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%TCIDATA{Created=Thu Nov 30 11:28:07 2000}
%TCIDATA{LastRevised=Mon Feb 12 12:33:08 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}


\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ 

\bigskip

\bigskip

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 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*}

\bigskip

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*}

\bigskip

Misalkan $u$\ panharmonik pada disk $(x-x_{0})^{2}+$\ $(y-y_{0})^{2}<R^{2},$%
\ 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*}

Untuk $0<r<R,$ persamaan di atas dapat ditulis sebagai,

\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*}

\bigskip Selanjutnya, 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\text{ dengan }\overline{x}=(x_{0},y_{0})
\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*}

\bigskip

\begin{theorem}
1. Misalkan $u$\ panharmonik pada disk $(x-x_{0})^{2}+$\ $(y-y_{0})^{2}<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 2. 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 dari fungsi panharmonik adalah
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 \bigskip

Selanjutnya akan kita lihat representasi integral fungsi panharmonik
dikaitkan dengan teorema Green. Perhatikan fungsi panharmonik pada disk $%
(x-x_{0})^{2}+$\ $(y-y_{0})^{2}<R^{2}$\ 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}
3. 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

PRINSIP MAKSIMUM DAN MINIMUM FUNGSI PANHARMONIK

\begin{theorem}
4. Misalkan $u$ fungsi panharmonik tak negatif pada daerah kompak terbatas $%
\Omega .$ Maka $u$ akan mencapai nilai maksimum di batas $\Omega .$

\begin{proof}
Misalkan $u$ fungsi panharmonik tak negatif pada daerah kompak terbatas $%
\Omega .$ Andaikan ada titik $\overline{x}$ pada interrior $\Omega $
sehingga $\nabla u(\overline{x})=0,$ $u_{xx}($ $\overline{x})u_{yy}($ $%
\overline{x})-u_{xy}^{2}($ $\overline{x})>0$ dan $u_{xx}($ $\overline{x})<0.$
Maka $u_{xx}($ $\overline{x})u_{yy}($ $\overline{x})>u_{xy}^{2}($ $\overline{%
x})\geq 0.$ Tetapi $u_{xx}($ $\overline{x})<0,$ akibatnya haruslah $u_{yy}($ 
$\overline{x})<0.$ Karena itu, $\mu ^{2}u(\overline{x})=$ $u_{xx}($ $%
\overline{x})+u_{yy}($ $\overline{x})<0.$ Ini bertentangan dengan yang
diketahui.
\end{proof}
\end{theorem}

\bigskip

\begin{theorem}
5. Misalkan $u$ fungsi panharmonik tak positif pada daerah kompak terbatas $%
\Omega .$ Maka $u$ akan mencapai nilai maksimum di batas $\Omega .$

\begin{proof}
Misalkan $u$ fungsi panharmonik tak positif pada daerah kompak terbatas $%
\Omega .$ Andaikan ada titik $\overline{x}$ pada interrior $\Omega $
sehingga $\nabla u(\overline{x})=0,$ $u_{xx}($ $\overline{x})u_{yy}($ $%
\overline{x})-u_{xy}^{2}($ $\overline{x})>0$ dan $u_{xx}($ $\overline{x})>0.$
Maka $u_{xx}($ $\overline{x})u_{yy}($ $\overline{x})>u_{xy}^{2}($ $\overline{%
x})\geq 0.$ Tetapi $u_{xx}($ $\overline{x})>0,$ akibatnya haruslah $u_{yy}($ 
$\overline{x})>0.$ Karena itu, $\mu ^{2}u(\overline{x})=$ $u_{xx}($ $%
\overline{x})+u_{yy}($ $\overline{x})>0.$ Ini bertentangan dengan yang
diketahui.
\end{proof}
\end{theorem}

\bigskip

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

\begin{proof}
Misalkan $\overline{y}=\left( x_{0},y_{0}\right) $\ sembarang titik pada
interior $\Omega .$\ Buat disk $\left( x-x_{0}\right) ^{2}+\left(
y-y_{0}\right) ^{2}\leq a^{2},$\ dimana $a$\ merupakan jarak terdekat dari $%
\overline{y}$\ terhadap batas $\Omega $, 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 $\Omega $ maka juga berlaku pada disk $%
\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2},$\ karena
itu 
\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}
7. Misalkan $u$\ fungsi panharmonik tak positif pada daerah kompak $\Omega $%
. Jika $u\geq M$\ pada batas $\Omega $\ untuk suatu $M<0$, dan $\overline{y}$%
\ suatu titik pada interior $\Omega $, maka 
\begin{equation*}
u(\overline{y}\ )\geq \frac{M}{\Psi _{0}\left( \mu a\right) },
\end{equation*}
dimana $a$\ jarak terdekat dari $\overline{y}$\ \ ke batas $\Omega .$
\end{theorem}

\begin{proof}
Misalkan $\overline{y}=$\ $\left( x_{0},y_{0}\right) $\ sembarang titik pada
interior $\Omega .$\ Buat disk $\left( x-x_{0}\right) ^{2}+\left(
y-y_{0}\right) ^{2}\leq a^{2},$\ dimana $a$\ merupakan jarak terdekat dari $%
\overline{y}$\ \ terhadap batas $\Omega $, 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 $\Omega ,$ maka juga berlaku pada disk $%
\left( x-x_{0}\right) ^{2}+\left( y-y_{0}\right) ^{2}\leq a^{2},$\ karena
itu 
\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}

\begin{corollary}
\bigskip 8. Misalkan $u$\ fungsi panharmonik tak negatif pada suatu domain
terbatas $\Omega ,$\ dan misalkan 
\begin{equation*}
lim_{k\rightarrow \infty }\sup u(a_{k})\leq M
\end{equation*}
untuk setiap barisan $\left( a_{k}\right) $\ di dalam $\Omega $\ yang
konvergen ke salah satu titik di batas $\Omega .$\ Maka $u\leq M$\ pada $%
\Omega .$

\begin{proof}
Misalkan $M^{\prime }=\sup \left\{ u(x):x\in \Omega \right\} ,$ \ dan
barisan $\left( b_{k}\right) $\ di $\Omega $\ sehingga $u\left( b_{k}\right)
\rightarrow M^{\prime }.$ Ada dua kasus yang akan ditinjau, pertama jika $%
\left( b_{k}\right) $\ memiliki sub barisan yang konvergen ke suatu titik $b$
di dalam $\Omega ,$ maka \smallskip\ $u(b)=M^{\prime }.$ Berdasarkan prinsip
maksimum fungsi panharmonik, maka $u=M^{\prime }$\ pada $\Omega .$\ Yang
kedua, jika $\left( b_{k}\right) \bigskip $\ tidak memiliki sub barisan yang
konvergen \ ke suatu titik di dalam $\Omega ,$\ maka $\left( b_{k}\right) $
memiliki sub barisan $\left( a_{k}\right) $\ yang konvergen ke suatu titik
di batas $\Omega $, sebut $a.$\ Maka $u(a)=M^{\prime }.$\ Ini berakibat $%
u\leq M^{\prime }.$\ Dari kedua kasus di atas, disimpulkan bahwa $M^{\prime
}\leq M.$
\end{proof}
\end{corollary}

\bigskip

Akibat yang sama seperti di atas, juga akan berlaku untuk fungsi panharmonik
tak positif.

\begin{corollary}
\bigskip 9. Misalkan $u$\ fungsi panharmonik tak positif pada suatu domain $%
\Omega ,$\ dan misalkan 
\begin{equation*}
lim_{k\rightarrow \infty }\inf (a_{k})\geq M
\end{equation*}
untuk setiap barisan $\left( a_{k}\right) $\ di dalam $\Omega $\ yang
konvergen ke salah satu titik di batas $\Omega .$\ Maka $u\geq M$\ pada $%
\Omega .$
\end{corollary}

\bigskip

\end{document}
