t,x,η(t,x)) The stability of the waves is dictated by the Taylor sign condition, which is the assumption that there exists a positive constant c such that (18) a(t,x) c > 0 This assumption is now classical and we refer to [11, 20, 21, 37, 52, 53] for various comments Here we only recall some basic facts First of all, as proved by Wu ([52, 53]), this assumption is automatically satisfied in the infinite depth case (that is when Γ = ) or for flat bottoms (when Γ = {y = k}) Notice that the proof remains valid for any C 1,α -domain, 0 < α < 1 (by using the fact that the Hopf Lemma is true for such domains, see [45] and the references therein) There are two other cases where this assumption is known to be satisfied For instance under a smallness assumption Indeed, if t φ = O(ε 2 ) and x,y φ = O(ε) then directly from the definition of the pressure we have P +gy = O(ε 2 ) Secondly, it was proved by Lannes ([37]) that the Taylor s assumption is satisfied under a smallness assumption on the curvature of the bottom (provided that the bottom is at least C 2 ) However, for general bottom we will assume that (18) is satisfied at time t = 0 14 Main result We work below with the vertical and horizontal traces of the velocity on the free boundary, namely B := ( y φ) y=η, V := ( x φ) y=η These can be defined only in terms of η and ψ by means of the formulas (19) B = η ψ +G(η)ψ 1+ η 2, V = ψ B η Also, recall that the Taylor coefficient a defined in (17) can be defined in terms of η,v,b,ψ only (see Section 15 below) Theorem 12 Let d 1, s > 1+d/2 and consider (η 0,ψ 0 ) such that (1) η 0 H s+1 2(R d ), ψ 0 H s+1 2(R d ), V 0 H s (R d ), B 0 H s (R d ), (2) there exists h > 0 such that condition (12) holds initially for t = 0, (3) there exists a positive constant c such that, for all x R d, a 0 (x) c Then there exists T > 0 such that the Cauchy problem for (16) with initial data (η 0,ψ 0 ) has a unique solution (η,ψ) C 0( [0,T];H s+1 2(R d ) H s+1 2(R d ) ), such that (1) we have (V,B) C 0( [0,T];H s (R d ) H s (R d ) ), (2) the condition (12) holds for 0 t T, with h replaced by h/2, (3) for all 0 t T and for all x R d, a(t,x) c/2 4

5 Remark 13 The main novelty is that, in view of Sobolev embeddings, the initial surfaces we consider turn out to be only of C 3/2+ǫ -class for some ǫ > 0 and consequently have unbounded curvature Remark 14 Assumption 1 in the above theorem is automatically satisfied if η 0 H s+1 2(R d ), ψ 0 H 1 2(R d ), V 0 H s (R d ), B 0 H 1 2(R d ) The only point where the estimates depend on ψ (and not only on η,v,b) come from the fact that we consider a general domain without assumption on the bottom Otherwise, we shall prove a priori estimates for the fluid velocity and not for the fluid potential (notice that the fluid potential is defined up to a constant) More precisely, the H s -norm of ψ is used only in the proof of an estimate (cf (410)) which holds trivially, say, in infinite depth (since the quantity γ which is estimated in (410) is 0 in infinite depth) We also refer the reader to Theorem 435 of Lannes [36] for a well-posedness result requiring only that ψ belongs to an homogeneous Sobolev space 15 The pressure The purpose of this paragraph is to clarify, for low regularity solutions of the water waves system in rough domains, the definition of the pressure which is required if one wants to come back from solutions to the Zakharov system to solutions to the free boundary Euler equation This definition will also provide the basic a priori estimates which will be later the starting point when establishing higher order elliptic regularity estimates required when studying the Taylor coefficient a = y P Σ On a physics point of view, the pressure is the Lagrange multiplier which is required by the incompressibility of the fluid (preservation of the null divergence condition) As a consequence, taking the divergence in (13), it is natural to define the pressure as a solution of (110) x,y P = div x,y (v x,y v), P y=η = 0 Notice however that the solution of such a problem may not beuniqueas can beseen in the simple case when Ω = (,0) R d Indeed, if P is a solution, then P +cy is another Notice also that if P satisfies (110), then x,y (P +gy v 2) = 0 Definition 15 Let (η,ψ) (W 1, H 1/2 (R d )) H 1/2 (R d ) Assume that the variational solution (as defined in 31) of the equation (111) x,y φ = 0, φ y=η = ψ, satisfies Let R be the variational solution of x,y φ 2 (x,η(x)) H 1/2 (R d ) x,y R = 0 in Ω, R y=η = gη x,yφ 2 y=η We define the pressure P in the domain Ω by P(x,y) := R(x,y) gy 1 2 x,yφ(x,y) 2 Remark 16 The main advantage of defining the pressure as the solution of a variational problem is that it will satisfy automatically an a priori estimate (the estimate given by the variational theory) 5

6 It remains to link the solutions to the Zakharov system to solutions of the free boundary Euler system (13) with boundary conditions (14) To do so, we proved in [3] that if (η,ψ) is a solution of thezakharov system, if we consider thevariational solution to (111), then the velocity field v = x,y φ satisfies (13), which is of course equivalent to (112) P = t φ gy 1 2 x,yφ 2 Theorem 17 (from [3]) Assume that (η,ψ) C([0,T];H s+1 2(R d ) H s+1 2(R d )), with s > 1/2 + d/2, is a solution of the Zakharov/Craig-Sulem system (16) Then the assumptions required to define the pressure are satisfied, and (112) is satisfied, and the distribution t φ is well defined for fixed t and belongs to the space H 1,0 (Ω(t)) (see Definition 33) 16 Plan of the paper At first glance, Theorem 12 looks very similar to our previous result in presence of surface tension [1, Theorem 11] Indeed, the regularity threshold exhibited by the velocity field (namely V,B H s (R d ),s > 1+d/2) is the same in both results and (as explained above) appears to be the natural one However, an important difference between both cases is that the algebraic nature of (16) (and its counter-part in presence of surface tension) requires that the free domain is 3/2 smoother than the velocity field in presence of surface tension and only 1/2 smoother without surface tension This algebraic rigidity of the system implies that in order to lower the regularity threshold to the natural one (Lipschitz velocities), we are forced to work with C 3/2 domains (compared to the much smoother C 5/2 regularity in [1]) This in turn poses new challenging questions in the study of the Dirichlet Neumann operator Indeed, at this level of regularity the regularity of the remainder term in the paradifferential description of the Dirichlet-Neumann operator G(η)ψ is not given by the regularity of the function ψ itself, but rather by the regularity of the domain This is this phenomenon which forces us to work with the new unknowns V,B rather than with ψ In Section 2, we wrote a review of paradifferential calculus and proved various technical results useful in the article In Section 3 we study the Dirichlet-Neumann operator In Section 4, we symmetrize the system and prove a priori estimates In Section 5 we prove the contraction estimates required to show uniqueness and stability of solutions In particular we prove a contraction estimate for the difference of two Dirichlet-Neumann operators, involving only (in terms of Sobolev embedding) the C 1 2-norm of the difference of the functions defining the domains (see Theorem 52), while in Section 6 we prove the existence of solutions by a regularization process Acknowledgements We would like to thank the referees for their comments which led to a better version of this paper 2 Paradifferential calculus Let us review notations and results about Bony s paradifferential calculus We refer to [13, 34, 42, 43] for the general theory Here we follow the presentation by Métivier in [42] 21 Paradifferential operators For k N, we denote by W k, (R d ) the usual Sobolev spaces For ρ = k + σ, k N,σ (0,1) denote by W ρ, (R d ) the space of functions whose derivatives up to order k are bounded and uniformly Hölder continuous with exponent σ 6

7 Definition 21 Given ρ [0,1] and m R, Γ m ρ (Rd ) denotes the space of locally bounded functions a(x,ξ) on R d (R d \0), which are C with respect to ξ for ξ 0 and such that, for all α N d and all ξ 0, the function x ξ α a(x,ξ) belongs to W ρ, (R d ) and there exists a constant C α such that, ξ 1 2, α ξ a(,ξ) W ρ, (R d ) C α(1+ ξ ) m α Given a symbol a, we define the paradifferential operator T a by (21) Ta u(ξ) = (2π) d χ(ξ η,η)â(ξ η,η)ψ(η)û(η)dη, where â(θ,ξ) = e ix θ a(x,ξ)dx is the Fourier transform of a with respect to the first variable; χ and ψ are two fixed C functions such that: (22) ψ(η) = 0 for η 1 5, ψ(η) = 1 for η 1 4, and χ(θ,η) satisfies, for 0 < ε 1 < ε 2 small enough, and such that χ(θ,η) = 1 if θ ε 1 η, (θ,η) : χ(θ,η) = 0 if θ ε 2 η, α θ β η χ(θ,η) Cα,β (1+ η ) α β Since we shall need to work with paraproducts, we chose a cut-off function χ such that when a = a(x), T a is given by the usual expression in terms of the Littlewood- Paley operators Namely, the function χ can be constructed as follows Let κ C 0 (Rd ) be such that κ(θ) = 1 for θ 11, κ(θ) = 0 for θ 19 Then we define χ(θ,η) = + k=0 κ k 3(θ)ϕ k (η), where κ k (θ) = κ(2 k θ) for k Z, ϕ 0 = κ 0, and ϕ k = κ k κ k 1 for k 1 Given a temperate distribution u and an integer k in N we also introduce S k u and k u by S k u = κ k (D x )u and k u = S k u S k 1 u for k 1 and 0 u = S 0 u 22 Symbolic calculus We shall use quantitative results from [42] about operator norms estimates in symbolic calculus Introduce the following semi-norms Definition 22 For m R, ρ [0,1] and a Γ m ρ (R d ), we set (23) M m ρ (a) = sup α 2(d+2)+ρ sup ξ 1/2 (1+ ξ ) α m α ξ a(,ξ) W ρ, (R d ) Definition 23 (Zygmund spaces) Consider a Littlewood-Paley decomposition: u = q=0 qu (with the notations introduced above) If s is any real number, we define the Zygmund class C s (R d ) as the space of tempered distributions u such that u C s := sup2 qs q u L < + q Remark 24 Recall that C s (R d ) is the Hölder space W s, (R d ) if s (0,+ )\N Definition 25 Let m R An operator T is said to be of order m if, for all µ R, it is bounded from H µ to H µ m The main features of symbolic calculus for paradifferential operators are given by the following theorem 7

8 Theorem 26 Let m R and ρ [0,1] (i) If a Γ m 0 (Rd ), then T a is of order m Moreover, for all µ R there exists a constant K such that (24) T a H µ H µ m KMm 0 (a) (ii) If a Γ m ρ (R d ),b Γ m ρ (R d ) then T a T b T ab is of order m+m ρ Moreover, for all µ R there exists a constant K such that (25) T a T b T ab H µ H µ m m +ρ KM m ρ (a)mm 0 (b)+kmm 0 (a)mm ρ (b) (iii) Let a Γ m ρ (Rd ) Denote by (T a ) the adjoint operator of T a and by a the complex conjugate of a Then (T a ) T a is of order m ρ Moreover, for all µ there exists a constant K such that (26) (T a ) T a H µ H µ m+ρ KMm ρ (a) We shall need in this article to consider paradifferential operators with negative regularity As a consequence, we need to extend our previous definition Definition 27 For m R and ρ (,0), Γ m ρ (R d ) denotes the space of distributions a(x,ξ) on R d (R d \0), which are C with respect to ξ and such that, for all α N d and all ξ 0, the function x α ξ a(x,ξ) belongs to Cρ (R d ) and there exists a constant C α such that, (27) ξ 1 2, α ξ a(,ξ) C ρ C α (1+ ξ ) m α For a Γ m ρ, we define (28) M m ρ (a) = sup α 2(d+2)+ ρ sup ξ 1/2 (1+ ξ ) α m α ξ a(,ξ) C ρ (R d ) 23 Paraproducts and product rules Ifa = a(x)isafunctionofxonly, the paradifferential operator T a is called a paraproduct A key feature of paraproducts is that one can replace nonlinear expressions by paradifferential expressions up to smoothing operators Also, one can define paraproducts T a for rough functions a which do not belong to L (R d ) but merely to C m (R d ) with m > 0 Definition 28 Given two functions a,b defined on R d we define the remainder R(a,u) = au T a u T u a We record here various estimates about paraproducts (see chapter 2 in [10] or [18]) Theorem 29 i) Let α,β R If α+β > 0 then (29) R(a,u) H α+β d K a 2(R d ) H α (R d ) u H β (R d ), (210) R(a,u) Hα+β(Rd) K a Cα (Rd ) u H β (R d ) ii) Let m > 0 and s R Then (211) T a u H s m K a C m u H s iii) Let s 0,s 1,s 2 be such that s 0 s 2 and s 0 < s 1 +s 2 d 2, then (212) T a u H s 0 K a H s 1 u H s 2 By combining the two previous points with the embedding H µ (R d ) C µ d/2 (R d ) (for any µ R) we immediately obtain the following results 8

9 Proposition 210 Let r,µ R be such that r +µ > 0 If γ R satisfies γ r and γ < r +µ d 2, then there exists a constant K such that, for all a H r (R d ) and all u H µ (R d ), au T a u H γ K a H r u H µ Corollary 211 i) If u j H s j (R d ) (j = 1,2) with s 1 +s 2 > 0 then (213) u 1 u 2 H s 0 K u 1 H s 1 u 2 H s 2, if s 0 s j, j = 1,2, and s 0 < s 1 +s 2 d/2 ii) (Tame estimate in Sobolev spaces) If s 0 then (214) u 1 u 2 H s K ( u 1 H s u 2 L + u 1 L u 2 H s) iii) Let µ,m R be such that µ,m > 0 and m N Then (215) u 1 u 2 H µ K ( ) u 1 L u 2 H µ + u 2 C m u 1 H µ+m iv) Let s > d/2 and consider F C (C N ) such that F(0) = 0 Then there exists a non-decreasing function F: R + R + such that, for any U H s (R d ) N, (216) F(U) H s F ( U L ) U H s Proof The first two estimates are well-known, see Hörmander [34] or Chemin [18] To prove iii) we write u 1 u 2 = T u1 u 2 +T u2 u 1 +R(u 1,u 2 ) and use that T u1 u 2 H µ u 1 L u 2 H µ (see (24)), T u2 u 1 H µ u 2 C m u 1 H µ+m (see (211)), R(u 1,u 2 ) H µ u 2 C m u 1 H µ+m (see (210)) Finally, iv) is due to Meyer [43, Théorème 25 and remarque] Finally, let us finish this section with a generalization of (211) Proposition 212 Let ρ < 0, m R and a Γ m ρ Then the operator T a is of order m ρ: (217) T a H s H s (m ρ) CMm ρ (a), T a C s C s (m ρ) CM m ρ (a) Proof Let us prove the first estimate The proof of the second is similar Notice that if m = 0 and a(x,ξ) = a(x), then (217) is simply (211) Furthermore, if a(x,ξ) = b(x)p(ξ), then T a = T b p(d) As a consequence, using the first step and the fact that p(ξ) = ξ m p ( ξ ξ ) we obtain T a H s H s (m ρ) C b C ρ p S d 1 L In the general case, we can expand, for fixed x, a(x,ξ) in terms of spherical harmonics Let ( h ν ) ν N be an orthonormal basis of L 2 (S d 1 ) consisting of eigenfunctions of the (self-adjoint) Laplace Beltrami operator, ω = S d 1 on L 2 (S d 1 ), ie ω hν = λ 2 ν h ν By the Weyl formula, we know that λ ν cν 1 d Setting h ν (ξ) = ξ m hν (ω), ω = ξ ξ, ξ 0, we can write a(x,ξ) = a ν (x)h ν (ξ) where a ν (x) = a(x,ω) h ν (ω)dω ν N S d 1 9

10 Then T a = ν N T a νh ν Now since we have λ 2k ν a ν (x) = k wa(x,ω) h ν (ω)dω, S d 1 taking k = d we obtain for all ν 1, a ν C ρ λν 2d d wa(x,ω) h ν (ω) dω C 1 λ 2d ν Mρ m (a) h ν L 2 (S d 1 ), S d 1 C 2 ν 2 Mρ m (a) Moreover by the Sobolev embedding we have, with ε > 0 small (218) hν L (S d 1 ) C 3λ d 1 2 +ε ν C 4 ν d +ε d C4 ν 1 2 Therefore by the second step we can conclude that T a H s H s (m ρ) C 5 This completes the proof ν 3 ν N 2M m ρ (a) We shall also need the following technical result Proposition 213 Set D x = (I ) 1/2 i) Let s > d 2 and σ R be such that σ s Then there exists K > 0 such that for all V W 1, (R d ) H s (R d ) and u H σ 1 2(R d ) one has [ D x σ,v]u L 2 (R d ) K { V W 1, (R d ) + V H s (R d )} u H σ 1 2(R d ) ii) Let s > 1+ d 2 and σ R be such that σ s Then there exists K > 0 such that for all V H s (R d ) and u H σ 1 (R d ) one has [ D x σ,v]u L 2 (R d ) K V H s (R d ) u H σ 1 (R d ) Proof To prove i) we write [ D x σ,v]u L 2 A+B, A = [ D x σ,t V ]u L 2, B = [ D x σ,v T V ]u L 2 By (25) we have A K V W 1, u H σ 1 On the other hand one can write B D x σ (V T V )u L 2 + (V T V ) D x σ u L 2 = B 1 +B 2 We use Proposition 210 two times To estimate B 1 we take γ = σ,r = s,µ = σ 1 2 To estimate B 2 we take γ = 0,r = s,µ = 1 2 and we obtain, B K V H s u H σ 1 2 To prove ii), to estimate B 1 (resp B 2 ) we useagain Proposition 210 with γ = σ,r = s,µ = σ 1 (resp γ = 0,r = s,µ = 1) We shall need well-known estimates on the solutions of transport equations Proposition 214 Let I = [0,T], s > 1+ d 2 and consider the Cauchy problem { t u+v u = f, t I, (219) u t=0 = u 0 We have the following estimates t (220) u(t) L (R d ) u 0 L (R d ) + f(σ, ) L (R d ) dσ 10 0

11 There exists a non decreasing function F : R + R + such that (221) u(t) L 2 (R d ) F ( V L 1 (I;W 1, (R d )))( u0 L 2 (R d ) + t 0 f(t, ) L 2 (R d )dt ) and for any σ [0,s] there exists a non decreasing function F : R + R + such that (222) u(t) H σ (R d ) F ( V L 1 (I;H s (R d )))( u0 H σ (R d ) + t 0 f(t, ) H σ (R d )dt ) 24 Commutation with a vector field We prove in this paragraph a commutator estimate between a paradifferential operator T p and the convective derivative t + V Inspired by Chemin [17] and Alinhac [6], we prove an estimate which depends on estimates on t p+v p and not on t,x p When a and u are symbols and functions depending on t I, we still denote by T a u the spatial paradifferential operator (or paraproduct) such that for all t I, (T a u)(t) = T a(t) u(t) Given a symbol a = a(t;x,ξ) depending on time, we use the notation M m 0 (a) := sup sup sup (1+ ξ ) α m ξ α a(t;,ξ) L (R d ) t [0,T] α 2(d+2)+ ρ ξ 1/2 Given a scalar symbol p = p(t,x,ξ) of order m, it follows directly from the symbolic calculus rules for paradifferential operators (see (24) and (25)) that, [ T p, t +T V ] u H µ K{M m 0 ( t p)+m m 0 ( p) V W 1, } u H µ+m A technical key point in our analysis is that one can replace this estimate by a tame estimate which does not involve the first order derivatives of p, but instead t p+v p Lemma 215 Let V C 0 ([0,T];C 1+ε (R d )) for some ε > 0 and consider a symbol p = p(t,x,ξ) which is homogeneous in ξ of order m Then there exists K > 0 (independent of p,v) such that for any t [0,T] and any u C 0 ([0,T];H m (R d )), (223) [ T p, t +T V ] u(t) L 2 (R d ) { } K M m 0 (p) V(t) C 1+ε +M m 0 ( tp+v p) u(t) H m (R d ) Proof Set I = [0,T] and denote by R the set of continuous operators R(t) from H m (R d ) to L 2 (R d ) with norm satisfying } R(t) L(H m (R d ),L 2 (R d )) {M K m 0 (p) V(t) C 1+ε +M m 0 ( tp+v p) We begin by noticing that it is sufficient to prove that ( (224) t +V ) ( T p = T p t +T V ) +R, R R Indeed, by Theorem 529 in [42], we have (for fixed t) (V T V ) T p u L 2 V W 1, T p u L 2 V W 1, M m 0 (p) u H m by using the operator norm estimate (24) This implies that ( V T V ) Tp R We split the proof of (224) into three steps By decomposing p into a sum of spherical harmonics, we shall reduce the analysis to establishing (224) for the special case when T p is a paraproduct In the first step we prove (224) for m = 0 and p = p(t,x) Inthesecondstepweprove(224)forp = a(t,x)h(ξ) wherehishomogeneous in ξ of order m Then we consider the general case 11

12 (225) Step 1: Paraproduct, m = 0, p = p(t,x) In this case M 0 0 (p) = p L We have { t T p u = T tpu+t p t u, V T p u = V T p u+vt p u =: A+B Decompose V = S j 3 (V)+S j 3 (V), with S j 3 (V) = k V, S j 3 (V) = k V k j 3 k j 2 With the choice of cut-off function χ made in 21, given two functions u and a we have T a u = j S j 3(a) j u and hence A = A 1 +A 2, (226) A 1 := S j 3 (V)S j 3 ( p) j u, A 2 := S j 3 (V)S j 3 ( p) j u j j Let us consider the term A 2 Since S j 3 (V) L k V L k j 2 k j 2 and S j 3 ( p) L 2 j p L, we obtain (227) A 2 L 2 j 2 jε V C 1+ε 2 k(1+ε) V C 1+ε p L u L 2 M 0 0(p) V C 1+ε We now estimate A 1 = A 11 +A 12, with A 11 := { S j 3 Sj 3 (V) p } j u, j (228) A 12 := {[ S j 3 ] } (V),S j 3 p j u j Write S j 3 (V) = V S j 3 (V), to obtain A 11 = j where Then I = j S j 3 ( V p ) j u j 2 j(1+ε) V C 1+ε u L 2 S j 3 { S j 3 (V) p } j u = T V p u+i +II ( { Sj 3 S j 3 (V)p }) j u, II = { S j 3 S j 3 ( V)p } j u j I L 2 j j Moreover, II L 2 j 2 j S j 3 (V)p L j u L 2 2 j 2 j(1+ε) V C 1+ε p L u L 2 V C 1+ε p L u L 2 S j 3 ( V) L p L j u L 2 V C 1+ε p L u L 2 Therefore (229) A 11 = T V p u+ru, R R In order to estimate A 12 we note that one can write [ S j 3 (V),S j 3 ] p = [ S j 3 (V),S j 3 ] p+s j 3 ( S j 3 ( V)p ) 12

13 Since S j 3 k = 2 j ψ 1 (2 j 3 D) where ψ 1 C 0 (Rd ) we have [ S j 3 (V),S j 3 ] p L C2 jε V C 1+ε p L Moreover we have ( S j 3 S j 3 ( V)p ) L C2 jε V C 1+ε p L It follows that A 12 = Ru with R R Consequently, we deduce from (228) and (229) that A 1 = T V p u + Ru for some R in R It thus follows from (226) and (227) that A = T V p u+ru with R R We estimate now the term B introduced in (225) We split this term as follows: B = V (T p u) = V S j 3 (p) j u = j S j 3 (V)S j 3 (p) j u+ j j S j 3 (V)S j 3 (p) j u =: B 1 +B 2 We have B 2 L 2 j j S j 3 (V) L S j 3 (p) L j u L 2 2 j(1+ε) V C 1+ε 2 j p L u L 2 and hence B 2 = Ru with R R To deal with the term B 1, let us introduce (230) C := T p T V u = S j 3 (p) j S k 3 (V) k u j Since the spectrum of S k 3 (V) k u is contained in {(3/8)2 k ξ (2+1/8)2 k }, the term j (S k 3 (V) k u) vanishes unless k j 3 On the other hand, for k j 3, S k 3 (V) S j 3 (V) = ± 3 l= 3 l+jv, and hence we can write C under the form C = C 1 +C 2 = C 1 + { S j 3 (p) j Sj 3 (V) k u } j where C 1 is given by so that C 1 = j k k j 3 3 i 2 { S j 3 (p) j l+j (V) i+j (u) l+j i (V) j i (u) }, i=1 l= 1 C 1 L 2 j p L 2 j(1+ε) V C 1+ε 2 j u L 2, which implies that C 1 = Ru with R R To estimate C 2, as before we write C 2 = C 21 +C 22 where C 21 := S j 3 (p) [ j,s j 3 (V) ] k u, j k j 3 C 22 := S j 3 (p)s j 3 (V) j k u, j k j 3 where (using frequency localization in dyadic annuli and Plancherel formula) C 21 2 L 2 p 2 L 2 2j V 2 W 1, 22j k u 2 L 2 p 2 L V 2 C 1+ε u 2 L 2 j k j 3 13

14 On the other hand, since j k j 3 k = j, we have C 22 = j S j 3 (V)S j 3 (p) j u = B 1 We thus end up with (231) B = T p T V u+ru, R R It follows from (225) and (231) that (232) ( t +V )T p u = T p ( t +T V )u+t tp+v pu+ru, R R The symbolic calculus shows that T tp+v p R, which proves (224) and concludes the proof of the first step Step 2 : Higher order paraproducts We now assume that p(t,x,ξ) = a(t,x)h(ξ) where h(ξ) = ξ m h( ξ ξ ) with h C (S d 1 ) Then, directly from the definition (21), we have T p = T a ψ(d x )h(d x ) where ψ satisfies (22) We have [ Tp, t +T V ] = [ T a, t +T V ] [ ψ(d x )h(d x )+T a ψ(dx )h(d x ),T V ] We claim that the norm from H m to L 2 of the operator in the left hand side is bounded by (233) C ( a L V C 1+ε + t a+v a L ) h H d+2 (S d 1 ) To obtain this claim, notice that the norm of the first term in the right-hand side is estimated by means of the previous step by C( a L V C 1+ε + t a+v a L ) h L (S d 1 ) To estimate the second term in the right-hand side we use a sharp version of the symbolic calculus estimate (25), see [42, Theorem 614], which implies that the norm of the second term is bounded by C a L V L h H d+2 (S d 1 ) The sharp version of symbolic result alluded to above asserts that (25) holds with the seminorm Mρ m (cf (23)) replaced by other semi-norms where the supremum on the multi-index α is taken for α smaller than d/2 plus an explicit number In particular we have the following lemma which suffices to complete the proof of the claim (233) Lemma 216 Set b(ξ) = ξ m ψ(ξ) h ( ξ ξ ) There exists C > 0 independent of V and h such that [b(d),tv]u L2(Rd) C V L (Rd) h Hd+2(Sd 1) u Hm 1(Rd) for every u H m 1 (R d ) Step 3 : Paradifferential operators Consider an orthonormal basis ( h ν ) ν N of L 2 (S d 1 ) consisting of eigenfunctions of the (self-adjoint) Laplace Beltrami operator, ω = S d 1 on L 2 (S d 1 ), ie ω hν = λ 2 ν h ν By the Weyl formula, we know that λ ν cν d 1 Setting h ν (ξ) = ξ m hν (ω), ω = ξ/ ξ, ξ 0, we can write p(t,x,ξ) = a ν (t,x)h ν (ξ) where a ν (t,x) = p(t,x,ω) h ν (ω)dω ν N S d 1 Since λ 2k ν a ν (t,x) = k wp(t,x,ω) h ν (ω)dω, S d 1 λ 2k ν ( t +V )a ν (t,x) = k w ( t +V )p(t,x,ω) h ν (ω)dω S d 1 14

15 taking k = d+2 we deduce (234) sup t I sup t I a ν (t, ) L Cλν 2(d+2) M m 0 (p) ( t +V )a ν (t, ) L Cλν 2(d+2) M m 0 ( tp+v p) Moreover, there exists a positive constant K such that, for all ν 1, (235) hν H d+2 (S d 1 ) Cλd+2 ν Now we can write [ t +T V,T p ]u L 2 ν N [ t +T V,T aνh ν ]u L 2 So using (233) for every ν 1 and the estimates (234) (235), we obtain (224), since the sum λ d+2 ν λν 2(d+2) ν 1 2 d ν ν is finite This completes the proof of the lemma We have also a Sobolev analogue of Lemma 215 which can be proved similarly Lemma 217 Let s > 1+d/2 and V C 0 ([0,T];H s (R d )) There exists a positive constant K such that for any symbol p = p(t,x,ξ) which is homogeneous in ξ of order m R and all u C 0 ([0,T];H s+m (R d )), [ T p, t +T V ] u(t) H s (R d ) K{M m 0 (p) V(t) H s +M m 0 ( t p+v p)} u(t) H s+m (R d ) 25 Parabolic evolution equation Consider the evolution equation z w + D x w = 0, where z R and x R d By using the Fourier transform, one easily checks that (236) sup w(z) H r + ( 1 z [0,1] 0 w(z) 2 H r+1 2 dz)1 2 K w(0) H r The purpose of this section is to prove similar results when the constant coefficient operator D x is replaced by an elliptic paradifferential operator GivenI R,z 0 I andafunctionϕ = ϕ(x,z) definedonr d I,wedenotebyϕ(z 0 ) the function x ϕ(x,z 0 ) For I R and a normed space E, ϕ Cz 0 (I;E) means that z ϕ(z) is a continuous function from I to E Similarly, for 1 p +, ϕ L p z(i;e) means that z ϕ(z) E belongs to the Lebesgue space L p (I) In this section, when a and u are symbols and functions depending on z, we still denote by T a u the function defined by (T a u)(z) = T a(z) u(z) where z I is seen as a parameter We denote by Γ m ρ (Rd I) the space of symbols a = a(z;x,ξ) such that z a(z; ) is bounded from I into Γ m ρ (R d ) (see Definition 22), with the semi-norm (237) M m ρ (a) = sup z I sup α 2(d+2)+ρ sup ξ 1/2 (1+ ξ ) α m α ξ a(z;,ξ) W ρ, (R d ) Given µ R we define the spaces X µ (I) = Cz 0 (238) (I;Hµ (R d )) L 2 z 2(R d (I;Hµ+1 )), Y µ (I) = L 1 z (I;Hµ (R d ))+L 2 z 2(R d (I;Hµ 1 )) 15

16 Proposition 218 Let r R, ρ (0,1), J = [z 0,z 1 ] R and let p Γ 1 ρ (Rd J) satisfying Rep(z;x,ξ) c ξ, for some positive constant c Then for any f Y r (J) and w 0 H r (R d ), there exists w X r (J) solution of the parabolic evolution equation (239) z w +T p w = f, w z=z0 = w 0, satisfying w X r (J) K { w 0 H r + f Y r (J) for some positive constant K depending only on r,ρ,c and M 1 ρ (p) Furthermore, this solution is unique in X s (J) for any s R Proof Letr R Denoteby, H r thescalarproductinh r (R d )andchosef 1 and F 2 such that f = F 1 +F 2 with }, F 1 L 1 z (J;H r ) + F 2 f L 2 (J;H r 1 2) Y r (J) +δ, δ > 0 Let us consider for ε > 0 the equation (240) z w ε +ε( +Id)w ε +T p w ε = f, w ε z=z0 = w 0 Then standard methods in parabolic equations show that for any z 1 > z 0, this equation have a unique solution in C 0 ([z 0,z 1 ];H r (R d )) L 2 ((z 0,z 1 );H r+2 (R d )) (here we only used that T p is a Sobolev first order operator) To let ε go to zero we need to establish uniform estimates with respect to ε Taking the scalar product in H r, directly from (240), we obtain 1 d 2dz w ε(z) 2 H r +ε ( +Id)w ε(z),w ε (z) H r +Re T p(z) w ε (z),w ε (z) H r F 1 (z) H r w ε (z) H r + F 2 (z) H r 2 1 w ε(z) H r+ 1 2 It follows from Gårding s inequality (see [42, Section 632]) that there exist two constants C 1,C 2 > 0 depending only on M 1 ρ (p) such that for any u Hr, Re T p(z) u(z),u(z) H r C 1 u(z) 2 C 2 u(z) 2 H r+1 2 H r+1 ρ 2 for each fixed z J Therefore, we obtain 1 d 2dz w ε(z) 2 H r +ε ( +Id)w ε(z),w ε (z) H r +C 1 w ε (z) 2 H r+1 2 F 1 (z) H r w ε (z) H r + F 2 (z) H r 2 1 w ε(z) H r C 2 w ε (z) 2 H r+1 ρ 2 Integrating in z we obtain that, for all z [z 0,z 1 ], A(z) := 1 } z { w ε (z) 2 H 2 r w ε(z 0 ) 2 H r +ε w ε (z ) 2 H dz r+1 z 0 z +C 1 is bounded by, z 0 wε (z ) 2 H r+1 2 dz B := F 1 L 1 (J;H r ) w ε L (J;H r ) + F 2 L 2 (J;H r 1 2) w ε L 2 (J;H r+1 2) 16 +C 2 w ε 2 L 2 (J;H r+1 ρ 2 )

17 By standard arguments, it follows that (241) w ε 2 L (J;H r ) +C 1 w ε 2 w 0 2 L 2 (J;H r+1 2) H r +C ( F 1 2 L 1 (J;H r ) + F w ε 2 ) L 2 (J;H r 1 2) L 2 (J;H r+1 ρ 2 ) Finally, to eliminate the last term in the right hand side of (241), one notices that the left hand side controls by interpolation c w ε 2, for some p > 2, L p (J;H r+1 ρ 2 ) hence by Hölder in the z variable, there exists κ > 0 (depending only on p) such that if z 0 z 1 κ, we have C w ε 2 1 ( wε 2 L 2 (J;H r+1 ρ L 2 ) 2 (J;H r ) +C 1 w ε 2 ) L 2 (J;H r+1 2) We consequently obtain w ε 2 L (J;H r ) +C 1 w ε 2 L 2 (J;H r+1 2) 2 w 0 2 H r +C( F 1 2 L 1 (J;H r ) + F 2 2 L 2 (J;H r 1 2)) We can now iterate the estimate between z 0 + κ and z 0 + 2κ, to get rid of the assumption z 1 z 0 κ (and of course the constants will depend on z 1 ) By using the equation, we obtain now that (w ε ) is bounded in X r (J) C 1 (J;H r 2 ) It follows from the Banach Alaoglu theorem that, up to a subsequence, (w ε ) converges in the sense of distributions to w L (J,H r ) L 2 (J,H r+1 2), which satisfies theequation z w+t p w = f Then,usingtheequation, weobtain z w Y r (J)which implies that w C 0 ([z 0,z 1 ];H r (R d )) thanks to the Lemma 219 below Moreover, by the Ascoli theorem, up to a subsequence, (w ε ) converges in C 0 ([z 0,z 1 ];H r µ loc ) for some µ > 0 Since w ε z=0 = w 0 we obtain that w z=0 = w 0, which completes the existence part The proof of uniqueness follows the same steps and we omit it We recall now the following classical lemma (see [40] th31) Lemma 219 Let I = ( 1,0) and s R Let u L 2 z 2(R (I,Hs+1 d )) such that z u L 2 z 2(R (I,Hs 1 d )) Then u C 0 ([ 1,0],H s (R d )) and there exists an absolute constant C > 0 such that sup z [ 1,0] u(z, ) H s (R d )) C ( u L 2 (I,H s+1 2(R d )) + zu L 2 (I,H s 1 2(R d )) ) 3 The Dirichlet-Neumann operator We shall prove some elliptic regularity results using a paradifferential approach 31 Definition and continuity We begin by recalling from [1] the definition of the Dirichlet Neumann operator under general assumptions on the bottom One of the novelties with respect to our previous work is that we clarify the regularity assumptions: assuming only that η W 1, (R d ) and ψ H 1 2(R d ), we show how to define G(η)ψ and prove that the map ψ H 1 2(R d ) G(η)ψ H 1 2(R d ) is continuous Our second contribution is to prove that the map η G(η) is Lipschitz (in a proper topology) The goal is to study the boundary value problem (31) x,y φ = 0 in Ω, φ Σ = f, n φ Γ = 0 17

18 See 11 for the definitions of Ω,Σ,Γ Since we make no assumption on Γ, the definition of φ requires some care We recall here the definition of φ as given in [1] Notation 31 Denote by D the space of functions u C (Ω) such that x,y u L 2 (Ω) We then define D 0 as the subspace of functions u D such that u is equal to 0 in a neighborhood of the top boundary Σ Proposition 32 ([1, Proposition 22]) There exists a positive weight g L loc (Ω), equal to 1 near the top boundary of Ω and a constant C > 0 such that for all u D 0, (32) g(x,y) u(x,y) 2 dxdy C x,y u(x,y) 2 dxdy Ω Definition 33 Denote by H 1,0 (Ω) the space of functions u on Ω such that there exists a sequence (u n ) D 0 such that, x,y u n x,y u in L 2 (Ω,dxdy), Ω u n u in L 2 (Ω,g(x,y)dxdy) We endow the space H 1,0 (Ω) with the norm u = x,y u L 2 (Ω) Let us recall that the space H 1,0 (Ω) is a Hilbert space (see [1]) We are able now to define the Dirichlet-Neumann operator Let f H 1 2(R d ) We first define an H 1 lifting of f in Ω To do so let χ 0 C (R) be such that χ 0 (z) = 1 if z 1 2 and χ 0(z) = 0 if z 1 We set ψ 1 (x,z) = χ 0 (z)e z Dx f(x), x R d,z 0 By the usual property of the Poisson kernel we have x,z ψ 1 L 2 ([ 1,0] R d ) C f H 1 2(R d ) Then we set ( y η(x) ) ψ(x,y) = ψ 1 x,, (x,y) Ω h This is well defined since Ω {(x,y) : y < η(x)} Moreover since the bottom Γ is contained in {(x,y) : y < η(x) h}, we see that ψ vanishes identically near Γ Now we have obviously ψ Σ = f and since η L (R d ), an easy computation shows that ψ H 1 (Ω) and (33) ψ H 1 (Ω) K(1+ η W 1, ) f H 1 2(R d ) Then the map v x,y ψ x,y vdxdy Ω is a bounded linear form on H 1,0 (Ω) It follows from the Riesz theorem that there exists a unique u H 1,0 (Ω) such that (34) v H 1,0 (Ω), x,y u x,y vdxdy = x,y ψ x,y vdxdy Then u is the variational solution to the problem Ω x,y u = x,y ψ in D (Ω), u Σ = 0, n u Γ = 0, the latter condition being justified as soon as the bottom Γ is regular enough Lemma 34 The function φ = u + ψ constructed by this procedure is independent on the choice of the lifting function ψ as long as it remains bounded in H 1 (Ω) and vanishes near the bottom 18 Ω

19 Proof Considertwofunctionsconstructedbythisprocedure,φ k = u k +ψ k,k = 1,2 Then, by standard density arguments, since ψ 1 ψ 2 vanishes at the top boundary Σ and in a neighborhood of the bottom Γ, there exists a sequence of functions ψ n C0 (Ω) supported in a fixed Lipschitz domain Ω Ω tending to ψ 1 ψ 2 in H0 1( Ω) and hence also in H 1,0 (Ω) As a consequence, ψ 1 ψ 2 H 1,0 (Ω) and the function φ = φ 1 φ 2 is the unique (trivial) solution in H 1,0 (Ω) of the equation x,y φ = 0 given by the Riesz Theorem Definition 35 We shall say that the function φ = u+ψ constructed by the above procedure is the variational solution of (31) It satisfies (35) x,y φ 2 dxdy K f 2 H 2(R 1, d ) Ω for some constant K depending only on the Lipschitz norm of η Formally the Dirichlet-Neumann operator is defined by (36) G(η)ψ = 1+ η 2 n φ y=η(x) = [ y φ η φ ] y=η(x) 311 Straightening the free boundary In what follows we shall set Ω 1 = {(x,y) : x R d,η(x) h < y < η(x)}, (37) Ω 2 = {(x,y) O : y η(x) h}, Ω = Ω 1 Ω 2, where O has been defined in section 11, and Ω 1 = {(x,z) : x R d,z I}, I = ( 1,0), (38) Ω 2 = {(x,z) R d (, 1] : (x,z +1+η(x) h) Ω 2 }, Ω = Ω 1 Ω 2 Guided by Lannes ([37]), we consider the map (x,z) ρ(x,z) from Ω to R defined as follows { ρ(x,z) = (1+z)e δz D x η(x) z { e (1+z)δ Dx η(x) h } if (x,z) Ω 1, (39) ρ(x,z) = z +1+η(x) h if (x,z) Ω 2 where δ > 0 will be chosen later on Lemma 36 Assume η W 1, (R d ) (1) There exists C > 0 such that for every (x,z) Ω we have x ρ(x,z) C η W 1, (R d ) (2) There exists K > 0 such that, if δ η W 1, (R d ) h 2K we have (310) min ( 1, h 2) z ρ(x,z) max (1, 3h 2 ), (x,z) Ω (3) The map (x,z) (x,ρ(x,z)) is a Lipschitz diffeomorphism from Ω 1 to Ω 1 Proof (1) For any λ < 0 the symbol a(ξ) = e λ ξ satisfies the estimate α ξ a(ξ) C α ξ α where C α is independent on λ Therefore its Fourier transform in an L 1 (R d ) function whose norm is uniformly bounded This implies that a(d)f L (R d ) M f L (R d ) with M independent of λ This proves our claim 19

20 (2) We have in Ω 1 (311) z ρ = h+ ( e δz Dx η η ) +(1+z)δe δz Dx D x η ( e δ(1+z) Dx η η ) +zδe δ(1+z) Dx D x η Since e δλ Dx η η = δλ 1 0 eδtλ Dx D x ηdt we can write e δz Dx η η L (R d ) +δ e δz Dx D x η L (R d ) + e δ(1+z) Dx η η L (R d ) which proves (2) +δ e δ(1+z) Dx D x η L (R d ) Kδ η W 1, (R d ) h 2 (3) Since ρ(x,0) = η(x),ρ(x, 1) = η(x) h our claim follows from (1) For later use we state the following result whose proof is straightforward (recall that the spaces X s (I) are defined in (238)) Lemma 37 Let I = ( 1,0) Assume s > d 2 There exists C > 0 such that for every η H s+1 2(R d ) we have (312) z ρ h X s 1 2(I) C δ η H s+ 1 2(R d ), xρ X s 1 2(I) C δ η H s+ 1 2(R d ) Coming back to the case where η is a Lipschitz function, Lemma 36 shows that the map (x,z) (x,ρ(x,z)) is a Lipschitz-diffeomorphism from Ω to Ω We denote by κ the inverse map of ρ: (x,z) Ω, (x,ρ(x,z)) = (x,y) (x,z) = (x,κ(x,y)), (x,y) Ω Let φ(x,z) = φ(x,ρ(x,z)) Then we have ( y φ)(x,ρ(x,z)) = (Λ 1 φ)(x,z), ( x φ)(x,ρ(x,z)) = (Λ 2 φ)(x,z), (313) Λ 1 = 1 z ρ z Λ 2 = x xρ z ρ z If φ is a solution of x,y φ = 0 in Ω then φ satisfies This yields (Λ 2 1 +Λ2 2 ) φ = 0 in Ω (314) (a 2 z + x +b x z c z ) φ = 0, where (315) a := 1+ xρ 2 ( z ρ) 2, b := 2 xρ z ρ, c := 1 ( a 2 z ρ z ρ+ x ρ+b x z ρ ) It will be convenient to have a constant coefficient in front of 2 z φ Dividing (314) by a we obtain (316) ( 2 z +α x +β x z γ z ) φ = 0, where (317) α := ( zρ) 2 1+ x ρ 2, β := 2 zρ x ρ 1 ( 1+ x ρ 2, γ := 2 z ρ z ρ+α x ρ+β x z ρ ) In the coordinates (x,z), according to (36) we have (318) G(η)ψ = Ũ z=0, Ũ = Λ 1 φ x ρ Λ 2 φ 20

21 The following remark will be useful in the sequel We have (319) z Ũ = x (( z ρ)λ 2 φ) Indeed we can write z Ũ = z Λ 1 φ x z ρ Λ 2 φ x ρ z Λ 2 φ = ( z ρ)λ 2 1 φ x z ρ Λ 2 φ+( z ρ)(λ 2 x )Λ 2 φ = ( z ρ)(λ 2 1 +Λ 2 2) φ x (( z ρ)λ 2 φ) Since (Λ 2 1 +Λ2 2 ) φ = 0 we obtain (319) 312 Continuity of the Dirichlet-Neumann operator Theorem 38 Let η W 1, (R d ) Then for all f H 2(R 1 d ), G(η)f = 1+ η 2 n φ y=η(x) = [ y η φ] y=η(x) is well defined in H 1 2(R d ) Furthermore there exists F : R + R + non decreasing independent of η,f such that G(η)f H 1 2(R d ) F( η W 1, (R d )) f H 1 2(R d ) Proof With the preceding notations we have G(η)f = ( Λ 1 φ x ρ Λ 2 φ) z=0 Let us set Ũ = Λ 1 φ x ρ Λ 2 φ Then the theorem will follow from Lemma 219 with s = 1 2 and the following inequality where I = ( 1,0) Ũ L 2 (I,L 2 (R d )) + z Ũ L 2 (I,H 1 (R d )) F( η W 1, (R d )) f H 1 2(R d ) TheestimateonŨ isaconsequenceof (35),(313)andtheestimate(1)inLemma36 on x ρ To estimate z Ũ we use (319) and the estimate on z ρ in Lemma 36 Notice that, asabyproductofthepreviousproof, inthesystemofcoordinates(x,z), the variational solution of (31), φ, satisfies (320) φ C 0 z ([ 1,0];H 1 2(R d )) C 1 z ([ 1,0];H 1 2(R d )) We also state a second basic strong continuity of the Dirichlet-Neumann operator Theorem 39 There exists a non decreasing function F: R + R + such that, for all η j W 1, (R d ), j = 1,2 and all f H 1 2(R d ), ( G(η 1 ) G(η 2 ) ) f H 1 2 F( (η 1,η 2 ) W 1, W 1, ) η1 η 2 W 1, f H 1 2 Proof We use the notations introduced in 311 Namely, for j = 1,2, x R d,z I := ( 1,0) we introduce ρ j (x,z) and v j (x,z) defined by (39), ρ j (x,z) = (1+z)(e δz Dx η j )(x) z { e (1+z)δ Dx η j (x) h }, ρ j (x,z) = z +1+η j h, if (x,z) Ω 2, where δ η j W 1, (R d ) is small enough for j = 1,2 Notice that we have the following estimates (see Lemma 36) (i) z ρ j min(1, h ), (x,z) Ω, 2 (321) (ii) x,z ρ j L ( Ω) C(1+ η j W 1, (R d ) ) (iii) x,z (ρ 1 ρ 2 ) L (I,L (R d )) C η 1 η 2 W 1, (R d ) 21

22 Recall also that we have set (322) Λ i 1 = 1 z ρ i z, Λ i 2 = x xρ i z ρ i z It follows from (321) that for k = 1,2 we have with W 1, = W 1, (R d ), (323) { (i) Λ 1 k Λ 2 k = β k z, with suppβ k R d I, (ii) β k L (I R d ) F( (η 1,η 2 ) W 1, W 1, ) η 1 η 2 W 1, Then we set φ j (x,z) = φ j (x,ρ j (x,z)) (where x,y φ j = 0 in Ω j,φ j Σj = f) and we recall (see (318)) that (324) G(η j )f = U j z=0, U j = Λ j 1 φ j x ρ j Λ j 2 φ j Lemma 310 Set I = ( 1,0), v = φ 1 φ 2, and Λ j = (Λ j 1,Λj 2 ) There exists a non decreasing function F : R + R + such that (325) Λ j v L 2 (I;L 2 (R d )) F( (η 1,η 2 ) W 1, W 1, ) η 1 η 2 W 1, f H 1 2 Let us show how this Lemma implies Theorem 39 According to (324) we have (326) U 1 U 2 = (1)+(2)+(3)+(4)+(5) where (1) = Λ 1 1 v, (2) = (Λ1 1 Λ2 1 ) φ 2, (3) = x (ρ 1 ρ 2 )Λ 1 2 φ 1 (4) = ( x ρ 2 )Λ 1 2v, (5) = ( x ρ 2 )(Λ 1 2 Λ 2 2) φ 2 The L 2 (I,L 2 (R d )) norms of (1) and (4) are estimated using (325) and (321) Also, the L 2 (I,L 2 (R d )) norms of (2) and (5) are estimated by the right hand side of (325) using (323) and (35) Eventually the L 2 (I,L 2 (R d )) norm of (3) is also estimated by the right hand side of (325) using (321) (iii) and (35) It follows that (327) U 1 U 2 L 2 (I,L 2 ) F( (η 1,η 2 ) W 1, W 1, ) η 1 η 2 W 1, f H 1 2 Now according to (319) we have (328) z (U 1 U 2 ) = x ( z (ρ 1 ρ 2 )Λ 1 2 φ 1 +( z ρ 2 )(Λ 1 2 Λ2 2 ) φ 1 +( z ρ 2 )Λ 2 2 v) Therefore using the same estimates as above we see easily that (329) z (U 1 U 2 ) L 2 (I,H 1 ) F( (η 1,η 2 ) W 1, W 1, ) η 1 η 2 W 1, f H 1 2 Then Theorem 39 follows from (327), (329) and Lemma 219 Proof of Lemma 310 We use the variational characterization of the solutions u i First of all we notice that φ 1 φ 2 = ũ 1 ũ 2 =: v Now setting X = (x,z) and recalling that Λ i = (Λ i 1,Λi 2 ), we have (330) Λ i ũ i Λ i θj i dx = Λ i f Λ i θj i dx Ω Ω for all θ H 1,0 ( Ω), where J i = z ρ i Taking the difference between the two equations (330), using (321) and setting θ = v = ũ 1 ũ 2 one can find a positive constant C such that 6 Λ 1 v 2 dx C A k, Ω 22 k=1

23 where A 1 = A 3 = A 5 = (Λ 1 Λ 2 )ũ 2 Λ 1 v J 1 dx, A 2 = Ω Λ 2 ũ 2 Λ 2 v J 1 J 2 dx, A 4 = Ω (Λ 1 Λ 2 )v Λ 2 f J1 dx, A 6 = Ω Using (323), (35), (321) we can write (Λ 1 Λ 2 )v Λ 2 ũ 2 J 1 dx, Ω (Λ 1 Λ 2 ) f Λ 1 v J 1 dx, Ω Λ 2 f Λ 2 v J 1 J 2 dx Ω (331) A 1 β L (I R d ) J 1 L (I R d ) z ũ 2 L 2 (I R d ) Λ 1 v L 2 ( Ω) F( (η 1,η 2 ) W 1, W 1, ) η 1 η 2 W 1, f H 1 2 Λ 1 v L 2 ( Ω) Since Λ 1 j Λ2 j = β j zρ 1 Λ 1 1 the term A 2 can beboundedby the right hand sideof (331) Now we have J 1 J 2 L (I R d ) C η 1 η 2 W 1, (R d ) and Λ 2 v L 2 ( Ω) F( (η 1,η 2 ) W 1, W 1, ) Λ1 v L 2 ( Ω) So using (35) we see that the term A 3 can be also estimated by the right hand side of (331) To estimate the terms A 4 to A 6 we use the same arguments and also (33) This completes the proof Let us finish this definition section by recalling also the following result which is a consequence of [1, Lemma 29] Lemma 311 Assume that 1 2 a < b 1 5 then the strip S a,b = {(x,y) R d+1 : ah < y η(x) < bh} is included in Ω and for any k 1, there exists C > 0 such that φ H k (S a,b ) C ψ H 1 2(R d ) 32 Paralinearization of the Dirichlet-Neumann operator In the case ofsmoothdomains, itisknownthat, moduloasmoothingoperator, G(η)isapseudodifferential operator with principal symbol given by (332) λ(x, ξ) := (1+ η(x) 2 ) ξ 2 ( η(x) ξ) 2 Notice that λ is well-defined for any C 1 function η The main result of this section allow to compare G(η) to the paradifferential operator T λ when η has limited regularity Namely we want to estimate the operator R(η) = G(η) T λ Such an analysis was at the heart of our previous work [4] [1, Proposition 314] for smooth domains (η H s+1 2,s > 2 + d 2 ) Here we are able to lower the regularity thresholds down to s > d 2 The following results, which we think are of independent interest, complement previous estimates about the Dirichlet- Neumann operator by Craig-Schanz-Sulem [27], Beyer Günther [12], Wu [52, 53], Lannes [37] Theorem 312 Let d 1, s > d 2 and 1 2 σ s Then there exists a non-decreasing function F: R + R + such that, for all η H s+1 2(R d ) and all f H σ (R d ), we have G(η)f H σ 1 (R d ), together with the estimate (333) G(η)f H σ 1 (R d ) F( η H s+ 1 2(R d )) f H σ (R d ) 23