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

Boundary Value Problems20142014:223

https://doi.org/10.1186/s13661-014-0223-2

  • Received: 9 July 2014
  • Accepted: 19 September 2014
  • Published:

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.

Keywords

  • homogenization
  • harmonic capacity
  • extension operator
  • Ginzburg-Landau equations

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 n 3 ( 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 n 3 ( 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 σ 2 ;

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 1 σ < 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 n 3 ( 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 , 2 l ] × [ 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 R Q R .
(2.1)

Let us observe that Q ( 3 4 R ) R . Let us pose Y c = Y Q . 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 = ε Q 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 σ 1 ; 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
Figure 1

The cell εY : case τ = 0 .

Figure 2
Figure 2

The cell εY : case τ 0 .

Let us define T ε = T ε ( Ω ) = ( { Q ε + ε k : 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 + ε k ) 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 + ε k : 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
Figure 3

The domain Ω ε τ , σ : case τ = 0 .

Figure 4
Figure 4

The domain Ω ε τ , σ : case τ 0 .

Definition 2.1

Let K be a compact subset of R n and Ω an open set such that K Ω . 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
A λ = m Y ( w λ ˜ ) = 1 | Y | Y c w λ d y , 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 , for n = 2 and n = 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)
where A is 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)
if n = 3 , and
μ τ , σ = { 4 π , if  σ < 2 , 2 π , if  σ 2 ,
(2.9)

if n = 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 ( t E , t Ω ) = 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 and C R be a compact set with Lipschitz boundary ∂C and 1 p < . Then there exists a linear bounded extension operator Φ : W 1 , p ( R C ) W 1 , p ( R ) such that
( Φ v ) L P ( R ) c v L P ( R C ) , for every  v W 1 , p ( R C ) .

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

Theorem 3.1

Let Ω 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 every u H 1 ( Ω ε ) and for every ( x 1 , x 2 , , x n ) l Y and k such that ε Y + ε k Ω . Eventually if u = 0 on Ω, then P ε u = 0 on Ω.

As a consequence we get the following result.

Corollary 3.1

Let Ω R n a bounded open set, ε > 0 and 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 = 0 on 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 ( 2 ε k 1 x 1 , x 2 , , x n ) ,
(3.3)

for a.e. x ( { ε ( l Y + 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 + ε k : k Z n } ) Y ε if 0 < ε < ε τ , D ε τ , σ = otherwise.

Let B Y 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 ε Y ).

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 1 σ < 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 1 weakly in  H 1 ( Ω ) ,
(3.9)
ψ ε τ , σ 1 strongly in  L p ( Ω ) , p [ 1 , + [ .
(3.10)
Moreover, for ν 2 2 p , we have
ψ ε τ , σ 0 strongly 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 ( 2 ε k 1 x 1 , x 2 , , x n )
(3.12)

for a.e. x ( { ε ( l Y + 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)
if n 3 , and
μ τ , σ = { 4 π , if  σ < 2 , 2 π , if  σ 2 ,
(3.15)

if n = 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 , φ D ( Ω ) .

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

Step 1. Let n 3 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 d x = 0 , φ D ( Ω ) .
(3.17)
Let us consider φ D ( Ω ) . Let us fix δ > 0 and consider η > 0 such that
| x y | < 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 ε Y + ε k Ω .

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 n 3 , 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 ε Y + ε k Ω , 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 ε 0 , we obtain
lim ε 0 Ω | ψ ε τ , σ | 2 φ = cap ( ( K + τ ) ( K τ ) ) Ω φ ( x ) d x , φ D ( 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 ) d x , φ D ( Ω ) .
Step 2. Let n 3 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 r B 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 δ 0 . 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 ) d x , φ D ( Ω ) .

Step 3. Let n 3 and let us examine the case 1 σ < 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 = 2 cap ( K ) Ω φ ( x ) d x , φ D ( 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 = 0 and lim ε 0 a ε = 0 ,
so that
1 ε 2 cap ( K , exp ( 1 ε σ + 1 ε 2 ) B ) 2 π .
(3.34)
Hence by (3.18) and (3.34) one has
lim ε 0 Ω | ψ ε τ , σ | 2 φ d x = 4 π Ω φ ( x ) d x .
Then we obtain
μ τ , σ , φ = 4 π Ω φ ( x ) d x , φ D ( Ω ) .
Step 5. Finally let n = 2 and let us examine the case σ 2 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 = 0 and lim ε 0 a ε = 0 .
So
1 ε 2 cap ( K ε , σ K ε , σ , exp ( 1 ε α ) B ) 2 π .
Indeed as in the previous case, by Lemma 3.3 of [10], we obtain
μ τ , σ , φ = 2 π Ω φ ( x ) d x , φ D ( 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 that C Y R n , a 1 , and
Ω ε ′′ = Ω ( { ε a C + ε k : k Z n } Y ε ) .
(3.36)

Let Ω 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 0 weakly 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, and φ = ( 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 fixed i = 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

Let n = 2 and n = 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 z weakly 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 h weakly 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  x l Y Q ,
where Φ is the extension operator defined by Lemma 3.1 with C = Q l Y and R = Y . From problem (2.4) we can note that
W λ ( y ) = λ y + v λ ( y ) , for  y Y ,
where v λ is an Y-periodic function. Let us denote by
P W λ ( y + k ) = λ ( y + k ) + v λ ( y + k ) = λ y + v λ ( y ) + λ k = W λ ( y ) + λ k
for y Y 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 ) λ x = w λ strongly in  L ( Ω ) ,
(4.29)
w ε , λ ( x ) λ x = 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 φ d σ = 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 0 strongly 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 φ D ( 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 ε 0 .

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 ε 0 ,
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 A λ φ 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 φ strongly in  L ( Ω ) ,
(4.53)

and ϕ ε is constant on ε ν B + ε k , 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 φ D ( 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 n 3 , 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 ) Ω ψ ε σ w ε , λ v ε , i σ φ .
(4.62)
By the equiboundedness of w ε σ in L ( Ω ) , by (4.13) and since, by (3.9), ψ ε σ 0 weakly in L 2 ( Ω ) , we see that Ω φ ψ ε σ ( v ε , i σ w ε , λ ) tends to zero. We can observe that
v ε , i σ w ε , λ  is equibounded in  L s ( Ω )  with  s > 2 .
(4.63)

In fact if n = 2 , by (4.32) we have the result that w ε , λ is equibounded in L r ( Ω ) with r > 2 . By (4.12) v ε , i σ is equibounded in L q ( Ω ) for 1 q < + and so by (4.63) there exists s > 2 such that v ε σ w ε , λ is equibounded in L s ( Ω ) . By (3.11), ψ ε σ 0 strongly in L s ( Ω ) , with 1 s + 1 s = 1 .

If n = 3 , by (4.32) there exists r such that 3 < r 6 3 2 α ( α ] 1 2 ,