Steady flows of an Oldroyd fluid with threshold slip

We consider a mathematical model that describes 3D steady flows of an incompressible viscoelastic fluid of Oldroyd type in a bounded domain under mixed boundary conditions, including a threshold-slip boundary condition. Using the concept of weak solutions, we reduce the original slip problem to a coupled system of variational inequalities and equations for the velocity field and stresses. For arbitrary large data (forcing and boundary data) and suitable material constants, we prove the existence of weak solutions and establish some of their properties.


1.
Introduction. It is well known that, under certain circumstances, many non-Newtonian fluids do not satisfy the classical no-slip boundary condition. For instance, polymeric flows can slip over solid surfaces when the shear stress exceeds a critical value. As the departure from the no-slip condition occurs in various ways, numerous mathematical models have been proposed to describe slip effects (see, e.g., the short survey [34]).
In this paper, we investigate a model that describes internal steady flows of an incompressible viscoelastic fluid of Oldroyd type [32,33] in a bounded domain Ω ⊂ R 3 subject to mixed boundary conditions, including a threshold-slip boundary condition [19]. Namely, the problem under consideration takes the form v · n = 0 on Γ,

EVGENII S. BARANOVSKII
Here Ω is the domain occupied by the fluid, Γ denotes the boundary of Ω, v is the velocity, S is the extra-stress tensor, p is the pressure, f denotes the body force, E is the elastic part of the extra-stress tensor, D(v) is the strain-rate tensor, the operator D ρ /Dt is the regularized Jaumann derivative [42]. In the stationary case, this operator is defined by the formula W(v) and W ρ (v) denote the vorticity tensor and the regularized vorticity tensor, respectively, where ρ : R 3 → R is a smooth function with compact support such that R 3 ρ(x) dx = 1 and ρ(x) = ρ(y) whenever |x| = |y|. In formula (10), we set W(v)(y) = 0 if y ∈ R 3 \ Ω. The material constants η (the viscosity coefficient) and λ (the stress relaxation time) are assumed to be positive, and α ∈ (0, 1) is a dimensionless parameter.
Finally, n denotes the unit outer normal to Γ, ( · ) tan stands for the tangential component, i.e., v tan = v − (v · n)n, and q is a non-negative function defined on the subset Γ 0 ⊂ Γ.
The symbol ↑↓ is used to denote oppositely directed vectors. In other words, for any vectors a, b ∈ R 3 a ↑↓ b ⇐⇒ a · b + |a||b| = 0.
In system (1)-(9), the unknowns are v, S, E, and p, while all other quantities are assumed to be given.

Remark 1.
Starting with the pioneering works of Renardy [35] and Guillopé & Saut [20], mathematical models of viscoelastic fluids of Oldroyd type have been studied by many authors. We mention here only the papers [2,3,5,9,10,12,13,14,15,16,17,18,21,22,26,28,30,39,40,41]. A detailed analysis of different problems and results related to the Oldroyd model and other similar non-Newtonian models can be found in the review article [37]. The present work extends the results of the conference paper [8], in which a slip problem for Oldroyd fluids is studied using the substantial derivative in the constitutive law instead of the Jaumann derivative. Namely, in [8], we deal with the following rheological equation of state, which is a simplified version of (4): In contrast to (11), the constitutive law (4) is frame-indifferent (see [42]), i.e., the form of (4) does not change after a change of spatial variables. This means that the model considered here does not violate the principle of material objectivity [38]. Note also that, for the case of flows of viscoelastic fluids subject to Navier's slip boundary condition, which does not take into account the effect of threshold slip, the corresponding boundary and initial-boundary value problems are examined in the recent papers [4,6,7,25].
Remark 2. Let us explain the reasons for using the regularized vorticity tensor W ρ instead of W. The goal of this work is to construct weak solutions of boundary value problem (1)-(9) with arbitrary large data. The main difficulty arises in the passageto-limit procedure in the terms E n W(v n ) and W(v n )E n as n → ∞. To overcome this difficulty, it is convenient to use the following property: if v n v 0 weakly in the Sobolev space This property is one of the key points in the proof of the existence theorem (see Sect. 4).
2. Weak formulation of the problem. We are interested in the existence of weak (generalized) solutions to problem (1)- (9). Before performing our study, let us introduce certain function spaces and notations.
We denote by |x| the Euclidean norm of a vector x and by |G| the Frobenius norm of a tensor G: By G : F denote the scalar product of two tensors G and F: We shall employ the following notations for the Lebesgue and Sobolev spaces of functions defined on Ω and with values in R d . Let R 3×3 sym be the space symmetric matrices of size 3 × 3. Denote by C ∞ 0 (Ω, R 3×3 sym ) the space of infinitely differentiable functions with compact support contained in Ω and with values in R 3×3 sym . By definition, put Assuming that Ω is a bounded domain in R 3 and Γ = ∂Ω is of class C 2 , we equip this space with the scalar product It follows from properties of the Laplace operator ∆ (see, e.g., [36,Chapter 9]) that this scalar product is well defined and the norm is equivalent to the norm induced from the Sobolev space H 2 (Ω, R 3×3 sym ). Introduce the spaces: and define the scalar product in X(Ω, R 3 ) by the formula assuming that at least one of the following two conditions holds: (A1) the domain Ω ⊂ R 3 is not a body of revolution, i.e., Ω is not a body obtained by rotating a plane curve around some straight line that lies on the same plane; (A2) the 2-dimensional Lebesgue measure of the set Γ \ Γ 0 is positive.
As the next proposition shows, the scalar product (14) is well defined and the norm v X(Ω, is equivalent to the norm induced from the space H 1 (Ω, R 3 ). Let U be a bounded locally Lipschitz domain in R 3 . The following statements hold.
Definition 2.1. We shall say that a triplet Remark 5. Let us compare weak and classical solutions to problem (1)- (9). Suppose that (v, S, E, p) is a classical solution to problem (1)- (9). Following exactly the same steps as in [8,Remark 3], we obtain relations (15) and (16). On the other hand, it can be proved that if a weak solution ( v, S, E) of problem (1)-(9) is sufficiently smooth, then there exists a function p such that ( v, S, E, p) is a classical solution to (1)-(9).
3. Main results. The following theorem gives our main results.
Theorem 3.1. Assume that Ω is a bounded domain in R 3 with boundary Γ ∈ C 2 and at least one of conditions (A1), (A2) holds. Furthermore, let f ∈ L 2 (Ω, R 3 ) and q ∈ L 2 + (Γ 0 , R). Then (i) boundary value problem (1)-(9) has at least one weak solution (v, S, E) such that The proof of this theorem uses methods for solving variational inequalities with pseudo-monotone operators and convex functionals [11], the method of introduction of auxiliary viscosity [27], and a passage-to-limit procedure based on energy estimates of approximate solutions and compactness arguments. 4. Proof of the theorem 3.1. First we establish the existence result (i). The proof of this statement is derived in six steps.
Step 2. Now let us estimate the H 2 0 -norm of the vector function R ε (u). Setting In particular, we have Using this inequality, we deduce from (20) that or equivalently, Step 3. We shall show that the operator R ε : X(Ω, R 3 ) → H 2 0 (Ω, R 3×3 sym ) is completely continuous, i.e., if u n u 0 weakly in X(Ω, R 3 ) as n → ∞, then as n → ∞.
Since the embedding H 1 (Ω, as n → ∞. Moreover, since the trace operator γ 0 : as n → ∞. In equation (19), we replace u by u n , E by R ε (u n ), and F by R ε (u n ) − R ε (u 0 ); this gives Further, if we take u = u 0 , E = R ε (u 0 ), and

EVGENII S. BARANOVSKII
Now, summing equalities (25) and (26), we get Moreover, in view of (21), we easily derive that where C denotes positive constant, which is independent of n.
Using the Hölder inequalities and the estimates (29) and (30), we deduce from equality (27) that Here and in the succeeding discussion, the symbols C 1 , C 2 , . . . denote positive constants that are independent of n. Further, taking into account the estimate (28) and the strong convergence (23) as well as (24), we derive from (31) that as n → ∞. This means that (22) holds. Thus, the operator R ε is completely continuous.
Step 4. Consider one more auxiliary problem:

EVGENII S. BARANOVSKII
Find v ∈ X(Ω, R 3 ) such that By [X(Ω, R 3 )] * denote the dual space of X(Ω, R 3 ) and introduce the operators: Then problem (32) can be written as the following variational inequality Next, we observe that the operator A is monotone and for any u ∈ X(Ω, R 3 ). Moreover, since the operator R ε is completely continuous, it can easily be checked that K ε is completely continuous too, i.e., if u n u 0 weakly in X(Ω, R 3 ) as n → ∞, then K ε (u n ) → K ε (u 0 ) strongly in [X(Ω, R 3 )] * as n → ∞.
Using the above-mentioned properties of the operators A and K ε , we deduce that the sum A + K ε is a pseudo-monotone operator and as u X(Ω,R 3 ) → +∞. Taking into account the existence results for variational inequalities with pseudo-monotone operators and convex functionals (see [11, Corollary 30, p. 138]), we see that, for any ε > 0, problem (32) possesses a solution v ε ∈ X(Ω, R 3 ).
Step 5. We want to obtain estimates that are independent of ε. Let ε n > 0 be a sequence such that ε n → 0 as n → ∞. Denote by E εn the vector function R εn (v εn ). We clearly have Putting F = E εn in (34) and using the following equalities: Furthermore, putting 2v εn in place of ϕ into (35), we get where we used the equality On the other hand, by setting ϕ = 0 in (35), we obtain

EVGENII S. BARANOVSKII
Combining (37) and (38), we obviously have Now we multiply (36) by (αη) −1 and add it to (39); this gives In particular, we have Besides, it follows from (40) and (41) This yields that Step 6. We shall construct a weak solution of the original boundary value problem by passing to the limit n → ∞. In view of the estimates (41) and (42), we can assume without loss of generality that, for some v * ∈ X(Ω, R 3 ) and E * ∈ L 2 (Ω, E εn E * weakly in L 2 (Ω, R 3×3 sym ) (44) as n → ∞. Moreover, the weak convergence (43) implies that From (36) it follows that If we combine this inequality with (35), we get Using (43), (44), (45), (46), and the relations we pass to the lower limit in (47). The result is Let Φ be an arbitrary vector function from the space C ∞ 0 (Ω, R 3×3 sym ). Putting F = Φ in (34) and integrating by parts the first term, we get By (43), (44), (45), we pass to the limit n → ∞ in the last equality and obtain Since the set C ∞ 0 (Ω, R 3×3 sym ) is dense in the space H 2 0 (Ω, R 3×3 sym ), the last equality remains valid if we replace Φ with an arbitrary vector function F ∈ H 2 0 (Ω, R 3×3 sym ). Thus, we have for any F ∈ H 2 0 (Ω, R 3×3 sym ). Let us define S * by the formula Clearly, relations (50) and (51) together with (52) mean that the triplet (v * , S * , E * ) is a weak solution of problem (1)- (9). It follows from (40) that