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%TCIDATA{Created=Mon Oct 08 12:54:33 2001}
%TCIDATA{LastRevised=Mon Oct 08 14:46:26 2001}
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\begin{document}


Pada bagian ini akan dicari seluruh transformasi yang mengawetkan fungsi
panharmonik melalui pencarian grup simetri dari persamaan diferensial
parsial Yukawa $\triangle u=u_{xx}+u_{yy}-\mu ^{2}u=0.$ Misalkan $u=f(x,y)$
solusi persamaan Yukawa, dan 
\begin{equation*}
V=\xi (x,y,u)\partial x+\Gamma (x,y,u)\partial y+\Phi (x,y,u)\partial u
\end{equation*}
adalah medan vektor pada $XxU,$ dengan prolongasi kedua dari medan vektor $V,
$ yaitu 
\begin{equation*}
\Pr \text{ }^{2}V=V+\Phi ^{x}\partial u_{x}+\Phi ^{y}\partial u_{y}+\Phi
^{xx}\partial u_{xx}+\Phi ^{xy}\partial u_{xy}+\Phi ^{yy}\partial u_{yy}
\end{equation*}
sehingga 
\begin{equation*}
\Pr \text{ }^{2}V(\triangle u)=\Phi ^{xx}+\Phi ^{yy}-\mu ^{2}\Phi =0.
\end{equation*}
Dimana

$\Phi ^{xx}=\Phi _{xx}+(2\Phi _{ux}-\xi _{xx})u_{x}+(\Phi _{uu}-2\xi
_{xu})u_{x}^{2}-\Gamma _{xx}u_{y}-\xi _{uu}u_{x}^{3}-2\Gamma _{ux}u_{x}u_{y}$

$\ \ \ \ \ \ \ \ \ \ -\Gamma _{uu}u_{x}^{2}u_{y}$ $+(\Phi _{u}-2\xi
_{x})u_{xx}-2\Gamma _{x}u_{yx}-3\xi _{u}u_{x}u_{xx}-\Gamma _{u}u_{y}u_{xx}$

$\ \ \ \ \ \ \ \ \ \ -2\Gamma _{u}u_{x}u_{yx}$

$\Phi ^{yy}=\Phi _{yy}+(\Phi _{uy}-\Gamma _{yy})u_{y}+(\Phi _{uu}-\Gamma
_{yu})u_{y}^{2}-\Gamma _{uu}u_{y}^{3}-\xi _{yy}u_{x}-2\xi _{uy}u_{x}u_{y}+$

$\ \ \ \ \ \ \ \ \ \ (\Phi _{u}-2\Gamma _{y})u_{yy}$ $-2\xi
_{y}u_{yx}-3\Gamma _{u}u_{y}u_{yy}-2\xi _{u}u_{yx}u_{y}-\xi
_{u}u_{x}u_{yy}-\xi _{uu}u_{x}u_{y}^{2}.$

Dengan menyamakan koefisien dari masing-masing turunan parsial $u,$
diperoleh $\Gamma =\Gamma (x,y)$ dan $\xi =\xi (x,y),$ dan $\Phi $ memenuhi
persamaan Yukawa, dan 
\begin{equation*}
\xi =\frac{1}{2}c_{1}x-c_{2}y+c_{4}
\end{equation*}
\begin{equation*}
\Gamma =\frac{1}{2}c_{1}y+c_{2}x+c_{3}
\end{equation*}
\begin{equation*}
\Phi =c_{1}u+\alpha (x,y).
\end{equation*}
Karena $\Phi $ memenuhi persamaan Yukawa, maka $\alpha $ juga harus memenuhi
persamaan Yukawa. Aljabar Lie dari simetri infinitesimal dari persamaan
Yukawa dibangun oleh empat buah medan vektor, yaitu

$V_{4}=\frac{1}{2}x\partial x+\frac{1}{2}y\partial y+u\partial u$

$V_{3}=-y\partial x+x\partial y$

$V_{2}=\partial y$

$V_{1}=\partial x$ 

Khusus $V_{\alpha }=\alpha (x,y)\partial u,$ ini merupakan sub aljabar
berdimensi tak hingga dimana $\alpha $ sembarang solusi persamaan Yukawa.
Grup berparameter-1 $G_{i}$ yang dibangun oleh $V_{i}$ adalah

$G_{1}:(x,y,u)\mapsto (x+\varepsilon ,y,u)$

$G_{2}:(x,y,u)\mapsto (x,y+\varepsilon ,u)$

$G_{3}:(x,y,u)\mapsto (x-\varepsilon y,y+\varepsilon x,u)$

$G_{4}:(x,y,u)\mapsto (xe^{\frac{1}{2}\varepsilon },ye^{\frac{1}{2}%
\varepsilon },e^{\varepsilon }u)$

$G_{\alpha }:(x,y,u)\mapsto (x,y,\alpha \varepsilon +u).$

Masing-masing $G_{i}$ grup simetri untuk persamaan Yukawa, karena itu jika $%
u=f(x,y)$ solusi persamaan Yukawa, maka 

$u_{1}=f(x-\varepsilon ,y)$

$u_{2}=f(x,y-\varepsilon )$

$u_{3}=f(\frac{x+\varepsilon y}{1+\varepsilon ^{2}},\frac{y-\varepsilon x}{%
1+\varepsilon ^{2}})$

$u_{4}=e^{\varepsilon }f(e^{\frac{-1}{2}\varepsilon }x,e^{\frac{-1}{2}%
\varepsilon }y)$

$u_{\alpha }=\alpha (x,y)\varepsilon +f(x,y)$

\end{document}
