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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 254, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/254\hfil Existence and uniqueness of solutions]
{Existence and uniqueness of solutions for miscible liquids model
 in porous media}

\author[K. Allali \hfil EJDE-2013/254\hfilneg]
{Karam Allali} 

\address{Karam Allali \newline
Department of Mathematics,
Faculty of Sciences and Technologies,
University Hassan II, PO Box 146, Mohammedia,
Morocco}
\email{allali@fstm.ac.ma}

\thanks{Submitted September 30, 2013. Published November 20, 2013.}
\subjclass[2000]{35A01, 35A02, 76D03, 76S05}
\keywords{Darcy approximation; Korteweg stress; miscible liquids; porous media}

\begin{abstract}
 In this article, we study the existence and uniqueness of
 solutions for  miscible liquids model in porous media.
 The model describing the phenomenon is a system of equations
 coupling hydrodynamic equations with concentration equation
 taking into account the Korteweg stress. We  assume that the
 fluid is incompressible and its motion is described by the
 Darcy law. We prove the existence and uniqueness of  global
 solutions for the initial boundary value problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\allowdisplaybreaks

\section{Introduction}

Two liquids are miscible if the molecules of the one liquid can
mix freely with the molecules of the other liquid. There is no
sharp interface between miscible liquids, but rather a transition
zone. An example of such phenomenon is the mixing of water and
glycerin \cite{Jos,PetMax}. It is possible that two liquids are
not completely miscible. They may mix until the concentration
reaches a certain saturation value. This saturation may be
affected by the temperature and pressure of the system \cite{Abr}.

There exists a transient capillary phenomena between two miscible
liquids. Theoretical and experimental studies of such phenomena
are reviewed in \cite{Jos}. The change of concentration gradients
near the transition zone causes capillary forces between the two
miscible liquids \cite{Kor}. So we need to take into account some
additional terms in the equation of motion due to the
concentration inhomogeneities called Korteweg stress
\cite{Bes1,Che}.

The existence and uniqueness of the solutions of miscible liquids
model with fully incompressible Navier-Stokes equations are
studied in \cite{Kos}. In this paper, we are interested in
studying porous media. The study of miscible liquids in porous
media is motivated by enhanced oil recovery, hydrology, frontal
polymerization, groundwater pollution and filtration
\cite{Bes,Hut,Off,Sch,Sch1}.

The paper is organized as follows. The next section introduces the
model, while section 3 deals with the existence of the model
solutions and we establish the uniqueness of solutions in section 4.


\subsection{The model}

The model describing the interaction between two miscible liquids
in porous media is given by the following system of equations in
the bounded open domain $\Omega \in \mathbb{R}^2$ with Lipschitz
continuous boundary \cite{All, Nie}:
\begin{gather}\label{prob1}
\frac{\partial C}{\partial t} + u\cdot \nabla C = d \Delta C,\\
\label{u}
\frac{\partial u}{\partial t} + \frac{\mu}{K}u = - \nabla p +
\nabla \cdot T(C), \\
\operatorname{div}(u) = 0,
\end{gather}
with the  boundary conditions
\begin{equation}
\frac{\partial C}{\partial n} =0, \quad u\cdot n=0, \quad\text{on } \Gamma,
\end{equation}
and the initial conditions
\begin{equation}\label{prob2}
C(x,0) = C_0(x),\quad  u(x,0) = u_0(x),\quad  x \in \Omega.
\end{equation}
 Here $u$ is the velocity, $p$ is the pressure, $C$ is
the concentration, $d$ is the coefficient of mass diffusion, $\mu$
is the viscosity, $K$ is the permeability of the medium, $\Gamma$
is the boundary of $\Omega$, $n$ is the unit outward normal vector
to $\Gamma$, the additional stress tensor term is
\begin{equation}
T_{11} = k \big(\frac{\partial C}{\partial x_2}\big)^2, \quad
T_{12} = T_{21} = -k \frac{\partial C}{\partial x_1}
 \frac{\partial C}{\partial x_2}, \quad
T_{22} = k\big(\frac{\partial C}{\partial x_1}\big)^2,
\end{equation}
where $k$ is a nonnegative constant. The gradient, divergence and
Laplace operators can be defined as follows
\[
  \nabla  v =\big(\frac{\partial v}{\partial
x_1},\frac{\partial v}{\partial x_2}\big), \quad
\operatorname{div}
\overrightarrow{\mathbf{v}} = \sum_{i=1}^{2} \frac{\partial
\mathbf{v}_i}{\partial x_i}, \quad
\Delta  v = \sum_{i=1}^{2}\frac{\partial^2 v}{\partial^2 x_i},
\]
the divergence of the additional tensor term will be in the
form
\begin{equation}
 \nabla \cdot T(C) = \begin{pmatrix}
 \frac{\partial T_{11}}{\partial x_1}+\frac{\partial
T_{12}}{\partial x_2}\\ \\  \frac{\partial
T_{21}}{\partial x_1}+\frac{\partial T_{22}}{\partial x_2}
\end{pmatrix}.
\end{equation}


\subsection{The problem in variational form}

We specify now the functional framework in which we  carry out
our analysis of the problem.
The velocity space $S_u$ is
\[
S_u=\{u\in H(\operatorname{div};\Omega); \operatorname{div}(u)=0,
u\cdot n=0\text{ on } \Gamma\}.
\]
The concentration space $S_C$ is
\[
 S_C=\{C\in H^2(\Omega); \frac{\partial C}{\partial n} =0\text{ on } \Gamma\}.
\]
By the Green formula, the variational form of problem
\eqref{prob1}-\eqref{prob2} is:
For each $B$ and $v$, find $C$ and $u$ such that
\begin{gather}\label{1}
\big(\frac{\partial C}{\partial t},B\big) + d (\nabla C,\nabla B) +
(u\cdot \nabla C,B)=0, \\
\label{2}
(\frac{\partial u}{\partial t},v)+\mu_p(u,v)-(div\ T(C),v)=0.
\end{gather}
Here $ \mu_p = \mu/K$. We will assume that $d> 0$ and $\mu_p > 0$.


\section{Existence of global solutions}

To prove the existence of global solutions, we  need the
following lemmas.

\begin{lemma} \label{lem3.1}
The concentration $C$ is bounded in the $L^{\infty}(0,t;L^2)$
space.
\end{lemma}

\begin{proof}
Choosing $C$ as test function in \eqref{1}, we will have:
\[
\frac{1}{2} \frac{\partial}{\partial t} (C,C) + d (\nabla C,
\nabla C) + (u\cdot \nabla C,C) = 0.
\]
Since $u\in S_u$ the last term vanishes. The second term is
positive, so by integrating over time:
\[
\|C(t=s)\|_{L^2}^2 \le \|C_0\|_{L^2}^2,
\]
from the inequality, it follows that $C$ is bounded in
$L^{\infty}(0,t;L^2)$.
\end{proof}

\begin{lemma} \label{lem3.2}
The concentration $C$ is bounded in $L^{\infty}(0,t;H^1)$ and the
velocity $u$ is bounded in $L^{\infty}(0,t;L^2)$.
\end{lemma}

\begin{proof}
By choosing  $-k \Delta C$ as test function in \eqref{1}, we have
\[
(\frac{\partial C}{\partial t},-k \Delta C) + (u\cdot \nabla C,-k
\Delta C) = d(\Delta C,-k \Delta C);
\]
therefore,
\[
\frac{k}{2}\frac{\partial}{\partial t}(\nabla C,\nabla C) +
dk(\Delta C,\Delta C) -k(u\cdot \nabla C,\Delta C) = 0,
\]
then
\begin{equation}\label{3}
\frac{1}{2}\frac{\partial}{\partial t}(\nabla C,\nabla C) +
d(\Delta C,\Delta C) -(u\cdot \nabla C,\Delta C) = 0.
\end{equation}
Also, by choosing  $u$ as test function in \eqref{2},
\begin{equation}\label{4}
\frac{1}{2} \frac{\partial}{\partial t} (u,u) + \mu_p (u,u)  -
(\nabla\cdot T(C), u)=0.
\end{equation}
To have an explicit expression of $\nabla\cdot T(C)$, we
calculate its first component:
\begin{equation}\label{star}
 \frac{\partial T_{11}}{\partial x_1}+\frac{\partial
T_{12}}{\partial x_2}=2k\frac{\partial C}{\partial
x_2}\frac{\partial^2 C}{\partial x_1\partial
x_2}-k\frac{\partial^2 C}{\partial x_1\partial x_2}\frac{\partial
C}{\partial x_2}-k\frac{\partial C}{\partial x_1}\frac{\partial^2
C}{\partial x_2^2},
\end{equation}
hence
\[
\frac{\partial T_{11}}{\partial x_1}+\frac{\partial
T_{12}}{\partial x_2}=k\frac{\partial C}{\partial
x_1}\frac{\partial^2 C}{\partial x_1\partial x_2}+k\frac{\partial
C}{\partial x_1}\frac{\partial^2 C}{\partial
x_1^2}-k\frac{\partial C}{\partial x_1}\Delta C,
\]
then
\[
\frac{\partial T_{11}}{\partial x_1}+\frac{\partial
T_{12}}{\partial x_2}=\frac{k}{2}\frac{\partial}{\partial
x_1}(\nabla C)^2 -k\frac{\partial C}{\partial x_1}\Delta C.
\]
Following the same steps for the second component, we have
\[
\nabla\cdot T = \frac{k}{2}\nabla(\nabla C)^2 -k\Delta C\nabla C.
\]
Replacing this last equality in \eqref{4} and since $u\in S_u$, we
have
\begin{equation}\label{5}
\frac{1}{2} \frac{\partial}{\partial t} (u,u) + \mu_p (u,u) -
k(\Delta C\nabla C, u)=0.
\end{equation}
Adding \eqref{3} and \eqref{5}, and with the fact $u \in S_u$ and
$C \in S_c$, we have
\[
\frac{1}{2}\frac{\partial}{\partial t} \big((u,u)+(\nabla
C,\nabla C)\big)+\mu_p (u,u)+dk(\Delta C,\Delta C)=0.
\]
Since the second and the third terms are positive, by integrating
over time, we have
\[
k\|C(t=s)\|_{H^1}^2+\|u(t=s)\|_{L^2}^2\le
k\|C(t=0)\|_{H^1}^2+\|u(t=0)\|_{L^2}^2.
\]
We conclude that $C$ is bounded in $L^{\infty}(0,t;H^1)$ and $u$
is bounded in $L^{\infty}(0,t;L^2)$.
\end{proof}

Now, we look for the estimates over the time derivatives of $u$
and $C$. To this end we will need the following Lemmas.

\begin{lemma} \label{lem3.3}
The  derivative 
$\frac{\partial C}{\partial t}$ of the concentration
is bounded in $L^2(0,t;L^2)$.
\end{lemma}

\begin{proof}
From  \eqref{1} and by the triangular inequality, we have
\[
\|\frac{\partial C}{\partial t}\|_{L^2} \le d\|\Delta C\|_{L^2} +
\|u\cdot \nabla C\|_{L^2}.
\]
Using H\"{o}lder inequality, we obtain
\[
\|\frac{\partial C}{\partial t}\|_{L^2} \le d\|\Delta C\|_{L^2} +
\|u\|_{L^4}\|\nabla C\|_{L^4},
\]
and by the Gagliardo-Nirenberg inequality, it follows that there exists
$N>0$ such that
\[
\|\frac{\partial C}{\partial t}\|_{L^2} \le d\|\Delta C\|_{L^2} +
N\|u\|_{L^2}^{1/2}\|\nabla u\|_{L^2}^{1/2}\|\nabla
C\|_{L^2}^{1/2}\|\nabla C\|_{H^1}^{1/2}.
\]
We conclude that $\frac{\partial C}{\partial t}$ is
bounded in $L^2(0,t;L^2)$.
\end{proof}

\begin{lemma} \label{lem3.4}
The time derivative of the velocity $ \frac{\partial
u}{\partial t}$ is bounded in $L^2(0,t;L^2)$.
\end{lemma}

\begin{proof}
To prove this lemma, it is sufficient to remark that $\nabla\cdot T(C)$
is sum of expressions of the form $\lambda D_i(D_j C D_l C)$.
Where $ D_i=\frac{\partial}{\partial x_i}$, $i=1,2$,
and $\lambda$ depends on $i,j$ and $l$ (see for example
\eqref{star}). Using the problem in its variational form and
using the same technics as for the previous lemmas, it follows
that $ \frac{\partial u}{\partial t}$ is bounded in
$L^2(0,t;L^2)$.
\end{proof}

Now, we can give our main result as follows:

\begin{theorem} \label{thm2.5}
The problem \eqref{prob1}-\eqref{prob2} admits a global solution.
\end{theorem}

\begin{proof}
A priori error estimates over concentration and speed (see all
the previous Lemmas) allow us to deduce that our finite
dimensional solution is global in time.
In addition, applying some classical compactness theorems (see for
example \cite{lio,tem}), it follows the existence of our
continuous problem.
\end{proof}


\section{Uniqueness of solutions}

To prove the uniqueness, we  assume that
\eqref{prob1}-\eqref{prob2} has two different solutions
$(C_1,u_1)$ and $(C_2,u_2)$.
From \eqref{prob1}, we have:
\begin{equation}\label{6}
\frac{\partial}{\partial t}(C_1-C_2) - d \Delta (C_1-C_2) +
u_1\nabla C_1 - u_2\nabla C_2 =0,
\end{equation}
and from \eqref{u}, we have also:
\begin{equation}\label{7}
\begin{aligned}
&\frac{\partial}{\partial t}(u_1-u_2) + \mu_p (u_1-u_2)
+ \nabla (p_1-p_2)\\
&= \frac{k}{2} \nabla\left((\nabla
C_1)^2-(\nabla C_2)^2\right)   
-k(\Delta C_1 \nabla C_1 - \Delta C_2 \nabla C_2).
\end{aligned}
\end{equation}
Multiplying \eqref{6} by $-k\Delta(C_1-C_2)$ and integrating, we
obtain
\begin{align*}
&(\frac{\partial}{\partial
t}(C_1-C_2),-k\Delta(C_1-C_2)) + dk(\Delta (C_1-C_2),\Delta
(C_1-C_2)) \\
&+ (u_1\nabla C_1,-k\Delta(C_1-C_2))+(u_2\nabla
C_2,k\Delta(C_1-C_2)) =0.
\end{align*}
Similarly, multiplying \eqref{7} by $u_1-u_2$ and integrating, we
have
\begin{align*}
&(\frac{\partial}{\partial t}(u_1-u_2),u_1-u_2) + \mu_p (u_1-u_2,u_1-u_2)\\
&= \frac{k}{2}(\nabla\left((\nabla C_1)^2-(\nabla
C_2)^2\right),u_1-u_2) -k(\Delta C_1 \nabla C_1 - \Delta C_2
\nabla C_2,u_1-u_2).
\end{align*}
Adding the two last equalities, using the Green formula and the fact
that $u_i \in S_u$, it follows that
\begin{align*}
&\frac{1}{2}\frac{\partial}{\partial t}
(\|u_1-u_2\|_{L^2}^2+k\|\nabla C_1- \nabla C_2\|_{L^2}^2) + \mu_p
\|u_1-u_2\|_{L^2}^2+kd \|\Delta(C_1-C_2)\|_{L^2}^2\\
&= k(u_1\nabla(C_1-C_2),\Delta (C_1-C_2))
 +k((u_1-u_2)\nabla C_2,\Delta(C_1-C_2))\\
&\quad -k(\Delta C_1\nabla(C_1-C_2),u_1-u_2)+k(-\Delta C_1 \nabla C_2+\Delta
C_2\nabla C_2,u_1-u_2);
\end{align*}
therefore,
\begin{equation} \label{8}
\begin{aligned}
&\frac{1}{2}\frac{\partial}{\partial t}
(\|u_1-u_2\|_{L^2}^2+k\|\nabla C_1- \nabla C_2\|_{L^2}^2) + \mu_p
\|u_1-u_2\|_{L^2}^2+kd \|\Delta(C_1-C_2)\|_{L^2}^2\\
&= k(u_1\nabla(C_1-C_2),\Delta (C_1-C_2))-k(\Delta
C_1\nabla(C_1-C_2),u_1-u_2).
\end{aligned}
\end{equation}
Now we look for estimates of the right-hand terms. We put
$C=C_1-C_2$ and $u=u_1-u_2$, by H\"{o}lder inequality, it follows
that
\begin{align*}
|(\Delta C_1\nabla C,u)|
&\le \|\Delta C_1\|_{L^2}\|\nabla C\cdot u\|_{L^2} \\
&\le \|\Delta C_1\|_{L^2}\|\nabla C\|_{L^4}\|u\|_{(L^4)^2}.
\end{align*}
Also, by Gagliardo-Nirenberg inequality, it follows that
\[
|(\Delta C_1\nabla C,u)| \le N_1 \|\Delta C_1\|_{L^2}\|\nabla
C\|_{L^2}^{1/2}\|\Delta C\|_{L^2}^{1/2}\|u\|_{L^2}^{1/2}\|\nabla
u\|_{L^2}^{1/2}.
\]
By applying Young inequality, we obtain
\[
|(\Delta C_1\nabla C,u)| \le \frac{N_1}{4}\|\Delta
C\|_{L^2}^{2}+\frac{3N_1}{4}\|\Delta C_1\|_{L^2}^{4/3}\|\nabla
C\|_{L^2}^{2/3}\|u\|_{L^2}^{2/3}\|\nabla u\|_{L^2}^{2/3}.
\]
Using that same technics, we obtain the following inequalities:
first,
\begin{align*}
|(u_1\nabla C,\Delta C)|
&\le \|\Delta C\|_{L^2}\|\nabla C.u_1\|_{L^2}\\
&\le \|\Delta C\|_{L^2}\|\nabla C\|_{L^4}\|u_1\|_{(L^4)^2};
\end{align*}
therefore,
\[
|(u_1\nabla C,\Delta C)| \le N_2 \|\Delta C\|_{L^2}^{3/2}\|\nabla
C\|_{L^2}^{1/2}\|u_1\|_{L^2}^{1/2}\|\nabla u_1\|_{L^2}^{1/2}.
\]
Finally,
\[
|(u_1\nabla C,\Delta C)| \le \frac{3N_2}{4} \|\Delta
C\|_{L^2}^{2}+\frac{N_2}{4}\|\nabla
C\|_{L^2}^{2}\|u_1\|_{L^2}^{2}\|\nabla u_1\|_{L^2}^{2}.
\]
From \eqref{8} and assuming that $N_1+3N_2\le 4d$, we have
\begin{align*}
&\frac{1}{2}\frac{\partial}{\partial t} (\|u\|_{L^2}^2
 +k\|\nabla C\|_{L^2}^2) \\
&\le \frac{3N_1k}{2}\|\Delta C_1\|_{L^2}^{4/3}\|\nabla C\|_{L^2}^{2/3}
  \|u\|_{L^2}^{2/3}\|\nabla u\|_{L^2}^{2/3}
  +\frac{N_2k}{2}\|\nabla C\|_{L^2}^{2}\|u_1\|_{L^2}^{2}\|\nabla
u_1\|_{L^2}^{2}
\\
&\le (\|u\|_{L^2}^2+k\|\nabla C\|_{L^2}^2)
\Big(\frac{N_2}{2}\|\nabla C\|_{L^2}^{2}\|u_1\|_{L^2}^{2}\|\nabla
u_1\|_{L^2}^{2}\\
&\quad +\frac{3N_1k}{2}\|\Delta C_1\|_{L^2}^{4/3}\|\nabla
C\|_{L^2}^{2/3}\|u\|_{L^2}^{-4/3}\|\nabla u\|_{L^2}^{2/3}\Big).
\end{align*}
If we denote
\begin{gather*}
\phi(t)=\|\nabla C\|_{L^2}^{2}\|u_1\|_{L^2}^{2}\|\nabla u_1\|_{L^2}^{2}
+\|\Delta C_1\|_{L^2}^{4/3}\|\nabla C\|_{L^2}^{2/3}\|u\|_{L^2}^{-4/3}
\|\nabla u\|_{L^2}^{2/3},
\\
M=\max(\frac{N_2}{2},\frac{3N_1k}{2}),
\end{gather*}
we have
\[
\frac{d}{dt}(exp(M\int_0^t \phi(s)ds)(\|u\|_{L^2}^2+k\|\nabla
C\|_{L^2}^2))\le 0 quad \forall t \ge 0\,.
\]
We deduce:
\[
\exp(M\int_0^t \phi(s)ds)(\|u\|_{L^2}^2+k\|\nabla C\|_{L^2}^2)\le
\|u(0)\|_{L^2}^2+k\|\nabla C(0)\|_{L^2}^2.
\]
Since $u(0)=C(0)=0$, we conclude the uniqueness of the solution.
\

Now, we can state our second theorem as follows.

\begin{theorem}
Problem \eqref{prob1}-\eqref{prob2} admits a unique solution.
\end{theorem}

\subsection*{Concluding remarks}

The interaction between two miscible liquids is modelled by a
system of equations, coupling hydrodynamic equations and the
species conservation equation. Due to the interfacial interaction
between the two liquids, we have taken into account two additional
terms, called Korteweg stress tensor. We have chosen the
appropriate functional framework for our variational problem. We
have established a priori estimates over the concentration and
speed which allow us to establish the existence of the solution.
Furthermore, we have also proved the uniqueness of solution.

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\end{document}