Recovering the full Navier Stokes equations with lattice...

fluid-d1q3 thermal-d1q3 NS-BGK (ρ, J, ζ) NS-ρJζ-d1q3q3 numerical experiments conclusion

Recovering the full Navier Stokes equationswith lattice Boltzmann schemes

Francois Dubois∗†, Benjamin Graille† and Pierre Lallemand‡§

NUMKIN 2016: International Workshopon Numerical Methods for Kinetic Equations

Strasbourg, 20 october 2016

∗ Conservatoire National des Arts et Metiers, Paris, France† Universite Paris-Sud, France‡ Centre National de la Recherche Scientifique, Paris, France§ Beijing Computational Science Research Center, China

Jean-Louis Loday (1946 - 2012)

From Boltzmann equation to lattice Boltzmann schemes

Boltzmann equation ∂t f + v•∇f = Q(f )Linearization Q(f ) ' Q(f eq) + dQ(f eq)•(f − f eq)Boltzmann BGK ∂t f + v•∇f = dQ(f eq)•(f − f eq)Discrete velocities

∂t(Mf ) + M•v•∇f =(M dQ(f eq)M−1

)•(M (f − f eq)

)Moments m = M f , meq = M f eq

Multiple Relaxation Times hypothesis :the matrix M dQ(f eq)M−1 is diagonal and real

Time discretization: alternate directions, or “collide-stream”Zero eigenvalues dmk

dt = 0, 0 ≤ k < N: conserved moments WRelaxation of the nonconserved moments mk , k ≥ Ndmkdt = − 1

(mk −meq

), m∗k = mk − ∆t

τk(mk −meq

k (W )), sk = ∆tτk

Particle densities f ∗j (x , t) =(M−1 m∗

)j(x , t)

x ∈ Lattice and x + vj ∆t ∈ Lattice, t = n∆tSolve the advection equation ∂t fj + vj•∇fj = 0 with CFL=1!

fj(x + vj ∆t, t + ∆t) = f ∗j (x , t)Lattice Boltzmann scheme DdQq

Outline

• Fluid flow with a Multi Relaxation TimesD1Q3 lattice Boltzmann scheme

Advection-diffusion with a Multi Relaxation TimesD1Q3 lattice Boltzmann scheme

• Double distribution for Navier StokesNavier Stokes with mass, momentum and total energyDouble distribution with the Bhatnagar Gross Krook

lattice Boltzmann framework

• Navier Stokes with mass, momentum and volumic entropyDouble distribution for a Multi Relaxation Times

D1Q3Q3 lattice Boltzmann schemeLinearized study with a focus on stability

• Definition of a nonlinear schemeFirst numerical experiments

Fluid flow with a MRT D1Q3 lattice Boltzmann scheme

f0(x , t) : motionless particles at x during the time step ∆t

f+(x , t) : density of particles going from x to x + ∆x during ∆t

f−(x , t) : particles going from x to x −∆x during the time step ∆t

numerical velocity λ = ∆x∆t fixed

discrete velocities v0 = 0, v+ = +1, v− = −1

Mass density ρ ≡ ρ0

∑j=0,+,−

fj = ρ0 (f+ + f− + f0)

Momentum J ≡ ρ0 λ∑

j=0,+,−vj fj = ρ0 λ (f+ − f−)

Energy e ≡ ρ0 λ2 (f+ + f− − 2 f0) third momentum

Fluid flow with MRT D1Q3 lattice Boltzmann scheme (ii)

Invertible matrix M between the particles and the moments:ρJe

= MD1Q3

f0f+f−

, MD1Q3 = ρ0

1 1 10 λ −λ−2λ2 λ2 λ2

• RelaxationThe density and momentum remain at equilibrium

and do not change: ρ∗ = ρ , J∗ = J

The equilibrium energy is a given function of the two momentsat equilibrium: eeq = eeq(ρ, J)

(the Gaussian distribution for the BGK model)

The energy after relaxation is obtained by a simple evolution:e∗ = e + se (eeq − e) with 0 < se < 2

The particle distribution after relaxation:f ∗0f ∗+f ∗−

=(MD1Q3

ρJe∗

MRT D1Q3 fluid lattice Boltzmann scheme (iii)

• Advectionf0(x , t + ∆t) = f ∗0 (x , t)f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t)

• Analysis with the Taylor expansion method[FD, ESAIM ProcS, 18, 2007]

A numerical Chapman-Enskog expansion withthe small parameter ε ≡ λ

D replaced by ε ≡ ∆xL

Equivalent partial differential equations up to order 2:

∂tρ+ ∂xJ = O(∆x2)

∂tJ + ∂x(

2 ρ+ 13e

eq)− 1

3 σe ∆t ∂xθe = O(∆x2)

with σe = 1se− 1

2 , θe ≡ ∂teeq + λ2 ∂xJ ' 3λ2ρ ∂xu

MRT D1Q3 fluid lattice Boltzmann scheme (iv)

1D “isentropic” Navier Stokes equationsmass ∂tρ+ ∂xJ = 0

momentum ∂tJ + ∂x(ρ u2 + p

)− ∂x

(ρ ν ∂xu

D1Q3 fluid lattice Boltzmann schememass ∂tρ+ ∂xJ = O(∆x2)

momentum

∂tJ + ∂x(

2 ρ+ 13e

eq)− ∂x

(σe ∆x λ ρ ∂xu

)= O(∆x2)

Energy at equilibrium: eeq = 3(ρ u2 + p

)− 2λ2 ρ

Relaxation parameter and kinematic viscosity:1

se− 1

2= σe =

λ∆x

Some algebraic tools...

Moments m ≡ M f

Momentum-velocity tensor Λk` ≡ λ∑j

Mkj vj (M−1)j`

ΛD1Q3 =

0 1 023λ

2 0 13

0 λ2 0

with the ordering (ρ , J , e) of the moments

Defect of conservation θ` ≡ ∂tmeq` +

Λ`p ∂xmeqp

Second order partial differential equationsatisfied by the ko conserved moment Wk

∂tWk +∑`

Λk` ∂xmeq` −∆t

σ` Λk` ∂xθ` = O(∆x2)

Straitforward generalization to two and three dimensions

Advection-diffusion with a MRT D1Q3 LB scheme

g0(x , t) : motionless particles at x during the time step ∆t

g+(x , t) : density of particles going from x to x + ∆x during ∆t

g−(x , t) : particles going from x to x −∆x during the time step

numerical velocity λ = ∆x∆t fixed

discrete velocities v0 = 0, v+ = +1, v− = −1

Scalar ζ ≡ ζ0

∑j=0,+,−

gj = ζ0 (g+ + g− + g0)

Momentum ψ ≡ ζ0 λ∑

j=0,+,−vj gj = ζ0 λ (g+ − g−)

Associated energy ε ≡ ζ0 λ2 (g+ + g− − 2 g0)

third momentum

Advection-diffusion with a MRT D1Q3 LB scheme (ii)

Invertible matrix M between the particles and the moments:ζψε

= MD1Q3

, MD1Q3 = ζ0

1 1 10 λ −λ−2λ2 λ2 λ2

• RelaxationThe scalar ζ remains at equilibrium : ζ∗ = ζ

The momentum ψ and associated energy ε at equilibriumare given functions of the scalar ζ :

ψeq = ψeq(ζ) , εeq = εeq(ζ)

The momentum and energy after relaxation are obtained bysimple evolutions: ψ∗ = ψ + sψ (ψeq − ψ) , 0 < sψ < 2

ε∗ = ε+ sε (εeq − ε) , 0 < sε < 2

Particle distribution g after relaxation:g∗0g∗+g∗−

=(MD1Q3

ζψ∗

Advection-diffusion with a MRT D1Q3 LB scheme (iii)

• Advectiong0(x , t + ∆t) = g∗0 (x , t)g+(x , t + ∆t) = g∗+(x −∆x , t)g−(x , t + ∆t) = g∗−(x + ∆x , t)

• Analysis with the Taylor expansion method

ΛD1Q3 =

0 1 023λ

2 0 13

0 λ2 0

Equivalent partial differential equation up to order 2:

∂tζ + ∂xψeq − σψ ∆t ∂xθψ = O(∆x2)

with σψ = 1sψ− 1

2 , θψ ≡ ∂tψeq + ∂x(

2 ζ + 13ε

Advection-diffusion with a MRT D1Q3 LB scheme (iv)

Linear advection-diffusion equation

∂tζ + ∂x(u0 ζ

)− ∂x

(κ ∂xζ

Thermal D1Q3 lattice Boltzmann scheme∂tζ + ∂xψ

eq − ∂x(σψ ∆t θψ

)= O(∆x2)

Momentum at equilibrium ψeq = u0 ζ

Defect of equilibrium θψ =(

2 − u20

)∂xζ + 1

3 ∂xεeq + O(∆x)

Energy at equilibrium εeq = α λ2 ζ

Thermal conductivity(

2+α3 λ2 − u2

) (1sψ− 1

)∆t = κ

Constraint for numerical stability −2 < α < 1

1D Navier Stokes equations

Conserved variables

ρ , J ≡ ρ u , ρE ≡ ρ(i +

State equation: polytropic perfect gasp = (γ − 1) ρ i = ρ r T , i = cv T

Sound velocityc2 = γ p

ρ = γ (γ − 1) i = (γ − 1) cp T , γ =cpcv

Kinematic viscosity, thermal conductivity and Prandtl numberPr =

ρ ν cpκ

Mass conservation∂tρ+ ∂xJ = 0

Momentum conservation∂tJ + ∂x

(ρ u2 + p

)− ∂x

(ρ ν ∂xu

Conservation of energy∂t(ρE)

+ ∂x(ρE u + p u

)−∂x

(ρ ν u ∂xu

)− ∂x

(κ ∂xT

A first and natural idea

Two distributions: fj for mass and momentumgj for total energy

Conserved momentsρ = ρ0

fj , J = ρ0 λ∑j

vj fj , ρE = ρ0 λ2∑j

Non conserved momentse = ρ0 λ

2(f+ + f− − 2 f0

), ψ = ρ0 λ

j vj gj ,

ε = ρ0 λ4(g+ + g− − 2 g0

)Particle representation

f ≡(f0 , f+ , f− , g0 , g+ , g−

)tMomentum representation

m ≡(ρ , J , ρE , e , ψ , ε

)t, m = M f

Previous work with the BGK framework

Two particle distributions for the full Navier Stokes equations

∂t fj + vj•∇fj = − 1τf

(fj − f eqj

)∂tgj + vj•∇gj = − 1

(gj − g eq

(fj − f eqj

)see Guo, Zheng, Shi, Phys. Rev. E, 75, 2007

Li, He, Wang, Tao Phys. Rev. E, 76, 2007Li, Luo, He, Gao, Tao, Phys. Rev. E, 85, 2012Karlin, Sichau, Chikatamarla Phys. Rev. E, 88, 2013. . .

Adapt this idea for the Multi Relaxation Times approach ?Change the relaxation step !

i conserved moments W ∗k = Wk (k < N)

ii first non-conserved moments m∗k = (1− sk)mk + sk meqk

iii second non-conserved momentsm∗` = m` − s`

(m` −meq

)+ K`k

(mk −meq

)with mk in the first family of nonconserved moments. . .

Add a source term !

Two distributions. . .

∂t fj + vj•∇fj + 1τf

(fj − f eqj

∂tgj + vj•∇gj + 1τg

(gj − g eq

(fj − f eqj

)Try to approach with a lattice Boltzmann scheme

the full Navier Stokes equations asa system of conservative partial differential equationswith a source term

∂tW + ∂xF (W )− ∂x(Φ(W ,∇W )

Present choice W = (ρ , J , ζ)ζ ≡ ρ s the volumic entropy

Navier Stokes with mass, momentum and volumic entropy

“Conserved” variablesρ , J ≡ ρ u , ζ ≡ ρ s

State equation: polytropic perfect gasp = (γ − 1) ρ i = ρ r T = p0

( ρρ0

)γexp

(γ (s−s0)cp

)Sound velocity

c2 = γ pρ = γ (γ − 1) i = (γ − 1) cp T , γ =

Kinematic viscosity, thermal conductivity and Prandtl numberPr =

ρ ν cpκ

Mass conservation∂tρ+ ∂xJ = 0 , J ≡ ρ u

Momentum conservation∂tJ + ∂x

(ρ u2 + p

)− ∂x

(ρ ν ∂xu

Production of entropy∂tζ + ∂x

(ζ u)− ∂x

(κT ∂xT

)= ρ ν

(∂xu)2

+ κT 2

(∂xT

D1Q3Q3 for the (ρ, J , ζ) formulation

Two distributions: fj for mass and momentumgj for volumic entropy

Conserved momentsρ = ρ0

fj , J = ρ0 λ∑j

vj fj , ζ = ρ0 cp∑j

Non conserved momentse = ρ0 λ

2(f+ + f− − 2 f0

), ψ = ρ0 cp λ

(g+ − g−

)ε = ρ0 cp λ

2(g+ + g− − 2 g0

)Particle representation

f ≡(f0 , f+ , f− , g0 , g+ , g−

)tMomentum representation

m ≡(ρ , J , ζ , e , ψ , ε

)t, m = M f

D1Q3Q3 for the (ρ, J , ζ) formulation (ii)

Matrix MD1Q3Q3 between particles and moments

MD1Q3Q3 = ρ0

1 1 1 0 0 00 λ −λ 0 0 00 0 0 cp cp cp−2λ2 λ2 λ2 0 0 0

0 0 0 0 cp λ −cp λ0 0 0 −2 cp λ

2 cp λ2 cpλ

Equilibrium of non-conserved momentaeeq = eeq

(ρ, J, ζ

), ψeq = ψeq

(ρ, J, ζ

), εeq = εeq

(ρ, J, ζ

)Relaxation of non-conserved momenta

e∗ = e + se(eeq − e

), ψ∗ = ψ + sψ

(ψeq − ψ

ε∗ = ε+ sε(εeq − ε

)Time iteration of the D1Q3Q3 scheme

f0(x , t + ∆t) = f ∗0 (x , t), f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t), g0(x , t + ∆t) = g∗0 (x , t)

g+(x , t + ∆t) = g∗+(x −∆x , t), g−(x , t + ∆t) = g∗−(x + ∆x , t) .

Linearization

Reference state W0 = (ρ0 , ρ u0 , ρ s0) , c20 = γ p0

Mass conservation∂tρ+ ∂xJ = 0

Momentum conservation∂tJ +

0 − u20 −

s0 c20

)∂xρ+ 2 u0 ∂xJ +

c20cp∂xζ

−ν0 ∂2xρ+ ν0 u0 ∂

2xJ = 0

Production of entropy∂tζ − u0 s0 ∂xρ+ s0 ∂xJ + u0 ∂xζ

− ν0Pr

((γ − 1) cp − γ s0

)∂2xρ− γ ν0

Pr ∂2x ζ = 0

• Decoupled simplified system

Momentum conservation∂tJ + (c2

0 − u20) ∂xρ+ 2 u0 ∂xJ − ν0 ∂

2xρ+ ν0 u0 ∂

2xJ = 0

Production of entropy∂tζ + u0 ∂xζ − γ ν0

Pr ∂2x ζ = 0

two systems analogous to the two first simple ones !

D1Q3Q3 for the decoupled simplified system

• Fluid systemequilibrium moment

2 ρ+ 13e

eq = (c20 − u2

0) ρ+ 2 u0 J

defect of conservationθe = 3 (λ2 − 3 u2

0 − c20 ) ∂xJ − 6 u0 (c2

0 − u20) ∂xρ ' 3λ2∂xJ

coefficient of relaxation se ν0 = σe λ∆x

• Scalar “thermal” equationequilibrium moment ψeq = u0 ζ

defect of conservation θψ =(

2 − u20

)∂xζ + 1

3 ∂xεeq

compatibility at second order[(23λ

2 − u20

)ζ + 1

3 εeq]σψ ∆t = γ ν0

then εeq = 3( γPr

σeσψ− 2

)λ2 ζ ≡ αλ2 ζ

• Proposed link between two relaxations: σψ = 32γPr σe

D1Q3Q3 for the linearized Navier Stokes equations

Coupled linearized Navier Stokes equations∂tρ+ ∂xJ = 0

∂tJ +(c2

0 − u20 −

s0 c20

)∂xρ+ 2 u0 ∂xJ +

c20cp∂xζ

−ν0 ∂2xρ+ ν0 u0 ∂

2xJ = 0

∂tζ − u0 s0 ∂xρ+ s0 ∂xJ + u0 ∂xζ− ν0

((γ − 1) cp − γ s0

)∂2xρ− γ ν0

Pr ∂2x ζ = 0

Maintain the previous relations for relaxation coefficientsν0 = σe λ∆x , σψ = 3

2γPr σe , σε = 1.5

Linear equilibria eeq = c41 ρ+ c42 J + c43 ζψeq = c51 ρ+ c52 J + c53 ζεeq = c61 ρ+ c62 J + c63 ζ

c41 = 3(1− s0

0 − 3 u20 − 2λ2 , c42 = 6 u0 , c43 = 3

cpc51 = −u0 s0 , c52 = s0 , c53 = u0 ,c61 = −3 (s0 c0)2 − 6 s0 u

20 + 3 s0 c

20 + 2− 2 s0 − 2

c62 = 6 u0 s0 , c63 = 3 cp u20 + 3 s0 c

D1Q3Q3 for the linearized Navier Stokes equations (ii)

simulation of a simple linear wave: γ = 1.4 , Pr = 1 , c0 = λ2

u0 = 0 , s0 = 0 , ∆x = 140 , se = 1.9 , ν = 6.579 10−4

numerical stability γ = 1.4 , Pr = 1 , c0 = λ2

u0 = 0.15λ , s0 = 0.2 cp , ∆x = 140 , se = 1.9 , Tf = 120 ∆t

D1Q3Q3 for the (ρ, J , ζ) full Navier Stokes equations

Matrix MD1Q3Q3 between particles and moments

MD1Q3Q3 = ρ0

1 1 1 0 0 00 λ −λ 0 0 00 0 0 cp cp cp−2λ2 λ2 λ2 0 0 0

0 0 0 0 cp λ −cp λ0 0 0 −2 cp λ

2 cp λ2 cpλ

Equilibrium of non-conserved momentaeeq = eeq

(ρ, J, ζ

), ψeq = ψeq

(ρ, J, ζ

), εeq = εeq

(ρ, J, ζ

)Relaxation of non-conserved momenta

e∗ = e + se(eeq − e

), ψ∗ = ψ + sψ

(ψeq − ψ

ε∗ = ε+ sε(εeq − ε

)Time iteration of the D1Q3Q3 scheme

f0(x , t + ∆t) = f ∗0 (x , t), f+(x , t + ∆t) = f ∗+(x −∆x , t)f−(x , t + ∆t) = f ∗−(x + ∆x , t), g0(x , t + ∆t) = g∗0 (x , t)

g+(x , t + ∆t) = g∗+(x −∆x , t), g−(x , t + ∆t) = g∗−(x + ∆x , t) .

First order analysis

Taylor expansion method: partial differential equationsfor the conserved variables W =

(ρ, J, ζ ≡ ρ s

)∂tWk + Λk` ∂xm

eq` = O(∆t)

with Λk` = λ∑j

Mkj vj M−1j`

Momentum-velocity tensor

0 1 0 0 0 02λ2

0 0 0 0 1 00 λ2 0 0 0 0

0 02λ2

30 0 0 0 λ2 0

ρ J ζ e ψ ε

First order analysis (ii)

D1Q3Q3 lattice Boltzmann scheme

Equivalent equations at first order∂tρ+ ∂xJ = O(∆t)

∂tJ + ∂x

3 ρ+ 13 e

= O(∆t)

∂tζ + ∂xψeq = O(∆t)

Navier-Stokes equations of gas dynamics∂tρ+ ∂xJ = 0∂tJ + ∂x

(ρ u2 + p

)= ∂x

(ρ ν ∂xu

)∂tζ + ∂x

(ζ u)− ∂x

(κT ∂xT

)= ρ ν

(∂xu)2

+ κT 2

(∂xT

By identificationeeq = 3

(ρ u2 + p

)− 2λ2 ρ

ψeq = ζ u.

Second order analysis

Henon’s coefficients

σe =1

se− 1

2, σψ =

sψ− 1

2, σε =

sε− 1

Defects of conservation : θ` ≡ ∂tmeq` + Λ`p ∂xm

Second order equivalent partial differential equations

∂tWk + Λk` ∂xmeq` = ∆t Λk` σ` ∂xθ` + O(∆t2)

For the D1Q3Q3 lattice Boltzmann scheme:∂tρ+ ∂xJ = O(∆t2)∂tJ + ∂x

(ρ u2 + p

)= ∆t

3 σe ∂xθe + O(∆t2)∂tζ + ∂x

(ζ u)

= ∆t σψ ∂xθψ + O(∆x2)

with θe = ∂teeq + λ2 ∂xJ ' 3λ2 ρ ∂xu Then ν = σe λ∆x

θψ = ∂tψeq + 2

3 λ2 ∂xζ + 1

3 ∂xεeq

= p ∂xs + ∂x[

(2λ2 ζ + εeq

)−((ρ u2 + p) s

)]' ∂x

(2λ2 ζ + εeq

)−((ρ u2 + p) s

Second order analysis (ii)

Entropy equation at second order∂tζ + ∂x

(ζ u)− ∂x

(κT ∂xT

)= ρ ν

(∂xu)2

+ κT 2

(∂xT

to compare with

∂tζ + ∂x(ζ u)

−∂2x

{σψ ∆t

(2λ2 ζ + εeq

)−((ρ u2 + p) s

)]}= O(∆x2)

then σψ ∆t[

(2λ2 ζ + εeq

)−((ρ u2 + p) s

)]= κ log

)Prandtl number Pr =

ρ ν cpκ , ν = σe λ∆x

Relation between the relaxation coefficients σψ = 32γPr σe

Equilibrium value εeq of the third momentum for entropy

εeq =2 ρ cp λ2

γ log(TT0

)+ eeq s a non-trivial relation !

Perfect gas p =ρ0 c2

( ρρ0

)γexp

(γ s−s0

)=(1− 1

)cp ρT

εeq = 2λ2[ρ (s − s0) +

(1− 1

)cp ρ log

( ρρ0

)]+ eeq s

First numerical experiments, δρ = 0.001 ρ0

progressive linear wave, periodic boundary conditionsγ = 1.4 , Pr = 1 , c0 = λ