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Comparative analysis of phenomenological growth models applied to epidemic outbreaks

  • Phenomenological models are particularly useful for characterizing epidemic trajectories because they often offer a simple mathematical form defined through ordinary differential equations (ODEs) that in many cases can be solved explicitly. Such models avoid the description of biological mechanisms that may be difficult to identify, are based on a small number of model parameters that can be calibrated easily, and can be utilized for efficient and rapid forecasts with quantified uncertainty. These advantages motivate an in-depth examination of 37 data sets of epidemic outbreaks, with the aim to identify for each case the best suited model to describe epidemiological growth. Four parametric ODE-based models are chosen for study, namely the logistic and Gompertz model with their respective generalizations that in each case consists in elevating the cumulative incidence function to a power p[0,1]. This parameter within the generalized models provides a criterion on the early growth behavior of the epidemic between constant incidence for p=0, sub-exponential growth for 0<p<1 and exponential growth for p=1. Our systematic comparison of a number of epidemic outbreaks using phenomenological growth models indicates that the GLM model outperformed the other models in describing the great majority of the epidemic trajectories. In contrast, the errors of the GoM and GGoM models stay fairly close to each other and the contribution of the adjustment of p remains subtle in some cases. More generally, we also discuss how this methodology could be extended to assess the "distance" between models irrespective of their complexity.

    Citation: Raimund Bürger, Gerardo Chowell, Leidy Yissedt Lara-Díıaz. Comparative analysis of phenomenological growth models applied to epidemic outbreaks[J]. Mathematical Biosciences and Engineering, 2019, 16(5): 4250-4273. doi: 10.3934/mbe.2019212

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  • Phenomenological models are particularly useful for characterizing epidemic trajectories because they often offer a simple mathematical form defined through ordinary differential equations (ODEs) that in many cases can be solved explicitly. Such models avoid the description of biological mechanisms that may be difficult to identify, are based on a small number of model parameters that can be calibrated easily, and can be utilized for efficient and rapid forecasts with quantified uncertainty. These advantages motivate an in-depth examination of 37 data sets of epidemic outbreaks, with the aim to identify for each case the best suited model to describe epidemiological growth. Four parametric ODE-based models are chosen for study, namely the logistic and Gompertz model with their respective generalizations that in each case consists in elevating the cumulative incidence function to a power p[0,1]. This parameter within the generalized models provides a criterion on the early growth behavior of the epidemic between constant incidence for p=0, sub-exponential growth for 0<p<1 and exponential growth for p=1. Our systematic comparison of a number of epidemic outbreaks using phenomenological growth models indicates that the GLM model outperformed the other models in describing the great majority of the epidemic trajectories. In contrast, the errors of the GoM and GGoM models stay fairly close to each other and the contribution of the adjustment of p remains subtle in some cases. More generally, we also discuss how this methodology could be extended to assess the "distance" between models irrespective of their complexity.


    In this paper, for a bounded domain Ω of Rd, we study the homogenization through Γ-convergence of the conductivity energy with a zero-order term of the type

    Fε(u):={Ω{A(xε)uu+|u|2}dx,if  uH10(Ω),,if  uL2(Ω)H10(Ω). (1)

    The conductivity A is a Yd-periodic, symmetric and non-negative matrix-valued function in L(Rd)d×d, denoted by Lper(Yd)d×d, which is not strongly elliptic, i.e.

    ess-infyYd(min{A(y)ξξ:ξRd,|ξ|=1})0. (2)

    This condition holds true when the conductivity energy density has missing derivatives. This occurs, for example, when the quadratic form associated to A is given by

    Aξξ:=Aξξforξ=(ξ,ξd)Rd1×R,

    where ALper(Yd)(d1)×(d1) is symmetric and non-negative matrix. It is known (see e.g. [13,Chapters 24 and 25]) that the strongly ellipticity of the matrix A, i.e.

    ess-infyYd(min{A(y)ξξ:ξRd,|ξ|=1})>0, (3)

    combined with the boundedness implies a compactness result of the conductivity functional

    uH10(Ω)ΩA(xε)uudx

    for the L2(Ω)-strong topology. The Γ-limit is given by

    ΩAuudx,

    where the matrix-valued function A is defined by the classical homogenization formula

    Aλλ:=min{YdA(y)(λ+v(y))(λ+v(y))dy:vH1per(Yd)}. (4)

    The Γ-convergence for the Lp(Ω)-strong topology, for p>1, for the class of integral functionals Fε of the form

    Fε(u)=Ωf(xε,Du)dx,foruW1,p(Ω;Rm), (5)

    where f:Ω×Rm×dR is a Borel function, 1-periodic in the first variable satisfying the standard growth conditions of order p, namely c1|M|pf(x,M)c2(|M|p+1) for any xΩ and for any real (m×d)-matrix M, has been widely studied and it is a classical subject (see e.g. [4,Chapter 12] and [13,Chapter 24]). On the contrary, the Γ-convergence of oscillating functionals for the weak topology on bounded sets of Lp(Ω) has been very few analysed. An example of the study of Γ-convergence for the Lp(Ω)-weak topology can be found in the paper [6] where, in the context of double-porosity, the authors compare the Γ-limit for non-linear functionals analogous to 5 computed with respect to different topologies and in particular with respect to Lp(Ω)-weak topology.

    In this paper, we investigate the Γ-convergence for the weak topology on bounded sets (a metrizable topology) of L2(Ω) of the conductivity functional under condition 2. In this case, one has no a priori L2(Ω)-bound on the sequence of gradients, which implies a loss of coerciveness of the investigated energy. To overcome this difficulty, we add a quadratic zeroth-order term of the form u2L2(Ω), so that we immediately obtain the coerciveness in the weak topology of L2(Ω) of Fε, namely, for uH10(Ω),

    Fε(u)Ω|u|2dx.

    This estimate guarantees that Γ-limit for the weak topology on bounded sets of L2(Ω) is characterized by conditions (i) and (ii) of the Definition 1.1 below (see [13,Proposition 8.10]), as well as, thanks to a compactness result (see [13,Corollary 8.12]), Fε Γ-converges for the weak topology of L2(Ω), up to subsequences, to some functional. We will show that, under the following assumptions:

    (H1) any two-scale limit u0(x,y) of a sequence uε of functions in L2(Ω) with bounded energy Fε(uε) does not depend on y (see [1,Theorem 1.2]);

    (H2) the space V defined by

    V:={YdA1/2(y)Φ(y)dy:ΦL2per(Yd;Rd)withdiv(A1/2(y)Φ(y))=0inD(Rd)}

    agrees with the space Rd,

    the Γ-limit is given by

    F0(u):={Ω{Auu+|u|2}dx,if  uH10(Ω),,if  uL2(Ω)H10(Ω), (6)

    where the homogenized matrix A is given through the expected homogenization formula

    Aλλ:=inf{YdA(y)(λ+v(y))(λ+v(y))dy:vH1per(Yd)}. (7)

    We need to make assumption (H1) since for any sequence uε with bounded energy, i.e. supε>0Fε(uε)<, the sequence uε in L2(Ω;Rd) is not bounded due to the lack of ellipticity of the matrix-valued conductivity A(y). Assumption (H2) turns out to be equivalent to the positive definiteness of the homogenized matrix (see Proposition 1).

    In the 2D isotropic elasticity setting of [11], the authors make use of similar conditions as (H1) and (H2) in the proof of the main results (see [11,Theorems 3.3 and 3.4]). They investigate the limit in the sense of Γ-convergence for the L2(Ω)-weak topology of the elasticity functional with a zeroth-order term in the case of two-phase isotropic laminate materials where the phase 1 is very strongly elliptic, while the phase 2 is only strongly elliptic. The strong ellipticity of the effective tensor is preserved through a homogenization process expect in the case when the volume fraction of each phase is 1/2, as first evidenced by Gutiérrez [14]. Indeed, Gutiérrez has provided two and three dimensional examples of 1-periodic rank-one laminates such that the homogenized tensor induced by a homogenization process, labelled 1-convergence, is not strongly elliptic. These examples have been revisited by means of a homogenization process using Γ-convergence in the two-dimensional case of [10] and in the three-dimensional case of [12].

    In the present scalar case, we enlighten assumptions (H1) and (H2) which are the key ingredients to obtain the general Γ-convergence result Theorem 2.1. Using Nguetseng-Allaire [1,16] two-scale convergence, we prove that for any dimension d2, the Γ-limit F0 6 for the weak topology of L2(Ω) actually agrees with the one obtained for the L2(Ω)-strong topology under uniformly ellipticity 3, replacing the minimum in 4 by the infimum in 7. Assumption (H2) implies the coerciveness of the functional F0 showing that its domain is H10(Ω) and that the homogenized matrix A is positive definite. More precisely, the positive definiteness of A turns out to be equivalent to assumption (H2) (see Proposition 1). We also provide two and three dimensional 1-periodic rank-one laminates which satisfy assumptions (H1) and (H2) (see Proposition 2 for the two-dimensional case and Proposition 3 for the three-dimensional case). Thanks to Theorem 2.1, the corresponding homogenized matrix A is positive definite. For this class of laminates, an alternative and independent proof of positive definiteness of A is performed using an explicit expression of A (see Proposition 5). This expression generalizes the classical laminate formula for non-degenerate phases (see [17] and also [2,Lemma 1.3.32], [8]) to the case of two-phase rank-one laminates with degenerate and anisotropic phases.

    The lack of assumption (H1) may induce a degenerate asymptotic behaviour of the functional Fε 1. We provide a two-dimensional rank-one laminate with two degenerate phases for which the functional Fε does Γ-converge for the L2(Ω)-weak topology to a functional F which differs from the one given by 6 (see Proposition 4). In this example, any two-scale limit u0(x,y) of a sequence with bounded energy Fε(uε), depends on the variable y. Moreover, we give two quite different expressions of the Γ-limit F which seem to be original up to the best of our knowledge. The energy density of the first expression is written with Fourier transform of the target function. The second expression appears as a non-local functional due to the presence of a convolution term. However, we do not know if the Γ-limit F is a Dirichlet form in the sense of Beurling-Deny [3], since the Markovian property is not stable by the L2(Ω)-weak topology (see Remark 2).

    The paper is organized as follows. In Section 2, we prove a general Γ-convergence result (see Theorem 2.1) for the functional Fε 1 with any non-uniformly elliptic matrix-valued function A, under assumptions (H1) and (H2). In Section 3 we illustrate the general result of Section 2 by periodic two-phase rank-one laminates with two (possibly) degenerate and anisotropic phases in dimension two and three. We provide algebraic conditions so that assumptions (H1) and (H2) are satisfied (see Propositions 2 and 3). In Section 4 we exhibit a two-dimensional counter-example where assumption (H1) fails, which leads us to a degenerate Γ-limit F involving a convolution term (see Proposition 4). Finally, in the Appendix we give an explicit formula for the homogenized matrix A for any two-phase rank-one laminates with (possibly) degenerate phases. We also provide an alternative proof of the positive definiteness of A using an explicit expression of A for the class of two-phase rank-one laminates introduced in Section 3 (see Proposition 5).

    Notation.

    ● For i=1,,d, ei denotes the i-th vector of the canonical basis in Rd;

    Id denotes the unit matrix of Rd×d;

    H1per(Yd;Rn) (resp. L2per(Yd;Rn), Cper(Yd;Rn)) is the space of those functions in H1loc(Rd;Rn) (resp. L2loc(Rd;Rn), Cloc(Rd;Rn)) that are Yd-periodic;

    ● Throughout, the variable x will refer to running point in a bounded open domain ΩRd, while the variable y will refer to a running point in Yd (or k+Yd, kZd);

    ● We write

    uεu0

    with uεL2(Ω) and u0L2(Ω×Yd) if uε two-scale converges to u0 in the sense of Nguetseng-Allaire (see [1,16])

    F1 and F2 denote the Fourier transform defined on L1(R) and L2(R) respectively. For fL1(R)L2(R), the Fourier transform F1 of f is defined by

    F1(f)(λ):=Re2πiλxf(x)dx.

    Definition 1.1. Let X be a reflexive and separable Banach space endowed with the weak topology σ(X,X), and let Fε:XR be a ε-indexed sequence of functionals. The sequence Fε Γ-converges to the functional F0:XR for the weak topology of X, and we write FεΓ(X)wF0, if for any uX,

    i) uεu, F0(u)lim infε0Fε(uε),

    ii) ¯uεu such that limε0Fε(¯uε)=F0(u).

    Such a sequence ¯uε is called a recovery sequence.

    Recall that the weak topology of L2(Ω) is metrizable on bounded sets, i.e. there exists a metric d on L2(Ω) such that on every norm bounded subset B of L2(Ω) the weak topology coincides with the topology induced on B by the metric d (see e.g. [13,Proposition 8.7]).

    In this section, we will prove the main result of this paper. As previously announced, up to a subsequence, the sequence of functionals Fε, given by 1 with non-uniformly elliptic matrix-valued conductivity A(y), Γ-converges for the weak topology on bounded sets of L2(Ω) to some functional. Our aim is to show that Γ-limit is exactly F0 when uH10(Ω).

    Theorem 2.1. Let Fε be functionals given by 1 with A(y) a Yd-periodic, symmetric, non-negative matrix-valued function in L(Rd)d×d satisfying 2. Assume the following assumptions

    (H1) any two-scale limit u0(x,y) of a sequence uε of functions in L2(Ω) with bounded energy Fε(uε) does not depend on y;

    (H2) the space V defined by

    V:={YdA1/2(y)Φ(y)dy:ΦL2per(Yd;Rd)withdiv(A1/2(y)Φ(y))=0inD(Rd)} (8)

    agrees with the space Rd.

    Then, Fε Γ-converges for the weak topology of L2(Ω) to F0, i.e.

    FεΓ(L2)wF0,

    where F0 is defined by 6 and A is given by 7.

    Proof. We split the proof into two steps which are an adaptation of [11,Theorem 3.3] using the sole assumptions (H1) and (H2) in the general setting of conductivity.

    Step 1 - Γ-lim inf inequality.

    Consider a sequence {uε}ε converging weakly in L2(Ω) to uL2(Ω). We want to prove that

    lim infε0Fε(uε)F0(u). (9)

    If the lower limit is then 9 is trivial. Up to a subsequence, still indexed by ε, we may assume that lim infFε(uε) is a limit and we can also assume henceforth that, for some 0<C<,

    Fε(uε)C. (10)

    As uε is bounded in L2(Ω), there exists a subsequence, still indexed by ε, which two-scale converges to a function u0(x,y)L2(Ω×Yd) (see e.g. [1,Theorem 1.2]). In other words,

    uεu0. (11)

    Assumption (H1) ensures that

    u0(x,y)=u(x)is independent ofy, (12)

    where, according to the link between two-scale and weak L2(Ω)-convergences (see [1,Proposition 1.6]), u is the weak limit of uε, i.e.

    uεuweakly inL2(Ω).

    Since all the components of the matrix A(y) are bounded and A(y) is non-negative as a quadratic form, in view of 10, for another subsequence (not relabeled), we have

    A(xε)uεσ0(x,y)withσ0L2(Ω×Yd;Rd),

    and also

    A1/2(xε)uεΘ0(x,y)withΘ0L2(Ω×Yd;Rd). (13)

    In particular

    εA(xε)uε0. (14)

    Consider ΦL2per(Yd;Rd) such that

    div(A1/2(y)Φ(y))=0inD(Rd), (15)

    or equivalently,

    YdA1/2(y)Φ(y)ψ(y)dy=0ψH1per(Yd).

    Take also φC(¯Ω). Since uεH10(Ω) and in view of 15, an integration by parts yields

    ΩA1/2(xε)uεΦ(xε)φ(x)dx=ΩuεA1/2(xε)Φ(xε)φ(x)dx.

    By using [1,Lemma 5.7], A1/2(y)Φ(y)φ(x) is an admissible test function for the two-scale convergence. Then, we can pass to the two-scale limit in the previous expression with the help of the convergences 11 and 13 along with 12, and we obtain

    Ω×YdΘ0(x,y)Φ(y)φ(x)dxdy=Ω×Ydu(x)A1/2(y)Φ(y)φ(x)dxdy. (16)

    We prove that the target function u is in H1(Ω). Setting

    N:=YdA1/2(y)Φ(y)dy, (17)

    and varying φ in Cc(Ω), the equality 16 reads as

    Ω×YdΘ0(x,y)Φ(y)φ(x)dxdy=Ωu(x)Nφ(x)dx

    Since the integral in the left-hand side is bounded by a constant times φL2(Ω), the right-hand side is a linear and continuous map in φL2(Ω). By the Riesz representation theorem, there exists gL2(Ω) such that, for any φCc(Ω),

    Ωu(x)Nφ(x)dx=Ωg(x)φ(x)dx,

    which implies that

    NuL2(Ω). (18)

    In view of assumption (H2), N is an arbitrary vector in Rd so that we infer from 18 that

    uH1(Ω). (19)

    This combined with equality 16 leads us to

    Ω×YdΘ0(x,y)Φ(y)φ(x)dxdy=Ω×YdA1/2(y)u(x)Φ(y)φ(x)dxdy. (20)

    By density, the last equality holds if the test functions Φ(y)φ(x) are replaced by the set of ψ(x,y)L2(Ω;L2per(Yd;Rd)) such that

    divy(A1/2(y)ψ(x,y))=0inD(Rd),

    or equivalently,

    Ω×Ydψ(x,y)A1/2(y)yv(x,y)dxdy=0vL2(Ω;H1per(Yd)).

    The L2(Ω;L2per(Yd;Rd))-orthogonal to that set is the L2-closure of

    K:={A1/2(y)yv(x,y):vL2(Ω;H1per(Yd))}.

    Thus, the equality 20 yields

    Θ0(x,y)=A1/2(y)u(x)+S(x,y)

    for some S in the closure of K, i.e. there exists a sequence vnL2(Ω;H1per(Yd)) such that

    A1/2(y)yvn(x,y)S(x,y)strongly inL2(Ω;L2per(Yd;Rd)).

    Due to the lower semi-continuity property of two-scale convergence (see [1,Proposition 1.6]), we get

    lim infε0A1/2(x/ε)uε2L2(Ω;Rd)Θ02L2(Ω×Yd;Rd)=limnA1/2(y)(xu(x)+yvn)2L2(Ω×Yd;Rd).

    Then, by the weak L2-lower semi-continuity of uεL2(Ω), we have

    lim infε0Fε(uε)limnΩ×YdA(y)(xu(x)+yvn(x,y))(xu(x)+yvn(x,y))dxdy+Ω|u|2dxΩinf{YdA(y)(xu(x)+yv(y))(xu(x)+yv(y))dy:vH1per(Yd)}dx+Ω|u|2dx.

    Recalling the definition 7, we immediately conclude that

    lim infε0Fε(uε)Ω{Auu+|u|2}dx,

    provided that uH10(Ω).

    It remains to prove that the target function u is actually in H10(Ω), giving a complete characterization of Γ-limit. To this end, take x0Ω a Lebesgue point for uΩ and for ν(x0), the exterior normal to Ω at point x0. Thanks to 19, we know that uH1(Ω), hence, after an integration by parts of the right-hand side of 16, we obtain, for φC(¯Ω),

    Ω×YdΘ0(x,y)Φ(y)φ(x)dxdy=ΩNu(x)φ(x)dxΩNν(x)u(x)φ(x)dH, (21)

    where N is given by 17. Varying φ in Cc(Ω), the first two integrals in 21 are equal and bounded by a constant times φL2(Ω). It follows that, for any φC(¯Ω),

    ΩNν(x)u(x)φ(x)dH=0,

    which leads to Nν(x)u(x)=0 H-a.e. on Ω. Since x0 is a Lebesgue point, we have

    Nν(x0)u(x0)=0. (22)

    In view of assumption (H2) and the arbitrariness of N, we can choose N such that N=ν(x0) so that from 22 we get u(x0)=0. Hence,

    uH10(Ω).

    This concludes the proof of the Γ-lim inf inequality.

    Step 2 - Γ-lim sup inequality.

    We use the same arguments of [12,Theorem 2.4] which can easily extend to the conductivity setting. We just give an idea of the proof, which is based on a perturbation argument. For δ>0, let Aδ be the perturbed matrix of Rd×d defined by

    Aδ:=A+δId,

    where Id is the unit matrix of Rd×d. Since the matrix A is non-negative, Aδ turns out to be positive definite, hence, the functional Fεδ, defined by 1 with Aδ in place of A, Γ-converges to the functional Fδ given by

    Fδ(u):={Ω{Aδuu+|u|2}dx,ifuH10(Ω),,ifuL2(Ω)H10(Ω),

    for the strong topology of L2(Ω) (see e.g. [13,Corollary 24.5]). Thanks to the compactness result of Γ-convergence (see e.g. [4,Proposition 1.42]), there exists a subsequence εj such that Fεj Γ-converges for the L2(Ω)-strong topology to some functional F0. Let uH10(Ω) and let uεj be a recovery sequence for Fεj which converges to u for the H1(Ω)-weak topology on bounded sets. Since FεjFδεj and since uεj belongs to some bounded set of H1(Ω), from [13,Propositions 6.7 and 8.10] we deduce that

    F0(u)Fδ(u)lim infεj0Ω{Aδ(xεj)uεjuεj+|uεj|2}dxlim infεj0Ω{A(xεj)uεjuεj+|uεj|2}dx+O(δ)=F0(u)+O(δ).

    It follows that Fδ converges to F0 as δ0. Then, the Γ-limit F0 of Fεj is independent on the subsequence εj. Repeating the same arguments, any subsequence of Fε has a further subsequence which Γ-converges for the strong topology of L2(Ω) to F0=limδ0Fδ. Thanks to the Urysohn property (see e.g. [4,Proposition 1.44]), the whole sequence Fε Γ-converges to the functional F0 for the strong topology of L2(Ω). On the other hand, in light of the definition 7 of A, we get that Aδ converges to A as δ0, i.e.

    limδ0Aδ=A. (23)

    Thanks to the Lebesgue dominated convergence theorem and in view of 23, we get that F0=limδ0Fδ is exactly F0 given by 6. Therefore, Fε Γ-converges to F0 for the L2(Ω)-strong topology.

    Now, let us show that Fε Γ-converges to F0 for the weak topology of L2(Ω). Recall that the L2(Ω)-weak topology is metrizable on the closed ball of L2(Ω). Fix nN and let dBn be any metric inducing the L2(Ω)-weak topology on the ball Bn centered on 0 and of radius n. Let uH10(Ω) and let ¯uε be a recovery sequence for Fε for the L2(Ω)-strong topology. Since the topology induced by the metric dBn on Bn is weaker than the L2(Ω)-strong topology, ¯uε is also a recovery sequence for Fε for the L2(Ω)-weak topology on Bn. Hence,

    limε0Fε(¯uε)=F0(u),

    which proves the Γ-lim sup inequality in Bn. Finally, since any sequence converging weakly in L2(Ω) belongs to some ball BnL2(Ω), as well as its limit, it follows that the Γ-lim sup inequality holds true for Fε for L2(Ω)-weak topology, which concludes the proof.

    The next proposition provides a characterization of Assumption (H2) in terms of homogenized matrix A.

    Proposition 1. Assumption (H2) is equivalent to the positive definiteness of A, or equivalently,

    Ker(A)=V.

    Proof. Consider λKer(A). Define

    H1λ(Yd):={uH1loc(Rd):uisYd-periodic andYdu(y)dy=λ}.

    Recall that uH1λ(Yd) if and only if there exists vH1per(Yd) such that u(y)=v(y)+λy (see e.g. [13,Lemma 25.2]). Since A is non-negative and symmetric, from 7 it follows that

    0=Aλλ=inf{YdA(y)u(y)u(y)dy:uH1λ(Yd)}.

    Then, there exists a sequence un of functions in H1λ(Yd) such that

    limnYdA(y)un(y)un(y)dy=0,

    which implies that

    A1/2un0strongly inL2(Yd;Rd). (24)

    Now, take ΦL2per(Yd;Rd) such that A1/2Φ is a divergence free field in Rd. Recall that, since unH1λ(Yd), we have that un(y)=vn(y)+λ, for some vnH1per(Yd). This implies that

    YdA1/2(y)un(y)Φ(y)dy=Ydun(y)A1/2(y)Φ(y)dy=λYdA1/2(y)Φ(y)dy+Ydvn(y)A1/2(y)Φ(y)dy=λYdA1/2(y)Φ(y)dy, (25)

    where the last equality is obtained by integrating by parts the second integral combined with the fact that A1/2Φ is a divergence free field in Rd. In view of convergence 24, the integral on the left-hand side of 25 converges to 0. Hence, passing to the limit as n in 25 yields

    0=λ(YdA1/2(y)Φ(y)dy),

    for any ΦL2per(Yd;Rd) such that A1/2Φ is a divergence free field in Rd. Therefore λV which implies that

    Ker(A)V.

    Conversely, by 23 we already know that

    limδ0Aδ=A,

    where Aδ is the homogenized matrix associated with Aδ=A+δId. Since Aδ is strongly elliptic, the homogenized matrix Aδ is given by

    Aδλλ=min{YdAδ(y)uδ(y)uδ(y)dy:uδH1λ(Yd)}. (26)

    Let ¯uδ be the minimizer of problem 26. Therefore, there exists a constant C>0 such that

    Aδλλ=YdAδ(y)¯uδ(y)¯uδ(y)dy=Yd|A1/2δ(y)¯uδ(y)|2dyC,

    which implies that the sequence Φδ(y):=A1/2δ(y)¯uδ(y) is bounded in L2per(Yd;Rd). Then, up to extract a subsequence, we can assume that Φδ converges weakly to some Φ in L2per(Yd;Rd).

    Now, we show that A1/2δ converges strongly to A1/2 in Lper(Yd)d×d. Since A(y) is a symmetric matrix, there exists an orthogonal matrix-valued function R in Lper(Yd)d×d such that

    A(y)=R(y)D(y)RT(y)for a.e. yYd,

    where D is a diagonal non-negative matrix-valued function in Lper(Yd)d×d and RT denotes the transpose of R. It follows that Aδ(y)=A(y)+δId=R(y)(D(y)+δId)RT(y), for a.e. yYd. Hence,

    A1/2δ(y)=R(y)(D(y)+δId)1/2RT(y)for a.e. yYd,

    which implies that A1/2δ converges strongly to A1/2=RD1/2RT in Lper(Yd)d×d.

    Now, passing to the limit as δ0 in

    div(A1/2δΦδ)=div(Aδ¯uδ)=0in D(Rd),

    we have

    div(A1/2Φ)=0in D(Rd).

    This along with ΦL2per(Yd;Rd) implies that Φ is a test function for the set V given by 8. From 26 it follows that

    Aδλ=YdAδ(y)¯uδ(y)dy=YdA1/2δ(y)Φδ(y)dy.

    Hence, taking into account the strong convergence of A1/2δ in Lper(Yd)d×d and the weak convergence of Φδ in L2per(Yd;Rd), we have

    Aλ=limδ0Aδλ=limδ0YdA1/2δ(y)Φδ(y)dy=YdA1/2(y)Φ(y)dy,

    which implies that AλV since Φ is a suitable test function for the set V. Therefore, for λV,

    Aλλ=0,

    so that, since A is a non-negative matrix, we deduce that λKer(A). In other words,

    VKer(A),

    which concludes the proof.

    In this section we provide a geometric setting for which assumptions (H1) and (H2) are fulfilled. We focus on a 1-periodic rank-one laminates of direction e1 with two phases in Rd, d=2,3. Specifically, we assume the existence of two anisotropic phases Z1 and Z2 of Yd given by

    Z1=(0,θ)×(0,1)d1andZ2=(θ,1)×(0,1)d1,

    where θ denotes the volume fraction of the phase Z1. Let Z#1 and Z#2 be the associated subsets of Rd, i.e. the open periodic sets

    Z#i:=Int(kZd(¯Zi+k))fori=1,2.

    Let X1 and X2 be unbounded connected components of Z#1 and Z#2 in Rd given by

    X1:=(0,θ)×Rd1andX2:=(θ,1)×Rd1,

    and we denote by Z the interface {y1=0}.

    The anisotropic phases are described by two constant, symmetric and non-negative matrices A1 and A2 of Rd×d which are possibly not positive definite. Hence, the conductivity matrix-valued function ALper(Yd)d×d, given by

    A(y1):=χ(y1)A1+(1χ(y1))A2fory1R, (27)

    where χ is the 1-periodic characteristic function of the phase Z1, is not strongly elliptic, i.e. 2 is satisfied.

    We are interested in two-phase mixtures in R2 with one degenerate phase. We specialize to the case where the non-negative and symmetric matrices A1 and A2 of R2×2 are such that

    A1=ξξandA2is positive definite, (28)

    for some ξR2. The next proposition establishes the algebraic conditions which provide assumptions (H1) and (H2) of Theorem 2.1.

    Proposition 2. Let A1 and A2 be the matrices defined by 28. Assume that ξe10 and the vectors ξ and A2e1 are linearly independent in R2. Then, assumptions (H1) and (H2) are satisfied. In particular, the homogenized matrix A, given by 7, associated to the matrix A defined by 27 and 28 is positive definite.

    From Theorem 2.1, we easily deduce that the energy Fε defined by 1 with A given by 27 and 28 Γ-converges to the functional F0 given by 6 with conductivity matrix A defined by 7. In the present case, the homogenized matrix A has an explicit expression given in Proposition 5 in the Appendix.

    Proof. Firstly, let us prove assumption (H1). We adapt the proof of Step 1 of [11,Theorem 3.3] to two-dimensional laminates. In our context, the algebra involved is different due to the scalar setting.

    Denote by ui0 the restriction of the two-scale limit u0 in phase Zi or Z#i for i=1,2. In view of 14, for any Φ(x,y)Cc(Ω×R2;R2) with compact support in Ω×Z#1, or due to periodicity in Ω×X1, we deduce that

    0=limε0εΩA(xε)uεΦ(x,xε)dx=limε0Ωuεdivy(A1Φ(x,y))(x,xε)dx=Ω×Z#1u10(x,y)divy(A1Φ(x,y))dxdy=Ω×Z#1A1yu10(x,y)Φ(x,y)dxdy,

    so that

    A1yu10(x,y)0inΩ×Z#1. (29)

    Similarly, taking Φ(x,y)Cc(Ω×R2;R2) with compact support in Ω×Z#2, or equivalently in Ω×X2, as test function and repeating the same arguments, we obtain

    A2yu20(x,y)0inΩ×Z#2. (30)

    Due to 29, in phase Z#1 we have

    yu10Ker(A1)=Span(ξ),

    where ξ=(ξ2,ξ1)R2 is perpendicular to ξ=(ξ1,ξ2). Hence, u10 reads as

    u10(x,y)=θ1(x,ξy)a.e.(x,y)Ω×X1, (31)

    for some function θ1L2(Ω×R). On the other hand, since the matrix A2 is positive definite, in phase Z#2 the relation 30 implies that

    u20(x,y)=θ2(x)a.e.(x,y)Ω×X2, (32)

    for some function θ2L2(Ω). Now, consider a constant vector-valued function Φ defined on Y2 such that

    (A1A2)Φe1=0onZ#1. (33)

    Note that condition 33 is necessary for divy(A(y)Φ) to be an admissible test function for two-scale convergence. In view of 14 and 32, for any φCc(Ω;Cper(Y2)), we obtain

    0=limε0εΩA(y)uεΦφ(x,xε)dx=limε0Ωuεdivy(A(y)Φφ(x,y))(x,xε)dx=Ω×Z1u10(x,y)divy(A1Φφ(x,y))dxdy+Ω×Z2θ2(x)divy(A2Φφ(x,y))dxdy.

    Take now φCc(Ω×R2) and use the periodized function

    φ#(x,y):=kZ2φ(x,y+k)

    as new test function. Then, we obtain

    0=Ω×Z1u10(x,y)divy(A1Φφ#(x,y))dxdy+Ω×Z2θ2(x)divy(A2Φφ#(x,y))dxdy=kZ2Ω×(Z1+k)u10(x,y)divy(A1Φφ(x,y))dxdy+kZ2Ω×(Z2+k)θ2(x)divy(A2Φφ(x,y))dxdy=Ω×Z#1u10(x,y)divy(A1Φφ(x,y))dxdy+Ω×Z#2θ2(x)divy(A2Φφ(x,y))dxdy. (34)

    Recall that A1=ξξ, where ξ is such that ξe10. This combined with the linear independence of the vectors ξ and A2e1 implies that the linear map

    ΦR2(A1e1Φ,A2e1Φ)R2

    is one-to-one. Hence, for any fR, there exists a unique ΦR2 such that

    A1Φe1=A2Φe1=f. (35)

    In view of the arbitrariness of f in 35, we can choose Φ such that

    A1e1Φ=A2e1Φ=1onZ#1. (36)

    Since A1yu10=0 in the distributional sense and A1=ξξ, we deduce that u10 is constant along the direction ξ. Using Fubini's theorem, we may integrate along straight lines parallel to the vector ξ where integration by parts is allowed. Therefore, performing an integration by parts in 34 combined with 36, it follows that for any φCc(Ω×R2),

    0=Ω×Zv0(x,y)φ(x,y)dxdHy,

    where we have set v0(x,y):=u10(x,y)θ2(x). We conclude that v0(x,) has a trace on Z for a.e. xΩ satisfying

    v0(x,)=0onZ. (37)

    Recall that Z={y1=0}. Fix xΩ. Taking into account 31 and 32, the equality 37 reads as

    θ1(x,ξ1y2)=θ2(x)onZ.

    Since ξe10, it follows that θ1 only depends on x so that u10(x,y) agrees with θ2(x). Finally, we conclude that u0(x,y):=χ(y1)u10(x,y)+(1χ(y1))u20(x,y) is independent of y and hence (H1) is satisfied.

    We prove assumption (H2). The proof is a variant of the Step 2 of [11,Theorem 3.4]. For arbitrary α,βR, let Φ be a vector-valued function given by

    A1/2(y)Φ(y):=χ(y1)αξ+(1χ(y1))(αξ+βe2)for a.e.yR2. (38)

    Such a vector field Φ does exist, since ξ is in the range of A1 and thus the right-hand side of 38 belongs pointwise to the range of A, or equivalently to the range of A1/2. Moreover, the difference of two constant phases in 38 is orthogonal to the laminate direction e1, so that A1/2Φ is a laminate divergence free periodic field in R2. Its average value is given by

    N:=Y2A1/2(y)Φ(y)dy=αξ+(1θ)βe2.

    Hence, due to ξe10 and the arbitrariness of α,β, the set of the vectors N spans R2, which yields assumption (H2).

    From Proposition 1, it immediately follows that the homogenized matrix A is positive definite. For the reader's convenience, the proof of explicit formula of A is postponed to Proposition 5 in the Appendix.

    We are going to deal with three-dimensional laminates where both phases are degenerate. We assume that the symmetric and non-negative matrices A1 and A2 of R3×3 have rank two, hence, there exist η1,η2R3 such that

    Ker(Ai)=Span(ηi)fori=1,2. (39)

    The following proposition gives the algebraic conditions so that assumptions required by Theorem 2.1 are satisfied.

    Proposition 3. Let η1 and η2 be the vectors in R3 defined by 39. Assume that the vectors {e1,η1,η2} as well as {A1e1,A2e1} are linearly independent in R3. Then, assumptions (H1) and (H2) are satisfied. In particular, the homogenized matrix A given by 7 and associated to the conductivity matrix A given by 27 and 39 is positive definite.

    Invoking again Theorem 2.1, the energy Fε defined by 1 with A given by 27 and 39, Γ-converges for the weak topology of L2(Ω) to F0 where the effective conductivity A is given by 7. As in two-dimensional laminate materials, A has an explicit expression (see Proposition 5 in the Appendix).

    Proof. We first show assumption (H1). The proof is an adaptation of the first step of [11,Theorem 3.3]. Same arguments as in the proof of Proposition 2 show that

    Aiyui0(x,y)0inΩ×Z#ifori=1,2. (40)

    In view of 39 and 40, in phase Z#i, ui0 reads as

    ui0(x,y)=θi(x,ηiy)a.e.(x,y)Ω×Xi, (41)

    for some function θiL2(Ω×R) and i=1,2. Now, consider a constant vector-valued function Φ on Y3 such that the transmission condition 33 holds. In view of 14, for any φCc(Ω,Cper(Y3)), we obtain

    0=limε0εΩA(y)uεΦφ(x,xε)dx=Ω×Z1u10(x,y)divy(A1Φφ(x,y))dxdy+Ω×Z2u20(x,y)divy(A2Φφ(x,y))dxdy. (42)

    Take φCc(Ω×R3). Putting the periodized function

    φ#(x,y):=kZ3φ(x,y+k)

    as test function in 42, we get

    Ω×Z#1u10(x,y)divy(A1Φφ(x,y))dxdy+Ω×Z#2u20(x,y)divy(A2Φφ(x,y))dxdy=0. (43)

    Since the vectors A1e1 and A2e1 are independent in R3, the linear map

    ΦR3(A1e1Φ,A2e1Φ)R2

    is surjective. In particular, for any fR, there exists ΦR3 such that

    A1Φe1=A2Φe1=f. (44)

    In view of the arbitrariness of f in 44, we can choose Φ such that 36 is satisfied. Due to 40 and 39, we deduce that ui0 is constant along the plane Πi perpendicular to ηi, for i=1,2. This implies that, thanks to Fubini's theorem, we may integrate along the plane Πi where an integration by part may be performed. Hence, an integration by parts in 43 combined with 36, yields for any φCc(Ω×R3),

    Ω×Z[u10(x,y)u20(x,y)]φ(x,y)dxdHy=0,

    which implies that

    u10(x,)=u20(x,)onZ. (45)

    Fix xΩ and recall that Z={y1=0}. In view of 41, the relation 45 reads as

    θ1(x,b1y2+c1y3)=θ2(x,b2y2+c2y3)onZ, (46)

    with ηi=(ai,bi,ci) for i=1,2. Due to the independence of {e1,η1,η2} in R3, the linear map (y2,y3)R2(z1,z2)R2 defined by

    z1:=b1y2+c1y3,z2:=b2y2+c2y3,

    is a change of variables so that 46 becomes

    θ1(x,z1)=θ2(x,z2)a.e.z1,z2R.

    This implies that θ1 and θ2 depend only on x and thus u10 and u20 agree with some function uL2(Ω). Finally, we conclude that u0(x,y)=χ(y1)u10(x,y)+(1χ(y1))u20(x,y) is independent of y and hence (H1) is satisfied.

    It remains to prove assumption (H2). To this end, let E be the subset of R3×R3 defined by

    E:={(ξ1,ξ2)R3×R3:(ξ1ξ2)e1=0,ξ1η1=0,ξ2η2=0}. (47)

    For (ξ1,ξ2)E, let Φ be the vector-valued function defined by

    A1/2(y)Φ(y):=χ(y1)ξ1+(1χ(y1))ξ2a.e.yR3. (48)

    The existence of such a vector field Φ is guaranteed by the conditions ξiηi=0, for i=1,2, which imply that ξi belongs to the range of Ai and hence the right-hand side of 48 belongs pointwise to the range of A, or equivalently to the range of A1/2. Moreover, since the difference of the phases ξ1 and ξ2 is orthogonal to the laminate direction e1, A1/2Φ is a laminate divergence free periodic field in R3. Its average value is given by

    N:=Y3A1/2(y)Φ(y)dy=θξ1+(1θ)ξ2.

    Note that E is a linear subspace of R3×R3 whose dimension is three. Indeed, let f be the linear map defined by

    (ξ1,ξ2)R3×R3((ξ1ξ2)e1,ξ1η1,ξ2η2)R3.

    If we identify the pair (ξ1,ξ2)R3×R3 with the vector (x1,y1,z1,x2,y2,z2)R6, with ξi=(xi,yi,zi), for i=1,2, the associated matrix MfR3×6 of f is given by

    Mf:=(100100a1b1c1000000a2b2c2),

    with ηi=(ai,bi,ci), i=1,2. In view of the linear independence of {e1,η1,η2}, the rank of Mf is three, which implies that the dimension of kernel Ker(f) is also three. Since the kernel Ker(f) agrees with E, we conclude that the dimension of E is three.

    Now, let g be the linear map defined by

    (ξ1,ξ2)Eθξ1+(1θ)ξ2R3.

    Let us show that g is invertible. To this end, consider (ξ1,ξ2)Ker(g). From the definition of the map g, Ker(g) consists of all vectors (ξ1,ξ2)E of the form

    (ξ1,θθ1ξ1). (49)

    In view of the definition of E given by 47, the vector 49 satisfies the conditions

    (ξ1θθ1ξ1)e1=0,ξ1η1=0,θθ1ξ1η2=0.

    This combined with the linear independence of {e1,η1,η2} implies that

    ξ1{e1,η1,η2}={0}.

    Hence, Ker(g)={(0,0)} which implies along with the fact that the dimension of E is three that g is invertible. This proves that all the vectors of R3 can be attained through the map g so that assumption (H2) is satisfied.

    Thanks to Proposition 1, the homogenized matrix A turns out to be positive definite. The proof of the explicit expression of A is given in Proposition 5 in the Appendix.

    In this section we are going to construct a counter-example of two-dimensional laminates with two degenerate phases, where the lack of assumption (H1) provides an anomalous asymptotic behaviour of the functional Fε 1.

    Let Ω:=(0,1)2 and let e2 be the laminate direction. We assume that the non-negative and symmetric matrices A1 and A2 of R2×2 are given by

    A1=e1e1andA2=ce1e1,

    for some positive constant c>1. The presence of c1 is essential to have oscillation in the conductivity matrix A. In the present case, the matrix-valued conductivity A is given by

    A(y2):=χ(y2)A1+(1χ(y2))A2=a(y2)e1e1fory2R, (50)

    with

    a(y2):=χ(y2)+c(1χ(y2))1. (51)

    Thus, the energy Fε, defined by 1 with A(y) given by 50 and 51 becomes

    Fε(u)={Ω[a(x2ε)(ux1)2+|u|2]dx, if uH10((0,1)x1;L2(0,1)x2),, if uL2(Ω)H10((0,1)x1;L2(0,1)x2). (52)

    We denote by 1 the convolution with respect to the variable x1, i.e. for fL1(R2) and gL2(R2)

    (f1g)(x1,x2)=Rf(x1t,x2)g(t,x2)dt.

    Throughout this section, cθ denotes the positive constant given by

    cθ:=cθ+1θ, (53)

    where θ(0,1) is the volume fraction of the phase Z1 in Y2. The following result proves the Γ-convergence of Fε for the weak topology of L2(Ω) and provides two alternative expressions of the Γ-limit, one of that seems nonlocal due to presence of convolution term (see Remark 2 below).

    Proposition 4. Let Fε be the functional defined by 52. Then, Fε Γ-converges for the weak topology of L2(Ω) to the functional defined by

    F(u):={10dx2R1ˆk0(λ1)|F2(u)(λ1,x2)|2dλ1,if  uH10((0,1)x1;L2(0,1)x2),,if  uL2(Ω)H10((0,1)x1;L2(0,1)x2),

    where F2(u)(λ1,) denotes the Fourier transform on L2(R) of parameter λ1 with respect to the variable x1 of the function x1u(x1,) extended by zero outside (0,1) and

    ˆk0(λ1):=1014π2a(y2)λ21+1dy2. (54)

    The Γ-limit F can be also expressed as

    F(u):={10dx2R{ccθ(ux1)2(x1,x2)+[αu(x1,x2)+(h1u)(x1,x2)]2}dx1, if uH10((0,1)x1;L2(0,1)x2),, if uL2(Ω)H10((0,1)x1;L2(0,1)x2), (55)

    where cθ is given by 53 and h is a real-valued function in L2(R) defined by means of its Fourier transform F2 on L2(R)

    F2(h)(λ1):=α+f(λ1)α, (56)

    where α and f are given by

    α:=c2θ+1θc2θ>0,f(λ1):=(c1)2θ(θ1)c2θ1cθ4π2λ21+1. (57)

    Moreover, any two-scale limit u0(x,y) of a sequence uε with bounded energy Fε depends on the variable y2Y1.

    Remark 1. From 57, we can deduce that

    α+f(λ1)=1c2θ(cθ4π2λ21+1){(c2θ+1θ)cθ4π2λ21+[(c1)θ+1]2}>0,

    for any λ1R, so that the Fourier transform of h is well-defined.

    Proof. We divide the proof into three steps.

    Step 1 - Γ-lim inf inequality.

    Consider a sequence {uε}ε converging weakly in L2(Ω) to uL2(Ω). Our aim is to prove that

    lim infε0Fε(uε)F(u). (58)

    If the lower limit is then 58 is trivial. Up to a subsequence, still indexed by ε, we may assume that lim infFε(uε) is a limit and we may assume henceforth that, for some 0<C<,

    Fε(uε)C. (59)

    It follows that the sequence uε is bounded in L2(Ω) and according to [1,Theorem 1.2], a subsequence, still indexed by ε, of that sequence two-scale converges to some u0(x,y)L2(Ω×Y2). In other words,

    uεu0. (60)

    In view of 51, we know that a1 so that, thanks to 59, for another subsequence (not relabeled) we have

    uεx1σ0(x,y)withσ0L2(Ω×Y2). (61)

    In particular,

    εuεx10. (62)

    Take φCc(Ω;Cper(Y2)). By integration by parts, we obtain

    εΩuεx1φ(x,xε)dx=Ωuε(εφx1(x,xε)+φy1(x,xε))dx.

    Passing to the limit in both terms with the help of 60 and 62 leads to

    0=Ω×Y2u0(x,y)φy1(x,y)dxdy,

    which implies that

    u0(x,y)is independent ofy1. (63)

    Due to the link between two-scale and weak L2-convergences (see [1,Proposition 1.6]), we have

    uεu(x)=Y1u0(x,y2)dy2weakly in   L2(Ω). (64)

    Now consider φC(¯Ω;Cper(Y2)) such that

    φy1(x,y)=0. (65)

    Since uεH10((0,1)x1;L2(0,1)x2), an integration by parts leads us to

    Ωuεx1φ(x,y)dx=Ωuεφx1(x,y)dx.

    In view of the convergences 60 and 61 together with 63, we can pass to the two-scale limit in the previous expression and we obtain

    Ω×Y2σ0(x,y)φ(x,y)dxdy=Ω×Y2u0(x,y2)φx1(x,y)dxdy. (66)

    Varying φCc(Ω;Cper(Y2)), the left-hand side of 66 is bounded by a constant times φL2(Ω×[0,1)) so that the right-hand side is a linear and continuous form in φL2(Ω×Y2). By the Riesz representation theorem, there exists gL2(Ω×Y2) such that, for any φCc(Ω;Cper(Y2)),

    Ω×Y2u0(x,y2)φx1(x,y)dxdy=Ω×Y2g(x,y)φ(x,y)dxdy,

    which yields

    u0x1(x,y2)L2(Ω×Y1). (67)

    Then, an integration by parts with respect to x1 of the right-hand side of 66 yields, for any φC(¯Ω;Cper(Y2)) satisfying 65,

    Ω×Y2σ0(x,y)φ(x,y)dxdy=Ω×Y2u0x1(x,y2)φ(x,y)dxdy10dx2Y2[u0(1,x2,y2)φ(1,x2,y)u0(0,x2,y2)φ(0,x2,y)]dy.

    Since for any φCc(Ω;Cper(Y2)) the first two integrals are equal and bounded by a constant times φL2(Ω×[0,1), we conclude that, for any φC(¯Ω;Cper(Y2)) satisfying 65,

    10dx2Y2[u0(1,x2,y2)φ(1,x2,y)u0(0,x2,y2)φ(0,x2,y)]dy=0,

    which implies that

    u0(1,x2,y2)=u0(0,x2,y2)=0a.e.(x2,y2)(0,1)×Y1.

    This combined with 67 yields

    u0(x1,x2,y2)H10((0,1)x1;L2((0,1)x2×Y1)).

    Finally, an integration by parts with respect to x1 of the right-hand side of 66 implies that, for any φC(¯Ω;Cper(Y2)) satisfying 65,

    Ω×Y2(σ0(x,y)u0x1(x,y2))φ(x,y)dxdy=0.

    Since the orthogonal of divergence-free functions is the gradients, from the previous equality we deduce that there exists ˜uH1per(Y1;L2(Ω×Y1)) such that

    σ0(x,y)=u0x1(x,y2)+˜uy1(x,y). (68)

    Now, we show that

    lim infε0Ωa(x2ε)(uεx1)2dxΩ×Y2a(y2)(u0x1(x,y2)+˜uy1(x,y))2dxdy. (69)

    To this end, set

    σε:=uεx1.

    Since aLper(Y1)L2per(Y1), there exists a sequence ak of functions in Cper(Y1) such that

    aakL2(Y1)0as  k, (70)

    hence, by periodicity, we also have

    a(x2ε)ak(x2ε)L2(Ω)CaakL2(Y1), (71)

    for some positive constant C>0. On the other hand, since σ0 given by 68 is in L2(Ω×Y2), there exists a sequence ψn of functions in Cc(Ω;Cper(Y2)) such that

    ψn(x,y)σ0(x,y)strongly in   L2(Ω×Y2). (72)

    From the inequality

    Ωa(x2ε)(σεψn(x,xε))2dx0,

    we get

    Ωa(x2ε)σ2εdx2Ωa(x2ε)σεψn(x,xε)dxΩa(x2ε)ψ2n(x,xε)dx=2Ω(a(x2ε)ak(x2ε))σεψn(x,xε)dx+2Ωak(x2ε)σεψn(x,xε)dxΩa(x2ε)ψ2n(x,xε)dx. (73)

    In view of 71, the first integral on the right-hand side of 73 can be estimated as

    |Ω(a(x2ε)ak(x2ε))σεψn(x,xε)dx|CaakL2(Y1)ψnL(Ω)σεL2(Ω)CaakL2(Y1).

    Hence, passing to the limit as ε0 in 73 with the help of 61 leads to

    lim infε0Ωa(x2ε)σ2εdxCaakL2(Y1)+2limε0Ωak(x2ε)σεψn(x,xε)dxlimε0Ωa(x2ε)ψ2n(x,xε)dxdy=2Ω×Y2ak(y2)σ0(x,y)ψn(x,y)dxdyCaakL2(Y1)Ω×Y2a(y2)ψ2n(x,y)dxdy.

    Thanks to 70, we take the limit as k\to\infty in the previous inequality and we obtain

    \begin{align*} \liminf\limits_{\varepsilon\to 0} \int_{{\Omega}} a\left ({\frac{x_2}{\varepsilon}}\right)\sigma^2_\varepsilon dx &\geq 2\int_{{\Omega}\times Y_2}a(y_2)\sigma_0(x, y)\psi_n(x, y)dxdy\notag\\ &\quad-\int_{{\Omega}\times Y_2}a(y_2)\psi_n^2(x, y)dxdy, \end{align*}

    so that in view of 72, passing to the limit as n\to\infty leads to

    \begin{align*} \liminf\limits_{\varepsilon\to 0} \int_{{\Omega}} a\left ({\frac{x_2}{\varepsilon}}\right)\sigma^2_\varepsilon dx &\geq \int_{{\Omega}\times Y_2}a(y_2)\sigma_0^2(x, y)dxdy. \end{align*}

    This combined with 68 proves 69.

    By 63, we already know that u_0 does not depend on y_1 . In view of the periodicity of \tilde{u} with respect to y_1 , an application of Jensen's inequality leads us to

    \begin{align} \int_{{\Omega}\times Y_2} a(y_2)&\left (\frac{\partial u_0}{\partial x_1}(x, y_2) +\frac{\partial \tilde{u}}{\partial y_1} (x, y)\right)^2dxdy\\ & = \int_{{\Omega}}dx\int_{Y_1}a(y_2)dy_2\int_{Y_1}\left (\frac{\partial u_0}{\partial x_1}(x, y_2) +\frac{\partial \tilde{u}}{\partial y_1} (x, y)\right)^2dy_1\\ &\geq\int_{{\Omega}}dx\int_{Y_1}a(y_2)dy_2\left (\int_{Y_1} \left [\frac{\partial u_0}{\partial x_1}(x, y_2) +\frac{\partial \tilde{u}}{\partial y_1} (x, y)\right] dy_1 \right)^2\\ & = \int_{{\Omega}}dx\int_{Y_1}a(y_2)\left (\frac{\partial u_0}{\partial x_1}\right)^2(x, y_2)dy_2. \end{align}

    This combined with 69 implies that

    \begin{align} \liminf\limits_{\varepsilon\to 0}\int_{{\Omega}}a\left (\frac{x_2}{\varepsilon}\right)\left (\frac{\partial{u_\varepsilon}}{\partial x_1}\right)^2dx &\geq\int_{{\Omega}}dx\int_{Y_1}a(y_2)\left (\frac{\partial u_0}{\partial x_1}\right)^2(x, y_2)dy_2. \end{align} (74)

    Now, we extend the functions in L^2({\Omega}) by zero with respect to x_1 outside (0, 1) so that functions in H^1_0((0, 1)_{x_1};L^2(0, 1)_{x_2}) can be regarded as functions in H^1(\mathbb{R}_{x_1}; L^2(0, 1)_{x_2}) . Due to the weak L^2 -lower semi-continuity of \|{u_\varepsilon}\|_{L^2({\Omega})} along with 74, we have

    \begin{align} &\liminf\limits_{\varepsilon\to 0} {\mathscr{F}_\varepsilon}({u_\varepsilon}) \\ &\quad \geq \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}}\biggl[a(y_2)\left (\frac{\partial u_0}{\partial x_1}\right)^2(x_1, x_2, y_2)+ |u_0|^2(x_1, x_2, y_2)\biggr]dx_1. \end{align} (75)

    We minimize the right-hand side with respect to u_0(x_1, x_2, y_2)\in H^1(\mathbb{R}_{x_1}; \quad L^2((0, 1)_{x_2} \times Y_1)) satisfying 64 where the weak limit u of {u_\varepsilon} in L^2({\Omega}) is fixed. The minimizer, still denoted by u_0 , satisfies the Euler equation

    \begin{align*} \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}}&\biggl[a(y_2)\frac{\partial u_0}{\partial x_1}(x_1, x_2, y_2)\frac{\partial v}{\partial x_1}(x_1, x_2, y_2)\notag\\ &\quad + u_0(x_1, x_2, y_2)v(x_1, x_2, y_2)\biggr]dx_1 = 0 \end{align*}

    for any v(x_1, x_2, y_2)\in H^1(\mathbb{R}_{x_1}; \quad L^2((0, 1)_{x_2}\times Y_1)) such that \int_{Y_1} v(x, y_2)dy_2 = 0 . Then, there exists b(x_1, x_2)\in H^{-1}(\mathbb{R}_{x_1}; \quad L^2(\mathbb{R})_{x_2}) independent of y_2 such that in distributions sense with respect to the variable x_1 ,

    \begin{equation} -a(y_2)\frac{\partial^2 u_0}{\partial x_1^2}(x_1, x_2 , y_2) + u_0(x_1, x_2, y_2) = b(x_1, x_2) \quad \text{in} \quad \mathscr{D}'({\mathbb{R}}) \end{equation} (76)

    for a.e. (x_2, y_2)\in (0, 1)\times Y_1 . Taking the Fourier transform {\mathcal{F}}_2 on L^2({\mathbb{R}}) of parameter \lambda_1 with respect to the variables x_1 , the equation 76 becomes

    \begin{equation} {\mathcal{F}}_2(u_0)(\lambda_1, x_2, y_2) = \frac{{\mathcal{F}}_2(b)(\lambda_1, x_2)}{4\pi^2a(y_2)\lambda_1^2+1} \qquad\text{a.e.} \quad (\lambda_1, x_2, y_2)\in {\mathbb{R}}\times (0, 1)\times Y_1. \end{equation} (77)

    Note that 77 proves in particular that the two-scale limit u_0 does depend on the variable y_2 , since its Fourier transform with respect to the variable x_1 depends on y_2 through the function a(y_2) .

    In light of the definition 54 of \hat{k}_0 and due to 64, integrating 77 with respect to y_2\in Y_1 yields

    \begin{equation} {\mathcal{F}}_2(u)(\lambda_1, x_2) = \hat{k}_0(\lambda_1){\mathcal{F}}_2 (b)(\lambda_1, x_2)\qquad\text{a.e.} \quad (\lambda_1, x_2)\in {\mathbb{R}}\times (0, 1). \end{equation} (78)

    By using Plancherel's identity with respect to the variable x_1 in the right-hand side of 75 and in view of 77 and 78, we obtain

    \begin{align} \liminf\limits_{\varepsilon\to 0} {\mathscr{F}_\varepsilon}({u_\varepsilon})&\geq \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}} (4\pi^2a(y_2)\lambda_1^2 +1)|{\mathcal{F}}_2(u_0)(\lambda_1, x_2, y_2)|^2d\lambda_1\\ & = \int_{0}^{1}dx_2\int_{\mathbb{R}} \frac{1}{\hat{k}_0(\lambda_1)} |{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1, \end{align}

    which proves the \Gamma - \liminf inequality.

    Step 2- \Gamma - \limsup inequality.

    For the proof of the \Gamma - \limsup inequality, we need the following lemma whose proof will be given later.

    Lemma 4.1. Let u\in{C^\infty_{\mathit{{c}}}}({\Omega}) . For fixed x_2\in (0, 1) and y_2\in Y_1 , let b(\cdot, x_2) be the distribution (parameterized by x_2 ) defined by

    \begin{equation} {\mathcal{F}}_2(b)(\lambda_1, x_2): = \frac{1}{\hat{k}_0(\lambda_1)} {\mathcal{F}}_2(u)(\lambda_1, x_2), \end{equation} (79)

    where u(\cdot, x_2) is extended by zero outside (0, 1) . Let u_0(\cdot, x_2, y_2) be the unique solution to problem

    \begin{align} \label{stuliopb} \left\{ \begin{array}{l} -a(y_2)\frac{\partial^2 u_0}{\partial x_1^2}(x_1, x_2, y_2) + u_0(x_1, x_2, y_2) = b(x_1, x_2), & x_1\in (0, 1), \\ u_0(0, x_2, y_2) = u_0(1, x_2, y_2) = 0, & \end{array} \right. \end{align} (80)

    with a(y_2) given by 51. Then b(x_1, x_2) is in C([0, 1]_{x_2}; \quad L^2(0, 1)_{x_1}) and u_0(x_1, x_2, y_2) is in C^1([0, 1]^2; \quad L^\infty_{\rm per}(Y_1)) .

    Let u\in{C^\infty_{\text{c}}}({\Omega}) . Thanks to Lemma 4.1, there exists a unique solution

    \begin{equation} u_0(x_1, x_2, y_2)\in C^1([0, 1]^2; \quad L^\infty_{\rm per}(Y_1)) \end{equation} (81)

    to the problem 80. Taking the Fourier transform {\mathcal{F}}_2 on L^2({\mathbb{R}}) of parameter \lambda_1 with respect to x_1 of the equation in 80 and taking into account 79, we get

    \begin{equation} {\mathcal{F}}_2(u_0)(\lambda_1, x_2, y_2) = \frac{{\mathcal{F}}_2(u)(\lambda_1, x_2) }{(4\pi^2a(y_2)\lambda_1^2+1)\hat{k}_0(\lambda_1)}\qquad \text{for} \quad (\lambda_1, x_2, y_2)\in{\mathbb{R}}\times [0, 1]\times Y_1, \end{equation} (82)

    where u_0(\cdot, x_2, y_2) and u(\cdot, x_2) are extended by zero outside (0, 1) . Integrating 82 over y_2\in Y_1 , we obtain

    \begin{equation} u(x_1, x_2) = \int_{Y_1} u_0(x_1, x_2, y_2)dy_2 \qquad \text{for} \quad (x_1, x_2)\in\mathbb{R}\times (0, 1). \end{equation} (83)

    Let \{{u_\varepsilon}\}_\varepsilon be the sequence in L^2({\Omega}) defined by

    \begin{equation*} \label{defuesp} u_\varepsilon(x_1, x_2) : = u_0\left (x_1, x_2, {\frac{x_2}{\varepsilon}}\right). \end{equation*}

    Recall that rapidly oscillating Y_1 -periodic function {u_\varepsilon} weakly converges in L^2({\Omega}) to the mean value of {u_\varepsilon} over Y_1 . This combined with 83 implies that {u_\varepsilon} weakly converges in L^2({\Omega}) to u . In other words,

    \begin{equation*} {u_\varepsilon} \rightharpoonup u \quad \text{weakly in} \quad L^2({\Omega}). \end{equation*}

    Due to 81, we can apply [1,Lemma 5.5] so that u_0(x_1, x_2, y_2) and {\partial u_0\over \partial x_1} are admissible test functions for the two-scale convergence. Hence,

    \begin{align} \lim\limits_{\varepsilon\to 0}&{\mathscr{F}_\varepsilon}({u_\varepsilon}) = \lim\limits_{\varepsilon\to 0} \int_{{\Omega}}\left [a\left ({\frac{x_2}{\varepsilon}}\right)\left (\frac{\partial u_0 }{\partial x_1} \right)^2\left (x_1, x_2, \frac{x_2}{\varepsilon}\right) +\left |u_0\left (x_1, x_2, {\frac{x_2}{\varepsilon}}\right)\right|^2 \right]dx\\ & = \int_{{\Omega}}dx\int_{Y_1}\left [ a(y_2)\left (\frac{\partial u_0}{\partial x_1} \right)^2(x_1, x_2, y_2) +\left |u_0(x_1, x_2, y_2)\right|^2\right]dy_2\\\ & = \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}}\left [ a(y_2)\left (\frac{\partial u_0}{\partial x_1} \right)^2(x_1, x_2, y_2)+\left |u_0(x_1, x_2, y_2)\right|^2\right]dx_1, \end{align} (84)

    where the function x_1\mapsto u_0(x_1, \cdot, \cdot) is extended by zero outside (0, 1) . In view of the definition 54 of \hat{k}_0 and due to 82, the Plancherel identity with respect to the variable x_1 and the Fubini theorem yield

    \begin{align*} \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}}&\left [ a(y_2)\left (\frac{\partial u_0}{\partial x_1} \right)^2(x_1, x_2, y_2)+\left |u_0(x_1, x_2, y_2)\right|^2\right]dx_1\notag\\ & = \int_{0}^{1}dx_2\int_{Y_1}dy_2\int_{\mathbb{R}}(4\pi^2a(y_2)\lambda^2_1+1)|{\mathcal{F}}_2(u_0)(\lambda_1, x_2, y_2)|^2d\lambda_1\notag\\ & = \int_{0}^{1}dx_2\int_{\mathbb{R}} \frac{1}{\hat{k}_0(\lambda_1)}|{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1. \label{limsup2} \end{align*}

    This together with 84 implies that, for u\in{C^\infty_{\text{c}}}({\Omega}) ,

    \begin{equation*} \lim\limits_{\varepsilon\to 0} {\mathscr{F}_\varepsilon}({u_\varepsilon}) = \int_{0}^{1}dx_2\int_{\mathbb{R}} \frac{1}{\hat{k}_0(\lambda_1)}|{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1, \end{equation*}

    which proves the \Gamma - \limsup inequality on {C^\infty_{\text{c}}}({\Omega}) .

    Now, we extend the previous result to any u\in H^1_0((0, 1)_{x_1}; \quad L^2(0, 1)_{x_2}) . To this end, we use a density argument (see e.g. [5,Remark 2.8]). Recall that the weak topology of L^2({\Omega}) is metrizable on the closed balls of L^2({\Omega}) . Fix n\in\mathbb{N} and denote d_{B_n} any metric inducing the L^2({\Omega}) -weak topology on the ball B_n centered on 0 and of radius n . Then, H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) can be regarded as a subspace of L^2({\Omega}) endowed with the metric d_{B_n} . On the other hand, H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) is a Hilbert space endowed with the norm

    \begin{equation*} \|u\|_{H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2})} : = \left ( \left \|\frac{\partial u}{\partial x_1} \right\|^2_{L^2({\Omega})} + \|u\|^2_{L^2({\Omega})}\right)^{1/2}. \end{equation*}

    The associated metric d_{H^1_0} on H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) induces a topology which is not weaker than that induced by d_{B_n} , i.e.

    \begin{equation} d_{H^1_0}(u_k, u)\to 0 \quad \text{implies } \quad d_{B_n}(u_k, u)\to 0. \end{equation} (85)

    Recall that {C^\infty_{\text{c}}}({\Omega}) is a dense subspace of H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) for the metric d_{H^1_0} and that the \Gamma - \limsup inequality holds on {C^\infty_{\text{c}}}({\Omega}) for the L^2({\Omega}) -weak topology, i.e. for any u\in{C^\infty_{\text{c}}}({\Omega}) ,

    \begin{equation} \Gamma\text{-}\limsup\limits_{\varepsilon\to 0}{\mathscr{F}_\varepsilon}(u)\leq \mathscr{F}(u). \end{equation} (86)

    A direct computation of \hat{k}_0 , given by 54, shows that

    \begin{align*} \hat{k}_0(\lambda_1) & = \frac{c_\theta4\pi^2\lambda_1^2+1}{(4\pi^2\lambda_1^2+ 1)(c4\pi^2\lambda_1^2+ 1)}, \end{align*}

    which implies that

    \begin{align} \frac{1}{\hat{k}_0(\lambda_1) } = \frac{c}{c_\theta}4\pi^2\lambda_1^2 + f(\lambda_1) + \alpha, \end{align} (87)

    where f(\lambda_1) and \alpha are given by 57. Hence, there exists a positive constant C such that

    \begin{equation} \frac{1}{\hat{k}_0(\lambda_1)} \leq C(4\pi^2\lambda_1^2 + 1). \end{equation} (88)

    This combined with the Plancherel identity yields

    \begin{align} \mathscr{F}(u)&\leq C\int_{0}^{1}dx_2\int_{{\mathbb{R}}} (4\pi^2\lambda_1^2 + 1) |{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1\\ & = C\int_{0}^{1}dx_2\int_{{\mathbb{R}}}\left [ \left ( \frac{\partial u}{\partial x_1} \right)^2(x_1, x_2)+ |u(x_1, x_2)|^2\right]dx_1\\ & = C \|u\|^2_{H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2})}, \end{align} (89)

    where u(\cdot, x_2) is extended by zero outside (0, 1) . Since \mathscr{F} is a non-negative quadratic form, from 89 we conclude that \mathscr{F} is continuous with respect to the metric d_{H^1_0} .

    Now, take u\in H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) . By density, there exists a sequence u_k in {C^\infty_{\text{c}}}({\Omega}) such that

    \begin{equation} d_{H^1_0} (u_k, u)\to 0\qquad\text{as} \quad k\to\infty. \end{equation} (90)

    In particular, due to 85, we also have that d_{B_n} (u_k, u)\to 0 as k\to\infty . In view of the weakly lower semi-continuity of \Gamma - \limsup and the continuity of \mathscr{F} , we deduce from 86 that

    \begin{align*} \Gamma\text{-}\limsup\limits_{\varepsilon\to 0}{\mathscr{F}_\varepsilon}(u)&\leq \liminf\limits_{k\to\infty} (\Gamma\text{-}\limsup\limits_{\varepsilon\to 0}{\mathscr{F}_\varepsilon}(u_k) )\\ &\leq \liminf\limits_{k\to\infty}\mathscr{F}(u_k)\\ & = \mathscr{F}(u), \end{align*}

    which proves the \Gamma - \limsup inequality in B_n . Since for any u\in H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) the sequence u_k of functions in {C^\infty_{\text{c}}}({\Omega}) satisfying 90 belongs to some ball B_n of L^2({\Omega}) , as well as its limit, the \Gamma - \limsup property holds true for the sequence {\mathscr{F}_\varepsilon} on H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2}) , which concludes the proof of \Gamma - \limsup inequality.

    Step 3 - Alternative expression of \Gamma -limit.

    The proof of the equality between the two expressions of the \Gamma -limit \mathscr{F} relies on the following lemma whose proof will be given later.

    Lemma 4.2. Let h \in L^2(\mathbb{R}) and u\in L^1(\mathbb{R})\cap L^2(\mathbb{R}) . Then, h\ast u\in L^2({\mathbb{R}}) and

    \begin{equation} {\mathcal{F}}_2 (h\ast u) = {\mathcal{F}}_2(h){\mathcal{F}}_2(u)\quad \mathit{{a.e. ~in }}~~ \mathbb{R} . \end{equation} (91)

    By applying Plancherel's identity with respect to x_1 , for any u\in H^1_0({\mathbb{R}}_{x_1}; L^2(0, 1)_{x_2}) extended by zero with respect to the variable x_1 outside (0, 1) , we get

    \begin{align} \int_{\mathbb{R}}&\left |\sqrt{\alpha}u(x_1, x_2) + (h\ast_1u)(x_1, x_2)\right|^2dx_1 \\ & = \int_{\mathbb{R}}\left |\sqrt{\alpha}{\mathcal{F}}_2(u)(\lambda_1, x_2) + {\mathcal{F}}_2(h\ast_1u)(\lambda_1, x_2)\right|^2d\lambda_1 \\ & = \int_{\mathbb{R}} \biggl[\alpha \left |{\mathcal{F}}_2 (u)(\lambda_1, x_2)\right|^2 + 2\sqrt{\alpha}{\rm Re}\left ({\mathcal{F}}_2(u)(\lambda_1, x_2) \overline{{\mathcal{F}}_2(h\ast_1u)}(\lambda_1, x_2)\right)\\ & \quad +\left |{\mathcal{F}}_2(h\ast_1u)(\lambda_1, x_2)\right|^2\biggr]d\lambda_1. \end{align} (92)

    Recall that the Fourier transform of h , given by 56, is real. From 92, an application of Lemma 4.2 leads us to

    \begin{align} \int_{\mathbb{R}} &\left [\alpha \biggl|{\mathcal{F}}_2 (u)(\lambda_1, x_2)\right|^2 + 2\sqrt{\alpha}{\rm Re}\left ({\mathcal{F}}_2(u)(\lambda_1, x_2) \overline{{\mathcal{F}}_2(h\ast_1u)}(\lambda_1, x_2)\right)\\ & \quad +\left |{\mathcal{F}}_2(h\ast_1u)(\lambda_1, x_2)\right|^2\biggr]d\lambda_1\\ & = \int_{\mathbb{R}} \left [\alpha+2\sqrt{\alpha}{\mathcal{F}}_2(h)(\lambda_1) + \left ({\mathcal{F}}_2(h)(\lambda_1)\right)^2\right]\left |{\mathcal{F}}_2(u)(\lambda_1, x_2)\right|^2d\lambda_1\\ & = \int_{\mathbb{R}} \left [\sqrt{\alpha}+ {\mathcal{F}}_2(h)(\lambda_1)\right]^2|{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1\\ & = \int_{\mathbb{R}} \left [\alpha+ f(\lambda_1)\right]|{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1. \end{align} (93)

    On the other hand, by applying Plancherel's identity with respect to x_1 , we obtain

    \begin{equation*} \int_{{\mathbb{R}}}\frac{c}{c_\theta}4\pi^2\lambda_1^2|{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1 = \int_{{\mathbb{R}}}\frac{c}{c_\theta}\left (\frac{\partial u}{\partial x_1}\right)^2(x_1, x_2)dx_1. \end{equation*}

    In view of the expansion of 1/\hat{k}_0(\lambda_1) given by 87, the previous equality combined with 92 and 93 implies that, for u\in H^1_0((0, 1)_{x_1}; \quad L^2(0, 1)_{x_2}) extended by zero with respect to x_1 outside (0, 1) ,

    \begin{align*} &\int_{0}^{1}dx_2\int_{\mathbb{R}} \frac{1}{\hat{k}_0(\lambda_1)} |{\mathcal{F}}_2(u)(\lambda_1, x_2)|^2d\lambda_1 \\ &\quad = \int_{0}^{1}dx_2 \int_{\mathbb{R}}\left \{\frac{c}{c_\theta}\left (\frac{\partial u}{\partial x_1}\right)^2(x_1, x_2)+[\sqrt{\alpha}u(x_1, x_2) + (h\ast_1u)(x_1, x_2)]^2\right\}dx_1, \end{align*}

    which concludes the proof.

    Proof of Lemma 4.1. In view of 87, the equality 79 becomes

    \begin{align} {\mathcal{F}}_2(b)(\lambda_1, x_2 ) & = \left (\frac{c}{c_\theta}4\pi^2\lambda_1^2+\alpha + f(\lambda_1)\right){\mathcal{F}}_2(u)(\lambda_1, x_2)\\ & = {\mathcal{F}}_2 \left (-\frac{c}{c_\theta}\frac{\partial^2 u}{\partial x_1^2} +\alpha u\right)(\lambda_1, x_2) + f(\lambda_1){\mathcal{F}}_2(u)(\lambda_1, x_2). \end{align} (94)

    Since

    \begin{equation*} \label{fCL1} f(\lambda_1) = \frac{(c-1)^2\theta(\theta-1)}{c^2_\theta}\frac{1}{c_\theta4\pi^2\lambda_1^2 + 1} = O(\lambda_1^{-2})\in C_0(\mathbb{R})\cap L^1(\mathbb{R}), \end{equation*}

    the right-hand side of 94 belongs to L^2({\mathbb{R}}) with respect to \lambda_1 , which implies that

    \begin{equation*} {\mathcal{F}}_2(b)(\cdot, x_2)\in L^2({\mathbb{R}}). \end{equation*}

    Applying the Plancherel identity, we obtain that b(\cdot, x_2)\in L^2({\mathbb{R}}) with respect to x_1 . Since u(\cdot, x_2) is extended by zero outside (0, 1) , b(\cdot, x_2) is also equal to zero outside (0, 1) so that

    \begin{equation} b(\cdot, x_2)\in L^2(0, 1). \end{equation} (95)

    We show that b(x_1, \cdot) is a continuous function with respect to x_2\in [0, 1] . Recall that the continuity of x_2\in [0, 1]\mapsto b(x_1, x_2) \in L^2(0, 1)_{x_1} is equivalent to

    \begin{equation*} \lim\limits_{t\to 0} \|b(\cdot, x_2+t) - b(\cdot, x_2)\|_{L^2(0, 1)_{x_1}} = 0. \end{equation*}

    Thanks to Plancherel's identity, we infer from 79 that

    \begin{align*} \|b(\cdot, x_2+t) -& b(\cdot, x_2)\|^2_{L^2(0, 1)_{x_1}} = \|{\mathcal{F}}_2(b)(\cdot, x_2+t) - {\mathcal{F}}_2(b)(\cdot, x_2)\|^2_{L^2({\mathbb{R}})_{\lambda_1}}\\ & = \int_{{\mathbb{R}}}\left |\frac{1}{\hat{k}_0(\lambda_1)}\left [{\mathcal{F}}_2(u)(\lambda_1, x_2+t) - {\mathcal{F}}_2(u)(\lambda_1, x_2)\right] \right|^2d\lambda_1. \end{align*}

    In view of 88 and thanks to the Plancherel identity, we obtain

    \begin{align} \|b(\cdot, x_2+t) &- b(\cdot, x_2)\|^2_{L^2(0, 1)_{x_1}}\\ & \leq C^2 \int_{{\mathbb{R}}}\left |(4\pi^2\lambda_1^2+1) ({\mathcal{F}}_2(u)(\lambda_1, x_2+t) - {\mathcal{F}}_2(u)(\lambda_1, x_2)) \right|^2d\lambda_1\\ &\leq C^2\left \|{\mathcal{F}}_2 \left (\frac{\partial u}{\partial x_1}\right) (\cdot, x_2+t ) -{\mathcal{F}}_2\left (\frac{\partial u}{\partial x_1}\right) (\cdot, x_2 ) \right\|^2_{L^2(0, 1)_{\lambda_1}} \\ &\quad+ C^2 \|{\mathcal{F}}_2(u)(\cdot, x_2+t)-{\mathcal{F}}_2(u)(\cdot, x_2)\|^2_{L^2(0, 1)_{\lambda_1}} \\ & = C^2\left \|\frac{\partial u}{\partial x_1}(\cdot, x_2+t ) -\frac{\partial u}{\partial x_1} (\cdot, x_2 ) \right\|^2_{L^2(0, 1)_{x_1}} \\ &\quad +C^2 \|u(\cdot, x_2+t)-u(\cdot, x_2)\|^2_{L^2(0, 1)_{x_1}}. \end{align}

    By the Lebesgue dominated convergence theorem and since u\in{C^\infty_{\text{c}}}([0, 1]^2) , from the previous inequality we conclude that the map x_2\in [0, 1]\mapsto b(x_1, x_2)\in L^2(0, 1)_{x_1} is continuous. Hence,

    \begin{equation} b(x_1, x_2)\in C([0, 1]_{x_2}; \quad L^2(0, 1)_{x_1}). \end{equation} (96)

    To conclude the proof, it remains to show the regularity of u_0 . Note that 80 is a Sturm-Liouville problem with constant coefficient with respect to x_1 , since x_2\in (0, 1) and y_2\in Y_1 play the role of parameters. By 95, we already know that b(\cdot, x_2)\in L^2(0, 1) , so that thanks to a classical regularity result (see e.g. [7] pp. 223-224), the problem 80 admits a unique solution u_0(\cdot, x_2, y_2) in H^2(0, 1) . Since H^2(0, 1) is embedding into C^1([0, 1]) , we have

    \begin{equation*} u_0(\cdot, x_2, y_2)\in C^1([0, 1])\qquad\text{a.e.} \quad (x_2, y_2)\in (0, 1)\times Y_1. \end{equation*}

    On the other hand, the solution u_0(x_1, x_2, y_2) to the Sturm-Liouville problem 80 is explicitly given by

    \begin{equation} u_0(x_1, x_2, y_2) : = \int_{0}^{1} G_{y_2}(x_1, s) b(s, x_2)ds, \end{equation} (97)

    where b(x_1, x_2) is defined by 79 and 96 and the kernel G_{y_2}(x_1, s) is given by

    \begin{equation*} G_{y_2}(x_1, s) : = \frac{1}{\sqrt{a(y_2)}\sinh\left (\frac{1}{\sqrt{a(y_2)}}\right) }\sinh\left (\frac{x_1\wedge s}{\sqrt{a(y_2)}}\right)\sinh\left (\frac{1-x_1\vee s}{\sqrt{a(y_2)}}\right). \end{equation*}

    This combined with 96 and 97 proves that

    \begin{equation*} \label{u0isregular} u_0(x_1, x_2, y_2) \in C^1([0, 1]^2, L^\infty_{\rm per}(Y_1)), \end{equation*}

    which concludes the proof.

    We prove now the Lemma 4.2 that we used in Step 3 of Proposition 4.

    Proof of Lemma 4.2. By the convolution property of the Fourier transform on L^2({\mathbb{R}}) , we have

    \begin{equation} h\ast u = \overline{{\mathcal{F}}_2}({\mathcal{F}}_2(h)) \ast \overline{{\mathcal{F}}_2}({\mathcal{F}}_2(h)) = \overline{{\mathcal{F}}_1}({\mathcal{F}}_2(h){\mathcal{F}}_2(u)), \end{equation} (98)

    where \overline{{\mathcal{F}}_i} denotes the conjugate Fourier transform for i = 1, 2 . On the other hand, since u\in L^1(\mathbb{R})\cap L^2({\mathbb{R}}) and due to Riemann-Lebesgue's lemma, we deduce that {\mathcal{F}}_2(u) = {\mathcal{F}}_1(u)\in C_0(\mathbb{R})\cap L^2({\mathbb{R}}) . This combined with {\mathcal{F}}_2(h)\in L^2({\mathbb{R}}) implies that

    \begin{equation*} {\mathcal{F}}_2(h){\mathcal{F}}_2(u) = {\mathcal{F}}_2(h){\mathcal{F}}_1(u)\in L^2(\mathbb{R})\cap L^1({\mathbb{R}}). \end{equation*}

    Since \overline{{\mathcal{F}}_1} = \overline{{\mathcal{F}}_2} on L^1({\mathbb{R}})\cap L^2({\mathbb{R}}) , from 98 we deduce that

    \begin{equation*} h\ast u = \overline{{\mathcal{F}}_2}({\mathcal{F}}_2(h){\mathcal{F}}_2(u))\in L^2(\mathbb{R}), \end{equation*}

    which yields 91. This concludes the proof.

    Remark 2. Thanks to the Beurling-Deny theory of Dirichlet forms [3], Mosco [15, Theorem 4.1.2] has proved that the \Gamma -limit F of a family of Markovian form for the L^2({\Omega}) -strong topology is a Dirichlet form which can be split into a sum of three forms: a strongly local form F_d , a local form and nonlocal one. More precisely, for u\in L^2({\Omega}) with F(u)<\infty , we have

    \begin{equation} F(u) = F_d(u) + \int_{{\Omega}}u^2k(dx) + \int_{({\Omega}\times{\Omega})\setminus{\rm diag}} (u(x)-u(y))^2j(dx, dy), \end{equation} (99)

    where F_d is called the diffusion part of F , k is a positive Radon measure on {\Omega} , called the killing measure, and j is a positive Radon measure on ({\Omega}\times{\Omega})\setminus{\rm diag} , called the jumping measure. Recall that a Dirichlet form F is a closed form which satisfies the Markovian property, i.e. for any contraction T:{\mathbb{R}}\to{\mathbb{R}} , such that

    \begin{equation*} T(0) = 0, \qquad\text{and}\qquad \forall x, y\in{\mathbb{R}}, \quad |T(x)-T(y)|\leq |x-y|, \end{equation*}

    we have F\circ T\leq F . A \Gamma -limit form obtained with the L^2({\Omega}) -weak topology does not a priori satisfy the Markovian property, since the L^2({\Omega}) -weak convergence does not commute with all contractions T . An example of a sequence of Markovian forms whose \Gamma -limit for the L^2({\Omega}) -weak topology does not satisfy the Markovian property is provided in [9, Theorem 3.1]. Hence, the representation formula 99 does not hold in general when the L^2({\Omega}) -strong topology is replaced by the L^2({\Omega}) -weak topology.

    In the present context, we do not know if the \Gamma -limit \mathscr{F} 55 is a Dirichlet form since the presence of the convolution term makes difficult to prove the Markovian property.

    We are going to give an explicit expression of the homogenized matrix A^\ast defined by 7, which extends the rank-one laminate formula in the case of a rank-one laminates with degenerate phases. We will recover directly from this expression the positive definiteness of A^* for the class of rank-one laminates introduced in Section 3. Indeed, by virtue of Theorem 2.1 the positive definiteness of A^* also follows from assumption (H2) which is established in Proposition 2 and Proposition 3.

    Set

    \begin{equation} a: = (1-\theta) A_1 e_1\cdot e_1 + \theta A_2e_1\cdot e_1, \end{equation} (100)

    with \theta\in (0, 1) being the volume fraction of phase Z_1 .

    Proposition 5. Let A_1 and A_2 be two symmetric and non-negative matrices of {\mathbb{R}}^{d\times d} , d\geq 2 . If a given by 100 is positive, the homogenized matrix A^\ast is given by

    \begin{equation} A^\ast = \theta A_1+(1-\theta)A_2 -\frac{\theta(1-\theta)}{a} (A_2-A_1)e_1\otimes (A_2-A_1)e_1. \end{equation} (101)

    If a = 0 , the homogenized matrix A^\ast is the arithmetic average of the matrices A_1 and A_2 , i.e.

    \begin{equation} A^\ast = \theta A_1+(1-\theta)A_2. \end{equation} (102)

    Furthermore, if one of the following conditions is satisfied:

    i) in two dimensions, a>0 and the matrices A_1 and A_2 are given by 28 with \xi\cdot e_1\neq 0 ,

    ii) in three dimensions, a>0 , the matrices A_1 and A_2 are given by 39 and the vectors \{e_1, \eta_1, \eta_2\} are independent in {\mathbb{R}}^3 ,

    then A^\ast is positive definite.

    Remark 3. The condition a>0 agrees with the \Gamma -convergence results of Propositions 2 and 3. In the two-dimensional framework, the degenerate case a = 0 does not agree with Propositions 2. Indeed, a = 0 implies that A_1 e_1\cdot e_1 = A_2 e_1\cdot e_1 = 0 in contradiction to positive definiteness of A_2 . Similar in the three-dimensional setting, where the independence of \{e_1, \eta_1, \eta_2 \} is not compatible with a = 0 . Indeed, a = 0 implies that A_i e_1 = A_i \eta_i = 0 , for i = 1, 2 , which contradicts the fact that A_1 and A_2 have rank two.

    Proof. Assume that a>0 . In view of the convergence 23, we already know that

    \begin{equation} \lim\limits_{\delta\to 0} A^\ast_\delta = A^\ast, \end{equation} (103)

    where, for \delta>0 , A^\ast_\delta is the homogenized matrix associated to conductivity matrix A_\delta given by

    \begin{equation*} A_\delta(y_1) = \chi(y_1) A_1^\delta + (1-\chi(y_1)) A_2 ^\delta\qquad\text{for} \quad y_1\in{\mathbb{R}}, \end{equation*}

    with A_i^\delta = A_i +\delta I_d . Since A_1 and A_2 are non-negative matrices, A_\delta is positive definite and thus the homogenized matrix A^\ast_\delta is given by the lamination formula (see [17] and also [2,Lemma 1.3.32])

    \begin{equation} A^\ast_\delta = \theta A_1^\delta+(1-\theta)A_2^\delta -\frac{\theta(1-\theta)}{(1-\theta)A_1^\delta e_1\cdot e_1 + \theta A_2^\delta e_1\cdot e_1 } (A_2^\delta-A_1^\delta)e_1\otimes (A_2^\delta-A_1^\delta)e_1. \end{equation} (104)

    If a>0 , we easily infer from the convergence 103 combined with the lamination formula 104 the expression 101 for A^\ast .

    We prove that A^\ast x\cdot x\geq 0 for any x\in{\mathbb{R}^d} . From the Cauchy-Schwarz inequality, we deduce that

    \begin{align} |(A_2-A_1)e_1\cdot x| &\leq |A_2e_1\cdot x|+ |A_1e_1\cdot x|\\ &\leq (A_2e_1\cdot e_1)^{1/2}(A_2x\cdot x)^{1/2} + (A_1e_1\cdot e_1)^{1/2}(A_1x\cdot x)^{1/2}. \end{align} (105)

    This combined with the definition 101 of A^\ast implies that, for any x\in{\mathbb{R}^d} ,

    \begin{align} A^\ast x\cdot x& = \theta (A_1 x\cdot x) + (1-\theta)(A_2 x\cdot x) - \theta(1-\theta)a^{-1}\left |(A_2-A_1)e_1\cdot x\right|^2\\ &\geq \theta (A_1x\cdot x) +(1-\theta) (A_2 x\cdot x)\\ &\quad-\theta(1-\theta)a^{-1}[(A_2e_1\cdot e_1)^{1/2} (A_2x\cdot x)^{1/2}+ (A_1e_1\cdot e_1)^{1/2} (A_1x\cdot x)^{1/2} ]^2\\ & = a^{-1} [a \theta (A_1x\cdot x) +a (1-\theta) (A_2 x\cdot x) -\theta(1-\theta)(A_2e_1\cdot e_1)( A_2x\cdot x)\\ & \quad -\theta(1-\theta)(A_1e_1\cdot e_1)(A_1x\cdot x)\\ & \quad - 2\theta(1-\theta)(A_2e_1\cdot e_1)^{1/2}( A_2x\cdot x)^{1/2}(A_1e_1\cdot e_1)^{1/2}(A_1x\cdot x)^{1/2}]. \end{align} (106)

    In view of definition 100 of a , we have that

    \begin{align} &a \theta( A_1x\cdot x) +a (1-\theta)( A_2 x\cdot x)\\& = \theta (1-\theta)(A_1 e_1\cdot e_1)(A_1 x\cdot x) + \theta^2(A_2 e_1\cdot e_1)(A_1 x\cdot x)\\ &\quad+ (1-\theta)^2 (A_1 e_1\cdot e_1)(A_2 x\cdot x)\\ &\quad + \theta(1-\theta)(A_2 e_1\cdot e_1)(A_2 x\cdot x). \end{align}

    Plugging this equality in 106, we deduce that

    \begin{align} A^\ast x\cdot x &\geq a^{-1}[\theta^2(A_2 e_1\cdot e_1)(A_1 x\cdot x)+ (1-\theta)^2(A_1 e_1\cdot e_1)(A_2 x\cdot x)\\ & \quad -2\theta(1-\theta)(A_2e_1\cdot e_1)^{1/2}(A_1x\cdot x)^{1/2}(A_1e_1\cdot e_1)^{1/2}( A_2x\cdot x)^{1/2}]\\ & = a^{-1}[\theta (A_2 e_1\cdot e_1)^{1/2}(A_1x\cdot x)^{1/2} -(1-\theta) (A_1 e_1\cdot e_1)^{1/2}(A_2x\cdot x)^{1/2} ]^2\geq 0, \end{align} (107)

    which proves that A^\ast is a non-negative definite matrix.

    Now, assume a = 0 . Since A_1 and A_2 are non-negative matrices, the condition a = 0 implies A_1e_1\cdot e_1 = A_2e_1\cdot e_1 = 0 or equivalently A_1e_1 = A_2e_1 = 0 . Hence,

    \begin{equation*} (A_2^\delta -A_1^\delta)e_1 = (A_2-A_1)e_1 = 0, \end{equation*}

    which implies that the lamination formula 104 becomes

    \begin{equation*} A^\ast_\delta = \theta A_1^\delta + (1-\theta)A_2^\delta. \end{equation*}

    This combined with the convergence 103 yields to the expression 102 for A^\ast .

    To conclude the proof, it remains to prove the positive definiteness of A^\ast under the above conditions i) and ii).

    Case (i). d = 2 , a>0 and A_1, A_2 given by 28.

    Assume A^* x\cdot x = 0 . Then, the inequality 107 is an equality, which yields in turn equalities in 105. In particular, we have

    \begin{equation} |A_2e_1\cdot x| = (A_2e_1\cdot e_1)^{1/2}(A_2x\cdot x)^{1/2} = \|A^{1/2}_2e_1\|\|A^{1/2}_2x\|. \end{equation} (108)

    Recall that the Cauchy-Schwarz inequality is an equality if and only if one of vectors is a scalar multiple of the other. This combined with 108 leads to A^{1/2}_2x = \alpha A^{1/2}_2e_1 for some \alpha\in{\mathbb{R}} , so that, since A_2 is positive definite or equivalently A^{1/2}_2 , we have

    \begin{equation} x = \alpha e_1 \qquad\text{for some} \quad \alpha\in{\mathbb{R}}. \end{equation} (109)

    From the definition 101 of A^\ast and due to the assumption \xi\cdot e_1\neq 0 , we get

    \begin{equation} A^\ast e_1\cdot e_1 = \frac{1}{a}(A_2 e_1\cdot e_1) (\xi\cdot e_1)^2 > 0. \end{equation} (110)

    Recall that A^\ast x\cdot x = 0 . This combined with 109 and 110 implies that x = 0 , which proves that A^\ast is positive definite.

    Case (ii). d = 3 , a>0 and A_1, A_2 given by 39.

    Assume that A^\ast x\cdot x = 0 . As in Case (i), we have equalities in 105. In other words,

    \begin{align} |A_1e_1\cdot x| & = (A_1e_1\cdot e_1)^{1/2}(A_1x\cdot x)^{1/2}, \end{align} (111)
    \begin{align} |A_2e_1\cdot x| & = (A_2e_1\cdot e_1)^{1/2}(A_2x\cdot x)^{1/2}. \end{align} (112)

    Let p_i(t) be the non-negative polynomials of degree 2 defined by

    \begin{equation*} p_i(t) : = A_i(x+te_1)\cdot (x+te_1) \qquad\text{for} \quad i = 1, 2. \end{equation*}

    In view of 111, the discriminant of p_1(t) is zero, so that there exists t_1\in{\mathbb{R}} such that

    \begin{equation} p_1(t_1) = A_1(x+t_1e_1)\cdot (x+t_1e_1) = 0. \end{equation} (113)

    Recall that {\text{Ker}} (A_1) = {\rm Span}(\eta_1) . Since A_1 is non-negative matrix, we deduce from 113 that x+t_1e_1 belongs to {\text{Ker}} (A_1) , so that

    \begin{equation} x\in{\rm Span}(e_1, \eta_1). \end{equation} (114)

    Similarly, recalling that {\text{Ker}} (A_2) = {\rm Span}(\eta_2) and using 112, we have

    \begin{equation} x\in{\rm Span}(e_1, \eta_2). \end{equation} (115)

    Since the vectors \{e_1, \eta_1, \eta_2\} are independent in {\mathbb{R}}^3 , 114 and 115 imply that

    \begin{equation*} x = \alpha e_1 \qquad\text{for some} \quad \alpha\in{\mathbb{R}}. \end{equation*}

    In light of definition 101 of A^\ast , we have

    \begin{equation*} A^\ast e_1\cdot e_1 = \frac{1}{a} (A_1e_1\cdot e_1)(A_2e_1\cdot e_1) > 0, \end{equation*}

    which implies that x = 0 , since A^\ast x\cdot x = 0 . This establishes that A^\ast is positive definite and concludes the proof.

    Note that when d = 2 and a>0 the assumption \xi\cdot e_1\neq 0 is essential to obtain that A^\ast is positive definite. Otherwise, the homogenized matrix A^\ast is just non-negative definite as shown by the following counter-example. Let A_1 and A_2 be symmetric and non-negative matrices of {\mathbb{R}}^{2\times 2} defined by

    \begin{equation*} A_1 = e_2\otimes e_2 \quad\text{and}\quad A_2 = I_2. \end{equation*}

    Then, it is easy to check that a = \theta>0 and A^\ast e_1\cdot e_1 = 0 .

    This problem was pointed out to me by Marc Briane during my stay at Institut National des Sciences Appliquées de Rennes. I thank him for the countless fruitful discussions. My thanks also extend to Valeria Chiadò Piat for useful remarks. The author is also a member of the INdAM-GNAMPA project "Analisi variazionale di modelli non-locali nelle scienze applicate".



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