New estimates for the div-curl-grad operators and elliptic problems with L1-data in the whole space and in the half-space

Transcription

1 New estimates for the div-curl-grad operators and elliptic problems with L1-data in the whole space and in the half-space Chérif Amrouche, Huy Hoang Nguyen To cite this version: Chérif Amrouche, Huy Hoang Nguyen. New estimates for the div-curl-grad operators and elliptic problems with L1-data in the whole space and in the half-space. 49 pages <hal > HAL Id: hal Submitted on 26 Jan 2011 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 New estimates for the div-curl-grad operators and elliptic problems with L 1 -data in the whole space and in the half-space Chérif Amrouche a, Huy Hoang Nguyen b a Laboratoire de Mathématiques Appliquées, UMR CNRS 5142 Université de Pau et des Pays de l Adour Pau France b Departamento de Matematica, IMECC, Universidade Estadual de Campinas Caixa Postal 6065, Campinas, SP , Brazil Abstract In this paper, we study the div-curl-grad operators and some elliptic problems in the whole space R n and in the half-space, with n 2. We consider data in weighted Sobolev spaces and in L 1. Keywords: div-curl-grad operators, elliptic, half-space, weighted Sobolev spaces 2000 MSC: 35J25, 35J47 1. Introduction The purpose of this paper is to present new results concerning the div-curlgrad operators and some elliptic problems in the whole space and in the halfspace with data and solutions which live in L 1 or in weighted Sobolev spaces, expressing at the same time their regularity and their behavior at infinity. Recently, new estimates for L 1 -vector field have been discovered by Bourgain, Brézis and Van Schaftingen (see [23], [10], [11], [12], [13], [15]) which yield in particular improved estimates for the solutions of elliptic systems in R n or in a bounded domain Ω R n. Our work presented in this paper is naturally based on these very interesting results and our approach rests on the use of weighted Sobolev spaces. This paper is organised as follows. In this section, we introduce some notations and the functional framework. Some results concerning the weighted Sobolev spaces and the spaces of traces are recalled. In Section 2 and Section 3, our work is focused on the div-grad-curl operators and elliptic problems in the whole space. After the case of the whole space, we then pass to the one of the halfspace. Results in the half-space are presented in Section 4 (The div-grad opera- addresses: cherif.amrouche@univ-pau.fr (Chérif Amrouche a ), nguyen@ime.unicamp.br (Huy Hoang Nguyen b ) Preprint submitted to Journal of Differential Equations December 28, 2010

3 tors), Section 5 (Vector potentials) and in the last section of this paper (Elliptic problems). In this paper, we use bold type characters to denote vector distributions or spaces of vector distributions with n components and C > 0 usually denotes a generic positive constant that may depends on the dimension n, the exponent p and possibly other parameters, but never on the functions under consideration. For any real number 1 < p <, we take p to be the Hölder conjugate of p. Let Ω be an open subset in the n-dimensional real euclidean space. A typical point x R n is denoted by x = (x, x n ), where x = (x 1, x 2,..., x n 1 ) R n 1 and x n R. Its distance to the origine is denoted by r = x = (x x 2 n) 1/2. Let denote the closure of the upper half-space = {x R n ; x n > 0}. In the half-space, its boundary is defined by Γ = { x R n ; x n = 0 } R n 1. In order to control the behavior at infinity of our functions and distributions, we use for basic weight the quantity ρ = ρ(r) = 1 + r, which is equivalent to r at infinity. We define D(Ω) to be the linear space of infinite differentiable functions with compact support on Ω. Now, let D (Ω) denote the dual space of D(Ω), the space of distributions on Ω. For any q N, P q stands for the space of polynomials of degree q. If q is strictly negative integer, we set by convention P q = {0}. Given a Banach space B, with dual space B and a closed subspace X of B, we denote by B X (or more simply X, if there is no ambiguity as to the duality product) the subspace of B orthogonal to X, i.e. B X = X = {f B v X, < f, v > = 0} = (B/X). The space X is called the polar space of X in B and is also denoted by X. We also introduce the space V(Ω) = { ϕ D(Ω), div ϕ = 0 }. In this paper, we want to consider some particular weighted Sobolev spaces (see [4], [5]). The open set Ω will be denote the whole space or the half-space. We begin by defining the space where W 1,p 0 (Ω) = {u D u (Ω), L p (Ω), u L p (Ω)}, w 1 w 1 = 1 + r if p n and w 1 = (1 + r) ln(2 + r) if p = n. This space is a reflexive Banach space when endowed with the norm: We also introduce the space u W 1,p 0 (Ω) = ( u w 1 p L p (Ω) + u p L p (Ω) )1/p. W 2,p 0 (Ω) = {u D (Ω), u w 2 L p (Ω), 2 u w 1 L p (Ω), D 2 u L p (Ω)},

4 where w 2 = (1 + r) 2 if p / { n 2, n} and w 2 = (1 + r) 2 ln(2 + r), otherwise, which is a Banach space endowed with its natural norm given by u W 2,p 0 (Ω) = ( u w 2 p L p (Ω) + u w 1 p L p (Ω) + D2 u p L p (Ω) )1/p. We need to give also the definition of the following space where W 1,p 1 (Ω) = {u D (Ω), u w 3 L p (Ω), u 1 + r Lp (Ω)}, w 3 = (1 + r) 2 if p n/2 and w 3 = (1 + r) 2 ln(2 + r), otherwise. This space is also a reflexive Banach space and we can show that W 2,p 0 (Ω) W 1,p 1 (Ω). Note that the logarithmic weight only appears if p = n or p = n 2. From now on, when we write Wα m,p (Ω), it means that m, p and α are taken as in these above definitions of the weighted Sobolev spaces. It is also true for the generalized case of the weighted Sobolev spaces. The weights in these above definitions are chosen so that the corresponding space satisfies two properties. On the one hand, the space D(Ω) is dense in Wα m,p (Ω). On the other hand, the following Poincar-type inequality holds in Wα m,p (Ω) (see [4], [5] and [6]). The semi-norm. W m,p α (Ω) = ( (1 + r) α λ u p L p (Ω) )1/p λ =m defines on W m,p α (Ω)/P q a norm which is equivalent to the quotient norm, u W m,p α (Ω), u W m,p α (Ω)/P q C u W m,p α (Ω), (1) with q = inf(q, m 1), where q is the highest degree of the polynomials contained in Wα m,p (Ω). We define the space W m,p α (Ω) = D(Ω). W m,p α (Ω), and its dual space, W m,p α (Ω), is a space of distributions. In addition, the semi-norm. W m,p α (Ω) is a norm on W m,p α (Ω) that is equivalent to the full norm. W m,p α (Ω): u W m,p α (Ω), u W m,p (Ω) C u W m,p α α (Ω). (2) When Ω = R n, we have Wα m,p (R n ) = W m,p α (R n ). We will now recall some properties of the weighted Sobolev spaces Wα m,p (Ω). All the local properties of Wα m,p (Ω) coincide with those of the classical Sobolev space W m,p (Ω). A quick computation shows that for m 0 and if n + α does not belong to p 3

5 {i Z; i m}, then P [m n/p α] is the space of all polynomials included in Wα m,p (R n ) (for s R, [s] stands for the integer part of s). For all λ N n where 0 λ 2m with m = 1 or m = 2, the mapping u W m,p α (Ω) λ u W m λ,p (Ω) is continuous. Recall the following Sobolev embeddings (see [1]): Also recall W 1,p 0 (Ω) L p (Ω) where p = np n p W 0 (R n ) BMO(R n ). α and 1 < p < n. The space BM O is defined as follows: A locally integrable function f belongs to BMO if 1 f BMO =: sup f(x ) f Q dx <, Q Q where the supremum is taken on all the cubes and f Q = 1 Q f(x ) dx is the Q average of f on Q. In the literature, we also find studies using the following spaces: Ŵ 1,p (Ω) = D(Ω). L p and Ŵ 2,p (Ω) = D(Ω) 2. Lp. In fact, when Ω = R n we can prove that Ŵ 1,p (R n ) = W 1,p 0 (R n ) if 1 < p < n and Ŵ 2,p (R n ) = W 2,p 0 (R n ) if 1 < p < n/2. When n/p m 0 with m = 1 or m = 2, Ŵ m,p (R n ) is not a space of distributions. For instance, in J. Deny, J. L. Lions [16], they show that Ŵ 1,2 (R 2 ) is not a space of distributions. Without going into details, let (ϕ ν ) be a sequence of functions of D(R 2 ) such that ϕ ν L 2 (R 2 ) is a Cauchy sequence. Applying Proposition 9.3 [4], there exists a constant c ν such that inf c R ϕ ν + c W 0 1, 1 (R 2 ) = ϕ ν + c ν W 0 1, 1 (R 2 ) is also a Cauchy sequence and therefore converges. But this does not mean that ϕ ν alone converges and from [4], ϕ ν L2 (R 2 ) tends to zero while < ϕ ν, ψ > tends to infinity for many ψ of D(R 2 ) instead of converging to a constant times the mean value of ψ. These considerations suggest that the space Ŵ 1,2 (R 2 ) lacks the constant functions and is not a space of distributions. In order to define the traces of functions of Wα m,p (), we introduce for any σ (0, 1) the space W σ,p 0 (R n ) = { u D (R n ); w σ u L p (R n ) and Rn Rn u(x) u(y) p dx dy < }, x y n+σp Q where w = ρ if n p σ and w = ρ(ln (1 + ρ))1/σ if n p = σ. 4

6 It is a reflexive Banach space equipped with its natural norm u W σ,p 0 (R n ) = ( u w σ p L p (R n ) Rn Rn u(x) u(y) p ) 1/p + dx dy. x y n+σp Similarly, for any real number α R, we define the space W σ,p α (R n ) = { u D (R n ); w α σ u L p (R n ) and Rn Rn ρ α (x)u(x) ρ α (y)u(y) p x y n+σp dx dy < }, where w = ρ if n p + α σ and w = ρ(ln (1 + ρ))1/(σ α) if n p + α = σ. For any s R +, we set W s,p α (R n ) = { u D (R n ); 0 λ k, ρ α s+ λ (ln(1 + ρ)) 1 λ u L p (R n ); k + 1 λ [s] 1, ρ α s+ λ λ u L p (R n ); λ = [s], λ u Wα σ,p (R n ) }, where s n k = k(s, n, p, α) = p α if n + α { σ,..., σ + [s] }, p 1, otherwise, with σ = s [s]. It is a reflexive Banach space equipped with the norm u W s,p α (R n ) = ( ρ α s+ λ (ln(1 + ρ)) 1 λ u p L p (R n ) 0 λ k + ρ α s+ λ λ u p L p (R n ) )1/p + λ u W σ,p α (R ). n k+1 λ [s] 1 λ =[s] We notice that this definition coincides with the previous definition of the weighted Sobolev spaces when s = m is a nonnegative integer. If u is a function on, we denote its trace of order j on the hyperplane Γ by j N, γ j u : x i u x j (x, 0). n Finally, we recall the following traces lemma due to Hanouzet [19] and extended by Amrouche-Nečasová [6] to this class of weighted Sobolev spaces. Lemma 1.1. The mapping γ = (γ 0, γ 1,..., γ m 1 ) : D() m 1 j=0 D(R n 1 ), 5

7 can be extended to a linear continuous mapping, still denoted by γ, m 1 γ : Wα m,p () Wα m j 1/p,p (R n 1 ). j=0 Moreover, γ is surjective and Ker γ = W m,p α (). In order to finish this section, we recall the definition of curl operator. When n = 2, we define the curl operator for distributions ϕ D (Ω) and v D (Ω) by curl ϕ = ( ϕ, ϕ ) and curl v = v 2 v 1. x 2 x 1 x 1 x 2 When n = 3, we define the curl operator of a distribution v D (Ω) as follows curl v = ( v 3 x 2 v 2 x 3, v 1 x 3 v 3 x 1, v 2 x 1 v 1 x 2 ). The following properties are easily obtained. curl(curl ϕ) = ϕ with n = 2, { curl (curl v) when n = 3, v + (div v) = curl(curl v) when n = The div-grad operators in the whole space The following proposition was given by J. Bourgain and H. Brézis [10]. We give here a detailed proof. Proposition 2.1. Let f L n (R n ). Then there exists u L (R n ) such that div u = f with u L (R n ) C n f Ln (R n ). (3) Proof. We consider the following unbounded operator with A = : L n/(n 1) (R n ) L 1 (R n ), D(A) = W 1,1 0 (R n ) = {v L n/(n 1) (R n ), v L 1 (R n )}. It is easy to see that A is closed: if v n v in L n/(n 1) and v n z in L 1 (R n ), then we have z = v. On the other hand, D(A) is dense in L n/(n 1) (R n ). We know that for all u D(A), u L n/(n 1) (R n ) C u L1 (R n ). (4) Thanks to Theorem II.20 [14], the adjoint operator A = div : L (R n ) L n (R n ) 6

8 is surjective, i.e., for all f L n (R n ), there exists u L (R n ) such that div u = f. Then from Theorem II.5 [14], there exists c > 0 such that div B L (0, 1) B L n(0, c), (5) where B E (0, α) is the open ball in E of radius α > 0 centered at origin. Let now f L n (R n ) satisfying f 0 and set f h = c. f L n (R n ) Therefore, we can deduce from (5) the existence of v n v n L (R n ) 1 such that in L (R n ) satisfying div v n h in L n (R n ). We can then extract a subsequence (v nk ) such that v nk v in L (R n ) with v L (R n ) 1 and h = div v. Hence, we obtain the property (3) with u = 1 c f L n (R n )v. Remark 1. Proposition 2.1 can be improved by showing that u actually belongs to W 0 (Rn ) L (R n ) (see Theorem 2.3). Note that W 0 (R n ) BMO(R n ), but the corresponding embedding in L (R n ) does not take place. Recall now De Rham s Theorem: let Ω be any open subset of R n and let f be a distribution of D (Ω) that satisfies: v V(Ω), < f, v > D (Ω) D(Ω) = 0. (6) Then there exists π in D (Ω) such that f = π. In particular, if f W 1,p 0 (R n ) with 1 < p < and satisfies v V(R n ), < f, v > W 1,p = 0, (7) 0 (R n ) W 1,p 0 (R n ) then there exists a unique π L p (R n ) such that f = π and the following estimate holds π L p (R n ) C f W 1,p 0 (R n ). Similarly, if f L p (R n ), with 1 < p <, and satisfies v V(R n ), f v = 0, (8) then f = π with π W 1,p 0 (R n ). Note that V(R n ) is dense in H p (R n ) for all 1 p < (see Alliot-Amrouche [2] for p > 1 and Miyakawa [20] for p = 1), but is not dense in H (R n ), where for any 1 p, H p (R n ) = {v L p (R n ); div v = 0}. Ω 7

9 As V(R n ) is dense in then (7) is equivalent to V 1,p 0 (R n ) = } {v W 1,p 0 (R n ); div v = 0, v V 1,p 0 (R n ), < f, v > W 1,p = 0. (9) 0 (R n ) W 1,p 0 (R n ) The same property holds for the relation (8) with V(R n ) replaced by H p (R n ). The following corollary gives an answer when f L 1 (R n ). Corollary 2.2. Assume that f L 1 (R n ) satisfying v H (R n ), f v dx = 0. R n Then there exists a unique π L n/(n 1) (R n ) such that f = π with the estimate π L n/(n 1) (R n ) C f L1 (R n ). (10) Proof. This corollary can be proved because from (4), we have Im is closed subspace of L 1 (R n ), so that [H (R n )] o = (Ker div) o = Im, where { } [H (R n )] o = f L 1 (R n ), f v dx = 0, v H (R n ). R n Recall that if E is a Banach space and M a subspace of the dual E, then the polar (or the orthogonal) of M is defined as follows M o = { f E; < f, v > = 0, v M }. Remark 2. Observe first that the hypothesis of Corollary 2.2 implies that f L 1 0(R n ), i.e., f L 1 (R n ) and f R n = 0. Next, note that the conclusion of the above corollary shows that f W /(n 1) 0 (R n ) and then div f W 2,n/(n 1) 0 (R n ). Moreover, we have λ P 1, < div f, λ > 2,n/(n 1) W = 0. (11) 0 (R n ) W 2,n 0 (R n ) Now recall a result in J. Bourgain - H. Brézis (cf. [11] or [12]). Theorem 2.3. For all f L n (R n ), there exists w W 0 (R n ) L (R n ) such that div w = f and w W 0 (R n ) + w L (R n ) C f L n (R n ). 8

10 Remark 3. Note that J. Bourgain and H. Brézis use the space Ŵ (R n ) (respectively, Ŵ 2,n (R n )), which is defined by the adherence of D(R n ) for the norm. L n (R n ) (respectively, the adherence of D(R n ) for the norm 2. L n (R n )) as their functional framework. Our choice is the weighted Sobolev space and it seems to us more adaptive (see the introduction in Section 1 for the explanation). The second point of the following theorem is an extension of Corollary 2.2 and (7) with p = n/(n 1). Corollary 2.4. i) There exists C > 0 such that for all have the following inequality u L n/(n 1) (R n ), we u L n/(n 1) (R n ) C inf g+h = u ( g L 1 (R n ) + h W /(n 1) 0 (R n ) ) (12) with g L 1 (R n ) and h W /(n 1) 0 (R n ). ii) Let f L 1 (R n ) + W /(n 1) 0 (R n ) satisfying the following compatibility condition v V 0 (R n ) L (R n ), < f, v > = 0. (13) Then there exists a unique π L n/(n 1) (R n ) such that f = π. Proof. i) We consider two following operators A = : L n/(n 1) (R n ) L 1 (R n ) + W /(n 1) 0 (R n ), A = div : W 0 (R n ) L (R n ) L n (R n ). The rest of this proof is similar to the one of Proposition 2.1. ii) The second point is a consequence of the first one. Remark 4. Remark that for all u W 1,1 0 (R n ), and for all u L n/(n 1) (R n ), u L n/(n 1) (R n ) C u L 1 (R n ), (14) u L n/(n 1) (R n ) C u W /(n 1) 0 (R n ). (15) The inequality (14) is well-known. We now consider (15). It is shown in [4] that is an isomorphism from W 2,n 0 (R n )/P 1 into L n (R n ). By duality, we have : L n/(n 1) (R n ) W 2,n/(n 1) 0 (R n ) P 1 is also an isomorphism. Then for all u L n/(n 1) (R n ), we can deduce u L n/(n 1) (R n ) C u W 2,n/(n 1) 0 (R n ). (16) 9

11 Moreover, we can also see immediately that and the following operator is continuous. Then u W 2,n/(n 1) 0 (R n ) = div u W 2,n/(n 1) 0 (R n ) div : W /(n 1) 0 (R n ) W 2,n/(n 1) 0 (R n ) u W 2,n/(n 1) 0 (R n ) C u W /(n 1) 0 (R n ) and we deduce easily (15). The inequality (12), stronger than (14) or (15) is especially interesting if u L n/(n 1) (R n ) and u / L 1 (R n ). Recall now the definition of Riesz transforms ( R j f = c n p.v. f x j x n+1 ), j = 1,..., n, where c n = Γ ( ) n+1 n+1 2 /π 2. Also recall that their Fourier transforms satisfy R j f = i ξ j ξ f. The following corollary is proved in [11]. We give here a little different proof. Corollary 2.5. Assume that F W 0 (R n ). Then there exists Y W 0 (R n ) L (R n ) such that n F = R j Y j. j=1 Proof. Let F W 0 (R n ) and define f by f(ξ) = ξ F (ξ). Then we have R j f(ξ) = F (ξ). Therefore, R j f = F L n (R n ) for all j = 1,..., n. Hence, x j x j n R j R j f L n (R n ) and we deduce f = R j R j f L n (R n ). Thanks to Theorem 2.3, there exists Y W 0 (R n ) L (R n ) such that f = div Y, i.e., n f = iξ j Y j. Then, F n = i ξ n j ξ Ŷj, that means that F = R j Y j. j=1 j=1 An another result was established by J. Bourgain and H. Brézis [11] (see also H. Brézis and J. Van Schaftingen [15]). Theorem 2.6. Let Ω be a Lipschitz bounded open domain in R n. i) For all ϕ W 0 (Ω), there exist ψ W 0 (Ω) L (Ω) and η W 2,n 0 (Ω) such that ϕ = ψ + η, (17) j=1 j=1 10

12 with the following estimate ψ L n (Ω) + ψ L (Ω) + D 2 η L n (Ω) C ϕ L n (Ω), (18) where C only depends on Ω. ii) For all ϕ W (Ω), there exist ψ W (Ω) L (Ω) and η W 2,n (Ω) such that (17) holds with ψ n = 0 on Γ and satisfying the following estimate ψ W (Ω) + ψ L (Ω) + η W 2,n (Ω) C ϕ W (Ω). (19) In the above theorem, W 0 (Ω) is the classical Sobolev space of functions in W (Ω) vanishing on the boundary of Ω and W 2,n 0 (Ω) is the one of functions in W 2,n (Ω) whose traces and the normal derivative are vanished on the boundary of Ω. We now prove a similar result corresponding to weighted Sobolev spaces. Corollary 2.7. For all ϕ W 0 (R n ), there exist ψ W 0 (R n ) L (R n ) and η W 2,n 0 (R n ) such that ϕ = ψ + η, with the following estimate ψ W 0 (R n ) + ψ L (R n ) + D 2 η L n (R n ) C ϕ L n (R n ). (20) Proof. Thanks to the density of D(R n ) in W 0 (R n ), there exists a sequence (ϕ k ) k N in D(R n ) that converges toward ϕ in W 0 (R n ). Let B rk be a ball such that supp ϕ k B rk and we set ϕ k (x ) = ϕ k(r k x ). Then we deduce ϕ k W 0 (B 1 ). Applying Theorem 2.6, there exist ψ k W 0 (B 1 ) L (B 1 ) and η k W 2,n 0 (B 1 ) such that ϕ k = ψ k + η k, with the following estimate ψ k Ln (B 1) + ψ k L (B 1) + D 2 η k L n (B 1) C ϕ k L n (B 1). We now set Then we have Moreover, since ψ k (x ) = ψ k( x r k ) and η k (x ) = r k η k( x r k ). ϕ k = ψ k + η k. (21) ψ k L n (R n ) = ψ k L n (B 1), ψ k L (R n ) = ψ k L (B 1), ϕ k L n (R n ) = ϕ k L n (B 1), we then have that ψ k Ln (R n ) + ψ k L (R n ) C ϕ k Ln (R n ) C ϕ Ln (R n ). (22) 11

13 Then there exists a sequence (a k ) in R n such that ψ k + a k is bounded in W 0 (R n ) and ψ k + a k W 0 (R n ) C ϕ L n (R n ). (23) As ψ k L (R n ) is also bounded, then the sequence (a k ) is bounded in R n. Then we can extract a subsequence, again denoted by (a k ), such that lim a k = a. k We know that there exists ψ 0 in W 0 (R n ) such that ψ k +a k ψ 0 in W 0 (R n ) and ψ 0 W 0 (R n ) C ϕ L n (R n ). Then, we have ψ k ψ 0 a in W 0 (R n ). Moreover, ψ k ψ in L (R n ), then it implies that ψ = ψ 0 a. In addition, we have the following estimate From (21), we can deduce that ψ Ln (R n ) + ψ L (R n ) C ϕ Ln (R n ). i.e., there exists α k P 1 such that D 2 η k Ln (R n ) C ϕ Ln (R n ), η k + α k is bounded in W 2,n 0 (R n ), (24) and there exists η 0 in W 2,n 0 (R n ) such that η k +α k η 0 in W 2,n 0 (R n ). As ϕ k and ψ k are bounded in W 0 (R n ), it is also true for η k, then from (24), we deduce that α k is bounded in W 0 (R n ). Therefore, there exists a real sequence b k such that α k + b k is bounded in W 2,n 0 (R n ). Consequently, the sequence η k b k is bounded in W 2,n 0 (R n ) and we can extract a subsequence, denoted in the same way, such that η k b k η in W 2,n 0 (R n ). Then we have with the estimate ϕ k = ψ k + (η k b k ) D 2 (η k b k ) Ln (R n ) C ϕ Ln (R n ). We pass to limite in the above decomposition, we shall obtain ϕ = ψ + η with ψ Ln (R n ) + ψ L (R n ) + D 2 η Ln (R n ) C ϕ Ln (R n ), and then we deduce (20). We have another version of Corollary 2.7. Theorem 2.8. For all ϕ W 0 (R n ), there exist ψ W 0 (R n ) L (R n ) and η W 2,n 0 (R n ) such that ϕ = ψ + η, with the following estimate ψ W 0 (R n ) + ψ L (R n ) + η W 2,n 0 (R n ) C ϕ W 0 (R n ). 12

14 Proof. Let ϕ W 0 (R n ). We resume the obtained functions a k, ψ k, a, ψ 0, ψ and η in the proof of Corollary 2.7. We have a k = 1 a k 1 ( ) 1/n a k n B 1 (n 1)/n. B 1 B 1 B 1 B 1 Then we deduce from (22) and (23) that and a k C a k L n (B 1) C a k + ψ k L n (B 1) + C ψ k L n (B 1) C ϕ L n (R n ). As ψ = ψ 0 a, then a L (R n ) C ϕ L n (R n ). ψ W 0 (R n ) C ϕ L n (R n ) C ϕ W 0 (R n ). As ϕ = ψ + η, then we have Therefore, there exists b R such that η W 0 (R n ) C ϕ W 0 (R n ). η + b W 2,n 0 (R n ) C η W 0 (R n ) C ϕ W 0 (R n ), and the proof is complete. Remark 5. This theorem suggests this open question: if ϕ W 2,n/2 0 (R n ), are there a function ψ W 2,n/2 0 (R n ) L (R n ) and a function η W 3,n/2 0 (R n ) such that ϕ = ψ + η, with the corresponding estimate? We define now the space X(R n ) = { f L 1 (R n ), div f W 2,n/(n 1) 0 (R n )}, which is Banach space endowed with the following norm f X(R n ) = f L 1 (R n ) + div f W 2,n/(n 1) 0 (R n ). (25) Theorem 2.9. Let f X(R n ). Then f W /(n 1) 0 (R n ) and we have the following inequality ϕ W 0 (R n ) L (R n ), R n f ϕ C f X(R n ) ϕ W 0 (R n ). (26) 13

15 Proof. We consider the following linear operator ϕ F f ϕ defined on R n D(R n ). Thanks to Theorem 2.8, we have < F, ϕ > = R n f (ψ + η) = R n f ψ < div f, η > W 2,n/(n 1) 0 (R n ) W 2,n f L1 (R n ) ψ L (R n ) + div f W 2,n/(n 1) 0 C f X(R n ) ϕ W 0 (R n ). 0 (R n ) η W 2,n 0 As D(R n ) is dense in W 0 (R n ) and by applying Hahn-Banach Theorem, we can uniquely extend F by an element F W /(n 1) 0 (R n ) satisfying F W /(n 1) 0 (R n ) C f X(R n ). Besides, the linear operator f F from X(R n ) into W /(n 1) 0 (R n ) is continuous and injective. Therefore, X(R n ) can be identified to a subspace of W /(n 1) 0 (R n ) with continuous and dense embedding. Remark 6. i) Let f X(R n ). Then f = 0 if and only if R n λ P 1, < div f, λ > 2,n/(n 1) W 0 (R n ) W 2,n 0 (R n ) = 0. Note that f i = < f i, 1 > /(n 1) W R n 0 (R n ) W 0 (R n ). ii) Let f X(R n ) satisfying f = 0. Then we have the following inequality: R n for every ϕ W 0 (R n ), < f, ϕ > W /(n 1) 0 (R n ) W 0 (R n ) C f X(R n ) ϕ L n (R n ). Actually, we observe that for any a R n, < f, ϕ > = < f, ϕ + a > f X(R n ) ϕ + a W 0 (R n ). Consequently, taking the infinum, we have for every ϕ W 0 (R n ) (see [4]): < f, ϕ > C f X(Rn ) ϕ Ln (R n ). (27) It is then easy to deduce the following corollary. Corollary Let f L 1 (R n ) and div f = 0. Then f = 0 and for every R n ϕ W 0 (R n ) L (R n ), we have f ϕ C f L1 (R n ) ϕ Ln (R n ). R n 14

16 Corollary Let f L 1 (R 3 ) and curl f W 2,3/2 0 (R 3 ). Then we have f W 1,3/2 0 (R 3 ) and the following estimate: for every ϕ W 1,3 0 (R3 ) L (R 3 ), f ϕ C ( f L1 (R 3 ) + curl f 2,3/2 W R 3 0 (R 3 ) ) ϕ W 1,3 0 (R3 ). If moreover following estimate R 3 f = 0, then for every ϕ W 1,3 0 (R3 ) L (R 3 ), we have the R 3 f ϕ C ( f L1 (R 3 ) + curl f W 2,3/2 0 (R 3 ) ) ϕ L 3 (R 3 ). Proof. Using Proposition 2.13, the proof is similar as the one of Theorem 2.9 and Remark 6. When f X(R n ), we can improve Corollary 2.2 as follows (recall that V(R n ) is not dense in H (R n )). Proposition Assume that f X(R n ) satisfying v V(R n ), f v dx = 0. (28) R n Then there exists a unique π L n/(n 1) (R n ) such that f = π and the following estimate holds π L n/(n 1) (R n ) C f W /(n 1) 0 (R n ). Proof. This proposition is an immediate consequence of the embedding X(R n ) W /(n 1) 0 (R n ) and De Rham s Theorem (see Alliot - Amrouche [2]). Remark 7. i) First remark that the hypothesis div f W 2,n/(n 1) 0 (R n ) of Proposition 2.12 is necessary because if f = π with π L n/(n 1) (R n ), then f W /(n 1) 0 (R n ) and div f = π W 2,n/(n 1) 0 (R n ). ii) Also note that (28) is equivalent to v V(R n ), < f, v > /(n 1) W = 0. (29) 0 (R n ) W 0 (R n ) As V(R n ) is dense in V 0 (R n ), (29) is also equivalent to v V 0 (R n ), < f, v > /(n 1) W = 0. (30) 0 (R n ) W 0 (R n ) Since vectors of the canonical basis of R n belong to V 0 (R n ), we deduce that if (28) holds, then f = 0. R n 15

17 Proposition Let ϕ W 1,3 0 (R3 ). Then there exist ψ W 1,3 0 (R3 ) L (R 3 ) and η W 2,3 0 (R3 ) such that and we have the following estimate ϕ = ψ + curl η ψ L 3 (R 3 ) + ψ L (R 3 ) + D 2 η L 3 (R 3 ) C ϕ L 3 (R 3 ). Moreover, ψ and η can be chosen such that ψ W 1,3 0 (R3 ) + ψ L (R 3 ) + η W 2,3 0 (R3 ) C ϕ W 1,3 0 (R3 ). Proof. Let ϕ W 1,3 0 (R3 ), then div ϕ L 3 (R 3 ). Thanks to Theorem 2.3, there exists ψ L (R 3 ) W 1,3 0 (R3 ) such that div ψ = div ϕ and ψ L (R 3 ) + ψ W 1,3 0 (R3 ) C div ϕ L 3 (R 3 ). Setting z = ϕ ψ. We know that there exists η 0 W 2,3 0 (R3 ) such that η 0 = curl z satisfying the following estimate η 0 W 2,3 0 (R3 ) C ϕ L3 (R 3 ). However, div η 0 W 1,3 0 (R 3 ) is harmonic, then we deduce div η 0 = a with a R. Therefore, we have curl (curl η 0 ) = curl z. We set y = z curl η 0. Then y W 1,3 0 (R3 ), div y = 0 and curl y = 0, i.e., y = 0. Then we can deduce y = b R 3. Let q P 1 such that b = curl q. We now set η = η 0 + q. Then and we have the following estimate ϕ = ψ + curl η ψ W 1,3 0 (R3 ) + ψ L (R 3 ) + D 2 η L 3 (R 3 ) C ϕ L 3 (R 3 ). We now introduce the following proposition. Proposition Assume that u L 1 loc (Rn ) satisfying u L 1 (R n ). Then there exists a unique constant K R such that u+k L n/(n 1) (R n ). Moreover, we have u + K L n/(n 1) (R n ) C u L1 (R n ) (31) and K = lim x where S n 1 is the unit sphere of R n and ω n its surface. 1 ω n S n 1 u(σ x )dσ, (32) 16

18 Proof. From Proposition 2.7 [21], it is easy to prove that there exists a unique constant K R verifying u + K L n/(n 1) (R n ) and we have (31). Thanks to Lemma 1.3 [2], we have lim x x n 1 u(σ x ) + K dσ = 0. (33) S n 1 We set D R (r) = {x R n,r< x <R} By proceeding similarly as in [22], we have D R (r) R ( S n 1 r u ρ ρ2 )dρ + R r u dx. ρ u dσdρ, S n 1 where u is the projection of gradient of u on the unit sphere S n 1. u 2 = r 2 [ u 2 u ρ 2 ]. By Hölder s and Wirtinger s inequalities D R (r) +C S n 1 R r R r ( u ρ dρ R r ρ 2 dρ S n 1 u( x σ) 1 ω n S n 1 u( x σ)dσ dσ)ρdρ. By consequence, we have D R (r) Cr u(rσ) u(rσ) dσ S n 1 R + C ( u( x σ) 1 u( x σ)dσ dσ)ρdρ. r S n 1 ω n S n 1 (34) Since the both integrals on the right are non-negative, each is separately bounded by D R (r). Then, there exists a function u L 1 (S n 1 ) such that lim u( x σ) u ( x σ ) dσ = 0, x S n 1 lim u( x σ)dσ = u (σ)dσ. x S n 1 S n 1 Thanks to (33), we deduce that u = K and from (34) we have (32). Remark 8. We have a similar result as the above proposition in the case u D (R n ) and u L p (R n ) for all p > 1 (see Payne and Weinberger [22], Amrouche and Razafison [9], for example). 17

19 3. Vector potentials and elliptic problems in the whole space Proposition 3.1. There exists C > 0 such that for any u L 3/2 (R 3 ) satisfying curl u L 1 (R 3 ) and div u = 0, we have the following estimate u L 3/2 (R 3 ) C curl u L1 (R 3 ). (35) Proof. Setting f = curl u. Then f belongs to W 1,3/2 0 (R 3 ) and for all i = 1, 2, 3, < f i, 1 > 1,3/2 W 0 (R 3 ) W 1,3 0 (R 3 ) = 0. Therefore, there exists a unique solution z W 1,3/2 0 (R 3 ) of z = f in R 3 and satisfying z W 1,3/2 0 (R 3 ) C curl u W 1,3/2 0 (R 3 ) C curl u L 1 (R 3 ). The last inequality is consequence of the embedding X(R 3 ) W 1,3/2 0 (R 3 ) and because div f = 0 and f = curl u X(R 3 ). Moreover, it is easy to see that div z = 0 in R 3. By setting w = u curl z, we can easily deduce that w = 0 in R 3. Then w = 0, u = curl z and we obtain the estimate (35). Proceeding similarly as in the proof of Proposition 2.1, we can show the following corollary. Corollary 3.2. Let f L 3 (R 3 ) such that div f = 0. L (R 3 ), with div u = 0 and such that Then there exists u curl u = f, with u L (R 3 ) C curl u L 3 (R 3 ). In three-dimensional space, thanks to Theorem 2.8, we can deduce the following proposition. Proposition 3.3. Let f L 3 (R 3 ) such that div f = 0. Then there exist ϕ W 1,3 0 (R3 ), unique up to a constant vector, and ψ W 1,3 0 (R3 ) L (R 3 ) such that curl ϕ = curl ψ = f and div ϕ = 0, satisfying the following estimate ϕ L 3 (R 3 ) + ψ W 1,3 0 (R3 ) + ψ L (R 3 ) C f L 3 (R 3 ). Proof. From the hypothesis, we deduce curl f W 1,3 0 (R 3 ). Then, from [4], there exists ϕ W 1,3 0 (R3 ), unique up to a constant vector, such that ϕ = curl f in Ω and satisfying the following estimate inf a R 3 ϕ + a W 1,3 0 (R3 ) C f L 3 (R 3 ). As div ϕ L 3 (R 3 ) is harmonic, then we deduce div ϕ = 0. Consequently, we have ϕ = curl curl ϕ div ϕ = curl curl ϕ. 18

20 Therefore, we obtain curl ( curl ϕ f ) = 0 in Ω. Setting z = curl ϕ f. Then z L 3 (R 3 ), div z = 0 and curl z = 0. Hence, we deduce z = 0 and z = 0, i.e., curl ϕ = f. Applying Theorem 2.8, there exist ψ W 1,3 0 (R3 ) L (R 3 ) and η W 2,3 0 (R 3 ) such that ϕ = ψ + η in R 3, with the following estimate ψ W 1,3 0 (R3 ) + ψ L (R 3 ) C ϕ L3 (R 3 ) C f L3 (R 3 ). The function ψ is the required function, then the proof is finished. Remark 9. From the previous proposition, we have the following Helmholtz decomposition: for all f L 3 (R 3 ), we have f = curl ψ + p (36) with ψ W 1,3 0 (R3 ) L (R 3 ) and p W 1,3 0 (R 3 ) and the following estimate ψ W 1,3 0 (R3 ) + ψ L (R 3 ) + p L 3 (R 3 ) C f L 3 (R 3 ). (37) Indeed, we have div f W 1,3 0 (R 3 ) R because of f L 3 (R 3 ). Then there exists p W 1,3 0 (R 3 ), unique up to a constant, such that p = div f and satisfying the following estimate p L 3 (R 3 ) C f L 3 (R 3 ). The function f p satisfies the hypothesis of Proposition 3.3, then we can decompose f as in (36) and we have the estimate (37). Corollary 3.4. There exists C > 0 such that for all div u = 0, we have the following inequality u L 3/2 (R 3 ) satisfying u L 3/2 (R n ) C inf f+g = curl u ( f L 1 (R 3 ) + g W 1,3/2 0 (R 3 ) ) (38) with f L 1 (R 3 ) and g W 1,3/2 0 (R 3 ). Proof. We consider two following operators A = curl : H 3/2 (R 3 ) L 1 (R 3 ) + W 1,3/2 0 (R 3 ), A = curl : W 1,3 0 (R3 ) L (R 3 ) H 3 (R 3 ). The rest of this proof is similar to the one of Proposition 2.1. The following corollary improves Corollary Corollary 3.5. Let f L 1 (R 3 ) such that div f = 0. Then for all ϕ W 1,3 0 (R3 ), we have the following estimate < f, ϕ > W 1,3/2 0 (R 3 ) W 1,3 0 (R3 ) C f L 1 (R 3 ) curl ϕ L3 (R 3 ). (39) 19

21 Proof. First remark that from the hypothesis, we deduce f W 1,3/2 0 (R 3 ). Let ϕ W 1,3 0 (R3 ). Then we have curl ϕ L 3 (R 3 ). Thanks to Proposition 3.3, there exists ψ W 1,3 0 (R3 ) L (R 3 ) such that curl ψ = curl ϕ with the following estimate ψ W 1,3 0 (R3 ) + ψ L (R 3 ) C curl ϕ L 3 (R 3 ). (40) Besides, there exists η W 2,3 0 (R 3 ) such that ϕ = ψ + η in R 3. Then we have < f, ϕ > 1,3/2 W = f ψ + < f, η > 0 (R 3 ) W 1,3 0 (R3 1,3/2 ) W R 3 0 (R 3 ) W 1,3 0 (R3 ) = f ψ. R 3 Therefore, the estimate (39) is deduced from the estimate (40). Remark 10. We have another proof for the above corollary as follows: We can write that f = u = curl curl u, with u W 1,3/2 0 (R 3 ) satisfying the following estimate Then we deduce u W 1,3/2 0 (R 3 ) C f W 1,3/2 0 (R 3 ) C f L 1 (R 3 ). < f, ϕ > W 1,3/2 0 (R 3 ) W 1,3 0 (R3 ) = < curl u, curl ϕ > L 3/2 (R 3 ) L 3 (R 3 ) curl u L 3/2 (R 3 ) curl ϕ L3 (R 3 ) C f L 1 (R 3 ) curl ϕ L 3 (R 3 ). We now prove the following proposition. Proposition 3.6. Let f L 1 (R 3 ) such that div f = 0. Then there exists a unique ϕ L 3/2 (R 3 ) such that curl ϕ = f and div ϕ = 0 in R 3 satisfying the following estimate ϕ L 3/2 (R 3 ) C f L 1 (R 3 ). Proof. From the definition of X(R 3 ), we have f W 1,3/2 0 (R 3 ). As : W 1,3/2 0 (R 3 ) W 1,3/2 0 (R 3 ) R is an isomorphism (see [4]), then there exists a unique h W 1,3/2 0 (R 3 ) such that h = f and we have the following estimate h W 1,3/2 0 (R 3 ) C f L 1 (R 3 ). Moreover, we can see that div h = 0 and then h = curl curl h. The proposition can be easily obtained by setting ϕ = curl h. We have the following Helmholtz decomposition. 20

22 Corollary 3.7. Let f L 1 0(R 3 ) such that div f W 2,3/2 0 (R 3 ). Then there exists a unique ϕ L 3/2 (R 3 ) such that div ϕ = 0 and a unique π L 3/2 (R 3 ) satisfying f = curl ϕ + π and the following estimate holds Proof. ϕ L 3/2 (R 3 ) + π L 3/2 (R 3 ) C It is clear that ( ) f L 1 (R 3 ) + div f 2,3/2 W. 0 (R 3 ) λ P 1, < div f, λ > = 0. From the hypothesis and [4], there exists a unique π L 3/2 (R 3 ) such that π = div f and π L 3/2 (R 3 ) C div f W 2,3/2 0 (R 3 ). Then, we deduce that f π W 1,3/2 0 (R 3 ) R 3. Therefore, there exists a unique z W 1,3/2 0 (R 3 ) such that z = f π. Moreover, we see that div z = 0. Then, f π = curl curlz. The proof is complete by setting ϕ = curl z. The following proposition is an extension of Proposition 3.6. Proposition 3.8. Let f L 1 0(R 3 ) + W 1,3/2 0 (R 3 ) such that div f = 0 and satisfying the following compatibility condition i = 1, 2, 3, < f i, 1 > = 0. Then there exists a unique ϕ L 3/2 (R 3 ) such that curl ϕ = f and div ϕ = 0 in R 3 satisfying the following estimate ϕ L 3/2 (R 3 ) C f L 1 (R 3 )+W 1,3/2 0 (R 3 ). Proof. Let f = g +h with g L 1 0(R 3 ), h W 1,3/2 0 (R 3 ) and div f = 0. Then div g = div h W 2,3/2 0 (R 3 ). Therefore we deduce g W 1,3/2 0 (R 3 ) and g R 3. As f W 1,3/2 0 (R 3 ) R 3, then there exists a unique z W 1,3/2 0 (R 3 ) such that z = f and divz = 0. The proof is finished by choosing ϕ = curl z. In two-dimensional space, we have a similar result as Proposition 3.8. Proposition 3.9. Assume that f L 1 0(R 2 ) + W 1,2 0 (R 2 ) such that div f = 0. Then there exists ϕ L 2 (R 2 ) such that curl ϕ = f and satisfying the following estimate ϕ L2 (R 2 ) C f L 1 (R 2 )+W 1,2 0 (R 2 ). 21

23 Corollary Let f X(R n ) and f = 0. Then the following problem R n u = f in R n has a unique solution u W /(n 1) 0 (R n ). Moreover, we have u = E n f and u satisfies the following estimate u W /(n 1) 0 (R n ) C f X(R n ). Proof. This corollary is an immediate consequence of Theorem 2.9, Remark 6 and the fact that : W /(n 1) 0 (R n ) W /(n 1) 0 (R n ) R if n 3 : W 1,2 0 (R 2 ) W 1,2 0 (R 2 ) R if n = 2 are isomorphisms (cf. [4]). We recall that W /(n 1) 0 (R n ) R = { f W /(n 1) 0 (R n ); < f, 1 > = 0 }. Remark 11. In particular, when n = 2, f L 1 (R 2 ) and div f = 0, the solution given in Corollary 3.10 belongs to L (R n ) C 0 (R n ). The reader can find this result in H. Brézis, J. Van Schaftingen [15] and J. Bourgain, H. Brézis [11]. Corollary Let f L 1 (R n ) such that n f W 2,n/(n 1) 0 (R n ). i) Then we have f W /(n 1) 0 (R n ) and the following estimate holds f W /(n 1) 0 (R n ) C ii) Furthermore if satisfying the following problem and we have the following estimate u W /(n 1) 0 (R n ) C ( f L1 (R n ) + n f W 2,n/(n 1) 0 (R n ) R n f = 0, then there exists a unique u W /(n 1) 0 (R n ) u = f in R n, ( f L 1 (R n ) + n f W 2,n/(n 1) 0 (R n ) Proof. This corollary can be obtained by applying Theorem 2.9 and Corollary 3.10 with f = (0,..., 0, f). ). ). 22

24 Remark 12. We know that if f L 1 (R n ), then E n f L n/(n 2) w (R n ) if n 3 and (E n f) L n/(n 1) w (R n ). As n f W 2,n/(n 1) 0 (R n ) with n 2, we deduce that n (E n f) L n/(n 1) (R n ), where { } L p w(r n ) = f measurable on R n ; sup t( {x R n ; f(x) > t} ) 1/p <. t>0 However, in Corollary 3.11, (E n f) L n/(n 1) (R n ) and E n f L n/(n 2) (R n ). Then there is an anisotropic phenomenon. Recall now a result in [2] concerning the Stokes problem in R n. Theorem Let (f, g) W 1,p 0 (R n ) L p (R n ) satisfying the compatibility condition as follows λ P [1 n/p ], < f, λ > W 1,p = 0. (41) 0 (R n ) W 1,p 0 (R n ) Then the Stokes system (S) u + π = f and div u = g in R n, has a unique solution (u, π) W 1,p 0 (Rn )/P [1 n/p] L p (R n ). Moreover, we have the estimate inf λ P [1 n/p] u+λ W 1,p 0 (Rn ) + π L p (R n ) C ( ) f W 1,p 0 (R n ) + g L p (R n ). Corollary Let (f, g) X(R n ) L n/(n 1) (R n ) satisfying the compatibility condition as follows < f i, 1 > = 0 for all i = 1, 2, 3. Then the Stokes system (S) has a unique solution (u, π) W /(n 1) 0 (R n ) L n/(n 1) (R n ) and the following estimate holds u W /(n 1) 0 + π L n/(n 1) C ( f /(n 1) W + g L n/(n 1) 0 Proof. This corollary is a consequence of X(R n ) W /(n 1) 0 (R n ) and Theorem ). 4. The div-grad operators in the half-space First of all, we introduce the following notations. If v is a function defined on, we set { v (x v(x, x n ) if x n > 0,, x n ) = v(x (42), x n ) if x n < 0, 23

25 and v (x, x n ) = { v(x, x n ) if x n > 0, v(x, x n ) if x n < 0. (43) We also set W 0,p 1 (div; Rn +) = { v L p (); div v W 0,p 1 () }, where W 0,p 1 () is the subspace of functions u in L p () which satisfy x u in L p (), and their normal traces are described in the following lemma (see C. Amrouche, S. Nečasová and Y. Raudin [7]): Lemma 4.1. The linear mapping γ en : D() D(R n 1 ), v v n Γ, can be extended to a linear continuous mapping γ en : W 0,p 1 (div; Rn +) W 1/p,p 0 (R n 1 ), for any 1 < p <. Moreover, for all v W 0,p 1 (div; Rn +) and for all ϕ W 1,p 0 (), we have the following Green formula v ϕ dx + ϕ div v dx = < v n, ϕ >. 1/p,p W 0 (Γ) W 1/p,p 0 (Γ) Define now the following spaces H p (div; ) = { v L p (); div v L p ()}, Hp () = { v L p (); div v = 0 in ; v n = 0 on Γ }, V 1,p 0 (Rn +) = { v W 1,p 0 (Rn +); div v = 0 in }, where W 1,p 0 (Rn +) is the subspace of functions of W 1,p 0 (Rn +) which are equal to zero on the boundary of. Remark that from Lemma 4.1, if v L p () and div v = 0 in, then v n W 1/p,p 0 (R n 1 ). We can show classically the following lemma. Lemma 4.2. For any 1 p <, we have that D() is dense in H p (div; ). Moreover, for any 1 < p <, the following linear mapping H p (div; ) v W 1/p,p 1 (Γ) v n is continuous and surjective. 24

26 Lemma 4.3. For any 1 p <, we have 1) V() is dense in Hp (), 2) V() is dense in V 1,p 0 (Rn +). Proof. 1) We give the proof for the case p = 1. With similar arguments, it is then easy to consider the case p > 1. The idea consists in using Hahn-Banach Theorem and showing that f = 0 if f [ H1 ()] satisfying v V(), < f, v > = 0. We know that there exists π D () such that f = π. Thanks to [4], we have π W 1,p loc (Rn +) for any p 1. On the other hand, π C 0 () and we can suppose π(0) = 0. So that x, π(x ) x. π L ( ). Let now ψ C () with 0 ψ 1 satisfying ψ(x ) = 1 if x 1 and ψ(x ) = 0 if x 2. We set ψ k (x ) = ψ( x k ) and π k = ψ k π. Then π k L () and supp π k B(0, 2k). Moreover, if x 2k, then π k (x ) C k π(x ) + C π(x ) 3C π L ( ). Therefore, we have that ( π k ) k is bounded in L (). In fact, we show that π k π in L (), i.e., ϕ L 1 () and j = 1,..., n, ( π k π )ϕ 0 x j x j when k. Let now v H1 (). Thanks to Lemma 4.1, we have π k v = π k div v = 0. By passing to the limit in the above equation, we obtain π v = 0 = < f, v >. This ends the proof of the case p = 1. 2) We content ourselves here with the case n = 3 and p > 1. Proceeding similarly as in the proof of Lemma 5.6, we can show that if f W 1,p 0 (R3 +) such that div f = 0, then f = curl ϕ with ϕ W 2,p 0 (R3 +) and ϕ W 2,p 0 (R3 + ) C f W 1,p 0 (R3 + ). 25

27 As D(R 3 +) is dense in W 2,p 0 (R3 +), there exists ϕ k D(R 3 +) such that ϕ k ϕ in W 2,p 0 (R3 +). The sequence f k = curl ϕ k answers this question. Lemma 4.4. Let 1 < p <. The following properties are satisfied: i) The mapping : L p () [ V 1,p 0 () ] 0 is an isomorphism. ii) The mapping is an isomorphism. div : W 1,p 0 ()/V 1,p 0 () L p () Proof. It suffices to show that the second operator is surjective. More generally, let ϕ L p () and g W 1 1/p,p 0 (R n 1 ) (instead of g = 0). We know that there exists u g W 1,p 0 () such that u g = g on Γ. This shows that we can assume g = 0. Then let π W 2,p 0 () be one solution of the following equation π = ϕ in and π x n = 0 on Γ. Let ψ i W 2,p 0 () with i = 1,..., n 1 such that ψ i = 0 and ψ i = π x n x i Γ. We set z = ( ψ 1,..., ψ n 1 x n x n n 1, k=1 ψ k x k ). The function u = π z satisfies u W 1,p 0 (), div u = ϕ in and u = 0 on Γ. Remark 13. The property i) of Lemma 4.4 can be rewritten as follows: for any f W 1,p 0 () such that for any v V 1,p 0 () satisfying < f, v > = 0, there exists a unique π L p () such that f = π with the following estimate We can improve this result as follows. π L p ( ) C f W 1,p 0 ( ). Theorem 4.5. Assume n 3 and 1 < p <. Let f W 1,p 0 () such that Then f = π, with π L p (). v V(), < f, v > = 0. on 26

28 Proof. Let ϕ W 1,p 0 (R n ). We define the operator P ϕ(x, x n ) = ϕ(x, x n ) ϕ(x, x n ), x n > 0. By duality, we also define the operator It is clear to see that < P f, ϕ > := < f, P ϕ > W 1,p 0 ( ) W 1,p 0 ( ). P : W 1,p 0 () W 1,p 0 (R n ) is continuous. Remark that, thanks to De Rham s Theorem, there exists θ D () such that f = θ. On the other hand, as θ D () and θ W 1,p 0 (), then θ L p loc (Rn +) and P θ = θ. Then we have θ W 1,p 0 (R n ) and θ W 2,p 0 (R n ). Moreover, < θ, 1 > = 0. Thus there exists a unique λ L p (R n ) such that (λ θ ) = 0 in R n. However, (λ θ ) is harmonic and belongs to W 1,p 0 (R n ). Consequently, θ = λ and there exists C R such that θ +C = λ. The function π = θ +C L p () is the required solution. Remark 14. In Lemma 4.3, we have given a constructive proof for the density of the space V() in V 1,p 0 (Rn +). Using Hahn-Banach Theorem and Theorem 4.5, a second proof of this result can be given. We introduce the following proposition. Proposition 4.6. Let f L 1 () such that v L () with div v = 0, f v = 0. Then there exists a unique π L n/(n 1) () satisfying f = π and the following estimate holds π L n/(n 1) ( ) C f L1 ( ). Proof. Let u L n/(n 1) () satisfying u L 1 (). Then u L n/(n 1) (R n ), u L 1 (R n ) and we have the following estimate u L n/(n 1) ( ) u L n/(n 1) (R n ) C u L1 (R n ) 2C u L1 ( ). The remains of this proof is identical to the one of Corollary 2.2. We set B a + = {x, x < a} with a R and a > 0. We introduce the following lemma. 27

29 Lemma 4.7. Let f L n 0 (B + a ). Then there exists u W 0 (B + a ) L (B + a ) such that div u = f and we have the following estimate where C does not depend on a. u L (B + a ) + u L n (B + a ) C f Ln (B + a ) (44) Proof. Let f L n 0 (B a + ) and we set g(x ) = a f(a x ) with x B 1 +. Then we can deduce g L n 0 (B 1 + ). Thanks to Theorem 3 [10], there exists v W 0 (B 1 + ) L (B 1 + ) such that div v = g and we have the following estimate v L (B + 1 ) + v L n (B + 1 ) C g L n (B + 1 ). (45) We now set u(x ) = v( x a ) with x B+ a. Then div u = f and we have B + a u(x ) n dx = B + 1 Besides, we can similarly prove that 1 a n v(t) n a n dt = B + 1 v(t) n dt. (46) u L (B a + ) = v L (B + 1 ) and f L n (B a + ) = g L n (B + 1 ). (47) The estimate (44) is deduced from (45), (46) and (47) and the proof is finished. We now give a similar result of Corollary 4.7 but the one is considered in the half-space. Theorem 4.8. Let f L n (). Then there exists u W 0 (Rn +) L () such that div u = f. However, we have the following estimate u L ( ) + u W 0 ( ) C f L n ( ). (48) Proof. It is easy to see that there exists a sequence (f k ) k N D() converging towards f L n () because D() is dense in L n (). Let B r + k such that supp f k B r + k. Applying Lemma 4.7, there exists u k W 0 (B r + k ) L (B r + k ) such that div u k = f k and we have the following estimate u k L (B + r k ) + u k L n (B + r k ) C f k L n (B + r k ) C f k Ln ( ) where C only depends on n. By extending u k in by zero outside B r + k denoting ũ k its extended function, we have and ũ k L ( ) = u k L (B + r k ) and ũ k W 0 ( ) = u k W 0 (B + r k ). Then (ũ k ) k is bounded in W 0 (Rn +) L () and we can deduce that there exists a subsequence, again denoted by (ũ k ) k such that ũ k u in W 0 (Rn +) and ũ k u in L (). Hence, we have div u = f and the estimate (48). Similarly to Corollary 2.4, we have the following result. 28

30 Corollary 4.9. i) There exists C > 0 such that for all u L n/(n 1) (), we have the following estimate u L n/(n 1) ( ) C inf f+g = u ( f L 1 ( ) + g W /(n 1) 0 ( ) ) (49) with f L 1 () and g W /(n 1) 0 (). ii) Let f L 1 () + W /(n 1) 0 () satisfying the following compatibility condition v V 0 () L (), < f, v > = 0. (50) Then there exists a unique π L n/(n 1) () such that f = π. We define now the space X() = { f L 1 (), div f W 2,n/(n 1) 0 ()}. Theorem Let f X(). Then f W /(n 1) 0 () and the following estimate holds f W /(n 1) 0 ( ) C f X(R n + ). Proof. The proof is similar to the one of Theorem 2.9. Proposition Assume that f X() satisfying v V(), f v dx = 0. Then there exists a unique π L n/(n 1) () such that f = π and the following estimate holds π L n/(n 1) ( ) C f W /(n 1) 0 ( ). Proof. This proposition is an immediate consequence of the embedding X() W /(n 1) 0 () and Theorem 4.5. We now introduce the following theorem. Theorem Let ϕ W 0 (Rn +). Then there exist ψ W 1, n 0 () L () and η W 2,n 0 (Rn +) such that ϕ = ψ + η. Moreover, we have the following estimate ψ W 0 ( ) + ψ L ( ) + η W 2,n 0 ( ) C ϕ W 0 ( ). (51) 29

31 Proof. We know that D(R n +) is dense in W 0 (Rn +), then there exists a sequence (ϕ k ) k N D() which converges towards ϕ W 0 (Rn +). Let now B r + k such that supp ϕ k B r + k and we set ϕ k (x ) = ϕ k(r k x ). Then we deduce ϕ k W 0 (B 1 + ). Thanks to Theorem 2.6, there exist ψ k W 0 (B 1 + ) L (B 1 + ) and η k W 2,n 0 (B 1 + ) such that ϕ k = ψ k + η k, with the following estimate ψ k L n (B + 1 ) + ψ k L (B + 1 ) + D2 η k L n (B + 1 ) C ϕ k L n (B + 1 ). We now set ψ k (x ) = ψ k( x r k ) and η k (x ) = r k η k( x r k ). Then we have ϕ k = ψ k + η k. Proceeding similarly as in the proof of Corollary 2.7 and by passing to limit, we can show ϕ = ψ+ η with the following estimate ψ L n ( ) + ψ L ( ) + D 2 η L n ( ) C ϕ L n ( ). The estimate (51) follows from the fact that the semi-norm. L n ( ) (respectively, D 2. L n ( ) ) defines on W 0 (Rn +) (respectively, W 2,n 0 (Rn +)) a norm which is equivalent to the norm. W 0 ( ) (respectively,. W 2,n 0 (R n )). + Proposition Let ϕ W 0 (). Then there exists ψ W 0 () L () and η W 2,n 0 () such that ψ n = 0 on Γ satisfying and the following estimate holds ϕ = ψ + η ψ L ( ) + ψ W 0 ( ) + η W 2,n 0 ( ) C ϕ W 0 ( ). (52) Proof. As in Theorem 4.12, we can prove that ϕ = ψ 0 + η 0 such that ψ 0 W 0 () L () and η 0 W 2,n 0 () with the following estimate ψ 0 L ( ) + ψ 0 W 0 ( ) + η 0 W 2,n 0 ( ) C ϕ W 0 ( ). Setting µ = ψ 0n on Γ, then µ L (Γ) W 1 1/n,n 0 (Γ). We can prove similarly as in Lemma 3.10 [15] that there exists v W 2,n 0 () W 1, () such that v = 0 and v = µ on Γ with the following estimates x n and v W 2,n 0 ( ) C µ W 1 1/n,n 0 (Γ) C ψ 0 W 0 ( ) v L ( ) C µ L (Γ) C ψ 0 L ( ). The proof is complete by setting ψ = ψ 0 v and η = η 0 + v. 30

32 Remark 15. We can give another proof of the existence of ϕ. We know that f W 1,3/2 0 (R 3 +) R 3. Then, there exists a unique z W 1,3/2 0 (R 3 +) satisfying z = f, with div z = 0 in R 3 +. The function ϕ = curl z is the required function. Proposition Let ϕ W 1,3 0 (R3 +). Then there exist ψ W 1,3 0 (R3 +) L (R 3 +) and η W 2,3 0 (R3 +) such that and we have the following estimate ϕ = ψ + curl η with ψ = 0 on Γ ψ W 1,3 0 (R3 + ) + ψ L (R 3 + ) + η W 2,3 0 (R3 + ) C ϕ W 1,3 0 (R3 + ). Proof. Let ϕ W 1,3 0 (R3 +). Then ϕ W 1,3 0 (R3 ) (see at the beginning of this section for the notations), we can use Proposition 2.13: there exist ψ 0 W 1,3 0 (R3 ) L (R 3 ) and η 0 W 2,3 0 (R3 ) such that ϕ = ψ 0 + curl η 0 in R 3, with the following estimate ψ 0 W 1,3 0 (R3 ) + ψ 0 L (R 3 ) + η 0 W 2,3 0 (R3 ) C ϕ W 1,3 0 (R3 + ). As in the proof of Lemma 3.10 of [15], we can prove that there exist α W 2,3 0 (R3 +) with α L (R 3 +) satisfying α = 0 and α x 3 = ψ 0 on Γ. Moreover, we have the estimate α W 2,3 0 (R3 + ) + α L (R 3 + ) C ( ψ 0 W 1,3 0 (R3 ) + ψ 0 L (R 3 )). One has thus the conclusion with ψ = ψ 0 curl α and η = η 0 + α. Remark 16. We have another result with the data ϕ W 1,3 0 (R3 +) (see Proposition 5.7). Proposition i) If f L 1 (R 3 +) with div f = 0, then f 3 = 0. R 3 + ii) If moreover f 3 = 0 on Γ, then f = 0. R 3 + Proof. i) Setting f = (f, f3 ). Then f L 1 (R 3 ), div f = 0 in R 3 and therefore f = 0, i.e. f 3 = 0. R 3 R 3 + ii) We have known f 3 = 0. Setting f = (f, f 3 ). Then f L 1 (R 3 ), R 3 + div f = 0 in R 3 and therefore f = 0 that implies R 3 f = 0. R 3 31

33 Remark 17. i) Let f L 1 (R 3 +) that satisfies the property: For any η W 2,3 0 (R 3 +) such that η L (R 3 +) and η = 0 on Γ, we have f η = 0. R 3 + By taking η = x 3, then we obtain f 3 = 0. ii) Let f L 1 (R 3 +) that satisfies the property: For any η W 2,3 0 (R 3 +) such that η L (R 3 ), we have taking η = x i with i = 1, 2 and after taking η = x 3, we find R 3 + R 3 + R 3 + f η = 0. By f = Vector potentials in the half-space Proposition 5.1. Let f L 3 (R 3 +) such that div f = 0 in R 3 +. Then there exists ϕ W 1,3 0 (R3 +) such that f = curl ϕ with div ϕ = 0 in R 3 +, ϕ 3 = 0 on Γ and we have the following estimate ϕ W 1,3 0 (R3 + ) C f L3 (R 3 + ). Proof. i) Setting f = (f, f3 ). It is easy to show f L 3 (R 3 ) and div f = 0 in R 3. Let θ W 1,3 0 (R3 ) such that θ = curl f. Then div θ = 0 and curl (curl θ f ) = 0 in R 3. We now set z = curl θ f in R 3 and can deduce z L 3 (R 3 ), div z = 0 and curl z = 0 in R 3. As z = 0, then z = 0 in R 3. It means that f = curl θ in R 3 and f = curl θ R 3 + with div θ = 0 in R 3 +. ii) Let h W 2,3 0 (R 3 +) be a solution of the following equation h = 0 in R 3 + and h x 3 = θ 3 on Γ. Setting ϕ = θ h. Then we have ϕ W 1,3 0 (R3 +), div ϕ = 0 in R 3 + and ϕ 3 = 0 on Γ. Theorem 5.2. Let f L 3 (R 3 +) such that div f = 0 in R 3 + and f 3 = 0 on Γ. Then there exists a unique ϕ W 1,3 0 (R3 +) such that f = curl ϕ with div ϕ = 0 in R 3 + and ϕ = 0 on Γ. Moreover, we have the following estimate ϕ W 1,3 0 (R3 + ) C f L3 (R 3 + ). Proof. i) We start this proof by showing the uniqueness of ϕ. Indeed, if ϕ W 1,3 0 (R3 +) such that curl ϕ = 0, div ϕ = 0 in R 3 + and ϕ = 0 on Γ, then there exists q W 2,3 0 (R 3 +) such that ϕ = q in R 3 + with q = 0 in R 3 + and q is a constant on Γ. Consequently, q is a constant in R 3 + and we then deduce ϕ = 0 in R 3 +. ii) We now consider the existence of ϕ. Setting f = (f, f 3 ). Then we have 32