Comparative analysis of phenomenological growth models applied to epidemic outbreaks

1.   Introduction

In this paper, for a bounded domain $\Omega$ of ${\mathbb{R}}^d$, we study the homogenization through $\Gamma$-convergence of the conductivity energy with a zero-order term of the type

The conductivity $A$ is a $Y_d$-periodic, symmetric and non-negative matrix-valued function in $L^\infty({\mathbb{R}^d})^{d\times d}$, denoted by $L^\infty_\text{per}(Y_d)^{d\times d}$, which is not strongly elliptic, i.e.

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

where $A'\in L^\infty_\text{per}(Y_d)^{(d-1)\times(d-1)}$ 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.

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

for the $L^2({\Omega})$-strong topology. The $\Gamma$-limit is given by

where the matrix-valued function $A^\ast$ is defined by the classical homogenization formula

The $\Gamma$-convergence for the $L^p({\Omega})$-strong topology, for $p>1$, for the class of integral functionals $F_\varepsilon$ of the form

where $f: {\Omega}\times \mathbb{R}^{m\times d}\to {\mathbb{R}}$ is a Borel function, $1$-periodic in the first variable satisfying the standard growth conditions of order $p$, namely $c_1|M|^p \leq f(x, M)\leq c_2(|M|^p +1)$ for any $x\in{\Omega}$ and for any real $(m\times d)$-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 $\Gamma$-convergence of oscillating functionals for the weak topology on bounded sets of $L^p({\Omega})$ has been very few analysed. An example of the study of $\Gamma$-convergence for the $L^p({\Omega})$-weak topology can be found in the paper [6] where, in the context of double-porosity, the authors compare the $\Gamma$-limit for non-linear functionals analogous to 5 computed with respect to different topologies and in particular with respect to $L^p({\Omega})$-weak topology.

In this paper, we investigate the $\Gamma$-convergence for the weak topology on bounded sets (a metrizable topology) of $L^2({\Omega})$ of the conductivity functional under condition 2. In this case, one has no a priori $L^2({\Omega})$-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 $\|u\|^2_{L^2({\Omega})}$, so that we immediately obtain the coerciveness in the weak topology of $L^2({\Omega})$ of ${\mathscr{F}_\varepsilon}$, namely, for $u\in H^1_0({\Omega})$,

This estimate guarantees that $\Gamma$-limit for the weak topology on bounded sets of $L^2({\Omega})$ 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]), ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges for the weak topology of $L^2({\Omega})$, up to subsequences, to some functional. We will show that, under the following assumptions:

$\rm (H1)$ any two-scale limit $u_0(x, y)$ of a sequence ${u_\varepsilon}$ of functions in $L^2({\Omega})$ with bounded energy ${\mathscr{F}_\varepsilon}({u_\varepsilon})$ does not depend on $y$ (see [1,Theorem 1.2]);

$\rm (H2)$ the space $V$ defined by

agrees with the space ${\mathbb{R}^d}$,

the $\Gamma$-limit is given by

where the homogenized matrix $A^\ast$ is given through the expected homogenization formula

We need to make assumption (H1) since for any sequence ${u_\varepsilon}$ with bounded energy, i.e. $\sup_{\varepsilon>0}{\mathscr{F}_\varepsilon}({u_\varepsilon})<\infty$, the sequence ${\nabla}{u_\varepsilon}$ in $L^2({\Omega}; \quad {\mathbb{R}}^d)$ 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 $2$D 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 $\Gamma$-convergence for the $L^2({\Omega})$-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 $1^\ast$-convergence, is not strongly elliptic. These examples have been revisited by means of a homogenization process using $\Gamma$-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 $\Gamma$-convergence result Theorem 2.1. Using Nguetseng-Allaire [1,16] two-scale convergence, we prove that for any dimension $d\geq 2$, the $\Gamma$-limit ${\mathscr{F}_0}$ 6 for the weak topology of $L^2({\Omega})$ actually agrees with the one obtained for the $L^2({\Omega})$-strong topology under uniformly ellipticity 3, replacing the minimum in 4 by the infimum in 7. Assumption (H2) implies the coerciveness of the functional ${\mathscr{F}_0}$ showing that its domain is $H^1_0({\Omega})$ and that the homogenized matrix $A^\ast$ is positive definite. More precisely, the positive definiteness of $A^\ast$ 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 $A^\ast$ is positive definite. For this class of laminates, an alternative and independent proof of positive definiteness of $A^\ast$ is performed using an explicit expression of $A^\ast$ (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 ${\mathscr{F}_\varepsilon}$ 1. We provide a two-dimensional rank-one laminate with two degenerate phases for which the functional ${\mathscr{F}_\varepsilon}$ does $\Gamma$-converge for the $L^2({\Omega})$-weak topology to a functional $\mathscr{F}$ which differs from the one given by 6 (see Proposition 4). In this example, any two-scale limit $u_0(x, y)$ of a sequence with bounded energy ${\mathscr{F}_\varepsilon}({u_\varepsilon})$, depends on the variable $y$. Moreover, we give two quite different expressions of the $\Gamma$-limit $\mathscr{F}$ 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 $\Gamma$-limit $\mathscr{F}$ is a Dirichlet form in the sense of Beurling-Deny [3], since the Markovian property is not stable by the $L^2({\Omega})$-weak topology (see Remark 2).

The paper is organized as follows. In Section 2, we prove a general $\Gamma$-convergence result (see Theorem 2.1) for the functional ${\mathscr{F}_\varepsilon}$ 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 $\Gamma$-limit $\mathscr{F}$ involving a convolution term (see Proposition 4). Finally, in the Appendix we give an explicit formula for the homogenized matrix $A^\ast$ for any two-phase rank-one laminates with (possibly) degenerate phases. We also provide an alternative proof of the positive definiteness of $A^\ast$ using an explicit expression of $A^\ast$ for the class of two-phase rank-one laminates introduced in Section 3 (see Proposition 5).

Notation.

● For $i = 1, \dots, d$, $e_i$ denotes the $i$-th vector of the canonical basis in ${\mathbb{R}^d}$;

● $I_d$ denotes the unit matrix of ${\mathbb{R}}^{d\times d}$;

● ${H^1_{\text{per}}}(Y_d; {\mathbb{R}^n})$ (resp. ${L^2_{\text{per}}}(Y_d; {\mathbb{R}^n})$, ${C^\infty_{\text{per}}}(Y_d; {\mathbb{R}^n})$) is the space of those functions in $H^1_{\text{loc}}({\mathbb{R}^d}; {\mathbb{R}^n})$ (resp. $L^2_{\text{loc}}({\mathbb{R}^d}; {\mathbb{R}^n})$, $C^\infty_{\text{loc}}({\mathbb{R}^d}; {\mathbb{R}^n})$) that are $Y_d$-periodic;

● Throughout, the variable $x$ will refer to running point in a bounded open domain ${\Omega}\subset{\mathbb{R}^d}$, while the variable $y$ will refer to a running point in $Y_d$ (or $k+Y_d$, $k\in\mathbb{Z}^d$);

● We write

with ${u_\varepsilon}\in L^2({\Omega})$ and $u_0\in L^2({\Omega}\times Y_d)$ if ${u_\varepsilon}$ two-scale converges to $u^0$ in the sense of Nguetseng-Allaire (see [1,16])

● ${\mathcal{F}}_1$ and ${\mathcal{F}}_2$ denote the Fourier transform defined on $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$ respectively. For $f\in L^1({\mathbb{R}})\cap L^2({\mathbb{R}})$, the Fourier transform ${\mathcal{F}}_1$ of $f$ is defined by

Definition 1.1. Let $X$ be a reflexive and separable Banach space endowed with the weak topology $\sigma(X, X')$, and let ${\mathscr{F}_\varepsilon}: X\to{\mathbb{R}}$ be a $\varepsilon$-indexed sequence of functionals. The sequence ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges to the functional ${\mathscr{F}_0}:X\to{\mathbb{R}}$ for the weak topology of $X$, and we write ${\mathscr{F}_\varepsilon}{{\stackrel{\Gamma(X)-w}{\rightharpoonup}}}{\mathscr{F}_0}$, if for any $u\in X$,

i) $\forall {u_\varepsilon}\rightharpoonup u$, ${\mathscr{F}_0}(u)\leq \liminf_{\varepsilon\to 0} {\mathscr{F}_\varepsilon}({u_\varepsilon})$,

ii) $\exists\overline{u}_{\varepsilon}\rightharpoonup u$ such that $\lim_{\varepsilon\to 0}{\mathscr{F}_\varepsilon}(\overline{u}_{\varepsilon}) = {\mathscr{F}_0}(u)$.

Such a sequence $\overline{u}_{\varepsilon}$ is called a recovery sequence.

Recall that the weak topology of $L^2({\Omega})$ is metrizable on bounded sets, i.e. there exists a metric $d$ on $L^2({\Omega})$ such that on every norm bounded subset $B$ of $L^2({\Omega})$ the weak topology coincides with the topology induced on $B$ by the metric $d$ (see e.g. [13,Proposition 8.7]).

2.   A preliminary general $\boldsymbol{\Gamma}$-result

In this section, we will prove the main result of this paper. As previously announced, up to a subsequence, the sequence of functionals ${\mathscr{F}_\varepsilon}$, given by 1 with non-uniformly elliptic matrix-valued conductivity $A(y)$, $\Gamma$-converges for the weak topology on bounded sets of $L^2({\Omega})$ to some functional. Our aim is to show that $\Gamma$-limit is exactly ${\mathscr{F}_0}$ when $u\in H^1_0({\Omega})$.

Theorem 2.1. Let ${\mathscr{F}_\varepsilon}$ be functionals given by 1 with $A(y)$ a $Y_d$-periodic, symmetric, non-negative matrix-valued function in $L^\infty({\mathbb{R}}^d)^{d\times d}$ satisfying 2. Assume the following assumptions

$\rm (H1)$ any two-scale limit $u_0(x, y)$ of a sequence ${u_\varepsilon}$ of functions in $L^2({\Omega})$ with bounded energy ${\mathscr{F}_\varepsilon}({u_\varepsilon})$ does not depend on $y$;

$\rm (H2)$ the space $V$ defined by

agrees with the space ${\mathbb{R}^d}$.

Then, ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges for the weak topology of $L^2({\Omega})$ to ${\mathscr{F}_0}$, i.e.

where ${\mathscr{F}_0}$ is defined by 6 and $A^\ast$ 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 - $\Gamma$-$\liminf$ inequality.

Consider a sequence $\{{u_\varepsilon} \}_\varepsilon$ converging weakly in $L^2({\Omega})$ to $u\in L^2({\Omega})$. We want to prove that

If the lower limit is $\infty$ then 9 is trivial. Up to a subsequence, still indexed by $\varepsilon$, we may assume that $\liminf{\mathscr{F}_\varepsilon}({u_\varepsilon})$ is a limit and we can also assume henceforth that, for some $0<C<\infty$,

As ${u_\varepsilon}$ is bounded in $L^2({\Omega})$, there exists a subsequence, still indexed by $\varepsilon$, which two-scale converges to a function $u_0(x, y)\in L^2({\Omega}\times Y_d)$ (see e.g. [1,Theorem 1.2]). In other words,

Assumption (H1) ensures that

where, according to the link between two-scale and weak $L^2({\Omega})$-convergences (see [1,Proposition 1.6]), $u$ is the weak limit of ${u_\varepsilon}$, i.e.

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

and also

In particular

Consider $\Phi\in{L^2_{\text{per}}}(Y_d; {\mathbb{R}^d})$ such that

or equivalently,

Take also $\varphi\in C^{\infty}(\overline{{\Omega}})$. Since ${u_\varepsilon}\in H^1_0({\Omega})$ and in view of 15, an integration by parts yields

By using [1,Lemma 5.7], $A^{1/2}(y)\Phi(y)\cdot{\nabla} \varphi(x)$ 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

We prove that the target function $u$ is in $H^1(\Omega)$. Setting

and varying $\varphi$ in ${C^\infty_{\text{c}}}({\Omega})$, the equality 16 reads as

Since the integral in the left-hand side is bounded by a constant times $\|\varphi\|_{L^2({\Omega})}$, the right-hand side is a linear and continuous map in $\varphi\in L^2({\Omega})$. By the Riesz representation theorem, there exists $g\in L^2({\Omega})$ such that, for any $\varphi\in{C^\infty_{\text{c}}}({\Omega})$,

which implies that

In view of assumption (H2), $N$ is an arbitrary vector in ${\mathbb{R}^d}$ so that we infer from 18 that

This combined with equality 16 leads us to

By density, the last equality holds if the test functions $\Phi(y)\varphi(x)$ are replaced by the set of $\psi(x, y)\in L^2({\Omega}; {L^2_{\text{per}}}(Y_d;{\mathbb{R}^d}))$ such that

or equivalently,

The $L^2({\Omega}; {L^2_{\text{per}}}(Y_d;{\mathbb{R}^d}))$-orthogonal to that set is the $L^2$-closure of

Thus, the equality 20 yields

for some $S$ in the closure of $\mathscr{K}$, i.e. there exists a sequence $v_n\in L^2({\Omega};{H^1_{\text{per}}}(Y_d))$ such that

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

Then, by the weak $L^2$-lower semi-continuity of $\|{u_\varepsilon}\|_{L^2({\Omega})}$, we have

Recalling the definition 7, we immediately conclude that

provided that $u\in H^1_0({\Omega})$.

It remains to prove that the target function $u$ is actually in $H^1_0({\Omega})$, giving a complete characterization of $\Gamma$-limit. To this end, take $x_0\in\partial{\Omega}$ a Lebesgue point for $u\lfloor\partial{\Omega}$ and for $\nu (x_0)$, the exterior normal to ${\Omega}$ at point $x_0$. Thanks to 19, we know that $u\in H^1({\Omega})$, hence, after an integration by parts of the right-hand side of 16, we obtain, for $\varphi\in C^\infty(\overline{{\Omega}})$,

where $N$ is given by 17. Varying $\varphi$ in ${C^\infty_{\text{c}}}({\Omega})$, the first two integrals in 21 are equal and bounded by a constant times $\|\varphi\|_{L^2({\Omega})}$. It follows that, for any $\varphi\in C^\infty(\overline{{\Omega}})$,

which leads to $N\cdot \nu(x) u(x) = 0$ $\mathscr{H}$-a.e. on $\partial{\Omega}$. Since $x_0$ is a Lebesgue point, we have

In view of assumption (H2) and the arbitrariness of $N$, we can choose $N$ such that $N = \nu(x_0)$ so that from 22 we get $u(x_0) = 0$. Hence,

This concludes the proof of the $\Gamma$-$\liminf$ inequality.

Step 2 - $\Gamma$-$\limsup$ 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 $\delta>0$, let $A_\delta$ be the perturbed matrix of ${\mathbb{R}}^{d\times d}$ defined by

where $I_d$ is the unit matrix of ${\mathbb{R}}^{d\times d}$. Since the matrix $A$ is non-negative, $A_\delta$ turns out to be positive definite, hence, the functional ${\mathscr{F}_\varepsilon}^\delta$, defined by 1 with $A_\delta$ in place of $A$, $\Gamma$-converges to the functional $\mathscr{F}^\delta$ given by

for the strong topology of $L^2({\Omega})$ (see e.g. [13,Corollary 24.5]). Thanks to the compactness result of $\Gamma$-convergence (see e.g. [4,Proposition 1.42]), there exists a subsequence $\varepsilon_j$ such that $\mathscr{F}_{\varepsilon_j}$ $\Gamma$-converges for the $L^2({\Omega})$-strong topology to some functional $F^0$. Let $u\in H^1_0({\Omega})$ and let $u_{\varepsilon_j}$ be a recovery sequence for $\mathscr{F}_{\varepsilon_j}$ which converges to $u$ for the $H^1({\Omega})$-weak topology on bounded sets. Since $\mathscr{F}_{\varepsilon_j}\leq \mathscr{F}_{\varepsilon_j}^\delta$ and since $u_{\varepsilon_j}$ belongs to some bounded set of $H^1(\Omega)$, from [13,Propositions 6.7 and 8.10] we deduce that

It follows that $\mathscr{F}^\delta$ converges to $F^0$ as $\delta\to 0$. Then, the $\Gamma$-limit $F^0$ of $\mathscr{F}_{\varepsilon_j}$ is independent on the subsequence $\varepsilon_j$. Repeating the same arguments, any subsequence of ${\mathscr{F}_\varepsilon}$ has a further subsequence which $\Gamma$-converges for the strong topology of $L^2({\Omega})$ to $F^0 = \lim_{\delta\to 0}\mathscr{F}^\delta$. Thanks to the Urysohn property (see e.g. [4,Proposition 1.44]), the whole sequence ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges to the functional $F^0$ for the strong topology of $L^2({\Omega})$. On the other hand, in light of the definition 7 of $A^\ast$, we get that $A^\ast_\delta$ converges to $A^\ast$ as $\delta\to 0$, i.e.

Thanks to the Lebesgue dominated convergence theorem and in view of 23, we get that $F^0 = \lim_{\delta\to 0}\mathscr{F}^\delta$ is exactly ${\mathscr{F}_0}$ given by 6. Therefore, ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges to ${\mathscr{F}_0}$ for the $L^2({\Omega})$-strong topology.

Now, let us show that ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges to ${\mathscr{F}_0}$ for the weak topology of $L^2({\Omega})$. Recall that the $L^2({\Omega})$-weak topology is metrizable on the closed ball of $L^2({\Omega})$. Fix $n\in\mathbb{N}$ and let $d_{B_n}$ be any metric inducing the $L^2({\Omega})$-weak topology on the ball $B_n$ centered on $0$ and of radius $n$. Let $u\in H^1_0({\Omega})$ and let $\overline{u}_{\varepsilon}$ be a recovery sequence for ${\mathscr{F}_\varepsilon}$ for the $L^2({\Omega})$-strong topology. Since the topology induced by the metric $d_{B_n}$ on $B_n$ is weaker than the $L^2({\Omega})$-strong topology, $\overline{u}_{\varepsilon}$ is also a recovery sequence for ${\mathscr{F}_\varepsilon}$ for the $L^2({\Omega})$-weak topology on $B_n$. Hence,

which proves the $\Gamma$-$\limsup$ inequality in $B_n$. Finally, since any sequence converging weakly in $L^2({\Omega})$ belongs to some ball $B_n\subset L^2({\Omega})$, as well as its limit, it follows that the $\Gamma$-$\limsup$ inequality holds true for ${\mathscr{F}_\varepsilon}$ for $L^2({\Omega})$-weak topology, which concludes the proof.

The next proposition provides a characterization of Assumption $\rm(H2)$ in terms of homogenized matrix $A^\ast$.

Proposition 1. Assumption $\rm(H2)$ is equivalent to the positive definiteness of $A^\ast$, or equivalently,

Proof. Consider $\lambda\in{\text{Ker}}(A^\ast)$. Define

Recall that $u\in H^1_\lambda(Y_d)$ if and only if there exists $v\in{H^1_{\text{per}}}(Y_d)$ such that $u(y) = v(y) + \lambda\cdot y$ (see e.g. [13,Lemma 25.2]). Since $A^\ast$ is non-negative and symmetric, from 7 it follows that

Then, there exists a sequence $u_n$ of functions in $H^1_\lambda(Y_d)$ such that

which implies that

Now, take $\Phi\in L^2_{\rm per}(Y_d; \quad {\mathbb{R}}^d)$ such that $A^{1/2}\Phi$ is a divergence free field in ${\mathbb{R}}^d$. Recall that, since $u_n\in H^1_\lambda(Y_d)$, we have that ${\nabla} u_n(y) = {\nabla} v_n(y) + \lambda$, for some $v_n\in H^1_{\rm per} (Y_d)$. This implies that

where the last equality is obtained by integrating by parts the second integral combined with the fact that $A^{1/2}\Phi$ is a divergence free field in ${\mathbb{R}}^d$. In view of convergence 24, the integral on the left-hand side of 25 converges to $0$. Hence, passing to the limit as $n\to\infty$ in 25 yields

for any $\Phi\in L^2_{\rm per}(Y_d; \quad {\mathbb{R}}^d)$ such that $A^{1/2}\Phi$ is a divergence free field in ${\mathbb{R}}^d$. Therefore $\lambda\in V^\perp$ which implies that

Conversely, by 23 we already know that

where $A^\ast_\delta$ is the homogenized matrix associated with $A_\delta = A+\delta I_d$. Since $A_\delta$ is strongly elliptic, the homogenized matrix $A^\ast_\delta$ is given by

Let $\overline{u}_\delta$ be the minimizer of problem 26. Therefore, there exists a constant $C>0$ such that

which implies that the sequence $\Phi_\delta(y): = A^{1/2}_\delta(y){\nabla} \overline{u}_\delta(y)$ is bounded in $L^2_{\rm per}(Y_d; {\mathbb{R}}^d)$. Then, up to extract a subsequence, we can assume that $\Phi_\delta$ converges weakly to some $\Phi$ in $L^2_{\rm per} (Y_d; {\mathbb{R}}^d)$.

Now, we show that $A^{1/2}_\delta$ converges strongly to $A^{1/2}$ in $L^\infty_{\rm per}(Y_d)^{d\times d}$. Since $A(y)$ is a symmetric matrix, there exists an orthogonal matrix-valued function $R$ in $L^\infty_{\rm per}(Y_d)^{d\times d}$ such that

where $D$ is a diagonal non-negative matrix-valued function in $L^\infty_{\rm per}(Y_d)^{d\times d}$ and $R^T$ denotes the transpose of $R$. It follows that $A_\delta (y) = A(y)+\delta I_d = R(y)(D(y)+\delta I_d)R^T(y)$, for a.e. $y\in Y_d$. Hence,

which implies that $A^{1/2}_\delta$ converges strongly to $A^{1/2} = RD^{1/2}R^T$ in $L^\infty_{\rm per} (Y_d)^{d\times d}$.

Now, passing to the limit as $\delta\to 0$ in

we have

This along with $\Phi\in L^2_{\rm per}(Y_d; {\mathbb{R}}^d)$ implies that $\Phi$ is a test function for the set $V$ given by 8. From 26 it follows that

Hence, taking into account the strong convergence of $A^{1/2}_\delta$ in $L^\infty_{\rm per}(Y_d)^{d\times d}$ and the weak convergence of $\Phi_\delta$ in $L^2_{\rm per}(Y_d; {\mathbb{R}}^d)$, we have

which implies that $A^\ast\lambda\in V$ since $\Phi$ is a suitable test function for the set $V$. Therefore, for $\lambda\in V^\perp$,

so that, since $A^\ast$ is a non-negative matrix, we deduce that $\lambda\in{\text{Ker}}(A^\ast)$. In other words,

which concludes the proof.

3.   Two-dimensional and three-dimensional examples

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 $e_1$ with two phases in ${\mathbb{R}}^d$, $d = 2, 3$. Specifically, we assume the existence of two anisotropic phases $Z_1$ and $Z_2$ of $Y_d$ given by

where $\theta$ denotes the volume fraction of the phase $Z_1$. Let $Z_1^\#$ and $Z_2^\#$ be the associated subsets of ${\mathbb{R}^d}$, i.e. the open periodic sets

Let $X_1$ and $X_2$ be unbounded connected components of $Z_1^\#$ and $Z_2^\#$ in ${\mathbb{R}^d}$ given by

and we denote by $\partial Z$ the interface $\{y_1 = 0\}$.

The anisotropic phases are described by two constant, symmetric and non-negative matrices $A_1$ and $A_2$ of ${\mathbb{R}}^{d\times d}$ which are possibly not positive definite. Hence, the conductivity matrix-valued function $A\in L^\infty_{\text{per}}(Y_d)^{d\times d}$, given by

where $\chi$ is the $1$-periodic characteristic function of the phase $Z_1$, is not strongly elliptic, i.e. 2 is satisfied.

3.1.   The two-dimensional case with one degenerate phase

We are interested in two-phase mixtures in ${\mathbb{R}^2}$ with one degenerate phase. We specialize to the case where the non-negative and symmetric matrices $A_1$ and $A_2$ of ${\mathbb{R}}^{2\times 2}$ are such that

for some $\xi\in{\mathbb{R}^2}$. The next proposition establishes the algebraic conditions which provide assumptions (H1) and (H2) of Theorem 2.1.

Proposition 2. Let $A_1$ and $A_2$ be the matrices defined by 28. Assume that $\xi\cdot e_1\neq 0$ and the vectors $\xi$ and $A_2e_1$ are linearly independent in ${\mathbb{R}^2}$. Then, assumptions $\rm(H1)$ and $\rm(H2)$ are satisfied. In particular, the homogenized matrix $A^\ast$, 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 ${\mathscr{F}_\varepsilon}$ defined by 1 with $A$ given by 27 and 28 $\Gamma$-converges to the functional ${\mathscr{F}_0}$ given by 6 with conductivity matrix $A^\ast$ defined by 7. In the present case, the homogenized matrix $A^\ast$ 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 $u^i_0$ the restriction of the two-scale limit $u_0$ in phase $Z_i$ or $Z_i^\#$ for $i = 1, 2$. In view of 14, for any $\Phi(x, y)\in{C^\infty_{\text{c}}}({\Omega}\times{\mathbb{R}^2}; \quad {\mathbb{R}^2})$ with compact support in ${\Omega}\times Z^\#_1$, or due to periodicity in ${\Omega}\times X_1$, we deduce that

so that

Similarly, taking $\Phi(x, y)\in{C^\infty_{\text{c}}}({\Omega}\times{\mathbb{R}^2}; \quad {\mathbb{R}^2})$ with compact support in ${\Omega}\times Z_2^\#$, or equivalently in ${\Omega}\times X_2$, as test function and repeating the same arguments, we obtain

Due to 29, in phase $Z_1^\#$ we have

where $\xi^\perp = (-\xi_2, \xi_1)\in{\mathbb{R}^2}$ is perpendicular to $\xi = (\xi_1, \xi_2)$. Hence, $u^1_0$ reads as

for some function $\theta^1\in L^2({\Omega}\times {\mathbb{R}})$. On the other hand, since the matrix $A_2$ is positive definite, in phase $Z_2^\#$ the relation 30 implies that

for some function $\theta^2\in L^2({\Omega})$. Now, consider a constant vector-valued function $\Phi$ defined on $Y_2$ such that

Note that condition 33 is necessary for ${\mathrm{div}}_y(A(y)\Phi)$ to be an admissible test function for two-scale convergence. In view of 14 and 32, for any $\varphi\in{C^\infty_{\text{c}}}({\Omega}; {C^\infty_{\text{per}}}(Y_2))$, we obtain

Take now $\varphi\in{C^\infty_{\text{c}}}({\Omega}\times{\mathbb{R}^2})$ and use the periodized function

as new test function. Then, we obtain

Recall that $A_1 = \xi\otimes \xi$, where $\xi$ is such that $\xi\cdot e_1\neq0$. This combined with the linear independence of the vectors $\xi$ and $A_2e_1$ implies that the linear map

is one-to-one. Hence, for any $f\in{\mathbb{R}}$, there exists a unique $\Phi\in{\mathbb{R}^2}$ such that

In view of the arbitrariness of $f$ in 35, we can choose $\Phi$ such that

Since $A_1 {\nabla}_y u^1_0 = 0$ in the distributional sense and $A_1 = \xi\otimes\xi$, we deduce that $u^1_0$ is constant along the direction $\xi$. Using Fubini's theorem, we may integrate along straight lines parallel to the vector $\xi$ where integration by parts is allowed. Therefore, performing an integration by parts in 34 combined with 36, it follows that for any $\varphi\in C^\infty_{\rm c}(\Omega\times\mathbb{R}^2)$,

where we have set $v_0(x, y): = u^1_0(x, y) - \theta^2(x)$. We conclude that $v_0(x, \cdot)$ has a trace on $\partial Z$ for a.e. $x\in\Omega$ satisfying

Recall that $\partial Z = \{y_1 = 0\}$. Fix $x\in{\Omega}$. Taking into account 31 and 32, the equality 37 reads as

Since $\xi\cdot e_1 \neq 0$, it follows that $\theta^1$ only depends on $x$ so that $u^1_0(x, y)$ agrees with $\theta^2(x)$. Finally, we conclude that $u_0(x, y) : = \chi(y_1)u_0^1(x, y) + (1-\chi(y_1))u^2_0(x, y)$ 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 $\alpha, \beta\in{\mathbb{R}}$, let $\Phi$ be a vector-valued function given by

Such a vector field $\Phi$ does exist, since $\xi$ is in the range of $A_1$ and thus the right-hand side of 38 belongs pointwise to the range of $A$, or equivalently to the range of $A^{1/2}$. Moreover, the difference of two constant phases in 38 is orthogonal to the laminate direction $e_1$, so that $A^{1/2}\Phi$ is a laminate divergence free periodic field in ${\mathbb{R}^2}$. Its average value is given by

Hence, due to $\xi\cdot e_1\neq 0$ and the arbitrariness of $\alpha, \beta$, the set of the vectors $N$ spans ${\mathbb{R}^2}$, which yields assumption (H2).

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

3.2.   The three-dimensional case with both degenerate phases

We are going to deal with three-dimensional laminates where both phases are degenerate. We assume that the symmetric and non-negative matrices $A_1$ and $A_2$ of ${\mathbb{R}}^{3\times 3}$ have rank two, hence, there exist $\eta_1, \eta_2\in{\mathbb{R}}^3$ such that

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

Proposition 3. Let $\eta_1$ and $\eta_2$ be the vectors in ${\mathbb{R}}^3$ defined by 39. Assume that the vectors $\{e_1, \eta_1, \eta_2\}$ as well as $\{A_1e_1, A_2e_1\}$ are linearly independent in ${\mathbb{R}}^3$. Then, assumptions $\rm(H1)$ and $\rm(H2)$ are satisfied. In particular, the homogenized matrix $A^\ast$ 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 ${\mathscr{F}_\varepsilon}$ defined by 1 with $A$ given by 27 and 39, $\Gamma$-converges for the weak topology of $L^2({\Omega})$ to ${\mathscr{F}_0}$ where the effective conductivity $A^\ast$ is given by 7. As in two-dimensional laminate materials, $A^\ast$ 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

In view of 39 and 40, in phase $Z_i^\#$, $u^i_0$ reads as

for some function $\theta^i\in L^2({\Omega}\times{\mathbb{R}})$ and $i = 1, 2$. Now, consider a constant vector-valued function $\Phi$ on $Y_3$ such that the transmission condition 33 holds. In view of 14, for any $\varphi\in{C^\infty_{\text{c}}}({\Omega}, \quad {C^\infty_{\text{per}}}(Y_3))$, we obtain

Take $\varphi\in{C^\infty_{\text{c}}}({\Omega}\times{\mathbb{R}}^3)$. Putting the periodized function

as test function in 42, we get

Since the vectors $A_1e_1$ and $A_2e_1$ are independent in ${\mathbb{R}}^3$, the linear map

is surjective. In particular, for any $f\in{\mathbb{R}}$, there exists $\Phi\in{\mathbb{R}}^3$ such that

In view of the arbitrariness of $f$ in 44, we can choose $\Phi$ such that 36 is satisfied. Due to 40 and 39, we deduce that $u^i_0$ is constant along the plane $\Pi_i$ perpendicular to $\eta_i$, for $i = 1, 2$. This implies that, thanks to Fubini's theorem, we may integrate along the plane $\Pi_i$ where an integration by part may be performed. Hence, an integration by parts in 43 combined with 36, yields for any $\varphi\in{C^\infty_{\text{c}}}({\Omega}\times {\mathbb{R}}^3)$,

which implies that

Fix $x\in{\Omega}$ and recall that $\partial Z = \{y_1 = 0\}$. In view of 41, the relation 45 reads as

with $\eta_i = (a_i, b_i, c_i)$ for $i = 1, 2$. Due to the independence of $\{e_1, \eta_1, \eta_2\}$ in $\mathbb{R}^3$, the linear map $( y_2, y_3)\in{\mathbb{R}}^2\mapsto (z_1, z_2)\in{\mathbb{R}}^2$ defined by

is a change of variables so that 46 becomes

This implies that $\theta^1$ and $\theta^2$ depend only on $x$ and thus $u^1_0$ and $u^2_0$ agree with some function $u\in L^2({\Omega})$. Finally, we conclude that $u_0(x, y) = \chi(y_1)u^1_0(x, y) + (1-\chi(y_1))u^2_0(x, y)$ is independent of $y$ and hence (H1) is satisfied.

It remains to prove assumption (H2). To this end, let $E$ be the subset of $\mathbb{R}^3\times\mathbb{R}^3$ defined by

For $(\xi_1, \xi_2)\in E$, let $\Phi$ be the vector-valued function defined by

The existence of such a vector field $\Phi$ is guaranteed by the conditions $\xi_i\cdot\eta_i = 0$, for $i = 1, 2$, which imply that $\xi_i$ belongs to the range of $A_i$ and hence the right-hand side of 48 belongs pointwise to the range of $A$, or equivalently to the range of $A^{1/2}$. Moreover, since the difference of the phases $\xi_1$ and $\xi_2$ is orthogonal to the laminate direction $e_1$, $A^{1/2}\Phi$ is a laminate divergence free periodic field in ${\mathbb{R}}^3$. Its average value is given by

Note that $E$ is a linear subspace of ${\mathbb{R}}^3\times{\mathbb{R}}^3$ whose dimension is three. Indeed, let $f$ be the linear map defined by

If we identify the pair $(\xi_1, \xi_2)\in{\mathbb{R}}^3\times {\mathbb{R}}^3$ with the vector $(x_1, y_1, z_1, x_2, y_2, z_2)\in{\mathbb{R}}^6$, with $\xi_i = (x_i, y_i, z_i)$, for $i = 1, 2$, the associated matrix $M_f\in{\mathbb{R}}^{3\times 6}$ of $f$ is given by

with $\eta_i = (a_i, b_i, c_i)$, $i = 1, 2$. In view of the linear independence of $\{ e_1, \eta_1, \eta_2 \}$, the rank of $M_f$ is three, which implies that the dimension of kernel ${\text{Ker}}(f)$ is also three. Since the kernel ${\text{Ker}}(f)$ agrees with $E$, we conclude that the dimension of $E$ is three.

Now, let $g$ be the linear map defined by

Let us show that $g$ is invertible. To this end, consider $(\xi_1, \xi_2)\in{\text{Ker}}(g)$. From the definition of the map $g$, ${\text{Ker}}(g)$ consists of all vectors $(\xi_1, \xi_2)\in E$ of the form

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

This combined with the linear independence of $\{e_1, \eta_1, \eta_2 \}$ implies that

Hence, ${\text{Ker}}(g) = \{(0, 0)\}$ which implies along with the fact that the dimension of $E$ is three that $g$ is invertible. This proves that all the vectors of ${\mathbb{R}}^3$ can be attained through the map $g$ so that assumption (H2) is satisfied.

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

4.   A two-dimensional counter-example

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 ${\mathscr{F}_\varepsilon}$ 1.

Let ${\Omega}: = (0, 1)^2$ and let $e_2$ be the laminate direction. We assume that the non-negative and symmetric matrices $A_1$ and $A_2$ of ${\mathbb{R}}^{2\times 2}$ are given by

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

with

Thus, the energy ${\mathscr{F}_\varepsilon}$, defined by 1 with $A(y)$ given by 50 and 51 becomes

We denote by $\ast_1$ the convolution with respect to the variable $x_1$, i.e. for $f\in L^1({\mathbb{R}}^2)$ and $g\in L^2({\mathbb{R}}^2)$

Throughout this section, $c_{\theta}$ denotes the positive constant given by

where $\theta\in (0, 1)$ is the volume fraction of the phase $Z_1$ in $Y_2$. The following result proves the $\Gamma$-convergence of ${\mathscr{F}_\varepsilon}$ for the weak topology of $L^2({\Omega})$ and provides two alternative expressions of the $\Gamma$-limit, one of that seems nonlocal due to presence of convolution term (see Remark 2 below).

Proposition 4. Let ${\mathscr{F}_\varepsilon}$ be the functional defined by 52. Then, ${\mathscr{F}_\varepsilon}$ $\Gamma$-converges for the weak topology of $L^2({\Omega})$ to the functional defined by

where ${\mathcal{F}}_2 (u)(\lambda_1, \cdot)$ denotes the Fourier transform on $L^2({\mathbb{R}})$ of parameter $\lambda_1$ with respect to the variable $x_1$ of the function $x_1\mapsto u(x_1, \cdot)$ extended by zero outside $(0, 1)$ and

The $\Gamma$-limit $\mathscr{F}$ can be also expressed as

where $c_\theta$ is given by 53 and $h$ is a real-valued function in $L^2({\mathbb{R}})$ defined by means of its Fourier transform ${\mathcal{F}}_2$ on $L^2({\mathbb{R}})$

where $\alpha$ and $f$ are given by

Moreover, any two-scale limit $u_0(x, y)$ of a sequence ${u_\varepsilon}$ with bounded energy ${\mathscr{F}_\varepsilon}$ depends on the variable $y_2\in Y_1$.

Remark 1. From 57, we can deduce that

for any $\lambda_1\in{\mathbb{R}}$, so that the Fourier transform of $h$ is well-defined.

Proof. We divide the proof into three steps.

Step 1 - $\Gamma$-$\liminf$ inequality.

Consider a sequence $\{{u_\varepsilon} \}_\varepsilon$ converging weakly in $L^2({\Omega})$ to $u\in L^2({\Omega})$. Our aim is to prove that

If the lower limit is $\infty$ then 58 is trivial. Up to a subsequence, still indexed by $\varepsilon$, we may assume that $\liminf{\mathscr{F}_\varepsilon}({u_\varepsilon})$ is a limit and we may assume henceforth that, for some $0<C<\infty$,

It follows that the sequence ${u_\varepsilon}$ is bounded in $L^2({\Omega})$ and according to [1,Theorem 1.2], a subsequence, still indexed by $\varepsilon$, of that sequence two-scale converges to some $u_0(x, y)\in L^2({\Omega}\times Y_2)$. In other words,

In view of 51, we know that $a\geq 1$ so that, thanks to 59, for another subsequence (not relabeled) we have

In particular,

Take $\varphi\in {C^\infty_{\text{c}}}({\Omega}; \quad {C^\infty_{\text{per}}}(Y_2))$. By integration by parts, we obtain

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

which implies that

Due to the link between two-scale and weak $L^2$-convergences (see [1,Proposition 1.6]), we have

Now consider $\varphi\in C^\infty(\overline{{\Omega}}; \quad C^\infty_{\text{per}}(Y_2))$ such that

Since ${u_\varepsilon}\in H^1_0((0, 1)_{x_1}; L^2(0, 1)_{x_2})$, an integration by parts leads us to

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

Varying $\varphi\in{C^\infty_{\text{c}}}({\Omega}; \quad {C^\infty_{\text{per}}}(Y_2))$, the left-hand side of 66 is bounded by a constant times $\|\varphi\|_{L^2({\Omega}\times [0, 1))}$ so that the right-hand side is a linear and continuous form in $\varphi\in L^2({\Omega}\times Y_2)$. By the Riesz representation theorem, there exists $g\in L^2({\Omega}\times Y_2)$ such that, for any $\varphi\in {C^\infty_{\text{c}}}({\Omega}; \quad {C^\infty_{\text{per}}}(Y_2))$,

which yields

Then, an integration by parts with respect to $x_1$ of the right-hand side of 66 yields, for any $\varphi\in C^\infty(\overline{{\Omega}}; \quad {C^\infty_{\text{per}}}(Y_2))$ satisfying 65,

Since for any $\varphi\in {C^\infty_{\text{c}}}({\Omega}; \quad {C^\infty_{\text{per}}}(Y_2))$ the first two integrals are equal and bounded by a constant times $\|\varphi\|_{L^2(\Omega\times [0, 1)}$, we conclude that, for any $\varphi\in C^\infty(\overline{{\Omega}}; \quad {C^\infty_{\text{per}}}(Y_2))$ satisfying 65,

which implies that

This combined with 67 yields

Finally, an integration by parts with respect to $x_1$ of the right-hand side of 66 implies that, for any $\varphi\in C^\infty(\overline{{\Omega}}; \quad {C^\infty_{\text{per}}}(Y_2))$ satisfying 65,

Since the orthogonal of divergence-free functions is the gradients, from the previous equality we deduce that there exists $\tilde{u}\in H^1_{\text{per}}(Y_1; L^2({\Omega}\times Y_1))$ such that

Now, we show that

To this end, set

Since $a\in L^\infty_{\rm per}(Y_1)\subset L^2_{\rm per}(Y_1)$, there exists a sequence $a_k$ of functions in ${C^\infty_{\text{per}}}(Y_1)$ such that

hence, by periodicity, we also have

for some positive constant $C>0$. On the other hand, since $\sigma_0$ given by 68 is in $L^2({\Omega}\times Y_2)$, there exists a sequence $\psi_n$ of functions in ${C^\infty_{\text{c}}}({\Omega}; \quad {C^\infty_{\text{per}}}(Y_2))$ such that

From the inequality

we get

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

Hence, passing to the limit as $\varepsilon\to 0$ in 73 with the help of 61 leads to

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

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

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

This combined with 69 implies that

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

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

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 $x_1$,

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

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

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

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

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

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

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

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

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

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,

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,

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

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

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

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.

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 $u\in{C^\infty_{\text{c}}}({\Omega})$,

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

which implies that

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

This combined with the Plancherel identity yields

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

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

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

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

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

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

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 $(0, 1)$,

which concludes the proof.

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

Since

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

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

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

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

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

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,

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

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

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

This combined with 96 and 97 proves that

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

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

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

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

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

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.

Appendix A.   Homogenized formula for a rank-one laminate

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

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

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

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

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

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])

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

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

In view of definition 100 of $a$, we have that

Plugging this equality in 106, we deduce that

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,

which implies that the lamination formula 104 becomes

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

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

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

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,

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

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

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

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

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

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

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

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

Acknowledgments

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".