\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 16, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/16\hfil Darcy-type law]
{Darcy-type law associated to an optimal control problem}

\author[T. Muthukumar, A. K. Nandakumaran\hfil EJDE-2008/16\hfilneg]
{T. Muthukumar, A. K. Nandakumaran}  % in alphabetical order

\address{T. Muthukumar \newline
Department of Mathematics\\
Indian Institute of Science \\
Bangalore-560012, India}
\email{tmk@math.iisc.ernet.in}

\address{ A. K. Nandakumaran \newline
Department of Mathematics\\
Indian Institute of Science \\
Bangalore-560012, India}
\email{nands@math.iisc.ernet.in}

\thanks{Submitted August 24, 2007. Published February 1, 2008.}
\subjclass[2000]{35B27, 49J20, 76D07}
\keywords{Homogenization; two-scale convergence; stokes
equation; \hfill\break\indent optimal control; porous medium}

\begin{abstract}
 The aim of this paper is to study the asymptotic behaviour
 (homogenization) of an optimal control problem in a periodically
 perforated domain with Dirichlet condition on the boundary of the
 holes. The optimal control problem considered here is governed by
 the Stokes system.  The holes are assumed to be of the same order
 as that of the period. The homogenized limit of the Stokes system
 as well as its adjoint system arising from the optimal control
 problem is obtained. The convergence of the optimal control and
 cost functional is obtained on some specific control sets.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

It is now well known that the fluid flow (governed by Stokes or
Navier-Stokes equations) in a periodically perforated domain (with a large
number of small holes) behaves differently according to the size
of the holes (say radius) compared to the period of the
distribution. In fact, there is a critical size of the holes,
where system behaves like a `Brinkman-type' flow. Further,
when the holes are much smaller, in order, than the critical one,
the flow is the standard Stokes or Navier-Stokes. Lastly, when the
holes are comparable to the period, then the system tends to a
`Darcy-type' law situation (cf.
\cite{allairestokes,allairens1,allairens2,tartarstokes,akn}).

In this article, we consider the Stokes system (it can be
carried to non-linear Navier-Stokes system as well) when the
holes and period are of the same order with a control acting on
the system and an associated cost functional. Some results
regarding the critical and subcritical case were proved in
\cite{sjpzou}. Substantial study had been carried out for optimal
control problems under various situation where the state is
determined by second order elliptic operators (cf .
\cite{sksjphocp,sksjppd,sksjplccp,rajesh,tmk,sktmk2,tmkakn}).

We shall now give some preliminaries and set the environment of
the paper.

Let $n\ge 2$ and $\Omega$ be a bounded open set in
$\mathbb{R}^n$. Let $\{e_1, e_2, \ldots, e_n\}$ be the canonical
basis of $\mathbb{R}^n$. Let $Y=(-1,1)^n$ be the reference cell
and $S$ be an open subset of $Y$, such that $S$ is contained in an
open ball of radius $\alpha$ ($0 < \alpha <1$) centered at the
origin. Let $Y' = Y\setminus \bar{S}$ be the fluid part and
$\bar{S}$ be the solid (or obstacle) part.  If $S+\mathbb{Z}^n =
\{x+k\mid x\in S$ and $k\in\mathbb{Z}^n\}$, then, for a
parameter $\varepsilon>0$ tending to zero, we set $S_\varepsilon =
\varepsilon(S+\mathbb{Z}^n)$. We define
\[
\Omega_\varepsilon = \Omega\cap(\mathbb{R}^n\setminus\bar{S_\varepsilon}).
\]
Observe that $\Omega_\varepsilon$ is bounded and we assume that $\Omega_\varepsilon$ is
connected and its boundary, $\partial\Omega_\varepsilon $, is of Lipschitz type.

In a more general situation, if $a_\varepsilon$ denotes the size (say,
diameter) of the obstacle $S_\varepsilon$ distributed periodically, then
observe that under the above setting $S_\varepsilon$ is exactly of order
$\varepsilon$, \emph{i.e.}, $\lim_{\varepsilon\to 0}(a_\varepsilon/\varepsilon)>0$.
Following the convention of Allaire \cite{allairens2}, we define
$\sigma_\varepsilon$ as the ratio between the actual size of the obstacles
and the critical size:
\[
\sigma_\varepsilon =
\begin{cases}
\varepsilon\left|\log\left(\frac{a_\varepsilon}{\varepsilon}\right)\right|^{1/2} & \text{for } n=2,\\
\left(\frac{\varepsilon^n}{a_\varepsilon^{n-2}}\right)^{1/2} & \text{for } n\ge 3.
\end{cases}
\]
Then, in our setting, we have $\sigma_\varepsilon \to 0$.

The inner-product in $(L^2(\Omega))^n$ is given as:
\[
\langle \boldsymbol{u},\boldsymbol{v}\rangle = \int_\Omega
\boldsymbol{u}.\boldsymbol{v}\,dx = \sum_{i=1}^n
\int_\Omega u_i v_i\,dx,\quad \forall
\boldsymbol{u},\boldsymbol{v}\in (L^2(\Omega))^n.
\]
We shall denote the norm in $(L^2(\Omega))^n$ by $\|\cdot\|_2$ and
the norm in $(L^2(\Omega_\varepsilon))^n$ by $\|\cdot\|_{2,\Omega_\varepsilon}$. For a function $g$
defined on $\Omega_\varepsilon$, we shall denote by $\tilde{g}$ its extension
by zero on $\Omega\cap S_\varepsilon$. The symbol $C$ will always denote a generic positive
constant independent of $\varepsilon$.

Let $a,b$ be given constants such that $0 < a \le b$.  Let
$B=B(y)$ be a $n\times n$ matrix with entries from $L^\infty(Y)$,
$Y$-periodic in $y$ satisfying
\[
a|\xi|^2 \leq B(y)\xi.\xi \leq b|\xi|^2 \quad \text{ a.e. in } y,
\quad \forall \xi=(\xi_i)\in \mathbb{R}^n.
\]
In addition, we assume that $B$ is symmetric. The symmetry assumption will
not play any role in the homogenization process and is inherited
from the optimal control problem.

Let $U_\varepsilon$ be a closed convex subset of $(L^2(\Omega_\varepsilon))^n$, called
the admissible control set and $\boldsymbol{f}\in(L^2(\Omega))^n$. Let $\nu > 0$ be
the cost of the control independent of $\varepsilon$. Given
$\boldsymbol{\theta}_\varepsilon\in U_\varepsilon$, let the cost functional $J_\varepsilon$ be defined as,
\begin{equation}\label{eq:gencstfn}
J_\varepsilon(\boldsymbol{\theta}_\varepsilon)=
\frac{1}{2}\int_{\Omega_\varepsilon}B\left(\frac{x}{\varepsilon}\right)\nabla\boldsymbol{u}_\varepsilon.\nabla\boldsymbol{u}_\varepsilon\,dx
+ \frac{\nu}{2}\|\boldsymbol{\theta}_\varepsilon\|^2_{2,\Omega_\varepsilon}
\end{equation}
where $\boldsymbol{u}_\varepsilon=\boldsymbol{u}_\varepsilon(\boldsymbol{\theta}_\varepsilon)\in (H_0^1(\Omega_\varepsilon))^n$ is the state in
the unique pair, $(\boldsymbol{u}_\varepsilon, p_\varepsilon)$, of solution of the Stokes
equation
\begin{equation}\label{eq:gensteqn}
 \begin{gathered}
\nabla p_\varepsilon -\Delta\boldsymbol{u}_\varepsilon   =  \boldsymbol{f} + \boldsymbol{\theta}_\varepsilon \quad
\text{in $\Omega_\varepsilon$},\\
\mathop{\rm div}(\boldsymbol{u}_\varepsilon)   =  0  \quad \text{in $\Omega_\varepsilon$},\\
\boldsymbol{u}_\varepsilon   =  0  \quad \text{on $\partial\Omega_\varepsilon $}.
\end{gathered}
\end{equation}
The pressure $p_\varepsilon$ is unique up to an additive constant and thus belongs
to $L^2(\Omega_\varepsilon)/\mathbb{R}$.
It is a classical result from the calculus of variations that
there exists a unique $\boldsymbol{\theta}_\varepsilon^*\in U_\varepsilon$ such that
\begin{equation}\label{eq:min}
J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*)
=\min_{\boldsymbol{\theta}_\varepsilon\in U_\varepsilon}J_\varepsilon(\boldsymbol{\theta}_\varepsilon).
\end{equation}

The system \eqref{eq:gencstfn}--\eqref{eq:gensteqn} was considered
by Saint Jean Paulin and Zoubairi (cf. \cite{sjpzou}) in the
abstract framework introduced by Allaire in \cite{allairens1,
allairens2}. They had considered the admissible control sets
$U_\varepsilon$ to be of obstacle-type. They studied the cases when
$\lim_{\varepsilon\to 0} \sigma_\varepsilon > 0$ and $\lim_{\varepsilon\to
0} \sigma_\varepsilon = \infty$. This, precisely, represents the critical
and subcritical case, where the holes are much smaller.
However, they were unable to conclude concerning the case
$\lim_{\varepsilon\to 0} \sigma_\varepsilon= 0$. We remark that the
homogenization in the comparable case is always different even in
Dirichlet problem of Laplacian.

In this article, we study the asymptotic behaviour of the optimal
control problem \eqref{eq:gencstfn}--\eqref{eq:gensteqn}, when
$\lim_{\varepsilon\to 0} \sigma_\varepsilon = 0$. Our method is based on
two-scale convergence. We end this section by recalling the notion
of two-scale convergence. We refer to
\cite{guetseng,allaire2s,akn,lukguet} for a detailed study of the
same and certain applications.

\begin{definition} \label{def1.1} \rm
A sequence of functions $\{v_\varepsilon\}$ in $L^2(\Omega)$ is said to
\textit{two-scale converge} to a limit $v\in L^2(\Omega\times Y)$
(denoted as $v_\varepsilon \stackrel{2\text{s}}{\rightharpoonup} v$) if
\[
\int_\Omega v_\varepsilon \phi\left(x,\frac{x}{\varepsilon}\right)\,dx \to
\int_\Omega \int_Y v(x,y)\phi(x,y)\,dy\,dx, \quad \forall \phi\in
L^2[\Omega; C_{\text{per}}(Y)].
\]
\end{definition}

The most interesting property of two-scale convergence is the
following compactness result.

\begin{theorem}\label{thm:2s}
For any bounded sequence $v_\varepsilon$ in $L^2(\Omega)$, there exists a
subsequence and $v\in L^2(\Omega\times Y)$ such that, $v_\varepsilon$
two-scale converges to $v$ along the subsequence.
\end{theorem}

The approach of the article is as follows: In the next section,
we introduce the adjoint problem associated to the optimal
control problem \eqref{eq:gencstfn}--\eqref{eq:gensteqn} and
homogenize the state-adjoint system. This yields a Darcy-type
result for the adjoint state as well. Finally, we consider
special situations where the optimal control problem can be
homogenized.

\section{Homogenization Process}\label{sec:stokes}

We begin by stating a lemma on the Poincar\'{e} inequality
proved by Tartar (cf. \cite{tartarstokes}) when the size of the
obstacles $a_\varepsilon$ are exactly of the order of $\varepsilon$.

\begin{lemma}\label{lm:tar}
There exists a positive constant $C$, independent of $\varepsilon$, such
that \[
\|v\|_{2,\Omega_\varepsilon} \le C\varepsilon \|\nabla v\|_{2,\Omega_\varepsilon}, \quad \forall
v\in H^1_0(\Omega_\varepsilon).
\]
\end{lemma}

Let $\boldsymbol{\theta}_\varepsilon^*$ be the unique optimal control of the system
\eqref{eq:gencstfn}--\eqref{eq:gensteqn}. Let $(\boldsymbol{u}_\varepsilon^*,
p_\varepsilon^*)\in (H^1_0(\Omega_\varepsilon))^n\times L^2(\Omega_\varepsilon)/\mathbb{R}$ be the
state and pressure corresponding to $\boldsymbol{\theta}_\varepsilon^*$ given by the system
of equations:
\begin{equation}\label{eq:optsteqn}
\begin{gathered}
\nabla p_\varepsilon^* -\Delta\boldsymbol{u}_\varepsilon^*   =  \boldsymbol{f} + \boldsymbol{\theta}_\varepsilon^* \quad \text{in
$\Omega_\varepsilon$},\\
\mathop{\rm div}(\boldsymbol{u}_\varepsilon^*)   =  0  \quad \text{in $\Omega_\varepsilon$},\\
\boldsymbol{u}_\varepsilon^*  =  0  \quad \text{on $\partial\Omega_\varepsilon $}.
\end{gathered}
\end{equation}
We introduce the adjoint optimal state associated to the optimal
control problem. Let
$(\boldsymbol{v}_\varepsilon^*, q_\varepsilon^*)\in (H^1_0(\Omega_\varepsilon))^n\times L^2(\Omega_\varepsilon)/\mathbb{R}$
 be the solution of
\begin{equation}\label{eq:adeqn}
 \begin{gathered}
\nabla q_\varepsilon^* -\Delta\boldsymbol{v}_\varepsilon^*   =
-\mathop{\rm div}\left(B\left(\frac{x}{\varepsilon}\right)\nabla\boldsymbol{u}_\varepsilon^*\right) \quad \text{in
$\Omega_\varepsilon$},\\
\mathop{\rm div}(\boldsymbol{v}_\varepsilon^*)  =  0  \quad \text{in $\Omega_\varepsilon$},\\
\boldsymbol{v}_\varepsilon^*   =  0  \quad \text{on $\partial\Omega_\varepsilon $}.
\end{gathered}
\end{equation}
Then the optimality condition, in terms of the adjoint optimal
state, is
\begin{equation}\label{eq:optcdn}
\int_{\Omega_\varepsilon} (\boldsymbol{v}_\varepsilon^* + \nu\boldsymbol{\theta}_\varepsilon^*).(\boldsymbol{\theta}_\varepsilon - \boldsymbol{\theta}_\varepsilon^*)\,dx \ge 0
\quad \forall \boldsymbol{\theta}_\varepsilon\in U_\varepsilon.
\end{equation}
Note that the symmetry hypothesis on $B$ comes in hand to
derive the optimality condition \eqref{eq:optcdn}.

\begin{lemma}\label{lm:bds}
Let $0\in U_\varepsilon$ for all $\varepsilon$, then
$\{\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*}\},
\{\varepsilon^{-2}\widetilde{\boldsymbol{u}_\varepsilon^*}\},
\{\varepsilon^{-2}\widetilde{\boldsymbol{v}_\varepsilon^*}\},
\{\varepsilon^{-1}\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}\}$ and
$\{\varepsilon^{-1}\nabla\widetilde{\boldsymbol{v}_\varepsilon^*}\}$ are bounded in
$(L^2(\Omega))^n$.
\end{lemma}

\begin{proof}
Let $\boldsymbol{w}_\varepsilon$ be the state corresponding to the control
$\boldsymbol{\theta}_\varepsilon=0$ in \eqref{eq:gensteqn}. Then it is easy to observe,
using Lemma~\ref{lm:tar}, that
\[
\|\nabla \boldsymbol{w}_\varepsilon\|_{2, \Omega_\varepsilon} \le C\varepsilon \|f\|_2.
\]
Since $J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*) \le J_\varepsilon(0)$, we deduce that
\[
\|\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*}\|^2_2 \le C \|f\|^2_2
\]
and similarly, we
also deduce that
\[
\|\varepsilon^{-1}\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}\|^2_2 \le C \|f\|^2_2.
\]
Now using, Lemma~\ref{lm:tar}, we get
$\|\varepsilon^{-2}\widetilde{\boldsymbol{u}_\varepsilon^*}\|^2_2$ is bounded. Hence,
using $\varepsilon^{-2}\boldsymbol{v}_\varepsilon^*$ as a test function in the adjoint state
\eqref{eq:adeqn}, we deduce the respective bounds of the adjoint
state.
\end{proof}

It follows from the above lemma that the optimal controls
$\boldsymbol{\theta}_\varepsilon^*$ converge to zero strongly in 
$(L^2(\Omega))^n$, $\boldsymbol{u}_\varepsilon^*$
converge to zero strongly in $(H^1_0(\Omega))^n$ and the cost
functional $J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*) \to 0$. Our interest is to get
information on the further terms on the asymptotic expansion of
these quantities. In fact, there exists
$\boldsymbol{\theta}^*\in(L^2(\Omega))^n$ such that
$\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \rightharpoonup
\boldsymbol{\theta}^*$ weakly in $(L^2(\Omega))^n$. More precisely, our
objective is to identify the role of $\boldsymbol{\theta}^*$ in
the non-zero homogenized limit of the system
\eqref{eq:gencstfn}--\eqref{eq:gensteqn}.

We now state a lemma proved in \cite{tartarstokes,allairestokes}.

\begin{lemma}\label{lm:ext}
There exists a restriction operator $R_\varepsilon: (H^1_0(\Omega))^n
\to (H^1_0(\Omega_\varepsilon))^n$ such that
\begin{itemize}
\item[(i)] $u\in (H^1_0(\Omega_\varepsilon))^n  \Rightarrow R_\varepsilon\widetilde{u} = u$ in
$\Omega_\varepsilon$
\item[(ii)] $\mathop{\rm div}(u) = 0 \Rightarrow \mathop{\rm div}(R_\varepsilon u) = 0$
in $\Omega_\varepsilon$
\item[(iii)] $\|\nabla(R_\varepsilon u)\|_{2,\Omega_\varepsilon} \le C \left[\frac{1}{\varepsilon}
\|u\|_2 + \|\nabla u\|_2\right]$.
\end{itemize}
\end{lemma}

The above lemma is used to prove the following result (cf.
\cite{tartarstokes,allairens2}).

\begin{lemma}
Let $0\in U_\varepsilon$. Then there exists $P^*_\varepsilon$ and $Q^*_\varepsilon$ in
$L^2(\Omega)/\mathbb{R}$ such that $P^*_\varepsilon = p^*_\varepsilon$ and $Q^*_\varepsilon =
q^*_\varepsilon$ in $\Omega_\varepsilon$,  and both $P^*_\varepsilon$ and $Q^*_\varepsilon$ are bounded
in $L^2(\Omega)/\mathbb{R}$.
\end{lemma}

We deduce from the \emph{a priori} estimates obtained in above
lemmas that there exists $\boldsymbol{u}_0^*(x,y),
\boldsymbol{v}^*_0(x,y)$ in $(L^2(\Omega\times Y))^n$,
$\boldsymbol{\xi}^*_0(x,y), \boldsymbol{\zeta}^*_0(x,y)$ in
$(L^2(\Omega\times Y))^{n\times n}$ and  $p_0^*(x,y), q_0^*(x,y)$
in $L^2(\Omega\times Y)/\mathbb{R}$ such that, up to a subsequence (cf.~
Theorem~\ref{thm:2s}),
\begin{gather*}
\varepsilon^{-2}\widetilde{\boldsymbol{u}_\varepsilon^*} \stackrel{2\text{s}}{\rightharpoonup}
\boldsymbol{u}_0^*(x,y),\quad  \varepsilon^{-2}\widetilde{\boldsymbol{v}_\varepsilon^*}
\stackrel{2\text{s}}{\rightharpoonup} \boldsymbol{v}_0^*(x,y),\\
\varepsilon^{-1}\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}
\stackrel{2\text{s}}{\rightharpoonup}
\boldsymbol{\xi}^*_0(x,y),\quad \varepsilon^{-1}\nabla\widetilde{\boldsymbol{v}_\varepsilon^*}
\stackrel{2\text{s}}{\rightharpoonup}  \boldsymbol{\zeta}^*_0(x,y),\\
P^*_\varepsilon \stackrel{2\text{s}}{\rightharpoonup}
p_0^*(x,y),\quad  Q^*_\varepsilon \stackrel{2\text{s}}{\rightharpoonup}
q_0^*(x,y).
\end{gather*}

Let $\boldsymbol{u}^*(x) = \frac{1}{|Y|}\int_Y
\boldsymbol{u}^*_0(x,y)\,dy$ and $\boldsymbol{v}^*(x) =\frac{1}{|Y|}\int_Y \boldsymbol{v}^*_0(x,y)\,dy$. It is a known
fact from the two-scale convergence theory that, for the same
subsequence,
\begin{gather*}
\varepsilon^{-2}\widetilde{\boldsymbol{u}_\varepsilon^*} \rightharpoonup \boldsymbol{u}^*
\text{ weakly in } (L^2(\Omega))^n,\\
\varepsilon^{-2}\widetilde{\boldsymbol{v}_\varepsilon^*} \rightharpoonup \boldsymbol{v}^*
\text{ weakly in } (L^2(\Omega))^n.
\end{gather*}
The extension of the pressures are not the trivial extension by
zero and it was observed in \cite{tartarstokes,allairens2} that, for
the same subsequence, $P^*_\varepsilon$, $Q_\varepsilon^*$ converges strongly in
$L^2(\Omega)/\mathbb{R}$ and, in fact, we get those limits as
\begin{gather*}
P^*_\varepsilon \to \frac{1}{|Y|}\int_Y p_0^*(x,y)\,dy \quad\text{
strongly in } L^2(\Omega)/\mathbb{R},\\
Q^*_\varepsilon \to \frac{1}{|Y|}\int_Y q_0^*(x,y)\,dy \quad\text{
strongly in } L^2(\Omega)/\mathbb{R}.
\end{gather*}

\begin{remark}\label{rmrk:div} \rm
Given $\mathop{\rm div}(\boldsymbol{u}_\varepsilon^*)=0$ in $\Omega_\varepsilon$ implies that
$\mathop{\rm div}(\widetilde{\boldsymbol{u}_\varepsilon^*})=0$ in $\Omega$, and hence
$\mathop{\rm div}(\boldsymbol{u}^*)=0$ in $\Omega$ and
$\boldsymbol{u}^*\cdot n = 0$ in $\partial\Omega$,
where $n$ is the unit outward normal. Similarly,
$\mathop{\rm div}(\boldsymbol{v}^*)=0$ in $\Omega$ and
$\boldsymbol{v}^*\cdot n = 0$ in $\partial\Omega$ (cf.
\cite{sanpalen,allairestokes}).
\end{remark}

We shall now define some cell problems which will be used in the
sequel to identify the limit problem. For $1\le i \le n$, let the
function $(\boldsymbol{\mu}_i,\rho_i)\in
(H^1_{\mathrm{per}}(Y'))^n\times
L_{\mathrm{per}}^2(Y')/\mathbb{R}$ be the solution of the cell problem
\begin{equation}\label{eq:percelleqn}
\begin{gathered}
\nabla_y\rho_i - \Delta_y\boldsymbol{\mu}_i  = e_i \quad \text{in $Y'$}\\
%\mathop{\rm div}_y(\boldsymbol{\mu}_i) = 0 \quad \text{ in } Y'\\
\mathrm{div}_y(\boldsymbol{\mu}_i) = 0 \quad \text{ in } Y'\\
\boldsymbol{\mu_i}  = 0 \quad \text{ on }
\partial Y'\setminus\partial Y\\
\boldsymbol{\mu}_i \text{ and  $\rho_i$  are  $Y$-periodic}.
\end{gathered}
\end{equation}

Let $(\boldsymbol{\chi}_i,\lambda_i)\in
(H^1_{\mathrm{per}}(Y'))^n\times
L_{\mathrm{per}}^2(Y')/\mathbb{R}$ be the solution of the adjoint cell problem
\begin{equation}\label{eq:adpercelleqn}
 \begin{gathered}
\nabla_y\lambda_i - \Delta_y\boldsymbol{\chi}_i  =
-\mathrm{div}_y\left(B(y)\nabla_y\boldsymbol{\mu}_i\right)
\quad  \text{in $Y'$}\\
\mathrm{div}_y(\boldsymbol{\chi}_i)  = 0 \quad \text{ in } Y'\\
\boldsymbol{\chi_i}  = 0 \quad \text{on } \partial Y'\setminus\partial Y\\
\boldsymbol{\chi}_i \text{ and  $\lambda_i$   are $Y$-periodic}.
\end{gathered}
\end{equation}

We extend $\boldsymbol{\mu}_i$ and $\boldsymbol{\chi}_i$ by zero
to $Y\setminus Y'$ and use the same notation for the extension.
Let $M$ and $N$ be the $n\times n$ matrices defined as follows:
\begin{gather*}
M e_i = \frac{1}{|Y|}\int_Y \boldsymbol{\mu}_i(y)\,dy, \\
N e_i = \frac{1}{|Y|}\int_Y \boldsymbol{\chi}_i(y)\,dy.
\end{gather*}
The matrix $M$ is standard in the homogenization of Stokes system
in perforated domain and it is known that $M$ is both symmetric
and positive definite (cf. \cite{sanpalen}). We have to establish
similar results for $N$.

\begin{lemma} \label{lem2.6}
If $B$ is symmetric, then the matrix $N$ is symmetric.
\end{lemma}

\begin{proof}
Using $\boldsymbol{\chi}_j$ as a test function in \eqref{eq:percelleqn}, we
get
\[
\int_Y \nabla_y \boldsymbol{\mu}_i\cdot \nabla_y
\boldsymbol{\chi}_j\,dy = \int_Y e_i. \boldsymbol{\chi}_j\,dy
= |Y|\langle e_i, N e_j\rangle.
\]
Now, using $\boldsymbol{\mu}_i$ as a test function in \eqref{eq:adpercelleqn}
corresponding to the index $j$,
we get
\[
\int_Y \nabla_y \boldsymbol{\mu}_i. \nabla_y
\boldsymbol{\chi}_j\,dy = \int_Y B(y)\nabla_y \boldsymbol{\mu}_j.
\nabla_y\boldsymbol{\mu}_i\,dy.
\]
Thus,
\[
|Y|\langle e_i, N e_j\rangle = \int_Y B(y)\nabla_y
\boldsymbol{\mu}_j.
\nabla_y\boldsymbol{\mu}_i\,dy.
\]
Similarly, interchanging the role of $i$ and $j$ in the above
argument and using the fact that scalar product commutes, we deduce
\[
|Y|\langle e_j, N e_i\rangle = \int_Y B(y)\nabla_y
\boldsymbol{\mu}_i.
\nabla_y\boldsymbol{\mu}_j\,dy.
\]
Hence, if $B$ is symmetric, then $N$ is symmetric.
\end{proof}

\begin{lemma} \label{lem2.7}
If $B$ is positive definite, then $N$ is positive definite.
\end{lemma}

\begin{proof}
Let $\xi\in\mathbb{R}^n$. Define
$(\boldsymbol{\alpha}_\xi,\gamma_1)\in
(H^1_{\mathrm{per}}(Y'))^n\times
L_{\mathrm{per}}^2(Y')/\mathbb{R}$ as the solution of the problem
\begin{equation}\label{eq:pdeqn}
 \begin{gathered}
\nabla_y\gamma_1 - \Delta_y\boldsymbol{\alpha}_\xi  = \xi \quad
 \text{in $Y'$}\\
\mathrm{div}_y(\boldsymbol{\alpha}_\xi)  = 0 \quad \text{in } Y'\\
\boldsymbol{\alpha_\xi}  = 0 \quad \text{on }
\partial Y'\setminus\partial Y\\
\boldsymbol{\alpha}_\xi \text{ and $\gamma_1$  are  $Y$-periodic}.
\end{gathered}
\end{equation}
Also, let $(\boldsymbol{\beta}_\xi,\gamma_2)\in
(H^1_{\mathrm{per}}(Y'))^n\times
L_{\mathrm{per}}^2(Y')/\mathbb{R}$ be the solution of the
corresponding adjoint problem
\begin{equation}\label{eq:adpdeqn}
 \begin{gathered}
\nabla_y\gamma_2 - \Delta_y\boldsymbol{\beta}_\xi  =
-\mathrm{div}_y\left(B(y)\nabla_y\boldsymbol{\alpha}_\xi\right) \quad
 \text{in $Y'$}\\
\mathrm{div}_y(\boldsymbol{\beta}_\xi)  = 0 \quad \text{in } Y'\\
\boldsymbol{\beta_\xi}  = 0 \quad \text{on } \partial Y'\setminus\partial Y\\
\boldsymbol{\beta}_\xi \text{ and $\gamma_2$  are $Y$-periodic}.
\end{gathered}
\end{equation}
We extend $\boldsymbol{\alpha}_\xi$ and $\boldsymbol{\beta}_\xi$ by zero
to $Y\setminus Y'$ and use the same notation for the extension.
Observe that $\boldsymbol{\alpha}_\xi = \sum_{i=1}^n \xi_i
\boldsymbol{\mu}_i$ and $\boldsymbol{\beta}_\xi = \sum_{i=1}^n
\xi_i \boldsymbol{\chi}_i$. Therefore, $\langle N\xi,\xi\rangle =
\frac{1}{|Y|}\int_Y \boldsymbol{\beta}_\xi(y) . \xi\,dy$. Using
$\boldsymbol{\beta}_\xi$ as a test function in \eqref{eq:pdeqn}
and $\boldsymbol{\alpha}_\xi$ as a test function in \eqref{eq:adpdeqn},
we deduce,
\begin{align*}
\langle N\xi,\xi\rangle & =  \frac{1}{|Y|}\int_Y
\boldsymbol{\beta}_\xi(y) . \xi\,dy\\
& =  \frac{1}{|Y|}\int_Y \nabla_y \boldsymbol{\beta}_\xi(y)
.\nabla_y \boldsymbol{\alpha}_\xi(y) \,dy\\
& =  \frac{1}{|Y|}\int_Y B(y)\nabla_y \boldsymbol{\alpha}_\xi(y)
.\nabla_y \boldsymbol{\alpha}_\xi(y) \,dy\\
& \ge  \frac{a}{|Y|}\|\nabla_y \boldsymbol{\alpha}_\xi(y)\|^2_2
\quad \ge 0.
\end{align*}
Thus, we have shown that $N$ is positive. It now remains to show
the positive definiteness of $N$. Suppose that $\langle
N\xi,\xi\rangle = 0$. Hence, $\|\nabla_y
\boldsymbol{\alpha}_\xi(y)\|^2_2 = 0$ and consequently
$\boldsymbol{\alpha}_\xi(y)=0$. This implies that
$\nabla_y\gamma_1 = \xi$, but since $\gamma_1$ is $Y$-periodic,
we have $\xi =0$.
\end{proof}

We now provide
the homogenization theorem for the state and adjoint-state
equations. Before doing so, we pause to remark that
$\text{div}_y$ will denote the divergence w.r.t the $y$ variable
and a `$\text{div}$' without subscript will denote the divergence
w.r.t the $x$ variable.

\begin{theorem}\label{thm:bstar}
Let $0\in U_\varepsilon$. If $(\boldsymbol{u}_\varepsilon^*,p^*_\varepsilon)$ and $(\boldsymbol{v}_\varepsilon^*,q^*_\varepsilon)$
are the solution of \eqref{eq:optsteqn} and \eqref{eq:adeqn},
respectively, then there exists $p^*$ and $q^*$ in
$L^2(\Omega)/\mathbb{R}$ such that
\begin{equation}\label{eq:limsteqn}
 \begin{gathered}
\boldsymbol{u}^* = M (\boldsymbol{f}-\nabla p^*) \quad \text{in $\Omega$},\\
\mathop{\rm div}(\boldsymbol{u}^*) =  0  \quad \text{in $\Omega$},\\
\boldsymbol{u}^*\cdot n  =  0  \quad \text{on $\partial\Omega$}
\end{gathered}
\end{equation}
and
\begin{equation}\label{eq:limadeqn}
 \begin{gathered}
\boldsymbol{v}^* = {N^t} (\boldsymbol{f}-\nabla p^*)- M\nabla q^* \quad
 \text{in $\Omega$},\\
\mathop{\rm div}(\boldsymbol{v}^*) =  0  \quad \text{in $\Omega$},\\
\boldsymbol{v}^*\cdot n  =  0  \quad \text{on $\partial\Omega$},
\end{gathered}
\end{equation}
where $N^t$ denotes the transpose of $N$. Further, the following
convergence hold for the entire sequence:
\begin{equation}\label{eq:con}
\begin{gathered}
\varepsilon^{-2}\widetilde{\boldsymbol{u}_\varepsilon^*} \rightharpoonup \boldsymbol{u}^*
\quad \text{weakly in } (L^2(\Omega))^n,\\
\varepsilon^{-2}\widetilde{\boldsymbol{v}_\varepsilon^*} \rightharpoonup \boldsymbol{v}^*
\quad \text{weakly in } (L^2(\Omega))^n,\\
P^*_\varepsilon \to p^* \quad \text{strongly in } L^2(\Omega)/\mathbb{R},\\
Q^*_\varepsilon \to q^* \quad \text{strongly in } L^2(\Omega)/\mathbb{R}
\end{gathered}
\end{equation}
and
\begin{equation}\label{eq:encon}
\varepsilon^{-2}\int_\Omega B\left(\frac{x}{\varepsilon}\right)
\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}\cdot\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}\,dx
\to \int_\Omega
N\left(M^{-1}\boldsymbol{u}^*\right)\cdot
M^{-1}\boldsymbol{u}^*\,dx
\end{equation}
\end{theorem}

\begin{proof}
Let $\Phi(x,y)\in
[\mathcal{D}(\Omega;C^\infty_{\mathrm{per}}(Y))]^{n\times n}$
be such that $\Phi(\cdot, y) = 0$ for all $y\in Y\setminus Y'$.
Then integration by parts will yield,
\[
\int_\Omega
\nabla\widetilde{\boldsymbol{u}_\varepsilon^*}\cdot\Phi\left(x,\frac{x}{\varepsilon}\right)\,dx =
-\int_\Omega
\widetilde{\boldsymbol{u}_\varepsilon^*}\Big[\mathop{\rm
div}\Phi\left(x,\frac{x}{\varepsilon}\right) +
\frac{1}{\varepsilon}
\mathrm{div}_y\Phi(x,\frac{x}{\varepsilon})\Big]\,dx.
\]
Multiplying by $\varepsilon^{-1}$ on both sides of the equality and then
passing to the limit, as $\varepsilon \to 0$, we get,
\[
\int_\Omega \int_Y \boldsymbol{\xi}^*_0(x,y)\cdot\Phi(x,y)\,dx\,dy =
-\int_\Omega \int_Y \boldsymbol{u}^*_0(x,y)\,
\mathrm{div}_y\Phi(x,y)\,dx\,dy.
\]
Since $\Phi$ is arbitrary, we have
$\boldsymbol{\xi}_0^*(x,y)=\nabla_y \boldsymbol{u}_0^*(x,y)$. A similar
argument for the adjoint-state $\boldsymbol{v}_\varepsilon^*$ will yield
$\boldsymbol{\zeta}_0^*(x,y)=\nabla_y \boldsymbol{v}_0^*(x,y)$.

Let $\boldsymbol{\phi}_1,\boldsymbol{\phi}_2\in
[\mathcal{D}(\Omega;\mathcal{D}(Y'))]^{n\times n}$ be such that
$\mathop{\rm div}_y(\boldsymbol{\phi}_2) = 0$. Using
$\varepsilon\boldsymbol{\phi}_1(x,\frac{x}{\varepsilon})
+\boldsymbol{\phi}_2(x,\frac{x}{\varepsilon})$ as a two-scale
test function in \eqref{eq:optsteqn}, we get,
\begin{align*}
&- \int_{\Omega}\int_Y
p^*_0(x,y)\Big[\mathop{\rm div}(\boldsymbol{\phi}_2(x,y)) +
\mathrm{div}_y(\boldsymbol{\phi}_1(x,y))\Big]\,dx\,dy\\
&+ \int_\Omega \int_Y \boldsymbol{\xi}_0^*(x,y)
\nabla_y\boldsymbol{\phi}_2(x,y)\,dx\,dy\\
&= \int_\Omega \int_Y \boldsymbol{f} \boldsymbol{\phi}_2(x,y)\,dx\,dy.
\end{align*}
By putting, $\boldsymbol{\phi}_2 \equiv 0$, we get
\[
- \int_{\Omega}\int_Y p^*_0(x,y)
\mathrm{div}_y(\boldsymbol{\phi}_1(x,y))\,dx\,dy = 0.
\]
Hence, $\nabla_y p^*_0(x,y) = 0$ a.e. and thus there exists a
$p^*\in L^2(\Omega)/\mathbb{R}$ such that $p^*_0(x,y) = p^*(x)$ in
$\Omega \times Y$.

By putting, $\boldsymbol{\phi}_1 \equiv 0$, we get
\begin{align*}
&- \int_{\Omega}\int_Y p^*(x)\mathop{\rm div}(\boldsymbol{\phi}_2(x,y))\,dx\,dy\\
&+ \int_\Omega \int_Y \nabla_y \boldsymbol{u}_0^*(x,y)
\nabla_y\boldsymbol{\phi}_2(x,y)\,dx\,dy\\
&= \int_\Omega \int_Y \boldsymbol{f}
\boldsymbol{\phi}_2(x,y)\,dx\,dy.
\end{align*}
Since $\boldsymbol{\phi}_2$ is such that
$\mathop{\rm div}_y(\boldsymbol{\phi}_2) = 0$, there
exists $p^*_1(x,y)\in L^2(\Omega \times Y)/\mathbb{R}$ such that
\[
-\nabla_y p_1^*(x,y) -\Delta_y \boldsymbol{u}_0^*(x,y) = \boldsymbol{f} -\nabla
p^* \text{ in } \Omega \times Y.
\]
By using the cell problem \eqref{eq:percelleqn} and from the
uniqueness of solution for the Stokes system, we derive
\[
\frac{\partial p_1^*}{\partial y_i} = \nabla_y\rho_i\cdot(\boldsymbol{f} -
\nabla p^*) \text{ in } \Omega\times Y.
\]
and
$\boldsymbol{u}^* = {M^t}(\boldsymbol{f}-\nabla p^*)$   in $\Omega$,
where $M^t$ is the transpose of $M$. But since $M$ is symmetric,
we have
\[
\boldsymbol{u}^* = M(\boldsymbol{f}-\nabla p^*) \textrm{ in $\Omega$},
\]
This combined with the facts mentioned in Remark~\ref{rmrk:div}
gives \eqref{eq:limsteqn}.

Now, using $\varepsilon\boldsymbol{\phi}_1(x,\frac{x}{\varepsilon}) +
\boldsymbol{\phi}_2(x,\frac{x}{\varepsilon})$ as a two-scale test
function in \eqref{eq:adeqn} and following a similar analysis as
before, we deduce that there exists a $q^*\in L^2(\Omega)/\mathbb{R}$
such that $q^*_0(x,y) = q^*(x)$ in $\Omega \times Y$ and there
exists $q^*_1(x,y)\in L^2(\Omega \times Y)/\mathbb{R}$ such that
\[
-\nabla_y q_1^*(x,y) -\Delta_y \boldsymbol{v}_0^*(x,y) =
-\mathrm{div}_y\left(B(y)\nabla_y \boldsymbol{u}_0^*(x,y)\right) -\nabla q^* \text{ in } \Omega \times Y.
\]
By using the cell problem \eqref{eq:adpercelleqn} and from the
uniqueness of solution for the Stokes system, we derive
\[
\frac{\partial q_1^*}{\partial y_i} = \nabla_y\lambda_i\cdot(\boldsymbol{f} -
\nabla p^*) - \nabla_y\rho_i\cdot\nabla q^* \text{ in } \Omega\times Y.
\]
and
\[
\boldsymbol{v}^* = {N^t}(\boldsymbol{f}-\nabla p^*) - {M^t}\nabla q^*
\textrm{ in }\Omega,
\]
where $M^t$ is the transpose of $M$. But since $M$ is symmetric,
we have,
\[
\boldsymbol{v}^* = {N^t}(\boldsymbol{f}-\nabla p^*) - M\nabla q^*
\textrm{ in }\Omega\,.
\]
This combined with the facts mentioned in Remark~\ref{rmrk:div}
gives \eqref{eq:limadeqn}. The uniqueness of $p^*$ and $q^*$,
up to additive constants, can be obtained by solving the Neumann
problem hidden in \eqref{eq:limsteqn} and \eqref{eq:limadeqn},
respectively. Hence the uniqueness of
$\boldsymbol{u}^*$ and $\boldsymbol{v}^*$, and we deduce the
convergence \eqref{eq:con} for the entire sequence.

It now remains to prove \eqref{eq:encon}. Using $\varepsilon^{-2}\boldsymbol{v}_\varepsilon^*$
as a test function in \eqref{eq:optsteqn} and $\varepsilon^{-2}\boldsymbol{u}_\varepsilon^*$
as a test function in \eqref{eq:adeqn}, we deduce that
\[
\varepsilon^{-2}\int_{\Omega_\varepsilon}B\left(\frac{x}{\varepsilon}\right) \nabla\boldsymbol{u}_\varepsilon^*\cdot\nabla\boldsymbol{u}_\varepsilon^*\,dx =
\int_{\Omega_\varepsilon} (\boldsymbol{f} + \boldsymbol{\theta}_\varepsilon^*).(\varepsilon^{-2}\boldsymbol{v}_\varepsilon^*)\,dx.
\]
Now, passing to the limit on the right-hand side of the above
equality, we get
\[
\varepsilon^{-2}\int_{\Omega_\varepsilon}B\left(\frac{x}{\varepsilon}\right) \nabla\boldsymbol{u}_\varepsilon^*\cdot\nabla\boldsymbol{u}_\varepsilon^*\,dx
\to \int_\Omega \boldsymbol{f}. \boldsymbol{v}^*\,dx.
\]
We deduce from the equation of $\boldsymbol{u}^*$ and
$\boldsymbol{v}^*$ in \eqref{eq:limsteqn} and
\eqref{eq:limadeqn}, respectively, that $\boldsymbol{f} =
M^{-1}\boldsymbol{u}^* + \nabla p^*$ and $\boldsymbol{v}^* =
{N^t}M^{-1}\boldsymbol{u}^* - M \nabla q^*$. Thus,
\begin{align*}
\int_\Omega \boldsymbol{f} \boldsymbol{v}^*\,dx & =  \int_\Omega
M^{-1}\boldsymbol{u}^*\cdot\boldsymbol{v}^*\,dx - \int_\Omega p^*
\mathop{\rm div}(\boldsymbol{v}^*)\,dx + \int_{\partial\Omega} p^*
(\boldsymbol{v}^*.n)\,d\sigma\\
& =  \int_\Omega M^{-1}\boldsymbol{u}^*\cdot
{N^t}(M^{-1}\boldsymbol{u}^*)\,dx - \langle
M^{-1}\boldsymbol{u}^*, M \nabla q^*\rangle\\
& =  \int_\Omega M^{-1}\boldsymbol{u}^*\cdot
{N^t}(M^{-1}\boldsymbol{u}^*)\,dx + \int_\Omega q^*
\mathop{\rm div}(\boldsymbol{u}^*)\,dx
 + \int_{\partial\Omega} q^* (\boldsymbol{u}^*.n)\,d\sigma\\
& =  \int_\Omega N(M^{-1}\boldsymbol{u}^*)\cdot
M^{-1}\boldsymbol{u}^*\,dx.
\end{align*}
Thus, we have shown \eqref{eq:encon}.
\end{proof}

We do not have the symmetry hypothesis on $B$ for the above
theorem. This hypothesis will only affect the symmetry property
of $N$ in the above proof.

\begin{remark} \label{rmk2.9} \rm
The limit pressure terms $p^*$ and $q^*$, in fact, are in
$H^1(\Omega)/\mathbb{R}$ since they solve the Neumann problem
hidden in \eqref{eq:limsteqn} and \eqref{eq:limadeqn},
respectively. Moreover, in particular, if $B$ is the identity
matrix, then \eqref{eq:encon} gives back the usual energy
convergence.\qed
\end{remark}

\begin{remark} \label{rmk2.10} \rm
In the above theorem, we have concluded regarding the limit
behaviour of the state-adjoint system. Equation
\eqref{eq:limsteqn} is called the Darcy law and, along with
convergences \eqref{eq:con}, is well-known in the literature.
However, the above theorem is original in the conclusion of
\eqref{eq:limadeqn} and the convergence \eqref{eq:encon} which
generalises the notion of energy convergence. It is an easy
exercise to note that when $B$ is the identity matrix, then
$M=N$, $\boldsymbol{u}_\varepsilon^*=\boldsymbol{v}_\varepsilon^*$ and
$\boldsymbol{u}^*=\boldsymbol{v}^*$. Also,
\[
\|\varepsilon^{-1}\nabla\boldsymbol{u}_\varepsilon^*\|^2_{2,\Omega_\varepsilon}\to \int_\Omega
\boldsymbol{u}^*\cdot M^{-1}\boldsymbol{u}^*\,dx.
\]\qed
\end{remark}

Recall that $\boldsymbol{\theta}^*$ is the weak limit of a
subsequence of $\varepsilon^{-1}\tilde{\boldsymbol{\theta}_\varepsilon^*}$ 
in $(L^2(\Omega))^n$. We
have no means of concluding that $\boldsymbol{\theta}^*$ is the
optimal control of an appropriate limit optimal control problem.
Moreover, we have no result on the convergence of
$\left\{\varepsilon^{-2}J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*)\right\}$, in the general case when
$0\in U_\varepsilon$. However, we do note that $\varepsilon^{-1}J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*)
\to 0$ and
\[
\liminf_{\varepsilon\to 0}\varepsilon^{-2} J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*)
\ge \frac{1}{2}\int_\Omega N\left(M^{-1}\boldsymbol{u}^*\right)\cdot
M^{-1}\boldsymbol{u}^*\,dx +
\frac{\nu}{2}\|\boldsymbol{\theta^*}\|^2_2.
\]

The main difficulty in concluding regarding the optimal control
problem is the disappearance of $\boldsymbol{\theta^*}$ in the
limit state equation \eqref{eq:limsteqn}. We give two situations
in which the convergence of the optimal controls will be obtained
due to extra regularity available on the optimal controls.

\begin{theorem}[unconstrained case]\label{thm:uncons}
Let $U_\varepsilon = (L^2(\Omega_\varepsilon))^n$ and $\boldsymbol{\theta}_\varepsilon^*$ be the minimiser of
the optimal control problem \eqref{eq:gencstfn}--\eqref{eq:gensteqn}
and $(\boldsymbol{u}_\varepsilon^*,p^*_\varepsilon)$ be the corresponding state and pressure.
Then \eqref{eq:limsteqn}, \eqref{eq:limadeqn}, \eqref{eq:con} and
\eqref{eq:encon} holds and, in addition, we have for the entire
sequence,
\begin{gather*}
\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \to 0 \quad \text{strongly in } (L^2(\Omega))^n,\\
\varepsilon^{-2}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \rightharpoonup
\frac{-\boldsymbol{v}^*}{\nu} \quad \text{weakly in } (L^2(\Omega))^n,
\end{gather*}
and
\begin{equation*}
\varepsilon^{-2}J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*) \to \frac{1}{2}\int_\Omega N
\left(M^{-1}\boldsymbol{u}^*\right)\cdot
M^{-1}\boldsymbol{u}^*\,dx.
\end{equation*}
\end{theorem}

\begin{proof}
Since $0\in U_\varepsilon$, by Theorem~\ref{thm:bstar},
\eqref{eq:limsteqn}, \eqref{eq:limadeqn}, \eqref{eq:con} and
\eqref{eq:encon} holds. It follows from \eqref{eq:optcdn} that
$\boldsymbol{\theta}_\varepsilon^*$ is the projection of $\frac{-\boldsymbol{v}_\varepsilon^*}{\nu}$ in
$U_\varepsilon$. Therefore, $\boldsymbol{\theta}_\varepsilon^* = \frac{-\boldsymbol{v}_\varepsilon^*}{\nu}$. Hence,
from the bounds on $\boldsymbol{v}_\varepsilon^*$, we deduce that
$\boldsymbol{\theta^*}=0$. Thus, using Lemma~\ref{lm:bds} (also
the arguments below it), we get for a subsequence,
\begin{gather*}
\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \rightharpoonup 0 \quad
\text{weakly in } (L^2(\Omega))^n,\\
\varepsilon^{-2}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \rightharpoonup
\frac{-\boldsymbol{v}^*}{\nu} \quad \text{weakly in } (L^2(\Omega))^n.
\end{gather*}
In fact, the first convergence is strong, since
$\varepsilon^{-1}\widetilde{\boldsymbol{v}_\varepsilon^*}$ converge to zero strongly in
$(L^2(\Omega))^n$. Moreover, from the uniqueness of \eqref{eq:limsteqn} and
\eqref{eq:limadeqn}, the convergences hold for the entire
sequence. The convergence of $\varepsilon^{-2}J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*)$ follows
from \eqref{eq:encon} and the strong convergence of
$\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*}$.
\end{proof}

\begin{theorem}[Constrained case]
Let $U_\varepsilon$ be the positive cone of $ (L^2(\Omega_\varepsilon))^n$,
\emph{i.e.}, $U_\varepsilon = \left\{\boldsymbol{\theta}\in (L^2(\Omega_\varepsilon))^n \mid
\boldsymbol{\theta} \ge 0 \text{ a.e. in }\Omega_\varepsilon\right\}$ and
$\boldsymbol{\theta}_\varepsilon^*$ be the minimiser of
the optimal control problem \eqref{eq:gencstfn}--\eqref{eq:gensteqn}
and $(\boldsymbol{u}_\varepsilon^*,p^*_\varepsilon)$ be the corresponding state and pressure.
Then \eqref{eq:limsteqn}, \eqref{eq:limadeqn}, \eqref{eq:con} and
\eqref{eq:encon} holds and, in addition, we have for the entire
sequence,
\begin{gather*}
\varepsilon^{-1}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \to 0 \quad\text{strongly in } (L^2(\Omega))^n,\\
\varepsilon^{-2}\widetilde{\boldsymbol{\theta}_\varepsilon^*} \rightharpoonup
\frac{(\boldsymbol{v}^*)^-}{\nu} \quad\text{weakly in } (L^2(\Omega))^n,
\varepsilon^{-2}J_\varepsilon(\boldsymbol{\theta}_\varepsilon^*) \to \frac{1}{2}\int_\Omega N\left(M^{-1}\boldsymbol{u}^*\right)\cdot
M^{-1}\boldsymbol{u}^*\,dx
\end{gather*}
where $(\boldsymbol{v}^*)^-$ is the negative part of
$\boldsymbol{v}^*$.
\end{theorem}

\begin{proof}
The proof is similar to that of Theorem~\ref{thm:uncons}, except
that we now note that $\boldsymbol{\theta}_\varepsilon^* = \frac{(\boldsymbol{v}_\varepsilon^*)^-}{\nu}$ where
$(\boldsymbol{v}_\varepsilon^*)^-$ is the negative part of $\boldsymbol{v}_\varepsilon^*$.
\end{proof}

\begin{remark} \label{rmk2.13} \rm
When $n=2$ or $3$, the results of \S\ref{sec:stokes} easily
carry forward to the situation when the state variable in the
cost functional is determined by the Navier-Stokes equations:
\begin{equation}\label{eq:nssteqn}
 \begin{gathered}
\nabla p_\varepsilon^* + \boldsymbol{u}_\varepsilon^*.\nabla\boldsymbol{u}_\varepsilon^* -\Delta\boldsymbol{u}_\varepsilon^*
 =  \boldsymbol{f} + \boldsymbol{\theta}_\varepsilon^* \quad \text{in $\Omega_\varepsilon$},\\
\mathop{\rm div}(\boldsymbol{u}_\varepsilon^*)   =  0  \quad \text{in $\Omega_\varepsilon$},\\
\boldsymbol{u}_\varepsilon^*   =  0  \quad \text{on $\partial\Omega_\varepsilon $}.
\end{gathered}
\end{equation}
The existence of solution for the above system is known when
$n=2$ or $3$. The {\it a priori} bounds obtained in
Lemma~\ref{lm:bds} remain valid, since
\[
\int_{\Omega_\varepsilon}[(\boldsymbol{u}_\varepsilon^*.\nabla)\boldsymbol{u}_\varepsilon^*].\boldsymbol{u}_\varepsilon^*\,dx =
-\frac{1}{2}\sum_{i=1}^3\int_{\Omega_\varepsilon}(\nabla.\boldsymbol{u}_\varepsilon^*)(u^*_\varepsilon)^2_i\,dx
= 0.
\]
The result of Theorem~\ref{thm:bstar} also remains valid, since
\[
\int_\Omega
[(\tilde{\boldsymbol{u}_\varepsilon^*}.\nabla)\tilde{\boldsymbol{u}_\varepsilon^*}].\left(\varepsilon\boldsymbol{\phi}_1\left(x,\frac{x}{\varepsilon}\right)
+ \boldsymbol{\phi}_2\left(x,\frac{x}{\varepsilon}\right)\right)\,dx
\to 0
\]
as $\varepsilon \to 0$. However, we note that the matrices $M$
and $N$ are the same and are defined as in \eqref{eq:percelleqn}
and \eqref{eq:adpercelleqn}, respectively. In other words, we do
not have the non-linear terms in the definition of the cell
problems \eqref{eq:percelleqn} and \eqref{eq:adpercelleqn}.
\end{remark}

\subsection*{Conclusion}

We have studied the asymptotic behaviour of the optimal control
problem \eqref{eq:gencstfn}--\eqref{eq:gensteqn} when the holes
are large, a situation left open in \cite{sjpzou}. We have
employed the two-scale method to achieve our result. We are
successful in computing the limit of the state-adjoint pair
\eqref{eq:optsteqn}--\eqref{eq:adeqn}, in the general case when
$0\in U_\varepsilon$. However, we are unable to conclude
anything about the optimal control problem in this generality.
Nevertheless, the optimal control problem is completely settled in
the unconstrained case and the positive cone case. It would be
interesting to see other non-trivial admissible control sets in
which the problem could be settled.

\subsection*{Acknowledgement}
The authors wish to thank the University Grants Commission (UGC),
India for their support. The first author would like to thank the
National Board of Higher Mathematics (NBHM), India, for their
financial support. This work was also partially supported by the
CEFIPRA Project No. 3701-1.

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\end{document}