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Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon with C. Besse, I. Violet and F. Charles, N. Vauchelet (CEMRACS’10) and J.P. Dudon (ThalesAleniaSpace) and S. Borghol (PhD) 1 / 31

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Page 1: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Hydrodynamic limits,Knudsen layers and

numerical fluxes

Thierry Goudonwith C. Besse, I. Violet

and F. Charles, N. Vauchelet(CEMRACS’10)and J.P. Dudon

(ThalesAleniaSpace)and S. Borghol (PhD)

1 / 31

Page 2: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Motivation from Spacecraft engineering

Models for Spacecraft Charging (that can generate electrical arcingproducing severe damages...)Vlasov-Poisson-Boltzmann eq...with a highly complicated Boundary Condition for the potentialBut the characteristics of the plasmas depends on the conditions offlight (GEO, LEO, PEO...)In some circumstances, it makes sense to make use of fluid models

2 / 31

Page 3: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Hydrodynamic limits

From BGK...

∂tF + v∂xF =1

τ(M[F ]− F ),

M[F ](v) = Mρ,u,θ(v) =ρ√2πθ

exp(− |v − u|2

),

(ρ, j ,E ) = (ρ, ρu, ρu2/2 + ρθ/2)(t, x) =

∫R

pF dv , p = (1, v , v2/2)

... to Euler

As τ → 0, we have F → Mρ,u,θ(v) with

∂t

ρρuE

+ ∂x

ρuρu2 + ρθ(ρu2 + 3ρθ)/2

= 0.

3 / 31

Page 4: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

From BGK to Euler

We have

∫R

p(M[F ]− F ) dv = 0 hence the conservation laws

∂t

∫R

pF dv + ∂x

∫R

v pF dv = 0

Replacing F by the equilibrium Mρ,u,θ(v) yields the Euler system...

∂t

ρρuρ

2(u2 + θ)

+ ∂x

ρuρu2 + ρθ1

2(ρu2 + 3ρθ)

= 0.

Warning : two sets of unknowns

U = (ρ, u, θ) 7→ U = (ρ, j = ρu,E = ρu2/2 + ρθ/2).

4 / 31

Page 5: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Knudsen Layers : x ∈ (−ω, ω)

But the Boundary Condition for the kinetic equation might be notcompatible with the thermodynamical equilibrium Mρ,u,θ... The differencecreates Boundary Layers.

Questions :

Q1 What is the boundary condition for the hydrodynamic equations(that is to be satisfied by (ρ, u, θ)) ?

Q2 In a numerical (FV) scheme for the hydrodynamic system, how do wedefine the boundary fluxes ?

A few references

Q1 leads to the (delicate !) analysis of half-space problems :Coron-Golse-Sulem’88 (linearized), Ukai-Yong-Yu’03

Similar questions arise for kinetic/fluid coupling : Vasseur ’09

Numerical investigations of boundary layers : Sone & Aoki

Q2 : Golse-Jin-Levermore’03 (diffusion limits), Dellacherie’03

5 / 31

Page 6: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Boundary Conditions, Boundary Fluxes

Kinetic vs. hydro. BC

One has to prescribe incoming fieds at x = ±ω. For the kinetic equation,this is quite clear :

f (t, x = −ω, v > 0) = ΦL(t, v), f (t, x = +ω, v < 0) = ΦR(t, v).

For the Euler system it becomes a nightmare... (the number of quantitiesto be prescribed depends on the unknown itself, and it might change withtime !).

Numerical scheme and boundary fluxes

FV discretisation : U n+1j = U n

j − ∆t

∆x(F n

j+1/2 −F nj−1/2) with, say,

F nj+1/2 = F(U n

j+1,Unj ) (e.g. Godunov fluxes...)

How do we define the boundary flux F n−1/2 (resp. F n

J+1/2) ? (that

certainly depends on the kinetic data ΦL/R).

6 / 31

Page 7: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Entropy, did you say “Entropy” ?

Entropy is dissipated by the BGK eq.

∂t

∫R

F ln(F ) dv +∂x

∫R

vF ln(F ) dv = −∫

R(M[F ]−F ) ln

(M[F ]

F

)dv ≤ 0.

It provides a natural entropy for the Euler system

U = (ρ, j ,E ), U = (ρ, u, θ)

η(U ) =

∫R

F ln(F ) dv∣∣∣F=MU

is (strictly) convex

∂tη(U ) + ∂xq(U ) = 0

Similar relations with the relative entropy

η(U |U?) =

∫R

[MU ln

( MU

MU?

)−MU + MU?

]dv ≥ 0.

But entropy is also the tool that allows to understand what needs to beprescribed in the hydrodynamic quantities. 7 / 31

Page 8: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Linearized Theory M? = M(ρ?>0,u?∈R,θ?>0)

Expand F = M?(1 + δf ), with 0 < δ � 1. The leading terms lead to

∂t f + v∂x f =1

τ(m[f ]− f ),

m[f ](v) = meU(v) =ρ

ρ?+

v − u?

θ?u +

θ

2θ?

( |v − u?|2

θ?− 1

),

U(t, x) = (ρ, u, θ)(t, x) =1

ρ?

∫R(1, v − u?, |v − u?|2 − θ?) f M? dv

Linearized Euler equations

As τ → 0, we have f ' meU(v) with

∂tU + A?∂x U = 0, A? =

ρ? u? θ?

θ?/ρ? u? 10 2θ? u?

Eigenvalues of A? : (u? −

√3θ?, u?, u? +

√3θ?)

8 / 31

Page 9: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Linearized equations and Entropy

Conserved vs. physical quantities

(ρ, u, θ) 7→ (ρ, j = ρu,E = ρu2/2 + ρθ/2) yields for fluctuations

U = (ρ, j , E ) = P?U ∂tU + A?∂xU = 0, A? = P?A?P−1? .

Entropy and Energy estimates

Expanding ∂tη(U |U?) + ∂xη(U |U?) = 0 yields

S?∂tU + Q?∂xU = 0, Q? = S?A?

with S? = D2η(U |U?)∣∣U =U?

symmetric positive definite and Q?

symmetric.

ddt

∫ +ω

−ωS?U · U dx + Q?U · U

∣∣∣+ω

−ω= 0

9 / 31

Page 10: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Entropy/Energy estimates

S?=entropy matrix, sym., positive definite, Q? = S?A?

0 ≤∫ +ω

−ωS?U · U (t, x) dx

+

∫ t

0

[Q?U · U (s, ω)

]+

ds +

∫ t

0

[Q?U · U (s,−ω)

]− ds

=

∫ +ω

−ωS?U · U (0, x) dx

+

∫ t

0

[Q?U · U (s,+ω)

]− ds +

∫ t

0

[Q?U · U (s,−ω)

]+

ds

One has to prescribe[Q?U · U (∓ω)

10 / 31

Page 11: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Coming back to the kinetic equation

Entropy dissipation

∂t

∫R

f 2M? dv + ∂x

∫R

vf 2M? dv ≤ 0

Set Q(U) = Entropy Flux∣∣f =meU = Σ?U · U.

Incoming/Outgoing Characteristics

Let n± =number of ≷ 0 eigenvalues of A? (or A?)

The signature of Q?U · U is (n+, n−),

Σ? = PT? Q?P?

Ker(m[f ]− f ) = Λ+ ⊕ Λ− ⊕ Λ0, with Q|Λ+ is positive definite, Q|Λ−

is negative definite, Λ0 ={U, Q(U) = 0

}, n± = dim(Λ±).

11 / 31

Page 12: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Summary

For the left hand side :

If u? >√

3θ? : 3 incoming characteristics, 3 quantities to beprescribed,

If√

3θ? > u? > 0 : 2 incoming characteristics, 2 quantities to beprescribed,

If 0 > u? > −√

3θ? : 1 incoming characteristics, 1 quantity to beprescribed,

If −√

3θ? > u? : all characteristics are outgoing, nothing to prescribe

... and reverse the conclusions at the right hand side

12 / 31

Page 13: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

An adapted basis

We remind that

Ker(m[f ]− f ) ={

m(eρ,eu.eθ) =

ρ

ρ?+

v − u?

θ?u +

θ

2θ?

( |v − u?|2

θ?− 1

)}We set E F (g) =

∫R

vg2M? dv and Q(U) = E F (meU) = Σ?U · U.

We introduce χ1(v) =1√6

(√3v − u?√

θ?+|v − u?|2

θ?

),

χ2(v) =1√6

(√3v − u?√

θ?− |v − u?|2

θ?

), χ0(v) =

1√6

( |v − u?|2

θ?− 3

).

Then Ker(m[f ]− f ) = Span{χ0, χ1, χ2} with E F (χ0) = ρ?u?,

E F (χ1) = ρ?(u? +√

3θ?), E F (χ2) = ρ?(u? −√

3θ?)

A useful decomposition is Ker(m[f ]− f ) = Λ+ ⊕ Λ0 ⊕ Λ− withΛ± = Span{χk , E F (χk) ≷ 0}, Λ0 = Span{χk , E F (χk) = 0}.

13 / 31

Page 14: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Half-Space Problem

Derivation of the Half-Space Problem

Start from ∂t f + v∂x f =1

τ(m[f ]− f ) with BC

f (t,−ω, v > 0) = ΦL(t, v), f (t,+ω, v < 0) = ΦR(t, v)

Expansion with Boundary Layer Corrections

f (t, x , v) = meU(t,x)(v) + GL

(t,

x + ω

τ, v

)+ GR

(t,

x − ω

τ, v

)+ ...

yields for z ≥ 0, v ∈ R

v∂zG = m[G ]− G , G (t, x = 0, v > 0) = Φ(t, v)−meU(t,−ω)(v)

14 / 31

Page 15: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Coron-Golse-Sulem’s Analysis

Theorem

Let V + be a subspace of Ker(m[f ]− f ) satisfying

i) For any U ∈ RN+2 \ {0} such that meU ∈ V + we have Q(U) ≥ 0,

ii) V + is maximal in the sense that any subspace V ⊂ Ker(m[f ]− f )verifying i) is included in V +.

Let Φ ∈ L2(RN , (1 + |v |)M?(v) dv). Then, for any m ∈ Ker(m[f ]− f )there exists a unique m∞ ∈ V + and a unique solutionG ∈ L∞

(0,∞; L2(RN ,M? dv)

)such that for any 0 < γ � 1, we have

eγz(G (z , v)−m −m∞(v)

)∈ L∞

(0,∞; L2(RN , (1 + |v |)M?(v) dv)

).

Consequences

The theorem defines a linear mapping C? : Φ−m 7→ m∞

The Boundary Layer Correction is expected to vanish far from theboundary : m∞ = 0 that is C?(Φ) = C?(meU(t,x)

).15 / 31

Page 16: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Boundary Conditions vs Boundary Fluxes...

The relation C?(Φ) = C?(meU(t,x)) defines the boundary condition for U.

We can think of this relation as follows :

split meU(t,x)= m+ + m− with m± ∈ Λ±.

m− corresponds to “outgoing” characteristics (known)

m+ corresponds to “incoming” characteristics (to be determined)

then m+ is the asymptotic state for the solution of the half-spaceproblem with incoming data Φ−m−.

Remarks :

when n+ = 0, there is no incoming characteristics Λ+ = {0} andm+ = 0... there is nothing to impose.

The incoming characteristics are defined as a functional of the kineticboundary data Φ and the outgoing characteristics.

How can we use this for defining numerical fluxes ?

16 / 31

Page 17: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Towards Boundary Fluxes...

FV scheme : U n+1j = U n

j − ∆t

∆x(F n

j+1/2 −F nj−1/2)

Idea : define the boundary flux as F−1/2 =

∫R

vpmbd M? dv .

Definition of mbd

Set mfluc =ρfluc

ρ?+

v − u?

θ?ufluc +

θfluc

2θ?

( |v − u?|2

θ?− 1

), with

(ρfluc , ufluc , θfluc)=unknowns in the boundary cell (i. e. P−1? U n

0 )

Project on Λ− to define m−(v) =∑k∈I−

α−k χk(v) (“outgoing part”)

Define m+ =∑k∈I+

α+k χk(v) “from the half-space problem” (...) with

data Φ−m−

mbd = m− + m+.

... but solving the half-space problem numerically looks as difficult assolving the original kinetic equation.

17 / 31

Page 18: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Approximation of the half space problem

Conservation Laws : p = (1, v , v2/2)

ddz

∫R

vpGM? dv = 0

Since G (∞, v) = 0 we get∫v>0

vpGinc(0, v)M? dv +

∫v<0

vpGout(0, v)M? dv = 0

with Ginc = Φ−m− −m+

Maxwell Approximation

Identify the outgoing distribution with the asymptotic state :Gout(0, v) = 0. It yields∫

v>0vp(Φ−m−)(0, v)M? dv =

∫v>0

vpm+(0, v)M? dv

18 / 31

Page 19: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Construction of the boundary fluxes

We seek m+ =ρ+

ρ?+

v − u?

θ?u+ +

θ+

2θ?

( |v − u?|2

θ?− 1

)∈ Λ+,

If n+ < 3 the Conservation Laws and the Maxwell Approximation providetoo many relations...

Computation of the “incoming state”

Pick the necessary number of relations (cf. : Golse-Klar’95,Arnold-Giering’97)

Least Square Approximation

minm+∈Λ+

∣∣∣ ∫v>0

vp(Φ−m−)(0, v)M? dv −∫

v>0vpm+(0, v)M? dv

∣∣∣2Numerical Flux

F−1/2 =

∫R

vpmbd M? dv =

∫R

vp(m− + m+) M? dv . It corresponds to

UpWind Fluxes on the diagonalized system ∂tV + D∂xV = 0,D = diag(u? −

√3θ?, u?, u? +

√3θ?).

19 / 31

Page 20: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Some results

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.50.4

0.5

0.6

0.7

0.8

0.9

1

1.1

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.50.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Comparison Hydro (dashed)/Kinetic (solid)

Top : Density, Bottom left : Velocity, Bottom right : Temperature, Φ = 0,n+ = 2, n− = 1

20 / 31

Page 21: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Extension to Non–Linear Problems

A naive attempt

Consider an initial data “close to” an absolute Maxwellian M?,(ρ?, u?, θ?) =constant state.

We expect to remain quite close to the linearized situation :U = U? + U (i. e. U is approximately a solution of the linearizedEuler system). We define numerical fluxes by

Fbd =

∫R

vp(1 + mbd)M? dv

with mbd = m− + m+, m−(v) =∑

k∈I− α−k χk(v)projection on Λ− ofmUn

0−U? ...

21 / 31

Page 22: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

It does not work...

Comparison Hydro (solid)/Kinetic (Dashed)

(ρ?, u?, θ?) = (1, 0.1, 1), Φdata = 0, final time t = 0.1.

22 / 31

Page 23: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

It does not work...

... because the method cannot take into account possible change of typeof the flow

Evolution of the eigenvalues on right and left boundaries.

(ρ?, u?, θ?) = (1, 0.1, 1), Φdata = 0

23 / 31

Page 24: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Local linearization

Definition of the Boundary Fluxes

• The hydro. field in the first cell becomes the reference state : Un1 7→ U?,

• See the hydro. field in the boundary cell as a perturbation of thereference state Un

0 7→ U? + Ufluc ,

• Repeat the construction Fbd =

∫R

vp(1 + mbd)M? dv with

mbd = m− + m+, m− =projection on Λ− of mUfluc...

Remark

The determination of m+(v) =∑

k∈I+ α+k χk(v) from the conservation law

minm+∈Λ+

∣∣∣ ∫v>0

vp(Φ−m−)(0, v)M? dv −∫

v>0vpm+(0, v)M? dv

∣∣∣2leads to a linear system

(ATA BBT 0

) (α+

q

)=

(b0

)but now the

matrices are computed as each time step, since they depend on Un1 .

24 / 31

Page 25: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Not that bad

Comparison Hydro (solid)/Kinetic (Dashed)

(ρ?, u?, θ?) = (1, 0.1, 1), Φdata = 0, final time t = 0.1.

25 / 31

Page 26: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Evaporation/Condensation problem

Boundary Data Mw (v) =ρw√2πθw

exp(− |v − uw |2

2θw

)

Kinetic (dashed) vs. hydro. (solid)

ρLw = 2/1.2, θL

w = 1.2/2, and final time t = 0.1. 26 / 31

Page 27: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Evaporation/Condensation problem

Kinetic (dashed) vs. hydro. (solid)

ρLw = 1.2/1.1, θL

w = 1.1/2 and final time t = 0.1.

27 / 31

Page 28: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Evaporation/Condensation problem

Kinetic (dashed) vs. hydro. (solid)

ρLw = 10/1.1, θL

w = 1.1/2 and final time t = 0.1.

28 / 31

Page 29: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Evaporation/Condensation problem

Convergence towards a stationary state

long time behaviour for ρLw = 2/1.2, θL

w = 1.2/2

29 / 31

Page 30: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

A couple of remarks

The scheme preserves equilibrium (initial and boundary data=thesame absolute Maxwellian)... up to the accuracy of the computationof the v−integrals

Comparison with kinetic runs could be a bit unfair... since the cost ofsuch computations becomes prohibitive as τ → 0, requiring small ∆xand ∆t... cf. : Sone & Aoki

Dellacherie’s boundary fluxes

Fbd =

∫v>0

vpΦ dv +

∫v<0

vpM(ρn0,un

0 ,θn0) dv

can be interpreted as a version of Enquist-Osher’s flux, cf. Vasseur’09.

30 / 31

Page 31: Hydrodynamic limits, Knudsen layers and numerical … · Hydrodynamic limits, Knudsen layers and numerical fluxes Thierry Goudon ... Kinetic vs. hydro. BC ... Ue (t,x) = (ρ,e eu,θe

Perspectives

Kinetic boundary condition involving a reflection operator can bedealt with

The Maxwell approximation looks a bit crude... in particular it doesnot contain much information on the details of the collision operator.Better approximations of the half-space problem can be used :Chapman-Enskog & duality relation : Golse-Klar’95,Spectral methods : Coron’88

Possible improvements based on recent progress of the analysis of thehalf-space problem (Ukai-Yong-Yu’03) ?

Multi-D simulations

Fluid/Kinetic matching conditions : same framework

Models for charged particles : incorporation of the electric field(collaboration with Thales)

31 / 31