A Carbuncle-free Roe-Type Solver for the Euler Equations Friedemann - - PowerPoint PPT Presentation

A Carbuncle-free Roe-Type Solver for the Euler Equations

Friedemann Kemm BTU Cottbus kemm@math.tu-cottbus.de

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Friedemann Kemm BTU Cottbus

Stability of Discrete Shock Profiles

1d:

• Post-shock oscillations (Quirk 1994; Jin & Liu 1996; Arora & Roe 1997, . . . )
• Godunov scheme: unstable discrete profiles (Bultelle, Grassin, Serre 1998)

⇒ tend to neighbouring stable profiles 2d:

• High resolution Riemann solvers produce unstable profiles

(Dumbser, Moschetta, Gressier 2004)

• Same mechanism in Carbuncle and Odd-Even-Decoupling

(Chauvat, Moschetta, Gressier 2005)

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Friedemann Kemm BTU Cottbus

1d-Stability ↔ 2d-Stability

entropy transport jump backwards jump forward

• riginal shock location

y x shear wave

⇒ Stabilization by viscosity on linear waves parallel to shock front

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Friedemann Kemm BTU Cottbus

HLLE Solver

x t Sl Sr ql qr qHLL

• HLL: Constant intermediate state according to conservation
• HLLE: Natural choice of bounding speeds:

SL = min{˜ u − ˜ a, ul − al, 0} SR = max{˜ u + ˜ a, ur + ar, 0}

• High viscosity on shear and entropy waves

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HLLEM

• Comparison of viscosity matrices
• With Roe eigenvalues as SL and SR

gRoe(qr, ql) = gHLL(qr, ql) − ˜ u2 − ˜ a2 4˜ a κ [˜ lT

2 ∆q ˜

r2 +˜ lT

3 ∆q ˜

r3] . with κ = 2˜ a ˜ a + |˜ u|

• Now for gHLL take SL, SR like for HLLE → HLLEM.
• Exact resolution of entropy- and shear waves (Park, Kwon 2002)

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Friedemann Kemm BTU Cottbus

HLL as Modification of Roe

x t ˜ u − ˜ a ˜ u + ˜ a ˜ u − φ(θ) ˜ a ˜ u + φ(θ) ˜ a ˜ u

• Harten Hyman type splitting of contact wave
• HLL for φ(θ) = 1
• Same flux with HLLEM and κ replaced by (1 − φ(θ)) κ

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Friedemann Kemm BTU Cottbus

Desirable Properties of the new Solver

• Exact Resolution of single discontinuities
• No carbuncle
• No information from neighbouring Riemann problems needed

strong shock? strong shock?

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Friedemann Kemm BTU Cottbus

Indicator for Entropy- and Shear Waves

Rankine-Hugoniot condition for single contact or shear wave: f(qr) − f(ql) = ˜ u(qr − ql) Idea: Residual in Rankine-Hugoniot condition as indicator: R := f(qr) − f(ql) − ˜ u(qr − ql) Relate to flow magnitudes: θ =

• R

˜ a

• 2

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Friedemann Kemm BTU Cottbus

Completing the Switching Function

Viscosity bounded by HLL(E): φ(θ) = min{1, θ} Relax by some parameter: φ(θ) = min{1, ε θ} Less dangerous when flow component parallel to shock: φ(θ) = min{1, ε θ max{0, 1 − M α

u }} ,

α > 0 Make φ concave (experimental): φ(θ) = min{1, (ε θ max{0, 1 − M α

u })β} ,

0 < β < 1

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Friedemann Kemm BTU Cottbus

Application of the Switch

Roe:

• Split wave with ˜

u into waves with ˜ u − φ(θ)˜ a and ˜ u + φ(θ)˜ a ⇒ RoeCC HLLEM:

• Multiply anti-diffusion coefficient κ by 1 − φ(θ)

⇒ HLLEMCC Both fluxes identical apart from entropy fix Reasonable setting: ε = 1 100 , α = β = 1 3

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Friedemann Kemm BTU Cottbus

Quirk Test

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 200 400 600 800 1000 1200 1400 1600 Quirk test: Godunov 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 200 400 600 800 1000 1200 1400 1600 Quirk test: HLLEMCC, eps=0.01

Godunov HLLEMCC

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 200 400 600 800 1000 1200 1400 1600 Quirk test: HLLEM 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 200 400 600 800 1000 1200 1400 1600 Quirk test: HLLE

HLLEM HLLE

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Friedemann Kemm BTU Cottbus

Steady Shock

1 2 3 4 5 6 7 10 20 30 40 50 60 70 80 90 100 steady shock, Godunov 1 2 3 4 5 6 7 10 20 30 40 50 60 70 80 90 100 steady shock, HLLEMCC

Godunov HLLEMCC

1 2 3 4 5 6 7 8 10 20 30 40 50 60 70 80 90 100 steady shock, HLLEM 1 2 3 4 5 6 7 10 20 30 40 50 60 70 80 90 100 steady shock, HLLE

HLLEM HLLE

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Friedemann Kemm BTU Cottbus

Colliding Flow (2nd-Order)

10 20 30 40 50 60 70 10 20 30 40 50 60 Colliding flow: Godunov 10 20 30 40 50 60 70 10 20 30 40 50 60 Colliding flow: HLLEMCC

Godunov HLLEMCC

10 20 30 40 50 60 70 10 20 30 40 50 60 Colliding flow: HLLEM 10 20 30 40 50 60 70 10 20 30 40 50 60 Colliding flow: HLLE

HLLEM HLLE

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Friedemann Kemm BTU Cottbus

Colliding Flow (2nd-Order)

10 20 30 40 50 60 70 10 20 30 40 50 60 5 10 15 20 25 30 10 20 30 40 50 60 70 10 20 30 40 50 60 70 10 20 30 40 50 60 5 10 15 20 25 30 10 20 30 40 50 60 70

Godunov HLLEMCC

10 20 30 40 50 60 70 10 20 30 40 50 60 5 10 15 20 25 30 10 20 30 40 50 60 70 10 20 30 40 50 60 70 10 20 30 40 50 60 5 10 15 20 25 30 10 20 30 40 50 60 70

HLLEM HLLE

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Friedemann Kemm BTU Cottbus

Sod Problem: Contact Discontinuity

0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.6 0.65 0.7 0.75 0.8 0.85 HLLEMCC HLLEM HLLE

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Conclusions

• No complete analysis available
• Possible to avoid carbuncle while retaining exact resolution of contact waves
• No information on neighbouring Riemann problems needed (efficiency)
• Steady profiles replaced by stable neighbouring profiles

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Appendix

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Friedemann Kemm BTU Cottbus

Comparison of Roe and HLLE Flux

Roe Flux gRoe(qr, ql) = 1 2[f(qr) + f(ql)] − 1 2| ˜ A(qr, ql)|(qr − ql) HLLE Flux gHLL(qr, ql) = 1 2[f(qr) + f(ql)] − 1 2 SR + SL SR − SL [f(qr) − f(ql)] + SRSL SR − SL (qr − ql) If Roe Matrix exists gHLL(qr, ql) = 1 2[f(qr) + f(ql)] − 1 2 SR + SL SR − SL ˜ A(qr, ql)(qr − ql) + SRSL SR − SL (qr − ql)

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Friedemann Kemm BTU Cottbus

Comparison of Viscosity Matrices

Roe VRoe = | ˜ A(qr, ql)| HLLE VHLL = SR + SL SR − SL ˜ A(qr, ql) − 2 SRSL SR − SL I If SL eigenvalue of ˜ A with eigenvector rl VHLL˜ rL = −SL˜ rL If SR eigenvalue of ˜ A with eigenvector rr VHLL˜ rr = SR˜ rr

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Friedemann Kemm BTU Cottbus

Idea of HLLEM: Write Roe as Correction of HLL

• Choose SL, SR to be the Roe eigenvalues for the outer waves
• ˜

L and ˜ R Matrices with left/right eigenvectors of Roe matrix as rows/columns and ˜ L ˜ R = I

• Find diagonal matrix K such that

VRoe = VHLL + SRSL SR − SL ˜ RK ˜ L

• Only entries for inner waves in K

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