Random homogenization of an obstacle problem

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1 Ann. I. H. Poincaré AN 26 (2009) Random homogenization of an obstacle problem L.A. Caffarelli a, A. Mellet b, a epartment of Mathematics and ICES, University of Texas at Austin, Austin, TX 78712, USA b epartment of Mathematics, University of British Columbia, Vancouver, BC V6T 1Z2, Canada Received 4 March 2007; received in revised form 9 September 2007; accepted 9 September 2007 Available online 26 October 2007 Abstract We study the homogenization of an obstacle problem in a perforated domain, when the holes are periodically distributed and have random shape and size. The main assumption concerns the capacity of the holes which is assumed to be stationary ergodic. Crown Copyright 2007 Published by Elsevier Masson SAS. All rights reserved. Keywords: Homogenization; Obstacle problem; Free boundary problem 1. Introduction Let (Ω, F, P) be a given probability space. For every ω Ω and every ε>0, we consider a domain ε (ω) obtained by perforating holes from a bounded domain of R n. We denote by T ε (ω) the set of the holes (we then have ε (ω) = \ T ε (ω)). The goal of this paper is to study the asymptotic behavior as ε 0 of the solution of the following obstacle problem: min 1 2 u 2 dx } fudx; u H0 1 (), u 0a.e.inT ε(ω) for some given function f(x)in L 2 (). This is a classical homogenization problem and the asymptotic behavior of the solutions strongly depends on the properties of the set T ε (ω). This type of problems was first studied by L. Carbone and F. Colombini [2] in periodic settings and then in more general frameworks by E. e Giorgi, G. al Maso and P. Longo [9], G. al Maso and P. Longo [7] and G. al Maso [6]. Our main reference will be the papers of. Cioranescu and F. Murat [4,5], in which the particular case of a periodic repartition of holes is studied. More precisely, they take T ε (ω) = T ε (independent of ω) given by T ε = B a ε(εk). (2) k Z n (1) The first author is partially supported by NSF grants. The second author is partially supported by NSERC grants. * Corresponding author. address: mellet@math.ubc.ca (A. Mellet) /$ see front matter Crown Copyright 2007 Published by Elsevier Masson SAS. All rights reserved. doi: /j.anihpc

2 376 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) It is then proved that there exists a critical radius a ε ε such that the limiting problem is no longer an obstacle problem, but a simple elliptic boundary value problem with a new term that takes into account the effect of the holes. More precisely, in dimension n 3, when a ε = r 0 ε n 2 n then there exists a constant α 0 > 0 such that u = lim u ε solves 1 min 2 u α 0u 2 dx fudx; u H0 }, 1 () where u = max(0, u). Our goal is to generalize this result to the case where the holes are still located in small neighborhoods of lattice points εz n but have random size and shape. More precisely, we assume that for any ε and ω, we can write ( ) T ε (ω) = S ε (k, ω), k Z n where the holes S ε (k, ω) satisfy, in particular, S ε (k, ω) B ε/2 (εk) for all k Z n. Further assumptions concerning the sets S ε (k, ω) will be made in the next section (we can already point out the fact that we will not exclude the case where S ε (k, ω) = for some k, thus allowing the fact that no holes may be present at some lattice points). The proof of the main theorem, which is detailed in Section 3, relies on the construction of an appropriate corrector. This construction is detailed in Sections 4 and 5, first in the case where the holes are balls of random radii in dimension n 2, then when no assumptions are made on the shape of the holes (in dimension n 3 only). 2. Assumptions and main result Let us now make precise the assumptions on the holes S ε (k, ω). The first assumption is mainly technical: Assumption 1. There exists a (large) constant M such that for all k Z n and a.e. ω Ω we have S ε (k, ω) B Mε n/(n 2)(εk) if n 3, S ε (k, ω) B exp( Mε 2 )(εk) if n = 2 for ε small. As mentioned in the introduction, the asymptotic behavior of the u ε strongly depends on the size of the holes. The critical size for which interesting phenomena is observed corresponds to finite, nontrivial capacity of the set T ε.more precisely, we assume: Assumption 2. For all k Z n and a.e. ω Ω, there exists γ(k,ω)(independent of ε) such that cap ( S ε (k, ω) ) = ε n γ(k,ω), where cap(a) denote the capacity of a subset A of R n (as defined below). Moreover, we assume that there exists a constant γ >0: γ(k,ω) γ for all k Z n and a.e. ω Ω. (3) We take the following definitions for the capacity of a subset A of R n : } cap(a) = inf h 2 dx; h H 1 (R n ), h 1inA, lim h(x) = 0, x R n

3 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) in dimension n 3 and } cap(a) = inf h 2 dx; h H0 1 (B 1), h 1inA, B 1 in dimension n = 2 and for sets A B 1. Finally, we need to make an assumption to ensure that some averaging process occurs as ε goes to zero: Assumption 3. The process γ : Z n Ω [0, ) is stationary ergodic: There exists a family of measure-preserving transformations τ k : Ω Ω satisfying γ(k+ k,ω)= γ(k,τ k ω) for all k,k Z n and ω Ω, and such that if A Ω and τ k A = A for all k Z n, then P(A)= 0orP(A)= 1 (the only invariant set of positive measure is the whole set). Let us make a few remarks concerning those assumptions: First of all, we stress out the fact that the shape of the holes S ε is left unspecified and may change with ε; Only the rescaled capacity γ(k,ω)is independent of ε. The first assumption, which implies that the diameters of the holes decrease faster than ε, guarantees that the capacities of neighboring sets separate at the limit (i.e. that cap( S ε ) cap(s ε )). And the choice of scaling for the capacity guarantee that cap(t ε ) remains bounded as ε goes to zero (since #Z n ε 1 } Cε n ). When n 3, we have cap(b r ) = c n r n 2, and so Assumption 2 implies that if S ε (k, ω) is a ball, then its radius is of the form r(k,ω)εn 2 n. In particular, we recover Cioranescu Murat s result in the periodic case. Finally, the hypothesis of stationarity is the most general extension of the notions of periodicity and almost periodicity for a function to have some self-averaging behavior. Under those assumptions, we prove the following result: Theorem 2.1. Assume that n 3 and that T ε (ω) = k Z n S ε(k, ω) satisfies Assumptions 1, 2 and 3 listed above. Then there exists α 0 0 such that when ε goes to zero, the solution u ε (x, ω) of } 1 min 2 u 2 dx fudx; u H0 1 (), u 0 a.e. in T ε(ω) converges weakly in H 1 () and almost surely ω Ω to the solution ū(x) of the following minimization problem: 1 min 2 u } 2 α 0u 2 fudx; u H 0 1 (), where u (x) = max(0, u(x)). Moreover, if there exists γ > 0 such that γ(k,ω) γ for all k Z n and a.e. ω Ω, then α 0 > 0. The same result holds when n = 2 in the case where the sets S ε (k, ω) are balls. The general result (with holes of random shape) holds also in dimension n = 2. However, because the fundamental solution of Laplace s equation is different in that case, the proof is slightly different and more technical. Note that the Euler Lagrange equations for the minimization problems yield u ε = f in ε, u ε (x) 0 for x T ε, (4) u ε (x) = 0 for x \ T ε

4 and ū α0 ū = f in, 378 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) ū(x) = 0 forx. As in Cioranescu Murat [4,5], the proof of this result relies on the construction of an appropriate corrector. More precisely, the key is the following result: Proposition 2.2. Assume that n 3 and that T ε (ω) = k Z n S ε(k, ω) satisfies Assumptions 1, 2 and 3 listed above. Then, there exists a nonnegative real number α 0 and a function w ε (x, ω) such that w ε (x, ω) = 1 for x T ε (ω), w ε (x, ω) = 0 for x \ T ε (ω), w ε bounded in L () w ε (,ω) 0 H 1 ()-weak for almost all ω Ω, and For all sequences v ε (x) satisfying: v ε (x) 0 for x T ε, v ε bounded in L (), v ε v in H 1 ()-weak and for any φ () such that φ 0, we have: lim φ w ε v ε dx α 0 φv dx with equality if v ε (x) = 0 for x T ε. The same result holds when n = 2 in the case where the sets S ε (k, ω) are balls. Condition (5) may seem rather complicated, but it is exactly the property that will be needed to prove Theorem 2.1. This condition is satisfied in particular if the Laplacian of w ε is equal to α 0 outside the holes as in the following lemma: (5) Lemma 2.3. Let w ε (x, ω) be a function satisfying w ε = α 0 in ε (ω), w ε (x, ω) = 1 for x T ε (ω), w ε (x, ω) = 0 for x \ T ε (ω) for almost all ω Ω, and w ε (,ω) 0 H 1 ()-weak a.s. ω Ω. Then w ε satisfies (5). The same conclusion holds if we only have ε w ε α 0 dx 0 a.s. ω Ω. (6) We will strongly rely on this lemma in the sequel. In particular, in the case where the holes are all balls of random radii, we will construct w ε by showing that for a critical α 0, the unique solution of (6) converges to 0 in H 1 weak for almost every ω.

5 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Proof of Lemma 2.3. Since ϕ = 0on,wehave: w ε ϕv ε dx = ϕ w ε v ε dx + ϕ w ε v ε dx ε ε ε T ε w ε ν ϕvε dσ (x). Since w ε goes to zero in H 1 ()-weak and v ε converges to v in L 2 ()-strong, it is readily seen that the second term in the right-hand side goes to zero as ε 0. Furthermore, it is readily seen that w ε 1 and w ε = 1onT ε, and so wν ε 0on T ε. Since v ε 0onT ε and ϕ 0, we deduce that lim w ε ϕv ε dx lim ϕ w ε v ε dx. ε ε ε ε Finally, we have lim w ε ϕv ε dx = lim α 0 ϕv ε dx + lim w ε α 0 ϕv ε dx lim α 0 ϕv ε dx ε using the fact that v ε is bounded in L. Hence we have lim ϕ w ε v ε dx α 0 ϕv dx. ε It is now easy to check that all the inequalities becomes equalities whenever v ε = 0onT ε. The proof of Proposition 2.2 will occupy most of this paper. It will be split into two parts: In Section 4, we consider the (simpler) case when the holes S ε (k, ω) are all balls of random radius. In Section 5, we will use this first result to treat the general case (when the holes have unspecified shapes). Before turning to this proof, we briefly give, in the next section, the proof of the main theorem. 3. Proof of Theorem 2.1 and We introduce the initial and limit energies: 1 J(v) = 2 v 2 fvdx J α (v) = 1 2 v α 0v 2 fvdx. With theses notations, we have that u ε (x, ω) satisfies J(u ε ) = inf J(v) v K ε with K ε =v H0 1(); v 0a.e.inT ε} and standard estimates give the existence of a function ū(x, ω) such that u ε (,ω) ū(,ω) in H0 1 ()-weak. We now have to show that for almost every ω, ū(,ω)satisfies: J α (ū) = inf J α (v). v H0 1() This will follow from two lemmas:

6 380 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Lemma 3.1. For any test function ϕ in () we have lim w ε 2 ϕdx= α 0 ϕdx. Lemma 3.2. Let u ε be any bounded sequence in H 1 0 () such that uε 0 in T ε.if u ε ū in H 1 ()-weak, then lim inf J(uε ) J α (ū). Proof of Theorem 2.1. For any v (), the function v + v w ε is nonnegative on T ε (ω) and is thus admissible for the initial obstacle problem. In particular by definition of u ε,wehave J(u ε ) J(v + v w ε ). Next, we can expand J(v + v w ε ) as follows: 1 J(v + v w ε [ ) = v 2 + v 2 w ε2 + v 2 w ε 2] dx 2 + [v v w ε w ε + v v w ε + vv w ε ] dx [fv+ fv w ε ] dx and it is readily check that Lemma 3.1 and the weak convergence of w ε to 0 in H 1 () implies lim J(v + v w ε ) = J α (v), as soon as v 2 (). We deduce J α (v) lim sup J(u ε ) and so, using Lemma 3.2, we get: J α (v) J α (ū) for all v () such that v 2 (). Finally, by approximating separately the positive and the negative parts, one can show that every function v H0 1() is the limit of a sequence of functions v k () such that (v k ) (). Theorem 2.1 follows. Proof of Lemma 3.1. This is an immediate consequence of (5): Since 1 w ε = 0inT ε we have lim ϕ w ε (1 w ε )dx = lim ϕ + w ε (1 w ε )dx lim ϕ w ε (1 w ε )dx = α 0 ϕ + dx + α 0 ϕ dx and so lim ϕ w ε w ε dx = α 0 ϕdx. Proof of Lemma 3.2. See Cioranescu and Murat [5], Proposition 3.1.

7 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Proof of Proposition 2.2: Balls of random radius Throughout this section, we assume that the sets S ε (k, ω) are balls centered at εk. Since n(n 2)ω n r n 2 if n 3, cap(b r ) = 2π if n = 2. log r Assumption 2 becomes in this framework: with and S ε (k, ω) = B a ε (r(k,ω))(εk) for all k Z n rε n/(n 2) a ε if n 3, (r) = exp( r 1 ε 2 ) if n = 2, ( ) γ(k,ω) 1/(n 2) r(k,ω) = if n 3, n(n 2)ω n γ(k,ω)/2π if n = 2. In particular that the process r : Z n Ω [0, ) is stationary ergodic and satisfies r(k,ω) r for all k Z n and a.e. ω Ω (7) for some constant r >0. Without loss of generality, we can always assume that r <1/2 (so that there is no overlapping of the holes for any ε<1): 4.1. The auxiliary obstacle problem After rescaling, we look for the corrector w ε (x, ω) in the form w ε (x, ω) = ε 2 v ε (x/ε, ω) with v ε (y, ω) solution to v = α, in ε 1 ε, a.e. ω Ω, v = ε 2 on k Z n B ā ε (k,ω)(k), where rε 2/(n 2) ā ε if n 3, (r) = ε 1 exp( r 1 ε 2 ) if n = 2, and such that ε 2 v ε (x/ε) 0 inh 1 -weak. One of the main tool in the proof is the fundamental solution of the Laplace equation, given by: 1 1 h(x) = n(n 2)ω n x n 2 if n 3, 1 log x if n = 2. 2π

8 382 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) In particular, we note that 1 h Bāε (r(k,ω)) (0) = n(n 2)ω n r n 2 ε 2 if n 3, 1 2π (log(ε) + r 1 ε 2 ) if n = 2, so we expect the rescaled corrector v ε (x, ω) to behave, near the hole Bā(k,ω) (k), like the function γ(k,ω)h(x k) = r(k,ω)n 2, if n 3, h k (x) := x k n 2 γ(k,ω)h(x k) = r(k,ω)log x k if n = 2, where ( ) n 2n(n r(k,ω) 2)ωn if n 3, γ(k,ω)= 2πr(k,ω) if n = 2. Since h k satisfies h k = γ(k,ω)δ(x k), we will construct v ε (x, ω) by solving v = α k Z n A γ(k,ω)δ(x k) in ε 1, v = 0 on ε 1. The main issue is thus to find the critical α for which the solution of the above equation has the appropriate behavior near x = k. Following [3], this will be done by introducing the following obstacle problem, for every open set A R n and α R: v α,a (x, ω) = inf v(x); v α Clearly, the function v α,a is solution of v = α γ(k,ω)δ(x k) k Z n A k Z n A whenever it is positive. Note that the function h α,k (x) := α 2n x k 2 + h k (x k) also satisfies α = 2n x k 2 + r(k,ω)n 2, if n 3, x k n 2 α 2n x k 2 r(k,ω)log x k if n = 2, h α,k (x) = α γ(k,ω)δ(x k). } γ(k,ω)δ(x k),v 0inA,v = 0on A. (8) It follows from (9) and the maximum principle that if B 1 (k) A, then, for all x in B 1 (k) and for almost every ω in Ω, we have h α,k (x) α v α,a (x, ω) 2n rn 2 if n 3, h α,k (x) α (11) if n = 2. 2n (9) (10)

9 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Critical α The purpose of this section is to prove that for a critical α, v α,a behaves like h α,k near S ε (k, ω). For that purpose, we introduce the following quantity, which measures the size of the contact set: m α (A, ω) = x A; v α,a (x, ω) = 0 }, where A denotes the Lebesgue measure of a set A. The starting point of the proof is the following lemma: Lemma 4.1. The random variable m α is subadditive, and the process T k m(a, ω) = m(k + A,ω) has the same distribution for all k Z n. Proof of Lemma 4.1. Assume that the finite family of sets (A i ) i I is such that A i A for all i I, A i A j = for all i j, A A i = 0 i I then v α,a is admissible for each A i, and so v α,ai v α,a. It follows that v α,a = 0} A i v α,ai = 0} and so m α (A, ω) = vα,a = 0} A i vα,ai = 0} = m α (A i,ω), i I i I i I which gives the subadditive property. Assumption 3 then yields T k m(a, ω) = m(a, τ k ω) which gives the last assertion of the lemma. Since m α (A, ω) A, and thanks to the ergodicity of the transformations τ k, it follows from the subadditive ergodic theorem (see [8] and [1]) that for each α, there exists a constant l(α) such that lim t m α (B t (0), ω) B t (0) = l(α) a.s., where B t (0) denotes the ball centered at the origin with radius t. Note that the limit exists and is the same if instead of B t (0), we use cubes or balls centered at tx 0 for some x 0. If we scale back and consider the function w ε α (y, ω) = ε2 v α,bε 1 (ε 1 x 0 ) (y/ε, ω) in B 1(x 0 ), we deduce y; w α ε (y, ω) = 0} lim = l(α) a.s. B 1 The next lemma summarizes the properties of l(α): Lemma 4.2. (i) l(α) is a nondecreasing functions of α.

10 384 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) (ii) If α<0, then l(α) = 0. Moreover, if the radii r(k,ω) are bounded from below, then l(α) = 0 for any α such that α<n(n 2) inf k Z n r(k,ω) n 2 almost surely. (iii) If α 2 n n(n 2) sup k Z n r(k,ω) n 2 (or α 8r for n = 2) almost surely, then l(α) > 0. Proof. (i) The proof follows immediately from the inequality v α,a v α,a for any α, α such that α α. (ii) If α is negative, then the function 2n α x x 0 2 2n α (tr)2, which is a sub-solution of (9), is positive in tb r (x 0 ) and vanishes along (tb r (x 0 )) for any ball B r (x 0 ) and for any t>0. We deduce: v α,tb > α 2n x x 0 2 α 2n (tr)2 > 0 intb r (x 0 ) for all t>0. Therefore m α (tb, ω) = 0 for all t>0, so l(α) = 0 for all α<0. Furthermore, if r(k,ω) is bounded below: r(k,ω) r > 0 for all k Z n, a.e. ω Ω, then, the function 2n α x k 2 + rn 2 α x k n 2 2n rn 2 is a sub-solution of (9) in B 1 (k) which vanishes on B 1 (k) and is strictly positive in B 1 (k) as long as α<n(n 2)r n 2. As above, we deduce that m α (tb, ω) = 0 for all t>0and for all α<n(n 2)r n 2. (iii) The function h α,k (x) = 2n α x k 2 + rn 2 is radially symmetric and reaches its minimum when x k n 2 ( n(n 2)r(k, ω) n 2 ) 1/n when n 3, x k =R(α,k) := α ( ) 2r(k,ω) 1/2 (12) when n = 2. α In particular, for α>2 n n(n 2)r(k, ω) n 2 (or n 8r(k,ω) when n = 2), we have R(α,k) < 1/2 and so the function hα,k (x) k in B R(α,k) (k), g k (x) = 0 inr n \ B R(α,k) (k) satisfies g k α γ(k,ω)δ(x k) in C 1 (k), and g k = 0 inc 1 (k) \ B 1/2 (k), where C 1 (k) denotes the cube of size 1 centered at k, and the constant C k is chosen in such a way that g k and g k vanish along B R(α,k) : ( ) n 2 ( ) 2 ( ) α n 2(n 2) n 2 n n r n when n 3, (α,k) := ( 2n ( )) 2 n 2 (13) r 2r 1 log when n = 2. 2 α By definition of v α,tb, we deduce that v α,tb (x) g k (x) in tb a.s. k Z n tb In particular, this implies that v α,tb vanishes in tb \ k Z n B 1/2(k), and so m α (tb, ω) tb C 1 B 1/2 C 1 = 1 ω n 2 n a.s.

11 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) We conclude l(α) 1 ω n 2 n > 0. Using Lemma 4.2, we can define α 0 = sup α; l(α) = 0 }. Note that α 0 is finite under Assumption 3 (Lemma 4.2(iii)) and that α 0 0 is strictly positive as soon as the r(k,ω) are bounded from below almost surely by a positive constant (Lemma 4.2(ii)). Our goal, in the rest of this section, is to show that the function w ε (x, ω) = inf w(x); w α 0 in \ T ε, w 1onT ε w = 0on \ T ε satisfies all the conditions of Proposition 2.2. For that purpose, we will introduce a series of intermediate functions Behavior of v α,ε 1 A near the holes We fix a bounded subset A of R n and we denote by }, v α ε (x, ω) = v α,ε 1 A (x, ω) (14) the solutions of (8) defined in ε 1 A. We also introduce the rescaled function w α ε (y, ω) = ε2 v α ε (y/ε, ω), defined in A. The key properties of v α ε are given by the following lemma: Lemma 4.3. (i) For every α and for every k Z n, we have h α,k (x) α v α ε (x) 2n rn 2 if n 3, h α,k (x) α if n = 2 2n for all x B 1 (k) and almost everywhere ω Ω (where h α,k is defined by (10)). (ii) For every α>α 0, we have v ε α (x) h α,k(x) + o(ε 2 ) for all x B 1/2 (k) and almost everywhere ω Ω. Since ε 2 + α 0 ā ε( r(k,ω) ) 2 if n 3, h α,k Bāε (r(k,ω)) (0) = 2n ε 2 + α 0 āε ( r(k,ω) ) 2 + r(k,ω)log ε if n = 2, 4 we deduce the following corollary: Corollary 4.4. (i) For every α and every k Z n such that r(k,ω) > 0, we have v ε α (x) ε 2 + o(1) on Bāε (r(k,ω))(k) a.e. ω Ω

12 386 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) and so w α ε (x) 1 + o(ε2 ) on T ε (ω) a.e. ω Ω for all α. (ii) For every α>α 0 and every k Z n, we have v α ε (x) ε 2 + o(ε 2 ) on Bāε (r(k,ω))(k) a.e. ω Ω and so w α ε (x) 1 + o(1) on T ε(ω) a.e. ω Ω. Proof of Lemma 4.3. (i) Immediate consequence of (11). (ii) Preliminary: First of all since A is bounded, we have A B R (x 0 ). Without loss of generality, we can always assume that B R (x 0 ) = B 1 (0). We then introduce vα ε (x, ω) = v α,ε 1 B 1 (x, ω), the solutions of (8) in B ε 1(0). It is readily seen that vα ε is admissible for (8) and thus v α ε (x, ω) vε α (x, ω) for all x ε 1 A a.e. ω Ω. It is thus enough to prove (ii) for vα ε. We will need the following consequence of Lemma 4.1 (see [3] for the proof): Lemma 4.5. For any ball B r (x 0 ) B 1 (0), the following limit holds, a.s. in ω vα ε lim (x, ω) = 0} B ε 1 r (ε 1 x 1 ) B ε 1 r = l(α). Step 1: We can now start the proof: For any δ>0, we can cover B ε 1 by a finite number N ( Cδ n )ofballsb i with radius δε 1 and center ε 1 x i. Since α>α 0,wehave l(α) > 0. By Lemma 4.5, we deduce that for every i, there exists ε i such that if ε ε i, then vα ε (x, ω) = 0} B i > 0 a.s. ω. In particular, if ε inf ε i, then vα ε (y i) = 0forsomey i in B i a.s. ω Ω. We now have to show that this implies that vα ε remains small in each B i as long as we stay away from the lattice points k Z n. More precisely, we want to show that sup B i \ v k Z n B α ε Cδ2 ε 2. 1/4(k) Step 2: Let η be a nonnegative function such that 0 η(x) 1 for all x, η(x) = 1inB 1/8 and η = 0inR n \ B 1/4. Then the function u = vα ε ηis nonnegative on 2B i and satisfies C u C, where C is a universal constant depending only on n and r. In particular, since B i has radius δε 1, Harnack inequality yields: sup u C inf u + Cα(δε 1 ) 2. B i B i Step 3: We need the following lemma: Lemma 4.6. If v α in B r (y 0 ), then 1 v(x)dx v(y 0 ) + αc(n)r 2, B r B r (y 0 ) where C(n) is a universal constant.

13 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Proof. We note that the function v(x) 2n α x y 0 2 is super-harmonic in B r (y 0 ). The lemma follows from the mean value formula. Now, we recall that vα ε (y i) = 0 and vα ε α in B 1/4(y i ).So 1 vα ε (x) dx vε α (y i) + αc(n). B 1/4 B 1/4 (y i ) In particular, we have u(y i ) vα ε (x) dx C(α,n). B 1/4 (y i ) Step 4: Steps 2 and 3 yield sup u C(α,n) ( 1 + α(δε 1 ) 2) B i and since vα ε 0inB i \ k Z nk},wehave: vα ε 1 (y) vα ε (x) dx Cu(y) B 1/8 B 1/8 (y) for all y B i \ k Z n B 1/4(k). It follows that for every δ and for ε small enough, we have: sup B ε 1 \ v k Z n B α ε Cδ2 ε 2. 1/4(k) The definition of vα ε and the fact that h α,k 0on B 1/2 implies that vα ε (x) h α,k(x) + Cδ 2 ε 2 in B 1/2 (k) for all k Z n Approximated corrector We now want to use the function v α ε, solution of the obstacle problem (8), to study the properties of the free solution w0 ε of w0 ε = α 0 k Z n γ(k,ω)δ(x εk) in, w0 ε = 0 on. We define h ε α,k (x) = ε2 h α,k (x/ε) for n 3 and h ε α,k (x) = ε2 h α,k (x/ε) + rε 2 log ε for n = 2. We have: α 0 h ε α,k (x) := 2n x εk 2 + εn r(k,ω) n 2 x εk n 2 if n 3, α 0 2n x εk 2 r(k,ω)ε 2 log x εk if n = 2. We then prove: Lemma 4.7. For every k Z n, w0 ε satisfies h ε α,k (x) o(1) wε 0 (x) hε α,k (x) + o(1) x B ε/2(εk) a.e. ω Ω. (15) In particular: w0 ε (x) = 1 + o(1) on T ε. (16)

14 388 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Proof. For every α, we denote by w α ε the function w ε α (x) = ε2 v α,ε 1 (x/ε), defined in and satisfying w ε α = 0on. 1. For every α>α 0,wehave (w ε 0 wε α ) α 0 α and w0 ε wε α = 0on. This implies w0 ε (x 0) wα ε (x 0) G(x 0, x)(α 0 α)dx, where G(, ) is the Green function on ( G = δ x0 and G = 0on ). Note that we have G(x 0,x) h(x x 0 ) x,x 0, and so w0 ε (x 0) wα ε (x 0) (α α 0 ) h(x x 0 )dx. We deduce with C 1/(n 1) sup(w0 ε wε α ) ρ α α 0 if n 3, C ρ log ρ α α 0 if n = 2, ρ = infρ; B ρ }. Hence we have w ε 0 wε α + O(α α 0). Using Lemma 4.3(ii) (since α>α 0 ), we deduce: w ε 0 hε α,k (x) + O(α α 0) + o(1) x B ε/2 (εk) a.e. ω Ω which gives the second inequality in (15). 2. Similarly, we observe that for every α α 0,wehave (w ε α wε 0 ) α α 0 α1 w ε α =0}. Proceeding as before, we deduce that for n 3, sup(wα ε wε 0 ) Cρ [ 1/(n 1) (α 0 α) + Cα w ε α = 0} 1/(n 1) ] and a similar inequality for n = 2. Using Lemma 4.3(i), we get w ε 0 hε α,k o(ε2 ) O(α 0 α) Cα w ε α = 0} 1/(n 1). Finally, since lim wα ε = 0} = 0 for all α α 0, and (15) follows.

15 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Proof of Proposition 2.2 We are now in position to complete the proof of Proposition 2.2: We define w ε (x, ω) = inf w(x); w α 0 in \ T ε, w 1onT } ε, w = 0on \ T ε it is readily seen that w ε (x, ω) = 1 for x T ε, w ε (x, ω) = α 0 for x \ T ε, w ε (x, ω) = 0 for x \ T ε. So in view of Lemma 2.3, we only have to show that w ε 0inH 1 ()-weak as ε goes to zero. More precisely, we will show that w ε converges to zero in L p strong and is bounded in H 1. Strong convergence in L p : First of all, (16) yields w0 ε (x) o(1) wε (x, ω) w0 ε (x) + o(1) x ε a.e. ω Ω, which in turns imply (using Lemma 4.7 again): h ε α,k (x) o(1) wε (x, ω) h ε α,k (x) + o(1) x B ε/2(εk) a.e. ω Ω. (17) Next, a simple computation shows that h ε Cε n (ε n 2 2n + ε 2p ) if n 3, α,k p dx Cε 2 ε 2p (log ε) p if n = 2. B ε \B a ε Since #εz n } Cε n for all n, we deduce from (17) that w ε C(ε p(n 2) 2n + ε 2 ) if n 3, L p Cε 2 (log ε) if n = 2. In particular w ε 0 inl p -strong, for all p [1, ). Bound in H 1 : First of all, a simple integration by parts together with the fact that w ε = 1on T ε yields w ε 2 dx α 0 + w ε dσ (x), ε T ε where T ε = S ε (k, ω). So we need an estimate in w ε along S ε (k, ω) = B a ε (r(k,ω)). We consider the function w ε (x) h ε α,k (x) + α 0 2n r2 ε n/(n 2) when n 3, z(x) = w ε (x) h ε α,k (x) + α 0 2n r2 e 2 ε 2 r when n = 2. It satisfies z = 0 inb 1/2 (εk) \ B a ε (r(k,ω))(εk), z(x) = o(1) in B 1/2 (εk) \ B a ε (r(k,ω))(εk), z(x) = 0 along B a ε (r(k,ω))(εk), and so z(x) o(r n 2 ε n ε n(n 1) n 2 ) = o ( ε n a ε (r) (n 1)) if n 3, o(ε 2 e r 1 ε 2 ) = o ( ε n a ε (r) (n 1)) if n = 2. (18)

16 390 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) on B a ε (r(k,ω))(εk). It follows that w ε h ε α,k (x) + z(x) Cε n a ε( r(k,ω) ) (n 1) along B a ε (r(k,ω))(εk), We deduce w ε 2 dx α 0 + w ε dσ(x) ε T ε α 0 + C, and the proof is complete. k Z n ε 1 B a ε (r(k,ω)) (εk) α 0 +Cε n a ε ( r) n 1 ε n a ε ( r) (n 1) 5. Proof of Proposition 2.2: General case w ε dσ(x) In this section, we treat the case where the sets S ε (k, ω) have unspecified shape, but satisfy Assumption 2: cap ( S ε (k, ω) ) = ε n γ(k,ω). Throughout this section we assume n 3. The proof makes use of the result of the previous section, after noticing that away from εk, the hole S ε (k, ω) is equivalent to a ball of radius a ε (r(k, ω)), where ( ) γ(k,ω) 1/(n 2) a ε (r) = rε n/(n 2), r(k,ω)=. n(n 2)ω n More precisely, we will rely on the following lemma: Lemma 5.1. For any k Z n and ω Ω,letϕk ε (x, ω) be defined by } v(x) 1, x Sε ϕk v(x); ε (x, ω) = inf (k, ω) v 0,. lim x v(x) = 0 Then for any δ>0, there exists R δ such that ϕ ε k (x, ω) ε n γ(k,ω)h(x εk) δε n h(x εk) for all x such that x εk a ε (R δ ) and for all ε>0. Moreover, R δ depends only on the constant M appearing in Assumption 1. In particular, R δ is independent of k and ω. 1. For a given δ>0, Lemma 5.1 implies that for every k Z n and ω Ω there exists a constant R δ (k, ω) such that ϕε k (x, ω) εn r(k,ω) n 2 ( ) r n 2 x εk n 2 δ in B 2a R ε (R δ ) \ B a ε (R δ )(εk) (19) δ for all ε>0. Moreover, it is readily seen that for any R there exists ε 1 (R) such that a ε (R) ε σ /4 for all ε ε 1 (20) for some σ>1. Finally, we note that by definition of ϕk ε,wehave ϕ ε ( k 2 dx = cap Sε (k) ) = ε n γ(k,ω). (21) R n

17 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) Next, let α 0 and w ε be the coefficient and corresponding corrector constructed in the previous section, and associated with holes S ε of radius r(k,ω). Lemma 4.7 implies that for δ and R given, there exists ε 2 (δ, R) < ε 1 (R) such that for all ε ε 2 (δ, R), wehave wε (x) εn r(k,ω) n 2 x εk n 2 δ R n 2 in B ε/2 (εk), (22) in dimension n 3. Note that thanks to (20), Inequality (22) holds in particular in B 2a ε (R) \ B a ε (R)(εk). The corrector given by Proposition 2.2 will be constructed by gluing together the functions ϕk ε (near the holes S ε (k)) and the function w ε (away from the holes). The gluing will have to be done in a very careful way so that the corrector satisfies all the properties listed in Proposition 2.2: For a given ε, we define δ ε to be the smallest positive number such that (20) and (22) hold with δ = δ ε and R = R δε. From the remarks above, we see that δ ε is well defined as soon as ε is small enough (say smaller than ε 2 (1,R 1 )). Moreover, for any δ>0, there exists ε 0 = ε 2 (δ, R δ ) such that δ ε δ ε ε 0. In particular lim δ ε = 0. From now on, we write R ε = R δε. We are now ready to define the corrector w ε :Letη ε (x) be a function defined on such that η ε (x) = 1 on \ B2aε(Rε)(εk), η ε (x) = 0 on k Zn B a ε (R ε )(εk) k Z n and satisfying η ε Ca ε (R ε ) 1 and η ε Ca ε (R ε ) 2 in B 2a ε (R ε ) \ B a ε (R ε )(k). We then define w ε (x, ω) in by: w ε (x, ω) = η ε (x)w ε (x, ω) + ( 1 η ε (x) ) ϕk ε (x, ω)1 B ε/2 (εk)(x). k Z n It satisfies ϕ w ε k ε(x) in B 2a ε (R ε )(k) \ S ε (k) k Z n, (x, ω) = w ε (x) in \ B a ε (R ε ). k Z n To simplify the notations in the sequel, we denote ϕ ε (x) := ϕk ε (x, ω)1 B ε/2 (εk)(x). k Z n The properties of w ε are summarize in the following lemma, which implies Proposition 2.2 with (5) instead of the first equation: Lemma 5.2. The function w ε satisfies (i) (ii) w ε = 1 on S ε for any ε>0. w ε converges to zero as ε goes to zero in L p () strong for all p [2, ) and w ε L p Cε p(n 2) p 2. 2n

18 392 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) (iii) w ε is bounded in H 1 (). (iv) w ε satisfies (5). Proof. (i) Immediate consequence of the definition of w ε since ϕk ε = 1onS ε(k, ω). (ii) Assumption 1 yields ϕk ε (x, ω) Cεn γ(k,ω)h(x εk) for all x such that x εk a ε (M). Since ϕk ε 1inB a ε (M)(εk), we deduce: (1 ηε )ϕ ε p L p (R n ) ϕk ε 1 B ε/2 (εk) p L p ( B R(k)a(ε) ) k Z n ε 1 k Z n ε 1 B a ε (M)(εk) k Z n ε 1 a ε (M) n + C γ Using (20) and the definition of a(ε), we deduce: (1 ηε )ϕ ε p L p (R n ) Cε n M n ε n 2 n2 + C γ CM n ε 2n n 2 + C γ ( ϕ ε k (x) ) p dx + C k Z n CM n ε n 2 2n + C γε 2p, where 2p n 2 2n if p 2 and n 3. Using (18), it follows that w ε L p () w ε L p () + C ( ε n 2 2n ) 1/p Cε p(n 2) 2n k Z n k Z n ε 1 ε n+2p ε pn ε n p(n 2) k Z n ε 1 B 2a ε (Rε)(εk) ε pn( a ε (R ε ) ) n p(n 2). ( ε n γ(k)h(x εk) ) p dx for all p 2. (iii) Next, we want to show that w ε is bounded in H 1 ( ε ). First, we note that in B ε/2 (εk),wehave: w ε = η ε (w ε ϕk ε ) + η ε w 0 ε + (1 η ε) ϕk ε, (23) where the function η ε is supported in B 2a ε (R ε )(εk) \ B a ε (R ε )(εk) and satisfies η ε C ( a ε (R) ) 1. Since w ε ϕk ε C δ ε in B Rε n 2 2a ε (R ε )(εk) \ B a ε (R ε )(εk), we deduce η ε (w ε ϕ ε ) 2 dx η ε (w ε ϕk ε ) 2 dx k εz n B 2a ε (Rε) (εk) k εz n k εz n ( a ε (R ε ) ) n( a ε (R ε ) ) 2 δ 2 ε Rε (n 2) ε n δε 2 R 2(n 2) ε Cε n ε n δ ε = Cδ ε, where we used the fact if ε is small enough, then δ ε < 1 and R ε 1. Finally, since w ε and ϕ ε are both bounded in H 1 (thanks to (21)), (23) implies w ε L 2 C.

19 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) (iv) Using Lemma 2.3, we only have to prove that lim w ε α 0 dx = 0. ε We have: w ε = α (1 η ε )α + 2 η ε (w ε ϕ ε ) + η ε (w ε ϕ ε ) in ε. Moreover, (22) and (19) yield w ε ϕk ε δ ε Rε n 2 in B 2a ε (R ε ) \ B a ε (R ε ), and by definition of w ε and ϕk ε,wehave ( w ε ϕk ε α 0 x εk 2 2n Interior gradient estimates thus imply (w ε ϕ ε k ) δ ε Rε n 2 ) = 0 inb 4a ε (R) \ B a ε (R ε )/2. a ε (R ε ) 1 + Ca ε (R ε ) in B 2a ε (R ε ) \ B a ε (R ε ). We deduce (using (20)): w ε α dx (1 η ε )α dx + η ε (w ε ϕ ε ) dx + η ε w ε ϕ ε dx ε ε ε k εz n ε 1 + C k εz n ε 1 k εz n ε 1 a ε (R ε ) n + a ε (R ε ) 2 a ε (R ε ) n + C Cε n a ε (R ε ) n + Cδ ε Cε (σ 1)n + Cδ ε. It follows that lim w ε α 0 dx Cδ ε for any δ>0, which gives the result. Appendix A. Proof of Lemma 5.1 k εz n ε 1 a ε (R ε ) 1 ε B 2a ε (Rε) \B a ε (R) B 2a ε (Rε) \B a ε (R) w ε 0 ϕε k dx k εz n ε 1 k εz n ε 1 ( a ε (R ε ) R ε δ ε R n 2 aε (R ε ) 2( a ε (R) ) n ) n 2 (w ε ϕ ε k ) dx We recall that n 3 in this section. For any k Z n, we define S ε (k) = ε n n 2 S ε (k). Then Assumption 2 yields: cap ( S ε (k) ) = γ(k) γ. and Assumption 1 gives S ε (k) B M (k). (24)

20 394 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) For the sake of simplicity, we take k = 0. We recall that h is defined by 1 1 h(x) = n(n 2)ω n x n 2. Lemma 5.1 will be a consequence of the following lemma: Lemma 5.3. Let ϕ be defined by v(x) 1, x S ε (k, ω) ϕ(x) = inf v(x); v 0, lim x v(x) = 0 Then for any δ>0, there exists R, depending only on δ and M such that ϕ(x,ω) γh(x) δh(x) for all x such that x R. Proof. We recall that ϕ solves ϕ(x) = 0 for all x R n \ S, ϕ(x) = 1 for all x S, lim x ϕ(x) = 0. In particular, (24) and the maximum principle imply ϕ(x) M n 2 n(n 2)ω n h(x) = Mn 2 x n 2 in R n \ B M (0). (25) Next, we observe that 0 = ϕ ϕ dx = and so R n \S R n \S ϕ 2 dx = S R n \S ϕ 2 dx ϕϕ ν dσ(x) = S S ϕ ν dσ (x). ϕϕ ν dσ(x) Moreover, for any R M,wehave 0 = ϕ dx = ϕ ν dσ(x)+ ϕ ν dσ (x). B R \S S B R We deduce: γ = R n \S ϕ 2 dx = }. B R ϕ ν dσ(x) for all R M. (26) We now introduce the function ( ) x 1 ( ) ( ) x 1 x Θ(x) = h x 2 ϕ x 2 = n(n 2)ω n x n 2 ϕ x 2 defined for x B 1/M (0). A straightforward computation yields Θ = 0 inb 1/M (0) and (25) implies Θ(x) M n 2 n(n 2)ω n in B 1/M (0).

21 L.A. Caffarelli, A. Mellet / Ann. I. H. Poincaré AN 26 (2009) A more delicate computation, making use of the mean formula for harmonic functions, gives ϕ ν dσ(x) = Θ(0). B R Hence (26) yields Θ(0) = cap( S ε ) = γ. To conclude, we note that interior gradient estimates for harmonic functions imply the existence of a universal C (depending only on M) such that Θ(x) γ C x for all x 1/(2M). Inverting back, we deduce ϕ(x) γh(x) C h(x) for all x 2M, x which yields the result. References [1] M.A. Akcoglu, U. Krengel, Ergodic theorems for superadditive processes, J. Reine Angew. Math. 323 (1981) [2] L. Carbone, F. Colombini, On convergence of functionals with unilateral constraints, J. Math. Pures Appl. (9) 59 (4) (1980) [3] L.A. Caffarelli, P.E. Souganidis, L. Wang, Homogenization of fully nonlinear, uniformly elliptic and parabolic partial differential equations in stationary ergodic media, Comm. Pure Appl. Math. 58 (3) (2005) [4]. Cioranescu, F. Murat, Un terme étrange venu d ailleurs, in: Nonlinear Partial ifferential Equations and their Applications. Collège de France Seminar, vol. II, Paris, 1979/1980, in: Res. Notes in Math, vol. 60, Pitman, Boston, MA, 1982, pp , [5]. Cioranescu, F. Murat, Un terme étrange venu d ailleurs. II, in: Nonlinear Partial ifferential Equations and their Applications. Collège de France Seminar, vol. III, Paris, 1980/1981, in: Res. Notes in Math., vol. 70, Pitman, Boston, MA, 1982, pp , [6] G. al Maso, Asymptotic behaviour of minimum problems with bilateral obstacles, Ann. Mat. Pura Appl. (4) 129 (1981) [7] G. al Maso, P. Longo, Γ -limits of obstacles, Ann. Mat. Pura Appl. (4) 128 (1981) [8] G. al Maso, L. Modica, Nonlinear stochastic homogenization and ergodic theory, J. Reine Angew. Math. 368 (1986) [9] E. e Giorgi, G. al Maso, P. Longo, Γ -limits of obstacles, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 68 (6) (1980)