Msc (2000): 35j25, 35j55, 35q30

L p estimates of solutions to mixed boundary value problems for the Stokes system in polyhedral domains. Abstract A mixed boundary value problem for the Stokes system in a polyhedral domain is considered. Here different boundary conditions (in particular, Dirichlet, Neumann, free surface conditions) are prescribed on the sides of the polyhedron. The authors prove the existence of solutions in (weighted and non-weighted) Lp Sobolev spaces and obtain regularity assertions for weak solutions. The results are based on point estimates of Green's matrix. 0 Introduction Steady-state flows of incompressible viscous Newtonian fluids are modelled by the Navier-Stokes equations −ν ∆u + (u · ∇) u + ∇p = f, ∇ · u = 0 (0.1) for the velocity u and the pressure p. To this system, one may add a variety of boundary conditions on different parts of the boundary (see e.g. [12]). For example, there is the Dirichlet condition u = 0 on solid walls. On other parts of the boundary (an artificial boundary such as the exit of a canal, or a free surface) a no-friction condition 2νε(u) n − pn = 0 may be useful. Here ε(u) denotes the matrix with the components 1 2 (∂ xi u j + ∂ xj u i), and n is the outward normal. It is also of interest to consider boundary conditions containing components of the velocity and of the friction. Frequently used combinations are the normal component of the velocity and the tangential component of the friction (slip condition for uncovered fluid surfaces) or the tangential component of the velocity and the normal component of the friction (condition for in/out-stream surfaces). In the present paper, we consider a mixed boundary value problem for the linear Stokes system −∆u + ∇p = f, −∇ · u = g (0.2) in a three-dimensional domain of polyhedral type, where components of the velocity and/or the friction are given on the boundary. To be more precise, we have one of the following boundary conditions on each side Γ j : (i) u = h, (iv) −pn + 2ε n (u) = φ, where u n = u · n denotes the normal and u τ = u − u n n the tangential component of u, ε n (u) is the vector ε(u) n, ε n,n (u) is the normal component and ε n,τ (u) the tangential component of ε n (u). In …