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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 111, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/111\hfil The Poisson equation]
{The Poisson equation on Klein surfaces}

\author[M. Ro\c{s}iu \hfil EJDE-2016/111\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}

\thanks{Submitted January 27, 2016. Published April 28, 2016.}
\subjclass[2010]{30F50, 35J05, 35J08}
\keywords{Klein surface; Poisson equation; Green's function}

\begin{abstract}
 We obtain a formula for the solution of the Poisson equation with
 Dirichlet boundary condition on a region of a Klein surface.
 This formula reveals the symmetric character of the solution.
\end{abstract}

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


\section{Introduction}

In this article we solve a boundary value problem involving the Poisson
equation on Klein surfaces. Our technique is based on the fact that
according to a classical result due to  Klein, the boundary value problems
on Klein surfaces can be reduced to similar problems on symmetric Riemann
surfaces. On Klein surfaces the formula for the solution is expressed in
terms of an analogue of the Green function, which has the symmetry in
argument and parameter. The extensive study of the Klein surfaces is due to
Schiffer and Spencer \cite{s1}. Other useful results about this topic are
the formulas for the Green function on the M\"obius strip expressed in 
\cite{b2} and the Dirichlet problem for harmonic functions treated
in \cite{b1}.

\section{Preliminaries}

A compact Klein surface is a pair $(X,A)$, consisting of a compact surface 
$X $ and a maximal dianalytic atlas $A$ on $X$, such that $A$ does not
contain any analytic subatlas.

It is known, see \cite{s1}, that given a compact Klein surface $(X,A)$,
its orientable double covering $O_2$ admits a fixed point free symmetry $k$,
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. We denote the canonical projection of $O_2$ onto $O_2/H$ by
$\pi$. By Klein's definition, the pair $(O_2,k)$ is a $k$-symmetric
compact Riemann surface. Forwards, we identify $X$ with the orbit space 
$O_2/H$.

A set $G$ of $O_2$ is called $k$-symmetric if $k(G)=G$. Thus, given $D$ a
subset of $X$, then $q^{-1}(D)=G$ is a $k$-symmetric subset of $O_2$.

A function $f$ defined on a $k$-symmetric set is called a $k$-symmetric
function if $f=f\circ k$.

Let $\widetilde{\gamma }:[0,1]\to D$ be a piecewise smooth Jordan
curve of $D$. Then the arc $\widetilde{\gamma }$ has exactly two liftings at
$\pi ^{-1}(D)$. If $\widetilde{\gamma }(0)=\widetilde{z}_{0}=\{z_{0},k(z_{0})\}$ 
and if $\gamma $ is the lifting of $\widetilde{\gamma }$
at $z_{0}$, then $k\circ \gamma $ is the lifting of $\widetilde{\gamma }$ at
$k(z_{0}).$Then $\pi ^{-1}(\widetilde{\gamma })=\gamma \cup k(\gamma )$ is a
$k$-symmetric curve of $O_2$. For any real valued function $F$ defined on $
\widetilde{\gamma }$, the function $f=F\circ \pi $ is a $k$-symmetric
function on $\gamma \cup k(\gamma )$, see \cite{r1}.

The Euclidean lengths of the two curves $\gamma $ and its symmetric, $k\circ
\gamma $, that is their lengths with respect to the metric $ds=|
dz| $ may be different. We modify this metric and get a new metric $
d\sigma $ on $O_2$, such that the lengths of $\gamma $ and $k\circ \gamma
, $ with respect to the metric $d\sigma $, will be the same. We define\ a $
k- $symmetric metric $d\sigma =\frac{1}{2}( ds+ds\circ k) $. Then
the $d\sigma $ -lengths of $\gamma $ and $k\circ \gamma $ are equal. By
definition, the length of $\widetilde{\gamma }$ is the common $d\sigma $
-length of $\gamma $ and $k\circ \gamma $.

Let $X$ be a compact Klein surface and let $D$ be a region bounded by a
finite number of $\sigma $-rectifiable Jordan curves. Given $F$ a continuous
real-valued function on $D$ and $H$ a continuous real-valued function on $
\partial D$, we study the problem defined by Poisson's equation \emph{\ }
\begin{equation}
\Delta U=F\quad \text{on }D  \label{e1}
\end{equation}
and the Dirichlet boundary condition
\begin{equation}
U=H\quad\text{on }\partial D.  \label{e2}
\end{equation}

Because the Klein surface $X$ is dianalytically equivalent to $O_2/H$,
problem \eqref{e1}--\eqref{e2} on a region $D$ of the Klein surface $X$, can be
replaced by an equivalent problem on a $k$-symmetric region $G$ of its
double $O_2$ as follows.

We define $G=\pi ^{-1}(D)$, $f=F\circ \pi $ on $G$ and $h=H\circ \pi $ on 
$ \partial G$. Then, $G$ is a $k$-symmetric region bounded by a finite number
of $\sigma$-rectifiable Jordan curves on $O_2$. Since $\pi \circ k=\pi $,
we obtain $f=f\circ k$ on $G$ and $h=h\circ k$ on the boundary $\partial G$,
thus $f$ and $h$ are $k$-symmetric, continuous real-valued functions. 
Problem \eqref{e1}--\eqref{e2} is equivalent to the problem
\begin{gather*}
\Delta u=f\quad \text{on  }G \\
u=h\quad\text{on }\partial G.
\end{gather*}

\section{Poisson's equation on the double cover}

Let $X$ be a compact Klein surface and let $\widehat{X}$ be the universal
covering surface. Then $\widehat{X}$ has a unique (up to conjugation)
analytic structure making the canonical projection dianalytic. 
Let $\mathcal{G}$ be the group of covering transformations and let 
$\mathcal{G}_1$ be the subgroup of conformal elements of $\mathcal{G}$. 
Then $\widehat{X}/\mathcal{G}$ is canonically identified with $X$ and 
$\widehat{X}/\mathcal{G}_1$ is canonically identified with $O_2$,
 see \cite{l1}.

Whence, a model of the bordered Riemann surface $O_2$ is a region of the
complex plane having a finite number of $\sigma$-rectifiable Jordan
boundary curves. The arc length parameter $s$ is a boundary uniformizer near
every boundary point. A half-neighborhood of a boundary point is mapped by 
$s $ onto a half-neighborhood bounded by a segment of the real $s$-axis, see
\cite{s1}.

Let $d\sigma =\lambda (z)| dz| $ be the $k$-symmetric metric on 
$G$, where $\lambda $ is a nonnegative continuous function on $\overline{G}$.

Let $G$ be a $k$-symmetric region in the complex plane, where $\partial G$
consists of $\sigma$-rectifiable Jordan curves $\Gamma $ ( exterior
boundary, positively oriented) and $C_1,\dots,C_n$ (interior boundary,
negatively oriented). Given a $k$-symmetric, continuous, real-valued
function $f$ on $G$ and a $k$- symmetric, continuous, real-valued function
$h $ on $\partial G$, then \eqref{e1}--\eqref{e2} can be
reduced to the problem consisting of the Poisson equation
\begin{equation}
\Delta u=f\quad\text{on }G  \label{e1p}
\end{equation}
and the Dirichlet boundary condition
\begin{equation}
u=h\quad\text{on }\partial G.  \label{e2p}
\end{equation}
To solve  \eqref{e1p}--\eqref{e2p} we combine the
solution of the Dirichlet problem for harmonic functions
\begin{gather*}
\Delta u=0\quad\text{on }G \\
u=h\quad\text{on }\partial G
\end{gather*}
and the solution of the Poisson equation with zero boundary values
\begin{equation}
\begin{gathered}
\Delta u=f\quad\text{on }G \\
u=0\quad\text{on }\partial G
\end{gathered} \label{e3}
\end{equation}
In this paper we only consider solutions which are in the class 
$C^{2}(G)\cap C^{1}(\partial G)$.

\begin{remark} \label{rmk1} \rm
From the maximum principle, it follows that a solution of 
\eqref{e1p}--\eqref{e2p}, if it exists, is necessarily uniquely
determined.
\end{remark}

For the general existence proof of a solution to \eqref{e1p}--\eqref{e2p} 
we refer to \cite{f1}. Moreover, in this case, the solution has the following
property.

\begin{proposition} \label{prop2}
A solution $u$ of \eqref{e1p}--\eqref{e2p} is a 
$k$-symmetric function in $G$.
\end{proposition}

\begin{proof}
Let $u$ be a solution of \eqref{e1p}--\eqref{e2p}. We
define $u_k:\overline{G}\to \mathbb{R}$ by
$u_k=\frac{1}{2}( u+u\circ k)$. The hypothesis, $f=f\circ k$, involves
$\Delta u_k$ $=\frac{1}{2}( f+f\circ k) =f$ on $G$ and $u_k=0$ on 
$\partial G$. Thus $u_k$ is also a solution of  \eqref{e1p}--\eqref{e2p}.
 Uniqueness of the solution yields $u_k=u$ on $G$,
therefore $u=u\circ k$ on $G$.
\end{proof}

The Dirichlet problem for harmonic functions on $G$ was solved in \cite{b1}.
To complete the solution we solve the Poisson equation with zero boundary
values for the $k$-symmetric region $G$.

In solving \eqref{e3} we  use the Green function of a region $G$. For the
existence of a harmonic function which vanishes on the boundary and has a
finite number of isolated singularities, with given singular parts, in a
relatively compact region of a Riemann surface, we refer to \cite{a1}.

First we will derive two important formulas. We recall the meaning of the
normal derivative with respect to the $k$-symmetric metric $d\sigma $. 
Let $u $ be a $C^{1}$-function defined on a $\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 $u$, denoted by 
$\frac{\partial u}{\partial n_{\sigma }}$, is the directional derivative of $u$ in
the direction of the unit normal vector $n_{\sigma }$, see \cite{b2}. For
any point of the curve $\gamma $, it follows that
\[
\frac{\partial u}{\partial n_{\sigma }}d\sigma
=-\frac{\partial u}{\partial y}dx+\frac{\partial u}{\partial x}dy
\]
and, in polar coordinates
\begin{equation}
\frac{\partial u}{\partial n_{\sigma }}d\sigma
=-\frac{1}{\rho }\frac{\partial u}{\partial \theta }d\rho
 +\rho \frac{\partial u}{\partial \rho }d\theta .  \label{e4}
\end{equation}

Let $G$ be a $k$-symmetric region bounded by a finite number of $\sigma $
-rectifiable Jordan curves. Suppose that $p$ and $q$ are continuous with
continuous partial derivatives functions on $\overline{G}$. By Green's
theorem
\[
\int_{\partial G} p(x,y)dx+q(x,y)dy=\iint_G
\Big(\frac{\partial q}{\partial x}-\frac{\partial p}{\partial y}\Big) \,dx\,dy.
\]

Given $u$ and $v$ two functions in the class $C^{2}(G)\cap C^{1}(\partial G)$, 
parameterized by $x(s)$ and $y(s)$, where $s$ is the arc $\sigma $-length,
applying Green's theorem, with $p=-u\frac{\partial v}{\partial y}$,
 $q=u\frac{\partial v}{\partial x}$ we obtain Green's first identity for the
symmetric region $G$,
\begin{equation}
\iint_G ( u\Delta v+( \operatorname{grad}u)\cdot ( \operatorname{grad}v) ) \,dx\,dy
=\int_{\partial G}u \frac{\partial v}{\partial n_{\sigma }}d\sigma .  \label{e5}
\end{equation}

Reversing the roles of $u$ and $v$ in Green's first identity and subtracting
the new identity from \eqref{e5} we arrive at Green's second identity for the
symmetric region $G$,
\[
\iint_G ( u\text{ }\Delta v-v\Delta u) \,dx\,dy=
\int_{\partial G}\Big( u\frac{\partial v}{\partial n_{\sigma }}-v
\frac{\partial u}{\partial n_{\sigma }}\Big) d\sigma .
\]

Let $\zeta $ be a point inside $G$. The function 
$\Phi ( z,\zeta ) =\ln | z-\zeta | $ is harmonic at all points 
$z\neq \zeta$. Let $w$ be the solution of the Dirichlet boundary-value
 problem on $G$, with the boundary condition $w(z)=\Phi (z,\zeta )$ on 
$\partial G$. The unique function $g_G(z;\zeta )=-\Phi (z,\zeta )+w(z)$ defined on
$\overline{G}\backslash \{ \zeta\} $ is called the Green's
function for the region $G$, with respect to the point $\zeta $, see
 \cite{a1}.

Let $g_G^{(k)}(z,\widetilde{\zeta })$ be the $k$-invariant Green's
function for the region $G$, with singularities at $\zeta $ and $k(\zeta )$,
defined by
\[
g_G^{(k)}(z,\widetilde{\zeta })=\frac{1}{2}[ g_G(z,\zeta
)+g_G(z,k(\zeta ))]
\]
on $\overline{G}\backslash \{ \zeta ,k(\zeta )\} $. For
additional information on this topic we refer to \cite{b2} and to the
original source \cite{s1}.

Let $w_{s}$ be the solution of the Dirichlet boundary-value problem on $G$,
with the boundary condition 
$w_{s}(z)=\frac{1}{2}[ \Phi (z,\zeta )+\Phi(z,k(\zeta )] $ on $\partial G$. 
From the definition of the Green function and the definition of 
$g_G^{(k)}(z,\widetilde{\zeta })$, it
follows that 
$$
g_G^{(k)}(z,\widetilde{\zeta })
=-\frac{1}{2}[ \Phi (z,\zeta )+\Phi (z,k(\zeta )] +w_{s}(z).
$$
Therefore, $g_G^{(k)}(z,\widetilde{\zeta })$ is a harmonic function of 
$z$ in $G\backslash \{\zeta ,k(\zeta )\} $, with singularities 
$-\frac{1}{2}\ln |z-\zeta | $ and 
$-\frac{1}{2}\ln | z-k(\zeta )| $ at 
$\zeta$ and $k(\zeta )$, respectively. Also, $g_G^{(k)}(z,\widetilde{\zeta })=0$
for all $z$ on $\partial G$.

\begin{theorem} \label{thm3}
Let $G$ be a $k$-symmetric region in the complex plane, where $\partial G$
consists of a finite number of $\sigma $-rectifiable Jordan curves. 
Let $u$ be a $C^{2}$-function on $G$, such that $u=0$ on $\partial G$. Then, for
all $\zeta $ in $G$,
\begin{equation}
u(\zeta )=\frac{1}{2\pi } \iint_G \Delta u(z)g_G(z;\zeta )\,dx\,dy.
\label{e6}
\end{equation}
\end{theorem}

\begin{proof}
Let $C_{\varepsilon }$ be a negatively oriented circle of radius 
$\varepsilon $, centered at $\zeta $ and let $G_{\varepsilon }$ be $G$ minus 
$D_{\varepsilon }$, the closed disk bounded by $C_{\varepsilon }$. The
boundary of $G_{\varepsilon }$ is the union of $\partial G$ and 
$C_{\varepsilon }$. Applying Green's second identity for $G_{\varepsilon }$,
with $v=g_G(\cdot,\zeta )$ and using that $u=0$ on $\partial G$, $g_G$ is
harmonic in $G_{\varepsilon }$ and $g_G=0$ on $\partial G$, we get
\[
-\iint_{G_{\varepsilon }} \Delta u(z)g_G(z,\zeta )\,dx\,dy 
=\int_{C_{\varepsilon }} u(z)\frac{\partial g_G}{\partial
n_{\sigma }}(z,\zeta )d\sigma 
-\int_{C_{\varepsilon }} g_G(z,\zeta )\frac{\partial u}{\partial n_{\sigma }}
(z)d\sigma .
\]
Next, we let $\varepsilon $ tends to zero, taking into account that the
outward normal derivative (with respect to the region $G_{\varepsilon }$) on
$C_{\varepsilon }$ is the inner radial derivative pointing towards the pole 
$\zeta $.

(1) First we prove that
\[
\lim_{\varepsilon \to 0} \int_{C_{\varepsilon }}
g_G(z,\zeta )\frac{\partial u}{\partial n_{\sigma }}(z)d\sigma =0.
\]
The curve $-C_{\varepsilon }$ is parameterized by
$z=z(\theta )=\zeta +\varepsilon e^{i\theta }$, $0\leq \theta \leq 2\pi $
and using \eqref{e4}, we obtain
$$
-\int_{C_{\varepsilon}} g_G(z;\zeta )\frac{\partial u}{
\partial n_{\sigma }}(z)d\sigma
=\int_{-C_{\varepsilon}} g_G(z;\zeta )\frac{\partial u}{\partial n_{\sigma }}(z)
d\sigma
=\varepsilon \underset{0}{\overset{2\pi }{\int }}g_G(z(\theta );\zeta )
\frac{\partial u}{\partial \rho }(z(\theta ))d\theta .
$$
By definition, $g_G(z;\zeta )=-\ln | z-\zeta | +w(z)$, where $w$ is harmonic
in $G$. Then $w$ is continuous in $D_{\varepsilon }$ and thus $w$ is bounded
on $C_{\varepsilon }$. As the function $u$ has continuous partial
derivatives in $G$, $\frac{\partial u}{\partial \rho }$ is continuous on
$D_{\varepsilon }$, hence is bounded on $C_{\varepsilon }$. Therefore there is
a constant $m$, such that $| w| \leq m$ and
$| \frac{\partial u}{\partial \rho }| \leq m$ on $C_{\varepsilon }$.
Since on $C_{\varepsilon }$, $\ln | z-\zeta | =\ln \varepsilon $, we arrive
at $| g_G(z,\zeta )| \leq m+| \ln \varepsilon | $,
for $z\in C_{\varepsilon }$. Therefore
\[
\big| \int_{C_{\varepsilon}} g_G(z,\zeta )\frac{\partial u}{
\partial n_{\sigma }}(z)d\sigma \big|
\leq 2\pi \varepsilon (m+| \ln \varepsilon | ) m
\]
and the right side of the last inequality tends to zero as
$\varepsilon $ tends to zero. Therefore,
\[
\lim_{\varepsilon \to 0}\int_{C_{\varepsilon}} g_G(z,\zeta )
\frac{\partial u}{\partial n_{\sigma }}(z)d\sigma =0.
\]

(2) We prove that
\[
\lim_{\varepsilon \to 0} \int_{C_{\varepsilon}}
u(z)\frac{\partial g_G}{\partial n_{\sigma }}(z,\zeta )d\sigma =-2\pi
u( \zeta ) .
\]
The definition of the Green function yields
\[
\int_{C_{\varepsilon}} u(z)\frac{\partial g_G}{\partial
n_{\sigma }}(z,\zeta )d\sigma
=-\int_{C_{\varepsilon}} u(z)\frac{
\partial \Phi }{\partial n_{\sigma }}(z)d\sigma
+\int_{C_{\varepsilon }} u(z)\frac{\partial w}{\partial n_{\sigma }}(z)d\sigma .
\]
Using \eqref{e4} and the mean value property, we have
$$
-\int_{C_{\varepsilon }} u(z)\frac{\partial \Phi }{\partial n_{\sigma }}(z)d\sigma
=- \int_0 ^{2\pi} u(z(\theta ))\frac{1}{\varepsilon }\varepsilon d\theta
=-2\pi u(\zeta ).
$$
Using again \eqref{e4}, we get
$$
\int_{C_{\varepsilon }} u(z)\frac{\partial w}{\partial n_{\sigma }}(z)d\sigma
=\varepsilon \int_0^{2\pi } u(z(\theta ))
\frac{\partial w}{\partial \rho }(z(\theta ))d\theta .
$$
As the function $w$ is harmonic, $w$ has continuous partial derivatives in $G$,
then $\frac{\partial w}{\partial \rho }$ is continuous on $D_{\varepsilon }$,
hence is bounded on $C_{\varepsilon }$. The function $u$ is also bounded on
$C_{\varepsilon }$. Therefore there is a constant $M$, such that
$|u| \leq M$ and $| \frac{\partial w}{\partial \rho }| \leq M$
on $C_{\varepsilon }$. So
$$
\big| \int_{C_{\varepsilon}} u(z) \frac{\partial w}{\partial n_{\sigma }}(z)d\sigma
\big|  \leq 2\pi \varepsilon M^{2},
$$
which tends to zero as $\varepsilon $ tends to zero.
Thus
\begin{gather*}
\lim_{\varepsilon \to 0}\int_{C_{\varepsilon }}
u(z)\frac{\partial w}{\partial n_{\sigma }}(z)d\sigma =0, \\
\lim_{\varepsilon \to 0}\int_{C_{\varepsilon}}
u(z)\frac{\partial g_G}{\partial n_{\sigma }}(z,\zeta )d\sigma =-2\pi
u( \zeta ) .
\end{gather*}

(3) Since $u$ is a $C^{2}$-function on $G$, $\Delta
u$ is continuous on $D_{\varepsilon }$, then $\Delta u$ is bounded on 
$D_{\varepsilon }$. Also, as we argued above, $w$ is bounded on 
$D_{\varepsilon }.$Thus there is a constant $M_{0}$, such that 
$| \Delta u| \leq M_{0}$ and $| w| \leq M_{0}$ on $D_{\varepsilon }$.
Then
\begin{align*}
&\big| \iint_G \Delta u(z)g_G(z,\zeta )\,dx\,dy-
\iint_{G_{\varepsilon }} \Delta u(z)g_G(z,\zeta )\,dx\,dy| \\
&=| \iint_{D_{\varepsilon }} \Delta u(z)g_G(z,\zeta ) \,dx\,dy| \\
&\leq \iint_{D_{\varepsilon }} | \Delta u(z)\ln |z-\zeta | | \,dx\,dy
+\iint_{D_{\varepsilon }} |\Delta u(z)w(z)| \,dx\,dy \\
&\leq M_{0}\iint_{D_{\varepsilon }}
| \ln | z-\zeta | | \,dx\,dy+M_{0}^{2}\iint_{D_{\varepsilon }} \,dx\,dy \\
&=M_{0}^{2}\pi \varepsilon ^{2}+M_{0}\int_0^{2\pi }
\int_0^{\varepsilon } \rho | \ln \rho | d\rho d\theta
\end{align*}
which tends to zero as $\varepsilon $ tends to zero. Thus 
$$
\lim_{\varepsilon \to 0} \iint_{G_{\varepsilon }}
\Delta u(z)g_G(z,\zeta )\,dx\,dy
=\iint_G \Delta u(z)g_G(z,\zeta )\,dx\,dy.
$$ 
By (1), (2) and (3) it follows \eqref{e6}.
\end{proof}

The next theorem yields the formula for the solution of the Poisson problem
with zero boundary values on a $k$-symmetric region $G$.

\begin{theorem} \label{thm4}
Let $G$ be a $k$-symmetric region bounded by a finite number of
 $\sigma $-rectifiable Jordan curves. Let $f$ be a $k$-symmetric, continuous 
function on $G$. There is a unique $k$-symmetric, $C^{2}$-function $u$ on $G$, 
such that $\Delta u=f$ on $G$ and $u=0$ on $\partial G$. For all
 $\zeta $ in $G$,
\begin{equation}
u(\zeta )=\frac{1}{4\pi }\iint_G f(z)[ g_G(z,\zeta
)+g_G(z;k(\zeta ))] \,dx\,dy.  \label{e7}
\end{equation}
\end{theorem}

\begin{proof}
Since $k$ is an involution of $G$, the function 
$\frac{u(\zeta )+u(k(\zeta ))}{2}$ is a $k$-symmetric function on $G$. 
By Theorem \ref{thm3}, 
\begin{gather*}
u(\zeta )=\frac{1}{2\pi }\iint_G f(z)g_G(z,\zeta )\,dx\,dy,\\
u(k(\zeta )) =\frac{1}{2\pi }\iint_G f(k(z))g_G(z;k(\zeta ))\,dx\,dy.
\end{gather*} 
The $k$-symmetry of $f$ implies 
$$
\frac{u(\zeta )+u(k(\zeta ))}{2}
=\frac{1}{ 2\pi }\iint_G f(z)\frac{g_G(z,\zeta )+g_G(z;k(\zeta ))}{2}\,dx\,dy.
$$ 
By Proposition \ref{prop2}, $u$ is a $k$-symmetric function on $G$, then the
left side of the last equality is $u(\zeta )$ and we conclude that 
$$
u(\zeta) =\frac{1}{4\pi }\iint_G f(z)[ g_G(z,\zeta
)+g_G(z;k(\zeta ))] \,dx\,dy.
$$ 
By Remark \ref{rmk1}, it follows \eqref{e7}.
\end{proof}

\section{Poisson's equation on the orbit space}

Let $X$ be compact Klein surface and let $D$ be a region bounded by a finite
number of $\sigma $-rectifiable Jordan curves. The Klein surface $X$ is the
factor manifold of the $k$-symmetric Riemann surface $O_2$ with respect to
the group $H$. Then, $D$ is obtained from the $k$-symmetric region $G$ by
identifying the $k$-symmetric points. Therefore, the $k$-symmetric Green's
function $g_G^{(k)}(z,\widetilde{\zeta })$, where 
$\widetilde{\zeta }=\pi(\zeta )$, is continuous on $\overline{D}$, 
harmonic on $\overline{D}\backslash\{ \widetilde{\zeta }\} $, 
$g_G^{(k)}(z,\widetilde{\zeta })=0$ for all $z$ on $\partial D$ and
 has the singularity $-\frac{1}{2}\ln | z-\widetilde{\zeta }| $ at 
$\widetilde{\zeta }$. Then, $g_G^{(k)}(z,\widetilde{\zeta })$ is the Green
function of $D$ with singularity at $\widetilde{\zeta }=\pi (\zeta )$.

We obtain the formula for the solution of \eqref{e1p}--\eqref{e2p} on a 
$k$-symmetric region $G$.

\begin{theorem} \label{thm5}
Let $G$ be a $k$-symmetric region bounded by a finite number of 
$\sigma $-rectifiable Jordan curves. Let $f$ be a $k$-symmetric, continuous 
function on $G$ and $h$ be a $k$-symmetric, continuous function on $\partial G$. 
There is a unique function $u$ on $\overline{G}$, such that $\Delta u=f$ on $G$
and $u=h$ on $\partial G$. For all $\zeta $ in $G$,
\begin{equation}
u(\zeta )=\frac{1}{2\pi }\iint_G f(z)g_G^{(k)}(z,
\widetilde{\zeta })\,dx\,dy+\frac{1}{2\pi }\int_{\partial G}h(z)\frac{
\partial g_G^{( k) }(z,\widetilde{\zeta })}{\partial n_{\sigma }
}d\sigma .  \label{e8}
\end{equation}
\end{theorem}

\begin{proof}
By definition, 
\[
\frac{1}{2}\big[ g_G(z,\zeta )+g_G(z,k(\zeta ))\big] 
=g_G^{(k)}(z,\widetilde{\zeta })
\]
 is the $k$-invariant Green function
for the region $G$, with singularities 
$-\frac{1}{2}\ln | z-\zeta | $ and $-\frac{1}{2}\ln | z-k(\zeta )| $ at 
$\zeta $ and $k(\zeta )$, respectively. We combine the solution of the 
Dirichlet problem for harmonic functions given by
\[
u(\zeta )=\frac{1}{2\pi }\int_{\partial G}h(z)\frac{\partial
g_G^{(k)}(z,\widetilde{\zeta })}{\partial n_{\sigma }}d\sigma
\]
for $\zeta \in $ $G$, see \cite{b2}, with the solution of the Poisson's
equation which is zero on the boundary given by Theorem \ref{thm4}.
\end{proof}

Next we derive the solution of \eqref{e1}--\eqref{e2} on the region $D$.

\begin{proposition} \label{prop6}
Let $F$ be the continuous real-valued function on $D$, defined by the
relation $f=F\circ \pi $ and let $H$ be the continuous real-valued function
on $\partial D,$defined by the relation $h=H\circ \pi $. The solution of 
\eqref{e1}--\eqref{e2} is the function $U$ defined on $\overline{D}$, 
by the relation $u=U\circ \pi $, where $\pi $ is the canonical projection of $O_2$
on $X$ and $u$ is the solution \eqref{e8}  of  \eqref{e1p}--\eqref{e2p} 
on the $k$-symmetric region $G$ of $O_2$.
\end{proposition}

\begin{proof}
The $k$-symmetry of the function $f$ on $G$, yields 
\[
\Delta U(\widetilde{\zeta })=\Delta u(\zeta )
=f(\zeta )=f(k(\zeta ))=F(\widetilde{\zeta}),
\]
 for all $\widetilde{\zeta }\in D$, where 
$\widetilde{\zeta }=\pi (\zeta)$. Also, the $k$-symmetry of the function $h$ 
on $\partial G$, yields 
\[
U(\widetilde{\zeta })=u(\zeta )=h(\zeta )=h(k(\zeta ))=H(\widetilde{
\zeta }), 
\]
for all $\widetilde{\zeta }\in \partial D.$Due to the uniqueness,
the function $U$ defined on $\overline{D}$ by
\[
U(\widetilde{\zeta })=u(\zeta ),
\]
for all $\widetilde{\zeta }$ in $\overline{D}$, where 
$\widetilde{\zeta } =\pi (\zeta )$, is the solution of  \eqref{e1}--\eqref{e2}.
\end{proof}

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\end{document}