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Original Articles

Two-dimensional Stokes–Brinkman cell model – a boundary integral formulation

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Pages 251-275 | Accepted 01 Aug 2011, Published online: 19 Oct 2011
 

Abstract

The purpose of this article is to prove the existence and uniqueness of the solution to a two-dimensional cell model problem, which describes the Stokes flow of a viscous incompressible fluid in a bounded Lipschitz region past a porous medium and in the presence of a solid core. The flow within the porous medium is described by the Brinkman equation. One uses the continuity of the velocity and traction fields at the fluid–porous interface, while on the exterior boundary of the fluid envelope, as well as on the boundary of the solid core the velocity field satisfies the prescribed Dirichlet conditions. In order to show the desired existence and uniqueness in certain Sobolev spaces, we develop a layer potential approach based on the potential theory for the Stokes and Brinkman equations. In addition, some particular cases are also analysed.

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Acknowledgements

M. Kohr and E.M. Ului have been supported by the UEFISCSU-CNCSIS Grant PN-II-ID-525/2007. G.P. Raja Sekhar acknowledges the support by Alexander von Humboldt Foundation for the Fellowship during his visit at the Institute of Applied Analysis and Numerical Simulation, University of Stuttgart, and the support by the Department of Science and Technology, Govt. of India, for the project, grant No. SR/S4/MS:405/07. The authors also acknowledge support by the Cluster of Excellence SimTech of the University of Stuttgart.

Notes

Notes

1. If Ω ⊂ ℝ2, the spaces H s (Ω; ℝ2), s ∈ ℝ, are simply denoted by H s (Ω), but their meaning is understood from the context.

2. Hereafter, ⟨·, ·⟩Γ denotes the pairing between two dual spaces H r (Γ) and H r (Γ), r ∈ (0, 1).

3. The Green formulas (Equation2.7) and (Equation2.9) can be naturally extended to the Brinkman operator, by replacing the spaces and with and , where ℒBr(u, π) = ∇π − (Δ − χ2𝕀)u.

4. For simplicity, one uses the notation (n Γ) ≔ n , ℓ = 1, 2.

5. Recall that and .

6. Note that in ℝ3, Ker(𝒱Γ : L 2(Γ) → H 1(Γ)) = ℝn Γ.

7. Recall that

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