Fourier Law and Non-Isothermal Boundary in the Boltzmann Theory - - PowerPoint PPT Presentation

Fourier Law and Non-Isothermal Boundary in the Boltzmann Theory

Joint work with Raffaele Esposito, Yan Guo, Rossana Marra Chanwoo Kim

DPMMS, University of Cambridge

ICERM November 8, 2011

Steady Boltzmann Equation

Steady Boltzmann Equation v · ∇xF = Q(F, F) F(x, v) : density distribution of rarefied gas 3 D velocity space v ∈ R3 Ω : bounded, connected domain in Rd for d = 1, 2, 3 nonlinear Boltzmann operator Q(F1, F2) :

quadratic, bilinear non-local in v ∈ R3 hard potential 0 ≤ γ ≤ 1 with angular cut-off collision invariant :

• R3{1, v, |v|2}Q(F, F)(v)dv = 0

Knudsen number ∼ 1 regime

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Non-Isothermal Boundary and Diffusive BC

Wall temperature θ(x) = θ0 + δϑ(x)

• n x ∈ ∂Ω

Diffusive boundary condition on x ∈ ∂Ω, n(x) · v < 0 F(x, v) = µθ(x, v)

• n(x)·u>0

F(x, u){n(x) · u}du Wall Maxwellian µθ(x, v) = 1 2πθ(x)2 exp

• − |v|2

2θ(x)

• with
• n(x)·v>0 µθ(x, v){n(x) · v}dv = 0

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Purpose of This Work

Analyze the thermal conduction phenomena in the kinetic regime(Knudsen number ∼ 1)

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Purpose of This Work

Analyze the thermal conduction phenomena in the kinetic regime(Knudsen number ∼ 1) when the wall temperature do not oscillate too much! ⇓ Steady Boltzmann equation with Duffuse BC |θ(x) − θ0| ≪ 1, |ϑ(x)| ≤ 1 and δ ≪ 1 ⇓ Fs ∼ µ Regime

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Natural Questions and Previous Works

Existence, Uniqueness, Non-Negativity for Steady Solution

S.-H.Yu : existence and stability, Ω is slab (length≪ 1), ARMA 2009 L.Arkeryd, A.Nouri : Ann. Fac. Sci. Toulouse, Math. 2000 : Existence in L1−space, Ω is slab

Regularity (Continuity and Singularity)

Y.Guo : for IBVP, Ω convex, continuity away from γ0 : ARMA 2010 C.K : for IBVP, Ω non-convex, singularity formation and propagation : CMP 2011

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Natural Questions and Previous Works

Dynamical Stability

L.Desvillettes, C.Villani : polynomial decay in Hk for some BCs : Invent. Math. 2005 C.Villani : polynomial decay in Hk, diffusive BC, θ ≡ θ0 :

• Mem. AMS 2009

Y.Guo : θ ≡ θ0, e−λt−decay in L∞ to µ: ARMA 2010 S.-H.Yu : e−λt−decay in L∞ to the steady solution : ARMA 2009

Hydrodynamic Limit

• R. Esposito, Lebowitz, R.Marra : CMP 1994, J.Stat.Phys.

1995

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Existence, Uniqueness and Non-Negativity

Let Ω ⊂ Rd, d = 1, 2, 3. For all M > 0,

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Existence, Uniqueness and Non-Negativity

Let Ω ⊂ Rd, d = 1, 2, 3. For all M > 0, there exists δ0 > 0 such that for 0 < δ < δ0 in |θ(x) − θ0| ≤ δ,

• n x ∈ ∂Ω,

then there exists a non-negative solution Fs = Mµ + √µfs ≥ 0 with

• Ω×R3 fs√µ = 0 to the problem

v · ∇xFs = Q(Fs, Fs), Fs|γ− = µθ

• γ+

Fsdγ,

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Existence, Uniqueness and Non-Negativity

Let Ω ⊂ Rd, d = 1, 2, 3. For all M > 0, there exists δ0 > 0 such that for 0 < δ < δ0 in |θ(x) − θ0| ≤ δ,

• n x ∈ ∂Ω,

then there exists a non-negative solution Fs = Mµ + √µfs ≥ 0 with

• Ω×R3 fs√µ = 0 to the problem

v · ∇xFs = Q(Fs, Fs), Fs|γ− = µθ

• γ+

Fsdγ, such that, for all 0 ≦ ζ <

1 4+2δ, β > 4,

||vβeζ|v|2fs||∞ + |vβeζ|v|2fs|∞ δ. If Mµ + √µgs is an another solution with

• Ω×R3 gs√µ = 0 such

that, for β > 4 ||vβgs||∞ + |vβgs|∞ ≪ 1, then fs ≡ gs.

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Continuity and Singularity

If θ(x) is continuous on ∂Ω then Fs is continuous away from D. In particular, if Ω is convex then D = γ0. On the other hand, if Ω is not convex then we can construct a continuous function θ(x) on ∂Ω such that the corresponding solution Fs in not continuous.

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Dynamical Stability

Let 0 ≦ ζ <

1 4+2δ, β > 4. There exists ε0 > 0, depends on δ0, and

λ > 0 such that if ||vβeζ|v|2[f (0) − fs]||∞ ≤ ε0 then there exists a unique non-negative dynamic solution F(t) = µ + fs√µ + f (t)√µ ≥ 0 to the dynamical problem ∂tF + v · ∇xF = Q(F, F), F(x, v) = µθ

• n(x)·v>0

F n · v for x ∈ ∂Ω and n(x) · v < 0 such that ||vβeζ|v|2[f (t) − fs]||∞ e−λt||vβeζ|v|2[f (0) − fs]||∞

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Why δ−Expansion ?

Fourier Law : a relation between the temperature and the heat flux qs = −κ(θs)∂xθs for suitable positive smooth function κ. Let Fs be the solution to the steady Boltzmann equation θs(x) = 1 3ρs

• R3 |v − us|2Fs(x, v)dv

us(x) = 1 ρs

• R3 vFs(x, v)dv

ρs(x) =

• R3 Fs(x, v)dv

qs(x) = 1 2

• R3(v − us(x))|v − u(x)|2Fs(x, v)dv.

Purpose : See the first order characterization of Fs

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

What is δ−Expansion ? : µδ−Expansion

Wall Temperature θ(x) = θ0 + δϑ(x), |ϑ(x)| ≤ 1, x ∈ ∂Ω. Wall Maxwellian µδ(x, v) = 1 2π[θ0 + δϑ(x)]2 exp

• −

|v|2 2[θ0 + δϑ(x)]

• Taylor Expansion in δ (µδ is analytic in δ )

µδ = µ + δµ1 + δ2µ2 + · · · + δmµm + · · ·

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

What is δ−Expansion ? : fs ∼ δf1 + δ2f2 + · · ·

Formal Expansion : Fs = µ + √µ

• δf1 + δ2f2 + · · ·
• fs = δf1 + δ2f2 + · · ·

Plug in v · ∇xFs = Q(Fs, Fs) with Diffusive Bounadary Condition to get the linear equation for fi (comparing the coefficients of power of δ) Once we solve fi, define the Remainder f δ

m such that

fs = δf1 + δ2f2 + · · · + δmf δ

m

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : δ−Expansion

δ−Expansion is valid ! There exist f1, f2, · · · , fm−1 and for all i = 1, 2, · · · m − 1 ||vβeζ|v|2fi||∞ 1 for all 0 ≦ ζ < 1

4, β > 4

and the remainder f δ

m exits and

||vβeζ|v|2f δ

m||∞ 1

for all 0 ≦ ζ <

1 4+2δ, β > 4

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Theorem : Criterion for Fourier Law

Let Ω = [0, 1]. If the Fourier Law holds for Fs = µ + √µfs, Fs = µ + δf1 √µ + O(δ2)√µ θs = θ0 + δθ1 + O(δ2) then θ1(x) is a linear function on [0, 1]

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Prediction

From an available numeric simulation (Ohwada, Aoki, Sone, 1989) θ1 is not linear! ⇓ Fourier Law is not valid at the kinetic regime !

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Hydrodynamic Part : Pf

Linearized Boltzmann operator Lf = − 1

√µ[Q(µ, √µf ) + Q(√µf , µ)] = ν(v)f − Kf

semi-positive : Lg, g ||{I − P}g||2

ν

kernel = ‘hydrodynamic part’ Pg ≡

• ag(t, x) + v · bg(t, x) + |v|2 − 3

2 cg(t, x) √µ Boltzmann equation = ⇒ macroscopic equation for bf ∆xbf = ∂2

x{I − P}f + · · ·

ellipticity in Hk = ⇒ Guo:VMB(Invent.Math.2003), VPL(JAMS2011); Gressman-Strain:BE without angular cut-off(JAMS2011)

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Difficulties with boundary conditions

Pf ≡

• af (t, x) + v · bf (t, x) + |v|2 − 3

2 cf (t, x) √µ Pf and {I − P}f do not make sense at the boundary no boundary condition for af , cf , only bf · n(x) = 0 on ∂Ω

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Mathematical Framework : L2 − L∞ Frame

Y.Guo : Initial Boundary Value Problem of BE, ARMA 2010 L2 Posivity : We Need A New Method ! L∞ Bound : We Need A New Method !

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

New L2 Positivity Estimate

v · ∇xf + Lf = g, fγ− = Pγf + r with

• Ω×R3 f √µ = 0 =
• Ω×R3 g√µ =
• n·v<0

r = ⇒ ||Pf ||ν ≤ M{||(I − P)f ||ν + |(1 − Pγ)f |2,+

• Good Terms !

} + · · · weak formulation(Green’s identity) + test functions constructive estimate with an explicit M dimension of Ω = 1, 2, 3

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

New L2 Positivity Estimate

Weak formulation (Green’s identity)

• γ

ψf −

• Ω×R3 v · ∇ψf = −
• Ω×R3 ψL(I − P)f +
• Ω×R3 ψg

bulk f = {af + v · bf + |v|2 − 3 2 cf }√µ + (I − P)f boundary fγ = Pγf + (1 − Pγ)f 1γ+ + r1γ−

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

New L2 Positivity Estimate

Test functions for cf : ψc = (|v|2 − βc)√µ{v · ∇x}(−∆0)−1cf with

• R3(|v|2 − βc)v2

i µ(v)dv = 0

for bf :

ψi,j

b = (v 2 i − βb)√µ∂j(−∆0)−1(bf )j for all i, j = 1, 2, · · · d

with

• R3(v 2

i − βb)µ(v)dv = 0, for all i

φi,j

b = vivj|v|2√µ∂j(−∆0)−1(bf )i for all i = j

for af : ψa = (|v|2 − βa){v · ∇x}(−∆N)−1af with

• R3(|v|2 − βa)( |v|2

2 − 3 2)(vi)2µ(v)dv = 0 for all i

Chanwoo Kim Fourier Law and Non-Isothermal Boundary

Future

Thanks !

Chanwoo Kim Fourier Law and Non-Isothermal Boundary