Lattice Boltzmann methods Summer term Cumulant-based LBM. 25. July 2017

Transcription

1 Cumulant-based LBM Prof. B. Wohlmuth Markus Muhr 25. July 2017

2 1. Repetition: LBM and notation Discretization SRT-equation Content 2. MRT-model Basic idea Moments and moment-transformation 3. Cumulant-based LBM Basic idea and technical tools Derivation of the cumulants Cumulant-transformation Cumulant collision operator 4. Validation a.k.a. colorful pictures Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 1/60

3 Repetition: LBM and notation Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 2/60

4 Computational Fluid Dynamics Numerical simulation of flows Calculation of the velocity field Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 3/60

5 Computational Fluid Dynamics Numerical simulation of flows Calculation of the velocity field Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 4/60

6 Discretization In space In velocity (D2Q9-model) c 6 c 2 c 5 c 3 c 1 c 7 c 4 c 8 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 5/60

7 (Discrete) Boltzmann-equation Considering probabilities of an encounter fi f 6 f 2 f 5 f 3 f 1 f 7 f 4 f 8 Derivation of a (discrete) transport-equation for the distributions f i : f i (t, x) t + c i x f i (t, x) = Q(f i ) 1 τ C (f i (t, x) f eq i (t, x) ) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 6/60

8 (Discrete) Boltzmann-equation Considering probabilities of an encounter fi f 6 f 2 f 5 f 3 f 1 f 7 f 4 f 8 Derivation of a (discrete) transport-equation for the distributions f i : f i (t, x) t + c i x f i (t, x) = Q(f i ) 1 τ C (f i (t, x) f eq i (t, x) ) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 6/60

9 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 7/60 Equilibrium distributions f eq i Maxwell-distribution as natural velocity-distribution of an ideal gas: ( ) ) m 3/2 m ξ u 2 M(ξ, ρ, u, T ) = ρ exp ( 2π k B T 2k B T Taylor expansion up to second order ξ = ci : ( f eq i = ρ ω i c i u c 2 3 u 2 2c ( ci c 4 u ) ) 2

10 SRT-equation Discretization of the derivative by finite differences f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Interpretation as collision and streaming: f 6 f 2 f 5 f 3 f 1 f 7 f 4 f 8

11 SRT-equation Discretization of the derivative by finite differences f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Interpretation as collision and streaming: f 6 f 2 f 5 f 6 f 2 f 5 f 3 f 1 f 3 f 1 f 7 f 4 f 8 f7 f4 f8

12 SRT-equation Discretization of the derivative by finite differences f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Interpretation as collision and streaming: f 6 f 2 f 5 f 6 f 2 f 5 f 6 f 2 f 5 f 3 f 1 f 3 f 1 f 3 f 1 f 7 f 4 f 8 f7 f4 f8 f7 f4 f8 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 8/60

13 SRT-equation Discretization of the derivative by finite differences f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Interpretation as collision and streaming: f 6 f 2 f 5 f 6 f 2 f 5 f 6 f 2 f 5 f 3 f 1 f 3 f 1 f 3 f 1 f 7 f 4 f 8 f7 f4 f8 f7 f4 f8 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 8/60

14 In three dimensions Analogously in three dimensions (D3Q27-model) z y x Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 9/60

15 Notation Convenient notation by index-tupel: f 6 f 2 f 5 f 11 f 01 f 11 f 3 f 1 f 10 f 10 f 7 f 4 f 8 f 1 1 f 0 1 f 1 1 f i f ij with i, j { 1, 0, 1} Analogously in three dimensions: fi f ijk with i, j, k { 1, 0, 1} Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 10/60

16 MRT-model Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 11/60

17 Basic idea of MRT (1) The SRT-equation reads: f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Idea: Mix the relaxation terms f i (t + t, x + c i t) = f i (t, x) n V j=0 1 ( ) f j (t, x) f eq j (t, x) τ ij Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 12/60

18 Basic idea of MRT (1) The SRT-equation reads: f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Idea: Mix the relaxation terms f i (t + t, x + c i t) = f i (t, x) n V j=0 1 ( ) f j (t, x) f eq j (t, x) τ ij In matrix-vector-form. f i (t + t, x + c i t). =. f i (t, x). S. ( fi (t, x) f eq i (t, x) ). Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 12/60

19 Basic idea of MRT (1) The SRT-equation reads: f i (t + t, x + c i t) = f i (t, x) 1 τ (f i (t, x) f eq i (t, x) ) Idea: Mix the relaxation terms f i (t + t, x + c i t) = f i (t, x) n V j=0 1 ( ) f j (t, x) f eq j (t, x) τ ij In matrix-vector-form. f i (t + t, x + c i t). =. f i (t, x). S. ( fi (t, x) f eq i (t, x) ). Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 12/60

20 Basic idea of MRT (2) In matrix-vector-form f (t + t, x + c t) = f (t, x) S (f (t, x) f eq (t, x)) Diagonalize the matrix S: f (t+ t, x +c t) = f (t, x) M 1Ŝ (Mf (t, x) Mf eq (t, x)) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 13/60

21 Basic idea of MRT (2) In matrix-vector-form f (t + t, x + c t) = f (t, x) S (f (t, x) f eq (t, x)) Diagonalize the matrix S: f (t+ t, x +c t) = f (t, x) M 1Ŝ (Mf (t, x) Mf eq (t, x)) Interpretation as transformation M : f m f (t + t, x + c t) = f (t, x) M 1Ŝ (m(t, x) meq (t, x)) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 13/60

22 Basic idea of MRT (2) In matrix-vector-form f (t + t, x + c t) = f (t, x) S (f (t, x) f eq (t, x)) Diagonalize the matrix S: f (t+ t, x +c t) = f (t, x) M 1Ŝ (Mf (t, x) Mf eq (t, x)) Interpretation as transformation M : f m f (t + t, x + c t) = f (t, x) M 1Ŝ (m(t, x) meq (t, x)) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 13/60

23 Definition: (Raw)-moments Moments For α, β, γ = 0, 1, 2,... one defined the (raw)-moments of order α + β + γ as follows: m αβγ = 1 i,j,k= 1 In matrix-vector-form (in 2D) m 00 m 10 m 01 m 20 m 02 m 11 m 21 m 12 m 22 = ( ) α ( ) β ( ) γ c ijk c ijk c ijk f ijk = x y z i,j,k= 1 i α j β k γ f ijk Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 14/60 f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8

24 Definition: (Raw)-moments Moments For α, β, γ = 0, 1, 2,... one defined the (raw)-moments of order α + β + γ as follows: m αβγ = 1 i,j,k= 1 In matrix-vector-form (in 2D) m 00 m 10 m 01 m 20 m 02 m 11 m 21 m 12 m 22 = ( ) α ( ) β ( ) γ c ijk c ijk c ijk f ijk = x y z i,j,k= 1 i α j β k γ f ijk Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 14/60 f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8

25 Interpretation of the moments The moments correspond to different physical quantities like: m 00 = 1 i,j= 1 f ij = ρ density c 6 c 2 c 5 c 3 c 1 m 10 = 1 ) (c ij i,j= 1 m01 = 1 i,j= 1 (c ij ) x y c 7 c 4 c 8 f ij = ρ u x x-momentum f ij = ρ u y y-momentum Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 15/60

26 Interpretation of the moments The moments correspond to different physical quantities like: m 00 = 1 i,j= 1 f ij = ρ density c 6 c 2 c 5 c 3 c 1 m 10 = 1 i,j= 1 m01 = 1 i,j= 1 (c ij ) (c ij ) x y c 7 c 4 c 8 f ij = ρ u x x-momentum f ij = ρ u y y-momentum Some moments have to be combined for a reasonable interpretation ( m20 + m 02 = 1 ( ) 2 ( ) ) 2 c ij + c ij f ij E kin x y i,j= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 15/60

27 Interpretation of the moments The moments correspond to different physical quantities like: m 00 = 1 i,j= 1 f ij = ρ density c 6 c 2 c 5 c 3 c 1 m 10 = 1 i,j= 1 m01 = 1 i,j= 1 (c ij ) (c ij ) x y c 7 c 4 c 8 f ij = ρ u x x-momentum f ij = ρ u y y-momentum Some moments have to be combined for a reasonable interpretation ( m20 + m 02 = 1 ( ) 2 ( ) ) 2 c ij + c ij f ij E kin x y i,j= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 15/60

28 Moment-transformation The respective linear combination of matrix-lines yields the moment-transformationmatrix M 0 M 1 M 2 M 3 M 4 M 5 M 6 M 7 M 8 = m 00 m 10 m 01 m 20 + m 02 m 20 m 02 m 11 m 21 m 12 m 22 = f 0 f 1 f 2 f 3 f 4 f 5 f 6 f 7 f 8 Further linear combinations of (raw)-moments are possible as for example row-wise orthogonalization. Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 16/60

29 MRT-algorithm Transform the given distributions f into moments M = M f und M eq = M f eq Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 17/60

30 MRT-algorithm Transform the given distributions f into moments M = M f und M eq = M f eq Relax the moment Mi with frequencies ω i M 0. M 8 = ω 0... ω 8 M 0 M eq 0. M 8 M eq 8 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 17/60

31 MRT-algorithm Transform the given distributions f into moments M = M f und M eq = M f eq Relax the moment Mi with frequencies ω i M 0. M 8 = ω 0... ω 8 M 0 M eq 0. M 8 M eq 8 Transform the relaxed moments back into distribution space f = M 1 M Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 17/60

32 MRT-algorithm Transform the given distributions f into moments M = M f und M eq = M f eq Relax the moment Mi with frequencies ω i M 0. M 8 = ω 0... ω 8 M 0 M eq 0. M 8 M eq 8 Transform the relaxed moments back into distribution space f = M 1 M Collision update and streaming step f (t+ t, x +c t) = f (t, x) M 1Ŝ (Mf (t, x) Mf eq (t, x)) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 17/60

33 MRT-algorithm Transform the given distributions f into moments M = M f und M eq = M f eq Relax the moment Mi with frequencies ω i M 0. M 8 = ω 0... ω 8 M 0 M eq 0. M 8 M eq 8 Transform the relaxed moments back into distribution space f = M 1 M Collision update and streaming step f (t+ t, x +c t) = f (t, x) M 1Ŝ (Mf (t, x) Mf eq (t, x)) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 17/60

34 Pro/Con MRT-model More tuning-parameter ωi as for SRT Better insight into the model by identification with physical quantities Linear transformations statistical dependencies among the moments ( dependencies between ω i ) Physics gets lost by (wrong) variation of parameters. How to choose them? Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 18/60

35 Pro/Con MRT-model More tuning-parameter ωi as for SRT Better insight into the model by identification with physical quantities Linear transformations statistical dependencies among the moments ( dependencies between ω i ) Physics gets lost by (wrong) variation of parameters. How to choose them? Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 18/60

36 Cumulant-based LBM Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 19/60

37 Basic idea of cumulant-based LBM Goal: Find a transformation to statistically independent quantities C m Mathematically this is described by a factorization of the probability-density f ijk = n V m=0 F m (C m ) Efficient implementation of that (nonlinear) transformation Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 20/60

38 Basic idea of cumulant-based LBM Goal: Find a transformation to statistically independent quantities C m Mathematically this is described by a factorization of the probability-density f ijk = n V m=0 F m (C m ) Efficient implementation of that (nonlinear) transformation Main tools: Multivariate Taylor-expansion, Laplace-transformation and distribution-theory Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 20/60

39 Basic idea of cumulant-based LBM Goal: Find a transformation to statistically independent quantities C m Mathematically this is described by a factorization of the probability-density f ijk = n V m=0 F m (C m ) Efficient implementation of that (nonlinear) transformation Main tools: Multivariate Taylor-expansion, Laplace-transformation and distribution-theory Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 20/60

40 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 21/60

41 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 21/60

42 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx One searches for a g L 2 (R) satisfying: G(f ) = g(x) f (x) dx f L 2 (R) R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60

43 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx One searches for a g L 2 (R) satisfying: G(f ) = g(x) f (x) dx f L 2 (R) R Solution: g(x) = 1[ 1,1] (x) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60

44 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx One searches for a g L 2 (R) satisfying: G(f ) = g(x) f (x) dx f L 2 (R) R Solution: g(x) = 1[ 1,1] (x) Interpretation of g as functional (Riesz representation theorem) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60

45 Distributions Recall from measure theory: Lebesgue-space L 2 : L 2 (R) := { f : R R R f (x) 2 dx < } Consider the following functional G : L 2 (R) R, G(f ) := 1 1 f (x) dx One searches for a g L 2 (R) satisfying: G(f ) = g(x) f (x) dx f L 2 (R) R Solution: g(x) = 1[ 1,1] (x) Interpretation of g as functional (Riesz representation theorem) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60

46 Dirac-sequence Interprete the functions gn (x) = n 2 1 [ 1 n, n] 1 (x), n N as functionals Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 23/60

47 Delta-distribution δ For continuous functions f there holds: R g n (x) f (x) dx = n 1 2 n f (x) dx 1 n = n ( ( ) ( 1 2 F F 1 )) n n Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 24/60

48 Delta-distribution δ For continuous functions f there holds: R g n (x) f (x) dx = n 1 2 n f (x) dx 1 n = n ( ( ) ( 1 2 F F 1 )) n n = F ( 1 n ) ( ) F 1 n 2 1 n n F (0) = f (0) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 24/60

49 Delta-distribution δ For continuous functions f there holds: R g n (x) f (x) dx = n 1 2 n f (x) dx 1 n = n ( ( ) ( 1 2 F F 1 )) n n = F ( 1 n ) ( ) F 1 n 2 1 n n F (0) = f (0) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 24/60

50 Delta-distribution δ For continuous functions f there holds: R g n (x) f (x) dx = n 1 2 n f (x) dx 1 n = n ( ( ) ( 1 2 F F 1 )) n n = F ( 1 n ) ( ) F 1 n 2 1 n Define the delta-distribution δ(x) := lim g n(x) = g (x) n by: δ(x) f (x) dx = f (0) R n F (0) = f (0) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 24/60

51 Delta-distribution δ For continuous functions f there holds: R g n (x) f (x) dx = n 1 2 n f (x) dx 1 n = n ( ( ) ( 1 2 F F 1 )) n n = F ( 1 n ) ( ) F 1 n 2 1 n Define the delta-distribution δ(x) := lim g n(x) = g (x) n by: δ(x) f (x) dx = f (0) R n F (0) = f (0) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 24/60

52 Delta-function Considered as a function there holds: δ(x) = { 0 if x 0 if x = 0 Which often gets normed to: δ(x) = { 0 if x 0 1 if x = 0 What you should remember for sure: δ(x) f (x) dx = f (0) R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 25/60

53 Delta-function Considered as a function there holds: δ(x) = { 0 if x 0 if x = 0 Which often gets normed to: δ(x) = { 0 if x 0 1 if x = 0 What you should remember for sure: δ(x) f (x) dx = f (0) R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 25/60

54 Repetition: Fourier-transform The Fourier-transform of f : R C is (modulo scaling) defined by: f (ω) = F[f (x)](ω) := f (x) e iωx dx R f (ω) tells, how strong e iω is present in the frequency-spectrum of f Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 26/60

55 Repetition: Fourier-transform The Fourier-transform of f : R C is (modulo scaling) defined by: f (ω) = F[f (x)](ω) := f (x) e iωx dx R f (ω) tells, how strong e iω is present in the frequency-spectrum of f δ(ω 1) δ(ω + 1) Example: F[sin(x)](ω) 2i Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 27/60

56 Laplace-transformation Instead of projection onto oscillations {e iω } ω R one projects onto damping-functions {e s } s R + F (s) := L[f (x)](s) := 0 f (x) e sx dx Laplace-transformations of a few common functions: L[x n ](s) = n! s n+1 L[sin(ax)](s) = a s 2 +a 2 L[ln(ax)](s) = 1 s (ln(s/a) + γ) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 28/60

57 Laplace-transformation Instead of projection onto oscillations {e iω } ω R one projects onto damping-functions {e s } s R + F (s) := L[f (x)](s) := 0 f (x) e sx dx Laplace-transformations of a few common functions: L[x n ](s) = n! s n+1 L[sin(ax)](s) = a s 2 +a 2 L[ln(ax)](s) = 1 s (ln(s/a) + γ) Certain features as for example the shift-property hold true: L[f (x a)](s) = e as F (s) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 28/60

58 Laplace-transformation Instead of projection onto oscillations {e iω } ω R one projects onto damping-functions {e s } s R + F (s) := L[f (x)](s) := 0 f (x) e sx dx Laplace-transformations of a few common functions: L[x n ](s) = n! s n+1 L[sin(ax)](s) = a s 2 +a 2 L[ln(ax)](s) = 1 s (ln(s/a) + γ) Certain features as for example the shift-property hold true: L[f (x a)](s) = e as F (s) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 28/60

59 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 29/60

60 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 29/60

61 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx = e s 0 = 1 R Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 30/60

62 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx = e s 0 = 1 R Transformation of the shifted delta-distribution by the shift-property: L[δ(x a)](s) = e as L[δ(x)](s) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 30/60

63 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx = e s 0 = 1 R Transformation of the shifted delta-distribution by the shift-property: L[δ(x a)](s) = e as L[δ(x)](s) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 30/60

64 Laplace-transformation of the delta-distribution Apply the definition of the (two-sided) Laplace-transform: L[δ(x)](s) = δ(x) e sx dx = e s 0 = 1 R Transformation of the shifted delta-distribution by the shift-property: L[δ(x a)](s) = e as L[δ(x)](s) = e as Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 31/60

65 Derivation of the cumulants (1) - Analytization Want to capture the discrete velocity-distributions analytically f 11 f 01 f 11 f 10 f 10 f 1 1 f 0 1 f 1 1 This can be done by the peak-function f (ξ): 1 f (ξ) = f (ξ x, ξ y, ξ z ) = f ijk δ(ic ξ x ) δ(jc ξ y ) δ(kc ξ z ) i,j,k= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 32/60

66 Derivation of the cumulants (1) - Analytization Want to capture the discrete velocity-distributions analytically f 11 f 01 f 11 f 10 f 10 f 1 1 f 0 1 f 1 1 This can be done by the peak-function f (ξ): 1 f (ξ) = f (ξ x, ξ y, ξ z ) = f ijk δ(ic ξ x ) δ(jc ξ y ) δ(kc ξ z ) i,j,k= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 32/60

67 Derivation of the cumulants (1) - Analytization Want to capture the discrete velocity-distributions analytically f 11 f 01 f 11 f 10 f 10 f 1 1 f 0 1 f 1 1 This can be done by the peak-function f (ξ): 1 f (ξ) = f (ξ x, ξ y, ξ z ) = f ijk δ(ic ξ x ) δ(jc ξ y ) δ(kc ξ z ) i,j,k= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 33/60

68 Derivation (2) - Laplace-transform To increase regularity one performs the two-sided Laplace-transform of f (ξ): 1 L[f (ξ)](ξ) = f ijk L[δ(ic ξ x )]L[δ(jc ξ y )]L[δ(kc ξ z )] i,j,k= 1 1 = f ijk e Ξx ic e Ξy jc e Ξz kc =: F (Ξ) i,j,k= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 34/60

69 Derivation (2) - Laplace-transform To increase regularity one performs the two-sided Laplace-transform of f (ξ): 1 L[f (ξ)](ξ) = f ijk L[δ(ic ξ x )]L[δ(jc ξ y )]L[δ(kc ξ z )] i,j,k= 1 1 = f ijk e Ξx ic e Ξy jc e Ξz kc =: F (Ξ) i,j,k= 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 34/60

70 Derivation (3) - statistical independence Applying the statistical independence (which one aims to have) in the frequency space: F (Ξ x, Ξ y, Ξ z ) = n V m=0 F m (C m ) And using the logarithm ln (F (Ξ x, Ξ y, Ξ z )) = n V m=0 ln (F m (C m )) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 35/60

71 Derivation (4) - Taylor-expansion The multivariate Taylor-expansion of ln (F (Ξx, Ξ y, Ξ z )) around Ξ = 0 is: α+β+γ ln(f (Ξ x, Ξ y, Ξ z )) = α+β+γ 0 1 α!β!γ! Ξ α x Ξ β y Ξ γ z ln(f (Ξ x, Ξ y, Ξ z )) Ξ=0 Definition: Cumulants For α, β, γ = 0, 1, 2,... one defines the cumulants of order α + β + γ as follows: c αβγ := c (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ ln(f (Ξ x, Ξ y, Ξ z )) z Ξ α x Ξβ y Ξ γ z Ξ=0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 36/60

72 Derivation (4) - Taylor-expansion The multivariate Taylor-expansion of ln (F (Ξx, Ξ y, Ξ z )) around Ξ = 0 is: α+β+γ ln(f (Ξ x, Ξ y, Ξ z )) = α+β+γ 0 1 α!β!γ! Ξ α x Ξ β y Ξ γ z ln(f (Ξ x, Ξ y, Ξ z )) Ξ=0 Definition: Cumulants For α, β, γ = 0, 1, 2,... one defines the cumulants of order α + β + γ as follows: c αβγ := c (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ ln(f (Ξ x, Ξ y, Ξ z )) z Ξ α x Ξβ y Ξ γ z Ξ=0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 36/60

73 Efficient calculation via moments Conduct the Taylor-expansion without the logarithm: α+β+γ F (Ξ x, Ξ y, Ξ z ) = α+β+γ 0 1 α!β!γ! Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) z Analogously consider the coefficients: α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) z Ξ=0 Ξ=0 Ξ α x Ξ β y Ξ γ z Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 37/60

74 Where the magic happens Explicit calculation of these coefficients yields: α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) 1 α+β+γ = z Ξ Ξ=0 α x Ξ β y Ξ γ z i,j,k= 1 f ijk e Ξx ic e Ξy jc e Ξz kc Ξ=0 = 1 i,j,k= 1 f ijk ( ic) α e Ξx ic ( jc) β e Ξy jc ( kc) γ e Ξz kc Ξ=0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 38/60

75 Where the magic happens Explicit calculation of these coefficients yields: α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) 1 α+β+γ = z Ξ Ξ=0 α x Ξ β y Ξ γ z i,j,k= 1 f ijk e Ξx ic e Ξy jc e Ξz kc Ξ=0 = 1 i,j,k= 1 f ijk ( ic) α e Ξx ic ( jc) β e Ξy jc ( kc) γ e Ξz kc Ξ=0 = ( c) α+β+γ 1 i,j,k= 1 i α j β k γ f ijk Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 38/60

76 Where the magic happens Explicit calculation of these coefficients yields: α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) 1 α+β+γ = z Ξ Ξ=0 α x Ξ β y Ξ γ z i,j,k= 1 f ijk e Ξx ic e Ξy jc e Ξz kc Ξ=0 = 1 i,j,k= 1 f ijk ( ic) α e Ξx ic ( jc) β e Ξy jc ( kc) γ e Ξz kc Ξ=0 = ( c) α+β+γ 1 i,j,k= 1 i α j β k γ f ijk = ( c) α+β+γ m αβγ Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 38/60

77 Where the magic happens Explicit calculation of these coefficients yields: α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) 1 α+β+γ = z Ξ Ξ=0 α x Ξ β y Ξ γ z i,j,k= 1 f ijk e Ξx ic e Ξy jc e Ξz kc Ξ=0 = 1 i,j,k= 1 f ijk ( ic) α e Ξx ic ( jc) β e Ξy jc ( kc) γ e Ξz kc Ξ=0 = ( c) α+β+γ 1 i,j,k= 1 i α j β k γ f ijk = ( c) α+β+γ m αβγ Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 38/60

78 Cumulants vs. Moments The (raw)-moments were shown to be: m αβγ = ( c) (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) z Ξ=0 The cumulants: c αβγ = c (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ z ln(f (Ξ x, Ξ y, Ξ z )) Ξ=0 Idea: By applying the chain rule one should be able to trace back the cumulants to the moments... Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 39/60

79 Cumulants vs. Moments The (raw)-moments were shown to be: m αβγ = ( c) (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ F (Ξ x, Ξ y, Ξ z ) z Ξ=0 The cumulants: c αβγ = c (α+β+γ) α+β+γ Ξ α x Ξ β y Ξ γ z ln(f (Ξ x, Ξ y, Ξ z )) Ξ=0 Idea: By applying the chain rule one should be able to trace back the cumulants to the moments... Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 39/60

80 Cumulant-transformation (1) We conduct the idea on the example c100 : c 100 = c (1+0+0) 1 Ξ 1 x Ξ 0 y Ξ 0 ln(f (Ξ x, Ξ y, Ξ z )) z = c 1 1 F (0, 0, 0) 1 F (Ξ x, Ξ y, Ξ z ) Ξ 1 x Ξ=0 Ξ=0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 40/60

81 Cumulant-transformation (1) We conduct the idea on the example c100 : c 100 = c (1+0+0) 1 Ξ 1 x Ξ 0 y Ξ 0 ln(f (Ξ x, Ξ y, Ξ z )) z = c 1 1 F (0, 0, 0) 1 F (Ξ x, Ξ y, Ξ z ) Ξ 1 x Ξ=0 = c 1 1 F (0, 0, 0) m 100 ( c) 1 = 1 F (0, 0, 0) m 100 Ξ=0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 40/60

82 Cumulant-transformation (1) We conduct the idea on the example c100 : c 100 = c (1+0+0) 1 Ξ 1 x Ξ 0 y Ξ 0 ln(f (Ξ x, Ξ y, Ξ z )) z = c 1 1 F (0, 0, 0) 1 F (Ξ x, Ξ y, Ξ z ) Ξ 1 x Ξ=0 = c 1 1 F (0, 0, 0) m 100 ( c) 1 = 1 F (0, 0, 0) m 100 With F (0, 0, 0) = 1 i,j,k= 1 Ξ=0 f ijk e 0 e 0 e 0 = ρ also: c 100 = 1 ρ m 100 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 40/60

83 Cumulant-transformation (1) We conduct the idea on the example c100 : c 100 = c (1+0+0) 1 Ξ 1 x Ξ 0 y Ξ 0 ln(f (Ξ x, Ξ y, Ξ z )) z = c 1 1 F (0, 0, 0) 1 F (Ξ x, Ξ y, Ξ z ) Ξ 1 x Ξ=0 = c 1 1 F (0, 0, 0) m 100 ( c) 1 = 1 F (0, 0, 0) m 100 With F (0, 0, 0) = 1 i,j,k= 1 Ξ=0 f ijk e 0 e 0 e 0 = ρ also: c 100 = 1 ρ m 100 Define: Cαβγ = ρc αβγ, then = C 100 = m 100 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 40/60

84 Cumulant-transformation (1) We conduct the idea on the example c100 : c 100 = c (1+0+0) 1 Ξ 1 x Ξ 0 y Ξ 0 ln(f (Ξ x, Ξ y, Ξ z )) z = c 1 1 F (0, 0, 0) 1 F (Ξ x, Ξ y, Ξ z ) Ξ 1 x Ξ=0 = c 1 1 F (0, 0, 0) m 100 ( c) 1 = 1 F (0, 0, 0) m 100 With F (0, 0, 0) = 1 i,j,k= 1 Ξ=0 f ijk e 0 e 0 e 0 = ρ also: c 100 = 1 ρ m 100 Define: Cαβγ = ρc αβγ, then = C 100 = m 100 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 40/60

85 Cumulant-transformation (2) Analogous procedure for the remaining cumulants: C 100 = m 100 C 010 = m 010 C 001 = m 001 C 200 = m ρ m2 100 C 020 = m ρ m2 010 C 002 = m ρ m2 001 C 110 = m ρ m 100m 010 C 101 = m ρ m 100m C 210 = m ρ 2 m2 100m ρ m 110m ρ m 200m 010. Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 41/60

86 Half-time score What do we have so far? How far do we come with that, starting with the current distributions? Transform distributions f to (raw)-moments m = M f Transform (raw)-moments m to cumulants C = K(m) = K(M f ) = C(f ) [Ŝ ] f (t+ t, x +c t) = f (t, x) C 1 (C(f (t, x)) C(f eq (t, x))) Efficient transformation of equilibria Choice of relaxation parameters ωi Inverse mapping C 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 42/60

87 Cumulant equilibria - Derivation The equilibria can be calculated in advance Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60

88 Cumulant equilibria - Derivation The equilibria can be calculated in advance Laplace-transformation of Maxwell-distribution: F eq (Ξ) = L [ M(ξ) ] (Ξ) = M(ξ) e Ξ ξ dξ x dξ y dξ z R = ρ ( ) 1 exp C 4C (Ξ2 x + Ξ 2 y + Ξ 2 z) Ξ x u x Ξ y u y Ξ z u z Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60

89 Cumulant equilibria - Derivation The equilibria can be calculated in advance Laplace-transformation of Maxwell-distribution: F eq (Ξ) = L [ M(ξ) ] (Ξ) = M(ξ) e Ξ ξ dξ x dξ y dξ z R = ρ ( ) 1 exp C 4C (Ξ2 x + Ξ 2 y + Ξ 2 z) Ξ x u x Ξ y u y Ξ z u z Apply the logarithm: ( ) ρ ln(f eq (Ξ)) = ln Ξ x u x Ξ y u y Ξ z u z + c2 ( s Ξ 2 2 x + Ξ 2 y + Ξ 2 z) ρ 0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60

90 Cumulant equilibria - Derivation The equilibria can be calculated in advance Laplace-transformation of Maxwell-distribution: F eq (Ξ) = L [ M(ξ) ] (Ξ) = M(ξ) e Ξ ξ dξ x dξ y dξ z R = ρ ( ) 1 exp C 4C (Ξ2 x + Ξ 2 y + Ξ 2 z) Ξ x u x Ξ y u y Ξ z u z Apply the logarithm: ( ) ρ ln(f eq (Ξ)) = ln Ξ x u x Ξ y u y Ξ z u z + c2 ( s Ξ 2 2 x + Ξ 2 y + Ξ 2 z) ρ 0 Taylor-expansion can be skipped, since this is a polynomial anyway Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60

91 Cumulant equilibria - Derivation The equilibria can be calculated in advance Laplace-transformation of Maxwell-distribution: F eq (Ξ) = L [ M(ξ) ] (Ξ) = M(ξ) e Ξ ξ dξ x dξ y dξ z R = ρ ( ) 1 exp C 4C (Ξ2 x + Ξ 2 y + Ξ 2 z) Ξ x u x Ξ y u y Ξ z u z Apply the logarithm: ( ) ρ ln(f eq (Ξ)) = ln Ξ x u x Ξ y u y Ξ z u z + c2 ( s Ξ 2 2 x + Ξ 2 y + Ξ 2 z) ρ 0 Taylor-expansion can be skipped, since this is a polynomial anyway Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60

92 Cumulant equilibria Only a certain few cumulants have an equilibrium not equal 0: c eq 000 = ln(ρ/ρ 0) c eq 100 = c 1 u x c eq 010 = c 1 u y c eq 001 = c 1 u z c eq 200 = c 2 c 2 s c eq 020 = c 2 c 2 s c eq 002 = c 2 c 2 s For all others, there holds: c eq αβγ = 0 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 44/60

93 Half-time score What do we have so far? How far do we come with that, starting with the current distributions? Transform distributions f to (raw)-moments m = M f Transform (raw)-moments m to cumulants C = K(m) = K(M f ) = C(f ) [Ŝ ] f (t+ t, x +c t) = f (t, x) C 1 (C(f (t, x)) C(f eq (t, x))) Efficient transformation of equilibria Choice of relaxation parameters ωi Inverse mapping C 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 45/60

94 Choice of relaxation parameters For the most cumulants the equilibrium is 0, hence one gets for example: C 222 = ω 10 (C 222 0) C 110 = ω 1 C 110 C 101 = ω 1 C 101 C 011 = ω 1 C 011 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 46/60

95 Choice of relaxation parameters For the most cumulants the equilibrium is 0, hence one gets for example: C 222 = ω 10 (C 222 0) C 110 = ω 1 C 110 C 101 = ω 1 C 101 C 011 = ω 1 C 011 C 211 = ω 8 C 211 C 121 = ω 8 C 121 C 112 = ω 8 C 112 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 46/60

96 Choice of relaxation parameters For the most cumulants the equilibrium is 0, hence one gets for example: C 222 = ω 10 (C 222 0) C 110 = ω 1 C 110 C 101 = ω 1 C 101 C 011 = ω 1 C 011 C 211 = ω 8 C 211 C 121 = ω 8 C 121 C 112 = ω 8 C 112 Those with non vanishing equilibrium one can combine: C 200 C 020 = ω 1 (C 200 C 020 ) C 200 C 002 = ω 1 (C 200 C 002 ) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 46/60

97 Choice of relaxation parameters For the most cumulants the equilibrium is 0, hence one gets for example: C 222 = ω 10 (C 222 0) C 110 = ω 1 C 110 C 101 = ω 1 C 101 C 011 = ω 1 C 011 C 211 = ω 8 C 211 C 121 = ω 8 C 121 C 112 = ω 8 C 112 Those with non vanishing equilibrium one can combine: C 200 C 020 = ω 1 (C 200 C 020 ) C 200 C 002 = ω 1 (C 200 C 002 ) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 46/60

98 Half-time score What do we have so far? How far do we come with that, starting with the current distributions? Transform distributions f to (raw)-moments m = M f Transform (raw)-moments m to cumulants C = K(m) = K(M f ) = C(f ) [Ŝ ] f (t+ t, x +c t) = f (t, x) C 1 (C(f (t, x)) C(f eq (t, x))) Efficient transformation of equilibria Choice of relaxation parameters ωi Inverse mapping C 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 47/60

99 Half-time score What do we have so far? How far do we come with that, starting with the current distributions? Transform distributions f to (raw)-moments m = M f Transform (raw)-moments m to cumulants C = K(m) = K(M f ) = C(f ) [Ŝ ] f (t+ t, x +c t) = f (t, x) C 1 (C(f (t, x)) C(f eq (t, x))) Efficient transformation of equilibria Choice of relaxation parameters ωi Inverse mapping C 1 Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 48/60

100 Advantages of Cumulant-based LBM Better representation of physical processes due to statistical independence Equilibria of cumulants are almost all 0 More (real) degrees of freedom than in classical MRT Gradual improvement of the method by asymptotic analysis and elimination of error-terms Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 49/60

101 Synthesis of Navier-Stokes-equation By a Chapman-Enskog expansion one can derive the Navier-Stokes equations: u = 0 u t + (u ) u = 1 ρ p + 1 ( 1 1 ) u + f 3 ω 1 2 ρ }{{} =ν With ω1 one can simulate fluids of different viscosity Only ω1 directly influences the solution Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 50/60

102 Synthesis of Navier-Stokes-equation By a Chapman-Enskog expansion one can derive the Navier-Stokes equations: u = 0 u t + (u ) u = 1 ρ p + 1 ( 1 1 ) u + f 3 ω 1 2 ρ }{{} =ν With ω1 one can simulate fluids of different viscosity Only ω1 directly influences the solution Further parameters influence higher order error terms (set them to 1 in the following) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 50/60

103 Synthesis of Navier-Stokes-equation By a Chapman-Enskog expansion one can derive the Navier-Stokes equations: u = 0 u t + (u ) u = 1 ρ p + 1 ( 1 1 ) u + f 3 ω 1 2 ρ }{{} =ν With ω1 one can simulate fluids of different viscosity Only ω1 directly influences the solution Further parameters influence higher order error terms (set them to 1 in the following) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 50/60

104 Validation a.k.a. colorful pictures Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 51/60

105 Turbulence measure The Reynolds-number is dimensionless parameter of a flow Re = U L ν At some specific threshold Re the laminar-turbulent transition starts Traditional LBM methods have problems with high Reynolds-numbers: Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 52/60

106 Comparison: BGK Kumulanten Comparison SRT- and cumulant-based LBM for a cylinder flow at Re = 8000 a) BGK-SRT-LBM (Picture source [5]) b) Cumulant-based LBM (Picture source [5]) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 53/60

107 Turbulence resolution in 3D Representation of (turbulence) vortices around a sphere flow: (Picture source [5]) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 54/60

108 Turbulence resolution in 3D (Picture source [5]) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 55/60

109 Turbulence resolution in 3D (Picture source [5]) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 56/60

110 The highlights MRT-like-method with nonlinear transformation to statistical independent quantities Mathematical methods: Distributions as generalized functions (Analytization) Laplace-transform Comparison of Taylor-coefficients Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 57/60

111 The highlights MRT-like-method with nonlinear transformation to statistical independent quantities Mathematical methods: Distributions as generalized functions (Analytization) Laplace-transform Comparison of Taylor-coefficients Advantages especially at high Reynolds-numbers (stability and adaptivity) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 57/60

112 The highlights MRT-like-method with nonlinear transformation to statistical independent quantities Mathematical methods: Distributions as generalized functions (Analytization) Laplace-transform Comparison of Taylor-coefficients Advantages especially at high Reynolds-numbers (stability and adaptivity) Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 57/60

113 Literature I d Humières, Dominique: Multiple relaxation time lattice Boltzmann models in three dimensions. Philosophical Transactions of the Royal Society of London A: Mathematical, Physical and Engineering Sciences, 360(1792): , Dubois, François: Equivalent partial differential equations of a lattice Boltzmann scheme. Computers & Mathematics with Applications, 55(7): , Geier, M: Ab initio derivation of the cascaded Lattice Boltzmann Automaton. University of Freiburg IMTEK, Geier, Martin, Andreas Greiner und Jan G Korvink: Cascaded digital lattice Boltzmann automata for high Reynolds number flow. Physical Review E, 73(6):066705, Geier, Martin, Martin Schönherr, Andrea Pasquali und Manfred Krafczyk: The cumulant lattice Boltzmann equation in three dimensions: Theory and validation. Computers & Mathematics with Applications, 70(4): , Hänel, Dieter: Molekulare Gasdynamik: Einführung in die kinetische Theorie der Gase und Lattice-Boltzmann-Methoden. Springer-Verlag, Lukacs, Eugene: Characteristics functions. Griffin, London, Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 58/60

114 Literature II Ning, Yang und Kannan N Premnath: Numerical Study of the Properties of the Central Moment Lattice Boltzmann Method. arxiv preprint arxiv: , Rohm, Florian: A commented Python Script for LBM-Flow-Simulations. Lecture on, Wolf-Gladrow, Dieter A: Lattice-gas cellular automata and lattice Boltzmann models: An Introduction. Nummer Springer Science & Business Media, Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 59/60