Regularity Structure, Vorticity Layer and Convergence Rates of Inviscid Limit of Free Surface Navier-Stokes Equations with or without Surface Tension
aa r X i v : . [ m a t h . A P ] A ug Regularity Structure, Vorticity Layer and ConvergenceRates of Inviscid Limit of Free Surface Navier-StokesEquations with or without Surface Tension
Fuzhou Wu ∗ Yau Mathematical Sciences Center, Tsinghua UniversityBeijing 100084, ChinaCenter of Mathematical Sciences and Applications, Harvard UniversityCambridge, Massachusetts 02138, USA
Abstract
In this paper, we study the inviscid limit of the free surface in-compressible Navier-Stokes equations with or without surface ten-sion. By delicate estimates, we prove the weak boundary layer ofthe velocity of the free surface Navier-Stokes equations and the exis-tence of strong or weak vorticity layer for different conditions. Whenthe limit of the difference between the initial Navier-Stokes vorticityand the initial Euler vorticity is nonzero, or the tangential projec-tion on the free surface of the Euler strain tensor multiplying bynormal vector is nonzero, there exists a strong vorticity layer. Oth-erwise, the vorticity layer is weak. We estimate convergence ratesof tangential derivatives and the first order standard normal deriva-tive in energy norms, we show that not only tangential derivativesand standard normal derivative have different convergence rates, butalso their convergence rates are different for different Euler bound-ary data. Moreover, we determine regularity structure of the freesurface Navier-Stokes solutions with or without surface tension, sur-face tension changes regularity structure of the solutions.
Keywords : free surface Navier-Stokes equations, free surface Eulerequations, inviscid limit, strong vorticity layer, weak vorticity layer, reg-ularity structure
Contents ∗ E-mail: [email protected]; [email protected]; [email protected] σ = 0 L ∞ Estimate of Strong Vorticity Layer . . . . . . . . . . . . . . . 354 Strong Vorticity Layer Caused by the Discrepancy between BoundaryValues of Vorticities L ∞ Estimate of Strong Vorticity Layer . . . . . . . . . . . . . . . 415 Convergence Rates of Inviscid Limit for σ = 0 S ϕ v n | z =0 = 0 . . . . . 505.4 Estimates for Normal Derivatives when Π S ϕ v n | z =0 = 0 . . . . . 565.5 Convergence Rates of the Inviscid Limit . . . . . . . . . . . . . . 576 Regularity Structure of Navier-Stokes Solutions for Fixed σ > σ > A Derivation of the Equations and Boundary Conditions B Derivation of the Equations for the Surface Tension References In this paper, we study the inviscid limit of the free surface incompressibleNavier-Stokes equations with or without surface tension (see [31, 41, 14]): u t + u · ∇ u + ∇ p = ǫ △ u, x ∈ Ω t , ∇ · u = 0 , x ∈ Ω t ,∂ t h = u · N , x ∈ Σ t ,p n − ǫ S u n = gh n − σH n , x ∈ Σ t , ( u, h ) | t =0 = ( u ǫ , h ǫ ) . (1.1)where x = ( y, z ), y is the horizontal variable, z is the vertical variable, thenormalized pressure p = p F + gz , p F is the hydrodynamical pressure of the fluid, gz corresponds to the gravitational force. The surface tension in the dynamicalboundary condition (1 . , namely H = −∇ x · (cid:0) ( −∇ y h, √ |∇ y h | (cid:1) = ∇ y · (cid:0) ∇ y h √ |∇ y h | (cid:1) ,is twice the mean curvature of the free surface Σ t . The initial data satisfies the2ompatibility condition Π S u ǫ n | z =0 = 0. Some notations are defined as follows:Ω t = { x ∈ R | − ∞ < z < h ( t, y ) } , Σ t = { x ∈ R | z = h ( t, y ) } , N = ( −∇ h, ⊤ , n = N | N | , S u = ( ∇ u + ( ∇ u ) ⊤ ) , (1.2)where the symbol ⊤ means the transposition of matrices or vectors. We suppose h ( t, y ) → | y | → + ∞ for any t ≥ t , thus we simply assume −∞ < z < h ( t, y ).Also, we neglect the Coriolis effect generated by the planetary rotation, thenthere is no Ekman layer near the free surface even if Rossby number is small.Let ǫ → . u t + u · ∇ u + ∇ p = 0 , x ∈ Ω t , ∇ · u = 0 , x ∈ Ω t ,∂ t h = u · N , x ∈ Σ t ,p = gh − σH, x ∈ Σ t , ( u, h ) | t =0 = ( u , h ) := lim ǫ → ( u ǫ , h ǫ ) , (1.3)where ( u , h ) = lim ǫ → ( u ǫ , h ǫ ) is in the pointwise sense or even the L sense(see [31, 14, 34] for the sufficient conditions of the inviscid limit), ( u , h ) areindependent of ǫ . Note that except for ( u , h ) = lim ǫ → ( u ǫ , h ǫ ), we do not restricttheir derivatives, especially normal derivatives. Furtherly, note that the Navier-slip boundary case requires u = lim ǫ → u ǫ (see [22]), while the Dirichlet boundarycase requires u ǫ ( y , y ) ∼ u ( y , y ) + u P ( y , y √ ǫ ) + o ( ǫ ), where u P is the initialdata of Prandtl equations with Dirichlet boundary condition (see [35]).The following Taylor sign condition should be imposed on (1 .
3) if σ = 0 , g − ∂ z p | z =0 ≥ δ p > . (1.4)In this paper, either σ = 0 or σ > .
1) and (1 . σ > σ = 0 case.In order to describe the strength of the initial vorticity layer, we define ̟ bl = ∇ × u ǫ − ∇ × u = ∇ × u ǫ − ∇ × lim ǫ → u ǫ . (1.5)We emphasize that the initial vorticity layer means a boundary layer at theinitial time rather than a time layer in the vicinity of the initial time.If u ǫ has a profile u ǫ ( y, z ) ∼ u ( y, z ) + √ ǫu bl ( y, z √ ǫ ) in its asymptotic ex-pansion, then ∂ z u ǫ does not converge uniformly to ∂ z u , and thenlim ǫ → ̟ bl = ( − ∂ z u bl, , ∂ z u bl, , ⊤ = 0 , (1.6)3hich means the initial vorticity layer is strong.For strong initial vorticity layer, there is a special case: if the Euler bound-ary data satisfies Π S u n | z =0 = 0, then lim ǫ → ̟ bl | z =0 = 0 on the free surface dueto the compatibility condition Π S u ǫ n | z =0 = 0. However, it can not prevent(1 .
6) from holding in the vicinity of the free surface. For example, we choosethe boundary layer profile to be u bl ( y, z √ ǫ ) = exp {− ( z √ ǫ ) } (1 , , ⊤ , for whichlim ǫ → ̟ bl (cid:12)(cid:12) z =0 = 0 , lim ǫ → ̟ bl (cid:12)(cid:12) z = −√ ǫ = 2 e − ( − , , ⊤ = 0 . (1.7)On the contrary, if lim ǫ → ̟ bl = 0, then lim ǫ → ̟ bl | z =0 = 0 due to its continuity, andthen Π S u n | z =0 = 0 at the initial time.If u ǫ has a profile u ǫ ( y, z ) ∼ u ( y, z ) + ǫ + δ ubl u bl ( y, z √ ǫ ) in its asymptoticexpansion, where δ ubl >
0, then lim ǫ → ̟ bl = 0, which means the initial vorticitylayer is weak.In order to describe the discrepancy between boundary value of Navier-Stokes vorticity and that of Euler vorticity, we investigate whether the Eulerboundary data satisfies Π S u n | Σ t = 0. If Π S u n | Σ t = 0, the boundary value ofNavier-Stokes vorticity converges to that of Euler vorticity; otherwise there is adiscrepancy.It is easy to have Π S u n | Σ t = 0 in (0 , T ], because it satisfies the forcedtransport equation. While Π S u n | Σ t = 0 in [0 , T ] is nontrivial. However, we canconstruct the Euler velocity field satisfying Π S u n | Σ t = 0 and finite energy. Thescenario of our problem is as follows: construct a Euler velocity field satisfyingΠ S u n | Σ t = 0, let Navier-Stokes initial data is a small perturbation of the Eulerinitial data, then we study the inviscid limit of Navier-Stokes solutions.One example of Π S u n | Σ t = 0 is that u = ( − y e − y − y − z , y e − y − y − z , , h = 0 , (1.8)and the pressure p is the solution of the Poisson equation: ( −△ p = e − y + y + z )( − y +4 y − y y ) ,p | z =0 = 0 . (1.9)Then the Euler boundary data satisfies S u n | z =0 = [ y ze − y − y − z , − y ze − y − y − z , ⊤ (cid:12)(cid:12) z =0 = 0 . (1.10)By deforming symmetrically the velocity field (1 .
8) where h is also symmetric,one may construct infinitely many velocity fields satisfying Π S u n | Σ t = 0. In this survey, we introduce the previous results on the well-posedness andinviscid limits.As to the irrotational fluids, refer to S. Wu [43, 44, 45, 46], Germain,Masmoudi and Shatah [15], Ionescu and Pusateri [24], Alazard and Delort [1] for4he water waves without surface tension, refer to K. Beyer and M. G¨ u nther [8],Germain, Masmoudi and Shatah [16] for the water waves with surface tension.Before introducing previous results on the boundary layer and inviscid limitproblem, we survery there some well-posedness results. The free surface Navier-Stokes equations have both local and global well-posedness results, while thefree surface Euler equations only have local well-posedness results.As to the free surface Navier-Stokes equations, refer to Beale [6], Hataya[20], Guo and Tice [17, 18, 19] for the zero surface tension, refer to Beale [7],Tani [38], Tanaka and Tani [39] for the surface tension case. Especially, [7, 39,20, 17, 18] proved the global in time results for the small initial data. Note thatthe viscosity is capable of producing the global well-posedness, while the surfacetension only provides the regularizing effect on the free surface and enhance thedecay rates of the solutions (see [17]).The general free surface Euler equations which are much more difficultand only have local results. Refer to Lindblad [26], Coutand and Shkiller [10],Shatah and Zeng [36], Zhang and Zhang [50] for the zero surface tension case,refer to Coutand and Shkiller [10], Shatah and Zeng [36] for the surface tensioncase.As the viscosity approaches zero, we hope that the solutions of Navier-Stokes equations converge to the solutions of Euler equations. However, this isonly proved in the whole spaces where there are no boundary conditions, see[37, 25, 12, 13, 9, 30]. However, in the presence of boundaries, the inviscid limitproblem will be challenging due to the formation of boundary layers.For Navier-Stokes equations with Dirichlet boundary condition in the fixeddomain, u | ∂ Ω = 0, strong boundary layer whose width is O ( √ ǫ ) and amplitude is O (1) forms near the boundary. Namely, the Navier-Stokes solution is expectedto behave like u ǫ ∼ u + u bl ( t, y, z/ √ ǫ ) where u is the Euler solution satisfyingcharacteristic boundary condition u · n | ∂ Ω = 0, u bl ( t, y, z/ √ ǫ ) is the boundarylayer profile. The inviscid limit is not rigorously verified except for the followingtwo cases, i. e., the analytic setting (see [5, 35]) and the case where the vorticityis located away from the boundary (see [28, 29]).For Navier-Stokes equations with Navier-slip boundary condition in thefixed domain, Π(2 S u n + γ s u ) | ∂ Ω = 0 , u · n | ∂ Ω = 0, weak boundary layer whosewidth and amplitude are O ( √ ǫ ) forms near the boundary. Namely, the Navier-Stokes solution is expected to behave like u ǫ ∼ u + √ ǫu bl ( t, y, z/ √ ǫ ), where u is the Euler solution satisfying characteristic boundary condition u · n | ∂ Ω = 0.For the inviscid limit, refer to Iftimie and Planas [22], Iftimie and Sueur [23],Masmoudi and Rousset [32], Xiao and Xin [49]. Note that H convergence issatisfied for general Navier-slip boundary condition or curved boundary, while H convergence happens for complete slip boundary condition ω × n | ∂ Ω = 0 , u · n | ∂ Ω = 0 and flat boundary (see [47, 11]).For the free surface Navier-Stokes equations with kinetical and dynamicalboundary conditions in the moving domain, the recent works on the inviscidlimit are studied in conormal Sobolev spaces for which the normal differentialoperators vanish on the free surface. Masmoudi and Rousset [31] proved theuniform estimates and inviscid limit of the free surface incompressible Navier-Stokes equations without surface tension in conormal Sobolev spaces. By ex-5ending this conormal analysis framework, Wang and Xin [41], Elgindi and Lee[14] proved the inviscid limit of the free surface incompressible Navier-Stokesequations with surface tension, Mei, Wang and Xin [34] proved the inviscid limitof the free surface compressible Navier-Stokes equations with or without surfacetension. [31] pointed out the free surface Navier-Stokes solutions are expectedto behave like u ǫ ∼ u + √ ǫu bl ( t, y, z/ √ ǫ ), where u is the free surface Eulersolutions. We first study N-S (abbreviation of Navier-Stokes) equations (1 .
1) with σ = 0. In this subsection, we formulate the free boundary problem into thefixed coordinates domain R − . Similar to [31], we define the diffeomorphismbetween R − and the moving domain Ω t :Φ( t, · ) : R − = R × ( −∞ , → Ω t ,x = ( y, z ) → ( y, ϕ ( t, y, z )) , (1.11)and define ϕ as ϕ ( t, y, z ) = Az + η ( t, y, z ) , (1.12)where A > η is defined as η ( t, y, z ) = ψ ∗ y h ( t, y ) , (1.13)here the symbol ∗ y is a convolution in the y variable and ψ decays sufficientlyfast in z such that (1 − z ) ψ, ψ, ∂ z ψ, · · · , ∂ m +1 z ψ ∈ L (d z ). For example, ψ = F − [ − z ) e − (1 − z ) (1+ | ξ | ) ] where F − is the inverse Fourier transformationwith respect to ξ ∈ R .The constant A > ∂ z ϕ (0 , y, z ) ≥ , ∀ x ∈ R − . (1.14)By the diffeomorphism (1 . v ( t, x ) = u ( t, y, ϕ ( t, y, z )) , q ( t, x ) = p ( t, y, ϕ ( t, y, z )) , ∀ x ∈ R − ,∂ ϕi v ( t, x ) = ∂ i u ( t, y, ϕ ( t, y, z )) , ∂ ϕi q ( t, x ) = ∂ i p ( t, y, ϕ ( t, y, z )) , i = t, , , , (1.15)while h ( t, y ) does not change.Then the free surface Navier-Stokes equations (1 .
1) with σ = 0 are equiva-lent to the following system: ∂ ϕt v + v · ∇ ϕ v + ∇ ϕ q = ǫ △ ϕ v, x ∈ R − , ∇ ϕ · v = 0 , x ∈ R − ,∂ t h = v ( t, y, · N, z = 0 ,q n − ǫ S ϕ v n = gh n , z = 0 , ( v, h ) | t =0 = ( v ǫ , h ǫ ) , (1.16)6here N = ( −∇ h ( t, y ) , ⊤ , n = N | N | , S ϕ v = ( ∇ ϕ v + ∇ ϕ v ⊤ ) . (1.17)Obviously, let ǫ → . ∂ ϕt v + v · ∇ ϕ v + ∇ ϕ q = 0 , x ∈ R − , ∇ ϕ · v = 0 , x ∈ R − ,∂ t h = v ( t, y, · N, z = 0 ,q = gh, z = 0 , ( v, h ) | t =0 = ( v , h ) , (1.18)where v is the limit of v ǫ in the L sense, h is the limit of h ǫ in the L sensefor σ = 0 and in the H sense for σ >
0, ( v , h ) is independent of ǫ . Thefollowing Taylor sign condition should be imposed on (1 .
18) when σ = 0, g − ∂ ϕz q | z =0 ≥ δ q > . (1.19)D. Coutand and S. Shkoller (see [10]) proved the well-posedness of the freesurface incompressible Euler equations (1 .
18) without surface tension. We statetheir results in our formulation as follows:
Suppose the Taylor sign condition (1 . holds at t = 0 , h ∈ H ( R ) , v ∈ H ( R − ) , then there exists T > and a unique solution ( v, q, h ) of (1 . with v ∈ L ∞ ([0 , T ] , H ( R − )) , ∇ q ∈ L ∞ ([0 , T ] , H ( R − )) , h ∈ L ∞ ([0 , T ] , H ( R )) . Though conormal derivatives of the Navier-Stokes solutions and conormalderivatives of Euler solutions vanish on the free boundary, their differences os-cillate dramatically in the vicinity of the free boundary, thus the conormalfunctional spaces are not suitable for studying the convergence rates of inviscidlimit. Thus, we define the following functional spaces: k v k X m,s := P ℓ ≤ m, | α |≤ m + s − ℓ k ∂ ℓt Z α v k L ( R − ) , k v k X m := k v k X m, , k v k X m,stan := P ℓ ≤ m, | α |≤ m + s − ℓ k ∂ ℓt ∂ αy v k L ( R − ) , k v k X mtan := k v k X m, tan , | h | X m,s := P ℓ ≤ m, | α |≤ m + s − ℓ | ∂ ℓt ∂ αy h | L ( R ) , | h | X m := | h | X m, , k v k Y m,stan := P ℓ ≤ m, | α |≤ m + s − ℓ k ∂ ℓt ∂ αy v k L ∞ ( R − ) , k v k Y mtan := k v k Y m, tan , | h | Y m,s := P ℓ ≤ m, | α |≤ m + s − ℓ | ∂ ℓt ∂ αy h | L ∞ ( R ) , | h | Y m := | h | Y m, , (1.20)where the differential operators Z = ∂ y , Z = ∂ y , Z = z − z ∂ z (see [31, 14,41, 34]). Also, we use | · | m to denote the standard Sobolev norm defined in thehorizontal space R . 7ssume ω ǫ = ∇ ϕ ǫ × v ǫ , ω = ∇ ϕ × v are Navier-Stokes vorticity, Eulervorticity respectively, ˆ ω = ω ǫ − ω . In this paper, bounded variables or quantitiesmean that they are bounded by O (1), small variables or quantities mean thatthey are bounded by O ( ǫ β ) for some β >
0. Now we state our motivations ofthis paper.1. As ǫ →
0, [31] showed that the velocity converges in L and L ∞ norms,the height function converges in L and W , ∞ norms. We can expect thattheir tangential derivatives converges, but we still do not know whether thevorticity and normal derivatives of the velocity converge in L ∞ norm. If theydo not converge in the L ∞ norm, there are must be a strong vorticity layer inthe vicinity of the free surface. [31] pointed the N-S solution is expected tobehave like u ǫ ∼ u + √ ǫu bl ( t, y, z/ √ ǫ ), however, this is not rigorously proved.It is expected that the velocity of the free surface N-S equations has a weakboundary layer, we have to prove the existence of strong vorticity layer for somesufficient conditions. Note that the energy norms are too weak, thus we use the L ∞ norm to describe the existence of strong boundary layers.2. We want to know the sufficient and necessary conditions for the exis-tence of strong vorticity layer, we also want to know these conditions for theweak vorticity layer. We show that there are two sufficient conditions for thestrong vorticity layer, note that these two conditions are almost independent.One condition is that the initial vorticity layer is strong, then it is transportedby the velocity field for any small ǫ , and then we get a strong vorticity layerwhen t ∈ (0 , T ]. Another condition is that the Euler boundary data satisfiesΠ S ϕ v n | z =0 = 0 in (0 , T ], then there is a discrepancy between N-S vorticity andEuler vorticity, and then we have a strong vorticity layer. When neither of twosufficient conditions is satisfied, we show that the vorticity layer is weak.3. [31, 41, 14] proved the uniform regularity and inviscid limit of the freesurface N-S equations with or without surface tension. In order to prove theuniform regularities, [31, 14, 41, 34] controlled the bounded quantities in conor-mal functional spaces and applied the following integration by parts formula tothe a priori estimates: dd t R R − f d V t = R R − ∂ ϕt f d V t + R { z =0 } f v · N d y, R R − ~a · ∇ ϕ f d V t = R { z =0 } ~a · N f d y − R R − ∇ ϕ · ~a f d V t , R R − ~a · ( ∇ ϕ × ~b ) d V t = R { z =0 } ~a · ( N × ~b ) d y + R R − ( ∇ ϕ × ~a ) · ~b d V t , (1.21)where d V t = ∂ z ϕ d y d z is defined on R − but measures the volume element of Ω t .Refer to [31] for the first and second formulae in (1 . . used in the fixed domain, refer to [40, 42, 49].Motivated by [31, 41, 14], we want to know convergence rates of the inviscidlimit, which involves two moving domain, we denote Navier-Stokes domain andEuler domain by Ω ǫ and Ω respectively. In general, Ω ǫ and Ω do not coincide,we can not compare these two velocity fields. Thus, we have to map Ω ǫ andΩ to the common fixed coordinate domain R − , namely Ω ǫ = Φ ǫ ( R − ) , Ω =Φ( R − ). For any x ∈ R − , two points Φ ǫ ( x ) and Φ( x ) do not coincide in general,However, Φ ǫ ( x ) converges to Φ( x ) pointwisely as ǫ →
0, thus | v ǫ ( x ) − v ( x ) | | ∂ ℓt Z α v ǫ ( x ) − ∂ ℓt Z α v ( x ) | must be small quantities. We have to overcomemany difficulties involving two different moving domains to close the estimatesof | ∂ ℓt Z α v ǫ ( x ) − ∂ ℓt Z α v ( x ) | .4. [47, 48, 49, 40] studied the inviscid limit of the incompressible or com-pressible N-S equations with Navier-slip boundary condition, where the initialNavier-Stokes data and initial Euler data are exactly the same and independentof ǫ . If the Navier-Stokes boundary condition satisfies ω ǫ × n | z =0 = 0 and theboundary is flat, then the Euler boundary data also satisfies ω × n | z =0 = 0, k u ǫ − u k L . O ( ǫ ) , k ω ǫ − ω k L + k u ǫ − u k H . O ( ǫ ), [47] proved the H convergence. While if the Euler boundary data is general or the boundary iscurved, then k u ǫ − u k L . O ( ǫ ) , k ω ǫ − ω k L + k u ǫ − u k H . O ( ǫ ). [23] showedthat it is impossible to prove H convergence.We are also interested in the convergence rates of the inviscid limit of thefree boundary problem for Navier-Stokes equations. However, in our formulationof the free boundary problem, the diffeomorphism between the fixed coordinates R − and two moving domains are twisted, the differential operators in N-S andEuler equations are also twisted, then the estimates of tangential derivativesand the estimates of normal derivatives can not be decoupled, we even cannot develop the L estimate of ( v ǫ − v, h ǫ − h ) themselves without involvingthe normal derivative ∂ z v ǫ − ∂ z v . Thus, we want to know whether tangentialderivatives and normal derivatives have different convergence rates.If the Euler boundary data satisfies Π S ϕ v | z =0 = 0, ω ǫ | z =0 does not convergeto ω | z =0 , we want to know how to calculate convergence rates of the vorticityin the energy norm. If Π S ϕ v | z =0 = 0, ω ǫ | z =0 → ω | z =0 , we want to know howto improve the convergence rates.5. To estimate the convergence rate of the inviscid limit, we need to usethe time derivatives. However, time derivatives can not be expressed in termsof space derivatives by using the equations, since we work on conormal spacesinstead of standard Sobolev spaces. Thus, we prove the uniform regularity con-cluding time derivatives and determine the regularity structure of N-S solutionsand Euler solutions in conormal functional spaces. When time derivatives areincluded, uniform estimates of tangential derivatives will be different from [31].Moreover, our estimates of normal derivatives are based on the estimates ofvorticity rather than those of Π S ϕ v n (see [31, 41, 14]). [31] proved the uniform regularity of space derivatives of the free surfaceNavier-Stokes equations (1.16), while the following proposition concerns theuniform regularity of time derivatives. Proposition 1.1.
For m ≥ , assume the initial data ( v ǫ , h ǫ ) satisfy the com-patibility condition Π S ϕ v ǫ n | z =0 = 0 and the regularities: sup ǫ ∈ (0 , (cid:0) | h ǫ | X m − , + ǫ | h ǫ | X m − , + k v ǫ k X m − , + k ω ǫ k X m − + k ω ǫ k , ∞ + ǫ k ∂ z ω ǫ k L ∞ (cid:1) ≤ C , (1.22)9 here C > is suitably small such that the Taylor sign condition g − ∂ ϕ ǫ z q ǫ | z =0 ≥ c > , then the unique Navier-Stokes solution to (1 . satisfies sup t ∈ [0 ,T ] (cid:0) | h ǫ | X m − , + ǫ | h ǫ | X m − , + k v ǫ k X m − , + k ∂ z v ǫ k X m − + k ω ǫ k X m − + k ∂ z v ǫ k , ∞ + ǫ k ∂ zz v ǫ k L ∞ (cid:1) + k ∂ mt h k L ([0 ,T ] ,L ) + ǫ k ∂ mt h k L ([0 ,T ] ,H ) + ǫ T R k∇ v ǫ k X m − , + k∇ ∂ z v ǫ k X m − d t ≤ C. (1.23) As ǫ → , the Euler solution to (1 . satisfies the following regularities: sup t ∈ [0 ,T ] (cid:0) | h | X m − , + k v k X m − , + k ∂ z v k X m − + k ω k X m − + k ∂ z v k , ∞ (cid:1) + k ∂ mt h k L ([0 ,T ] ,L ) ≤ C, (1.24) where the Taylor sign condition g − ∂ ϕz q | z =0 ≥ c > holds. For the initial regularities (1 . k ∂ mt v ǫ k L ([0 ,T ] ,L ) . Toprove k ∂ mt v ǫ k L ([0 ,T ] ,L ) , it requires (1 .
22) as well as ∂ mt v ǫ , ∂ mt h ǫ ∈ L ( R − ).Note that when σ = 0, we must use the following Alinhac’s good unknown(see [2, 31]) to estimate tangential derivatives: V ℓ,α = ∂ ℓt Z α v − ∂ ϕz v∂ ℓt Z α η, < ℓ + | α | ≤ m, ℓ ≤ m − ,Q ℓ,α = ∂ ℓt Z α q − ∂ ϕz q∂ ℓt Z α η, < ℓ + | α | ≤ m, ℓ ≤ m − . (1.25)Our proof of Proposition 1 . k ∂ ℓt q k L has no boundin general. When | α | = 0, we estimate V ℓ, and ∇ ∂ ℓt q where 0 ≤ ℓ ≤ m −
1, thedynamical boundary condition can not be used.(ii) [31] as well as [41, 14, 34] estimated normal derivatives by using Π S ϕ n and its evolution equations. While in this paper, we estimate normal derivativesby using the vorticity and the following equations: ∂ ϕt ω h + v · ∇ ϕ ω h − ǫ △ ϕ ω h = ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) ,ω | z =0 = F [ ∇ ϕ ]( ∂ j v i ) ,ω | z =0 = F [ ∇ ϕ ]( ∂ j v i ) , (1.26)where j = 1 , , i = 1 , , ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) is a quadratic polynomial vectorwith respect to ω h and ∂ j v i , F [ ∇ ϕ ]( ∂ j v i ), F [ ∇ ϕ ]( ∂ j v i ) are polynomials withrespect to ∂ j v i , all the coefficients are fractions of ∇ ϕ .(iii) In [31], the Taylor sign condition is g − ∂ ϕ ǫ z q ǫ,E | z =0 ≥ c >
0, thatis imposed on the Euler part of the pressure q ǫ . q ǫ has a decomposition q ǫ = q ǫ,E + q ǫ,NS which satisfy ( △ ϕ ǫ q ǫ,E = − ∂ ϕ ǫ i v ǫ,j ∂ ϕ ǫ j v ǫ,i ,q ǫ,E | z =0 = gh ǫ . ( △ ϕ ǫ q ǫ,NS = 0 ,q ǫ,NS | z =0 = 2 ǫ S ϕ ǫ v n · n . (1.27)10owever, the force term of q ǫ,E has boundary layer in the vicinity of the freeboundary in general, thus ∂ ϕ ǫ z q ǫ,E | z =0 may also have boundary layer, it is un-known whether ∂ ϕ ǫ z q ǫ,E | z =0 converges pointwisely to ∂ ϕz q | z =0 or not. Differ-ent from [31], our Taylor sign condition is g − ∂ ϕ ǫ z q ǫ | z =0 ≥ c >
0. Since ∂ ϕ ǫ z q ǫ | z =0 = ǫ △ ϕ ǫ v − ∂ t v − v ǫy ·∇ y v ǫ, and k ∂ zz v k L ∞ , √ ǫ k ∂ zz v k L ∞ are bounded,thus ∂ ϕ ǫ z q ǫ | z =0 converges to ∂ ϕz q | z =0 pointwisely.For classical solutions to the free surface Navier-Stokes equations (1 . σ = 0, we will estimate the convergence rates of the velocity later, whichimplies the weak boundary layer of the velocity. Before estimating the conver-gence rates, we show the following theorem which states the existence of strongvorticity layer. Theorem 1.2.
Assume
T > is finite, fixed and independent of ǫ , ( v ǫ , h ǫ ) isthe solution in [0 , T ] of Navier-Stokes equations (1 . with initial data ( v ǫ , h ǫ ) satisfying (1 . , ω ǫ is its vorticity. ( v, h ) is the solution in [0 , T ] of Eulerequations (1 . with initial data ( v , h ) ∈ X m − , ( R − ) × X m − , ( R ) , ω is itsvorticity.(1) If the initial Navier-Stokes velocity satisfies lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0 in the initial set A , the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , then the Navier-Stokes solution of (1 . has a strong vorticity layersatisfying lim ǫ → k ω ǫ − ω k L ∞ ( X ( A ) × (0 ,T ]) = 0 , lim ǫ → k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k L ∞ ( X ( A ) × (0 ,T ]) = 0 , lim ǫ → kS ϕ ǫ v ǫ − S ϕ v k L ∞ ( X ( A ) × (0 ,T ]) = 0 , lim ǫ → k∇ ϕ ǫ q ǫ − ∇ ϕ q k L ∞ ( X ( A ) × (0 ,T ]) = 0 . (1.28) where X ( A ) = {X ( t, x ) (cid:12)(cid:12) X (0 , x ) ∈ A , ∂ t X ( t, x ) = v ( t, Φ − ◦ X ) } .(2) If lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0 , the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in (0 , T ] , then the Navier-Stokes solution of (1 . has a strongvorticity layer satisfying lim ǫ → (cid:12)(cid:12) ω ǫ | z =0 − ω | z =0 (cid:12)(cid:12) L ∞ ( R × (0 ,T ]) = 0 , lim ǫ → k ω ǫ − ω k L ∞ ( R × [0 ,O ( ǫ − δz )) × (0 ,T ]) = 0 , lim ǫ → k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k L ∞ ( R × [0 ,O ( ǫ − δz )) × (0 ,T ]) = 0 , lim ǫ → kS ϕ ǫ v ǫ − S ϕ v k L ∞ ( R × [0 ,O ( ǫ − δz )) × (0 ,T ]) = 0 , lim ǫ → k∇ ϕ ǫ q ǫ − ∇ ϕ q k L ∞ ( R × [0 ,O ( ǫ − δz )) × (0 ,T ]) = 0 , (1.29) for some constant δ z ≥ .(3) lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0 and Π S ϕ v n | z =0 = 0 in [0 , T ] arenecessary conditions for the Navier-Stokes solution of (1 . to have a weak orticity layer satisfying lim ǫ → k ω ǫ − ω k L ∞ ( Cl ( R − ) × (0 ,T ]) = 0 , lim ǫ → k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k L ∞ ( Cl ( R − ) × (0 ,T ]) = 0 , lim ǫ → kS ϕ ǫ v ǫ − S ϕ v k L ∞ ( Cl ( R − ) × (0 ,T ]) = 0 , lim ǫ → k∇ ϕ ǫ q ǫ − ∇ ϕ q k L ∞ ( Cl ( R − ) × (0 ,T ]) = 0 , (1.30) where Cl ( R − ) = R − ∪ { x | z = 0 } is the closure of R − . We give some remarks on Theorem 1 . Remark 1.3. (i) To represent ∂ ϕ ǫ z v ǫ − ∂ ϕz v is more natural than ∂ z v ǫ − ∂ z v .However, lim ǫ → k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k L ∞ = 0 results from lim ǫ → k ∂ z v ǫ − ∂ z v k L ∞ = 0 and lim ǫ → k ∂ z ( η ǫ − η ) k L ∞ = 0 , due to the formula: ∂ ϕ ǫ z v ǫ − ∂ ϕz v = ∂ ϕ ǫ z ( v ǫ − v ) − ∂ ϕz v ∂ ϕ ǫ z ( η ǫ − η )= ∂ z ϕ ǫ · ∂ z ( v ǫ − v ) − ∂ ϕz v ∂ z ϕ ǫ · ∂ z ( η ǫ − η ) . (1.31) (ii) The energy norm k · k L is weaker than the L ∞ norm, because k ω ǫ − ω k L ( R − ) = 0 , even though we have the profile ω ǫ ( t, y, z ) ∼ ω ( t, y, z )+ ω bl ( t, y, z √ ǫ ) .While k ω ǫ − ω k L ∞ ( R − ) = 0 . Thus, we use the L ∞ norm to describe the strongvorticity layer.(iii) S n = Π S ϕ v n satisfies the forced transport equations: ∂ ϕt S n + v · ∇ ϕ S n = − Π (cid:0) ( ∇ ϕ v ) + (( ∇ ϕ v ) ⊤ ) (cid:1) n − Π(( D ϕ ) q ) n +( ∂ ϕt Π + v · ∇ ϕ Π) S ϕ v n + Π S ϕ v ( ∂ ϕt n + v · ∇ ϕ n ) , (1.32) where (cid:0) ( D ϕ ) q (cid:1) is the Hessian matrix of q . The equation (1 . implies thateven if S n | t =0 = 0 , then S n = 0 in (0 , T ] is possible due to the force terms of (1 . .However, S n | z =0 ≡ in [0 , T ] can be constructed, see an example con-structed in (1 . , (1 . , (1 . .(iv) Π S ϕ v n | z =0 = 0 at t = 0 implies that lim ǫ → ( ∇ ϕ ǫ × v ǫ ) | z =0 − ∇ ϕ × lim ǫ → v ǫ | z =0 = 0 . But it does not contradict with lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0 in the initial set A , see (1 . where A = { x | z = −√ ǫ } in local coordinates. If Π S ϕ v n | z =0 = 0 in [0 , T ] and lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0 in the initial set A , then it is easy to know the results are the union of (1 . and (1 . .(v) N · ∂ ϕz v and N · ∂ z v do not have boundary layer, but ∂ z v has boundarylayer in general. Similarly, N · ω does not have boundary layer, but ω hasboundary layer in general. The reason is that both v | z =0 and ω | z =0 are notperpendicular to the free surface in general. The proof of Theorem 1 . ω = ω ǫ − ω ∂ ϕ ǫ t ˆ ω h + v ǫ · ∇ ϕ ǫ ˆ ω h − ǫ △ ϕ ǫ ˆ ω h = ~ F [ ∇ ϕ ǫ ]( ω ǫh , ∂ j v ǫ,i ) − ~ F [ ∇ ϕ ]( ω h , ∂ j v i )+ ǫ △ ϕ ǫ ω h + ∂ ϕz ω h ∂ ϕ ǫ t ˆ η + ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ η − ˆ v · ∇ ϕ ω h , ˆ ω h | z =0 = F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − ω bh , ˆ ω h | t =0 = (ˆ ω , ˆ ω ) ⊤ , (1.33)where ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) and F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) = ( F [ ∇ ϕ ]( ∂ j v i ) , F [ ∇ ϕ ]( ∂ j v i )) ⊤ are defined in (1 . ~ F [ ∇ ϕ ]( ω h , ∂ j v i ), ω h has degree one.By introducing Lagrangian coordinates (3 . .
33) can betransformed into the heat equation with damping and force terms.By splitting (3 .
9) and estimating (3 .
10) and (3 . ǫ → (cid:13)(cid:13) ˆ ω h | t =0 (cid:13)(cid:13) L ∞ ( A ) = 0, we prove that the limitof ˆ ω h is equal to that of the initial vorticity layer in Lagrangian coordinates,thus the limit of the initial vorticity layer is transported in Eulerian coordinates.Namely, lim ǫ → k ˆ ω k L ∞ ( X ( A ) × (0 ,T ]) = 0.By splitting (4 .
16) and estimating (4 .
17) and (4 . ω h | z =0 = 0 and lim ǫ → (cid:13)(cid:13) ˆ ω h | t =0 (cid:13)(cid:13) L ∞ = 0, there is a discrepancy between N-S vor-ticity and Euler vorticity, we prove that lim ǫ → k ω ǫ − ω k L ∞ ( R × [0 ,O ( ǫ − δz )) × (0 ,T ]) = 0 by using symbolic analysis.The following theorem concerns the convergence rates of the inviscid limitsof (1 . Theorem 1.4.
Assume
T > is finite, fixed and independent of ǫ , ( v ǫ , h ǫ ) isthe solution in [0 , T ] of Navier-Stokes equations (1 . with initial data ( v ǫ , h ǫ ) satisfying (1 . , ω ǫ is its vorticity. ( v, h ) is the solution in [0 , T ] of Eulerequations (1 . with initial data ( v , h ) ∈ X m − , ( R − ) × X m − , ( R ) , ω is itsvorticity. g − ∂ ϕ ǫ z q ǫ | z =0 ≥ c > , g − ∂ ϕz q | z =0 ≥ c > . Assume there existsan integer k where ≤ k ≤ m − , such that k v ǫ − v k X k − , ( R − ) = O ( ǫ λ v ) , | h ǫ − h | X k − , ( R ) = O ( ǫ λ h ) , k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) , where λ v > , λ h > , λ ω > .If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , then theconvergence rates of the inviscid limit satisfy k v ǫ − v k X k − , tan + | h ǫ − h | X k − , = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k X k − tan + k N ǫ · ω ǫ − N · ω k X k − tan = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan + k ω ǫ − ω k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k∇ ϕ ǫ q ǫ − ∇ ϕ q k X k − tan + k△ ϕ ǫ q ǫ − △ ϕ q k X k − tan = O ( ǫ min { , λv , λh , λω } ) , (1.34)13 v ǫ − v k Y k − tan + | h ǫ − h | Y k − = O ( ǫ min { , λv , λh , λω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k Y k − tan + k N ǫ · ω ǫ − N · ω k Y k − tan = O ( ǫ min { , λv , λh , λω } ) . If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , assume k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) where λ ω > , then the convergence rates of the inviscidlimit satisfy k v ǫ − v k X k − , tan + | h ǫ − h | X k − , = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k X k − tan + k N ǫ · ω ǫ − N · ω k X k − tan = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan + k ω ǫ − ω k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k∇ ϕ ǫ q ǫ − ∇ ϕ q k X k − tan + k△ ϕ ǫ q ǫ − △ ϕ q k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k v ǫ − v k Y k − tan + | h ǫ − h | Y k − = O ( ǫ min { , λv , λh , λω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k Y k − tan + k N ǫ · ω ǫ − N · ω k Y k − tan = O ( ǫ min { , λv , λh , λω } ) . (1.35)We give some remarks on Theorem 1 . Remark 1.5. (i) Convergence rates are represented in functional spaces con-taining only tangential derivatives, such as X k − , tan and Y k − tan , because we are ac-tually interested in standard derivatives rather than conormal derivatives. The-orem . has already described the behaviors of standard normal derivatives.However, we have to use conormal Sobolev spaces to estimate convergence rates,the initial data is also required to converge in conormal Sobolev spaces, thus theindex k depends on the strength of the initial vorticity layer. The weaker theinitial vorticity layer is, the larger k is.(ii) In general, ∇ · ( v ǫ − v ) = 0 , q ǫ − q is infinite for the infinite fluiddepth. Thus, we even can not obtain the L estimate of ( v ǫ − v, h ǫ − h ) withoutinvolving ∂ z v ǫ − ∂ z v . The estimates of tangential derivatives and the estimates ofnormal derivatives can not be decoupled, but tangential derivatives and normalderivatives have different convergence rates.(iii) The convergence rate of the initial vorticity is related to whether theEuler boundary data satisfies Π S ϕ v n | z =0 ,t =0 = 0 or not. In general, λ ω = λ ω .The convergence rates of the inviscid limit for the free boundary problem areslower than those of the Navier-slip boundary case.(iv) To represent ∂ ϕ ǫ z v ǫ − ∂ ϕz v is more natural than ∂ z v ǫ − ∂ z v . However, theestimate of k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan results from the estimate of k ∂ z v ǫ − ∂ z v k X k − tan and k ∂ z ( η ǫ − η ) k X k − tan , due to the formula (1 . . The L ∞ type estimates in (1 . are based on the formula k f k L ∞ . k f k H s tan k ∂ z f k H s tan where s + s > (see [31]). k v ǫ − v k X ktan and | h ǫ − h | X k can not be estimated because we can notcontrol k ∂ kt ( q ǫ − q ) k L . v) For the finite fluid depth R × [ − L, and L > can be very small, if theinitial data satisfy k v ǫ − v k X m − , ( R × [ − L, = O ( ǫ λ v , L ) , | h ǫ − h | X m − , ( R ) = O ( ǫ λ h ) . k ω ǫ − ω k X m − ( R × [ − L, = O ( ǫ λ ω , L ) . If Π S ϕ v n | z =0 = 0 in [0 , T ] ,then the pressure itself has L type estimates: k q ǫ ( · , z ) − q ( · , z ) k X k − tan ( R ) . (cid:12)(cid:12) q e | z =0 − q | z =0 (cid:12)(cid:12) X k − tan + k ∂ z q ǫ − ∂ z q k X k − tan . O ( ǫ min { , λv , λh , λω } ) , k q ǫ − q k X k − tan ( R × [ − L, . O ( ǫ min { , λv , λh , λω } , L ) , k ω ǫ − ω k X k − tan ( R × [ − L, . O ( ǫ min { , λv , λh , λω } ) , k v ǫ − v k X k − tan ( R × [ − L, . O ( ǫ min { ,λ v ,λ h ,λ ω } , L ) , k h ǫ − h k X k − tan ( R ) . O ( ǫ min { ,λ v ,λ h ,λ ω } ) . If Π S ϕ v n | z =0 = 0 in [0 , T ] , we only adjust the indices of ǫ in the aboveconvergence rates, the results are similar. Now we show our strategies of the proofs. Denote ˆ v = v ǫ − v , ˆ h = h ǫ − h ,ˆ q = q ǫ − q , then ˆ v, ˆ h, ˆ q satisfy the following equations ∂ ϕ ǫ t ˆ v − ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η = 2 ǫ ∇ ϕ ǫ · S ϕ ǫ ˆ v + ǫ △ ϕ ǫ v − ˆ v · ∇ ϕ v, x ∈ R − , ∇ ϕ ǫ · ˆ v = ∂ ϕz v · ∇ ϕ ǫ ˆ η, x ∈ R − ,∂ t ˆ h + v y · ∇ y ˆ h = ˆ v · N ǫ , { z = 0 } , (ˆ q − g ˆ h ) N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = 2 ǫ S ϕ ǫ v N ǫ , { z = 0 } , (ˆ v, ˆ h ) | t =0 = ( v ǫ − v , h ǫ − h ) . (1.36)In order to close the estimates for (1 . V ℓ,α = ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ℓt Z α ˆ η, ˆ Q ℓ,α = ∂ ℓt Z α ˆ q − ∂ ϕz q∂ ℓt Z α ˆ η. (1.37)When | α | >
0, we study the equations (5 .
15) and estimate ˆ V ℓ,α , ˆ Q ℓ,α . When | α | = 0, we study the equations (5 .
21) and estimate ˆ V ℓ, and ∂ ℓt ∇ ˆ q .When we prove the estimates, we always rewrite the viscous terms by usingthe formula △ ϕ v = 2 ∇ ϕ · S ϕ v − ∇ ϕ ( ∇ ϕ · v ) = ∇ ϕ ( ∇ ϕ · v ) − ∇ ϕ × ( ∇ ϕ × v ). Thedissipation of the velocity and the vorticity are controlled by using inequalities: k∇ v k L . R R − |S ϕ v | d V t + k v k L , k∇ ω k L . R R − |∇ ϕ × ω | d V t + k ω k L + | ω · n | , (1.38)where (1 . is Korn’s inequality (see [31, 21]), (1 . can be proved by usingHodge decomposition and ∇ ϕ · ω = 0. 15 .4 Main Results for N-S Equations with Surface Tension We studied the free surface Navier-Stokes equations with surface tension,the free surface Navier-Stokes equations (1 .
1) with fixed σ > ∂ ϕt v + v · ∇ ϕ v + ∇ ϕ q = ǫ △ ϕ v, x ∈ R − , ∇ ϕ · v = 0 , x ∈ R − ,∂ t h = v ( t, y, · N, z = 0 ,q n − ǫ S ϕ v n = gh n − σH n , z = 0 , ( v, h ) | t =0 = ( v ǫ , h ǫ ) , (1.39)where H = −∇ x · (cid:0) ( −∇ y h, √ |∇ y h | (cid:1) = ∇ y · (cid:0) ∇ y h √ |∇ y h | (cid:1) .Let ǫ →
0, we formally get the free surface Euler equations with surfacetension: ∂ ϕt v + v · ∇ ϕ v + ∇ ϕ q = 0 , x ∈ R − , ∇ ϕ · v = 0 , x ∈ R − ,∂ t h = v ( t, y, · N, z = 0 ,q = gh − σH, z = 0 , ( v, h ) | t =0 = ( v , h ) , (1.40)where ( v , h ) = lim ǫ → ( v ǫ , h ǫ ), we do not need the Taylor sign condition (1 . .
40) when σ > . .
39) and (1 .
40) with σ > Proposition 1.6.
Fix σ > . For m ≥ , assume the initial data ( v ǫ , h ǫ ) satisfy the compatibility condition Π S ϕ v ǫ n | z =0 = 0 and the regularities: sup ǫ ∈ (0 , (cid:0) | h ǫ | X m + ǫ | h ǫ | X m, + σ | h ǫ | X m, + k v ǫ k X m + k ω ǫ k X m − + ǫ k∇ v k X m − , + ǫ k∇ ω k X m − + k ω ǫ k , ∞ + ǫ k ∂ z ω ǫ k L ∞ (cid:1) ≤ C , (1.41) then the unique Navier-Stokes solution to (1 . satisfies sup t ∈ [0 ,T ] (cid:0) | h ǫ | X m − , + ǫ | h ǫ | X m − , + σ | h ǫ | X m − , + k v ǫ k X m − , + k ∂ z v ǫ k X m − + k ω ǫ k X m − + k ∂ z v ǫ k , ∞ + ǫ k ∂ zz v ǫ k L ∞ (cid:1) + k ∂ z v k L ([0 ,T ] ,X m − ) + k ∂ mt v k L ([0 ,T ] ,L ) + k ∂ mt h k L ([0 ,T ] ,L ) + ǫ k ∂ mt h k L ([0 ,T ] ,X , ) + σ k ∂ mt h k L ([0 ,T ] ,X , ) + ǫ T R k∇ v ǫ k X m − , + k∇ ∂ z v ǫ k X m − d t ≤ C. (1.42) As ǫ → , the Euler solution to (1 . with the initial data ( v , h ) satisfies he following regularities: sup t ∈ [0 ,T ] (cid:0) | h | X m − , + σ | h | X m − , + k v k X m − , + k ∂ z v k X m − + k ω k X m − (cid:1) + k ∂ z v k L ([0 ,T ] ,X m − ) + k ∂ mt v k L ([0 ,T ] ,L ) + k ∂ mt h k L ([0 ,T ] ,L ) + σ k ∂ mt h k L ([0 ,T ] ,X , ) ≤ C. (1.43)For any σ ≥
0, the equation of vorticity and its equivalent boundary con-dition Π S ϕ v n | z =0 = 0 are the same, thus the estimates of normal derivativesare also the same. Thus, our proof of Proposition 1 . σ = 0case. Note that our proof is different from [31, 41].However, the estimates of the pressure are very different from the σ =0 case. If we couple △ ϕ q = − ∂ ϕj v i ∂ ϕi v j with its nonhomogeneous Dirichletboundary condition q | z =0 = gh − σH + 2 ǫ S ϕ v n · n , the estimates can not beclosed due to the less regularity of h .The elliptic equation of the pressure coupled with its Neumann boundarycondition is as follows (see [41, 14, 32]): ( △ ϕ q = − ∂ ϕj v i ∂ ϕi v j , ∇ ϕ q · N | z =0 = − ∂ ϕt v · N − v · ∇ ϕ v · N + ǫ △ ϕ v · N , (1.44)Using (1 .
44) to estimate the pressure, we have to prove ∂ mt v ∈ L ([0 , T ] , L ).When we estimate ∂ mt v , we have to overcome the difficulties generated by ∂ mt q .Besides integrating the energy estimates in time twice (see [41]), we apply the fol-lowing Hardy’s inequality (see [33]) to the terms ∂ z ( ∂ m − t q · f ) where k f k L ( R − )and k (1 − z ) ∂ z f k L ( R − ) are bounded. k − z ∂ ℓt q k L ( R − ) . (cid:12)(cid:12) ∂ ℓt q | z =0 (cid:12)(cid:12) L ( R ) + k ∂ z ∂ ℓt q k L ( R − ) , ≤ ℓ ≤ m − . (1.45)Note that [41] considered the finite fluid depth, for which k ∂ ℓt q k L ( R × [ − L, is bounded. While we consider the infinite fluid depth in this paper, k ∂ ℓt q k L hasno bound in general. Another difference is that [41] needs Taylor sign conditionand Alinhac’s good unknown (1 .
25) to estimate tangential derivatives. Whilewe do not need them for the fixed σ > .
40) with surface tension. We statetheir results in our formulation as follows:
Suppose that σ > is fixed, h ∈ H . ( R ) , v ∈ H . ( R − ) , then there exists T > and a solution ( v, q, h ) of (1 . with v ∈ L ∞ ([0 , T ] , H . ( R − )) , ∇ q ∈ L ∞ ([0 , T ] , H ( R − )) , h ∈ L ∞ ([0 , T ] , H . ( R )) . The solution is unique if h ∈ H . ( R ) , v ∈ H . ( R − ) . Since the equations of the vorticity and its equivalent boundary conditionΠ S ϕ v n | z =0 = 0 are the same as the σ = 0 case, Theorem 1 . .
16) with σ >
0. Thus, we are mainly concerned with convergencerates of the inviscid limit. The results are stated in the following theorem:17 heorem 1.7.
Assume
T > is finite, fixed and independent of ǫ , ( v ǫ , h ǫ ) isthe solution in [0 , T ] of Navier-Stokes equations (1 . with initial data ( v ǫ , h ǫ ) satisfying (1 . , ω ǫ is its vorticity. ( v, h ) is the solution in [0 , T ] of Eulerequations (1 . with initial data ( v , h ) ∈ X m − , ( R − ) × X m − , ( R ) , ω isits vorticity. Assume there exists an integer k where ≤ k ≤ m − , suchthat k v ǫ − v k X k ( R − ) = O ( ǫ λ v ) , | h ǫ − h | X k ( R ) = O ( ǫ λ h ) , k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) , where λ v > , λ h > , λ ω > .If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , then theconvergence rates of the inviscid limit satisfy k v ǫ − v k X k − , tan + | h ǫ − h | X k − , = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k X k − tan + k N ǫ · ω ǫ − N · ω k X k − tan = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan + k ω ǫ − ω k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k∇ ϕ ǫ q ǫ − ∇ ϕ q k X k − tan + k△ ϕ ǫ q ǫ − △ ϕ q k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k v ǫ − v k Y k − tan + | h ǫ − h | Y k − = O ( ǫ min { , λv , λh , λω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k Y k − tan + k N ǫ · ω ǫ − N · ω k Y k − tan = O ( ǫ min { , λv , λh , λω } ) . (1.46) If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , assume k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) where λ ω > , then the convergence rates of the inviscidlimit satisfy k v ǫ − v k X k − , tan + | h ǫ − h | X k − , = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k X k − tan + k N ǫ · ω ǫ − N · ω k X k − tan = O ( ǫ min { ,λ v ,λ h ,λ ω } ) , k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan + k ω ǫ − ω k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k∇ ϕ ǫ q ǫ − ∇ ϕ q k X k − tan + k△ ϕ ǫ q ǫ − △ ϕ q k X k − tan = O ( ǫ min { , λv , λh , λω } ) , k v ǫ − v k Y k − tan + | h ǫ − h | Y k − = O ( ǫ min { , λv , λh , λω } ) , k N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v k Y k − tan + k N ǫ · ω ǫ − N · ω k Y k − tan = O ( ǫ min { , λv , λh , λω } ) . (1.47)Beside Remark 1 .
5, we supplement the following remarks:
Remark 1.8. (i) When σ > , the surface tension changes the regularity struc-ture of Navier-Stokes solutions and Euler solutions, it does not change conver-gence rates of the inviscid limit. For the fixed σ > , we need neither Taylorsign condition nor Alinhac’s good unknown. However, if σ → is allowed, weneed both Taylor sign condition and Alinhac’s good unknown to close a prioriestimates.(ii). [3, 4, 41] studied the zero surface tension limit of water waves or thefree surface N-S equations, but convergence rates of the zero surface tensionlimit are unknown. For the equations (1 . , σ → is very different from → , because ǫ → implies some boundary layers generate, σ → implies theheight function h loses some regularities. By using the variables (1 . , it is notdifficult for extending our estimates to provide the convergence rates.Assume ≤ k ≤ m − , k v ǫ − v k X k ( R − ) = O ( σ µ v , ǫ λ v ) , | h ǫ − h | X k ( R ) = O ( σ µ h , ǫ λ h ) , k ω ǫ − ω k X k − ( R − ) = O ( σ µ ω , ǫ λ ω ) , g − ∂ ϕ ǫ z q ǫ ≥ c > . If Π S ϕ v n | z =0 = 0 in [0 , T ] , we have the convergence rates of the inviscid limit: k v ǫ − v k X k − , tan + | h ǫ − h | X k − , = O ( σ min { µ v ,µ h ,µ ω } , ǫ min { ,λ v ,λ h ,λ ω } ) , k ∂ ϕ ǫ z v ǫ − ∂ ϕz v k X k − tan + k ω ǫ − ω k X k − tan = O ( σ min { µ v ,µ h ,µ ω } , ǫ min { , λv , λh , λω } ) , k∇ ϕ ǫ q ǫ − ∇ ϕ q k X k − tan + k△ ϕ ǫ q ǫ − △ ϕ q k X k − tan = O ( σ min { µ v ,µ h ,µ ω } , ǫ min { , λv , λh , λω } ) , k v ǫ − v k Y k − tan + | h ǫ − h | Y k − = O ( σ min { µ v ,µ h ,µ ω } , ǫ min { , λv , λh , λω } ) . If Π S ϕ v n | z =0 = 0 in [0 , T ] , we only adjust the indices of ǫ in the aboveconvergence rates, the results are similar. The rest of the paper is organized as follows: In Section 2, we study theboundary value of the vorticity, determine the regularity structure of N-S so-lutions with σ = 0. In Section 3, we study the strong vorticity layer causedby the strong initial vorticity layer. In Section 4, we study the strong vorticitylayer caused by the discrepancy between boundary values of the vorticities. InSection 5, we estimate the convergence rates of the inviscid limit for σ = 0. InSection 6, we determine the regularity structure of N-S solutions with σ >
0. InSection 7, we estimate the convergence rates of the inviscid limit for σ >
0. Inthe Appendices A and B, we derive the equations and their boundary conditionswhich are useful for a priori estimates. σ = 0 In this section, we determine the relationship between the vorticity on thefree boundary and normal derivatives of the velocity on the free boundary, andderive the equations of ω h = ( ω , ω ) and their boundary conditions. When σ = 0, we prove Proposition 1 . ǫ in this section,which represents Navier-Stokes solutions. The following lemma states that the normal derivatives ( ∂ z v , ∂ z v ) can beestimated by the tangential vorticity ω h .19 emma 2.1. Assume v and ω are the velocity and vorticity of the free surfaceNavier-Stokes equations (1 . respectively, k v k X m − , + | h | X m − , < + ∞ , then k ∂ z v k X k + k ∂ z v k X k . k ω h k X k + k v k X k, + | h | X k, , k ≤ m − . (2.1) Proof.
We calculate the vorticity: ω = ∂ ϕ v − ∂ ϕz v = ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ z ϕ ∂ z v ,ω = ∂ ϕz v − ∂ ϕ v = − ∂ v + ∂ ϕ∂ z ϕ ∂ z v + ∂ z ϕ ∂ z v ,ω = ∂ ϕ v − ∂ ϕ v = ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ v + ∂ ϕ∂ z ϕ ∂ z v . (2.2)Plug the following divergence free condition ∂ z v = ∂ ϕ∂ z v + ∂ ϕ∂ z v − ∂ z ϕ ( ∂ v + ∂ v ) , (2.3)into (2 . ω = − ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ v + ∂ ϕ ( ∂ v + ∂ v ) ,ω = ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ v − ∂ ϕ ( ∂ v + ∂ v ) . (2.4)It follows from (2 .
4) that ∂ ϕ∂ ϕ∂ z ϕ ∂ z v + ∂ ϕ ) ∂ z ϕ ∂ z v = − ω + ∂ v + ∂ ϕ ( ∂ v + ∂ v ) , ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ ϕ∂ ϕ∂ z ϕ ∂ z v = ω + ∂ v + ∂ ϕ ( ∂ v + ∂ v ) . (2.5)For (2 . ∂ z v , ∂ z v ) ⊤ is (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ∂ ϕ∂ ϕ∂ z ϕ ∂ ϕ ) ∂ z ϕ ∂ ϕ ) ∂ z ϕ ∂ ϕ∂ ϕ∂ z ϕ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = − ∂ ϕ ) +( ∂ ϕ ) ( ∂ z ϕ ) = 0 , (2.6)thus we can solve ∂ z v and ∂ z v from (2 . f k [ ∇ ϕ ]( ∂ j v i ) , k = 1 , , ,
4, which are one order with respect to ∂ j v i , the coefficients are fractions of ∇ ϕ . ( ∂ z v = f [ ∇ ϕ ]( ω , ω ) + f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,∂ z v = f [ ∇ ϕ ]( ω , ω ) + f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , , (2.7)then we have the estimates: k ∂ z v k X k + k ∂ z v k X k . k ω h k X k + P i,j k ∂ j v i k X k + k∇ ϕ k X k . (2.8)Thus, Lemma 2 . S ϕ v n | z =0 = 0, thefollowing lemma claims that the boundary value of normal derivatives and tan-gential vorticity can be expressed in terms of that of tangential derivatives, thetangential vorticity ω h satisfies (1 . emma 2.2. Assume v and ω are the velocity and vorticity of the free surfaceNavier-Stokes equations (1 . respectively. If ǫ > , then there exist polynomi-als ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) , F [ ∇ ϕ ]( ∂ j v i ) , F [ ∇ ϕ ]( ∂ j v i ) such that ω h satisfies (1 . ,where ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) is a quadratic polynomial vector with respect to ω h and ∂ j v i , F [ ∇ ϕ ]( ∂ j v i ) , F [ ∇ ϕ ]( ∂ j v i ) are polynomials with respect to ∂ j v i , all thecoefficients are fractions of ∇ ϕ .Proof. Firstly, we investigate the following quantity on the free boundary: S ϕ v n = n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕ v ) + n ( ∂ ϕ v + ∂ ϕz v ) n ( ∂ ϕ v + ∂ ϕ v ) + n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕz v ) n ( ∂ ϕ v + ∂ ϕz v ) + n ( ∂ ϕ v + ∂ ϕz v ) − n ∂ ϕ v − n ∂ ϕ v . (2.9)Since Π S ϕ v n = 0, S ϕ v n = ( S ϕ v n · n ) n , then S ϕ v n is parallel to n . Byusing S ϕ v n × n = 0, we have n [ n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕ v ) + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕz v ) + n ( ∂ ϕ v + ∂ ϕz v ) − n ∂ ϕ v − n ∂ ϕ v ] ,n [ n ( ∂ ϕ v + ∂ ϕ v ) + n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕz v ) + n ( ∂ ϕ v + ∂ ϕz v ) − n ∂ ϕ v − n ∂ ϕ v ] ,n [ n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕ v ) + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕ v ) + n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕz v )] . (2.10)Firstly, we solve ∂ ϕz v from (2 . ( n ) − ( n ) − ( n ) ] ∂ ϕz v = − [ ( n ) − ( n ) − ( n ) ] ∂ ϕ v +( n ( n ) n − n n ) ∂ ϕ v − ( n n + n ( n ) n ) ∂ ϕ v +[ ( n ) n n − n n n n − n n ]( ∂ ϕ v + ∂ ϕ v ) , (2.11)Secondly, we solve ∂ ϕz v from (2 . ( n ) − ( n ) − ( n ) ] ∂ ϕz v = − [ ( n ) − ( n ) − ( n ) ] ∂ ϕ v − ( n n + ( n ) n n ) ∂ ϕ v + ( ( n ) n n − n n ) ∂ ϕ v +( ( n ) n n − n n − n n n n )( ∂ ϕ v + ∂ ϕ v ) . (2.12)It follows from (2 .
11) and (2 .
12) that (cid:2) ( n ) + ( n ) +
12 ( n ) ( n ) −
12 ( n ) ( n ) (cid:3) ∂ z v + (cid:2) n n + ( n ) n ( n ) + n ( n ) ( n ) (cid:3) ∂ z v = − [ ( n ) − ( n ) − ( n ) ] (cid:2) ∂ z ϕ∂ v − ∂ ϕ [ − ∂ z ϕ ( ∂ v + ∂ v )] (cid:3) +( n ( n ) n − n n )( ∂ z ϕ∂ v ) − ( n n + n ( n ) n )( ∂ z ϕ∂ v )+[ ( n ) n n − n n n n − n n ]( ∂ z ϕ∂ v + ∂ z ϕ∂ v ) , n n + ( n ) n ( n ) + n ( n ) ( n ) (cid:3) ∂ z v + (cid:2) ( n ) + ( n ) + ( n ) n ) − ( n ) n ) (cid:3) ∂ z v = − [ ( n ) − ( n ) − ( n ) ] (cid:2) ∂ z ϕ∂ v − ∂ z ϕ [ − ∂ z ϕ ( ∂ v + ∂ v )] (cid:3) − ( n n + ( n ) n n )( ∂ z ϕ∂ v ) + ( ( n ) n n − n n )( ∂ z ϕ∂ v )+( ( n ) n n − n n − n n n n )( ∂ z ϕ∂ v + ∂ z ϕ∂ v ) . where the coefficient matrix of ( ∂ z v , ∂ z v ) ⊤ is M = ( n ) + ( n ) +
12 ( n ) ( n ) −
12 ( n ) ( n ) n n + ( n ) n ( n ) + n ( n ) ( n ) n n + ( n ) n ( n ) + n ( n ) ( n ) ( n ) + ( n ) + ( n ) n ) − ( n ) n ) . (2.13)Assume |∇ h | ∞ is suitably small, then n is suitably large and | n | + | n | is suitably small such that M is strictly diagonally dominant matrix. By Levy-Desplanques theorem, M is nondegenerate, thus we can solve ∂ z v and ∂ z v ,namely there exist two homogeneous polynomials f [ ∇ ϕ ]( ∂ j v i ) and f [ ∇ ϕ ]( ∂ j v i ),which are one order with respect to ∂ j v i , the coefficients are fractions of ∇ ϕ . ( ∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , . (2.14)By (2 . ω h = ( ω , ω ): ω = − ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ v + ∂ ϕ ( ∂ v + ∂ v )= − ∂ ϕ∂ ϕ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i ) − ∂ ϕ ) ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i ) + ∂ v + ∂ ϕ ( ∂ v + ∂ v ):= F [ ∇ ϕ ]( ∂ j v i ) ,ω = ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ v − ∂ ϕ ( ∂ v + ∂ v )= ∂ ϕ ) ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i ) + ∂ ϕ∂ ϕ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i ) − ∂ v − ∂ ϕ ( ∂ v + ∂ v ):= F [ ∇ ϕ ]( ∂ j v i ) (2.15)Since ω h satisfies the equation: ∂ ϕt ω h + v · ∇ ϕ ω h − ǫ △ ϕ ω h = ω h · ∇ ϕh v h + ω ∂ ϕz v h , (2.16)where the force term can be transformed as follows: ω h · ∇ ϕh v h + ω ∂ ϕz v h = ω ( ∂ v h − ∂ ϕ∂ z ϕ ∂ z v h ) + ω ( ∂ v h − ∂ ϕ∂ z ϕ ∂ z v h )+( ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ v + ∂ ϕ∂ z ϕ ∂ z v ) ∂ z ϕ ∂ z v h = ω ∂ v h + ω ∂ v h − ω ∂ ϕ∂ z ϕ ~f , [ ∇ ϕ ]( ∂ j v i ) − ω ∂ ϕ∂ z ϕ ~f , [ ∇ ϕ ]( ∂ j v i )+[ ∂ v − ∂ ϕ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i ) − ∂ v + ∂ ϕ∂ z ϕ f [ ∇ ϕ ]( ∂ j v i )] ∂ z ϕ ~f , [ ∇ ϕ ]( ∂ j v i )= ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) , j = 1 , , i = 1 , , , (2.17)22here ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) is a quadratic polynomial vector with ω h and ∂ j v i , thecoefficients are fractions of ∇ ϕ , ω h has degree one.Namely, ω h satisfies the equation (1 . . Thus, Lemma 2 . For the free surface N-S equations (1 . S n = Π S ϕ v n ,while we investigate the vorticity in this paper. q satisfies the elliptic equation with its nonhomogeneous Dirichlet boundarycondition ( △ ϕ q = − ∂ ϕj v i ∂ ϕi v j ,q | z =0 = gh + 2 ǫ S ϕ v n · n , (2.18)then it is standard to prove the gradient estimate of q : k∇ q k X m − . k ∂ ϕi v j ∂ ϕj v i k X m − + (cid:12)(cid:12) q | z =0 (cid:12)(cid:12) X m − , . k v k X m − , + k ∂ z v k X m − + g | h | X m − , + ǫ | v z =0 | X m − , + ǫ | h | X m − , . (2.19)Note that [31] estimated the pressure by decomposition q ǫ = q ǫ,E + q ǫ,NS , whichsatisfy two systems (1 . v , that is tobound k ∂ ℓt v k L and √ ǫ k∇ ∂ ℓt Z α v k L , we must prove two preliminary lemmas of h by using the kinetical boundary condition (1 . .The first preliminary lemma concerns | ∂ ℓt h | L where 0 ≤ ℓ ≤ m −
1. Notethat the estimates of mix derivatives ∂ ℓt Z α h will be obtained when we estimatemix derivatives ∂ ℓt Z α v , where | α | > Lemma 2.3.
Assume ≤ ℓ ≤ m − , | ∂ ℓt h | L have the estimates: | ∂ ℓt h | L . | h | X m − + t R | h | X m − , + k v k X m − , d t + k ∂ z v k L ([0 ,T ] ,X m − ) . (2.20) Proof.
By the kinetical boundary condition (1 . , we have ∂ t h + v y ·∇ y h = v ,apply ∂ ℓt to the above equation, we get ∂ t ∂ ℓt h + v y · ∇ y ∂ ℓt h = ∂ ℓt v − [ ∂ ℓt , v y · ∇ y ] h. (2.21)Multiply (2 .
21) with ∂ ℓt h , integrate in R , we have dd t R R | ∂ ℓt h | d y = 2 R R (cid:0) ∂ ℓt v − [ ∂ ℓt , v y · ∇ y ] h (cid:1) ∂ ℓt h d y + R R | ∂ ℓt h | ∇ y · v y d y . | ∂ ℓt h | L + | h | X ℓ, + (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) X ℓ . | ∂ ℓt h | L + | h | X k − , + k v k X k − , + k ∂ z v k X k − . (2.22)23um ℓ , integrate (2 .
22) in time and apply the integral form of Gronwall’sinequality, we have R R | ∂ ℓt h | d y . | h | X m − + t R | h | X m − , + k v k X m − , + k ∂ z v k X m − d t . | h | X m − + t R | h | X m − , + k v k X m − , d t + k ∂ z v k L ([0 ,T ] ,X m − ) . (2.23)Thus, Lemma 2 . √ ǫ | ∂ ℓt Z α h | , by which we bound √ ǫ kS ϕ ∂ ℓt Z α η k L and then we can bound √ ǫ kS ϕ ∂ ℓt Z α v k L . Lemma 2.4.
Assume ≤ ℓ ≤ m − , √ ǫ | ∂ ℓt Z α h | have the estimates: ǫ | h | X m − , ≤ ǫ | h | X m − , + t R | h | X m − , + ǫ P ℓ ≤ m − ,ℓ + | α |≤ m |∇ V ℓ,α | L d t. (2.24) Proof.
Let Λ be a differential operator with respect to y , whose Fourier multi-plier is (1 + | ξ | ) , so | Λ h | L = | h | .By the kinetical boundary condition (1 . , we have ∂ t h + v y · ∇ y h = v ,apply ∂ ℓt Z α Λ to the above equation, we get ∂ t ∂ ℓt Z α Λ h + v y · ∇ y ∂ ℓt Z α Λ h = ∂ ℓt Z α Λ v − [ ∂ ℓt Z α Λ , v y · ∇ y ] h. (2.25)Multiply (2 .
25) with ǫ∂ ℓt Z α Λ h , integrate in R , we have ǫ dd t R R | ∂ ℓt Z α Λ h | d y = 2 ǫ R R (cid:0) ∂ ℓt Z α Λ v − [ ∂ ℓt Z α Λ , v y · ∇ y ] h (cid:1) ∂ ℓt Z α Λ h d y + ǫ R R | ∂ ℓt Z α Λ h | ∇ y · v y d y . ǫ | h | X m − , + ǫ (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) X m − , + ǫ | ∂ ℓt Z α Λ h | L . ǫ | h | X m − , + ǫ k v k X m − , tan + ǫ k ∂ z v k X m − , tan + ǫ | ∂ ℓt Z α Λ h | L . ǫ | h | X m − , + P ℓ ≤ m − ,ℓ + | α |≤ m ( ǫ k∇ y V ℓ,α k L + ǫ k∇ y ∂ ℓt Z α η k L )+ P ℓ ≤ m − ,ℓ + | α |≤ m (cid:2) ǫ k ∂ z V ℓ,α k L + k ∂ ϕz v k L ∞ · ǫ k ∂ z ∂ ℓt Z α η k L +( √ ǫ k ∂ ϕzz v k L ∞ ) k ∂ ℓt Z α η k L + ǫ | ∂ ℓt Z α Λ h | L (cid:3) . ǫ | h | X m − , + | h | X m − , + ǫ P ℓ ≤ m − ,ℓ + | α |≤ m |∇ V ℓ,α | L . (2.26)Sum ℓ, α , integrate (2 .
26) in time and apply the integral form of Gronwall’sinequality, we have (2 . . ∂ ℓt v where 1 ≤ ℓ ≤ m −
1, since k ∂ ℓt q k has no bound for infinite fluid depth.24 emma 2.5. Assume the conditions are the same with those of Proposition . ,then v and h satisfy the a priori estimate: k v k X m − , + | h | X m − , + ǫ | h | X m − , + ǫ t R k∇ v k X m − , d t . k v k X m − , + | h | X m − , + ǫ | h | X m − , + k ∂ z v k L ([0 ,T ] ,X m − ) . (2.27) Proof.
Apply ∂ ℓt Z α to (1 . ∂ ℓt Z α , ∂ ϕi ] f = − ∂ ϕz f ∂ ϕi ( ∂ ℓt Z α η ) + b.t. , i = t, , , . (2.28)where the abbreviation b.t. represents bounded terms in this paper. Note that ∂ mt ϕ and ∂ mt h are bounded in L ([0 , T ] , L ), thus they are also represented by b.t. in (2 . .
25) as our vari-able, then V ℓ,α and Q ℓ,α satisfies ∂ ϕt V ℓ,α + v · ∇ ϕ V ℓ,α + ∇ ϕ Q ℓ,α − ǫ ∇ ϕ · S ϕ V ℓ,α = − ∂ ϕt ∂ ϕz v ∂ ℓt Z α η − v · ∇ ϕ ∂ ϕz v ∂ ℓt Z α η − ∇ ϕ ∂ ϕz q ∂ ℓt Z α η +2 ǫ ∇ ϕ · ( S ϕ ∂ ϕz v ∂ ℓt Z α η ) − ǫ∂ ϕz S ϕ v y · ∇ y ∂ ℓt Z α η + b.t. , ∇ ϕ · V ℓ,α = − ( ∇ ϕ · ∂ ϕz v ) ∂ ℓt Z α η + b.t. = b.t. ,∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h = V ℓ,α · N + b.t. ,Q ℓ,α N − ǫ S ϕ V ℓ,α N = ( g − ∂ ϕz q ) ∂ ℓt Z α h N + 2 ǫ ( S ϕ ∂ ϕz v N ) ∂ ℓt Z α h − [ ∂ ℓt Z α , ǫ S ϕ v n · n , N ]+(2 ǫ S ϕ v − ǫ S ϕ v n · n ) ∂ ℓt Z α N + 2 ǫ [ ∂ ℓt Z α , S ϕ v, N ] + b.t. , ( ∂ ℓt Z α v, ∂ ℓt Z α h ) | t =0 = ( ∂ ℓt Z α v , ∂ ℓt Z α h ) . (2.29)When | α | ≥ , ≤ ℓ + | α | ≤ m , we develop the L estimate of V ℓ,α . Theestimates are similar to [31], but we do not use g − ∂ ϕz q E .
12 dd t R R − | V ℓ,α | d V t − R R − Q ℓ,α ∇ ϕ · V ℓ,α d V t + 2 ǫ R R − |S ϕ V ℓ,α | d V t ≤ R { z =0 } (2 ǫ S ϕ V ℓ,α N − Q ℓ,α N ) · V ℓ,α d y + k ∂ z v k X m − + k∇ q k X m − + b.t. ≤ − R { z =0 } ( g − ∂ ϕz q ) ∂ ℓt Z α h N · V ℓ,α d y + k ∂ z v k X m − + k∇ q k X m − + b.t. ≤ − R { z =0 } ( g − ∂ ϕz q ) ∂ ℓt Z α h ( ∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h )d y + k ∂ z v k X m − + k∇ q k X m − + b.t. ≤ −
12 dd t R { z =0 } ( g − ∂ ϕz q ) | ∂ ℓt Z α h | d y + k ∂ z v k X m − + k∇ q k X m − + b.t. , (2.30)25hen dd t R R − | V ℓ,α | d V t + dd t R { z =0 } ( g − ∂ ϕz q ) | ∂ ℓt Z α h | d y + ǫ R R − |S ϕ V ℓ,α | d V t . k ∂ z v k X m − + k∇ q k X m − + b.t. (2.31)Since ( g − ∂ ϕz q ) | z =0 ≥ c >
0, a priori estimates can be closed. Thus, k ∂ ℓt Z α v k + | ∂ ℓt Z α h | + ǫ | ∂ ℓt Z α h | + ǫ t R k∇ ∂ ℓt Z α v k d t . k v k X m − , + | h | X m − , + ǫ | h | X m − , + T R k ∂ z v k X m − + k∇ q k X m − d t. (2.32)where we used the estimate of ǫ | ∂ ℓt Z α h | that is proved by Lemma 2 . | α | = 0 and 0 ≤ ℓ ≤ m −
1, we have no bounds of q and ∂ ℓt q .Without using Hardy’s inequality (1 . V ℓ, , we neither use the variable Q ℓ,α and nor apply the integration by partsto the pressure terms. Also, the divergence free condition and the dynamicalboundary condition will not be used here. Then
12 dd t R R − | V ℓ, | d V t + 2 ǫ R R − |S ϕ V ℓ, | d V t ≤ − R R − ∂ ℓt ∇ ϕ q · V ℓ, d V t + R { z =0 } ǫ S ϕ V ℓ, N · V ℓ, d y + k ∂ z v k X m − + b.t. . k ∂ ℓt ∇ q k L + ǫ R { z =0 } | V ℓ, | d y + 4 ǫ R { z =0 } |S ϕ V ℓ, | d y + k ∂ z v k X m − + b.t. . k ∂ ℓt ∇ q k L + ǫ k ∂ ℓt v | z =0 k L + ǫ | ∂ ℓt h | L + ǫ (cid:12)(cid:12) ∂ ℓt ∂ y v | z =0 (cid:12)(cid:12) L + ǫ (cid:12)(cid:12) ∂ ℓt ∂ z v | z =0 (cid:12)(cid:12) L + ǫ | ∂ ℓt h | X , + k ∂ z v k X m − + b.t. . (2.33)Since ∂ z v | z =0 can be expressed in terms of tangential derivatives, see (2 . ∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,∂ z v = ∂ ϕ∂ z v + ∂ ϕ∂ z v − ∂ z ϕ ( ∂ v + ∂ v ) , (2.34)then dd t R R − | V ℓ, | d V t + ǫ R R − |S ϕ V ℓ, | d V t . k ∂ ℓt ∇ q k L + ǫ (cid:12)(cid:12) ∂ ℓt ∂ y v | z =0 (cid:12)(cid:12) L + ǫ | ∂ ℓt h | X , + k ∂ z v k X m − + b.t. . k ∂ ℓt ∇ q k L + ǫ k∇ v k X m − , + ǫ | ∂ ℓt h | X , + k ∂ z v k X m − + b.t. . (2.35)Similarly, we have k V ℓ, k + ǫ t R k∇ V ℓ, k d t . k v k X m − , + | h | X m − , + ǫ | h | X m − , + k∇ q k L ([0 ,T ] ,X m − ) + k ∂ z v k L ([0 ,T ] ,X m − ) + ǫ T R k∇ v k X m − , d t. (2.36)26ombining (2 .
36) and Lemma 2 .
3, we obtain the estimate of ∂ ℓt v : R R | ∂ ℓt v | d y . k v k X m − , + | h | X m − , + ǫ | h | X m − , + k ∂ z v k L ([0 ,T ] ,X m − ) + k∇ q k L ([0 ,T ] ,X m − ) + t R | h | X m − , + k v k X m − , d t + ǫ T R k∇ v k X m − , d t. (2.37)Sum ℓ and α in (2 . , (2 .
37) and Lemma 2 .
3, we get the estimate (2 . . ∂ z v , [32] estimated the quantity ω h − αu ⊥ h and got k ∂ z v k L ∞ ([0 ,T ] ,H m − co ) . However, for the free surface Navier-Stokes equations(1 . ∂ z v . Similar to[31], we estimate k ∂ z v k L ([0 ,T ] ,X m − ) to close energy estimates. Lemma 2.6.
Assume v and ω are the velocity and vorticity of the free surfaceNavier-Stokes equations (1 . respectively. ω h satisfies the following estimate: k ω h k L ([0 ,T ] ,X m − ) + k ∂ z v k L ([0 ,T ] ,X m − ) . (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + T R k v k X m − , + | h | X m − , d t + ǫ t R k ∂ z v k X m − , d t. (2.38) Proof.
By Lemma 2 .
2, we have the equations of ω h : ∂ ϕt ω h + v · ∇ ϕ ω h − ǫ △ ϕ ω h = ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) ,ω h | z =0 = ~ F , [ ∇ ϕ ]( ∂ j v i ) ,ω h | t =0 = ( ω , ω ) ⊤ . (2.39)where j = 1 , , i = 1 , , ω h = ω nhomh + ω homh , such that ω nhomh satisfiesthe nonhomogeneous equations: ∂ ϕt ω nhomh + v · ∇ ϕ ω nhomh − ǫ △ ϕ ω nhomh = ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) ,ω nhomh | z =0 = 0 ,ω nhomh | t =0 = ( ω , ω ) ⊤ , (2.40)and ω homh satisfies the homogeneous equations: ∂ ϕt ω homh + v · ∇ ϕ ω homh − ǫ △ ϕ ω homh = 0 ,ω homh | z =0 = ~ F , [ ∇ ϕ ]( ∂ j v i ) ,ω homh | t =0 = 0 . (2.41)(2 . is equivalent to ∂ t ω nhomh + v y · ∇ y ω nhomh + V z ∂ z ω nhomh − ǫ △ ϕ ω nhomh = ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) . (2.42)27here V z = ∂ z ϕ ( v · N − ∂ t ϕ ) = ∂ z ϕ ( v − ∂ t η − v y · ∇ y η ), see [31].Apply ∂ ℓt Z α , where ℓ + | α | ≤ m −
1, to the equations (2 . ∂ t ∂ ℓt Z α ω nhomh + v y · ∇ y ∂ ℓt Z α ω nhomh + V z ∂ z ∂ ℓt Z α ω nhomh − ǫ △ ϕ ∂ ℓt Z α ω nhomh = ∂ ℓt Z α ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) − [ ∂ ℓt Z α , v y · ∇ y ] ω nhomh − [ ∂ ℓt Z α , V z ∂ z ] ω nhomh + ǫ ∇ ϕ · [ ∂ ℓt Z α , ∇ ϕ ] ω nhomh + ǫ [ ∂ ℓt Z α , ∇ ϕ · ] ∇ ϕ ω nhomh ,∂ ℓt Z α ω nhomh | z =0 = 0 ,∂ ℓt Z α ω nhomh | t =0 = ( ∂ ℓt Z α ω , ∂ ℓt Z α ω ) ⊤ . (2.43)Develop the L estimate of (2 . dd t k ∂ ℓt Z α ω nhomh k L + 2 ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L . k ∂ ℓt Z α ω h k L + k ∂ ℓt Z α ∂ j v i k L + k ∂ ℓt Z α ∇ ϕ k L + (cid:13)(cid:13) [ ∂ ℓt Z α , V z ∂ z ] ω nhomh (cid:13)(cid:13) L + ǫ R R − ∇ ϕ · [ ∂ ℓt Z α , ∇ ϕ ] ω nhomh · ∂ ℓt Z α ω nhomh d V t + ǫ R R − [ ∂ ℓt Z α , ∇ ϕ · ] ∇ ϕ ω nhomh · ∂ ℓt Z α ω nhomh d V t + b.t. . (2.44)Now we estimate the last three terms on the right hand of (2 . (cid:13)(cid:13) [ ∂ ℓt Z α , V z ∂ z ] ω nhomh (cid:13)(cid:13) L = P ℓ + | α | > (cid:13)(cid:13) − zz ∂ ℓ t Z α [ ∂ z ϕ ( v − η t − v · ∇ y η )] · z − z ∂ ℓ t Z α ∂ z ω nhomh (cid:13)(cid:13) L . (cid:13)(cid:13) ∂ z ∂ ℓt Z α [ A + ∂ z ( ψ ∗ h ) ( v − ψ ∗ h t − v · ∇ y ( ψ ∗ h )) (cid:13)(cid:13) L + b.t. . (cid:13)(cid:13) ∂ ℓt Z α ∂ z [ A + ∂ z ( ψ ∗ h ) ( v − ψ ∗ ( v − v y · ∇ y h ) − v · ∇ y ( ψ ∗ h )) (cid:13)(cid:13) L + b.t. . | ∂ m +1 z ψ | L (d z ) | ∂ ℓt ∂ α y y h | L + k ∂ ℓt Z α ∂ z ( ∂ z ϕ v ) k L + | ∂ mz ψ k L (d z ) k ∂ ℓt ∂ α y y ∂ y h k L + b.t. . | h | X m − , + k ∂ z v k X m − + b.t. , (2.45)the second term is ǫ R R − ∇ ϕ · [ ∂ ℓt Z α , ∇ ϕ ] ω nhomh · ∂ ℓt Z α ω nhomh d V t = P i =1 ǫ R R − [ ∂ ℓt Z α , ∂ ϕi ] ω nhomh · ∂ ϕi ∂ ℓt Z α ω nhomh d V t = P i =1 ǫ R R − ∂ ϕi ∂ ℓt Z α η ∂ ϕz ω nhomh · ∂ ϕi ∂ ℓt Z α ω nhomh d V t + b.t. (2.46)28 P i =1 ǫ R R − z ∂ ϕi ∂ ℓt Z α ( ψ ∗ h ) ∂ z ϕ Z ω nhomh · ∂ ϕi ∂ ℓt Z α ω nhomh d V t + b.t. . | ∂ m +1 z ψ | L (d z ) | h | X m − , + ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L + b.t. . | h | X m − , + ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L + b.t. . and the third term is ǫ R R − [ ∂ ℓt Z α , ∇ ϕ · ] ∇ ϕ ω nhomh · ∂ ℓt Z α ω nhomh d V t . ǫ P i =1 R R − ∂ ϕi ∂ ℓt Z α η∂ ϕz ∂ ϕi ω nhomh · ∂ ℓt Z α ω nhomh d V t + b.t. . − ǫ P i =1 R R − ∂ ϕz ∂ ϕi ∂ ℓt Z α η∂ ϕi ω nhomh · ∂ ℓt Z α ω nhomh d V t − ǫ P i =1 R R − ∂ ϕi ∂ ℓt Z α η∂ ϕi ω nhomh · ∂ ϕz ∂ ℓt Z α ω nhomh d V t + b.t. . | ∂ m +1 z ψ | L (d z ) | h | X m − , + ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L + b.t. . | h | X m − , + ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L + b.t.. (2.47)Plug (2 . , (2 . , (2 .
47) into (2 . dd t k ∂ ℓt Z α ω nhomh k L + 2 ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L . k ∂ ℓt Z α ω h k L + k ∂ z v k X m − + | h | X m − , + ǫ k∇ ϕ ∂ ℓt Z α ω nhomh k L + b.t. . (2.48)Sum ℓ and α in (2 . .
48) from 0 to t , we get k ω nhomh k X m − + ǫ t R k∇ ω nhomh k X m − d t . (cid:13)(cid:13) ω nhomh | t =0 (cid:13)(cid:13) X m − + t R k ω h k X m − d t + t R k v k X m − , + | h | X m − , d t . (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + √ t k ω h k L ([0 ,t ] ,X m − ) + t R k v k X m − , d t + t R | h | X m − , d t. (2.49)It follows from (2 .
49) that k ω nhomh k X m − . (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + T k ω h k L ([0 ,t ] ,X m − ) + (cid:0) T R k v k X m − , d t (cid:1) + (cid:0) T R | h | X m − , d t (cid:1) , t R k ω nhomh k X m − d t . T (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + T t R k ω h k L ([0 ,t ] ,X m − ) d t + T (cid:0) T R k v k X m − , d t (cid:1) + T (cid:0) T R | h | X m − , d t (cid:1) . (2.50)29or the homogeneous equations (2 . L ([0 , T ] , L ) es-timate in [31] and paradifferential calculus (see Theorem 10.6 in [31]), when ℓ + | α | ≤ m −
1, we have k ∂ ℓt Z α ω homh k L ([0 ,T ] ,L ( R − )) . k ∂ ℓt Z α ω homh k H ([0 ,T ] ,L ( R − )) . √ ǫ T R (cid:12)(cid:12) ∂ ℓt Z α ω homh | z =0 (cid:12)(cid:12) L ( R ) d t . √ ǫ T R (cid:12)(cid:12) ∂ ℓt Z α (cid:0) ~ F , [ ∇ ϕ ]( ∂ j v i ) (cid:1) | z =0 (cid:12)(cid:12) L ( R ) d t . √ ǫ T R | h | X m − , d t + √ ǫ T R (cid:12)(cid:12) ∂ j v i | z =0 (cid:12)(cid:12) X m − d t . √ ǫ T R | h | X m − , d t + √ ǫ T R (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) X m − , tan d t . √ ǫ T R | h | X m − , d t + √ ǫ T R k ∂ z v k X m − , tan k v k X m − , tan d t . √ ǫ T R | h | X m − , d t + T R k v k X m − , d t + ǫ T R k ∂ z v k X m − , d t, (2.51)where (cid:12)(cid:12) ∂ j v i | z =0 (cid:12)(cid:12) X m − = (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) H m since j = 1 , α in (2 . k ω homh k L ([0 ,T ] ,X m − ) . T R | h | X m − , + k v k X m − , d t + ǫ T R k ∂ z v k X m − , d t. (2.52)Square (2 . k ω homh k L ([0 ,t ] ,X m − ) . k ω homh k L ([0 ,T ] ,X m − ) . (cid:0) T R | h | X m − , d t (cid:1) + (cid:0) T R k v k X m − , d t (cid:1) + (cid:0) ǫ T R k ∂ z v k X m − , d t (cid:1) . (2.53)By (2 .
50) and (2 . k ω h k L ([0 ,t ] ,X m − ) . k ω nhomh k L ([0 ,t ] ,X m − ) + k ω homh k L ([0 ,t ] ,X m − ) . (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + t R k ω h k L ([0 ,t ] ,X m − ) d t + (cid:0) T R k v k X m − , d t (cid:1) + (cid:0) T R | h | X m − , d t (cid:1) + (cid:0) ǫ T R k ∂ z v k X m − , d t (cid:1) . (2.54)By the integral form of Gronwall’s inequality, it is easy to have k ω h k L ([0 ,T ] ,X m − ) . (cid:13)(cid:13) ω h | t =0 (cid:13)(cid:13) X m − + T R k v k X m − , d t + T R | h | X m − , d t + ǫ T R k ∂ z v k X m − , d t. (2.55)30hile by (2 . k ∂ z v h k L ([0 ,T ] ,X m − ) . k ω h k L ([0 ,T ] ,X m − ) + | h | L ([0 ,T ] ,X m − , ) + k v k L ([0 ,T ] ,X m − , ) . (2.56)By the divergence free condition (2 . k ∂ z v k L ([0 ,T ] ,X m − ) . k ∂ z v h k L ([0 ,T ] ,X m − ) + k∇ ϕ k L ([0 ,T ] ,X m − ) + k ∂ j v i k L ([0 ,T ] ,X m − ) . k ω h k L ([0 ,T ] ,X m − ) + | h | L ([0 ,T ] ,X m − , ) + k v k L ([0 ,T ] ,X m − , ) . (2.57)Thus, Lemma 2 . L ∞ estimates which imply ∂ z v, Z ∂ z v, √ ǫ∂ zz v ∈ L ∞ .The argument is based on analyzing 1D Fokker Planck equation which hasexplicit Green function.In the following lemma, we estimate k ∂ z v k L ∞ ([0 ,T ] ,X m − ) . Note that we cannot have ∂ z v ∈ L ∞ ([0 , T ] , X m − ) due to ω | z =0 = F , [ ∇ ϕ ]( ∂ j v i ), see (1 . Lemma 2.7.
Assume v and ω are the velocity and vorticity of the free surfaceNavier-Stokes equations (1 . respectively. ω h satisfies the following estimate: k ∂ z v k X m − + ǫ t R k ∂ zz v k X m − d V t d t + k ω k X m − + ǫ t R k∇ ω k X m − d V t d t . k ∂ z v k X m − + t R k v k X m − , + | h | X m − d t + k ∂ z v k L ([0 ,T ] ,X m − ) . (2.58) Proof.
By the divergence free condition ∇ ϕ · v = 0, we have △ ϕ v = ∇ ϕ ( ∇ ϕ · v ) − ∇ ϕ × ( ∇ ϕ × v ) = −∇ ϕ × ω. (2.59)Firstly, we develop the L estimate of ω . Multiple (1 .
16) with ∇ ϕ × ω ,integrate in R − , use the integration by parts formula (1 . , we get R R − (cid:0) ∂ ϕt v + v · ∇ ϕ v + ∇ ϕ q + ǫ ∇ ϕ × ω (cid:1) · ∇ ϕ × ω d V t = 0 , R R − (cid:0) ∂ ϕt ω + v · ∇ ϕ ω + ∇ ϕ × ∇ ϕ q (cid:1) · ω d V t + ǫ R R − |∇ ϕ × ω | d V t = − R z =0 (cid:0) ∂ t v + v y · ∇ y v + ∇ ϕ q (cid:1) · N × ω d V t − R R − ( P i =1 ∇ ϕ v i ∂ ϕi v · ω d V t , k ω k L + ǫ t R k∇ ω k L d t ≤ (cid:13)(cid:13) ω | t =0 (cid:13)(cid:13) L + b.t. . (2.60)Note that R z =0 ∇ ϕ q · N × ω d V t . (cid:12)(cid:12) ∇ ϕ q | z =0 (cid:12)(cid:12) − + (cid:12)(cid:12) N × ω | z =0 (cid:12)(cid:12) = b.t. . ≤ ℓ + | α | ≤ m −
2, apply ∂ ℓt Z α to (1 .
16) and rewrite the viscousterms, we have ∂ ϕt ∂ ℓt Z α v + v · ∇ ϕ ∂ ℓt Z α v + ∇ ϕ ∂ ℓt Z α q + ǫ ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) = ǫ I , + I , , (2.61)where I , = − [ ∂ ℓt Z α , ∇ ϕ × ] ω − ∇ ϕ × [ ∂ ℓt Z α , ∇ ϕ × ] v, I , = − [ ∂ ℓt Z α , v y · ∇ y + V z ∂ z ] v − [ ∂ ℓt Z α , N ∂ ϕz ] q. Multiple (2 .
61) with ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ), integrate in R − , we have R R − ∂ ϕt ∂ ℓt Z α v · ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v )d V t + R R − v · ∇ ϕ ∂ ℓt Z α v · ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v )d V t + R R − ∇ ϕ ∂ ℓt Z α q · ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v )d V t + ǫ R R − |∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) | d V t = R R − ( ǫ I , + I , ) · ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v )d V t , (2.62)Use the integration by parts formula (1 . and note that [ ∂ ϕt , ∇ ϕ ] = 0,we have R R − ∂ ϕt |∇ ϕ × ∂ ℓt Z α v | d V t + R R − v · ∇ ϕ |∇ ϕ × ∂ ℓt Z α v | d V t + R R − ( ∇ ϕ × ∇ ϕ ∂ ℓt Z α q ) · ( ∇ ϕ × ∂ ℓt Z α v )d V t + ǫ R R − |∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) | d V t = − R z =0 ( ∂ ϕt ∂ ℓt Z α v + v · ∇ ϕ ∂ ℓt Z α v ) · N × ( ∇ ϕ × ∂ ℓt Z α v )d y − R R − [( P i =1 ∇ ϕ v i · ∂ ϕi ) × ∂ ℓt Z α v ] · ( ∇ ϕ × ∂ ℓt Z α v )d V t − R z =0 ∇ ϕ ∂ ℓt Z α q · N × ( ∇ ϕ × ∂ ℓt Z α v )d y + R R − ǫ I , · ∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v )d V t + R z =0 I , · N × ( ∇ ϕ × ∂ ℓt Z α v )d y + R R − ∇ ϕ × I , · ( ∇ ϕ × ∂ ℓt Z α v )d V t . (2.63)By ∇ ϕ × ∇ ϕ ∂ ℓt Z α q = 0, we have dd t R R − |∇ ϕ × ∂ ℓt Z α v | d V t + 2 ǫ R R − |∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) | d V t . k∇ ϕ × ∂ ℓt Z α v k L + ǫ k∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) k L + (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) X m − tan + (cid:12)(cid:12) ∇ ϕ × ∂ ℓt Z α v | z =0 (cid:12)(cid:12) + (cid:12)(cid:12) ∇ ϕ ∂ ℓt Z α q | z =0 (cid:12)(cid:12) − + k∇ ∂ ℓt Z α v k L + P ℓ + | α | ≤ m − (cid:12)(cid:12) ∇ ϕ ∂ ℓ t Z α q | z =0 (cid:12)(cid:12) − + ǫ kI , k L + k∇ ϕ × I , k L . (2.64)It is easy to prove that kI , k L . k∇ ω k X m − . Then we estimate ∇ ϕ ×I , . k∇ ϕ × [ ∂ ℓt Z α , V z ∂ z ] v k L . P ℓ + | α | > (cid:0) k − zz ∂ ℓ t Z α V z · z − z ∇ ϕ × ∂ ℓ t Z α ∂ z v k L + k∇ ϕ ∂ ℓ t Z α V z × ∂ ℓ t Z α ∂ z v k L (cid:1) (2.65)32 P ℓ + | α | > (cid:0) k∇ ∂ ℓ t Z α V z k L + k∇ ϕ × ∂ ℓ t Z α Z v k L + k ∂ ℓ t Z α ∂ z v k L (cid:1) . k ω k X m − + k v k X m − + k ∂ z v k X m − + k ∂ zz η k X m − + k ∂ zt η k X m − . k ω k X m − + k v k X m − + | h | X m − . k∇ ϕ × [ ∂ ℓt Z α , N ∂ ϕz ] q k L . k ( ∇ y , ⊤ × [ ∂ ℓt Z α , N ∂ ϕz ] q k L + k N ∂ ϕz × [ ∂ ℓt Z α , N ∂ ϕz ] q k L . k∇ q k X m − + P ℓ + | α | > (cid:0) k N ∂ ϕz × ∂ ℓ t Z α N · ∂ ℓ t Z α ∂ ϕz q k L + k N × ∂ ℓ t Z α N · ∂ ϕz ∂ ℓ t Z α ∂ ϕz q k L . k∇ q k X m − + k ∂ ϕzz q k X m − . k∇ q k X m − + k△ ϕ q k X m − . k∇ q k X m − + k ∂ ϕi v j ∂ ϕj v i k X m − . k∇ q k X m − + k v k X m − + k ω k X m − . Plug (2 .
65) into (2 . k∇ ϕ × ∂ ℓt Z α v k L + ǫ t R k∇ ϕ × ( ∇ ϕ × ∂ ℓt Z α v ) k L d t . (cid:13)(cid:13) ∇ ϕ × ∂ ℓt Z α v | t =0 (cid:13)(cid:13) L + t R k ω k X m − + k v k X m − , + | h | X m − + k ∂ z v k X m − + k∇ q k X m − d t + b.t. (2.66) ∂ ℓt Z α ω is equivalent to ∇ ϕ × ∂ ℓt Z α v , due to ℓ + | α | ≤ m − ∂ ℓt Z α ω − ∂ ℓt Z α ( ∇ ϕ × v ) = P ℓ + | α | > ∂ ℓ t Z α ( N ∂ z ϕ ) ∂ z × ∂ ℓ t Z α v, k ∂ ℓt Z α ω − ∂ ℓt Z α ( ∇ ϕ × v ) k L . k ∂ z v k X m − + | h | X m − , . (2.67)Then we have the estimate of the vorticity: k ω k X m − + ǫ t R k∇ ω k X m − d V t d t . k ω k X m − + t R k v k X m − , + | h | X m − + k ∂ z v k X m − + k∇ q k X m − d t. (2.68)Thus, Lemma 2 . Remark 2.8.
There is another approach to estimate k ω h k X m − and k ∂ z v k X m − ,that is, we define variables: ζ = ω − F [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,ζ = ω − F [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , , (2.69)33 hen ζ satisfies the following equation: ∂ t ζ + v y · ∇ y ζ + V z ∂ z ζ − ǫ △ ϕ ζ = ~ F [ ∇ ϕ ]( ζ + ~ F , [ ∇ ϕ ]( ∂ j v i ) , ∂ j v i )+ ~ F , [ ∇ ϕ ]( ∂ j ∂ ϕi q + [ ∂ j , ∂ ϕt + v · ∇ ϕ − ǫ △ ϕ ] v i ) − ∂ j v i ( ∂ ϕt ~ F , [ ∇ ϕ ]+ v · ∇ ϕ ~ F , [ ∇ ϕ ] − ǫ △ ϕ ~ F , [ ∇ ϕ ]) + ǫ ∇ ϕ ~ F , [ ∇ ϕ ] · ∇ ϕ ∂ j v i ,ζ | z =0 = 0 ,ζ | t =0 = ω h, − F , [ ∇ ϕ ]( ∂ j v i ) | t =0 . (2.70) Then we have the estimate of ζ : k ζ k X m − + ǫ t R k∇ ζ k X m − d t . (cid:13)(cid:13) ζ | t =0 (cid:13)(cid:13) X m − + t R k v k X m − , + k∇ q k X m − , + | h | X m d t . (cid:13)(cid:13) ω | t =0 (cid:13)(cid:13) X m − + (cid:13)(cid:13) h | t =0 (cid:13)(cid:13) X m − , + (cid:13)(cid:13) v | t =0 (cid:13)(cid:13) X m − , + t R · · · d t. (2.71) By using (2 . and (2 . , we can estimate k ω k X m − . However, we cannot use this method and variables (2 . to estimate convergence rates of theinviscid limit, see Remark . . By the estimates proved in Lemmas 2 . , . , .
7, it is standard to proveProposition 1 . In this section, we study the strong vorticity layer for the free surface N-Sequations (1 . In this preliminaries, we derive the evolution equations of ˆ ω h = ω ǫh − ω h ,and construct a variable which satisfies the heat equations with damping.ˆ ω satisfies the equations (1 . . ~ F [ ∇ ϕ ǫ ]( ω ǫh , ∂ j v ǫ,i ) − ~ F [ ∇ ϕ ]( ω h , ∂ j v i )= f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ]ˆ ω h + f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∂ j ˆ v i + f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∇ ˆ ϕ, (3.1)34here these coefficients f [ · · · ] , f [ · · · ] , f [ · · · ] are uniformly bounded with re-spect to ǫ . Then we obtain the following system of ˆ ω h : ∂ ϕ ǫ t ˆ ω h + v ǫ · ∇ ϕ ǫ ˆ ω h − ǫ △ ϕ ǫ ˆ ω h − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ]ˆ ω h = f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∂ j ˆ v i + f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∇ ˆ ϕ + ǫ △ ϕ ǫ ω h + ∂ ϕz ω h ∂ ϕ ǫ t ˆ ϕ + ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ ϕ − ˆ v · ∇ ϕ ω h , ˆ ω h | z =0 = F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − ω bh := ˆ ω bh , ˆ ω h | t =0 = ω ǫh, − ω h, := ˆ ω h, . (3.2)Similar to [31], we eliminate the convection term by using the Lagrangianparametrization of Ω t : ∂ t X ( t, x ) = u ǫ ( t, X ( t, x )) = v ǫ ( t, Φ − ◦ X ) , X (0 , x ) = Φ(0 , x ) . (3.3)Define the Jacobian of the change of variables J ( t, x ) = | det ∇X ( t, x ) | , then J ( t, x ) = J (0 , x ) := J ( x ) due to the divergence free condition. Denote a = | J ( x ) | , define the matrix ( a ij ) = | J | P − , where the matrix P satisfies P ij = ∂ i X · ∂ j X .Define W = e − γt ˆ ω h ( t, Φ − ◦ X ), then W satisfies the equation: a ∂ t W − ǫ∂ i ( a ij ∂ j W ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W = ǫ e − γt △ ϕ ǫ ω h + e − γt ∂ ϕz ω h ∂ ϕ ǫ t ˆ ϕ + e − γt ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ ϕ − e − γt ˆ v · ∇ ϕ ω h + e − γt f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∂ j ˆ v i + e − γt f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∇ ˆ ϕ := I , (3.4)where kI k L ∞ → ǫ → a >
0, we can choose suitably large γ > γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] > , (3.5)then (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W is a damping term. Since the matrix( a ij ) is definitely positive, − ǫ∂ i ( a ij ∂ j W ) is the diffusion term. L ∞ Estimate of Strong Vorticity Layer
In this subsection, we prove that if the initial vorticity layer is strong, thenthe vorticity layer is strong.Before proving our results, let us investigate the simplest model by usingthe heat kernel, that is the following heat equation with damping, Dirichletboundary condition and constant coefficients.
Proposition 3.1.
Assume k W k X tan < + ∞ , w ini = 0 , γ > is constant, W isthe solution of the following heat equation with damping: ∂ t W − ǫ △ W + γW = 0 ,W | z =0 = 0 ,W | t =0 = w ini , (3.6)35 hen lim ǫ → k W k L ∞ ( R − × (0 ,T ]) = 0 .Proof. Define ˜ W = e γt W , the equations (3 .
6) are rewritten as ∂ t ˜ W − ǫ △ ˜ W = 0 , ˜ W | z =0 = 0 , ˜ W | t =0 = w ini = 0 , (3.7)Note that ǫ △ ˜ W ǫ →
0. Otherwise, we have ∂ t ˜ W = 0, then ˜ W = w ini ( y, z √ ǫ ). However, ǫ∂ zz ˜ W = ∂ zz w ini ( y, z √ ǫ ) = 0. This is a contradiction. ǫ → ǫt →
0, then the limit of the solution ˜ W satisfies˜ W ( t, x ) = √ πǫt R R − w ini ( y ) (cid:0) exp {− | x − y | ǫt } − exp {− | x + y | ǫt } (cid:1) d y → w ini ( x ) , as ǫt → . (3.8)The convergence (3 .
8) is strong. Because lim ǫ → k w ini ( x ) k L ∞ ( R − ) = 0, then wehave lim ǫ → k W k L ∞ ( R − × (0 ,T ]) = 0. Thus, Proposition 3 . S ϕ v n | z =0 = 0. Theorem 3.2.
Assume ω ǫ , v ǫ are the vorticity, velocity of Navier-Stokes equa-tions (1 . , ω, v, n are the vorticity, velocity, normal vector of Euler equations (1 . . If the initial Navier-Stokes velocity satisfies lim ǫ → ( ∇ ϕ ǫ × v ǫ ) −∇ ϕ × lim ǫ → v ǫ =0 in the initial set A , the Euler solution satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , then lim ǫ → k ω ǫ − ω k L ∞ ( X ( A ) × (0 ,T ]) = 0 .Proof. Since Π S ϕ v n | z =0 = 0 in [0 , T ], | ω ǫ | z =0 − ω | z =0 | ∞ → ǫ →
0. thenthere exist a set A ∩ { x | z < } 6 = ∅ such that lim ǫ → ( ∇ ϕ ǫ × v ǫ ) − ∇ ϕ × lim ǫ → v ǫ = 0in the initial set A .We study the equations (3 .
2) in the Lagrangian coordinates: a ∂ t W − ǫ∂ i ( a ij ∂ j W ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W = I ,W | z =0 = ˆ ω bh := ω ǫ | z =0 − ω | z =0 → ,W | t =0 = ˆ ω h, . (3.9)We decompose W = W fo + W bdy + W ini , such that W fo satisfies thenonhomogeneous equations: a ∂ t W fo − ǫ∂ i ( a ij ∂ j W fo ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W fo = I ,W fo | z =0 = 0 ,W fo | t =0 = 0 , (3.10)36 bdy satisfies the following equations: a ∂ t W bdy − ǫ∂ i ( a ij ∂ j W bdy ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W bdy = 0 ,W bdy | z =0 = ˆ ω bh ,W bdy | t =0 = 0 , (3.11)and W ini satisfies the homogeneous equations: a ∂ t W ini − ǫ∂ i ( a ij ∂ j W ini ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W ini = 0 ,W ini | z =0 = 0 ,W ini | t =0 = ˆ ω h, , (3.12)where lim ǫ → k ˆ ω h, k L ∞ ( A ) = 0.Note the diffusion term and damping term in (3 . k W fo k L ∞ → ǫ →
0, due to the force term thatvanishes when ǫ → . φ = W bdy − (cid:0) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:1) , (3.13)then φ satisfies the following equations: a ∂ t W bdy − ǫ∂ i ( a ij ∂ j W bdy ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W bdy = − a ∂ t (cid:0) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:1) + ǫ∂ i (cid:2) a ij ∂ j (cid:0) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:1)(cid:3) − (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1)(cid:0) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:1) ,φ | z =0 = 0 ,φ | t =0 = − F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) | t =0 + F , [ ∇ ϕ ]( ∂ j v i ) | t =0 . (3.14)It is easy to prove k φ k L ∞ → ǫ →
0, due to the force term and theinitial data φ | t =0 that vanish when ǫ →
0. Thus, it follows from (3 .
13) that k W bdy k L ∞ → ǫ → .
12) which are already expressed in theLagrangian coordinates. In order to prove lim ǫ → k ω ǫ − ω k L ∞ ( X ( A ) × (0 ,T ]) = 0, wehave to prove lim ǫ → k W ini k L ∞ ( A × (0 ,T ]) = 0.By defining the variable˜ W ini = W ini exp { R t a ( γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ]) d t } , (3.15)(3 .
12) can be rewritten as ∂ t ˜ W ini − √ ǫa exp {− R t a ( γa − f [ · · · ]) d t } · a ij ∂ ij ˜ W ini = √ ǫ I , ˜ W | z =0 = 0 , ˜ W | t =0 = ˆ ω h, , (3.16)37here I = √ ǫa P i,j =1 ∂ i a ij · ∂ j (cid:0) ˜ W ini exp {− R t a ( γa − f [ · · · ]) d t } (cid:1) + √ ǫa P i,j =1 a ij ∂ i ˜ W ini · ∂ j (cid:0) exp {− R t a ( γa − f [ · · · ]) d t } (cid:1) . (3.17)Note that kI k L ∞ < + ∞ , because I contains normal differential operator ∂ z of order at most one.(3 .
16) is uniformly parabolic, which has fundament solution satisfying theparabolic scaling. Let H ( x √ ǫt ) to denote the fundament solution of the followinghomogeneous parabolic equation in R : ∂ t f − ǫa a ij exp {− R t a ( γa − f [ · · · ]) d t } · ∂ ij f = 0 , (3.18)by using the fundament solution H , the equations (3 .
16) have the explicit for-mula by using Duhamel’s principle,˜ W ini ( t, x ) = R R − ˆ ω h, ( y )( H ( x − y √ ǫt ) − H ( x + y √ ǫt )) d y + t R R R − √ ǫ I ( t − s, y )( H ( x − y √ ǫs ) − H ( x + y √ ǫs )) d y d s, (3.19)then ˜ W ini ( t, x ) → ˆ ω h, + √ ǫO (1) → ˆ ω h, pointwisely, as ǫ →
0, lim ǫ → ˜ W ini ( t, x )and lim ǫ → ˆ ω h, ( x ) have the same support. The limit of the solution is equal to thatof the initial data in Lagrangian coordinates, namely the limit of the vorticityis also transported by the velocity field in Eulerian coordinates.By (3 . ǫ → k W ini k L ∞ ( A ) = lim ǫ → k ˜ W ini exp {− R t a ( γa − f [ · · · ]) d t }k L ∞ ( A ) = 0 . (3.20)So lim ǫ → k ˆ ω h k L ∞ ( X ( A ) × (0 ,T ]) = 0. Thus, Theorem 3 . ǫ → k ∂ z v ǫ − ∂ z v k L ∞ ( X ( A ) × (0 ,T ]) = 0 , lim ǫ → kS v ǫ − S v k L ∞ ( X ( A ) × (0 ,T ]) = 0 , lim ǫ → k∇ q ǫ − ∇ q k L ∞ ( X ( A ) × (0 ,T ]) = 0 . (3.21) In this section, we study the strong vorticity layer for the free surface N-Sequations (1 . .1 Discrepancy of the Vorticity on the Free Boundary The following lemma shows that if the tangential projection on the freesurface of the Euler strain tensor multiply by normal vector does not vanishes,then there is a discrepancy between Navier-Stokes vorticity and Euler vorticity.
Lemma 4.1.
Assume ω ǫ , v ǫ , N ǫ are the vorticity, velocity, normal vector ofNavier-Stokes equations (1 . , ω, v, N are the vorticity, velocity, normal vectorof Euler equations (1 . , ω ǫ,b and ω b are boundary values of ω ǫ , ω respectively.If Π S ϕ v n | z =0 = 0 in (0 , T ] , then lim ǫ → | ω ǫ,b − ω b | L ∞ ( R × (0 ,T ]) = 0 .Proof. We denote S n = Π S ϕ v n . Since S n = 0, S ϕ v n = ( S ϕ v n · n ) n + Π S ϕ v n ,then S ϕ v n is not parallel to n , namely, S ϕ v n × n = ( S ϕ v n · n ) n × n + Π S ϕ v n × n = Π S ϕ v n × n = 0 . (4.1)Denote Π S ϕ v n × n := (Θ , Θ , Θ ) ⊤ , which is a nonzero vector.By S ϕ v n × n = (Θ , Θ , Θ ) ⊤ , similar to (2 . n [ n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕ v ) + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕz v ) + n ( ∂ ϕ v + ∂ ϕz v ) − n ∂ ϕ v − n ∂ ϕ v ] − Θ ,n [ n ( ∂ ϕ v + ∂ ϕ v ) + n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕz v ) + n ( ∂ ϕ v + ∂ ϕz v ) − n ∂ ϕ v − n ∂ ϕ v ] + Θ ,n [ n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕ v ) + n ( ∂ ϕ v + ∂ ϕz v )]= n [ n ( ∂ ϕ v + ∂ ϕ v ) + n ∂ ϕ v + n ( ∂ ϕ v + ∂ ϕz v )] + Θ . (4.2)Then we have the following two equations involving ∂ z v and ∂ z v : (cid:2) ( n ) + ( n ) +
12 ( n ) ( n ) −
12 ( n ) ( n ) (cid:3) ∂ z v + (cid:2) n n + ( n ) n ( n ) + n ( n ) ( n ) (cid:3) ∂ z v = − [ ( n ) − ( n ) − ( n ) ] (cid:2) ∂ z ϕ∂ v − ∂ ϕ [ − ∂ z ϕ ( ∂ v + ∂ v )] (cid:3) +( n ( n ) n − n n )( ∂ z ϕ∂ v ) − ( n n + n ( n ) n )( ∂ z ϕ∂ v )+[ ( n ) n n − n n n n − n n ]( ∂ z ϕ∂ v + ∂ z ϕ∂ v ) − ∂ z ϕ Θ − n n ∂ z ϕ Θ , (cid:2) n n + ( n ) n ( n ) + n ( n ) ( n ) (cid:3) ∂ z v + (cid:2) ( n ) + ( n ) + ( n ) n ) − ( n ) n ) (cid:3) ∂ z v = − [ ( n ) − ( n ) − ( n ) ] (cid:2) ∂ z ϕ∂ v − ∂ z ϕ [ − ∂ z ϕ ( ∂ v + ∂ v )] (cid:3) − ( n n + ( n ) n n )( ∂ z ϕ∂ v ) + ( ( n ) n n − n n )( ∂ z ϕ∂ v )+( ( n ) n n − n n − n n n n )( ∂ z ϕ∂ v + ∂ z ϕ∂ v ) + ∂ z ϕ Θ + n n ∂ z ϕ Θ . (4.3)When |∇ h | ∞ is suitably small, the coefficient matrix of ( ∂ z v , ∂ z v ) ⊤ is39ondegenerate, then we solve ∂ z v = f [ ∇ ϕ ]( ∂ j v i ) − M ( ∂ z ϕ Θ + n n ∂ z ϕ Θ ) + M ( ∂ z ϕ Θ + n n ∂ z ϕ Θ ) ,j = 1 , , i = 1 , , ,∂ z v = f [ ∇ ϕ ]( ∂ j v i ) − M ( ∂ z ϕ Θ + n n ∂ z ϕ Θ ) + M ( ∂ z ϕ Θ + n n ∂ z ϕ Θ ) ,j = 1 , , i = 1 , , , (4.4)where the matrix M = ( M ij ) is defined in (2 . M ij ) = ( M ij ) − .By (2 .
4) and (4 . ω h = ( ω , ω ): ω = − ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ v + ∂ ϕ ( ∂ v + ∂ v ):= F [ ∇ ϕ ]( ∂ j v i ) + ς Θ + ς Θ + ς Θ ,ω = ∂ ϕ ) ∂ z ϕ ∂ z v + ∂ ϕ∂ ϕ∂ z ϕ ∂ z v − ∂ v − ∂ ϕ ( ∂ v + ∂ v ):= F [ ∇ ϕ ]( ∂ j v i ) + ς Θ + ς Θ + ς Θ , (4.5)where the coefficients ς i are as follows: ς = ∂ z ϕ [ ∂ ϕ∂ ϕ M + (1 + ( ∂ ϕ ) ) M ] ,ς = ∂ ϕ∂ ϕ M + (1 + ( ∂ ϕ ) ) M ,ς = (cid:2) (1 + ( ∂ ϕ ) ) (cid:0) − M n n + M n n ∂ z ϕ (cid:1) + ∂ ϕ∂ ϕ (cid:0) − M n n + M n n ∂ z ϕ (cid:1)(cid:3) ,ς = ∂ z ϕ [(1 + ( ∂ ϕ ) ) M + ∂ ϕ∂ ϕ M ] ,ς = − (1 + ( ∂ ϕ ) ) M − ∂ ϕ∂ ϕ M ,ς = (1 + ( ∂ ϕ ) ) (cid:0) − M n n + M n n ∂ z ϕ (cid:1) + ∂ ϕ∂ ϕ (cid:0) − M n n + M n n ∂ z ϕ (cid:1) . (4.6)If | ς Θ + ς Θ + ς Θ | ∞ = | ς Θ + ς Θ + ς Θ | ∞ = 0, then ( ∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , ,∂ z v = f [ ∇ ϕ ]( ∂ j v i ) , j = 1 , , i = 1 , , . (4.7)Since the proof of Lemma 2 . .
7) implies S ϕ v n × n = 0. Thisstrongly contradicts with (4 . | ς Θ + ς Θ + ς Θ | ∞ = 0 or | ς Θ + ς Θ + ς Θ | ∞ = 0.Without lose of generality, we assume the former holds.As ǫ → | F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F [ ∇ ϕ ]( ∂ j v i ) | L ∞ →
0, this convergence isstrong due to enough uniform regularities in conormal Sobolev space of Navier-Stokes solutions and its tangential derivatives (see [31]). Thus, when ǫ is suffi-ciently small, | F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F [ ∇ ϕ ]( ∂ j v i ) | ∞ ≤ | ς Θ + ς Θ + ς Θ | ∞ . It40ollows from (4 .
5) that | ω ǫ, − ω | ∞ ≥ | ς Θ + ς Θ + ς Θ | ∞ − (cid:12)(cid:12) F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F [ ∇ ϕ ]( ∂ j v i ) (cid:12)(cid:12) ∞ ≥ | ς Θ + ς Θ + ς Θ | ∞ . (4.8)Then | ω ǫ,bh − ω bh | L ∞ ( R × (0 ,T ]) ≥ max {| ω ǫ, − ω | ∞ , | ω ǫ, − ω | ∞ }≥ max {| ς Θ + ς Θ + ς Θ | ∞ , | ς Θ + ς Θ + ς Θ | ∞ } > . (4.9)Thus, Lemma 4 . L ∞ Estimate of Strong Vorticity Layer
In this subsection, we prove the existence of strong vorticity layer when theEuler boundary data satisfies Π S ϕ v n | z =0 = 0 in (0 , T ].Before proving our results, let us investigate the simplest model, that is thefollowing heat equation with damping and constant coefficients. Proposition 4.2.
Assume w b ∈ H ( R × [0 , T ]) , w b , γ > is constant, W is the solution of the following heat equation with damping: ∂ t W − ǫ △ W + γW = 0 ,W | z =0 = w b ,W | t =0 = 0 , (4.10) then lim ǫ → k W k L ∞ ( R − × (0 ,T ]) = 0 .Proof. We define the following Fourier transformation with respect to ( t, y ) ∈ R + × R , F [ W ]( τ, ξ, z ) = + ∞ R R R − e − iτt − iξ · y W ( t, y, z ) d t d y, (4.11)Note that W | t =0 = 0, there is no term involving W | t =0 appears as a force term.By applying Fourier transformation (4 .
11) to (4 . iτ F [ W ] − ǫ∂ zz F [ W ] + ǫ | ξ | F [ W ] + γ F [ W ] = 0 ,∂ zz F [ W ] − ǫ ( iτ + ǫ | ξ | + γ ) F [ W ] = 0 , F [ W ]( τ, ξ, z ) = exp { ( iτ + ǫ | ξ | + γ ) z √ ǫ }F [ w b ]( τ, ξ ) , (4.12)where the complex root ( iτ + ǫ | ξ | + γ ) has two branches, one of which alwayshas a positive real part due to ǫ | ξ | + γ >
0, then we choose this branch.41f | z | = O ( ǫ − δ z ) where δ z >
0, we simply assume z = − ǫ − δ z , then as ǫ → | exp { ( iτ + ǫ | ξ | + γ ) z √ ǫ }| = exp {− ( ǫ | ξ | + γ ) ǫ − δ z } ≤ exp {− γ ǫ − δ z } → , kF [ W ]( τ, ξ, z ) k L (d τ d ξ ) = k exp {− ( ǫ | ξ | + γ ) ǫ − δ z }F [ w b ]( τ, ξ ) k L (d τ d ξ ) → , (4.13)note that F [ w b ] ∈ L (d τ d ξ ) requires w b ∈ H ( R × [0 , T ]), thenlim ǫ → k W k L ∞ ( t,y,z ) = lim ǫ → kF − [ F [ W ]] k L ∞ ( t,y,z ) . kF [ W ]( τ, ξ, z ) k L (d τ d ξ ) → . (4.14)If | z | = O ( ǫ + δ z ) where δ z ≥
0, we simply assume z = − ǫ + δ z , then as ǫ → | exp { ( iτ + ǫ | ξ | + γ ) z √ ǫ }| = exp {− ( ǫ | ξ | + γ ) ǫ δ z } → e −√ γ , (4.15)for any finite ξ ∈ R , then kF [ W ]( τ, ξ, z ) k L ∞ (d τ d ξ ) = 0.We use the proof by contradiction. Assume that k W k L ∞ ( t,y,z ) = 0, then k W k L ( t,y,z ) = 0, and then kF [ W ]( τ, ξ, z ) k L ∞ = 0, this is a contradiction. Thus,if z = O ( ǫ + δ z ), lim ǫ → k W k L ∞ ( t,y,z ) = 0. Thus, Proposition 4 . .
2, since we have to use the symbolic analysis and paradifferentialcalculus for our problem.In order to prove that the discrepancy of boundary values of vorticities isone of sufficient conditions for the existence of strong vorticity layer, we assumethat the initial vorticity layer is weak.
Theorem 4.3.
Assume the conditions are the same with Lemma . . If theEuler solution satisfies Π S ϕ v n | z =0 = 0 in (0 , T ] , and Π S ϕ v n | z =0 ∈ H ( R × [0 , T ]) , the initial Navier-Stokes velocity satisfies lim ǫ → ( ∇ ϕ ǫ × v ǫ ) −∇ ϕ × lim ǫ → v ǫ = 0 ,then lim ǫ → k ω ǫ − ω k L ∞ ( R × [0 ,O ( √ ǫ )) × (0 ,T ]) = 0 .Proof. Since the initial Navier-Stokes velocity satisfies lim ǫ → ( ∇× u ǫ ) −∇× lim ǫ → u ǫ =0, lim ǫ → | ω ǫ − ω | L ∞ = 0, then Π S ϕ v n | z =0 ,t =0 = 0, that does not contradict withΠ S ϕ v n | z =0 = 0 in (0 , T ].We study the equations (3 .
2) with small initial data: a ∂ t W − ǫ∂ i ( a ij ∂ j W ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W = I ,W | z =0 = e − γt ˆ ω bh ,W | t =0 = ˆ ω h, → . (4.16)We decompose W = W bdy + W fo , such that W fo satisfies the nonhomoge-42eous equations: a ∂ t W fo − ǫ∂ i ( a ij ∂ j W fo ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W fo = I ,W fo | z =0 = 0 ,W fo | t =0 = ˆ ω h, → . (4.17)and W bdy satisfies the homogeneous equations: a ∂ t W bdy − ǫ∂ i ( a ij ∂ j W bdy ) + (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W bdy = 0 ,W bdy | z =0 = e − γt ˆ ω bh ,W bdy | t =0 = 0 . (4.18)Note the diffusion term and damping term of (4 . k W fo k L ∞ ≤ k W fo | t =0 k L ∞ + t R kI k ∞ d t → . Next, we study the homogeneous equations (4 .
18) with variable coefficients,which differs from the equations (4 .
10) up to coefficients. By using symbolicanalysis, it is standard to prove that the limit of the solution of (4 .
18) behavessimilarly to that of (4 . .
18) in the following form: ǫ∂ zz W bdy + ǫ ( ∂ z a a + P j =1 , ∂ j a j a ) ∂ z W bdy + 2 ǫ P j =1 , a j a ∂ jz W bdy + ǫ P j =1 , ∂ z a j a ∂ j W bdy + ǫ P j =1 , ∂ i a ij a ∂ j W bdy + ǫ P j =1 , a ij a ∂ ij W bdy − a a ∂ t W bdy − a (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) W bdy = 0 ,W bdy | z =0 = e − γt ˆ ω bh ,W bdy | t =0 = 0 . (4.19)Take z as a parameter, then the symbolic version of (4 .
19) is ǫ∂ zz ˜ W bdy + A √ ǫ∂ z ˜ W bdy + A ˜ W bdy = 0 , ˜ W bdy | z =0 = F [ e − γt ˆ ω bh ] , ˜ W bdy | t =0 = 0 . (4.20)where the Fourier multipliers are as follows: A = √ ǫ ( ∂ z a a + P j =1 , ∂ j a j a + 2 i P j =1 , a j a ξ j ) A = iǫ P j =1 , ∂ z a j a ξ j + iǫ P j =1 , ∂ i a ij a ξ j − ǫ P j =1 , a ij a ξ i ξ j − iτ a a − a (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) , (4.21)43ue to | a | + | a ij | + √ ǫ | ∂ z a ij | ≤ C for some C > ǫ → A → √ ǫ ∂ z a a , − A → a (cid:0) γa − f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ] (cid:1) + iτ a a − iǫ P j =1 , ∂ z a j a ξ j . When ǫ > A and A are around their limits.The solution of the ODE (4 .
20) is that˜ W bdy = exp { − A + √ A − A z √ ǫ } ˜ W bdy | z =0 . (4.22)The complex root p A − A has two branches, but one of which alwayshas positive real part, since ℜ ( A − A ) > ǫ is sufficiently small, where ℜ represents the real part. Then we choose this branch. Since ℜ ( − A ) > ǫ is sufficiently small, then |ℜ p A − A | > |− A | , and then ℜ − A + √ A − A > (cid:13)(cid:13) − A + √ A − A (cid:13)(cid:13) L ∞ < + ∞ .Define T [ W bdy ] = F − [ ˜ W bdy ]. Note that (4 .
21) has the same form with(4 . , apply the same argument in Proposition 4 . . z = O ( ǫ + δ z ) where δ z > k ˜ W bdy k L ǫ →
0, then kT [ W bdy ] k L ∞ = 0.If z = O ( ǫ − δ z ) where δ z ≥ k ˜ W bdy k L → ǫ →
0, then kT [ W bdy ] k L ∞ → T [ W bdy ] and W bdy is bounded by W bdy (refer to theresults of paradifferential calculus in [31, 27]). If we assume lim ǫ → k W bdy k L ∞ = 0,then lim ǫ → kT [ W bdy ] k L ∞ = 0. This is a contradiction. So lim ǫ → k W bdy k L ∞ = 0 insome set located in the interior. Thus, Theorem 4 . σ = 0 In this section, we estimate convergence rates of the inviscid limit when σ = 0. We denote ˆ v = v ǫ − v, ˆ q = q ǫ − q, ˆ h = h ǫ − h , we denote the i − thcomponents of v ǫ and v by v ǫ,i and v i respectively. Lemma 5.1.
Assume ≤ s ≤ k − , k ≤ m − , the difference of the pressure ˆ q has the following gradient estimate: k∇ ˆ q k X s . k ˆ v k X s, + k ∂ z ˆ v k X s + | ˆ h | X s, + O ( ǫ ) . (5.1) Proof. [31] introduced the following matrices E and P satisfying E = ∂ z ϕ PP ⊤ : E = ∂ z ϕ − ∂ ϕ ∂ z ϕ − ∂ ϕ − ∂ ϕ − ∂ ϕ ∂ ϕ ) +( ∂ ϕ ) ∂ z ϕ , P = ∂ z ϕ ∂ z ϕ − ∂ ϕ − ∂ ϕ . ∇ ϕ ǫ · to (1 . and apply the divergenceoperator ∇ ϕ · to (1 . , then we get ( ∇ · ( E ǫ ∇ q ǫ ) = ∂ z ϕ ǫ △ ϕ ǫ q ǫ = − ∂ z ϕ ǫ ∇ ϕ ǫ · ( v ǫ · ∇ ϕ ǫ v ǫ ) , ∇ · ( E ∇ q ) = ∂ z ϕ △ ϕ q = − ∂ z ϕ ∇ ϕ · ( v · ∇ ϕ v ) . (5.2)It follows from (5 .
2) and ( A.
2) that ∇ · ( E ǫ ∇ ˆ q ) + ∇ · (( E ǫ − E ) ∇ q ) = ∇ · ( E ǫ ∇ q ǫ ) − ∇ · ( E ∇ q )= − ∂ z ϕ ǫ ∇ ϕ ǫ · ( v ǫ · ∇ ϕ ǫ v ǫ ) + ∂ z ϕ ∇ ϕ · ( v · ∇ ϕ v )= −∇ · (cid:2) P ǫ ( v ǫ · ∇ ϕ ǫ v ǫ ) (cid:3) + ∇ · (cid:2) P ( v · ∇ ϕ v ) (cid:3) = −∇ · (cid:2) P ǫ ( v ǫ · ∇ ϕ ǫ v ǫ − v · ∇ ϕ v ) (cid:3) − ∇ · (cid:2) ( P ǫ − P )( v · ∇ ϕ v ) (cid:3) = −∇ · (cid:2) P ǫ ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ ϕ∂ ϕz v + ˆ v · ∇ ϕ v ) (cid:3) − ∇ · (cid:2) ( P ǫ − P )( v · ∇ ϕ v ) (cid:3) . (5.3)Namely, ˆ q satisfies the following elliptic equation: ∇ · ( E ǫ ∇ ˆ q ) = −∇ · (( E ǫ − E ) ∇ q ) − ∇ · [( P ǫ − P )( v · ∇ ϕ v )] −∇ · [ P ǫ ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ ϕ∂ ϕz v + ˆ v · ∇ ϕ v )] ,q | z =0 = g ˆ h + 2 ǫ S ϕ ǫ v ǫ n ǫ · n ǫ . (5.4)The matrix E ǫ is definitely positive, then it is standard to prove that ˆ q satisfies the following gradient estimate: k∇ ˆ q k X s . k ( E ǫ − E ) ∇ q k X s + k ( P ǫ − P )( v · ∇ ϕ v ) k X s + k P ǫ ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ ϕ∂ ϕz v + ˆ v · ∇ ϕ v ) k X s + | g ˆ h + 2 ǫ S ϕ ǫ v ǫ n ǫ · n ǫ | X s, . k E ǫ − E k X s + k P ǫ − P k X s + k ˆ v k X s + k∇ ˆ v k X s + k∇ ˆ ϕ k X s + g | ˆ h | X s, + 2 ǫ (cid:12)(cid:12) S ϕ ǫ v ǫ | z =0 (cid:12)(cid:12) X s, . k ˆ v k X s, + k ∂ z ˆ v k X s + k∇ ˆ η k X s + g | ˆ h | X s, + 2 ǫ (cid:12)(cid:12) S ϕ ǫ v ǫ | z =0 (cid:12)(cid:12) X s, . k ˆ v k X s, + k ∂ z ˆ v k X s + | ˆ h | X s, + O ( ǫ ) , (5.5)where (cid:12)(cid:12) S ϕ ǫ v ǫ | z =0 (cid:12)(cid:12) X s, . k ∂ z ∂ j v ǫ k X s k ∂ j v ǫ k X s +1 < + ∞ .Thus, Lemma 5 . In order to close the estimates of tangential derivatives of ˆ v , that is tobound k ∂ ℓt ˆ v k L and √ ǫ k∇ ∂ ℓt Z α ˆ v k L , we must prove two preliminary lemmas ofˆ h by using the kinetical boundary condition (1 . .45he first preliminary lemma concerns | ∂ ℓt ˆ h | L where 0 ≤ ℓ ≤ k −
1. Notethat the estimates of mix derivatives ∂ ℓt Z α ˆ h will be obtained when we estimatemix derivatives ∂ ℓt Z α ˆ v , where | α | > Lemma 5.2.
Assume ≤ k ≤ m − , ≤ ℓ ≤ k − , | ∂ ℓt ˆ h | L has the estimates: | ∂ ℓt ˆ h | L . | ˆ h | X k − + t R | ˆ h | X k − , + k ˆ v k X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) . (5.6) Proof.
Apply ∂ ℓt to the kinetical boundary condition (1 . , we get ∂ t ∂ ℓt ˆ h + v y · ∇ y ∂ ℓt ˆ h = ∂ ℓt ˆ v · N ǫ + [ ∂ ℓt , N ǫ · ]ˆ v − [ ∂ ℓt , v y · ∇ y ]ˆ h. (5.7)Multiply (5 .
7) with ∂ ℓt ˆ h , integrate in R , we have dd t R R | ∂ ℓt ˆ h | d y = 2 R R (cid:0) ∂ ℓt ˆ v · N ǫ + [ ∂ ℓt , N ǫ · ]ˆ v − [ ∂ ℓt , v y · ∇ y ]ˆ h (cid:1) ∂ ℓt ˆ h d y + R R | ∂ ℓt ˆ h | ∇ y · v y d y . | ∂ ℓt ˆ h | L + | ˆ h | X ℓ, + (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X ℓ . | ∂ ℓt ˆ h | L + | ˆ h | X k − , + k ˆ v k X k − , + k ∂ z ˆ v k X k − . (5.8)Sum ℓ , integrate (5 .
8) in time and apply the integral form of Gronwall’sinequality, we have R R | ∂ ℓt ˆ h | d y . | ˆ h | X k − + t R | ˆ h | X k − , + k ˆ v k X k − , + k ∂ z ˆ v k X k − d t . | ˆ h | X k − + t R | ˆ h | X k − , + k ˆ v k X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) . (5.9)Thus, Lemma 5 . √ ǫ | ∂ ℓt Z α ˆ h | , by which we bound √ ǫ kS ϕ ∂ ℓt Z α ˆ η k L and then we can bound √ ǫ kS ϕ ∂ ℓt Z α ˆ v k L . Lemma 5.3.
Assume ≤ k ≤ m − , ≤ ℓ ≤ k − , ℓ + | α | ≤ k , √ ǫ | ∂ ℓt Z α ˆ h | has the estimates: ǫ | ˆ h | X k − , ≤ ǫ | ˆ h | X k − , + t R | ˆ h | X k − , + ǫ k∇ ˆ v k X k − , d t. (5.10) Proof.
The differential operator Λ is defined in the proof of Lemma 2 .
4. Apply ∂ ℓt Z α Λ to the kinetical boundary condition (1 . , we get ∂ t ∂ ℓt Z α Λ ˆ h + v y · ∇ y ∂ ℓt Z α Λ ˆ h = ∂ ℓt Z α Λ ˆ v · N ǫ + [ ∂ ℓt Z α Λ , N ǫ · ]ˆ v − [ ∂ ℓt Z α Λ , v y · ∇ y ]ˆ h. (5.11)Multiply (5 .
11) with ǫ∂ ℓt Z α Λ ˆ h , integrate in R , we have ǫ dd t R R | ∂ ℓt Z α Λ ˆ h | d y = ǫ R R | ∂ ℓt Z α Λ ˆ h | ∇ y · v y d y +2 ǫ R R (cid:0) ∂ ℓt Z α Λ ˆ v · N ǫ + [ ∂ ℓt Z α Λ , N ǫ · ]ˆ v − [ ∂ ℓt Z α Λ , v y · ∇ y ]ˆ h (cid:1) ∂ ℓt Z α Λ ˆ h d y (5.12)46 ǫ | ˆ h | X k − , + ǫ (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X k − , + ǫ | ∂ ℓt Z α Λ ˆ h | L . ǫ | ˆ h | X k − , + ǫ k ˆ v k X k − , tan + ǫ k ∂ z ˆ v k X k − , tan + ǫ | ∂ ℓt Z α Λ ˆ h | L . ǫ | ˆ h | X k − , + | ˆ h | X k − , + ǫ k∇ ˆ v k X k − , . Sum ℓ, α , integrate (5 .
12) in time and apply the integral form of Gronwall’sinequality, we have (5 . . ∂ z ˆ v can be estimated by ∂ z ˆ v h , that is k ∂ z ˆ v k X s . k ˆ v h k X s, + k ∂ z ˆ v h k X s + | ˆ h | X s, . The proof is based on the following equality that followsfrom the divergence free condition (2 . ∂ z ˆ v = − ∂ z ϕ ǫ ( ∂ ˆ v + ∂ ˆ v ) − ∂ z ˆ ϕ ( ∂ v + ∂ v )+ ∂ ϕ ǫ ∂ z ˆ v + ∂ ˆ ϕ∂ z v + ∂ ϕ ǫ ∂ z ˆ v + ∂ ˆ ϕ∂ z v . (5.13)Now we develop the estimates for tangential derivatives. Lemma 5.4.
Assume ≤ k ≤ m − , ∂ ℓt Z α ˆ v and ∂ ℓt Z α ˆ h have the estimates: k ˆ v k X k − , + | ˆ h | X k − , + ǫ | ˆ h | X k − , + ǫ t R k∇ ˆ v k X k − , d t . k ˆ v k X k − , + | ˆ h | X k − , + ǫ | ˆ h | X k − , + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + | ∂ kt ˆ h | L ([0 ,T ] ,L ) + k∇ ˆ q k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.14) Proof. ( ˆ V ℓ,α , ˆ Q ℓ,α ) satisfy the following equations: ∂ ϕ ǫ t ˆ V ℓ,α + v ǫ · ∇ ϕ ǫ ˆ V ℓ,α + ∇ ϕ ǫ ˆ Q ℓ,α − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt Z α ˆ v = 2 ǫ [ ∂ ℓt Z α , ∇ ϕ ǫ · ] S ϕ ǫ ˆ v + 2 ǫ ∇ ϕ ǫ · [ ∂ ℓt Z α , S ϕ ǫ ]ˆ v + ǫ∂ ℓt Z α △ ϕ ǫ v − ∂ ℓt Z α ˆ ϕ∂ ϕ ǫ t ∂ ϕz v − ∂ ℓt Z α ˆ ϕ v ǫ · ∇ ϕ ǫ ∂ ϕz v − ∂ ℓt Z α ˆ v · ∇ ϕ v − ∂ ℓt Z α ˆ ϕ ∇ ϕ ǫ ∂ ϕz q − [ ∂ ℓt Z α , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt Z α , v ǫ · ∇ ϕ ǫ ]ˆ v − [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v +[ ∂ ℓt Z α , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ := I , ∇ ϕ ǫ · ˆ V ℓ,α = − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η − ∂ ℓt Z α ˆ η ∇ ϕ ǫ · ∂ ϕz v,∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h − N ǫ · ˆ V ℓ,α = N ǫ · ∂ ϕz v∂ ℓt Z α ˆ η − ˆ v y · ∇ y ∂ ℓt Z α h − ∂ y ˆ h · ∂ ℓt Z α v y + [ ∂ ℓt Z α , ˆ v, N ǫ ] − [ ∂ ℓt Z α , v y , ∂ y ˆ h ] , ˆ Q ℓ,α N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ − ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ǫ S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , ǫ S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] + 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ , ( ∂ ℓt Z α ˆ v, ∂ ℓt Z α ˆ h ) | t =0 = ( ∂ ℓt Z α v ǫ − ∂ ℓt Z α v , ∂ ℓt Z α h ǫ − ∂ ℓt Z α h ) , (5.15)47hen | α | ≥ , ℓ ≤ k − , ≤ ℓ + | α | ≤ k , we develop the L estimate ofˆ V ℓ,α , we have
12 dd t R R − | ˆ V ℓ,α | d V t − R R − ˆ Q ℓ,α ∇ ϕ ǫ · ˆ V ℓ,α d V t + 2 ǫ R R − |S ϕ ǫ ∂ ℓt Z α ˆ v | d V t = − R { z =0 } (cid:0) ˆ Q ℓ,α N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ (cid:1) · ˆ V ℓ,α d y + R R − I · V ℓ,α d V t +2 ǫ R R − S ϕ ǫ ∂ ℓt Z α ˆ v · S ϕ ǫ ( ∂ ϕz v∂ ℓt Z α ˆ η ) d V t . − R { z =0 } (cid:0) ˆ Q ℓ,α N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ (cid:1) · ˆ V ℓ,α d y + k ˆ V ℓ,α k L + k ∂ z ˆ v k X k − + k ˆ v k X k − , + k ˆ η k X k − , + ǫ | ˆ h | X k − , + k ∂ kt ˆ η k L + k∇ ˆ q k X k − + O ( ǫ ) . (5.16)We develop the boundary estimates in (5 . − R { z =0 } (cid:0) ˆ Q ℓ,α N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ (cid:1) · ˆ V ℓ,α d y = R { z =0 } − ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h N ǫ · ˆ V ℓ,α − (cid:0) ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ +(2 ǫ S ϕ ǫ v ǫ − ǫ S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , ǫ S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ]+2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] + 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ (cid:1) · ˆ V ℓ,α d y . R { z =0 } − ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h N ǫ · ˆ V ℓ,α d y + O ( ǫ ) , (5.17)note that normal derivatives ∂ z v ǫ on the boundary can be expressed in termsof tangential derivatives of v ǫ , thus we get O ( ǫ ) rather than O ( √ ǫ ). Namely, − R { z =0 } (cid:0) ˆ Q ℓ,α N ǫ − ǫ∂ ℓt Z α S ϕ ǫ ˆ v N ǫ (cid:1) · ˆ V ℓ,α d y . R { z =0 } − ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h (cid:0) ∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h − N ǫ · ∂ ϕz v∂ ℓt Z α ˆ η +ˆ v y · ∇ y ∂ ℓt Z α h + ∂ y ˆ h · ∂ ℓt Z α v y − [ ∂ ℓt Z α , ˆ v, N ǫ ] + [ ∂ ℓt Z α , v y , ∂ y ˆ h ] (cid:1) d y + O ( ǫ ) . −
12 dd t R { z =0 } ( g − ∂ ϕz q ) | ∂ ℓt Z α ˆ h | d y + R { z =0 } ∇ · ( gv y − v y ∂ ϕz q ) | ∂ ℓt Z α ˆ h | d y − R { z =0 } ∂ t ∂ ϕz v | ∂ ℓt Z α ˆ h | d y − R { z =0 } ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h (cid:0) − N ǫ · ∂ ϕz v∂ ℓt Z α ˆ η +ˆ v y · ∇ y ∂ ℓt Z α h + ∂ y ˆ h · ∂ ℓt Z α v y − [ ∂ ℓt Z α , ˆ v, N ǫ ] + [ ∂ ℓt Z α , v y , ∂ y ˆ h ] (cid:1) d y + O ( ǫ ) . −
12 dd t R { z =0 } ( g − ∂ ϕz q ) | ∂ ℓt Z α ˆ h | d y + k ˆ v k X k − + k ∂ z ˆ v k X k − + | ˆ h | X k − , + O ( ǫ ) . (5.18)By (5 .
16) and (5 . dd t R R − | ˆ V ℓ,α | d V t + dd t R { z =0 } ( g − ∂ ϕz q ) | ∂ ℓt Z α ˆ h | d y + ǫ R R − | ∂ ℓt Z α S ϕ ǫ ˆ v | d V t . k ˆ V ℓ,α k L + k ∂ z ˆ v k X k − + k ˆ v k X k − , + | ˆ h | X k − , + | ∂ kt ˆ h | L + ǫ | ˆ h | X k − , + k∇ ˆ q k X k − + O ( ǫ ) . (5.19)48ince g − ∂ ϕz q ≥ c >
0, integrate (5 .
19) in time, apply the integral form ofGronwall’s inequality, we get k ∂ ℓt Z α ˆ v k L + | ∂ ℓt Z α ˆ h | + ǫ | ˆ h | X k − , + ǫ t R k∇ ∂ ℓt Z α ˆ v k L d t . k ˆ v k X k − , + | ˆ h | X k − , + t R k ˆ v k X k − , + | ˆ h | X k − , + ǫ | ˆ h | X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + | ∂ kt ˆ h | L ([0 ,T ] ,L ) + k∇ ˆ q k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.20)When | α | = 0 , ≤ ℓ ≤ k −
1, we have no bounds of ˆ q and ∂ ℓt ˆ q , so weneither use the variable ˆ Q ℓ,α and nor apply the integration by parts to thepressure terms. Also, the dynamical boundary condition will not be used. Sincethe main equation of ˆ V ℓ, and its kinetical boundary condition satisfy ∂ ϕ ǫ t ˆ V ℓ, + v ǫ · ∇ ϕ ǫ ˆ V ℓ, − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt ˆ v = ǫ∂ ℓt △ ϕ ǫ v ǫ + 2 ǫ [ ∂ ℓt , ∇ ϕ ǫ · ] S ϕ ǫ ˆ v + 2 ǫ ∇ ϕ ǫ · [ ∂ ℓt , S ϕ ǫ ]ˆ v − ∂ ℓt ∇ ϕ ǫ ˆ q + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt ˆ ϕ − ∂ ℓt ˆ ϕ∂ ϕ ǫ t ∂ ϕz v − ∂ ℓt ˆ v · ∇ ϕ v − ∂ ℓt ˆ ϕ v ǫ · ∇ ϕ ǫ ∂ ϕz v − [ ∂ ℓt , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt , v ǫ · ∇ ϕ ǫ ]ˆ v + [ ∂ ℓt , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt , ∇ ϕ v · ]ˆ v + [ ∂ ℓt , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ := I ,∂ t ∂ ℓt ˆ h + v ǫy · ∇ y ∂ ℓt ˆ h − N ǫ · ˆ V ℓ, = N ǫ · ∂ ϕz v∂ ℓt ˆ η − ˆ v y · ∇ y ∂ ℓt h − ∂ y ˆ h · ∂ ℓt v y + [ ∂ ℓt , ˆ v, N ǫ ] − [ ∂ ℓt , v y , ∂ y ˆ h ] , ( ∂ ℓt ˆ v, ∂ ℓt ˆ h ) | t =0 = ( ∂ ℓt v ǫ − ∂ ℓt v , ∂ ℓt h ǫ − ∂ ℓt h ) , (5.21)then we have L estimate of ˆ V ℓ, :
12 dd t R R − | ˆ V ℓ, | d V t + 2 ǫ R R − |S ϕ ǫ ∂ ℓt ˆ v | d V t = 2 ǫ R { z =0 } S ϕ ǫ ∂ ℓt ˆ v N ǫ · ˆ V ℓ, d y +2 ǫ R R − S ϕ ǫ ∂ ℓt ˆ v · S ϕ ǫ ( ∂ ϕz v∂ ℓt ˆ η ) d V t + R R − I · ˆ V ℓ, d V t . (5.22)Now we estimate the right hand side of (5 . ǫ R { z =0 } S ϕ ǫ ∂ ℓt v ǫ N ǫ · ˆ V ℓ, d y = 2 ǫ R { z =0 } S ϕ ǫ ∂ ℓt v ǫ N ǫ · ( ∂ ℓt ˆ v − ∂ ϕz v∂ ℓt ˆ η ) d y . (cid:12)(cid:12) ∂ ℓt ˆ v | z =0 (cid:12)(cid:12) L + | ∂ ℓt ˆ h | L + O ( ǫ ) . k ∂ ℓt ˆ v k L + k ∂ ℓt ∂ z ˆ v k L + | ∂ ℓt ˆ h | L + O ( ǫ ) . (5.23)It is easy to check that R R − I · ˆ V ℓ, d V t . k ˆ V ℓ, k L + k ∂ ℓ − t ∂ z ˆ v k L + k ∂ ℓ − t ∂ y ˆ v k L + k ∂ ℓ − t ˆ v k L + k ∂ ℓ − t ∇ ˆ η k L + k ∂ ℓt ˆ η k L + k ∂ ℓt ∇ ˆ η k L + k ∂ ℓt ∇ ˆ q k L + O ( ǫ ) . (5.24)49y (5 . .
23) and (5 . dd t R R − | ˆ V ℓ, | d V t + ǫ R R − |S ϕ ǫ ∂ ℓt ˆ v | d V t . k ˆ V ℓ, k L + k ∂ z ˆ v k X k − + k ˆ v k X k − , + | ˆ h | X k − , + ǫ | ˆ h | X k − , + k∇ ˆ q k X k − + O ( ǫ ) . (5.25)Integrate (5 .
25) in time, apply the integral form of Gronwall’s inequality,we have k ˆ V ℓ, k L + ǫ k∇ ∂ ℓt ˆ v k L . k ˆ v k X k − + | ˆ h | X k − + t R k ∂ z ˆ v k X k − + k ˆ v k X k − , + | ˆ h | X k − , + k∇ ˆ q k X k − + ǫ | ˆ h | X k − , d t + O ( ǫ ) . k ˆ v k X k − + | ˆ h | X k − + t R k ˆ v k X k − , + | ˆ h | X k − , + ǫ | ˆ h | X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + k∇ ˆ q k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.26)Combining (5 .
26) and Lemma 5 .
2, we have k ∂ ℓt ˆ v k L + | ∂ ℓt ˆ h | L + ǫ | ˆ h | X k − , + ǫ t R k∇ ∂ ℓt ˆ v k L d t . k ˆ v k X k − + | ˆ h | X k − + t R k ˆ v k X k − , + | ˆ h k X k − , + ǫ | ˆ h | X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + k∇ ˆ q k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.27)Apply the integral form of Gronwall’s inequality to (5 .
20) and (5 . . . Π S ϕ v n | z =0 = 0 In this subsection, we develop the estimates for normal derivatives ∂ z ˆ v . Inthe following lemma, we estimate k ∂ z ˆ v k L ([0 ,T ] ,X k − ) by studying the equationsof ˆ ω h . Lemma 5.5.
Assume k ≤ m − , if Π S ϕ v n | z =0 = 0 , then the vorticity has thefollowing estimate: k ∂ z ˆ v h k L ([0 ,T ] ,X k − ) + k ˆ ω h k L ([0 ,T ] ,X k − ) . (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + T R k ˆ v k X k − , d t + T R | ˆ h | X k − , d t + k ∂ kt ˆ h k L ([0 ,T ] ,L ) + O ( √ ǫ ) . (5.28) Proof.
Assume ℓ + | α | ≤ k −
1, we study the equations (1 .
33) and decomposeˆ ω h = ˆ ω nhomh + ˆ ω homh , such that ˆ ω nhomh satisfies the following nonhomogeneous50quations: ∂ ϕ ǫ t ˆ ω nhomh + v ǫ · ∇ ϕ ǫ ˆ ω nhomh − ǫ △ ϕ ǫ ˆ ω nhomh = ~ F [ ∇ ϕ ǫ ]( ω ǫh , ∂ j v ǫ,i ) − ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) + ǫ △ ϕ ǫ ω h + ∂ ϕz ω h ∂ ϕ ǫ t ˆ η + ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ η − ˆ v · ∇ ϕ ω h , ˆ ω nhomh | z =0 = 0 , ˆ ω nhomh | t =0 = (ˆ ω , ˆ ω ) ⊤ , (5.29)and ˆ ω homh satisfies the following homogeneous equations: ∂ ϕ ǫ t ˆ ω homh + v ǫ · ∇ ϕ ǫ ˆ ω homh − ǫ △ ϕ ǫ ˆ ω homh = 0 , ˆ ω homh | z =0 = F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − ω b , ˆ ω homh | t =0 = 0 , (5.30)By using (3 .
1) and ∂ ϕ ǫ t + v ǫ · ∇ ϕ ǫ = ∂ t + v ǫy · ∇ y + V ǫz ∂ z , (5 .
29) is equivalentto the following equations: ∂ t ˆ ω nhomh + v ǫy · ∇ y ˆ ω nhomh + V ǫz ∂ z ˆ ω nhomh − ǫ △ ϕ ǫ ˆ ω nhomh = f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i ]ˆ ω h + f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∂ j ˆ v i + f [ ∇ ϕ ǫ , ∇ ϕ, ∂ j v ǫ,i , ∂ j v i , ω ǫh , ω h ] ∇ ˆ ϕ + ǫ △ ϕ ǫ ω h + ∂ ϕz ω h ∂ ϕ ǫ t ˆ η + ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ η − ˆ v · ∇ ϕ ω h := I , ˆ ω nhomh | z =0 = 0 , ˆ ω nhomh | t =0 = (ˆ ω , ˆ ω ) ⊤ , (5.31)Apply ∂ ℓt Z α to (5 . ∂ t ∂ ℓt Z α ˆ ω nhomh + v ǫy · ∇ y ∂ ℓt Z α ˆ ω nhomh + V ǫz ∂ z ∂ ℓt Z α ˆ ω nhomh − ǫ △ ϕ ǫ ∂ ℓt Z α ˆ ω nhomh = ∂ ℓt Z α I − [ ∂ ℓt Z α , v ǫy · ∇ y ]ˆ ω nhomh − [ ∂ ℓt Z α , V z ∂ z ]ˆ ω nhomh + ǫ ∇ ϕ ǫ · [ ∂ ℓt Z α , ∇ ϕ ]ˆ ω nhomh + ǫ [ ∂ ℓt Z α , ∇ ϕ · ] ∇ ϕ ǫ ˆ ω nhomh ,∂ ℓt Z α ˆ ω nhomh | z =0 = 0 ,∂ ℓt Z α ˆ ω nhomh | t =0 = ( ∂ ℓt Z α ˆ ω , ∂ ℓt Z α ˆ ω ) ⊤ , (5.32)Develop the L estimate of ∂ ℓt Z α ˆ ω nhomh , we get dd t k ∂ ℓt Z α ˆ ω nhomh k L + 2 ǫ k∇ ϕ ǫ ∂ ℓt Z α ˆ ω nhomh k L . k ∂ ℓt Z α ˆ ω nhomh k L + k ∂ ℓt Z α I k L + k [ ∂ ℓt Z α , V z ∂ z ]ˆ ω nhomh k L + ǫ R R − ∇ ϕ ǫ · [ ∂ ℓt Z α , ∇ ϕ ]ˆ ω nhomh ∂ ℓt Z α ˆ ω nhomh d V t + ǫ R R − [ ∂ ℓt Z α , ∇ ϕ · ] ∇ ϕ ǫ ˆ ω nhomh ∂ ℓt Z α ˆ ω nhomh d V t (5.33)51 k ∂ ℓt Z α ˆ ω nhomh k L + k ˆ ω h k X k − + k ˆ v k X k − , + k∇ ˆ η k X k − + k ∂ kt ˆ η k L + P ℓ + | α | > k − zz ∂ ℓt Z α V z · ∂ ℓt Z α z − z ∂ z ˆ ω nhomh k L − ǫ R R − [ ∂ ℓt Z α , N ∂ ϕz ]ˆ ω nhomh · ∇ ϕ ǫ ∂ ℓt Z α ˆ ω nhomh d V t + ǫ R R − P ℓ + | α | > (cid:2) ( ∂ ϕz ) − ∂ ℓ t Z α ( N ∂ z ϕ ) ∂ ℓ t Z α ∂ z (cid:3) · ∇ ϕ ǫ ˆ ω nhomh ∂ ϕz ∂ ℓt Z α ˆ ω nhomh d V t . where the notation ( ∂ ϕz ) − means a cancellation such that ( ∂ ϕz ) − ( ∂ ϕz ) = 1.Integrate (5 .
33) in time, apply the integral form of Gronwall’s inequality,it is easy to have k ˆ ω nhomh k X k − + 2 ǫ t R k∇ ˆ ω nhomh k X k − d t ≤ k ˆ ω ,h k X k − + t R k ˆ ω h k X k − d t + k ˆ h k X k − , d t + k ∂ kt ˆ h k L d t + O ( ǫ ) . (5.34)Similar to (2 . k ˆ ω nhomh k L ([0 ,T ] ,X k − ) . √ T (cid:13)(cid:13) ˆ ω ,h (cid:13)(cid:13) X k − + T k ˆ ω h k L ([0 ,T ] ,X k − ) + √ T T R k ˆ v k X k − , d t + √ T T R | ˆ h | X k − , d t + √ T | ∂ kt ˆ h | L ([0 ,T ] ,L ) + O ( ǫ ) . (5.35)For the homogeneous equations (5 . .
41) or [31], we have k ∂ ℓt Z α ˆ ω homh k L ([0 ,T ] ,L ( R − )) . k ∂ ℓt Z α ˆ ω homh k H ([0 ,T ] ,L ( R − )) . √ ǫ T R (cid:12)(cid:12) ˆ ω homh | z =0 (cid:12)(cid:12) X k − ( R ) d t . √ ǫ T R (cid:12)(cid:12) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:12)(cid:12) X k − ( R ) d t + √ ǫ T R (cid:12)(cid:12) ς Θ + ς Θ + ς Θ (cid:12)(cid:12) X k − ( R ) d t + √ ǫ T R (cid:12)(cid:12) ς Θ + ς Θ + ς Θ (cid:12)(cid:12) X k − ( R ) d t . O ( √ ǫ ) , (5.36)where ς i and Θ i are defined in the proof of Lemma 4 . .
35) and (5 . k ˆ ω h k L ([0 ,T ] ,X k − ) . k ˆ ω nhomh k L ([0 ,T ] ,X k − ) + k ˆ ω homh k L ([0 ,T ] ,X k − ) . (cid:13)(cid:13) ˆ ω ,h (cid:13)(cid:13) X k − + | ∂ kt ˆ h | L ([0 ,T ] ,L ) + T R k ˆ v k X k − , d t + T R | ˆ h | X k − , d t + O ( √ ǫ ) . (5.37)Thus, Lemma 5 . emark 5.6. If Π S ϕ v n | z =0 = 0 , then Θ i = 0 where i = 1 , · · · , , and then theestimate (5 . is reduced into the following estimate: k ∂ ℓt Z α ˆ ω homh k L ([0 ,T ] ,L ( R − )) . √ ǫ T R (cid:12)(cid:12) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:12)(cid:12) X k − ( R ) d t . O ( √ ǫ ) , (5.38) Since we do not have the convergence rates of | ∂ j v ǫ,i − ∂ j v i | X k − ( R ) , thus wecan not improve the convergence rates of k ω k L ([0 ,T ] ,X k − ) . However, we canimprove the convergence rates of k ω k L ([0 ,T ] ,X k − ) , see subsection 5.4. If Π S ϕ v n | z =0 = 0, we estimate k ∂ z ˆ v k L ∞ ([0 ,T ] ,X m − ) and k ˆ ω k L ∞ ([0 ,T ] ,X m − ) .Note that when Π S ϕ v n | z =0 = 0, not only (cid:12)(cid:12) ∇ ϕ ǫ × ∂ ℓt Z α ( v ǫ − v ) | z =0 (cid:12)(cid:12) L = 0 butalso (cid:12)(cid:12) N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ( v ǫ − v )) | z =0 (cid:12)(cid:12) L = 0. Lemma 5.7.
Assume ≤ k ≤ m − , ˆ ω h = ω ǫh − ω h , ∂ z ˆ v = ∂ z v ǫ − ∂ z v , then ˆ ω h and ∂ z ˆ v satisfy the following estimate: k ˆ ω k X k − + k ∂ z ˆ v k X k − . k ˆ ω k X k − + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + k ˆ h k X k − d t + O ( ǫ ) . (5.39) Proof.
By using (2 . . as ∂ ϕ ǫ t ˆ v − ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ˆ v · ∇ ϕ v + ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η = − ǫ ∇ ϕ ǫ × ˆ ω − ǫ ∇ ϕ ǫ × ω (5.40)Firstly, we develop L estimate of ˆ ω . Multiple (5 .
40) with ∇ ϕ ǫ × ( ∇ ϕ ǫ × ˆ v ),integrate in R − , use the integration by parts formula (1 . , we get R R − (cid:0) ∂ ϕ ǫ t ˆ v − ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ˆ v · ∇ ϕ v + ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η + ǫ ∇ ϕ ǫ × ( ∇ ϕ ǫ × ˆ v ) + ǫ △ ϕ ǫ v (cid:1) · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ˆ v ) d V t = 0 , R R − ∇ ϕ ǫ × (cid:0) ∂ ϕ ǫ t ˆ v + v ǫ · ∇ ϕ ǫ ˆ v + ∇ ϕ ǫ ˆ q (cid:1) d V t + ǫ R R − |∇ ϕ ǫ × ( ∇ ϕ ǫ × ˆ v ) | d V t = R R − ∇ ϕ ǫ × (cid:0) ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v − ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ˆ η (cid:1) · ˆ ω d V t − R z =0 (cid:0) ∂ t ˆ v + v ǫy · ∇ y ˆ v + ˆ v · ∇ ϕ v − ∂ ϕz v∂ t ˆ η − v ǫy · ∇ y ˆ η ∂ ϕz v + ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η (cid:1) · N ǫ × ( ∇ ϕ ǫ × ˆ v ) d y − ǫ R R − ∇ ϕ ǫ × ω · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ˆ v ) d V t . k ˆ ω k L + | ˆ h | X , + k ˆ v k X + k ∂ z ˆ v k L + ǫ R R − |∇ ϕ ǫ × ˆ ω | d V t + O ( ǫ )+ (cid:12)(cid:12) N ǫ × ( ∇ ϕ ǫ × ˆ v ) | z =0 (cid:12)(cid:12) L (cid:0)(cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X tan + (cid:12)(cid:12) ˆ h | z =0 (cid:12)(cid:12) X (cid:1) + (cid:12)(cid:12) N ǫ × ( ∇ ϕ ǫ × ˆ v ) | z =0 (cid:12)(cid:12) (cid:12)(cid:12) ∇ ˆ q | z =0 (cid:12)(cid:12) − . (5.41)53ince ∇ ϕ ǫ × ∇ ϕ ǫ ˆ q = 0, ∇ ϕ ǫ × ω is bounded, (cid:12)(cid:12) N ǫ × ( ∇ ϕ ǫ × ˆ v ) | z =0 (cid:12)(cid:12) L = 0, k ˆ ω | L + ǫ t R k∇ ˆ ω k d t . k ˆ ω k L + t R | ˆ h | X , + k ˆ v k X + k ∂ z ˆ v k L + (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X tan + (cid:12)(cid:12) ˆ h | z =0 (cid:12)(cid:12) X + (cid:12)(cid:12) ∇ ˆ q | z =0 (cid:12)(cid:12) − d t + O ( ǫ ) . k ˆ ω k L + t R | ˆ h | X , + k ˆ v k X + k ∂ z ˆ v k L + k ˆ v k X tan + k ∂ z ˆ v k X tan + k ˆ h k X + k∇ ˆ q k L d t + O ( ǫ ) . (5.42)When ℓ + | α | ≤ k −
2, we study the quantity ∇ ϕ ǫ × ∂ ℓt Z α ˆ v .The equations ( A. is rewritten as ∂ ϕ ǫ t ∂ ℓt Z α ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q + ǫ ∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v = ǫ I , + I , , (5.43)where I , = − [ ∂ ℓt Z α , ∇ ϕ ǫ × ] ∇ ϕ ǫ × ˆ v − ∇ ϕ ǫ × [ ∂ ℓt Z α , ∇ ϕ ǫ × ]ˆ v + ∂ ℓt Z α △ ϕ ǫ v, I , := ∂ ϕz v ( ∂ t + v ǫy · ∇ y + V ǫz ∂ z ) ∂ ℓt Z α ˆ ϕ − ∂ ℓt Z α ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − [ ∂ ℓt Z α , ∂ t + v ǫ ∂ y + V ǫz ∂ z ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v ( ∂ t + v ǫy · ∇ y + V ǫz ∂ z ] ˆ ϕ − [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v − [ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ. (5.44)Multiply (5 .
43) with ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ), integrate in R − , we get R R − ∂ ϕ ǫ t ∂ ℓt Z α ˆ v · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R R − v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R R − ∇ ϕ ǫ ∂ ℓt Z α ˆ q · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + ǫ R R − |∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt = R R − ( ǫ I , + I , ) · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt . (5.45)Use the integration by parts formula (1 . and note that [ ∂ ϕ ǫ t , ∇ ϕ ǫ ] = 0,we have R z =0 ∂ ϕ ǫ t ∂ ℓt Z α ˆ v · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + R R − ∂ ϕ ǫ t ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R z =0 v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + R R − v ǫ · ∇ ϕ ǫ ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt (5.46)54 R R − [( P i =1 ∇ ϕ ǫ v ǫ,i · ∂ ϕ ǫ i ) × ∂ ℓt Z α ˆ v ] · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R z =0 ∇ ϕ ǫ ∂ ℓt Z α ˆ q · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + R R − ∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ q · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + ǫ R R − |∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt = ǫ R R − I , · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R z =0 I , · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + R R − ∇ ϕ ǫ × I , · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt . Note that ∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ q = 0 and ( ∂ ϕ ǫ t + v ǫ · ∇ ϕ ǫ ) | z =0 = ( ∂ t + v ǫy · ∇ y ),we have
12 dd t R R − |∇ ϕ ǫ × ∂ ℓt Z α ˆ v | d V ǫt + ǫ R R − |∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt = − R z =0 ( ∂ t + v ǫy · ∇ y ) ∂ ℓt Z α ˆ v · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y − R R − [( P i =1 ∇ ϕ ǫ v ǫ,i · ∂ ϕ ǫ i ) × ∂ ℓt Z α ˆ v ] · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt − R z =0 ∇ ϕ ǫ ∂ ℓt Z α ˆ q · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + ǫ R R − I , · ∇ ϕ ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt + R z =0 I , · N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d y + R R − ∇ ϕ ǫ × I , · ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) d V ǫt . k∇ ϕ ǫ × ∂ ℓt Z α ˆ v k L + | N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) | (cid:12)(cid:12) ∇ ϕ ǫ ∂ ℓt Z α ˆ q | z =0 (cid:12)(cid:12) − + | N ǫ × ( ∇ ϕ ǫ × ∂ ℓt Z α ˆ v ) | L (cid:0)(cid:12)(cid:12) I , | z =0 (cid:12)(cid:12) L + (cid:12)(cid:12) ∂ ℓt Z α ˆ v | z =0 (cid:12)(cid:12) X tan (cid:1) + k ∂ ℓt Z α ˆ v k X + k ∂ z ∂ ℓt Z α ˆ v k L + ǫ kI , k L + k∇ ϕ ǫ × I , k L . (5.47)It is easy to prove that (cid:12)(cid:12) I , | z =0 (cid:12)(cid:12) L . | ˆ h | X k − + (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X k − + (cid:12)(cid:12) ∇ ˆ q | z =0 (cid:12)(cid:12) X k − . | ˆ h | X k − + k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − ,ǫ kI , k L . ǫ P ℓ + | α |≤ k − k∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v k L + O ( ǫ ) , k∇ ϕ ǫ × I , k L . k ˆ η k X k − , + k∇ ϕ ǫ × ˆ v k X k − tan + k∇ ϕ ǫ × [ ∂ ℓt Z α , V ǫz ∂ z ]ˆ v k L + k∇ ϕ ǫ × [ ∂ ℓt Z α , N ǫ ∂ ϕ ǫ z ]ˆ q k L , (5.48)where the estimates for the last two terms are similar to (2 . , (2 . .
47) in time, apply the integral form of Gronwall’s inequality,55e have k∇ ϕ ǫ × ∂ ℓt Z α ˆ v k L + ǫ R R − |∇ ϕ ǫ × ∇ ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt . (cid:13)(cid:13) ∇ ϕ ǫ × ∂ ℓt Z α ˆ v | t =0 (cid:13)(cid:13) L + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + k ˆ h k X k − d t + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k ˆ h k X k − , d t + O ( ǫ ) . (cid:13)(cid:13) ∂ ℓt Z α ˆ ω | t =0 (cid:13)(cid:13) L + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + k ˆ h k X k − d t + O ( ǫ ) . (5.49)Since ˆ ω = ∇ ϕ ǫ × ˆ v − ∇ ϕ ǫ ˆ η × ∂ ϕz v , we have k ˆ ω k X k − + k ∂ z ˆ v k X k − . k ˆ ω | t =0 k X k − + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + k ˆ h k X k − d t + O ( ǫ ) . (5.50)Thus, Lemma 5 . Remark 5.8.
We can not use the following variables to estimate ˆ ω h : ( ˆ ζ = ˆ ω − F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) + ω b, , j = 1 , , i = 1 , , , ˆ ζ = ˆ ω − F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) + ω b, , j = 1 , , i = 1 , , , (5.51) because F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) may not converge to the extension of ω b . That is, ˆ ζ may not be a small quantity. Π S ϕ v n | z =0 = 0 When Π S ϕ v n | z =0 = 0, the boundary value of Navier-Stokes vorticity con-verges to that of Euler vorticity, the convergence rates of the inviscid limits canbe improved. In the following lemma, we estimate normal derivatives for thespecial Euler boundary data. Lemma 5.9.
Assume k ≤ m − , if Π S ϕ v n | z =0 = 0 , then the vorticity has thefollowing estimate: k ∂ z ˆ v h k L ([0 ,T ] ,X k − ) + k ˆ ω h k L ([0 ,T ] ,X k − ) . (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + T R k ˆ v k X k − , d t + T R | ˆ h | X k − , d t + k ∂ k − t ˆ h k L ([0 ,T ] ,L ) + √ ǫ k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.52) Proof.
If Π S ϕ v n | z =0 = 0, Θ i = 0 where i = 1 , · · · ,
6, See Remark 5 . ℓ + | α | ≤ k −
2, we study the equations (1 .
33) and decompose ˆ ω h =ˆ ω nhomh + ˆ ω homh , such that ˆ ω nhomh satisfies the nonhomogeneous equations (5 . ω homh satisfies the homogeneous equations (5 . ω nhomh satisfies the following estimate: k ˆ ω nhomh k L ([0 ,T ] ,X k − ) . √ T (cid:13)(cid:13) ˆ ω ,h (cid:13)(cid:13) X k − + T k ˆ ω h k L ([0 ,T ] ,X k − ) + √ T T R k ˆ v k X k − , d t + √ T T R | ˆ h | X k − , d t + √ T | ∂ k − t ˆ h | L ([0 ,T ] ,L ) + O ( ǫ ) . (5.53)When ℓ + | α | ≤ k −
2, the estimate (5 .
36) is reduced as follows: k ∂ ℓt Z α ˆ ω homh k L ([0 ,T ] ,L ( R − )) . √ ǫ T R (cid:12)(cid:12) F , [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − F , [ ∇ ϕ ]( ∂ j v i ) (cid:12)(cid:12) X k − ( R ) d t . √ ǫ T R (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X k − ( R ) d t + √ ǫ T R | ˆ h | X k − , ( R ) d t . √ ǫ T R k ˆ v k X k − , ( R ) d t + √ ǫ √ T k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + √ ǫ T R | ˆ h | X k − , ( R ) d t. (5.54)By (5 .
53) and (5 . k ∂ z ˆ v h k L ([0 ,T ] ,X k − ) + k ˆ ω h k L ([0 ,T ] ,X k − ) . k ˆ ω nhomh k L ([0 ,T ] ,X k − ) + k ∂ ℓt Z α ˆ ω homh k L ([0 ,T ] ,L ( R − )) . (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + T R k ˆ v k X k − , d t + T R | ˆ h | X k − , d t + k ∂ k − t ˆ h k L ([0 ,T ] ,L ) + √ ǫ k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (5.55)Thus, Lemma 5 . In this subsection, we calculate convergence rates of the inviscid limit.
Theorem 5.10.
Assume
T > is finite, fixed and independent of ǫ , ( v ǫ , h ǫ ) isthe solution in [0 , T ] of Navier-Stokes equations (1 . with initial data ( v ǫ , h ǫ ) satisfying (1 . , ω ǫ is its vorticity. ( v, h ) is the solution in [0 , T ] of Eulerequations (1 . with initial data ( v , h ) ∈ X m − , ( R − ) × X m − , ( R ) , ω isits vorticity. Assume there exists an integer k where ≤ k ≤ m − , such that k v ǫ − v k X k − , ( R − ) = O ( ǫ λ v ) , | h ǫ − h | X k − , ( R ) = O ( ǫ λ h ) , k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) , where λ v > , λ h > , λ ω > .If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , then theconvergence rates of the inviscid limit satisfy (1 . .If the Euler boundary data satisfies Π S ϕ v n | z =0 = 0 in [0 , T ] , assume k ω ǫ − ω k X k − ( R − ) = O ( ǫ λ ω ) where λ ω > , then the convergence rates of the inviscidlimit satisfy (1 . .Proof. If Π S ϕ v n | z =0 = 0, we prove the converge rates of the inviscid limit:57y Lemmas 5 . , . , .
5, we have k ˆ v k X k − , + | ˆ h | X k − , . k ˆ v k X k − , + | ˆ h | X k − , + | ˆ ω | X k − + t R k ˆ v k X k − , + t R k ˆ h k X k − , d t + O ( √ ǫ ) . (5.56)Apply the integral form of Gronwall’s inequality to (5 . k ˆ v k X k − , + | ˆ h | X k − , . k ˆ v k X k − , + | ˆ h | X k − , + (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + O ( √ ǫ ) . O ( ǫ min { , λ v , λ h , λ ω } ) . (5.57)By Lemma 5 .
5, we have k ∂ z ˆ v h k L ([0 ,T ] ,X k − ) + k ˆ ω h k L ([0 ,T ] ,X k − ) . O ( ǫ min { , λ v , λ h , λ ω } ) . (5.58)By Lemmas 5 . , . , .
7, we have k ˆ ω k X k − + k ∂ z ˆ v k X k − . k ˆ ω k X k − + t R k ˆ v k X k − + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + k ˆ h k X k − d t + O ( ǫ ) . O ( ǫ min { ,λ v ,λ h ,λ ω } ) . (5.59)If Π S ϕ v n | z =0 = 0, we prove the converge rates of the inviscid limit:By Lemma 5 .
9, we have k ∂ z ˆ v h k L ([0 ,T ] ,X k − ) + k ˆ ω h k L ([0 ,T ] ,X k − ) . (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + T R k ˆ v k X k − , d t + T R | ˆ h | X k − , d t + √ ǫO ( ǫ min { , λ v , λ h , λ ω } ) . (5.60)Couple (5 .
60) with the following tangential estimates, k ˆ v k X k − , + | ˆ h | X k − , . k ˆ v k X k − , + | ˆ h | X k − , + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + t R k ˆ v k X k − , + t R k ˆ h k X k − , d t + O ( ǫ ) , (5.61)apply the integral form of Gronwall’s inequality, then we get k ˆ v k X k − , + | ˆ h | X k − , . (cid:13)(cid:13) ˆ ω (cid:13)(cid:13) X k − + k ˆ v k X k − , + | ˆ h | X k − , + √ ǫO ( ǫ min { , λ v , λ h , λ ω } ) . O ( ǫ min { , λ v , λ h , λ ω , λ ω +1 } ) = O ( ǫ min { , λ v , λ h , λ ω } ) . (5.62)Similar to (5 . k ˆ ω k X k − + k ∂ z ˆ v k X k − . O ( ǫ min { ,λ v ,λ h ,λ ω } ) . (5.63)Thus, Theorem 5 .
10 is proved. 58o estimate N ǫ · ∂ ϕ ǫ z v ǫ − N · ∂ ϕz v , we use the equality N · ∂ ϕz v = − ( ∂ v + ∂ v ).To estimate N ǫ · ω ǫ − N · ω , we use the following equality: N · ω = − ∂ ϕ ( ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ z ϕ ∂ z v ) − ∂ ϕ ( − ∂ v + ∂ ϕ∂ z ϕ ∂ z v + ∂ z ϕ ∂ z v )+ ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ v + ∂ ϕ∂ z ϕ ∂ z v = − ∂ ϕ∂ v + ∂ ϕ∂ v + ∂ v − ∂ v . (5.64) σ > In this section, σ >
0, we prove Proposition 1 . ǫ in this section, which represents Navier-Stokes solutions.Since the estimates of normal derivatives are the same as the σ = 0 case, weonly focus on the estimates of the pressure gradient and tangential derivativeswhen σ > Lemma 6.1.
Assume the pressure q satisfies the elliptic equation with Neumannboundary condition (1 . , then q has the following gradient estimate: k∇ q k X m − . k ∂ mt v k X m + k v k X m − , + k ∂ z v k X m − + | h | X m, + ǫ k∇ y v k X m − , + ǫ k ∂ z v k X m − + ǫ | h | X m − , . (6.1) Proof.
The L estimate of the elliptic equation with its Neumann boundarycondition (1 .
44) is standard, that is k∇ q k L . k v · ∇ ϕ v k L + (cid:12)(cid:12) ∇ ϕ q · N | z =0 (cid:12)(cid:12) − . k v k X , + k ∂ z v k L + | h | X , + (cid:12)(cid:12) ∂ ϕt v · N | z =0 (cid:12)(cid:12) − + (cid:12)(cid:12) v · ∇ ϕ v · N | z =0 (cid:12)(cid:12) − + ǫ (cid:12)(cid:12) △ ϕ v · N | z =0 (cid:12)(cid:12) − . k v k X , + k ∂ z v k L + | h | X , + k ∂ ϕt v k L + k∇ ϕ · ∂ ϕt v k L + k v · ∇ ϕ v k L + k∇ · ( v · ∇ ϕ v ) k L + ǫ (cid:12)(cid:12) v | z =0 (cid:12)(cid:12) + ǫ | h | . (6.2)where we used the inequality | v · N | − . k v k + k∇ ϕ · v k (see [41]).Similar to [41, 32], we have higher order estimates for (1 . k∇ q k X m − . k v · ∇ ϕ v k X m − + (cid:12)(cid:12) ∇ ϕ q · N | z =0 (cid:12)(cid:12) X m − , − . k v k X m − , + k ∂ z v k X m − + | h | X m − , + (cid:12)(cid:12) ∇ ϕ q · N | z =0 (cid:12)(cid:12) X m − , − . (6.3)Next, we estimate (cid:12)(cid:12) ∇ ϕ q · N | z =0 (cid:12)(cid:12) X m − , − .59irstly, it is easy to estimate ǫ (cid:12)(cid:12) △ ϕ v · N | z =0 (cid:12)(cid:12) X m − , − . It follows from thedivergence free condition ∇ ϕ · v = 0 that ∂ z v · N = − ∂ z ϕ ∇ y · v y , z ≤ ,∂ zz v · N | z =0 = ∂ z ( ∂ z v · N ) − ∂ z v · ∂ z N = − ∂ z ( ∂ z ϕ ∇ y · v y ) + ∂ z v y · ∂ z ∂ y ϕ = − ∂ z ϕ ∇ y · v y − ∂ z ϕ ∇ y · ∂ z v y + ∂ z v y · ∂ z ∂ y ϕ = − ∂ z ϕ ∇ y · v y − ∂ z ϕ ∇ y · [ f , [ ∇ ϕ ]( ∂ j v i )] + [ f , [ ∇ ϕ ]( ∂ j v i )] · ∂ z ∂ y ϕ, (6.4)where ∂ z v h = f , [ ∇ ϕ ]( ∂ j v i ) is proved in (2 . ǫ (cid:12)(cid:12) △ ϕ v · N | z =0 (cid:12)(cid:12) X m − , − . ǫ (cid:12)(cid:12) ∇ y v | z =0 (cid:12)(cid:12) X m − , + ǫ | h | X m − , . ǫ k∇ y v k X m − , + ǫ k ∂ z v k X m − + ǫ | h | X m − , . (6.5)Secondly, we estimate (cid:12)(cid:12) ( ∂ ϕt v + v · ∇ ϕ v ) · N | z =0 (cid:12)(cid:12) X m − , − . (cid:12)(cid:12) ( ∂ ϕt v + v · ∇ ϕ v ) · N | z =0 (cid:12)(cid:12) X m − , − = (cid:12)(cid:12) ( ∂ t v + v y · ∇ y v ) · N | z =0 (cid:12)(cid:12) X m − , − . P ℓ + | α |≤ m − (cid:0)(cid:12)(cid:12) ∂ ℓt Z α ( ∂ t v + v y · ∇ y v ) · N | z =0 (cid:12)(cid:12) − + (cid:12)(cid:12) ∂ ℓt Z α N | z =0 (cid:12)(cid:12) − (cid:1) . P ℓ + | α |≤ m − (cid:0) k ∂ ℓt Z α ∂ t v k L + k∇ ϕ · ∂ ℓt Z α ∂ t v k L (cid:1) + P ℓ + | α |≤ m − (cid:0) k ∂ ℓt Z α ∇ y v k L + k∇ ϕ · ∂ ℓt Z α ∇ y v k L (cid:1) + | h | X m − , . k v k X m + k ∂ z v k X m − + | h | X m, . (6.6)Thus, Lemma 6 . v , we have the estimate of ∂ ℓt h byusing the kinetical boundary condition (1 . , which is the same with (1 . ,we give the following lemma without proof, which is the same with Lemma 2 . Lemma 6.2.
Assume ≤ ℓ ≤ m − , ∂ ℓt h have the estimates: R R | ∂ ℓt h | d y . | h | X m − + t R | h | X m − , + k v k X m − , d t + k ∂ z v k L ([0 ,T ] ,X m − ) . (6.7)Now, we develop a priori estimates for tangential derivatives including timederivatives. Our equations and variables are different from [41] which usedAlinhac’s good unknown. Lemma 6.3.
Assume the conditions are the same with those of Proposition . ,then v and h satisfy the a priori estimate: k v k X m − , + | h | X m − , + ǫ | h | X m − , + σ | h | X m − , + ǫ t R k∇ v k X m − , d t . k v k X m − , + | h | X m − , + ǫ | h | X m − , + σ | h | X m − , + k ∂ z v k L ([0 ,T ] ,X m − ) + k ∂ mt v k L ([0 ,T ] ,L ) + k ∂ mt h k L ([0 ,T ] ,X , ) . (6.8)60 roof. For the fixed σ >
0, we do not need Alinhac’s good unknown (1 . ∂ ℓt Z α to (1 . ∂ ℓt Z α v and ∂ ℓt Z α q satisfy the following equations: ∂ ϕt ∂ ℓt Z α v + v · ∇ ϕ ∂ ℓt Z α v + ∇ ϕ ∂ ℓt Z α q − ǫ ∇ ϕ · S ϕ ∂ ℓt Z α v = ∂ ℓ +1 t Z α η∂ ϕz v + ∂ ℓt Z α ∇ η · v∂ ϕz v + ∂ ℓt Z α ∇ η · ∂ ϕz q +2 ǫ ∇ ϕ · [ ∂ ℓt Z α , S ϕ ] v + 2 ǫ [ ∂ ℓt Z α , ∇ ϕ · ] S ϕ v + b.t., ∇ ϕ · ∂ ℓt Z α v = ∂ ℓt Z α ∇ η · ∂ ϕz v + b.t.,∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h = ∂ ℓt Z α v · N + [ ∂ ℓt Z α , v, N ] ,∂ ℓt Z α q N − ǫ S ϕ ∂ ℓt Z α v N = g∂ ℓt Z α h N − σ ∇ y · √ |∇ y h | (cid:0) ∇ y ∂ ℓt Z α h − ∇ y h ( ∇ y h ·∇ y ∂ ℓt Z α h )1+ |∇ y h | (cid:1) N +2 ǫ [ ∂ ℓt Z α , S ϕ ] v N + (2 ǫ S ϕ v − ( q − gh )) ∂ ℓt Z α N − [ ∂ ℓt Z α , q − gh, N ] + 2 ǫ [ ∂ ℓt Z α , S ϕ v, N ] − σ [ ∂ ℓt Z α , N ] H − σ ∇ y · [ ∂ ℓt Z α , ∇ y h, √ |∇ y h | ] N , ( ∂ ℓt Z α v, ∂ ℓt Z α h ) | t =0 = ( ∂ ℓt Z α v , ∂ ℓt Z α h ) . (6.9)When | α | ≥ , ≤ ℓ + | α | ≤ m, ≤ ℓ ≤ m −
1, we develop the L estimate ∂ ℓt Z α v and ∂ ℓt Z α h , we get
12 dd t R R − | ∂ ℓt Z α v | d V t − R R − ∂ ℓt Z α q ∇ ϕ · ∂ ℓt Z α v d V t + 2 ǫ R R − |S ϕ ∂ ℓt Z α v | d V t ≤ R { z =0 } (2 ǫ S ϕ ∂ ℓt Z α v N − ∂ ℓt Z α q N ) · ∂ ℓt Z α v d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t. ≤ − R { z =0 } (cid:2) g∂ ℓt Z α h − σ ∇ y · √ |∇ y h | (cid:0) ∇ y ∂ ℓt Z α h − ∇ y h ( ∇ y h ·∇ y ∂ ℓt Z α h )1+ |∇ y h | (cid:1)(cid:3) N · ∂ ℓt Z α v d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t. ≤ σ R { z =0 } ∇ y · √ |∇ y h | (cid:0) ∇ y ∂ ℓt Z α h − ∇ y h ( ∇ y h ·∇ y ∂ ℓt Z α h )1+ |∇ y h | (cid:1) · ( ∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h )d y − R { z =0 } g∂ ℓt Z α h · ( ∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h )d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t. ≤ − σ R { z =0 } √ |∇ y h | (cid:0) ∇ y ∂ ℓt Z α h − ∇ y h ( ∇ y h ·∇ y ∂ ℓt Z α h )1+ |∇ h | (cid:1) · ( ∂ t ∇ y ∂ ℓt Z α h + v y · ∇ y ∇ y ∂ ℓt Z α h )d y − g t R { z =0 } | ∂ ℓt Z α h | d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t. (6.10)61 − g t R { z =0 } | ∂ ℓt Z α h | d y − σ t R { z =0 } √ |∇ y h | (cid:0) |∇ y ∂ ℓt Z α h | − |∇ y h ·∇ y ∂ ℓt Z α h | |∇ y h | (cid:1) d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t.then
12 dd t R R − | ∂ ℓt Z α v | d V t + g t R { z =0 } | ∂ ℓt Z α h | d y + 2 ǫ R R − |S ϕ ∂ ℓt Z α v | d V t + σ t R { z =0 } √ |∇ y h | (cid:0) |∇ y ∂ ℓt Z α h | − |∇ y h ·∇ y ∂ ℓt Z α h | |∇ y h | (cid:1) d y ≤ k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | L + b.t. (6.11)Integrate (6 .
11) in time, apply the integral form of Gronwall’s inequality,we have k ∂ ℓt Z α v k + | ∂ ℓt Z α h | + ǫ | ∂ ℓt Z α h | + σ | ∂ ℓt Z α h | + ǫ t R k∇ ∂ ℓt Z α v k d t . k v k X m − , + | h | X m − , + ǫ | h | X m − , + σ | h | X m − , + T R k ∂ z v k X m − , + k∇ q k X m − , + | ∂ mt h | L d t. (6.12)Note that we use the following inequality to control the surface tension term: σ R { z =0 } √ |∇ y h | (cid:0) |∇ y ∂ ℓt Z α h | − |∇ y h ·∇ y ∂ ℓt Z α h | |∇ y h | (cid:1) d y ≥ σ R { z =0 } |∇ y h | ) |∇ y ∂ ℓt Z α h | d y ≥ σ R { z =0 } |∇ y ∂ ℓt Z α h | d y. (6.13)When | α | = 0 and 0 ≤ ℓ ≤ m −
1, we have no bounds of q and ∂ ℓt q , so we cannot apply the integration by parts to the pressure terms. The divergence freecondition and the dynamical boundary condition will not be used here. Then
12 dd t R R − | ∂ ℓt v | d V t + 2 ǫ R R − |S ϕ ∂ ℓt v | d V t ≤ − R R − ∂ ℓt ∇ ϕ q · ∂ ℓt v d V t + R { z =0 } ǫ S ϕ ∂ ℓt v N · ∂ ℓt v d y + k ∂ z v k X m − + k ∂ z q k X m − + m − P ℓ =0 | ∂ ℓ +1 t h | L + b.t. . k∇ q k X m − + k ∂ ℓt v k L + k ∂ z v k X m − + ǫ k∇ y ∂ ℓt v k X , + ǫ k ∂ z ∇ y ∂ ℓt v k L + m P ℓ =0 | ∂ ℓt h | L + b.t. (6.14)Combining (6 .
14) and (6 . k ∂ ℓt v k + k ∂ ℓt h k + ǫ t R k∇ ∂ ℓt v k d t . k v k X m − + k∇ q k L ([0 ,T ] ,X m − ) + k ∂ z v k L ([0 ,T ] ,X m − ) + | ∂ mt h | L ([0 ,T ] ,L ) + b.t. (6.15)62um ℓ and α . By (6 . , (6 .
15) and Lemma 6 .
2, we get the estimate (6 . . k ∂ mt v k L ([0 ,T ] ,L ) and k ∂ mt h k L ([0 ,T ] ,X , ) , which appear in Lemma 6 .
3. Thus,we estimate ∂ mt v and ∂ mt h . Lemma 6.4. ∂ mt v, ∂ mt h, ∂ m +1 t h satisfies the following estimate: k ∂ mt v k L ([0 ,T ] ,L ) + | ∂ mt h | L ([0 ,T ] ,X , ) + | ∂ m +1 t ∇ h | L ([0 ,T ] ,L ) . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ h | L + k ∂ z ∂ m − t v k L + k ∂ z v k L ([0 ,T ] ,X m − ) + b.t. (6.16) Proof.
In (6 . α = 0 and ℓ = m . Then multiply with ∂ mt v , integrate in R − ,then we get
12 dd t R R − | ∂ mt v | d V t − R R − ∂ mt q ∇ ϕ · ∂ mt v d V t + 2 ǫ R R − |S ϕ ∂ mt v | d V t ≤ R { z =0 } (2 ǫ S ϕ ∂ mt v N − ∂ mt q N ) · ∂ mt v d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L + b.t. ≤ − R { z =0 } (cid:2) g∂ mt h − σ ∇ y · √ |∇ y h | (cid:0) ∇ y ∂ mt h − ∇ y h ( ∇ y h ·∇ y ∂ mt h )1+ |∇ y h | (cid:1)(cid:3) N · ∂ mt v d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L + b.t. ≤ σ R { z =0 } ∇ y · √ |∇ y h | (cid:0) ∇ y ∂ mt h − ∇ y h ( ∇ y h ·∇ y ∂ mt h )1+ |∇ y h | (cid:1) · ( ∂ t ∂ mt h + v y · ∇ y ∂ mt h )d y − R { z =0 } g∂ mt h · ( ∂ t ∂ mt h + v y · ∇ y ∂ mt h )d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L + b.t. ≤ − σ R { z =0 } √ |∇ y h | (cid:0) ∇ y ∂ mt h − ∇ y h ( ∇ y h ·∇ y ∂ mt h )1+ |∇ y h | (cid:1) · ( ∂ t ∇ y ∂ mt h + v y · ∇ y ∇ y ∂ mt h )d y − g t R { z =0 } | ∂ mt h | d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L + b.t. ≤ − g t R { z =0 } | ∂ mt h | d y − σ t R { z =0 } √ |∇ y h | (cid:0) |∇ y ∂ mt h | − |∇ y h ·∇ y ∂ mt h | |∇ y h | (cid:1) d y + k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L + b.t.(6.17)The same as [41], we will integrate in time twice, we get the L ([0 , T ] , L )type estimate. After the first integration in time, we have k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ h | L + 4 ǫ R R − |S ϕ ∂ mt v | d V t . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ h | L + t R R R − ∂ mt q ∇ ϕ · ∂ mt v d V t d t (6.18)63 t R k ∂ z v k X m − + k∇ q k X m − + | h | X m − , + | ∂ mt h | X , + | ∂ m +1 t h | L d t + b.t. . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ h | L + t R R R − ∂ mt q ∇ ϕ · ∂ mt v d V t d t + k ∂ z v k L ([0 ,T ] ,X m − ) + | ∂ mt h | L ([0 ,T ] ,X , ) + | ∂ mt v | L ([0 ,T ] ,L ) + b.t. . Now we deal with the pressure term: t R R R − ∂ mt q ∇ ϕ · ∂ mt v d V t d t = − t R R R − ∂ mt q [ ∂ mt , ∇ ϕ · ] v d V t d t = P ℓ > t R R R − ∂ mt q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ t ∂ z v d x d t = P ℓ > t R t R R − ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ t ∂ z v d x d t − P ℓ > t R R R − ∂ m − t q ∂ t (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ t ∂ z v d x d t − P ℓ > t R R R − ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ +1 t ∂ z v d x d t = P ℓ > R { z =0 } ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ t v d y − P ℓ > R R − ∂ z (cid:2) ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1)(cid:3) · ∂ ℓ t v d x − P ℓ > R { z =0 } ∂ m − t q | t =0 (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) | t =0 (cid:1) · ∂ ℓ t v | t =0 d y + P ℓ > R R − ∂ z (cid:2) ∂ m − t q | t =0 (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) | t =0 (cid:3) · ∂ ℓ t v | t =0 d x − P ℓ > t R R { z =0 } ∂ m − t q ∂ t (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ t v d y d t + P ℓ > t R R R − ∂ z (cid:2) ∂ m − t q ∂ t (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1)(cid:3) · ∂ ℓ t v d x d t − P ℓ > t R R { z =0 } ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) · ∂ ℓ +1 t v d y d t + P ℓ > t R R R − ∂ z (cid:2) ∂ m − t q (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1)(cid:3) · ∂ ℓ +1 t v d x d t. (6.19)(6 .
19) contains ∂ z (cid:2) ∂ m − t qf (cid:3) , where f represents the terms (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) , (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) | t =0 , ∂ t (cid:0) ∂ z ϕ∂ ℓ t ( N ∂ z ϕ ) (cid:1) . By using Hardy’s inequality (1 . R R − ∂ z (cid:2) ∂ m − t qf (cid:3) · ∂ ℓ t v d x = R R − ( ∂ z ∂ m − t q ) f · ∂ ℓ t v d x + R R − − z ∂ m − t q [(1 − z ) ∂ z f ] · ∂ ℓ t v d x (6.20)64 k ∂ z ∂ m − t q k L + k − z ∂ m − t q k L + k (1 − z ) ∂ z f k L + k ∂ ℓ t v k L . k ∂ z ∂ m − t q k L + (cid:12)(cid:12) ∂ m − t q | z =0 (cid:12)(cid:12) L + | ∂ mt h | X , + k ∂ m − t v k L , where (1 − z ) ∂ z f ∼ (1 − z ) ∂ z ψ ∗ ∂ ℓt h and (1 − z ) ∂ z ψ ∈ L (d z ).Denote I := k ∂ z ∂ m − t q k L + (cid:12)(cid:12) ∂ m − t q | z =0 (cid:12)(cid:12) L + | ∂ mt h | X , + k ∂ m − t v k L + (cid:12)(cid:12) ∂ m − t v | z =0 (cid:12)(cid:12) L . k ∂ mt v k L + k ∂ z ∂ m − t v k L + | ∂ mt h | X , + b.t. (6.21)Plug (6 .
20) into (6 . t R R R − ∂ mt q ∇ ϕ · ∂ mt v d V t d t . I | t =0 + I + T R I d s . k ∂ mt v k L + k ∂ z ∂ m − t v k L + | ∂ mt h | X , + k ∂ mt v k L + k ∂ z ∂ m − t v k L + | ∂ mt h | X , + k ∂ mt v k L ([0 ,T ] ,L ) + k ∂ z ∂ m − t v k L ([0 ,T ] ,L ) + | ∂ mt h | L ([0 ,T ] ,L ) . (6.22)By (6 .
18) and (6 . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ y h | L . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ y h | L + k ∂ z ∂ m − t v k L + k ∂ mt v k L + k ∂ z ∂ m − t v k L + | ∂ mt h | X , + k ∂ mt v k L ([0 ,T ] ,L ) + k ∂ z v k L ([0 ,T ] ,X m − ) + | ∂ mt h | L ([0 ,T ] ,X , ) + b.t. (6.23)Square (6 .
23) and integrate in time again, apply the integral form of Gron-wall’s inequality, we have k ∂ mt v k L ([0 ,T ] ,L ) + g | ∂ mt h | L ([0 ,T ] ,L ) + σ | ∂ mt ∇ y h | L ([0 ,T ] ,L ) . k ∂ mt v k L + g | ∂ mt h | L + σ | ∂ mt ∇ y h | L + k ∂ z ∂ m − t v k L + k ∂ z v k L ([0 ,T ] ,X m − ) + b.t. (6.24)Thus, Lemma 6 . σ = 0 case, we have the estimates of normal derivatives: ∂ z v, ω ∈ L ([0 , T ] , X m − ) ∩ L ∞ ([0 , T ] , X m − ) . (6.25)Couple Lemmas 6 . , . , . . .
6. 65
Convergence Rates of Inviscid Limit for Fixed σ > In this section, we estimate convergence rates of the inviscid limit for the σ > v = v ǫ − v, ˆ q = q ǫ − q, ˆ h = h ǫ − h , we denote the i − th componentsof v ǫ and v by v ǫ,i and v i respectively. ˆ v, ˆ h, ˆ q satisfy the following equations ∂ ϕ ǫ t ˆ v + v ǫ · ∇ ϕ ǫ ˆ v + ∇ ϕ ǫ ˆ q − ǫ ∇ ϕ ǫ · S ϕ ǫ ˆ v = ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v − ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ˆ η + ǫ △ ϕ ǫ v, x ∈ R − , ∇ ϕ ǫ · ˆ v = ∂ ϕz v · ∇ ϕ ǫ ˆ η, x ∈ R − ,∂ t ˆ h + v y · ∇ ˆ h = ˆ v · N ǫ , { z = 0 } , ˆ q N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = g ˆ h N ǫ − σ ∇ y · (cid:0) H ∇ y ˆ h + H ∇ y ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) N ǫ + 2 ǫ S ϕ ǫ v N ǫ , { z = 0 } , (ˆ v, ˆ h ) | t =0 = ( v ǫ − v , h ǫ − h ) , (7.1)where the quantities H and H are defined as H = √ |∇ y h ǫ | + √ |∇ y h | , H = − √ |∇ y h ǫ | √ |∇ y h | ( √ |∇ y h ǫ | + √ |∇ y h | ) . (7.2)Since the estimates for normal derivatives are the same as the σ > ∇ ˆ q = ∇ q ǫ − ∇ q : Lemma 7.1.
Assume ≤ s ≤ k − , k ≤ m − , the difference of the pressure ˆ q has the following gradient estimate: k∇ ˆ q k X s . k ˆ v k X s, + k ∂ z ˆ v k X s + k ∂ s +1 t ˆ v k L + | ∂ st ˆ h k X , + | ˆ h | X s, + O ( ǫ ) . (7.3) Proof.
The Navier-Stokes pressure q ǫ satisfies the elliptic equations (1 . ( △ ϕ q = − ∂ ϕj v i ∂ ϕi v j , ∇ ϕ q · N | z =0 = − ∂ ϕt v · N − v · ∇ ϕ v · N , (7.4)Then the difference between boundary values is ∇ ϕ ǫ q ǫ · N ǫ | z =0 − ∇ ϕ q · N | z =0 = ǫ △ ϕ ǫ v ǫ · N ǫ − ( ∂ ϕ ǫ t v ǫ · N ǫ − ∂ ϕt v · N ) − ( v ǫ · ∇ ϕ ǫ v ǫ · N ǫ − v · ∇ ϕ v · N ) , (7.5)66 ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η ) · N ǫ | z =0 + ∇ ϕ q · ˆ N | z =0 = ǫ △ ϕ ǫ v ǫ · N ǫ − ( ∂ ϕ ǫ t ˆ v − ∂ ϕz v∂ ϕ ǫ t ˆ η ) · N ǫ − ∂ ϕt v · ˆ N − ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ˆ v · ∇ ϕ v ) · N ǫ − v · ∇ ϕ v · ˆ N , ∇ ϕ ǫ ˆ q · N ǫ | z =0 = ∂ ϕz q ∇ ϕ ǫ ˆ η · N ǫ | z =0 − ∇ ϕ q · ˆ N | z =0 + ǫ △ ϕ ǫ v ǫ · N ǫ − ( ∂ t ˆ v + v ǫy · ∇ y ˆ v ) · N ǫ − ( ∂ t v + v y · ∇ y v ) · ˆ N +[( ∂ t ˆ η + v ǫy · ∇ y ˆ η ) ∂ ϕz v − ˆ v · ∇ ϕ v ] · N ǫ := I . Similar to (5 . q satisfies the following elliptic equation: ∇ · ( E ǫ ∇ ˆ q ) = −∇ · (( E ǫ − E ) ∇ q ) − ∇ · [( P ǫ − P )( v · ∇ ϕ v )] −∇ · [ P ǫ ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ ϕ∂ ϕz v + ˆ v · ∇ ϕ v )] , ∇ ϕ ǫ ˆ q · N ǫ | z =0 = I . (7.6)The matrix E ǫ is definitely positive, then it is standard to prove that ˆ q satisfies the following gradient estimate: k∇ ˆ q k X s . k ( E ǫ − E ) ∇ q k X s + k ( P ǫ − P )( v · ∇ ϕ v ) k X s + k P ǫ ( v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ ϕ∂ ϕz v + ˆ v · ∇ ϕ v ) k X s + |I | X s, − . k E ǫ − E k X s + k P ǫ − P k X s + k ˆ v k X s + k∇ ˆ v k X s + k∇ ˆ ϕ k X s + |I | X s, − . k ˆ v k X s, + k ∂ z ˆ v k X s + | ˆ h | X s, + |I | X s, − . (7.7)Now we estimate the boundary terms. |I | X s, − . (cid:12)(cid:12) ∂ t ˆ v · N ǫ | z =0 (cid:12)(cid:12) X s, − + (cid:12)(cid:12) v ǫy · ∇ y ˆ v · N ǫ | z =0 (cid:12)(cid:12) X s, − + (cid:12)(cid:12) ˆ v · N ǫ | z =0 (cid:12)(cid:12) X s, − + (cid:12)(cid:12) ∂ t ˆ η · N ǫ | z =0 (cid:12)(cid:12) X s, − + (cid:12)(cid:12) ∇ y ˆ η · N ǫ | z =0 (cid:12)(cid:12) X s, − + | ˆ h | X s, + O ( ǫ ) . | ˆ h | X s, + k ∂ t ˆ v k X s + k ˆ v k X s, + k ∂ t ˆ η k X s + k ∂ t ˆ η k X s, + k∇ ˆ η k X s, + O ( ǫ ) . k ∂ s +1 t ˆ v k L + k ˆ v k X s, + | ∂ t ˆ h k X s, + | ˆ h | X s, + O ( ǫ ) , (7.8)refer to (6 .
4) for the estimate of ǫ △ ϕ ǫ · N ǫ .By (7 .
7) and (7 . . . v , we have the estimate of ∂ ℓt ˆ h by using the kinetical boundary condition (7 . , which is the same with (1 . ,we give the following lemma without proof, which is the same with Lemma 5 . Lemma 7.2.
Assume ≤ k ≤ m − , ≤ ℓ ≤ k − , ∂ ℓt ˆ h have the estimates: R R | ∂ ℓt ˆ h | d y . | ˆ h | X k − + t R | ˆ h | X k − , + k ˆ v k X k − , d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) . (7.9)67e develop the estimates for tangential derivatives. Lemma 7.3.
Assume ≤ k ≤ m − , ∂ ℓt Z α ˆ v and ∂ ℓt Z α ˆ h have the estimates: k ˆ v k X k − , + | ˆ h | X k − , . k ˆ v k X k − , + | ˆ h | X k − , + t R k ∂ z ˆ v k X k − + k ˆ v k X k − , + k ˆ h k X k − , d t + O ( ǫ ) . (7.10) Proof. ( ∂ ℓt Z α ˆ v, ∂ ℓt Z α ˆ h, ∂ ℓt Z α ˆ q ) satisfy the following equations: ∂ ϕ ǫ t ∂ ℓt Z α ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt Z α ˆ v = ǫ∂ ℓt Z α △ ϕ ǫ v + 2 ǫ [ ∂ ℓt Z α , ∇ ϕ ǫ · ] S ϕ ǫ ˆ v + 2 ǫ ∇ ϕ ǫ · [ ∂ ℓt Z α , S ϕ ǫ ]ˆ v + ∂ ϕz v∂ ϕ ǫ t ∂ ℓt Z α ˆ ϕ + ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − ∂ ℓt Z α ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − [ ∂ ℓt Z α , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt Z α , v ǫ · ∇ ϕ ǫ ]ˆ v − [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v +[ ∂ ℓt Z α , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ := I , ∇ ϕ ǫ · ∂ ℓt Z α ˆ v = ∂ ϕz v · ∇ ϕ ǫ ∂ ℓt Z α ˆ η − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η,∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h − N ǫ · ∂ ℓt Z α ˆ v = − ˆ v y · ∇ y ∂ ℓt Z α h − ∂ y ˆ h · ∂ ℓt Z α v y +[ ∂ ℓt Z α , ˆ v, N ǫ ] − [ ∂ ℓt Z α , v y , ∂ y ˆ h ] ,∂ ℓt Z α ˆ q N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ − g∂ ℓt Z α ˆ h N ǫ + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h (cid:1) N ǫ + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) N ǫ = I , + I , , ( ∂ ℓt Z α ˆ v, ∂ ℓt Z α ˆ h ) | t =0 = ( ∂ ℓt Z α v ǫ − ∂ ℓt Z α v , ∂ ℓt Z α h ǫ − ∂ ℓt Z α h ) , (7.11)where I , := 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ + 2 ǫ ( S ϕ ǫ v ǫ − S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ +2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ − S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ] − ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ , I , := − σ ∇ y · (cid:0) [ ∂ ℓt Z α , H ∇ y ]ˆ h (cid:1) N ǫ − σ ∇ y · (cid:0) [ ∂ ℓt Z α , H ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) · ∇ y ]ˆ h (cid:1) N ǫ . (7.12)When | α | ≥ ≤ ℓ + | α | ≤ k , we develop the L estimate of ∂ ℓt Z α ˆ v ,we have
12 dd t R R − | ∂ ℓt Z α ˆ v | d V ǫt − R R − ∂ ℓt Z α ˆ q ∇ ϕ ǫ · ∂ ℓt Z α ˆ v d V ǫt + 2 ǫ R R − |S ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt . − R { z =0 } (cid:0) ∂ ℓt Z α ˆ q N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ (cid:1) · ∂ ℓt Z α ˆ v d y + R R − I · ∂ ℓt Z α ˆ v d V ǫt . R { z =0 } (cid:2) − g∂ ℓt Z α ˆ h + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h (cid:1) + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1)(cid:3) N ǫ · ∂ ℓt Z α ˆ v d y − R { z =0 } ( I , + I , ) · ∂ ℓt Z α ˆ v d y + R R − I · ∂ ℓt Z α ˆ v d V ǫt + O ( ǫ ) (7.13)68 R { z =0 } (cid:2) − g∂ ℓt Z α ˆ h + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h (cid:1) + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1)(cid:3) · (cid:16) ∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h + ˆ v y · ∇ y ∂ ℓt Z α h + ∂ y ˆ h · ∂ ℓt Z α v y − [ ∂ ℓt Z α , ˆ v, N ǫ ] + [ ∂ ℓt Z α , v y , ∂ y ˆ h ] (cid:17) d y + R R − I · ∂ ℓt Z α ˆ v d V ǫt − R { z =0 } I , · ∂ ℓt Z α ˆ v d y − R { z =0 } I , · ∂ ℓt Z α ˆ v d y + O ( ǫ ) . We develop the following boundary estimates in (7 . R { z =0 } (cid:0) ∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h (cid:1)(cid:0) − g∂ ℓt Z α ˆ h + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h (cid:1) + σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1)(cid:1) d y = − g t R { z =0 } | ∂ ℓt Z α ˆ h | d y + g R { z =0 } | ∂ ℓt Z α ˆ h | ∇ y · v ǫy d y − σ R { z =0 } (cid:0) ∂ t ∇ y ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∇ y ∂ ℓt Z α ˆ h + ∇ y v ǫ,jy · ∂ j ∂ ℓt Z α ˆ h (cid:1) · (cid:2) H ∇ y ∂ ℓt Z α ˆ h + H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:3) d y . − g t R { z =0 } | ∂ ℓt Z α ˆ h | d y − σ t R { z =0 } H |∇ y ∂ ℓt Z α ˆ h | + H |∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) | d y + k∇ y ∂ ℓt Z α ˆ h k L + k ∂ ℓt Z α ˆ h k L . (7.14)It is easy to check that − R { z =0 } I , · ∂ ℓt Z α ˆ v d y = O ( ǫ ) . (7.15)Another boundary estimate is that − R { z =0 } I , · ∂ ℓt Z α ˆ v d y = σ R { z =0 } ∇ y · (cid:16) [ ∂ ℓt Z α , H ∇ y ]ˆ h +[ ∂ ℓt Z α , H ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) · ∇ y ]ˆ h (cid:17) N ǫ · ∂ ℓt Z α ˆ v d y = σ R { z =0 } ∇ y · (cid:16) [ ∂ ℓt Z α , H ∇ y ]ˆ h + [ ∂ ℓt Z α , H ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) · ∇ y ]ˆ h (cid:17) · (cid:16) ∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h + ˆ v y · ∇ y ∂ ℓt Z α h + ∂ y ˆ h · ∂ ℓt Z α v y − [ ∂ ℓt Z α , ˆ v, N ǫ ] + [ ∂ ℓt Z α , v y , ∂ y ˆ h ] (cid:17) d y . σ | ˆ h | X k − , + σ | ∂ kt ˆ h | X , + (cid:12)(cid:12) ˆ v | z =0 (cid:12)(cid:12) X k − . σ | ˆ h | X k − , + σ | ∂ kt ˆ h | X , + k ˆ v (cid:12)(cid:12) X k − , + k ∂ z ˆ v (cid:12)(cid:12) X k − . (7.16)69lug (7 . , (7 . , (7 .
16) into (7 .
12 dd t R R − | ∂ ℓt Z α ˆ v | d V ǫt + 2 ǫ R R − |S ϕ ǫ ∂ ℓt Z α ˆ v | d V ǫt + g t R { z =0 } | ∂ ℓt Z α ˆ h | d y + σ t R { z =0 } H |∇ y ∂ ℓt Z α ˆ h | + H |∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) | (cid:3) d y . k ˆ v k X k − , + k ∂ kt ˆ v k L + | ˆ h | X k − , + | ∂ kt ˆ h | X , + k ˆ q k X k − + O ( ǫ ) . (7.17)Since R { z =0 } H |∇ y ∂ ℓt Z α ˆ h | + H |∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) | (cid:3) d y ≥ R { z =0 } |∇ y ∂ ℓt Z α ˆ h | ( H − | H ||∇ y ( h ǫ + h ) | ) d y ≥ R { z =0 } | H ||∇ y ∂ ℓt Z α ˆ h | d y. (7.18)where 4 | H | ≥ δ σ > .
17) in time, apply the integral form of Gronwall’s inequality,note that (7 . R R − | ∂ ℓt Z α ˆ v | d V ǫt + g R { z =0 } | ∂ ℓt Z α ˆ h | d y + σ R { z =0 } |∇ y ∂ ℓt Z α ˆ h | d y . k ˆ v k X k − , + | ˆ h | X k − , + t R k ∂ kt ˆ v k L + | ∂ kt ˆ h | X , + k ˆ q k X k − d t + O ( ǫ ) . (7.19)When | α | = 0 , ≤ ℓ ≤ k −
1, we have no bounds of ˆ q and ∂ ℓt ˆ q , so we cannot apply the integration by parts to the pressure terms. Also, the dynamicalboundary condition will not be used. Since the main equation of ∂ ℓt ˆ v and itskinetical boundary condition satisfy ∂ ϕ ǫ t ∂ ℓt ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt ˆ v − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt ˆ v = −∇ ϕ ǫ ∂ ℓt ˆ q + 2 ǫ [ ∂ ℓt , ∇ ϕ ǫ · ] S ϕ ǫ ˆ v + 2 ǫ ∇ ϕ ǫ · [ ∂ ℓt , S ϕ ǫ ]ˆ v + ǫ∂ ℓt △ ϕ ǫ v + ∂ ϕz v∂ ϕ ǫ t ∂ ℓt ˆ ϕ + ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt ˆ ϕ − ∂ ℓt ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt ˆ ϕ − [ ∂ ℓt , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt , v ǫ · ∇ ϕ ǫ ]ˆ v − [ ∂ ℓt , ∇ ϕ v · ]ˆ v +[ ∂ ℓt , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ := I ,∂ t ∂ ℓt ˆ h + v ǫy · ∇ y ∂ ℓt ˆ h − N ǫ · ∂ ℓt ˆ v = − ˆ v y · ∇ y ∂ ℓt h − ∂ y ˆ h · ∂ ℓt v y +[ ∂ ℓt , ˆ v, N ǫ ] − [ ∂ ℓt , v y , ∂ y ˆ h ] , ( ∂ ℓt ˆ v, ∂ ℓt ˆ h ) | t =0 = ( ∂ ℓt v ǫ − ∂ ℓt v , ∂ ℓt h ǫ − ∂ ℓt h ) , (7.20)then we have L estimate of ∂ ℓt ˆ v :
12 dd t R R − | ∂ ℓt ˆ v | d V ǫt + 2 ǫ R R − |S ϕ ǫ ∂ ℓt ˆ v | d V ǫt . ǫ R { z =0 } S ϕ ǫ ∂ ℓt ˆ v N ǫ · ∂ ℓt ˆ v d y + R R − I · ∂ ℓt ˆ v d V ǫt . kI k L + k ∂ ℓt ˆ v k L + O ( ǫ ) . (7.21)70t is easy to check the last term of (7 .
21) satisfies kI k L . k ∂ z ˆ v k X k − + k ˆ v k X k − , + k ∂ kt ˆ v k L + | ∂ kt ˆ h | L + | ˆ h | X k − , + k∇ ˆ q k X k − + O ( ǫ ) . (7.22)Plug (7 .
22) into (7 . k ∂ ℓt ˆ v k L + ǫ t R k∇ ∂ ℓt ˆ v k L . k ∂ ℓt ˆ v k L + t R k ∂ z ˆ v k X k − + k ˆ v k X k − , + k ∂ kt ˆ v k L + | ∂ kt ˆ h | L + | ˆ h | X k − , + k∇ ˆ q k X k − d t + O ( ǫ ) . k ˆ v k X k − + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + | ∂ kt ˆ h | L ([0 ,T ] ,L ) + k∇ ˆ q k L ([0 ,T ] ,X k − ) + t R k ˆ v k X k − + | ˆ h | X k − , d t + O ( ǫ ) . (7.23)Sum ℓ and α . By (7 . .
23) and Lemma 7 .
2, we have (7 . . k ∂ kt ˆ v k L ([0 ,T ] ,L ) and k ∂ kt ˆ h k L ([0 ,T ] ,X , ) , which appear in Lemma 7 .
3. Thus, weestimate ∂ kt ˆ v and ∂ kt ˆ h . Lemma 7.4. ∂ kt ˆ v, ∂ kt ˆ h, ∂ k +1 t ˆ h satisfies the following estimate: k ∂ kt ˆ v k L ([0 ,T ] ,L ) + | ∂ kt ˆ h | L ([0 ,T ] ,X , ) + | ∂ k +1 t ∇ ˆ h | L ([0 ,T ] ,L ) . k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ ˆ h | L + k ∂ z ∂ k − t ˆ v k L + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (7.24) Proof. ( ∂ kt ˆ v, ∂ kt ˆ h, ∂ kt ˆ q ) satisfy the following equations: ∂ ϕ ǫ t ∂ kt ˆ v + v ǫ · ∇ ϕ ǫ ∂ kt ˆ v + ∇ ϕ ǫ ∂ kt ˆ q − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ kt ˆ v = I | ℓ = k, | α | =0 , ∇ ϕ ǫ · ∂ kt ˆ v = ∂ ϕz v · ∇ ϕ ǫ ∂ kt ˆ η − [ ∂ kt , ∇ ϕ ǫ · ]ˆ v + [ ∂ kt , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η,∂ t ∂ kt ˆ h + v ǫy · ∇ y ∂ kt ˆ h − N ǫ · ∂ kt ˆ v = − ˆ v y · ∇ y ∂ kt h − ∂ y ˆ h · ∂ kt v y +[ ∂ kt , ˆ v, N ǫ ] − [ ∂ kt , v y , ∂ y ˆ h ] ,∂ kt ˆ q N ǫ − ǫ S ϕ ǫ ∂ kt ˆ v N ǫ − g∂ kt ˆ h N ǫ + σ ∇ y · (cid:0) H ∇ y ∂ kt ˆ h (cid:1) N ǫ + σ ∇ y · (cid:0) H ∇ y ∂ kt ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) N ǫ = I , | ℓ = k, | α | =0 + I , | ℓ = k, | α | =0 , ( ∂ kt ˆ v, ∂ kt ˆ h ) | t =0 = ( ∂ kt v ǫ − ∂ kt v , ∂ kt h ǫ − ∂ kt h ) , (7.25)71hen multiply (7 .
25) with ∂ kt ˆ v , integrate in R − , then we get
12 dd t R R − | ∂ kt ˆ v | d V ǫt − R R − ∂ kt ˆ q ∇ ϕ ǫ · ∂ kt ˆ v d V ǫt + 2 ǫ R R − |S ϕ ǫ ∂ kt ˆ v | d V ǫt ≤ R { z =0 } (2 ǫ S ϕ ∂ kt ˆ v N ǫ − ∂ kt ˆ q N ǫ ) · ∂ kt ˆ v d y + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L + O ( ǫ ) ≤ − R { z =0 } h g∂ ℓt ˆ h − σ ∇ y · (cid:0) H ∇ y ∂ ℓt ˆ h (cid:1) − σ ∇ y · (cid:0) H ∇ y ∂ ℓt ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1)i N ǫ · ∂ kt ˆ v d y + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L + O ( ǫ ) ≤ σ R { z =0 } ∇ y · (cid:0) H ∇ y ∂ ℓt ˆ h + H ∇ y ∂ ℓt ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) · ( ∂ t ∂ kt ˆ h + v y · ∇ y ∂ kt ˆ h )d y − R { z =0 } g∂ kt ˆ h · ( ∂ t ∂ kt ˆ h + v y · ∇ y ∂ kt ˆ h )d y + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L + O ( ǫ ) ≤ − σ R { z =0 } (cid:0) H ∇ y ∂ ℓt ˆ h + H ∇ y ∂ ℓt ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) · ( ∂ t ∇ y ∂ kt ˆ h + v y · ∇ y ∇ y ∂ kt ˆ h )d y − g t R { z =0 } | ∂ kt ˆ h | d y + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L + O ( ǫ ) ≤ − g t R { z =0 } | ∂ kt ˆ h | d y − σ t R { z =0 } (cid:0) H |∇ y ∂ kt ˆ h | + H |∇ y ∂ ℓt ˆ h · ∇ y ( h ǫ + h ) | (cid:1) d y + k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L + O ( ǫ ) . (7.26)The same as (7 . R { z =0 } H |∇ y ∂ kt ˆ h | + H |∇ y ∂ kt ˆ h · ∇ y ( h ǫ + h ) | (cid:3) d y ≥ R { z =0 } |∇ y ∂ kt ˆ h | ( H − | H ||∇ y ( h ǫ + h ) | ) d y ≥ R { z =0 } | H ||∇ y ∂ kt ˆ h | d y. (7.27)where 4 | H | ≥ δ σ >
0, since |∇ y h ǫ | ∞ and |∇ y h | ∞ have their upper bounds.The same as [41], we will integrate in time twice, we get the L ([0 , T ] , L )type estimate. After the first integration in time, we have k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ y ˆ h | L + ǫ t R k∇ ∂ kt ˆ v k L d t . k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ y ˆ h | L + t R R R − ∂ kt ˆ q ∇ ϕ ǫ · ∂ kt ˆ v d V ǫt d t + t R k ∂ z ˆ v k X k − + k∇ ˆ q k X k − + | ˆ h | X k − , + | ∂ kt ˆ h | X , + | ∂ k +1 t ˆ h | L d t + O ( ǫ )(7.28)72 k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ ˆ h | L + t R R R − ∂ kt ˆ q ∇ ϕ ǫ · ∂ kt ˆ v d V ǫt d t + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + | ∂ kt ˆ h | L ([0 ,T ] ,X , ) + | ∂ kt ˆ v | L ([0 ,T ] ,L ) + O ( ǫ ) . Similar to the procedures (6 . . . . t R R R − ∂ kt ˆ q ∇ ϕ ǫ · ∂ kt ˆ v d V ǫt d t by using Hardy’s inequality. Denote I := k ∂ z ∂ k − t ˆ q k L + (cid:12)(cid:12) ∂ k − t ˆ q | z =0 (cid:12)(cid:12) L + | ∂ kt ˆ h | X , + k ∂ k − t ˆ v k L + (cid:12)(cid:12) ∂ k − t ˆ v | z =0 (cid:12)(cid:12) L . k ∂ kt ˆ v k L + k ∂ z ∂ k − t ˆ v k L + | ∂ kt ˆ h | X , + O ( ǫ ) , (7.29)then we have t R R R − ∂ kt ˆ q ∇ ϕ · ∂ kt ˆ v d V ǫt d t . I | t =0 + I + T R I d s . k ∂ kt ˆ v k L + k ∂ z ∂ k − t ˆ v k L + | ∂ kt ˆ h | X , + k ∂ kt ˆ v k L + k ∂ z ∂ k − t ˆ v k L + | ∂ kt ˆ h | X , + k ∂ kt ˆ v k L ([0 ,T ] ,L ) + k ∂ z ∂ k − t ˆ v k L ([0 ,T ] ,L ) + | ∂ kt ˆ h | L ([0 ,T ] ,L ) . (7.30)By (7 .
28) and (7 . k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ y ˆ h | L . k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ y ˆ h | L + k ∂ z ∂ k − t ˆ v k L + k ∂ kt ˆ v k L + k ∂ z ∂ k − t ˆ v k L + | ∂ kt ˆ h | X , + k ∂ kt ˆ v k L ([0 ,T ] ,L ) + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + | ∂ kt ˆ h | L ([0 ,T ] ,X , ) + O ( ǫ ) . (7.31)Square (7 .
31) and integrate in time again (see [41]), apply the integral formof Gronwall’s inequality, we have k ∂ kt ˆ v k L ([0 ,T ] ,L ) + g | ∂ kt ˆ h | L ([0 ,T ] ,L ) + σ | ∂ kt ∇ y ˆ h | L ([0 ,T ] ,L ) . k ∂ kt ˆ v k L + g | ∂ kt ˆ h | L + σ | ∂ kt ∇ y ˆ h | L + k ∂ z ∂ k − t ˆ v k L + k ∂ z ˆ v k L ([0 ,T ] ,X k − ) + O ( ǫ ) . (7.32)Thus, Lemma 7 . .
2, 7 . .
4, the estimates of tangential derivativescan be closed. The estimates of normal derivatives are the same as the σ = 0case. Finally, it is standard to estimates (1 .
46) and (1 .
47) in Theorem 1 . Derivation of the Equations and BoundaryConditions
In this appendix, we derive the equations and their boundary conditionsfor the σ = 0 case.Since ∂ ϕ ǫ i ϕ ǫ = ∂ i ϕ ǫ − ∂ i ϕ ǫ ∂ z ϕ ǫ ∂ z ϕ ǫ = 0 and ∂ ϕ ǫ z ϕ ǫ = ∂ z ϕ ǫ ∂ z ϕ ǫ = 1, ∂ ϕ ǫ i v ǫ − ∂ ϕi v = ∂ ϕ ǫ i ˆ v − ( ∂ i ϕ ǫ ∂ z ϕ ǫ − ∂ i ϕ∂ z ϕ ) ∂ z v = ∂ ϕ ǫ i ˆ v + ( ∂ i ϕ − ∂ i ϕ ǫ ∂ z ϕ ǫ ∂ z ϕ ) ∂ z ϕ ∂ z v = ∂ ϕ ǫ i ˆ v + ∂ ϕz v∂ ϕ ǫ i ϕ = ∂ ϕ ǫ i ˆ v + ∂ ϕz v∂ ϕ ǫ i ϕ − ∂ ϕz v∂ ϕ ǫ i ϕ ǫ = ∂ ϕ ǫ i ˆ v − ∂ ϕz v∂ ϕ ǫ i ˆ ϕ = ∂ ϕ ǫ i ˆ v − ∂ ϕ ǫ i ˆ η ∂ ϕz v, i = t, , ,∂ ϕ ǫ z v ǫ − ∂ ϕz v = ∂ ϕ ǫ z ˆ v + ( ∂ z ϕ ǫ − ∂ z ϕ ) ∂ z v = ∂ ϕ ǫ z ˆ v + ( ∂ z ϕ ǫ ∂ z ϕ − ∂ z ϕ ∂ z v = ∂ ϕ ǫ z ˆ v + ( ∂ ϕ ǫ z ϕ − ∂ z ϕ ∂ z v = ∂ ϕ ǫ z ˆ v + ( ∂ ϕ ǫ z ϕ − ∂ ϕ ǫ z ϕ ǫ ) ∂ ϕz v = ∂ ϕ ǫ z ˆ v − ∂ ϕz v∂ ϕ ǫ z ˆ ϕ = ∂ ϕ ǫ z ˆ v − ∂ ϕ ǫ z ˆ η ∂ ϕz v. (A.1)Similarly, we have ∂ ϕ ǫ i q ǫ − ∂ ϕi q = ∂ ϕ ǫ i ˆ q − ∂ ϕ ǫ i ˆ η ∂ ϕz q, i = t, , , v ǫ · ∇ ϕ ǫ v ǫ − v · ∇ ϕ v = v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ˆ v · ∇ ϕ v,ω ǫ − ω = ∇ ϕ ǫ × v ǫ − ∇ ϕ × v = ∇ ϕ ǫ × ˆ v − ∇ ϕ ǫ ˆ η × ∂ ϕz v. (A.2) Lemma A.1. (ˆ v = v ǫ − v, ˆ h = h ǫ − h, ˆ q = q ǫ − q ) satisfy the equations (1 . .Proof. Plug ( A. , ( A.
2) into ∂ ϕ ǫ t v ǫ − ∂ ϕt v + v ǫ · ∇ ϕ ǫ v ǫ − v · ∇ ϕ v + ∇ ϕ ǫ q ǫ − ∇ ϕ q = 2 ǫ ∇ ϕ ǫ · S ϕ ǫ ( v ǫ − v ) + ǫ △ ϕ ǫ v, (A.3)then we get the equation (1 . .It follows from the divergence free condition that0 = ∇ ϕ ǫ · v ǫ − ∇ ϕ · v = P i =1 ( ∂ ϕ ǫ i ˆ v i − ∂ ϕz v i ∂ ϕ ǫ i ˆ η ) = ∇ ϕ ǫ · ˆ v − ∂ ϕz v · ∇ ϕ ǫ ˆ η. (A.4)It follows from the kinetical boundary condition that ∂ t ˆ h = ∂ t h ǫ − ∂ t h = v ǫ ( t, y, · N ǫ − v ( t, y, · N ,v ǫ · N ǫ − v · N = ˆ v · N ǫ + v · ˆ N = v · ( −∇ y ˆ h,
0) + ˆ v · N ǫ ,∂ t ˆ h + v y · ∇ y ˆ h = ˆ v · N ǫ . (A.5)74he dynamical boundary condition for the Euler equation with σ = 0 is ascalar equation, that is q = gh . For any vector such as N ǫ , q N ǫ = gh N ǫ makessense. It follows from the dynamical boundary condition that q ǫ N ǫ − q N ǫ − ǫ S ϕ ǫ ( v ǫ − v ) N ǫ = gh ǫ N ǫ − gh N ǫ + 2 ǫ S ϕ ǫ v N ǫ , ˆ q N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = g ˆ h N ǫ + 2 ǫ S ϕ ǫ v N ǫ , (ˆ q − g ˆ h ) N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = 2 ǫ S ϕ ǫ v N ǫ . (A.6)Thus, Lemma A. Lemma A.2.
Assume ≤ ℓ + | α | ≤ k, ≤ ℓ ≤ k − , | α | ≥ , let ˆ V ℓ,α = ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ℓt Z α ˆ ϕ , ˆ Q ℓ,α = ∂ ℓt Z α ˆ q − ∂ ϕz q∂ ℓt Z α ˆ ϕ , then ˆ V ℓ,α , ˆ Q ℓ,α satisfy theequations (5 . .Proof. Apply ∂ ℓt Z α to the equations (1 . . . is as follows: ∂ ϕ ǫ t ∂ ℓt Z α ˆ v + [ ∂ ℓt Z α , ∂ ϕ ǫ t ]ˆ v − ∂ ϕz v∂ ϕ ǫ t ∂ ℓt Z α ˆ ϕ − [ ∂ ℓt Z α , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v + [ ∂ ℓt Z α , v ǫ · ∇ ϕ ǫ ]ˆ v − ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − [ ∂ ℓt Z α , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ + ∂ ℓt Z α ˆ v · ∇ ϕ v + [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q +[ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q − ∂ ϕz q ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ = ǫ∂ ℓt Z α △ ϕ ǫ ˆ v + ǫ∂ ℓt Z α △ ϕ ǫ v,∂ ϕ ǫ t ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ϕ ǫ t ∂ ℓt Z α ˆ ϕ + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v − ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ + ∂ ℓt Z α ˆ v · ∇ ϕ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q − ∂ ϕz q ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − ǫ∂ ℓt Z α ∇ ϕ ǫ · S ϕ ǫ ˆ v = ǫ∂ ℓt Z α △ ϕ ǫ v − [ ∂ ℓt Z α , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt Z α , v ǫ · ∇ ϕ ǫ ]ˆ v +[ ∂ ℓt Z α , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v − [ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ,∂ ϕ ǫ t ( ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ℓt Z α ˆ ϕ ) + v ǫ · ∇ ϕ ǫ ( ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ℓt Z α ˆ ϕ )+ ∇ ϕ ǫ ( ∂ ℓt Z α ˆ q − ∂ ϕz q∂ ℓt Z α ˆ ϕ ) − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt Z α ˆ v = I ,∂ ϕ ǫ t ˆ V ℓ,α + v ǫ · ∇ ϕ ǫ ˆ V ℓ,α + ∇ ϕ ǫ ˆ Q ℓ,α − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt Z α ˆ v = I . (A.7)The derivation of the divergence free condition (5 . is as follows: ∇ ϕ ǫ · ∂ ℓt Z α ˆ v + [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v − ∂ ϕz v · ∇ ϕ ǫ ∂ ℓt Z α ˆ η − [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η = 0 , ∇ ϕ ǫ · ∂ ℓt Z α ˆ v − ∂ ϕz v · ∇ ϕ ǫ ∂ ℓt Z α ˆ η = − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η, ∇ ϕ ǫ · ( ∂ ℓt Z α ˆ v − ∂ ϕz v∂ ℓt Z α ˆ η ) + ∂ ℓt Z α ˆ η ∇ ϕ ǫ · ∂ ϕz v = − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η, (A.8)75 ϕ ǫ · ˆ V ℓ,α = − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η − ∂ ℓt Z α ˆ η ∇ ϕ ǫ · ∂ ϕz v, Next, we derive the kinetical boundary condition (5 . . Apply ∂ ℓt Z α toNavier-Stokes and Euler kinetical boundary conditions, we get ∂ t ∂ ℓt Z α h ǫ + v ǫy · ∇ y ∂ ℓt Z α h ǫ = ∂ ℓt Z α v ǫ · N ǫ + [ ∂ ℓt Z α , v ǫ , N ǫ ] ,∂ t ∂ ℓt Z α h + v y · ∇ y ∂ ℓt Z α h = ∂ ℓt Z α v · N + [ ∂ ℓt Z α , v, N ] , (A.9)then the kinetical boundary condition (5 . is derived as follows: ∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h + ˆ v y · ∇ y ∂ ℓt Z α h = N ǫ · ∂ ℓt Z α ˆ v − ∂ y ˆ h · ∂ ℓt Z α v y +[ ∂ ℓt Z α , ˆ v, N ǫ ] − [ ∂ ℓt Z α , v y , ∂ y ˆ h ] ,∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h − N ǫ · ∂ ℓt Z α ˆ v = − ˆ v y · ∇ y ∂ ℓt Z α h − ∂ y ˆ h · ∂ ℓt Z α v y +[ ∂ ℓt Z α , ˆ v, N ǫ ] − [ ∂ ℓt Z α , v y , ∂ y ˆ h ] ,∂ t ∂ ℓt Z α ˆ h + v ǫy · ∇ y ∂ ℓt Z α ˆ h − N ǫ · ˆ V ℓ,α = N ǫ · ∂ ϕz v∂ ℓt Z α ˆ η − ˆ v y · ∇ y ∂ ℓt Z α h − ∂ y ˆ h · ∂ ℓt Z α v y + [ ∂ ℓt Z α , ˆ v, N ǫ ] − [ ∂ ℓt Z α , v y , ∂ y ˆ h ] , (A.10)Finally, we derive the dynamical boundary condition (5 . . Apply ∂ ℓt Z α to Navier-Stokes and Euler dynamical boundary conditions, we get( ∂ ℓt Z α q ǫ − g∂ ℓt Z α h ǫ ) N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α v ǫ N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ( q ǫ − gh ǫ )) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , q ǫ − gh ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] ,∂ ℓt Z α q = g∂ ℓt Z α h, (A.11)then the dynamical boundary condition (5 . is derived as follows:( ∂ ℓt Z α ˆ q − g∂ ℓt Z α ˆ h ) N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ǫ S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , ǫ S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] + 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ , ˆ Q ℓ,α N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ − ( g − ∂ ϕz q ) ∂ ℓt Z α ˆ h N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ǫ S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , ǫ S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] + 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ , (A.12)Thus, Lemma A. Lemma A.3.
Assume ≤ ℓ ≤ k − , | α | = 0 , let ˆ V ℓ, = ∂ ℓt ˆ v − ∂ ϕz v∂ ℓt ˆ ϕ , thenthe main equation of ˆ V ℓ, and its kinetical boundary condition satisfy (5 . . roof. The derivation of the main equation of ˆ V ℓ, is as follows: ∂ ϕ ǫ t ( ∂ ℓt ˆ v − ∂ ϕz v∂ ℓt ˆ ϕ ) + ∂ ℓt ˆ ϕ∂ ϕ ǫ t ∂ ϕz v + v ǫ · ∇ ϕ ǫ ( ∂ ℓt ˆ v − ∂ ϕz v∂ ℓt ˆ ϕ ) + ∂ ℓt ˆ v · ∇ ϕ v + ∂ ℓt ˆ ϕ v ǫ · ∇ ϕ ǫ ∂ ϕz v + ∂ ℓt ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ∂ ℓt ˆ ϕ − ǫ∂ ℓt ∇ ϕ ǫ · S ϕ ǫ ˆ v = ǫ∂ ℓt △ ϕ ǫ v ǫ − [ ∂ ℓt , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt , v ǫ · ∇ ϕ ǫ ]ˆ v +[ ∂ ℓt , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt , ∇ ϕ v · ]ˆ v + [ ∂ ℓt , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ,∂ ϕ ǫ t ˆ V ℓ, + v ǫ · ∇ ϕ ǫ ˆ V ℓ, − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt ˆ v = I . (A.13)The derivation of the kinetical boundary condition is the same as Lemma A.
2, but let α = 0 in (5 . . Thus, Lemma A. Lemma A.4. ˆ ω h = ω ǫh − ω h satisfies the equations (1 . .Proof. By Lemma 2 . . ω ǫh satisfies ∂ ϕ ǫ t ω ǫh + v ǫ · ∇ ϕ ǫ ω ǫh − ǫ △ ϕ ǫ ω ǫh = ~ F [ ∇ ϕ ǫ ]( ω ǫh , ∂ j v ǫ,i ) ,ω ǫ, | z =0 = F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) ,ω ǫ, | z =0 = F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) , (A.14)Similar to the arguments in (2 . , (2 . ω ǫh satisfies ∂ ϕt ω h + v · ∇ ϕ ω h = ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) ,ω | z =0 = ∂ ϕ v − ∂ ϕz v = ∂ v − ∂ ϕ∂ z ϕ ∂ z v − ∂ z ϕ ∂ z v := ω b, ,ω | z =0 = ∂ ϕz v − ∂ ϕ v = ∂ z ϕ ∂ z v − ∂ v + ∂ ϕ∂ z ϕ ∂ z v := ω b, , (A.15)By ( A. − ( A. ∂ ϕ ǫ t ˆ ω h − ∂ ϕz ω h ∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ ω h − ∂ ϕz ω h v ǫ · ∇ ϕ ǫ ˆ η + ˆ v · ∇ ϕ ω h − ǫ △ ϕ ǫ ˆ ω h = ~ F [ ∇ ϕ ǫ ]( ω ǫh , ∂ j v ǫ,i ) − ~ F [ ∇ ϕ ]( ω h , ∂ j v i ) + ǫ △ ϕ ǫ ω h , ˆ ω | z =0 = F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − ω b, , ˆ ω | z =0 = F [ ∇ ϕ ǫ ]( ∂ j v ǫ,i ) − ω b, . (A.16)Thus, Lemma A. B Derivation of the Equations for the SurfaceTension
In this appendix, we derive the equations and their boundary conditionsfor the σ > emma B.1. (ˆ v = v ǫ − v, ˆ h = h ǫ − h, ˆ q = q ǫ − q ) satisfy the equations (7 . .Proof. The surface tension term appear in the dynamical boundary condition,thus we only need to derive the difference equation of the dynamical boundarycondition, other equations and the kinetical boundary condition are the samewith the σ = 0 case, see (1 . σ > q = gh − σH . For any vector such as N ǫ , q N ǫ = gh N ǫ − σH N ǫ makes sense.Denote ˆ H = H ǫ − H , it follows from the dynamical boundary conditionthat q ǫ N ǫ − q N ǫ − ǫ S ϕ ǫ v ǫ N ǫ = gh ǫ N ǫ − gh N ǫ − σH ǫ N ǫ + σH N ǫ , ˆ q N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = ( g ˆ h − σ ˆ H ) N ǫ + 2 ǫ S ϕ ǫ v N ǫ . (B.1)ˆ v, ˆ h, ˆ q satisfy the following equations ∂ ϕ ǫ t ˆ v − ∂ ϕz v∂ ϕ ǫ t ˆ η + v ǫ · ∇ ϕ ǫ ˆ v − v ǫ · ∇ ϕ ǫ ˆ η ∂ ϕz v + ˆ v · ∇ ϕ v + ∇ ϕ ǫ ˆ q − ∂ ϕz q ∇ ϕ ǫ ˆ η = 2 ǫ ∇ ϕ ǫ · S ϕ ǫ ˆ v + ǫ △ ϕ ǫ v, x ∈ R − , ∇ ϕ ǫ · ˆ v − ∂ ϕz v · ∇ ϕ ǫ ˆ η = 0 , x ∈ R − ,∂ t ˆ h + v y · ∇ ˆ h = ˆ v · N ǫ , { z = 0 } , ˆ q N ǫ − ǫ S ϕ ǫ ˆ v N ǫ = ( g ˆ h − σ ˆ H ) N ǫ + 2 ǫ S ϕ ǫ v N ǫ , { z = 0 } , (ˆ v, ˆ h ) | t =0 = ( v ǫ − v , h ǫ − h ) , (B.2)whereˆ H = ∇ y · (cid:16) ∇ y h ǫ √ |∇ y h ǫ | − ∇ y h √ |∇ y h | (cid:17) = ∇ y · (cid:16) ∇ y ˆ h (cid:0) √ |∇ y h ǫ | + √ |∇ y h | (cid:1)(cid:17) + ∇ y · (cid:16) √ |∇ y h | − √ |∇ y h ǫ | √ |∇ y h ǫ | √ |∇ y h | · ∇ y h ǫ + ∇ y h (cid:17) = ∇ y · (cid:16) ∇ y ˆ h (cid:0) √ |∇ y h ǫ | + √ |∇ y h | (cid:1)(cid:17) −∇ y · (cid:16)(cid:0) ∇ y ˆ h ·∇ y ( h ǫ + h )2 √ |∇ y h ǫ | √ |∇ y h | ( √ |∇ y h ǫ | + √ |∇ y h | ) (cid:1) ∇ y ( h ǫ + h ) (cid:17) . (B.3)Plug ( B.
3) into ( B. .
1) and (7 . B. Lemma B.2.
Assume ≤ ℓ + | α | ≤ k, ≤ ℓ ≤ k − , | α | ≥ , then ( ∂ ℓt Z α ˆ v, ∂ ℓt Z α ˆ h, ∂ ℓt Z α ˆ q ) satisfy the equations (7 . . roof. Apply ∂ ℓt Z α to the equations (7 . . . follows from ( A. , ∂ ϕ ǫ t ∂ ℓt Z α ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q − ǫ∂ ℓt Z α ∇ ϕ ǫ · S ϕ ǫ ˆ v = ∂ ϕz v∂ ϕ ǫ t ∂ ℓt Z α ˆ ϕ + ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ − ∂ ℓt Z α ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt Z α ˆ ϕ + ǫ∂ ℓt Z α △ ϕ ǫ v − [ ∂ ℓt Z α , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt Z α , v ǫ · ∇ ϕ ǫ ]ˆ v +[ ∂ ℓt Z α , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt Z α , ∇ ϕ v · ]ˆ v − [ ∂ ℓt Z α , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt Z α , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ,∂ ϕ ǫ t ∂ ℓt Z α ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt Z α ˆ v + ∇ ϕ ǫ ∂ ℓt Z α ˆ q − ǫ ∇ ϕ ǫ · S ϕ ǫ ∂ ℓt Z α ˆ v = I . (B.4)The divergence free condition (7 . follows from ( A. ∇ ϕ ǫ · ∂ ℓt Z α ˆ v = ∂ ϕz v · ∇ ϕ ǫ ∂ ℓt Z α ˆ η − [ ∂ ℓt Z α , ∇ ϕ ǫ · ]ˆ v + [ ∂ ℓt Z α , ∂ ϕz v · ∇ ϕ ǫ ]ˆ η. (B.5)Next, the kinetical boundary condition (7 . is exactly ( A. .Finally, we derive the dynamical boundary condition (7 . . Apply ∂ ℓt Z α to Navier-Stokes and Euler dynamical boundary conditions, we get( ∂ ℓt Z α q ǫ − g∂ ℓt Z α h ǫ + σ∂ ℓt Z α H ǫ ) N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α v ǫ N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ( q ǫ − gh ǫ + σH ǫ )) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , q ǫ − gh ǫ + σH ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] , ( ∂ ℓt Z α q − g∂ ℓt Z α h + σ∂ ℓt Z α H ) N ǫ = 0 . (B.6)By ( B. − ( B. , we get( ∂ ℓt Z α ˆ q − g∂ ℓt Z α ˆ h + σ∂ ℓt Z α ˆ H ) N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ = 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ + (2 ǫ S ϕ ǫ v ǫ − ( q ǫ − gh ǫ + σH ǫ )) ∂ ℓt Z α N ǫ − [ ∂ ℓt Z α , q ǫ − gh ǫ + σH ǫ , N ǫ ] + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ , N ǫ ] + 2 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ , (B.7)and then we calculate ∂ ℓt Z α ˆ H , ∂ ℓt Z α ˆ H = ∂ ℓt Z α ∇ y · (cid:2) H ∇ y ˆ h + H ∇ y ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:3) = ∇ y · (cid:2) H ∇ y ∂ ℓt Z α ˆ h + H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:3) + ∇ y · (cid:2) [ ∂ ℓt Z α , H ∇ y ]ˆ h + [ ∂ ℓt Z α , H ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) · ∇ y ]ˆ h (cid:3) . (B.8)Plug ( B.
8) into ( B. ∂ ℓt Z α ˆ q N ǫ − ǫ S ϕ ǫ ∂ ℓt Z α ˆ v N ǫ = g∂ ℓt Z α ˆ h N ǫ − σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h (cid:1) N ǫ − σ ∇ y · (cid:0) H ∇ y ∂ ℓt Z α ˆ h · ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) (cid:1) N ǫ + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ ] v ǫ N ǫ +2 ǫ ( S ϕ ǫ v ǫ − S ϕ ǫ v ǫ n ǫ · n ǫ ) ∂ ℓt Z α N ǫ + 2 ǫ [ ∂ ℓt Z α , S ϕ ǫ v ǫ − S ϕ ǫ v ǫ n ǫ · n ǫ , N ǫ ](B.9)792 ǫ S ϕ ǫ ∂ ℓt Z α v N ǫ − σ ∇ y · (cid:2) [ ∂ ℓt Z α , H ∇ y ] (cid:3) ˆ h N ǫ − σ ∇ y · (cid:2) [ ∂ ℓt Z α , H ∇ y ( h ǫ + h ) ∇ y ( h ǫ + h ) · ∇ y ]ˆ h (cid:3) N ǫ . Thus, Lemma B. Lemma B.3.
Assume ≤ ℓ ≤ k − , | α | = 0 , then the main equation of ∂ ℓt ˆ v and its kinetical boundary condition satisfy the equations (7 . .Proof. The derivation of the main equation of ∂ ℓt ˆ v is as follows: ∂ ϕ ǫ t ∂ ℓt ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt ˆ v + ∇ ϕ ǫ ∂ ℓt ˆ q − ǫ∂ ℓt ∇ ϕ ǫ · S ϕ ǫ ˆ v = ∂ ϕz v∂ ϕ ǫ t ∂ ℓt ˆ ϕ + ∂ ϕz v v ǫ · ∇ ϕ ǫ ∂ ℓt ˆ ϕ − ∂ ℓt ˆ v · ∇ ϕ v + ∂ ϕz q ∇ ϕ ǫ ∂ ℓt ˆ ϕ + ǫ∂ ℓt △ ϕ ǫ v − [ ∂ ℓt , ∂ ϕ ǫ t ]ˆ v + [ ∂ ℓt , ∂ ϕz v∂ ϕ ǫ t ] ˆ ϕ − [ ∂ ℓt , v ǫ · ∇ ϕ ǫ ]ˆ v +[ ∂ ℓt , ∂ ϕz v v ǫ · ∇ ϕ ǫ ] ˆ ϕ − [ ∂ ℓt , ∇ ϕ v · ]ˆ v − [ ∂ ℓt , ∇ ϕ ǫ ]ˆ q + [ ∂ ℓt , ∂ ϕz q ∇ ϕ ǫ ] ˆ ϕ,∂ ϕ ǫ t ∂ ℓt ˆ v + v ǫ · ∇ ϕ ǫ ∂ ℓt ˆ v − ǫ∂ ℓt ∇ ϕ ǫ · S ϕ ǫ ˆ v = I . (B.10)The derivation of the kinetical boundary condition is the same as Lemma A.
2, but let α = 0 in (7 . . Thus, Lemma B. Acknowledgements
The author is extremely grateful to Prof. S. T. Yau for all his help. Thisresearch is conducted under the supervision of Prof. S. T. Yau and supportedby the Center of Mathematical Sciences and Applications, Harvard University.This research is also supported by the scholarship of Chinese Scholarship Council(No. 201500090074).
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