Numerical simulation of a viscous Oldroyd-B model Bangwei She, M - PowerPoint PPT Presentation

Numerical simulation of a viscous Oldroyd-B model Bangwei She, M´ aria Luk´ aˇ cov´ a JGU-Mainz cooperation with Prof. M.Tabata, H. Notsu, A. Tezuka Waseda University IRTG 1529 Mathematical fluid dynamics Sino-German Symposium on advanced numerical methods for compressible fluid mechanics and related problems

Oldroyd-B model  Re ( ∂ u β ∂ t + u · ∇ u ) = −∇ p + α ∆ u + We ∇ · σ  ∇ · u = 0 (1) ∂ t + ( u · ∇ ) σ − ∇ u · σ − σ · ( ∇ u ) T = 1 ∂ σ We ( I − σ )  u : velocity, p : pressure Re : Reynolds number σ : conformation tensor, symmetric positive definite, elastic We : Weissenberg number, relaxation time over characteristic time α : ratio of Newtonian viscocity in total viscocity. β = 1 − α . B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 2 / 20

Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 3 / 20

Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 3 / 20

Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 regularized model: regularity in 2D, Constantin,2012 global existence of regularized model: Barrett, 2014 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 3 / 20

Existence results local in time existence: Guillope and Saut, 1990 Renardy 1991 Jourdain, Lelivre and Bris, 2004 Li and Zhang,2004 global existence with small initial data: Guillope and Saut, 1990 Lin, Liu and Zhang, 2005 regularized model: regularity in 2D, Constantin,2012 global existence of regularized model: Barrett, 2014 global existence and uniqueness in the discrete scheme: Lee, Xu, and Zhang, 2011 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 3 / 20

Numerical results High Weissenberg number problem. 201 Re = 1, We = 1 x 10 3.5 3 2.5 2 1.5 1 0.5 0 0 5 10 15 20 25 30 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 4 / 20

Positivity-preserving method: B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 5 / 20

Positivity-preserving method: Log-transformation: ψ = log σ Fattal and Kupfermann, 2005 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 5 / 20

Positivity-preserving method: Log-transformation: ψ = log σ Fattal and Kupfermann, 2005 Square-root: 1 ψ = ( σ ) 2 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 5 / 20

Positivity-preserving method: Log-transformation: ψ = log σ Fattal and Kupfermann, 2005 Square-root: 1 ψ = ( σ ) 2 Euler-Lagrangian method: δ σ δ t = ∂ σ ∂ t + ( u · ∇ ) σ − ∇ u · σ − σ · ( ∇ u ) T is discretized as σ k +1 − F ( σ k ◦ X ) F T ∆ t d F dt = ∇ uF Trebotich, 2005 Lee and Xu, 2006 Lee, Xu, and Zhang, 2011 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 5 / 20

A cartoon model: ∂ t + a ( x ) ∂φ ∂φ ∂ x − b ( x ) φ = − 1 We φ φ ( x , 0) = 0. a ( x ) , b ( x ) > 0 play the role as u , ∇ u . � x 0 exp( b ( x ′ ) − We − 1 ) dx ′ is exponential. Steady state φ ( x ) = a ( x ′ ) B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

A cartoon model: ∂ t + a ( x ) ∂φ ∂φ ∂ x − b ( x ) φ = − 1 We φ φ ( x , 0) = 0. a ( x ) , b ( x ) > 0 play the role as u , ∇ u . � x 0 exp( b ( x ′ ) − We − 1 ) dx ′ is exponential. Steady state φ ( x ) = a ( x ′ ) What would happen if we do logarithm transformation? ψ = log φ . B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

A cartoon model: ∂φ ∂ t + a ( x ) ∂φ ∂ x − b ( x ) φ = − 1 We φ φ ( x , 0) = 0. a ( x ) , b ( x ) > 0 play the role as u , ∇ u . � x 0 exp( b ( x ′ ) − We − 1 ) dx ′ is exponential. Steady state φ ( x ) = a ( x ′ ) What would happen if we do logarithm transformation? ψ = log φ . ∂ x − b ( x ) = − 1 ∂ψ ∂ t + a ( x ) ∂ψ We B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

A cartoon model: ∂φ ∂ t + a ( x ) ∂φ ∂ x − b ( x ) φ = − 1 We φ φ ( x , 0) = 0. a ( x ) , b ( x ) > 0 play the role as u , ∇ u . � x 0 exp( b ( x ′ ) − We − 1 ) dx ′ is exponential. Steady state φ ( x ) = a ( x ′ ) What would happen if we do logarithm transformation? ψ = log φ . ∂ x − b ( x ) = − 1 ∂ψ ∂ t + a ( x ) ∂ψ We Numerical methods based on polynomials is difficult to catch the exponential growth!! Re = 1, We = 2, t = 2 1 250 300 0.9 250 0.8 200 0.7 200 0.6 150 0.5 τ 11 150 0.4 100 100 0.3 0.2 50 50 0.1 0 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 y (x=0.5) B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

some stability technique For any matrix ∇ u and symmetric positive definite matrix σ : ∇ u = B + Ω + N σ − 1 . N , Ω are anti-symmetric, B is symmetric and commutes with τ . B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

some stability technique For any matrix ∇ u and symmetric positive definite matrix σ : ∇ u = B + Ω + N σ − 1 . N , Ω are anti-symmetric, B is symmetric and commutes with τ . ∂ t + ( u · ∇ ) σ − ∇ u · σ − σ · ( ∇ u ) T = Eq.(1) 3 ∂ σ 1 We ( I − σ ) can be written as 1 ∂ σ ∂ t + ( u · ∇ ) σ = Ω σ − σ Ω + 2 B σ + We ( I − σ ) B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

some stability technique For any matrix ∇ u and symmetric positive definite matrix σ : ∇ u = B + Ω + N σ − 1 . N , Ω are anti-symmetric, B is symmetric and commutes with τ . ∂ t + ( u · ∇ ) σ − ∇ u · σ − σ · ( ∇ u ) T = Eq.(1) 3 ∂ σ 1 We ( I − σ ) can be written as 1 ∂ σ ∂ t + ( u · ∇ ) σ = Ω σ − σ Ω + 2 B σ + We ( I − σ ) � Log-transformation ψ = log( σ ): diag ( λ i ) = R T σ R , ψ = R diag ( log λ i ) R T . B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

some stability technique For any matrix ∇ u and symmetric positive definite matrix σ : ∇ u = B + Ω + N σ − 1 . N , Ω are anti-symmetric, B is symmetric and commutes with τ . ∂ t + ( u · ∇ ) σ − ∇ u · σ − σ · ( ∇ u ) T = Eq.(1) 3 ∂ σ 1 We ( I − σ ) can be written as 1 ∂ σ ∂ t + ( u · ∇ ) σ = Ω σ − σ Ω + 2 B σ + We ( I − σ ) � Log-transformation ψ = log( σ ): diag ( λ i ) = R T σ R , ψ = R diag ( log λ i ) R T . ∂ ψ 1 We ( e − ψ − I ) ∂ t + u · ∇ ψ = Ω ψ − ψ Ω + 2 B + B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

kinetic energy 0.018 We=0.5 We=1 0.016 We=3 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 0 5 10 15 20 25 30 time No blow up!! B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 8 / 20

kinetic energy 0.018 We=0.5 We=1 0.016 We=3 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 0 5 10 15 20 25 30 time No blow up!! Table: L 2 error || σ h − σ h / 2 || h We = 0 . 5 We = 1 We = 3 1/32 0.3502 1.5846 4.8967 1/64 0.5006 3.3141 10.8242 1/128 0.7181 5.3517 18.5389 B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 8 / 20 Results do not converge.

A viscous Oldroyd-B model in log-transformation: ∂ u β  We ∇ · e ψ ∂ t + u · ∇ u = −∇ p + α ∆ u +  ∇ · u = 0 We ( e − ψ − I ) + ε ∆ ψ ∂ ψ 1  ∂ t + u · ∇ ψ = Ω ψ − ψ Ω + 2 B + (2) B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 9 / 20

A viscous Oldroyd-B model in log-transformation: ∂ u β  We ∇ · e ψ ∂ t + u · ∇ u = −∇ p + α ∆ u +  ∇ · u = 0 We ( e − ψ − I ) + ε ∆ ψ ∂ ψ 1  ∂ t + u · ∇ ψ = Ω ψ − ψ Ω + 2 B + (2) The viscous model satisfies the inequality: d � β � |∇ u | 2 + tr ( σ + σ − 1 − 2 I ) ≤ 0 dt F ( u , σ ) + α (3) 2 We 2 D D and the free-energy F ( u , σ ) decreases exponentially fast to zero in time. B. She, M. Luk´ aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 9 / 20