A bi-projection method for Bingham type flows Laurent Chupin, - PowerPoint PPT Presentation

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

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

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

Bingham model Conservation equations (incompressible and isothermal case): � � � ρ 0 ∂ t u + u · ∇ u + ∇ p = div τ , div u = 0 . The stress is given by: ✘ ✘✘✘✘✘✘✘✘✘✘ D u τ = 2 µ 0 D u + σ 0 τ = 2 µ 0 D u + σ 0 σ , | D u | where the extra-stress satisfies: σ = D u if | D u | � = 0 , | D u | | σ | ≤ 1 if | D u | = 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 )+ 2 µ 0 D u : ( D v − D u )+ σ 0 | D v |−| D u | ≥ 0 Ω Ω Ω � Well suited for finite volume methods 2 Regularization D u D u ε τ = 2 µ 0 D u + σ 0 � τ ε = 2 µ 0 D u ε + σ 0 | D u | | D u ε | + ε � Use results of the quasi-Newtonian models � No stop in finite time! Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 5 / 20

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

Bingham model Conservation equations (incompressible and isothermal case): � � � ρ 0 ∂ t u + u · ∇ u + ∇ p = div τ , div u = 0 . The stress is given by: ✘ ✘✘✘✘✘✘✘✘✘✘ D u τ = 2 µ 0 D u + σ 0 τ = 2 µ 0 D u + σ 0 σ , | D u | where the extra-stress satisfies: ✘ ✘✘✘✘✘✘✘✘✘✘✘✘ σ = D u si | D u | � = 0 , | D u | σ = P ( σ + r D u ) , | σ | ≤ 1 if | D u | = 0 . where P is the projector on the following convex closed set: � � σ ∈ L 2 (Ω) d × d Λ := ; | 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: R e = ρ 0 VL B i = σ 0 L and µ 0 V . µ 0 2 The model becomes ∂ t u + u · ∇ u + ∇ p − 1 R e ∆ u = B i R e div σ , under the two constraints ( r being any positif real) div u = 0 and σ = P ( σ + r D u ) . Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 7 / 20

Algorithm: time discretization and p n being given we obtain � u n u n + 1 solving  u n + 1 − u n � + ∇ p n − 1  u n + 1 = 0  R e ∆ �    δ t (1)   u n + 1 �  �   � ∂ Ω = 0 . � We obtain ( u n + 1 , p n + 1 ) using the free divergence constraint:  u n + 1 − � u n + 1  + ∇ ( p n + 1 − p n ) = 0     δ t div u n + 1 = 0 (2)  �    � u n + 1 · n  ∂ Ω = 0 . � 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 � u n + 1 , σ n + 1 solving  u n + 1 − u n � + ∇ p n − 1 u n + 1 = B i   R e div σ n + 1 R e ∆ �    δ t σ n + 1 = P ( σ n + 1 + r D � u n + 1 + θ ( σ n − σ n + 1 )) (1)   u n + 1 �  �   � ∂ Ω = 0 . � We obtain ( u n + 1 , p n + 1 ) using the free divergence constraint:  u n + 1 − � u n + 1  + ∇ ( p n + 1 − p n ) = 0     δ t div u n + 1 = 0 (2)  �    � u n + 1 · n  ∂ Ω = 0 . � Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 8 / 20

Algorithm: time discretization Fixed point to solve (1) :  u n + 1 − u n � + ∇ p n − 1 u n + 1 = B i   R e div σ n + 1 R e ∆ �    δ t σ n + 1 = P ( σ n + 1 + r D � u n + 1 + θ ( σ n − σ n + 1 )) (1)   u n + 1 �  �   � ∂ Ω = 0 . � σ n , k being given, we obtain � u n , k then σ n , k + 1 :  u n , k − u n � + ∇ p n − 1 u n , k = B i   R e div σ n , k , R e ∆ �    δ t σ n , k + 1 = P ( σ n , k + r D � u n , k + θ ( σ n − σ n , k )) , (3)   u n , k �   �  � � ∂ Ω = 0 . Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 9 / 20

Numerical study Theorem 1 Assume that 2 θ + r B i ≤ 2 . u n , k , σ n , k ) k solution of system (3) For each integer n ∈ N , the sequence ( � u n + 1 , σ n + 1 ) , solution of system (1) . converges to ( � 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 = � u n , k − � u n + 1 and σ k = σ n , k − σ n + 1  δ t u k − 1 1 R e ∆ u k = B i  R e div σ k ,     u k �   � ∂ Ω = 0 , �   σ k + 1 = P ( σ n , k + r D � u n , k + θ ( σ n − σ n , k ))      − P ( σ n + 1 + r D � u n + 1 + θ ( σ n − σ n + 1 )) . 2 Energy estimate: 1 L 2 (Ω) + 1 L 2 (Ω) = − B i δ t � u k � 2 R e �∇ u k � 2 R e � σ k , D u 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) + 2 r ( 1 − θ ) � σ k , D u 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 B i ≤ 1 et θ ≤ 1 / 2 The sequence ( u n , � u n , p n , σ n ) n solution of system (1) – (2) is bounded.  u n + 1 − u n � + ∇ p n − 1 u n + 1 = 0 B i   R e div σ n + 1  R e ∆ �   δ t    σ n + 1 = P ( σ n + 1 + r D � u n + 1 + θ ( σ n − σ n + 1 ))     u n + 1 − � u n + 1 + ∇ ( p n + 1 − p n ) = 0   δ t    div u n + 1 = 0     � u n + 1 �  � �  u n + 1 · n  ∂ Ω = 0 and � ∂ Ω = 0 . � � Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 12 / 20

Numerical study - Proof of theorem 2 1 Energy estimate: L 2 (Ω) + 2 δ t u n + 1 − u n � 2 u n + 1 � 2 u n + 1 � 2 L 2 (Ω) − � u n � 2 � � L 2 (Ω) + � � R e �∇ � L 2 (Ω) u n + 1 � − 2 δ t B i = − 2 δ t �∇ p n , � � σ n + 1 , D � u n + 1 � . R e u n + 1 � taking δ ( u n + 1 + � u n + 1 ) + δ t 2 ∇ ( p n + 1 + p n ) as 2 Control of �∇ p n , � test function: � u n + 1 � 2 L 2 (Ω) + δ t 2 �∇ p n + 1 � 2 L 2 (Ω) = 2 δ t �∇ p n , � u n + 1 � + � � u n + 1 � 2 L 2 (Ω) + δ t 2 �∇ p n � 2 L 2 (Ω) . 3 Control of � σ n + 1 , D � u n + 1 � : L 2 (Ω) + ( 1 − 2 θ ) θ � σ n + 1 − σ n � 2 θ � σ n + 1 � 2 L 2 (Ω) ≤ 2 r 2 �∇ � u n + 1 � 2 L 2 (Ω) + 2 r � σ n + 1 , D � u 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 B i ≤ 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 n � � u ( t n ) − u n � 2 �∇ u ( t k ) − ∇ u k � 2 L 2 ≤ C ( θ δ t + δ t 2 ) . L 2 + δ t k = 0 Laurent Chupin (Clermont-Ferrand) Bingham type flows GdR EGRIN - Juin 2015 14 / 20

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