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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 93, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/93\hfil Analytic solution]
{Analytic solution of an initial-value problem
from Stokes flow with free boundary}

\author[X. Xie\hfil EJDE-2008/93\hfilneg]
{Xuming Xie}

\address{Xuming Xie \newline
Department of Mathematics\\
Morgan State University\\
Baltimore, MD 21251, USA}
\email{xuming.xie@morgan.edu}

\thanks{Submitted March 5, 2008. Published July 2, 2008.}
\thanks{Supported by grant DMS-0500642 from National Science Foundation}
\subjclass[2000]{35Q72, 74F05, 80A22}
\keywords{Stokes flow; free boundary problem; analytic solution;
 \hfill\break\indent abstract Cauchy-Kovalevsky problem;
 initial-value problem}

\begin{abstract}
 We study an initial-value problem arising from Stokes flow
 with free boundary. If the initial data is analytic in disk
 $\mathcal{R}_r$ containing the unit disk, it is proved that
 unique  solution, which is  analytic in $\mathcal{R}_s$ for
 $s\in (1,r)$, exists locally in time.
\end{abstract}

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

\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\norm}[1]{\lVert#1\rVert}

\section{Introduction}

The study of the deformation and breakup of bubbles in a slow
viscous flow is of importance in many practical applications such
as the  rheology and mixing in multiphase viscous system. There
has been a lot of research on this subject, the review article by
Stone \cite{s1} summarizes the state of affair in the early
nineties. This problem has been recently studied by some
investigators. Tanveer and Vasconcelos \cite{t1,t2} obtained polynomial
exact solutions; Cummings et al \cite{c5} also obtained explicit
solutions and some conserved quantities. Crowdy and Siegel \cite{c2}
obtained new conserved quantities and exact solution based on
Cauchy transform approach. Nie et al \cite{n1} numerically studied the
singularity formation of the Stokes flow. Prokert \cite{p1} obtained
existence result of solutions in Sobolev space for a similar
problem.

 In this paper, we are going to establish a local
existence result for an initial-value problem arising from free
evolving bubble in Stokes flow. We first derive the initial-value
problem using complex variables theory \cite{c1}, then obtain the local
existence result based on a Nirenberg theorem \cite{n2,n3} on abstract
Cauchy-Kowalewski problem in properly chosen Banach spaces. The
same techniques have been used for other problems \cite{x1,x2}.

\section{Stokes flow with free boundary}

We consider the quasi-steady evolution of a bubble in an ambient
Stokes flow \cite{t1,t2}. The fluid inside the bubble has a negligible
viscosity and is at a constant pressure, which is set to be zero.
The fluid outside the bubble has a viscosity $\mu$ and is
incompressible. Under the assumption of no inertial effects,
gravitational or other body forces, the fluid motion is governed
by Stokes equation and the incompressibility condition
\begin{gather}\label{e1}
\mu \Delta u= \nabla p,\\
\label{e2} \nabla\cdot u=0
\end{gather}
The above equations hold in the fluid region outside of the
bubble.

On the bubble boundary, we have stress condition
\begin{equation}\label{e3}
-p n_j +2 \mu e_{jk} n_k = \tau \kappa n_j
\end{equation}
where $n=(n_1,n_2) $ is a unit normal vector pointing outward from
the bubble, $\tau$ is the surface tension coefficient, $\kappa$
is the curvature and
\begin{equation}\label{e4}
e_{jk}=\frac{1}{2}\big( \frac{\partial u_j}{\partial x_k}
+\frac{\partial u_k}{\partial x_j}\big)
\end{equation}
is the rate of strain tensor.

The kinematic condition on the free boundary is
\begin{equation}\label{e5}
u\cdot n=V_n,
\end{equation}
where $V_n$ is the normal component of the free surface motion.
Introducing streaming function $\psi(x,y)$ such that
\begin{equation}\label{e6}
u=\nabla^{\perp }\psi
\end{equation}
then $\psi (x,y)$ satisfies the biharmonic equation
\begin{equation}\label{e7}
\nabla^4 \psi =0\,.
\end{equation}
Here $\psi(x,y)$ can be expressed as
\begin{equation}\label{e8}
\psi=\mathop{\rm Im} [z^*f(z,t)+g(z,t)]
\end{equation}
where $z=x+iy$ and $*$ denotes complex conjugate. Here Goursat functions
$f(z,t)$ and $g(z,t)$ are analytic functions in the fluid region.

In terms of Goursat functions, the physical quantities are established
\begin{equation}\label{e9}
\begin{gathered}
\frac{p}{\mu}-i\omega =4 f'(z,t), \\
u=u_1+iu_2=-f(z,t)+z[f'(z)]^*+[g( z)]^*,\\
 e_{11}+ie_{12}=z[f''(z)]^*+[g''(z)]^*.
\end{gathered}
\end{equation}
where $*$ denotes the complex conjugate and $\omega$ is the vorticity.

Defining $s $ to be the arc-length traversed in a counterclockwise direction
around the bubble boundary, then the stress condition can be written as
\begin{equation}\label{e10}
f(z,t)+z[f'(z,t)]^*+[g'(z,t)]^*=-i\frac{z_s}{2},
\end{equation}
the kinematic equation can be written as
\begin{equation}\label{e11}
\mathop{\rm Im} [(z_t+2f)z^*_s] =-\frac{1}{2}.
\end{equation}
Equations \eqref{e10} and \eqref{e1} will be supplemented by
far field conditions on $f$ and $g$ at infinity.

\section{An initial-value problem}

We consider the conformal mapping $z(\xi,t)$ that maps the interior of
the unit circle $|\xi|<1$  in the $\xi$ plane to the fluid region in
$z$-plane such that the $\xi=0$ is mapped to the point $z=\infty$.
So $\xi z(\xi,t)$ is analytic in $|\xi|<1$. The kinematic condition can
be written as
\begin{equation}\label{e12}
\mathop{\rm Re} \big[\frac{z_t+2f(z,t)}{\xi z_\xi}\big]=\frac{\tau}{2|z_\xi|}.
\end{equation}
as in Tanveer and Vasconcelos \cite{t1}, we use the far field condition
\begin{equation}\label{e13}
f(z)\sim az+b+O(1/z) \quad \text{as } |z|\to \infty
\end{equation}
where $a$ and $b$ are functions of $t$ only. In particular,
we choose $f(z)= az +b$ where $a$ and $b$ are constants.
From Poisson's formula, \eqref{e12} becomes
\begin{equation}\label{e14}
z_t+2(az+b)=\xi z_\xi I_-(\xi,t) \text{ for } |\xi|<1;
\end{equation}
where $I_-(\xi,t)$ is defined by
\begin{equation}\label{e15}
I_-(\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'}
[ \frac{\xi'+\xi}{\xi'-\xi}] \frac{1}{|z_\xi|}.
\end{equation}
Let $h(\xi,t)=\xi z(\xi,t)$, then $h(\xi,t)$ is analytic in
$|\xi|<1$. If $h(\xi,t)$ can be analytically extended to some region where
$|\xi|>1$, then
\begin{equation}\label{e16}
h_t+2ah+2b\xi =[\xi h_\xi-h]I_+[h](\xi,t)
+\frac{\tau (\xi h_\xi-h)^{1/2}}{(\xi \bar h_\xi -\bar h)^{1/2}},
\end{equation}
where $\bar h(\xi,t)$ is defined as
\begin{equation}\label{e17}
\bar h(\xi,t)=[ h(\frac{1}{\xi^*})]^*,
\end{equation}
and
\begin{equation}\label{e18}
I_+[h](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'}
[ \frac{\xi'+\xi}{\xi'-\xi}] \frac{1}{|\xi h_\xi- h|} \quad
\text{for  } |\xi|>1.
\end{equation}
Making the change of variable,
\begin{equation}\label{e19}
v(\xi,t)=\frac{1}{(\xi h_\xi -h)^{1/2}},
\end{equation}
and using \eqref{e16}, we obtain
\begin{equation}\label{e20}
v_t=\xi v_\xi I_+[v] -\frac{1}{2}\xi v [I_+[v]]_\xi
+\frac{1}{2}\tau \xi v \bar v v_\xi
-\frac{1}{2}\tau \xi v^2 \bar v_\xi +\frac{1}{2}v I_+[v]
+ \tau v^2\bar v +av,
\end{equation}
where
\begin{equation}\label{e21}
I_+[v](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'}
\big[ \frac{\xi'+\xi}{\xi'-\xi}\big] v(\xi',t)\bar v(\xi',t) \quad
\text{for  } |\xi|>1.
\end{equation}
The analytic continuation of \eqref{e20} to $|\xi|<1$ is
\begin{equation}\label{e22}
v_t= \xi v_\xi I_-[v]-\frac{1}{2}\xi v [I_-[v]]_\xi +\frac{1}{2}v I_-[v] +a v,
\end{equation}
where
\begin{equation}\label{e23}
I_-[v](\xi,t)=\frac{\tau}{4\pi i}\int_{|\xi|=1}\frac{d\xi'}{\xi'}
[ \frac{\xi'+\xi}{\xi'-\xi}] v(\xi',t)\bar v(\xi',t)\quad
 \text{for  } |\xi|<1.
\end{equation}
We will consider  equation \eqref{e20} \eqref{e22} with the initial
condition
\begin{equation}\label{e24}
v(\xi,0)=v_0(\xi).
\end{equation}
Let us first introduce a scale of Banach spaces which are spaces of
bounded analytic functions in disks.

\begin{definition} \label{de3.1} \rm
Let $\mathcal{R}_s$ be the disk in complex $\xi$ plane with radius $s$;
 i.e., $\mathcal{R}_s=\{\xi,|\xi|<s\}$; we define function space
$\mathbf{B}_s$ consisting of functions
$f(\xi)$ is analytic in
$\mathcal{R}_s$ and continuous on $\overline{\mathcal{R}_s}$
with norm $\norm{f}_s=\sup_{\mathcal{R}_s} |f(\xi)|$.
\end{definition}

Also we define the constant
\begin{equation}\label{e25}
M=\norm{v_0}_r\,.
\end{equation}
We will obtain the following local existence result.

\begin{theorem} \label{mainthm}
If $v_0\in \mathbf{B}_r$ with $r>1$, then there exists one and only
one solution $v\in C^1([0,T),\mathcal{B}_s)$, $1<s<r,\norm{v}\le 2M$
to  \eqref{e20} and $v|_{t=0}=v_0$, where $T=a_0(r-s)$, $a_0$ is
a suitable positive constant independent of $s$.
\end{theorem}

The proof of above theorem will be based on  Nirenberg-Nishida
theorem \cite{n2,n3}.

\begin{theorem}[Nirenberg-Nishida] \label{niremberg}
Let  $\{\mathbf{B_s}\}_{r_1\le s\le r}$ be a scale of Banach spaces
satisfying that $\mathbf{B}_s\subset\mathbf{B}_{s'}$,
$\norm{\cdot}_{s'}\le\norm{\cdot}_{s} $ for any $r_1<s'<s< r$.
Consider the abstract Cauchy-Kowalewski problem
\begin{equation}
\label{e35}
\frac{du}{dt}=\mathcal{L}(u(t),t), \quad u(0)=0
\end{equation}
Assume the following conditions on $\mathcal{L}$:
\begin{itemize}
\item[(i)] For some constants $M>0,\delta>0$ and every pair of numbers
$s,s'$ such that $r_1< s'<s< r$, $(u,t)\to \mathcal{L}(u,t)$
is a continuous mapping of
\begin{equation}
\label{e36}
\{u\in \mathbf{B}_s:\norm{u}_s<M\}\times \{t; |t|<\delta \} \text{ into  } \mathbf{B}_{s'}
\end{equation}

\item[(ii)] For any $r_1\le s'<s\le r$ and all $u,v\in\mathbf{B}_s$
with $\norm{u}_s<M, \norm{v}_s<M$ and for any t, $|t|<\delta$,
$\mathcal{L}$ satisfies
\begin{equation} \label{e37}
\norm{\mathcal{L}(u,t)-\mathcal{L}(v,t)}_{s'}\leq C \frac{\norm{u-v}_s}{s-s'}
\end{equation}
where $C$ is some positive constant independent of $t,u,v,s,s'$.

\item[(iii)] $\mathcal{L}(0,t)$ is a continuous function of t, $|t|<\delta$
with values in $\mathbf{B}_s$ for every $r_1< s<r$ and satisfies, with
some positive constant $K$,
\begin{equation} \label{e38}
\norm{\mathcal{L}(0,t)}_s\le \frac{K}{(r-s)}
\end{equation}
\end{itemize}
Under the preceding assumptions there is a positive constant $a_0$ such
that there exists a unique function $u(t)$ which, for every $r_1<s<r$
and $|t|<a_0(r-s)$, is a continuously differentiable function of $t$
with values in $\mathbf{B}_s, \norm{u}_s<M$, and satisfies \eqref{e35}.
\end{theorem}

\section{Properties of Banach space $\mathbf{B}_s$}

Let $r_0$ and $r_1$ be two fixed numbers so that $r>r_1>r_0>1$,
then $\mathcal{R}_{r_0}\subset\mathcal{R}_{r_1}\subset\mathcal{R}_{r}$
and $\mathbf{B}_{r}\subset\mathbf{B}_{r_1}\subset\mathbf{B}_{r_0}$.

In this and the following sections, $C>0$ represents a generic constant,
it may vary from line to line. $C$ may depend on $r_0, r_1$ and $r$;
but it is always independent of $s$ and $s'$.

\begin{lemma}\label{lem:3.1}
If $f\in \mathbf{B}_s, r_1<s<r$, then $\norm{f_\xi}_{r_0}\leq K_1\norm{f}_s$
where $K_1$ is a positive constant independent of $s$ and $f$.
\end{lemma}

\begin{proof}
For $\xi\in\mathcal{R}_{r_0}$ and $t\in \{ |t|=r_1\}$, we have
$|t-\xi|\ge |r_1-r_0|$. From Cauchy Integral Formula, we have
\begin{equation*}
f_\xi(\xi)=\frac{1}{2\pi i}\int_{|t|=r_1}\frac{f(t)}{(t-\xi)^2}dt
\end{equation*}
so
\begin{equation*}
|f(\xi)|\le\frac{1}{2\pi }\int_{|t|=r_1}\frac{|f(t)|}{|t-\xi|^2}|dt|
         \le \frac{r_1\norm{f}_s}{2\pi (r_1-r_0)^2},
\end{equation*}
which completes the lemma.
\end{proof}

\begin{definition}\label{def:3.1} \rm
We define the function
\begin{equation}\label{3.1}
\bar{f}(\xi)=[f(\frac{1}{\xi^*})]^*
\end{equation}
where $ ~^*$ denotes complex conjugate.
\end{definition}

\begin{remark}\label{rem:3.3} \rm
For $s>0$, if $f\in\mathbf{B}_s$, then $\bar{f}$ is analytic in
$|\xi|>\frac{1}{s}$ and $|\bar{f}|\leq \norm{f}_s$ for $|\xi|>\frac{1}{s}$.
\end{remark}

\begin{lemma}\label{lem:3.4}
If $f\in \mathbf{B}_s, r_1<s<r$, then $|\bar{f}_\xi(\xi)|\leq K_2\norm{f}_s$
for $r\ge |\xi|\ge \frac{1}{r_0}$, where $K_2$ is a positive constant
independent of $s$ and $f$.
\end{lemma}

\begin{proof}
Due to Remark \ref{rem:3.3}, $\bar{f}$ is analytic in
$|\xi|\ge \frac{1}{r_0}$, by Cauchy Integral Formula, we have for
$1\le |\xi|\le s$
\begin{equation*}
\bar{f}_\xi(\xi,t)=\frac{1}{2\pi i}
\int_{|\xi'|=2r}\frac{\bar{f}(\xi',t)}{(\xi'-\xi)^2}d\xi'
-\frac{1}{2\pi i}\int_{|\xi'|
=\frac{1}{r_1}}\frac{\bar{f}(\xi',t)}{(\xi'-\xi)^2}d\xi'
\end{equation*}
For $\frac{1}{r_0}\le |\xi|\le r, \xi'\in \{|\xi'|
=\frac{1}{r_1}\}\cup\{ |\xi'|=2r\}$, we have $|\xi-\xi|\ge C$,
where $C$ depends only on $r_0$, $r_1$ and $r$; hence each integral
in the above equation can be bounded by $C\norm{f}_s$.
This completes the proof.
\end{proof}

The following lemma is essential for applying the Nirenberg-Nishida Theorem.

\begin{lemma}\label{lem:3.5}
If $f\in \mathbf{B}_{s}$, $r_1 <s'<s\le r$; then $f_\xi\in \mathbf{B}_{s'}$
and
\begin{equation}\label{3.2}
\norm{f_\xi}_{s'}\leq \frac{K_3}{s-s'}\norm{f}_{s},
\end{equation}
where $K_3>0$ is independent of $s,s'$ and $f$.
\end{lemma}

\begin{proof}
Since
$\mathop{\rm dist}(\partial\mathbf{B}_{s'},\partial\mathbf{B}_{s} )=s-s'$,
for $\xi\in\mathbf{B}_{s'}$, we are able to find a disk $D(\xi)$ centered
at $\xi$ with radius $s-s'$ such that $D(\xi)$ is contained in
$\mathcal{R}_s$. Using Cauchy integral formula, we have
\begin{equation*}
f_\xi(\xi)=\frac{1}{2\pi i}\int_{|t-\xi|=s-s'}\frac{f(t)}{(t-\xi)^2}dt
\end{equation*}
so
\begin{equation*}
\begin{split}
|f_\xi(\xi)|
&\le\frac{1}{2\pi }\int_{|t-\xi|=s-s'}\frac{|f(t)|}{|t-\xi|^2}|dt|\\
&\le \frac{1}{2\pi }\int_0^{2\pi}\frac{|f(\xi+|s-s'| e^{i\theta})|}{s-s'}d\theta\\
&\le \frac{K_3\norm{f}_s}{s-s'},
\end{split}
\end{equation*}
which proves the lemma.
\end{proof}

\begin{definition} \label{def4.6} \rm
We define the function
\begin{equation}\label{3.4}
G_1[v](\xi,t)=v(\xi,t)\bar{v}(\xi,t).
\end{equation}
\end{definition}

\begin{remark}\label{rem:3.7} \rm
If $v\in\mathbf{B}_s, s>1$, then $G_1[v]$ are analytic in
$\frac{1}{s}\le |\xi|\le s$.
\end{remark}

\begin{lemma}\label{lem:3.8}
If $v\in\mathbf{B}_s$ and $ 1<s<r$, then
$|G_1[v](\xi,t)|\le \norm{v}_s^2$ for $ \frac{1}{s}\le |\xi|\le s$.
\end{lemma}

The above lemma follows from equation (\ref{3.4}).

\begin{lemma}\label{lem:3.9}
If $v\in\mathbf{B}_s$ and $ r_1<s<r$, then $I_+[v](\xi,t)\le C\norm{v}^2$
for $|\xi|\ge 1$, where $C>0$ is a constant independent of $s, g $ and $h$.
\end{lemma}

\begin{proof}
Due to Remark \ref{rem:3.7}, the integrands of $I_+$ are analytic in
$\frac{1}{r_0}\le |\xi|\le 1$, changing the contour of integration in
the definitions of $I^+$ from $|\xi|=1$ to $|\xi|=\frac{1}{r_0}$ gives
\begin{equation}\label{3.5}
I_+[v](\xi,t)=\frac{1}{4\pi i}\int_{|\xi '|
=\frac{1}{r_0}}\frac{d\xi'}{\xi'}[\frac{\xi+\xi'}{\xi'-\xi}]G_1[v](\xi',t)
\end{equation}
For $|\xi|>1$ and $|\xi '|=\frac{1}{r_0}$, from simple geometry, we have
\begin{equation}\label{3.7}
|\frac{\xi+\xi'}{\xi'-\xi}|\le C
\end{equation}
where $C$ depends on only $r_0$.
The lemma now follows from (\ref{3.5}) and Lemma \ref{lem:3.8}.
\end{proof}

Similarly we have
\begin{lemma}\label{lem:3.10}
If $v\in\mathbf{B}_s$ and $ 1<s<r$, then $(I^+[v])_\xi(\xi,t)\le C\norm{v}^2$
for $|\xi|\ge 1$.
\end{lemma}

\begin{proof}
Taking derivative in (\ref{3.5}),
\begin{equation}\label{3.9}
(I_+[v])_\xi (\xi,t)=\frac{1}{4\pi i}\int_{|\xi '|
=\frac{1}{r_0}}\frac{d\xi'}{\xi'}[\frac{\xi+\xi'}{(\xi'-\xi)^2}]G_1[v](\xi',t)
\end{equation}
For $|\xi|>1$ and $|\xi '|=\frac{1}{r_0}$, from simple geometry, we have
\begin{equation}\label{3.10}
|\frac{\xi+\xi'}{(\xi'-\xi)^2}|\le C
\end{equation}
where $C$ depends on only $r_0$.
The lemma now follows from (\ref{3.10}) and Lemma \ref{lem:3.8}.
\end{proof}

\begin{lemma}\label{lem:3.11}
If $v\in\mathbf{B}_s$ and $ 1<s<r$, then $I_-[v](\xi,t)\le C\norm{v}^2$
for $|\xi|\le 1$.
\end{lemma}

\begin{proof}
Due to Remark \ref{rem:3.7}, the integrands of $I_-$ is analytic in
$r_0\ge |\xi|\ge 1$, changing the contour of integration in the definitions
of $I_-$ from $|\xi|=1$ to $|\xi|=r_0$ gives
\begin{equation}\label{3.12}
I_-[v](\xi,t)=\frac{1}{4\pi i}\int_{|\xi '|=r_0}\frac{d\xi'}{\xi'}
[\frac{\xi+\xi'}{\xi'-\xi}]G_1[v](\xi',t)
\end{equation}
For $|\xi|<1$ and $|\xi '|=r_0$, from simple geometry, we have
\begin{equation}\label{3.13}
|\frac{\xi+\xi'}{\xi'-\xi}|\le C
\end{equation}
where $C$ depends on only $r_0$.
The lemma now follows from(\ref{3.12}), (\ref{3.13}) and Lemma \ref{lem:3.8}.
\end{proof}

\begin{lemma}\label{lem:3.12}
If $v\in\mathbf{B}_s$ and $ 1<s<r$, then $(I_-)_\xi[v](\xi,t)\le C\norm{v}^2$
for $|\xi|\le 1$.
\end{lemma}

\begin{proof}
Taking derivatives in (\ref{3.12}),
\begin{equation}\label{3.15}
(I_-[v])_\xi (\xi,t)=\frac{1}{4\pi i}
\int_{|\xi '|=r_0}\frac{d\xi'}{\xi'}[\frac{\xi+\xi'}{(\xi'-\xi)^2}]
G_1[v](\xi,t)
\end{equation}
For $|\xi|<1$ and $|\xi '|=r_0$, from simple geometry, we have
\begin{equation}\label{3.16}
|\frac{\xi+\xi'}{(\xi'-\xi)^2}|\le C
\end{equation}
where $C$ depends on only $r_0$.
The lemma now follows from (\ref{3.15}), (\ref{3.16}) and Lemma \ref{lem:3.8}.
\end{proof}

\section{Proof of the main theorem}

In this section,  to prove the main theorem, we
apply Nirenberg's theorem to initial problem \eqref{e20}, \eqref{e22}
and \eqref{e24}. To this end, we need more estimates of the type
as in \eqref{e37}.

\begin{lemma}\label{lem:4.1}
If $u\in \mathbf{B}_s,v\in \mathbf{B}_s, r_1<s'<s<r$, then
$\norm{u_\xi-v_\xi}_{s'}\le \frac{C}{s-s'}\norm{u-v}_s$,
where $C>0$ is independent of $s,s'$ and $u$ and $v$.
\end{lemma}

The above lemma follows from applying Lemma \ref{lem:3.5} with $f=u-v$.

\begin{lemma}\label{lem:4.2}
If $u\in \mathbf{B}_s,v\in \mathbf{B}_s, r_1<s<r$, then
$|\bar{u}_\xi-\bar{v}_\xi|\le C\norm{u-v}_s$ for $\frac{1}{r_1}\le|\xi|$,
where $C>0$ is independent of $s$, $u$ and $v$.
\end{lemma}

The above lemma follows from
applying Lemma \ref{lem:3.4} with $f=u-v$.

\begin{lemma}\label{lem:4.2.5}
If $u\in \mathbf{B}_s,v\in \mathbf{B}_s,\norm{u}_s\le M,\norm{v}_s\le M$,
$r_1<s<r$, then for $|\xi|>\frac{1}{r_0}$,
$$
|G_1[v](\xi,t)-G_1[u](\xi,t)|\le C\norm{v-u}_s .
$$
\end{lemma}

\begin{proof}
From  (\ref{3.4}), we have
$$
G_1[v]-G_1[u]=(v-u)\bar{v}+u(\bar{v}-\bar{u})
$$
which proves the lemma by using Remark \ref{rem:3.3}.
\end{proof}

\begin{lemma}\label{lem:4.3}
If $u\in \mathbf{B}_s,v\in \mathbf{B}_s,\norm{u}_s\le M,\norm{v}_s\le M$,
$r_1<s<r$, then for $|\xi|\le 1$,
\begin{gather*}
|I_-[v](\xi,t)-I_-[u](\xi,t)|\le C \norm{v-u}_s;\\
|(I_-[v])_\xi (\xi,t)-(I_-[u])_\xi (\xi,t)|\le C \norm{v-u}_s.
\end{gather*}
\end{lemma}

\begin{proof}
From (\ref{3.12}), we have
\begin{equation*}
I_-[v](\xi,t)-I_-[u](\xi,t)=\frac{1}{4\pi i}\int_{|\xi '|=r_0}
\frac{d\xi'}{\xi'}[\frac{\xi+\xi'}{\xi'-\xi}](G_2[v](\xi',t)-G_2[u](\xi',t))
\end{equation*}
Now the lemma can be proved in the same fashion as
Lemma \ref{lem:3.11} and Lemma \ref{3.12} in light of Lemma \ref{lem:4.2.5}.
\end{proof}

\begin{lemma}\label{lem:4.4}
If $u\in \mathbf{B}_s,v\in \mathbf{B}_s,\norm{u}_s\le M,\norm{v}_s\le M$,
then for $|\xi|\ge 1$,
\begin{gather*}
|I_+[v](\xi,t)-I_+[u](\xi,t)|\le C \norm{v-u}_s, \\
|(I_+[v])_\xi (\xi,t)-(I_+[u])_\xi (\xi,t)|\le C \norm{v-u}_s.
\end{gather*}
\end{lemma}

The above lemma can be proved in the same fashion as Lemma \ref{lem:3.9} and
Lemma \ref{3.10}, in light of Lemma \ref{lem:4.2.5}.

Let $v\in \mathcal{B}_s$. We define the following operator;
for $|\xi|>1$, $L[v]$ is defined by
\begin{equation}\label{4.1}
L[v](\xi,t)=
\xi v_\xi I_+[v] -\frac{1}{2}\xi v [I_+[v]]_\xi +\frac{1}{2}\tau \xi v \bar v v_\xi
-\frac{1}{2}\tau \xi v^2 \bar v_\xi +\frac{1}{2}v I_+[v] + \tau v^2\bar v +av,
\end{equation}
The analytic continuation of $L[v]$ to $|\xi|<1$ is
\begin{equation}\label{4.2}
L[v](\xi,t)=
\xi v_\xi I_-[v] -\frac{1}{2}\xi v [I_-[v]]_\xi +\frac{1}{2}v I_-[v] +av\,.
\end{equation}

\begin{lemma}\label{lem:4.7}
If $u\in \mathbf{B}_s$, $v\in \mathbf{B}_s$, $\norm{u}_s\le M$,
$\norm{v}_s\le M$, then for $|\xi|\ge 1$ and $r_1<s'<s<r$, we have
$$
\norm{L[u]-L[v]}_{s'}\le \frac{C}{s-s'}\norm{v-u}_s,
$$
where $C>0$ is independent of $s$ and $s'$.
\end{lemma}

\begin{proof}
By (\ref{4.2}), for $|\xi|<1$,
\begin{equation}\label{4.5}
\begin{aligned}
L[v](\xi,t)-L[u](\xi,t)
&=\xi(v_\xi-u_\xi)I_-[v]+\xi u_\xi \{I_-[v]-I_-[u]\}+a(v-u)\\
&\quad -\frac{1}{2}\xi(v-u)(I_-[v])_\xi-\frac{1}{2}
 \xi u\{(I_-[v])_\xi -(I_-[u])_\xi\},
\end{aligned}
\end{equation}
and for $|\xi|>1$,
\begin{equation}\label{4.6}
\begin{aligned}
L[v](\xi,t)-L[u](\xi,t)
&=\xi(v_\xi-u_\xi)I_+[v]+\xi u_\xi \{I_+[v]-I_+[u]\}\\
&\quad +a(v-u)-\frac{1}{2}\xi(v-u)(I_+[v])_\xi-\frac{1}{2}
  \xi u\{(I_+[v])_\xi -(I_+[u])_\xi\}\\
&\quad +\frac{1}{2}\tau \xi \{(v-u)\bar{v}v_\xi+u(\bar{v}
 -\bar{u} v_\xi + u\bar{u} (v_\xi-u_\xi)\}\\
&\quad -\frac{1}{2}\tau \xi \{ (v-u)(v+u)\bar{v}_\xi
 + u^2(\bar{v}_\xi+\bar{u}_\xi)(\bar{v}_\xi-\bar{u}_\xi)\}\\
&\quad +\tau (v-u)(v+u)\bar{v}+\tau u^2(\bar{v}-\bar{u}),
\end{aligned}
\end{equation}
By Lemmas \ref{lem:3.4}--\ref{lem:3.12} and  \ref{lem:4.1}-- \ref{lem:4.4},
each term in equations (\ref{4.5}) and (\ref{4.6}) can bounded by
$\frac{C}{s-s'}\norm{v-u}_s$; hence the proof is complete.
\end{proof}

Let $p=v-v_0$, then $v$ is a solution of initial problem
\eqref{e20}, \eqref{e22} and \eqref{e24} if and only if $p$
solves the following initial problem
\begin{equation}\label{4.9}
p_t=\mathcal{L}[p], p|_{t=0}=0.
\end{equation}
where the operator $\mathcal{L}$ is defined by
\begin{equation}\label{4.10}
\mathcal{L}[p]=L[p+v_0]
\end{equation}

\begin{lemma}\label{lem:4.9}
If $p\in B_s, u\in B_s, \norm{p}_s\le M$ and
$\norm{u}_s\le M, r_1<s'<s<r$, then
$$
\norm{\mathcal{L}[p]-\mathcal{L}[u]}_{s'}\le \frac{C}{s-s'}\norm{p-u}_s
$$
\end{lemma}

The proof of the above lemma follows from Lemma \ref{lem:4.7}
and (\ref{4.10}).

\begin{lemma}\label{lem:4.10}
If $r_1<s'<r$, then $\norm{\mathcal{L}[0]}_{s'}\le \frac{K}{r-s'}$.
\end{lemma}

\begin{proof}
From (\ref{4.10}), we have
$\mathcal{L}[0]=L[v_0])=L[v_0]-L[0]$,
for any $s$ such that $ s'<s<r$, using Lemma \ref{lem:4.7}
 with $v=v_0,u=0$, we obtain
\begin{equation*}
\norm{\mathcal{L}[0]}_{s'}\le \frac{C}{s-s'}\norm{v_0}_s,
\end{equation*}
Letting $s\to r$ in the above equation, we complete the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{mainthm}]
We first apply Nirenberg theorem to system (\ref{4.9}).
For $p\in B_s$, by Lemma \ref{lem:4.9} with $p=p, u=0$, we have
$\mathcal{L}[p]\in B_{s'}$. Since the system (\ref{4.9}) is autonomous,
 the continuity of the operator $\mathcal{L}$ is implied by
Lemma \ref{lem:4.9}; hence \eqref{e36} holds. \eqref{e37} and \eqref{e38}
are given by Lemma \ref{lem:4.9} and Lemma \ref{lem:4.10} respectively.
Therefore, there exists unique solution $p\in B_s,\norm{p}_s\le M$,
so $v=p+v_0$ is the unique solution of the problem \eqref{e20}, \eqref{e22}
and \eqref{e24}.
\end{proof}

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