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%TCIDATA{Created=Tue Nov 07 08:40:25 2000}
%TCIDATA{LastRevised=Mon Feb 12 12:42:42 2001}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}


Tulisan ini merupakan cara penulisan lain dari suatu fungsi panharmonik pada
suatu disk yang telah dikemukakan oleh Duffin, dalam Yukawa Potential
Theory. Untuk keperluan penulisan yang akan dikemukakan disini, terlebih
dulu akan dikemukakan cara penulisan fungsi Bessel yang dimodifikasi.

\begin{equation*}
I_{n}(x)=\frac{1}{n!}\left( \frac{x}{2}\right) ^{n}\Psi _{n}(x)
\end{equation*}

\begin{equation*}
\Psi _{n}(x)=1+\frac{\left( \frac{x}{2}\right) ^{2}}{1.(n+1)}+\frac{\left( 
\frac{x}{2}\right) ^{4}}{1.2(n+1)(n+2)}+\frac{\left( \frac{x}{2}\right) ^{6}%
}{1.2.3(n+1)(n+2)(n+3)}+...
\end{equation*}
Secara khusus,

\begin{equation*}
I_{0}(x)=\Psi _{0}(x)=\sum_{n=0}^{\infty }\frac{\left( \frac{x}{2}\right)
^{2n}}{n!.n!}
\end{equation*}

\begin{equation*}
I_{1}(x)=\left( \frac{x}{2}\right) \Psi _{1}(x)=\left( \frac{x}{2}\right)
\sum_{n=0}^{\infty }\frac{\left( \frac{x}{2}\right) ^{2n}}{n!(n+1)!}.
\end{equation*}

\begin{eqnarray*}
\int_{0}^{R}rI_{0}(\mu r)dr &=&\int_{0}^{R}\sum_{n=0}^{\infty }\frac{\left( 
\frac{\mu r}{2}\right) ^{2n}}{n!.n!}rdr \\
&=&\sum_{n=0}^{\infty }\left( \int_{0}^{R}\frac{\left( \frac{\mu r}{2}%
\right) ^{2n}}{n!.n!}rdr\right) \\
&=&\frac{R^{2}}{2}\sum_{n=0}^{\infty }\frac{\left( \frac{\mu R}{2}\right)
^{2n}}{n!(n+1)!} \\
&=&\frac{R}{\mu }I_{1}(\mu R) \\
&=&\frac{R^{2}}{2}\Psi _{1}(\mu R).
\end{eqnarray*}
Misalkan $u$ panharmonik pada disk $(x-x_{0})^{2}+$ $(y-y_{0})^{2}\leq
R^{2}, $ atau biasa ditulis $D(\overline{x},R)$\ maka

\begin{equation*}
u(x_{0},y_{0})=\frac{1}{2\pi I_{0}(\mu R)}\int_{0}^{2\pi }u\left(
x_{0}+R\cos \theta ,y_{0}+R\sin \theta \right) d\theta .
\end{equation*}
Selanjutnya gerakan $r$, mulai $0$ sampai dengan $R$ , maka

\begin{equation*}
u(x_{0},y_{0})=\frac{1}{2\pi I_{0}(\mu r)}\int_{0}^{2\pi }u\left(
x_{0}+r\cos \theta ,y_{0}+r\sin \theta \right) d\theta
\end{equation*}
Kalikan kedua ruas dengan $\mu rI_{0}(\mu r)$, dan integralkan terhadap $r$
dari $0$ sampai $R$, diperoleh

\begin{equation*}
\int_{0}^{R}\mu rI_{0}(\mu r)u(x_{0},y_{0})dr=\int_{0}^{R}\left( \frac{1}{%
2\pi }\int_{0}^{2\pi }u\left( x_{0}+r\cos \theta ,y_{0}+r\sin \theta \right)
d\theta \right) \mu rdr
\end{equation*}

\begin{equation*}
u(x_{0},y_{0})RI_{1}(\mu R)=\frac{\mu }{2\pi }\int_{D(\overline{x}%
,R)}u\left( x,y\right) dxdy
\end{equation*}

\begin{eqnarray*}
u(x_{0},y_{0}) &=&\frac{\mu }{2\pi RI_{1}(\mu R)}\int_{D(\overline{x}%
,R)}u\left( x,y\right) dxdy \\
&=&\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x},R)}u\left(
x,y\right) dxdy
\end{eqnarray*}
Jadi kita punya,

\begin{theorem}
Misalkan $u$ panharmonik pada disk $(x-x_{0})^{2}+$ $(y-y_{0})^{2}\leq R^{2}$
maka 
\begin{equation*}
u(x_{0},y_{0})=\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x}%
,R)}u\left( x,y\right) dxdy,
\end{equation*}
dimana $\Psi _{1}(x)=1+\frac{\left( \frac{x}{2}\right) ^{2}}{1!.2!}+\frac{%
\left( \frac{x}{2}\right) ^{4}}{2!.3!}+\frac{\left( \frac{x}{2}\right) ^{6}}{%
3!.4!}+\frac{\left( \frac{x}{2}\right) ^{8}}{4!.5!}+...$
\end{theorem}

\bigskip

\begin{theorem}
\bigskip Misalkan $u$ panharmonik tak negatif pada disk $(x-x_{0})^{2}+$ $%
(y-y_{0})^{2}\leq R^{2}$ maka 
\begin{equation*}
\left| \frac{\partial u}{\partial x}\right| _{\left( x_{0},y_{0}\right)
}\leq \frac{2}{R}\frac{\Psi _{0}(\mu R)}{\Psi _{1}(\mu R)}u(x_{0},y_{0}).
\end{equation*}
\end{theorem}

\begin{proof}
\bigskip Karena masing-masing turunan parsial fungsi panharmonik juga
panharmonik maka  
\begin{eqnarray*}
\frac{\partial u}{\partial x}\left| _{\left( x_{0},y_{0}\right) }\right.  &=&%
\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x},R)}\frac{\partial
u\left( x,y\right) }{\partial x}dxdy \\
&=&\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{\partial D}u\left( x,y\right) dy
\\
&=&\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{0}^{2\pi }uR\cos \theta d\theta .
\end{eqnarray*}
Jadi kita punya, 
\begin{eqnarray*}
\left| \frac{\partial u}{\partial x}\right| _{\left( x_{0},y_{0}\right) }
&=&\left| \frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{0}^{2\pi }uR\cos \theta
d\theta .\right|  \\
&\leq &\frac{1}{\pi R\Psi _{1}(\mu R)}\int_{0}^{2\pi }ud\theta  \\
&=&\frac{1}{\pi R\Psi _{1}(\mu R)}2\pi I_{0}(\mu R)u(x_{0},y_{0}) \\
&=&\frac{2}{R}\frac{\Psi _{0}(\mu R)}{\Psi _{1}(\mu R)}u(x_{0},y_{0}).
\end{eqnarray*}
\end{proof}

\bigskip

Dengan cara yang sama, juga diperoleh $\left| \frac{\partial u}{\partial y}%
\right| _{\left( x_{0},y_{0}\right) }\leq \frac{2}{R}\frac{\Psi _{0}(\mu R)}{%
\Psi _{1}(\mu R)}u(x_{0},y_{0}).$ Untuk $\mu \rightarrow 0,$ maka kita
peroleh ketaksamaan Harnack untuk fungsi harmonik, $\left| \frac{\partial u}{%
\partial x}\right| _{\left( x_{0},y_{0}\right) }\leq \frac{2}{R}%
u(x_{0},y_{0}).$ Untuk fungsi panharmonik tak positif diperoleh $\left| 
\frac{\partial u}{\partial x}\right| _{\left( x_{0},y_{0}\right) }\leq \frac{%
2}{R}\frac{\Psi _{0}(\mu R)}{\Psi _{1}(\mu R)}\left| u(x_{0},y_{0})\right| .$

\bigskip

Selanjutnya akan kita lihat representasi integral fungsi panharmonik
dikaitkan dengan teorema Green. Perhatikan fungsi panharmonik pada disk $%
(x-x_{0})^{2}+$ $(y-y_{0})^{2}\leq R^{2}$ yaitu 
\begin{equation*}
u(x_{0},y_{0})=\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x}%
,R)}u\left( x,y\right) dxdy.
\end{equation*}
Selanjutnya tulis, 
\begin{eqnarray*}
u(x_{0},y_{0}) &=&\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x}%
,R)}u\left( x,y\right) dxdy \\
&=&\frac{1}{\pi \mu ^{2}R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x},R)}\mu
^{2}u\left( x,y\right) dxdy \\
&=&\frac{1}{\pi \mu ^{2}R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x}%
,R)}\triangle u\left( x,y\right) dxdy \\
&=&\frac{1}{\pi \mu ^{2}R^{2}\Psi _{1}(\mu R)}\int_{\partial D(\overline{x}%
,R)}\frac{\partial u}{\partial n}ds
\end{eqnarray*}
dimana $n$ adalah vektor normal satuan terhadap disk, dan $\frac{\partial u}{%
\partial n}$ merupakan turunan berarah $u$ dalam arah $n$. Untuk menunjukkan
ketunggalan \ representasi di atas, dapat dilihat dari teorema berikut.

\bigskip

\begin{theorem}
Misalkan $u$ dan $v$ fungsi panharmonik pada $\Omega $ dengan syarat batas
yang sama $\frac{\partial u}{\partial n}=\frac{\partial v}{\partial n}=g,$ $g
$ fungsi kontinu pada $\partial \Omega $, maka $u=v$ pada $\overline{\Omega 
\text{.}}$

\begin{proof}
Dengan menggunakan teorema green, 
\begin{equation*}
\int_{\Omega }\left( u-v\right) \triangle \left( u-v\right) dV+\int_{\Omega
}\left| \nabla \left( u-v\right) \right| ^{2}dV=\int_{\partial \Omega
}\left( u-v\right) \frac{\partial (u-v)}{\partial n}
\end{equation*}
\begin{equation*}
\int_{\Omega }\mu ^{2}\left( u-v\right) ^{2}dV+\int_{\Omega }\left| \nabla
\left( u-v\right) \right| ^{2}dV=0.
\end{equation*}
Karena $\left( u-v\right) ^{2}$ dan $\left| \nabla \left( u-v\right) \right|
^{2}$ terintegralgan dan positif, maka haruslah $u=v.$
\end{proof}
\end{theorem}

\bigskip\ 

Selanjutnya, akan dilihat nilai fungsi panharmonik di suatu titik dalam
daerah kompak melalui representasi integral fungsi panharmonik di atas
khususnya untuk fungsi panharmonik tak negatif dan juga untuk fungsi
panharmonik tak positif. Ini akan dikaji melalui akibat hasil terdahulu
mengenai prinsip maksimum dan minimum fungsi panharmonik tak negatif dan
juga fungsi panharmonik tak positif.

\begin{theorem}
Misalkan $u$ fungsi panharmonik tak negatif pada daerah kompak $R$. Jika $%
u\leq M$ pada batas $R$, dan $y$ suatu titik pada interior $R$, maka 
\begin{equation*}
u(y)\leq \frac{M}{\Psi _{0}\left( \mu a\right) },
\end{equation*}
dimana $a$ jarak terdekat dari $y$ ke batas $R.$
\end{theorem}

\begin{proof}
Misalkan $\left( x_{0},y_{0}\right) $ sembarang titik pada interior $R,$ dan
disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2},$
dimana $a$ merupakan jarak terpendek dari $y$ terhadap batas $R$, maka 
\begin{equation*}
u\left( x_{0},y_{0}\right) =\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u(x_{0}+a\cos \phi ,y_{0}+a\sin \phi )d\phi .
\end{equation*}
Karena $0\leq u\leq M$ pada batas $R$ mengakibatkan $u<M$ pada $R,$ dan disk 
$\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2}$ di dalam
interior $R,$ maka 
\begin{eqnarray*}
u\left( x_{0},y_{0}\right)  &=&\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u(x_{0}+a\cos \phi ,y_{0}+a\sin \phi )d\phi  \\
&\leq &\frac{M}{\Psi _{0}\left( \mu a\right) }.
\end{eqnarray*}
\end{proof}

\bigskip

\begin{theorem}
Misalkan $u$ fungsi panharmonik tak positif pada daerah kompak $R$. Jika $%
u\geq M$ pada batas $R$ untuk suatu $M<0$, dan $y$ suatu titik pada interior 
$R$, maka 
\begin{equation*}
u(y)\geq \frac{M}{\Psi _{0}\left( \mu a\right) },
\end{equation*}
dimana $a$ jarak terdekat dari $y$ ke batas $R.$
\end{theorem}

\begin{proof}
Misalkan $\left( x_{0},y_{0}\right) $ sembarang titik pada interior $R,$ dan
disk $\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2},$
dimana $a$ merupakan jarak terpendek dari $y$ terhadap batas $R$, maka 
\begin{equation*}
u\left( x_{0},y_{0}\right) =\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u(x_{0}+a\cos \phi ,y_{0}+a\sin \phi )d\phi .
\end{equation*}
Karena $0\geq u\geq M$ pada batas $R$ mengakibatkan $u>M$ pada $R,$ dan disk 
$\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2}$ di dalam
interior $R,$ maka 
\begin{eqnarray*}
u\left( x_{0},y_{0}\right)  &=&\frac{1}{2\pi \Psi _{0}\left( \mu a\right) }%
\int_{0}^{2\pi }u(x_{0}+a\cos \phi ,y_{0}+a\sin \phi )d\phi  \\
&\geq &\frac{M}{\Psi _{0}\left( \mu a\right) }.
\end{eqnarray*}
\end{proof}

\bigskip

Selanjutnya, bila kita punya fungsi panharmonik $u$ dan $v$ yang saling
konjugat pada disk $\left| z-\overline{x}\right| \leq R^{2},$ maka $%
f(z)=u(x,y)+iv(x,y)$ \ $\mu $ regular, dan karena setiap fungsi panharmonik
pada disk memiliki representasi integral seperti pada teorema di \ atas,
maka 
\begin{eqnarray*}
f(\overline{x}) &=&u(x_{0},y_{0})+iv(x_{0},y_{0}) \\
&=&\frac{1}{\pi R^{2}\Psi _{1}(\mu R)}\left( \int_{D(\overline{x},R)}u\left(
x,y\right) dxdy+i\int_{D(\overline{x},R)}v\left( x,y\right) dxdy\right) \\
&=&\frac{1}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x},R)}\left( \mu
u\left( x,y\right) +i\mu v\left( x,y\right) \right) dxdy \\
&=&\frac{1}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{D(\overline{x},R)}\left(
u_{x}-v_{y}\right) +i\left( -u_{y}-v_{x}\right) dxdy \\
&=&\frac{1}{\mu \pi R^{2}\Psi _{1}(\mu R)}\left[ \int_{\partial D(\overline{x%
},R)}\left( vdx+udy\right) +i\int_{\partial D(\overline{x},R)}\left(
udx-vdy\right) \right] \\
&=&\frac{1}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{\partial D(\overline{x}%
,R)}i\left( u-iv\right) \left( dx-idy\right) \\
&=&\frac{i}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{\partial D(\overline{x},R)}%
\overline{f(z)}d\overline{z} \\
&=&\frac{i}{2\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x},R)}\overline{%
f(z)}d\overline{z}.
\end{eqnarray*}
Karena 
\begin{equation*}
\oint_{\partial \Omega }f(z)dz=\iint_{\Omega }\frac{-1}{2}\mu \overline{f(z)}%
dz\wedge d\overline{z}
\end{equation*}
maka 
\begin{equation*}
\overline{\oint_{\partial \Omega }f(z)dz}=\iint_{\Omega }\frac{1}{2}\mu
f(z)dz\wedge d\overline{z}
\end{equation*}
dan 
\begin{eqnarray*}
f(\overline{x}) &=&\frac{i}{2\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x}%
,R)}\overline{f(z)}d\overline{z} \\
&=&\frac{i}{2\pi RI_{1}(\mu R)}\iint_{\Omega }\frac{1}{2}\mu f(z)dz\wedge d%
\overline{z} \\
&=&\frac{\mu i}{4\pi R^{2}\Psi _{1}(\mu R)}\iint_{\Omega }f(z)dz\wedge d%
\overline{z}.
\end{eqnarray*}
Tulis $z-z_{0}=R$ $e^{i\theta },$ maka $dz=iR$ $e^{i\theta }d\theta $ dan \ $%
d\overline{z}=-iR$ $e^{-i\theta }d\theta =\frac{-1}{e^{2i\theta }}iR$ $%
e^{i\theta }d\theta =\frac{-1}{e^{2i\theta }}dz,$ sehingga bentuk di atas
menjadi 
\begin{eqnarray*}
f(\overline{x}) &=&\frac{i}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{\partial D(%
\overline{x},R)}\overline{f(z)}d\overline{z} \\
&=&\frac{i}{\mu \pi R^{2}\Psi _{1}(\mu R)}\int_{\partial D(\overline{x},R)}%
\overline{f(z_{0}+R\text{ }e^{i\theta })}\frac{-1}{e^{2i\theta }}dz \\
&=&\frac{-i}{\mu \pi \Psi _{1}(\mu R)}\int_{\partial D(\overline{x},R)}%
\overline{f(z_{0}+R\text{ }e^{i\theta })}\frac{-1}{R^{2}e^{2i\theta }}dz \\
&=&\frac{-i}{\mu \pi \Psi _{1}(\mu R)}\int_{\partial D(\overline{x},R)}\frac{%
\overline{f(z_{0}+R\text{ }e^{i\theta })}}{\left( z-z_{0}\right) ^{2}}dz \\
&=&\frac{-i}{\mu \pi \Psi _{1}(\mu R)}\int_{\partial D(\overline{x},R)}\frac{%
\overline{f(z)}}{\left( z-z_{0}\right) ^{2}}dz
\end{eqnarray*}

Selanjutnya perhatikan turunan fungsi $f$ terhadap $z$ di $z=\overline{x},$%
\begin{eqnarray*}
\frac{\partial f}{\partial z}(\overline{x}) &=&\frac{i}{2\pi RI_{1}(\mu R)}%
\int_{\partial D(\overline{x},R)}\frac{\partial \overline{f(z)}}{\partial z}d%
\overline{z} \\
&=&\frac{i}{2\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x},R)}\frac{1}{2}%
\mu f(z)d\overline{z} \\
&=&\frac{\mu i}{4\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x},R)}f(z)d%
\overline{z}
\end{eqnarray*}

\bigskip Catatan $f^{\prime }(z_{0})=\lim_{h\rightarrow 0}\frac{%
f(z_{0}+h)-f(z_{0})}{h}.$

Selanjutnya misalkan $z=z_{0}+$ $R$ $e^{i\theta }$ maka $d\overline{z}%
=-iRe^{-i\theta }d\theta =-i\left( \overline{z}-\overline{z_{0}}\right)
d\theta $, maka 
\begin{eqnarray*}
f(z_{0}) &=&\frac{i}{2\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x},R)}%
\overline{f(z)}d\overline{z} \\
&=&\frac{i}{2\pi RI_{1}(\mu R)}\int_{\partial D(\overline{x},R)}\overline{%
f(z_{0}+Re^{i\theta })}.-iRe^{-i\theta }d\theta \\
&=&\frac{1}{2\pi I_{1}(\mu R)}\int_{\partial D(\overline{x},R)}\overline{%
f(z_{0}+Re^{i\theta })}.e^{-i\theta }d\theta
\end{eqnarray*}

Panharmonik pada $x^{2}+y^{2}\leq 1.$

Untuk $0\leq r<1,$ diperoleh fungsi panharmonik 
\begin{equation*}
u(re^{i\theta })=\sum_{-\infty }^{\infty }c_{n}I_{\left| n\right| }(\mu
r)e^{in\theta }
\end{equation*}
dengan 
\begin{equation*}
c_{n}=\frac{1}{2\pi I_{\left| n\right| }(\mu )}\int_{0}^{2\pi }u(e^{i\theta
})e^{-in\theta }d\theta .
\end{equation*}
Misalkan $\left| x\right| =\left| re^{i\theta }\right| =r<1,$ maka $\ $%
\begin{eqnarray*}
u(x) &=&\sum_{-\infty }^{\infty }c_{n}I_{\left| n\right| }(\mu r)e^{in\theta
} \\
&=&\sum_{-\infty }^{\infty }\frac{1}{n!}\left( \frac{\mu }{2}\right)
^{n}c_{n}\Psi _{\left| n\right| }(\mu \left| x\right| )x^{n}
\end{eqnarray*}
dengan 
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi I_{\left| n\right| }(\mu )}\int_{0}^{2\pi
}u(e^{i\theta })e^{-in\theta }d\theta \\
&=&\frac{1}{2\pi I_{\left| n\right| }(\mu )}\int_{S}f(\eta )\eta ^{-n}dS,%
\text{ \ }\left| \eta \right| =1.
\end{eqnarray*}
Selanjutnya perhatikan 
\begin{eqnarray*}
u(x) &=&\sum_{-\infty }^{\infty }c_{n}I_{\left| n\right| }(\mu \left|
x\right| )x^{n} \\
&=&\sum_{-\infty }^{\infty }\frac{1}{n!}\left( \frac{\mu }{2}\right)
^{n}\left( \frac{1}{2\pi I_{\left| n\right| }(\mu )}\int_{S}f(\eta )\eta
^{-n}dS\right) \Psi _{\left| n\right| }(\mu \left| x\right| )x^{n} \\
&=&\int_{S}\sum_{-\infty }^{\infty }\frac{1}{n!}\left( \frac{\mu }{2}\right)
^{n}\left( \frac{1}{2\pi I_{\left| n\right| }(\mu )}\right) \Psi _{\left|
n\right| }(\mu \left| x\right| )x^{n}f(\eta )\eta ^{-n}dS \\
&=&\int_{S}\left( \sum_{-\infty }^{\infty }\frac{1}{2\pi }\frac{\Psi
_{\left| n\right| }(\mu \left| x\right| )}{\Psi _{\left| n\right| }(\mu )}%
\left( x\eta ^{-1}\right) ^{n}\right) f\left( \eta \right) dS.
\end{eqnarray*}
Tulis $u$ sebagai 
\begin{equation*}
u(x)=\int_{S}P(x,\eta )f(\eta )dS,\text{ dimana}
\end{equation*}
\begin{equation*}
P(x,\eta )=\frac{1}{2\pi }\sum_{-\infty }^{\infty }\frac{\Psi _{\left|
n\right| }(\mu \left| x\right| )}{\Psi _{\left| n\right| }(\mu )}\left(
x\eta ^{-1}\right) ^{n},\text{ }\left| \eta \right| =1,\text{ }\left|
x\right| <1,\text{ dan }f(\eta )=u(1.e^{i\theta }).
\end{equation*}
Selanjutnya perhatikan 
\begin{eqnarray*}
\int_{S}P(x,\eta )dS &=&\int_{0}^{2\pi }\frac{1}{2\pi }\sum_{-\infty
}^{\infty }\frac{\Psi _{\left| n\right| }(\mu \left| x\right| )}{\Psi
_{\left| n\right| }(\mu )}x^{n}e^{-in\theta }d\theta \\
&=&\frac{\Psi _{0}(\mu \left| x\right| )}{\Psi _{0}(\mu )} \\
&\leq &1
\end{eqnarray*}
dan untuk $\left| x\right| \rightarrow 1$ maka $\int_{S}P(x,\eta
)dS\rightarrow 1.$

\end{document}
