Convergence of monotone operators with respect to measures

Transcription

1 Convergence of monotone operators with respect to measures Department of Mathematics, Faculty of Sciences University Abou Bakr Belkaïd Tlemcen, Algeria m mamchaoui@mail.univ-tlemcen.dz Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 1/ SciencesUniv 1 / 31

2 Table of Contents 1 References 2 Homogenization of Monotone Operators Setting of the Homogenization Problem The Homogenized Problem 3 Convergence in the variable L p 4 Our Goal 5 Monotonicity and convergence First Case : Second Case : 6 Diagram 7 Conclusion Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 2/ SciencesUniv 2 / 31

3 References References 1 G. A. Chechkin, V. V. Jikov, D. Lukassen and A. L. Piatniski, On homogenization of networks and junctions, Asymptotic Analysis, 30 (2002) A. Defranceschi. An Introduction to Homogenization and G-convergence, School on Homogenization, at the ICTP, Trieste. September 6-8, V. V. Jikov, S. M. Kozlov and O. A. Oleinik, Homogenization of Differential Operators and Integral Functionals, Springer. Berlin, V. V. Jikov(Zhikov): Weighted Sobolev spaces, Russian Acad. Sci. Sb. Math. 189(8) V. V. Zhikov: On two-scale convergence, Journal of Mathematical Sciences, Vol 120, No.3, D. Lukkassen and P. Wall, Two-scale convergence with respect to measures and homogenization of monotone operators, (Function Spaces and Applications), Volume 3, Number 2 (2005) Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 3/ SciencesUniv 3 / 31

4 Homogenization of Monotone Operators Setting of the Homogenization Problem Setting of the Homogenization Problem Let be a bounded open set of R n and Y = ]0, 1[ n the unit cube. For a fixed ɛ > 0, let us consider the nonlinear monotone operator: A ɛ (u) := div(a( x ɛ, Du)), u H1 0 () where a(x,.) is Y -periodic and satisfies the following properties : 1 a : R n R n R n such that : ξ R n, a(., ξ) is Lebesgue mesurable and Y -periodic. 2 (a(x, ξ 1 ) a(x, ξ 2 ), ξ 1 ξ 2 ) α ξ 1 ξ 2 2 a.e x R n and ξ 1, ξ 2 R n. 3 a(x, ξ 1 ) a(x, ξ 2 ) β ξ 1 ξ 2 a.e x R n and ξ 1, ξ 2 R n. 4 a(x, 0) = 0 a.e x R n. If a N, then for all real positif ɛ and f ɛ L 2 (), we consider the following problem : { div(a( x ɛ, Du ɛ)) = f ɛ in u ɛ H 1 0 () (1) Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 4/ SciencesUniv 4 / 31

5 Homogenization of Monotone Operators The Homogenized Problem The Homogenized Problem Theorem : Let a N and let (ɛ h ) h be a sequence of positive real numbers converging to 0. Assume that f ɛ f strongly in H 1 (). ɛ 0 Let (u ɛ) ɛ be the solution to (1). Then, u ɛ ɛ 0 u in H 1 0 () where u is the unique solution to the homogenized problem : a( x ɛ, Duɛ) ɛ 0 b(du ) in L2 (; R n ) { div(b(du )) = f in u H 1 0 () (2) The operator b is defined by : b : R n R n ξ b(ξ) = a(y, ξ + Dw ξ (y))dy (3) Y where w ξ is the unique solution to the local problem : { Y (a(y, ξ + Dw ξ (y)), Dv(y))dy = 0 v H 1 # (Y ) w ξ H 1 # (Y ) (4) Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 5/ SciencesUniv 5 / 31

6 Homogenization of Monotone Operators The Homogenized Problem The Homogenized Problem Proposition : The homogenized operator b has the following properties: 1 b is strictly monotone 2 There exists a constant γ = β2 α > 0 such that: b(ξ 1 ) b(ξ 2 ) γ ξ 1 ξ 2 ξ 1, ξ 2 R n 3 b(0) = 0. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 6/ SciencesUniv 6 / 31

7 Convergence in the variable L p Let be a bounded Lipschitz domain in R N. It will always assumed that p and q are conjugate indexes, i.e. 1/p + 1/q = 1 and that 1 < p <. Let µ δ and µ be Randon measures in and µ δ µ. We say that a sequence u δ is bounded in L p (, dµ δ ) if sup u δ p dµ δ <. Definition A bounded sequence u δ L p (, dµ δ ) is weakly convergent to u L p (, dµ), u δ u if u δ ϕdµ δ = uϕdµ. (5) for any test function ϕ C 0 (). We have the following results: each bounded sequence has a weakly convergent subsequence. if the sequence u δ weakly converges to u, then inf u δ p dµ δ u p dµ. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 7/ SciencesUniv 7 / 31

8 Convergence in the variable L p Definition A bounded sequence u δ L p (, dµ δ ) is called strongly convergent to u L p (, dµ), u δ u if We have strong convergence implies weak convergence. u δ v δ dµ δ = uvdµ, if v δ v in L q (, dµ δ ). the weak convergence u δ u, together with the relation u δ p dµ δ = u p dµ, is equivalent to the strong convergence u δ u. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 8/ SciencesUniv 8 / 31

9 Our Goal Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty Octoberof 11, 9/ SciencesUniv 9 / 31

10 Monotonicity and convergence Let be a bounded Lipschitz domain in R N, where the Lebesgue measure of the boundary is zero, Y = [0, 1) N the semi-open cube in R N. It will always assumed that p and q are conjugate indexes, i.e. 1/p + 1/q = 1 and that 1 < p <. Let µ δ and µ be Randon measures in and µ δ µ. Consider a family of sequence (u δ ) such that ( uδ p + Du δ p) dµ δ c. Let a : R N R N R N be a function such that a(., ξ) is Y periodic and µ δ measurable for every ξ R N. Moreover, assume that there exists constants c 1, c 2 > 0 and two more constants α and β, with 0 α min {1, p 1} and max {p, 2} β < such that a satisfies the following continuity and monotonicity assumptions: a(y, 0) = 0 (6) a(y, ξ 1 ) a(y, ξ 2 ) c 1 (1 + ξ 1 + ξ 2 ) p 1 α ξ 1 ξ 2 α (7) (a(y, ξ 1 ) a(y, ξ 2 ), ξ 1 ξ 2 ) c 2 (1 + ξ 1 + ξ 2 ) p β ξ 1 ξ 2 β, (8) for µ δ a.e. y R N and any ξ 1, ξ 2 R N. We will treat two cases: a(x, Du δ ) and a δ (x, Du δ ). We Suppose that a(x, ϕ) ( C 0 () ) N for all ϕ ( C 0 () ) N. As δ is small enough, we obtain the following results: Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of10/ SciencesUniv 10 / 31

11 Monotonicity and convergence First Case : Lemma There exist a subseqence of u δ still denoted by u δ and a (L q (, dµ)) N such that a(x, Du δ ) a. Proof. a(x, Du δ ) c 1 (1 + Du δ ) p 1 α Du δ α c 1 (1 + Du δ ) p 1 2 p 1 c 1 (1 + Du δ p 1 ) which gives that a(x, Du δ ) is bounded in (L q (, dµ δ )) N since (Du δ ) is bounded in (L p (, dµ δ )) N. By proposition 3 there exist a subsequence (still denoted by δ) such that: there exists a a (L q (, dµ)) N such that a(x, Du δ ) a i.e : (a(x, Du δ ), ϕ)dµ δ = (a, ϕ)dµ for all ϕ (C 0 ()) N. Lemma Assume that (a a(x, ϕ), z ϕ)dµ 0, ϕ ( C 0 () ) N for some a and z belong to (L q (, dµ)) N and (L p (, dµ)) N respectivelly. Then, the inequality holds for every ϕ (L p (, dµ)) N. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of11/ SciencesUniv 11 / 31

12 Monotonicity and convergence First Case : Theorem 1: Assume that a satisfies the conditions (6)-(7)-(8). Let (Du δ ) be a bounded sequence in (L p (, dµ δ )) N which converges weakly to z (L p (, dµ)) N and assume that a(x, Du δ ) converges weakly to a (L q (, dµ)) N. Then inf (a(x, Du δ ), Du δ )dµ δ (a, z)dµ. If equality holds, i.e : then inf (a(x, Du δ ), Du δ )dµ δ = (a, z)dµ, a = a(x, z) for µ a.e x. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of12/ SciencesUniv 12 / 31

13 Assume the contradiction, i.e. Monotonicity and convergence First Case : inf (a(x, Du δ ), Du δ )dµ δ < (a, z)dµ. Then there exists a positive constant C > 0 such that inf (a(x, Du δ ), Du δ )dµ δ = (a, z)dµ C. (9) By monotonicity we have that (a(x, Du δ ) a(x, ϕ), Du δ ϕ)dµ δ 0, ϕ ( C 0 () ) N. Using (9) on the corresponding term and passing to the it when δ 0 in this inequality. Its left-handside contains four terms. We have (a a(x, ϕ), z ϕ)dµ C, ϕ ( C 0 () ) N. By density and continuity (see lemma 4) this inequality holds also for any ϕ (L p (, dµ)) N. Let tφ = z ϕ, with t > 0. Then after dividing by t we get (a a(x, z tφ), φ)dµ C t. Letting t 0 gives (a a(x, z), φ)dµ. Repeating the procedure for t < 0 implies that (a a(x, z), φ)dµ. We have clearly reached a contradiction. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of13/ SciencesUniv 13 / 31

14 Monotonicity and convergence First Case : If equality holds we can repeat the procedure above : (a(x, Du δ ) a(x, ϕ), Du δ ϕ)dµ δ 0, for every ϕ ( C 0 () ) N. Let us pass to the it in this inequality. Its left-hand side contains four terms. We have : As a result, we obtain the inequality : (a(x, Du δ ), ϕ)dµ δ = (a(x, ϕ), ϕ)dµ δ = (a(x, ϕ), Du δ )dµ δ = (a, ϕ)dµ. (a(x, ϕ), ϕ)dµ. (a(x, ϕ), z)dµ. inf (a(x, Du δ ), Du δ )dµ δ = (a, z)dµ. (a a(x, ϕ), z ϕ)dµ 0, ϕ ( C 0 () ) N. By density and continuity this inequality holds for any ϕ (L p (, dµ)) N. Put ϕ = z tφ, φ (L p (, dµ)) N. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of14/ SciencesUniv 14 / 31

15 Monotonicity and convergence First Case : For t > 0 we get that For t < 0 we obtain that (a a(x, z tφ), φ)dµ 0. (10) (a a(x, z tφ), φ)dµ 0. (11) By taking (10) and (11) into account and letting t tend to 0 we find that (a a(x, z), φ)dµ = 0, φ ( L p (, dµ) ) N. This implies that a = a(x, z) for µ a.e. x. This ends the proof. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of15/ SciencesUniv 15 / 31

16 Monotonicity and convergence First Case : Theorem 2: Assume that a satisfies the conditions (6)-(7)-(8). Let (Du δ ) be a bounded sequence in (L p (, dµ δ )) N which converges weakly to z (L p (, dµ)) N, a(x, Du δ ) converges weakly to a (L q (, dµ)) N, and Du δ p 2 Du δ converges weakly to v (L q (, dµ)) N. If then the sequence Du δ is strongly convergent to z. For k > 0, we have : This together with (12) give = (a(x, Du δ ), Du δ )dµ δ = (a(x, Du δ ), Du δ )dµ δ = sup sup (a, z)dµ, (12) k Du δ p dµ δ sup k Du δ p dµ δ + (a(x, Du δ ), Du δ )dµ δ = sup + (a(x, Du δ ), Du δ )dµ δ. k Du δ p dµ δ + inf (kv, z)dµ + (a kv, z)dµ. k Du δ p dµ δ + inf k Du δ p dµ δ (a(x, Du δ ) kdu δ Du δ p 2, Du δ )dµ δ Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of16/ SciencesUniv 16 / 31

17 Monotonicity and convergence First Case : For k > 0 sufficiently small The function a(x, Du δ ) kdu δ Du δ p 2 satisfies the conditions in Theorem 1 therefore inf (a(x, Du δ ) kdu δ Du δ p 2, Du δ )dµ δ (a kv, z)dµ. (13) This and (12) imply sup Du δ p dµ δ (v, z)dµ. (14) The function Du δ satisfies the conditions in Theorem 1 which implies that inf Du δ p dµ δ = inf (Du δ Du δ p 2, Du δ )dµ δ (v, z)dµ. (15) By (14) and (15) it follows that Du δ p dµ δ = (Du δ Du δ p 2, Du δ )dµ δ = (v, z)dµ Hence if follows by Theorem 1 that v = z p 2 z. Then, Du δ converges strongly to z. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of17/ SciencesUniv 17 / 31

18 Monotonicity and convergence First Case : Theorem 3: Assume that a satisfies the conditions (6)-(7)-(8). Let (Du δ ) be a bounded sequence in (L p (, dµ δ )) N which converges strongly to z (L p (, dµ)) N and assume that a(x, Du δ ) converges weakly to a (L q (, dµ)) N. Then a = a(x, z) for µ a.e x. Proof. In the same spirit of the proof of Theorem 1 we show that a = a(x, z) for µ a.e x. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of18/ SciencesUniv 18 / 31

19 Monotonicity and convergence Second Case : Lemma There exists a subseqence and a 0 (L q (, dµ)) N such that a δ (x, Du) a 0. Proof. a δ (x, Du) c 1 (1 + Du ) p 1 α Du α c 1 (1 + Du ) p 1 2 p 1 c 1 (1 + Du p 1 ) which gives that a δ (x, Du) is bounded in (L q (, dµ δ )) N since (Du) is bounded in (L p (, dµ δ )) N. By proposition 3 there exist a subsequence (still denoted by δ) such that: There exist a a 0 (L q (, dµ)) N such that a δ (x, Du) a 0 i.e : (a δ (x, Du), ϕ)dµ δ = (a 0, ϕ)dµ for all ϕ (C 0 ()) N. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of19/ SciencesUniv 19 / 31

20 Monotonicity and convergence Second Case : Theorem 4: Assume that a δ satisfies the conditions (6)-(7)-(8) and converges weakly to a 0 in sens (5). Then the ite a 0 also satisfies the conditions (6)-(7)-(8). For all ϕ ( C 0 () ) N : 0 = (a δ (x, 0), ϕ)dµ δ = (a 0 (x, 0), ϕ)dµ. For ϕ C 0 () and for all ξ 1, ξ 2 R n, we have: (a δ (x, ξ 1 ) a δ (x, ξ 2 ), ϕ(ξ 1 ξ 2 ))dµ δ (a δ (x, ξ 1 ) a δ (x, ξ 2 ) ϕ(ξ 1 ξ 2 ) dµ δ c 1 (1 + ξ 1 + ξ 2 ) p 1 α ξ 1 ξ 2 α ϕ(ξ 1 ξ 2 ) dµ δ Passing here to the it we obtain the inequality (a 0 (x, ξ 1 ) a 0 (x, ξ 2 ), ϕ(ξ 1 ξ 2 ))dµ c 1 (1 + ξ 1 + ξ 2 ) p 1 α ξ 1 ξ 2 α ϕ(ξ 1 ξ 2 ) dµ Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of20/ SciencesUniv 20 / 31

21 Monotonicity and convergence Second Case : which shows that (a 0 (x, ξ 1 ) a 0 (x, ξ 2 ) c 1 (1 + ξ 1 + ξ 2 ) p 1 α ξ 1 ξ 2 α, for µ a.e. x and any ξ 1, ξ 2 R N. Finally, we have for all ϕ C 0 (), ϕ 0 : (a δ (x, ξ 1 ) a δ (x, ξ 2 ), ϕ(ξ 1 ξ 2 ))dµ δ = c 2 (a δ (x, ξ 1 ) a δ (x, ξ 2 ), (ξ 1 ξ 2 ))ϕdµ δ (1 + ξ 1 + ξ 2 ) p β ξ 1 ξ 2 β ϕdµ δ Passing here to the it we obtain inequality (a 0 (x, ξ 1 ) a 0 (x, ξ 2 ), (ξ 1 ξ 2 ))ϕdµ c 2 (1 + ξ 1 + ξ 2 ) p β ξ 1 ξ 2 β ϕdµ for every ϕ C 0 (), ϕ 0, which complete the proof. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of21/ SciencesUniv 21 / 31

22 Monotonicity and convergence Second Case : Lemma There exists a subsequence and a 1 (L q (, dµ)) N such that a δ (x, Du δ ) a 1. Proof. a δ (x, Du δ ) c 1 (1 + Du δ ) p 1 α Du δ α c 1 (1 + Du δ ) p 1 2 p 1 c 1 (1 + Du δ p 1 ) which gives that a δ (x, Du δ ) is bounded in (L q (, dµ δ )) N since (Du δ ) is bounded in (L p (, dµ δ )) N. By proposition 3 there exists a subsequence (still denoted by δ) such that: There exist a a 1 (L q (, dµ)) N such that a δ (x, Du δ ) a 1 i.e : (a δ (x, Du δ ), ϕ)dµ δ = (a 1, ϕ)dµ for all ϕ (C 0 ()) N. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of22/ SciencesUniv 22 / 31

23 Monotonicity and convergence Second Case : Theorem 5: Assume that a δ satisfies the conditions (6)-(7)-(8). Let (Du δ ) be a bounded sequence in (L p (, dµ δ )) N which converges strongly to z (L p (, dµ)) N and assume that a δ (x, Du δ ) converges weakly to a 1 (L q (, dµ)) N. Then, a 1 = a 0 (x, z) where a 0 is the it of a δ in sens (5). By the monotonicity of a δ we have (a δ (x, Du δ ) a δ (x, ϕ), Du δ ϕ)dµ δ 0, for every ϕ ( C 0 () ) N. Let us pass to the it in this inequality. Its left-hand side contains four terms. We have : and (a δ (x, Du δ ), ϕ)dµ δ = (a 1, ϕ)dµ, (a δ (x, ϕ), ϕ)dµ δ = (a 0 (x, ϕ), ϕ)dµ, (a δ (x, ϕ), Du δ )dµ δ = (a 0 (x, ϕ), z)dµ, (a δ (x, Du δ ), Du δ )dµ δ = (a 1, z)dµ. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of23/ SciencesUniv 23 / 31

24 Monotonicity and convergence Second Case : As a result, we obtain the inequality (a 1 a 0 (x, ϕ), z ϕ)dµ 0, ϕ ( C 0 () ) N. By density and continuity this inequality holds for any ϕ (L p (, dµ)) N. Put ϕ = z tφ, φ (L p (, dµ)) N. For t > 0 we get that (a 1 a 0 (x, z tφ), φ)dµ 0. (16) For t < 0 we obtain that (a 1 a 0 (x, z tφ), φ)dµ 0 (17) By taking (16) and (17) into account and letting t tend to 0 we find that (a 1 a 0 (x, z), φ)dµ = 0, φ ( L p (, dµ) ) N. Which implies that a 1 = a 0 (x, z) for µ a.e. x. The proof is complete. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of24/ SciencesUniv 24 / 31

25 Monotonicity and convergence Second Case : Example N = p = 2. Let I = {x = (x 1, x 2 ) / a x 1 b ; x 2 = 0} be a segment in R 2, and suppose that a bounded domain contains I. For any sufficiently small δ > 0 consider the bar I δ := {x = (x 1, x 2 ) / a < x 1 < b ; δ < x 2 < δ}. Denote by µ δ the measure in, concentrated and uniformly distributed on I δ : µ δ (dx) = χ I δ (x) 2δ(b a) dx 1dx 2. The family µ δ converges weakly, as δ 0, to a measure µ (dx) = 1 (b a) dx 1 δ(x 2 ) where δ(z) stands for the Dirac mass at zero. Consider the operator a δ with the following form : ( ) 1 + δ x a δ (x, Du δ ) = 2 + δ x Du δ. We can prove that a δ satisfies suitable assumptions of uniform strict monotonicity and uniform Lipschitz- continuity and a δ (x, 0) = 0 for a.e. x R 2. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of25/ SciencesUniv 25 / 31

26 Example Monotonicity and convergence Second Case : Suppose that Du δ is strongly convergent to z. (a δ (x, ϕ), ϕ)dµ δ = (( 1 + δ x 2 + δ x )ϕ, ϕ)dµ δ = ( 1 2 ϕ, ϕ)dµ, ϕ ( C 0 () ) 2. This implies that and a 0 (x, ξ) = 1 2 ξ a 0 (x, ξ 1 ) a 0 (x, ξ 2 ) = 1 2 ξ 1 ξ 2, for a.e. x R 2 and for every ξ 1, ξ 2 R 2. Finally, the hypothesis of Theorem 5 are satisfied then (a δ (x, Du δ ), ϕ)dµ δ = = = (a 1, ϕ)dµ (a 0 (x, z), ϕ)dµ ( 1 2 z, ϕ)dµ, ϕ ( C 0 () ) 2. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of26/ SciencesUniv 26 / 31

27 Diagram Theorem 6: Assume that a satisfies the conditions (6)-(7)-(8). Let (Du ɛ,δ ) be a bounded sequence in ( L p (, dµ ɛ,δ ) ) N which converges weakly to z ɛ (L p (, dµ ɛ)) N and assume that a( x ɛ, Du ɛ,δ) converges weakly to a ɛ (L q (, dµ ɛ)) N. If inf (a( x ɛ, Du ɛ,δ), Du ɛ,δ )dµ ɛ,δ = (a ɛ, z ɛ)dµ ɛ, then the diagram is commutative. a( x ɛ, Du ɛ,δ) δ 0 ɛ 0 b(du 0,δ ) δ 0 a( x ɛ, Duɛ) ɛ 0 b(du 0 ) Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of27/ SciencesUniv 27 / 31

28 Diagram Proof. From Theorem 1 it follows that a ɛ = a( x, zɛ) for µ a.e x. ɛ We have for all ϕ ( C 0 () ) N (a( x ɛ 0 ɛ, Du ɛ,δ), ϕ)dµ ɛ,δ = (b(du 0,δ ), ϕ)dx. Since z ɛ = Du ɛ, we obtain : (a( x ɛ 0 ɛ, Du ɛ,δ), ϕ)dµ ɛ,δ = (a ɛ, ϕ)dµ ɛ. ɛ 0 = (a( x, Duɛ), ϕ)dµɛ. ɛ 0 ɛ = (b(du 0 ), ϕ)dx. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of28/ SciencesUniv 28 / 31

29 Diagram Theorem 7: Assume that a δ satisfies the conditions (6)-(7)-(8). Let (Du ɛ,δ ) be a bounded sequence in ( L p (, dµ ɛ,δ ) ) N which converges strongly to z ɛ (L p (, dµ ɛ)) N and assume that a δ ( x ɛ, Du ɛ,δ) converges weakly to a ɛ 1 (Lq (, dµ ɛ)) N. Then the diagram a δ ( x ɛ, Du ɛ,δ) ɛ 0 b(du 0,δ ) δ 0 δ 0 a 0 ( x ɛ, Duɛ)) ɛ 0 b(du 0 ) is commutative. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of29/ SciencesUniv 29 / 31

30 Diagram Proof. From Theorem 5 it follows that a ɛ 1 = a 0( x, zɛ) for µ a.e x. ɛ We have for all ϕ ( C 0 () ) N (a δ ( x ɛ 0 ɛ, Du ɛ,δ), ϕ)dµ ɛ,δ = (b(du 0,δ ), ϕ)dx. Since z ɛ = Du ɛ, we obtain : (a δ ( x ɛ 0 ɛ, Du ɛ,δ), ϕ)dµ ɛ,δ = (a ɛ ɛ 0 1, ϕ)dµɛ. = (a 0 ( x, Duɛ), ϕ)dµɛ. ɛ 0 ɛ = (b(du 0 ), ϕ)dx. Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of30/ SciencesUniv 30 / 31

31 Conclusion Conclusion a( x ɛ, Du ɛ,δ) a( x ɛ, z ɛ) ɛ 0 b(du,δ ) ɛ 0 b(z ) Basque Center Department for Applied Mathematics,Bilbao,Spain. of Faculty October 11, of31/ SciencesUniv 31 / 31