Magic. The Lattice Boltzmann Method

Transcription

1 Magic The Lattice Boltzmann Method

2 Lattice Boltzmann: secret ingredients Kn(Ma t f + v x x f + v y y f + v z z f ) = Ω( f ) Analysis Modeling Kn(Ma t π α + x π αx + y π αy + z π αz ) = ω α ( α π α ) ω α : magic relaxation rates α : magic attractors

3 Cumulant method: a magic example 10-1 D3Q27 lattice Magic attractors: relax cumulants Magic relaxation rate: Relative error ER BGK ELB CUM II ω xyz = 2 ω xy L x =L z Error in viscosity Magic cumulant method O(k 4 ) BGK/ELB O(k 2 )

4 From Magic to Science Step 1: Galilean invariance Central moments or cumulants Step 2: Optimize relaxation rates Errors from Taylor analysis still to Find roots of error pre-factors be done

5 Galilean invariance in weak sense Problem: relaxation of (raw) moments with different rates like in MRT method breaks Galilean invariance

6 Galilean invariance in weak sense Regard MRT raw moment equations: Kn(Ma t π xy + x π xxy + y π xyy )=ω xy (π eq xy π xy ) Kn(Ma t π xx + x π xxx + y π xxy )=ω xx (π eq xx π xx ) Kn(Ma t π xxy + x π xxxy + y π xxyy )=ω xxy (π eq xxy π xxy ) Apply Galilean transformation to 3rd eq. π eq xx=π 2 x + 1/3 π eq xy=π x π y π eq xxy=π y /3 π x π x + a π y π y + b π xxy π xxy + bπ xx + 2aπ xy + 2abπ x + a 2 π y + a 2 b rhs xxy +b rhs xx + 2a rhs xy + 2ab rhs x + a 2 rhs y + a 2 b rhs 0 = lhs xxy + b ω xxy ω xx lhs xx + 2a ω xxy ω xy lhs xy Failed!

7 Scattering cascade Non-linear transformation to Galilean invariant variables Central moments Cumulants Back to raw moments n x k x n y k y n z k z vf { f } n x k x n y k y n z k z log(f { f }) n x k x n y k y n z k z F { f } Example 3rd moment ( xxy = 2π x π xy + ω ) xy ( xy π xy ) 2π 2xπ ( y + π y π xx + ω ) xx ( xx π xx ) ω xxy ω xxy

8 in weak sense Galilean invariance at 3rd order moments Perform the Galilean transformation π x π x + a π y π y + b π xxy π xxy + bπ xx + 2aπ xy + 2abπ x + a 2 π y + a 2 b rhs xxy +b rhs xx + 2a rhs xy + 2ab rhs x + a 2 rhs y + a 2 b rhs 0 = lhs xxy + b lhs xx + 2a lhs xy Correct! 3rd order moment now as Galilean invariant as they would be with BGK ansatz

9 Central moments versus cumulants Raw moments: orthogonal decomposition in lattice (=arbitrary) frame of reference Central moments: orthogonal decomposition in frame moving with flow Cumulants or irreducible moments orthogonal decomposition in any frame of reference

10 Equilibrium central moments Transform to raw moments Central moment equilibria polynomials in first order raw moments and monomials in raw moments advection for nonconserved moment! xxyz=2π xyz π x π yz π 2 x + π xxz π y 2π xz π x π y +π xxy π z 2π xy π x π z π xx π y π z + 3π 2 xπ y π z

11 Equilibrium cumulants Transform to raw moments diffusion for nonconserved moment! xxyz= 4π y π x π xz + 2π xz π xy + 2π x π xyz + π yz (π xx 2π 2 x) +π y π xxz + π z (π y (6π 2 x 2π xx ) + π xxy 4π x π xy )) Cumulant equilibria polynomial in all lower order raw moments

12 Test case: 3rd order O(Ma 2 ) Let us investigate: Kn(Ma t π xyy + x π xxyy + y π xyyy + z π xyyz ) = ω xyy ( xyy π xyy ) Assumptions: truncated ( equilibrium state) at 4th order Expansion to low order using π yy = 1/3 +O(Ma 2 ) π xyy =π x /3 +O(Ma 3 ) xyy π xyy =O(Ma 3 )

13 Test case: 3rd order O(Ma 2 ) Note that: Ma t π x = x π xx y π xy z π xz So: x (π xxyy π xx /3) + y (π xyyy π xy /3) + z (π xyyz π xz /3) = O(Ma 4 ) Hence: π xxyy π xx /3 +O(Ma 4 ) π xyyy π xy /3 +O(Ma 4 ) π xyyz π xz /3 +O(Ma 4 )

14 Test case: compare Central moments Cumulants Failed! xxyy=1/9 + π 2 x/3 + π 2 z/3 +O(Ma 4 ) xyyy=π x π y +O(Ma 4 ) xyyz=π x π z /3 +O(Ma 4 ) xxyy=π xx /3 +O(Ma 4 ) xyyy=π xy + π x π y +O(Ma 4 ) xyyz=π xz /3 +O(Ma 4 ) Failed! Failed! Failed!

15 Understand the errors (a research plan) Taylor method (Dubois) a little modified π α + MaKn t π α + = π α Kn( x π αx + y π αy + z π αz) Pre-collision expansion in time = Post-collision expansion in space Expansion in moment form Two scale parameters No generic (acoustic or diffusive) scaling several possibilities Ma & Kn suggested by Asinari

16 Some pitfalls on our way... Expansion parameters ought to be small Not clear for time step expansion Asymptotic order ought to increase with expansion True for Knudsen but false for Mach!

17 Do not forget to scale relaxation rates ν = c 2 s ( 1 ω xy 1 2 O(Ma 2 ) π xy = π eq xy ) O(Kn/Ma) ν Re 1 = Kn Ma 1 = Kn ω xy Ma ( ) 1 1 (Kn x π ω xxy + Kn y π xyy + ) xy O(Ma)!!! O(Ma)!!! diffusive scaling: Ma=Kn better? Be careful using this: π xy = π eq xy +O(Kn) equilibrium as small as reminder!!!

18 Equilibria and their scale in Mach number π eq xy=o(ma 2 ) π eq xx=o(1) π eq xyy=o(ma) π eq xyz=o(ma 3 ) π eq xxyy=o(1) π eq xxyz=o(ma 2 ) π eq xxyyz=o(ma) cyclic repetition of O(Ma), O(Ma 2 ), & O(Ma 3 ) for all higher order moments O(Ma 3 ) is highest order for any moment! no 4th order moment higher than O(Ma 2 ) all odd order moments are at least O(Ma) Convergence completely up to Knudsen number!

19 Magic parameters Analytic solution NSE u y (x,y,z,t = 0)=u 0 sin(2x/l x π)cos(2z/l z π) [ u y (x,y,z,t)=u y (x,y,z,t = 0)e νt ( 2π Lx) 2 +( 2π Lz) 2] L z (" # ") / " ω 4 = ω 5 = ω ! 1 ω 1 = 2 ω 1 = 1 ω 4 = ω 5 = 2 ω 1 ω 1 :viscosity ω 3 = ω 1 :π xxy + π yzz ω 4 :π xxy π yzz ω 5 :π xyz BGK ELB MRT II CLB II CUM II MRT III CLB III CUM III Relative error ER BGK ELB CUM II L x u y BGK & ELB magic cumulants ν = O(h 2 ) O(h 4 ) L x =L z

20 Magic MRT versus magic cumulants 10-1 MRT needs magic parameter for 4th order moment central moments & cumulants Relative error ER BGK ELB CLB * CLB I CLB II MRT I MRT II CUM I CUM II best result: cumulants truncated at 4th order L x =L z use cumulant equilibrium!! strong violations of Galilean invariance in MRT CLB II CUM II MRT II u x O(Ma 2 )!!! this makes the difference! xxyz= 4π y π x π xz + 2π xz π xy + 2π x π xyz + π yz (π xx 2π 2 x) +π y π xxz + π z (π y (6π 2 x 2π xx ) + π xxy 4π x π xy ))

21 Questionnaire Are parameters magic or quadric? Magic! I do not understand them! Do they help us? If they work under general condition, then they are a big leap forward. If not, we might still learn something important from them. We have to understand them! Are cumulants better than moments? Yes! Galilean invariance and stability is improved. Quadric behavior even if we equilibrate 4th cumulants. But they do not solve all problems...