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1 COMMUN. MATH. SCI. Vol. 8, No. 4, pp c 00 International Press BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS WITH NAVIER BOUNDARY CONDITIONS FOR THE VANISHING VISCOSITY LIMIT XIAO-PING WANG, YA-GUANG WANG, AND ZHOUPING XIN Abstract. In this paper, we study the vanishing viscosity limit for the incompressible Navier- Stokes equations with the Navier friction boundary condition. To simplify the expansion of solutions in terms of the viscosity, we shall only consider the case that the slip length α in the Navier boundary condition is a power of the viscosity ǫ, α=ǫ γ. First, by multi-scale analysis we formally deduce that γ= is critical in determining the boundary layer behavior. When γ>, the boundary layer appears in the zero-th order terms of the expansion of solutions, and satisfies the same boundary value problem for the nonlinear Prandtl equations as in the non-slip case, when γ=, the boundary layer also appears in the zero-th order terms of solutions, and satisfies the nonlinear Prandtl equations but with a Robin boundary condition for the tangential velocity profile, and when γ<, the boundary layer appears in the order O(ǫ γ terms of solutions, and satisfies a boundary value problem for the linearized Prandtl equations. Secondly, we justify rigorously the asymptotic behavior of the vanishing viscosity limit for the incompressible Navier-Stokes equations with anisotropic viscosities by using the energy method, when the slip length is larger than the square root of the vertical viscosity. Even though the boundary layer appears in the lower order terms of solutions and satisfies a linear problem, the vorticity of flow is unbounded in the vanishing viscosity limit. Key words. Incompressible Navier-Stokes equations, Navier friction boundary condition, boundary layers, anisotropic viscosities. AMS subject classifications. 76D05, 76D0, 35K65.. Introduction In this paper, we consider the vanishing viscosity limit for the following incompressible Navier-Stokes equations with the Navier boundary condition in {t > 0,x } with being a domain of IR n (n= or 3: t u ǫ +(u ǫ uǫ + p ǫ =ǫ u ǫ, t>0,x u ǫ =0, t>0,x u ǫ n=0, (D(u ǫ n τ+ηu ǫ τ =0, on u ǫ t=0 =u ǫ 0(x,y, (. where ǫ is the viscosity and D(u ǫ = ( uǫ +( u ǫ T is the rate of the strain tensor, with n and τ being unit normal and tangent vectors on the boundary. The boundary condition given in (. is the so-called Navier friction condition, which was first proposed by Navier [0] and derived for gases by Maxwell [9]. It means that the rate of strain on the boundary is proportional to the tangential slip velocity. This friction boundary condition was also justified rigorously as an effective boundary condition for flows over rough boundaries; see [5]. Received: October 4, 009; accepted: February, 00. Communicated by Song Jiang. Hong Kong University of Science and Technology, Clear Water Bay, Kowloon, Hong Kong, China (mawang@ust.hk. Corresponding author. Department of Mathematics, Shanghai Jiao Tong University, Shanghai 0040, China (ygwang@sjtu.edu.cn. The Institute of Mathematical Sciences, The Chinese University of Hong Kong, Shatin, Hong Kong, China (zpxin@ims.cuhk.edu.hk. 965

2 966 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS The asymptotic behavior of solutions to the incompressible Navier-Stokes equations in the vanishing viscosity limit, in the case where there are physical boundaries, is a challenging problem due to the formation of boundary layers. The problem with the non-slip boundary condition was formally studied by Prandtl in [], in which it was derived that the boundary layer can be described by an initial-boundary problem for a nonlinear degenerate parabolic-elliptic coupled system, which is now called the Prandtl equations. Under the monotonic assumption on the velocity of the outflow, Oleinik and her collaborators established the local existence of smooth solutions for boundary value problems of the Prandtl equations in the 960 s, and their works were surveyed in the monograph []. The existence of global weak solutions to the Prandtl equations was obtained by Xin and Zhang in [8]. Recently, it was announced that such a solution is in fact unique and is classical by Xin, Zhang and Zhao in [9]. In [5], Sammartino and Caflisch obtained the local existence of analytic solutions to the Prandtl equations, and a rigorous theory on the stability of boundary layers in incompressible fluids with analytic data in the frame of the abstract Cauchy-Kowaleskaya theory. Rather recently, a rigorous theory was obtained in [7] for the behavior of boundary layers in a circularly symmetric flow with non-slip boundary conditions in two space variables. As in [], by a simple computation it is known that the Navier friction boundary condition given in (. can be rewritten as curl u ǫ =(κ ηu ǫ τ, on (. in two space variables, where curl u ǫ is the vorticity, and κ is the curvature of. The problem of the vanishing viscosity limit when the non-slip boundary condition is replaced by the Navier friction condition has been studied by many mathematicians since the 960 s. Yodovich [0] and Lions [6] studied the vanishing viscosity limit for the incompressible Navier-Stokes equations in two space variables with a free boundary condition, u ǫ n=0, and curl u ǫ =0 on. For the two-dimensional Navier-Stokes equations with the Navier friction condition, Clopeau, Mikelic, and Robert([], Lopes Filho, Nussenzveig Lopes, and Planas [8] obtained that the solution u ǫ to (. converges to the solution of the corresponding Euler equations in L ([0,T],L ( under certain boundedness assumptions on the initial vorticity when the slip length η is a constant. Recently, Xiao and Xin [7] studied the vanishing viscosity limit from the Navier-Stokes equations to the Euler equations in three space variables for the slip case, u ǫ n=0, and curl u ǫ τ =0 on. Almost all of these results do not have any detail description of the boundary layer behavior when the viscosity goes to zero. Certainly, this is a very interesting problem from the physical point of view. Only recently, Iftimie and Sueur [4] investigated the boundary layer behavior for the problem (. when the slip length η is independent of the viscosity ǫ. As mentioned in [4], many interesting physical phenomena show that the slip length should depend on viscosities in general. The main proposal of this work is to describe the asymptotic behavior of solutions to (. in the vanishing viscosity limit, especially the behavior of boundary layers when the slip length η depends on the viscosity ǫ. In this paper, we shall first study the asymptotic behavior of solutions to the problem (. when the viscosity ǫ goes to zero for different dependencies of η on the viscosity, and derive problems of boundary layer profiles. Then, we study rigorously the stability of boundary layers. For simplicity of presentation, we shall only consider the problem (. in the half plane ={x IR,y>0}. In the following sections, one will see that it is not difficult

3 X.P. WANG, Y.G. WANG AND Z.P. XIN 967 to generalize our discussion to multi-dimensional problems in an arbitrary bounded domain. In order to have a complete expansion of solutions in terms of the viscosity, we shall only consider the case that the slip length in the Navier boundary condition is a power of the viscosity. Let η= β α ǫ with β independent of ǫ and α ǫ =ǫ γ for an index γ IR. Then the Navier boundary condition given in (. can be simplified as u ǫ =0, βu ǫ α ǫ uǫ y =0, on y=0. (.3 Obviously, when α ǫ 0 the boundary conditions in (.3 formally tend to the nonslip case, u ǫ y=0 =0, while if α ǫ + the boundary conditions in (.3 tend to the complete slip case, u ǫ y=0 =0 and y u ǫ y=0 =0. From the above discussion, we already knew that the behavior of boundary layers has completely different phenomena for the non-slip and slip boundary condition cases. Therefore, the behavior of the vanishing viscosity limit for the problem (. with the boundary conditions (.3 should be clearly influenced by the amplitude of the slip length. Indeed, in the following sections, by multi-scale analysis we shall deduce that γ= is critical in determining the boundary layer behavior. When γ is super-critical, the leading boundary layer profile satisfies the same boundary problem for the nonlinear Prandtl equations as in the non-slip case, in the critical case γ=, the boundary layer profile also satisfies the nonlinear Prandtl equations but with a Robin boundary condition for the tangential velocity profile, and when γ is subcritical, the boundary layer appears in the order O(ǫ γ terms of solutions, and satisfies a boundary value problem for linearized Prandtl equations. The second goal of this paper is to study the stability of boundary layers rigorously. We shall justify the asymptotic behavior of the vanishing viscosity limit for the incompressible Navier-Stokes equations with anisotropic viscosities by using the energy method, when the slip length is larger than the square root of the vertical viscosity, in which even though the boundary layer appears in the lower order terms of solutions and obeys a linear law but it still produces an unbounded vorticity of flow in the vanishing viscosity limit. From the approach of this paper, one can easily deduce Iftimie and Sueur s results on the leading profile expansion of boundary layers hold not only in L (0,T,L ( as given in [4], but even in L ([0,T], moreover we have a complete expansion of u ǫ with respect to the viscosity ǫ. The remainder of this paper is arranged as follows: In section, we study the asymptotic behavior of solutions to the problem (. (.3 in the vanishing viscosity limit by multi-scale analysis, from which we observe that the power γ= of the slip length α ǫ =ǫ γ is critical for the behavior of boundary layers. In section 3 and section 4, we justify rigorously the asymptotic behavior of the vanishing viscosity limit for the anisotropic Navier-Stokes equations by using the energy method, and obtain that boundary layer is stable when the slip length is larger than the square root of the vertical viscosity. The preliminary version of the results given in this paper was announced in [6].

4 968 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS. Formal asymptotic analysis In this section, we study the vanishing viscosity limit for the following initial boundary value problem: t u ǫ +(u ǫ uǫ + p ǫ =ǫ u ǫ, t>0,(x,y IR + u ǫ =0, t>0,(x,y IR + u =0, βu ǫ α ǫ u ǫ y =0, on y=0 u ǫ t=0 =u 0 (x,y (. by multi-scale analysis for different dependencies of α ǫ on the viscosity. In order to simplify the presentation, we shall only consider the case where the slip length is a power of the viscosity, α ǫ =ǫ γ... The cases α ǫ =ǫ and ǫ. In these cases, we take the following ansatz: u ǫ (t,x,y= ǫ (u j I,j (t,x,y+u B,j (t,x, y ǫ p ǫ (t,x,y= (. ǫ (p j I,j (t,x,y+p B,j (t,x, y ǫ for the solutions of (., where u B,j (t,x,z and p B,j (t,x,z are rapidly decreasing when z= y ǫ +. Plugging (. into the divergence free condition given in (., it follows and which implies u I,j =0,, (.3 z u B,0 =0, x u B,j + z u B,j+ =0, (.4 u B,0 0, (.5 by noting that u B,0 (t,x,z is fast decay when z +. Plugging (. into the equations given in (., it follows ǫ j t (u I,j +u B,j + ǫ j +ǫ (u I,0 +u B,0 z u B,0 + + ǫ p j I,j +ǫ ( 0 z p B,0 j ((u I,k +u B,k u I,j k +(u I,k j+ ǫ j (u I,k +u B,k + ǫ j ( x p B,j z u B,j+ k z p B,j+ +u B,k x u B,j k = zu B,0 +ǫ zu B, + ǫ + ( u j I,j + xu B,j + zu B,j+. (.6 Letting z go to + in (.6, it gives t u I,0 +(u I,0 u I,0 + p I,0 =0, (.7

5 and X.P. WANG, Y.G. WANG AND Z.P. XIN 969 t u I,j +(u I,0 u I,j +(u I,j u I,0 + p I,j = u I,j k j I,j k (u I,k u (.8 for all j, where we denote by u I, =0. In the following discussion, we shall always denote the trace of a function u(t,x,y on {y=0} by u(t,x = u(t,x,0. The vanishing of the order O(ǫ terms in (.6 implies that ( (u I,0 +u B,0 z u B,0 0 + z p B,0 =0 (.9 which yields by using (.5. From the order O(ǫ 0 terms of (.6 we obtain p B,0 0 (.0 t (u I,0 +u B,0 +(u I,0 +u B,0 u I,0 +(u I,0 +u B,0 x u B,0 +z y u I,0 zu B,0 whose second component reads as implying +(u I, +u B, z u B,0 + p I,0 + by noting that y p I,0 =0 from (.7. For any, define with 0!=. y p I,0 + z p B, =0 u p,j (t,x,z=u B,j (t,x,z+ ( x p B,0 z p B, = zu B,0, (. p B, 0 (. j z k k! k yu I,j k (t,x (.3 From the first component of (. and (.4, (.5, we know that u p,0 =u I,0 (t,x,0+ub,0 (t,x,z u p, =u I, (t,x,0+ub, (t,x,z+z y u I,0 (t,x,0 satisfy the following Prandtl equations: t u p,0 +u p,0 xu p,0 +u p, zu p,0 + x p I,0 = zu p,0 x u p,0 + z u p, =0. (.4

6 970 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS The vanishing of the order O(ǫ terms in (.6 implies that ( t +(u I,0 +u B,0 x (u I, +u B, +z y u I,0 +(u I, +u B, +z y u I,0 x(u I,0 +u B,0 +(u I, +u B, +z y u I, + z yu I,0 zu B,0 +(u I, +u B, +z y u I,0 ( zu B, + y u I,0 ( + p I, +z y p I,0 x p + B, z p B, = zu B,. (.5 Obviously, the second component of (.5 can be written as t u p, +u p,0 xu p, +u p, zu p, + y p I, +z yp I,0 + z p B, = zu B,, (.6 which is equivalent to z p B, = zu B, ( t +u I,0 x+u p, z+ y u I,0 ub, u B,0 x u p,, (.7 by using the facts and ( t +u I,0 x+ y u I,0 yu I,0 + yp I,0 =0 ( t +u I,0 x+ y u I,0 ui, + y p I, =0 derived directly from (.7 and (.8. The unknown p B, (t,x,z rapidly decreasing in z + can be easily determined uniquely from (.7. From the first component of (.5 and (.4, (.5, (. we know that u p, =u I, (t,x,0+ub, (t,x,z+z y u I,0 (t,x,0 u p, =u I, (t,x,0+ub, (t,x,z+z y u I, (t,x,0+ z yu I,0 (t,x,0 satisfy the following linearized Prandtl equations: ( t +u p,0 x+u p, z+ x u p,0 up, +u p, zu p,0 + x (p I,0 +z y p I,0 = zu p, x u p, + z u p, =0. (.8 Similarly, for any j, from the O(ǫ order j terms of (.6 one can determine p B,j+ (t,x,z uniquely provided that {u p,k (t,x,z} k j and {u p,k (t,x,z} k j are known already. From the O(ǫ order j terms of (.6 and (.4, (.5 we deduce that (u p,j,up,j+ satisfy a linearized Prandtl system similar to (.8. To solve the boundary layer profiles from equation(.4 and(.8, the boundary conditions must be determined. First, from the first condition given in (. 3 we have for all j. u p,j z=0=0 (.9

7 X.P. WANG, Y.G. WANG AND Z.P. XIN 97 Substituting the ansatz (. into the Navier boundary condition given in (. 3, it follows that β ǫ j (u I,j +ub,j =α ǫ {ǫ z u B,0 + ǫ j ( y u I,j + zu B,j+ } (.0 on {y=z=0}. Now, we study (.0 for two cases. and Case : α ǫ =ǫ. In this case, from (.0 we immediately obtain u p,0 z=0=0 lim z + (u p,0 u I,0 =0 exponentially, (. u p,j = β zu p,j, on z=0 lim z + (u p,j j for all j. Therefore, one concludes z k k! k yu I,j k =0 exponentially (. Conclusion.. The solutions (u ǫ,p ǫ to the problem (. with α ǫ =ǫ formally have the following asymptotic expansions: u ǫ (t,x,y= ǫ(u j I,j (t,x,y+ub,j (t,x, y ǫ u ǫ (t,x,y=u I,0 (t,x,y+ ǫ(u j I,j (t,x,y+ub,j (t,x, y ǫ j p ǫ (t,x,y= ǫp j I,j (t,x,y+ ǫp j B,j (t,x, y ǫ j (.3 for rapidly decreasing (u B,j,p B,j (t,x,z in z +, where ( (u I,0,p I,0 are solutions to the following problem for the Euler equations: t u I,0 +(u I,0 ui,0 + p I,0 =0 u I,0 =0 (.4 u I,0 y=0=0 and for all j, (u I,j,p I,j are solutions to problems for the linearized Euler equations (.8 and (.3, ( the leading boundary layer profiles (u p,0,up, =(ub,0 +u I,0,uB, +u I, + z y u I,0 satisfy the following problem for the Prandtl equations: t u p,0 +u p,0 xu p,0 +u p, zu p,0 + x p I,0 = zu p,0 x u p,0 + z u p, =0 (.5 u p,0 z=0=u p, z=0=0 lim z + (u p,0 =0 exponentially, u I,0

8 97 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS p B, (t,x,z is uniquely determined by the equation (.7, for all j, (u p,j,up,j+ with u p,j =u B,j + j z k k! k yu I,j k satisfy problems for the linearized Prandtl equations similar to (.8 with boundary conditions given in (.9 and (., and p B,j+ (t,x,z are uniquely determined by equations similar to (.7. Case : α ǫ =ǫ. In this case, from (.0 we immediately obtain z u p,j βup,j =0, on z=0 lim z + (u p,j j z k k! k yu I,j k =0 exponentially (.6 for all. Therefore, we deduce Conclusion.. The solutions (u ǫ,p ǫ to the problem (. with α ǫ =ǫ formally have the same expansions as given in Conclusion. except that the leading boundary layer profiles (u p,0,up, satisfy the nonlinear Prandtl equations as given in (.5 but with the boundary conditions u p, =0, z u p,0 βu p,0 =0, on z=0 (.7 lim z + (u p,0 u I,0 =0 exponentially and the lower order boundary layer profiles (u p,j,up,j+ (j satisfy the linearized Prandtl equations as given in (.8 with the boundary conditions (.9 and (.6. Remark.3. From the above discussion, in general when the slip length α ǫ =ǫ γ for a fixed γ>, for the solution uǫ to the problem (. we can deduce that the boundary layer appears in the zero-th order terms of the expansion of u ǫ, and the leading boundary layer profiles satisfy the boundary value problem (.5, which is the same as in the non-slip case []. Complete expansions of solutions can be derived as well in a way similar to the one given in sections 4. and The case α ǫ =ǫ4. When α ǫ =ǫ4, we take the following ansatz: u ǫ (t,x,y= ǫ4(u j I,j (t,x,y+u B,j (t,x, y ǫ p ǫ (t,x,y= ǫ4(p j I,j (t,x,y+p B,j (t,x, y ǫ (.8 for solutions of (., where u B,j (t,x,z and p B,j (t,x,z are rapidly decreasing when z= y ǫ +. Plugging (.8 into (., we obtain and u I,j =0,, (.9 z u B,0 = z u B, =0, x u B,j + z u B,j+ =0,, (.30

9 X.P. WANG, Y.G. WANG AND Z.P. XIN 973 which implies u B,0 =u B, 0, (.3 yielding from the boundary condition u ǫ y=0 =0. Plugging (.8 into (., it follows that ǫ4 j t (u I,j +u B,j + + j=0 ǫ j 4 j (u I,k +u B,k + ǫ4 p j I,j +ǫ u I,0 =u I, =0, on y=0 (.3 ǫ j 4 ( 0 z p B,0 j ((u I,k +u B,k u I,j k +(u I,k z u B,j k + ( +ǫ 4 0 j+ ǫ j 4 (u I,k +u B,k z p B, + ǫ j 4 = 3 ǫ k 4 zu B,k + ǫ + 4( u j I,j + xu B,j + zu B,j+4. +u B,k z u B,j+ k ( x p B,j z p B,j+ Letting z + in (.33, this yields t u I,0 +(u I,0 ui,0 + p I,0 =0 u I,0 =0 u I,0 y=0=0 and { t u I,j +(u I,0 ui,j +(u I,j ui,0 + p I,j = u I,j 4 k j x u B,j k (.33 (.34 I,j k (ui,k u u I,j =0 (.35 for all j, where we set u I,k =0 when k. By using (.3, the vanishing of the order O(ǫ and O(ǫ 4 terms in (.33 implies that which yields From the order O(ǫ 0 terms of (.33, we obtain z p B,0 = z p B, =0, (.36 p B,0 =p B, 0. (.37 t (u I,0 +u B,0 +(u I,0 +u B,0 u I,0 +(u I,0 +u B,0 x u B,0 +z y u I,0 zu B,0 + (u I,k +u B,k ( z u B, k + p I,0 x p + B,0 z p B, = zu B,0, (.38

10 974 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS whose second component reads as implying by noting y p I,0 =0 from (.34. For any, define y p I,0 + z p B, =0 p B, 0 (.39 [ j ] u q,j (t,x,z=u B,j z k (t,x,z+ k! k yu I,j k (t,x (.40 with 0!=. From the first component of (.38 and (.9, (.30, we know that (u q,0,uq, satisfy the following Prandtl equations: t u q,0 +uq,0 xu q,0 +uq, zu q,0 + xp I,0 = zu q,0 (.4 x u q,0 + zu q, =0. The vanishing of the order O(ǫ 4 terms in (.33 implies that t (u I, +u B, + + z y u I,k zu B, k + 3 { (u I,k +u B,k u I, k +(u I,k +u B,k x u B, k } (u I,k +u B,k ( z u B,3 k + p I, x p + B, z p B,3 Obviously, the second component of (.4 can be written as which implies that = zu B,. (.4 y p I, + z p B,3 =0, (.43 p B,3 (t,x,z 0 (.44 by using the fact y p I, =0 from (.35 with j=. From the first component of (.4 and (.9, (.30 we know that (u q,,uq,3 satisfy the following linearized Prandtl equations: t u q, +uq,0 xu q, +uq, xu q,0 +uq, zu q, +uq,3 zu q,0 + xp I, = zu q, (.45 x u q, + zu q,3 =0, From the order O(ǫ terms of (.33, we deduce ( t +u q,0 xu q, +u q, xu q, +u q, xu B,0 +u q, u I,0 + 4 k= u q,k zu B,4 k ( + p I, +z y p I,0 x p + B, z p B,4 = zu B,. (.46

11 X.P. WANG, Y.G. WANG AND Z.P. XIN 975 By using (.34 and (.35, the second component of (.46 can be written as z p B,4 = zu B, ( t +u q,0 x+u p, z+ y u I,0 ub, u B,0 x (u I, +z y u I,0, (.47 whichdeterminesp B,4 (t,x,zuniquelyprovidedthatu B,0 andu B, are known already. To solve (u q,0,uq, and (uq,,uq,3 from equations (.4 and (.45 respectively, we need to study their boundary conditions. First, from (. 3 we immediately have for all j. u q,j z=0=0 (.48 Substituting the ansatz (.8 into the Navier boundary condition, it follows that β ǫ j 4 (u I,j +ub,j =ǫ 4 z u B,0 + z u B, + ǫ j+ 4 ( y u I,j + zu B,j+ (.49 on {y=z=0}, which implies that z u B,0 z=0 =0 z u B, z=0 =β(u I,0 +u B,0 y=z=0 z u B,j =β(u I,j +u B,j y u I,j on y=z=0, j. Therefore, from (.4 and (.50 we know that u q,0 (t,x,z=ui,0 +u B,0 u q, (t,x,z=ui, +u B, +z y u I,0 satisfy the following problem: t u q,0 +uq,0 xu q,0 +uq, zu q,0 + xp I,0 = zu q,0 x u q,0 + zu q, =0 z u q,0 z=0=u q, z=0=0 lim z + (u q,0 =0 exponentially. ui,0 (.50 (.5 By uniqueness of classical solutions to (.5, it follows that (t,x,z=ui,0 (t,x, i.e. ub,0 0, u B, 0. (.5 u q,0 Substituting (.5 into (.47 and (.45, it follows immediately that p B,4 (t,x,z 0 (.53 and u q, (t,x,z=ui, (t,x+ub, (t,x,z u q,3 (t,x,z=z yu I, (t,x+ui,3 (t,x+ub,3 (t,x,z (.54

12 976 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS satisfy the following problem for the linearized Prandtl equations: t u q, +ui,0 xu q, +z yu I,0 zu q, +uq, xu I,0 + x p I, = zu q, x u q, + zu q,3 =0 u q,3 z=0=0, z u q, z=0=βu I,0 (t,x lim z + u q, (t,x exponentially. =u I, (.55 FromtheorderO(ǫ 3 4termsof(.33wegetthefollowingequationfordetermining p B,5 (t,x,z: z p B,5 = zu B,3 ( t +u I,0 x+(u I, +z y u I,0 z+ y u I,0 ub,3. (.56 Similar to the above discussion, for any j, from the O(ǫ4 order j terms of (.33 and (.9, (.30 we deduce that (u q,j,uq,j+ satisfy a problem for the linearized Prandtl equations as given in (.55 with the boundary conditions u q,j+ =0, z u q,j =βuq,j on z=0 ( lim z + u q,j [j/] (.57 z k k! k yu I,j k = 0 exponentially. From the O(ǫ4 order j terms of (.33 one can determine p B,j+ (t,x,z uniquely provided that {u q,k (t,x,z} k j and {u q,k (t,x,z} k j are known already. Therefore, we conclude: Conclusion.4. The solutions (u ǫ,p ǫ to the problem (. with α ǫ =ǫ 4 formally have the following asymptotic expansions: u ǫ (t,x,y=u I,0 (t,x,y+ ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ j u ǫ (t,x,y= ǫ4u j I,0 (t,x,y+ ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ (.58 j=0 j 3 p ǫ (t,x,y= 4 ǫ4p j I,j + ǫ4(p j I,j (t,x,y+p B,j (t,x, y ǫ j=0 j 5 for rapidly decreasing (u B,j,p B,j (t,x,z in z +, where (u I,0,p I,0 are solutions to the problem (.34 for the Euler equations, for all j, (u I,j,p I,j are solutions to the linearized Euler equations (.35, for all j, u q,j (t,x,z=ub,j (t,x,z + [j/] u q,j+ (t,x,z=u B,j+ (t,x,z + z k k! k yu I,j k (t,x,0 [ j+ ] z k k! k yu I,j+ k (t,x,0 (.59 satisfy a boundary value problem for linearized Prandtl equations similar to (.55, and for all j 5, p B,j (t,x,z are given by equations similar to (.56 directly.

13 X.P. WANG, Y.G. WANG AND Z.P. XIN 977 Remark.5. As in the above discussion, in general when the slip length α ǫ =ǫ γ for a fixed 0<γ<, for the solution uǫ to the problem (. we can deduce that the boundary layer appears in the order O(ǫ γ terms of solutions, and satisfies a boundary value problem for linearized Prandtl equations, but it still yields the vorticity of flow being unbounded in the vanishing viscosity limit..3. The case α ǫ =. In the case α ǫ =, we take the same ansatz as in (.. From the boundary condition (.0 with α ǫ =, we obtain z u p,0 z=0=0 z u p,j βup,j =0 on z=0, j lim z + ( u p,j j z k k! k yu I,j k Thus, from (.4, (.9, and (.60 we know that =0 exponentially,. u p,0 (t,x,z=ub,0 (t,x,z+u I,0 (t,x u p, (t,x,z=ub, (t,x,z+u I, (t,x+z yu I,0 (t,x (.60 satisfy the following problem: t u p,0 +u p,0 xu p,0 +u p, zu p,0 + x p I,0 = zu p,0 x u p,0 + z u p, =0 z u p,0 z=0=u p, z=0=0 lim z + ( u p,0 u I,0 = 0 exponentially. (.6 On the other hand, from (.7 and u I,0 y=0=0, we have ( t +u I,0 xu I,0 + x p I,0 =0. (.6 So, by uniqueness of solutions to (.6, we deduce u B,0 (t,x,z=u B, (t,x,z 0. (.63 Substituting (.63 into (.7 and (.8 respectively, it follows that p B, (t,x,z 0 (.64 and u p, (t,x,z=ub, (t,x,z+u I, (t,x+z yu I,0 (t,x u p, (t,x,z=ub, (t,x,z+u I, (t,x+z yu I, z (t,x+ yu I,0 (t,x

14 978 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS satisfy the following problem for the linearized Prandtl equations: ( t u p, +u I,0 xu p, +z y u I,0 zu p, +u p, xu I,0 + x p I, +z y p I,0 = zu p, x u p, + z u p, =0 z u p, z=0=βu I,0 (t,x, up, ( lim z + u p, u I, z y u I,0 implies z=0=0 = 0 exponentially. (.65 The vanishing of the second component of the O(ǫ order terms in that (.6 ( ( t +u I,0 x+z y u I,0 z+ y u I,0 u B, +u I, +z y u I, + z yu I,0 ( +z xyu I,0 u I, +z y u I,0 + y p I, +z yp I, + z 3 yp I,0 + z p B,3 = yu I,0 + zu B, (.66 which gives rise to z p B,3 = ( z t u I,0 x z y u I,0 z y u I,0 u B, (.67 by using and ( yp 3 I,0 + ( yp I, + t +3 y u I,0 t + y u I,0 yu I,0 +u I,0 x yu I,0 + y u I,0 xyu I,0 =0, y u I, +u I,0 xyu I, +u I, xyu I,0 =0 ( y p I, + t +u I,0 x+ y u I,0 u I, = yu I,0 derived immediately from (.7 and (.8. Therefore, one concludes Conclusion.6. The solutions (u ǫ,p ǫ to the problem (. with α ǫ = formally have the following asymptotic expansions: u ǫ (t,x,y=u I,0 (t,x,y+ j u ǫ (t,x,y=(u I,0 +ǫ u I,0 (t,x,y+ ǫ j (u I,j (t,x,y+ub,j (t,x, y ǫ p ǫ (t,x,y= ǫp j I,j (t,x,y+ ǫ(p j I,j (t,x,y+p B,j (t,x, y ǫ j=0 j 3 j ǫ j (u I,j (t,x,y+ub,j (t,x, y ǫ (.68 for ( rapidly decreasing (u B,j,p B,j (t,x,z in z +, where (u p,,up, = u B, +u I, +z y u I,0,uB, + z k k! k yu I, k satisfy the problem (.65 for the

15 X.P. WANG, Y.G. WANG AND Z.P. XIN 979 linearized Prandtl equations, and p B,3 (t,x,z is uniquely determined by equation (.67. Remark.7. Similar to the above discussion, one can deduce that when γ 0, the amplitude of the boundary layer is at most of the order O(ǫ, which yields that theconvectionterm(u ǫ u ǫ isuniformlyboundedinǫ. Byusinganapproachsimilar to the one presented in section 3, one can justify rigorously the asymptotic expansions of solutions (u ǫ,p ǫ given in (.68 in the vanishing viscosity limit. Rather recently, in [4] Iftimie and Sueur studied this expansion up to the order o(ǫ in L (0,T,L (. 3. Stability of boundary layers with unbounded vorticity In this section, we study rigorously the asymptotic behavior of solutions to the initial-boundary value problem for anisotropic Navier-Stokes equations with the Navier friction boundary condition for the vanishing viscosity limit. Due to the degeneracy of the Prandtl equations, it is a challenging problem to rigorously justify the formal asymptotic expansions of solutions obtained in section for the vanishing viscosity limit in the Sobolev spaces, except that one can verify these expansions when the data are analytic in the frame of the abstract Cauchy- Kowaleskaya theory as done by Sammartino and Caflisch in [5] in the case where the velocity field satisfies the non-slip condition on the boundary. As we shall see, the crucial point in rigorously justifying the formal expansions of solutions obtained in section in the Sobolev norms is estimating the convection term u ǫ uǫ by the viscous term in the Navier-Stokes equations. To do so, in this section, we shall first study a problem similar to (. in the case where the slip length α ǫ =ǫ 4 with ǫ being the vertical viscosity, and the horizontal viscosity vanishes as well when ǫ goes to zero. As we have seen, from the formal analysis given in section, even though in this case the boundary layer profiles satisfy a linearized Prandtl system, but the vorticity of flow in the layer is not uniformly bounded in ǫ, and the convection term is unbounded as well. In order to control the convection term by the viscous term, instead of (. we study this problem for the anisotropic Navier-Stokes equations with the horizontal viscosity being ǫ. The vanishing viscosity limit problem for the Navier-Stokes equation with the Navier boundary condition for general anisotropic viscosities will be considered later. The anisotropic Navier-Stokes equations are widely used in geophysical fluid dynamics as a mathematical model for water flows in lakes and oceans, and also in the study of the Ekman boundary layers for rotating fluids; see [, 3]. Similar to that mentioned in section, for simplicity of presentation we shall mainly study the problem in the two-dimensional half space, though it is not difficult to generalize our discussion to the problem in a smooth bounded domain in IR n (n= or 3. Consider the following problem: t u ǫ +(u ǫ uǫ + p ǫ =ǫ xu ǫ +ǫ yu ǫ, t>0,(x,y IR + u ǫ =0, t>0,(x,y IR + u ǫ =0, βu ǫ ǫ 4 uǫ y =0, on y=0 u ǫ t=0 =u 0 (x,y, (3. where β is a positive constant. For the solutions (u ǫ,p ǫ of (3., by taking the same ansatz as (.8 one can

16 980 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS formally derive problems for all order profiles in a way similar to that given in section., and conclude that Conclusion 3.. The solutions (u ǫ,p ǫ to the problem (3. formally have the following asymptotic expansions: u ǫ (t,x,y=u I,0 4u B, (t,x,y+ǫ (t,x, y ǫ + ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ u ǫ (t,x,y=u I,0 (t,x,y+ǫ u I, (t,x,y+ p ǫ (t,x,y= 4 ǫ4p j I,j (t,x,y+ ǫ4(p j I,j (t,x,y+p B,j (t,x, y ǫ j=0 j 5 j 3 j ǫ j 4(u I,j (t,x,y+ub,j (t,x, y ǫ (3. for rapidly decreasing (u B,j,p B,j (t,x,z in z +, where (u I,0,p I,0 are solutions to the problem (.34 for the Euler equations with u I,0 t=0 =u 0 (x,y, p I, a constant, for all j, (u I,j,p I,j are solutions to the following problem for the linearized Euler equations: t u I,j +(u I,0 ui,j +(u I,j ui,0 + p I,j = xu I,j + yu I,j 4 I,j k (ui,k u k j u I,j =0 (3.3 u I,j y=0= + 0 x u B,j (t,x,ξdξ u I,j t=0 =0, and for all j, u B,j satisfies the boundary value problem for a linear degenerate parabolic equation: t u B,j +u I,0 xu B,j +z y u I,0 zu B,j + x u I,0 ub,j zu B,j =f j where z u B,j =β(u I,j +u B,j y u I,j, on{z=0} lim z + u B,j (t,x,z = 0 exponentially u B,j t=0 =0 f j = xu B,j j j 3 k= ( [ k ] n=0 ( z n n! x yu n I,k n u B,j k +u B,k x u B,j k x p B,j [ k ] z n n! n+ y u I,k n u B,j k j [ k ]+ n= k= n= u B,j+ is given explicitly by u B,j+ (t,x,z = and for all j 5, p B,j (t,x,z is given by p B,j = + z + z z n n! n yu I,k+ n z u B,j k. (3.4 x u B,j (t,x,ξdξ (3.5 f j (t,x,ξdξ, (3.6

17 where f j = zu B,j + xu B,j 4 t u B,j j 5 j 3 [ k ] k=n=0 j 3 k= ( [ k ] n=0 X.P. WANG, Y.G. WANG AND Z.P. XIN 98 z n n! x yu n I,k n u B,j k j 5 z n n! n yu I,k n +u B,k ( [ k ] n=0 z n n! n yu I,k n +u B,k [ k ] z n n! n+ y u I,k n u B,j k n=0 z u B,j k. Now, let us justify rigorously the above expansions (3.. First as in [4], for the problem (3. we have x u B,j k Proposition 3.. Assume that u 0 L ( with u 0 =0. There exists a global weak solution u ǫ Cw([0,+,L 0 ( L loc ([0,+,H ( to (3., moreover, we have the estimate u ǫ (t L +ǫ t 0 xu ǫ (τ L dτ+ǫ t 0 yu ǫ (τ L dτ (3.7 +ǫ4 3 t 0 β uǫ dxdτ u 0 L for all t 0, where L denotes the L norm on ={x IR,y>0} with the boundary ={y=0}. Remark 3.3. The estimate (3.7 can easily be obtained by multiplying equation (3. by u ǫ and integrating over. When β is non-negative, from (3.7 the a priori boundfollowsforu ǫ inthespacel ([0,+,L ( L loc ([0,+,H (. Asnoted in [4], even if β is not non-negative (3.7 also implies an a priori bound for u ǫ in the space L loc ([0,+,L ( L loc ([0,+,H (. Indeed, the boundary term in (3.7 can be estimated as ǫ 3 4 β u ǫ dx= ǫ 3 4 y (β u ǫ dxdy Cǫ u ǫ (t L ǫ yu ǫ (t L. Substituting this estimate into (3.7, it follows that t u ǫ (t L + e C ǫ(t τ (ǫ x u ǫ (τ L +ǫ yu ǫ (τ L dτ ec ǫt u 0 L (3.8 0 for all t 0. The existence of weak solutions to (3. can be obtained by using the argument of Theorem 3. in [4]. For the asymptotic behavior of the solution u ǫ when ǫ goes to zero, first, similar to [4], we have: Proposition 3.4. Assume that u 0 H s ( with u 0 =0 for a fixed s> and let u 0 C([0,T],H s ( C ([0,T],H s ( be a unique solution to the initial-boundary problem for the Euler equations (.34. Then, for the solutions u ǫ to (3., we have when ǫ goes to zero. u ǫ u 0 L (0,T;L (=O(ǫ 4 (3.9 Proof. The proof of this proposition is similar to the one given in [4]. For completeness, let us sketch the main ideas.

18 98 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS Set ũ ǫ =u ǫ u 0 and p ǫ =p ǫ p 0. From (3. and (.34 we know that (ũ ǫ, p ǫ satisfy the following problem: t ũ ǫ +(u ǫ ũǫ +(ũ ǫ u0 + p ǫ ǫ xu ǫ ǫ yu ǫ =0 ũ ǫ =0 ũ ǫ y=0 =0 ũ ǫ t=0 =0. (3.0 Multiplying the equations in (3.0 by ũ ǫ and integrating in space variables, it follows that d dt ũǫ (t L + ũ ǫ (ũ ǫ u 0 dxdy= (ǫ ũǫ xu ǫ +ǫũ ǫ yu ǫ dxdy. (3. and Obviously, we have ũ ǫ xu ǫ dxdy= x ũ ǫ x u ǫ dxdy ( xũ ǫ (t L xu 0 (t L ũǫ yu ǫ dxdy = yũ ǫ y u ǫ dxdy ǫ 4 βũǫ u ǫ dx ( yũ ǫ (t L y u 0 (t L ǫ 4 βũǫ u ǫ dx by using (βu ǫ ǫ 4 uǫ y y=0=0. On the other hand, we have ǫ 4 βũǫ u ǫ dx = ǫ 4 So, from (3. we deduce d dt ũǫ (t L +ǫ x ũ ǫ (t L +ǫ y ũ ǫ (t L y (βũǫ u ǫ dxdy C ǫ 4 ũ ǫ (t H ( ũ ǫ (t L + u 0 (t H. C ǫ 3 4 ũ ǫ (t H ( ũ ǫ (t L + u 0 (t H + u 0 (t L ũ ǫ (t L +ǫ u 0 (t H which implies ũ ǫ (t L C ǫ t 0 t u 0 (τ H dτexp(c 3t+C 3 u 0 (τ L dτ (3. for all 0 t T. From (3. we immediately conclude the estimate (3.9. Remark 3.5. If the slip length and horizontal viscosity in the problem (3. are generalized as ǫ γ and ǫ δ, with 0 γ< and δ>0, respectively, then by the same approach as above we can obtain u ǫ u 0 L (0,T;L (=O(ǫ min( γ,δ (3.3 0

19 X.P. WANG, Y.G. WANG AND Z.P. XIN 983 under the same assumption as in Proposition 3.3. In particular, the case γ=0, δ= is the one studied by Iftimie and Sueur in [4]. We are going to justify rigorously the asymptotic expansions (3.. For simplicity, let β be a positive constant in (3.. Suppose that for a fixed s>8, u 0 H s ( with u 0 =0 and u 0, =0 on {y=0}. Then, from the problems of profiles given in Conclusion 3., it is easy to have u I,k j=0 Cj ([0,T],H s k j (, 0 k 0 u B,k j=0 Cj ([0,T],H s k j (, k 0 u B,k j=0 Cj ([0,T],H s k j (, 3 k. Denote (u ǫ,a,p ǫ,a to be u ǫ,a 0 (t,x,y= ǫ k 4u I,k 0 (t,x,y+ ǫ k 4u B,k (t,x, y ǫ k= u ǫ,a 0 (t,x,y= ǫ k 4u I,k (t,x,y+ ǫ k 4u B,k (t,x, y ǫ k=3 p ǫ,a (t,x,y= 0 ǫ k 4p I,k (t,x,y+ 0 ǫ k 4p B,k (t,x, y ǫ k=5 (3.4 (3.5 the approximate solutions to (3., where all profiles are given in Conclusion 3., and let the solutions of the problem (3. have the expansions: u ǫ (t,x,y=u ǫ,a (t,x,y+ǫ 4 R ǫ (t,x,y (3.6 p ǫ (t,x,y=p ǫ,a (t,x,y+ǫ 4 π ǫ (t,x,y. Then, from Conclusion 3., we know that (R ǫ,π ǫ satisfy the following problem: t R ǫ +(u ǫ Rǫ + π ǫ (ǫ x+ǫ yr ǫ +(R ǫ uǫ,a =F ǫ R ǫ =0 R ǫ y=0 = (u B, +ǫ 4u B, (t,x,0 βr ǫ ǫ 4 Rǫ y = y(u I,0 +ǫ 4u I,9 4(u I,0 βǫ +u B,0, on {y=0} R ǫ t=0 =0, (3.7 where F ǫ is bounded in the space L (0,T;Hǫ s 4 (IR + W, (0,T;Hǫ s 6 (IR + with the norm F ǫ L (0,T;Hǫ s(ir + = max x α ( ǫ 0 t T y α F ǫ (t L (IR +. α s By constructing R ǫ j=0 Cj ([0,T],H s j (IR + satisfying R ǫ =0 R ǫ y=0 = (u B, +ǫ 4u B, (t,x,0,

20 984 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS we know that R ǫ =R ǫ R ǫ satisfies the following problem: t R ǫ +(u ǫ Rǫ + π ǫ (ǫ x+ǫ yr ǫ +(R ǫ uǫ,a =F ǫ R ǫ =0 R=0, ǫ βr ǫ ǫ 4 Rǫ y =rǫ (t,x, on {y=0} R ǫ t 0 =0, (3.8 where we have dropped tilde notation for simplicity, r ǫ =O(ǫ 4 in L (0,T;H s 7 ǫ (IR W, (0,T;H s 3 ǫ L (0,T;H s 4 ǫ ( W, (0,T;H s 6 ǫ (IR, and F ǫ =O(ǫ 4 in the space (. Proposition 3.6. For the solution R ǫ to the problem (3.8, we have the following estimate: sup 0 t T T R ǫ (t L + ( ǫ x R ǫ (t L +ǫ yr ǫ (t L dt Cǫ. (3.9 0 Proof. Multiplying the equations in (3.8 by R ǫ and integrating on, it follows that d dt Rǫ (t L + R ǫ (R ǫ u ǫ,a dxdy R ǫ (ǫ x +ǫ yr ǫ dxdy= R ǫ F ǫ dxdy. (3.0 Obviously, by using the boundary conditions given in (3.8, we have R ǫ (ǫ x +ǫ yr ǫ dxdy= ǫ x R ǫ (t L ǫ yr ǫ (t L ǫ 3 4 β R(t ǫ dx+ǫ 3 4 Rr ǫ ǫ dx. (3. It is easy to show that ǫ 3 4 β R(t ǫ dx+ǫ 3 4 R ǫ r ǫ dx ǫ yr ǫ (t L +C Rǫ (t L +Cǫ r ǫ (t L (. (3. On the other hand, by using (3.4, the divergence free part of R ǫ, and R ǫ y=0 =0, we obtain R ǫ (R ǫ u ǫ,a dxdy C R ǫ (t L + ǫ 4 RR ǫ ǫ z u B, dxdy C R ǫ (t L +C ǫ 4 R ǫ (t L x R ǫ L. (3.3 Plugging (3., (3., and (3.3 into (3.0, it follows that d dt Rǫ (t L +ǫ x R ǫ (t L +ǫ yr ǫ (t L C( R ǫ (t L + Fǫ (t L +ǫ r ǫ (t L (, (3.4

21 X.P. WANG, Y.G. WANG AND Z.P. XIN 985 which implies the estimate (3.9 by using the Gronwall inequality. Proposition 3.7. For the problem (3.8, we have the following estimate for t R ǫ : sup 0 t T T t R ǫ (t L + ( ǫ x t R ǫ (t L +ǫ y t R ǫ (t L dt Cǫ 3. (3.5 0 Proof. From (3.8, we know that t R ǫ satisfies the following problem t ( t R ǫ +(u ǫ ( t R ǫ + ( t π ǫ (ǫ x+ǫ y t R ǫ +(( t R ǫ u ǫ,a ( t R ǫ =0 +( t u ǫ R ǫ +(R ǫ t u ǫ,a = t F ǫ (3.6 t R=0, ǫ β t R ǫ ǫ 4 ( trǫ y = t r ǫ (t,x, on {y=0} t R ǫ t=0 =0. Multiplying the equations in (3.6 by t R ǫ and integrating on, it follows that d dt tr ǫ (t L + t R ǫ (( t R ǫ u ǫ,a +( t u ǫ R ǫ +(R ǫ t u ǫ,a dxdy t R ǫ (ǫ x +ǫ y t R ǫ dxdy= t R ǫ t F ǫ dxdy. (3.7 As in (3., by using the boundary conditions given in (3.6, we have and t R ǫ (ǫ x +ǫ y t R ǫ dxdy= ǫ x t R ǫ (t L ǫ y t R ǫ (t L β t R(t ǫ dx+ǫ 3 4 t R ǫ t r ǫ dx. ǫ 3 4 ǫ 3 4 β t R(t ǫ dx+ǫ 3 4 t R ǫ t r ǫ dx (3.8 ǫ y t R ǫ (t L +C tr ǫ (t L +Cǫ t r ǫ (t L (. (3.9 By using (3.4, the divergence free of R ǫ and R ǫ y=0 = t R ǫ y=0 =0, we obtain tr ǫ ( t R ǫ u ǫ,a dxdy C t R ǫ (t L + ǫ 4 tr ǫ t R ǫ z u B, dxdy and t R ǫ (R ǫ t u ǫ,a dxdy C 3 t R ǫ (t L R ǫ (t L + ǫ 4 C t R ǫ (t L +C ǫ 4 t R ǫ (t L x t R ǫ L (3.30 t R ǫ R ǫ z t u B, dxdy C 3 t R ǫ (t L R ǫ (t L +C 4 ǫ 4 t R ǫ (t L x R ǫ L. (3.3

22 986 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS On the other hand, we have = t R ǫ ( t u ǫ R ǫ dxdy t R ǫ ( t (u ǫ,a +ǫ 4 R ǫ R ǫ dxdy C 5 R ǫ (t L t R ǫ (t L +ǫ 4 R ǫ L t R ǫ L 4 C 5 R ǫ (t L t R ǫ (t L +ǫ 4 R ǫ L t R ǫ L t R ǫ H ǫ 4 tr ǫ L +C 6(ǫ 3 R ǫ L + tr ǫ L + Rǫ L. (3.3 Plugging (3.8 (3.3 into (3.7, it follows that which implies that sup 0 t T T 0 d dt tr ǫ (t L +ǫ x t R ǫ (t L +ǫ y t R ǫ (t L C 7 (ǫ 3 R ǫ L + tr ǫ L + Rǫ L + Rǫ (t L + t F ǫ (t L +ǫ t r ǫ (t L (, (3.33 T t R ǫ (t L + ( ǫ x t R ǫ (t L +ǫ y t R ǫ (t L dt 0 e T t C7(ǫ3 R ǫ (s L +ds ( R ǫ (t L + Rǫ (t L + t F ǫ (t L +ǫ t r ǫ (t L ( dt. (3.34 By using (3.9 in (3.34, the estimate (3.5 follows immediately. Similarly, by differentiating the problem (3.8 with respect to the x-variable, and using the same argument as above, we can conclude Proposition 3.8. For the problem (3.8, the following estimate holds for x R ǫ : sup 0 t T T x R ǫ (t L + ( ǫ xr ǫ (t L +ǫ y x R ǫ (t L dt Cǫ 3. ( Finally, let us study the estimate for txr ǫ. For this, we obtain Proposition 3.9. For the problem (3.8, we have the following estimate: sup 0 t T T txr ǫ (t L + ( ǫ x txr ǫ (t L +ǫ y txr ǫ (t L dt Cǫ 5. (3.36 0

23 X.P. WANG, Y.G. WANG AND Z.P. XIN 987 Proof. From (3.8, we know that txr ǫ satisfies the following problem: t ( txr ǫ +(u ǫ ( txr ǫ + ( txπ ǫ (ǫ x+ǫ y txr ǫ +(( txr ǫ u ǫ,a +( txu ǫ Rǫ +( t u ǫ x R ǫ +( x u ǫ t R ǫ +(R ǫ txu ǫ,a +( x R ǫ t u ǫ,a +( t R ǫ x u ǫ,a = txf ǫ ( txr ǫ =0 txr =0, ǫ β txr ǫ ǫ 4 ( tx Rǫ y = txr ǫ (t,x, on {y=0} txr ǫ t=0 =0. (3.37 Multiplying the equations in (3.37 by txr ǫ and integrating on, it follows that d dt txr ǫ (t L txr ǫ (ǫ x +ǫ y txr ǫ dxdy = txr ǫ txf ǫ dxdy txr ǫ {( txr ǫ u ǫ,a +( t R ǫ x u ǫ,a +( x R ǫ t u ǫ,a +(R ǫ txu ǫ,a +( txu ǫ R ǫ +( t u ǫ x R ǫ +( x u ǫ t R ǫ} dxdy. (3.38 As in (3., by using the boundary conditions given in (3.37, we have txr ǫ (ǫ x+ǫ y txr ǫ dxdy= ǫ x txr ǫ (t L ǫ y txr ǫ (t L ǫ 3 4 β txr(t ǫ dx+ǫ 3 4 txr ǫ txr ǫ dx (3.39 and ǫ 3 4 β txr (t ǫ dx+ǫ 3 4 txr ǫ txr ǫ dx ǫ y txr ǫ (t L +C txr ǫ (t L +C ǫ tx r ǫ (t L (. (3.40 Now, let us estimate each term in the last integral of (3.38. By using (3.4, the divergence free part of R ǫ, and R ǫ y=0 =0, we obtain txr ǫ ( txr ǫ u ǫ,a dxdy C txr ǫ (t L + ǫ 4 and txr ǫ txr ǫ z u B, dxdy C txr ǫ (t L +C 3 ǫ 4 txr ǫ (t L x txr ǫ L, (3.4 txr ǫ (( t R ǫ x u ǫ,a +( x R ǫ t u ǫ,a +(R ǫ txu ǫ,a dxdy C 4 txr ǫ (t L ( t R ǫ (t L + x R ǫ (t L + R ǫ (t L +ǫ 4 txr ( ǫ t R ǫ zxu B, + x R ǫ ztu B, +R ǫ z txu B, dxdy C 4 txr ǫ (t L ( t R ǫ (t L + x R ǫ (t L + R ǫ (t L +C 5 ǫ 4 tx R ǫ (t L ( txr ǫ (t L + xr ǫ (t L + x R ǫ (t L. (3.4

24 988 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS On the other hand, we have = txr ǫ ( txu ǫ R ǫ dxdy txr ǫ ( tx(u ǫ,a +ǫ 4 R ǫ R ǫ dxdy C 6 R ǫ (t L txr ǫ (t L +ǫ 4 R ǫ L txr ǫ L 4 C 6 R ǫ (t L txr ǫ (t L +ǫ 4 R ǫ L txr ǫ L txr ǫ H ǫ 8 txr ǫ L +C 7(ǫ 3 R ǫ L + txr ǫ L + Rǫ L, (3.43 = txr ǫ ( x u ǫ t R ǫ dxdy txr ǫ ( x (u ǫ,a +ǫ 4 R ǫ t R ǫ dxdy C 8 t R ǫ (t L txr ǫ (t L +ǫ 4 t R ǫ L txr ǫ L 4 x R ǫ L 4 C 8 t R ǫ (t L txr ǫ (t L +C 9 ǫ 4 t R ǫ L txr ǫ L txr ǫ H x R ǫ L x R ǫ H C 8 t R ǫ (t L txr ǫ (t L +C 0 ǫ 9 8 t R ǫ L txr ǫ L txr ǫ H x R ǫ H ǫ 8 txr ǫ L +C (ǫ 5 x R ǫ H + txr ǫ L + tr ǫ L, (3.44 and = txr ǫ ( t u ǫ x R ǫ dxdy txr ǫ ( t (u ǫ,a +ǫ 4 R ǫ x R ǫ dxdy C x R ǫ (t L txr ǫ (t L +ǫ 4 x R ǫ L txr ǫ L 4 t R ǫ L 4 C x R ǫ (t L txr ǫ (t L +C 3 ǫ 4 x R ǫ L txr ǫ L txr ǫ H t R ǫ L t R ǫ H C x R ǫ (t L txr ǫ (t L +C 4 ǫ 7 8 x R ǫ L txr ǫ L txr ǫ H t R ǫ H ǫ 8 txr ǫ L +C 5(ǫ 5 t R ǫ H + txr ǫ L + xr ǫ L. (3.45

25 X.P. WANG, Y.G. WANG AND Z.P. XIN 989 Plugging (3.39 (3.45 into (3.38, it follows that d dt txr ǫ (t L +ǫ x txr ǫ (t L +ǫ y txr ǫ (t L C 6 (ǫ 3 R ǫ L +ǫ5 t,x R ǫ H + txr ǫ L + Rǫ, t R ǫ L + tr ǫ L + x R ǫ L + Rǫ (t L + txf ǫ (t L +ǫ tx r ǫ (t L (, (3.46 which implies that sup 0 t T T 0 T txr ǫ (t L + ( ǫ x txr ǫ (t L +ǫ y txr ǫ (t L dt 0 e T t C6(ǫ3 R ǫ (s L +ǫ 5 t,xr ǫ (s H +ds ( R ǫ (t, t R ǫ (t L + tr ǫ (t L + x R ǫ (t L + Rǫ (t L + txf ǫ (t L +ǫ tx r ǫ (t L ( dt. (3.47 By using (3.9, (3.5, and (3.35 in (3.47, the estimate (3.36 follows immediately. Finally, by combining the estimates given in Propositions 3,6, 3.7, 3.8, and 3.9, with the classical Sobolev embedding theorem, R ǫ L C( Rǫ L + tr ǫ, R ǫ L + txr ǫ, tyr ǫ, xyr ǫ L + t xyr ǫ L, and with L being the norm in L ([0,T] IR +, we obtain Proposition 3.0. For the solution R ǫ to the problem (3.8, we have the estimate R ǫ L ((0,T IR + Cǫ 7 4. (3.48 Thus, from the representation of solutions given in (3.5, we conclude: Theorem 3.. For the problem (3., suppose that the initial data u 0 of (3. belongs to H s ( for a fixed s>8, with u 0 =0 and u 0, =0 on {y=0}. Then, in L ((0,T IR + we have u ǫ (t,x,y=u I,0 (t,x,y+ 3 ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ +O(ǫ u ǫ (t,x,y= 3 j=0 j= ǫ j 4u I,j (t,x,y+ǫ3 4u B,3 (t,x, y ǫ +O(ǫ p ǫ (t,x,y= 3 ǫ4p j I,j (t,x,y+o(ǫ j=0 (3.49 with all profiles being given in Conclusion 3.. Remark 3.. For any fixed J 4, under certain compatibility and regularity assumptions on initial data one can similarly conclude u ǫ (t,x,y=u I,0 (t,x,y+ J ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ +O(ǫ J+ 4 j= u ǫ (t,x,y= ǫ4u j I,j (t,x,y+ J j=0 j=3 p ǫ (t,x,y= 4 ǫ4p j I,j (t,x,y+ J j=0 j=5 ǫ4(u j I,j (t,x,y+ub,j (t,x, y ǫ +O(ǫ J+ 4 ǫ j 4(p I,j (t,x,y+p B,j (t,x, y ǫ +O(ǫ J+ 4 (3.50

26 990 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS in L ((0,T IR Remarks on some general cases In this section, we are going to sketch the main idea for generalizing the above discussion to the case that the slip length ǫ 4 in the problem (. is replaced by ǫ γ for a fixed 0<γ<. 4.. The case ǫ γ for a rational number 0<γ<. Consider the following problem in {t,y>0,x IR}: t u ǫ +(u ǫ uǫ + p ǫ =ǫ γ xu ǫ +ǫ yu ǫ u ǫ =0 u ǫ y=0 =0 (4. (βu ǫ ǫ γ u ǫ y y=0=0 u ǫ t=0 =u 0 (x,y, for a rational number 0<γ<. Let r>0 be such that a= γ r and b= r are co-prime integers. Take the following ansatz for the solutions to (4.: u ǫ (t,x,y= ǫ rj (u I,j (t,x,y+u B,j (t,x, y ǫ p ǫ (t,x,y= ǫ rj (p I,j (t,x,y+p B,j (t,x, y (4. ǫ, where u B,j (t,x,z and p B,j (t,x,z are rapidly decreasing when z= y ǫ +. Plugging (4. into (4., it follows that u I,j =0, which implies that z u B,j =0, j b x u B,j + z u B,j+b =0, (4.3 u B,j 0, u I,j y=0=0, j b. (4.4 Plugging (4. into (4. and (4. 4 respectively, it follows that ǫ rj t (u I,j +u B,j + b + ǫ rj j=0 j (u I,k +u B,k + b ǫ rj p I,j + b = j=0 ǫ rj j ǫ rj ((u I,k +u B,k u I,j k +(u I,k +u B,k z u B,j k + j+b ǫ rj (u I,k +u B,k z u B,j+b k ( 0 z p B,j + ( ǫ rj x p B,j z p B,j+b x u B,j k ǫ rk zu B,k + ǫ +r(j a x(u I,j +u B,j + ǫ +rj ( yu I,j + zu B,j+b (4.5

27 X.P. WANG, Y.G. WANG AND Z.P. XIN 99 and β ǫ rj (u I,j +ub,j =ǫ ar ǫ rj y u I,j + ǫ rj z u B,j (4.6 on {y=z=0}. Set [j/b] u p,j (t,x,z=u B,j z k (t,x,z+ k! k yu I,j bk. (4.7 From (4.6 we immediately obtain { z u B,j z=0 =0, j b a ( z u p,j βup,j b+a z=0 =0, j b a. From (4.5 and (4.8, we can deduce that (4.8 u B,j (t,x,z 0, j b a (4.9 which implies u B,j (t,x,z 0, j b a (4.0 from ( The vanishing of the order O(ǫ r(b a terms in the first component of (4.5 implies that u p,b a (t,x,z satisfies the following problem for a linear degenerate parabolic equation: t u p,b a +u I,0 xu p,b a +u p,b a x u I,0 +z y u I,0 zu p,b a + b a z u p,b a z=0 =βu I,0 k= u I,k xu I,b a k + x p I,b a = zu p,b a lim z + u p,b a (t,x,z=u I,b a (t,x exponentially. (4. From (4.3, we immediately get that u p,b a (t,x,z satisfies z u p,b a + x u p,b a =0 u p,b a z=0 =0. (4. The vanishing of the order O(ǫ r(b a terms in the second component of (4.5 gives rise to an equation for determining p B,3b a (t,x,z by using u B,b a and u B,b a. Similarly, foranyj b a+, fromtheo(ǫ rj ordertermsofthefirstcomponent of (4.5 we deduce an equation for solving u p,j. Then, we can determine up,j+b from ( The vanishing of the O(ǫ r(j+b order terms of the second component of (4.5 gives an explicit formula for determining p B,j+b (t,x,z by using {u B,k } k j and {u B,k } k j+b.

28 99 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS In the same way as in the proof of Theorem 3., for the problem (4., we can conclude Theorem 4.. For the problem (4., under certain assumptions on the regularity and compatibility conditions on the initial data u 0, the solutions to (4. have the following expansions: u ǫ (t,x,y= b a ǫ rk u I,k (t,x,y+ J j=b a u ǫ (t,x,y= b a ǫ rk u I,k (t,x,y+ J p ǫ (t,x,y= 3b a ǫ rk p I,k (t,x,y+ J j b a j 3b a ǫ rj (u I,j (t,x,y+ub,j (t,x, y ǫ +o(ǫ rj ǫ rj (u I,j (t,x,y+ub,j (t,x, y ǫ +o(ǫ rj ǫ rj (p I,j (t,x,y+p B,j (t,x, y ǫ +o(ǫ rj (4.3 in L ((0,T IR + for a fixed J. Moreover, the first components of the boundary layer profiles [j/b] u p,j (t,x,z=u B,j z k (t,x,z+ k! k yu I,j bk satisfy linear degenerate parabolic problems similar to (4. when j b a, the second components are explicitly given by the problem (4. for all j b a, and for all j 3b a, p B,j (t,x,z are given explicitly by {u p,k (t,x,z} k j b and {u p,k (t,x,z} k j b as in ( The case ǫ γ for a irrational number 0<γ<. Consider the following problem in {t,y>0,x IR}: t u ǫ +(u ǫ uǫ + p ǫ =ǫ γ xu ǫ +ǫ yu ǫ u ǫ =0 u ǫ y=0 =0 (4.4 (βu ǫ ǫ γ u ǫ y y=0=0 u ǫ t=0 =u 0 (x,y, for a irrational number 0<γ<. For the solutions to the problem (4.4, we take the following ansatz: u ǫ (t,x,y= u 0 (t,x,y+ k ǫ k (v I,k (t,x,y+v B,k (t,x, y ǫ + k ǫ ( γk [(u I,k (t,x,y+u B,k (t,x, y ǫ + ǫ(w j I,k,j (t,x,y+w B,k,j (t,x, y ǫ ] j p ǫ (t,x,y= p 0 (t,x,y+ k ǫ k (p I,k v (t,x,y+p B,k v (t,x, y ǫ (4.5 + k ǫ ( γk [(p I,k u (t,x,y+p B,k u (t,x, y ǫ + ǫ(p j I,k,j w (t,x,y+p B,k,j w (t,x, y ǫ ], j

29 X.P. WANG, Y.G. WANG AND Z.P. XIN 993 where all profiles with the index B decay exponentially when z= y ǫ +. From the following discussion, one can see the motivation for taking the ansatz (4.5 is that first the term u B, (t,x, y ǫ is to cancel βu 0 in the Navier boundary condition, secondly, the term v B, (t,x, y ǫ is to cancel y u 0 in the Navier boundary condition, thirdly, the term w B,, (t,x, y ǫ is for satisfying the divergence free condition associated with u B, (t,x, y ǫ, fourthly, the term p B,, w (t,x, y ǫ in the pressure is to to cancel the second component of the equations for w B,, (t,x, y ǫ, and finally all higher order terms come from the nonlinear interaction in the Navier-Stokes equations. Now, we are going to describe how to determine each order profile in (4.5 step by step. First, by plugging (4.5 into the divergence free relation, we get that the vanishing of the order O(ǫ ( γk terms for each k implies that yielding z u B,k =0, (4.6 u B,k (t,x,z 0, for all k. (4.7 Therefore, the vanishing of the order O(ǫ ( γk terms for the second component of the equations in (4.4 gives which implies that z p B,k, w =0, (4.8 p B,k, w (t,x,z 0, for all k. (4.9 Obviously, the vanishing of the order O(ǫ ( γk terms for the second component of the equations in (4.4 gives which implies that The divergence free condition implies that z p B,k u =0, (4.0 p B,k u (t,x,z 0, for all k. (4. divu 0 =divv I,k =divu I,k =divw I,k,j =0 ( k,j z v B, =0 x v B,k + z v B,k+ =0 ( k (4. x u B,k + z w B,k, =0 ( k x w B,k,j + z w B,k,j+ =0 ( k,j,

30 994 BOUNDARY LAYERS IN INCOMPRESSIBLE NAVIER-STOKES EQUATIONS which leads to v B, 0 v B,k+ (t,x,z= z xv B,k (t,x,ξdξ ( k w B,k, (t,x,z= z xu B,k (t,x,ξdξ ( k w B,k,j+ (t,x,z= z xw B,k,j (t,x,ξdξ ( k,j. By a direct computation, we have (4.3 βu ǫ ǫ γ y u ǫ =βu 0 z u B, ǫ γ ( y u 0 + z v B, + k ǫ ( γk (β(u I,k +u B,k z u B,k+ +ǫ (β(v I, +v B, y u I, z w B,, + ǫ γ+k ( y v I,k + z v B,k+ k + k + k + k j ǫ k (β(v I,k +v B,k y w I,,k z w B,,k ǫ ( γk+ (β(w I,k, +w B,k, y u I,k+ z w B,k+, ǫ ( γk+ j (β(w I,k,j +w B,k,j y w I,k+,j z w B,k+,j. (4.4 So, the Navier boundary condition implies that z u B, z=0 =βu 0 y=0 z v B, z=0 = y u 0 y=0 z w B,, z=0 =(β(v I, +v B, y u I, y=z=0 z u B,k z=0 =β(u I,k +u B,k y=z=0 ( k z v B,k z=0 = y v I,k y=0 ( k z w B,,k z=0 =(β(v I,k +v B,k y w I,,k y=z=0 ( k z w B,k, z=0 =(β(w I,k, +w B,k, y u I,k y=z=0 ( k z w B,k,j z=0 =(β(w I,k,j +w B,k,j y w I,k,j y=z=0 ( k,j. Obviously, the boundary condition u ǫ y=0 =0 implies that u 0 y=0 =0 (4.5 v I,k y=0= v B,k z=0 ( k u I,k y=0= u B,k z=0 =0 ( k w I,k,j y=0 = w B,k,j z=0 ( k,j. (4.6

31 X.P. WANG, Y.G. WANG AND Z.P. XIN 995 Now, we determine each order profile in the expansions (4.5 as the following steps. ( First, we solve (u 0,p 0 from the Euler equations (.7. ( Secondly, the vanishing of the order O(ǫ γ terms in the first component of the equations (4.4 gives that u B, (t,x,z satisfies a linear degenerate parabolic equation with the boundary condition (4.5, which has a unique solution u B, (t,x,z which decays quickly when z +. The observation (4.7givesthatu I, =0on{y=0}, sowecandetermine(u I,,p I, u bysolving the linearized Euler equations. From the divergence free condition (4. 4 with k=, we get w B,, (t,x,z, from which one can determine p B,, w by using the vanishing of the order O(ǫ γ terms in the second component of the equations (4.4. This also gives the boundary condition of w I,, (t,x,z, which can be used to determine (w I,,,pw I,, by solving the linearized Euler equations. (3 Thirdly, the vanishing of the order O(ǫ terms in the first component of the equations (4.4 gives that v B, (t,x,z satisfies a linear degenerate parabolic equation with the boundary condition (4.5, which has a unique solution v B, (t,x,z which decays quickly when z +. From (4.3 and the second component of the equations (4.4 at the order O(ǫ, we get p B, v 0. The fact (4.3 also gives that v I, =0 on {y=0}, so we can determine (v I,,p I, v by solving the linearized Euler equations. From the divergence free condition (4. 3 with k=, we get v B, (t,x,z, from which one can determine p B,3 v by using the vanishing of the order O(ǫ terms in the second component of the equations (4.4. This also gives the boundary condition of v I, (t,x,z, which can be used to determine (v I,,p I, v by solving the linearized Euler equations. (4 Fourthly, the vanishing of the order O(ǫ γ terms in the first component of the equations (4.4 gives that u B, (t,x,z satisfies a linear degenerate parabolic equation with the boundary condition (4.5 4 with k=, which has a unique solution u B, (t,x,z which decays quickly when z +. From the divergence free condition (4. 4 with k=, we get w B,, (t,x,z, from which onecandeterminep B,, w byusingthevanishingoftheordero(ǫ 3 γ termsin the second component of the equations (4.4. This also gives the boundary condition of w I,, (t,x,z, which can be used to determine (w I,,,pw I,, by solving the linearized Euler equations. (5 Fifthly, the vanishing of the order O(ǫ γ terms in the first component of the equations (4.4 gives that w B,, (t,x,z satisfies a linear degenerate parabolic equation with the boundary condition (4.5 3, which has a unique solution w B,, (t,x,z which decays quickly when z +. From the divergence free condition (4. 5 with k=j=, we get w B,, (t,x,z, from which one can determine p B,,3 w by using the vanishing of the order O(ǫ 3 γ terms in the second component of the equations (4.4. This also gives the boundary condition of w I,, (t,x,z, which can be used to determine (w I,,,pw I,, by solving the linearized Euler equations. (6 Similarly, for any fixed k, the vanishing of the order O(ǫ k terms in the firstcomponentoftheequations(4.4 givesthatv B,k (t,x,z satisfies a linear