Navier–Stokes equations with periodic boundary conditions and pressure loss

Abstract

We present in this note the existence and uniqueness results for the Stokes and Navier-Stokes equations which model the laminar flow of an incompressible fluid inside a two-dimensional channel of periodic sections. The data of the pressure loss coefficient enables us to establish a relation on the pressure and to thus formulate an equivalent problem.

The camera method, or how to track numerically a deformable particle moving in a fluid network

“…Mathematically, this succession of "cells" can be represented with periodic boundary conditions for the fluid on I and O. The fluid equations are (see for example [2]):…”

Stokes problem with the possibility of controlling the velocity in a L-shaped domain

“…Domain Ω Figure 1: Vertical plane from the channel As applications of this problem, we cite the different types of flows for example see [2,3]. This problem was studied by C. Amrouche, M. Batchi and J. Batina in the stationary case with only one parameter see [1], we extend the preceding work to a problem of evolution with four parameters. Our aim is, in first time to prove the existence, uniqueness and regularity of the solution, second time we show the equivalence between the variational problem, where the notion of pressure does not appear explicitly, and classic problem which highlights the pressure and these differences between the opposite sides of our field.…”

“…Let Ω ⊂ R 2 be a bounded domain with boundary ∂Ω = Γ = 8 i=1 Γ i , where Γ 1 = {0}×[0, 1], Γ 2 = {0}× [1,3], Γ 3 = [0, 1]×{3}, Γ 4 = {1}× [1,3], Γ 5 = [1,5]×{1}, Γ 6 = {5} × [0, 1], Γ 7 = [1,5] × {0} and Γ 8 = [0, 1] × {0}, see Figure 1. Given four real numbers λ 1 , λ 2 , λ 3 and λ 4 , we consider the problem:…”

“…We consider the space 11 (Γ 1 ), we put ν = (0, µ 1 ) t where µ 1 = µ on Γ 1 ∪ Γ 6 0 on Γ 2 ∪ Γ 3 ∪ Γ 4 ∪ Γ 5 ∪ Γ 7 ∪ Γ 8 . It is clear that ν ∈ (H 1 2 (Γ)) 2 and ∂Ω ν. ηdσ = 0, so there exists v ∈ (H 1 (Ω)) 2 such that div v = 0 in Ω and v = ν on Γ (see [1,5]), therefore v ∈ V 1 . According to ( On the other hand, let…”

“…ηdσ = 0. So, there exists v ∈ (H 1 (Ω)) 2 such that div v = 0 in Ω and v = τ on Γ (see [1,5]). In particular v ∈ V 1 .…”