Existence Results for Regularized Equations of Second-grade Fluids with Wall Slip

In this paper, we study the equations governing the steady motion of a class of second-grade fluids in a bounded domain of R n , n = 2, 3, with the nonlinear slip boundary condition. We prove the existence of a weak solution without assuming smallness of the data. Moreover, we give estimates for weak solutions and show that the solution set is sequentially weakly closed.


Introduction
Experimental and theoretical investigations show that non-Newtonian fluids often exhibit wall slip, and that this is governed by nonlinear relation between the slip velocity and the shear stress (see, e.g., [10,14] and the references therein).In this paper, we investigate the boundary value problem for the regularized stationary equations of motion of second-grade fluids with the nonlinear slip boundary condition.
Steady flows of an incompressible fluid are described by the system of differential equations in the Cauchy form [13]: where Ω ⊂ R n , n = 2, 3, is the flow domain, v = v(x) is the velocity vector of the particle at a point x ∈ Ω, p = p(x) is the pressure, ∇ denotes Eulerian spatial gradient, f = f (x) is the external body force, ρ is the positive constant density (without loss of generality it can be assumed that ρ = 1), S = S(x) is the extra stress that is specified constitutively.The divergence Div S is the vector with coordinates ∂S ij ∂x i .

E. S. Baranovskii
For fluids of second-grade (see [23]) the extra stress S is given by Here, µ is the viscosity, α 1 and α 2 are the normal stress moduli, A 1 and A 2 are the first two Rivlin-Ericksen tensors: It was shown in [11,12] that Let us denote The tensor W is called the vorticity tensor.
Combining (1.3), (1.4), and (1.5), we get Following [28], we consider the regularized constitutive law Here, W ρ is the regularized vorticity tensor: ) Note that the constitutive law (1.6) is frame-indifferent.This means (see e.g.[27]) that the form of (1.6) does not change after a change of spatial variables.This can be proved by methods of [28].
We will use general Navier-type slip boundary conditions (see [21] for the original reference): where Γ is the boundary of the flow region, n is the outer unit normal on Γ, v • n is the scalar product of the vectors v and n in space R 3 , u τ denotes the tangential component of any vector field u defined on Γ, i.e., and the slip coefficient λ is a given function λ : Boundary value problems for the equations of motion of second-grade fluids have been studied by several authors.Existence results for the corresponding no-slip problems have been established under various restrictions on the data in [5,[7][8][9]20].In [2], the optimal flow control problem for second-grade fluids was studied.The slip problem for time-dependent flows of second-grade fluids was considered by Le Roux in [17].He showed the existence of a unique classical solution under suitable regularity and growth restrictions on the data and the body and surface forces.For small initial data, the global existence of H 3 solutions is shown in Busuioc and Ratiu [6].Moreover, Tani and Le Roux [25] proved the unique solvability in Hölder spaces of the stationary slip problem with a sufficiently small body force.
In this paper, we consider the slip problem without restrictions on the data.Instead of assuming smallness of the data, we use the regularization of constitutive law.We show the existence of a weak solution and establish some estimates for the flow velocity.We also show that the solution set is sequentially weakly closed.
To prove the existence of weak solutions, we give the operator treatment for the slip problem and investigate the corresponding operator equation.First we construct a suitable system of functional spaces with a special basis.Then we construct approximate solutions by the Galerkin method.Using the special properties of the basis, we show that the limit of a sequence of approximate solutions is a solution of the original problem.
Note that the proposed approach is adapted for the study of non-homogeneous boundary value problems.The classical methods are not always effective for such problems.For example, Oskolkov [22] (see also [29]) proved the existence of a weak solution for a simplified version of (1.1)-(1.3)with homogeneous boundary data.He used the method of introduction of auxiliary viscosity.However, the variant with Navier slip boundary condition produces significant technical obstacles.The integration by parts of auxiliary viscosity terms produces additional terms in the motion equations.These terms do not vanish when the regularization parameter tends to zero, and one has to find new methods for the study of non-homogeneous boundary value problems.One of such methods is presented in this article.
We set

Suppose the function λ
where λ 0 and λ 1 are constants.Define the scalar product in X(Ω) as The scalar product in well defined.This follows from Korn's inequality (see, e.g., [19, Chapter II, Theorems 2.2 and 2.3]): where C is a constant.Suppose that f ∈ L 2 (Ω).
holds for any ϕ ∈ X (Ω). ) where Taking the L 2 -scalar product of equality (2.6) with ϕ ∈ X (Ω), we obtain (2.9) Integrating by parts the terms of equality (2.9), we have (2.10) Since the matrix S is symmetric, it follows that Furthermore, taking into account Using (2.11) and (2.12), we can rewrite (2.10) in the following form: with S defined in (1.6).This yields that for any ψ ∈ H 1 (Ω) such that div ψ = 0 and ψ| Γ = 0. Hence (see, e.g., [16]), there exists a function p such that for any ϕ ∈ X(Ω).Integrating by parts, we obtain that the left-hand side of (2.16) is equal to zero.Therefore, we have The main result of this paper is the following.
1) Boundary value problem (2.1)-(2.4)has a weak solution, which satisfies the estimate 2) The weak solution set is sequentially weakly closed.
The proof of Theorem 2.5 is given in Section 4.

Auxiliary results
To prove Theorem 2.5, we study a class of operator equations.Let F 0 , F 1 , Z 1 , . . ., Z m be separable real Hilbert spaces, and X be a linear subspace of F 0 .Suppose the embedding F 0 ⊂ F 1 is continuous.Let symbols X and Y denote the closures of X in F 1 and F 0 , respectively.
Let Z * i be the set of all continuous linear functionals on Z i .The value of a functional from Z * i on an element from Z i is denoted by We assume the following: (iv) for any sequence {v k } ⊂ X such that v k → v 0 weakly in X, it follows that Consider the following problem: for any ϕ ∈ X .
Theorem 3.1.Under conditions (i)-(iv), there exist a solution to problem (A) in the ball {v ∈ X : v X ≤ r 0 }.
Proof.Consider a sequence (x 1 , y 1 ), (x 2 , y 2 ), . . ., (x k , y k ), . . .such that (x i , y i ) ∈ X × Y for any i ∈ N, the set {x 1 , x 2 , . . ., x k , . . .} is everywhere dense in the space X, and the set {y 1 , y 2 , . . ., y k , . . .} is everywhere dense in the space Y. Since the set X × X is everywhere dense in X × Y, we see that there exist a pair (ϕ 11 , ψ 11 ) ∈ X × X such that Moreover, there are pairs (ϕ 21 , ψ 21 ), (ϕ 22 , ψ 22 ) ∈ X × X such that Similarly, for each k ∈ N, there are k pairs Consider the following sequence It is clear that system (3.1) is complete in the space X, i.e., the smallest closed subspace containing (3.1) is the whole space X.Moreover, system (3.1) is complete in Y.
We make the following transformations of sequence (3.1): 1) we eliminate from sequence (3.1) all elements which can be written as linear combinations of elements with smaller indices; 2) we apply the orthogonalization process (in X) to the obtained system.Denote the resulting system as u 1 , u 2 , . . ., u j , . . .We get an orthonormal basis in X such that u j ∈ X for any j ∈ N and the system {u j } ∞ j=1 is complete in Y.We fix a number k ∈ N. Consider the auxiliary problem: It is clear that problem (3.2), (3.3) is equivalent to the operator equation Observe that for all ξ ∈ R k such that ξ R k = r 0 .In fact, using (i)-(iii), we get Then, by [ Since r 0 does not depend on k, there exist a element v * ∈ X such that v k j → v * weakly in X for some subsequence {k j }.We can assume without loss of generality that v k → v * weakly in X.Using (3.2) and (iv), we conclude that for any ϕ ∈ X .Thus, the element v * is a solution to problem (A).Let us recall that v k → v * weakly in X. Taking into account (3.5), we obtain v * X ≤ r 0 .This completes the proof.

Proof of Theorem 2.5
The proof of Theorem 2.5 is based on Theorem 3.1.We set Let Y be the closure of X in the space F 0 .Define the operators T i : X → Z * i , Q i : X × Y → Z i by the following formulas: It is clear that the weak statement of problem (2.1)-(2.4) is equivalent to the following equation Trivially, the operators Q 1 , . . ., Q 5 satisfy condition (i).
Using the equalities for any v ∈ X .Let the function a be given by r → a(r) = r 2 .Obviously, the function a satisfies condition (iii).Taking into account (4.1), we obtain 5 Now we must only prove that condition (iv) holds.Let v k → v 0 weakly in X.Since the embedding H 1 (Ω) ⊂ L 4 (Ω) is compact (see e.g.[1]), we see that v k → v 0 strongly in L 4 (Ω).Therefore, for i = 1, 2, 3, condition (iv) holds.