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Homogenization and corrector for the wave equation with discontinuous coefficients in time

Casado Díaz, Juan; Couce Calvo, Julio; Maestre, Faustino; Martín Gómez, José Domingo

Abstract

In this paper we analyze the homogenization of the wave equation with bounded variation coefficients in time, generalizing the classical result, which assumes Lipschitz-continuity. We start showing a general existence and uniqueness result for a general sort of hyperbolic equations. Then, we obtain our homogenization result comparing the solution of a sequence of wave equations to the solution of a sequence of elliptic ones. We conclude the paper making an analysis of the corrector. Firstly, we obtain a corrector result assuming that the derivative of the coefficients in the time variable is equicontinuous. This result was known for non-time dependent coefficients. After, we show, with a counterexample, that the regularity hypothesis for the corrector theorem is optimal in the sense that it does not hold if the time derivative of the coefficients is just bounded.

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J. Math. Anal. Appl. 379 (2011) 664–681 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa Homogenization and corrector for the wave equation with discontinuous coefficients in time Juan Casado-Díaz∗, Julio Couce-Calvo, Faustino Maestre, José D. Martín-Gómez Dpto. de Ecuaciones Diferenciales y Análisis Numérico, Universidad de Sevilla, Spain article info abstract Article history: Received 2 November 2010 Available online 28 January 2011 Submitted by Steven G. Krantz Keywords: Homogenization Corrector Wave equation In this paper we analyze the homogenization of the wave equation with bounded variation coefficients in time, generalizing the classical result, which assumes Lipschitz-continuity. We start showing a general existence and uniqueness result for a general sort of hyperbolic equations. Then, we obtain our homogenization result comparing the solution of a sequence of wave equations to the solution of a sequence of elliptic ones. We conclude the paper making an analysis of the corrector. Firstly, we obtain a corrector result assuming that the derivative of the coefficients in the time variable is equicontinuous. This result was known for non-time dependent coefficients. After, we show, with a counterexample, that the regularity hypothesis for the corrector theorem is optimal in the sense that it does not hold if the time derivative of the coefficients is just bounded. ©2011 Elsevier Inc. All rights reserved. 1. Introduction For a bounded open set Ω⊂RNand a positive number T, we are interested in the present paper in the homogenization and corrector for the wave problem ⎧ ⎨ ⎩ ∂tρn(t,x)∂tun−divxAn(t,x)∇xun=Fnin (0,T)×Ω, un=0on(0,T)×∂Ω, un(0,x)=u0 n(x), (ρn∂tun)(0,x)=ϑ1 n(x)in Ω. (1.1) The homogenization of (1.1) has been carried out in [6] (see also [2] for the case of periodic coefficients) assuming ρn≡1 and the symmetric matrix functions Anuniformly elliptic, bounded and Lipschitz with respect to the time variable. The construction of correctors for problem (1.1) can be found in [3] (see also [9]) for the case where the coefficients do not depend on t. Our purpose here is to extend these results to more general coefficients and second members. We remark that some smoothness in the time variable is needed in order to assure the existence and uniqueness of solution for problem (1.1). In this way, we recall the following results: It is proved in [11] that (even with zero second member) problem (1.1) has not a solution in general for ρn≡1 and Anconstant in the two sides of a hyperplane not parallel to {t=0}.In[8]it is obtained a non-existence result for coefficients in C0,α(¯ Ω×[0,T]),foreveryα∈(0,1). Moreover, if the coefficients are rapidly oscillating then even if there exists a solution, it can be not bounded. An example of such phenomenon is considered in [7] where the matrices Anare supposed of the form An(t,x)=A(nt,x)with Asmooth and periodic in the time variable (and ρn≡1). Then, it is proved the existence of very smooth initial conditions such that the solutions of (1.1) with vanishing second member are not bounded in the space of distributions. In [19] (see also [13]), it is considered the homogenization *Corresponding author. E-mail addresses: [email protected] (J. Casado-Díaz), [email protected] (J. Couce-Calvo), [email protected] (F. Maestre), [email protected] (J.D. Martín-Gómez). 0022-247X/$ – see front matter ©2011 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2011.01.054 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 665 of problem (1.1) for coefficients of the form A(nt,nx),withAperiodic, but the existence and boundness of the solutions is assumed by hypothesis. Taking into account the above remarks, let us assume in the present paper that ρnand Anare bounded in BV(0,T;L∞(Ω)) and BV(0,T;L∞(Ω;Ms N)) respectively (so, in particular, they are not continuous in general with respect to t), the initial conditions u0 nand ϑ1 nare bounded in H1 0(Ω) and L2(Ω) respectively and the second member Fnis of the form Fn=fn+gnwith fnbounded in the space of measures M([0,T]; L2(Ω)) and gnbounded in BV(0,T;H−1(Ω)) (and satisfying the compact assumption in the spatial variable given by (3.9)). With these assumptions it is possible to prove that there exists a unique solution unof problem (1.1). Moreover unand ∂tunare bounded in L∞(0,T;H1 0(Ω)) and L∞(0,T;L2(Ω)) respectively. Although the ideas to prove this existence and uniqueness result are classical (see e.g. [10–12,14] for related results), we have not found it in the literature in all of its generality. Thus, we give in Section 2 a sketch of the proof. In Section 3 we carry out the homogenization of problem (1.1). Our main result (Theorem 3.4) establishes that, for a subsequence, the solution unof (1.1) converges weakly-∗in L∞(0,T;H1 0(Ω)) to the solution of a similar problem where ρn is replaced by its weak-∗limit ρin L∞((0,T)×Ω) and Anby the matrix Asuch that A(t,.) is the H-limit of An(t,.) for every t∈(0,T)up to a countable set. This theorem generalizes the results obtained in [3] (where it is considered the case ρn(t,x)and An(t,x)independent of tand gn=0) and in [6] (where it is considered the case ρn=1, An(t,x)uniformly Lipschitz in tand gn=0). Section 4 is devoted to give a corrector result for the solution of (1.1), i.e. an approximation of unin the strong topology of H1((0,T)×Ω). Our aim is to generalize the following corrector result proved in [3]: Assume that ρnand Ando not depend on the time variable, the second member Fnconverges weakly in L2((0,T)×Ω),thesequenceϑ1 nconverges strongly in L2(Ω) and the sequence u0 nconverges weakly in H1 0(Ω) and it is such that −divxAn∇xu0 nconverges strongly in H−1(Ω). Then the corrector for ∇xunis given by the corrector corresponding to the elliptic operators −divxAn∇x, while ∂tunconverges strongly in L2(0,T;L2(Ω)). Here, we generalize this result for coefficients depending on t, but satisfying smoothness in the time variable than in Section 3. Namely, we assume that ∂tρnand ∂tAnare continuous on [0,T]with values in L∞(Ω) and L∞(Ω;Ms N)respectively, with a modulus of continuity uniform in n. This smoothness hypothesis may seem very restrictive, but it is optimal such as we will see in Section 5. There, we prove that even for ρn≡1, Anbounded in C1([0,T]; L∞(Ω;Ms N)) and converging strongly in C0([0,T]; L∞(Ω;Ms N)),Fn≡0, u0 n≡0 and ϑ1 nequals to a function in C∞(Ω) independent of n,wehavethatthesolutionunof (1.1) does not converge in the strong topology of H1((0,T)×Ω). In particular this shows that the corrector for the elliptic operators −divxAn∇xdoes not give a corrector for the space derivatives of un, and so that the corrector result proved in [3] for the case of coefficients independent of the time variable cannot be generalized to the framework considered in [6], where the coefficients are supposed uniformly Lipschitz in the time variable. 1.1. Notations and recalls •For a Banach space X, and T>0, we denote by M([0,T]; X)the space of bounded Borel measures from [0,T]into X. In the particular case X=Rwe just denote M([0,T]; R)as M([0,T]). •For a Banach space X, and T>0, we define BV(0,T;X)as the space of functions ζ:[0,T]→Xsuch that VT(ζ) =sup {t0=0<t1<···<tm=T} m  i=1 ζ(ti)−ζ(ti−1) X<+∞. This implies in particular that ζis continuous up to a countable set. Changing the values of ζin this set, we can always assume that ζis right-continuous of [0,T)and left-continuous on {T}. This gives a unique representative for a function in BV(0,T;X). Along the paper we will usually consider this representative. It is also known (it follows for example from the structure theorem for BV functions given in [4]) that for every ζ∈BV(0,T;X), there exists a nonnegative measure μ∈M([0,T]),withμM([0,T];X)=VT(ζ), such that  ζ(t)−ζ(ˆ t) X⩽μ[t,ˆ t],∀t,ˆ t∈[0,T],t<ˆ t. Moreover, if Xis reflexive, the distributional derivative ∂tζbelongs to M([0,T]; X)and satisfies ζ(t)=ζ(0)+∂tζ(0,t],∀t∈(0,T). •We denote by MNand Ms Nthe spaces of squared matrices of order Nand symmetric matrices of order Nrespectively. •Sometimes, for functions depending of the time and space variables (t,x)we only specify the dependence in tin order to write shorter expressions. •We will denote by Ca nonnegative generic constant which can change from line to line. 666 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 2. Existence and uniqueness of weak solution In this section we prove an abstract result for the existence and uniqueness of solution of a hyperbolic equation, which in particular can be used for the hyperbolic problem whose homogenization is the goal of this paper. As usual, we consider two separable Hilbert spaces Vand H, such that V⊂H, with continuous injection and Vdense in H. Identifying Hwith its dual Hwe obtain V⊂H⊂V. The scalar product in His denoted by (·,·)=(·,·)Hand the duality between Vand Vis denoted by ·,· = ·,·V,V. For T>0 we consider two operators: R∈BV0,T;L(H,H),A∈BV0,T;LV,V (2.1) satisfying the symmetry assumption: R(t)h1,h2=R(t)h2,h1,∀h1,h2∈H,a.e. t∈(0,T), A(t)v1,v2=A(t)v2,v1,∀v1,v2∈V,a.e. t∈(0,T), (2.2) and the hyperbolic assumption: There exists α>0 such that R(t)h,h⩾αh2 H,∀h∈H,a.e. t∈(0,T), A(t)v,v⩾αv2 V,∀v∈V,a.e. t∈(0,T). (2.3) For f∈M[0,T]; H,g∈BV0,T;V,u0∈V,ϑ 1∈H,(2.4) we will study the initial boundary value problem for the hyperbolic equation R(t)u(t)+A(t)u(t)=f(t)+g(t)in (0,T), u(0)=u0,Ru0+=ϑ1.(2.5) Theorem 2.1. Under the above assumptions (2.1),(2.2),(2.3),(2.4), there exists a unique u ∈L∞(0,T,V)with u∈L∞(0,T,H), solution of (2.5) in the sense that R(t)u(t), v+A(t)u(t), v=f(t), v+g(t), vin D(0,T), ∀v∈V, u(0)=u0,Ru0+=ϑ1.(2.6) Moreover,wehavethefollowingestimate  u(t)  2 H+ u(t)  2 V⩽C u0  2 V+ ϑ1  2 H+f2 M([0,T];H)+g2 BV(0,T;V),a.e. t ∈(0,T), (2.7) where the constant C depends continuously on α,RBV(0,T;L(H;H)) and ABV(0,T;L(V,V)). Remark 2.2. Since uis in L∞(0,T;H), the initial condition u(0)=u0has a sense at least in H. Moreover, from (2.6) (or (2.5)) we have that Rubelongs to BV(0,T;V)and therefore the initial condition (Ru)(0+)=ϑ1has a sense at least in V. If in Theorem 2.1 we assume f∈L1(0,T;H)then Ruis in W1,1(0,T;V)and therefore we can just write (Ru)(0)=ϑ1 at the place of (Ru)(0+)=ϑ1. Proof of Theorem 2.1. When Ris the identity operator, Theorem 2.1 is proved in [1] (see also [10–12,14] for related results). For the sake of completeness we give here a sketch of the proof of Theorem 2.1 which is valid for a general operator R. Part I: Existence. Taking into account the existence of Rn∈W1,1(0,T;L(H,H)),An∈W1,1(0,T;L(V,V)),fn∈L1(0,T;H) and gn∈W1,1(0,T;V)such that Rn→Rin L10,T;L(H,H),RnW1,10,T;L(H,H)→RBV(0,T;L(H,H)), An→Ain L10,T;LV,V,AnW1,10,T;L(V,V)→ABV(0,T;L(V,V)), fn ∗ fin M[0,T]; H,fnL1(0,T;H)→fM([0,T];H), gn→gin L10,T;V,gnW1,1(0,T;V)→gBV(0,T;V), J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 667 with Rn,Ansatisfying (2.2) and (2.3) (with αindependent of n), we can always assume in the following that R∈ W1,1(0,T;L(H,H)),A∈W1,1(0,T;L(V,V)),f∈L1(0,T;H)and g∈W1,1(0,T;V). Along the proof, we denote by Ca generic nonnegative constant which depends continuously on α,RBV(0,T;L(H,H)) and ABV(0,T;L(V,V)), and can change from line to line. Step 1. Galerkin approximations. Let {wi,i=1,2,...}be a basis of Vand therefore of H. For a positive integer kwe take Wk=span{w1,...,wk}, and we consider the problem R(t)u k(t), w+A(t)uk(t), w=f(t), w+g(t), win D(0,T), ∀w∈Wk, uk(0)=u0 k,R(0)u k(0)=ϑ1 k.(2.8) where u0 k,ϑ1 k∈Wkconverge to u0,ϑ1in Vand Hrespectively. Thanks to (2.2) and (2.3), the standard theory of ODE provides a unique solution uk∈W2,1(0,T;Wk). Step 2.Energyestimate.We write uk=k j=1dkjwj,withdkj ∈W2,1(0,T),j=1,...,k. Then, taking in (2.8) w=wj,multiplying by d kj(t)and adding in jwe get R(t)u k(t),u k(t)+A(t)uk(t), u k(t)=f(t), u k(t)+g(t), u k(t),t∈(0,T), (2.9) which implies E k(t)=f(t), u k(t)+g(t), u k(t)−1 2R(t)u k(t), u k(t)+1 2A(t)uk(t), uk(t),(2.10) with Ek(t)=1 2R(t)u k(t), u k(t)+A(t)uk(t), uk(t),t∈[0,T].(2.11) Using (2.3) and the inequality f(t), u k(t)⩽C f(t) HEk(t)⩽CFδ f(t) H+C Fδ f(t) HEk(t), ∀t∈[0,T],(2.12) in (2.10), with Fδ=fL1(0,T;H)+δ,δ>0, we get E k(t)⩽CFδ f(t) H+g(t), u k(t)+Cf(t)H Fδ +μ(t)Ek(t), (2.13) with μ(t)= R(t) L(H,H)+ A(t) L(V,V).(2.14) Applying Gronwall’s inequality to (2.13) we deduce e−mδ(t)Ek(t)⩽Ek(0)+ t  0CFδ f(s) H+g(s), u k(s)e−mδ(s)ds,∀t∈(0,T)(2.15) with mδdefined as mδ(s)=C s  0f(r)H Fδ +μ(r)dr,∀r∈[0,T].(2.16) If g≡0, inequality (2.15) shows that ukand u kare bounded in L∞(0,T;V)and L∞(0,T;H)respectively. In the general case (g≡ 0), we use the following estimate for the last term in (2.15). We denote Gδ=gL1(0,T;V)+δ. Integrating by parts, taking into account that e−mδ(t)⩽1, using g(t)V⩽ CgW1,1(0,T;V),foreveryt∈[0,T], and reasoning similarly to (2.12) with g(s), uk(s),wehave t  0g(s), u k(s)e−mδ(s)ds =g(t), uk(t)e−mδ(t)−g(0), uk(0)− t  0g(s)−Cg(s)f(s)H Fδ +μ(s),uk(s)e−mδ(s)ds 668 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 ⩽1 2Ek(t)e−mδ(s)−g(0), uk(0)+Cg2 W1,1(0,T;V)+CGδ t  0 g(s) Vds +Cg2 W1,1(0,T;V) t  0f(s)H Fδ +μ(s)ds +C t  0g(s)V Gδ +f(s)H Fδ +μ(s)Ek(s)ds. Using in (2.15) the above estimate, applying Gronwall’s inequality and making δconverging to zero we conclude  u k(t)  2 H+ uk(t)  2 V⩽CeC(T+T 0μ(s)ds) u0 k  2 V+ ϑ1 k  2 H+f2 L1(0,T;H)+g2 W1,1(0,T;V),(2.17) for every t∈[0,T], where Cdepends continuously on α. Step 3. Passing to the limit. Thanks to (2.17), up to a subsequence, there exists u∈L∞(0,T;V),withu∈L∞(0,T;H)such that uk ∗ uin L∞(0,T;V), u k ∗ uin L∞(0,T;H). Thanks to the linearity of the problems satisfied by ukit is easy to show that uis a solution of (2.6), which satisfies (2.7). Part II: Uniqueness. By linearity, it is enough to prove that the problem of finding u∈L∞(0,T;V)with u∈L∞(0,T;H) solution of R(t)u(t)+A(t)u(t)=0in(0,T), u(0)=0,Ru0+=0,(2.18) has the unique solution u≡0. For this purpose, given h∈L1(0,T,H),wetakevh∈L∞(0,T,V),withv h∈L∞(0,T,H) solution of R(t)v h(t)+A(t)vh(t)=h(t)in (0,T), vh(T)=0,Rv hT−=0. This function exists by the first part of the proof, using the change of variables s=T−t.Takingvhas a test function (2.18) we get (the integrations by parts can be easily justified) T  0u(t),h(t)dt =0,∀h∈L1(0,T,H), and thus u≡0. 2 3. Homogenization In this section we analyze the homogenization of a wave equation with BV coefficients in time. The main strategy in order to compute the homogenized problem consists in an appropriated application of results of homogenization for elliptic problems. As we said in the introduction, let us consider the following wave equation with Dirichlet boundary condition: ⎧ ⎨ ⎩ ∂tρn(t,x)∂tun−divxAn(t,x)∇xun=fn+gnin (0,T)×Ω, un=0on(0,T)×∂Ω, un(0)=u0 n,(ρn∂tun)0+=ϑ1 nin Ω, (3.1) where ρn∈BV(0,T;L∞(Ω)) and An∈BV(0,T;L∞(Ω;Ms N)) satisfy the following hypotheses ρnis bounded in BV0,T;L∞(Ω),Anis bounded in BV0,T;L∞Ω;Ms N.(3.2) There exists α>0 such that ρn(t,x)⩾α,a.e. (t,x)∈(0,T)×Ω, (3.3) An(t,x)ξ ·ξ⩾α|ξ|2,∀ξ∈RN,a.e. (t,x)∈(0,T)×Ω. (3.4) Let us prove in the present section that the limit problem of (3.1) has the same structure with ρnand Anrespectively replaced by the weak-∗limit of ρnin L∞((0,T)×Ω) and the H-limit of Anrespectively. J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 669 We recall the definition of H-limit: Definition 3.1. We consider a bounded sequence Mnin L∞(Ω;MN)such that there exists α>0 satisfying Mn(x)ξ ·ξ⩾α|ξ|2,∀ξ∈RN,a.e. x∈Ω. (3.5) We say that MnH-converges to M∈L∞(Ω;MN), which also satisfies (3.5), if for every F∈H−1(Ω),thesolutionznof −divx(Mn∇xzn)=Fin Ω, zn=0on∂Ω, satisfies znzin H1 0(Ω), Mn∇xznM∇xzin L2Ω;RN, where zis the solution of −divx(M∇xz)=Fin Ω, z=0on∂Ω. It is proved in [15] (and [18] for the case of symmetric matrices) the following compactness theorem for the Hconvergence: Every sequence of matrices Mnwhich is bounded in L∞(Ω;MN)and satisfies (3.5) admits a subsequence which H-converges to some M∈L∞(Ω;MN). InthecaseofthesequenceofmatricesAnwhich we consider in (3.1), we have Proposition 3.2. For every sequence Anwhich is bounded in BV(0,T;L∞(Ω;Ms N)) and satisfies (3.4), there exist a subsequence of n, still denoted by n, and a matrix function A ∈BV(0,T;L∞(Ω;Ms N)) such that An(t,.) H A(t,.), ∀t∈(0,T)\N,(3.6) with N⊂(0,T)a countable subset. Remark 3.3. The properties of the H-limit [15,18] imply that (3.4) is still satisfied with Anreplaced by A. Since ρnis bounded in BV(0,T;L∞(Ω)) and satisfies (3.3), there exist a subsequence of nstill denoted by nand a function ρ∈BV(0,T;L∞(Ω)) satisfying (3.3), with ρnreplaced by ρsuch that ρn ∗ ρin L∞(0,T)×Ω.(3.7) Taking into account Proposition 3.2 we can also assume that this sequence is chosen in such way that (3.6) is satisfied. Therefore, to assume that Anand ρnsatisfy (3.6) and (3.7) respectively, is not a restriction because it always holds for a subsequence. The main result of the present section is the following homogenization result for problem (3.1). Theorem 3.4. We consider Anand ρnwhich satisfy (3.2),(3.4),(3.3),(3.6) and (3.7).Then,forevery f n∈M([0,T]; L2(Ω)),g n∈ BV(0,T;H−1(Ω)),u 0 n∈H1 0(Ω),ϑ1 n∈L2(Ω) such that there exist f ∈M([0,T]; L2(Ω)),g∈BV(0,T;H−1(Ω)),u 0∈H1 0(Ω) and ϑ1∈L2(Ω) satisfying fn ∗ finM[0,T]; L2(Ω),(3.8) gnis bounded in BV0,T;H−1(Ω), s r gn(t)dt → s r g(t)dt in H−1(Ω), ∀r,s∈(0,T), (3.9) u0 nu0in H1 0(Ω), ϑ1 nϑ1in L2(Ω), (3.10) we have that the unique solution unof (3.1) satisfies un ∗ uinL ∞0,T;H1 0(Ω), ∂tun ∗ ∂ tuinL ∞0,T;L2(Ω), where u is the unique solution of ⎧ ⎨ ⎩ ∂tρ(t,x)∂tu−divxA(t,x)∇xu=f+gin(0,T)×Ω, u=0on (0,T)×∂Ω, u(0)=u0,(ρ∂tu)0+=ϑ1in Ω. (3.11) 670 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 Proof of Proposition 3.2. We take μn∈M([0,T])such that μnM([0,T])=VT(An)and  An(t)−An(ˆ t) L∞(0,T;Ms N(Ω)) ⩽μn[t,ˆ t],∀t,ˆ t∈[0,T],with t<ˆ t. Since μnis bounded, up to a subsequence, there exists μ∈M([0,T])such that μnconverges to μweakly-∗in the measures. On the other hand, since An∈BV(0,T;L∞(Ω;Ms N)), we can always assume Ancontinuous on the right in [0,T)and on the left in T, and then that Anis well defined in every point t∈[0,T]. We consider a countable dense set {tk}k∈Nin (0,T)such that μ({tk})=0, for every k∈N.UsingtheH-convergence compactness theorem and reasoning by a diagonal argument we can extract a subsequence of n,stilldenotedbyn, such that there exists A:{tk}k∈N→L∞(Ω;Ms N)satisfying An(tk,.) H A(tk,.), ∀k∈N. This subsequence of nwill be the subsequence which appears in the statement of Proposition 3.2. By Lemma 3.5 below, for ti<tj,wehave  A(ti,.)−A(tj,.)  L∞(Ω;Ms N)⩽Climinf n→∞  An(ti,.)−An(tj,.)  L∞(Ω;Ms N) ⩽Climinf n→∞ μn[ti,tj]⩽Cμ[ti,tj].(3.12) Using this property, we define A∈L∞((0,T)×Ω;Ms N)by A(t,x)=lim st s∈{tk} A(s,x). (3.13) Let us see that Asatisfies the thesis of Proposition 3.2. First, we prove that the limit on the right-hand side of (3.13) exists. This is a simple consequence of the fact that thanks to (3.12), for every ti,tj,witht<ti<tj,wehave  A(ti,.)−A(tj,.)  L∞(Ω;Ms N)⩽Cμ[ti,tj]⩽Cμ(t,tj], where the right-hand side tends to zero when tjtends to t. On the other hand, (3.12) easily implies  A(t,.)−A(ˆ t,.)  L∞(Ω;Ms N)⩽Cμ[t,ˆ t], for every t,ˆ t∈(0,T),witht<ˆ tand therefore Abelongs to BV(0,T;L∞(Ω;Ms N)). In order to finish the proof of Proposition 3.2, it only remains to show that (3.6) is satisfied. For this purpose we take N as the countable set of t∈(0,T)such that μ({t})>0. For t∈(0,T)\Nand f∈H−1(Ω),wedefinezn,z∈H1 0(Ω) as the solutions of −divxAn(t,x)∇xzn=fin Ω, zn=0on∂Ω, −divxA(t,x)∇xz=fin Ω, z=0on∂Ω. We must show that znconverges weakly to zin H1 0(Ω).Sinceznis bounded in H1 0(Ω), it is enough to check that zn converges to zin L2(Ω).Forti>t,wedefinezi n,zi∈H1 0(Ω) as the solutions of −divxAn(ti,x)∇xzi n=fin Ω, zi n=0on∂Ω, −divxA(ti,x)∇xzi=fin Ω, zi=0on∂Ω. Taking zi n−znas a test function in the difference of the equations satisfied by zi nand zn,wehave  Ω An(t,x)∇xzn−zi n·∇ xzn−zi ndx = ΩAn(ti,x)−An(t,x)∇xzi n·∇ xzn−zi ndx ⩽μn[t,ti] zi n H1 0(Ω) zn−zi n H1 0(Ω), which, using that zi nH1 0(Ω) is bounded and the uniform ellipticity of An,implies limsup n→∞  zn−zi n H1 0(Ω) ⩽Cμ[t,ti]. Analogously, we have  z−zi H1 0(Ω) ⩽Cμ[t,ti]. J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 671 Therefore, using that zi nconverges to zin L2(Ω),weget limsup n→∞ zn−zL2(Ω) ⩽limsup n→∞  zn−zi n L2(Ω) +limsup n→∞  zi n−zi L2(Ω) + zi−z L2(Ω) ⩽Cμ[t,ti]. Since μ({t})=0wehavethatμ([t,ti])tends to zero when ticonverges to tand so znconverges to zin L2(Ω).2 Lemma 3.5. We consider two sequences of matrix functions M1 n,M 2 nin L∞(Ω;MN),suchthat  Mi n L∞(Ω;MN)⩽β, Mi n(x)ξ ·ξ⩾α|ξ|2,∀ξ∈RN,a.e. x ∈Ω, i=1,2, which H-converge to M1and M2respectively. Then, for a constant C >0, which only depends on β/α,wehave  M1−M2 L∞(Ω;MN)⩽Climinf n→∞  M1 n−M2 n L∞(Ω;MN).(3.14) Proof. For the case where Mnare symmetric this lemma can be found in [5]. We present here a more direct proof which also does not need to assume Mnsymmetric. Extracting a subsequence if necessary, we can always assume that the liminf in (3.14) is a limit. We consider ξ∈RNand ui n,i=1,2, the solutions of −divxMi n∇xui n=−divxMiξin Ω, ui n=ξ·xon ∂Ω, then (see e.g. [15]) ∇xui nand Mi n∇xui nconverge respectively to ξand Miξin L2(Ω;RN)weakly, for i=1,2. Now, for ϕ∈C∞ c(Ω),ϕ⩾0inΩ, the div-curl lemma [16] shows lim n→∞  ΩM1 n∇xu1 n−M2 n∇xu2 n·∇ xu1 n−u2 nϕdx =0,(3.15) lim n→∞  Ω M1 n∇xu1 n·∇ xu1 nϕdx = Ω M1ξ·ξϕdx.(3.16) Thanks to (3.15), we have limsup n→∞  Ω M2 n∇xu1 n−u2 n·∇ xu1 n−u2 nϕdx ⩽limsup n→∞  ΩM2 n−M1 n∇xu1 n·∇ xu1 n−u2 nϕdx +lim n→∞  ΩM1 n∇xu1 n−M2 n∇xu2 n·∇ xu1 n−u2 nϕdx =limsup n→∞  ΩM2 n−M1 n∇xu1 n·∇ xu1 n−u2 nϕdx, and hence, by (3.16), we get limsup n→∞  Ω∇xu1 n−u2 n 2ϕdx ⩽β α3|ξ|2lim n→∞ M1 n−M2 n  2 L∞(Ω;Ms N) Ω ϕdx.(3.17) By the semicontinuity of the norm for the weak convergence in L2(Ω;RN), (3.16) and (3.17) we get  ΩM1−M2ξ 2ϕdx ⩽liminf n→∞  ΩM1 n∇xu1 n−M2 n∇xu2 n 2ϕdx ⩽2liminf n→∞  ΩM1 n−M2 n∇xu1 n 2ϕdx + ΩM2 n∇xu1 n−u2 n 2ϕdx ⩽2β α1+β2 α2|ξ|2lim n→∞ M1 n−M2 n  2 L∞(Ω;Ms N) Ω ϕdx,∀ϕ∈C∞ c(Ω), and therefore (3.14). 2 672 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 We are now in position to prove Theorem 3.4. Proof of Theorem 3.4. A simple application of Theorem 2.1 with V=H1 0(Ω),H=L2(Ω) proves that there exists a unique solution of (3.1), which is bounded in L∞(0,T;H1 0(Ω)) and is such that ∂tunis bounded in L∞(0,T;L2(Ω)). Therefore, up to a subsequence there exists u∈L∞(0,T;H1 0(Ω)),with∂tu∈L∞(0,T;L2(Ω)), such that un ∗ uin L∞0,T;H1 0(Ω), ∂tun ∗ ∂ tuin L∞0,T;L2(Ω). Let us prove that uis the unique solution of (3.11). For this purpose, we need to compute the limits of the products ρn∂tun and An∇xun. We take μn∈M([0,T])such that μnM([0,T])⩽VT(ρn)+VT(An),  ρn(t)−ρn(ˆ t) L∞(Ω) + An(t)−An(ˆ t) L∞(Ω;Ms N)⩽μn[t,ˆ t], for every t,ˆ t∈[0,T]with t<ˆ t.Sinceμnis bounded in M([0,T]), extracting a subsequence if necessary, we can assume that there exists the weak-∗limit μof μnin M([0,T]). Since ρn∂tunis bounded in L2(0,T;L2(Ω)), we can assume that there exists the weak limit zof ρn∂tunin L2(0,T;L2(Ω)). In order to characterize z, we consider τ∈(0,T),h∈(0,T−τ).Forϕ∈C∞ c(τ,τ+h),ϕ⩾0, we have, in the sense of L2(Ω) τ+h  τ ρn(t)∂tun(t)ϕ(t)dt = τ+h  τρn(t)−1 h τ+h  τ ρn(s)ds∂tun(t)ϕ(t)dt −1 h τ+h  τ ρn(s)dsτ+h  τ un(t)ϕ(t)dt.(3.18) Using that ∂tunis bounded in L∞(0,T;L2(Ω)), the first term on the right-hand side of the above equality can be estimated by      τ+h  τρn(t)−1 h τ+h  τ ρn(s)ds∂tun(t)ϕ(t)dt    L2(Ω) ⩽Cμn[τ,τ+h] τ+h  τ ϕ(t)dt, while for the second one, using the weak-∗convergence of ρnin L∞((0,T)×Ω), the strong convergence of unto uin L2(0,T;L2(Ω)), and an integration by parts, we get 1 h τ+h  τ ρn(s)dsτ+h  τ un(t)ϕ(t)dt −1 h τ+h  τ ρ(s)dsτ+h  τ ∂tu(t)ϕ(t)dt in L2(Ω). Therefore, using the semicontinuity of the norm for the weak convergence, we deduce from (3.18)      τ+h  τ z(t)ϕ(t)dt −1 h τ+h  τ ρ(s)dsτ+h  τ ∂tu(t)ϕ(t)dt    L2(Ω) ⩽Cμ[τ,τ+h] τ+h  τ ϕ(t)dt, which implies z(t)=ρ(t)∂tu(t)for a.e. t∈(0,T). (3.19) This characterizes the weak limit in L2(0,T;L2(Ω)) of ρn∂tun. In order to characterize the weak limit in L2(0,T;L2(Ω;RN)) of An∇xun, we first remark that (3.19), ∂t(ρn∂tun)bounded in M([0,T]; H−1(Ω)) and Lemma A.1 in Appendix A prove that ρn(t)∂tun(t)converges to ρ(t)∂tu(t)in H−1(Ω), ∀t∈(0,T)\N,(3.20) with Nthe countable set of t∈(0,T)such that μ({t})>0. We take τ∈(0,T)and has above such that τ,τ+h/∈N. Integrating Eq. (3.1) in (τ,τ+h]and dividing by h,wehave, in the sense of H−1(Ω) J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 679 5. A counterexample In Section 4, we have obtained a corrector result for problem (3.1) assuming that ∂tρn,∂tAnare uniformly continuous from [0,T]into L∞(Ω) and L∞(Ω;MN)respectively, with a continuity modulus independent of n.WegiveinProposition 5.1 a counterexample showing that Theorem 4.1 is optimal in the sense that it does not hold if we just assume Anand ρnbounded in C1([0,T]; L∞(Ω)) and C1([0,T]; L∞(Ω;MN)) respectively. In particular the result is not true in the general framework of Section 3 and neither in the framework of [6], where it is considered the case of Lipschitz functions in the time variable. Proposition 5.1. For bn∈C∞([0,T]×[0,π]),definedby bn(t,x)=1 nsin(nt)cos(nx)cos(x), ∀(t,x)∈[0,T]×[0,π],(5.1) we take unas the unique solution of ⎧ ⎨ ⎩ ∂2 ttun−∂2 xxun−∂x(bn∂xun)=0in (0,T)×(0,π), un(t,0)=un(t,π)=0, un(0,x)=0,∂ tun(0,x)=sin(x), a.e. in (0,π). (5.2) Then, un ∗ uinL ∞0,T;H1 0(0,π),(5.3) ∂tun ∗ ∂ tuinL ∞0,T;L2(0,π),(5.4) with u(t,x)=sin(t)sin(x), the unique solution of ⎧ ⎨ ⎩ ∂2 ttu−∂2 xxu=0in (0,T)×(0,π), u(t,0)=u(t,π)=0, u(0,x)=0,∂ tu(0,x)=sin(x), a.e. in (0,π), (5.5) but un−uL2(0,T;H1 0(Ω)) 0, ∂t(un−u) L2(0,T;L2(Ω)) 0.(5.6) Remark 5.2. The sequence bndefined by (5.1) is bounded in C1([0,T]×[0,π])and converges strongly to zero in C0([0,T]× [0,π]). This last assertion implies in particular that An(t)=1+bn(t)H-converges to A(t)=1in(0,π)for every t∈[0,T], ∂x(An(0)∂xun(0)) ≡∂x(A(0)∂xu(0)) ∈C∞([0,π]),∂tun(0)=∂tu(0)∈C∞([0,π]). Therefore, if Theorem 4.1 were true for An just bounded in W1,∞((0,T)×Ω) we would get that un−uL∞(0,T;H1 0(Ω)) →0, ∂t(un−u) L∞(0,T;L2(Ω)) →0, in contradiction with (5.6). Proof of Proposition 5.1. Statements (5.3) and (5.4) are a simple consequence of Theorem 3.4. On the other hand, using un−uas a test function in the difference of (5.2) and (5.5), we have π  0un(T)−u(T)∂tun(T)−u(T)dx − T  0 π  0∂t(un−u) 2dxdt + T  0 π  0∂x(un−u) 2dxdt + T  0 π  0 bn∂xun∂x(un−u)dxdt =0. Since unand ∂tunare bounded in L∞(0,T;H1 0(Ω)) and L∞(0,T;L2(Ω)) respectively, we have that (see e.g. [17]) un(T) converges strongly to u(T)in L2(Ω). Therefore the first term in the above equality tends to zero. Using also that bntends to zero in C0([0,T]×[0,π])we conclude that lim n→∞T  0 π  0∂t(un−u) 2dxdt − T  0 π  0∂x(un−u) 2dxdt=0.(5.7) 680 J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 In order to prove (5.6) let us reason by contradiction. Thus, we assume that one of the assertions in (5.6) is not true and then by (5.7), that none of these assertions are satisfied. Using the Fourier expansion of unin the space variable given by un(t,x)= ∞  k=1 ϕk n(t)sin(kx), with ϕk n(t)=2 π π  0 un(t,y)sin(ky)dy, we then have that T  0ϕ1 n−sin(t) 2dt + ∞  k=2 k2 T  0ϕk n 2dt →0,(5.8) T  0ϕ1 n−cos(t) 2dt + ∞  k=2 T  0ϕk n 2dt →0.(5.9) But taking into account the equation satisfied by un(5.2), we easily have, for n⩾2, ϕn n(t)=−2 π t  0 π  0 bn(s,y)∂xun(s,y)cos(ny)sinn(t−s)dyds,∀t∈[0,π]. Using here the expression (5.1) of bnand that, by the contradiction assumption, ∂xunconverges strongly to ∂xuin L2(0,T;L2(0,π)),weget nϕn n+2 π t  0 π  0 sin(ns)sinn(t−s)cos2(ny)cos2(y)sin(s)dyds =−2n π t  0 π  0 bn(s,y)∂xun(s,y)−∂xu(s,y)cos(ny)sinn(t−s)dyds →0inL∞(0,T), where a simple calculus shows 2 π t  0 π  0 sin(ns)sinn(t−s)cos2(ny)cos2(y)sin(s)dyds +1 4cos(nt)1−cos(t)→0inL∞(0,T). Therefore lim n→∞ n2 T  0ϕn n(t) 2dt =1 16 lim n→∞ T  0 cos2(nt)1−cos(t)2dt =1 32 T  01−cos(t)2dt = 0, in contradiction with (5.8). 2 Acknowledgments The authors have been partially supported by the projects MTM2008-00306 of the “Ministerio de Ciencia e Innovación” of Spain and FQM-309 of the “Junta de Andalucía”. Appendix A This appendix is devoted to prove the following Lemma A.1 which was used in the proof of Theorem 3.4. Lemma A.1. We consider two Banach spaces X, Y , with X compactly embedded in Y . Let φnbe a bounded sequence in L1(0,T;X)∩ BV(0,T;Y), which converges weakly in L1(0,T;Y)toafunctionφ.Thesequenceφnis assumed to be continuous on the right in (0,T) with values in Y . Taking μn∈M([0,T])such that μnM([0,T])=VT(φn),  φn(t)−φn(ˆ t) Y⩽μn[t,ˆ t],∀t,ˆ t∈[0,T],t<ˆ t, we denote by μthe weak-∗limit of μnin M([0,T]), which exists up to a subsequence. Then, φn(t)converges strongly to φ(t)in Y for every t ∈(0,T)\N,withNthe countable set of t ∈(0,T)such that μ({t})>0. J. Casado-Díaz et al. / J. Math. Anal. Appl. 379 (2011) 664–681 681 Remark A.2. Assuming in Lemma A.1 that the sequence φnis bounded in L1(0,T;X)∩W1,1(0,T;Y)at the place of L1(0,T;X)∩BV(0,T;Y), the result is a simple consequence of [17], where it is proved that in these conditions φnconverges strongly to φin C0([0,T]; Y). Proof of Lemma A.1. We define Nas the countable set of t∈(0,T)such that μ({t})= 0. For every t∈(0,T)\Nand h∈(0,T−t),wehave  φn(t)−φ(t) Y⩽     φn(t)−1 h t+h t φn(s)ds    Y +     1 h t+h tφn(s)−φ(s)ds    Y +     1 h t+h t φ(s)ds −φ(t)    Y .(A.1) Taking into account that  φn(t)−φn(ˆ t) Y⩽μn[t,ˆ t], φ(t)−φ(ˆ t) ⩽μ[t,ˆ t],∀t,ˆ t∈[0,T],t<ˆ t, the weak-∗convergence of μnto μin M([0,T])and that the weak convergence of φnto φin L1(0,T;X)joining to the compact embedding of Xin Yimplies 1 h t+h tφn(s)−φ(s)ds →0inY, we deduce from (A.1) limsup n→∞  φn(t)−φ(t) Y⩽2μ[t,t+h],∀h∈(0,T−t), and thus, taking the limit in this inequality for htending to zero, we get limsup n→∞  φn(t)−φ(t) Y⩽2μ{t}=0. This finishes the proof of Lemma A.1. 2 References [1] A. Arosio, Linear second order differential equations in Hilbert spaces. The Cauchy problem and asymptotic behaviour for large time, Arch. Rat. Anal. 86 (1984) 147–180. [2] A. Bensoussan, J.L. Lions, G. 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