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\begin{document}

\title{PANHARMONIC FUNCTIONS}
\author{Endang Cahya M.A.\thanks{%
Ph.D student of Mathematics Department, Bandung Institute of Technology. Jl.
Ganesha 10 Bandung 40132, Indonesia.} \\
%EndAName
{\small Department of Mathematics Education, }\\
{\small \ Indonesia University of \ Education.}\\
{\small Jalan Setyabudhi 229 Bandung 40141, Indonesia}\\
{\small E-mail : ecma@dns.math.itb.ac.id}}
\maketitle

\bigskip \textbf{Abstract}

{\small This paper we establish many properties of panharmonic functions in
a circular disk. Integral representation can be used to aproximate
derivative valued of panharmonic functioncs. Furthermore, we discuss the
maximum and the minimum principle of real and complex valued panharmonic
functions.\bigskip }

\section{Introduction}

In this paper we establish some results to the solutions of the Yukawa
equation, i.e. :

\begin{equation}
\triangle u=\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{%
\partial y^{2}}=\mu ^{2}u\text{ }  \label{yukawa}
\end{equation}
\newline
where $\mu $ is a positive constant. A $C^{2}$-solution of $\left( \ref
{yukawa}\right) $ in a domain $\Omega \subseteq R^{2}$ is called
panharmonic. A comprehensive panharmonic theory has subsequently develoved
by Duffin, J.L. Schiff \& W.J. Walker, E. Cahya, W. Setyabudhi, to which we
refer to the reader.

This paper will inform many other properties of panharmonic functions. Many
of properties will be proved by using integral representation of panharmonic
function, and another properties by using correspondence properties between
panharmonics functions in two space and a subclass of harmonic functions in
three space.

In this paper, the function $u$ is said to be panharmonic in a closed domain 
$\Omega $, if it is panharmonic in the interior and continuous on the
boundary of $\Omega .$

\section{Integral Representation}

The following result is the analog of the Gauss mean value theorem. For the
proof see \cite[p. 111]{duffin}$.$

\begin{theorem}
\label{repere1}Let u be panharmonic in the circular disk $\left(
x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2}.$ Then 
\begin{equation}
u\left( x_{0},y_{0}\right) =\frac{1}{2\pi I_{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta   \label{rep1}
\end{equation}
where $I_{0}(x)$ is the modified Bessel function, and 
\begin{equation}
I_{n}(x)=\frac{1}{n!}\left( \frac{x}{2}\right) ^{n}\left( 1+\frac{\left( 
\frac{x}{2}\right) ^{2}}{1.\left( n+1\right) }+\frac{\left( \frac{x}{2}%
\right) ^{4}}{1.2\left( n+1\right) \left( n+2\right) }+...\right) .
\label{bessel}
\end{equation}
We can see this theorem in another version. Let 
\begin{equation*}
I_{n}(x)=\frac{1}{n!}\left( \frac{x}{2}\right) ^{n}\Psi _{n}\left( x\right) ,
\end{equation*}
where 
\begin{equation}
\Psi _{n}\left( x\right) =1+\frac{\left( \frac{x}{2}\right) ^{2}}{1.\left(
n+1\right) }+\frac{\left( \frac{x}{2}\right) ^{4}}{1.2\left( n+1\right)
\left( n+2\right) }+...  \label{bessel1}
\end{equation}
and we can see that 
\begin{equation}
\int_{0}^{R}r\Psi _{0}\left( \mu r\right) dr=\frac{R^{2}}{2}\Psi _{1}\left(
\mu R\right) .  \label{bessel2}
\end{equation}
Write that 
\begin{equation*}
\Psi _{0}\left( \mu r\right) u\left( x_{0},y_{0}\right) =\frac{1}{2\pi }%
\int_{0}^{2\pi }u\left( x_{0}+r\cos \theta ,y_{0}+r\sin \theta \right)
d\theta 
\end{equation*}
for $0<r<a,$ multiplied the both side by $\mu r$, and then integrating that
equation with respect to $r$, gives 
\begin{equation*}
\int_{0}^{R}\mu r\Psi _{0}\left( \mu r\right) u\left( x_{0},y_{0}\right) dr=%
\frac{\mu }{2\pi }\int_{0}^{R}\int_{0}^{2\pi }u\left( x_{0}+r\cos \theta
,y_{0}+r\sin \theta \right) rdrd\theta ,
\end{equation*}
so 
\begin{equation*}
u\left( x_{0},y_{0}\right) =\frac{1}{\pi R^{2}\Psi _{1}\left( \mu R\right) }%
\int_{D(\overline{x},R)}u\left( x,y\right) dxdy
\end{equation*}
where $D(\overline{x},R)$ is the circular disk on above, and $\overline{x}%
=\left( x_{0},y_{0}\right) .$

\begin{theorem}
Let u be panharmonic in the circular disk $D(\overline{x},R)$. Then 
\begin{equation}
u\left( \overline{x}\right) =\frac{1}{\pi R^{2}\Psi _{1}\left( \mu R\right) }%
\int_{D(\overline{x},R)}udA,  \label{ec}
\end{equation}
where $\Psi _{1}\left( x\right) =1+\frac{\left( \frac{x}{2}\right) ^{2}}{1!2!%
}+\frac{\left( \frac{x}{2}\right) ^{4}}{2!3!}+\frac{\left( \frac{x}{2}%
\right) ^{6}}{3!4!}+\frac{\left( \frac{x}{2}\right) ^{8}}{4!5!}+...$
\end{theorem}

\begin{theorem}
Let u be panharmonic and non negative in the circular disk $D(\overline{x},R)
$. Then 
\begin{equation*}
\left| \frac{\partial u}{\partial x}\right| _{\left( x_{0},y_{0}\right)
}\leq u\left( x_{0},y_{0}\right) \frac{2}{R}\frac{\Psi _{0}\left( \mu
R\right) }{\Psi _{1}\left( \mu R\right) }
\end{equation*}
\end{theorem}
\end{theorem}

\begin{proof}
The idea of proof, see $\left[ 3,p.121\right] .$ Since the derivative of a
panharmonic functions is a panharmonic function, then we have 
\begin{equation*}
\frac{\partial u}{\partial x}\left( x_{0},y_{0}\right) =\frac{1}{\pi
R^{2}\Psi _{1}\left( \mu R\right) }\int_{D(\overline{x},R)}\frac{\partial u}{%
\partial x}dxdy.
\end{equation*}
By using \ Green's theorem, we have 
\begin{eqnarray*}
\frac{\partial u}{\partial x}\left( x_{0},y_{0}\right)  &=&\frac{1}{\pi
R^{2}\Psi _{1}\left( \mu R\right) }\int_{\partial D(\overline{x},R)}udy \\
&=&\frac{1}{\pi R^{2}\Psi _{1}\left( \mu R\right) }\int_{0}^{2\pi }u\left(
x_{0}+R\cos \theta ,y_{0}+R\sin \theta \right) R\cos \theta d\theta .
\end{eqnarray*}
So that, 
\begin{eqnarray*}
\left| \frac{\partial u}{\partial x}\right| _{\left( x_{0},y_{0}\right) }
&\leq &\frac{1}{\pi R\Psi _{1}\left( \mu R\right) }\int_{0}^{2\pi }ud\theta 
\\
&\leq &u\left( x_{0},y_{0}\right) \frac{2}{R}\frac{\Psi _{0}\left( \mu
R\right) }{\Psi _{1}\left( \mu R\right) }.
\end{eqnarray*}
\end{proof}

By the same way, we can prove for $\left| \frac{\partial u}{\partial y}%
\right| .$

\begin{corollary}
Let u be panharmonic and non negative in the circular disk $D(\overline{x},R)
$. Then 
\begin{equation*}
\left| div\text{ }u\right| _{\left( \overline{x}\right) }\leq u\left( 
\overline{x}\right) \frac{4}{R}\frac{\Psi _{0}\left( \mu R\right) }{\Psi
_{1}\left( \mu R\right) }.
\end{equation*}
\end{corollary}

\begin{proof}
Since $div$ $u=\nabla \cdot u=\frac{\partial u}{\partial x}+\frac{\partial u%
}{\partial y},$ then $\left| div\text{ }u\right| _{\left( \overline{x}%
\right) }\leq \left| \frac{\partial u}{\partial x}\right| _{_{\left( 
\overline{x}\right) }}+\left| \frac{\partial u}{\partial y}\right|
_{_{\left( \overline{x}\right) }}\leq u\left( \overline{x}\right) \frac{4}{R}%
\frac{\Psi _{0}\left( \mu R\right) }{\Psi _{1}\left( \mu R\right) }$
\end{proof}

\begin{corollary}
Let u be panharmonic and non negative in the circular disk $D(\overline{x},R)
$. Then 
\begin{equation*}
\left| grad\text{ }u\right| _{\left( \overline{x}\right) }\leq \frac{2\sqrt{2%
}}{R}\frac{\Psi _{0}\left( \mu R\right) }{\Psi _{1}\left( \mu R\right) }%
u\left( \overline{x}\right) .
\end{equation*}
\end{corollary}

\begin{proof}
Since $grad$ $u=\nabla u=\frac{\partial u}{\partial x}i+\frac{\partial u}{%
\partial y}j,$ then 
\begin{eqnarray*}
\left| grad\text{ }u\right| _{\left( \overline{x}\right) } &=&\left| \frac{%
\partial u}{\partial x}i+\frac{\partial u}{\partial y}j\right| _{\left( 
\overline{x}\right) } \\
&=&\left( \sqrt{\left( \frac{\partial u}{\partial x}\right) ^{2}+\left( 
\frac{\partial u}{\partial y}\right) ^{2}}\right) _{\left( \overline{x}%
\right) } \\
&=&\sqrt{2\left( \frac{2}{R}\frac{\Psi _{0}\left( \mu R\right) }{\Psi
_{1}\left( \mu R\right) }u\left( \overline{x}\right) \right) ^{2}} \\
&\leq &\frac{2\sqrt{2}}{R}\frac{\Psi _{0}\left( \mu R\right) }{\Psi
_{1}\left( \mu R\right) }u\left( \overline{x}\right) .
\end{eqnarray*}
\end{proof}

Next, the following theorems are consequences of the first Green Identity.
Let $\Omega $ be a bounded domain with boundary $\partial \Omega $ and $n$
is the outward unit normal vector to $\partial \Omega .$ Let $u,v$ are $%
C^{2}\left( \Omega \right) $ functions. We obtain the first Green's identity 
\begin{equation*}
\int_{\Omega }v\triangle udV+\int_{\Omega }\nabla u.\nabla
vdV=\int_{\partial \Omega }v\frac{\partial u}{\partial n}dS
\end{equation*}
Replacing $v$ by 1$\in C^{2}\left( \Omega \right) ,$ then we have 
\begin{equation*}
\int_{\Omega }\triangle udV=\int_{\partial \Omega }\frac{\partial u}{%
\partial n}dS.
\end{equation*}

\begin{theorem}
Let $u$ be panharmonic in the circular disk $D(\overline{x},R)$. Then 
\begin{equation*}
u(\overline{x})=\frac{1}{\pi \left( \mu R\right) ^{2}\Psi _{1}\left( \mu
R\right) }\int_{\partial D(\overline{x},R)}\frac{\partial u}{\partial n}dS.
\end{equation*}
\end{theorem}

\begin{proof}
Write that 
\begin{eqnarray*}
u\left( \overline{x}\right)  &=&\frac{1}{\pi R^{2}\Psi _{1}\left( \mu
R\right) }\int_{D(\overline{x},R)}udA \\
&=&\frac{1}{\pi \left( \mu R\right) ^{2}\Psi _{1}\left( \mu R\right) }%
\int_{D(\overline{x},R)}\mu ^{2}udA \\
&=&\frac{1}{\pi \left( \mu R\right) ^{2}\Psi _{1}\left( \mu R\right) }%
\int_{D(\overline{x},R)}\triangle udA \\
&=&\frac{1}{\pi \left( \mu R\right) ^{2}\Psi _{1}\left( \mu R\right) }%
\int_{\partial D(\overline{x},R)}\frac{\partial u}{\partial n}dS
\end{eqnarray*}
\end{proof}

\begin{corollary}
Let $u$ be panharmonic in the circular disk $D(\overline{x},R)$. Then $u(%
\overline{x})\leq \frac{4\sqrt{2}}{\mu ^{2}R^{3}}\frac{\Psi _{0}\left( \mu
R\right) }{\Psi _{1}^{2}\left( \mu R\right) }u\left( \overline{x}\right) .$
\end{corollary}

\begin{proof}
Since $\left| \frac{\partial u}{\partial n}\right| =\left| \nabla u\cdot
n\right| \leq $ $\left| \frac{\partial u}{\partial x}\right| \left|
n_{1}\right| +\left| \frac{\partial u}{\partial y}\right| \left|
n_{2}\right| \leq \frac{2\sqrt{2}}{R}\frac{\Psi _{0}\left( \mu R\right) }{%
\Psi _{1}\left( \mu R\right) }u\left( \overline{x}\right) ,$ then 
\begin{eqnarray*}
u(\overline{x}) &\leq &\frac{1}{\pi \left( \mu R\right) ^{2}\Psi _{1}\left(
\mu R\right) }\int_{\partial D(\overline{x},R)}\left| \frac{\partial u}{%
\partial n}\right| dS \\
&\leq &\frac{4\sqrt{2}}{\mu ^{2}R^{3}}\frac{\Psi _{0}\left( \mu R\right) }{%
\Psi _{1}^{2}\left( \mu R\right) }u\left( \overline{x}\right) 
\end{eqnarray*}
\end{proof}

For the uniqueness of this integral value, we can see the following theorem.
For the complete proof, see \cite[p. 98]{cahya}

\begin{theorem}
Let $u$ be panharmonic in $\Omega $ and satisfies the condition on the
boundary $\frac{\partial u}{\partial n}=g,$ where $g$ is continuous function
on $\partial \Omega .$ Then there is no different panharmonic function
satisfying the same condition.
\end{theorem}

\begin{proof}
Let us $u$ and $v$ are both panharmonic in $\Omega $ and satisfy the same
condition on the boundary i.e: $\frac{\partial u}{\partial n}=\frac{\partial
v}{\partial n}=g,$ where $g$ is a continuous function. By the first Green's
Identity, we have $\int_{\Omega }\mu ^{2}(u-v)^{2}dV+\int_{\Omega }\left|
\nabla (u-v)\right| ^{2}dV=0.$ Since $(u-v)^{2}$and $\left| \nabla
(u-v)\right| ^{2}$ are integrable and positive, then it must $u=v$ in $%
\Omega .$
\end{proof}

\section{The Maximum and The Minimum Principle}

Panharmonic functions in two space are in one to one correspondence with a
sub class of harmonic functions in three space. Thus given that $u(x,y)$ is
panharmonic, this correspondence is defined by the mapping 
\begin{equation*}
v(x,y,z)=u(x,y)\cos \mu z.
\end{equation*}
So, 
\begin{equation*}
\frac{\partial ^{2}v}{\partial x^{2}}+\frac{\partial ^{2}v}{\partial y^{2}}+%
\frac{\partial ^{2}v}{\partial z^{2}}=0
\end{equation*}
therefore $v$ is a harmonic function. Conversely, if $v(x,y)$ is a harmonic
function, then 
\begin{equation*}
U(x,y,z)=v(x,y)\cosh \mu z
\end{equation*}
is a panharmonic function, because 
\begin{equation*}
\frac{\partial ^{2}U}{\partial x^{2}}+\frac{\partial ^{2}U}{\partial y^{2}}+%
\frac{\partial ^{2}U}{\partial z^{2}}=\mu ^{2}U.
\end{equation*}

The virtue of this correspondence is that panharmonic functions may inherit
many of the well-known properties of harmonic functions.

\subsection{The Maximum Principle}

The following theorem is an important consequence of correspondence
principle.

\begin{theorem}
Let $u$ be a panharmonic function and non constant in a compact region $%
\Omega $, and suppose $u$ has a positive maximum values on $\Omega $. Then $u
$ attains its maximum values over $\Omega $ on $\partial \Omega .$

\begin{proof}
Suppose $u$ attains a maximum at $p$ interior point of $\Omega $. Consider
the cylindrical region $\Re $ of three space, defined as $\left( x,y\right)
\subseteq \Omega $ and $\frac{-\pi }{2\mu }\leq z\leq \frac{-\pi }{2\mu }.$
Let's define $V(x,y,z)=u(x,y)\cos \mu z$ on $\Re .$ It is clear that $V$ is
a harmonic function, and $V$ vanishes on the top and on the bottom of this
cylinder. On the sides and on all interior points of cylinder, $%
V(p,0)=u(p)\geq u(p)\cos \mu z>u(x,y)\cos \mu z.$ Thus, $V$ attains a
maximum at $(p,z)$ interior point of $\Re $ . This is contradict with the
maximum principle of harmonic function.
\end{proof}
\end{theorem}

The following corollary, which the proof is immediately follows from the
theorem 9, is frequently useful.

\begin{corollary}
If $u$ is a panharmonic function and non negative in a compact region $%
\Omega $, then $u$ attains it's maximum values over $\Omega $ on $\partial
\Omega .$

\begin{proof}
Since $u$ is a panharmonic function in $\Omega ,$ then $u$ continuous in $%
\Omega .$ Therefore, $u$ attains a maximum in $\Omega $, and it has a
positive maximum values over $\Omega .$ By applying theorem 9, it's maximum
values over $\Omega $ occurs on $\partial \Omega $.
\end{proof}
\end{corollary}

\begin{corollary}
Let $u$ be a panharmonic function and non negative in a domain $\Omega .$ If 
$u$ has a maximum in $\Omega $ then $u$ is a constant function. (in this
case $u=0$ on $\Omega ).$

\begin{proof}
Let $u$ be a panharmonic function in the domain $\Omega $ and assume that u
attains it's maximum valued in $\Omega .$ Let $\Omega _{M}=\left\{ x\in
\Omega :u(x)=M\right\} $ where $M:=\sup_{x\in \Omega }u(x),$ then $\Omega
_{M}$ not empty. Since $u$ is continuous function, $\Omega _{M}$ is closed
relative to $\Omega .$ Let $\overline{x}$ be any point in $\Omega _{M}$ and
apply the theorema 2, to the panharmonic function $u$ in an open ball $D(%
\overline{x},R)$ with $D(\overline{x},R)\subset \Omega .$ Then 
\begin{equation*}
0=M-u\left( \overline{x}\right) =\frac{1}{\pi R^{2}\Psi _{1}\left( \mu
R\right) }\int_{D(\overline{x},R)}\left( \Psi _{1}\left( \mu R\right) M-\mu
u\right) dA.
\end{equation*}
So, that $u=M$ in an open ball $D(\overline{x},R)$ therefore $\Omega _{M}$
is open relative to $\Omega .$ By the connectivity properties $\Omega
=\Omega _{M}.$ Which means $u$ is constant in $\Omega .$ But there is no
constant panharmonic functions exept nol function. (note that the
connectivity of $\Omega $ is needed here). So, $u\equiv M$ on $\Omega .$
\end{proof}
\end{corollary}

The next version of the maximum principle can be applied even when $u$ is
not continuous on $\overline{\Omega }$.

\begin{corollary}
Let $u$ be a panharmonic function and non negative in a bounded domain $%
\Omega $, and suppose there is a constant $M>0$ such that 
\begin{equation}
\lim_{k\rightarrow \infty }\sup u(b_{k})\leq M
\end{equation}
for every sequence $\left( b_{k}\right) $ in $\Omega $ converging to a point
in $\partial \Omega .$ Then $u\leq M$ on $\Omega .$
\end{corollary}

\begin{proof}
Let $M^{\prime }=\sup \left\{ u(x):x\in \Omega \right\} ,$and choose a
sequence $\left( b_{k}\right) $ in $\Omega $ such that $u(b_{k})\rightarrow
M^{\prime }.$ There are two cases to be consider. The first, if $\left(
b_{k}\right) $ has a sub sequence converging to some point $b\in \Omega ,$
then $u(b)=M\prime $. By the maximum principle panhamonic function, then $%
u=M^{\prime }$ on $\Omega .$ The second case, if no subsequence of $\left(
b_{k}\right) $ converges to a point in $\Omega ,$ then $\left( b_{k}\right) $
has a sub sequence $\left( a_{k}\right) $ converging to a boundary point of $%
\Omega ,$ called $a,$ then $u(a)=M^{\prime }.$ This implies $u\leq M^{\prime
}.$ Thus in all cases, we obtain $M^{\prime }\leq M.$
\end{proof}

\subsection{\bigskip The Minimum Principle}

The same as above, we have the following theorem.

\begin{theorem}
let $u$ be a panharmonic function and non constant in a compact region $%
\Omega $, and suppose $u$ has a negative minimum values on $\Omega $. Then $%
u $ attains its minimum values over $\Omega $ on $\partial \Omega .$
\end{theorem}

The following corollary is a consequence of the theorem.

\begin{corollary}
If $u$ is a panharmonic function and non positive in a compact region $%
\Omega $, then $u$ attains it's minimum values over $\Omega $ on $\partial
\Omega .$
\end{corollary}

\begin{corollary}
Let $u$ be a non positive panharmonic function in a domain $\Omega .$ If $u$
has a minimum in $\Omega $ then $u$ is a constant function. (in this case $%
u=0$ on $\Omega ).$
\end{corollary}

\begin{example}
Let $u(x,y)=\cosh \left( \mu \frac{x+y}{\sqrt{2}}\right) $ in the circular
disk $x^{2}+y^{2}\leq a^{2}.$ It is clear that $u$ is a panharmonic function
on the disk. We can see that the maximum of $u$ occurs on the boundary, but
the minimum of $u$ occurs on the straight line $x+y=0.$ But, if $%
v(x,y)=-u(x,y)$, maximum of $v$ over disk occurs on the straight line $x+y=0$
and its minimum occurs on the boundary.
\end{example}

\begin{example}
Let $u(x,y)=\sinh \left( \mu \frac{x+y}{\sqrt{2}}\right) $ in the circular
disk $x^{2}+y^{2}\leq a^{2}.$ It is clear that $u$ is a panharmonic function
on the disk. We can see that the maximum and the minimum of $u$ occurs on
the boundary.
\end{example}

The next version of the minimum principle can be applied even when $u$ is
not continuous on $\overline{\Omega }$.

\begin{corollary}
\bigskip Let $u$ be a panharmonic function and non positive in a bounded
domain $\Omega $, and suppose there is a constant $M<0$ such that 
\begin{equation*}
\lim_{k\rightarrow \infty }\inf u(b_{k})\geq M
\end{equation*}
for every sequence $\left( b_{k}\right) $ in $\Omega $ converging to a point
in $\partial \Omega .$ Then $u\geq M$ on $\Omega .$
\end{corollary}

\section{Complex -Valued Panharmonic Functions}

The all results in above apply to real valued functions only. The next, we
want to see the integral representation and the maximum valued for complex
valued panharmonic function.

\begin{theorem}
Let $f(z)=u(x,y)+iv(x,y)$ be a complex valued function in $C^{2}\left(
\Omega \right) .$ $f$ is a panharmonic if only if $u$ and $v$ are both
panharmonic.
\end{theorem}

The theorem in above implies that the values of $f$ on the center of disk
can be represented in the form of integral representation, by using the
theorem 1, we have the following theorem.

\begin{theorem}
Let $f$ be a complex valued panharmonic function in the circular disk $%
D(z_{0},R).$ Then $f(z_{0})=\frac{1}{2\pi \Psi _{0}\left( \mu R\right) }%
\int_{0}^{2\pi }f(z_{0}+R$ $e^{i\theta })d\theta .$
\end{theorem}

\begin{proof}
Let $f$ be a complex valued panharmonic functions. Write that $%
f(z)=u(x,y)+iv(x,y),$ where $u,v$ are both panharmonic functions. Since $u$
and $v$ can be represented as (2), then 
\begin{eqnarray*}
f(z_{0}) &=&u\left( x_{0},y_{0}\right) +iv\left( x_{0},y_{0}\right) \\
&=&\frac{1}{2\pi \Psi _{0}\left( \mu R\right) }\int_{0}^{2\pi }(u+iv)\left(
x_{0}+R\cos \theta ,y_{0}+R\sin \theta \right) d\theta \\
&=&\frac{1}{2\pi \Psi _{0}\left( \mu R\right) }\int_{0}^{2\pi
}f(z_{0}+Re^{i\theta })d\theta .
\end{eqnarray*}
\end{proof}

Now we want to see the integral representation of a complex valued
panharmonic function in a disk and in another form. Let's $z=z_{0}+R$ $%
e^{i\theta },$ then $dz=i(z-z_{0})d\theta .$ Therefore, we have 
\begin{equation*}
f(z_{0})=\frac{1}{2\pi i\Psi _{0}(\mu R)}\int_{\partial D(z_{0},R)}\frac{f(z)%
}{z-z_{0}}dz
\end{equation*}

\begin{corollary}
Let $f$ be a complex valued panharmonic functions in a disk $\left|
z-z_{0}\right| <R.$ Then 
\begin{equation*}
f(z_{0})=\frac{1}{2\pi i\Psi _{0}(\mu R)}\int_{\partial D(z_{0},R)}\frac{f(z)%
}{z-z_{0}}dz
\end{equation*}
\end{corollary}

\begin{theorem}
Let $f$ be a complex valued panharmonic functions and $c$ is a complex
constant. Then $cf$ is a panharmonic functions
\end{theorem}

\begin{proof}
Let's $f(z)=u(x,y)+iv(x,y),$ where $u,v$ are real panharmonic functions.
Let's $c\in C,$ where $c=a+ib.$ We have $h(z)=cf(z)=(au-bv)+i(av+bu),$ and $%
h_{xx}=(au-bv)_{xx}+i(av+bu)_{yy},$ $h_{yy}=(au-bv)_{yy}+i(av+bu)_{yy}.$
Then $h_{xx}+h_{yy}=\mu ^{2}cf(z).$
\end{proof}

\begin{corollary}
Let $f$ be a complex valued panharmonic function. Then $f^{\prime }(z)$ is a
panharmonic function.
\end{corollary}

\begin{proof}
Let's $h(z)=2\frac{\partial f}{\partial z}=\left( \frac{\partial }{\partial x%
}-i\frac{\partial }{\partial y}\right) \left( u+iv\right) =\left(
u_{x}+v_{y}\right) +i\left( v_{x}-u_{y}\right) .$ Then $h_{xx}(z)=\left(
u_{x}+v_{y}\right) _{xx}+i\left( v_{x}-u_{y}\right) _{xx}$ and $%
h_{yy}(z)=\left( u_{x}+v_{y}\right) _{yy}+i\left( v_{x}-u_{y}\right) _{yy}.$
We obtain $h_{xx}+h_{yy}=\mu ^{2}(\left( u_{x}+v_{y}\right) +i\left(
v_{x}-u_{y}\right) )=\mu ^{2}h.$
\end{proof}

\begin{corollary}
Let $f$ be a complex valued panharmonic functions in a disk $\left|
z-z_{0}\right| <R.$ Then 
\begin{equation*}
f^{\prime }(z_{0})=\frac{1}{2\pi i\Psi _{0}(\mu R)}\int_{\partial D(z_{0},R)}%
\frac{f^{\prime }(z)}{z-z_{0}}dz
\end{equation*}
\end{corollary}

\begin{proof}
See corollary 21 and combine with corollary 23.
\end{proof}

We can conclude that if $f\in C^{\infty }(D),$ and $f$ is a complex valued
panharmonic functions then we can show that $f^{n}(z)$ is a panharmonic
functions. And then, by using the corollary 24, we have 
\begin{equation*}
f^{n}(z_{0})=\frac{1}{2\pi i\Psi _{0}(\mu R)}\int_{\partial D(z_{0},R)}\frac{%
f^{n}(z)}{z-z_{0}}dz.
\end{equation*}

\subsection{The Maximum Principle of a Complex -Valued Panharmonic Functions}

\begin{theorem}
If $u$ is a complex valued panharmonic function in a domain $\Omega ,$ then $%
\left| u\right| $ attains a maximum valued in $\overline{\Omega }$ on the
boundary $\partial \Omega .$
\end{theorem}

\begin{proof}
Suppose $\left| u\right| $ attains a maximum value of $M,$ at some point $%
a\in \Omega .$ Choose $b\in C$ such that $\left| b\right| =1$ and $bu(a)=M.$
Then the real valued panharmonic function $\func{Re}bu$ attains its maximum
valued is $M$ at $a$ $\in \Omega ..$ It is contradict with theorem 9. So, $%
\left| u\right| $ must be attains a maximum value on the boundary $\partial
\Omega .$
\end{proof}

No minimum principle hold for $\left| u\right| $ as an example, let $%
u=e^{\mu \frac{x+y}{\sqrt{2}}}+ie^{-\mu \frac{x+y}{\sqrt{2}}}$ on the
circular disk $D(0,R).$ The minimum value of $\left| u\right| $ occurs on
the straight line $x+y=0,$ but the maximum value of $\left| u\right| $
occurs on the boundary, where $x+y$ attains a maximum on the circular disk.

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\bibitem{wono}  W.Setyabudhi, Fungsi Panharmonik di Cakram, MIHMI
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\bibitem{axler}  S.Axler, P.Bourdon, W.Ramey, ''Harmonic Function Theory'',
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\end{document}
