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Homogenization of a Ginzburg-Landau problem in a perforated domain with mixed boundary conditions
Boundary Value Problems volume 2014, Article number: 223 (2014)
Abstract
In this paper we study the asymptotic behavior of a Ginzburg-Landau problem in a ε-periodically perforated domain of 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 , with Lipschitz boundary ∂ Ω and Y be . 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 , a piece of which is on the hyperplane . The latter kind is of size if ( if ) and it is obtained by rescaling a reference hole K strictly contained in Y and in the half plane (and containing a segment on the line if ). Moreover, every element of the second family moves perpendicularly with respect to the -axis towards an element of the first family with an approaching speed of order if ( if ) where , . 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 and we take 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 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:
Let us observe that the equation in (1.1) is known as the Ginzburg-Landau equation.
By using the energy method (see [1]), for and we prove the weakly convergence in of a suitable extension of the sequence of solutions of (1.1) to a function , unique solution of the following limit problem:
where 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 depends both on σ or on the dimension n. More precisely
if , is equal to 4π if and 2π if ;
if , is equal to
-
(a)
the double of the capacity in of the Dirichlet reference hole if ;
-
(b)
the capacity in of a set obtained by perpendicularly translating the Dirichlet reference hole to a distance , from the hyperplane and after doubling it by reflection with respect to the same hyperplane if ;
-
(c)
the capacity in of a set obtained by doubling the Dirichlet reference hole by reflection with respect to the hyperplane if .
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 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 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 obtained by removing from Ω closed smoothed cubes well contained in Ω (the holes) of diameter periodically distributed with period ε in . 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 if ( if ), the Neumann condition is non-homogeneous so the reference hole has to be rescaled with . 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 . On the other side, an extensive study of Ginzburg-Landau equation in a bounded domain Ω of 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 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 with oscillating boundary.
2 Statement of the problem and main result
Let Ω be an open bounded subset of with Lipschitz boundary. Let and such that the cube . For , or we take . For we consider a domain Q which has boundary such that
Let us observe that . Let us pose . Let K be a compact subset of , contained in the half plane and in Y; moreover, if let contain a segment.
Let and let be a null average function.
Let and . Let and
where , , and ; let us observe that there exists such that if then is well contained in εY (see Figure 1 and Figure 2).
Let us define and .
Let be the function defined on by if and .
Let .
Let if , otherwise.
Let . Let , (see Figure 3 and Figure 4).
Definition 2.1
Let K be a compact subset of and Ω an open set such that . We define the (harmonic) capacity of K with respect to Ω, and we will denote by the following quantity:
We will, moreover, denote by the quantity .
Let us consider, for every , the following problem:
whose variational formulation is
where denote the closure of in .
For any , let be the solution of the following problem:
Since is linear in λ and the extension operator to zero is linear, we can consider the matrix given by
where denotes the extension to zero of on the whole Y. In what follows, with we will denote the average of the function u over the subset .
We give the following result.
Theorem 2.1
Let ε be a parameter taking values in a sequence going to zero and letbe the sequence of solutions of problem (2.2). Then there exists a bounded sequenceinextendingand weakly converging in, forand, to the solutionof the following homogenized problem:
or, in the variational formulation,
whereis the constant matrix defined in (2.5), and
if, and
if.
Moreover, any sequence of functions bounded inand extendingconverges to.
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 ofand E, , subsets of Ω. Then
-
(i)
;
-
(ii)
(monotonicity);
-
(iii)
(strong subadditivity);
-
(iv)
if is an increasing sequence of subsets of Ω and , then ;
-
(v)
if is a sequence of subsets of Ω and , then ;
-
(vi)
if and are open subsets of and , then ;
-
(vii)
if is an increasing sequence of open sets such that , then ;
-
(viii)
if , then ;
-
(ix)
if is a decreasing sequence of compact subsets of Ω with , then .
Now we recall two results of [5].
Lemma 3.1
Let R a cube inandbe a compact set with Lipschitz boundary ∂C and. Then there exists a linear bounded extension operatorsuch that
Let , and .
Theorem 3.1
Letbe a bounded open set, , and let Y, l, Q, , , be defined as in Section 2.
Then there exists a familyof uniform extension operators (i.e. in) fromto, such that
for everyand for everyand k such that. Eventually if u = 0 on ∂ Ω, thenon ∂ Ω.
As a consequence we get the following result.
Corollary 3.1
Leta bounded open set, and let Y, l, Q, , , , be defined as in Section 2.
Letsuch thaton.
Then there exists a sequence of uniform extensions of and a constant c independent of ε such that
and
for a.e. and for every k of this kind and related to x.
Proof
Let and let us define
We observe that and therefore . Then the sequence given by meets the requirements by Theorem 3.1. □
Let , and if , otherwise.
Let an open ball centered at the origin, (, if ) and construct the following families of sets:
if and otherwise (let us observe that there exists such that if then ).
Let be the set of functions v∈ such that in a neighborhood of .
Let be the closure of in . Let be the unique solution of the problem
Let us pose . We can consider as a function of , and observe that
Using a result proved in Lemma 3.5 of [5], we can study the behavior, and explicitly calculate, the ‘strange term’ , when σ is equal to , and σ is different from , i.e., when the distance between the small hole and the big one is different from the size of the latter. In fact if the distance between the small hole and the big one is bigger and bigger than the size of the latter; if the condition is just the opposite.
Theorem 3.2
For any fixed ε letthe unique solution of (3.6). Then
Moreover, for, we have
Furthermore, if we restrictto, we get
for a.e. and for every k of this kind and related to x.
Moreover, there existandand a unique distributionsuch that
where
if, and
if.
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 and such that (3.13) holds up to a subsequence.
Now we identify the measure .
Let us recall that by virtue of Proposition 1.1 in [3] up to a subsequence
The proof of (3.14) and (3.15) will be performed in five steps.
Step 1. Let and . At first we observe that
Let us consider the first term in the right-hand side of (3.16); by (3.10) and (3.11), one has
Let us consider . Let us fix and consider such that
Then if is small enough
for every such that .
Moreover, by definition of , taking into account its εY periodicity in and by (viii) of Proposition 3.1, if , we have
Now if is small enough, then and by the last expression and summing up relations (3.18), for every such that , we have
Then by (vii) of Proposition 3.1 and the arbitrariness of δ, passing to the limit for , we obtain
Indeed, using condition (i) in (3.13), one has
Consequently by combining (3.13)(iii), with (3.16), (3.17), and (3.21), we obtain
Step 2. Let and let us consider the case . We set
where . As in the previous case
Let us fix , and . If ε is small enough, we have and . Then
for ε is small enough (dependent on r).
Consequently
On the other hand
Let us observe that is an increasing sequence converging to , and since K is a perfect set. Consequently, by (iv) of Proposition 3.1, it follows that
Therefore by combining (3.24) with (3.25) and (3.26) we get
Since K is a compact set, decreases to as . Then, by (ix) of Proposition 3.1, one has
By passing to the limit, as in (3.27), by (3.28), it follows that
By passing to the limit as in (3.29) we have the result
By arguing as in the previous case, (3.30) provides
Step 3. Let and let us examine the case .
By recalling the expression of the in (3.5), and by following the same arguments as above, one has
By repeating the same steps (3.16)-(3.22), we have
Step 4. Let . Let , and 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
Let us observe that contains a segment and that K is well contained in another ball centered at the origin, say it . So, by (ii) of Proposition 3.1 we have
By Lemma 3.3 of [10] we obtain
where
so that
Hence by (3.18) and (3.34) one has
Then we obtain
Step 5. Finally let and let us examine the case setting
and
Let us suppose that contains a segment. Let us denote by and , respectively, the following subsets:
where with we denote the symmetry of subset E with respect to the hyperplane .
By (ii) and (viii) of Proposition 3.1, and by Lemma 3.3 of [10] we have
Let be another ball centered at the origin and containing
then on the one hand we have
on the other hand
where
So
Indeed as in the previous case, by Lemma 3.3 of [10], we obtain
□
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, , and
Leta bounded open set, , and letdefined as in (3.36).
Letandtwo bounded sequences insuch thata.e. in. Then
Lemma 3.3
Let Ω, Y, l, Q, , , , , be defined as in Section 2, anda function in. Then there exists a constant C, depending on, such that for every fixed
Proof
For every i fixed, let us consider the following problem:
Set where , so that problem (3.39) becomes
whose variational formulation is
By Hölder’s inequality we have
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 the zero extension to the whole Ω of a vector function u defined on a subset of Ω and by the periodic extension to of a vector function u Y-periodic in . Moreover, let be defined as in (3.4).
By Corollary 3.1 there exists a sequence satisfying (3.2). So by Poincaré’s inequality and (3.2) in Corollary 3.1 we get
where is the Poincaré constant of Ω and c is the constant given by Corollary 3.1.
By the Hölder inequality we get
By Lemma 3.3 and (4.2), inequality (4.1) becomes
Now let us consider the following quantity:
Let us suppose
Then (4.3) becomes
which is
where is a constant independent of ε.
Now let us suppose
By (4.2), inequality (4.3) becomes
Then
i.e. again
where is a constant independent of ε.
So there exists a subsequence, still denoted by ε, such that
Obviously, since on , we deduce, by (4.6), that
where is a constant independent of ε.
By the Rellich theorem,
and
up to a subsequence still denoted by ε. Let us observe that
Now we are able to prove the following result.
Lemma 4.1
Letand. Letbe the sequence of solutions of problem (2.2), then
up to a subsequence still denoted by ε.
Proof
By (4.11) we get
Then
up to a subsequence still denoted by ε. Since
by (4.13) and (4.14) we obtain
and then (4.15) holds. In order to prove (4.16) we observe that
where C is a constant independent of ε. Indeed by (4.11) we have also
where is a constant independent of ε, which implies
Multiplying by in (2.2) we get
By (4.20) and (3.2) in Corollary 3.1 we get
which easily implies
where C is a constant independent of ε and then (4.18) holds. As a consequence
up to a subsequence still denoted by ε. In order to identify h we need to prove
To this aim we can write
By (4.25) and (4.13) we get (4.24) for and .
Now, since
by (4.24) and (4.15) we get and then (4.16). So the lemma is completely proved. □
Let us consider the function defined as
where is the solution of problem (2.4).
Let us denote
where Φ is the extension operator defined by Lemma 3.1 with and . From problem (2.4) we can note that
where is an Y-periodic function. Let us denote by
for and for the periodic extension of on . Finally let us set
It can be proved that (see (4.6)-(4.19) of [5])
where c is a constant independent of ε, and
for p such that
with .
It is easy to see also that (see (4.19) of [5])
Let us pose in . This function satisfies the problem
whose variational formulation, by the periodicity of , is
Moreover, we have
4.2 Identification of the limit problem
Let us pose where for and observe that by (2.2) we get
for . Moreover, by (4.11) we obtain
for .
Now, let us fix i and observe that problem (4.35) implies
Let us determine . For let be the function such that
and let us pose
Let be the function defined as
for . Using (4.38), verifies the following problem:
Moreover, by the periodicity of , we have
Let us pose . The function , where is the sequence of extension operators given by Theorem 3.1, is bounded in .
In particular, is a null average function, so we obtain
Instead of problem (4.37) we can consider the following one:
If we take as a test function in (4.43), where is given in Lemma 3.2, and , using the extension to zero on the whole Ω, we obtain
Moreover, if we take as test function in problem (4.33), we have
By (4.44) and (4.45), we obtain
Since , and the support of is included in , the second and the third term in our formulation can be majorized.
In fact
By the absolute continuity of the integral and as , we get tending to zero as .
By (3.10) and by (4.9) we obtain the following convergence result:
Then by (4.36), (4.41), and (4.47), we have, as ,
By (4.13), (4.34), and (4.42) we get
Moreover, for and , by (4.10), (4.14), (4.15), and (4.47) we have
By (3.10), (4.16), and (4.29) we get
for and .
Let us consider the fourth term in (4.46).
Let be a sequence in such that
and is constant on , for every such that .
Since in and in , we have
Denoting by the extension operators given by Theorem 3.6, by the properties of the symmetry of given by (3.12), , and since is constant on , we have
By (4.54) and (4.55) we obtain