THE DIFFUSION APPROXIMATION FOR THE LINEAR BOLTZMANN ... · the diffusion approximation for the...

THE DIFFUSION APPROXIMATION FOR THE LINEAR

BOLTZMANN EQUATION WITH VANISHING ABSORPTION

CLAUDE BARDOS, ETIENNE BERNARD, FRANCOIS GOLSE, AND REMI SENTIS

Abstract. The present paper discusses the di↵usion approximation of thelinear Boltzmann equation in cases where the collision frequency is not uni-formly large in the spatial domain. Our result applies for instance to the caseof radiative transfer in a composite medium with optically thin inclusions inan optically thick background medium. The equation governing the evolutionof the approximate particle density coincides with the limit of the di↵usionequation with infinite di↵usion coe�cient 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 ofidentical point particles interacting with some background material. In other words,f(t, x, v) is the number density of particles located at the position x 2 ⌦, where ⌦is a domain of RN , with velocity v ⇢ RN at time t � 0.

The notation Lx designates a linear integral operator acting on the v variable inf , i.e.

(2) Lxf(t, x, v) =

Z

RN

k(x, v, w)(f(t, x, v)� f(t, x, w))dµ(w)

where µ is a Borel probability measure on RN , while k is a nonnegative functiondefined µ⌦ µ-a.e. on RN ⇥RN that measures the probability of a transition fromvelocity v to velocity w for particles located at the position x.

Henceforth we denote

(3) h�i =Z

RN

�(v)dµ(v) and⌦⌦�↵↵=

ZZ

RN⇥RN

�(v, w)dµ(v)dµ(w)

for all � 2 L1(RN , dµ) and � 2 L1(RNv ⇥RN

w ; dµ(v)dµ(w)).We assume that k satisfies 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)

1991 Mathematics 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

2 C. BARDOS, E. BERNARD, F. GOLSE, AND R. SENTIS

for the absorption rate, so that

Lxf(t, x, v) = a(x, v)f(t, x, v)�Kxf(t, x, v)

where Kx designates 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 discussionare introduced later.

We further assume that ⌦ is a bounded domain of RN with C1 boundary @⌦and denote by nx the unit outward normal field at x 2 @⌦. 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 bediscussed later.

We are concerned with the di↵usion approximation of the linear Boltzmannequation (1) — see [7] and the references therein for a general presentation of thisapproximation. We briefly recall its main features below.

Set L to be a length scale that measures the size of ⌦ while V is the averageparticle velocity; this defines a time scale T := L/V . The di↵usion limit of (1) isbased on the assumption that the dimensionless quantity Ta(x, v) is large. Thuswe introduce a scaling parameter 0 < ✏⌧ 1 and set

k✏(x, v, w) := ✏k(x, v, w)

so that k✏(x, v, w) is of order unity. Accordingly, we define

a✏(x, v) := ✏a✏(x, v) , Lx = ✏Lx , and Kx = ✏Kx .

(For notational simplicity, we do not mention explicitly the dependence of Lx andKx in ✏.) Assume further that variations of order unity of the boundary data drivingthe solution of (1) do not occur on time scales shorter than T/✏.

In that case, the solution f of (1) is sought in the form

f(t, x, v) = f✏(✏t, x, v)

with the notation t = ✏t for the rescaled time variable. Thus (1) takes the form

✏@tf✏(t, x, v) + v ·rxf✏(t, x, v) +1

✏Lxf✏(t, x, v) = 0 .

DIFFUSION APPROXIMATION WITH VANISHING ABSORPTION 3

Henceforth we drop hats on rescaled variables and consider the initial-boundaryvalue problem for the scaled linear Boltzmann equation

(8)

8>><

>>:

(✏@t + v ·rx)f✏(t, x, v) +1

✏Lxf✏(t, x, v) = 0 , x 2 ⌦ , v 2 RN , t > 0 ,

f✏(t, x, v) = 0 , (x, v) 2 �� , t > 0 ,

f✏(0, x, v) = f in(x, v) x 2 ⌦ , v 2 RN ,

in the limit as ✏! 0.

2. Existence, uniqueness and a priori estimates

Assume that k is a nonnegative measurable function on RN ⇥RN satisfying thesemi-detailed balance assumption (4) and the condition

(9) a✏ 2 L1(⌦⇥RN , dxdµ) .

Lemma 2.1. Under assumption (9)

a) the integral operators Lx and Kx are bounded on L2(RN , µ) for a.e. x 2 ⌦, with

kKxkL(L2(RN ,µ)) ka✏(x, ·)kL1(RN ,µ) ;

b) the adjoints of Kx and Lx 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. x 2 ⌦;

c) for a.e. x 2 ⌦

{ functions a.e. constant on RN} = R ⇢ Ker(Lx) and Ker(L⇤x) ;

d) for each � 2 L2(RN ; dµ),

h�Lx�i = 12

ZZ

RN⇥RN

k✏(x, v, w)(�(v)� �(w))2dµ(v)dµ(w) ,

for a.e. x 2 ⌦;

e) if in addition k✏(x, v, w) > 0 for dµ(v)dµ(w)-a.e. (v, w) 2 RN ⇥RN, then

Ker(Lx) = Ker(L⇤x) = { functions a.e. constant on RN} = R .

Remark. The discussion of the properties of the operator Lx di↵ers from [3]. In[3], it is assumed that the measure µ is the uniform probability measure on the setV of admissible velocities, that can be a ball, or a sphere, or a spherical annuluscentered at the origin in RN . The scattering kernel k✏(x, v, w) is of the form

k✏(x, v, w) = �✏(x)f(v, w)

where f(v, w) = f(w, v) a.e. on V ⇥ V is positive and such thatZ

V

f(v, w)dw = 1 for a.e. v 2 V .

Thus Lx = �✏(x)(I � F ), where F is the integral operator defined by

F�(v) :=

Z

V

f(v, w)�(w)dw for a.e. v 2 V .


Then Ker(Lx) = Ker(I � F ) whenever �✏(x) > 0, and R ⇢ Ker(I � F ). On theother hand, it is assumed in [3] that f is chosen so that F is a compact operator onL2(V ). Then 1 is the principal eigenvalue of F and Ker(I � F ) is one-dimensionalby 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 Lemma1 in §2 of chapter XXI in [7]). The formula for Kx in statement b) and statementc) are obvious.

As for the formula for L⇤x in 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)

for dxdµ-a.e. (x, v) 2 ⌦⇥RN by the semi-detailed balance assumption (4).For each � 2 L2(RN ; dµ) and a.e. in x 2 ⌦

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) + 1

2

Z

RN

a✏(x,w)�(w)2dµ(v)

�ZZ

RN⇥RN


=

ZZ

RN⇥RN

k✏(x, v, w)12 (�(v)

2 + �(w)2)dµ(v)dµ(w)

�ZZ

RN⇥RN


= 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 provesstatement d).

By statement d), if � 2 L2(RN ; dµ) satisfies Lx� = 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 for dµ(v)dµ(w)� a.e. (v, w) 2 RN ⇥RN

so that

�(v)� �(w) = 0 for dµ(v)dµ(w)� a.e. (v, w) 2 RN ⇥RN .

Averaging in w shows that

�(v) = h�i for dµ(v)� a.e. v 2 RN ,

so thatKer(Lx) ⇢ { functions a.e. constant on RN} = R .


With statement c), this shows that

Ker(Lx) = { functions a.e. constant on RN} = R .

Since the function (v, w) 7! k✏(x,w, v) satisfies the same properties as k✏,

Ker(Lx) = { functions a.e. constant on RN} = R .

⇤

Proposition 2.2. Assume that k✏ is a nonnegative measurable function defined

dxd(µ⌦µ)-a.e. on ⌦⇥RN ⇥RNsatisfying (4) and (9) with a✏ defined by (5). For

each ✏ > 0 and each f in 2 L2(⌦⇥RN ; dxdµ(v)), there exists a unique weak solution

of the initial-boundary value problem (8) in the space Cb(R+;L2(⌦⇥RN ; dxdµ(v))).This solution satisfies

a) the continuity equation

@thf✏i+ divx1

✏hvf✏i = 0

in the sense of distributions on R⇤+ ⇥ ⌦;

b) the “entropy inequality”

Z

⌦hf✏(t, x, ·)2idx+

Z t

0

Z

⌦

⌦⌦k✏(x, ·, ·)q✏(s, x, ·, ·)2

↵↵dxds

Z

⌦hf in(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 advectionoperator �v ·rx with absorbing boundary condition (7) that is the generator of astrongly continuous contraction semigroup on L2(⌦⇥RN ; dxdµ(v)).

This implies the existence and uniqueness of the weak solution f✏ of the initial-boundary value problem (8) in the functional space Cb(R+;L2(⌦⇥RN ; dxdµ(v))).

Statement a) follows from the inclusion R ⇢ Ker(L⇤x) in Lemma 2.1. Statement

b) follows from Lemma 2.1 d) and Lemma 2.3 below. ⇤

Lemma 2.3. Let f in 2 L2(⌦⇥RN ; dxdµ(v)) and S 2 L2([0, T ]⇥⌦⇥RN ; dtdxdµ(v)).For each ✏ > 0, let f✏ be the weak solution in Cb(R+;L2(⌦⇥RN ; dxdµ(v))) of

8>><

>>:

✏@tf✏ + v ·rxf✏ = S , x 2 ⌦ , v 2 RN , t > 0 ,

f✏��

= 0 ,

f✏��t=0

= f in .

Then

12

Z

⌦hf(t, x, ·)2idx 1

✏

Z t

0

Z

⌦hS(s, x, ·)f✏(s, x, ·)idxds+ 1

2

Z

⌦hf in(x, ·)2idx

The proof of this lemma is classical; we give it in the appendix for the sake ofbeing self-contained.


3. Diffusion approximation with vanishing absorption rate:

main results

Assume that the spatial domain ⌦ = A [ B, where A is open and B is closedin RN (i.e. B \ @⌦ = ?), with finitely many connected components denoted Bl,for l = 1, . . . ,m. We further assume that Bl has piecewise C1 boundary, that Bl islocally on one side of its boundary @Bl. Finally, we denote by nx the unit normalfield at x 2 @A, oriented towards the exterior of A.

Henceforth, it is assumed that the measure µ satisfies

(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 position x with the velocity v; in other words

(12) ⌧l(x, v) := inf{t > 0 s.t. x+ tv 2 @Bl} .We assume that, for each l = 1, . . . ,m and for each g 2 L2(@Bl),

(13)

g(x+ ⌧l(x, v)v) = g(x) for d�(x)dµ(v)� a.e. (x, v) 2 @Bl ⇥RN

) g(x) =1

|@Bl|Z

@Bl

g(y)d�(y) for a.e. x 2 @Bl .

We further assume that the scattering kernel k✏ in the linear Boltzmann equationis a dxdµ(v)dµ(w)-a.e. nonnegative measurable function on ⌦⇥RN⇥RN satisfyingthe following assumptions, in addition to (4):

(a) the absorption rate a✏ is uniformly small on B ⇥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 ✏ anddenoted kA ⌘ 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) , for dxdµ(v)� a.e. (x, v) 2 A⇥RN ;

(c) there exists a RN -valued vector field b ⌘ b(x, v) defined dxdµ-a.e. on A⇥RN

such that

(17) b(x, ·) 2 L2(RN , dµ) , hb(x, ·)i = 0 and Lxb(x, ·) = L⇤xb(x, ·) = v

for a.e. x 2 A.

With the vector field b, we defined the MN (R)- valued matrix field

(18) M(x) = hb(x, ·)⌦ vi , for a.e. x 2 A .


3.1. Coercivity properties of Lx. The coercivity properties of Lx on the orthog-onal of its null-space for a.e. x 2 A are crucial in order to study the matrix fieldM . 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↵usionmatrix is (a.e.) defined on A only, and not in all of ⌦.

Lemma 3.1. Assume that k✏ satisfies the assumptions (4)-(14)-(15)-(17) while the

probability measure µ satisfies (11). Then

a) one has

hvi = 0 ;

b) for a.e. x 2 A and each � 2 L2(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 field M satisfies

M(x) = M(x)T for a.e. x 2 A ;

d) the matrix field M satisfies the bound

|Mij(x)| 2CKkvikL2(RN ,dµ)kvjkL2(RN ,dµ) for a.e. x 2 A ;

e) denoting by � > 0 the smallest eigenvalue of the positive matrix hv⌦2i, one has

⇠ ·M(x)⇠ � �

2CK|⇠|2 for all ⇠ 2 RN , for a.e. x 2 A .

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 coe�-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 ✏ ! 0in the subregion B of the spatial domain ⌦. This suggest that the limit equationshould be a di↵usion equation with infinite di↵usion constant in B.

Before going further, we recall the precise statement of this problem and itsvariational formulation.

Define

H :=

⇢u 2 L2(⌦) s.t. u(x) =

1

|Bl|Z

Bl

u(y)dy for a.e. x 2 Bl , l = 1, . . . , n

�,

and

V := H \H10 (⌦) = {u 2 H1

0 (⌦) s.t. ru(x) = 0 for a.e. x 2 Bl , l = 1, . . . , n} .For each ⇢in 2 H, consider the following variational problem

(19)8>>><

>>>:

⇢ 2 Cb(R+;H) \ L2(R+;V) , @t⇢ 2 L2(R+;V 0) , and ⇢��t=0

= ⇢in ,

d

dt

Z

⌦⇢(t, x)w(x)dx+

Z

A

rw(x) ·M(x)rx⇢(t, x)dx = 0 , for a.e. t � 0 ,

for all w 2 V .


Proposition 3.2. Assume that x 7! M(x) is an MN (R)-valued measurable matrix

field on A satisfying

Mij 2 L1(A) for all i, j = 1, . . . , N , and there exists ↵ > 0 s.t.

⇠ ·M(x)⇠ � ↵|⇠|2 for a.e. x 2 A and all ⇠ 2 RN .

For each ⇢in 2 H, the variational problem (19) has a unique solution. This solution

satisfies the “energy” identity

12

Z

⌦⇢(t, x)2dx+

Z t

0

Z

A

rx⇢(s, x) ·M(x)rx⇢(s, x)dxds =12

Z

⌦⇢in(x)2dx

for each t � 0.

Next we state the PDE formulation equivalent to (19).

Proposition 3.3. Assume that x 7! M(x) is an MN (R)-valued measurable matrix

field on A such that Mij 2 L1(A) for all i, j = 1, . . . , N . Let ⇢ 2 C([0, T ];H) \L2([0, T ];V) with @t⇢ 2 L2([0, T ];V 0).

Then ⇢ satisfies

8>>><

>>>:

d

dt

Z

⌦⇢(t, x)w(x)dx+

Z

A

rw(x) ·M(x)rx⇢(t, x)dx = 0 , for all w 2 V ,

⇢��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) x 2 A , t > 0 ,

⇢(t, x) = 0 x 2 @⌦ , t > 0 ,

⇢l(t) =1

�l

Z

@Bl

@⇢

@nM(t, x)d�(x) l = 1, . . . ,m , t > 0 ,

⇢(0, x) = ⇢in(x) x 2 ⌦ .

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

⇢ 2 Cb(R+;H) \ L2(R+;V) and @t⇢ 2 L2(R+;V 0)


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 wherethe di↵usion coe�cient tends to +1 in B: see for instance Theorem 2.4 in [8] fora 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 withabsorption rate vanishing in B.

Theorem 3.5. Assume that µ satisfies (11) and that k✏ satisfies the assumptions

(4)-(14)-(15)-(17). Let M be the matrix field defined on A by (18).

Let ⇢in 2 H, and for each ✏ > 0, let f✏ be the unique weak solution of the initial-

boundary value problem (8) in the space Cb(R+;L2(⌦⇥RN ; dxdµ(v))). Then

f✏(t, ·, ·) ! ⇢(t, ·) strongly in L2(⌦⇥RN ; dxdµ)

uniformly in t 2 [0, T ] for all T > 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 of L2([0, T ]⇥A⇥RN ⇥RN ; kA(x, v, w)dtdxdµ(v)dµ(w)) forall T > 0 as ✏! 0.

Remark. The last convergence statement above is equivalent to the following:

(21)1

✏(f✏ � hf✏i) ! �b ·rx⇢ strongly in L2([0, T ]⇥A⇥RN ; dtdxdµ)

as ✏! 0. This statement is the analogue of formula (37) in [3] giving the O(✏) termin the Hilbert expansion of the solution f✏ 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 from L2([0, T ]⇥A⇥RN ⇥RN ; kA(x, v, w)dtdxdµ(v)dµ(w))to L2([0, T ]⇥A⇥RN ; dtdxdµ(v)). Indeed

�(t, x, v)2 CK

Z

RN

kA(x, v, w)�(t, x, v, w)2dµ(w)

by the Cauchy-Schwarz inequality and assumption (15). The conclusion followsfrom integrating both sides of this inequality in t, 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)

in L2([0, T ]⇥A⇥RN ; dtdxdµ(v)) as ✏! 0, since hb(x, ·)i = 0 for a.e. x 2 A. ⇤


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 thedi↵usion matrix M . In the present formulation, the existence of b is postulated in(17) and the condition hvi = 0 in Lemma 3.1 deduced from the existence of b.

Conversely, one can assume that the transition kernel kA is chosen so that Kx isa compact operator on L2(RN ; dµ) for a.e. x 2 A. In that case, Lx is a Fredholmoperator on L2(RN ; dµ) for a.e. x 2 A, because the multiplication by aA(x, v) isan invertible operator on L2(RN ; dµ) for a.e. x 2 A — 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

for dxdµ(v)-a.e. (x, v) 2 A⇥RN . Applying then Lemma 2.1 e) shows that

Im(Lx) = Ker(L⇤x)

? = {� 2 L2(RN ; dµ(v)) s.t. h�i = 0} .Thus

hvi = 0 , v 2 Im(Lx)

for a.e. x 2 A. If in addition kA is chosen such that

kA(x, v, w) = kA(x,w, v) for dµ(v)dµ(w)� a.e. (v, w) 2 RN ⇥RN

for a.e. x 2 A, then L⇤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 in RN centered at the origin,(13) is obviously satisfied. Indeed, for each x, y 2 @Bl, the segment [x, y] is includedin Bl, so that g(x) = g(y) for a.e. x, y 2 @Bl.

But even when Bl is convex, the assumption (13) may fail to be satisfied for somemeasures µ. For instance, assume that N = 2, and take Bl = {x 2 R2 s.t. |x| 1}.Denote by (e1, e2) the canonical basis of R2, and let

µ = 14 (�e1 + ��e1 + �e2 + ��e2) .

For x = (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 thecondition

g(x+ ⌧l(x, v)v) = g(x) for dxdµ(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) 2 B ⇥RN ⇥RN , or if k✏(x, v, w) = O(✏) on B. The assumption used inthe present paper is obviously much more general. For instance, it is satisfied if onehas k✏(x, v, w) = O(| ln ✏|��l) on Bl with �l > 0 for each l = 1, . . . ,m. The Hilbertexpansion method used in [3] does not apply to this situation, and therefore cannotbe used on the problem considered here in its fullest generality.

Even in the nondegenerate case where B = ?, observe that our assumptionson the transition kernel k✏ do not imply that the vector field b in (17) dependssmoothly on x. This again excludes the possibility of using the Hilbert expansionas in [3] to establish the validity of the di↵usion limit. Accordingly, the di↵usionmatrix field M obtained in Theorem 3.5 is in general not even continuous. In thisregularity 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 forinstance section 5.1 and Remark 5.1.6 in [14]). One should bear in mind that adi↵usion equation with di↵usion matrix that has a type I discontinuity across somesmooth surface is equivalent to a transmission problem for two di↵usion equationson each side of the discontinuity surface, with continuity of the solution and of thenormal 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 each i = 1, . . . , N and a.e. x 2 A, one has

hvii = hLxbi(x, ·)i = h(L⇤x1)bi(x, ·)i = 0

since L⇤x1 = 0 by Lemma 2.1 c), which proves statement a).

Set Lx� = ; 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. x 2 A,

|�(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

since L⇤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, weobtain 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 all i, j = 1, . . . , N and a.e. x 2 A. 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. x 2 A, 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. x 2 A. 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

A

ru(x) ·M(x)rv(x)dx , u, v 2 V .

Indeed, this bilinear form satisfies the assumptions of the Lions-Magenes theoremsince 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 each u, v 2 V.Consider the linear functional

L(t) : V 3 w 7! h@t⇢, wiV0,V + a(⇢(t, ·), w)defined for a.e. t � 0.

Since L(t) = 0 for a.e. t 2 R, one has

hL(t), ⇢(t, ·)iV0,V = 0 for a.e. t � 0 ,

for each w 2 V. By Lemma B.2, one has

L(t) = 0 in V 0 for a.e. t 2 R+ .

In particular, for a.e. s � 0, one has

0 = hL(s), ⇢(s, ·)iV0,V = h@t⇢(s, ·), ⇢(s, ·)iV0,V +

Z

A

rx⇢(s, x) ·M(x)rx⇢(s, x)dx ,

and one concludes by integrating in s 2 [0, t] and applying Lemma B.1 b). ⇤


Proof of Proposition 3.3. Specializing (19) to the case where w 2 C1c (A) is equiv-

alent to

@t⇢� divx(Mrx⇢) = 0 in D0(R⇤+ ⇥A) .

In particular, the (a.e. defined) vector field

(0, ⌧)⇥A 3 (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

A

rw(x) ·M(x)rx⇢(t, x)dx

=d

dt

Z

A

⇢(t, x)w(x)dx+mX

l=1

�lwl⇢l(t) +

Z

A

rw(x) ·M(x)rx⇢(t, x)dx

=mX

l=1

wl

�l⇢l(t)�

⌧@⇢

@nM(t, ·)

��@Bl

, 1

�

H�1/2(@Bl),H1/2(@Bl)

!

for each w 2 V, 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, oneconcludes that

�l⇢i �⌧

@⇢

@nM

��@Bl

, 1

�

H�1/2(@Bl),H1/2(@Bl)

= 0

in H�1((0, ⌧)) for all l = 1, . . . ,m, which is precisely the transmission condition on@Bl. Finally, the Dirichlet condition on @⌦ comes from the condition ⇢ 2 L2(R+;V)since V ⇢ H1

0 (⌦).Conversely, if ⇢ 2 Cb(R+,H) \ L2(R+,V) s.t. @t⇢ 2 L2(R+,V 0) satisfies the

initial condition and the di↵usion equation in (20) in the sense of distributions onR⇤

+ ⇥ A, together with the transmission condition on @Bl for each l = 1, . . . ,m, itfollows 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

kpk✏q✏kL2(R+⇥⌦⇥RN⇥RN ;dtdxd(µ⌦µ) k⇢inkL2(⌦)

By the Banach-Alaoglu theorem, the families f✏ andpk✏q✏ are relatively compact

in L1(R+;L2(⌦⇥RN ; dxdµ)) weak-* and L2(R+ ⇥⌦⇥RN ⇥RN ; dtdxd(µ⌦ µ))weak respectively. Extracting subsequences if needed, one has

(23) f✏ ! f in L1(R+;L2(⌦⇥RN ; dxdµ)) weak-*

while

(24)pk✏q✏ ! r in L2(R+ ⇥ ⌦⇥RN ⇥RN ; dtdxd(µ⌦ µ)) weak.


In particular

(25) q✏ ! q in L2(R+ ⇥A⇥RN ⇥RN ; kA(x, v, w)dtdxd(µ⌦ µ)) weak,

where

(26) q(t, x, v, w) := r(t, x, v, w)/pkA(x, v, w) ,

for dtdxdµ(vdµ(w)-a.e. (t, x, v, w) 2 R+ ⇥A⇥RN ⇥RN .

Step 2: asymptotic form of the linear Boltzmann equation

One has1

✏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 to L2(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 of L2(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 of L2(R+ ⇥ ⌦⇥RN ; dxdµ) as ✏! 0. Passing to the limit inthe scaled Boltzmann equation (8) we see that

(28)

v ·rxf 2 L2(R+ ⇥ ⌦⇥RN , dtdxdµ) andZ

RN

kA(·, ·, w)q(·, ·, ·, w)dµ(w) 2 L2(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 ,

for dtdxdµ-a.e. (t, x, v) 2 R+ ⇥ ⌦⇥RN .

Step 3: asymptotic form of f✏.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) 2 R⇤+ ⇥ ⌦⇥RN .

By Lemma 2.1 e), this implies that f(t, x, v) is independent of v for a.e. x 2 A, i.e.is of the form

(30) f(t, x, v) = ⇢(t, x) for a.e. (t, x, v) 2 R⇤+ ⇥A⇥RN .


By (23) and (28)

(31) ⇢ 2 L1(R+;L2(A)) and rx⇢ 2 L2(R+ ⇥A) ,

since(v ·rxf)v = (v ⌦ v) ·rx⇢ 2 L2(R+ ⇥A;L1(RN , dµ))

so thathv ⌦ vi ·rx⇢ 2 L2(R+ ⇥A) ;

one concludes since det(hv ⌦ vi) 6= 0 by assumption (11).In particular

⇢��@Bi

2 L2([0, T ];H1/2(@Bi))

for each T > 0 and each i = 1, . . . , n.In particular, the first condition in (28) and (10) imply that s 7! f(t, x+ sv, v)

is continuous in s for dtdxdµ-a.e. (t, x, v) 2 R+ ⇥ ⌦ ⇥RN . Therefore, we deducefrom (29) and (30) that, for each l = 1, . . . ,m

(v ·rxf(t, x, v) = 0 , x 2 Bl , v 2 RN , t > 0 ,

f(t, x, v) = ⇢(t, x) , x 2 @Bl , v 2 RN , t > 0 .

Henced

dsf(t, x+ sv, v) = 0 for all s s.t. x+ sv 2 Bl

for dtdxdµ(v)-a.e. (t, x, v) 2 R+ ⇥ Bl ⇥ RN . By assumption (10), one concludesthat

⇢(t, x+ ⌧l(x, v)v) = ⇢(t, x) for d�(x)dµ(v)� a.e. (x, v) 2 @Bl ⇥RN

by solving the boundary value problem above by the method of characteristics. Byassumption (13)

⇢(t, x) =1

|@Bl|Z

@Bl

⇢(t, y)d�(y) =: ⇢l(t) for a.e. x 2 @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)

for dtdxdµ-a.e. (t, x, v) 2 R+ ⇥Bl ⇥RN , for l = 1, . . . ,m.Summarizing, we have proved that

(32)f(t, x, v) = ⇢(t, x) for dtdxdµ� a.e. (t, x, v) 2 ⌦

with ⇢ 2 L1(R+;H) and rx⇢ 2 L2(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) 2 R+ ⇥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 to L1(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 of L2(R+ ⇥A) as ✏! 0, on account of (29).Therefore, for each w 2 V, one has

d

dt

Z

⌦hf✏(t, x, ·)iw(x)dx+

Z

A

1

✏hvf✏(t, x, ·)i ·rw(x)dx = 0 ,

(since rw = 0 on B) and passing to the limit in each side of this identity as ✏! 0shows that

(35)d

dt

Z

⌦⇢(t, x)w(x)dx+

Z

A

rw(x) ·M(x)rx⇢(t, x)dx = 0

in the sense of distributions on R⇤+.

Step 5: limiting entropy production

By definition of q✏, one has

q✏(t, x, v, w) = �q✏(t, x, w, v)

for dtdxdµ(vdµ(w)-a.e. (t, x, v, w) 2 R+⇥A⇥RN ⇥RN and each ✏ > 0; by passingto the limit as ✏! 0

q(t, x, v, w) = �q(t, x, w, v)

for dtdxdµ(vdµ(w)-a.e. (t, x, v, w) 2 R+ ⇥A⇥RN ⇥RN . Defining

ksA(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) 2 R+ ⇥A. LikewiseZZ

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) 2 R+ ⇥ ⌦. With �(v) = ⇠ · b(x, v) for some ⇠ 2 RN to be chosenlater, and applying the Cauchy-Schwarz inequality, one finds that(36) ✓ZZ

RN⇥RN

ksA(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 of ksAZZ

RN⇥RN

ksA(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) 2 R+ ⇥ A where the last equality follows from (17). Passing to thelimit 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) 2 R+ ⇥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. x 2 A, 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) 2 R+ ⇥A. By convexity and weak convergence(38)Z 1

0

Z

A

⌦⌦kA(x, ·, ·)q(t, x, ·, ·)2

↵↵dxdt lim

✏!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

0krx⇢(t, ·)k2L2(A)dt2

Z 1

0

Z

A

rx⇢(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

⇥ supessx2A

ZZ

RN⇥RN

kA(x, v, w)|b(x, v)|2dµ(v)dµ(w)

8C3Kh|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 each w 2 V, 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 each w 2 V(41)

Z

⌦(hf✏(t, x, ·)i � ⇢(t, x))w(x)dx ! 0 uniformly in t 2 [0, T ]

for all T > 0. In particularZ

⌦⇢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 each w 2 V .

Returning to (40), we have proved that @thf✏i is bounded in L2(R+,V 0) for eachT > 0, so that

(43) @t⇢ 2 L2(R+;V 0) .

Step 7: Dirichlet condition

Next we establish the Dirichlet condition on @⌦ for the di↵usion equation. Thescaled linear Boltzmann equation implies that, for each � 2 C1

c (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 in L2(⌦⇥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 thenotation ⌧(x, v) designates the forward exit time from ⌦ starting from x withvelocity v, i.e.

⌧(x, v) := inf{t > 0 s.t. x+ tv 2 @⌦} , x 2 ⌦ , v 2 RN .


In particularZ 1

0�(t)⇢(t, ·)dt

��@⌦

= 0 .

By (32), we already know that the limiting density ⇢ 2 L2([0, T ];H1(⌦)). Therefore

(44) ⇢(t, ·)��@⌦

= 0 in L2([0, T ];H1/2(@⌦))

for each T > 0.

Step 8: convergence to the di↵usion equation

Summarizing, we have proved that

f✏ is relatively compact in L1(R+;L2(⌦⇥RN , dxdµ(v))) weak-*

and that, if f is a limit point of f✏ as ✏! 0, it is of the form

f(t, x, v) = ⇢(t, x) dtdxdµ(v)� a.e. in (t, x, v) 2 R+ ⇥ ⌦⇥RN

where

(45)⇢ 2 L1(R+;H) \ L2(R+;H

10 (⌦)) = L1(R+;H) \ L2(R+;V)

and @t⇢ 2 L2(R+;V 0)

since ⇢ satisfies the Dirichlet boundary condition (44) and rx⇢ 2 L2(R+ ⇥ ⌦) by(32), together with (43). In particular, this implies that

(46) ⇢ 2 Cb(R+;H) .

Besides ⇢ satisfies (35) for each test function w 2 V , together with the initialcondition (42). Therefore ⇢ is the unique solution of the Dirichlet problem for thedi↵usion equation with di↵usion matrixM(x) defined in (18) with infinite di↵usivityin B, with initial data ⇢in. By compactness and uniqueness of the limit point, weconclude that

f✏ ! ⇢ in L1(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 in B is known to satisfy the identity

12

Z

⌦⇢(t, x)2dx+

Z t

0

Z

A

rx⇢(s, x) ·M(x)rx⇢(s, x)dxds =12

Z

⌦⇢in(x)2dx

for each t � 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 in t 2 [0, T ] for each T > 0 by (41).With the entropy identity satisfied by f✏ (see Proposition 2.2) and the inequality

(38), the two inequalities above imply thatZ

⌦hf✏(t, x, ·)2idx !

Z

⌦⇢(t, x)2dx


uniformly in t 2 [0, T ] for each T > 0, so that

f✏(t, ·, ·) ! ⇢(t, ·) strongly in L2(⌦⇥RN , dxdµ)

uniformly in t 2 [0, T ] for each T > 0.By the same token

12

Z t

0

Z

A

⌦⌦kA(x, ·, ·)q✏(s, x, ·, ·)2

↵↵dxds ! 1

2

Z t

0

Z

A

⌦⌦kA(x, ·, ·)q✏(s, x, ·, ·)2

↵↵dxds

=

Z t

0

Z

A

rx⇢(s, x) ·M(x)rx⇢(s, x)dxds .

Therefore

q✏ ! q strongly in L2([0, t]⇥A⇥RN ⇥RN ; kA(x, v, w)dsdxdµ(v)dµ(w)))

as ✏! 0. Besides q satisfies the equality in the Cauchy-Schwarz inequality (36) sothat q 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) 2 R+ ⇥A, which concludes the proof.

Remark. The proof of Theorem 3.5 is inspired from the discussion of the Stokesand of the Navier-Stokes limit of the Boltzmann equation initiated in [1]. Theprocedure in step 9 for obtaining strong convergence is a simplified analogue of theproof 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 step5 is a simplified version of Lemma 4.7 in [1]. Notice also the role of the limitinglinearized Boltzmann equation (29) and of the limiting collision integrand q, thatis reminiscent of the analogous objects considered in [1] (see formula (4.3) andProposition 4.1 there) in the case of the Boltzmann equation of the kinetic theoryof 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 analogywith 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 inseveral directions.

First, our method obviously applies to a scaled linear Boltzmann equation of theform

(✏@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 anamplifying or damping coe�cient, i.e. Bf✏(t, x, v) = �(x)f✏(t, x, v) as in [3]. In otherwords, problems where the collision process is nearly, but not exactly conservativecan be treated as above.

More general boundary conditions than the absorbing condition on @⌦ can alsobe considered. For instance, imposing a specular or di↵use reflection condition at


the boundary, or a convex combination thereof, i.e. assuming that

f✏(t, x, v) = (1�✓(x))f✏(t, x, v�2v·nxnx)+✓(x)

h(w · nx)+iZ

RN

f✏(t, x, w)(w·nx)+dµ(w)

with ✓ 2 C(@⌦) satisfying 0 ✓(x) 1 for all x 2 @⌦ and with a measure µinvariant under all transformations of the form v 7! Qv for Q 2 ON (R) leads tothe same result as in Theorem 3.5, except that the homogeneous Dirichlet conditionon @⌦ should be replaced with the homogeneous Neuman condition.

The methods presented in this paper should also apply to some nonlinear prob-lems, such as the radiative transfer equations. In fact the compactness methodused in the proof of Theorem 3.5 finds its origin in [2].

Comparing Theorem 3.5 with the result in [9] is a more delicate issue. We recallthat the problem considered in [9] involves the juxtaposition of a medium wherethe collision cross-section is of order 1 and a highly collisional medium, where thecollision cross section is of order 1/✏. The setting is one dimensional, but extensionsto higher dimensions are possible and discussed in [9]. The main result in [9] is aproof of the validity of a domain decomposition strategy where the highly collisionalmedium is treated by the di↵usion equation, with a boundary layer term that isthe solution of a Milne problem (see [3]) to accurately describe the interface. Theinterested reader is referred to [9] for a more accurate description of this domaindecomposition algorithm.

At first sight, the situation considered in the present paper is of the same type,as the case of a transition kernel k✏ such that k✏(x, v, w) = O(1) for a.e. x 2 Bis covered by our assumptions. Yet the result in Theorem 3.5 obviously does notinvolve any sophisticated treatment of the interface between A and B that wouldrequire solving a Milne problem. The di↵erence between both results comes fromthe type of boundary data considered in [9] and here. In the situation consideredin Theorem 3.5, the distribution function of particles entering each connected com-ponent of B, i.e. of the region where the collision cross-section is of order 1, isindependent of the variable v. For such boundary data, one easily verifies that theboundary layer matching the kinetic and the di↵usion domain in [9] is trivial toleading order.

Appendix A. Proof of Lemma 2.3

Let �in and ⌃ be the extensions of f in and S respectively by 0 for x /2 ⌦. Let� be the solution of the Cauchy problem

(✏@t�✏ + v ·rx�✏ = ⌃ , x, v 2 RN , t > 0 ,

�✏��t=0

= �in .

Then, for all t 2 [0, T ]

(47) f✏(t, x, v) = �✏(t, x, v) for dxdµ(v) a.e. (x, v) 2 ⌦⇥RN .

Denoting by �✏, �in and ⌃ the partial Fourier transforms of �✏, �in and ⌃✏

respectively, one finds that

✏d

dt

⇣eitv·⇠/✏�✏(t, ⇠, v)

⌘= eitv·⇠/✏⌃(t, ⇠, v)

so that the functiont 7! ✏(t, ⇠, v) := eitv·⇠/✏�✏


belongs to C1([0, T ];L2(RN ⇥RN ; d⇠dµ(v))) and one has

✏ 12d

dt| ✏(t, ⇠, v)|2 = <

⇣eitv·⇠/✏⌃(t, ⇠, v) ✏(t, ⇠, v)

⌘

or equivalently

✏ 12d

dt|�✏(t, ⇠, v)|2 = <

⇣⌃(t, ⇠, v)�✏(t, ⇠, v)

⌘

Integrating in t 2 [0, T ] and ⇠, v 2 RN and applying Plancherel’s theorem showsthat

12

ZZ

⌦⇥RN

�✏(T, x, v)2dxdv 1

2

ZZ

RN⇥RN

�✏(T, x, v)2dxdv

=1

✏

Z T

0

ZZ

RN⇥RN

⌃(t, x, v)�✏(t, x, v)2dxdvdt+ 1

2

ZZ

RN⇥RN

�(0, x, v)2dxdv

=1

✏

Z T

0

ZZ

⌦⇥RN

S(t, x, v)�✏(t, x, v)2dxdvdt+ 1

2

ZZ

⌦⇥RN

f in(x, v)2dxdv

and the result follows from (47).

Appendix B. Auxiliary Lemmas on Evolution Equations

Let V and H be two separable Hilbert spaces such that V ⇢ H with continuousinclusion and V is dense in H. The Hilbert space H is identified with its dual andthe map

H 3 u 7! Lu 2 V 0 ,

where Lu is the linear functional

Lu : V 3 v 7! (u|v)H 2 R ,

identifies H with a dense subspace of V 0.

Lemma B.1. Assume that

v 2 L2(0, T ;V) and

dLv

dt2 L2(0, T ;V 0) .

Then

a) the function v is a.e. equal to a unique element of C([0, T ],H) still denoted v;

b) this function v 2 C([0, T ],H) satisfies

12 |v(t2)|2H � 1

2 |v(t1)|2H =

Z t2

t1

⌧dLv

dt(t), v(t)

�

V0,Vdt

for all t1, t2 2 [0, T ]

Statement a) follows from Proposition 2.1 and Theorem 3.1 in chapter 1 of [13],and statement b) from Theorem II.5.12 of [4].

Lemma B.2. Let L 2 L2(0, T ;V 0) satisfy

hL(t), wiV0,V = 0 for a.e. t 2 [0, T ]

for all w 2 V. Then

L(t) = 0 for a.e. t 2 [0, T ] .


Proof. Pick Nw ⇢ [0, T ] negligible such that L is defined on [0, T ] \ Nw and

hL(t), wiV0,V = 0 for all t 2 [0, T ] \ Nw .

Let D be a dense countable subset of V and let

N :=[

w2DNw .

For all t 2 [0, T ] \ N , one has

hL(t), wiV0,V = 0 for all w 2 D so that L(t) = 0

because L(t) is a continuous linear functional on V and D is dense in V. ⇤

The next lemma recalls the functional background for Green’s formula in thecontext of evolution equations.

Lemma B.3. Let ⌦ be an open subset of RNwith smooth boundary, and let T > 0.

Denote by n the unit outward normal field on @⌦. Let ⇢ 2 C([0, T ];L2(⌦)) and

m 2 L2((0, T )⇥ ⌦,RN ). Assume that

@t⇢+ divx m = 0 in the sense of distributions in (0, T )⇥ ⌦ .

Then

a) the vector field m has a normal trace m · n��(0,T )⇥@⌦

2 H1/200 ((0, T )⇥ @⌦)0;

b) for each 2 H1(⌦)

d

dt

Z

⌦⇢(·, x) (x)dx�

Z

⌦m(·, x) ·rx (x)dx

= �hm · n��@⌦

, ��@⌦

iH�1/2(@⌦),H1/2(@⌦)

in H�1(0, T ).

Proof. Let � 2 C1c (R) be such that

�(t) = 1 for t 2 [�1, T + 1] and supp(�) ⇢ [�2, T + 2] .

Define

⇢(t, x) :=

8><

>:

⇢(t, x) if 0 t T

�(t)⇢(0, x) if t < 0

�(t)⇢(T, x) if t > T

and

m(t, x) :=

(m(t, x) if 0 t T

0 if t /2 [0, T ]

so that the vector field X := (⇢, m) is an extension of (⇢,m) to R⇥ ⌦ satisfying

X 2 L2(R⇥ ⌦;RN+1) .

Besides

(@t⇢+ divx m)(t, x) = �0(t)(1t<0⇢(0, x) + 1t>T ⇢(T, x)) =: S(t, x)

with S 2 L2(R⇥ ⌦) so that

divt,x X = S 2 L2(R⇥ ⌦) .

Therefore X has a normal trace on the boundary @(R ⇥ ⌦) = R ⇥ @⌦, denotedX · n��

R⇥@⌦2 H�1/2(R⇥ @⌦).


Let � 2 H1/200 ((0, T ) ⇥ @⌦); denote by � its extension by 0 to R ⇥ @⌦. Thus

� 2 H1/2(R ⇥ @⌦) and there exists � 2 H1(R ⇥ ⌦) such that � = ��R⇥@⌦

. Thenormal trace of m is then defined as follows: by Green’s formula

hm · n��R⇥@⌦

,�iH

1/200 ((0,T )⇥@⌦)0H1/2

00 ((0,T )⇥@⌦)

:= hX · n��R⇥@⌦

, �iH1/2((0,T )⇥@⌦)0H1/2((0,T )⇥@⌦)

=

ZZ

R⇥⌦(⇢@t�+ m ·rx�+ S�)(t, x)dxdt .

Applying Green’s formula on (0, T )⇥ ⌦ shows that two di↵erent extensions of thevector field (⇢,m) define the same distribution m · n��

(0,T )⇥ @⌦) on (0, T ) ⇥ @⌦.

This completes the proof of statement a).As for statement b), let 2 H1

0 (0, T ) and 2 H1(⌦), define �(t, x) := (t) (x)and let � be the extension of � by 0 to R ⇥ ⌦, so that � 2 H1(R ⇥ ⌦). Thus

� = ��(0,T )⇥@⌦

2 H1/200 ((0, T )⇥ @⌦) and

hhm · n��@⌦

, ��@⌦

iH�1/2(@⌦),H1/2(@⌦),iH�1(0,T ),H10 (0,T )

:= hm · n��R⇥@⌦

,�iH

1/200 ((0,T )⇥@⌦)0H1/2

00 ((0,T )⇥@⌦)

=

ZZ

R⇥⌦(⇢@t�+ m ·rx�+ S�)(t, x)dxdt

=

Z T

0

Z

⌦(⇢(t, x)0(t) (x) +m(t, x) ·r (x)(t))dxdt

= �⌧

d

dt

Z

⌦⇢(t, x) (x)dx,

�

H�1(0,T ),H10 (0,T )

+

Z T

0

Z

⌦m(t, x) ·r (x)(t))dxdt

which is precisely the identity in statement b). ⇤

References

[1] C. Bardos, F. Golse, C.D. Levermore: Fluid Dynamic Limits of Kinetic Equations II: Con-vergence Proofs for the Boltzmann Equation, Comm. on Pure and Appl. Math. 46 (1993),667–753.

[2] C. Bardos, F. Golse, B. Perthame, R. Sentis: The nonaccretive radiative transfer equations:existence of solutions and Rosseland approximation, J. Funct. Anal. 77 (1988), 434–460.

[3] C. Bardos, R. Santos, R. Sentis: Di↵usion approximation and computation of the critical

size, Trans. Amer. Math. Soc. 284 (1984), 617–649.[4] F. Boyer, P. Fabrie: “Elements d’analyse pour l’etude de quelques modeles d’ecoulements de

fluides visqueux incompressibles”. Mathematiques et Applications Vol. 52. Springer Verlag,Berlin, Heidelberg 2006.

[5] H. Brezis: “Analyse Fonctionnelle. Theorie et Applications.” Masson, Paris 1987.[6] M. Cessenat: Theoremes de trace Lp pour des espaces de fonctions de la neutronique, C. R.

Acad. Sci. Paris Ser. I Math. 299 (1984), 831–834.[7] R. Dautray, J.-L. Lions: “Analyse mathematique et calcul numerique pour les sciences et les

techniques”, Masson, Paris 1985.[8] L. Desvillettes, F. Golse, V. Ricci: Homogenization of the heat equation with infinitely con-

ducting inclusions, manuscript.[9] F. Golse, S. Jin, C.D. Levermore: A domain decomposition analysis for a two-scale linear

transport problem, ESAIM Math. Model. Numer. Anal. 37 (2003), 869–892.


[10] L. Hormander: “The Analysis of Linear Partial Di↵erential Operators III”. Springer-Verlag,Berlin, Heidelberg, 1985.

[11] L. Landau, E. Lifshitz: “Physical Kinetics”. Pergamon Press, Oxford, 1981.[12] J.-L. Lions: “Problmes aux limites et equations aux derivees partielles”. Seminaire de

Mathematiques Superieures, no. 1 (ete 1962), Les Presses de l’Universite de Montreal,Montreal,1965.

[13] J.-L. Lions, E. Magenes: “Problemes aux limites non homogenes et applications. Volume 1”.Dunod, Paris 1968.

[14] D. Stroock, S.R.S. Varadhan: “Multidimensional Di↵usion Processes”. Springer-Verlag,Berlin, Heidelberg, 2006.

.

(C.B.) Universit

´

e Paris-Diderot, Laboratoire J.-L. Lions, BP187, 4 place Jussieu,

75252 Paris Cedex 05 France

E-mail address: [email protected]

(E.B.) Ecole Polytechnique, Centre de Math

´

ematiques L. Schwartz, 91128 Palaiseau

Cedex, France & Universit

´

e Paris-Diderot, Laboratoire J.-L. Lions, BP187, 4 place

Jussieu, 75252 Paris Cedex 05 France


(F.G.) Ecole Polytechnique, Centre de Math

´

ematiques L. Schwartz, 91128 Palaiseau

Cedex, France & Universit

´

e Paris-Diderot, Laboratoire J.-L. Lions, BP187, 4 place

Jussieu, 75252 Paris Cedex 05 France


(R.S.) CEA-DIF Bruy

`

eres-le-Ch

ˆ

atel & Universit

´

e Pierre-et-Marie Curie, Labora-

toire J.-L. Lions, BP187, 4 place Jussieu, 75252 Paris Cedex 05 France


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