Lagrangian approach to global well-posedness of the viscous surface wave equations without surface tension
aa r X i v : . [ m a t h . A P ] D ec Lagrangian approach to global well-posedness of the viscoussurface wave equations without surface tension
Guilong Gui ∗ Abstract
In this paper, we revisit the global well-posedness of the classical viscous surface wavesin the absence of surface tension effect with the reference domain being the horizontalinfinite slab, for which the first complete proof was given in Guo-Tice (2013, Analysis andPDE) via a hybrid of Eulerian and Lagrangian schemes. The fluid dynamics are governedby the gravity-driven incompressible Navier-Stokes equations. Even though Lagrangianformulation is most natural to study free boundary value problems for incompressibleflows, few mathematical works for global existence are based on such an approach in theabsence of surface tension effect, due to breakdown of Beale’s transformation. We developa mathematical approach to establish global well-posedness based on the Lagrangianframework by analyzing suitable ”good unknowns” associated with the problem, whichrequires no nonlinear compatibility conditions on the initial data.
Keywords:
Viscous surface waves; Lagrangian coordinates; Global well-posedness
AMS Subject Classification (2010):
We consider in this paper the global existence of time-dependent flows of an viscous incom-pressible fluid in a moving domain Ω( t ) with an upper free surface Σ F ( t ) and a fixed bottomΣ B ∂ t u + ( u · ∇ ) u + ∇ p − ν ∆ u = − g e in Ω( t ) , ∇ · u = 0 in Ω( t ) , ( p I − ν D ( u )) n ( t ) = p atm n ( t ) on Σ F ( t ) , V (Σ F ( t )) = u · n ( t ) on Σ F ( t ) ,u | Σ B = 0 , (1.1)where we denote n ( t ) the outward-pointing unit normal on Σ F ( t ), I the 3 × D u ) ij = ∂ i u j + ∂ j u i the symmetric gradient of u , the constant g > g = 1 without loss of generality), and ν > pI − ν D ( u )) is known as the viscous stress tensor. Equation (1.1) is the conservation ∗ Center for Nonlinear Studies, School of Mathematics, Northwest University, Xi’an 710069, China. Email: [email protected] .
1f momentum, where gravity is the only external force; the second equation in (1.1) meansthe fluid is incompressible; Equation (1.1) means the fluid satisfies the kinetic boundary con-dition on the free boundary Σ F ( t ), where p atm stands for the atmospheric pressure, assumedto be constant. the kinematic boundary condition (1.1) states that the free boundary Σ F ( t )is moving with speed equal to the normal component of the fluid velocity; (1.1) implies thatthe fluid is no-slip, no-penetrated on the fixed bottom boundary. Here the effect of surfacetension is neglected on the free surface.The problem can be equivalently stated as follows. Given an initial domain Ω ⊂ R bounded by a bottom surface Σ B , and a top surface Σ F (0), as well as an initial velocity field u , where the upper boundary does not touch the bottom, we wish to find for each t ∈ [0 , T ]a domain Ω( t ), a velocity field u ( t, · ) and pressure p ( t, · ) on Ω( t ), and a transformation¯ η ( t, · ) : Ω → R so that Ω( t ) = ¯ η ( t, Ω ) , ¯ η ( t, Σ B ) = Σ B ,∂ t ¯ η = u ◦ ¯ η,∂ t u + ( u · ∇ ) u + ∇ p − ν ∆ u = − g e in Ω( t ) , ∇ · u = 0 in Ω( t ) , ( p I − ν D ( u )) n ( t ) = p atm n ( t ) on Σ F ( t ) ,u | Σ B = 0 ,u | t =0 = u , ¯ η | t =0 = x (1.2)For convenience, it is natural to subtract the hydrostatic pressure from p in the usual wayby adjusting the actual pressure p according to e p = p + g x − p atm , and still denote the newpressure e p by p for simplicity, so that after substitution the gravity term in (1.2) and theatmospheric pressure term in (1.2) are eliminated. A gravity term appears in (1.2) .The conditions on the initial domain Ω are as follows: Let the equilibrium domain Ω ⊂ R be the horizontal infinite slabΩ = { x = ( x , x h ) | − b < x < , x h ∈ R } (1.3)with the bottom Σ b = { x = − b } and the top surface Σ = { x = 0 } , where b will be thedepth of the fluid at infinity. We assume that Ω is the image of Ω under a diffeomorphism σ : Ω → Ω , where σ (Σ b ) = Σ B , σ (Σ ) = Σ F (0), σ is of the form σ ( x ) = x + ξ ( x ), ξ ∈ ˚ H , Σ tan ,N − (Ω) (to be defined later) with N ≥ There are two methods used to solve this viscous surface problem: the first one is Lagrangiancoordinates transformation, and the other is the flattening transformation introduced byBeale [6].In Lagrangian coordinates, the geometry of the domain is encoded in the flow map η :(0 , T ) × Ω → Ω( t ) satisfying ∂ t ¯ η ( t, x ) = v ( t, x ) which gives the trajectory of a particlelocated at x ∈ Ω = Ω(0) at t = 0, where v = u ◦ ¯ η is the Lagrangian velocity field in thefixed domain Ω with u being the Eulerian velocity field in the moving domain Ω( t ). TheLagrangian pressure is q = p ◦ ¯ η for p the pressure in Eulerian coordinates.2y using Lagrangian coordinates transformation, Solonnikov [13] proved the local well-posedness of this viscous surface problem in H¨older spaces for the fluid motion in a boundeddomain whose entire boundary is a free surface. For the fluid motion in a horizontal infinitedomain in this paper, Beale [5] obtained the local well-posedness in L -based space-timeSobolev spaces under the assumption with necessary compatibility condition, which wasextended to L p -based space-time Sobolev spaces by Abels [1]. The local well-posedness in [5]showed that, given v = u ∈ H r − (Ω ) for r ∈ (3 , / , T ), so that u ∈ K r ((0 , T ) × Ω ), where K r ((0 , T ) × Ω ) := H ((0 , T ); H r (Ω )) ∩ H r/ ((0 , T ); H (Ω )) . (1.4)It also showed [5] that, if the initial domain Ω is the image of the equilibrium domain Ωgiven by (1.3) under a diffeomorphism σ : Ω → Ω with σ ( x ) = x + ξ ( x ), then, for any fixed0 < T < ∞ , there exists a collection of sufficiently small data so that a unique solution existson (0 , T ) and so that the solutions depend analytically on the data, where the flow map ismodified by η = Id + ξ def = ¯ η ◦ ¯ σ . This result suggests that solutions should exist globallyin time for small data. If global solutions do exist, it is natural to expect that the solutionsshould approach equilibrium as t → + ∞ , and therefore ξ ( t ) should have a limiting value suchthat ξ ( ∞ ) = 0 on Σ . (1.5)However, Beale gave the negative answer to the global well-posedness in [5]. As a mat-ter of fact, he proved a non-decay theorem which showed that a reasonable extension tosmall-data global well-posedness with decay of the free surface fails. More precisely, for r ∈ (3 , / K r ( R + , Ω) the space of ( v, q ) such that v ∈ K r ( R + , Ω), ξ ∈ K r − ( R + , Σ ), ∇ q ∈ K r − ( R + , Ω), and also v ∈ L ( R + ; H r (Ω)), there exists certain θ ∈ H r (Ω) so that therecannot exist a curve ( v ( ǫ ) , q ( ǫ )) ∈ K r ( R + , Ω) , defined for ǫ near 0, such that the system (1.1)hold with ξ ( ǫ ) = ǫθ, v = 0, ξ ( t = + ∞ ) = 0 on Σ holds for each ǫ , and v ( ǫ ) is of the form v ( ǫ ) = ǫv (1) + ǫ v (2) + O ( ǫ ). Let us mention that the extra condition v ∈ L ( R + ; H r (Ω)) in( v, q ) ∈ K r ( R + , Ω) is included so that ξ ( ∞ ) is meaningful. In the discussion of this result,Beale pointed out that it does not imply the non-existence of global-in-time solutions, inparticular by using Lagrangian approach, but rather that establishing global-in-time resultsrequires weaker or different hypotheses than those imposed in the non-decay theorem.In order to get the global well-posedness of the viscous surface problem, Beale [6] turnedto introduce another method—the flattening transformation to successful solve the viscoussurface problem with surface tension in the space-time Sobolev spaces. Such a viscous surfaceproblem (without surface tension) was studied by Sylvester [14] and Tani-Tanaka [16] inthe Beale-Solonnikov functional framework, where higher regularity and more compatibilityconditions were required.Recently, by using the flattening transformation, Guo and Tice [9, 10] employed the geo-metric structure in the Eulerian coordinates to study the local well-posedness of the system(1.1) in Sobolev spaces for small initial data, and then introduced the two-tiered energymethod to get the algebraic decay rate of the solutions, which leads to the construction ofglobal-in-time solutions to the surface wave problem (1.1). We mention that, in this newframework, the two-tiered energy method couples the decay of low-order energy and the3oundedness of high-order energy, where higher regularity and more compatibility condi-tions, as well as low horizontal regularity of the solutions were needed, which also avoids tocontradict Beale’s non-decay theorem. Wang, Tice, and Kim [17] also considered the globalwell-posedness and decay for two layers compressible fluid by using this two-tiered energymethod. Wu [18] extended their local well-posedness result from small data to general data.More recently, Ren, Xiang and Zhang [15] proved the local well-posedness of the viscous sur-face wave equation in low regularity Sobolev spaces, where only the first-order time-derivativeof the velocity is involved in the compatibility condition, but the global existence of the so-lution is left open.Beale and Nishida studied the decay properties of solutions to viscous surface waves withsurface tension in [7], and then Hataya [11] gave a complete proof to their decay estimates: if ξ ∈ L (Ω) and initial data is small, the global smooth solution to the viscous surface waveswith surface tension has the optimal (upper) decay rate k u ( t ) k H . (1 + t ) − , which meansthe decay of k u ( t ) k H is not fast enough to guarantee that u ∈ L ([0 , ∞ ); H (Ω)). This resultalso suggests that, for the solution u to the viscous surface waves without surface tension,its norm k u k L ([0 , ∞ ); H (Ω)) may still blow-up. In effect, with the additional low horizontalregularity assumption on the small initial data, Guo and Tice [10] got the global smoothsolution with the decay rate k u k C (Ω) . (1 + t ) − − ǫ for some positive constant ǫ . From this,the norm k u k L ([0 , ∞ ); C (Ω)) does not blow up, but the quantity k u k L ([0 , ∞ ); H (Ω)) may. Motivated by Beale’s remark on the non-decay theorem in [5], it is of interest to know whetherthe Lagrangian approach works on the study of the global well-posedness of the viscous surfacewaves without surface tension. So far as I know, even though Lagrangian formulation is quitedirect approach to study free boundary value problems for incompressible flows, there arefew mathematical works for global existence based on such an approach. On the other hand,in previous works, the global well-posedness was established for the initial data which hashigh normal regularity and some compatibility conditions in terms of the time-derivatives ofthe velocity on the initial data are needed. Usually, it is difficult to verify the valid of thecompatibility condition in terms of the time-derivatives of the velocity, which is essentiallysome type of the nonlinear compatibility conditions from the momentum equation in (1.1).Suppose we considered the Navier-Stokes equations in the fixed domain Ω with the condition u = 0 on ∂ Ω, for small initial velocity u , Heywood [12] investigated the unique, globalsolvability to this system without any compatibility conditions in terms of ∂ t u on the initialdata. In the present case, a natural and important question is whether a corresponding well-posedness result can be obtained with low normal regularity and without any compatibilityconditions in terms of the time-derivatives of the velocity (nonlinear compatibility conditions)on the initial data.The aim of this paper is to employ the Lagrangian approach to study the global well-posedness of the viscous surface waves under the assumptions of low normal regularity andno any compatibility conditions in terms of the time-derivatives of the velocity on the initialdata.Let us explain our main ideas to get the global well-posedness of the viscous surface waveswithout surface tension.As mentioned above, in the proof of Beale’s non-decay theorem [5], to make η ( ∞ ) | Σ = 04eaningful, Beale introduced the space-time Sobolev space K r ( R + , Ω) in which the additionalcondition v ∈ L ( R + ; H (Ω)) is required, in particular, the quantity k v ( t ) k L (Ω) should beglobally integral with respect to the time t ∈ R + . While inspired by the optimal decay rate k v ( t ) k H . (1 + t ) − for the solution u to the the viscous surface waves with surface tensionunder the assumption η ∈ L (Ω), it is natural to expect that the solution v to the the viscoussurface waves without surface tension also satisfies the optimal decay rate k v ( t ) k L . (1+ t ) − if η ∈ L (Ω), which means that the condition v ∈ L ( R + ; L (Ω)) will be not satisfied in anycase. Therefore, the condition v ∈ L ( R + ; L (Ω)) is not essential to get Beale’s non-decaytheorem. In effect, we will give a modified non-decay theorem (Theorem 3.2 below) where theconditions ˙Λ σ h v ∈ L ([0 , + ∞ ); H , (Ω)) and ˙Λ h v ∈ L ([0 , + ∞ ); H , (Ω)) for some positive σ ∈ (0 ,
1) are required, which makes η ( ∞ ) | Σ = 0 meaningful in (1.5), while η ( ∞ ) = 0 maynot. Due to the Sobolev embedding theorem, the condition ˙Λ σ h v ∈ L ([0 , + ∞ ); H , (Ω))ensures that k∇ v ( t ) k L ∞ (Ω) belongs to L ( R + t ), from this, it may guarantee the propagationof horizontal regularities of the velocity as well as its smallness, which plays a key role in thestudy of the global well-posedness of the the viscous surface waves. Proceed analogously to theproof of the optimal decay k v ( t ) k L . (1 + t ) − whence η ∈ L (Ω) in [7, 11], we may expectto get that k ˙Λ σ h v k L (Ω)) . (1 + t ) − ( σ + λ ) if ˙Λ − λh η ∈ L (Ω) with λ ∈ (0 , − λh η ∈ L (Ω) with λ > − σ ∈ (0 , σ h v ∈ L ([0 , + ∞ ); H , (Ω)), which also impliesthat the initial condition ˙Λ − λh η ∈ L (Ω) contradicts the choice of the initial data in Beale’snon-decay theorem. With low horizontal regularity assumptions on the initial data, we willemploy Guo-Tice’s two-tiered energy method to study the decay estimates of the solution.Without any compatibility conditions in terms of the time-derivatives of the velocity on theinitial data, we should avoid using the energy in terms of ∂ t v and its derivatives. In this case,to close the nonlinear energy estimates, we will first introduce four good unknowns replacing ∇ v and the pressure q , which plays a crucial role in the proof of our main theorem about theglobal well-posedness of the system (1.1). The rest of the paper is organized as follows. Section 2 gives some equivalent formulations ofthe viscous surface waves (1.1) in Lagrangian coordinates, and introduces four new variablesaccording to the kinetic boundary condition (1.1) in (1.1). In Section 3, we define theenergy and the dissipation, and state the main result of our paper. We first recall, in Section4, some necessary estimates, and then derive the basic estimates in terms of the flow map.Then, in Section 5, we apply the four new good unknowns introduced first in the paper to getenergy estimates of the horizontal derivatives of the velocity. Second normal derivatives of thevelocity as well as their more horizontal regularities are showed by using the Stokes estimatesin Section 6. As a consequence, the total energy estimates are stated in Section 7. Finally,in Section 8, we complete the proof of our main theorem about the global well-posedness ofthe system (1.2). In Appendix, we give the proof of the modified non-decay theorem. Let us end this introduction by some notations that will be used in all that follows.For operators
A, B, we denote [ A ; B ] = AB − BA to be the commutator of A and B.
5e denote 1 + t by h t i . For a . b , we mean that there is a uniform constant C, which maybe different on different lines, such that a ≤ Cb . The notation a ∼ b means both a . b and b . a . Throughout the paper, the subscript notation for vectors and tensors as well asthe Einstein summation convention has been adopted unless otherwise specified, Einsteinssummation convention means that repeated Latin indices i, j, k , etc., are summed from 1 to3, and repeated Greek indices α, β, γ , etc., are summed from 2 to 3.We introduce the operator P ( ∂ h ) to denote the horizontal derivatives of some functions,here P ( ∂ h ) is a pseudo-differential operator with the symbol P ( ς h ) depending only on thehorizontal frequency ς h = ( ς , ς ) T , that is, P ( ∂ h ) f ( x , x h ) := F − ς h → x h (cid:18) P ( ς h ) F x h → ς h ( f ( x , x h )) (cid:19) . (1.6)In particular, we denote ˙Λ sh (or Λ sh ) the homogeneous (or nonhomogeneous) horizontal differ-ential operator with the symbol | ς h | s (or h ς h i s ) respectively, where s ∈ R , h ς h i def = (1+ | ς h | ) / . The quite direct approach used to solve the system (1.1) is Lagrangian. Let us now introducethe Lagrangian coordinates in which the free boundary becomes fixed.
1. The flow map :Let η def = ¯ η ◦ ¯ σ be a position of the fluid particle x in the equilibrium domain Ω at time t so that ( ddt η ( t, x ) = u ( t, η ( t, x )) , t > , x ∈ Ω ,η | t =0 = x + ξ ( x ) , x ∈ Ω , (2.1)then the displacement ξ ( t, x ) def = η ( t, x ) − x satisfies ( ddt ξ ( t, x ) = u ( t, x + ξ ( t, x )) ,ξ | t =0 = ξ . (2.2)We define Lagrangian quantities the velocity v and the pressure q in fluid as (where x =( x , x , x ) T ∈ Ω): v ( t, x ) def = u ( t, η ( t, x )) , q ( t, x ) def = p ( t, η ( t, x )) . Denote the Jacobian of the flow map η by J def = det( Dη ) , where Dη = ∂ ξ ∂ ξ ∂ ξ ∂ ξ ∂ ξ ∂ ξ ∂ ξ ∂ ξ ∂ ξ . Define A def = ( Dη ) − T , then according to definitions of the flow map η and the displacement ξ , we may get the identities: A ki ∂ k η j = A jk ∂ i η k = δ ji , ∂ k ( J A ki ) = 0 , ∂ i η j = δ ji + ∂ i ξ j , A ji = δ ji − A ki ∂ k ξ j . (2.3)6et a ij def = J A ji , the explicit form of the entries of the matrix J A can be found from thedefinition of A that a = (1 + ∂ ξ )(1 + ∂ ξ ) − ∂ ξ ∂ ξ , a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) ,a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) , a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) ,a = (1 + ∂ ξ )(1 + ∂ ξ ) − ∂ ξ ∂ ξ , a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) ,a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) , a = − ( ∂ ξ + ∂ ξ ∂ ξ − ∂ ξ ∂ ξ ) ,a = (1 + ∂ ξ )(1 + ∂ ξ ) − ∂ ξ ∂ ξ . (2.4)If the displacement ξ is sufficiently small in an appropriate Sobolev space, then the flowmapping η is a diffeomorphism from Ω to Ω( t ), which makes us to switch back and forth fromLagrangian to Eulerian coordinates.Simple computation implies that J = 1 + ∇ · ξ + B + B with B := ∂ ξ ∇ h · ξ h − ∂ ξ h · ∇ h ξ + ∇ ⊥ h ξ · ∇ h ξ , B := ∂ ξ ∇ ⊥ h ξ · ∇ h ξ + ∂ ξ ∇ ⊥ h ξ · ∇ h ξ + ∂ ξ ∇ ⊥ h ξ · ∇ h ξ with ∇ h def = ( ∂ x , ∂ x ) T , ∇ ⊥ h def = ( − ∂ x , ∂ x ) T and then ∇ ⊥ h f · ∇ h g = −∇ ⊥ h g · ∇ h f .
2. Derivatives of J and A in Lagrangian coordinates Next, we give some useful equations which we often use in what follows.Since A ( Dη ) T = I , differentiating it with respect to t and x once yields ∂ t A ji = −A jk A mi ∂ m v k , ∂ s A ji = −A jk A mi ∂ m ∂ s ξ k , (2.5)where we used the fact ∂ t η = v in the first equation in (2.5). Whence differentiate theJacobian determinant J , we get ∂ t J = J A ji ∂ j v i , ∂ k J = J A ji ∂ j ∂ k ξ i . (2.6)Moreover, it is easy to verify the following Piola identity: ∂ j ( J A ji ) = 0 ∀ i = 1 , , . (2.7)Here and in what follows, the subscript notation for vectors and tensors as well as the Einsteinsummation convention has been adopted unless otherwise specified.
3. Navier-Stokes equations in Lagrangian coordinates
Under Lagrangian coordinates, we may introduce the differential operators with theiractions given by ( ∇ A f ) i = A ji ∂ j f , D A ( v ) = ∇ A v + ( ∇ A v ) T , ∆ A f = ∇ A · ∇ A f , so theLagrangian version of the system (1.1) can be written on the fixed reference domain Ω as ∂ t ξ = v in Ω ,∂ t v + ∇ A q − ν ∇ A · D A ( v ) = 0 in Ω , ∇ A · v = 0 in Ω , ( q − ξ ) N − ν D A ( v ) N = 0 on Σ ,v | Σ b = 0 ,ξ | t =0 = ξ , v | t =0 = v , (2.8)7here n = (1 , , T is the outward-pointing unit normal vector on the interface Σ , N := J A n stands for the outward-pointing normal vector on the moving interface Σ F ( t ). In the fluid, under the assumption J = 0, the incompressibility condition ∇ A · v = 0 isequivalent to the equation ∇ J A · v = 0, which implies that the incompressibility condition ∇ J A · v = 0 implies a ∂ v + a ∂ v + a ∂ v = −∇ h · v h + B α ,i ∂ α v i with B α ,i ∂ α v i def = − a β ∂ β v − ( a − ∂ v − a ∂ v − a ∂ v − ( a − ∂ v . (2.9)Hence, the incompressibility condition ∇ A · v = 0 is equivalent to the linearized form of thedivergence-free condition that ∇ · v = − g a α ∂ v α + B α ,i ∂ α v i (2.10)with f a := a − a , f a := a − a , B α ,i ∂ α v i def = − a − a β ∂ β v − a − ( a − a ) ∂ v − a − a ∂ v − a − a ∂ v − a − ( a − a ) ∂ v . In this section, we want to introduce four new good unknowns related to the vertical deriva-tives ∂ v and the pressure q , which plays a significant role for the proof of the main result ofthe paper.Notice that N = J A n is the outer normal vector field on the interface Σ , N = J A n = a j e j = a e + a e + a e . Let τ and τ be two independent tangential vector fields on Σ τ = − a e + a e , τ = − a e + a e . It is easy to see that
N · τ α = 0 with α = 2 , (cid:0) D J A ( v ) N (cid:1) i = ( D J A ( v )) ij N j = ( a ik ∂ k v j + a jk ∂ k v i ) a j with i, j, k = 1 , ,
3, weseparate all the components of D J A ( v ) N into two types of the terms: the one is of thevertical derivative of v , and the other is of the horizontal derivative of v (cid:0) D J A ( v ) N (cid:1) = ( a + |−→ a | ) ∂ v + a a α ∂ v α + B α ,i ∂ α v i , (cid:0) D J A ( v ) N (cid:1) = a a ∂ v + ( |−→ a | + a ) ∂ v + a a ∂ v + ∂ v + B α ,i ∂ α v i , (cid:0) D J A ( v ) N (cid:1) = a a ∂ v + a a ∂ v + ( |−→ a | + a ) ∂ v + ∂ v + B α ,i ∂ α v i , (2.11)8here −→ a := ( a , a , a ) T , |−→ a | := ( P i =1 a i ) / , and B α ,i ∂ α v i def = a iβ a i ∂ β v + a β a i ∂ β v i , B α ,i ∂ α v i def = ( a a − ∂ v + a a ∂ v + a iβ a i ∂ β v + a β a γ ∂ β v γ , B α ,i ∂ α v i def = a a ∂ v + a β a γ ∂ β v γ + a iβ a i ∂ β v + ( a a − ∂ v . Hence, multiplying the boundary equation on Σ in (2.8) by the tangential vectors τ α (with α = 2 ,
3) implies0 = τ · ( D J A ( v ) N ) = − a |−→ a | ∂ v + a |−→ a | ∂ v + a ∂ v + ( − a B α ,i + a B α ,i ) ∂ α v i , and 0 = τ · ( D J A ( v ) N ) = − a |−→ a | ∂ v + a |−→ a | ∂ v + a ∂ v + ( − a B α ,i + a B α ,i ) ∂ α v i , which along with the incompressibility condition (2.10) gives rise to − a ∂ v + a ∂ v = − a |−→ a | − ∂ v + ( a B α ,i − a B α ,i ) |−→ a | − ∂ α v i , − a ∂ v + a ∂ v = − a |−→ a | − ∂ v + ( a B α ,i − a B α ,i ) |−→ a | − ∂ α v i ,a ∂ v + a ∂ v + a ∂ v = −∇ h · v h + B α ,i ∂ α v i . (2.12)Solving (2.12) in terms of the vertical derivative of v , we conclude that on the interface Σ ∂ v = −∇ h · v h + B α ,i ∂ α v i , ∂ v = − ∂ v + B α ,i ∂ α v i , ∂ v = − ∂ v + B α ,i ∂ α v i , (2.13)where B α ,i ∂ α v i def = − f a ∂ v − f a ∂ v + ( f a B α ,i + f a B α ,i + B α ,i ) ∂ α v i , B α ,i ∂ α v i def = a − |−→ a | − (cid:20) − a ( a + a − |−→ a | ) ∂ v + ( a + a ) (cid:18) ( a B α ,i − a B α ,i ) ∂ α v i + |−→ a | a ( −∇ h · v h + B α ,i ∂ α v i ) (cid:19) − a a (cid:18) − a ∂ v + ( a B α ,i − a B α ,i ) ∂ α v i + |−→ a | a ( −∇ h · v h + B α ,i ∂ α v i ) (cid:19)(cid:21) , B α ,i ∂ α v i def = a − |−→ a | − (cid:20) − a ( a + a − |−→ a | ) ∂ v + ( a + a ) (cid:18) ( a B α ,i − a B α ,i ) ∂ α v i + |−→ a | a ( −∇ h · v h + B α ,i ∂ α v i ) (cid:19) − a a (cid:18) − a ∂ v + a ( a B α ,i − a B α ,i ) ∂ α v i + |−→ a | a ( −∇ h · v h + B α ,i ∂ α v i ) (cid:19)(cid:21) .
9n the other hand, taking the dot product between the boundary condition on Σ in (2.8)and N , it immediately follows( q − ξ ) N · N − ν N · ( D A ( v ) N ) = 0 , which, together with N · N = |−→ a | , (2.2), (2.11), (2.12), and N · (cid:0) D J A ( v ) N (cid:1) = 2 |−→ a | a i ∂ v i + a β ∂ β v + a j B αj,i ∂ α v i = 2 |−→ a | ( −∇ h · v h + B α ,i ∂ α v i ) + a β ∂ β v + a j B αj,i ∂ α v i , implies J ( q − ξ ) − ν (2( −∇ h · v h + B α ,i ∂ α v i ) + |−→ a | − ( a β ∂ β v + a j B αj,i ∂ α v i )) = 0 on Σ . We thus conclude that q = ξ − ν ∇ h · v h + ν B α ,i ∂ α v i on Σ , (2.14)where the nonlinear term B α ,i ∂ α v i def = − J − − ∇ h · v h + 2 J − B α ,i ∂ α v i + J − |−→ a | − ( a β ∂ β v + a j B αj,i ∂ α v i ) . Combining (2.13) with (2.14), the boundary condition on Σ in (2.8) can be written as thelinearized form q e − ν D ( v ) e = ξ + ν ( B α ,i − B α ,i ) ∂ α v i − ν B α ,i ∂ α v i − ν B α ,i ∂ α v i . (2.15)In order to extend the interface boundary forms of ∂ v and q in (2.13) and (2.14) to theinterior domain of the fluid, let us first introduce H ( f ) as the harmonic extension of f | Σ intoΩ: ( ∆ H ( f ) = 0 in Ω , H ( f ) | Σ = f | Σ , H ( f ) | Σ b = 0 . We state the classical result on such an operator, which proof is left to the reader.
Lemma 2.1. kH ( f ) k H (Ω) . k f k H (Σ ) , kH ( f ) k H r (Ω) . k f k H r − (Σ ) ( r ≥ , k ∂ kh H ( f ) k H (Ω) . k ∂ kh f k H (Σ ) , k ∂ kh H ( f ) k H (Ω) . k ∂ kh f k H (Σ ) ( k ∈ N ) . Based on (2.13) and (2.14), we introduce four new good variables in the fluid G def = ∂ v + ∇ h · v h − H ( B α ,i ) ∂ α v i , G def = ∂ v + ∂ v − H ( B α ,i ) ∂ α v i , G def = ∂ v + ∂ v − H ( B α ,i ) ∂ α v i , Q def = q − H ( ξ ) + 2 ν ∇ h · v h − ν H ( B α ,i ) ∂ α v i in Ω . (2.16)10t is easy to obtain from (2.13) and (2.14) that G = 0 , G = 0 , G = 0 , Q = 0 on Σ , (2.17)and ∂ v and q satisfy ∂ v = G − ∇ h · v h + H ( B α ,i ) ∂ α v i ,∂ v = G − ∂ v + H ( B α ,i ) ∂ α v i ,∂ v = G z − ∂ v + H ( B α ,i ) ∂ α v i ,q = Q + H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i in Ω . (2.18)For convenience, we briefly write q = Q + e Q in Ωwith e Q def = H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i . These four new unknowns G i with i = 1 , , Q play an important role in the estimateof the L ∞ t L norms of ∇ v as well as its tangential derivatives, as it is possible to close all theestimates by using integration in parts and the relations (2.18) with (2.17) when we encounterthe integrals including the vertical derivative of ∂ v and q . More details will be showed inthe proof of Lemma 5.8. This makes it possible to avoid using the compatibility conditionsin terms of the acceleration ∂ t v and its derivatives. Let’s now derive the linearized form of the system (2.8) in the Lagrangian coordinates, whichis crucially used to recover the bounds for high order derivatives.Under the Lagrangian coordinates, we first divide the dissipative term ∇ A · D A ( v ) intolinear and nonlinear parts ( ∇ A · D A ( v )) k = ( ∇ · D ( v )) k + g kk (2.19)with k = 1 , ,
3, where nonlinear parts g kk have the forms g := [ f , + ( A ) − ∂ v + 2( A β A αβ + 2 A A α ) ∂ α ∂ v + ( A α A ) ∂ v α + B α ,β ∂ α ∂ v β + B αβ ,i ∂ α ∂ β v i + ( f ,j + A i ∂ i A j ) ∂ j v + A iα ∂ i A ∂ v α + A iβ ∂ i A α ∂ α v β ,g := ( A A ) ∂ v + A i ∂ i A ∂ v + [ f , + ( A ) ] ∂ v + ( A A ) ∂ v + f , ∂ v + A iβ ∂ i A ∂ v β + B α ∂ α ∂ v + B αβ ,i ∂ α ∂ β v i + B αβ ,i ∂ α ∂ β v i + B α ,β ∂ α ∂ v β ,g := A i ∂ i A ∂ v + ( A A ) ∂ v + B αβ ,i ∂ α ∂ β v i + B α ,β ∂ α ∂ v β + B α ∂ α ∂ v + B α ∂ α v i + B ,α ∂ v α + B ,α ∂ v α , B α ,β ∂ α ∂ v β := ( A A + A A ) ∂ ∂ v + ( A A + A A ) ∂ ∂ v + ( A A + A A − ∂ ∂ v + ( A A + A A − ∂ ∂ v ,B αβ ,i ∂ α ∂ β v i := ( A βi A α ) ∂ β ∂ α v i + 2( A i A i ) ∂ ∂ v + X α =2 f ,αα ∂ α v ,B α ∂ α ∂ v := ( A A + A A − ∂ ∂ v + ( A A + A A ) ∂ ∂ v ,B αβ ,i ∂ α ∂ β v i := X α =2 f ,αα ∂ α v + (2 A i A i + A A ) ∂ ∂ v + ( A A − ∂ v + ( A A − ∂ ∂ v + ( A A ) ∂ v + ( A β A ) ∂ β ∂ v + ( A βi A ) ∂ β ∂ v i ,B αβ ,i ∂ α ∂ β v i := A k ∂ k A α ∂ α v + f ,α ∂ α v + A kβ ∂ k A α ∂ α v β ,B α ,β ∂ α ∂ v β := 2 A i A αi ∂ α ∂ v + ( A β A α + A αβ A ) ∂ α ∂ v β ,B αβ ,i ∂ α ∂ β v i := X α =2 f ,αα ∂ α v + ( A β A α ) ∂ β ∂ α v + ( A A ) ∂ v + ( A A ) ∂ v + ( A A + A A − ∂ ∂ v + ( A A ) ∂ v + ( A A − ∂ v + 2( A i A i + A A ) ∂ ∂ v ,B α ,β ∂ α ∂ v β := 2( A i A αi + A α A ) ∂ α ∂ v + ( A A α + A α A ) ∂ α ∂ v ,B α ∂ α ∂ v := ( A A + A A ) ∂ ∂ v + ( A A + A A − ∂ ∂ v ,B α ∂ α v i := f ,α ∂ α v + A k ∂ k A α ∂ α v + A kβ ∂ k A α ∂ α v β ,B ,α ∂ v α := ( A A ) ∂ v + [ f , + ( A ) ] ∂ v , B ,α ∂ v α := f , ∂ v + A kβ ∂ k A ∂ v β , and f , := P i =1 ( A i ) − f ,αα := P i =1 ( A αi ) − f ,j := A ki ∂ k A ji .Combining (2.10), (2.15), (2.19), the system (2.8) can be written as the linearized form ∂ t ξ = v,∂ t v + ∇ q − ν ∇ · D ( v ) = g, ∇ · v = − g a α ∂ v α + B α ,i ∂ α v i in Ω ,q e − ν D ( v ) e = ξ + ν ( B α ,i − B α ,i ) ∂ α v i − ν B α ,i ∂ α v i − ν B α ,i ∂ α v i on Σ ,v | Σ b = 0 . (2.20)where g = ( g , g , g ) T with g := −A h ∂ h q − ( A − ∂ q + ν g ,g := −A ∂ q − ( A − ∂ q − A ∂ q + ν g ,g := −A ∂ q − ( A − ∂ q − A ∂ q + ν g . (2.21)Using the incompressibility condition (2.10), one has ∂ k ∇ · v = e g kk with k = 1 , ,
3, and e g kk = − f a ∂ k ∂ v − f a ∂ k ∂ v + B α ,i ∂ k ∂ α v i − ∂ k f a ∂ v − ∂ k f a ∂ v + ∂ k B α ,i ∂ α v i .
12n particular, for ∂ v , there holds ∂ v = − ∂ ∇ h · v h + e g . (2.22)From this, the momentum equations in (2.20) are equivalent to the equations ∂ t v + ∂ q − ν ∆ v = g + ν e g ,∂ t v + ∂ q − ν ∆ v = g + ν e g ,∂ t v + ∂ q − ν ∆ v = g + ν e g , which follows ( ∂ q = − ∂ t v + ν (∆ h v − ∂ ∇ h · v h ) + g + 2 ν e g , − ν∂ v β + ∂ β q = − ∂ t v β + ν ∆ h v β + g β + ν e g ββ for β = 2 , . (2.23)Thanks to (2.22) and (2.23), it is easy to get, for any smooth sub-channel domain Ω ⊆ Ω, kP ( ∂ h ) ∂ v k L (Ω ) . kP ( ∂ h ) ∂ h ∇ v k L (Ω ) + kP ( ∂ h ) e g k L (Ω ) , kP ( ∂ h ) ∂ q k L (Ω ) . kP ( ∂ h ) ∂ t v k L (Ω ) + kP ( ∂ h ) ∂ h ∇ v k L (Ω ) + kP ( ∂ h ) g k L (Ω ) + kP ( ∂ h ) e g k L (Ω ) , kP ( ∂ h )( − ν∂ v β + ∂ β q ) k L (Ω ) . kP ( ∂ h ) ∂ t v β k L (Ω ) + kP ( ∂ h )∆ h v β k L (Ω ) + kP ( ∂ h ) g β k L (Ω ) + kP ( ∂ h ) e g ββ k L (Ω ) for β = 2 , . (2.24) Let us explain that, at the beginning of this section, how to define the energy and dissipationfunctions. Usually, in order to get the global well-posedness of the nonlinear system, welook at it as a perturbation of its linearized equation, which relies on the global control ofthe perturbed nonlinear terms. In our case, it depends on t the uniformly global (in time)estimates of the L ∞ (Ω)-norms of A − I (or equivalently a ij − δ ji ) as well as its derivatives,and at least, it requires that kA − I k L ∞ (Ω) <
1. Take the (1 , A of A − I as anexample, its mainly linear part is ∂ ξ . Since there is no information of ∂ ξ in the energybased on the momentum equations, we turn to use the flow map equation ∂ t ξ = v to get ∂ ξ ( t ) = ∂ ξ (0) + R t ∂ v ( τ ) dτ . Hence, in order to get the control of k ∂ ξ ( t ) k L ∞ (Ω) uniformly in terms of time t , it requires the global L t integrability of k ∂ v ( t ) k L ∞ (Ω) on ∈ [0 , + ∞ ). Because of the Sobolev embedding theorem, k ∂ v ( t ) k L ∞ ≤ C k ˙Λ σ h Λ h ∂ v k H ,we turn to find the global L t integrability of k ˙Λ σ h Λ h ∂ v k H on ∈ [0 , + ∞ ), which dependson the decay (enough) of k ˙Λ σ h Λ h ∂ v k L according to the energy estimates arising from themomentum equations. We first define the decay instantaneous energy and dissipation ˙ E N = ˙ E N ( ξ, v ) and ˙ D N =˙ D N ( ξ, v ) with N ≥ E N includes three parts:˙ E N def = ˙ E ℓ,N + ˙ E ℓ,σ + ˙ E ℓ, σ (for N ≥ , E ℓ,σ def = k ˙Λ σh v k L (Ω) + k ˙Λ σh ξ k L (Σ ) (for σ ∈ ( − , E ℓ, def = k ∂ h v k L (Ω) + k ∂ h ξ k L (Σ ) , ˙ E ℓ, σ def = k ˙Λ σ h ∇ v k L (Ω) , ˙ E ℓ,N def = N − X i =1 ( k ∂ ih v k H (Ω) + k ∂ ih ξ k H (Σ ) ) (for N ≥ , and the decay instantaneous dissipation ˙ D N (with N ≥
2) is defined as˙ D def = ˙ D ℓ,σ + ˙ D ℓ, σ + ˙ D ℓ, , ˙ D N def = ˙ D + ˙ D ℓ,N + ¨ D ℓ, σ + ¨ D ℓ,N (for N ≥ , where˙ D ℓ,σ := k ˙Λ σ h ∇ v k L (Ω) , ˙ D ℓ, σ := k ˙Λ σ h ∂ t v k L , ˙ D ℓ,N def = N X i =1 k ∂ ih ∇ v k L (Ω) + N − X i =1 k ∂ ih ∂ t v k L (for N ≥ , ¨ D ℓ, σ := k ˙Λ σ h ( ∇ v, ∇ p ) k L (Ω) + k ˙Λ σ +1 h ξ k L (Σ ) , ¨ D ℓ, := k ∂ h ( ∇ v, ∇ p ) k L (Ω) + k ∂ h ξ k L (Σ ) , ¨ D ℓ, := X i =1 k ∂ ih ( ∇ v, ∇ p ) k L (Ω) + k ∂ h ξ k H (Σ ) , ¨ D ℓ,N := ¨ D ℓ, + N − X i =3 ( k ∂ ih ( ∇ v, ∇ p ) k L (Ω) + k ∂ ih ξ k H (Σ ) ) (for N ≥ . Remark 3.1.
We mention that in the notations of the above energy and dissipation, theenergy ˙ E ℓ,s and dissipation ˙ D ℓ,s with s ∈ R come from the tangential derivatives estimates ofthe system (2.20) , and the dissipation ¨ D ℓ,s with s ∈ R comes from the Stokes estimates of thesystem (2.20) . Remark 3.2.
It is easy to check that for any N ≥ D N = ˙ D N − + ˚ D N , where ˚ D N def = k ∂ N − h ( ∇ v, ∇ p, ∂ t v ) k L (Ω) + k ∂ N − h ξ k H (Σ ) . On the other hand, we define the bounded instantaneous energy E N and dissipation D N with N ≥ E N def = k ˙Λ − λh v k H (Ω) + k ( ˙Λ − λh ξ , ξ ) k L (Σ ) + ˙ E N , D N def = k ( ˙Λ − λh ∇ v, ˙Λ − λh ∂ t v, ∇ v, ∂ t v ) k L (Ω) + ˙ D N . Remark 3.3.
Set ˚ E N def = k ∂ N − h ∇ v k L (Ω) + k ∂ Nh ξ k L (Σ ) , we have E N = ˚ E N + E N − .
14e also define instantaneous energy quantities E N = E N ( ξ, v ) (with N ≥
3) in terms ofthe velocity v and the flow map η : E N def = E h ,N + E N (for N ≥ , with E h ,N def = k ˙Λ σ − h ∂ ξ k H + k ˙Λ σ h ∂ ξ h k H + k ˙Λ σ h ξ k L + N − X i =1 k ∂ ih ξ k H , which is corresponding to the energy space F N def = { ( ξ, v ) | E N < + ∞} (for N ≥ k ( ξ, v ) k F N def = E N ( ξ, v ) . If we set˚ E N def = k ∂ N − h ∇ ξ k L (Ω) + k ∂ N − h ∇ v k L (Ω) + k ∂ Nh ξ k L (Σ ) = k ∂ N − h ∇ ξ k L (Ω) + ˚ E N , then E N = ˚ E N + E N − ∼ k ∂ N − h ξ k H + E N + E N − . Let E ℓ,N = E ℓ,N ( ξ, v ) def = N − X i =0 ( k ∂ ih v k H (Ω) + k ∂ ih ξ k H (Σ ) + k ∂ ih ∇ ξ k H (Ω) ) . We define H s tan ,N (Ω) def = { f ∈ H s (Ω) | N X i =0 k ∂ ih f k H s (Ω) < + ∞} , H s , tan ,N (Ω) def = { f ∈ H s (Ω) | N X i =0 k ∂ ih f k H s (Ω) < + ∞ , f | Σ b = 0 } , ˚ H , Σ tan ,N (Ω) def = { f ∈ H , (Ω) | N X i =0 ( k ∂ ih f k H (Σ ) + k ∂ ih ∇ f k H (Ω) ) < + ∞} (for N ∈ N )equipped with the norms respectively k f k H s tan ,N (Ω) def = ( N X i =0 k ∂ ih f k H s (Ω) ) / , k f k H s , tan ,N (Ω) def = ( N X i =0 k ∂ ih f k H s (Ω) ) / k f k ˚ H , Σ0tan ,N (Ω) def = ( N X i =0 ( k ∂ ih f k H (Σ ) + k ∂ ih ∇ f k H (Ω) )) . Inspired by the works in [8] and [9], we may get the following local well-posedness of thesystem (2.8), which proof is similar to the one in [8], and we leave it to the reader.15 heorem 3.1 (Local well-posedness) . Let N ≥ be an integer. Assume ( ξ , v ) ∈ ˚ H , Σ tan ,N − (Ω) ×H , tan ,N − (Ω) satisfies ∇ J A · v = 0 . There exists a positive constant ǫ such that the viscoussurface wave problem (2.8) with initial data ( ξ , v ) has a unique solution ( ξ, v, p ) (dependingcontinuously on the initial data) in C ([0 , T ]; ˚ H , Σ tan ,N − (Ω)) × ( C ([0 , T ]; H , tan ,N − (Ω)) ∩ L ([0 , T ]; H tan ,N − (Ω)) × L ([0 , T ]; H tan ,N − (Ω)) , with T = min { , ( k ξ k ˚ H , Σ0 tan ,N − (Ω)) ) − } provided k ξ k ˚ H , Σ0 tan ,N − (Ω)) + k v k H , tan ,N − (Ω) < ǫ , andthe solution satisfies the estimate sup t ∈ [0 ,T ] ( k ξ ( t ) k H , Σ0 tan ,N − (Ω) + k v ( t ) k H , tan ,N − (Ω) ) + k ( ∇ v, p ) k L ([0 ,T ]; H tan ,N − (Ω) ≤ C ( k ξ k H , Σ0 tan ,N − (Ω)) + k v k H , tan ,N − (Ω) + T k ξ k H , Σ0 tan ,N − (Ω)) ) , (3.1) And if the maximal existence time T ∗ < + ∞ , then lim t ր T ∗ ( k ξ k ˚ H , Σ0 tan ,N − (Ω)) + k v k H , tan ,N − (Ω) + k J − k L ∞ ) = + ∞ . (3.2) Moreover, if in addition ( ξ , v ) ∈ F N +1 , then ( ξ, v ) ∈ C ([0 , T ]; F N +1 ) . Remark 3.4.
According to (3.1) , we see that, if ǫ in Theorem 3.1 is small enough, there is apositive δ ∈ (0 , ) such that − δ ≤ sup ( t,x ) ∈ [0 ,T ] × Ω J ( t, x ) ≤ , which implies that the flow-map η ( t, x ) defines a diffeomorphism from the equilibrium domain Ω to the moving domain Ω( t ) with the boundary Σ F ( t ) . From this, together with the fact that η is a diffeomorphismfrom the equilibrium domain Ω to the initial domain Ω(0) , we deduce a diffeomorphism fromthe initial domain
Ω(0) to the evolving domain Ω( t ) for any t ∈ [0 , T ] . Denote the inverse ofthe flow map η ( t, x ) by η − ( t, y ) for t ∈ [0 , T ] so that if y = η ( t, x ) for y ∈ Ω( t ) and t ∈ [0 , T ] ,then x = η − ( t, y ) ∈ Ω . Along the proof of Beale’s non-decay theorem in [5], we give the modified the non-decaytheorem as follows, which proof will be given in Appendix.
Theorem 3.2 (Non-decay theorem) . Suppose thet σ ∈ ( , . For certain θ ∈ ˚ H , Σ tan , (Ω) ,there cannot exist a curve ( v ( ε ) , q ( ε ) , ξ ( ε )) with v ( ε ) ∈C b ([0 , + ∞ ); H , tan , (Ω)) ∩ L loc ([0 , + ∞ ); H tan , (Ω)) ∩ L ([0 , + ∞ ); ˙Λ − σ h H tan , (Ω)) ∩ L ([0 , + ∞ ); ˙Λ − h H tan , (Ω)) ,q ( ε ) ∈ L loc ([0 , + ∞ ); H tan , (Ω)) ,ξ ( ε )) ∈C ([0 , + ∞ ); ˚ H , Σ tan , (Ω)) ∩ C b ([0 , + ∞ ); H (Σ )) ∩ C b ([0 , + ∞ ); ˚ H , Σ tan , (Ω)) , defined for ε near , such that viscous surface wave problem (2.8) hold with the initial data ξ = ε θ , v = 0 , and k ξ ( t ) k H (Σ ) → as t → + ∞ ) (3.3)16 olds for each ε , and v ( ε ) is of the form v ( ε ) = ε v (1) + ε v (2) + ε v (3) (3.4) in the space E := C b ([0 , + ∞ ); H , tan , (Ω)) ∩ L loc ([0 , + ∞ ); H tan , (Ω)) ∩ L ([0 , + ∞ ); ˙Λ − σ h H tan , (Ω)) ∩ L ([0 , + ∞ ); ˙Λ − h H tan , (Ω)) . (3.5) Remark 3.5.
We call v ∈ L ([0 , + ∞ ); ˙Λ − σ h H tan , (Ω)) and v ∈ L ([0 , + ∞ ); ˙Λ − h H tan , (Ω)) if and only if ˙Λ σ h v ∈ L ([0 , + ∞ ); H tan , (Ω)) and ˙Λ h v ∈ L ([0 , + ∞ ); H tan , (Ω)) respectively. The main result of this paper states as follows, the proof of which will be presented inSection 8.
Theorem 3.3 (Global well-posedness) . Let N ≥ be an integer, ( λ, σ ) ∈ (0 , satisfy − λ < σ ≤ − λ . If ( ξ , v ) ∈ F N +1 satisfies ∇ J A · v = 0 , then there exists a smallpositive constant ǫ , such that, if E N +1 (0) ≤ ǫ , then the viscous surface wave problem (2.8) is globally well-posed in F N +1 . Moreover, for any t > , there hold E N ( t ) . E N (0) + E ℓ,N +1 (0) , E N +1 ( t ) . E N +1 (0) + h t i E N +1 (0) , sup τ ∈ [0 ,t ] (cid:18) ˙ E N ( τ ) h τ i λ + σ + E N +1 ( τ ) (cid:19) + Z t (cid:18) h τ i ( λ + σ +1) / ˙ D N ( τ ) + D N +1 ( τ ) (cid:19) dτ . E N +1 (0) . Remark 3.6.
Notice that the global solution v stated in Theorem 3.3 belongs to the space E in (3.5) and the free surface ξ satisfies (3.3) in the non-decay theorem (Theorem 3.2), weknow that the assumption of the low horizontal regularity on the initial data is the essentialelement to prevent from occurring the statement in Theorem 3.2. Remark 3.7.
Compared to the work of Guo-Tice [10] about the global well-posedness ofthe system (1.1) , there is no requirement about the compatibility conditions in terms of thematerial derivative of the velocity in Theorem 3.3, which plays a significant role in the studyof the vacuum free boundary problem of the viscous compressible Navier-Stokes system (see [8]for example). On the other hand, the Lagrangian method used in the proof of Theorem 3.3 is anew approach to study the global well-posedness of the incompressible Navier-Stokes equationswith free boundary problem, which can be applied to investigate the well-posedness of manytypes of free boundary problem of the incompressible or compressible fluid dynamical system.Finally, some anisotropic algebraic decay estimates in terms of the energy and dissipation areuncovered in Theorem 3.3, and this kind of result may be helpful to understand the possibleblow-up or decay mechanism of the solution.
Let us first recall some basic estimates, which will be heavily used throughout the paper.17 emma 4.1 ([3], Theorem 2.61 in [4]) . Let f be a smooth function on R vanishing at , s , s be two positive real number, s ∈ (0 , , s > . If u belongs to ˙ H s ( R ) ∩ ˙ H s ( R ) ∩ L ∞ ( R ) ,then so does f ◦ u , and we have k f ◦ u k ˙ H si ≤ C ( s, f ′ , k u k L ∞ ) k u k ˙ H si for i = 1 , . Lemma 4.2 (Classical product laws in Sobolev spaces [4]) . k f g k ˙ H s s − ( R ) . k f k ˙ H s ( R ) k g k ˙ H s ( R ) for | s | , | s | < , s + s > , k ˙Λ sh ( f g ) k L ( R h ) . k ˙Λ sh f k L ( R h ) k g k L ∞ ( R h ) + k ˙Λ sh g k L ( R h ) k f k L ∞ ( R h ) for s > . (4.1) Lemma 4.3 (Embedding inequality [4]) . For any σ ∈ (0 , , there holds k f k L ∞ ( R h ) . k ˙Λ σh Λ h f k L ( R h ) . (4.2)The following result gives a version of Korn’s type inequality for the equilibrium domainΩ, which proof can be found in Beale’s paper [5]. Lemma 4.4 (Korn’s lemma, Lemma 2.7 in [5]) . Let Ω be the equilibrium domain given in (1.3) , then there exists a positive constant C , independent of u , such that k u k H (Ω) ≤ C k D ( u ) k L (Ω) for all u ∈ H (Ω) with u | Σ b = 0 . The entries of the matrices
A − I , J A − I , as well as the Jacobian J of the flow map andits inverse J − , are anisotropic when we consider their low horizontal regularities, which isstated as follows. Lemma 4.5.
Let i, j = 1 , , , ≤ k ∈ N , if E ( t ) is uniformly bounded for all existencetime t , say E ( t ) ≤ , then there hold (1) . k ˙Λ σh ( a , a , A , A ) k H + k ( a , a , A , A ) k L ∞ . E ∀ σ ∈ [ σ , . k ˙Λ σh ( J − , J − − k H + k J − k L ∞ . E ∀ σ ∈ [ σ − , . k ˙Λ σh ( a ij , A ji ) k H + k ( a ij , A ji ) k L ∞ . E ∀ σ ∈ [ σ − , with ( i, j ) = (2 , , (2 , , (3 , , (3 , . k ˙Λ σh ( a ii − , A ii − k H + k ( a ii − , A ii − k L ∞ . E ∀ σ ∈ [ σ − , . k ˙Λ kh ( J − , J − − , a ij i = j , A ji i = j , a jj − , A jj − k H . E k +1 . (4.3) Proof.
Due to the expression of a ij in (2.4), we obtain from Lemma 4.2 that k ˙Λ σ h a k L . k ˙Λ σ h ∂ ξ k L + k ˙Λ σ h ( ∂ ξ ∂ ξ ) k L + k ˙Λ σ h ( ∂ ξ ∂ ξ ) k L . k ˙Λ σ h ∂ ξ k L + k ˙Λ σ h ∂ ξ h k L k ∂ ξ h k L ∞ + k ∂ ξ h k L x L ∞ h k ˙Λ σ h ∂ ξ h k L ∞ x L h , which follows from (4.2) that k ˙Λ σ h a k L . k ˙Λ σ h ∂ ξ k L + k ˙Λ σ h ∂ ξ h k L k ˙Λ σ h Λ h ∂ ξ h k H + k ˙Λ σ h Λ h ∂ ξ h k L k ˙Λ σ h ∂ ξ h k H . E . (4.4)18hile from ∇ a = − ( ∇ ∂ ξ + ∇ ∂ ξ ∂ ξ + ∂ ξ ∇ ∂ ξ − ∇ ∂ ξ ∂ ξ − ∂ ξ ∇ ∂ ξ ), applyingLemma 4.2 implies k ˙Λ σ h ∇ a k L . k ˙Λ σ h ∇ ∂ ξ k L + k ˙Λ (1+ σ ) / h ∇ ∂ ξ h k L k ˙Λ (1+ σ ) / h ∂ ξ h k L ∞ x L h + k ˙Λ (1+ σ ) / h ∂ ξ h k L ∞ x L h k ˙Λ (1+ σ ) / h ∂ ∇ ξ h k L . E . (4.5)Combining (4.4) with (4.5) leads to k ˙Λ σ h a k H . E .The same conclusion can be drawn for k ˙Λ h a k H , that is, k ˙Λ h a k H . E .Combining the above two inequalities, we thus use the interpolation inequality to find k ˙Λ σh a k H . E ∀ σ ∈ [ σ , , and then, from (4.2), it follows k a k L ∞ . k ˙Λ σ h Λ h a k L ∞ x L h . k ˙Λ σ h Λ h a k H . E . By a similar argument, one can see k ˙Λ σh a k H + k a k L ∞ . E ∀ σ ∈ [ σ , . For a −
1, it immediately shows k ˙Λ σ − h ( a − k H . k ˙Λ σ − h ( a − k L + k ˙Λ σ − h ∇ a k L . k ˙Λ σ − h ( ∂ ξ , ∂ ξ ) k H + k ˙Λ σ − h ( ∂ ξ ∂ ξ , ∂ ξ ∂ ξ ) k L + k ˙Λ σ − h ( ∂ ξ ∇ ∂ ξ , ∇ ∂ ξ ∂ ξ , ∇ ∂ ξ ∂ ξ , ∂ ξ ∇ ∂ ξ ) k L . (4.6)which along with (4.2) yields k ˙Λ σ − h ( ∇ ∂ ξ ∂ ξ ) k L . k ˙Λ σ h ∇ ∂ ξ k L k ∂ ξ k L ∞ x L h . E . The same estimate holds for other nonlinear terms in the right hand side of (4.6), which,combining k ˙Λ σ − h ( ∂ ξ , ∂ ξ ) k H . E , implies k ˙Λ σ − h ( a − k H . E .Similarly, for k = 1 , ,
3, ( i, j ) = (2 , , (2 , , (3 , , (3 , k ˙Λ σ − h ( a kk − , J − , a ij ) k H . E and k ˙Λ h ( a kk − , J − , a ij ) k H . E , so, applying the interpolationinequality yields k ˙Λ σh ( a kk − , J − , a ij ) k H + k ( a kk − , J − , a ij ) k L ∞ . E ∀ σ ∈ [ σ − , . (4.7)Next, applying Lemma 4.1, it produces from (4.7) that k ˙Λ σh ( J − − k H . E for σ ∈ [ σ , , which ensures that A ji shares the same estimate with a ij with i, j = 1 , , a −
1, andthe same proof remains valid for the others. Indeed, for any k ∈ N , it is easy to verify that k ˙Λ kh ( a − k H . k ˙Λ kh ( a − k L + k ˙Λ kh ∇ a k L . k ˙Λ kh ( ∂ ξ , ∂ ξ ) k H + k ˙Λ kh ( ∂ ξ ∂ ξ , ∂ ξ ∂ ξ ) k L + k ˙Λ kh ( ∂ ξ ∇ ∂ ξ , ∇ ∂ ξ ∂ ξ , ∇ ∂ ξ ∂ ξ , ∂ ξ ∇ ∂ ξ ) k L . (4.8)19ue to (4.2), we may get k ˙Λ kh ( ∇ ∂ ξ ∂ ξ ) k L . k ˙Λ kh ∇ ∂ ξ k L k ∂ ξ k L ∞ + k∇ ∂ ξ k L x L ∞ h k ˙Λ kh ∂ ξ k L ∞ x L h . E k +1 E . The same estimate holds for other nonlinear terms in the right hand side of (4.8), which,together with k ˙Λ kh ( ∂ ξ , ∂ ξ ) k H . E k +1 , follows k ˙Λ kh ( a − k H . E k +1 .The proof of the lemma is thus completed.With Lemma 4.5 in hand, let us now estimate B αj,i and B αj,i ∂ α v i with j = 1 , · · · , . Lemma 4.6.
Let j = 1 , · · · , . , i = 1 , , , α = 2 , , σ ∈ [ σ − , , ≤ k ∈ N , if E ( t ) ≤ ,then there hold that k ˙Λ σh B αj,i k H . E , k ˙Λ kh B αj,i k H . E k +1 , k ˙Λ σh ( B αj,i ∂ α v i ) k H . E k ∂ h Λ h v k H , (4.9) k ˙Λ kh ( B αj,i ∂ α v i ) k L . E k +1 k ∂ h Λ h v k L + E k ∂ h Λ kh v k L , (4.10) k ˙Λ kh ( B αj,i ∂ α v i ) k H . E k +1 k ∂ h Λ h v k H + E k ∂ h Λ kh v k H , (4.11) k ˙Λ σh ( f a ∂ v , f a ∂ v ) k H . E ( k ˙Λ σ +1 h ∂ v h k H + k ∂ h Λ h ∇ v k L ) , (4.12) k ˙Λ kh ( f a ∂ v , f a ∂ v ) k H . E ˙ D k +1 + E k +1 ˙ D . (4.13) Proof.
In view of Lemma 4.5, it is easy to get the first two inequalities in (4.9). Let’s nowfocus on the estimates of B αj,i ∂ α v i with j = 1 , · · · , . We first directly use (4.2) to get k ˙Λ σ − h ( B αj,i ∂ α v i ) k H . k ˙Λ σ − h ( B αj,i ∂ α v i ) k L + k ˙Λ σ − h ( ∇ B αj,i ∂ α v i + B αj,i ∇ ∂ α v i ) k L . k ˙Λ σ h B αj,i k L k ∂ α v i k L ∞ x L h + k ˙Λ σ h ∇ B αj,i k L k ∂ α v i k L ∞ x L h + k ˙Λ σ h B αj,i k L ∞ x L h k∇ ∂ α v i k L , then applying the Sobolev embedding theorem shows k ˙Λ σ − h ( B αj,i ∂ α v i ) k H . k ˙Λ σ h B αj,i k H k ∂ α v i k H . E k ∂ α v i k H . Similar estimates hold for k ˙Λ σh ( B αj,i ∂ α v i ) k H with σ ∈ [ σ − , k ˙Λ kh ( B αj,i ∂ α v i ) k L . k ˙Λ kh B αj,i k L ∞ x L h k ∂ α v i k L x L ∞ h + kB αj,i k L ∞ k ˙Λ kh ∂ α v i k L . E k +1 k ∂ h Λ h v k L + E k ∂ h Λ kh v k L , and k ˙Λ kh ∇ ( B αj,i ∂ α v i ) k L . k ˙Λ kh ( ∇B αj,i ∂ α v i + B αj,i ∇ ∂ α v i ) k L . k ˙Λ kh ∇ B αj,i k L k ∂ α v i k L ∞ + k ˙Λ kh ∂ α v i k L ∞ x L h k∇ B αj,i k L x L ∞ h + k ˙Λ kh B αj,i k L ∞ x L h k∇ ∂ α v i k L x L ∞ h + kB αj,i k L ∞ k ˙Λ kh ∇ ∂ α v i k L , so using the Sobolev embedding theorem yields (4.11).20inally, for (4.12), due to the same argument above, we need only prove it in the case of σ = σ −
1. Notice that from the product law (4.1) k ˙Λ σ − h ( f a ∂ v , f a ∂ v ) k L . k ( f a , f a ) k L ∞ x L h k ˙Λ σ h ( ∂ v , ∂ v ) k L , and k ˙Λ σ − h ∇ ( f a ∂ v , f a ∂ v ) k L . k ( f a , f a ) k L ∞ x L h k ˙Λ σ h ∇ ( ∂ v , ∂ v ) k L + k∇ ( f a , f a ) k L k ˙Λ σ h ( ∂ v , ∂ v ) k L ∞ x L h , it follows from Lemma 4.5 that(4.12). The estimate in (4.13) can be showed by using thesame argument in the proof of (4.10).The lemma is therefore completely proved.Due to the incompressibility condition (2.10), the vertical derivatives of the vertical com-ponent v of the velocity gains the − Lemma 4.7.
Let ( v, ξ ) be smooth solution to the system (2.8) , if E ( t ) ≤ for all existencetime t , then there holds ¨ D ℓ, σ + ˙ D ℓ,σ + ¨ D ℓ, ∼ k ( ˙Λ σ − h ∇ v , ˙Λ σ h ( ∇ v h , ∇ p )) k L (Ω) + k ˙Λ σ +1 h ξ k L (Σ ) + ¨ D ℓ,σ + ˙ D ℓ, , Proof.
By the definitions of ¨ D ℓ, σ , ˙ D ℓ,σ and ¨ D ℓ, , it suffices to show that k ˙Λ σ − h ∂ v k L (Ω) . ¨ D ℓ, σ + ˙ D ℓ,σ + ¨ D ℓ, . (4.14)Indeed, thanks to the incompressibility condition (2.10), one can see˙Λ σ − h ∂ v = − ˙Λ σ − h ∂ ∇ h · v h − ˙Λ σ − h ∂ ( g a α ∂ v α ) + ˙Λ σ − h ∂ ( B α ,i ∂ α v i ) , and then k ˙Λ σ − h ∂ v k L . k ˙Λ σ h ∂ v h k L + k ˙Λ σ − h ∂ ( g a α ∂ v α ) k L + k ˙Λ σ − h ∂ ( B α ,i ∂ α v i ) k L . Hence, by virtue of (4.9) and (4.12), we deduce that k ˙Λ σ − h ∂ v k L . k ˙Λ σ h ∂ v h k L + E ( k ∂ h Λ h v k H + k ˙Λ σ h ∂ v h k H + k ∂ h Λ h ∇ v k L ) . ( ¨ D ℓ, σ + ˙ D ℓ,σ + ¨ D ℓ, ) , which results in (4.14), and ends the proof of Lemma 4.7. In this section, we will derive some global energy estimates. In view of the local well-posednesstheorem (Theorem 3.1), it suffices to get necessary a priori estimates. Here and in whatfollows, all the C -forms, such as C j , c j , C j , e C j , c j , and e c j , are generic positive constants,which may be different on different lines. 21 .1 Energy estimates in L We begin with the energy identity in L based on the nonlinear structure of the system (2.8). Proposition 5.1.
Let ( v, ξ ) be smooth solution to the system (2.8) , then there holds ddt (cid:18) Z Ω | v | J dx + Z Σ a | ξ | dS (cid:19) + ν Z Ω | D J A ( v ) | J − dx = 0 . (5.1) Proof.
We multiply the i -th component of the momentum equations of (2.8) by J v i , sumover i , and integrate over Ω to find12 ddt Z Ω J | v | dx − Z Ω | v | ∂ t J dx + I + II = 0 (5.2)with I = R Ω ∇ J A q · v dx, II = R Ω ( − ν ∇ J A · D A ( v )) · v dx. Due to the incompressibility condition ∇ J A · v = 0, from (2.6), it provides ∂ t J = 0 . (5.3)Integrating by parts in I , II , and utilizing the Piola identity (2.7), yields I = Z Σ v · ( q N ) dS − Z Ω q ∇ J A · v dx = Z Σ v · ( q N ) dS , (5.4)and II = Z Σ ( − ν D A ( v ) N ) · v dS + ν Z Ω D A ( v ) : ∇ J A v dx = Z Σ ( − ν D A ( v ) N ) · v dS + Z Ω ν | D J A ( v ) | J − dx, (5.5)where the identity D A ( v ) : ∇ A u = D A ( v ) : D A ( u ) has been used in the second equality in(5.5).Plugging (5.3)-(5.5) into (5.2) implies12 ddt Z Ω | v | J dx + Z Ω ν | D J A ( v ) | J − dx + Z Σ v · (cid:18) q N − ν D A ( v ) N (cid:19) dS = 0 . (5.6)For the boundary integral in (5.6), making use of the interface boundary condition in (2.8)yields Z Σ v · (cid:18) q N − ν D A ( v ) N (cid:19) dS = Z Σ ξ N · v dS = Z Σ ξ a i v i dS = Z Σ a ∂ t | ξ | dS + Z Σ ξ ( a v + a v ) dS , which together with the expression of a ij in (2.4) follows Z Σ v · (cid:18) q N − ν D A ( v ) N (cid:19) dS = 12 ddt Z Σ a | ξ | dS + Z Σ (cid:18) ξ ( a v + a v ) − | ξ | ∂ t a (cid:19) dS = 12 ddt Z Σ a | ξ | dS . (5.7)Inserting (5.7) into (5.6) ensures (5.1), which ends the proof of Lemma 5.1.22 .2 Energy estimates of the horizontal derivatives of the velocity In general, since the equations of the derivatives of the solution ( v, ξ ) to the system (2.8) lackthe nonlinear symmetric structure, we have no idea to get the high-order energy identity. Forthis reason, we turn to employ the linearized form (2.20) of the system (2.8) to study thehigh-order energy estimates.For the general horizontal derivatives of the velocity, we first derive
Lemma 5.1.
Let ( v, ξ ) be smooth solution to the system (2.8) , then there holds ddt ( kP ( ∂ h ) v k L (Ω) + kP ( ∂ h ) ξ k L (Σ ) ) + ν k D ( P ( ∂ h ) v ) k L (Ω) = X j =1 K j (5.8) with K := ν Z Σ (cid:18) P ( ∂ h )(( B α ,i − B α ,i ) ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v (cid:19) dS , K := Z Ω P ( ∂ h ) q P ( ∂ h )( B α ,i ∂ γ v i − g a α ∂ v α ) dx, K := Z Ω P ( ∂ h ) g · P ( ∂ h ) v dx. (5.9) Proof.
Acting the operator P ( ∂ h ) to both the momentum equations and the incompressibilityequation in (2.20) implies ( ∂ t P ( ∂ h ) v + ∇ P ( ∂ h ) q − ν ∇ · D ( P ( ∂ h ) v ) = P ( ∂ h ) g, ∇ · P ( ∂ h ) v = P ( ∂ h )( − g a α ∂ v α + B α ,i ∂ α v i ) in Ω . (5.10)We multiply the i -th component of the momentum equations of (5.10) by P ( ∂ h ) v i , sum over i , and integrate over Ω to find12 ddt Z Ω |P ( ∂ h ) v | dx + I + II = III (5.11)with I = R Ω ∇P ( ∂ h ) q ·P ( ∂ h ) v dx, II = − ν R Ω [ ∇ · D ( P ( ∂ h ) v )] ·P ( ∂ h ) v dx , III = R Ω P ( ∂ h ) g ·P ( ∂ h ) v dx .In view of the boundary condition v | Σ b = 0, integrating by parts in I and II shows I = Z Σ P ( ∂ h ) v · ( P ( ∂ h ) q e ) dS − Z Ω P ( ∂ h ) q ∇ · P ( ∂ h ) v dx, (5.12)and II = − ν Z Σ ( D ( P ( ∂ h ) v ) e ) · P ( ∂ h ) v dS + ν Z Ω | D ( P ( ∂ h ) v ) | dx, (5.13)so summing up (5.12) and (5.13) provides I + II = Z Σ P ( ∂ h ) v · P ( ∂ h ) (cid:18) q e − ν D ( v ) e (cid:19) dS − Z Ω P ( ∂ h ) q ∇ · P ( ∂ h ) v dx + ν Z Ω | D ( P ( ∂ h ) v ) | dx. (5.14)23or the first boundary integral of the right hand side in (5.14), applying the interface boundarycondition in (2.20) leads to Z Σ P ( ∂ h ) v · P ( ∂ h ) (cid:18) q e − ν D ( v ) e (cid:19) dS = 12 ddt Z Σ |P ( ∂ h ) ξ | dS − ν Z Σ (cid:18) P ( ∂ h )(( B α ,i − B α ,i ) ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v (cid:19) dS . (5.15)On the other hand, thanks to the second equation in (5.10), one may see that Z Ω P ( ∂ h ) q ∇ · P ( ∂ h ) v dx = Z Ω P ( ∂ h ) q P ( ∂ h )( − g a α ∂ v α + B α ,i ∂ α v i ) dx. (5.16)Hence, plugging (5.15) and (5.16) into (5.14) gives rise to I + II = 12 ddt kP ( ∂ h ) ξ k L (Σ ) − ν Z Σ (cid:18) P ( ∂ h )(( B α ,i − B α ,i ) ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v + P ( ∂ h )( B α ,i ∂ α v i ) P ( ∂ h ) v (cid:19) dS + ν k D ( P ( ∂ h ) v ) k L (Ω) + Z Ω P ( ∂ h ) q P ( ∂ h )( − g a α ∂ v α + B α ,i ∂ α v i ) dx. (5.17)Combining (5.17) with (5.11) yields (5.8), which completes the proof of Lemma 5.1.In order to get the energy estimates of the tangential derivatives of the velocity, we willdeal with the remainder terms on the right hand side in (5.8). For this, let us first estimatethe terms with respect to g in (5.8). Lemma 5.2.
Under the assumption in Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ , then k ˙Λ σ h g k L . E ˙ D , k ˙Λ − λh g jj k L + k ˙Λ − λh e g jj k L . E ˙ D ( with j = 1 , , , k ˙Λ − λh g k L . E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D , k g k L . E k ˙Λ − σ h ξ k L (Σ ) + E ˙ D , k ∂ N − h g k L . E N ˙ D + E ˙ D N ( for N ≥ . (5.18) Proof.
We first bound the term g jj with j = 1 , , g in (5.18). For this, we split the termsof g jj in (2.21) into two types: the one is of the second-order derivative terms of v and theother is of the first-order derivative terms of v .For the first one, we use the product law (4.1) to get k ˙Λ σ h ( A α A ∂ v α ) k L . k ˙Λ (1+ σ ) / h ( A α A ) k L ∞ x L h k ˙Λ (1+ σ ) / h ∂ v α k L . ( k ˙Λ (1+ σ ) / h A α k L ∞ x L h + k ˙Λ (3+ σ ) / h A α k L ∞ x L h k ˙Λ (3+ σ ) / h ( A − k L ∞ x L h ) k ˙Λ (1+ σ ) / h ∂ v α k L , k ˙Λ σ h ( A α A ∂ v α ) k L . k ˙Λ (1+ σ ) / h ∂ v α k L (cid:18) k ˙Λ (1+ σ ) / h A α k H + k ˙Λ (3+ σ ) / h A α k H k ˙Λ (3+ σ ) / h ( A − k H (cid:19) . E ˙ D . The same estimate holds for any second-order derivative terms of v in g jj with j = 1 , , v in g jj with j = 1 , ,
3, we employ the productlaw (4.1) once again to yield k ˙Λ σ h ( A i ∂ A ∂ v i ) k L . k ˙Λ (1+ σ ) / h ( A i ∂ A ) k L k ˙Λ (1+ σ ) / h ∂ v k L ∞ x L h . ( k ˙Λ (3+ σ ) / h ( A α , A − k L ∞ x L h ( k ˙Λ (1+ σ ) / h ∂ A k L + k ˙Λ (3+ σ ) / h ∂ A k L ) × ( k ˙Λ σ h ∂ v k L + k ˙Λ σ h ∂ v k L ) , from which, applying Lemma 4.5 implies k ˙Λ σ h ( A i ∂ A ∂ v i ) k L . E ˙ D . This bound also holds true for any first-order derivative term of v in g jj and e g jj with j =1 , , j = 1 , , k ˙Λ σ h g jj k L + k ˙Λ σ h e g jj k L . E ˙ D . (5.19)For the pressure term in g , making use of the product law (4.1) again yields k ˙Λ σ h ( A ∂ q ) k L . k ˙Λ (1+ σ ) / h ∂ q k L k ˙Λ (1+ σ ) / h A k L ∞ x L h . E ˙ D , (5.20)where we used the estimate k ˙Λ (1+ σ ) / h A k L ∞ x L h . k ˙Λ (1+ σ ) / h A k H . E in Lemma 4.5.In the same manner it is easy to get k ˙Λ σ h ( A h ∂ h q ) k L + k ˙Λ σ h (( A − ∂ q ) k L + k ˙Λ σ h (( A − ∂ q ) k L + k ˙Λ σ h ( A ∂ q ) k L + k ˙Λ σ h ( A ∂ q ) k L + k ˙Λ σ h (( A − ∂ q ) k L . E ˙ D . (5.21)Combing (5.19)-(5.21) yields the first inequality in (5.18).Let’s now estimate k ˙Λ − λh g k L including the L norms of the negative horizontal derivativeof g jj , e g jj and the pressure terms.Firstly, the product law (4.1) ensures k ˙Λ − λh ( A α A ∂ v α ) k L . k ˙Λ σ − h ( A α A ) k L ∞ x L h k ˙Λ − σ − λh ∂ v α k L . (5.22)While from Lemma 4.5 and E ( t ) ≤
1, there holds k ˙Λ σ − h ( A α A ) k L ∞ x L h . k ˙Λ σ − h A α k L ∞ x L h + k ˙Λ σ h A α k L ∞ x L h kA − k L ∞ x L h . k ˙Λ σ − h A α k H + k ˙Λ σ h A α k H kA − k H . E . (5.23)25ubstituting (5.23) into (5.22) yields k ˙Λ − λh ( A α A ∂ v α ) k L . E k ˙Λ − σ − λh ∂ v α k L . E ˙ D , (5.24)where the fact 2 − σ − λ ≥ σ (since σ ≤ − λ ) has been used in the last inequality.For A A ∂ ∂ v , thanks to the fact 2 − σ − λ ≥ σ again, we have k ˙Λ − λh ( A A ∂ ∂ v ) k L . k ˙Λ − σ − λh ( A A ) k L k ˙Λ σ − h ∂ ∂ v k L ∞ x L h . E ˙ D . (5.25)This estimate also holds for any second-order derivative terms of v in g jj and e g jj with j = 1 , , v in g jj and e g jj with j = 1 , ,
3, applying the productlaw (4.1), and then using Lemma 4.5 ensures k ˙Λ − λh ( A i ∂ A ∂ v i ) k L . k ˙Λ σ − h ( A i ∂ A ) k L k ˙Λ − σ − λh ∂ v k L ∞ x L h . k ˙Λ σ − h ( A i ∂ A ) k L k ˙Λ − σ − λh ∂ v k H . E ˙ D . (5.26)This estimate remains valid for any other first-order derivative terms of v in g jj and e g jj with j = 1 , , j = 1 , , k ˙Λ − λh g jj k L + k ˙Λ − λh e g jj k L . E ˙ D . For the pressure terms in g , from (4.1), it can be obtained that k ˙Λ − λh ( A ∂ q ) k L . k ˙Λ σ − h A k L ∞ x L h k ˙Λ − σ − λh ∂ q k L . E k ˙Λ − σ − λh ∂ q k L . E ˙ D . While for the term about ∂ h q , we have k ˙Λ − λh ( A h ∂ h q ) k L . k ˙Λ σ h A k L ∞ x L h k ˙Λ − σ − λh ∂ h q k L . E k ˙Λ − σ − λh ∂ h q k L . In order to control k ˙Λ − σ − λh ∂ h q k L , we employ the Poincare’s inequality and the boundarycondition q = ξ − ν ∇ h · v h + ν B α ,i ∂ α v i on Σ to get k ˙Λ − σ − λh ∂ h q k L (Ω) . k ˙Λ − σ − λh ∂ h ( ξ − ν ∇ h · v h + ν B α ,i ∂ α v i ) k L (Σ ) + k ˙Λ − σ − λh ∂ h ∂ q k L (Ω) . k ˙Λ − σ − λh ξ k L (Σ ) + k ˙Λ − σ − λh ∂ h ( − ν ∇ h · v h + ν B α ,i ∂ α v i ) k H (Ω) + k ˙Λ − σ − λh ∂ q k L (Ω) . k ˙Λ − σ − λh ξ k L (Σ ) + ˙ D , where we have used the assumption 2 − σ − λ ≥ σ in the last inequality. It thus follows k ˙Λ − λh ( A h ∂ h q ) k L . E ( k ˙Λ − σ − λh ξ k L (Σ ) + ˙ D ) . Repeating the above argument to the other pressure terms in g ensures that k ˙Λ − λh ( A ∂ q ) k L + k ˙Λ − λh ( A h ∂ h q ) k L + k ˙Λ − λh (( A − ∂ q ) k L + k ˙Λ − λh (( A − ∂ q ) k L + k ˙Λ − λh ( A ∂ q ) k L + k ˙Λ − λh ( A ∂ q ) k L + k ˙Λ − λh (( A − ∂ q ) k L . E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D . k ˙Λ − λh g k L . E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D , this is, the third inequality in (5.18) holds true.The forth inequality can be proved by repeating the same line in the proof of the thirdinequality in (5.18).Finally, let us now turn to derive the last inequality in (5.18).For this, we first apply the product law (4.1) to get k ∂ N − h ( A α A ∂ v α ) k L . k ∂ N − h ( A α A ) k L ∞ x ( L h ) k ∂ v α k L x L ∞ h + k ∂ N − h ∂ v α k L kA α A k L ∞ . It is easy to check k ∂ N − h ( A α A ) k L ∞ x ( L h ) . k ∂ N − h ( A α A ) k H . E N and kA α A k L ∞ . k ˙Λ σ h Λ h ( A α A ) k H . E , and by using the Sobolev embedding theorem k ∂ v α k L x L ∞ h . k ˙Λ σ h Λ h ∂ v α k L . ˙ D , we obtain k ∂ N − h ( A α A ∂ v α ) k L . E N ˙ D + E ˙ D N . The same conclusion can be drawn for any other terms of v in g jj with j = 1 , , j = 1 , , k ∂ N − h g jj k L . E N ˙ D + E ˙ D N . Similarly, thanks to the product law (4.1) again, it can be established that k ∂ N − h ( A ∂ q ) k L . k ∂ N − h A k L ∞ x L h k ∂ q k L x ( L ∞ h ) + k ∂ N − h ∂ q k L kA k L ∞ . E N ˙ D + E ˙ D N , and also k ∂ N − h ( A h ∂ h q ) k L + k ∂ N − h (( A − ∂ q ) k L + k ∂ N − h (( A − ∂ q ) k L + k ∂ N − h ( A ∂ q ) k L + k ∂ N − h ( A ∂ q ) k L + k ∂ N − h (( A − ∂ q ) k L . E N ˙ D + E ˙ D N . We thus prove that k ∂ N − h g k L . E N ˙ D + E ˙ D N . The proof of the lemma is thereforeaccomplished.With Lemmas 5.1 and 5.2 in hand, we will deal with the estimates k ˙Λ sh v k L ∞ t L (Ω) + k ˙Λ sh ξ k L ∞ t L (Σ ) with s = σ , N, − λ . k ˙Λ σ h v k L ∞ t L (Ω) + k ˙Λ σ h ξ k L ∞ t L (Σ ) Lemma 5.3.
Under the assumption of Lemma 5.1, if E ( t ) ≤ for all the existence times t , then there holds ddt ( k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) ) + ν k D ( ˙Λ σ h v ) k L (Ω) . E ˙ D . (5.27)27 roof. Taking the horizontal derivative operator P ( ∂ h ) = ˙Λ σ h in (5.8) yields12 ddt (cid:18) k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) (cid:19) + ν k D ( ˙Λ σ h v ) k L (Ω) = X j =1 K j , (5.28)where the remainder terms K j with j = 1 , , P ( ∂ h ) =˙Λ σ h .For the first integral K , we first observe | Z Σ ˙Λ σ h ( B α ,i ∂ α v i ) ˙Λ σ h v dS | . k ˙Λ σ h ( B α ,i ∂ α v i k L (Σ ) ) k ˙Λ σ h v k L (Σ ) . k ˙Λ σ h ( B α ,i ∂ α v i k H (Ω) k ˙Λ σ h v k H (Ω) , which follows from Lemma 4.6 that | R Σ ˙Λ σ h ( B α ,i ∂ α v i ) ˙Λ σ h v dS | . E ˙ D . Along the sameline, it can be obtained that | Z Σ ˙Λ σ h ( B α ,i ∂ α v i ) ˙Λ σ h v dS | + | Z Σ ˙Λ σ h ( B α ,i ∂ α v i ) ˙Λ σ h v dS | + | Z Σ ˙Λ σ h ( B α ,i ∂ α v i ) ˙Λ σ h v dS | . E ˙ D . We thus get | K | . E ˙ D . (5.29)On the other hand, due to the product law (4.1), one can see that | K | . k ˙Λ σ +1 h q k L ( k ˙Λ σ − h ( g a α ∂ v α ) k L + k ˙Λ σ − h ( B α ,i ∂ α v i ) k L ) . k ˙Λ σ h ∂ h q k L (cid:18) k g a α k L ∞ x L h k ˙Λ σ h ∂ v h k L + k ˙Λ σ h B α ,i k L k ∂ α v i k L ∞ x L h (cid:19) , which follows from (4.3) and (4.9) that | K | . k ˙Λ σ h ∂ h q k L (cid:18) k g a α k H k ˙Λ σ h ∂ v h k L + k ˙Λ σ h B α ,i k L k ∂ α v i k H (cid:19) . E ˙ D . (5.30)For the term K = R Ω ˙Λ σ h g · ˙Λ σ h v dx , from (5.18), it follows that | K | . k ˙Λ σ h g k L k ˙Λ σ h v k L . k ˙Λ σ h g k L k ˙Λ σ h v k H . E ˙ D . (5.31)Substituting (5.29)-(5.31) into (5.28) leads to (5.27), which furnishes the proof of Lemma5.3. k ∂ Nh v k L ∞ t ( L ) Lemma 5.4.
Let N ≥ , under the assumption of Lemma 5.1, if E ( t ) ≤ for all theexistence times t , there holds ddt (cid:18) k ∂ Nh v k L (Ω) + k ∂ Nh ξ k L (Σ ) (cid:19) + ν k D ( ∂ Nh v ) k L (Ω) . ˙ D N ( E N ˙ D + E ˙ D N ) . (5.32)28 roof. Similar to the proof of Lemma 5.3, taking the operator P ( ∂ h ) = ∂ Nh in (5.8), it canbe easily obtained that12 ddt ( k ∂ Nh v k L (Ω) + k ∂ Nh ξ k L (Σ ) ) + ν k D ( ∂ Nh v ) k L (Ω) = X j =1 K j , (5.33)where K j with j = 1 , , P ( ∂ h ) = ∂ Nh .For the boundary integral K , making use of the trace theorem ensures | Z Σ ∂ Nh ( B α ,i ∂ α v i ) ∂ Nh v dS | . k ˙Λ N − h ( B α ,i ∂ α v i ) k L (Σ ) k ˙Λ N + h v k L (Σ ) . k ˙Λ N − h ( B α ,i ∂ α v i ) k H (Ω) k ˙Λ Nh v k H (Ω) . We thus obtain from Lemma 4.6 that | Z Σ ∂ Nh ( B α ,i ∂ α v i ) ∂ Nh v dS | . (cid:18) E N ˙ D + E ˙ D N (cid:19) ˙ D N . (5.34)It can be similarly proved that | Z Σ ∂ Nh ( B α ,i ) ∂ α v i ) ∂ Nh v dS | + | Z Σ ∂ Nh ( B α ,i ∂ α v i ) ∂ Nh v dS | + | Z Σ ∂ Nh ( B α ,i ∂ α v i ) ∂ Nh v dS | . (cid:18) E N ˙ D + E ˙ D N (cid:19) ˙ D N , which along with (5.34) leads to | K | . (cid:18) E N ˙ D + E ˙ D N (cid:19) ˙ D N . (5.35)On the other hand, for K , it is easy to produce | K | . k ∂ Nh q k L ( k ∂ Nh ( g a α ∂ v α ) k L + k ∂ Nh ( B α ,i ∂ γ v i ) k L ) . (5.36)Thanks to Lemma 4.6 again, one has k ∂ Nh ( g a α ∂ v α ) k L + k ∂ Nh ( B α ,i ∂ α v i ) k L . E N ˙ D + E ˙ D N , which along with (5.36) follows | K | . ( E N ˙ D + E ˙ D N ) ˙ D N . (5.37)Finally, for K = R Ω ∂ Nh g · ∂ Nh v dx , we use (5.18) to get | K | . k ∂ N − h g k L k ∂ N +1 h v k L . ( E N ˙ D + E ˙ D N ) ˙ D N . (5.38)Therefore, inserting (5.35), (5.37), (5.38) into (5.33) yields (5.32), which gives the desiredresult. 29hanks to Lemmas 5.3, 5.4, and the Korn’s inequality, we obtain Lemma 5.5.
Let N ≥ , under the assumption of Lemma 5.1, if E ( t ) ≤ for all theexistence times t , then there holds ddt (cid:18) k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) + N X i =1 ( k ∂ ih v k L (Ω) + k ∂ ih ξ k L (Σ ) ) (cid:19) + c ( k ˙Λ σ h ∇ v k L (Ω) + N X i =1 k ∂ ih ∇ v k L (Ω) ) . ˙ D N ( E N ˙ D + E ˙ D N ) . k ˙Λ − λh v k L ∞ t ( L ) Lemma 5.6.
Under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt (cid:18) k ˙Λ − λh v k L (Ω) + k ˙Λ − λh ξ k L (Σ ) (cid:19) + c k ˙Λ − λh ∇ v k L (Ω) . k ˙Λ − λh ξ k L (Σ ) E ˙ D + k ˙Λ − λh ∂ t v k L E ˙ D + E ˙ D + E k ˙Λ σ h ξ k L (Σ ) . (5.39) Proof.
First, taking the operator P ( ∂ h ) = ˙Λ − λh in (5.8) shows12 ddt (cid:18) Z Ω | ˙Λ − λh v | dx + Z Σ | ˙Λ − λh ξ | dS (cid:19) + ν Z Ω | D ( ˙Λ − λh v ) | dx = I + I + I (5.40)with I := R Ω ˙Λ − λh q ˙Λ − λh ( B α ,i ∂ α v i − g a α ∂ v α ) dx , I := ν Z Σ (cid:18) ˙Λ − λh (( B α ,i − B α ,i ) ∂ α v i ) ˙Λ − λh v + ˙Λ − λh ( B α ,i ∂ α v i ) ˙Λ − λh v + ˙Λ − λh ( B α ,i ∂ α v i ) ˙Λ − λh v (cid:19) dS , I := Z Ω ˙Λ − λh g · ˙Λ − λh v dx. Using the relation q = Q + H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i , we split I into two parts: I = Z Ω ˙Λ − λh Q ˙Λ − λh ( B α ,i ∂ α v i − g a α ∂ v α ) dx + Z Ω ˙Λ − λh ( H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i ) ˙Λ − λh ( B α ,i ∂ α v i − g a α ∂ v α ) dx =: I , + I , . For I , , we have | I , | = | Z Ω ˙Λ − λh Q ˙Λ − λh ( B α ,i ∂ α v i − g a α ∂ v α ) dx | . k ˙Λ − σ − λ ) h Qk L ∞ x L h k ˙Λ σ − h ( B α ,i ∂ α v i − g a α ∂ v α ) k L x L h . (5.41)30y using the product law (4.1), it follows k ˙Λ σ − h ( B α ,i ∂ α v i − g a α ∂ v α ) k L x L h . k ˙Λ σ h B α ,i k L k ˙Λ σ − h ∂ α v i k L + k ˙Λ σ h ∂ v α k L k ˙Λ σ − h g a α k L . E k ˙Λ σ h ∇ v k L , (5.42)which along with Q| Σ = 0 and (5.41) implies | I , | . k ˙Λ − σ − λ ) h ∂ Qk L E k ˙Λ σ h ∇ v k L . The estimate of k ˙Λ − σ − λ ) h ∂ Qk L can be treated by k ˙Λ − σ − λ ) h ∂ Qk L . k ˙Λ − σ − λ ) h ∂ q k L + k ˙Λ − σ − λ ) h ∂ H ( ξ ) k L + k ˙Λ − σ − λ ) h ∂ ∇ h · v h k L + k ˙Λ − σ − λ ) h ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ˙Λ − λh ∂ q k L + k ˙Λ − λh ξ k L (Σ ) + k ˙Λ σ +2 h ∂ q k L + k ˙Λ σ +1 h ξ k H (Σ ) + k ˙Λ − λh ∇ v k L + k ˙Λ σ +1 h ∇ v k L + k ˙Λ − σ − λ ) h ∂ ( H ( B α ,i ) ∂ α v i ) k L . (5.43)For k ˙Λ − λh ∂ q k L in (5.43), thanks to (2.24) and (5.18), one can see that k ˙Λ − λh ∂ q k L . k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + k ˙Λ − λh g k L + k ˙Λ − λh e g k L . k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D , which, together with (5.43), 1 − λ < σ ≤ − λ , and the estimate k ˙Λ − λh ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ˙Λ − λh ( ∂ H ( B α ,i ) ∂ α v i ) k L + k ˙Λ − λh ( H ( B α ,i ) ∂ ∂ α v i ) k L . k ˙Λ σ h ∂ H ( B α ,i ) k L k ˙Λ − σ − λh ∂ α v i k L ∞ x L h + k ˙Λ σ h H ( B α ,i ) k L ∞ x L h k ˙Λ − σ − λh ∂ ∂ α v i ) k L . E k ˙Λ − σ − λh v k H . E ˙ D by using (4.1), ensures that k ˙Λ − σ − λ ) h ∂ Qk L . k ˙Λ − σ − λ ) h ∂ q k L + k ˙Λ − σ − λ ) h ∂ H ( ξ ) k L + k ˙Λ − σ − λ ) h ∂ ∇ h · v h k L + k ˙Λ − σ − λ ) h ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ˙Λ − λh ξ k L (Σ ) + k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∇ v k L + ˙ D . Hence, we conclude that | I , | . k ˙Λ − λh ξ k L (Σ ) E ˙ D + ( k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∇ v k L ) E ˙ D + E ˙ D . (5.44)For I , , from (5.42) and (5.44), we find | I , | . k ˙Λ − σ − λ ) h ( H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i ) k L ∞ x L h × k ˙Λ − σ ) h ( B α ,i ∂ α v i − g a α ∂ v α ) k L x L h . ( k ˙Λ − λh ξ k L (Σ ) + k ˙Λ − λh ∇ v k L + ˙ D ) E k ˙Λ σ h ∇ v k L . (5.45)31ombining (5.44) with (5.45) yields | I | . k ˙Λ − λh ξ k L (Σ ) E ˙ D + ( k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∇ v k L ) E ˙ D + E ˙ D . (5.46)For I , from (4.1), it is easy to get | Z Σ ˙Λ − λh ( B α ,i ∂ α v i ) ˙Λ − λh v dS | . k ˙Λ − λh ( B α ,i ∂ α v i ) k L (Σ ) k ˙Λ − λh v k L (Σ ) . k ˙Λ σ h B α ,i k L (Σ ) k ˙Λ − σ − λh ∂ α v i k L (Σ ) k ˙Λ − λh v h k L (Σ ) . E k ˙Λ − σ − λh ∇ v k L k ˙Λ − λh ∇ v h k L . Similarly, one may see | Z Σ (cid:18) ˙Λ − λh (( B α ,i − B α ,i ) ∂ α v i ) ˙Λ − λh v + ˙Λ − λh ( B α ,i ∂ α v i ) ˙Λ − λh v (cid:19) dS | . E k ˙Λ − σ − λh ∇ v k L k ˙Λ − λh ∇ v h k L . Hence, from the condition 1 − λ < σ ≤ − λ , it follows | I | . E k ˙Λ − σ − λh ∇ v k L k ˙Λ − λh ∇ v k L . E ˙ D k ˙Λ − λh ∇ v k L . (5.47)Finally, to deal with I , we use (5.18) to deduce | I | . k ˙Λ − λh g k L k ˙Λ − λh v k L . ( E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D ) k ˙Λ − λh ∇ v k L . ( E k ˙Λ σ h ξ k L (Σ ) + E ˙ D ) k ˙Λ − λh ∇ v k L . (5.48)Therefore, substituting (5.46), (5.47) and (5.48) into (5.40) yields12 ddt (cid:18) Z Ω | ˙Λ − λh v | dx + Z Σ | ˙Λ − λh ξ | dS (cid:19) + ν Z Ω | D ( ˙Λ − λh v ) | dx . k ˙Λ − λh ξ k L (Σ ) E ˙ D + k ˙Λ − λh ∂ t v k L E ˙ D + E ˙ D + ( E k ˙Λ σ h ξ k L (Σ ) + E ˙ D ) k ˙Λ − λh ∇ v k L . Applying Young’s inequality and the Korn inequality implies (5.39), which is what we wantedto prove.Thanks to (5.1), Lemmas 5.5 and 5.6, we have
Lemma 5.7.
Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt (cid:20) k ˙Λ − λh v k L (Ω) + k ˙Λ − λh ξ k L (Σ ) + k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) + N +1 X i =0 ( k ∂ ih v k L (Ω) + k ∂ ih ξ k L (Σ ) (cid:21) + c ( k ˙Λ − λh ∇ v k L (Ω) + k∇ v k L (Ω) + k ˙Λ σ h ∇ v k L (Ω) + N +1 X i =0 k ∂ ih ∇ v k L (Ω) ) . ˙ D N +1 ( E N +1 ˙ D + E ˙ D N +1 ) + k ˙Λ − λh ξ k L (Σ ) E ˙ D + E k ˙Λ σ h ξ k L (Σ ) + k ˙Λ − λh ∂ t v k L E ˙ D . In this subsection, we derive energy estimates in terms of ∇ v as well as its horizontal deriva-tives. For this, we still use the linearized form (2.20) of the system (2.8) as in Section 5.2.Set ˚ E ( P ( ∂ h ) ∇ v ) def = kP ( ∂ h ) ∇ v k L + 4 kP ( ∂ h ) ∇ h · v h k L + 2 Z Ω ( P ( ∂ h ) ∂ v h · P ( ∂ h )( ∇ h v ) + P ( ∂ h ) ∂ v P ( ∂ h ) ∇ h · v h ) dx + 2 ν Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ H ( ξ ) dx − Z Ω 3 X j =1 P ( ∂ h ) ∂ v j P ( ∂ h )( H ( B αj +5 ,i ) ∂ α v i ) dx. The standard procedure for getting the energy estimates can be adopted to get
Lemma 5.8.
Let ( v, ξ ) be smooth solution to the system (2.8) , there holds that ν ddt ˚ E ( P ( ∂ h ) ∇ v ) + kP ( ∂ h ) ∂ t v k L = X j =1 J j , (5.49) where J := Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ ∂ t H ( ξ ) dx, J := − Z Ω P ( ∂ h ) ∂ QP ( ∂ h )( g a α ∂ t v α ) dx, J := 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h (( g a α ∂ v α ) − B α ,i ∂ α v i ) dx, J := − ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ (cid:18)g a α ∂ v α + ( H ( B α ,i ) − B α ,i ) ∂ α v i (cid:19) dx, J := − Z Ω P ( ∂ h ) QP ( ∂ h ) (cid:18) ∂ g a α ∂ t v α − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i (cid:19) dx, := Z Ω (cid:18) P ( ∂ h ) ∂ α Q P ( ∂ h )( B α ,i ∂ t v i ) − P ( ∂ h ) Q P ( ∂ h )( ∂ α B α ,i ∂ t v i ) (cid:19) dx, J := ν Z Ω 3 X j =1 (cid:18) P ( ∂ h ) ∂ ∂ α v j P ( ∂ h ) ∂ t ( H ( B αj +5 ,i ) v i ) + P ( ∂ h ) ∂ v j P ( ∂ h ) ∂ t ( ∂ α H ( B αj +5 ,i ) v i )+ P ( ∂ h ) ∂ t v j P ( ∂ h ) ∂ ( H ( B αj +5 ,i ) ∂ α v i ) (cid:19) dx, J := Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) g dx. Proof.
We multiply the i -th component of the momentum equations of (5.10) by P ( ∂ h ) ∂ t v i ,sum over i , and integrate over Ω to find that kP ( ∂ h ) ∂ t v k L + I + II = Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) g dx, (5.50)where I := R Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ q dx , II := − ν R Ω P ( ∂ h ) ∂ t v · ( ∇ · D ( P ( ∂ h ) v )) dx .Using q = Q + H ( ξ ) − ν ∇ h · v h + ν H ( B α ,i ) ∂ α v i , the integral I can be split into fourparts I = Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇Q dx + Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ H ( ξ ) dx − ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇∇ h · v h dx + ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ ( H ( B α ,i ) ∂ α v i ) dx =: X i =1 I i . (5.51)In view of the boundary conditions v | Σ b = 0 and Q| Σ = 0, integrating by parts in I shows I = − Z Ω P ( ∂ h ) Q ∇ · P ( ∂ h ) ∂ t v dx = Z Ω P ( ∂ h ) Q P ( ∂ h )( g a α ∂ ∂ t v α ) dx − Z Ω P ( ∂ h ) Q P ( ∂ h ) (cid:18) B α ,i ∂ α ∂ t v i − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i (cid:19) dx, (5.52)where we used the fact ∇ · ∂ t v = − g a α ∂ ∂ t v α + ( B α ,i ∂ α ∂ t v i − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i ), whichcomes from the divergence-free condition (2.10).Applying integration by parts ensures Z Ω P ( ∂ h ) Q P ( ∂ h )( g a α ∂ ∂ t v α ) dx = − Z Ω P ( ∂ h ) ∂ Q P ( ∂ h )( g a α ∂ t v α ) dx − Z Ω P ( ∂ h ) Q P ( ∂ h )( ∂ g a α ∂ t v α ) dx, and − Z Ω P ( ∂ h ) Q P ( ∂ h )( B α ,i ∂ α ∂ t v i ) dx = Z Ω P ( ∂ h ) ∂ α Q P ( ∂ h )( B α ,i ∂ t v i ) dx + Z Ω P ( ∂ h ) Q P ( ∂ h )( ∂ α B α ,i ∂ t v i ) dx, I = − Z Ω P ( ∂ h ) ∂ Q P ( ∂ h )( g a α ∂ t v α ) dx + Z Ω P ( ∂ h ) ∂ α Q P ( ∂ h )( B α ,i ∂ t v i ) dx − Z Ω P ( ∂ h ) Q P ( ∂ h ) (cid:18) ∂ g a α ∂ t v α − ∂ α B α ,i ∂ t v i − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i (cid:19) dx. (5.53)For I , a direct computation gives rise to I = Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ H ( ξ ) dx = ddt Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ H ( ξ ) dx − Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ H ( v ) dx. (5.54)While for I , it is easy to get I = − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ∇ h · v h dx − ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ∇ h · v h dx = − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ∇ h · v h dx + ν ddt kP ( ∂ h ) ∇ h · v h k L . (5.55)Thus, plugging (5.53)-(5.55) into (5.51), one has I = ddt (cid:18) Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ H ( ξ ) dx + ν kP ( ∂ h ) ∇ h · v h k L (cid:19) − Z Ω P ( ∂ h ) v · P ( ∂ h ) ∇ H ( v ) dx − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ∇ h · v h dx − Z Ω P ( ∂ h ) ∂ QP ( ∂ h )( g a α ∂ t v α ) dx − Z Ω P ( ∂ h ) Q P ( ∂ h ) (cid:18) ∂ g a α ∂ t v α − ∂ α B α ,i ∂ t v i − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i (cid:19) dx + Z Ω P ( ∂ h ) ∂ α Q P ( ∂ h )( B α ,i ∂ t v i ) dx + ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∇ ( H ( B α ,i ) ∂ α v i ) dx. For II , we split it into two parts: II = − ν Z Ω P ( ∂ h ) ∂ t v · ∆( P ( ∂ h ) v ) dx − ν Z Ω P ( ∂ h ) ∂ t v · ∇ ( ∇ · P ( ∂ h ) v ) dx =: II + II . Due to the second equation in (5.10), one can see II = − ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h )( −∇ ( g a α ∂ v α ) + ∇ ( B α ,i ∂ α v i )) dx. (5.56)The calculation of II should be more delicate. In fact, it can be immediately verified that II = − ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h )∆ h v dx − ν Z Ω P ( ∂ h ) ∂ t v · P ( ∂ h ) ∂ v dx = ν ddt Z Ω |P ( ∂ h ) ∂ h v | dx + X j =1 II ,j , (5.57)35here II ,j := − ν R Ω P ( ∂ h ) ∂ t v j ∂ v j dx with j = 1 , , II , , using the equation ∂ v = G − ∇ h · v h + H ( B α ,i ) ∂ α v i in (2.18) produces that II , = − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ G dx − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ( −∇ h · v h + H ( B α ,i ) ∂ α v i ) dx. (5.58)For the first integral of the right hand side in (5.58), thanks to the boundary conditions v | Σ b = 0 and G | Σ = 0, integrating by parts yields − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ G dx = ν Z Ω P ( ∂ h ) ∂ t ∂ v P ( ∂ h ) G dx = ν Z Ω P ( ∂ h ) ∂ t ∂ v P ( ∂ h )( ∂ v + ∇ h · v h − H ( B α ,i ) ∂ α v i ) dx = ν ddt Z Ω |P ( ∂ h ) ∂ v | dx + ν ddt Z Ω P ( ∂ h ) ∂ v P ( ∂ h )( ∇ h · v h − H ( B α ,i ) ∂ α v i ) dx − ν Z Ω P ( ∂ h ) ∂ v P ( ∂ h ) ∂ t ( ∇ h · v h − H ( B α ,i ) ∂ α v i ) dx, which follows − ν Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ G dx = − ν Z Ω P ( ∂ h ) ∂ ∂ α v P ( ∂ h ) ∂ t ( H ( B α ,i ) v i ) dx − ν Z Ω P ( ∂ h ) ∂ v P ( ∂ h ) ∂ t ( ∂ α H ( B α ,i ) v i ) dx + ν Z Ω P ( ∂ h ) ∇ h ∂ v · P ( ∂ h ) ∂ t v h dx + ν ddt Z Ω (cid:18) |P ( ∂ h ) ∂ v | + 2 P ( ∂ h ) ∂ t ∂ v P ( ∂ h )( ∇ h · v h − H ( B α ,i ) ∂ α v i ) (cid:19) dx. (5.59)Plugging (5.59) into (5.58), we get II , = ν ddt (cid:20) kP ( ∂ h ) ∂ v k L + 2 Z Ω P ( ∂ h ) ∂ v P ( ∂ h )( ∇ h · v h − H ( B α ,i ) ∂ α v i ) dx (cid:21) + ν (cid:18) Z Ω P ( ∂ h ) ∇ h ∂ v · P ( ∂ h ) ∂ t v h dx + Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ∇ h · v h dx (cid:19) − ν (cid:18) Z Ω P ( ∂ h ) ∂ ∂ α v P ( ∂ h ) ∂ t ( H ( B α ,i ) v i ) dx + Z Ω P ( ∂ h ) ∂ v P ( ∂ h ) ∂ t ( ∂ α H ( B α ,i ) v i ) dx + Z Ω P ( ∂ h ) ∂ t v P ( ∂ h ) ∂ ( H ( B α ,i ) ∂ α v i ) dx (cid:19) . (5.60)Following our analysis of the term II , , we use ∂ v = G − ∂ v + H ( B α ,i ) ∂ α v i , ∂ v =36 − ∂ v + H ( B α ,i ) ∂ α v i in (2.18) to II , , II , respectively to obtain II , + II , = ν ddt (cid:20) kP ( ∂ h ) ∂ v h k L + 2 Z Ω P ( ∂ h ) ∂ v h · P ( ∂ h )( ∇ h v ) dx − Z Ω 3 X β =2 P ( ∂ h ) ∂ v β P ( ∂ h )( H ( B αβ +5 ,i ) ∂ α v i ) dx (cid:21) − ν X β =2 Z Ω (cid:18) P ( ∂ h ) ∂ ∂ α v β P ( ∂ h ) ∂ t ( H ( B αβ +5 ,i ) v i ) + P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ t ( ∂ α H ( B αβ +5 ,i ) v i )+ P ( ∂ h ) ∂ t v β P ( ∂ h ) ∂ ( H ( B αβ +5 ,i ) ∂ α v i ) (cid:19) dx + ν Z Ω P ( ∂ h ) ∂ ∇ h · v h P ( ∂ h ) ∂ t v dx + ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ∂ v dx. (5.61)Combining (5.60) with (5.61) leads to II , + II , + II , = ν ddt (cid:20) kP ( ∂ h ) ∂ v k L + 2 Z Ω P ( ∂ h ) ∂ v h · P ( ∂ h )( ∇ h v ) dx + 2 Z Ω P ( ∂ h ) ∂ v P ( ∂ h ) ∇ h · v h dx − Z Ω 3 X j =1 P ( ∂ h ) ∂ v j P ( ∂ h )( H ( B αj +5 ,i ) ∂ α v i ) dx (cid:21) − ν Z Ω 3 X j =1 (cid:18) P ( ∂ h ) ∂ ∂ α v j P ( ∂ h ) ∂ t ( H ( B αj +5 ,i ) v i ) + P ( ∂ h ) ∂ v j P ( ∂ h ) ∂ t ( ∂ α H ( B αj +5 ,i ) v i )+ P ( ∂ h ) ∂ t v j ∂ ( H ( B αj +5 ,i ) ∂ α v i ) (cid:19) dx + 2 ν Z Ω P ( ∂ h ) ∂ ∇ h · v h P ( ∂ h ) ∂ t v dx + 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ∂ v dx. (5.62)Using the incompressible condition ∂ v = −∇ h · v h − g a α ∂ v α + B α ,i ∂ α v i in (2.10), we infer2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ∂ v dx = 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ( −∇ h · v h ) dx + 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ( − g a α ∂ v α + B α ,i ∂ α v i ) dx = ν ddt kP ( ∂ h ) ∇ h · v h k L + 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ( − g a α ∂ v α + B α ,i ∂ α v i ) dx, II = ν ddt (cid:20) kP ( ∂ h ) ∇ v k L + kP ( ∂ h ) ∇ h · v h k L + 2 Z Ω P ( ∂ h ) ∂ v h · P ( ∂ h )( ∇ h v ) dx + 2 Z Ω P ( ∂ h ) ∂ v P ( ∂ h ) ∇ h · v h dx − Z Ω 3 X j =1 P ( ∂ h ) ∂ v j P ( ∂ h )( H ( B αj +5 ,i ) ∂ α v i ) dx (cid:21) + 2 ν Z Ω P ( ∂ h ) ∂ ∇ h · v h P ( ∂ h ) ∂ t v dx − ν Z Ω 3 X j =1 (cid:18) P ( ∂ h ) ∂ ∂ α v j P ( ∂ h ) ∂ t ( H ( B αj +5 ,i ) v i )+ P ( ∂ h ) ∂ v j P ( ∂ h ) ∂ t ( ∂ α H ( B αj +5 ,i ) v i ) + P ( ∂ h ) ∂ t v j ∂ ( H ( B αj +5 ,i ) ∂ α v i ) (cid:19) dx + 2 ν Z Ω P ( ∂ h ) ∂ t v h · P ( ∂ h ) ∇ h ( − g a α ∂ v α + B α ,i ∂ α v i ) dx. (5.63)Combining (5.50), (5.53), (5.56), (5.63) gives rise to I + II = I + II + II = ν ddt ˚ E ( P ( ∂ h ) ∇ v ) − X j =1 J j . (5.64)Inserting (5.64) into (5.50) yields (5.49), which is the desired result. k ˙Λ σ h ∇ v k L Lemma 5.9.
Under the assumption of Lemma 5.8, if E ( t ) ≤ for all the existence times t , then there holds ν ddt ˚ E ( ˙Λ σ h ∇ v ) + k ˙Λ σ h ∂ t v k L . k ˙Λ σ h ∇ v k L + E ˙ D . (5.65) Proof.
Taking P ( ∂ h ) = ˙Λ σ h in (5.49), we will estimate all the integrals in the right hand sideof (5.49) one by one.For J , we directly bound it by | J | . k ˙Λ σ h v k L k ˙Λ σ h ∇ ∂ t H ( ξ ) k L . k ˙Λ σ h v k L k ˙Λ σ h ∇ v k L . k ˙Λ σ h ∇ v k L , While for J , the product law (4.1) ensures that | J | . k ˙Λ σ +1 h ∂ Qk L k ˙Λ σ − h ( g a α ∂ t v α ) k L . k ˙Λ σ +1 h ∂ Qk L k g a α k L ∞ x L h k ˙Λ σ h ∂ t v h k L . Since Q = q − H ( ξ ) + 2 ν ∇ h · v h − ν H ( B α ,i ) ∂ α v i , we have k ˙Λ σ h ∂ Qk L . k ˙Λ σ h ∂ q k L + k ˙Λ σ h ∂ H ( ξ ) k L + k ˙Λ σ h ∂ ∇ h · v h k L + k ˙Λ σ h ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ˙Λ σ h ∂ q k L + k ˙Λ σ h ξ k H (Σ ) + k ˙Λ σ h ∂ ∇ h · v h k L + k ˙Λ σ h ∂ ( H ( B α ,i ) ∂ α v i ) k L . ˙ D , (5.66)38hich, along with k ( f a , f a ) k L ∞ x L h . E , follows that | J | . E ˙ D . For J , one has | J | . k ˙Λ σ h ∂ t v k L ( k ˙Λ σ h ∇ ( g a α ∂ v α ) k L + k ˙Λ σ h ∇ ( B α ,i ∂ α v i ) k L ) , Notice that k ˙Λ σ h ∇ ( g a α ∂ v α ) k L . k ˙Λ σ h ( g a α ∇ ∂ v α ) k L + k ˙Λ σ h ( ∇ g a α ∂ v α ) k L . k ˙Λ (1+ σ ) / h g a α k L ∞ x ( L h ) k ˙Λ (1+ σ ) / h ∇ ∂ v α k L + kk ˙Λ (1+ σ ) / h ∇ g a α k L k ˙Λ (1+ σ ) / h ∂ v α k L ∞ x L h . E ˙ D . Similarly, it can be checked that k ˙Λ σ h ∇ ( B α ,i ∂ α v i ) k L ) . E ˙ D , which ensures that | J | . E ˙ D . The same proof remains valid for J , and one has | J | . E ˙ D . We control J by | J | . k ˙Λ σ +1 h Qk L ∞ x L h ( k ˙Λ σ − h ( ∂ f a ∂ t v + ∂ f a ∂ t v ) k L x L h + k ˙Λ σ − h ( − ∂ t f a ∂ v − ∂ t f a ∂ v + ∂ t B α ,i ∂ α v i ) k L x L h ) . k ˙Λ σ +1 h ∂ Qk L k ( ∂ f a , ∂ f a , ∂ t f a , ∂ t f a , ∂ t B α ,i ) k L k ˙Λ σ h ( ∂ t v h , ∂ v h , ∂ α v i ) k L , which along with (5.66) yields that | J | . E ˙ D . For J , in the same manner, it can be obtained that | J | . k ˙Λ σ h ∂ α Qk L ∞ x L h k ˙Λ σ h ( B α ,i ∂ t v i ) k L x L h + k ˙Λ σ +1 h Qk L ∞ x L h k ˙Λ σ − h ( ∂ α B α ,i ∂ t v i ) k L x L h . k ˙Λ σ +1 h ∂ Qk L ( k ˙Λ ( σ +1) / h B α ,i k L k ˙Λ ( σ +1) / h ∂ t v i k L + k ∂ α B α ,i k L k ˙Λ σ h ∂ t v i k L ) , which implies | J | . E ˙ D . (5.67)For J , we deal with the integral R Ω ˙Λ σ h ∂ ∂ α v j ˙Λ σ h ∂ t ( H ( B αj +5 ,i ) v i ) dx in it by | Z Ω ˙Λ σ h ∂ ∂ α v j ˙Λ σ h ∂ t ( H ( B αj +5 ,i ) v i ) dx | . k ˙Λ σ h ∂ ∂ α v j k L k ˙Λ σ h ∂ t ( H ( B αj +5 ,i ) v i ) k L . k ˙Λ σ h ∂ ∂ α v j k L ( k ˙Λ σ h ( H ( B αj +5 ,i ) ∂ t v i ) k L + k ˙Λ σ h ( ∂ t H ( B αj +5 ,i ) v i ) k L ) , | Z Ω ˙Λ σ h ∂ ∂ α v j ˙Λ σ h ∂ t ( H ( B αj +5 ,i ) v i ) dx | . k ˙Λ σ h ∂ ∂ α v j k L ( k ˙Λ ( σ +1) / h H ( B αj +5 ,i ) k L ∞ x ( L h ) k ˙Λ ( σ +1) / h ∂ t v i k L + k ˙Λ ( σ +1) / h ∂ t H ( B αj +5 ,i ) k L k ˙Λ ( σ +1) / h v i k L ∞ x L h ) . ˙ D ( E ˙ D + ˙ D ˙ E ) . E ˙ D , (5.68)where we have used that k ˙Λ σ h ∂ ∂ α v j k L . ˙ D , k ˙Λ ( σ +1) / h H ( B αj +5 ,i ) k L ∞ x ( L h ) . k ˙Λ ( σ +1) / h H ( B αj +5 ,i ) k H . E , k ˙Λ ( σ +1) / h ∂ t H ( B αj +5 ,i ) k L . ˙ D , k ˙Λ ( σ +1) / h v i k L ∞ x L h . k ˙Λ ( σ +1) / h ∇ v k L . ˙ E . The following results may be proved in much the same way as in the proof of (5.68) | Z Ω ˙Λ σ h ∂ v j ˙Λ σ h ∂ t ( ∂ α H ( B αj +5 ,i ) v i ) dx | . E ˙ D , | Z Ω ˙Λ σ h ∂ t v j ˙Λ σ h ∂ ( H ( B αj +5 ,i ) ∂ α v i ) dx | . E ˙ D . Hence, it follows | J | . E ˙ D . Finally, making directly use of (5.18) implies | J | . k ˙Λ σ h ∂ t v k L k ˙Λ σ h g k L . E ˙ D . Therefore, we conclude that X j =1 | J j | . k ˙Λ σ h ∇ v k L + E ˙ D , which leads to (5.65), and we complete the proof of Lemma 5.9. k ∂ N − h ∇ v k L ∞ t ( L ) Lemma 5.10.
Let N ≥ , under the assumption of Lemma 5.8, if E ( t ) ≤ for all theexistence times t , then there holds ν ddt ˚ E ( ∂ N − h ∇ v ) + k ∂ N − h ∂ t v k L . k ∂ N − h ∇ v k L + ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D . (5.69) Proof.
We estimate all the integrals in the right hand side of (5.49), where we take P ( ∂ h ) = ∂ N − h . We first estimate J to get | J | . k ∂ N − h v k L k ∂ N − h ∇ ∂ t H ( ξ ) k L . k ∂ N − h v k L k ∂ N − h ∇ v k L . k ∂ N − h ∇ v k L . | J | . k ∂ N − h ∂ Qk L k ∂ N − h ( g a α ∂ t v α ) k L . k ∂ N − h ∂ Qk L ( k ∂ N − h g a α k L ∞ x L h k ∂ t v h k L x L ∞ h + k g a α k L ∞ k ∂ N − h ∂ t v h k L ) . k ∂ N − h ∂ Qk L ( E ˙ D N + E N ˙ D ) , Since Q = q − H ( ξ ) + 2 ν ∇ h · v h − ν H ( B α ,i ) ∂ α v i , we have k ∂ N − h ∂ Qk L . k ∂ N − h ∂ q k L + k ∂ N − h ∂ H ( ξ ) k L + k ∂ N − h ∂ ∇ h · v h k L + k ∂ N − h ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ∂ N − h ∂ q k L + k ∂ N − h ξ k H (Σ ) + k ∂ N − h ∂ ∇ h · v h k L + k ∂ N − h ∂ ( H ( B α ,i ) ∂ α v i ) k L . ˙ D N + E N ˙ D , (5.70)where we used the estimate (4.10) k ∂ N − h ( B α ,i ∂ α v i ) k H . E ˙ D N + E N ˙ D and k ∂ N − h ξ k H (Σ ) . k ∂ N − h q k H (Σ ) + k ∂ N − h ∇ h · v h k H (Σ ) + k ∂ N − h ( B α ,i ∂ α v i ) k H (Σ ) . ˙ D N + E N ˙ D . Hence, one has | J | . ( ˙ D N + E N ˙ D )( E ˙ D N + E N ˙ D ) . ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D . For J , we get | J | . k ∂ N − h ∂ t v k L ( k ∂ N − h ∇ ( g a α ∂ v α ) k L + k ∂ N − h ∇ ( B α ,i ∂ α v i ) k L ) , which along with Lemma 4.6 leads to | J | . ˙ D N ( E ˙ D N + E N ˙ D ) . This estimate also holds true for J : | J | . ˙ D N ( E ˙ D N + E N ˙ D ) . For J , we first get | Z Ω ∂ N − h Q ∂ N − h ( ∂ g a α ∂ t v α ) dx | . k ∂ N − h Qk L ∞ x L h k ∂ N − h ( ∂ g a α ∂ t v α ) k L x L h . k ∂ N − h ∂ Qk L ( k ∂ N − h ∂ g a α k L k ∂ t v α k L x L ∞ h + k ∂ N − h ∂ t v α k L k ∂ g a α k L x L ∞ h ) , which along with the estimate (5.70) yields that | Z Ω ∂ N − h Q ∂ N − h ( ∂ g a α ∂ t v α ) dx | . ( ˙ D N + E N ˙ D )( E ˙ D N + E N ˙ D ) . ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D , (5.71)41epeating the argument in the proof of (5.71), one can immediately obtain | Z Ω ∂ N − h Q ∂ N − h ( − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i ) dx | . k ∂ Nh Qk L k ∂ N − h ( − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i ) k L . k ∂ Nh Qk L ( k ∂ N − h ( ∂ t g a α , ∂ t B α ,i ) k L k ( ∂ v h , ∂ α v i ) k L ∞ + k ∂ N − h ( ∂ v h , ∂ α v i ) k L k ( ∂ t f a , ∂ t f a , ∂ t B α ,i ) k L ∞ ) . ( ˙ D N + E N ˙ D )( E ˙ D N + E N ˙ D ) . ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D . Therefore, we have | J | . ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D . The same conclusion can be drawn for J , J , and J | J | . ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D , | J | + | J | . ˙ D N ( E ˙ D N + E N ˙ D ) . Therefore, we obtain X j =1 | J j | . k ∂ N − h ∇ v k L + ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D , and then reach (5.69), which completes the proof of Lemma 5.10.With Lemmas 5.9 and 5.10 in hand, we have Lemma 5.11.
Let N ≥ , under the assumption of Lemma 5.1, if E ( t ) ≤ for all theexistence times t , then there holds ddt (cid:18) ˚ E ( ˙Λ σ h ∇ v ) + N − X k =1 ˚ E ( ∂ kh ∇ v ) (cid:19) + c ( k ˙Λ σ h ∂ t v k L + N − X k =1 k ∂ kh ∂ t v k L ) ≤ C ( k ˙Λ σ h ∇ v k L + N − X k =1 k ∂ kh ∇ v k L + E N ˙ D ˙ D N + E ˙ D N + E N ˙ D ) . Thanks to Lemmas 5.5 and 5.11, for any small positive constant δ ≤ min { , c C } (whichwill be determined later), we have Lemma 5.12.
Let N ≥ , under the assumption of Lemma 5.1, if E ( t ) ≤ for all theexistence times t , then there holds ddt (cid:20) k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) + N X i =1 ( k ∂ ih v k L (Ω) + k ∂ ih ξ k L (Σ ) + δ (cid:18) ˚ E ( ˙Λ σ h ∇ v ) + N − X k =1 ˚ E ( ∂ kh ∇ v ) (cid:19)(cid:21) + c k ˙Λ σ h ∇ v k L (Ω) + N X i =1 k ∂ ih ∇ v k L (Ω) ) + δ c ( k ˙Λ σ h ∂ t v k L + N − X k =1 k ∂ kh ∂ t v k L ) . E N ˙ D ˙ D N + E ˙ D N + E N ˙ D . .3.3 Estimate of k ˙Λ − λh ∇ v k L ∞ t ( L ) Lemma 5.13.
Under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt ˚ E ( ˙Λ − λh ∇ v ) + c k ˙Λ − λh ∂ t v k L ≤ C (cid:18) k ˙Λ − λh ∇ v k L + E ˙ D k ˙Λ − λh ξ k H (Σ ) + E k ˙Λ σ h ξ k L (Σ ) + E ˙ D (cid:19) . (5.72) Proof.
Taking P ( ∂ h ) = ˙Λ − λh in (5.49), we will estimate all the integrals in the right hand sideof (5.49).For J := R Ω ˙Λ − λh v · ∇ ˙Λ − λh ∂ t H ( ξ ) dx , it is easy to see | J | . k ˙Λ − λh v k L k ˙Λ − λh ∇ ∂ t H ( ξ ) k L . k ˙Λ − λh v k L k ˙Λ − λh ∇ v k L . k ˙Λ − λh ∇ v k L . While the product law (4.1) ensures that | J | . k ˙Λ − λh ∂ Qk L k ˙Λ − λh ( g a α ∂ t v α ) k L . k ˙Λ − λh ∂ Qk L k ˙Λ − λ − σ h ( g a α ) k L ∞ x L h k ˙Λ σ h ∂ t v h k L . Thanks to Q = q − H ( ξ ) + 2 ν ∇ h · v h − ν H ( B α ,i ) ∂ α v i , we have k ˙Λ − λh ∂ Qk L . k ˙Λ − λh ∂ q k L + k ˙Λ − λh ∂ H ( ξ ) k L + k ˙Λ − λh ∂ ∇ h · v h k L + k ˙Λ − λh ∂ ( H ( B α ,i ) ∂ α v i ) k L . According to the equation of ∂ q , we get from (5.18) that k ˙Λ − λh ∂ q k L . k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + k ˙Λ − λh g k L + k ˙Λ − λh e g k L . k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + E k ˙Λ − σ − λh ξ k L (Σ ) + E ˙ D , which, together with k ˙Λ − λh ∂ ( H ( B α ,i ) ∂ α v i ) k L . k ˙Λ σ h v k H k ˙Λ − σ − λh B α ,i k H . E ˙ D (where we used the fact 2 − λ − σ ≥ σ since σ ≤ − λ ), follows that k ˙Λ − λh ∂ Qk L . k ˙Λ − λh ξ k H (Σ ) + E k ˙Λ − σ − λh ξ k L (Σ ) + k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + E ˙ D . k ˙Λ − λh ξ k H (Σ ) + k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + E ˙ D , (5.73)where the interpolation inequality k ˙Λ − σ − λh ξ k L (Σ ) . k ˙Λ − λh ξ k L (Σ ) + k ˙Λ σ h ξ k L (Σ ) isused in the last inequality. Therefore, we obtain, from k ˙Λ − λ − σ h ( f a , f a ) k L ∞ x L h . k ˙Λ − λ − σ h ˙Λ σ − h ( f a , f a ) k H . E − λ − σ ≥ σ ≤ − λ ), that | J | . E ˙ D (cid:18) k ˙Λ − λh ξ k H (Σ ) + k ˙Λ − λh ∂ t v k L + k ˙Λ − λh ∂ h ∇ v k L + E ˙ D (cid:19) . E ˙ D k ˙Λ − λh ξ k H (Σ ) + E ˙ D k ˙Λ − λh ∂ t v k L + E ˙ D k ˙Λ − λh ∂ h ∇ v k L + E ˙ D . For J , one has | J | . k ˙Λ − λh ∂ t v k L ( k ˙Λ − λh ∇ ( g a α ∂ v α ) k L + k ˙Λ − λh ∇ ( B α ,i ∂ α v i ) k L ) , Notice that k ˙Λ − λh ∇ ( g a α ∂ v α ) k L . k ˙Λ − λh ( g a α ∇ ∂ v α ) k L + k ˙Λ − λh ( ∇ g a α ∂ v α ) k L . k ˙Λ σ − h g a α k L ∞ x L h k ˙Λ − σ − λh ∇ ∂ v α k L + kk ˙Λ σ − h ∇ g a α k L k ˙Λ − σ − λh ∂ v h k L ∞ x − L h . E ˙ D . Similarly, we have k ˙Λ − λh ∇ ( B α ,i ∂ α v i ) k L ) . E ˙ D , which ensures that | J | . k ˙Λ − λh ∂ t v k L E ˙ D . (5.74)Similar to the proof of (5.74), one can obtain | J | . k ˙Λ − λh ∂ t v k L E ˙ D . We control J by | J | . k ˙Λ − λh Qk L ∞ x L h ( k ˙Λ − λh ( ∂ g a α ∂ t v α − ∂ t g a α ∂ v α + ∂ t B α ,i ∂ α v i ) k L x L h ) . k ˙Λ − λh ∂ Qk L (cid:18) k ˙Λ − σ − λh ∂ g a α k L ∞ x L h k ˙Λ σ h ∂ t v h k L + k ˙Λ − σ − λh ∂ t g a α k L k ˙Λ σ h ∂ v h k L ∞ x L h + k ˙Λ − σ − λh ∂ t B α ,i k L k ˙Λ σ − h ∂ h v k L ∞ x L h (cid:19) . k ˙Λ − λh ∂ Qk L E ˙ D , which along with (5.73) follows that | J | . E ˙ D k ˙Λ − λh ξ k H (Σ ) + E ˙ D k ˙Λ − λh ∂ t v k L + E ˙ D ( k ˙Λ − λh ∂ h ∇ v k L + ˙ D ) . Similarly, for J , we have | J | . E ˙ D k ˙Λ − λh ξ k H (Σ ) + E ˙ D k ˙Λ − λh ∂ t v k L + E ˙ D ( k ˙Λ − λh ∂ h ∇ v k L + ˙ D ) . J , we bound the integral R Ω ˙Λ − λh ∂ ∂ α v j ˙Λ − λh ∂ t ( H ( B αj +5 ,i ) v i ) dx in it by | Z Ω ˙Λ − λh ∂ ∂ α v j ˙Λ − λh ∂ t ( H ( B αj +5 ,i ) v i ) dx | . k ˙Λ σ − h ∂ ∂ α v j k L ∞ x L h k ˙Λ − σ − λh ∂ t ( H ( B αj +5 ,i ) v i ) k L x L h . k ˙Λ σ h ∂ v k H ( k ˙Λ − σ − λh ( H ( B αj +5 ,i ) ∂ t v i ) k L x L h + k ˙Λ − σ − λh ( ∂ t H ( B αj +5 ,i ) v i ) k L x L h ) , which, together with 2 − σ − λ ≥ − λ and k ˙Λ − σ − λh ( H ( B αj +5 ,i ) ∂ t v i ) k L x L h . k ˙Λ − σ − λh H ( B αj +5 ,i ) k L k ˙Λ − λh ∂ t v k L . E k ˙Λ − λh ∂ t v k L , k ˙Λ − σ − λh ( ∂ t H ( B αj +5 ,i ) v i ) k L x L h . k ˙Λ σ h ∂ t H ( B αj +5 ,i ) k L k ˙Λ − σ − λh v k L . ˙ D E , follows that | Z Ω ∂ ∂ α v j ∂ t ( H ( B αj +5 ,i ) v i ) dx | . ˙ D E k ˙Λ − λh ∂ t v k L + ˙ D E . Similarly, we have | Z Ω ˙Λ σ h ∂ v j ˙Λ σ h ∂ t ( ∂ α H ( B αj +5 ,i ) v i ) dx | . ˙ D E k ˙Λ − λh ∂ t v k L + ˙ D E , | Z Ω ˙Λ σ h ∂ t v j ˙Λ σ h ∂ ( H ( B αj +5 ,i ) ∂ α v i ) dx | . ˙ D E k ˙Λ − λh ∂ t v k L + ˙ D E . Hence, we have | J | . ˙ D E k ˙Λ − λh ∂ t v k L + ˙ D E . Finally, due to (5.18), we find | J | . k ˙Λ − λh ∂ t v k L k ˙Λ − λh g k L . E k ˙Λ − λh ∂ t v k L ( k ˙Λ − σ − λh ξ k L (Σ ) + ˙ D ) . Therefore, we obtain that X j =1 | J j | . k ˙Λ − λh ∇ v k L + E ˙ D k ˙Λ − λh ξ k H (Σ ) + E ˙ D k ˙Λ − λh ∂ t v k L + E ˙ D k ˙Λ − λh ∂ h ∇ v k L + E ˙ D + E k ˙Λ − λh ∂ t v k L k ˙Λ − σ − λh ξ k L (Σ ) , which together with Young’s inequality applied leads to (5.72), and the desired result isproved. k∇ v k L ∞ t ( L ) Lemma 5.14.
Under the assumption of Lemma 5.8, if E ( t ) ≤ for all the existence times t , then there holds ddt ˚ E ( ∇ v ) + c k ∂ t v k L . k∇ v k L + E ˙ D k ˙Λ − σ ) h ξ k H (Σ ) + E ˙ D + E k ˙Λ σ h ξ k H (Σ ) . (5.75)45ith Lemmas 5.11, 5.13 and 5.14 in hand, we have Lemma 5.15.
Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt (cid:18) ˚ E ( ˙Λ − λh ∇ v ) + ˚ E ( ˙Λ σ h ∇ v ) + N X k =1 ˚ E ( ∂ kh ∇ v ) (cid:19) + c (cid:18) k ˙Λ − λh ∂ t v k L + k ˙Λ σ h ∂ t v k L + N X k =0 k ∂ kh ∂ t v k L (cid:19) ≤ C (cid:18) k ˙Λ − λh ∇ v k L + k ˙Λ σ h ∇ v k L + N X k =0 k ∂ kh ∇ v k L + E N +1 ˙ D ˙ D N +1 + E ˙ D N +1 + E N +1 ˙ D + E ˙ D k ˙Λ − λh ξ k H (Σ ) + E k ˙Λ σ h ξ k H (Σ ) (cid:19) . (5.76)Thanks to Lemmas 5.7 and 5.15, for any small positive constant δ ≤ min { , c C } (whichwill be determined later on), we have Lemma 5.16.
Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt b E N +1 ,tan,δ + b D N +1 ,tan,δ ≤ C (cid:18) E N +1 ˙ D ˙ D N +1 + E ˙ D N +1 + E N +1 ˙ D + E ˙ D k ˙Λ − λh ξ k H (Σ ) + E k ˙Λ σ h ξ k H (Σ ) (cid:19) + C δ − E ˙ D . (5.77) with b E N +1 ,tan,δ := k ˙Λ − λh v k L (Ω) + k ˙Λ − λh ξ k L (Σ ) + k ˙Λ σ h v k L (Ω) + k ˙Λ σ h ξ k L (Σ ) + N +1 X i =0 ( k ∂ ih v k L (Ω) + k ∂ ih ξ k L (Σ ) + δ (cid:18) ˚ E ( ˙Λ − λh ∇ v ) + ˚ E ( ˙Λ σ h ∇ v ) + N X k =0 ˚ E ( ∂ kh ∇ v ) (cid:19) (5.78) and b D N +1 ,tan,δ := c k ˙Λ − λh ∇ v k L (Ω) + k ˙Λ σ h ∇ v k L (Ω) + N +1 X i =0 k ∂ ih ∇ v k L (Ω) )+ δ c (cid:18) k ˙Λ − λh ∂ t v k L + k ˙Λ σ h ∂ t v k L + N X k =0 k ∂ kh ∂ t v k L (cid:19) . (5.79) In this section, we will investigate the dissipative estimates in terms of ∇ v and ∇ q as wellas their horizontal derivatives in the L framework, which is based on the Stokes estimates.We first recall the classical regularity theory for the Stokes problem with mixed boundaryconditions on the boundary stated in [2]. 46 emma 6.1 ([2]) . Suppose v , q solve −∇ · D ( v ) + ∇ q = φ ∈ H r − (Ω) , div v = ψ ∈ H r − (Ω) , ( q I − D ( v )) e = k ∈ H r − (Σ ) ,v | Σ b = 0 . Then, for r ≥ , k v k H r (Ω) + k q k H r − (Ω) . k φ k H r − (Ω) + k ψ k H r − (Ω) + k k k H r − (Σ ) . Denote the cut-off function near the bottom by the smooth function χ in ( x ) = , x ∈ [ − b, − b );smooth ∈ [0 , , r ∈ ( − b, − b ];0 , x ∈ ( − b, , with | d n dr n χ in ( r ) | . b − n for any n ∈ N , and the cut-off function near the free surface by thesmooth function χ f ( x ) = , x ∈ ( − b, ∈ [0 , , r ∈ ( − b, − b ];0 , x ∈ [ − b, − b ) , with | d n dr n χ f ( r ) | . b − n for any n ∈ N . Notice that χ f (Supp χ ′ in ) ≡
1. We also denoteΩ in := Ω χ in with the measure χ in dx and Ω f := Ω χ f with the measure χ f dx .Since there are different boundary conditions on the top boundary Σ and the bottomboundary Σ b in (2.20), we need to deal with these two different situations separately. Define v in def = v χ in , q in def = q χ in , then, according to (2.20), ( v in , q in ) solves − ν ∇ · D ( v in ) + ∇ q in = − ∂ t v χ in + g in , div v in = ψ in , ( q in I − D ( v in )) e | Σ b = 0 ,v in | Σ = 0 , where g in ∼ gχ in + χ ′ in ∇ v + χ ′′ in v + χ ′ in q, ψ in ∼ ψχ in + χ ′ in v , ψ := − g a α ∂ v α + B α ,i ∂ α v i . Thanks to Lemma 6.1, we obtain kP ( ∂ h ) v in k H (Ω) + kP ( ∂ h ) q in k H (Ω) . kP ( ∂ h ) ∂ t v χ in k L (Ω) + kP ( ∂ h ) g in k L (Ω) + kP ( ∂ h ) ψ in k H (Ω) , which implies kP ( ∂ h ) v k H (Ω in ) + kP ( ∂ h ) q k H (Ω in ) . kP ( ∂ h ) v k H (Ω f ) + kP ( ∂ h ) q k L (Ω f ) + kP ( ∂ h ) ∂ t v k L (Ω in ) + kP ( ∂ h ) g k L (Ω) + kP ( ∂ h ) ψ k H (Ω) , (6.1)where we used the fact χ f (Supp χ ′ in ) ≡
1. 47 .2 Estimates near the free boundary
Consider the equation of ∂ v β (with β = 2 , − ν∂ v β + ∂ β q = − ∂ t v β + ν ∆ h v β + g β + ν e g ββ . (6.2) Lemma 6.2.
Let ( v, ξ ) be smooth solution to the system (2.8) , then there holds that ν kP ( ∂ h ) ∂ v h k L (Ω f ) + 12 ddt kP ( ∂ h ) ∂ h ξ k L (Σ ) = X j =1 I j , (6.3) where I := 2 ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ∇ h · v h dS , I := − Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β ∂ q χ f dx, I := − Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q χ ′ f dx, I := Z Ω P ( ∂ h ) ∂ v β P ( ∂ h )( − ∂ t v β + ν ∆ h v β ) χ f dx, I := − ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ( B α ,i ∂ α v i ) dS , I := Z Σ P ( ∂ h )( B αβ +5 ,i ∂ α v i ) P ( ∂ h ) ∂ β q dS , I := Z Ω P ( ∂ h ) ∂ v β P ( ∂ h )( g β + ν e g ββ ) χ f dx. Proof.
Applying the operator P ( ∂ h ) to (6.2) yields − ν P ( ∂ h ) ∂ v β + ∂ β P ( ∂ h ) q = − ∂ t P ( ∂ h ) v β + ν ∆ h P ( ∂ h ) v β + P ( ∂ h )( g β + ν e g ββ ) . (6.4)Multiplying (6.4) by −P ( ∂ h ) ∂ v β χ f and integrating it in Ω, we get ν Z Ω |P ( ∂ h ) ∂ v β | χ f dx + I = Z Ω P ( ∂ h ) ∂ v β P ( ∂ h )( − ∂ t v β + ν ∆ h v β ) χ f dx + Z Ω P ( ∂ h ) ∂ v β P ( ∂ h )( g β + ν e g ββ ) χ f dx (6.5)with I = − R Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q χ f dx .Integrating by parts in I gives rise to I = − Z Σ P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q dS + Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β ∂ q χ f dx + Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q χ ′ f dx. Thanks to the boundary conditions of ∂ v β and q on Σ , we have − Z Σ P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q dS = − Z Σ P ( ∂ h ) (cid:18) − ∂ β v + B αβ +5 ,i ∂ α v i (cid:19) P ( ∂ h ) ∂ β q dS = Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β q dS − Z Σ P ( ∂ h )( B αβ +5 ,i ∂ α v i ) P ( ∂ h ) ∂ β q dS = Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) (cid:18) ∂ β ξ − ν∂ β ∇ h · v h + ν∂ β ( B α ,i ∂ α v i ) (cid:19) dS − Z Σ P ( ∂ h )( B αβ +5 ,i ∂ α v i ) P ( ∂ h ) ∂ β q dS , − Z Σ P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q dS = 12 ddt Z Σ |P ( ∂ h ) ∂ β ξ | dS − ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ∇ h · v h dS + ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ( B α ,i ∂ α v i ) dS − Z Σ P ( ∂ h )( B αβ +5 ,i ∂ α v i ) P ( ∂ h ) ∂ β q dS . Therefore, we obtain I = 12 ddt Z Σ |P ( ∂ h ) ∂ β ξ | dS − ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ∇ h · v h dS + ν Z Σ P ( ∂ h ) ∂ β v P ( ∂ h ) ∂ β ( B α ,i ∂ α v i ) dS − Z Σ P ( ∂ h )( B ,β +5 ,α,i ∂ α v i ) P ( ∂ h ) ∂ β q dS + Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β ∂ q χ f dx + Z Ω P ( ∂ h ) ∂ v β P ( ∂ h ) ∂ β q χ ′ f dx. (6.6)Plugging (6.6) into (6.5), we get (6.3). k ( ˙Λ σ h ∇ v, ˙Λ σ h ∇ q ) k L t ( L ) Lemma 6.3.
Under the assumption of Lemma 5.8, if E ( t ) ≤ for all the existence times t , then there holds ddt k ˙Λ σ h ∂ h ξ k L (Σ ) + c ( k ˙Λ σ h ∇ v k L (Ω) + k ˙Λ σ h ∇ q k L (Ω) ) . k ˙Λ σ h ∇ v k L + k ∂ h ∇ v k L + ( k ˙Λ σ h ∂ h ∂ t v k L + k ˙Λ σ h ∂ t v k L ) + E ˙ D . (6.7) Proof.
First, taking P ( ∂ h ) = ˙Λ σ h in (2.24), we get k ˙Λ σ h ∂ q k L . k ˙Λ σ h ∂ t v k L + k ˙Λ σ h ∂ h v k H + E ˙ D , k ˙Λ σ h ∂ v k L . k ˙Λ σ h ∂ h v k H + E ˙ D , k ˙Λ σ − h ∂ v k L . k ˙Λ σ h v k H + E ˙ D , k − ν ˙Λ σ h ∂ v h + ˙Λ σ h ∂ h q k L . k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D , k − ν ˙Λ σ h ∂ v h + ˙Λ σ h ∂ h q k L (Ω f ) . k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D , (6.8)and in particular, k ˙Λ σ h ∂ h q k L (Ω f ) . k ˙Λ σ h ∂ v h k L (Ω f ) + k ˙Λ σ h ∂ t v k L + k ˙Λ σ h ∂ h v k H + E ˙ D . (6.9)While taking P ( ∂ h ) = ˙Λ σ h ∂ h in (6.1), we have k ˙Λ σ h ∂ h v k H (Ω in ) + k ˙Λ σ h ∂ h q k H (Ω in ) . k ˙Λ σ h ∂ h v k H (Ω f ) + k ˙Λ σ h ∂ h q k L (Ω f ) + k ˙Λ σ h ∂ h ∂ t v k L (Ω in ) + k ˙Λ σ h ∂ h g k L (Ω) + kP ( ∂ h ) ψ k H (Ω) with ψ = − g a α ∂ v α + B α ,i ∂ α v i . 49hanks to (5.18), we find k ˙Λ σ h ∂ h g k L (Ω) + k ˙Λ σ h ∂ h ψ k H (Ω) . E ˙ D , so we get from (6.9) that k ˙Λ σ h ∂ h v k H (Ω in ) + k ˙Λ σ h ∂ h q k H (Ω in ) . k ˙Λ σ h ∂ h v k H + k ˙Λ σ h ∂ h ∂ t v k L + E ˙ D + (cid:18) k ˙Λ σ h ∂ v h k L (Ω f ) + k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D (cid:19) . Hence, we get k ˙Λ σ h ∂ h q k L (Ω in ) . k ˙Λ σ h ∂ h v k H + k ˙Λ σ h ∂ h ∂ t v k L + k ˙Λ σ h ∂ t v h k L + E ˙ D + k ˙Λ σ h ∂ v h k L (Ω f ) , which along with (6.9) and the first inequality in (6.8) ensures k ˙Λ σ h ∇ q k L (Ω) . k ˙Λ σ h ∂ q k L (Ω) + k ˙Λ σ h ∂ h q k L (Ω f ) + k ˙Λ σ h ∂ h q k L (Ω in ) . k ˙Λ σ h ∂ v h k L (Ω f ) + k ˙Λ σ h ∂ h ∂ t v k L + k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D , (6.10)and then, due to (6.8), we have k ˙Λ σ h ∂ v h k L . k ˙Λ σ h ∂ h q k L + k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D . k ˙Λ σ h ∂ v h k L (Ω f ) + k ˙Λ σ h ∂ h ∂ t v k L + k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D . (6.11)On the other hand, taking P ( ∂ h ) = ˙Λ σ h in (6.3) gives ν k ˙Λ σ h ∂ v h k L (Ω f ) + 12 ddt k ˙Λ σ h ∂ h ξ k L (Σ ) = X j =1 I j . For the remainder terms I j with j = 1 , , ,
4, from the linear terms, we have | I | . k ˙Λ σ h ∂ β v k H (Σ ) k ˙Λ σ h ∇ h · v h k H (Σ ) . k ˙Λ σ h ∂ h v k H (Ω) , | I | . k ˙Λ σ +1 h ∂ v β k L k ˙Λ σ h ∂ q k L (Ω f ) . k ˙Λ σ +1 h ∇ v k L k ˙Λ σ h ∂ q k L , | I | . k ˙Λ σ h ∂ v β k L k ˙Λ σ +1 h q k L , and | I | . k ˙Λ σ h ∂ v h k L (Ω f ) ( k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v h k L ) . For the nonlinear remainder terms I , I , and I , there hold | I | . k ˙Λ σ h ∂ β v k L (Σ ) k ˙Λ σ h ∂ β ( B α ,i ∂ α v i ) k L (Σ ) . k ˙Λ σ h ∂ β v k H (Ω) k ˙Λ σ h ∂ β ( B α ,i ∂ α v i ) k H (Ω) . E ˙ D . I | . k ˙Λ σ h ( B αβ +5 ,i ∂ α v i ) k L (Σ ) k ˙Λ σ h ∂ β q k L (Σ ) . E ˙ D , and | I | . k ˙Λ σ h ∂ v β k L (Ω f ) ( k ˙Λ σ h g β k L + k ˙Λ σ h e g ββ k L ) . E ˙ D , Hence, we get ν k ˙Λ σ h ∂ v h k L (Ω f ) + 12 ddt k ˙Λ σ h ∂ h ξ k L (Σ ) . k ˙Λ σ h ∂ h v k H (Ω) + k ˙Λ σ +1 h ∇ v k L k ˙Λ σ h ∂ q k L + k ˙Λ σ h ∂ v h k L k ˙Λ σ h ∂ h q k L + k ˙Λ σ h ∂ v h k L (Ω f ) ( k ˙Λ σ h ∂ t v k L + k ˙Λ σ h ∂ h v k L ) + E ˙ D , which follows from (6.10) that ν k ˙Λ σ h ∂ v h k L (Ω f ) + ddt k ˙Λ σ h ∂ h ξ k L (Σ ) . k ˙Λ σ h ∂ h ∇ v k L + k ˙Λ σ h ∂ t v k L + E ˙ D + k ˙Λ σ h ( ∇ v, ∂ h ∇ v ) k L k ˙Λ σ h ∂ v h k L (Ω f ) + k ˙Λ σ h ( ∇ v, ∂ h ∇ v ) k L ( k ˙Λ σ h ∂ h ∂ t v k L + k ˙Λ σ h ∂ t v h k L + k ˙Λ σ h ∂ h v k H + E ˙ D ) . Hence, one has c k ˙Λ σ h ∂ v h k L (Ω f ) + ddt k ˙Λ σ h ∂ h ξ k L (Σ ) . E ˙ D + k ˙Λ σ h ( ∇ v, ∂ h ∇ v ) k L + k ˙Λ σ h ( ∂ h ∂ t v, ∂ t v h ) k L . (6.12)Combining (6.12),(6.10),(6.11), we obtain ddt k ˙Λ σ h ∂ h ξ k L (Σ ) + c ( k ˙Λ σ h ∂ v h k L (Ω) + k ˙Λ σ h ∇ q k L (Ω) ) . E ˙ D + k ˙Λ σ h ( ∇ v, ∂ h ∇ v ) k L + k ˙Λ σ h ( ∂ h ∂ t v, ∂ t v h ) k L , which along with the second inequality in (6.8) follows (6.7). The proof of the lemma isaccomplished. k ∂ N − h ∂ v h k L t L Lemma 6.4.
Under the assumption of Lemma 5.8, if E ( t ) ≤ for all the existence times t , then there holds ddt k ∂ Nh ξ k L (Σ ) + c ( k ∂ N − h ( ∇ v, ∇ q ) k L ) . k ∂ N − h ( ∂ t v, ∂ h ∂ t v ) k L + k ∂ N − h ( ∂ h v, v ) k H + k ˙Λ σ h ∂ v h k L (Ω) + ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D (6.13)51 roof. First, taking P ( ∂ h ) = ∂ N − h in (2.24), we get k ∂ N − h ∂ q k L . k ∂ N − h ∂ t v k L + k ∂ N − h ∂ h v k H + ( E ˙ D N + E N ˙ D ) , k ∂ N − h ∂ v k L . k ∂ N − h ∂ h v k H + ( E ˙ D N + E N ˙ D ) , k ∂ N − h ( ∂ h q − ν∂ v h ) k L . k ∂ N − h ∂ t v k L + k ∂ N − h ∂ h v k H + ( E ˙ D N + E N ˙ D ) , k ∂ N − h ( ∂ h q − ν∂ v h ) k L (Ω f ) . k ∂ N − h ∂ t v k L + k ∂ N − h ∂ h v k H + ( E ˙ D N + E N ˙ D ) , (6.14)and in particular, k ∂ k − h ∂ h q k L (Ω f ) . k ∂ k − h ∂ v h k L (Ω f ) + k ∂ k − h ∂ t v k L + k ∂ k − h ∂ h v k H + ( E ˙ D k + E k ˙ D ) (6.15)for any k ≥
3. While taking P ( ∂ h ) = ∂ N − h in (6.1), we have k ∂ N − h v k H (Ω in ) + k ∂ N − h q k H (Ω in ) . k ∂ N − h v k H (Ω f ) + k ∂ N − h q k L (Ω f ) + k ∂ N − h ∂ t v k L (Ω in ) + k ∂ N − h g k L (Ω) + k ∂ N − h ψ k H (Ω) with ψ = − ( f a ∂ v + f a ∂ v ) + B α ,i ∂ α v i .Thanks to (5.18), we find k ∂ N − h g k L (Ω) + k ∂ N − h ψ k H (Ω in ) . E ˙ D N + E N ˙ D . so we get from (6.15) that k ∂ N − h v k H (Ω in ) + k ∂ N − h q k H (Ω in ) . k ∂ N − h v k H + k ∂ N − h ∂ t v k L + k ∂ N − h ∂ t v k L + E ˙ D N + E N ˙ D + k ∂ N − h ∂ v h k L (Ω f ) . In particular, we have k ∂ N − h ∇ q k L (Ω in ) . k ∂ N − h v k H + k ∂ N − h ∂ t v k L + k ∂ N − h ∂ t v k L + E ˙ D N + E N ˙ D + k ∂ N − h ∂ v h k L (Ω f ) . Thanks to (6.14), we get k ∂ N − h ∇ q k L (Ω f ) . k ∂ N − h ∂ q k L (Ω f ) + k ∂ Nh q k L (Ω f ) . k ∂ N − h ∂ t v k L + k ∂ N − h ∂ h v k H + ( E ˙ D N + E N ˙ D ) + k ∂ N − h ∂ v h k L (Ω f ) . Hence, one has k ∂ N − h ∇ q k L (Ω) . k ∂ N − h ∇ q k L (Ω f ) + k ∂ N − h ∇ q k L (Ω in ) . k ∂ N − h ( ∂ t v, ∂ h ∂ t v ) k L + k ∂ N − h ( ∂ h v, v ) k H + ( E ˙ D N + E N ˙ D )+ k ˙Λ σ h ∂ v h k L (Ω f ) + k ∂ N − h ∂ v h k L (Ω f ) , k ∂ N − h ( ∂ v h , ∇ q ) k L (Ω) . k ∂ N − h ( ∂ t v, ∂ h ∂ t v ) k L + k ∂ N − h ( ∂ h v, v ) k H + ( E ˙ D N + E N ˙ D ) + k ˙Λ σ h ∂ v h k L (Ω f ) + k ∂ N − h ∂ v h k L (Ω f ) . (6.16)On the other hand, taking P ( ∂ h ) = ∂ N − h in (6.3) gives ν k ∂ N − h ∂ v h k L (Ω f ) + 12 ddt k ∂ N − h ∂ h ξ k L (Σ ) = X j =1 I j . For the remainder terms I j with j = 1 , , ,
4, from the linear terms, we have | I | . k ∂ N − h ∂ β v k H (Σ ) k ∂ N − h ∇ h · v h k H (Σ ) . k ∂ N − h ∂ β v k H k ∂ N − h ∇ h · v h k H . k ∂ Nh ∇ v k L , | I | + | I | . ( k ∂ Nh ∂ v h k L + k ∂ N − h ∂ v h k L ) k ∂ N − h ∇ q k L , and | I | . k ∂ N − h ∂ v h k L (Ω f ) ( k ∂ N − h ∂ t v h k L + k ∂ N +1 h v h k L ) . For the nonlinear remainder terms I , I , and I , there hold | I | . k ∂ N − h ∂ β v k H (Σ ) k ∂ N − h ( B α ,i ∂ α v i ) k H (Σ ) . k ∂ N − h ∂ β v k H k ∂ N − h ( B α ,i ∂ α v i ) k H . ˙ D N ( E ˙ D N + E N ˙ D ) , | I | . k ∂ N − h ( B αβ +5 ,i ∂ α v i ) k H (Σ ) k ∂ N − h q k H (Σ ) . k ∂ N − h ( B αβ +5 ,i ∂ α v i ) k H k ∂ N − h q k H . ˙ D N ( E ˙ D N + E N ˙ D ) , and | I | . k ∂ N − h ∂ v h k L (Ω f ) ( k ∂ N − h g β k L + k ∂ N − h e g ββ k L ) . k ∂ N − h ∂ v h k L (Ω f ) ( E ˙ D N + E N ˙ D ) , where we used the inequality in (5.18) in the last inequality.Hence, we have ν k ∂ N − h ∂ v β k L (Ω f ) + 12 ddt k ∂ N − h ∂ β ξ k L (Σ ) . k ∂ Nh ∇ v k L + ˙ D N ( E ˙ D N + E N ˙ D ) + ( k ∂ Nh ∂ v h k L + k ∂ N − h ∂ v h k L ) k ∂ N − h ∇ q k L + k ∂ N − h ∂ v h k L (Ω f ) ( k ∂ N − h ∂ t v h k L + k ∂ N +1 h v h k L + E ˙ D N + E N ˙ D ) , which follows that ν k ∂ N − h ∂ v β k L (Ω f ) + ddt k ∂ N − h ∂ β ξ k L (Σ ) . k ∂ Nh ∇ v k L + k ∂ N − h ∂ t v h k L + ( k ∂ Nh ∂ v h k L + k ∂ N − h ∂ v h k L ) k ∂ N − h ∇ q k L + ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D . (6.17)53ombining (6.17) with (6.16) yields ddt k ∂ N − h ∂ β ξ k L (Σ ) + c k ∂ N − h ( ∂ v h , ∇ q ) k L (Ω) . k ∂ N − h ( ∂ t v, ∂ h ∂ t v ) k L + k ∂ N − h ( ∂ h v, v ) k H + k ˙Λ σ h ∂ v h k L (Ω) + ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D , which proves (6.13), and completes the proof of Lemma 6.4.With Lemmas 6.3, 6.4 in hand, we get Lemma 6.5.
Under the assumption of Lemma 5.8, if E ( t ) ≤ for all the existence times t , then there holds ddt ( k ˙Λ σ h ∂ h ξ k L (Σ ) + N − X k =1 k ∂ k +1 h ξ k L (Σ ) ) + c (cid:18) k ˙Λ σ h ( ∇ v, ∇ q ) k L + k ˙Λ σ +1 h ξ k H (Σ ) + N − X k =1 k ∂ kh ( ∇ v, ∇ q ) k L + N − X k =2 k ∂ kh ξ k H (Σ ) (cid:19) ≤ C ( k ˙Λ σ h ∇ v k L (Ω) + N X i =1 k ∂ ih ∇ v k L (Ω) + k ˙Λ σ h ∂ t v k L + N − X k =1 k ∂ kh ∂ t v k L )+ C ( ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D ) . (6.18) Proof.
Thanks to Lemmas 6.3 and 6.4, we get ddt ( k ˙Λ σ h ∂ h ξ k L (Σ ) + N − X k =1 k ∂ k +1 h ξ k L (Σ ) )+ c (cid:18) k ˙Λ σ h ( ∇ v, ∇ q ) k L + N − X k =1 k ∂ kh ( ∇ v, ∇ q ) k L (cid:19) ≤ C ( k ˙Λ σ h ∇ v k L (Ω) + N X i =1 k ∂ ih ∇ v k L (Ω) + k ˙Λ σ h ∂ t v k L + N − X k =1 k ∂ kh ∂ t v k L )+ C ( ˙ D N ( E ˙ D N + E N ˙ D ) + E N ˙ D ) . (6.19)On the other hand, due to (2.14), one has k ˙Λ σ +1 h ξ k H (Σ ) . k ˙Λ σ +1 h q k H (Σ ) + kk ˙Λ σ +1 h ∇ h · v h k H (Σ ) + k ˙Λ σ +1 h ( B α ,i ∂ α v i ) k H (Σ ) . k ˙Λ σ h ∂ h q k H (Ω) + k ˙Λ σ h ∂ h v k H (Ω) + k ˙Λ σ h ∂ h ( B α ,i ∂ α v i ) k H (Ω) , k ∂ kh ξ k H (Σ ) . k ∂ kh q k H (Σ ) + k ∂ kh ∇ h · v h k H (Σ ) + k ∂ kh ( B α ,i ∂ α v i ) k H (Σ ) . k ∂ kh q k H (Ω) + k ∂ k +1 h v k H (Ω) + k ∂ kh ( B α ,i ∂ α v i ) k H (Ω) (for k = 2 , ..., N − , which follows that k ˙Λ σ +1 h ξ k H (Σ ) . k ˙Λ σ h ∂ h q k H (Ω) + k ˙Λ σ h ∂ h v k H (Ω) + k ∂ h v k H (Ω) , k ∂ kh ξ k H (Σ ) . k ∂ kh q k H (Ω) + k ∂ kh ∂ h v k H (Ω) + E ˙ D k +1 + E k +1 ˙ D k = 2 , ..., N − k ˙Λ σ +1 h ξ k H (Σ ) + N − X k =2 k ∂ kh ξ k H (Σ ) ≤ C (cid:18) k ˙Λ σ h ∂ h q k H (Ω) + k ˙Λ σ h ∂ h v k H (Ω) + k ∂ h v k H (Ω) + N − X k =2 ( k ∂ kh q k H (Ω) + k ∂ kh ∂ h v k H (Ω) ) + E ˙ D N + E N ˙ D (cid:19) , which along with (6.19) ensures (6.18). Combining Lemma 6.5 with Lemma 5.12, and taking the positive δ in Lemma 5.12 so smallthat δ ≤ min { , c C , q c C , c C } , we immediately get the following energy estimate. Lemma 7.1.
Let N ≥ , under the assumption of Lemma 5.1, if E ( t ) ≤ for all theexistence times t , then there holds ddt b ˙ E N,δ + b ˙ D N,δ ≤ C ( E N ˙ D ˙ D N + E ˙ D N + E N ˙ D ) with b ˙ E N,δ := k ˙Λ σ h v k L (Ω) + k ˙Λ σh ξ k L (Σ ) + N X i =1 ( k ∂ ih v k L (Ω) + k ∂ ih ξ k L (Σ ) )+ δ (cid:18) ˚ E ( ˙Λ σ h ∇ v ) + N − X k =1 ˚ E ( ∂ kh ∇ v ) (cid:19) + δ ( k ˙Λ σ h ∂ h ξ k L (Σ ) + N − X k =1 k ∂ k +1 h ξ k L (Σ ) ) and b ˙ D N,δ := c k ˙Λ σ h ∇ v k L (Ω) + N X i =1 k ∂ ih ∇ v k L (Ω) ) + c δ ( k ˙Λ σ h ∂ t v k L + N − X k =1 k ∂ kh ∂ t v k L )+ δ c (cid:18) k ˙Λ σ h ( ∇ v, ∇ q ) k L + N − X k =1 k ∂ kh ( ∇ v, ∇ q ) k L + k ˙Λ σ +1 h ξ k H (Σ ) + N − X k =2 k ∂ kh ξ k H (Σ ) (cid:19) . From this, together with Lemma 5.16, It is convenient to derive55 emma 7.2.
Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E ( t ) ≤ for all the existence times t , then there holds ddt ( b E N +1 ,tan,δ + b ˙ E N +1 ,δ ) + ( b D N +1 ,tan,δ + b ˙ D N +1 ,δ ) ≤ C (cid:18) E N +1 ˙ D ˙ D N +1 + E ˙ D N +1 + E N +1 ˙ D + E ˙ D k ˙Λ − λh ξ k H (Σ ) + E k ˙Λ σ h ξ k H (Σ ) (cid:19) + C δ − E ˙ D . The following comparison lemma discovers the equivalence of two types of the energy.
Lemma 7.3 (Comparison lemma) . Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E N ( t ) ≤ for all the existence times t , then thereexists a genius positive constant δ such that ˙ E N ∼ b ˙ E N,δ , ˙ D N ∼ b ˙ D N,δ , E N +1 ∼ b E N +1 ,tan,δ + b ˙ E N +1 ,δ , D N +1 ∼ b D N +1 ,tan,δ + b ˙ D N +1 ,δ . (7.1) Proof.
In order to prove (7.1), we need only to estimate the nonlinear energy ˚ E ( P ( ∂ h ) ∇ v )from upper and low ones. In fact, due to the definition of ˚ E ( P ( ∂ h ) ∇ v ), we obtain˚ E ( ˙Λ σ h ∇ v ) ≤ k ˙Λ σ h ∇ v k L + 4 k ˙Λ σ h ∇ h · v h k L + k ˙Λ σ h ∂ v k L + k ˙Λ σ h ∇ h v k L + 2 ν k ˙Λ σ h v k L k ˙Λ σ h ∇ H ( ξ ) k L + 4 X j =1 k ˙Λ σ h ∂ v k L k ˙Λ σ h ( H ( B αj +5 ,i ) ∂ α v i ) k L ≤ k ˙Λ σ h ∇ v k L + C ( k ˙Λ σ h ∂ h v k L + k ˙Λ σ h v k L + k ˙Λ σ h ξ k H (Σ ) ) + C E k ˙Λ σ h Λ h ∇ v k L , N − X k =1 ˚ E ( ∂ kh ∇ v ) ≤ N − X k =1 (cid:18) k ∂ kh ∇ v k L + 4 k ∂ kh ∇ h · v h k L + 2 k ∂ kh ∂ v k L k ∂ kh ∇ h v k L + C k ∂ kh v k L k ∂ kh ξ k H (Σ ) + 2 k ∂ kh ∂ v k L X j =1 k ∂ kh ( H ( B αj +5 ,i ) ∂ α v i ) k L ≤ N − X k =1 (cid:18) k ∂ kh ∇ v k L + C ( k ∂ kh v k L + k ∂ kh ξ k H (Σ ) + E N k ∂ kh ∇ v k L ) (cid:19) , and ˚ E ( ˙Λ σ h ∇ v ) ≥ k ˙Λ σ h ∇ v k L − C ( k ˙Λ σ h ∂ h v k L + k ˙Λ σ h v k L + k ˙Λ σ h ξ k H (Σ ) + E k ˙Λ σ h Λ h ∇ v k L ) , N − X k =1 ˚ E ( ∂ kh ∇ v ) ≥ N X k =1 (cid:18) k ∂ kh ∇ v k L − C ( k ∂ kh Λ h v k L + k ∂ kh ξ k H (Σ ) ) − C ( E k +1 k ∂ h Λ h v k L + E k ∂ h Λ kh v k L ) (cid:19) . δ . Thisfinishes the proof of Lemma 7.3.With Lemma 7.3 in hand, for simplicity, we denote b ˙ E N,δ , b ˙ D N,δ , b E N +1 ,tan,δ + b ˙ E N +1 ,δ , b D N +1 ,tan,δ + b ˙ D N +1 ,δ by b ˙ E N , b ˙ D N , b E N +1 ,tan + b ˙ E N +1 , b D N +1 ,tan + b ˙ D N +1 respectively, and b E N +1 := b E N +1 ,tan + b ˙ E N +1 , b D N +1 := b D N +1 ,tan + b ˙ D N +1 . And we also denote the positiveconstant C ≥ C − b ˙ E N ≤ ˙ E N ≤ C b ˙ E N , C − b ˙ D N ≤ ˙ D N ≤ C b ˙ D N , C − b E N +1 ≤ E N +1 ≤ C b E N +1 , C − b D N +1 ≤ D N +1 ≤ C b D N +1 . (7.2)Therefore, we restate Lemmas 7.1 and 7.2 to get the total energy estimates. Lemma 7.4.
Let N ≥ , under the assumption of Lemma 5.1, if ( λ, σ ) ∈ (0 , satisfies − λ < σ ≤ − λ , and E N ( t ) ≤ for all the existence times t , then there hold ddt b ˙ E N + 2 c ˙ D N ≤ C E N ˙ D N , (7.3) and ddt b E N +1 + 2 c D N +1 ≤ C (cid:18) E ˙ D N +1 + E N +1 ˙ D + E ˙ D + E ˙ E (cid:19) . (7.4) Thanks to (7.3) and (7.4) in Lemma 7.4, we will show that the low order energy E N isuniformly bounded, while the low order energy E N +1 grows algebraicly as the time t goes toinfinity by an inductive argument. Hence, inspired by the two-tiered energy method [10], wemay find the decay estimate of the low order energy ˙ E N . For this, we first prove the followingdecay estimate. Lemma 8.1 (Decay estimate) . If the non-negative function f satisfies the differential in-equality ( ddt f + c f s ≤ , ∀ t > ,f | t =0 = f (8.1) for two positive constants s > and c > , then there holds for any t > f ( t ) ≤ ( c s ) − s (( c s ) − f − s + t ) − s . (8.2) Proof.
Due to the differential inequality in (8.1), we have f − (1+ s ) ddt f + c ≤ . Hence, one has ddt f − s ≥ c s. (8.3)57ntegrating (8.3) on [0 , t ] implies f − s ( t ) ≥ f − s + c s t, which follows (8.2).We are now in a position to complete the proof of Theorem 3.3. Proof of Theorem 3.3.
Thanks to the local well-posedness theorem (Theorem 3.1), there ex-ists a positive time T ∗ such that the system (2.8) with initial data ( ξ , v ) has a uniquesolution ( ξ, v ) with ( ξ, v ) ∈ C ([0 , T ∗ ); F N +1 ) . Without loss of generality, we may assume that T ∗ maximal time of the existence to thissolution. The aim of what follows is to prove that T ∗ = + ∞ .Suppose, by way of contradiction, that T ∗ < + ∞ . We will show that solutions can actuallybe extended past T ∗ and satisfy E N +1 ( T ∗∗ ) < + ∞ for some T ∗∗ > T ∗ , which contradicts thedefinition of T ∗ . It suffices to show that there is a positive constant C such thatlim t ր T ∗ E N ( t ) ≤ C E N (0) , lim t ր T ∗ ( h t i − E N +1 ( t )) ≤ C E N +1 (0) . (8.4)Indeed, if (8.4) holds, then k J − k L ∞ ≤ C E ( t ) ≤ C C E (0) ≤ C C ǫ , so we take ǫ ≤ C C so that k J − k L ∞ ≤ , which ensures that J ( t ) ≥ for any existence time t .While the second inequality implies lim t ր T ∗ E N +1 ( t ) ≤ C h T ∗ i E N +1 (0), which contradictsthe fact that T ∗ < + ∞ is the maximal existence time by applying Theorem 3.1, and we canextend the solution past T ∗ to some T ∗∗ > T ∗ .Let’s now focus on the proof of (8.4), which is reduced to claim that T ∗ = sup (cid:26) T ∈ [0 , T ∗ ] | ( ξ, v ) ∈ C ([0 , T ); F N +1 ) ,E N ( t ) ≤ C ( E N (0) + E N +1 (0)) , E N +1 ( t ) ≤ C (1 + t ) E N +1 (0) (cid:27) , where the constant C will be determined later.Indeed, setting T def = sup (cid:26) T ∈ [0 , T ∗ ] | ( ξ, v ) ∈ C ([0 , T ); F N +1 ) , E N +1 ( t ) ≤ C E N +1 (0) ,E N ( t ) ≤ C ( E N (0) + E N +1 (0)) , E N +1 ( t ) ≤ C (1 + t ) E N +1 (0) ∀ t ∈ [0 , T ) (cid:27) , (8.5)we first get that T ≤ T ∗ . On the other hand, we will verify that T ≥ T ∗ . Otherwise, wesuppose that T < T ∗ by a contradiction argument.From this, take ǫ ∈ (0 , c C C ] small enough, then C E N ≤ C C E N (0) ≤ C C ǫ ≤ c ,
58e get from (7.3) that ddt b ˙ E N + c ˙ D N ≤ . (8.6)Hence, thanks to the definitions of ˙ E N , E N , and ˙ D N , we get from the interpolation inequality k ˙Λ σ h ξ k L (Σ ) . k ˙Λ σ h ξ k λ + σ λ + σ L (Σ ) k ˙Λ − λh ξ k λ + σ L (Σ ) , k ˙Λ Nh ξ k L (Σ ) . k ˙Λ N − h ξ k L (Σ ) k ˙Λ N +1 h ξ k L (Σ ) that ˙ E N ≤ C ( ˙ D N ) λ + σ λ + σ ( E N +1 ) λ + σ , where C is the interpolation constant independent of C .Hence, one has b ˙ E N ≤ C ˙ E N ≤ C C ( ˙ D N ) λ + σ λ + σ ( E N +1 ) λ + σ ≤ C C (2 C ) λ + σ ( ˙ D N ) λ + σ λ + σ E N +1 (0) λ + σ . Denote e C := C C (2 C ) λ + σ , then there holds b ˙ E N ≤ e C ( ˙ D N ) λ + σ λ + σ E N +1 (0) λ + σ , which implies [ e C − b ˙ E N ] λ + σ λ + σ E N +1 (0) − λ + σ ≤ ˙ D N . Therefore, from (8.6), we get ddt b ˙ E N + c [ e C − b ˙ E N ] λ + σ λ + σ E N +1 (0) − λ + σ ≤ , that is, ddt b ˙ E N + c E N +1 (0) − λ + σ b ˙ E λ + σ λ + σ N ≤ c := c [ e C − ] λ + σ λ + σ = c [ C C ] − λ + σ λ + σ (2 C ) − λ + σ .Applying Lemma 7.4, where we take c := c E N +1 (0) − λ + σ , s := 1 λ + σ , f ( t ) := b ˙ E N ( t ) , we infer that f ( t ) ≤ ( c λ + σ ) − ( λ + σ ) E N +1 (0) (cid:18) c − ( λ + σ )( E N +1 (0)[ b ˙ E N (0)] − ) λ + σ + t (cid:19) − ( λ + σ ) . b ˙ E N (0) and E N +1 (0), we denote that b ˙ E N (0) ≤ C ( N ) E N +1 (0)with the positive constant C ( N ) depending only on N , so one can see that c − ( λ + σ )( E N +1 (0)[ b ˙ E N (0)] − ) λ + σ ≥ c − ( λ + σ ) C ( N ) − λ + σ , which leads to f ( t ) ≤ ( c λ + σ ) − ( λ + σ ) E N +1 (0) (cid:18) c − ( λ + σ ) C ( N ) − λ + σ + t (cid:19) − ( λ + σ ) . Therefore, we obtain b ˙ E N ( t ) ≤ C E N +1 (0) h t i − ( λ + σ ) with C = C ( λ + σ , C ( N ) , c ), and then˙ E N ( t ) ≤ C C E N +1 (0) h t i − ( λ + σ ) , Z t ˙ E N dτ ≤ e C E N +1 (0)with e C = C C R t h τ i − ( λ + σ ) dτ .Since ddt ( h t i (1+ λ + σ ) / b ˙ E N ) + c h t i (1+ λ + σ ) / ˙ D N = h t i (1+ λ + σ ) / ddt ( b ˙ E N ) + c h t i (1+ λ + σ ) / ˙ D N + b ˙ E N h t i (1+ λ + σ ) / − (1 + λ + σ ) / , from (8.6), we get ddt ( h t i (1+ λ + σ ) / b ˙ E N ) + c h t i (1+ λ + σ ) / ˙ D N ≤ b ˙ E N h t i (1+ λ + σ ) / − (1 + λ + σ ) / , and then ddt ( h t i (1+ λ + σ ) / b ˙ E N ) + c h t i (1+ λ + σ ) / ˙ D N ≤ C E N +1 (0) h t i − (1+ λ + σ ) / (1 + λ + σ ) / . It follows that Z t h τ i (1+ λ + σ ) / ˙ D N ( τ ) dτ ≤ C E N +1 (0) , Z t ˙ D N ( τ ) dτ ≤ e C E N +1 (0) . (8.7)With this in hand, we are ready to estimate E N .Since k ˙Λ σ − h ∂ ξ ( t ) k H + k ˙Λ σ h ∂ ξ h ( t ) k H + k ˙Λ σ h ξ ( t ) k L + N − X i =1 k ∂ ih ξ ( t ) k H ≤ k ˙Λ σ − h ∂ ξ (0) k H + k ˙Λ σ h ∂ ξ h (0) k H + k ˙Λ σ h ξ (0) k L + N − X i =1 k ∂ ih ξ (0) k H + Z t ( k ˙Λ σ − h ∂ v k H + k ˙Λ σ h ∂ v h k H + k ˙Λ σ h v k L + N − X i =1 k ∂ ih v k H ) dτ, E h ,N ( t ) ≤ C ( N ) E h ,N (0) + C ( N )( Z t ˙ D N dτ ) . Due to (8.7), we find that E N ( t ) = E h ,N ( t ) + E N ( t ) ≤ C ( N ) E h ,N (0) + C ( N )( Z t ˙ D N dτ ) + E N ( t ) ≤ C ( N ) E h ,N (0) + ( C ( N ) e C + 2 C ) E N +1 (0) , and then E N ( t ) ≤ C ( E N (0) + E N +1 (0)) with C := C ( N )( e C + 1) + 2 C . (8.8)Notice that k ∂ Nh ∇ ξ k L ( t ) ≤ k ∂ Nh ∇ ξ (0) k L + Z t k ∂ Nh ∇ v k L ( τ ) dτ ≤ k ∂ Nh ∇ ξ k L (0) + Z t D N +1 dτ, which along with (8.8) and the definition of T leads to E N +1 ( t ) ≤ k ∂ Nh ∇ ξ k L + 2( E N +1 + E N ) ≤ k ∂ Nh ∇ ξ (0) k L + 2 t Z t D N +1 dτ + 4 C E N +1 (0) + 2 C ( E N (0) + E N +1 (0)) ≤ t Z t D N +1 dτ + e C E N +1 (0) (8.9)with e C := 2 + 4 C + 2 C .On the other hand, thanks to (7.4), we get ddt b E N +1 + c D N +1 ≤ C (cid:18) E N +1 ˙ D + E ( ˙ D + ˙ E ) (cid:19) . Hence, from (8.9), one has ddt ( b E N +1 + c Z t D N +1 dτ ) ≤ C t ˙ D Z t D N +1 dτ + 2 C C E N +1 (0)( ˙ D + ˙ E ) + C e C E N +1 (0) ˙ D . Thanks to the Gronwall inequality, we get b E N +1 ( t ) + c Z t D N +1 dτ ≤ exp { C c − Z t τ ˙ D dτ }× (cid:20) b E N +1 (0) + 2 C C E N +1 (0) Z t ( ˙ D + ˙ E ) dτ + C e C E N +1 (0) Z t ˙ D dτ (cid:21) , which follows that b E N +1 ( t ) + c Z t D N +1 dτ ≤ exp { C c − C E N +1 (0) }× (cid:20) b E N +1 (0) + 2 C C E N +1 (0) ( e C E N +1 (0) + e C E N +1 (0)) + C e C E N +1 (0) C E N +1 (0) (cid:21) . b E N +1 ( t ) + c Z t D N +1 dτ ≤ exp { C c − C ǫ } b E N +1 (0)(1 + C ǫ )with C = 2 C C ( e C + e C ) + C e C C C .Taking ǫ ≤ min { c C C log( ) , C } , we obtainsup τ ∈ [0 ,t ] b E N +1 ( τ ) + c Z t D N +1 ( τ ) dτ ≤ b E N +1 (0) , and then E N +1 ( t ) ≤ C E N +1 (0) . Therefore, we get E N +1 ( t ) ≤ t Z t D N +1 dτ + C E N +1 (0) ≤ t c C E N +1 (0) + C E N +1 (0) ≤ e C ( t + 1) E N +1 (0))with e C = c C + C .Notice that all the constants C i , e C j (with i = 0 , , ..., j = 1 , ...,
6) above are independentof C .We take ǫ small enough (satisfying above conditions) and the constant C = 2( C + e C )in (8.5), then, according to the above estimates, for any t ∈ [0 , T ), there hold that E N +1 ( t ) ≤ C E N +1 (0) , E N ( t ) ≤ C ( E N (0) + E N +1 (0)) ,E N +1 ( t ) ≤ C (1 + t ) E N +1 (0) , which contradicts the fact that T is maximal by the definition. This ends the proof ofTheorem 3.3. A Proof of Theorem 3.2
Proof of Theorem 3.2.
Motivated by the method in [5], we shall suppose that v (1) is knownand show that a contradiction arises in solving for v (2) .We first assume that ∇ · θ = 0. Because of the assumption about ( v ( ε ) , q ( ε )), the flowmap ξ ( ε ) satisfies ξ ( t ) = ξ (0) + Z t v dτ = ε ( θ + Z t v (1) dτ ) + ε Z t v (2) dτ + ε Z t v (3) dτ, and then ξ ( ε ) has an expansion ξ ( ε ) = ε ξ (1) + ε ξ (2) + ε ξ (3) , (A.1)62nd also J ( ε ) = 1 + ε ∇ · ξ (1) + O ( ε ) , A ji ( ε ) = δ ji − ε ∂ i ξ (1) j + O ( ε ) . (A.2)Since ∇ J A · v = 0, vanishing the order ε of it, we get from (3.4) and (A.2) that ∇ · v (1) = 0 . (A.3)While from (A.1) and the fact ∇ · θ = 0, it follows ε ∇ · ξ (1) + ε ( ∇ · ξ (2) + ε ∇ · ξ (3) )= ε ( Z t ∇ · v (1) dτ ) + ε Z t ( ∇ · v (2) + ε ∇ · v (3) ) dτ, (A.4)then from (A.3), vanishing the order ε in (A.4) implies ∇ · ξ (1) = 0 , (A.5)The next divergence term v (2) is determined by v (1) from ∇ A · v = 0. Indeed, thanks to(A.3) and (A.2), one has0 = ∇ A · v = A ji ∂ j v i = ( δ ji − ε ∂ i ξ (1) j + O ( ε ))( ε ∂ j v (1) i + ε ∂ j v (2) i + O ( ε ))= ε ( ∇ · v (2) − ∂ i ξ (1) j ∂ j v (1) i ) + O ( ε ) . (A.6)Vanishing the order ε in (A.6) leads to ∇ · v (2) = ∂ i ξ (1) j ∂ j v (1) i , (A.7)and then ∇ · v (2) = 12 ∂ j ( ∂ i ξ (1) j v (1) i + ξ (1) i ∂ i v (1) j ) = 12 ∂ t ∇ · ( ξ (1) · ∇ ξ (1) ) , (A.8)where we used (A.3) and (A.5) in the second equality.Set ~ ( t, x ) def = Z t ∇ · v (2) dτ, l ( t ) def = Z Ω ~ ( t, x ) dx, we obtain from (A.8) that ~ ( t, x ) = 12 ∇ · ( ξ (1)1 ∂ ξ (1) + ξ (1) h · ∇ h ξ (1) )( t ) − ∇ · ( θ ∂ θ + θ h · ∇ h θ ) , and l ( t ) = 12 Z Σ ( ξ (1)1 ∂ ξ (1)1 + ξ (1) h · ∇ h ξ (1)1 )( t ) dS − Z Σ ( θ ∂ θ + θ h · ∇ h θ ) dS = 12 Z Σ ( − ξ (1)1 ∇ h · ξ (1) h + ξ (1) h · ∇ h ξ (1)1 )( t ) dS − Z Σ ( θ ∂ θ − ∇ h · θ h θ ) dS = Z Σ ( ξ (1) h · ∇ h ξ (1)1 )( t ) dS − Z Σ θ ∂ θ dS. | Z Σ ( ξ (1) h · ∇ h ξ (1)1 )( t ) dS | . k ξ (1) h ( t ) k L (Σ ) k∇ h ξ (1)1 ( t ) k L (Σ ) → t → + ∞ ) , applying Lebesgue Dominated Convergence Theorem yields Z ∞ Z Ω ∇ · v (2) dxdτ = l ( ∞ ) = Z Σ ( ξ (1) h · ∇ h ξ (1)1 )( ∞ ) dS − Z Σ θ ∂ θ dS = − Z Σ θ ∂ θ dS. On the other hand, set a ( t, x h ) def = R t R − b ∇ · v (2) dx dτ , one can see a ( t, x h ) = Z t v (2)1 | Σ dτ + Z t Z − b ∇ h · v (2) h dx dτ = η (2)1 | Σ ( t ) − η (2)1 | Σ (0) + Z t Z − b ∇ h · v (2) h dx dτ, then a ( ∞ , x h ) = η (2)1 | Σ ( ∞ ) − η (2)1 | Σ (0) + ∇ h · Z ∞ Z − b v (2) h dx dτ = Z ∞ Z − b ∇ h · v (2) h dx dτ. (A.9)While from (A.7), a direct computation implies a ( t, x h ) = Z t Z − b ∂ i ξ (1) j ∂ j v (1) i dx dτ = Z t Z − b ∂ γ ξ (1)1 ∂ v (1) γ dx dτ + Z t Z − b ( ∂ ξ (1)1 ∂ v (1)1 + ∂ ξ (1) α ∂ α v (1)1 + ∂ γ ξ (1) α ∂ α v (1) γ ) dx dτ, where Z t Z − b ∂ γ ξ (1)1 ∂ v (1) γ dx dτ = Z t Z − b ∂ γ ξ (1)1 ∂ ∂ t ξ (1) γ dx dτ = Z − b ( ∂ γ ξ (1)1 ∂ ξ (1) γ ) | tτ =0 dx − Z t Z − b ∂ γ v (1)1 ∂ ξ (1) γ dx dτ = Z − b ∂ γ ξ (1)1 ( t ) ∂ ξ (1) γ ( t )) dx − Z − b ( ∂ γ θ (1)1 ∂ θ (1) γ ) dx − Z t Z − b ∂ γ v (1)1 ∂ ξ (1) γ dx dτ, so it can be deduced that a ( t, x h ) = Z − b ∂ γ ξ (1)1 ( t ) ∂ ξ (1) γ ( t )) dx − Z − b ( ∂ γ θ (1)1 ∂ θ (1) γ ) dx + Z t Z − b ( ∇ h · ξ (1) h ∇ h · v (1) h + ∂ ξ (1) α ∂ α v (1)1 + ∂ γ ξ (1) α ∂ α v (1) γ − ∂ γ v (1)1 ∂ ξ (1) γ ) dx dτ.
64e thus prove that k a ( t, x h ) k L ( R h ) . k ∂ γ ξ (1)1 ( t ) k L (Ω) k ∂ ξ (1) γ ( t ) k L (Ω) + k ∂ γ θ (1)1 k L (Ω) k ∂ θ (1) γ k L (Ω) + Z t k∇ ξ (1) k L (Ω) k∇ h · v (1) k L (Ω) dτ, which follows k a ( ∞ , x h ) k L ( R h ) . k ∂ h ξ (1)1 k L ∞ ([0 , ∞ ); L (Ω)) k ∂ ξ (1) h k L ∞ ([0 , ∞ ); L (Ω)) + k ∂ h θ (1)1 k L (Ω) k ∂ θ (1) h k L (Ω) + k∇ ξ (1) k L ∞ ([0 , ∞ ); L (Ω)) Z ∞ k ∂ h v (1) k L (Ω) dτ < + ∞ , namely, a ( ∞ , x h ) ∈ L ( R h ). From this, integrating a ( ∞ , x h ) of (A.9) with respect to x h ∈ R h yields Z R h a ( ∞ , x h ) dx h = Z ∞ Z − b Z R h ∇ h · v (2) h dx h dx dτ = 0 , that is 0 = Z Σ a ( ∞ , x h ) dS = l ( ∞ ) . (A.10)Aiming for a contradiction by choosing θ so that l ( ∞ ) is nonzero, suppose that a ∈ C ∞ c (Σ ; R ) is nonzero. Let w ∈ H , tan , be some vector field such that w = a , ∂ w = a onΣ , w = ∂ w = 0 on Σ . If θ = ∇× w , we have ∇· θ = 0 and θ = ∂ w − ∂ w = ∂ w = ∂ a , ∂ θ = ∂ ∂ w − ∂ ∂ w = ∂ ∂ w = ∂ a . Hence, one has l ( ∞ ) = − Z Σ θ ∂ θ dS = − Z Σ ( ∂ a ) dS < , which contradicts (A.10). This ends the proof of Theorem 3.2. Acknowledgments.
The author would like to thank Professor J. Thomas Beale for hisvaluable suggestions and constructive comments. This work is supported in part by theNational Natural Science Foundation of China under the Grant 11571279.
References [1] H. Abels, The initial-value problem for the Navier-Stokes equations with a free surfacein L q -Sobolev spaces. Adv. Differential Equations , (2005), 45–64.[2] S. Agmon, A. Douglis, and L. Nirenberg, Estimates near the boundary for solutions ofellipic partial differential equations satisfying general boundary conditions II, Comm.Pure Appl. Math. , XVII (1964), 35–92.[3] S. Alinhac, Paracomposition et op´erateurs paradiff´erentiels,
Communications in PartialDifferential Equations , (1986), 87–121.654] H. Bahouri, J. Y. Chemin, and R. Danchin, Fourier analysis and nonlinear partial differ-ential equations , Grundlehren der mathematischen Wissenschaften 343, Springer-VerlagBerlin Heidelberg, 2011.[5] J. T. Beale, The initial value problem for the Navier-Stokes equations with a free surface,
Comm. Pure Appl. Math. , (3) (1981), 359–392.[6] J. T. Beale, Large-time regularity of viscous surface waves, Arch Rational. Mech.Anal. , Recent topicsin nonlinear PDE , II (Sendai, 1984), 1-14, North-Holland Math. Stud., 128, North-Holland, Amsterdam, 1985.[8] G. Gui, C. Wang, and Y. Wang, Local well-posedness of the vacuum free boundaryof 3-D compressible NavierCStokes equations,
Calculus of Variations and PDE , 2019,58:166.[9] Y. Guo and I. Tice, Local well-posedness of the viscous surface wave problem withoutsurface tension,
Analysis and PDE , Analysis and PDE , Bulletin of the Institute of Mathematics,Academia Sinica (New Series) , (2011), 293–303.[12] J.G. Heywood, The Navier-Stokes Equations: On the Existence, Regularity and Decayof Solutions, Indiana Univ. Math. J. , (1980), 639–681.[13] V. A. Solonnikov, Solvability of the problem of the motion of a viscous incompressible uidthat is bounded by a free surface, Izv Akad Nauk SSSR Ser. Mat. , Comm. Partial Differential Equations , Science China Mathematics , (2019), 1887–1924.[16] A. Tani and N. Tanaka, Large-time existence of surface waves in incompressible viscousuids with or without surface tension, Arch. Rational Mech. Anal. , (1995), 303–314.[17] Y. Wang, I. Tice, and C. Kim, The viscous surface-internal wave problem: Global well-posedness and decay, Arch. Rational Mech. Anal. , (2014), 1–92.[18] L. Wu, Well-Posedness and decay of the viscous surface wave, SIAM J. Math. Anal. ,46