mellet/publi/slc.pdf

Asymptotic Analysis 33 (2003) 337–361 337IOS Press

Macroscopic model for coupled surfaceand volume collisions in semiconductorsuperlattices

A. MelletMathématiques pour l’Industrie et la Physique, UMR CNRS 5640, Université Paul Sabatier,118, route de Narbonne, 31062 Toulouse cedex, FranceE-mail: [email protected]

Abstract. This paper follows [7], by N. Ben Abdallah, P. Degond, F. Poupaud and the author, in which a diffusion model forsemiconductor superlattices was derived. At the starting point of our study, the device consists of a periodic array of localizedscatters (the heterojunction between two semiconductor materials), and electron motion is described by the Boltzmann equation.Assuming that both the collisions and the scattering processes preserve the energy of the particles, we prove the convergenceof the electron distribution function to the solution of the SHE model equation (for Spherical Harmonics Expansion). Since thedevice has microscopic periodicity, the proof relies on the recently developed tool of two-scale convergence.

Keywords: Boltzmann equation, Spherical Harmonics Expansion model, semiconductor superlattices, diffusion approximation,homogenization, interface operators, two-scale limit

1. Introduction and mathematic model

The present paper is devoted to the modeling of electron transport in a semiconductor superlattice. Thedevice consists of a periodic superposition of layers of two different semiconductor materials. Model-ing of semiconductor superlattices is mostly due to Esaki and Tsu’s [14]. We shall not detail here thephysical background of such devices; we refer to [7] and reference therein for further details. Basically,we describe the cloud of electrons by its distribution function, which is solution of some transport equa-tion within each layer, and satisfies some reflexion–transmission conditions along the heterojunctions(see [13] for the modeling of such conditions).

As in [7], we consider a situation in which one of the semiconductor layers is much narrower than theother one, so that it can be described as a single plane. Taking the microscopic period equal to 1, thesuperlattice is therefore reduced to a periodic array of cells separated by interfaces located atx = n forn ∈ Z (we avoid boundary problems by considering an infinite device).

Within each cell, we shall describe the electron motion by the Boltzmann equation for semiconduc-tor. Assuming that the electric potential only depends on the one-dimensional space variablex, somesymmetry considerations allow us to make use of the following one-dimensional Boltzmann equation:

∂tf + vx∂xf + ∂xV ∂kxf = L(f ), x ∈ R \ Z, k ∈ B, t 0.

The unknownf (x,k, t) depends on the space variablex, the time variablet, and the momentum variable(or wave-vector)k, lying in the first Brillouin zoneB, which can be identified with the torusR3/L

0921-7134/03/$8.00 2003 – IOS Press. All rights reserved

338 A. Mellet / Macroscopic model for semiconductor superlattices

(L being the dual lattice of the crystal lattice).vx = vx(k) is thex-component of the particle velocityv(k) defined by

v(k) = ∇kε(k), k ∈ B,

whereε(k) is a given smooth periodic function ofk which gives the energy-wave-vector relationship inthe semiconductor cell (the so-called band diagram). It is assumed to be even (this assumption is theexpression of the microscopic time-reversibility). The electrostatic potentialV = V (x) is due to chargesand externally applied biases (∂xV is the corresponding electric field). It will be assumed given, timeindependent and as regular as necessary.

In [7], we neglected the effect of the collisions that electrons may undergo within the cell; it is the goalof this article to extend the result when collisions are taken into account through the operatorL, whichis a linear integral operator given by:

L(x)(f )(k) =∫

Bs(x,k,k′)

(f (k′) − f (k)

)dk.

Moreover, we assume that the energy of the particles is preserved during collisions. Therefore, we dealwith the elastic cross-sections(x,k,k′) = λ(x,k,k′)δ(ε(k′) − ε(k)). This leads to the following elasticoperator:

L(x)(f )(k) =∫

Bλ(x,k,k′)

(f (k′) − f (k)

)δ(ε(k′) − ε(k)

)dk′. (1)

We refer to [6,11] for further details about such collision operator. Its main properties will be summarizedin Section 3.3.

The presence of the second material is described by the scattering process that the particles goingthrough the interfaces undergo. For mathematical convenience we need to define the trace along theseinterfaces. For a functionϕ(x,k) defined onR\Z, we denote byγ±n (ϕ) the following limits (if they exist):

γ±n (ϕ)(k) = limx→n±0

ϕ(x,k),

and we define the outgoing and incoming traces ofϕ at pointn respectively by:

γoutn (ϕ)(k) =

γ−n (ϕ), vx(k) > 0,γ+

n (ϕ), vx(k) < 0,γ inc

n (ϕ)(k) =γ+

n (ϕ), vx(k) > 0,γ−n (ϕ), vx(k) < 0.

(2)

The outgoing (respectively incoming) trace is the distribution of particles leaving (respectively enter-ing) the semiconductor material at the interface located atx = n. The scattering process through theinterfaces is described by the relation

γ incn (f ) = Bn

(γout

n (f )),

where the scattering operatorBn is an integral operator given for any functionϕ(k) defined onB by

Bnϕ(k) =∫

Bσn(k′,k)ϕ(k′)

∣∣vx(k)′∣∣δ(ε(k′) − ε(k)

)dk′.

Note that once again the energy of the particles is preserved.

A. Mellet / Macroscopic model for semiconductor superlattices 339

Introducing the ratioα of the superlattice period to the typical macroscopic lengh scale, it was shownin [7] that the large time behaviour (i.e., after a time rescaling inα−2) is asymptoticaly given by the SHEmodel (for Spherical Harmonics Expansion), asα goes to zero. There exist a lot of papers concernedwith the approximation of diffusion models (drift-diffusion model, energy-transport model. . . ) by kineticequations. We refer to [3,8,5,19] for detailed review on this topic. The SHE model, which was firstderived from kinetic equations by Ben Abdallah and Degond in [6] (see also [11]), is a diffusion modelacting on both the position variablex, and the energy variableε. This model has been proved to be veryrelevant in many situation (see [12] for an application to plasma thruster).

In [7], the convergence of the electron distribution function to a function depending onk through theenergy variable only (equilibrium state for elastic interactions) was ensured by the interface operator. Wenow deal with two scattering sources (a boundary one and a volume one), both of them mixing electronshaving the same energy. However, it is well known (see [6,11]) that, under appropriate assumption,elastic collision operators, likeL, satisfy aH-theorem, which gives the convergence offα to an isotropicequilibrium (a function depending onk through the energy only). This property will be enough to derivethe continuity equation, and the interface operator will not play any role in this first part (this has tobe compared with [7,12] where relaxation was provided by the boundary operator). The coupling ofthe two scattering sources clearly appears in the current equation, and the diffusion constant will takeinto account the two scattering phenomenons. In order to derive the current equation, we shall use thetwo-scale convergence method introduced by Nguetseng [22] and Allaire [1].

The main theorem we wish to prove, is stated in the next section.

2. Rescaling and main theorem

As anounced in the introduction, the passage to the macroscopic scale is provided by the diffusionrescaling

x′ = αx, t′ = α2t,

whereα 1 is a small parameter which is the ratio of the microscopic unit length (the superlatticeperiod) to the macroscopic unit length (the typical size of the device),x′ being the macroscopic spacevariable. The square ofα appears in the time rescaling because we are aiming at a diffusion model at themacroscopic scale (it amounts to study the long time behaviour of the solution). Setting

fα(x′,k, t′) = f (x,k, t), V (x′) = V (x),

we obtain the following model, which is the starting point of our analysis (we have dropped the primesfor clarity):

α2∂tfα + α

(vx(k)∂xf

α + ∂xV ∂kxfα)

= Lα(fα)

, x ∈ R \ αZ, k ∈ B, t 0, (3)

γ incnα

(fα)

= B(nα)(γout

nα

(fα))

, (4)

fα(x,k, t = 0) = fαI (x,k), x ∈ R \ αZ, k ∈ B. (5)


Note that the electrostatic potential is assumed to be independent ofα, which amounts to supposingthat it varies over the macroscopic scale only. Similarly, Eq. (4) is obtained by assuming that there existsa smooth functionσ(x,k′,k), depending on the macroscopic variablex, such that

σn(k′,k) = σ(nα,k′,k).

The operatorB(x) appearing in (4) is then defined by:

B(x)ϕ(k) =∫

Bσ(x,k′,k)ϕ(k′)

∣∣vx(k′)∣∣δ(ε(k′) − ε(k)

)dk′.

Concerning the initial data, we shall assume that there exists a smooth functionFI (x, ε) defined onR×R,whereR is the closure of the numerical range of the functionε(k), satisfying:

FI

(x, ε(k)

)∈ L2(R ×B), vx∂xFI

(x, ε(k)

)∈ L2(R ×B),

and such that

fαI (x,k) = FI

(x, ε(k)

)|R\αZ×B . (6)

Eq. (6) states thatfαI is the restriction toR \ αZ of an everywhere defined function independent ofα.

SinceFI depends onk through the energyε(k) only, we shall see that the initial datum (6) satisfies theinterface condition (4). This allows us to avoid the treatment of initial layers.

In this paper, we are concerned with the limitα → 0 of the kinetic model (3) and (4), with initialdata (6). More precisely we shall prove the following result:

Theorem 2.1.

(i) Asα goes to zero, the solutionfα to the problem(3), (4), formally converges to an equilibriumstateF (x, ε(k), t) solution to the following SHE model:

N (ε)∂tF +(∂x + ∂xV ∂ε

)J = 0, (7)

J(x, ε, t) = −D(x, ε)(∂x + ∂xV ∂ε

)F , (8)

F (x, ε, t = 0) = FI (x, ε), (9)

J(x, ε, t) = 0, ε ∈ ∂R. (10)

N (ε) is the density-of-states of material, the definition of which is given in(12). The diffusionconstantD(x, ε) is given by

D(x, ε) =∫

B

∫ 1

0vx(k)χ(x,y,k)δ

(ε(k) − ε

)dy dk, (11)

whereχ(x,y,k) is solution of the cell equation(34).Moreover the diffusion constantD(x, ε) is strictly positive for(x, ε) ∈ R × R whereR denotesthe interior ofR.


(ii) When the electric field vanishes,fα converges tof0 in the weak star topology ofL∞([0,T ],L2(R ×B)) for anyT > 0, wheref0(x,k, t) = F (x, ε(k), t) andF (x, ε, t) is the weak solution ofthe problem(7)–(10).

The result is very close to [7]. Nevertheless, the proof will be very different, though we shall use a lotof results developed in [7]. These results are essentially recalled in the next section, in which we list themain properties of the collision, interface and evolution operators.

In Section 4, we shall formally derive the asymptotic model (7), (8), using a formal Hilbert expansion,the rigorous proof (including the proof of the existence of a solution to the kinetic model) being developedin Section 5.

As in many other papers concerned with the SHE model, the rigorous study is achieved assumingthat the electric field∂xV vanishes. The proof of the existence and the uniqueness of the solution ofEqs (3), (4) relies on semi-group theory and the Hille–Yosida theorem. We shall give the proof of thisresult when the electric field vanishes. Note, however, that the existence and uniqueness of a solution inL1(R ×B) ∩ L2(R ×B) is proved in [21] by S. Mischler and the author, for nonvanishing electric field.

The main contribution of this paper is the derivation of the current equation in Section 5.4. As usualwhen dealing with periodic problems, we shall use the two-scale convergence theory (see [22,1]), intro-ducing a cell problem, that will be studied in detail in Section 5.5.

3. Mathematical preliminaries

In this section, we state the properties of the operators appearing in Eqs (3), (4).

3.1. Notations

We introduceΩα = R \ αZ, Oα = Ωα × B, andΓα = αZ × B. Oα is equipped with the usualL2

norm and inner product. Note thatL2(Oα) = L2(R ×B) sinceαZ ×B is a zero measure set. Then

|f |2L2(Oα) =∫

R×B

∣∣f (x,k)∣∣2 dxdk.

The energy functionε(k) is assumed to beC2 onB with values inR. We denote bySε the manifold ink-spaceSε = k ∈ B, ε(k) = ε. Let dSε(k) be the Euclidean surface element onSε andR the closureof the numerical range ofε(k). We also denote by dNε(k) the coarea and byN (ε) the energy-density ofstates:

dNε(k) =dSε(k)|∇kε(k)| , N (ε) =

∫Sε

dNε(k). (12)

Then we have the coarea formula [15]: for any continuous functionψ(k) defined onB, we have

∫Bψ(k) dk =

∫R

(∫Sε

ψ(k)dSε(k)|∇kε(k)|

)dε :=

∫R

(∫Sε

ψ(k) dNε(k))

dε. (13)


We define the space of tracesL2(Γα) as the set of functionsu = (uαn)n∈Z such thatuαn ∈ L2(B) forall n ∈ Z and

|u|2L2(Γ α) :=∑n∈Z

α|uαn|2L2(B),

whereL2(B) is equipped with the weighted norm

|u|L2(B) =∫

B

∣∣vx(k)∣∣∣∣u(k)

∣∣2 dk.

In the same way, we denote byL2(Sε) the space associated with the weight|vx(k)| (in such a way wehaveL2(B) = L2(R;L2(Sε))).

In the sequel we shall need the following geometrical hypothesis, the validity of which is discussedin [7]:

Hypothesis 3.1. N (ε) lies in L∞(R), and for all compact subsetsK ⊂ R, there exist two positiveconstantsµK , andνK such that

N (ε) =∫Sε

dNε(k) µK , ∀ε ∈ K,∫Sε

∣∣vx(k)∣∣2 dNε(k) νK , ∀ε ∈ K.

3.2. The interface operator

This section is devoted to the operatorB, which can be rewritten as follows (using (13)):

B(x)ϕ(k) =∫Sε

σ(x,k′,k)ϕ(k′)∣∣vx(k′)

∣∣ dNε(k)(k′).

All results stated in this section have been proved in [7]. We first specify the required assumptionsconcerning the cross-sectionσ(x,k,k′).

Hypothesis 3.2.

(i) Positivity: σ(x,k,k′) 0.(ii) Particle conservation:∫

Sε(k′)

σ(x,k′,k)∣∣vx(k)

∣∣ dNε(k′)(k) = 1. (14)

(iii) Reciprocity:

σ(x,k′,k) = σ(x,−k,−k′). (15)


Equality (14), together with (15) yields the following normalization condition:∫Sε(k)

σ(x,k′,k)∣∣vx(k′)

∣∣ dNε(k)(k′) = 1. (16)

From (16) and the Cauchy–Schwartz inequality, we easily deduce the following inequality

∣∣B(x)u∣∣2L2(Sε) |u|2L2(Sε),

which plays a similar role as the Darrozes–Guiraud inequality in gas–surface interaction [10].Since, by the normalization condition (16), constant functions are fixed points ofB(x), we deduce that

for eachε ∈ R, the operatorB(x) is a continuous linear operator onL2(Sε) of norm 1:∥∥B(x)∥∥L(L2(Sε)) = 1, ∀ε ∈ R. (17)

Note that, ifB(x) is a compact operator (for instance ifσ is bounded), then we can make precise theproperties ofB(x):

Proposition 3.3.

(i) Considering the operator as acting onL2(Sε), we have,

ker(I − B(x)

)= R.

(ii) Denoting byQ the orthogonal projector ofL2(Sε) onto the space of constant functions, andintroducing P = I − Q, whereI is the identity onL2(Sε), we havePB(x) = B(x)P , andQB(x) = B(x)Q = Q.

Note also that the adjointB∗(x) of B(x) is given by

B∗(x)u(k) =∫Sε

σ(x,k,k′)u(k′)∣∣vx(k′)

∣∣ dNε(k)(k′). (18)

It obviously satisfies‖B∗(x)‖L(L2(Sε)) = 1, ∀ε ∈ R.We end this section by introducingBα the operator acting onL2(Γα), which coincides withB(αn) on

each interface, namely:

u = (uαn)n∈Z ∈ L2(Γα)→ Bαu =

(B(αn)uαn

)n∈Z

∈ L2(Γα).

Obviously,Bα is a continuous operator onL2(Γα).

3.3. The collision operator

We now recall the properties of the collision operator (1). Many papers have already dealt with suchan operator, and we quote [6] for details.


Using the coarea formula, we rewriteLα as follows:

L(f )(k) =∫Sε

λα(x,k,k′)[f (k′) − f (k)

]dNε(k

′),

and we shall assume that the cross-sectionλα satisfies:

Hypothesis 3.4.

(i) λα(x,k,k′) = λ(x,k,k′) 0 does not depend onα.(ii) λ(x,k,k′) satisfies the so-called detailed balance principle (or micro-reversibility condition)

which in the case of elastic operator reads as:

λ(x,k,k′) = λ(x,k′,k), ∀x ∈ R, ∀k,k′ ∈ B.

(iii) There exist two positive constantsλ1 andλ2 such that

0<λ1

N (ε) λ(x,k,k′) λ2

N (ε).

Let us discuss the relevance of these hypotheses. First of all, in view of the periodicity of the device, itseems reasonable to assume that the following development holds:

λα(x,k,k′) = λ0(x,x

α,k,k′

)+ αλα(x,k,k′).

Such a situation has been considered in [17] by Goudon and the author in the framework of neutrontransport. Nevertheless, in our paper, the dependence ofλ0 on the fast variablex/α would not introducesignificant modifications of the result. We therefore restrict ourselves to the framework of hypothesis (i).

Hypothesis (ii) is related to physical considerations (it ensures relaxation to an isotropic equilibrium),and is well justified in such a situation. Hypothesis (iii) is purely technical; it will provide the relaxationof fα to an equilibrium state.

The following proposition summarize the main properties ofL:

Proposition 3.5.

(i) L is a self-adjoint, bounded operator onL2(B).(ii) Denoting〈f〉(ε) = (1/N (ε))

∫Sεf (k) dNε(k), we have

−∫Sε

L(f )f dNε(k) λ1

∫Sε

∣∣f − 〈f〉∣∣2 dNε(k).

It yieldsker(L) = f ∈ L2(B); f (k) = g(ε(k)) .


3.4. The transport operator

We end the presentation of the mathematical framework by introducing the transport operatorAα:

Aαf = vx(k)∂xf + ∂xV ∂kxf , (19)

with domain

Hα(A,B) =f ∈ Hα(A) | γout

α (f ) ∈ L2(Γα), γ inc

α (f ) = Bαγoutα (f )

,

where the spaceHα(A) is given by

Hα(A) =f ∈ L2(Oα)

, Aαf ∈ L2(Oα),

and

γoutα (f ) =

(γout

nα(f ))n∈Z

, γ incα (f ) =

(γ inc

nα(f ))n∈Z

.

We recall that the outcoming and incoming traces are defined by (2). The spacesHα(A) andHα(A,B)are equipped with the graph norm

|f |2Hα(A) = |f |2L2(Oα) +∣∣Aαf

∣∣2L2(Oα).

We shall denote byA the bare differential operator (19), when no indication of the domain is needed.It is well known [4,24] that the regularityu ∈ Hα(A) is not sufficient to guarantee that the traces

γoutα (u) andγ inc

α (u) are square integrable for the measure|vx(k)|dk. But if one of these traces is inte-grable, the other one is also integrable. Therefore, defining

Hα0 (A) =

f ∈ Hα(A), γout

α (f ) ∈ L2(Γα)=

f ∈ Hα(A), γ inc

α (f ) ∈ L2(Γα),

we have the following Green’s formula, [4,24]:

Lemma 3.6 (Green’s formula). For f , g inHα0 (A), we have

(Af ,g)L2(Oα) + (f ,Ag)L2(Oα) =1α

((γout

α (f ),γoutα (g)

)L2(Γ α) −

(γ inc

α (f ),γ incα (g)

)L2(Γ α)

), (20)

where(·, ·)L2(Oα) and (·, ·)L2(Γ α) stand for the inner products associated with the norms ofL2(Oα) andL2(Γα).

Applying Green’s formula tou ∈ Hα(A,B), and thanks to (17), we get

2α(Af ,f )L2(Oα) =∣∣γout

α (f )∣∣2L2(Γ α) −

∣∣γ incα (f )

∣∣2L2(Γ α) =

∣∣γoutα (f )

∣∣2L2(Γ α) −

∣∣Bγoutα (f )

∣∣2L2(Γ α) 0,

which yields the accretivity of operatorAα.


OperatorAα has been studied in [7]. In particular, it has been proved that in the general case,Aα

is not maximal on the domainHα(A,B). Therefore, the existence of a solution for the evolution prob-lem (3), (4) has been proved under weaker hypotheses concerning the trace (the projectionQγout

α (fα) wassupposed to belong to someL2

loc(Γα) space). However, we shall see in Section 5 that the case∂xV = 0

is less technical. We shall therefore restrict ourselves to this case for the rigorous derivation.In the next section, we formally study the asymptotic behaviour offα asα goes to zero, without

restriction concerning the electric field.

4. The formal derivation of the SHE model

The formal asymptotic approach relies on the following Hilbert expansion:

fα(x) = f0(x,x

α

)+ αf1

(x,x

α

)+ α2f2

(x,x

α

)+ · · · , (21)

with f i(x,y) having a periodic behaviour with respect to the fast variabley. Inserting (21) in (3), andidentifying the terms having the same order with respect toα, we respectively obtain:

vx∂yf0 − L

(f0) = 0,

vx∂yf1 − L

(f1) = −

(vx∂xf

0 + ∂xV ∂kxf0),

vx∂yf2 − L

(f2) = −

(∂tf

0 + vx∂xf1 + ∂xV ∂kxf

1), (22)

and the boundary condition (4) yields:

γ inc#

(f i)(x) = B(x)γout

#(f i)(x), ∀x ∈ R, ∀i = 0, 1, 2, (23)

where incoming and outcoming traces are defined as follows:

γout# (ψ) =

lim

y→1−ψ(y), vx(k) > 0,

limy→0+

ψ(y), vx(k) < 0,γ inc

# (ψ) =

lim

y→0+ψ(y), vx(k) > 0,

limy→1−

ψ(y), vx(k) < 0.(24)

The formal expansion (21) therefore reduces the transport equations (3), (4) to three cell problems onY ×B.

We introduce the asymptotic operatorT 0 = A0 − L = vx∂y − L with domain

D(A0) =

f , vx∂yf ∈ L2(Y × Sε); γ

out# (f ) ∈ L2(Sε), γ

inc# (f ) = Bγout

# (f ).

Anticipating Section 5.5, in which the formal adjoint operatorT = −vx∂y − L is studied, we list belowthe properties ofT 0:

Proposition 4.1.

(i) The kernelker(T 0) is composed of functions independent ofy, and depending onk through εonly.


(ii) Im(T 0) = (ker(T 0))⊥ = f ∈ L2;∫Sε

∫ 10 f (y,k) dy dNε(k) = 0 .

(iii) For all h ∈ Im(T 0) there exists a uniquef ∈ (ker(T 0))⊥

solvingT 0(f ) = h. We shall denote byT 0−1

(h) this function. Moreover, for allg ∈ ker(T 0) we haveT 0−1(gh) = gT 0−1

(h).

We now investigate the consequence for Eqs (22), (23):

– The zeroth order equation yieldsf0 ∈ ker(T 0) and therefore:

f0(x,y,k, t) = F (x, ε, t),

which gives the (formal) convergence to a function depending onk through the energy variable.– Noticing that∂kxF = vx∂εF , the first order equation in (22) becomes:

T 0f1 = −vx(∂xF + ∂xV ∂εF ).

Since∫Sεvx(k) dNε(k) = 0 (see (38)), we can define:

χ0(x,y,k) = T 0−1(vx), (25)

and since∂xF + ∂xV ∂εF lies in ker(T 0), we deduce:

f1 = −χ0(x,y,k)(∂xF + ∂xV ∂εF ). (26)

– The solvability condition for the second order equation in (22) (obtained by integrating with respectto y,k ∈ Sε) yields:

N (ε)∂tF + ∂x

∫ 1

0

∫Sε

vxf1 dNε(k) dy + ∂xV

∂

∂ε

∫Sε

vxf1 dNε(k) dy = 0,

and defining the current

J(x, ε, t) =∫ 1

0

∫Sε

vx(k)f1(x,y,k, t) dNε(k) dy,

we obtain the continuity equation (7).A straightforward computation using (26) also implies

J(x, ε, t) = −[∫ 1

0

∫Sε

vx(k)χ0(x,y,k) dNε(k) dy](∂xF + ∂xV ∂εF ),

which is the current equation (8), with a diffusion constant defined by

D(x, ε) =∫ 1

0

∫Sε

vx(k)χ0(x,y,k) dNε(k) dy.


The positivity of this constant is provided by the accretivity of the operatorT 0, since we can write:

D(x, ε) =∫ 1

0

∫Sε

T 0(χ0)χ0 dNε(k) dy.

We have therefore formally derived the asymptotic SHE model. This approach could be the startingpoint of the proof of the convergence using the Chapman–Enskog method [5]. Nevertheless, such amethod needs a lot of regularity for the solution of the cell problemT 0−1

(h), and it is now well known(see [7,20]) that this operator introduces some singularities with respect to the energy variable close tothe critical values of the energy functionk → ε(k).

Anyway, we shall develop a direct proof, relying on the moment method, and the use of the two-scaleconvergence. In order to bypass the difficulty due to the regularity with respect tok, we shall develop therigorous approach when the electric potential vanishes only.

5. The rigorous derivation of the SHE model

In this section we are aiming at the proof of Theorem 2.1. As announced in the introduction, we shallassume throughout this section that∂xV = 0, the transport operator being therefore reduced to

Aαf = vx∂xf.

The first step will be devoted to the existence and the uniqueness of the solution of the Boltzmannequations (3), (4). Then, we shall prove the convergence to the asymptotic SHE equation (part (ii) ofTheorem 2.1).

5.1. The Boltzmann equations(3), (4)

The existence and the uniqueness offα are given by the following proposition and Hille–Yosidatheorem:

Proposition 5.1. OperatorAα, with domainHα(A,B) is maximal and accretive.

This proposition has to be compared with [7], in which we failed to obtain the maximality and theaccretivity of operatorA0 on its domain. We shall see that the difficulties met in [7] break down whenthe electric potential vanishes.

The outline of the proof is the same as in [7]:(1) We introduce the perturbed interface operator:

Bη =1

1 + ηB, η > 0,

and the associated operatorAαη = A with domainHα(A,Bη):

Hα(A,Bη) =u ∈ Hα(A), γout

α (u) ∈ L2(Γα), γ inc

α (u) = Bηγoutα (u)

.

Then we have:


Lemma 5.2. Aαη , with domainHα(A,Bη), is a maximal accretive operator forη > 0.

This lemma is a consequence of the following inequality:‖Bη‖L(L2(B)) 1/(1 + η) < 1,∀η > 0, anda contraction mapping argument (see [7] for details).

(2) We now take the limitη → 0, and we get the maximality ofAα by proving that we can control thetrace in terms of the graph norm uniformly with respect toη (in [7], we do not control the whole tracebut only theP -projection):

Let f be inL2(Oα), sinceAαη is maximal for allη > 0 (Lemma 5.2), there existsuη ∈ Hα(A,Bη)

such that

uη + Auη = f.

Multiplying the previous expression byuη, and integrating with respect tox, k, we get (using the accre-tivity of Aα

η ):

|uη|L2(Oα) |f |L2(Oα),

|Auη|L2(Oα) |f − uη|L2(Oα) 2|f |L2(Oα).

This yields the weak convergence ofuη in L2(Oα) to a functionu ∈ L2(Oα) and that ofAuη to Au. Itfollows thatu ∈ Hα(A) andu+ Aαu = f .

It remains to check that the so-obtained functionu lies inHα(A,B): to that purpose, we establish thefollowing estimate, which holds for allu ∈ Hα(A,Bη):

∣∣γ incα (u)

∣∣2L2(Γ α) +

∣∣γoutα (u)

∣∣2L2(Γ α) C1α|u|2Hα(A) + C2|u|2L2(Oα). (27)

Before we prove this estimate, we end the proof of Proposition 5.1: by (27), we can assume thatγoutα (uη),

γ incα (uη) converge inL2(Γα) for the weak topology, and by a classical result [16], their limits are re-

spectivelyγoutα (u), γ inc

α (u), in the distributional sense (and even inH−1/2(B)). We deduce that the tracessatisfy:

γ incα (u),γout

α (u) ∈ L2(Γα)and γ inc

α (u) = Bγoutα (u),

and therefore thatu ∈ Hα(A,B). This achieves the proof of the maximality ofAα, sinceu satisfies:

u ∈ Hα(A,B), u+ Aαu = f.

Proof of estimate (27). We introduce:

φα(x) =2α

(x−

(n− 1

2

)α

), x ∈

((n− 1)α,nα

),

and the sign function denoted by sgn(vx). Note that|φα|L∞(Ωα) = 1, thatγ−nα(φα) = 1, γ+(n−1)α(φα) =

−1 and that∂φα/∂x = 2/α.


Then, using Green’s formula (20), we get:

2(Aαu,u sgn(vx)φα)

L2(Oα) +2α

∫Oα

|u|2∣∣vx(k)

∣∣ dxdk

=∑n∈Z

∫B

(∣∣γ−nα(u)∣∣2 +

∣∣γ+(n−1)α(u)

∣∣2)∣∣vx(k)∣∣ dk =

∑n∈Z

∫B

(∣∣γoutnα(u)

∣∣2 +∣∣γ inc

nα(u)∣∣2)∣∣vx(k)

∣∣ dk

which gives (27).

Let the evolution operator be defined byTα = Aα − 1αL, with domainHα(A,B); it is a perturbation

of a maximal accretive operator by a bounded operator. Hence, we have (see [23]):

Corollary 5.3. OperatorTα = Aα − 1αL with domainHα(A,B) generates a strongly continuous semi-

group of contractions.

We deduce:

Proposition 5.4. Under hypotheses listed in Theorem2.1, there exists a unique solutionfα to prob-lem(3)–(5), such that

fα ∈ C1(0,T ;L2(Oα))∩ C0(0,T ;Hα(A,B)

).

Note thatfα satisfies the boundary conditions(4) sincefα(t) ∈ Hα(A,B).

5.2. The two-scale convergence

We can now investigate the convergence offα: we have to justify the two-scale asymptotic devel-opment (21) introduced in Section 4. It is now usual (see [1]) in such a situation to use two-scale (ordouble-scale) convergence.

Before we get into the very heart of the proof, we setup some notation, and briefly recall the mainresults that we shall use (the proof of which can be found in [1]):

We define the space domainΩ = R \ Z, and the elementary cellY = (0, 1). We denote byC∞# (Ω) the

space ofY -periodic smooth functions, whose restriction toY is compactly supported, and byC∞# (Ω) the

space of smooth functions inΩ that areY -periodic (Ω denotes the disjoint union of the closed intervals[n,n + 1]). L2

#(Y ) denotes the completion ofC∞# (Ω) under theL2 norm (it coincides with the space of

functions inL2(Y ), extended byY -periodicity toR).Then we have the following definition–theorem:

Theorem 5.5. Letuα(x,k) be a bounded sequence inL2(R×B), then there existsu0(x,y,k) in L2#(R×

Y ×B) and a subsequence still denoteduα such that for allψ(x,y,k) ∈ L2(R ×B; C#(Ω)), we have

limα→0

∫R×B

uα(x,k)ψ(x,x

α,k

)dxdk =

∫R×B

∫Yu0(x,y,k)ψ(x,y,k) dy dxdk. (28)

A sequence satisfying(28) is said to two-scale converge tou0(x,y,k). Moreover(28)still holds with lessregular test function(e.g.,ψ ∈ L2

#(Y ×B; Cc(R))).


We refer to [1,2] and references therein for the proof and more details about admissible test functions.

Remark 5.6. It is also known that if the sequenceuα two-scale converges to a functionu0(x,y), then itweakly converges inL2(R) to

u(x) =∫ 1

0u0(x,y) dy.

Moreover, we have limα→0 |uα|L2(R) |u0|L2(R×Y ) |u|L2(R).

Finally, we recall that in Section 4 we have defined for functionsψ in C#(Ω), the incoming and out-coming traces on the cellY : γout

# (ψ), andγ inc# (ψ) (see (24)).

We end this section by writing the following weak form of the evolution problem (3), (4):

Lemma 5.7. Let fα be the solution of problem(3), (4), defined by Proposition5.4. Then, for any testfunctionψ in C1

c ([0,T [×R ×B; C1#(Ω)), we have∫ T

0

∫Oαfα

(α2∂tψ

(x,x

α

)+ α

(Aαψ

)(x,x

α

)+ vx∂yψ

(x,x

α

))dxdk dt

+ α2∫Oαfα

I ψ|t=0

(x,x

α

)dxdk = −

∫ T

0

∫OαfαL(ψ)

(x,x

α

)dxdk dt

+∑n∈Z

∫ T

0

(γout

nα

(fα(t)

),γout

# (ψ)(nα,k, t) − B∗(nα)γ inc# (ψ)(nα,k, t)

)L2(B) dt. (29)

Proof. Lemma 5.7 is nothing but Eq. (3) in weak form, withϕα(x,k, t) = ψ(x, xα ,k, t) as a test function.

Sinceγoutnα(fα(t)) ∈ L2

loc(Γα), Green’s formula yields:∫ T

0

∫Oαfα(

α2∂tϕα + αAαϕα)

dxdk dt+ α2∫Oαfα

I ϕα|t=0 dxdk = −

∫ T

0

∫OαfαL

(ϕα)

dxdk dt

+∑n∈Z

∫ T

0

(γout

nα

(fα(t)

),γout

nα

(ϕα)

(k, t) − B∗(nα)γ incnα

(ϕα)

(k, t))L2(B) dt.

We obviously haveAαϕα = (Aαψ)(x, xα )+ 1

αvx∂yψ(x, xα ). Moreover, sinceψ is continuous with respect

to x, we have

γoutnα

(ϕα)

= γout# (ψ)(nα,k, t),

and (29) follows straightforwardly.

5.3. The continuity equation

We define the density and current of particles of energyε by:

Fα(x, ε, t) =1

N (ε)

∫Bfα(x,k, t)δ

(ε(k) − ε

)dk =

⟨fα⟩

,

Jα(x, ε, t) =1α

∫Bfα(x,k, t)vx(k)δ

(ε(k) − ε

)dk =

N (ε)α

⟨vxf

α⟩.


Then we have:

Lemma 5.8. For any test functionφ(x, ε, t) in C1c ([0,T ] × R ×R) (i.e., continuously differentiable and

compactly supported in[0,T ] × R ×R), such thatφ(·, ·,T ) ≡ 0, we have

∫ T

0

∫Ωα×R

[N (ε)Fα∂tφ+ Jα∂xφ

]dxdεdt +

∫R×R

FIφ|t=0 dxdk = 0. (30)

Proof. Sinceφ only depends onk throughε, it is easy to check that the right-hand side in (29) vanishes,and the result obviously follows from the coarea formula and (29).

The continuity equation (7) follows from (30) by taking the limitα → 0 in (30). To this purpose, weprove the following lemma:

Lemma 5.9. There exists two functionsF (x, ε, t) andJ(x, ε, t) such that

fα F , L∞(0,T ;L2(R ×B)

)two scale, and weak star,

Jα J , L2([0,T ] × R ×R)

weak star.

SinceF depends onk throughε only, we have

N (ε)Fα N (ε)F , L∞(0,T ;L2(R ×R)

)weak star.

This leads to:

Corollary 5.10. For any test functionφ(x, ε, t) in C1c ([0,T ] ×R×R) (twice continuously differentiable

with compact support inR ×R), such thatφ(·, ·,T ) ≡ 0, we have

∫ T

0

∫Ωα×R

[N (ε)F∂tφ+ J∂xφ

]dxdεdt +

∫R×R

FIφ|t=0 dxdk = 0. (31)

Eq. (31) is nothing else but the weak form of the continuity equation (7) with initial condition (9).Moreover, we obtain the zero flux boundary condition (10) by means of an integration by parts in (31).

The remainder of this section is devoted to the proof of Lemma 5.9 which is divided in three steps:

First step: Convergence to a function depending onk throughε(k) only. Multiplying (3) by fα andintegrating with respect tox, k, we get

∂t

∣∣fα(t)∣∣2L2(R×B) +

1α

(Aαfα,fα)

L2(R×B) =1α2

(Lfα,fα)

L2(R×B).

Propositions 5.1 (accretivity ofAα) and 3.5 (coercivity ofL) imply:

∂t

∣∣fα(t)∣∣2L2(R×B) −λ1

α2

∣∣fα −⟨fα⟩∣∣2

L2(R×B) 0,


and integrating with respect tot, we get∣∣fα(t)∣∣2L2(R×B) |FI |2L2(R×B), (32)∫ t

0

∣∣fα(s) −⟨fα(s)

⟩∣∣2L2(R×B) ds α2

λ1|FI |2L2(R×B). (33)

From (32), we deduce thatfα is bounded inL∞(0,T ;L2(R × B)), and thus two-scale converges asαtends to zero (Theorem 5.5) to a functionf0(x,y,k, t).

Moreoverfα − 〈fα〉 obviously two-scale converges tof0 − 〈f0〉, and Remark 5.6 yields:∣∣f0 −⟨f0⟩∣∣

L2(Y ×R×B) limα→0

∣∣fα −⟨fα⟩∣∣

L2(R×B).

The right-hand side vanishes thanks to (33), and thereforef0 = 〈f0〉 which means

f0(t,x,y,k) = F(t,x,y, ε(k)

).

Second step:f0 does not depend ony. We now consider Eq. (29) with a test functionψ(x,y,k, t) inC1

c ([0,T ] × R × B; C1#(Ω)). Such a function being compactly supported inΩ =

⋃n∈N ]n,n + 1[ with

respect toy, ψ satisfiesγout# (ψ)(x) = γ inc

# (ψ)(x) = 0 for all x ∈ R. Therefore equation (29) gives thelimit asα→ 0:∫ T

0

∫R×B

∫ 1

0f0vx∂yψ(x,y) dy dxdk dt+

∫ T

0

∫R×B

∫ 1

0L

(f0)ψ(x,y) dy dxdk dt = 0,

which formally meansT 0f0 = vx∂yf0 − L(f0) = 0 (note that this is the zeroth order equation (22)).

SinceL(f0) = L(F ) = 0 we deduce:∫ T

0

∫R×B

∫ 1

0F (x,y,k, t)vx∂yψ(x,y,k, t) dy dxdk dt = 0.

HenceF does not depend ony, and Remark 5.6 gives the convergence offα to F for the weak startopology.

Third step: Convergence of the current.Since∫Sεvx(k) dNε(k) = 0 (see (38)),Jα can be rewritten as:

Jα =1α

∫Sε

(fα(x,k, t) −

⟨fα⟩

(x, ε, t))vx(k) dNε(k),

and (33) yields the boundedness ofJα in L2([0,T ] × R ×R)). This achieves the proof.

Remark 5.11. The shortness of this proof has to be compared with the length of the correspondingproof in [7]. Actually the energy-dependence off0, and the boundedness of the currentJα are given bythe coercivity of the collision operatorL, while in [7] these results followed from the properties of thetraceγ(fα).

But if the proof of the continuity equation is much easier in this case, it is not the same for the currentequation, that will be investigated in the next section.


5.4. Equation for the current

The last point that remains to be proved concerns the current equation. To that purpose, we introducethe following cell problem (which is nothing but the dual problem to Eq. (25)):

−vx∂yχ(x,y,k) − L(x)(χ(x,y)

)= vx(k),

γout# (χ)(x, t) = B(x)γ inc

# (χ)(x, t).(34)

The aim of the next section to investigate the existence of a solution of the cell-problem. Let us summarizethe results:

First notice that operators−vx∂y, B andL only act ony,k ∈ [0, 1] × Sε, and leavex and ε asparameters. We can thus omit thex-dependence. But we do not omit the energy dependence, sinceestimates will not be uniform with respect toε (we shall see that some singularities occur forε close tothe boundary∂R). We introduce the spaceL∞

loc(R,L2(Y × Sε)) of functions satisfying for all compactsubsetsK ⊂ R:

supε∈K

1N (ε)

∫ 1

0

∫Sε

∣∣f (y,k)∣∣2 dNε(k) dy < +∞.

The existence ofχ is provided by the following proposition the proof of which is given in the nextsection:

Proposition 5.12. Under Hypothesis3.4, we have: for all h ∈ L∞loc(R,L2(Y × Sε)), such that

1N (ε)

∫ 1

0

∫Sε

h(y,k) dNε(k) dy = 0, ∀ε ∈ R,

there exists a functiong in L∞loc(R,L2(Y × Sε)) solving

−vx∂yg − L(g) = h,

γout# (g) = Bγ inc

# (g),(35)

and such that

vx∂yg ∈ L∞loc

(R,L2(Y × Sε)

), γ inc

# (g) ∈ L∞loc

(R,L2(Sε)

). (36)

Moreover, there exists a unique such function satisfying

1N (ε)

∫ 1

0

∫Sε

g(y,k) dNε(k) dy = 0, ∀ε ∈ R. (37)

We now note that the coarea formula gives∫Sε

vx(k) dNε(k) =∫

B∂kxε(k)δ

(ε(k) − ε

)dk =

∫B

∂kx

(H

(ε(k) − ε

))dk = 0, (38)


becauseH(ε(k)−ε) (whereH denotes the Heaviside function) is a periodic function ofk and the integralover a period of the derivative of a periodic function is zero. Proposition 5.12 therefore gives the existenceand the uniqueness ofχ(x,y,k) satisfying conditions (36) and (37), and solving (34).

Moreover, we have the following lemma, the proof of which is postponed to the end of the next section:

Lemma 5.13. The auxiliary functionχ belongs toL∞(R;L∞loc(R;L2(Y × Sε))). Furthermore,χ has

the same regularity with respect tox as the cross-sectionsσ and λ. In particular under appropriateassumption,∂xχ is continuous with respect tox.

We are now in position to establish the current equation. Actually we shall prove the following:

Lemma 5.14. For any test functionϕ(x, ε, t) in C1c (0,T × R × R) (i.e., continuously differentiable and

compactly supported in]0,T [ × R × R), we have

∫ T

0

∫R×R

J(x, ε, t)ϕ(x, ε, t) dεdxdt =∫ T

0

∫R×R

F (x, ε, t)∂x(D(x, ε)ϕ(x, ε, t)

)dk dxdt,

with the diffusion constant being defined by

D(x, ε) =∫Sε

∫ 1

0vx(k)χ(x,y,k) dy dNε(k). (39)

Proof. Letϕ(t,x, ε) be a test function inC1c (]0,T [ × R × R), we define

ψ(x,y,k, t) = χ(x,y,k)ϕ(x, ε(k), t

).

Sinceϕ does not depend ony, and only depends onk throughε, we have:

γout# (ψ)(x,k) = ϕ(x, ε)γout

# (χ)(x,k) = ϕ(x, ε)Bγ inc# (χ) = B(ϕ(x, ε)γ inc

# (χ))

= B(γ inc# (ψ)

).

Insertingψ as a test function in (29), we get:

α

∫ T

0

∫R×B

fαχ

(x,x

α,k

)∂tϕ(x, ε, t) dk dxdt+

∫ T

0

∫R×B

fα(Aα(χϕ)

)(x,x

α,k, t

)dk dxdt

=1α

∫ T

0

∫R×B

fα(−vx∂yχ− L(χ)

)(x,x

α,k

)ϕ(x, ε, t) dk dxdt.

The first term on the left-hand side obviously vanishes whenα tends to zero, sincefα is bounded inL∞(0,T ;L2(R × B)). In view of the definition ofχ and the coarea formula, the right-hand side can berewritten∫ T

0

∫R×R

Jα(x, ε, t)ϕ(x, ε, t) dεdxdt,

which converges asα tends to zero to∫ T

0

∫R×R J(x, ε, t)ϕ(x, ε, t) dεdxdt.


We now investigate the second term on the left-hand side: Since∂xV = 0, we have

Aα(χϕ)(x,y,k, t) = vx∂x(χϕ)(x,y,k, t).

Lemma 5.13 yieldsAα(χϕ) ∈ L2#(Y × B; Cc(R))), and thus is an admissible test function (we point out

that no derivative ofχ with respect toy arises, so that it does not matter thatvx∂yχ does not belong toL2

#(Y )). Recalling thatfα two-scale converges toF , (28) yields:

∫ T

0

∫R×B

fα(Aαψ

)(x,x

α

)dk dxdt α→0−→

∫ T

0

∫R×B

∫ 1

0FA(χϕ) dy dk dxdt

=∫ T

0

∫R×R

F (x, ε, t)∫Sε

∫ 1

0A

(χ(x,y,k)ϕ(x, ε, t)

)dy dNε(k) dεdxdt.

We can write

∫Sε

∫ 1

0A


)dy dNε(k) =

∫Sε

∫ 1

0vx(k)∂x


)dy dNε(k)

= ∂x

(∫Sε

∫ 1

0vx(k)χ(x,y,k) dy dNε(k)ϕ(x, ε, t)

).

Defining the diffusion constant by (39), we eventually get

∫ T

0

∫R×R

J(x, ε, t)ϕ(x, ε, t) dεdxdt =∫ T

0

∫R×R

F (x, ε, t)∂x(D(x, ε)ϕ(x, ε, t)

)dεdxdt,

and the proof is complete.

We end this section with the following lemma:

Lemma 5.15. The diffusion constantD(x, ε) is positive.

Proof. Inserting (34) in (39), we have:

D(x, ε) =∫Sε

∫ 1

0vx(k)χ(x,y,k) dy dNε(k) =

∫Sε

∫ 1

0

(−vx∂yχ− L(χ)

)χ(x,y,k) dy dNε(k)

=12

∫Sε

∣∣vx(k)∣∣(∣∣γ inc

# (χ)∣∣2 − ∣∣γout

# (χ)∣∣2) dNε(k) +

∫ 1

0

∫Sε

−L(χ)χdNε(k) dy.

The second term in the last expression is positive by Proposition 3.5(ii), and the first one is positive sinceγout

# (χ) = Bγ inc# (χ) and‖B∗‖L(L2(Sε)) = 1.


5.5. The auxiliary equation

This section is devoted to the proof of Proposition 5.12. In a first step,ε is assumed to be fixed inR,and we prove the existence ofg|Sε in L2(Y × Sε).

We introduce the cell-operatorTε = −vx∂y − L, with domain

D(Tε) =f ∈ L2(Y × Sε); vx∂yf ∈ L2(Y × Sε), γ

inc# (f ) ∈ L2(Sε), γ

out# (f ) = Bγ inc

# (f ).

We recall thatD(Tε) is closed for the graph norm|f |L2 + |Tε(f )|L2 (see estimate (27)), and we state thefollowing lemma, the proof of which relies on the same ideas as Lemma 5.1:

Lemma 5.16. The operator−vx∂y is maximal accretive on its domainD(Tε).

We can now get to the heart of the proof of Proposition 5.12, which will be divided in two steps: thenext lemma determines the closure of the range of the operatorTε, and the closeness of Im(Tε) will beproved afterward.

Lemma 5.17. Im(Tε) =g ∈ L2(Y × Sε) |

∫ 10

∫Sεg(y,k) dNε(k) dy = 0

.

Proof. The proof relies on functional analytic arguments, which can be found in [9]. SinceIm(Tε) =ker(T

ε )⊥, it is equivalent to show that the kernel ker(T ε ) is reduced to constant functions onL2(Sε×Y ).

L being a self adjoint bounded operator, one has

T ε = (−vx∂y) − L = vx∂y − L,

with D(T ε ) = D((−vx∂y)), and since−vx∂y is maximal accretive (Lemma 5.16), its formal adjoint

vx∂y is also maximal accretive onD(T ε ). (Note that we do not need to characterizeD(T

ε ); it is enoughto know thatvx∂y is accretive on it.) It follows (by Proposition 3.5) that for allf in D(T

ε ) we have

(T

ε f ,f)L2(Y ×Sε)

((−vx∂y)f ,f

)L2(Y ×Sε) −

(Lf ,f

)L2(Y ×Sε) λ1

∣∣f − 〈f〉∣∣2L2(Y ×Sε).

Let now f belong to ker(T ε ). Then the previous inequality yieldsf (k,y) = 〈f〉(ε,y). The equality

T ε (f ) = 0 becomesvx∂yf = 0, and finallyf is a constant onSε × Y . Moreover, it is readily seen that

the constants lie in ker(T ε ). Hence we have ker(T

ε ) = R, which concludes the proof.

Note that in the same way, we prove that ker(Tε) = R. It gives the uniqueness condition (37). The nextstep states the closeness of Im(Tε):

Lemma 5.18. Im(Tε) is closed inL2(Y × Sε).

Proof. Let (gn)n∈N be a sequence in Im(Tε) such that

gnn→∞−→ g in L2(Y × Sε) strong,

and let (fn)n∈N be the associated sequence inD(Tε) ∩ N (Tε)⊥ such thatTεfn = gn. If we prove that|fn|L2(Y ×Sε) is bounded, then there will exist a functionf ∈ L2(Y × Sε) such thatfn f weakly inL2; In the same way as we proved the closeness ofD(Tε), we deducef ∈ D(Tε) and thusg ∈ Im(Tε).


To show that|fn|L2(Y ×Sε) is bounded, we proceed by contradiction. First notice that, up to a subse-quence, we have|fn|L2(Y ×Sε) → ∞. Then, settingFn = fn/|fn|L2(Y ×Sε), we have

TεFn → 0 inL2(Y × Sε) sinceTεfn is bounded,

|Fn|L2(Y ×Sε) = 1,∫Sε

∫ 1

0Fn(y,k) dy dNε(k) = 0,

(40)

the last condition following from the choice offn in ker(Tε)⊥. This implies thatFn weakly converges inL2(Y × Sε). Furthermore, since we have

(TεFn,Fn)L2(Y ×Sε) λ1∣∣Fn − 〈Fn〉

∣∣2L2(Y ×Sε),

(40) yieldsFn − 〈Fn〉 → 0 inL2(Y × Sε) strongly.Writing Fn = 〈Fn〉 + (Fn − 〈Fn〉), we readily obtain

|Fn|2L2(Y ×Sε) = 1 =∣∣Fn − 〈Fn〉

∣∣2L2(Y ×Sε) +

∣∣〈Fn〉∣∣2L2(Y ×Sε),

and ∣∣〈Fn〉∣∣L2(Y ×Sε) → 1. (41)

We will have a contradiction if we prove

〈Fn〉 → 0 L2(Y × Sε) strong.

To this purpose, we write, following Goudon and Poupaud in [18]

vx∂y〈Fn〉 = −vx∂y

(Fn − 〈Fn〉

)+ vx∂yFn = −vx∂y

(Fn − 〈Fn〉

)− Tε(Fn) − L(Fn).

Multiplying the equality byvx(k), and integrating onSε (we recall that〈Fn〉 is constant onSε), we get(∫Sε

|vx|2 dNε(k))

∂y〈Fn〉

= −∫Sε

|vx|2∂y

(Fn − 〈Fn〉

)dNε(k) −

∫Sε

vx

(Tε(Fn) − L(Fn)

)dNε(k).

SinceFn − 〈Fn〉 converges to 0 inL2(Y × Sε), the first term in the right-hand side converges to 0 inH−1(Y ), and the second one obviously strongly converges to 0 inL2(Y ). Hypothesis 3.1 implies that|∂y〈Fn〉|H−1(Y ) tends to 0. Definingϕ(y) =

∫ y0 〈Fn〉(z) dz, the condition

∫ 10

∫SεFn dNε(k) dy = 0 gives

ϕ ∈ H10(Y ), hence we have

−(∂y〈Fn〉,ϕ

)H−1,H1

0=

∣∣〈Fn〉∣∣2L2(Y )

∣∣∂y〈Fn〉∣∣H−1|ϕ|H1

0,


and since|ϕ|H10

= |∂yϕ|L2(Y ) = |〈Fn〉|L2(Y ), it yields

∣∣〈Fn〉∣∣L2(Y ) → 0,

which contradicts (41).

Forg ∈ L2(Y × Sε) satisfying∫ 1

0

∫Sεg(y,k) dNε(k) dy = 0, we denote byT−1

ε (g) the unique function

f ∈D(Tε) such thatTε(f ) = g and∫ 1

0

∫Sεf (y,k) dNε(k) dy = 0 (the uniqueness condition on the van-

ishing integral amounts to choosingf ∈ ker(Tε)⊥).In order to complete the proof of Proposition 5.12, we need the following lemma:

Lemma 5.19. For all compact subsetK ⊂ R, there exists a constantCK such that

∀ε ∈ K, ∀g ∈ Im(Tε),∣∣T−1

ε (g)∣∣L2(Y ×Sε) CK |g|L2(Y ×Sε). (42)

Proof. LetK be a fixed compact subset inR. We proceed as in the previous lemma: assuming that (42)is false, we construct sequencesεn ∈ K, gn ∈ Im(Tεn), such that

|gn|L2(Y ×Sεn ) = 1 and∣∣T−1

ε (gn)∣∣L2(Y ×Sεn ) → +∞.

Settingfn = T−1εn

(gn), andFn = fn/|fn|, we have

|Fn| = 1 and∣∣Tε(Fn)

∣∣ → 0,

and the coercivity of operatorL gives once again∣∣Fn − 〈Fn〉∣∣L2(Y ×Sεn ) → 0.

The equality|Fn|2 = 1 = |Fn − 〈Fn〉|2 + |〈Fn〉|2, yields∣∣〈Fn〉∣∣L2(Y ×Sεn ) → 1.

We now write(∫Sεn

|vx|2 dNε(k))

∂y〈Fn〉

= −∫Sεn

|vx|2∂y(Fn − 〈Fn〉

)dNε(k) −

∫Sεn

vx(Tε(Fn) − L(Fn)

)dNε(k).

It is readily seen that the right-hand side tends to 0 inH−1(Y ), and sinceεn ∈ K, one has, usingHypothesis 3.1,∫

Sεn

|vx|2 dNε(k) νK > 0.

We conclude in the same way as in the proof of the previous lemma.


Note that, since we have∫Sε

|vx|2 dNε(k) → 0 asε→ ∂R,

we cannot expect to obtain better estimates thanL∞loc with respect to the energy variable.

5.6. Proof of Lemma 5.13

We end this paper by proving Lemma 5.13, which states the regularity ofχ(x,y,k) with respect tox.Actually, we shall prove under the following hypothesis that∂xχ belongs toL∞

loc(R × R;L2(Y × Sε)):

Hypothesis 5.20. Denote by∂xB∗ the operators defined by (18) withσ replaced by∂xσ. Then, weassume that∂xB∗ is a bounded operators onL2(Sε), uniformly with respect to (x, ε) ∈ R ×R.

In the same way,∂xL is assumed to be a bounded operator onL2(B).

The formal derivative ofχ with respect tox solves:−vx∂y

(∂xχ(x,y,k)

)− L(x)

(∂xχ(x,y)

)= (∂xL)(χ),

γout# (∂xχ) = B(x)γ inc

# (∂xχ) + ∂xB(x)γ inc# (χ),

and satisfies the condition∫ 1

0

∫Sε

∂xχdNε(k) dy = 0. We defineΦ0 as follows:

Φ0(y,k) = y∂xB(x)γ inc

# (χ), vx(k) > 0,

Φ0(y,k) = (1− y)∂xB(x)γ inc# (χ), vx(k) < 0.

Since∂xB(x)γ inc# (χ) lies inL∞

loc(R × R;L2(Sε)), Φ0 satisfies:

φ0 ∈ L∞

loc

(R × R;L2(Y × Sε)

),

vx∂yφ0 ∈ L∞loc

(R × R;L2(Y × Sε)

),

γ inc# (φ0) = 0,

γout# (φ0) = ∂xB(x)γ inc

# (χ),

and the functionφ = ∂xχ− φ0 +∫ 1

0

∫SεΦ0 dNε(k) dy solves

−vx∂y

(φ(x,y,k)

)− L(x)

(φ(x,y)

)= (∂xL)(χ) − vx∂yφ0 − L(x)(φ0),

γout# (φ) = B(x)γ inc

# (φ).

Therefore,φ is the unique solution of the cell problem (35), with right-hand side

h(y,k) = (∂xL)(χ) − vx∂yφ0 − L(x)(φ0) ∈ L∞loc

(R × R;L2(Y × Sε)

).

By Proposition 5.12, it follows thatφ ∈ L∞loc(R × R;L2(Y × Sε)).


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