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

Bol. Soc. Paran. Mat. c SPM –ISSN-2175-1188 on line SPM: www.spm.uem.br/bspm (3s.) v. 32 2 (2014): 119–131. ISSN-00378712 in press doi:10.5269/bspm.v32i2.21085 Stokes problem with the possibility of controlling the velocity in a L-shaped domain Omar Chakrone, Okacha Diyer, Driss Sbibih abstract: The movement is studied from a viscous and incompressible homogeneous fluid which crosses a field of the channel in the form of L, with the possibility to exert pressure of known difference between two opposite edges. We extend previous work in [1] which studies a problem of Stokes in the stationary case and with one parameter that characterizes the pressure difference between two sides in a specific domain (symmetric channel). We show existence, unicity and regularity of the solution of an evolution problem with four parameters that characterize the pressure difference between two opposite sides of our field. Key Words: Stokes problem, Regularity of the solution, weak solution. Contents 1 Introduction 119 2 Preliminaries 120 3 Existence, uniqueness and regularity of solution 121 4 Classical problem 126 1. Introduction S8 Let Ω ⊂ R be a bounded domain with boundary ∂Ω = Γ = i=1 Γi , where Γ1 = {0}×[0, 1], Γ2 = {0}×[1, 3], Γ3 = [0, 1]×{3}, Γ4 = {1}×[1, 3], Γ5 = [1, 5]×{1}, Γ6 = {5} × [0, 1], Γ7 = [1, 5] × {0} and Γ8 = [0, 1] × {0}, see Figure 1. Given four real numbers λ1 , λ2 , λ3 and λ4 , we consider the problem:  u = (u1 , u2 ) ∈ V1 such that   Find  2 R 2 R R3 R1  P P  ∂ui   Ω ∇ui ∇vi = λ1 0 v1 (5, y)dy + λ2 1 v1 (1, y)dy Ω ∂t vi + i=1 i=1 R R5 (S1 ) 1  ∀v = (v1 , v2 ) ∈ V1 , v (x, 3)dx + λ4 1 v2 (x, 1)dx +λ 2 3  0    u (x, 0) = a (x) a.e. x inΩ, 1 01   u2 (x, 0) = a02 (x) a.e. x inΩ, 2 ∂v2 1 where V1 is the closing of {v = (v1 , v2 ) ∈ C 1 ([0, T ]; H); div v = ∂v ∂x + ∂y = 0, vi |Γ1 = vi |Γ6 , vi |Γ2 = vi |Γ4 , vi |Γ3 = vi |Γ8 , vi |Γ5 = vi |Γ7 for i = 1, 2} in C([0, T ]; H), H is the closing of ϑ = {u ∈ (D(Ω))2 ; div v = 0} in (L2 (Ω))2 , where D consist of all functions in C ∞ (Ω) which have compact support in Ω and 2000 Mathematics Subject Classification: 76D05, 35Q30 119 Typeset by BSP style. M c Soc. Paran. de Mat. 120 Omar Chakrone, Okacha Diyer, Driss Sbibih a0 = (a01 , a02 ) ∈ H. Γ3 Γ2 Γ4 Γ5 Domain Ω Γ1 Γ6 Γ8 Γ7 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. We cite also a variety of works in the stationary case see [2,4]. This paper is organized as follows. In Section 2, we give preliminaries. In Section 3, we establish existence, unicity and regularity of the solution. In Section 4, we prove the equivalence between our variational problem and the classical problem associated. 2. Preliminaries Let us denote by V the closing of ϑ in (H 1 (Ω))2 . We consider the following Banach spaces L2 (0, T ; V ) and C([0, T ]; H) with the norms kukL2(0,T ;V ) = R  21 T 2 and kukC([0,T ];H) = sup ku(t)kH respectively. 0 ku(t)kV dt t∈[0,T ] Proposition 2.1. If u ∈ W (a, b; V, V ′ ), then for all v in V , we have d (u(.), v)V,V = hu′ (.), viD′ (]a,b[),V in D′ (]a, b[), dt 121 Stokes problem where W (a, b; X, Y ) = {u ∈ L2 (a, b; X); u′ ∈ L2 (a, b; Y )} is a Hilbert space with R  21 b the norm kukW = a (|u(t)|2X + |u′ (t)|2Y )dt (see [3]). Rb Rb Proof: Let ϕ ∈ D(]a, b[), we have a hu′ (t), viϕ(t)dt = a hu′ (t)ϕ(t), vidt. Since u′ ∈ L2 (a, b; V ′ ), we deduce that the function t → hu′ (t), vi is in L2 (a, b) for all v ∈V. In the same way, we have Z b Z b Z b hu′ (t)ϕ(t), vidt = h u′ (t)ϕ(t)dt, vi = −h u(t)ϕ′ (t)dt, vi a a = − a Z b hu(t)ϕ′ (t), vidt = − a Rb Z b (u(t), v)ϕ′ (t)dt. a Rb Thus a hu′ (t), viϕ(t)dt = − a (u(t), v)ϕ′ (t)dt = d hu′ (.), vi = dt (u(.), v). Rb d a dt (u(t), v)ϕ(t)dt, and therefore ✷ 3. Existence, uniqueness and regularity of solution Theorem 3.1. If the solution of the problem (S1 ) exists, it is necessarily unique. Proof: Let u1 and u2 be two solutions of the problem (S1 ). Put w = u1 − u2 , (∂t w, v) + ((w, v)) = 0 ∀v ∈ V1 and w(0) = 0, where (∂t w, v) = 2 R P 2 R P ∂wi Ω ∂t vi , ((w, v)) = Ω ∇wi ∇vi . Then i=1 i=1 Z T (∂t w, v)dt + 0 Z T Rs (∂t w, w)dt + 0 ((w, w))dt = 0. Z s ((w, w))dt = − kw(t)k2V dt ≤ 0. Let us take v = wχ(0,s) (t), s ∈ (0, T ), thus Hence Z s Z s (∂t w, w)dt = − 0 Consequently ((w, v))dt = 0. 0 0 Rs 0 0 1 1 1 kw(s)k2H − kw(0)k2H = kw(s)k2H ≤ 0, 2 2 2 and this shows that w = 0. The problem (S1 ) becomes    ∂Find u ∈ V1 such that  ∂t (u, v) + ((u, v)) = hλ1 e1 , viΓ6 + hλ2 e1 , viΓ4 +hλ3 e2 , viΓ3 + hλ4 e2 , viΓ5    u(x, 0) = a0 (x) ∀v ∈ V1 , a.e. x in Ω, ✷ 122 Omar Chakrone, Okacha Diyer, Driss Sbibih where e1 = (1, 0) and e2 = (0, 1). We are looking for an approximate solution (um ) of the form: um (t) = m X (3.1) gi (t)wi , i=1 where {wi } is a Hilbertienne basis of H and gi ∈ C([0, T ]), gi (0) = gi0 = (u0 , wi ). We have ∀j = 1, . . . , m ∂ (um (t), wj ) + ((um (t), wj )) = hλ1 e1 , wj iΓ6 + hλ2 e1 , wj iΓ4 ∂t + hλ3 e2 , wj iΓ3 + hλ4 e2 , wj iΓ5 . (3.2) Put Vm = {w1 , w2 , . . . , wm }. By replacing (3.1) in (3.2), we obtain m m X ∂ X gi (t)wi , wj )) = γ j , gi (t)wi , wj ) + (( ( ∂t i=1 i=1 where γ j = hλ1 e1 , wj iΓ6 + hλ2 e1 , wj iΓ4 + hλ3 e2 , wj iΓ3 + hλ4 e2 , wj iΓ5 . Thus # "m m X ∂ X gi ((wi , wj )) = γ j . gi (t)(wi , wj ) + ∂t i=1 i=1 (3.3) (3.4) So gj′ (t) + α1j gj (t) = γ j , where αj are the eigenvalues of ψ : H → V, f 7→ u, with u is the unique solution of the problem u ∈ V ; ((u, v)) = (f, v) ∀v ∈ V. We have the following results: ψ(wi ) = αi wi for all i ≥ 1, all the eigenvalues are strictly positive and αi → 0 when i → ∞, {wi } is an orthogonal system in V and we have ((wi , wj )) = α1i δ ij , where δ ij is the kronocker symbol. We obtain  ′ gj (t) + α1j gj (t) = γ j ∀j = 1, . . . , m a.e t ∈ (0, T ), (E) gj (0) = gj0 which admits a unique solution. Consequently um = m P gi (t)wi is the solution of i=1 the approximate problem and in the same way as previously, we show the unicity of the solution um . Theorem 3.2. If u0 ∈ V1 , the approximate solution um satisfies there exist c, c′ > 0 such that ku′m kL2 (0,T ;H) ≤ c and kum kC([0,T ];V ) ≤ c′ . Proof: The function um (x, t) = m P gi (t)wi satisfies i=1 ∂ ∂t (um , v) + ((um , v)) = hλ1 e1 , viΓ6 + hλ2 e1 , viΓ4 + hλ3 e2 , viΓ3 + hλ4 e2 , viΓ5 := b(λ1 , λ2 , λ3 , λ4 , v, Γ) ∀v ∈ V. 123 Stokes problem Then (u′m , v) + ((um , v)) = b(λ1 , λ2 (...truncated)