INCOMPRESSIBLE FLUIDS IN THIN DOMAINS WITH NAVIER FRICTION BOUNDARY CONDITIONS (II) Luan Thach Hoang. IMA Preprint Series #2406.

INCOMPRESSIBLE FLUIDS IN THIN DOMAINS WITH NAVIER FRICTION BOUNDARY CONDITIONS II By Luan Thach Hoang IMA Preprint Series #2406 August 2012 INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS UNIVERSITY OF MINNESOTA 400 Lind Hall 207 Church Street S.E. Minneapolis, Minnesota 55455-0436 Phone: 612-624-6066 Fax: 612-626-7370 URL: http://www.ima.umn.edu

INCOMPRESSIBLE FLUIDS IN THIN DOMAINS WITH NAVIER FRICTION BOUNDARY CONDITIONS II LUAN THACH HOANG ABSTRACT. We study the Navier-Stokes equations for incompressible fluids in a three-dimensional thin two-layer domain whose top, bottom and interface boundaries are not flat. In addition to the Navier friction boundary conditions on the top and bottom boundaries of the domain, and the periodicity condition on the sides, the fluid velocities are subject to an interface boundary condition which relates the normal stress of each fluid to the relative velocity between them on the common boundary. We prove that the strong solutions exist for all time if the initial data and body force, measured in relevant norms, are appropriately large as the domain becomes very thin. In our analysis, the interface boundary condition is interpreted as a variation of the Navier boundary conditions containing an interaction part. The effect of that interaction on the Stokes operator and the nonlinear term of the Navier-Stokes equations is expressed and carefully estimated in different ways in order to obtain suitable estimates. CONTENTS 1. Introduction 1 2. Problem formulation 3 3. Consequences of the boundary conditions 7 4. Stokes and averaging operators 16 4.1. Stokes operator and functional form of NSE 16 4.2. Averaging operators 22 5. Estimate of a nonlinear term 24 6. Global solutions 30 Appendix A. Auxiliary Inequalities 36 References 37 1. INTRODUCTION This is a continuation of our previous work [8] in which the global existence in time of strong solutions to incompressible Navier-Stokes equations NSE was proved for a three-dimensional single layer thin domain under large data conditions, that is, when the initial data and body Date: July 9, 2012. Key words and phrases. Navier-Stokes equations, strong solution, Navier boundary conditions, interface boundary value problem, thin domain. 1

2 Luan Thach Hoang force are very large, but not arbitrarily large. The fluid is subject to the Navier friction boundary conditions on the top and bottom boundaries, and to the periodicity condition on the sides. In the current paper we study the NSE in thin two-layer domains with, in addition the mentioned Navier boundary conditions, an interface boundary condition for the stress tensors. Such an interface boundary condition is used, for example, in the coupled atmosphere-ocean models [19,20], see also the survey [30]. The reader is referred to [1,8 11,15,16,25] and references therein for NSE with Navier boundary conditions, and to [13, 14, 17, 18, 21 23, 28, 29] among many others for NSE in thin domains. Previous works on the problem are [4] and [12] which deal with domains with flat boundaries. Thanks to the simple geometry of the boundaries, all or some explicit solutions of the corresponding eigenvalue problems are found. Based on those, an appropriate closed subspace in which important commutativity properties of involved operators Leray projection, or special projections, or Laplacian, or Stokes operator, or averaging operators are available. In contrast, the domain considered in this paper has non-flat top, bottom and interface boundaries. Therefore it is impossible to find and take advantage of those explicit solutions. Also, while [4, 12] have free boundary conditions on the top and bottom boundaries, which correspond to the Navier conditions without friction, here we consider the Navier conditions with positive friction coefficients. We adopt the approach in [7 9, 15] which does not rely on any special spectral spaces. Instead, certain properties of the velocity field satisfying those boundary conditions need be realized in order to obtain suitable estimates. On one hand, the interface boundary condition is a variation of the Navier boundary conditions see 3.5, hence the treatment of the normal stress on the boundary resembles that in the Navier conditions. On the other hand, it can be specifically interpreted as having a Navier part and an interaction part see 3.6 and 3.3. The Navier part can be treated together with the conditions of the same type on the top and bottom boundaries, while the interaction part requires extra work. Thanks to such coherent treatments, we can take advantage of our previous works on the Navier conditions and prove the global existence of strong solutions for this involved case within a reasonable length. Our result Theorems 6.2 and 6.3 is comparable to those obtained in [7,9,15] for single layer domains. It only differs from [4, 12] for two-layer domains only on the technical dependence on the domain s thickness ε. Briefly speaking, the global-in-time strong solutions exist even when the initial data has large H 1 -norm and the body force has large L 2 -norm, both expressed in terms of ε. Moreover, thanks to strictly positive friction coefficients, none of the zero average condition in [4, 12], or the orthogonality condition in [15], or the generic domain condition in [8, 9], is imposed. It is noteworthy that our result shows that the larger friction we have on the top and bottom boundaries, the larger in L 2 -norm initial and body force are allowed. The reader is referred to Remark 6.5 for more discussions on the result. The article is organized as follows. In section 2, we introduce the problem, formulation, notation and basic assumptions. In section 3, properties of vector fields satisfying the boundary conditions are presented. These include both their behavior on the boundaries and their norm

Incompressible fluids in thin two-layer domains 3 estimates on the whole domain. In section 4, we define the Stokes operator A corresponding to our boundary conditions, and derive the functional form of the NSE. Relations between Lebesgue and Sobolev norms with A-related norms are obtained. Particular is the H 2 -norm relation and comparison estimate in Propositions 4.3, 4.4 and Corollary 4.5. We also define the averaging operators M 0 and M in subsection 4.2 which play crucial roles in our result s formulation and in obtaining various estimates. Inequalities of Sobolev, Poincaré and Agmon types for these averages and their corresponding oscillations are derived with constants depending explicitly on the domain s thickness and fluids viscosities. In section 5, the key nonlinear estimate is proved under the main assumption 5.1 on the viscosities and friction coefficients. It is derived with the desired boundedness of the term ε A 1/2 u 2 L 2 kept in mind. In section 6, we state and prove the main theorem of the paper Theorem 6.2. Its argument is arranged for clarity with the use of Lemma 6.4. This theorem is restated in the form of Theorem 6.3 without using the Stokes operator. Final comments are given in Remark 6.5. The Appendix recalls some auxiliary inequalities for thin domains. 2. PROBLEM FORMULATION Let T 2 be the two-dimensional torus R/Z 2 and ε 0, 1]. Let Ω ε + and Ω ε be two layers Ω ε + = {x, x 3 : x T 2, h ε 0x < x 3 < h ε +x }, Ω ε = {x, x 3 : x T 2, h ε x < x 3 < h ε 0x }, where h ε 0 = εg 0, h ε + = εg 0 + g + and h ε = εg 0 g, with the functions g 0 x, g + x and g x belonging to C 3 T 2, and g +, g positive everywhere. Hence there is a positive constant c 0 > 0 such that c 1 0 g 0, g +, g c 0 in T 2, g 0 C 3 T 2, g + C 3 T 2, g C 3 T 2 c 0. We will use the same notation c 0 to denote positive constants in our assumptions below. We define our two-layer domain by Ω ε = Ω ε + Ω ε. The top and bottom boundaries of Ω ε are while the interface boundary is Γ ε + = {x, x 3 : x T 2, x 3 = h ε +x }, Γ ε = {x, x 3 : x T 2, x 3 = h ε x }, Γ ε 0 = {x, x 3 : x T 2, x 3 = h ε 0x }. Each layer Ω ε +, resp. Ω ε, is occupied with an incompressible viscous fluid of velocity u + = u ε +x, t for x Ω ε + and t R, resp. u = u ε x, t for x Ω ε + and t R, pressure p + = p ε +x, t, resp. p = p ε x, t, having kinetic viscosity ν + = ν+, ε resp. ν = ν. ε Each u + x, t, resp. u x, t satisfies the corresponding NSE on Ω ε +, resp. Ω ε with an outer body force f + = f+x, ε t, resp. f = f x, ε t.

4 Luan Thach Hoang For each vector field u, we denote by u the 3 3 matrix of its partial derivatives, that is, u ij = u i x j, for i, j = 1, 23. We also denote Du = u + u 2 where u is the transpose matrix of u. Below, we will drop the superscript ε except for the domain notation Ω ε, Ω ε +, Ω ε. The velocity fields u + and u satisfy the following boundary conditions. The Navier friction conditions on the top and bottom boundaries: u + N + = 0, [ν + Du + N + ] tan + γ + u + = 0 on Γ +, 2.1 u N = 0, [ν Du N ] tan + γ u = 0 on Γ, 2.2 where γ + = γ ε + and γ = γ ε are positive constants. Above, N +, resp. N, is the outward normal vector to the boundary of Ω ε +, resp. Ω ε. The subscript tan denotes the tangential component of a vector on the boundary. The interface boundary condition on : u + N + = 0, [ν + Du + N + ] tan + γ 0 u + u = 0, 2.3 u N = 0, [ν Du N ] tan + γ 0 u u + = 0, 2.4 where γ 0 = γ0 ε is a positive constant. For a unified treatment of equations on both Ω ε + and Ω ε, we introduce the index notation ι {+, } and set if ι = +, ι = 2.5 + if ι =. In calculations, when not used as an index, i.e., not a subscript, ι = +,, is interpreted as +1, 1, respectively. Therefore the previous convention 2.5 for ι agrees with these numeric values. With the notation ι, we rewrite h ι = εg 0 + ιg ι and have Ω ε ι = {x, x 3 : x T 2, x 3 is between h 0 x and h ι x }, Γ ι = {x, x 3 : x T 2, x 3 = h ι x }. Note that Ω ε ι = Γ ι and N ι = N ι on. We also rewrite the Navier boundary conditions 2.1 and 2.2 as u ι N ι = 0, [ν ι Du ι N ι ] tan + γ ι u ι = 0 on Γ ι for ι = +,, 2.6 and rewrite the boundary conditions 2.3 and 2.4 as u ι N ι = 0, [ν ι Du ι N ι ] tan + γ 0 u ι u ι = 0 on for ι = +,. 2.7

Incompressible fluids in thin two-layer domains 5 To avoid writing the + and components repeatedly like in 2.6 and 2.7, we pair them together into vectors: Ω ε = [Ω ε +, Ω ε ], Ω ε = [ Ω ε +, Ω ε ], Γ = [Γ +, Γ ] and = [, ], u = [u +, u ], p = [p +, p ], f = [f +, f ] and ν = [ν +, ν ]. The initial value problem for the NSE in Ω ε is written in the form: u + u u ν u = p + f t on Ωε 0,, div u = 0 on Ω ε 0,, ux, 0 = u 0 x on Ω ε, 2.8 2.9 where u 0 = [u 0 +, u 0 ] is the known initial velocity field. We will write u + = u +,1, u +,2, u +,3, u = u,1, u,2, u,3 and u = u 1, u 2, u 3 where u i = [u +,i, u,i ] for each i = 1, 2, 3. Denoting N = [N +, N ] and γ = [γ +, γ ], we rewrite the set of boundary conditions 2.6 and 2.7 in the following compact forms: u N = 0 on Γ, 2.10 and [νdun] tan + γu = 0 on Γ, 2.11 u N = 0 on, 2.12 where [νdun] tan + γ 0 u Su = 0 on, 2.13 Su = S[u +, u ] def == [u, u + ]. The slip conditions 2.10 and 2.12 can be combined into u N = 0 on Ω ε. 2.14 We now discuss the relations between viscosities ν ε +, ν ε and friction coefficients γ ε +, γ ε, γ ε 0 as ε 0. We remind the reader that the superscript ε here is an index, not an exponent. First we assume that ν + and ν are comparable to each other as ε 0, that is, 0 < lim inf ε 0 νι ε lim sup ν ε + 2 + ν ε 2 ε 0 Assume that there exist δ +, δ, δ 0 [0, 1] such that νι ε <, ι = +,. ν ε + 2 + ν ε 2 and 0 < lim inf ε 0 ε δι γε ι ν ε ι 0 < lim inf ε 0 lim sup ε δι γε ι ε 0 νι ε ε δ 0 γε 0 ν ε ι lim sup ε δ γε 0 0 ε 0 νι ε < for ι = +,, <.

6 Luan Thach Hoang For the sake of simplicity, we consider through out that 0 < δ + = δ = δ δ 0 1. 2.15 Taking into account the above conditions, without loss of generality, we assume ν +, ν c 1 0 ν, 2.16 c 1 0 ε δ ν γ, γ + c 0 ε δ ν and c 1 0 ε δ 0 ν γ 0 c 0 ε δ 0 ν, 2.17 for all ε 0, 1], where ν = ν 2 + + ν 2 1/2. Notation. Concerning the vectors in 2.8, for any f, f = [f +, f ] and u = [u +, u ] we denote fu = [fu +, fu ] and fu = [f + u +, f u ]. For example, u = [ u +, u ] and ν u v = [ ν + u + v +, ν u v ]. The usual Euclidean length of vector u is denoted by Obviously, while For integrals, we use the notation [ fudx = Ω ε fudx = Ω ε u = u + 2 + u 2 1/2. Su = u. 2.18 Ω ε + Ω ε + ] f + u + dx, f u dx, Ω ε f + u + dx + Ω ε f u dx. Similar notation is used for surface integrals on Ω ε, Γ and. The Lebesgue and Sobolev spaces on Ω ε are naturally defined as follows. Let L 2 Ω ε = L 2 Ω ε + L 2 Ω ε = {u = [u +, u ] : u ι L 2 Ω ε ι, ι = +, }. 2.19 This space is equipped with the inner product u, v = u +, v + L 2 Ω ε + + u, v L 2 Ω ε = Ω ε u v dx. We denote the corresponding norm by u L 2, while denoting [[ u ]] L 2 = [ u + L 2 Ω ε +, u L 2 Ω ε ] u L 2 = u + 2 L 2 Ω ε + + u + 2 L 2 Ω ε 1/2. The spaces H 1 Ω ε and H 2 Ω ε with their norms are defined similarly. Same definitions and notations apply to the spaces L 2 Ω ε k, H 1 Ω ε k and H 2 Ω ε k, for k = 1, 2, 3,... The norms in those spaces will be denoted by the same notations L 2, H 1 and H 2, disregarding k. Such disregard of k will be used also for pairs of norms [[ u ]], [[ u ]] and [[ ]] L 2 H 1 u H 2.

Incompressible fluids in thin two-layer domains 7 Throughout, the expression [u +, u ] =, [v +, v ] + C means u + =, v + + C and u =, v + C, where C is a number. Also, we write f 1 = f 2 + Oε α f 3 where f 3 0, to indicate that f 1 f 2 = f 4 with f 4 Cε α f 3, when ε > 0 is sufficiently small. 3. CONSEQUENCES OF THE BOUNDARY CONDITIONS We present in this section some properties of the vector fields satisfying the Navier and/or interface boundary conditions. They extend the similar ones for Navier friction boundary conditions in [2, 7 9]. For a function ux and a vector τ R 3 not necessarily having unit length we denote ux ux + hτ ux = lim. τ h 0 h We restate Lemma 4.1 of [8] for our two-layer domain see also [2]: Lemma 3.1 [8]. Let u satisfy 2.10 and τ be a tangential vector field to Γ. Then u τ N + N u = 0 on Γ. 3.1 τ If, in addition, u satisfies 2.11 then { } u N N τ = u τ 2γ ν τ on Γ 3.2 and N u = 2N { n n u γ ν N u} on Γ, 3.3 where n is any C 1 -extension of one of the four vector fields [±N +, ±N ] from Γ to its neighborhood. For the interface boundary, we obtain similar properties: Lemma 3.2. Let u satisfy 2.12 and τ be a tangential vector field to. Then u τ N + N τ u = 0 on. 3.4 If, in addition, u satisfies 2.13, then and u N τ = u N τ 2γ 0 ν u Su τ on 3.5 N u = 2N { n n u γ 0 ν N u} 2 γ 0 ν Su on, 3.6 where n is any C 1 -extension of one of the four vector fields [±N +, ±N ] from to its neighborhood.

8 Luan Thach Hoang Proof. Identity 3.4 is the same as 3.1. For 3.5, we observe that on, the relation 2.13 implies ν[dun] τ = γ 0 u Su τ, which in turn yields un τ + uτ N = 2 γ 0 u Su τ, ν hence u N τ + u τ N = 2γ 0 u Su τ. 3.7 ν Therefore 3.5 follows from 3.7 and 3.4. We now prove 3.6. Let ω = u. Let τ 1 and τ 2 be two unit tangential vectors on which are orthogonal to each other. Then 0 = u n τ 1 τ 2 u n τ 1 = τ 1 τ 2 u n. τ 2 Since n = ±τ 1 τ 2, we have on that n u n = 0, hence where Ku = 1 2 { u u }. Note that with and, thanks to 2.13, 0 = n [ u n] + n [ n u] = n [Dun Kun] + n [ n u], n [Dun Kun] = N [DuN KuN], KuN = 1 2 ω N, N [DuN] = N [DuN] tan = γ 0 N u Su. ν Thus we have from 3.8 that Therefore 0 = γ 0 ν N u Su 1 2 N ω N + n [ n u]. 3.8 N N ω = 2n [ n u] 2 γ 0 N u Su. 3.9 ν Applying N to 3.9 and using the identity a a a b = a 2 a b, we obtain N ω = 2N { n n u γ 0 ν N u Su} on. 3.10 Note that N Su = SN Su = 0 on. Then N N Su = Su on. Therefore 3.6 follows 3.10. To make use of properties 3.3 and 3.6, we specify below extensions of N ι from Γ ι to the whole domain T 2 R. We define for x = x, x 3 = x 1, x 2, x 3 T 2 R that N ι x, x 3 = ι 1h ι x, 2 h ι x, 1 1 + 2 h ι x 2, 3.11

Incompressible fluids in thin two-layer domains 9 where 2 = / x 1, / x 2. Similarly, for as part of the boundary of Ω ε ι, we define N 0,ι x, x 3 = ι 1h 0 x, 2 h 0 x, 1 1 + 2 h 0 x 2. 3.12 Then N ι and N 0,ι, for ι = +,, are x 3 -independent vector fields on T 2 R that satisfy N ι Γι = N ι and N 0,ι Γ0 = N ι. For specific tangential vector fields on Ω ε and their extensions, we define for x = x, x 3 T 2 R: / τι j x = e j + j h ι x e 3 1 + j h ι x 2, ι =, +, j = 1, 2, 3.13 / τ0x j = e j + j h 0 x e 3 1 + j h 0 x 2, j = 1, 2, 3.14 where {e 1, e 2, e 3 } is the standard canonical basis of R 3. On each boundary Γ ι, for ι = +,, the two vectors τ 1 ι and τ 2 ι form a basis of the tangent space to Γ ι. Similarly, the two vectors τ 1 0 and τ 2 0 form a basis of the tangent space to. They are all defined on T 2 R and are independent of x 3. For these vector fields, we have the following estimates in T 2 R: N ι, N 0,ι, τ j ι, τ j 0 Cε, 3.15 e 3 ιn ι, e 3 + ιn 0,ι, e j τ j ι, e j τ j 0 Cε 3.16 for ι = +, and j = 1, 2. For vector forms, we denote N = [N +, N ] and N 0 = [N 0,+, N 0, ], 3.17 τ j = [τ j +, τ j ] and τ j 0 = [τ j 0, τ j 0] for j = 1, 2. 3.18 For the orthonormal frames on the boundary, we define By 3.15 we have for j = 1, 2 that τ 1 = τ 1, τ 2 = N τ 1 and τ 1 = τ 1 0, τ 2 = N 0 τ 1. 3.19 τ j, τ j Cε in T 2 R. 3.20 As seen in [7 9, 15], boundary conditions affect estimates of different Sobolev norms. For instance, the following special Poincaré-like inequality for the third component of a vector field is proved in [9]. Lemma 3.3 [9], Lemma 4.7. If u is in H 1 Ω ε 3 and satisfies 2.14, then [[ u3 ]] L 2 Cε [[ u ]] H 1, 3.21 where C > 0 does not depend on ε. For the partial derivative in x 3, we have the following version of Lemma 4.8 in [9].

10 Luan Thach Hoang Lemma 3.4. Let u belong to H 2 Ω ε 3 and satisfy 2.10 and 2.11. Then [[ ]] 3 u j Cε [[ u ]] + C ε γ [[ ]] L 2 H u, j = 1, 2, 3.22 2 ν L 2 Γ where C > 0 is independent of ε. Proof. The proof is the same as that of Lemma 4.2 in [8], using the boundary Γ ι in place of there. Let ι = +, and j = 1. Using trace inequality A.1, we have On Γ ι : 3 u ι,1 L 2 Ω ι C ε 3 u ι,1 L 2 Γ ι + Cε 3 3 u ι,1 L 2 Ω ι. 3.23 3 u ι,1 = u ι,1 e 3 ιn ι + ι u ι,1 N ι = u ι,1 e 3 ιn ι + ι u ι e 1 τ 1 ι N ι + ι u ι τ 1 ι N ι = u ι,1 e 3 ιn ι + ι u ι e 1 τ 1 ι N ι + ι u ι N ι τ 1 ι. Take τ in Lemma 3.1 to be τ 1 defined by 3.18. Since N = N on Γ, and we have from 3.2 that u ι N ι τι 1 = u ι N ι τι 1 2 γ ι u ι τι 1 on Γ ι. ν ι Then thanks to 3.15 and 3.16, we obtain 3 u ι,1 Cε u ι + C ε + γ ι ν ι uι Cε u ι + C γ ι ν ι u ι on Γ ι. Applying trace inequality A.2 to u ι, we have Together with 3.23, 3 u ι,1 L 2 Γ ι Cε 1 ε u ι H 1 Ω ι + ε u ι H 2 Ω ι + γ ι ν ι u ι L 2 Γ ι. 3 u ι,1 L 2 Ω ι Cε u ι H 2 Ω ι + C ε γ ι ν ι u ι L 2 Γ ι, ι = +,. Therefore we obtain 3.22 for j = 1. The case j = 2 is similar. Comparing 3.22 to the Poincaré inequality A.1, we not only replace 3 u j by u for the L 2 -norm on Γ but also gain the factor γ/ν ε δ. Using estimates 3.15 and 3.20, we restate the integration by parts result of Lemma A.1 in [8] as follows. Lemma 3.5 [8]. Let Σ =, resp. Γ +, Γ. Suppose ux and vx are two functions on Σ and τx is τ0x j or τ j± x, resp. τ+x j or τ +x j, τ x j or τ x j, with j = 1 or 2. Then u τ v dσ = v u dσ + Oε u v dσ. 3.24 τ Σ Σ Next, we compare the norm 2 u L 2 with u L 2. We obtain a version of Lemma 5.4 in [8]. Σ

Incompressible fluids in thin two-layer domains 11 Lemma 3.6. There is ε 1 0, 1] such that if ε 0, ε 1 ] and u H 2 Ω ε 3 satisfies the boundary conditions 2.10, 2.11 and 2.12, 2.13, then ν 1/2 2 u 2 L 2 C ν1/2 u 2 L + ν 2 u 2 H1 3.25 and, consequently, 2 u L 2 C u L 2 + u H 1. 3.26 Proof. We follow the proof in Appendix A of [8]. Integration by parts twice gives ν 2 u 2 dx = ν u 2 dx + I 0, 3.27 Ω ε Ω ε where 1 u 2 I 0 = ν Ω ε 2 N u N u dσ. 3.28 We write I 0 = I 0,+ + I 0, + I 0,0 where I 0,+, I 0, are the integrals on Γ + and Γ, respectively, and 1 u 2 I 0,0 = 2 N u N u dσ. 3.29 ν Exactly as in [8], by using extensions τ 1, τ 2 and N and working only on the separate components Γ + and Γ of Γ, we have hence I 0,ι = Oεν ι Γ ι u ι 2 + u ι 2 dσ, 3.30 I 0,+ + I 0, = Oε Γ ν u 2 + u 2 dσ. 3.31 Many arguments leading to 3.30 are repeated below while we estimate I 0,0. Let τ 1, τ 2 be defined by 3.19. We have τ 1, τ 2, 2 τ 1, 2 τ 2 Cε on T 2 R. 3.32 In this proof, for calculations below, we use the extension N 0 defined in 3.17 for N on. Therefore, regarding estimates of derivatives of N on we will use the following N 0, 2 N 0 Cε on T 2 R. 3.33 Recall that { τ 1, τ 2, N 0 } restricted to forms an orthonormal frame on. For I 0,0 we have similar calculations up to A.9 of Appendix A in [8]: I 0,0 = 2 u ν τ 1 N u dσ + τ 1 2 u ν + Γ τ 1 τ 1 = I 1 + I 2 + I 3 + J = Γ ν 2 u τ 2 N u dσ τ 2 2 u u τ 2 τ 2 N dσ + J ι=+, I 1,ι + I 2,ι + I 3,ι + J ι,

12 Luan Thach Hoang where J ι = Oεν ι u ι 2 dσ. Same as in [8], I 3,ι = I 1,ι + I 2,ι + Oεν ι u ι 2 dσ, ι = +,. Therefore Combining 3.34 with 3.31 gives I 0,0 = 2I 1 + I 2 + Oε ν u 2 dσ. 3.34 I 0 = 2I 1 + I 2 + Oε ν u 2 + u 2 dσ. Ω ε 3.35 We now estimate I 1 and I 2. For τ, τ { τ 1, τ 2 }, taking the derivative / τ of identity 3.5 on gives 2 u τ N τ + u N N u = τ τ N τ + u 2 N τ τ 2 γ 0 u Su ν τ 2 γ 0 τ ν u Su τ τ. 3.36 Using 3.33 to estimate the first and second order derivatives of N on we can bound the second term on the left-hand side and the first two terms on the right-hand side by Oε u + u. Using properties 2.17, 2.18 and 3.32, we bound the last term of 3.36 by Oε u. Thus, we have 2 u τ N τ = 2γ 0 u Su τ + Oε u + u ν τ. 3.37 For the integrand of I 1, writing the dot product in the orthonormal basis { τ 1, τ 2, N} gives ν 2 u τ 1 N u + ν 2 u = ν τ 1 u 2 u τ 1 N τ 2 u τ 1 N τ 1 τ 1 τ 1 τ 2 + ν τ 1 2 u τ 1 N N u τ 1 N. 3.38 By 3.37, the first term on the right-hand side of 3.38 is 2 u u ν τ 1 N τ 1 τ 1 τ 1 { u Su } u = 2γ 0 τ 1 + Oεν u + u τ τ 1 1 τ 1 = 2γ 0 u Su τ 1 τ 1 u τ 1 τ 1 + Oεν u 2 + u u.

Incompressible fluids in thin two-layer domains 13 The second term on the right-hand side of 3.38 is treated similarly. Also, we use 3.4 for the last factor of 3.38. We obtain ν 2 u τ 1 N u τ 1 u Su = 2γ 0 τ 1 ν 2 u τ 1 N N N τ 1 u u u Su τ 1 τ 1 2γ 0 τ 1 τ 1 + Oεν u 2 + u u. τ 2 u τ 1 τ 2 3.39 We have on that SN = N, S τ 1 = τ 1, hence S τ 2 = τ 2. Using these properties and writing the following sums explicitly, we obtain 2 [ u Su 2 u τ 1 τ 1 = u Su τ ι=+, 1 τ 1 τ 1 ]ι τ 1 0, 2 [ u Su 2 u τ 2 τ 2 = u Su τ 1 τ 1 τ 2 ]ι τ 1 0. ν ι=+, Therefore, summing the + and components of 3.39 and integrating over yields I 1 = [ 2 u ν ι ι=+, τ 1 N u dσ τ 1 ]ι { 2 u τ 1 N N u N } 3.40 dσ + Oε ν τ u 1 Γ 2 + u u dσ. 0 We apply Lemma 3.5 to the middle integral of 3.40: ν u { N u N } dσ τ 1 N τ 1 = ν u { N N u N } dσ + Oε τ 1 τ 1 = Oε ν u u + u dσ + Oε ν u N ν u u dσ. { N u N } dσ τ 1 For the last identity, we use 3.33 again to estimate N/ τ 1. Therefore, it follows 3.40 and Cauchy s inequality that I 1 Cε ν u 2 + u 2 dσ Cε ν u 2 + u 2 dσ. Similarly, I 2 Cε ν u 2 + u 2 dσ. Then combining 3.35 with above estimates of I 1, I 2 and using the trace inequality A.2 we have I 0 Cε ν u 2 + u 2 dσ C ν u 2 Ω ε H + 1 ε2 2 u 2 L2. 3.41

14 Luan Thach Hoang By 2.16, it follows 3.27 and 3.41 that ν 1/2 2 u 2 L 2 ν1/2 u 2 L + C ν 2 u 2 H + 1 Cε2 ν 1/2 2 u 2 L2. 3.42 For ε > 0 sufficiently small, the term Cε 2 ν 1/2 2 u 2 on the right-hand side of 3.42 can L 2 be absorbed into the left-hand side, hence we obtain 3.25. Then by 2.16 again, inequality 3.26 follows 3.25. The proof is complete. For further estimates in sections 4 and 5 below, we need to apply 3.3 and 3.6 with the same extension n. The above extensions N and N 0, however, cannot be matched on Γ and in general. Therefore we introduce new extension vector fields in the following. For x = x, x 3 T 2 R, define and its related function Ñ ι x = x 3 h 0 x ιεg ι x N ι x h ιx x 3 N ιεg ι x 0,ι x 3.43 Ñ ι x = x 3 h 0 x γι N ιεg ι x ι x + h ιx x 3 γ0 N ν ι ιεg ι x 0,ι x. 3.44 ν ι Then Ñι Γι = N ι and Ñι Γ0 = N ι. We have on T 2 R: Ñι,j, j Ñ ι Cε, for j = 1, 2, Ñι,3, 3 Ñ ι, 2 Ñ ι C, 3.45 Also, Ñ ι Γι = γι ν ι N ι and Ñ ι Γ0 = γ 0 ν ι N ι and Ñι + Ñ0,ι Cε, Ñι Ñ0,ι C. 3.46 Ñ ι C γ 0 + γ ι ν ι Cε δ, 2 Ñ ι Cε δ, 3 Ñ ι Cε δ 1 on T 2 R. 3.47 Above, we used the facts 0 < δ δ 0 and 0 < ε 1. Therefore Ñ ι Cε δ 1 on T 2 R. 3.48 Let Ñ = [Ñ+, Ñ ] and Ñ = [Ñ +, Ñ ]. Then Ñ is an extension of the upward/downward normal vectors on the boundary Ω ε. Lemma 3.7. Suppose u H 2 Ω ε 3 satisfies the boundary conditions 2.10, 2.11 and 2.12, 2.13. Then N u = N Gu on Γ, 3.49 N u = N Gu 2 γ 0 ν Su on, 3.50 where Gu is a vector field on Ω ε defined by 3.53 below and satisfies the following estimates in Ω ε : Gu Cε δ u, 3.51 Gu Cε δ u + ε δ 1 u. 3.52

Incompressible fluids in thin two-layer domains 15 Proof. Using the identity 3.3 and 3.6 with n = Ñ, we have 3.49 and 3.50, where the vector field Gu is defined on Ω ε by with As in [9, 15], we have Gu = G 1 u G 2 u, 3.53 G 1 u = 2Ñ [ Ñ u] and G 2 u = 2Ñ u. 3.54 G 1 u Cε u and G 1 u Cε u + C u on Ω ε. 3.55 By 3.47 and 3.48, one obtains G 2 u 2 Ñ u Cε δ u, 3.56 G 2 u C u Ñ + u Ñ Cε δ u + ε δ 1 u. 3.57 Combining 3.53, 3.55, 3.56 and 3.57 yields 3.51 and 3.52. The NSE has an important curl structure. The following lemma connects this curl structure to the boundary conditions, especially on the interface. Lemma 3.8. If u H 2 Ω ε 3 satisfies 2.10, 2.11 and 2.12, 2.13, and Φ H 1 Ω ε 3, then Ω ε ν u Φ dx = where Gu is defined in Lemma 3.7. Ω ε ν Φ u Gudx + Ω ε νφ Gudx 2γ 0 Proof. Let ω = u, one has ν ω Φ dx = νω Φdx + Ω ε Ω ε For the boundary integral, by 3.49 and 3.50, = Ω ε νn ω Φ dσ = = Γ νn Gu Φ dσ 2γ 0 Ω ε νφ Gu N dσ 2γ 0 Γ νn ω Φ dσ + Su Φ dσ u SΦ dσ, 3.58 Ω ε νn ω Φ dσ. u SΦ dσ. Applying the Divergence Theorem to the integral on Ω ε gives = Ω ε νφ Gu N dσ = νn ω Φ dσ Ω ε ν Φ Gudx Ω ε ν Φ Gu νφ Gudx.

16 Luan Thach Hoang Combining the above calculations, we obtain 3.58. Comparing Lemma 3.8 with its counterpart for a single layer domain Lemma 5.2 in [8] the additional boundary integrals on in 3.58 shows the interaction between u ι and Φ ι on for ι = +,. 4. STOKES AND AVERAGING OPERATORS 4.1. Stokes operator and functional form of NSE. For u = [u +, u ] H 2 Ω ε 3 and v = [v +, v ] H 1 Ω ε 3, we have the following Green s formula c.f. [15, 25]: ν u v dx = 2νDu : Dv + ν u vdx Ω ε Ω ε + {2νDuN v ν uv N}dσ, 4.1 Ω ε where the symbol : denotes the usual scalar product of two matrices. If u is divergence-free and satisfies the boundary conditions 2.10, 2.11 and 2.12, 2.13, and v is tangential to the boundary Ω ε then ν u v dx = 2 νdu : Dvdx Ω ε Ω ε 4.2 + 2 γu v dσ + 2γ 0 u Su v dσ. Consequently, we obtain where Eu, v = 2 In case u = v we have where Eu is defined by Eu = Eu, u = 2 Ω ε νdu : Dv dx + 2 Γ ν u v dx = Eu, v, Ω ε 4.3 Γ γu v dσ + γ 0 u Su v Svdσ. 4.4 ν u u dx = Eu. Ω ε 4.5 Ω ε ν Du 2 dx + 2 Γ γ u 2 dσ + γ 0 As with [9, 15, 25], a Korn s inequality is needed for the coercivity of E,. u Su 2 dσ. 4.6 Lemma 4.1. There is ε 2 0, 1] such that if ε 0, ε 2 ] and u H 1 Ω ε 3 satisfying 2.14 then ν 1/2 u L 2 c 1 ε 1 δ/2 Eu 1/2, 4.7 ν 1/2 u H 1 c 1 Eu 1/2, 4.8 Eu 1/2 c 1 ν 1/2 u L 2 + ε δ 1/2 u L 2 2c 1 ε δ 1/2 ν 1/2 u H 1, 4.9

where c 1 is a positive constant independent of ε and ν. Incompressible fluids in thin two-layer domains 17 Proof. By Lemma 4.13 of [9] applied to each domain Ω ι, we have ν 1/2 u 2 L 2 4 ν1/2 Du 2 L + 2 C ν1/2 u 2 L 2Eu + 2 C ν1/2 u 2 L2. 4.10 By Poincaré s inequality A.1, ν 1/2 u 2 L 2 Cε2 ν 1/2 u 2 L 2 + Cε ν1/2 u 2 L 2 Γ. By relation 2.17, the last L 2 -norm can be bounded by Hence Using this in 4.10 yields ν 1/2 u 2 L 2 Γ C ε δ/2 γ 1/2 u 2 L 2 Γ Cε δ Eu. ν 1/2 u 2 L 2 Cε2 ν 1/2 u 2 L 2 + Cε1 δ Eu. 4.11 ν 1/2 u 2 L 2 2Eu + Cε2 ν 1/2 u 2 L 2 + Cε1 δ Eu CEu + Cε 2 ν 1/2 u 2 L 2. For sufficiently small ε we obtain ν 1/2 u 2 L2 CEu. 4.12 By 4.11 and 4.12, we obtain 4.7. In turn, 4.7 and 4.12 imply 4.8. By definition 4.6, property 2.18, relations 2.15 and 2.17 and the trace inequality A.2, Eu 2 ν u 2 L 2 + Cεδ ν u 2 L 2 Ω ε 2 ν u 2 L 2 + C ν εδ ε u 2 L 2 + ε 1 u 2 L 2 C ν u 2 L 2 + C ν εδ 1 u 2 L 2. Thus one obtains the first inequality in 4.9. The second one is obvious. We now present the functional formulation for the NSE in the context of boundary conditions 2.14, 2.11 and 2.13. First, we recall the Helmholtz-Leray decomposition: where L 2 Ω ε ι 3 = H ι H ι, 4.13 H ι = {u L 2 Ω ε ι 3 : u = 0 on Ω ε, u N ι = 0 on Ω ε ι}, H ι Then we have for Ω ε : where = { φ : φ H 1 Ω ε ι}. L 2 Ω ε 3 = H H, 4.14 H = H + H = {u L 2 Ω ε 3 : u = 0 on Ω ε, u N = 0 on Ω ε }, H = H + H = { φ : φ H 1 Ω ε }.

18 Luan Thach Hoang Let P ι be the orthogonal projection from L 2 Ω ε ι 3 onto H ι. Define the Leray-Helmholtz projection for the two-layer domain Ω ε by P = [P +, P ] = P + P, i.e., P is the orthogonal projection from L 2 Ω ε 3 onto H. Define V = H 1 Ω ε 3 H. Consider ε ε 0, where ε 0 is from Lemma 4.1. By the same classical arguments as in [4,12] and by Lemma 4.1, there exist a bounded operator A : V V such that Au, v V,V = Eu, v, u, v V, and a bi-linear form B, from V V to the dual space V, such that Bu, v, w V,V = u v, w for all u, v, w V. 4.15 The domain of A, considered as an unbounded operator on H, is a subspace D A such that Au H for all u D A. Then A and D A can be characterized by D A = {u H 2 Ω ε : u = 0 and u satisfies 2.10, 2.11 and 2.12, 2.13}. 4.16 and Au = νp u for u D A. 4.17 We also have V = D A 1/2 the domain of the fractional power A 1/2. The Navier Stokes equations can now be written in the following variational form d u, v + Eu, v + Bu, u, v = f, v dt for all v V, 4.18 or, equivalently, in the functional form as in the space V. For more details, see e.g. [5, 27]. We have du dt + Au + Bu, u = P f, 4.19 Au, v = Eu, v, u D A, v V, 4.20 A 1/2 u 2 L2 = Eu, u V. 4.21 Below, we find relations between the Lebesgue and Sobolev norms of u with the A-related norms A 1/2 u L 2, Au L 2. First, a simple interpolation inequality: A 1/2 u 2 L 2 = Au, u u L 2 Au L 2, u D A. 4.22 Lemma 4.2. There is a positive number c 2 such that for all ε 0, ε 2, one has the following: i If u V then ν 1/2 u L 2 c 2 ε 1 δ/2 A 1/2 u L 2, 4.23 ν 1/2 u H 1 c 2 A 1/2 u L 2. 4.24

Incompressible fluids in thin two-layer domains 19 ii If u D A then ν u L 2 c 2 ε 1 δ Au L 2, 4.25 ν u H 1 c 2 ε 1 δ/2 Au L 2, 4.26 ν 1/2 A 1/2 u L 2 c 2 ε 1 δ/2 Au L 2. 4.27 Proof. Thanks to 4.21, the estimates in 4.23 and 4.24 are restatements of 4.7 and 4.8, respectively. For u D A and u 0, thanks to 4.22 and 4.23, one has A 1/2 u 2 L 2 u L 2 Au L 2 c 2ε 1 δ/2 ν 1/2 A 1/2 u L 2 Au L 2, hence one obtains 4.27. Inequalities 4.24 and 4.27 then imply 4.26. Applying 4.23 and then 4.27, we have 4.25. Proposition 4.3. If u D A then Au + ν u L 2 C ν ε δ 0 2 u L 2 + ε δ 0 1 u L 2 + ε δ 1 u L 2. 4.28 Proof. Let ω = u and Φ = Au + ν u. Assume Φ 0. By the Helmholtz-Leray decomposition 4.14, Φ = q where q = [q +, q ] H 1 Ω ε. Since Au H and Φ H are orthogonal in L 2 Ω ε 3, we have Φ 2 dx = Au + ν u Φ dx = ν u Φ dx = Ω ε Ω ε Ω ε By virtue of Lemma 3.8 noting that Φ = 0, we obtain Φ 2 dx Ω ε νφ Gudx + 2γ 0 Ω ε Ω ε ν ω Φ dx. [N N u] SΦ dσ = I 1 + I 2. 4.29 By Lemma 3.7, I 1 = νφ Gudx C ν Φ ε δ u + ε δ 1 u dx Ω ε Ω ε C ν Φ L 2{ε δ u L 2 + ε δ 1 u L 2}. To estimate I 2, we use an idea in [4,15]. Shifting q ι by a constant we can assume that q ι has zero average on Ω ι. Using a dilation argument c.f. Lemma 3.8 of [15], we have q L 2 C q L 2. Combining this with the trace inequality A.2 gives q L 2 C ε q L 2 + C ε q L 2 C ε q L 2, and hence q L 2 C ε Φ L 2. 4.30

20 Luan Thach Hoang Let v = N u. Then Noticing SN = N, we have I 2 = 2γ 0 J, where J = [ N SΦ] v dσ. J = [SN SΦ] v dσ = [SN Φ] v dσ = N Φ Sv dσ. Let τ 1 and τ 2 be defined in 3.19. We have N = τ 1 τ 2 on, thus q J = N q Sv dσ = τ 2 q τ 1 Sv dσ τ 1 τ 2 = Γ 0 q τ 2 Sv q τ 1 Svdσ. τ 1 τ 2 Applying Lemma 3.5 to this boundary integral on yields J = q τ 2 Sv + q τ 1 Sv dσ + Oε q τ 1 Sv + q τ 2 Sv dσ. τ 1 τ 2 Note Sv = v = u. Then J = q τ 2 Sv τ 1 Γ 0 Since S τ 1 = τ 1 and S τ 2 = τ 2, we obtain J = q S τ 2 Sv S τ 1 We have = + q τ 1 Sv dσ + Oε q τ u dσ. 2 + q S τ 1 Sv S τ 2 Sq τ 2 v Sq τ 1 v dσ + Oε τ 1 τ 2 dσ + Oε q u dσ. q u dσ τ 1 v = u τ 1 N = u τ 2 and τ 2 v = u τ 2 N = u τ 1. It follows that u τ 1 J = Sq + u τ 2 dσ + Oε q τ 1 τ u dσ 2 Γ 0 u = Sq τ 1 + u τ 2 dσ + Sq u τ 1 + u τ 2 dσ τ 1 τ 2 τ 1 τ 2 + Oε q u dσ Γ 0 u = Sq τ 1 + u τ 2 dσ + Oε q τ 1 τ u dσ. 2 Therefore, J q u dσ + Oε q u dσ C q L 2 u L 2 + ε u L 2. 4.31

Incompressible fluids in thin two-layer domains 21 Using 4.30 to estimate q L 2 and applying the trace inequality A.2 to the remaining terms in 4.31, we have J C 1 ε Φ L 2 u L 2 + ε 2 u L 2 + ε u L 2 + ε ε u L 2 ε ε C Φ L 2 2 u L 2 + ε 1 u L 2 + ε u L 2. Thus, by 2.17, I 2 = 2γ 0 J Cε δ 0 ν Φ L 2 Combining above estimates for I 1 and I 2 with 4.29 gives 2 u L 2 + ε 1 u L 2 + ε u L 2. Φ 2 L I 2 1 + I 2 C ν Φ L 2{ε δ u L 2 + ε δ 1 u L 2} + C ν Φ L 2 ε δ 0 2 u L 2 + ε δ0 1 u L 2 + ε δ0+1 u L 2. Then dividing both sides of this inequality by Φ L 2 0 yields Au + ν u L 2 = Φ L 2 C ν ε δ 0 2 u L 2 + ε δ 0 1 u L 2 + ε δ 1 u L 2, hence we obtain 4.28. Proposition 4.3 above is a counterpart of Corollary 5.3 in [8] and Theorem 2.1 in [7]. However, its proof is more involved due to the interface boundary condition expressed in I 2, which results in the term 2 u L 2 of higher derivatives. See also [4, 15] for similar estimates. The equivalence between two norms Au L 2 and u H 2 then follows. Proposition 4.4. There is 0 < ε min{ε 1, ε 2 } such that if ε 0, ε ] and u D A, then C 1 Au L 2 ν u H 2 C1 + ε δ 0 1+δ/2 Au L 2. 4.32 Proof. Let 0 < ε min{ε 1, ε 2 } and u D A. On one hand, Au L 2 = νp u L 2 ν u L 2 C ν u H 2. On the other hand, it follows from Lemma 3.6 and Proposition 4.3 that u H 2 C u H 1 + 2 u L 2 C u H 1 + C u L 2 + C u H 1 C u H 1 + C u + ν 1 Au L 2 + ν 1 Au L 2 C u H 1 + Cε δ 0 2 u L 2 + ε δ 0 1 u L 2 + ε δ 1 u L 2 + C ν 1 Au L 2 Cε δ 0 u H 2 + C ν 1 Au L 2 + Cε δ 0 1 u H 1 + Cε δ 1 u L 2. Since δ 0 > 0, for ε sufficiently small, absorbing Cε δ 0 u H 2 into the left-hand side, using 4.25 and 4.27 we obtain u H 2 C ν 1 Au L 2 + Cε δ0 1 ε 1 δ/2 ν 1 A 1/2 u L 2 + Cε δ 1 ε 1 δ ν 1 Au L 2 C1 + ε δ 0 1+δ/2 ν 1 Au L 2. Thus the second inequality in 4.32 follows.

22 Luan Thach Hoang Corollary 4.5. If δ 0 = 1, ε 0, ε ] and u D A then and C 1 Au L 2 ν u H 2 C Au L 2 4.33 Au + ν u L 2 Cε Au L 2 + ν 1/2 A 1/2 u L 2 + ν ε δ 1 u L 2. 4.34 Proof. In this case δ 0 1 + δ/2 = 1 δ/2 0, hence 4.33 follows 4.32. By 4.28, Au + ν u L 2 C ν ε 2 u L 2 + u L 2 + ε δ 1 u L 2. Applying 4.33 and 4.24 to estimate the first two norms on the right-hand side, we obtain 4.34. 4.2. Averaging operators. We extend the averaging operators M 0 and M in [8] see also [9, 15] for single layer domains to the two-layer ones. Their properties are stated below c.f. [9] for the proofs. First, we define an averaging operator M 0 on L 2 Ω ε by M 0 = [M 0 +, M0 ] where M ι 0φx = 1 ειg ι x for ι = +, and φ L 2 Ω ε ι. Then hιx h 0 x φx, y 3 dy 3, x = x, x 3 T 2 R, 4.35 M 0 φ = [M + 0 φ +, M 0 φ ] for φ = [φ +, φ ] L 2 Ω ε 3. Note that M 0 φx is independent of x 3. Also, M 0 is an orthogonal projection on L 2 Ω ε, hence M 0 φ 2 L + φ M 0φ 2 2 L = 2 φ 2 L2. 4.36 Define a function ψ = [ψ +, ψ ] = ψ 1, ψ 2 on T 2 R by ψ ι x = ψ ι,1, ψ ι,2 x def == 1 ειg ι x { x 3 h 0 2 h ι x + h ι x x 3 2 h 0 x } 4.37 for ι = +,, and x = x, x 3 = x 1, x 2, x 3 T 2 R, where we recall 2 = / x 1, / x 2. Note that ψ ι Γ0 = 2 h 0, ψ ι Γι = 2 h ι, 3 ψ ι = 1/g ι 2 g ι, ι = +,, 4.38 ψ Cε, 3 ψ C, 2 ψ Cε, 2 ψ C. 4.39 Let φ H 1 Ω ε and j = 1, 2, then Consequently, for m = 0, 1, 2, and φ H m Ω ε, where positive number Cm is independent of ε. j M 0 φ = M 0 j φ + M 0 ψ j 3 φ. 4.40 M 0 φ H m Ω ε Cm φ H m Ω ε, 4.41

Let u H and v = M 0 u 1, M 0 u 2. Then Incompressible fluids in thin two-layer domains 23 2 v = v 1 g 2g, where g = [g +, g ]. 4.42 Next, we define the operator M. For u = u 1, u 2, u 3 L 2 Ω ε 3, let and define v = M 0 u 1, M 0 u 2 Mu = v, v ψ = M 0 u 1, M 0 u 2, ψ 1 M 0 u 1 + ψ 2 M 0 u 2, 4.43 Same as in [9] we have that Mu H provided u H. Let u H 1 Ω ε 3 and v = Mu = v, v 3. Then we have and, consequently, v 3 Cε v, 3 v 3 C v, 2 v 3 Cε v + 2 v in Ω ε 4.44 v C v, v C v + 2 v in Ω ε. 4.45 Combining 4.45 with 4.40 and 4.41, we have for m = 0, 1, 2, and u H m Ω ε that v H m Ω ε Cm u H m Ω ε. 4.46 Applying 2D Ladyzhenskaya inequality A.6 to v, we derive from 4.45 that v L 4 Cε 1/4 u 1/2 L 2 u 1/2 H 1 for u H 1 Ω ε 3, 4.47 v L 4 Cε 1/4 u 1/2 H 1 u 1/2 H 2 for u H 2 Ω ε 3. 4.48 Now, let w = u Mu. The following Poincaré s inequality for w is from Lemma 3.5 of [9]: Lemma 4.6. If ε 0, ε ] and u H 1 Ω ε 3 satisfies the boundary condition 2.14 then w L 2 Cε w L 2. 4.49 Consequently, w L 2 Cε ν 1/2 A 1/2 u L 2, u V. 4.50 Above, inequality 4.50 results from 4.49 and 4.24. Lemma 4.7. Assume δ 0 = 1 and ε 0, ε ]. i Poincaré inequalities. If u D A then w L 2 Cε ν 1 Au L 2 Cε u H 2, 4.51 and, consequently, w L 2 Cε 2 ν 1 Au L 2 Cε 2 u H 2 4.52 w H 1 Cε ν 1 Au L 2 Cε u H 2. 4.53

24 Luan Thach Hoang ii Sobolev inequalities. If u V then If u D A then iii Agmon inequality. If u D A then Proof. i Same as Lemma 4.3 in [8], By Lemma 3.4, 2.16 and 2.17, w L 6 C w H 1 C ν 1 A 1/2 u L 2. 4.54 w L 6 C ν 1 Au L 2 C u H 2. 4.55 w L Cε 1/2 ν 1 Au L 2. 4.56 i w j L 2, 3 w 3 L 2, i w 3 L 2 Cε u H 2, i, j = 1, 2. 3 w j L 2 = 3 u j L 2 C ε u H 2 + ε 1/2 1 ν 1/2 γ ν 1/2 γ u 1/2 L 2 Γ Cε u H 2 + ε 1/2 ε δ/2 ν 1/2 A 1/2 u L 2. Summing up the above estimates for i w j L 2, for i, j = 1, 2, 3, and using 4.33 and 4.27, we obtain w L 2 Cε ν 1 Au L 2 + Cε 1+δ/2 ν 1/2 ε 1 δ/2 ν 1/2 Au L 2 Cε ν 1 Au L 2, hence inequality 4.51 follows. This and 4.49 imply 4.52. ii and iii Now that we have i, the proofs of ii and iii can be proceeded the same way as in Lemmas 4.4 and 4.5 of [8]. 5. ESTIMATE OF A NONLINEAR TERM Regarding the nonlinear term in the NSE, here is the main estimate of the paper. Proposition 5.1. Assume 2/3 δ 1 and δ 0 = 1. 5.1 Given α > 0, there is C α > 0 such that for any ε 0, ε ] and u D A, we have Bu, u, Au = [u u] Au dx Ω ε α Au 2 L + c 3ε 1/2 ν 3/2 2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L + u 2 L A 1/2 2 u 2 L 2. where ε is from Proposition 4.4, and c 3 > 0 is independent of ε, ν, α and u. 5.2

Incompressible fluids in thin two-layer domains 25 Proof. The equality in 5.2 comes from the fact that Bu, u = P [u u] and P Au = Au. We now prove the inequality. The proof follows the key steps in Proposition 6.1 of [8] and uses many calculations there. Let v = Mu and w = u v, also ω = u and Φ = v ω. We first rewrite where hence J 1 = I def == Ω ε u u Au dx = = u ω Au dx = Ω ε = J 1 + J 2 J 3, Ω ε w ω Au dx, J 2 = Ω ε {u u 1 2 u 2 } Au dx Ω ε w ω Au dx + Ω ε Φ Au + ν udx, J 3 = Ω ε v ω Au dx Ω ε Φ ν u dx. Estimate of J 1. By Agmon s inequality 4.56, Hölder s inequality and relation 4.24 J 1 w L ω L 2 Au L 2 Cε 1/2 ν 1 Au L 2 ν 1/2 A 1/2 u L 2 Au L 2, J 1 Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L2. 5.3 Estimate of J 2. First, we apply 4.45 and Cao-Titi inequality A.8 to have hence by Lemma 4.2 and Corollary 4.5, By Hölder s inequality and 4.34, Using 5.4, we have where Φ L 2 Cε 1/2 v 1/2 L v 1/2 2 H ω 1/2 1 L ω 1/2 2 H 1 Cε 1/2 u 1/2 L 2 u H 1 u 1/2 H 2, Φ L 2 Cε 1/2 ν 1 u 1/2 L 2 A 1/2 u L 2 Au 1/2 L 2. 5.4 J 2 Φ L 2 Au + ν u L 2 C Φ L 2ε Au L 2 + C ν 1/2 A 1/2 u L 2 + ε δ 1 ν u L 2. J 2 CJ 2,1 + J 2,2 + J 2,3, J 2,1 = ε 1/2 ν 1 u 1/2 L 2 A 1/2 u L 2 Au 3/2 L 2, J 2,2 = ε 1/2 ν 1/2 u 1/2 L A 1/2 u 2 2 L 2 Au 1/2 L, 2 J 2,3 = ε δ 3/2 u 3/2 L 2 A 1/2 u L 2 Au 1/2 L 2.

26 Luan Thach Hoang Using 4.25 to estimate u L 2 in J 2,1 gives J 2,1 Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. We define J 0 = ε 1/2 ν 1/2 u L 2 A 1/2 u L 2 Au L 2 and use it as an intermediate term aiding other estimates. By Young s inequality, J 0 α Au 2 L 2 + C α ε 1 ν 1 u 2 L 2 A1/2 u 2 L 2 for all α > 0. 5.5 Using interpolation inequality 4.22 to estimate A 1/2 u L 2, but not A 1/2 u 2 L 2, in J 2,2 and then applying 5.5, we have J 2,2 J 0 α Au 2 L 2 + C αε 1 ν 1 u 2 L 2 A1/2 u 2 L 2. For J 2,3, applying Young s inequality yields Using 4.23 to estimate u 2/3 L 2 J 2,3 α Au 2 L 2 + C αε 4δ/3 2 u 2 L 2 A1/2 u 4/3 L 2. in the last product gives J 2,3 α Au 2 L 2 + C αε 4δ/3 2 u 4/3 L 2 ε 1 δ/2 ν 1/2 A 1/2 u L 2 Since δ 2/3, Therefore we obtain = α Au 2 L 2 + C αε δ 5/3 ν 1/3 u 4/3 L 2 A 1/2 u 2 L 2. J 2,3 α Au 2 L 2 + C αε 1 ν 1/3 u 4/3 L 2 A 1/2 u 2 L 2. 2/3 A 1/2 u 4/3 L 2 J 2 2α Au 2 L + 2 Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 5.6 + C α ε 1 ν 1/3 u 4/3 L + ν 1 2 u 2 L A 1/2 u 2 2 L 2. Estimate of J 3. We have from Lemma 3.8 that J 3 = + Ω ε νφ ωdx = Ω ε νφ Gudx 2γ 0 = J 3,1 + J 3,2 + J 3,3 + J 3,4. Ω ε ν Φ ω dx u SΦ dσ Ω ε ν Φ Gudx We estimate J 3,3 first. By Lemma 3.7, Hölder s inequality, estimate 5.4 and Lemma 4.2, J 3,3 C ν Φ L 2ε δ u H 1 + ε δ 1 u L 2 C ν ε 1/2 ν 1 u 1/2 L A 1/2 u 2 L 2 Au 1/2 L ε δ ν 1/2 2 A 1/2 u L 2 + ε δ 1 u L 2 ε δ 1/2 ν 1/2 u 1/2 L A 1/2 u 2 2 L 2 Au 1/2 L + ε δ 3/2 u 3/2 2 L A 1/2 u 2 L 2 Au 1/2 L 2 CJ 2,2 + J 2,3.

Incompressible fluids in thin two-layer domains 27 Using above estimates of J 2,2 and J 2,3 with an adjustment of α and C α to compensate the factor C, we obtain J 3,3 2α Au 2 L + C αε 1 ν 1/3 2 u 4/3 L + ν 1 2 u 2 L A 1/2 u 2 2 L2. 5.7 We estimate J 3,2 next. This is similar to the corresponding estimate in [8]. Note that Φ = ω v v ω + ω v v ω. Since v and ω are divergence-free, we have Φ = ω v v ω C v u + v 2 u. 5.8 It follows from 3.51 and 5.8 that J 3,2 C ν Φ ε δ u dx Cε δ ν v u + v 2 u u dx. Ω ε Ω ε For the last integral, we apply 4.45, Hölder s inequality and A.8 to obtain ε δ ν Ω ε v u u dx Cε δ ν Cε δ ν 2 v + v u L 2 u L 2 Cε δ ν ε 1/2 v 1/2 H 1 v 1/2 H 2 u 1/2 L 2 u 1/2 H 1 u H 1 Cε δ 1/2 ν u 1/2 L 2 u 2 H 1 u 1/2 H 2, and then by Lemma 4.2i and Corollary 4.5, Similarly, ε δ ν ε δ ν Cε δ ν Ω ε 2 v + v u u dx v u u dx Cε δ 1/2 ν 1/2 u 1/2 Ω ε L A 1/2 u 2 2 L 2 Au 1/2 L 2 Ω ε v u 2 u dx CJ 2,2 CJ 0. Ω ε v u 2 u dx Cε δ ν v u L 2 2 u L 2 Cε δ 1/2 ν u L 2 u H 1 u H 2 Cε δ 1/2 ν 1/2 u L 2 A 1/2 u L 2 Au L 2 CJ 0. Therefore, by 5.5, J 3,2 2α Au 2 L + C αε 1 ν 1 2 u 2 L 2 A1/2 u 2 L2. 5.9 We estimate J 3,1 by following calculations for the same term in Proposition 6.1 of [8] with some alterations. We have J 3,1 = Ω ε νω v ω dx Ω ε νv ω ω dx.

28 Luan Thach Hoang It is well-known that the second integral vanishes. Hence J 3,1 = = νω v ω dx Ω ε v v ω dx + Ω ε ν = J 1 3,1 + J 2 3,1. Ω ε ν w v ω dx For J 1 3,1, we have exactly the same as in [8] and then use relations 4.24 and 4.33, Thus, by 5.5, J 1 3,1 Cε 1/2 ν u L 2 u H 1 u H 2 + Cε 1/2 ν u 2 H 1 u H 2 Cε 1/2 ν 1/2 u L 2 A 1/2 u L 2 Au L 2 + Cε 1/2 ν 1 A 1/2 u 2 L 2 Au L 2 CJ 0 + Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. J 1 3,1 α Au 2 L 2 + C αε 1 ν 1 u 2 L 2 A1/2 u 2 L 2 + Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. For J 2 3,1, letting v = v + v 3 e 3, we have J 2 3,1 = ν w v v + v 3 e 3 + wdx + Ω ε hence = ν w 2 v v 3 e 3 dx + Ω ε + w v 3 e 3 ω dx Ω ε ν = K 1 + K 2 + K 3. As in [8], Ω ε ν w v 3 e 3 ω dx Ω ε ν w 2 v w dx K 1 Cε 1/2 ν u H 1 u H 2 w L 2 Cε 1/2 ν u H 1 u 2 H 2, K 1 Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. Applying Hölder s inequality and A.8 yields K 2 C ν 2 v w L 2 w L 2 C ν ε 1/2 v 1/2 H 1 v 1/2 H 2 w 1/2 L 2 w 1/2 H 1 w L 2 Cε 1/2 ν u 1/2 H 1 w 3/2 L 2 u H 2 Cε 1/2 ν u H 1 w L 2 u H 2.

Incompressible fluids in thin two-layer domains 29 Estimating w L 2 by 4.51, and also using 4.24 and 4.33 for the H 1 and H 2 norms, we obtain K 2 Cε 1/2 ν ν 1/2 Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. For K 3, by estimates in 4.44, Using A.8, we estimate ε ν K 3 C ν A 1/2 u L 2ε ν 1 Au L 2 ν 1 Au L 2 Ω ε w ε 2 v + v u dx. Ω ε w 2 v u dx Cε ν ε 1/2 w 1/2 L 2 w 1/2 H 1 v 1/2 H 1 v 1/2 H 2 u H 1 Cε 1/2 ν u 2 H 1 u H 2 Cε1/2 ν u H 1 u 2 H 2 Cε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2. By the 2D Agmon inequality A.7 applied to v, we have ν Combining with 5.5, we obtain Ω ε w v u dx ν v L w L 2 u L 2 C ν ε 1/2 u 1/2 L 2 u 1/2 H 2 u 2 H 1 Cε 1/2 ν 1/2 u 1/2 L 2 A 1/2 u 2 L 2 Au 1/2 L 2 CJ 2,2 CJ 0. K 3 α Au 2 L 2 + Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L 2 A1/2 u 2 L 2. Gathering the estimates of K 1, K 2 and K 3, we have Summing up, J 2 3,1 K 1 + K 2 + K 3 α Au 2 L 2 + Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L 2 A1/2 u 2 L 2. J 3,1 J 1 3,1 + J 2 3,1 2α Au 2 L 2 + Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L 2 A1/2 u 2 L 2. 5.10

30 Luan Thach Hoang Finally, we estimate the boundary term J 3,4. By 2.17, 5.1 and 2.18, we have J 3,4 2γ 0 u Sv Sω dσ Cε ν u v ω dσ Cε ν v u u dσ. For an estimation of ε v u u dσ, we use the calculations in estimating the term I 3 on page 595, from line 7 to line 23, of [9]. We obtain J 3,4 C ν ε 1/2 u L 2 u 2 H + C ν 1 ε u 1/2 L u 3/2 2 H u 1 H 2 Hence, by 4.24 and 4.33, + C ν ε 1/2 u L 2 u H 1 u H 2 + C ν ε 2 u 1/2 L 2 u 1/2 H 1 u 2 H 2 C ν ε 1/2 u L 2 u 2 H 1 + C ν ε1/2 u H 1 u 2 H 2. J 3,4 Cε 1 u L 2 A 1/2 u 2 L + 2 Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L2. 5.11 It follows from estimates 5.10, 5.9, 5.7 and 5.11 that J 3 J 3,1 + J 3,2 + J 3,3 + J 3,4 6α Au 2 L + 2 Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L + 2 ν 1/3 u 4/3 L + u 2 L 2 A 1/2 u 2 L 2. 5.12 Summing up the estimates 5.3, 5.6 and 5.12 for J 1, J 2 and J 3, respectively, we obtain I 8α Au 2 L + 2 Cε1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + C α ε 1 ν 1 u 2 L + 5.13 2 ν 1/3 u 4/3 L + u 2 L 2 A 1/2 u 2L 2. Using Young s inequality to estimate ν 1/3 u 4/3 L 2 by α/8, we obtain the assertion 5.2 from 5.13. ν 1 u 2 L + u 2 L 2, and replacing α 6. GLOBAL SOLUTIONS First, we state the standard local existence theorem for strong solutions of the Navier Stokes equations. Below, we use the conventional notation ut = u, t and ft = f, t. We recall that ε is defined in Proposition 4.4. Theorem 6.1. Assume 5.1. Suppose ε 0, ε ] and u 0 V and f L 0,, L 2 Ω ε 3. Then there exist T > 0 and a unique solution ut C[0, T, V L 2 0, T ; D A of the equation 4.19 on 0, T satisfying u0 = u 0.

Incompressible fluids in thin two-layer domains 31 Furthermore, if the maximal time interval of the above existence is [0, T max and T max is finite, then lim A 1/2 ut L 2 =. 6.1 t Tmax We call the above solution ut a strong solution on [0, T of the IBVP 2.9, 2.10, 2.11, 2.12, 2.13. We have a few observations before stating our global existence result. Let u = u 1, u 2, u 3 L 2 Ω ε 3, we define Mu = M 0 u 1, M 0 u 2, 0. 6.2 One can verify that M is an orthogonal projection on L 2 Ω ε 3, hence For u V, it follows from A.4 and 3.21 that and, consequently, by 4.24 u 2 L = 2 Mu 2 L + I 2 Mu 2 L2. 6.3 I Mu L 2 I Mu 1, u 2 L 2 + u 3 L 2 Cε u H 1, For a function f L 0, ; L 2 Ω ε, denote I Mu L 2 Cε ν 1/2 A 1/2 u L 2. 6.4 Our main result is: f L L2 = ess sup ft L 2. t 0, Theorem 6.2. Assume 5.1. There is a positive number κ > 0 such that if ε 0, ε ] and u 0 V, f L 0,, L 2 Ω ε 3 satisfy that def U 0 == ν 2 Mu 0 2 L 2 κ2 def, U 1 == ε ν 3 A 1/2 u 0 2 L 2 κ2, 6.5 def F 0 == ε 2 2δ ν 4 MP f 2 L L 2 κ2 def, F 1 == ε 2 δ ν 4 I MP f 2 L L 2 κ2 then the strong solution ut in Theorem 6.1 exists for all t 0, i.e., T max =. Moreover, and where ν 2 ut 2 L 2 c Λ 1 e c 2 2 εδ 1 ν t + Λ 2, t [0,, 6.6 ε ν 3 A 1/2 ut 2 L 2 c Λ 3 + Λ 4, t [0, ε 1 δ / ν, 6.7 ε ν 3 A 1/2 ut 2 L 2 c Λ 1 e c 2 2 εδ 1 ν t + Λ 4, t [ε 1 δ / ν,, 6.8 Λ 1 = U 0 + εu 1, Λ 2 = F 0 + εf 1, Λ 3 = U 0 + U 1, Λ 4 = F 0 + F 1, 6.9 the number c 2 > 0 is defined in Lemma 4.2, and c > 0 is independent of ε, ν, κ, u 0, f.

32 Luan Thach Hoang For u V, let E u = Ω ε Du 2 dx + ε δ Still considering δ 0 = 1, we then have or, equivalently, Γ C 1 E u ν 1 Eu CE u, u 2 dσ + ε u Su 2 dσ. 6.10 C 1 E u ν 1 A 1/2 u 2 L 2 CE u. 6.11 Therefore, Theorem 6.2 can be rewritten without using the Stokes operator as: Theorem 6.3. Suppose the assumptions in Theorem 6.2 hold with U 1 in 6.5 being redefined by def U 1 == ε ν 2 E u 0. 6.12 Then one has 6.6 and, in place of 6.7 and 6.8, ε ν 2 E ut c Λ 3 + Λ 4, t [0, ε 1 δ / ν, 6.13 ε ν 2 E ut c Λ 1 e c 2 2 εδ 1 ν t + Λ 4, t [ε 1 δ / ν,. 6.14 Proof of Theorem 6.2. Note from 6.3 and 6.4 that Also, u 0 2 L 2 Mu0 2 L 2 + c2 4ε 2 ν 1 A 1/2 u 0 2 L 2 C ν 2 U 0 + εu 1 = C ν 2 Λ 1. 6.15 P f 2 L 2 = MP f 2 L 2 + I MP f 2 L 2 ν 4 ε2δ 2 F 0 + ε δ 2 F 1 ε δ 2 ν 4 Λ 4. 6.16 Obviously, from 6.9 and 6.5: Set d = 1 16c 2 3 Λ 1, Λ 2, Λ 3, Λ 4 2κ 2. 6.17 and { κ 2 = min 1, d } 2, d, 6.18 2C 0 where c 3 > 0 is from Proposition 5.1 and C 0 > 0 is the constant in Lemma 6.4 below. Let 0 < ε < ε 1. We estimate ut L 2 first. Taking the inner product of equation 4.19 with ut for t 0, T max, and applying inequalities 4.23 and 4.50, we have 1 d 2 dt u 2 L + 2 A1/2 u 2 L2 u, f Mu, MP f + I Mu, I MP f Cε 1 δ/2 ν 1/2 A 1/2 u L 2 MP f L 2 + Cε ν 1/2 A 1/2 u L 2 I MP f L 2 1 2 A1/2 u 2 L + 2 Cε1 δ ν 1 MP f 2 L + 2 Cε2 ν 1 I MP f 2 L 2 1 2 A1/2 u 2 L 2 + Cεδ 1 ν 3 F 0 + εf 1.

Incompressible fluids in thin two-layer domains 33 Hence d dt u 2 L + 2 A1/2 u 2 L 2 Cεδ 1 ν 3 Λ 2. 6.19 Using 4.23 in 6.19 gives Applying Gronwall s inequality, we obtain d dt u 2 L + ν 2 c 2 2ε 1 δ u 2 L 2 Cεδ 1 ν 3 Λ 2. ut 2 L 2 u0 2 L 2e c 2 2 εδ 1 ν t + C ν 2 Λ 2, t [0, T max. Using 6.15 to estimate u 0 2 L 2, we have and consequently, by 6.17 ut 2 L 2 C 1 ν 2 Λ 1e c 2 2 εδ 1 ν t + Λ 2, t [0, T max, 6.20 ut 2 L 2 C 1 ν 2 Λ 1 + Λ 2 4C 1 ν 2 κ2, t [0, T max. 6.21 For any number θ [0, ε 1 δ / ν ] [0, T max and t [0, T max θ, integrating 6.19 from t to t + θ and using 6.20 yield t+θ Hence we have t+θ t t A 1/2 uτ 2 L 2dτ ut 2 L 2 + Cεδ 1 ν 3 t+θ t Λ 2 dτ C ν 2 Λ 1e c 2 2 εδ 1 ν t + Λ 2 + Cε δ 1 ν 3 θλ 2. A 1/2 uτ 2 L 2dτ C 2 ν 2 Λ 1e c 2 2 εδ 1 ν t + Λ 2, t [0, T max θ. 6.22 and, again, by 6.17 t+θ t A 1/2 uτ 2 L 2dτ 4C 2 ν 2 κ2, t [0, T max θ. 6.23 Next, we estimate A 1/2 ut L 2. The initial data satisfies We claim that ε ν 3 A 1/2 u 0 2 L 2 = U 1 κ 2 d 2. ε ν 3 A 1/2 ut 2 L 2 d for all t [0, T max. 6.24 Suppose 6.24 is false. By the continuity of ut in V for t [0, T, there is T 0, T max such that ε ν 3 A 1/2 ut 2 L2 < d for all t < T, and ε ν 3 A 1/2 ut 2 L2 = d. 6.25 We apply Lemma 6.4 below and have from its conclusion 6.27 that ε ν 3 A 1/2 ut 2 L 2 C 0κ 2 d 2 for all t [0, T.

34 Luan Thach Hoang Taking t T, we have ε ν 3 A 1/2 ut 2 L 2 d 2 < d, which contradicts 6.25. Therefore 6.24 must hold true. As a consequence of 6.24, the norm A 1/2 ut L 2 is bounded on [0, T max, which implies T max =, by virtue of Theorem 6.1. The estimate 6.6 follows 6.20. Since 6.29 and 6.32 in Lemma 6.4 now hold with arbitrarily large T > 0, we obtain 6.7 and 6.8, respectively. Lemma 6.4. Let d be defined in 6.18. There is C 0 > 0 such that for any ε 0, ε ], κ 0, 1] and T > 0, if u 0 V is such that U 1, U 2, F 1 and F 2 defined in Theorem 6.2 are less than κ 2, and ut is a strong solution on [0, T satisfying u0 = u 0 and ε ν 3 A 1/2 ut 2 L2 d for all t [0, T, 6.26 then ε ν 3 A 1/2 ut 2 L 2 C 0κ 2 for all t [0, T. 6.27 Proof. For t 0, T, taking the inner product of equation 4.19 with Aut and applying the estimate in Proposition 5.1 with α = 1/4, we have 1 d 2 dt A1/2 u 2 L + 2 Au 2 L2 Bu, u, Au + Au, P f 1 4 Au 2 L 2 + c 3ε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 + Cε 1 ν 1 u 2 L 2 + u L 2 A 1/2 u 2 L 2 + 1 4 Au 2 L 2 + P f 2 L 2, hence d { } dt A1/2 u 2 L + 1 2c 2 3 ε 1/2 ν 3/2 A 1/2 u L 2 Au 2 L 2 Cε 1 ν 1 u 2 L + u 2 L A 1/2 2 u 2 L + 2 P 2 f 2 L 2. By assumption 6.26 and the L 2 -estimate 6.21, it follows that d dt A1/2 u 2 L + { } 1 2c 2 3 d Au 2 L 2 C 3 ε 1 κ 2 + κ ν A 1/2 u 2 L + 2 Cεδ 2 Λ 4, where C 3 and C do not depend on κ. Since 2c 3 d = 1/2 and κ 1, we obtain d dt A1/2 u 2 L 2 + 1 2 Au 2 L 2 2C 3ε 1 ν A 1/2 u 2 L 2 + Cεδ 2 ν 4 Λ 4. 6.28 Consider the case t < min{ε 1 δ / ν, T }. By 6.28 and 4.27, d dt A1/2 u 2 L + ν 2 2c 2 2ε 1 δ A1/2 u 2 L 2 Cε 1 ν A 1/2 u 2 L + 2 Cεδ 2 ν 4 Λ 4.