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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 82, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/82\hfil Existence and uniqueness]
{Existence and uniqueness of classical solutions to  second-order
quasilinear elliptic equations}

\author[D. Denny\hfil EJDE-2010/82\hfilneg]
{Diane Denny} 

\address{Diane Denny \newline
Department of Mathematics and Statistics\\
Texas A\&M University - Corpus Christi \\
Corpus Christi,   TX 78412, USA}
\email{diane.denny@tamucc.edu}

\thanks{Submitted April 13, 2010. Published June 18, 2010.}
\subjclass[2000]{35A05}
\keywords{Existence; uniqueness; quasilinear; elliptic}

\begin{abstract}
 This article studies the existence of solutions to the
 second-order quasilinear elliptic equation
 $$
 -\nabla \cdot(a(u) \nabla u) +\mathbf{v}\cdot \nabla u=f
 $$
 with the condition $u(\mathbf{x}_0)=u_0$ at a certain point
 in the domain, which is the 2 or the 3 dimensional torus.
 We prove that if the functions $a$, $f$, $\mathbf{v}$ satisfy
 certain conditions, then there exists a unique classical solution.
 Applications of our results include stationary heat/diffusion
 problems with convection and with a source/sink, when
 the value of the solution is known at a certain location.
\end{abstract}

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


\section{Introduction}

In this article, we consider the following quasilinear elliptic
equation  for $u(\mathbf{x})$ under periodic boundary
conditions:
\begin{gather}
-\nabla \cdot(a(u) \nabla u)+\mathbf{v}\cdot \nabla u =f,  \label{1.1} \\
u(\mathbf{x}_0)=u_0,  \label{1.2}
\end{gather}
where $u_0$ is a given constant and $\mathbf{x}_0$ a given point
in the domain  $\Omega$. Here, $f(\mathbf{x})$ and
$\mathbf{v}(\mathbf{x})$ are given smooth functions for
$\mathbf{x}\in \Omega$, where the domain $\Omega =\mathbb{T}^N$,
the $N$-dimensional torus, with $N=2,3$. We assume that $a(u)$ is
a smooth, positive function of $u$ for $u \in \bar{G}$, where
$G\subset \mathbb{R}$ is a bounded interval.

The purpose of this article is to prove the existence of a unique
classical solution $u(\mathbf{x})$ to \eqref{1.1}-\eqref{1.2}.
What is new in this paper is the requirement that condition
\eqref{1.2} holds for a quasilinear elliptic equation of the form
\eqref{1.1} which includes a convection term $\mathbf{v}\cdot
\nabla u$. The proof of the existence theorem uses the method of
successive approximations in which an iteration scheme, based on
solving a linearized version of the equation, will be defined and
then convergence of the sequence of approximating solutions to a
unique solution satisfying the quasilinear equation will be
proven. It will be shown that there exist positive constants
$\delta_0$, $\delta_1$, and $\delta_2$ such that if
$\big| \frac{d a}{d u} \big|_{s,\bar{G}_1}^2 \| f\|_{s-1}^2
\leq \delta_0$, and
$|\nabla \cdot \mathbf{v}|_{L^\infty} \leq \delta_1$, and
$\max\{1,|\mathbf{v}|_{L^\infty}^2\}\| f\|_{s-1}^2 \leq \delta_2$,
and $\| D \mathbf{v}\|_{s} \leq \frac 12$,  where
$s> \frac N2 +1$, and where $G_1\subset G$, then there exists
a classical solution $u(\mathbf{x})$ to \eqref{1.1}-\eqref{1.2}.
Here we
define $ |\frac{da}{du} |_{s,\bar{G}_1}=\max\{\big|\frac{d^{j+1}
a}{d u^{j+1}} (u_{*}) \big|: u_{*} \in \bar{G}_1, 0 \leq j \leq s
\}$. And $u(\mathbf{x}) \in \bar{G_1}$ for all $\mathbf{x}\in
\Omega$. The solution $u(\mathbf{x}) \in \bar{G_1}$ will be unique
if $a''(u_{*})\leq \frac{1}{a(u_{*})}(a'(u_{*}))^2$ for all $u_{*} \in \bar{G_1}$. The key to the proof
lies in obtaining a priori estimates for $u$.

Applications of the existence of a unique solution to \eqref{1.1}-\eqref{1.2} include stationary heat/diffusion problems with
convection and with a source/sink. Solutions could be obtained for
problems in which, for example, the temperature or the
concentration of a substance in a fluid is monitored at a given
spatial location $\mathbf{x}_0 \in \Omega$ .

This article is organized as follows. First, the main result
is presented and proved as Theorem \ref{T3.1} in the next section.
Then lemmas supporting the proof of the theorem are proven in Appendix
A (which proves the existence of a solution to the linearized
equation used in the iteration scheme) and in Appendix B (which
presents proofs of the a priori estimates used in the proof of the
theorem).

\section{Existence theorem}

We  use the Sobolev space $H^s(\Omega )$ (where $s$ is a non-negative
integer) of real-valued functions in $L^2(\Omega )$ whose
distribution derivatives up to order $s$ are in $L^2(\Omega )$,
with norm given by $\|g\|_s^2=\sum_{0 \leq |\alpha |\leq
s}\int_\Omega |D^\alpha g|^2d \mathbf{x}$ and inner product
$(g,h)_s=\sum_{0 \leq |\alpha |\leq s}\int_\Omega (D^\alpha
g)\cdot (D^\alpha h)d \mathbf{x}$. We  use the notation
$\|g\|_s^2=\sum_{0 \leq r \leq s} \int_\Omega |D^r g|^2 d
\mathbf{x}$, where $D^r g$ is the set of  all space derivatives
$D^{\alpha}g$ with $|\alpha|=r$, and $|D^r g|^2=
\sum_{|\alpha|=r}|D^{\alpha}g|^2$, where $r \geq 0$ is an integer.
Also, $C(\Omega )$ is the space of real-valued, continuous
functions with domain $\Omega$. Here, we are using the standard
multi-index notation. Also, we let both $\nabla g$ and $Dg$ denote
the gradient of $g$.

\begin{theorem} \label{T3.1}
Let $f(\mathbf{x}) \in C(\Omega)\cap
H^{s-1}(\Omega)$, $\mathbf{v}(\mathbf{x}) \in C^2(\Omega)\cap
H^{s+1}(\Omega)$, and let $a(u)$ be a smooth, positive function
of $u$ for $u \in \bar{G}$, where $G \subset \mathbb{R}$ is a
bounded interval. We require that the given data $u(\mathbf{x}_0)$
satisfy $u(\mathbf{x}_0)\in G$, where $\mathbf{x}_0\in\Omega$ and
where $\Omega =\mathbb{T}^N$, the $N$-dimensional torus, with
$N=2,3$. There exist positive constants $\delta_0$, $\delta_1$,
and $\delta_2$, and an interval $G_1\subset G$, such that if
$\big| \frac{d a}{d u} \big|_{s,\bar{G}_1}^2 \| f\|_{s-1}^2 \leq
\delta_0$, and $|\nabla \cdot \mathbf{v}|_{L^\infty} \leq
\delta_1$,  and $\max\{1,|\mathbf{v}|_{L^\infty}^2\}\| f\|_{s-1}^2
\leq \delta_2$, and $\| D\mathbf{v}\|_{s} \leq 1/2$, then
there exists a classical solution $u(\mathbf{x})$ to \eqref{1.1}-\eqref{1.2}. And $u(\mathbf{x}) \in \bar{G_1}$ for all
$\mathbf{x}\in \Omega$. Here, we define  $ |\frac{da}{du}
|_{s,\bar{G}_1}=\max\{\big|\frac{d^{j+1} a}{d u^{j+1}} (u_{*})
\big|: u_{*} \in \bar{G}_1, 0 \leq j \leq s \}$, where $s>\frac
N2+1$. The solution $u(\mathbf{x}) \in \bar{G_1}$ will be unique
if $a''(u_{*})\leq \frac{1}{a(u_{*})}(a'(u_{*}))^2$ for all $u_{*} \in \bar{G_1}$. The regularity of the
solution is $u \in C^2(\Omega)\cap H^{s+1}(\Omega)$.
\end{theorem}

\begin{proof}
 We will construct the solution of the problem for \eqref{1.1}-\eqref{1.2} through an iteration scheme. To define the iteration
scheme, we will let the sequence of approximate solutions be
$\{u_k\}_{k=1}^{\infty}$. Set $u_0=u(\mathbf{x}_0)$. For
$k=0,1,2,\dots $, construct $ u_{k+1}$ from the previous iterate
$u _k$ by solving
\begin{gather}
-\nabla \cdot( a(u_k)\nabla u _{k+1})+\mathbf{v}\cdot \nabla u^{k+1}
 = f, \label{3.1} \\
u_{k+1}(\mathbf{x}_0)=u(\mathbf{x}_0), \label{3.2}
\end{gather}
Existence of a sufficiently smooth solution to \eqref{3.1},
\eqref{3.2} for fixed $k$ is proven in Appendix A. The a priori
estimates used in the proof are proven in Appendix B. We proceed
now to prove convergence of the iterates as $k\to \infty $ to a
unique classical solution of \eqref{1.1}-\eqref{1.2}.

We fix an interval $G_1 \subset G$ by defining
$G_1 =\{u_{*} \in G : |u_{*} -u _0|_{L^\infty } < R \}$,
where $R = \mathop{\rm dist}(u_0,\partial{G})$.
We fix a positive constant $c_1$
such that $a(u_{*} )> c_1$ for all $u_{*}\in \bar{G}_1$. Using a
proof by induction on $k$, we assume that $u_{k}(\mathbf{x})\in
\bar{G_1}$ for all $\mathbf{x}\in \Omega$, and then later we will
show that $u_{k+1}(\mathbf{x})\in \bar{G_1}$ for all
$\mathbf{x}\in \Omega$.
\end{proof}

\begin{proposition} \label{P3.1}
Assume that the hypotheses of Theorem
\ref{T3.1} hold. Assume that $|\nabla \cdot \mathbf{v}|_{L^\infty}
\leq \frac{c_1}{C_*}$, where $C_*$ is the constant from
Poincar\'{e}'s inequality $\|\bar{u} \|_0^2\leq C_*\|\nabla u
\|_0^2$, and where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac{1}{|\Omega|}
\int_{\Omega} u(\mathbf{x})  d\mathbf{x}$. There exist constants
$C_4$, $C_5$, $C_1$, $L$ such that if $\big| \frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2 \leq \frac{1}{C_4}$, and if
$\max\{1,|\mathbf{v}|_{L^\infty}^2\}\| f\|_{s-1}^2 \leq \frac
{R^2}{C_5^2}$, and if $\| D\mathbf{v}\|_{s} \leq \frac 12$, where
$s> \frac N2 +1$, then the following hold for $k=1,2,3\dots$
\begin{gather}
\|\nabla u_k\|_{s}^2 \leq 2 C_1\| f\|_{s-1}^2, \label{e3.3}\\
|u_{k} - u_0|_{L^{\infty}} \leq R ,\label{e3.5} \\
\|u_k\|_{s+1}^2 \leq L,  \label{e3.4} \\
\sum_{k=0}^{\infty}| |u _{k+1}-u _k\|_{s+1}^2 < \infty
\label{e3.6}
\end{gather}
Here, $R = \mathop{\rm dist}(u_0,\partial{G})$ and $C_1$ is the constant
in \eqref{B.27} from Lemma \ref{LB.6} in Appendix B .
\end{proposition}

\begin{proof}
The proof is done by induction on $k$. We show only the inductive
step. We will derive estimates for $u _{k+1}$, and then use these
estimates to show that if $u _k$ satisfies the estimates
\eqref{e3.3}, \eqref{e3.5}, \eqref{e3.4} then $u _{k+1}$ also
satisfies the same estimates. We will prescribe $L$ a priori,
independent of $k$ so that \eqref{e3.4} holds for all $k\geq 1$.
We assume by the induction hypothesis that $u_k(\mathbf{x}) \in
\bar{G_1}$, and then we will show that $u_{k+1}(\mathbf{x}) \in
\bar{G_1}$, for all $\mathbf{x}\in \mathbb{T}^N$. In the estimates
below, we use $C$ to denote a generic constant whose value may
change from one relation to the next. Recall that we let both
$\nabla g$ and $Dg$ denote the gradient of $g$.
\smallskip

\noindent\textbf{Estimate for $\|\nabla u _{k+1}\|_s^2$:} We begin
by applying estimate \eqref{B.27} from Lemma \ref{LB.6} in
Appendix B to equation \eqref{3.1}, which yields
\begin{equation}
\|\nabla u _{k+1}\|_{s}^2  \leq C_1
\Big[\sum_{j=0}^{s}(\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\})^{j}\Big] \|f\|_{s-1}^2   \label{e3.7}
\end{equation}
where $s_1=\max\{s-1,s_0\}$, and $s_0=[ \frac N2]+1=2$, and
$s>\frac N2+1$, for $N=2,3$, so $s\geq 3$ and $s_1=s-1$.

We consider two cases: when $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _k))\|_{s_1}^2$, and when
$\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D\mathbf{v}\|_{s_1}$.

\textbf{Case 1}: Suppose that $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _k))\|_{s_1}^2$ in \eqref{e3.7}.

To estimate the term $\|D(a(u _k))\|_{s_1}^2$, we  apply the
Sobolev space inequality \eqref{3.3} from  Lemma \ref{LB.1} in
Appendix B, which yields the following:
\begin{align}
\|D(a(u _k))\|_{s_1}^2&= \sum_{0 \leq r\leq
s_1}\|D^r(D(a(u_k)))\|_0^2
 = \sum_{0 \leq r\leq s_1}\|D^{r+1}(a(u_k))\|_0^2  \nonumber \\
&\leq  \sum_{0 \leq r\leq s_1} \Big [C \big|\frac{d a}{d
u}\big|_{r,\bar{G}_1}^2(1+|u_k|_{L^\infty})^{2r}\|\nabla
u_k\|_{r}^2\Big]
\nonumber  \\
&\leq  C \big|\frac{d a}{d
u}\big|_{s_1,\bar{G}_1}^2(1+|u_k|_{L^\infty})^{2s_1}\|\nabla
u_k\|_{s_1}^2
\nonumber  \\
&\leq  C \big|\frac{d a}{d u}
\big|_{s_1,\bar{G}_1}^2(1+|u_k-u_0|_{L^\infty}+|u_0|_{L^\infty})^{2s_1}\|\nabla
u_k\|_{s_1}^2
\nonumber  \\
&\leq   C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_0)|)^{2s}\|\nabla
u_k\|_{s_1}^2
 \label{e3.8} \\
&\leq   C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+(1+|\Omega|^{1/2})
|u(\mathbf{x}_0)|)^{2s}\|\nabla u_k\|_{s_1}^2
\nonumber  \\
&= C_2 \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2\|\nabla
u_k\|_{s_1}^2  \nonumber \\
&\leq  C_3 \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2\|\nabla
u_k\|_{s_1}^2 \nonumber
\end{align}
where  $ |\frac{da}{du} |_{s,\bar{G}_1}=\max\{\big|\frac{d^{j+1}
a}{d u^{j+1}} (u_{*}) \Big|: u_{*} \in \bar{G}_1, 0 \leq j \leq s
\}$, from  \eqref{3.3} in  Lemma \ref{LB.1}. Here $C_2=
C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$, and we
define $C_3=M C_2$, where $M$ is a constant to be defined later
and $M\geq 1$. We can assume that $C_2 \geq 1$, so that $C_3 \geq
1$. And we used the fact that $\big|\frac{d a}{d u}
\big|_{r,\bar{G}_1} \leq \big|\frac{d a}{d u}
\big|_{s_1,\bar{G}_1} \leq \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}$ for $r\leq s_1$ and $s_1 \leq s$. We also
used the fact that $|u_k-u_0|_{L^\infty}\leq R$ holds by
\eqref{e3.5}, since $u_k(\mathbf{x}) \in \bar{G_1}$ for all
$\mathbf{x}\in \mathbb{T}^N$ by the induction hypothesis.

We now define the constant $C_4$ to be $C_4= 4 C_3^2 C_1^2$, where
$C_1$ is the constant in \eqref{e3.3} and in estimate \eqref{B.27}
from Lemma \ref{LB.6} in Appendix B, and where we may assume that
$C_1 \geq 1$. We assume that $\Big| \frac{d a}{d u}
\big|_{s,\bar{G}_1}^2 \| f\|_{s-1}^2 \leq \frac{1}{C_4}$.
Substituting \eqref{e3.8} into \eqref{e3.7}, and using estimate
\eqref{e3.3}, namely $\|\nabla u_k\|_{s}^2 \leq 2 C_1\|
f\|_{s-1}^2$, which holds by the induction hypothesis for $u_k$,
and using the fact that $s_1 \leq s$, yields
\begin{equation}
\begin{aligned}
\|\nabla u _{k+1}\|_{s}^2
&\leq C_1 \Big[\sum_{j=0}^{s} C_3^j
\big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^{2j}\|\nabla
u_k\|_{s_1}^{2j}\Big] \|f\|_{s-1}^2  \\
&\leq C_1\Big[\sum_{j=0}^{s}C_3^j (2C_1)^{j}
\big|\frac{d a}{d u}\big|_{s,\bar{G}_1}^{2j}\|f\|_{s-1}^{2j}\Big] \|f\|_{s-1}^2 \\
&\leq  C_1\Big[\sum_{j=0}^{s}C_3^j(2C_1)^j
\Big(\frac{1}{C_4} \Big)^j\Big] \|f\|_{s-1}^2 \\
&\leq  C_1\Big[\sum_{j=0}^{s} \big(\frac 12\big)^j\Big] \|f\|_{s-1}^2 \\
&\leq  2 C_1 \|f\|_{s-1}^2
\end{aligned}\label{e3.9}
\end{equation}
where we used the fact that $|\frac{d a}{d u}|_{s,\bar{G}_1}^{2}\|
f\|_{s-1}^2 \leq \frac{1}{C_4}$. And we used the fact that $\frac{
2 C_3 C_1}{C_4} \leq \frac 12$ and $\frac{1}{C_3 C_1}\leq 1$,
since $C_4= 4 C_3^2 C_1^2$ and $C_3 C_1 \geq 1$. Therefore
\eqref{e3.3} holds for $\|\nabla u _{k+1}\|_s^2$ when
$\max\{\|D(a(u _k))\|_{s_1}^2, \|D\mathbf{v}\|_{s_1}\}=\|D(a(u
_k))\|_{s_1}^2$.


\textbf{Case 2}: Suppose that $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D \mathbf{v}\|_{s_1}$ in \eqref{e3.7}.
 From \eqref{e3.7}, we obtain
\begin{equation}
\|\nabla u _{k+1}\|_{s}^2  \leq C_1
\Big[\sum_{j=0}^{s}\|D\mathbf{v}\|_{s_1}^{j}\Big] \|f\|_{s-1}^2
 \leq C_1 \Big[\sum_{j=0}^{s} \big(\frac 12 \big)^{j}\Big]
\|f\|_{s-1}^2  \leq  2 C_1\|f\|_{s-1}^2 \label{e3.10}
\end{equation}
where we used  the fact that $\|D\mathbf{v}\|_{s_1} \leq
\|D\mathbf{v}\|_{s} \leq 1/2$. This is the same result as
inequality \eqref{e3.9}, and therefore \eqref{e3.3} holds for
$\|\nabla u _{k+1}\|_s^2$.
\smallskip

\noindent\textbf{Estimate for $|u _{k+1} -u _{0}|_{L^{\infty}}$:}
To obtain an estimate for $|u _{k+1} -u _{0}|_{L^{\infty}}$, we
will use Sobolev's inequality $|h|_{L^{\infty}}^2 \leq
C\|h\|_{s_0}^2$ (see, e.g.,  \cite{a1}, \cite{e1}), where $s_0=[
\frac N2]+1=2$. We will also apply inequality \eqref{B.55} from
Lemma \ref{LB.1} in Appendix B, which yields the estimate $\| u
_{k+1}-u _0\|_{0}^2 \leq C \|\nabla( u _{k+1}-u _0)\|_{2}^2$. And
we will use the estimate \eqref{e3.9}, \eqref{e3.10} just proven
for $\|\nabla u_{k+1}\|_{s}^2$ . We then obtain the following
inequality:
\begin{equation}
\begin{aligned}
|u _{k+1}-u _0|_{L^\infty}^2 &\leq  C \| u _{k+1}-u _0\|_{s_0}^2
\leq  C \| u _{k+1}-u _0\|_{s+1}^2
 \\
&= C\|u _{k+1}-u_0\|_0^2+C \sum_{1 \leq |\alpha|\leq
s+1}\|D^{\alpha}( u _{k+1}-u_0)\|_{0 }^2    \\
&\leq  C\| u _{k+1}-u _0\|_{0}^2+C\| \nabla (u _{k+1}-u
_0)\|_{s}^2 \\
&\leq  C\|\nabla(u_{k+1}-u_0)\|_{2}^2 + C\|\nabla(u_{k+1}-u_0)\|_{s}^2   \\
&\leq  C\|\nabla(u_{k+1}-u_0)\|_{s}^2   \\
&=  C\|\nabla u_{k+1}\|_{s}^2   \\
&\leq   2 C C_1 \|f\|_{s-1}^2
\end{aligned} \label{e3.19}
\end{equation}
Therefore, from \eqref{e3.19} we have $|u _{k+1}-u _0|_{L^\infty}
\leq C_5 \|f\|_{s-1}$, where we define $C_5 = ( 2 C C_1)^{1/2}$ from \eqref{e3.19}. We will assume that
$\max\{1,|\mathbf{v}|_{L^\infty}^2\}\|f\|_{s-1}^2$ $\leq \frac{
R^2}{C_5^2} $. It follows that $\|f\|_{s-1} \leq \frac{ R}{C_5}$,
and therefore $|u _{k+1}-u _0|_{L^{\infty}}\leq R $. And so
\eqref{e3.5} holds for $u_{k+1}$, and $u_{k+1}(\mathbf{x}) \in
\bar{G_1}$ for all $\mathbf{x}\in \mathbb{T}^N$.
\smallskip

\noindent\textbf{Estimate for $\|u _{k+1}\|_{0}^2$ and $\|u
_{k+1}\|_{s+1}^2$:} To obtain an $L^2$ estimate for $u _{k+1}$ we
apply inequality \eqref{B.56} from Lemma \ref{LB.1} in Appendix B,
which yields
\begin{equation}
\begin{aligned}
\|u _{k+1}\|_0^2 &\leq  C\|u _0\|_0^2+C\|\nabla u
_0\|_{2 }^2+C\|\nabla u _{k+1}\|_{2}^2    \\
&\leq C\|u _0\|_0^2+C\|\nabla u _{k+1}\|_{s}^2    \\
&\leq  C |\Omega||u (\mathbf{x}_0)|^2+2 C C_1 \|f\|_{s-1}^2
\end{aligned} \label{e3.11}
\end{equation}
Here we used the estimate for $\| \nabla u _{k+1}\|_s^2$ from the
result just proven in \eqref{e3.9}, \eqref{e3.10}. And we used the
fact that $u_0$ is a constant.  From the estimates \eqref{e3.9},
\eqref{e3.10}, \eqref{e3.11} and using the fact that
$\|f\|_{s-1}^2 \leq \frac{ R^2}{C_5^2}$ where $C_5^2 = 2 C C_1$,
yields
\begin{equation}
\begin{aligned}
\| u _{k+1}\|_{s+1}^2
&=   \|u _{k+1}\|_0^2+\sum_{1 \leq
|\alpha|\leq s+1}\|D^{\alpha} u_{k+1}\|_{0 }^2    \\
&\leq   \|u _{k+1}\|_0^2+C\|\nabla u
_{k+1}\|_{s }^2  \\
&\leq  C |\Omega||u (\mathbf{x}_0)|^2+2 C C_1 \|f\|_{s-1}^2  \\
&\leq  C |\Omega||u (\mathbf{x}_0)|^2+C R^2
\end{aligned} \label{e3.12}
\end{equation}
We now define the constant $L$ to be $L = C |\Omega||u
(\mathbf{x}_0)|^2+C R^2$ from \eqref{e3.12}. Then we have $\|u
_{k+1}\|_{s+1}^2\leq L$, and so \eqref{e3.4} holds for $\|u
_{k+1}\|_{s+1}^2$. Therefore \eqref{e3.3}, \eqref{e3.5},
\eqref{e3.4} hold for all $k\geq 1$, and $u_k(\mathbf{x}) \in
\bar{G_1}$ for all $\mathbf{x}\in \mathbb{T}^N$ and for all $k\geq
1$.
\smallskip

\noindent\textbf{Estimate for $\|u _{k+1} -u _{k}\|_{s+1}^2$:}
Subtracting the equation \eqref{3.1} for $u_k$ from the equation
\eqref{3.1} for $u_{k+1}$ yields the following equation
\begin{equation}
-\nabla \cdot(a(u_k)\nabla (u _{k+1}-u _k))+\mathbf{v}\cdot \nabla
(u^{k+1}-u^k)  =\nabla \cdot(( a(u_k)-a(u_{k-1}))\nabla u_k)
\label{e3.20}
\end{equation}
We consider two cases: when $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _k))\|_{s_1}^2$, and when
$\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D\mathbf{v}\|_{s_1}$.

\textbf{Case 1}: Suppose that $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _k))\|_{s_1}^2$.
Applying estimate \eqref{B.27} from Lemma \ref{LB.6} in
Appendix B to equation \eqref{e3.20}, and using estimate
\eqref{e3.8} for
$\|D(a(u _k))\|_{s_1}^2$, and using estimate \eqref{e3.3} for
$\|\nabla u_k\|_{s}^2$, yields the following:
\begin{equation}
\begin{aligned}
&\|\nabla (u _{k+1}-u _k)\|_{s}^2 \\
& \leq C_1\Big[\sum_{j=0}^s\|D(a(u _k))\|_{s_1}^{2j}\Big] \|\nabla
\cdot((a(u _k)-a(u
_{k-1}))\nabla u _k)\|_{s-1}^2    \\
&\leq  C C_1\Big[\sum_{j=0}^sC_3^j \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^{2j}\|\nabla u_k\|_{s_1}^{2j}\Big] \|(a(u
_k)-a(u
_{k-1}))\nabla u _k\|_{s}^2   \\
&\leq  C  C_1\Big[\sum_{j=0}^s C_3^j \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^{2j}\|\nabla u_k\|_{s}^{2j}\Big] \|a(u _k)-a(u
_{k-1})\|_s^2\|\nabla u _k\|_{s}^2    \\
&\leq  C  C_1\Big[\sum_{j=0}^s C_3^j(2 C_1)^{j} \big|\frac{d a}{d
u} \big|_{s,\bar{G}_1}^{2j}\|f\|_{s-1}^{2j}\Big] \|a(u _k)-a(u
_{k-1})\|_s^2 (2 C_1)\|f\|_{s-1}^2    \\
&\leq  C C_1\Big[\sum_{j=0}^s C_3^j( 2 C_1)^j \Big(\frac{1}{C_4}
\Big)^j \Big] \|a(u _k)-a(u
_{k-1})\|_s^2 (2 C_1)\|f\|_{s-1}^2    \\
&\leq  C C_1\Big[\sum_{j=0}^s \big(\frac 12 \big)^{j}\Big]
\|a(u _k)-a(u
_{k-1})\|_s^2(2 C_1)\|f\|_{s-1}^2    \\
&\leq  C (2 C_1)^2 \|a(u _k)-a(u _{k-1})\|_s^2\|f\|_{s-1}^2
\end{aligned}\label{e3.13}
\end{equation}
where we used the Sobolev calculus inequality
$\|g h \|_r^2 \leq C\| g\|_r^2\|h\|_r^2$ for $r>\frac N2$,
where $C$ is a constant
which depends on $r$ (see, e.g.,  \cite{e1}, \cite{m1}), and we
let $r=s$ where $s>\frac N2 +1$. We also used the fact that
$|\frac{d a}{d u}|_{s,\bar{G}_1}^{2}\| f\|_{s-1}^2 \leq
\frac{1}{C_4}$. And we used the fact that $\frac{2 C_3 C_1}{C_4}
\leq \frac 12$ and $\frac{1}{C_3 C_1}\leq 1$, since $C_4= 4 C_3^2
C_1^2$ and $C_3 C_1 \geq 1$.

To estimate the term $\|a(u _k)-a(u _{k-1})\|_s^2$, we will apply
the Sobolev space inquality \eqref{3.98} from Lemma \ref{LB.1} in
Appendix B, which yields
\begin{equation}
\begin{aligned}
&\|a(u _k)-a(u _{k-1})\|_s^2\\
& \leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u_k|_{L^\infty}+|u_{k-1}|_{L^\infty})^2(\|u_k\|_s+\|u_{k-1}\|_s)^2
\|u_k-u_{k-1}\|_{s}^2  \\
&\leq   C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u_k-u_0|_{L^\infty}
+|u_{k-1}-u_0|_{L^\infty}+2|u_{0}|_{L^\infty})^2\\
&\quad\times (\|u_k\|_s^2+\|u_{k-1}\|_s^2)
\|u_k-u_{k-1}\|_{s}^2  \\
&\leq  C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(2+2R+2|u_{0}|_{L^\infty})^2
(\|u_k\|_{s+1}^2+\|u_{k-1}\|_{s+1}^2)
\|u_k-u_{k-1}\|_{s}^2  \\
&\leq   C L \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_{0})|)^{2s}
\|u_k-u_{k-1}\|_{s}^2  \\
&\leq   C L C_2  \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2
\|u_k-u_{k-1}\|_{s}^2   \\
&\leq   C L C_3  \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2
\|u_k-u_{k-1}\|_{s}^2
\end{aligned}\label{e3.90}
\end{equation}
where $C$ depends on $s$, and we used \eqref{e3.4} to estimate
$\|u_k\|_{s+1}^2\leq L$ and $\|u_{k-1}\|_{s+1}^2\leq L$. We also
used Cauchy's inequality $gh \leq \frac 12 g^2 + \frac 12 h^2$.
Here, $C_2$, $C_3$ are the same constants as in \eqref{e3.8}. Then
from \eqref{e3.13} and \eqref{e3.90} we obtain
\begin{equation}
\begin{aligned}
\|\nabla (u _{k+1}-u _k)\|_{s}^2 &\leq  C L C_3(2 C_1)^2
\big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2
\|f\|_{s-1}^2 \|u_k-u_{k-1}\|_{s}^2   \\
&\leq  C L C_3(2 C_1)^2\Big(\frac{1}{C_4}\Big)
\|u _k-u _{k-1}\|_s^2  \\
&=  \frac{C L }{C_3} \|u _k-u _{k-1}\|_{s+1}^2
\end{aligned} \label{e3.14}
\end{equation}
Here we used the fact that $C_4 =4 C_3^2 C_1^2$, and
 that $\big|\frac{d a}{d u}\big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2
\leq \frac{1}{C_4}$.

\textbf{Case 2}: Suppose that $\max\{\|D(a(u _k))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D \mathbf{v}\|_{s_1}$.
Applying estimate \eqref{B.27} from Lemma \ref{LB.6} in Appendix B
to equation \eqref{e3.20}, and using \eqref{e3.3}, \eqref{e3.90},
and using the proof of \eqref{e3.14}, yields the inequality
\begin{equation}
\begin{aligned}
&\|\nabla (u _{k+1}-u _k)\|_{s}^2 \\
& \leq
C_1\Big[\sum_{j=0}^s\|D\mathbf{v}\|_{s_1}^{j}\Big] \|\nabla
\cdot((a(u _k)-a(u _{k-1}))\nabla u _k)\|_{s-1}^2    \\
&\leq C C_1\Big[\sum_{j=0}^s \big(\frac 12\big)^j \Big] \|a(u
_k)-a(u
_{k-1})\|_s^2\|\nabla u _k\|_{s}^2    \\
&\leq C  ( 2 C_1) \|a(u _k)-a(u
_{k-1})\|_s^2 (2 C_1)\|f\|_{s-1}^2    \\
&\leq C L C_3(2 C_1)^2 \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2
\|f\|_{s-1}^2 \|u_k-u_{k-1}\|_{s}^2  \\
&\leq C L C_3(2 C_1)^2 \Big(\frac{1}{C_4}\Big)\|u_k-u_{k-1}\|_{s}^2  \\
&\leq  \frac{C L }{C_3} \|u _k-u _{k-1}\|_{s+1}^2
\end{aligned}\label{e3.91}
\end{equation}
which is the same result as \eqref {e3.14}. Here, we used the
facts that $\big|\frac{d a}{d u}\big|_{s,\bar{G}_1}^2\|
f\|_{s-1}^2 \leq \frac{1}{C_4}$ where $C_4= 4 C_3^2 C_1^2$, and
that $\|D\mathbf{v}\|_{s_1}\leq \|D\mathbf{v}\|_{s} \leq \frac
12$.

To obtain an $L^2$ estimate for $u _{k+1}-u_k$, we apply
inequality \eqref{B.55} from Lemma \ref{LB.1} in Appendix B, which
yields
\begin{equation}
\|u _{k+1}-u _k\|_0^2 \leq C\|\nabla (u_{k+1}-u _k)\|_2^2 \leq
C\|\nabla (u_{k+1}-u _k)\|_s^2
 \label{e3.15}
\end{equation}
 From \eqref{e3.14}--\eqref{e3.15}, we obtain
\begin{equation}
\begin{aligned}
\|u _{k+1}-u _k\|_{s+1}^2
&=  \|u _{k+1}-u_k\|_0^2+\sum_{1 \leq
|\alpha|\leq s+1}\|D^{\alpha}( u
_{k+1}-u_k)\|_{0 }^2    \\
&\leq   \|u _{k+1}-u _k\|_0^2+C\|\nabla
(u _{k+1}-u _k)\|_s^2  \\
&\leq   C\|\nabla(u _{k+1}-u _{k})\|_s^2    \\
&\leq   \frac{C L }{C_3} \|u _k-u _{k-1}\|_{s+1}^2
\end{aligned} \label{e3.16}
\end{equation}
where $L = C |\Omega||u (\mathbf{x}_0)|^2+C R^2$ from
\eqref{e3.12} , and $C_2= C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$ and $C_3=MC_2$ from \eqref{e3.8}. It
follows that $\frac{C L }{ C_3}\leq \frac{C}{M}$, where $C$
depends on $s$ and $s\geq 3$. We now define the constant $M$ to be
large enough so that $\frac{C}{M}<1$. Then from \eqref{e3.16}, we
have
\begin{equation}
\sum_{k=0}^{\infty}\|u _{k+1}-u _k\|_{s+1}^2< \infty
\label{e3.18}
\end{equation}
which is the inequality \eqref{e3.6} to be proven. This completes
the proof of Proposition \ref{P3.1}.
\end{proof}


We now complete the proof of Theorem \ref{T3.1}.  From Lemma
\ref{LA.4} in Appendix A, we know that  $u_k \in C^2(\Omega)\cap
H^{s+1}(\Omega)$ for each $k\geq 1$, where $s> \frac N2 +1$.  From
\eqref{e3.4} in Proposition \ref{P3.1} and from Sobolev's
inequality $|h|_{L^{\infty}}^2 \leq C\|h\|_{s_0}^2$ (see, e.g.,
 \cite{a1}, \cite{e1}), where $s_0=[ \frac N2]+1=2$, we
know that $\{u_k\}_{k=1}^{\infty}$ is bounded in $ C^2(\Omega)\cap
H^{s+1}(\Omega)$. And from \eqref{e3.6} in Proposition \ref{P3.1},
it follows that $\|u _{k+1} -u _{k}\|_{s+1} \rightarrow 0 $ as
$k\to \infty$. We conclude that there exists $u \in
C^2(\Omega)\cap H^{s+1}(\Omega)$ such that $\|u _{k}-u
\|_{s+1}\rightarrow 0$ as $k\to \infty $.  From Lemma \ref{LA.4} in
Appendix A,  we know that $u_{k+1}$ is a solution of the linear
equation \eqref{3.1} for each $k\geq 0$, and
$u_{k+1}(\mathbf{x}_0)=u(\mathbf{x}_0)$ for each $k\geq 0$, and so
it follows that $u$ is a solution of the quasilinear equation
\eqref{1.1}, and $u$ satisfies \eqref{1.2}.


To prove uniqueness of the solution, let us assume that
$u_1(\mathbf{x})$, $u_2(\mathbf{x})$ are two solutions of
\eqref{1.1}-\eqref{1.2}, and $u_1(\mathbf{x})\in \bar{G_1}$ and
$u_2(\mathbf{x})\in \bar{G_1}$ for all $\mathbf{x}\in
\mathbb{T}^N$. We will show that there exists a constant $C_7$,
such that if $\Big| \frac{d a}{d u} \big|_{s,\bar{G}_1}^2\|
f\|_{s-1}^2 \leq \frac{1}{C_7}$, and if $\|D\mathbf{v}\|_{s_1}\leq
\frac 12$, and if $a''(u_{*})\leq
\frac{1}{a(u_{*})}(a'(u_{*}))^2$ for all $u_{*} \in
\bar{G_1}$, and if
$\max\{1,|\mathbf{v}|_{L^\infty}^2\}\|f\|_{s-1}^2$ $ \leq
\frac{R^2}{C_5^2} $, and if $|\nabla \cdot \mathbf{v}|_{L^\infty}
\leq \frac{c_1}{C_*}$, where $C_5$, $c_1$, $C_*$ are the constants
from the proof of Proposition \ref{P3.1}, then $u_1=u_2$.

Note that since $u_1(\mathbf{x})\in \bar{G_1}$ and
$u_2(\mathbf{x})\in \bar{G_1}$, it  follows that
$|u_1-u_0|_{L^\infty}\leq R$ and $|u_2-u_0|_{L^\infty} \leq R$,
and $a(u_1)
> c_1$ and $a(u_2) > c_1$,  and $a''(u_{1})\leq \frac{1}{a(u_{1})}(a'(u_{1}))^2$ and
$a''(u_{2})\leq \frac{1}{a(u_{2})}(a'(u_{2}))^2$. By Lemma \ref{LB.8} from Appendix B applied to
equation \eqref{1.1} for $u_1$ and $u_2$, there exist constants
$C_7$, $C_8$, such that if $\Big| \frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2 \leq \frac{1}{C_7}$, then
$u_1$, $u_2$ satisfy
\begin{equation}
\begin{gathered}
\|\nabla u _1\|_{s}^2 \leq  2 C_8\|f\|_{s-1}^2, \\
\|\nabla u _2\|_{s}^2 \leq  2 C_8\|f\|_{s-1}^2
\end{gathered}\label{e3.82}
\end{equation}
 From Lemma \ref{LB.8} in  Appendix B, the constant $C_7= 4 C_0^2
C_3^2 C_1^2 K_1^2$, and the constant $C_8=C_0 C_1 K_1$ so that we
have $C_7=4 C_3^2 C_8^2 $, and $C_0$ is a constant which depends
on $s$, $c_1$, and the constant
$K_1=\max\{1,|\mathbf{v}|_{L^\infty}^2 \}$. We may assume that
$C_0\geq 1$, so that $C_1 \leq C_8$.

And we have $\| u _1\|_{0}^2 \leq |\Omega| |u_1|_{L^\infty}^2$
$\leq 2|\Omega|( |u_1-u_0|_{L^\infty}^2+ |u_0|_{L^\infty}^2)$
$\leq 2|\Omega|(R^2+ |u(\mathbf{x}_0)|^2)$. So $\| u _1\|_{s+1}^2
\leq \| u _1\|_{0}^2 +C\|\nabla u _1\|_{s}^2 $ $\leq
2|\Omega|(R^2+ |u(\mathbf{x}_0)|^2)+ 2 C C_8\|f\|_{s-1}^2$. It
follows that $u_1 \in C^2(\Omega)\cap H^{s+1}(\Omega)$. Similarly,
$u_2 \in C^2(\Omega)\cap H^{s+1}(\Omega)$. Here, we used Sobolev's
inequality $|h|_{L^{\infty}}^2 \leq C\|h\|_{s_0}^2$, where $s_0=[
\frac N2]+1=2$.

Subtracting  \eqref{1.1} for $u_1$ from
\eqref{1.1} for $u_{2}$ yields the equation
\begin{equation}
-\nabla \cdot(a(u_1)\nabla (u _{2}-u_1))+\mathbf{v}\cdot \nabla
(u_{2}-u_{1})  =\nabla \cdot(( a(u_2)-a(u_{1}))\nabla u_2)
\label{e3.83}
\end{equation}

To obtain an estimate for $\|u_{2}-u_1\|_{s+1}^2$, we repeat the
proof of the estimate for $\|u_{k+1}-u_k\|_{s+1}^2$ from
\eqref{e3.13}-\eqref{e3.16}, and apply this proof to
\eqref{e3.83}. We use inequality \eqref{B.55} from Lemma
\ref{LB.1} in Appendix B, which yields $\|u_2-u _1\|_0^2 \leq C
\|\nabla (u_2-u _1)\|_2^2$, and we use inequality \eqref{B.27}
from Lemma \ref{LB.6} in Appendix B to estimate
$\|\nabla(u_2-u_1)\|_s^2$, and we use inequality \eqref{e3.8} to
estimate $\|D(a(u _1))\|_{s_1}^2$ and we use inequality
\eqref{e3.82} to estimate $\|\nabla u_1\|_{s}^2$ and $\|\nabla
u_2\|_{s}^2$. We also use the inequality $\Big| \frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2 \leq \frac{1}{C_7}$ and the
inequality $\|D\mathbf{v}\|_{s_1}\leq \frac 12 $.


We consider two cases: when $\max\{\|D(a(u _1))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _1))\|_{s_1}^2$, and when
$\max\{\|D(a(u _1))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D\mathbf{v}\|_{s_1}$.

\textbf{Case 1}: Suppose that $\max\{\|D(a(u _1))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D(a(u _1))\|_{s_1}^2$.
We obtain
\begin{equation}
\begin{aligned}
&\|u_{2}-u_1\|_{s+1}^2\\
&=  \|u _{2}-u_1\|_0^2+\sum_{1 \leq
|\alpha|\leq s+1}\|D^{\alpha}( u_2-u_1)\|_{0 }^2    \\
&\leq   \|u_2-u _1\|_0^2+C\|\nabla
(u_2-u_1)\|_s^2  \\
&\leq   C\|\nabla(u_2-u_1)\|_2^2+ C\|\nabla(u_2-u_1)\|_s^2    \\
&\leq   C\|\nabla(u_2-u_1)\|_s^2    \\
&\leq  C C_1\Big[\sum_{j=0}^s
(\max\{\|D(a(u_1))\|_{s_1}^2,\|D\mathbf{v}\|_{s_1}\})^{j}\Big]
\|\nabla \cdot((a(u_2)-a(u_1))\nabla u_2)\|_{s-1}^2    \\
&\leq  C C_8\Big[\sum_{j=0}^s \|D(a(u_1))\|_{s_1}^{2j}\Big]
\|a(u_2)-a(u_1)\|_s^2\|\nabla u_2\|_{s}^2    \\
&\leq  C  C_8\Big[\sum_{j=0}^s C_3^j \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^{2j}\|\nabla u_1\|_{s}^{2j}\Big] \|a(u_2)-a(u_1)\|_s^2\|\nabla u_2\|_{s}^2    \\
&\leq  C  C_8\Big[\sum_{j=0}^s C_3^j(2 C_8)^{j} \big|\frac{d a}{d
u} \big|_{s,\bar{G}_1}^{2j}\|f\|_{s-1}^{2j}\Big]
\|a(u_2)-a(u_1)\|_s^2(2 C_8)\|f\|_{s-1}^2    \\
&\leq  C C_8\Big[\sum_{j=0}^s C_3^j(2 C_8)^j \Big(\frac{1}{C_7}
\Big)^j \Big] \|a(u
_2)-a(u_1)\|_s^2 (2 C_8)\|f\|_{s-1}^2    \\
&\leq  C C_8 \Big[\sum_{j=0}^s \big(\frac 12\big)^j\Big]
\|a(u_2)-a(u_1)\|_s^2
(2C_8)\|f\|_{s-1}^2    \\
&\leq  C (2C_8)^2 \|a(u_2)-a(u_1)\|_s^2\|f\|_{s-1}^2
\end{aligned}\label{e3.84}
\end{equation}
where we used  that  $C_1 \leq C_8$, and $C_7=  4 C_3^2
C_8^2$, and $\frac{ 2 C_3 C_8}{C_7}\leq \frac 12$, since $C_3 C_8
\geq 1$.

 From the third line in the proof of estimate \eqref{e3.90}, we
have the inequality
\begin{equation}
\begin{aligned}
&\|a(u _2)-a(u _{1})\|_s^2\\
& \leq  C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(2+2R+2|u(\mathbf{x}_{0})|)^2
(\|u_2\|_{s+1}^2+\|u_1\|_{s+1}^2) \|u_2-u_1\|_{s}^2
 \\
&\leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_{0})|)^2(\|u_2\|_{0}^2
+C\|\nabla u_2\|_{s}^2+\|u_1\|_{0}^2\\
&\quad +C\|\nabla u_1\|_{s}^2) \|u_2-u_1\|_{s}^2
\end{aligned}\label{e3.85}
\end{equation}
By inequality \eqref{B.56} from Lemma \ref{LB.1} in Appendix B, we
have the estimate $ \|u _{2}\|_0^2 \leq C\|u _0\|_0^2+C\|\nabla u
_0\|_{2 }^2+C\|\nabla u _{2}\|_{2}^2 $ $\leq C
|\Omega||u(\mathbf{x}_{0})|^2 +C\|\nabla u _{2}\|_{s}^2$. And a
similar inequality holds for $\|u_1\|_{0}^2$. Substituting these
$L^2$ estimates into \eqref{e3.85} yields
\begin{equation}
\begin{aligned}
&\|a(u _2)-a(u _{1})\|_s^2\\
&\leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_{0})|)^2(|\Omega||u(\mathbf{x}_{0})|^2
+\|\nabla u_2\|_{s}^2+\|\nabla u_1\|_{s}^2)
\|u_2-u_1\|_{s}^2  \\
&\leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_{0})|)^2(|\Omega||u(\mathbf{x}_{0})|^2
+4 C_8\|f\|_{s-1}^2)
\|u_2-u_1\|_{s}^2  \\
&\leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_{0})|)^2
\Big(|\Omega||u(\mathbf{x}_{0})|^2
+4 C_8\Big( \frac{R^2}{K_1 C_5^2}\Big)\Big)
\|u_2-u_1\|_{s}^2  \\
&= C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_{0})|)^2
\Big(|\Omega||u(\mathbf{x}_{0})|^2
+\Big(\frac{ 4 C_0 C_1 K_1 R^2}{2 K_1 C C_1}\Big)\Big)
\|u_2-u_1\|_{s}^2  \\
&\leq C \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_{0})|)^{2s}(|\Omega||u(\mathbf{x}_{0})|^2 + R^2)
\|u_2-u_1\|_{s}^2  \\
&\leq C C_3 L \big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2
\|u_2-u_1\|_{s}^2
\end{aligned}\label{e3.86}
\end{equation}
where we used  $\|f\|_{s-1}^2 \leq \frac{ R^2}{K_1
C_5^2}$, where $K_1=\max\{1,|\mathbf{v}|_{L^\infty}^2\}$ and $C_5
= (2 C C_1)^{1/2}$. And we used the fact that $C_8=C_0 C_1
K_1$, where $C_0$ depends on $s$, $c_1$. And we used inequality
\eqref{e3.82} to estimate $\|\nabla u_1\|_{s}^2$ and $\|\nabla
u_2\|_{s}^2$. Also, $L = C |\Omega||u (\mathbf{x}_0)|^2+C R^2$
from \eqref{e3.12}, and
$C_2= C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$ and
$C_3=M C_2$ from \eqref{e3.8},
where $M \geq 1$. Substituting \eqref{e3.86} into \eqref{e3.84}
yields
\begin{equation}
\begin{aligned}
\|u_{2}-u_1\|_{s+1}^2
&\leq   C C_3 L (2 C_8)^2 \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\|f\|_{s-1}^2 \|u_2-u_1\|_{s}^2 \\
&\leq  C C_3 L (2 C_8)^2 \Big(\frac {1}{C_7}\Big)
\|u_2-u_1\|_{s}^2 \\
&=  \Big(\frac{C L}{C_3 }\Big)\|u_2-u_1\|_{s+1}^2
\end{aligned}\label{e3.87}
\end{equation}
where we used the fact that $C_7= 4 C_3^2 C_8^2$  and
$\big|\frac{d a}{d u} \big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2 \leq
\frac{1}{C_7}$ .

\textbf{Case 2}: Suppose that $\max\{\|D(a(u _1))\|_{s_1}^2,
\|D\mathbf{v}\|_{s_1}\}=\|D \mathbf{v}\|_{s_1}$.
Repeating the proof of \eqref{e3.84}--\eqref{e3.87}  yields the
following:
\begin{align*}
&\|u_{2}-u_1\|_{s+1}^2\\
&\leq  C C_1\Big[\sum_{j=0}^s
(\max\{\|D(a(u_1))\|_{s_1}^2,\|D\mathbf{v}\|_{s_1}\})^{j}\Big]
\|\nabla \cdot((a(u_2)-a(u_1))\nabla u_2)\|_{s-1}^2    \\
&\leq  C  C_8\Big[\sum_{j=0}^s \|D \mathbf{v}\|_{s_1}^{j}\Big]
\|a(u_2)-a(u_1)\|_s^2\|\nabla u_2\|_{s}^2    \\
&\leq    C C_8\Big[\sum_{j=0}^s \big(\frac 12\big)^j \Big]
\|a(u_2)-a(u_1)\|_s^2
(2 C_8)\|f\|_{s-1}^2    \\
&\leq  C (2 C_8)^2
\|a(u_2)-a(u_1)\|_s^2\|f\|_{s-1}^2  \\
&\leq  \Big(\frac{C L}{ C_3 }\Big) \|u_2-u_1\|_{s+1}^2
\end{align*}
which is the same estimate as \eqref{e3.87}.

Recall that $L = C |\Omega||u (\mathbf{x}_0)|^2+C R^2$ from
\eqref{e3.12} , and $C_2= C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$ and $C_3=M C_2$ from \eqref{e3.8}. It
follows that $\frac{C L }{C_3}\leq \frac{C}{M}$, where $C$ depends
on $s$. As in the proof of \eqref{e3.18}, we define the constant
$M$ to be large enough so that $\frac{C}{M}<1$. It follows that
$| |u_{2}-u_1\|_{s+1}^2=0$, and therefore $u_1=u_2$ and the
solution is unique.

This completes the proof of Theorem \ref{T3.1}. Note that
$\delta_0= \min\{\frac{1}{C_4}, \frac{1}{C_7}\}$, and
$\delta_1=\frac{c_1}{C_*}$, and $\delta_2=\frac { R^2}{C_5^2}$ in
the statement of Theorem \ref{T3.1} in which we assume that $\Big|
\frac{d a}{d u} \big|_{s,\bar{G}_1}^2 \| f\|_{s-1}^2 \leq
\delta_0$, and $|\nabla \cdot \mathbf{v}|_{L^\infty} \leq \delta_1
$ , and $\max\{1,|\mathbf{v}|_{L^\infty}^2\}\| f\|_{s-1}^2 \leq
\delta_2$, and $\|D\mathbf{v}\|_s \leq \frac 12$. And $\delta_0$,
$\delta_1$, $\delta_2$, $C_4$, $C_5$, $C_7$ depend on $s$, $c_1$,
$R$, $|\Omega|$, and $|u (\mathbf{x}_0)|$.
%\end{proof}


\appendix

\section{Existence for the linear equation}


In this section, we present the proof of the existence of a
solution to the linear problem \eqref{3.1}, \eqref{3.2}.

\begin{lemma}\label{LA.4}
Let $a_1 \in C^1(\Omega)\cap H^{s}(\Omega)$, $f \in C(\Omega)\cap
H^{s-1}(\Omega)$, $\mathbf{v} \in C^1(\Omega)\cap H^{s}(\Omega)$
be given functions, where $a_1(\mathbf{x})> c_1$ for some positive
constant $c_1$, for $\mathbf{x}\in\Omega$, $\Omega= \mathbb{T}^N$,
$N=2$ or $N=3$. We assume that $|\nabla \cdot
\mathbf{v}|_{L^\infty} \leq  \frac{c_1}{ C_*}$, where $C_*$ is the
constant from Poincar\'{e}'s inequality $\|\bar{u} \|_0^2\leq
C_*\|\nabla u \|_0^2$, and where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac{1}{|\Omega|}\int_{\Omega}
u(\mathbf{x}) d\mathbf{x}$. Then there is a unique classical
solution $u \in C^2(\Omega)\cap H^{s+1}(\Omega)$ of
\begin{gather}
-\nabla\cdot (a_1\nabla u)+\mathbf{v}\cdot \nabla u= f,  \label{A.1}\\
u(\mathbf{x}_0)=u_0, \label{A.2}
\end{gather}
where $u_0$ is a given constant and $\mathbf{x}_0 \in \Omega$ is a
given point, and where $s>\frac N2+1$.
\end{lemma}

\begin{proof}
The operator in \eqref{A.1} is linear with $a_1(\mathbf{x})> c_1$
for $\mathbf{x}\in\Omega$. The existence of a zero-mean solution
$\bar{u}(\mathbf{x})$ of equation \eqref{A.1} follows from the
standard theory for elliptic equations, specifically, the
Lax-Milgram Lemma (see, e.g., \cite{e2}). We then define the
chosen solution $u(\mathbf{x})$ to \eqref{A.1}, \eqref{A.2} to be
$u(\mathbf{x})= \bar{u}(\mathbf{x})-\bar{u}(\mathbf{x}_0)+u_0$.

We remark that the condition for the Lax-Milgram Lemma that
$\|\bar{u}\|_1^2 \leq C B[\bar{u},\bar{u}]$, where
$B[\bar{u},\bar{u}]=(a_1\nabla \bar{u},\nabla
\bar{u})+(\mathbf{v}\cdot \nabla \bar{u},\bar{u})$, and where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac{1}{|\Omega|}\int_{\Omega}
u(\mathbf{x})  d\mathbf{x}$, follows from the following
inequality:
\begin{align*}
(c_1\nabla u,\nabla u )&\leq (a_1\nabla u,\nabla u ) \\
& = -(\mathbf{v}\cdot \nabla \bar{u},\bar{u})
+B[\bar{u},\bar{u}] \\
 &= \frac 12(\nabla \cdot\mathbf{v}\cdot \bar{ u},\bar{u})
+B[\bar{u},\bar{u}]
 \\
& \leq \frac 12|\nabla \cdot \mathbf{v}|_{L^\infty}\|\bar{ u}
\|_0^2+B[\bar{u},\bar{u}]  \\
&\leq  \frac 12 C_*|\nabla \cdot \mathbf{v}|_{L^\infty}\|\nabla{
u}
\|_0^2+B[\bar{u},\bar{u}]  \\
&\leq  \frac {c_1}{2} \|\nabla{ u} \|_0^2+B[\bar{u},\bar{u}]
\end{align*} % A3
where we used the fact that $|\nabla \cdot \mathbf{v}|_{L^\infty}
\leq  \frac{c_1}{ C_*}$. And so $\frac 12(c_1\nabla u,\nabla u )
\leq B[\bar{u},\bar{u}]$.  From Poincar\'{e}'s inequality
$\|\bar{u} \|_0^2\leq C_*\|\nabla u \|_0^2$, we obtain the desired
inequality $ \|\bar{u}\|_1^2=\|\bar{u}\|_0^2+\|\nabla u\|_0^2 \leq
(C_*+1)\|\nabla u\|_0^2\leq
\frac{2(C_*+1)}{c_1}B[\bar{u},\bar{u}]$.

The regularity of the chosen solution $u(\mathbf{x})$ follows from
the estimates \eqref{B.56} from Lemma \ref{LB.1} and \eqref{B.27}
from Lemma \ref{LB.6} in Appendix B, applied to equation
\eqref{A.1}, which yield:
\begin{gather*}
\|u \|_{0}^2 \leq
C\|u(\mathbf{x}_0)\|_{0}^2+C\|\nabla(u(\mathbf{x}_0))\|_{2
}^2+C\|\nabla u \|_{2 }^2  \leq C |\Omega||u (\mathbf{x}_0)|^2+
C\|\nabla u
\|_{s}^2 \\
\|\nabla u \|_{s}^2 \leq  C_1\Big[\sum_{j=0}^{s}(\max\{\|Da_1\|
_{s_1}^2, \|D\mathbf{v}\|_{s_1}\})^{j}\Big]\|f\|_{s-1}^2
\end{gather*}
where $s_1=\max\{s-1,s_0\}=s-1$, and $s_0=[ \frac N2]+1=2$, and
$s>\frac N2+1$, so $s\geq 3$. It follows that  $u \in
C^2(\Omega)\cap H^{s+1}(\Omega)$ by the above estimates and by
Sobolev's inequality $|h|_{L^{\infty}}^2 \leq C\|h\|_{s_0}^2$
(see, e.g., \cite{a1,e1}).
\end{proof}

\section{A priori estimates}

Recall that we will be using the Sobolev space $H^s(\Omega )$
(where $s\geq 0$ is an integer) of real-valued functions in
$L^2(\Omega )$ whose distribution derivatives up to order $s$ are
in $L^2(\Omega )$, with norm given by $\|g\|_s^2=\sum_{|\alpha
|\leq s}\int_\Omega |D^\alpha g|^2d\mathbf{x}$ and inner product
$(g,h)_s=\sum_{|\alpha |\leq s}\int_\Omega (D^\alpha g)\cdot
(D^\alpha h)d\mathbf{x}$. The domain $\Omega$ is the N-dimensional
torus $\mathbb{T}^N$, where $N=2$ or $N=3$. Here, we are using the
standard multi-index notation. For convenience, we are going to
denote derivatives by $g_\alpha =D^\alpha g$. And we will denote
the $L^2$ inner product by $(g,h)=\int_\Omega g\cdot h$
$d\mathbf{x}$. We will use $C$ to denote a generic constant whose
value may change from one relation to the next. Recall that we let
both $\nabla g$ and $Dg$ denote the gradient of $g$.

We begin by listing several standard Sobolev space inequalities.

\begin{lemma}[Calculus Inequalities]  \label{LB.1}
\quad

(a) Let $g(u)$ be a smooth function on $G$, where $u(\mathbf{x})$
is a continuous function and where $u(\mathbf{x}) \in G_1$ for
$\mathbf{x} \in \Omega$ and $G_1 \subset G$ and $u \in
H^{r}(\Omega )\cap L^{\infty}(\Omega)$. Then for $r \geq 1$,
\begin{equation}
\|D^r(g(u))\|_0 \leq  C \big|\frac{d g}{d u}
\big|_{r-1,\bar{G}_1}(1+|u|_{L^\infty})^{r-1}\|Du\|_{r-1},\label{3.3}
\end{equation}
where $ |h |_{r,\bar{G}_1}=\max\{\big|\frac{d^j h}{d u^j} (u_{*})
\Big|: u_{*} \in \bar{G}_1, 0 \leq j \leq r \}$, and where $C$
depends on $r$, $\Omega$.

(b) And
\begin{equation}
\|g(u)-g(v)\|_r \leq  C \big|\frac{d g}{d u}
\big|_{r,\bar{G}_1}(1+|u|_{L^\infty}+|v|_{L^\infty})
(\|u\|_r+\|v\|_r)\|u-v\|_{r},\label{3.98}
\end{equation}
 where $C$ depends on $r$, $\Omega$.

(c) If $Dg\in H^{r_1}(\Omega )$, $h\in H^{r-1}(\Omega )$, where
$r_1=\max\{r-1,s_0\}$, $s_0=[ \frac N2]+1$, then for any $r\geq
1$, $g,h$ satisfy the estimate
\begin{equation}
\|D^\alpha (gh)-gD^\alpha h\|_0\leq C\|Dg\|_{r_1}\|h\|_{r-1},
\label{B.75}
\end{equation}
where $r=|\alpha |$, and the constant $C$ depends on $r$,
$\Omega$.


(d) Let $v$, $w$ be $C^1(\Omega)\cap H^{3}(\Omega)$ functions on a
bounded, open, connected, convex domain $\Omega$. And let
$v(\mathbf{x}_0)=w(\mathbf{x}_0)$ at a point $\mathbf{x}_0 \in
\Omega$. Then $v-w$ and $v$ satisfy the estimates
\begin{gather}
\|v -w\|_{0}^2\leq C\|\nabla (v -w)\|_{2}^2, \label{B.55}\\
\|v \|_{0}^2 \leq C\|w\|_{0}^2+C\|\nabla w\|_{2 }^2+C\|\nabla v
\|_{2 }^2 \label{B.56}
\end{gather}
Here $C$ is a constant which depends on $\Omega$.
\end{lemma}

Proofs of the inequalities (a), (b) may be found, for example, in
\cite{k1}, \cite{m2}. Proof of inequalities (c), (d) may be found
in \cite{DD1}. Inequalities (a), (b) also appear in \cite{e1}.


Lemmas \ref{LB.6} and  \ref{LB.8} provide the key a priori
estimates used in the proof of the theorem.

\begin{lemma} \label{LB.6}
Let $a_1(\mathbf{x})$,
$\mathbf{v}(\mathbf{x})$, and $f(\mathbf{x})$ be sufficiently
smooth functions in the following equation
\begin{equation}
-\nabla \cdot (a_1 \nabla u)+\mathbf{v}\cdot \nabla u = f,
\label{B.5}
\end{equation}
where $a_1(\mathbf{x})>c_1$, for some positive constant $c_1$, and
for all $\mathbf{x}\in \Omega$, with $\Omega=\mathbb{T}^N$, and
$N=2$ or $N=3$. We assume that $|\nabla \cdot
\mathbf{v}|_{L^\infty} \leq \frac{c_1}{ C_*}$, where $C_*$ is the
constant from Poincar\'{e}'s inequality $\|\bar{u} \|_0^2\leq
C_*\|\nabla u \|_0^2$, and where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac{1}{|\Omega|}\int_{\Omega}
u(\mathbf{x}) d\mathbf{x}$. Then we obtain the inequalities:
\begin{gather}
\|\nabla u\|_0^2 \leq  C\|f\|_{0}^2, \label{B.30}\\
\|\nabla u\|_{r}^2  \leq  C
\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\} \|\nabla u
\|_{r-1}^2 +C\|f\|_{r-1}^2, \label{B.31} \\
\|\nabla u \|_{r}^2  \leq
C_1\Big[\sum_{j=0}^{r}(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{j}\Big]\|f\|_{r-1}^2
\label{B.27}
\end{gather}
where $r \geq 1$, where $r_1=\max \{r-1,s_0\}$, and where $s_0=[
\frac N2]+1=2$. Here constant $C$ in \eqref{B.30} depends on
$c_1$, and the constant $C$ in \eqref{B.31} depends on $r$, $c_1$,
and the constant $C_1$ in \eqref{B.27} depends on $r$, $c_1$.
\end{lemma}

\begin{proof}
First, we obtain an $L^2$ estimate. Integrating equation
\eqref{B.5} by parts with $\bar{u} $, where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac 1{|\Omega |}\int_\Omega
u(\mathbf{x}) d \mathbf{x}$, yields
\begin{equation}
\begin{aligned}
(c_1\nabla u,\nabla u )
&\leq (a_1\nabla u,\nabla u ) = -(\nabla
\cdot (a_1\nabla u),\bar{u} ) =-(\mathbf{v}\cdot \nabla u,\bar{u})
+(f,\bar{u}) \\
 &= \frac 12(\nabla \cdot\mathbf{v}\cdot \bar{ u},\bar{u})
+(f,\bar{u}) \\
& \leq \frac 12|\nabla \cdot \mathbf{v}|_{L^\infty}\|\bar{ u}
\|_0^2+\frac{1}{4\epsilon}\|f\|_0^2+\epsilon\|\bar{ u} \|_0^2  \\
&\leq  \frac 12 C_*|\nabla \cdot \mathbf{v}|_{L^\infty}\|\nabla{
u} \|_0^2+\frac{1}{4\epsilon}\|f\|_0^2+\epsilon C_*\|\nabla u
\|_0^2
\end{aligned}\label{B.9}
\end{equation}
where we used Cauchy's inequality with $\epsilon$, namely
$gh \leq \frac{1}{4\epsilon}g^2+\epsilon h^2$, and
where we used the fact
that $a_1(\mathbf{x}) > c_1$. We also used Poincar\'{e}'s
inequality (see, e.g., \cite{e1}, \cite{e2}) to estimate
$\|\bar{u} \|_0^2\leq C_*\|\nabla u \|_0^2$, where $C_*$ is a
constant. We assume that $|\nabla \cdot \mathbf{v}|_{L^\infty}
\leq  \frac{c_1}{ C_*}$. And we let $\epsilon = \frac{c_1}{4
C_*}$. Then from \eqref{B.9}, we obtain
\begin{equation} \label{B.10}
\|\nabla u\|_0^2\leq C\|f\|_{0}^2
\end{equation}
where $C$ depends on $c_1$. This is the desired inequality
\eqref{B.30}.

Next, after applying $D^\alpha $ to the equation \eqref{B.5}, we
obtain the equation:
\begin{equation} \label{B.11}
-\nabla \cdot (a_1\nabla u_\alpha)+\mathbf{v}\cdot \nabla
u_{\alpha}=F_\alpha
\end{equation}
where $F_\alpha =f_{\alpha}+[\nabla \cdot (a_1\nabla u)_\alpha
-\nabla \cdot (a_1\nabla u_\alpha )]$ $-[(\mathbf{v}\cdot \nabla
u)_{\alpha}-\mathbf{v}\cdot \nabla u_{\alpha}]$.

 From \eqref{B.11} we obtain
\begin{equation}
\begin{aligned}
c_1(\nabla u_\alpha, \nabla u _\alpha )
&\leq(a_1\nabla u_\alpha, \nabla u _\alpha )   \\
&= -(\nabla \cdot(a_1\nabla u_\alpha), u _\alpha) \\
&= -(\mathbf{v}\cdot \nabla u_\alpha,u_\alpha)+(F_\alpha ,u_\alpha)  \\
&\leq  \frac 12 |\nabla \cdot \mathbf{v}|_{L^\infty}\| u_\alpha
\|_0^2+|(F_{\alpha} ,u_{\alpha} )|
 \\
&\leq  C|D \mathbf{v}|_{L^\infty}\|\nabla{ u} \|_{k-1}^2
+|(F_{\alpha} ,u_{\alpha} )|
\end{aligned} \label{B.12}
\end{equation}
where $|\alpha|=k$.

Next, we estimate $|(F_{\alpha} ,u _{\alpha} )|$. We use
integration by parts, and then apply inequality \eqref{B.75} from
Lemma \ref{LB.1}, to obtain the following inequality:
\begin{equation}
\begin{aligned}
&|(F_{\alpha} ,u _{\alpha} )|\\
&\leq  |( f_{\alpha},u_{\alpha})|
+|([\nabla \cdot (a_1\nabla u)_{\alpha} -\nabla \cdot (a_1\nabla
u_{\alpha} )],u _{\alpha} )|+|((\mathbf{v}\cdot \nabla
u)_{\alpha}-\mathbf{v}\cdot \nabla u_{\alpha},u_{\alpha})|
   \\
&=  |(f_{\alpha-\beta},u_{\alpha+\beta})|+|([(a_1\nabla
u)_{\alpha} -a_1\nabla u_{\alpha} ],\nabla u _{\alpha}
)|+|((\mathbf{v}\cdot \nabla
u)_{\alpha}-\mathbf{v}\cdot \nabla u_{\alpha},u_{\alpha})|  \\
&\leq  \|f_{\alpha-\beta}\|_0\|u_{\alpha+\beta}\|_0+\|(a_1\nabla
u)_{\alpha} -a_1\nabla u_{\alpha} \|_0 \|\nabla u _{\alpha}
\|_0\\
&\quad +\|(\mathbf{v}\cdot \nabla
u)_{\alpha}-\mathbf{v}\cdot \nabla u_{\alpha}\|_0\|u_{\alpha}\|_{0}  \\
&\leq   C\|f\|_{k-1}\|\nabla u\|_{k}+ C\|Da_1\|_{k_1}\|\nabla u
\|_{k-1}\|\nabla u \|_k + C\|D\mathbf{v}\|_{k_1}\|\nabla u
\|_{k-1}^2   \\
&\leq  \frac{C}{4\epsilon}\|f\|_{k-1}^2+\epsilon\| \nabla u
\|_{k}^2+ \frac{C}{4\epsilon}\|Da_1\| _{k_1}^2\|\nabla u
\|_{k-1}^2+\epsilon\| \nabla u \|_k^2+
C\|D\mathbf{v}\|_{k_1}\|\nabla u \|_{k-1}^2
\end{aligned}\label{B.16}
\end{equation}
where $|\beta|  = 1$, $k=|\alpha |$, and $k_1=\max \{k-1,s_0\}$,
with  $s_0=[\frac N2 ]+1$. Again, we used Cauchy's inequality with
$\epsilon$. Substituting \eqref{B.16} into \eqref{B.12}, and
adding \eqref{B.12} over $|\alpha |=k\leq r$, including the
estimate \eqref{B.9}, we obtain for $r\geq 1$ the estimate
\begin{equation}
\|\nabla u \|_r^2 \leq  \frac{C}{4\epsilon}(\|Da_1\|_{r_1}^2+
\|D\mathbf{v}\|_{r_1}) \|\nabla u \|_{r-1}^2
+\frac{C}{4\epsilon}\|f\|_{r-1}^2 +\epsilon C\|\nabla u\|_{r}^2
\label{B.17}
\end{equation}
where $r_1=\max \{r-1,s_0\}$, with $s_0=[\frac N2 ]+1$, and where
$C$ depends on $r$, $c_1$. Here we used Sobolev's lemma to obtain
$|D \mathbf{v}|_{L^\infty}\leq C\|D\mathbf{v}\|_{s_0}$. Choosing
$\epsilon$ sufficiently small yields
\begin{equation}
\|\nabla u \|_r^2 \leq
C\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\} \|\nabla u \|
_{r-1}^2 +C\|f\|_{r-1}^2 \label{B.18}
\end{equation}
where $C$ depends on $r$, $c_1$. This is the desired inequality
\eqref{B.31}.

Applying the inequality \eqref{B.18} to $ \|\nabla u \| _{r-1}^2$
which appears on the right-hand side of \eqref{B.18} yields
\begin{equation}
\begin{aligned}
\|\nabla u \|_r^2
&\leq C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})\\
&\quad\times \Big[C(\max\{\|Da_1\|_{r_2}^2,\|D\mathbf{v}\|_{r_2}\})
 \|\nabla u \|_{r-2}^2 +C\|f\|_{r-2}^2 \Big]
+C\|f\|_{r-1}^2   \\
&\leq  C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^2
\|\nabla u \|_{r-2}^2\\
&\quad +C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})
\|f\|_{r-2}^2 +C\|f\|_{r-1}^2
\end{aligned}\label{B.23}
\end{equation}
where $r_1=\max \{r-1,s_0\}$, $r_2=\max \{r-2,s_0\}$, $ r_2 \leq
r_1$, with $s_0=[\frac N2 ]+1=2$ for $N=2,3$.

Similarly, by  applying the estimate \eqref{B.18} to
$\|\nabla u \| _{r-j}^2$ for $j=2,3,\dots ,r-1$, which will appear
in the term $
C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{j}\|\nabla u \|
_{r-j}^2$ on the right-hand side of \eqref{B.23}, we obtain
\begin{equation}
\begin{aligned}
\|\nabla u \|_r^2
&\leq C\sum_{j=1}^{r-1}(\max\{\|Da_1\|_{r_1}^2,
\|D\mathbf{v}\|_{r_1}\})^{j} \|f\|_{r-1-j}^2  \\
&\quad +C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{r}
\|\nabla u \|_{0}^2 +C\|f\|_{r-1}^2    \\
&\leq C\Big[\sum_{j=0}^{r-1}(\max\{\|Da_1\|_{r_1}^2,
\|D\mathbf{v}\|_{r_1}\})^{j}\Big]\|f\|_{r-1}^2
 \\
&\quad +C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{r}
\|\nabla u \|_{0}^2
\end{aligned} \label{B.24}
\end{equation}
Substituting the estimate  $\|\nabla u\|_0^2 \leq C\|f\|_{0}^2$
into \eqref{B.24} yields
\begin{align*}
&\|\nabla u \|_r^2 \\
&\leq C
\Big[\sum_{j=0}^{r-1}(\max\{\|Da_1\|_{r_1}^2,
 \|D\mathbf{v}\|_{r_1}\})^{j}\Big]\|f\|_{r-1}^2
+C(\max\{\|Da_1\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{r}\|f\|_0^2 \\
&\leq C_1\Big[\sum_{j=0}^{r}(\max\{\|Da_1\|_{r_1}^2,
\|D\mathbf{v}\|_{r_1}\})^{j}\Big] \|f\|_{r-1}^2
\end{align*}
where $C_1$ depends on $r$, $c_1$. This completes the proof.
\end{proof}

\begin{lemma} \label{LB.8}
 Let $a(u)$ be a smooth function of $u$, and
let $\mathbf{v}(\mathbf{x})$ and $f(\mathbf{x})$ be sufficiently
smooth functions in the  equation
\begin{equation} \label{B.99}
-\nabla \cdot (a(u)\nabla u)+\mathbf{v}\cdot \nabla u =f
\end{equation}
for $\mathbf{x} \in \Omega$, where $\Omega = \mathbb{T}^N$,
$N=2,3$, where $a(u) > c_1$, for some positive constant $c_1$, and
where $|u-u_0|_{L^\infty}\leq R$, where $u_0$, $R$ are given
constants. We assume that $|\nabla \cdot \mathbf{v}|_{L^\infty}
\leq \frac{c_1}{ C_*}$, where $C_*$ is the constant from
Poincar\'{e}'s inequality $\|\bar{u} \|_0^2\leq C_*\|\nabla u
\|_0^2$, and where
$\bar{u}(\mathbf{x})=u(\mathbf{x})-\frac{1}{|\Omega|}\int_{\Omega}
u(\mathbf{x})  d\mathbf{x}$. Then there exist constants $C_7$,
$C_8$, such that if $\Big| \frac{d a}{d u} \big|_{s,\bar{G}_1}^2\|
f\|_{s-1}^2 \leq \frac{1}{C_7}$, and if $\|D\mathbf{v}\|_{s} \leq
\frac 12$ , and if $a''(u)\leq
\frac{1}{a(u)}(a'(u))^2$, then $u$ satisfies the
 inequality
\begin{equation}
\|\nabla u\|_{s}^2 \leq 2 C_8 \| f\|_{s-1}^2 \label{B.88}
\end{equation}
We define $C_7= 4 C_3^2 C_8^2$ and $C_8=C_0 C_1 K_1$, where $C_1$
is the constant from estimate \eqref{B.27} in Lemma \ref{LB.6},
and where $C_2=C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$ and $C_3 = M C_2$ are the same
constants as in \eqref{e3.8} from Proposition \ref{P3.1}, and
$C_0$ is a constant which depends on $s$, $c_1$, where $s
> \frac N2+ 1$. We define the constant $K_1=\max\{1,|\mathbf{v}|_{L^\infty}^2 \}$.
And we define $ |\frac{da}{du}
|_{s,\bar{G}_1}=\max\{\big|\frac{d^{j+1} a}{d u^{j+1}} (u_{*})
\Big|: u_{*} \in \bar{G}_1, 0 \leq j \leq s \}$.
\end{lemma}

\begin{proof}
First we obtain estimates for $\|\nabla u\|_{0}^2$, $\|\nabla
u\|_{1}^2$, $\|\nabla u\|_{2}^2$, and $\|\nabla u\|_{r}^2$, where
$3 \leq r \leq s$. It is necessary to have an estimate for
$\|\nabla u\|_{j}^2$ in order to obtain an estimate for $\|\nabla
u\|_{j+1}^2$, for $j=0,1,2,\dots ,s-1 $. We will apply estimate
\eqref{B.27} from Lemma \ref{LB.6} to obtain an estimate for
$\|\nabla u\|_{r}^2$, when $3 \leq r \leq s$.

 From inequality \eqref{B.30} in Lemma \ref{LB.6} applied to
equation \eqref{B.99}, we obtain
\begin{equation} \label{e3.24}
\|\nabla u\|_0^2\leq C\|f\|_{0}^2
\end{equation}
where $C$ depends on $c_1$. We now obtain an estimate for
$\|\nabla u\|_{1}^2$.
Applying $D^\alpha$ to equation \eqref{B.99}, with $|\alpha|=1$,
yields
\begin{equation}
-\nabla \cdot(a(u)\nabla u_\alpha)  =  \nabla \cdot
((a(u))_\alpha \nabla u)-(\mathbf{v}\cdot \nabla
u)_\alpha+f_\alpha \label{e3.26}
\end{equation}
Integrating \eqref{e3.26} by parts with $u_\alpha$, where
$|\alpha|=1$, and using the fact from equation \eqref{B.99} that
$\nabla a(u) \cdot \nabla u =$ $-a(u)\Delta u+\mathbf{v}\cdot
\nabla u -f $, yields
\begin{equation}
\begin{aligned}
(a(u)\nabla u_\alpha,\nabla u_\alpha)
&= -(\nabla \cdot(a(u)\nabla u_\alpha),u_\alpha)  \\
&=  (\nabla \cdot ((a(u))_\alpha \nabla
u),u_\alpha)-((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha)+
(f_\alpha,u_\alpha)  \\
&=  -((a(u))_\alpha \nabla u,\nabla u_\alpha) -((\mathbf{v}\cdot
\nabla u)_\alpha, u_\alpha)+
(f_\alpha,u_{\alpha})  \\
&=   -\frac 12((a(u))_\alpha, (\nabla u \cdot \nabla
u)_\alpha)-((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha) +
(f_\alpha,u_{\alpha})  \\
&=  -\frac 12(a^{\prime}(u) u_\alpha, (\nabla u \cdot \nabla
u)_\alpha) -((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha)+
(f_\alpha,u_{\alpha})  \\
&=  -\frac 12(u_\alpha, ( a^{\prime}(u)\nabla u \cdot \nabla
u)_\alpha)  +\frac 12( u_\alpha, (a^{\prime}(u))_\alpha(\nabla u
\cdot \nabla u)) \\
&\quad -((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha)+
(f_\alpha,u_{\alpha})  \\
&=  -\frac 12(u_\alpha, (\nabla a(u) \cdot \nabla u)_\alpha)
+\frac 12( u_\alpha , a''(u) u_\alpha(\nabla u \cdot
\nabla u))  \\
&\quad -((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha)+
(f_\alpha,u_{\alpha})  \\
&=  \frac 12(u_\alpha, (a(u)\Delta u-\mathbf{v}\cdot \nabla
u+f)_\alpha) +\frac 12((u_\alpha)^2, a''(u)(\nabla u
\cdot \nabla u)) \\
&\quad -((\mathbf{v}\cdot \nabla u)_\alpha, u_\alpha) +
(f_\alpha,u_{\alpha})  \\
&=  -\frac 12(u_{\alpha+\alpha},a(u) \Delta u)+\frac
32((\mathbf{v}\cdot \nabla u), u_{\alpha+\alpha})-\frac 32
(f,u_{\alpha+\alpha}) \\
&\quad +\frac 12((u_\alpha)^2,a''(u) (\nabla u \cdot
\nabla u))
\end{aligned}\label{e3.27}
\end{equation}
Adding \eqref{e3.27} over $|\alpha|=1$ yields
\begin{equation}
\begin{aligned}
\sum_{|\alpha|=1} (a(u)\nabla u_\alpha,\nabla u_\alpha)
&=  -\frac 12\sum_{|\alpha|=1}(u_{\alpha+\alpha},a(u) \Delta u)
 +\frac 32 \sum_{|\alpha|=1}((\mathbf{v}\cdot \nabla u),
 u_{\alpha+\alpha})\\
&\quad -\frac 32 \sum_{|\alpha|=1} (f,u_{\alpha+\alpha})
 + \frac 12\sum_{|\alpha|=1} ((u_\alpha)^2, a''(u)(\nabla u
  \cdot \nabla u)) \\
&= -\frac 12 ( \Delta u,a(u) \Delta u) +\frac 32((\mathbf{v}\cdot
\nabla u), \Delta u)-\frac 32 (f,\Delta
u)  \\
&\quad +\frac 12((\nabla u \cdot \nabla u),a''(u) (\nabla
u \cdot \nabla u))
\end{aligned} \label{e3.28}
\end{equation}
Next, we  estimate the term $\frac 12((\nabla u \cdot \nabla
u),a''(u) (\nabla u \cdot \nabla u))$ in
\eqref{e3.28}. We assume that $a''(u)\leq
\frac{1}{a(u)}(a'(u))^2$. We then obtain the inequality
\begin{align}
&\frac 12((\nabla u \cdot \nabla u),a''(u) (\nabla u
\cdot \nabla u)) \nonumber \\
&\leq  \frac 12
(\frac{1}{a(u)}(a^{\prime}(u))^2(\nabla u \cdot \nabla
u), (\nabla u \cdot \nabla u)) \nonumber\\
&=  \frac 12((\nabla a(u) \cdot \nabla
u), \frac{1}{a(u)}(\nabla a(u) \cdot \nabla u)) \\
&= \frac 12((a(u)\Delta u -\mathbf{v}\cdot \nabla u+f),
\frac{1}{a(u)} (a(u)\Delta u-\mathbf{v}\cdot \nabla u+f)) \nonumber\\
&= \frac 12(\Delta u,a(u) \Delta u)+\frac
12(f,\frac{1}{a(u)}f)+\frac 12(\mathbf{v}\cdot \nabla
u,\frac{1}{a(u)}\mathbf{v}\cdot \nabla u)
\label{e3.29} \\
&\quad +(\Delta u, f)-(\mathbf{v}\cdot \nabla u, \Delta
u)-(\mathbf{v}\cdot \nabla u,\frac{1}{a(u)} f) \nonumber
\end{align}
Substituting \eqref{e3.29} into \eqref{e3.28} yields
\begin{equation}
\begin{aligned}
\sum_{|\alpha|=1} (a(u)\nabla u_\alpha,\nabla u_\alpha)
&=  \frac
12(f,\frac{1}{a(u)}f) +\frac 12(\mathbf{v}\cdot \nabla
u,\frac{1}{a(u)}\mathbf{v}\cdot \nabla u) \\
&\quad -\frac 12(\Delta u, f)+\frac 12(\mathbf{v}\cdot \nabla u, \Delta
u)-(\mathbf{v}\cdot
\nabla u,\frac{1}{a(u)} f)  \\
& \leq  \frac {1}{c_1}(f,f) +
\frac{1}{c_1}|\mathbf{v}|_{L^\infty}^2\|\nabla
u\|_0^2+\epsilon(\Delta u, \Delta u) \\
&\quad +\frac {1}{16\epsilon}(f, f)
+\frac{1}{16\epsilon}|\mathbf{v}|_{L^\infty}^2\|\nabla u\|_0^2
+\epsilon(\Delta u, \Delta u)
\end{aligned}\label{e3.30}
\end{equation}
where we used Cauchy's inequality with $\epsilon$, and we used the
fact that $a(u)>c_1$.

We now use the fact that $\sum_{|\alpha|=1} (\nabla
u_\alpha,\nabla u_\alpha)$ $= \sum_{|\alpha|=1}
((u_{\alpha+\alpha},\Delta u)$ $=(\Delta u,\Delta u)$. We also use
the fact that $a(u )>c_1$, and we define $\epsilon
=\frac{c_1}{4}$. Then \eqref{e3.30} becomes
\begin{equation}
\begin{aligned}
&\sum_{|\alpha|=1} (c_1\nabla u_\alpha,\nabla u_\alpha)\leq
\sum_{|\alpha|=1} (a(u)\nabla
u_\alpha,\nabla u_\alpha)  \\
& \leq  \frac{c_1}{2}(\Delta u, \Delta u)+\frac {5}{ 4c_1}(f, f)+
\frac {5}{4c_1}|\mathbf{v}|_{L^\infty}^2\|\nabla
u\|_0^2 \\
&= \frac{c_1}{2}\sum_{|\alpha|=1} ( \nabla u_\alpha,\nabla
u_\alpha)+\frac {5}{4 c_1}(f, f)+ \frac{5}{4
c_1}|\mathbf{v}|_{L^\infty}^2\|\nabla u\|_0^2 \label{e3.31}
\end{aligned}
\end{equation}
Subtracting the term $\frac {c_1}{2}\sum_{|\alpha|=1}(\nabla
u_\alpha, \nabla u_\alpha) $ on both sides of \eqref{e3.31}, and
using inequality \eqref{e3.24}, namely $\|\nabla u\|_0^2\leq
C\|f\|_{0}^2$, yields the estimate
\begin{equation}
\sum_{|\alpha|=1} (\nabla u_\alpha,\nabla u_\alpha) \leq
C\max\{1,|\mathbf{v}|_{L^\infty}^2\}\|f\|_0^2 \label{e3.32}
\end{equation}
where $C$ depends on $c_1$. Adding the inequalities \eqref{e3.32},
\eqref{e3.24}  yields
\begin{equation}
\|\nabla u\|_{1}^2=\|\nabla u\|_{0}^2 +\sum_{|\alpha|=1} (\nabla
u_\alpha,\nabla u_\alpha) \leq C \max\{1,
|\mathbf{v}|_{L^\infty}^2 \}\|f\|_0^2 = C K_1 \|f\|_0^2
\label{e3.33}
\end{equation}
where $C$ depends on $c_1$, and where we define the constant
$K_1=\max\{1, |\mathbf{v}|_{L^\infty}^2 \}$.

We now obtain an estimate for $\|\nabla u\|_{2}^2$. Applying
$D^\alpha$ to  equation \eqref{B.99}, with $|\alpha|=2$, yields
\begin{equation}
\begin{aligned}
-\nabla \cdot(a(u)\nabla u_\alpha)
&= \nabla \cdot ((a(u))_\alpha
\nabla u)+\nabla \cdot ((a(u))_{\alpha-\beta} \nabla
u_{\beta}) +\nabla \cdot ((a(u))_{\beta} \nabla u_{\alpha-\beta})\\
&\quad -\mathbf{v}\cdot \nabla
u_\alpha-\mathbf{v}_\alpha\cdot \nabla u-\mathbf{v}_{\beta}\cdot
\nabla u_{\alpha-\beta}-\mathbf{v}_{\alpha-\beta}\cdot \nabla
u_{\beta}+f_\alpha
\end{aligned} \label{e3.34}
\end{equation}
where $|\beta| =1$. Integrating by parts with $u_\alpha$, where
$|\alpha| =2$ and $|\beta| =1$, and using inequality \eqref{3.3}
from Lemma \ref{LB.1}, yields
\begin{align}
&((a(u)\nabla u_\alpha,\nabla u_\alpha) \nonumber \\
&=-( \nabla
\cdot (a(u)\nabla u_\alpha),u_\alpha) \nonumber \\
&= (\nabla \cdot ((a(u))_\alpha \nabla u),u_\alpha)+(\nabla \cdot
((a(u))_{\alpha-\beta} \nabla u_{\beta}),u_\alpha) \nonumber \\
&\quad + (\nabla \cdot ((a(u))_{\beta} \nabla
u_{\alpha-\beta}),u_\alpha)-(\mathbf{v}\cdot \nabla u_\alpha,
u_\alpha) \nonumber \\
&\quad- (\mathbf{v}_\alpha\cdot \nabla u,
u_\alpha)-(\mathbf{v}_{\beta}\cdot \nabla
u_{\alpha-\beta},u_\alpha)-(\mathbf{v}_{\alpha-\beta}\cdot \nabla
u_{\beta},u_{\alpha})
+(f_\alpha,u_\alpha) \nonumber \\
&= -((a(u))_\alpha \nabla u,\nabla u_\alpha)
-((a(u))_{\alpha-\beta} \nabla u_{\beta},\nabla u_\alpha) \nonumber\\
&\quad- ((a(u))_{\beta} \nabla u_{\alpha-\beta},\nabla u_\alpha)+\frac
12 ((\nabla \cdot\mathbf{v}) u_\alpha, u_\alpha)
-(\mathbf{v}_\alpha\cdot \nabla u,
u_\alpha) \nonumber\\
&\quad- (\mathbf{v}_{\beta}\cdot \nabla
u_{\alpha-\beta},u_\alpha)-(\mathbf{v}_{\alpha-\beta}\cdot \nabla
u_{\beta},u_{\alpha})
-(f_{\alpha-\beta},u_{\alpha+\beta}) \nonumber\\
&\leq  \|(a(u))_\alpha\|_0|\nabla u|_{L^\infty}\|\nabla
 u_\alpha\|_0+|(a(u))_{\alpha-\beta}|_{L^\infty}\|\nabla
u_{\beta}\|_0\|\nabla u_\alpha\|_0  \nonumber\\
&\quad + |(a(u))_\beta|_{L^\infty}\|\nabla u_{\alpha-\beta}\|_0\|\nabla
u_\alpha\|_0+\frac 12|\nabla \cdot \mathbf{v}|_{L^\infty}\|
u_\alpha\|_0^2 +|\mathbf{v}_\alpha|_{L^\infty}\|
\nabla u\|_0\| u_\alpha \|_0 \nonumber \\
&\quad + |\mathbf{v}_\beta|_{L^\infty}\|\nabla
u_{\alpha-\beta}\|_0\|u_{\alpha}\|_0
+|\mathbf{v}_{\alpha-\beta}|_{L^\infty}\| \nabla u_\beta\|_0\|
u_{\alpha}\|_0 + \|f_{\alpha-\beta}\|_{0}\|u_{\alpha+\beta}\|_{0}
\nonumber \\
&\leq  C\|D^2(a(u))\|_0\|\nabla u\|_{2}\|\nabla
u_\alpha\|_0+C\|D(a(u))\|_2\|\nabla u\|_{1}\|\nabla u_\alpha\|_0
+C|\nabla \cdot \mathbf{v}|_{L^\infty}\| \nabla u\|_1^2  \nonumber\\
&\quad + C|D^2\mathbf{v}|_{L^\infty}\| \nabla u\|_0\|\nabla u\|_1 +C|D
\mathbf{v}|_{L^\infty}\|\nabla u\|_1^2 +C \|\nabla f\|_{0}\|\nabla
u_{\alpha}\|_{0}
\nonumber \\
&\leq  \frac{C}{4\epsilon}\|D^2(a(u))\|_0^2\|\nabla
u\|_{2}^2+\epsilon\|\nabla
u_\alpha\|_0^2+\frac{C}{4\epsilon}\|D(a(u))\|_2^2\|\nabla
u\|_{1}^2+\epsilon\|\nabla u_\alpha\|_0^2 \nonumber\\
&\quad +C|D \mathbf{v}|_{L^\infty}\| \nabla u\|_1^2
 + C|D^2\mathbf{v}|_{L^\infty}\| \nabla
 u\|_0^2+C|D^2\mathbf{v}|_{L^\infty}\|\nabla u\|_1^2 \nonumber\\
&\quad +C|D
 \mathbf{v}|_{L^\infty}\|\nabla u\|_1^2
 + \frac{C}{4\epsilon}\|\nabla f\|_{0}^2+\epsilon\|\nabla
u_{\alpha}\|_{0}^2  \nonumber \\
&\leq  \frac{C}{4\epsilon}\|D^2(a(u))\|_0^2\|\nabla u\|_{2}^2
 +\frac{C}{4\epsilon}\Big[\sum_{0\leq r \leq 2}\|D^{r+1}(a(u))\|_0^2
\Big]\|\nabla u\|_{1}^2  \nonumber \\
&\quad + C(|D\mathbf{v}|_{L^\infty}+|D^2\mathbf{v}|_{L^\infty})\| \nabla
u\|_1^2 +C|D^2\mathbf{v}|_{L^\infty}\|\nabla u\|_0^2
+\frac{C}{4\epsilon}\|\nabla f\|_{0}^2+3\epsilon\|\nabla
u_{\alpha}\|_{0}^2  \nonumber\\
&\leq  \frac{C}{4\epsilon} \big|\frac{d a}{d u}
\big|_{1,\bar{G}_1}^2(1+|u|_{L^\infty})^2\|\nabla u\|_1^2 \|\nabla
u\|_{2}^2  \nonumber \\
&\quad + \frac{C}{4\epsilon}\Big[\sum_{0\leq r \leq 2
}\big|\frac{d a}{d u}
\big|_{r,\bar{G}_1}^2(1+|u|_{L^\infty})^{2r}\|\nabla
u\|_r^2 \Big]\|\nabla u\|_{1}^2  \nonumber \\
&\quad + C(|D\mathbf{v}|_{L^\infty}+|D^2\mathbf{v}|_{L^\infty})\| \nabla
u\|_1^2 +C|D^2\mathbf{v}|_{L^\infty}\|\nabla u\|_0^2 \nonumber\\
&\quad + \frac{C}{4\epsilon}\|\nabla f\|_{0}^2+3\epsilon\|\nabla
u_{\alpha}\|_{0}^2
  \label{e3.35} \\
&\leq  \frac{C}{4\epsilon} \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u|_{L^\infty})^{2s}\|\nabla u\|_1^2
\|\nabla u\|_{2}^2 \nonumber\\
&\quad + \frac{C}{4\epsilon}\big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u|_{L^\infty})^{2s}\Big[\sum_{0\leq r
\leq 2 }\|\nabla u\|_r^2 \Big]\|\nabla u\|_{1}^2  \nonumber\\
&\quad + C(\|D\mathbf{v}\|_{s_0}+\|D^2\mathbf{v}\|_{s_0})\| \nabla
u\|_1^2 +C\|D^2\mathbf{v}\|_{s_0}\|\nabla u\|_0^2+
\frac{C}{4\epsilon}\|\nabla f\|_{0}^2+3\epsilon\|\nabla
u_{\alpha}\|_{0}^2  \nonumber \\
&\leq  \frac{C}{4\epsilon} \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u|_{L^\infty})^{2s}\|\nabla u\|_1^2
\|\nabla u\|_{2}^2 + \frac{C}{4\epsilon}\big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u|_{L^\infty})^{2s}\|\nabla u\|_2^2
\|\nabla u\|_{1}^2  \nonumber \\
&\quad + C\|D\mathbf{v}\|_{s}\| \nabla u\|_1^2
+C\|D\mathbf{v}\|_{s}\|\nabla u\|_0^2 +
\frac{C}{4\epsilon}\|\nabla f\|_{0}^2+3\epsilon\|\nabla
u_{\alpha}\|_{0}^2  \nonumber \\
&\leq  \frac{C}{2\epsilon} \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+|u-u_0|_{L^\infty}+|u_0|_{L^\infty})^{2s}\|\nabla
u\|_1^2 \|\nabla u\|_{2}^2  \nonumber \\
&\quad + C\|D\mathbf{v}\|_{s}\| \nabla u\|_1^2
+C\|D\mathbf{v}\|_{s}\|\nabla u\|_0^2 +
\frac{C}{4\epsilon}\|\nabla f\|_{0}^2+3\epsilon\|\nabla
u_{\alpha}\|_{0}^2  \nonumber \\
&\leq  \frac{C}{2\epsilon} \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_0)|)^{2s}\|\nabla u\|_1^2
\|\nabla u\|_{2}^2  \nonumber \\
&\quad + C\|D\mathbf{v}\|_{s}\| \nabla u\|_1^2
+C\|D\mathbf{v}\|_{s}\|\nabla u\|_0^2+ \frac{C}{4\epsilon}\|\nabla
f\|_{0}^2+3\epsilon\|\nabla u_{\alpha}\|_{0}^2 \nonumber
\end{align}
where we used inequality \eqref{3.3} from Lemma \ref{LB.1}. We
also used Sobolev's lemma to obtain $|D \mathbf{v}|_{L^\infty}\leq
C\|D\mathbf{v}\|_{s_0}$ and $|D^2 \mathbf{v}|_{L^\infty}\leq
C\|D^2\mathbf{v}\|_{s_0}$, where $s_0=[\frac N2 ]+1=2$ and $s\geq
3$. We also used Cauchy's inequality with $\epsilon$.

We assume that $\Big| \frac{d a}{d u} \big|_{s,\bar{G}_1}^2\|
f\|_{s-1}^2 \leq  \frac{1}{C_7}$, where the constant $C_7$ will be
defined later. And we assume that $\|D\mathbf{v}\|_{s} \leq \frac
12$. Substituting estimates \eqref{e3.33}, \eqref{e3.24} for
$\|\nabla u\|_1^2$, $\|\nabla u\|_0^2$ into \eqref{e3.35}, and
using the fact that $a(u )> c_1$, and letting
$\epsilon=\frac{c_1}{6}$, yields
\begin{equation}
\begin{aligned}
&(c_1\nabla u_\alpha,\nabla u_\alpha)\\
&\leq ((a(u)\nabla u_\alpha,\nabla u_\alpha) \\
& \leq  C K_1\big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_0)|)^{2s}\|f\|_0^2
\|\nabla u\|_{2}^2+ C K_1\|D\mathbf{v}\|_{s}\|
f\|_{0}^2 \\
&\quad +  C \|D\mathbf{v}\|_{s}\|
f\|_{0}^2+C\|f\|_{1}^2+\frac{c_1}{2}\|\nabla
 u_{\alpha}\|_{0}^2  \\
& \leq  C K_1\big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2(1+R+|u(\mathbf{x}_0)|)^{2s}\|f\|_{s-1}^2
\|\nabla u\|_{2}^2 + C K_1\| f\|_{1}^2+\frac{c_1}{2}\|\nabla
 u_{\alpha}\|_{0}^2  \\
&\leq  C
K_1\Big(\frac{1}{C_7}\Big)(1+R+|u(\mathbf{x}_0)|)^{2s}\|\nabla
u\|_{2}^2+ C K_1\| f\|_{1}^2+\frac{c_1}{2}\|\nabla
 u_{\alpha}\|_{0}^2
\end{aligned} \label{e3.36}
\end{equation}
where $C$ depends on $s$, $c_1$, and where we used the facts that
$\|D\mathbf{v}\|_{s} <1$, and that $K_1=\max\{1,
|\mathbf{v}|_{L^\infty}^2 \}$.

Adding \eqref{e3.36}  over all $|\alpha|=2$ after moving the term
$\frac{c_1}{2}\|\nabla u_{\alpha}\|_{0}^2 $ to the left-hand side,
and adding the estimate \eqref{e3.33} for $\|\nabla u\|_1^2$
yields
\begin{equation}
\begin{aligned}
\|\nabla u\|_{2}^2
&=  \|\nabla u\|_1^2+\sum_{|\alpha|=2}(\nabla
 u_\alpha,\nabla  u_\alpha)  \\
& \leq  C
K_1\Big(\frac{1}{C_7}\Big)(1+R+|u(\mathbf{x}_0)|)^{2s}\|\nabla
u\|_{2}^2 + CK_1\|
f\|_{1}^2 \\
& \leq C K_1\Big(\frac{1}{C_7}\Big)(1+R+(1+|\Omega|^{1/2}) |u(\mathbf{x}_0)|)^{2s}\|\nabla u\|_{2}^2 + CK_1\|
f\|_{1}^2 \\
& \leq \Big(\frac{C_0 C_2 K_1}{C_7}\Big)\|\nabla u\|_{2}^2 +
C_0 K_1\| f\|_{1}^2  \\
& \leq \Big(\frac{C_0 C_3 K_1}{C_7}\Big)\|\nabla u\|_{2}^2 +
C_0 K_1\| f\|_{1}^2
\end{aligned}\label{e3.37}
\end{equation}
where the constant $C_0$ depends on $s$, $c_1$, and
$C_2=C(1+R+(1+|\Omega|^{1/2})|u(\mathbf{x}_0)|)^{2s}$ and
$C_3=M C_2$ are the same constants as in \eqref{e3.8} from
Proposition \ref{P3.1}, and $M\geq 1$.

We now define $C_7= 4 C_0^2 C_3^2 C_1^2 K_1^2$, where $C_0$ is the
constant from \eqref{e3.37}, and where $C_1$ is the constant from
estimate \eqref{B.27} in Lemma \ref{LB.6}, and we may assume that
$C_1 \geq 1$, $C_0 \geq 1$, and $C_3 \geq 1$. Substituting the
definition of $C_7$ into \eqref{e3.37} yields
\begin{equation}
\|\nabla u\|_{2}^2 \leq   \frac{1}{4 C_0 C_1^2 C_3 K_1}\|\nabla
u\|_{2}^2 + C_0 K_1\| f\|_{1}^2  \leq \frac 12 \|\nabla u\|_{2}^2+
C_0 K_1\| f\|_{1}^2 \label{e3.38}
\end{equation}
where we used  that $K_1 \geq 1$. We define $C_8=C_0 C_1
K_1$. It follows from \eqref{e3.38} that
\begin{equation}
 \|\nabla u\|_{2}^2 \leq 2 C_0 K_1\|f\|_{1}^2
 \leq  2 C_8\|f\|_{1}^2\,.
 \label{e3.39}
\end{equation}
Note that since $C_8=C_0 C_1 K_1$ we have $C_7= 4 C_3^2 C_8^2$.

Next we estimate $\|\nabla u\|_{r}^2$, where $3 \leq r \leq s$.
Using estimate \eqref{B.27} from Lemma \ref{LB.6} in Appendix B
applied to equation \eqref{B.99} yields
\begin{equation}
\begin{aligned}
\|\nabla u\|_{r}^2  &\leq  C_1 \Big[\sum_{j=0}^{r}(\max \{\|D(a(u
))\|_{r_1}^2,\|D\mathbf{v}\|_{r_1}\})^{j}\Big] \|f\|_{r-1}^2
 \\
&\leq  C_8\Big[\sum_{j=0}^{r}(\max \{\|D(a(u
))\|_{r-1}^2,\|D\mathbf{v}\|_{r-1}\})^{j}\Big]
\|f\|_{r-1}^2,
\end{aligned}\label{e3.40}
\end{equation}
where $r_1=\max\{r-1,s_0\}=r-1$,  and $s_0=[ \frac N2]+1=2$.


We consider two cases: when $\max\{\|D(a(u ))\|_{r-1}^2,
\|D\mathbf{v}\|_{r-1}\}=\|D(a(u ))\|_{r-1}^2$, and when
$\max\{\|D(a(u ))\|_{r-1}^2,
\|D\mathbf{v}\|_{r-1}\}=\|D\mathbf{v}\|_{r-1}$.

\textbf{Case 1}: Suppose that $\max\{\|D(a(u))\|_{r-1}^2,
\|D\mathbf{v}\|_{r-1}\}=\|D(a(u ))\|_{r-1}^2$.
 From \eqref{e3.8} in Proposition \ref{P3.1}, we have
 $ \|D(a(u))\|_{r-1}^2 \leq C_3 \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\|\nabla u\|_{r-1}^2$. Repeatedly applying
estimate \eqref{e3.40}, letting $r=3,4,\dots ,s$ and using the fact
that $\|\nabla u\|_{r-1}^2 \leq 2C_8\|f\|_{r-2}^2$, and using
estimate \eqref{e3.8} for $\|D(a(u ))\|_{r-1}^2$, and using the
fact that $r_1=\max\{r-1,s_0\}=r-1$ when $r\geq 3$, we obtain
\begin{equation}
\begin{aligned}
\|\nabla u \|_{r}^2
&\leq  C_8\Big[\sum_{j=0}^{r}\|D(a(u ))\|_{r-1}^{2j}\Big]
\|f\|_{r-1}^2  \\
&\leq  C_8\Big[\sum_{j=0}^{r} C_3^j \big|\frac{d a}{d u}
\big|_{s,\bar{G}_1}^{2j}\|\nabla u\|_{r-1}^{2j}\Big]\|f\|_{r-1}^2  \\
&\leq  C_8 \Big[\sum_{j=0}^{r}C_3^j( 2 C_8)^{j} \big|\frac{d a}{d
u}\big|_{s,\bar{G}_1}^{2j}
\|f\|_{r-2}^{2j}\Big]\|f\|_{r-1}^2  \\
&\leq  C_8 \Big[\sum_{j=0}^{s}C_3^j(2 C_8)^{j} \big|\frac{d a}{d
u}\big|_{s,\bar{G}_1}^{2j}
\|f\|_{s-1}^{2j}\Big]\|f\|_{r-1}^2  \\
&\leq   C_8 \Big[\sum_{j=0}^{s}C_3^j(2 C_8)^j \Big(\frac{1}{C_7}
\Big)^j\Big]\|f\|_{r-1}^2  \\
&\leq  C_8\Big[\sum_{j=0}^{s} \big(\frac 12\big)^j\Big] \|f\|_{r-1}^2 \\
&\leq  2 C_8\|f\|_{r-1}^2
\end{aligned} \label{e3.80}
\end{equation}
where we used the fact that $\big| \frac{d a}{d u}
\big|_{s,\bar{G}_1}^2\| f\|_{s-1}^2 \leq \frac{1}{C_7}$, and $C_7=
4 C_3^2 C_8^2$, and $C_3 C_8 \geq 1$.

\textbf{Case 2}: Suppose that $\max\{\|D(a(u))\|_{r-1}^2,
\|D\mathbf{v}\|_{r-1}\}=\|D \mathbf{v}\|_{r-1}$.
 From \eqref{e3.40}, and using the fact that
$\|D\mathbf{v}\|_{s}\leq \frac 12$, we obtain the following:
\[
\|\nabla u \|_{r}^2  \leq
C_8\Big[\sum_{j=0}^{r}\|D\mathbf{v}\|_{r-1}^{j}\Big] \|f\|_{r-1}^2
\leq C_8\Big[\sum_{j=0}^{s} \big(\frac 12\big)^j\Big]
\|f\|_{r-1}^2 \leq 2 C_8\|f\|_{r-1}^2
\]
for $3 \leq r \leq s$, which is the same estimate as
\eqref{e3.80}. Therefore we have $\|\nabla u \|_{r}^2 \leq 2
C_8\|f\|_{r-1}^2$ for $3 \leq r \leq s$. It follows that $\|\nabla
u\|_{s}^2 \leq 2 C_8 \|f\|_{s-1}^2$. This completes the proof.
\end{proof}

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\end{document}