Problem of Second grade fluids in convex polyhedrons

Problem of Second grade fluids in convex polyhedrons J. M. Bernard* Abstract This article studies the solutions of a three-dimensional grade-two fluid model with a tangential boundary condition, in a polyhedron. We begin to split the problem into a system with a generalized Stokes problem and a transport equation, as V. Girault and L.R. Scott in the two dimensions case. But, compared to the two-dimensional problem, we have an additional term difficult to bound, which requests regularity of the solutions, and we have to prove that the solutions of the transport equation, which is no longer scalar, are divergence-free. In order to deal with these specific difficulties of the three dimensions, we establish a new system which implies the previous one without being equivalent. A subtantial part of the article is devoted to prove that any solution in L 2 (Ω) 3 of the transport equation is divergence-free if the right-hand side is divergence-free and the velocity small enough in W 1, (Ω) 3. Existence is proven in a convex polyhedron, with adequate restrictions on the size of the data and parameters. Uniqueness requires, in addition, inner angles smaller than 3π 4. *Université d Evry Val d Essonne, Boulevard F. Mitterand. 91025 Evry Cedex, France. 0

1 Introduction This paper studies the stationary problem of a class of grade two fluids in three dimensions. The equation governing the motion of these fluids is ν u + curl(u α u) u + p = f, (1.1) where the constant ν is the kinematic viscosity and α is a normal stress moduli (cf. W. Noll & C. Truesdell [19]). The incompressibility requires that: and we impose the Dirichlet tangential boundary condition: div u = 0 (1.2) u = g on Ω with g. n = 0, (1.3) where n denotes the unit exterior normal to the boundary Ω of Ω. The thermodynamics of fluids of grade 2 entail that ν and α be nonnegative (cf. J. E. Dunn & R. L. Fosdick [11]), but, from a strictly mathematical point view, since the sign of α is unimportant, we shall assume: ν > 0, α 0. (1.4) Concerning fluids of grade n, we refer to W. Noll & C. Truesdell [19] and R. L. Fosdick & K. R. Rajagopal [12], [13]. Problem (1.1) is difficult because of its nonlinear terms involve third order derivatives. A closer look shows that it behaves somewhat like an Euler equation. Indeed, by taking the Helmoltz decomposition of u α 1 u: u α 1 u = w + q, where w and q are uniquely determined by the conditions div w = 0, w. n = 0 on the boundary of the domain. With w as an auxiliary variable, problem (1.1) can be written formally: αu. w + νw + (αp + νq) = νu + αf + N(u), (1.5) where N(u) is a function of u and its partial derivatives. This is similar to an Euler equation with a damping effect arising from the term νw. This observation suggests to take the curl of equation (1.1). Thus by setting z = curl(u α 1 u), 1

we shall obtain formally from (1.1) the following transport equation νz + α(u. z z. u) = ν curl u + α curl f. (1.6) D. Cioranescu and E. H. Ouazar in [8], [9] were the first ones to solve the problem (1.1)-(1.3) successfully. They used Galerkin s method with the special basis of eigenfunctions corresponding to the scalar product associated with the operator curl(u α 1 u). This basis was designed to split the discrete Galerkin problem into a continuous Stokes problem and a discrete version of the transport equation (1.6), from which they recovered sharp energy estimates. In two dimensions, Ouazar proved in [20] that the time-dependent version of problem (1.1)-(1.3) with a homogeneous Dirichlet boundary condition always has a unique global solution, in a simply-connected domain with sufficiently smooth boundary, for initial data with H 3 regularity. The same approach allows to prove existence of the solutions of the homogeneous stationary problem (1.1), (1.2) in two dimensions without restriction on the data. In three dimensions, they obtained existence, uniqueness and stability of the variational solution during some time interval, again without restriction on the data. In 1997, D. Cioranescu and V. Girault in [7] extended the results of [8] to prove global existence in time for the 3-D problem (1.1) and additional regularity of the solution. Some years thereafter, J. M. Bernard in [3], [4] and [5] proved the same types of results for the general problem of grade two on decreasing the required regularity of the boundary for existence, precisely with a boundary of class C 1,1, and extended the study to the non-simply-connected domain. Moreover, applying a special method, using together a H 1 bound of the velocity, a pseudo continuous dependence with respect to the data and a polynomial inequality (verified by an adequate norm of the velocity), he showed existence of solutions, uniqueness and continuous dependence with respect to the data, with small data, for the stationary problem. In 1999, V. Girault and L.R. Scott in [15] extended the results of the steady-state grade-two fluid models in two dimensions to the problem with the non-homogeneous Dirichlet tangential boundary condition (1.3) and reduced the regularity of the boundary to Lipschitz-continuity. Moreover, they established that any solution of (1.1)-(1.3) converges to a solution of the Navier-Stokes equations when α tends to zero. This new important result relies on Green s formula in a Lipschitz-continuous domain, which follows from a sharp analysis of the steady transport equation in arbitrary dimension. As a result of the weak regularity of the boundary, V. Girault and L.R. Scott proposed finite element discretizations of a two dimenional grade-two fluid model in [16]. In 2005, Another finiteelement discretizations were studied in [1] by M. Amara, C. bernardi, V. Girault and F. Hecht. Other publications devoted to grade-two fluids have used another approach : each one decomposed the original system of equations in their own way, but all applied a Schauder fixed point argument. If we compare the method of energy estimates of D. Cioranescu and E. H. Ouazar, extended by J.M. Bernard, V. Girault and L.R. Scott, to those methods, first we can note that this method is the only one that gives global existence of solutions in dimension two for the second grade fluids, without restriction on the size of the data. Second, in three dimensions, if the two approaches prove the existence only for sufficiently 2

small data, the method of energy estimates leads to conditions of existence that are more precise and more explicit. The reason could be due to the fact that, in those methods using a Schauder fixed point argument, the non-linear terms are placed straight without conversions on the right hand side. In this article, we propose to study the stationary grade-two fluids models, in an analogous way as V. Girault and L.R. Scott in [15], but in three dimensions, with a non-homogeneous Dirichlet tangential boundary condition, reducing the regularity of the boundary to allow corners. The extension to three-dimensional problem is not straightforward for several reasons. First, there is an additional term, which is difficult to bound and which requests sharp regularity results and a new formulation of the problem. For a few years now, V Maj ya and J. Rossmann obtained new regularity results for the Stokes and Navier-Stokes problem in polyhedral domains, a regularity W 1, (Ω) 3 for example, which allows us to bound this additional term. These authors extended the regularity results in the Hilbert spaces of M. Dauge in [10] to the Banach spaces. The second difficulty is to prove that the solution in L 2 (Ω) 3 of the transport equation (1.6) is divergence-free. We shall establish that any solution in L 2 (Ω) 3 of such transport equation is divergence-free if the velocity is small enough in W 1, (Ω) 3 and if the right hand side is divergence-free. By splitting the problem into a generalized Stokes problem and a transport equation the solutions of which are assumed divergence-free, in the same way as V. Girault and L.R. Scott in [15] and next by using a new formulation, which frees us from the divergence-free condition for the solutions of the transport equation, we shall prove the existence of solution for problem (1.1)-(1.3) in convex polyhedrons with adequate restrictions on the size of the data and parameters. Uniqueness will require more regularity and we shall be led to assume that the inner angle of Ω are smaller than 3π. Note that expanding on this 4 article, finite-element discretizations should be possible in three dimensions. After this introduction, this article is organized as follows. In Section 2, problem (1.1)- (1.3) is split into an equivalent coupled system consisting of a generalized Stokes problem and a transport equation in the way of V. Girault and L.R. Scott in [15]. Section 3 is devoted to a third formulation of the problem without divergence-free condition for the solution of the transport equation, to a sharp study of this transport equation when the right hand side is divergence-free, and to establishing regularity results. In section 4, we construct a solution of the third formulation s problem by Galerkin s method when Ω is a convex polyhedron with adequate restrictions on the size of the data and parameters, which also is a solution of the initial problem (1.1)-(1.3). Finally, in Section 5, we prove uniqueness of the solution when the inner angles are smaller than 3π and the data and 4 parametres are small enough. In order to set this problem into adequate spaces, recall the definition of the following standard Sobolev spaces (cf. J. Nečas [18]). For any multi-index k = (k 1, k 2, k 3 ) with k i 0, set k = k 1 + k 2 + k 3 and denote k v = k v x k 1 1 x k 2 2 x k 3 3 Then for any integer m 0 and number p with 1 p, we define: 3.

W m,p (Ω) = {v L p (Ω); k v L p (Ω) for 1 k m}, which is a Banach space equipped with the norm m v W m,p (Ω) = ( k =0 k k v p L p (Ω) )1/p, with the usual modification when p =. We denote H m (Ω) = W m,2 (Ω). Let l be nonnegative integer, we define the space C l (Ω) as the set of all functions on Ω having bounded and continuous derivatives up to order l on Ω. The norm in C l (Ω) is defined as v C l (Ω) = sup k v(x). k l x Ω Let σ be a real number, 0 < σ 1. A function v is called Hölder-continuous with exponent σ (Lipschitz-continuous if σ = 1) (cf V. Maz ya and J. Rossmann [17]) if there exists a constant C < such that for all x, y Ω, x y, v(x) v(y) x y σ C. The space C l,σ (Ω) is defined as the set of all functions v on Ω having bounded and continuous derivatives up to order l and Hölder-continuous with exponent σ derivatives of order l. The norm in the Hölder space C l,σ (Ω) is defined as v C l,σ (Ω) = v C l (Ω) + sup k =l x,y Ω, x y k v(x) k v(y) x y σ. In the same way, we define the space C l,σ ( Ω) for a regular enough boundary Ω, for example in the case where Ω is a polyhedron of IR 3. We define C 1,σ (Ω) as the set of the distributions v of the form (see [17], page 517) v = v (0) + 3 j=1 v (j) x j, where v (j) C 0,σ (Ω), j = 0, 1, 2, 3. (1.7) For vector-valued functions v = (v 1, v 2,..., v N ), we use special norms: if 1 p, we set v L p (Ω) N = v L p (Ω), (1.8) where. is the euclidian norm in IR N. The advantage of using this definition is that it leads to smaller constants in applying Hölder s inequality to trilinear terms. To simplify, we shall denote v L p (Ω) instead of v L p (Ω) N. We consider the matrices 3 3 as elements of L p (Ω) 9 and we define their norms L p by using (1.8) with N = 9. In the same way, we define the norms of tensors. 4

We shall frequently use the scalar product of L 2 (Ω) (f, g) = f(x)g(x) dx, the semi-norm of H 1 (Ω) Ω v H 1 (Ω) = v L 2 (Ω), and the subspaces of L 2 (Ω), H 1 (Ω) and L 2 (Ω) 3 : L 2 0(Ω) = {q L 2 (Ω); Ω q dx = 0}, H 1 0(Ω) = {v H 1 (Ω); v = 0 on Ω}, H(div, Ω) = {v L 2 (Ω) 3 ; div v L 2 (Ω)}, H(curl, Ω) = {v L 2 (Ω) 3 ; curl v L 2 (Ω) 3 }. We shall often use Sobolev s imbeddings : for any real number p, 1 p 6, there exists a constant S p such that v H 1 0(Ω), v L p (Ω) S p v H 1 (Ω). (1.9) When p = 2, this reduces to Poincaré s inequality and S 2 is the Poincaré s constant. For tangential boundary values, we define : H 1 T (Ω) = {v H 1 (Ω) 3 ; v. n = 0 on Ω}. (1.10) A straightforward application of Peetre-Tartar s Theorem (see [14]) shows that the analogue of Sobolev s imbeddings holds in H 1 T (Ω) for any real number p, 1 p 6 : v H 1 T (Ω), v L p (Ω) S p v H 1 (Ω). (1.11) In particular, for p = 2, the mapping v v H 1 (Ω) is a norm on H 1 T (Ω), equivalent to the H 1 norm and S 2 is the analogue of Poincaré s constant. 2 An equivalent formulation 2.1 The second formulation Following the approach of [15], we shall establish a mixed formulation of the problem. In this subsection, the assumptions on the data are: Ω is a bounded domain in IR 3, with lipschitz-continuous boundary Ω, f is a given function in H(curl; Ω), g is a given vector field in H 1 2 ( Ω) 3 such that g. n = 0 and ν > 0 and α 0 are two given real constants. We introduce the space W = {v H 1 T (Ω); curl v L 2 (Ω) 3 }, (2.1) 5

in which we shall look for the velocity u. Let (u, p) W L 2 0(Ω) be a solution of (1.1)-(1.3) and introduce the auxiliary variable: z = curl(u α u). (2.2) We set Z = {v L 2 (Ω) 3, div v = 0}. (2.3) Note that z belongs to Z. With this notation, we write (1.1) as: ν u + z u + p = f in Ω, (2.4) that is with (1.2) and (1.3) a generalized Stokes equation. Taking the curl of (2.4) in the sense of distributions, we obtain ν curl u + u. z z. u = curl f, that we can write as the transport equation (1.6). Conversely, let (u, p, z) H 1 T (Ω) L 2 0(Ω) Z be a solution of (2.4), (1.2), (1.3) and (1.6). Then, since div z = 0, taking the curl of (2.4) in the sense of distributions yields: ν curl u + u. z z. u = curl f. Next, multiplying by α and comparing with (1.6), we obtain: z = curl(u α u). Therefore u belongs to W and substituting the expression of z into (2.4) shows that (u, p) is a solution of the original equations (1.1)-(1.3). This is summarised in the following lemma. Lemma 2.1 Problem (1.1-(1.3) with (u, p) in W L 2 0(Ω) is equivalent to: Find (u, p, z) in H 1 T (Ω) L 2 0(Ω) Z solution of the generalized Stokes problem (2.4), (1.2), (1.3) and the transport equation (1.6), namely: ν u + z u + p = f in Ω, div u = 0 in Ω, u = g on Ω with g. n = 0 Ω, (2.5) νz + α (u. z z. u) = ν curl u + α curl f in Ω. 2.2 Estimates for the solution of the generalized Stokes problem In this subsection, the assumptions on the data are the same as the previous subsection. For a given z in L 2 (Ω) 3, the generalized Stokes problem (2.4), (1.2), (1.3) has the following variational formulation (see [15]): Find (u(z), p(z)) in H 1 (Ω) 3 L 2 0(Ω), such that v H 1 0(Ω) 3, a z (u(z), v) + b(v, p(z)) = (f, v), (2.6) q L 2 0(Ω), b(u(z), q) = 0, (2.7) u(z) = g on Ω with g. n = 0 on Ω, (2.8) 6

where a z (w, v) = ν( w, v) + (z w, v), b(v, q) = (q, div v). In the same way as in [15], we define the following lifting of g: let w g be the solution in H 1 (Ω) 3 of the non-homogeneous Stokes problem: w g + p g = 0 and div w g = 0 in Ω, w g = g on Ω. (2.9) This problem has a unique solution and it satisfies the bound (see [14]): w g H 1 (Ω) T g H 1 2 ( Ω). (2.10) By the same proof as in [15], substituting the three dimensions into the two dimensions, this lifting allows us to show the following lemma, the proof of which we give for the reader convenience. Lemma 2.2 Let Ω be Lipchitz-continuous, ν > 0, f L 2 (Ω) 3 and g H 1 2 ( Ω) 3 satisfying the second part of (1.3). For any z in L 2 (Ω) 3, the generalized Stokes problem (2.6)-(2.8) has a unique solution (u(z), p(z)) in HT 1 (Ω) L 2 0(Ω). This solution satisfies the following bounds: u(z) H 1 (Ω) S ( 2 ν f L 2 (Ω) + T g 1 H 2 ( Ω) 1 + S S ) 4 4 z L ν 2 (Ω), (2.11) p(z) L 2 (Ω) 1 β (S 2 f L 2 (Ω) + νt g H 1 2 ( Ω) + S 4 S4 u(z) H 1 (Ω) z L 2 (Ω)), (2.12) where β > 0 is the isomorphism constant of the divergence operator, as given in formula (2.14), S p and S p are defined in (1.9) and (1.11) respectively and T is defined in (2.10). Proof. We set u 0 = u w g. Then (2.6)-(2.8) is equivalent to: Find u 0 V such that : v V, a z (u 0, v) = (f, v) a z (w g, v). (2.13) For fixed z in L 2 (Ω) 3, the bilinear form a z is elliptic on H 1 0(Ω) 3 H 1 0(Ω) 3 since (z v, v) = 0, and it is continuous on H 1 (Ω) 3 H 1 (Ω) 3 since L 4 (Ω) H 1 (Ω) and u, v H 1 (Ω) 3, (z u, v) z L 2 (Ω) u L 4 (Ω) v L 4 (Ω). Therefore (2.13) has a unique solution u 0 V and in turn this implies that (2.6)-(2.8) has a unique solution (u(z), p(z)) in H 1 T (Ω) L 2 0(Ω). Next, the choice v = u 0 in (2.13) yields ν u 0 2 H 1 (Ω) f L 2 (Ω) u 0 L 2 (Ω) + z L 2 (Ω) u 0 L 4 (Ω) w g L 4 (Ω), which implies u 0 H 1 (Ω) 1 ν (S 2 f L 2 (Ω) + S 4 S4 z L 2 (Ω) w g H 1 (Ω)). 7

With (2.10) and the triangle inequality, this gives (2.11). Let us recall that it follows from the isomorphism properties of the divergence (cf for instance [6] or [14]) that there exists a unique v p in H 1 0(Ω) 3 such that div v p = p(z) in Ω, v p H 1 (Ω) 1 β p(z) L 2 (Ω), (2.14) w V, ( v p, w) = 0. Then taking v p for test function in (2.6), we obtain : p(z) 2 L 2 (Ω) = (z u(z), v p ) + ν( w g, v p ) (f, v p ). Therefore, applying (2.14), we derive (2.12). 3 A third formulation without divergence-free condition In this section we assume that Ω is a convex polyhedron. 3.1 A third formulation Let w be in Z which is defined by (2.3). Since Ω is connected, we have < w. n, 1 > Ω = 0 and since, in addition, Ω is simply connected, there exists an unique function ϕ H 1 (Ω) 3 (see [2] or [14]) such that curl ϕ = w in Ω, div ϕ = 0 in Ω and ϕ. n = 0 on Ω. (3.1) Moreover, there exists a constant C > 0 such that Let Φ the map from Z to H 1 (Ω) 3 defined by ϕ H 1 (Ω) C w L 2 (Ω). (3.2) Φ : Z H 1 (Ω) 3 w ϕ, (3.3) where ϕ is the unique solution of (3.3). Since Z is a closed subspace of L 2 (Ω) 3, let P Z the orthogonal projection from L 2 (Ω) 3 on Z. Then, we define the following problem: find (u, p, v, z) in H 1 T (Ω) L 2 0(Ω) H 1 T (Ω) L 2 (Ω) 3 such that ν u + z u + p = f in Ω, div u = 0 in Ω, u = g on Ω with g. n = 0 Ω, 8

where Φ is defined by (3.3). ν v + q = Φ( ν α (P Z(z) curl u)) in Ω, (3.4) div v = 0 in Ω, v = g on Ω with g. n = 0 Ω, ν z + α (v. z z. v) = ν curl u + α curl f in Ω, 3.2 Transport equation with right hand side in Z Let l be given in Z and v be given in H 1 T (Ω) with div v = 0, we consider the transport problem: find z L 2 (Ω) 3, such that νz + α(v. z z. v) = l. (3.5) Lemma 3.1 Let Ω be a convex polyhedron, ν > 0, l Z and v H 1 T (Ω) W 1, (Ω) 3, with div v = 0, satisfying α v L (Ω) < ν. (3.6) If z is a solution in L 2 (Ω) 3 of the transport equation (3.5), then z belongs to Z. Proof. Let us prove first that, if l L 2 (Ω) 3 and under the condition (3.6), the transport equation (3.5) has at most one solution in L 2 (Ω) 3. Let z = (z 1, z 2, z 3 ) L 2 (Ω) 3 be a solution of: νz + α(v. z z. v) = 0 in Ω, with v verifying the assumptions of Lemma 3.1. We must show that necessarily z = 0. For fixed u in H 1 T (Ω), with div u = 0, we introduce the Hilbert space (see [15]): X u = {z L 2 (Ω), u. z L 2 (Ω)}. (3.7) Since, for j = 1, 2, 3, z j belongs to X v, applying Green formula (3.14) page 998 of [15] yields (v. z j, z j ) = 0, (3.8) which implies ν(z, z) α(z. v, z) = 0 and next (ν α v L (Ω)) z 2 L 2 (Ω) 0. Then, in view of (3.6), we derive z = 0, thus proving uniqueness. Let us introduce the following spaces X N (Ω) = {v H(curl, Ω) H(div, Ω), v n = 0 on Ω}, (3.9) X 0 N(Ω) = {v X N (Ω), < v. n, 1 > Ω = 0}, V N (Ω) = {v X N (Ω), div v = 0}. (3.10) We recall two results of [2]. First (Theorem 2.17), if the domain Ω is convex, then X N (Ω) is continuously imbedded in H 1 (Ω) 3. Second (Corollary 3.19), v ( curl v 2 L 2 (Ω) + div v 2 L 2 (Ω)) 1/2 9

is a norm in X 0 N(Ω) equivalent to the norm in H(curl, Ω) H(div, Ω) and, therefore, (v, w) (curl v, curl w) + (div v, div w) (3.11) is a scalar product in X 0 N(Ω). Since V N (Ω) is a closed subspace of X 0 N(Ω), v curl v L 2 (Ω) is a norm in V N (Ω) and we have v V N (Ω), v H 1 (Ω) C curl v L 2 (Ω). (3.12) Let V N (Ω) be the orthogonal of V N (Ω) in X 0 N(Ω) for the scalar product (3.11). Let v V N (Ω) and let w H 1 0(Ω) 3. Using the Helmoltz decomposition [14, Corollary 3.4, page 50] and [14, Theorem 3.6, page 48], w has the orthogonal decomposition in L 2 (Ω) 3 : w = curl ϕ + q, with ϕ V N (Ω) and q solution in H 2 (Ω) of the homogeneous Neumann problem: Then we have { q = div w, q = 0. n Ω (curl (v ), w) = (curl (v ), curl ϕ) + (curl (v ), q) = (curl (v ), q), since v V N (Ω) and ϕ V N (Ω). Applying Green formula [14, Theorem 2.11 page 34] with q in H 1 (Ω) 3 and v n Ω = 0, we obtain w H 1 0(Ω) 3, (curl (v ), w) = 0, which imply curl (v ) = 0. Conversely, if v XN(Ω) 0 verifies curl v = 0, then v V N (Ω). Therefore, we have the following characterization V N (Ω) = {v X 0 N(Ω), curl v = 0}. (3.13) Consider the following variational problem. For each ε > 0, find z ε V N (Ω) solving : ζ V N (Ω), ε(curl z ε, curl ζ) + ν(z ε, ζ) + α(v. z ε z ε. v, ζ) = (l, ζ). (3.14) The bilinear form associated with the left-hand side of (3.14) is defined by a ε (z, ζ) = ε(curl z, curl ζ) + ν(z, ζ) + α(v. z z. v, ζ). Owing to (3.8) and (3.6), we have z V N (Ω), a ε (z, z) ε curl z 2 L 2 (Ω) + (ν α v L ) z 2 L 2 (Ω) ε curl z 2 L 2 (Ω), which implies that a ε is V N (Ω) elliptic. Moreover, a ε is continuous on V N (Ω) 2. Indeed, owing to (3.12), we have z V N (Ω), ζ V N (Ω), a ε (z, ζ) (ε + (ν + α ( v L (Ω) + v L (Ω)))C 2 ) curl z L 2 (Ω) curl ζ L 2 (Ω). 10

Therefore, by Lax-Milgram s Lemma, (3.14) defines a unique function z ε V N (Ω), and z ε 1 L 2 (Ω) l L ν α v 2 (Ω), ε curl z ε 1 L 2 (Ω) L (Ω) 2 ν α v L (Ω) Let us recall that z ε H 1 (Ω) 3 and, since div z ε = div v = 0, we have v. z ε z ε. v = curl(z ε v) l L 2 (Ω). (3.15) and then div (v. z ε z ε. v) = 0. Since z ε, v. z ε z ε. v and l belong to Z, in view of (3.1), there exist functions ϕ ε H 1 (Ω) 3, ψ ε H 1 (Ω) 3 and ξ H 1 (Ω) 3 such that z ε = curl ϕ ε, v. z ε z ε. v = curl ψ ε, l = curl ξ. Next, applying Green formula [14, Theorem 2.11 page 34] and considering (3.13), we obtain In the same way, v V N (Ω), v V N (Ω), (z ε, v ) = (curl ϕ ε, v ) = 0. (v. z ε z ε. v, v ) = 0 and (l, v ) = 0. Finally, since X 0 N(Ω) = V N (Ω) + V N (Ω), (3.14) becomes ζ X 0 N(Ω), ε(curl z ε, curl ζ) + ν(z ε, ζ) + α(v. z ε z ε. v, ζ) = (l, ζ). (3.16) From (3.15), we derive that there exists a function z such that Finally, we have lim ε 0 zε = z weakly in L 2 (Ω) 3. ϕ D(Ω) 3, ε(curl z ε, curl ϕ) + ν(z ε, ϕ) + α(v. z ε, ϕ) α(z ε. v, ϕ) = (l, ϕ). Since (v. z ε, ϕ = (v. ϕ, z ε ) the above convergence and (3.15) allow us to pass to the limit and we obtain which implies ν( z, ϕ) α(v. ϕ, z) + ( z. v, ϕ)) = (l, ϕ), ν z + α(v. z z. v) = l. Since this tranport equation has a unique solution, we derive z = z. In view of div z ε = 0 and lim div ε 0 zε = div z in D (Ω), we obtain div z = 0, that is to say z Z. Now we can prove that Problem (3.4) implies Problem (2.5) in the following way. 11

Lemma 3.2 Let Ω be a convex polyhedron of IR 3 and ν > 0. If (u, p, v, z) is a solution of Problem (3.4) in H 1 T (Ω) L 2 0(Ω) H 1 T (Ω) (L 2 (Ω) 3 ) with v W 1, (Ω) 3, satisfying (3.6), then (u, p, z) is a solution of Problem (2.5) in H 1 T (Ω) L 2 0(Ω) Z. Moreover, v = u in Ω. Proof. Let (u, p, v, z) H 1 T (Ω) L 2 0(Ω) H 1 T (Ω) L 2 (Ω) 3 be a solution of problem (3.4), with v verifying (3.6). From Lemma 3.1, we derive that z belongs to Z, that is to say P Z (z) = z. Thus, we can write curl( ν v) = ν (z curl u) = curl f (v. z z. v) = curl(f z v), α which implies that there exists p L 2 0(Ω) such that ν v + z v + p = f. Thus v and u are solutions of the generalized Stokes problem (2.6)-(2.8). In view of Lemma 2.2, the uniqueness of the solution yields v = u in Ω and, therefore, (u, p, z) is a solution in HT 1 (Ω) L 2 0(Ω) Z of problem (2.5). 3.3 A regularity result Lemma 3.3 Let Ω be a convex polyhedron of IR 3 and ν > 0. We suppose that there exist real numbers σ 0, 0 < σ 0 1 2, T > 0 and a function v g C 1,σ 0 (Ω) 3 such that v g = g on Ω, v g C 1,σ 0 (Ω) T g C 1,σ 0 ( Ω), div v g = 0 on Ω, (3.17) where g. n = 0 on Ω. Then, for any z in L 2 (Ω) 3, the solution v(z) of the Stokes problem ν v + q = Φ( ν α (P Z(z) curl u(z))) in Ω, v = g on Ω with g. n = 0 Ω, div v = 0 in Ω, (3.18) where u(z) is the solution of the generalized Stokes problem (2.6)-(2.8), belongs to W 1, (Ω) 3 and α v(z) L (Ω) C CK( z L 2 (Ω) + 2 u(z) H 1 (Ω))+( 3K +1) α T g C 1,σ 0 ( Ω), (3.19) where C is defined in (3.20), K is defined in (3.21) and C is defined in (3.2). Proof. Let us set v 0 (z) = v(z) v g, where v g is defined by (3.17). Then v 0 (z) is the solution in H 1 (Ω) 3 of the homogeneous Stokes s problem with Φ( ν α (P Z(z) curl u(z))) + ν v g 12

as right-hand side. On the one hand, owing to Lemma 2.2 and (3.3), Φ( ν (z curl u(z))) α belongs to H 1 (Ω) 3. On the other hand, considering the Sobolev imbedding from H 1 (Ω) into C 1, 1 2 (Ω), there exist a strictly positive constant C such that v H 1 (Ω), v C 1,σ 0 (Ω) C v H 1 (Ω). (3.20) Therefore, Φ( ν α (P Z(z) curl u(z))) + ν v g belongs to C 1,σ 0 (Ω) 3. Then, in view of Theorem 11.3.2 of [17], there exist a real number σ, 0 < σ σ 0, such that v 0 (z) belongs to C 1,σ (Ω) 3 and there exists a constant K such that v 0 (z) C 1 (Ω) K ν Φ( ν α (z curl u(z))) + ν v g C 1,σ 0 (Ω). (3.21) Finally, we derive that v(z) belongs to C 1 (Ω) 3 and verifies α v(z) C 1 (Ω) C CK( z L 2 (Ω) + 2 u(z) H 1 (Ω)) + ( 3K + 1) α T g C 1,σ 0 ( Ω), (3.22) which implies the estimate (3.19). 4 Existence of a solution The solution of the mixed formulation (3.4) is constucted by means of Galerkin s discretization of z. Let {w j } j 1 be a basis of H 2 (Ω) 3. For any positive integer m, we denote by X m the vector space spanned by the first m eigenfunctions {w j } m j=1. For each z m in X m, we denote by u(z m ) the unique solution of the generalized Stokes problem (2.6)-(2.8), by v(z m ) the unique solution of the Stokes problem (3.18) and we define an approximate solution of problem (1.6) by: Find solution for 1 i m, of m z m = c j,m w j X m, j=1 ν(z m, w i ) + α (v(z m ). z m z m. v(z m ), w i ) = ν (curl u(z m ), w i ) + α (curl f, w i ). (4.1) We recall the following classic lemma, which is a variant of the Brouwer Theorem, that we shall use to prove the existence of approximate solutions. Lemma 4.1 Let X be a finite-dimensional Hilbert space whose scalar product is denoted by (.,.) and corresponding norm by.. Let P be a continuous mapping from X into X. Suppose that there exists a real number ρ > 0 such that, for all ξ in X, with ξ = ρ, we have (P (ξ).ξ) 0. Then there exists ξ X, with ξ ρ, such that P (ξ ) = 0. 13

Lemma 4.2 Let Ω be a convex polyhedron of IR 3 and ν > 0. Let f belong to H(curl; Ω) and g belong to C 1,σ 0 ( Ω), verifying (3.17). Let us assume that the data f and g are small enough so that they satisfy 2C CK( α curl f L 2 (Ω) + 2 2(S 2 f L 2 (Ω) + νt g 1 H 2 ( Ω) ))(1 + ν T S S 4 4 g 1 H 2 ( Ω) ) +ν 2T (C CK+S4 S4 ) g H 1 2 ( Ω) + 2C CKS2 f L 2 (Ω)+ α ν( 3K+1) T g C 1,σ 0 ( Ω) ν2 2. (4.2) Then, for all integers m 1, the discrete problem (4.1) has at least one solution z m in X m and z m satisfies the uniform estimate with respect to m: z m L 2 (Ω) 2 ν ( α curl f L 2 (Ω) + 2S 2 f L 2 (Ω) + 2νT g H 1 2 ( Ω) ), (4.3) where T, C, K, C are defined in Lemma 3.3, Sp and S p are defined by (1.9) and (1.11) respectively and T is defined in (2.10). Proof. We construct a mapping P from X m to X m as follows: for all τ m Xm, we define P (τ m ) in X m by : for 1 i m, (P (τ m ), w i ) = ν(τ m, w i ) + α (v(τ m ). τ m τ m. v(τ m ), w i ) α(curl f, w i ) ν(curl u(τ m ), w i ). (4.4) This square system of linear equations defines P (τ m ) X m uniquely. 1) First, owing to the finite dimension, P is a continuous mapping from X m into X m. 2) Next, let us show that for τ m L 2 (Ω) = ρ large enough, (P (τ m ), τ m ) 0. Let us compute (P (τ m ), τ m ). (P (τ m ), τ m ) = ν τ m 2 L 2 (Ω) + α (v(τ m ). τ m, τ m ) α (τ m. v(τ m ), τ m ) α(curl f, τ m ) ν(curl u(τ m ), τ m ). Since div v = 0, v H 1 T (Ω) and τ m H 2 (Ω) 3, Green s formula yields Therefore, we obtain: From (3.19, we derive (v(τ m ). τ m, τ m ) = 1 2 Ω τ m 2 v(τ m ). n ds = 0. (P (τ m ), τ m ) ν τ m 2 L 2 (Ω) α v(τ m ) L (Ω) τ m 2 L 2 (Ω) α curl f L 2 (Ω) τ m L 2 (Ω) ν 2 u(τ m ) H 1 (Ω) τ m L 2 (Ω). (P (τ m ), τ m ) ν τ m 2 L 2 (Ω) α curl f L 2 (Ω) τ m L 2 (Ω) ν 2 u(τ m ) H 1 (Ω) τ m L 2 (Ω) (C CK( τ m L 2 (Ω) + 2 u(τ m ) H 1 (Ω)) + ( 3K + 1) α T g C 1,σ 0 ( Ω) ) τ m 2 L 2 (Ω). 14

Substituting (2.11) in the previous inequality yields with (P (τ m ), τ m ) τ m L 2 (Ω)((ν A B τ m L 2 (Ω)) τ m L 2 (Ω) C), A = 2 T (S 4 S4 + C CK) g H 1 2 ( Ω) + 2C CK S 2 ν f L 2 (Ω) + α ( 3K + 1) T g C 1,σ 0 ( Ω), 2 B = C CK(1+ ν T S S 4 4 g 1 H 2 ( Ω) ), C = α curl f L 2 (Ω)+ 2(S 2 f L 2 (Ω)+νT g 1 H 2 ( Ω) ). If we suppose A + B τ m L 2 (Ω) ν 2 ν A B τ m L 2 (Ω) ν (4.5) 2 and ν 2 τ m L 2 (Ω) = C τ m L 2 (Ω) = 2 ν ( α curl f L 2 (Ω) + 2S 2 f L 2 (Ω) +ν 2T g H 1 2 ( Ω) ), (4.6) we obtain (P (τ m ), τ m ) 0. Substituting (4.6) in the inequality A + B τ m L 2 (Ω) ν 2 yields the condition (4.2). Let us note that (4.6), (4.5) are equivalent to (4.6), (4.2). Thus, if we suppose (4.2), (4.6) implies (P (τ m ), τ m ) 0. Hence, Lemma 4.1 implies that P (τ m ) has at least one zero in the sphere τ m L 2 (Ω) 2 ν ( α curl f L 2 (Ω) + 2 S 2 f L 2 (Ω) + ν 2T g H 1 2 ( Ω) ). Therefore (4.1) has at least one solution z m and it satisfies the uniform bound (4.3). Now we can pass to the limit and prove existence of a solution. Theorem 4.3 Let Ω be a convex polyhedron of IR 3 and ν > 0. Let f belong to H(curl; Ω) and g belong to C 1,σ 0 ( Ω) 3, verifying (3.17). Let us assume that the data f and g are small enough so that they satisfy the condition (4.2) of Lemma 4.2. Then problem (2.5) has at least one solution (u, p, z) in H 1 T (Ω) L 2 0(Ω) Z and this solution satisfies the following estimates: z L 2 (Ω) 2 ν ( α curl f L 2 (Ω) + 2S 2 f L 2 (Ω) + 2νT g H 1 2 ( Ω) ), (4.7) u H 1 (Ω) S ( 2 ν f L 2 (Ω) + T g 1 H 2 ( Ω) 1 + S S ) 4 4 z L ν 2 (Ω), (4.8) p L 2 (Ω) 1 β (S 2 f L 2 (Ω) + νt g H 1 2 ( Ω) + S 4 S4 u(z) H 1 (Ω) z L 2 (Ω)), (4.9) where β > 0 is the isomorphism constant of the divergence operator, as given in formula (2.14), S p and S p are defined in (1.9) and (1.11) respectively and T is defined in (2.10). 15

Proof. In view of (4.3), z m is uniformly bounded in L 2 (Ω). Hence, (2.11), (2.12) and (3.22) imply that u(z m ), p(z m ) and v(z m ) are uniformly bounded in H 1 (Ω) 3, L 2 (Ω) and W 1, (Ω) 3, respectively. Therefore, there exists a subsequence, still denoted by the index m, and four functions z L 2 (Ω) 3, u H 1 (Ω) 3, p L 2 (Ω) and v W 1, (Ω) 3 such that lim z m = z weakly in L 2 (Ω) 3, (4.10) m lim u(z m) = u weakly in H 1 (Ω) 3, m lim p(z m) = p m weakly in L 2 (Ω), These convergences imply lim v(z m) = v weak in L (Ω) 9, (4.11) m lim v(z m) = v weakly in W 1,3 (Ω) 3. m div u = div v = 0 and u Ω = v Ω = g. (4.12) The weak convergence of u(z m ) in H 1 (Ω) 3 and the weak convergence of v(z m ) in W 1,3 (Ω) 3 imply that for all real p, 1 p < 6 and all real q 1 and lim u(z m) = u strongly in L p (Ω) 3 m lim v(z m) = v strongly in L q (Ω) 3. (4.13) m From the Green s formula, owing to v(z m ). n Ω = 0 and div v(z m ) = 0 in Ω, we derive (v(z m ). z m, w i ) = (v(z m ), (z m. w i ) (v(z m ). w i, z m ) = (v(z m ). w i, z m ). (4.14) Since w i belongs to L 6 (Ω) 9, (4.13) yields that lim v(z m). w i = v. w i strongly in L 2 (Ω) 3, m which implies, in view of (4.14) and (4.10), Next, (4.10) and (3.19) imply that so we obtain Passing to the limit in (4.1) yields lim (v(z m). z m, w i ) = (v. w i, z). (4.15) m lim (z m. v(z m )) = z. v weakly in L 2 (Ω) 3, m lim (z m. v(z m ), w i ) = (z. v, w i ). (4.16) m ν(z, w i ) α((v. w i, z) + (z. v, w i )) = ν(curl u, w i ) + α(curl f, w i ). 16

Then the density of the finite linear combinations of the functions w i in H 2 (Ω) 3 implies : ϕ H 2 (Ω) 3, ν(z, ϕ) α((v. ϕ, z) + (z. v, ϕ)) = ν(curlu, ϕ) + α(curlf, ϕ). But we have for all ϕ in D(Ω) 3, 3 3 < v. z, ϕ > D(Ω) 3= < v. z j, ϕ j > D(Ω) = j=1 j=1 Hence, we obtain νz + α(v. z z. v) = νcurl u + αcurl f. < z j, ϕ j v > H 1 0 (Ω) 3= (v. ϕ, z). Then the transport equation of problem (3.4) is verified. Moreover, the above convergences allow to pass to the limit in the other equations of problem (3.4). Therefore, we obtain a solution (u, p, v, z) of problem (3.4) in H 1 T (Ω) L 2 0(Ω) H 1 T (Ω) L 2 (Ω) 3. From (4.3) and the lower semi-continuity of the norm for the weak topology, we derive the bound (4.7). In the same way of the proof of Lemma 2.2, passing to the limit yields (4.8) and (4.9). Finally, substituting (4.8) and (4.7) into the bound (3.19) and owing to the condition (4.2), we derive α v L (Ω) ν 2 < ν and Lemma 3.2 implies that (u, p, z) is a solution in H 1 T (Ω) L 2 0(Ω) Z of problem (2.5). 5 Uniqueness Uniqueness of the solution of problem (2.5) is proved in [5], but with a boundary of class C 1,1 and with the homogeneous condition g = 0. Let us review this proof in order to relax its assumptions. We propose first to establish further regularity properties of the solution of (2.5), while assuming that Ω is a convex polyhedron of IR 3 with inner angle of Ω strictly smaller than 3π 4 solution. and next use this regularity to prove uniqueness of the Lemma 5.1 Let Ω be a convex polyhedron of IR 3 such that all its inner angles ω i satisfy 0 < ω i < 3π 4 and ν > 0. We suppose that there exist real number σ 0, 0 < σ 0 1 2 and a function v g C 1,σ 0 (Ω) 3 W 2,3 (Ω) 3 such that v g = g on Ω and div v g = 0 on Ω, (5.1) where g. n = 0 on Ω. Then, for any z in L 2 (Ω) 3, the solution v(z) of the Stokes problem (3.18) belongs to W 1, (Ω) 3 W 2,3 (Ω) 3 with the estimates α v(z) L (Ω) C CK( z L 2 (Ω) + 2 u(z) H 1 (Ω)) + ( 3K + 1) α v g C 1,σ 0 (Ω) (5.2) 17

and α 2 v(z) L 3 (Ω) C CL( z L 2 (Ω) + 2 u(z) H 1 (Ω)) + ( 3L + 1) α 2 v g L 3 (Ω), (5.3) where C is defined in (3.20), K is defined in (3.21), C is defined in (5.4), L is defined in (5.5) and C is defined in (3.2). Proof. In the same way as in the proof of Lemma 3.3, we prove that v(z) belongs to W 1, (Ω) 3 with the estimate (5.2) by subtituting v g C 1,σ 0 (Ω) in the place of T g C 1,σ 0 ( Ω). Let us recall that v 0 (z) = v(z) v g is the solution in H 1 (Ω) 3 of the homogeneous Stokes s problem with Φ( ν α (P Z(z) curl u(z))) + ν v g as right-hand side and that this right-hand side belongs to H 1 (Ω) 3. Considering the Sobolev imbedding from H 1 (Ω) into L 3 (Ω), there exists a strictly positive constant C such that v H 1 (Ω), v L 3 (Ω) C v H 1 (Ω). (5.4) Since, in addition, v g belongs to W 2,3 (Ω) 3, we derive that Φ( ν α (P Z(z) curl u(z)))+ν v g belongs to L 3 (Ω) 3. Then, in view of Theorem 11.3.1 of [17], since the inner angles of Ω are strictly smaller than 3π 4, v 0(z) belongs to W 2,3 (Ω) 3 and there exists a constant L such that v 0 (z) W 2,3 (Ω) L ν Φ( ν α (P Z(z) curl u(z))) + ν v g L 3 (Ω). (5.5) Hence, we derive α 2 v 0 (z) L 3 (Ω) C CL( z L 2 (Ω) + 2 u(z) H 1 (Ω)) + 3 α 2 v g L 3 (Ω) and with the triangle inequality, this gives (5.3). Let (u, p) be a solution of (1.1)-(1.3) in W L 2 0(Ω). We introduce the space V = {v H 1 0(Ω), div v = 0 in Ω}. (5.6) Then u is a solution of the following variational equation v V, ν( u, v) + (curl(u α u) u, v) = (f, v). (5.7) Let us suppose in addition that u belongs to H 2 (Ω) 3. Owing to the Green formula, since u v belongs to H 1 0(Ω) 3, we obtain (curl u u, v) = (u v, curl u) = (curl(u v), u). Since div u = div v = 0, we have curl(u v) = v. u u. v. Hence, we derive Applying Green s formula yields (curl u u, v) = (v. u, u) (u. v, u). (v. u, u) = (v, 1 2 ( u 2 )) = 1 2 (div v, u 2 ) + 1 2 < v. n, u 2 > Ω = 0. 18

In the same way, we have (u. v, u) = (u, (v. u)) (u. u, v) = (u. u, v). Finally, we obtain (curl u u, v) = (u. u, v). (5.8) Since u belongs to H(curl, Ω) and u v belongs to H 1 0(Ω) 3, Green s formula yields (curl( u) u, v) = (curl( u), u v) = (curl(u v), u) = (v. u u. v, u). Finally, we rewrite (5.7) as v V, ν( u, v) + (u. u, v) + α((u. v, u) (v. u, u)) = (f, v). (5.9) Let (u 1, p 1 ) and (u 2, p 2 ) be two solutions of (1.1)-(1.3) in (W H 2 (Ω) 3 ) L 2 0(Ω). Since u 1 and u 2 satisfy (5.9), we obtain v H 1 0(Ω) 3, ν(u 1 u 2, v) + (u 1. u 1 u 2. u 2, v) +α((u 1. v, u 1 ) (u 2. v, u 2 ) (v. u 1, u 1 ) + (v. u 2, u 2 )) = 0. Let w = u 1 u 2 and let us set v = w. From Green s formula, we derive (u 1. u 1 u 2. u 2, w) = (w. u 1, w) + (u 2. w, w) = (w. u 1, w). Finally, w satisfies: ν w 2 H 1 (Ω) + (w. u 1, w) + α((u 1. w, w) (w. u 1, w)) = 0. (5.10) Let us bound these terms. First, Hölder s and Poincaré s inequalities, with the special norm (1.8), yield (w. u 1, w) S 2 2 u 1 L (Ω) w 2 H 1 (Ω). (5.11) Next, Green s formula gives 3 (w. u 1, w) = ( w. u 1, w 3 ) (w. ( u1 ), w ). k=1 x k x k k=1 x k x k Hence, from (1.9), we derive (w. u 1, w) ( u 1 L (Ω) + S 6 2 u 1 L 3 (Ω)) w 2 H 1 (Ω). (5.12) The bound of the last term (u 1. w, w) is more difficult to obtain because u 1 does not belong to H 1 0(Ω) 3. So, let (w n ) be a sequence of functions of D(Ω) 3 such that lim w n = w in H 1 n + 0(Ω) 3. 19

From Green s formula and div u 1 = 0, we derive 3 (u 1. w n, w) = ( u1. w n, w 3 ) (u 1. ( w n ), w ) k=1 x k x k k=1 x k x k 3 = ( u1. w n, w 3 ) + (u 1. ( w ), w n ). k=1 x k x k k=1 x k x k Passing to the limit yields 3 (u 1. w, w) = ( u1. w, w 3 ) + (u 1. ( w ), w ), k=1 x k x k k=1 x k x k which implies, with again Green s formula, div u 1 = 0 and u 1. n Ω = 0, Hence, we obtain the following bound 3 (u 1. w, w) = ( u1. w, w ). k=1 x k x k Finally from (5.11), (5.12) and (5.13), we derive (u 1. w, w) u 1 L (Ω) w 2 H 1 (Ω). (5.13) (ν (2 + S2 2 α ) α u1 L (Ω) α S 6 2 u 1 L 3 (Ω)) w 2 H 1 (Ω) 0. Therefore, if we want to prove uniqueness of the solution of problem (2.5) from this last estimate, we must be able to apply Lemma 5.1, which gives bounds for u 1 L (Ω) and 2 u 1 L 3 (Ω). Thus, we assume that the domain verifies the assumptions of Lemmas 3.2 and 5.1 and Theorem 4.3. Subtituting the bounds (4.7) and (4.8) in the estimates (5.2) and (5.3) yields a condition on the data f and g, which provides uniqueness of the solution of problem (2.5). Note that, in order to have existence in addition, we add the term ν 2T S 4 S4 g H 1/2 ( Ω) in this condition. Hence, we derive the following theorem. Theorem 5.2 Under the assumptions of Lemma 5.1, the solution of problem (2.5) is unique, if C((2+ S2 2 α )(C K+S 6 C L)( α curl f L 2+ 2 2(S 2 f L 2+νT g 1 H 2 ))(1+ ν T S S 4 4 g 1 H 2 ) 2 C + 2 ((2 + S2 2 α )(C K + S 6 C L)(S 2 f L 2 + νt g 1 H 2 ) + 2νT S 4 S4 g 1 H 2 +ν( α + S2 2 2 )( 3K + 1) v g C 1,σ 0 + ν S 6 2 ( 3L + 1) α 2 v g L 3 < ν2 2, (5.14) where C is defined in (3.20), K is defined in (3.21), C is defined in (5.4), L is defined in (5.5), T is defined in (2.10), S p is defined in (1.9), S p is defined in (1.11) and C is defined in (3.2). 20

Remark 5.3 When g = 0, the sufficient condition for uniqueness (5.14) substantially simplifies and becomes : C α ((2+ S 2 α )C L+S 6 C L) curl f L 2+3 2S 2 C((1+ S 2 2 α )C K+ S 6 2 C L) f L 2 < ν2 2. (5.15) References [1] M. Amara, C. Bernardi, V. Girault and F. Hecht, Regularized finite element discretizations of a grade-two fluid model, Int. J. for Numerical Methods in Fluids 48, 1375-1414 (2005). [2] C. Amrouche, C. Bernardi, M. Dauge and V. Girault, Vector potentials in Three- Dimensional Nonsmooth Domains, Math. Meth. in the Appl. Sci. (21), 823-864 (1998). [3] J.M. Bernard, Fluides de second et troisième grade en dimension trois: solution globale et régularité, Thèse de doctorat, Université Pierre-et-Marie-Curie (1998). [4] J.M. Bernard,Stationary problem of second-grade fluids in three dimensions: existence, uniqueness and regularity, Math. Meth. Appl. Sci., 22, 655-687 (1999). [5] J.M. Bernard, Solutions W 2,p, p > 3, for second grade fluid equations with a boundary of class C 1,1, Com. on Appl. Nonlinear Anal. 9, 1, 1-29 (2002). [6] S.C. Brenner and L.R. Scott, The Mathematical Theory of Finite Element Methods, TAM 15, Springer-Verlag, Berlin 1994. [7] D. Cioranescu and V. Girault, Weak and classical solutions of a family of second grade fluids, Int. J. Non-linear Mechanics, 32, 2, 317-335 (1997). [8] D. Cioranescu and E.H. Ouazar, Existence et unicité pour les fluides de second grade, Note CRAS 298 Série I, 285-287 (1984). [9] D. Cioranescu and E.H. Ouazar, Existence and uniqueness for fluids of second grade, in Nonlinear Partial Differential Equations, Collège de France Seminar, Pitman 109 178-197 (1984). [10] M. Dauge, Stationary Stokes and Navier-Stokes Systems on two or three dimensonal Domains with Corners. Part I: Linearized Equations, SIAM J. Math. Anal. 20, 1, 1989. [11] J.E. Dunn and R.L. Fosdick, Thermodynamics, stability and boundedness of fluids of complexity two and fluids of second grade, Arch. Rat. Mech. Anal. 56,3, 191-252 (1974). [12] R.L. Fosdick and K.R. Rajagopal, Anomalous features in the model of second order fluids, Arch. Rat. Mech. Anal. 70, 3, 1-46 (1979). [13] R.L. Fosdick and K.R. Rajagopal, Thermodynamics and stability of fluids of third grade, Proc. Royal Soc. London A 339, 351-377 (1980). [14] V. Girault and P.A. Raviart, Finite Element Approximation for Navier-Stokes Equations. Theory and Algorithms, SMC 5, Springer-Verlag, Berlin, 1986. [15] V. Girault and L.R. Scott, Analysis of two-dimentional grade-two fluid model with a 21

tangential boundary condition, J. Math. Pures Appl., 78, 981-1011 (1999). [16] V. Girault and L.R. Scott, Finite-element discretizations of a two-dimensional gradetwo fluid model, Modél. Math. et Anal. Numér. 35, 1007-1053 (2001). [17] V. Maz ya and J. Rossmann, Elliptc Equations in Polyhedral Domains, Mathematical Surveys and Monographs, Vol. 162, American Mathematical Society (2010). [18] J. Nečas, Les Méthodes Directes en Théorie des Equations Elliptiques, Masson, Paris (1967). [19] W. Noll and C. Truesdell, The Nonlinear Field Theory of Mechanics. Handbuch of Physik, Vol. III, Springer-Verlag, Berlin (1975). [20] E.H. Ouazar, Sur les Fluides de Second Grade, Thèse 3ème Cycle, Université Pierreet-Marie-Curie (1981). 22