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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 132, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/132\hfil Kernel function and integral representations]
{Kernel function and integral representations on Klein surfaces}

\author[M. Ro\c{s}iu\hfil EJDE-2017/132\hfilneg]
{Monica Ro\c{s}iu}

\address{Monica Ro\c{s}iu \newline
Department of Mathematics,
University of Craiova, Street A.I. Cuza No 13,
Craiova 200585, Romania}
\email{monica\_rosiu@yahoo.com} 

\dedicatory{Communicated by Giovanni Molica Bisci}

\thanks{Submitted  April 5, 2017. Published May 16, 2017.}
\subjclass[2010]{30F50, 35J25}
\keywords{Klein surface; harmonic kernel function; Green's function;
\hfill\break\indent Neumann's function}

\begin{abstract}
 Some representation theorems for the solutions of the Dirichlet problem and
 the Neumann problem on Klein surfaces are proved by using an analogue of the
 harmonic kernel function on symmetric Riemann surfaces.
\end{abstract}

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


\section{Introduction}

In this article we study boundary-value problems for harmonic functions on
Klein surfaces, through their double covers by symmetric Riemann surfaces in
the sense of Klein, that is, Riemann surfaces endowed with fixed point free
antianalytic involutions. The harmonic kernel function is related to the
classical domain functions, such as the Green function and the Neumann
function on a Klein surface, introduced earlier, see \cite{b1,r1}. Thus, it is
possible to solve both the boundary value problems of potential theory on a
Klein surface, once the harmonic kernel function on a symmetric Riemann
surface is known. We develop a symmetrization technique based on the
correspondence between  Klein surfaces and symmetric Riemann surfaces. The
idea of using the double cover has been successfully used to study objects
on a Klein surface by Alling and Greenleaf \cite{a2}, Andreian Cazacu \cite{a3}, 
B\^{a}rz\u{a} and Ghi\c{s}a \cite{b1,b2}.

\section{Preliminaries}

Klein surfaces are the most general two-dimensional real manifolds that
support harmonic functions. In this paper, the methods introduced in 
\cite{a2,b1,r1} are used to extend results about boundary value problems for harmonic
functions on Riemann surfaces to Klein surfaces. The extension required new
concepts and techniques such as the well known relationship between Riemann
surfaces and Klein surfaces, see \cite{a2,s1}. 
Namely, given a compact Klein surface $(X,A)$, there exists a double
 cover $f:O_2\to X$ of $X$
by a Riemann surface $O_2$, such that $O_2$ has an antianalytic
involution $k$ (called symmetry in the sense of Klein) with $f\circ k=f$. 
Moreover, $X$ is dianalytically equivalent with $O_2/H$, where $H$ is the group
generated by $k$, with respect to the usual composition of functions. Also,
if $O_2$ is a Riemann surface on which a fixed point free antianalytic
involution $k$ exists, then $O_2/H$ carries a unique dianalytic structure,
which makes the canonical projection $\pi :$ $O_2\to O_2/H$ a
dianalytic function. Throughout this paper, we identify $X$ with 
the orbit space $O_2/H$.

By Klein's definition, the pair $(O_2,k)$ is a symmetric Riemann surface.

Any compact Klein surface can be conceived as a region $X$ of the complex
plane bounded by a finite number of analytic Jordan curves, see \cite{s1}.

Let $F:X\to \mathbf{R}$ be a function on $X$. Its lifting $f$ to 
$O_2$ is given by
\begin{equation} \label{e2.1}
F(\widetilde{z})=f(z)=f(k(z)),\quad z\in O_2,\; \widetilde{z}=\pi (z).
\end{equation}
A function $f$ on $O_2$ with the property \eqref{e2.1} is called a symmetric
function.

Also, if $g:O_2\to \mathbf{R}$ is a function on $O_2$, then the
function $f=g+g\circ k$ is a symmetric function on $O_2$. 
Thus, relation \eqref{e2.1} defines a function $F$ on $X$.

We consider the symmetric metric on $O_2$, defined by 
$d\sigma =\frac{1}{2}( | dz| +| dw|) $, where $
w=k(z)$, $z\in O_2$. Then
\begin{equation*}
d\Sigma (\widetilde{z})=d\sigma (z)=d\sigma (k(z)),\ z\in O_2,
\end{equation*}
is a metric on $X$. The metric $d\Sigma $ is invariant with respect to the
group of conformal or anticonformal transition functions of $X$.

Let $\widetilde{\gamma }$ be a piecewise smooth Jordan curve on $X$. 
Then $\widetilde{\gamma }$ has exactly two liftings $\gamma $ and 
$k\circ \gamma $ on $O_2$, see \cite{f1}, and by definition
\begin{equation*}
\int_{\widetilde{\gamma }} Fd\Sigma =\int_{\gamma }
fd\sigma =\int_{k\circ \gamma } fd\sigma .
\end{equation*}

For more details about measure and integration on Klein surfaces, 
see \cite{a2}.

Let $u$ be a $C^{1}$-function defined in a neighborhood of the 
$\sigma $-rectifiable Jordan curve $\gamma $, parameterized in 
terms of the arc $\sigma $-length. Therefore, 
$\gamma:z=z(s)=x(s)+iy(s)$, 
$s\in [0,l]$, where $l$ is the $\sigma $-length of $\gamma$.
Then the normal derivative of $u$ on $\gamma $ with respect to $d\sigma $, 
denoted by $\frac{\partial u}{\partial n_{\sigma }}$, is the directional 
derivative of $u$ in the direction of the unit normal vector 
$n_{\sigma }=( \frac{dy}{d\sigma },-\frac{dx}{d\sigma }) $.

 Given $\Omega $ a region of $X$ bounded by a finite number of 
$\sigma $-rectifiable Jordan curves, then $\pi ^{-1}(\Omega )=D$ is a
symmetric subset of $O_2$, since $k$ is an antianalytic involution,
without fixed points and $\pi \circ k=\pi $. For details about Green's
identities for the symmetric region $D$ in terms of $d\sigma $, see \cite{b1}.

Let $F$ be a continuous real-valued function on $\partial \Omega $. The
Dirichlet problem on $X$ for the region $\Omega $, consists in finding a
harmonic function $U$ in $\Omega $ with prescribed values $F$ on 
$\partial \Omega $. We define $f=F\circ \pi $ on $\partial D$. 
Then $f=f\circ k$ on $\partial D$, thus $f$ is a symmetric, 
continuous real-valued functions on $\partial D$. The Dirichlet problem on $X$,
\begin{equation} \label{e2.2}
\begin{gathered}
\Delta U=0 \quad\text{in } \Omega \\
U=F \quad \text{on }\partial \Omega
\end{gathered}
\end{equation}
is equivalent with the Dirichlet problem on $O_2$
\begin{equation} \label{e2.3}
\begin{gathered}
\Delta u=0\quad \text{in }D \\
u=f\quad\text{on }\partial D,
\end{gathered}
\end{equation}
see \cite{b1,r1}.

Let $G$ be a continuous real-valued function on $\partial \Omega $. The
Neumann problem on $X$ for the region $\Omega $, consists in finding a
harmonic function $U$ in $\Omega $ with prescribed normal derivatives values
$G$ on $\partial \Omega $. We define $g=G\circ \pi $ on $\partial D$. 
Then $g=g\circ k$ on $\partial D$, thus $g$ is a symmetric, 
continuous real-valued functions on $\partial D$. By the symmetry of $g$,
 we obtain the compatibility condition $\int_{\partial D} gd\sigma =0$, for the
existence of a solution to the Neumann problem on $O_2$ for the symmetric
region $D$, see \cite{n1}. The Neumann problem on $X$
\begin{equation} \label{e2.4}
\begin{gathered}
\Delta U=0\quad \text{in }\Omega \\
\frac{\partial U}{\partial n_{\Sigma }}=G\quad\text{on }\partial \Omega
\end{gathered}
\end{equation}
is equivalent with the Neumann problem on $O_2$
\begin{equation} \label{e2.5}
\begin{gathered}
\Delta u=0\quad\text{in }D \\
\frac{\partial u}{\partial n_{\sigma }}=g\quad\text{on }\partial D,
\end{gathered}
\end{equation}
see \cite{b1}.

The Dirichlet problem on $O_2$ for the region $D$ and the boundary
function $f$ has a unique solution, provided that $\partial D$ has only regular
points, see \cite{a1}. If the Neumann problem \eqref{e2.5} has a solution, then it is
unique up to an additive constant, see \cite{n1}.

The symmetric conditions on the boundary imply symmetric solutions for the
problems \eqref{e2.3} and \eqref{e2.5}, for details see \cite{b1} and the 
original source \cite{s1}.

\begin{proposition} \label{prop2.1}
The solution $u$ of  problem \eqref{e2.3} is a symmetric function in $D$.
\end{proposition}

\begin{proposition} \label{prop2.2}
A solution $u$ of  problem \eqref{e2.5} is a symmetric function in $D$.
\end{proposition}

\section{Symmetric harmonic kernel function}

Let $D$ be a symmetric region in the complex plane, bounded by a finite
number of $\sigma $-rectifiable Jordan curves. In this section we introduce
closed systems $(\varphi _{i})_{i\in I\text{ }}$ of harmonic functions in 
$D$, which are orthonormal with respect to the Dirichlet integral
\begin{equation*}
D\{\varphi _{i},\varphi _{j}\}
=\iint_D \Big( \frac{\partial
\varphi _{i}}{\partial x}\frac{\partial \varphi _{j}}{\partial y}+\frac{
\partial \varphi _{i}}{\partial y}\frac{\partial \varphi _{j}}{\partial x}
\Big) \,dx\,dy.
\end{equation*}
We recall some notions and results about orthogonal harmonic functions, see
\cite{b3}.

Let $\Lambda ^{2}(D)$ be the set of harmonic functions $\varphi (z)$ in $D$
with a finite Dirichlet integral
\begin{equation} \label{e3.1}
D\{\varphi \}=D\{\varphi ,\varphi \}<\infty
\end{equation}
such that
\begin{equation} \label{e3.2}
D\{N_{D}(z;\zeta ),\varphi \}=-2\pi \varphi (\zeta ),
\end{equation}
where $N_{D}(z;\zeta )$ is the Neumann's function of $D$ with its
singularity at the fixed point $\zeta $, $\zeta \in D$.

\begin{remark} \label{rmk3.1} \rm
If $\varphi $ has continuous boundary values, $\varphi $ will be normalized
by the condition
\begin{equation} \label{e3.3}
\int_{\partial D}\varphi d\sigma =0.
\end{equation}
\end{remark}

\begin{proposition} \label{prop3.2}
There exists a closed system $(\varphi _{i})_{i\in I}$ for $\Lambda ^{2}(D)$,
which is orthonormal with respect to the Dirichlet integral, i.e.
\begin{equation*}
D\{\varphi _{i},\varphi _{j}\}=\delta _{ij},\quad \delta _{ii}=1,\;
\delta_{ij}=0,\;i\neq j.
\end{equation*}
\end{proposition}

Let $\zeta $ be a point inside $D$. The harmonic kernel function 
$K_{D}(z;\zeta )$ of the closed orthonormal system $(\varphi _{i})_{i\in I}$,
for the region $D$, with respect to the point $\zeta $, is the function
defined by
\begin{equation*}
K_{D}(z;\zeta )=\sum_{i=1}^{\infty } \varphi_{i}(z)\varphi _{i}(\zeta ),\quad
 z\in \overline{D}.
\end{equation*}

\begin{remark} \label{rmk3.3} \rm
The harmonic kernel function depends on the domain $D$.
\end{remark}

An extensive study of the harmonic kernel function can be found in 
Bergman \cite{b3}.
It is known, see \cite{b3}, that the harmonic kernel function $K_{D}(z;\zeta )$,
the Green function $G_{D}(z;\zeta )$ and the Neumann function $
N_{D}(z;\zeta )$ satisfy the relation
\begin{equation} \label{e3.4}
K_{G}(z;\zeta )=\frac{1}{2\pi }[ N_{D}(z;\zeta )-G_{D}(z;\zeta )],\quad 
z\in \overline{D}.
\end{equation}

First, we  derive formulas that solve problem \eqref{e2.3}. We
prove that if $u$ is harmonic inside a region $D$ and continuous on 
$\partial D$, then we can determine the values of $u$ inside of $D$ by
integrating on $\partial D$ the product of $u$ times the normal derivative\
of the harmonic kernel function for the region $D$, which is a fixed
function that depends only on $D$.

\begin{proposition} \label{prop3.4}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. If $u$ is harmonic in $D$ and continuous 
on $\overline{D}$, then for all $\zeta $ in $D$,
\begin{equation} \label{e3.5}
u(\zeta )=-\int_{\partial D}u(z)\frac{\partial K_{D}(z;\zeta )}{
\partial n_{\sigma }}d\sigma .
\end{equation}
\end{proposition}

\begin{proof}
From \cite{b1,p1}, the solution of the Dirichlet problem \eqref{e2.3} is
\begin{equation} \label{e3.6}
u(\zeta )=\frac{1}{2\pi }\int_{\partial D}u(z)\frac{\partial
G_{D}(z;\zeta )}{\partial n_{\sigma }}d\sigma ,\quad  \zeta \in D.
\end{equation}
Using \eqref{e3.4}, we obtain
$$
\frac{\partial K_{D}(z;\zeta )}{\partial n_{\sigma }}
=\frac{1}{2\pi } \frac{\partial N_{D}(z;\zeta )}{\partial n_{\sigma }}
-\frac{1}{2\pi }\frac{\partial g_{D}(z;\zeta )}{\partial n_{\sigma }}
=-\frac{1}{L}-\frac{1}{2\pi }\frac{\partial G_{D}(z;\zeta )}{\partial n_{\sigma }}, 
$$
for $z\in \partial D$, where $L$ is the length of $\partial D$, see \cite{p1}.
Combining this with \eqref{e3.6}, we find that
\begin{equation*}
u(\zeta )
=-\int_{\partial D}u(z)\frac{\partial K_{D}(z;\zeta )}{
\partial n_{\sigma }}d\sigma -\frac{1}{L}\int_{\partial D}
u(z)d\sigma .
\end{equation*}
Since the boundary values satisfy the normalization condition 
$\int_{\partial D} u(z)d\sigma =0$, we obtain relation \eqref{e3.5}.
\end{proof}

Next, we  derive formulas that solve problem \eqref{e2.5}. From Green's
formula for the Laplacian, in terms of $d\sigma $, the prescribed values of
the normal derivatives must satisfy the compatibility condition
\begin{equation*}
\int_{\partial D}\frac{\partial u(z)}{\partial n_{\sigma }} d\sigma =0.
\end{equation*}

\begin{proposition} \label{prop3.5}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. If $u$ is harmonic in $D$, 
then, up to an additive constant,
\begin{equation} \label{e3.7}
u(\zeta )=\int_{\partial D}\frac{\partial u(z)}{\partial
n_{\sigma }}K_{D}(z;\zeta )d\sigma ,\quad \zeta \in D.
\end{equation}
\end{proposition}

\begin{proof}
Using Green's second identity, it follows that, up to an additive constant,
a solution of the Neumann problem is given by
\begin{equation} \label{e3.8}
u(\zeta )=\frac{1}{2\pi }\int_{\partial D}\frac{\partial u(z)}{
\partial n_{\sigma }}N_{D}(z;\zeta )d\sigma ,\quad \zeta \in D.
\end{equation}
The constant is chosen such that $u(z)$ is in $\Lambda ^{2}(D)$.
By \eqref{e3.4}, for $\zeta \in \partial D$, we obtain
\begin{equation*}
K_{D}(z;\zeta )=\frac{1}{2\pi }N_{D}(z;\zeta ).
\end{equation*}
Substituting this in \eqref{e3.8}, we obtain \eqref{e3.7}.
\end{proof}

Let $K_{D}^{(k)}(z;\widetilde{\zeta })$ be the function defined by
\begin{equation*}
K_{D}^{(k)}(z;\widetilde{\zeta })
=\frac{1}{2} [ K_{D}(z;\zeta)+K_{D}(z;k(\zeta ))] ,\quad  z\in \overline{D},
\end{equation*}
where $K_{D}(z;k(\zeta ))$ is the harmonic kernel function of the closed
orthonormal system $(\varphi _{i})_{i\in I}$, for the region $D$, with
respect to the point $k(\zeta )$. The function 
$K_{D}^{(k)}(z;\widetilde{\zeta })$ is in $\Lambda ^{2}(D)$.

\begin{proposition} \label{prop3.6}
If  $D$ is a symmetric region, then the function 
$K_{D}^{(k)}(z;\widetilde{\zeta })$ is symmetric with respect to $z$ 
on $\overline{D}$ i.e. for every $z\in \overline{D}$,
\begin{equation*}
K_{D}^{(k)}(z;\widetilde{\zeta })=K_{D}^{(k)}(k(z);\widetilde{\zeta }).
\end{equation*}
\end{proposition}

\begin{proof}
We  use \eqref{e3.4} and the symmetric properties of the corresponding
symmetric Green's function and symmetric Neumann's function, see \cite{b1,r1}.
\end{proof}

The function $K_{D}^{(k)}(z;\widetilde{\zeta })$ is called the symmetric
harmonic kernel function of the closed orthonormal system 
$(\varphi_{i})_{i\in I}$, for the region $D$, with respect to the point
 $\widetilde{\zeta }=\{\zeta ,k(\zeta )\}$.

\section{Integral representations on the double cover}

The next theorem yields a formula for the symmetric solution of the problem
\eqref{e2.3}.

\begin{theorem} \label{thm4.1}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. Let $f$ be a symmetric, continuous 
function on $\partial D$. There exists a unique symmetric function $u$ 
on $\overline{D}$, which is harmonic on $D$, continuous on $\overline{D}$, 
such that $u=f$ on $\partial D$. For all $\zeta $ in $D$,
\begin{equation} \label{e4.1}
u(\zeta )=-\frac{1}{2}\int_{\partial D}f(z)\big[ \frac{\partial
K_{D}(z;\zeta )}{\partial n_{\sigma }}+\frac{\partial K_{D}(z;k(\zeta ))}{
\partial n_{\sigma }}\big] d\sigma .
\end{equation}
\end{theorem}

\begin{proof}
Since $k$ is an involution of $D$, the function 
$\frac{u(\zeta )+u(k(\zeta ))}{2}$ is a symmetric function on $D$.
 By Proposition 3.4,
\begin{equation} \label{e4.2}
u(\zeta )=-\int_{\partial D}u(z)
\frac{\partial K_{D}(z;\zeta )}{\partial n_{\sigma }}d\sigma ,\quad \zeta \in D.
\end{equation}
Replacing $\zeta $ with $k(\zeta )$ in \eqref{e4.2}, we obtain
\begin{equation} \label{e4.3}
u(k(\zeta ))=-\int_{\partial D}u(z)\frac{\partial K_{D}(z;k(\zeta
))}{\partial n_{\sigma }}d\sigma ,\quad \zeta \in D.
\end{equation}
Adding \eqref{e4.2} to \eqref{e4.3} and dividing by 2, it follows that
\begin{equation*}
\frac{u(\zeta )+u(k(\zeta ))}{2}=-\frac{1}{2}\int_{\partial D}u(z)
\big[ \frac{\partial K_{D}(z;\zeta )}{\partial n_{\sigma }}+\frac{\partial
K_{D}(z;k(\zeta ))}{\partial n_{\sigma }}\big] d\sigma ,\quad  \zeta \in D.
\end{equation*}
By Proposition 2.1, $u$ is a symmetric function on $D$, then the left-hand side
of the last equality is $u(\zeta )$ and we conclude that for all $\zeta $ in
$D$,
\begin{equation*}
u(\zeta )=-\frac{1}{2}\int_{\partial D}u(z)\big[ \frac{\partial
K_{D}(z;\zeta )}{\partial n_{\sigma }}+\frac{\partial K_{D}(z;k(\zeta ))}{
\partial n_{\sigma }}\big] d\sigma .
\end{equation*}
The uniqueness of the solution of the Dirichlet problem for harmonic functions
implies relation \eqref{e4.1}.
\end{proof}

Next we obtain a formula for the symmetric solution of the problem \eqref{e2.5}.

\begin{theorem} \label{thm4.2}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. Let $g$ be a symmetric, continuous function 
on $\partial D$. If $u$ is harmonic in $D$ and $g$ is its normal derivative on 
$\partial D$, then up to an additive constant,
\begin{equation} \label{e4.4}
u(\zeta )=\frac{1}{2}\int_{\partial D}g(z)[ K_{D}(z;\zeta
)+K_{D}(z;k(\zeta ))] d\sigma ,\zeta \in D.
\end{equation}
\end{theorem}

\begin{proof}
By analogy with the proof of the Theorem \ref{thm4.1}, we are using Proposition 3.5
instead of Proposition 3.4.
\end{proof}

\section{Integral representations on a Klein surface}

Let $X$ be compact Klein surface and let $\Omega $ be a region bounded by a
finite number of $\sigma $-rectifiable Jordan curves. Then there exists a
symmetric Riemann surface $(O_2,k)$ such that $X$ is dianalytically
equivalent with $O_2/H$, where $H$ is the group generated by $k$, with
respect to the usual composition of functions. Then, $\Omega $ is obtained
from the symmetric region $D$ by identifying the corresponding symmetric
points.

Let $\widetilde{\zeta }$ be a point inside $\Omega $. The harmonic kernel
function $K_{\Omega }(\widetilde{z};\widetilde{\zeta })$ of the closed
orthonormal system $(\varphi _{i})_{i\in I}$, for the region $\Omega $, with
respect to the point $\widetilde{\zeta }=\{\zeta ,k(\zeta )\}$ is defined by
\begin{equation*}
K_{\Omega }(\widetilde{z};\widetilde{\zeta })=K_{D}^{(k)}(z;\widetilde{\zeta
})=K_{D}^{(k)}(k(z);\widetilde{\zeta }),\quad \widetilde{z}=\pi (z)\in \Omega .
\end{equation*}

\begin{remark} \label{rmk5.1} \rm
From Proposition \eqref{e3.6}, it follows that $K_{\Omega }(\widetilde{z};
\widetilde{\zeta })$ is well defined on $\Omega $.
\end{remark}

By Theorem \ref{thm4.1}, we obtain the following representation of the solution of
 problem \eqref{e2.3} on a symmetric region $D$, in terms of the symmetric
harmonic kernel function.

\begin{theorem} \label{thm5.2}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. Let $f$ be a symmetric, continuous 
function on $\partial D$. There exists a unique symmetric function 
$u$ on $\overline{D}$, which is harmonic on $D$, continuous on $\overline{D}$, 
such that $u=f$ on $\partial D$. For all $\zeta $ in $D$ we have
\begin{equation} \label{e5.1}
u(\zeta )=-\int_{\partial D}f(z)\frac{\partial K_{D}^{(k)}(z;
\widetilde{\zeta })}{\partial n_{\sigma }}d\sigma .
\end{equation}
\end{theorem}

By Theorem \ref{thm4.2}, we obtain the following representation of the solution of
 \eqref{e2.5} on a symmetric region $D$, in terms of the symmetric
harmonic kernel function.

\begin{theorem} \label{thm5.3}
Let $D$ be a symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. Let $g$ be a symmetric, continuous function 
on $\partial D$. If $u$ is harmonic in $D$ and $g$ is its normal derivative on 
$\partial D$, then up to an additive constant,
\begin{equation} \label{e5.2}
u(\zeta )=\int_{\partial D}g(z)K_{D}^{(k)}(z;\widetilde{\zeta }
)d\sigma ,\zeta \in D.
\end{equation}
\end{theorem}

The symmetric solutions on $O_2$ determine the solutions of the similar
problems on the Klein surface $X$.

We obtain the solution of  \eqref{e2.2} on the region $\Omega $, with
respect to the harmonic kernel function, for the region $\Omega $.

\begin{theorem} \label{thm5.4}
Let $F$ be a continuous real-valued function on the border
 $\partial \Omega$. The solution of  \eqref{e2.2} with the boundary function 
$F$ is the function $U$ defined on $\overline{\Omega }$, by the relation 
$u=U\circ \pi$, where $\pi $ is the canonical projection of $O_2$ on 
$X$ and $u$ is the solution \eqref{e5.1} of the problem \eqref{e2.3} 
on the symmetric region $D$, with the boundary function $f=F\circ \pi $.
\end{theorem}

\begin{proof}
By definition, $\Delta U(\widetilde{\zeta })=\Delta u(\zeta )$ $=0$, for all
$\widetilde{\zeta }\in \Omega $, where $\widetilde{\zeta }=\pi (\zeta )$,
thus $U$ is a harmonic function. The symmetry of the function $f$ on 
$\partial D$, implies
$$
U(\widetilde{\zeta })=u(\zeta )=f(\zeta )
=f(k(\zeta ))
=F(\widetilde{\zeta }),\quad \text{for all }\widetilde{\zeta }\in
\partial \Omega .
$$
By the uniqueness of the solution, the function $U$
defined on $\overline{\Omega }$ by
\begin{equation*}
U(\widetilde{\zeta })=u(\zeta )=u(k(\zeta )),
\end{equation*}
for all $\widetilde{\zeta }$ in $\overline{\Omega }$, where 
$\widetilde{\zeta }=\pi (\zeta )$, is the solution of the problem \eqref{e2.2} 
on $\Omega $.
\end{proof}

The next theorem gives a solution of the problem \eqref{e2.4} on the region 
$\Omega$, with respect to the harmonic kernel function, for the region $\Omega $.

\begin{theorem} \label{thm5.5}
Let $G$ be a continuous real-valued function on the border $\partial \Omega$.
Then, up to an additive constant, the solution of  \eqref{e2.4} with
the normal derivative $G$ on $\partial \Omega $ is the function $U$ defined
on $\overline{\Omega }$, by the relation $u=U\circ \pi $, where $\pi $ is
the canonical projection of $O_2$ on $X$ and $u$ is the solution \eqref{e5.2} of
the problem \eqref{e2.5} on the symmetric region $D$, with the normal derivative
function $g=G\circ \pi $ on $\partial D$.
\end{theorem}

\begin{proof}
By definition, $\Delta U(\widetilde{\zeta })=\Delta u(\zeta )$ $=0$, for all
$\widetilde{\zeta }\in \Omega $, where $\widetilde{\zeta }=\pi (\zeta )$,
thus $U$ is a harmonic function. The symmetry of the function $g$ on $
\partial D$, implies 
$$
\frac{\partial U(\widetilde{\zeta })}{\partial
n_{\sigma }}=\frac{\partial u(\zeta )}{\partial n_{\sigma }}=g(\zeta
)=g(k(\zeta ))=G(\widetilde{\zeta }),\quad \text{for all }
\widetilde{\zeta }\in \partial \Omega .
$$
Thus, up to an additive constant, the function $U$ defined
on $\overline{\Omega }$ by
\begin{equation*}
U(\widetilde{\zeta })=u(\zeta )=u(k(\zeta )),
\end{equation*}
is the solution of  problem \eqref{e2.4} on $\Omega $.
\end{proof}

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\bibitem{a2} N. Alling, N. Greenleaf;
\emph{The Foundation of the Theory of Klein Surfaces,} 
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