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Homogenization of a Ginzburg-Landau problem in a perforated domain with mixed boundary conditions

Abstract

In this paper we study the asymptotic behavior of a Ginzburg-Landau problem in a ε-periodically perforated domain of R n with mixed Dirichlet-Neumann conditions. The holes can verify two different situations. In the first one they have size ε and a homogeneous Dirichlet condition is posed on a flat portion of each hole, whose size is an order smaller than ε, the Neumann condition being posed on the remaining part. In the second situation, we consider two kinds of ε-periodic holes, one of size of order smaller than ε, where a homogeneous Dirichlet condition is prescribed and the other one of size ε, on which a non-homogeneous Neumann condition is given. Moreover, in this case as ε goes to zero, the two families of holes approach each other. In both situations a homogeneous Dirichlet condition is also prescribed on the whole exterior boundary of the domain.

MSC: 35J20, 35J25, 35B25, 35J55, 35B40.

1 Introduction

Let Ω be a bounded set in R n , with Lipschitz boundary Ω and Y be [ 1 2 , 1 2 ] n . We consider two kinds of holes removed from Ω, both periodically distributed. The first kind is of size ε, ε being a positive parameter. It is obtained by rescaling a reference hole Q (a cube or a smoothed one) contained in Y and in the half plane { x 1 0}, a piece of which is on the hyperplane { x 1 =0}. The latter kind is of size ε n n 2 if n3 (exp( ε 2 ) if n=2) and it is obtained by rescaling a reference hole K strictly contained in Y and in the half plane x 1 0 (and containing a segment on the line x 1 =0 if n=2). Moreover, every element of the second family moves perpendicularly with respect to the x n -axis towards an element of the first family with an approaching speed of order τ ε σ if n3 (exp(1/ ε σ )τ if n=2) where τ=( τ 1 ,0,,0) R n , τ 1 0. Then we assign a non-homogeneous Neumann condition on the first family so that there is no total flow in Ω through these holes and a homogeneous Dirichlet condition on the latter family. We observe explicitly that if K is contained in the set {x R n : x 1 =0} and we take τ=0 we obtain the case when the zone where homogeneous Dirichlet condition is imposed, lies exactly on the boundary of the holes of the first family. Let Ω ε τ , σ be the set obtained by removing from Ω the two families of holes previously described. In this paper we study the homogenization process of the following vectorial nonlinear problem with mixed boundary conditions:

{ Δ u ε τ , σ u ε τ , σ + | u ε τ , σ | 2 u ε τ , σ = f , in  Ω ε τ , σ , u ε τ , σ = 0 , on  Ω , described conditions on  Ω ε τ , σ Ω .
(1.1)

Let us observe that the equation in (1.1) is known as the Ginzburg-Landau equation.

By using the energy method (see [1]), for n=2 and n=3 we prove the weakly convergence in H 0 1 (Ω) of a suitable extension of the sequence of solutions u ε τ , σ of (1.1) to a function u τ , σ , unique solution of the following limit problem:

{ div A ( u τ , σ ) + 1 2 μ τ , σ u τ , σ u τ , σ + | u τ , σ | 2 u τ , σ = θ f , in  Ω , u τ , σ = 0 , on  Ω ,
(1.2)

where A is the standard homogenized matrix which appears in [2] and θ is, roughly speaking, the volume of the ‘material’ in Ω. Furthermore, the definition of μ τ , σ depends both on σ or on the dimension n. More precisely

if n=2, μ τ , σ is equal to 4π if σ<2 and 2π if σ2;

if n=3, μ τ , σ is equal to

  1. (a)

    the double of the capacity in R 3 of the Dirichlet reference hole if 1σ<3;

  2. (b)

    the capacity in R 3 of a set obtained by perpendicularly translating the Dirichlet reference hole to a distance τ=( τ 1 ,0,0) R 3 , τ 1 0 from the hyperplane { x 1 =0} and after doubling it by reflection with respect to the same hyperplane if σ=3;

  3. (c)

    the capacity in R 3 of a set obtained by doubling the Dirichlet reference hole by reflection with respect to the hyperplane { x 1 =0} if σ>3.

We observe explicitly that no additional term of flow appears in (1.2). This is a consequence of the fact that there is no total flow. Eventually we observe that in the case (a) the term 1 2 μ τ , σ is exactly the ‘strange term’ of [3] and in the case (b) it is exactly the half of this quantity.

The paper is organized as follows. In Section 2, we describe the domain with appropriate spaces required and we give the main result. In Section 3, we recall some preliminary lemmas and prove an important result involving the capacity in R n of the Dirichlet reference hole. Finally, Section 4 is devoted to the proof of the main theorem, by establishing some a priori norm estimates for the sequence of solutions and convergence results.

Many authors (see for example [2]–[10]) studied the asymptotic behavior, as ε tends to zero, of solutions of scalar boundary value problems defined in a domain Ω ε obtained by removing from Ω closed smoothed cubes well contained in Ω (the holes) of diameter r(ε)ε periodically distributed with period ε in R n . In particular in [10] is studied, perhaps for the first time, a problem in which both Neumann and Dirichlet conditions are present on the boundary of the holes. The Dirichlet condition is given on a flat portion of diameter ε n n 2 if n3 (exp( ε 2 ) if n=2), the Neumann condition is non-homogeneous so the reference hole has to be rescaled with r(ε)= ε n n 1 . In [5] a problem where both Neumann and Dirichlet conditions are present on the boundary of the holes is also studied, where, the Neumann condition being homogeneous, the reference hole is rescaled with r(ε)=ε. On the other side, an extensive study of Ginzburg-Landau equation in a bounded domain Ω of R 2 is performed by several authors starting from the pioneering papers of Bethuel, Brezis and Hélein (see for example [11]–[18]). The limit behavior of the Ginzburg-Landau equation in a perforated domain in R 3 with holes along a plane is studied in [19] while in [20] is studied the homogenization of the Ginzburg-Landau equation in a domain of R 2 with oscillating boundary.

2 Statement of the problem and main result

Let Ω be an open bounded subset of R n with Lipschitz boundary. Let Y= [ 1 / 2 , 1 / 2 ] n R n and l>0 such that the cube R=[0,2l]× [ l , l ] n 1 Y . For n=2, or n=3 we take Q=R. For n>3 we consider a domain Q which has C boundary such that

3 4 RQR.
(2.1)

Let us observe that Q( 3 4 R)R. Let us pose Y c =YQ. Let K be a compact subset of R n , contained in the half plane x 1 0 and in Y; moreover, if n=2 let K{x R n : x 1 =0} contain a segment.

Let f=( f 1 , f 2 ) ( L 2 ( Ω ) ) 2 and let g=( g 1 , g 2 ) ( L 2 ( Q ) ) 2 be a null average function.

Let ε>0 and Y ε = Y ε (Ω)={ε(Y+k):k Z n  and ε(Y+k)Ω}. Let Q ε =εQ and

S ε τ , σ = { ε n n 2 K + ε σ τ , n 3 , exp ( 1 / ε 2 ) K + exp ( 1 / ε σ ) τ , n = 2 ,

where τ=( τ 1 ,0,,0) R n , τ 1 0, and σ1; let us observe that there exists ε τ >0 such that if 0<ε< ε τ then S ε τ , σ is well contained in εY (see Figure 1 and Figure 2).

Figure 1
figure1

The cell εY : case τ=0 .

Figure 2
figure2

The cell εY : case τ0 .

Let us define T ε = T ε (Ω)=({ Q ε +εk:k Z n }) Y ε and Ω ε =Ω T ε .

Let g ε be the function defined on T ε by g ε (x)=g( x ε ) if x( Q ε +εk) and k Z n .

Let K ={x R n :( x 1 , x 2 ,, x n )K}.

Let D ε τ , σ = D ε τ , σ (Ω)=({ S ε τ , σ +εk:k Z n }) Y ε if 0<ε< ε τ , D ε τ , σ = otherwise.

Let Ω ε τ , σ =Ω( T ε D ε τ , σ ). Let Γ ε D , τ , σ = D ε τ , σ , Γ ε N , τ , σ = T ε D ε τ , σ (see Figure 3 and Figure 4).

Figure 3
figure3

The domain Ω ε τ , σ : case τ=0 .

Figure 4
figure4

The domain Ω ε τ , σ : case τ0 .

Definition 2.1

Let K be a compact subset of R n and Ω an open set such that KΩ. We define the (harmonic) capacity of K with respect to Ω, and we will denote by cap(K,Ω) the following quantity:

cap(K,Ω)=inf { Ω | φ | 2 d x : φ C c 1 ( Ω ) , 0 φ 1  and  φ 1  on  K } .

We will, moreover, denote by cap(K) the quantity cap(K, R n ).

Let us consider, for every ε>0, the following problem:

{ Δ u ε τ , σ u ε τ , σ + | u ε τ , σ | 2 u ε τ , σ = f , in  Ω ε τ , σ , u ε τ , σ = 0 , on  Ω Γ ε D , τ , σ , u ε τ , σ n = g ε , on  Γ ε N , τ , σ ,
(2.2)

whose variational formulation is

{ u ε τ , σ ( V ε τ , σ ) 2 , Ω ε τ , σ u ε τ , σ , φ + Ω ε τ , σ | u ε τ , σ | 2 u ε τ , σ φ Ω ε τ , σ u ε τ , σ φ = Ω ε τ , σ f φ + Γ ε N , τ , σ g ε φ , φ ( V ε τ , σ ) 2 ,
(2.3)

where V ε τ , σ denote the closure of C 0 1 (Ω Γ ε D , τ , σ ) in H 1 ( Ω ε τ , σ ).

For any λ R n , let w λ H 1 ( Y c ) be the solution of the following problem:

{ Δ w λ = 0 , in  Y c , w λ ( y ) λ y , Y -periodic , w λ n = 0 , on  Q .
(2.4)

Since w λ is linear in λ and the extension operator to zero is linear, we can consider the matrix A given by

Aλ= m Y ( w λ ˜ )= 1 | Y | Y c w λ dy,λ R n ,
(2.5)

where w λ ˜ denotes the extension to zero of w λ on the whole Y. In what follows, with m S (u) we will denote the average of the function u over the subset S R n .

We give the following result.

Theorem 2.1

Let ε be a parameter taking values in a sequence going to zero and let { u ε τ , σ } ε be the sequence of solutions of problem (2.2). Then there exists a bounded sequence { v ε τ , σ } ε in ( H 0 1 ( Ω ) ) 2 extending { u ε τ , σ } ε and weakly converging in ( H 0 1 ( Ω ) ) 2 , forn=2andn=3, to the solution u τ , σ of the following homogenized problem:

{ div ( A u τ , σ ) + 1 2 μ τ , σ u τ , σ u τ , σ + | u τ , σ | 2 u τ , σ = θ f in  Ω , u τ , σ = 0 on  Ω
(2.6)

or, in the variational formulation,

{ u τ , σ ( H 0 1 ( Ω ) ) 2 , Ω A u τ , σ , φ + 1 2 μ τ , σ Ω u τ , σ φ Ω u τ , σ φ + Ω | u τ , σ | 2 u τ , σ φ = Ω θ f φ , φ ( D ( Ω ) ) 2 ,
(2.7)

whereAis the constant matrix defined in (2.5), and

μ τ , σ = { 2 cap ( K ) , if  1 σ < 3  for every  τ , cap ( ( K + τ ) ( K τ ) ) , if  σ = 3 , cap ( K K ) , if  σ > 3  for every  τ ,
(2.8)

ifn=3, and

μ τ , σ = { 4 π , if  σ < 2 , 2 π , if  σ 2 ,
(2.9)

ifn=2.

Moreover, any sequence of functions bounded in ( H 0 1 ( Ω ) ) 2 and extending { u ε τ , σ } ε converges to u τ , σ .

3 Preliminary results

Let us recall some properties of the capacity of general dimension n (see [21] and [22]).

Proposition 3.1

Let Ω be an open subset of R n and E, E 1 , E 2 subsets of Ω. Then

  1. (i)

    cap(,Ω)=0;

  2. (ii)

    E 1 E 2 cap( E 1 ,Ω)cap( E 2 ,Ω) (monotonicity);

  3. (iii)

    cap( E 1 E 2 ,Ω)+cap( E 1 E 2 ,Ω)cap( E 1 ,Ω)+cap( E 2 ,Ω) (strong subadditivity);

  4. (iv)

    if { E h } h is an increasing sequence of subsets of Ω and E= h E h Ω, then cap(E,Ω)= lim h cap( E h ,Ω);

  5. (v)

    if { E h } h is a sequence of subsets of Ω and E h E h , then cap(E,Ω) h cap( E h ,Ω);

  6. (vi)

    if Ω 1 and Ω 2 are open subsets of R n and E Ω 1 Ω 2 , then cap(E, Ω 2 )cap(E, Ω 1 );

  7. (vii)

    if { Ω h } h is an increasing sequence of open sets such that h N Ω h =Ω, then lim h cap(E, Ω h )=cap(E,Ω);

  8. (viii)

    if t>0, then cap(tE,tΩ)= t n 2 cap(E,Ω);

  9. (ix)

    if { E h } h is a decreasing sequence of compact subsets of Ω with E= h E h , then cap(E,Ω)= lim h cap( E h ,Ω).

Now we recall two results of [5].

Lemma 3.1

Let R a cube in R n andCRbe a compact set with Lipschitz boundary ∂C and1p<. Then there exists a linear bounded extension operatorΦ: W 1 , p (RC) W 1 , p (R)such that

( Φ v ) L P ( R ) c v L P ( R C ) ,for every v W 1 , p (RC).

Let T =Q(l Y ), T ε ={ε T +εk,k Z n  s.t. ε(Y+k)Ω} and Ω ε =Ω T ε .

Theorem 3.1

LetΩ R n be a bounded open set, ε>0, and let Y, l, Q, Y ε , T ε , Ω ε be defined as in Section  2.

Then there exists a family { P ε } ε of uniform extension operators (i.e. P ε u= u ε in Ω ε ) from H 1 ( Ω ε )to H 1 (Ω), such that

( P ε u ) L 2 ( Ω ) c u L 2 ( Ω ε ) , for every  u H 1 ( Ω ε ) , P ε u ( ε ( x 1 , x 2 , , x n ) + ε k ) = P ε u ( ε ( x 1 , x 2 , , x n ) + ε k ) ,

for everyu H 1 ( Ω ε )and for every( x 1 , x 2 ,, x n )lYand k such thatεY+εkΩ. Eventually if u = 0 on Ω, then P ε u=0on Ω.

As a consequence we get the following result.

Corollary 3.1

LetΩ R n a bounded open set, ε>0and let Y, l, Q, Y ε , Q ε , T ε , Ω ε τ , σ be defined as in Section  2.

Let { u ε τ , σ } ε ( H 1 ( Ω ε τ , σ ) ) 2 such that u ε τ , σ =0on Γ ε D , τ , σ .

Then there exists a sequence { v ε τ , σ } ε ( H 1 ( Ω ) ) 2 of uniform extensions of { u ε τ , σ } ε and a constant c independent of ε such that

v ε τ , σ = u ε τ , σ , in  Ω ε τ , σ , v ε τ , σ = 0 , on  Γ ε D , τ , σ ,
(3.1)
v ε τ , σ L 2 ( Ω ) c u ε τ , σ L 2 ( Ω ε τ , σ ) ,
(3.2)

and

v ε τ , σ ( x 1 , x 2 ,, x n )= v ε τ , σ (2ε k 1 x 1 , x 2 ,, x n ),
(3.3)

for a.e. x({ε(lY+ k ) , k Z n  s.t. ε(Y+ k ) Ω} Ω ε ) Ω ε and for every k of this kind and related to x.

Proof

Let u ε τ , σ =( u ε , 1 τ , σ , , u ε , 2 τ , σ ) ( H 1 ( Ω ε τ , σ ) ) 2 and let us define

u ¯ ε τ , σ = { u ε τ , σ , in  Ω ε τ , σ , 0 , in  Ω ε Ω ε τ , σ .
(3.4)

We observe that Ω ε τ , σ ( Ω ε Ω ε τ , σ )= Γ ε D , τ , σ and therefore u ¯ ε τ , σ ( H 1 ( Ω ε ) ) 2 . Then the sequence { v ε τ , σ } ε given by v ε τ , σ =( P ε u ¯ ε , 1 τ , σ , P ε u ¯ ε , 2 τ , σ ) meets the requirements by Theorem 3.1. □

Let ( S ε τ , σ ) ={x R n :( x 1 , x 2 ,, x n ) S ε τ , σ }, S ε τ , σ = S ε τ , σ ( S ε τ , σ ) and D ε τ , σ =({ S ε τ , σ +εk:k Z n }) Y ε if 0<ε< ε τ , D ε τ , σ = otherwise.

Let BY an open ball centered at the origin, 1<ν< n n 2 (1<ν<+, if n=2) and construct the following families of sets:

A ε σ = { ( { ε ν B + ε k : k Z n } ) Y ε , if  σ n n 2 ( n 3 ) , ( { ε σ ( ( B + τ ) ( B τ ) ) } : k Z n ) Y ε , if  1 σ < n n 2 ( n 3 ) , ( { exp ( 1 ε σ ) ( ( B + τ ) ( B τ ) ) } : k Z n ) Y ε , if  σ < 2 ( n = 2 ) , ( { exp ( 1 ε α ) B + ε k } : k Z n ) Y ε , with  α < 2  if  σ 2 ( n = 2 ) ,
(3.5)

if 0<ε< ε 1 and A ε σ = otherwise (let us observe that there exists ε 1 >0 such that if 0<ε< ε 1 then S ε τ , σ B ε εY).

Let C 0 (Ω, D ε τ σ ,1) be the set of functions v C 0 (Ω) such that v=1 in a neighborhood of D ε τ , σ .

Let H 0 1 (Ω, D ε τ , σ ,1) be the closure of C 0 (Ω, D ε τ , σ ,1) in H 0 1 (Ω). Let Ψ ε τ , σ be the unique solution of the problem

min { Ω | φ | 2 d x : φ H 0 1 ( Ω , D ε τ , σ , 1 ) , φ = 0  on  Ω A ε σ } .
(3.6)

Let us pose ψ ε τ , σ =1 Ψ ε τ , σ . We can consider ψ ε τ , σ as a function of H 0 1 (Ω), and observe that

ψ ε τ , σ = 0 , on  D ε τ , σ , ψ ε τ , σ = 1 , in  Ω A ε σ .
(3.7)

Using a result proved in Lemma 3.5 of [5], we can study the behavior, and explicitly calculate, the ‘strange term’ μ τ , σ , when σ is equal to n n 2 , and σ is different from n n 2 , i.e., when the distance between the small hole and the big one is different from the size of the latter. In fact if 1σ< n n 2 the distance between the small hole and the big one is bigger and bigger than the size of the latter; if σ> n n 2 the condition is just the opposite.

Theorem 3.2

For any fixed ε let ψ ε τ , σ the unique solution of (3.6). Then

0 ψ ε τ , σ 1,
(3.8)
ψ ε τ , σ 1weakly in  H 1 (Ω),
(3.9)
ψ ε τ , σ 1strongly in  L p (Ω),p[1,+[.
(3.10)

Moreover, forν 2 2 p , we have

ψ ε τ , σ 0strongly in  L p (Ω),p[1,2[.
(3.11)

Furthermore, if we restrict ψ ε τ , σ to Ω ε , we get

ψ ε τ , σ ( x 1 , x 2 ,, x n )= ψ ε τ , σ (2ε k 1 x 1 , x 2 ,, x n )
(3.12)

for a.e. x({ε(lY+k):k Z n  s.t. ε(Y+k)Ω} Ω ε ) Ω ε and for every k of this kind and related to x.

Moreover, there exist μ ε τ , σ and γ ε τ , σ H 1 (Ω)and a unique distribution μ τ , σ W 1 , such that

i i ( i ) i ( ii ) ( iii ) { Δ ψ ε τ , σ = μ ε τ , σ γ ε τ , σ , μ ε τ , σ μ τ , σ , strongly in  H 1 ( Ω ) , γ ε τ , σ , v ε = 0 , v ε H 0 1 ( Ω D ε τ , σ ) ,
(3.13)

where

μ τ , σ = { 2 cap ( K ) , if  1 σ < n n 2  for every  τ , cap ( ( K + τ ) ( K τ ) ) , if  σ = n n 2 , cap ( K K ) , if  σ > n n 2  for every  τ ,
(3.14)

ifn3, and

μ τ , σ = { 4 π , if  σ < 2 , 2 π , if  σ 2 ,
(3.15)

ifn=2.

Proof

Properties (3.8)-(3.10) are proved in [3].

By arguing as in Lemma 3.5 of [5] we obtain conditions (3.11) and (3.12). Moreover, Theorem 2.7 and Lemma 2.8 of [3] provide the existence of two sequences { μ ε τ , σ } ε , { γ ε τ , σ } ε H 1 (Ω) and μ τ , σ W 1 , such that (3.13) holds up to a subsequence.

Now we identify the measure μ τ , σ .

Let us recall that by virtue of Proposition 1.1 in [3] up to a subsequence

μ ε τ , σ , φ = lim ε | ψ ε τ , σ | 2 φ,φD(Ω).

The proof of (3.14) and (3.15) will be performed in five steps.

Step 1. Let n3 and σ= n n 2 . At first we observe that

Δ ψ ε τ , σ , φ ψ ε τ , σ = Ω ψ ε τ , σ ( φ ψ ε τ , σ ) d x = Ω ψ ε τ , σ ( φ ) ψ ε τ , σ d x + Ω ψ ε τ , σ ( ψ ε τ , σ ) φ d x = Ω ψ ε τ , σ ( φ ) ψ ε τ , σ d x + Ω | ψ ε τ , σ | 2 φ .
(3.16)

Let us consider the first term in the right-hand side of (3.16); by (3.10) and (3.11), one has

lim ε 0 Ω ψ ε τ , σ (φ) ψ ε τ , σ dx=0,φD(Ω).
(3.17)

Let us consider φD(Ω). Let us fix δ>0 and consider η>0 such that

|xy|<η | φ ( x ) φ ( y ) | <δ.

Then if ε>0 is small enough

ε n ε Y + ε k ( φ ( y ) δ ) d y ε Y + ε k | ψ ε τ , σ | 2 d x ε Y + ε k | ψ ε τ , σ | 2 φ d x ε n ε Y + ε k ( φ ( y ) + δ ) d y ε Y + ε k | ψ ε τ , σ | 2 d x
(3.18)

for every k Z n such that εY+εkΩ.

Moreover, by definition of ψ ε τ , σ , taking into account its εY periodicity in Y ε and by (viii) of Proposition 3.1, if n3, we have

ε Y + ε k | ψ ε τ , σ | 2 = ε Y + ε k | Ψ ε τ , σ | 2 = ε ν B + ε k | Ψ ε τ , σ | 2 = cap ( ε n n 2 ( ( K + τ ) ( K τ ) ) , ε ν B ) = ε n cap ( ( K + τ ) ( K τ ) , ε ν n n 2 B ) .
(3.19)

Now if ε>0 is small enough, then supp(φ) Y ε and by the last expression and summing up relations (3.18), for every k Z n such that εY+εkΩ, we have

cap ( ( K + τ ) ( K τ ) , ε ν n n 2 B ) Ω ( φ ( y ) δ ) Ω | ψ ε τ , σ | 2 φ cap ( ( K + τ ) ( K τ ) , ε ν n n 2 B ) Ω ( φ ( y ) + δ ) , φ D ( Ω ) .
(3.20)

Then by (vii) of Proposition 3.1 and the arbitrariness of δ, passing to the limit for ε0, we obtain

lim ε 0 Ω | ψ ε τ , σ | 2 φ=cap ( ( K + τ ) ( K τ ) ) Ω φ(x)dx,φD(Ω).
(3.21)

Indeed, using condition (i) in (3.13), one has

Δ ψ ε τ , σ , φ ψ ε τ , σ = μ ε τ , σ γ ε τ , σ , φ ψ ε τ , σ = μ ε τ , σ , φ ψ ε τ , σ γ ε τ , σ , φ ψ ε τ , σ , φ D ( Ω ) .
(3.22)

Consequently by combining (3.13)(iii), with (3.16), (3.17), and (3.21), we obtain

μ τ , σ , φ =cap ( ( K + τ ) ( K τ ) ) Ω φ(x)dx,φD(Ω).

Step 2. Let n3 and let us consider the case σ> n n 2 . We set

K = K K , K ε γ = ( ( K + ε γ τ ) ( K ε γ τ ) ) , ( K ) δ = { x R n : dist ( x , K ) < δ } ,

where γ=σ n n 2 . As in the previous case

ε Y + ε k | ψ ε τ , σ | 2 = ε ν B + ε τ | Ψ ε τ , σ | 2 = cap ( ( ε n n 2 K + ε σ τ ) ( ε n n 2 K ε σ τ ) , ε ν B ) = ε n cap ( ( K + ε γ τ ) ( K ε γ τ ) , ε ν n n 2 B ) = ε n cap ( K ε γ , ε ν n n 2 B ) .

Let us fix δ]0,1[, and r>dist( K )+1. If ε is small enough, we have K ε γ ( K ) δ and rB ε ν n n 2 B. Then

cap ( K ε γ , ε ν n n 2 B ) cap ( ( K ) δ , ε ν n n 2 B ) cap ( ( K ) δ , r B ) ,
(3.23)

for ε is small enough (dependent on r).

Consequently

lim sup ε 0 cap ( K ε γ , ε ν n n 2 B ) cap ( ( K ) δ , r B ) .
(3.24)

On the other hand

cap ( K ε γ , ε ν n n 2 B ) cap ( K ε γ K , ε ν n n 2 B ) cap ( K ε γ K ) ,ε>0.
(3.25)

Let us observe that K ε γ K is an increasing sequence converging to K , and K ε γ K since K is a perfect set. Consequently, by (iv) of Proposition 3.1, it follows that

lim ε 0 + cap ( K ε γ K ) =cap ( K ) .
(3.26)

Therefore by combining (3.24) with (3.25) and (3.26) we get

cap ( K ) = lim ε 0 + cap ( K ε γ K ) lim inf ε 0 + cap ( K ε γ , ε ν n n 2 B ) lim sup ε 0 + cap ( K ε γ , ε ν n n 2 B ) cap ( ( K ) δ , r B ) .
(3.27)

Since K is a compact set, ( K ) δ decreases to K as δ0. Then, by (ix) of Proposition 3.1, one has

lim δ 0 + cap ( ( K ) δ , r B ) =cap ( K , r B ) .
(3.28)

By passing to the limit, as δ 0 + in (3.27), by (3.28), it follows that

cap ( K ) lim inf ε 0 + cap ( K ε γ , ε ν n n 2 B ) lim sup ε 0 + cap ( K ε γ , ε ν n n 2 B ) cap ( K , r B ) .
(3.29)

By passing to the limit as r in (3.29) we have the result

lim ε 0 + cap ( K ε γ , ε ν n n 2 B ) =cap ( K ) .
(3.30)

By arguing as in the previous case, (3.30) provides

μ τ , σ , φ =cap ( K ) Ω φ(x)dx,φD(Ω).

Step 3. Let n3 and let us examine the case 1σ< n n 2 .

By recalling the expression of the A ε σ in (3.5), and by following the same arguments as above, one has

ε Y + ε k | ψ ε τ , σ | 2 = ε σ ( B + τ ) ε σ ( B τ ) | Ψ ε τ , σ | 2 = 2 ε σ ( B + τ ) | Ψ ε τ , σ | 2 = 2 cap ( ( ε n n 2 K + ε σ τ ) , ε σ ( B + τ ) ) = 2 cap ( ε n n 2 K , ε σ B ) = 2 ε n cap ( K , ε σ n n 2 B ) .
(3.31)

By repeating the same steps (3.16)-(3.22), we have

μ τ , σ , φ =2cap(K) Ω φ(x)dx,φD(Ω).

Step 4. Let n=2. Let σ<2, and A ε σ be the respective family sets (see (3.5)).

In a similar way as in the previous case, by (viii) of Proposition 3.1 we have

ε Y + ε k | ψ ε τ , σ | 2 d x = exp ( ε σ ) ( ( B + τ ) ( B τ ) ) | Ψ ε τ , σ | 2 = 2 exp ( ε σ ) ( ( B + τ ) ) | Ψ ε τ , σ | 2 = 2 cap ( exp ( 1 ε 2 ) K , exp ( 1 ε σ ) B ) = 2 cap ( K , exp ( 1 ε σ + 1 ε 2 ) B ) .
(3.32)

Let us observe that E=K{x R n : x 1 =0} contains a segment and that K is well contained in another ball centered at the origin, say it B 1 . So, by (ii) of Proposition 3.1 we have

1 ε 2 cap ( E , exp ( 1 ε σ + 1 ε 2 ) B ) 1 ε 2 cap ( K , exp ( 1 ε σ + 1 ε 2 ) B ) 1 ε 2 cap ( B 1 , exp ( 1 ε σ + 1 ε 2 ) B ) .

By Lemma 3.3 of [10] we obtain

1 ε 2 2 π ε σ + ε 2 ( 1 + b ε ) 1 ε 2 cap ( K , exp ( 1 ε σ + 1 ε 2 ) B ) 1 ε 2 2 π ε σ + ε 2 ( 1 + a ε ) ,
(3.33)

where

lim ε 0 b ε =0and lim ε 0 a ε =0,

so that

1 ε 2 cap ( K , exp ( 1 ε σ + 1 ε 2 ) B ) 2π.
(3.34)

Hence by (3.18) and (3.34) one has

lim ε 0 Ω | ψ ε τ , σ | 2 φdx=4π Ω φ(x)dx.

Then we obtain

μ τ , σ , φ =4π Ω φ(x)dx,φD(Ω).

Step 5. Finally let n=2 and let us examine the case σ2 setting

K ε , σ =exp ( 1 ε 2 ) K+exp ( 1 ε σ ) τ

and

K ε , σ =exp ( 1 ε 2 ) K exp ( 1 ε σ ) τ.

Let us suppose that E=K{x R n : x 1 =0} contains a segment. Let us denote by E ε , σ and E ε , σ , respectively, the following subsets:

E ε , σ = exp ( 1 ε 2 ) E + exp ( 1 ε σ ) τ , E ε , σ = exp ( 1 ε 2 ) E exp ( 1 ε σ ) τ ,

where with E we denote the symmetry of subset E with respect to the hyperplane { x 1 =0}.

By (ii) and (viii) of Proposition 3.1, and by Lemma 3.3 of [10] we have

ε Y + ε k | ψ ε τ , σ | 2 d x = exp ( ε α ) ( B + τ ) | Ψ ε τ , σ | 2 = cap ( K ε , σ K ε , σ , exp ( 1 ε α ) B ) = cap ( ( K + exp ( 1 ε 2 1 ε σ ) τ ) ( K exp ( 1 ε 2 1 ε σ ) τ ) , exp ( 1 ε 2 1 ε α ) B ) .

Let B 2 be another ball centered at the origin and containing

( K + exp ( 1 ε 2 1 ε σ ) τ ) ( K exp ( 1 ε 2 1 ε σ ) τ )

then on the one hand we have

1 ε 2 cap ( ( K + exp ( 1 ε 2 1 ε σ ) τ ) ( K exp ( 1 ε 2 1 ε σ ) τ ) , exp ( 1 ε 2 1 ε α ) B ) 1 ε 2 cap ( B 2 , exp ( 1 ε 2 1 ε α ) B ) = 1 ε 2 2 π ε α + ε 2 ( 1 + a ε ) ,

on the other hand

1 ε 2 cap ( K ε , σ K ε , σ , exp ( 1 ε α ) B ) 1 ε 2 cap ( E ε , σ E ε , σ , exp ( 1 ε α ) B ) 1 ε 2 cap ( E ε , σ , exp ( 1 ε α ) B ) cap ( exp ( 1 ε 2 ) E , exp ( 1 ε α ) B ) = 1 ε 2 cap ( E , exp ( 1 ε 2 1 ε α ) B ) = 1 ε 2 2 π ε α + ε 2 ( 1 + b ε ) ,
(3.35)

where

lim ε 0 b ε =0and lim ε 0 a ε =0.

So

1 ε 2 cap ( K ε , σ K ε , σ , exp ( 1 ε α ) B ) 2π.

Indeed as in the previous case, by Lemma 3.3 of [10], we obtain

μ τ , σ , φ =2π Ω φ(x)dx,φD(Ω).

 □

The following lemma is a consequence of [23] and Lemma 3.11 of [10].

Lemma 3.2

Let C be a compact set such thatCY R n , a1, and

Ω ε ′′ =Ω ( { ε a C + ε k : k Z n } Y ε ) .
(3.36)

LetΩ R n a bounded open set, ε>0, and let Ω ε ′′ defined as in (3.36).

Let { u ε } ε and { v ε } ε two bounded sequences in ( H 0 1 ( Ω ) ) 2 such that u ε = v ε a.e. in Ω ε ′′ . Then

u ε v ε 0weakly in  ( H 0 1 ( Ω ) ) 2  as ε 0 + .
(3.37)

Lemma 3.3

Let Ω, Y, l, Q, Y ε , Q ε , T ε , Ω ε , g ε be defined as in Section  2, andφ=( φ 1 , φ 2 )a function in ( H 1 ( Ω ε ) ) 2 . Then there exists a constant C, depending on|Ω|, such that for every fixedi=1,2

T ε g ε , i φ i d H n 1 C φ i L 2 ( Ω ε ) .
(3.38)

Proof

For every i fixed, let us consider the following problem:

{ Δ β i = 0 , in  Y c , β i n = g i , on  Q , β  is  Y -periodic s.t.  m Q ( β ) = 0 .
(3.39)

Set β ε , i (x)=ε β i ( x ε ) where x Y c , so that problem (3.39) becomes

{ Δ β ε , i = 0 , in  Ω ε , β ε , i n = g ε , i , on  T ε ,

whose variational formulation is

Ω ε β ε , i φ i = T ε g ε , i φ i d H n 1 , φ i H 1 ( Ω ε ).

By Hölder’s inequality we have

T ε g ε , i φ i d H n 1 = Ω ε β i ( x ε ) φ i ( x ) d x ( Ω ε | β i ( x ε ) | 2 d x ) 1 2 φ i L 2 ( Ω ε ) ( c | Ω | ε 2 ) ( ε Y c | β i ( x ε ) | 2 d x ) 1 2 φ i L 2 ( Ω ε ) = c | Ω | ( Y c | β i ( y ) | 2 d y ) 1 2 φ i L 2 ( Ω ε ) = c | Ω | β i L 2 ( Y c ) φ i L 2 ( Ω ε ) ,

and the lemma is proved. □

4 Proof of Theorem 2.1

4.1 Compactness and convergence results

To simplify the notation we omit the explicit dependence on the parameter τ.

We denote by u ˜ the zero extension to the whole Ω of a vector function u defined on a subset of Ω and by u ˆ the periodic extension to R n k Z 2 (Q+k) of a vector function u Y-periodic in H 1 ( Y c ). Moreover, let u ¯ ε σ be defined as in (3.4).

By Corollary 3.1 there exists a sequence ( v ε σ ) ε satisfying (3.2). So by Poincaré’s inequality and (3.2) in Corollary 3.1 we get

v ε σ H 0 1 ( Ω ) 2 c Ω 2 Ω | v ε σ | 2 c Ω 2 c 2 Ω ε σ | u ε σ | 2 = c Ω 2 c 2 [ Ω ε σ f u ε σ Ω ε σ | u ε σ | 4 + Ω ε σ | u ε σ | 2 + Γ ε N , σ g ε u ε σ ] = c Ω 2 c 2 [ Ω ε f u ¯ ε σ Ω ε | u ¯ ε σ | 4 + Ω ε | u ¯ ε | 2 + T ε g ε u ¯ ε σ ] ,
(4.1)

where c Ω is the Poincaré constant of Ω and c is the constant given by Corollary 3.1.

By the Hölder inequality we get

u ¯ ε σ L 2 ( Ω ε ) | Ω | 1 4 u ¯ ε σ L 4 ( Ω ε ) .
(4.2)

By Lemma 3.3 and (4.2), inequality (4.1) becomes

v ε σ H 0 1 ( Ω ) 2 c Ω 2 c 2 [ f L 2 ( Ω ) u ¯ ε σ L 2 ( Ω ε ) u ¯ ε σ L 4 ( Ω ε ) 4 + | Ω | 1 2 u ¯ ε σ L 4 ( Ω ε ) 2 + C u ¯ ε σ L 2 ( Ω ε ) ] .
(4.3)

Now let us consider the following quantity:

u ¯ ε σ L 4 ( Ω ε ) 4 | Ω | 1 2 u ¯ ε σ L 4 ( Ω ε ) 2 .
(4.4)

Let us suppose

u ¯ ε σ L 4 ( Ω ε ) | Ω | 1 4 .

Then (4.3) becomes

v ε σ H 0 1 ( Ω ) 2 c Ω 2 c 2 [ f L 2 ( Ω ) u ¯ ε σ L 2 ( Ω ε ) + C u ¯ ε σ L 2 ( Ω ε ) ] c Ω 2 c 2 max { f L 2 , C } u ¯ ε σ H 1 ( Ω ε ) C Ω v ε σ H 0 1 ( Ω ) ,
(4.5)

which is

v ε σ H 0 1 ( Ω ) C Ω ,
(4.6)

where C Ω is a constant independent of ε.

Now let us suppose

0 u ¯ ε σ L 4 ( Ω ε ) < | Ω | 1 4 .
(4.7)

By (4.2), inequality (4.3) becomes

v ε σ H 0 1 ( Ω ) 2 c Ω 2 c 2 [ f L 2 ( Ω ) u ¯ ε σ L 2 ( Ω ε ) + | Ω | + C u ¯ ε σ L 2 ( Ω ε ) ] .
(4.8)

Then

v ε σ H 0 1 ( Ω ) 2 c Ω 2 c 2 [ max { f L 2 , C } u ¯ ε σ H 1 ( Ω ε ) + | Ω | ] ,
(4.9)

i.e. again

v ε σ H 0 1 ( Ω ) C Ω ,

where C Ω is a constant independent of ε.

So there exists a subsequence, still denoted by ε, such that

v ε σ u σ weakly in  H 0 1 (Ω).
(4.10)

Obviously, since v ε σ = u ε σ on Ω ε σ , we deduce, by (4.6), that

u ε σ H 0 1 ( Ω ε σ ) C Ω ,
(4.11)

where C Ω is a constant independent of ε.

By the Rellich theorem,

if  n = 2 , v ε σ L q , for  q [ 1 , + [ , if  n 3 , v ε σ L q , for  q [ 1 , 2 n n 2 ] ,
(4.12)

and

v ε σ u σ strongly in  L p (Ω), { for  1 p < 2 = 2 n n 2 , if  n > 2 , for  2 p < , if  n = 2 ,
(4.13)

up to a subsequence still denoted by ε. Let us observe that

χ Ω ε σ θ= | Y c | | Y | weakly  in  L (Ω).
(4.14)

Now we are able to prove the following result.

Lemma 4.1

Letn=2andn=3. Let ( u ε σ ) ε be the sequence of solutions of problem (2.2), then

u ˜ ε σ θ u σ weakly in  L 2 (Ω),
(4.15)
u ˜ ε σ | u ˜ ε σ | 2 θ u σ | u σ | 2 weakly in  L 4 3 (Ω),
(4.16)

up to a subsequence still denoted by ε.

Proof

By (4.11) we get

u ˜ ε σ L 2 ( Ω ) = u ε σ L 2 ( Ω ε σ ) C Ω .

Then

u ˜ ε σ zweakly in  L 2 (Ω),

up to a subsequence still denoted by ε. Since

u ˜ ε σ = χ Ω ε σ v ε σ ,
(4.17)

by (4.13) and (4.14) we obtain

z=θ u σ ,

and then (4.15) holds. In order to prove (4.16) we observe that

u ˜ ε σ | u ˜ ε σ | 2 L 4 3 ( Ω ) C,
(4.18)

where C is a constant independent of ε. Indeed by (4.11) we have also

u ¯ ε σ H 1 ( Ω ε ) C Ω ,
(4.19)

where C Ω is a constant independent of ε, which implies

u ¯ ε σ L 2 ( Ω ε ) C Ω .
(4.20)

Multiplying by u ε σ in (2.2) we get

Ω | u ˜ ε σ | 2 + Ω | u ˜ ε σ | 4 = Ω | u ˜ ε σ | 2 + Ω f u ˜ ε σ + T ε g ε u ¯ ε σ .
(4.21)

By (4.20) and (3.2) in Corollary 3.1 we get

Ω | u ˜ ε σ | 2 + Ω | u ˜ ε σ | 4 Ω | u ˜ ε σ | 2 + Ω f u ˜ ε σ + C Ω ,
(4.22)

which easily implies

Ω | u ˜ ε σ | 4 C,
(4.23)

where C is a constant independent of ε and then (4.18) holds. As a consequence

u ˜ ε σ | u ˜ ε σ | 2 hweakly in  L 4 3 (Ω),

up to a subsequence still denoted by ε. In order to identify h we need to prove

| v ε σ | 2 | u σ | 2 strongly in  L 2 (Ω).
(4.24)

To this aim we can write

| v ε σ | 2 | u σ | 2 L 2 ( Ω ) 2 Ω | v ε σ u σ | 2 ( | v ε σ | 2 + | u σ | 2 ) d x 2 v ε σ u σ L 4 ( Ω ) 2 ( v ε σ L 4 ( Ω ) 2 + u σ L 4 ( Ω ) 2 ) .
(4.25)

By (4.25) and (4.13) we get (4.24) for n=2 and n=3.

Now, since

u ˜ ε σ | u ˜ ε σ | 2 = u ˜ ε σ | v ε σ | 2 ,

by (4.24) and (4.15) we get h=θ u σ | u σ | 2 and then (4.16). So the lemma is completely proved. □

Let us consider the function w λ defined as

w λ ( x 1 , x 2 ,, x n )= { w λ ( x 1 , x 2 , , x n ) , for  x Y c , w λ ( x 1 , x 2 , , x n ) , for a.e.  x l Y Q ,
(4.26)

where w λ is the solution of problem (2.4).

Let us denote

W λ =Φ w λ ( x 1 , x 2 ,, x n ),for xlYQ,

where Φ is the extension operator defined by Lemma 3.1 with C=QlY and R=Y. From problem (2.4) we can note that

W λ (y)=λy+ v λ (y),for yY,

where v λ is an Y-periodic function. Let us denote by

P W λ (y+k)=λ(y+k)+ v λ (y+k)=λy+ v λ (y)+λk= W λ (y)+λk

for yY and for k Z n the periodic extension of W λ (y) on R n . Finally let us set

w ε , λ =ε ( P W λ ) ( x ε ) ,for x R n .
(4.27)

It can be proved that (see (4.6)-(4.19) of [5])

w ε , λ L ( Ω ) c,
(4.28)

where c is a constant independent of ε, and

w ε , λ (x)λx= w λ strongly in  L (Ω),
(4.29)
w ε , λ (x)λx= w λ weakly in  W 1 , p (Ω),
(4.30)

for p such that

2 < p 4 2 2 α , if  n = 2 , 3 < p 6 3 2 α , if  n = 3 , 1 p < + , if  n > 3 ,
(4.31)

with α] 1 2 ,1].

It is easy to see also that (see (4.19) of [5])

w ε , λ W 1 , p ( Ω ) C , 2 < p 4 2 2 α ( 4 2 2 α > 4 ) ,  if  n = 2 , w ε , λ W 1 , p ( Ω ) C , 3 < p 6 3 2 α ( 6 3 2 α > 3 ) ,  if  n = 3 , w ε , λ W 1 , p ( Ω ) C , 1 p < + ,  if  n > 3 .
(4.32)

Let us pose η ε , λ =P W λ ( x ε ) in Ω ε . This function satisfies the problem

{ div η ε , λ = 0 , in  Ω ε , η ε , λ n = 0 , on  T ε ,

whose variational formulation, by the periodicity of w ε , λ , is

Ω η ˜ ε , λ φ=0,φ H 1 (Ω).
(4.33)

Moreover, we have

η ˜ ε , λ (x) A λ weakly  in  ( L 2 ( Ω ) ) n .
(4.34)

4.2 Identification of the limit problem

Let us pose ξ ε σ =( ξ ε , 1 σ , ξ ε , 2 σ ) where ξ ε , i σ = u ε , i σ for i=1,2 and observe that by (2.2) we get

{ div ξ ε , i σ u ε , i σ + | u ε σ | 2 u ε , i σ = f i , in  Ω ε σ , u ε σ = 0 , on  Ω Γ ε D , σ , ξ ε , i σ n = g ε , i , on  Γ ε N , σ ,
(4.35)

for i=1,2. Moreover, by (4.11) we obtain

ξ ˜ ε , i σ ξ , i σ weakly in  ( L 2 ( Ω ) ) n ,
(4.36)

for i=1,2.

Now, let us fix i and observe that problem (4.35) implies

Ω ξ ˜ ε , i σ , φ Γ ε N , σ g ε , i φdσ= Ω u ˜ ε , i σ φ Ω | u ˜ ε , i σ | 2 u ˜ ε σ φ+ Ω χ Ω ε σ f i φ,φ V ε σ .
(4.37)

Let us determine ξ , i σ . For i=1,2 let ϑ g i H 1 ( Y c ) be the function such that

{ Δ ϑ g i = 0 , in  Y c , ϑ g i n = g i , on  Q , ϑ g i , Y -periodic s.t.  m Y c ( ϑ g i ) = 0 ,
(4.38)

and let us pose

γ i = y ϑ g i ,in  Y c .
(4.39)

Let γ ε , i be the function defined as

γ ε , i (x)= γ ˆ i ( x ε ) ,in  R n k Z n ε(Q+k)

for i=1,2. Using (4.38), γ ε , i verifies the following problem:

{ div γ ε , i = 0 , in  k ε ( Y c + k ) , γ ε , i n = g ε , i , on  k ( ε ( Y c + k ) ) .
(4.40)

Moreover, by the periodicity of γ ε , i , we have

γ ˜ ε , i m Y c ( γ i )weakly in  ( L 2 ( Ω ) ) n .
(4.41)

Let us pose ϑ g i ε =ε ϑ ˆ i ( x ε ). The function P ε ϑ g i ε , where { P ε } ε is the sequence of extension operators given by Theorem 3.1, is bounded in H 1 (Ω).

In particular, ϑ g i ε is a null average function, so we obtain

ϑ ˜ g i ε 0strongly in  L 2 (Ω).
(4.42)

Instead of problem (4.37) we can consider the following one:

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) φ = Ω ε σ ( ξ ˜ ε , i σ γ ε , i ) φ = Ω u ˜ ε , i σ φ Ω | u ˜ ε σ | 2 u ˜ ε , i σ φ + Ω χ Ω ε σ f i φ , φ V ε σ .
(4.43)

If we take φ w ε , λ ψ ε σ as a test function in (4.43), where ψ ε σ is given in Lemma 3.2, and φD(Ω), using the extension to zero on the whole Ω, we obtain

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) φ ( w ε , λ ψ ε σ ) + Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ ( φ ψ ε σ ) + Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ ( w ε , λ φ ) = Ω u ˜ ε , i σ ( φ w ε , λ ψ ε σ ) Ω | u ˜ ε σ | 2 u ˜ ε , i σ ( φ w ε , λ ψ ε σ ) + Ω χ Ω ε σ f i ( φ w ε , λ ψ ε σ ) .
(4.44)

Moreover, if we take φ( v ε , i σ ϑ ˜ g i ε ) as test function in problem (4.33), we have

Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) + Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) =0.
(4.45)

By (4.44) and (4.45), we obtain

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) φ ( w ε , λ ψ ε σ ) + Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ ( φ ψ ε σ ) Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) + Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ ( w ε , λ φ ) Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) = Ω u ˜ ε , i ( φ w ε , λ ψ ε σ ) Ω | u ˜ ε σ | 2 u ˜ ε , i σ ( φ w ε , λ ψ ε σ ) + Ω χ Ω ε σ f i ( φ w ε , λ ψ ε σ ) .
(4.46)

Since 0 ψ ε σ 1, and the support of ( ψ ε σ 1) is included in A ε , the second and the third term in our formulation can be majorized.

In fact

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ ( φ ψ ε σ ) Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) = Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ ( φ ψ ε σ ) Ω η ˜ ε , λ φ ( ξ ˜ ε , i σ γ ˜ ε , i ) | Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ φ ( ψ ε σ 1 ) | = | Ω ε σ ( ξ ˜ ε , i σ γ ˜ ε , i ) η ˜ ε , λ φ ( ψ ε σ 1 ) | φ L ( Ω ) ( ξ ε , i σ γ ˜ ε , i ) ( L 2 ( Ω ε σ ) ) n η ε , λ ( L 2 ( A ε Ω ε σ ) ) n C 1 ε n Φ w λ ( L 2 ( A ε ε ) ) n C 2 | Ω | Φ w λ ( L 2 ( A ε ε Y ) ) n .

By the absolute continuity of the integral and as | A ε ε |0, we get Φ w λ ( L 2 ( A ε ε Y ) ) n tending to zero as ε0.

By (3.10) and by (4.9) we obtain the following convergence result:

w ε , λ ψ ε σ w λ strongly in  L 2 (Ω).
(4.47)

Then by (4.36), (4.41), and (4.47), we have, as ε0,

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) φ ( w ε , λ ψ ε σ ) Ω ( ξ , i σ m Y c ( γ i ) ) φ w λ .
(4.48)

By (4.13), (4.34), and (4.42) we get

Ω η ˜ ε , λ φ ( v ε , i σ ϑ ˜ g i ε ) Ω Aλφ u i σ .
(4.49)

Moreover, for n=2 and n=3, by (4.10), (4.14), (4.15), and (4.47) we have

Ω χ Ω ε σ f i ( φ w ε , λ ψ ε σ ) Ω θ f i φ w λ ,
(4.50)
Ω u ˜ ε , i σ ( φ w ε , λ ψ ε σ ) Ω θ u i σ φ w λ .
(4.51)

By (3.10), (4.16), and (4.29) we get

Ω | u ˜ ε σ | 2 u ˜ ε , i σ ( φ w ε , λ ψ ε σ ) Ω θ | u σ | 2 u i σ φ w λ ,
(4.52)

for n=2 and n=3.

Let us consider the fourth term in (4.46).

Let ( ϕ ε ) ε be a sequence in L (Ω) such that

ϕ ε φstrongly in  L (Ω),
(4.53)

and ϕ ε is constant on ε ν B+εk, for every ε,k Z n such that ε(Y+k)Ω.

Since ψ ε σ =1 in Ω A ε and ψ ε σ =0 in Ω A ε , we have

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ ( w ε , λ φ ) = Ω ε ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ ( w ε , λ φ ) = Ω ε φ w ε , λ ( ξ ˜ ε , i σ γ ε , i ) ψ ε σ Ω ε A ε ϕ ε w ε , λ ( ξ ˜ ε , i σ γ ˜ ε , i ) w ε , λ ψ ε σ + Ω ε A ε ( ξ ˜ ε , i σ γ ˜ ε , i ) w ε , λ ψ ε σ ϕ ε = Ω ε A ε ( ξ ˜ ε , i σ γ ˜ ε , i ) w ε , λ ψ ε σ ( φ ϕ ε ) + Ω ε A ε ϕ ε ( ξ ˜ ε , i σ γ ˜ ε , i ) w ε , λ ψ ε σ .
(4.54)

Denoting by P ε the extension operators given by Theorem 3.6, by the properties of the symmetry of ψ ε σ given by (3.12), w ε , λ , and since ϕ ε is constant on Ω ε A ε , we have

Ω ε A ε ϕ ε w ε , λ ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ = Ω ε A ε ϕ ε w ε , λ ( u ε , i σ P ε ϑ g i ε ) ψ ε σ = 1 2 A ε ϕ ε w ε , λ ( P ε u ε , i σ P ε ϑ g i ε ) ψ ε σ .
(4.55)

By (4.54) and (4.55) we obtain

Ω ( ξ ˜ ε , i σ γ ˜ ε , i ) ψ ε σ ( w ε , λ φ ) = Ω ε ( ξ ˜ ε , i σ γ ε , i ) w ε , λ ψ ε σ ( φ ϕ ε ) + 1 2 A ε ϕ ε w ε , λ ( P ε u ε , i σ P ε ϑ g i ε ) ψ ε σ 1 2 A ε φ w ε , λ ( P ε u ε , i σ P ε ϑ g i ε ) ψ ε σ + 1 2 A ε φ w ε , λ ( P ε u ε , i σ P ε ϑ g i ε ) ψ ε σ = Ω ε ( ξ ˜ ε , i σ γ ε , i ) w ε , λ ψ ε σ ( φ ϕ ε ) + 1 2 A ε ( ϕ ε φ ) ( P ε u ε , i σ P ε ϑ g i ε ) w ε , λ ψ ε σ + 1 2 A ε φ ( P ε u ε , i σ P ε ϑ g i ε ) w ε , λ ψ ε σ .
(4.56)

By (3.9), (4.27), (4.36), (4.41), (4.53) we obtain

| Ω ε ( ξ ˜ ε , i σ γ ε , i ) w ε , λ ψ ε σ ( φ ϕ ε ) | φ ϕ ε L ( Ω ) w ε , λ L ( Ω ) ( ξ ˜ ε , i σ γ ˜ ε , i ) ( L 2 ( Ω ) ) n ψ ε σ L 2 ( Ω ) .
(4.57)

The last expression tends to zero as the first term of the product tends to zero and the other ones are bounded. In a similar way we obtain

1 2 A ε (φ ϕ ε ) ( P ε u ε , i σ P ε ϑ g i ε ) w ε , λ ψ ε σ 0.
(4.58)

Let us consider the last term in (4.56) and observe that

1 2 A ε φ ( P ε u ε , i σ P ε ϑ g i ε ) w ε , λ ψ ε σ = 1 2 Ω φ P ε u ε , i σ w ε , λ ψ ε σ 1 2 A ε φ P ε ϑ g i ε w ε , λ ψ ε σ = 1 2 Ω φ v ε , i σ w ε , λ ψ ε σ 1 2 A ε φ P ε ϑ g i ε w ε , λ ψ ε σ .
(4.59)

Let us analyze these two terms separately. We observe that

Ω φ v ε , i σ w ε , λ ψ ε σ = A ε φ v ε , i σ w ε , λ ψ ε σ
(4.60)

and

w ε , λ v ε , i σ φ H 0 1 ( Ω D ε σ ) .
(4.61)

In fact since v ε , i σ =0 on D ε σ and φD(Ω), by Corollary 3.1 we must only prove that w ε , λ v ε , i σ H 1 (Ω). Obviously ( w ε , λ v ε , i σ )= w ε , λ v ε , i σ + v ε , i σ w ε , λ . By (4.28) and since v ε , i σ L 2 (Ω) we have w ε , λ v ε , i σ L 2 (Ω) If n3, by (4.12) and (4.32) it follows that w ε , λ v ε , i σ L r (Ω)r<p. So we have

Ω φ v ε , i σ w ε , λ ψ ε σ = Δ ψ ε σ , v ε , i σ w ε , λ φ Ω φ ψ ε τ ( v ε , i σ w ε , λ )