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Unfolding Method for Diffusion Process in a Rarefied Binary Structure
Georges Griso, Louiza Merzougui
To cite this version:
Georges Griso, Louiza Merzougui. Unfolding Method for Diffusion Process in a Rarefied Binary
Structure. Applicable Analysis, Taylor & Francis, 2017. �hal-01449413�
Unfolding Method for Diffusion Process in a Rarefied Binary Structure.
Georges Griso
a, Louiza Merzougui
ba
Laboratoire J.-L. Lions–CNRS, Boˆıte courrier 187, Universit´e Pierre et Marie Curie, 4 place Jussieu, 75005 Paris, France, Email: griso@ljll.math.upmc.fr
b
Laboratory of EDP’s and their applications, Faculty of mathematics and computer science, University of Batna2, Algeria, Email: merzougui louiza@yahoo.fr
Abstract
The aim of this paper is to study the homogenization of a diffusion process which takes place in a binary structure made by an ambient connected phase surrounding the suspensions (very small particles of diameter of order εδ) distributed in an ε-periodic network. Using the periodic unfolding method introduced in [4], in the critical case, when ε and δ go to 0 we determine the asymptotic behavior of the solution of an evolution problem.
KEY WORDS: evolution problem, homogenization, strange term, binary structure Mathematics Subject Classification (2010): 35A01, 74Q10, 74Q15, 76M50.
1 Introduction
This article is devoted to describing the asymptotic behavior of an evolution problem which governs the diffusion process of a rarefied binary structure. The problem is given by (4.1) with assumptions (4.2)-(4.3) and (5.7). In this problem, the first constant a
ε,δrepre- sents the mass density of the suspensions while the second one b
ε,δis their diffusivity. The parameter δ is the size order of the suspensions in the reference cell Y and ε is the periodicity length. The critical case lim
(ε,δ)→(0,0)δ
N−2/ε
2∈ (0, +∞) is considered.
This problem was first treated in a paper of Bentalha and al. (see [6]) in 2006 . In this ar-
ticle, the authors investigate the case of a diffusivity b
ε,δgoing to infinity, the suspensions
being spheres. They use the control zone method. One year later, Gruais (see [7]) study the
stationary problem making the assumption lim
(ε,δ)→(0,0)b
ε,δ= lim
(ε,δ)→(0,0)δ
Na
ε,δ= 1. She
use a multiple scale method. The same year, using again the control zone method Gruais and
Poliˇ sevski find the limit problem in the case lim
(ε,δ)→(0,0)b
ε,δ< +∞ (see [8]) .
In the present paper, we revisit the evolution problem (4.1) with the periodic unfolding method introduced in [4], but we focus our attention in the most interesting case: namely the critical case with lim
(ε,δ)→(0,0)b
ε,δ∈ (0, +∞) and lim
(ε,δ)→(0,0)δ
Na
ε,δ∈ (0, +∞) (see assumptions (5.7)).
The periodic unfolding method is a general tool to study periodic homogenization. Nowa- days, this method is used in a lot of new papers since it is a simple way to replace the two-scale convergence by a convergence in a fixed domain: via T
εin Ω × Y (where Y is the reference periodic cell) or via T
ε,δin Ω × R
N.
The paper is organized as follows. Section 2 contains the main notations regarding the structure. In Section 3, we recall the definitions of some unfolding operators (T
ε, T
ε,δand M
ε) and we also recall the main results concerning T
ε,δ(Lemmas 3.1 and 3.2).
In Section 4, we present the evolution problem in its variational formulation (4.4). Then, the total energy estimate (4.5) allows to introduce the assumptions on the data (4.6)-(4.7)-(5.8).
In Section 5, transforming the solution of the evolution problem with the operator T
ε,δleads to the assumptions on the two small parameters ε and δ and on a
ε,δ, b
ε,δ(see (5.7)).
Theorem (6.1) in Section 6 gives the unfolded limit problem. Here, it is worth to note that for the first time the limit evolution problem in its variational form is first posed in Ω × R
N.
In the last section, using the Laplace transform we give the homogenized limit problem (Theorem 7.2). The limit equation differs from the initial one (4.1) regarding the convolution term, this new term accounting for the memory effects is here in fact a ”strange term” (see [2]).
2 Notations
Let Ω and B be two bounded domains in R
Nwith Lipschitz boundary.
Denote
• Y = (−1/2, 1/2)
Nthe reference cell,
• Ω
T= (0, T ) × Ω,
• Y
δ= Y \ δB,
• Ξ
ε= n
ξ ∈ ZZ
N|ε(ξ + Y) ⊂ Ω o ,
• Ω b
ε= interior n S
ξ∈Ξε
ε(ξ + Y ) o
, Λ
ε= Ω \ Ω b
ε,
• B
ε,δ= Ω ∩ S
ξ∈Ξε
ε(ξ + δB),
• Ω
∗ε,δ= Ω \ B
ε,δ,
• 1
Ω∗ε,δis the characteristic function of the set Ω
∗ε,δ.
Figure 1. Integer and fractional part of z.
For a.e z ∈ R
N, one has
z = [z] + {z}
where [z] ∈ ZZ
Nis the integer part of z while {z} ∈ Y denotes the fractional part of z.
Ωԑ
ʌ
ԑFigure 2. The sets Ω, Ω b
εand Λ
ε.
3 Some recalls on the unfolding operators T ε , M ε and T ε,δ
• For φ measurable function on Ω, the unfolding operator T
εis defined by T
(φ)(x, y) =
(
φ(ε h x ε i
+ εy) (x, y) ∈ Ω b
ε× Y,
0 (x, y) ∈ Λ
ε× Y. (3.1)
The properties of T
εare given in [4].
• The local average operator M
ε: L
1(Ω) −→ L
1(Ω) is defined as follows:
∀φ ∈ L
1(Ω), M
ε(φ)(x) = Z
Y
T
ε(φ)(x, y )dy, for a.e. x ∈ Ω.
For any sequence {φ
ε}
εsuch that φ
ε* φ weakly in L
2(Ω), we have
M
ε(φ
ε) * φ weakly in L
2(Ω). (3.2)
• For φ measurable function on Ω, the unfolding operator T
ε,δis defined by T
,δ(φ)(x, z) =
(
T
ε(φ)(x, δz) if (x, z) ∈ Ω b
ε× 1 δ Y,
0 otherwise.
(3.3)
Ω
*𝜀, 𝛿Y
𝐵 𝛿𝐵
Figure 3. The sets Ω
∗ε,δand B.
Below, we recall the main properties of the operator T
ε,δ(see Theorem 2.11 of [5]).
Lemma 3.1. The operator T
ε,δis linear and continuous from L
p(Ω) into L
p(Ω × R
N), p ∈ [1, +∞].
1. For any measurable functions ψ, φ one has T
ε,δ(ψφ) = T
ε,δ(ψ )T
ε,δ(φ).
2. For any φ in L
1(Ω) δ
NZ
Ω×RN
T
ε,δ(φ)(x, z)dxdz = Z
Ω×Y
T
ε(φ)(x, y)dxdy = Z
Ωbε
φdx,
Z
Ω
φ(x) dx − δ
NZ
Ω×RN
T
ε,δ(φ)(x, z)dxdz ≤
Z
Λε
|φ| dx.
3. For any φ ∈ L
2(Ω)
kT
ε,δ(φ)k
2L2(Ω×RN)≤ 1
δ
Nkφk
2L2(Ω), kT
ε,δ(φ)k
2L2(Ω×B)≤ 1
δ
Nkφk
2L2(Bε,δ). (3.4)
4. For any φ in H
1(Ω)
T
ε,δ(∇φ) = 1
εδ ∇
zT
ε,δ(φ)
in Ω × 1
δ Y (3.5)
and
k∇
z(T
ε,δ(φ))k
2L2(Ω×RN)≤ ε
2δ
N−2k∇φk
2L2(Ω), k∇
z(T
ε,δ(φ))k
2L2(Ω×B)≤ ε
2δ
N−2k∇φk
2L2(Bε,δ).
(3.6)
5. Denote 2
∗.
= 2N
N − 2 , N ≥ 3, the Sobolev exponent associated to 2. For any φ in H
1(Ω) kT
ε,δ(φ − M
ε(φ))k
2L2(Ω;L2∗(RN))≤ C ε
2δ
N−2k∇φk
2L2(Ω). (3.7) Lemma 3.2. For any w ∈ C(Ω)
T
ε,δ(w) −→ w strongly in L
2(Ω; L
2loc( R
N)). (3.8) Proof. Let O be a bounded open subset of R
N. There exists R > 0 suth that O ⊂ B(O, R) where B (O, R) is the ball of radius R and center O. For δ small enough O ⊂ B(0, R) ⊂ 1
δ Y . Hence
Z
Ω×O
|T
ε,δ(w)(x, z) − w(x)|
2dxdz
= Z
Λε×O
|w(x)|
2dxdz + Z
Ωbε×O
w
ε h x ε i
+ εδz
− w(x)
2
dxdz
≤ |O|
Z
Λε
|w(x)|
2dx + |Ω × O| sup
(x,z)∈Ω×O
w
ε h x ε i
+ εδz
− w(x)
2
. Let $ be the modulus of continuity of w
Z
Ω×O
|T
ε,δ(w)(x, z) − w(x)|
2dxdz ≤ C kwk
2L∞(Ω)|Λ
ε| + $(ε + εδR)
2. Passing to the limit gives (3.8).
4 The evolution problem
We consider the evolution problem which governs the diffusion process in our binary mixture
Find u
ε,δsuth that ρ
ε,δdu
ε,δdt − div
xk
ε,δ∇
xu
ε,δ= f
ε,δin Ω
T,
u
ε,δ= 0 on ∂Ω × (0, T ),
u
ε,δ(0) = u
0ε,δon Ω × {0},
(4.1)
where f
ε,δ∈ L
2(Ω
T) and ρ
ε,δ=
( 1 in Ω
∗ε,δ,
a
ε,δin B
ε,δ, k
ε,δ=
( 1 in Ω
∗ε,δ,
b
ε,δin B
ε,δ. (4.2) Here a
ε,δand b
ε,δare strictly positive constants bounded from below independently of ε and δ. Hence, there exists c
0> 0 such that for any (ε, δ)
Assumption A
1.
a
ε,δ≥ c
0, b
ε,δ≥ c
0. (4.3)
The first constant a
ε,δrepresents the density of mass of the suspensions while the second one is the diffusivity of suspensions.
Lemma 4.1. Problem (4.1) has the following variational formulation:
Find u
ε,δ∈ L
2(0, T ; H
01(Ω)) ∩ L
∞(0, T ; L
2(Ω)), ρ
ε,δu
ε,δ∈ H
1(0, T ; H
−1(Ω)),
< ρ
ε,δdu
ε,δdt (t), w >
H−1(Ω),H10(Ω)
+ Z
Ω
k
ε,δ(x)∇
xu
ε,δ(t, x)∇
xw(x)dx
= Z
Ω
f
ε,δ(t, x)w(x)dx for a.e. t ∈ (0, T ), ∀w ∈ H
01(Ω), u
ε,δ(0) = u
0ε,δ.
(4.4)
Furthermore u
ε,δsatisfies 1
2 Z
Ω
ρ
ε,δ(x)|u
ε,δ(t, x)|
2dx + Z
t0
Z
Ω
k
ε,δ(x)|∇
xu
ε,δ(s, x)|
2dsdx
= Z
t0
Z
Ω
f
ε,δ(s, x)u
ε,δ(s, x)dsdx + 1 2
Z
Ω
ρ
ε,δ(x)|u
0ε,δ(x)|
2dx
for all t ∈ [0, T ]. (4.5)
Proof. The existence and uniqueness of the solution of Problem (4.4) is given by the Faedo- Galerkin’s method (see e.g. [3] or [9]) which also gives the equality (4.5).
Assumption A
2. We assume that there exists a constant C
0, which does not depend on ε and δ, such that
1Z
Ω
ρ
ε,δ|u
0ε,δ(x)|
2dx + Z
ΩT
1
ρ
ε,δ|f
ε,δ(t, x)|
2dtdx ≤ C
0. (4.6) As a consequence of the above assumptions (4.3)
1and (4.6) the sequences {u
0ε,δ}, {f
ε,δ1
Ω∗ε,δ
} are bounded respectively in L
2(Ω), L
2(Ω
T).
Assumption A
3. We assume that there exist u
0∈ L
2(Ω) and f ∈ L
2(Ω
T) such that u
0ε,δ* u
0weakly in L
2(Ω),
f
ε,δ1
Ω∗ε,δ
* f weakly in L
2(Ω
T). (4.7)
1
We will see in Lemma 4.2 the reason of this assumption.
Lemma 4.2. There exists a positive constant C independent of ε and δ such that Z
Ω
ρ
ε,δ(x)|u
ε,δ(t, x)|
2dx ≤ C, ∀t ∈ [0, T ], Z
ΩT
k
ε,δ(x)|∇
xu
ε,δ(s, x)|
2dsdx ≤ C.
(4.8)
Moreover
ku
ε,δk
L∞(0,T;L2(Ω))+ ku
ε,δk
L2(0,T;H10(Ω))
≤ C. (4.9)
Proof. According to the Cauchy-Schwarz inequality we have Z
t0
Z
Ω
f
ε,δu
ε,δdsdx ≤ Z
t0
√ 1
ρ
ε,δf
ε,δL2(Ω)
k √
ρ
ε,δu
ε,δk
L2(Ω)ds
≤ 1 2
√ 1
ρ
ε,δf
ε,δ2
L2(ΩT)
+ 1 2
Z
t 0k √
ρ
ε,δu
ε,δk
2L2(Ω)ds.
(4.10)
The above inequality and (4.5) yield k √
ρ
ε,δu
ε,δ(t)k
2L2(Ω)≤ k √
ρ
ε,δu
0ε,δk
2L2(Ω)+
√ 1
ρ
ε,δf
ε,δ2 L2(ΩT)
+ Z
t0
k √
ρ
ε,δu
ε,δ(s)k
2L2(Ω)ds.
Then, the Gronwall’s lemma and (4.6) give k √
ρ
ε,δu
ε,δ(t)k
2L2(Ω)+ Z
t0
k √
ρ
ε,δu
ε,δ(s)k
2L2(Ω)ds
≤ e
Tk √
ρ
ε,δu
0ε,δk
2L2(Ω)+
√ 1
ρ
ε,δf
ε,δ2 L2(ΩT)
≤ C
0e
T(4.11)
which in turn leads to (4.8)
1. Again with (4.6) and (4.10)-(4.11) we get (4.8)
2. Then, the assumptions (4.3) yield
min{1, c
0} Z
Ω
|u
ε,δ(t, x)|
2dx ≤ C ∀t ∈ [0, T ], min{1, c
0}
Z
ΩT
|∇
xu
ε,δ(s, x)|
2dsdx ≤ C.
That gives (4.9).
As a consequence of the above lemma, one has
Corollary 4.3. There exist a subsequence of (ε, δ), still denoted (ε, δ), and u ∈ L
∞(0, T ; L
2(Ω))∩
L
2(0, T ; H
01(Ω)) such that
u
ε,δ* u weakly * in L
∞(0, T ; L
2(Ω)),
u
ε,δ* u weakly in L
2(0, T ; H
01(Ω)). (4.12)
5 Further estimates and convergences
Lemma 5.1. There exists a positive constant C independent of ε and δ, such that kT
ε,δ(u
ε,δ− M
ε(u
ε,δ))k
2L2(ΩT;L2∗(RN)≤ C ε
2δ
N−2, k∇
z(T
ε,δ(u
ε,δ))k
2L2(ΩT×RN)≤ C ε
2δ
N−2(5.1)
and
kT
ε,δ(u
ε,δ)k
2L2(ΩT×B)≤ C
δ
Na
ε,δ, k∇
z(T
ε,δ(u
ε,δ))k
2L2(ΩT×B)≤ C ε
2δ
N−2b
ε,δ. (5.2) Proof. From (3.4) and (4.8) we immediately deduce (5.1). From (3.4)
2we have
kT
ε,δ(u
ε,δ)k
2L2(ΩT×B)≤ 1
δ
Nku
ε,δk
2L2(Bε,δ)(5.3) and from (4.8)
1Z
Ω∗ε,δ
|u
ε,δ(t, x)|
2dx + Z
Bε,δ
a
ε,δ|u
ε,δ(t, x)|
2dx ≤ C. (5.4) Hence
Z
Bε,δ
|u
ε,δ(t, x)|
2dx ≤ C
a
ε,δ, (5.5)
We substitute (5.5) in (5.3) that gives (5.2)
1. Then (3.6)
2yields k∇
z(T
ε,δ(u
ε,δ))k
2L2(ΩT×B)≤ ε
2δ
N−2k∇
xu
ε,δk
L2(BTε,δ). (5.6) Estimate (4.8)
2leads to
Z
Bε,δ×(0,T)
|∇u
ε,δ(t, x)|
2dtdx ≤ C b
ε,δ. The above inequality and (5.6) give (5.2)
2.
So, the above estimates (5.2) highlight several cases. We will focus our attention on the most interesting one.
Assumption A
4. We assume that
(ε,δ)→(0,0)
lim δ
N−2ε
2= γ
0∈ (0, +∞), lim
(ε,δ)→(0,0)
δ
Na
ε,δ= a
0∈ (0, +∞),
(ε,δ)→(0,0)
lim b
ε,δ= b
0∈ (0, +∞).
(5.7)
Now, as a consequence of Assumptions A
1, A
2and the estimate (3.4)
2, the sequences {T
ε,δ(u
0ε,δ)},
{δ
NT
ε,δ(f
ε,δ)} are bounded respectively in L
2(Ω × B), L
2(Ω
T× B).
Assumption A
5. We assume that there exist U
0∈ L
2(Ω × B) and F ∈ L
2(Ω
T× B) such that T
ε,δ(u
0ε,δ) * U
0weakly in L
2(Ω × B),
δ
NT
ε,δ(f
ε,δ) * F weakly in L
2(Ω
T× B ). (5.8) Lemma 5.2. There exist a subsequence of (ε, δ), still denoted (ε, δ), and U ∈ L
2(Ω
T; H
loc1( R
N)) with ∇
zU ∈ L
2(Ω
T× R
N) such that
T
ε,δ(u
ε,δ) * U weakly in L
2(Ω
T; H
loc1( R
N)),
∇
zT
ε,δ(u
ε,δ)1
1/δY* ∇
zU weakly in L
2(Ω
T× R
N). (5.9) Furthermore, there also exists V ∈ L
2(Ω
T, L
2∗( R
N)) with ∇
zV ∈ L
2(Ω
T× R
N) such that
T
ε,δ(u
ε,δ− M
ε(u
ε,δ))1
1/δY* V weakly in L
2(Ω
T; L
2∗( R
N)),
∇
zT
ε,δ(u
ε,δ)1
1/δY* ∇
zV weakly in L
2(Ω
T× R
N) (5.10) and we have U = V + u.
Proof. From the estimates (5.1)-(5.2), there exist subsequences of (ε, δ) such that the conver- gences (5.9) and (5.10) hold. Convergences (4.12)
2and (3.2) yield
M
ε(u
ε,δ) * u weakly in L
2(Ω
T),
thus M
ε(u
ε,δ) * u weakly in L
2(Ω
T; L
2loc∗( R
N)). (5.11) Then, (5.9)
1, (5.10)
1and (5.11) lead to V = U − u.
Set
H =
Φ ∈ H
loc1( R
N) | ∇
zΦ ∈ [L
2( R
N)]
N= .
H
1( R
N) ⊕ R
(2), L =
Φ ∈ L
2(Ω; H) | Φ(·, ∞) ∈ H
01(Ω) . (5.12) For any Φ ∈ H, we know that there exists a constant denoted Φ(∞) such that
kΦ − Φ(∞)k
L2∗(RN)
≤ Ck∇Φk
L2(RN). (5.13) The constant C does not depend on Φ.
Observe that U ∈ L
2(0, T ; L) and U (t, x, ∞) = u(t, x) for a.e. (t, x) ∈ Ω
T.
2
Recall that the space .
H
1( R
N) is the completion of D( R
N) for the norm k∇φk
L2(RN). The Sobolev
imbedding theorem implies that for N ≥ 3, it is a subspace of L
2∗( R
N), where 2
∗is the Sobolev exponent
associated to 2. Therefore, all its elements admit 0 as limit at ∞ of R
N(in the weak sense of L
2∗( R
N)).
6 The unfolded limit problem
Denote
k
0(z) =
( 1 if z ∈ R
N\ B, ¯
b
0if z ∈ B. (6.1)
Theorem 6.1. Let u
ε,δbe the solution of the evolution Problem (4.5). For the whole sequence (ε, δ) we have the convergences (4.12), (5.9) and (5.10).
The limit function U is the solution of the unfolded evolution problem U ∈ L
2(0, T ; L), U(·, ·, ∞) ∈ H
1(0, T ; H
−1(Ω)),
U ∈ H
1(0, T ; L
2(Ω; (H
1(B))
0)),
< dU
dt (t, ·, ∞), Φ(·, ∞) >
H−1(Ω),H10(Ω)
+a
0< dU
dt (t), Φ >
L2(Ω;(H1(B))0),L2(Ω;H1(B))+
Z
Ω
∇
xU (t, ·, ∞) ∇
xΦ(·, ∞) dx + γ
0Z
Ω×RN
k
0∇
zU(t) ∇
zΦ dxdz
= Z
Ω
f (t)Φ(·, ∞) dx + Z
Ω×B
F (t)Φ dxdz, for a.e. t ∈ (0, T ), ∀Φ ∈ L, U (0, x, ∞) = u
0(x), U (0, x, z) = U
0(x, z) for a.e. (x, z) ∈ Ω × B.
(6.2)
Furthermore U satisfies 1
2 Z
Ω
|U (t, ·, ∞)|
2dx + a
02
Z
Ω×B
|U (t)|
2dxdz +
Z
t 0Z
Ω
|∇
xU (s, ·, ∞)|
2dsdx + γ
0Z
t0
Z
Ω×RN
k
0|∇
zU|
2dsdxdz
= Z
t0
Z
Ω
f(s)U (s, ·, ∞)dsdx + Z
t0
Z
Ω×B
F (t)U(t) dxdz + 1
2 Z
Ω
|u
0|
2dx + a
02
Z
Ω×B
|U
0|
2dxdz.
for all t ∈ [0, T ]. (6.3)
Before proving the unfolded limit problem, we introduce and prove three lemmas. The first one concerns a density result, the second one gives a convergence result for functions vanishing in B
ε,δand the last one introduces a test function.
Since the open set B is bounded, there exists η
0> 0 such that 2η
0B ⊂ Y .
Lemma 6.2. The set
[
η∈(0,η0]
n
ψ ∈ H
per1(Y ) | ψ
|ηB= 0 o
is dense in H
per1(Y ).
Proof. We prove the lemma in two steps. In the first one we choose an element in C
per∞(Y ) and in the second one we investigate the general case.
Step 1. We fix a function χ in C
∞( R
N) satisfying χ = 0 on B,
χ = 1 on R
N\ 2B,
χ(x) ∈ [0, 1] for all x ∈ R
N.
(6.4)
For η ∈ (0, η
0], set
χ
η= χ · η
.
For any φ ∈ C
per∞(Y ), denote φ
η= χ
ηφ. Since χ
η= 1 on ∂Y , the function φ
ηbelongs to H
per1(Y ). We have φ
η= φ on Y \ 2B and ∇φ
η= ∇φχ
η+ ∇χ
ηφ. Hence
kφ
η− φk
2L2(Y)+ k∇φ
η− ∇φk
2L2(Y)≤ Z
2ηB
|φ|
2dy + 2 Z
2ηB
|∇φ|
2dy + 2 Z
2ηB
|∇χ
η|
2|φ|
2dy.
One has
Z
2ηB
|∇χ
η|
2|φ|
2dy ≤ Cη
N−2k∇χk
[L∞(RN)]Nkφk
2L∞(Y).
Since φ belongs to C
per∞(Y ) and the measure of ηB goes to zero, one obtains the strong con- vergence of φ
ηto φ in H
per1(Y ).
Step 2. Let ψ be in H
per1(Y ). The space C
per∞(Y ) is dense in H
per1(Y ). Hence for any ε > 0 there exists ϕ ∈ C
per∞(Y ) such that
kψ − ϕk
H1(Y)≤ ε. (6.5)
We fix a function ϕ satisfying (6.5). From Step 1, there exists η
1> 0 such that
∀η ∈ (0, η
1] kϕ − ϕ
ηk
H1(Y)< ε. (6.6) As a consequence of (6.5) and (6.6) we get
∀η ∈ (0, η
1] kψ − ϕ
ηk
H1(Y)≤ 2ε.
The lemma is proved.
Lemma 6.3. Suppose p ∈ [1, +∞). For any w ∈ L
p(Ω) w1
Ω∗ε,δ
−→ w strongly in L
p(Ω). (6.7)
Proof. Step 1. We prove (6.7) for w ∈ L
∞(Ω).
Take w in L
∞(Ω) Z
Ω
|(w1
Ω∗ε,δ
− w)(x)|
pdx = Z
Bε,δ
|w(x)|
pdx
≤ |B
ε,δ|kwk
pL∞(Ω)≤ Cδ
Nkwk
pL∞(Ω). Passing to the limit gives (6.7) for any w ∈ L
∞(Ω).
Step 2. We prove (6.7) for w ∈ L
p(Ω).
Let w be in L
p(Ω). Since C(Ω) is dense in L
p(Ω), p ∈ [1, ∞), for any ε
1> 0 there exists W ∈ C (Ω) suth that
kw − W k
Lp(Ω)≤ ε
1. (6.8)
We fix W satisfying (6.8). Then kW 1
Ω∗ε,δ
− W k
Lp(Ω)≤ |B
ε,δ|kW k
L∞(Ω)≤ Cδ
N|kW k
L∞(Ω). Hence, there exists δ
1> 0 such that
∀δ ∈ (0, δ
1], kW 1
Ω∗ε,δ
− W k
Lp(Ω)≤ ε
1. So
kw1
Ω∗ε,δ
− wk
Lp(Ω)≤ kw1
Ω∗ε,δ
− W 1
Ω∗ε,δ
k
Lp(Ω)+ kW 1
Ω∗ε,δ
− W k
Lp(Ω)+ kW − wk
Lp(Ω)≤ kW 1
Ω∗ε,δ
− W k
Lp(Ω)+ 2kW − wk
Lp(Ω)≤ 3ε
1. As a result, the strong convergence (6.7) is proved.
Lemma 6.4. Let ψ be in H ∩ C
∞( R
N) such that the support of ∇
zψ is bounded. Set ψ
ε,δ(x) = ψ 1
δ n x
ε o
if x ∈ Ω b
ε, ψ
ε,δ(x) = ψ(∞) if x ∈ Λ
ε.
(6.9)
The function ψ
ε,δbelongs to C
∞(Ω) and
ψ
ε,δ−→ ψ(∞) strongly in L
2(Ω). (6.10)
Proof. There exists R > 0 such that the support of ∇
zψ is included in the ball B (O; R). For δ small enough B(O; δR) ⊂ Y . We perform a change of variables and we use the fact that ψ − ψ(∞) belongs to H
01(B(O; R)). That gives
Z
εξ+εY
|ψ
ε,δ(x) − ψ(∞)|
2dx = Z
εξ+εY
ψ 1
δ n x
ε
o − ψ(∞)
2
dx
= (εδ)
NZ
B(O;R)
|ψ(z) − ψ(∞)|
2dz
≤ C(εδ)
NR
2Z
B(O;R)
|∇
zψ(z)|
2dz.
The constant does not depend on ε and δ. Summing over Ξ
εyields Z
Ω
|ψ
ε,δ(x) − ψ(∞)|
2dx = X
ξ∈Ξε
Z
εξ+εY
|ψ
ε,δ(x) − ψ(∞)|
2dx ≤ Cδ
Nk∇
zψk
2L2(RN).
Hence, we get the strong convergence (6.10).
Proof of Theorem 6.1. First observe that Problem (6.2) has a unique solution U belonging to L
2(0, T ; L). It is a consequence of the Faedo-Galerkin’s method which also gives the equality (6.3). This will imply the convergence of the full sequences in (4.12), (5.9) and (5.10). It is therefore enough to obtain the limit Problem (6.2) for a subsequence, which we do below.
The aim of the three first steps is to obtain the variational formulation (6.20) of the limit problem. Then, in the last step we prove that U (·, ·, ∞) ∈ H
1(0, T ; H
−1(Ω)) and U ∈ H
1(0, T ; L
2(Ω; (H
1(B ))
0)).
Step 1. Let (φ, w) be in C
1([0, T ]) × D(Ω) such that φ(T ) = 0. Taking φ(t)w(x) as test function in (4.4) and then integrating by parts lead to
− Z
(0,T)×Ω∗ε,δ
u
ε,δφ
0w dtdx − Z
(0,T)×Bε,δ
a
ε,δu
ε,δφ
0w dtdx +
Z
(0,T)×Ω∗ε,δ
φ ∇u
ε,δ∇w dtdx + Z
(0,T)×Bε,δ
b
ε,δφ ∇u
ε,δ∇w dtdx
= Z
ΩT
f
ε,δφ w dtdx + Z
Ω
ρ
ε,δu
0ε,δφ(0)w dx.
(6.11)
Due to the convergences (4.12) and (6.7) we obtain Z
(0,T)×Ω∗ε,δ
u
ε,δφ
0w dtdx = Z
ΩT
u
ε,δφ
0w1
Ω∗ε,δdtdx −→
Z
ΩT
u φ
0w dtdx, Z
(0,T)×Ω∗ε,δ
φ∇
xu
ε,δ∇
xw dtdx = Z
ΩT
φ∇
xu
ε,δ∇
xw1
Ω∗ε,δdtdx −→
Z
ΩT
φ∇
xu∇w dtdx.
Now, unfolding the integral over (0, T ) × B
ε,δand using Lemma 3.2 and convergences (5.9) yield
Z
(0,T)×Bε,δ
a
ε,δu
ε,δφ
0w dtdx = Z
(0,T)×Ω×B
δ
Na
ε,δT
ε,δ(u
ε,δ) φ
0T
ε,δ(w) dtdxdz
−→
Z
(0,T)×Ω×B
a
0U φ
0w dtdxdz, Z
(0,T)×Bε,δ
b
ε,δφ ∇u
ε,δ∇w dtdx = Z
ΩT×B
εδ δ
N−2ε
2b
ε,δφ ∇
zT
ε,δ(u
ε,δ) T
ε,δ(∇w) dtdxdz −→ 0.
The convergences (3.8), (5.8), (4.7) and (6.7), lead to Z
Ω
ρ
ε,δu
0ε,δφ(0)w dx = Z
Ω
u
0ε,δφ(0) w 1
Ω∗ε,δdx + Z
Ω×B
δ
Na
ε,δT
ε,δ(u
0ε,δ) φ(0)T
ε,δ(w) dxdz
−→
Z
Ω
u
0φ(0)w dx + Z
Ω×B
a
0U
0φ(0)w dxdz, Z
ΩT
f
ε,δφ w dtdx = Z
ΩT
f
ε,δφ w 1
Ω∗ε,δdtdx + Z
ΩT×B
δ
NT
ε,δ(f
ε,δ)φ T
ε,δ(w) dtdxdz
−→
Z
ΩT
f φ w dtdx + Z
ΩT×B
F φ w dtdxdz.
Finally, summarizing the above limits give
− a
0Z
ΩT×B
U (t, x, z)φ
0(t)w(x)dtdxdz
− Z
ΩT
u(t, x)φ
0(t)w(x)dtdx + Z
ΩT
φ(t)∇
xu(t, x)∇w(x)dtdxdy
= Z
ΩT
f(t, x)φ(t)w(x)dtdx + Z
ΩT×B
F (t, x, z)φ(t)w(x)dtdxdz + φ(0)
Z
Ω
u
0(x) + a
0Z
B
U
0(x, z)dz
w(x)dx.
(6.12)
Step 2. Now, let (φ, w) be in C
1([0, T ]) × D(Ω) such that φ(T ) = 0 and let ψ ∈ H ∩ C
∞( R
N) such that the support of ∇
zψ is bounded. In (4.4) we take as test function φ(t)w(x)ψ
ε,δ(x), where ψ
ε,δis given by (6.9).
If δ is small enough the function ψ
ε,δis constant (and equal to ψ(∞)) on the boundary of the cells εξ +εY (ξ ∈ Ξ
ε). As a consequence, the function wψ
ε,δis an admissible test function belonging to H
01(Ω). Hence
− Z
ΩT
ρ
ε,δu
ε,δφ
0w ψ
ε,δdtdx + Z
ΩT
k
ε,δφ ∇u
ε,δ∇(w ψ
ε,δ) dtdx
= Z
ΩT
f
ε,δ(t, x)φ(t)w(x)ψ
ε,δ(x)dtdx + Z
Ω
ρ
ε,δu
0ε,δ(x)φ(0)w(x)ψ
ε,δ(x)dx.
(6.13)
Since ∇
zψ has a compact support in R
N, from the equality (3.5) and Lemma 3.2 we obtain εδT
ε,δ(w∇ψ
ε,δ) = T
ε,δ(w)∇
zψ −→ w∇
zψ strongly in L
2(Ω × R
N). (6.14) Now, due to the convergences (4.12)
2, (5.9)
1, (6.14) and (6.10) we can write
Z
ΩT
ρ
ε,δu
ε,δφ
0wψ
ε,δdtdx
= Z
ΩT
u
ε,δφ
0wψ
ε,δ1
Ω∗ε,δdtdx + Z
ΩT×B
δ
Na
ε,δT
ε,δ(u
ε,δ)φ
0T
ε,δ(w)ψdtdxdz
−→
Z
ΩT
uφ
0wψ(∞)dtdx + a
0Z
ΩT×B
U φ
0wψdtdxdz.
(6.15)
We have
Z
ΩT
k
ε,δφ∇
xu
ε,δ∇(wψ
ε,δ)dtdx
= Z
ΩT
k
ε,δφ ∇
xu
ε,δ∇w ψ
ε,δdtdx + Z
ΩT
k
ε,δφ ∇
xu
ε,δw ∇ψ
ε,δdtdx.
Convergences (5.9)
1, (5.10)
2and (6.10) lead to Z
ΩT
k
ε,δφ ∇
xu
ε,δ∇w ψ
ε,δdtdx = Z
ΩT
φ ∇
xu
ε,δ∇w ψ
ε,δ1
Ω∗εδ
dtdx +
Z
ΩT×B
b
ε,δφ δ
N−2ε
2∇
zT
ε,δ(u
ε,δ)
εδT
ε,δ(∇w)
ψ dtdxdz
−→
Z
ΩT
φ ∇
xu ∇w ψ(∞) dtdx + 0.
(6.16)
Again from (5.9)
1, (5.10)
2and now (6.14) we get Z
ΩT
k
ε,δφ u
ε,δw ∇ψ
ε,δdtdx
= Z
ΩT×RN
φ T
ε,δ(k
ε,δ) δ
N−2ε
2∇
zT
ε,δ(u
ε,δ)T
ε,δ(w) ∇
zψ dtdxdz
−→ γ
0Z
ΩT×(RN\B)¯
φ ∇
zV w ∇
zψ dtdxdz + b
0Z
ΩT×B
φ ∇
zU w ∇
zψ dtdxdz
.
(6.17)
In the same way, using (4.7)
2and (5.8)
2give Z
ΩT
f
ε,δφ w ψ
ε,δdtdx −→
Z
ΩT
f φ w ψ(∞) dtdx + Z
ΩT×B
F φ w ψ dtdxdz, Z
Ω
ρ
ε,δu
0ε,δφ(0) w ψ
ε,δdx −→
Z
Ω
u
0φ(0)w ψ(∞) dx + Z
Ω×B
a
0U
0φ(0)w ψ dxdz.
(6.18)
Finally, from the convergences (6.15), (6.16), (6.17) and (6.18) one has
− Z
ΩT
uφ
0wψ(∞)dtdx − a
0Z
ΩT×B
U φ
0w ψ dtdxdz + Z
ΩT
φ ∇u ∇w ψ(∞) dtdx + γ
0Z
ΩT×RN
k
0φ ∇
zU w ∇
zψ dtdxdz
= Z
ΩT
f φ w ψ(∞) dtdx + Z
ΩT×B
F φ w ψ dtdxdz +
Z
Ω
u
0φ(0)w ψ(∞) dx + a
0Z
Ω×B
U
0φ(0)w ψ dxdz.
(6.19)
Step 3. The space of functions belonging to H ∩ C
∞( R
N) and whose gradient support is
bounded is dense in H. Hence, equality (6.19) is satisfied for any ψ in H. Then, the density
of the tensor product C
1([0, T ]) ⊗ D(Ω) ⊗ H in H
1(0, T ; L) implies
− Z
ΩT
U(t, ·, ∞) ∂Φ
∂t (t, ·, ∞)dtdx − a
0Z
ΩT×B
U ∂ Φ
∂t dtdxdz +
Z
ΩT
∇
xU(·, ∞) ∇
xΦ(t, ·, ∞) dtdx + γ
0Z
ΩT×RN
k
0∇
zU ∇
zΦ dtdxdz
= Z
ΩT
f Φ(t, ·, ∞) dtdx + Z
ΩT×B
F Φ dtdxdz +
Z
Ω
u
0Φ(0, ·, ∞) dx + a
0Z
Ω×B
U
0Φ(0) dxdz
(6.20)
for any Φ ∈ H
1(0, T ; L) such that Φ(T, ·, ·) = 0. Hence (6.2).
Step 4. Let Θ be in L satisfying
Θ(∞) = 1 and Θ(z) = 0 for a.e. z ∈ B.
For any Ψ ∈ H
01(0, T ; H
01(Ω)) we choose Φ = ΨΘ as test function in (6.20). That gives
− Z
ΩT
U (t, ·, ∞) ∂ Ψ
∂t dtdx + Z
ΩT
∇
xU (·, ∞) ∇
xΨ dtdx + γ
0Z
ΩT×RN
k
0∇
zU Ψ ∇
zΘ dtdxdz = Z
ΩT
f Ψ dtdx.
(6.21)
Hence
Z
ΩT
U (t, ·, ∞) ∂Ψ
∂t dtdx
≤ k∇
xU(·, ∞)k
L2(ΩT)k∇
xΨk
L2(ΩT)+ Ck∇
zU k
L2(ΩT×RN)kΨk
L2(ΩT)k∇
zΘk
L2(RN)+ kf k
L2(ΩT)kΨk
L2(ΩT). Since U belongs to L
2(0, T ; L), we obtain
Z
ΩT
U(t, ·, ∞) ∂Ψ
∂t dtdx
≤ CkΨk
L2(0,T;H10(Ω))
.
The constant depends on kU k
L2(0,T;L), k∇
zΘk
L2(RN)and kf k
L2(ΩT). As a consequence of the above inequality we get U (·, ·, ∞) ∈ H
1(0, T ; H
−1(Ω)).
Now, let Φ be in H
01(0, T ; L
2(Ω; H
1(B))) we extend Φ as a function Φ belonging to H
01(0, T ; L
2(Ω; H
1( R
N))) satisfying
Φ(t, x, z) = Φ(t, x, z) for a.e. (t, x, z) ∈ Ω
T× B, kΦk
H10(0,T;L2(Ω;H1(RN)))
≤ CkΦk
H10(0,T;L2(Ω;H1(B)))
. We choose Φ as test function in (6.20). That leads to
−a
0Z
ΩT×B
U ∂Φ
∂t dtdxdz + γ
0Z
ΩT×RN
k
0∇
zU ∇
zΦ dtdxdz = Z
ΩT×B
F Φ dtdxdz.
Hence
a
0Z
ΩT×B
U ∂Φ
∂t dtdxdz
≤ Ck∇
zU k
L2(ΩT×B)k∇
zΦk
L2(ΩT×RN)+ kF k
L2(ΩT×B)kΦk
L2(ΩT×RN).
(6.22) Again, using the fact that U belongs to L
2(0, T ; L), we get
Z
ΩT×B
U ∂ Φ
∂t dtdxdz
≤ CkΦk
L2(ΩT;H1(B)).
The constant depends on kU k
L2(0,T;L)and on kF k
L2(ΩT×B). As a consequence of the above inequality we get U ∈ H
1(0, T ; L
2(Ω; (H
1(B ))
0)).
7 The homogenized limit problem
From now on, we extend f and F by zero on [T, +∞[.
For any φ ∈ L
2(0, +∞) and for any p ∈ R
∗+, the Laplace transform (see e.g. [10], [11] or [12]) is defined by
φ(p) = b Z
+∞0
φ(t)e
−ptdt.
Hence, the evolution Problem (6.2) becomes p
Z
Ω
U b (p, x, ∞)Φ(x, ∞)dx + a
0p Z
Ω×B
U b (p, x, z)Φ(x, z)dxdz +
Z
Ω
∇
xU b (p, x, ∞)∇
xΦ(x, ∞)dx + γ
0Z
Ω×RN
k
0(z)∇
zU b (p, x, z)∇
zΦ(x, z)dxdz
= Z
Ω
f b (p, x)Φ(x, ∞)dx + Z
Ω×B
F b (p, x, z)Φ(x, z)dxdz +
Z
Ω
u
0(x)Φ(x, ∞)dx + a
0Z
Ω×B
U
0(x, z)Φ(x, z)dxdz, ∀p > 0, ∀Φ ∈ L.
(7.1)
For any p > 0, the Lax-Milgram theorem gives the existence and uniqueness of U b (p) ∈ L.
The following estimates hold:
pk U b (p, ·, ∞)k
2L2(Ω)+ k∇
xU b (p, ·, ∞)k
2L2(Ω)+ pk U b (p)k
2L2(Ω×B)+ k∇
zU b (p)k
2L2(Ω×RN)≤ C p
k f(p)k b
2L2(Ω)+ k F b (p)k
2L2(Ω×B)+ ku
0k
2L2(Ω)+ kU
0k
2L2(Ω×B), ∀p > 0.
The constant does not depend on p. We recall that
∀p > 0, k f b (p)k
2L2(Ω)≤ 1
p kf k
2L2(ΩT), k F b (p)k
2L2(Ω×B)≤ 1
p kF k
2L2(ΩT×B).
Now, let us introduce the solutions θ b
0and θ b
1of the following cell problems:
θ b
0(p) ∈ L
2(Ω; .
H
1( R
N)), a
0p
Z
B
θ b
0(p, x)Φ dz + γ
0Z
RN
k
0∇
zθ b
0(p, x)∇
zΦ dz
= Z
B
F b (p, x)Φ dz + a
0Z
B
U
0(x, z)Φdxdz, for a.e. x ∈ Ω, ∀Φ ∈ .
H
1( R
N), ∀p > 0,
(7.2)
and
θ b
1(p) ∈ H, θ b
1(p, ∞) = 1, a
0p
Z
B
θ b
1(p) Φ dz + γ
0Z
RN
k
0∇
zθ b
1(p)∇
zΦ dz = 0,
∀Φ ∈ .
H
1( R
N), ∀p > 0.
(7.3)
Again, the Lax-Milgram theorem gives the existence of θ b
0, θ b
1and pkb θ
0(p)k
2L2(Ω×B)+ k∇
zθ b
0(p)k
2L2(Ω×RN)≤ C
p k F b (p)k
2L2(Ω×B)+ kU
0k
2L2(Ω×B)∀p > 0.
Now, in order to estimate θ b
1we consider a function ϑ ∈ H ∩ C
∞( R
N) satisfying
ϑ(∞) = 1, ϑ = 0 in B, ∇
zϑ has a compact support. (7.4) The function θ b
2= θ b
1− ϑ is the solution of
θ b
2(p) ∈ .
H
1( R
N), a
0p
Z
B
θ b
2(p) Φ dz + γ
0Z
RN
k
0∇
zθ b
2(p)∇
zΦ dz
= − γ
0Z
RN
k
0∇
zϑ ∇
zΦ dz = γ
0Z
RN
k
0∆
zϑ Φ dz,
∀Φ ∈ .
H
1( R
N), ∀p > 0.
(7.5)
Then, taking b θ
2(p) has test function in (7.5) leads to pkb θ
2(p)k
2L2(B)+ k∇
zθ b
2(p)k
2L2(RN)≤ C
p k∆
zϑk
2L2(RN), ∀p > 0.
We express U b (p) in terms of θ b
0(p), θ b
2(p), ϑ and u(p). To do that, observe that the Laplace b transform b u(p) of u is in fact U b (p, ·, ∞). Hence, we have
U(p, x, z) = b u(p, x)b b θ
1(p, z) + θ b
0(p, x, z) = b u(p, x)b θ
2(p, z) + θ b
0(p, x, z) + u(p, x)ϑ(z) b
for a.e. (x, z) ∈ Ω × R
N, ∀p > 0.
Let ψ be in H
01(Ω), we choose ψ as test-function in (7.1). That gives p
Z
Ω
u(p)ψ dx b + a
0p Z
Ω
Z
B
θ b
2(p, z)dz
b u(p)ψ dx +a
0p
Z
Ω
Z
B
θ b
0(p, ·, z)dz
ψ dx + Z
Ω
∇ u(p)∇ψ dx b
= Z
Ω
f(p) + b u
0ψ dx + Z
Ω
Z
B
F b (p, ·, z)dz ψ dx +a
0Z
Ω
Z
B
U
0(·, z)dz ψ dx.
(7.6)
Now, applying the inverse Laplace transform (see [1], [10] and [11]) in Problems (7.2), (7.3) and (7.6) give
θ
0∈ H
1(0, T ; L
2(Ω; (H
1(B))
0)), θ
0∈ L
2(Ω
T; .
H
1( R
N)), a
0< dθ
0dt (t, x), Φ >
(H1(B))0,H1(B)+γ
0Z
RN
k
0∇
zθ
0(t, x)∇
zΦ dz
= Z
B
F (t, x)Φ dz, for a.e. (t, x) ∈ (0, +∞) × Ω, θ
0(0) = U
0, ∀Φ ∈ .
H
1( R
N),
(7.7)
θ
2is the solution of the following evolution problem:
θ
2∈ H
1(0, T ; (H
1(B))
0), θ
2∈ L
2(0, T ; .
H
1( R
N)), a
0< dθ
2dt (t), Φ >
(H1(B))0,H1(B)+γ
0Z
RN
k
0∇
zθ
2(t)∇
zΦ dz = 0, θ
2(0) = γ
0k
0a
0∆
zϑ, for a.e. t ∈ (0, +∞), ∀Φ ∈ .
H
1( R
N).
(7.8)
Here again, the Faedo-Galerkin’s method gives the existence and uniqueness of the solutions of the two above evolution equations. Proceeding as in Theorem 6.1 we show that θ
0∈ H
1(0, T ; L
2(Ω; (H
1(B))
0)) and θ
2∈ H
1(0, T ; (H
1(B))
0).
Lemma 7.1. Set
Θ
0= Z
B
θ
0dz, Θ
2= Z
B
θ
2dz.
We have
Θ
0∈ H
1(0, T ; L
2(Ω)), Θ
2∈ H
1(0, T ).
Moreover, the function Θ
2does not depend on the choice of the function ϑ.
Proof. Since θ
0∈ L
2(Ω
T× B) we obtain Θ
0∈ L
2(Ω
T) (as a consequence of the Cauchy-
Schwarz inequality and the Fubini’s theorem). Choosing 1−ϑ as test function in (7.7) (observe
that 1 − ϑ ∈ .
H
1( R
N)) gives a
0d dt
< θ
0(t, x), 1 >
(H1(B))0,H1(B)= a
0< dθ
0dt (t, x), 1 >
(H1(B))0,H1(B)=a
0< dθ
0dt (t, x), 1 − ϑ >
(H1(B))0,H1(B)=γ
0Z
RN
k
0∇
zθ
0(t, x)∇
zϑ dz + Z
B
F (t, x) dz.
(7.9)
Now, the fact that
• < θ
0(t, x), 1 >
(H1(B))0,H1(B)= Z
B
θ
0(t, x) dz for a.e. x ∈ Ω,
• ∇
zθ
0∈ [L
2(Ω
T× R
N)]
N,
• ∇
zϑ ∈ [C
∞( R
N)]
N(with compact support)
• F ∈ L
2(Ω
T× B )
and the above equalities (7.9) yield dΘ
0dt ∈ L
2(Ω
T). As a consequence Θ
0belongs to H
1(0, T ; L
2(Ω)). Proceeding in the same way gives Θ
2∈ H
1(0, T ).
Denote Θ b
2the Laplace transform of Θ
2. One has Θ b
2=
Z
B
θ b
2dz = Z
B
(b θ
1− ϑ) dz = Z
B
θ b
1dz.
It follows that Θ
2does not depend of the choice of ϑ.
We transform Problem (7.6) by the inverse Laplace transform, that gives
< d dt
u + a
0Θ
2∗ u
, ψ >
H−1(Ω),H10(Ω)
+ Z
Ω
∇u(t)∇ψ dx
= −a
0< dΘ
0dt (t), ψ >
H−1(Ω),H10(Ω)
+ Z
Ω
f(t) ψ dx, +a
0Z
Ω
Z
B
F (t) dz ψ dx u(0) = u
0, Θ
0(0) =
Z
B
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