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A bi-projection method for Bingham type flows Laurent Chupin, Thierry Dubois Laboratoire de Mathématiques Université Blaise Pascal, Clermont-Ferrand Ecoulements Gravitaires et RIsques Naturels Juin 2015 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 1 / 20

Summary 1 Motivations 2 One difficulty: the presence of a threshold 3 Numerical simulations

Complex fluids Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 2 / 20

Rheology Rheology is the study of the flow of matter, primarily in a liquid state, but also as solids under conditions in which they respond with plastic flow rather than deforming elastically in response to an applied force. Force Huile Maizena Peinture Boue Eau Pente = viscosite Cisaillement Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 3 / 20

Summary 1 Motivations 2 One difficulty: the presence of a threshold 3 Numerical simulations

Bingham model Conservation equations (incompressible and isothermal case): { ρ0 ( t u + u u ) + p = div τ, The stress is given by: div u = 0. τ = 2µ 0 Du + σ 0 Du Du The deformation tensor and its Froebenius norm are defined by: Du = 1 2 ( u + T ( u) ) and Du 2 = 1 i,j d (Du) ij 2. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 4 / 20

Bingham model Conservation equations (incompressible and isothermal case): { ρ0 ( t u + u u ) + p = div τ, The stress is given by: div u = 0. Du τ = 2µ 0 Du + σ 0 Du τ = 2µ 0 Du + σ 0 σ, where the extra-stress satisfies: σ = Du Du if Du 0, σ 1 if Du = 0. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 4 / 20

Classical approaches 1 Variational inequalities ( ρ 0 t u+u u ) (v u)+2µ 0 Ω Well suited for finite volume methods 2 Regularization τ = 2µ 0 Du + σ 0 Du Du Ω Du : (Dv Du)+σ 0 Use results of the quasi-newtonian models No stop in finite time! Ω Dv Du 0 τ ε = 2µ 0 Du ε + σ 0 Du ε Du ε + ε Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 5 / 20

Bingham model Conservation equations (incompressible and isothermal case): { ρ0 ( t u + u u ) + p = div τ, The stress is given by: div u = 0. τ = 2µ Du 0 Du + σ 0 Du where the extra-stress satisfies: τ = 2µ 0 Du + σ 0 σ, σ = Du Du if Du 0, σ 1 if Du = 0. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 6 / 20

Bingham model Conservation equations (incompressible and isothermal case): { ρ0 ( t u + u u ) + p = div τ, The stress is given by: div u = 0. τ = 2µ Du 0 Du + σ 0 Du where the extra-stress satisfies: σ = Du si Du Du 0, σ 1 if Du = 0. τ = 2µ 0 Du + σ 0 σ, σ = P(σ + r Du), where P is the projector on the following convex closed set: { } Λ := σ L 2 (Ω) d d ; σ(x) 1, p.p.. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 6 / 20

Bingham model 1 We introduce two non-dimensional numbers: 2 The model becomes Re = ρ 0VL µ 0 and Bi = σ 0L µ 0 V. t u + u u + p 1 Bi u = div σ, Re Re under the two constraints (r being any positif real) div u = 0 and σ = P(σ + r Du). Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 7 / 20

Algorithm: time discretization u n and p n being given we obtain ũ n+1 solving ũ n+1 u n + p 1 δt Re ũn+1 = 0 ũ n+1 Ω = 0. We obtain (u n+1, p n+1 ) using the free divergence constraint: u n+1 ũ n+1 + (p n+1 p n ) = 0 δt div u n+1 = 0 u n+1 n = 0. Ω (1) (2) Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 8 / 20

Algorithm: time discretization u n, σ n and p n being given we obtain ũ n+1, σ n+1 solving ũ n+1 u n + p n 1 δt Re ũn+1 = Bi Re σ n+1 = P(σ n+1 + r Dũ n+1 + θ(σ n σ n+1 )) ũ n+1 Ω = 0. div σn+1 We obtain (u n+1, p n+1 ) using the free divergence constraint: u n+1 ũ n+1 + (p n+1 p n ) = 0 δt div u n+1 = 0 u n+1 n = 0. Ω (1) (2) Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 8 / 20

Algorithm: time discretization Fixed point to solve (1) : ũ n+1 u n + p n 1 δt Re ũn+1 = Bi Re σ n+1 = P(σ n+1 + r Dũ n+1 + θ(σ n σ n+1 )) ũ n+1 Ω = 0. div σn+1 σ n,k being given, we obtain ũ n,k then σ n,k+1 : ũ n,k u n + p n 1 δt Re ũn,k = Bi Re div σn,k, σ n,k+1 = P(σ n,k + r Dũ n,k + θ(σ n σ n,k )), ũ n,k Ω = 0. (1) (3) Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 9 / 20

Numerical study Theorem 1 Assume that 2θ + r Bi 2. For each integer n N, the sequence (ũ n,k, σ n,k ) k solution of system (3) converges to (ũ n+1, σ n+1 ), solution of system (1). Moreover the convergence is geometric with common ratio 1 θ. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 10 / 20

Numerical study - Proof of theorem 1 1 Equation on differences u k = ũ n,k ũ n+1 and σ k = σ n,k σ n+1 2 Energy estimate: 1 δt uk 1 Re uk = Bi Re div σk, u k Ω = 0, σ k+1 = P(σ n,k + r Dũ n,k + θ(σ n σ n,k )) P(σ n+1 + r Dũ n+1 + θ(σ n σ n+1 )). 1 δt uk 2 L 2 (Ω) + 1 Re uk 2 L 2 (Ω) = Bi Re σk, Du k. 3 The projection P is a contraction: σ k+1 2 L 2 (Ω) (1 θ)2 σ k 2 L 2 (Ω) +r 2 u k 2 L 2 (Ω) +2r(1 θ) σk, Du k. 4 Simple combination gives the result. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 11 / 20

Numerical study: stability result Theorem 2 (Stability) We assume that r Bi 1 et θ 1/2 The sequence (u n, ũ n, p n, σ n ) n solution of system (1) (2) is bounded. ũ n+1 u n + p n 1 δt Re ũn+1 = 0 Bi Re σ n+1 = P(σ n+1 + r Dũ n+1 + θ(σ n σ n+1 )) u n+1 ũ n+1 + (p n+1 p n ) = 0 δt div u n+1 = 0 u n+1 n = 0 and ũ n+1 Ω = 0. Ω div σn+1 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 12 / 20

Numerical study - Proof of theorem 2 1 Energy estimate: ũ n+1 2 L 2 (Ω) un 2 L 2 (Ω) + ũn+1 u n 2 L 2 (Ω) + 2δt Re ũn+1 2 L 2 (Ω) = 2δt p n, ũ n+1 2δt Bi σ n+1, Dũ n+1. Re 2 Control of p n, ũ n+1 taking δ(u n+1 + ũ n+1 ) + δt 2 (p n+1 + p n ) as test function: u n+1 2 L 2 (Ω) + δt2 p n+1 2 L 2 (Ω) 3 Control of σ n+1, Dũ n+1 : = 2δt p n, ũ n+1 + ũ n+1 2 L 2 (Ω) + δt2 p n 2 L 2 (Ω). θ σ n+1 2 L 2 (Ω) + (1 2θ) θ σn+1 σ n 2 L 2 (Ω) 2 r 2 ũ n+1 2 L 2 (Ω) + 2r σn+1, Dũ n+1 + θ σ n 2 L 2 (Ω). Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 13 / 20

Numerical study: convergence result Theorem 3 (Convergence) We assume that r Bi 1/3 et θ 1/3 If there exists a regular solution (u, p, σ) of continuous system then the sequence (u n ) n 0 issued from the previous algorithm converges to u as n tends to +. More precisely, there exists a constant C such that forall 0 n N, we have u(t n ) u n 2 L 2 n + δt u(t k ) u k 2 L C (θ δt + δt 2 ). 2 k=0 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 14 / 20

Summary 1 Motivations 2 One difficulty: the presence of a threshold 3 Numerical simulations

Numerical scheme and implementation Time discretization: second-order projection method based on the combination of the BDF2 (Backward Differentiation Formulae) and the AB2 (Adams-Bashforth) schemes. Spatial discretization: Ω = (0, L x ) (0, L y ) is discretized by using a Cartesian uniform mesh. v i,j+1 (x i, y j+1 ) (x i+1, y j+1 ) u i,j p ij σ ij u i+1,j (x i, y j ) v ij (x i+1, y j ) Implemented in a F90/MPI code. The PETSc library to solve the linear systems and to manage data on structured grids. Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 15 / 20

Stationary flows in the lid-driven cavity at Re = 1000 Bi = 0 Bi = 10 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 16 / 20

Stationary flows in the lid-driven cavity at Re = 1000 Bi = 0 Bi = 10 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 16 / 20

Stationary flows in the lid-driven cavity at Re = 1000 Bi = 0 Bi = 10 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 16 / 20

Numerical estimates of the convergence rate 1e-01 Data: Re = 1000 Bi = 10 r = 0.01 θ = δt N x = N y = 1024 1e-02 1e-03 Error graph: 1e-04 u u ref 2 = F (δt) 1e-05 1e-04 1e-03 1e-02 1e-01 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 18 / 20

Finite stopping times Physical data: Re = 1 Bi = 1 Stopped training velocity at time t = 0.05 Numerical data: r = 0.1 θ = 0.00125 δt = 0.00125 N x = N y = 128 0,2 0,15 Graph of the L 2 norm: u 2 = G(t) 0,1 0,05 0-0,05 0 0,025 0,05 0,075 0,1 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 19 / 20

T H E E N D