\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 269, pp. 1--7.\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-2017/269\hfil Harmonic measures ]
{Harmonic measures and Poisson kernels on \\ Klein surfaces}
\author[M. Ro\c{s}iu \hfil EJDE-2017/269\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 Vicentiu D. Radulescu}
\thanks{Submitted July 20, 2017. Published October 31, 2017.}
\subjclass[2010]{30F50, 35J25}
\keywords{Klein surface; Green's function; harmonic measure; Poisson kernel}
\begin{abstract}
We introduce harmonic measure on a Klein surface and obtain a
formula for the solution of the Dirichlet problem on a Klein surface,
which is an analogue for the Poisson integral. We rewrite the
Radon-Nikodym derivative of harmonic measure against the corresponding
arc length.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks
\section{Introduction}
We study the Dirichlet problem 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. We prove that symmetric conditions on the boundary determine
symmetric solutions which lead to solutions for the similar problems for
Klein surfaces. Thus, it is possible to solve boundary value problems on a
Klein surface, once the harmonic measure on the symmetric Riemann surface is
known. 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{b2,b3}.
In this paper, the methods introduced in \cite{a2,b1,r2} are used to extend the
use of the Green's function to the study of harmonic measure for a Klein
surface. Since dianalytic structures of Klein surfaces are related to
symmetric conformal metrics on their double covers, we represent the
harmonic measure in terms of such metrics, using the concept of normal
derivative of the $k$-invariant Green's function introduced in \cite{b3}. The
symmetric harmonic measure provides an explicit formula for the solution of
the Dirichlet problem on a Klein surface, that is an analogue for the
Poisson integral. Also, we rewrite the Radon-Nikodym derivative of harmonic
measure against $\sigma $-arc length, the symmetric arc length.
\section{Preliminaries}
We present some definitions and basic results about the relationship between
Klein surfaces and symmetric Riemann surfaces.
Let $O_{2}$ be a region in the complex plane, bounded by a finite number of
analytic Jordan curves. Then $\overline{O_{2}}=O_{2}\cup \partial O_{2}$ can
be conceived as a bordered Riemann surface, see \cite{a1}.
A symmetric Riemann
surface is a pair $(O_{2},k)$, where $O_{2}$ is a Riemann surface and $k$ is
an antianalytic involution without fixed points.
If $(X,A)$ is a a compact Klein surface, 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.Conversely, if $(O_{2},k)$ is a symmetric Riemann
surface, then on the orbit space $O_{2}/H$ there exists a dianalytic atlas
$A $, such that $(O_{2}/H,A)$ is a Klein surface, see \cite{a2}.
\begin{remark} \label{rmk2.1} \rm
In this paper, we identify $X$ with the orbit space $O_{2}/H$. The canonical
projection of $O_{2}$ onto $O_{2}/H$ is denoted by $\pi $. Also, we denote
with $\widetilde{z}$ the $H$ - orbit of $z\in O_{2}$, namely
$\widetilde{z}=\widetilde{k(z)}=\pi (z)=\pi (k(z))=\{z,k(z)\}$.
\end{remark}
Let $F:X\to \mathbb{R}$ be a function on $X$. Its lifting $f$ to
$O_{2}$ is given by
\begin{equation}
F(\widetilde{z})=f(z)=f(k(z)),z\in O_{2},\widetilde{z}=\pi (z). \label{e2.1}
\end{equation}
A function $f$ on $O_{2}$ with the property \eqref{e2.1} is called a symmetric
function.
Conversely, if $g:O_{2}\to \mathbb{R}$ 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)),\quad 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 $X$.
We denote by $\mathcal{B}(O_{2})$ and $\mathcal{B}(X)$ the $\sigma$-algebra
of Borel sets on $O_{2}$, respectively, on $X$. The $\sigma $-algebra of
symmetric Borel sets of $O_{2}$ is denoted by $\mathcal{B}_{s}(O_{2})$ and
$\mathcal{B}_{s}(O_{2})=\{U\cup k(U): U\in \mathcal{B}(O_{2})\}$, see
\cite{b2}.
Let $\widetilde{\gamma }$ be a piecewise smooth Jordan curve on $X$.
Then $\widetilde{\gamma }$ has exactly two liftings $\gamma $ and
$k\circ \gamma $ on $O_{2}$ 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{b1}.
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 $\gamma$.
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 relatively compact 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{b3}.
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}
\begin{gathered}
\Delta U=0,\quad \text{in }\Omega \\
U=F,\quad \text{on }\partial \Omega
\end{gathered} \label{e2.2}
\end{equation}
is equivalent with the Dirichlet problem on $O_{2}$
\begin{equation}
\begin{gathered}
\Delta u=0,\quad \text{in }D \\
u=f,\quad \text{on }\partial D,
\end{gathered} \label{e2.3}
\end{equation}
see \cite{b2,r2}.
The Dirichlet problem for the region $D$ and the boundary function $f$ has a
unique solution, provided that $\partial D$ has only regular points, see
\cite{s1}.
The symmetric conditions on the boundary imply symmetric solutions for the
problem \eqref{e2.3}, for details see \cite{b2} and the original source
\cite{s1}.
\begin{proposition} \label{prop2.2}
The solution $u$ of \eqref{e2.3} is a symmetric function in $D$.
\end{proposition}
\section{Symmetric harmonic measure}
First we recall some notions and results about harmonic measure. An
extensive study of the harmonic measure is developed in \cite{g1}.
Let $D$ be a symmetric region in the complex plane and
$\mathcal{B}_{s}(\partial D)$ the $\sigma $-algebra of symmetric Borel
sets of $\partial D$. The harmonic measure for $D$ is known (see \cite{r1})
to be a function $\omega_{D}:D\times \mathcal{B}_{s}(\partial D)\to [ 0,1]$
such that:
\begin{enumerate}
\item for each $\zeta \in D$, the map $B\to \omega _{D}(\zeta ,B)$
is a Borel probability measure on $\partial D$;
\item if $f:\partial D\to \mathbb{R}$
is a continuous function, then the solution of the Dirichlet problem, for
$D$ and the boundary function $f$, is the generalized Poisson integral of
$f $ on $D$, $P_{D}f(z)$, given by
\begin{equation}
P_{D}f(\zeta )=\int_{\partial D} f(z)d\omega _{D}(\zeta ,z),\zeta
\in D. \label{e3.1}
\end{equation}
\end{enumerate}
\begin{remark} \label{rmk3.1}\rm
The uniqueness of $\omega _{D}$ is a consequence of the Riesz representation
theorem.
\end{remark}
A method of determining the harmonic measure is given by the following
characterization (see \cite{r1}).
\begin{proposition} \label{prop3.2}
The function $\omega _{D}(\cdot ,B)$, is the solution of the generalized
Dirichlet problem with boundary function $f=1_{B}$.
\end{proposition}
\begin{remark} \label{rmk3.3}\rm
The function $\omega _{D}(\cdot ,B)$ is well defined on a compact Riemann
surface.
\end{remark}
The harmonic measure for $D$ is related to another conformal invariant, the
Green's function for the symmetric region $D$. We are using the following
integral Barza's representation \cite{b3}:
\begin{theorem} \label{thm3.4}
Let $D$ be a symmetric region, whose boundary $\partial D$ consists of a
finite number of pairwise disjoint $\sigma $-rectifiable Jordan curves . If
$u\in C(\overline{D})$ is harmonic on $D$, then\ for all $\zeta $ in $D$,
\begin{equation}
u(\zeta )=\frac{1}{2\pi }\int_{\partial D} u(z)\frac{\partial
g_{D}(z;\zeta )}{\partial n_{\sigma }}d\sigma (z). \label{e3.2}
\end{equation}
\end{theorem}
Because of \eqref{e3.2},
\begin{equation*}
P_{\zeta }(z)=\frac{1}{2\pi }\frac{\partial g_{D}(z;\zeta )}{\partial
n_{\sigma }}
\end{equation*}
is called the Poisson kernel for the region $D$.
Using the above theorem and the fact that Borel measures are determined by
their actions on continuous functions, we obtain a representation of the
harmonic measure in terms of the normal derivative of the Green's function
with respect to $d\sigma $.
\begin{proposition} \label{prop3.5}
Let $D$ be a symmetric region, whose boundary $\partial D$ consists of a
finite number of pairwise disjoint $\sigma $-rectifiable Jordan curves.
If $\zeta \in D$, then
\begin{equation*}
d\omega _{D}(\zeta ,z)
=\frac{\partial g_{D}(z;\zeta )}{\partial n_{\sigma }}
\cdot \frac{d\sigma (z)}{2\pi },\quad z\in \partial D.
\end{equation*}
Thus, harmonic measure for $\zeta \in D$ is absolutely continuous to arc
$\sigma $-length on $\partial D$ and, the density is
\begin{equation*}
\frac{d\omega _{D}}{d\sigma }
=\frac{1}{2\pi }\frac{\partial g_{D}(z;\zeta )}{\partial n_{\sigma }}
=P_{\zeta }(z),\quad\text{on }\partial D.
\end{equation*}
\end{proposition}
For $\zeta $, a point inside $D$, let $g_{D}^{(k)}(z;\widetilde{\zeta })$
be the $k$-invariant Green's function for the region $D$, with singularities
at $\zeta $ and $k(\zeta )$, defined by
\begin{equation*}
g_{D}^{(k)}(z;\widetilde{\zeta })
=\frac{1}{2}[ g_{D}(z;\zeta
)+g_{D}(z;k(\zeta ))] \quad\text{on }\overline{D}\backslash
\{\zeta ,k(\zeta )\}.
\end{equation*}
For additional information on this topic we refer to \cite{b3,r2}.
One can also derive the following statement (see \cite{b3}).
\begin{proposition} \label{prop3.6}
For every symmetric region $D$, the function
$g_{D}^{(k)}(\cdot ;\widetilde{\zeta })$ is $k$-invariant on
$\overline{D}$, i.e.
\begin{equation*}
g_{D}^{(k)}(z;\widetilde{\zeta })=g_{D}^{(k)}(k(z);\widetilde{\zeta }),
\quad\text{for every }z\in \overline{D}.
\end{equation*}
\end{proposition}
Let $\omega _{D}^{(k)}:D\times \mathcal{B}_{s}(\partial D)\to
[ 0,1]$ be the function defined by
\begin{equation*}
\omega _{D}^{(k)}(\widetilde{\zeta };B)=\frac{1}{2}[ \omega _{D}(\zeta
,B)+\omega _{D}(k(\zeta ),B)] ,\quad
\widetilde{\zeta }=\{\zeta ,k(\zeta)\},\quad
\zeta \in D,\; B\in \mathcal{B}_{s}(\partial D).
\end{equation*}
\begin{remark} \label{rmk3.7} \rm
The symmetry of the region $D$, implies that the function
$\omega _{D}^{(k)}(\widetilde{\zeta };B)$ is symmetric with respect to
$B$ on $\mathcal{B}_{s}(\partial D)$, i.e. for every
$B\in \mathcal{B}_{s}(\partial D)$:
\begin{equation*}
\omega _{D}^{(k)}(\widetilde{\zeta };B)=\omega _{D}^{(k)}(\widetilde{\zeta };k(B)).
\end{equation*}
\end{remark}
The function $\omega _{D}^{(k)}(\widetilde{\zeta };B)$ is called the
symmetric harmonic measure for $D$.
The function
\begin{equation*}
P_{\widetilde{\zeta }}^{(k)}(z)=\frac{1}{2\pi }
\frac{\partial g_{D}^{(k)}(z;\widetilde{\zeta })}{\partial n_{\sigma }},\quad
z\in D
\end{equation*}
is called the symmetric Poisson kernel for the region $D$.
\section{Integral representation on the double cover}
The next theorem yields a formula for the symmetric solution of problem
\eqref{e2.3}.
\begin{theorem} \label{thm4.1}
Let $D$ be a symmetric region bounded by a finite number of pairwise
disjoint $\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}
u(\zeta )=\frac{1}{2}\int_{\partial D} f(z)[ d\omega
_{D}(\zeta ,z)+d\omega _{D}(k(\zeta ),z)] . \label{e4.1}
\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 \eqref{e3.1},
\begin{equation}
u(\zeta )=\int_{\partial D} f(z)d\omega _{D}(\zeta ,z),\zeta \in D.
\label{e4.2}
\end{equation}
Replacing $\zeta $ with $k(\zeta )$ in \eqref{e4.2}, we get
\begin{equation}
u(k(\zeta ))=\int_{\partial D} f(z)d\omega _{D}(k(\zeta ),z),\zeta
\in D. \label{e4.3}
\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} f(z)
[ d\omega (\zeta ,z)+d\omega (k(\zeta ),z)] ,\ \zeta \in D.
\end{equation*}
By Proposition \ref{prop2.2}, $f$ 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} f(z)[ d\omega_{D}(\zeta ,z)
+d\omega _{D}(k(\zeta ),z)] .
\end{equation*}
The uniqueness of the solution of the Dirichlet problem for harmonic
functions implies \eqref{e4.1}.
\end{proof}
By Theorem \ref{thm4.1}, we obtain the Radon-Nikodym derivative of symmetric
harmonic measure for $D$ against $\sigma $-arc length.
\begin{proposition} \label{prop4.2}
Let $D$ be a symmetric region whose boundary $\partial D$ consists of a
finite number of pairwise disjoint $\sigma $- rectifiable Jordan curves.
If $\zeta \in D$, then
\begin{equation*}
d\omega _{D}^{(k)}(\widetilde{\zeta };z)
=\frac{\partial g_{D}^{(k)}(z;
\widetilde{\zeta })}{\partial n_{\sigma }}\cdot \frac{d\sigma (z)}{2\pi },
\quad z\in \partial D.
\end{equation*}
Thus, symmetric harmonic measure for $D$ is absolutely continuous to arc
$\sigma $-length on $\partial D$ and, the density is
\begin{equation*}
\frac{d\omega _{D}^{(k)}}{d\sigma }=\frac{1}{2\pi }\frac{\partial
g_{D}^{(k)}(z;\widetilde{\zeta })}{\partial n_{\sigma }}=P_{\widetilde{\zeta
}}^{(k)}(z),\quad\text{on }\partial D.
\end{equation*}
\end{proposition}
\section{Integral representations on a Klein surface}
Let $X$ be compact Klein surface and let $\Omega $ be a region of $X$
bounded by a finite number of pairwise disjoint $\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.
The harmonic measure for $\Omega $,
$\omega _{\Omega }:\Omega\times \mathcal{B} (\partial \Omega )\to [ 0,1]$, is defined by
\begin{equation*}
\omega _{\Omega }(\widetilde{\zeta },\widetilde{B})
=\omega _{D}^{(k)}(\widetilde{\zeta },B)
=\omega _{D}^{(k)}(\widetilde{\zeta },k(B)),\widetilde{
\zeta }\in \Omega ,\widetilde{B}
=\pi (B)\in \mathcal{B}(\partial \Omega )
\end{equation*}
The function
\begin{equation*}
P_{\widetilde{\zeta }} (\widetilde{z})
=P_{\widetilde{\zeta }}^{(k)}(z)
=P_{\widetilde{\zeta }}^{(k)}(k(z)),z\in D
\end{equation*}
is called the Poisson kernel for the region $\Omega $.
\begin{remark} \label{rmk5.1} \rm
From Remark \ref{rmk3.7}, it follows that the function $\omega _{\Omega }$
is well defined. From Proposition \ref{prop3.6}, it follows that the function
$P_{\widetilde{\zeta }}$ is well defined.
\end{remark}
By Theorem \ref{thm4.1}, we obtain the following representation of the solution of
the problem \eqref{e2.3} on a symmetric region $D$, in terms of the symmetric
harmonic measure.
\begin{theorem} \label{thm5.2}
Let $D$ be a symmetric region bounded by a finite number of disjoint $\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*}
u(\zeta )=\int_{\partial D} f(z)d\omega _{D}^{(k)}(\widetilde{
\zeta },z).
\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 the problem \eqref{e2.2} on the region $\Omega $, with
respect to the harmonic measure for the region $\Omega $.
\begin{theorem} \label{thm5.3}
Let $F$ be a continuous real-valued function on the border $\partial \Omega .
$ The solution of the problem \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{e4.1} 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
\begin{equation*}
U(\widetilde{\zeta })=u(\zeta )=f(\zeta )=f(k(\zeta ))=F(\widetilde{\zeta }
), \quad \text{for all }\widetilde{\zeta }\in \partial \Omega .\quad
\end{equation*}
Due to 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}
By Proposition \ref{prop4.2}, we obtain the Radon-Nikodym derivative of harmonic
measure for $\Omega $ against $\Sigma $-arc length.
\begin{proposition} \label{prop5.4}
Let $\Omega $ be a region bounded by a finite number of disjoint
$\sigma $-rectifiable Jordan curves. If $\widetilde{\zeta }\in \Omega $, then
\begin{equation*}
d\omega _{\Omega }(\widetilde{\zeta };\widetilde{z})=d\omega _{D}^{(k)}(
\widetilde{\zeta },z)=d\omega _{D}^{(k)}(\widetilde{\zeta },k(z)),\quad
z\in \partial D.
\end{equation*}
Thus, harmonic measure for $\Omega $ is absolutely continuous to arc $\Sigma
$-length on $\partial \Omega $ and, the density is
\begin{equation*}
\frac{d\omega _{\Omega }}{d\sigma }=P_{\widetilde{\zeta }} (\widetilde{z
}),\quad\text{on }\partial \Omega .
\end{equation*}
\end{proposition}
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\end{document}