Homogenization of multiscale Maxwell wave equations

TL;DR

This paper develops a homogenization theory for multiscale Maxwell wave equations, addressing the complexities due to non-compact embeddings and microscopic scale dependencies, and provides error estimates for different regularity levels.

Contribution

It introduces a homogenization framework for multiscale Maxwell equations, including explicit error estimates and correctors, especially for cases with lower regularity.

Findings

Derived explicit homogenization error estimates for two-scale Maxwell equations.

Extended homogenization analysis to cases with lower regularity solutions.

Developed correctors for numerical approximation of multiscale Maxwell wave solutions.

Abstract

We study homogenization of multiscale Maxwell wave equation that depends on $n$ separable microscopic scales in a domain $D\subset{\mathbb R}^d$ on a finite time interval $(0,T)$. Due to the non-compactness of the embedding of $H_0(\curl,D)$ in $L^2(D)^d$, homogenization of Maxwell wave equation can be significantly more complicated than that of scalar wave equations in the $H^1(D)$ setting, and requires analysis uniquely for Maxwell wave equations. We employ multiscale convergence. The homogenized Maxwell wave equation and the initial condition are deduced from the multiscale homogenized equation. When the coefficient of the second order time derivative in the multiscale equation depends on the microscopic scales, the derivation is significantly more complicated, comparing to scalar wave equations, due to the corrector terms for the solution $\ue$ of the multiscale equation in the $L^2(D)^d$ norm, which do not appear in the $H^1(D)$ setting. For two scale equations, we derive an explicit homogenization error estimate for the case where the solution $u_0$ of the homogenized equation belongs to $L^\infty((0,T);H^1(\curl,D))$. When $u_0$ only belongs to a weaker regularity space $L^\infty((0,T);H^s(\curl,D))$ for $0<s<1$, we contribute an approach to deduce a new homogenization error in this case, which depends on $s$. For general multiscale problems, a corrector is derived albeit without an explicit homogenization error estimate. These correctors and homogenization errors play an essential role in deriving numerical correctors for approximating the solutions to the multiscale problems numerically, as considered in our forthcoming publication.

Equations

$$\begin{array}[]{lr}\displaystyle{\alpha|\xi|^{2}\leq a_{ij}(x,\boldsymbol{y})\xi_{i}\xi_{j},\ \ \ a_{ij}\xi_{i}\zeta_{j}\leq\beta|\xi||\zeta|}\\ \displaystyle{\alpha|\xi|^{2}\leq b_{ij}(x,\boldsymbol{y})\xi_{i}\xi_{j},\ \ \ b_{ij}\xi_{i}\zeta_{j}\leq\beta|\xi||\zeta|}\end{array}$$

$$\underset{\epsilon \to 0}{\lim }\frac{{\epsilon }_{i+1}(\epsilon )}{{\epsilon }_i(\epsilon )}=0.$$

$$a^{\epsilon }(x)=a(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_n}),b^{\epsilon }(x)=b(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_n}).$$

$$W=H_0(\mathrm{curl},D)=\{u\in L^2(\Omega {)}^3,\mathrm{curl}u\in L^2(\Omega {)}^3,u\times \nu =0\},H=L^2(D{)}^3$$

$$W=H_0(\mathrm{curl},D)=\{u\in L^2(D{)}^2,\mathrm{curl}u\in L^2(D),u\times \nu =0\},H=L^2(D{)}^2$$

$$\{\begin{array}{ll}{b}^{\epsilon }(x)\frac{{\partial }^2{u}^{\epsilon }(t,x)}{\partial t^2}+\mathrm{curl}(a^{\epsilon }(x)\mathrm{curl}{u}^{\epsilon }(t,x))=f(t,x),& (t,x)\in (0,T)\times D\\ u^{\epsilon }(0,x)=g_0(x)& \\ u_t^{\epsilon }(0,x)=g_1(x)& \end{array}$$

$$H=L^2(D{)}^3,H_i=L^2(D\times {\mathbf{Y}}_i{)}^3,i=1,\dots ,n.$$

$${⟨b^{\epsilon }(x)\frac{{\partial }^2{u}^{\epsilon }}{\partial t^2},\varphi (x)⟩}_{W^{\prime },W}+{\int }_D{a}^{\epsilon }(x)\mathrm{curl}{u}^{\epsilon }(t,x)\cdot \mathrm{curl}\varphi (x)dx={\int }_{D}f(t,x)\cdot \varphi (x)dx$$

$$\Vert u^{\epsilon }{\Vert }_{L^2(0,T;W)}+\Vert u^{\epsilon }{\Vert }_{H^1(0,T;H)}+\Vert u^{\epsilon }{\Vert }_{H^2(0,T;W^{\prime })}\le c(\Vert f{\Vert }_{L^2(0,T;H)}+\Vert g^0{\Vert }_W+\Vert g^1{\Vert }_H)$$

$$\underset{\epsilon \to 0}{\lim }{\int }_0^T{\int }_D{w}^{\epsilon }(t,x)\varphi (t,x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_n})dxdt={\int }_0^T{\int }_D{\int }_{\mathbf{Y}}{w}_0(t,x,\mathbf{y})\varphi (t,x,\mathbf{y})d\mathbf{y}dxdt.$$

$$w^{\epsilon }\stackrel{(n+1)-\text{mbox}scale}{⟶}{w}_0+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{w}}_i.$$

$$\mathrm{curl}{w}^{\epsilon }\stackrel{(n+1)-\text{mbox}scale}{⟶}\mathrm{curl}{w}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{w}_i.$$

$$u^{\epsilon }\stackrel{(n+1)-\text{mbox}scale}{⟶}{u}_0+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i,$$

$$\mathrm{curl}{u}^{\epsilon }\stackrel{(n+1)-\text{mbox}scale}{⟶}\mathrm{curl}{u}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{u}_i.$$

$$\mathbf{V}=W\times W_1\times \dots \times W_n\times V_1\times \dots \times V_n.$$

$$|||\mathbf{v}|||=\Vert v_0{\Vert }_{H(\mathrm{curl},D)}+\sum _{i=1}^n\Vert v_i{\Vert }_{L^2(D\times {\mathbf{Y}}_{i-1},{\tilde{H}}_{\#}(\mathrm{curl},Y_i))}+\sum _{i=1}^n\Vert {\mathfrak{v}}_i{\Vert }_{L^2(D\times {\mathbf{Y}}_{i-1},H_{\#}^1(Y_i))}.$$

$${\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right)d\mathbf{y}$$

$${⟨{\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right)d\mathbf{y},v_0⟩}_{W^{\prime },W}={\int }_D{\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right)\cdot v_{0}d\mathbf{y}dx;$$

$${⟨b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right),{\mathfrak{v}}_j⟩}_{V_j^{\prime },V_j}={\int }_D{\int }_{\mathbf{Y}}b(x,\mathbf{y})(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y}))\cdot {\nabla }_{y_j}{\mathfrak{v}}_j(x,{\mathbf{y}}_j)d\mathbf{y}dx.$$

$${⟨\frac{{\partial }^2}{\partial t^2}{\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right)d\mathbf{y},v_0⟩}_{W^{\prime },W}=$$

$${\int }_{D}f(x)\cdot v_0(x)dx-{\int }_D{\int }_{\mathbf{Y}}a(x,\mathbf{y})\left(\mathrm{curl}{u}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{u}_i\right)\cdot \left(\mathrm{curl}{v}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{v}_i\right)d\mathbf{y}dx$$

$${⟨\frac{{\partial }^2}{\partial t^2}\left(b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right)\right),{\mathfrak{v}}_j⟩}_{W_j^{\prime },W_j}=0,j=1,\dots ,n,$$

$$\begin{gathered} {⟨\frac{{\partial }^2}{\partial t^2}{\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i\right)d\mathbf{y},v_0⟩}_{W^{\prime },W}+ \\ \sum _{j=1}^n{\left(\frac{{\partial }^2}{\partial t^2}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,\mathbf{y})\right),{\mathfrak{v}}_j\right)}_{V_j^{\prime },V_j} \\ +{\int }_D{\int }_{\mathbf{Y}}a(x,\mathbf{y})\left(\mathrm{curl}{u}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{u}_i\right)\cdot \left(\mathrm{curl}{v}_0+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{v}_i\right)d\mathbf{y}dx={\int }_{D}f(x)\cdot v_0(x)dx \end{gathered}$$

$$\varphi (t,x)=\left(v_0(x)+\sum _{i=1}^n{\epsilon }_i{v}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+\sum _{i=1}^n{\epsilon }_i\nabla {\mathfrak{v}}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})\right)q(t)$$

$${\int }_0^T{\int }_{D}b(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_n})u^{\epsilon }(t,x)\cdot (v_0(x)+\sum _{i=1}^n{\epsilon }_i{v}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+$$

$$\sum _{i=1}^n\left({\epsilon }_i{\nabla }_x{\mathfrak{v}}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+\sum _{j=1}^i\frac{{\epsilon }_i}{{\epsilon }_j}{\nabla }_{y_j}{\mathfrak{v}}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})\right))q^{\prime \prime }(t)dxdt$$

$$+{\int }_0^T\underset{D}{\int }a(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_n})\mathrm{curl}{u}^{\epsilon }(t,x)\cdot$$

$$\left(\mathrm{curl}{v}_0(x)+\sum _{i=1}^n\left({\epsilon }_i{\mathrm{curl}}_x{v}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+\sum _{j=1}^i\frac{{\epsilon }_i}{{\epsilon }_j}{\mathrm{curl}}_{y_j}{v}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})\right)\right)q(t)dxdt$$

$$={\int }_0^T{\int }_{D}f(t,x)\cdot (v_0(x)+\sum _{i=1}^n{\epsilon }_i{v}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+\sum _{i=1}^n{\epsilon }_i{\nabla }_x{\mathfrak{v}}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i})+$$

$$\sum _{i=1}^n\sum _{j=1}^i\frac{{\epsilon }_i}{{\epsilon }_j}{\nabla }_{y_j}{\mathfrak{v}}_i(x,\frac{x}{{\epsilon }_1},\dots ,\frac{x}{{\epsilon }_i}))q(t)dxdt$$

$${\int }_0^T{\int }_D{\int }_{\mathbf{Y}}b(x,\mathbf{y})\left(u_0(t,x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{u}}_i(t,x,{\mathbf{y}}_i)\right)\cdot \left(v_0(x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{v}}_i(x,{\mathbf{y}}_i)\right)q^{\prime \prime }(t)d\mathbf{y}dxdt$$

$$+{\int }_0^T{\int }_D{\int }_{\mathbf{Y}}a(x,\mathbf{y})\left(\mathrm{curl}{u}_0(t,x)+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{u}_i(t,x,y)\right)\cdot \left(\mathrm{curl}{v}_0(x)+\sum _{i=1}^n{\mathrm{curl}}_{y_i}{v}_i(x,{\mathbf{y}}_i)\right)q(t)d\mathbf{y}dxdt$$

$$={\int }_0^T{\int }_D{\int }_{\mathbf{Y}}f(t,x)\cdot \left(v_0(x)+\sum _{i=1}^n{\nabla }_{y_i}{\mathfrak{v}}_i(x,{\mathbf{y}}_i)\right)q(t)d\mathbf{y}dxdt$$

$$={\int }_0^T{\int }_{D}f(t,x)\cdot v_0(x)q(t)dxdt.$$

$$u_0(0,\cdot )=g_0,{\nabla }_{y_i}{\mathfrak{u}}_i(0,\cdot ,\cdot )=0,$$

Full text

1 Introduction

We consider multiscale Maxwell wave equation (2.3) in a domain $D\in\mathbb{R}^{d}$ which depends on $n$ microscopic scales. Using multiscale convergence ([1], [12], we deduce the multiscale homogenized equation, from which we deduce the homogenized equation. In comparison to multiscale wave equation in the $H^{1}(D)$ setting, homogenization of a multiscale Maxwell wave equation is quite different due to the embedding of the space $H_{0}({\rm curl\,},D)$ in $L^{2}(D)^{d}$ is not compact (see [3]). It is far more complicated to derive the homogenized equation, and the initial condition for the solution of the homogenized equation than for a multiscale scale wave equation when the coefficient $b^{\varepsilon}$ of the second order time derivative in (2.3) depends also on the microscopic scales. This is due to the correctors $\mathfrak{u}_{i}$ for $u^{\varepsilon}$ in $L^{2}(D)^{d}$ in (2.7), and the second order time derivative ${\partial^{2}\over\partial t^{2}}\nabla_{y_{i}}\mathfrak{u}_{i}$ in (2.9) which do not belong to $L^{2}(D\times Y_{1}\times\ldots\times Y_{i})$, and can only be understood in the distribution sense. Derivation of the initial condition requires representations of the time derivative ${\partial\over\partial t}\nabla_{y_{i}}\mathfrak{u}_{i}$ in terms of the solutions of the cell problems, which are quite non-trivial. The main contributions of the paper are the new correctors and homogenization errors in two sale problems, especially when the solution to the homogenized equation possesses low regularity, and a general corrector for multiscale problems. Their derivation are quite non trivial, comparing to elliptic and wave equations in the $H^{1}(D)$ setting.

As is well known, for two scale wave equations, a corrector similar to that of two scale elliptic problems does not hold in general, as the energy of the multiscale problem does not converge to the energy of the homogenized problem (see [4]). Therefore, to derive a corrector, we restrict our consideration to the case where the initial condition $g_{0}$ in (2.3) equals zero. For two scale elliptic problems in a domain $D\subset\mathbb{R}^{d}$, it is well known that a $O(\varepsilon^{1/2})$ homogenization rate of convergence in the $H^{1}(D)$ norm can be derived when the solution of the homogenized equation is smooth (see [3] and [10]). In polygonal domains, which are of interests in numerical discretization, this regularity may not hold. However, if the domain is convex, the solution to the homogenized equation belongs to $H^{2}(D)$; the $O(\varepsilon^{1/2})$ homogenization rate still holds with the same proof. For wave equation, the $O(\varepsilon^{1/2})$ homogenization rate is deduced in [15] for scalar wave equations and in [16] for elastic wave equations. In this paper, for two scale Maxwell wave equations, we derive the $O(\varepsilon^{1/2})$ homogenization rate when the solution $u_{0}$ of the homogenized equation belongs to $L^{\infty}(0,T;H^{1}({\rm curl\,},D))$. However, in polygonal domains, this regularity condition normally does not hold. The solution $u_{0}$ of the homogenized equation only belongs to a weaker regularity space $L^{\infty}(0,T;H^{s}({\rm curl\,},D))$ for $0<s<1$. We develop an approach to deduce a new homogenization error for this case of low regularity, using the ideas of [5]. For multiscale problems, an explicit homogenization error is not available. However, we can deduce a general corrector which requires a procedure quite different from that for multiscale wave equation in [15] due to the corrector terms $\mathfrak{u}_{i}$ in (2.7). The correctors and homogenization errors in this paper play a key role in establishing numerical correctors using finite element solutions for the multiscale problems as studied in Chu and Hoang [6].

The paper is organized as follows. In the next section, we define the multiscale Maxwell wave equation, we review the concept of multiscale convergence, extended to functions that depend on the time variable, and use it to derive the multiscale Maxwell wave equation. We then derive the initial condition for the multiscale homogenized Maxwell wave equation, and show that this problem has a unique solution. In Section 3, we derive the homogenized equation from the multicale homogenized equation, together with the initial condition. The derivation is quite non trivial in comparison to the multiscale wave equation in [4] and [15] due to the multiscale coefficient $b^{\varepsilon}$ in (2.3) and the corrector terms $\mathfrak{u}_{i}$ in (2.7). In Section 4, we study the regularity of the solution $u_{0}$ of the homogenized equation which is necessary for the derivation of the correctors and homogenization errors. We study correctors in Section 5. For two scale problems, when the solution $u_{0}$ is in $L^{\infty}(0,T;H^{1}({\rm curl\,},D))$, we derive the $O(\varepsilon^{1/2})$ homogenization error in the $H({\rm curl\,},D)$ norm. When the solution only belongs to a weaker regularity space $L^{\infty}(0,T;H^{s}({\rm curl\,},D))$ we deduce a weaker homogenization convergence rate. The proof requires substantial technical developments. For general multiscale problems, we derive in this section a general corrector, without an explicit convergence rate.

Throughout the paper, without indicating a variable, ${\rm curl\,}$ and $\nabla$ denote the ${\rm curl\,}$ and gradient of a function of the variable $x$, with respect to $x$, and by ${\rm curl\,}_{x}$ and $\nabla_{x}$, we denote partial ${\rm curl\,}$ and partial gradient with respect to $x$, of a function depending on $x$ and other variables. Repeated indices indicate summation.

2 Multiscale homogenization of multiscale Maxwell wave equation

2.1 Problem setting

Let $D$ be a bounded domain in $\mathbb{R}^{d}$ ($d=2,3$). Let $Y$ be the unit cube in $\mathbb{R}^{d}$. By $Y_{1},\ldots,Y_{n}$ we denote $n$ copies of $Y$. We denote by ${\bf Y}$ the product set $Y_{1}\times Y_{2}\times\ldots\times Y_{n}$ and by $\boldsymbol{y}=(y_{1},\ldots,y_{n})$. For $i=1,\ldots,n$, we denote by ${\bf Y}_{i}=Y_{1}\times\ldots\times Y_{i}$. Let $a$ and $b$ be functions from $D\times Y_{1}\times\ldots\times Y_{n}$ to $\mathbb{R}^{d\times d}_{sym}$. We assume that the symmetric matrix functions $a$ and $b$ satisfy the boundedness and coerciveness conditions: for all $x\in D$ and $\boldsymbol{y}\in{\bf Y}$, and all $\xi,\zeta\in\mathbb{R}^{d}$,

[TABLE]

where $\alpha$ and $\beta$ are positive numbers. Let $\varepsilon$ be a small positive value. Let $\varepsilon_{1},\ldots,\varepsilon_{n}$ be $n$ functions of $\varepsilon$ that denote the $n$ microscopic scales that the problem depends on. We assume the following scale separation properties: for all $i=1,\ldots,n-1$

[TABLE]

Without loss of generality, we assume that $\varepsilon_{1}(\varepsilon)=\varepsilon$. We define the multiscale coefficient of the Maxwell equation $a^{\varepsilon}$ and $b^{\varepsilon}$ which are functions from $D$ to $\mathbb{R}^{d\times d}_{sym}$ as

[TABLE]

When $d=3$ we define the spaces

[TABLE]

and when $d=2$

[TABLE]

where $\nu$ denotes the outward normal vector on the boundary $\partial D$. We have the Gelfand triple $W\subset H\subset W^{\prime}$. We denote by $\langle\cdot,\cdot\rangle$ the inner product in $H$, extending to the duality pairing between $W^{\prime}$ and $W$. We note that when $d=3$, ${\rm curl\,}u^{\varepsilon}$ is a vector function in $L^{2}(D)^{3}$ and when $d=2$, ${\rm curl\,}u^{\varepsilon}$ is a scalar function in $L^{2}(D)$. Let $f\in L^{2}(0,T;H)$, $g^{0}\in W$ and $g^{1}\in H$. We consider the problem: Find $u^{\varepsilon}(t,x)\in L^{2}(0,T;W)$

[TABLE]

with the boundary condition $u^{\varepsilon}\times\nu=0$ on $\partial D$. We will mostly present the analysis for the case $d=3$ and only discuss the case $d=2$ when there is significant difference. For notational conciseness, we denote by

[TABLE]

In variational form, this problem becomes: Find $u^{\varepsilon}\in L^{2}(0,T;W)\cap H^{1}(0,T;H)$ so that

[TABLE]

for all $\phi\in W$ when $d=3$; and when $d=2$ we need to replace the vector product for ${\rm curl\,}$ by the scalar multiplication. Problem (2.5) has a unique solution $u^{\varepsilon}\in L^{2}(0,T;W)\bigcap H^{1}(0,T;H)\bigcap H^{2}(0,T;W^{\prime})$ that satisfies

[TABLE]

where the constant $c$ only depends on the constants $\alpha$ and $\beta$ in (2.1) and $T$ (see Wloka [14]).

We will study this problem via multiscale convergence.

2.2 Multiscale convergence

We recall the definition of multiscale convergence (see Nguetseng [12], Allaire [1] and Allaire and Briane [2].

Definition 2.1

A sequence of functions $\{w^{\varepsilon}\}_{\varepsilon}\subset L^{2}(0,T;H)$ $(n+1)$-scale converges to a function $w_{0}\in L^{2}((0,T);D\times{\bf Y})$ if for all smooth functions $\phi(t,x,\boldsymbol{y})$ which are $Y$ periodic w.r.t. $y_{i}$ for all $i=1,\ldots,n$:

[TABLE]

We have the following result.

Proposition 2.2

From a bounded sequence in $L^{2}(0,T;H)$ we can extract an $(n+1)$-scale convergent subsequence.

We note that the definition above for functions which depend also on $t$ is slightly different from that in [12] and [1] as we take also the integral with respect to $t$. However, the proof of proposition 2.2 is similar.

For a bounded sequence in $L^{2}(0,T;W)$, we have the following results which are very similar to those in [5] for functions which do not depend on $t$. The proofs for these results follow the same lines of those in [5] so we do not present them here. As in [13] and [5], we denote by $\tilde{H}_{\#}({\rm curl\,},Y_{i})$ the space of equivalent classes of functions in $H_{\#}({\rm curl\,},Y_{i})$ of equal ${\rm curl\,}$.

Proposition 2.3

Let $\{w^{\varepsilon}\}_{\varepsilon}$ be a bounded sequence in $L^{2}(0,T;W)$. There is a subsequence (not renumbered), a function $w_{0}\in L^{2}(0,T;W)$, $n$ functions $\mathfrak{w}_{i}\in L^{2}((0,T)\times D\times Y_{1}\times\ldots\times Y_{i-1},H^{1}_{\#}(Y_{i})/\mathbb{R})$ such that

[TABLE]

Further, there are $n$ functions $w_{i}\in L^{2}((0,T)\times D\times\ldots\times Y_{i-1},\tilde{H}_{\#}({\rm curl\,},Y_{i}))$ such that

[TABLE]

2.3 Multiscale homogenized Maxwell wave problem

From (2.6) and Proposition (2.3), we can extract a subsequence (not renumbered), a function $u_{0}\in L^{2}(0,T;W)$, $n$ functions $\mathfrak{u}_{i}\in L^{2}(0,T;D\times Y_{1}\times\ldots\times Y_{i-1},H^{1}_{\#}(Y_{i})/\mathbb{R})$ and $n$ functions $u_{i}\in L^{2}(0,T;D\times Y_{1}\times\ldots\times Y_{i-1},\tilde{H}_{\#}({\rm curl\,},Y_{i}))$ such that

[TABLE]

and

[TABLE]

For $i=1,\ldots,n$, let $W_{i}=L^{2}(D\times Y_{1}\times\ldots\times Y_{i-1},\tilde{H}_{\#}({\rm curl\,},Y_{i}))$ and $V_{i}=L^{2}(D\times Y_{1}\times\ldots\times Y_{i-1},H^{1}_{\#}(Y_{i})/\mathbb{R})$. We define the space ${\bf V}$

[TABLE]

For $\boldsymbol{v}=(v_{0},\{v_{i}\},\{\mathfrak{v}_{i}\})\in{\bf V}$, we define the form

[TABLE]

Let $\boldsymbol{u}=(u_{0},\{u_{i}\},\{\mathfrak{u}_{i}\})\in{\bf V}$. We define the function

[TABLE]

in $W^{\prime}$ as

[TABLE]

and the function $b(x,\boldsymbol{y})(u_{0}(t,x)+\sum_{i=1}^{n}\nabla_{y_{i}}\mathfrak{u}_{i}(t,x,\boldsymbol{y}))$ in $V_{j}^{\prime}$ as

[TABLE]

We then have the following result.

Proposition 2.4

The function $\boldsymbol{u}=(u_{0},\{u_{i}\},\{\boldsymbol{u}_{i}\})$ satisfies

[TABLE]

and

[TABLE]

i.e. the function $\boldsymbol{u}$ satisfies the multiscale homogenized equation

[TABLE]

for all $\boldsymbol{v}=(v_{0},\{v_{i}\},\{\mathfrak{v}_{i}\})\in{\bf V}$.

Proof  Let $q\in{\cal D}(0,T)$. Let $v_{0}\in{\cal D}(D)$, $v_{i}\in{\cal D}(D,C^{\infty}_{\#}(Y_{1},\ldots,C^{\infty}_{\#}(Y_{i})\ldots))^{3}$ and $\mathfrak{v}_{i}\in{\cal D}(D,C^{\infty}_{\#}(Y_{1},\ldots,C^{\infty}_{\#}(Y_{i})\ldots))$ for $i=1,\ldots,n$. Choosing a test function of the form

[TABLE]

we obtain

[TABLE]

Passing to the two scale limit, using the scale separation (2.2), we have

[TABLE]

Using a density argument, we find that this equation holds for all $(v_{0},\{v_{i}\},\{\mathfrak{v}_{i}\})\in{\bf V}$. The conclusion then follows. $\Box$

We now establish the initial conditions.

Proposition 2.5

We have $u_{0}\in H^{1}(0,T;H)$, $\nabla_{y_{i}}\mathfrak{u}_{i}\in H^{1}(0,T;H_{i})$ for all $i=1,\ldots,n$. Further

[TABLE]

[TABLE]

and for $i=1,\ldots,n$

[TABLE]

Proof  As $u^{\varepsilon}$ is bounded in $H^{1}(0,T;H)$, $u_{0}$ belongs to $H^{1}(0,T;H)\subset C([0,T];H)$ and is the weak limit in this space of $u^{\varepsilon}$. Let $\phi\in C^{\infty}([0,T]\times D)$ with $\phi=0$ when $t=T$. We have

[TABLE]

On the other hand

[TABLE]

Thus $u_{0}(0,x)=g_{0}$.

As $\{{\partial u^{\varepsilon}\over\partial t}\}_{\varepsilon}$ is bounded in $L^{2}(0,T;H)$ so there is a subsequence that $(n+1)$-scale converges. Let $\xi\in L^{2}((0,T);L^{2}(D\times{\bf Y}))$ be the $(n+1)$-scale limit. Let $\phi(t,x,y_{1})\in C^{\infty}_{0}((0,T);C^{\infty}_{0}(D,C^{\infty}_{\#}(Y_{1})))$. We have that

[TABLE]

On the other hand

[TABLE]

which converges to

[TABLE]

Thus

[TABLE]

so

[TABLE]

(we refer to (2.4) for the definition of the spaces $H_{i}$). Similarly, using a function $\phi(t,x,y_{1},y_{2})\in C^{\infty}_{0}((0,T),C^{\infty}_{0}(D,C^{\infty}_{\#}(Y_{1},C^{\infty}_{\#}(Y_{2}))))$, we have

[TABLE]

Continuing this process, we have that for all $i=1,\ldots,n$,

[TABLE]

Finally, we have

[TABLE]

Let $q\in C^{\infty}([0,T])$ with $q(T)=0$. Let $\phi(x)=v_{0}(x)+\sum_{i=1}^{n}\varepsilon_{i}v_{i}(x,{x\over\varepsilon_{1}},\ldots,{x\over\varepsilon_{i}})+\sum_{i=1}^{n}\varepsilon_{i}\nabla\mathfrak{v}_{i}(x,{x\over\varepsilon_{1}},\ldots,{x\over\varepsilon_{i}})$. We have

[TABLE]

when $\varepsilon\to 0$. On the other hand, let $q_{n}$ be a sequence in $C^{\infty}_{0}(0,T)$ that converges to $q(t)$ in $L^{2}(0,T)$ when $n\to\infty$. As $b^{\varepsilon}{\partial^{2}\over\partial t^{2}}u^{\varepsilon}$ is bounded in $L^{2}(0,T;W^{\prime})$ so there is a constant $c>0$ such that

[TABLE]

As $q_{n}\in C^{\infty}_{0}(0,T)$, when $\varepsilon\to 0$,

[TABLE]

Passing to the limit when $n\to\infty$, we have

[TABLE]

From Proposition 2.4, as ${\partial^{2}\over\partial t^{2}}\int_{{\bf Y}}b(x,\boldsymbol{y})(u_{0}(t,x)+\sum_{j=1}^{n}\nabla_{y_{j}}\mathfrak{u}_{j}(t,x,\boldsymbol{y}_{i}))d\boldsymbol{y}\in L^{2}(0,T;W^{\prime})$, ${\partial\over\partial t}\int_{{\bf Y}}b(x,\boldsymbol{y})(u_{0}(t,x)+\sum_{i=1}^{n}\nabla_{y_{i}}\mathfrak{u}_{i}(t,x,\boldsymbol{y}_{i}))d\boldsymbol{y}\in C([0,T];W^{\prime})$ so the initial condition ${\partial\over\partial t}\int_{{\bf Y}}b(x,\boldsymbol{y})(u_{0}(t,x)+\sum_{j=1}^{n}\nabla_{y_{j}}\mathfrak{u}_{j}(t,x,\boldsymbol{y}_{j}))d\boldsymbol{y}$ at $t=0$ is well defined in $W^{\prime}$. Similarly, the initial condition ${\partial\over\partial t}b(x,\boldsymbol{y})(u_{0}+\sum_{j=1}^{n}\nabla_{y_{j}}\mathfrak{u}_{i}(t,x,\boldsymbol{y}_{j})$ at $t=0$ is well defined in $V_{i}^{\prime}$. The right hand side of (2.14) can be written as

[TABLE]

Comparing (2.13) and (2.15), we have

[TABLE]

and

[TABLE]

for all $i=1,\ldots,n$. We get the desired result.

$\Box$

Proposition 2.6

With the initial conditions (2.10), (2.11) and (2.12), problem (2.9) has a unique solution.

Proof  We show that when $f=0$, $g_{0}=0$ and $g_{1}=0$, the solution of (2.9) is $u_{0}=0$, $u_{i}=0$ and $\mathfrak{u}_{i}=0$ for all $i=1,\ldots,n$.

Following the procedure in [14] Theorem 19.1 for showing the uniqueness of a solution of a wave equation, fixing $s\in(0,T)$, we define

[TABLE]

We have

[TABLE]

Integrating over $(0,s)$, we get

[TABLE]

due to $u_{0}={\partial w_{0}\over\partial t}$, $u_{i}={\partial w_{i}\over\partial t}$ and $\mathfrak{u}_{i}={\partial\mathfrak{w}_{i}\over\partial t}$ for $i=1,\ldots,n$ from (2.16). Using (2.11) and (2.12), we have

[TABLE]

We thus deduce that $u_{0}(s)=0$, $\nabla_{y_{i}}\mathfrak{u}_{i}(s)=0$ $\forall\,s$, ${\rm curl\,}w_{0}(0)=0$ and ${\rm curl}_{y_{i}}w_{i}(0)=0$. This means that

[TABLE]

for all $s$. Thus for all $\sigma$, ${\rm curl\,}u_{0}(\sigma)=0$ and ${\rm curl}_{y_{i}}(\sigma)=0$. $\Box$

3 Homogenized equation

In this section, we use the multiscale homogenized problem (2.9) to deduce the homogenized equation. From (2.9), we have that for all $\mathfrak{v}_{n}\in V_{n}$ and all $q\in{\cal D}(0,T)$

[TABLE]

Thus

[TABLE]

where $w_{n}^{k}\in L^{2}(D\times{\bf Y}_{n-1},H^{1}_{\#}(Y_{n})/\mathbb{R})$ is the solution of the cell problem

[TABLE]

$e^{k}$ is the $k$-th unit vector in $\mathbb{R}^{d}$. Therefore

[TABLE]

so

[TABLE]

From this we have

[TABLE]

for a function $G_{n}(x,\boldsymbol{y}_{n})$ in $L^{2}(D\times{\bf Y}_{n})$. We then have $\forall\,\mathfrak{v}_{n-1}\in V_{n-1}$

[TABLE]

where $b^{n-1}(x,\boldsymbol{y}_{n-1})$ is the $(n-1)$th level homogenized coeffcient which is defined by

[TABLE]

Similarly we have

[TABLE]

where $G_{n-1}(x,\boldsymbol{y}_{n-1})\in L^{2}(D\times{\bf Y}_{n-1})$, and $w^{k}_{n-1}\in L^{2}(D\times{\bf Y}_{n-2},H^{1}_{\#}(Y_{n-1})/\mathbb{R})$ satisfies the cell problem

[TABLE]

Recursively, letting $b^{n}(x,\boldsymbol{y}_{n})=b(x,\boldsymbol{y})$, we have for all $i=0,\ldots,n-1$:

[TABLE]

where $w^{k}_{i}\in L^{2}(D\times{\bf Y}_{i-1},H^{1}_{\#}(Y_{i})/\mathbb{R})$ is the solution of the cell problem

[TABLE]

From an argument as above, we have

[TABLE]

for a function $G_{i}(x,\boldsymbol{y}_{i})\in L^{2}(D\times{\bf Y}_{i})$. The positive definite matrix function $b^{0}(x)$, which is defined as

[TABLE]

is the homogenized coefficient. It satisfies

[TABLE]

for all $v_{0}\in W$ and $\mathfrak{v}_{i}\in V_{i}$.

From (2.9)

[TABLE]

for all $v_{n}\in W_{n}$. For each $l=1,...,d$, let $N^{l}_{n}\in V_{n}$ be the solution of

[TABLE]

We can write $u_{n}$ as

[TABLE]

For all $v_{n-1}\in W_{n-1}$,

[TABLE]

i.e.

[TABLE]

for all $v_{n-1}\in W_{n-1}$. Let

[TABLE]

We have that

[TABLE]

where $N_{(n-1)}^{l}$ satisfies the cell problem

[TABLE]

Letting $a^{n}=a$, we then have, recursively,

[TABLE]

where $N_{i}^{l}\in W_{i}$ satisfies the cell problem

[TABLE]

i.e.

[TABLE]

for all $v_{i}\in W_{i}$. For $i=1,\ldots,n-1$, the $i$th level homogenized coefficient $a^{i}$ is defined as

[TABLE]

Continuing this process, we finally get the homogenized coefficient $a^{0}(x)$ as

[TABLE]

The homogenization equation is

[TABLE]

i.e.

[TABLE]

Now we derive the initial conditions. From (2.10), $u_{0}(0)=g_{0}$. As a distribution in $V_{n}^{\prime}$,

[TABLE]

so for all $t$

[TABLE]

i.e.

[TABLE]

From (3.2) we have

[TABLE]

due to (3.1). From (3.2) we have

[TABLE]

Thus there is a function $\tilde{G}_{n}\in L^{2}(D\times{\bf Y}_{n-1},H^{1}_{\#}(Y_{n})/\mathbb{R})$ such that $G_{n}(x,\boldsymbol{y}_{n})=\nabla_{y_{n}}\tilde{G}_{n}(x,\boldsymbol{y}_{n})$. From

[TABLE]

we deduce that

[TABLE]

where $w^{k}_{n}$ is the solution of the cell problem (3.1). As a function in $V_{n-1}^{\prime}$, ${\partial^{2}\over\partial t^{2}}b(x,\boldsymbol{y})(u_{0}+\sum_{i=1}^{n}\nabla_{y_{i}}\mathfrak{u}_{i})=0$ so

[TABLE]

From (2.12), we have

[TABLE]

From (3.2) and (3.9) we have

[TABLE]

Thus

[TABLE]

Continuing this, we find that

[TABLE]

Therefore, for all $t$

[TABLE]

Therefore for $q\in C^{\infty}([0,T])$ with $q(T)=0$, using the fact that ${\partial\over\partial t}b(x,\boldsymbol{y})(u_{0}+\sum_{i=1}^{n}\nabla_{y_{i}}\mathfrak{u}_{i})$ is continuous as a map from $[0,T]$ to $W^{\prime}$, we have

[TABLE]

From (2.13) and (2.14), we deduce

[TABLE]

We then have

[TABLE]

We note that

[TABLE]

Continuing this we have

[TABLE]

Thus as distribution in $V^{\prime}$

[TABLE]

Therefore, $u^{0}$ is the solution of the problem (3.8) with the initial condition $u^{0}(0)=g^{0}$ and (3.10) which has a unique solution.

The solution $\boldsymbol{u}$ is written in terms of $u_{0}$ as

[TABLE]

and

[TABLE]

Given that at $t=0$, $\nabla_{y_{i}}\mathfrak{u}_{i}=0$, we then have

[TABLE]

Without loss of generality, we let

[TABLE]

4 Regularity of the solution

To deduce the homogenization errors in the next section, we need regularity for the solution $u^{0}$ of the homogenized equation (3.8), and of the solutions of the cell problems (3.4) and (3.6). We assume:

Assumption 4.1

The matrix functions $a$ and $b$ belong to $C^{1}(\bar{D},C^{2}(\bar{Y}_{1},\ldots,C^{2}(\bar{Y}_{n})\ldots))^{d\times d}$.

With this assumption, we have

Proposition 4.2

Under Assumption 4.1, for all $i,r=1,\ldots,d$, ${\rm curl}_{y_{i}}N_{i}^{r}\in C^{1}(\bar{D},C^{2}(\bar{Y}_{1},\ldots,C^{2}(\bar{Y}_{i-1},H^{2}(Y_{i}))\ldots))$ and $w_{i}^{r}\in C^{1}(\bar{D},C^{2}(\bar{Y}_{1},\ldots,C^{2}(\bar{Y}_{i-1},H^{3}(Y_{i}))\ldots))$.

We refer to [5] for a proof of this proposition.

We have the following regularity results for the solution $u_{0}$ of the homogenized equation (3.8).

Proposition 4.3

Under Assumption 4.1, assume

[TABLE]

then

[TABLE]

Further, if

[TABLE]

then

[TABLE]

Proof  We use the regularity theory of general hyperbolic equations (see, e.g., Wloka [14], Chapter 5). From (4.1) we have that

[TABLE]

with compatibility initial conditions

[TABLE]

and

[TABLE]

with compatibility initial conditions

[TABLE]

We thus deduce that

[TABLE]

From (3.13) and Proposition 4.2, we deduce that

[TABLE]

Similarly, we deduce regularity (4.4) from (4.3). $\Box$

For the regularity of $u_{0}$, we have the following result.

Proposition 4.4

Under Assumption 4.1, if $D$ is a Lipschitz polygonal domain, $f\in H^{1}(0,T;H)$, $g_{0}\in H^{1}({\rm curl\,},D)$ and $g_{1}\in W$, ${\rm div}f\in L^{\infty}(0,T;L^{2}(D))$, ${\rm div}(b^{0}g_{0})\in L^{2}(D)$ and ${\rm div}(b^{0}g_{1})\in L^{2}(D)$, there is a constant $s\in(0,1]$ such that $u_{0}\in L^{\infty}(0,T;H^{s}({\rm curl\,},D))$.

Proof  Using Proposition 4.2, equations (3.7) and (3.5), we have that $a^{0},b^{0}\in C^{1}(\bar{D})^{d\times d}$. As $f\in H^{1}(0,T;H)$ and $g_{0}\in H^{1}({\rm curl\,},D)$, we have that $(b^{0})^{-1}[f-{\rm curl\,}(a^{0}{\rm curl\,}g^{0})]\in H$. The compatibility initial conditions hold so that ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H)$. Thus

[TABLE]

Let $U(t)=a^{0}{\rm curl\,}u_{0}(t)$. As ${\rm div}((a^{0})^{-1}U(t))=0$ and $(a^{0})^{-1}U(t)\cdot\nu=0$, there is a constant $c$ and a constant $s\in(0,1]$ which depend on $a^{0}$ and the domain $D$ so that

[TABLE]

so $U\in L^{\infty}(0,T;H^{s}(D)^{3})$. As ${\rm curl\,}u_{0}(t)=(a^{0})^{-1}U(t)$ and $(a^{0})^{-1}\in C^{1}(\bar{D})^{d\times d}$, ${\rm curl\,}u_{0}\in L^{\infty}(0,T;H^{s}(D))$. We note that

[TABLE]

so

[TABLE]

From Theorem 4.1 of Hiptmair [8], we deduce that there is a constant $s\in(0,1]$ (we take it as the same constant as above), so that

[TABLE]

Thus $u_{0}\in L^{\infty}(0,T;H^{s}({\rm curl\,},D))$. $\Box$

Similarly, we can deduce the regularity for ${\partial^{2}u_{0}\over\partial t^{2}}$.

Proposition 4.5

Under Assumption 4.1, if $D$ is a Lipschitz polygonal domain, if the comparibility conditions (4.3) hold, and if ${\rm div}f\in L^{\infty}(0,T;L^{2}(D))$, then there is a constant $s\in(0,1]$ such that ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{s}({\rm curl\,},D))$.

Proof  From equation (4.6), we have

[TABLE]

as ${\partial^{4}u_{0}\over\partial t^{4}}\in L^{\infty}(0,T;H)$ due to (4.3). Following a similar argument as in the proof of Proposition 4.4 we deduce that ${\rm curl\,}{\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{s}(D)^{3})$. We note that

[TABLE]

From Theorem 4.1 of [8], we deduce that ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{s}(D)^{3})$. $\Box$

5 Corrector for the homogenization problem

We derive homogenization correctors in the section. For two scale problems, with sufficient regularity for the solution of the homogenized equation and the cell problems, we derive explicit homogenization errors in terms of the microscopic scales. We consider both the cases where $u_{0}$ and ${\partial\over\partial t}u_{0}$ belong to $L^{\infty}((0,T);H^{1}({\rm curl\,},D))$ and where they only belong to a weaker regularity space $L^{\infty}((0,T);H^{s}({\rm curl\,},D))$ for $0<s<1$, which is normally the case in polygonal domains $D$. For multiscale problems, we are not able to prove an explicit homogenization error but a general corrector is derived.

5.1 Corrector for two-scale problem

For conciseness, we denote the coefficients $a(x,\boldsymbol{y})$ and $b(x,\boldsymbol{y})$ as $a(x,y)$ and $b(x,y)$. The cell problems become

[TABLE]

and

[TABLE]

The homogenized coefficient is determined by

[TABLE]

and

[TABLE]

We then have the following

Proposition 5.1

For two-scale problems, assume that $g_{0}=0$, $g_{1}\in H^{1}(D)^{3}\bigcap W$, $f\in H^{1}(0,T;H)$, $u_{0}\in L^{\infty}(0,T;H^{1}({\rm curl\,};D))$, ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;H^{1}({\rm curl\,},D))$, ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H({\rm curl\,},D)\bigcap H^{1}(D)^{3})$, $N^{r}\in C^{1}(\bar{D},C(\bar{Y}))^{3}$, ${\rm curl}_{y}N^{r}\in C^{1}(\bar{D},C(\bar{Y}))^{3}$ and $\omega^{r}\in C^{1}(\bar{D},C(\bar{Y}))$ for all $r=1,2,3$. There exists a constant $c$ that does not depend on $\varepsilon$ such that

[TABLE]

Proof  We note that ${\partial u^{\varepsilon}\over\partial t}$ satisfies

[TABLE]

with the initial condition ${\partial u^{\varepsilon}\over\partial t}(0)=g_{1}\in W$ and ${\partial^{2}u^{\varepsilon}\over\partial t^{2}}(0)=f(0)\in H$ (due to $g_{0}=0$). We therefore deduce that

[TABLE]

where $c$ only depends on the constants $\alpha$ and $\beta$ in (2.1). Thus ${\partial u^{\varepsilon}\over\partial t}$ is uniformly bounded in $L^{\infty}(0,T;W)$ for all $\varepsilon$.

We consider the function

[TABLE]

where we have from (3.14)

[TABLE]

We first show that

[TABLE]

We have

[TABLE]

Thus

[TABLE]

where the vector functions $G_{r}(x,y)$ are defined by

[TABLE]

and

[TABLE]

From (5.1), we have that ${\rm curl}_{y}G_{r}(x,y)=0$. Further from (5.3), $\int_{Y}G_{r}(x,y)dy=0$. From these we deduce that there are functions $\tilde{G}_{r}(x,y)$ such that $G_{r}(x,y)=\nabla_{y}\tilde{G}_{r}(x,y)$. Since $\nabla_{y}\tilde{G}_{r}(x,\cdot)=G_{r}(x,\cdot)\in H^{1}(Y)^{3}$, we deduce that $\Delta_{y}\tilde{G}_{r}(x,\cdot)\in L^{2}(Y)$. From elliptic regularity, $\tilde{G}_{r}(x,\cdot)\in H^{2}(Y)\subset C(\bar{Y})$. As $G_{r}(x,\cdot)\in C^{1}(\bar{D},H^{1}_{\#}(Y)^{3})$, $\tilde{G}_{r}(x,y)\in C^{1}(\bar{D},H^{2}(Y))\subset C^{1}(\bar{D},C(\bar{Y}))$. We note that

[TABLE]

Therefore, for all $\phi\in{\cal D}(D)^{d}$ we have

[TABLE]

We note that

[TABLE]

As $N^{j}\in C^{1}(\bar{D},C(\bar{Y}))$ and ${\rm curl\,}u_{0}\in L^{\infty}(0,T;H^{1}(D))$ we deduce that

[TABLE]

where $c$ is independent of $t$. From these we conclude that

[TABLE]

Using a density argument, we have that this holds for all $\phi\in W$. Thus

[TABLE]

Let $\tau^{\varepsilon}(x)$ be a function in ${\cal D}(D)$ such that $\tau^{\varepsilon}(x)=1$ outside an $\varepsilon$ neighbourhood of $\partial D$ and $\sup\limits_{x\in D}\varepsilon|\nabla\tau^{\varepsilon}(x)|<c$ where $c$ is independent of $\varepsilon$. Let

[TABLE]

The function $w_{1}^{\varepsilon}(x)$ belongs to $L^{2}(0,T;H_{0}({\rm curl\,},D))$. We note that

[TABLE]

From this,

[TABLE]

Since $\frac{\partial^{2}u_{0}}{\partial t^{2}}\in L^{\infty}(0,T;H^{1}(D)^{3})$, we have

[TABLE]

Therefore

[TABLE]

For $\phi\in W$, we have

[TABLE]

where the vector function $F_{r}$ is defined by

[TABLE]

Since ${\rm div}_{y}F_{r}(x,y)=0$, and $\int_{Y}F_{r}(x,y)dy=0$, there is a function $\tilde{F}_{r}(x,y)$ such that

[TABLE]

Detailed construction of the function $\tilde{F}_{r}$ is presented in [10] which implies that $F_{r}\in C^{1}(\bar{D},C(\bar{Y}))^{3}$. We note that

[TABLE]

Therefore

[TABLE]

due to the conditions ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{1}(D)^{d})$. Let $D^{\varepsilon}\subset D$ be the $\varepsilon$ neighbourhood of the boundary $\partial D$. We know that

[TABLE]

for $\phi\in H^{1}(D)$ (see Hoang and Schwab [9]). From the condition ${\rm curl\,}u_{0}\in L^{\infty}(0,T;H^{1}(D))$, ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;H^{1}({\rm curl\,};D))$ and ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{1}(D))$ we deduce that:

[TABLE]

From this and (5.13), we get

[TABLE]

From

[TABLE]

we have

[TABLE]

We note that

[TABLE]

and

[TABLE]

As ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;W\bigcap H^{1}({\rm curl\,},D))$, we deduce that ${\partial w^{\varepsilon}_{1}\over\partial t}$ is uniformly bounded in $L^{\infty}(0,T;W)$. Since ${\partial u^{\varepsilon}\over\partial t}(t)$ and ${\partial w_{1}^{\varepsilon}\over\partial t}(t)$ belong to $W$ and are uniformly bounded in the norm of $W$, from (5.18), we deduce that

[TABLE]

From (5.8) we deduce

[TABLE]

where $c$ is independent of $t$. From (5.11) we get

[TABLE]

Using the initial condition $u_{0}(0)=w_{1}^{\varepsilon}(0)=0$, from (5.17), we deduce that

[TABLE]

Thus

[TABLE]

The left hand side equals

[TABLE]

due to condition (2.1).

We also have

[TABLE]

Thus, we deduce that for all $t\in(0,T)$

[TABLE]

which implies

[TABLE]

Due to ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;H^{1}(D)^{3})$ and $g_{1}\in H^{1}(D)^{3}$,

[TABLE]

From (5.17) and the facts that $N^{r}\in C^{1}(\overline{D},C(\overline{Y})),w^{r}\in C^{1}(\overline{D},C(\overline{Y}))$ we have

[TABLE]

Similarly, from (5.11), we have

[TABLE]

We therefore have

[TABLE]

We note that

[TABLE]

Thus

[TABLE]

Similarly, from (5.6) we have

[TABLE]

We then get the conclusion. $\Box$

Next, we derive the homogenization error where $u_{0}$ only possesses the weaker regularity $W^{2,\infty}(0,T;H^{s}({\rm curl\,},D))$ for $0<s<1$.

Proposition 5.2

Assume that $g_{0}=0$, $g_{1}\in H^{1}(D)\bigcap W$, $f\in H^{1}(0,T;H)$, $u_{0}$ and ${\partial u_{0}\over\partial t}$ belong to $L^{\infty}((0,T);H^{s}({\rm curl\,},D))$, and ${\partial^{2}u_{0}\over\partial t^{2}}$ belong to $L^{\infty}(0,T;H^{s}(D))$ for $0<s\leq 1$, $N^{r}\in C^{1}(\overline{D},C(\overline{Y}))^{3}$, ${\rm curl\,}_{y}N^{r}\in C^{1}(\overline{D},C(\overline{Y}))^{3}$, $\omega^{r}\in C^{1}(\overline{D},C(\overline{Y}))$ for all $r=1,2,3$. There exists a constant $c$ that does not depend on $\varepsilon$ such that

[TABLE]

Proof  We consider a set of $M$ open cubes $Q_{i}~{}(i=1,..,M)$ of size $\varepsilon^{s_{1}},s_{1}>0$ to be chosen later such that $D\subset\bigcup_{i=1}^{M}Q_{i}$ and $Q_{i}\cap D\not=\emptyset$. Each cube $Q_{i}$ intersects with only a finite number, which does not depend on $\varepsilon$, of other cubes. We consider a partition of unity that consists of $M$ functions $\rho_{i}$ such that $\rho_{i}$ has support in $Q_{i}$, $\sum_{i=1}^{M}\rho_{i}(x)=1$ for all $x\in D$ and $|\nabla\rho_{i}|\leq c\varepsilon^{-s_{1}}$ for all $x$. For $r=1,2,3$ and $i=1,...,M$, we denote by

[TABLE]

and

[TABLE]

(as $u_{0}\in H^{s}(D)^{3}$ and ${\rm curl\,}u_{0}\in H^{s}(D)^{3}$, for the Lipschitz domain $D$, we can extend each of them, separately, continuously outside $D$ and understand $u_{0}$ and curl $u_{0}$ as these extensions (see Wloka [??] Theorem 5.6)). Let $U_{i}$ and $V_{i}$ denote the vector $(U_{i}^{1},U_{i}^{2},U_{i}^{3})$ and $(V_{i}^{1},V_{i}^{2},V_{i}^{3})$ respectively. Let $B$ be the unit cube in $\mathbb{R}^{3}$. From Poincare inequality, we have

[TABLE]

By translation and scaling, we deduce that

[TABLE]

i.e

[TABLE]

Together with

[TABLE]

we deduce from interpolation that

[TABLE]

Thus for $t$ fixed

[TABLE]

We consider the function

[TABLE]

We first show that

[TABLE]

We have

[TABLE]

Thus

[TABLE]

We have

[TABLE]

where the vector functions $G_{r}(x,y)$ are defined in (5.7) and

[TABLE]

For all $\phi\in{\cal D}(D)^{d}$ we have

[TABLE]

where $G_{r}(x,y)=\nabla_{y}\tilde{G}_{r}(x,y)$ as shown above. Further,

[TABLE]

We note that

[TABLE]

From

[TABLE]

and the fact that the support of each function $\rho_{i}$ intersects only with the support of a finite number (which does not depend on $\varepsilon$) of other functions $\rho_{j}$ in the partition of unity, we deduce

[TABLE]

Thus

[TABLE]

We also have

[TABLE]

As the support of each function $\rho_{i}$ intersects with the support of a finite number of other functions $\rho_{j}$ and $\|\nabla\rho_{j}\|_{L^{\infty}(D)}\leq c\varepsilon^{-s_{1}}$, we have

[TABLE]

so

[TABLE]

We have further that

[TABLE]

We note that, for $t$ fixed

[TABLE]

Using the support property of $\rho_{j}$, we have from (5.19)

[TABLE]

Thus

[TABLE]

From these we conclude that

[TABLE]

Using a density argument, we have that this holds for all $\phi\in W$. Thus

[TABLE]

Since ${\partial u_{0}\over\partial t}\in L^{\infty}((0,T);H^{s}(D))$, by an identical argument, we deduce that

[TABLE]

Choose $s_{1}=\frac{1}{s+1}$ we have

[TABLE]

and

[TABLE]

Let $\tau^{\varepsilon}(x)$ be a function in ${\cal D}(D)$ such that $\tau^{\varepsilon}(x)=1$ outside an $\varepsilon$ neighbourhood of $\partial D$ and $\sup\limits_{x\in D}\varepsilon|\nabla\tau^{\varepsilon}(x)|<c$ where $c$ is independent of $\varepsilon$. Let

[TABLE]

We then have

[TABLE]

From this,

[TABLE]

Let $\tilde{D}^{\varepsilon}$ be the $3\varepsilon^{s_{1}}$ neighbourhood of $\partial D$. We note that ${\rm curl\,}u_{0}$ is extended continously outside $D$. As shown in Hoang and Schwab [9], for $\phi\in H^{1}(\tilde{D}^{\varepsilon})$

[TABLE]

From this and

[TABLE]

using interpolation we get

[TABLE]

for all $\phi\in H^{s}(D)$ extended continuously outside $D$. We then have for $t$ fixed,

[TABLE]

As $\partial D$ is Lipschitz, $D^{\varepsilon}$ is the $\varepsilon$ neighbourhood of $\partial D$ and $Q_{j}$ has size $\varepsilon^{s_{1}}$, $|Q_{j}\cap D^{\varepsilon}|\leq c\varepsilon^{1+(d-1)s_{1}}$ so $\frac{|Q_{j}\cap D^{\varepsilon}|}{|Q_{j}|}\leq c\varepsilon^{1-s_{1}}$. When $Q_{j}\cap D^{\varepsilon}\not=\emptyset$ , $Q_{j}\subset\tilde{D}^{\varepsilon}.$ Thus

[TABLE]

Therefore

[TABLE]

and

[TABLE]

Similarly, we have

[TABLE]

Thus

[TABLE]

Therefore

[TABLE]

Arguing as above (with $t$ fixed) we deduce that

[TABLE]

We have

[TABLE]

Thus

[TABLE]

For $\phi\in W$, we consider

[TABLE]

where the function $F_{r}(x,y)$ is defined in (5.14). Using the function $\tilde{F}_{r}$ defined in (5.15), we have

[TABLE]

As ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{\infty}(0,T;H^{s}(D))$, there is a constant $c$ such that for all $t\in(0,T)$

[TABLE]

Thus

[TABLE]

when we choose $s_{1}=1/(1+s)$. Using

[TABLE]

we have

[TABLE]

As shown in the proof of Proposition 5.1, ${\partial u^{\varepsilon}\over\partial t}$ is uniformly bounded in $L^{\infty}(0,T;W)$ with respect to $\varepsilon$. For $w_{1}^{\varepsilon}$, we have

[TABLE]

and

[TABLE]

As ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;H({\rm curl\,},D))$, $\|{\partial U_{j}^{r}\over\partial t}\rho_{j}\|_{L^{2}(D)^{3}}\leq c$, $\|{\partial V_{j}^{r}\over\partial t}\rho_{j}\|_{L^{2}(D)^{3}}\leq c$, $\|{\partial U_{j}^{r}\over\partial t}\nabla\rho_{j}\|_{L^{2}(D)^{3}}\leq c\varepsilon^{-s_{1}}$ and $\|{\partial V_{j}^{r}\over\partial t}\nabla\rho_{j}\|_{L^{2}(D)^{3}}\leq c\varepsilon^{-s_{1}}$. Therefore ${\rm curl\,}{\partial w^{\varepsilon}_{1}\over\partial t}$ and ${\partial w^{\varepsilon}_{1}\over\partial t}$ are uniformly bounded in $H$ with respect to $\varepsilon$, i.e. ${\partial w^{\varepsilon}_{1}\over\partial t}$ is uniformly bounded in $L^{\infty}(0,T;H^{1}({\rm curl\,},D))$. We have

[TABLE]

We also have

[TABLE]

Since $u_{0}(0)=0$ and $u_{1}^{\varepsilon}(0)=0$, together with (5.24) we have that

[TABLE]

Now we estimate

[TABLE]

using (5.27). We have that

[TABLE]

and

[TABLE]

For the other two terms in (5.27), we have

[TABLE]

Thus

[TABLE]

as $u_{0}$ and ${\partial u_{0}\over\partial t}$ belong to $L^{\infty}(0,T;H^{s}(D)))$. Similarly, we have

[TABLE]

Therefore,

[TABLE]

We then deduce

[TABLE]

Therefore

[TABLE]

(note that $u^{\varepsilon}(0)=w^{\varepsilon}_{1}(0)=0$). We have

[TABLE]

As $g_{1}\in H^{1}(D)^{3}$, a similar argument as for showing (5.22) for $s=1$ shows that

[TABLE]

Further,

[TABLE]

Thus

[TABLE]

Using (2.1) we get

[TABLE]

From this we deduce that for all $t\in(0,T)$

[TABLE]

From (5.26) we have

[TABLE]

Therefore, using $g_{1}\in H^{1}(D)^{3}$ we get

[TABLE]

As ${\partial\over\partial t}{\rm curl\,}u_{0}\in L^{\infty}(0,T;H^{s}(D)^{3})$, ${\partial\over\partial t}u_{0}\in L^{\infty}(0,T;H^{s}(D)^{3})$ and $g_{1}\in H^{1}(D)^{3}$, we deduce that

[TABLE]

Therefore

[TABLE]

From (5.31), (5.34) and (5.35), we deduce that

[TABLE]

We note that

[TABLE]

so from (5.21)

[TABLE]

Thus we deduce from (5.22) that

[TABLE]

We further have

[TABLE]

As

[TABLE]

we have that

[TABLE]

Using

[TABLE]

we deduce that

[TABLE]

From (5.36), (5.37) and (5.38), we get

[TABLE]

$\Box$

5.2 Corrector for multiscale problem

For the case of more than two scales, we cannot deduce an explicit homogenization error. However, we can deduce correctors for the case where $\varepsilon_{i-1}/\varepsilon_{i}$ is an integer for all $i=2,\ldots,n$. We use the operator ${\cal T}_{n}^{\varepsilon}$ and ${\cal U}_{n}^{\varepsilon}$ defined as follows. We define the map

[TABLE]

for $\phi\in L^{1}(D)$ extended to 0 outside $D$. Letting $D^{\varepsilon_{1}}$ be the $2\varepsilon_{1}$ neighbourhood of $D$, we have

[TABLE]

for all $\phi\in L^{1}(D)$. If a sequence $\{\phi^{\varepsilon}\}_{\varepsilon}$ $(n+1)$-scale converges to $\phi(x,y_{1},\ldots,y_{n})$, then

[TABLE]

in $L^{2}(D\times Y_{1}\times\ldots\times Y_{n})$. We define the operator ${\cal U}_{n}^{\varepsilon}$

[TABLE]

for all functions $\Phi\in L^{1}(D\times{\bf Y})$. For each function $\Phi\in L^{1}(D\times{\bf Y})$ we have

[TABLE]

The proofs for (5.39) and (5.40) may be found in [7]. We then have:

[TABLE]

and

[TABLE]

in $L^{2}(D\times{\bf Y})$ when $\varepsilon\to 0$.

We have the following result.

Theorem 5.3

Assume that $g_{0}=0$, $g_{1}\in W$ and $f\in H^{1}(0,T;H)$. We have

[TABLE]

Proof  We consider

[TABLE]

We note that

[TABLE]

where we have used the energy formula for wave equation (see Lions and Magenes [11]) and the initial condition $g_{0}=0$. Using (5.41) and (5.42), we have

[TABLE]

From (3.2) and (3.9), we have

[TABLE]

From (3.1) and (3.3), this equals

[TABLE]

Continuing this process, this expression equals

[TABLE]

On the other hand we have

[TABLE]

Therefore

[TABLE]

From (3.8), we get

[TABLE]

From (3.4) we have

[TABLE]

Thus

[TABLE]

We consider

[TABLE]

Continuing this, we get

[TABLE]

Thus

[TABLE]

We show that the convergence is uniform. To make the notation concise, we denote by

[TABLE]

We consider

[TABLE]

We note that $\|{\partial u^{\varepsilon}\over\partial t}\|_{L^{\infty}(0,T;H)}$ is uniformly bounded for all $\varepsilon$, ${\partial u_{0}\over\partial t}\in L^{\infty}(0,T;H)$, and from (3.12), ${\partial\over\partial t}\nabla_{y_{i}}\mathfrak{u}_{i}\in L^{\infty}(0,T;L^{2}(D\times{\bf Y}_{i})^{3})$. Further

[TABLE]

so

[TABLE]

From (5.5), we have that ${\partial^{2}u^{\varepsilon}\over\partial t^{2}}$ is uniformly bounded in $L^{2}(0,T;H)$. By a similar argument using the compatible initial condition, we show that ${\partial^{2}u_{0}\over\partial t^{2}}\in L^{2}(0,T;H)$ which implies that ${\partial^{2}\over\partial t^{2}}\nabla_{y_{i}}\mathfrak{u}_{i}\in L^{2}(0,T;L^{2}(D\times{\bf Y}_{i})^{3})$. We then have

[TABLE]

By the same argument, we have similar estimates for other terms in (5.43). We can perform similarly for the other terms in $E^{\varepsilon}(t)$. From the Arzelà-Ascoli theorem, we deduce that $E^{\varepsilon}(t)$ converges to 0 when $\varepsilon\to 0$ uniformly for all $t\in[0,T]$. The conclusion of the proposition follows from the fact that

[TABLE]

and

[TABLE]

$\Box$

Acknowledgement The authors gratefully acknowledge a postgraduate scholarship of Nanyang Technological University, the AcRF Tier 1 grant RG30/16, the Singapore A*Star SERC grant 122-PSF-0007 and the AcRF Tier 2 grant MOE 2013-T2-1-095 ARC 44/13.