BOLTZMANN EQUATION WITH VANISHING ABSORPTION
CLAUDE BARDOS, ETIENNE BERNARD, FRANC¸ OIS GOLSE, AND R ´EMI SENTIS
Abstract. The present paper discusses the di↵usion approximation of the linear Boltzmann equation in cases where the collision frequency is not uni- formly large in the spatial domain. Our result applies for instance to the case of radiative transfer in a composite medium with optically thin inclusions in an optically thick background medium. The equation governing the evolution of the approximate particle density coincides with the limit of the di↵usion equation with infinite di↵usion coefficient in the optically thin inclusions.
1. Presentation of the Problem Consider the linear Boltzmann equation
(1) (@t+v·rx)f(t, x, v) +Lxf(t, x, v) = 0
for the unknown f ⌘ f(t, x, v) that is the distribution function for a system of identical point particles interacting with some background material. In other words, f(t, x, v) is the number density of particles located at the positionx2⌦, where⌦ is a domain ofRN, with velocityv⇢RN at timet 0.
The notationLxdesignates a linear integral operator acting on thevvariable in f, i.e.
(2) Lxf(t, x, v) = Z
RN
k(x, v, w)(f(t, x, v) f(t, x, w))dµ(w)
where µis a Borel probability measure on RN, whilek is a nonnegative function definedµ⌦µ-a.e. onRN ⇥RN that measures the probability of a transition from velocityv to velocityw for particles located at the positionx.
Henceforth we denote
(3) h i=
Z
RN
(v)dµ(v) and⌦⌦ ↵↵
= ZZ
RN⇥RN
(v, w)dµ(v)dµ(w) for all 2L1(RN, dµ) and 2L1(RNv ⇥RNw;dµ(v)dµ(w)).
We assume thatksatisfies the semi-detailed balance condition (4)
Z
RN
k(x, v, w)dµ(w) = Z
RN
k(x, w, v)dµ(w) and introduce the notation
(5) a(x, v) :=
Z
RN
k(x, v, w)dµ(w)
1991Mathematics Subject Classification. 45K05 (45M05, 82A70, 82C70, 85A25).
Key words and phrases. Linear Boltzmann equation, Di↵usion approximation, Neutron trans- port equation, Radiative transfer equation.
1
for the absorption rate, so that
Lxf(t, x, v) =a(x, v)f(t, x, v) Kxf(t, x, v) whereKxdesignates the integral operator
(6) Kxf(t, x, v) :=
Z
RN
k(x, v, w)f(t, x, w)dµ(w).
The semi-detailed balance assumption appears for instance in [11] — see formula (2.9) in§2.
The assumptions on the transition kernel other than (4) used in our discussion are introduced later.
We further assume that⌦ is a bounded domain of RN with C1 boundary @⌦
and denote bynxthe unit outward normal field atx2@⌦. Let
+:={(x, v)2@⌦⇥RN|v·nx>0},
0:={(x, v)2@⌦⇥RN|v·nx= 0}, :={(x, v)2@⌦⇥RN|v·nx<0}.
The linear Boltzmann equation is supplemented with the absorption boundary con- dition
(7) f(t, x, v) = 0, (x, v)2 , t >0.
(In other words, it is assumed that there are no particles entering the domain⌦.) This choice is made for the sake of simplicity; other boundary conditions will be discussed later.
We are concerned with the di↵usion approximation of the linear Boltzmann equation (1) — see [7] and the references therein for a general presentation of this approximation. We briefly recall its main features below.
Set L to be a length scale that measures the size of ⌦while V is the average particle velocity; this defines a time scaleT :=L/V. The di↵usion limit of (1) is based on the assumption that the dimensionless quantity T a(x, v) is large. Thus we introduce a scaling parameter 0<✏⌧1 and set
ˆk✏(x, v, w) :=✏k(x, v, w) so that ˆk✏(x, v, w) is of order unity. Accordingly, we define
ˆ
a✏(x, v) :=✏a✏(x, v), Lˆx=✏Lx, and ˆKx=✏Kx.
(For notational simplicity, we do not mention explicitly the dependence ofLx and Kxin✏.) Assume further that variations of order unity of the boundary data driving the solution of (1) do not occur on time scales shorter thanT /✏.
In that case, the solutionf of (1) is sought in the form f(t, x, v) = ˆf✏(✏t, x, v)
with the notation ˆt=✏tfor the rescaled time variable. Thus (1) takes the form
✏@ˆtf✏(ˆt, x, v) +v·rxf✏(ˆt, x, v) +1
✏Lˆxf✏(ˆt, x, v) = 0.
Henceforth we drop hats on rescaled variables and consider the initial-boundary value problem for the scaled linear Boltzmann equation
(8) 8>
><
>>
:
(✏@t+v·rx)f✏(t, x, v) +1
✏Lxf✏(t, x, v) = 0, x2⌦, v2RN, t >0,
f✏(t, x, v) = 0, (x, v)2 , t >0,
f✏(0, x, v) =fin(x, v) x2⌦, v2RN, in the limit as✏!0.
2. Existence, uniqueness and a priori estimates
Assume thatkis a nonnegative measurable function onRN⇥RN satisfying the semi-detailed balance assumption (4) and the condition
(9) a✏2L1(⌦⇥RN, dxdµ).
Lemma 2.1. Under assumption (9)
a) the integral operatorsLx andKx are bounded onL2(RN, µ)for a.e. x2⌦, with kKxkL(L2(RN,µ)) ka✏(x,·)kL1(RN,µ);
b) the adjoints of Kx andLx are given by the formulas K⇤x (v) =
Z
RN
k✏(x, w, v) (w)dµ(w) and
L⇤x (v) = Z
RN
k✏(x, w, v)( (v) (w))dµ(w) for a.e. x2⌦;
c) for a.e. x2⌦
{ functions a.e. constant onRN}=R⇢Ker(Lx)and Ker(L⇤x) ; d) for each 2L2(RN;dµ),
h Lx i= 12 ZZ
RN⇥RN
k✏(x, v, w)( (v) (w))2dµ(v)dµ(w), for a.e. x2⌦;
e) if in addition k✏(x, v, w)>0 fordµ(v)dµ(w)-a.e. (v, w)2RN ⇥RN, then Ker(Lx) = Ker(L⇤x) ={ functions a.e. constant on RN}=R.
Remark. The discussion of the properties of the operatorLx di↵ers from [3]. In [3], it is assumed that the measureµis the uniform probability measure on the set V of admissible velocities, that can be a ball, or a sphere, or a spherical annulus centered at the origin inRN. The scattering kernelk✏(x, v, w) is of the form
k✏(x, v, w) = ✏(x)f(v, w)
wheref(v, w) =f(w, v) a.e. onV ⇥V is positive and such that Z
V
f(v, w)dw= 1 for a.e. v2V .
ThusLx= ✏(x)(I F), whereF is the integral operator defined by F (v) :=
Z
V
f(v, w) (w)dw for a.e. v2V .
Then Ker(Lx) = Ker(I F) whenever ✏(x)>0, and R⇢ Ker(I F). On the other hand, it is assumed in [3] thatf is chosen so thatF is a compact operator on L2(V). Then 1 is the principal eigenvalue ofF and Ker(I F) is one-dimensional by the Krein-Rutman theorem, which implies that Ker(I F) =R.
Proof. Statement a) follows from Schur’s lemma (Lemma 18.1.12 in [10] or Lemma 1 in§2 of chapter XXI in [7]). The formula for Kx in statement b) and statement c) are obvious.
As for the formula forL⇤xin statement b), observe that Lx (v) +Kx (v) =
Z
RN
k✏(x, v, w) (v)dµ(w)
= Z
RN
k✏(x, w, v) (v)dµ(w) =a✏(x, v) (v) fordxdµ-a.e. (x, v)2⌦⇥RN by the semi-detailed balance assumption (4).
For each 2L2(RN;dµ) and a.e. inx2⌦ h Lx i=
ZZ
RN⇥RN
k✏(x, v, w)( (v)2 (v) (w))dµ(v)dµ(w)
= Z
RN
a✏(x, v) (v)2dµ(v) ZZ
RN⇥RN
k✏(x, v, w) (v) (w)dµ(v)dµ(w)
=12 Z
RN
a✏(x, v) (v)2dµ(v) +12 Z
RN
a✏(x, w) (w)2dµ(v) ZZ
RN⇥RN
k✏(x, v, w) (v) (w)dµ(v)dµ(w)
= ZZ
RN⇥RN
k✏(x, v, w)12( (v)2+ (w)2)dµ(v)dµ(w) ZZ
RN⇥RN
k✏(x, v, w) (v) (w)dµ(v)dµ(w)
=12 ZZ
RN⇥RN
k✏(x, v, w)( (v) (w))2dµ(v)dµ(w)
by Fubini’s theorem and the semi-detailed balance assumption (4). This proves statement d).
By statement d), if 2L2(RN;dµ) satisfiesLx = 0, then 0 =h Lx i= 12
ZZ
RN⇥RN
k✏(x, v, w)( (v) (w))2dµ(v)dµ(w). Therefore
k✏(x, v, w)( (v) (w)) = 0 fordµ(v)dµ(w) a.e. (v, w)2RN ⇥RN so that
(v) (w) = 0 fordµ(v)dµ(w) a.e. (v, w)2RN ⇥RN. Averaging inw shows that
(v) =h i fordµ(v) a.e. v2RN, so that
Ker(Lx)⇢{functions a.e. constant onRN}=R.
With statement c), this shows that
Ker(Lx) ={functions a.e. constant onRN}=R. Since the function (v, w)7!k✏(x, w, v) satisfies the same properties ask✏,
Ker(Lx) ={functions a.e. constant onRN}=R.
⇤ Proposition 2.2. Assume that k✏ is a nonnegative measurable function defined dxd(µ⌦µ)-a.e. on⌦⇥RN⇥RN satisfying (4) and (9) witha✏ defined by (5). For each✏>0and eachfin2L2(⌦⇥RN;dxdµ(v)), there exists a unique weak solution of the initial-boundary value problem (8) in the spaceCb(R+;L2(⌦⇥RN;dxdµ(v))).
This solution satisfies a) the continuity equation
@thf✏i+ divx1
✏hvf✏i= 0 in the sense of distributions onR⇤+⇥⌦;
b) the “entropy inequality”
Z
⌦hf✏(t, x,·)2idx+ Z t
0
Z
⌦
⌦⌦k✏(x,·,·)q✏(s, x,·,·)2↵↵
dxds Z
⌦hfin(x,·)2idx for each✏>0 and each t 0, where
q✏(t, x, v, w) = 1
✏(f✏(t, x, v) f✏(t, x, w)).
Proof. The operator (x, v)7!Lx (x, v) is a bounded perturbation of the advection operator v·rxwith absorbing boundary condition (7) that is the generator of a strongly continuous contraction semigroup onL2(⌦⇥RN;dxdµ(v)).
This implies the existence and uniqueness of the weak solutionf✏ of the initial- boundary value problem (8) in the functional spaceCb(R+;L2(⌦⇥RN;dxdµ(v))).
Statement a) follows from the inclusionR⇢Ker(L⇤x) in Lemma 2.1. Statement
b) follows from Lemma 2.1 d) and Lemma 2.3 below. ⇤
Lemma 2.3. Letfin2L2(⌦⇥RN;dxdµ(v))andS2L2([0, T]⇥⌦⇥RN;dtdxdµ(v)).
For each ✏>0, letf✏ be the weak solution inCb(R+;L2(⌦⇥RN;dxdµ(v))) of 8>
><
>>
:
✏@tf✏+v·rxf✏=S , x2⌦, v2RN, t >0, f✏ = 0,
f✏t=0=fin. Then
1 2
Z
⌦hf(t, x,·)2idx 1
✏ Z t
0
Z
⌦hS(s, x,·)f✏(s, x,·)idxds+12 Z
⌦hfin(x,·)2idx The proof of this lemma is classical; we give it in the appendix for the sake of being self-contained.
3. Diffusion approximation with vanishing absorption rate:
main results
Assume that the spatial domain ⌦=A[B, where A is open andB is closed in RN (i.e. B\@⌦=?), with finitely many connected components denotedBl, forl= 1, . . . , m. We further assume thatBlhas piecewiseC1boundary, thatBlis locally on one side of its boundary@Bl. Finally, we denote by nx the unit normal field atx2@A, oriented towards the exterior ofA.
Henceforth, it is assumed that the measureµsatisfies
(10) µ({0}) = 0.
We further assume that
(11) h|v|2i<1 and det(hv⌦vi)6= 0.
For each l = 1, . . . , m, denote by ⌧l ⌘ ⌧l(x, v) the forward exit time from Bl
starting from the positionxwith the velocityv; in other words (12) ⌧l(x, v) := inf{t >0 s.t. x+tv2@Bl}. We assume that, for eachl= 1, . . . , mand for eachg2L2(@Bl),
(13)
g(x+⌧l(x, v)v) =g(x) ford (x)dµ(v) a.e. (x, v)2@Bl⇥RN )g(x) = 1
|@Bl| Z
@Bl
g(y)d (y) for a.e. x2@Bl.
We further assume that the scattering kernelk✏in the linear Boltzmann equation is adxdµ(v)dµ(w)-a.e. nonnegative measurable function on⌦⇥RN⇥RN satisfying the following assumptions, in addition to (4):
(a) the absorption ratea✏ is uniformly small onB⇥RN as✏!0, i.e.
(14) ka✏kL1(B⇥RN,dxdµ)!0 as✏!0 ;
(b) the restriction of k✏ to A⇥RN ⇥RN is assumed to be independent of ✏ and denotedkA⌘kA(x, v, w); it satisfies
(15) CK := supess
(x,v)2A⇥RN
Z
RN
✓
kA(x, v, w) + 1 kA(x, v, w)
◆
dµ(w)<1; we henceforth denote
(16) aA(x, v) :=
Z
RN
kA(x, v, w)dµ(w), fordxdµ(v) a.e. (x, v)2A⇥RN; (c) there exists aRN-valued vector fieldb⌘b(x, v) defineddxdµ-a.e. on A⇥RN such that
(17) b(x,·)2L2(RN, dµ), hb(x,·)i= 0 andLxb(x,·) =L⇤xb(x,·) =v for a.e. x2A.
With the vector fieldb, we defined theMN(R)- valued matrix field (18) M(x) =hb(x,·)⌦vi, for a.e. x2A .
3.1. Coercivity properties ofLx. The coercivity properties ofLxon the orthog- onal of its null-space for a.e. x2A are crucial in order to study the matrix field M. This matrix field is of fundamental importance in the sequel as it is the di↵u- sion matrix that appears in the limit equation. Notice however that this di↵usion matrix is (a.e.) defined onAonly, and not in all of⌦.
Lemma 3.1. Assume thatk✏satisfies the assumptions (4)-(14)-(15)-(17) while the probability measure µsatisfies (11). Then
a) one has
hvi= 0 ; b) for a.e. x2Aand each 2L2(RN, dµ)
k h ikL2(RN;dµ)2CKkLx kL2(RN,dµ); in particular
kb(x,·)kL2(RN,dµ)2CKh|v|2i1/2; c) the matrix fieldM satisfies
M(x) =M(x)T for a.e. x2A; d) the matrix fieldM satisfies the bound
|Mij(x)|2CKkvikL2(RN,dµ)kvjkL2(RN,dµ) for a.e. x2A;
e) denoting by >0 the smallest eigenvalue of the positive matrixhv⌦2i, one has
⇠·M(x)⇠
2CK|⇠|2 for all ⇠2RN, for a.e. x2A .
3.2. The di↵usion equation on⌦with infinite di↵usivity in B. In the clas- sical di↵usion approximation of the linear Boltzmann equation, the di↵usion coeffi- cient is proportional to the reciprocal absorption rate (see for instance formula (47) in [3]). In the situation considered above, the absorption rate vanishes as ✏ ! 0 in the subregionB of the spatial domain⌦. This suggest that the limit equation should be a di↵usion equation with infinite di↵usion constant inB.
Before going further, we recall the precise statement of this problem and its variational formulation.
Define H:=
⇢
u2L2(⌦) s.t. u(x) = 1
|Bl| Z
Bl
u(y)dyfor a.e. x2Bl, l= 1, . . . , n , and
V :=H\H01(⌦) ={u2H01(⌦) s.t. ru(x) = 0 for a.e. x2Bl, l= 1, . . . , n}. For each⇢in2H, consider the following variational problem
(19)8
>>
><
>>
>:
⇢2Cb(R+;H)\L2(R+;V), @t⇢2L2(R+;V0), and⇢t=0=⇢in, d
dt Z
⌦
⇢(t, x)w(x)dx+ Z
Arw(x)·M(x)rx⇢(t, x)dx= 0, for a.e. t 0, for allw2V.
Proposition 3.2. Assume thatx7!M(x)is anMN(R)-valued measurable matrix field onA satisfying
Mij 2L1(A)for alli, j= 1, . . . , N , and there exists↵>0 s.t.
⇠·M(x)⇠ ↵|⇠|2 for a.e. x2Aand all ⇠2RN.
For each⇢in2H, the variational problem (19) has a unique solution. This solution satisfies the “energy” identity
1 2
Z
⌦
⇢(t, x)2dx+ Z t
0
Z
Arx⇢(s, x)·M(x)rx⇢(s, x)dxds= 12 Z
⌦
⇢in(x)2dx for eacht 0.
Next we state the PDE formulation equivalent to (19).
Proposition 3.3. Assume thatx7!M(x)is anMN(R)-valued measurable matrix field on A such that Mij 2 L1(A) for all i, j = 1, . . . , N. Let ⇢ 2C([0, T];H)\ L2([0, T];V)with @t⇢2L2([0, T];V0).
Then⇢satisfies 8>
>>
<
>>
>: d dt
Z
⌦
⇢(t, x)w(x)dx+ Z
Arw(x)·M(x)rx⇢(t, x)dx= 0, for all w2V,
⇢t=0=⇢in. if and only if
8>
>>
>>
><
>>
>>
>>
:
@t⇢ divx(Mrx⇢) = 0 in D0(R⇤+⇥A),
⇢(t,·)@⌦= 0 in L2([0, T];H1/2(@⌦)),
˙
⇢l= 1
l
⌧ @⇢
@nM
,1
H 1/2(@Bl),H1/2(@Bl)
in H 1((0, T)), l= 1, . . . , m
⇢t=0=⇢in, where
l=|Bl|, l= 1, . . . , m .
In the PDE formulation equivalent to (19), we have used the notation
@⇢
@nM
(t, x) :=nx·M(x)rx⇢(t, x). Consider therefore the problem
(20)
8>
>>
>>
><
>>
>>
>>
:
@t⇢(t, x) = divx(M(x)rx⇢(t, x) x2A , t >0,
⇢(t, x) = 0 x2@⌦, t >0,
˙
⇢l(t) = 1
l
Z
@Bl
@⇢
@nM
(t, x)d (x) l= 1, . . . , m , t >0,
⇢(0, x) =⇢in(x) x2⌦. The proposition above justifies the following definition.
Definition 3.4. For ⇢in 2 H, a weak solution of the problem (20) is a function
⇢⌘⇢(t, x)such that
⇢2Cb(R+;H)\L2(R+;V) and@t⇢2L2(R+;V0)
which satisfies the variational formulation and the initial condition in (19).
Remark. The problem (20) is the limit of the di↵usion equation in the case where the di↵usion coefficient tends to +1in B: see for instance Theorem 2.4 in [8] for a proof of this result.
3.3. The di↵usion approximation. With the preparations above, we can for- mulate the di↵usion approximation for the scaled linear Boltzmann equation with absorption rate vanishing inB.
Theorem 3.5. Assume thatµ satisfies (11) and thatk✏ satisfies the assumptions (4)-(14)-(15)-(17). Let M be the matrix field defined onA by (18).
Let⇢in2H, and for each✏>0, letf✏ be the unique weak solution of the initial- boundary value problem (8) in the spaceCb(R+;L2(⌦⇥RN;dxdµ(v))). Then
f✏(t,·,·)!⇢(t,·)strongly inL2(⌦⇥RN;dxdµ)
uniformly in t2[0, T] for allT >0, where⇢is the unique weak solution of (20).
In addition 1
✏(f✏(t, x, v) f✏(t, x, w))! (b(x, v) b(x, w))·rx⇢(t, x)
in the strong topology ofL2([0, T]⇥A⇥RN⇥RN;kA(x, v, w)dtdxdµ(v)dµ(w))for allT >0 as✏!0.
Remark. The last convergence statement above is equivalent to the following:
(21) 1
✏(f✏ hf✏i)! b·rx⇢ strongly inL2([0, T]⇥A⇥RN;dtdxdµ) as✏!0. This statement is the analogue of formula (37) in [3] giving theO(✏) term in the Hilbert expansion of the solutionf✏as a formal power series in✏.
Proof of formula (21). Observe that the linear map 7! defined by (t, x, v) =
Z
RN
(t, x, v, w)dµ(w)
is a bounded operator fromL2([0, T]⇥A⇥RN⇥RN;kA(x, v, w)dtdxdµ(v)dµ(w)) toL2([0, T]⇥A⇥RN;dtdxdµ(v)). Indeed
(t, x, v)2CK
Z
RN
kA(x, v, w) (t, x, v, w)2dµ(w)
by the Cauchy-Schwarz inequality and assumption (15). The conclusion follows from integrating both sides of this inequality int, x, v. Therefore
1
✏(f✏(t, x, v) hf✏i(t, x)) = Z
RN
1
✏(f✏(t, x, v) f✏(t, x, w))dµ(w)
! Z
RN
(b(x, v) b(x, w))·rx⇢(t, x)dµ(w)
= b(x, v)·rx⇢(t, x)
inL2([0, T]⇥A⇥RN;dtdxdµ(v)) as✏!0, sincehb(x,·)i= 0 for a.e. x2A. ⇤
3.4. Remarks on the assumptions of Theorem 3.5. The existence of the vec- tor field b is obviously an important assumption as it enters the definition of the di↵usion matrixM. In the present formulation, the existence ofb is postulated in (17) and the conditionhvi= 0 in Lemma 3.1 deduced from the existence ofb.
Conversely, one can assume that the transition kernelkAis chosen so thatKxis a compact operator onL2(RN;dµ) for a.e. x2A. In that case,Lxis a Fredholm operator onL2(RN;dµ) for a.e. x2A, because the multiplication by aA(x, v) is an invertible operator onL2(RN;dµ) for a.e. x2A— see Corollary 19.1.8 in [10]
or chapter 6 in [5]. Indeed, by (15) and Jensen’s inequality
|aA(x, v) 1|=hkA(x, v,·)i 1 hkA(x, v,·) 1i CK
fordxdµ(v)-a.e. (x, v)2A⇥RN. Applying then Lemma 2.1 e) shows that Im(Lx) = Ker(L⇤x)? ={ 2L2(RN;dµ(v)) s.t. h i= 0}. Thus
hvi= 0,v2Im(Lx) for a.e. x2A. If in additionkA is chosen such that
kA(x, v, w) =kA(x, w, v) fordµ(v)dµ(w) a.e. (v, w)2RN ⇥RN for a.e. x2A, thenL⇤x=Lx and assumption (17) holds.
Another key assumption is (13).
If Bl is convex for l = 1, . . . , m and µ is of the form dµ(v) = r(|v|)dv or µ is the uniform probability measure on a sphere included inRN centered at the origin, (13) is obviously satisfied. Indeed, for eachx, y2@Bl, the segment [x, y] is included inBl, so thatg(x) =g(y) for a.e. x, y2@Bl.
But even whenBlis convex, the assumption (13) may fail to be satisfied for some measuresµ. For instance, assume thatN = 2, and takeBl={x2R2s.t. |x|1}. Denote by (e1, e2) the canonical basis ofR2, and let
µ= 14( e1+ e1+ e2+ e2).
Forx= (x1, x2)2@Bl, one has⌧l(x,±e1) = 2|x1|and⌧l(x,±e2) = 2|x2|, so that ( x1, x2) +⌧l(x, e1)e1= (x1, x2), (x1, x2) ⌧l(x, e1)e1= ( x1, x2), (x1, x2) +⌧l(x, e2)e2= (x1, x2), (x1, x2) ⌧l(x, e2)e2= (x1, x2). Thus g(x) =|x1| or g(x) = |x2| are not a.e. constant on @Bl and yet satisfy the condition
g(x+⌧l(x, v)v) =g(x) fordxdµ(v) a.e. (x, v)2@Bl⇥R2.
Assumption (14) is obviously satisfied if k✏(x, v, w) = 0 for dxdµ(v)dµ(w)-a.e.
(x, v, w)2B⇥RN ⇥RN, or if k✏(x, v, w) =O(✏) onB. The assumption used in the present paper is obviously much more general. For instance, it is satisfied if one hask✏(x, v, w) =O(|ln✏| l) onBlwith l>0 for eachl= 1, . . . , m. The Hilbert expansion method used in [3] does not apply to this situation, and therefore cannot be used on the problem considered here in its fullest generality.
Even in the nondegenerate case where B = ?, observe that our assumptions on the transition kernel k✏ do not imply that the vector field b in (17) depends smoothly onx. This again excludes the possibility of using the Hilbert expansion as in [3] to establish the validity of the di↵usion limit. Accordingly, the di↵usion matrix fieldM obtained in Theorem 3.5 is in general not even continuous. In this regularity class, the classical interpretation of the di↵usion equation with di↵usion
matrix M in terms of the associated stochastic di↵erential equation fails (see for instance section 5.1 and Remark 5.1.6 in [14]). One should bear in mind that a di↵usion equation with di↵usion matrix that has a type I discontinuity across some smooth surface is equivalent to a transmission problem for two di↵usion equations on each side of the discontinuity surface, with continuity of the solution and of the normal component of the current across the discontinuity surface. See for instance [12] on p. 107 or Lemma 1.1 in [8] for a discussion of this well known issue.
4. Proof of Lemma 3.1 For eachi= 1, . . . , N and a.e. x2A, one has
hvii=hLxbi(x,·)i=h(L⇤x1)bi(x,·)i= 0 sinceL⇤x1 = 0 by Lemma 2.1 c), which proves statement a).
SetLx = ; by statement d) in Lemma 2.1 h i=h Lx i= 12
ZZ
RN⇥RN
k✏(x, v, w)( (v) (w))2dµ(v)dµ(w) 0. By the Cauchy-Schwarz inequality, for a.e. x2A,
| (v) h i|2=
✓Z
RN
( (v) (w))dµ(w)
◆2
Z
RN
dµ(w) kA(x, v, w)
Z
RN
kA(x, v, w)( (v) (w))2dµ(w) so that
k h ik2L2(RN;dµ)CK
ZZ
RN⇥RN
kA(x, v, w)( (v) (w))2dµ(v)dµ(w)
= 2CKh i. Next
h i=hLx i=h(L⇤x1) i= 0 sinceL⇤x1 = 0 by Lemma 2.1 c), so that
h i=h( h i) i k kL2(RN,dµ)k h ikL2(RN,dµ)
by the Cauchy-Schwarz inequality. Putting together the last two inequalities, we obtain the bound
k h ikL2(RN;dµ)2CKk kL2(RN,dµ)
which is statement b).
Next
Mij(x) =hbi(x,·)vji=hbi(x,·)Lxbj(x,·)i
=hbj(x,·)L⇤xbi(x,·)i=hbj(x,·)vii=Mji(x) for alli, j= 1, . . . , N and a.e. x2A. This proves statement c).
Statement d) follows from the identity
Mij(x) =hbi(x,·)vji,
from the Cauchy-Schwartz inequality and statement b) with =bi(x,·).
Applying again the Cauchy-Schwarz inequality with (x, v) := ⇠·b(x, v) and (v) :=⇠·v=Lx (x, v), one has
(v)2=
✓Z
RN
kA(x, v, w)( (x, v) (x, w))dµ(w)
◆2
aA(x, v) Z
RN
kA(x, v, w)( (x, v) (x, w))2dµ(w) for a.e. x2A, so that, by Lemma 2.1 d)
h 2i CK
ZZ
RN⇥RN
kA(x, v, w)( (x, v) (x, w))2dµ(v)dµ(w)
= 2CKh (x,·)i= 2CK⇠·M(x)⇠. for a.e. x2A. Obviously
h 2i=⇠·hv⌦2i⇠ |⇠|2 and statement e) follows.
5. Proofs of Propositions 3.2 and 3.3
Proof of Proposition 3.2. The existence and uniqueness of the solution of the varia- tional problem (19) is a straightforward consequence of the Lions-Magenes theorem, i.e. Theorem X.9 in [5], with the bilinear form
a(u, v) :=
Z
Aru(x)·M(x)rv(x)dx , u, v2V.
Indeed, this bilinear form satisfies the assumptions of the Lions-Magenes theorem since Lemma 3.1 d) implies that
|a(u, v)|2CKh|v|2ikrukL2(A)krvkL2(A)2CKh|v|2ikukVkvkV, while Lemma 3.1 e) implies that
a(u, u)
2CKkruk2L2(A)=
2CKkruk2L2(⌦)= 2CK
(kuk2V kuk2H) for eachu, v2V.
Consider the linear functional
L(t) : V 3w7! h@t⇢, wiV0,V+a(⇢(t,·), w) defined for a.e. t 0.
SinceL(t) = 0 for a.e. t2R, one has
hL(t),⇢(t,·)iV0,V= 0 for a.e. t 0, for eachw2V. By Lemma B.2, one has
L(t) = 0 inV0 for a.e. t2R+. In particular, for a.e. s 0, one has
0 =hL(s),⇢(s,·)iV0,V=h@t⇢(s,·),⇢(s,·)iV0,V+ Z
Arx⇢(s, x)·M(x)rx⇢(s, x)dx , and one concludes by integrating ins2[0, t] and applying Lemma B.1 b). ⇤
Proof of Proposition 3.3. Specializing (19) to the case wherew2Cc1(A) is equiv- alent to
@t⇢ divx(Mrx⇢) = 0 inD0(R⇤+⇥A). In particular, the (a.e. defined) vector field
(0,⌧)⇥A3(t, x)7!(⇢(t, x), M(x)rx⇢(t, x))
is divergence free in (0,⌧)⇥A. Applying statement b) in Lemma B.3 shows that 0 = d
dt Z
⌦
⇢(t, x)w(x)dx+ Z
Arw(x)·M(x)rx⇢(t, x)dx
= d dt
Z
A
⇢(t, x)w(x)dx+ Xm l=1
lwl⇢˙l(t) + Z
Arw(x)·M(x)rx⇢(t, x)dx
= Xm l=1
wl l⇢˙l(t)
⌧ @⇢
@nM
(t,·)
@Bl
,1
H 1/2(@Bl),H1/2(@Bl)
!
for eachw2V, where
wl:= 1
|Bl| Z
Bl
w(y)dy , l= 1, . . . , m .
Since this is true for all w 2 V, and therefore for all (w1, . . . , wm) 2 Rm, one concludes that
l⇢˙i
⌧ @⇢
@nM
@Bl
,1
H 1/2(@Bl),H1/2(@Bl)
= 0
inH 1((0,⌧)) for alll= 1, . . . , m, which is precisely the transmission condition on
@Bl. Finally, the Dirichlet condition on@⌦comes from the condition⇢2L2(R+;V) sinceV⇢H01(⌦).
Conversely, if ⇢ 2 Cb(R+,H)\L2(R+,V) s.t. @t⇢ 2 L2(R+,V0) satisfies the initial condition and the di↵usion equation in (20) in the sense of distributions on R⇤+⇥A, together with the transmission condition on @Blfor eachl= 1, . . . , m, it follows from the identity above that⇢must satisfy (19). ⇤
6. Proof of Theorem 3.5 The proof is split in several steps.
Step 1: uniform bounds and weak compactness.
By the entropy inequality (statement b) in Proposition 2.2), one has the bounds (22)
(kf✏(t,·,·)kL2(⌦⇥RN;dxdµ) k⇢inkL2(⌦) and kp
k✏q✏kL2(R+⇥⌦⇥RN⇥RN;dtdxd(µ⌦µ) k⇢inkL2(⌦)
By the Banach-Alaoglu theorem, the families f✏ andp
k✏q✏ are relatively compact inL1(R+;L2(⌦⇥RN;dxdµ)) weak-* andL2(R+⇥⌦⇥RN⇥RN;dtdxd(µ⌦µ)) weak respectively. Extracting subsequences if needed, one has
(23) f✏!f inL1(R+;L2(⌦⇥RN;dxdµ)) weak-*
while
(24) p
k✏q✏!rin L2(R+⇥⌦⇥RN ⇥RN;dtdxd(µ⌦µ)) weak.
In particular
(25) q✏!qinL2(R+⇥A⇥RN⇥RN;kA(x, v, w)dtdxd(µ⌦µ)) weak, where
(26) q(t, x, v, w) :=r(t, x, v, w)/p
kA(x, v, w), fordtdxdµ(vdµ(w)-a.e. (t, x, v, w)2R+⇥A⇥RN ⇥RN. Step 2: asymptotic form of the linear Boltzmann equation
One has 1
✏Lxf✏(t, x, v) = Z
RN
1A(x)kA(x, v, w)q✏(t, x, v, w)dµ(w) +
Z
RN
1B(x)k✏(x, v, w)q✏(t, x, v, w)dµ(w)
Since (x, v, w)7!1A(x) belongs toL2(A⇥RN⇥RN;k(x, v, w)dxd(µ⌦µ)) by (15) Z
RN
1A(x)kA(x, v, w)q✏(t, x, v, w)dµ(w)! Z
RN
1A(x)kA(x, v, w)q(t, x, v, w)dµ(w) in the weak topology ofL2(R+⇥A⇥RN;dtdxdµ) as✏!0. On the other hand, the Cauchy-Schwarz inequality and (15) imply that
Z
RN
k✏(·,·, w)q✏(·,·,·, w)dµ(w)
2
L2(R+⇥B⇥RN;dtdxdµ)
ka✏kL1(B⇥RN)
ZZ
R+⇥⌦
⌦⌦k✏(x,·,·)q✏(t, x,·,·)2↵↵
dtdx
ka✏kL1(B⇥RN)k⇢ink2L2(⌦)!0 as✏!0, by (14) and the entropy inequality in Proposition 2.2. Thus
(27) 1
✏Lxf✏(t, x, v)! Z
RN
1A(x)kA(x, v, w)q(t, x, v, w)dµ(w)
in the weak topology ofL2(R+⇥⌦⇥RN;dxdµ) as✏!0. Passing to the limit in the scaled Boltzmann equation (8) we see that
(28)
v·rxf 2L2(R+⇥⌦⇥RN, dtdxdµ) and Z
RN
kA(·,·, w)q(·,·,·, w)dµ(w)2L2(R+⇥A⇥RN, dtdxdµ), while
(29) v·rxf(t, x, v) +1A(x) Z
RN
kA(x, v, w)q(t, x, v, w)dµ(w) = 0, fordtdxdµ-a.e. (t, x, v)2R+⇥⌦⇥RN.
Step 3: asymptotic form off✏.
Multiplying both sides of the scaled linear Boltzmann equation (8) by ✏ and passing to the limit in the sense of distributions as✏!0, one finds that
Lxf(t, x, v) = 0 for a.e. (t, x, v)2R⇤+⇥⌦⇥RN.
By Lemma 2.1 e), this implies thatf(t, x, v) is independent ofv for a.e. x2A, i.e.
is of the form
(30) f(t, x, v) =⇢(t, x) for a.e. (t, x, v)2R⇤+⇥A⇥RN.
By (23) and (28)
(31) ⇢2L1(R+;L2(A)) andrx⇢2L2(R+⇥A), since
(v·rxf)v= (v⌦v)·rx⇢2L2(R+⇥A;L1(RN, dµ)) so that
hv⌦vi·rx⇢2L2(R+⇥A) ; one concludes since det(hv⌦vi)6= 0 by assumption (11).
In particular
⇢@B
i2L2([0, T];H1/2(@Bi)) for eachT >0 and eachi= 1, . . . , n.
In particular, the first condition in (28) and (10) imply thats7!f(t, x+sv, v) is continuous ins fordtdxdµ-a.e. (t, x, v)2R+⇥⌦⇥RN. Therefore, we deduce from (29) and (30) that, for eachl= 1, . . . , m
(v·rxf(t, x, v) = 0, x2Bl, v2RN, t >0, f(t, x, v) =⇢(t, x), x2@Bl, v2RN, t >0. Hence
d
dsf(t, x+sv, v) = 0 for allss.t. x+sv2Bl
for dtdxdµ(v)-a.e. (t, x, v)2 R+⇥Bl⇥RN. By assumption (10), one concludes that
⇢(t, x+⌧l(x, v)v) =⇢(t, x) ford (x)dµ(v) a.e. (x, v)2@Bl⇥RN by solving the boundary value problem above by the method of characteristics. By assumption (13)
⇢(t, x) = 1
|@Bl| Z
@Bl
⇢(t, y)d (y) =:⇢l(t) for a.e. x2@Bl, for a.e. t 0. In other words,⇢(t,·) is a.e. equal to a constant on@Bl.
Solving for f along characteristics, this implies that f(t,·,·) itself is a.e. equal to a constant on@Bl⇥RN, i.e.
f(t, x, v) = 1
|Bl| Z
Bl
hf(t, x,·)idx=:⇢l(t) fordtdxdµ-a.e. (t, x, v)2R+⇥Bl⇥RN, forl= 1, . . . , m.
Summarizing, we have proved that
(32) f(t, x, v) =⇢(t, x) fordtdxdµ a.e. (t, x, v)2⌦ with⇢2L1(R+;H) andrx⇢2L2(R+⇥⌦). Step 4: Fourier’s law and continuity equation
Observe that the flux satisfies
(33) 1
✏hvf✏(t, x,·)i=1
✏h(L⇤xb(x,·))f✏(t, x,·)i=
⌧ b(x,·)1
✏Lxf✏(t, x,·)
= ZZ
RN⇥RN
b(x, v)k(x, v, w)q✏(t, x, v, w)dµ(v)dµ(w) for a.e. (t, x)2R+⇥A and for all✏>0.
Since b 2 L1(A;L2(RN;dµ)) by statement b) in Lemma 3.1 , the function (x, v, w)7!p
kA(x, v, w)b(x, v) belongs toL1(A;L2(RN⇥RN;dµ(v)µ(w))). Thus
(34) 1
✏hvf✏(t, x,·)i= ZZ
RN⇥RN
b(x, v)k(x, v, w)q✏(t, x, v, w)dµ(v)dµ(w)
! ZZ
RN⇥RN
b(x, v)k(x, v, w)q(t, x, v, w)dµ(v)dµ(w)
=hb(x,·)v·rx⇢(t, x)i=M(x)rx⇢(t, x) in for the weak topology ofL2(R+⇥A) as✏!0, on account of (29).
Therefore, for eachw2V, one has d
dt Z
⌦hf✏(t, x,·)iw(x)dx+ Z
A
1
✏hvf✏(t, x,·)i·rw(x)dx= 0,
(sincerw= 0 on ˚B) and passing to the limit in each side of this identity as✏!0 shows that
(35) d
dt Z
⌦
⇢(t, x)w(x)dx+ Z
Arw(x)·M(x)rx⇢(t, x)dx= 0 in the sense of distributions onR⇤+.
Step 5: limiting entropy production By definition ofq✏, one has
q✏(t, x, v, w) = q✏(t, x, w, v)
fordtdxdµ(vdµ(w)-a.e. (t, x, v, w)2R+⇥A⇥RN⇥RN and each✏>0; by passing to the limit as✏!0
q(t, x, v, w) = q(t, x, w, v)
fordtdxdµ(vdµ(w)-a.e. (t, x, v, w)2R+⇥A⇥RN ⇥RN. Defining kAs(t, x, v, w) =12(kA(t, x, v, w) +kA(t, x, w, v)) one has
⌦⌦kA(x,·,·)q(t, x,·,·)2↵↵
=⌦⌦
ksA(x,·,·)q(t, x,·,·)2↵↵
for a.e. (t, x)2R+⇥A. Likewise ZZ
RN⇥RN
kA(x, v, w)( (v) (w))2dµ(v)dµ(w)
= ZZ
RN⇥RN
ksA(x, v, w)( (v) (w))2dµ(v)dµ(w)
and ZZ
RN⇥RN
kA(x, v, w)( (v) (w))q(t, x, v, w)dµ(v)dµ(w)
= ZZ
RN⇥RN
ksA(x, v, w)( (v) (w))q(t, x, v, w)dµ(v)dµ(w)
for a.e. (t, x) 2R+⇥⌦. With (v) = ⇠·b(x, v) for some ⇠ 2RN to be chosen later, and applying the Cauchy-Schwarz inequality, one finds that
(36) ✓ZZ
RN⇥RN
kAs(x, v, w)⇠·(b(x, v) b(x, w))q(t, x, v, w)dµ(v)dµ(w)
◆2
ZZ
RN⇥RN
kA(x, v, w)(⇠·(b(x, v) b(x, w)))2dµ(v)dµ(w)⌦⌦
kA(x,·,·)q(t, x,·,·)2↵↵
.
On the other hand, by definition ofkAs ZZ
RN⇥RN
kAs(x, v, w)⇠·(b(x, v) b(x, w))q✏(t, x, v, w)dµ(v)dµ(w)
= 2
✏ ZZ
RN⇥RN
ksA(x, v, w)⇠·(b(x, v) b(x, w))f✏(t, x, v)dµ(v)dµ(w)
= 1
✏hf✏(t, x,·)(Lx+L⇤x)⇠·b(x,·)i=2
✏h⇠·vf✏(t, x,·)i for a.e. (t, x)2R+⇥A where the last equality follows from (17). Passing to the limit as✏!0, one finds that
ZZ
RN⇥RN
ksA(x, v, w)⇠·(b(x, v) b(x, w))q(t, x, v, w)dµ(v)dµ(w)
= 2⇠·M(x)rx⇢(t, x) for a.e. (t, x)2R+⇥A. On the other hand
ZZ
RN⇥RN
kA(x, v, w)(⇠·(b(x, v) b(x, w)))2dµ(v)dµ(w)
= 2h⇠·b(x,·)Lx(⇠·b(x,·))i= 2h⇠·b(x,·)⇠·vi= 2⇠·M(x)⇠
for a.e. x2A, by Lemma 2.1 b). Hence
2(⇠·M(x)rx⇢(t, x))2⇠·M(x)⇠⌦⌦
kA(x,·,·)q(t, x,·,·)2↵↵
and choosing⇠=rx⇢(t, x), we find that (37) 2rx⇢(t, x)·M(x)rx⇢(t, x)⌦⌦
kA(x,·,·)q(t, x,·,·)2↵↵
for a.e. (t, x)2R+⇥A. By convexity and weak convergence (38)Z 1
0
Z
A
⌦⌦kA(x,·,·)q(t, x,·,·)2↵↵
dxdtlim
✏!0
Z 1 0
Z
A
⌦⌦kA(x,·,·)q✏(t, x,·,·)2↵↵
dxdt .
Using Lemma 3.1 e) and the entropy inequality (39)
CK
Z 1
0 krx⇢(t,·)k2L2(A)dt2 Z 1
0
Z
Arx⇢(t, x)·M(x)rx⇢(t, x)dxdt k⇢ink2L2(⌦). Step 6: limiting initial condition
By (33) and the Cauchy-Schwarz inequality 1
✏hvf✏i
2
L2([0,T]⇥A) Z
R+
Z
A
⌦⌦kA(x,·,·)q✏(t, x,·,·)2↵↵
dxdt
⇥supess
x2A
ZZ
RN⇥RN
kA(x, v, w)|b(x, v)|2dµ(v)dµ(w)
8CK3h|v|2ik⇢ink2L2(⌦)
using the entropy inequality in Proposition 2.2 and Lemma 3.1 b) and d). Since d
dt Z
⌦hf✏(t, x,·)iw(x)dx= Z
A
1
✏hvf✏(t, x,·)i·rw(x)dx for eachw2V, one has
(40) d
dt Z
⌦hf✏(·, x,·)iw(x)dx (2CK)3/2h|v|2i1/2k⇢inkL2(⌦)krwkL2(⌦). Applying the Ascoli-Arzela theorem shows that, for eachw2V
(41)
Z
⌦
(hf✏(t, x,·)i ⇢(t, x))w(x)dx!0 uniformly in t2[0, T] for allT >0. In particular
Z
⌦
⇢in(x)w(x)dx= Z
⌦hf✏(0, x,·)iw(x)dx! Z
⌦
⇢(0, x))w(x)dx so that
(42)
Z
⌦
⇢(0, x)w(x)dx= Z
⌦
⇢in(x)w(x)dx for eachw2V.
Returning to (40), we have proved that @thf✏i is bounded inL2(R+,V0) for each T >0, so that
(43) @t⇢2L2(R+;V0).
Step 7: Dirichlet condition
Next we establish the Dirichlet condition on@⌦for the di↵usion equation. The scaled linear Boltzmann equation implies that, for each 2Cc1(R⇤+),
v·rx Z 1
0
(t)f✏(t, x, v)dt= Z 1
0
(t)1
✏Lxf✏(t, x, v)dt+✏ Z 1
0
0(t)f✏(t, x, v)dt is bounded in L2(⌦⇥RN;dxdµ) by (27) and the uniform boundedness principle (Banach-Steinhaus’ theorem) and the entropy inequality in Proposition 2.2, while
Z 1
0
(t)f✏(t, x, v)dt
is bounded inL2(⌦⇥RN;dxdµ) by the same entropy inequality. Hence 0 =
Z 1 0
(t)f✏(t,·,·)dt ! Z 1
0
(t)⇢(t,·)dt
in L2( ;|v ·nx|⌧(x, v)^1d (x)dv) by Cessenat’s trace theorem [6], where the notation ⌧(x, v) designates the forward exit time from ⌦ starting from x with velocityv, i.e.
⌧(x, v) := inf{t >0 s.t. x+tv2@⌦}, x2⌦, v2RN.
In particular
Z 1 0
(t)⇢(t,·)dt
@⌦
= 0.
By (32), we already know that the limiting density⇢2L2([0, T];H1(⌦)). Therefore (44) ⇢(t,·)@⌦= 0 in L2([0, T];H1/2(@⌦))
for eachT >0.
Step 8: convergence to the di↵usion equation Summarizing, we have proved that
f✏ is relatively compact inL1(R+;L2(⌦⇥RN, dxdµ(v))) weak-*
and that, iff is a limit point off✏as ✏!0, it is of the form
f(t, x, v) =⇢(t, x) dtdxdµ(v) a.e. in (t, x, v)2R+⇥⌦⇥RN where
(45) ⇢2L1(R+;H)\L2(R+;H01(⌦)) =L1(R+;H)\L2(R+;V) and @t⇢2L2(R+;V0)
since⇢satisfies the Dirichlet boundary condition (44) and rx⇢2L2(R+⇥⌦) by (32), together with (43). In particular, this implies that
(46) ⇢2Cb(R+;H).
Besides ⇢ satisfies (35) for each test function w 2 V, together with the initial condition (42). Therefore ⇢is the unique solution of the Dirichlet problem for the di↵usion equation with di↵usion matrixM(x) defined in (18) with infinite di↵usivity in B, with initial data⇢in. By compactness and uniqueness of the limit point, we conclude that
f✏!⇢ inL1(R+;L2(⌦⇥RN, dxdµ(v))) weak-*
as✏!0.
Step 9: strong convergence
The weak solution⇢of the initial-boundary value problem for the di↵usion equa- tion with infinite di↵usivity inB is known to satisfy the identity
1 2
Z
⌦
⇢(t, x)2dx+ Z t
0
Z
Arx⇢(s, x)·M(x)rx⇢(s, x)dxds= 12 Z
⌦
⇢in(x)2dx for eacht 0.
By Jensen’s inequality Z
⌦hf✏(t, x,·)2idx Z
⌦hf✏(t, x,·)i2dx while, by convexity and weak convergence,
lim
✏!0
Z
⌦hf✏(t, x,·)i2dx Z
⌦
⇢(t, x)2dx uniformly int2[0, T] for eachT >0 by (41).
With the entropy identity satisfied byf✏(see Proposition 2.2) and the inequality (38), the two inequalities above imply that
Z
⌦hf✏(t, x,·)2idx! Z
⌦
⇢(t, x)2dx
uniformly int2[0, T] for eachT >0, so that
f✏(t,·,·)!⇢(t,·) strongly inL2(⌦⇥RN, dxdµ) uniformly int2[0, T] for eachT >0.
By the same token
1 2
Z t 0
Z
A
⌦⌦kA(x,·,·)q✏(s, x,·,·)2↵↵
dxds! 12
Z t 0
Z
A
⌦⌦kA(x,·,·)q✏(s, x,·,·)2↵↵
dxds
= Z t
0
Z
Arx⇢(s, x)·M(x)rx⇢(s, x)dxds . Therefore
q✏!q strongly inL2([0, t]⇥A⇥RN ⇥RN;kA(x, v, w)dsdxdµ(v)dµ(w))) as✏!0. Besidesqsatisfies the equality in the Cauchy-Schwarz inequality (36) so thatq is of the form
q(t, x, v, w) = (t, x)(b(x, v) b(x, w))·rx⇢(t, x).
for some measurable function defined a.e. on R+⇥A. Using (29) shows that (t, x) = 1 for a.e. (t, x)2R+⇥A, which concludes the proof.
Remark. The proof of Theorem 3.5 is inspired from the discussion of the Stokes and of the Navier-Stokes limit of the Boltzmann equation initiated in [1]. The procedure in step 9 for obtaining strong convergence is a simplified analogue of the proof of Theorem 6.2 in [1] using the notion of “entropic convergence” (see formula (4.32) in [1]). The discussion bearing on the limiting entropy production in step 5 is a simplified version of Lemma 4.7 in [1]. Notice also the role of the limiting linearized Boltzmann equation (29) and of the limiting collision integrand q, that is reminiscent of the analogous objects considered in [1] (see formula (4.3) and Proposition 4.1 there) in the case of the Boltzmann equation of the kinetic theory of gases. Notice in particular that the next to leading order in the Hilbert expansion, i.e. the second convergence statement in Theorem 3.5 is stated in complete analogy with formula (6.18) in [1] for the case of the Stokes limit of the Boltzmann equation.
7. Conclusions
The main result presented above (Theorem 3.5) can obviously be generalized in several directions.
First, our method obviously applies to a scaled linear Boltzmann equation of the form
(✏@t+v·rx)f✏(t, x, v) +1
✏Lxf✏(t, x, v) +✏Bf✏(t, x, v) =✏S(t, x, v)
where B is a bounded operator on L2(⌦⇥RN;dxdµ) and S 2 L1(R+;L2(⌦⇥ RN, dxdµ)) is a source term. For instance B could be the multiplication by an amplifying or damping coefficient, i.e. Bf✏(t, x, v) = (x)f✏(t, x, v) as in [3]. In other words, problems where the collision process is nearly, but not exactly conservative can be treated as above.
More general boundary conditions than the absorbing condition on@⌦can also be considered. For instance, imposing a specular or di↵use reflection condition at
