Cumulant-basedLBM - Technische Universität München€¦ · Cumulant-basedLBM Prof.B.Wohlmuth MarkusMuhr 25.July2017. Lattice Boltzmann methods Summer term 2017 Content 1.Repetition:LBMandnotation

Lattice Boltzmann methodsSummer term 2017

Cumulant-based LBMProf. B. Wohlmuth Markus Muhr

25. July 2017


Content1. Repetition: LBM and notation

DiscretizationSRT-equation

2. MRT-modelBasic ideaMoments and moment-transformation

3. Cumulant-based LBMBasic idea and technical toolsDerivation of the cumulantsCumulant-transformationCumulant collision operator

4. Validation a.k.a. colorful pictures

Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 1/60


Repetition: LBM and notation



Computational Fluid Dynamics

� Numerical simulation of flows

� Calculation of the velocity field



Computational Fluid Dynamics

� Numerical simulation of flows

� Calculation of the velocity field



Discretization

� In space

� In velocity (D2Q9-model)

c1c3

c2

c4

c5c6

c7 c8



(Discrete) Boltzmann-equation

� Considering „probabilities of an encounter“ fi

f1f3

f2

f4

f5f6

f7 f8

� Derivation of a (discrete) transport-equation for thedistributions fi :

∂fi(t, x)∂t + ci ·∇x fi(t, x) = Q(fi) ≈ −

1τC·(fi(t, x)− f eqi (t, x)

)



(Discrete) Boltzmann-equation

� Considering „probabilities of an encounter“ fi

f1f3

f2

f4

f5f6

f7 f8

� Derivation of a (discrete) transport-equation for thedistributions fi :

∂fi(t, x)∂t + ci ·∇x fi(t, x) = Q(fi) ≈ −

1τC·(fi(t, x)− f eqi (t, x)

)



Equilibrium distributions f eqi

� Maxwell-distribution as „natural“velocity-distribution of anideal gas:

M(ξ, ρ, u,T ) = ρ ·( m2π kBT

)3/2· exp

(−

m |ξ − u|2

2kBT

)

� Taylor expansion up to second order ξ = ci :

f eqi = ρωi ·(1 +

3 ci · uc2 − 3 |u|2

2c2 + 12

9c4(ci · u

)2)

CKumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 7/60


SRT-equation

� Discretization of the derivative by finite differencesfi(t + ∆t, x + ci∆t) = fi(t, x)− 1

τ·(fi(t, x)− f eqi (t, x)

)� Interpretation as collision and streaming:

f1f3

f2

f4

f5f6

f7 f8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8



SRT-equation




f1f3

f2

f4

f5f6

f7 f8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8



SRT-equation




f1f3

f2

f4

f5f6

f7 f8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8



SRT-equation




f1f3

f2

f4

f5f6

f7 f8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8

f ∗1f ∗3

f ∗2

f ∗4

f ∗5f ∗6

f ∗7 f ∗8



In three dimensions

� Analogously in three dimensions (D3Q27-model)

1

2

3

4

5

6

78

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

x

yz

1



Notation� Convenient notation by index-tupel:

f1f3

f2

f4

f5f6

f7 f8

f10f−10

f01

f0−1

f11f−11

f−1−1 f1−1

fi → fij with i , j ∈ {−1, 0, 1}

� Analogously in three dimensions: fi → fijk withi , j , k ∈ {−1, 0, 1}

�Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 10/60


MRT-model



Basic idea of MRT (1)

� The SRT-equation reads:

fi(t + ∆t, x + ci∆t) = fi(t, x)− 1τ·(fi(t, x)− f eqi (t, x)

)� Idea: Mix the relaxation terms

fi(t + ∆t, x + ci∆t) = fi(t, x)−nV∑j=0

1τij·(fj(t, x)− f eqj (t, x)

)

� In matrix-vector-form...

fi(t + ∆t, x + ci∆t)...

=

...

fi(t, x)...

−S·

...(

fi(t, x)− f eqi (t, x))

...









)� In matrix-vector-form

...fi(t + ∆t, x + ci∆t)

...

=

...

fi(t, x)...

−S·

...(


...









)� In matrix-vector-form

...fi(t + ∆t, x + ci∆t)

...

=

...

fi(t, x)...

−S·

...(


...




� 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−1S ·(Mf (t, x)−Mf eq(t, x))

� Interpretation as transformation M : f 7→ m

f (t + ∆t, x + c×∆t) = f (t, x)−M−1S · (m(t, x)−meq(t, x))





















Moments

Definition: (Raw)-momentsFor α, β, γ = 0, 1, 2, ... one defined the (raw)-moments of orderα + β + γ as follows:

mαβγ =1∑

i ,j,k=−1

(cijk)αx·(cijk)βy·(cijk)γz· fijk =

1∑i ,j,k=−1

iα · jβ ·kγ · fijk

� In matrix-vector-form (in 2D)

m00m10m01m20m02m11m21m12m22

=

1 1 1 1 1 1 1 1 10 1 0 −1 0 1 −1 −1 10 0 1 0 −1 1 1 −1 −10 1 0 1 0 1 1 1 10 0 1 0 1 1 1 1 10 0 0 0 0 1 −1 1 −10 0 0 0 0 1 1 −1 −10 0 0 0 0 1 −1 −1 10 0 0 0 0 1 1 1 1

·

f0f1f2f3f4f5f6f7f8



Moments

Definition: (Raw)-momentsFor α, β, γ = 0, 1, 2, ... one defined the (raw)-moments of orderα + β + γ as follows:

mαβγ =1∑

i ,j,k=−1

(cijk)αx·(cijk)βy·(cijk)γz· fijk =

1∑i ,j,k=−1

iα · jβ ·kγ · fijk

� In matrix-vector-form (in 2D)

m00m10m01m20m02m11m21m12m22

=

1 1 1 1 1 1 1 1 10 1 0 −1 0 1 −1 −1 10 0 1 0 −1 1 1 −1 −10 1 0 1 0 1 1 1 10 0 1 0 1 1 1 1 10 0 0 0 0 1 −1 1 −10 0 0 0 0 1 1 −1 −10 0 0 0 0 1 −1 −1 10 0 0 0 0 1 1 1 1

·

f0f1f2f3f4f5f6f7f8



Interpretation of the moments

� The moments correspond to different physical quantities like:

I m00 =1∑

i,j=−1fij = ρ −→ density

c1c3

c2

c4

c5c6

c7 c8

I m10 =1∑

i,j=−1

(cij)x· fij = ρ · ux −→ x-momentum

I m01 =1∑

i,j=−1

(cij)y· fij = ρ · uy −→ y-momentum

� Some moments have to be combined for a reasonableinterpretationI m20 + m02 =

1∑i,j=−1

((cij)2x

+(

cij)2y

)· fij ∼ Ekin





I m00 =1∑


c1c3

c2

c4

c5c6

c7 c8

I m10 =1∑

i,j=−1


I m01 =1∑

i,j=−1



1∑i,j=−1

((cij)2x

+(

cij)2y

)· fij ∼ Ekin





I m00 =1∑


c1c3

c2

c4

c5c6

c7 c8

I m10 =1∑

i,j=−1


I m01 =1∑

i,j=−1



1∑i,j=−1

((cij)2x

+(

cij)2y

)· fij ∼ Ekin



Moment-transformation

� The respective linear combination of matrix-lines yields themoment-transformationmatrix

M0M1M2M3M4M5M6M7M8

=

m00m10m01

m20 + m02m20 − m02

m11m21m12m22

=

1 1 1 1 1 1 1 1 10 1 0 −1 0 1 −1 −1 10 0 1 0 −1 1 1 −1 −10 1 1 1 1 2 2 2 20 1 −1 1 −1 0 0 0 00 0 0 0 0 1 −1 1 −10 0 0 0 0 1 1 −1 −10 0 0 0 0 1 −1 −1 10 0 0 0 0 1 1 1 1

·

f0f1f2f3f4f5f6f7f8

� Further linear combinations of (raw)-moments are possible asfor example row-wise orthogonalization.



MRT-algorithm

� Transform the given distributions f into moments M = M · fund Meq = M · f eq

� Relax the moment Mi with frequencies ωiM∗0...

M∗8

=

ω0

. . .ω8

·

M0 −Meq0

...M8 −Meq

8

� Transform the relaxed moments back into distribution space

f ∗ = M−1 ·M∗

� Collision update and streaming stepf (t+∆t, x +c×∆t) = f (t, x)−M−1S ·(Mf (t, x)−Mf eq(t, x))



MRT-algorithm



M∗8

=

ω0

. . .ω8

·

M0 −Meq0

...M8 −Meq

8

� Transform the relaxed moments back into distribution spacef ∗ = M−1 ·M∗




MRT-algorithm



M∗8

=

ω0

. . .ω8

·

M0 −Meq0

...M8 −Meq

8


f ∗ = M−1 ·M∗




MRT-algorithm



M∗8

=

ω0

. . .ω8

·

M0 −Meq0

...M8 −Meq

8


f ∗ = M−1 ·M∗




MRT-algorithm



M∗8

=

ω0

. . .ω8

·

M0 −Meq0

...M8 −Meq

8


f ∗ = M−1 ·M∗




Pro/Con MRT-model� More „tuning-parameter“ ωi as for SRT� Better insight into the model by identification with physicalquantities

� Linear transformations

� statistical dependencies among the moments(→ dependencies between ωi)

� Physics gets lost by (wrong) variation of parameters. How tochoose them?



Pro/Con MRT-model� More „tuning-parameter“ ωi as for SRT� Better insight into the model by identification with physicalquantities

� Linear transformations

� statistical dependencies among the moments(→ dependencies between ωi)

� Physics gets lost by (wrong) variation of parameters. How tochoose them?



Cumulant-based LBM



Basic idea of cumulant-based LBM

� Goal: Find a transformation to statistically independentquantities Cm

� Mathematically this is described by a factorization of theprobability-density

fijk =nV∏m=0

Fm(Cm)

� Efficient implementation of that (nonlinear) transformation

� Main tools: Multivariate Taylor-expansion,Laplace-transformation and distribution-theory






fijk =nV∏m=0

Fm(Cm)

� Efficient implementation of that (nonlinear) transformation� Main tools: Multivariate Taylor-expansion,

Laplace-transformation and distribution-theory






fijk =nV∏m=0

Fm(Cm)

� Efficient implementation of that (nonlinear) transformation� Main tools: Multivariate Taylor-expansion,

Laplace-transformation and distribution-theory



Distributions� Recall from measure theory: Lebesgue-space L2:

L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞

}� Consider the following functional

G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx




L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞


G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx




L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞


G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx

� One searches for a g ∈ L2(R) satisfying:

G(f ) =∫R

g(x) · f (x) dx ∀f ∈ L2(R)

� Solution: g(x) = 1[−1,1](x)

� Interpretation of g as functional (Riesz representationtheorem)




L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞


G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx


G(f ) =∫R

g(x) · f (x) dx ∀f ∈ L2(R)

� Solution: g(x) = 1[−1,1](x)

� Interpretation of g as functional (Riesz representationtheorem)




L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞


G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx


G(f ) =∫R

g(x) · f (x) dx ∀f ∈ L2(R)

� Solution: g(x) = 1[−1,1](x)

� Interpretation of g as functional (Riesz representationtheorem)Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60



L2(R) :={f : R→ R |

∫R |f (x)|2 dx <∞


G : L2(R)→ R, G(f ) :=∫ 1

−1f (x) dx


G(f ) =∫R

g(x) · f (x) dx ∀f ∈ L2(R)

� Solution: g(x) = 1[−1,1](x)

� Interpretation of g as functional (Riesz representationtheorem)Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 22/60


Dirac-sequence

� Interprete the functions gn(x) = n2 · 1[− 1

n ,1n ](x), n ∈ N as

functionals



Delta-distribution δ

� For continuous functions f there holds:∫R

gn(x) · f (x) dx = n2 ·∫ 1

n

− 1n

f (x) dx

= n2 ·(

F(1

n

)− F

(−1

n

))

=F(1n

)− F

(− 1

n

)2 · 1n

n→∞−→ F ′(0) = f (0)

� Define the delta-distribution δ(x) := limn→∞

gn(x) = „g∞(x)“by: ∫

Rδ(x) · f (x) dx = f (0)





gn(x) · f (x) dx = n2 ·∫ 1

n

− 1n

f (x) dx

= n2 ·(

F(1

n

)− F

(−1

n

))

=F(1n

)− F

(− 1

n

)2 · 1n

n→∞−→ F ′(0) = f (0)


gn(x) = „g∞(x)“by: ∫

Rδ(x) · f (x) dx = f (0)





gn(x) · f (x) dx = n2 ·∫ 1

n

− 1n

f (x) dx

= n2 ·(

F(1

n

)− F

(−1

n

))

=F(1n

)− F

(− 1

n

)2 · 1n

n→∞−→ F ′(0) = f (0)


gn(x) = „g∞(x)“by: ∫

Rδ(x) · f (x) dx = f (0)





gn(x) · f (x) dx = n2 ·∫ 1

n

− 1n

f (x) dx

= n2 ·(

F(1

n

)− F

(−1

n

))

=F(1n

)− F

(− 1

n

)2 · 1n

n→∞−→ F ′(0) = f (0)


gn(x) = „g∞(x)“by: ∫

Rδ(x) · f (x) dx = f (0)





gn(x) · f (x) dx = n2 ·∫ 1

n

− 1n

f (x) dx

= n2 ·(

F(1

n

)− F

(−1

n

))

=F(1n

)− F

(− 1

n

)2 · 1n

n→∞−→ F ′(0) = f (0)


gn(x) = „g∞(x)“by: ∫

Rδ(x) · f (x) dx = f (0)



Delta-function

� Considered as a function there holds: δ(x) ={0 if x 6= 0∞ if x = 0

� Which often gets normed to: δ(x) ={0 if x 6= 01 if x = 0

� What you should remember for sure:∫Rδ(x) · f (x) dx = f (0)



Delta-function

� Considered as a function there holds: δ(x) ={0 if x 6= 0∞ if x = 0

� Which often gets normed to: δ(x) ={0 if x 6= 01 if x = 0

� What you should remember for sure:∫Rδ(x) · f (x) dx = f (0)



Repetition: Fourier-transform� The Fourier-transform of f : R→ C is (modulo scaling)defined by:

f (ω) = F [f (x)](ω) :=∫R

f (x) · e−iωx dx

� f (ω) tells, how strong eiω is present in thefrequency-spectrum of f



Repetition: Fourier-transform� The Fourier-transform of f : R→ C is (modulo scaling)defined by:

f (ω) = F [f (x)](ω) :=∫R

f (x) · e−iωx dx

� f (ω) tells, how strong eiω is present in thefrequency-spectrum of f

� Example: F [sin(x)](ω) ∼ δ(ω − 1)− δ(ω + 1)2i



Laplace-transformation

� Instead of projection onto oscillations {eiω}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:I L[xn](s) = n!

sn+1

I L[sin(ax)](s) = as2+a2

I L[ln(ax)](s) = − 1s (ln(s/a) + γ)

� Certain features as for example the shift-property hold true:L[f (x − a)](s) = e−asF (s)





F (s) := L[f (x)](s) :=∫ ∞0

f (x) · e−sx dx


sn+1


I L[ln(ax)](s) = − 1s (ln(s/a) + γ)






F (s) := L[f (x)](s) :=∫ ∞0

f (x) · e−sx dx


sn+1


I L[ln(ax)](s) = − 1s (ln(s/a) + γ)




Laplace-transformation of the delta-distribution

� Apply the definition of the (two-sided) Laplace-transform:

L[δ(x)](s) =∫Rδ(x) · e−sx dx





L[δ(x)](s) =∫Rδ(x) · e−sx dx





L[δ(x)](s) =∫Rδ(x) · e−sx dx = e−s·0 = 1

� Transformation of the shifted delta-distribution by theshift-property:

L[δ(x − a)](s) = e−asL[δ(x)](s)





L[δ(x)](s) =∫Rδ(x) · e−sx dx = e−s·0 = 1


L[δ(x − a)](s) = e−asL[δ(x)](s)





L[δ(x)](s) =∫Rδ(x) · e−sx dx = e−s·0 = 1


L[δ(x − a)](s) = e−asL[δ(x)](s)





L[δ(x)](s) =∫Rδ(x) · e−sx dx = e−s·0 = 1


L[δ(x − a)](s) = e−asL[δ(x)](s) = e−as

♦Kumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 31/60


Derivation of the cumulants (1) - „Analytization“

� Want to capture the discrete velocity-distributions analytically

f10f−10

f01

f0−1

f11f−11

f−1−1 f1−1

� This can be done by the „peak-function“ f (ξ):

f (ξ) = f (ξx , ξy , ξz) =1∑

i ,j,k=−1fijk δ(ic−ξx ) δ(jc−ξy ) δ(kc−ξz)





f10f−10

f01

f0−1

f11f−11

f−1−1 f1−1


f (ξ) = f (ξx , ξy , ξz) =1∑






f10f−10

f01

f0−1

f11f−11

f−1−1 f1−1


f (ξ) = f (ξx , ξy , ξz) =1∑




Derivation (2) - Laplace-transform

� To increase regularity one performs the two-sidedLaplace-transform of f (ξ):

L[f (ξ)](Ξ) =1∑

i ,j,k=−1fijkL[δ(ic−ξx )]L[δ(jc−ξy )]L[δ(kc−ξz)]

=1∑

i ,j,k=−1fijk e−Ξx ic e−Ξy jc e−Ξzkc =: F (Ξ)



Derivation (2) - Laplace-transform

� To increase regularity one performs the two-sidedLaplace-transform of f (ξ):

L[f (ξ)](Ξ) =1∑

i ,j,k=−1fijkL[δ(ic−ξx )]L[δ(jc−ξy )]L[δ(kc−ξz)]

=1∑

i ,j,k=−1fijk e−Ξx ic e−Ξy jc e−Ξzkc =: F (Ξ)



Derivation (3) - statistical independence

� Applying the statistical independence (which one aims tohave) in the frequency space:

F (Ξx ,Ξy ,Ξz) =nV∏m=0

Fm(Cm)

� And using the logarithm

ln (F (Ξx ,Ξy ,Ξz)) =nV∑m=0

ln (Fm(Cm))



Derivation (4) - Taylor-expansion

� The multivariate Taylor-expansion of ln (F (Ξx ,Ξy ,Ξz))around Ξ = 0 is:

ln(F (Ξx ,Ξy ,Ξz )) =∑

α+β+γ≥0

1α!β!γ!

∂α+β+γ

∂Ξαx ∂Ξβy ∂Ξγzln(F (Ξx ,Ξy ,Ξz ))

∣∣∣∣Ξ=0

· Ξαx Ξβy Ξγz

Definition: CumulantsFor α, β, γ = 0, 1, 2, ... one defines the cumulants of orderα + β + γ as follows:

cαβγ := c−(α+β+γ) ∂α+β+γ

∂Ξαx ∂Ξβy ∂Ξγzln(F (Ξx ,Ξy ,Ξz))

∣∣∣∣∣Ξ=0



Derivation (4) - Taylor-expansion

� The multivariate Taylor-expansion of ln (F (Ξx ,Ξy ,Ξz))around Ξ = 0 is:

ln(F (Ξx ,Ξy ,Ξz )) =∑

α+β+γ≥0

1α!β!γ!

∂α+β+γ

∂Ξαx ∂Ξβy ∂Ξγzln(F (Ξx ,Ξy ,Ξz ))

∣∣∣∣Ξ=0


Definition: CumulantsFor α, β, γ = 0, 1, 2, ... one defines the cumulants of orderα + β + γ as follows:

cαβγ := c−(α+β+γ) ∂α+β+γ


∣∣∣∣∣Ξ=0



Efficient calculation via moments

� Conduct the Taylor-expansion without the logarithm:

F (Ξx ,Ξy ,Ξz ) =∑

α+β+γ≥0

1α!β!γ!

∂α+β+γ

∂Ξαx ∂Ξβy ∂ΞγzF (Ξx ,Ξy ,Ξz )

∣∣∣∣Ξ=0


� Analogously consider the coefficients:

∂α+β+γ

∂Ξαx ∂Ξβy ∂ΞγzF (Ξx ,Ξy ,Ξz)

∣∣∣∣∣Ξ=0



Where the magic happens

� Explicit calculation of these coefficients yields:

∂α+β+γ


∣∣∣∣Ξ=0

=∂α+β+γ

∂Ξαx ∂Ξβy ∂Ξγz

1∑i,j,k=−1

fijk e−Ξx ic e−Ξy jc e−Ξzkc

∣∣∣∣∣Ξ=0

=1∑

i,j,k=−1

fijk (−ic)αe−Ξx ic (−jc)βe−Ξy jc · (−kc)γe−Ξzkc∣∣

Ξ=0

= (−c)α+β+γ1∑

i,j,k=−1

iα jβ kγ fijk

= (−c)α+β+γ mαβγ





∂α+β+γ


∣∣∣∣Ξ=0

=∂α+β+γ


1∑i,j,k=−1


∣∣∣∣∣Ξ=0

=1∑

i,j,k=−1


Ξ=0

= (−c)α+β+γ1∑

i,j,k=−1

iα jβ kγ fijk






∂α+β+γ


∣∣∣∣Ξ=0

=∂α+β+γ


1∑i,j,k=−1


∣∣∣∣∣Ξ=0

=1∑

i,j,k=−1


Ξ=0

= (−c)α+β+γ1∑

i,j,k=−1

iα jβ kγ fijk






∂α+β+γ


∣∣∣∣Ξ=0

=∂α+β+γ


1∑i,j,k=−1


∣∣∣∣∣Ξ=0

=1∑

i,j,k=−1


Ξ=0

= (−c)α+β+γ1∑

i,j,k=−1

iα jβ kγ fijk




Cumulants vs. Moments

� The (raw)-moments were shown to be:

mαβγ = (−c)−(α+β+γ) ∂α+β+γ


∣∣∣∣∣Ξ=0

� The cumulants:

cαβγ = c−(α+β+γ) ∂α+β+γ


∣∣∣∣∣Ξ=0

� Idea: By applying the chain rule one should be able to traceback the cumulants to the moments...



Cumulants vs. Moments

� The (raw)-moments were shown to be:

mαβγ = (−c)−(α+β+γ) ∂α+β+γ


∣∣∣∣∣Ξ=0

� The cumulants:

cαβγ = c−(α+β+γ) ∂α+β+γ


∣∣∣∣∣Ξ=0

� Idea: By applying the chain rule one should be able to traceback the cumulants to the moments...



Cumulant-transformation (1)

� We conduct the idea on the example c100:

c100 = c−(1+0+0) ∂1

∂Ξ1x ∂Ξ0

y ∂Ξ0zln(F (Ξx ,Ξy ,Ξz))

∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

∂1

∂Ξ1x

F (Ξx ,Ξy ,Ξz)∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

m100(−c)−1 = − 1

F (0, 0, 0)m100

� With F (0, 0, 0) =1∑

i ,j,k=−1fijke0e0e0 = ρ also: c100 = −1

ρm100

� Define: Cαβγ = ρcαβγ , then =⇒ C100 = −m100





c100 = c−(1+0+0) ∂1

∂Ξ1x ∂Ξ0


∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

∂1

∂Ξ1x

F (Ξx ,Ξy ,Ξz)∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

m100(−c)−1 = − 1

F (0, 0, 0)m100

� With F (0, 0, 0) =1∑


ρm100






c100 = c−(1+0+0) ∂1

∂Ξ1x ∂Ξ0


∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

∂1

∂Ξ1x

F (Ξx ,Ξy ,Ξz)∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

m100(−c)−1 = − 1

F (0, 0, 0)m100

� With F (0, 0, 0) =1∑


ρm100






c100 = c−(1+0+0) ∂1

∂Ξ1x ∂Ξ0


∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

∂1

∂Ξ1x

F (Ξx ,Ξy ,Ξz)∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

m100(−c)−1 = − 1

F (0, 0, 0)m100

� With F (0, 0, 0) =1∑


ρm100






c100 = c−(1+0+0) ∂1

∂Ξ1x ∂Ξ0


∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

∂1

∂Ξ1x

F (Ξx ,Ξy ,Ξz)∣∣∣∣∣Ξ=0

= c−1 1F (0, 0, 0) ·

m100(−c)−1 = − 1

F (0, 0, 0)m100

� With F (0, 0, 0) =1∑


ρm100





� Analogous procedure for the remaining cumulants:

C100 = −m100 C010 = −m010 C001 = −m001

C200 = m200−1ρ

m2100 C020 = m020−

1ρ

m2010 C002 = m002−

1ρ

m2001

C110 = m110−1ρ

m100m010 C101 = m101−1ρ

m100m001 . . .

C210 = −m210 −2ρ2

m2100m010 + 2

ρm110m100 + 1

ρm200m010

...



Half-time score

� What do we have so far? How far do we come with that,starting with the current distributions?I Transform distributions f to (raw)-moments m = M · fI Transform (raw)-moments m to cumulants

C = K(m) = K(M · f ) = C(f )

f (t+∆t, x +c×∆t) = f (t, x)−C−1[S · (C(f (t, x))− C(f eq(t, x)))

]� Efficient transformation of equilibria� Choice of relaxation parameters ωi� Inverse mapping C−1



Cumulant equilibria - Derivation

� The equilibria can be calculated in advance

� Laplace-transformation of Maxwell-distribution:

F eq(Ξ) = L[M(ξ)

](Ξ) =

∫∫∫R

M(ξ) · e−Ξ·ξ dξx dξy dξz

= ρ√C· exp

(14C (Ξ2

x + Ξ2y + Ξ2

z)− Ξx ux − Ξy uy − Ξz uz)

� Apply the logarithm:

ln(F eq(Ξ)) = ln(ρ

ρ0

)−Ξx ux −Ξy uy −Ξz uz + c2s

2(Ξ2x + Ξ2

y + Ξ2z)

� Taylor-expansion can be skipped, since this is a polynomialanyway




� The equilibria can be calculated in advance� Laplace-transformation of Maxwell-distribution:

F eq(Ξ) = L[M(ξ)

](Ξ) =

∫∫∫R


= ρ√C· exp

(14C (Ξ2

x + Ξ2y + Ξ2




ρ0


2(Ξ2x + Ξ2

y + Ξ2z)






F eq(Ξ) = L[M(ξ)

](Ξ) =

∫∫∫R


= ρ√C· exp

(14C (Ξ2

x + Ξ2y + Ξ2




ρ0


2(Ξ2x + Ξ2

y + Ξ2z)






F eq(Ξ) = L[M(ξ)

](Ξ) =

∫∫∫R


= ρ√C· exp

(14C (Ξ2

x + Ξ2y + Ξ2




ρ0


2(Ξ2x + Ξ2

y + Ξ2z)

� Taylor-expansion can be skipped, since this is a polynomialanywayKumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60




F eq(Ξ) = L[M(ξ)

](Ξ) =

∫∫∫R


= ρ√C· exp

(14C (Ξ2

x + Ξ2y + Ξ2




ρ0


2(Ξ2x + Ξ2

y + Ξ2z)

� Taylor-expansion can be skipped, since this is a polynomialanywayKumulanten-basierte LBM Prof. B. Wohlmuth Markus Muhr 43/60


Cumulant equilibria

� Only a certain few cumulants have an equilibrium not equal 0:

ceq000 = ln(ρ/ρ0)ceq100 = −c−1uxceq010 = −c−1uyceq001 = −c−1uzceq200 = c−2c2sceq020 = c−2c2sceq002 = c−2c2s

� For all others, there holds: ceqαβγ = 0



Half-time score


C = K(m) = K(M · f ) = C(f )


]

� Efficient transformation of equilibria� Choice of relaxation parameters ωi� Inverse mapping C−1



Choice of relaxation parameters

� For the most cumulants the equilibrium is 0, hence one getsfor example:

C∗222 = ω10 (C222 − 0)

C∗110 = ω1 C110

C∗101 = ω1 C101

C∗011 = ω1 C011

C∗211 = ω8 C211

C∗121 = ω8 C121

C∗112 = ω8 C112

� Those with non vanishing equilibrium one can combine:

C∗200 − C∗020 = ω1 (C200 − C020)C∗200 − C∗002 = ω1 (C200 − C002)





C∗222 = ω10 (C222 − 0)

C∗110 = ω1 C110

C∗101 = ω1 C101

C∗011 = ω1 C011

C∗211 = ω8 C211

C∗121 = ω8 C121

C∗112 = ω8 C112


C∗200 − C∗020 = ω1 (C200 − C020)C∗200 − C∗002 = ω1 (C200 − C002)





C∗222 = ω10 (C222 − 0)

C∗110 = ω1 C110

C∗101 = ω1 C101

C∗011 = ω1 C011

C∗211 = ω8 C211

C∗121 = ω8 C121

C∗112 = ω8 C112


C∗200 − C∗020 = ω1 (C200 − C020)C∗200 − C∗002 = ω1 (C200 − C002)





C∗222 = ω10 (C222 − 0)

C∗110 = ω1 C110

C∗101 = ω1 C101

C∗011 = ω1 C011

C∗211 = ω8 C211

C∗121 = ω8 C121

C∗112 = ω8 C112


C∗200 − C∗020 = ω1 (C200 − C020)C∗200 − C∗002 = ω1 (C200 − C002)



Half-time score


C = K(m) = K(M · f ) = C(f )


]

� Efficient transformation of equilibria

� Choice of relaxation parameters ωi� Inverse mapping C−1



Half-time score


C = K(m) = K(M · f ) = C(f )


]

� Efficient transformation of equilibria

� Choice of relaxation parameters ωi� Inverse mapping C−1



Advantages of Cumulant-based LBM

� Better representation of physical processes due to statisticalindependence

� Equilibria of cumulants are almost all 0� More (real) degrees of freedom than in classical MRT� Gradual improvement of the method by asymptotic analysisand elimination of error-terms



Synthesis of Navier-Stokes-equation

� By a Chapman-Enskog expansion one can derive theNavier-Stokes equations:

∇ · u = 0∂u∂t + (u · ∇) u = −1

ρ∇p + 1

3

( 1ω1− 1

2

)︸︷︷︸

=ν

∆u + fρ

� With ω1 one can simulate fluids of different viscosity� Only ω1 directly influences the solution

� Further parameters influence higher order error terms (setthem to 1 in the following)





∇ · u = 0∂u∂t + (u · ∇) u = −1

ρ∇p + 1

3

( 1ω1− 1

2

)︸︷︷︸

=ν

∆u + fρ

� With ω1 one can simulate fluids of different viscosity� Only ω1 directly influences the solution� Further parameters influence higher order error terms (setthem to 1 in the following)





∇ · u = 0∂u∂t + (u · ∇) u = −1

ρ∇p + 1

3

( 1ω1− 1

2

)︸︷︷︸

=ν

∆u + fρ

� With ω1 one can simulate fluids of different viscosity� Only ω1 directly influences the solution� Further parameters influence higher order error terms (setthem to 1 in the following)



Validation a.k.a. colorfulpictures



Turbulence measure� The Reynolds-number is dimensionless parameter of a flow

Re = U · Lν

� At some specific threshold Re∗ the laminar-turbulenttransition starts

� Traditional LBM methods have problems with highReynolds-numbers:



Comparison: BGK ↔ Kumulanten

� Comparison SRT- and cumulant-based LBM for a cylinderflow at Re = 8000

a) BGK-SRT-LBM (Picture source [5]) b) Cumulant-based LBM (Picture source [5])



Turbulence resolution in 3D

� Representation of (turbulence) vortices around a sphere flow:

(Picture source [5])











The highlights

� MRT-like-method with nonlinear transformation to statisticalindependent quantities

� Mathematical methods:I Distributions as generalized functions (Analytization)I Laplace-transformI Comparison of Taylor-coefficients

� Advantages especially at high Reynolds-numbers (stability andadaptivity)



The highlights






The highlights






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 EngineeringSciences, 360(1792):437–451, 2002.

Dubois, François: Equivalent partial differential equations of a lattice Boltzmann scheme.Computers & Mathematics with Applications, 55(7):1441–1449, 2008.

Geier, M: Ab initio derivation of the cascaded Lattice Boltzmann Automaton.University of Freiburg–IMTEK, 2006.

Geier, Martin, Andreas Greiner und Jan G Korvink: Cascaded digital lattice Boltzmann automatafor high Reynolds number flow.Physical Review E, 73(6):066705, 2006.

Geier, Martin, Martin Schönherr, Andrea Pasquali und Manfred Krafczyk: The cumulantlattice Boltzmann equation in three dimensions: Theory and validation.Computers & Mathematics with Applications, 70(4):507–547, 2015.

Hänel, Dieter: Molekulare Gasdynamik: Einführung in die kinetische Theorie der Gase undLattice-Boltzmann-Methoden.Springer-Verlag, 2006.

Lukacs, Eugene: Characteristics functions.Griffin, London, 1970.



Literature II

Ning, Yang und Kannan N Premnath: Numerical Study of the Properties of the Central MomentLattice Boltzmann Method.arXiv preprint arXiv:1202.6351, 2012.

Rohm, Florian: A commented Python Script for LBM-Flow-Simulations.Lecture on Lattice Boltzmann methods, 2015.

Wolf-Gladrow, Dieter A: Lattice-gas cellular automata and lattice Boltzmann models: AnIntroduction.Nummer 1725. Springer Science & Business Media, 2000.


Documents

Cumulant-basedLBM - Technische Universität München€¦ · Cumulant-basedLBM Prof.B.Wohlmuth MarkusMuhr 25.July2017. Lattice Boltzmann methods Summer term 2017 Content 1.Repetition:LBMandnotation