Two-scale convergence

Transcription

1 Two-scale convergence Dag Lukkassen y, Gabriel Nguetseng z and Peter Wall x (Dedicated to the memory of Jacques-Louis Lions ) Abstract This paper is devoted to the properties and applications of two-scale convergence introduced by Nguetseng in In a self-contained way, we present the details of the basic ideas in this theory. Moreover, we give an overview of the main homogenization problems which have been studied by this technique. We also bridge gaps in previous presentations, make generalizations and give alternative proofs. AMS Subject Classi cation (2000): 35B27, 35B40 Keywords: Homogenization, two-scale convergence, multiscale convergence, reiterated homogenization. 1 Introduction This paper is devoted to a special type of convergence in L p spaces. Let be an open bounded subset in R N, the unit cube in R N and (") a sequence of positive numbers converging to 0. In 1989 G. Nguetseng, see [58], proved that for each bounded sequence (u " ) in L 2 () there exists a subsequence, still indexed by ", and a u 2 L 2 ( ) such that u " (x)(x; x " ) dx! u(x; y)(x; y) dydx; (1) for every su ciently smooth (x; y) which is -periodic in y. Nguetseng also proved that for a bounded sequence (u " ) in W 1;2 () there exist functions u 2 L 2 ( ) and u 1 2 L 2 (; Wper( 1;2 )) such that, up to a subsequence, u "! u weakly in W 1;2 (); Du " (x); (x; x " ) dx! (Du(x) + D y u 1 (x; y); (x; y)) dydx; Published in: Int. J. of Pure and Appl. Math. 2, 1, 35-86, y Narvik University College, P.O.B. 385 N-8505 Narvik, Norway, dl@hin.no, z Department of Mathematics, University of aounde1, P.O.B. 812 aounde, Cameroun, gnguets@uycdc.uninet.cm x Department of Mathematics, Luleå University, S Luleå, Sweden, wall@sm.luth.se, 1

2 for every su ciently smooth (x; y): The latter result made it possible for Nguetseng to make a new proof of the homogenization result corresponding to the linear elliptic problems div A x " Du" = f on ; (2) u " = 0 where A is a -periodic matrix. Later on G. Allaire, see [6], started to call the type of convergence de ned by (1) two-scale convergence. Allaire also developed the theory further by studying some general properties of two-scale convergence. Moreover he used two-scale convergence to analyze several homogenization problems, both linear and nonlinear. Two-scale convergence has also been generalized to n-scale convergence (or multi scale convergence) in the obvious way. In [26], [49] and [50] n-scale convergence were used to study so called reiterated homogenization of the type div a x " ; ; x " ; Du n " = f on ; (3) u " = 0 Two-scale convergence is now a well-known concept among people who work with homogenization. A lot of di erent homogenization problems have been analyzed by using this tool (see Section 8). The aim of this paper is to, in a self contained way, present the details of the basic ideas in this theory and give an overview of the main homogenization problems which have been studied by this technique. We believe that it is important both for researchers in homogenization theory and other elds to have access to a paper of this type. In several cases we bridge gaps in previous presentations, generalize known results and present di erent proofs. We also clear out a common mistake concerning the de nition of two-scale convergence. Moreover, we point out how two-scale convergence is related to strong and weak convergence in L p (). We demonstrate how n-scale convergence may be used in the homogenization of (2) and (3) (earlier results by the authors G. Nguetseng [58] and J.L. Lions et al. [49], [50]). In particular we give some existence and uniqueness proofs which were left out in [49] and [50]. The paper is organized in the following way: In Section 2 we present the basic properties of some function spaces which are frequently used throughout the paper. In Section 3 we de ne two-scale convergence and discuss some of its main properties. Section 4 is devoted to the homogenization of (2) and Section 5 concerns so-called corrector results related to (2). In Section 6 we give some existence and uniqueness results for monotone operators which are important in Section 7 where we consider n-scale convergence and the homogenization of (3). Finally in Section 8 we present a guide to the literature. 2 Some function spaces and their properties Let be an open bounded subset of R N, and are two copies of the unit cube in R N. C per ( ) denotes the set of -periodic continuous functions de ned on R N. By de nition L 1 (; C per ( )) is the space of functions f :! C per ( ) which are measurable and satis es that R kf(x)k C per( ) dx < 1. Each function 2

3 f in L 1 (; C per ( )) can be identi ed with a function f(x; y) de ned on R N via f(x; y) = f(x)(y). We have the following characterization of the functions in L 1 (; C per ( )): Theorem 1 A function f belongs to L 1 (; C per ( )) if and only if there exists a subset E of measure zero in such that (a) For any x 2 ne, the function y! f(x; y) is continuous and -periodic, (b) For any y 2, the function x! f(x; y) is measurable. (c) The function x! sup y2 jf(x; y)j has nite L 1 () norm. Proof. The proof relays on Pettis theorem, see e.g. [73] page 131: A function f de ned on with values in the separable Banach space C per ( ) is measurable if and only if the real valued function x! h ; f(x)i is measurable for each in Cper( 0 ). (i) Let f be a function in L 1 (; C per ( )). By de nition it is clear that (a) and (c) hold. It remains to prove (b). Let be a functional on C per ( ). By Riez representation theorem there exists a unique -periodic Radon measure such that h ; gi = g d(y); for every g in C per ( ). From Pettis theorem we know that h ; f(x)i = f(x; y) d(y); is measurable for each. By choosing as the Dirac measure concentrated at y 0 we obtain that f(x; y) d y0 (y) = f(x; y 0 ) is measurable for each y 0. (ii) Assume that f satis es (a)-(c). Then it is clear that f is a function from into C per ( ) and that R sup y2 jf(x; y)j dx. It remains to prove that f :! C per ( ) is measurable. By Pettis theorem we would be done if we could prove that h ; f(x)i is measurable for each in Cper( 0 ). We will do this by, for each, constructing a sequence of functionals n such that h n ; f(x)i is measurable and h n; f(x)i = h ; f(x)i : n!1 Then it follows that h ; f(x)i is measurable since the it of measurable functions is measurable. Choose in Cper( 0 ) arbitrarily. Then there are two positive functionals + and such that = +, see [62] theorem By Riez representation theorem, see [63] theorem 2.14, + and can be identi ed with a unique positive measures + and, respectively. Let f i g be a partition of into disjoint cubes of side length 1=n, + i and i denotes + ( i ) and ( i ), respectively. Moreover, let i (y) = (y y i ) where y i is a point in 3

4 i and denotes the Dirac measure. We now de ne n as n = + n n, where +n ; f(x) = n ; f(x) = f(x; y) d( + 1 1(y) + : : : + + n N n N (y)) = f(x; y) d( 1 1 (y) + : : : + n N n N (y)) = n N X i=1 n N X i=1 f(x; y i ) + i ; f(x; y i ) i : From (b) it is clear that h n ; f(x)i is measurable since the sum of measurable functions is measurable. It remains to prove Let us rst prove h n; f(x)i = h ; f(x)i : n!1 +n ; f(x) = + ; f(x) (4) n!1 We note that it is enough to prove the inequality +n ; f(x) + ; f(x) : (5) n!1 For once (5) is established, the linearity of + n and + shows that n!1 +n ; f(x) = +n ; f(x) + ; f(x) = + ; f(x) ; n!1 which together with (5), shows that equality holds in (5). Let A be the set of functions s(x; y) which are simple in y and of the form s(x; y) = P n N i=1 i(x) i (y) such that s(x; y) f(x; y): Moreover, let S be the set of all functions s(x; y) which are simple in y such that s(x; y) f(x; y). First assume that the function y! f(x; y) is positive. Then +n ; f(x) = f(x; y) d( + n!1 n!1 1 1(y) + : : : + + n N n N (y)) = sup s(x; y) d + (y) s2a sup s(x; y) d + (y) (6) s2s = f(x; y) d + (y) = + ; f(x) ; where the second equality follows by the continuity of y! f(x; y). If y! f(x; y) is not positive then there exists a positive function y! g(x; y) such that f(x; y) = g(x; y) K(x) and by the same arguments as in (6) we obtain that +n ; f(x) + ; f(x) ; n!1 for all f(x) in C per ( ). In the same way it follows that n ; f(x) = ; f(x) : (7) n!1 4

5 By (4) and (7) we get that h n; f(x)i = h ; f(x)i : n!1 Let (" h ) h be a x sequence of positive real numbers (when it is clear from the context we will omit the subscript h) converging to 0. Then we have the following convergence result for functions in L 1 (; C per ( )): Theorem 2 Let f(x; y) 2 L 1 (; C per ( )). Then f(x; x=") is a measurable function on such that f(x; x " ) L1 kf(x; y)k L1 (;C () per( )) := sup jf(x; y)j dx; (8) and y2 f(x; x " ) dx = f(x; y) dydx: (9) Proof. From (a) and (b) in Theorem 1 we have that f is of Carathedory type which assure the measurability of f(x; x="). The inequality (8) is obvious. It remains to prove (9). Fix f 2 L 1 (; C per ( )): Let n be a positive integer and f i g a partition of consisting of cubes with side lengths n N such that j i \ j j = 0 if i 6= j, j i j = 1 N and = [n nn i=1 i. We denote the characteristic function of i, extended by -periodicity to R N, by i and y i is an arbitrary point in i. First we prove (9) for step functions f n of the form f n (x; y) = n N X i=1 f(x; y i ) i (y): We note that by Theorem 1 the function x! f n (x; y i ) is in L 1 (). Since i is a periodic function in L 1 () we have that i (x=") converges weak- to its mean value, especially this means that f n (x; x N n " ) dx = X i=1 f(x; y i ) i ( x " ) dx = f n (x; y) dy: (10) By using (10) we obtain that f(x; x " ) dx f(x; y) dydx f(x; x " ) f n(x; x " ) dx + f n (x; x " ) f n (x; y) dydx + f n (x; y) dydx f(x; y) dydx 2 sup jf(x; y) f n (x; y)j dx = 2 kf n fk L1 (;C per( )) ; y2 5

6 for every n 2 N. We are done if we show that kf n to 0. Since f(x; y) is continuous in y we have that fk L 1 (;C per( )) converges Moreover, g n (x) = sup jf n (x; y) f(x; y)j! 0 for a.e. x 2 : y2 g n (x) 2 sup jf(x; y)j 2 L 1 (): y2 The Lebesgue dominated convergence theorem implies that kf n fk L1 (;C per( )) = sup jf n (x; y) f(x; y)j dx! 0: y2 Example 3 Let f(x; y) 2 L 2 (; C per ( )). Then f(x; x " ) 2 dx = jf(x; y)j 2 dydx: Results similar to Theorem 2 can be proved for other function spaces as well. For example we have the following result: Theorem 4 Let B p (; ), 1 p < 1, denote any of the spaces L p (; C per ( )), L p per( ; C()), C(; C per ( )). Then B p (; ) has the following properties: 1. B p (; ) is a separable Banach space. 2. B p (; ) is dense in L p ( ): 3. If f(x; y) 2 B p (; ). Then f(x; x=") is a measurable function on such that f(x; x " ) Lp kf(x; y)k B(; ) : () 4. For every f(x; y) 2 B p (; ), one has f(x; x " ) p dx = jf(x; y)j p dydx: We also have the following result concerning weak convergence, see [15]. Theorem 5 Suppose that (x; y) = 1 (x) 2 (y), 1 2 L sp (), 2 2 L tp per( ) with 1 s; t 1; 1 p < 1 and such that 1 s + 1 t = 1: Then (x; x=") 2 L p () and (x; x " )! 1(x) weakly in L p (): 2(y) dy; 6

7 3 Two-scale convergence This section is devoted to the de nition of the concept of two-scale convergence and some of its properties. We also point out how some results concerning twoscale convergence are related to classical results for weak and strong convergence. From now on p and q are real numbers such that 1 < p < 1 and 1=p + 1=q = 1: De nition 6 Let (" h ) h be a x sequence of positive real numbers (when it is clear from the context we will omit the subscript h) converging to 0. A sequence (u " ) of functions in L p () is said to two-scale converge to a it u 2 L p ( ) if u " (x)(x; x " ) dx! u(x; y)(x; y) dydx; (11) for every 2 L q (; C per ( )): We remark that if we instead of the unit cube consider a parallelepiped = Q N i=1 (0; a i) the right hand side of (11) should be multiplied by 1= j j, where jj denotes the Lebesgue measure of. The two scale it is unique. Indeed, assume that a sequence (u " ) in L p () two-scale converges to u and v in L p ( ). Then u " (x)(x; x " )dx! u(x; y)(x; y)dydx; (12) u " (x)(x; x " )dx! v(x; y)(x; y)dydx; (13) for every 2 L q (; C per ( )). By (12) and (13) we have that [u(x; y) v(x; y)] (x; y)dydx = 0: (14) This implies that u(x; y) v(x; y) = 0 a.e.. We now give some examples of two-scale its. Example 7 Assume that u " :! R admits an asymptotic expansion of the form u " (x) = u 0 (x; x " ) + "u 1(x; x " ) + "2 u 2 (x; x " ) + ; where u i are smooth functions which are -periodic in the second argument. Then u " two scale converges to u 0 since Theorem 2 implies that u i (x; x " )(x; x " ) dx! u i (x; y)(x; y) dydx: Example 8 Let u(x; y) be a smooth function de ned on R N, -periodic in y. Then the two scale it of u " (x) = u(x; x=" 2 ) is equal to the weak L p it, namely R u(x; z) dz. Indeed, let f(x; y; z) be a smooth function which is -periodic in y and -periodic in z. Then it is well known that f(x; x " ; x " 2 )! f(x; y; z) dzdy weakly in L p : 7

8 This implies that u(x; x=" 2 )(x; x " ) dx! u(x; z)dz(x; y) dydx Or in other words the two scale it v of u " is v(x) = R u(x; z) dz: Theorem 9 If (u " ) converges to u in L p () then (u " ) two-scale converges to u 1 (x; y) = u(x): Proof. Let be a function in L q (; C per ( )). Then we have that u " (x)(x; x " ) dx u 1 (x; y)(x; y) dydx (15) ku " uk Lp () (x; x " ) Lq + () u(x)(x; x " ) dx u(x)(x; y)dydx : We note that u(x)(x; y) belong to L 1 (; C per ( )) and by Theorem 4 k(x; x=")k L q () is bounded. Thus by assumption and Theorem 2 both terms in the right hand side of (15) converges to 0: Theorem 10 Let (u " ) be a sequence in L p () which two-scale converges to u 2 L p ( ). Then u "! v(x) = u(x; y) dy weakly in L p (): and (u " ) is bounded. Proof. By the de nition of two-scale convergence it follows that u " (x)(x; x " ) dx! u(x; y)(x; y) dydx; for every 2 L q (; C per ( )). For independent of y we obtain that u " (x)(x) dx! u(x; y)dy(x) dx: Since every function in L q () can be identi ed with 2 L q (; C per ( )) independent of y the result follows. That (u " ) is bounded in L p () follows from the well-known fact that every weakly convergent sequence is bounded. We remark that the choice of space of test functions, L q (; C per ( )), in De nition 6 is essential in Theorem 10. Unfortunately one often nd, in the literature available on two-scale convergence, the mistake that one claim Theorem 10 even though the space of test functions L q (; C per ( )) is replaced by D(; Cper( 1 )), i.e. the space of measurable functions on R N such that u(x; ) 2 Cper( 1 ) for any x 2 and the map x 2 7! u(x; ) 2 Cper( 1 ) is inde nitely di erentiable with compact support in. The following example exhibits a sequence (u " ) in L p () satisfying (11) for 2 D(; Cper( 1 )) but which is neither bounded nor weakly convergent in L p (): 8

9 Example 11 Let = (0; 1); u(x; y) = 0; and u " be de ned as 1=" if 0 < x "; u " (x) = 0 if " < x < 1: Then u " (x)(x; x " ) dx! u(x; y)(x; y) dydx; is satis ed for all (x; y) in D(; Cper( 1 )), but (u " ) does not converge to 0 weakly in L p (): This is easily seen by choosing the function g in the dual as g 1; 1 0 u " g dx = 1: We also note that (u " ) is not bounded in L p (). Even if D(; C 1 per( )) is replaced by C(; C 1 per( )) we are still in trouble, which is shown by the following example: Example 12 Let = (0; 1) and u " be de ned as u" (x=") if 1=4 < x 3=4; u " (x) = 0 elsewhere where u " is the (0; 1)-periodic extension to R of the function de ned in the Example 11, and 1 if 1=4 < x 3=4; u(x; y) = 0 elsewhere. Then u " (x)(x; x " ) dx! u(x; y)(x; y) dydx; is satis ed for all (x; y) in C(; C 1 per( ));but (u " ) does not converge to v(x) = R u(x; y) dy weakly in Lp () and is certainly not bounded. The following result shows that it is possible to replace L q (; C per ( )) by D(; C 1 per( )) in our de nition of two-scale convergence provided we add the assumption that (u " ) is bounded in L p (). Proposition 13 Let (u " ) be a bounded sequence in L p () such that u " (x) (x; x " ) dx! u(x; y) (x; y) dydx: for every 2 D(; C 1 per( )). Then u " two-scale converges to u. Proof. Let be an arbitrary function in L q (; C per ( )): Moreover, let ( m ) be a sequence of functions in D(; Cper( 1 )) which converges to in L q (; C per ( )) as m! 1. We have that = m!1 u " (x)(x; x ) dx (16) " u " (x)( m )(x; x " ) dx + u " (x) m (x; x " ) dx : 9

10 By assumption we can pass to the it in the second term on the right hand left hand side of (16). Indeed, u " (x) m (x; x ) dx = u(x; y) m!1 " m (x; y) dydx (17) m!1 = u(x; y) (x; y) dydx; where the last equality follows by noting that u(x; y) [ m (x; y) (x; y)] dydx c k m k Lq ( ) c k m k Lq (;C per( ))! 0: It remains to show that the rst term in the right hand side of (16) is equal to 0. By the Hölder inequality and the fact that (u " ) is bounded in L p () we have that m!1 u " (x)( m )(x; x ( ) dx " c m ) (; m!1 " ) L q () c k mk m!1 L q (;Cper 1 ( )) = 0 and the proof is complete. We now state a compactness result for two-scale convergence. Theorem 14 For each bounded sequence, (u " ), in L p () there exist a subsequence and a u 2 L p ( ) such that the subsequence two-scale converges to u. Proof. By Hölder s inequality and Theorem 2 it follows that: u " (x)(x; x " ) dx ku " k L p () (x; x " ) Lq c k(x; y)k L () q (;C per( )) : (18) This means that each u " can be identi ed with an element U " in the dual space, L q (; C per ( )) 0, of L q (; C per ( )), i.e. hu " ; i = u " (x)(x; x " ) dx: The sequence (U " ) is bounded in L q (; C per ( )) 0. Indeed, (18) implies that ku " k = sup jhu " ; ij = sup u " (x)(x; x kk=1 kk=1 " ) dx c. Next, we use the well-known result that it is possible to extract a subsequence from any bounded sequence in the dual space of a separable Banach space, which converges weak-*. In our case we obtain that there exists a U 2 L q (; C per ( )) 0 such that u " (x)(x; x " ) dx = hu "; i! hu; i, (19) 10

11 for every 2 L q (; C per ( )). The proof would be done if we show that U is on the form hu; i = u(x; y)(x; y) dydx: By (18) we have that jhu " ; ij c (x; x " ) Lq : () By taking (19) and Theorem 2 into account and passing to the it we get jhu; ij c kk Lq ( ) ; for each 2 L q (; C per ( )). Since L q (; C per ( )) is dense in L q ( ) we can for each 2 L q ( ) nd a sequence ( h ) in L q (; C per ( )) such that h! in L q ( ) this means that we can de ne the extension e U of U to L q ( ) such that for each D eu; E = hu; h i : Riez representation theorem guarantees the existence of an u 2 L p ( ) such that D E eu; = u(x; y)(x; y) dydx; for each in L q ( ). Especially, for in L q (; C per ( )) we have that D E hu; i = eu; = u(x; y)(x; y) dydx: Theorem 14 can be compared with the well-known fact that each bounded sequence, (u " ), in L p (); 1 < p < 1; contains a weakly convergent subsequence. The following two theorems shows that if (u " ) two-scale converges then by density arguments the relation (11) holds for test functions in other spaces than L q (; C per ( )): Theorem 15 Let (u " ) be a sequence in L p () which two-scale converges to u. Then u " (x) (x; x " ) dx! u(x; y) (x; y) dydx; for every in L q per( ; C()). Proof. Fix 2 L q per( ; C()). Let ( m ) be a sequence in D(; Cper( 1 )) such that m! in L q ( ). We have that u " (x) (x; x h ) dx = " u " (x) (x; x m!1 " ) m(x; x i " ) dx + u " (x) m (x; x ) dx (20) m!1 " 11

12 Since u " two-scale converges to u and m converges in L q ( ) we get that the second term in the right hand side of (20) is equal to u " (x) m (x; x ) dx = u(x; y) m!1 " m (x; y) dydx m!1 = u(x; y) (x; y) dydx: This means that we would be done if we prove that the rst term in the right hand side of (20) is equal to 0. By the Hölder inequality and the fact that (u " ) is bounded in L p () (since (u " ) two-scale converges) we get that m!1 c m!1 h u " (x) (x; x " ) m(x; x i " ) dx (x; x " ) m(x; x " ) q 1=q dx : First we note that (x; x=") m (x; x=") is in L q per( ; C()): Then by applying Theorem 4 and subsequently using the assumption that m! in L q ( ) implies that m!1 c m!1 h u " (x) (x; x " ) m(x; x i " ) dx j (x; y) m (x; y)j q dx 1=q = 0: Theorem 16 Let (u " ) be a sequence in L p () which two-scale converges to u. Then u " (x) (x; x " ) dx! u(x; y) (x; y) dydx; for every on the form (x; y) = 1 (x) 2 (y), 1 2 L sq (), 2 2 L tq per( ) with 1 s; t 1 and such that 1=s + 1=t = 1: Proof. Fix 1 ; 1 and 2. Let ( m ) be a sequence in D() which converges to 1 in L sq () and ( m ) be a sequence in Cper( 1 ) which converges to 2 in L tq ( ). We have that u " (x) (x; x ) dx = " m!1 + m!1 h u " (x) (x; x " ) m(x) m ( x i " ) dx u " (x) m (x) m ( x ) dx (21) " Since u " two-scale converges to u and m m converges in L q ( ) we get that the second term in the right hand side of (21) is equal to u " (x) m (x) m ( x ) dx = u(x; y) m (x) m!1 " m (y) dydx m!1 = u(x; y) (x; y) dydx: 12

13 This means that we would be done if we prove that the rst term in the right hand side of (21) is equal to 0. By the Hölder inequality and the fact that (u " ) is bounded in L p () (since (u " ) two-scale converges) we get that c h u " (x) (x; x " ) m(x) m ( x i " ) dx (x; x " ) m(x) m ( x " ) q 1=q dx c j 1 (x) m (x)j q 2 ( x " ) q + jm (x)j q 2 ( x " ) m( x " ) q 1=q dx ; observe that c denotes a constant that may di er from one line to another. By applying Theorem 5 we obtain h u " (x) (x; x " ) m(x) m ( x i " ) dx (22) 1=q c j 1 (x) m (x)j q j 2 (y)j q + j m (x)j q j 2 (y) m (y)j dydx q : By using Hölder s inequality in (22) together with the facts that m! 1 in L sq () and m! 2 in L tq ( ) we get h u " (x) (x; x m!1 " ) m(x) m ( x i " ) dx = 0: and we are done. Theorem 17 Let (u " ) be a sequence in L p () which two-scale converges to u 2 L p ( ). Then where v(x) = R u(x; y)dy. inf ku "k Lp () kuk L p ( ) kvk L p () ; Proof. Let ( m ) be a sequence in L q (; C per ( )) such that m converges to juj p 2 u strongly in L q ( ). The oung inequality for real numbers a and b states that ab jaj p =p + jbj q =q and implies that ju " j p dx p u " (x) m (x; x " )dx (p 1) m (x; x " ) q dx: By passing to the it in " we get inf ku "k p L p () p u(x; y) m (x; y) dydx (p 1) By passing to the it in m we obtain inf ku "k p L p () p ju(x; y)j p dydx (p 1) = kuk p L p ( ) : j m (x; y)j q dydx: ju(x; y)j p dydx 13

14 Next, by Jensen s inequality we have that kvk p p L p () = u(x; y)dy dx ju(x; y)j p dydx = kuk p L p ( ) ; which nish the proof. We remark that the result in Theorem 17 can be compared with the wellknown fact that if u "! u weakly in L p () then inf ku "k Lp () kuk L p () : Theorem 18 Let (u " ) be a sequence in L p () which two-scale converges to u 2 L p ( ) and assume that ku "k L p () = kuk L p ( ) : (23) Then, for any sequence (v " ) in L q () which two-scale converges to v 2 L q ( ), we have that u " (x)v " (x)(x) dx! u(x; y)v(x; y)(x) dydx ; (24) for every in C0 1 (): Moreover, if the -periodic extension of u belong to L p (; C per ( )), then u " (x) u(x; x " ) Lp = 0: (25) () Proof. Let ( m ) be a sequence in L p (; C per ( )) such that m converges to u strongly in L p ( ) and a function in C0 1 (). To prove (24) we rst note that by passing to the two scale it and the choice of ( m ) we have m (x; x m!1 " )v "(x)(x) dx = u(x; y)v(x; y)(x) dydx: (26) We also have that u " (x)v " (x)(x) dx u(x; y)v(x; y)(x) dydx hu " (x) m (x; x i " ) v " (x)(x) dx (27) + m (x; x " )v "(x)(x) dx u(x; y)v(x; y)(x) dydx : From (26) and (27) we obtain sup u " (x)v " (x)(x) dx u(x; y)v(x; y)(x) dydx (28) sup sup hu " (x) m (x; x i m!1 " ) v " (x)(x) dx : It is clear that (24) follows from (28) if we prove sup sup hu " (x) m (x; x i m!1 " ) v " (x)(x) dx = 0: (29) 14

15 The Hölder inequality and the facts that each two scale convergent sequence is bounded imply hu " (x) m (x; x i " ) v " (x)(x) dx (30) max (x) u " (x) m (x; x x2 " ) p 1=p 1=q dx jv " (x)j q dx c u " (x) m (x; x " ) L : p () The Clarksson inequalities give, respectively, for p 2 and for p 2 u " m (x; x " ) p L p () " 2 p 1 2 ku "k p L p () m (x; x " ) p u " + m (x; x " ) # p L p () 2 ; L p () u " m (x; x " ) q L p () 2 p 1 " 1 2 ku "k p L p () m (x; x " ) p L p () 1 p 1 u " + m (x; x " ) 2 q L p () By (23) and an application of Theorem 17 to the sequence u " (x) + m (x; x=") (which two-scale converges to u + m 2 L p ( )) it follows sup u " m (x; x " ) p L p () " 2 p 1 2 kukp L p ( ) k mk p L p ( ) sup u " m (x; x " ) q L p () 2 p 1 " 1 2 kukp L p ( ) k mk p L p ( ) u + m 2 1 p 1 p L p ( ) # u + m 2 ; q L p ( ) # (31) (32) # By the choice of m we have that (u + m ) =2 converges to u in L p ( ). Applying sup as m! 1 on both sides in (31) and (32) gives sup sup u " m (x; x m!1 " ) L = 0: (33) p () Now (29) follows from (30) and (33). Finally, we prove (25). Assume that u is smooth, e.g. u is a function in L p (; C per ( )), then the Clarksson inequalities with m replaced by u implies (33) which is exactly (25). From the proof it is clear that the function 2 C0 1 () in Theorem 18 may be replaced by a function (x; x=") where 2 D(; Cper( 1 )). Theorem 18 can be compared with the following well-known result: If (u " ) converges weakly to u in L p () and ku "k L p () = kuk L p () : : : 15

16 Then ku " uk Lp () = 0: In Theorem 18 some type of regularity of the two scale it was required so u(x; x=") in (25) is measurable. It is natural to ask whether such regularity always holds for the two-scale it or not. The answer to this question is given in the next theorem. Theorem 19 Any function u 2 L p ( ) is attained as a two-scale it. Proof. Fix u(x; y) in L p ( ). Let (u n ) be a sequence of functions in L p (; C per ( )) which converges to u in L p ( ). L q (; C per ( )) is a separable Banach space (see Theorem 4), i.e. we can nd a countable dense family ( k ) which is dense in L q (; C per ( )). Moreover, L q (; C per ( )) is dense in L q ( ), see Theorem 4. This means that for each in L q ( ) and every > 0 it is possible to rst nd in L q (; C per ( )) and subsequently a k such that k k k L q ( ) k k k L q ( ) + k k L q ( ) k k k Lq (;C per( )) + 2: This means that ( k ) is dense in L q ( ). For x n, the sequence (u n (x; x=" h )) two scale converges to u n (x; y), i.e. for every > 0 there exists a "(n) 2 (") such that " > "(n) implies that u n (x; x " )(x; x " ) dx u n (x; y)(x; y)dydx. Let us now extract a diagonal sequence in the following way: Let n = ku n uk Lp ( ) then there exists a subsequence ("(n)) n of (") such that "(n)! 0 and p u x n(x; "(n) ) dx ju n (x; y)j p dydx n ; (34) x u n (x; "(n) ) x k(x; "(n) ) dx u n (x; y) k (x; y) dydx n ; (35) for k = 1; : : : ; n. Let v "(n) be de ned as v "(n) (x) = u n (x; x="(n)). We have now constructed a sequence ("(n)) n such that "(n)! 0 as n! 1 and a sequence of functions v "(n) such that (v "(n) ) two scale converges to u. Indeed, for each > 0 16

17 and each in L q (; C per ( )) we can nd a k such that k k k L q ( ) < x v "(n) (x)(x; "(n) ) dx u(x; y)(x; y) dydx (36) = x u n (x; "(n) )(x; x "(n) ) dx u(x; y)(x; y) dydx x u n (x; "(n) )(x; x "(n) ) dx x u n (x; "(n) ) x k(x; "(n) ) dx + x u n (x; "(n) ) x k(x; "(n) ) dx u n (x; y) k (x; y) dydx + u n (x; y) k (x; y) dydx u(x; y) k (x; y) dydx + u(x; y) k (x; y) dydx u(x; y)(x; y) dydx : Let us know study the four terms in the right hand side of (36) term by term. Term 1: By (34) u n (x; x="(n)) is bounded in L p (). Hölder s inequality implies q x n!1 u n (x; "(n) )(x; x "(n) ) dx x u n (x; "(n) ) x k(x; "(n) ) dx q c n!1 (x; x "(n) ) x k(x; "(n) ) dx (37) = c j(x; y) k (x; y)j q dydx c Term 2: By (35) we have x u n (x; "(n) ) x k(x; "(n) ) dx n!1 Term 3: By the Hölder inequality we have that u n (x; y) k (x; y) dydx n!1 ku n uk n!1 Lp ( ) k kk Lp ( ) = 0 u n (x; y) k (x; y) dydx = 0 (38) u(x; y) k (x; y) dydx (39) Term 4: By the Hölder inequality we have that u(x; y) k (x; y) dydx u(x; y)(x; y) dydx (40) kuk Lp ( ) k k k Lp ( ) c By (36)-(40) and the fact that was arbitrary chosen the proof is nished. So far we have considered two-scale convergence in L p spaces. We now proceed by two-scale convergence in Sobolev spaces. By Wper( 1;p ) we denote the subset of W 1;p ( ) of all functions u with mean value zero which have the same trace on opposite faces of. We will not make any di erence in the notation of u 2 Wper( 1;p ) and it s -periodic extension to an element in W 1;p loc (RN ): Moreover, (x; y) denotes the usual scalar product between x and y in R N : 17

18 Theorem 20 Let (u " ) be a sequence in W 1;p () such that u "! u weakly in W 1;p (): Then (u " ) two-scale converges to u and there exist a subsequence " 0 and a u 1 2 L p (; W 1;p per( )) such that Du " 0! Du + D y u 1 two-scale. Proof. Since every weakly convergent sequence is bounded we conclude that (u " ) and (Du " ) are bounded in L p () and [L p ()] n, respectively. By Theorem 14 there exists a subsequence (still denoted by "), u 0 (x; y) in L p ( ) and U(x; y) in [L p ( )] n such that for every (x; y) in D(; Cper( 1 )) and (x; y) in D(; Cper( 1 )) n, we have u " (x)(x; x " ) dx = u 0 (x; y)(x; y) dydx; Du " (x); (x; x " ) dx = (U(x; y); (x; y)) dydx: (41) By Theorem 10 we know that u(x) = R u 0(x; y) dy. If we show that u 0 does not depend on y we would, by the uniqueness of the weak it, have that the whole sequence two-scale converges to u. Indeed, integration by parts gives us " Du " (x); (x; x " ) dx = In the it we obtain 0 = u " (x) h"div x (x; x " ) + div y(x; x " ) i dx: u 0 (x; y)div y (x; y) dydx: This implies that u 0 does not depend on y, see [2] page 178. Next, we want to show that U(x; y) is equal to Du(x)+D y u 1 (x; y). Let be a function such that div y (x; y) = 0. Then integration by parts in (41) gives (U(x; y); (x; y)) dydx = Du " (x); (x; x " ) dx = u " (x) = u(x)div x (x; y) dydx = (Du(x); (x; y)) dydx: div x (x; x" ) + 1" div y(x; x" ) dx Thus, for any function (x; y) in D(; Cper( 1 )) n such that divy (x; y) = 0, we have (U(x; y) Du(x); (x; y)) dydx = 0: This means that U(x; y) L p (; Wper( 1;p )) such that Du(x) is a gradient, i.e. there exists a u 1 (x; y) in D y u 1 (x; y) = U(x; y) Du(x): 18

19 4 Homogenization of a linear second order elliptic problem In this section we demonstrate how two-scale convergence can be used to study a classical homogenization problem. For more information concerning this problem and homogenization theory we refer to e.g. [29]. Let A(y) = (a ij (y)) be a -periodic and measurable N N matrix. Moreover, let f 2 L 2 () and assume that there exists two constants 0 < < 1 such that jaj jj and (A; ) jj 2 : (42) Let us now study the class of partial di erential equations given by 8 < div A x " Du" = f on ; : u " = 0 The corresponding weak formulation is 8 R < Du" ; Dw dx = R 1;2 fw dx for every w 2 W 0 () : A x " u " 2 W 1;2 0 (): (43) Since A is periodic the functions a ij (x=") will oscillate very rapidly for small values of ". The eld of mathematics which treat the asymptotic behavior of the sequence of solutions (u " ) is known as homogenization. In the theorem below we use two-scale convergence to prove that u " converges to u, where u is given as the solution of a variational problem. This it variational problem is called the homogenized problem since it not any longer involve rapidly oscillating functions. Theorem 21 The sequence of solutions of (43) converges weakly to u in W 1;2 0 () and the sequence Du " two-scale converges to Du(x) + D y u 1 (x; y), where (u; u 1 ) is the unique solution in W 1;2 0 ()L 2 (; Wper( 1;2 )) of the homogenized equation: (A(y)(Du(x) + D y u 1 (x; y)); Dv(x) + D y v 1 (x; y)) dydx = for every v 2 W 1;2 0 () and v 1 2 L 2 (; W 1;2 per( )). fv dx; (44) Proof. By the assumptions on A the Lax-Milgram lemma guarantees existence and uniqueness of the solution of (43) and we also have the a priori estimate ku " k W 1;2 0 () c. Since W 1;2 0 () is re exive we can use the well-known fact that from each bounded sequence in a re exive space one can extract a weakly convergent subsequence. Thus there exists a subsequence, still denoted by ", such that u "! u weakly in W 1;2 0 (): 19

20 From Theorem 20 it follows that there exists a subsequence, still denoted by ", and u 1 2 L 2 (; W 1;2 per( )) such that Du "! Du + D y u 1 two-scale. Let w " be on the form w " (x) = v(x) + "v 1 (x; x="), where v 2 C 1 0 () and v 1 2 D(; C 1 per( )). By using w " as a test function in (43) we obtain that or = = A +" f v(x) + "v 1 (x; x " ) dx x " Du " ; Dv(x) + D y v 1 (x; x " ) dx x A Du " ; D x v 1 (x; x " " ) dx Du " ; A t x " +" Du " ; A t x " f v(x) + "v 1 (x; x " ) dx. hdv(x) + D y v 1 (x; x " ) i dx D x v 1 (x; x " ) dx We can pass to the it in the left hand side by using Theorem 15. In the right hand side we can pass to the it by using Theorem 4. Indeed, Du(x) + D y u 1 (x; y); A t (y) [Dv(x) + D y v 1 (x; y)] dydx = fv dx or (A(y) [Du(x) + D y u 1 (x; y)] ; Dv(x) + D y v 1 (x; y)) dydx = fv dx: (45) By density (45) holds true for each (v; v 1 ) in W 1;2 0 () L 2 (; Wper( 1;2 )). So far we have proved the result for a subsequence. If we prove that the solution (u; u 1 ) of (45) is unique then it follows that it is true for the whole sequence. This will be done by an application of Lax-Milligram lemma. Let us rst de ne the Hilbert space H = W 1;2 0 () L 2 (; Wper( 1;2 )). Moreover, If U = (u; u 1 ) 2 H, then the norm of U is de ned as kuk 2 H = kduk2 L 2 (;R N ) + kd yu 1 k 2 L 2 ( ) : We also de ne the bilinear form b as b(u; V ) = (A(y) [Du(x) + D y u 1 (x; y)] ; Dv(x) + D y v 1 (x; y)) dydx and the functional fon e D E H as ef; V = R fv dx: The homogenized problem (45) then reads as follows: Find U in H such that D E b(u; V ) = ef; V for each V 2 H. 20

21 The Lax-Milligram lemma guarantees the existence an uniqueness of the solution U if b is coercive and continuous, i.e. that there exists constants, 0 < c 1 ; c 2 < 1 such that b(u; U) c 1 kuk 2 H and jb(u; V )j c 2 kuk H kv k H : Let us prove that b is coercive: The assumptions on A implies that b(u; V ) = (A(y) [Du(x) + D y u 1 (x; y)] ; Du(x) + D y u 1 (x; y)) dydx jdu(x) + D y u 1 (x; y)j 2 dydx = jduj 2 dydx + jd y u 1 j 2 dydx +2 (Du(x); D y u 1 ) dydx : By Greens formula and the -periodicity of u 1 (x; ) we obtain that the third integral in the right hand side is equal to 0. Thus we have that b(u; U) kuk 2 H : Let us no check that b is bounded. Schwartz inequality, (42) and Hölder s inequality implies that jb(u; V )j = (A(y) [Du(x) + D y u 1 (x; y)] ; Dv(x) + D y v 1 (x; y)) dydx ja(y) [Du(x) + D y u 1 (x; y)]j jdv(x) + D y v 1 (x; y)j dydx jdu(x) + D y u 1 (x; y)j jdv(x) + D y v 1 (x; y)j dydx kuk H kv k H ; and the proof is done. We remark that the homogenization problem just described is classical and is often analyzed by using Tartar s method of oscillating test functions, see e.g. [29]. Then the result is formulated in the following way: The solutions u " of (43) converges weakly in W 1;2 0 () to the solution u of the homogenized problem 8 < R (A homdu; Dv) dx = R 1;2 fv dx for every v 2 W 0 (); : u 2 W 1;2 0 (); where the constant matrix A hom is given by A hom e i = A(y) [e i + D y w i (y)] dy for i = 1; : : : ; N; where w i is the solution of the local problem 8 R < (A(y) [e i + D y w i (y)] ; v 1 ) dx = 0 for every v 1 2 W 1;2 : w i 2 W 1;2 per( ): 21 per( ); (46) (47)

22 By using the relation u 1 (x; y) = NX w i (y)d xi u(x): (48) i=1 we will see that the formulation (46) of the homogenized equation is equivalent to (44) in Theorem 21. By uniqueness of the solutions u; u 1 and w i it is su cient to prove that if u and w i are solutions of (46) and (47) then (u; u 1 ) is a solution of (44). Assume that (46) and (47) hold. We note that (A hom Du; Dv) dx " ( N )#! X = A(y) Du + D y w i (y)d xi u dy; Dv dx: Put u 1 (x; y) = P N i=1 w i(y)d xi u(x) and we get (A(y) [Du + D y u 1 ] ; Dv) dydx = hf; vi (49) By summing the equations in (47) we have that (A(y) [Du + D y u 1 ] ; D y v 1 ) dydx = 0; (50) for every v 1 2 L 2 (; W 1;2 per( )). Equation (44) now follows by (49) and (50). 5 Correctors By Theorem 21 the sequence of solutions of (43) converges weakly to u in W 1;2 0 () and the sequence Du " two-scale converges to Du(x)+D y u 1 (x; y), where (u; u 1 ) is the unique solution in W 1;2 0 () L 2 (; Wper( 1;2 )) of the homogenized equation: (A(y)(Du(x) + D y u 1 (x; y)); Dv(x) + D y v 1 (x; y)) dydx = fv dx; (51) for every v 2 W 1;2 0 () and v 1 2 L 2 (; Wper( 1;2 )). Since the imbedding of W 1;2 0 () in L 2 () is compact we have that u " converges to u strongly in L 2 (). For the gradients we only have weak convergence of Du " to Du in L 2 (; R N ). To improve this convergence we have to add an extra term (corrector) which take care of the oscillations.. We will now see that such corrector results can be obtained by using two-scale convergence. Theorem 22 Let u ", u; u 1 and w i be de ned as in (43), (51) and (48) and let 1 s; t 1 such that 1=s + 1=t = 1: Moreover, assume that D xi u 2 L 2s () and w i 2 L 2t per( ): Then, i=1 Du " Du Du 1 (; =")! 0 in L 2 () N : 22

23 Proof. By the ellipticity condition (42) on A and Hölder inequality we get that Du " Du D y u 1 (x; x " ) 2 dx (52) A( x hdu " ) " Du D y u 1 (x; x i " ) ; Du " Du D y u 1 (x; x " ) dx = A( x " )Du "; Du " dx Du " ; A + A t ( x hdu " ) + D y u 1 (x; x i " ) dx + A( x hdu " ) + D y u 1 (x; x i " ) ; Du + D y u 1 (x; x " ) dx: By (48) we also know that u 1 is of the form u 1 (x; y) = NX D xi u(x)w i (y); (53) i=1 where w i is the solution of the local problem (47). Let us now study the convergence as "! 0 for the three terms in the right hand side of (52) separately. Term 1: By (43) we have that A( x " )Du "; Du " dx = fu " dx! fu dx as "! 0: (54) Term 2: By (53) and Theorem 16 we have Du " ; A + A t ( x hdu " ) + D y u 1 (x; x i " ) dx (55) #! = Du " ; A + A t ( x NX "Du " ) + D xi u(x)d y w i ( x " ) dx! = 2 i=1 Du + Du 1 (x; y); A + A t (y) [Du + D y u 1 (x; y)] dydx (A(y) [Du + D y u 1 (x; y)] ; Du + Du 1 (x; y)) dydx: Term 3: By (53) and Theorem 5 we have A( x hdu " ) + D y u 1 (x; x i " ) ; Du + D y u 1 (x; x " ) dx # = A( x NX "Du " ) + D xi u(x)d y w i ( x " ) ;! Du + i=1! NX D xi u(x)d y w i ( x " ) dx (56) i=1 (A(y) [Du + D y u 1 (x; y)] ; Du + Du 1 (x; y)) dydx: 23

24 Finally by (52), (54), (55), and (56) we have Du " Du D y u 1 (x; x " ) 2 dx fu dx (A(y) [Du + D y u 1 (x; y)] ; Du + Du 1 (x; y)) dydx = 0; where we in the last equality used (51). 6 Monotone operators Let X be a real re exive Banach space, let A : X! X and let h; i denote the dual pairing between X and X : We recall the following de nitions: A is called monotone if A is called strictly monotone if A is called hemicontinuous if for all u; v 2 X A is called coercive if hau 1 Au 2 ; u 1 u 2 i 0 8u 1 ; u 2 2 X hau 1 Au 2 ; u 1 u 2 i > 0 whenever u 1 6= u 2 A(u + tv) = Au weakly in X t!0 hau; ui = +1 kuk!1 kuk Let us consider the operator equation of the form Au = b; u 2 X: (57) Theorem 23 (Browder (1963), Minty (1963)) Suppose that A is strictly monotone, hemicontinuous and coercive. Then (57) has a unique solution u 2 X for every b 2 X : Let a " : R N! R N ful lls the conditions: 1. a " is on the form a " (x; ) = a( x " ; x " 2 ; ); where a(y; z; ) is -periodic in y and -periodic in z. 2. a " (; ) is Lebesgue measurable for every 2 R N ; 24

25 3. There exists two constants c 1 ; c 2 > 0 and two constants and, with 0 min f1; p 1g and max fp; 2g < 1 such that a satis es the following boundedness, continuity and monotonicity assumptions: a " (x; 0) = 0 for a.e. x 2 ; ja " (x; 1 ) a " (x; 2 )j c 1 (1 + j 1 j + j 2 j) p 1 j 1 2 j ; (58) (a " (x; 1 ) a " (x; 2 ); 1 2 ) c 2 (1 + j 1 j + j 2 j) p j 1 2 j Fix " > 0. Let us consider a partial di erential equation of the form div(a " (x; Du " ) = f on ; u " 2 W 1;p 0 (): (59) The corresponding weak formulation (operator equation) can be written as T u " = b; u " 2 X = W 1;p 0 (); (60) where T is given by and where f 2 L q (): ht w; vi = (a " (x; Dw); Dv)dx hb; vi = fvdx; Lemma 24 The operator T maps X into X and is hemicontinuous, strictly monotone and coercive. Moreover, b 2 X (hence (60) has a unique solution). Proof. We put kvk X = kdvk L p (this is possible due to Friedrichs inequality). We rst prove that T w 2 X for all w 2 X: By Schwarz inequality, the growth conditions on a and the Hölder inequality and the fact that (1 + jdwj) p 1 jdwj (1 + jdwj) p 1 ; we have that for all v; w 2 X and all k 2 R jht w; vij c 1 (1 + jdwj) p 1 jdvj dx Hence, c 1 kjdwj + 1k 1=q L kdvk p L p = c 1 kjdwj + 1k 1=q L p kvk X : ht w; vi c kvk X ; where c is a real constant which is independent of v: This and the fact that T w is linear shows that T w 2 X : We proceed by showing that T is hemicontinuous. By similar arguments as above we have for all u; v; w 2 X and k 2 R that jht w T (w + ku); vij jkj c 1 (1+jDwj+jD(w + ku)j) p 1 jduj jdvj dx: 25

26 By letting k! 0 and using the Hölder inequality we se that the right side of this inequality vanishes. To see this put g = 1 + jdwj + jd(w + ku)j + jduj : It follows that (1 + jdwj + jd(w + ku)j) p 1 jduj jdvj dx jgj p 1 jgj jdvj dx = jgj p 1 jdvj dx kgk p 1 L kdvk Hölder p L : p Now, since kgk L p k1k L p + 2 kdwk L p + k kduk L p + kduk L p : we have that the integral is bounded as k! 0: Accordingly, the right side vanishes. We conclude that ht w T (w + ku); vi! 0 as k! 0; i.e. T (w + ku)! T (w) in X weakly, that is T is hemicontinuous. Finally we show that T is strictly monotone and coercive. We have that ht w T v; w vi = (a(x; Dw) a(x; Dv); Dw Dv)dx c 2 (1 + jdwj + jdvj) p jdw Dvj dx c 2 k1 + jdwj + jdvjk p L p kdw Dvk L p where we in the last inequality have used the reversed Hölder inequality with dual exponents p=(p ) and p=: This shows that T is strictly monotone. Moreover, by this inequality we have that which implies that ht w; wi c 2 k1 + jdwjk p L p kdwk L p c 2 (k1k L p + kdwk L p) p kdwk L p = c 2 (k1k L p + kwk X ) p kwk X = c 2 (k1k L p + kwk X ) p kwk X ; k1k L p + kwk X ht w; wi kwk X! 1; as kwk X! 1. Thus T is coercive. Moreover, By Hölder and Friedrich s inequality we obtain that jhb; vij = fvdx kfk L kvk q L c kfk p L kdvk q L = c kfk p L kvk q X ; 26

27 thus b 2 X and the proof is complete. Now, assume that the function z! a(y; z; ) belongs to L q per(; C per ( )) for every (note e.g. that this ensure that a " (; ) is measurable) and let X = X 0 X 1 X 2 = W 1;p 0 () L p (; W 1;p per( )) L p ( ; W 1;p per()) Consider the weak formulation (operator equation) Su = b; u 2 X; (61) where S is de ned as hsw; vi = (a(y; z; Dw 0 + D y w 1 + D z w 2 ); Dv 0 + D y v 1 + D z v 2 ) dz dy dx and hb; vi = fv 0 dx; where f 2 L q () for all w = (w 0 ; w 1 ; w 2 ) and v = (v 0 ; v 1 ; v 2 ) in X : Lemma 25 Let X be endowed with the norm : kvk X = kdv 0 + D y v 1 + D z v 2 k L p (where L p = L p ( )). Then X is a re exive Banach space and S maps X into X. Moreover, S is hemicontinuous, strictly monotone and coercive and b 2 X (hence (61) has a unique solution). Proof. Let us rst consider a more familiar norm kk on the product space X; namely kvk = kdv 0 k L p + kd y v 1 k L p + kd z v 2 k L p (actually we could have chosen any norm of the form kvk s = kdv 0 k s L p + kd yv 1 k s L p + kd z v 2 k s L p ; where 1 s < 1). Concerning the three norms on the right hand side we note the following: 1. kdv 0 k L p (= j j jj kdv 0 k L p () ) and kv 0k X0 are equivalent norms due to Friedrichs inequality. 2. kd y v 1 k L p (= jj kd y v 1 k L p ( ) ) and kv 1 k L p (;Wper( 1;p )) are equivalent norms due to the Poincaré inequality. Indeed, p jv 1 j p dydx = v 1 1 v 1 dy j j dydx c jd y v 1 j p dydx: X Thus, we obtain that kd y v 1 k Lp ( ) kv 1k Lp (;W 1;p per( )) k kd yv 1 k Lp ( ) ; and the equivalence follows. 3. It follows similarly that the norms kd z v 2 k L p and kv 2 k L p ( ;Wper()) 1;p are equivalent. Since all three components of X (endowed with these three norms, respectively) are re exive Banach spaces, it follows that (X; kk) also is a re exive Banach space. In order to see this observe that f 2 (X; kk) if and only if 27

28 fv = f 0 v 0 + f 1 v 1 + f 2 v 2 ; for all v 2 X; where f i 2 Xi : This gives that a sequence (v) " converges weakly in (X; kk) if and only if (v i ) " converges weakly in Xi : Moreover, (v) " is bounded in (X; kk) if and only if (v i) " is bounded in X i : Thus, by recalling that a Banach space is re exive if and only if its closed unit sphere is weakly compact, we conclude that (X; kk) is re exive. Hence, in order to prove that (X; kk X ) is re exive we only have to prove that the norms kk and kk X are equivalent. By Jensen inequality we observe that jdv 0 + D y v 1 + D z v 2 j p dz dy dx jj 1 = jj 1 p p p Dv 0 + D y v 1 + D z v 2 dz dy dx p jj Dv 0 + jj D y v 1 + D z v 2 dz dy dx (62) {z } 0 (jj j j) 1 p p jj Dv 0 + jj D y v 1 dy dx 0 = (jj j j) 1 p j j jj Dv 0 + jj j j D y v 1 dy {z } 0 = jj j j jdv 0 j p dx: Thus we have shown the following inequalities: p 1 dxc A kdv 0 + D y v 1 + D z v 2 k L p kdv 0 + D y v 1 k L p kdv 0 k L p ; (63) Using the rst inequality in (63) and next the Minkowski s inequality we get 2 kdv 0 + D y v 1 + D z v 2 k L p k (Dv 0 + D y v 1 )k L p + kdv 0 + D y v 1 + D z v 2 k L p kd z v 2 k L p (64) Similarly, using both inequalities in (63) and next the Minkowski s inequality we obtain 2 kdv 0 + D y v 1 + D z v 2 k L p 2 kdv 0 + D y v 1 k L p k Dv 0 k L p + kdv 0 + D y v 1 k L p kd y v 1 k L p (65) Adding the inequalities (63), (64) and (65) and next using the Minkowski s inequality, we obtain that 5 kdv 0 + D y v 1 + D z v 2 k L p kdv 0 k L p + kd y v 1 k L p + kd z v 2 k L p kdv 0 + D y v 1 + D z v 2 k L p : 28

29 Hence kk and kk X are equivalent norms. We continue by repeating the proof of Lemma 24 with replaced by, L p replaced by L p ( ) and Dw and Dv replaced by Dw 0 +D y w 1 +D z w 2 and Dv 0 + D y v 1 + D z v 2 respectively. The monotonicity, the coercivity, the fact that Sw 2 X and that S is hemicontinuity follows exactly as in that proof. Correspondingly, by using the additional information that kdv 0 k L p kvk X ; we see that b 2 X : 7 Multiscale convergence and reiterated homogenization of monotone operators Two-scale convergence can be generalized to three-scale convergence (or n-scale convergence). De nition 26 A sequence (u " ) of functions in L p () is said to three-scale converge to a it u in L p ( ) if u " (x)(x; x " ; x " 2 ) dx! u(x; y; z)(x; y; z) dzdydx; (66) for every 2 L q (; C per ( )). By same arguments as in Theorem 15 it follows that if (u " ) three scale converges then (66) also holds for every 2 L q per(; C per ( ), where C per ( ) is the set of functions (x; y) which are continuous on and -periodic in y. Also other results obtained for two-scale convergence have corresponding versions for three-scale convergence. Especially we mention the following: Theorem 27 For each bounded sequence, (u " ), in L p () there exist a subsequence and a u 2 L p ( ) such that the subsequence three-scale converges to u. Theorem 28 Let (u " ) be a sequence in W 1;p () such that u "! u weakly in W 1;p (): Then (u " ) three-scale converges to u and there exist a subsequence " 0 and u 1 2 L p (; Wper( 1;p )) and u 2 2 L p ( ; Wper()) 1;p such that Du " 0! Du + D y u 1 + D z u 2 three-scale. By studying the asymptotic behavior of the sequence of solutions of (59) by three-scale convergence we obtain the following homogenization result: Theorem 29 Let u " be a solution of (60). Then as "! 0; u " three-scale converges to u 0 2 W 1;p 0 () and Du " 3-scale converges to Du 0 + D y u 1 + D z u 2, where fu 0 ; u 1 ; u 2 g is the solution of (61) Remark 1 The theorem above can easily be extended to the case div (a(x; x="; : : : ; x=" n ; Du " )) = f: 29

30 Proof. By the assumptions on a we have the a priori estimate ku " k W 1;p 0 () c. Since W 1;p 0 () is re exive we can use the well-known fact that from each bounded sequence in a re exive space one can extract a weakly convergent subsequence. Thus there exists a subsequence, still denoted by ", such that u "! u weakly in W 1;p 0 (): From Theorem 28 it follows that there exists a subsequence, still denoted by ", and u 1 2 L p (; Wper 1;p ( )) and u 2 2 L p ( ; Wper 1;p ()) such that Du " 3-scale converges to Du 0 + D y u 1 + D z u 2. Let w " be de ned as w " (x) = w 0 (x) + "w 1 (x; x " ) + "2 w 2 (x; x " ; x " 2 ); where w 0 2 C0 1 (), w 1 2 D(; Cper( 1 )), w 2 2 D(; Cper( 1 )). Then the monotonicity assumption on a implies that (a( x " ; x " 2 ; Du ") a( x " ; x " 2 ; Dw "); D(u " w " )) dx 0; By taking the weak formulation of (59) into account we obtain. f(u " w " ) dx a( x " ; x " 2 ; Dw "); D(u " w " ) dx 0: (67) We note that a( x " ; x " 2 ; Dw "); D(u " w " ) dx (68) = (a( x " ; x h " 2 ; Dw ") a( x " ; x " 2 ; Dw 0(x) + + D y w 1 (x; x " ) + D zw 2 (x; x " ; x i " 2 )) ; D(u " w " ) )dx + (a( x " ; x " 2 ; Dw 0(x) + D y w 1 (x; x " ) + D zw 2 (x; x " ; x " 2 )); D(u " w " ) )dx We can pass to the three scale it in the second term in right hand side of (68) by noting that a(y; z; Dw 0 (x) + D y w 1 (x; y) + D z w 2 (x; y; z)) belong to L q per(; C per ( ). Indeed! a( x " ; x " 2 ; Dw 0(x) + D y w 1 (x; x " ) + D zw 2 (x; x " ; x " 2 )); D(u " w " ) dx (a(y; z; Dw 0 + D y w 1 + D z w 2 ); (69) D(u 0 w 0 ) + D y (u 1 w 1 ) + D z (u 2 w 2 )) dzdydx: 30

31 By using (58) and Hölder in the second term in right hand side of (68) we get that (a( x " ; x h " 2 ; Dw ") a( x " ; x " 2 ; Dw 0(x) + D y w 1 (x; x )+ (70) " D z w 2 (x; x " ; x i " 2 )) ; D(u " w " ) )dx c 1 ( (1 + jdw " j + jdw 0 + D y w 1 + D z w 2 j) (p 1 )q jdw " Dw 0 D y w 1 D z w 2 j q dx) 1=q jd(u " 1=p w " )j p dx = c 1 ( (1 + jdw " j + jdw 0 + D y w 1 + D z w 2 j) (p 1 )q j"d x w 1 + 1=p " 2 D x w 2 + "D y w 2 q dx) 1=q jd(u " w " )j dx p : Since w 0 (x); w 1 (x; y); w 2 (x; y; z) are given one can pass to the it in (67) and we obtain that f(u 0 w 0 ) dx (a(y; z; Dw 0 + D y w 1 + D z w 2 ); (71) D(u 0 w 0 ) + D y (u 1 w 1 ) + D z (u 2 w 2 )) dzdydx 0: We use then the standard trick of monotonicity, we take in (71), w i = u i v i ; > 0 where v 0 = v 0 (x); v 1 = v 1 (x; y) and v 2 = v 2 (x; y; z): After dividing by we obtain that fv 0 dx (a(y; z; Du 0 + D y u 1 + D z u 2 (Dv 0 + D y v 1 + D z v 2 )); By letting! 0 we obtain that fv 0 dx = Dv 0 + D y v 1 + D z v 2 )) dzdydx 0: (a(x; y; z; Du 0 + D y u 1 + D z u 2 ) ; Dv 0 + D y v 1 + D z v 2 ) dzdydx and the proof is completed. 8 A guide to the literature In addition to the papers referred to in the introduction many authors have studied di erent homogenization problems by means of two-scale convergence. Below we give a brief overview of these works. Using the two-scale convergence method Polisevski and Mascarenhas [60] studied the homogenized behavior of the warping and the torsion problems in a two-dimensional domain with a quasi-periodic perforation, as well as the relationship between the two homogenized problems. The Neumann problem were also analyzed. 31