Tamtam Proceedings - lamsin

232 Amoura et al.1. IntroductionFor a start, we consider the Stokes problem in the cylindrical coordinates. We write:⎧−ν(∂r ⎪⎨2 u r + 1 r ∂ ru r − 1ru 2 r + ∂r 2 u r + ∂zu 2 r ) + ∂ r p = f r , inΩ−ν(∂r 2 u z + 1 r ∂ ru z + ∂zu 2 z ) + ∂zu 2 z ) + ∂ z p = f z , inΩ−∂ r u r + ⎪⎩1 r u (1)r + ∂ z u z = 0, inΩu r = 0, u z = 0 on Γ.For the known f = (f r , f r ) ∈ (v 1 0) ′ × (H 1 0 ) ′ , there existsu = (u r , u z ) ∈ (v 1 0) × (H 1 0 ). The nul divergence can be written as:−∂ r (ru r ) + ∂ z (ru z ) = 0, in Ω. (2)Equation (2) show that the vector (ru r, ru z ) is of nul divergence. From Theorem 3.1 inChapter 1 of [3], there exists a function ϕ verifying( )( )rur∂z ϕ= ru = rot(ϕ) =.ru z −∂ r ϕBy taking ϕ = rψ, we have ∂ z ϕ = r∂ z ψ and −∂ r ϕ = −ψ − r∂ r ψ, then we set{u r = ∂ z ψu z = − 1 r (∂ .r(rψ))Now, let us denote (2) by div r u which can also be written as∂ r u r + 1 r u r + ∂ z u z = 0 (3)and by rot r ψ the vector (∂ z ψ, − 1 r ∂ r(rψ)). We also introduce the operator Rot given by:and the operator rot such that for any ψ , we have:(∂z ψrotψ =−∂ r ψWe can then writeRot(v) = ∂ r v z − ∂ z v r /v = (v r , v z ) (4)).Rot(rotψ) = −∆ψ. (5)Note that If div v = 0 we have rotRot(v) = −∆v. If we define the operator ∆ r by :∆ r ψ = ∂ 2 r ψ + 1 r ∂ rψ − 1 r 2 ψ + ∂2 zψ, (6)we can then show thatRot(rot r ψ) = −∆ r ψ. (7)TAMTAM –Tunis– 2005