On the Weak Solvability Via Lagrange Multipliers for a Bingham Model
- PDF / 448,795 Bytes
- 16 Pages / 439.37 x 666.142 pts Page_size
- 54 Downloads / 176 Views
On the Weak Solvability Via Lagrange Multipliers for a Bingham Model Mariana Chivu Cojocaru and Andaluzia Matei Abstract. We study the stationary flow of an incompressible non-Newtonian fluid of Bingham type, mathematically described by means of a nonlinear boundary value problem governed by PDEs. The variational formulation which we propose is a mixed variational problem with Lagrange multipliers. First, we obtain existence, uniqueness, and stability results into an abstract framework. Then, we discuss the well-posedness of the mechanical model based on the auxiliary abstract results. Mathematics Subject Classification. 35J66, 35Q35, 47J30, 49J40, 76D05. Keywords. Non-Newtonian fluid, Bingham constitutive law, mixed variational formulation, Lagrange multipliers, weak solution, fixed point, well-posedness.
1. Introduction The Bingham fluid is a medium who enjoys rigidity below a critical value; above the critical value, such a medium behaves like an incompressible viscous fluid. Some clay suspensions, blood in the capillaries, mayonnaise, mustard, or tooth pastes are only a few examples of materials having a Bingham medium behavior. The mathematical form of the Bingham fluid was proposed by Eugene C. Bingham, see [1]. The flow of the Bingham fluid was studied for many decades by applied mathematicians and numerical analysts. The main progress is due to [8]. For a broad discussion about the numerical simulation of Bingham flow, we send the reader to [6]. The literature contains very much references related to the modeling and simulation of Bingham fluid flow. In addition to [6, 8], we mention herein [4,5,10–12,16,17,19,21–23,26,29], to give only a few examples. In the present paper, we focus on a variational approach via Lagrange multipliers for a model describing the stationary flow of an incompressible non-Newtonian fluid of Bingham type, by means of the following boundary value problem. 0123456789().: V,-vol
164
Page 2 of 16
M. C. Cojocaru, A. Matei
MJOM
¯ → R3 , σ : Ω ¯ → S3 and p : Ω ¯ → R, such that: Problem 1. Find u : Ω − Div σ (x) + (u · ∇)u (x) + ∇p (x) = f 0 (x) σ (x) = 2ηDu(x) + σ (x)S3 ≤ g
(x) g DDuu(x) S3
if Du(x)S3 = 0 if Du(x)S3 = 0
div u(x) = 0 u(x) = 0 uν (x) = 0, σ τ (x) = 0
in Ω,
in Ω,
in Ω, on Γ1 , on Γ2 .
(1) (2)
(3) (4) (5)
Herein, Ω ⊂ R3 is a bounded domain with smooth boundary ∂Ω partitioned in two measurable parts Γ1 and Γ2 with positive measure, u is the velocity, σ is the stress tensor, and σ is the deviatoric stress tensor ) I, where I is the identity tensor), p is the pressure, (σ = σ − trace(σ 3 f 0 is the density of the volume force, ν is the outward unit normal vector at ∂Ω, η, g > 0 are parameters of material, Du = 12 (∇u + ∇uT ) where ∇uT is the transpose of the tensor ∇u, uν = u · ν and σ τ = σν − σν ν, where σν = (σν) · ν. The operator div is the divergence of a vector, e.g., 3 ∂v div v = j=1 ∂xjj for a vector v ∈ R3 , and Div is the divergence of a tensor, 3 ∂θ e.g., Div θ = ( j=1 ∂xijj ) for a tensor θ ∈ S3 , where S3 is the space of the second-order
Data Loading...
