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\begin{document}

\title{THE MAXIMUM AND MINIMUM\\
PRINCIPLE OF PANHARMONIC FUNCTIONS}
\author{Endang Cahya M. A. \\
%EndAName
Jurusan Pendidikan Matematika\\
FPMIPA\ UPI\ Bandung\\
Jl. Dr. Setyabudhi 229 Bandung 40141\\
E-mail ecma@dns.math.itb.ac.id}
\maketitle

\section{Introduction}

A function $u=u(x,y)$ in $C^{2}\left( \Omega \right) $ be a panharmonic
function if it satisfies the Yukawa equation, ie 
\begin{equation}
\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}%
=\mu ^{2}u,  \label{yukawa}
\end{equation}
for some constants real $\mu ,$ and $\Omega $ is a domain in $\Re ^{2}.$ The
panharmonic function $u$ is said to be panharmonic in a closed domain if it
is panharmonic in the interior of the domain and continuous on the boundary.

In this paper, we give some results of literatures survey and also give some
new results. The approach is similar to one that is used in the harmonic
functions. It can be used because there is a similarly between Laplacian
equation and Yukawan equation.

We firstly consider a maximum and minimum valued problem of $a$ panharmonic
function on a disk, which will be used as a first step of general conjecture
about maximum and minimum principle of panharmonic function. Secondly we
discussed the maximum valued of modulus of a complex valued panharmonic
functions.

\section{The Maximum and Minimum Principle}

To study maximum and minimum principle for panharmonic functions, let see
the idea from the following facts.The following theorem from Duffin $\left[ 2%
\right] $ page $115.$

\begin{theorem}
If $u(r,\theta )$ is a panharmonmic in the disk $x^{2}+y^{2}\leq a^{2},$
then for $0\leq r<a$%
\begin{equation}
u(r,\theta )=\sum_{-\infty }^{\infty }c_{n}I_{\left| n\right| }\left( \mu
r\right) e^{in\theta }  \label{panh}
\end{equation}
where 
\begin{equation*}
c_{n}=\frac{1}{2\pi I_{\left| n\right| }\left( \mu a\right) }\int_{0}^{2\pi
}u(a,\theta )e^{-in\theta }d\theta .
\end{equation*}
\end{theorem}

Here $I_{n}$ is the modified Bessel function of the first kind given by 
\begin{equation*}
I_{n}\left( x\right) =\frac{1}{n!}\left( \frac{x}{2}\right) ^{n}\Psi
_{n}\left( x\right)
\end{equation*}
and 
\begin{equation*}
\Psi _{n}\left( x\right) =1+\frac{\left( \frac{x}{2}\right) ^{2}}{1.(n+1)}+%
\frac{\left( \frac{x}{2}\right) ^{4}}{1.2(n+1)(n+2)}+...
\end{equation*}
We note that for each $n,I_{\left| n\right| }\left( \mu r\right) e^{in\theta
}$ is a solution and that in particular $I_{\left| 0\right| }\left( \mu
r\right) $ is another positive solution. From the above theorem, first we
see if the value of $u$ on the boundary is constant, say $u(a,\theta )=k$
for arbitrary real value of $k.$ Then each coefficient of $c_{n}$ becomes 
\begin{equation*}
c_{n}=\frac{k}{2\pi I_{\left| n\right| }\left( \mu a\right) }\int_{0}^{2\pi
}e^{-in\theta }d\theta .
\end{equation*}
But 
\begin{equation*}
\int_{0}^{2\pi }e^{-in\theta }d\theta =\left\{ 
\begin{array}{c}
2\pi ,\text{ }n=0 \\ 
0,\text{ }n\neq 0
\end{array}
\right.
\end{equation*}
then we obtain 
\begin{equation*}
c_{n}=\left\{ 
\begin{array}{c}
2\pi ,\text{ }n=0 \\ 
0,\text{ }n\neq 0
\end{array}
\right.
\end{equation*}
and 
\begin{equation*}
u(r,\theta )=\frac{k\Psi _{0}\left( \mu r\right) }{\Psi _{0}\left( \mu
a\right) }.
\end{equation*}
For this function $u,$ let see the following cases. First, if $k>0,$ then $%
u(r,\theta )>0$ for all $\theta $ and $r,$ with $0\leq r\leq a.$ The maximum
value of $u$ is achieved if $r=a,$ i.e $u(r,\theta )=k.$ The minimum value
of $u$ is achieved if $r=0.$ This condition says that the maximum value of $%
u $ is achieved at the boundary and the minimum value will be achieved at
the \ center of disk. Second, if $k<0,$ then $u(r,\theta )<0$ for all $%
\theta $ and $r,$ with $0\leq r\leq a.$ The maximum value of $u$ will be at
the center of the disk and the minimum value of $u$ will be achieved at the
boundary of disk. From the above cases, the point where $\max $imum and $%
\min $imum value of a panharmonic function $u$ are happened, are different
from harmonic function.

\subsection{The Maximum Principle}

The example in above, gives the idea for the following theorem,

\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 the boundary ($\partial \Omega ).$

\begin{proof}
Suppose that there is $\overline{x}$ in $\Omega ,$ such that $\nabla u(%
\overline{x})=0,u_{xx}(\overline{x})u_{yy}(\overline{x})-u_{xy}^{2}(%
\overline{x}\dot{)}>0,$ and $u_{xx}(\overline{x})<0.$ Then $u_{xx}(\overline{%
x})u_{yy}(\overline{x})>u_{xy}^{2}(\overline{x}\dot{)}\geq 0,$ but $u_{xx}(%
\overline{x})<0.$ As it's consequence $u_{yy}(\overline{x})<0.$ Therefore, $%
\mu ^{2}u(\overline{x})=u_{xx}(\overline{x})+u_{yy}(\overline{x})<0.$ This
is contradict with the fact that $u(\overline{x})>0.$ In other word, there
is no $\overline{x}$ in $\Omega $ that gives the maximum valued, therefore $%
\overline{x}$ mus be on the boundary of $\Omega .$
\end{proof}
\end{theorem}

The following corollary which the proof is immediately follows from the
theorem 2, is frequently useful.

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

\begin{proof}
Since $u$ is a panharmonic function in $\Omega ,$ then $u$ is continuous in $%
\Omega .$ Therefore, $u$ attains a maximum in $\Omega ,$ and it has a
positive maximum values over $\Omega .$ By applying theorem 2, it's maximum
values over $\Omega $ occurs on the $\partial \Omega .$
\end{proof}

\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 $\Omega ).$
\end{corollary}

For the proof, see $\left[ 1\right] $ and $\left[ 3\right] .$

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 $u<M$ 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 $u<M$ on $\Omega .$
\end{proof}

\subsection{\protect\bigskip The Minimum Principle}

The maximum principle in above, tend to used on a positive panharmonic
function. But the same as above, we have the following theorem for the
negative panharmonic functions.

\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}

\begin{proof}
Suppose that there is $\overline{x}$ in $\Omega ,$ such that $\nabla u(%
\overline{x})=0,u_{xx}(\overline{x})u_{yy}(\overline{x})-u_{xy}^{2}(%
\overline{x}\dot{)}>0,$ and $u_{xx}(\overline{x})>0.$ Then $u_{xx}(\overline{%
x})u_{yy}(\overline{x})>u_{xy}^{2}(\overline{x}\dot{)}\geq 0,$ but $u_{xx}(%
\overline{x})>0.$ As it's consequence $u_{yy}(\overline{x})>0.$ Therefore, $%
\mu ^{2}u(\overline{x})=u_{xx}(\overline{x})+u_{yy}(\overline{x})>0.$ This
is contradict with the fact that $u(\overline{x})<0.$ In other word, there
is no $\overline{x}$ in $\Omega $ that gives the minimum valued, therefore $%
\overline{x}$ mus be on the boundary of $\Omega .$
\end{proof}

The following corollary which the proof is immediately follows from the
theorem 6, is frequently useful.

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

\begin{proof}
Since $u$ is a negative panharmonic function, then the minimum value of $u$
is a negative too. By the above theorem, the minimum of $\ u$ must be occur
on the boundary.
\end{proof}

\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}

For the proof, see $\left[ 1\right] $ and $\left[ 3\right] .$

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 $u>M$ on $\Omega .$
\end{corollary}

\begin{proof}
Let $M^{\prime }=\inf \left\{ u(x):x\in \Omega \right\} ,$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 minimum 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 $u\geq M^{\prime
}.$ Thus in all cases, we obtain $u>M$ on $\Omega .$
\end{proof}

Furthermore, we will saw the value of a panharmonic function at some
interior points of a compact set through away integral representation of a
panharmonic function on a disk. This result can be obtained as implication
from the maximum principle in above. For this, let see the following
integral representation, and for the proof, see $\left[ 2\right] $ page 111.

\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 \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta .
\end{equation*}
\end{theorem}

\begin{corollary}
Let u be a non negative panharmonic function on a compact region $\Omega .$
If u$\leq M$ \ on the boundary of $\Omega $ for some M\TEXTsymbol{>}0, and $%
\overline{x}$ any an interior points then 
\begin{equation*}
u\left( \overline{x}\right) \leq \frac{M}{\Psi _{0}\left( \mu a\right) },
\end{equation*}
where $a$ is the shortest path from $\overline{x}$ to the boundary of $%
\Omega .$
\end{corollary}

\begin{proof}
Let's $\overline{x}=$ $\left( x_{0},y_{0}\right) $ any interior point of $%
\Omega .$ Make a disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right)
^{2}\leq a^{2}$ \ where $a$ is the shortest path from $\overline{x}$ to the
boundary of $\Omega .$ Then we have 
\begin{equation*}
u\left( x_{0},y_{0}\right) =\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta .
\end{equation*}
Since $0\leq u\leq M$ on the boundary of $\Omega $ then it is satisfied on
the disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2}$
in $\Omega ,$ therefore 
\begin{eqnarray*}
u\left( x_{0},y_{0}\right)  &=&\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta  \\
&\leq &\frac{M}{\Psi _{0}\left( \mu a\right) }.
\end{eqnarray*}
\end{proof}

\begin{corollary}
Let u be a non positive panharmonic function on a compact region $\Omega .$
If $u\geq M$ \ on the boundary of $\Omega $ for some M\TEXTsymbol{<}0, and $%
\overline{x}$ any an interior points then 
\begin{equation*}
u\left( \overline{x}\right) \geq \frac{M}{\Psi _{0}\left( \mu a\right) },
\end{equation*}
where $a$ is the shortest path from $\overline{x}$ to the boundary of $%
\Omega .$
\end{corollary}

\begin{proof}
Let's $\overline{x}=$ $\left( x_{0},y_{0}\right) $ any interior point of $%
\Omega .$ Make a disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right)
^{2}\leq a^{2}$ \ where $a$ is the shortest path from $\overline{x}$ to the
boundary of $\Omega .$ Then we have 
\begin{equation*}
u\left( x_{0},y_{0}\right) =\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta .
\end{equation*}
Since $0\geq u\geq M$ on the boundary of $\Omega $ then it is satisfied on
the disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2}$
in $\Omega ,$ therefore 
\begin{eqnarray*}
u\left( x_{0},y_{0}\right)  &=&\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u\left( x_{0}+a\cos \theta ,y_{0}+a\sin \theta \right)
d\theta  \\
&\geq &\frac{M}{\Psi _{0}\left( \mu a\right) }.
\end{eqnarray*}
\end{proof}

\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}

\subsection{Complex-Valued Panharmonic Functions}

The all results in above apply to real valued functions only. The next, we
want to see the maximum modulus 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}

\begin{proof}
Let $f(z)=u(x,y)+iv(x,y)$ is a complex valued panharmonic function in $%
\Omega ,$ then $f_{xx}+f_{yy}=u_{xx}+iv_{xx}+u_{yy}+iv_{yy}=\left(
u_{xx}+u_{yy}\right) +i\left( v_{xx}+v_{yy}\right) =\mu ^{2}\left(
u+iv\right) =\mu ^{2}f.$ Therefore, $u$ and $v$ must be satisfies $%
u_{xx}+u_{yy}=\mu ^{2}u$ and $v_{xx}+v_{yy}=\mu ^{2}v.$
\end{proof}

\begin{corollary}
Let $f(z)=u(x,y)+iv(x,y)$ be a complex valued panharmonic function and c any
complex constant. Then cf is a panharmonic too.
\end{corollary}

\begin{proof}
Let $f$ be a complex valued panharmonic function and c any complex constant.
Then $\left( cf\right) _{xx}+$ $\left( cf\right) _{yy}=cf_{xx}+$ $%
cf_{yy}=c(f_{xx}+f_{yy})=c\mu ^{2}f=\mu ^{2}(cf).$
\end{proof}

\subsubsection{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 $%
a\in \Omega .$ By the rotation principle, we can 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 2. 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.

\begin{thebibliography}{9}
\bibitem{cahya}  E.Cahya, Panharmonic Functions (to appear).

\bibitem{duffin}  R.J.Duffin,Yukawan Potential Theory, J.Math.Anal.Appl
35(1971).104-130.

\bibitem{wono}  W.Setyabudhi, Fungsi Panharmonik di Cakram, MIHMI
6(2000).-119-124.

\bibitem{axler}  S.Axler, P.Bourdon, W.Ramey, ''Harmonic Function Theory'',
Springer-Verlag, NewYork,1992.
\end{thebibliography}

\end{document}
