Reflection principles and kernels in R n _+ for the biharmonic and Stokes operators. Solutions in a large class of weighted Sobolev spaces

Transcription

1 Refection principes and kernes in R n _+ for the biharmonic and Stokes operators. Soutions in a arge cass of weighted Soboev spaces Chérif Amrouche, Yves Raudin To cite this version: Chérif Amrouche, Yves Raudin. Refection principes and kernes in R n _+ for the biharmonic and Stokes operators. Soutions in a arge cass of weighted Soboev spaces. 30 pages <ha v2> HAL Id: ha Submitted on 6 Jan 2010 HAL is a muti-discipinary open access archive for the deposit and dissemination of scientific research documents, whether they are pubished or not. The documents may come from teaching and research institutions in France or abroad, or from pubic or private research centers. L archive ouverte puridiscipinaire HAL, est destinée au dépôt et à a diffusion de documents scientifiques de niveau recherche, pubiés ou non, émanant des étabissements d enseignement et de recherche français ou étrangers, des aboratoires pubics ou privés.

2 Advances in Differentia Equations Voume 15, Numbers , REFLECTION PRINCIPLES AND KERNELS IN FOR THE BIHARMONIC AND STOKES OPERATORS. SOLUTIONS IN A LARGE CLASS OF WEIGHTED SOBOLEV SPACES Chérif Amrouche and Yves Raudin Laboratoire de Mathématiques Appiquées, CNRS UMR 5142 Université de Pau et des Pays de Adour IPRA, Avenue de Université, Pau, France Submitted by: Yoshigazu Giga Abstract. In this paper, we study the Stokes system in the haf-space, with n 2. We consider data and give soutions which ive in weighted Soboev spaces, for a whoe scae of weights. We start to study the kernes of the biharmonic and Stokes operators. After the centra case of the generaized soutions, we are interested in strong soutions and symmetricay in very weak soutions by means of a duaity argument. 1. Introduction The purpose of this paper is the resoution of the Stokes system u + π = f in, S + div u = h in R n +, u = g on Γ R n 1. Weighted Soboev spaces provide a functiona framework quite suitabe to express the reguarity and the behavior at infinity of data and soutions. This paper is the continuation of a previous work in which we ony deat with the basic weights see [8]. Here, we are interested in a arge cass of weights. This eads us to dea with the kerne of the operator associated to this probem and symmetricay with the compatibiity condition for the data. So, an important part of this work is devoted to the study of the refection principes for the biharmonic and Stokes operators. We give weak formuations of these principes with the aim of getting the kernes in some distribution spaces see Section 2. The main resuts of [8] wi be naturay incuded in this paper, but we wi not discuss again these particuar cases. Accepted for pubication: Juy AMS Subject Cassifications: 35J50, 35J55, 35Q30, 76D07, 76N

3 202 Chérif Amrouche and Yves Raudin We wi aso base our work on the previousy estabished resuts on the harmonic and biharmonic operators see [4], [5], [6], [7]. Among the first works on the Stokes probem in the haf-space, we can cite Cattabriga. In [10], he appeas to potentia theory to expicity get the veocity and pressure fieds. For the homogeneous probem f = 0 and h = 0, for instance, he shows that, if g L p Γ and the semi-norm g 1 1/p, p W 0 Γ <, then u Lp and π L p. We can find simiar resuts in Farwig-Sohr see [15] and Gadi see [16], who aso have chosen the setting of homogeneous Soboev spaces. On the other hand, Maz ya-pamenevskiĭ-stupyais see [18] work within the suitabe setting of weighted Soboev spaces and consider different sorts of boundary conditions. However, their resuts are imited to the dimension 3, to the weight zero and to the Hibertian framework, in which they give generaized and strong soutions. This is aso the case with Boumezaoud see [9], who ony gives strong soutions; however, he suggests an interesting characterization of the kerne that we wi get here in another way. Otherwise, aways in dimension 3, by Fourier anaysis techniques, we can find in Tanaka the case of very reguar data, corresponding to veocities which beong to W m+3, 2 2 R 3 +, with m 0 see [19]. For any integer n 2, writing a typica point R n as x = x, x n, we denote by the upper haf-space of R n and Γ its boundary. We sha use the two basic weights = 1 + x 2 1/2 and g = n2 + x 2, where x is the Eucidean norm of x. For any integer q, P q stands for the space of poynomias of degree smaer than or equa to q; Pq respectivey Pq 2 is the subspace of harmonic respectivey biharmonic poynomias of P q ; A q respectivey Nq is the subspace of poynomias of Pq, odd respectivey even with respect to x n, or equivaenty, which satisfy the condition ϕx,0 = 0 respectivey n ϕx, 0 = 0, with the convention that these spaces are reduced to {0} if q < 0. For any rea number s, we denote by [s] the integer part of s. Given a Banach space B, with dua space B and a cosed subspace X of B, we denote by B X the subspace of B orthogona to X. For any k Z, we sha denote by {1,...,k} the set of the first k positive integers, with the convention that this set is empty if k is nonpositive. In the whoe text, bod characters are used for the vector and matrix fieds. Let Ω be an open set of R n. For any m N, p 1,, α, β R 2, we define the foowing space: W m, p α, β Ω = {u D Ω : 0 λ k, α m+ λ g β 1 λ u L p Ω;

4 Refection principes and kernes in 203 } k + 1 λ m, α m+ λ g β λ u L p Ω, 1.1 where k = m n/p α if n/p + α {1,...,m}, and k = 1 otherwise. In the case β = 0, we simpy denote the space by Wα m, p Ω. Note that Ω is a refexive Banach space equipped with the graph-norm. Now, W m, p α, β we define the space W m, p α, β Rn + = DR n m, p W α, β Rn + +, and its dua is denoted m, p by W α, β Rn +. In order to define the traces of functions of Wα m, p here we don t consider the case β 0, for any σ 0,1, we introduce the space W σ, p α R n = { u D R n : w α σ u L p R n, Rn Rn αx ux αy uy p } x y n+σp dxdy <, where w = if n/p + α σ and w = g 1/σ α if n/p + α = σ. For any s R +, we set { Wα s, p R n = u D R n : 0 λ k, α s+ λ g 1 λ u L p R n ; } k + 1 λ [s] 1, α s+ λ λ u L p R n ; [s] u Wα σ, p R n, where k = s n/p α if n/p + α {σ,...,σ + [s]}, with σ = s [s] and k = 1 otherwise. In the same way, we aso define, for any rea number β, the space W s, p α, β Rn = { v D R n : g β v Wα s, p R n }. Let us reca, for any integer m 1 and any rea number α, the foowing trace emma. Lemma 1.1. For any integer m 1 and rea number α, we have the inear continuous mapping m 1 γ = γ 0, γ 1,...,γ m 1 : Wα m, p Wα m j 1/p, p R n 1. j=0 Moreover, γ is surjective and Kerγ = W m, p α. On the Stokes probem in R n S : u + π = f and div u = h in R n, et us reca the fundamenta resuts on which we are based in the seque. First, for any k Z, we introduce the space S k = { λ, µ P k P k 1 : div λ = 0, λ + µ = 0}.

5 204 Chérif Amrouche and Yves Raudin Theorem 1.2 Aiot-Amrouche [2]. Let Z and assume that n/p / {1,...,} and n/p / {1,..., }. For any f, h W 1, p R n W 0, p R n S [1+ n/p ], probem S admits a soution u, π W 1, p R n W 0, p R n, unique up to an eement of S [1 n/p], with the estimate inf u + λ λ, µ S W 1, p [1 n/p] R n + π + µ W 0, p R n C f W 1, p R n + h W 0, p. R n Theorem 1.3 Aiot-Amrouche [2]. Let Z and m 1 be two integers and assume that n/p / {1,..., + 1} and n/p / {1,..., m}. For any f, h W m 1, p m+ R n W m, p m+ Rn S [1+ n/p ], probem S admits a soution u, π W m+1, p m+ R n W m, p m+ Rn, unique up to an eement of S [1 n/p], with the estimate inf u + λ λ, µ S W m+1, p [1 n/p] m+ R n + π + µ W m, p m+ Rn C f W m 1, p m+ R n + h W m, p m+ Rn. 2. Refection principes and kernes in The aim of this section is to characterize the kerne of the Stokes operator with Dirichet boundary conditions in the haf-space. In this geometry, the natura way is to use a refection principe simiar to the we-known Schwarz refection principe for harmonic functions. Since, in the Stokes system, the veocity fied is biharmonic and the pressure is harmonic, it is reasonabe to start with the refection principe for the biharmonic functions. Let us notice that R. Farwig gives these continuation formuae in [14], referring to eiptic reguarity theory see Agmon, Dougis and Nirenberg, [1]. Let us especiay quote R. J. Duffin, who first estabished in [13] the continuation formua of biharmonic functions in the three-dimensiona case and then anaogous formuae for the Stokes fow equations. Next, A. Huber extended in [17] this principe to poyharmonic functions. From the cassica point of view, the ony serious difficuty is the argument at the boundary. Starting with the biharmonic operator, we wi give a weak formuation of the continuation formua, which wi aow us to characterize the kerne of this operator, even for very weak soutions. At first, et us introduce a usefu notation. For any function ϕ defined on an open set Ω of R n, we wi denote by ϕ, the composite function ϕ = ϕ r

6 Refection principes and kernes in 205 defined on Ω = rω, of ϕ with the C -diffeomorphism r : Ω Ω, x = x, x n x = x, x n. Thus, if ϕ D, then ϕ D and conversey. In the same way, if u D Ω, we wi denote by u the distribution in D Ω, defined for any ϕ DΩ by u, ϕ D Ω DΩ = u, ϕ D Ω DΩ. Thus, if u D, then u D and conversey. Now, for the convenience of the reader, et us reca the essentia too i.e., the Green formua in the study of singuar boundary conditions for the biharmonic probem see [7]. For any Z, we introduce the space Y p, 1 Rn + = { v W 0, p 2 Rn + : 2 v W 0, p +2, 1 Rn + }, which is a refexive Banach space equipped with its natura norm v Y p, 1 Rn + = v W 0, p 2 Rn v W 0, p +2, 1 Rn +. Then we proved in [7], Lemma 4.1, the foowing resut. Lemma 2.1. Let Z such that n p / {1,..., 2} and then the space DR n p + is dense in Y, 1 Rn +. n p / {1,..., + 2}; 2.1 Thanks to this density emma, we proved in [7], Lemma 4.2, the foowing resut of traces with the Green formua. Lemma 2.2. Let Z. Under hypothesis 2.1, the mapping γ 0, γ 1 : D DRn 1 2 can be extended to a inear continuous mapping γ 0, γ 1 : Y p, 1 Rn + W 1/p, p 2 Γ W 1 1/p, p 2 Γ, and we have the foowing Green formua: 4, p v Y p, 1 Rn +, ϕ W +2 Rn + such that ϕ = n ϕ = 0 on Γ, v, ϕ W 0, p +2, 1 Rn p v, 2 ϕ + W0, 2, 1 Rn + W 0, p 2 Rn p + W0, +2 Rn + = v, N ϕ 1/p, p 1/p, p W 2 Γ W nv, ϕ. +2 Γ 1 1/p, p 1+1/p, p W 2 Γ W +2 Γ Now, we can estabish the foowing resut. Lemma 2.3. Let Z with hypothesis 2.1 and u W 0, p 2 Rn + satisfying 2 u = 0 in, u = n u = 0 on Γ;

7 206 Chérif Amrouche and Yves Raudin then there exists a unique biharmonic extension ũ D R n of u, which is given for a ϕ DR n by ũ, ϕ D R n DR n = u ϕ 5 ϕ 6 x n n ϕ x 2 n ϕ dx. 2.3 Moreover, we have ũ W 2, p 4 Rn with the estimate ũ W 2, p 4 Rn + C u W 0, p 2 Rn Proof. 1 Let us notice an important point to start with. According to Wey s emma, since u is a biharmonic distribution in, we know that u C see e.g. Dautray-Lions [12], page 327, Proposition 1. Next, et us remark that the integra in 2.3 is we defined. Indeed, u W 0, p 2 Rn + and ϕ thus ϕ aso has compact support. Now, et us show that ũ beongs to D R n and more precisey the end of our statement. From 2.3 we get the foowing estimate: ũ, ϕ C u W 0, p 2 Rn + ϕ W 0, p + ϕ + ϕ +2 Rn 1, p W +3 Rn 2, p. W +4 Rn Since n 2, p p / { 3, 2}, we have W 2, p and then, for any ϕ W +4 Rn, ũ, ϕ C u W 0, p 2 Rn + ϕ W 1, p 0, p +4 Rn W +3 Rn W +2 Rn. 2, p +4 Rn So, we can deduce that ũ W 2, p 4 Rn and the estimate For the uniqueness, et us consider two biharmonic extensions ũ 1 and ũ 2 of u which beong to D R n and set U = ũ 2 ũ 1. Then we have 2 U = 0 in R n and we can deduce that U is anaytic in R n. Since U = 0 in, the anaytic continuation principe impies that in fact U = 0 in R n. 3 Evidenty, ũ is an extension of u. Indeed, et ϕ D and ϕ the zero extension of ϕ to R n, then ũ, ϕ D R n DR n = u ϕ dx; that is, ũ R n = u. + On the other hand, et ϕ D and ϕ the zero extension of ϕ to R n, then we get ũ, ϕ D R n DR n = u 5 ϕ 6 x n n ϕ x 2 n ϕ dx.

8 Refection principes and kernes in 207 Moreover, we can express ũ, ϕ by means of an integra in : I 1 = u ϕ dx = u ϕ dx, Hence, I 2 = = I 3 = = u x n n ϕ dx = n x n u ϕdx = u x 2 n ϕ dx = x n u n ϕ dx u + x n n u ϕ dx, x 2 n u ϕ dx = 2 u + 4 x n n u + x 2 n u ϕ dx. ũ, ϕ D R n DR n = x 2 n u ϕ dx u + 2 x n n u x 2 n u ϕ dx; that is, ũ R n = u 2 x n n u x 2 n u. So, ũ R n C and we find the cassica formuation obtained by R. J. Duffin see [13] in the three dimensiona case: for any x, ũx = u 2 x n n u x 2 n u x It remains to show that this extension is actuay biharmonic in R n. From the definition 2.3, we obtain the foowing expression: for a ϕ DR n, 2ũ, ϕ D R n DR n = ũ, 2 ϕ D R n DR n = u [ 2 ϕ 5 ϕ 6 x n n 2 ϕ x 2 n 3 ϕ ] dx, that we can rewrite as foows: 2ũ, ϕ = u 2 Φ dx, where Φ = ϕ ϕ x 2 n ϕ + 2 x n n ϕ. Besides, we have { Φ = ϕ ϕ = 0 on Γ, n Φ = n ϕ + n ϕ = n ϕ n ϕ = 0 on Γ.

9 208 Chérif Amrouche and Yves Raudin Then, according to Lemma 2.2, we get 2 ũ, ϕ = 0 for a ϕ DR n ; that is, 2 ũ = 0 in R n. As a main consequence of Lemma 2.3, we are going to characterize the kerne K of the biharmonic operator 2, γ 0, γ 1 in W 0, p 2 Rn + as a poynomia space. For any q Z, et us introduce B q as a subspace of Pq 2 : { } B q = u Pq 2 : u = n u = 0 on Γ. Coroary 2.4. Let Z with hypothesis 2.1, then K = B [2 n/p]. Proof. Given u K, thanks to Lemma 2.3, we know that ũ W 2, p 4 Rn S R n and 2 ũ = 0 in R n. We can deduce that ũ, and consequenty u, is a poynomia. In addition, u W 0, p 2 Rn + impies that u P [2 n/p] see [3]. Remark 2.5. Coming back to Lemma 2.3, since ũ P [2 n/p], we get in fact ũ W m+2, p m+ R n for any integer m 4. Indeed, under hypothesis R n W 2, p 4 Rn and 2.1, we have the imbedding chain W m+2, p m+ besides P [2 n/p] W m+2, p m+ R n. Better, we can see that this kerne does not reay depend on the reguarity according to the Soboev imbeddings. More precisey, if we denote by K m the kerne of 2, γ 0, γ 1 in W m+2, p m+, identica arguments ead us to the foowing resut. Coroary 2.6. Let Z and m 2 be two integers and assume that n p / {1,..., + min{m, 2}} then K m = B [2 n/p]. and n p / {1,..., m}; 2.6 Finay, we showed in [6] that we can ink this kerne to those of the Dirichet and Neumann probems for the Lapacian. With this intention, we defined the two operators Π D and Π N by r A k, Π Dr = 1 2 s N k, xn Π Ns = 1 2 x n 0 xn 0 t rx, t dt, sx, t dt,

10 satisfying the foowing properties: Refection principes and kernes in 209 r A k, Π Dr = r in, Π D r = n Π D r = 0 on Γ, s N k, Π Ns = s in, Π N s = n Π N s = 0 on Γ. So we got a second characterization of this kerne: 2.7 B [2 n/p] = Π D A [ n/p] Π NN [ n/p]. 2.8 Now, we can use these resuts in the study of the Stokes operator. But to begin with we must estabish a resut equivaent to Lemma 2.2. Let us denote by T : u, π u + π, div u the Stokes operator. For any Z, we introduce the space { T p, 1 Rn + = u, π W 0, p 1 Rn + W 1, p 1 Rn + : Tu, π W 0, p +1, 1 Rn + W 0, p, 1 Rn + which is a refexive Banach space equipped with the graph-norm. Then we have the foowing density resut. Lemma 2.7. Let Z such that }, n/p / {1,..., 1} and n/p / {1,..., + 1}; 2.9 then the space D DRn + is dense in T p, 1 Rn +. Proof. For every continuous inear form Λ T p,1 Rn +, there exists a unique 1,p f, ϕ,g, ψ W 0,p +1 Rn + W +1 W 0,p 1, 1 Rn + W 0,p, 1 Rn +, such that, for a u, π T p, 1 Rn +, Λ, u, π = f, ϕ, u, π + g, ψ, Tu, π Thanks to the Hahn-Banach theorem, it suffices to show that any Λ which vanishes on D DRn + is actuay zero on T p, 1 Rn +. Let us suppose that Λ = 0 on D DRn +, thus on DRn + D. Then we can deduce from 2.10 that f, ϕ + Tg, ψ = 0 in, 1, p 0, p hence, Tg, ψ W +1 Rn + W +1 Rn +. Let f, ϕ, g, ψ be respectivey the zero extensions of f, ϕ, g, ψ to R n. By 2.10, it is cear that we have

11 210 Chérif Amrouche and Yves Raudin f, ϕ + T g, ψ = 0 in R n, and thus T g, ψ 0, p W +1 Rn W +1 Rn. According to the resuts in the whoe space see Theorem 1.3, we can deduce that g, ψ 2, p W +1 Rn 1, p W +1 Rn. Since g and ψ are the zero extensions, it foows that g, ψ W 2, p +1 Rn + W 1, p +1 Rn +. Then, by density of D D in W 2, p +1 Rn + W 1, p +1 Rn +, we can construct a sequence g k, ψ k k N D D such that g k, ψ k g, ψ in 2, p 1, p W +1 Rn + W +1 Rn +. Thus, for any u, π T p, 1 Rn +, we have Λ, u, π = Tg, ψ, u, π + g, ψ, Tu, π = im k { Tg k, ψ k, u, π + g k, ψ k, Tu, π } = 0; i.e., Λ is identicay zero. Thanks to this density emma, we can prove the foowing resut. Lemma 2.8. Let Z. Under hypothesis 2.9, we can define the inear continuous mapping the trace of the veocity fied τ 0 : T p, 1 Rn + W 1/p, p 1 Γ, u, π u Γ = γ 0 u 1,..., γ 0 u n. Moreover, we have the foowing Green formua: u, π T p, 1 Rn 2, p +, ϕ, ψ W +1 Rn 1, p + W +1 Rn + such that ϕ = 0 and div ϕ = 0 on Γ, 1, p Tu, π, ϕ, ψ W 0, p +1, 1 Rn + W0, p, 1 Rn p +, W0, 1, 1 Rn p + W0,, 1 Rn = u, π, Tϕ, ψ W 0, p 1 Rn 1, p + W 1 Rn p +, W0, u, n ϕ, ψ W 1/p, p +1 Rn + W 1, p +1 Rn +. 1/p, p 1 Γ W +1 Γ Proof. Let us make three remarks to start. Firsty, the eft-hand term in 2.11 is nothing but the integra Tu, π ϕ, ψ dx. Secondy, the reason for the ogarithmic factor in the definition of T p, 1 Rn + is that the imbeddings W 2, p +1 Rn 0, p + W 1, 1 Rn 1, p + and W +1 Rn 0, p + W, 1 Rn + hod without suppementary critica vaues with respect to 2.9 whereas the 2, p imbedding W +1 Rn 0, p + W 1 Rn + fais if n/p {, +1}. Thirdy, for 2, p any ϕ W +1 Rn +, the boundary conditions ϕ = 0 and div ϕ = 0 on Γ are equivaent to ϕ = 0 and n ϕ n = 0 on Γ.

12 Refection principes and kernes in 211 So we can write the foowing Green formua: u, π D DRn 2, p +, ϕ, ψ W +1 Rn + W +1 Rn such that ϕ = 0 and div ϕ = 0 on Γ, Tu, π ϕ, ψ dx = u, π Tϕ, ψ dx u n ϕ, ψ dx. We can deduce the foowing estimate: u, n ϕ, ψ W 1/p, p 1/p, p 1 Γ W +1 Γ u, π T p, 1 Rn + ϕ, ψ W 2, p +1 Rn p + W1, +1 Rn. + By Lemma 1.1, for any g W 1 1/p, p Γ, there exists a ifting function 2, p 1, p +1 ϕ, ψ W +1 Rn + W +1 Rn + such that ϕ = 0, n ϕ = g, n ϕ n = 0 and ψ = g n on Γ, satisfying Γ 1, p ϕ, ψ W 2, p C g +1 Rn p + W1, +1 Rn + W 1 1/p, p, +1 Γ where C is a constant not depending on ϕ, ψ and g. Then we can deduce that u 1/p, p W 1 Γ C u, π T p, 1 Rn +. Thus, the inear mapping τ 0 : u, π u Γ defined on D DRn + is continuous for the norm of T p, 1 Rn +. Since D DRn + is dense in T p, 1 Rn +, τ 0 can be extended by continuity to a mapping sti caed τ 0 L T p, 1 Rn +; W 1/p, p 1 Γ. Moreover, we aso can deduce the formua 2.11 from 2.12 by density of D DRn + in T p, 1 Rn +. We now can give the continuation resut for the Stokes operator. Lemma 2.9. Let Z with hypothesis 2.9 and u, π W 0, p 1 Rn + W 1, p 1 Rn + satisfying u + π = 0 and div u = 0 in, u = 0 on Γ; then there exists a unique extension ũ, π D R n D R n of u, π satisfying ũ + π = 0 and div ũ = 0 in R n, 2.13

13 212 Chérif Amrouche and Yves Raudin which is given for a ϕ, ψ DR n DR n by ũ, ϕ = [u ϕ ϕ 2 u n ϕ n + 2 u n x n div ϕ ] dx and + π, 2 x n ϕ n x 2 n div ϕ W 1, p π, ψ = π, ψ ψ 2 x n n ψ W 1, p 1 Rn + W 1, p 1 Rn + W, 1, p +1 Rn Rn u n n ψ dx Moreover, we have ũ, π W 2, p 3 Rn W 2, p 2 Rn with the estimate ũ, π W 2, p 3 Rn W 2, p C u, π 2 Rn W 0, p 1 Rn 1, p + W 1 Rn Remark Knowing that u n satisfies the biharmonic probem, see 2.20, naturay we must find 2.3 from Indeed, if we take ϕ = 0 in 2.14, we get the foowing: for a ϕ n DR n, ũ n, ϕ n = [u n ϕ n ϕ n 2 u n ϕ n 2 u n x n n ϕ n] dx + π, 2 x n ϕ n + x 2 n n ϕ n Since u n = n π in, we can write π, 2 xn ϕ n + x 2 n n ϕ n Hence, ũ n, ϕ n = W 1, p 1 Rn + W 1, p +1 Rn + = π, n x 2 n ϕ n W 1, p = n π, x 2 n ϕ n W 2, p = u n, x 2 n ϕ n W 1, p 1 Rn + W. 1, p +1 Rn + 1 Rn + W 1, p +1 Rn + 1 Rn + W 2, p +1 Rn + W 2, p 1 Rn + W 2, p +1 Rn + = u n, x 2 n ϕ n W 0, p 1 Rn p + W0, = u n, 2 ϕ n + 4 x n n ϕ n x 2 n ϕ n which is exacty the formua 2.3 for u n. +1 Rn +. W 0, p 1 Rn p + W0, +1 Rn + u n ϕn 5 ϕ n 6 x n n ϕ n x 2 n ϕ n dx,

14 Refection principes and kernes in 213 Proof of Lemma As for the biharmonic operator, according to 2.14 and 2.15, we can readiy check that ũ, π W 2,p 3 Rn W 2,p 2 Rn with the estimate Besides, the argument for the uniqueness of the extension aso hods for the Stokes operator and it is cear that 2.14 and 2.15 define an extension of u, π to R n. Indeed, we have both for a ϕ D, ũ, ϕ = R u ϕ dx and for a ψ D, π, ψ = n + π, ψ W 1, p 1 Rn + W 1, p +1 Rn, where ϕ and ψ are respectivey the zero extensions of ϕ and ψ to R n. + 2 Now, we aso can give the functiona writing of this extension in. For a ϕ D, we have [ ũ, ϕ = u ϕ 3 u n ϕ n + 2 u n x n div ϕ ] dx + π, 2 x n ϕ n x 2 n div ϕ D DRn +. Breaking down this expression, we get u ϕ 3 u n ϕ n dx = u ϕ 3 u n ϕ n dx, = 2 u n x n div ϕ dx = 2 u n x n ϕ dx = 2 u n x n div ϕ dx 2 u n ϕ n + x n u n ϕ dx, π, 2 x n ϕ n D DRn + = 2 π, x n ϕ n D DRn = 2 x n π, ϕ n D DRn, π, x 2 n div ϕ D R n + DRn + = x 2 n π, div ϕ D DRn + = x 2 n π, div ϕ D DRn = x 2 n π, ϕ D DRn Hence, = 2 x n π, ϕ n D DRn + x 2 n π, ϕ D DRn. ũ, ϕ = u ϕ u n ϕ n + 2 x n u n ϕ dx + x 2 n π, ϕ D DRn.

15 214 Chérif Amrouche and Yves Raudin Let us notice that here we aways can repace the duaity brackets by integras. Indeed, ϕ has a compact support in, in addition to u and π beonging to C, thus to L 1 oc Rn. So, we get ũ, ϕ = u + 2 x n u n + x 2 n π ϕ dx; i.e., and i.e., x, ũ x = u + 2 x n u n + x 2 n π x ũ n, ϕ n = u n + 2 x n n u n + x 2 n n π ϕ n dx; x, ũ n x = u n 2 x n n u n x 2 n n π x. Likewise, for a ψ D, we have π, ψ = π, ψ 2 x n n ψ D R n+ DRn+ + 4 Separatey, we get π, ψ D DRn + = π, ψ D DRn, u n n ψ dx. π, 2 x n n ψ = 2 x n π, n ψ D DRn + = 2 x n π, n ψ D DRn = 2 n x n π, ψ D DRn = 2 π + x n n π, ψ D DRn, u n n ψ dx = u n n ψ dx = n u n ψ dx. Hence, π, ψ = D R n DR n π + 2 x n n π + 4 n u n, ψ D DRn ; i.e., x, πx = π 2 x n n π 4 n u n x. So, we find the cassica continuation formuae: for a x, ũ x = u + 2 x n u n + x 2 n π x, ũ n x = u n 2 x n n u n x 2 n n π x, πx = π 2 x n n π 4 n u n x.

16 Refection principes and kernes in Finay, it remains to show that this extension satisfies 2.13 that is the Stokes system in the whoe space. For a ϕ, ψ DR n DR n, we have R n Tũ, π ϕ, ψ dx = ũ, π, Tϕ, ψ D R n D R n, DR n DR n = ũ, ϕ + ψ D R n DR n π, div ϕ D R n DR n. Then, according to 2.14 and 2.15, we get Tũ, π ϕ, ψ dx = R n [u ϕ ϕ 2 u n ϕ n + 2 u n x n div ϕ ] dx π, 2 x n ϕ n x 2 n div ϕ W 1, p 1 Rn + W 1, p +1 Rn + + [u ψ ψ + 4 u n n ψ + 2 u n x n ψ ] dx + π, 2 x n n ψ x 2 n ψ W 1, p 1 Rn + W 1, p +1 Rn + π, divϕ ϕ + 2 n ϕ n 2 x n n div ϕ W 1, p 1 Rn + W 1, p +1 Rn + 4 u n n div ϕ dx. With the intention of showing that R Tũ, π ϕ, ψ dx = 0, we are going n to rewrite this expression as foows: u Φ + Ψ dx + π, divφ W 1, p 1, p 1 Rn + W +1 Rn + = u, π, TΦ, Ψ W 0, p 1 Rn 1, p + W 1 Rn p +, W0, 1, p +1 Rn + W, 1, p +1 Rn + 2, p where Φ, Ψ W +1 Rn + W +1 Rn +, with Φ = 0 and n Φ n = 0 on Γ. Then, the zero of u, π, TΦ, Ψ wi be a straightforward consequence of the Green formua Let us construct the functions Φ and Ψ. We wi start with the terms π,. Noticing that div ϕ = div ϕ, we find π, 2 xn ϕ n + x 2 n div ϕ 2 x n n ψ x 2 n ψ divϕ ϕ 2 n ϕ n + 2 x n n div ϕ

17 216 Chérif Amrouche and Yves Raudin = π, div [ ϕ ϕ + 2 x n ϕ n + x 2 n ψ div ϕ ], hence we put Φ = ϕ ϕ + 2 x n ϕ n + x 2 n ψ div ϕ. Next, we can group together the remaining terms in A+B dx, where A = u [ ϕ ϕ + ψ ψ ] B = u n [2 ϕ n 2 x n div ϕ + 4 n ψ + 2 x n ψ 4 n div ϕ ]. We can rewrite B by means of e n = 0,..., 0, 1 as foows: B = u 2 e n [ϕ n + x n ψ div ϕ ]. Using the identity e n ξ = x n ξ x n ξ + 2 n ξ, we get So, we find B = u [2 x n ϕ n + x n ψ div ϕ 2 x n ϕ n + x n ψ div ϕ + 4 n ϕ n + x n ψ div ϕ ], = u [ 2 x n ϕ n + x 2 n ψ div ϕ x 2 n ψ div ϕ + 2 x n ϕ n + x n ψ div ϕ + 4 n ϕ n + x n ψ div ϕ ], = u [ 2 x n ϕ n + x 2 n ψ div ϕ ] + u [ x 2 n ψ div ϕ + 2 x n ϕ n + x n ψ div ϕ + 4 n ϕ n + x n ψ div ϕ ], = u [ 2 x n ϕ n + x 2 n ψ div ϕ ] + u [ 2 x n ϕ n + x 2 n ψ div ϕ ]. A + B = u Φ + Ψ, where { Φ = ϕ ϕ + 2 x n ϕ n + x 2 n ψ div ϕ, Ψ = ψ ψ + 2 x n ϕ n + x 2 n ψ div ϕ. 4, p We can consider ϕ, ψ W +3 Rn 3, p W +3 Rn, then we can easiy 2, p check that Φ, Ψ W +1 Rn 1, p + W +1 Rn + under hypothesis 2.9. In addition, { Φ = ϕ ϕ = 0 on Γ, n Φ n = n ϕ n + n ϕ n = n ϕ n n ϕ n = 0 on Γ.

18 Then, Refection principes and kernes in 217 R n Tũ, π ϕ, ψ dx = 0, for a ϕ, ψ DR n DR n ; that is, Tũ, π = 0, 0. Now, we can characterize the Stokes kerne. For Z, et us denote by K S the kerne of the Stokes operator T, τ 0 in W 0, p 1 Rn + W 1, p 1 Rn + and for any k Z, introduce the poynomia space S + k = { λ, µ P 2 k P k 1 : λ + µ = 0 and div λ = 0 in, λ = 0 on Γ }. Let u, π K S. By Lemma 2.9, we can see that π and ũ are respectivey harmonic and biharmonic tempered distributions in R n, thus poynomias. Hence, we have the foowing resut. Coroary Let Z with hypothesis 2.9; then K S = S + [1 n/p]. Again, this kerne does not depend on the reguarity. That is, if we denote by KS m the kerne of the Stokes operator T, τ 0 in W m+1, p m+ W m, p m+ Rn +, we have the foowing resut. Coroary Let Z and m 1 be two integers and assume that n n p / {1,..., + min{m, 1}} and / {1,..., m}; 2.17 p then K m S = S+ [1 n/p]. We can be more specific about poynomias which buid up this kerne. The idea of this characterization is due to T.Z. Boumezaoud see [9]. We give it with a competey different proof, based on the kernes of the Dirichet and Neumann probems for the Lapacian and the one of the biharmonic probem with Dirichet boundary conditions in the haf-space. Lemma Let Z. Then u, π S + [1 n/p] if and ony if there exists ϕ A [1 n/p] such that u = ϕ Π D div ϕ + Π N n ϕ n, 2.18 π = div ϕ Proof. Given u, π S + [1 n/p], then we aso have div u = 0 on Γ and thus n u n = 0 on Γ. Moreover π = 0 in and thus 2 u n = 0 in. So we get the biharmonic probem 2 u n = 0 in and u n = n u n = 0 on Γ. 2.20

19 218 Chérif Amrouche and Yves Raudin Hence, u n B [1 n/p] and there exists r, s A [ 1 n/p] N [ 1 n/p] such that u n = Π D r + Π N s. We can deduce from 2.7 that n π = u n = r + s in and thus π satisfies π = 0 in and n π = s on Γ. Then, there exists ψ N[ n/p] see [6] such that where π = ψ + Ks in, 2.21 Ksx, x n = xn 0 sx, t dt. So, we have u n = r + s = n π = n ψ + s in, thus r = n ψ. Hence, u n = Π D n ψ + Π N s in From 2.21, we get, for every i {1,...,n 1}, u i = i π = i ψ + i Ks N [ 1 n/p] A [ 1 n/p] = Π N i ψ + Π D i Ks. Then, w i = u i Π N i ψ Π D i Ks satisfies w i = 0 in and w i = 0 on Γ. Hence, we have the existence of ϕ i A [1 n/p] see [4], such that w i = ϕ i ; i.e., u i = Π N i ψ + Π D i Ks + ϕ i. Thereby, writing ϕ = ϕ 1,...,ϕ n 1, we get div u = Π N ψ + Π D Ks + div ϕ In addition, by 2.22, we have n u n = n Π D n ψ + n Π N s = 1 2 x n n ψ x n s + = Π N 2 nψ Π D 2 nks + div ϕ = 1 2 x n n ψ 1 2 x n n Ks Ks + div ϕ = 1 2 x n n ψ 1 2 x n s Ks + div ϕ. xn 0 sx, t dt = 1 2 x n n ψ x n s + Ks. Since div u = 0, we can deduce that div ϕ = Ks and thus 2.21 can be rewritten as π = ψ div ϕ. Now, if we set ϕ n = x n 0 ψx, t dt, then

20 Refection principes and kernes in 219 we have ψ = n ϕ n and ϕ n A [1 n/p]. So, we obtain π = div ϕ; i.e., 2.19, with ϕ = ϕ, ϕ n A [1 n/p]. Coming back to the veocity fied, we get, for every i {1,...,n 1}, u i = ϕ i i Π N n ϕ n i Π D div ϕ Likewise, for the norma component, 2.22 yieds u n = Π D 2 nϕ n + Π N n Ks = 1 2 ϕ n x n n ϕ n x n Ks = ϕ n 1 2 x n n ϕ n 1 2 ϕ n 1 2 x n div ϕ = ϕ n n Π N n ϕ n n Π D div ϕ. So, combining this with 2.23, we get u = ϕ Π N n ϕ n + Π D div ϕ ; i.e., the statement Conversey, we can verify that such a pair u, π beongs to S + [1 n/p]. 3. Generaized soutions to the Stokes system In this section, we wi estabish the centra resut on the generaized soutions to the Stokes system in the haf-space, with Theorem 3.3. We wi be interested in the existence of a soution u, π W 1, p W 0, p to S +, for data f W 1, p, h W 0, p and g W 1 1/p, p Γ. To avoid troubes with the compatibiity conditions, we wi start with the study of the negative weights. For this, as for the weight = 0 in [8], we wi adapt a method used by Farwig-Sohr in [15]. Then, we get back the positive weights by a duaity argument, and the compatibiity condition naturay comes from the kerne of the dua case. First, we wi estabish the resut for the homogeneous probem in the case of negative weights. Lemma 3.1. Let be a negative integer and assume that n/p / {1,..., }. For any g W 1 1/p, p Γ, the homogeneous Stokes probem u + π = 0 in, 3.1 inf λ, µ S + [1 n/p] div u = 0 in, 3.2 u = g on Γ 3.3 has a soution u, π W 1, p W 0, p, unique up to an eement of S + [1 n/p], with the estimate u + λ W 1, p + π + µ W 0, p C g 1 1/p, p W Γ.

21 220 Chérif Amrouche and Yves Raudin Proof. The operator associated to this probem is ceary continuous, moreover its kerne is known. The ast point concerns its surjectivity, then the fina estimate wi be a straightforward consequence of the Banach theorem. So, we ony must prove the existence of a soution u, π. 1 Firsty, we wi show that system can be reduced to a set of three probems on the fundamenta operators 2 and. Appying the operator div to the first equation 3.1, we obtain π = 0 in. 3.4 Now, appying the operator to the same equation 3.1, we deduce From the boundary condition 3.3, we take out 2 u = 0 in. 3.5 u n = g n on Γ, 3.6 and moreover div u = div g on Γ, where div u = n 1 i=1 iu i. Since div u = 0 in, we aso have div u = 0 on Γ, then we can write n u n + div u = 0 on Γ, hence n u n = div g on Γ. 3.7 Combining 3.5, 3.6 and 3.7, we obtain the biharmonic probem P : 2 u n = 0 in, u n = g n and N u n = div g on Γ. Then, combining 3.4 with the trace on Γ of the n th component in the equations 3.1, we obtain the Neumann probem Q : π = 0 in and n π = u n on Γ. Lasty, if we consider the n 1 first components of the equations 3.1 and 3.3, we can write the Dirichet probem R : u = π in and u = g on Γ. 2 Next, we wi sove these three probems. Step 1: Probem P. Since g W 1 1/p, p Γ, we have g n W 1 1/p, p Γ and div g W 1/p, p Γ, so P is a homogeneous biharmonic probem with singuar boundary conditions. Since < 0, we know that probem P has a soution u n W 1, p, unique up to an eement of B [1 n/p] see [7], Theorem 4.5. Step 2: Probem Q. Since 2 u n = 0 in, according to an appropriate trace resut see [8], Lemma 3.7, we can deduce that u n Γ

22 W 1 1/p, p W 0, p Refection principes and kernes in 221 Γ. As < 0, we know that probem Q has a soution π, unique up to an eement of N[ n/p] see [5], Theorem 3.4. Step 3: Probem R. Thanks to the previous resut, we can deduce that π W 1, p and moreover g W 1 1/p, p Γ. Since < 0, we know that probem R has a soution u W 1, p, unique up to an eement of A [1 n/p] see [4], Theorem In order, we have found u n, π and u, non-unique, which satisfy 3.3 and partiay satisfy 3.1, more precisey such that u + π = 0 in. It remains to show we can choose them satisfying 3.2 and the n th component of 3.1; i.e., u n + n π = 0 in. Consider such a pair u, π satisfying probems P, Q and R. Thanks to the first equations of P and Q, we obtain u n n π = 2 u n = 0 in. Thus, with the boundary condition of Q, we can deduce that the distribution u n n π W 1, p satisfies the Dirichet probem u n n π = 0 in, u n n π = 0 on Γ. Then, we have u n n π = µ A [ 1 n/p] see [8], Theorem 3.5. Moreover, we can write µ = Π D µ, with Π D µ = q B [1 n/p]. Setting u n = u n q, we now get u n n π = 0 in, and u n is sti a soution to probem P. Note that π is unchanged with u n, because q = µ = 0 on Γ. Thus, if we set u = u, u n, the pair u, π competey satisfies 3.1. Next, as π = 0 in, we aso have div u = 0 in. Moreover, from the boundary condition in R, we obtain div u = div g on Γ. Then, with the boundary condition in P, we can write div u = div u + n u n = div g div g = 0 on Γ. So, we have div u W 0, p, which satisfies the Dirichet probem div u = 0 in, div u = 0 on Γ. Then, we have div u = ν A [ n/p] see [8], Theorem 3.8. If we take for instance rx = x 1 0 νt, x 2,...,x n dt, we have ν = 1 r and, thus, ν = div r, with r = r, 0,...,0. Setting u = u r, we get div u = 0 in and, as

23 222 Chérif Amrouche and Yves Raudin r A [1 n/p], we sti have u 1 = u 1 r a soution to the first component of the equations 3.1 and 3.3. Consequenty, the pair u, π now competey satisfies the probem Remark 3.2. If g is sufficienty smooth; i.e., g DΓ, using a potentiatheoretic method, it has been shown see [10], [11] that there exists a unique soution of with a finite Dirichet integra. In that case, we can see that this soution is naturay coming in the functiona setting of Lemma 3.1. Now, we can give the foowing genera resut. Theorem 3.3. Let Z and assume that n/p / {1,...,} and n/p / {1,..., }. 3.8 For any f W 1, p, h W 0, p and g W 1 1/p, p Γ, satisfying the compatibiity condition ϕ A [1+ n/p ], f h, ϕ W 1, p + div f, Π D div ϕ + Π N n ϕ n W 2, p W 1, p Rn + W 2, p Rn + + g, n ϕ W 1 1/p, p Γ W 1/p, p Γ = 0, 3.9 probem S + admits a soution u, π W 1, p W 0, p, unique up to an eement of S + [1 n/p], and there exists a constant C such that inf u + λ λ, µ S + W 1, p + π + µ W 0, p [1 n/p] C f W 1, p + h W 0, p + g W 1 1/p, p Proof. 1 First, we sti assume that < 0. We write f = div F, where F = F i 1 i n W 0, p, with the estimate F W 0, p C f W 1, p R n. + Let us respectivey denote by F = F i 1 i n W 0, p R n and h W 0, p R n the zero extensions of F and h to R n. By Theorem 1.2, we know that there exists a soution ũ, π W 1, p R n W 0, p R n to the probem S : ũ + π = div F and div ũ = h in R n. Consequenty, we can reduce the system S + to the homogeneous probem S : v + ϑ = 0 and div v = 0 in, v = g on Γ, Γ.

24 Refection principes and kernes in 223 where we have set g = g ũ Γ W 1 1/p, p Γ. Next, thanks to Lemma 3.1, we know that S admits a soution v, ϑ W 1, p W 0, p. Then, u, π = v + ũ R n +, ϑ + π R n + W 1, p W 0, p is a soution to S +. 2 We now assume that > 0. We wi reason by duaity from the case < 0. So, we have estabished that, under hypothesis 3.8, the Stokes operator T : W 1, p W 0, p /S + [1 n/p] 1, p W W 0, p u, π u + π, div u is an isomorphism for any integer < 0 and rea number p > 1. Thus, repacing p by p and by, we deduce that its adjoint operator T : W 1, p W 0, p W 1, p W 0, p S + [1+ n/p ] is an isomorphism for any integer > 0 and rea number p > 1, aways under hypothesis 3.8. Moreover, by a density argument, we can readiy show that T v, ϑ = v + ϑ, div v. So, we have proved that, for any > 0, probem S + with g = 0 admits a unique soution provided f, h S + [1+ n/p ]. Now, it remains to show that the genera probem S + can be reduced to the particuar case with g = 0, by means of a ifting function; and then that the orthogonaity condition on the ifted probem is equivaent to the compatibiity condition 3.9. First, by Lemma 1.1, there exists a ifting function u g W 1, p of g, i.e., u g = g on Γ, such that u g W 1, p C g 1 1/p, p W Γ. Set v = u u g ; then probem S + is equivaent to the foowing, with homogeneous boundary conditions: v + π = f + u g in, S div v = h div u g in R n +, v = 0 on Γ. So, provided f + u g, h+div u g S + [1+ n/p ], we know that S admits a unique soution. This condition is written in the foowing way: λ, µ S + [1+ n/p ], f, λ + u g, λ h, µ + div u g, µ = 0.

25 224 Chérif Amrouche and Yves Raudin Moreover, we have the Green formua u g, λ W 1, p W = 1, p Rn + = u g λ dx + g, n λ Γ, u g λ dx + g, n λ Γ, because n λ n = 0 on Γ, according to the definition of the kerne. Next, we have another Green formua div u g, µ = W 0, p R n p u + W0, Rn + g µdx g n, µ Γ. Finay, since λ + µ = 0, we have u g λ dx u g µdx = 0, we then get a first formuation for this compatibiity condition: λ, µ S + [1+ n/p ], f, λ h, µ + g, n λ Γ g n, µ Γ = 0. Now, according to the characterization , we can repace each pair λ, µ S + [1+ n/p ] by ϕ Π D div ϕ + Π N n ϕ n, div ϕ, where ϕ beongs to A [1+ n/p ]. Then we have f, λ W 1, p W 1, p Rn + = f, ϕ f, Π D div ϕ + Π N n ϕ n, because Π D div ϕ + Π N n ϕ n Γ = 0. Likewise, = f, ϕ + div f, Π D div ϕ + Π N n ϕ n, h, µ = h, div ϕ W 0, p R n p + W0, Rn + W 0, p R n p + W0, Rn + = h, ϕ W 1, p 1, p W. Rn + Moreover, we can remark that, on the one hand, µ = n ϕ n on Γ, and on the other hand, according to 2.7, we have n λ = n ϕ on Γ, hence the equivaent formuation ϕ A [1+ n/p ], f h, ϕ + div f, Π D div ϕ + Π N n ϕ n + g, n ϕ Γ = 0, i.e., the compatibiity condition 3.9.

26 Refection principes and kernes in Strong soutions and reguarity In this section, we are interested in the existence of strong soutions, i.e., of soutions u, π W 2, p +1 Rn + W 1, p +1 Rn +; and more generay, in the reguarity of soutions to the Stokes system S + according to the data. Theorem 4.1. Let Z and m 1 be two integers and assume that n/p / {1,..., + 1} and n/p / {1,..., m}. 4.1 For any f W m 1, p m+, h W m, p m+ Rn + and g W m+1 1/p, p m+ Γ, satisfying the compatibiity condition 3.9, probem S + admits a soution u, π W m+1, p m+ W m, p m+ Rn +, unique up to an eement of S + [1 n/p], and there exists a constant C such that inf λ, µ S + [1 n/p] u + λ W m+1, p m+ + π + µ W m, p C f W m 1, p m+ + h W m, p m+ Rn + + g m+1 1/p, p W m+ Rn + We have aready proved this resut for = 0 and = 1 in our previous work see [8], Coroaries 5.5 and 5.7. We wi use simiar arguments for the other negative weights, with the aim of minimizing the set of critica vaues, thanks to the known resuts on the harmonic and biharmonic operators in the haf-space. Then, for the positive weights, we wi use a reguarity argument to avoid the compatibiity conditions which woud naturay appear in the auxiiary probems with the previous method. At first, we adapt Lemma 3.1 and its proof for more reguar data. Lemma 4.2. Let 2 and m 1 be two integers and assume that m+ Γ. n/p / {1,..., m}. 4.2 For any g W m+1 1/p, p m+ Γ, the Stokes probem has a soution u, π W m+1, p m+ W m, p m+ Rn +, unique up to an eement of S + [1 n/p], with the corresponding estimate. Proof. Point 1 of Lemma 3.1 is ceary unchanged. Since g W m+1 1/p,p m+ Γ, under hypothesis 4.2, probem P has a soution u n W m+1, p m+, unique up to an eement of B [1 n/p] see [6], Lemma Hence we have u n Γ W m 1 1/p, p m+ Γ, and then probem Q has a soution π W m, p m+ Rn +, unique up to an eement of N[ n/p] see [8], Theorem 3.4, for m = 1; and [6], Theorem 2.8, for m 2. Hence, π

27 226 Chérif Amrouche and Yves Raudin W m 1, p m+, and then probem R has a soution u W m+1, p m+, unique up to an eement of A [1 n/p] see [4], Coroary 3.4. Likewise, point 3 is unchanged with respect to the proof of Lemma 3.1. Proof of Theorem Assume that 2. The proof is quite simiar to the one of Theorem 3.3. Here again, the ony question is the surjectivity of the Stokes operator for such data. For that, we must simpy repace Theorem 1.2 by Theorem 1.3 and Lemma 3.1 by Lemma 4.2 in the proof of the existence of a soution for negative weights in Theorem Assume that > 0. We simpy extend the reguarity argument used in [8] for the cases = 0 and = 1. Now, hypothesis 4.1 is reduced to n/p / {1,..., + 1}. 4.3 Since n/p + 1, we have the imbedding W m 1, p m+ W 1, p ; moreover, W m, p m+ Rn + W 0, p and W m+1 1/p, p m+ Γ W 1 1/p, p Γ hod. So, thanks to Theorem 3.3, we know that probem S + admits a unique soution u, π W 1, p W 0, p. We wi show by induction that f, h, g W m 1, p m+ W m, p m+ Rn + W m+1 1/p, p m+ Γ = u, π W m+1, p m+ W m, p m+ Rn For m = 0, 4.4 is true. Now, assume that 4.4 is true for 0, 1,..., m and suppose that f, h, g W m, p m++1 Rn + W m+1, p m++1 Rn + W m+2 1/p, p m++1 Γ. Let us prove that u, π W m+2, p m++1 Rn + W m+1, p m++1 Rn +. Since we aso have the imbeddings W m, p m++1 Rn + W m 1, p m+, W m+1, p m++1 Rn + W m, p m+ Rn + and W m+2 1/p, p m++1 Γ W m+1 1/p, p m+ Γ, according to the induction hypothesis, we can deduce that the soution u, π W m+1, p m+ W m, p m+ Rn +. Now, for any i {1,...,n 1}, we have i u + i π = i f + 2 x. n 1 iu i u + 1 x iπ. Thus, i u + i π W m 1, p m+. Moreover, div i u = 1 x iu + i h.

28 Refection principes and kernes in 227 Thus, div i u W m, p m+ Rn +. We aso have γ 0 i u = i γ 0 u = i g Γ. So, by the induction hypothesis, we can deduce that W m+1 1/p, p m+ i {1,...,n 1}, i u, i π W m+1, p m++1 Rn + W m, p m++1 Rn +. It remains to prove that n u, n π W m+1, p m++1 Rn + W m, p m++1 Rn +. For that, et us observe that, for any i {1,...,n 1}, we have i n u = n i u W m, p m++1 Rn +, nu 2 i = u i + i π f i W m, p m++1 Rn +, nu 2 n = n h n div u W m, p m++1 Rn +, n π = f n + u n W m, p m++1 Rn +. Hence, n u W m, p m++1 Rn + and, knowing that n u W m, p m+ Rn +, we can deduce that n u W m+1, p m++1 Rn +, according to definition 1.1. Consequenty, we have u W m+1, p m++1 Rn +. Likewise, π W m, p m++1 Rn + and, finay, we can concude that u, π W m+2, p m++1 Rn + W m+1, p m++1 Rn Very weak soutions The aim of this section is to come back to the homogeneous Stokes system , but with singuar data on the boundary. In [8], we soved it in the cases g W 1/p, p 1 Γ and g W 1/p, p 0 Γ. Here, we wi extend these resuts to the other weights, introducing the question of the kerne and, by duaity, the compatibiity condition. Thanks to Lemma 2.8, the proof wi be more direct. Theorem 5.1. Let Z with hypothesis 2.9. For any g W 1/p, p 1 Γ, satisfying the compatibiity condition ϕ A [1+ n/p ], g, nϕ W 1/p, p 1/p, p 1 Γ W +1 Γ = 0, 5.1 probem admits a soution u, π W 0, p 1 Rn + W 1, p 1 Rn +, unique up to an eement of S + [1 n/p], and there exists a constant C such that inf λ, µ S + [1 n/p] u + λ W 0, p 1 Rn + + π + µ W 1, p 1 Rn + C g W 1/p, p 1 Γ. Proof. To start with, et us observe that Lemma 2.8 gives a meaning to these boundary conditions. Besides, thanks to the Green formua 2.11, we get

29 228 Chérif Amrouche and Yves Raudin the equivaence between probem and the variationa formuation: Find u, π T p, 1 Rn + satisfying 1, p 2, p v, ϑ W +1 Rn + W +1 Rn +, 5.2 such that v = 0 and div v = 0 on Γ, u, π, Tv, ϑ W 0, p 1 Rn 1, p + W 1 Rn p +, W0, = g, n v, ϑ W 1/p, p 1/p, p +1 Rn + W 1, p +1 Rn +. 1 Γ W +1 Γ Now, et us sove probem 5.2. By Theorem 4.1, we know that under 0, p hypothesis 2.9, for a f, h W +1 Rn + W 1, p +1 Rn + S + [1 n/p], 2, p there exists a unique soution v, ϑ W +1 Rn 1, p + W +1 Rn +/S + [1+ n/p ] to v + ϑ = f and div v = h in, v = 0 on Γ, with the estimate v, ϑ W 2, p +1 Rn p C f + W1, +1 Rn + /S+ [1+ n/p ] W 0, p +1 Rn + h + W Consider the inear form Λ : f, h g, n v, ϑ W 1/p, p 1, p +1 Rn + 1/p, p 1 Γ W +1 Γ 0, p defined on W +1 Rn + W 1, p +1 Rn + S + [1 n/p]. By 5.1, we have for any ϕ A [1+ n/p ], or, equivaenty, for any λ, µ S+ [1+ n/p ], Λf, h = g, n v, ϑ + n ϕ W 1/p, p 1/p, p 1 Γ W +1 Γ = g, n [v + λ ], [ϑ + µ] W 1/p, p 1/p, p 1 Γ W +1 Γ. Thus, C g 1/p, p W 1 Γ v, ϑ + λ, µ W 2, p +1 Rn p + W1, +1 Rn. + Λf, h C g 1/p, p W 1 Γ v, ϑ W f W 0, p C g 1/p, p W 1 Γ 2, p +1 Rn p + W1, +1 Rn + /S+ [1+ n/p ] +1 Rn + + h W 1, p +1 Rn + 0, p In other words, Λ is continuous on W +1 Rn + W 1, p +1 Rn + S + [1 n/p], and according to the Riesz representation theorem, we can deduce that there.

30 Refection principes and kernes in 229 exists a unique u, π W 0, p 1 Rn + W 1, p 1 Rn +/S + [1 n/p], which is the 0, p dua space of W +1 Rn + W 1, p +1 Rn + S + [1 n/p], such that 0, p f, h W +1 Rn + W 1, p +1 Rn +, Λf, h = u,f + π, h W 0, p 1 Rn p + W0, +1 Rn + W 1, p 1 Rn + W ; 1, p +1 Rn + i.e., the pair u, π satisfies 5.2 and the kerne of the associated operator is S + [1 n/p]. References [1] S. Agmon, A. Dougis, and L. Nirenberg, Estimates near the boundary for soutions of eiptic partia differentia equations, I, Comm. Pure App. Math., , [2] F. Aiot and C. Amrouche, The Stokes probem in R n : an approach in weighted Soboev spaces, Math. Meth. App. Sci., , [3] C. Amrouche, V. Giraut, and J. Giroire, Weighted Soboev spaces for Lapace s equation in R n, J. Math. Pures App., , [4] C. Amrouche and S. Nečasová, Lapace equation in the haf-space with a nonhomogeneous Dirichet boundary condition, Mathematica Bohemica, , [5] C. Amrouche, The Neumann probem in the haf-space, C. R. Acad. Sci. Paris, Série I, , [6] C. Amrouche and Y. Raudin, Nonhomogeneous biharmonic probem in the haf-space, L p theory and generaized soutions, J. Differentia Equations, , [7] C. Amrouche and Y. Raudin, Singuar boundary conditions and reguarity for the biharmonic probem in the haf-space, Comm. Pure App. Ana., , [8] C. Amrouche, S. Nečasová, and Y. Raudin, Very weak, generaized and strong soutions to the Stokes system in the haf-space, J. Differentia Equations, , [9] T. Z. Boumezaoud, On the Stokes system and the biharmonic equation in the hafspace: an approach via weighted Soboev spaces, Math. Meth. App. Sci., , [10] L. Cattabriga, Su un probema a contorno reativo a sistema di equazioni di Stokes, Rend. Sem. Mat. Padova, , [11] H. Chang, The steady Navier-Stokes probem for ow Reynods number viscous jets into a haf-space, The Navier-Stokes equations II theory and numerica methods Oberwofach, 1991, 85 96, Lecture Notes in Math., 1530, Springer, Berin, [12] R. Dautray and J.-L. Lions, Anayse mathématique et cacu numérique, 9 vo., Masson, Paris, [13] R. J. Duffin, Continuation of biharmonic functions by refection, Duke Math. J., , [14] R. Farwig, A Note on the Refection Principe for the Biharmonic Equation and the Stokes system, Acta app. Math., , [15] R. Farwig and H. Sohr, On the Stokes and Navier-Stokes System for Domains with Noncompact Boudary in L q -spaces, Math. Nachr., ,

31 230 Chérif Amrouche and Yves Raudin [16] G. P. Gadi, An Introduction to the Mathematica Theory of the Navier-Stokes Equations, Voİ and VoİI, Springer tracts in natura phiosophy, [17] A. Huber, On the Refection Principe for Poyharmonic Functions, Comm. Pure App. Math., IX 1956, [18] V. G. Maz ya, B. A. Pamenevskiĭ, and L. I. Stupyais, The three-dimentiona probem of steady-state motion of a fuid with a free surface, Amer. Math. Soc. Trans., , [19] N. Tanaka, On the boundary vaue probem for the stationary Stokes system in the haf-space, J. Differentia Equations, ,