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

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

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Numerical simulation of a viscous Oldroyd-B model Bangwei She, M aria Luk a cov a JGU-Mainz cooperation with Prof. M.Tabata, H. Notsu, A. Tezuka Waseda University IRTG 1529 Mathematical fluid dynamics Sino-German Symposium on

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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

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Oldroyd-B model    Re( ∂u

∂t + u · ∇u) = −∇p + α∆u + β We ∇ · σ

∇ · u = 0

∂σ ∂t + (u · ∇)σ − ∇u · σ − σ · (∇u)T = 1 We (I − σ)

(1) 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

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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

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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

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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

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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

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Numerical results

High Weissenberg number problem.

5 10 15 20 25 30 0.5 1 1.5 2 2.5 3 3.5 x 10

201

Re = 1, We = 1

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 4 / 20

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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

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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

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Positivity-preserving method: Log-transformation: ψ = log σ Fattal and Kupfermann, 2005 Square-root: ψ = (σ)

1 2

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 5 / 20

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Positivity-preserving method: Log-transformation: ψ = log σ Fattal and Kupfermann, 2005 Square-root: ψ = (σ)

1 2

Euler-Lagrangian method: δσ δt = ∂σ ∂t + (u · ∇)σ − ∇u · σ − σ · (∇u)T is discretized as σk+1 − F(σk ◦ X)FT ∆t

dF 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

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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. Steady state φ(x) = x

0 exp( b(x′)−We−1 a(x′)

)dx′ is exponential.

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

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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. Steady state φ(x) = x

0 exp( b(x′)−We−1 a(x′)

)dx′ is exponential. 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

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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. Steady state φ(x) = x

0 exp( b(x′)−We−1 a(x′)

)dx′ is exponential. What would happen if we do logarithm transformation? ψ = log φ. ∂ψ ∂t + a(x)∂ψ ∂x − b(x) = − 1 We

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

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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. Steady state φ(x) = x

0 exp( b(x′)−We−1 a(x′)

)dx′ is exponential. What would happen if we do logarithm transformation? ψ = log φ. ∂ψ ∂t + a(x)∂ψ ∂x − b(x) = − 1 We Numerical methods based on polynomials is difficult to catch the exponential growth!!

0.2 0.4 0.6 0.8 1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 50 100 150 200 250 0.2 0.4 0.6 0.8 1 50 100 150 200 250 300 Re = 1, We = 2, t = 2 y (x=0.5) τ11

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 6 / 20

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some stability technique

For any matrix ∇u and symmetric positive definite matrix σ: ∇u = B + Ω + Nσ−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

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some stability technique

For any matrix ∇u and symmetric positive definite matrix σ: ∇u = B + Ω + Nσ−1. N, Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1)3 ∂σ

∂t + (u · ∇)σ − ∇u · σ − σ · (∇u)T = 1 We (I − σ)

can be written as ∂σ ∂t + (u · ∇)σ = Ωσ − σΩ + 2Bσ + 1 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

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some stability technique

For any matrix ∇u and symmetric positive definite matrix σ: ∇u = B + Ω + Nσ−1. N, Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1)3 ∂σ

∂t + (u · ∇)σ − ∇u · σ − σ · (∇u)T = 1 We (I − σ)

can be written as ∂σ ∂t + (u · ∇)σ = Ωσ − σΩ + 2Bσ + 1 We (I − σ) Log-transformation ψ = log(σ): diag(λi) = RT σR, ψ = Rdiag(logλi)RT .

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

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some stability technique

For any matrix ∇u and symmetric positive definite matrix σ: ∇u = B + Ω + Nσ−1. N, Ω are anti-symmetric, B is symmetric and commutes with τ. Eq.(1)3 ∂σ

∂t + (u · ∇)σ − ∇u · σ − σ · (∇u)T = 1 We (I − σ)

can be written as ∂σ ∂t + (u · ∇)σ = Ωσ − σΩ + 2Bσ + 1 We (I − σ) Log-transformation ψ = log(σ): diag(λi) = RT σR, ψ = Rdiag(logλi)RT . ∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 7 / 20

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5 10 15 20 25 30 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 time kinetic energy We=0.5 We=1 We=3

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

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5 10 15 20 25 30 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 time kinetic energy We=0.5 We=1 We=3

No blow up!!

Table: L2 error ||σh − σh/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 Results do not converge.

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 8 / 20

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A viscous Oldroyd-B model in log-transformation:   

∂u ∂t + u · ∇u = −∇p + α∆u + β We ∇ · eψ

∇ · u = 0

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

(2)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 9 / 20

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A viscous Oldroyd-B model in log-transformation:   

∂u ∂t + u · ∇u = −∇p + α∆u + β We ∇ · eψ

∇ · u = 0

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

(2) The viscous model satisfies the inequality: d dt F(u, σ) + α

  • D

|∇u|2 + β 2We2

  • D

tr(σ + σ−1 − 2I) ≤ 0 (3) 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

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A viscous Oldroyd-B model in log-transformation:   

∂u ∂t + u · ∇u = −∇p + α∆u + β We ∇ · eψ

∇ · u = 0

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

(2) The viscous model satisfies the inequality: d dt F(u, σ) + α

  • D

|∇u|2 + β 2We2

  • D

tr(σ + σ−1 − 2I) ≤ 0 (3) and the free-energy F(u, σ) decreases exponentially fast to zero in time. The free-energy F(u, σ) = Re 2

  • D

|u|2 + β 2We

  • D

tr(σ − log σ − I) (4) is found to be the entropy(Hu and Leli` evre, 2006).

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 9 / 20

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Inner product of the Navier-Stokes equation with the velocity: Re 2 d dt

  • D

|u|2 + α

  • D

|∇u|2 + β We

  • D

∇u : eψ = 0 (5)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 10 / 20

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Inner product of the Navier-Stokes equation with the velocity: Re 2 d dt

  • D

|u|2 + α

  • D

|∇u|2 + β We

  • D

∇u : eψ = 0 (5) For (2)3

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

we contract it with eψ: d dt ψ : eψ = d dt tr(eψ) (Ωψ − ψΩ) : eψ = tr(Ωψeψ) − tr(ψΩeψ) = 0. ∇u : eψ = Ω : eψ + B : eψ + Ne−ψ : eψ = B : eψ (e−ψ − I) : eψ = tr(I − eψ)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 10 / 20

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Inner product of the Navier-Stokes equation with the velocity: Re 2 d dt

  • D

|u|2 + α

  • D

|∇u|2 + β We

  • D

∇u : eψ = 0 (5) For (2)3

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

we contract it with eψ: d dt ψ : eψ = d dt tr(eψ) (Ωψ − ψΩ) : eψ = tr(Ωψeψ) − tr(ψΩeψ) = 0. ∇u : eψ = Ω : eψ + B : eψ + Ne−ψ : eψ = B : eψ (e−ψ − I) : eψ = tr(I − eψ)

d dt

  • D

tr(eψ) = 2

  • D

∇u : eψ + 1 We tr(I − eψ) + ε

  • D

∆ψ : eψ (6)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 10 / 20

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Inner product of the Navier-Stokes equation with the velocity: Re 2 d dt

  • D

|u|2 + α

  • D

|∇u|2 + β We

  • D

∇u : eψ = 0 (5) For (2)3

∂ψ ∂t + u · ∇ψ = Ωψ − ψΩ + 2B + 1 We (e−ψ − I) + ε∆ψ

we contract it with eψ: d dt ψ : eψ = d dt tr(eψ) (Ωψ − ψΩ) : eψ = tr(Ωψeψ) − tr(ψΩeψ) = 0. ∇u : eψ = Ω : eψ + B : eψ + Ne−ψ : eψ = B : eψ (e−ψ − I) : eψ = tr(I − eψ)

d dt

  • D

tr(eψ) = 2

  • D

∇u : eψ + 1 We tr(I − eψ) + ε

  • D

∆ψ : eψ (6)

(5) +

β 2We × (6) ⇒

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 10 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ − I)

  • =

εβ 2We

  • D

∆ψ : eψ = − εβ 2We

  • D

∇ψ : ∇eψ ≤ 0 (7)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 11 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ − I)

  • =

εβ 2We

  • D

∆ψ : eψ = − εβ 2We

  • D

∇ψ : ∇eψ ≤ 0 (7)

∂eψ ∂x : ∂ψ ∂x ≥ 0. ∂eψ ∂x = eψ ∂ψ ∂x .

ψ = Rdiag(λi)RT, σ = Rdiag(eλi)RT R =

  • cosθ

−sinθ sinθ cosθ

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 11 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ − I)

  • =

εβ 2We

  • D

∆ψ : eψ = − εβ 2We

  • D

∇ψ : ∇eψ ≤ 0 (7)

∂eψ ∂x : ∂ψ ∂x ≥ 0. ∂eψ ∂x = eψ ∂ψ ∂x .

ψ = Rdiag(λi)RT, σ = Rdiag(eλi)RT R =

  • cosθ

−sinθ sinθ cosθ

  • Taking the trace of (2)3 ⇒

d dt

  • D

tr(ψ) = 1 We tr(e−ψ − I) (8)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 11 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ − I)

  • =

εβ 2We

  • D

∆ψ : eψ = − εβ 2We

  • D

∇ψ : ∇eψ ≤ 0 (7)

∂eψ ∂x : ∂ψ ∂x ≥ 0. ∂eψ ∂x = eψ ∂ψ ∂x .

ψ = Rdiag(λi)RT, σ = Rdiag(eλi)RT R =

  • cosθ

−sinθ sinθ cosθ

  • Taking the trace of (2)3 ⇒

d dt

  • D

tr(ψ) = 1 We tr(e−ψ − I) (8) (7)-

β 2We × (8) ⇒

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 11 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − ψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ + e−ψ − 2I)

  • ≤ 0
  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 12 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − ψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ + e−ψ − 2I)

  • ≤ 0

Poincar´ e inequality:

  • D|u|2 ≤ Cp
  • D|∇u|2 .
  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 12 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − ψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ + e−ψ − 2I)

  • ≤ 0

Poincar´ e inequality:

  • D|u|2 ≤ Cp
  • D|∇u|2 .

tr(eψ + e−ψ − 2I) ≥ tr(eψ − ψ − I) ≥ 0.

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 12 / 20

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d dt

  • D

Re 2 |u|2 + β 2We tr(eψ − ψ − I)

  • +
  • D
  • α|∇u|2 +

β 2We2 tr(eψ + e−ψ − 2I)

  • ≤ 0

Poincar´ e inequality:

  • D|u|2 ≤ Cp
  • D|∇u|2 .

tr(eψ + e−ψ − 2I) ≥ tr(eψ − ψ − I) ≥ 0. d dt F(u, σ) ≤ −cF(u, σ) ≤ 0(c = min( 2α Re Cp , 1 We ) > 0). Apply the Gronwall inequality, F(u, σ) ≤ F(u(t = 0), σ(t = 0))e−ct (9)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 12 / 20

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characteristic method - time discretization

Let X be the position of a particle,

  • d

dt X = u(X, t), ∀t ∈ [tn, tn+1],

X(t; x) = x. Material derivative:

Dφ Dt = ∂φ ∂t + u · ∇φ

is discretized as

Dφ Dt ≈ φ−φ(X(t−∆t;x),t−∆t) ∆t

t n+1 t

n

t x=X(t,x)

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 13 / 20

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characteristic method - time discretization

Let X be the position of a particle,

  • d

dt X = u(X, t), ∀t ∈ [tn, tn+1],

X(t; x) = x. Material derivative:

Dφ Dt = ∂φ ∂t + u · ∇φ

is discretized as

Dφ Dt ≈ φ−φ(X(t−∆t;x),t−∆t) ∆t

t n+1 t

n

t x=X(t,x)

An implicit scheme which also satisfies the entropy inequality in the discrete level:          Re uk+1−uk◦X(uk,∆t)

∆t

+ ∇pk+1 − α∆uk+1 =

β We ∇ · eψk+1

∇ · uk+1 = 0

ψk+1−ψk◦X(uk,∆t) ∆t

= Ωk+1ψk+1 − ψk+1Ωk+1 + 2Bk+1 + 1

We (e−ψk+1 − I) + ε∆ψk+1

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 13 / 20

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A characteristic finite element scheme

For a given (un

h, pn h, ψn h), find

(un+1

h

, pn+1

h

, ψn+1

h

) ∈ (P1)d × P1 × (P1)

d(d+1) 2

, such that for any test function (v, q, φ) ∈ (P1)d × P1 × (P1)

d(d+1) 2

and for all n = 0, · · · , NT − 1:

Re ∆t (uk+1

h

, v) + 2α(D(uk+1

h

), D(v)) − (pk+1

h

, ∇ · v) − (∇ · uk+1

h

, q) +sh(pk+1

h

, q) = Re ∆t (uk

h ◦ X, v) +

β We (∇ · eψk+1

h

, v) 1 ∆t (ψk+1

h

, φ) + ε(∇ψk+1

h

, ∇φ) = (Ωk+1

h

ψk+1

h

− ψk+1

h

Ωk+1

h

+ 2Bk+1, φ) +( 1 ∆t ψk

h ◦ X1(uk h, ∆t), φ) +

1 We (e−ψk+1

h

− I, φ)

Penalty stabilization: sh(p, q) = −δ

κ h2 κ(∇p, ∇q)κ.

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 14 / 20

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A characteristic finite element scheme

For a given (un

h, pn h, ψn h), find

(un+1

h

, pn+1

h

, ψn+1

h

) ∈ (P1)d × P1 × (P1)

d(d+1) 2

, such that for any test function (v, q, φ) ∈ (P1)d × P1 × (P1)

d(d+1) 2

and for all n = 0, · · · , NT − 1:

Re ∆t (uk+1

h

, v) + 2α(D(uk+1

h

), D(v)) − (pk+1

h

, ∇ · v) − (∇ · uk+1

h

, q) +sh(pk+1

h

, q) = Re ∆t (uk

h ◦ X, v) +

β We (∇ · eψk+1

h

, v) 1 ∆t (ψk+1

h

, φ) + ε(∇ψk+1

h

, ∇φ) = (Ωk+1

h

ψk+1

h

− ψk+1

h

Ωk+1

h

+ 2Bk+1, φ) +( 1 ∆t ψk

h ◦ X1(uk h, ∆t), φ) +

1 We (e−ψk+1

h

− I, φ)

Penalty stabilization: sh(p, q) = −δ

κ h2 κ(∇p, ∇q)κ.

The discrete energy inequality can be derived by taking the test function as (v, q, φ) = (uk+1

h

, −pk+1

h

,

β 2 We (eψk+1

h

− I))

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 14 / 20

slide-41
SLIDE 41

Numerical Test

lid-driven cavity

Geometry: (0,1)x(0,1) Initial values: u = (0, 0), σ = I. Boundary condition for velocity: left, right and bottom, u = 0; top u = (16x2(1 − x)2, 0). Timestep dt = hk(average size of the mesh) Re=1, We=0.5,1,3,5. α = 0.5

u

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 15 / 20

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SLIDE 42

Velocity component and energy at We = 5

  • 0.169

Velocity X

0.75

0.5

0.25

1

  • 0.257

Velocity Y

0.2 0.1

  • 0.1
  • 0.2

0.254

5 10 15 20 25 30 0.02 0.04 0.06 0.08 0.1 0.12 0.14

t kinetic energy @ Re=1,We=5

5 10 15 20 25 30 1 2 3 4 5 6

tr(σ−ln(σ)−I) t

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 16 / 20

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SLIDE 43

Conformation tensor and streamline at We = 5

0.432 PHI 20 16 12 8 4 20.8

  • 1.83

PHI 6 4 2 7.95 0.236 PHI 30 20 10 33.7

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 17 / 20

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SLIDE 44

5 10 15 20 25 30 2 3 4 5 6 7 8

time trσ

we=0.5 we=1 we=3

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 18 / 20

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SLIDE 45

u v p σ1 σ2 σ3 h L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC 1.27e-1 9.62e-3 7.38e-3 1.03e-1 2.50e-1 6.36e-2 1.83e-1 6.37e-2 3.07e-3 1.65 2.03e-3 1.86 4.61e-2 1.16 5.76e-2 2.12 2.23e-2 1.51 3.62e-2 2.34 3.15e-2 6.03e-4 2.35 4.70e-4 2.11 1.67e-2 1.46 1.44e-2 2.00 6.68e-3 1.74 8.79e-3 2.04 1.58e-2 1.55e-4 1.96 1.21e-4 1.96 6.90e-3 1.28 5.04e-3 1.51 2.83e-3 1.24 2.74e-3 1.68 7.87e-2 3.79e-5 2.04 2.69e-5 2.17 3.06e-3 1.18 1.95e-3 1.37 1.34e-3 1.08 1.22e-3 1.16

Table: We=0.5

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 19 / 20

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SLIDE 46

u v p σ1 σ2 σ3 h L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC 1.27e-1 9.62e-3 7.38e-3 1.03e-1 2.50e-1 6.36e-2 1.83e-1 6.37e-2 3.07e-3 1.65 2.03e-3 1.86 4.61e-2 1.16 5.76e-2 2.12 2.23e-2 1.51 3.62e-2 2.34 3.15e-2 6.03e-4 2.35 4.70e-4 2.11 1.67e-2 1.46 1.44e-2 2.00 6.68e-3 1.74 8.79e-3 2.04 1.58e-2 1.55e-4 1.96 1.21e-4 1.96 6.90e-3 1.28 5.04e-3 1.51 2.83e-3 1.24 2.74e-3 1.68 7.87e-2 3.79e-5 2.04 2.69e-5 2.17 3.06e-3 1.18 1.95e-3 1.37 1.34e-3 1.08 1.22e-3 1.16

Table: We=0.5

u v p σ1 σ2 σ3 h L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC L2 EOC 1.27e-1 9.62e-3 7.38e-3 1.03e-1 1.25e-0 7.29e-1 9.01e-1 6.37e-2 3.07e-3 1.65 2.03e-3 1.86 4.61e-2 1.16 2.91e-1 2.11 3.18e-1 1.19 6.14e-1 0.55 3.15e-2 6.03e-4 2.35 4.70e-4 2.11 1.67e-2 1.46 8.74e-2 1.74 8.66e-2 1.88 1.47e-1 2.07 1.58e-2 1.55e-4 1.96 1.21e-4 1.96 6.90e-3 1.28 4.17e-2 1.07 3.25e-2 1.41 5.17e-2 1.50 7.87e-3 3.79e-5 2.04 2.69e-5 2.17 3.06e-3 1.18 2.06e-2 1.01 1.42e-2 1.20 2.20e-2 1.23

Table: We=5.0

Results are convergent!

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 19 / 20

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SLIDE 47

Conclusion: introduction of the model research state and challenge a viscous model based on the log-formula characteristic FEM, free energy dissipative scheme mesh convergence

Thank you for your attention!

  • B. She, M. Luk´

aˇ cov´ a (JGU-Mainz) Numerical tests of a viscous Oldroyd-B model 2014-05-24 20 / 20