NEW RESULTS ON THE ASYMPTOTIC BEHAVIOUR OF DIRICHLET PROBLEMS IN PERFORATED DOMAINS

Transcription

1 NEW RESULTS ON THE ASYMPTOTIC BEHAVIOUR OF DIRICHLET PROBLEMS IN PERFORATED DOMAINS GIANNI DAL MASO ad ADRIANA GARRONI S.I.S.S.A., via Beirut 4, Trieste, Italy Let A be a liear elliptic operator of the secod order with bouded measurable coefficiets o a bouded ope set of R, ad let ( h ) be a arbitrary sequece of ope subsets of. We prove the followig compactess result: there exist a subsequece, still deoted by ( h ), ad a positive Borel measure µ o, ot chargig polar sets, such that, for every f H 1 (), the solutios u h H0 1( h) of the equatios Au h = f i h, exteded to 0 o \ h, coverge weakly i H0 1 () to the uique solutio u H0 1() L2 µ () of the problem Au, v + uv dµ = f, v v H 1 0 () L2 µ(). Whe A is symmetric, this compactess result is already kow ad was obtaied by Γ-covergece techiques. Our ew proof, based o the method of oscillatig test fuctios, exteds the result to the o-symmetric case. The ew techique, which is completely idepedet of Γ -covergece, relies o the study of the behaviour of the solutios wh H1 0 ( h) of the equatios A wh = 1 i h, where A is the adjoit operator. We prove also that the limit measure µ does ot chage if A is replaced by A. Moreover, we prove that µ depeds oly o the symmetric part of the operator A, if the coefficiets of the skew-symmetric part are cotiuous, while a explicit example shows that µ may deped also o the skew-symmetric part of A, whe the coefficiets are discotiuous. 1. Itroductio I this paper we study the asymptotic behaviour of the solutios of elliptic equatios with Dirichlet boudary coditios i perforated domais. Amog the physical motivatios of the problem we metio the applicatios to scatterig theory (see [28], [46]), electrostatic screeig (see [47]), ad heat coductio i domais with a complicated boudary (see [46], [10]). A further motivatio for the study

2 of this problem i the most geeral case, without ay geometric assumptio o the domais, is give by the recet applicatios to a relaxed formulatio of some optimal desig problems (see [1], [6], [14], [5], [26]). Our problem ca be formulated as follows. Let A be a liear elliptic operator of the secod order with bouded measurable coefficiets o a bouded ope set of R, ad let ( h ) be a arbitrary sequece of ope subsets of. For every f H 1 () we cosider the sequece (u h ) of the solutios of the Dirichlet problems u h H 1 0 ( h ), Au h = f i h. (1.1) If we exted u h to by settig u h = 0 o \ h, the (u h ) ca be regarded as a sequece i H 1 0 (). The problem is to describe the asymptotic behaviour of (u h ) as h. The mai result of the paper is the followig compactess theorem (Theorem 4.6), which holds without ay further hypothesis o the geometry of the sets h. For every sequece ( h ) of ope subsets of there exist a subsequece, still deoted by ( h ), ad a positive Borel measure µ o, ot chargig polar sets, such that, for every f H 1 (), the solutios u h of (1.1) coverge weakly i H 1 0 () to the uique solutio u H 1 0 () L 2 µ() of the problem Au, v + uv dµ = f, v v H0 1 () L 2 µ(), (1.2) where, deotes the duality pairig betwee H 1 () ad H 1 0 (). To prove this compactess theorem we observe (Remark 2.3) that all problems of the form (1.1) ca be writte as problems of the form (1.2) for a suitable choice of the measure µ i a special class of positive measures, deoted by M 0 (), which icludes also measures which take the value + o a large family of sets. We prove (Theorem 4.5) that, for every sequece (µ h ) of measures of the class M 0 (), there exist a subsequece, still deoted by (µ h ), ad a measure µ M 0 () such that, for every f H 1 (), the solutios u h H 1 0 () L 2 µ h () of the problems Au h, v + u h v dµ h = f, v v H0 1 () L 2 µ h () (1.3) coverge weakly i H 1 0 () to the uique solutio u H 1 0 () L 2 µ() of (1.2). This more geeral formulatio of the compactess theorem icludes i our framework the problem of the asymptotic behaviour of the solutios of Schrödiger equatios with positive oscillatig potetials. Whe A is symmetric, this compactess result is already kow (see [2], [1], [23], [8], [38]), ad the origial proof is based o Γ-covergece techiques, for which we refer to [1] ad [20]. I this case it is also possible to costruct µ by usig the limits of the capacities of the sets U \ h, whe U varies i the class of all relatively compact ope subsets of (see [7] ad [19]).

3 Uder some special hypotheses o the sequece ( h ), which imply, i particular, that the limit measure µ belogs to H 1 (), the asymptotic behaviour of the solutios (u h ) of (1.1) was studied i [30], [34], [46], [47], [31], [35], [32] by a orthogoal projectio method, i [46], [12], [13] by Browia motio estimates, i [39], [40], [41] by Gree s fuctio estimates, i [17], [15], [16] by meas of oscillatig test fuctios, i [43], [25] by the poit iteractio approximatio, i [4] by capacitary methods. These papers provide also a descriptio of the limit measure µ i terms of the relevat properties of the sets h. The case of radom sets h was studied i [28], [45], [42], [44], [24], [11], [3]. Our ew proof of the compactess theorem holds i the geeral case, eve if the operator A is ot symmetric. The ew method, which is more direct tha the previous oe, ad is completely idepedet of Γ-covergece, is based o the origial techique of the oscillatig test fuctios, which was itroduced by Tartar [50] i the study of homogeizatio problems for elliptic operators, ad was adapted to the case of perforated domais by Cioraescu ad Murat [16]. However, our choice of the test fuctios is ew, ad allows us to avoid ay additioal hypothesis o the sequece ( h ). Our proof relies o the study of the behaviour of the solutios wh of the Dirichlet problems w h H 1 0 ( h ), A w h = 1 i h, (1.4) where A is the adjoit operator. For a complete study of the asymptotic behaviour of the solutios of (1.1) whe A is symmetric ad the sequece (wh ) coverges strogly i H0 1 () we refer to [49]. The mai difficulty of our result lies i the fact that (wh ) is compact oly i the weak topology of H1 0 (). I the geeral case (1.3) we cosider the solutios wh H1 0 () L 2 µ h () of the problems A wh, v + whv dµ h = v dx v H0 1 () L 2 µ h (). (1.5) By a elemetary variatioal estimate the sequece (wh ) is bouded i H1 0 (), ad so we may assume that (wh ) coverges weakly i H1 0 () to some fuctio w. We prove (Sectio 3) that ν = 1 A w is a positive Rado measure o, which belogs to H 1 (), ad thus we ca cosider the measure µ M 0 () defied by dν µ(b) = B w, if cap(b {w = 0}, ) = 0, +, if cap(b {w = 0}, ) > 0. This is the measure which appears i the limit problem (1.2). Sice, by a elemetary variatioal estimate, the sequece (u h ) of the solutios of (1.3) is bouded i H 1 0 (), we may assume also that (u h ) coverges weakly i H 1 0 () to a fuctio u. Moreover, if f L (), by the compariso priciple (Propositio 2.5) the sequece (u h ) is bouded i L (), ad thus u L ().

4 To prove that u is the solutio of (1.2), we show that u h satisfies the equatio Au h, whϕ A wh, u h ϕ = fwhϕ dx u h ϕ dx (1.6) for every ϕ H0 1 () L (). As the differece of the first two terms is cotiuous with respect to the weak covergece of (u h ) ad (wh ), it is easy to take the limit i (1.6) ad to show that Au, w ϕ A w, uϕ = fw ϕ dx uϕ dx (1.7) for every ϕ H 1 0 () L (). The we prove (Lemma 3.5) that (1.7) has a uique solutio u H 1 0 () L (), which coicides with the solutio u H 1 0 () L 2 µ() of (1.2). This cocludes the proof of our compactess result i the case f L (). The case f H 1 () ca be treated by a easy approximatio argumet. If we repeat the proof with A replaced by A, we obtai the same limit measure µ (Theorem 4.3). So far we have cosidered oly the problem of the weak covergece of (u h ) i H 1 0 (). I Sectio 5 we cosider also the problem of the strog covergece of the gradiets (Du h ) i L p (, R ). Usig Meyers estimate [36] ad a geeral result due to Murat [37], we prove (Theorem 5.1) that, without ay additioal hypothesis, the sequece (Du h ) coverges to Du strogly i L p (, R ) for every 1 p < 2. To obtai strog covergece of the gradiets i L 2 (, R ) we costruct a corrector term P h (x, s), x, s R, which depeds o the sequece (µ h ), but is idepedet of f, u, u h. We prove (Theorem 5.2) that for every f L () we have Du h (x) = Du(x) + P h (x, u(x)) + R h (x) a.e. i, where the remaiders R h ted to 0 strogly i L 2 (, R ). This improves the corrector results of [16] ad [29], which assume that µ H 1 (), ad those of [27], which assume that w > 0 a.e. i. The corrector P h (x, s) is costructed explicitly i terms of the solutios of (1.4) or (1.5), with A replaced by A. If these fuctios coverge strogly i H0 1 (), we recover (Corollary 5.8) the result of [49]. I the last sectio we study the problem of the depedece of µ o the skewsymmetric part of the operator A. Extedig a result of [16], we prove (Theorem 6.1) that the limit measure µ depeds oly o the symmetric part of the operator A, if the coefficiets of the skew-symmetric part are cotiuous. Fially, we costruct a explicit example, which shows that µ may deped also o the skew-symmetric part of A, whe the coefficiets are discotiuous.

5 2. Notatio ad Prelimiaries Let be a bouded ope subset of R, 2. We deote by H 1,p () ad H 1,p 0 (), 1 p < +, the usual Sobolev spaces, ad by H 1,q (), 1/q + 1/p = 1, the dual of H 1,p 0 (). Whe p = 2 we adopt the stadard otatio H1 (), H0 1 (), ad H 1 (). O H0 1 () we cosider the orm u 2 H 1 0 () = Du 2 dx. By L p µ(), 1 p +, we deote the usual Lebesgue space with respect to the measure µ. If µ is the Lebesgue measure, we use the stadard otatio L p (). For every subset E of the (harmoic) capacity of E i, deoted by cap(e, ), is defied as the ifimum of Du 2 dx over the set of all fuctios u H 1 0 () such that u 1 a.e. i a eighbourhood of E. We say that a property P(x) holds quasi everywhere (abbreviated as q.e.) i a set E if it holds for all x E except for a subset N of E with cap(n, ) = 0. The expressio almost everywhere (abbreviated as a.e.) refers, as usual, to the Lebesgue measure. A fuctio u: R is said to be quasi cotiuous if for every ε > 0 there exists a set A, with cap(a, ) < ε, such that the restrictio of u to \A is cotiuous. It is well kow that every u H 1 () has a quasi cotiuous represetative, which is uiquely defied up to a set of capacity zero. I the sequel we shall always idetify u with its quasi cotiuous represetative, so that the poitwise values of a fuctio u H 1 () are defied quasi everywhere. We recall that, if a sequece (u h ) coverges to u i H 1 0 (), the a subsequece of (u h ) coverges to u q.e. i. For all these properties of quasi cotiuous represetatives of Sobolev fuctios we refer to [51], Sectio 3. A subset A of is said to be a quasi ope if for every ε > 0 there exists a ope subset U ε of, with cap(u ε, ) < ε, such that A U ε is ope. Lemma 2.1. For every quasi ope subset A of there exists a icreasig sequece (v h ) of o-egative fuctios of H0 1 () covergig to 1 A poitwise q.e. i. Proof. This lemma is a easy cosequece of a more geeral result proved i [18], Lemma 1.5. For the reader s coveiece, we give here the easy proof i this particular case. Let A be a quasi ope subset of. The there exists a sequece (U k ) of ope subsets of, with cap(u k, ) < 1/k, such that the sets A k = A U k

6 are ope. Therefore, for every k N there exists a icreasig sequece (ϕ k h ) h of o-egative fuctios of C0 () covergig to 1 Ak poitwise q.e. i. Sice cap(u k, ) < 1/k, for every k N there exists u k H0 1 () such that u k 1 q.e. i U k, u k 0 q.e. i, ad Du k 2 dx < 1/k. This implies that a subsequece of (u k ) coverges to 0 q.e. i. Moreover, as ϕ k h 1 A k, we have (ϕ k h u k) + 1 A q.e. i. Let us defie v h = max 1 k h (ϕk h u k ) +, ψ = sup v h. h The v h H 1 0 (), v h 0 i, the sequece (v h ) is icreasig, ad ψ 1 A q.e. i. For every h k we have v h ϕ k h u k. As A A k, we get ψ 1 u k q.e. i A. Takig the limit as k alog a suitable subsequece, we obtai ψ 1 q.e. i A. This shows that ψ = 1 A ad cocludes the proof of the lemma. By a Borel measure o we mea a positive, coutably additive set fuctio defied i the Borel σ -field of ad with values i [0, + ]. By a Rado measure o we mea a Borel measure which is fiite o every compact subset of. We deote by M 0 () the set of all Borel measures µ o such that (i) µ(b) = 0 for every Borel set B with cap(b, ) = 0, (ii) µ(b) = if{µ(a) : A quasi ope, B A} for every Borel set B. This defiitio differs from the defiitio used i [22] ad [23], where coditio (ii) is ot preset. Our class M 0 () coicides with the class M 0() itroduced i [19] ad used i [8]. We refer to [19] for a compariso betwee these defiitios. It is well kow that every Rado measure satisfies (ii), while there are examples of Borel measures which satisfy (i), but do ot satisfy (ii). For every closed set E we deote by E the measure of the class M 0 () defied by { 0, if cap(b E, ) = 0, E (B) = (2.1) +, otherwise. We shall see i Theorem 4.6 that the measures E will be useful i the study of the asymptotic behaviour of sequeces of Dirichlet problems i varyig domais. Fially, we say that a Rado measure ν o belogs to H 1 () if there exists f H 1 () such that f, ϕ = ϕ dν ϕ C 0 (), (2.2) where, deotes the duality pairig betwee H 1 () ad H0 1 (). We shall always idetify f ad ν. Note that, by the Riesz theorem, for every positive fuctioal f H 1 (), there exists a Rado measure ν such that (2.2) holds. It is well kow that every Rado measure which belogs to H 1 () belogs also to M 0 () (see [51], Sectio 4.7).

7 Let A: H 1 () H 1 () be a elliptic operator of the form Au = D i (a ij D j u), (2.3) where (a ij ) is a matrix of fuctios of L () satisfyig, for a suitable costat α > 0, the ellipticity coditio a ij (x)ξ j ξ i α ξ 2 (2.4) for a.e. x ad for every ξ R. Let A : H 1 () H 1 () be the adjoit operator, defied by A u = D i (a ji D j u) for every u H 1 (). It is well kow that A u, v = Av, u for every u, v H0 1 (). Let µ M 0 () ad f H 1 (). We shall cosider the followig relaxed Dirichlet problem (see [22] ad [23]): fid u H0 1 () L 2 µ() such that Au, v + uv dµ = f, v v H0 1 () L 2 µ(). (2.5) The ame is motivated by Theorem 4.6 ad by the desity results proved i [22] ad [21]. Theorem 2.2. (2.5). For every f H 1 () there exists a uique solutio of problem Proof. The proof is a straightforward applicatio of the Lax-Milgram lemma, see, e.g., [22], Theorem 2.4. By the ellipticity coditio (2.4), if we take u as test fuctio i (2.5), we obtai the followig estimate u H 1 0 () 1 α f H 1 (). (2.6) A coectio betwee classical Dirichlet problems ad relaxed Dirichlet problems (2.5) is give by the followig remark. Remark 2.3. It is easy to see that, if E is a closed set ad µ = E, the u H0 1 () is the solutio of problem (2.5) if ad oly if u = 0 q.e. i E ad u is the solutio i \E of the classical boudary value problem u H 1 0 (\E), Au = f i \E. The solutios of relaxed Dirichlet problems satisfy a compariso priciple.

8 Propositio 2.4. Let µ M 0 (), let f H 1 (), ad let u H0 1 () L 2 µ() be the solutio of problem (2.5). If f 0 i, the u 0 q.e. i. Proof. The proof is give i [22], Propositio 2.9, i a more geeral cotext. For the sake of completeess we give the proof i this simple case. Let v = (u 0). The v is a o-egative fuctio of H0 1 () L 2 µ(). Sice uv 0 q.e. i ad f, v 0, takig v as test fuctio i (2.5) we obtai Au, v 0. As Dv = Du a.e. i {v > 0} ad Dv = 0 a.e. i {v = 0}, we have that Av, v = Au, v 0. By the ellipticity assumptio we obtai v = 0 q.e. i, hece u 0 q.e. i. Propositio 2.5. Let f 1, f 2 H 1 () ad let µ 1, µ 2 M 0 (). Let u 1, u 2 H 1 0 () be the solutios of problem (2.5) correspodig to f 1, µ 1 ad to f 2, µ 2. If 0 f 1 f 2 ad µ 2 µ 1 i, the 0 u 1 u 2 q.e. i. Proof. This result is proved i [22], Propositio For the reader s coveiece we give here the complete proof. By Propositio 2.4 we have that u 1 0 q.e. i ad u 2 0 q.e. i. Let v = (u 1 u 2 ) +. Sice 0 v u 1 ad µ 2 µ 1, we have v L 2 µ 1 () L 2 µ 2 (). As u 2v dµ 2 u 2v dµ 1, takig v as test fuctio i the problems solved by u 1 ad u 2 ad subtractig the correspodig equatios, we obtai A(u 1 u 2 ), v + (u 1 u 2 )v dµ 1 f 1 f 2, v 0. Sice (u 1 u 2 )v 0 q.e. i, by the ellipticity coditio (2.4) we have α v 2 H 1 0 () Av, v = A(u 1 u 2 ), v 0. Thus v = 0 q.e. i ad, cosequetly, u 1 u 2 q.e. i. The followig result will be useful i the sequel. Propositio 2.6. Let ν be a positive Rado measure o which belogs to H 1 () ad let u be the solutio i H0 1 () L 2 µ() of problem (2.5) correspodig to f = ν. The Au, v v dν for every v H 1 0 () with v 0 q.e. i. Proof. This propositio is proved i [22], Propositio 2.6, uder more geeral hypotheses. Here we sketch the proof oly i our particular case. Let v H0 1 () with v 0 q.e. i ad let v h = ( 1 hv) u. Sice u 0 (Propositio 2.4), we have that v h 0 q.e. i ad v h H0 1 () L 2 µ(). The, takig v h as test fuctio

9 i problem (2.5) with f = ν, we obtai Au, v h v h dν. Sice Dv h = 1 h Dv i {v < hu} ad Dv h = Du i {v hu}, we have 1 a h ij D j ud i v ) dx + {v<hu} v h dν 1 h {v hu} a ij D j ud i u ) dx v dν. By eglectig the secod term, which is o-egative by the ellipticity assumptio, we get a ij D j ud i v ) dx v dν. {v<hu} Takig the limit as h, we obtai {u>0} a ij D j ud i v ) dx v dν. Sice u 0 q.e. i ad D j u = 0 a.e. i {u = 0}, the coclusio follows. 3. A Covex Set I this sectio we shall study the properties of the set K() of all fuctios w H 1 0 () such that w 0 q.e. i ad Aw 1 i i the sese of H 1 (). It is easy to see that K() is a closed covex subset of H 1 0 (). Moreover, for every w K() we have α Dw 2 dx Aw, w w dx. This shows that K() is bouded, ad hece weakly compact, i H 1 0 (). Let w 0 be the solutio of the Dirichlet problem w 0 H 1 0 (), Aw 0 = 1. By the maximum priciple we have w w 0 q.e. i for every w K(). As w 0 L () (see [48]), the set K() is bouded i L (). Give w K(), let ν = 1 Aw. By the defiitio of K() we have ν 0 i i the sese of distributios, hece ν is a positive Rado measure. As Aw H 1 (), we have also ν H 1 (). We shall see that, if w K(), the w ca be characterized as the solutio of a particular relaxed Dirichlet problem. To this aim we eed some prelimiary results.

10 Propositio 3.1. Let µ M 0 () ad let u H0 1 () L 2 µ(). For every h N let u h H0 1 () L 2 µ() be the solutio of the problem Au h, v + u h v dµ + h (u h u)v dx = 0 v H0 1 () L 2 µ(). (3.1) The (u h ) coverges to u strogly i H0 1 () ad i L 2 µ(). Moreover ( ) lim Au h, u h + u 2 h dµ + h (u h u) 2 dx = Au, u + u 2 dµ. (3.2) h Proof. hece Takig v = u h u as test fuctio i (3.1) we obtai Au h, u h u + u h (u h u) dµ + h (u h u) 2 dx = 0, (3.3) A(u h u), u h u + = Au, u h u From the ellipticity coditio (2.4) we get (u h u) 2 dµ + h (u h u) 2 dx = u(u h u) dµ. hece α u h u 2 H0 1() + u h u 2 L 2 µ () + h u h u 2 L 2 () Au, u h u u(u h u) dµ, (3.4) α u h u 2 H 1 0 () + u h u 2 L 2 µ () + h u h u 2 L 2 () Au H 1 () u h u H 1 0 () + u L 2 µ () u h u L 2 µ (). By usig the Cauchy iequality we obtai α 2 u h u 2 H 1 0 () u h u 2 L 2 µ () + h u h u 2 L 2 () 1 2α Au 2 H 1 () u 2 L 2 µ (). This shows that (u h ) coverges to u weakly i H0 1 () ad i L 2 µ(). By (3.4) this implies that (u h ) coverges to u strogly i H0 1 () ad i L 2 µ(). Fially (3.3) gives Au h, u h + u 2 h dµ + h (u h u) 2 dx = Au h, u + u h u dµ, which proves (3.2).

11 Lemma 3.2. Let µ M 0 () ad let w H0 1 () L 2 µ() be the solutio of the problem Aw, v + wv dµ = v dx v H0 1 () L 2 µ(). The µ(b) = + for every Borel subset B of with cap(b {w = 0}, ) > 0. Proof. Let u H 1 0 () L 2 µ(), with 0 u 1 q.e. i, ad, for every h N, let u h H 1 0 () L 2 µ() be the solutio of the problem Au h, v + u h v dµ + h u h v dx = h uv dx v H0 1 () L 2 µ(). By the compariso priciple (Propositio 2.5) we have 0 u h hw q.e. i, hece u h = 0 q.e. i {w = 0}. Sice, by Propositio 3.1, (u h ) coverges to u i H0 1 (), we have u = 0 q.e. i {w = 0}. Let U be a quasi ope subset of such that µ(u) < +. By Lemma 2.1 there exists a icreasig sequece (z h ) i H0 1 () covergig to 1 U poitwise q.e. i ad such that 0 z h 1 U q.e. i for every h N. As µ(u) < +, each fuctio z h belogs to L 2 µ(), hece z h = 0 q.e. o {w = 0} by the previous step. This implies that cap(u {w = 0}, ) = 0. Let us cosider a Borel set B with cap(b {w = 0}, ) > 0. For every quasi ope set U cotaiig B we have cap(u {w = 0}, ) > 0, hece µ(u) = + by the previous step of the proof. By the defiitio of M 0 () we coclude that µ(b) = +. Lemma 3.3. Let λ ad µ be measures of M 0 (). Assume that there exists a fuctio w H0 1 () L 2 λ () L2 µ() such that The λ = µ. Aw, v + Aw, v + wv dλ = wv dµ = v dx v H 1 0 () L 2 λ(), (3.5) v dx v H 1 0 () L 2 µ(). (3.6) Proof. Let us cosider the measures λ 0 ad µ 0 defied for every Borel set B by λ 0 (B) = w dλ, µ 0 (B) = w dµ. B Let us prove that λ 0 = µ 0. For every ε > 0 let λ ε ad µ ε be the measures defied by λ ε (B) = w dλ, µ ε (B) = w dµ. B {w>ε} B B {w>ε}

12 To prove that λ 0 = µ 0 it is eough to show that λ ε = µ ε for every ε > 0. Let us fix ε > 0. As w L 2 λ () L2 µ(), λ ε ad µ ε are bouded measures. Therefore it is eough to show that λ ε (U) = µ ε (U) for every ope subset U of. Let us fix U ad let U ε = U {w > ε}. As U ε is quasi ope, by Lemma 2.1 there exists a icreasig sequece (z h ) i H0 1 () covergig to 1 Uε poitwise q.e. i ad such that 0 z h 1 Uε q.e. i. As w L 2 λ () L2 µ() ad w > ε q.e. i U ε, we have λ(u ε ) < + ad µ(u ε ) < +, hece z h L 2 λ () L2 µ(). From (3.5) ad (3.6) we get wz h dλ = wz h dµ. Takig the limit as h we obtai λ ε (U) = w dλ = U ε w dµ = µ ε (U). U ε This shows that λ ε = µ ε for every ε > 0, hece λ 0 = µ 0. For every Borel set B {w > 0} we have 1 λ(b) = w dλ 1 0 = B w dµ 0 = µ(b). B If B is Borel set cotaied i {w = 0} ad cap(b, ) > 0, the λ(b) = µ(b) = + by Lemma 3.2. If cap(b, ) = 0, the λ(b) = µ(b) = 0 by the defiitio of M 0 (). Therefore λ(b) = λ(b {w > 0})+λ(B {w = 0}) = µ(b {w > 0})+ µ(b {w = 0}) = µ(b) for every Borel set B. We give ow the characterizatio of K() i terms of relaxed Dirichlet problems. Propositio 3.4. A fuctio w H0 1 () belogs to K() if ad oly if there exists µ M 0 () such that w H0 1 () L 2 µ() ad Aw, v + wv dµ = v dx v H0 1 () L 2 µ(). (3.7) The measure µ M 0 () is uiquely determied by w K(). More precisely, for every w K() ad for every Borel set B we have dν, if cap(b {w = 0}, ) = 0, µ(b) = B w (3.8) +, if cap(b {w = 0}, ) > 0, where ν is the measure of H 1 () defied by ν = 1 Aw. Moreover, we have ν(b {w > 0}) = w dµ (3.9) for every Borel set B. B

13 Proof. We follow the lies of the proof of Theorem 1 of [14]. Let µ M 0 () ad let w H0 1 () L 2 µ() be a solutio of (3.7). The w 0 q.e. i by Propositio 2.4 ad Aw 1 i by Propositio 2.6, hece w K(). Coversely, assume that w K() ad let µ be the measure defied by (3.8). Let us prove that µ M 0 (). Sice ν H 1 (), we have µ(b) = 0 for every Borel set B with cap(b, ) = 0. It remais to prove that µ(b) = if{µ(a) : A quasi ope, B A} (3.10) for every Borel set B with µ(b) < +. For every h N let µ h measure o defied by µ h (B) = µ(b {w > 1 h }). Note that be the µ h () = µ({w > 1 h }) h ν({w > 1 h }) h2 w dν = h 2 1 Aw, w < +. Let us fix a Borel set B with µ(b) < +. By the defiitio of µ we have cap(b {w = 0}, ) = 0. For every h 2 let B h = B { 1 h < w 1 h 1 }, ad let B 1 = {1 < w}, so that µ(b) = h µ(b h). Sice µ h () < +, for every ε > 0 ad for every h N there exists a ope set U h, with B h U h, such that µ h (U h ) < µ h (B h ) + ε2 h = µ(b h ) + ε2 h. Let A h = U h {w > 1 h }. As w is quasi cotiuous, the set A h is quasi ope. Moreover B h A h ad µ(a h ) = µ h (U h ) < µ(b h ) + ε2 h. Let A 0 = B {w = 0} ad let A be the uio of all sets A h for h 0. The A is quasi ope, cotais B, ad µ(a) < µ(b) + ε. Sice ε > 0 is arbitrary, this proves (3.10). Let us prove that w is a solutio of (3.7). By (3.8) we have w 2 dµ = {w>0} w 2 dµ = {w>0} w dν = 1 Aw, w < +, hece w L 2 µ(). Let v H 1 0 () L 2 µ(). By (3.8) we have v = 0 q.e. i {w = 0}. By the defiitios of µ ad ν we have Aw, v + = Aw, v + wv dµ = Aw, v + {w>0} v dν = Aw, v + {w>0} wv dµ = v dν = which proves (3.7). The uiqueess of µ follows from Lemma 3.3. Property (3.9) is a easy cosequece of (3.8). v dx, The followig lemma will be crucial i the proof of Theorem 4.3.

14 Lemma 3.5. Let µ M 0 () ad let f L (). Let u H0 1 () L 2 µ() ad w H0 1 () L 2 µ() be the solutios of the problems Au, v + uv dµ = fv dx v H0 1 () L 2 µ(), (3.11) A w, v + w v dµ = v dx v H0 1 () L 2 µ(), (3.12) The u is the uique solutio i H0 1 () L () of the problem Au, w ϕ A w, uϕ = fw ϕ dx uϕ dx ϕ C0 (). (3.13) Proof. First of all, we ote that (3.13) ca be writte as a ij D j ud i ϕ ) w dx a ij D j ϕd i w ) u dx = = fw ϕ dx uϕ dx ϕ C 0 (). (3.14) Let w H0 1 () L 2 µ() be the solutio of (3.7). By the compariso priciple (Theorem 2.5) we have u c w q.e. i, with c = f L (). Sice w is bouded, this implies that u L (). Let ν = 1 A w. By Propositio 2.6 ν is a o-egative Rado measure. By Lemma 3.4 (applied to A ) we have that ν (B {w > 0}) = B w dµ (3.15) for every Borel set B. As w L 2 µ(), we have w ϕ H 1 0 () L 2 µ() for every ϕ C 0 (). As u L 2 µ(), by Lemma 3.2 (applied to A ) we have u = 0 q.e. i {w = 0}. Therefore (3.15) implies that uw ϕ dµ = {w >0} uϕ dν = uϕ dν. Takig v = w ϕ i (3.11) we obtai Au, w ϕ + uϕ dν = fw ϕ dx for every ϕ C 0 (). As ν = 1 A w, we coclude that u is a solutio i H 1 0 () L () of (3.13).

15 Let us prove that the solutio of (3.13) is uique. First of all we observe that, by a easy approximatio argumet, (3.13) holds for every ϕ H0 1 () L (). Sice the equatio is liear i u, it is eough to cosider the case f = 0. Let us fix a solutio z H0 1 () L () of (3.13) with f = 0. By (3.14) we have that a ij D j zd i v ) w dx a ij D j vd i w ) z dx + zv dx = 0 for every v H0 1 () L (). Takig v = z we obtai a ij D j zd i z ) w dx a ij D j zd i w ) z dx + As zd j z = 1 2 D j(z 2 ) ad ν 0 we have a ij D j zd i w ) z dx = 1 2 A w, z ad so (3.16) gives a ij D j zd i z ) w dx z 2 dx = 0. (3.16) z 2 dx, z 2 dx 0. (3.17) Sice w 0 q.e. i (Propositio 2.4), (3.17) ad the ellipticity coditio (2.4) imply that z = 0 a.e. i. This cocludes the proof of the uiqueess. 4. The γ A -Covergece ad the Compactess Theorem I this sectio we itroduce the otio of γ A -covergece i M 0 (), related to the covergece of the solutios of the correspodig relaxed Dirichlet problems. Whe A is the Laplace operator, this otio is defied i [23] i terms of the Γ-covergece of the fuctioals Du 2 dx + u2 dµ associated with the relaxed Dirichlet problems. For the extesio of this defiitio to the case of symmetric operators see [7] ad [19]. The defiitio give here ivolves oly the solutios of (2.5), ad coicides with the previous oes i the symmetric cases. Defiitio 4.1. Let (µ h ) be a sequece of measures of M 0 () ad let µ M 0 (). We say that (µ h ) γ A -coverges to µ (i ) if for every f H 1 () the solutios u h H0 1 () L 2 µ h () of the problems Au h, v + u h v dµ h = f, v v H0 1 () L 2 µ h () (4.1) coverge weakly i H0 1 (), as h, to the solutio u H0 1 () L 2 µ() of the problem Au, v + uv dµ = f, v v H0 1 () L 2 µ(). (4.2) We uderlie the fact that the γ A -limit depeds o the operator A. This fact will be discussed later i Sectio 6.

16 Remark 4.2. Sice A is liear ad the solutios of (4.1) deped cotiuously o the data, uiformly with respect to h (see the estimate (2.6)), a sequece (µ h ) γ A -coverges to µ if ad oly if the solutios of (4.1) coverge weakly i H0 1 () to the solutio of (4.2) for every f i a dese subset of H 1 (). Let (µ h ) be a sequece of measures of the class M 0 () ad let µ M 0 (). Let w h H 1 0 () L 2 µ h () ad w H 1 0 () L 2 µ() be the solutios of the problems Aw h, v + w h v dµ h = v dx v H0 1 () L 2 µ h (), (4.3) Aw, v + wv dµ = v dx v H0 1 () L 2 µ(), (4.4) ad let w h H1 0 () L 2 µ h () ad w H 1 0 () L 2 µ() be the solutios of the correspodig problems for the adjoit operator A. We are ow i a positio to characterize the γ A -covergece of a sequece of measures (µ h ) i terms of the weak covergece i H 1 0 () of the sequeces (w h ) ad (w h ). Theorem 4.3. Let (µ h ) be a sequece of measures of M 0 () ad let µ M 0 (). Let w h H0 1 () L 2 µ h () ad w H0 1 () L 2 µ() be the solutios of problems (4.3) ad (4.4), ad let wh H1 0 () L 2 µ h (), w H0 1 () L 2 µ() be the solutios of the correspodig problems for A. The followig coditios are equivalet: (a) (w h ) coverges to w weakly i H 1 0 (); (b) (w h ) coverges to w weakly i H 1 0 (); (c) (µ h ) γ A -coverges to µ; (d) (µ h ) γ A -coverges to µ. Proof. (b) (c). Give f L (), let u h H 1 0 () L 2 µ h () ad u H 1 0 () L 2 µ() be the solutios of the problems (4.1) ad (4.2). By Lemma 3.5 ad by (3.14) we have a ij D j u h D i ϕ ) wh dx a ij D j ϕd i wh) uh dx = = fwhϕ dx u h ϕ dx ϕ C 0 (). (4.5) By the estimate (2.6) the sequece (u h ) is bouded i H 1 0 (), so we may assume that (u h ) coverges weakly i H 1 0 () to some fuctio ũ. By the compariso priciple (Theorem 2.5) we have u h c w h q.e. i, with c = f L (). Takig the

17 limit as h we get ũ c w q.e. i, ad hece ũ L (). Moreover, takig the limit i (4.5) we obtai that ũ satisfies (3.14), ad so ũ = u by Lemma 3.5. Therefore (µ h ) γ A -covereges to µ by Remark 4.2. (c) (a). It is eough to take f = 1 i the defiitio of γ A -covergece. (a) (d). It is eough to replace A by A i the proof of (b) (c). (d) (b). It is eough to take f = 1 i the defiitio of γ A -covergece. Remark 4.4. The uiqueess of the γ A -limit is a easy cosequece of Theorem 4.3, which implies that, if (µ h ) γ A -coverges to λ ad µ, the w satisfies (3.5) ad (3.6), so that λ = µ by Lemma 3.3. The followig theorem proves the compactess of M 0 () with respect to γ A -covergece. Theorem 4.5. Every sequece of measures of M 0 () cotais a subsequece. γ A -coverget Proof. Let (µ h ) be a sequece of measures of M 0 () ad, for every h N, let w h H0 1 () L 2 µ h () be the solutio of problem (4.3). By Propositio 3.4 we have w h K(). Sice K() is compact i the weak topology of H0 1 (), a subsequece of (w h ) coverges weakly i H0 1 () to some fuctio w K(). By Propositio 3.4 there exists a measure µ M 0 () such that w is a solutio i H0 1 () L 2 µ() of problem (4.4). The coclusio follows ow from Theorem 4.3. The case of Dirichlet problems i perforated domais is cosidered i the followig theorem. Theorem 4.6. Let ( h ) be a arbitrary sequece of ope subsets of. The there exist a subsequece, still deoted by ( h ), ad a measure µ M 0 () such that, for every f H 1 (), the solutios u h H0 1 ( h ) of the equatios Au h = f i h, exteded to 0 o \ h, coverge weakly i H0 1 () to the uique solutio u H0 1 () L 2 µ() of problem (4.2). Proof. The coclusio follows easily from the compactess theorem (Theorem 4.5) ad from the fact that each fuctio u h ca be regarded as the solutio of problem (4.1) with µ h = \h (Remark 2.3). Usig Theorem 4.3 we ca prove the followig desity result i M 0 (). We shall see i Corollary 5.8 that the strog coveregece i H 1 0 () of the sequece (w h ) implies the strog coveregece i H 1 0 () of the sequece (u h ) of the solutios of (4.1) for every f H 1 ().

18 Propositio 4.7. Every measure µ M 0 () is the γ A -limit of a sequece (µ h ) of Rado measures of M 0 () such that the solutios w h of (4.3) coverge strogly i H 1 0 () to the solutio w of (4.4). Proof. By (3.8) a measure µ M 0 () is a Rado measure if the solutio w of (4.4) satisfies if w > 0 for every compact set K. (4.6) K Now let w 0 H0 1 () be the solutio of the equatio Aw 0 = 1 i. By the strog maximum priciple (see [48]) we have that w 0 satisfies (4.6). Let us fix µ M 0 () ad let w K() be the solutio of (4.4). For every h N let us defie w h = (1 1 h )w + 1 h w 0. It is easy to see that w h is a positive subsolutio of the equatio Au = 1, hece w h K(). Moreover the fuctios w h satisfy (4.6) ad coverge to w strogly i H0 1 (). Therefore the measures µ h M 0 () associated with w h by Propositio 3.4 are Rado measures ad γ A -coverge to µ by Theorem 4.3. The followig propositio deals with the case where also f varies. Propositio 4.8. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (). Let (f h ) be a sequece i H 1 () covergig strogly to f H 1 (). For every h N let v h H0 1 () L 2 µ h () be the solutio of the problem Av h, v + v h v dµ h = f h, v v H 1 0 () L 2 µ h () ad let u H 1 0 () L 2 µ() be the solutio of problem (4.2). The (v h ) coverges to u weakly i H 1 0 (). Proof. For every h N, let u h be the solutio i H 1 0 () L 2 µ h () of problem (4.1). By the estimate (2.6) ad by the liearity of the problem the sequece (v h u h ) coverges to 0 strogly i H 1 0 (). Moreover, by the defiitio of γ A -covergece, (u h ) coverges to u weakly i H 1 0 (). Therefore (v h ) coverges to u weakly i H 1 0 (). The followig results (Theorem 4.9, Theorem 4.10, Corollary 4.11) show the local character of the γ A -covergece. Let ω be a ope subset of. With a little abuse of otatio we still deote by A the operator defied by (2.3) o H 1 (ω), ad by, the duality pairig betwee H 1 (ω) ad H 1 0 (ω).

19 Theorem 4.9. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig i to a measure µ M 0 (). Let ω be a ope subset of, let (f h ) be a sequece i H 1 (ω) covergig to f strogly i H 1 (ω), ad let (u h ) be a sequece i H 1 (ω) covergig to u weakly i H 1 (ω). Suppose that u h L 2 µ h (ω ) for every ω ω ad that Au h, v + u h v dµ h = f h, v (4.7) ω for every v H0 1 (ω) L 2 µ h (ω) with supp(v) ω. The u L 2 µ(ω ) for every ω ω ad Au, v + uv dµ = f, v (4.8) ω for every v H0 1 (ω) L 2 µ(ω) with supp(v) ω. Proof. Let ϕ C0 (ω) ad let z h = ϕu h. Sice for every v H0 1 () we have a ij D j z h D i v ) dx = a ij D j ϕd i v ) u h dx + + a ij D j u h D i v ) ϕ dx = a ij D j ϕd i v ) u h dx + + a ij D j u h D i (vϕ) ) dx a ij D j u h D i ϕ ) v dx, from (4.7) we obtai Az h, v + z h v dµ h = g h, v v H 1 0 () L 2 µ h (), where g h, v = f h, vϕ + a ij D j ϕd i v ) u h dx a ij D j u h D i ϕ ) v dx for every v H0 1 (). Sice (u h ) coverges to u weakly i H0 1 (ω), (f h ) coverges to f strogly i H 1 (ω), ad ϕ has compact support i ω, it follows that (g h ) coverges strogly i H 1 () to the fuctioal g H 1 () defied by g, v = f, vϕ + a ij D j ϕd i v ) u dx a ij D j ud i ϕ ) v dx for every v H0 1 (). As (µ h ) γ A -coverges to µ ad (z h ) coverges to z = ϕu weakly i H0 1 (), by Propositio 4.8 the fuctio z = ϕu is the solutio i H0 1 () L 2 µ() of the problem Az, v + zv dµ = g, v v H0 1 () L 2 µ(). (4.9) Let us fix a ope set ω ad a fuctio v H0 1 (ω) L 2 µ(ω) with supp(v) ω ω. If we choose ϕ C0 (ω) such that ϕ = 1 i ω, the u = z q.e. i ω, hece u L 2 µ(ω ) ad (4.9) implies (4.8).

20 Theorem Let (µ h ) a sequece of measures of M 0 () γ A -covergig i to a measure µ M 0 (), ad let ω be a ope subset of. The (µ h ) γ A -coverges to µ i ω. Proof. Let us fix f H 1 (ω). For every h N let u h be the solutio i H0 1 (ω) L 2 µ h (ω) of problem (4.1), with replaced by ω. By the estimate (2.6) we kow that a subsequece, still deoted by (u h ), coverges weakly i H0 1 (ω) to a fuctio u H0 1 (ω). The, by Theorem 4.9, u L 2 µ(ω ) for every ope set ω ω ad u is a solutio of problem (4.8). It remais to prove that u L 2 µ(ω). Sice u H0 1 (ω) ad u L 2 µ(ω ) for every ope set ω ω, there exists a sequece (v h ) i H0 1 (ω) L 2 µ(ω), covergig to u weakly i H0 1 (ω), with supp(v h ) ω ad uv h 0 q.e. i ω, such that the sequece (uv h ) is icreasig ad coverges to u 2 poitwise q.e. i ω. Takig v = v h i (4.8) we get Au, v h + uv h dµ = f, v h. ω Takig the limit as h we obtai ω u2 dµ = f, u Au, u < +, ad thus u L 2 µ(ω). By a easy approximatio argumet we ca prove that u is the uique solutio i H 1 0 (ω) L 2 µ(ω) of the problem Au, v + ω uv dµ = f, v v H 1 0 (ω) L 2 µ(ω). Sice the limit does ot deped o the subsequece, the proof is complete. Corollary Let µ h, µ M 0 (). Let ( i ) i I be a family of ope subsets of which covers. The (µ h ) γ A -coverges to µ i if ad oly if (µ h ) γ A -coverges to µ i i for every i I. Proof. The coclusio follows easily from the compactess theorem (Theorem 4.5) ad from Theorem Strog Covergece Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (). Let f H 1 () ad let u h ad u be the solutios of problems (4.1) ad (4.2). By the defiitio of γ A -covergece the sequece (u h ) coverges to u weakly i H 1 0 (). I this sectio we study the strog covergece of the sequece of the gradiets (Du h ) i the space L p (, R ), 1 p 2. The followig theorem proves that (Du h ) coverges strogly to Du i L p (, R ) for every 1 p < 2.

21 Theorem 5.1. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (). Let f H 1 () ad let u h H0 1 () L 2 µ h () ad u H0 1 () L 2 µ() be the solutios of problems (4.1) ad (4.2). The (u h ) coverges to u strogly i H 1,p 0 () for every 1 p < 2. Proof. Sice A is liear ad the solutios of (4.1) deped cotiuously o the data, uiformly with respect to h (see the estimate (2.6)), it is ot restrictive to suppose that f L () ad f 0. By the defiitio of γ A -covergece the sequece (u h ) coverges to u weakly i H0 1 (), ad hece (Au h ) coverges to Au weakly i H 1 (). By Propositio 2.6 we have Au h f, ad so f Au h H+ 1 (), the positive coe of H 1 (). Sice H+ 1 () is compactly imbedded i H 1,p () for every 1 p < 2 (see [37]), the sequece (Au h ) coverges to Au strogly i H 1,p () for every 1 p < 2. If we apply Meyers estimate (see [36]) to the operator A, we fid that there exists a real umber s > 2 such that the operator A : H 1,q 0 () H 1,q () is a isomorphism for every 2 q s. Deote by r the expoet cojugate to s, i.e., 1/r + 1/s = 1. The A: H 1,p 0 () H 1,p () is a isomorphism for every r p 2. Sice (Au h ) coverges to Au strogly i H 1,p () for every r p < 2, the sequece (u h ) coverges to u strogly i H 1,p 0 () for every r p < 2, ad hece for every 1 p < 2. Let f L () ad let u h ad u be the solutios of problems (4.1) ad (4.2). By Theorem 5.1 the sequece (Du h ) coverges to Du weakly i L 2 (, R ) ad strogly i L p (, R ) for every 1 p < 2. To obtai strog covergece i L 2 (, R ) we eed a corrector term. This is a sequece of Borel fuctios P h from R to R, depedig o the sequece (µ h ), but idepedet of f, u, u h, such that Du h (x) = Du(x) + P h (x, u(x)) + R h (x) a.e. i, (5.1) where (R h ) teds to 0 strogly i L 2 (, R ). This coditio meas that the oscillatios of the sequece of the gradiets (Du h ) ear a poit x are determied, up to a term which is small i L 2 (, R ), oly by the values of the limit fuctio u ear x ad by the correctors P h, which deped oly o the sequece (µ h ). Let w h H 1 0 () L 2 µ h () ad w H 1 0 () L 2 µ() be the solutios of problems (4.3) ad (4.4). The fuctios P h : R R are defied by s ( Dwh (x) Dw(x) ), if w(x) > 0, P h (x, s) = w(x) 0, if w(x) = 0. (5.2) We are ow i a positio to state the mai theorem of this sectio.

22 Theorem 5.2. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (), ad let (P h ) be the sequece defied by (5.2). Let f L () ad let u h H0 1 () L 2 µ h () ad u H0 1 () L 2 µ() be the solutios of problems (4.1) ad (4.2). The (5.1) holds, with (R h ) covergig to 0 strogly i L 2 (, R ). Remark 5.3. Let w 0 be the uique fuctio of H0 1 () such that Aw 0 = 1 i. By the compariso priciple (Propositio 2.5) we have u h c w h c w 0 ad u c w c w 0 q.e. i, with c = f L (). As w 0 L () (see [48]), the fuctios u ad w belog to L (), ad the sequeces (u h ) ad (w h ) are bouded i L (). To prove Theorem 5.2 we eed the followig lemmas. ε = {w > ε}. For every ε > 0 we set Lemma 5.4. Assume that all hypotheses of Theorem 5.2 are satisfied. Let ε > 0 ad, for every h N, let r ε h = u h uw h w ε, where w h H0 1 () L 2 µ h () ad w H0 1 () L 2 µ() are the solutios of problems (4.3) ad (4.4). The rh ε H1 0 () L () ad (Drh ε ) coverges to 0 strogly i L 2 ( 2ε, R ). Proof. Sice the fuctios u ad 1 w ε belog to H1 0 () L (), ad, i additio, the sequeces (u h ) ad (w h ) are bouded i L () (Remark 5.3) ad coverge to u ad w weakly i H0 1 () (Defiitio 4.1), we coclude that rh ε H1 0 () L 2 µ h () ad that (rh ε uw ) coverges to u w ε weakly i H1 0 (). As u uw w ε = 0 a.e. i ε, we obtai that (rh ε) coverges to 0 strogly i L2 ( ε ) ad (Drh ε) coverges to 0 weakly i L 2 ( ε, R ). Let us fix a fuctio ϕ H0 1 () L () such that 0 ϕ 1 q.e. i, ϕ = 1 q.e. i 2ε, ad ϕ = 0 q.e. i \ ε. For istace, we ca take ϕ(x) = Φ ε (w(x)), where Φ ε : R R is the Lipschitz fuctio defied by Φ ε (t) = 0 for t ε, Φ ε (t) = t ε 1 for ε t 2ε, Φ ε(t) = 1 for t 2ε. To coclude the proof it is eough to show that lim h By the ellipticity coditio (2.4) we have Dr ε h 2 ϕ dx = 0. (5.3) α Drh ε 2 ϕ dx + (rh) ε 2 ϕ dµ h a ij D j rhd ε i r ε ) h ϕ dx + (rh) ε 2 ϕ dµ h = = a ij D j u h D i r ε ) h ϕ dx a ij D j w h D i r ε ) uϕ h w ε dx

23 ( a ij D j = + u w ε a ij D j u h D i (r ε hϕ) ) dx + ) Di rh ε ) wh ϕ dx + u h rhϕ ε dµ h w h ur ε h ϕ w ε dµ h ( uϕ )) a ij D j w h D i r ε w ε h dx u h r ε hϕdµ h ( a ij D j u h D i ϕ ) rh ε dx + ( u a ij D j w ε By (4.1) ad (4.3) we obtai α Drh ε 2 ϕ dx + (rh) ε 2 ϕ dµ h frhϕ ε dx ε a ij D j u h D i ϕ ) rh ε dx + ε ( ε ( ε ( ( u ) a ij D j Di r ε ) h wh ϕ dx. w ε uw h w ε rε hϕ dµ h = ( urh ε a ij D j w h D ϕ )) i dx w ε ε ) Di rh ε ) wh ϕ dx. ur ε h ϕ w ε dx ( uϕ )) a ij D j w h D i r ε w ε h dx Sice all terms i the right had side of the previous iequality ted to 0 as h, (5.3) holds ad the proof is complete. Lemma 5.5. Assume that all hypotheses of Theorem 5.2 are satisfied, ad let w H0 1 () L 2 µ() be the solutio of problem (4.4). The lim lim sup Du h 2 dx = 0. (5.4) ε 0 h {w<ε} Proof. For every ε > 0 let Φ ε : R R be the Lipschitz fuctio defied by Φ ε (t) = 1 for t ε, Φ ε (t) = 2 t ε for ε t 2ε, Φε (t) = 0 for t 2ε, ad let w ε H 1 () L () be the fuctio defied by w ε (x) = Φ ε (w(x)). As w ε 0 q.e. i ad w ε = 1 q.e. i {w < ε}, by the ellipticity coditio (2.4) ad by (4.1) we have α Du h 2 dx + (u h ) 2 ) dµ h a ij D j u h D i u h w ε dx + {w<ε} {w<ε} + (u h ) 2 w ε dµ h = a ij D j u h D i (u h w ε ) ) dx + (u h ) 2 w ε dµ h a ij D j u h D i w ε) u h dx = fu h w ε dx a ij D j u h D i w ε) u h dx.

24 Sice, by the defiitio of γ A -covergece, (u h ) coverges to u weakly i H 1 0 () ad strogly i L 2 (), we ca take the limit i the last two terms as h. Therefore we obtai α lim sup h {w<ε} Du h 2 dx fuw ε dx a ij D j ud i w ε) u dx. (5.5) As (w ε ) is bouded i L () ad coverges poitwise to the characteristic fuctio of {w = 0}, we have that (uw ε ) coverges to 0 strogly i L 2 () as ε 0 (recall that u c w q.e. i by Remark 5.3). Moreover, u 2 Dw ε 2 dx c2 ε 2 w 2 Dw 2 dx 4c 2 Dw 2 dx, {ε<w<2ε} {ε<w<2ε} ad thus (udw ε ) coverges to 0 strogly i L 2 (). Takig the limit i (5.5) as ε 0 we obtai (5.4). Proof of Theorem 5.2. Let us fix ε > 0, let rh ε = u h uw h w ε as i Lemma 5.4, ad let 2ε = {w > 2ε}. The R h = ( w h w 1)Du ( w h w 1) u w Dw+Drε h a.e. i 2ε. Sice (Drh ε) coverges to 0 strogly i L2 ( 2ε, R ) (Lemma 5.4) ad, i additio, ( w h w ) is bouded i L ( 2ε ) ad coverges to 1 strogly i L 2 ( 2ε ), we coclude that (R h ) coverges to 0 strogly i L 2 ( 2ε, R ). As R2 h dx = 2ε Rh 2dx+ {w 2ε} R2 h dx, it is eough to prove that lim lim sup Rhdx 2 = 0. (5.6) ε 0 h {w 2ε} Sice u c w q.e. i (Remark 5.3), we have R h Du h Du + c Dw h Dw a.e. i. Therefore lim sup h {w 2ε} Rhdx 2 4 lim sup + 4 c 2 lim sup h h {w 2ε} {w 2ε} Du h 2 dx + 4 Dw h 2 dx + 4 c 2 {w 2ε} {w 2ε} Dw 2 dx Du 2 dx + for every ε > 0. As u c w, we have Du = Dw = 0 a.e. i {w = 0}. Sice Lemma 5.5 ca be applied to the sequeces (u h ) ad (w h ), from the previous iequality we obtai (5.6), which cocludes the proof of the theorem. Lemmas 5.4 ad 5.5 eable us to prove the followig corrector result i H 1 0 ().

25 Theorem 5.6. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (), ad let w h H0 1 () L 2 µ h () ad w H0 1 () L 2 µ() be the solutios of problems (4.3) ad (4.4). Let f L () ad let u h H0 1 () L 2 µ h () ad u H0 1 () L 2 µ() be the solutios of problems (4.1) ad (4.2). The for every ε > 0 we have u h = uw h w ε + rε h, with lim lim sup r ε ε 0 h H 1 0 () = 0. h Proof. Settig 2ε = {w > 2ε}, we have Drh ε 2 dx = Drh ε 2 dx + 2ε {w 2ε} Dr ε h 2 dx. (5.7) Sice, by Lemma 5.4, (Dr ε h ) coverges to 0 strogly i L2 ( 2ε, R ) as h, we have oly to estimate the last term of (5.7). As Dr ε h = Du h u w ε Dw h ad u c w (Remark 5.3), we have w h w ε Du + uw h D(w ε), (w ε) Drε h 2 Du h 2 + c 2 Dw h 2 + ( w h ) 2 Du 2 + c 2( w h ) 2 Dw 2. w ε w ε Sice (w h ) is bouded i L () ad coverges to w weakly i H0 1 (), we obtai 1 lim sup Dr 4 h ε 2 dx lim sup Du h 2 dx + h {w 2ε} h {w 2ε} + c 2 lim sup Dw h 2 dx + Du 2 dx + c 2 Dw 2 dx. h {w 2ε} {w 2ε} {w 2ε} As u c w, we have Du = 0 a.e. i {w = 0}, ad so the last two terms ted to 0 as ε 0. The coclusio follows ow from Lemma 5.5. The case f / L () requires a further approximatio (see [9]). Theorem 5.7. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (), ad let (P h ) be the sequece of correctors defied by (5.2). Let f H 1 () ad let u h H0 1 () L 2 µ h () ad u H0 1 () L 2 µ() be the solutios of problems (4.1) ad (4.2). Fially, let (f λ ) be a sequece i L () covergig to f strogly i H 1 (), ad let u λ H0 1 () L 2 µ() be the solutios of the problems Au λ, v + u λ v dµ = f λ v dx v H0 1 () L 2 µ(). (5.8)

26 The Du h (x) = Du(x) + P h (x, u λ (x)) + Rh λ (x) a.e. i, with lim λ lim sup (Rh) λ 2 dx = 0. (5.9) h Proof. For every λ ad for every h let u λ h H1 0 () L 2 µ h () be the solutio of the problem Au λ h, v + u λ hv dµ h = f λ v dx v H 1 0 () L 2 µ h (). By Theorem 5.2 we have Du λ h (x) = Duλ (x) + P h (x, u λ (x)) + Sh λ (x) a.e. i, where (Sh λ) coverges to 0 strogly i L2 (, R ) for every λ. As Rh λ Sλ h = (Du h Du λ h ) (Du Duλ ), from the estimate (2.6) we obtai R λ h L 2 (,R ) S λ h L 2 (,R ) + 2 α f f λ H 1 (), which implies (5.9). Corollary 5.8. Let (µ h ) be a sequece of measures of M 0 () γ A -covergig to a measure µ M 0 (), ad let w h H0 1 () L 2 µ h () ad w H0 1 () L 2 µ() be the solutios of problems (4.3) ad (4.4). Let f H 1 () ad let u h H0 1 () L 2 µ h () ad u H0 1 () L 2 µ() be the solutios of problems (4.1) ad (4.2). If (w h ) coverges strogly i H0 1 (), the (u h ) coverges strogly i H0 1 (). Proof. Let (f λ ) be a sequece i L () covergig to f strogly i H 1 (), ad, for every λ, let u λ H0 1 () L 2 µ() be the solutio of problem (5.8). By Remark 5.3 each fuctio u λ /w is bouded o {w > 0}. Therefore, if (w h ) coverges strogly i H0 1 (), the ( P h (x, u λ (x)) ) coverges to 0 strogly i L 2 (, R ) for every λ, ad so the coclusio follows from Theorem The Rôle of the Skew-Symmetric Part of the Operator Let (a s ij ) ad (b ij ) be the symmetric ad the skew-symmetric part of the matrix (a ij ), ad let A s be the operator associated with the matrix (a s ij ) accordig to (2.3). I this sectio we shall study the depedece of the γ A -limit of a sequece (µ h ) o the skew-symmetric part (b ij ) of the matrix (a ij ). We begi by provig that, if the fuctios b ij are cotiuous, the the γ A -limit depeds oly o the symmetric part a s ij.

27 Theorem 6.1. Let µ, µ h M 0 (). If the fuctios b ij, i, j = 1,...,, are cotiuous, the (µ h ) γ A -coverges to µ if ad oly if (µ h ) γ As -coverges to µ. Proof. Sice the γ A -covergece ad the γ As -covergece are compact (Theorem 4.5), we may assume that (µ h ) γ As -coverges to a measure µ, ad we have oly to prove that (µ h ) γ A -coverges to µ. Suppose that b ij C 1 () for every i, j = 1,...,. The, for every pair of fuctios u, v H0 1 () H 2 (), we have Au, v = a s ijd j ud i v ) dx + b ij D j ud i v ) dx = = A s u, v D i (b ij D j u) ) v dx = A s u, v D i b ij D j u ) v dx, (6.1) where, i the last equality, we have used the fact that (b ij ) is skew-symmetric, while (D i D j u) is symmetric. By cotiuity, the same equality holds for every u, v H0 1 (). Therefore the solutio w h H0 1 () L 2 µ h () of problem (4.3) satisfies A s w h, v + w h v dµ h = f h, v v H0 1 () L 2 µ h (), with f h = 1 + D i b ij D j w h. By the estimate (2.6) the sequece (w h ) is bouded i H 1 0 (). Passig, if ecessary, to a subsequece, we may assume that (w h ) coverges weakly i H 1 0 () to a fuctio w. This implies that (f h ) coverges to f = 1 + D i b ij D j w weakly i L 2 (), ad hece strogly i H 1 (). Sice (µ h ) γ As -coverges to µ, by Propositio 4.8 the fuctio w is the solutio i H 1 0 () L 2 µ() of the problem A s w, v + wv dµ = 1 + D i b ij D j w ) v dx v H0 1 () L 2 µ(). By (6.1) w turs out to be the solutio i H 1 0 () L 2 µ() of (4.4), ad this implies that (µ h ) γ A -coverges to µ by Theorem 4.3. Sice the limit does ot deped o the subsequece, the whole sequece (µ h ) γ A -coverges to µ.

28 Let us cosider ow the more geeral hypothesis b ij C 0 (). Let (b ε ij ) be a sequece of skew-symmetric matrices of class C 1 covergig uiformly to (b ij ) as ε 0. Let a ε ij = as ij + bε ij ad let A ε be the correspodig elliptic operators o H 1 (). By the first step of the proof (µ h ) γ Aε -coverges to µ. Therefore, if wh ε H1 0 () L 2 µ h () ad w ε H0 1 () L 2 µ() are the solutios of the problems A ε wh, ε v + whv ε dµ h = v dx v H0 1 () L 2 µ h (), A ε w ε, v + w ε v dµ = v dx v H0 1 () L 2 µ(), the (w ε h ) coverges to wε weakly i H 1 0 () for every ε > 0. Let us prove that the solutios w h H 1 0 () L 2 µ h () of (4.3) coverge weakly i H 1 0 () to the solutio w H 1 0 () L 2 µ() of (4.4). For every ε > 0 we have w h w L2 () w h w ε h L2 () + w ε h w ε L2 () + w ε w L2 (). (6.2) We already proved that the secod term of the right had side teds to 0 as h. Let us estimate the first term. If we choose wh ε w h as test fuctios i the problems solved by wh ε ad w h, we obtai A ε wh, ε wh ε w h + wh(w ε h ε w h ) dµ h = (wh ε w h ) dx, Aw h, wh ε w h + w h (wh ε w h ) dµ h = (wh ε w h ) dx. By subtractig the secod equatio from the first oe we get A ε (wh ε w h ), wh ε w h + (b ε ij b ij )D j w h D i (wh ε w h ) dx + (w ε h w h ) 2 dµ h = 0. The, usig the ellipticity assumptio (2.4)(that depeds oly o the symmetric part of the matrix) ad the Hölder iequality, we obtai wh ε w h 2 H0 1() 1 α A ε(wh ε w h ), wh ε w h 1 (b ε ij b α ij )D j w h D i (wh ε w h ) dx 1 α b ε ij b ij L () w h H 1 0 () wh ε w h H 1 0 (). Sice (b ε ij ) coverges uiformly to b ij as ε 0, ad (w h) is bouded i H0 1 (), it follows that wh ε w h H 1 0 () teds to 0, as ε 0, uiformly with respect to h. To prove that w ε w H 1 0 () teds to zero we ca use the same argumets. Therefore (6.2) shows that (w h ) coverges to w strogly i L 2 (). As (w h ) is bouded i H0 1 (), we obtai that (w h ) coverges to w weakly i H0 1 (), ad, by Theorem 4.3, we coclude that (µ h ) γ A -coverges to µ.