# Sulfate attack in sewer pipes: Derivation of a concrete corrosion model via
two-scale convergence

Tasnim Fatima  Adrian Muntean  Centre for Analysis, Scientific computing and
Applications (CASA),
Department of Mathematics and Computer Science,
Technical University Eindhoven, Eindhoven, The Netherlands.
E-mail<EMAIL_ADDRESS>Centre for Analysis, Scientific computing and
Applications (CASA),
Institute for Complex Molecular Systems (ICMS),
Department of Mathematics and Computer Science,
Technical University Eindhoven, Eindhoven, The Netherlands.
E-mail<EMAIL_ADDRESS>
###### Abstract

We explore the homogenization limit and rigorously derive upscaled equations
for a microscopic reaction-diffusion system modeling sulfate corrosion in
sewer pipes made of concrete. The system, defined in a periodically-perforated
domain, is semi-linear, partially dissipative and weakly coupled via a non-
linear ordinary differential equation posed on the solid-water interface at
the pore level. Firstly, we show the well-posedness of the microscopic model.
We then apply homogenization techniques based on two-scale convergence for an
uniformly periodic domain and derive upscaled equations together with explicit
formulae for the effective diffusion coefficients and reaction constants. We
use a boundary unfolding method to pass to the homogenization limit in the
non-linear ordinary differential equation. Finally, besides giving its strong
formulation, we also prove that the upscaled two-scale model admits a unique
solution.

###### keywords:

Sulfate corrosion of concrete, periodic homogenization, semi-linear partially
dissipative system, two-scale convergence, periodic unfolding method,
multiscale system.

and

##  1 Introduction

This paper treats the periodic homogenization of a semi-linear reaction-
diffusion system coupled with a nonlinear differential equation arising in the
modeling of the sulfuric acid attack in sewer pipes made of concrete. The
concrete corrosion situation we are dealing with here strongly influences the
durability of cement-based materials especially in hot environments leading to
spalling of concrete and macroscopic fractures of sewer pipes. It is
financially important to have a good estimate on the moment in time when such
pipe systems need to be replaced, for instance, at the level of a city like
Los Angeles. To get good such practical estimates, one needs on one side easy-
to-use macroscopic corrosion models to be used for a numerical forecast of
corrosion, while on the other side one needs to ensure the reliability of the
averaged models by allowing them to incorporate a certain amount of
microstructure information. The relevant question is: How much of this
oscillatory-type information is needed to get a sufficiently accurate
description of the heterogeneous reality? Due to the complexity of possible
shapes of the microstructure, averaging concrete materials is far more
difficult than averaging metallic composites with rigorously defined well-
packed structure. In this paper, we imagine our concrete piece to be made of a
periodically-distributed microstructure. Based on this assumption, we provide
here a rigorous justification of the formal asymptotic expansion performed by
us (in [1]) for this reaction-diffusion scenario. Note that in [1] upscaled
models are derived for a more general situation involving a locally-periodic
distribution of perforations111The word ”perforation” is seen here as a
synonym for ”pore” or ”microstructure”.. Locally periodic geometries refer to
a special case of $x$-dependent microstructures, where, inherently, the outer
normals to (microscopic) inner interfaces are dependent on both spatial slow
variable, say $x$, and fast variable, say $y$.

In the framework of this paper, we combine two-scale convergence concepts with
the periodic unfolding of interfaces to pass to the homogenization limit (i.e.
to $\varepsilon\to 0$, where $\varepsilon$ is a small parameter linked to the
relative size of the perforation) for the uniformly periodic case. Here, the
outer normals to the inner interfaces are dependent only on the spatial fast
variable. For more details on the mathematical modeling of sulfate corrosion
of concrete, we refer the reader to [2, 3] (a moving-boundary approach:
numerics and formal matched asymptotics), [4] (a two-scale reaction-diffusion
system modeling sulfate corrosion), as well as to [5], where a nonlinear
Henry-law type transmission condition (modeling $H_{2}S$ transfer across all
air-water interfaces present in this sulfatation problem) is analyzed.
Mathematical background on periodic homogenization can be found in e.g., [6,
7, 8], while a few relevant (remotely resembling) worked-out examples of this
averaging methodology are explained, for instance, in [9, 10, 11, 12, 13, 14].
It is worth noting that, since it deals with the homogenization of a linear
Henry-law setting, the paper [11] is related to our approach. The major
novelty here compared to [11] is that we now need to pass to the limit in a
non-dissipative object, namely a nonlinear ordinary differential equation
(ode). The ode is describing sulfatation reaction at the inner water-solid
interface – place where corrosion localizes. This aspect makes a rigorous
averaging challenging. For instance, compactness-type methods do not work in
the case when the nonlinear ode is posed on $\epsilon$-dependent surfaces. We
circumvent this issue by ”boundary unfolding” the ode. Thus we fix, as
independent of $\epsilon$, the reaction interface similarly as in [15], and
only then we pass to the limit. Alternatively, one could use varifolds (cf.
e.g. [16]), since this seems to be the natural framework for the rigorous
passage to the limit when both the surface measure and the oscillating
sequences depend on $\epsilon$. However, we find the boundary unfolding
technique easier to adapt to our scenario than the varifolds.

Note that here we approach the corrosion problem deterministically. However,
we have reasons to expect that the uniform periodicity assumption can be
relaxed by assuming instead a Birkhoff-type ergodicity of the microstructure
shapes and positions, and hence, the natural averaging context seems to be the
one offered by random fields; see ch. 1, sect. 6 in [17], ch. 8 and 9 in [18],
or [19]. But, methodologically, how big is the overlap between homogenizing
deterministically locally-periodic distributions of microstructures compared
to working in the random fields context? We will treat these and related
aspects elsewhere.

The paper is organized as follows: We start off in section 2 (and continue in
section 3) with the analysis of the microscopic model. In section 4, we obtain
the $\varepsilon$-independent estimates needed for the passage to the limit
$\varepsilon\to 0$. Section 5 contains the main result of the paper: the set
of the upscaled two-scal equations.

##  2 The microscopic model

In this section, we describe the geometry of our array of periodic
microstructures and briefly indicate the most aggressive chemical reaction
mechanism typically active in sewer pipes. Finally, we list the set of
microscopic equations.

###  2.1 Basic geometry

Fig. 1 (i) shows a cross-section of a sewer pipe hosting corrosion. We assume
that the geometry of the porous medium in question consists of a system of
pores periodically distributed inside the three-dimensional cube
$\Omega:=[a,b]^{3}$ with $a,b\in\mathbb{R}$ and $b>a$. The exterior boundary
of $\Omega$ consists of two disjoint, sufficiently smooth parts:
${\Gamma^{N}}$ \- the Neumann boundary and ${\Gamma^{D}}$ \- the Dirichlet
boundary. The reference pore, say $Y:=[0,1]^{3}$, has three pairwise disjoint
connected domains $Y^{s}$, $Y^{w}$ and $Y^{a}$ with smooth boundaries
$\Gamma^{sw}$ and $\Gamma^{wa}$, as shown in Fig. 1 (iii). Moreover,
$Y:=\bar{Y}^{s}\cup\bar{Y}^{w}\cup\bar{Y}^{a}$.

Figure 1: Left: Cross-section of a sewer pipe pointing out one region. Middle:
Periodic approximation of the periodic rectangular domain. Right: Reference
pore configuration.

Let $\varepsilon$ be a sufficiently small scaling factor denoting the ratio
between the characteristic length of the pore $Y$ and the characteristic
length of the domain $\Omega$. Let $\chi^{w}$ and $\chi^{a}$ be the
characteristic functions of the sets $Y^{w}$ and $Y^{a}$, respectively. The
shifted set $Y^{w}_{k}$ is defined by

$Y^{w}_{k}:=Y+\Sigma_{j=0}^{3}k_{j}e_{j}\;\mbox{for}\;\;k=(k_{1},k_{2},k_{3})\in\mathbb{Z}^{3},$

where $e_{j}$ is the $j^{th}$ unit vector. The union of all shifted subsets of
$Y^{w}_{k}$ multiplied by $\varepsilon$ (and confined within $\Omega$) defines
the perforated domain $\Omega^{\varepsilon}$, namely

$\Omega^{\varepsilon}:=\cup_{k\in\mathbb{Z}^{3}}\\{\varepsilon
Y^{w}_{k}\;|\;\varepsilon Y^{w}_{k}\subset\Omega\\}.$

Similarly, $\Omega_{1}^{\varepsilon}$, $\Gamma^{sw}_{\varepsilon}$, and
$\Gamma^{wa}_{\varepsilon}$ denote the union of the shifted subsets (of
$\Omega$) $Y^{a}_{k}$, $\Gamma^{sw}_{k}$, and $\Gamma^{wa}_{k}$ scaled by
$\varepsilon$. Since usually the concrete in sewer pipes is not completely
dry, we decide to take into account a partially saturated porous
material222The solid, water and air parts corresponds to $Y^{s}$, $Y^{w}$ and
$Y^{a}$, respectively.. We assume that every pore has three distinct non-
overlapping parts: a solid part (grain) which is placed in the center of the
pore, the water film which surrounds the solid part, and an air layer bounding
the water film and filling the space of $Y$ as shown in Fig. 1. The air
connects neighboring pores to one another. The geometry defined above
satisfies the following assumptions:

  1. 1.

Neither solid nor water-filled parts touch the boundary of the pore.

  2. 2.

All internal (air-water and water-solid) interfaces are sufficiently smooth
and do not touch each other.

These geometrical restrictions imply that the pores are connected by air-
filled parts only which is needed not only to give a meaning to functions
defined across interfaces, but also to introduce the concept of extension as
given, for instance, in [20]. Furthermore, there are no solid-air interfaces.

###  2.2 Description of the chemistry

There are many variants of severe attack to concrete in sewer pipes, we focus
here on the most aggressive one – the sulfuric acid attack. The situation can
be described briefly as follows: (The anaerobic bacteria in the flowing waste
water release hydrogen sulfide gas ($H_{2}S$) within the air space of the
pipe. These bacteria are especially active in hot environments. From the air
space inside the pipe, $H_{2}S(g)$333$H_{2}S(g)$ and $H_{2}S(aq)$ refer to
gaseous, and respectively, aqueous $H_{2}S$. enters the pores of the concrete
matrix where it diffuses and then dissolves in the pore water. The aerobic
bacteria catalyze some of the $H_{2}S$ into sulfuric acid $H_{2}SO_{4}$.
$H_{2}S$ molecules can move between air-filled part and water-filled part the
water-air interfaces [21]. We model this microscopic interfacial transfer via
Henry’s law [22], (see the boundary conditions at $\Gamma^{wa}_{\varepsilon}$
in (3) and (4)). $H_{2}SO_{4}$ being an aggressive acid reacts with the solid
matrix444The solid matrix is assumed here to consist of $CaCO_{3}$ only. This
assumption can be removed in the favor of a more complex cement chemistry. at
the solid-water interface, which is made up of cement, sand, and aggregate,
and produces gypsum (i.e. $CaSO_{4}\cdot 2H_{2}O$). Here we restrict our
attention to a minimal set of chemical reactions mechanisms as suggested in
[2], namely.

$\left\\{\begin{aligned} 10H^{+}+SO_{4}^{-2}+\mbox{org.
matter}\;&\longrightarrow\;H_{2}S(aq)+4H_{2}O+\mbox{oxid. matter}\\\
H_{2}S(aq)+2O_{2}\;&{\longrightarrow}\;2H^{+}+SO_{4}^{-2}\\\
H_{2}S(aq)\;&\rightleftharpoons\;H_{2}S(g)\\\
2H_{2}O+H^{+}+SO_{4}^{-2}+CaCO_{3}\;&{\longrightarrow}\;CaSO_{4}\cdot
2H_{2}O+HCO_{3}^{-}\end{aligned}\right.$ (1)

We assume that reactions (1) do not interfere with the mechanics of the solid
part of the pores. This is a rather strong assumption since it is known that
(1) can actually produce local ruptures of the solid matrix [23]. For more
details on the involved cement chemistry and connections to acid corrosion, we
refer the reader to [24] (for a nice enumeration of the involved
physicochemical mechanisms), [23] (standard textbook on cement chemistry), as
well as to [25, 26, 27] and references cited therein. For a mathematical
approach of a similar theme related to the conservation and restoration of
historical monuments, we refer to the work by R. Natalini and co-workers (cf.
e.g. [28]).

###  2.3 Setting of the equations

The data and unknown are given by

$\displaystyle{u^{\varepsilon}_{1}}_{0}$
$\displaystyle:\Omega\longrightarrow\mathbb{R}_{+}\mbox{ - initial
concentration of }H_{2}SO_{4}(aq)$ $\displaystyle{u^{\varepsilon}_{2}}_{0}$
$\displaystyle:\Omega\longrightarrow\mathbb{R}_{+}\mbox{ - initial
concentration of }H_{2}S(aq)$ $\displaystyle{u^{\varepsilon}_{3}}_{0}$
$\displaystyle:\Omega\longrightarrow\mathbb{R}_{+}\mbox{ - initial
concentration of }H_{2}S(g)$ $\displaystyle{u^{\varepsilon}_{4}}_{0}$
$\displaystyle:\Omega\longrightarrow\mathbb{R}_{+}\mbox{ - initial
concentration of moisture}$ $\displaystyle{u^{\varepsilon}_{5}}_{0}$
$\displaystyle:\Omega\longrightarrow\mathbb{R}_{+}\mbox{ - initial
concentration of gypsum}$ $\displaystyle u_{3}^{D}$
$\displaystyle:\Gamma_{D}\times(0,T)\longrightarrow\mathbb{R}_{+}\mbox{ -
exterior concentration (Dirichlet data) of }H_{2}S(g)$ $\displaystyle
u^{\varepsilon}_{1}$
$\displaystyle:\Omega^{\varepsilon}\times(0,T)\longrightarrow\mathbb{R}\mbox{
- concentration of }H_{2}SO_{4}(aq)$ $\displaystyle u^{\varepsilon}_{2}$
$\displaystyle:\Omega^{\varepsilon}\times(0,T)\longrightarrow\mathbb{R}\mbox{
- concentration of }H_{2}S(aq)$ $\displaystyle u^{\varepsilon}_{3}$
$\displaystyle:\Omega^{\varepsilon}_{1}\times(0,T)\longrightarrow\mathbb{R}\mbox{
- concentration of }H_{2}S(g)$ $\displaystyle u^{\varepsilon}_{4}$
$\displaystyle:\Omega^{\varepsilon}\times(0,T)\longrightarrow\mathbb{R}\mbox{
- concentration of moisture}$ $\displaystyle u^{\varepsilon}_{5}$
$\displaystyle:\Gamma^{sw}_{\varepsilon}\times(0,T)\longrightarrow\mathbb{R}\mbox{
- concentration of gypsum }$

All concentrations are viewed as mass concentrations. We consider the
following system of mass-balance equations defined at the pore level. The
mass-balance equation for $H_{2}SO_{4}$ is

$\displaystyle\partial_{t}u^{\varepsilon}_{1}+div(-d^{\varepsilon}_{1}\nabla
u^{\varepsilon}_{1})$
$\displaystyle=-k^{\varepsilon}_{1}u^{\varepsilon}_{1}+k^{\varepsilon}_{2}u^{\varepsilon}_{2},\quad
x\in\Omega^{\varepsilon},\;t\in(0,T)$ (2) $\displaystyle
u^{\varepsilon}_{1}(x,0)$ $\displaystyle={u^{\varepsilon}_{1}}_{0}(x),\quad
x\in\Omega^{\varepsilon}$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{1}\nabla u^{\varepsilon}_{1}$ $\displaystyle=0,\quad
x\in\Gamma^{wa}_{\varepsilon},\;t\in(0,T)$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{1}\nabla u^{\varepsilon}_{1}$
$\displaystyle=\varepsilon\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{5}),\quad
x\in\Gamma^{sw}_{\varepsilon}\;t\in(0,T).$

The mass-balance equation for $H_{2}S(aq)$ is given by

$\displaystyle\partial_{t}u^{\varepsilon}_{2}+div(-d^{\varepsilon}_{2}\nabla
u^{\varepsilon}_{2})$
$\displaystyle=k^{\varepsilon}_{1}u^{\varepsilon}_{1}-k^{\varepsilon}_{2}u^{\varepsilon}_{2},\quad
x\in\Omega^{\varepsilon},\;t\in(0,T),$ (3) $\displaystyle
u^{\varepsilon}_{2}(x,0)$ $\displaystyle={u^{\varepsilon}_{2}}_{0}(x),\quad
x\in\Omega^{\varepsilon}$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{2}\nabla u^{\varepsilon}_{2}$
$\displaystyle=\varepsilon(a^{\varepsilon}(x)u^{\varepsilon}_{3}-b^{\varepsilon}(x)u^{\varepsilon}_{2}),\quad
x\in\Gamma^{wa}_{\varepsilon},\;t\in(0,T)$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{2}\nabla u^{\varepsilon}_{2}$ $\displaystyle=0,\quad
x\in\Gamma^{sw}_{\varepsilon},\;t\in(0,T).$

The mass-balance equation for $H_{2}S(g)$ reads

$\displaystyle\partial_{t}u^{\varepsilon}_{3}+div(-d^{\varepsilon}_{3}\nabla
u^{\varepsilon}_{3})$ $\displaystyle=0,\quad
x\in\Omega^{\varepsilon}_{1},\;t\in(0,T)$ (4) $\displaystyle
u^{\varepsilon}_{3}(x,0)$ $\displaystyle={u^{\varepsilon}_{3}}_{0}(x),\quad
x\in\Omega^{\varepsilon}_{1}$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{3}\nabla u^{\varepsilon}_{3}$ $\displaystyle=0,\quad
x\in\Gamma^{N},\;t\in(0,T)$ $\displaystyle u^{\varepsilon}_{3}(x,t)$
$\displaystyle=u^{D}_{3}(x,t),\quad x\in\Gamma^{D},\;t\in(0,T)$
$\displaystyle-n^{\varepsilon}\cdot d^{\varepsilon}_{3}\nabla
u^{\varepsilon}_{3}$
$\displaystyle=-\varepsilon(a^{\varepsilon}(x)u^{\varepsilon}_{3}-b^{\varepsilon}(x)u^{\varepsilon}_{2}),\quad
x\in\Gamma^{wa}_{\varepsilon},\;t\in(0,T).$

The mass-balance equation for moisture follows

$\displaystyle\partial_{t}u^{\varepsilon}_{4}+div(-d^{\varepsilon}_{4}\nabla
u^{\varepsilon}_{4})$
$\displaystyle=k^{\varepsilon}_{1}u^{\varepsilon}_{1},\quad
x\in\Omega^{\varepsilon},\;t\in(0,T)$ (5) $\displaystyle
u^{\varepsilon}_{4}(x,0)$ $\displaystyle={u^{\varepsilon}_{4}}_{0}(x),\quad
x\in\Omega^{\varepsilon}$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{4}\nabla u^{\varepsilon}_{4}$ $\displaystyle=0,\quad
x\in\Gamma^{wa}_{\varepsilon},\;t\in(0,T)$ $\displaystyle-n^{\varepsilon}\cdot
d^{\varepsilon}_{4}\nabla u^{\varepsilon}_{4}$ $\displaystyle=0,\quad
x\in\Gamma^{sw}_{\varepsilon},\;t\in(0,T).$

The mass-balance equation for the gypsum produced at the water-solid interface
is

$\displaystyle\partial_{t}u^{\varepsilon}_{5}$
$\displaystyle=\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{5}),\quad
x\in\Gamma^{sw}_{\varepsilon},\;t\in(0,T)$ (6) $\displaystyle
u^{\varepsilon}_{5}(x,0)$ $\displaystyle={u^{\varepsilon}_{5}}_{0}(x),\quad
x\in\Gamma^{sw}_{\varepsilon},\;t\in(0,T).$

##  3 Weak formulation and basic results

We begin this section with a list of notations and function spaces. Then we
indicate our working assumptions and give the weak formulation of the
microscopic problem; we bring reader’s attention to the well-posedness of the
system (2)–(6).

###  3.1 Notations and function spaces

We use
$(\alpha,\beta)_{(0,T)\times\Omega^{\varepsilon}}:=\int_{0}^{T}\int_{\Omega^{\varepsilon}}\alpha\beta
dxdt$,
$(\alpha,\beta)_{(0,T)\times\Gamma_{\varepsilon}}:=\int_{0}^{T}\int_{\Gamma_{\varepsilon}}\alpha\beta
d\sigma_{x}dt$. $\langle\cdot\rangle$, $|\cdot|$ and $\|\cdot\|$ denote the
dual pairing of $H^{1}(\Omega^{\varepsilon})$ and
$H^{-1}(\Omega^{\varepsilon})$, the norm in $L^{2}(\Omega^{\varepsilon})$, and
the norm in $H^{1}(\Omega^{\varepsilon}),$ respectively. $\varphi^{+}$ and
$\varphi^{-}$ will point out the positive and respectively the negative part
of the function $\varphi$. We denote by $C^{\infty}_{\\#}(Y)$,
$H^{1}_{\\#}(Y)$, and $H^{1}_{\\#}(Y)/\mathbb{R}$, the space of infinitely
differentiable functions in $\mathbb{R}^{n}$ that are periodic of period $Y$,
the completion of $C^{\infty}_{\\#}(Y)$ with respect to $H^{1}-$norm, and the
respective quotient space, respectively. Furthermore,
$H^{1}_{\Gamma^{D}}(\Omega):=\\{u\in H^{1}(\Omega)|u=0\mbox{ on
}\Gamma^{D}\\}$. The Sobolev space $H^{\beta}(\Omega)$ as a completion of
$C^{\infty}(\Omega)$ is a Hilbert space equipped with a norm

$\|\varphi\|_{H^{\beta}(\Omega)}=\|\varphi\|_{H^{[\beta]}(\Omega)}+\left(\int_{\Omega}\int_{\Omega}\frac{|\varphi(x)-\varphi(y)|^{2}}{|x-y|^{n+2(\beta-[\beta])}}dxdy\right)^{\frac{1}{2}}$

and (cf. Theorem 7.57 in [29]) the embedding $H^{\beta}(\Omega)\hookrightarrow
L^{2}(\Omega)$ is continuous. Since we deal with an evolution problem, we need
typical Bochner spaces like $L^{2}(0,T;H^{1}(\Omega))$,
$L^{2}(0,T;H^{-1}(\Omega))$, $L^{2}(0,T;H^{1}_{\Gamma^{D}}(\Omega))$, and
$L^{2}((0,T)\times\Omega;H^{1}_{\\#}(Y)/\mathbb{R})$. In the analysis of the
microscopic model, we use frequently the following trace inequality for
$\varepsilon-$dependent hypersurfaces $\Gamma^{wa}_{\varepsilon}$: For
$\varphi_{\varepsilon}\in H^{1}(\Omega^{\varepsilon})$, there exists a
constant $C^{*}$, which is independent of $\varepsilon$, such that

$\varepsilon|\varphi_{\varepsilon}|^{2}_{L^{2}(\Gamma_{\varepsilon})}\leq
C^{*}(|\varphi_{\varepsilon}|^{2}_{L^{2}(\Omega^{\varepsilon})}+\varepsilon^{2}|\nabla\varphi_{\varepsilon}|^{2}_{L^{2}(\Omega^{\varepsilon})}).$
(7)

The proof of (7) is given in Lemma 3 of [30]. For a function
$\varphi^{\varepsilon}\in H^{\beta}(\Omega^{\varepsilon})$ with
$\beta\in(\frac{1}{2},1)$, the inequality (7) refines into

$\varepsilon|\varphi_{\varepsilon}|^{2}_{L^{2}(\Gamma_{\varepsilon})}\leq
C^{*}_{0}(|\varphi_{\varepsilon}|^{2}_{L^{2}(\Omega^{\varepsilon})}+\varepsilon^{2\beta}\int_{\Omega}^{\varepsilon}\int_{\Omega}^{\varepsilon}\frac{|\varphi^{\varepsilon}(x)-\varphi^{\varepsilon}(y)|^{2}}{|x-y|^{n+2\beta}}dxdy),$
(8)

where $C^{*}_{0}$ is again a constant independent of $\varepsilon$. For proof
of (8), see [15]. To simplify the writing of some of the estimates, we employ
the next set of notations:

$\displaystyle d_{i}$ $\displaystyle:=\min_{[0,T]\times\bar{\Omega}}\mid
d_{i}^{\varepsilon}\mid,\;i\in\\{1,2,3,4\\},\quad\tilde{d}_{i}:=\min_{[0,T]\times\bar{\Omega}}\mid\tilde{d}_{i}^{\varepsilon}\mid,$
$\displaystyle D_{m}$
$\displaystyle:=\max_{[0,T]\times\Omega^{\varepsilon}}|\partial_{t}d^{\varepsilon}_{m}|,\;m\in\\{1,2,3\\},\quad
k_{j}:=\min_{[0,T]\times\bar{\Omega}}\mid
k_{j}^{\varepsilon}\mid,\;j\in\\{1,2\\}$ $\displaystyle K_{j}$
$\displaystyle:=\min_{[0,T]\times\bar{\Omega}}\mid\partial_{t}k_{j}^{\varepsilon}\mid,\quad\tilde{k}_{j}:=\min_{[0,T]\times\bar{\Omega}}\mid{\tilde{k}}_{j}^{\varepsilon}\mid,$
$\displaystyle k^{\infty}_{m}$ $\displaystyle:=\sup_{(0,T)\times\Omega}\mid
k_{m}^{\varepsilon}\mid,\quad\tilde{k}^{\infty}_{m}:=\sup_{(0,T)\times\Omega}\mid\tilde{k}_{m}^{\varepsilon}\mid,$
$\displaystyle K^{\infty}_{m}$
$\displaystyle:=\sup_{(0,T)\times\Omega}\mid\partial_{t}k_{m}^{\varepsilon}\mid,\quad
M_{i}:=\sup_{(0,T)\times\Omega}\mid
u_{i}^{\varepsilon}\mid,\;i\in\\{1,2,3,4,5\\},$ $\displaystyle A^{\infty}$
$\displaystyle:=\sup_{(0,T)\times\Gamma^{wa}_{\varepsilon}}|a_{\varepsilon}|,\;B^{\infty}:=\sup_{(0,T)\times\Gamma^{wa}_{\varepsilon}}|b_{\varepsilon}|,$
$\displaystyle A^{\infty}$
$\displaystyle:=\sup_{(0,T)\times\Gamma^{wa}_{\varepsilon}}|\partial_{t}a_{\varepsilon}|,\;B^{\infty}:=\sup_{(0,T)\times\Gamma^{wa}_{\varepsilon}}|\partial_{t}b_{\varepsilon}|,$
$\displaystyle\tilde{a}^{\infty}$
$\displaystyle:=\sup_{(0,T)\times\Gamma^{wa}}|\tilde{a}|,\;\tilde{b}^{\infty}:=\sup_{(0,T)\times\Gamma^{wa}}|\tilde{b}|,$
$\displaystyle Q^{\infty}$
$\displaystyle:=\sup_{s\in(0,T)\times\Gamma^{sw}_{\varepsilon}}|Q(s)|,\quad\bar{\eta}:=||\eta||_{\infty},\quad\hat{\eta}:=||\partial_{t}\eta||_{\infty}.$

###  3.2 Assumptions on the data and parameters

We consider the following restriction on the data and parameters:

  1. (A1)

$d_{i}\in L^{\infty}((0,T)\times Y)^{3\times 3}$, $\partial_{t}d_{i}\in
L^{\infty}((0,T)\times Y)^{3\times 3}$, $\partial_{tt}d_{i}\in
L^{\infty}((0,T)\times Y)^{3\times 3}$, $(d_{i}(t,x)\xi,\xi)\geq
d_{i0}\mid\xi\mid^{2}$ for $d_{i0}>0$, for every $\xi\in\mathbb{R}^{3}$,
$(t,x)\in(0,T)\times Y$, $i\in\\{1,2,3,4\\}$.

  2. (A2)

$\eta$ is measurable w.r.t. $t$ and $x$ and
$\eta(\alpha,\beta)=k_{3}^{\varepsilon}R(\alpha)Q(\beta)$, $R$ is sub-linear
and locally Lipschitz function and $Q$ is bounded and locally Lipschitz
function such that

$\displaystyle
R(\alpha)=\left\\{\begin{array}[]{ccc}\mbox{positive},\;\;\mbox{if}\;\;\alpha\geq
0,\\\ 0,\;\;\;\mbox{otherwise}\end{array}\right.\quad\quad
Q(\beta)=\left\\{\begin{array}[]{ccc}\mbox{positive},\;\;\mbox{if}\;\;\beta<\beta_{max},\\\
0,\;\;\;\mbox{otherwise}\end{array}\right.$

Additionally to (A2), we sometimes assume (A2)’, that is

  3. (A2)’

$\partial_{t}\eta\leq\hat{\eta}$.

  4. (A3)

$u^{\varepsilon}_{i0}\in L^{2}(\Omega^{\varepsilon})\cap
L_{+}^{\infty}(\Omega^{\varepsilon}),\;i\in\\{1,2,4\\}$,
$u^{\varepsilon}_{30}\in L^{2}(\Omega^{\varepsilon}_{1})\cap
L_{+}^{\infty}(\Omega^{\varepsilon}_{1})$, $u^{\varepsilon}_{50}\in
L^{2}(\Gamma^{sw}_{\varepsilon})\cap
L_{+}^{\infty}(\Gamma^{sw}_{\varepsilon})$.

  5. (A4)

$a^{\infty}M_{3}=b^{\infty}M_{2}$, $k^{\infty}_{1}M_{1}=M_{4}$,
$k_{1}M_{1}=k^{\infty}_{2}M_{2}$.

  6. (A5)

$a,b\in{C}^{1}([0,T];C^{0,\alpha}(\Gamma^{wa})),a,b\geq 0$ in
$[0,T]\times\Gamma^{wa},\partial_{t}a,\partial_{t}b\in
L^{\infty}((0,T)\times\Gamma^{wa})$.

  7. (A6)

$\partial_{t}u_{3}^{D}$, $\partial_{tt}u_{3}^{D}$ and
$\nabla\partial_{t}u_{3}^{D}$ are bounded.

  8. (A7)

$k_{3}\in C^{1}([0,T];C^{0,\alpha}(\Gamma^{sw}))$ and $k_{j}\in
C^{1}([0,T];C^{0,\alpha}(\bar{Y}))$ for any $j\in\\{1,2\\}$ and
$\alpha\in]0,1]$.

The assumptions (A1)–(A3), (A5), and (A6) are of technical nature. The first
equality in (A4) points out an infinitely fast (equilibrium) Henry law, while
the last two equalities remotely resemble a detailed balance in two of the
involved chemical reactions.

###  3.3 Weak formulation of the microscopic model

######  Definition 1.

Assume (A1) and (A3). We call the vector
$u^{\varepsilon}=(u^{\varepsilon}_{1},u^{\varepsilon}_{2},u^{\varepsilon}_{3},u^{\varepsilon}_{4},u^{\varepsilon}_{5})$,
a weak solution to (2)–(6) if $u^{\varepsilon}_{j}\in
L^{2}(0,T;H^{1}(\Omega^{\varepsilon})),\partial_{t}u^{\varepsilon}_{j}\in
L^{2}(0,T;H^{-1}(\Omega^{\varepsilon})),j\in\\{1,2,4\\}$,
$u^{\varepsilon}_{3}\in
u_{3}^{D}+L^{2}(0,T;H_{\Gamma^{D}}^{1}(\Omega_{1}^{\varepsilon})),\partial_{t}u^{\varepsilon}_{3}\in
u_{3}^{D}+L^{2}(0,T;H^{-1}(\Omega_{1}^{\varepsilon})),u^{\varepsilon}_{5}\in
L^{\infty}((0,T)\times\Gamma^{sw}_{\varepsilon}),\partial_{t}u^{\varepsilon}_{5}\in
L^{\infty}((0,T)\times\Gamma^{sw}_{\varepsilon})$ such that the following
identities hold

$\displaystyle\langle\partial_{t}u^{\varepsilon}_{1},\varphi_{1}\rangle_{(0,T)\times\Omega^{\varepsilon}}$
$\displaystyle+(d_{1}\nabla
u^{\varepsilon}_{1}),\nabla\varphi_{1})_{(0,T)\times\Omega^{\varepsilon}}$
(12)
$\displaystyle=-(k_{1}u^{\varepsilon}_{1},\varphi_{1})_{(0,T)\times\Omega^{\varepsilon}}+(k_{2}u^{\varepsilon}_{2},\varphi_{1})_{(0,T)\times\Omega^{\varepsilon}}$
$\displaystyle-\varepsilon(\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{4}),\varphi_{1})_{(0,T)\times\Gamma^{sw}_{\varepsilon}},$
$\displaystyle\langle\partial_{t}u^{\varepsilon}_{2},\varphi_{2}\rangle_{(0,T)\times\Omega^{\varepsilon}}$
$\displaystyle+(d^{\varepsilon}_{2}\nabla
u^{\varepsilon}_{2}),\nabla\varphi_{2})_{(0,T)\times\Omega^{\varepsilon}}$
(13)
$\displaystyle=(k^{\varepsilon}_{1}u^{\varepsilon}_{1},\varphi_{2})_{(0,T)\times\Omega^{\varepsilon}}-(k^{\varepsilon}_{2}u^{\varepsilon}_{2},\varphi_{2})_{(0,T)\times\Omega^{\varepsilon}}$
$\displaystyle+\varepsilon(a_{\varepsilon}u^{\varepsilon}_{3},\varphi_{2})_{(0,T)\times\Gamma^{wa}_{\varepsilon}}-\varepsilon(a_{\varepsilon}u^{\varepsilon}_{2},\varphi_{2})_{(0,T)\times\Gamma^{wa}_{\varepsilon}},$
$\displaystyle\langle\partial_{t}u^{\varepsilon}_{3},\varphi_{3}\rangle_{(0,T)\times\Omega_{1}^{\varepsilon}}$
$\displaystyle=-(d^{\varepsilon}_{3}\nabla
u^{\varepsilon}_{3}),\nabla\varphi_{3})_{(0,T)\times\Omega_{1}^{\varepsilon}}$
(14)
$\displaystyle-\varepsilon(a_{\varepsilon}u^{\varepsilon}_{3},\varphi_{3})_{(0,T)\times\Gamma^{wa}_{\varepsilon}}+\varepsilon(a_{\varepsilon}u^{\varepsilon}_{2},\varphi_{3})_{(0,T)\times\Gamma^{wa}_{\varepsilon}},$
$\displaystyle\langle\partial_{t}u^{\varepsilon}_{4},\varphi_{4}\rangle_{(0,T)\times\Omega^{\varepsilon}}$
$\displaystyle=-(d^{\varepsilon}_{4}\nabla
u^{\varepsilon}_{4}),\nabla\varphi_{4})_{(0,T)\times\Omega^{\varepsilon}}+(k^{\varepsilon}_{1}u^{\varepsilon}_{1},\varphi_{4})_{(0,T)\times\Omega^{\varepsilon}}$
(15)

for all $\varphi_{j}\in
L^{2}(0,T;H^{1}(\Omega^{\varepsilon})),j\in\\{1,2,4\\}$ and $\varphi_{3}\in
L^{2}(0,T;H_{\Gamma^{D}}^{1}(\Omega_{1}^{\varepsilon}))$ together with the ode

$\displaystyle\partial_{t}u^{\varepsilon}_{5}$ $\displaystyle=$
$\displaystyle\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{5})\;\;\mbox{a.e.
on}\;\;(0,T)\times{\Gamma^{ws}_{\varepsilon}}$ (16)

and the initial conditions

$\displaystyle u^{\varepsilon}_{i}(0,x)$
$\displaystyle=u^{\varepsilon}_{i0}(x)\;\;x\in\Omega^{\varepsilon}\mbox{ for
all }i\in\\{1,2,4\\},$ (17) $\displaystyle u^{\varepsilon}_{3}(0,x)$
$\displaystyle=u^{\varepsilon}_{30}(x)\;\;x\in\Omega_{1}^{\varepsilon},$
$\displaystyle u^{\varepsilon}_{5}(0,x)$
$\displaystyle=u^{\varepsilon}_{50}(x)\;\;x\in\Gamma^{ws}_{\varepsilon}.$

###  3.4 Basic results

######  Lemma 2.

(Positivity and $L^{\infty}$-estimates) Assume (A1)-(A6), and let $t\in[0,T]$
be arbitrarily chosen. Then the following estimates hold:

  * (i)

$u^{\varepsilon}_{i}(t)\geq 0,\;\;i\in\\{1,2,4\\}$ a.e. in
${\Omega^{\varepsilon}}$, $u^{\varepsilon}_{3}(t)\geq 0$ a.e.
${\Omega^{\varepsilon}_{1}}$ and $u^{\varepsilon}_{5}(t)\geq 0$ a.e. on
$\Gamma^{ws}_{\varepsilon}$.

  * (ii)

$u^{\varepsilon}_{i}(t)\leq M_{i}$, $i\in\\{1,2\\}$,
$u^{\varepsilon}_{4}(t)\leq(t+1)M_{4}$ a.e. in ${\Omega^{\varepsilon}}$ ,
$u^{\varepsilon}_{3}(t)\leq M_{3}$ a.e. in ${\Omega^{\varepsilon}_{1}}$ and
$u^{\varepsilon}_{5}(t)\leq M_{5}$ a.e. on $\Gamma^{ws}_{\varepsilon}$.

Proof (i). We test (12)-(15) with
$\varphi=(-{u^{\varepsilon}_{1}}^{-},-{u^{\varepsilon}_{2}}^{-},-{u^{\varepsilon}_{3}}^{-},-{u^{\varepsilon}_{4}}^{-})$
element of the space $[L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))]^{2}\times
L^{2}(0,T;H^{1}_{\Gamma^{D}}(\Omega_{1}^{\varepsilon})\times
L^{2}(0,T;H^{1}(\Omega^{\varepsilon})$. We obtain the following inequality

$\displaystyle\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{1}}^{-}|^{2}+d_{1}|\nabla{u^{\varepsilon}_{1}}^{-}|^{2}$
$\displaystyle\leq-
k_{1}|{u^{\varepsilon}_{1}}^{-}|^{2}+k^{\infty}_{2}({u^{\varepsilon}_{1}}^{-},{u^{\varepsilon}_{2}}^{-})$
(18)
$\displaystyle-\varepsilon(\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{5}),-{u^{\varepsilon}_{1}}^{-})_{\Gamma^{sw}_{\varepsilon}}.$

Note that the first term on the r.h.s of (18) is negative, while the third
term is zero because of (A2). We then get

$\partial_{t}|{u^{\varepsilon}_{1}}^{-}|^{2}+2d_{1}|\nabla{u^{\varepsilon}_{1}}^{-}|^{2}\leq
k^{\infty}_{2}\left(|{u^{\varepsilon}_{1}}^{-}|^{2}+|{u^{\varepsilon}_{2}}^{-}|^{2}\right).$
(19)

On the other hand, (13) leads to

$\displaystyle\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{2}}^{-}|^{2}+d_{2}|\nabla{u^{\varepsilon}_{2}}^{-}|^{2}$
$\displaystyle\leq\frac{k^{\infty}_{1}}{2}\left(|{u^{\varepsilon}_{1}}^{-}|^{2}+|{u^{\varepsilon}_{2}}^{-}|^{2}\right)$
$\displaystyle+\varepsilon
a^{\infty}({u^{\varepsilon}_{2}}^{-},{u^{\varepsilon}_{3}}^{-})_{\Gamma^{wa}_{\varepsilon}}+\varepsilon
b^{\infty}|{u^{\varepsilon}_{2}}^{-}|^{2}_{\Gamma^{wa}_{\varepsilon}}.$

By the trace inequality (7) (with $\varepsilon<1$), we get

$\displaystyle\partial_{t}|{u^{\varepsilon}_{2}}^{-}|^{2}$
$\displaystyle+2(d_{2}-C^{*}b^{\infty})|\nabla{u^{\varepsilon}_{2}}^{-}|^{2}\leq{k^{\infty}_{1}}\left(|{u^{\varepsilon}_{1}}^{-}|^{2}+|{u^{\varepsilon}_{2}}^{-}|^{2}\right)$
(20)
$\displaystyle+2C^{*}b^{\infty}|{u^{\varepsilon}_{2}}^{-}|^{2}+2\varepsilon
a^{\infty}({u^{\varepsilon}_{2}}^{-},{u^{\varepsilon}_{3}}^{-})_{\Gamma^{wa}_{\varepsilon}}.$

(14) leads to

$\partial_{t}|{u^{\varepsilon}_{3}}^{-}|^{2}+2(d_{3}-C^{*}a^{\infty})|\nabla{u^{\varepsilon}_{3}}^{-}|^{2}\leq
2\varepsilon
b^{\infty}({u^{\varepsilon}_{2}}^{-},{u^{\varepsilon}_{3}}^{-})_{\Gamma^{wa}_{\varepsilon}}+2C^{*}a^{\infty}|{u^{\varepsilon}_{3}}^{-}|^{2},$
(21)

while from (15), we see that

$\partial_{t}|{u^{\varepsilon}_{4}}^{-}|^{2}+2d_{4}|\nabla{u^{\varepsilon}_{5}}^{-}|^{2}\leq
k^{\infty}_{1}\left(|{u^{\varepsilon}_{1}}^{-}|^{2}+|{u^{\varepsilon}_{5}}^{-}|^{2}\right).$
(22)

Adding up inequalities (19)-(22) gives

$\displaystyle\partial_{t}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle+2d_{1}|\nabla{u^{\varepsilon}_{1}}^{-}|^{2}+2(d_{2}-C^{*}b^{\infty})|\nabla{u^{\varepsilon}_{2}}^{-}|^{2}$
(23)
$\displaystyle+2(d_{3}-C^{*}a^{\infty})|\nabla{u^{\varepsilon}_{3}}^{-}|^{2}+2d_{4}|\nabla{u^{\varepsilon}_{4}}^{-}|^{2}$
$\displaystyle\leq\left(2k^{\infty}_{1}+k^{\infty}_{2}+2C^{*}b^{\infty}+2C^{*}a^{\infty}\right)\sum_{{i=1}}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle+2\varepsilon(a^{\infty}+b^{\infty})({u^{\varepsilon}_{2}}^{-},{u^{\varepsilon}_{3}}^{-})_{\Gamma^{wa}_{\varepsilon}},$

and hence,

$\displaystyle\partial_{t}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle+2d_{1}|\nabla{u^{\varepsilon}_{1}}^{-}|^{2}+2(d_{2}-C^{*}b^{\infty})|\nabla{u^{\varepsilon}_{2}}^{-}|^{2}$
(24)
$\displaystyle+2(d_{3}-C^{*}a^{\infty})|\nabla{u^{\varepsilon}_{3}}^{-}|^{2}+2d_{4}|\nabla{u^{\varepsilon}_{5}}^{-}|^{2}$
$\displaystyle\leq\left(2k^{\infty}_{1}+k^{\infty}_{2}+C^{*}(a^{\infty}+b^{\infty})\right)\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle+\varepsilon\left(a^{\infty}+b^{\infty})(\delta|{u^{\varepsilon}_{2}}^{-}|_{\Gamma^{wa}_{\varepsilon}}^{2}+\frac{1}{\delta}|{u^{\varepsilon}_{3}}^{-}|^{2}_{\Gamma^{wa}_{\varepsilon}}\right).$

Applying the trace inequality (7) to estimate the last term on the right side
of (24), we finally get

$\displaystyle\partial_{t}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle+2d_{1}|\nabla{u^{\varepsilon}_{1}}^{-}|^{2}+(2d_{2}-2C^{*}b^{\infty}-C^{*}\delta(a^{\infty}+b^{\infty}))|\nabla{u^{\varepsilon}_{2}}^{-}|^{2}$
$\displaystyle+(2d_{3}-2C^{*}a^{\infty}-\frac{C^{*}2}{\delta}(a^{\infty}+b^{\infty}))|\nabla{u^{\varepsilon}_{3}}^{-}|^{2}+2d_{4}|\nabla{u^{\varepsilon}_{4}}^{-}|^{2}$
$\displaystyle\leq C_{1}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}.$

Thus, we have

$\displaystyle\partial_{t}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}$
$\displaystyle\leq$ $\displaystyle
C_{1}\sum_{i=1}^{4}|{u^{\varepsilon}_{i}}^{-}|^{2}.$

where
$C_{1}:=2k^{\infty}_{1}+k^{\infty}_{2}+C^{*}(a^{\infty}+b^{\infty})+C^{*}(\delta+\frac{1}{\delta})(a^{\infty}+b^{\infty})$
and $\delta$ is chosen conveniently. Gronwall’s inequality together with
$[u^{\varepsilon}_{i}(0)]^{-}=0$ gives now the desired result. Note that (A2)
ensures automatically the positivity of $u^{\varepsilon}_{5}$.

(ii). We consider the test function

$(\varphi_{1},\varphi_{2},\varphi_{3},\varphi_{4})=((u^{\varepsilon}_{1}-M_{1})^{+},(u^{\varepsilon}_{2}-M_{2})^{+},(u^{\varepsilon}_{3}-M_{3})^{+},(u^{\varepsilon}_{4}-(t+1)M_{4})^{+}).$

Obviously, $\varphi\in[L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))]^{2}\times
L^{2}(0,T;H^{1}_{\Gamma^{D}}(\Omega_{1}^{\varepsilon})\times
L^{2}(0,T;H^{1}(\Omega^{\varepsilon})$ is allowed as test function. We obtain
from (12) that

$\displaystyle\frac{1}{2}\partial_{t}|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}$
$\displaystyle+d_{1}|\nabla(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}\leq-
k_{1}|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}$
$\displaystyle-(k_{1}M_{1},(u^{\varepsilon}_{1}-M_{1})^{+})$
$\displaystyle+k^{\infty}_{2}((u^{\varepsilon}_{1}-M_{1})^{+},(u^{\varepsilon}_{2}-M_{2})^{+})$
$\displaystyle+(k^{\infty}_{2}M_{2},(u^{\varepsilon}_{1}-M_{1})^{+})$
$\displaystyle-\varepsilon(\eta(u^{\varepsilon}_{1},u^{\varepsilon}_{5}),(u^{\varepsilon}_{1}-M_{1})^{+})_{\Gamma^{sw}_{\varepsilon}}.$

Relying on (A4), we get the estimate

$\displaystyle\begin{array}[]{ccc}\partial_{t}|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}&\leq&k^{\infty}_{2}(|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}+|(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}).\end{array}$
(26)

(13) in combination with (A4) gives that

$\displaystyle\partial_{t}|(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}$
$\displaystyle+2(d_{2}-C^{*}b^{\infty})|\nabla(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}$
(27) $\displaystyle\leq
k^{\infty}_{1}(|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}+|(u^{\varepsilon}_{2}-M_{2})^{+}|^{2})$
$\displaystyle+2C^{*}b^{\infty}|(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}$
$\displaystyle+2\varepsilon
a^{\infty}((u^{\varepsilon}_{2}-M_{2})^{+},(u^{\varepsilon}_{3}-M_{3})^{+})_{\Gamma^{wa}_{\varepsilon}}.$

By (14), we obtain

$\displaystyle\partial_{t}|(u^{\varepsilon}_{3}-M_{3})^{+}|^{2}$
$\displaystyle+2(d_{3}-C^{*}a^{\infty})|\nabla(u^{\varepsilon}_{3}-M_{3})^{+}|^{2}$
(28) $\displaystyle\leq
2C^{*}a^{\infty}|\nabla(u^{\varepsilon}_{3}-M_{3})^{+}|^{2}$
$\displaystyle+2\varepsilon
b^{\infty}((u^{\varepsilon}_{2}-M_{2})^{+},(u^{\varepsilon}_{3}-M_{3})^{+})_{\Gamma^{wa}_{\varepsilon}}.$

Using again (A4), (15) yields

$\displaystyle\begin{array}[]{ccc}\partial_{t}|(u^{\varepsilon}_{4}-(t+1)M_{4})^{+}|^{2}&\leq&k^{\infty}_{1}(|(u^{\varepsilon}_{1}-M_{1})^{+}|^{2}+|(u^{\varepsilon}_{4}-(t+1)M_{4})^{+}|^{2}).\end{array}$
(30)

Adding up (26)–(30) side by side, we get

$\displaystyle\sum_{j=1}^{3}\partial_{t}|(u^{\varepsilon}_{j}-M_{j})^{+}|^{2}$
$\displaystyle+\partial_{t}|(u^{\varepsilon}_{4}-(t+1)M_{4})^{+}|^{2}+(2d_{2}-2C^{*}b^{\infty})|\nabla(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}$
$\displaystyle+(2d_{3}-2C^{*}a^{\infty})|\nabla(u^{\varepsilon}_{3}-M_{3})^{+}|^{2}$
$\displaystyle\leq(2k^{\infty}_{2}+k^{\infty}_{1}+2C^{*}a^{\infty}+2C^{*}b^{\infty})(\sum_{j=1}^{3}|(u^{\varepsilon}_{j}-M_{j})^{+}|^{2}$
$\displaystyle+|(u^{\varepsilon}_{4}-(t+1)M_{4})^{+}|^{2})+\varepsilon(a^{\infty}+b^{\infty})(\delta|(u^{\varepsilon}_{2}-M_{2})^{+}|^{2}_{\Gamma^{wa}_{\varepsilon}}$
$\displaystyle+\frac{1}{\delta}|(u^{\varepsilon}_{3}-M_{3})^{+}|^{2}_{\Gamma^{wa}_{\varepsilon}}).$

We use the trace inequality (7) (with $\varepsilon<1$) to deal with the
boundary terms in (3.4). Then Gronwall’s inequality yields for all $t\in(0,T)$
the following estimate

$\displaystyle u^{\varepsilon}_{j}(t)$ $\displaystyle\leq
M_{j},\;\;j\in\\{1,2,5\\}\;a.\;e.\;in\;\Omega^{\varepsilon}$ $\displaystyle
u^{\varepsilon}_{3}(t)$ $\displaystyle\leq
M_{3},\;a.\;e.\;in\;\Omega_{1}^{\varepsilon}$ $\displaystyle
u^{\varepsilon}_{4}$ $\displaystyle\leq(t+1)M_{4}\;\mbox{a.e.
in}\;\Omega^{\varepsilon}.$

Furthermore, by (A2) $u^{\varepsilon}_{5}$ is bounded.

######  Proposition 3.

(Uniqueness) Assume (A1)–(A4). Then there exists at most one weak solution in
the sense of Definition 1.

Proof. We assume that
${u}^{j,\varepsilon}=(u_{1}^{j,\varepsilon},u_{2}^{j,\varepsilon},u_{3}^{j,\varepsilon},u_{4}^{j,\varepsilon},u_{5}^{j,\varepsilon}),j\in\\{1,2\\}$
are two distinct weak solutions in the sense of Definition 1. We set
$u_{i}^{\varepsilon}:=u_{i}^{1,\varepsilon}-u_{i}^{2,\varepsilon}$ for all
$i\in\\{1,2,3,4\\}$. Firstly, we deal with (15). We obtain

$\displaystyle\partial_{t}{u^{1,\varepsilon}_{5}}-\partial_{t}{u^{2,\varepsilon}_{5}}=\eta(u^{1,\varepsilon}_{1},u^{1,\varepsilon}_{5})-\eta(u^{2,\varepsilon}_{1},u^{2,\varepsilon}_{5}).$
(31)

Integrating (31) along (0,T) and using (A2), we get

$\displaystyle|u^{1,\varepsilon}_{5}-u^{2,\varepsilon}_{5}|\leq
k_{3}^{\infty}c_{R}c_{Q}M_{1}\int_{0}^{t}|{u^{1,\varepsilon}_{5}}-{u^{2,\varepsilon}_{5}}|d\tau+k_{3}^{\infty}c_{R}Q^{\infty}\int_{0}^{t}|{u^{1,\varepsilon}_{1}}-{u^{2,\varepsilon}_{1}}|d\tau.$

Gronwall’s inequality implies

$\displaystyle|{u^{1,\varepsilon}_{5}}(t)-{u^{2,\varepsilon}_{5}(t)}|\leq
C_{2}\int_{0}^{t}|{u^{1,\varepsilon}_{1}}-{u^{2,\varepsilon}_{1}}|d\tau\;\;\mbox{for
a.e. }t\in(0,T),$ (32)

where $C_{2}:=k_{3}^{\infty}c_{R}Q^{\infty}(1+C_{3}te^{C_{3}t})$ and
$C_{3}:=k_{3}^{\infty}c_{R}c_{Q}M_{1}$. We calculate

$\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{1}}|^{2}+d_{1}|\nabla{u^{\varepsilon}_{1}}|^{2}\leq-
k_{1}|{u^{\varepsilon}_{1}}|^{2}+k^{\infty}_{2}({u^{\varepsilon}_{1}},{u^{\varepsilon}_{2}})+\varepsilon(\eta_{1}-\eta_{2},{u^{\varepsilon}_{1}})_{\Gamma^{sw}_{\varepsilon}},$
(33)

where we denote
$\eta_{1}-\eta_{2}:=\eta(u^{1,\varepsilon}_{1},u^{1,\varepsilon}_{5})-\eta(u^{2,\varepsilon}_{1},u^{2,\varepsilon}_{5})$.
We can write

$\displaystyle\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{1}}|^{2}+d_{1}|\nabla{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle\leq-
k_{1}|{u^{\varepsilon}_{1}}|^{2}+\frac{k^{\infty}_{2}}{2}(|{u^{\varepsilon}_{1}}|^{2}+|{u^{\varepsilon}_{2}}|^{2})$
(34) $\displaystyle+\varepsilon
C_{3}({u^{1,\varepsilon}_{5}}-{u^{2,\varepsilon}_{5}},{u^{\varepsilon}_{1}})_{\Gamma^{sw}_{\varepsilon}}$
$\displaystyle+\varepsilon
k_{3}^{\infty}c_{R}Q^{\infty}({u^{1,\varepsilon}_{1}}-{u^{2,\varepsilon}_{1}},{u^{\varepsilon}_{1}})_{\Gamma^{sw}_{\varepsilon}}.$

Now, inserting (32) in (34) yields

$\displaystyle\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{1}}|^{2}+d_{1}|\nabla{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle\leq-
k_{1}|{u^{\varepsilon}_{1}}|^{2}+\frac{k^{\infty}_{2}}{2}(|{u^{\varepsilon}_{1}}|^{2}+|{u^{\varepsilon}_{2}}|^{2})$
(35)
$\displaystyle+C_{4}\varepsilon|{u^{\varepsilon}_{1}}|^{2}_{\Gamma^{sw}_{\varepsilon}}+\frac{\varepsilon
C_{3}^{2}}{2\delta}\int_{0}^{t}|{u^{\varepsilon}_{1}}|^{2}_{\Gamma^{sw}_{\varepsilon}}d\tau,$

where $C_{4}:=k_{3}^{\infty}c_{R}Q^{\infty}+\frac{C_{3}}{2\delta}$. Using (7),
we estimate the last two terms in (35) to obtain the inequality

$\displaystyle\frac{1}{2}\partial_{t}|{u^{\varepsilon}_{1}}|^{2}+d_{1}|\nabla{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle\leq-
k_{1}|{u^{\varepsilon}_{1}}|^{2}+\frac{k^{\infty}_{2}}{2}(|{u^{\varepsilon}_{1}}|^{2}+|{u^{\varepsilon}_{2}}|^{2})+C^{*}C_{4}(|{u^{\varepsilon}_{1}}|^{2}$
(36)
$\displaystyle+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|)+C^{*}\frac{C_{3}^{2}}{2\delta}\int_{0}^{t}(|{u^{\varepsilon}_{1}}|^{2}+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|^{2})d\tau.$

Note that the constant $C^{*}$, arising from in (36), stems from (7).
Rearranging now the terms, we have

$\displaystyle\partial_{t}|{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle+(2d_{1}-2C^{*}C_{4}\varepsilon^{2})|\nabla{u^{\varepsilon}_{1}}|^{2}+2k_{1}|{u^{\varepsilon}_{1}}|^{2}\leq({k^{\infty}_{2}}+C^{*}C_{4})(|{u^{\varepsilon}_{1}}|^{2}$
(37)
$\displaystyle+|{u^{\varepsilon}_{2}}|^{2})+C^{*}\frac{C_{3}^{2}}{2\delta}\int_{0}^{t}(|{u^{\varepsilon}_{1}}|^{2}+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|^{2})d\tau.$

Following the same line of arguments as before, we obtain from (13) that

$\displaystyle\partial_{t}|{u^{\varepsilon}_{2}}|^{2}+2d_{2}|\nabla{u^{\varepsilon}_{2}}|^{2}$
$\displaystyle\leq-2k_{2}|{u^{\varepsilon}_{2}}|^{2}+{k^{\infty}_{1}}(|{u^{\varepsilon}_{1}}|^{2}+{u^{\varepsilon}_{2}}|^{2})$
(38) $\displaystyle+2\varepsilon
a^{\infty}({u^{\varepsilon}_{3}},{u^{\varepsilon}_{2}})_{\Gamma^{wa}_{\varepsilon}}+2\varepsilon
b^{\infty}|{u^{\varepsilon}_{2}}|^{2}_{\Gamma^{wa}_{\varepsilon}},$

while from (14), we deduce

$\displaystyle\partial_{t}|{u^{\varepsilon}_{3}}|^{2}+2d_{3}|\nabla{u^{\varepsilon}_{3}}|^{2}$
$\displaystyle\leq 2\varepsilon
b^{\infty}({u^{\varepsilon}_{2}},{u^{\varepsilon}_{3}})_{\Gamma^{wa}_{\varepsilon}}+2\varepsilon
a^{\infty}|{u^{\varepsilon}_{3}}|^{2}_{\Gamma^{wa}_{\varepsilon}}.$ (39)

Proceeding similarly, (15) yields

$\displaystyle\partial_{t}|{u^{\varepsilon}_{4}}|^{2}+2d_{4}|\nabla{u^{\varepsilon}_{4}}|^{2}$
$\displaystyle\leq{k^{\infty}_{2}}(|{u^{\varepsilon}_{1}}|^{2}+|{u^{\varepsilon}_{4}}|^{2}).$
(40)

Putting together (37)–(40), we get

$\displaystyle\partial_{t}\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}$
$\displaystyle+(2d_{1}-C^{*}C_{4}\varepsilon^{2})|\nabla{u^{\varepsilon}_{1}}|^{2}+2d_{2}|\nabla{u^{\varepsilon}_{2}}|^{2}+2d_{3}|\nabla{u^{\varepsilon}_{3}}|^{2}$
(41)
$\displaystyle+2d_{4}|\nabla{u^{\varepsilon}_{4}}|^{2}+2k_{1}|{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle\leq\left(2{k^{\infty}_{1}}+{k^{\infty}_{2}}+C^{*}C_{2}\right)\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}$
$\displaystyle+C^{*}\frac{C_{1}^{2}}{2\delta}\int_{0}^{t}(|{u^{\varepsilon}_{1}}|^{2}+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|^{2})d\tau$
$\displaystyle+2\varepsilon
b|{u^{\varepsilon}_{2}}|^{2}_{\Gamma^{wa}_{\varepsilon}}+2\varepsilon
a|{u^{\varepsilon}_{3}}|^{2}_{\Gamma^{wa}_{\varepsilon}}$
$\displaystyle+\varepsilon(a^{\infty}+b^{\infty})(\delta|{u^{\varepsilon}_{2}}|_{\Gamma^{wa}_{\varepsilon}}^{2}+\frac{1}{\delta}|{u^{\varepsilon}_{3}}|_{\Gamma^{wa}_{\varepsilon}}^{2}).$

Applying the trace inequality (7) to the boundary terms in (41), we get

$\displaystyle\partial_{t}\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}$
$\displaystyle+(2d_{1}-2C^{*}C_{4}\varepsilon^{2})|\nabla{u^{\varepsilon}_{1}}|^{2}$
$\displaystyle+(2d_{2}-2C^{*}b^{\infty}\varepsilon^{2}-C^{*}\delta\varepsilon^{2}(a^{\infty}+b^{\infty}))|\nabla{u^{\varepsilon}_{2}}|^{2}$
$\displaystyle+(2d_{3}-2C^{*}a^{\infty}\varepsilon^{2}-\frac{C^{*}\varepsilon^{2}}{\delta}(a^{\infty}+b^{\infty}))|\nabla{u^{\varepsilon}_{3}}|^{2}$
$\displaystyle+2d_{4}|\nabla{u^{\varepsilon}_{4}}|^{2}+2k_{1}|{u^{\varepsilon}_{1}}|^{2}\leq
C_{5}\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}$
$\displaystyle+C^{*}\frac{C_{1}^{2}}{2\delta}\int_{0}^{t}(|{u^{\varepsilon}_{1}}|^{2}+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|^{2})d\tau,$
(42)

where
$C_{5}:=2{k^{\infty}_{1}}+{k^{\infty}_{2}}+C^{*}C_{2}+2C^{*}(a^{\infty}+b^{\infty})+C^{*}(a^{\infty}+b^{\infty})(\delta+\frac{1}{\delta})$.
Let us choose $\varepsilon$ and $\delta$ such that

$\displaystyle\varepsilon$
$\displaystyle\in\left]0,\frac{2d_{1}}{C_{1}C^{*}}\right[$
$\displaystyle\delta$
$\displaystyle\in\left[\frac{C^{*}\varepsilon^{2}(a^{\infty}+b^{\infty})}{2d_{3}-C^{*}a^{\infty}\varepsilon^{2}},\frac{2d_{2}-C^{*}b^{\infty}\varepsilon^{2}}{C^{*}\varepsilon^{2}(a^{\infty}+b^{\infty})}\right].$

With this choice of $(\varepsilon,\delta)$, (42) takes the form

$\partial_{t}\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}+\bar{C}|\nabla{u^{\varepsilon}_{1}}|^{2}+\bar{C}|{u^{\varepsilon}_{1}}|^{2}\leq{C_{6}}{}(\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}+\int_{0}^{t}(|{u^{\varepsilon}_{1}}|^{2}+\varepsilon^{2}|\nabla{u^{\varepsilon}_{1}}|^{2})d\tau),$

where
$C_{6}:=2{k^{\infty}_{1}}+{k^{\infty}_{2}}+C^{*}C_{2}+C^{*}(a^{\infty}+b^{\infty})+C^{*}\frac{C_{1}^{2}}{2\delta}$
and $\bar{C}:=\min\\{2d_{1}-2C^{*}C_{2}\varepsilon^{2},2k_{1}\\}$. Taking in
(42) the supremum along $t\in(0,T)$ and applying Gronwall’s inequality, we
obtain the following estimate

$\Sigma^{4}_{i=1}|{u^{\varepsilon}_{i}}|^{2}+\bar{C}\int_{0}^{T}|\nabla{u^{\varepsilon}_{1}}|^{2}dt+\bar{C}\int_{0}^{T}|{u^{\varepsilon}_{1}}|^{2}dt\leq
0.$ (43)

Thus, the proof of Proposition 3 is completed.

######  Theorem 4.

(Global Existence) Assume $(A1)-(A3)$. Then there exists at least a global-in-
time weak solution in the sense of Definition 1.

Proof. The proof is based on the Galerkin argument. Since the proof is rather
standard, and here we wish to focus on the passage to the limit
$\varepsilon\to 0$, we omit it.

##  4 A priori estimates for microscopic solutions

This section includes the $\varepsilon-$ independent estimates.

######  Lemma 5.

Assume (A1)-(A6). Then the weak solution of the microscopic model (12)-(17)
satisfies the following a priori bounds:

$\displaystyle\parallel
u^{\varepsilon}_{j}\parallel_{L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))}\leq
C,\;j\in\\{1,2,3,4\\}$ (44)
$\displaystyle\parallel\nabla\partial_{t}u^{\varepsilon}_{2}\parallel_{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}))}\leq
C,$ (45)
$\displaystyle\parallel\partial_{t}u^{\varepsilon}_{j}\parallel_{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}))}\leq
C,$ (46) $\displaystyle\parallel
u^{\varepsilon}_{3}\parallel_{L^{2}(0,T;H^{1}(\Omega^{\varepsilon}_{1}))}\leq
C,$ (47)
$\displaystyle\parallel\nabla\partial_{t}u^{\varepsilon}_{3}\parallel_{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}_{1}))}\leq
C,$ (48)
$\displaystyle\parallel\partial_{t}u^{\varepsilon}_{3}\parallel_{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}_{1}))}\leq
C,$ (49) $\displaystyle\parallel
u^{\varepsilon}_{5}\parallel_{L^{\infty}((0,T)\times\Gamma^{sw}_{\varepsilon})}\leq
C,$ (50)
$\displaystyle\parallel\partial_{t}u^{\varepsilon}_{5}\parallel_{L^{2}((0,T)\times\Gamma^{sw}_{\varepsilon})}\leq
C.$ (51)

In (44)–(51), the generic constant $C$ is independent of $\varepsilon$.

Proof. We test (12) with $\varphi_{1}=u^{\varepsilon}_{1}$ to get

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{1}|^{2}+d_{1}|\nabla
u^{\varepsilon}_{1}|^{2}$ $\displaystyle\leq-
k_{1}|u^{\varepsilon}_{1}|^{2}+k_{2}^{\infty}(u^{\varepsilon}_{1},u^{\varepsilon}_{2})-\varepsilon(\eta,u^{\varepsilon}_{1})_{\Gamma^{sw}_{\varepsilon}},$
(52)
$\displaystyle\leq\frac{k_{2}^{\infty}}{2}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{2}|^{2})+\varepsilon
k^{\infty}_{3}Q^{\infty}c_{R}(u^{\varepsilon}_{1},u^{\varepsilon}_{1})_{\Gamma^{sw}_{\varepsilon}}.$

After applying the trace inequality to the last term on r.h.s of (52), we get

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{1}|^{2}+d_{1}|\nabla
u^{\varepsilon}_{1}|^{2}$
$\displaystyle\leq\frac{k_{2}^{\infty}}{2}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{2}|^{2})+C^{*}k^{\infty}_{3}Q^{\infty}c_{R}(|u^{\varepsilon}_{1}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{1}|^{2})_{\Gamma^{sw}_{\varepsilon}}.$
$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{1}|^{2}$
$\displaystyle+(d_{1}-\varepsilon^{2}C^{*}k^{\infty}_{3}Q^{\infty}c_{R})|\nabla
u^{\varepsilon}_{1}|^{2}\leq
C_{7}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{2}|^{2}),$ (53)

where $C_{7}:=\frac{k_{2}^{\infty}}{2}+C^{*}k^{\infty}_{3}Q^{\infty}c_{R}.$
Taking $\varphi_{2}=u^{\varepsilon}_{2}$ in (13), we get

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{2}|^{2}+d_{2}|\nabla
u^{\varepsilon}_{2}|^{2}$
$\displaystyle\leq\frac{k_{1}^{\infty}}{2}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{2}|^{2})-k_{2}|u^{\varepsilon}_{2}|^{2}$
$\displaystyle+\varepsilon
a^{\infty}(u^{\varepsilon}_{3},u^{\varepsilon}_{2})_{\Gamma^{wa}_{\varepsilon}}+\varepsilon
b^{\infty}|u^{\varepsilon}_{2}|^{2}_{\Gamma^{wa}_{\varepsilon}}.$

Application of the trace inequality (7) only to the last term leads to

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{2}|^{2}+(d_{2}-C^{*}b^{\infty}\varepsilon^{2})|\nabla
u^{\varepsilon}_{2}|^{2}$
$\displaystyle\leq\frac{k_{1}^{\infty}}{2}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{2}|^{2})+2$
(54) $\displaystyle+\varepsilon
a^{\infty}(u^{\varepsilon}_{3},u^{\varepsilon}_{2})_{\Gamma^{wa}_{\varepsilon}}.$

We choose $\varphi_{3}=u^{\varepsilon}_{3}$ as a test function in (14) to
calculate

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{3}|^{2}+(d_{3}-C^{*}a^{\infty}\varepsilon^{2})|\nabla
u^{\varepsilon}_{3}|^{2}$ $\displaystyle\leq\varepsilon
b^{\infty}(u^{\varepsilon}_{3},u^{\varepsilon}_{2})_{\Gamma^{wa}_{\varepsilon}}+C^{*}a^{\infty}|u^{\varepsilon}_{3}|^{2}.$
(55)

Setting $\varphi_{4}=u^{\varepsilon}_{4}$ in (15), we are led to

$\displaystyle\frac{1}{2}\partial_{t}|u^{\varepsilon}_{4}|^{2}+d_{4}|\nabla
u^{\varepsilon}_{4}|^{2}$ $\displaystyle\leq$
$\displaystyle\frac{k_{1}^{\infty}}{2}(|u^{\varepsilon}_{1}|^{2}+|u^{\varepsilon}_{4}|^{2}).$
(56)

Putting together (53)-(56), we obtain

$\displaystyle\frac{1}{2}\Sigma_{i=1}^{4}\partial_{t}|u^{\varepsilon}_{i}|^{2}$
$\displaystyle+(d_{1}-\varepsilon^{2}C^{*}k^{\infty}_{3}Q^{\infty}c_{R})|\nabla
u^{\varepsilon}_{1}|^{2}+d_{4}|\nabla u^{\varepsilon}_{4}|^{2}$ (57)
$\displaystyle+(d_{2}-C^{*}b^{\infty}\varepsilon^{2})|\nabla
u^{\varepsilon}_{2}|^{2}+(d_{3}-C^{*}a^{\infty}\varepsilon^{2})|\nabla
u^{\varepsilon}_{3}|^{2}$
$\displaystyle\leq({k_{1}^{\infty}}{}+\frac{k_{2}^{\infty}}{2}+C^{*}b^{\infty}+C^{*}a^{\infty})\Sigma_{i=1}^{4}|u^{\varepsilon}_{i}|^{2}$
$\displaystyle+\varepsilon(a^{\infty}+b^{\infty})(u^{\varepsilon}_{3},u^{\varepsilon}_{2})_{\Gamma^{wa}_{\varepsilon}}.$

Combing Young’s inequality and the trace inequality to the boundary term, (57)
turns out to be

$\displaystyle\frac{1}{2}\Sigma_{i=1}^{4}\partial_{t}|u^{\varepsilon}_{i}|^{2}$
$\displaystyle+(d_{1}-\varepsilon^{2}C^{*}k^{\infty}_{3}Q^{\infty}c_{R})|\nabla
u^{\varepsilon}_{1}|^{2}$
$\displaystyle+(d_{2}-C^{*}b\varepsilon^{2}-\frac{C^{*}\varepsilon^{2}\delta}{2}(a^{\infty}+b^{\infty}))|\nabla
u^{\varepsilon}_{2}|^{2}$
$\displaystyle+(d_{3}-C^{*}a\varepsilon^{2}-\frac{C^{*}\varepsilon^{2}}{2\delta}(a^{\infty}+b^{\infty}))|\nabla
u^{\varepsilon}_{3}|^{2}+d_{4}|\nabla u^{\varepsilon}_{4}|^{2}$
$\displaystyle\leq({k_{1}^{\infty}}{}+\frac{k_{2}^{\infty}}{2}+C^{*}(a^{\infty}+b^{\infty})(\delta+\frac{1}{\delta}))\Sigma_{i=1}^{4}|u^{\varepsilon}_{i}|^{2}.$

Choosing $\varepsilon$ small enough and $\delta$ conveniently such that the
coefficients of the terms involving $|\nabla u^{\varepsilon}_{2}|^{2}$ and
$|\nabla u^{\varepsilon}_{3}|^{2}$ stay positive, we are led to

$\displaystyle\Sigma_{i=1}^{4}\partial_{t}|u^{\varepsilon}_{i}|^{2}+d^{\prime}_{1}|\nabla
u^{\varepsilon}_{1}|^{2}+d^{\prime}_{2}|\nabla
u^{\varepsilon}_{2}|^{2}+d^{\prime}_{3}|\nabla u^{\varepsilon}_{3}|^{2}$
$\displaystyle+$ $\displaystyle 2d_{4}|\nabla u^{\varepsilon}_{4}|^{2}\leq
C_{7}\Sigma_{i=1}^{4}|u^{\varepsilon}_{i}|^{2},$

where

$\displaystyle d^{\prime}_{1}$
$\displaystyle:=2(d_{1}-\varepsilon^{2}C^{*}k^{\infty}_{3}Q^{\infty}c_{R}),$
$\displaystyle d^{\prime}_{2}$
$\displaystyle:=2(d_{2}-C^{*}b^{\infty}\varepsilon^{2}-\frac{C^{*}\varepsilon^{2}\delta}{2}(a^{\infty}+b^{\infty})),$
$\displaystyle d^{\prime}_{3}$
$\displaystyle:=2(d_{3}-C^{*}a^{\infty}\varepsilon^{2}-\frac{C^{*}b\varepsilon^{2}}{2\delta}(a^{\infty}+b^{\infty})),$

while the constant $C$ is given by

$C_{8}:=2{k_{1}^{\infty}}{}+\frac{k_{2}^{\infty}}{2}+C^{*}a^{\infty}+C^{*}b^{\infty}+C^{*}(a^{\infty}+b^{\infty})(\delta+\frac{1}{\delta}).$

Summarizing, we have

$\displaystyle\Sigma_{i=1}^{4}\partial_{t}|u^{\varepsilon}_{i}|^{2}+d_{0}\Sigma_{j=1}^{3}|\nabla
u^{\varepsilon}_{j}|^{2}+d_{0}|\nabla u^{\varepsilon}_{3}|^{2}\leq
C\Sigma_{i=1}^{4}|u^{\varepsilon}_{i}|^{2},$ (58)

where
$d_{0}:=min\\{d^{\prime}_{1},d^{\prime}_{2},d^{\prime}_{3},d^{\prime}_{4}\\}$.
By Gronwall’s inequality, we have

$\displaystyle\Sigma_{i=1}^{4}|u^{\varepsilon}_{i}|^{2}\leq
C\Sigma_{i=1}^{4}|u_{i}(0)|^{2},$

and hence,

$\displaystyle\parallel
u^{\varepsilon}_{j}\parallel_{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}))}\leq
C\mbox{ for all
}i\in\\{1,2,4\\}\;\mbox{and}\;\|u^{\varepsilon}_{3}\|_{{L^{2}(0,T;L^{2}(\Omega^{\varepsilon}_{1}))}}\leq
C,$ (59)

where $C$ depends on initial data and model parameters but is independent of
$\varepsilon$. Integrating (58) along $(0,T)$, we get

$\displaystyle\parallel
u^{\varepsilon}_{j}\parallel_{L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))}$
$\displaystyle\leq C,\;j\in\\{1,2,4\\},$ (60) $\displaystyle\parallel
u^{\varepsilon}_{3}\parallel_{L^{2}(0,T;H^{1}(\Omega^{\varepsilon}_{1}))}$
$\displaystyle\leq C.$

With the help of (A2) together with the boundedness of $u^{\varepsilon}_{1}$,
we conclude from (16) that

$\displaystyle\parallel
u^{\varepsilon}_{5}\parallel_{L^{\infty}((0,T)\times\Gamma^{sw}_{\varepsilon})}\leq
C.$

Multiplying (16) by $\partial_{t}u^{\varepsilon}_{5}$ and using (A2), we get

$\displaystyle\|\partial_{t}u^{\varepsilon}_{5}\|_{L^{2}((0,T)\times\Gamma^{sw}_{\varepsilon})}\leq
C.$

Now, we focus on obtaining $\varepsilon-$independent estimates on the time
derivative of the concentrations. Firstly, we choose
$\varphi_{1}=\partial_{t}u^{\varepsilon}_{1}$ and get

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}\partial_{t}u^{\varepsilon}_{1}\partial_{t}u^{\varepsilon}_{1}dxd\tau$
$\displaystyle+\int_{0}^{t}\int_{\Omega^{\varepsilon}}d_{1}^{\varepsilon}\nabla
u^{\varepsilon}_{1}\nabla\partial_{t}u^{\varepsilon}_{1}dxd\tau$ (61)
$\displaystyle=-\int_{0}^{t}\int_{\Omega^{\varepsilon}}k^{\varepsilon}_{1}u^{\varepsilon}_{1}\partial_{t}u^{\varepsilon}_{1}dxd\tau+\int_{0}^{t}\int_{\Omega^{\varepsilon}}k^{\varepsilon}_{2}u^{\varepsilon}_{2}\partial_{t}u^{\varepsilon}_{1}dxd\tau$
$\displaystyle-\varepsilon\int_{0}^{t}\int_{\Gamma^{sw}_{\varepsilon}}\eta\partial_{t}u^{\varepsilon}_{1}d\sigma_{x}d\tau.$

Consequently, it holds

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{1}|^{2}dxd\tau$
$\displaystyle+\int_{0}^{t}\int_{\Omega^{\varepsilon}}\left(\frac{1}{2}\partial_{t}(d_{1}^{\varepsilon}|\nabla
u^{\varepsilon}_{1}|^{2})-(\partial_{t}d_{1}^{\varepsilon})|\nabla
u^{\varepsilon}_{1}|^{2}\right)dxd\tau$
$\displaystyle\leq-\frac{k_{1}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}\partial_{t}|u^{\varepsilon}_{1}|^{2}dxd\tau$
$\displaystyle+\frac{k_{2}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}\left(\frac{1}{\delta}|u^{\varepsilon}_{2}|^{2}+\delta|\partial_{t}u^{\varepsilon}_{1}|^{2}\right)dxd\tau$
$\displaystyle-{\varepsilon}{}\int_{0}^{t}\int_{\Gamma^{sw}_{\varepsilon}}(\partial_{t}(\eta
u^{\varepsilon}_{1})-(\partial_{t}\eta)u^{\varepsilon}_{1})d\sigma_{x}d\tau,$
$\displaystyle(1-\frac{k_{2}^{\infty}\delta}{2})\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{1}|^{2}dxd\tau$
$\displaystyle\leq D_{1}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{1}|^{2}dxd\tau$ (62)
$\displaystyle+\frac{d_{1}^{\infty}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u_{1}}_{0}|^{2}dx+\frac{k_{2}^{\infty}}{2\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|u^{\varepsilon}_{2}|^{2}dxd\tau$
$\displaystyle+\frac{\varepsilon}{2}\int_{\Gamma^{sw}_{\varepsilon}}\left(|\eta|^{2}+|u^{\varepsilon}_{1}|^{2}+|\eta(0)|^{2}+|u^{\varepsilon}_{1}(0)|^{2}\right)d\sigma_{x}$
$\displaystyle+\frac{\varepsilon}{2}\int_{0}^{t}\int_{\Gamma^{sw}_{\varepsilon}}\left(|\partial_{t}\eta|^{2}+|u^{\varepsilon}_{1}|^{2}\right)d\sigma_{x}d\tau,$

where $\eta(0):=\eta(u^{\varepsilon}_{1}(0),u^{\varepsilon}_{5}(0))$. Applying
(7) and recalling (60), we have

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{1}|^{2}dxd\tau$
$\displaystyle\leq$ $\displaystyle C_{9},$ (63)

where

$\displaystyle C_{9}$
$\displaystyle:=D_{1}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{1}|^{2}dxd\tau+\frac{k_{1}}{2}\int_{\Omega^{\varepsilon}}|u^{\varepsilon}_{1}(0)|^{2}dx+\frac{d_{1}^{\infty}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u_{1}}_{0}|^{2}dx$
$\displaystyle+\frac{k_{2}^{\infty}}{2\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|u^{\varepsilon}_{2}|^{2}dxd\tau+\frac{\varepsilon}{2}\int_{\Gamma^{sw}_{\varepsilon}}(|\overline{\eta}|^{2}+|{\eta}(0)|^{2}+|\hat{\eta}|^{2})$
$\displaystyle+\frac{C^{*}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{1}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{1}|^{2})dxd\tau+\frac{C^{*}}{2}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{1}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{1}|^{2})dxd\tau,$

and $\delta\in\left]0,\frac{2}{k_{2}^{\infty}}\right[$. Testing (13) with
$\varphi_{2}=\partial_{t}u^{\varepsilon}_{2}$ gives

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dxd\tau$
$\displaystyle+$
$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}(\frac{1}{2}\partial_{t}(d_{2}^{\varepsilon}|\nabla
u^{\varepsilon}_{2}|^{2})-(\partial_{t}d_{2}^{\varepsilon})|\nabla
u^{\varepsilon}_{2}|^{2})dxd\tau$ $\displaystyle\leq$
$\displaystyle-\frac{k_{2}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}\partial_{t}|u^{\varepsilon}_{2}|^{2}dxd\tau+\frac{k_{1}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(\frac{1}{\delta}|u^{\varepsilon}_{1}|^{2}$
$\displaystyle+$
$\displaystyle\delta|\partial_{t}u^{\varepsilon}_{2}|^{2})dxd\tau+\frac{\varepsilon
a^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}\left(|u^{\varepsilon}_{3}|^{2}+|\partial_{t}u^{\varepsilon}_{2}|^{2}\right)d\sigma_{x}d\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
b^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}\partial_{t}|u^{\varepsilon}_{2}|^{2}d\sigma_{x}d\tau,$

and hence,

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dx\tau$
$\displaystyle+$
$\displaystyle\frac{d_{2}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u_{2}}^{\varepsilon}|^{2}dxd\tau$
$\displaystyle\leq$
$\displaystyle\frac{d_{2}^{\infty}}{2}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{2}(0)|^{2}dx+D_{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{2}|^{2}dxd\tau$ $\displaystyle+$
$\displaystyle\frac{k_{1}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(\frac{1}{\delta}|u^{\varepsilon}_{1}|^{2}+\delta|\partial_{t}u^{\varepsilon}_{2}|^{2})dxd\tau$
$\displaystyle+$
$\displaystyle\frac{C^{*}a^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}\left(|u^{\varepsilon}_{3}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{3}|^{2}+\varepsilon^{2}|\nabla\partial_{t}u^{\varepsilon}_{2}|^{2}\right)dxd\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
b^{\infty}}{2}\int_{\Gamma^{wa}_{\varepsilon}}(|u^{\varepsilon}_{2}|^{2}-|u^{\varepsilon}_{2}(0)|^{2})d\sigma_{x}.$

By (7) and (60), we get

$\displaystyle\left(1-\frac{C^{*}a^{\infty}}{2}-\frac{k_{1}^{\infty}\delta}{2}\right)\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dxd\tau$
$\displaystyle\leq$ $\displaystyle
C_{10}\left(1+\varepsilon^{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}u^{\varepsilon}_{2}|^{2}dxd\tau\right).$

Consequently, choosing
$\delta\in]0,\frac{2-C^{*}a^{\infty}}{k_{1}^{\infty}}[$, we are led to

$\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dxd\tau\leq
C_{10}(1+\varepsilon^{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}u^{\varepsilon}_{2}|^{2}dxd\tau),$
(64)

where

$\displaystyle C_{10}$ $\displaystyle:=$ $\displaystyle
D_{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{1}|^{2}dxd\tau+\frac{d_{2}^{\infty}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u^{\varepsilon}}_{2}(0)|^{2}dx+\frac{k_{1}^{\infty}}{2\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|u^{\varepsilon}_{2}|^{2}dxd\tau$
$\displaystyle+$
$\displaystyle\frac{C^{*}b^{\infty}}{2}\int_{\Omega^{\varepsilon}}\left(|u^{\varepsilon}_{2}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{2}|^{2}+|u^{\varepsilon}_{2}(0)|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{2}(0)|^{2}\right)dx$ $\displaystyle+$
$\displaystyle\frac{C^{*}a^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{3}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{3}|^{2})dxd\tau.$

The initial data $u^{\varepsilon}_{30}$ holding in $\Omega^{\varepsilon}_{1}$
and the Dirichlet data $u_{3}^{D}$ acting on the exterior boundary of
$\Omega_{1}^{\varepsilon}$ are considered here as restrictions of the
respective functions defined on whole of $\overline{\Omega}$. Testing now (14)
with $\varphi_{3}=\partial_{t}(u^{\varepsilon}_{3}-u_{3}^{D})$ leads to

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{3}|^{2}dxd\tau$
$\displaystyle+\frac{d_{3}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u^{\varepsilon}}_{3}|^{2}$
$\displaystyle\leq\frac{d_{3}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u^{\varepsilon}_{3}}(0)|^{2}+\frac{1}{2}(|\partial_{t}u^{\varepsilon}_{3}|^{2}+|\partial_{t}u^{D}_{3}|^{2})$
$\displaystyle+D_{3}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla{u^{\varepsilon}_{3}}|^{2}+\frac{d_{3}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\nabla{u^{\varepsilon}_{3}}|^{2}+|\nabla\partial_{t}u^{D}_{3}|^{2})$
$\displaystyle+\frac{{\varepsilon
a^{\infty}}}{\delta}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|u^{\varepsilon}_{3}|^{2}+\frac{\varepsilon\delta}{2}(a^{\infty}+a^{\infty})\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|\partial_{t}u^{\varepsilon}_{3}|^{2}$
$\displaystyle+\frac{\varepsilon}{2}(a^{\infty}+b^{\infty})\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|\partial_{t}u^{D}_{3}|^{2}+\frac{\varepsilon
b^{\infty}}{\delta}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|u^{\varepsilon}_{2}|^{2}.$

Using (7) and (A6), we obtain

$\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{3}|^{2}dxd\tau\leq
C_{11}(1+\varepsilon^{2}\delta\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}u^{\varepsilon}_{3}|^{2}dxd\tau),$
(65)

where $\delta\in]0,\frac{2}{C^{*}(a^{\infty}+b^{\infty})}[$ and

$\displaystyle C_{11}$
$\displaystyle:=D_{3}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{\varepsilon}_{3}|^{2}dxd\tau+\frac{d_{3}}{2}\int_{\Omega^{\varepsilon}}|\nabla{u^{\varepsilon}_{3}}(0)|^{2}dx+\frac{1}{2\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla
u^{D}_{3}|^{2}$
$\displaystyle+\frac{d^{\infty}_{3}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\nabla
u^{\varepsilon}_{3}|^{2}+|\nabla\partial_{t}u^{D}_{3}|^{2})+\frac{C^{*}a^{\infty}}{\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{3}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{3}|^{2})$
$\displaystyle+\frac{C^{*}b^{\infty}}{\delta}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{2}|^{2}+\varepsilon^{2}|\nabla
u^{\varepsilon}_{2}|^{2})dxd\tau$
$\displaystyle+\frac{C^{*}(a^{\infty}+b^{\infty})}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\partial_{t}u^{D}_{3}|^{2}+\varepsilon^{2}|\nabla\partial_{t}u^{D}_{3}|^{2})dxd\tau.$

From (15), we get

$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{4}|^{2}dxd\tau$
$\displaystyle\leq$ $\displaystyle C_{12}.$ (66)

In order to estimate (64) and (65), we proceed first with differentiating (13)
with respect to time and then testing the result with
$\partial_{t}u^{\varepsilon}_{2}$. Consequently, we derive

$\displaystyle\frac{1}{2}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dx$
$\displaystyle+$
$\displaystyle{d_{2}}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{2}}^{\varepsilon}|^{2}dxd\tau$
(67) $\displaystyle+$
$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}\left(\frac{1}{2}(\partial_{t}d_{2}|\nabla{u_{2}}^{\varepsilon}|^{2}-(\partial_{t}\partial_{t}d_{2})|\nabla{u_{2}}^{\varepsilon}|^{2}\right)$
$\displaystyle\leq$
$\displaystyle\frac{k_{1}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\partial_{t}u^{\varepsilon}_{1}|^{2}+|\partial_{t}u^{\varepsilon}_{2}|^{2})+\frac{K_{1}^{\infty}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|u^{\varepsilon}_{1}|^{2}+|\partial_{t}u^{\varepsilon}_{2}|^{2})$
$\displaystyle-$ $\displaystyle
k_{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}-\frac{K_{2}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}\partial_{t}|u^{\varepsilon}_{2}|^{2}$
$\displaystyle+$ $\displaystyle\frac{{\varepsilon
a^{\infty}}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}(\frac{1}{\delta}|\partial_{t}u^{\varepsilon}_{2}|^{2}+\delta|\partial_{t}u^{\varepsilon}_{3}|^{2})d\sigma_{x}d\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
b^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}d\sigma_{x}d\tau+\frac{\varepsilon
B}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}\partial_{t}|u^{\varepsilon}_{2}|^{2}d\sigma_{x}d\tau.$

Using (7), it yields

$\displaystyle\frac{1}{2}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2}dx$
$\displaystyle+\left({d_{2}}-\frac{C^{*}A^{\infty}\varepsilon^{2}}{2}-\frac{C^{*}B^{\infty}\varepsilon^{2}}{2}-\frac{C^{*}a^{\infty}\varepsilon^{2}}{2\delta}\right)\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{2}}^{\varepsilon}|^{2}dxd\tau$
(68) $\displaystyle\leq
C_{13}+\frac{C^{*}a^{\infty}\delta}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\partial_{t}u^{\varepsilon}_{3}|^{2}+\varepsilon^{2}|\nabla\partial_{t}u^{\varepsilon}_{3}|^{2})$
$\displaystyle+\left(\frac{k_{1}^{\infty}}{2}+\frac{K_{1}^{\infty}}{2}-k_{2}+\frac{C^{*}A^{\infty}}{2}+\frac{C^{*}B^{\infty}\varepsilon^{2}}{2}+\frac{C^{*}a^{\infty}}{2\delta}\right)\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{2}|^{2},$

where $C_{13}$ depends on the bounded terms of r.h.s of (67). Differentiating
now (14) with respect to time and then testing the result with
$\partial_{t}(u^{\varepsilon}_{3}-u^{D}_{3})$, we get

$\displaystyle\frac{1}{2}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{3}|^{2}dx$
$\displaystyle+$
$\displaystyle{d_{3}}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{3}}^{\varepsilon}|^{2}dxd\tau$
$\displaystyle+$
$\displaystyle\int_{0}^{t}\int_{\Omega^{\varepsilon}}\left(\frac{1}{2}(\partial_{t}d_{3}|\nabla{u_{3}}^{\varepsilon}|^{2}-(\partial_{t}\partial_{t}d_{3})|\nabla{u_{3}}^{\varepsilon}|^{2}\right)$
$\displaystyle\leq$
$\displaystyle\frac{D_{3}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla{u_{3}}^{\varepsilon}|^{2}dxd\tau+\frac{d^{\infty}_{3}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{3}}^{\varepsilon}|^{2}dxd\tau$
$\displaystyle+$
$\displaystyle\frac{d^{\infty}_{3}+D_{3}}{2}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{3}}^{D}|^{2}dxd\tau+\frac{\varepsilon
A^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}\partial_{t}|{u_{3}}^{\varepsilon}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle{\varepsilon
a^{\infty}}{}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}|\partial_{t}{u_{3}}^{\varepsilon}|^{2}dxd\tau+\frac{\varepsilon
A^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}(|{u_{3}}^{\varepsilon}|^{2}+|\partial_{t}{u_{3}}^{D}|^{2})dxd\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
a^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}(|\partial_{t}{u_{3}}^{\varepsilon}|^{2}+|\partial_{t}{u_{3}}^{D}|^{2})dxd\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
B^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}(|{u_{2}}^{\varepsilon}|^{2}+|\partial_{t}{u_{3}}^{\varepsilon}|^{2}+|{u_{2}}^{\varepsilon}|^{2}+|\partial_{t}{u_{3}}^{D}|^{2})dxd\tau$
$\displaystyle+$ $\displaystyle\frac{\varepsilon
b^{\infty}}{2}\int_{0}^{t}\int_{\Gamma^{wa}_{\varepsilon}}(\frac{1}{\delta}|\partial_{t}{u_{2}}^{\varepsilon}|^{2}+\delta|\partial_{t}{u_{3}}^{\varepsilon}|^{2}+|\partial_{t}{u_{2}}^{\varepsilon}|^{2}+|\partial_{t}{u_{3}}^{D}|^{2})dxd\tau.$

Using (7) to deal with the boundary terms, we obtain

$\displaystyle\frac{1}{2}\int_{\Omega^{\varepsilon}}|\partial_{t}u^{\varepsilon}_{3}|^{2}dx$
$\displaystyle+$
$\displaystyle\left({d_{3}}-\frac{d_{3}^{\infty}}{2}-\frac{C^{*}\varepsilon^{2}}{2}(3a^{\infty}+B^{\infty}+b^{\infty}+a^{\infty}\delta)\right)\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\nabla\partial_{t}{u_{3}}^{\varepsilon}|^{2}dxd\tau$
$\displaystyle\leq$ $\displaystyle
C_{14}+C_{15}\int_{0}^{t}\int_{\Omega^{\varepsilon}}|\partial_{t}{u_{3}}^{\varepsilon}|^{2}dxd\tau$
(69) $\displaystyle+$ $\displaystyle
C^{*}b^{\infty}\int_{0}^{t}\int_{\Omega^{\varepsilon}}(|\partial_{t}{u_{3}}^{\varepsilon}|^{2}+\varepsilon^{2}|\nabla\partial_{t}{u_{3}}^{\varepsilon}|^{2})dxd\tau$
(70)

Adding (68) and (4) and using (64) and (65) to get the desired result.

###  4.1 Extension step

Since we deal here with an oscillating system posed in a perforated domain,
the natural next step is to extend all concentrations to the whole $\Omega$.
We do this by following a two-steps procedure: In Step 1, we rely on the
standard extension results indicated in section 4.2 to extend all active
concentrations $u^{\varepsilon}_{\ell}$ ($\ell\in\\{1,\dots,4\\}$) to
$\Omega$. In step 2, we unfold the ode for $u^{\varepsilon}_{5}$ such that the
unfolded concentration is defined on the fixed boundary $\Gamma$; see section
5.1.

###  4.2 Extension lemmas

Since all the concentrations are defined in $\Omega^{\varepsilon}$ and
$\Omega^{\varepsilon}_{1}$, to get macroscopic equations we need to extend
them into $\Omega$.

######  Remark 6.

Take $\varphi^{\varepsilon}\in L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))$. Note
that since our microscopic geometry is sufficiently regular, we can speak in
terms of extensions. Recall the linearity of the extension operator

$\mathcal{P}^{\varepsilon}:L^{2}(0,T;H^{1}(\Omega^{\varepsilon}))\rightarrow
L^{2}(0,T;H^{1}(\Omega))$

defined by
$\mathcal{P}^{\varepsilon}\varphi^{\varepsilon}=\tilde{\varphi}^{\varepsilon}$.
To keep notation simple, we denote the extension
$\tilde{\varphi}^{\varepsilon}$ again by $\varphi^{\varepsilon}$.

######  Lemma 7.

(Extension) Consider the geometry described in Section 2.1. There exists an
extension $\tilde{u}^{\varepsilon}$ of $u^{\varepsilon}$ such that

  1. 1.

$\parallel\tilde{u}^{\varepsilon}\parallel_{L^{2}(Y)}\leq\hat{C}\parallel
u^{\varepsilon}\parallel_{L^{2}(Y^{w})},$ for
$u^{\varepsilon}\in{L^{2}(Y^{w})}$

  2. 2.

$\parallel\nabla\tilde{u}^{\varepsilon}\parallel_{L^{2}(Y)}\leq\hat{C}\parallel\nabla
u^{\varepsilon}\parallel_{L^{2}(Y^{w})},$ for $\nabla
u^{\varepsilon}\in{L^{2}(Y^{w})}$

  3. 3.

$\parallel\tilde{u}^{\varepsilon}\parallel_{H^{1}(\Omega)}\leq\hat{C}\parallel
u^{\varepsilon}\parallel_{H^{1}(\Omega^{\varepsilon})}$, for
$u^{\varepsilon}\in H^{1}(\Omega^{\varepsilon})$

Proof. For the proof of this Lemma, see Section 2 in [20] or compare Lemma 5,
p.214 in [30].

######  Definition 8.

(Two-scale convergence cf. [31, 32]) Let $\\{u^{\varepsilon}\\}$ be a sequence
of functions in $L^{2}((0,T)\times\Omega)$ ($\Omega$ being an open set of
$\mathbb{R}^{N}$) where $\varepsilon$ being a sequence of strictly positive
numbers that tends to zero. $\\{u^{\varepsilon}\\}$ is said to two-scale
converge to a unique function $u_{0}(t,x,y)\in L^{2}((0,T)\times\Omega\times
Y)$ if and only if for any $\psi\in
C^{\infty}_{0}((0,T)\times\Omega,C^{\infty}_{\\#}(Y))$, we have

$\displaystyle lim_{\varepsilon\rightarrow
0}\int_{0}^{T}\int_{\Omega}u^{\varepsilon}\psi(t,x,\frac{x}{\varepsilon})dxdt=\int_{\Omega}\int_{Y}u_{0}(t,x,y)\psi(t,x,y)dydxdt.$
(71)

We denote (71) by $u^{\varepsilon}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}u_{0}$.

######  Theorem 9.

  1. (i)

From each bounded sequence $\\{u^{\varepsilon}\\}$ in
$L^{2}((0,T)\times\Omega)$, one can extract a subsequence which two-scale
converges to $u_{0}(t,x,y)\in L^{2}((0,T)\times\Omega\times Y)$.

  2. (ii)

Let $\\{u^{\varepsilon}\\}$ be a bounded sequence in
$H^{1}((0,T)\times\Omega)$, which converges weakly to a limit function
$u_{0}(t,x,y)\in H^{1}((0,T)\times\Omega\times Y)$. Then there exists
$\tilde{u}\in L^{2}(\Omega;H^{1}_{\\#}(Y)/\mathbb{R})$ such that up to a
subsequence $\\{u^{\varepsilon}\\}$ two-scale converges to $u_{0}(t,x,y)$ and
$\nabla u^{\varepsilon}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}\nabla_{x}u_{0}+\nabla_{y}\tilde{u}.$

  3. (iii)

Let $\\{u^{\varepsilon}\\}$ and $\\{\varepsilon\nabla u^{\varepsilon}\\}$ be
bounded sequences in $L^{2}((0,T)\times\Omega)$, then there exists${u_{0}}\in
L^{2}((0,T)\times\Omega;H^{1}_{\\#}(Y))$ such that up to a subsequence
$u^{\varepsilon}$ and $\varepsilon\nabla u^{\varepsilon}$ two-scale converge
to $u_{0}(t,x,y)$ and $\nabla_{y}u_{0}(t,x,y)$ respectively.

######  Definition 10.

(Two-scale convergence for $\varepsilon-$periodic hypersurfaces [33]) A
sequence of functions $\\{u^{\varepsilon}\\}$ in
$L^{2}((0,T)\times\Gamma_{\varepsilon})$ is said to two-scale converge to a
limit $u_{0}\in L^{2}((0,T)\times\Omega\times\Gamma)$ if and only if for any
$\psi\in C^{\infty}_{0}((0,T)\times\Omega,C^{\infty}_{\\#}(\Gamma))$ we have

$\displaystyle lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma_{\varepsilon}}u^{\varepsilon}\psi(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt=\int_{\Omega}\int_{\Gamma}u_{0}(t,x,y)\psi(t,x,y)d\sigma_{y}dxdt.$

######  Theorem 11.

  1. (i)

From each bounded sequence $\\{u^{\varepsilon}\\}\in
L^{2}((0,T)\times\Gamma_{\varepsilon})$, one can extract a subsequence
$u^{\varepsilon}$ which two-scale converges to a function $u_{0}\in
L^{2}((0,T)\times\Omega\times\Gamma)$.

  2. (ii)

If a sequence of functions $\\{u^{\varepsilon}\\}$ is bounded in
$L^{\infty}((0,T)\times\Gamma_{\varepsilon})$, then $u^{\varepsilon}$ two-
scale converges to a function $u_{0}\in
L^{\infty}((0,T)\times\Omega\times\Gamma)$.

Proof. For proof of (i), see [33] and the one for (ii), see [15].

######  Lemma 12.

Assume the hypotheses of Lemma 5 and Lemma 7 to hold. The a priori estimates
lead to the following convergence results:

  1. (a)

$u^{\varepsilon}_{i}\rightharpoonup u_{i}$ in $L^{2}(0,T;H^{1}(\Omega)$ for
all $i\in\\{1,2,3,4\\}$,

  2. (b)

$u^{\varepsilon}_{i}\stackrel{{\scriptstyle\ast}}{{\rightharpoonup}}u_{i}$ in
$L^{\infty}((0,T)\times\Omega)$,

  3. (c)

$\partial_{t}u^{\varepsilon}_{i}\rightharpoonup\partial_{t}u_{i}$ in
$L^{2}((0,T)\times\Omega)$,

  4. (d)

$u^{\varepsilon}_{i}\rightarrow u_{i}$ in $L^{2}(0,T;H^{\beta}(\Omega))$ for
$\frac{1}{2}<\beta<1$, also $\parallel
u^{\varepsilon}_{i}-u_{i}\parallel_{L^{2}((0,T)\times\Gamma_{\varepsilon})}\rightarrow
0$ as $\varepsilon\rightarrow 0$,

  5. (e)

$u^{\varepsilon}_{i}\stackrel{{\scriptstyle 2}}{{\rightharpoonup}}u_{i},\nabla
u^{\varepsilon}_{i}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}\nabla_{x}u_{i}+\nabla_{y}u_{i1}$, $u_{i1}\in
L^{2}((0,T)\times\Omega;H^{1}_{\\#}(Y)/\mathbb{R})$,

  6. (f)

$u^{\varepsilon}_{5}\stackrel{{\scriptstyle 2}}{{\rightharpoonup}}u_{5}$, and
$u_{5}\in L^{\infty}((0,T)\times\Omega\times\Gamma^{sw})$,

  7. (g)

$\partial_{t}u^{\varepsilon}_{5}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}\partial_{t}u_{5}$, and $u_{5}\in
L^{2}((0,T)\times\Omega\times\Gamma^{sw})$.

Proof. (a) and (b) are obtained as a direct consequence of the fact that
$u^{\varepsilon}_{i}$ is bounded in $L^{2}(0,T;H^{1}(\Omega))\cap
L^{\infty}((0,T)\times\Omega)$; up to a subsequence (still denoted by
$u^{\varepsilon}_{i}$) $u^{\varepsilon}_{i}$ converges weakly to $u_{i}$ in
$L^{2}(0,T;H^{1}(\Omega))\cap L^{\infty}((0,T)\times\Omega)$. A similar
argument gives (c). To get (d), we use the compact embedding
$H^{\beta^{\prime}}(\Omega)\hookrightarrow H^{\beta}(\Omega)$, for
$\beta\in(\frac{1}{2},1)$ and $0<\beta<\beta^{\prime}\leq 1$ (since $\Omega$
has Lipschitz boundary). We have

$W:=\\{u_{i}\in L^{2}(0,T;H^{1}(\Omega))\;\mbox{and}\;\partial_{t}u_{i}\in
L^{2}((0,T)\times\Omega)\mbox{ for all}\;i\in\\{1,2,3,4\\}\\}$

For a fixed $\varepsilon$, $W$ is compactly embedded in
$L^{2}(0,T;H^{\beta}(\Omega))$ by the Lions-Aubin Lemma; cf. e.g. [34]. Using
the trace inequality (8)

$\displaystyle\parallel
u^{\varepsilon}_{i}-u_{i}\parallel_{L^{2}((0,T)\times\Gamma_{\varepsilon})}$
$\displaystyle\leq$ $\displaystyle{C^{*}_{0}}\parallel
u^{\varepsilon}_{i}-u_{i}\parallel_{L^{2}(0,T;H^{\beta}(\Omega^{\varepsilon}))},$
$\displaystyle\leq$ $\displaystyle C\parallel
u^{\varepsilon}_{i}-u_{i}\parallel_{L^{2}(0,T;H^{\beta}(\Omega))},$

where $\parallel
u^{\varepsilon}_{i}-u_{i}\parallel_{L^{2}(0,T;H^{\beta}(\Omega))}\rightarrow
0$ as $\varepsilon\rightarrow 0.$ To investigate (e), (f) and (g), we use the
notion of two-scale convergence as indicated in Definition 8 and 10. Since
$u^{\varepsilon}_{i}$ are bounded in $L^{2}(0,T;H^{1}(\Omega)$, up to a
subsequence $u^{\varepsilon}_{i}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}u_{i}$ in $L^{2}((0,T)\times\Omega\times Y)$, and
$\nabla u^{\varepsilon}_{i}\stackrel{{\scriptstyle
2}}{{\rightharpoonup}}\nabla_{x}u_{i}+\nabla_{y}\tilde{u}_{i}$,
$\tilde{u}_{i}\in L^{2}((0,T)\times\Omega;H^{1}_{\\#}(Y)/\mathbb{R})$. By
Theorem 11, $u^{\varepsilon}_{5}$ in
$L^{\infty}((0,T)\times\Omega\times\Gamma)$ converges two-scale to $u_{5}$ in
the same space and $\partial_{t}u^{\varepsilon}_{5}$ converges two-scale to
$\partial_{t}u_{5}$ in $L^{2}((0,T)\times\Omega\times\Gamma)$. Due to the
presence of the non-linear reaction rate on the interface
$\Gamma^{sw}_{\varepsilon}$, the convergences listed in Lemma 12 are still not
sufficient to pass to the limit $\varepsilon\rightarrow 0$ in the microscopic
model. To be more precise, we can pass to $\varepsilon\rightarrow 0$ in the
pde’s, but not in the ode.

###  4.3 Cell problems

In order to be able to formulate the upscaled equations, we define two classes
of cell problems very much in the spirit of [9]. One class of problems will
refer to the water-filled parts of the pore, while the second class will refer
to the air-filled part of the pores.

######  Definition 13.

(Cell problems) The cell problems in water-filled part are given by

$\displaystyle-\nabla_{y}.(D_{\ell}(t,y)\nabla_{y}\chi_{i})$
$\displaystyle=\sum_{k=1}^{3}\partial_{y_{k}}{D_{\ell}}_{ki}(t,y),\;\mbox{in}\;\;Y^{w},$
$\displaystyle-D_{\ell}(t,y)\frac{\partial\chi_{i}}{\partial n}$
$\displaystyle=\sum_{k=1}^{3}{D_{\ell}}_{ki}(t,y)n_{k}\;\;\mbox{on}\;\Gamma^{sw},$
$\displaystyle-D_{\ell}(t,y)\frac{\partial\chi_{i}}{\partial n}$
$\displaystyle=\sum_{k=1}^{3}{D_{\ell}}_{ki}(t,y)n_{k}\;\;\mbox{on}\;\Gamma^{wa},$

for all $i,\ell\in\\{1,2,4\\}$ and $\chi_{i}$ are Y-periodic in y. The cell
problems in air-filled part are given by

$\displaystyle-\nabla_{y}.(D_{3}(t,y)\nabla_{y}\varsigma_{i})$
$\displaystyle=\sum_{k=1}^{3}\partial_{y_{k}}{D_{3}}_{ki}(t,y),\;\mbox{in}\;\;Y^{a},$
$\displaystyle-D_{3}(t,y)\frac{\partial\varsigma_{i}}{\partial n}$
$\displaystyle=\sum_{k=1}^{3}{D_{3}}_{ki}(t,y)n_{k}\;\;\mbox{on}\;\Gamma^{wa},$
$\displaystyle-D_{3}(t,y)\frac{\partial\varsigma_{i}}{\partial n}$
$\displaystyle=\sum_{k=1}^{3}{D_{3}}_{ki}(t,y)n_{k}\;\;\mbox{on}\;\partial
Y^{a}-\Gamma^{wa},$

for all $i\in\\{1,2,3\\}$ and $\varsigma_{i}$ are $Y$-periodic in $y$.

##  5 Two-scale limit equations

######  Theorem 14.

The sequences of the solutions of the weak formulation (12)-(16) converges to
the weak solution $u_{i},i\in\\{1,2,3,4,5\\}$ as $\varepsilon\rightarrow 0$
such that $u_{i}\in H^{1}(0,T;L^{2}(\Omega))\cap L^{2}(0,T;H^{1}(\Omega))\cap
L^{\infty}((0,T)\times\Omega)$ and $u_{5}\in
H^{1}(0,T;L^{2}(\Omega\times\Gamma))\cap
L^{\infty}((0,T)\times\Omega\times\Gamma))$. The weak formulation of the two-
scale limit equations is given by

$\displaystyle\int_{0}^{T}\int_{\Omega}\partial_{t}u_{i}(t,x)\phi_{i}(t,x)dxdt$
$\displaystyle+$ $\displaystyle\int_{0}^{T}\int_{\Omega}\tilde{d}_{i}(t)\nabla
u_{i}(t,x)\nabla\phi_{i}dxdt$ $\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{\Omega}F_{i}(u)\phi_{i}dxdt\mbox{ for all
}i\in\\{1,2,3,4\\},$

where

$\displaystyle F_{1}(u)$ $\displaystyle:=$
$\displaystyle-\tilde{k}_{1}(t)u_{1}(t,x)+\tilde{k}_{2}(t)u_{2}(t,x)$
$\displaystyle\;-\frac{1}{|Y|}\int_{\Gamma}k_{3}(t,y)R(u_{1}(t,x))Q(u_{5}(t,x,y))d\sigma_{y},$
$\displaystyle F_{2}(u)$ $\displaystyle:=$
$\displaystyle\tilde{k}_{1}(t)u_{1}(t,x)-\tilde{k}_{2}(t)u_{2}(t,x)+\tilde{a}(t)u_{3}(t,x)-\tilde{b}(t)u_{2}(t,x),$
$\displaystyle F_{3}(u)$ $\displaystyle:=$
$\displaystyle-\tilde{a}(t)u_{3}(t,x)+\tilde{b}(t)u_{2}(t,x),$ $\displaystyle
F_{4}(u)$ $\displaystyle:=$ $\displaystyle\tilde{k}_{1}(t)u_{1}(t,x),$

with the initial values $u_{i}(0,x)={u_{i}}_{0}(x)$ for $x\in\Omega$, and

$\displaystyle\int_{0}^{T}\int_{\Omega\times\Gamma}$
$\displaystyle\partial_{t}u_{5}(t,x,y)\phi_{5}(t,x,y)dtdxd\sigma_{y}$ (73)
$\displaystyle=\int_{0}^{T}\int_{\Omega\times\Gamma}k_{3}(t,y)R(u_{1}(t,x))Q(u_{5}(t,x,y))\phi_{5}(t,x,y)dtdxd\sigma_{y},$

with $u_{5}(0,x,y)={u_{5}}_{0}(x,y)$ for $\;x\in\Omega,\;y\in\Gamma^{sw}$.
Also
$\phi:=(\phi_{1},\phi_{2},\phi_{3},\phi_{4})\in[C^{\infty}((0,T)\times\Omega)]^{4}$,
$\psi:=(\psi_{1},\psi_{2},\psi_{3},\psi_{4})\in[C^{\infty}((0,T)\times\Omega);C^{\infty}_{\\#}(Y)]^{4}$,

$\displaystyle\tilde{k}_{j}(t)$ $\displaystyle:=$
$\displaystyle\frac{1}{|Y|}\int_{Y}k_{j}(t,y)dy,\;j\in\\{1,2\\},$ (74)
$\displaystyle\tilde{a}(t)$ $\displaystyle:=$
$\displaystyle\frac{1}{|Y|}\int_{\Gamma^{wa}}a(t,y)d\sigma_{y},$ (75)
$\displaystyle\tilde{b}(t)$ $\displaystyle:=$
$\displaystyle\frac{1}{|Y|}\int_{\Gamma^{wa}}b(t,y)d\sigma_{y},$ (76)
$\displaystyle{\tilde{d}_{\ell ij}}$ $\displaystyle:=$
$\displaystyle\sum_{k=1}^{3}\int_{Y}(d_{\ell ij}(t,y)+d_{\ell
ik}(t,y)(\delta_{n\ell}\partial_{y_{k}}\chi_{j}+\delta_{3\ell}\partial_{y_{k}}\varsigma_{j})dy,$
$\displaystyle\ell\in\\{1,2,3,4\\},\,n\in\\{1,2,4\\}$

with $\chi_{j},\varsigma_{j}$ being solutions of the cell problems defined in
Definition 13, while $\delta$ denotes here the Kronecker’s symbol.

Proof. We apply two-scale convergence techniques together with Lemma 12 to get
macroscopic equations. We take test functions incorporating the following
oscillating behavior
$\varphi_{i}(t,x)=\phi_{i}(t,x)+\varepsilon\psi_{i}(t,x,\frac{x}{\varepsilon}),\phi_{i}\in
C^{\infty}((0,T)\times\Omega),\phi_{i}\in
C^{\infty}((0,T)\times\Omega,;C^{\infty}_{\\#}(Y)),i\in\\{1,2,3,4\\}$.
Applying two-scale convergence yields

$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}$
$\displaystyle\partial_{t}u_{i}\phi_{i}(t,x)dxdt+\int_{0}^{T}\int_{\Omega}\int_{Y}d_{i}(t,y)(\nabla_{x}u_{i}(t,x)$
(78) $\displaystyle+$
$\displaystyle\nabla_{y}\tilde{u}_{i}(t,x,y))(\nabla_{x}\phi_{i}(t,x)+\nabla_{y}\psi_{i}(t,x,\frac{x}{\varepsilon}))dydxdt$
$\displaystyle=$
$\displaystyle\int_{0}^{T}\int_{\Omega}f_{i}(u)\phi_{i}(t,x)dxdt.$
$\displaystyle\int_{0}^{T}\int_{\Omega}f_{1}(u)$
$\displaystyle\phi_{1}(t,x)dxdt=-\lim_{\varepsilon\rightarrow
0}\int_{0}^{T}\int_{\Omega^{\varepsilon}}k_{1}^{\varepsilon}u^{\varepsilon}_{1}(\phi_{1}(t,x)+\varepsilon\psi_{1}(t,x,\frac{x}{\varepsilon}))dxdt$
$\displaystyle+$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\int_{0}^{T}\int_{\Omega^{\varepsilon}}k_{2}^{\varepsilon}u^{\varepsilon}_{2}(\phi_{1}(t,x)+\varepsilon\psi_{1}(t,x,\frac{x}{\varepsilon}))dxdt$
$\displaystyle-$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma^{sw}_{\varepsilon}}\eta(R(u^{\varepsilon}_{1}),Q(u^{\varepsilon}_{5}))(\phi_{1}(t,x)+\varepsilon\psi_{1}(t,x,\frac{x}{\varepsilon}))d\sigma_{x}dt.$

Using Lemma 12, we have

$\displaystyle\int_{0}^{T}\int_{\Omega}f_{1}(u)\phi_{1}(t,x)dxdt$
$\displaystyle=$
$\displaystyle-\int_{0}^{T}\int_{\Omega}\int_{Y}k_{1}(t,y)u_{1}(t,x)\phi_{1}(t,x)dydxdt$
$\displaystyle+$
$\displaystyle\int_{0}^{T}\int_{\Omega}\int_{Y}k_{2}(t,y)u_{2}(t,x)\phi_{1}(t,x)dydxdt$
$\displaystyle-$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma^{sw}_{\varepsilon}}\partial_{t}u^{\varepsilon}_{5}(\phi_{1}(t,x)+\varepsilon\psi_{1}(t,x,\frac{x}{\varepsilon}))d\sigma_{x}dt.$
$\displaystyle\int_{0}^{T}\int_{\Omega}f_{1}(u)\phi_{1}(t,x)dxdt$
$\displaystyle=$
$\displaystyle-|Y|\int_{0}^{T}\int_{\Omega}\tilde{k}_{1}(t)u_{1}(t,x)\phi_{1}(t,x)dxdt$
$\displaystyle+$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{k}_{2}(t)u_{2}(t,x)\phi_{1}(t,x)dxdt$
$\displaystyle-$
$\displaystyle\int_{0}^{T}\int_{\Omega}\int_{\Gamma}\partial_{t}u_{5}\phi_{1}(t,x)d\sigma_{y}dxdt.$
$\displaystyle\int_{0}^{T}\int_{\Omega}f_{2}(u)\phi_{2}(t,x)dxdt$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\int_{0}^{T}\int_{\Omega^{\varepsilon}}k_{1}^{\varepsilon}u^{\varepsilon}_{1}(\phi_{2}(t,x)+\varepsilon\psi_{2}(t,x,\frac{x}{\varepsilon}))dxdt$
$\displaystyle-$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\int_{0}^{T}\int_{\Omega^{\varepsilon}}k_{2}^{\varepsilon}u^{\varepsilon}_{2}(\phi_{2}(t,x)+\varepsilon\psi_{2}(t,x,\frac{x}{\varepsilon}))dxdt$
$\displaystyle+$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma^{wa}_{\varepsilon}}a_{\varepsilon}u^{\varepsilon}_{3}(\phi_{2}(t,x)+\varepsilon\psi_{2}(t,x,\frac{x}{\varepsilon}))d\sigma_{x}dt$
$\displaystyle-$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma^{wa}_{\varepsilon}}b_{\varepsilon}u^{\varepsilon}_{2}(\phi_{2}(t,x)+\varepsilon\psi_{2}(t,x,\frac{x}{\varepsilon}))d\sigma_{x}dt.$
$\displaystyle\int_{0}^{T}\int_{\Omega}f_{2}(u)\phi_{2}(t,x)dxdt$
$\displaystyle=$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{k}_{1}(t)u_{1}(t,x)\phi_{2}(t,x)dxdt$
$\displaystyle-$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{k}_{2}(t)u_{2}(t,x)\phi_{2}(t,x)dxdt$
$\displaystyle+$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{a}(t)u_{3}(t,x)\phi_{2}(t,x)dxdt$
$\displaystyle-$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{b}(t)u_{2}(t,x)\phi_{2}(t,x)dxdt.$

We also have

$\displaystyle\int_{0}^{T}\int_{\Omega}f_{3}(u)\phi_{3}(t,x)dxdt$
$\displaystyle=$
$\displaystyle-|Y|\int_{0}^{T}\int_{\Omega}\tilde{a}(t)u_{3}(t,x)\phi_{3}(t,x)dxdt$
$\displaystyle+$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{b}(t)u_{2}(t,x)\phi_{3}(t,x)dxdt$

and

$\displaystyle\int_{0}^{T}\int_{\Omega}f_{4}(u)\phi_{4}(t,x)dxdt$
$\displaystyle=$
$\displaystyle|Y|\int_{0}^{T}\int_{\Omega}\tilde{k}_{1}(t)u_{1}(t,x)\phi_{4}(t,x)dxdt.$

We set $\phi_{i}=0,i\in\\{1,2,3,4\\}$ in (78) to calculate the expression of
the known function $\tilde{u}_{1}$ and obtain

$\int_{0}^{T}\int_{\Omega}\int_{Y}d_{i}(t,y)(\nabla_{x}u_{i}(t,x)+\nabla_{y}\tilde{u}_{i}(t,x,y))\nabla_{y}\psi_{i}(t,x,\frac{x}{\varepsilon})dydxdt=0,\mbox{
forall }\psi_{i}.$

Since $\tilde{u}_{1}$ depends linearly on $\nabla_{x}u_{1}$, it can be defined
as

$\tilde{u}_{i}:=\sum_{j=1}^{3}\partial_{x_{j}}u_{i}(\delta_{in}\chi_{j}(t,y)+\delta_{3i}\varsigma_{j}(t,y))\;\mbox{for}\;n\in\\{1,2,4\\}$

where the function $\chi_{j},\varsigma_{j}$ are the unique solutions of the
cell problems defined in Definition 13. Setting $\psi_{i}=0$ in (78), we get

$\displaystyle\int_{0}^{T}\int_{\Omega}\int_{Y}$
$\displaystyle\sum_{j,k=1}^{3}{d_{i}}_{jk}(t,y)(\partial_{x_{k}}u_{i}(t,x)$
$\displaystyle+\sum_{m=1}^{3}(\delta_{in}\partial_{y_{k}}\chi_{m}+\delta_{3i}\partial_{y_{k}}\varsigma_{m})\partial_{x_{m}}u_{i}(t,x))\partial_{x_{j}}\phi_{k}(t,x)dydxdt$
$\displaystyle=|Y|\int_{0}^{T}\int_{\Omega}\sum_{j,k=1}^{3}{\tilde{d}_{ijk}}\partial_{x_{k}}u_{i}(t,x)\partial_{x_{j}}\phi_{i}(t,x)dxdt.$

Hence, the coefficients (entering the effective diffusion tensor) are given by

${\tilde{d}_{ijk}}:=\frac{1}{|Y|}\sum_{k=1}^{3}\int_{Y}(d_{\ell
ij}(t,y)+d_{\ell
ik}(t,y)(\delta_{in}\partial_{y_{k}}\chi_{j}+\delta_{3i}\partial_{y_{k}}\varsigma_{j})dy.$
(79)

$i\in\\{1,2,3,4\\}$, $n\in\\{1,2,4\\}$ and $j,k\in\\{1,2,3\\}$.

###  5.1 Passing to the limit $\varepsilon\to 0$ in (16)

It is not yet possible to pass to the limit $\varepsilon\rightarrow 0$ with
the convergence results stated in Lemma 12. To overcome this difficulty, we
use the notion of periodic unfolding. It si worth mentioning that there is an
intimate link between the two-scale convergence and weak convergence of the
unfolded sequences; see [35, 15]. The key idea is: Instead of getting strong
convergence for $u^{\varepsilon}_{5}$, obtain strong convergence for the
periodic unfolding of $u^{\varepsilon}_{5}$.

######  Definition 15.

For $\varepsilon>0$, the boundary unfolding of a measurable function $\varphi$
posed on oscillating surface $\Gamma_{\varepsilon}$ is defined by

$T^{b}_{\varepsilon}\varphi(x,y)=\varphi(\varepsilon y+\varepsilon
k),\;\;y\in\Gamma,x\in\Omega$

where $k:=[\frac{x}{\varepsilon}]$ denotes the unique integer combination
$\Sigma_{j=1}^{3}k_{j}e_{j}$ of the periods such that
$x-[\frac{x}{\varepsilon}]$ belongs to $Y$. Note that the oscillation due to
the perforations are shifted into the second variable $y$ which belongs to
fixed surface $\Gamma$.

######  Lemma 16.

If $u_{\varepsilon}$ converges two-scale to $u$ and
$T_{b}^{\varepsilon}u_{\varepsilon}$ converges weakly to $u^{*}$ in
$L^{2}((0,T)\times\Omega;L^{2}_{\\#}(\Gamma))$, then $u=u^{*}$ a.e. in
$(0,T)\times\Omega\times\Gamma$.

Proof. The proof details for this statement can be found in Lemma 4.6 of [15].

######  Lemma 17.

If $\varphi\in L^{2}((0,T)\times\Gamma^{\varepsilon})$, then the following
identity holds

$\frac{1}{|Y|}\|T^{\varepsilon}_{b}\varphi\|_{L^{2}((0,T)\times\Omega\times\Gamma)}=\varepsilon\|\varphi\|_{L^{2}((0,T)\times\Gamma^{\varepsilon})}.$

Proof. Consider

$\displaystyle\frac{1}{|Y|}|T^{\varepsilon}_{b}\varphi|^{2}_{L^{2}(\Omega\times\Gamma)}$
$\displaystyle=\frac{1}{|Y|}\int_{\Omega\times\Gamma}|T^{\varepsilon}_{b}\varphi|^{2}dxd\sigma_{y}=\frac{1}{|Y|}\int_{\Omega\times\Gamma}T^{\varepsilon}_{b}\varphi^{2}dxd\sigma_{y},$
$\displaystyle=\frac{1}{|Y|}\Sigma^{3}_{k=1}\int_{\varepsilon(Y+k)}\int_{\Gamma}T^{\varepsilon}_{b}\varphi^{2}dxd\sigma_{y}=\frac{1}{|Y|}\Sigma^{3}_{k=1}\int_{\varepsilon(Y+k)}dx\int_{\Gamma}\varphi^{2}d\sigma_{y},$
$\displaystyle=\Sigma^{3}_{k=1}\varepsilon^{3}\int_{\Gamma}\varphi^{2}d\sigma_{y}.$

Changing variable $z=\varepsilon(y+k)$, where $k=[\frac{x}{\varepsilon}]$, we
get

$\displaystyle\frac{1}{|Y|}|T^{\varepsilon}_{b}\varphi|^{2}_{L^{2}(\Omega\times\Gamma)}$
$\displaystyle=$
$\displaystyle\Sigma^{3}_{k=1}\varepsilon^{3}\int_{\Gamma}\varphi^{2}d\sigma_{y}=\Sigma^{3}_{k=1}\varepsilon\int_{\varepsilon(\Gamma+k)}\varphi^{2}d\sigma_{z}=\varepsilon\int_{\Gamma^{\varepsilon}}\varphi^{2}d\sigma_{z}.$

This completes the proof of (17).

######  Lemma 18.

If $\varphi\in L^{2}(\Omega)$, then
$T^{\varepsilon}_{b}\varphi\rightarrow\varphi$ as $\varepsilon\rightarrow 0$
strongly in $L^{2}(\Omega\times\Gamma)$.

Proof. See in [36, 37] for proof details.

Using the boundary unfolding operator $T^{\epsilon}_{b}$, we unfold the ode
(16). Changing the variable, $x=\varepsilon y+\varepsilon k$ (for
$x\in\Gamma^{sw}_{\varepsilon}$) to the fixed domain
$(0,T)\times\Omega\times\Gamma$, we have

$\displaystyle\partial_{t}T^{\epsilon}_{b}u^{\epsilon}_{5}(t,x,y)=\eta(T^{\epsilon}_{b}u^{\epsilon}_{1}(t,x,y),T^{\epsilon}_{b}u^{\epsilon}_{5}(t,x,y)).$
(80)

In the remainder of this section, we prove that
$T^{\varepsilon}_{b}u^{\varepsilon}_{5}$ converges strongly to $u_{5}$ in
$L^{2}(\Omega\times\Gamma)$. From the two-scale convergence of
$u^{\epsilon}_{5}$, we obtain weak convergence of
$T^{\epsilon}u^{\epsilon}_{5}$ to $u_{5}$ in
$L^{\infty}((0,T)\times\Omega;L^{2}_{per}(\Gamma))$. We start with showing
that $\\{T^{\varepsilon}_{b}u^{\varepsilon}_{5}\\}$ is a Cauchy sequence in
$L^{2}(\Omega\times\Gamma)$. To this end, we choose $m,n\in\mathbb{N}$ with
$n>m$ arbitrary. Writing down (80) for the two different choices of
$\varepsilon$ (i.e. $\varepsilon_{i}=\varepsilon_{n}$ and
$\varepsilon_{i}=\varepsilon_{m}$), we obtain after subtracting the
corresponding equations that

$\displaystyle\partial_{t}$
$\displaystyle\int_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{5}-T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{5}|^{2}d\sigma_{y}dx$
(81) $\displaystyle=$
$\displaystyle\int_{\Omega\times{\Gamma}}[k^{\varepsilon}_{3}R(T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1})Q(T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{5})-k^{\varepsilon}_{3}R(T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{1})Q(T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{5}))$
$\displaystyle\times$
$\displaystyle(T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{5}-T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{5})d\sigma_{y}dx,$
$\displaystyle\leq$ $\displaystyle
k^{\infty}_{3}c_{R}(\frac{Q^{\infty}}{2}+c_{Q}sup_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1}|)\int_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{5}-T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{5}|^{2}d\sigma_{y}dx$
$\displaystyle+$
$\displaystyle\frac{k^{\infty}_{3}c_{R}Q^{\infty}}{2}\int_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1}-T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{1}|^{2}d\sigma_{y}dx.$

To get (81), we have used the uniform boundedness of
$T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1}$. We consider now

$\displaystyle\int_{\Omega\times{\Gamma}}$ $\displaystyle|T^{\epsilon
n}_{b}u^{\epsilon n}_{1}-T^{\epsilon n}_{b}u^{\epsilon
m}_{1}|^{2}d\sigma_{y}dx$ (82) $\displaystyle\leq$
$\displaystyle\int_{\Omega\times{\Gamma}}(|T^{\epsilon n}_{b}u^{\epsilon
n}_{1}-T^{\epsilon m}_{b}u_{1}|^{2}+|T^{\epsilon
n}_{b}u_{1}-u_{1}|^{2})d\sigma_{y}dx$ $\displaystyle+$
$\displaystyle\int_{\Omega\times{\Gamma}}(|T^{\epsilon
m}_{b}u_{1}-u_{1}|^{2}+|T^{\epsilon m}_{b}u^{\varepsilon m}_{1}-T^{\epsilon
m}_{b}u_{1}|^{2})d\sigma_{y}dx.$

Since $u_{1}$ is constant w.r.t. $y$, we have that
$T^{\epsilon_{m}}_{b}u_{1}\rightarrow u_{1}$ strongly in
$L^{2}((0,T)\times\Omega\times\Gamma)$ as $\varepsilon\rightarrow 0$. From
Lemma 17, we conclude that

$\int_{\Omega\times{\Gamma}}|T^{\epsilon}_{b}u^{\epsilon}_{1}-T^{\epsilon}_{b}u_{1}|^{2}d\sigma_{y}dx\leq\varepsilon\int_{{\Gamma}^{\varepsilon}}|u^{\epsilon}_{1}-u_{1}|^{2}d\sigma_{y}dx\leq\varepsilon
C.$

(82) turns out to be

$\displaystyle\int_{\Omega\times{\Gamma}}$
$\displaystyle|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1}-T^{\epsilon_{n}}_{b}u^{\epsilon_{m}}_{1}|^{2}d\sigma_{y}dxdt\leq
C(\varepsilon_{n}+\varepsilon_{m}),$

while (81) becomes

$\partial_{t}\int_{\Omega\times{\Gamma}}|T^{\epsilon n}_{b}u^{\epsilon
n}_{5}-T^{\epsilon m}_{b}u^{\epsilon m}_{5}|^{2}d\sigma_{y}dx\leq
C_{15}\int_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{5}-T^{\epsilon_{m}}_{b}u^{\epsilon_{m}}_{5}|^{2}d\sigma_{y}dx+\frac{C_{16}}{n},$

where
$C_{15}:=k^{\infty}_{3}c_{R}(\frac{Q^{\infty}}{2}+c_{Q}sup_{\Omega\times{\Gamma}}|T^{\epsilon_{n}}_{b}u^{\epsilon_{n}}_{1}|)$
and $C_{16}:=\frac{k^{\infty}_{3}c_{R}Q^{\infty}}{2}C$. The Gronwall’s
inequality gives

$\displaystyle\parallel T^{\epsilon n}_{b}u^{\epsilon n}_{5}-T^{\epsilon
m}_{b}u^{\epsilon
m}_{5}\parallel_{L^{2}(\Omega\times\Gamma)}\leq\frac{C_{16}}{n}.$ (83)

By (83), $\\{T^{\epsilon}_{b}u^{\epsilon}_{5}\\}$ is a Cauchy sequence. Now,
we take the two-scale limit in the ode (80) to get

$\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}\partial_{t}T^{\epsilon}_{b}u^{\varepsilon}_{5}\phi_{1}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt=\lim_{\varepsilon\rightarrow
0}\varepsilon\int_{0}^{T}\int_{\Gamma^{sw}_{\varepsilon}}\eta(T^{\epsilon}_{b}u^{\varepsilon}_{1},T^{\epsilon}_{b}u^{\varepsilon}_{5})\phi_{1}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt.$

Consequently, we have

$\displaystyle\int^{T}_{0}\int_{\Omega\times\Gamma^{sw}}$
$\displaystyle\partial_{t}u_{5}\phi_{5}(t,x,y)dxd\sigma_{y}dt$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}T^{\epsilon}_{b}k^{\varepsilon}_{3}R(T^{\epsilon}_{b}u^{\varepsilon}_{1})Q(u^{\varepsilon}_{5})\phi_{5}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt,$
$\displaystyle=$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}T^{\epsilon}_{b}k^{\varepsilon}_{3}R(T^{\epsilon}_{b}u^{\varepsilon}_{1})Q(u_{5})\phi_{5}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt$
$\displaystyle+$ $\displaystyle\lim_{\varepsilon\rightarrow
0}\varepsilon\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}T^{\epsilon}_{b}k^{\varepsilon}_{3}R(T^{\epsilon}_{b}u^{\varepsilon}_{1})(Q(T^{\epsilon}_{b}u^{\varepsilon}_{5})-Q(u_{5}))\phi_{5}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt.$

By (A2) and the strong convergence of $u^{\varepsilon}_{1}$, the first term on
the right hand side of (5.1) converges two-scale to

$\int^{T}_{0}\int_{\Omega}\int_{\Gamma^{sw}}k_{3}(t,y)R(u_{1})Q(u_{5})\phi_{5}(t,x,y)d\sigma_{y}dxdt,$

while the second integral of (5.1)

$\displaystyle\varepsilon$
$\displaystyle\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}T^{\epsilon}_{b}k^{\varepsilon}_{3}R(T^{\epsilon}_{b}u^{\varepsilon}_{1})(Q(T^{\epsilon}_{b}u^{\varepsilon}_{5})-Q(u_{5}))\phi_{5}(t,x,\frac{x}{\varepsilon})d\sigma_{x}dt$
$\displaystyle\leq$
$\displaystyle\varepsilon\left(\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}|T^{\epsilon}_{b}k^{\varepsilon}_{3}R(T^{\epsilon}_{b}u^{\varepsilon}_{1})\phi_{5}(t,x,\frac{x}{\varepsilon})|^{2}d\sigma_{x}dt\right)^{\frac{1}{2}}\cdot$
$\displaystyle\cdot$
$\displaystyle\left(\int^{T}_{0}\int_{\Gamma^{sw}_{\varepsilon}}|Q(T^{\epsilon}_{b}u^{\varepsilon}_{5})-Q(u_{5})|^{2}d\sigma_{x}dt\right)^{\frac{1}{2}},$
$\displaystyle\rightarrow$ $\displaystyle 0\;\mbox{as}\;\varepsilon\rightarrow
0.$

At this point, we have used again (A2) in combination with the strong
convergence of $T^{\epsilon}_{b}u^{\varepsilon}_{5}$. So, as result of passing
to the limit $\varepsilon\rightarrow 0$ in (16) we get (73).

It is worth noting that the weak solution to the two-scale model inherits a.e.
the positivity and boundedness properties from the corresponding properties of
the weak solution of the microscopic model. Now, it only remains to ensure the
uniqueness of weak solutions to the upscaled model.

######  Lemma 19.

(Uniqueness of solutions of (14)-(73) Assume (A1)-(A6). There exists at most
one weak solution to the two-scale limit problem (14) and (73).

Proof. Suppose there are two weak solutions to the two-scale limit problem
$(u^{j}_{1},u^{j}_{2},u^{j}_{3},u^{j}_{4},u^{j}_{5})$ with $j\in\\{1,2\\}$. We
denote $u_{\ell}=u^{1}_{\ell}-u^{2}_{\ell},\;\ell\in\\{1,2,3,4\\}$ and choose
as test function $\phi_{\ell}=u_{\ell}$. After straightforward calculations,
we have from (73)

$\displaystyle|u^{1}_{5}-u^{2}_{5}|\leq
C\int_{0}^{t}|u^{1}_{1}-u^{2}_{1}|d\tau.$ (85)

Take $\phi_{1}=u_{1}$ in (14) to obtain

$\displaystyle\frac{1}{2}\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxdt$
$\displaystyle+$ $\displaystyle\tilde{d}_{1}\int^{t}_{0}\int_{\Omega}|\nabla
u_{1}|^{2}dxdt$ (86) $\displaystyle\leq$
$\displaystyle-\tilde{k}_{1}\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxdt+\frac{\hat{k}_{2}^{\infty}}{2}\int^{t}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{2}|^{2})dxdt$
$\displaystyle+$
$\displaystyle\frac{k_{3}^{\infty}}{|Y|}c_{R}c_{Q}M_{1}\int^{t}_{0}\int_{\Omega\times\Gamma^{sw}}(u^{1}_{5}-u^{2}_{5})u_{1}dxd\sigma_{y}dt$
$\displaystyle+$
$\displaystyle\frac{k_{3}^{\infty}}{|Y|}c_{R}Q^{\infty}\int^{t}_{0}\int_{\Omega\times\Gamma^{sw}}|u_{1}|^{2}dxd\sigma_{y}dt.$

Using (85) together with the trace inequality for fixed domains, see section
5.5 Theorem 1 in [38] and also the fact that $u_{1}$ is independent of y in
(86), we get

$\displaystyle\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxdt$
$\displaystyle+$
$\displaystyle(2\tilde{d}_{1}-{k_{3}^{\infty}c_{R}}{}C^{*}(\delta
M_{1}+Q^{\infty}))\int^{t}_{0}\int_{\Omega}|\nabla u_{1}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle
2\tilde{k}_{1}\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxd\tau$
$\displaystyle\leq$
$\displaystyle({\tilde{k}_{2}^{\infty}}+{k_{3}^{\infty}c_{R}}{}C^{*}(\delta
M_{1}+Q^{\infty}))\int^{t}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{2}|^{2})dxd\tau$
$\displaystyle+$
$\displaystyle{k_{3}^{\infty}}{\delta}c_{R}c_{Q}M_{1}C^{*}\int^{t}_{0}\int_{\Omega}\int^{\tau}_{0}(|u_{1}|^{2}+|\nabla
u_{1}|^{2})dsdxd\tau.$

For suitable choice of
$\delta\in]0,\frac{2d_{1}-k_{3}^{\infty}c_{R}C^{*}Q^{\infty}}{k_{3}^{\infty}c_{R}C^{*}M_{1}}[$,
we have

$\displaystyle\int^{T}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxdt$
$\displaystyle+$ $\displaystyle\tilde{d}_{1}\int^{T}_{0}\int_{\Omega}|\nabla
u_{1}|^{2}dxd\tau+2\tilde{k}_{1}\int^{T}_{0}\int_{\Omega}\partial_{t}|u_{1}|^{2}dxd\tau$
(87) $\displaystyle\leq$
$\displaystyle({\tilde{k}_{2}^{\infty}}+k_{3}^{\infty}c_{R}C^{*}(\delta
M_{1}+Q^{\infty}))\int^{T}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{2}|^{2})dxd\tau$
$\displaystyle+$
$\displaystyle\frac{k_{3}^{\infty}}{\delta}c_{R}c_{Q}M_{1}C^{*}\int^{T}_{0}\int_{\Omega}\int^{\tau}_{0}(|u_{1}|^{2}+|\nabla
u_{1}|^{2})dsdxd\tau.$

Take $\phi_{2}=u_{2}$ in (14), we get

$\displaystyle\frac{1}{2}\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{2}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle\tilde{d}_{2}\int^{t}_{0}\int_{\Omega}|\nabla
u_{2}|^{2}dxd\tau$ $\displaystyle\leq$
$\displaystyle-\tilde{k}_{2}\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{2}|^{2}dxd\tau+\frac{\tilde{k}_{1}^{\infty}}{2}\int^{t}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{2}|^{2})dxd\tau$
$\displaystyle+$
$\displaystyle{\tilde{a}^{\infty}}\int^{t}_{0}\int_{\Omega}u_{2}u_{3}dxdt-{\tilde{b}}\int^{T}_{0}\int_{\Omega}|u_{2}|^{2}dxd\tau.$
$\displaystyle\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{2}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle\tilde{d}_{2}\int^{t}_{0}\int_{\Omega}|\nabla
u_{2}|^{2}dxd\tau$ (88) $\displaystyle\leq$
$\displaystyle({\tilde{k}_{1}^{\infty}}+\tilde{a}^{\infty})\int^{t}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{2}|^{2}+|u_{3}|^{2})dxd\tau.$

Similarly, we obtain from (14)

$\displaystyle\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{3}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle\tilde{d}_{3}\int^{t}_{0}\int_{\Omega}|\nabla
u_{3}|^{2}dxd\tau\leq\tilde{b}^{\infty}\int^{t}_{0}\int_{\Omega}(|u_{2}|^{2}+|u_{3}|^{2})dxd\tau,$
(89) $\displaystyle\int^{t}_{0}\int_{\Omega}\partial_{t}|u_{4}|^{2}dxd\tau$
$\displaystyle+$ $\displaystyle\tilde{d}_{4}\int^{t}_{0}\int_{\Omega}|\nabla
u_{4}|^{2}dxd\tau\leq\tilde{k}^{\infty}_{1}\int^{T}_{0}\int_{\Omega}(|u_{1}|^{2}+|u_{3}|^{2})dxd\tau.$
(90)

Adding side by side (87)-(90) and applying Gronwall’s inequality to the
corresponding result, we receive

$\displaystyle\Sigma_{i=1}^{4}\int_{\Omega}|u_{i}|^{2}dx+\hat{d}\Sigma_{i=1}^{4}\int^{t}_{0}\int_{\Omega}|\nabla
u_{i}|^{2}dxd\tau+\hat{d}\int^{t}_{0}\int_{\Omega}|u_{1}|^{2}dxd\tau\leq 0.$
(91)

In (91), we have
$\hat{d}:=min\\{\tilde{d}_{1},\tilde{d}_{2},\tilde{d}_{3},\tilde{d}_{4},\tilde{k}_{1}\\}>0$.
Taking in (92) supremum over $(0,T)$, we obtain

$\displaystyle\Sigma_{i=1}^{4}\int_{\Omega}|u_{i}|^{2}dx+\hat{d}\Sigma_{i=1}^{4}\int^{T}_{0}\int_{\Omega}|\nabla
u_{i}|^{2}dxd\tau\leq 0,$ (92)

which concludes the proof of the Lemma.

######  Lemma 20.

(Strong formulation of the two-scale limit equations) Assume the hypothesis of
Lemma 12 to hold. Then the strong formulation of the two-scale limit equations
(for all $t\in(0,T)$) reads

$\displaystyle\partial_{t}u_{1}(t,x)$
$\displaystyle+\nabla\cdot(-\tilde{d}_{1}\nabla u_{1}(t,x))$ (93)
$\displaystyle=-\tilde{k}_{1}(t)u_{1}(t,x)+\tilde{k}_{2}(t)u_{2}(t,x)$
$\displaystyle-\frac{1}{|Y|}\int_{\Gamma^{sw}}k_{3}(t,y)R(u_{1}(t,x))Q(u_{5}(t,x,y))d\sigma_{y},\;x\in\Omega$
$\displaystyle u_{1}(0,x)$ $\displaystyle=u_{10}(x),\;x\in\bar{\Omega},$
$\displaystyle n\cdot(-\tilde{d}_{1}\nabla u_{1}(t,x))$
$\displaystyle=0,\;x\in\partial\Omega$ $\displaystyle\partial_{t}u_{2}(t,x)$
$\displaystyle+\nabla\cdot(-\tilde{d}_{2}\nabla
u_{2}(t,x))=\tilde{k}_{1}(t)u_{1}(t,x)-\tilde{k}_{2}(t)u_{2}(t,x)$ (94)
$\displaystyle+\tilde{a}(t)u_{3}(t,x)-\tilde{b}(t)u_{2}(t,x),\;x\in\Omega,$
$\displaystyle u_{2}(0,x)$ $\displaystyle=u_{20}(x),\;x\in\bar{\Omega},$
$\displaystyle n\cdot(-\tilde{d}_{2}\nabla u_{2}(t,x))$
$\displaystyle=0,\;x\in\partial\Omega,$ $\displaystyle\partial_{t}u_{3}(t,x)$
$\displaystyle+\nabla\cdot(-\tilde{d}_{3}\nabla
u_{3}(t,x))=-\tilde{a}(t)u_{3}(t,x)+\tilde{b}(t)u_{2}(t,x),\;x\in\Omega,$ (95)
$\displaystyle u_{3}(0,x)$ $\displaystyle=u_{30}(x),\;x\in\bar{\Omega},$
$\displaystyle u_{3}(t,x)$ $\displaystyle=u^{D}_{3}(x),\;x\in\Gamma^{D},$
$\displaystyle n\cdot(-\tilde{d}_{3}\nabla u_{3}(t,x))$
$\displaystyle=0,\;x\in\Gamma^{N}$ $\displaystyle\partial_{t}u_{4}(t,x)$
$\displaystyle+\nabla\cdot(-\tilde{d}_{4}\nabla
u_{4}(t,x))=\tilde{k}_{1}(t)u_{1}(t,x),\;x\in\Omega,$ (96) $\displaystyle
u_{4}(0,x)$ $\displaystyle=u_{40}(x),\;x\in\bar{\Omega},$ $\displaystyle
n\cdot(-\tilde{d}_{4}\nabla u_{4}(t,x))$
$\displaystyle=0,\;x\in\partial\Omega$ $\displaystyle\partial_{t}u_{5}(t,x,y)$
$\displaystyle=k_{3}(t,y)R(u_{1}(t,x))Q(u_{5}(t,x,y)),\;x\in\Omega,y\in\Gamma^{sw},$
(97) $\displaystyle u_{5}(0,x,y)$
$\displaystyle=u_{50}(x,y)\;x\in\bar{\Omega},y\in\Gamma^{sw},$

where $\tilde{d}_{i},i\in\\{1,2,3,4\\}$ and $\tilde{k}_{j},j\in\\{1,2\\}$ are
defined in Theorem 14.

### Acknowledgements

We would like to thank M. Ptashnyk (RWTH Aachen) and M. A. Peletier (TU
Eindhoven) for fruitful discussions on this subject.

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