\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small {\em Electronic Journal of
Differential Equations}, Vol. 2003(2003), No. 33, pp. 1--25.
\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/33\hfil Analytic solution to integro-differential equations]
{Analytic solution to a class of integro-differential equations}

\author[Xuming Xie \hfil EJDE--2003/33\hfilneg]
{Xuming Xie}

\address{Xuming Xie\hfill\break
Department of Mathematical Sciences, University of Delaware, \hfill\break
501 Ewing Hall, Newark, DE 19716, USA}
\email{xie@math.udel.edu}

\date{}
\thanks{Submitted August 13, 2002. Published March 28, 2003.}
\subjclass[2000]{34A20, 45E05}
\keywords{Analytic solution, singular integro-differential equation}

\begin{abstract}
 In this paper, we consider the integro-differential equation
 $$
 \epsilon^2 y''(x)+L(x)\mathcal{H}(y)=N(\epsilon,x,y,\mathcal{H}(y)),
 $$
 where $\mathcal{H}(y)[x]=\frac{1}{\pi}(P)\int_{-\infty}^{\infty}
 \frac{y(t)}{t-x}dt$ is the Hilbert transform.
 The existence and uniqueness of analytic solution in appropriately
 chosen space is proved. Our method consists of extending the
 equation to an appropriately chosen region in the complex plane,
 then use the Contraction Mapping Theorem.
\end{abstract}

\maketitle

\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
%\newtheorem{xca}[theorem]{Exercise}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}
\newcommand{\abs}[1]{|#1|}
\newcommand{\norm}[1]{\|#1\|}

\section{Introduction}

The second order ordinary differential equations with singular
perturbation have been discussed in works such as \cite{Olver74,Wasow68}.
While singular integral equations have also been studied systematically
in \cite{Musk77}.  Many physical problems can be modelled by
 singular integro-differential equation
\begin{equation}
\label{eq:1.1} \epsilon^2 y''(x)+Q(x)y(x)+L(x)\mathcal{H}(y)=
N(\epsilon^2,x,y,y',\mathcal{H}(y),\mathcal{H}(y')),
\end{equation}
for $x\in (-\infty,+\infty)$, where
$\mathcal{H}(y)[x]=\frac{1}{\pi}(P)\int_{-\infty}^{\infty}\frac{y(t)}{t-x}dt$
is the Hilbert transform.

Saffman and Taylor \cite{SaffmanTaylor58} studied  the
displacement of a viscous fluid by a less viscous  fluid in a
Hele-Shaw cell. It was noted that a single finger of the less  viscous
fluid is eventually formed and propagates at constant velocity keeping a
steady shape. In the absence of surface tension ($\epsilon =0$),
Saffman and Taylor obtained a family of exact solutions. When the surface
tension is non-zero, by conformal
mapping, Maclean and Saffman \cite{McleanSaffman81} have reduced
the determination of the finger to the solution of two coupled
nonlinear integrodifferential equations. In Maclean-Saffman
equations, the integral term is a Cauchy type singular integral
over the interval [0,1]. By a transformation of the independent
variable, Combescot et al \cite{Combescot88} and Chapman
\cite{Chapman} cast the finger problem with small surface tension
as an integrodifferntial equation on the whole real line. By using
Saffman-Taylor exact solutions, the integrodifferntial equations
of Combescot et al and Chapman can be reduced to equation
(\ref{eq:1.1}) with $Q(x)\ne 0$ and $L(x)\ne 0$. The works
mentioned above include numerical and asymptotic studies.
Recently, Xie and Tanveer \cite{XieTanveer01, TanveerXie02}
 reformulated the Saffman-Taylor finger problem as solving the
 integro-differential equation
\begin{equation}
\label{eq:1.1.5} \epsilon^2 y''(x)+Q(x)y'(x)= N(\epsilon,x,y,y',
\mathcal{H}(\mathcal{G}[y]))
\end{equation}
where $\mathcal{G}[y]$ is an operator. Existence of steadily
translating finger solutions (i.e analytic solution of
(\ref{eq:1.1.5})) were proved rigorously if relative finger width
is in a set of infinite discrete values
and surface tension is sufficiently small.

Dendritic crystal growth has long been a subject of
interest to both physicists and Mathematicians. The simplest
example of dentrite growth is the growth of needle crysyal in
solidification from a pure undercooled melt. The growth of a
steadily moving interface between solid and liquid is the ultimate
evolution of the Mullins-Sekerka instability. When surface tension
is neglected, Ivanstov \cite{Ivans} found an infinite continuous
family of parabolic crystal interfaces. When surface tension is
taken into account, in the limit of small Peclet number, Pelce and
Pomeau \cite {Pelce,BenAmar86} reduced the Nash-Gicksman
\cite{Nash} equation to a simpler set of integrodifferential
equations in which the singular integral terms are no longer
of Cauchy type. In a recent work, Xie \cite{Xie03} reduced the
one-sided needle crystal growth problem to solving an
integrodifferential equation of form (\ref{eq:1.1.5}); symmetric
analytic solutions are obtained
if the surface tension is small and the crystalline anisotropy is in a set
of  infinite discrete values.

In this paper, we consider (\ref{eq:1.1}) with $L(x)\ne 0$, $Q(x)=0$,
and $N$ does not depend on $y'$. i.e., equations of form
\begin{equation}
\label{eq:1.2} \epsilon^2 y''(x)+L(x)\mathcal{H}(y)=
N(\epsilon,x,y,\mathcal{H}(y))\,.
\end{equation}
We believe that the method developed in this paper will be useful for other
type of integro-differential equations such as Nash-Glicksman equations
\cite{Nash,Pelce,Tanveer89} for the two-sided steady needle crystal growth
problem.

Although equation (\ref{eq:1.2}) is given on the real $x$ axis, we
will extend the equation to some domains in the complex plane as
in the viscous fingering case (\cite{XieTanveer01,TanveerXie02}).
The main reason to go to complex plane is
that it is possible to control the nonlocal integral terms in
(\ref{eq:1.2}) and estimate the decay rate of the derivatives.

\subsection*{Notation and Main result}
We define regions in complex z-plane:
\begin{definition} \label{def:1.1}  \rm
Let $\mathcal{R}_{\alpha,\varphi}$ be an open
connected region on the complex plane bounded by the lines
\[
r_u=r_{u,1}\cup r_{u,2}\cup r_{u,3},\quad
r_l=r_{l,1}\cup r_{l,2}\cup r_{l,3}
\]
where
\begin{gather*}
r_{u,1}=\{z: z=\alpha i-R +re^{(\pi-\varphi)i},0\le r\le\infty \},\\
r_{u,2}=\{z: z=\alpha i+r,-R\le r\le R\},\\
r_{u,3}=\{z: z=\alpha i-R +re^{\varphi i},0\le r\le\infty \},\\
r_{l,1}=\{z: z=-\frac{\alpha}{2} i-R +re^{(\pi+\frac{\varphi i}{2})},0\le
r\le\infty \},\\
r_{l,2}=\{z: z=-\frac{\alpha}{2} i+r,-R\le r\le R\},\\
r_{l,3}=\{z: z=-\frac{\alpha}{2}i-R +re^{-\frac{\varphi
i}{2}},0\le r\le\infty \},
\end{gather*}
where $1>\alpha>0$, $0<\varphi <\pi/2$ and $R>0$.
\end{definition}

\begin{figure} \label{fig:1}
\begin{center}
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\caption{Region $\mathcal{R}_{\alpha,\varphi}$ in the complex plane}
\end{figure}

Denoting by  $*$ the complex conjugate, we define
\begin{equation}
\label{eq:1.3}
\tilde{\mathcal{R}}_{\alpha,\varphi}=\mathcal{R}_{\alpha,\varphi}^*
\equiv\{z^*: z\in\mathcal{R}\}
\end{equation}


\begin{definition}
\label{def:1.2} We define function
\begin{equation}
\label{eq:1.4} P(z)=\int_0^z \sqrt{iL(t)}dt,\quad \tilde
P(z)=\int_0^z \sqrt{-iL(t)}dt
\end{equation}
\end{definition}
We assume that there exist numbers $1>\alpha_0>0$, $R>0$ and
$\frac{\pi}{2}>\varphi_0>0$, so that $L(z), P(z)$, and
$\tilde P(z)$ satisfy the following properties:

\noindent\textbf{Property 1.}  $L(z)$ is
analytic in $\mathcal{R}_{\alpha_0,\varphi_0}\cup
\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$,
and $L(z)\ne 0$  in $\mathcal{R}_{\alpha_0,\varphi_0}\cup
\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$ and
\begin{equation}
\label{eq:1.5} |L(z)|= C|z|^{\gamma}(1+o(1)), |L'(z)|=
C|z-2i|^{\gamma-1}(1+o(1)),
\end{equation}
\begin{equation}
\label{eq:1.6} |L''(z)|= C|z-2i|^{\gamma -2}(1+o(1))\text{ as
}|z|\to
\infty,z\in\mathcal{R}_{\alpha_0,\varphi_0}\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}
\end{equation}
where $\gamma >-2$ and $C$ independent of $\epsilon$.

\noindent\textbf{Property 2.} A branch of $\sqrt{iL}$ and $\sqrt{-iL}$ in
(\ref{eq:1.4}) can be chosen so that $P(z)$ and $\tilde P(z)$ are
analytic in $\mathcal{R}_{\alpha_0,\varphi_0}
\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$ and
$\mathop{\rm Re}P(-\infty)=-\infty$ , $\mathop{\rm Re}P(\infty )=\infty$,$\mathop{\rm Re}\tilde P(-\infty)=-\infty$ , $\mathop{\rm Re} \tilde P(\infty )=\infty$ .

\noindent\textbf{Property 3.} For $\mathop{\rm Re}z\geq R$, $0<\alpha\leq\alpha_0,~0<\varphi\leq\varphi_0$,
$\mathop{\rm Re}P(t)$ (resp. $\mathop{\rm Re}\tilde P(t)$) is increasing with increasing
$s$ along any ray
$r=\{t:t=z+se^{i\theta},0<s<\infty,-\varphi<\theta<\varphi\}$ in
$\mathcal{R}_{\alpha,\varphi}$ (resp. in $\tilde
{\mathcal{R}}_{\alpha,\varphi}$) from $z$ to $z+\infty
e^{i\theta}$ and  $C_1|t-2i|^{\gamma/2}\leq \frac{d}{ds}\text{ Re
}P(t(s))$ $(~resp. ~C_1|t-2i|^{\gamma/2}\leq
\frac{d}{ds}\text{ Re }\tilde P(t(s)))$ where $C_1$ is a
constant which can be made independent of $\epsilon$ and $C_1>0$.

\noindent\textbf{Property 4.} For $\mathop{\rm Re}z\le -R$, $0<\alpha\leq\alpha_0,~0<\varphi\leq\varphi_0$,
$\mathop{\rm Re}P(t)$ (resp. $\mathop{\rm Re}\tilde P(t)$ ) is decreasing with increasing
$s$ along  ray
$r=\{t:t=z+se^{i(\pi-\theta)},0<s<\infty,-\varphi<\theta<\varphi\}$
in $\mathcal{R}_{\alpha,\varphi}$ (resp.
$\tilde{\mathcal{R}}_{\alpha,\varphi}$) from $z$ to $z+\infty
e^{i(\pi -\varphi)}$ and  $\frac{d}{ds}\text{ Re }P(t(s))\leq -C_2
|t-2i|^{\gamma/2}$,(resp. $\frac{d}{ds}\text{ Re }\tilde
P(t(s))|\leq - C_2 |t-2i|^{\gamma/2}$) where $C_2$ is constant
which can be made independent of $\epsilon$ and $C_2>0$.

\noindent\textbf{Property 5.} For any $z\in \mathcal{R}_{\alpha,\varphi}$ (resp. $z\in\tilde{\mathcal{R}}_{\alpha,\varphi}$), $0<\alpha\leq\alpha_0,~ 0<\varphi\leq\varphi_0$, there is a path $\mathcal{P}(z,\infty)$ in $\mathcal{R}_{\alpha,\varphi}$  (resp. $\tilde {\mathcal{P}}(z,\infty)$ in $\tilde{\mathcal{R}}_{\alpha,\varphi}$) which is $\mathbf{C}^1$ curve
connecting $z$ to $\infty$  so that $\frac{d}{ds}~[Re ~P(t(s))]\ge C|t-2i|^{\gamma/2}>0$ for $t(s) \in \mathcal{P}(z,\infty)$ \\
$(~ resp.~\frac{d}{ds}~[Re ~\tilde P(t(s))]\ge
C|t-2i|^{\gamma/2}>0\text{ for }~t(s) \in \tilde
{\mathcal{P}}(z,\infty))$, $s$ being an arc length of
$\mathcal{P}(z,\infty)$ (resp. $\tilde{\mathcal{P}}(z,\infty)$),
which increases toward $t=\infty$.

\noindent\textbf{Property 6.}
 For any $z\in \mathcal{R}_{\alpha,\varphi}$,(resp. $z\in\tilde{\mathcal{R}}_{\alpha,\varphi}$), $0<\alpha\leq\alpha_0,~0<\varphi\leq\varphi_0$, there is a path $\mathcal{P}(z,-\infty)$ in $\mathcal{R}_{\alpha,\varphi}$  (resp. $\tilde {\mathcal{P}}(z,-\infty)$ in $\tilde{\mathcal{R}}_{\alpha,\varphi}$) which is $\mathbf{C}^1$ curve
connecting $z$ to $-\infty$  so that $\frac{d}{ds}~[Re ~P(t(s))]\le -C|t-2i|^{\gamma/2}<0$ for $t(s) \in \mathcal{P}(z,\infty)$ \\
$(~ resp.~\frac{d}{ds} [\mathop{\rm Re}\tilde P(t(s))]\le
-C|t-2i|^{\gamma/2}<0\text{ for }~t(s) \in \tilde
{\mathcal{P}}(z,\infty))$, $s$ being an arc length of
$\mathcal{P}(z,-\infty)$ (resp. $\tilde{\mathcal{P}}(z,-\infty)$),
which increases toward $t=-\infty$.


\begin{remark}\label{rem:1} \rm
In section 4, we are going to give two explicit
functions of $L(z)$ and region $\mathcal{R}_{\alpha_0,\varphi_0}$
so that Property 1-6 hold. Note that Property 1-6 are crucial to
prove Lemmas \ref{lem:2.7} and \ref{lem:2.8}.
\end{remark}

We assume that $N(\epsilon, z,u,v)$  can be written as
\begin{equation}
\label{eq:1.7} N(\epsilon, z,u,v)=\sum_{k=2}^np_k(z)T_k(u,v) +
\epsilon^2\sum_{k=0}^{l}f_k(z)Q_k(u,v)
\end{equation}
where $Q_k(u,v),T_k(u,v)$ is analytic in $\{(u,v):
|u|<\frac{1}{\rho},|v|<\frac{1}{\rho}\}$ and
\begin{equation} \label{eq:1.8}
\begin{gathered}
Q_k(u,v)=\sum_{\alpha_k+\beta_k\ge k}q_{\alpha_k,\beta_k}
u^{\alpha_k}v^{\beta_k},\quad
T_k(u,v)=\sum_{\alpha_k+\beta_k\ge k}t_{\alpha_k,\beta_k}u^{\alpha_k}
v^{ \beta_k},\\
 |q_{\alpha_k,\beta_k}|\leq
A\rho^{\alpha_k+\beta_k},\quad |t_{\alpha_k,\beta_k}|\leq
A\rho^{\alpha_k+\beta_k},
\end{gathered}
\end{equation}
where $A$ and $\rho$ are some positive constants. Then
$f_k(z)$ and $p_k(z)$ are analytic in
$\mathcal{R}_{\alpha_0,\varphi_0}\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$
and for
$z\in\mathcal{R}_{\alpha_0,\varphi_0}\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$,
\begin{equation}
\label{eq:1.9} |f_k(z)|\leq C|z-2i|^{-\tau+\gamma+k\tau},\quad
|p_k(z)|\leq C|z-2i|^{-\tau+\gamma+k\tau}
\end{equation}

Let $0 <\tau <1$ be fixed and independent of $\epsilon$. Let
$\mathcal{D}$ be any connected open set in complex $z$-plane.
 We introduce the following function spaces:
\begin{align*}
\mathbf{A}_k(\mathcal{D})
=&\big\{y(z):y(z)\text{ is analytic in } \mathcal{D}\text{ and continuous in }
\overline{\mathcal{D}},\\
&\text{ with }\sup_{z\in\overline{\mathcal{D}}}
\abs{(z-2i)^{k+\tau}y(z)} <\infty\big\}
\end{align*}
for $k=0,1$, with
$\norm{y}_{k,\mathcal{D}}:=\sup_{z\in\overline{\mathcal{D}}}
\abs{(z-2i)^{k+\tau}y(z)}$

Clearly, $\mathbf{A}_k(\mathcal{D})$ are Banach spaces, and
$\mathbf{A}_2(\mathcal{D})\subset\mathbf{A}_1(\mathcal{D})
\subset\mathbf{A}_0(\mathcal{D})\equiv\mathbf{A}(\mathcal{D})$.
\begin{gather*}
\mathbf{A}_k\equiv\mathbf{A}_k(\mathcal{R}_{\alpha_0,\varphi_0}),\quad
\norm{y}_k=\norm{\cdot}_{k,\mathcal{R}_{\alpha_0,\varphi_0}}
\text{ for }y\in\mathbf{A}_k \\
\tilde{\mathbf{A}}_k\equiv\mathbf{A}_k(\tilde{\mathcal{R}}_{\alpha_0,
\varphi_0}),\quad \norm{\tilde
y}_k=\norm{\cdot}_{k,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}}
\text{ for }\tilde y\in\tilde{\mathbf{A}}_k
\end{gather*}
Let $\delta$ be a constant such that $0<\delta<1$,
We define
\begin{equation} \label{eq:1.10}
\mathbf{A}_{0,\delta}=\{y: y\in\mathbf{A}_0,\norm{y}_0\leq \delta\},\quad
\tilde{\mathbf{A}}_{0,\delta}=\{\tilde y: \tilde
y\in\tilde{\mathbf{A}}_0,\norm{\tilde y}_0\leq \delta\}
\end{equation}

We will prove the following result.

\begin{theorem}
\label{lem:1.6} For sufficiently small $\epsilon$ and $\delta$,
there exists a unique solution $y\in\mathbf{A}_{0,\delta}(
\mathcal{R}_{\alpha_0,\varphi_0}\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0})$
to equation (\ref{eq:1.2}). Furthermore
$\norm{y}_{0,\mathcal{R}_{\alpha_0,\varphi_0}\cup\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}}\sim
O(\epsilon^2)$.
\end{theorem}

The proof of this Theorem will be given at the end of \S 3,
after some preliminary results. The solution strategy followed in
this paper is as follows: In \S 2, we first derive two
integro-differential equations. One is  the extension of equation
(\ref{eq:1.2}) to the upper part of $\mathcal{R}$ and the other is
the extension of equation (\ref{eq:1.2}) to lower part of
$\tilde{\mathcal{R}}$. Since the integral terms $I_{\pm}$ derived
from the Hilbert transform $\mathcal{H}$ (see Definition
\ref{def:2.1} in the sequel)
 are not  contraction terms (or small terms), the classical contraction argument does not work for both equations. By integration by parts and Hilbert Inverse transform, we can change the integral term into a small term. However in doing so, we get derivatives of $I_{\pm}$, the classical contraction argument
still fails. To circumvent this difficulty, we formulate a coupled
system of integral equations using the equations in $\{ Im
~z>0\}\cap \mathcal{R}$ and in $\{ Im ~z<0\}\cap
\tilde{\mathcal{R}}$ at the same time. In \S 3, we use contraction
argument to show the existence and uniqueness of solution to the
coupled system of integral equations. Then, we further show the
solution to the coupled system is actually the solution to
(\ref{eq:1.2}). In \S 4, in order to demonstrate the relevance of
the method, we give two explicit functions for $L(x)$ in
(\ref{eq:1.2}) and show that there are constants $\alpha_0$ and
$\varphi_0$ so that property 1-6 hold in
$\mathcal{R}_{\alpha_0,\varphi_0}$. Therefore, Theorem
\ref{lem:1.6} can be applied to these two examples. These simple
model problems are derived from more complex and physically sound
problem \cite{McleanSaffman81,Tanveer89,BenAmar88}.

\section{Formulation of equivalent integral equations in complex regions}

In this section, for simple notation, we use $\mathcal{R}$ to
denote $\mathcal{R}_{\alpha_0,\varphi_0}$ and
$\tilde{\mathcal{R}}$ to denote
$\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$ respectively. We will
use $C$ (and sometimes $C_1,C_2$) as generic constant, whose value
is allowed to differ from Lemma to Lemma  and from line to line.
However $C$ does not depend on $\epsilon$.

\begin{lemma} \label{lem:2.1}
Let $\Gamma =\{t,t=\xi_0+\rho e^{i\varphi},0\leq\rho<\infty\}$ be a ray,
with $2i$ not in $\Gamma$.
$\mathcal{D}$ is a region with $\mathop{\rm dist} (2i,\mathcal{D})>0$ and
\begin{equation} \label{eq:2.1}
\mathop{\rm dist}(\xi,\Gamma)\geq m|\xi-\xi_0|>0;\text{ for
}\xi\in\mathcal{D};
\end{equation}
\begin{equation}
\label{eq:2.2} \mathop{\rm dist}(t,\mathcal{D})\geq m|t-\xi_0|;\text{ for
}t\in\Gamma;
\end{equation}
\begin{equation}
\label{eq:2.3} \mathop{\rm dist}(t,2i)\geq m|t-\xi_0|;\text{ for }t\in\Gamma;
\end{equation}
for some constant $m>0$ independent of $\epsilon$. Assume $g(\xi)$
to be a continuous function on $\Gamma$ with
$\norm{g}_{0,\Gamma}<\infty$, then for $k =0,1,2$,
\begin{equation}
\label{eq:2.4}
\sup_{\mathcal{D}} \big|(\xi-2i)^{k+\tau}\int_{\Gamma}
\frac{g(t)}{(t-\xi)^{k+1}}dt\big| \leq C \norm{g}_{0,\Gamma};
\end{equation}
where constant $C$ that depends on $\varphi$ and $m$ only .
\end{lemma}

\begin{proof}
This lemma was proved in \cite{XieTanveer01}, we give the proof
here for completeness.
\begin{equation}
\label{eq:2.5}
\big\lvert(\xi-2i)^{k+\tau}\int_{\Gamma}\frac{g(t)}{(
t-\xi)^{k+1}}dt\big\rvert\leq
\norm{g}_{0,\gamma}|\xi-2i|^{k+\tau}\int_{\Gamma}\frac{|dt|}{|(t-\xi)|^{k+1}|t-2i|^\tau};
\end{equation}
On $\Gamma$,$t-\xi_0=\rho e^{i\varphi},|dt|=d\rho$. Breaking up
the integral in (\ref{eq:2.5}) into two parts:
\begin{multline}
\label{eq:2.6} \int_\Gamma\frac{|dt|}{|(t-\xi)|^{k+1}|t-2i|^\tau}\\
=\int_0^{|\xi-\xi_0|}\frac{d\rho}{|(t-\xi)|^{k+1}|t-2i|^\tau}
+\int_{|\xi-\xi_0|}^\infty\frac{d\rho}{|(t-\xi)|^{k+1}|t-2i|^\tau};
\end{multline}
for the first integral in (\ref{eq:2.6}),we use (2.1) and (2.3)
and for the second, we use (2.2) and (2.3) to obtain:(on scaling
$\rho$ by $|\xi-\xi_0|$)
\[
\int_\Gamma\frac{|dt|}{|(t-\xi)|^{k+1}|t-2i|^\tau}=\frac{C}{|\xi-\xi_0|^{k+\tau}}(
\int_0^{1}\frac{d\rho}{\rho^\tau}
+\int_{1}^\infty\frac{d\rho}{\rho^{k+\tau+1}});
\]
\end{proof}

\begin{figure} \label{fig:2}
\begin{center}
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\caption{Angular subset $\mathcal{D}'$ of $\mathcal{D}$}
\end{figure}

\begin{definition}
\label{def:2.0} Let $\mathcal{D}$ be an open connected set in
complex plane with one or more straight line boundaries.
$\mathcal{D}'$ is defined as an angular subset of $\mathcal{D}$ if
$\mathcal{D}'\subset\mathcal{D}$, {\bf $dist (\mathcal{D}',
\partial\mathcal{D}) >0$}
and $\mathcal{D}'$ has straight line boundaries that make a
nonzero angle with respect to $\partial\mathcal{D}$ asymptotically
at large distances from the origin (see Figure \ref{fig:2}). This
means that if $z'\in\mathcal{D}'$ and
$z\in\partial\mathcal{D}$,then $dist (z,\partial \mathcal{D}')\geq
C|z|\sin \theta_0$, as $|z|\to\infty$; $\mathop{\rm dist}(z',\partial
\mathcal{D})\geq C|z'|\sin \theta_0$, as $|z'|\to\infty$,where $C$
is some positive constant and $0<\theta_0\le \pi/2$.
\end{definition}

\begin{remark} \label{rem:2} \rm
 Note if $\Gamma$ is in $\mathcal{D}_c'$,an angular subset of $\mathcal{D}_c$
(complement of $\mathcal{D}$),then (\ref{eq:2.1}) and (\ref{eq:2.2})
 hold (see Figure \ref{fig:3} ).
 Also note (\ref{eq:2.3}) is valid for any $\Gamma$ in  $\mathcal{R}$.
\end{remark}
\begin{figure}
\label{fig:3}
\begin{center}
\includegraphics{figure3.eps}
\end{center}
\caption{Relevant to Lemma \ref{lem:2.1} and  Remark \ref{rem:2}}
\end{figure}


\begin{definition}\label{def:2.1} Let $y$ be continuous on $(-\infty,\infty)$ and
and $\norm{y}_{0,(-\infty,\infty)}<\infty$, we define $I_+(y)(z)$
and $I_-(y)(z)$ as
\begin{gather}
\label{2.7} I_+(y)(z)=\frac{1}{\pi}\int_{-\infty}^{\infty}
\frac{y(t)}{t-z}dt \text{ for Im }z>0; \\
\label{eq:2.8} I_-(y)(z)=\frac{1}{\pi}\int_{-\infty}^{\infty}
\frac{y(t)}{t-z}dt \text{ for Im }z<0;
\end{gather}
\end{definition}

\begin{lemma} \label{lem:2.2}
If $y\in\mathbf{A}_0$ and $\tilde y\in\tilde{\mathbf{A}}_0$, then \\
(1) $I_+(y)\in \mathbf{A}_0,\norm{I_+(y)}_0\leq C \norm{y}_0$.\\
(2) $\frac{d^k}{dz^k}I_+(\tilde y)\in
\mathbf{A}_k,\norm{\frac{d^k}{dz^k}I_+(\tilde y)}_k\leq C
\norm{\tilde y}_0$ for any integer $k\ge 1$
\end{lemma}

\begin{proof}
(1) For $z\in \{ \mathop{\rm Im}z\geq 0\}$, By Cauchy integral formula, we have
\begin{equation}
\label{eq:2.9} I_+(y)(z)=\frac{1}{\pi}\int_{r_l}
\frac{y(t)}{t-z}dt
\end{equation}
Using Lemma \ref{lem:2.1} with $\mathcal{D}=\{ \mathop{\rm Im}z>0\}$,we have $\underset{\mathcal{D}}{\sup}|(z-2i|^{\tau}|I_+(y)|\leq C\norm{y}_0$.\\
For $z\in \{ \mathop{\rm Im}z <0\}\cap\mathcal{R}$, by Plemelj formula and
Cauchy integral formula, $I_+(y)$ can be analytically continued to
low part of $\mathcal{R}$,
\begin{equation}
\label{eq:2.10} I_+(y)(z)=I_-(y)(z)+2iy(z)=\frac{1}{\pi}\int_{r_u}
\frac{y(t)}{t-z}dt+2iy(z);
\end{equation}
Using Lemma \ref{lem:2.1} with $\mathcal{D}=\{ \mathop{\rm Im}z<0\}\cap\mathcal{R}$, we have $\underset{\mathcal{D}}{\sup}|(z-2i|^{\tau}|I_-(y)(z)|\leq C\norm{y}_0$. Using (\ref{eq:2.10}) and $y\in \mathbf{A}_0$, we get the lemma.\\
(2) For $z\in \{ \mathop{\rm Im}z\geq 0\}$,By Cauchy integral formula,we have
\begin{equation}
\label{eq:2.11} \frac{d^k}{dz^k}I_+(\tilde
y)=\frac{1}{k!\pi}\int_{\tilde r_u} \frac{\tilde
y(t)}{(t-z)^{(k+1)}}dt
\end{equation}
where $\tilde r_u=[r_u]^*$.
Using Lemma \ref{lem:2.1} with $\mathcal{D}=\{ \mathop{\rm Im}z>0\}$,
we have $\sup_{\mathcal{D}}|z-2i|^{k+\tau}|\frac{d^k}{dz^k}I_+(\tilde y)|
\leq C\norm{\tilde y}_0$.
For $z\in \{ \mathop{\rm Im}z <0\}\cap\mathcal{R}$, by (\ref{eq:2.8}), Cauchy
integral formula and Plemelj formula, $I_+(\tilde y)$ can be
analytically continued to low part of $\tilde{\mathcal{R}}$,
\begin{equation}
\label{eq:2.12} I_+(\tilde y)=I_-(\tilde y)+2i\tilde
y=\frac{1}{\pi}\int_{\tilde r_l} \frac{\tilde
y(t)}{t-z}dt+2i\tilde y;
\end{equation}
So
\begin{equation}
\label{eq:2.13} \frac{d^k}{dz^k}I_+(\tilde
y)=\frac{d^k}{dz^k}I_-(\tilde y)+2i\frac{d^k}{dz^k}\tilde
y=\frac{1}{k!\pi}\int_{\tilde r_l} \frac{\tilde
y(t)}{(t-z)^k}dt+2i\frac{d^k}{dz^k}\tilde y;
\end{equation}
Using Lemma \ref{lem:2.1} with $\mathcal{D}=\{ \mathop{\rm Im}z<0\}
\cap\mathcal{R}$, we have $\underset{\mathcal{D}}{\sup}|z-2i|^{k+\tau}|
\frac{d^k}{dz^k}I_-(\tilde y)|\leq C\norm{\tilde y}_0$.
Now by the Cauchy integral formula: for $z\in \{ \mathop{\rm Im}z
<0\}\cap\mathcal{R}$
\begin{equation}
\label{eq:2.14} \tilde y(z)=\frac{1}{2\pi i}\int_{\tilde
r_u}-\int_{\tilde r_l}\frac{\tilde y}{(t-z)}dt;
\end{equation}
where $\tilde r_l=[r_l]^*,\tilde r_u=[r_u]^*$. So
\begin{equation}
\label{eq:2.15} \frac{d^k}{dz^k}\tilde y(z)=\frac{1}{2k!\pi
i}\int_{\tilde r_u}-\int_{\tilde y_l}\frac{\tilde
y}{(t-z)^{k+1}}dt;
\end{equation}
Using Lemma \ref{lem:2.1} with $\mathcal{D}=\{
\mathop{\rm Im}z<0\}\cap\mathcal{R}$ which is an angular subset of
$\tilde{\mathcal{R}}$, we have
$\underset{\mathcal{D}}{\sup}|(z-2i|^{k+\tau}|\frac{d^k}{dz^k}(\tilde
y)|\leq C\norm{\tilde y}_0$.
\end{proof}

\begin{remark} \label{rem:2.0} \rm
In general, from Cauchy integral formula and Lemma \ref{lem:2.1},
the following statements are true:
\begin{enumerate}
\item If $y\in\mathbf{A}_0(\mathcal{D})$ and $\mathcal{D}'$ is an angular
subset of $\mathcal{D}$, then $\frac{d^ky}{dz^k}\in \mathbf{A}_k(\mathcal{D}')$
and $\norm{\frac{d^ky}{dz^k}}_{k,\mathcal{D}'}\leq C \norm{y}_{0,\mathcal{D}}$.

\item If $y\in\mathbf{A}_0(\mathcal{D})$,
$(-\infty,\infty )\subset\mathcal{D}$, then
$I_{\pm}(y)\in\mathbf{A}_0(\mathcal{D})$ and
$\norm{I_{\pm}(y)}_{0,\mathcal{D}}\leq C \norm{y}_{0,\mathcal{D}}$.

\item If $y\in\mathbf{A}_0(\mathcal{D})$,
$\mathcal{D}'\cap \{\mathop{\rm Im}z<0\}$ is an angular subset of
$\mathcal{D}$, and $(-\infty,\infty )\subset\mathcal{D}$, then for any
integer $k\ge 1$, $\frac{d^kI_{+}(y)}{dz^k}\in \mathbf{A}_k(\mathcal{D}')$
and $\norm{\frac{d^kI_{+}(y)}{dz^k}}_{k,\mathcal{D}'}\leq C
\norm{y}_{0,\mathcal{D}}$.

\item If $y\in\mathbf{A}_0(\mathcal{D})$, $\mathcal{D}'\cap
\{\mathop{\rm Im}z>0\}$ is an angular subset of $\mathcal{D}$, and
$(-\infty,\infty )\subset\mathcal{D}$, then for any integer $k\ge
1$, $\frac{d^kI_{-}(y)}{dz^k}\in \mathbf{A}_k(\mathcal{D}')$ and
$\norm{\frac{d^kI_{-}(y)}{dz^k}}_{k,\mathcal{D}'}\leq C
\norm{y}_{0,\mathcal{D}}$.
\end{enumerate}
\end{remark}

\begin{lemma}
\label{lem:2.3}
If $y\in\mathbf{A}_0$ and $\tilde y\in\tilde{\mathbf{A}}$, then
 \begin{enumerate}
\item $I_-(\tilde y)\in \tilde{\mathbf{A}}_0,\norm{I_-(\tilde y)}_0
\leq C \norm{\tilde y}_0$

\item $\frac{d^k}{dz^k}I_-(y)\in
\tilde{\mathbf{A}}_k,\norm{\frac{d^k}{dz^k}I_-(y)}_k\leq C
\norm{y}_0$ for any integer $k\ge 1$
\end{enumerate}
\end{lemma}

The proof of this lemma is parallel to that of Lemma \ref{lem:2.2}.

\begin{lemma} \label{lem:2.4}
If $y\in\mathbf{A}_0$ and $\tilde y\in\tilde{\mathbf{A}}_0$, then
\begin{enumerate}
\item The Hilbert transform
$\mathcal{H}(y)(x)=(P)\int_{-\infty}^{\infty}\frac{y(t)}{t-x}dt$
can be extended to region $\mathcal{R}$ and
$\mathcal{H}(y)(z)\in\mathbf{A}_0$,$\norm{\mathcal{H}(y)}_0\leq C\norm{y}_0$.

\item The Hilbert transform $\mathcal{H}(\tilde
y)(x)=(P)\int_{-\infty}^{\infty}\frac{\tilde y(t)}{t-x}dt$ can be
extended to region $\tilde{\mathcal{R}}$ and  $\mathcal{H}(\tilde
y)(z)\in\tilde{\mathbf{A}}_0$,$\norm{\mathcal{H}(\tilde y)}_0\leq
C\norm{\tilde y}_0$.
\end{enumerate}
\end{lemma}

\begin{proof}
We prove only (1). The proof of (2) is similar.
Using Plemelj formula, we have
\begin{equation}
\label{eq:2.16}
\begin{split}
&\mathcal{H}(y)(z)=I_+(y)(z)-iy(z); \text{ for } z\in \{ \mathop{\rm Im}z>0\}\cap\mathcal{R}\\
&\mathcal{H}(y)(z)=I_-(y)(z)+iy(z);\text{ for } z\in \{
\mathop{\rm Im}z<0\}\cap\mathcal{R}
\end{split}
\end{equation}
the lemma follows from Lemma \ref{lem:2.2}.
\end{proof}

\subsection*{Formulation of Equivalent integral equations}

\begin{lemma}
\label{lem:2.5} Let $ y(z)\in \mathbf{A}_0$, then $y(z)$ is a
solution of (\ref{eq:1.2}) on real axis, if and only if  $y(z)$
satisfies
\begin{equation}
\label{eq:2.17} \epsilon^2 y''(z)-iL(z)y(z)=-L(z)I_+(y)(z)+
N(\epsilon , z, y, I_+(y)-iy),
\end{equation}
$\text {for }z\in \mathcal{R}\cap \{ \mathop{\rm Im}z>0\}$.
\end{lemma}

\begin{proof}
If $y\in\mathbf{A}_0$ satisfies (\ref{eq:1.2}), extending
(\ref{eq:1.2}) to upper half complex plane Im $z>0$ and using
Plemelj formulae and Lemma \ref{lem:2.4}, we get the equation
(\ref{eq:2.17}). Conversely, in (\ref{eq:2.17}) , let z go to to
real axis from above, using Plemelj formulae, we get
(\ref{eq:1.2}).
\end{proof}

\begin{lemma}
\label{lem:2.6} If $ \tilde y(z)\in \tilde{\mathbf{A}}_0$ is a
solution of (\ref{eq:1.2}) on real axis, then for $z\in
\tilde{\mathcal{R}}\cap \{ \mathop{\rm Im}z<0\}$, $\tilde y(z)$ satisfies
\begin{equation}
\label{eq:2.18} \epsilon^2 \tilde y''(z)+iL(z)\tilde
y(z)=-L(z)I_-(\tilde y)(z)+ N(\epsilon, z,\tilde y,
I_-(\tilde y)+i\tilde y),
\end{equation}
\end{lemma}

\begin{proof}
If $y\in\mathbf{A}_0$ satisfies (\ref{eq:1.2}), extending
(\ref{eq:1.2}) to lower half complex plane Im $z<0$ and using
Plemelj formulae and Lemma \ref{lem:2.4}, we get the equation
(\ref{eq:2.18}). Conversely, in (\ref{eq:2.18}) , let z go to to
real axis from below, using Plemelj formulae, we get
(\ref{eq:1.2}).
\end{proof}

Note that the equation  $\epsilon^2 \phi''-iL\phi=0$ has the following
two Wentzel-Kramers-Brillouin solutions (i.e WKB solutions):
\begin{gather}
\label{eq:2.19}
Y_1(z)=L^{-1/4}(z)\exp \{\frac{1}{\epsilon}P(z)\}\\
\label{eq:2.20} Y_2(z)=L^{-1/4}(z)\exp
\{-\frac{1}{\epsilon}P(z)\}
\end{gather}
The Wronskian of these two solutions is
\begin{equation}
\label{eq:2.21} W=-\frac{\sqrt{2}(1+i)}{\epsilon},
\end{equation}
While two WKB solutions to $\epsilon^2 \tilde \phi''+iL\tilde\phi
=0$ are
\begin{align}
\label{eq:2.22}
\tilde Y_1(z)&=L^{-1/4}(z)\exp \{\frac{1}{\epsilon}\tilde P(z)\}\\
\label{eq:2.23} \tilde Y_2(z)&=L^{-1/4}(z)\exp
\{-\frac{1}{\epsilon}\tilde P(z)\}
\end{align}
The Wronskian of $\tilde Y_1(z)$ and $\tilde Y_2(z)$ is
\begin{equation}
\label{2.24} \widetilde W=-\frac{\sqrt{2}(1-i)}{\epsilon},
\end{equation}
and $Y_1(z),Y_2(z)$ satisfy the equation
\begin{equation}
\label{eq:2.25} \epsilon^2 \phi''(z)-iL(z)\phi(z)+\epsilon^2
L_1(z)\phi(z)=0
\end{equation}
where
\begin{equation}
\label{eq:2.26} L_1(z)=\frac{L''(z)}{4L(z)}-\frac{5(L'(z))^2}{16 L^2(z)}
\end{equation}

\begin{remark}\label{rem:3} \rm
 From Property 1, $L_1(z)$ is analytic in $\mathcal{R}\cup\tilde{\mathcal{R}}$,
 and  $L_1(z)\sim O(|z-2i|^{-2})$.
\end{remark}

The functions $\tilde Y_1(z),\tilde Y_2(z)$ satisfy the equation
\begin{equation}
\label{eq:2.27} \epsilon^2 \tilde \phi''(z)+iL(z)\tilde
\phi(z)+\epsilon^2 L_1(z)\tilde\phi(z)=0
\end{equation}
so (\ref{eq:2.17}) can be written as
\begin{equation}
\label{eq:2.28} \epsilon^2 \phi ''-iL(z)\phi
(z)+L_1(z)\epsilon^2\phi (z) =-L(z)I_+(\phi
)(z)+N_1(\epsilon,\phi)(z)
\end{equation}
where $N_1$ is an operator
\begin{equation}
\label{eq:2.29} N_1(\epsilon, \phi)(z)=\epsilon^2
L_1(z)\phi (z)+N(\epsilon , z,\phi(z), I_+(\phi
)(z)-i\phi(z)),
\end{equation}
while  (\ref{eq:2.18}) can be written as
\begin{equation}
\label{eq:2.30} \epsilon^2 \tilde\phi ''+iL(z)\tilde \phi
(z)+L_1(z)\epsilon^2\tilde\phi (z)
=-L(z)I_-(\tilde\phi)+\tilde{N}_1(\epsilon,\tilde\phi)(z),
\end{equation}
where $\tilde{N}_1$ is an operator defined by
\begin{equation}
\label{eq:2.31} \tilde{N}_1(\epsilon,
\tilde{\phi})=\epsilon^2 L_1(z)\tilde{\phi}
(z)+N(\epsilon , z,\tilde{\phi}(z), I_-(\tilde{\phi}
)(z)+i\tilde{\phi}(z)),
\end{equation}

\begin{lemma} \label{lem:2.7}
If $\sup_{\mathcal{R}} \abs{y(z)(z-2i)^m}<\infty$, for $m\geq 0$ ,then
\[
| Y_1(z)\int_{\infty}^z y(t)Y_2(t)dt |  \leq
\epsilon\frac{C\sup \abs{(z-2i)^m y(z)}}{|z-2i|^{\gamma +m}}
\quad\text{for } z\in\mathcal{R},
\]
where $C$ is a constant independent of $\epsilon$ and $y(z)$.
\end{lemma}

\begin{proof}
{\bf Case 1}: when $\mathop{\rm Re}z> 2R$, on path $\mathcal{P}=\{t: t=z+s,
0<s<\infty\}$, $\mathop{\rm Re}(P(t)-P(z))$ goes from 0 to $\infty $ as $s\to
\infty$.
\begin{align*}
&| Y_1(z)\int_{\infty}^z y(t)Y_2(t)dt | \\
&=|L^{-1/4}\int^{\infty}_zy(t)L(t)^{-1/4}e^{-\frac{1}{\sqrt{2\epsilon}}
(P(t)-P(z))}dt | \\
&\leq
\sup\abs{(z-2i)^m y(z)}\,|L(z)|^{-1/4}\int_z^{\infty}|L(t)|^{-1/4}|t-2i|^{-m}e^{-\frac{1}
{\epsilon}\mathop{\rm Re }(P(t)-P(z))}dt\\
&\leq\epsilon \sup\abs{(z-2i)^my(z)}\,|L(z)|^{-1/4}\int_0^\infty
\frac{|L(t(s))|^{-1/4}|t(s)-2i|^{-m}}{\mathop{\rm Re}P'(t(s))}
d[e^{-\frac{1} {\epsilon}\mathop{\rm Re}(P(t)-P(z))}]
\end{align*}
Note that $|L^{-1/4}(z)|\sim C |z-2i|^{-\gamma /4}, |z-2i|\leq
|t-2i|$ for $t$ on the integral range, and we have
$\mathop{\rm Re}P'(t(s))|\geq C |t-2i|^{\gamma /2}$, so we have
$$
\big| Y_1(z)\int_{\infty}^z y(t)Y_2(t)dt\big|
\leq C \epsilon\sup\abs{(z-2i)^my(z)}|z-2i|^{-m-\gamma}
$$
{\bf Case 2:} when $|\mathop{\rm Re}z|\leq 2R$, by property 5, there is a path
$\mathcal{P}(z,\infty)$ on which $\mathop{\rm Re}P(t)$ increases as $t$
goes from $z$ to $\infty$. Using the same steps as in Case 1,
we can get the estimate.

\noindent{\bf Case 3:} when $\mathop{\rm Re}z\leq -2R$, we choose paths connecting $z$
to $\infty$ , $\mathcal{P}=\mathcal{P}_1\cup [\mathop{\rm Re}z/2,\infty)$,
where
$$\mathcal{P}_1=\{t: t=\mathop{\rm Re}z/2 +\rho e^{i\arg (z-\frac{\mathop{\rm Re}z}{2})}, 0\leq\rho\leq |z-\frac{\mathop{\rm Re}z}{2}|\}$$
Note on $\mathcal{P}_1$,by Property 4, we have
$$
\mathop{\rm Re}(P(z)-P(t))\leq -C_1\int_{|\mathop{\rm Re}t|}^{|
\mathop{\rm Re}z|}r^{\gamma/2}dr
\leq -C_1 (|\mathop{\rm Re}z|^{1+\gamma/2}-|\mathop{\rm Re}t|^{1+\gamma/2})
$$
So
\begin{align*}
&\big| Y_1(z)\int_{\mathcal{P}_1}y(t)Y_2(t)dt\big| \\
&=\big| L^{-1/4}\int_{\mathcal{P}_1}y(t)
L(t)^{-1/4}e^{-\frac{1}{\epsilon}(P(t)-P(z))}dt\big| \\
&\leq \sup\abs{(z+i)^my(z)}|L(z)^{-1/4}|\int^{|\mathop{\rm Re}
z|}_{|\mathop{\rm Re}z/2|}|L(t)|^{-1/4}|t-2i|^{-m}e^{-\frac{1}
{\epsilon}\mathop{\rm Re}(P(t)-P(z))}dt\\
&\leq C\sup\abs{(z+i)^my(z)} |z|^{-m-\gamma/2}\int^{|
\mathop{\rm Re}z|}_{|\mathop{\rm Re}z|/2}e^{-C_1\frac{1}{\epsilon}
(|\mathop{\rm Re}z|^{1+\gamma/2}-t^{1+\gamma/2})}dt\\
&\leq C\epsilon\sup\abs{(z+i)^my(z)} |z|^{-m-\gamma}
\end{align*}
and
\begin{align*}
&\big| Y_1(z)\int_{\mathop{\rm Re}z/2}^\infty y(t)Y_2(t)dt\big|\\
& =\big| L^{-1/4}\int_{\mathop{\rm Re}z/2}^\infty y(t)L(t)^{-1/4}
e^{-\frac{1}{\epsilon}(P(t)-P(z))}dt\big| \\
&\leq \sup\abs{(z-2i)^my(z)}|L(z)|^{-1/4}\int_{\mathop{\rm Re}
z/2}^{\infty}|L(t)|^{-1/4}|t-2i|^{-m}e^{-\frac{1}
{\epsilon}\mathop{\rm Re}(P(t)-P(z))}dt\\
&\leq\epsilon\sup\abs{(z-2i)^my(z)}|L(z)^{-1/4}|\\
&\quad\times \int_{\mathop{\rm Re}z/2}^\infty
\frac{|L(t(s))|^{-1/4}|t(s)-2i|^{-m}}{\mathop{\rm Re}P'(t(s))}d\big[
e^{-\frac{1} {\epsilon}\mathop{\rm Re}(P(t)-P(z))}\big]\\
&\leq C\epsilon\sup\abs{(z-2i)^my(z)} |L|^{-1/4}e^{-\frac{1}
{\epsilon}\mathop{\rm Re}(P(\mathop{\rm Re}z/2)-P(z))}\\
&\leq C\epsilon\sup\abs{(z-2i)^my(z)} |z-2i|^{-m-\gamma}
\end{align*}
Note that
$$e^{-\frac{1}
{\epsilon}\mathop{\rm Re }(P(\mathop{\rm Re}z/2)-P(z))}\leq
e^{-\frac{C_1}{\epsilon}|\mathop{\rm Re}z|^{1+\gamma}}\leq C|z-2i|^{-l}
$$
for any integer $l>0$, which completes the proof.
\end{proof}

\begin{lemma} \label{lem:2.8}
If
$\sup_{\mathbf{R}}\abs{y(z)|z-2i|^m}<\infty$, for
$m\geq 0$, then $z\in\mathcal{R}$,
\[
\big| Y_2(z)\int_{-\infty}^z y(t)Y_1(t)dt\big|
 \leq \epsilon\frac{C\sup \abs{(z+i)^m y(z)}}{|z-2i|^{\gamma +m}}\,,
\]
where $C$ is a constant independent of $\epsilon$ and $y(z)$.
\end{lemma}
The proof is similar to the proof of Lemma \ref{lem:2.7}.

\begin{lemma} \label{lem:2.9}
If
$\sup_{\tilde{\mathcal{R}}} \abs{\tilde y(z)|z-2i|^m}<\infty$, for
$ m\geq 0$, then
\begin{gather*}
\big| \tilde Y_1(z)\int_{\infty }^z \tilde y(t)\tilde Y_2(t)dt\big|
\leq \epsilon\frac{C\max\abs{(z-2i)^m\tilde y(z)}}{|z-i|^{\gamma+m}}\\
\big|\tilde Y_2(z)\int_{-\infty}^z \tilde y(t)\tilde Y_1(t)dt\big|
\leq \epsilon\frac{C\max\abs{(z-2i)^m\tilde y(z)}}{|z-i|^{\gamma+m}}
\end{gather*}
where $C$ is a constant independent of $\epsilon$ and $\tilde
y(z)$.
\end{lemma}

The proof is similar  to the proof of Lemmas \ref{lem:2.7} and
 \ref{lem:2.8}.

\begin{lemma} \label{lem:2.10}
Let $u,v\in \mathbf{A}_{0,\delta}$, then for $\delta$ sufficiently small
and $z\in \mathcal{R}$
\begin{equation}
\label{eq:2.32} |N(\epsilon,z,u,v)|\leq
C|z-2i|^{-\tau+\gamma}(\epsilon^2+\delta
(\norm{u}_0+\norm{v}_0))
\end{equation}
\end{lemma}

\begin{proof}
By (\ref{eq:1.7}), (\ref{eq:1.8}) and (\ref{eq:1.9}):
\begin{equation}
\label{eq:2.33}
\begin{split}
&|N(\epsilon,z,u,v)|\\
&\leq\sum_{k=2}^{n}|p_k(z)||T_k(u,v)|+\epsilon^2\sum_{k=0}^l|f_k(z)||Q_k(u,v)|\\
&\leq C\sum_{k=2}^n|z-2i|^{-\tau +\gamma+k\tau}\sum_{\alpha_k+\beta_k\ge k}|t_{\alpha_k,\beta_k}||u|^{\alpha_k}|v|^{\beta_k}\\
& \quad +\epsilon^2C\sum_{k=0}^l|z-2i|^{-\tau +\gamma+k\tau}\sum_{\alpha_k+\beta_k\ge k}|q_{\alpha_k,\beta_k}||u|^{\alpha_k}|v|^{\beta_k}\\
&\leq C\sum_{k=2}^n|z-2i|^{-\tau +\gamma+k\tau}\sum_{\alpha_k+\beta_k\ge k}A\rho^{\alpha_k+\beta_k}(|z-2i|^{-\tau} \norm{u})^{\alpha_k}(|z-2i|^{-\tau} \norm{v})^{\beta_k}\\
& +\epsilon^2C\sum_{k=0}^l|z-2i|^{-\tau +\gamma+k\tau}\sum_{\alpha_k+\beta_k\ge k}A\rho^{\alpha_k+\beta_k}(|z-2i|^{-\tau} \norm{u}_0)^{\alpha_k}(|z-2i|^\tau \norm{v}_0)^{\beta_k}\\
&\leq C|z-2i|^{-\tau +\gamma}\sum_{k=2}^n\sum_{\alpha_k+\beta_k\ge k}A(\rho (\norm{u}_0+\norm{v}_0))^{\alpha_k+\beta_k}\\
&\quad +C\epsilon^2|z-2i|^{-\tau
+\gamma}\sum_{k=0}^l\sum_{\alpha_k+\beta_k\ge
k}A(\rho\delta)^{\alpha_k+\beta_k}
\end{split}
\end{equation}
If $\delta$ is less than $\frac{1}{4\rho}$, then
\begin{gather*}
\sum_{k=2}^n\sum_{\alpha_k+\beta_k\ge k}A
(\rho(\norm{u}+\norm{v}))^{\alpha_k+\beta_k}
\leq C\delta (\norm{u}_0+\norm{v}_0),\\
\sum_{k=0}^l\sum_{\alpha_k+\beta_k\ge k}A(\rho\delta)^{\alpha_k+\beta_k}\leq C
\end{gather*}
hence, the Lemma follows from (\ref{eq:2.33}).
\end{proof}

\begin{lemma} \label{lem:2.11}
Let $\tilde{u},\tilde{v}\in \tilde{\mathbf{A}}_{0,\delta}$,
then for $\delta$ sufficiently enough and $z\in\tilde{\mathcal{R}}$
\begin{equation}
\label{eq:2.34} |N(\epsilon,z,\tilde u,\tilde v)|\leq
C|z-2i|^{-\tau+\gamma}(\epsilon^2+\delta (\norm{\tilde
u}_0+\norm{\tilde v}_0) )
\end{equation}
\end{lemma}

The proof is similar to that of Lemma \ref{lem:2.10}.


We want to convert (\ref{eq:2.17}) and (\ref{eq:2.18}) into
integral equations by using variation of parameters.

\begin{definition}
\label{def:2.3} Define operator $\mathcal{U}$ so that for function
$N(z)$ satisfying
$$\sup_{z\in \mathcal{R}}|z-2i|^m |N(z)|<\infty, m+\gamma>0$$
\begin{equation}
\label{eq:2.35}
\mathcal{U}(N)[z]:=\frac{Y_1(z)}{\sqrt{2}(1+i)\epsilon}
\int_{\infty}^zN(t)Y_2(t) dt-\frac{Y_2(z)}{\sqrt{2}(1+i)\epsilon}
\int_{-\infty}^zN(t)Y_1(t) dt
\end{equation}
\end{definition}

\begin{remark} \label{rem:4}
In light of Lemma \ref{lem:2.7} and Lemma
\ref{lem:2.8}, we have
$$\sup_{z\in \mathcal{R}}|z-2i|^{m+\gamma}|\mathcal{U} N[z]|<C\sup_{z\in \mathcal{R}}|z-2i|^m |N(z)| $$
\end{remark}

\begin{definition} \label{def:2.4}
Let  $\phi\in\mathbf{A}_{0,\delta}$, we define
operator $G(\epsilon,\phi)$ so that
\begin{equation} \label{eq:2.36}
G(\epsilon,\phi)[z]:=\mathcal{U}(N_1(\epsilon,\phi))[z]
\end{equation}
\end{definition}

\begin{lemma}
\label{lem:2.12} Let  $\phi(z)\in\mathbf{A}_{0,\delta}$, then for
sufficiently small $\delta$, $G(\epsilon,\phi)\in\mathbf{A}_0$ and
\begin{equation}
\label{eq:2.37} \norm{G(\epsilon,\phi)}_0\leq
C(\delta\norm{\phi}_0+\epsilon^2)
\end{equation}
\end{lemma}

\begin{proof}
Since $\phi(z)\in\mathbf{A}_{0,\delta}$, by lemma \ref{lem:2.2},
$I_+(\phi)\in\mathbf{A}_0,\norm{I_+(\phi)}\leq C\norm{\phi}_0$. By
(\ref{eq:2.29}) and Lemma \ref{lem:2.10}, we have
$$|N_1(\epsilon,t,\phi)[z]|\leq C|z-2i|^{-\tau+\gamma}(\epsilon^2+\delta\norm{\phi}_0)$$
Then the lemma follows from Remark \ref{rem:4}.
\end{proof}

\begin{lemma} \label{lem:2.13}
Let  $\phi(z)\in\mathbf{A}_{0,\delta}$, $\phi$ is
a solution to (\ref{eq:2.17}) if and only if it is a solution to
the following integral equation:
\begin{equation}
\label{eq:2.38}
\phi(z)=\mathcal{U}(-LI_+(\phi))[z]+G(\epsilon,\phi)[z]
\end{equation}
\end{lemma}

\begin{proof}
Using  variation of parameters, the equation (\ref{eq:2.17}) is
equivalent to the integral equation
\[
\phi
(z)=C_1Y_1(z)+C_2Y_2(z)+\mathcal{U}(-LI_+(\phi))[z]+G(\epsilon,\phi)[z]
\]
since $\phi(z)\in\mathbf{A}_{0,\delta}$,by (\ref{eq:1.5}) and
Lemma \ref{lem:2.2}, we have $|L(z)I_+(\phi)(z)|\leq
C|z-2i|^{-\tau+\gamma}$. By Remark \ref{rem:4},
$\mathcal{U}(LI_+(\phi))\in\mathbf{A}_0$. Note $Y_1(z)\to\infty$
as $z\to -\infty$ and $Y_2(z)\to\infty$ as $z\to \infty$, we must
have $C_1=C_2=0$.
\end{proof}

By the same method, we convert (\ref{eq:2.18}) into integral
equations.

\begin{definition} \label{def:2.5}
Define operator $\tilde{\mathcal{U}}$ so that for
function $\tilde N(z)$ satisfying
$$\underset{z\in\tilde{\mathcal{R}}}{\sup}|z-2i|^m |\tilde N(z)| <\infty, m+\gamma>0$$
\begin{equation}
\label{eq:2.39} \tilde{\mathcal{U}}(\tilde N)[z]:=\frac{\tilde
Y_1(z)}{\sqrt{2}(1-i)\epsilon}\int_{\infty}^z\tilde N(t)\tilde
Y_2(t) dt-\frac{\tilde
Y_2(z)}{\sqrt{2}(1-i)\epsilon}\int_{-\infty}^z\tilde N(t)\tilde
Y_1(t) dt
\end{equation}
\end{definition}

\begin{remark} \label{rem:5}
In light of Lemma \ref{lem:2.9}, we have
$$\sup_{z\in\tilde{\mathcal{R}}}
|z-2i|^{m+\gamma}\tilde{\mathcal{U}}(\tilde N)[z]
<\underset{z\in\tilde{\mathcal{R}}}{\sup}|z-2i|^m |\tilde N(z)|
$$
\end{remark}

\begin{definition} \label{def:2.6}
Let $\tilde\phi(z)\in\tilde{\mathbf{A}}_{0,\delta}$, we define
operator $\tilde{G}(\epsilon,\tilde\phi)$ so that
\begin{equation}
\label{eq:2.40}
\tilde{G}(\epsilon,\tilde\phi)[z]:=\mathcal{U}(\tilde{N}_1(\epsilon,\tilde\phi))[z]
\end{equation}
\end{definition}

\begin{lemma} \label{lem:2.14}
Let $\tilde\phi(z)\in\tilde{\mathbf{A}}_{0,\delta}$, then for
sufficiently small $\delta$,
$G(\epsilon,\tilde\phi)\in\tilde{\mathbf{A}}_0$ and
\begin{equation}
\label{eq:2.41} \norm{\tilde{G}(\epsilon,\tilde\phi)}_0\leq
C(\delta\norm{\tilde\phi}_0 +\epsilon^2)
\end{equation}
\end{lemma}

The proof is similar to that of Lemma \ref{lem:2.12}.

\begin{lemma} \label{lem:2.15}
Let $\tilde\phi(z)\in\tilde{\mathbf{A}}_{0,\delta}$, $\tilde\phi$ is a
solution to (\ref{eq:2.18}) if and only if it is a solution to the
following integral equation:
\begin{equation} \label{eq:2.42}
\tilde\phi(z)=\tilde{\mathcal{U}}(-LI_-(\tilde\phi))[z]
+\tilde{G}(\epsilon,\tilde\phi)[z]
\end{equation}
\end{lemma}

The proof is similar to the proof of Lemma \ref{lem:2.13}.

We would  not be able to use contraction argument in integral
equation (\ref{eq:2.38}) or (\ref{eq:2.42}) to get existence of
solution, since the linear part of the left hand side of
(\ref{eq:2.38}) or (\ref{eq:2.42}), $\mathcal{U}(-LI_{\pm}(\phi))$,
is not a contraction. In the following lemma, we are going to deal
with this linear term.

\begin{lemma} \label{lem:2.16}
Let  $\phi(z)\in\mathbf{A}_{0,\delta}$, $\phi$ is
a solution to (\ref{eq:2.38}) if and only if it is a solution to
the following integral equation:
\begin{equation} \label{eq:2.43}
 \phi (z)=iI_+(G(\epsilon,\phi))(z) +i\epsilon
I_+(G_1(\phi,\phi))(z)
+G(\epsilon,\phi)[z] +\epsilon
G_1(\phi,\phi)[z]
\end{equation}
where $G_1$ is an operator acting on two functions $u$ and $v$
\begin{multline} \label{eq:2.44}
G_1(u,v)[z]=\frac{iY_1(z)}{2\epsilon}\int_{\infty}^z
\biggl(\frac{1}{4}L^{-1/2}(t)L'(t)I_+(u
)(t)+L^{1/2}(t)\frac{d}{dt}\{I_+(v)(t) \}
\biggr)Y_2(t)dt \\
+\frac{iY_2(z)}{2\epsilon}\int_{-\infty}^z\biggl(\frac{1}{4}L^{-1/2}(t)L'(t)I_+(u
)+L^{1/2}(t)\frac{d}{dt}\{I_+(v)(t) \} \biggr) Y_1(t)dt
\end{multline}
\end{lemma}
\begin{proof}
Integrating by parts,
\begin{equation}
\label{eq:2.45} \mathcal{U}(-LI_+(\phi)) =-iI_+(\phi
)+\epsilon G_1(\phi,\phi)
\end{equation}
From (\ref{eq:2.38}),(\ref{eq:2.45}):
\begin{equation}
\label{eq:2.46} \phi(z)=-iI_+(\phi )(z)+
G(\epsilon,\phi)[z] +\epsilon G_1(\phi.\phi))[z]
\end{equation}
In above equation,Let $\mathop{\rm Im}z\to 0^+$, i.e $z$ goes to $x=\mathop{\rm Re}z$ on
the real axis, using Plemelj formulae:
\begin{equation}
\label{eq:2.47} \mathcal{H}(\phi)(x)=-i
G(\epsilon,\phi)[x] -i\epsilon G_1(\phi,\phi)[x]
\end{equation}
Using Hilbert inverse formulae:
\begin{equation}
\label{eq:2.48} \phi (x)=i
\mathcal{H}(G(\epsilon,\phi))(x) +i\epsilon
\mathcal{H}(G_1(\phi,\phi))(x)
\end{equation}
Extending (\ref{eq:2.48}) to upper half z-plane, we get (\ref{eq:2.43}).\\
Conversely,since the above steps are reversible,it can be seen
that a solution $\phi\in\mathbf{A}$ to (\ref{eq:2.43}) satisfies
(\ref{eq:2.38}).
\end{proof}

\begin{lemma} \label{lem:2.17}
Let $\tilde\phi(z)\in\tilde{\mathbf{A}}_{0,\delta}$, $\tilde\phi$ is a
solution to (\ref{eq:2.42}) if and only if  it is a solution to the following
integral equation:
\begin{equation} \label{eq:2.49}
\tilde\phi =-iI_-(\tilde G(\epsilon,\tilde\phi))
-i\epsilon I_-(\tilde G_1(\tilde\phi,\tilde\phi)) +
\tilde G(\epsilon,\tilde\phi) +\epsilon \tilde
G_1(\tilde\phi,\tilde \phi)
\end{equation}
where $\tilde G_1$ is an operator acting on $\tilde u$ and $\tilde v$
\begin{multline}
\label{eq:2.50} \tilde G_1(\tilde u,\tilde v)
=-\frac{i\tilde Y_1(z)}{2\epsilon}\int_{\infty}^z
\biggl(\frac{1}{4}L^{-1/2}(t)L'(t)I_-(\tilde u )+L^{1/2}(t)\frac{d}{dt}\{I_-(\tilde v)(t)\} \biggr)\tilde Y_2(t)dt \\
-\frac{i\tilde
Y_2(z)}{2\epsilon}\int_{-\infty}^z\biggl(\frac{1}{4}L^{-1/2}(t)L'(t)I_-(\tilde
u )+L^{1/2}(t)\frac{d}{dt} \{I_-(\tilde v)(t)\} \biggr) \tilde
Y_1(t)dt
\end{multline}
\end{lemma}

The proof is parallel to that of Lemma \ref{lem:2.16}.


To get the small $\epsilon$ factor in front of
$G_1(\phi,\phi)$ terms in equation (\ref{eq:2.43}), we paid a
price. Instead we get the derivative term $\frac{d}{dt}I_+(\phi)$
in the expression of $G_1(\phi,\phi)$. Since
$\frac{d}{dt}I_+(\phi)$ can not be estimated in terms of
$\norm{\phi}_0$ (see Remark \ref{rem:2.0}),we are not be able to
use contraction argument to prove existence in (\ref{eq:2.43}).
Same situation is true for (\ref{eq:2.49}). To circumvent this
difficulty, we replace the derivative term $\frac{d}{dt}I_+(\phi)$
in $G_1(\phi,\phi)$ with $\frac{d}{dt}I_+(\tilde\phi)$ to get
$G_1(\phi,\tilde\phi)$, and $\frac{d}{dt}I_-(\tilde\phi)$ in
$\tilde G_1(\tilde\phi,\tilde\phi)$ with $\frac{d}{dt}I_-(\phi)$
to get $\tilde G_1(\tilde\phi,\phi)$ . That is, we consider a
{\bf coupled system }of integral equations:
\begin{gather}
\label{eq:2.51} \phi
(z)=iI_+(G(\epsilon\phi)) +i\epsilon
I_+(G_1(\phi,\tilde\phi)) +G(\epsilon,\phi) +\epsilon G_1(\phi,\tilde\phi)\\
\label{eq:2.52} \tilde\phi (z)=-i I_-(\tilde
G(\epsilon,\tilde\phi)) -i\epsilon
I_-(\tilde G_1(\tilde\phi ,\phi)) +\tilde
G(\epsilon,\tilde\phi) +\epsilon \tilde G_1(\tilde\phi,\phi)
\end{gather}

\begin{remark} \label{rem:6}
If $\phi \in \mathbf{A}_{0,\delta},\tilde\phi\in\tilde{\mathbf{A}}_{0,\delta}$
are solutions to the coupled system (\ref{eq:2.51}) and
(\ref{eq:2.52}), and $\phi(x)=\tilde \phi (x)$ for real $x$,then
(\ref{eq:2.51}) is equivalent to (\ref{eq:2.43}) in
$\mathcal{R}\cup\{\mathop{\rm Im}z>0\}$,(\ref{eq:2.52}) is equivalent to
(\ref{eq:2.49}) in $\tilde{\mathcal{R}}\cup\{\mathop{\rm Im}z<0\}$. Therefore
$\phi\equiv \tilde\phi$ in $\mathcal{R}\cup\tilde{\mathcal{R}}$,
and by lemma \ref{lem:2.5}, Lemma \ref{lem:2.13} and Lemma
\ref{lem:2.16}, either $\phi$ or $\tilde\phi$ is a solution to
(\ref{eq:1.2}).
\end{remark}

\section{Existence and Uniqueness of Analytic Solutions }

In this section, we use a contraction argument to prove
the existence and uniqueness of solutions of a coupled system.
Then we argue that solutions to the coupled system are
solutions of (\ref{eq:1.2}).

\begin{lemma} \label{lem:3.1}
If $y\in\mathbf{A}_{0,\delta},\tilde y(z) \in \mathbf{\tilde A}_{0,\delta}$,
then $G_1(y,\tilde y)\in \mathbf{A}_0$ and
\begin{equation} \label{eq:3.1}
\| G_1(y,\tilde y)\|_{0}\leq C(\norm{y}_0+\| \tilde y\|_{0})
\end{equation}
where  $C$ is independent of $\epsilon$ and $y,\tilde y$.
\end{lemma}
\begin{proof}
By Lemma \ref{lem:2.2}, $\frac{d}{dz}I_+(\tilde y)\in \mathbf{A}_1$,
and $\norm{\frac{d}{dz}I_+(\tilde y)}_1\leq C\norm{\tilde y}_0$.
Since $|L^{1/2}|\sim O(|z-2i|^{\gamma /2})$, we have
\begin{equation}
\label{eq:3.2}
\sup_{z\in \mathcal{R}}|z-2i|^{1+\tau-\gamma/2}|L^{1/2}(z)\frac{d}{dz}I_+
(\tilde y)|\leq C\norm{\tilde y}_0
\end{equation}
By Lemma \ref{lem:2.2}, $I_+(y)\in \mathbf{A}_0,
\norm{I_+(y)}_0\leq C\norm{y}_0$.Using (\ref{eq:1.5}), we have
\begin{equation}
\label{eq:3.3}
\sup_{z\in \mathcal{R}}|z-2i|^{1+\tau-\gamma/2}|L^{-1/2}(z)L'(z)I_+(y)|\leq
C\norm{y}_0
\end{equation}
The Lemma follows from Lemma \ref{lem:2.7}, Lemma \ref{lem:2.8}
and (\ref{eq:2.44}).
\end{proof}

\begin{lemma} \label{lem:3.2}
If $y\in\mathbf{A}_{0,\delta},\tilde y(z) \in \mathbf{\tilde A}_{0,\delta}$,
then $\tilde G_1(\tilde y,y)\in \mathbf{\tilde A}_0$ and
\begin{equation} \label{eq:3.4}
\| \tilde G_1(\tilde y,y)\|_{0}\leq C(\norm{y}_0+\| \tilde y\|_{0}).
\end{equation}
where  $C$ is independent of $\epsilon$ and $y,\tilde y$.
\end{lemma}

The proof is similar to the proof of Lemma \ref{lem:3.1}.

\begin{lemma} \label{lem:3.3}
If $y_k\in\mathbf{A}_{0,\delta},\tilde y_k(z) \in \mathbf{\tilde A}_{0,\delta}$,
then
\begin{equation} \label{eq:3.5}
\| G_1( y_1,\tilde y_1)-G_1(y_1,\tilde y_2)\|_{0}
\leq C(\norm{y_1-y_2}_0+\| \tilde y_1-\tilde
y_2\|_{0})
\end{equation}
where  $C$ is independent of $\epsilon$, $y_k$, and $\tilde y_k$.
\end{lemma}

\begin{proof} By (\ref{eq:2.44}),
$G_1(y,\tilde y)$ is linear in $y$ and $\tilde
y$, the lemma follows from lemma \ref{lem:3.1}.
\end{proof}

\begin{lemma} \label{lem:3.4}
If $y_k\in\mathbf{A}_{0,\delta},\tilde y_k(z) \in \mathbf{\tilde A}_{0,\delta}$,
then
\begin{equation} \label{eq:3.6}
\| \tilde G_1(\tilde y_1,y_1)-\tilde G_1(\tilde y_2,y_2)\|_{0}
\leq C(\norm{y_1-y_2}_0+\| \tilde y_1-\tilde
y_2\|_{0})
\end{equation}
where  $C$ is independent of $\epsilon$, $y_k$, and $\tilde y_k$.
\end{lemma}

The proof is similar to the proof of Lemma \ref{lem:3.3}.

\begin{lemma} \label{lem:3.5}
Let $u_k\in\mathbf{A}_{0,\delta},v_k(z) \in \mathbf{A}_{0,\delta}$,
then for sufficiently small $\delta$,
\begin{equation} \label{eq:3.7}
|N(\epsilon,z,u_1,v_1)-N(\epsilon,z,u_2,v_2)|\leq
C|z-2i|^{-\tau+\gamma}(\epsilon^2+\delta)(\norm{u_1-u_2}_0
+\norm{v_1-v_2}_0)
\end{equation}
where  $C$ is independent of $\epsilon$ and $u_k,v_k$.
\end{lemma}

\begin{proof}
For $\alpha_k\ge 1,\beta_k\ge 1$,
\begin{equation}
\label{eq:3.8}
\begin{split}
&|u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}|\\
&\le |(u_1^{\alpha_k}-u_2^{\alpha_k})| |v_1|^{\beta_k} +|u_2|^{\alpha_k}|v_1^{\beta_k}-v_2^{\beta_k}|\\
&\leq
\alpha_k(|u_1|^{\alpha_k-1}+|u_2|^{\alpha_k-1})|u_1-u_2||v_1|^{\beta_k}
+\beta_k(|v_1|^{\beta_k-1}+|v_2|^{\beta_k-1})|v_1-v_2||u_2|^{\alpha_k}\\
&\leq |z-2i|^{-(\alpha_k+\beta_k)\tau}
\big\{ 2\alpha_k\delta^{\alpha_k+\beta_k-1}\norm{u_1-u_2}_0+2\beta_k
\delta^{\alpha_k+\beta_k-1}\norm{v_1-v_2}_0\big\}\\
&\le |z-2i|^{-(\alpha_k+\beta_k)\tau}\big\{
2(\alpha_k+\beta_k)\delta^{\alpha_k+\beta_k-1}(\norm{u_1-u_2}_0
+\norm{v_1-v_2}_0)\big\}
\end{split}
\end{equation}
By (\ref{eq:1.7}) and (\ref{eq:1.8}):
\begin{equation}
\label{eq:3.9}
\begin{split}
&|N(\epsilon,z,u_1,v_1)-N(\epsilon,z,u_2,v_2)|\\
&\leq \sum_{k=2}^n|p_k(z)|\sum_{\alpha_k+\beta_k\ge
k}|t_{\alpha_k,\beta_k}|
|u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}|\\
& \quad+\epsilon^2\sum_{k=0}^l|f_k(z)|\sum_{\alpha_k+\beta_k\ge
1}|q_{\alpha_k,\beta_k}|
|u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}|\\
&\le C |z-2i|^{-\tau+\gamma}(\norm{u_1-u_2}_0+\norm{v_1-v_2}_0)\\
&\quad\times \sum_{k=2}^n\sum_{\alpha_k+\beta_k\ge k}A\rho^{\alpha_k+\beta_k}
\{ 2(\alpha_k+\beta_k)\delta^{\alpha_k+\beta_k-1}\}\\
&\quad +C\epsilon^2|z-2i|^{-\tau+\gamma}(\norm{u_1-u_2}_0
+\norm{v_1-v_2}_0)\\
&\quad \times\sum_{k=1}^l\sum_{\alpha_k+\beta_k\ge
k}A\rho^{\alpha_k+\beta_k}\{2(\alpha_k+\beta_k)\delta^{\alpha_k+\beta_k-1}\}
\end{split}
\end{equation}
For $\delta\leq \frac{1}{4\rho}$,
\begin{gather*}
\sum_{\alpha_k+\beta_k\ge 2}A\rho^{\alpha_k+\beta_k}
\{ 2(\alpha_k+\beta_k)\delta^{\alpha_k+\beta_k-1}\}\leq C\delta\\
\sum_{\alpha_k+\beta_k\ge 1}A\rho^{\alpha_k
+\beta_k}\{ 2(\alpha_k+\beta_k)\delta^{\alpha_k+\beta_k-1}\}\leq C
\end{gather*}
Equation (\ref{eq:3.7}) follows from (\ref{eq:3.9}) and the inequalities
above.
\end{proof}

\begin{lemma} \label{lem:3.6}
If $y_k\in\mathbf{A}_{0,\delta}$, then
\begin{equation}\label{eq:3.10}
\| G(\epsilon,y_1)-G(\epsilon,y_2)\|_{0}\leq C(\epsilon^2
+\delta)\norm{y_1-y_2}_0
\end{equation}
where  $C$ is independent of $\epsilon$ .
\end{lemma}
\begin{proof}
By lemma \ref{lem:2.2}, we have
$I_+(y_k)\in\mathbf{A}_0, \norm{I_+(y_k)}\leq C\norm{y_k}_0$.
Replacing $u_k=y_k,v_k=I_+(y_k)-iy_k$ in Lemma \ref{lem:3.5}, we
have
\begin{equation}
\label{eq:3.11} |N(\epsilon,z,u_1,v_1)-N(\epsilon,z,u_2,v_2)|\leq
C|z-2i|^{-\tau+\gamma}(\epsilon^2+\delta)\norm{y_1-y_2}_0
\end{equation}
By Remark \ref{rem:3},
\begin{equation}
\label{eq:3.12} |\epsilon^2L_1(z)(y_1-y_2)|\leq
\epsilon^2|z-2i|^{-\tau-2}\norm{y_1-y_2}_0\,.
\end{equation}
Then the  lemma follows from (\ref{eq:2.29}), Definition \ref{def:2.4}
and Remark \ref{rem:4}.
\end{proof}

\begin{lemma}\label{lem:3.7}
If $\tilde y_k\in\mathbf{\tilde A}_{0,\delta}, k=1,2$ , then
\begin{equation} \label{eq:3.13}
\| \tilde G(\epsilon,\tilde y_1)-\tilde G(\epsilon,
\tilde y_2)\|_{0}\leq C(\epsilon^2
+\delta )\norm{\tilde y_1-\tilde y_2}_0
\end{equation}
where  $C$ is independent of $\epsilon$.
\end{lemma}

The proof is simimlar to the proof of Lemma \ref{lem:3.6}.

\begin{lemma} \label{lem:3.8}
If $\phi\in\mathbf{A}_{0,\delta},\tilde\phi\in\mathbf{A}_{0,\delta}$
is a solution of equations (\ref{eq:2.51}) and (\ref{eq:2.52}) for
sufficiently small $\delta$ but independent of $\epsilon$, then
$\norm{\phi}_0\leq K_0\epsilon^2, \norm{\tilde \phi}_0\leq
K_0\epsilon^2$. where $K_0$ is some constant independent of
$\epsilon$.
\end{lemma}

\begin{proof} By (\ref{eq:2.51}) and  Lemmas \ref{lem:2.2}, \ref{lem:2.12}, \ref{lem:3.1}:
\begin{equation} \label{eq:3.14}
\begin{split}
\norm{\phi}_0&\leq \norm{I_+(G(\epsilon,\phi))}_0+\epsilon
\norm{I_+(G_1(\phi,\tilde\phi))}+\norm{G(\epsilon,\phi)}_0
+\epsilon\norm{G_1(\phi,\tilde\phi)}_0\\
&\leq C_1\norm{G(\epsilon,\phi)}_0+C_1\epsilon\norm{G_1(\phi,\tilde\phi)}_0\\
&\leq C_2(\epsilon^2+\delta\norm
{\phi}_0)+C_2\epsilon(\norm{\phi}_0+\norm{\tilde\phi}_0)
\end{split}
\end{equation}
By (\ref{eq:2.52}), Lemmas \ref{lem:2.3}, \ref{lem:2.14}, \ref{lem:3.2}:
\begin{equation} \label{eq:3.15}
\begin{split}
\norm{\tilde\phi}_0&\leq \norm{I_-(\tilde G(\epsilon,\tilde \phi))}_0
+\epsilon\norm{I_-(\tilde G_1(\tilde\phi,\phi))}
+\norm{\tilde G(\epsilon,\tilde\phi)}_0
+\epsilon\norm{\tilde G_1(\tilde\phi,\phi)}_0\\
&\leq C_1\norm{\tilde G(\epsilon,\tilde\phi)}_0+C_1\epsilon\norm{\tilde G_1
(\tilde \phi,\phi)}_0\\
&\leq C_2(\epsilon^2+\delta\norm
{\tilde\phi}_0)+C_2\epsilon(\norm{\phi}_0+\norm{\tilde\phi}_0)\,.
\end{split}
\end{equation}
Adding (\ref{eq:3.15}) to (\ref{eq:3.14}),
\begin{equation}
\label{eq:3.16} \norm{\phi}_0+\norm{\tilde\phi}_0\leq
C_3(\epsilon+\delta)(\norm{\phi}_0+\norm{\tilde\phi}_0)+C_3\epsilon^2
\end{equation}
Choosing $\delta$ small enough so that $C_3(\epsilon+\delta)\leq
\frac{1}{2}$, the lemma follows from (\ref{eq:3.16}).
\end{proof}

\begin{definition} \label{def:3.1}
\begin{equation}
\label{eq:3.17} \mathbf{E}=\mathbf{A}_{0,\delta}\oplus
\mathbf{\tilde A}_{0,\delta}
\end{equation}
\begin{equation}
\label{eq:3.18} \norm{(y,\tilde y)}_{\mathbf{E}}
=\norm{y}_{\mathbf{A}_0}+\norm{\tilde y}_{\mathbf{\tilde A}_0}
\end{equation}
\end{definition}

It is clear that $\mathbf{E}$ is a Banach space.

\begin{definition} \label{def:3.2} Let
\begin{equation}
\label{eq:3.19} \mathbf{E}_\epsilon =\{(y,\tilde
y)\in\mathbf{E}~~|~~\norm{y}_0\leq K\epsilon^2, \norm{\tilde
y}_0\leq K\epsilon^2\},
\end{equation}
\end{definition}
\begin{definition}
\label{def:3.3} For $\mathbf{e}=(\phi,\tilde \phi)\in \mathbf{E}$,
we define operator $\mathop{O}(\mathbf{e})$ as follows:
\begin{equation}
\label{eq:3.20} O(\mathbf{e})=(O_1(\mathbf{e}),O_2(\mathbf{e}))
\end{equation}
where
\begin{gather}
\label{eq:3.21} O_1(\mathbf{e})=i
I_+(G(\epsilon,\phi)) +i\epsilon I_+(G_1(\phi,\tilde\phi))
+G(\epsilon,\phi) +\epsilon G_1(\phi,\tilde\phi) \\
\label{eq:3.22} O_2(\mathbf{e})=-iI_-(\tilde
G(\epsilon,\tilde\phi)) -i\epsilon
I_-(\tilde G_1(\tilde\phi,\phi))) +\tilde
G(\epsilon,\tilde\phi) +\epsilon \tilde
G_1(\tilde\phi,\phi)
\end{gather}
\end{definition}

\begin{theorem}\label{lem:3.9}
For sufficiently small $\epsilon$ and properly
chosen $K$,the operator $O(\mathbf{e})$ is a contraction mapping
from $\mathbf{E}_\epsilon$ to $\mathbf{E}_\epsilon$; therefore
there exists a unique solution $(\phi,\tilde\phi ) \in
\mathbf{E}_\epsilon$ to (\ref{eq:2.51}),(\ref{eq:2.52}).
\end{theorem}

\begin{proof} By (\ref{eq:3.21}) and replacing $\delta = K\epsilon^2$ in
(\ref{eq:3.14}), we have $\norm{O_1(\mathbf{e})}_0\leq
C\epsilon^2+CK\epsilon^3+CK\epsilon^4$.By (\ref{eq:3.22}) and
replacing $\delta = K\epsilon^2$ in (\ref{eq:3.15}), we have
$\norm{O_2(\mathbf{e})}_0\leq
C\epsilon^2+CK\epsilon^3+CK\epsilon^4$. Hence for $K$ suitably
chosen, we have
$\mathop{O}(\mathbf{E}_\epsilon)\subset \mathbf{E}_\epsilon$ .\\
Let $(\phi_1,\tilde \phi_1)\in \mathbf{E}_\epsilon, (\phi_2,\tilde
\phi_2)\in \mathbf{E}_\epsilon$. By (\ref{eq:3.21}), Lemmas \ref{lem:2.2},
\ref{lem:3.3}, \ref{lem:3.5}:
\[
\begin{split}
&\|\mathop{O_1}(\phi_1,\tilde \phi_1)- \mathop{O_1}(\phi_2,\tilde \phi_2)\|_{0}\\
&\leq
\norm{I_+(G(\epsilon,\phi_1)-G(\epsilon,\phi_2))}
+\epsilon \norm{I_+(G_1(\phi_1,\tilde\phi_1)-G_1(\phi_2,\tilde\phi_2))}\\
&quad+\norm{(G(\epsilon,\phi_1)-G(\epsilon,\phi_2))}
+\epsilon \norm{(G_1(\phi_1,\tilde\phi_1)-G_1(\phi_2,\tilde\phi_2))}\\
&\leq C\norm{(G(\epsilon,\phi_1)-G(\epsilon,\phi_2))}
+\epsilon C\norm{(G_1(\phi_1,\tilde\phi_1)-G_1(\phi_2,\tilde\phi_2))}\\
&\leq \epsilon
\| (y_1-y_2,\tilde y_1-\tilde y_2)\|_{\mathbf{E}}\,.
\end{split}
\]
By (\ref{eq:3.22}), Lemmas \ref{lem:2.3}, \ref{lem:3.4}, \ref{lem:3.6}, we have
\[
\begin{split}
&\|\mathop{O_2}(\phi_1,\tilde \phi_1)- \mathop{O_1}(\phi_2,\tilde \phi_2)\|_{0}\\
&\leq \norm{I_-(\tilde
G(\epsilon,\tilde\phi_1)-G(\epsilon,\tilde\phi_2))}
+\epsilon \norm{I_-(\tilde G_1(\tilde\phi_1,\phi_1)-\tilde G_1(\tilde\phi_2,\phi_2))}\\
&\quad+\norm{(\tilde G(\epsilon,\tilde\phi_1)-\tilde
G(\epsilon,\tilde \phi_2))}
+\epsilon \norm{(\tilde G_1(\tilde \phi_1,\phi_1)-\tilde G_1(\tilde\phi_2,\phi_2))}\\
&\overset{}{\leq} C\norm{(\tilde
G(\epsilon,\tilde\phi_1)-\tilde
G(\epsilon,\phi_2))}
+\epsilon C\norm{(\tilde G_1(\tilde \phi_1,\phi_1)-\tilde G_1(\tilde\phi_2,\phi_2))}\\
&\leq C \epsilon\| (\phi_1-\phi_2,\tilde \phi_1-\tilde
\phi_2)\|_{\mathbf{E}}
\end{split}
\]
\end{proof}

Next, we show that $\phi\equiv\tilde\phi$ in the
above Theorem. First, we show that $u=\phi-\tilde\phi$ satisfies a
homogeneous differential equation on real axis. Then we use the a priori
estimates obtained in \S 2 and \S 3 to prove $u\equiv 0$.

\begin{definition} \label{def:3.4}
We define the differential operators
\begin{gather} \label{eq:3.23}
\mathcal{V}\phi=\epsilon^2\phi''+(-iL+\epsilon^2L_1)\phi \,,\\
\label{eq:3.24}
\tilde{\mathcal{V}}\tilde\phi=\epsilon^2\tilde\phi''
+(iL+\epsilon^2L_1)\tilde\phi\,.
\end{gather}
\end{definition}

\begin{remark} \label{rem:7}
By (\ref{eq:2.35}) and (\ref{eq:2.39}), we have
$\mathcal{V}\mathcal{U}N=N$ and
$\tilde{\mathcal{V}}\tilde{\mathcal{U}}\tilde N=\tilde N$.
\end{remark}

\begin{lemma} \label{lem:3.10}
Let $(\phi,\tilde\phi)$ be as in Theorem
\ref{lem:3.9},then $(\phi,\tilde\phi)$ satisfies the following
equation on real axis:
\begin{equation}
\label{eq:3.25}
\epsilon^2\tilde\phi''+L\mathcal{H}(\phi)=\epsilon^2i(\mathcal{H}(\tilde\phi-\phi))''+\epsilon^2L_1[(\phi-\tilde\phi)+i\mathcal{H}(\tilde\phi-\phi)]
+ N(\epsilon,x,\phi,\mathcal{H}(\phi))
\end{equation}
\end{lemma}

\begin{proof}
In equation (\ref{eq:2.51}), let $\mathop{\rm Im} z\to 0^+$. Using Plemelj
formula,
\begin{equation}
\label{eq:3.26} \phi=i\mathcal{H}(G(\epsilon,\phi)) +
i\epsilon\mathcal{H}(G_1(\phi,\tilde \phi)),
\end{equation}
On the real axis. Applying Hilbert inverse Transform:
\begin{equation}
\label{eq:3.27} \mathcal{H}(\phi)=-iG(\epsilon,\phi) - i\epsilon
G_1(\phi,\tilde \phi)
\end{equation}
on real axis. Extending above to $\mathop{\rm Im} z>0$, and using Plemelj
formula,
\begin{equation}
\label{eq:3.28} \phi=-iI_+(\phi)+G(\epsilon,\phi) +\epsilon
G_1(\phi,\tilde \phi).
\end{equation}
Using (\ref{eq:2.45}), we have
\begin{equation}
\label{eq:3.29} -iI_+(\phi)=\mathcal{U}(
-LI_+(\phi))-\epsilon G_1(\phi,\phi).
\end{equation}
Substituting  (\ref{eq:3.29}) in (\ref{eq:3.28}), we have
\begin{equation}
\label{eq:3.30} \phi=G(\epsilon,\phi) +\epsilon G_1(\phi,\tilde
\phi-\phi)+\mathcal{U}( -LI_+(\phi) )\,.
\end{equation}
Using (\ref{eq:2.45}), we have
\begin{equation}
\label{eq:3.31} \epsilon
G_1(\phi,\tilde\phi-\phi)=iI_+(\tilde\phi-\phi)+\mathcal{U}(
-LI_+(\tilde\phi-\phi)).
\end{equation}
Substituting (\ref{eq:3.31}) in (\ref{eq:3.30}), we have
\begin{equation}
\label{eq:3.32} \phi=G(\epsilon,\phi) +\mathcal{U}(
-LI_+(\tilde\phi) )+iI_+(\tilde\phi-\phi).
\end{equation}
Applying $\mathcal{V}$ to (\ref{eq:3.30}) and Using Remark
\ref{rem:7},
\begin{equation} \label{eq:3.33}
\mathcal{V}\phi=R_1(\epsilon,\phi)-LI_+(\tilde\phi)
+\mathcal{V}(iI_+(\tilde\phi-\phi))
\end{equation}
Let $\mathop{\rm Im}z\to 0^+$ in (\ref{eq:2.33}) and using Plemelj formula, we
get the lemma.
\end{proof}

\begin{lemma} \label{lem:3.11}
Let $(\phi,\tilde\phi)$ be as in Theorem \ref{lem:3.9},
then $(\phi,\tilde\phi)$ satisfies the following
equation on real axis:
\begin{equation} \label{eq:3.34}
\epsilon^2\phi''+L\mathcal{H}(\tilde\phi)=\epsilon^2i(\mathcal{H}(\tilde\phi-\phi))''+\epsilon^2L_1[(\tilde\phi-\phi)+i\mathcal{H}(\tilde\phi-\phi)]
+ N(\epsilon,x,\tilde\phi,\mathcal{H}(\tilde\phi))
\end{equation}
\end{lemma}

\begin{proof}
Starting with (\ref{eq:2.52}), using the same steps as in the
proof of Lemma \ref{lem:3.10}, we get the lemma.
\end{proof}

From now on in this section, we are going to work with different
domains. We use notation $\mathbf{A}(\mathcal{D})$ to indicate the
dependence of function space on domain $\mathcal{D}$. Let
$\mathcal{R}_1=\mathcal{R}_{\alpha_0,\varphi_0}\cap\tilde{\mathcal{R}}_{\alpha_0,\varphi_0}$.
By Definition
\ref{def:1.1},$\mathcal{R}_1=\mathcal{R}_{\alpha_0/2,\varphi_0/2}\cup\tilde{\mathcal{R}}_{\alpha_0/2,\varphi_0/2}$.

\begin{lemma} \label{lem:3.12}
Let $(\phi,\tilde\phi)$ be as in Theorem \ref{lem:3.9},then
\begin{equation} \label{eq:3.35}
N(\epsilon,x,\tilde\phi,\mathcal{H}(\tilde\phi))-N(\epsilon,x,\phi,\mathcal{H}(\phi))=B_1(\epsilon,x)(\tilde\phi-\phi)
+B_2(\epsilon,x)\mathcal{H}(\tilde\phi-\phi)
\end{equation}
where $B_1(\epsilon,x)$ and $B_2(\epsilon,x)$ can be extended to
$\mathcal{R}_1$ and
\begin{equation} \label{eq:3.36}
\underset{z\in\mathcal{R}_1}{\sup}|z-2i|^{\tau-\gamma}|B_j(\epsilon,z)|\leq
C\epsilon^2,\quad j=1,2
\end{equation}
\end{lemma}

\begin{proof}
By Theorem \ref{lem:3.9},
$\phi\in\mathbf{A}(\mathcal{R}_1),\tilde\phi\in\mathbf{A}(\mathcal{R}_1)$
and
\begin{equation}
\label{eq:3.37} \norm{\phi}_{0,\mathcal{R}_1}\leq
K\epsilon^2,\norm{\tilde\phi}_{0,\mathcal{R}_1}\leq K\epsilon^2
\end{equation}
Let $u_1=\tilde\phi,v_1=\mathcal{H}(\tilde\phi)$ and
$u_2=\phi,v_2=\mathcal{H}(\phi)$. By Lemma \ref{lem:2.2} and Lemma
\ref{lem:2.3}:
\begin{equation}
\label{eq:3.38} \norm{u_j}_{0,\mathcal{R}_1}\leq
K\epsilon^2,\quad \norm{v_j}_{0,\mathcal{R}_1}\leq K\epsilon^2,\quad
j=1,2
\end{equation}
For any integer $m\ge 1$, define
\begin{equation}
\label{eq:3.39}
g_m(x):=\frac{[u_1(x)]^m-[u_2(x)]^m}{u_1(x)-u_2(x)}=\sum_{k=0}^{m-1}
u_1^ku_2^{m-1-k}
\end{equation}
\begin{equation}
\label{eq:3.40}
h_m(x):=\frac{[v_1(x)]^m-[v_2(x)]^m}{v_1(x)-v_2(x)}=\sum_{k=0}^{m-1}
v_1^kv_2^{m-1-k}
\end{equation}
Then $g_m$ and $h_m$ can be extended to $\mathcal{R}_1$ and for
$z\in \mathcal{R}_1$,
\begin{equation} \label{eq:3.41}
\begin{aligned}
|g_m(z)|&\leq \sum_{j+k=m-1}|u_1^{j}||u_2^{k}|\\
&\leq \sum_{j+k=m-1}\{|z-2i|^{-\tau}\norm{u_1}_0\}^{j}\{|z-2i|^{-\tau}
\norm{u_2}_0\}^{k}\\
&\leq Km\epsilon^{2(m-1)}|z-2i|^{-\tau(m-1)}
\end{aligned}
\end{equation}
Similarly
\begin{equation}
\label{eq:3.42} |h_m(z)|\leq
Km\epsilon^{2(m-1)}|z-2i|^{-\tau(m-1)}\,.
\end{equation}
For $\alpha_k+\beta_k\ge 1$,
\begin{equation} \label{eq:3.43}
\begin{aligned}
u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}
&=(u_1^{\alpha_k}-u_2^{\alpha_k})v_1^{\beta_k}
+u_2^{\alpha_k}(v_1^{\beta_k}-v_2^{\beta_k})\\
&=(u_1-u_2)g_{\alpha_k}v_1^{\beta_k}+(v_1-v_2)u_2^{\alpha_k}h_{\beta_k}
\end{aligned}
\end{equation}
By (\ref{eq:1.7}) and (\ref{eq:1.8}), we have
\begin{equation} \label{eq:3.44}
\begin{aligned}
N(\epsilon,x,u_1,v_1)-N(\epsilon,x,u_2,v_2)
&=\sum_{k=2}^np_k(x)\sum_{\alpha_k+\beta_k\ge k}t_{\alpha_k,\beta_k}
[u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}]\\
&\quad +\epsilon^2\sum_{k=1}^nf_k(x)\sum_{\alpha_k+\beta_k\ge
k}q_{\alpha_k,\beta_k}[u_1^{\alpha_k}v_1^{\beta_k}-u_2^{\alpha_k}v_2^{\beta_k}]
\end{aligned}
\end{equation}
Substituting (\ref{eq:3.43}) in (\ref{eq:3.44}), we have
(\ref{eq:3.35}) with $B_1$ and $B_2$ given by:
\begin{equation}
\label{eq:3.45}
B_1(\epsilon,x)=\sum_{k=2}^np_k(x)\sum_{\alpha_k+\beta_k\ge
k}t_{\alpha_k,\beta_k}g_{\alpha_k}(x)v_1^{\beta_k}+\epsilon^2\sum_{k=1}^nf_k(x)\sum_{\alpha_k+\beta_k\ge
k}q_{\alpha_k,\beta_k}g_{\alpha_k}(x)v_1^{\beta_k}
\end{equation}
\begin{equation}
\label{eq:3.46}
B_2(\epsilon,x)=\sum_{k=2}^np_k(x)\sum_{\alpha_k+\beta_k\ge
k}t_{\alpha_k,\beta_k}h_{\alpha_k}(x)v_1^{\beta_k}+\epsilon^2\sum_{k=1}^nf_k(x)\sum_{\alpha_k+\beta_k\ge
k}q_{\alpha_k,\beta_k}h_{\alpha_k}(x)v_1^{\beta_k}
\end{equation}
To obtain (\ref{eq:3.36}) and to show the convergence of the series in
$B_1$ and $B_2$, we consider
\begin{equation} \label{eq:3.47}
\begin{split}
&\sum_{k=2}^n|p_k(x)|\sum_{\alpha_k+\beta_k\ge k}|t_{\alpha_k,\beta_k}||g_{\alpha_k}(x)||v_1|^{\beta_k}\\
&\leq \sum_{k=2}^n|z-2i|^{-\tau+\gamma+k\tau} \\
&\quad\times\sum_{\alpha_k+\beta_k\ge k}A\rho^{\alpha_k+\beta_k}( \alpha_k(K\epsilon^2)^{\alpha_k-1}|z-2i|^{-\tau(\alpha_k-1)})(|z-2i|^{-\tau}\norm{v_1})^{\beta_k}\\
&\leq |z-2i|^{\gamma-\tau}\sum_{k=2}^n\sum_{\alpha_k+\beta_k\ge k}A\rho\alpha_k(K\epsilon^2\rho)^{\alpha_k+\beta_k -1}\\
&\leq C\epsilon^2|z-2i|^{\gamma-\tau}
\end{split}
\end{equation}
The other series can be estimated similarly. Then the proof is complete.
\end{proof}

\begin{lemma}\label{lem:3.13}
Let $u=\phi-\tilde\phi$, then $u$ satisfies the
following homogeneous equation on real axis:
\begin{equation}
\label{eq:3.48}
\epsilon^2u''-L\mathcal{H}(u)=-2\epsilon^2L_1u+B_1(\epsilon,x)u+B_2(\epsilon,x)\mathcal{H}(u)
\end{equation}
\end{lemma}

This lemma follows from Lemmas \ref{lem:3.11} and \ref{lem:3.12}.

\begin{lemma}\label{lem:3.14}
The function $u$ satisfies the  homogeneous
equations:
\begin{equation}
\label{eq:3.49}
\epsilon^2u''+(iL+\epsilon^2L_1)u=LI_+(u)-\epsilon^2L_1u+B_1(\epsilon,z)u+B_2(\epsilon,z)(
I_+(u)-iu)
\end{equation}
for $z\in\mathcal{R}_{\alpha_0/2,\varphi_0/2}$, and
\begin{equation}
\label{eq:3.50}
\epsilon^2u''+(-iL+\epsilon^2L_1)u=LI_-(u)-\epsilon^2L_1u+B_1(\epsilon,z)u+B_2(\epsilon,z)(
I_-(u)+iu)
\end{equation}
for $z\in\tilde{\mathcal{R}}_{\alpha_0/2,\varphi_0/2}$.
\end{lemma}

\begin{proof}
Extending (\ref{eq:3.48}) to $\{ \mathop{\rm Im}z>0\}$, using Plemelj
formula, we obtain (\ref{eq:3.49}). While extending
(\ref{eq:3.48}) to $\{ \mathop{\rm Im}z<0\}$, using Plemelj formula, we
obtain (\ref{eq:3.50}).
\end{proof}

\begin{definition}
We define operator
\begin{equation}
\label{eq:3.51} \tilde
G_2(\epsilon,u)[z]:=\tilde{\mathcal{U}}(-\epsilon^2L_1u+B_1(\epsilon,z)u+B_2(\epsilon,z)(
I_+(u)-iu))[z]
\end{equation}
where $\tilde{\mathcal{U}}$ is given by (\ref{eq:2.39}).
\end{definition}

\begin{lemma} \label{lem:3.15} If
$u\in\mathbf{A}(\mathcal{R}_{\alpha_0/2,\varphi_0/2})$, then
$\tilde G_2(\epsilon,u)\in\mathbf{A}(\mathcal{R}_{\alpha_0/2,\varphi_0/2})$
and
\begin{equation}
\label{eq:3.52} \norm{\tilde
G_2(\epsilon,u)}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}\leq
C\epsilon^2 \norm{u}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}
\end{equation}
\end{lemma}

\begin{proof}
 By Remark \ref{rem:2.0},
$\norm{I_+(u)}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}\leq
C\norm{u}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}$. Using
(\ref{eq:3.36}), we have
\begin{equation}
\label{eq:3.53}
\left\lvert-\epsilon^2L_1u+B_1(\epsilon,z)u+B_2(\epsilon,z)(
I_+(u)-iu)\right\rvert\leq C\epsilon^2|z-2i|^{\gamma
-\tau}\norm{u}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}
\end{equation}
We note that Lemma \ref{lem:2.9} still hold if we replace
$\tilde{\mathcal{R}}$ by $\mathcal{R}_{\alpha_0/2,\varphi_0/2}$,
the lemma follows from Lemma \ref{lem:2.9} , equation
(\ref{eq:3.51}) and (\ref{eq:3.53}).
\end{proof}

\begin{definition}
We define operator $\tilde G_3(u)$ by
\begin{multline}
\label{eq:3.54}
\tilde G_3(u)[z]:=\frac{i\tilde Y_1(z)}{2}\int_{\infty}^z[L^{1/2}(t)\frac{d}{dt}\{I_+(u)(t)\}+\frac{1}{4}L^{-1/2}(t)L'(t)I_+(u)(t)]\tilde Y_2(t) dt\\
+\frac{i\tilde
Y_2(z)}{2}\int_{-\infty}^z[L^{1/2}(t)\frac{d}{dt}\{I_+(u)(t)\}+\frac{1}{4}L^{-1/2}(t)L'(t)I_+(u)(t)]\tilde
Y_1(t) dt
\end{multline}
\end{definition}

\begin{lemma}\label{lem:3.16}
If $u\in\mathbf{A}(\mathcal{R}_{1})$, then
$\tilde G_3(u)\in\mathbf{A}(\mathcal{R}_{\alpha_0/2,\varphi_0/2})$
and
\begin{equation} \label{eq:3.55}
\norm{\tilde G_3(u)}_{0,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}
\leq C \norm{u}_{0,\mathcal{R}_{1}}
\end{equation}
\end{lemma}

\begin{proof}
Since $\mathcal{R}_{\alpha_0/2,\varphi_0/2}\cap \{ ~\mathop{\rm Im}z<0\}$ is an angular subset of $\mathcal{R}_1$,from Remark \ref{rem:2.0}, $ \norm{\frac{d}{dz}I_+(u)}_{1,\mathcal{R}_{\alpha_0/2,\varphi_0/2}}\leq C\norm{u}_{0,\mathcal{R}_1}$.\\
Since $|L^{1/2}|\sim O(|z-2i|^{\gamma /2})$, we have
\begin{equation} \label{eq:3.56}
\sup_{z\in\mathcal{R}_{\alpha_0/2,\varphi_0/2}}
|z-2i|^{1+\tau-\gamma/2}|L^{1/2}(z)\frac{d}{dz}I_+(u)|
\leq C\norm{u}_{0,\mathcal{R}_1}
\end{equation}
By Lemma \ref{lem:2.2}, $I_+(u)\in \mathbf{A}_{0,\mathcal{R}_1},
\norm{I_+(u)}_{0,\mathcal{R}_1}\leq
C\norm{u}_{0,\mathcal{R}_1}$.Using (\ref{eq:1.5}), we have
\begin{equation}
\label{eq:3.57}
\underset{z\in\mathcal{R}_{\alpha_0/2,\varphi_0/2}}{\sup}|z-2i|^{1+\tau-\gamma/2}|L^{-1/2}(z)L'(z)I_+(u)|\leq
C\norm{u}_{0,\mathcal{R}_1}
\end{equation}
The proof follows from Lemma \ref{lem:2.9}, (\ref{eq:3.54}),
(\ref{eq:3.56}) and (\ref{eq:3.57}).
\end{proof}

\begin{lemma} \label{lem:3.17}
The function $u$ satisfies the following integral equation
for $z\in\mathcal{R}_{\alpha_0/2,\varphi_0/2}$:
\begin{equation}
\label{eq:3.58} u=i[I_+(\tilde G_2(\epsilon,u))-i\tilde
G_2(\epsilon,u)]+i\epsilon [I_+(\tilde G_3(u))-i\tilde
G_3(u)]
\end{equation}
\end{lemma}

\begin{proof}
Using variation of parameters in (\ref{eq:3.49}):
\begin{equation}
\label{eq:3.59} u=\tilde{\mathcal{U}}(LI_+(u))+\tilde
G_2(\epsilon, u)
\end{equation}
Integration by parts,
\begin{equation}
\label{eq:3.60} \tilde{\mathcal{U}}(LI_+(u))
=-iI_+(u)+\epsilon\tilde G_3(u)
\end{equation}
From (\ref{eq:3.59}) and (\ref{eq:3.60}),
\begin{equation}
\label{eq:3.61} u=-iI_+(u)+\epsilon\tilde G_3(u)+\tilde G_2\,.
\end{equation}
In the above equation, let $\mathop{\rm Im}z\to 0^+$, using Plemelj formula:
\begin{equation}
\label{eq:3.62} \mathcal{H}(u)=-i\epsilon\tilde G_3(u)-i\tilde G_2
\end{equation}
Taking inverse Hilbert Transforms,
\begin{equation}
\label{eq:3.63} u=i\epsilon\mathcal{H}(\tilde
G_3(u))+i\mathcal{H}(\tilde G_2).
\end{equation}
Extending (\ref{eq:3.63}) to $\mathop{\rm Im}z>0$, we complete the proof.
\end{proof}

\begin{definition}
We define operator
\begin{equation} \label{eq:3.64}
G_2(\epsilon,u)[z]:=\mathcal{U}(-\epsilon^2L_1u+B_1(\epsilon,z)u
+B_2(\epsilon,z)(I_-(u)+iu))
\end{equation}
where $\mathcal{U}$ is given by (\ref{eq:2.35}).
\end{definition}

\begin{lemma} \label{lem:3.18}
If $u\in\mathbf{A}(\tilde{\mathcal{R}}_{\alpha_0/2,\varphi_0/2})$,
then $G_2(\epsilon,u)\in\mathbf{A}(\tilde{\mathcal{R}}_{\alpha_0/2,
\varphi_0/2})$
and
\begin{equation} \label{eq:3.65}
\norm{G_2(\epsilon,u)}_{0,\tilde{\mathcal{R}}_{\alpha_0/2,\varphi_0/2}}\leq
C\epsilon^2
\norm{u}_{0,\tilde{\mathcal{R}}_{\alpha_0/2,\varphi_0/2}}
\end{equation}
\end{lemma}

 The proof is similar to the proof of Lemma \ref{lem:3.15}.

\begin{definition}
We define operator
\begin{multline}
\label{eq:3.66}
G_3(u)[z]:=-\frac{iY_1(z)}{2}\int_{\infty}^z[L^{1/2}(t)\frac{d}{dt}\{I_-(u)(t)\}+\frac{1}{4}L^{-1/2}(t)L'(t)I_-(u)(t)] Y_2(t) dt\\
-\frac{iY_2(z)}{2}\int_{-\infty}^z[L^{1/2}(t)\frac{d}{dt}\{I_+(u)(t)\}+\frac{1}{4}L^{-1/2}(t)L'(t)I_-(u)(t)]
Y_1(t) dt
\end{multline}
\end{definition}

\begin{lemma} \label{lem:3.19}
If $u\in\mathbf{A}(\mathcal{R}_{1})$, then
$G_3(u)\in\mathbf{A}(\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2})$
and
\begin{equation} \label{eq:3.67}
\norm{G_3(u)}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}\leq C
\norm{u}_{0,\mathcal{R}_{1}}
\end{equation}
\end{lemma}

The proof is similar to the proof of Lemma \ref{lem:3.16}.

\begin{lemma} \label{lem:3.20}
the function $u$ satisfies the following integral equations
for $z\in\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}$,
\begin{equation}
\label{eq:3.68}
u=-i[I_-(G_2(\epsilon,u)+iG_2(\epsilon,u))]-i\epsilon
[I_-(G_3(u))+iG_3(u)]
\end{equation}
\end{lemma}

The proof is parallel to that of Lemma \ref{lem:3.17}.


\begin{lemma} \label{lem:3.21}
Let $(\phi,\tilde \phi)$ be as in Theorem \ref{lem:3.9},
then $\phi\equiv\tilde\phi$ in $\mathcal{R}_1$
\end{lemma}

\begin{proof} By (\ref{eq:3.58}), Lemmas \ref{lem:2.2}, \ref{lem:3.15},
\ref{lem:3.16}, we have
\begin{equation}
\label{eq:3.69}
\begin{split}
\norm{u}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}
&\leq \norm{I_+(\tilde G_2(\epsilon,u))}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}+\epsilon\norm{I_+(\tilde G_3(u))}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}\\
&\quad+\norm{\tilde G_2(\epsilon,\phi)}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}+\epsilon\norm{\tilde G_3(u)}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}\\
&\leq C_1\norm{\tilde G_2(\epsilon,u)}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}+C\epsilon\norm{\tilde G_3(u)}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}\\
&\leq C(\epsilon+\epsilon^2)\norm{u}_{0,\mathcal{R}_1}\,.
\end{split}
\end{equation}
By (\ref{eq:3.68}), Lemmas \ref{lem:2.3}, \ref{lem:3.18}, \ref{lem:3.19},
we have
\begin{equation} \label{eq:3.70}
\begin{split}
\norm{u}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}&\leq \norm{I_-(G_2(\epsilon,u))}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}+\epsilon\norm{I_-(G_3(u))}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}\\
&\quad+\norm{G_2(\epsilon,\phi)}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}
+\epsilon\norm{G_3(u)}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}\\
&\leq C_1\norm{G_2(\epsilon,u)}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}+C\epsilon\norm{G_3(u)}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}\\
&\leq C(\epsilon+\epsilon^2)\norm{u}_{0,\mathcal{R}_1}\,.
\end{split}
\end{equation}
Adding (\ref{eq:3.69}) to (\ref{eq:3.70}),
\begin{equation}
\label{eq:3.71} \norm{u}_{0,\mathcal{R}_1}\leq
\norm{u}_{0,\tilde{\mathcal{R}}_{\alpha_0,\varphi_0/2}}+\norm{u}_{0,\mathcal{R}_{\alpha_0,\varphi_0/2}}\leq
C(\epsilon+\epsilon^2)\norm{u}_{0,\mathcal{R}_1}\,.
\end{equation}
For sufficiently small $\epsilon$, (\ref{eq:3.71}) implies
$\norm{u}_{0,\mathcal{R}_1}=0$. So $u\equiv 0 $ in
$\mathcal{R}_1$.
\end{proof}

The proof of Theorem \ref{lem:1.6} follows from
Remark \ref{rem:6}, Lemma \ref{lem:3.8}, Lemma \ref{lem:3.9} and
Lemma \ref{lem:3.21}.

\section{Some Explicit Examples}
In this section  we give explicit functions of $L(x)$ so that
Properties 1--6 hold.

\noindent{\bf Example 1} Let $L(x)$ be constant.
Without loss of generality, we assume $L(x)\equiv 1$, then
$\gamma=0$, $P(z)=e^{i\frac{\pi}{4}}z$,
$\tilde P(z)=e^{-i\frac{\pi}{4}}z$. Let
$r=\{t: z+se^{i\theta},0\le s<\infty \}$ be a ray, and $z$ be a complex number.
Then along ray $r$,
$$ \mathop{\rm Re} P(t(s))=s\cos (\frac{\pi}{4}+\theta)
+\mathop{\rm Re}(e^{i\frac{\pi}{4}}z), \quad
\mathop{\rm Re} \tilde P(z)=t\cos (-\frac{\pi}{4}+\theta)
+ \mathop{\rm Re}(e^{-i\frac{\pi}{4}}z).
$$
Therefore,
\begin{gather*}
\frac{d\mathop{\rm Re}P(t(s)}{ds}=\cos (\frac{\pi}{4}+\theta)>0,
\text{ for } -\frac{\pi}{4}<\theta <\frac{\pi}{4},\\
\frac{d\mathop{\rm Re}P(t(s)}{ds}=\cos (\frac{\pi}{4}+\theta)<0,
\text{ for } \frac{3\pi}{4}<\theta <\frac{5\pi}{4},\\
\frac{d\mathop{\rm Re}\tilde P(t(s)}{ds}=\cos (\frac{\pi}{4}+\theta)>0,
\text{ for } -\frac{\pi}{4}<\theta <\frac{\pi}{4},\\
\frac{d\mathop{\rm Re}\tilde P(t(s)}{ds}=\cos (\frac{\pi}{4}+\theta)<0,
\text{ for } \frac{3\pi}{4}<\theta <\frac{5\pi}{4},
\end{gather*}
Also for  $R>0$, on the line $ \{\mathop{\rm Im}t=\text{constant},
-R\leq s=\mathop{\rm Re}t\leq R\}$,
$$
\mathop{\rm Re}P(t(s))=s\cos\frac{\pi}{4}+\text{costant}, \quad
\mathop{\rm Re}\tilde P(t(s))=s\cos\frac{\pi}{4}+\text{costant},
$$
So
$$
\frac{d\mathop{\rm Re}P(t(s)}{ds}=\cos \frac{\pi}{4}>0, \quad
\frac{d\mathop{\rm Re}\tilde P(t(s)}{ds}=\cos \frac{\pi}{4}>0 ,
$$
Therefore, if we choose $\alpha_0$ and $\varphi_0$ so that
$0<\alpha_0<1,~0<\varphi_0<\frac{\pi}{4}$, $R>0$ can be chosen
arbitrarily, then Property 1-6 hold from above equations.\\

\noindent{\bf Example 2} Let $L(x)= x^2+a^2$ with $a\ge 1$.
Without loss of
generality, we assume $L(x)=x^2+1$, then $\gamma=2$,
$P(z)=e^{i\frac{\pi}{4}}\int_0^z\sqrt{1+t^2}dt$, $\tilde
P(z)=e^{-i\frac{\pi}{4}}\int_0^z\sqrt{1+t^2}dt $, where we use
lines $\{ti:t\geq 1 \}$ and $\{ti:t\leq -1\}$ as the branch cut of
$\sqrt{t^2+1}$ and
\[
\sqrt{t^2+1}=\sqrt{|t^2+1|}e^{1/2[\arg (t+i)+\arg (t-i)]}
\]
where $-\frac{3\pi}{2}\leq \arg (t-i)\leq \frac{\pi}{2},-\frac{\pi}{2}\leq \arg(t+i)\leq \frac{3\pi}{2} .$\\

\begin{lemma} \label{lem:4.1}
 For $\theta \in (-\frac{\pi}{2},\frac{\pi}{2})$,
 then  as $|z|\to \infty$  and
 $t\in \{t(s):t(s) = z+se^{i\theta},0<s<\infty\}$,
\begin{equation} \label{eq:4.1}
\begin{gathered}
\mathop{\rm Re}P(t(s))=|t|^2\cos(\frac{\pi}{4}+2\theta)(1+o(1)),  \\
\frac{d\mathop{\rm Re}P(t(s))}{ds}=2|t|\cos(\frac{\pi}{4}+2\theta)(1+o(1)),
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
Note that
\begin{equation} \label{eq:4.2}
P(t(s))=e^{i\frac{\pi}{4}}\int_0^z\sqrt{1+\xi^2}d\xi +
e^{i\frac{\pi}{4}}\int_z^t(s)\sqrt{1+\xi^2}d\xi\,.
\end{equation}
The first integral of (\ref{eq:4.2}) is independent of $s$, and
can be treated  as a constant.
We choose the path of integration in the second integral of
(\ref{eq:4.2}) to be the ray segment $ \{\xi:\xi=z+\rho
e^{i\theta},0<\rho <s\}$, then
\begin{equation}
\label{eq:4.3}
\begin{split}
P(t(s))&=e^{i\frac{\pi}{4}}\int_0^s\sqrt{(z+\rho e^{i\theta})+i}
\sqrt{(z+\rho e^{i\theta})-i}e^{i\theta}d\rho + \text{constant}\\
&=se^{i(\theta+\frac{\pi}{4})}\int_0^1\sqrt{(z+s\omega
e^{i\theta})+i}\sqrt{(z+s\omega e^{i\theta})-i}d\omega
\end{split}
\end{equation}
The proof follows from the above  equation and the fact that
$\arg (z+s\omega e^{i\theta}+i)\to \theta$,
$\arg (z+s\omega e^{i\theta}-i)\to \theta$, as $|z|\to\infty$.
\end{proof}

\begin{lemma} \label{lem:4.2}
 For $\theta \in (-\frac{\pi}{2},\frac{\pi}{2})$, if $|z|\to \infty$
  and $t\in \{t(s):t=z+se^{i\theta},0<s<\infty\}$, then
\begin{equation}
\label{eq:4.4}
\begin{gathered}
\mathop{\rm Re}\tilde P(t(s))=|t|^2\cos(-\frac{\pi}{4}+2\theta)(1+o(1)),  \\
\frac{d\mathop{\rm Re}\tilde
P(t(s))}{ds}=2|t|\cos(-\frac{\pi}{4}+2\theta)(1+o(1)),
\end{gathered}
\end{equation}
\end{lemma}

The proof is similar to the proof of Lemma \ref{lem:4.1}.

\begin{lemma} \label{lem:4.3}
For $\theta\in (\frac{\pi}{2},\frac{3\pi}{2})$, if $|z|\to \infty$
and $t\in \{t(s):t=z+se^{i\theta},0<s<\infty\}$, then
\begin{equation}
\label{eq:4.5}
\begin{gathered}
\mathop{\rm Re}P(t)=|t|^2\cos(-\frac{3\pi}{4}+2\theta)(1+o(1)), \\
\frac{d\mathop{\rm Re}P(t(s))}{ds}=2|t|\cos(-\frac{3\pi}{4}+2\theta)(1+o(1)),
\end{gathered}
\end{equation}
\end{lemma}

\begin{proof}
The proof follows from (\ref{eq:4.3}) and the fact that
$\arg (z+s\omega e^{i\theta}+i)\to \theta$, $\arg (z+s\omega
e^{i\theta}-i)\to \theta-2\pi$, as $|z|\to\infty$.
\end{proof}

\begin{lemma} \label{lem:4.4}
For $\theta\in (\frac{\pi}{2},\frac{3\pi}{2})$, if $|z|\to \infty$
and  $t\in \{t(s):t=z+se^{i\theta},0<s<\infty\}$, then
\begin{equation}
\label{eq:4.6}
\begin{gathered}
\mathop{\rm Re}\tilde P(t(s))=|t|^2\cos(-\frac{5\pi}{4}+2\theta)(1+o(1)), \\
\frac{d\mathop{\rm Re}\tilde
P(t(s))}{ds}=2|t|\cos(-\frac{5\pi}{4}+2\theta)(1+o(1)),
\end{gathered}
\end{equation}
\end{lemma}

\begin{lemma} \label{lem:4.5}
Let $R>0$ be any fixed number, there exists a
number $0<\alpha_0<1$ so that $\frac{d \mathop{\rm Re}P(t(s))}{ds}>1$ on line
segment $\{ z: z=s+id,-R<s<R\}$ where $-\alpha_0\le d \le
\alpha_0$.
\end{lemma}

\begin{proof}
When $t(s)$ is on the real axis, i.e., $d=0$, we have
\[
\mathop{\rm Re}P(t(s))=\cos(\pi/4)\int_0^s\sqrt{1+t^2}dt
\]
so $\frac{d \mathop{\rm Re}P(z)}{ds}=1+s^2>1$.
Since $\frac{d \mathop{\rm Re}P(t)}{ds}$
is continuous with respect to $d$, we get the lemma.
\end{proof}

Now by Lemmas \ref{lem:4.1}--\ref{lem:4.4}, there exist $R$
large enough and $\varphi_0=\pi/8$ so that
\begin{equation}
\label{eq:4.7}
\frac{d \mathop{\rm Re}P(t)}{ds}>C|t|>0,\frac{d \mathop{\rm Re}
\tilde P(t)}{ds}>C|t|>0
\end{equation}
for $z>R$, $t\in\{t(s)=z+se^{i\theta}$, $0<s<\infty\}$,
$-\varphi_0<\theta<\varphi_0$. Also
\begin{equation} \label{eq:4.8}
\frac{d \mathop{\rm Re}P(t)}{ds}<-C|t|<0,\frac{d \mathop{\rm Re}
\tilde P(t)}{ds}<-C|t|<0
\end{equation}
for $z<-R$, $t\in\{t(s)=z+se^{i(\pi-\theta)}$, $0<s<\infty\}$,
$-\varphi_0<\theta<\varphi_0$.

Choose $\alpha_0$ so that Lemma \ref{lem:4.5} holds. It can
be checked easily that Properties 1--6 hold in
$\mathcal{R}_{\alpha_0,\varphi_0}$, by (\ref{eq:4.7}),
(\ref{eq:4.8}) and Lemma \ref{lem:4.5}.

\subsection*{Acknowlegment.}
The author wants to thank Professor Saleh Tanveer for suggesting this
problem and for having helpful discussions.

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\end{document}