Schauder estimates for solutions to a mixed boundary value problems for Stokes system in polyhedral domains

Transcription

1 Schauder estimates for solutions to a mixed boundary value problems for Stokes system in polyhedral domains by V. Mazya and J. Rossmann Abstract A mixed boundary value problems for the Stokes system in a polyhedral domain is considered. The authors prove the existence of solutions in weighted and non-weighted Hölder spaces and obtain regularity assertions for the solutions. The results are essentially based on estimates of the Green s matrix. eywords: Stokes system, nonsmooth domains MSC 1991: 35J25, 35J55, 35C15, 35Q30 0 Introduction Schauder estimates, i.e. coercive estimates of Hölder norms, for solutions to linear elliptic equations and systems in domains with smooth boundaries have important applications to linear and especially nonlinear boundary value problems see, e.g., Agmon, Douglis, Nirenberg [1] and Gilbarg, Trudinger [8]. In the present paper, we consider a mixed boundary value problem for the linear Stokes system u + p = f, u = g 0.1 in a three-dimensional domain of polyhedral type, where components of the velocity and/or the friction are given on the boundary. To be more precise, we have one of the following boundary conditions on each side Γ j : i u = h, ii u τ = h, p + 2ε n,n u = φ, iii u n = h, ε n,τ u = φ, iv pn + 2ε n u = φ, where u n = u n denotes the normal and u τ = u u n n the tangential component of u, ε n u is the vector εu n, ε n,n u is the normal component and ε n,τ u the tangential component of ε n u. In our previous papers [22, 23] we proved estimates for the Green s matrix and estimates of solutions in weighted and nonweighted Sobolev spaces. The goal of the present paper is to prove the existence of solutions in weighted Hölder spaces. Furthermore, we will obtain regularity results for the solutions. There is an extensive bibliography concerning elliptic boundary value problems in domains with edges see e.g. the references in the books of Grisvard [10], Dauge [3], Nazarov and Plamenevskiĭ [25]. Moreover, many works deal with boundary value problems in Lipschitz domains. We mention here the papers of Jerison and enig [11], enig [12] and for the Stokes system the papers of Fabes, enig and Verchota [7], Brown and Shen [2], Deuring and von Wahl [5], Dindos and Mitrea [6]. However, most of the works in this field deal with solutions in Sobolev spaces with or without weight. Concerning Schauder estimates for solutions of boundary value problems in domains with edges, we mention the papers by Maz ya and Plamenevskiĭ [16, 17], where boundary value problems for elliptic differential equations of arbitrary order were studied. The results obtained in [16, 17] are applicable, e.g., to the Dirichlet problem but not to the Stokes system with boundary conditions i iv. In [19], weighted L p and Schauder estimates were obtained for solutions of the Stokes system with Dirichlet condition i and 1

2 free surface condition iii on parts of the boundary. Weighted L p and Schauder estimates for solutions to the Neumann problem in a domain with nonintersecting edges were proved in the preprint [26] by Solonnikov. For the Neumann problem to second order systems A i,j xi xj u = f i, we refer to our paper [21]. The boundary value problems considered in [16, 17, 19] allow to use weighted Hölder spaces N with homogeneous norms. However, the Neumann problem or in general the mixed problem with boundary conditions i iv requires the use of weighted spaces C with inhomogeneous norms. This makes the consideration of the boundary value problem more difficult. On the other hand, in some cases e.g. the Dirichlet problem in convex polyhedral domains, the results can be improved when considering solutions in weighted spaces with inhomogeneous norms. We also note that on the set of functions vanishing in a neighborhood of the corners of the domain, the C norm with = 0 is equivalent to the norm in the non-weighted Hölder space C. The largest part of the paper Sections 2 4 concerns the boundary value problem for the Stokes system in a dihedron D and in polyhedral cone with sides Γ 1,..., Γ N and edges M 1,..., M N. We do not a priori pose in this paper that the cone is Lipschitz. Section 3 deals with the existence of solutions u, p C 2,σ 3 C 1,σ of the boundary value problem if f C0,σ 3, g C 1,σ, and the boundary data h j, φ j are from the corresponding trace spaces. Here, C is a weighted Hölder space with weight parameters β R, = 1,..., N [0, N. In the special case 1 = = N =, the norm in this space is given by u C = + x α l x y </2 α =l k β l σ+ α rx max0, l σ+ α α x ux β α x ux α y uy x y k+σ + x y <rx/2 α =l β rx α x ux α y uy x y σ, where rx denotes the distance of x to the set S = {0} M 1 M N and k is the integral part of σ + 1. It is shown in Section 3 that there is a uniquely determined solution if g and the boundary data satisfy certain compatibility conditions on the edges, the line Re λ = 2 + σ β is free of eigenvalues of a certain operator pencil Aλ, and the components k of are such that k 0, k σ not integer, and 2 µ k < k σ < 2 for k = 1,..., N. Here µ k are certain positive numbers depending on the angle θ k at the edge M k. For example, in the case of the Dirichlet problem, we have µ k = π/θ k if θ k < π, while µ k is the smallest positive solution of the equation sinµθ k + µ sin θ k = 0 if θ k > π. Estimates for the eigenvalues of the pencil Aλ can be found e.g. in [4, 13, 14, 18]. It is further shown that the solution u, p C 2,σ 3 C 1,σ belongs to C β, 3 C l 1,σ β, if f C l 2,σ β, 3, g C 1,σ β,, the boundary data are from the corresponding trace spaces, the closed strip between the lines Re λ = 2+σ β and Re λ = l + σ β is free of eigenvalues of the pencil Aλ, and the components of are such that k 0, k σ not integer, and l µ k < k σ < l. In Section 4, we deal with weak solutions of the boundary value problem. We prove that under conditions analogous to those in Section 3, there exists a unique weak solution u, p C 1,σ 3 C 0,σ. The boundary value problem for the Stokes system in a bounded polyhedral domain G is studied in the last Section 5. As an example, we consider the weak solution u, p W 1 G 3 L 2 G of the Dirichlet problem u + p = f, u = g in G, u = 0 on Γ j, j = 1,..., N, in a polyhedron G with sides Γ j and edges M k. We denote by θ k the angle at the edge M k. As a consequence of our results, we obtain for example the following regularity assertion in a neighborhood of an edge point ξ M k. If f C l 2,σ, g C l 1,σ in a neighborhood of ξ, g Mk = 0, and l + σ < π/θ k, then u C and p C l 1,σ in a neighborhood of ξ. 2

3 This result is also true for l = 1. This means that in the case θ k < π, we obtain u C 1,σ, p C 0,σ in a neighborhood of ξ provided σ < π θ k /θ k and f, g satisfy the above conditions. Another result is the following. If f C l 2,σ l,l G 3 and g C l 1,σ l,l G, where 0 < σ < minre Λ j, µ k, then u, p C l,l G3 C l 1,σ l,l G. In particular, we have u C 0,σ G 3. If the polyhedron G is convex, then this result is true for arbitrary σ 0, 1. Here Λ j denotes the eigenvalue of the pencil A j λ with smallest positive real part. For convex polyhedrons, we have Λ j = 1. Then there is also the following result. Let f C 1,σ G 3, g C 0,σ G, where ε is a sufficiently small positive number, and let g Mk =0 for all k. Then the solution u, p belongs to C 1,σ G 3 C 0,σ G. Other examples are given at the end of Section 5. In a forthcoming paper, we will apply the results to the nonlinear Navier-Stokes system. 1 Weighted Hölder spaces 1.1 Weighted Hölder spaces in an angle an in a dihedron Let be the angle {x 1, x 2 R 2 : 0 < r <, θ/2 < ϕ < θ/2}, where r, ϕ are the polar coordinates of x = x 1, x 2, and let γ ± : ϕ = ±θ/2 be the sides of. Furthermore, let D = R = {x = x, x 3 : x = x 1, x 2, x 3 R} be a dihedron with sides Γ ± = γ ± R and edge M. For arbitrary integer l 0 and real, σ, 0 < σ < 1, we define N as the space of all functions with continuous derivatives up to order l on \{0} such that u N = x l σ+ α x α ux + x α x ux y α uy x x,y x y σ <. α l α =l x y < x /2 Analogously, we define the weighted Hölder space N D as the set of all functions with continuous derivatives up to order l on D\M such that u N D = x l σ+ α x α ux + u,;d <, 1.1 where x D α l u,;d = x,y D α =l x y < x /2 u C D = α ux + x D α l x α x ux α y uy x y σ. Note that N D is continuously imbedded into C l k,k +σ D if k 1 σ < k, where k is a nonnegative integer, k l see [21]. Here C D denotes the nonweighted Hölder space with the norm α ux α uy An equivalent norm is given in the following lemma. Lemma 1.1 The norm in C D is equivalent to u = α ux + x D α l α =l + x 3 x x,y D α =l x y <1 x y σ α ux, x 3 α ux, y 3 x 3 y 3 <1 x 3 y 3 σ α ux, x 3 α uy, x 3 x y σ. x y < x /2 3

4 P r o o f. It suffices to prove the lemma for l = 0. Obviously, the norm in C 0,σ D is equivalent to ux, x 3 ux, y 3 u = ux + x D x x 3 y 3 <1 + x 3 x 3 y 3 σ ux, x 3 uy, x 3 x y σ. x y <1 We show, that there exists a constant c independent of u such that x 3 x,y ux, x 3 uy, x 3 x y σ c x 3 x,y x y < x /2 ux, x 3 uy, x 3 x y σ 1.2 We denote the right-hand side of 1.2 by A σ u. Let x, y be arbitrary points in such that x y > x /2, and let x 3 R. We put ξ n = 2 n x. Then uξ n, x 3 uξ n+1, x 3 c 0 ξ n ξ n+1 σ = c 0 x σ 2 n+1σ, where c 0 = A σ u. Consequently, ux, x 3 u0, x 3 uξ n, x 3 uξ n+1, x 3 c 0 x σ n=0 and, analogously, uy, x 3 u0, x 3 c 0 2 σ 1 y σ Since x < 2 x y and y < 3 x y, it follows that n=0 2 n+1σ = c 0 2 σ 1 x σ ux, x 3 uy, x 3 c 0 x 2 σ σ + y σ 2σ + 3 σ 1 2 σ 1 A σu x y σ what proves 1.2. The result follows. Let 0 < l + σ and 0 < σ 1. Then by C D, we denote the weighted Hölder space with the norm l u C D = u C l k,k +σ D + x l σ+ α x α ux + u,;d, x D α =l k+1 where k = [ σ] + 1, [s] denotes the greatest integer less or equal to s. In the case l + σ, we set C D = N D. Every function u C D, 0 < l + σ, is continuous in D. The restriction of u to M belongs to the Hölder space C l k,k +σ M, where k = [ σ] + 1. Conversely, every function f C l k,k +σ M can be extended to a function u C D. For this, we consider the operator Efx, x 3 def = 1 0 fx 3 + tr ψt dt, 1.3 where ψ C 0 R is a given function with port in 0, 1 satisfying the condition 1 0 ψt dt = 1, 1 0 t j ψt dt = 0 for j = 1, 2,..., l k. In [21, Le.2.7] it is shown that E realizes a continuous mapping C l k,k +σ M C D, and Ef M = f for arbitrary f C l k,k +σ M, k = [ σ] + 1 l xi Ef N +1 D, x i Ef N +1 D c f C l k,k +σ M for i = 1, 2. For the following lemma we refer to [21, Le.2.8,Le.2.9]. 4

5 Lemma 1.2 Let u C D, where 0, k 1 < σ < k, k {0, 1,..., l}. 1 Furthermore, let f i,j = x i 1 x j 2 u M for i + j l k and u i,j x = χ x Ef i,j x, where χ is a smooth cut-off function on [0,, p χ [0, 2, χ = 1 on [0, 1]. Then u admits the decomposition u = v + w, where v N D, w = i+j l k 1 i! j! u i,j x i 1x j 2 Cl+m,σ +m D, m = 0, 1, 2, For the inclusion u N D it is necessary and sufficient that α u = 0 on M for α l k. 1.2 Weighted Hölder spaces in a polyhedral cone Let = {x R 3 : x/ Ω} 1.4 be a polyhedral cone in R 3 whose boundary consists of plane sides Γ k and edges M k, j = 1,..., N. We denote by r k x the distance to the edge M k and by rx the distance to the set S = M 1 M N {0}. The subset {x : r j x < 3rx/2} is denoted by j. Note that there are the inequalities c 1 r k x rx c 2 with positive constants c 1, c 2 independent of x. Let l be a nonnegative integer, β R, and = 1,..., N R N. We define the space N l as the set of all l times continuously differentiable functions in \S such that u N l = N β l+ α x α l r k x rk x k l+ α α x ux dx <. The weighted Hölder space N is defined as the set of all l times continuously differentiable functions on \S with finite norm u N = N β l σ+ α x α l + x y <rx/2 α =l β rk x k l σ+ α x α ux 1.5 rk x k x α ux y α uy x y σ. 1.6 Furthermore, the space C is defined for nonnnegative k, k = 1,..., N, as the set of all l times continuously differentiable functions on \S with finite norm u C = N β l σ+ α x α l + j: k j l + β j x,y α =l k j j x y </2 x,y α =l x y <rx/2 β rk x max0,k l σ+ α x α ux α x ux α y uy x y k j+σ j rk x k x α ux y α uy x y σ, 1.7 where k j = [ j σ] + 1, [s] denotes the greatest integer less or equal to s. Replacing N α ux α uy x,y α =l k j j x y </2 β j x y kj+σ j 5

6 in 1.6 by N α =l k j x,y j x y <rx/2 β j α ux α uy x y kj+σ j + x j 1/2<t<3/2 β j α ux α utx x tx kj+σ j, we obtain an equivalent norm in C cf. Lemma 1.1. Obviously, N is a subset of C. If k l + σ for k = 1,..., N, then both spaces coincide. Furthermore, there are the following continuous imbeddings: N l+1 β σ+1, σ+1 N β, N l β σ, σ, C Cl,σ β, if l + σ l + σ, β l σ = β l σ, k l σ k l σ. The trace spaces for N and C on Γ j are denoted by N Γ j and C Finally, we introduce the following notation. If R N and s R, then by N +s we mean the spaces N and C with = 1 + s,..., N + s. 2 The problem in a dihedron Γ j, respectively. and C +s Let D be the dihedron introduced in the previous section. We consider a boundary value problem for the Stokes system, where on each of the sides Γ ± one of the boundary conditions i iv is given. Let n ± = n ± 1, n± 2, 0 be the exterior normal to Γ±, ε ± n u = εu n ± and ε ± nnu = ε ± n u n ±. Furthermore, let d ± {0, 1, 2, 3} be integer numbers characterizing the boundary conditions on Γ + and Γ, respectively. We put S ± u = u for d ± = 0, S ± u = u u n ± n ±, N ± u, p = p + 2ε ± nnu for d ± = 1, S ± u = u n ±, N ± u, p = ε ± n u ε ± nnu n ± for d ± = 2 N ± u, p = pn ± + 2ε ± n u for d ± = 3 and consider the boundary value problem u + p = f, u = g in D, 2.8 S ± u = h ±, N ± u, p = φ ± on Γ ±. 2.9 Here the condition N ± u, p = φ ± is absent in the case d ± = 0, while the condition S ± u = h ± is absent in the case d ± = Reduction to homogeneous boundary conditions Lemma 2.1 Let h ± N Γ ± 3 d±, φ ± N l 1,σ Γ ± d±, l 1, be given. Then there exists a vector function u N D 3 such that S ± u = h ± and N ± u, 0 = φ ± on Γ ±. The norm of u can be estimated by the norms of h ± and φ ±. The analogous result in the space C holds only under additional assumptions on the boundary data. If u C D 3, < l+σ, then there exists the trace u M C l k,k +σ M 3, k = [ σ]+1, and from the boundary conditions 2.9 it follows that S ± u M = h ± M. Here S + and S are considered as operators on C l k,k +σ M 3. Consequently, the boundary data h + and h must satisfy the compatibility condition h + M, h M RT, where RT is the range of the operator T = S +, S. This condition can be also written in the form A + h + M = A h M, 2.10 where A +, A are certain constant matrices. For example in the case of the Dirichlet problem, A + and A are the identity matrices. 6

7 Lemma 2.2 Let h ± C Γ ± 3 d± and φ ± C l 1,σ Γ ± d±, l 1, l + σ 1 < < l + σ. Suppose that h + and h satisfy the compatibility condition 2.10 on M. Then there exists a vector function u C D 3 such that S ± u = h ±, N ± u, 0 = φ ± on Γ ±. The norm of u can be estimated by the norms of h ± and φ ±. P r o o f. By 2.10, there exists a vector function ψ C 0,l+σ M 3 such that S ± ψ = h ± M. Let v C D 3 be an extension of ψ. Then the trace of h ± S ± v Γ ± on M is equal to zero and, consequently, h ± S ± v Γ ± N Γ ± 3 d± cf. Lemma 1.1. Furthermore, φ ± N ± v, 0 Γ ± C l 1,σ Γ ± d± N l 1,σ Γ ± d±. Thus, according to Lemma 2.1, there exists a function w N D 3 such that S ± w = h ± S ± v and N ± w, 0 = φ ± N ± v, 0 on Γ ±. Then u = v + w has the desired properties. Now let g C l 1,σ D, h ± C Γ ± 3 d±, and φ C Γ ± 3 d±, l+σ 2 < < l+σ 1. Then the traces of g, h ±, r h ± and φ ± on M exist. We pose that there is a pair u, p C D 3 C l 1,σ D such that S ± u = h ±, N ± u, p = φ ± on Γ ± and u = g on M 2.11 We put b = u M, c = x1 u M, d = x2 u M and q = p M. Then from the equations S ± u = h ± on Γ ± it follows that S ± r u = r h ± on Γ ±, and therefore, Moreover u = g on M if and only if S ± b = h ± M, 2.12 S ± c cos θ 2 ± d sin θ 2 = r h ± M c 1 + d 2 + x3 b 3 = g M Obviously, the trace of N ± u, p on M can be written as a linear form M ± c, d, x3 b, q. Thus, from N ± u, p = φ ± on Γ ± it follows that M ± c, d, x3 b, q = φ ± M Lemma 2.3 Suppose that g C l 1,σ D, h ± C Γ ± 3 d±, and φ C l 1,σ Γ ± 3 d±, l 1, l+σ 2 < < l + σ 1, are such that the system with the unknowns b, c, d, q is solvable. Then there exists a pair u, p C D 3 C l 1,σ D satisfying P r o o f. Let b C 1,l 1 +σ M 3, c, d C 0,l 1 +σ M 3, q C 0,l 1 +σ M solve the linear system We set v = Eb + x 1 Ec + x 2 Ed, p = Eq, where E is the extension operator 1.3. Then v C D 3, p C l 1,σ D, v M = b, x1 v M = c, x 2 v M = d, and p M = q. Consequently, S ± v M = h ± M, r S ± v M = r h ± M, v M = g M, N ± v, p M = φ ± M. From this and from Lemma 1.1 we conclude that S ± v h ± N Γ ± 3 d± and N ± v, p φ ± N l 1,σ Γ ± d±. Thus, by Lemma 2.1, there exists a vector function w N D 3 such that S ± w = h ± S ± v, N ± w, 0 = φ ± on Γ ±. Then u, p = v + w, p satisfies Remark 2.1 The condition of Lemma 2.3 is always satisfied if d + + d is odd, h + and h satisfy the compatibility condition 2.10 and sin 2θ 0 for d + + d = 3, cos θ cos 2θ 0 for d + + d {1, 5}. If d + + d is even, then additionally to 2.10 another compatibility condition must be satisfied. For 7

8 example, in the case of the Dirichlet problem, θ π, θ 2π, for the validity of Lemma 2.3 it is necessary and sufficient that h + M = h M, n r h + M + n + r h M = g M + x3 h + 3 M sin θ. In the case d = 0, d + = 2, θ π/2, θ 3π/2, the data h +, h, φ + and g must satisfy the compatibility conditions h n + = h + and r h + cos 2θ 2n + cos θ + n r h + 2 sin 2 θ φ + 1 cos θ/2 + φ+ 2 sin θ/ g + x 3 h 3 sin 2θ = 0, and in the case of the Neumann problem, θ π, θ 2π, the compatibility condition φ + n = φ n + on M is necessary and sufficient for the validity of Lemma 2.3, see [22]. Lemma 2.4 Let f C l 2,σ D 3, g C l 1,σ D, h ± C Γ ± 3 d±, and φ C l 1,σ Γ ± 3 d±, l 2, 0 < l + σ 1, σ not integer. Suppose that g, h ± and φ ± are such that the system with the unknowns b, c, d, q is solvable. In the case l + σ > 2 we assume furthermore that for k = 2, 3,..., [l + σ ] there do not exist homogeneous polynomials u = c i,j x i 1 x j 2, p = d i,j x i 1 x j 2 i+j=k i+j=k 1 satisfying u + p = 0, u = 0 in D and the homogeneous boundary conditions S ± u = 0, N ± u, p = 0 on Γ ±. Then there exists a pair u, p C D 3 C l 1,σ D such that u p + f N l 2,σ D 3, u + g N l 1,σ D, S ± u Γ ± = h ±, N ± u, p Γ ± = φ ±. P r o o f. Let s < l + σ < s + 1, where s {1,..., l}, and let k be an integer, 1 k s. We show by induction in k that there exists a pair u, p C D 3 C l 1,σ D satisfying the following condition C k : i x1 j x 2 u p + f M = 0 for i + j k 2, i x1 j x 2 u + g M = 0 for i + j k 1, j r S ± u Γ ± h ± M = 0 for j k, j r N ± u, p Γ ± φ ± M = 0 for j k 1. For k = 1 condition C k means that u = g, S ± u = h ±, r S ± u Γ ± = r h ±, N ± u, p = φ ± on M. Suppose that 2 k s and the assertion is true for k 1, i.e. there exists a pair u, p C D 3 D satisfying condition C k 1. We put C l 1,σ v ν = α =k Ev ν α x α, ν = 1, 2, 3, q = Eq γ x γ α! γ! γ =k 1 and show that the functions v α ν, q γ C s k,l s+σ M can be chosen such that i x 1 j x 2 v q = i x 1 j x 2 u p + f on M for i + j = k 2, 2.16 x i 1 x j 2 v = x i 1 x j 2 u + g on M for i + j = k 1, 2.17 r k S ± v Γ ± = r k h ± S ± u Γ ±, r k 1 N ± v, q Γ ± = k 1 r φ ± N ± v, q Γ ± on M Equation 2.16 is equivalent to v ν i+2,j + vν i,j+2 + ν,1 q i+1,j + ν,2 q i,j+1 = i x 1 j x 2 u p + f M for i + j = k 2, ν = 1, 2, 3, while equation 2.16 is equivalent to v 1 i+1,j + v2 i,j+1 = i x 1 j x 2 u + g M for i + j = k 1. 8

9 Analogously, 2.18 can be written in the form S ± {v α ν }, {q γ } = r k h ± S ± u Γ ± M, N ± {v α ν }, {q γ } = r k 1 φ ± N ± v, q Γ ± M, where S ±, N ± are linear forms. Thus, is equivalent to a system of 4k + 3 linear equations with constant coefficients and 4k + 3 unknowns v α ν, q γ, ν = 1, 2, 3, α = k, γ = k 1. This system is uniquely solvable. Otherwise, the corresponding homogeneous system has a nontrivial solution ν {c α }, {d γ }, and the functions U ν = α =k c ν α x α, ν = 1, 2, 3, P = x γ d γ α! γ! γ =k 1 satisfy the homogeneous equations 2.8, 2.9 what contradicts the assumptions of the lemma. This proves that v α ν and q γ can be chosen such that v, q satisfies Obviously, x i 1 x j 2 v q = 0 on M for i + j k 3, x i 1 x j 2 v = 0 on M for i + j k 2, rs j ± v Γ ± = 0, N ± v, q Γ ± on M for j k 1. k 1 r Consequently, the pair u + v, p + q satisfy condition C k. In particular, it follows that there exists a pair u, p C for k = s. This means that u p+f N l 2,σ and N ± u, p φ ± N l 1,σ the lemma. D 3, u+g N l 1,σ D 3 C l 1,σ D satisfying condition C k D, S ± u h ± N Γ± 3 d±, Γ ± d± see Lemma 1.2. Applying Lemma 2.1, we obtain the assertion of Remark 2.2 The last assumptions in Lemma 2.4 the nonexistence of homogeneous polynomials of degrees k and k 1, respectively, satisfying the homogeneous equations 2.8 and 2.9 is satisfied, e.g., if λ = k is not an eigenvalue of the pencil Aλ introduced below. 2.2 Regularity assertions for the solution The next lemma follows from [24, Th.6.3.7] and [1, Th.9.3]. Lemma 2.5 Let G 1, G 2 be bounded subdomains of R 3 such that G 1 G 2, G 1 D = and G 1 M =. If u, p is a solution of 2.8, 2.9, u W 2,s D G 2 3, p W 1,s D G 2, f C l 2,σ D G 2 3, g C l 2,σ D G 2, h ± C Γ ± G 2 3 d±, φ ± C l 1,σ Γ ± G 2 d±, l 2, 0 < σ < 1, then u C D G 1 3 C l 1,σ D G 2, and u C D G 1 + p l C l 1,σ D G 2 c f C l 2,σ D G 2 + g l C l 2,σ D G 2 + h ± C Γ ± G 2 ± φ ± C l 1,σ Γ ± G 2 + u CD G2 3 + p CD G 2 with a constant c independent of u and p. + ± Let W l,s loc D\M be the set of all functions u such that ζu W l,s D for all ζ C 0 D\M. Lemma 2.6 Let u, p W 2,s loc D\M 3 W 1,s loc D\M be a solution of problem 2.8, 2.9 such that If f N l 2,σ p N D, and D 3, l 2, g N l 1,σ u N D + p 3 N l 1,σ D c x l σ ux + x l σ+1 px <. D, h ± N Γ ± 3 d±, φ ± N l 1,σ Γ ± d±, then u N D 3, f N l 2,σ D + g 3 N l 1,σ D + ± + ± h ± N Γ ± φ ± N l 1,σ Γ ± + u N 0 l σ D3 + p N 0 l+1 σ D

10 P r o o f. Due to Lemma 2.1, we may restrict ourselves to the case h ± = 0, φ ± = 0. For an arbitrary point y D we denote by B y the ball with center y and radius y /2 and by B y the ball with center y and radius 3 y /4. For an arbitrary subdomain U D let the norm in N U be defined by 1.1, where D is replaced by U. For y = 1 this norm is equivalent to the C norm, and Lemma 2.5 implies u N B y D + p 3 N l 1,σ B y D c f N l 2,σ B + g y D3 N l 1,σ B + u y D N l σ 0 B y D3 + p N 0 l+1 σ B y D 2.20 with a constant c independent of y. Let y = 1 and z = y 1 y. We introduce the functions ũξ = u y ξ, pξ = y p y ξ, fξ = y 2 f y ξ, and gξ = y g y ξ. Then Therefore, by 2.20, we have ũ + p = f, ũ = g in D, S ± ũ = 0, N ± ũ, p = 0 on Γ ±. ũ N B z D + p 3 N l 1,σ B z D c Using the inequalities f N l 2,σ B + g z D3 N l 1,σ B z D + ũ N l σ 0 B z D3 + p N 0 l+1 σ B z D. c 1 y l+σ u N B y D 3 ũ N B z D 3 c 2 y l+σ u N B y D 3 and the analogous inequalities for the norms of p, f and g, we obtain estimate 2.20 for arbitrary y D. This proves the lemma. We further need the following modification of Lemma 2.6 Lemma 2.7 Let ζ, η be smooth functions with compact ports, η = 1 in a neighborhood of p ζ. Furthermore, let u, p W 2,s loc D\M 3 W 1,s loc D\M be a solution of problem 2.8, 2.9 such that If ηf N l 2,σ N D 3, ζp N x l σ ηx ux + x l σ+1 ηx px <. D 3, l 2, ηg N l 1,σ, and ζu N D + ζp 3 N l 1,σ D c D, ηh ± N ηf N l 2,σ + ± Γ ± 3 d±, ηφ ± N l 1,σ Γ ± d±, then ζu D + ηg 3 N l 1,σ D + ηh ± N Γ ± ± ηφ ± N l 1,σ Γ ± + ηu N l σ 0 + ηp D3 N l+1 σ 0 D. P r o o f. We may again restrict ourselves in the proof to the case h ± = 0, φ ± = 0. Let U be a neighborhood of p ζ such that η = 1 in a neighborhood of U. Obviously, we obtain an equivalent norm in N D if we replace the expression u,;d in 1.1 by u,;d = x,y D α =l x y <ε x x α x ux α y uy x y σ, where ε is an arbitrarily small positive number. Using this norm with sufficiently small ε, then we have ζu N D + ζp 3 N l 1,σ D c u N U D + p 3 N l 1,σ U D. 10

11 Here, we used the same notation as in the proof of Lemma 2.6. Furthermore, estimate 2.20 is also valid if we denote by B y and B y the balls centered at y with radii ε y and 2ε y, respectively. From this it follows if ε is sufficiently small that u N U D + p 3 N l 1,σ U D c f N l 2,σ U D + g 3 N l 1,σ U D + u N 0 l σ U D 3 + p N 0 l+1 σ U D, where U = {x D : ηx = 1}. This proves the lemma. Next, we prove a regularity assertion for the solution in the class of the spaces C, Lemma 2.8 Let u, p W 2,s loc D\M 3 W 1,s loc D\M be a solution of problem 2.8, 2.9, and let ζ, η be smooth functions with compact ports, η = 1 in a neighborhood of p ζ. Suppose that ηu C l 1,σ 1 D3, ηp C l 2,σ 1 D, ηf Cl 2,σ D 3, ηg C l 1,σ D, ηh ± C Γ ± 3 d±, ηφ ± C l 1,σ Γ ± d±, where l 2, 1, 0 < σ < 1. Then ζu, p C D 3 C l 1,σ D. P r o o f. By Lemma 1.2, there are the representations ηu = u + u, ηp = p + p, where u N l 1,σ 1 D3, p N l 2,σ 1 D, u C D 3, and p C l 1,σ D. Let χ be a smooth cut-off function equal to one in a neighborhood of p ζ such that η = 1 in a neighborhood of p χ. Then χ u + p = χf + χ u p C l 2,σ D 3, χ u = χg + χ u C l 1 D. In the case l 2+σ, the space C l 2,σ D coincides with N l 2,σ D, and in the case 1 < l 2+σ we have l 3 and χ u + p N l 3,σ 1 D3. From Lemma 1.2 it follows that N l 3,σ 1 D Cl 2,σ D N l 2,σ D. Therefore in both cases, we obtain χ u + p N l 2,σ D 3. Analogously, χ u N l 1,σ D, χs ± u N Γ ± 3 d±, and χn ± u, p N l 1,σ Γ ± d±. This together with Lemma 2.7 implies ζu N D 3, ζp N l 1,σ D. The result follows. We define the operator Aλ as follows Aλ Uϕ, P ϕ = r 2 λ u + p, r 1 λ u, r λ S ± u ϕ=±θ/2, r 1 λ N ± u, p ϕ=±θ/2, where u = r λ Uϕ, p = r λ 1 P ϕ, λ C, r, ϕ are the polar coordinates of the point x = x 1, x 2. The operator Aλ depends quadratically on the parameter λ and realizes a continuous mapping W 2,s θ 2, + θ 2 3 W 1,s θ 2, + θ 2 W 1,s θ 2, + θ 2 3 L s θ 2, θ 2 C3 C 3 for every λ C. In [22] a description of the spectrum of the pencil Aλ is given for different d and d +. For example, in the cases of the Dirichlet problem d + = d = 0 and Neumann problem d + = d = 3, the spectrum of Aλ consists of the solutions of the equation sinλθ λ 2 sin 2 θ sin 2 λθ = 0, λ 0 for d + = d = 0. In the case d = 0, d + = 1, the eigenvalues of Aλ are the nonzero solutions of the equation sinλθ λ sin2θ + sin2λθ = 0. If d = 0, d + = 2, then the eigenvalues are the nonzero solutions of the equation while the nonzero solutions of the equation are eigenvalues of Aλ if d = 0 and d + = 3. sin2λθ λ sin2θ sin2λθ = 0, sin2λθ λ 2 sin 2 θ cos 2 λθ = 0 11

12 Lemma 2.9 Let ζ, η be the same functions as in Lemma 2.8, and let u, p be a solution of problem 2.8, 2.9 such that η x j 3 u, p N D 3 N l 1,σ D for j = 0 and j = 1, where l 2, 0 < σ < 1. If ηf N l 1,σ D 3, ηg N D, ηh ± N l+1,σ Γ ± 3 d±, ηφ ± N Γ ± d±, and the strip l + σ Re λ l σ does not contain eigenvalues of the pencil Aλ, then ζu, p N l+1,σ D 3 N D. P r o o f. Let χ be a smooth cut-off function such that χ = 1 in a neighborhood of p ζ and η = 1 in a neighborhood of p χ. We denote by x, x the Laplace and Nabla operators in the coordinates x = x 1, x 2. Then x χu 3 = F 3 = χ f 3 + x 2 3 u 3 x3 p 2 x χ x u 3 u 3 x χ N l 1,σ D. Furthermore, χu 3 satisfies the boundary conditions χu 3 Γ ± = H ± 3 N l+1,σ Γ ± for d ± 1, χu 3 n ± Γ ± = Φ± 3 N Γ ± for d ± 2 on Γ ±, where H ± 3 = χh± 3 and Φ± 3 = χ 4 d ± φ ± 3 n± x3 u +u 3 χ/ n ±. Analogously for u = u 1, u 2 and p, we obtain the equations x χu + x χp = F, x χu = G, S± χu Γ ± = H ±, Ñ ± χu, χp Γ ± = Φ ± with certain functions F N l 1,σ D 2, G N D, H ± N l+1,σ Γ ±, Φ ± N Γ ±, where S ± u = S ± u, 0 and Ñ ± u, p = N ± u, 0, p. Consequently, by [?, Th.8.4], we obtain χu, x 3 N l+1,σ 3, χp, x 3 N, and χu, x 3 N l+1,σ + χp, x 3 3 N l+1,σ c F, x 3 N l 1,σ + ± + G, x 3 3 N H N l+1,σ γ ± + ± Φ ± N γ ± with a constant c independent of x 3. From this and from the inclusions η x3 u N D 3, η x3 p N l 1,σ D we conclude that ζu, p N l+1,σ D 3 N D. We prove the analogous result for the spaces C. Lemma 2.10 Let ζ, η be the same functions as in Lemma 2.8, and let u, p be a solution of problem 2.8, 2.9 such that η x j 3 u, p C D 3 C l 1,σ D for j = 0 and j = 1, where l 2, 0 < σ < 1, σ is not integer. If ηf C l 1,σ D 3, ηg C D, ηh ± C l+1,σ Γ ± 3 d±, ηφ ± C Γ ± d±, and the strip l + σ Re λ l σ does not contain eigenvalues of the pencil Aλ, then ζu, p C l+1,σ D 3 C D. P r o o f. Suppose that k 1 < σ < k, where k is an integer, k l. Then both ηu and x3 ηu belong to C l k,k +σ D 3. Consequently, the traces u i,j of i x 1 j x 2 ηu on M are from C l k i j+1,k +σ M 3 for i + j l k. Analogously, the traces p i,j of i x 1 j x 2 ηp on M are from C l k i j,k +σ M for i + j l k 1. By Lemma 1.2, there are the representations ζu = ζ i+j l k Eu i,j i!j! x i 1 x j 2 + v, ζp = ζ i+j l k 1 Ep i,j i!j! x i 1 x j 2 + q, where E is the extension operator 1.3, x j 3 v N D 3, and x j 3 q N l 1,σ D for j = 0, 1. From the properties of the extension operator E it follows that ζu v C l+1,σ D 3, ζp q C D, 12

13 Therefore, v + q C l 1,σ C l+1,σ Γ ± d±. Since v N D 3, v C l 1,σ D 3 and q N l 1,σ D, S ± v C l+1,σ Γ ± 3 d±, and N ± v, q D, we have x α q v = 0 on M for α l k 2 and α x v = 0 on M for α l k 1. Furthermore, j rs ± v = 0 on M for j l k and j rn ± v, q = 0 on M for j l k 1. We put f i,j = i x 1 j x 2 q v M for i + j = l k 1, g i,j = i x 1 j x 2 v M for i + j = l k, H ± = r l k+1 S ± v M, and Φ ± = r l k N ± v, q M. Obviously f i,j, g i,j, H ±, Φ ± belong to the space C 0,k+σ M. By virtue of Lemma 1.2, we have η q v η S ± v i+j=l k 1 Ef i,j i! j! x i 1 x j 2 N l 1,σ D 3, η v + EH ± l k + 1! rl k+1 N l+1,σ Γ ± 3 d±, η i+j=l k Eg i,j i! j! x i 1 x j 2 N D, N ± v, q EΦ± l k! rl k N Γ ± d±. Since λ = l k + 1 is not an eigenvalue of the pencil Aλ, there exist homogeneous vector-valued polynomials U i,j,µ x 1, x 2 of degree l k + 1 and homogeneous polynomials P i,j,µ x 1, x 2 of degree l k, µ = 1, 2, 3, 4, i + j = l k 1 µ,4, satisfying the equations U i,j,µ ν + xν P i,j,µ = µ,ν x i 1 x j 2 i! j! for ν = 1, 2, 2, U i,j,µ = µ,4 x i 1 x j 2 i! j! and the boundary conditions S ± U i,j,µ = 0, N ± U i,j,µ, P i,j,µ = 0 on Γ ± see Remark 2.2. We define Ux = P x = µ=1 µ=1 i+j=l k 1 i+j=l k 1 U i,j,µ x Ef i,j µ x + P i,j,µ x Ef i,j µ x + i+j=l k 2 i+j=l k 2 U i,j,4 x Eg i,j x P i,j,4 x Eg i,j x Then ηu, P C l+1,σ D 3 C D, η x j 3 U, P N D 3 N l 1,σ D for j = 0, 1, S ± U = 0 on Γ ± η U v P q N l 1,σ D 3, η U v N D, η N ± U, P N Γ ±. For the last, we used the fact that η xν Ef i,j N +l k+1 D3 and η xν Eg i,j N ν = 1, 2, 3. Analogously there exist functions V and Q such that ηv, Q C l+1,σ N D 3 N l 1,σ D for j = 0, 1, η V Q N l 1,σ D 3, η V N D η S ± V v N l+1,σ Γ ±, η N ± V v, Q q N Γ ±. +l k+1 D for D 3 C D, η x j 3 V, Q Applying Lemma 2.9 to the vector-function U + V v, P + Q q, we obtain χu + V v, P + Q q N l+1,σ D 3 N D, where χ is the same cut-off function as in the proof of Lemma 2.9. The result follows. 3 Solvability of the problem in a cone Let d j {0, 1, 2, 3} for j = 1,..., N. We consider the boundary value problem u + p = f u = g in, 3.1 S j u = h j, N j u, p = φ j on Γ j, j = 1,..., N,

14 where u for d j = 0, S j u = u u n n for d j = 1, u n for d j = 2, p + 2ε nn u for d j = 1, N j u, p = ε n u ε nn u n for d j = 2, pn + 2ε n u for d j = 3. We will prove that this problem is uniquely solvable in C 2,σ 3 C 1,σ under certain conditions on the data g, h j, φ j and on β and. 3.1 Operator pencils 1 Let Γ k± be the sides of adjacent to the edge M k, and let θ k be the angle at the edge M k. We consider the Stokes system in the dihedron D k bounded by the half-planes Γ k ± Γ k± with the boundary conditions S k± u = h ±, N k± u, p = φ ± on Γ k ±. By A k λ we denote the operator pencil introduced before Lemma 2.9 for this problem. Furthermore, let λ k 1 denote the eigenvalue with smallest positive real part of this pencil, while λ k 2 is the eigenvalue with smallest real part greater than 1. Finally, we define { Re λ k µ k = 1 if d k+ + d k is odd or d k+ + d k is even and α k π/m k, Re λ k if d k+ + d k is even and θ k < π/m k, where m k = 1 if d k+ = d k, m k = 2 if d k+ d k. 2 Let ρ =, ω = x/, V Ω = {u W 1 Ω 3 : S j u = 0 on γ j for j = 1,..., n}, and u a p v, q ; λ = 1 log 2 1<<2 2 i, ε i,j U ε i,j V P V U Q dx, where U = ρ λ uω, V = ρ 1 λ vω, P = ρ λ 1 pω, Q = ρ 2 λ qω, u, v V Ω, p, q L 2 Ω, and λ C, ε i,j U = 1 2 xi U j + xj U i. The bilinear form a, ; λ generates the linear and continuous operator Aλ : V Ω L 2 Ω V Ω L 2 Ω by Ω u Aλ p v q u dω = a p v, q ; λ, u, v V Ω, p, q L 2 Ω. 3.2 Reduction to homogeneous boundary conditions Lemma 3.1 Let h j N Γ j 3 d j, φ j N l 1,σ Γ j d j, j = 1,..., N, l 1, be given. Then there exists a vector function u N 3 such that S j u = h j and N j u, 0 = φ j on Γ j. The norm of u can be estimated by the norms of h j and φ j. P r o o f. Let ζ k be smooth functions on 0, such that We set p ζ k 2 k 1, 2 k+1, h k,j x = ζ k 2 k h j 2 k x, j ρζ k ρ c 2 kj, and + k= φ k,j x = 2 k ζ k 2 k φ j 2 k x. ζ k = The ports of h k,j and φ k,j are contained in {x : 1/2 < < 2}. Consequently, by Lemma 2.1, there exists a vector function v k N 3 such that v k x = 0 for < 1/4 and > 4, S j v k = h k,j on N j v k, 0 = h k,j on Γ j, j = 1,..., N, v k N c 3 N h k,j N Γj3 d j + φ k,j N l 1,σ Γ j d j

15 where c is independent of k. From this we conclude that the functions u k x = v k 2 k x satisfy S j u k x = ζ k g j x, N j u k, 0 = ζ k φ j x on Γ j, and the estimate 3.5 with ζ k h j, ζ k φ j instead of h k,j and φ k,j, respectively. Thus, u = u k has the desired properties. An analogous result in C is only valid under additional compatibility conditions on the boundary data. Denote by Γ k+ and Γ k the sides of the cone adjacent to the edge M k and by θ k the inner angle at M k. If u C and k < l + σ, then the trace of u on M k exists and from the equations S j u = h j on Γ j it follows that the pair h k+ Mj, h k Mj belongs to the range of the matrix operator Sk+, S k. This condition can be also written in the form A + k h k + Mk = A k h k Mk, 3.6 where A + k, A k are certain constant matrices see Section 2.1. Using Lemma 2.4, one can prove the following result analogously to Lemma 3.1. Lemma 3.2 Let h j C Γ j 3 dj, φ j C l 1,σ Γ j dj, f C l 2,σ 3, and g C l 1,σ, where l 1, 0 k < l + σ, k σ not integer for k = 1,..., N in the case l = 1 the condition on f can be omitted. Suppose that the boundary data h j satisfy the compatibility condition 3.6 and that in the case k < l 1 + σ the functions g, h k±, φ k± satisfy the compatibility conditions given in Lemma 2.3. Furthermore, we assume that the numbers 2, 3,..., [l + σ k ] do not belong to the spectrum of the pencil A k λ if k < l 2 + σ. Then there exists a vector function u, p C 3 C l 1,σ satisfying S j u = h j, N j u, p = φ j on Γ j, j = 1,..., n, u p+f N l 2,σ 3, u+g N l 1,σ 3.7 The norms of u and p can be estimated by the norms of f, g, h j and φ j. Note again that the condition of Lemma 2.3 is always satisfied if d k+ + d k = 3 and sin2θ k 0 or d k+ + d k {1, 5} and cos θ k cos2θ k 0. For even d k+ + d k, one can find explicit conditions on g Mk, h k± Mk, and φ k± Mk for different combinations of boundary conditions on Γ k+ and Γ k, see Remark Regularity result for solutions of the boundary value problem Lemma 3.3 Let u, p W 2,s loc \S 3 W 1,s loc \S be a solution of problem 3.1, 3.2 such that l σ rk x k l σ ux + l+1 σ rk x k l+1 σ px <. If f N l 2,σ 3, l 2, g N l 1,σ, h j N Γ j 3 d j, φ j N l 1,σ Γ j d j, j = 1,..., N, then u N 3 and p N l 1,σ. P r o o f. Due to Lemma 3.1, we may assume without loss of generality that h j = 0 and φ j = 0. From Lemma 2.7 it follows that ζu N 3 and ζp N l 1,σ for every smooth function ζ with compact port vanishing in a neighborhood of the origin. Let ρ be a positive integer, ρ = {x : ρ/2 < < 2ρ}, and ρ = {x : ρ/4 < < 4ρ}. Furthermore, let ũx = uρx, px = ρ pρx, fx = ρ 2 fρx, and gx = ρ gρx. Then ũ+ p = f and ũ = g in. Moreover ũ and p satisfy the homogeneous boundary conditions 3.2. Consequently, by Lemma 2.7, we have ũ N + p 13 N l 1,σ 1 c f N l 2,σ + g 1 3 N l 1,σ 1 + ũ N 0 β l σ, l σ 1 + p 3 Nβ l+1 σ, l+1 σ with a constant c independent of u, p and ρ. Here the norm in N ρ is defined by 1.6, where has to be replaced by ρ. Since ũ N = ρ l+σ β u 1 3 N, ρ 3 15

16 we obtain an analogous estimate for the norms of u and p in N ρ 3 and N l 1,σ ρ, respectively. Here, the constant c is the same as in 3.8. The result follows. In the same way, the following two lemmas can be proved using Lemmas 2.8 and 2.9. Lemma 3.4 Let u, p W 2,s loc \S 3 W 1,s loc \S be a solution of problem 3.1, 3.2. Suppose that u C l 1,σ β 1, 1 3, p C l 2,σ β 1, 1, f Cl 2,σ 3, g C l 1,σ, h j C Γ j 3 dj, φ j C l 1,σ Γ j dj, where l 2, k 1 for k = 1,..., N, 0 < σ < 1. Then u, p C 3 C l 1,σ. Lemma 3.5 Let u, p be a solution of problem 3.1, 3.2 such that ρ ρ j u, p C 3 C l 1,σ for j = 0, 1, where l 2, 0 < σ < 1. If f C l 1,σ 3, g C, h j C l+1,σ Γ j 3 d j, φ j C Γ j d j, j = 1,..., N, and the strip l + σ k Re λ l σ k does not contain eigenvalues of the pencil A k λ, k = 1,..., n, then u, p C l+1,σ 3 C. 3.4 Estimates of Green s matrix We denote by Vκ l the weighted Sobolev space wit the norm 1/2. u V l κ = 2κ l+ α x α ux dx 2 α l A matrix Gx, ξ = G i,j x, ξ 4 is called Green s matrix for problem 3.1, 3.2 if i, x Gj x, ξ + x G 4,j x, ξ = x ξ 1,j, 2,j, 3,j t for x, ξ, 3.9 x G j x, ξ = 4,j x ξ for x, ξ, 3.10 S k Gj x, ξ = 0, N k x Gj x, ξ, G 4,j x, ξ = 0 for x Γ k, ξ, k = 1,..., N Here G j denotes the vector with the components G 1,j, G 2,j, G 3,j. Suppose that the line Re λ = κ 1/2 is free of eigenvalues of the pencil Aλ. Then, by [22, Th.4.5], there exists a unique Green matrix Gx, ξ such that the function x ζ x ξ /rξ G i,j x, ξ belongs to V 1 κ for i = 1, 2, 3 and to V 0 κ for i = 4, where ζ is an arbitrary smooth function on 0, equal to one in 1, and to zero in 0, 1 2. We denote by Λ < Re λ < Λ + the widest strip in the complex plane which is free of eigenvalues of the pencil Aλ and which contains the line Re λ = κ 1/2. Furthermore, we introduce the following notation. σ k,i,α = min0, µ k α i,4 ε, σ i,α x = min0, µ x α i,4 ε Here ε is an arbitrarily small positive real number, µ x = µ kx, and kx is the smallest integer k such that rx = r k x. For the following theorem we refer to [22, Th.4.5,Th.4.6]. Theorem 3.1 Let Gx, ξ be the above introduced Green matrix. Then for > 2 ξ there is the estimate α x γ ξ G i,jx, ξ c Λ i,4 α +ε ξ Λ 1 j,4 γ ε For ξ > 2 we have α x γ ξ G i,jx, ξ c Λ + i,4 α ε ξ Λ + 1 j,4 γ +ε rk x σk,i,α N rk ξ σk,j,γ, ξ rk x σk,i,α N rk ξ σk,j,γ, ξ 16

17 while for /2 < ξ < 2 the estimates x α γ ξ G i,jx, ξ T α γ c x ξ if x ξ < minrx, rξ, x α γ ξ G i,jx, ξ c x ξ T α γ rx σi,α x rξ x ξ x ξ σj,γ ξ if x ξ > minrx, rξ are valid, where T = 1 + i,4 + j,4. Furthermore, for i = 1,..., 4 there is the representation G i,4 x, ξ = ξ P i x, ξ + Q i x, ξ, where P i x, ξ n = 0 for ξ Γ k, x D, and P i and Q i satisfy the estimates α x γ ξ P i x, ξ c α,γ x ξ 1 i,4 α γ, α x γ ξ Q ix, ξ c α,γ rξ 2 i,4 α γ for x ξ < minrx, rξ. Remark 3.1 For derivatives with respect to ρ = there are the sharper estimates ρ α x γ ξ G i,jx, ξ c Λ 1 i,4 α +ε ξ Λ 1 j,4 γ ε ρ α x γ ξ G i,jx, ξ c Λ + i,4 α ε ξ Λ + 1 j,4 γ +ε rk x σk,i,α N rk ξ σk,j,γ, ξ σk,i,α N σk,j,γ rk x rk ξ ξ if ξ < /2 and ξ > 2, respectively. For /2 < ξ < 2, x ξ > minrx, rξ, the estimate is valid. ρ x α γ ξ G i,jx, ξ c x ξ T 1 α γ rx σi,α x rξ σj,γ ξ x ξ x ξ 3.5 Representation of the solution by Green s matrix Suppose that f N 0,σ 1,σ, g N, where β R and = 1,..., N [0, N are such that 2 µ k < k σ < 2, k σ not integer for k = 1,..., N, 3.12 Λ < 2 + σ β < Λ Here Λ +, Λ are the same numbers as in Theorem 3.1. We consider the functions u i x = px = gx + fj ξ + ξj gξ G i,j x, ξ dξ + gξ G i,4 x, ξ dξ, i = 1, 2, 3, 3.14 fj ξ + ξj gξ G 4,j x, ξ dξ + gξ G 4,4 x, ξ dξ The vector-function u, p is a solution of problem 3.1, 3.2 with h j = 0, φ j = 0 see [22, Th.4.5]. Let χ be an arbitrary smooth cut-off function on [0,, χt = 1 for t < 1/4, χt = 0 for t > 1/2. We put x ξ χ + x, ξ = χ, χ x, ξ = 1 χ + x, ξ. rx Then χ + x, ξ = 0 for x ξ > rx/2, χ x, ξ = 0 for x ξ < rx/4, and α x χ ± x, ξ c rx α with a constant c independent of x and ξ. We write u nd p in the form u = u + + u, p = p + + p, 17

18 where u ± i x = p ± x = gx 2 fj ξ + ξj gξ χ ± x, ξ G i,j x, ξ dξ + gξ χ ± x, ξ G i,4 x, ξ dξ, Weighted L estimates for u +, p + fj ξ + ξj gξ χ ± x, ξ G 4,j x, ξ dξ + gξ χ ± x, ξ G 4,4 x, ξ dξ We consider the functions u +, p + defined by 3.16, 3.17, where Gx, ξ is the Green matrix introduced in Section 3.4. Lemma 3.6 Suppose conditions 3.12, 3.13 are satisfied. Then for arbitrary f N 0,σ 3 and g N 1,σ, there are the estimates β 2 σ N β 1 σ rk x k 2 σ rk x i=1 k 1 σ i,ν=1 u + i f x c N 0,σ + g 3 N 1,σ, 3.18 xν u + i x + p+ x c f N 0,σ + g 3 N 1,σ.3.19 P r o o f. We have xν u + i x A + B, where A = fj ξ + ξj gξ xν χ + x, ξ G i,j x, ξ dξ B = xν gξ χ + x, ξ G i,4 x, ξ dξ On the port of χ + there are the inequalities /2 ξ 3/2, r k x/2 r k ξ 3r k x/2. This together with Theorem 3.1 implies N A c σ β rk x σ k f N 0β σ, σ + g 3 N 0β σ, σ x ξ 2 dξ x ξ <rx/2 c σ β+1 rk x σ k +1 f N 0β σ, σ + g 3 N 0β σ, σ Using the representation G i,4 x, ξ = ξ P i x, ξ+q i x, ξ in Theorem 3.1 and the properties of P i, Q i, we obtain B = xν gξ ξ χ + x, ξ P i x, ξ + Q i x, ξ + χ + x, ξ ξ gξ P i x, ξ dξ c σ β+1 + c σ β c σ β+1 N rk x σ k +1 g N 0 β 1 σ, 1 σ rk x σ k g N 0 β σ, σ x ξ <rx/2 x ξ <rx/2 and rx 3 dξ x ξ 2 dξ rk x σ k +1 g N 0β 1 σ, 1 σ + g 3 N 0β σ, σ. This proves the desired estimate for u. Analogously, the estimates for u and p hold. 18

19 3.7 Weighted L estimates for the derivatives of u, p Next we show that and β 2 σ+ α β 1 σ+ α rk x max0,k 2 σ+ α x α u x c f N 0,σ + g 3 N 1,σ rk x max0,k 1 σ+ α x α p x c f N 0,σ + g 3 N 1,σ for an arbitrary multi-index α. For this we need the following lemmas Lemma 3.7 Let f N t,σ and vx = ξ </2 x, ξ fξ dξ, where the kernel satisfies the estimate x, ξ c Λ s+ε ξ Λ 1 t ε rk x min0,µk s ε N rk ξ min0,µk t ε. ξ with nonnegative integers s, t. Suppose that β, satisfy conditions 3.12, Then β 2+s σ with a constant c independent of x. P r o o f. Obviously vx c Λ s+ε rk x max0,k 2+s σ vx c f N 0 β t σ, t σ 3.22 ξ </2 rk x min0,µk s ε f N 0 β t σ, t σ ξ Λ 1 β+σ ε rk ξ min0,µk t ε k +t+σ dξ 3.23 ξ From the conditions on β and it follows that Λ 1 β+σ > 3 and min0, µ k t ε k +t+σ > 2. Hence the integral on the right-hand side of 3.23 is equal to c Λ +2 β+σ ε. Therefore, vx c β+2 s+σ rk x min0,µk s ε f N 0 β t σ, t σ Using the inequality min0, µ k s ε min0, 2 k + σ s = max0, k 2 + s σ, we obtain the desired estimate for v. Analogously, the following lemma can be proved. Lemma 3.8 Let f N t,σ and vx = ξ >2 x, ξ fξ dξ, 19

20 where the kernel satisfies the estimate x, ξ c Λ + s+ε ξ Λ + 1 t ε rk x min0,µk s ε N rk ξ min0,µk t ε. ξ with nonnegative integers s, t. Suppose that β, satisfy conditions 3.12, Then 3.22 is valid with a constant c independent of x. Note that the functions α x G i,j x, ξ satisfy the assumption on the kernel in Lemmas 3.7 and 3.8 with s = α + i,4, t = j,4. For the proof of an analogous estimate for the integral over the set {x : /2 < ξ < 2}, we need the following lemma. Lemma 3.9 Let D be a dihedron, x D and R rx/4. Then x ξ α rξ dξ c R 3 α if α + < 3, < 2, 3.24 D rx/4< x ξ <R D x ξ >R Here the constant c is independent of x and R. x ξ α rξ dξ c R 3 α if α + > 3, < P r o o f. 1 We denote the left-hand side of 3.24 by A. Substituting x/rx = y and ξ/rx = η, we obtain A = rx 3 α y η α rη dη, where ry = 1. Obviously the integral D 1/4< y η <R/rx D 1/4< y η <2 y η α rη dη is finite and can be estimated by a constant independent of y. We denote by y = 0, 0, y 3 the orthogonal projection of y onto the edge M. Then, for 2 < y η < R/rx, we have 1 < η y < 1 + R/rx, 2 η y /3 < y η < 2 η y and, therefore, 2< y η <R/rx y η α rη β dη c 1< η y <1+R/rx R 3 α. η y α rη y dη c rx This proves If 0, then α > 3 and rξ c rx + x ξ c R + x ξ, and x ξ α rξ dξ c x ξ α R + x ξ α dξ c R 3 α. x ξ >R x ξ >R Let 0. Then D rξ> x ξ >R x ξ α rξ dξ c D x ξ >R x ξ α dξ c R 3 α. 20

21 If x ξ > rξ > R, then ξ x rξ > R and ξ x ξ ξ + ξ x rξ+ ξ x < 2 ξ x. Here again x = 0, 0, x 3, ξ = 0, 0, ξ 3 denote the nearest points to x and ξ on M. Therefore, x ξ α rξ dξ c ξ x α rξ x dξ c R 3 α. D x ξ >rξ>r D ξ x >R Finally, since 2 x ξ > R + x 3 ξ 3 for x ξ > R, we obtain D x ξ >R>rξ x ξ α rξ dξ c + R + x3 ξ 3 R α dξ3 r 1 dr = c R 3 α. 0 This proves Lemma 3.10 Let f N t,σ and vx = /2< ξ <2 x, ξ fξ dξ, where x, ξ vanishes for x ξ < rx/4 and satisfies the estimate x, ξ c x ξ 1 s t rx min0,µx s ε rξ min0,µξ t ε. x ξ x ξ with nonnegative integers s, t. Suppose that satisfy condition Then 3.22 is valid with a constant c independent of x. P r o o f. Obviously, vx c σ+t β f /2< ξ <2 x, ξ rk ξ σ+t k dξ ξ here by f, we mean the Nβ t σ, t σ 0 -norm of f. If rx > /2, then /2 < r kx < for k = 1,..., N and /8 < x ξ < 3 for all ξ satisfying the inequalities /2 < ξ < 2, x ξ > rx/4. This implies vx c 1 β s+σ f /2< ξ <2 Since σ + t k + min0, µ k t ε σ k > 2, it follows that vx c 2 β s+σ f c 2 β s+σ f rk ξ σ+t k +min0,µ k t ε dξ ξ rk x max0,k 2+s σ. This proves 3.22 for the case rx > /2. Suppose now that rx = r 1 x /2 and fξ = 0 for r 1 ξ < 2rξ. Then there exist positive constants c 1, c 2 such that c 1 < x ξ < 3, rx min0,µx s ε N rk x min0,µk s ε c2 x ξ 21

22 for ξ p x, f, /2 < ξ < 2. From this and from the inequality µ k > 2 k + σ we conclude that vx c 1 β s+σ rk x min0,µk s ε f /2< ξ <2 c 2 β s+σ f rk ξ ξ σ+t k +min0,µ k t ε dξ rk x min0,2 s k +σ what implies Next we pose that rx = r 1 x /2 and fξ = 0 for r 1 ξ > 3rξ. Then there exist positive constants c 1, c 2 such that rξ min0,µξ t ε r1 ξ min0,µ1 t ε N c1, x ξ x ξ for ξ p x, f, /2 < ξ < 2. Thus, rk ξ ξ σ+t k r1 ξ σ+t 1 c ξ vx c 1 β r 1 x min0,µ 1 s ε f x ξ 1 s t min0,µ 1 s ε min0,µ 1 t ε r 1 ξ σ+t 1+min0,µ 1 t ε dξ, 3.26 x where x denotes the domain {ξ : /2 < ξ < 2, r 1 ξ < 3rξ, x ξ > rx/4}. Let 1 σ > 2 s. Then according to 3.12, we have 1 s min0, µ 1 s ε + σ 1 = 3 + max2 s, 2 µ 1 + ε + σ 1 < 3. Hence, using 3.25 with R = r 1 x/4, we can estimate the integral on the right of 3.26 by what implies c r 1 x 2 s min0,µ 1 s ε+σ 1 vx c 1 β r 1 x 2 1 s+σ f = c 2 β s+σ r 1 x 2 1 s+σ f. If 2 µ 1 < 1 σ < 2 s, then 1 s min0, µ 1 s ε + σ 1 = 1 s + σ 1 > 3. Consequently, using 3.24 with R = 3, the integral on the right of 3.26 can be estimated by c 2 1 s+σ from what we obtain vx c 2 β s+σ f = c 2 β s+σ r 1 x min0,2 1 s+σ f Thus, both in the cases 1 σ > 2 s and 1 σ < 2 s, the estimate 3.22 follows. It remains to note that every f N t,σ can be written as a sum f = f 1 + f 2 such that f 1 ξ = 0 for r 1 ξ < 2rξ, f 2 ξ = 0 for r 1 ξ > 3rξ and f 1 N 0 β t σ, t σ + f 2 N 0 β t σ, t σ c f N 0 β t σ, t σ This completes the proof. Theorem 3.2 Let f N 0,σ 3, g N 1,σ, and let u, p be defined by 3.16, 3.17, where Gx, ξ is the Green matrix introduced in Section 3.4. Suppose that β, satisfy conditions 3.12 and Then u, p satisfy 3.20, P r o o f. It is sufficient to note that x, ξ = α x χ x, ξg i,j x, ξ satisfies the conditions of Lemmas 3.7, 3.8 and 3.10 with s = α + i,4, t = j,4 for ξ < /2, ξ > 2 and /2 < ξ < 2, respectively. Hence, the result follows immediately from the representation of u, p and from Lemmas 3.7, 3.8 and

23 3.8 Hölder estimates for u, p Let ν = {x : r ν x < 3rx/2} for ν = 1,..., N. Lemma 3.11 Let f N t,σ and vx = x, ξ fξ dξ. We assume that β, satisfy conditions 3.12, 3.13, x, ξ vanishes for x ξ > rx/4 and that the estimates s ρx, ξ c x ξ 3 s t+kν rξ x ξ ρ x, ξ c Λ 3+k ν +ε ξ Λ 1 t ε ρ x, ξ c Λ + 3+k ν ε ξ Λ + 1 t+ε min0,µξ t ε for x ν, /4 < ξ < 2, s = 0, 1, rk ξ min0,µk t ε ξ rk ξ min0,µk t ε ξ are valid, where k ν = 1 + [ ν σ] and ρ =. Then there ist the estimate β ν for x ν, ξ < 3/4, for x ν, ξ > 3/2 vx vy x y σ ν+kν c f N 0 β t σ, t σ 3.27 for x ν, y = τx, 4/5 < τ < 5/4, x y > rx/4, where c is independent of f, x, τ. P r o o f. Let x ν and y = τx, 4/5 < τ < 5/4, x y > rx/4. Then 4 x y < min, y. We have vx vy A 1 + A 2 + A 3, where A 1 = x, ξ fξ dξ, A 2 = y, ξ fξ dξ, A 3 = ξ x <2 x y ξ x >2 x y x, ξ y, ξ fξ dξ, ξ x <2 x y From x ξ < 2 x y < /2 it follows that /2 < ξ < 3/2. Therefore, A 1 c f N 0 β t σ, t σ t+σ β rξ min0,µξ t ε x ξ 3 t+kν x ξ rk ξ t+σ k dξ, ξ where the domain of integration is contained in the set of all ξ satisfying the inequalities /2 < ξ < 2, rx/4 < x ξ < 2 x y. Since k ν + σ ν > 0 and min0, µ k t ε + t + σ k > 2, by virtue of 3.24, we obtain A 1 c f N 0 β t σ, t σ ν β x y σ ν+k ν. Analogously, this estimate holds for A 2. For the proof, one can use the fact that ξ x < 2 x y implies ξ y < 3 x y and y /4 < ξ < 2 y. We consider the expression A 3. By the mean value theorem, there is the inequality x, ξ y, ξ ρ x, ξ x y, where x is a certain point on the line between x and y, i.e. 4/5 < x < 5/4. Hence, A 3 x y ρ x, ξ fξ dξ. ξ x >2 x y 23

24 Here, for the integral over the set of all ξ such that ξ < /2, we obtain ρ x, ξ fξ dξ c f N 0 β t σ, t σ Λ 3+k ν +ε c f N 0 β t σ, t σ σ β+kν 1. ξ σ β Λ 1 ε n rk ξ t+σ k +min0,µ k t ε dξ ξ The same estimate holds for the integral over the set of all ξ such that ξ > 2, while for the integral over the set of all ξ satisfying /2 < ξ < 2 and ξ x > 2 x y we have ρ x, ξ fξ dξ c f N 0 β t σ, t σ t+σ β c f N 0 β t σ, t σ ν β x y σ ν+k ν 1. x ξ 4 t+k ν rξ min0,µξ t ε x ξ n rk ξ t+σ k dξ ξ Here, we used the inequality x ξ /2 < x ξ < 3 x ξ /2 and the estimate 3.25 with R = 2 x y. Thus, we obtain A 3 c f N 0 β t σ, t σ x y σ β+kν 1 + ν β x y σ ν+kν 1 c f N 0 β t σ, t σ ν β x y σ ν+kν. This proves the lemma. Theorem 3.3 Let the condition of Theorem 3.2 be satisfied. Then u C 2,σ 3, p C 1,σ, and there is the inequality u C 2,σ + p 3 C 1,σ c f N 0,σ + g 3 N 1,σ 3.28 with a constant c independent of f and g. P r o o f. According to Theorem 3.2, the vector function u satisfies We show that β rk x k x α u x y α u y x y σ c f N 0,σ + g 3 N 1,σ 3.29 for α = 2, x y < rx/2. By the mean value theorem, we have α ux α uy = x y α u x, where x = x + ty x, t 0, 1. Furthermore, for x y < rx/2, there are the inequalities /2 < x < 3/2, r j x/2 < r j x < 3r j x/2, j = 1,..., N. From this and from 1.5 it follows that β rk x k x α u x y α u y x y σ β c x β+1 σ rk x krx 1 σ α u x rk x k 1+σ α u x x for α = 2. This together with 3.20 implies Analogously, the estimate α β ν x u x y α u y x y k ν+σ ν c f N 0,σ + g 3 N 1,σ