# A free boundary problem describing migration into rubbers – quest of the
large time behavior

Kota Kumazaki
Faculty of Education, Nagasaki University,
1-14 Bunkyo-cho, Nagasaki-city, Nagasaki, 851-8521, Japan
<EMAIL_ADDRESS>
Toyohiko Aiki
Department of Mathematics Faculty of Sciences, Japan Women’s University,
2-8-1, Mejirodai, Bunkyo-ku, Tokyo, 112-8681, Japan
<EMAIL_ADDRESS>
Adrian Muntean
Department of Mathematics and Computer Science, Karlstad University,
Universitetsgatan 2, 651 88 Karlstad, Sweden
<EMAIL_ADDRESS>
###### Abstract

In many industrial applications, rubber-based materials are routinely used in
conjunction with various penetrants or diluents in gaseous or liquid form. It
is of interest to estimate theoretically the penetration depth as well as the
amount of diffusants stored inside the material. In this framework, we prove
the global solvability and explore the large time-behavior of solutions to a
one-phase free boundary problem with nonlinear kinetic condition that is able
to describe the migration of diffusants into rubber. The key idea in the proof
of the large time behavior is to benefit of a contradiction argument, since it
is difficult to obtain uniform estimates for the growth rate of the free
boundary due to the use of a Robin boundary condition posed at the fixed
boundary.

##  1 Introduction

In many industrial applications, the behavior of rubber-based materials is
difficult to predict theoretically. This intriguing fact is especially due to
their internal structure which allows for unexpected local changes
(deformations, concentration localization, network entaglement, etc.)
typically facilitated by the absorption and migration of diffusants into the
material; this is from where our motivation stems. There is a variety of
possible modeling approaches for such scenarios. Motivated by our recent work
[17], where solutions to our free boundary model did recover experimental
data, we choose to follow a macroscopic modeling approach with kinetically-
driven interfaces capturing the penetration of diffusants into the material.
We refer the reader, for instance, to [8, 9, 11, 7] for closely related work
especially what concerns the mathematics of Case II diffusion as it arises for
some classes of polymers, but not directly applicable to the rubber case.

In this paper we consider the mathematical analysis of the following free
boundary problem which was discussed in [17] in connection with the
absorption, penetration and diffusion-induced swelling in dense and foamed
rubbers. Let $[0,s(t)]$ be a region occupied by a solvent (e.g. water, tea)
occupying the one-dimensional pore $[0,\infty)$, where $t$ is the time
variable, $s=s(t)$ is the position of the moving interface, while $u=u(t,z)$
is the content of the diffusant situated at the position $z\in[0,s(t)]$. The
function $u(t,z)$ acts in the non-cylindrical region $Q_{s}(T)$ given by

$\displaystyle Q_{s}(T):=\\{(t,z)|0<t<T,\ 0<z<s(t)\\}.$

Our free boundary problem $(\mbox{P})(u_{0},s_{0},b)$ reads: Find the pair
$(u,s)$ satisfying

$\displaystyle u_{t}-u_{zz}=0\mbox{ for }(t,z)\in Q_{s}(T),$ (1.1)
$\displaystyle-u_{z}(t,0)=\beta(b(t)-\gamma u(t,0))\mbox{ for }t\in(0,T),$
(1.2) $\displaystyle-u_{z}(t,s(t))=u(t,s(t))s_{t}(t)\mbox{ for }t\in(0,T),$
(1.3) $\displaystyle s_{t}(t)=a_{0}\sigma(u(t,s(t)))\mbox{ for }t\in(0,T),$
(1.4) $\displaystyle s(0)=s_{0},u(0,z)=u_{0}(z)\mbox{ for }z\in[0,s_{0}],$
(1.5)

where $\beta$, $\gamma$ and $a_{0}$ are given positive constants, $b$ is a
given threshold function defined on $[0,T]$, while $s_{0}$ and $u_{0}$ are the
corresponding initial data. In (1.4), $\sigma$ is a function on $\mathbb{R}$
given by

$\sigma(r)=\begin{cases}r\quad\mbox{ if }r\geq 0,\\\ 0\quad\mbox{ if
}r<0.\end{cases}$

In [18], A. Visintin refers to this type of problems as free or moving
boundary problems with kinetic boundary condition. The reason for calling this
way is linked to the fact that relation (1.4) is an explicit description of
the speed of the free boundary. Note also that, in Refs. [15, 16], the authors
have considered the mathematical analysis of a similar problem to
$(\mbox{P})(u_{0},s_{0},b)$ related to water-induced swelling in porous
materials, viz.

$\displaystyle u_{t}-u_{zz}=0\mbox{ for }(t,z)\in(0,T)\times(a,s(t)),$ (1.6)
$\displaystyle-u_{z}(t,a)=\hat{\beta}(h(t)-\gamma u(t,a))\mbox{ for
}t\in(0,T),$ (1.7) $\displaystyle-u_{z}(t,s(t))=u(t,s(t))s_{t}(t)\mbox{ for
}t\in(0,T),$ (1.8) $\displaystyle
s_{t}(t)=a_{0}(u(t,s(t))-\varphi(s(t)))\mbox{ for }t\in(0,T),$ (1.9)
$\displaystyle s(0)=\hat{s}_{0},u(0,z)=\hat{u}_{0}(z)\mbox{ for
}z\in[a,\hat{s}_{0}].$ (1.10)

In this context, $a$ is a positive constant, $h$ is a given non-negative
function on $[0,T]$, while $\hat{s}_{0}$ and $\hat{u}_{0}$ are the initial
data such that $\hat{s}_{0}>a$. Also, $\hat{\beta}$ and $\varphi$ are
continuous functions on $\mathbb{R}$ such that $\hat{\beta}(r)>0$ and
$\varphi(r)>0$ for $r>0$, and $\hat{\beta}(r)=\varphi(r)=0$ for $r\leq 0$.
Denote the above problem $\\{(\ref{1-6})-(\ref{1-10})\\}$ by
$(\hat{\mbox{P}})(\hat{u}_{0},\hat{s}_{0},h)$. In [15] in was assumed that
$\varphi$ is conveniently small and $\hat{u}_{0}\in H^{1}(a,\hat{s}_{0})$ such
that $\varphi(a)\leq\hat{u}_{0}\leq h^{*}/\gamma$ on $[a,\hat{s}_{0}]$. Such
conditions ensure the existence of a locally-in-time solution $(u,s)$ to
$(\hat{\mbox{P}})(\hat{u}_{0},\hat{s}_{0},h)$ on $[0,T_{0}]$ such that
$\varphi(a)\leq u\leq h^{*}/\gamma$ on $Q^{a}_{s}(T_{0})$ for some
$0<T_{0}\leq T$, where $h^{*}$ is a upper bound of $h$. In [16], relying on
the same assumptions as in [15], the authors have constructed a globally-in-
time solution $(u,s)$ to $(\hat{\mbox{P}})(\hat{u}_{0},\hat{s}_{0},h)$ on
$[0,T]$ such that $\varphi(a)\leq u\leq h^{*}/\gamma$ on $Q^{a}_{s}(T)$. What
concerns the large time behavior of solutions to
$(\hat{\mbox{P}})(\hat{u}_{0},\hat{s}_{0},h)$, one reports in [16] that the
following situation holds:

$\displaystyle\mbox{ if
}\lim_{t\to\infty}\int_{0}^{t}\hat{\beta}(h(\tau)-\gamma
u(\tau,a))d\tau=\infty,\mbox{ then}\lim_{t\to\infty}s(t)=\infty.$

One of our concrete aims here is to construct a global-in-time solution of
$(\mbox{P})(u_{0},s_{0},b)$. As anticipated, the key to the proof is to
establish the strictly positivity for the free boundary. To this end, we
consider the free boundary condition (1.4), which should be seen as
$\varphi\equiv 0$ in (1.9) in the models proposed in [15, 16]. This is a
simplification of the modeling setting which is convenient for mathematical
analysis purposes. Furthermore, we adopt the positive part in (1.4), and
hence, we can easily show that the free boundary $s(t)$ is indeed strictly
positive and the expected global existence is now reachable. We will
investigate elsewhere to which extent such structural restrictions can be
relaxed.

Moreover, we establish that our free boundary grows up, namely, it is
unbounded. In order to obtain a control on the growth of the free boundary,
the mass conservation law (respectively, the momentum balance law) are
effective ingredients in case the boundary condition at the fixed boundary of
Neumann (respectively, Dirichlet) type, see for instance [5]. In the present
setting, we impose a Robin boundary condition at the fixed boundary and the
usual approach does not work well. Hence, the rationale beyond showing that
$s(t)\to\infty$ as $t\to\infty$ is as follows: If the free boundary is
bounded, then we can obtain some uniform-in-time estimates for the target
solution, and consequently, this solution $u(t)$ converges towards the
stationary solution of our problem. It is worthwhile to note here that the
stationary problem still contains a free boundary condition. However, as a
consequence of our uniform estimates, the solution never satisfies the
stationary free boundary condition. Thus we can prove the large time behavior
by detecting a contradiction. The idea of applying a contradiction argument
concerned with a stationary solution was already applied in [1]. The quest for
growth (convergence) rates of this kind of kinetically-driven free boundaries
was completed in the series of papers [2, 3, 4]. For the problem at hand,
proving quantitative estimates on the growth rate of the free boundary is
currently an open problem.

##  2 Notation, assumptions and results

In this paper, we use the following notations. We denote by $|\cdot|_{X}$ the
norm for a Banach space $X$. The norm and the inner product of a Hilbert space
$H$ are denoted by $|\cdot|_{H}$ and $(\cdot,\cdot)_{H}$, respectively.
Particularly, for $\Omega\subset{\mathbb{R}}$, we use the notation of the
usual Hilbert spaces $L^{2}(\Omega)$, $H^{1}(\Omega)$ and $H^{2}(\Omega)$.
Throughout this paper, we assume the following parameters and functions:

(A1) $a_{0}$, $\gamma$, $\beta$ and $T$ are positive constants.

(A2) $b\in W^{1,2}(0,T)$ with $b_{*}\leq b\leq b^{*}$ on $(0,T)$, where
$b_{*}$ and $b^{*}$ are positive constants.

(A3) $s_{0}>0$ and $u_{0}\in H^{1}(0,s_{0})$ such that $0\leq u_{0}\leq
b^{*}/\gamma$ on $[0,s_{0}]$.

Next, we define our concept of solution to (P)$(u_{0},s_{0},b)$ on $[0,T]$ in
the following way:

######  Definition 2.1.

For $T>0$, let $s$ be a function on $[0,T]$ and $u$ be a function on
$Q_{s}(T)$. We call the pair $(s,u)$ a solution to (P)$(u_{0},s_{0},b)$ on
$[0,T]$ if the following conditions (S1)-(S6) hold:

(S1) $s\in W^{1,\infty}(0,T)$, $0<s$ on $[0,T]$, $u\in L^{\infty}(Q_{s}(T))$,
$u_{t}$, $u_{zz}\in L^{2}(Q_{s}(T))$ and
$t\in[0,T]\to|u_{z}(t,\cdot)|_{L^{2}(0,s(t))}$ is bounded;

(S2) $u_{t}-u_{zz}=0$ on $Q_{s}(T)$;

(S3) $-u_{z}(t,0)=\beta(b(t)-\gamma u(t,0))$ for a.e. $t\in[0,T]$;

(S4) $-u_{z}(t,s(t))=u(t,s(t))s_{t}(t)$ for a.e. $t\in[0,T]$;

(S5) $s_{t}(t)=a_{0}\sigma(u(t,s(t)))$ for a.e. $t\in[0,T]$;

(S6) $s(0)=s_{0}$ and $u(0,z)=u_{0}(z)$ for $z\in[0,s_{0}]$.

The first result of this paper is concerned with the existence and uniqueness
of a locally-in-time solution in the sense of Definition 2.1 to the problem
(P)$(u_{0},s_{0},b)$.

######  Theorem 2.2.

Let $T>0$. If (A1)-(A3) hold, then there exists $T^{*}\in(0,T]$ such that
(P)$(u_{0},s_{0},b)$ has a unique solution $(s,u)$ on $[0,T^{*}]$ satisfying
$0\leq u\leq b^{*}/\gamma$ on $Q_{s}(T^{*})$.

To prove Theorem 2.2, we transform (P)$(u_{0},s_{0},b)$, initially posed in a
non-cylindrical domain, to a cylindrical domain. Let $T>0$. For given $s\in
W^{1,2}(0,T)$ with $s(t)>0$ on $[0,T]$, we introduce the following new
function obtained by the change of variables and fix the moving domain:

$\displaystyle\tilde{u}(t,y)=u(t,ys(t))\mbox{ for }(t,y)\in
Q(T):=(0,T)\times(0,1)$ (2.1)

By using the function $\tilde{u}$, (P)$(u_{0},s_{0},b)$ becomes the following
problem (PC)$(\tilde{u}_{0},s_{0},b)$ on the cylindrical domain $Q(T)$:

$\displaystyle\tilde{u}_{t}(t,y)-\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t,y)=\frac{ys_{t}(t)}{s(t)}\tilde{u}_{y}(t,y)\mbox{
for }(t,y)\in Q(T),$ (2.2)
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,0)=\beta(b(t)-\gamma\tilde{u}(t,0))\mbox{
for }t\in(0,T),$ (2.3)
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,1)=\tilde{u}(t,1)s_{t}(t)\mbox{
for }t\in(0,T),$ (2.4) $\displaystyle
s_{t}(t)=a_{0}\sigma(\tilde{u}(t,1))\mbox{ for }t\in(0,T),$ (2.5)
$\displaystyle s(0)=s_{0},$ (2.6)
$\displaystyle\tilde{u}(0,y)=u_{0}(ys(0))(:=\tilde{u}_{0}(y))\mbox{ for
}y\in[0,1].$ (2.7)

######  Definition 2.3.

For $T>0$, let $s$ be a function on $[0,T]$ and $\tilde{u}$ be a function on
$Q(T)$, respectively. We call that a pair $(s,\tilde{u})$ is a solution of
$(\mbox{P})(\tilde{u}_{0},s_{0},b)$ on $[0,T]$ if the conditions (S’1)-(S’2)
hold:

(S’1) $s\in W^{1,\infty}(0,T)$, $s>0$ on $[0,T]$, $\tilde{u}\in
W^{1,2}(Q(T))\cap L^{\infty}(0,T;H^{1}(0,1))\cap L^{2}(0,T;H^{2}(0,1))$.

(S’2) (2.2)–(2.7) hold.

Here, we introduce the following function space: For $T>0$, we put
$V(T)=L^{\infty}(0,T;L^{2}(0,1))\cap L^{2}(0,T;H^{1}(0,1))$ and
$|z|_{V(T)}=|z|_{L^{\infty}(0,T;L^{2}(0,1))}+|z_{y}|_{L^{2}(0,T;L^{2}(0,1))}$
for $z\in V(T)$. Note that $V(T)$ is a Banach space with the norm
$|\cdot|_{V(T)}$.

Now, we state the existence and uniqueness of a locally-in-time solution of
$(\mbox{PC})(\tilde{u}_{0},s_{0},b)$.

######  Theorem 2.4.

Let $T>0$. If (A1)-(A3) hold, then there exists $T^{*}\in(0,T]$ such that
$(\mbox{PC})(\tilde{u}_{0},s_{0},b)$ has a unique solution $(s,\tilde{u})$ on
$[0,T^{*}]$.

By Theorem 2.4, we see that for a solution $(s,\tilde{u})$ of
$(\mbox{PC})(\tilde{u}_{0},s_{0},b)$ on $[0,T^{*}]$, a pair of the function
$(s,u)$ with the variable

$\displaystyle u(t,z):=\tilde{u}\left(t,\frac{z}{s(t)}\right)\mbox{ for
}z\in[0,s(t)]$ (2.8)

is a solution of $(\mbox{P})(u_{0},s_{0},b)$ on $[0,T^{*}]$. Moreover, by
proving that $(s,u)$ satisfies $0\leq u\leq b^{*}/\gamma$ on $Q_{s}(T^{*})$,
the pair $(s,u)$ is a desired solution of $(\mbox{P})(u_{0},s_{0},b)$ on
$[0,T^{*}]$ which leads to Theorem 2.2.

The second result of this paper is the existence and uniqueness of a globally-
in-time solution in the sense of Definition 2.1 to the problem
(P)$(u_{0},s_{0},b)$.

######  Theorem 2.5.

Let $T>0$. If (A1)-(A3) hold, then $(\mbox{P})(u_{0},s_{0},b)$ has a unique
solution $(s,u)$ on $[0,T]$ satisfying $0\leq u\leq b^{*}/\gamma$ on
$Q_{s}(T)$.

Throughout Sections 3 and 4, we show Theorem 2.2 by proving Theorem 2.4 and
the boundedness of a solution of $(\mbox{P})(u_{0},s_{0},b)$. In Section 5, we
give a proof of Theorem 2.5. In the last section, we discuss the large time
behavior of a solution of $(\mbox{P})(u_{0},s_{0},b)$ as $t\to\infty$. In
fact, we obtain the result that $s\to\infty$ as $t\to\infty$. The precise
statement is stated as Theorem 6.2.

##  3 Auxiliary Problem

In this section, we prove Theorem 2.4 on the existence and uniqueness of a
locally-in-time solution of $(\mbox{PC})(\tilde{u}_{0},s,b)$. To do so, we
introduce the following auxiliary problem $(\mbox{AP})(\tilde{u}_{0},s,b)$:
For $T>0$, $s_{0}>0$ and given $s\in W^{1,2}(0,T)$ with $s(0)=s_{0}$ and
$s\geq s_{0}$ on $[0,T]$,

$\displaystyle\tilde{u}_{t}(t,y)-\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t,y)=\frac{ys_{t}(t)}{s(t)}\tilde{u}_{y}(t,y)\mbox{
for }(t,y)\in Q(T),$ (3.1)
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,0)=\beta(b(t)-\gamma\tilde{u}(t,0))\mbox{
for }t\in(0,T),$ (3.2)
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,1)=a_{0}\tilde{u}(t,1)\sigma(\tilde{u}(t,1))\mbox{
for }t\in(0,T),$ (3.3) $\displaystyle\tilde{u}(0,y)=\tilde{u}_{0}(y)\mbox{ for
}y\in[0,1],$ (3.4)

where $\sigma$ is the same function as in (1.4).

First of all, to solve $(\mbox{AP})(\tilde{u}_{0},s,b)$, for given $s\in
W^{1,\infty}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on $[0,T]$ and $f\in
L^{2}(0,T;H^{1}(0,1))$, we consider the problem
$(\mbox{AP}1)(\tilde{u}_{0},s,f,b)$:

$\displaystyle\tilde{u}_{t}(t,y)-\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t,y)=\frac{ys_{t}(t)}{s(t)}f_{y}(t,y)\mbox{
for }(t,y)\in Q(T),$
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,0)=\beta(b(t)-\gamma\tilde{u}(t,0))\mbox{
for }t\in(0,T),$
$\displaystyle-\frac{1}{s(t)}\tilde{u}_{y}(t,1)=a_{0}\tilde{u}(t,1)\sigma(\tilde{u}(t,1))\mbox{
for }t\in(0,T),$ $\displaystyle\tilde{u}(0,y)=\tilde{u}_{0}(y)\mbox{ for
}y\in[0,1].$

Now, we define a family $\\{\psi^{t}\\}_{t\in[0,T]}$ of time-dependent
functionals $\psi^{t}:L^{2}(0,1)\to\mathbb{R}\cup\\{+\infty\\}$ for
$t\in[0,T]$ as follows:

$\psi^{t}(u):=\begin{cases}\displaystyle{\frac{1}{2s^{2}(t)}\int_{0}^{1}|u_{y}(y)|^{2}dy}+\displaystyle{\frac{1}{s(t)}\int_{0}^{u(1)}a_{0}\xi\sigma(\xi)d\xi}\\\
-\displaystyle{\frac{1}{s(t)}\int_{0}^{u(0)}\beta(b(t)-\gamma\xi)d\xi}\mbox{
if }u\in D(\psi^{t}),\\\ +\infty\quad\rm{otherwise},\end{cases}$

where $D(\psi^{t})=\\{z\in H^{1}(0,1)|z\geq 0\mbox{ on }[0,1]\\}$ for
$t\in[0,T]$. Here, we show the property of $\psi^{t}$.

######  Lemma 3.1.

Let $s\in W^{1,2}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on $[0,T]$ and
assume (A1)-(A3). Then the following statements hold:

  * (1)

There exists positive constants $C_{0}$ and $C_{1}$ such that the following
inequalities hold:

$\displaystyle(i)$ $\displaystyle\ |u(y)|^{2}\leq C_{0}\psi^{t}(u)+C_{1}\mbox{
for }u\in D(\psi^{t}),y=0,1\mbox{ and }t\in[0,T],$ $\displaystyle(ii)$
$\displaystyle\ \frac{1}{2s^{2}(t)}|u_{y}|^{2}_{L^{2}(0,1)}\leq
C_{0}\psi^{t}(u)+C_{1}\mbox{ for }u\in D(\psi^{t})\mbox{ and }t\in[0,T].$

  * (2)

For $t\in[0,T]$, the functional $\psi^{t}$ is proper, lower semi-continuous,
and convex on $L^{2}(0,1)$.

###### Proof.

First, we note that for $t\in[0,T]$ if $u\in D(\psi^{t})$, then $u(0)$ is non-
negative. Let $t\in[0,T]$ and $u\in D(\psi^{t})$. Then, it holds

$\displaystyle-\frac{1}{s(t)}\int_{0}^{u(0)}\beta(b(t)-\gamma\xi)d\xi$
$\displaystyle=\frac{\beta}{s(t)}\left[\frac{\gamma}{2}u^{2}(0)-b(t)u(0)\right]$
$\displaystyle\geq\frac{\beta\gamma}{2l}u^{2}(0)-\frac{\beta
b^{*}}{s_{0}}u(0)$
$\displaystyle\geq\frac{\beta\gamma}{4l}u^{2}(0)-\frac{\beta
l}{\gamma}\left(\frac{b^{*}}{s_{0}}\right)^{2}.$ (3.5)

Since the second term of the right-hand side of $\psi^{t}$ is positive, by
(3.5), we have that

$\displaystyle\psi^{t}(u)$
$\displaystyle\geq\frac{1}{2s^{2}(t)}\int_{0}^{1}|u_{y}(y)|^{2}dy+\frac{\beta\gamma}{4l}u^{2}(0)-\frac{\beta
l}{\gamma}\left(\frac{b^{*}}{s_{0}}\right)^{2}.$ (3.6)

Also, it holds that

$\displaystyle|u(1)|^{2}$
$\displaystyle=\biggl{|}\int_{0}^{1}u_{y}(y)dy+u(0)\biggr{|}^{2}\leq
2\left(\int_{0}^{1}|u_{y}(y)|^{2}dy+|u(0)|^{2}\right)$ $\displaystyle\leq
2\left(\frac{2l^{2}}{2s^{2}(t)}\int_{0}^{1}|u_{y}(y)|^{2}dy+|u(0)|^{2}\right).$
(3.7)

Therefore, by (3.6) and (3.7) we see that the statement (1) of Lemma 3.1
holds.

Next, we prove statement (2). For $t\in[0,T]$ and $r\in\mathbb{R}$, put

$\displaystyle
g_{1}(s(t),r)=\frac{1}{s(t)}\int_{0}^{r}a_{0}\xi\sigma(\xi)d\xi,$
$\displaystyle
g_{2}(s(t),b(t),r)=-\frac{1}{s(t)}\int_{0}^{r}\beta(b(t)-\gamma\xi)d\xi.$

Then, by $0<s(t)$ we see that

$\displaystyle\frac{\partial^{2}}{\partial
r^{2}}g_{1}(s(t),r)=\frac{2a_{0}}{s(t)}r>0\mbox{ for }r>0,$
$\displaystyle\frac{\partial^{2}}{\partial
r^{2}}g_{2}(s(t),b(t),r)=\frac{\beta\gamma}{s(t)}>0\mbox{ for
}r\in\mathbb{R}.$

This means that $\psi^{t}$ is convex on $L^{2}(0,1)$. Also, by using Lemma 3.1
and Sobolev’s embedding $H^{1}(0,1)\hookrightarrow C([0,1])$ in one
dimensional case, it is easy to prove that the level set of $\psi^{t}$ is
closed in $L^{2}(0,1)$ which leads to the lower semi-continuity of $\psi^{t}$.
Thus, we see that statement (2) holds. ∎

By Lemma 3.1 we obtain the existence of a solution to
$(\mbox{AP}1)(\tilde{u}_{0},s,f,b)$.

######  Lemma 3.2.

Let $T>0$ and $s_{0}>0$. If (A1)-(A3) hold, then, for given $s\in
W^{1,2}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on $[0,T]$ and $f\in
L^{2}(0,T;H^{1}(0,1))$, the problem $(\mbox{AP}1)(\tilde{u}_{0},s,f,b)$ admits
a unique solution $\tilde{u}$ on $[0,T]$ such that $\tilde{u}\in
W^{1,2}(Q(T))\cap L^{\infty}(0,T;H^{1}(0,1))$ with $\tilde{u}\geq 0$ on
$Q(T)$. Moreover, the function $t\to\psi^{t}(\tilde{u}(t))$ is absolutely
continuous on $[0,T]$.

###### Proof.

By Lemma 3.1, for $t\in[0,T]$ $\psi^{t}$ is a proper lower semi-continuous
convex function on $L^{2}(0,1)$. From the definition of the subdifferential of
$\psi^{t}$, for $t\in[0,T]$, $z^{*}\in\partial\psi^{t}(u)$ is characterized by
$u$, $z^{*}\in L^{2}(0,1)$,

$\displaystyle z^{*}=-\frac{1}{s^{2}(t)}u_{yy}\mbox{ on
}(0,1),\quad-\frac{1}{s(t)}u_{y}(0)=\beta(b(t)-\gamma
u(0)),\quad-\frac{1}{s(t)}u_{y}(1)=a_{0}u(1)\sigma(u(1)).$

Namely, $\partial\psi^{t}$ is single-valued. Also, we see that there exists a
positive constant $C_{3}$ such that for each $t_{1}$, $t_{2}\in[0,T]$ with
$t_{1}\leq t_{2}$, and for any $u\in D(\psi^{t_{1}})$, there exists
$\bar{u}\in D(\psi^{t_{2}})$ such that

$\displaystyle|\psi^{t_{2}}(\bar{u})-\psi^{t_{1}}(u)|\leq
C_{3}(|s(t_{1})-s(t_{2})|+|b(t_{1})-b(t_{2})|)(1+|\psi^{t_{1}}(u)|).$ (3.8)

Indeed, by taking $\bar{u}:=u$ and using (i) and (ii) of Lemma 3.1, we can
find $C_{3}>0$ such that (3.8) holds. Now, we consider the following Cauchy
problem (CP):

$\begin{cases}\tilde{u}_{t}+\partial\psi^{t}(\tilde{u}(t))=\frac{ys_{t}(t)}{s(t)}f_{y}(t)\mbox{
in }L^{2}(0,1),\\\ \tilde{u}(0,y)=\tilde{u}_{0}(y)\mbox{ for
}y\in[0,1].\end{cases}$

Since $\frac{ys_{t}}{s}f_{y}\in L^{2}(0,T;L^{2}(0,1))$, the general theory of
evolution equations governed by time dependent subdifferentials (cf. [14])
guarantees that (CP) has a non-negative solution $\tilde{u}$ on $[0,T]$ such
that $\tilde{u}\in W^{1,2}(Q(T))$, $\psi^{t}(\tilde{u}(t))\in L^{\infty}(0,T)$
and $t\to\psi^{t}(\tilde{u}(t))$ is absolutely continuous on $[0,T]$. This
implies that $\tilde{u}$ is a unique solution of
$(\mbox{AP}1)(\tilde{u}_{0},s,f,b)$ on $[0,T]$. ∎

######  Lemma 3.3.

Let $T>0$, $s_{0}>0$ and $s\in W^{1,\infty}(0,T)$ with $s(0)=s_{0}$ and $s\geq
s_{0}$ on $[0,T]$. If (A1)-(A3) hold, then, $(\mbox{AP})(\tilde{u}_{0},s,b)$
has a unique solution $\tilde{u}$ on $[0,T]$ such that $\tilde{u}\in
W^{1,2}(Q(T))\cap L^{\infty}(0,T;H^{1}(0,1))$.

###### Proof.

First, we define a solution operator $\Gamma_{T}(f)=\tilde{u}$, where
$\tilde{u}$ is the unique solution of $(\mbox{AP}1)(\tilde{u}_{0},s,f,b)$ for
given $f\in V(T)$. Now, for $i=1,2$ we put $\Gamma_{T}(f_{i})=\tilde{u}_{i}$
and $f=f_{1}-f_{2}$ and $\tilde{u}=\tilde{u}_{1}-\tilde{u}_{2}$. Then, we have
that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}-\int_{0}^{1}\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t)\tilde{u}(t)dy=\int_{0}^{1}\frac{ys_{t}(t)}{s(t)}f_{y}(t)\tilde{u}(t)dy.$
(3.9)

Using the boundary condition, it holds that

$\displaystyle-\int_{0}^{1}\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t)\tilde{u}(t)dy$
$\displaystyle=\frac{1}{s^{2}(t)}\left(-\tilde{u}_{y}(t,1)\tilde{u}(t,1)+\tilde{u}_{y}(t,0)\tilde{u}(t,0)+\int_{0}^{1}|\tilde{u}_{y}(t)|^{2}dy\right)$
$\displaystyle=\frac{a_{0}}{s(t)}\biggl{(}\tilde{u}_{1}(t,1)\sigma(\tilde{u}_{1}(t,1))-\tilde{u}_{2}(t,1)\sigma(\tilde{u}_{2}(t,1))\biggr{)}\tilde{u}(t,1)$
$\displaystyle-\frac{1}{s(t)}\biggl{(}\beta(b(t)-\gamma\tilde{u}_{1}(t,0))-\beta(b(t)-\gamma\tilde{u}_{2}(t,0))\biggr{)}\tilde{u}(t,0)+\frac{1}{s^{2}(t)}\int_{0}^{1}|\tilde{u}_{y}(t)|^{2}dy.$
(3.10)

Since the function $r\sigma(r)$ is monotone for $r\in\mathbb{R}$, the first
term of the right-hand side of $(\ref{3-10})$ is non-negative. The second term
of the right-hand side of (3.10) is also non-negative, hence we see that

$\displaystyle-\int_{0}^{1}\frac{1}{s^{2}(t)}\tilde{u}_{yy}(t)\tilde{u}(t)dy\geq\frac{1}{s^{2}(t)}\int_{0}^{1}|\tilde{u}_{y}(t)|^{2}dy.$
(3.11)

Accordingly, by (3.9)-(3.11), we have that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{s^{2}(t)}\int_{0}^{1}|\tilde{u}_{y}(t)|^{2}dy\leq\int_{0}^{1}\frac{ys_{t}(t)}{s(t)}f_{y}(t)\tilde{u}(t)dy.$
(3.12)

Here, using Hölder’s inequality, it holds that

$\displaystyle\int_{0}^{1}\frac{ys_{t}(t)}{s(t)}f_{y}(t)\tilde{u}(t)dy\leq\frac{|s_{t}|_{L^{\infty}(0,T)}}{s_{0}}|\tilde{u}(t)|_{L^{2}(0,1)}|f_{y}(t)|_{L^{2}(0,1)}.$
(3.13)

Let $T_{1}\in(0,T]$. Then, by putting $l=\mbox{max}_{0\leq t\leq T}|s(t)|$ and
integrating (3.12) with (3.13) over $[0,t]$ for any $t\in[0,T_{1}]$ we obtain
that

$\displaystyle\frac{1}{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2l^{2}}\int_{0}^{t}\int_{0}^{1}|\tilde{u}_{y}(\tau)|^{2}dyd\tau$
$\displaystyle\leq$
$\displaystyle\frac{|s_{t}|_{L^{\infty}(0,T)}}{s_{0}}|\tilde{u}|_{L^{\infty}(0,T_{1};L^{2}(0,1))}T^{1/2}_{1}\left(\int_{0}^{T_{1}}|f_{y}(\tau)|^{2}_{L^{2}(0,1)}d\tau\right)^{1/2}$
$\displaystyle\leq$
$\displaystyle\frac{|s_{t}|_{L^{\infty}(0,T)}}{s_{0}}|\tilde{u}|_{V(T_{1})}T^{1/2}_{1}|f|_{V(T_{1})}$
(3.14)

Therefore, by putting $\delta=\mbox{min}\\{1/2,1/2l^{2}\\}$ in (3.14) we have
that

$\displaystyle\delta|\tilde{u}|_{V(T_{1})}\leq\frac{|s_{t}|_{L^{\infty}(0,T)}}{s_{0}}T^{1/2}_{1}|f|_{V(T_{1})}\mbox{
for }T_{1}\in(0,T].$

From this result, we infer that for some $T_{1}\leq T$ such that
$\Gamma_{T_{1}}$ is a contraction mapping in $V(T_{1})$. Therefore, by
Banach’s fixed point theorem there exists $\tilde{u}\in V(T_{1})$ such that
$\Gamma_{T_{1}}(\tilde{u})=\tilde{u}$ which implies $\tilde{u}$ is a solution
of $(\mbox{AP})(\tilde{u}_{0},s,b)$ on $[0,T_{1}]$. Since $T_{1}$ is
independent of the choice of initial data of $\tilde{u}$, by repeating the
argument of the local existence result, we can extend the solution $\tilde{u}$
beyond $T_{1}$. Thus, we prove that Lemma 3.3 holds. ∎

Next, for given $s\in W^{1,2}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on
$[0,T]$, we construct a solution to $(\mbox{AP})(\tilde{u}_{0},s,b)$.

######  Lemma 3.4.

Let $T>0$ and $s_{0}>0$. If (A1)-(A3) hold, then, for given $s\in
W^{1,2}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on $[0,T]$,
$(\mbox{AP})(\tilde{u}_{0},s,b)$ has a unique solution $\tilde{u}$ on $[0,T]$.

###### Proof.

For given $s\in W^{1,2}(0,T)$ with $s(0)=s_{0}$ and $s\geq s_{0}$ on $[0,T]$,
we choose a sequence $\\{s_{n}\\}\subset W^{1,\infty}(0,T)$ and $l>0$
satisfying $s_{0}\leq s_{n}\leq l$ on $[0,T]$ for each $n\in\mathbb{N}$,
$s_{n}\to s$ in $W^{1,2}(0,T)$ as $n\to\infty$. By Lemma 3.3 we can take a
sequence $\\{\tilde{u}_{n}\\}$ of solutions to
$(\mbox{AP})(\tilde{u}_{0},s_{n},b)$ on $[0,T]$. Then, we see that
$t\to\psi^{t}(\tilde{u}_{n}(t))$ is absolutely continuous on $[0,T]$ so that
$t\to\frac{1}{s^{2}_{n}(t)}|\tilde{u}_{ny}(t)|^{2}_{L^{2}(0,1)}$ is continuous
on $[0,T]$. First, it holds that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}-\int_{0}^{1}\frac{1}{s^{2}_{n}(t)}\tilde{u}_{nyy}(t)\tilde{u}_{n}(t)dy=\int_{0}^{1}\frac{ys_{nt}(t)}{s_{n}(t)}\tilde{u}_{ny}(t)\tilde{u}_{n}(t)dy.$

For the second term in the left-hand side, we have that

$\displaystyle-\int_{0}^{1}\frac{1}{s^{2}_{n}(t)}\tilde{u}_{nyy}(t)\tilde{u}_{n}(t)dy$
$\displaystyle=$
$\displaystyle\frac{a_{0}}{s_{n}(t)}\tilde{u}_{n}(t,1)\sigma(\tilde{u}_{n}(t,1))\tilde{u}_{n}(t,1)-\frac{1}{s_{n}(t)}\beta(b(t)-\gamma\tilde{u}_{n}(t,0))\tilde{u}_{n}(t,0)+\frac{1}{s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy$
$\displaystyle\geq$
$\displaystyle-\frac{1}{s_{n}(t)}\beta(b(t)-\gamma\tilde{u}_{n}(t,0))\tilde{u}_{n}(t,0)+\frac{1}{s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy.$

Hence, we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}\frac{ys_{nt}(t)}{s_{n}(t)}\tilde{u}_{ny}(t)\tilde{u}_{n}(t)dy+\frac{1}{s_{n}(t)}\beta(b(t)-\gamma\tilde{u}_{n}(t,0))\tilde{u}_{n}(t,0)\mbox{
for }t\in[0,T].$ (3.15)

Using Young’s inequality, we have that

$\displaystyle\int_{0}^{1}\frac{ys_{nt}(t)}{s_{n}(t)}\tilde{u}_{ny}(t)\tilde{u}_{n}(t)dy\leq\frac{1}{2s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy+\frac{|s_{nt}(t)|^{2}}{{2}}\int_{0}^{1}|\tilde{u}_{n}(t)|^{2}dy.$
(3.16)

Here, by Sobolev’s embedding theorem in one dimension, we note that it holds
that

$\displaystyle|v(t,y)|^{2}\leq
C_{e}|v(t)|_{H^{1}(0,1)}|v(t)|_{L^{2}(0,1)}\mbox{ for }v\in H^{1}(0,1)\mbox{
and }y\in[0,1],$ (3.17)

where $C_{e}$ is a positive constant defined from Sobolev’s embedding theorem.
By (3.17) and $s_{0}\leq s_{n}$ on $[0,T]$ we get

$\displaystyle\frac{1}{s_{n}(t)}\beta(b(t)-\gamma\tilde{u}_{n}(t,0))\tilde{u}_{n}(t,0)\leq\frac{\beta
b^{*}}{s_{n}(t)}|\tilde{u}_{n}(t,0)|$ $\displaystyle\leq$
$\displaystyle\frac{\beta
b^{*}C_{e}}{2s_{n}(t)}\biggl{(}|\tilde{u}_{ny}(t)|_{L^{2}(0,1)}|\tilde{u}_{n}(t)|_{L^{2}(0,1)}+|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}\biggr{)}+\frac{\beta
b^{*}}{2s_{n}(t)}$ $\displaystyle\leq$
$\displaystyle\frac{1}{4s^{2}_{n}(t)}|\tilde{u}_{ny}(t)|^{2}_{L^{2}(0,1)}+\left((\beta
b^{*}C_{e})^{2}+\frac{\beta
b^{*}C_{e}}{2s_{0}}\right)|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}+\frac{\beta
b^{*}}{2s_{0}}.$ (3.18)

As a result, we see from (3.15)-(3) that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{4s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(\tau)|^{2}dy$
$\displaystyle\leq$ $\displaystyle\left(\frac{|s_{nt}(t)|^{2}}{{2}}+(\beta
b^{*}C_{e})^{2}+\frac{\beta
b^{*}C_{e}}{2s_{0}}\right)|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}+\frac{\beta
b^{*}}{2s_{0}}\mbox{ for }t\in[0,T].$

Now, we denote $F_{n}(t)$ the coefficient of
$|\tilde{u}_{n}|^{2}_{L^{2}(0,1)}$ in the right-hand side. Then, by $s_{n}\leq
l$ on $[0,T]$, the boundedness of $\\{F_{n}\\}$ in $L^{1}(0,T)$ and Gronwall’s
inequality we obtain that

$\displaystyle\frac{1}{2}|\tilde{u}_{n}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{4l^{2}}\int_{0}^{t}|\tilde{u}_{ny}(t)|^{2}_{L^{2}(0,1)}d\tau$
$\displaystyle\leq\left(\frac{1}{2}|\tilde{u}_{0}|^{2}_{L^{2}(0,1)}+\left(\frac{\beta
b^{*}}{2s_{0}}\right)T\right)e^{C}\mbox{ for }t\in[0,T].$ (3.19)

Next, we put $\tilde{u}_{n}(t)=u_{0}$ for $t<0$. For each $n\in\mathbb{N}$ and
$h>0$, it holds

$\displaystyle\int_{0}^{1}\tilde{u}_{nt}(t)\frac{\tilde{u}_{n}(t)-\tilde{u}_{n}(t-h)}{h}dy-\int_{0}^{1}\frac{1}{s^{2}_{n}(t)}\tilde{u}_{nyy}(t)\frac{\tilde{u}_{n}(t)-\tilde{u}_{n}(t-h)}{h}dy$
$\displaystyle=\int_{0}^{1}\frac{ys_{nt}(t)}{s_{n}(t)}\tilde{u}_{ny}(t)\frac{\tilde{u}_{n}(t)-\tilde{u}_{n}(t-h)}{h}dy.$
(3.20)

The second term of (3.20) can be dealt as follows:

$\displaystyle-\int_{0}^{1}\frac{1}{s^{2}(t)}\tilde{u}_{nyy}(t)\frac{\tilde{u}_{n}(t)-\tilde{u}_{n}(t-h)}{h}dy$
$\displaystyle=$
$\displaystyle-\frac{\tilde{u}_{ny}(t,1)}{s^{2}_{n}(t)}\frac{\tilde{u}_{n}(t,1)-\tilde{u}_{n}(t-h,1)}{h}+\frac{\tilde{u}_{ny}(t,0)}{s^{2}_{n}(t)}\frac{\tilde{u}_{n}(t,0)-\tilde{u}_{n}(t-h,0)}{h}$
$\displaystyle+\int_{0}^{1}\frac{\tilde{u}_{ny}(t)}{s^{2}_{n}(t)}\frac{\tilde{u}_{ny}(t)-\tilde{u}_{ny}(t-h)}{h}dy.$

We denote $I_{1}$, $I_{2}$ and $I_{3}$ the three terms in the last identity
and estimate three terms separately. For $I_{1}$, using the same notation
$g_{1}$ and $g_{2}$ in the proof of Lemma 3.1, it follows that

$\displaystyle
I_{1}\geq\frac{1}{h}\frac{1}{s_{n}(t)}\left(\int_{0}^{\tilde{u}_{n}(t,1)}a_{0}\xi\sigma(\xi)d\xi-\int_{0}^{\tilde{u}_{n}(t-h,1)}a_{0}\xi\sigma(\xi)d\xi\right)$
$\displaystyle=$
$\displaystyle\frac{g_{1}(s_{n}(t),\tilde{u}_{n}(t,1))-g_{1}(s_{n}(t-h),\tilde{u}_{n}(t-h,1))}{h}+\frac{1}{h}\left(\frac{1}{s_{n}(t-h)}-\frac{1}{s_{n}(t)}\right)\int_{0}^{\tilde{u}_{n}(t-h,1)}a_{0}\xi\sigma(\xi)d\xi.$

Next, for $I_{2}$ and $I_{3}$ we have that

$\displaystyle I_{2}$
$\displaystyle\geq\frac{1}{h}\frac{1}{s_{n}(t)}\left(-\int_{0}^{\tilde{u}_{n}(t,0)}\beta(b(t)-\gamma\xi)d\xi+\int_{0}^{\tilde{u}_{n}(t-h,0)}\beta(b(t)-\gamma\xi)d\xi\right)$
$\displaystyle=$
$\displaystyle\frac{g_{2}(s_{n}(t),b(t),\tilde{u}_{n}(t,0))-g_{2}(s_{n}(t-h),b(t-h),\tilde{u}_{n}(t-h,0))}{h}$
$\displaystyle+\frac{1}{h}\left(-\frac{1}{s_{n}(t-h)}+\frac{1}{s_{n}(t)}\right)\int_{0}^{\tilde{u}_{n}(t-h,0)}\beta(b(t-h)-\gamma\xi)d\xi$
$\displaystyle-\frac{1}{h}\frac{1}{s_{n}(t)}\int_{0}^{\tilde{u}_{n}(t-h,0)}\biggl{(}\beta(b(t-h)-\gamma\xi)-\beta(b(t)-\gamma\xi)\biggr{)}d\xi,$

and

$\displaystyle I_{3}$
$\displaystyle\geq\frac{1}{h}\frac{1}{2s^{2}_{n}(t)}\left(\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy-\int_{0}^{1}|\tilde{u}_{ny}(t-h)|^{2}dy\right)$
$\displaystyle=\frac{1}{h}\left(\frac{1}{2s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy-\frac{1}{2s^{2}_{n}(t-h)}\int_{0}^{1}|\tilde{u}_{ny}(t-h)|^{2}dy\right)$
$\displaystyle+\frac{1}{h}\left(\frac{1}{2s^{2}_{n}(t-h)}-\frac{1}{2s^{2}_{n}(t)}\right)\int_{0}^{1}|\tilde{u}_{ny}(t-h)|^{2}dy.$

Combining the above three estimates and using the fact that $t\to
1/s^{2}_{n}(t)|\tilde{u}_{ny}(t)|^{2}$ is continuous on $[0,T]$, we obtain

$\displaystyle\liminf_{h\to 0}(I_{1}+I_{2}+I_{3})$ $\displaystyle\geq$
$\displaystyle\frac{d}{dt}\psi^{t}(\tilde{u}_{n}(t))+\frac{s_{nt}(t)}{s^{2}_{n}(t)}\int_{0}^{\tilde{u}_{n}(t,1)}a_{0}\xi\sigma(\xi)d\xi+\frac{s_{nt}(t)}{s^{2}_{n}(t)}\int_{0}^{\tilde{u}_{n}(t,0)}\beta(b(t)-\gamma\xi)d\xi$
$\displaystyle+\frac{1}{s_{n}(t)}\int_{0}^{\tilde{u}_{n}(t,0)}\beta
b_{t}(t)d\xi+\frac{s_{nt}(t)}{s^{3}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy.$

Applying this result to (3.20) and letting $h\to 0$, we observe that

$\displaystyle|\tilde{u}_{nt}(t)|^{2}_{L^{2}(0,1)}+\frac{d}{dt}\psi^{t}(\tilde{u}_{n}(t))$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}\frac{ys_{nt}(t)}{s_{n}(t)}\tilde{u}_{ny}(t)\tilde{u}_{nt}(t)dy+\frac{|s_{nt}(t)|}{s^{2}_{n}(t)}\int_{0}^{\tilde{u}_{n}(t,1)}a_{0}\xi\sigma(\xi)d\xi+\frac{|s_{nt}(t)|}{s^{2}_{n}(t)}\biggl{|}\int_{0}^{\tilde{u}_{n}(t,0)}\beta(b(t)-\gamma\xi)d\xi\biggr{|}$
$\displaystyle+\frac{1}{s_{n}(t)}\left|\int_{0}^{\tilde{u}_{n}(t,0)}\beta
b_{t}(t)d\xi\right|+\frac{|s_{nt}(t)|}{s^{3}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy.$
(3.21)

Denote $J_{i}(1\leq i\leq 5)$ each terms in the right-hand side of (3.21).
Using Lemma 3.1 and $s_{n}\geq s_{0}$ on $[0,T]$, we estimate each terms
$J_{i}$ except for $i=2$ as follows:

$\displaystyle J_{1}$
$\displaystyle\leq\frac{1}{2}|\tilde{u}_{nt}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2}\frac{|s_{nt}(t)|^{2}}{s^{2}_{n}(t)}|\tilde{u}_{ny}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle\leq\frac{1}{2}|\tilde{u}_{nt}(t)|^{2}_{L^{2}(0,1)}+|s_{nt}(t)|^{2}\left(C_{0}\psi^{t}(\tilde{u}_{n}(t))+C_{1}\right),$
$\displaystyle J_{3}$
$\displaystyle\leq\frac{|s_{nt}(t)|\beta}{s^{2}_{0}}\left(b^{*}|\tilde{u}_{n}(t,0)|+\gamma\frac{|\tilde{u}_{n}(t,0)|^{2}}{2}\right)$
$\displaystyle\leq\frac{\beta
b^{*}}{s^{2}_{0}}\left(\frac{|s_{nt}(t)|^{2}}{2}+\frac{|\tilde{u}_{n}(t,0)|^{2}}{2}\right)+\frac{\beta\gamma|s_{nt}(t)|}{2s^{2}_{0}}|\tilde{u}_{n}(t,0)|^{2},$
$\displaystyle J_{4}$
$\displaystyle\leq\frac{\beta}{s_{0}}|b_{t}(t)|\tilde{u}_{n}(t,0)\leq\frac{\beta}{s_{0}}\left(\frac{|b_{t}(t)|^{2}}{2}+\frac{|\tilde{u}_{n}(t,0)|^{2}}{2}\right),$
$\displaystyle J_{5}$
$\displaystyle\leq\frac{|s_{nt}(t)|}{s^{3}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t)|^{2}dy\leq\frac{2|s_{nt}(t)|}{s_{0}}\left(C_{0}\psi^{t}(\tilde{u}_{n}(t))+C_{1}\right).$

For $J_{2}$, by the definition of $\psi^{t}$, we have that

$\displaystyle J_{2}$ $\displaystyle=$
$\displaystyle\frac{|s_{nt}(t)|}{s_{n}(t)}\left(\psi^{t}(\tilde{u}_{n}(t))-\frac{1}{2s^{2}_{n}(t)}\int_{0}^{1}|\tilde{u}_{ny}(t,y)|^{2}dy+\frac{1}{s_{n}(t)}\int_{0}^{\tilde{u}_{n}(t,0)}\beta(b(t)-\gamma\xi)d\xi\right)$
$\displaystyle\leq$
$\displaystyle\frac{|s_{nt}(t)|}{s_{n}(t)}\left(\psi^{t}(\tilde{u}_{n}(t))+\frac{1}{s_{n}(t)}\beta
b^{*}\tilde{u}_{n}(t,0)\right)$ $\displaystyle\leq$
$\displaystyle\frac{|s_{nt}(t)|}{s_{0}}\left(\psi^{t}(\tilde{u}_{n}(t))+\frac{\beta
b^{*}}{2s_{n}(t)}(1+\tilde{u}^{2}_{n}(t,0))\right)$

Hence, by the estimates for each $J_{i}$ and (3.21), we obtain that

$\displaystyle\frac{1}{2}|\tilde{u}_{nt}(t)|^{2}_{L^{2}(0,1)}+\frac{d}{dt}\psi^{t}(\tilde{u}_{n}(t))$
$\displaystyle\leq$
$\displaystyle\frac{|s_{nt}(t)|}{s_{0}}\psi^{t}(\tilde{u}_{n}(t))+\left(|s_{nt}(t)|^{2}+\frac{2|s_{nt}(t)|}{s_{0}}\right)(C_{0}\psi^{t}(\tilde{u}_{n}(t)+C_{1})$
$\displaystyle+\frac{\beta
b^{*}}{s^{2}_{0}}\frac{|s_{nt}(t)|^{2}}{2}+\left(\frac{\beta
b^{*}}{2s^{2}_{0}}+\frac{\beta\gamma|s_{nt}(t)|}{2s^{2}_{0}}+\frac{\beta}{2s_{0}}+\frac{|s_{nt}(t)|}{s_{0}}\frac{\beta
b^{*}}{2s_{0}}\right)\tilde{u}^{2}_{n}(t,0)$
$\displaystyle+\frac{\beta}{2s_{0}}|b_{t}(t)|^{2}+\frac{\beta
b^{*}}{2s_{0}}\frac{|s_{nt}(t)|}{s_{0}}\mbox{ for a.e. }t\in[0,T].$ (3.22)

Here, by using (i) of Lemma 3.1 we put the coefficients of
$\psi^{t}(\tilde{u}_{n}(t))$ by $l_{1}(t)$ and otherwise by $l_{2}(t)$. Then,
by the fact that $\\{s_{n}\\}$ is bounded in $W^{1,2}(0,T)$ and (A2), $l_{1}$,
$l_{2}\in L^{1}(0,T)$. Now, we see from (3) that

$\displaystyle\frac{1}{2}|\tilde{u}_{nt}(t)|^{2}_{L^{2}(0,1)}+\frac{d}{dt}\psi^{t}(\tilde{u}_{n}(t))\leq
l_{1}(t)\psi^{t}(\tilde{u}_{n}(t))+l_{2}(t)\mbox{ for a.e. }t\in[0,T].$

Therefore, by using Gronwall’s lemma, we have that

$\displaystyle\frac{1}{2}\int_{0}^{t}|\tilde{u}_{nt}(\tau)|^{2}_{L^{2}(0,1)}d\tau+\psi^{t}(\tilde{u}_{n}(t))\leq\biggl{[}\psi^{0}({\tilde{u}_{0}})+\int_{0}^{t}l_{2}(\tau)d\tau\biggr{]}e^{\int_{0}^{t}l_{1}(\tau)d\tau}\mbox{
for }t\in[0,T].$

From this result, we infer that the sequence $\\{\tilde{u}_{n}\\}$ is bounded
in $W^{1,2}(0,T;L^{2}(0,1))$ and the sequence
$\\{\psi^{(\cdot)}(\tilde{u}_{n}(\cdot))\\}$ is bounded in $L^{\infty}(0,T)$.
By these boundedness results and Lemma 3.1, we can take a sequence
$\\{n_{k}\\}\subset\\{n\\}$ such that for some $\tilde{u}\in
W^{1,2}(0,T;L^{2}(0,1))\cap L^{\infty}(0,T;H^{1}(0,1))$,
$\tilde{u}_{n_{k}}\to\tilde{u}$ weakly in $W^{1,2}(0,T;L^{2}(0,1))$, weakly -*
in $L^{\infty}(0,T;H^{1}(0,1))$ and in $C(\overline{Q(T)})$ as $k\to\infty$.
Finally, by letting $k\to\infty$, we see that $\tilde{u}$ is a solution of
$(\mbox{AP})(\tilde{u}_{0},s,b)$ on $[0,T]$.

To complete the proof, we show the uniqueness of a solution
$(\mbox{AP})(\tilde{u}_{0},s,b)$. Let $s\in W^{1,2}(0,T)$ with $s(0)=s_{0}$
and $s\geq s_{0}$ on $[0,T]$ and $\tilde{u}_{1}$ and $\tilde{u}_{2}$ be
solutions of $(\mbox{AP})(\tilde{u}_{0},s,b)$ on $[0,T]$. Put
$\tilde{u}=\tilde{u}_{1}-\tilde{u}_{2}$. Then, by (3.1) and the same argument
of the derivation of (3.12), we have that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{s^{2}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}\leq\int_{0}^{1}\frac{ys_{t}(t)}{s(t)}\tilde{u}_{y}(t)\tilde{u}(t)dy.$
(3.23)

For the right-hand side of (3.23), we deal as follows:

$\displaystyle\int_{0}^{1}\frac{ys_{t}(t)}{s(t)}\tilde{u}_{y}(t)\tilde{u}(t)dy\leq\frac{1}{2s^{2}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}+\frac{|s_{t}(t)|^{2}}{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}.$

From the above result and (3.23) we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2l^{2}}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}\leq\frac{|s_{t}(t)|^{2}}{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)},$

where $\displaystyle{l=\mbox{max}_{0\leq t\leq T}|s(t)|}$. Therefore, by
Gronwall’s lemma we have that $|\tilde{u}(t)|_{L^{2}(0,1)}=0$ for $t\in[0,T]$.
This implies that $\tilde{u}_{1}=\tilde{u}_{2}$ on $[0,T]$. Thus, Lemma 3.4 is
proved. ∎

##  4 Proof of Theorem 2.2

In this section, using the results obtained in Section 3, we establish the
existence of a locally-in-time solution $(\mbox{PC})(\tilde{u}_{0},s_{0},b)$.
In the rest of this section, we assume (A1)-(A3). For $T>0$ and $l>0$ such
that $s_{0}<l$ we set

$\displaystyle M(T,s_{0},l):=\\{s\in W^{1,2}(0,T)|s_{0}\leq s\leq l\mbox{ on
}[0,T],s(0)=s_{0}\\}.$

Also, for given $s\in M(T,s_{0},l)$, we define two solution mappings as
follows: $\Psi:M(T,s_{0},l)\to W^{1,2}(0,T;L^{2}(0,1))\cap
L^{\infty}(0,T;H^{1}(0,1))$ by $\Psi(s)=\tilde{u}$, where $\tilde{u}$ is a
unique solution of $(\mbox{AP})(\tilde{u}_{0},s,b)$, and
$\Gamma_{T}:M(T,s_{0},l)\to W^{1,2}(0,T)$ by
$\Gamma_{T}(s)=s_{0}+\int_{0}^{t}a_{0}\sigma(\Psi(s)(\tau,1))d\tau$ for
$t\in[0,T]$. Moreover, for any $K>0$ we put

$\displaystyle M_{K}(T):=\\{s\in M(T,s_{0},l)|\ |s|_{W^{1,2}(0,T)}\leq K\\}.$

Now, we show that for some $T>0$, $\Gamma_{T}$ is a contraction mapping on the
closed set of $M_{K}(T)$ for any $K>0$.

######  Lemma 4.1.

Let $K>0$. Then, there exists a positive constant $T^{*}\leq T$ such that the
mapping $\Gamma_{T^{*}}$ is a contraction on the closed set $M_{K}(T^{*})$ in
$W^{1,2}(0,T^{*})$.

###### Proof.

For $T>0$ and $l>0$ such that $s_{0}<l$, let $s\in M(T,s_{0},l)$ and
$\tilde{u}=\Psi(s)$. First, we note that it holds

$\displaystyle|\Psi(s)|_{W^{1,2}(0,T;L^{2}(0,1))}+|\Psi(s)|_{L^{\infty}(0,T;H^{1}(0,1))}\leq
C\mbox{ for }s\in M_{K}(T),$ (4.1)

where $C=C(T,\tilde{u}_{0},K,l,b^{*},\beta,s_{0})$ is a positive constant
depending on $T$, $\tilde{u}_{0}$, $K$, $l$, $b^{*}$, $\beta$ and $s_{0}$.

Next, we show that there exists $T_{0}\leq T$ such that
$\Gamma_{T_{0}}:M_{K}(T_{0})\to M_{K}(T_{0})$ is well-defined. Let $K>0$ and
$s\in M_{K}(T)$. First, by the definition of $\sigma$, we see that

$\displaystyle\Gamma_{T}(s)(t)$
$\displaystyle=s_{0}+\int_{0}^{t}a_{0}\sigma(\Psi(s)(\tau,1))d\tau\geq
s_{0}\mbox{ for }t\in[0,T].$ (4.2)

Also, by (3.17) and (4.1), we have that
$|\tilde{u}(t,1)|\leq\sqrt{C_{e}}|\tilde{u}(t)|_{H^{1}(0,1)}\leq\sqrt{C_{e}}C$
for a.e. $t\in(0,T)$. Hence, by $\sigma(r)\leq r$ for $r\in\mathbb{R}$ we
obtain that

$\displaystyle\Gamma_{T}(s)(t)\leq
s_{0}+a_{0}\sqrt{C_{e}}CT,\quad\int_{0}^{t}|\Gamma_{T}(s)(\tau)|^{2}d\tau\leq
2s^{2}_{0}T+2a^{2}_{0}T^{3}(C_{e}C^{2}),$ (4.3)

and

$\displaystyle\int_{0}^{t}|\Gamma^{\prime}_{T}(s)(\tau)|^{2}d\tau\leq
a^{2}_{0}\int_{0}^{t}|\Psi(s)(\tau,1)|^{2}d\tau\leq a^{2}_{0}TC_{e}C^{2}.$
(4.4)

Therefore, by (4.3) and (4.4) we see that there exists $T_{0}\leq T$ such that
$\Gamma_{T_{0}}(s)\in M_{K}(T_{0})$.

Next, for $s_{1}$ and $s_{2}\in M_{K}(T_{0})$, let $\tilde{u}_{1}=\Psi(s_{1})$
and $\tilde{u}_{2}=\Psi(s_{2})$ and set
$\tilde{u}=\tilde{u}_{1}-\tilde{u}_{2}$, $s=s_{1}-s_{2}$. Then, it holds that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{H}-\int_{0}^{1}\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1yy}(t)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2yy}(t)\right)\tilde{u}(t)dy$
$\displaystyle=\int_{0}^{1}\left(\frac{ys_{1t}(t)}{s_{1}(t)}\tilde{u}_{1y}(t)-\frac{ys_{2t}(t)}{s_{2}(t)}\tilde{u}_{2y}(t)\right)\tilde{u}(t)dy.$
(4.5)

For the second term of the left-hand side of (4.5), we observe that

$\displaystyle-\int_{0}^{1}\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1yy}(t)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2yy}(t)\right)\tilde{u}(t)dy$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1y}(t)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2y}(t)\right)\tilde{u}_{y}(t)dy$
$\displaystyle-\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1y}(t,1)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2y}(t,1)\right)\tilde{u}(t,1)+\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1y}(t,0)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2y}(t,0)\right)\tilde{u}(t,0)$
$\displaystyle=:$ $\displaystyle I_{1}+I_{2}+I_{3}.$

For the term $I_{1}$, the following estimate below holds:

$\displaystyle
I_{1}=\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}+\int_{0}^{1}\left(\frac{1}{s^{2}_{1}(t)}-\frac{1}{s^{2}_{2}(t)}\right)\tilde{u}_{2y}(t)\tilde{u}_{y}(t)dy$
$\displaystyle\geq$
$\displaystyle\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}-\frac{2l|s(t)|}{s^{3}_{0}s_{1}(t)}|\tilde{u}_{2y}(t)|_{L^{2}(0,1)}|\tilde{u}_{y}(t)|_{L^{2}(0,1)}$
$\displaystyle\geq$
$\displaystyle\left(1-\frac{\eta}{2}\right)\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}-\frac{1}{2\eta}\left(\frac{2l}{s^{3}_{0}}\right)^{2}|s(t)|^{2}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)},$

where $\eta$ is arbitrary positive number. For $I_{2}$, we separate in the
following way:

$\displaystyle-\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1y}(t,1)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2y}(t,1)\right)\tilde{u}(t,1)$
$\displaystyle=$ $\displaystyle
a_{0}\left(\frac{\tilde{u}_{1}(t,1)\sigma(\tilde{u}_{1}(t,1))}{s_{1}(t)}-\frac{\tilde{u}_{2}(t,1)\sigma(\tilde{u}_{2}(t,1))}{s_{2}(t)}\right)\tilde{u}(t,1)$
$\displaystyle=$ $\displaystyle
a_{0}\left(\frac{1}{s_{1}(t)}\biggl{(}\tilde{u}_{1}(t,1)\sigma(\tilde{u}_{1}(t,1))-\tilde{u}_{2}(t,1)\sigma(\tilde{u}_{2}(t,1))\biggr{)}+\left(\frac{1}{s_{1}(t)}-\frac{1}{s_{2}(t)}\right)\tilde{u}_{2}(t,1)\sigma(\tilde{u}_{2}(t,1))\right)\tilde{u}(t,1)$
$\displaystyle=$ $\displaystyle:I_{21}+I_{22}.$

Similarly to (3.10), the term $I_{21}$ is non-positive because the function
$r\sigma(r)$ is monotone for $r\in\mathbb{R}$. For $I_{22}$, using the fact
that $\sigma(r)\leq|r|$ for $r\in\mathbb{R}$ and (3.17), the following
inequalities hold:

$\displaystyle|I_{22}|$
$\displaystyle=\left(\frac{s(t)}{s_{1}(t)s_{2}(t)}\right)a_{0}\tilde{u}_{2}(t,1)\sigma(\tilde{u}_{2}(t,1))\tilde{u}(t,1)$
$\displaystyle\leq\frac{C_{e}(a_{0}\tilde{u}^{2}_{2}(t,1))^{2}}{2s^{2}_{0}s^{2}_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}+\frac{1}{2}|s(t)|^{2}.$
(4.6)

Put $L^{(1)}_{s_{2}}(t)=C_{e}a^{2}_{0}|\tilde{u}_{2}(t,1)|^{4}/2s^{2}_{0}$. As
for $I_{2}$, we consider the term $I_{3}$ as follows:

$\displaystyle\left(\frac{1}{s^{2}_{1}(t)}\tilde{u}_{1y}(t,0)-\frac{1}{s^{2}_{2}(t)}\tilde{u}_{2y}(t,0)\right)\tilde{u}(t,0)$
$\displaystyle=$
$\displaystyle-\left(\frac{1}{s_{1}(t)}\beta(b(t)-\gamma\tilde{u}_{1}(t,0))-\frac{1}{s_{2}(t)}\beta(b(t)-\gamma\tilde{u}_{2}(t,0))\right)\tilde{u}(t,0)$
$\displaystyle=$
$\displaystyle-\frac{1}{s_{1}(t)}\biggl{(}\beta(b(t)-\gamma\tilde{u}_{1}(t,0))-\beta(b(t)-\gamma\tilde{u}_{2}(t,0))\biggr{)}\tilde{u}(t,0)$
$\displaystyle-\left(\frac{1}{s_{1}(t)}-\frac{1}{s_{2}(t)}\right)\beta(b(t)-\gamma\tilde{u}_{2}(t,0))\tilde{u}(t,0).$

Then, by using (3.17) and (A3), we have that

$\displaystyle|I_{3}|$ $\displaystyle\leq\frac{\beta
C_{e}\gamma}{s_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}$
$\displaystyle+\frac{(\beta(b^{*}+\gamma|\tilde{u}_{2}(t,0)|)^{2}C_{e}}{2s^{2}_{0}s^{2}_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}+\frac{1}{2}|s(t)|^{2}\mbox{
for }t\in[0,T_{0}].$ (4.7)

For the right-hand side of (4.5), we can write as follows:

$\displaystyle\int_{0}^{1}\left(\frac{ys_{1t}(t)}{s_{1}(t)}\tilde{u}_{1y}(t)-\frac{ys_{2t}(t)}{s_{2}(t)}\tilde{u}_{2y}(t)\right)\tilde{u}(t)dy$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{ys_{1t}(t)}{s_{1}(t)}\tilde{u}_{y}(t)\tilde{u}(t)dy+\int_{0}^{1}\frac{ys_{t}(t)}{s_{1}(t)}\tilde{u}_{2y}(t)\tilde{u}(t)dy+\int_{0}^{1}\left(\frac{1}{s_{1}(t)}-\frac{1}{s_{2}(t)}\right)ys_{2t}(t)\tilde{u}_{2y}(t)\tilde{u}(t)dy$
$\displaystyle:=I_{41}+I_{42}+I_{43}.$

The three terms are estimated in the following way:

$\displaystyle I_{41}$
$\displaystyle\leq\frac{\eta}{2s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2\eta}|s_{1t}(t)|^{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)},$
$\displaystyle I_{42}$
$\displaystyle\leq\frac{1}{2s_{0}}\biggl{(}|s_{t}(t)|^{2}+|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}\biggr{)},$
$\displaystyle I_{43}$
$\displaystyle\leq\frac{1}{2s^{2}_{0}}\biggl{(}|s(t)|^{2}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+|s_{2t}(t)|^{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}\biggr{)}.$

Then, by (4.5)-(4.7) we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+(1-\eta)\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle\leq$ $\displaystyle\frac{\beta
C_{e}\gamma}{s_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}$
$\displaystyle+\frac{1}{s^{2}_{1}(t)}\biggl{(}L^{(1)}_{s_{2}}(t)+\frac{(\beta(b^{*}+\gamma|\tilde{u}_{2}(t,0)|)^{2}C_{e}}{2s^{2}_{0}}\biggr{)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}$
$\displaystyle+\left(\frac{1}{2\eta}|s_{1t}(t)|^{2}+\frac{1}{2s_{0}}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2s^{2}_{0}}|s_{2t}(t)|^{2}\right)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle+\left(\frac{1}{2s^{2}_{0}}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2\eta}\left(\frac{2l}{s^{3}_{0}}\right)^{2}|\tilde{u}_{2y}|^{2}_{L^{2}(0,1)}+1\right)|s(t)|^{2}+\frac{1}{2s_{0}}|s_{t}(t)|^{2}.$
(4.8)

Here, by (3.17) and (4.1), we see that

$\displaystyle|\tilde{u}_{i}(t,0)|^{2}$ $\displaystyle\leq
C_{e}(|\tilde{u}_{iy}(t)|_{L^{2}(0,1)}|\tilde{u}_{i}(t)|_{L^{2}(0,1)}+|\tilde{u}_{i}(t)|^{2}_{L^{2}(0,1)})$
$\displaystyle\leq 2C_{e}C^{2}\mbox{ for }t\in[0,T_{0}],$ (4.9)

where $C$ is the same constant as in (4.1). Then, by (4.9) we note that
$\\{L^{(1)}_{s_{2}}|s_{2}\in M_{k}(T)\\}$ is bounded in $L^{\infty}(0,T_{0})$.
Also, by putting $C_{5}=(\beta(b^{*}+2\gamma C_{e}C^{2}))^{2}C_{e}$ and
Young’s inequality it follows that

$\displaystyle\frac{\beta
C_{e}\gamma}{s_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}$
$\displaystyle\leq$ $\displaystyle\frac{\beta
C_{e}\gamma}{s_{1}(t)}\biggl{(}|\tilde{u}_{y}(t)|_{L^{2}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}+|\tilde{u}(t)|^{2}_{L^{2}(0,1)}\biggr{)}$
$\displaystyle\leq$ $\displaystyle\beta
C_{e}\gamma\left(\frac{\eta}{2s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}+(\frac{1}{2\eta}+\frac{1}{s_{0}})|\tilde{u}(t)|^{2}_{L^{2}(0,1)}\right),$

and

$\displaystyle\left(L^{(1)}_{s_{2}}(t)+\frac{C_{5}}{2s^{2}_{0}}\right)\frac{1}{s^{2}_{1}(t)}|\tilde{u}(t)|_{H^{1}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}$
$\displaystyle\leq$
$\displaystyle\left(L^{(1)}_{s_{2}}(t)+\frac{C_{5}}{2s^{2}_{0}}\right)\frac{1}{s^{2}_{1}(t)}(|\tilde{u}_{y}(t)|_{L^{2}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}+|\tilde{u}(t)|^{2}_{L^{2}(0,1)})$
$\displaystyle\leq$
$\displaystyle\left(L^{(1)}_{s_{2}}(t)+\frac{C_{5}}{2s^{2}_{0}}\right)\left[\frac{1}{s^{2}_{1}(t)}\frac{\eta}{2}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{s^{2}_{0}}(\frac{1}{2\eta}+1)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}\right].$

Accordingly, by applying these results to (4.8) and taking a suitable
$\eta=\eta_{0}$, we have

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2}\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle\leq$ $\displaystyle\beta
C_{e}\gamma\left(\frac{1}{2\eta_{0}}+\frac{1}{s_{0}}\right)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle+\left(L^{(1)}_{s_{2}}(t)+\frac{C_{5}}{2s^{2}_{0}}\right)\frac{1}{s^{2}_{0}}\left(\frac{1}{2\eta_{0}}+1\right)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle+\left(\frac{1}{2\eta_{0}}|s_{1t}(t)|^{2}+\frac{1}{2s_{0}}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2s^{2}_{0}}|s_{2t}(t)|^{2}\right)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}$
$\displaystyle+\left(\frac{1}{2s^{2}_{0}}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2\eta_{0}}\left(\frac{2l}{s^{3}_{0}}\right)^{2}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}+1\right)|s(t)|^{2}+\frac{1}{2s_{0}}|s_{t}(t)|^{2}.$
(4.10)

Now, we put the summation of all coefficients of
$|\tilde{u}(t)|^{2}_{L^{2}(0,1)}$ by $L^{(2)}_{s}(t)$ for $t\in[0,T_{0}]$ and
$L^{(3)}_{s_{2}}(t)=|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)}/2s^{2}_{0}+(4l^{2}|\tilde{u}_{2y}(t)|^{2}_{L^{2}(0,1)})/2\eta_{0}s^{6}_{0}+1+1/2s_{0}$.
Then, we have

$\displaystyle\frac{1}{2}\frac{d}{dt}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2}\frac{1}{s^{2}_{1}(t)}|\tilde{u}_{y}(\tau)|^{2}_{L^{2}(0,1)}$
$\displaystyle\leq$ $\displaystyle
L^{(2)}_{s}(t)|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+L^{(3)}_{s_{2}}(t)(|s(t)|^{2}+|s_{t}(t)|^{2})\mbox{
for }t\in[0,T_{0}].$ (4.11)

Here, using (4.1) and the fact that $s_{i}\in M_{K}(T_{0})$ for $i=1,2$, we
see that $L^{(2)}_{s}\in L^{1}(0,T_{0})$ and $L^{(3)}_{s_{2}}\in
L^{\infty}(0,T_{0})$. Therefore, Gronwall’s inequality guarantees that

$\displaystyle\frac{1}{2}|\tilde{u}(t)|^{2}_{L^{2}(0,1)}+\frac{1}{2}\frac{1}{s^{2}_{1}(t)}\int_{0}^{t}|\tilde{u}_{y}(\tau)|^{2}_{L^{2}(0,1)}d\tau$
$\displaystyle\leq$
$\displaystyle\left(|L^{(3)}_{s_{2}}|_{L^{\infty}(0,T_{0})}|s|^{2}_{W^{1,2}(0,T)}\right)e^{2\int_{0}^{t}L^{(2)}_{s}(\tau)d\tau}\mbox{
for }t\in[0,T_{0}].$ (4.12)

By using (4.12) we show that there exists $T^{*}\leq T_{0}$ such that
$\Gamma_{T^{*}}$ is a contraction mapping on the closed subset of
$M_{K}(T^{*})$. To do so, from the subtraction of the time derivatives of
$\Gamma_{T_{0}}(s_{1})$ and $\Gamma_{T_{0}}(s_{2})$ and relying on (3.17) and
(4.12), we have for $T_{1}\leq T_{0}$ the following estimate:

$\displaystyle|(\Gamma_{T_{1}}(s_{1}))_{t}-(\Gamma_{T_{1}}(s_{2}))_{t}|_{L^{2}(0,T_{1})}$
$\displaystyle\leq$ $\displaystyle
a_{0}\biggl{(}|\sigma(\tilde{u}_{1}(\cdot,1))-\sigma(\tilde{u}_{2}(\cdot,1))|_{L^{2}(0,T_{1})}\biggr{)}$
$\displaystyle\leq$ $\displaystyle
a_{0}\sqrt{C_{e}}\biggl{(}\int_{0}^{T_{1}}(|\tilde{u}_{y}(t)|_{L^{2}(0,1)}|\tilde{u}(t)|_{L^{2}(0,1)}+|\tilde{u}(t)|^{2}_{L^{2}(0,1)})dt\biggr{)}^{1/2}$
$\displaystyle\leq$ $\displaystyle
a_{0}\sqrt{C_{e}}\left(|\tilde{u}|^{\frac{1}{2}}_{L^{\infty}(0,T;L^{2}(0,1))}\left(\int_{0}^{T_{1}}|\tilde{u}_{y}(t)|_{L^{2}(0,1)}dt\right)^{\frac{1}{2}}+\sqrt{T_{1}}|\tilde{u}|_{L^{\infty}(0,T;L^{2}(0,1))}\right).$
(4.13)

Using (4.12), we obtain

$\displaystyle|\Gamma_{T_{1}}(s_{1})-\Gamma_{T_{1}}(s_{2})|_{W^{1,2}(0,T_{1})}\leq
T_{1}C_{6}\biggl{(}T^{\frac{1}{4}}_{1}|s|_{W^{1,2}(0,T_{1})}+\sqrt{T_{1}}|s|_{W^{1,2}(0,T_{1})}\biggr{)},$
(4.14)

where $C_{6}$ is a positive constant obtained by (4.12). Therefore, by (4.13)
and (4.14) we see that there exists $T^{*}\leq T_{0}$ such that
$\Gamma_{T^{*}}$ is a contraction mapping on a closed subset of
$M_{K}(T^{*})$. ∎

From Lemma 4.1, by applying Banach’s fixed point theorem, there exists $s\in
M_{K}(T^{*})$, where $T^{*}$ is the same as in Lemma 4.1 such that
$\Gamma_{T^{*}}(s)=s$. This implies that $(\mbox{PC})(\tilde{u}_{0},s_{0},b)$
has a unique solution $(s,\tilde{u})$ on $[0,T^{*}]$. Thus, we can prove
Theorem 2.4. Moreover, this shows that by the change of variables (2.8) a pair
of the function $(s,u)$ is a solution of $(\mbox{P})(u_{0},s_{0},b)$ on
$[0,T^{*}]$.

At the end of this section, we show the boundedness of the solution to
$(\mbox{P})(u_{0},s_{0},b)$.

######  Lemma 4.2.

Let $T>0$ and $(s,u)$ be a solution of $(\mbox{P})(u_{0},s_{0},b)$ on $[0,T]$.
Then, $0\leq u(t)\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$.

###### Proof.

First, we show that $u(t)\geq 0$ on $[0,s(t)]$ for $t\in[0,T]$. From (1.1), we
have that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}|[-u(t)]^{+}|^{2}dz-\frac{s_{t}(t)}{2}|[-u(t,s(t))]^{+}|^{2}$
$\displaystyle+\int_{0}^{s(t)}u_{zz}(t)[-u(t)]^{+}dz=0\mbox{ for a.e.
}t\in[0,T].$ (4.15)

By the boundary conditions (1.2) and (1.3) it follows that

$\displaystyle u_{z}(t,s(t))[-u(t,s(t))]^{+}$
$\displaystyle=-u(t,s(t))s_{t}(t)[-u(t,s(t))]^{+}=s_{t}(t)|[-u(t,s(t))]^{+}|^{2}$

and

$\displaystyle-u_{z}(t,0)[-u(t,0)]^{+}=\beta(b(t)-\gamma
u(t,0))[-u(t,0)]^{+}\geq 0.$

Therefore, we derive that

$\displaystyle\frac{d}{dt}\int_{0}^{s(t)}|[-u(t)]^{+}|^{2}dz+\frac{s_{t}(t)}{2}|[-u(t,s(t))]^{+}|^{2}+\int_{0}^{s(t)}|[-u(t)]_{z}^{+}|^{2}dz\leq
0\mbox{ for a.e. }t\in[0,T].$ (4.16)

Note that by $s_{t}(t)=a_{0}\sigma(u(t,s(t))$, the second term in the left-
hand side of (4.16) is equal to 0. Therefore, by integrating (4.16) over
$[0,T]$ we conclude that $u\geq 0$ on $[0,s(t)]$ for $t\in[0,T]$.

Next, we show that $u(t)\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$. Put
$U(t,z)=[u(t,z)-b^{*}/\gamma]^{+}$ for $z\in[0,s(t)]$ and $t\in[0,T]$. Then,
we have that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}|U(t,z)|^{2}dz-\frac{s_{t}(t)}{2}|U(t,s(t))|^{2}-\int_{0}^{s(t)}u_{zz}(t)U(t,z)dz=0\mbox{
for a.e. }t\in[0,T].$ (4.17)

Using the boundary condition (1.2), it holds that

$\displaystyle-u_{z}(t,s(t))U(t,s(t))=u(t,s(t))s_{t}(t)U(t,s(t))$
$\displaystyle=$ $\displaystyle
s_{t}(t)\left(u(t,s(t))-\frac{b^{*}}{\gamma}\right)U(t,s(t))+s_{t}(t)\frac{b^{*}}{\gamma}U(t,s(t))$
$\displaystyle=$ $\displaystyle
s_{t}(t)|U(t,s(t))|^{2}+s_{t}(t)\frac{b^{*}}{\gamma}U(t,s(t)).$

Also, by (1.3) and $b\leq b^{*}$, we observe that

$\displaystyle u_{z}(t,0)U(t,0)=-\beta(b(t)-\gamma u(t,0))U(t,0)$
$\displaystyle=$ $\displaystyle\beta(\gamma u(t,0)-b^{*}+b^{*}-b(t))U(t,0)$
$\displaystyle=$
$\displaystyle\beta\gamma|U(t,0)|^{2}+\beta(b^{*}-b(t))U(t,0)\geq 0.$

By applying the above two results to (4.17) we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}|U(t,z)|^{2}dz+\int_{0}^{s(t)}|U_{z}(t,z)|^{2}dz$
$\displaystyle+\frac{s_{t}(t)}{2}|U(t,s(t))|^{2}+s_{t}(t)\frac{b^{*}}{\gamma}U(t,s(t))\leq
0\mbox{ for a.e. }t\in[0,T].$ (4.18)

Here, by $s_{t}(t)=a_{0}\sigma(u(t,s(t)))$ we notice that $s_{t}(t)\geq 0$ on
$[0,T]$, and the third and forth terms in the left-hand side of (4.18) are
non-negative. Therefore, we have that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}|U(t,z)|^{2}dz+\int_{0}^{s(t)}|U_{z}(t,z)|^{2}dz\leq
0\mbox{ for a.e. }t\in[0,T].$ (4.19)

Finally, by integrating (4.19) over $[0,t]$ for $t\in[0,T]$ and using (A3), we
see that $u(t)\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$. Thus, Lemma
4.2 is proven. ∎

By Lemma 4.2, we can conclude that Theorem 2.2 holds.

##  5 Proof of Theorem 2.5

In this section, we prove Theorem 2.5 which ensure the existence and
uniqueness of a globally-in-time solution of (P)$(u_{0},s_{0},b)$. First, we
provide uniform estimates of a solution of (P)$(u_{0},s_{0},b)$.

######  Lemma 5.1.

Let $(s,u)$ be a solution of (P)$(u_{0},s_{0},b)$ on $[0,T]$ satisfying $0\leq
u\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$. Then, there exists a
positive constant $\tilde{C}$ which is independent of $T$ such that

$\displaystyle\int_{0}^{t}|u_{t}(\tau)|^{2}_{L^{2}(0,s(\tau))}d\tau+|u_{z}(t)|^{2}_{L^{2}(0,s(t))}\leq\tilde{C}\mbox{
for all }t\in(0,T).$ (5.1)

###### Proof.

Let $(s,u)$ be a solution of (P)$(u_{0},s_{0},b)$ on $[0,T]$ such that $0\leq
u\leq b^{*}/\gamma$ on $Q_{s}(T)$. Then, by the change of variables (2.1) we
see that $(s,\tilde{u}$) is a solution of (PC)$(\tilde{u}_{0},s_{0},b)$ on
$[0,T]$ in the sense of Definition 2.3 and satisfies that $0\leq\tilde{u}\leq
b^{*}/\gamma$ on $Q(T)$. Now, we put
$v_{h}(t)=\frac{\tilde{u}(t)-\tilde{u}(t-h)}{h}$ for $h>0$ and
$u(t)=u(0)=u_{0}$ and $b(t)=b(0)$ for $t<0$. By (1.1), it holds that

$\displaystyle\int_{0}^{1}\tilde{u}_{t}(t)s(t)v_{h}(t)dy-\int_{0}^{1}\frac{1}{s(t)}\tilde{u}_{yy}(t)v_{h}(t)dy=\int_{0}^{1}\frac{ys_{t}(t)\tilde{u}_{y}(t)}{s(t)}s(t)v_{h}(t)dy\quad\mbox{
for }t\in[0,T].$ (5.2)

Then, using (1.2)-(1.4) and
$s_{t}(t)=a_{0}\sigma(\tilde{u}(t,1))=a_{0}\tilde{u}(t,1)$ for $t\in[0,T]$, we
observe that

$\displaystyle-\int_{0}^{1}\frac{1}{s(t)}\tilde{u}_{yy}(t)\frac{\tilde{u}(t)-\tilde{u}(t-h)}{h}dy$
$\displaystyle=$ $\displaystyle
a_{0}\tilde{u}^{2}(t,1)v_{h}(t,1)-\beta(b(t)-\gamma\tilde{u}(t,0))v_{h}(t,0)+\int_{0}^{1}\frac{1}{s(t)}\tilde{u}_{y}(t)v_{hy}(t)dy,$
(5.3)

and

$\displaystyle\int_{0}^{1}\frac{1}{s(t)}\tilde{u}_{y}(t)v_{hy}(t)dy$
$\displaystyle\geq$
$\displaystyle\frac{1}{2h}\int_{0}^{1}\frac{1}{s(t)}(|\tilde{u}_{y}(t)|^{2}-|\tilde{u}_{y}(t-h)|^{2})dy$
$\displaystyle=$
$\displaystyle\frac{1}{2h}\biggl{[}\int_{0}^{s(t)}|u_{z}(t)|^{2}dz-\int_{0}^{s(t-h)}\frac{s(t-h)}{s(t)}|u_{z}(t-h)|^{2}dz\biggr{]}$
$\displaystyle=$
$\displaystyle\frac{1}{2h}\biggl{[}\int_{0}^{s(t)}|u_{z}(t)|^{2}dz-\int_{0}^{s(t-h)}|u_{z}(t-h)|^{2}dz+\int_{0}^{s(t-h)}\frac{s(t)-s(t-h)}{s(t)}|u_{z}(t-h)|^{2}dz\biggr{]}.$
(5.4)

Here, for $t\in[0,T]$ the following inequality holds:

$\displaystyle a_{0}\tilde{u}^{2}(t,1)v_{h}(t,1)$
$\displaystyle=a_{0}\frac{\tilde{u}^{3}(t,1)-\tilde{u}^{2}(t,1)\tilde{u}(t-h,1)}{h}$
$\displaystyle\geq a_{0}\frac{\tilde{u}^{3}(t,1)-\tilde{u}^{3}(t-h,1)}{3h}.$
(5.5)

Also, by introducing $\Phi(b(t),r)=-\beta(b(t)r-\frac{\gamma}{2}r^{2})$ for
$r\in\mathbb{R}$, it is easy to see that $\frac{\partial^{2}}{\partial
r^{2}}\Phi(b(t),r)=\beta\gamma\geq 0$ for $r\in\mathbb{R}$. Hence, for
$t\in[0,T]$, $\Phi(b(t),\tilde{u}(t,0))$ is convex with respect to the second
component so that we can see that the following inequality holds.

$\displaystyle-\beta(b(t)-\gamma\tilde{u}(t,0))v_{h}(t,0)\geq\frac{\Phi(b(t),\tilde{u}(t,0))-\Phi(b(t),\tilde{u}(t-h,0))}{h}\mbox{
for }t\in[0,T].$ (5.6)

Combining (5.2)-(5.6) with (5.1) , we have

$\displaystyle\int_{0}^{1}\tilde{u}_{t}(t)s(t)v_{h}(t)dy$
$\displaystyle+\frac{1}{2h}\biggl{[}\int_{0}^{s(t)}|u_{z}(t)|^{2}dz-\int_{0}^{s(t-h)}|u_{z}(t-h)|^{2}dz+\int_{0}^{s(t-h)}\frac{s(t)-s(t-h)}{s(t)}|u_{z}(t-h)|^{2}dz\biggr{]}$
$\displaystyle+a_{0}\frac{\tilde{u}^{3}(t,1)-\tilde{u}^{3}(t-h,1)}{3h}+\frac{\Phi(b(t),\tilde{u}(t,0))-\Phi(b(t),\tilde{u}(t-h,0))}{h}$
$\displaystyle\leq$
$\displaystyle\int_{0}^{1}ys_{t}(t)\tilde{u}_{y}(t)v_{h}(t)dy\mbox{ for
}t\in[0,T].$ (5.7)

Now, we integrate (5) over $[0,t_{1}]$ for $t_{1}\in(0,T]$ and take the limit
as $h\to 0$. Then, by the change of variables (2.8) the first term of the
left-hand side of (5) is as follows:

$\displaystyle\lim_{h\to
0}\int_{0}^{t_{1}}\int_{0}^{1}\tilde{u}_{t}(t)s(t)v_{h}(t)dydt=\int_{0}^{t_{1}}\int_{0}^{1}|\tilde{u}_{t}(t)|^{2}s(t)dydt$
$\displaystyle=$
$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}\biggl{(}|u_{t}(t)|^{2}+2u_{t}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)+\left(u_{z}(t)\frac{z}{s(t)}s_{t}(t)\right)^{2}\biggr{)}dzdt.$
(5.8)

As arguing the local existence, the function
$t\to\int_{0}^{s(t)}|u_{z}(t)|^{2}dz$ is absolutely continuous on $[0,T]$.
Then, the second and third terms of the left-hand side of (5) can be dealt
with as

$\displaystyle\lim_{h\to
0}\frac{1}{2h}\int_{0}^{t_{1}}\biggl{(}\int_{0}^{s(t)}|u_{z}(t)|^{2}dz-\int_{0}^{s(t-h)}|u_{z}(t-h)|^{2}dz\biggr{)}d\tau$
$\displaystyle=$
$\displaystyle\frac{1}{2}\biggl{(}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz-\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz\biggr{)},$
(5.9)

and

$\displaystyle\lim_{h\to
0}\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t-h)}\frac{1}{s(t)}\frac{s(t)-s(t-h)}{h}|u_{z}(t-h)|^{2}dzdt=\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt.$
(5.10)

Moreover, since $\tilde{u}$ is continuous on $\overline{Q(T)}$ we have that

$\displaystyle\lim_{h\to
0}\frac{a_{0}}{3h}\int_{0}^{t_{1}}\biggl{(}\tilde{u}^{3}(t,1)-\tilde{u}^{3}(t-h,1)\biggr{)}dt=$
$\displaystyle\lim_{h\to
0}\biggl{(}\frac{a_{0}}{3h}\int_{t_{1}-h}^{t_{1}}\tilde{u}^{3}(t,1)dt\biggr{)}-\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1)$
$\displaystyle=$
$\displaystyle\frac{a_{0}}{3}\tilde{u}^{3}(t_{1},1)-\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1).$
(5.11)

Similarly to the derivation of (5),

$\displaystyle\lim_{h\to
0}\frac{1}{h}\int_{0}^{t_{1}}\biggl{(}\Phi(b(t),\tilde{u}(t,0))-\Phi(b(t),\tilde{u}(t-h,0))\biggr{)}dt$
$\displaystyle=$
$\displaystyle\Phi(b(t_{1}),\tilde{u}(t_{1},0))-\Phi(b(0),\tilde{u}_{0}(0))$
$\displaystyle+\lim_{h\to
0}\biggl{(}-\frac{1}{h}\int_{0}^{t_{1}}\biggl{[}\Phi(b(t),\tilde{u}(t-h,0))-\Phi(b(t-h),\tilde{u}(t-h,0))\biggr{]}dt\biggr{)}.$
(5.12)

For the last term of the right-hand side of (5) we observe that

$\displaystyle\lim_{h\to
0}\biggl{(}-\frac{1}{h}\int_{0}^{t_{1}}\biggl{[}\Phi(b(t),\tilde{u}(t-h,0))-\Phi(b(t-h),\tilde{u}(t-h,0))\biggr{]}dt\biggr{)}$
$\displaystyle\geq$ $\displaystyle\lim_{h\to
0}\biggl{(}-\frac{1}{h}\int_{0}^{t_{1}}\beta|b(t)-b(t-h)||\tilde{u}(t-h,0)|dt\biggr{)}$
$\displaystyle\geq$ $\displaystyle\lim_{h\to
0}\biggl{(}-\frac{\beta}{h}\int_{0}^{t_{1}}\biggl{(}\int_{t-h}^{t}|b_{t}(\tau)|d\tau\biggr{)}|\tilde{u}(t-h,0)|dt\biggr{)}$
$\displaystyle\geq$ $\displaystyle-\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt.$ (5.13)

From (5) and the estimates (5)-(5), we obtain that

$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}\biggl{(}|u_{t}(t)|^{2}+2u_{t}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)+\left(u_{z}(t)\frac{z}{s(t)}s_{t}(t)\right)^{2}\biggr{)}dzdt$
$\displaystyle+\frac{1}{2}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz-\frac{1}{2}\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz+\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt$
$\displaystyle+\frac{a_{0}}{3}\tilde{u}^{3}(t_{1},1)-\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1)+\Phi(b(t_{1}),\tilde{u}(t_{1},0))-\Phi(b(0),\tilde{u}_{0}(0))-\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt$ $\displaystyle\leq$
$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}\biggl{(}u_{t}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)+\left(u_{z}(t)\frac{z}{s(t)}s_{t}(t)\right)^{2}\biggr{)}dzdt\
\mbox{ for }t_{1}\in[0,T].$ (5.14)

Then, we see that the third term of the left-hand side of (5) is same to the
second term of the right-hand side of (5) . Then, by moving the second term of
the left-hand side of (5) to the right-hand side we have that

$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}|u_{t}(t)|^{2}dzdt$
$\displaystyle+\frac{1}{2}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz-\frac{1}{2}\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz+\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt$
$\displaystyle+\frac{a_{0}}{3}\tilde{u}^{3}(t_{1},1)-\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1)+\Phi(b(t_{1}),\tilde{u}(t_{1},0))-\Phi(b(0),\tilde{u}_{0}(0))-\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt$ $\displaystyle\leq$
$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}-u_{t}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)dzdt\
\mbox{ for }t_{1}\in[0,T].$ (5.15)

Using (1.1) and the fact that $s_{t}(t)\geq 0$ for $t\in[0,T]$ we obtain the
following inequality:

$\displaystyle-\int_{0}^{t_{1}}\int_{0}^{s(t)}u_{t}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)dzdt$
$\displaystyle=$
$\displaystyle-\int_{0}^{t_{1}}\int_{0}^{s(t)}u_{zz}(t)u_{z}(t)\frac{z}{s(t)}s_{t}(t)dzdt$
$\displaystyle=$
$\displaystyle-\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{1}{2}\biggl{(}\frac{\partial}{\partial
z}|u_{z}(t)|^{2}\biggr{)}\frac{z}{s(t)}s_{t}(t)dzdt$ $\displaystyle=$
$\displaystyle-\int_{0}^{t_{1}}\frac{1}{2}|u_{z}(t,s(t))|^{2}s_{t}(t)dt+\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt.$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt.$
(5.16)

Hence, by (5) and (5) we have that

$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}|u_{t}(t)|^{2}dzdt$
$\displaystyle+\frac{1}{2}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz-\frac{1}{2}\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz+\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt$
$\displaystyle+\frac{a_{0}}{3}\tilde{u}^{3}(t_{1},1)-\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1)+\Phi(b(t_{1}),\tilde{u}(t_{1},0))-\Phi(b(0),\tilde{u}_{0}(0))-\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt$ $\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{t_{1}}\int_{0}^{s(t)}\frac{s_{t}(t)}{s(t)}|u_{z}(t)|^{2}dzdt\
\mbox{ for }t_{1}\in[0,T].$ (5.17)

The forth term in the left-hand side and the right-hand side are canceled out
and the fifth and eighth terms in the left-hand side are positive. Therefore,
we finally obtain that

$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}|u_{t}(t)|^{2}+\frac{1}{2}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz+\frac{a_{0}}{3}\tilde{u}^{3}_{0}(1)-\Phi(b(t_{1}),\tilde{u}(t_{1},0))+\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt\mbox{ for }t_{1}\in[0,T].$ (5.18)

In the right-hand side of (5), by (A2) and $0\leq\tilde{u}(t)\leq
b^{*}/\gamma$ on $[0,1]$ for $t\in[0,T]$ and the definition of $\Phi$, we can
estimate as follows:

$\displaystyle-\Phi(b(t_{1}),\tilde{u}(t_{1},0))=-\beta
b(t_{1})\tilde{u}(t_{1},0)+\frac{\beta\gamma}{2}\tilde{u}^{2}(t_{1},0)\leq\frac{\beta\gamma}{2}\left(\frac{b^{*}}{\gamma}\right)^{2}.$
(5.19)

Finally, by (5.19), $b\in W^{1,2}(0,T)$ as in (A2) and (A3) we see that there
exists $\tilde{C}$ which depends on $b^{*}$, $a_{0}$, $\gamma$, $\beta$ such
that (5.1) holds. Thus, Lemma 5.1 is proved. ∎

At the end of this section, we prove Theorem 2.5. Let $T>0$. By the local
existence result there exists $T_{1}<T$ such that (P)$(u_{0},s_{0},b)$ has a
unique solution $(s,u)$ on $[0,T_{1}]$ satisfying $0\leq u\leq b^{*}/\gamma$
on $Q_{s}(T_{1})$. Then, the pair $(s,\tilde{u})$ with the variable (2.1) is a
solution of $(\mbox{PC})(\tilde{u}_{0},s_{0},b)$ satisfying
$0\leq\tilde{u}\leq b^{*}/\gamma$ on $Q(T_{1})$. Let put

$\displaystyle\tilde{T}:=\mbox{sup}\\{T_{1}>0|(\mbox{PC})(\tilde{u}_{0},s_{0},b)\mbox{
has a solution }(s,\tilde{u})\mbox{ on }[0,T_{1}]\\}.$

From the local existence result, we deduce that $\tilde{T}>0$. Now, we assume
$\tilde{T}<T$. First, by (2.5) and the result that $\tilde{u}(t)\geq 0$ on
$[0,1]$ for $t\in[0,\tilde{T})$ we see that $s_{t}(t)\geq 0$ for
$t\in[0,\tilde{T})$, and therefore $s(t)\geq s_{0}$ for $t\in[0,\tilde{T})$.
Also, by putting $L(t)=a_{0}\frac{b^{*}}{\gamma}t+s_{0}$ for $t\in[0,T]$ we
have that

$\displaystyle s(t)$
$\displaystyle=s_{0}+\int_{0}^{t}a_{0}\sigma(\tilde{u}(\tau,1))d\tau$
$\displaystyle=s_{0}+\int_{0}^{t}a_{0}\tilde{u}(\tau,1)d\tau$
$\displaystyle\leq
s_{0}+a_{0}\frac{b^{*}}{\gamma}\tilde{T}=L(\tilde{T})<L(T)\mbox{ for
}t\in[0,\tilde{T}).$ (5.20)

Next, by using the change of the variable (2.1), it holds that

$\displaystyle\int_{0}^{1}|\tilde{u}_{y}(t)|^{2}dy=\int_{0}^{s(t)}\frac{1}{s(t)}|u_{z}(t)s(t)|^{2}dzdt.$

Therefore, from (5) and Lemma 5.1, we obtain that

$\displaystyle|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}\leq L(T)\tilde{C}\mbox{ for
all }t<\tilde{T},$ (5.21)

where $\tilde{C}$ is the same constant as in Lemma 5.1. By (5.21) we see that
for some $\tilde{u}_{\tilde{T}}\in H^{1}(0,1)$,
$\tilde{u}(t)\to\tilde{u}_{\tilde{T}}$ strongly in $L^{2}(0,1)$ and weakly in
$H^{1}(0,1)$ as $t\to\tilde{T}$ and $0\leq\tilde{u}_{\tilde{T}}\leq
b^{*}/\gamma$ on $(0,1)$. Also, by $|s_{t}(t)|\leq a_{0}b^{*}/\gamma$ for
$t\in[0,\tilde{T})$, $\\{s(t)\\}_{t\in[0,\tilde{T})}$ is a Cauchy sequence in
$\mathbb{R}$ so that for some $s_{\tilde{T}}\in\mathbb{R}$, $s(t)\to
s_{\tilde{T}}$ in $\mathbb{R}$ as $t\to\tilde{T}$. Moreover, by (5),
$s_{\tilde{T}}$ satisfies that $0<s_{0}\leq s_{\tilde{T}}\leq L(\tilde{T})$.
Now, we put $u_{\tilde{T}}(z)=\tilde{u}_{\tilde{T}}(\frac{z}{s_{\tilde{T}}})$
for $z\in[0,s_{\tilde{T}}]$. Then, we see that $u_{\tilde{T}}\in
H^{1}(0,s_{\tilde{T}})$ and $0\leq u_{\tilde{T}}\leq b^{*}/\gamma$ on
$(0,s_{\tilde{T}})$ and we can consider $(s_{\tilde{T}},u_{\tilde{T}})$ as a
initial data. Therefore, by repeating the argument of the local existence we
can extend a solution beyond $\tilde{T}$. This is a contradiction for the
definition of $\tilde{T}$ and we have a solution on the whole interval
$[0,T]$. Thus Theorem 2.5 is proved.

##  6 Large time behavior of the free boundary

In this section, we discuss the large time behavior of a solution to
(P)$(u_{0},s_{0},b)$ as $t\to\infty$. First, we assume (A2)’ replaced by (A2):
(A2)’: $b\in W^{1,2}_{loc}([0,\infty))$, $b_{t}\in L^{1}(0,\infty)$,
lim${}_{t\to\infty}b(t)=b_{\infty}$, $b-b_{\infty}\in L^{1}(0,\infty)$ and
$b_{*}\leq b\leq b^{*}$ on $(0,\infty)$, where $b_{*}$ and $b^{*}$ are
positive constants as in (A2).
Clearly, we see that $b_{*}\leq b_{\infty}\leq b^{*}$. Next, we consider the
following stationary problem (P)∞: find a pair $(u_{\infty},s_{\infty})\in
L^{2}(0,s_{\infty})\times\mathbb{R}$ satisfying

$\begin{cases}-u_{\infty zz}=0\mbox{ on }(0,s_{\infty}),\\\ -u_{\infty
z}(0)=\beta(b_{\infty}-\gamma u_{\infty}(0)),\ \ -u_{\infty
z}(s_{\infty})=0,\\\ u_{\infty}(s_{\infty})=0.\end{cases}$

By using the change of variables
$\tilde{u}_{\infty}(y)=u_{\infty}(ys_{\infty})$ for $y\in(0,1)$, (P)∞ can be
written in the following problem $(\tilde{\mbox{P}})_{\infty}$:

$\begin{cases}-\displaystyle{\frac{1}{s^{2}_{\infty}}}\tilde{u}_{\infty
yy}=0\mbox{ on }(0,1),\\\\[8.53581pt]
-\displaystyle{\frac{1}{s_{\infty}}}\tilde{u}_{\infty
y}(0)=\beta(b_{\infty}-\gamma\tilde{u}_{\infty}(0)),\ \
-\displaystyle{\frac{1}{s_{\infty}}}\tilde{u}_{\infty y}(1)=0,\\\\[5.69054pt]
\tilde{u}_{\infty}(1)=0.\end{cases}$

The next lemma is concerned with non-existence of a solution
$(s_{\infty},\tilde{u}_{\infty})$ of the problem
$(\tilde{\mbox{P}})_{\infty}$.

######  Lemma 6.1.

A solution $(s_{\infty},\tilde{u}_{\infty})$ of $(\tilde{\mbox{P}})_{\infty}$
satisfying $0<s_{\infty}<+\infty$ and $\tilde{u}_{\infty}\in H^{2}(0,1)$ does
not exist.

###### Proof.

Let $(s_{\infty},\tilde{u}_{\infty})$ be a solution of
$(\tilde{\mbox{P}})_{\infty}$ such that $0<s_{\infty}<+\infty$ and
$\tilde{u}_{\infty}\in H^{2}(0,1)$. Then, it holds that

$\displaystyle-\frac{1}{s_{\infty}}\tilde{u}_{\infty
y}(1)+\frac{1}{s_{\infty}}\tilde{u}_{\infty y}(0)=0.$

Then, we see that $\tilde{u}_{\infty}(0)=b_{\infty}/\gamma$. Hence,
$\tilde{u}_{\infty}\in H^{2}(0,1)$ satisfies $-\tilde{u}_{\infty yy}=0$ on
$(0,1)$ with $\tilde{u}_{\infty y}(1)=\tilde{u}_{\infty y}(0)=0$ and
$\tilde{u}_{\infty}(1)=0$ so that $\tilde{u}_{\infty}\equiv 0$ on $[0,1]$.
This is a contradiction to $\tilde{u}_{\infty}(0)\neq 0$. Thus, we conclude
that Lemma 6.1 holds. ∎

Now, we state the result on the large time behavior of a solution as
$t\to\infty$.

######  Theorem 6.2.

Assume (A1), (A2)’ and (A3) and let (P)$(u_{0},s_{0},b)$ be a solution $(s,u)$
on $[0,\infty)$. Then, $s\to\infty$ as $t\to\infty$.

We prove this result in the rest of the section.

###  6.1 Global estimates

To prove Theorem 6.2, we provide some uniform estimates for the solution with
respect to time $t$. We assume (A1), (A2)’ and (A3). Then, by Theorem 2.5,
(P)$(u_{0},s_{0},b)$ has a solution $(s,u)$ on $[0,T]$ for $T>0$ satisfying
$0\leq u\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$.

######  Lemma 6.3.

Let $(s,u)$ be a solution of (P)$(u_{0},s_{0},b)$ on $[0,\infty)$. If there
exists a constant $C>0$ such that $s(t)\leq C$ for $t>0$, then it holds

$\displaystyle(i)$
$\displaystyle\int_{0}^{t}|s_{t}(\tau)|^{2}d\tau+\int_{0}^{t}|u_{z}(\tau)|^{2}_{L^{2}(0,s(\tau))}d\tau+\int_{0}^{t}\left|u(\tau,0)-\frac{b_{\infty}}{\gamma}\right|^{2}d\tau\leq
C_{1}\mbox{ for }t>0,$ (6.1) $\displaystyle(ii)$
$\displaystyle\int_{0}^{t}|u_{t}(\tau)|^{2}_{L^{2}(0,s(\tau))}d\tau+|u_{z}(t)|^{2}_{L^{2}(0,s(t))}\leq
C_{2}\mbox{ for }t>0,$ (6.2)

where $C_{1}$ and $C_{2}$ are positive constants which is independent of time
$t$.

###### Proof.

First, we prove that (6.1) holds. By (1.1) we have that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}\left|u(t)-\frac{b_{\infty}}{\gamma}\right|^{2}dz-\frac{s_{t}(t)}{2}\left|u(t,s(t))-\frac{b_{\infty}}{\gamma}\right|^{2}-\int_{0}^{s(t)}u_{zz}(t)\left(u(t)-\frac{b_{\infty}}{\gamma}\right)dz=0.$
(6.3)

For the third term of the left-hand side of (6.3), it holds that

$\displaystyle-\int_{0}^{s(t)}u_{zz}(t)\left(u(t)-\frac{b_{\infty}}{\gamma}\right)$
$\displaystyle=$ $\displaystyle-
u_{z}(t,s(t))\left(u(t,s(t))-\frac{b_{\infty}}{\gamma}\right)+u_{z}(t,0)\left(u(t,0)-\frac{b_{\infty}}{\gamma}\right)+\int_{0}^{s(t)}|u_{z}(t)|^{2}dz$
$\displaystyle=$ $\displaystyle
s_{t}(t)\left|u(t,s(t))-\frac{b_{\infty}}{\gamma}\right|^{2}+\frac{b_{\infty}}{\gamma}s_{t}(t)\left(u(t,s(t))-\frac{b_{\infty}}{\gamma}\right)$
$\displaystyle-\beta(b(t)-\gamma
u(t,0))\left(u(0)-\frac{b_{\infty}}{\gamma}\right)+\int_{0}^{s(t)}|u_{z}(t)|^{2}dz.$
(6.4)

By (6.3) with (6.1) it follows that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}\left|u(t)-\frac{b_{\infty}}{\gamma}\right|^{2}dz+\frac{s_{t}(t)}{2}\left|u(t,s(t))-\frac{b_{\infty}}{\gamma}\right|^{2}+\int_{0}^{s(t)}|u_{z}(t)|^{2}dz$
$\displaystyle+\frac{b_{\infty}}{\gamma}s_{t}(t)\left(u(t,s(t))-\frac{b_{\infty}}{\gamma}\right)-\beta(b(t)-\gamma
u(t,0))\left(u(0)-\frac{b_{\infty}}{\gamma}\right)=0.$ (6.5)

Since $s_{t}(t)=a_{0}\sigma(u(t,s(t))=a_{0}u(t,s(t))$, we have that

$\displaystyle\frac{b_{\infty}}{\gamma}s_{t}(t)\left(u(t,s(t))-\frac{b_{\infty}}{\gamma}\right)=\frac{b_{\infty}}{\gamma}\frac{|s_{t}(t)|^{2}}{a_{0}}-\left(\frac{b_{\infty}}{\gamma}\right)^{2}s_{t}(t).$
(6.6)

Also, it holds that

$\displaystyle-\beta(b(t)-\gamma
u(t,0))\left(u(0)-\frac{b_{\infty}}{\gamma}\right)$ $\displaystyle=$
$\displaystyle\beta(\gamma
u(t,0)-b_{\infty}+b_{\infty}-b(t))\left(u(t,0)-\frac{b_{\infty}}{\gamma}\right)$
$\displaystyle=$
$\displaystyle\beta\gamma\left|u(t,0)-\frac{b_{\infty}}{\gamma}\right|^{2}+\beta(b_{\infty}-b(t))\left(u(t,0)-\frac{b_{\infty}}{\gamma}\right).$
(6.7)

Combining with (6.1)-(6.1), we obtain that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}\left|u(t)-\frac{b_{\infty}}{\gamma}\right|^{2}dz+\frac{s_{t}(t)}{2}\left|u(t,s(t))-\frac{b_{\infty}}{\gamma}\right|^{2}+\int_{0}^{s(t)}|u_{z}(t)|^{2}dz$
$\displaystyle+\frac{b_{\infty}}{\gamma}\frac{|s_{t}(t)|^{2}}{a_{0}}+\beta\gamma\left|u(t,0)-\frac{b_{\infty}}{\gamma}\right|^{2}$
$\displaystyle=$
$\displaystyle\left(\frac{b_{\infty}}{\gamma}\right)^{2}s_{t}(t)+\beta(b(t)-b_{\infty})\left(u(t,0)-\frac{b_{\infty}}{\gamma}\right).$
(6.8)

Here, by the fact that $u(t)\geq 0$ on $[0,s(t)]$ for $t\in[0,T]$ we note that
$s_{t}(t)\geq 0$ for $t\in[0,T]$ and the second term of the left-hand side of
(6.1) is non-negative. Hence, we derive that

$\displaystyle\frac{1}{2}\frac{d}{dt}\int_{0}^{s(t)}\left|u(t)-\frac{b_{\infty}}{\gamma}\right|^{2}dz+\int_{0}^{s(t)}|u_{z}(t)|^{2}dz+\frac{b_{\infty}}{\gamma}\frac{|s_{t}(t)|^{2}}{a_{0}}+\beta\gamma\left|u(t,0)-\frac{b_{\infty}}{\gamma}\right|^{2}$
$\displaystyle\leq$
$\displaystyle\left(\frac{b_{\infty}}{\gamma}\right)^{2}s_{t}(t)+\beta(b(t)-b_{\infty})\left(u(t,0)-\frac{b_{\infty}}{\gamma}\right).$
(6.9)

By using $u(t)\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t\in[0,T]$ and integrating
over $[0,t_{1}]$ for $t_{1}\in[0,T]$ we obtain that

$\displaystyle\frac{1}{2}\int_{0}^{s(t_{1})}\left|u(t_{1})-\frac{b_{\infty}}{\gamma}\right|^{2}dz+\int_{0}^{t_{1}}\int_{0}^{s(t)}|u_{z}(t)|^{2}dzdt$
$\displaystyle+\frac{b_{\infty}}{a_{0}\gamma}\int_{0}^{t_{1}}|s_{t}(t)|^{2}dt+\beta\gamma\int_{0}^{t_{1}}\left|u(t,0)-\frac{b_{\infty}}{\gamma}\right|^{2}dt$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{s_{0}}\left|u_{0}-\frac{b_{\infty}}{\gamma}\right|^{2}dz+\left(\frac{b_{\infty}}{\gamma}\right)^{2}(s(t_{1})-s(0))+\frac{\beta(b^{*}+b_{\infty})}{\gamma}\int_{0}^{t_{1}}|b_{\infty}-b(t)|dt.$
(6.10)

Hence, from $b-b_{\infty}\in L^{1}(0,\infty)$ in (A2)’ and the assumption that
$s(t)\leq C$ for $t\in[0,T]$, we conclude that (6.1) holds.

Also, for the estimate (6.2), by repeating the proof of Lemma 5.1 we infer
that it holds that

$\displaystyle\int_{0}^{t_{1}}\int_{0}^{s(t)}|u_{t}(t)|^{2}+\frac{1}{2}\int_{0}^{s(t_{1})}|u_{z}(t_{1})|^{2}dz$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\int_{0}^{s_{0}}|u_{z}(0)|^{2}dz+|\tilde{u}^{3}_{0}(1)|+|\Phi(b(t_{1}),\tilde{u}(t_{1},0))|+\frac{\beta
b^{*}}{\gamma}\int_{0}^{t_{1}}|b_{t}(t)|dt\mbox{ for }t_{1}\in[0,T].$ (6.11)

Therefore, by (5.19) and $b_{t}\in L^{1}(0,\infty)$ we can find a positive
constant $C_{2}$ which is independent of $t$ such that (6.2) holds. This
completes the proof of this lemma. ∎

###  6.2 Proof of Theorem 6.2

At the end of the paper, by using the uniform estimate obtained in previous
subsection, we complete the proof of Theorem 6.2 concerning the large-time
behavior of solutions to (P)$(u_{0},s_{0},b)$.

Let us assume (A1), (A2)’ and (A3). Then, by Theorem 2.5, we have a solution
$(s,u)$ of (P)$(u_{0},s_{0},b)$ on $[0,T]$ for any $T>0$ such that $0\leq
u\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t>0$.

Now, we show Theorem 6.2 by contradiction. Let us assume that there exists a
constant $C>0$ such that $s(t)\leq C$ for $t\in[0,T]$. Then, by Lemma 6.3, we
have that

$\displaystyle\int_{0}^{t}|s_{t}(\tau)|^{2}d\tau+\int_{0}^{t}|u_{z}(\tau)|^{2}_{L^{2}(0,s(\tau))}d\tau+\int_{0}^{t}\left|u(\tau,0)-\frac{b_{\infty}}{\gamma}\right|^{2}d\tau\leq
C_{1}\mbox{ for }t>0,$ (6.12)
$\displaystyle\int_{0}^{t}|u_{t}(\tau)|^{2}_{L^{2}(0,s(\tau))}d\tau+|u_{z}(t)|^{2}_{L^{2}(0,s(t))}\leq
C_{2}\mbox{ for }t>0.$ (6.13)

Here, by (1.4), we see that $s_{t}(t)\geq 0$, and hence $s(t)\geq s_{0}$ for
$t>0$. Also, $u\leq b^{*}/\gamma$ on $[0,s(t)]$ for $t>0$ so that it holds
that $|s_{t}(t)|\leq a_{0}b^{*}/\gamma$ for $t>0$. From these results and the
change of variables (2.1) we obtain that

$\displaystyle\int_{0}^{t}|\tilde{u}_{t}(\tau)|^{2}_{L^{2}(0,1)}d\tau$
$\displaystyle=$
$\displaystyle\int_{0}^{t}\int_{0}^{s(\tau)}\frac{1}{s(\tau)}|u_{t}(\tau,z)+u_{z}(\tau,z)\frac{z}{s(\tau)}s_{t}(\tau)|^{2}dzd\tau$
$\displaystyle\leq$
$\displaystyle\int_{0}^{t}\int_{0}^{s(\tau)}\frac{1}{s(\tau)}|u_{t}(t,z)|^{2}dzd\tau+\int_{0}^{t}\int_{0}^{s(\tau)}\frac{2}{s(\tau)}|u_{t}(\tau,z)||u_{z}(\tau,z)||s_{t}(\tau)|dzd\tau$
$\displaystyle+\int_{0}^{t}\int_{0}^{s(\tau)}\frac{1}{s(\tau)}|u_{z}(\tau,z)|^{2}|s_{t}(\tau)|^{2}dzd\tau$
$\displaystyle\leq$
$\displaystyle\frac{1}{s_{0}}\biggl{(}C_{2}+\frac{2a_{0}b^{*}}{\gamma}C^{1/2}_{1}C^{1/2}_{2}+\left(\frac{a_{0}b^{*}}{\gamma}\right)^{2}C_{1}\biggr{)}\mbox{
for }t>0,$ (6.14)

and

$\displaystyle|\tilde{u}_{y}(t)|^{2}_{L^{2}(0,1)}=\int_{0}^{s(t)}\frac{1}{s(t)}|u_{z}(t)s(t)|^{2}dzdt\leq
CC_{2}\mbox{ for }t>0,$ (6.15)

where $C_{1}$ and $C_{2}$ are positive constants as in (6.12) and (6.13).

Here, for $\\{t_{n}\\}$ such that $t_{n}\to\infty$ as $n\to\infty$, we put
$\tilde{u}_{n}(t,y):=\tilde{u}(t+t_{n},y)$ for $(t,y)\in[0,1]\times[0,1]$,
$s_{n}(t):=s(t+t_{n})$, $b_{n}(t):=b(t+t_{n})$ for $t\in[0,1]$. By (A2)’, it
is clear that $b_{n}\to b_{\infty}$ in $L^{1}(0,1)$ as $n\to\infty$. Also, by
(6.12), (6.2) and (6.15) we see that $\\{s_{nt}\\}$ is bounded in $L^{2}(0,1)$
and $\\{\tilde{u}_{n}\\}$ is bounded in $W^{1,2}(0,1;L^{2}(0,1))\cap
L^{\infty}(0,1;H^{1}(0,1))$. Therefore, we can take a subsequence
$\\{n_{j}\\}\subset\\{n\\}$ such that the following convergences holds for
some $\tilde{u}_{\infty}\in H^{1}(0,1)$ and $s_{\infty}\in\mathbb{R}$
satisfying $s_{0}\leq s_{\infty}<+\infty$:

$\begin{cases}\tilde{u}_{nj}(0)=\tilde{u}(t_{nj})\to\tilde{u}_{\infty}\mbox{
in }C([0,1]),\mbox{ weakly in }H^{1}(0,1),\\\ s_{nj}(0)=s(t_{nj})\to
s_{\infty}\mbox{ in }\mathbb{R},\\\ \tilde{u}_{njt}\to 0\mbox{ in
}L^{2}(0,1;L^{2}(0,1))\mbox{ and }s_{njt}\to 0\mbox{ in }L^{2}(0,1),\\\
\tilde{u}_{nj}\to\tilde{u}_{\infty}\mbox{ in }C([0,1];L^{2}(0,1)),\\\ \hskip
51.21504pt\mbox{ weakly in }W^{1,2}(0,1;L^{2}(0,1)),\\\ \hskip
51.21504pt\mbox{ weakly -* in }L^{\infty}(0,1;H^{1}(0,1)),\\\ s_{nj}\to
s_{\infty}\mbox{ in }C([0,1]),\mbox{ weakly in }W^{1,2}(0,1)\end{cases}$

as $j\to\infty$. Also, by Sobolev’s embedding theorem in one dimension (3.17),
it holds that for $y\in[0,1]$,

$\displaystyle\int_{0}^{1}|\tilde{u}_{nj}(t,y)-\tilde{u}_{\infty}(y)|^{2}dt\leq
C_{e}\int_{0}^{1}|\tilde{u}_{nj}(t)-\tilde{u}_{\infty}|_{H^{1}(0,1)}|\tilde{u}_{nj}(t)-\tilde{u}_{\infty}|_{L^{2}(0,1)}dt.$
(6.16)

Hence, by the strong convergence of $\tilde{u}_{nj}$ and (6.16) we see that

$\displaystyle\tilde{u}_{nj}(y)\to\tilde{u}_{\infty}(y)\mbox{ in
}L^{2}(0,1)\mbox{ at }y=0,1\mbox{ as }j\to\infty.$ (6.17)

Now, for each $j$, $(s_{nj},\tilde{u}_{nj})$ satisfies

$\begin{cases}\tilde{u}_{njt}(t,y)-\frac{1}{s^{2}_{nj}(t)}\tilde{u}_{njyy}(t,y)=\frac{ys_{njt}(t)}{s_{nj}(t)}\tilde{u}_{njy}(t,y)\mbox{
for }(t,y)\in Q(1),\\\
-\frac{1}{s_{nj}(t)}\tilde{u}_{njy}(t,0)=\beta(b_{nj}(t)-\gamma\tilde{u}_{nj}(t,0))\mbox{
for }t\in(0,1),\\\
-\frac{1}{s_{nj}(t)}\tilde{u}_{njy}(t,1)=\tilde{u}_{nj}(t,1)s_{njt}(t)\mbox{
for }t\in(0,1),\\\ s_{njt}(t)=a_{0}\tilde{u}_{nj}(t,1)\mbox{ for
}t\in(0,1).\end{cases}$

By letting $j\to\infty$ in the above system and using the strong convergences
of $\tilde{u}_{nj}$ and $s_{nj}$, we see that $\tilde{u}_{\infty}\in
H^{2}(0,1)$ and

$\displaystyle-\displaystyle{\frac{1}{s^{2}_{\infty}}}\tilde{u}_{\infty
yy}=0\mbox{ on }(0,1).$ (6.18)

Hence, by using the above convergences of $\tilde{u}_{nj}$ and $s_{nj}$,
(6.17) and (6.18) we infer that $(s_{\infty},\tilde{u}_{\infty})$ satisfies

$\displaystyle-\displaystyle{\frac{1}{s_{\infty}}}\tilde{u}_{\infty
y}(0)=\beta(b_{\infty}-\gamma\tilde{u}_{\infty}(0)),\quad-\displaystyle{\frac{1}{s_{\infty}}}\tilde{u}_{\infty
y}(1)=0,\quad\tilde{u}_{\infty}(1)=0.$

Therefore, we see that $(s_{\infty},\tilde{u}_{\infty})$ is a solution of
$(\tilde{\mbox{P}})_{\infty}$ such that $\tilde{u}_{\infty}\in H^{2}(0,1)$ and
$s_{0}\leq s_{\infty}<+\infty$. This contradicts that
$(\tilde{\mbox{P}})_{\infty}$ does not have a solution (see Lemma 6.1). Thus,
we conclude that $s$ goes to $\infty$ as $t\to\infty$ and Theorem 6.2 holds.

##  7 Numerical illustration

In this section, we use our free boundary model to approximate numerically the
diffusion of a population of solvent molecules (cyclohexane) into a piece of
material made of ethylene propylene diene monomer rubber (EPDM). The actual
migration experiment and the set of basic parameters are reported in [17].

In this framework, we take the effective diffusivity with an order of
magnitude higher and explore briefly of the depth of the penetration front
depending on variations in the kinetic parameter $a_{0}$ arising in (1.4). In
fact, we look only at a particular instance of the large-time behavior of our
problem and point out that, depending on the choice of model parameters, the
free boundary position $s(t)$ behaves like a power law of type $t^{\beta}$,
where $\beta$ is typically different than $\frac{1}{2}$ or $1$ as expected for
the classical diffusion and for the Case II diffusion, respectively; see [12]
for a detailed discussion based on first principles on the large time behavior
of sharp diffusion fronts in the transition from glassy to rubbery polymers.

(a) Comparison to experimental data.

(b) Penetration fronts for various values $a_{0}$.

Figure 1: Approximation of the large-time behavior of solutions to
$(P)(u_{0},s_{0},b)$ on $[0,T]$ with $T=5000$ minutes.

As shown in Figure 1 (a), the behavior of our free boundary seems to be
different from the real experimental result. From a phenomenological point of
view, a more realistic behavior of the free boundary is obtained in [17]. On
the other hand, Theorem 6.2 guarantees that the growth observed in numerical
results is correct, theoretically. Moreover, in order to measure the growth
rate for the free boundary, we show numerical results for varying positive
constants $a_{0}$ in Figure 1 (b). From these results we conjecture that the
free boundary position corresponding to Figure 1 (a) behaves like $t^{0.41}$.
This is a sub-diffusive regime. However, other parameters can bring the front
in a super-diffusive regime. Based on our current simulation and mathematical
analysis results, we can only state that we expect the free boundary position
to follow a power law for large times, but we are, for the moment, unable to
establish rigorously quantitative upper and lower bounds on $s(t)$.
Nevertheless, relying also on results from [10], we hope to be able to adapt
some parts of our working technique developed in [3] to handle this case. The
main difficulty lies on the fact that it seems that, for a large region in the
parameter spaces, our sharp diffusion fronts tend to deviate from
$t^{\frac{1}{2}}$. This makes us wonder what is the most relevant exponent
$\beta$ and also for which parameter case and type(s) of rubber-like materials
this corresponds.

##  8 Discussion

We were able to prove the global solvability for a one-phase free boundary
problem with nonlinear kinetic condition that is meant to describe the
migration of diffusants into rubber. Despite of its apparently simple one-
dimensional structure, our free boundary model brings in a number of open
questions. The most important ones include the identification of an asymptotic
dependence of type $s(t)\sim\mathcal{O}(t^{\beta})$ as $t\to+\infty$ and its
rigorous mathematical justification. Also, capturing numerically the large
time behavior so that a certain power law is preserved requires a special
care; compare e.g. the ideas from [6, 19] to be adapted for the finite element
method used here; see [17] for a detailed description of the numerical scheme
used in this context. Of course, to bring the one-dimensional model equations
to describe better the physical scenario of diffusants migrating into rubbers,
more modeling components must be added, viz. macroscopic swelling, capillarity
transport. The case of more space dimensions is out of reach as it is not at
all clear how the kinetic condition on the moving sharp diffusion front should
be formulated especially close to corners or other singularities of the
geometry.

## Acknowledgments

T. A. and A. M. thank the KK Foundation for financial support (project nr.
2019-0213). The work of T. A. is partially supported also by JSPS KAKENHI
Grant Number JP19K03572, while the one by K.K. is partially supported by JSPS
KAKENHI Grant Numbers JP16K17636, JP19K03572 and JP20K03704. Fruitful
discussions with U. Giese, N. Kröger, R. Meyer (Deutsches Institut für
Kautschuktechnologie, Hannover, Germany) and S. Nepal, Y. Wondmagegne
(Karlstad, Sweden) concerning the potential applicability of this research in
the case of diffusants migration into polymers have greatly influenced our
work.

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