BOUNDARY LAYERS AND HYDRODYNAMIC LIMITS OF
BOLTZMANN EQUATION (I):
INCOMPRESSIBLE NAVIER-STOKES-FOURIER LIMIT
NING JIANG AND NADER MASMOUDI
Abstract. We establish an incompressible Navier-Stokes-Fourier limit for solutions to
the Boltzmann equation (with a general collision kernel) considered over a bounded
domain. Appropriately scaled families of DiPerna-Lions renormalized solutions with
Maxwell reflection boundary conditions are shown to have fluctuations that converge as
the Knudsen number ε goes to zero. Every limit point is a weak solution to a NavierStokes-Fourier system with different types of fluid boundary conditions depending on the
ratio between the accommodation coefficient and the Knudsen number. The main new
result of the paper is that this convergence is strong in the case of Dirichlet boundary
condition. Indeed, we prove that the acoustic waves are damped immediately which
means that they are damped in a boundary layer in time. This damping is due to the
presence of viscous and kinetic boundary layers in space. These acoustic waves are known
to oscillate faster and faster when the Knudsen number ε goes to zero and to prevent
strong convergence in the periodic or the whole space case.
This extends the work of Golse and Saint-Raymond [18, 19] and Levermore and Masmoudi [27] to the case of a bounded domain. The case of a bounded domain was already
considered by Masmoudi and Saint-Raymond [33] for the (linear) Stokes-Fourier limit
and without studying the damping of the acoustic waves.
Contents
1. Introduction
2. Boltzmann Equation in Bounded Domain
2.1. Maxwell Boundary Condition
2.2. Nondimensionalized Form of the Boltzmann Equation
2.3. Navier-Stokes Scaling
2.4. A Priori Estimates
2.5. DiPerna-Lions Solutions
3. Statement of the Main Results
3.1. Dirichlet Boundary Condition
3.2. Navier Boundary Condition
4. Geometry of the boundary ∂Ω
5. Acoustic Modes
5.1. Acoustic Operator A
5.2. Eigenspaces of A
5.3. Conditions on Φk
5.4. the operator A − iλτ,k
6. Analysis of the Kinetic Boundary Layer Equation
7. Approximate Eigenfunctions-Eigenvalues
Date: March 22, 2010.
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N. JIANG AND N. MASMOUDI
7.1. Motivation
7.2. Statement of the Proposition
7.3. Main Idea of the Proof
8. Proof of Proposition 7.1: the case 0 < β < 21
8.1. Double Induction
8.2. (m, n) = (0, 0)
8.3. (m, n) = (1, 0)
8.4. (m, 0) for all m ≥ 2.
8.5. General Process
9. Proof of Proposition 7.1: the case β = 0
10. Proof of Proposition 7.1: the case 21 < β < 1
10.1. (0, n) for n ≥ 0
10.2. (1, n) for n ≥ −1
10.3. General process
11. Proof of Proposition 7.1: the case β = 12
12. Proof of Proposition 7.1: Truncation Error Estimates
τ,k
12.1. Estimates of Rε,N
τ,k
τ,k,int
12.2. Estimates of gε,N
− g0,0
12.3. Boundary error estimate
13. Proof of the Weak Convergence
14. Proof of the Strong Convergence
14.1. Strong Convergence in L1
14.2. Proof of Proposition 14.1
14.3. Estimates of c±
k,ε
14.4. Estimates of b̃±
k,ε
14.5. Proof of Lemma 14.1
15. Conclusion
References
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1. Introduction
The hydrodynamic limit of the Boltzmann equation got a lot of interest in the previous
decade. The main contribution is the rigorous derivation of the Navier-Stokes-Fourier
system from the Boltzmann equation by Golse and Saint-Raymond [18, 19] for the case
of hard potentials. This was then extended by Levermore and Masmoudi [27] to include
soft potentials. One of the ingredients of the proof of convergence is the treatment of
the acoustic waves which are highly oscillating. A compensated compactness type argument was used by Lions and Masmoudi [30] to prove that they have no contribution
on the equation satisfied by the weak limit. This argument was previously used in the
compressible incompressible limit [29].
In this paper, we prove that in a bounded domain and with an accommodation coefficient bigger than the Knudsen number, there is no need for the argument in [29] since we
can prove that the acoustic waves are damped instantaneously. Our work
√ is based on the
construction of viscous and kinetic (Knudsen) boundary layers of size ε and ε. A similar
property was used by Desjardins, Grenier, Lions, and Masmoudi [10] in the compressible
NAVIER-STOKES-FOURIER
3
incompressible limit. However, there are two main differences with [10]. The first one
is that a generic assumption on the domain has to be made in [10] (in particular there
are modes which are not damped in the disc). Here, we do not need this assumption.
The reason is that we are dealing with the whole acoustic system, namely including the
temperature and that we are including a kinetic effect. The second difference, is that we
need to build a kinetic boundary layer. The solvability condition of this layer gives the
boundary data for the viscous boundary layer.
We consider a bounded domain Ω ⊂ RD , D ≥ 2, with boundary ∂Ω ∈ C 2 . The
Navier-Stokes-Fourier system governs the fluctuations of mass density, bulk velocity, and
temperature (ρ, u, θ) about their spatially homogeneous equilibrium values in a Boussinesq regime. Specifically, after a suitable choice of units, these fluctuations satisfy the
incompressibility and Boussinesq relations
∇x ·u = 0 ,
ρ + θ = 0;
(1.1)
while their evolution is determined by the Navier-Stokes and heat equations
∂t u + u·∇x u + ∇x p = ν∆x u , u(x, 0) = u0 (x) ,
∂t θ + u·∇x θ =
2
κ∆x θ ,
D+2
θ(x, 0) = θ0 (x) ,
(1.2)
where ν > 0 is the kinematic viscosity and κ > 0 is the heat thermal conductivity.
We will derive two types of natural physical boundary conditions for the incompressible
Navier-Stokes-Fourier system (1.2). The first is the (homogeneous) Dirichlet boundary
condition, namely,
u = 0 , θ = 0 on R+ × ∂Ω .
(1.3)
The other is the so-called Navier slip boundary condition, which was proposed by Navier
[38]:
[2νd(u)·n + χu]tan = 0 , u·n = 0 on R+ × ∂Ω ,
(1.4)
D+1
κ∂n θ + χ D+2
θ = 0 on R+ × ∂Ω ,
where d(u) denotes the symmetric part of the stress tensor dij (u) = 21 (∂i uj + ∂j ui ), and ∂n
denotes the directional derivative along the outer normal n. In the above Navier boundary
condition, χ > 0 is the reciprocal of the slip length which depends on the material of the
container.
Fluid regimes are those where the mean free path is small compared to the macroscopic
length scales−i.e. where the Knudsen number ε is small. The incompressible NavierStokes-Fourier system can be formally derived from the Boltzmann equation through a
scaling in which the fluctuations of the density F about an absolute Maxwellian M are
scaled to be on an order ε, see [2].
The program that justifies the fluid dynamic limits from the Boltzmann equation in the
framework of DiPerna-Lions [11] was initiated by Bardos-Golse-Levermore [2, 3]. Since
then, there has been lots of contributions to this program [4, 12, 18, 19, 23, 27, 30, 31,
33, 39] among which we particularly mention the result of Golse and Saint-Raymond [18]
which is the first complete justification of Navier-Stokes-Fourier limit from Boltzmann
equation in a class of bounded collision kernels, without making any nonlinear weak
compactness hypothesis. They have recently extended their result to the case of hard
potentials [19]. With some new nonlinear estimates, Levermore and Masmoudi [27] treated
a broader class of collision kernels which includes all hard potential cases and, for the first
time in this program, soft potential cases.
4
N. JIANG AND N. MASMOUDI
However, all of the above works are carried out in either the periodic spatial domain
or the whole space, except for [33] in which the (linear) Stokes-Fourier is recovered with
the same collision kernels assumption as in [12]. In [33], the renormalized solutions to
the Boltzmann equation in bounded domain (see [37]) was proved to pass to the limit
in the different types of kinetic boundary conditions and recovered different fluid boundary conditions, either Dirichlet condition (1.3), or Navier slip boundary condition (1.4),
depending on the relative sizes of the accommodation coefficient and Knudsen number.
Y. Sone and his collaborators observed that formally the boundary conditions of the
fluid equations derived from the kinetic equation with Maxwell type boundary conditions,
depend on the relative importance of the accommodation coefficient and the Knudsen
number, see [43] Chapter 3 and 4.
The current work, generalizes the previous work of Masmoudi and Saint-Raymond [33]
in three ways: First, it works for general collision kernels, namely in the framework of
[27]. Second, it extends the result of [33] from linear Stokes-Fourier system to nonlinear
Navier-Stokes-Fourier system.
The third extension which is the main novelty of the current work is in the treatment
of the Dirichlet boundary condition case. Indeed, we prove, in the Dirichlet boundary
condition case, that the convergence is strong. We point out that in all the previous works
mentioned above, the convergence is in w-L1 , unless the initial data is well-prepared, i.e.
is hydrodynamic and satisfies the Boussinesq and incompressibility relations. This weak
convergence is caused by the persistence of fast acoustic waves. In the Navier-Stokes
regime, the Reynold number Re is unit order, then the von Kámán relation ε = Ma
Re
implies that in the fluid limit ε → 0, the Mach number Ma must go to zero. As is well
know physically, one expects that as Ma → 0, fast acoustic waves are generated carrying
the energy of the potential part of the flow. For the periodic flows, or for some particular
boundary conditions such as Navier condition (1.4), these waves subsist forever and their
frequency grows with ε. Mathematically, this means that the convergence is only weak.
This phenomena happens in many cases among which we only mention [28, 29].
In [10], a striking phenomena (damping of acoustic waves) caused by the Dirichlet
boundary condition was found when Desjardins, Grenier, Lions, and Masmoudi considered the incompressible limit of the isentropic Navier-Stokes equations. In the case of a
viscous flow in a bounded domain with Dirichlet boundary condition, and under a generic
assumption on the domain (related to the so-called Schiffer’s conjecture and the Pompeiu
problem [9]), they showed that the acoustic waves are instantaneously (asymptotically)
damped, due to the formation of a thin boundary layer in time. This layer is caused by
a boundary layer in space and dissipates the energy carried by the acoustic waves. From
a mathematical point of view, strong convergence was obtained.
The current work can be seen as an extension of [10] to the kinetic case. The main
idea is to use a family of test functions which solve the linearized Boltzmann equation
and which can capture the propagation of the fast acoustic waves. These test functions
are constructed through considering a family of approximate eigenfunctions of a dual
operator with a dual kinetic boundary condition with respect to the original Boltzmann
equation. The leading term of the approximate eigenvalue is purely imaginary describing
the acoustic mode, the real part of the next order term is strictly negative which gives
the strict dissipation when applying the test functions to the renormalized Boltzmann
equation.
NAVIER-STOKES-FOURIER
5
In contrast to [10], the family of the approximate eigenfunctions includes two boundary
layers: viscous fluid boundary layer and kinetic Knudsen layer, while in [10], only a fluid
boundary layer was necessary. An other important difference is the fact that here, we do
not need any assumption on the domain. The reason behind it is that the Navier-StokesFourier system has also some dissipation in the temperature equation which is ignored
in the isentropic model. (in particular this dissipation property holds in the case of the
ball).
In the current paper, the kinetic-fluid coupled boundary layers are constructed for the
case that the accommodation coefficient is of the form α = χεβ with 0 ≤ β < 1. We
found that β = 12 is a threshold.
A key role is played by the linearized kinetic boundary layer equation which has been
studied extensively (see [1, 8, 15, 14, 44]) and recently extended by the authors to a
broader class of the collision kernel [24]. Applying the linearized kinetic boundary layer
equation to construct the two layer eigenfunctions is the main novelty of the current
paper. To the best of our knowledge these two layer eigenfunctions are new even in the
applied literature. For instance, Y. Sone, only construct Knudsen layers relative to the
incompressible flow (which then does not require the fluid boundary layer).
The paper is organized as follows: the next section contains preliminary material regarding the Boltzmann equation in a bounded domain. We state the main theorems in
Section 3 which include the weak convergence for the Navier slip boundary and strong
convergence for the Dirichlet boundary. In Section 4, we list some differential geometry
properties of the boundary ∂Ω as a submanifold of RD . Section 5 provides an introduction to the acoustic modes while Section 6 is about the analysis of the kinetic boundary
layer equation whose solvability provides the boundary conditions of the fluid variables.
In Section 7, we presents the constructions of the test functions used in the proof of the
Main Theorem. The proof of the main proposition on the boundary layers is given in
Section 8, 9 and 10. In Section 11, we establish the weak convergence results in the main
theorem. Section 12 contains a proof of the strong convergence for the Dirichlet boundary
case. In the last section, we make some concluding remarks.
2. Boltzmann Equation in Bounded Domain
Here we will introduce the Boltzmann equation in a bounded domain, only so far as to
set our notations, which is essentially that of [3] and [33]. More complete introduction to
the Boltzmann equation can be found in [6, 7, 16, 43].
2.1. Maxwell Boundary Condition. We take Ω, a smooth bounded domain of RD ,
and O = Ω × RD , the space-velocity domain. Let n(x) be the outward unit normal vector
at x ∈ ∂Ω. We denote by dσx the Lebesgue measure on the boundary ∂Ω. We define the
outgoing and incoming sets Σ+ and Σ− at the boundary ∂Ω by
Σ± = {(x, v) ∈ Σ : ±n(x)·v > 0} where Σ = ∂Ω × RD .
(2.1)
Denoted by γF the trace of F over Σ. The boundary condition takes the form of a
balance between the values of the outgoing and incoming parts of the trace of F , namely
γ± F = 1Σ± γF . In order to describe the interaction between particles and the wall, J.-C.
Maxwell [34] proposed in 1879 the following phenomenological law which splits into a
local reflection and a diffuse (or Maxwell) reflection
γ− F = (1 − α)L(γ+ F ) + αK(γ+ F ) on Σ−
(2.2)
6
N. JIANG AND N. MASMOUDI
where α ∈ [0 , 1] is a constant, also called the “accommodation coefficient.” The local
reflection operator L is given by
Lφ(x , v) = φ(x , Rx v) ,
(2.3)
where Rx v = v − 2 [n(x)·v] n(x) is the velocity before the collision with the wall. The
diffuse reflection operator K is given by
√
Kφ(x , v) = 2π φ̃(x)M (v)
(2.4)
where φ̃ is the outgoing mass flux
Z
φ̃(x) =
φ(x , v)v·n(x) dv ,
(2.5)
v·n(x)>0
and M is the following absolute Maxwellian
M (v) =
1
(2π)D/2
exp − 12 |v|2 ,
(2.6)
that corresponds to the spatially homogeneous fluid state with density and temperature
equal to 1 and bulk velocity equals to 0.
We notice that
Z
Z
√
√
v·n(x) 2πM (v) dv =
|v·n(x)| 2πM (v) dv = 1 ,
(2.7)
v·n(x)>0
v·n(x)<0
which expresses the conservation of mass at the boundary. Here we take the temperature
of the wall to be constant and equal to 1.
2.2. Nondimensionalized Form of the Boltzmann Equation. We consider a sequence of renormalized solutions to the rescaled Boltzmann equation
1
ε∂t F + v · ∇x F = B(F , F ) on R+ × O ,
ε
(2.8)
F (0 , x , v) = Fin (x , v) on O ,
γ− F = (1 − α)L(γ+ F ) + αK(γ+ F ) on R+ × Σ− ,
where the measurable function Fin ≥ 0 is the initial distrubution funtion. The positive,
nondimensional parameter ε is the Knudsen number that is the ratio of the mean free
path to the macroscopic length scale. it is supposed to be small, which justifies studying
the limit ε going to 0. The Boltzmann collision operator B acts only on the v argument
of F and is formally given by
ZZ
B(F , F ) =
(2.9)
(F10 F 0 − F1 F )b(ω , v1 − v) dω dv1 ,
SD−1 ×RD
where v1 ranges over RD endowed with its Lebesgue measure dv1 , while ω ranges over
the unit sphere SD−1 = {ω ∈ RD : |ω| = 1} endowed with its rotationally invariant
unit measure dω. The F10 , F 0 , F1 , and F appearing in the integrand designate F (t , x , ·)
evaluated at the velocities v10 , v 0 , v1 and v, respectively, where the primed velocities are
defined by
v10 = v1 − ω[ω · (v1 − v)] , v 0 = v + ω[ω · (v1 − v)] ,
(2.10)
NAVIER-STOKES-FOURIER
7
for any given (ω , v1 , v) ∈ SD−1 × RD × RD . This expresses the conservation of momentum
and energy for particle pairs after a collision, namely,
v + v1 = v 0 + v10 ,
|v|2 + |v1 |2 = |v 0 |2 + |v10 |2 .
(2.11)
The collision kernel b is a positive, locally integrable function. The Galilean invariance of
the collision kernel implies that b has the classical form
b(ω , v) = |v|Σ(|ω · v̂| , |v|) ,
(2.12)
where v̂ = v/|v| and
R Σ is the specific differential cross section. This symmetry implies
that the quantity b(ω , v) dω is a function of |v| only. The DiPerna-Lions theory that
provides the global existence of renormalized solutions requires that b satisfies
ZZ
1
b(ω , v1 − v) dω dv1 = 0
(2.13)
lim
|v|→∞ 1 + |v|2
SD−1 ×K
for any compact set K ⊂ RD . There are some additional assumptions on b needed in
the work of Levermore-Masmoudi [27]. For the convenience of the reader, we list these
assumptions here.
A major role will be played by the attenuation coefficient a, which is defined as
Z
ZZ
a(v) = b̄(v1 − v)M1 dv1 =
b(ω, v1 − v) dωM1 dv1 .
(2.14)
RD
SD−1 ×RD
A few facts about a are readily evident from what we have already assumed. Because
(2.13) holds, one can show that
a(v)
= 0.
(2.15)
lim
|v|→∞ 1 + |v|2
Our second assumption regarding the collision kernel b is that a satisfies a lower bound
of the form
Ca (1 + |v|)α ≤ a(v) ,
(2.16)
for some constant Ca > 0 and α ∈ R. The third assumption is that there exists s ∈ (1 , ∞]
and Cb ∈ (0 , ∞) such that
 1s

s
Z  b̄(v1 − v) a(v1 )M1 dv1  ≤ Cb .
(2.17)
a(v1 )a(v) RD
Another major role in what follows will be played by the linearized around the global
Maxwellian M collision operator L, which is defined by
ZZ
Lg̃ =
(g̃ + g̃1 − g̃ 0 − g̃10 )b(ω , v1 − v) dωM1 dv1 .
(2.18)
SD−1 ×RD
One has the decomposition
1
L = I + K− − 2K+ ,
a
where the loss operator K− and the gain operator K+ are defined by
Z
1
−
g̃1 b̄(v1 − v)M1 dv1 ,
K g̃ =
a
RD
(2.19)
(2.20)
8
N. JIANG AND N. MASMOUDI
1
K g̃ =
a
+
Z
(g̃ 0 + g̃10 )b(ω, v1 − v) dωM1 dv1 .
(2.21)
RD
The fourth assumption regarding the collision kernel b is that
K+ : L2 (aM dv) → L2 (aM dv) is compact .
(2.22)
Combining the gain operator assumption (2.22) and the loss operator assumption (2.17),
we conclude that
1
L : Lp (aM dv) → Lp (aM dv) is Fredholm
(2.23)
a
for every p ∈ (1 , ∞). From this Fredholm property we can define the psuedo-inverse of
L, called L−1 :
L−1 : Lp (a1−p M dv) ∩ Null⊥ (L) → Lp (aM dv) .
(2.24)
Moreover, L−1 is a bounded operator.
The fifth assumption regarding b is that for every δ > 0 there exists Cδ such that b̄
satisfies
b̄(v1 − v)
≤ Cδ (1 + a(v1 ))(1 + a(v)) for every v1 , v ∈ RD .
(2.25)
b̄(v1 −v)
1 + δ 1+|v1 −v|2
It is well known that the null space of the linearized Boltzmann operator L is given by
Null(L) ≡ span{1, v1 , · · · , vD , |v|2 }. Define P as the orthogonal projection from L2 (M dv)
onto Null(L), which is given by
2
P g̃ = hg̃i + v · hg̃i + ( |v|2 − D2 )h( D1 |v|2 − 1)g̃i .
(2.26)
Futhermore, we define P ⊥ = I − P.
The matrix-valued function A(v) and the vector-valued function B(v) are defined by
A(v) = v ⊗ v − D1 |v|2 I ,
B(v) = 12 |v|2 v −
D+2
v
2
.
(2.27)
It is easy to see that each entry of A and B is in L2 (a−1 M dv) ∩ Null⊥ (L). We also
b ∈ L2 (aM dv; RD×D ) and B
b ∈ L2 (aM dv; RD ) which are given by
introduce A
b = L−1 A ,
A
b = L−1 B .
B
(2.28)
For the sake of simplicity, we take the following normalizations
ZZZ
b(ω , v1 − v) dωM1 dv1 dv = 1 ,
SD−1 ×RD ×RD
Z
Z
dω = 1 ,
SD−1
M dv = 1 ,
RD
(2.29)
Z
dx = 1 ,
Ω
associated with the domains SD−1 × RD × RD , SD−1 , RD and Ω respectively, and
ZZ
Fεin dx dv = 1 ,
(2.30)
Ω×RD
associated with the initial data Fεin .
NAVIER-STOKES-FOURIER
9
Because M dv is a positive unit measure on RD , we denote by hξi the average over this
measure of any integrable function ξ = ξ(v),
Z
hξi = ξ(v)M dv .
(2.31)
RD
Moreover, we also use the following average on the boundary
Z
√
hξi∂Ω = ξ(v) [n(x)·v] 2πM dv .
(2.32)
RD
Notice that
h1Σ+ i∂Ω = −h1Σ− i∂Ω = 1 .
(2.33)
Because dµ = b(ω , v1 −v) dωM1 dv1 M dv is a positive unit measure on SD−1 ×RD ×RD , we
denote by hhΞii the average over this measure of any integrable function Ξ = Ξ(ω , v1 , v)
ZZZ
Ξ(ω , v1 , v) dµ .
(2.34)
hhΞii =
SD−1 ×RD ×RD
The measure dµ is invariant under the coordinate transformations
(ω , v1 , v) 7→ (ω , v , v1 ) ,
(ω , v1 , v) 7→ (ω , v10 , v 0 ) .
(2.35)
These, and compositions of these, are called dµ-symmetries.
2.3. Navier-Stokes Scaling. The Navier-Stokes-Fourier system can be formally derived
from the Boltzmann equation through a scaling in which the fluctuations of the kinetic
densities Fε about the absolute Maxwellian M are scaled to be of order ε. More precisely,
we take
Fε = M Gε = M (1 + εgε ) .
(2.36)
Rewriting equation (2.8) for Gε yields
1
ε∂t G + v · ∇x G = Q(G , G ) on R+ × O ,
ε
G (0 , x , v) = Gin
(x , v) on O ,
γ− G = (1 − α)L(γ+ G ) + αhγ+ Gε i∂Ω
(2.37)
on R+ × Σ− ,
where the collision kernel Q is now given by
ZZ
Q(G , G) =
(G01 G0 − G1 G)b(ω , v1 − v)dωM1 dv1 .
(2.38)
SD−1 ×RD
In terms of gε and with the previous notations, the system (2.8) finally reads
ε∂t gε + v · ∇x gε + 1ε Lgε = Q(gε , gε ) on R+ × O ,
gε (0 , x , v) = gεin (x , v) on O ,
γ− gε = (1 − α)L(γ+ gε ) + αhγ+ gε i∂Ω
+
on R × Σ− .
(2.39)
10
N. JIANG AND N. MASMOUDI
2.4. A Priori Estimates. Due to the presence of the boundary, the classical a priori
estimates for the Boltzmann equation, namely the entropy and energy bounds, are modified. First, because all particles arriving at the boundary are reflected or diffused, we
have conservation of mass, which can be written as
Z
Z
hG i dx = hGin
(2.40)
i dx = 1 .
Ω
Ω
Mulplying the equation (2.37) by log(G ) and integrating in x and v, we get formally
Z
ZZ
ε∂t hG log(G ) − G + 1i dx +
(G log(G ) − G + 1)v·n(x) dσx M dv
Ω
Σ
Z
(2.41)
1
=
hlog(G )Q(G , G )i dx .
ε Ω
By denoting h(z) = (1+z) log(1+z)−z, for z > −1 and using that it is a convex function,
we can compute the boundary term in the following way:
ZZ
e
[G log(G ) − G + 1]v·n(x) dσx M dv
Eε (γ+ G ) =
Σ
ZZ
=
[h(εγ+ g ) − h ((1 − α )εγ+ g + α hεγ+ g i∂Ω )] v·n(x) M dvdσx
Σ+
(2.42)
ZZ
≥
[h(εγ+ g ) − (1 − α )h(εγ+ g ) − α h(hεγ+ g i∂Ω )] v·n(x) M dvdσx
Σ+
α
= √ E(γ+ G ) ,
2π
where E(γ+ G ), the so-called Darrozès-Guiraud information, is given by
Z
E(γ+ G ) =
[hh(εγ+ g )i∂Ω − h(εhγ+ g i∂Ω )] dσx .
(2.43)
∂Ω
E(γ+ G ), we
Jensen’s inequality implies that E(γ+ G ) ≥ 0. Noticing that Eeε (γ+ G ) ≥ √α2π
get the entropy inequality
Z t
1e
1
(2.44)
H(G (t)) +
R(G (s)) + Eε (γ+ G (s)) ds ≤ H(Gin
),
2
ε
ε
0
where H(G) is the relative entropy functional
Z
H(G) = hG log(G) − G + 1i dx ,
(2.45)
Ω
and R(G) is the entropy dissipation rate functional
0 0
Z 1
G1 G
0 0
R(G) =
log
(G1 G − G1 G) dx .
4
G1 G
Ω
(2.46)
NAVIER-STOKES-FOURIER
11
2.5. DiPerna-Lions Solutions. We will work in the setting of DiPerna-Lions renormalized solutions. These solutions were initially constructed by DiPerna and Lions [11] over
the whole space RD for any initial data satisfying natural physical bounds. Their result
was then extended to the case of a bounded domain by Hamdache [21] for the case of
specular reflection at the boundary. Mischler [35, 36, 37] developed a general theory to
treat other boundary conditions, namely the case of general Maxwell boundary conditions
(2.2).
The DiPerna-Lions theory does not yield solutions that are known to solve the Boltzmann equation in the usual weak sense. Rather, it gives the existence of a global weak
solution to a class of formally equivalent initial value problems that are obtained by
multiplying (2.37) by Γ0 (Gε ), where Γ0 is the derivative of an admissible function Γ:
1
(ε∂t + v · ∇x )Γ(G ) = Γ0 (G )Q(G , G ) on R+ × O ,
ε
(2.47)
in
G (0 , · , ·) = G ≥ 0 on O .
A function Γ : [0 , ∞) → R is called admissible if it is continuously differentiable and for
some constant CΓ < ∞ its derivative satisfies
√
(2.48)
|Γ0 (z)| 1 + z ≤ CΓ .
Mischler [37] extended DiPerna-Lions theory to domains with a boundary on which the
Maxwell reflection boundary condition is imposed. This required the proof of a so-called
trace theorem that shows thatγGε makes sense. In particular, Mischler showed that γGε
lies in the set of all |v ·n|M dv dσx dt-measurable functions over R+ × ∂Ω that are finite
almost everywhere, which we denote L0 (|v·n|M dv dσx dt).
The weak formulation of the renormalized Boltzmann equation (2.47) is given by
Z
Z
ε hΓ(G (t2 ))Y i dx − ε hΓ(G (t1 ))Y i dx
Ω
Ω
Z
t2
Z
−
t2
Z
hΓ(G )v · ∇x Y i dx dt +
t1
=
Z
1
ε
Z
Ω
t2
t1
hΓ(γG )Y (v · n(x))i dσx dt
t1
Z
(2.49)
∂Ω
hΓ0 (G )Q(G , G )Y i dx dt ,
Ω
for every Y ∈ C ∩ L∞ (Ω̄ × RD ) and every [t1 , t2 ] ⊂ [0 , ∞].
Moreover, the boundary condition is also understood in the renormalized sense:
e
Γ(γ− G ) = Γ (1 − α)L(γ+ G ) + αF
on R+ × Σ− ,
(2.50)
1
where the equality holds almost everywhere and in the sense of distribution.
Proposition 2.1. (Renormalized solutions [37]) Let b satisfy the condition (2.13). Given
any initial data Gin
satisfying
ZZ
2
in
Gin
(2.51)
ε (1 + |v| + | log Gε |) M dv dx < +∞ ,
O
there exists at least one G ≥ 0 in C([0 , ∞); L1 ( M dv dx)) such that (2.49) and (2.50)
hold for all admissible functions Γ. Moreover, G satisfies the following global entropy
12
N. JIANG AND N. MASMOUDI
inequality for all t > 0:
1
H(G (t)) + 2
ε
Z
0
t
1
R(G (s)) ds +
ε
Z
t
Eeε (γ+ G (s)) ≤ H(Gin
).
(2.52)
0
DiPerna-Lions renormalized solutions are not known to satisfy many properties that
one would formally expect to be satisfied by solutions to the Boltzmannn equation. In
particular, the theory does not assert either the local conservation of momentum, the
global conservation of energy, the global entropy equality, or even a local entropy inequality; none does it assert the uniqueness of the solution. Nevertheless, as will be stated
below, it provides enough control to establish the Navier-Stokes limit.
3. Statement of the Main Results
In this section we state our main results. Specifically, we derive the incompressible
Navier-Stokes-Fourier limits with different boundary conditions depending on the quotient
between the accommodation coefficients αε and the Knudsen number ε. More importantly,
the limit in the case of Dirichlet boundary condition is strong because of the instantaneous
damping from the boundary layers.
3.1. Dirichlet Boundary Condition. The main theorem of this paper is the following
strong convergence to the Navier-Stokes-Fourier system with Dirichlet boundary condition
when the accommodation coefficient α is much larger than the Knudsen number ε, i.e.
α
→ ∞ as ε → 0.
ε
Theorem 3.1. (Dirichlet Boundary Condition) Let b be a collision kernel that satisfies
conditions (2.15), (2.16), (2.17), and (2.22). Let Gin
ε be any family satisfying (2.51)
and the renormalization (2.30). Let gεin be the associated family of fluctuations given by
in
in
Gin
ε = 1 + εgε . Assume that the family gε satisfies
and
in 2
H(Gin
ε ) ≤ C ε ,
(3.1)
2
lim hgεin i , hvgεin i , h( |v|D − 1)gεin i = (ρin , uin , θin ) ,
(3.2)
ε→0
in the sense of distributions for some (ρin , uin , θin ) ∈ L2 (dx ; R × RD × R).
Let Gε be any family of DiPerna-Lions renormalized solutions to the Boltzmann equation (2.37) that have Gin
ε as initial values, and the accommodation coefficient α satisfies
α
→ ∞ as ε → 0 .
(3.3)
ε
Then the family of fluctuations gε given by (2.36) is relatively compact in L1loc (dt; L1 (σM dvdx)) .
Every limit point g of gε has the infinitesimal Maxwellian form
g = v·u + 21 |v|2 − D+2
θ,
(3.4)
2
where (u, θ) ∈ C([0, ∞); L2 (dx ; RD ×R))∩L2 (dt ; H 1 (dx ; RD ×R)) with mean zero over Ω,
and it satisfies the Navier-Stokes-Fourier system with Dirichlet boundary condition (1.1),
(1.2), and (1.3), where kinematic viscosity ν and thermal conductivity κ are given by
1
b : LAi
b , κ = 1 hB·L
b Bi
b .
ν=
hA
(3.5)
(D−1)(D+2)
D
The initial data is given by
u0 = Puin ,
θ0 =
D
θin
D+2
−
2
ρin
D+2
.
(3.6)
NAVIER-STOKES-FOURIER
13
Moreover, every subsequence gεk of gε that converges to g as εk → 0 also satisfies
hvgεk i → u in Lploc (dt; L1 (dx; RD )) ,
h( D1 |v|2 − 1)gεk i → θ
in Lploc (dt; L1 (dx; R)) for every
1 ≤ p < ∞.
(3.7)
Furthermore, 1ε P ⊥ gε is relatively compact in w-L1loc (dt; w-L1 (σM dvdx)). For every subsequence εk so that gεk converges to g,
1 ⊥
P gε → 12 A : u ⊗ u + B·uθ + 12 Cθ2
ε
(3.8)
1
1
b
b
− A : ∇x u − B·∇x θ , in w-Lloc (dt; w-L (σM dvdx)) ,
as εk → 0.
Remark: In previous works [18, 19, 27], under the assumptions (3.1) and (3.2), the
convergence to (3.4) and (3.7) are only in w-L1 . As a consequence, the convergence to the
quadratic term (3.8), which is the first correction to the infinitesimal Maxwellian that one
obtains from the Chapman-Enskog expansion with the Navier-Stokes scaling, could not
be obtained. In Theorem 3.1, by showing the acoustic waves are instantaneously damped,
we justify not only the strong convergence to the leading order term g, but also weak
convergence to the next order corrector (3.8).
3.2. Navier Boundary Condition. The second result treats the case of Navier boundary condition. We will make the same assumption as in the previous theorem, but instead
of αε → ∞, we assume that α = χε with χ ≥ 0 or, more generally,
α
→ χ , as ε → 0 .
(3.9)
ε
For this case, although the coupled viscous boundary layer and the Knudsen layer still
have dissipative effect, however, the damping happen in a longer time scale O(1). So,
unlike the Dirichlet boundary condition case, the fast acoustic waves can be damped,
but not instantaneously. Nevertheless, we can show the weak convergence result, thus
generalize the Stokes-Fourier limit of [33] to Navier-Stokes-Fourier limit with slip Navier
boundary condition.
Theorem 3.2. (Navier Boundary Condition) Let b be a collision kernel that satisfies
conditions (2.15), (2.16), (2.17), and (2.22). Let Gin
ε be any family satisfying (2.51)
in
and the renormalization (2.30) and gε be the associated family of fluctuations given by
in
Gin
ε = 1 + εgε . Assume that
in 2
H(Gin
(3.10)
ε ) ≤ C ε ,
and
|v|2
in
in
in
(3.11)
lim hgε i , hvgε i , h( D − 1)gε i = (ρin , uin , θin )
ε→0
in the sense of distributions for some (ρin , uin , θin ) ∈ L2 (dx ; R × RD × R). Let Gε be any
corresponding family of renormalized solutions to the Boltzmann equation (2.37), where
the accommodation coefficients satisfy
α
→ χ , as ε → 0 .
(3.12)
ε
Then the family gε is relatively compact in w-L1loc (dt; w-L1 (σM dvdx)) . Every limit
point g of gε in w-L1loc (dt; w-L1 (σM dvdx)) has the infinitesimal Maxwellian form
g = v·u + 12 |v|2 − D+2
θ,
(3.13)
2
14
N. JIANG AND N. MASMOUDI
where (u, θ) ∈ C([0, ∞); L2 (dx ; RD ×R))∩L2 (dt; H 1 (dx ; RD ×R)) and satisfies the NavierStokes-Fourier system with Navier boundary condition (1.1), (1.2), and (1.4), where kinematic viscosity ν and thermal conductivity κ are given by (3.5), the initial data is given
by (3.6).
Moreover, every subsequence gεk of gε that converges to g as εk → 0 also satisfies
Phvgεk i → u in L1loc (dt ; L1 (dx ; RD )) ,
1
|v|2 − 1)gεk i → θ
h( D+2
in L1loc (dt ; L1 (dx ; R)) .
(3.14)
4. Geometry of the boundary ∂Ω
In this section, we collect some differential geometry properties related to the boundary
∂Ω which can be considered as a (D − 1) dimension compact Riemannian manifold with a
metric induced from the standard Euclidian metric of RD . Let T (∂Ω) and T (∂Ω)⊥ denote
the tangent and normal bundles of ∂Ω in RD respectively. Furthermore, let ∇v be the
covariant derivative along the vector v ∈ RD , and An : T (∂Ω) → T (∂Ω) be the Wengarten
map which is defined as:
∇v n = −An (v) + (∇v n)⊥ ,
for any v ∈ T (∂Ω) .
(4.1)
Define h : T (∂Ω) × T (∂Ω) → T (∂Ω)⊥ the second fundamental form of ∂Ω:
∇v u = (∇v u)tan + h(v, u) ,
for any v, u ∈ T (∂Ω) .
(4.2)
The Wengarten map and the second fundamental form are related by the following
formula:
hAn (u) , vi = hh(u , v) , ni , for any v, u ∈ T (∂Ω) .
(4.3)
T
tan
For the vector field [(∇x u + (∇x u) ) · n] , we can give a more geometric representation.
We have the following formula:
[(∇x u + (∇x u)T ) · n]tan = [∇n u]tan + An (u) .
Proof. Take any a = ai ∂i ∈ T (∂Ω), where ∂i =
∂
∂xi
(4.4)
is the coordinate base of RD . Then
i
i
∂u j
ai ∂x
j n = h∇nj ∂j u , a ∂i i = h∇n u , ai .
(4.5)
ai ∂u
nj = h∇ai ∂i u , ni = hh(a , u) , ui = hAn (u) , ai .
∂xi
(4.6)
j
There is a tubular neighborhood U∂Ω = {x ∈ Ω : dist(x, ∂Ω) < δ} of ∂Ω such that
the nearest point projection map is well defined. More precisely, we have the following
lemma:
Lemma 4.1. If ∂Ω is a compact C k submanifold of dimension D − 1 embedded in RD ,
then there is δ = δ∂Ω > 0 and a map π ∈ C k−1 ({x ∈ Ω : dist(x, ∂Ω) < δ} ; RD ) such that
the following properties hold:
(i): for all x ∈ Ω ⊂ RD with dist(x , ∂Ω) < δ ;
π(x) ∈ ∂Ω ,
⊥
x − π(x) ∈ TΠ(x)
(∂Ω) ,
|x − π(x)| = dist(x , ∂Ω) , and
|z − x| > dist(x , ∂Ω) for any z ∈ ∂Ω \ {π(x)} ;
(ii):
π(x + z) ≡ x ,
for x ∈ ∂Ω , z ∈ Tx (∂Ω)⊥ , |z| < δ ,
NAVIER-STOKES-FOURIER
15
(iii): Let HessΠx denote the Hessian of π at x, then
Hessπ x (V1 , V2 ) = hx (V1 , V2 ) ,
for x ∈ ∂Ω V1 , V2 ∈ Tx (∂Ω) ,
where hx is the second fundamental form of ∂Ω at x.
For the proof of this lemma for even more general submanifold, see [41].
Because ∂Ω is a (D − 1) dimensional manifold, so locally π(x) can be represented as
π(x) = (π 1 (x) , · · · , π D−1 (x)) .
(4.7)
Let d(x) be the distance function to the boundary ∂Ω, i.e.
d(x) = dist(x , ∂Ω) = |x − π(x)| .
(4.8)
It is easy to see that ∇x d is perpendicular to the level surface of the distance function d,
i.e. the set S z = {x ∈ Ω : d(x) = z}. In particular, on the boundary, ∇x d is perpendicular
to S0 = ∂Ω. Without loss of generality, we can normalize the distance function so that
∇x d(x) = −n(x) when x ∈ ∂Ω. By the definition of the projection Π, we have
π(x + t∇x d(x)) = π(x) for t small ,
(4.9)
and consequently, ∇x π α · ∇x d = 0, for α = 1 , · · · , D − 1. In particular, for t small enough,
∇x π α (x) ∈ Tx (∂Ω) when x ∈ ∂Ω.
Next, we calculate the induced Riemannian metric from RD on ∂Ω. In a local coordinate
system, these Riemannian metric can be represented as
g = gαβ dπ α ⊗ dπ β ,
(4.10)
∂π α
where gαβ = h ∂π∂ α , ∂π∂ β i. Noticing that ∂x∂ i = ∂xi ∂π∂ α , and h ∂x∂ i , ∂x∂ j i = δij , the metric gαβ
can be determined by
α ∂π β
gαβ ∂π
= 1.
(4.11)
∂xi ∂xi
5. Acoustic Modes
5.1. Acoustic Operator A. Recall that the Leray’s projection P on the space of divergencefree vector fields and Q on the space of gradients are defined by
P = I − Q,
(5.1)
where Qu = ∇x q and q solves
∆x q = ∇x ·u in Ω ,
Z
∇x q·n = u·n on ∂Ω ,
q dx = 0 .
and
(5.2)
Ω
We define Hilbert spaces
H = U = (ρ , u , θ) ∈ L2 (dx; C × CD × C) ,
Z
2
V= U ∈H:
|∇x U | dx < ∞ ,
(5.3)
Ω
endowed with inner product
Z
hU1 | U2 iH =
Ω
(ρ1 ρ2 + u1 · u2 + D2 θ1 θ2 ) dx ,
(5.4)
16
N. JIANG AND N. MASMOUDI
where f means the complex conjugate of f . Next, we define the acoustic operator A:

  
∇x ·u
ρ
(5.5)
A u = ∇x (ρ + θ) ,
2
θ
∇
·u
D x
over the domain
D(A) = {U = (ρ , u , θ) ∈ V : u·n = 0 on ∂Ω} .
(5.6)
The null space of A and its orthogonal with respect to the inner product (5.4) are
characterized as
Null(A) = {(−ϕ , w , ϕ) ∈ V : ∇x ·w = 0 and w·n = 0 on ∂Ω} ,
(5.7)
and
Null(A)⊥ = {(ρ , u , θ) ∈ V : θ =
2
ρ,u
D
= ∇x φ , for some φ ∈ H 1 (Ω)} ,
(5.8)
respectively. Because Null(A) includes the incompressibility and Boussinesq relations, we
call it incompressible regime. We will see in the next subsection that Null(A)⊥ is spanned
by the eigenspaces of the acoustic operator A, so we call it acoustic regime.
For any U = (ρ , u , θ) ∈ H, we can define Π and Π⊥ the projections to the incompressible
regime Null(A) and acoustic regime Null(A)⊥ respectively as follows:
D
D
2
2
ρ − D+2
θ , Pu , D+2
θ − D+2
ρ ,
ΠU = D+2
(5.9)
2
D
Π⊥ U = D+2
(ρ + θ) , Qu , D+2
(ρ + θ) .
5.2. Eigenspaces of A. The eigenvalues and eigenvectors of the acoustic operator A in
a bounded domain can be constructed from those of the Laplace operator with Neumann
boundary condition in the following way.
D
[λk ]2 , λk > 0 , k ∈ N be the nondecreasing sequence of eigenvalues and Φk the
Let D+2
orthonormal basis of L2 (Ω) functions with eigenvectors of the Laplace operator ∆N with
homogeneous Neumann boundary condition:
−∆x Φk =
D
[λk ]2 Φk
D+2
in Ω,
∇x Φk ·n = 0
on ∂Ω .
(5.10)
More specifically,
0 < λ1 ≤ λ2 ≤ · · · ≤ λk → +∞ ,
as k → ∞ .
The (non-zero) eigenvalues of A read as follows
T
Φk
2
D
Φk , ∇iλxτ,k
, D+2
Φk
,
U τ,k = D+2
(5.11)
(5.12)
and corresponding eigenvectors are U τ,k such that
AU τ,k = iλτ,k U τ,k ,
(5.13)
where τ denotes either + or −, and λτ,k = τ λk .
2D
Note that U τ,k have a norm D+2
under the inner product (5.4), then we have an orthonormal basis of the acoustic modes, i.e.
q
L2
⊥
D+2 τ,k
Null(A) = Span
.
(5.14)
U |k ∈ N , τ = ±
2D
NAVIER-STOKES-FOURIER
17
Furthermore, from U τ,k we can construct infinitesimal Maxwellians g τ,k which are in the
null space of L:
2
Φk
2
D
+ D+2
g τ,k = D+2
Φk + v· ∇iλxτ,k
Φk |v|2 − D2 .
(5.15)
These infinitesimal Maxwellians will be the building blocks of the approximate eigenfunctions of Lε = 1ε L − v·∇x .
5.3. Conditions on Φk . Note that Φk , k ≥ 1 are determined up to a constant. Some
orthogonality condition is required for eigenvalues with multiplicity greater than 1. If
λk = λl and k 6= l, we require that Φk and Φl satisfy one of the following two conditions:
Condition 1:
Z
√
√
2
ν∇xΦk ·∇xΦl + D+2
[λk ]2 κΦk Φl dσx = 0 ,
(5.16)
∂Ω
and
Z
√
k 2
ν|∇xΦ | +
∂Ω
2
[λk ]2
D+2
√
k 2
κ|Φ |
√
Z
dσx 6=
ν|∇xΦl |2 +
∂Ω
2
[λk ]2
D+2
√
κ|Φl |2 dσx
(5.17)
Condition 2:
Z
∇x Φk ·∇x Φl +
∂Ω
and
Z
∂Ω
k 2
|∇xΦ | +
2(D+1) k 2
[λ ] |Φk |2
(D+2)2
2(D+1) k 2 k l
[λ ] Φ Φ
(D+2)2
Z
dσx 6=
∂Ω
|∇xΦl |2 +
dσx = 0 ,
2(D+1) k 2
[λ ] |Φl |2
(D+2)2
(5.18)
dσx .
(5.19)
The above two conditions should be compared with condition (10) in [10]. When the
eigenvalue of −∆N is not simple, these orthogonality conditions will be needed in the
construction of the kinetic-fluid coupled boundary layers. That Φk satisfies conditions
(5.16)-(5.17) or (5.18)-(5.19) depends on the values of β which appears in the accommodation coefficients α = χεβ . When 0 ≤ β < 12 , the condition (5.16) is required, while
when 21 ≤ β < 1, the condition (5.18) is required.
5.4. the operator A − iλτ,k . Next, for each acoustic mode k ≥ 1 and τ = + or −, we
consider operator A−iλτ,k . This operator, especially its pseudo inverse, will be important
in the construction of the boundary layers. Then main difficulty is that the kernel is
nontrivial when then eigenvalues have multiplicity greater than 1. More specifically,
Ker A − iλτ,k = Span U τ,l : λl = λk ,
(5.20)
while its orthogonal space with respect to the inner product (5.4) is
⊥
Ker A − iλτ,k =Span U +,l , U −,l : λl 6= λk ⊕ Span U −τ,l : λl = λk
⊕ Null(A) .
We can define a bounded pseudo inverse of A − iλτ,k by
−1
⊥
⊥
A − iλτ,k
: Ker A − iλτ,k
−→ Ker A − iλτ,k
,
0,0
0,0
0,0
(5.21)
(5.22)
by
−1
δ,l
1
A − iλτ,k U δ,l = iλδ,l −iλ
, for any U δ,l with λl 6= λk ,
τ,k U
−1
−τ,l
1
A − iλτ,k U −τ,l = −2iλ
, for any U −τ,l with λl = λk ,
τ,k U
(5.23)
(5.24)
18
N. JIANG AND N. MASMOUDI
and
−1
A − iλτ,k (ρ, v, −ρ)T = iλ1τ,k (ρ, v, −ρ)T ,
(5.25)
T
for any (ρ, v, −ρ) ∈ Null(A) and τ, δ ∈ {+, −}. It is obvious that this pseudo inverse is
a bounded operator.
D
[λk ]2 ) be the
Lemma 5.1. For each acoustic modes k ≥ 1 and τ ∈ {+, −}, let (Φk , D+2
eigenfunctions and eigenvalues of the Laplace operator ∆N with homogeneous Neumann
boundary condition defined in (5.10). Let U τ,k be defined in (5.12). For given µτ,k , F τ,k
and g τ,k , consider the system for V τ,k = (ρk , vτ,k , θk )T :
(A − iλτ,k )V τ,k = iµτ,k U τ,k + F τ,k ,
(5.26)
vτ,k ·n = g τ,k on ∂Ω .
V τ,k can be determined modulo Ker A − iλτ,k
0,0 under the following two conditions:
τ,k
(i) iµ must satisfy
Z
τ,k
D+2
iµ = 2D
hF τ,k |U τ,k i .
(5.27)
g τ,k Φk dσx − D+2
2D
∂Ω
(ii) F
τ,k
must satisfy the compatibility condition:
Z
g τ,k Φl dσx = hF τ,k |U τ,l i , for λl = λk
with k 6= l .
(5.28)
∂Ω
Remark: Strictly speaking, (5.26) is not rigorous because vτ,k ·n is non-zero, so V τ,k is
not in the domain of A. For notational simplicity, we still use A in (5.26) and later, just
mean the expression of A in (5.5) regardless of the domain.
1
Proof. For any g τ,k ∈ H 2 (∂Ω), there exists ṽ τ,k ∈ H 1 (Ω ; RD ), such that γṽτ,k ·n = g τ,k ,
1
where γ is the usual trace operator from H 1 (Ω ; RD ) to H 2 (∂Ω). We define
Ve τ,k = V τ,k − (0, ṽτ,k , 0)T .
(5.29)
Then Ve τ,k has zero the normal velocity on the boundary ∂Ω, thus is in the domain of A.
From (5.26), Ve τ,k satisfies
(A − iλτ,k )Ve τ,k = −(A − iλτ,k )(0, ṽτ,k , 0) + iµτ,k U τ,k + F τ,k .
(5.30)
⊥
The solvability of (5.30) is that the righthand side is in Ker A − iλτ,k . Inner product
with U τ,k gives (5.27) while inner product with U τ,l with λk = λl , k 6= l gives (5.28).
−1
Under these conditions, by applying the psedu inverse operator A − iλτ,k , we can
⊥
uniquely solve Ve τ,k in Ker A − iλτ,k , denoted by Ve1τ,k . Finally, V τ,k can be represented
as
X
2D
V τ,k = D+2
hV τ,k | U τ,l iU τ,l
λk =λl
+ Ve1τ,k + (0, ṽτ,k , 0)T −
2D
D+2
X
h(0, ṽ τ,k , 0)T | U τ,l iU τ,l .
λk =λl
τ,k
(5.31)
At this stage, the projection of V τ,k on Ker A − iλ , i.e. the first term in the righthand
side of (5.31), can not be determined. It is easy to see that the rest part of V τ,k , which
is the last three terms in (5.31), is uniquely determined, although the lifting of the trace
g τ,k is not unique.
NAVIER-STOKES-FOURIER
19
6. Analysis of the Kinetic Boundary Layer Equation
In this section, we collect three results in kinetic equations which will be frequently
used later in this paper. The first one is a standard result in kinetic theory:
Lemma 6.1. The solvability condition for the linear kinetic equation Lg = f is
for ζ ∈ Span{1 , v , |v|2 } .
hf , ζ(v)i = 0 ,
(6.1)
This result will be used to derive the equations and algebraic relations satisfied by the
fluid variables. The second result we will use is quoted from [3] Lemma 4.4.
b and hB · Bi
b satisfy the following identities:
Lemma 6.2. The components of hA ⊗ Ai
b kl i = ν δik δjl + δil δjk − 2 δij δkl ,
hAij ⊗ A
D
(6.2)
D+2
b
κδij ,
hBi Bj i =
2
where ν and κ are given by (3.5).
The next result is about the linear boundary layer equation which will be used to
determine the boundary conditions of the fluid variables.
Lemma 6.3. Considering the following linear kinetic boundary layer equation in half
space:
(v·n)∂ξ g bb + Lg bb = K bb , in ξ > 0 ,
(6.3)
g bb −→ 0 , as ξ → ∞ ,
with boundary condition at ξ = 0
γ+ g bb = Lγ− g bb + hbb ,
on
ξ = 0,
v · n > 0,
(6.4)
In the above equations, the boundary source term hbb is taken the following form:
hbb = − [γ+ g − Lγ− g] + χ [hγ− f i∂Ω − Lγ− f ] ,
(6.5)
where g and f are of the forms:
g =ρg + ug · v + θg
|v|2
2
−
D
2
(6.6)
b + ∂ζ θb n · B)
b + (∇x uint : A
b + ∇x θint · B)
b + Sg ,
−(∂ζ ub ⊗ n : A
and
f = ρf + uf · v + θf
|v|2
2
−
D
2
+ Sf ,
(6.7)
and where Sg ∈ Null(L)⊥ and Sf are source terms.
Then there exists a solution g bb of the equation (6.3) if the following the boundary
conditions are satisfied by the fluid variables:
(i) On the boundary ∂Ω, the normal components of velocity is
Z ∞
ug · n =
hK bb i dξ .
(6.8)
0
20
N. JIANG AND N. MASMOUDI
(ii) On the boundary ∂Ω, the tangential components of velocities and temperature satisfy
Z
tan
b tan
tan
tan
int
ν ∂ζ u
=χ [uf ] + 2νd(u ) · n
−χ
[hγ− Sf i∂Ω − Lγ− Sf ] (v·n)vM dv
v·n>0
Z ∞
tan
+ h(v·n)vSg i −
hK bb vitan dξ ,
0
κ∂ζ θ
b
χ
+ κ∂n θ − 2(D+2)
uf · n
Z
1
[hγ− Sf i∂Ω − Lγ− Sf ] (v·n)|v|2 M dv +
− D+2
Z ∞ v·n>0
|v|2
− 1)i dξ ,
−
hK bb ( D+2
int
χθf
= D+1
D+2
1
h(v·n)|v|2 Sg i
D+2
0
tan
where kinematic viscosity ν and thermal conductivity κ are given by (3.5), u
the tangential components of the vector u.
(6.9)
denotes
Proof. The solvability conditions of the linear boundary layer equation (6.3) with boundary condition (6.4) are given by
Z
Z ∞
bb
h η(v · n)M dv = −
hK bb ηi dξ ,
(6.10)
v·n>0
0
for all η(v) ∈Null(L) satisfying the condition:
η(Rx v) = η(v) .
(6.11)
It is obvious that 1 and |v|2 satisfy (6.11). We assume η(v) =
also satisfy (6.11) which requires that
(v · n)
D
X
PD
1
ai ni = 0 ,
ai vi with v ·n > 0
(6.12)
i=1
which implies that the vector a = (a1 , · · · , aD )T is perpendicular to the outer normal
vector n.
The formula (6.8) can be derived by taking η = 1 in (6.10). Simple calculations show
that
Z
hbb (v)(v · n)M dv = −hvγgi · n
v·n>0
(6.13)
= −ug · n − hvSg i · n .
Note that Sg ∈Null(L)⊥ , hence (6.8) follows.
P
To prove (6.9), take η = D
1 ai vi in (6.10).
Z
Z
bb
h (ai vi )(nj vj )χM dv = −
v·n>0
∞
hK bb (ai vi )i dξ ,
(6.14)
[γ+ g − Lγ− g] (ai vi )(nj vj )M dv = hγgAij iai nj .
(6.15)
0
which implies that
Z
v·n>0
NAVIER-STOKES-FOURIER
Using the definition of the viscosity ν in (3.5) and Lemma 6.2, we have
Z
[γ+ g − Lγ− g] (ai vi )(nj vj )M dv
v·n>0
= − ν[∂ζ ub ] · a + ν (∇x uint + (∇x uint )T ) · n · a + hSg Ani · a .
Take η = |v|2 , we have
Z
[γ+ g − Lγ− g] (v · n)|v|2 M dv = 2hγgBi · n + (D + 2)h(v · n)γgi .
21
(6.16)
(6.17)
v·n>0
Using the definition of the thermal conductivity κ in (3.5) and Lemma 6.2, we have
Z
[γ+ g − Lγ− g] (v · n)|v|2 M dv
(6.18)
v·n>0
b
int
2
= − (D + 2)κ∂ζ θ + (D + 2)κ∇x θ · n + (D + 2)ug · n + h(v · n)|v| Sg i .
It can be calculated that
hγ− f i∂Ω − Lγ− f =
D+1
θf
2
√
2π
uf
2
−
· n − v·(Rx uf ) − 12 θf |v|2
+ hγ− Sf i∂Ω − Lγ− Sf ,
(6.19)
Where Rx uf = uf − 2(uf · n)n is the reflection of uf with respect to the normal n(x).
Thus
Z
[hγ− f i∂Ω − Lγ− f ] (ai vi )(v · n)M dv
v·n>0
Z
(6.20)
1
= − √2π (Rx uf ) · a +
[hγ− Sf i∂Ω − Lγ− Sf ] A · nM dv · a ,
v·n>0
and
Z
[hγ− f i∂Ω − Lγ− f ] v1 |v|2 M dv
v·n>0
Z
1
D+1
[hγ− Sf i∂Ω − Lγ− Sf ] (v · n)|v|2 M dv .
= 2 uf · n − √2π θf +
(6.21)
v·n>0
As showed in (14.2), the vector a is perpendicular to n. Thus the resulting vector of
inner product with a is the tangential part. Finally, noticing [Rx uf ]tan = [uf ]tan , we then
finish the proof of the Lemma 6.3.
Corollary 6.1. Considering the same linear kinetic boundary layer equation (6.3), with
the boundary condition:
γ+ g bb − Lγ− g bb = χ hγ− g bb i∂Ω − Lγ− g bb + hbb , on ξ = 0 , v · n > 0 ,
(6.22)
where hbb is given the same as in (6.5), (6.6) and (6.7). Then to solve g bb , only one
solvability condition is needed: On the boundary ∂Ω, the normal components of velocity is
Z ∞
hK bb i dξ ,
(6.23)
ug · n =
0
which is the same as (6.8). Furthermore, under this condition, on the boundary ∂Ω,
the tangential components of velocities and temperature satisfy (6.9) with replacing f by
f + g bb .
22
N. JIANG AND N. MASMOUDI
Proof. For equation (6.3) with boundary condition (6.22) we need only one solvability
condition, i.e. η = 1 in (6.10). Thus, the boundary condition for the normal direction
(6.8) isderived as before. Under
this solvability condition, g bb is solved. Note that the
bb
bb
term χ hγ− g i∂Ω − Lγ− g
has the same form as in the source term χ [hγ− f i∂Ω − Lγ− f ].
Thus, we can put g bb into the term Sf . Thus, for this case, the fluid boundary conditions
are the same as before, except that the source term Sf includes g bb .
7. Approximate Eigenfunctions-Eigenvalues
7.1. Motivation. We define the operators Lε and L∗ε as
1
1
Lε = L − v·∇x ,
L∗ε = L + v·∇x .
ε
ε
∗
Formally, Lε and Lε are “dual” in the following sense:
hL∗ε g ∗ |gi = hg ∗ |Lε gi ,
(7.1)
(7.2)
provided that g ∗ satisfies the Maxwell reflect boundary condition
γ− g ∗ = (1 − α)L(γ+ g ∗ ) + αhγ+ g ∗ i∂Ω
on Σ− ,
(7.3)
and g satisfies the dual boundary condition
γ+ g = (1 − α)L(γ− g) + αhγ− gi∂Ω
on Σ+ .
(7.4)
If gε is the fluctuation defined in (2.36), then gε obeys the scaled Boltzmann equation
(2.39) in which L∗ε gε appears with gε satisfies the boundary condition (7.3). Then from
(7.2), Lε gεBL appears in the weak formulation of the Boltzmann equation if gεBL is a test
function. Thus, it is natural to construct approximate eigenfunctions and eigenvalues of
Lε , satisfying the dual boundary condition (7.4). Specifically, we consider the kinetic
eigenvalue problem:
BL
Lε gεBL = −iλBL
(7.5)
ε gε ,
BL
with gε satisfying the dual Maxwell boundary condition (7.4), where the accommodation
coefficient α takes the value α = χεβ , with β ∈ [0, 1).
To solve the eigenvalue problem (7.5) and (7.4), we have two key observations. First,
the solutions must include interior term and
√ two boundary layer terms: the fluid viscous boundary layer terms with thickness ε, and the kinetic Knudsen layer term with
thickness ε. Second, the fluid-kinetic coupled boundary layers depend on the values of β.
Specifically, the constructions of the boundary layers are different for the cases 0 ≤ β < 21
and 12 ≤ β < 1. Thus, we make the ansatz for gεBL and λBL
ε :
i m 1
Xh
d(x)
d(x)
int
b
bb
BL
√
gε (x, v) =
gm,n (x, v) + gm,n (π(x), ε , v) + gm,n (π(x), ε , v) ε 2 +| 2 −β|n , (7.6)
m,n
and
=
λBL
ε
X
m
1
λm,n ε 2 +| 2 −β|n ,
(7.7)
m,n
where the summations are taken over the values of m and n such that m2 + | 12 − β|n ≥ 0.
int
In the above ansatz, gεBL is consisted of three types of terms: the interior terms gm,n
b
bb
, the fluid viscous boundary layer terms gm,n
, and the kinetic Knudsen layer terms gm,n
.
They are coupled through the boundary condition (7.4), which depends on the value of
β, and β = 21 is the threshold.
NAVIER-STOKES-FOURIER
23
b
bb
Each gm,n
and gm,n
are defined in U∂Ω , the δ∂Ω tubular neighborhood of ∂Ω in Ω, where
δ∂Ω is the small
√ number defined in Lemma 4.1, the projection π is defined in (4.7). After
rescaling by ε and ε respectively,
b
bb
gm,n
, gm,n
: ∂Ω × R+ × RD −→ R .
(7.8)
b
bb
b
bb
Both gm,n
and gm,n
will vanish in the outside of U∂Ω . Thus gm,n
and gm,n
are required to
√ and
be rapidly decreasing to 0 in the ζ and ξ respectively, which are defined by ζ = d(x)
ε
ξ=
d(x)
.
ε
7.2. Statement of the Proposition. Now we state the proposition which is the key
part of this paper. It can be considered as a kinetic analogue of the Proposition 2 in [10].
Proposition 7.1. Let Ω be a C 2 bounded domain of RD . Then, for every α = χεβ ,
β ∈ [0, 1), every acoustic mode k ≥ 1, non-negative integer N , and each τ ∈ {+ , −},
τ,k
there exists approximate eigenfunctions gε,N
and eigenvalues −iλτ,k
ε,N of Lε , and error
τ,k
τ,k
terms Rε,N and rε,N respectively, such that
τ,k
τ,k
τ,k
Lε gε,N
= −iλτ,k
ε,N gε,N + Rε,N ,
(7.9)
τ,k
and gε,N
satisfy the approximate dual Maxwell boundary condition:
h
i
τ,k
τ,k
τ,k
τ,k
τ,k
γ+ gε,N − Lγ− gε,N = α hγ− gε,N i∂Ω − Lγ− gε,N + rε,N
on Σ+ .
(7.10)
Moreover, iλτ,k
ε,N has the following expansions (depending on the value of β):
τ,k √
τ,k
iλτ,k
=
iλ
+
iλ
ε + O(ε1−β ) ,
1
ε,N
with Re(iλτ,k
1 ) < 0
2
(7.11)
2
when 0 ≤ β < 12 ,
τ,k
β
iλτ,k
+ iλτ,k
ε,N = iλ
β ε + O(ε) ,
with Re(iλτ,k
β ) < 0
when 21 ≤ β < 1.
Furthermore, for all 1 < r, p ≤ ∞, we have error estimates:
√
N −1
τ,k
ε
kRε,N
kLr (dx,Lp (a1−p M dv)) = O
,
(7.12)
(7.13)
and
τ,k
kgε,N
−
τ,k,int
g0,0
kLr (dx,Lp (a1−p M dv))
τ,k
τ,k,int
kgε,N
− g0,0
kLr (dx,Lp (a1−p M dv))
1
2r
, for 0 ≤ β < 12 ,
=O ε
1
1
= O εβ− 2 + 2r , for 12 ≤ β < 1 ,
τ,k,int
where g0,0
is defined in (5.15). We also have the boundary error estimates:
N 1 τ,k
krε,N
kLr (dσx ,Lp (a1−p M dv)) = O ε 2 +| 2 −β| , for 0 < β < 1 ,
τ,k
and rε,N
= 0, for β = 0.
(7.14)
(7.15)
24
N. JIANG AND N. MASMOUDI
int
7.3. Main Idea of the Proof. For each (m, n), gm,n
is the summation of the hydrodyint
int
namic parts Pgm,n , i.e. their projections on Null(L), and the kinetic parts P⊥ gm,n
, the
⊥
projections on Null(L) . The hydrodynamic part is denoted by
2
|v|
int
int
int
int
D
Pgm,n = ρm,n + v·um,n + 2 − 2 θm,n
,
(7.16)
int
int
int T
int
and we call Um,n
= (ρint
m,n , um,n , θm,n ) the fluid variables of gm,n . It can be shown that
0
0
int
int
the coefficients of P⊥ gm,n
are in terms of Um
0 ,n0 and their derivatives for (m , n ) < (m, n)
int
which means m0 ≤ m, n0 ≤ n and m0 + n0 < m + n. Thus, we need only to solve Um,n
int
b
for all (m, n) to determine gm,n
. Similar notations will be used for gm,n
, and also we need
b
only to solve Um,n .
We put the ansatz (7.6) and (7.7) into the equation (7.5), then collect the same order
int
terms. For all 0 ≤ β < 1, the leading order of the kinetic-fluid boundary layer g0,0
is
int
int
hydrodynamic, which means g0,0 is completely determined by U0,0 . we can derive that
int
U0,0
satisfies the equation
int
int
AU0,0
= iλ0,0 U0,0
.
(7.17)
For (7.17) we can consider the following two cases:
int
int
• Case 1: λ0,0 = 0, which implies that ρint
0,0 + θ0,0 = 0 and ∇x ·u0,0 = 0. Starting from
here, we can construct the boundary layer gεBL which we call the boundary layer
in the incompressible modes;
• Case 2: by comparing the equations (7.17) and (5.13), iλ0,0 is an eigenvalue of the
int
acoustic operator A, i.e. λ0,0 = λτ,k and U0,0
= U τ,k , where k ≥ 0 is the acoustic
modes, and τ denotes either + or −. Starting from here, we can construct the
boundary layer gεBL which we call the boundary layer in the acoustic modes.
Because the main goal of the current paper is about the interaction between the acoustic
waves and the boundary layers, we only consider the kinetic-fluid boundary layers in the
acoustic modes for each k ≥ 0 and each τ ∈ {+, −}. Consequently, we add superscript τ, k
for each term in the ansatz. We will consider the boundary layers in the incompressible
modes in the forthcoming paper [25].
The basic strategy to solve all terms in the ansatz (7.6) and (7.7) is the following:
τ,k,bb
(1) gm,n
satisfies the linear kinetic boundary layer equation (6.3). Applying Lemma
τ,k,bb
6.3, the solvability conditions for gm,n
give the boundary conditions (both normal
τ,k,int
τ,k,b
and tangential directions) of Um0 ,n0 and Um
00 ,n00 . The relations between the pairs
0
0
00
00
(m, n), (m , n ) and (m , n ) depend on the values of β;
τ,k,int
(2) Um,n
and iλτ,k
m,n can be solved by applying Lemma 5.1, where only the boundary
condition in the normal direction uτ,k,int
m,n ·n is needed;
(3) If the multiplicity of the eigenvalue λτ,k
0,0 is greater than 1, applying Lemma 5.1,
τ,k,int
τ,k,int
Um,n can be only solved modulo Ker(A − iλτ,k
in
0,0 ). The components of Um,n
τ,k
Ker(A − iλ0,0 ) will be solved by applying Lemma (5.1) again in later steps;
τ,k,b
τ,k,b
(4) uτ,k,int
m,n ·n = −um,n ·n + some known terms, while um,n ·n can be derived from the
τ,k,b
ordinary differential system of Um−1,n
;
(5) To solve the ordinary differential system in (3), the boundary conditions on the
τ,k,b
tangent direction for Um,n
are needed. These boundary conditions are coupled
τ,k,int
with Um0 ,n0 , and these coupling relations are given in (1);
NAVIER-STOKES-FOURIER
25
Remark: For the cases 0 ≤ β < 12 and 21 ≤ β < 1, the constructions of the boundary
layers have some difference. That is because that in (1) and (5) above, the dependance
of (m, n), (m0 , n0 ) and (m00 , n00 ) are different for these two cases. We will explain in more
details in the later sections.
τ,k,int
Remark: To apply Lemma 5.1 Um,n
, we require a priori the following orthogonality for
1
1
both cases 0 ≤ β < 2 and 2 ≤ β < 1.
τ,l,int
τ,k,int
hUm,n
|U0,0
i = 0,
for all (m, n) 6= (0, 0) or k 6= l .
8. Proof of Proposition 7.1: the case 0 < β <
(7.18)
1
2
In this section, we construct the kinetic-fluid boundary layers corresponding to the
the
accommodation coefficient α = χεβ for 0 < β < 12 . We make for gεBL and λBL
ε
following ansatz:
i m 1
X h
d(x)
τ,k,int
τ,k,b
√ , v) + g τ,k,bb (π(x),
gεBL =
,
v)
ε 2 +( 2 −β)n ,
gm,n
(x, v) + gm,n
(π(x), d(x)
m,n
ε
ε
m≥0,n≥−1
(8.1)
and
λBL
=
ε
X
m
1
+( 2 −β)n
2
λτ,k
.
m,n ε
(8.2)
m,n≥0
In the above ansatz, we require that
τ,k,b
τ,k,bb
τ,k,int
= 0 for m ≥ 1, g0,n
= 0 for all n ≥ 0;
= 0, gm,−1
• gm,−1
τ,k,b
τ,k,b
• g0,n are hydrodynamic, i.e. Lg0,n = 0;
• The orthogonality condition (7.18) holds.
We can use the following infinite matrix to represent the index
(8.1):

.
(0, 0) −→ (0, 1) −→ (0, 2) −→ (0, 3) −→

↓
↓
↓
↓

(1, −1)bb −→ (1, 0) −→ (1, 1) −→ (1, 2) −→ (1, 3) −→

↓
↓
↓
↓
↓


(2, −1)bb −→ (2, 0) −→ (2, 1) −→ (2, 2) −→ (2, 3) −→

↓
↓
↓
↓
↓

(3, −1)bb −→ (3, 0) −→ (3, 1) −→ (3, 2) −→ (3, 3) −→


↓
↓
↓
↓
↓

 · · · −→
· · · −→
· · · −→
· · · −→
· · · −→
↓
↓
↓
↓
↓
(m, n) in the ansatz

· · · −→
↓ 

· · · −→

↓ 

· · · −→
.
↓ 
· · · −→

↓ 

· · · −→
↓
(8.3)
τ,k,int
In the above matrix, (m, −1)bb means that for the index (m, −1), there is no gm,−1
and
τ,k,b
gm,−1 terms. The element (m, n) without super-index denotes that there are corresponding
τ,k,int
τ,k,b
τ,k,bb
gm,n
, gm,n
and gm,n
in the ansatz (8.1).
Remark: For the case 0 < β < 21 , the subscript (m, n) stands for m2 + ( 12 − β)n in the
ansatz (8.1). If β is a rational number, then for a given rational number q, it is possible
that there are different pairs (m, n) so that q = m2 + ( 21 − β)n. But it does not affect
the construction of the approximate eigenfunction-eigenvalues gετ,k , λτ,k
ε if they are in the
q
same order ε .
26
N. JIANG AND N. MASMOUDI
8.1. Double Induction. To solve all terms in the ansatz (8.1) and (8.2), we use a double
induction argument, i.e. we do induction for both m and n, which we call m-th row and
n-th column. For the case 0 < β < 21 , the order of induction for m and n is the following:
we start from n = 0, make induction for m. After we finish the induction for the 0-th
column, we move to the next column, i.e. n = 1, then make induction for m, and so on,
thus solve all terms. When we solve n-th column, we need the information from (n − 1)-th
column, and save some information to be used for the (n + 1)-th column. In particular,
when the multiplicity of the eigenvalues iλk0,0 is greater than 1, the induction will be even
more complex. The details will be given in the following subsections.
8.2. (m, n) = (0, 0). The inductions start from (m, n) = (0, 0). We formally plug the
ansatz (8.1) into the equation (7.5) and then collect the terms of the same order of ε in
the interior, the viscous boundary layer, and the Knudsen layer respectively.
We will frequently use the following calculations:
√ ) = (v·∇x π α )∂Πα g b +
v·∇x g b (π(x), d(x)
ε
√1 (v·∇x d)∂ζ g b
ε
.
(8.4)
τ,k,int
τ,k,b
First we solve g0,0
, g0,0
and iλτ,k
0,0 . It includes the following five steps:
τ,k,b
k,b
Step 0: The values of u0,0 · ∇x d and ρk,b
0,0 + θ0,0 .
The terms of order O(ε−1 ) in the interior and viscous boundary layers in the equation
(7.5) yield
τ,k,int
τ,k,b
Lg0,0
= 0 and Lg0,0
= 0,
τ,k,int
τ,k,b
which imply that g0,0
and g0,0
are hydrodynamic, i.e.
τ,k,int
τ,k,int
g0,0
= ρk,int
+
0,0 + v·u0,0
and
τ,k,b
τ,k,b
g0,0
= ρk,b
0,0 + v·u0,0 +
|v|2
2
|v|2
2
−
−
D
2
D
2
k,int
,
θ0,0
k,b
θ0,0
.
τ,k,int
τ,k,b
In the above expressions of g0,0
and g0,0
, the fluid variables are to be determined.
We need to know the equations satisfied by the corresponding fluid variables and their
boundary conditions.
1
The order O(ε− 2 ) in the viscous boundary layer reads
τ,k,b
τ,k,b
Lg1,0
= (v·∇x d)∂ζ g0,0
|v|2
k,b
τ,k,b
k,b
k,b
=∇x d⊗∂ζ uτ,k,b
0,0 : A + ∂ζ θ0,0 ∇x d·B + ∂ζ u0,0 ·∇x d D + ∂ζ (ρ(0,0) + θ0,0 )∇x d·v .
(8.5)
Lemma 6.1 implies that the solvability conditions for equation (8.5) are:
k,b
∂ζ (ρk,b
0,0 + θ0,0 )∇x d = 0 ,
(8.6)
∂ζ (uτ,k,b
0,0 ·∇x d) = 0 , .
(8.7)
k,b
k,b
k,b
From (8.6), we deduce that ρk,b
0,0 + θ0,0 is constant in ζ. Since ρ0,0 + θ0,0 → 0 as ζ → ∞,
then
k,b
ρk,b
(8.8)
0,0 + θ0,0 = 0 .
Similarly, we have
uτ,k,b
(8.9)
0,0 ·∇x d = 0 ,
which implies that on the boundary ∂Ω,
uτ,k,b
0,0 (x, ζ = 0)·n = 0 .
(8.10)
NAVIER-STOKES-FOURIER
27
τ,k,b
Under these conditions, g1,0
can be expressed as
τ,k,b
k,b
b
b
g1,0
= ∇x d⊗∂ζ uτ,k,b
0,0 : A + ∂ζ θ0,0 ∇x d· B
h
τ,k,b
+ ρk,b
1,0 + v·u1,0 +
|v|2
2
−
D
2
k,b
θ1,0
i
.
(8.11)
Step 1: The boundary value of uτ,k,int
· n.
0,0
k,bb
0
Since g0,0 = 0, the order O(ε ) in (7.4) gives the boundary conditions
τ,k
τ,k
γ+ g̃0,0
− γ− Lg̃0,0
= 0,
(8.12)
τ,k,int
τ,k,b
τ,k
τ,k,int
τ,k,b
where we use the notation g̃m,n
= gm,n
+ gm,n
. Because g0,0
and g0,0
are hydrodynamic, then from then formula (6.8) in Lemma 6.3, (8.12) implies that on the boundary
∂Ω,
τ,k,int
u0,0 + uτ,k,b
· n = 0.
(8.13)
0,0
Combining (8.13) with the condition (8.10), we deduce that on the boundary ∂Ω,
uτ,k,int
·n = 0 .
0,0
(8.14)
τ,k,int
Step 2: Solve iλτ,k
.
0,0 and U0,0
τ,k,int
τ,k,int k,int T
To find the equation satisfied by U0,0
= (ρk,int
, θ0,0 ) , we consider the order
0,0 , u0,0
0
O(ε ) in the interior part in (7.4):
τ,k,int
τ,k,int
τ,k,int
Lg2,0
= v·∇x g0,0
− iλτ,k
.
0,0 g0,0
(8.15)
τ,k,int
Using the boundary condition (8.14) which means that U0,0
is in the domain of the
acoustic operator A, the solvability of (8.15) gives
τ,k,int
τ,k,int
AU0,0
= iλτ,k
.
0,0 U0,0
(8.16)
τ,k,int
For the system (8.16), if iλτ,k
is in the incompressible regime. This
0,0 = 0, then U0,0
τ,k
case will be treated in a separate paper [25]. If iλ0,0 6= 0, then from (5.13), we know that
τ,k,int
U0,0
are eigenvectors of A, and can be constructed from the eigenvectors of the Laplace
operator with Neumann boundary condition, i.e.
τ,k,int
τ,k
,
= U0,0
U0,0
k
and λτ,k
0,0 = τ λ ,
k = 1, 2, · · ·
(8.17)
where τ denotes either + or −, see the discussion of the spectrum of A in the section 5,
in particular (5.10), (5.12) and (5.13)
Remark: (8.17) is the building-block of the construction of the approximate eigenvector
τ,k
gε,N
and eigenvalues λτ,k
ε,N for any N ≥ 1 and any fixed acoustic mode k ≥ 1.
τ,k,int
Under solvability conditions (8.17), we can represent g2,0
as
τ,k,int
k,int b
b
g2,0
= ∇x uτ,k,int
: A(v)
+ ∇x θ0,0
· B(v)
0,0
h
i
τ,k,int
k,int |v|2
D
+ ρk,int
+
u
·v
+
θ
(
−
)
,
2,0
2,0
2,0
2
2
where uτ,k,int
=
0,0
∇x Φk
iλτ,k
0,0
k,int
and θ0,0
=
2
Φk , and Φk is defined
D+2
k,b
tan
.
values of [uτ,k,b
and θ0,0
0,0 ]
(8.18)
in (5.10).
Step 3: The boundary
k,b
The boundary conditions for the tangential components of uτ,k,b
0,0 and θ0,0 can be derived
τ,k,bb
from the solvability conditions of g1,−1
which satisfies the linear boundary layer equation
28
N. JIANG AND N. MASMOUDI
τ,k
(6.3) with g = 0, f = g̃0,0
, Sf = 0 and K bb = 0. Then from (6.9) of Lemma 6.3, we
deduce that on the boundary ∂Ω,
tan
[uτ,k,b
= −[uτ,k,int
]tan
0,0 ]
0,0
k,b
k,int
and θ0,0
= −θ0,0
.
(8.19)
k,b
tan
From (8.19), we can deduce the boundary conditions for [uτ,k,b
and θ0,0
because
0,0 ]
τ,k,int
k,int
u0,0 and θ0,0 are already solved in Step 2, thus their boundary conditions are known.
k,b
tan
Step 4: Solve [uτ,k,b
and θ0,0
.
0,0 ]
τ,k,b tan
k,b
The equations satisfied by [u0,0 ] and θ0,0
can be found by considering the order
0
O(ε ) of the viscous boundary layer in (8.1):
τ,k,b
τ,k,b
τ,k,b
τ,k,b
Lg2,0
= v·∇x d∂ζ g1,0
+ (v·∇x Πα )∂Πα g0,0
− iλτ,k
0,0 g0,0 .
(8.20)
Lemma 6.1 implies that the projection of the righthand side of (8.20) must be in the
τ,k,b
null space of L. We first use the expression (8.11) of g1,0
to calculate the projection of
τ,k,b
v·∇x d∂ζ g1,0 onto Null(L).
|v|2
D
b
It is easy to see that components of P(∂i d∂m d∂ζ2 (uτ,k,b
0,0 )l vi Aml ) on 1 and 2 − 2 are
zeros. Applying Lemma 6.2, we get
2 τ,k,b
b
b
P ∂i d∂m d∂ζ2 (uτ,k,b
)
v
A
l i ml = ∂i d∂m d∂ζ (u0,0 )l hvi vj Aml ivj
0,0
2
= ν∂i d∂m d∂ζ2 (uτ,k,b
0,0 )l δim δjl + δil δjm − D δij δml vj
i
h
(8.21)
τ,k,b
2
2 2 τ,k,b
2
= ν|∇x d| ∂ζ u0,0 + ν(1 − D )∂ζ u0,0 · ∇x d ∇x d · v
= ν|∇x d|2 ∂ζ2 uτ,k,b
0,0 · v .
k,b b
Similarly, the components of P(∂i d∂j d∂ζ2 θ0,0
vi Bj ) on 1 and v are zeros. Then applying
Lemma 6.2 again, we get again
2
k,b b
k,b
b j i |v|2 − D
P ∂i d∂j d∂ζ2 θ0,0
vi Bj = ∂i d∂j d∂ζ2 θ0,0
hvi ( |v|D − 1)B
2
2
(8.22)
2 k,b
2 |v|2
D+2
D
= D κ∂ζ θ0,0 |∇x d| 2 − 2 ,
where kinematic viscosity ν and thermal conductivity κ are given by (3.5). The above
calculations give the solvability conditions for (8.20):
k,b
τ,b
τ,k,b
α
iλτ,k
0,0 ρ0,0 =∂ζ (u1,0 ·∇x d) + ∂Πα u0,0 ·∇x Π ,
τ,k,b
k,b
k,b
2 τ,k,b
2
iλτ,k
0,0 u0,0 =∂ζ (ρ1,0 + θ1,0 )∇x d + ν∂ζ u0,0 |∇x d| ,
τ,b
τ,k,b
k,b
α
2
2
iλτ,k
0,0 θ0,0 = D ∂ζ (u1,0 ·∇x d) + D ∂Πα u0,0 ·∇x Π +
(8.23)
k,b
D+2
κ∂ζ2 θ0,0
|∇x d|2
D
.
The above equations can be written as
τ,k,b
τ,k,b
τ,k,b
τ,k,b
iλτ,k
= Ab U1,0
+ AΠ U0,0
+ Db U0,0
,
0,0 U0,0
where the operators Ab , AΠ and Db are defined as




∂ζ (ũ·∇
d)
x
∂Πα ũ·∇x Πα


Ab Ũ = ∇x d∂ζ ρ̃ + θ̃  , AΠ Ũ = ∂Πα (ρ̃ + θ̃)∇x Πα 
2
2
∂ α ũ·∇x Πα
∂ (ũ·∇x d)
D Π
D ζ
(8.24)
(8.25)
NAVIER-STOKES-FOURIER
29
and

0
2
Db Ũ = ν∂ζ2 ũ|∇x d|2 + ν(1 − D )∇x d·∂ζ2 ũ∇x d ,
D+2
κ∂ζ2 θ̃|∇x d|2
D

(8.26)
where Ũ = (ρ̃ , ũ , θ̃)T . Using (8.9) and taking the inner product of the second equation
in (8.23) with ∇x d, we have
k,b
∂ζ (ρk,b
1,0 + θ1,0 ) = 0 ,
k,b
hence ρk,b
1,0 + θ1,0 = 0 .
(8.27)
Thus taking the inner product of the second equation in (8.23) with ∇x π α gives the
α
equation for uτ,k,b
0,0 ·∇x π :
τ,k
τ,k,b
α
α
ν|∇x d|2 ∂ζ2 uτ,k,b
·∇
π
=
iλ
u
·∇
π
,
x
x
0,0
0,0
0,0
τ,k,int
α
uτ,k,b
(x)·∇x π α ,
0,0 (x, ζ = 0)·∇x π = −u0,0
lim
ζ→∞
α
uτ,k,b
0,0 ·∇x π
(8.28)
= 0.
α
The boundary condition in the second line of (8.28) follows from the fact that uτ,k,b
0,0 ·∇x π
α
is the tangential components of uτ,k,b
0,0 because ∇x π is tangential to ∂Ω, see the arguments
after (4.9).
Using the relation (8.8), the third equation minus D2 of the first equation in (8.23) yields
k,b
the equation for θ0,0
:
k,b
k,b
κ|∇x d|2 ∂ζ2 θ0,0
= iλτ,k
0,0 θ0,0 ,
k,b
k,int
θ0,0
(x, ζ = 0) = −θ0,0
(x) ,
lim
ζ→∞
k,b
θ0,0
(8.29)
= 0.
The equation (8.29) and the first (or equivalently the third) equation in (8.23) yield
τ,k,b
τ,k k,b
α
α
∂ζ (uτ,k,b
(1,0) ·∇x d) = −∂π (u0,0 ·∇x π ) − iλ0,0 θ0,0 .
(8.30)
The solutions to the ordinary differential equations (8.28) and (8.29) are
q
λk0,0 1
τ,k,b
τ,k,int
α
α
u0,0 (x, ζ)·∇x π = −(u0,0 (x)·∇x π ) exp −(1 + τ i) 2ν |∇x d| ζ ,
k,b
θ0,0
(x, ζ)
=
k,int
−θ0,0
(x) exp
q
λk0,0 1
−(1 + τ i) 2κ |∇x d| ζ ,
(8.31)
(8.32)
Since we already know that uτ,k,b
0,0 · ∇x d = 0 from (8.9), then coupled with (8.31), we
τ,k,b
know u0,0 completely. By integrating ζ from 0 to ∞, (8.30) gives uτ,k,b
1,0 ·∇x d, in particular
τ,k,b
the value of u1,0 ·n on the boundary ∂Ω:
q
q
∂πα (∇x Φk ·∇x π α )
τ,k,b
k
k
ν
2
κ
+ τ iλ0,0 D+2
.
(8.33)
u1,0 ·n = (1 − τ i)
Φ
2λk
2λk
iλτ,k
0,0
0,0
0,0
30
N. JIANG AND N. MASMOUDI
τ,k,b
Finally we can represent g2,0
from (8.20):
τ,k,b
k,b
−1 ⊥
2 τ,k,b
2 k,b b
b
b
b
g2,0
= ∇x uτ,k,b
:
A
+
∇
θ
·
B
+
L
P
∂
d∂
d∂
(u
)
v
A
+
∂
d∂
d∂
θ
v
B
i
m
x
l
i
ml
i
l
i
l
0,0
0,0
ζ
ζ 0,0
(0,0)
i
h
2
|v|
k,b
k,b
k,b
τ,k,b
b
b
− D2 θ2,0
.
+ ∇x d ⊗ ∂ζ uτ,k,b
(1,0) : A + ∂ζ θ(1,0) ∇x d· B + ρ2,0 + v·u2,0 +
2
(8.34)
In summary, from the equation (8.20), we know the following facts:
k,b
τ,k,b
k,b
k,b
α
• The equations of uτ,k,b
0,0 ·∇x π and θ0,0 (using u0,0 · ∇x d and ρ0,0 + θ0,0 ) then we
k,b
τ,k,bb
completely solve uτ,k,b
0,0 and θ0,0 . Furthermore, g1,−1 can now be solved;
k,b
k,b
• The relations uτ,k,b
1,0 · ∇x d and ρ1,0 + θ1,0 which will be used later to start the
induction argument for m = 1. In particular, we determine the value of uτ,k,b
1,0 · n
τ,k,int
on the boundary ∂Ω which later will give the boundary value of u1,0 · n to solve
τ,k,int
g1,0
;
τ,k,b
• The expression of g2,0
modulo the hydrodynamic part.
8.3. (m, n) = (1, 0). In this subsection, we continue to solving terms involving (m, n) =
(1, 0). We basically following the same strategy as in the previous subsection. However,
unlike in the case (m, n) = (0, 0), if iλτ,k
0,0 is not a simple eigenvalue of the acoustic operator
τ,k,int
τ,k,b
A, g1,0 and g1,0 can be only partially determined in the current row, and will be fully
determined in the next row, i.e. (m, n) = (2, 0).
τ,k
We also emphasize that among all iλτ,k
m,n , the value of iλ1,0 plays a particularly important
τ,k,int
τ,k,b
role in the rest of paper, not only in solving the general gm,n
, gm,n
and iλτ,k
m,n for the
1
case 0 < β < 2 , but also in proving the damping of the acoustic waves in the justification
of the Navier-Stokes limit. More specifically, we will show the key property of iλτ,k
1,0 : the
real part is strictly negative.
τ,k,int τ,k,b
τ,k,bb
The process of solving g1,0
, g1,0 , iλτ,k
1,0 and g2,−1 also includes the following 5 steps:
k,b
k,b
Step 0: The values of uτ,k,b
1,0 · ∇x d and ρ1,0 + θ1,0 .
From Step 4 of the last row (but in the same column), i.e. (m, n) = (0, 0), we already
k,b
k,b
− 12
) in the interior parts in
know the values of uτ,k,b
1,0 · ∇x d and ρ1,0 + θ1,0 . The order O(ε
(8.1) yields
τ,k,int
Lg1,0
= 0,
τ,k,int
which means g1,0
is hydrodynamic, i.e.
τ,k,int
τ,k,int
g1,0
= ρk,int
+
1,0 + v·u1,0
|v|2
2
−
D
2
k,int
θ1,0
.
(8.35)
Step 1: The boundary value of uτ,k,int
(1,0) · n.
1
τ,k,bb
satisfies the boundary layer equation (6.3) with
From the order O(ε 2 ) in (7.4), g1,0
τ,k
τ,k
bb
g = g̃1,0 , f = g̃0,1 and K = 0. Then (6.8) in Lemma 6.3 implies that
uτ,k,int
· n = −uτ,k,b
1,0
1,0 ·n on ∂Ω .
(8.36)
Thus from (8.33), we get that
uτ,k,int
1,0
· n = (−1 + τ i)
∂πα (∇x Φk ·∇x π α )
iλτ,k
0,0
q
ν
2λk0,0
+
k
2
iλτ,k
0,0 D+2 Φ
q
κ
2λk0,0
.
(8.37)
NAVIER-STOKES-FOURIER
31
k,b
tan
Furthermore, (6.9) in Lemma 6.3 gives the boundary conditions of [uτ,k,b
and θ0,1
in
0,1 ]
τ,k,b tan
k,b
terms of [∂ζ u0,0 ] and ∂ζ θ0,0 . More specifically,
τ,k,b tan
tan
[uτ,k,b
= −[uτ,k,int
]tan + χν [∂ζ u0,0
] ,
0,1 ]
0,1
k,b
k,int
+
θ0,1
= −θ0,1
D+2 κ
∂ θk,b
D+1 χ ζ 0,0
.
(8.38)
These conditions will be saved and will be used in the next column, i.e. n = 1, to
τ,k,b
determine g0,1
.
τ,k,int
Step 2: Solve iλτ,k
modulo Ker A − iλτ,k
1,0 and U1,0
0,0 .
1
The order O(ε 2 ) in the interior part of (8.1) yields
τ,k,int
τ,k,int
τ,k,int
τ,k,int
Lg3,0
= v · ∇x g1,0
− iλτ,k
− iλτ,k
.
0,0 g1,0
1,0 g0,0
(8.39)
The solvability condition for (8.39) is
τ,k,int
τ,k,int
(A − iλτ,k
= iλτ,k
.
0,0 )U1,0
1,0 U0,0
(8.40)
Applying Lemma 5.1 to (8.39), the formula (5.27) gives
Z
τ,k
D+2
iλ1,0 = 2D
[uτ,k,int
· n]Φk dσx .
1,0
(8.41)
∂Ω
k
∂
= ∇x Φk = g γβ ∂Φ
, and ∇x π α = g αδ ∂π∂ δ , we have
From (8.37) and notice that uτ,k,int
0,0
∂π β ∂π γ
Z
Z
k ∂Φk
k
α
k
dσx
∂πα (∇x Φ · ∇x π )Φ dσx = −
gγδ g αδ g βγ ∂Φ
∂π α ∂π β
∂Ω
∂Ω
Z
=−
|∇x Φk |2 dσx .
∂Ω
Thus
iλτ,k
1,0
=
√1+τki 3
− D+2
2D
2[λ0,0 ]
Z
√
ν|∇x Φk |2 +
∂Ω
2
[λk ]2
D+2 0,0
√
κ|Φk |2 dσx ,
(8.42)
which has the important property:
Re(iλτ,k
1,0 ) < 0 .
(8.43)
Remark: From (8.42), we can only conclude Re(iλτ,k
1,0 ) ≤ 0. The strict negativity comes
from the following argument. Indeed, assume that Re(iλτ,k
1,0 ) = 0. This would imply that
∇x Φk = 0 and Φk = 0 on the boundary ∂Ω. Hence, extending Φk by 0 outside of Ω and
denoting by Φ̃k this extension, we see that Φ̃k is an eigenvector of −∆x on the whole space
with compact support which is impossible. Hence (8.43) holds.
Remark: The above formula (8.42) is crucial in this paper, because it gives dissipativity,
which will be used later in proving the damping of the acoustic waves in the Navier-Stokes
limit.
The compatibility condition (5.28) gives
Z
[uτ,k,int
·n]Φl dσx = 0 if λk0,0 = λl0,0 and k 6= l .
(8.44)
1,0
∂Ω
Specifically,
Z
∂Ω
√
ν∇xΦk ·∇xΦl +
2
[λk ]2
D+2 0,0
√
κΦk Φl dσx = 0 ,
(8.45)
32
N. JIANG AND N. MASMOUDI
if λk0,0 = λl0,0 and k 6= l. This is exactly the orthogonality condition (5.16). Under these
τ,k,int
conditions, applying Lemma 5.1, we solve U1,0
modulo Ker A − iλτ,k
0,0 , i.e.
X
τ,k,int
τ,k,int
τ,l
τ,l
U1,0
= D+2
hU1,0
| U0,0
iU0,0
+
2D
l6=k ,λl =λk
=
X
(8.46)
τ,l
Aτ,k,l
1,0 U0,0 + ,
l6=k ,λl =λk
which implies that
X
uτ,k,int
=
1,0
l
k,int
∇x Φ
Aτ,k,l
1,0 iλk + and θ1,0 =
X
0,0
l
2
Aτ,k,l
1,0 D+2 Φ + ,
(8.47)
l6=k ,λl =λk
l6=k ,λl =λk
where the coefficients Aτ,k,l
m,n is defined as
Aτ,k,l
m,n =
τ,k,int
D+2
hUm,n
2D
τ,l
| U0,0
i.
(8.48)
From now on and in the rest of the paper, denotes terms already known from the
previous steps.
The values of Aτ,k,l
1,0 will be determined in the next row (but the same column), i.e. when
(m, 0) = (2, 0).
k,b
τ,k
tan
Step 3: The boundary values of [uτ,k,b
]
and
θ
modulo
Ker
A
−
iλ
1,0
1,0
0,0 .
1
τ,k,bb
+β
The order O(ε 2 ) in the kinetic boundary layer which implies that g(2,−1) satisfies
τ,k
τ,k,b
k
(6.3) with g = 0, f = g̃1,0
. So (ρf , uf , θf ) = (ρ̃k1,0 , ũτ,k
:
1,0 , θ̃1,0 ), and Sf = −(n ⊗ ∂ζ u0,0
k,b
b
b
A + ∂ζ θ0,0 n · B). Thus (6.9) in Lemma 6.3 gives the boundary conditions:
h
itan
h
itan
τ,k,b
τ,k,int
u1,0
= − u1,0
+ on ∂Ω ,
(8.49)
k,b
k,int
θ1,0 = −θ1,0 + on ∂Ω .
k,int
From (8.47) we know that [uτ,k,int
]tan and θ1,0
have been solved in the Step 2 modulo
1,0
τ,k
τ,k,l
k,b
Ker A − iλ0,0 , i.e. with A1,0 unknown. Thus the boundary values of uτ,k,b
1,0 and θ1,0 are
only known partially.
k,b
τ,k
tan
Step 4: Solve [uτ,k,b
]
and
θ
modulo
Ker
A
−
iλ
1,0
1,0
0,0 .
1
τ,k,b
k,b
The equations of u1,0 and θ1,0 can be found by analyzing the order O(ε 2 ) of the viscous
boundary layer in (8.1) which gives
τ,k,b
τ,k,b
τ,k,b
τ,k,b
τ,k τ,k,b
+ (v·∇x π α )∂πα g1,0
− iλτ,k
Lg3,0
= v·∇x d∂ζ g2,0
0,0 g1,0 − iλ1,0 g0,0 .
(8.50)
Noticing that iλτ,k
1,0 is known, the solvability condition for (8.50) yields that
τ,k,b
τ,k,b
τ,k,b
τ,k,b
iλτ,k
= Ab U2,0
+ AU1,0
+ Db U1,0
+ ,
0,0 U1,0
(8.51)
which has the same structure with the system (8.23) except some known source terms, so
we can treat (8.51) in the same way from which we can deduce (the details will be given
in the next subsection when we describe the more general case):
k,b
τ,k,b
k,b
k,b
α
• The equations of uτ,k,b
1,0 ·∇x π and θ1,0 (using the relations u1,0 · ∇x d and ρ1,0 + θ1,0
which are known in the last round). Then using the boundary conditions (8.49),
k,b
τ,k,bb
τ,k
uτ,k,b
1,0 , θ1,0 and thus g2,−1 can be solved modulo Ker A − iλ0,0 ;
NAVIER-STOKES-FOURIER
33
k,b
k,b
• The relations uτ,k,b
2,0 · ∇x d and ρ2,0 + θ2,0 which will be used later to start the
induction argument for m = 2. In particular, we determine the value of uτ,k,b
2,0 ·n
τ,k,int
on the boundary ∂Ω which later will give the boundary value of u2,0 · n, (again,
τ,k
modulo Ker A − iλτ,k
0,0 ) which will be used to determine the value of iλ2,0 ;
τ,k,b
• The kinetic part of g3,0
modulo the hydrodynamic part.
τ,k,int
Remark: From Step 2, U1,0
is only partially solved. The components in the kernel of
τ,k
A − iλ0,0 will be determined in the next row. However, a key point is that to determine
τ,k
the value of iλτ,k
2,0 , the components in Ker A − iλ0,0 will not be needed.
8.4. (m, 0) for all m ≥ 2. We can solve terms for all m ≥ 2 by an induction argument
for m.
k,b
k,b
Step 0: The values of uτ,k,b
m,0 · ∇x d and ρm,0 + θm,0 .
τ,k,b
k,b
k,b
We already know uτ,k,b
m,0 · ∇x d then um,0 · n on ∂Ω and ρm,0 + θm,0 from the last row, i.e.
the Step 0 is already done. Note that they are only partially determined at this stage.
Specifically, uτ,k,b
m,0 ·n can be calculated through
τ,k k,b
α
(8.52)
= −∂πα (uτ,k,b
·∇
d
∂ζ uτ,k,b
x
m−1,0 ·∇x π ) − iλ0,0 θm−1,0 + .
m,0
Thus
−1+τ i
√
uτ,k,b
m,0 ·n = −
k
√
2λ0,0
τ,k √
k,int
α
·∇
π
)
+
iλ
ν∂πα (uτ,k,int
κθ
x
0,0
m−1,0
m−1,0 + .
(8.53)
τ,k,int
From (5.31) in Lemma 5.1, we know that Um−1,0
can be written as
X
τ,k,int
τ,k,int
τ,l
τ,l
Um−1,0
= D+2
hUm−1,0
| U0,0
iU0,0
+
2D
k6=l ,λk0,0 =λl0,0
X
=
τ,l
Aτ,k,l
m−1,0 U0,0 + ,
(8.54)
k6=l ,λk0,0 =λl0,0
where is already solved in the last round. Thus, from (8.53),
X
√ ∆g Φl
τ,k,b
τ,k,l
τ,k 2 √
l
−1+τ
i
um,0 ·n = − √ k
Am−1,0
ν iλτ,k + iλ0,0 D+2 κΦ + ,
2λ0,0
(8.55)
0,0
k6=l ,λk0,0 =λl0,0
and Aτ,k,l
m−1,0 are still to be computed.
Step 1: The boundary value of uτ,k,int
· n.
m,0
m
2
Considering the order O(ε ) in the Knudsen layer and in the Maxwell boundary condiτ,k,bb
τ,k
τ,k
and f = g̃m−1,1
.
tion (7.4), gm,0
satisfies the boundary layer equation (6.3) with g = g̃m,0
Then (6.8) in Lemma 6.3 gives
uτ,k,int
· n = −uτ,k,b
m,0
m,0 · n + on ∂Ω .
Furthermore, (6.9) in Lemma 6.3 gives that on the boundary ∂Ω,
h
h
itan
h
itan
itan
τ,k,b
τ,k,int
τ,k,b
ν
+ ,
um−1,1
= − um−1,1
+ χ ∂ζ um−1,0
k,b
θm−1,1
=
k,int
−θm−1,1
+
D+2 κ
∂ θk,b
D+1 χ ζ m−1,0
τ,k,b
These conditions will be used for solving gm−1,1
later.
+ ,
(8.56)
(8.57)
34
N. JIANG AND N. MASMOUDI
τ,k,l
τ,k,int
Step 2: iλτ,k
modulo Ker A − iλτ,k
m,0 , Am−1,0 and Um,0
0,0 .
τ,k,int
Um−1,0
is only solved modulo Ker A − iλτ,k
0,0 in the last round. The components in
τ,k,l
τ,k,int
Ker A − iλτ,k
will
0,0 , i.e. Am−1,0 , will be solved in this round. At the same time, Um,0
τ,k
be determined, again, modulo Ker A − iλ0,0 .
m
The order O(ε 2 ) in the interior part of (8.1) yields
τ,k,int
τ,k,int
τ,k,int
τ,k,int
τ,k,int
Lgm+2,0
= v · ∇x gm,0
− iλτ,k
− iλτ,k
− iλτ,k
0,0 gm,0
m,0 g0,0
1,0 gm−1,0 + .
The solvability condition for (8.58) is
τ,k,int
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
+ iλτ,k
0,0 Um,0
m,0 U0,0
1,0 Um−1,0 + .
τ,k,int
τ,k,int
Noticing that hUm−1,0
| U0,0
i = 0 for all m ≥ 2, (5.27) in Lemma 5.1 gives
Z
τ,k
D+2
iλm,0 = 2D
[uτ,k,int
· n]Φk dσx + m,0
∂Ω
Z
k
D+2
= − 2D
[uτ,k,b
m,0 · n]Φ dσx + .
(8.58)
(8.59)
(8.60)
∂Ω
Here we applied the formula (8.56) and put all the known terms in .
Using the formula (8.55),
iλτ,k
m,0 =
D+2 √
−1+τ i 1
τ,k
2D
2λk0,0 iλ0,0
X
×
Aτ,k,l
m−1,0
Z
√
ν∇xΦk ·∇xΦl +
∂Ω
k6=l ,λk0,0 =λl0,0
2
[λk ]2
D+2 0,0
√
κΦk Φl dσx + (8.61)
= .
Here the key point is that although Aτ,k,l
m−1,0 are unknown so far, they do not have contriτ,k
bution in determining iλm,0 because of the compatibility condition (5.16). Thus iλτ,k
m,0 is
completely solved.
Furthermore, the condition (5.28) in Lemma 5.1 gives
X
0
τ,k,l0
τ,k τ,k,l
al,l
(8.62)
k Am−1,0 = iλ1,0 Am−1,0 + ,
k6=l0 ,λk0,0 =λl0,0
0
where the coefficients al,l
k is given by
Z h
√
0
l,l0
D+2 √1+τ i
ak = − 2D
ν∇xΦl ·∇xΦl +
k 3
2[λ0,0 ]
∂Ω
2
[λk ]2
D+2 0,0
√
i
0
κΦl Φl dσx .
(8.63)
0
0
Again, from the compatibility condition (5.16), al,l
k = 0 when l 6= l. Thus, from (8.62),
τ,l
τ,k,l
iλτ,k
−
iλ
(8.64)
1,0
1,0 Am−1,0 = .
τ,l
k
l
Note that because of the condition (5.17), iλτ,k
1,0 6= iλ1,0 for all k 6= l, λ0,0 = λ0,0 . Then,
τ,k,int
τ,k,b
Aτ,k,l
m−1,0 and consequently gm−1,0 and gm−1,0 are completely determined.
τ,k,int
Under the conditions (8.60) and (8.62), applying Lemma 5.1 to (8.59), Um,0
is solved
τ,k
modulo Ker A − iλ0,0 . Thus we finish Step 2.
k,b
tan
Step 3: The boundary values of [uτ,k,b
and θm,0
modulo Ker A − iλτ,k
m,0 ]
0,0 .
NAVIER-STOKES-FOURIER
35
m
τ,k,bb
with g = 0,
Next, the order O(ε 2 +β ) in the kinetic boundary layer implies that gm+1,−1
τ,k
f = g̃m,0 . Thus (6.9) in Lemma 6.3 gives the boundary conditions:
h
itan
h
itan
τ,k,b
τ,k,int
um,0
= − um,0
+ ,
(8.65)
k,b
k,int
θm,0 = −θm,0 + .
τ,k,int
τ,k,b tan
Again, Um,0
in Ker A − iλτ,k
and
0,0 is not determined until the next round. So [um,0 ]
k,b
θm,0 are only partially determined at this level..
k,b
tan
Step 4: Solve [uτ,k,b
and θm,0
modulo Ker A − iλτ,k
m,0 ]
0,0 .
m
Finally, the order O(ε 2 ) of the viscous boundary layer in (8.1) gives:
τ,k,b
τ,k,b
τ,k,b
τ,k,b
Lgm+2,0
= v·∇x d∂ζ gm+1,0
+ (v·∇x π α )∂πα gm,0
− iλτ,k
0,0 gm,0 + .
(8.66)
Here we used the fact that iλ±
m,0 is now known. The solvability of (8.66) yields that
τ,k,b
τ,k,b
b τ,k,b
b τ,k,b
iλτ,k
0,0 Um,0 = A Um+1,0 + AUm,0 + D Um,0 + .
(8.67)
From the equation (8.66) we know:
k,b
τ,k,b
k,b
k,b
α
• The equations of uτ,k,b
m,0 ·∇x π and θm,0 (using um,0 · ∇x d and ρm,0 + θm,0 ) then using
k,int
the boundary conditions (8.65), they can be solved in terms of uτ,k,int
and θm,0
;
m,0
τ,k,b
k,b
k,b
• The relations um+1,0 · ∇x d and ρm+1,0 + θm+1,0 which will be used later to start the
induction argument for m + 1. In particular, we determine the value of uτ,k,b
m+1,0 · n
τ,k,int
on the boundary ∂Ω which later will give the boundary value of um+1,0 · n;
τ,k,b
• The kinetic part of gm+2,0
modulo the hydrodynamic part.
τ,k,int
All above are solved with the components of Um,0
in Ker A − iλτ,k
0,0 to be determined in
τ,k,int
τ,k,b
the next round. Thus we close the induction process and can determine gm,0
and gm,0
for all m ≥ 2.
8.5. General Process. In the previous subsections, we already finish the induction for
the 0-th column. For every fixed n ≥ 1, the induction for each m ≥ 0 includes the
following 5 steps:
k,b
k,b
Step 0: The values of uτ,k,b
m,n · ∇x d and ρm,n + θm,n ;
τ,k,b
For m = 0, because g0,n
is hydrodynamic, we can deduce that uτ,k,b
0,n · ∇x d = 0 and
k,b
k,b
ρ0,n + θ0,n = 0, as in (m, n) = (0, 0). Thus we can start the induction. For general m ≥ 1,
k,b
k,b
uτ,k,b
m,n · ∇x d and ρm,n + θm,n are from the Step 4 of the previous round, i.e. (m − 1)-th row.
In particular, similar to (8.55), we have
X
√ ∆g Φl
τ,k 2
τ,k,l
τ,k,b
l
−1+τ
i
um,n ·n = − √ k
ν iλτ,k + iλ0,0 D+2 Φ + ,
(8.68)
Am−1,n
2λ0,0
k6=l ,λk0,0 =λl0,0
0,0
where Aτ,k,l
m−1,n is defined in (8.48), and will be determined in the following Step 2.
Step 1: The boundary value of uτ,k,int
m,n · n.
m
1
+n(
−β)
2
) of the kinetic boundary layer in the ansatz (8.1) and the
The order O(ε 2
τ,k,bb
Maxwell condition (7.4) implies that gm,n
satisfies the linear boundary layer equation
36
N. JIANG AND N. MASMOUDI
τ,k
τ,k
(6.3) with g = g̃m,n
and f = g̃m−1,n+1
. Then (6.8) in Lemma 6.3 gives that
τ,k,b
uτ,k,int
m,n · n = −um,n · n + for m ≥ 1 .
Furthermore, (6.9) in Lemma 6.3 gives that on the boundary ∂Ω,
h
itan
h
itan
h
itan
τ,k,int
τ,k,b
ν
uτ,k,b
=
−
u
+
∂
u
+ ,
ζ
m−1,n+1
m−1,n+1
m−1,n
χ
k,b
θm−1,n+1
=
k,int
−θm−1,n+1
+
D+2 κ
∂ θk,b
D+1 χ ζ m−1,n
(8.69)
(8.70)
+ ,
These conditions will be used for (n + 1)-th column.
k,bb
The case m = 0 is special. Indeed, similar as in (m, n) = (0, 0), since g0,n
= 0, the
( 12 −β)n
order O(ε
) in (7.4) gives the boundary conditions
τ,k
τ,k
γ+ g̃0,n
− γ− Lg̃0,n
= 0,
(8.71)
which implies that on the boundary ∂Ω,
τ,k,int
u0,n + uτ,k,b
· n = 0.
0,n
(8.72)
Combining with the boundary condition uτ,k,b ·n = 0, we deduce that on the boundary,
uτ,k,int
·n = 0 .
0,n
(8.73)
τ,k,l
τ,k,int
Step 2: Solve iλτ,k
modulo Ker A − iλ0,0 .
m,n , Am−1,n and Um,n
1
m
+n(
−β)
2
) in the interior part of (8.1) yields
The order O(ε 2
τ,k,int
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
+ iλτ,k
0,0 Um,n
1,0 Um−1,n + .
m,n U0,0
τ,k
(8.74)
Then the solvability conditions of (8.74) are given by (5.27)
and (5.28) in Lemma 5.1,
τ,k,int
τ,k
τ,k
which give the value of iλm,n and Um−1,n in Ker A − iλ0,0 , just as we did in (8.61) and
τ,k,int
(8.62). Under these conditions, we can solve Um,n
modulo Ker A − iλτ,k
0,0 , again by
Lemma 5.1, as
X
τ,l
τ,k,int
Um,n
Aτ,k,l
(8.75)
=
m,n U0,0 + .
k6=l ,λk0,0 =λl0,0
We specifically treat the case m = 0, we have
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
.
0,0 U0,n
0,n U0,0
(8.76)
Thus applying Lemma 5.1 to the equation (8.76), the formula (5.27) implies that
iλτ,k
0,n = 0 ,
(8.77)
τ,k
τ,k,int
τ,k,int
which means U0,n
∈ Ker A − iλ0,0 . We set U0,n
as
τ,k,int
U0,n
= 0,
if iλτ,k
0,0 is a simple eigenvalue of A ;
(8.78)
Otherwise,
τ,k,int
U0,n
=
X
τ,l
Aτ,k,l
0,n U0,0 .
(8.79)
k6=l ,λk0,0 =λl0,0
tan
k,b
Step 3: The boundary values of [uτ,b
and θm,n
modulo Ker A − iλτ,k
0,0 .
m,n ]
τ,k,int tan
k,b
k,int
Using the boundary values of [uτ,k,b
and θm,n
+θm,n
which are from the Step
m,n +um,n ]
1 of the (n − 1)-th column, see (8.70) (replacing n + 1 by n), together with the boundary
tan
int
values of [uτ,k,int
and θm,n
modulo Ker A − iλτ,k
0,0 known in the Step 2, we can get the
m,n ]
NAVIER-STOKES-FOURIER
37
tan
k,b
boundary values of [uτ,k,b
and θm,n
with parts in Ker A − iλτ,k
0,0 to be determined in
m,n ]
the next column;
tan
k,b
Step 4: Solve [uτ,k,b
and θm,n
modulo Ker A − iλτ,k
0,0 .
m,n ]
tan
k,b
Find the ordinary differential equations for [uτ,k,b
and θm,n
then together with the
m,n ]
boundary values known in the Step 3, we can solve the equations. Thus we can determine
τ,k,b
τ,k,b
gm,n
. This is achieved from the solvability conditions of the equation of gm+2,n
which at
τ,k,b
k,b
k,b
the same time also gives the values of um+1,n · ∇x d and ρm+1,n + θm+1,n . Then we can go
back to Step 0 to start the (m + 1)-th row so close the induction argument for m.
After finishing the n-th column, we move to (n + 1)-th column using the data saved
from the Step 1 in the n-th column to close the induction for n.
Remark: When 0 < β < 12 , the leading term of iλBL
is
ε
1
τ,k 1−β
2
iλBL
= iλτ,k + iλτ,k
+ ···
1,0 ε + iλ1,1 ε
ε
with Re(iλτ,k
1,0 ) < 0 .
(8.80)
1 −β
τ,k,int
τ,k,b
τ,k,bb β
τ,k,int
τ,k,b
gεBL = g0,0
+ g0,0
+ g1,−1
ε + g0,1
+ g0,1
ε2 + · · ·
(8.81)
The leading term of gεBL is
9. Proof of Proposition 7.1: the case β = 0
We treat the case β = 0 separately in this section. For this case, the accommodation
coefficient α is a constant χ ∈ (0, 1). The construction of the kinetic-fluid boundary
layers is the same as the case 0 < β < 21 in the last section. However, the ansatz of the
eigenfunction and eigenvalue only includes the terms with subscript (m, 0) in (8.1) and
(8.2), i.e. we make for gεBL and λBL
the following ansatz:
ε
i m
X h τ,k,int
d(x)
τ,k,b
τ,k,bb
√ , v) + g
(π(x), d(x)
(π(x),
(9.1)
,
v)
ε2 ,
gεBL =
gm,0 (x, v) + gm,0
m,0
ε
ε
m≥0
λBL
=
ε
X
m
2
λτ,k
m,0 ε .
(9.2)
m≥0
τ,k,bb
g0,0
Note that
= 0 in the above ansatz. The only difference of the construction of the
boundary layers between the cases β = 0 and 0 < β < 21 is on the boundary conditions
which we describe bellow.
The most important boundary conditions are for (m, 0) = (0, 0), the normal and tank,int
k,b
gential components of uτ,k,int
+ uτ,k,b
0,0
0,0 , and θ0,0 + θ0,0 , i.e. (8.13) thus (8.14), and (8.19).
For the case 0 < β < 12 , (8.13) was from the kinetic condition (8.12), while (8.19) was
τ,k,bb
deduced from considering the kinetic boundary layer equation satisfied by g1,−1
.
For the case β = 0, the analogue of (8.12) is the kinetic boundary condition
τ,k
τ,k
τ,k
τ,k
= χ[hγ− g̃0,0
i∂Ω − Lγ− g̃0,0
].
γ+ g̃0,0
− Lγ− g̃0,0
(9.3)
τ,k,bb
Note that for β = 0, there is no boundary layer term g1,−1
as in the case 0 < β < 21 .
τ,k,int
τ,k,b
Because both g0,0
and g0,0 are hydrodynamic, then straightforward calculations
show that (9.3) implies that on the boundary ∂Ω,
2[uτ,k,int
+ uτ,k,b
0,0
0,0 ]·n
√
τ,k,int
k,int
k,b
=χ 22π [uτ,k,int
+ uτ,k,b
+ uτ,k,b
0,0
0,0 ]·n + 2 [u0,0
0,0 ]·n n·v − θ0,0 + θ0,0
|v|2
2
−
D
2
.
(9.4)
38
N. JIANG AND N. MASMOUDI
2
Comparing the coefficients of 1, v, and |v|2 − D2 in (9.4), we deduce (8.13) thus (8.14), and
(8.19).
When m > 0, for the case 0 < β < 21 , the boundary conditions (8.56) and (8.65) are
τ,k,bb
deduced from considering the kinetic boundary layer equations satisfied by gm,0
and
τ,k,bb
τ,k,bb
gm+1,−1 respectively. For β = 0, these are no gm+1,−1 , these boundary conditions can be
τ,k,bb
deduced by considering gm,0
, and applying Corollary 6.1.
Except the difference we described above, the other process is exactly the same as
determining the terms with subscript (m, 0) in the case 0 < β < 21 .
Remark: When β = 0, the leading term of iλBL
is
ε
1
τ,k
2
iλBL
= iλτ,k + iλτ,k
1,0 ε + iλ2,0 ε + · · ·
ε
with Re(iλτ,k
1,0 ) < 0 .
(9.5)
1
τ,k,int
τ,k,b
τ,k,int
τ,k,b
τ,k,bb
gεBL = g0,0
+ g0,0
+ g1,0
+ g1,0
+ g1,0
ε2 + · · ·
(9.6)
The leading term of gεBL is
10. Proof of Proposition 7.1: the case
1
2
<β<1
In this section, we construct the kinetic-fluid boundary layers corresponding to the
accommodation coefficient α = χεβ for 12 < β < 1. In contrast to the case 0 ≤ β < 21 ,
we make for the gεBL and λBL
the following ansatz:
ε
X m +(β− 1 )n
τ,k,int
τ,k,b
τ,k,bb
2
,
(10.1)
gεBL =
gm,n
+ gm,n
+ gm,n
ε2
m≥0 ,n≥−m
and
λBL
=
ε
X
m
1
λm,n ε 2 +n(β− 2 ) .
(10.2)
m≥0,n≥−m
In the above ansatz, we require that
τ,k,b
τ,k,bb
τ,k,bb
τ,k,bb
• gm,−m
= 0 for all m ≥ 0, g0,n
= 0, for all n ≥ 0, gm,−m
= gm,−m+1
= 0 for all
m ≥ 1;
τ,k,int
τ,k,b
• g0,n
and g0,n
for n ≥ 1 are hydrodynamic, for n ≥ 1;
• The orthogonality condition (7.18) holds.
We can also use the following infinite matrix to represent the index (m, n) in the ansatz
(10.1):
(0, 0) −→

↓


.
.
.
(1, −1) −→ (1, 0) −→

↓
↓


.
.
(2, −2) −→ (2, −1) −→ (2, 0) −→


↓
↓
↓


.
· · · −→
· · · −→
· · · −→
· · · −→


↓
↓
↓
↓

(m, −m) −→
−→ (m, −2) −→ (m, −1) −→ (m, 0) −→
↓
↓
↓
↓
↓

.
.
.
.
(0, 1) −→
↓
(1, 1) −→
↓
(2, 1) −→
↓
· · · −→
↓
(m, 1) −→
↓

· · · −→
↓ 

· · · −→

↓ 

· · · −→
.
↓ 
· · · −→



· · · −→
(10.3)
NAVIER-STOKES-FOURIER
39
The process to determine all the terms in the ansatz (10.1) is similar to the case
0 ≤ β < 12 . We will still use double induction argument both for m and n which are still
called m-th row and n-th column.
However, different with the case 0 ≤ β < 12 , the order of induction for m and n is the
following: we start from m = 0, make induction for n. After we finish the induction for
the 0-th row, we move to the next row, i.e. m = 1, then make induction for n, and so on,
thus solve all terms. When we solve n-th row, we need the information from (n − 1)-th
row, and save some information to be used for the (n + 1)-th row.
10.1. (0, n) for n ≥ 0. Similar as before, for each (0, n), we divide the process into the
following five steps.
k,b
k,b
Step 0: The values of uτ,k,b
0,n · ∇x d and ρ0,n + θ0,n ;
τ,k,b
By the requirements on the ansatz (10.1), all g0,n
for n ≥ 1 are hydrodynamic. Thus,
the same arguments before (8.8) and (8.9) yield the relations
k,b
k,b
uτ,k,b
0,n ·∇x d = 0 and ρ0,n + θ0,n = 0 .
(10.4)
In particular, we have the boundary condition
uτ,k,b
0,n ·n = 0 on ∂Ω .
(10.5)
· n.
Step 1: The boundary value of uτ,k,int
0,n
· n = 0 for all
Following the same arguments before (8.73), we can deduce that uτ,k,int
0,n
n ≥ 0.
τ,k,int
τ,k
and
U
modulo
Ker
A
−
iλ
Step 2: Solve iλτ,k
0,n
0,n
0,0 .
τ,k,int
This step is also exactly the same as before. For n = 0, iλτ,k
are the
0,0 and U0,0
eigenvalue and eigenfunctions of the acoustic operator A. For n ≥ 1, the conclusions are
the same as (8.77), (8.78) and (8.79), where Aτ,k,l
0,n will be determined in the next row, i.e.
when m = 1.
k,b
tan
Step 3: The boundary values of [uτ,k,b
and θ0,n
.
0,n ]
τ,k,bb
We consider the linear kinetic boundary equation (6.3)-(6.5) satisfied by g1,n
with
τ,k,int
τ,k,b
τ,k,int
τ,k,b
g = g1,n + g1,n and f = g0,n−1 + g0,n−1 in (6.5). The formula (6.9) in Lemma 6.3 gives
h
itan
itan
h
τ,k,int
τ,k,b
χ
+
u
on ∂Ω ,
=
∂ζ uτ,k,b
u
0,n
0,n−1
0,n−1
ν
(10.6)
k,b
k,int
k,b
D+1 χ
∂ζ θ0,n = D+2 κ θ0,n−1 + θ0,n−1 on ∂Ω .
Furthermore the formula (6.8) in Lemma 6.3 gives
uτ,k,int
·n = −uτ,k,b
1,n
1,n ·n on ∂Ω ,
(10.7)
which will be used in the next row.
k,b
tan
Step 4: Solve [uτ,k,b
and θ0,n
.
0,n ]
τ,k,b tan
k,b
The equations satisfied by [u0,n ] and θ0,n
can be found by considering the order
( 12 −β)n
) of the viscous boundary layer in (8.1):
O(ε
τ,k,b
τ,k,b
τ,k,b
τ,k,b
Lg2,n
= v·∇x d∂ζ g1,n
+ (v·∇x Πα )∂Πα g0,n
− iλτ,k
0,0 g0,n ,
(10.8)
for which the solvability condition is an ordinary differential system like (8.23). From this
k,b
k,b
system, using the relations uτ,k,b
0,n · ∇x d and ρ0,n + θ0,n , we can also derive as before the
following:
40
N. JIANG AND N. MASMOUDI
k,b
k,b
• The value of ρ1,n
+ θ1,n
;
τ,k,b
k,b
α
• u0,n ·∇x π and θ0,n satisfy the ordinary differential systems (8.28) and (8.29) with
vanishing boundary conditions at ζ → ∞ respectively. However, the boundary
conditions at ζ = 0 are (10.6), which are different as before. Furthermore, because
k,int
the boundary conditions depend on uτ,k,int
0,n−1 and θ0,n−1 which are only solved partially
k,b
α
(except n = 1!) from the Step 2, then for n > 1, uτ,k,b
0,n · ∇x π and θ0,n , and
consequently uτ,k,b
1,n · ∇x d can be only solved partially.
k,b
α
For n = 1, the boundary conditions for uτ,k,b
0,1 ·∇x π and θ0,1 are given completely by
τ,k,b
(10.6) because there is no g0,0
, i.e.
h
itan
h
itan
τ,k,int
χ
u
on ∂Ω ,
∂ζ uτ,k,b
=
0,0
0,1
ν
(10.9)
k,b
χ k,int
∂ζ θ0,1
= D+1
θ
on
∂Ω
.
D+2 κ 0,0
Thus they can be completely solved by
α
uτ,k,b
0,1 ·∇x π
k,b
θ0,1
= (−1 +
= (−1 +
τ i) √ χ k (uτ,k,int
·∇x π α ) exp
0,0
2νλ0,0
D+1 √ χ
τ i) D+2
2κλk
k,int
θ0,0
0,0
q
λk0,0 1
−(1 + τ i) 2ν |∇x d| ζ ,
q
λk0,0 1
exp −(1 + τ i) 2κ |∇x d| ζ .
(10.10)
(10.11)
Consequently, we have
τ,k k,b
α
∂ζ uτ,k,b
·∇
d
= −∂πα (uτ,k,b
x
1,1
0,1 ·∇x π ) − iλ0,0 θ0,1 ,
from which we derive that on the boundary
Φk·∇x π α )
τ,k,b
+
u1,1 · n = χ ∂πα (∇[λxτ,k
]2
0,0
2(D+1) k
Φ
(D+2)2
(10.12)
.
(10.13)
k,b
α
For n = 2, since uτ,k,b
0,1 ·∇x π and θ0,1 are known now, then from the boundary condition
(10.6) we can get
α
uτ,k,b
0,2 ·∇x π
=(−1 +
τ i) √ χ k
2νλ0,0
X
τ,k,int
Aτ,k,l
·∇x π α ) exp
0,1 (u0,0
q
λk0,0 1
−(1 + τ i) 2ν |∇x d| ζ
(10.14)
k6=l ,λk0,0 =λl0,0
+ ,
k,b
θ0,2
= (−1 +
D+1 √ χ
τ i) D+2
2κλk
0,0
X
l,int
Aτ,k,l
0,1 θ0,0
q
λk0,0 1
exp −(1 + τ i) 2κ |∇x d| ζ + ,
k6=l ,λk0,0 =λl0,0
(10.15)
k,b
k,b
and the values of uτ,k,b
1,2 · ∇x d and ρ1,2 + θ1,2 , and consequently,
X
2(D+1) l
∂πα (∇x Φl·∇x π α )
τ,k,b
τ,k,l
+ (D+2)2 Φ + on ∂Ω .
u1,2 · n = χ
A0,1
[λτ,k ]2
k6=l ,λk0,0 =λl0,0
(10.16)
0,0
For n > 2, the boundary condition (10.6) is given recursively. For example, if (0, 3),
k,b
tan
from Step 0 to Step 3, the process is the same. For Step 4, [uτ,k,b
and θ0,3
thus,
0,3 ]
NAVIER-STOKES-FOURIER
41
τ,k,l
τ,k,l
uτ,k,b
1,3 · ∇x d depend on the unknown coefficients both A0,2 and A0,1 . This is because
tan
tan
from the boundary condition (10.6), [uτ,k,b
depends on both [uτ,k,int
]tan and [uτ,k,b
.
0,3 ]
0,2
0,2 ]
τ,k,l
τ,k,l
The former is involved A0,2 , the latter depends on A0,1 because of the solution formula
(10.15). Then we can move to (0, 4), and so on.
Summary: For the 0-th row, i.e. m = 0, after finish the induction for n, we solve the
following:
τ,k,int
τ,k,int
• Solve g0,0
completely, but g0,n
for n ≥ 1 partially, with Aτ,k,l
0,n to be determined;
τ,k,b
τ,k,b
• Solve g0,1 completely, but g0,n for n ≥ 1 partially, which depends on Aτ,k,l
0,i for
1 ≤ i ≤ n − 1 to be determined;
• Solve iλτ,k
0,n completely;
τ,k,b
k,b
τ,k,l
• Solve u1,n · ∇x d and ρk,b
1,n + θ1,n partially, which depend on A0,i for 1 ≤ i ≤ n − 1
to be determined.
τ,k,int
10.2. (1, n) for n ≥ −1. We start from n = −1. It is easy to deduce that g1,−1
is
τ,k,int
τ,k
hydrodynamic and U1,−1 = 0 if iλ0,0 is a simple eigenvalue of A. Otherwise,
X
τ,k,int
τ,l
U1,−1
=
(10.17)
Aτ,k,l
1,−1 U0,0 .
k6=l ,λk0,0 =λl0,0
For n = 0, we can follow the same 5 steps as before:
k,b
k,b
Step 0: The values of uτ,k,b
1,0 · ∇x d and ρ1,0 + θ1,0 ;
τ,k,b
k,b
k,b
Since g1,0
is hydrodynamic, we can deduce uτ,k,b
1,0 · ∇x d = 0 and ρ1,0 + θ1,0 = 0.
τ,k,int
Step 1: The boundary value of u1,0
· n.
τ,k
τ,k
τ,k,b
From γ+ g̃1,0 − Lγ− g̃1,0 = 0 and u1,0 · n = 0, we have uτ,k,int
· n = 0 on ∂Ω.
1,0
τ,k
τ,k,int
τ,k
Step 2: Solve iλ1,0 and U1,0
modulo Ker A − iλ0,0 .
τ,k
τ,k,int
It is easy to deduce that iλ1,0 = 0 and U1,0
= 0 if iλτ,k
0,0 is a simple eigenvalue of A.
Otherwise,
X
τ,k,int
τ,l
U1,0
=
Aτ,k,l
(10.18)
1,0 U0,0 .
k6=l ,λk0,0 =λl0,0
k,b
tan
.
Step 3: The boundary values of [uτ,k,b
and θ1,0
1,0 ]
τ,k,bb
We consider the linear kinetic boundary equation (6.3)-(6.5) satisfied by g2,0
with
τ,k,int
τ,k,b
τ,k,int
g = g2,0 + g2,0 and f = g1,−1 in (6.5). The formula (6.9) in Lemma 6.3 gives
h
itan
h
itan
h
itan
τ,k,int
τ,k,int
χ
∂ζ uτ,k,b
=
)·n
on ∂Ω ,
u
+
2d(u
1,0
1,−1
0,0
ν
(10.19)
k,b
D+1 χ k,int
∂ζ θ1,0 = D+2 κ θ1,−1 on ∂Ω .
Furthermore the formula (6.8) in Lemma 6.3 gives
uτ,k,int
·n = −uτ,k,b
2,0
2,0 ·n on ∂Ω ,
(10.20)
which will be used in the next row.
k,b
tan
Step 4: Solve [uτ,k,b
and θ1,0
.
1,0 ]
k,b
tan
Applying the same method, we can have [uτ,k,b
, θ1,0
and the boundary value of
1,0 ]
τ,k,b
τ,k,l
u2,0 · n similar to (10.14), (10.15) and (10.16), with A1,−1 to be determined in the next
row.
42
N. JIANG AND N. MASMOUDI
For n = 1, we can follow the same 5 steps. We omit the details here except the Step
τ,k,int
2 which is particularly important for the later inductions. We can derive that U1,1
satisfies the system
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
.
(10.21)
0,0 U1,1
1,1 U0,0
Applying the formula (5.27) in Lemma 5.1, iλτ,k
1,1 can be represented as
Z
D+2
iλτ,k
[uτ,k,int
·n]Φk dσx ,
1,1
1,1 = 2D
(10.22)
The formula (5.28) in Lemma 5.1 yields the compatibility condition
Z
τ,l
[uτ,k,int
·n]Φl dσx = 0 , for all l 6= k , λτ,k
1,1
0,0 = λ0,0 .
(10.23)
∂Ω
∂Ω
Furthermore, Lemma 5.1 gives the “partial” solution to the system (10.21):
X
τ,l
τ,k,int
Aτ,k,l
U1,1
=
1,1 U0,0 + ,
(10.24)
k6=l ,λk0,0 =λl0,0
where Aτ,k,l
1,1 is to be determined in the next row, i.e. when m = 2.
From (10.7), (10.13) and (10.22) we can get
Z 2(D+1)
τ,k
k 2
k 2
1
D+2
|∇x Φ | + (D+2)2 |Φ | dσx .
iλ1,1 = −χ 2D
[λk ]2
∂Ω
(10.25)
0,0
Thus, we have
iλτ,k
1,1 < 0 .
Furthermore, the compatibility condition (10.23) reads
Z 2(D+1) k l
k
l
1
Φ
Φ
dσx = 0 ,
∇
Φ
·∇
Φ
+
x
(D+2)2
[λk ]2 x
∂Ω
(10.26)
(10.27)
0,0
for all k 6= l and λk0,0 = λl0,0 , which is the orthogonality condition (5.18).
τ,k
1
Remark: iλτ,k
1,1 plays the same role as iλ1,0 in the case 0 < β < 2 . Specifically, to
τ,k
τ,k,int
τ,k,int
solve the gm,n
, gm,n
and iλτ,k
m,n inductively when the eigenvalue iλ0,0 is not simple,
the negativity of iλτ,k
1,1 , together with the compatibility condition (10.27) are crucially
needed. Furthermore, iλτ,k
1,1 is strictly negative, this will damp the fast acoustic waves in
the Navier-Stokes limit.
To continue the induction argument for n ≥ 2, we next consider n = 2. For general
n ≥ 2, the process is the same.
k,b
k,b
Step 0: The values of uτ,k,b
1,2 · ∇x d and ρ1,2 + θ1,2 ;
This step has been done partially in the Step 4 of the inductions in the last row on
(0, 2), see (10.14)-(10.16). However, all the values include Aτ,k,l
0,1 which is to be determined
in the Step 2 below.
τ,k,int
Step 1: The boundary value of u1,2
· n.
It can be deduced from (10.7) and (10.16), again, with Aτ,k,l
0,1 undetermined.
τ,k
τ,k,l
τ,k,int
Step 2: Solve iλ1,2 , A0,1 and U1,2
modulo Ker A − iλτ,k
0,0 .
τ,k,int
U1,2
satisfies the system
τ,k,int
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
+ iλτ,k
+ .
(10.28)
0,0 U1,2
1,2 U0,0
1,1 U0,1
NAVIER-STOKES-FOURIER
43
τ,k,int
τ,k,int
Noticing that hU0,1
|U0,0
i = 0 which is from the orthogonality condition (7.18), we
apply Lemma 5.1 to (10.28). The formula (5.27) gives
Z
τ,k
k
2D
iλ1,2 = −
[uτ,k,b
1,2 ·n]Φ dσx + D+2
∂Ω
Z X
2(D+1) k l
τ,k,l
k
l
1
∇
Φ
·∇
Φ
+
=−χ
A0,1
Φ
Φ
dσx + (10.29)
x
(D+2)2
[λk ]2 x
k6=l ,λk0,0 =λl0,0
∂Ω
0,0
= .
Here we used the compatibility condition (10.27). The formula (5.28) in Lemma 5.1 yields
Z
τ,k τ,k,l
l
2D
−
for all l 6= k , λk0,0 = λl0,0 .
(10.30)
[uτ,k,b
1,2 ·n]Φ dσx = D+2 iλ1,1 A0,1 + ,
∂Ω
Applying the compatibility condition (10.27) again yields
τ,k,l
τ,k
2D
iλτ,l
1,1 − iλ1,1 A0,1 = ,
D+2
(10.31)
τ,l
τ,k,l
Note the condition (5.19) implies iλτ,k
1,1 6= iλ1,1 . Then we determine A0,1 for all l 6=
τ,l
k , λτ,k
0,0 = λ0,0 .
Furthermore, from Lemma 5.1 again, the partial solution to the system (10.28) is
X
τ,l
τ,k,int
(10.32)
=
Aτ,k,l
U1,2
1,2 U0,0 + .
τ,l
k6=l ,λτ,k
0,0 =λ0,0
After we know Aτ,k,l
0,1 , the following quantities are completely determined:
τ,k,int
;
• g0,1
τ,k,b
• g0,2 , from (10.14) and (10.15);
k,b
k,b
• The values of uτ,k,b
1,2 · ∇x d and ρ1,2 + θ1,2 in the Step 0.
k,b
tan
Step 3: The boundary values of [uτ,k,b
and θ1,2
.
1,2 ]
τ,k,bb
We consider the linear kinetic boundary equation (6.3)-(6.5) satisfied by g2,2
with
τ,k,int
τ,k,b
τ,k,int
τ,k,b
g = g2,2 + g2,2 and f = g1,1 + g1,1 in (6.5). The formula (6.9) in Lemma 6.3 gives
h
itan
itan h
itan
h
τ,k,int
τ,k,b
τ,k,int
χ
∂ζ uτ,k,b
=
)·n
on ∂Ω ,
+
2d(u
u
+
u
1,2
0,2
1,1
1,1
ν
(10.33)
k,b
k,b
D+1 χ k,int
∂ζ θ1,2 = D+2 κ (θ1,1 + θ1,1 ) on ∂Ω .
Furthermore the formula (6.8) in Lemma 6.3 gives
uτ,k,int
·n = −uτ,k,b
2,2
2,2 ·n + on ∂Ω ,
(10.34)
which will be used in the next row.
k,b
τ,k,l
tan
Remark: From (10.33), the boundary values of [uτ,k,b
and θ1,2
depend on Aτ,k,l
1,2 ]
1,1 , A1,0
k,b
τ,k,l
τ,k,l
(because uτ,k,b
and θ1,1
depends on Aτ,k,l
1,1
1,0 ) and A0,2 which are unknown. A0,2 will be
determined in the the induction of the same row and next column, i.e. the induction on
(2, 3). However, Aτ,k,l
1,1 will be determined in the next row, i.e. m = 2.
τ,k,b tan
k,b
Step 4: Solve [u1,2 ] and θ1,2
.
τ,k,b
k,b
Using the values of u1,2 · ∇x d and ρk,b
1,2 + θ1,2 which are known now, we can derive
k,b
α
the equations of uτ,k,b
1,2 · ∇x π and θ1,2 then solve them, and consequently the values of
44
N. JIANG AND N. MASMOUDI
k,b
k,b
uτ,k,b
2,2 · ∇x d and ρ2,2 + θ2,2 which will be used in the next row m = 2. All the quantities are
τ,k,l
τ,k,l
included the unknown Aτ,k,l
1,1 and A0,2 . However as we remarked, A0,2 will be determined
in the next column in the same row m = 1.
After finish these 5 steps for (1, 2), we can move to the next column, i.e (1, 3), and
continue the induction for n. For each (1, n), n ≥ 3, the 5 steps are the same as in (1, 2),
thus we omit the details here.
Summary: For m = 1, after finish the induction for n, we solve the following:
• Solve all Aτ,k,l
0,n , thus all the terms in (0, n) are completely determined;
τ,k,int
• Solve g1,n partially, with Aτ,k,l
1,n to be determined;
τ,k,b
• Solve g1,n partially, which depends on Aτ,k,l
1,i for −1 ≤ i ≤ n − 1 to be determined;
τ,k
• Solve iλ1,n completely;
k,b
k,b
τ,k,l
• Solve uτ,k,b
2,n · ∇x d and ρ2,n + θ2,n partially, which depend on A1,i for −1 ≤ i ≤ n − 1
to be determined.
After we finish the inductions on n ≥ −1 for all (1, n), we move to the next row, i,e,
(2, n) for n ≥ −2), then continue the induction for n, and so on. We describe the general
process briefly below.
10.3. General process. Suppose we have finished the induction process in (m − 1)-th
row, the following facts are already known:
τ,k,int
• gm−1,n
modulo Ker A − iλτ,k
0,0 , i.e.
X
τ,k,int
τ,l
Um−1,n
=
Aτ,k,l
for n ≥ −(m − 1) ,
(10.35)
m−1,n U0,0 + ,
k6=l ,λk0,0 =λl0,0
with the coefficients Aτ,k,l
m−1,n to be determined ;
τ,k,b
τ,k,b
α
• gm−1,n modulo Ker A − iλτ,k
0,0 , specifically, we have the equations for um−1,n · ∇x π
k,b
and θm−1,n
, n ≥ −(m − 1) with boundary conditions
h
itan
itan
h
itan h
τ,k,int
τ,k,b
τ,k,int
χ
u
+
u
+
2d(u
)·n
,
=
∂ζ uτ,k,b
m−1,n
m−1,n−1
m−1,n−1
m−2,n
ν
(10.36)
k,int
k,b
k,b
k,b
D+1 χ
∂ζ θm−1,n = D+2 κ θm−1,n−1 + θ1,n−1 + ∂n θm−2,n + .
τ,k,b
Thus, we can solve gm−1,n
which depends on Aτ,k,l
m−1,i , −(m − 1) ≤ i ≤ n − 1.
k,b
k,b
τ,k,b
• um−1,n · ∇x d and ρm−1,n + θm−1,n which depends on Aτ,k,l
m−1,i , −(m − 1) ≤ i ≤ n − 1.
τ,k,int
τ,k,int
First, it is easy to derive that both gm,−m
and gm,−m+1
are hydrodynamic and in Ker A−
τ,k
τ,k,l
τ,k,l
iλ0,0 , thus will be determined after Am,−m and Am,−m+1 are known.
The induction start from n = −m + 2. At each (m, n), the 5 steps are the same as
before, we omit the details here. We only emphasize that the key is the system
τ,k,int
τ,k,int
τ,k,int
A − iλτ,k
= iλτ,k
+ iλτ,k
(10.37)
0,0 Um,n
1,1 Um−1,n−1 + ,
m,n U0,0
τ,k,l
τ,k,int
from which we can solve iλτ,k
modulo Ker A − iλτ,k
0,0 , with
m,n , Am−1,n−1 , and solve Um,n
the coefficient Aτ,k,l
m,n to be determined in the next row.
The induction process for general n at m-th row is the same as we did in m = 1.
We omit the details here. After we finish the induction for n at m-th row, we solve the
following:
NAVIER-STOKES-FOURIER
•
•
•
•
•
45
Solve all Aτ,k,l
m−1,n , thus all the terms in (m − 1, n) are completely determined;
τ,k,int
Solve gm,n partially, with Aτ,k,l
m,n to be determined;
τ,k,b
Solve gm,n partially, which depends on Aτ,k,l
m,i for −m ≤ i ≤ n−1 to be determined;
τ,k
Solve iλm,n completely;
k,b
k,b
τ,k,l
Solve uτ,k,b
m+1,n · ∇x d and ρm+1,n + θm+1,n partially, which depend on Am,i for −m ≤
i ≤ n − 1 to be determined.
Then we move to m + 1-th row, and continue. Finally we can solve for all m and n.
Remark: For the case 12 < β < 1, the leading term of iλτ,k
ε is
τ,k
τ,k
β
iλτ,k
+ iλτ,k
1,1 ε + iλ2,0 ε + · · ·
ε = iλ
with iλτ,k
1,1 < 0 .
(10.38)
The leading term of gετ,k is
τ,k,int
τ,k,int
τ,k,b β− 12
τ,k,bb β
ε
+ g1,1
ε + ···
gετ,k = g0,0
+ g0,1
+ g0,1
11. Proof of Proposition 7.1: the case β =
(10.39)
1
2
We treat the threshold case β = 21 separately in this section. The ansatz for gεBL and
λBL
are the same as the case β = 0. The process to determine all terms are also the same
ε
as before except the boundary condition. The key difference is that for the case β = 12
k,b
tan
the boundary condition of [uτ,k,b
and θ0,0
:
0,0 ]
itan
h
itan
h
τ,k,int
τ,k,b
χ
+
u
on ∂Ω ,
=
∂ζ uτ,k,b
u
0,0
0,0
0,0
ν
(11.1)
k,b
k,int
k,b
D+1 χ
∂ζ θ0,0 = D+2 κ θ0,0 + θ0,0
on ∂Ω .
k,b
tan
Using the boundary condition (11.1), we can solve [uτ,k,b
and θ0,0
, thus the boundary
0,0 ]
τ,k,int
value of u1,0 · n:
q
q
−1+τ (2b+1)i
−1+τ (2a+1)i ∂πα (∇x Φk ·∇x π α )
τ,k 2
k
ν
κ
+
iλ
,
(11.2)
Φ
uτ,k,int
·
n
=
0,0 D+2
1,0
(a+1)2 +a2
(b+1)2 +b2
2λk
2λk
iλτ,k
0,0
0,0
0,0
where
q
a=
λk0,0 ν 1
2 χ
,
b=
D+2
D+1
q
λk0,0 κ 1
2 χ
.
Following the same process as (8.39), (8.40) and (8.41), we can deduce that
Z
√
(2a+1)+i
τ,k
D+2 √ 1
iλ1,0 = − 2D
ν|∇x Φk |2 dσx
2 +a2
k
3
(a+1)
2[λ0,0 ]
Z ∂Ω
√
(2b+1)+i
2
√ 1k 3 (b+1)
− D+2
[λk ]2 κ|Φk |2 dσx .
2 +b2
2D
D+2 0,0
2[λ0,0 ]
(11.3)
(11.4)
∂Ω
Note: If a = 0, b = 0, then (11.2) and (11.4) reduce to the corresponding formulas (8.37)
and (8.41) in the case 0 ≤ β < 12 .
Apparently, we can deduce that
Re(iλτ,k
1,0 ) < 0 .
The rest of the process is similar as before, we omit the details here.
(11.5)
46
N. JIANG AND N. MASMOUDI
12. Proof of Proposition 7.1: Truncation Error Estimates
In the previous sections, we construct the kinetic-fluid boundary layers up to any order
for different values of β. For notational simplicity, now for different β, the powers m2 +( 12 −
β)n, m2 +(β − 21 )n and m2 +(β −1)n can be uniformly written as 2i +j with 0 ≤ j < 21 . Now
τ,k
we defined the approximated eigenfunction and eigenvalues gε,N
and λτ,k
ε,N by truncation
in the corresponding ansatzs. More specifically,
i
X τ,k,int
τ,k
τ,k,b
τ,k,bb
gε,N =
g i +j + g i +j + g i +j ε 2 +j ,
2
i
+j≤ N
2
2
X
λτ,k
ε,N =
2
2
(12.1)
i
λτ,k
ε 2 +j ,
i
+j
2
i
+j≤ N
2
2
τ,k
12.1. Estimates of Rε,N
. Using the eigen-equation (7.5), we can easily find that the
τ,k
error term Rε,N has the form of
N 1
N
τ,k,b
τ,k,bb
τ,k
− 12
2
Rε,N
= −L g τ,k,int
ε 2 − 2 + (v·∇x d)∂ξ g τ,k,bb
N
1 + gN
1 + gN
1
1 ε
N
+
+
+
+
2
2
2
2
2
2
2
2
(12.2)
+ higher order terms .
From the construction of the interior parts which is based on the solutions of equation
(5.26), and viscous fluid boundary layers, it is easy to know that
kg τ,k,int
kLr (dx ;Lp (aM dv)) ≤ C ,
i
+j
kLr (dx ;Lp (aM dv)) ≤ C ,
and kg τ,k,int
i
+j
2
(12.3)
2
for all i, j, and 1 < r, p < ∞. Moreover, the solutions of the linear kinetic boundary
layer equation (6.3) for g τ,k,bb
are also bounded in Lr (dx, Lp (aM dv)). Note that for every
i
+j
2
1 < p < ∞,
L : Lp (aM dv) → Lp (a1−p M dv) is bounded .
(12.4)
Thus we have the estimate
τ,k,int
τ,k,b
τ,k,bb ≤C.
(12.5)
L g N + 1 + g N + 1 + g N + 1 2
2
2
2
2
2
Lr (dx ;Lp (a1−p M dv))
and (v ·∇x d)∂ξ g τ,k,bb
, we integrate over Ω × RD and use
For the term (v ·∇x d)∂ζ g τ,k,bb
i
i
+j
+j
2
2
d(x)
√ or yD =
simple change of variable (y1 , y2 , · · · , yD−1 ) = π(x) , yD = d(x)
, we can have
ε
ε
√
extra ε and ε, so all will be in the higher order terms. Thus, we have the error estimate
(7.13).
τ,k
τ,k,int
12.2. Estimates of gε,N
− g0,0
. From (8.81), (10.39),, we know that the leading order
τ,k,b β− 1
τ,k
τ,k,int
τ,k,b
1
2 whenever
term of gε,N − g0,0 is g0
whenever 0 ≤ β < 12 ; gβ−
≤ β < 1 and λk
1 ε
2
1
τ,k,int β−
2 whenever
is simple, gβ−
1 ε
2
2
1
2
≤ β < 1 and λk is not simple.
k,b
Using the expressions above for uτ,k,b
, θ0k,b and uτ,k,b
, θβ−
1 , and a simple change of
0
β− 21
2
√
variable which will an extra ε, we have
1
τ,k
±,k,int
kgε,N
− g0,0
kLr (dx,Lp (aM dv)) ≤ Cε 2r ,
1
for 0 ≤ β < 12 ,
1
τ,k
±,k,int
kgε,N
− g0,0
kLr (dx,Lp (aM dv)) ≤ Cεβ− 2 + 2r ,
for
1
2
≤ β < 1.
(12.6)
NAVIER-STOKES-FOURIER
47
Thus we get (7.14).
τ,k
12.3. Boundary error estimate. Finally, we prove the boundary error term rε,N
which
is different for different values of β. When β = 0, it is obvious that there is no boundary
τ,k
τ,k
error, i.e. rε,N
= 0. The leading term of rε,N
is, when 0 < β < 12 ,
i N 1
h
τ,k,bb
τ,k
τ,k,bb
+ −β
+
γ
g
i
−
Lγ
g̃
−
Lγ
g
.
(12.7)
hγ− g̃ τ,k
− N + 1 −2β ∂Ω
− N + 1 −2β
− N + 1 −2β ε 2 2
N
+ 1 −2β
2
When
1
2
2
2
2
2
2
2
2
≤ β < 1, it is
h
i N 1
τ,k,bb
τ,k
τ,k,bb
hγ− g̃ τ,k
ε 2 − 2 +β ;
N
1 + γ− g N
1 i∂Ω − Lγ− g̃ N
1 − Lγ− g N
1
−
−
−
−
2
2
2
2
2
2
2
(12.8)
2
from which we can get the estimate (7.15). Thus we finish the proof of the Proposition
7.1.
13. Proof of the Weak Convergence
In order to derive the fluid equation with the boundary conditions, we need to pass to
the limit in approximate local conservation laws built from the renormalized Boltzmann
equation (2.8). We choose the renormalization used in [27]:
Γ(Z) =
Z −1
.
1 + (Z − 1)2
After dividing by ε, equation (2.8) becomes
ZZ
1 0
1
∂t g̃ + v · ∇x g̃ = Γ (G )
ε
ε
qε b(ω , v1 − v) dωM1 dv1 ,
(13.1)
(13.2)
SD−1 ×RD
where g̃ = Γ(G )/ε can be considered as the L2 part of the fluctuations gε . By introducing
Nε = 1 + ε2 gε2 , we can write
g̃ =
gε
,
Nε
Γ0 (G ) =
1
2
−
.
2
Nε
Nε
(13.3)
When moments of the renormalized Boltzmann equation (13.2) are formally taken with
respect to any ζ ∈ span{1 , v1 , · · · , vD , |v|2 }, one obtains the local conservation laws with
defects
1
1
∂t ρ̃ε + ∇x · ũε = hhΓ0 (G )qε ii ,
ε
ε
1
1
1
(13.4)
∂t ũε + ∇x (ρ̃ε + θ̃ε ) + ∇x ·hA(v)g̃ i = hhvΓ0 (G )qε ii ,
ε
ε
ε
EE
12
21
1 DD |v|2
0
∂t θ̃ε +
∇x · ũε +
∇x ·hB(v)g̃ i =
,
−
1
Γ
(G
)q
ε
D
εD
Dε
ε
which can be written as
eε + 1 AU
eε + Q
eε = R
eε ,
∂t U
(13.5)
ε
where
eε = ρ̃ε , ũε , θ̃ε = hg̃ i , hvg̃ i , h( |v|2 − 1)g̃ i ,
U
(13.6)
D
eε = 0 , 1 ∇x ·hA(v)g̃ i , 1 ∇x ·hB(v)g̃ i ,
Q
(13.7)
ε
ε
48
N. JIANG AND N. MASMOUDI
and the local conservation defect
EE
1 DD
|v|2
0
e
Rε =
(13.8)
1 , v , D − 1 Γ (G )qε .
ε
eε is not necessary in the domain of
Notice that because we don’t know if ũε · n = 0, so U
A for every ε > 0. However we can show that the weak limit of ũ, u, satisfies u · n = 0 on
the boundary, also see [23]. From the local conservation laws with defect (13.5), formally
eε will be in the null space of the acoustic operator A.
the limit of U
eε in (13.5) describes the acoustic waves with propagation speed 1 . As ε
The term 1ε AU
ε
goes to zero, the sound waves propagate faster and faster to make the fluid limit singular.
To derive the incompressible fluid equations, a natural way is to project the local consereε can be decomposed
vation laws (13.5) onto Null(A) and Null(A)⊥ respectively. First U
as
eε = ΠU
eε + Π⊥ U
eε
U
|v|2
|v|2
)g̃ i , Phvg̃ i , h( D+2
− 1)g̃ i
= h(1 − D+2
(13.9)
2
|v|
2|v|2
+ h D+2
g̃ i , Qhvg̃ i , h D(D+2)
g̃ i .
By definition of Leray projection in a bounded domain (5.2), the boundary conditions
of Phvg̃ i and Qhvg̃ i are
Phvg̃ i · n = 0 and Qhvg̃ i · n = ũε · n on ∂Ω .
(13.10)
eε , we take the test function
To derive the weak form of the evolution equations of ΠU
Y in (2.49) as special infinitesimal Maxwellian in the incompressible mode:
2
Y incom (x , v) = −χ + w·v + χ |v|2 − D2 ,
(13.11)
where (χ , w) ∈ C ∞ (Ω , RD × R) with ∇x ·w = 0 in Ω and w·n = 0 on ∂Ω. Because χ and
w are independent, the weak form of (13.5) can be written separately as:
Z
Z
Phvg̃ (t2 )i · w dx − Phvg̃ (t1 )i · w dx
Ω
Ω
1
−
ε
1
=
ε
t2
Z
t1
Z
Z
1
hAg̃ i : ∇x w dx dt + √
2πε
t1
Z
Z
hγg̃ (w·v)i∂Ω dσx dt
t1
Ω
t2
t2
Z
(13.12)
∂Ω
w·hhvΓ0 (G )qε ii dx dt ,
Ω
and
D+2
2
Z D
|v|2
D+2
E
− 1 g̃ (t2 ) χ dx −
D+2
2
Ω
1
−
ε
1
=
ε
Z D
|v|2
D+2
E
− 1 g̃ (t1 ) χ dx
Ω
t2
Z
t1
Z
t2
Z
1
hBg̃ i · ∇x χ dx dt + √
2πε
Ω
Z
χ
t1
Ω
DD
|v|2
D+2
Z
t2
t1
Z D E
|v|2
χ D+2
− 1 γg̃
∂Ω
EE
− 1 Γ0 (G )qε dx dt .
∂Ω
dσx dt
(13.13)
NAVIER-STOKES-FOURIER
49
Identities (13.12) and (13.13) are the local conservation laws in the incompressible
modes. It is the starting point of the proof of the weak convergence to the incompressible
Navier-Stokes equations with boundary conditions in the Main Theorems 3.1 and 3.2. It
has been proved in [27] the convergence of the interior terms of (13.12) and (13.13) as
ε → 0 to recover the weak form of incompressible Navier-Stokes equations. It is only left
to derive the boundary conditions of the limiting equations.
The strategy to recover the boundary conditions in the limit is basically the same
as Masmoudi and Saint-Raymond [33] except some necessary modifications. For the
convenience of the readers, we briefly go through the proof here. In [27], the author proved
that inside the domain R+ × Ω × RD , the family of fluctuations gε is relatively compact in
w-L1loc (dt; w-L1 (σM dvdx)), and that every limit point g has the form (3.4). Lemma 5.1
of [33] showed that the trace of the limit point γg belongs to L1loc (dt; L1 (M |v·n(x)|dσx ))
and satisfies
γg = v·γu +
|v|2
2
−
D+2
2
γθ
(13.14)
where γu and γθ denote the fluid traces of u and θ.
We list some key a priori estimates from [33] on γgε . The first one is from the inside,
we generalize it to the more general collision kernel case considered in this paper.
Lemma 13.1. For all p > 0, as ε → 0,
γg̃ → γg in w-L1loc (dt; w-L1 (M (1 + |v|p )|v·n(x)|dvdσx )) .
(13.15)
Proof. The proof is essentially the same as Lemma 5.2 of [33], except for some new argument to treat the soft potential collision kernel case.
Using the function
5/3
Z−1
Γ(Z) = 1+(Z−1)
(13.16)
2
in the renormalized formulation (2.47) gives
ZZ
2
1
5
2/3
5/3
g̃ qε
(ε∂t + v·∇x )g̃ =
−
b(ω, v − v1 )dωM1 dv1 .
3
Nε2 Nε
(13.17)
SD−1 ×RD
To estimate the right-hand side in (13.17), we apply the classical Young’s inequality,
namely,
pz ≤ r∗ (p) + r(z) ,
(13.18)
for every p and z in the domains of r∗ and r. Here the function r is defined over z > −1
by r(z) = z log(1 + z) and is strictly convex. r∗ is the Legendre dual of r.
!
2 2 2/3
qε
1
ε
q
1
ε
g̃
ε
|g̃ |2/3 ≤ G Gε1 r
+ 4 G Gε1 r∗
N2
ε4
G
G
ε
Nε2
ε1
ε
(13.19)
2 1
ε qε
|g̃ |4/3 ∗
≤ 4 G Gε1 r
+ G Gε1
r (1) ,
ε
G Gε1
Nε4
The second inequality above used the superquadratic homogeneity of r∗ . By the entropy
dissipation rate bound, the first term on the right-hand
side of (13.19) is bounded in
√
1
1
Lloc (dt , L (dν dx)). Noting that Nε ≥ 1 and G ≤ 2Nε , the integral of the second term
50
N. JIANG AND N. MASMOUDI
can be bounded as follows:
√ ZZ
b(v1 − v)
|g̃ |4/3 Gε1
a1 M1 dv1 aM dv
2
a(v1 )a(v)
RD ×RD
√ ZZ
≤ 2
RD ×RD
|g̃ |4/3
b(v1 − v)
√ (1 + ε|g̃ε1 |)
a1 M1 dv1 aM dv
3
a(v1 )a(v)
Nε Nε
(13.20)
√ ZZ
b(v1 − v)
|g̃ |4/3 ε|gε1 − g̃ε1 |
a1 M1 dv1 aM dv .
+ 2
a(v1 )a(v)
RD ×RD
Using the assumption 3 (2.17) and |εg̃ | ≤ 21 , the first term in (13.20) is bounded. Indeed,
it is bounded by
Z
Z
|g̃ |4/3
gε2
√
√
C
aM
dv
≤
C
aM dv ≤ C .
(13.21)
3 N
Nε
ε
R D Nε
RD
The second term in (13.20) is bounded as
2
√ ZZ
|εgε1 | gε1
b(v1 − v)
√
2
|εg̃ |4/3 ε2/3 √
aM dv a1 M1 dv1
Nε1 Nε1 a(v1 )a(v)
RD ×RD
(13.22)
4/3 Z
1
g2
2/3
√ ε1 a1 M1 dv1 .
≤ε 2
2
Nε1
RD
5/3
Notice that (13.22) vanishes as ε goes to zero. Thus we deduce that (ε∂t + v·∇x )g̃ε is
uniformly bounded in L1loc (dt , L1 (M dvdx)). The rest of the proof of (13.15) is the same
as that of Lemma 5.2 in [33].
The next lemma is the a priori estimate of γgε from the boundary term in the entropy inequality (2.52). The proof is the same as Lemma 6.1 in [33] with some trivial
modification. So we just state the lemma without giving the proof.
Lemma 13.2. Define γε = γ+ gε − hγ+ gε i∂Ω and
γε(1) = γε 1γ+ Gε ≤2hGε i∂Ω ≤4γ+ Gε ,
γε(2) = γε − γε(1) .
(13.23)
Then each of these is bounded as follows:
r
(1)
α
γε
in L2loc (dt; L2 (M |v·n(x)|dvdσx )) ,
(13.24)
ε [1 + ε2 (γ+ gε )2 ] 41
r
(1)
α
γε
in L2loc (dt; L2 (M |v·n(x)|dvdσx )) ,
(13.25)
ε [1 + ε2 hγ+ gε i2∂Ω ] 41
α (2)
in L1loc (dt; L1 (M |v·n(x)|dvdσx )) .
(13.26)
γε
2
ε
Using Lemma 13.1 and Lemma 13.2, we can prove the following lemma which describes
how to define the renormalized outgoing mass flux 1Σ+ ρ.
NAVIER-STOKES-FOURIER
51
√
Lemma 13.3. Assume that α /( 2πε) → χ ∈ (0 , +∞]. Then up to the extraction of a
subsequence,
γε
γε
and
(13.27)
2
2
2
1 + ε γ+ gε
1 + ε hγ+ gε i2∂Ω
converge in w-L1loc (dt; w-L1 (M |v ·n(x)|dvdx)) and have the same weak limit. Moreover,
there exists ρ ∈ L1loc (dt; L1 (dσx )) such that, up to the extraction of a subsequence,
1Σ+ hγ+ gε i∂Ω
→ 1 Σ+ ρ
1 + ε2 γ+ gε2
in w-L1loc (dt; w-L1 (M |v·n(x)|dvdx)) .
(13.28)
Furthermore,
ρ = hγ+ gi∂Ω .
(13.29)
Lemma 13.3 is Lemma 6.2 and Lemma 6.3 in [33]. The proof is basically the same
except some trivial modifications because we use some different renormalizations. We
skip the proof here.
Now it is ready to recover the Dirichlet boundary condition. For the case α /ε → ∞,
by (13.24) and (13.26), we deduce that
(1)
γε
→ 0 strongly in L2loc (dt; L2 (M |v·n(x)|dvdx)),
1 + ε2 hγ+ gε i2∂Ω
(2)
γε
→ 0 strongly in L1loc (dt; L1 (M |v·n(x)|dvdx));
2
2
1 + ε hγ+ gε i∂Ω
hence, we get
1+
γε
2
ε hγ+ gε i2∂Ω
→ 0 strongly in L1loc (dt; L1 (M |v·n(x)|dvdx)) .
(13.30)
On the other hand,
1Σ+ hγ+ gε i∂Ω
γε
=
γ
g̃
−
→ γ+ g − 1Σ+ ρ ,
+
1 + ε2 γ+ gε2
1 + ε2 γ+ gε2
(13.31)
in w-L1loc (dt; w-L1 (M |v·n(x)|dvdx)). Then (13.30) and (13.31) imply that
γ+ g = 1Σ+ ρ
(13.32)
where ρ depends only on (t, x). Thus, by (13.14), we get the Dirichlet boundary condition
γu = 0 and γθ = 0 .
(13.33)
Now, we concentrate on the Navier boundary condition case. Using the previous convergence results, we can take limits in the conservation laws (13.12) and (13.13) to get the
weak form of the boundary conditions. In the weak forms (13.12) and (13.13), the limits
of the interior terms have been carried in Levermore-Masmoudi [27], which is stated in
the following lemma:
Lemma 13.4. Assume that
|v|2
√α
2πε
→ χ ∈ [0, ∞), then up to the extraction of a sequence,
Phvg̃ i and h( D+2 − 1)g̃ i converge to u and θ in C([0, ∞); w-L1 (dx)) such that, for all
52
N. JIANG AND N. MASMOUDI
w ∈ C ∞ (Ω; RD ) with ∇x ·w = 0 in Ω and w·n = 0 on ∂Ω, for all χ ∈ C ∞ (Ω; R), and for
all t1 , t2 > 0,
Z
Z
Z t2 Z X
u(t2 )·w dx − u(t1 )·w dx −
ui uj ∂i wj dxdt
Ω
Ω
Z
+ν
t2 Z X
t1
t1
Ω i,j
(∂i uj + ∂j ui )∂i wj dxdt
(13.34)
Ω i,j
+
(1)
2
γ
(w·v)1
α
ε
|v| ≤20| log ε|
dσx dt ,
= − lim √
2
2
ε→0
2πε t1 ∂Ω (1 + ε γ+ gε )(1 + ε2 γ+ ĝε2 ) ∂Ω
Z t2 Z
Z
Z
Z t2 Z
2
∇x θ·∇x χ dxdt
θ(t2 )·χ dx − θ(t1 )·χ dx −
θu·∇x χ dxdt + D+2 κ
t1
t1
Ω
Ω
Ω
Ω
+
Z t2 Z *
(1)
2
α
γε
|v|
= − lim √
dσx dt .
χ D+2 − 1 1|v|2 ≤20| log ε|
ε→0
2πε t1 ∂Ω (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
∂Ω
(13.35)
where γ+ ĝε = (1 − α )γ+ gε + α hγ+ gε i∂Ω .
Z
t2
Z
*
Proof. It is the analogue of Lemma 7.1 and Lemma 7.2 of [33] and the idea of the proof
|v|2
is the same. Denote by Yε the test function (w ·v)1|v|2 ≤20| log ε| or χ( D+2
− 1)1|v|2 ≤20| log ε| .
Then Yε has the property: Yε = LYε . From (2.50), the renormalized form of the Maxwell
boundary condition reads
Lγ+ g
hγ+ g i∂Ω
+ α
,
(13.36)
γ− g̃ = (1 − α )
2
2
1 + ε (Lγ+ ĝε )
1 + ε2 (Lγ+ ĝε )2
where
γ+ ĝε = (1 − α )γ+ gε + α hγ+ gε i∂Ω ,
(13.37)
= γ+ g − α γε .
Then
Z
Z 1
1
ε2 γ+ gε (γ+ ĝε2 − γ+ gε2 )
dσx
hγg̃ Yε i∂Ω dσx =
Yε 1 Σ +
ε ∂Ω
ε ∂Ω (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
∂Ω
Z α
γε
+
Yε 1 Σ +
dσx
ε ∂Ω 1 + ε2 γ+ ĝε2
∂Ω
+
Z *
(1)
(2)
(13.38)
α
γε + γε
=
Y
1
dσ
ε
Σ
x
+
ε ∂Ω (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
+∂Ω
Z * (1)
(2) 2
α
(γε + γε )ε (γ+ gε γ+ ĝε )
−
Yε 1 Σ +
dσx .
ε ∂Ω (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
∂Ω
By (13.26),
Z *
+
(2)
α
γ
ε
Y
1
dσ
x dt
2 γ g 2 )(1 + ε2 γ ĝ 2 ) ε Σ+
ε
(1
+
ε
+ ε
+ ε
t1
∂Ω
∂Ω
Yε
≤ Cε (1 + ε2 γ+ g 2 )(1 + ε2 γ+ ĝ 2 ) ≤ Cε| log ε| .
ε
ε
∞
Z
t2
(13.39)
NAVIER-STOKES-FOURIER
53
(2)
(1)
The γε part in the last term of (13.38) can be estimated as (13.39). For the γε part,
from (13.24),
r
(1)
γε
α
p
is relatively compact in w-L1loc (dt; w-L1 (M |v·n|dvdσx )) (13.40)
2
2
ε 1 + ε γ+ gε
Use the fact that
r
α
εγ+ gε εγ+ ĝε
p
ε 1 + ε2 γ+ gε2 (1 + ε2 γ+ ĝε2 )
(13.41)
is bounded in L∞ and goes to 0 a.e. Then by the Product Limit Theorem of [3], the
product of (13.40) and (13.41) goes to 0 in L1loc (dt) as ε → 0. Thus we finish the proof of
the lemma.
Now, it is ready to recover the Navier boundary condition by taking limit in the last
terms in (13.34) and (13.35). As in [33], we can deduce that
+
* (1)
γε (w·v)1|v|2 ≤20| log ε|
α
√
→ λh(γ+ g − 1Σ+ hγ+ gi∂Ω )(w·v)i∂Ω ,
(13.42)
2πε (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 ) ∂Ω
*
+
(1)
2
α
γε
|v|
√
χ D+2
− 1 1|v|2 ≤20| log ε|
2 γ g 2 )(1 + ε2 γ ĝ 2 )
(1
+
ε
2πε
+ ε
+ ε
(13.43)
∂Ω
2
|v|
− 1)i∂Ω
→ λh(γ+ g − 1Σ+ hγ+ gi∂Ω )χ( D+2
in w-L1loc (dt; w-L1 (dσx )). Use (13.14), we finally prove the weak form of the incompressible
Navier-Stokes equations with Navier boundary conditions:
Z
Z
Z t2 Z X
ui uj ∂i wj dxdt
u(t2 )·w dx − u(t1 )·w dx −
Ω
Ω
t2 Z X
Z
+ν
t1
Z
Ω i,j
(13.44)
(∂i uj + ∂j ui )∂i wj dxdt
Ω i,j
t2
Z
=λ
γu·wdσx dt ,
t1
∂Ω
Z
Z
θ(t2 )·χ dx −
Ω
+
t1
Z
Z
t2
Z
∇x θ·∇x χ dxdt =
t1
Ω
Z
θ(t1 )·χ dx −
Ω
2
κ
D+2
t2
θu·∇x χ dxdt
t1
Ω
D+1
α
D+2
Z
t2
(13.45)
Z
γθχσx dt .
t1
∂Ω
Thus we finish the proof of the weak convergence results in the Main Theorem.
14. Proof of the Strong Convergence
eε converge only weakly
We proved that the projection on the incompressible mode ΠU
to solutions to the incompressible Navier-Stokes-Fourier equations, see (3.14). This weak
eε , as for the periodic doconvergence is caused by the persistence of fast acoustic part Π⊥ U
⊥e
main [27]. If Π Uε vanishes in the limit as ε goes to zero, we can improve the convergence
results from weak to strong. The main novelty of this paper is that when √α2πε
→ ∞, the
acoustic modes will be damped.
54
N. JIANG AND N. MASMOUDI
eε be defined as (13.9). If α /ε → ∞ as ε → 0, then
Proposition 14.1. Let Π⊥ U
eε → 0 in L2loc (dt; L2 (dx)) ,
Π⊥ U
(14.1)
as ε → 0.
We will prove Proposition 14.1 later. Now we apply it to prove the Main Theorem 3.1.
14.1. Strong Convergence in L1 . We first show that we can improve the relative compactness of the family of fluctuations gε from weak to strong in L1loc (dt; L1 (σM dvdx)).
Indeed, gε can be decomposed as
ε2 gε3
gε = P g̃ + P ⊥ g̃ +
Nε
2
|v|
D+2
2
D
θ̃ε − D+2
ρ̃ε
−
= v·Pũε + D+2
2
2
(14.2)
|v|2
+ v·P⊥ ũε + D+2
ρ̃ε + θ̃ε
ε2 gε g 2
+ P ⊥ g̃ + √ √ ε ,
Nε Nε
where P is the projection to Null(L), i.e.
2
P g̃ = hg̃i + hvg̃i·v + h( |v|D − 1)g̃i
|v|2
2
−
D
2
.
(14.3)
It is proved in [27] that P ⊥ g̃ → 0 in L2loc (dt; L2 (aM dvdx)) and
Pũε → u ,
D
θ̃
D+2 ε
−
2
ρ̃
D+2 ε
→ θ,
in L2loc (dt; L2 (dx)) .
(14.4)
The Proposition 14.1 yields that
|v|2
⊥
(14.5)
v·P ũε + D+2 ρ̃ε + θ̃ε → 0 in L2loc (dt; L2 (M dvdx)) .
Thus P g̃ → g = v · u + 12 |v|2 − D+2
θ in L2loc (dt; L2 (M dvdx)), as ε → 0. The key
2
nonlinear estimate in [3] claims that
g2
σ √ ε = O(| log ε|) in L∞ (dt; L1 (aM dvdx)) .
Nε
It is easy to see that
√εg̃
Nε
(14.6)
is bounded, thus
ε2 gε3
→ 0 in L1loc (dt; L1 (σM dvdx)) .
(14.7)
Nε
We deduce that gε is relatively compact in L1loc (dt; L1 (σM dvdx)) and that every limit g
has the form (3.4), Combining the above estimates.
Next, we can also improve the convergence of the moment of gε . In [27], the author
1
|v|2 − 1)gε i converge to u and θ in
proved that the incompressible part Phvgε i and h( D+2
1
C([0, ∞); w-L1 (dx)), respectively. We also have Phvg̃ i and h( D+2
|v|2 − 1)g̃ i converge to
2
2
u and θ in Lloc (dt; L (dx)). Now, from Proposition 14.1, we know that the acoustic part
1
P⊥ hvg̃ i and h( D+2
|v|2 g̃ i converge strongly to 0 in L2loc (dt; L2 (dx)). So combining this
with (14.7), we get
hvgε i → u in L1loc (dt; L1 (dx; RD )) ∩ C([0, ∞); w-L1 (dx; RD )) ,
h( D1 |v|2 − 1)gε i → θ
in L1loc (dt; L1 (dx; R)) ∩ C([0, ∞); w-L1 (dx; R)) .
(14.8)
NAVIER-STOKES-FOURIER
55
Furthermore, since now we have ũε → u and θ̃ → θ in L2loc (dt; L2 (dx)), we can improve
the Quadratic Limit Theorem 13.1 in [27] to
ũε ⊗ ũε → u ⊗ u ,
θ̃ε ũε → uθ ,
θ̃ε2 → θ2
in L1loc (dt; L1 (dx)) ,
(14.9)
as ε → 0.
1
Let s ∈ (0, ∞] be from the assumed bound (2.17) on b. Let p = 2 + s−1
, so that p = 2
ˆ hence,
when s = ∞. Let ξˆ ∈ Lp (aM dv) be such that P ξˆ = 0 and set ξ = Lξ,
1
1
ˆ
ˆ
ˆ
hξg̃ i = hξP ⊥ g̃ i = hξQ(g̃
, g̃ )i − hhξ q̃ε ii + hhξTε ii .
ε
ε
We know from in [27] that
ˆ ε ii → 0 in L1 (dt; L1 (dx)) ,
hhξT
loc
(14.10)
(14.11)
and
b : ∇x u + hξ Bi·∇
b
hhξˆq̃ε ii → hξ Ai
xθ
in w-L2loc (dt; w-L2 (dx)) .
(14.12)
Note that
⊥
ˆ
ˆ
ˆ
hξQ(g̃
, g̃ )i = hξQ(P g̃ , P g̃ )i + 2hξQ(P g̃ , P g̃ )i
⊥
ˆ
+ hξQ(P
g̃ , P g̃ )i .
(14.13)
It is easy to show that the last two terms above vanish as ε → 0. For the first term,
ˆ
hξQ(P
g̃ , P g̃ )i = 12 hξP ⊥ (P g̃ )2 i
= 21 hξAi : (ũε ⊗ ũε ) + hξBi· ũε θ̃ε + 12 hξCiθ̃ε2 .
(14.14)
Applying the quadratic limit (14.9), (14.14) can be taken limit in L1loc (dt; L1 (dx)) strongly.
Combining with convergence (14.11) and (14.12), we get
D E
1
b : ∇x u − B·∇
b xθ
hξP ⊥ g̃ i → ξ 21 A : u ⊗ u + B·uθ + 21 Cθ2 − A
(14.15)
ε
in w-L1loc (dt; w-L1 (dx)). Since gε − g̃ → 0 in L∞ (dt; L1 (σM dvdx)), the convergence above
implies (3.8). Thus we finish the proof of the Main Theorem 3.1.
14.2. Proof of Proposition 14.1. We will reduce the proof of the Proposition 14.1 to
eε on each fixed acoustic mode goes to zero in L2 (dt; L2 (dx)).
show that the projection of U
loc
⊥e
∞
2
We know that Π Uε is uniformly bounded in L (dt; L (dx)), so it can be represented as
q
X
eε = D+2
eε , U + U + + U
eε , U − U −
Π⊥ U
U
k,0
k,0
k,0
k,0
2D
k∈N
2D|v|2
k
h (D+2)
2 g̃ iΦk,0 dx Φ
Ω
X R
k
D+2
∇x Φk 
xΦ
 Ω hvg̃ i· ∇iλ

k dx
2D
iλk
R 4|v|2
k
k∈N
h
g̃ iΦk,0 dx Φ
Ω (D+2)2 R
=
q

(14.16)
The above summation includes infinitely many terms. To reduce the problem to a
finite number of modes, we need some regularity in x of hvg̃ i and h|v|2 g̃ i. The tool
adapted to investigating this property is the velocity averaging theorem given in [13] and
the improvement to L1 averaging in [17].
56
N. JIANG AND N. MASMOUDI
Following the similar argument in the proof of Proposition 11.2 in [27], and apply to
(13.2), we can show that for each ζ ∈ Span{1 , v , |v|2 } and T > 0, there exists a function
η : R+ → R+ such that lim+ η(z) = 0,
z→0
khζ g̃ (t, x + y, v) − ζ g̃ (t, x, v)ikL2 ([0,T ]×Ω) ≤ η(|y|) ,
(14.17)
for every y ∈ Ω such that |y| ≤ 1, uniformly in ε ∈ [0, 1]. From the classical criterion
of compactness in L2 , hvg̃ i and |v|2 g̃ are relatively compact in L2loc (dt , L2 (dx)) which
implies that
2
X Z T τ,k eε , U0,0
eε kL2 ([t ,t ];L2 (dx)) → 0 as N → ∞ ,
(14.18)
dt ≤ CN kΠ⊥ U
U
1 2
k>N
0
⊥e
recalling from (14.17) that CN → 0 as N → ∞. (14.18) implies that,
to show Π Uε → 0
τ,k
eε , U0,0
strongly in L2loc (dt , L2 (dx)), we need only to prove that U
converges strongly to
0 in L2 (0, T ) for any fixed acoustic mode k. Furthermore, the relation
Z D
E
τ,k
±,k,int
e
Uε , U0,0 =
g̃ , g0,0
dx ,
(14.19)
Ω
implies that the proof of Proposition 14.1 is reduced to showing that :
Proposition 14.2. Assume that α /ε → ∞ as ε → 0 and let g̃ be the renormalized
±,k,int
fluctuation defined in (13.3), satisfying the scaled Boltzmann equation (13.2), and g0,0
be the infinitesimal Maxwellian of acoustic mode k ≥ 1:
±,k,int
=
g0,0
D
Φk
D+2
+
∇x Φk
±iλk
Then, for any fixed mode k,
Z D
E
±,k,int
dx → 0
g̃ g0,0
in
·v+
2
2
Φk ( |v|2
D+2
L2 (0, T ) ,
as
− D2 ) .
ε → 0.
(14.20)
(14.21)
Ω
Proof. We start from the weak formulation of the rescaled Boltzmann equation (2.49) with
the test function Y taken to be one of the approximate eigenfunctions of Lε constructed
τ,k
in Proposition 7.1 to the order N = 4, namely Y = gε,4
:
Z
Z
τ,k
τ,k
hg̃ (t2 )gε,4
i dx − hg̃ (t1 )gε,4
i dx
Ω
Ω
Z Z
Z Z
1 t2
1 t2
τ,k
τ,k
hg̃ Lε gε,4 i dx dt +
hγg̃ γgε,4
(v·n)i dσx dt
+
(14.22)
ε t1 Ω
ε t1 ∂Ω
Z Z
1 t2
τ,k
=
hhRε gε,4
ii dx dt ,
ε t1 Ω
where
and
1
Lε = L − v·∇x ,
ε
0
1 gε1
gε
gε1
gε0
0
Rε = Γ (Gε )qε +
+
− 0 − 0 .
ε Nε1 Nε Nε1
Nε
(14.23)
(14.24)
NAVIER-STOKES-FOURIER
Define
b̃±
k,ε (t)
Z
57
τ,k
hg̃ (t)gε,4
i dx ,
=
(14.25)
Ω
then from (14.22) b̃±
k,ε (t) satisfies
b̃±
k,ε (t2 )
−
b̃±
k,ε (t1 )
1
− iλτ,k
ε ε,4
Z
t2
b̃±
k,ε (t) dt
Z
t2
=
t1
c±
k,ε (t) dt ,
(14.26)
t1
where c±
k,ε (t) is:
c±
k,ε (t)
Z
Z
1
1
τ,k
τ,k
hg̃ (t)Rε,4 i dx −
hγg̃ γgε,4
i dσx
=−
ε Ω
ε ∂Ω
Z
1
τ,k
hhRε gε,4
ii dx .
+
ε Ω
(14.27)
We claim that the boundary contribution in (14.27) is zero as ε → 0, i.e.
τ,k
Lemma 14.1. Let gε,4
be the approximate eigenfunction of Lε constructed in Proposition
7.1. Then,
Z
1
τ,k
hγg̃ γgε,4
(v·n)i dσx
(14.28)
ε
∂Ω
is the summation of two terms, one is bounded in Lploc (dt) for p > 1, the other vanishes
in L1loc (dt) as ε → 0.
We leave the proof of Lemma 14.1 to the section 14.5.
±
14.3. Estimates of c±
k,ε . We will decompose ck,ε (t) into two parts: one is vanishing in
p
L1loc (dt), the other is bounded in Lloc (dt) for some p > 1. First, taking N = 4, r = p = 2
2
in (7.13) and noticing the L∞
loc (dt , L (aM dvdx)) boundedness of g̃ , we have the estimate
of the first term in (14.27):
1 Z
1
τ,k
τ,k
kL2 ( 1 M dvdx) kg̃ kL2 (aM dvdx)
hg̃ (t)Rε,4 i dx ≤ kRε,4
a
ε Ω
ε
(14.29)
√
≤ C ε.
Second, Lemma 14.1 implies that the boundary term in (14.27) will be vanishing in L1loc (dt)
as ε goes to zero. The third term in (14.27) is estimated as follows:
Z
1
τ,k
hhRε gε,4
ii dx
ε Ω
Z
Z
(14.30)
1
1
τ,k
⊥ τ,k
hhRε Pgε,4 ii dx +
hhRε P gε,4 ii dx .
=
ε Ω
ε Ω
τ,k
Because Pgε,4
is in Null(L), so the first term in the right-hand side of (14.30) has the
form of
Z
Z
1
1
0
hhΓ (Gε )qε ζii dx + 2 hLg̃ ζidx , for some ζ ∈ Null(L) .
(14.31)
ε Ω
ε Ω
The second term above is zero, the first term converges to zero strongly in L1loc (dt) as
ε → 0 by the Conservation Defect Theorem (Proposition 8.1) in Levermore-Masmoudi
58
N. JIANG AND N. MASMOUDI
[27]. For the second term in the right-hand side of (14.30), from the calculations in
Proposition 7.1 again, we have
√
1 ⊥ τ,k ∇x2 Φk b 2∇x Φk b −1
k,b
b
b
P gε,4 = ±iλk : A + D+2 · B + ∇x d⊗∂ζ uτ,k,b
:
A
+
∂
θ
∇
d·
B
ε
1
x
ζ 1
2
2
ε
(14.32)
+P ⊥ gβτ,k,bb εβ−1 + higher order terms .
We decompose Rε into
Rε = Tε +
0 0
(g̃ε1
g̃ε
− g̃ε1 g̃ε1 ) + qε
1
2
1
− 0 0
−
2
Nε
Nε Nε1 Nε Nε1 Nε
,
(14.33)
1 0
0 0
− (g̃ε1
+ g̃ε0 − g̃ε1 − g̃ε ) − (g̃ε1
g̃ε − g̃ε1 g̃ε1 ) .
ε
(14.34)
where Tε is defined as
Tε =
qε
0
Nε1 Nε0 Nε1 Nε
When we integrate (14.32) over Ω, for the second term of (14.32) which is a function of
√ ), we make the change of variables:
(x , d(x)
ε
√ .
(14.35)
y1 = Π(x)1 , · · · , yD−1 = Π(x)D−1 , yD = d(x)
ε
√
√
√ −1
Then dx = ε|∇x d|dy. This extra ε cancels with the ε in the second term of
(14.32). Similarly, for the third term of (14.32), we make the change of variables y1 =
Π(x)1 , · · · , yD−1 = Π(x)D−1 , yD = d(x)
, and consequently dx = ε|∇x d|dy . Thus, for the
ε
integral of (14.32), the first two terms are the same order, while the third term is of higher
order εβ with β > 0. By the Flux Remainder Theorem (Proposition 10.1) of [27], we have
Z
1
τ,k
hhTε P ⊥ gε,4
iidx → 0 , in L1loc (dt) ,
(14.36)
ε Ω
as ε → 0. Furthermore, from the Bilinear estimates (Lemma 9.1) of [27], we have
Z
Z
1
0 0
⊥ τ,k
hh(g̃ε1 g̃ε − g̃ε1 g̃ε1 )P gε,4 iidx ≤ C hag̃2 i dx ≤ C .
(14.37)
ε Ω
Ω
√
that
The third term in (14.33) can be written as qε / Nε times a bounded sequence √
2
vanishes almost everywhere as ε → 0. Then the L (dνdxdt) boundedness of qε / Nε
implies that it times (14.32) is relatively compact in w-L1loc (dt; w-L1 (dx)), then following
from the Product Limit Theorem of [3],
Z
1
τ,k
2
1
1
hhqε N 2 − Nε − N 0 N 0 Nε1 Nε P ⊥ gε,4
iidx
ε
ε
ε1
ε Ω
(14.38)
1
→ 0 , in Lloc (dt) , as ε → 0 .
Based on the above arguments, we now can decompose c±
k,ε (t) into
Z
Z
1
1
τ,k
τ,k
0 0
c±,1
hh(g̃ε1
g̃ε − g̃ε1 g̃ε1 )P ⊥ gε,4
iidx +
hγg̃ γgε,4
(v·n)i dσx ,
k,ε (t) =
ε Ω
ε
(14.39)
∂Ω
c±,2
k,ε (t)
and
= c±
k,ε (t) −
p
Lloc (dt) for p > 1.
c±,1
k,ε (t),
where
c±,2
k,ε (t)
→ 0, in L1loc (dt), and c±,1
k,ε (t) is bounded in
NAVIER-STOKES-FOURIER
59
±
14.4. Estimates of b̃±
k,ε . From (14.26), b̃k,ε satisfies the ordinary differential equation
d ±
1
±,2
b̃± = c±,1
b̃k,ε − iλτ,k
k,ε (t) + ck,ε (t) .
dt
ε ε,4 k,ε
The solution to (14.40) is
Z t
1
±,2
iλτ,k
t
− 1ε iλτ,k
±
±
ε,4
ε,4 (s−t) ds .
ε
b̃k,ε (t) = b̃k,ε (0)e
+
[c±,1
k,ε (s) + ck,ε (s)]e
(14.40)
(14.41)
0
From the Proposition 7.1, when 0 ≤ β < 21 ,
i
1
1 τ,k
1 h
−β
2
iλε,4 t = √ Re(iλτ,k
Re(iλ̃τ,k
)
t
1 ) + ε
1−β
2
ε
ε
1
1 k
1
τ,k
τ,k
− i ± λ + √ Im(iλ 1 ) + β Im(iλ̃1−β ) t ,
2
ε
ε
ε
while when
1
2
≤ β < 1,
i
1 h
1 τ,k
τ,k
1−β
)
+
ε
Re(i
λ̃
)
t
iλε,4 t = 1−β Re(iλτ,k
1
β
ε
ε
1 k
1
τ,k
τ,k
− i ± λ + 1−β Im(iλβ ) + Im(iλ̃1 ) t .
ε
ε
(14.42)
(14.43)
Using (14.42) and (14.43), the first term in (14.41) is estimated as follows: when 0 ≤ β <
1
,
2
1
τ,k
iλε,4 t
ε
kL2 (0,T )
kb̃±
k,ε (0)e
=|b̃±
k,ε (0)|
h
while when
1
2
τ,k
−2 Re(iλ 1 ) + ε
2
1
−β
2
i− 12 Re(iλ̃τ,k
1−β )
1−e
1 −β
1
2
√
[Re(iλτ,k
Re(iλ̃τ,k
1 )+ε
1−β )]T
ε
2
12
1
ε4 ,
(14.44)
1
≤ β < 1,
τ,k
iλε,4 t
ε
kL2 (0,T )
kb̃±
k,ε (0)e
h i− 12 12 1
1
1−β Re(iλ̃τ,k )]T
√
[Re(iλτ,k
τ,k
τ,k
1−β
1
β )+ε
ε
=|b̃±
(0)|
−2
Re(iλ
)
+
ε
Re(i
λ̃
)
1
−
e
ε 2 (1−β) .
1
k,ε
β
(14.45)
±
To estimate |b̃k,ε (0)|, from
Z
Z
τ,k,int
τ,k
τ,k,int
±
in
− g0,0
)i dx ,
(14.46)
b̃k,ε (0) = hg̃ , g0,0 i dx + hg̃in (gε,4
Ω
Ω
±,k,int
g0,0
noticing that
∈Null(L) and khζ(v)g̃in ikL2 (dx) is bounded for every ζ(v) ∈Null(L),
τ,k
±,k,int
and the error estimate for gε,4
− g0,0
in (7.14), we deduce that |b̃±
k,ε (0)| is bounded.
τ,k
τ,k
1
Using the fact that Re(iλ 1 ) < 0 when 0 ≤ β < 2 , and Re(iλβ ) < 0 when 21 ≤ β < 1,
2
we deduce that for any 0 < T < ∞, sufficiently small ε:
1
1
− 12 iλτ,k
ε,4 t k 2
ε
kb̃±
when 0 ≤ β < ,
L (0,T ) ≤ Cε 4 ,
k,ε (0)e
2
1
1
(1−β)
≤ Cε 2
≤β<1 .
, when
2
(14.47)
60
N. JIANG AND N. MASMOUDI
In order to estimate the remaining term in (14.41), we observe that for any a ∈ Lp (0, t)
and 1 ≤ p, r ≤ ∞, such that p−1 + r−1 = 1, we have
Z t
Z t
− √1 Re(iλτ,k
1 )(s−t)
− 1ε iλτ,k
(s−t)
ε,4
2
a(s)e
ds ≤ C
e ε
|a(s)| ds ,
0
when 0 ≤ β <
0
1
. (14.48)
2
Direct calculations show that

1
− √ Re(iλτ,k
1 )(t−s)
e
2
1
= ε 2r 
Lr (0,t)
1
−rRe(iλτ,k
1 )
− √r Re(iλ± 1 )t
k, 2
e
1
r
− √1 Re(iλτ,k
1 )t
2
−1  e ε
.
2
(14.49)
Using the fact −Re(iλτ,k
1 ) < 0 again, we have
2
Z t
1
− 1ε iλτ,k
ε,4 (s−t) ds ≤ Ckak p
a(s)e
L (0,t) ε 2r ,
for 0 ≤ β <
0
1
.
2
(14.50)
Similarly, we also have the estimate
Z t
1
− 1ε iλτ,k
(s−t)
ε,4
a(s)e
ds ≤ CkakLp (0,t) ε r (1−β) ,
for
0
1
≤ β < 1.
2
(14.51)
±,2
Now applying a(t) in (14.50) and (14.51) to c±,1
k,ε and ck,ε , finally we get:
b̃±
k,ε → 0 ,
strongly in L2loc (dt) .
(14.52)
To finish the proof of the Proposition, we notice that
Z D
τ,k,int
g̃ g0,0
Ω
E
dx =
b̃±
k,ε
Z
+
τ,k,int
τ,k
hg̃ (g0,0
− gε,4
)i dx ,
Ω
Applying the error estimate (7.14) in Proposition 7.1, we finish the proof.
14.5. Proof of Lemma 14.1.
(14.53)
NAVIER-STOKES-FOURIER
61
τ,k
Proof. Using the boundary condition of gε,4
(7.4), simple calculations yields that
1
ε
Z
τ,k
hγg̃ γgε,4
(v·n)i dσx
1
=
ε
ZZ
τ,k
γ− g̃ γ− gε,4
(v·n)M dv dσx
Σ−
∂Ω
ZZ
i
1
τ,k
τ,k
τ,k
γ+ g̃ (1 − α )Lγ− gε,4
+ α hγ− gε,4
i∂Ω ) + rε,4
(v·n)M dv dσx
ε
Σ+
ZZ
1
τ,k
=
γ− gε,4
[γ− g̃ − (1 − αε )Lγ+ g̃ − αε hγ+ g̃ i∂Ω ] (v·n)M dv dσx
ε
Σ−
ZZ
1
τ,k
+
γ+ g̃ rε,4
(v·n)M dv dσx ,
ε
Σ+
ZZ
ZZ
1
α
=
(γ+ g̃ − Lγ− g̃ )dν̃ε −
(γ+ g̃ − hγ+ g̃ i∂Ω )dν̃ε
ε
ε
Σ+
Σ+
ZZ
1
τ,k
γ+ g̃ rε,4
(v·n)M dv dσx ,
+
ε
h
+
(14.54)
Σ+
τ,k
(v ·n)M dv dσx . From the boundary error estimate (7.15)
where the measure dν̃ε = Lγ− gε,4
in Proposition 7.1 (letting r = ∞, p = 2), γ+ g̃ is bounded in L1 (dσx , L2 (aM dv)) it is easy
to see that
ZZ
1
τ,k
γ+ g̃ rε,4
(v·n)M dv dσx → 0 , in L1loc (dt) ,
(14.55)
ε
Σ+
as ε → 0.
It remains to show that the first two terms in the righthand side of (14.54) go to zero
as ε → 0. It again follows from the a priori estimates Lemma 13.1 and 13.2. The main
difficulty is α /ε → ∞.
=
1
ε
ZZ
α
(γ+ g̃ − Lγ− g̃ )dν̃ε −
ε
ZZ
Σ+
Σ+
ZZ
α
ε
ZZ
1
ε
(γ+ g̃ − hγ+ g̃ i∂Ω )dν̃ε
γε
dν̃ε
1 + ε2 γ+ gε2
Σ+
Σ+
ZZ α
hγ+ gε i∂Ω
hγ+ gε i∂Ω
+
−
dν̃ε
ε
1 + ε2 hγ+ gε i2∂Ω 1 + ε2 γ+ gε2
Σ+
Z Z α
γ+ gε
hγ+ gε i∂Ω
+
−
dν̃ε .
ε
1 + ε2 γ+ gε2 ∂Ω 1 + ε2 hγ+ gε i2∂Ω
(γ+ g̃ − Lγ− g̃ )dν̃ε −
Σ+
(14.56)
62
N. JIANG AND N. MASMOUDI
The renormalized boundary condition (13.36) yields that
1
α
γε
(γ+ g̃ − Lγ− g̃ ) −
ε
ε 1 + ε2 γ+ gε2
α γε ε2 γ+ gε (γ+ gε + γ+ ĝε )
α
γε
γε
=−
+
−
.
ε (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
ε 1 + ε2 γ+ ĝε2 1 + ε2 γ+ gε2
Thus, after simple calculations, we have
Z
1
τ,k
hγg̃ γgε,4
(v·n)i dσx
ε
∂Ω
ZZ
α
γε ε2 γ+ ĝε (γ+ gε + γ+ ĝε )
±
=−
Lγ− gk,,2
(v·n)M dv dσx
2
2
2
2
ε
(1 + ε γ+ gε )(1 + ε γ+ ĝε )
Σ+
γε ε2 hγ+ gε i∂Ω (γ+ gε + hγ+ gε i∂Ω )
±
Lγ− gk,,2
(v·n)M dv dσx
2
2
2
2
(1 + ε γ+ gε )(1 + ε hγ+ gε i∂Ω )
Σ+
ZZ α
γε ε2 γ+ gε (γ+ gε + hγ+ gε i∂Ω )
±
Lγ− gk,,2
(v·n)M dv dσx .
−
ε
(1 + ε2 γ+ gε2 )(1 + ε2 hγ+ gε i2∂Ω ) ∂Ω
+
α
ε
ZZ
(14.57)
(14.58)
Σ+
The a priori estimates from boundary yields that all the three terms on the right-hand
side of (14.58) are bounded in in Lploc (dt) for p > 1. In deed, the integral of the first term
over [t1 , t2 ] is bounded by
√
√
Z t2 Z Z r
√
(1)
α (εγ+ ĝε ) ( εγ+ g + εγ+ ĝε )
α
γε
±
Lγ− gk,,2
(v·n)M dv dσx
2
2
1/4
2
2
2
2
3/4
ε (1 + ε γ+ g ) 1 + ε γ+ ĝε (1 + ε γ+ g )
t1
Σ+
Z
t2
ZZ
−ε
t1
Σ+
α (2) εγ+ ĝε (εγ+ gε + εγ+ ĝε )
±
Lγ− gk,,2
(v·n)M dv dσx
γ
ε2 ε (1 + ε2 γ+ gε2 )(1 + ε2 γ+ ĝε2 )
(14.59)
Note that
√
α (εγ+ ĝε )
1 + ε2 γ+ ĝε2
√
√
( εγ+ g + εγ+ ĝε )
and
1γ+ Gε ≤2hGε i∂Ω ≤4γ+ Gε
(1 + ε2 γ+ g2 )3/4
(14.60)
±
are bounded. Furthermore,Lγ− gk,,2
(v · n) is bounded in Lq ((v · n)M dv dσx ) for q ≥ 2.
Now, the estimates (13.24), (13.26) of Lemma 13.2 imply that the first term in (14.59) is
bounded in Lplox (dt), the second term vanishes as ε → 0. Similarly, using Lemma 13.1,
Lemma 13.2, we can prove that the integrals over [t1 , t2 ] of the second and third terms of
(14.58) can be decomposed into two terms, one is bounded in Lplox (dt), the other vanishes
in L1lox (dt). Thus we proved the Lemma 14.1.
15. Conclusion
In the current paper, we construct the kinetic-fluid boundary layers for the case β ∈
[0, 1) which appears in the accommodation coefficient α = χεβ , and in the acoustic
regime. The reason to do so is that our main goal here is to damp the fast acoustic waves
in the Dirichlet boundary condition which corresponds to β ∈ [0, 1). Indeed, we can also
NAVIER-STOKES-FOURIER
63
construct kinetic-fluid boundary layers for β ≥ 1 in the acoustic regime. Furthermore,
if in (8.16), we let iλτ,k
0,0 = 0, then the boundary layers will be completely different. For
that case, the leading orders terms will be in incompressible regime and no viscous fluid
boundary layers appears. In the constructions of the boundary layers for the cases β ≥ 1
in the acoustic regime and for all β ≥ 0 for the incompressible regime, the geometry of
the boundary ∂Ω as a submanifold of RD , will play a significant role. We will discuss
these cases in the forthcoming paper [25], and we will apply the boundary layers to the
higher order acoustic approximation to the linear Boltzmann equation.
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Courant Institute of Mathematical Sciences, 251 Mercer Street, New York, NY 10012
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Courant Institute of Mathematical Sciences, 251 Mercer Street, New York, NY 10012
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