hydrodynamic limits, knudsen layers and numerical … · hydrodynamic limits, knudsen layers and...

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)

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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

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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

2θ

),

(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.

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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).

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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

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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).

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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

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?)

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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

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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)

]±

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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±).

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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

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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}.

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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)

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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

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

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.

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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

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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?).

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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

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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? ...

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It does not work...

Comparison Hydro (solid)/Kinetic (Dashed)

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

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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

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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 .

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Not that bad

Comparison Hydro (solid)/Kinetic (Dashed)

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

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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

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

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

Evaporation/Condensation problem

Convergence towards a stationary state

long time behaviour for ρLw = 2/1.2, θL

w = 1.2/2

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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.

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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)

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