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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 36, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/36\hfil
Hyperbolic equations with degenerate curve]
{Oblique derivative problems for second-order hyperbolic equations
with degenerate curve}

\author[G.-C. Wen\hfil EJDE-2011/36\hfilneg]
{Guo-Chun Wen}

\address{Guo-Chun Wen \newline
LMAM, School of Mathematical Sciences,
Peking University, Beijing 100871, China}
\email{Wengc@math.pku.edu.cn}

\thanks{Submitted December 22, 2010. Published March 3, 2011.}
\subjclass[2000]{35L20, 35L80}
\keywords{Oblique derivative problem; hyperbolic equations; degenerate curve}

\begin{abstract}
 The present article concerns the oblique derivative problem
 for second order hyperbolic equations with degenerate circle arc.
 Firstly the formulation of the oblique derivative problem for
 the equations is given, next the representation and estimates
 of solutions for the above problem are obtained, moreover the
 existence of solutions for the problem is proved by the successive
 iteration of solutions of the equations. In this article, we
 use the complex analytic method, namely the new partial
 derivative notations, hyperbolic complex functions are
 introduced, such that the second order hyperbolic equations with
 degenerate curve are reduced to the first order hyperbolic complex
 equations with singular coefficients, then the advantage of complex
 analytic method can be applied.
\end{abstract}

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

\section{Formulation of the oblique derivative problem}


 In \cite{b1,b2,r1,s1,s2,w3,w4,w5},
 the authors posed and discussed the Cauchy
problem, Dirichlet problem and oblique derivative boundary value
problem of second order hyperbolic equations and mixed equations
with parabolic degenerate straight lines by using the methods of
integral equations, functional analysis, energy integrals, complex
analysis and so on, the obtained results possess the important
applications. Here we generalize the above results to the oblique
derivative problem of hyperbolic equations with degenerate circle
arc. In this article, the used notations are the same as in
\cite{w1,w2,w3,w4,w5}.

  Let $D$ be a simply connected bounded domain $D$ in the
hyperbolic complex plane $\mathbb{C}$ with the boundary
$\partial D=L\cup L_0$, where $L=L_1\cup L_2$.
Herein and later on, denote $\hat
y=y - \sqrt{R^2-x^2}$, and
\begin{gather*}
L_1 = \{x + G(\hat y) = R_*,x \in [R_*,0]\},\quad
L_2 = \{x - G(\hat y) = R^*,x \in [0,R^*]\}, \\
 L_0 = \{R_*\le x\le R^*, \hat y=0\},
\end{gather*}
in which $K(\hat y)=-|\hat y|^m$, $m, R$ are positive numbers,
$R_*=-R$, $R^*=R$,
$z_0=z_1=jy_0=jy_1$ the intersection of $L_1, L_2$,
$G(\hat y)=\int_0^{\hat y} \sqrt{|K(t)|}dt$,
$H(\hat y)=|K(\hat y)|^{1/2}$. In this article
we use the hyperbolic unit $j$ with the condition $j^2=1$ in
$\overline{D}$, and $x+jy, w(z)=U(z)+jV(z)=[H(\hat y)u_x-ju_y]/2$ are
called the hyperbolic number and hyperbolic complex function in $D$.
Consider the second-order linear equation of hyperbolic type with degenerate
circle arc
\begin{equation}
Lu=K(\hat y)u_{xx}+u_{yy}+au_x+bu_y+cu=-d\quad\text{in }D,\label{e1}
\end{equation}
where $a,b,c,d$ are real functions of $z$ ($z\in \overline D$), and suppose
that the equation \eqref{e1} satisfies the following conditions:
\subsection*{Condition C}
The coefficients $a,b,c,d$ in $\overline{D}$ satisfy
\begin{equation}
\begin{gathered}
 \tilde C[d,\overline{D}]=C[d,\overline{D}]+C[d_x,\overline{D}] \le k_1, \quad
\tilde C[\eta,\overline{D}]\le k_0,\quad  \eta=a,b,c, \\
|a(x,y)|{|\hat y|^{1-m/2}}=\varepsilon_1(\hat y)\quad\text{as }
\hat y\to0,\;m\ge2, \;z\in\overline{D},
\end{gathered}\label{e2}
\end{equation}
in which $\varepsilon_1(\hat y)$ is a non-negative function satisfying
the condition: $\varepsilon_1(\hat y)\to 0$ as $\hat y\to0$.

  To write the complex form of the above equation,
denote $Y=G(\hat y)$, $\hat y=y-\sqrt{R^2-x^2}$,
$\hat x=x$, and
\begin{gather*}
 W(z)=U+jV = \frac12[H(\hat y)u_x -ju_y]=\frac{H(\hat y)}
2[u_x  - ju_Y] = H(\hat y)u_Z, \\
 H(\hat y)W_{\overline Z}=\frac{H(\hat y)}2[W_x+jW_Y]
=\frac12[H(\hat y)u_x+jW_y]=W_{\overline{\tilde z}}\quad\text{in }\overline{D},
\end{gather*}
where
$Z = Z(z)  = x + jY = x + jG(\hat y)$ in $\overline{D}$,
$G(\hat y)=\int_0^{\hat y}H(t)dt$,
$H(\hat y)=\sqrt{|K(\hat y)|}$. Moreover,
\begin{equation}
\begin{gathered}
\begin{aligned}
-K(\hat y)u_{xx}-u_{yy}
&=H(\hat y)[H(\hat y)u_x-ju_y]_x 
+j[H(\hat y)u_x-ju_y]_y-[jH_y+HH_x]u_x \\
&=4H(\hat y)W_{\overline Z} - [jH_y/H + H_x]Hu_x \\
&= au_x + bu_y + cu+d,
\end{aligned} \\
\begin{aligned}
&H(\hat y)W_{\overline Z}\\
&= H[W_x + jW_Y]/2 \\
&= H[(U + jV)_x + j(U + jV)_Y]/2 \\
&=\frac14(e_1 - e_2)(H_{\hat y}/H)Hu_x + (e_1 + e_2)
[(H_x + a/H)Hu_x  + bu_y + cu + d],
\end{aligned}
\\
(U+V)_\mu=\frac1{4H}\{2[H_{\hat y}/H + H_x+a/H]U-2bV+cu+d\},
\quad \text{in }D,\\
(U-V)_\nu=\frac1{4H}\{-2[H_{\hat y}/H - H_x - a/H]U - 2bV
 +cu+d\}, \quad \text{in }D,
\end{gathered} \label{e3}
\end{equation}
where
$e_1 = (1+j)/2$, $e_2 = (1-j)/2$, $x=\mu+\nu$, $Y=\mu-\nu$,
$\partial x/\partial\mu=1/2=\partial Y/\partial\mu$,
$\partial x/\partial\nu=1/2=-\partial Y/\partial\nu$.
Hence the
complex form of \eqref{e1} can be written as
\begin{equation}
\begin{gathered}
W_{\overline{\tilde z}}\, =A_1W+A_2\overline W+A_3u+A_4\quad\text{in }\overline{D}, \\
 u(z) = 2\operatorname{Re}  \int_{z_0}^z [\frac{U(z)}{H(\hat
y)} - jV(z)]dz + b_0 \quad\text{in }\overline{D},
\end{gathered}\label{e4}
\end{equation}
where $b_0=u(z_0)$, $z_0=jy_0$, and the coefficients
$A_l=A_l(z)$ ($l=1,2,3,4$) are as follows
\begin{gather*}
A_1 = \frac14[\frac a{H}+\frac{jH_{\hat y}}{H} + H_x -jb],\quad
A_2 = \frac14[\frac a{H}+\frac{jH_{\hat y}}{H} + H_x +jb],\\
A_3=\frac c4,\quad
A_4=\frac d4\quad\text{in }\overline{D}.
\end{gather*}
For convenience, sometimes the hyperbolic complex number
$\hat z=\hat x+j\hat y=x+j(y-\sqrt{R^2-x^2})$ and the function
$F[z(Z)]$ are simply written as $z=x+j\hat y$ and $F(Z)$ respectively.
We mention that in
this article, three domains; i.e., the original domain $D$, the
characteristic domain $D_{\hat z}$ and the image domain $D_Z$ are
used, and the corresponding characteristic domain $D_{\hat z}$
almost is written as the original domain $D$.

  The oblique derivative problem for \eqref{e1} may be
formulated as follows.

\subsection*{Problem  O}
 Find a continuous solution
$u(z)$ of \eqref{e1} in $\overline D\backslash L_0$, which satisfies the boundary
conditions
\begin{equation}
\begin{gathered}
\frac12\frac{\partial u}{\partial l} = \frac1{H(y)}\operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}]
 = \operatorname{Re}[\overline{\Lambda(z)}u_z] = r(z),\quad  z \in L = L_1 \cup L_2,\\
 u(z_0)=b_0,\quad
\frac1{H(\hat y)}\operatorname{Im}[\overline{\lambda(z)}u_{\tilde z}]|_{z=z_0}
=\operatorname{Im}[\overline{\Lambda(z)} u_z]|_{z=z_0}=b_{1},
\end{gathered}\label{e5}
\end{equation}
in which $l$ is a given vector at every point
$z\in L$, $u_{\tilde z}=[H(\hat y)u_x -ju_{y}]/2$,
$u_{\bar{\tilde z}}=[H(\hat y)u_x+ju_{y}]/2, b_0,b_1$
are real constants,
$\lambda(z)=\lambda_1(x)+j\lambda_2(x)$,
$\Lambda(z) = \cos(l,x) + j\cos(l,y)$,
$R(z) = H(\hat y)r(z)$, $z \in  L$,
$b'_1 = H(\hat y_1)b_1$, $\lambda_1(z)$ and
$\lambda_2(x)$ are real functions, $\lambda(z),r(z),b_0,b_1$ satisfy the
conditions
\begin{equation}
\begin{gathered}
C^1[\lambda(z),L] \le k_0, \quad
C^1[r(z),L] \le k_2, \quad
|b_0|,|b_1| \le k_2,\\
\max_{z\in L_1}\frac1{|\lambda_1(x) - \lambda_2(x)|},\quad
\max_{z\in L_2}\frac1{|\lambda_1(x) + \lambda_2(x)|}\le k_0,
\end{gathered}\label{e6}
\end{equation}
in which $k_0, k_2$ are positive constants.

  For the Dirichlet problem (Problem D) with the boundary condition:
\begin{equation}
u(z)=\phi(x)\quad\text{on }L=L_1\cup L_2,\label{e7}
\end{equation}
where $L_1, L_2$ are as stated before, we find the derivative for
\eqref{e7} according to the parameter $s=x$ on $L_1,L_2$,
and obtain
\begin{gather*}
 u_s=u_x+u_yy_x=u_x-\frac{u_y}{H(\hat y)}=\phi'(x)\quad\text{on }L_1,
\\
 u_s=u_x+u_yy_x=u_x+\frac{u_y}{H(\hat y)}=\phi'(x)
\quad\text{on }L_2;
\end{gather*}
i. e.,
\begin{gather*}
 U(z)+V(z)=\frac12H(\hat y)\phi'(x)=R(z)\quad\text{on }L_1,\\
 U(z)-V(z)=\frac12H(\hat y)\phi'(x)=R(z)\quad\text{on }L_2;
\end{gather*}
i. e.,
\begin{gather*}
\operatorname{Re}[(1+j)(U+jV)]=U(z)+V(z)=R(z)\quad\text{on }L_1, \\
\operatorname{Im}[(1+j)(U+jV)]|_{z=z_0-0}=[U(z)+V(z)]|_{z=z_0-0}=R(z_0-0), \\
\operatorname{Re}[(1-j)(U+jV)]=U(z)-V(z)=R(z)\quad\text{on }L_2, \\
\operatorname{Im}[(1-j)(U +jV)]|_{z=z_0+0} = [-U(z) + V(z)]|_{z=z_0+0}
= -R(z_0 + 0),
\end{gather*}
where
\begin{gather*}
 U(z)=\frac12H(\hat y)u_x,\quad V(z)=-\frac{u_y}2, \\
\lambda_1+j\lambda_2=1-j,\quad \lambda_1=1\not=\lambda_2=-1\quad\text{on }L_1,\\
\lambda_1+j\lambda_2=1+j,\quad \lambda_1=1\not= -\lambda_2=-1\quad\text{on }L_2.
\end{gather*}
From the above formulas, we can write the complex forms of boundary
conditions of $U+jV$:
\begin{equation}
\begin{gathered}
\operatorname{Re}[\overline{\lambda(z)}(U+jV)]=R(z)\quad\text{on }L, \\
\operatorname{Im}[\overline{\lambda(z)}(U+jV)]|_{z=z_0-0}=R(z_0-0)=b'_1, \\
\lambda(z) = \begin{cases} 1 - j = \lambda_1 + j\lambda_2, \\
1 + j = \lambda_1 + j\lambda_2,
\end{cases} \quad
R(z) = \begin{cases}
 H(\hat y)\phi'(x)/2 &\text{on }L_1, \\
 H(\hat y)\phi'(x)/2 &\text{on }L_2,
\end{cases}\\
 u(z)=2\operatorname{Re}\int_{z_0}^z[\frac{U(z)}{H(\hat
y)}-jV(z)]dz+\phi(z_0)\quad\text{in } D.
\end{gathered}\label{e8}
\end{equation}
 Hence Problem D is a special case of Problem O.

  Noting that the condition \eqref{e6}, we can find a twice
continuously differentiable functions $u_0(z)$ in $\overline{D}$,
for instance, which
is a solution of the oblique derivative problem with the boundary
condition in \eqref{e5} for harmonic equations in $D$
(see \cite{w1,w2}), thus
the functions $v(z)=u(z)-u_0(z)$ in $D$ is the solution of the
following boundary value problem in the form
\begin{gather}
K(\hat y)v_{xx}+v_{yy}+av_x+bv_y+cv
=-\hat d\quad\text{in }D,\label{e9}\\
\begin{gathered}
\operatorname{Re}[\overline{\lambda(z)}v_{\tilde z}(z)] = r(z)\quad\text{on }L,\\
v(z_0)=b_0, \quad
\operatorname{Im}[\overline{\lambda(z_0)}v_{\tilde z}(z_0)]=b'_1,
\end{gathered} \label{e10}
\end{gather}
where
$W(z)=U+jV=v_{\tilde z}$ in $\overline{D}$,
$r(z)=0$ on $L$, $b_0=b'_1=0$.
Hence later on we only discuss the case of the homogeneous boundary
condition. From $v(z)=u(z)-u_0(z)$ in $\overline{D}$, we have
$u(z)=v(z)+u_0(z)$ in $\overline{D}$, and $v_y=2\tilde R_0(x)$ on
$L_0=D_{\hat z}\cap\{\hat y=0\}$, in which $\tilde R_0(x)$ is an
undermined real function. The boundary vale problem
\eqref{e9}, \eqref{e10} is
called Problem $\tilde O$.

\section{Properties of solutions to the oblique derivative problem}

 In this section, we consider the special mixed equation
\begin{equation}
\begin{gathered}
 u_{\tilde z\overline{\tilde z}}=W_{\overline{\tilde z}}=0,\quad\text{i.e.,}\\
(U+V)_\mu=0,\quad (U-V)_\nu=0\quad\text{in }\overline D,
\end{gathered}\label{e11}
\end{equation}
where $U(z)=\operatorname{Re} W(z)$, $V(z)=\operatorname{Im} W(z)$.

\begin{theorem} \label{thm1}
Any solution $u(z)$ of Problem O
for the hyperbolic equation \eqref{e11} can be expressed as
\begin{equation}
u(z) =  u(x) -  2  \int_0^{\hat y}  V(\hat y)d\hat y = 2\operatorname{Re}
\int_{z_0}^z[\frac{\operatorname{Re} W(z)}{H(\hat y)} - j\operatorname{Im}
W(z)]dz + b_0\quad\text{in }\overline{D},\label{e12}
\end{equation}
where
\begin{equation}
\begin{aligned}
W(z)&= U + jV = f(x - Y)e_1 + g(x + Y)e_2\\
&= f(\nu)e_1 + g(\mu)e_2 \\
&= \frac12\{f(x-Y) + g(x+Y) + j[f(x-Y) - g(x+Y)]\},
\end{aligned}\label{e13}
\end{equation}
in which $Y=G(\hat y)$. For convenience denote by the
functions $\lambda_1(x),\lambda_2(x),r(x)$ of $x$ the functions
$\lambda_1(z),\lambda_2(z),r(z)$ of $z$ in \eqref{e10},
and $f(x-Y)=f(\nu)$, $g(x+Y)=g(\mu)$ possess the forms
\begin{equation}
\begin{gathered}
\begin{aligned}
f(\nu) = f(x - Y)
&= \frac{2r((x-Y+R_*)/2)}{\lambda_1((x-Y +R_*)/2)-\lambda_2((x-Y +R_*)/2)}\\
&\quad -\frac{[\lambda_1((x - Y+R_*)/2)+\lambda_2((x-Y+R_*)/2)]
g(R^*)}{\lambda_1((x-Y+R_*)/2) - \lambda_2((x-Y+R_*)/2)},
\end{aligned}\\
R_*\le x-Y\le R^*, \\
(\lambda_1(0) + \lambda_2(0))g(R_*)
= (\lambda_1(0) + \lambda_2(0))(U(z_1) - V(z_1))
= r(0) - b_1\quad\text{or }0, \\
\begin{aligned}
g(\mu)&=g(x+Y)= \\
&=\frac{2r((x + Y  + R^*)/2)
 - [\lambda_1((x + Y  + R^*)/2) - \lambda_2((x +
 Y  + R^*)/2)]f(R^*)}{\lambda_1((x+Y+R^*)/2)+\lambda_2((x+Y+R^*)/2)},
\end{aligned} \\
R_*\le x+Y\le R^*, \\
(\lambda_1(0) - \lambda_2(0))f(R^*)
= (\lambda_1(0) - \lambda_2(0))(U(z_1) + V(z_1))
= r(0) + b_1\quad\text{or }0.
\end{gathered}\label{e14}
\end{equation}
Moreover $u(z)$ satisfies the estimate
\begin{equation}
C^1_\delta[u(z),\overline D]\le M_1,\quad
C^1_\delta[u(z),\overline D]\le M_2k_1,\label{e15}
\end{equation}
where
$\delta = \delta(\alpha,k_0,k_1,D)<1$, $M_1 = M_1(\alpha,k_0,k_1,D)$,
$M_2 = M_2 (\alpha,k_0,D)$ are positive constants.
\end{theorem}

\begin{proof}
Let the general solution
$$
W(z)=u_{\tilde z}=\frac12\{f(x-Y)+g(x+Y)+j[f(x-Y)-g(x+Y)]\}
$$
of \eqref{e11} be substituted in the boundary condition \eqref{e10},
thus \eqref{e10} can be rewritten as
\begin{gather*}
\lambda_1(x)U(z) -\lambda_2(x)V(z) = r(z)\quad\text{on }L, \\
\overline{\lambda(z_1)}W(z_1) = r(z_1)+jb_1;
\end{gather*}
i.e.,
\begin{gather*}
{[\lambda_1(x)-\lambda_2(x)]f(2x- R_*)+[\lambda_1(x)+\lambda_2(x)]g(R_*)
=2r(x)\quad\text{on }L_1,} \\
{[\lambda_1(x)-\lambda_2(x)]f(R^*)+[\lambda_1(x)+\lambda_2(x)]g(2x-R^*)
=2r(x)\quad\text{on }L_2,}
\end{gather*}
the above formulas can be rewritten as
\begin{gather*}
\begin{aligned}
&\Big[\lambda_1\Big(\frac{t+ R_*}2\Big) - \lambda_2\Big(\frac{t+ R_*}2
\Big)\Big]f(t)+
\Big[\lambda_1\Big(\frac{t+ R_*}2\Big)
 + \lambda_2\Big(\frac{t+ R_*}2\Big)\Big]g(R_*) \\
&=2r\Big(\frac{t+R_*}2\Big),\quad t\in[R_*,R^*],
\end{aligned}\\
(\lambda_1(0) + \lambda_2(0))g(R_*)
= (\lambda_1(0) + \lambda_2(0))(U(z_1) - V(z_1))
= r(0) - b_1\quad\text{or}\quad 0,\\
\begin{aligned}
&\Big[\lambda_1\Big(\frac{t+ R^*}2\Big) - \lambda_2\Big(\frac{t+R^*}2\Big)
\Big]f(R^*) + \Big[\lambda_1\Big(\frac{t+ R^*}2)
+\lambda_2\Big(\frac{t+ R^*}2\Big)\Big]g(t) \\
&=2r(\frac{t+R^*}2),\quad t\in[R_*,R^*],
\end{aligned} \\
(\lambda_1(0) - \lambda_2(0))f(R^*) = (\lambda_1(0) - \lambda_2(0))(U(z_1) + V(z_1))
= r(0) + b_1\quad\text{or} \quad 0,
\end{gather*}
thus the solution $W(z)$ can be expressed as \eqref{e13}. Here
we mention that for the oblique derivative boundary condition,
by \eqref{e10}, we have
$(\lambda_1(0) + \lambda_2(0))g(R_*) = 0$,
$(\lambda_1(0) - \lambda_2(0))f(R^*) = 0$. If
$(\lambda_1(x) + \lambda_2(x))g(R_*)$ on $L_1$ and
$(\lambda_1(x) - \lambda_2(x))f(R^*)$ on $L_2$ are known. From the
condition \eqref{e6} and the relation \eqref{e12},
we see that the estimate \eqref{e15}
of the solution $u(z)$ for \eqref{e11}, \eqref{e12}
is obviously true.
\end{proof}

\section{Uniqueness of solutions to the oblique derivative problem}

  The representation of solutions of Problem O for equation \eqref{e1}
is as follows.

\begin{theorem} \label{thm2}
Under Condition C, any solution $u(z)$ of Problem O for equation
\eqref{e1} in $D$ can be expressed as
\begin{equation}
\begin{gathered}
u(z)  =  2\operatorname{Re} \int_{z_0}^z[\frac{\operatorname{Re} W}{H(\hat y)}-j\operatorname{Im} W]{\rm d}z+b_0, \\
W(z)  =  w(z)+{\Phi}(z)+{\Psi}(z)\quad\text{in }D, \\
w(z)  =  f(\nu)e_1 + g(\mu)e_2, {\Phi}(z)
 = \tilde f(\nu)e_1 + \tilde g(\mu)e_2, \\
{\Psi}(z) =  \int^\mu_{R_*}  g_1(z)e_1{\rm d}\mu
+ \int_{R^*}^{\nu}  g_2(z)e_2d\nu, \\
g_l(z)  = A_l\xi+B_l\eta+Cu+D,\quad
l=1,2.\end{gathered}
\label{e16}
\end{equation}
Here
\begin{equation}
\begin{gathered}
A_1 = \frac1{4H}[\frac aH + H_x + \frac{H_y}H - b],\quad
B_1 = \frac1{4H}[\frac aH + H_x + \frac{H_y}H + b], \quad
C = \frac c{4H},\\
A_2 = \frac1{4H}[\frac aH + H_x - \frac{H_y}H - b], \quad
B_2 = \frac1{4H}[\frac aH + H_x - \frac{H_y}H + b],\quad
D = \frac d{4H},
\end{gathered}\label{e17}
\end{equation}
where $f(\nu), g(\mu)$ are as stated in \eqref{e14}, and
$\tilde f(\nu), \tilde g(\mu)$ are similar to $f(\nu), g(\mu)$,
and $\Phi(z)$ satisfy the boundary condition
\begin{equation}
\begin{gathered}
\operatorname{Re}[\overline{\lambda(z)}(\Phi(z) + {\Psi}(z))] = 0, \quad z\in L,\\
\operatorname{Im}[\overline{\lambda(z_0)}({\Phi}(z_0)
 + {\Psi}(z_0))] = 0.
\end{gathered}\label{e18}
\end{equation}
\end{theorem}

\begin{proof}
 Since Problem O is equivalent to
the Problem A for \eqref{e4}, from Theorem \ref{thm1} and \eqref{e3},
it is not difficult to see that the function ${\Psi}(z)$ satisfies
the complex equation
\begin{equation}
[{\Psi}]_{\bar{\tilde z}} = H\{[A_1\xi + B_1\eta + Cu + D]e_1 + [A_2\xi
+ B_2\eta + Cu + D]e_2\}\quad\text{in } D,\label{e19}
\end{equation}
and ${\Phi}(z)=W(z)-w(z)-{\Psi}(z)$ satisfies  \eqref{e11} and the
boundary conditions
\begin{equation}
\begin{gathered}
 \operatorname{Re}[\overline{\lambda(z)}{\Phi}(z)] = -\operatorname{Re}[\overline{\lambda(z)}{\Psi}(z)]\quad\text{on }L, \\
\operatorname{Im}[\overline{\lambda(z_0)}{\Phi}(z_0)]=-\operatorname{Im}
[\overline{\lambda(z_0)}{\Psi}(z_0)].
\end{gathered} \label{e20}
\end{equation}
 By the representation of solutions of Problem A for \eqref{e4}
as stated in the final four formulas of \eqref{e16}, we can
obtain the representation of solutions of Problem O for \eqref{e1} as
stated in the first formula of \eqref{e16}.
\end{proof}

  Next, we prove the uniqueness of solutions of Problem O for
equation \eqref{e1}.

\begin{theorem} \label{thm3}
 Suppose that \eqref{e1} satisfies the Condition C.
Then Problem O for \eqref{e1} in $D$ has a
unique solution.
\end{theorem}

\begin{proof}
 Let $u_1(z), u_2(z)$ be two
solutions of Problem O for \eqref{e1}. Then $u(z)=u_1(z)-u_2(z)$ is a
solution of the homogeneous equation
\begin{equation}
K(\hat y)u_{xx} + u_{yy} + au_x + bu_y + cu=0\quad\text{in }D,
\label{e21}
\end{equation}
satisfying the boundary conditions
\begin{equation}
\begin{gathered}
u(z) = 0 ;\quad\text{i.e., } \operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}(z)] = 0
\quad\text{on }L, \\
u(z_0)=0,\quad \operatorname{Im}[\overline{\lambda(z_0)}u_{\tilde z}(z_0)]=0,
\end{gathered}\label{e22}
\end{equation}
where the function $W(z)=[H(\hat y)u_x-ju_y]/2$ is a solution of the
homogeneous problem of Problem A; namely $W(z)$ satisfies the
homogeneous equation and boundary conditions
\begin{equation}
\begin{gathered}
W_{\bar z}=A_1W+A_2\overline W+A_3u\quad\text{in }D, \\
 u(z)=2\operatorname{Re} \int_{z_0}^z[\frac{\operatorname{Re} W}{H(\hat y)}-j\operatorname{Im} W]dz,\\
\operatorname{Re}[\overline{\lambda(z)}W(z)] = 0\quad \text{on }L, \quad
\operatorname{Im}[\overline{\lambda(z_0)}W(z_0)] = 0.
\end{gathered}\label{e23}
\end{equation}
On the basis of Theorem \ref{thm2}, the function $W(z)$ can be expressed in
the form
\begin{equation}
\begin{gathered}
W(z)={\Phi}(z)+{\Psi}(z),\\
\begin{aligned}
{\Psi}(z) &=  \int^{\mu}_{R_*}
[A_1\xi + B_1\eta + Cu]e_1d\mu +  \int^{\nu}_{R^*}
[A_2\xi + B_2\eta + Cu]e_2d\nu\\
&=  \int^{\hat y}_{y_1'} 2H(\hat
y)[A_1\xi + B_1\eta + Cu]e_1dy -   \int^{\hat
y}_{y_1''}  2H(\hat
y)[A_2\xi + B_2\eta + Cu]e_2dy
\end{aligned}
\end{gathered}\label{e24}
\end{equation}
in $D$, where $z_1'=x_1'+j\hat y'_1, z_1''=x_1''+j\hat y''_1$ are two
intersection points of $L_1,L_2$ and two families of characteristics
lines
\begin{equation}
s_1 : \frac{{\rm d}x}{{\rm d}y}={\sqrt{|K(\hat y)|}}=H(\hat y),\quad
s_2 :\frac{{\rm d}x}{{\rm d}y}= -{\sqrt{|K(\hat y)|}}=-H(\hat
y)\label{e25}
\end{equation}
passing through $z=x+\hat y\in\overline D$ respectively.
Suppose $w(z)\not\equiv0$ in the neighborhood of the point $z_1$. We
may choose a sufficiently small positive number $R_0$, such that
$8M_2MR_0 < 1$, where $M_2 = \max\{C[A_1,Q_0]$,
$C[B_1,Q_0],C[A_2,Q_0]$, $C[B_2,Q_0],C[C,Q_0]\}$,
$M = 1 + 4k_0^2(1 + 2k_0^2)$ is a positive constant, and
$M_0 = C[W(z),\overline{Q_0}]+C[u(z),\overline{Q_0}] > 0$. Herein
\[
\|W(z)\|=\hat C[W(z),\overline{Q_0}]=C[\operatorname{Re} W(z)/H(\hat y)
+j\operatorname{Im} W(z),\overline{Q_0}],
\]
$Q_0 = \{R_* \le \mu \le R_*+R_0\}
\cap\{R^*-R_0 \le \nu \le R^*\}$.
From \eqref{e14}--\eqref{e18}, \eqref{e24} and
Condition C, we have
$$
\|{\Psi}(z)\|\le8M_2M_0R_0,\quad
\|{\Phi}(z)\|\le32M_2k_0^2(1+2k_0^2)M_0R_0,
$$
thus an absurd inequality $M_0\le 8M_2MM_0R_0<M_0$ is derived. It
shows $W(z)=0$, $(x,\hat y)\in Q_0$. Moreover, we extend along the
positive direction of $\mu=x+Y$ and the negative direction of
$\nu=x-Y$ successively, and finally obtain $W(z)=0$ in $D$. This
proves the uniqueness of solutions of Problem $A$ for \eqref{e23},
 and then $u(z)=u_1(z)-u_2(z)=0$ in $D$, this shows that
Problem O for \eqref{e1} has a unique solution.
\end{proof}

\section{Solvability of the oblique derivative problem}

In this section, we prove the existence of solutions of Problem O
for  \eqref{e1} by the method of the successive approximations.

\begin{theorem} \label{thm4}
 If \eqref{e1} satisfies Condition C, then Problem O
for \eqref{e1} has a solution.
\end{theorem}

\begin{proof}
To find a solution $u(z)$
of Problem O in $D$, we first find a solution $[W(z),u(z)]$ of
Problem A for \eqref{e4} in the closed domain
$D_\delta=\overline D\cap\{\hat y\le-\delta\}$, where $\delta$ is a small
positive constant. In the following, a solution of Problem $A$
for the equation \eqref{e1} in $D_\delta$
can be found by using successive approximations. First of all,
substituting the solution
$[W_0(z),u_0(z)]=[\xi_0 e_1+\eta_0 e_2,u_0(z)]$
of Problem A for \eqref{e4} into the position of
$W=\xi e_1+\eta e_2$ on the right-hand side of \eqref{e16},
the functions
\begin{equation}
\begin{gathered}
W_1(z)= W_0(z)+{\Phi}_1(z)+{\Psi}_1(z),\\
\begin{aligned}
\Psi_1(z)&= \int^\mu_{R_*} [A_1\xi_0 + B_1\eta_0 + Cu_0 + D]e_1d\mu\\
&\quad +\int^\nu_{R^*} [A_2\xi_0 + B_2\eta_0 + Cu_0 + D]e_2d\nu,
\end{aligned}\\
 u_1(z)= 2\operatorname{Re} \int_{z_0}^z [\frac{\operatorname{Re}
W_1}{H(\hat y)} - j\operatorname{Im} W_1]dz + b_0\quad\text{in } D_\delta,
\end{gathered}\label{e26}
\end{equation}
are determined, where $\mu=x+Y, \nu=x-Y, {\Phi}_1(z)$ is a solution of
\eqref{e11} in $D_\delta$ satisfying the boundary conditions
\begin{equation}
\begin{gathered}
\operatorname{Re}[\overline{\lambda(z)}{\Phi}_1(z)] = -\operatorname{Re}[\overline{\lambda(z)}{\Psi}_1(z)]\quad\text{on }
L,\\
\operatorname{Im}[\overline{\lambda(z_0)}{\Phi}_1(z_0)]=-\operatorname{Im}[\overline{\lambda(z_0)}{\Psi}_1(z_0)].
\end{gathered}\label{e27}
\end{equation}
Thus from \eqref{e26}, we have
\begin{equation}
\begin{split}
\|W_1(z) - W_0(z)\|
& = C[W_1(z) - W_0(z),D_\delta] + C[u_1(z) - u_0(z),D_\delta] \\
& \le  2M_3M(4M_0 + 1)R',
\end{split}\label{e28}
\end{equation}
where $M_3=\max_{z\in D_\delta}(|A_1|,|B_1|,|A_2|,|B_2|,|C|)$,
$M_0 = C[w_0(z),D_\delta]+C[u_0(z),D_\delta]$,
$R'=\max(R^*,|R_*|), M=1+4k_0^2(1+2k_0^2)$ is a
positive constant similar to the one in the proof of Theorem \ref{thm3}.
Moreover, we substitute $W_1(z)=W_0(z) +{\Phi}_1(z)+{\Psi}_1(z)$ and
the corresponding functions
$\xi_1(z)=\operatorname{Re} W_1(z)+\operatorname{Im} W_1(z)$,
$\eta_1(z)=\operatorname{Re} W_1(z)-\operatorname{Im} W_1(z), u_1(z)$ into the positions
of $W(z), \xi(z), \eta(z), u(z)$ in \eqref{e16}, and similarly to
\eqref{e26}--\eqref{e28}, we can find the corresponding functions
${\Psi}_2(z),{\Phi}_2(z),u_2(z)$ in $\overline D$ and the function
\begin{gather*}
W_2(z)  =  W_0(z)+{\Phi}_2(z)+{\Psi}_2(z)\quad\text{in }D_\delta, \\
 u_2(z) =  2\operatorname{Re}  \int_{z_0}^z [\frac{\operatorname{Re}
W_2}{H(\hat y)}-j\operatorname{Im} W_2]dz+b_0.
\end{gather*}
It is clear that the function
$W_2(z)-W_1(z)$ satisfies the equality
\begin{gather*}
\begin{aligned}
W_2(z) - W_1(z)
& = \Phi_2(z)-\Phi_1(z)+\Psi_2(z) - \Psi_1(z)=\Phi_2(z) - \Phi_1(z) \\
&\quad + \int^\mu_{R_*} [A_1(\xi_1 - \xi_0) +
B_1(\eta_1 - \eta_0) + C(u_1 - u_0)]e_1d\mu\\
&\quad + \int^\nu_{R^*} [A_2(\xi_1 - \xi_0) +
B_2(\eta_1 - \eta_0) + C(u_1 - u_0)]e_2d\nu,
\end{aligned}\\
 u_2(z) - u_1(z) = 2\operatorname{Re}\int_{z_0}^z[\frac{\operatorname{Re}
W_1}{H(\hat y)}-j\operatorname{Im} W_1]dz\quad\text{in }D_\delta,
\end{gather*}
and then
$$
\|W_2 - W_1\| \le [2M_3M(4M_0 + 1)]^2 \int_0^{R'} R'{\rm
d}R' \le  \frac{[2M_3M(4M_0 + 1)R']^2}{2\,!},
$$
where $M_3$ is a constant as stated in \eqref{e28}. Thus we can find a
sequence of functions $\{W_n(z)\}$ satisfying
\begin{equation}
\begin{gathered}
W_{n}(z)  =  W_0(z)+{\Phi}_n(z)+{\Psi}_n(z), \\
{\Psi}_n(z)  =  \int^\mu_{R_*}[A_1\xi_n+B_1\eta_n+Cu_n] e_1d\mu
 + \int^\nu_{R^*}[A_2\xi_n + B_2\eta_n + Cu_n]e_2d\nu,\\
 u_n(z)   =   2\operatorname{Re} \int_{z_0}^z [\frac{\operatorname{Re}
W_n}{H(\hat y)} - j\operatorname{Im} W_n]dz + b_0,
\end{gathered}\label{e29}
\end{equation}
and
$W_{n}(z)-W_{n-1}(z)$ satisfies
\begin{equation}
\begin{gathered}
W_{n}(z) - W_{n-1}(z)={\Phi}_n(z) - {\Phi}_{n-1}(z)+{\Psi}_n(z)
- {\Psi}_{n-1}(z), \\
\begin{aligned}
&\Phi_n(z) - {\Phi}_{n-1}(z)\\
&= \int^\mu_{R_*} [A_1(\xi_{n-1} - \xi_{n-2})
+ B_1(\eta_{n-1} - \eta_{n-2})
+ C[u_{n-1} - u_{n-2})]e_1d\mu\\
&\quad+ \int^\nu_{R^*}
[A_2(\xi_{n-1} - \xi_{n-2}) +
B_2(\eta_{n-1} - \eta_{n-2})]e_2d\nu,
\end{aligned}\\
u_n(z) - u_{n-1}(z) = 2\operatorname{Re}  \int_{z_0}^z
[\frac{\operatorname{Re}(W_n - W_{n-1})}{H(\hat y)}  - j\operatorname{Im}(W_n
- W_{n-1})]dz\quad\text{in } D_\delta,
\end{gathered}\label{e30}
\end{equation}
and then
\begin{align*}
\|W_n - W_{n-1}\|
&\le   [2M_3M(4M_0 + 1)]^n \int_0^{R'}   \frac{R'^{n -1}}{(n-1)!}
{\rm d}R'\\
& \le \frac{[2M_3M(4M_0 + 1)R']^n}{n\,!}\quad\text{in }D_\delta.
\end{align*}
 From the above inequality, we see that the sequences of the
functions $\{W_n(z)\}$, $\{u_n(z)\}$; i.e.,
\begin{gather*}
W_n(z)= W_0(z)+[W_1(z)-W_0(z)]+\dots+[W_n(z)-W_{n-1}(z)],\\
u_n(z)= u_0(z) + [u_1(z) - u_0(z)] + \dots + [u_n(z) - u_{n-1}(z)],
\quad n = 1,2,\dots
\end{gather*}
converges uniformly  to a function $[W_*(z),u^*(z)]$ and
$[W_*(z),u_*(z)]$ satisfies
\begin{equation}
\begin{gathered}
W_*(z) = W_0(z)+{\Phi}_*(z)+{\Psi}_*(z), \\
\begin{aligned}
{\Psi}_n(z)& = \int^\mu_{R_*}[A_1\xi_* + B_1\eta_* + Cu_*
+ D]e_1d\mu \\
&\quad +\int^\nu_{R^*}[A_2\xi_*+B_2\eta_*+Cu_*+D]e_2d\nu,
\end{aligned}\\
 u_*(z) = u_0(z) + 2\operatorname{Re} \int_{z_0}^z [\frac{\operatorname{Re} W_*}{H(\hat y)} -  j\operatorname{Im}
W_*]dz + b_0\quad\text{in } D_\delta.
\end{gathered}\label{e31}
\end{equation}
It is easy to see that $[W_{*}(z),u_*(z)]$ satisfies
 \eqref{e4} in $D_\delta$ and the
boundary condition \eqref{e10}, hence $u_*(z)$ is just a solution of
Problem O for  \eqref{e1} in the domain $D_\delta$. Finally
letting $\delta\to 0$, we can choose a limit function $u(z)$, which is a
solution of Problem O for \eqref{e1} in $D$.
\end{proof}

\section{Oblique derivative problem in general domains}

  Now we consider some general domains with non-characteristic
boundary and prove the unique solvability of Problem O for
 \eqref{e1}. Denote by $D$ a simply connected bounded domain $D$ in
the hyperbolic complex plane ${\mathbb{C}}$ with the boundary
$\partial D =L_0\cup L$, where $L_0, L=L_1\cup L_2$ are as stated
in Section 1.

  (1) We consider the domain $D'$ with the boundary
$L_0\cup L', L'=$ $L'_1\cup L'_2$, where the parameter equations
of the curves $L'_1, L'_2$ are as follows:
\begin{equation}
L'_1=\{\hat y=-\gamma_1(s), 0\le s\le s_0\}, \quad
L_2'=\{x-G(\hat y)=R^*, 0\le x\le R^*\}.\label{e32}
\end{equation}
Herein $Y=G(\hat y)=\int_0^{\hat y}\sqrt{K(t)}dt$,
$s$ is the parameter of arc length of $L'_1$,
 $\gamma_1(s)$ on $\{0\le s\le s_0\}$ is continuously differentiable,
$\gamma_1(0)=0,\gamma_1(s)>0$
on $\{0<s\le s_0\}$, and the slope of curve $L'_1$ at a point $z^*$
is not equal to $dy/dx= -1/H(\hat y)$ of the characteristic curve
$s_2: dy/dx=-1/H(\hat y)$ at the point, where $z^*$ is an
intersection point of $L'_1$ and the characteristic curve of $s_2$,
and $z'_0=x'_0-j\gamma_1(s_0)$ is the intersection point of $L'_1$ and
$L'_2$.

  The boundary conditions of the oblique derivative problem
(Problem O') for \eqref{e1} in $D'$ are as follows:
\begin{equation}
\begin{gathered}
\frac12\frac{\partial u}{\partial\nu} = \frac1{H(\hat y)}\operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}]
 = r(z), \quad z \in L'=L_1' \cup L'_2,\\
 u(z'_0)=b_0,\quad \frac1{H(\hat y)}\operatorname{Im}[\overline{\lambda(z)}u_{\tilde
z}]|_{z=z'_0}=b_1,
\end{gathered} \label{e33}
\end{equation}
where $\lambda(z)=\lambda_1(x)+j\lambda_2(x)$,
$R(z)=H(\hat y)r(z)$ on
$L', b'_1=H(\hat y'_0)b_1=H(\operatorname{Im} z_0')b_1$, and $\lambda(z)$, $r(z)$,
$b_1'$ satisfy the conditions
\begin{equation}
\begin{gathered}
C^1[\lambda(z),L']\le k_{0}, \quad
C^1[r(z),L']\le k_2, \quad |b_0|,|b_1|\le k_2,\\
\max_{z\in L'_1}\frac1{|\lambda_1(x) - \lambda_2(x)|}\le k_0, \quad
\max_{z\in L'_2}\frac1{|\lambda_1(x) + \lambda_2(x)|}\le k_0,
\end{gathered}
\label{e34}
\end{equation}
in which $k_0, k_2$ are positive constants.

  Set $Y=G(\hat y)=\int_0^{\hat y}\sqrt{K(t)}dt$. By the
conditions in \eqref{e32}, the inverse function
$x=\sigma(\nu)=(\mu+\nu)/2$ of
$\nu=x-G(\hat y)$ can be found, and then
$\mu=2\sigma(\nu)-\nu$,
$R_*\le\nu\le R^*$. We make a transformation
\begin{equation}
\begin{gathered}
\tilde\mu=R_*[\mu-2\sigma(\nu)+\nu]/[2\sigma(\nu)-\nu]+R_*, \quad \tilde\nu=\nu, \\
2\sigma(\nu)-\nu\le\mu\le 0,\quad
R_*\le\nu\le R^*,
\end{gathered}\label{e35}
\end{equation}
where $\mu, \nu$ are real variables, its inverse transformation is
\begin{equation}
\begin{gathered}
\mu=[2\sigma(\nu)-\nu](\tilde\mu-R_*)/R_*+2\sigma(\nu)-\nu, \quad \nu=\tilde\nu, \\
R_*\le\tilde\mu\le R^*,\quad
R_*\le\tilde\nu\le R^*.
\end{gathered}\label{e36}
\end{equation}
It is not difficult to see that the transformation in \eqref{e36}
 maps the domain $D'$ onto $D$,
$\tilde x= (\tilde\mu+\tilde\nu)/2, \tilde
Y=(\tilde\mu-\tilde\nu)/2$, and $x = (\mu+\nu)/2$, $Y = (\mu-\nu)/2$.
Denote by $\tilde Z=\tilde x+j\tilde Y=f(Z)$,
$Z=x+jY=f^{-1}(\tilde Z)$ the
transformation \eqref{e35} and the inverse transformation \eqref{e36}
respectively. In this case, the system  \eqref{e3} can be
rewritten as
\begin{equation}
\begin{gathered}
\xi_\mu=A_1\xi+B_1\eta+C_1(\xi+\eta)+Du+E, \quad z \in  D',\\
\eta_\nu=A_2\xi+B_2\eta+C_2(\xi+\eta)+Du+E, \quad z \in  D'.
\end{gathered}\label{e37}
\end{equation}
Suppose that \eqref{e1} in $D'$ satisfies Condition $C$, through the
transformation \eqref{e36}, we obtain
$\xi_{\tilde\mu}= [2\sigma(\nu)-\nu]\xi_\mu/R_*$,
$\eta_{\tilde\nu}=\eta_\nu$, in $D'$, where
$\xi=U+V$, $\eta=U-V$, and then
\begin{equation}
\begin{gathered}
\xi_{\tilde\mu} = [2\sigma(\nu) - \nu][A_1\xi + B_1\eta + C_1(\xi
 + \eta) + Du + E]/R_*, \\
\eta_{\tilde\nu}=A_2\xi+B_2\eta+C_2(\xi+\eta)+Du+E\quad\text{in } D,
\end{gathered} \label{e38}
\end{equation}
and through the transformation \eqref{e36}, the boundary condition
\eqref{e33} is reduced to
\begin{equation}
\begin{gathered}
\operatorname{Re}[\overline{\lambda(f^{-1}(\tilde Z))}W(f^{-1}(\tilde Z))]
= H[\hat y(Y)]r(f^{-1}(\tilde Z)), \quad
\tilde Z = \tilde x + j\tilde Y \in L = L_1 \cup L_2, \\
\operatorname{Im}[\overline{\lambda(f^{-1}(\tilde Z'_0))}W(f^{-1}(\tilde
Z_0')]=b_1, \quad u(z_0)=b_0,
\end{gathered}\label{e39}
\end{equation}
in which $Z=f^{-1}(\tilde Z)$, $\tilde Z_0'=f(Z'_0)$,
$Z'_0=x_0'+jG[-\gamma_1(s_0)]$. Therefore, the boundary
value problem \eqref{e37}, \eqref{e33} (Problem A') is
transformed into the
boundary value problem \eqref{e38}, \eqref{e39}; i.e.,
 the corresponding Problem A in $D$.
On the basis of Theorem \ref{thm4}, we see that the boundary
value problem \eqref{e38}-\eqref{e39} has a unique solution
$w(\tilde Z)$, and
\begin{equation}
u(z) = 2\operatorname{Re} \int_{z_0'}^z[\frac{\operatorname{Re} W}{H(\hat y)}
 - j\operatorname{Im} W]dz+b_0\quad\text{in }
\begin{pmatrix} D^+\\\overline{D^-}\end{pmatrix}
\label{e40}
\end{equation}
is just a solution of Problem O' for
\eqref{e1} in $D'$ with the boundary conditions \eqref{e33},
where $W=W(\tilde Z(z)]$.

\begin{theorem} \label{thm5}
 If  \eqref{e1} in $D'$ satisfies Condition $C$ in the domain $D'$
with the boundary $L_0\cup L'_1\cup L'_2$, where $L'_1,L'_2$ are
as stated in \eqref{e32},
then Problem O' for $\eqref{e1}$ with the boundary conditions
\eqref{e33} has a unique solution $u(z)$.
\end{theorem}

(2)  Next let the domain $D''$ be a simply
connected domain with the boundary $L_0\cup$ $L''_1\cup L''_2$,
where $L_0$ is as stated before and
\begin{equation}
L''_1 = \{\hat y = \gamma_1(s),0 \le s \le s_0\},\quad
L''_2 = \{\hat y = \gamma_2(s), 0 \le x \le s'_0\},\label{e41}
\end{equation}
in which $s$ is the parameter of arc
length of $L''_1$ or $L''_2$, $\gamma_1(0)=0$,
$\gamma_2(0)=0$, $\gamma_1(s)>0$, $0<s\le s_0$, $\gamma_2(s)>0$,
$0<x \le s'_0$, and
$\gamma_1(s)$ on $0\le x\le s_0$ and $\gamma_2(s)$ on $0\le s\le s'_0$ are
continuously differentiable,
$z''_0=x''_0-j\gamma_1(s_0)=x_0''-j\gamma_2(s'_0)$.
Denote by two points
$z^*_1, z^*_2$ the intersection points of $L''_1, L''_2$ and the
characteristic curves $s_2 : dy/dx=-1/H(\hat y)$,
$s_1 : dy/dx=1/H(\hat y)$ respectively, we require that the slopes of
curves $L''_1, L''_2$ at $z^*_1, z^*_2$ are not equal to those at
the characteristic curves $s_2, s_1$ at the corresponding points,
hence $\gamma_1(s),\gamma_2(s)$ can be expressed by
$\gamma_1[s(\mu)]$ ($R_*\le\mu\le R^*$),
$\gamma_2[s(\nu)]$ ($R_*\le\nu\le R^*$).
We consider the oblique derivative problem (Problem O'')
for  \eqref{e1} in $D''$ with the boundary conditions
\begin{equation}
\begin{gathered}
\operatorname{Re}[\overline{\lambda(z)}u_{\tilde z}] = R(z), \quad
z \in L'' = L''_1 \cup  L''_2, \\
u(z''_0)=b_0,\quad \operatorname{Im}[\overline{\lambda(z)}u_{\tilde
z}]\,|_{z=z''_0}=b_1,
\end{gathered}\label{e42}
\end{equation}
where
$\lambda(z)=\lambda_1(x)+j\lambda_2(x)$, $r(z)$ satisfy the corresponding
conditions
\begin{equation}
\begin{gathered} C^1[\lambda(z),L''] \le k_{0}, \quad
C^1[r(z),L''] \le k_2, \quad |b_0|,|b_1| \le k_2,\\
\max_{z\in L''_1}\frac1{|\lambda_1(x)-\lambda_2(x)|}, \quad
\max_{z\in L''_2}\frac1{|\lambda_1(x)+\lambda_2(x)|}\le k_0,
\end{gathered}\label{e43}
\end{equation}
 in which $k_0, k_2$ are positive constants.
By the conditions in \eqref{e41}, the
inverse function $x=(\mu+\nu)/2=\tau(\mu)$,
$x=(\mu+\nu)/2= \sigma(\nu)$
of $\mu=x+G(y),\nu=x-G(y)$ can be found, namely
\begin{equation}
\mu=2\sigma(\nu)-\nu, \quad \nu=2\tau(\mu)-\mu,\quad
R_*\le\mu\le R^*,\quad  R_*\le\nu\le R^*.\label{e44}
\end{equation}
We make the transformation
\begin{equation}
\begin{gathered}
\tilde\mu=\mu, \quad
\tilde\nu=R^*[\nu-2\tau(\mu)+\mu]/[2\tau(\mu)-\mu] +R^*,\\
R_*\le \mu\le R^*,\quad 0\le\nu\le2\tau(\mu) - \mu.
\end{gathered}\label{e45}
\end{equation}
It is clear that its inverse transformation is
\begin{equation}
\begin{gathered}
\mu=\tilde\mu, \quad
\nu = \frac{[\tilde\nu-R^*][2\tau(\mu)-\mu]}{ R^*}+2\tau(\mu)-\mu, \\
R_*\le \tilde\mu\le R^*,\quad
R_*\le\tilde\nu\le R^*.
\end{gathered}\label{e46}
\end{equation}
Hence $\tilde x=(\tilde\mu+\tilde\nu)/2$,
$\tilde Y=(\tilde\mu-\tilde\nu)/2$,
$x=(\mu+\nu)/2$, $Y=(\mu-\nu)/2$.
Denote by $\tilde Z=\tilde x+j\tilde Y=g(z)$,
$Z=x+jY=g^{-1}(\tilde Z)$ the transformation \eqref{e45} and its
inverse transformation in \eqref{e46} respectively. Through the
transformation \eqref{e46}, we obtain
$(u+v)_{\tilde\mu}=(u+v)_\mu$,
$(u - v)_{\tilde\nu} = [2\tau(\mu)-\mu](u - v)_\nu /R^*$ in $D''$.
Thus the system \eqref{e37} in $D''$ is reduced to
\begin{equation}
\begin{gathered}
\xi_{\tilde\mu}=A_1\xi+B_1\eta+C_1(\xi+\eta)+Du+E \quad\text{in }D',\\
\eta_{\tilde\nu} = [2\tau(\mu)-\mu][A_2\xi + B_2\eta + C_2(\xi
+ \eta) + Du + E]/R^*\quad\text{in }D'.
\end{gathered}\label{e47}
\end{equation}
Moreover, through
the transformation \eqref{e46}, the boundary condition \eqref{e42}
 on $L''_1,L''_2$ is reduced to
\begin{equation}
\begin{gathered}
 \operatorname{Re}[\overline{\lambda(g^{-1}(\tilde Z))}W(g^{-1}(\tilde Z))] = H_1[\hat y(Y)]r[g^{-1}
(\tilde Z)], \quad z = x + jy \in  L_1' \cup L_2',\\
\operatorname{Im}[\overline{\lambda(g^{-1}(Z_0'))}W(g^{-1}(Z'_0)]=b_1',\quad
u(z_0')=b_0,
\end{gathered}\label{e48}
\end{equation}
in which $Z=g^{-1}(\tilde Z)$, $\tilde Z'_0=g(Z''_0)$,
$Z''_0=l_0'+jG[-\gamma_2(s_0')]$. Therefore the
boundary-value problem \eqref{e37}, \eqref{e42} in $D''$ is
transformed into the boundary-value problem
\eqref{e47}, \eqref{e48}, where we require that the boundaries
$L'_1, L'_2$ satisfy the similar conditions in \eqref{e32}.
According to the method in the proof of Theorem \ref{thm5}, we can see
that the boundary-value problem \eqref{e37}, \eqref{e42} has a
unique solution $u(\tilde Z)$, and then
the corresponding $u=u(z)$ is a solution of the oblique derivative
problem (Problem O'') of  \eqref{e1}.

\begin{theorem} \label{thm6}
 If \eqref{e1} satisfies Condition C in the domain $D''$
bounded by the boundary $L_0\cup L''_1\cup L''_2$, where
$L''_1, L''_2$ are as stated in \eqref{e41},
then Problem O'' for \eqref{e1} in $D''$ with the boundary condition
\eqref{e42} on $L''$ has a unique solution $u(z)$.
\end{theorem}


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