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%TCIDATA{Created=Wed May 23 09:10:38 2001}
%TCIDATA{LastRevised=Tue Jun 12 08:43:21 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}}
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\begin{document}


\section{Invariance In Two Dimensions}

The Yukawa equation is invariant under all rigid motion. A rigid motion in
the plane consists of tranlations and rotations. A translation in the plane
is a transformation 
\begin{equation*}
x^{\prime }=x+a\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ }y^{\prime }=y+b.
\end{equation*}
Invariance under translations means simply that 
\begin{equation*}
u_{xx}+u_{yy}=u_{x^{\prime }x^{\prime }}+u_{y^{\prime }y^{\prime }.}
\end{equation*}
A rotation in the plane through the angle $\alpha $ is given by 
\begin{eqnarray*}
x^{\prime } &=&x\cos \alpha +y\sin \alpha \\
y^{\prime } &=&-x\sin \alpha +y\cos \alpha .
\end{eqnarray*}
By the chain rule we calculate 
\begin{eqnarray*}
u_{x} &=&u_{x^{\prime }}\cos \alpha -u_{y^{\prime }}\sin \alpha \\
u_{y} &=&u_{x^{\prime }}\sin \alpha +u_{y^{\prime }}\cos \alpha
\end{eqnarray*}
and 
\begin{eqnarray*}
u_{xx} &=&\left( u_{x^{\prime }}\cos \alpha -u_{y^{\prime }}\sin \alpha
\right) _{x^{\prime }}\cos \alpha -\left( u_{x^{\prime }}\cos \alpha
-u_{y^{\prime }}\sin \alpha \right) _{y^{\prime }}\sin \alpha \\
u_{yy} &=&\left( u_{x^{\prime }}\sin \alpha +u_{y^{\prime }}\cos \alpha
\right) _{x^{\prime }}\sin \alpha +\left( u_{x^{\prime }}\sin \alpha
+u_{y^{\prime }}\cos \alpha \right) _{y^{\prime }}\cos \alpha .
\end{eqnarray*}
Adding, we have 
\begin{eqnarray*}
u_{xx}+u_{yy} &=&u_{x^{\prime }x^{\prime }}\cos ^{2}\alpha -u_{y^{\prime
}x^{\prime }}\sin \alpha \cos \alpha -u_{x^{\prime }y^{\prime }}\cos \alpha
\sin \alpha +u_{y^{\prime }y^{\prime }}\sin ^{2}\alpha + \\
&&u_{x^{\prime }x^{\prime }}\sin ^{2}\alpha +u_{y^{\prime }x^{\prime }}\sin
\alpha \cos \alpha +u_{x^{\prime }y^{\prime }}\cos \alpha \sin \alpha
+u_{y^{\prime }y^{\prime }}\cos ^{2}\alpha \\
&=&u_{x^{\prime }x^{\prime }}(\cos ^{2}\alpha +\sin ^{2}\alpha
)+u_{y^{\prime }y^{\prime }}(\sin ^{2}\alpha +\cos ^{2}\alpha ) \\
&=&u_{x^{\prime }x^{\prime }}+u_{y^{\prime }y^{\prime }}.
\end{eqnarray*}
This proves the invariance of the Laplace operator under translation.

\begin{theorem}
Let u(x,y) is a panharmonic function. Then the translate u(x-a,y-b) is a
panharmonic function for any fixed a and b.
\end{theorem}

The rotational invariance suggests that the two-dimensional laplacian 
\begin{equation*}
\triangle =\frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{%
\partial y^{2}}
\end{equation*}
should take a particularly simple form in polar coordinates. The
transformation has the jacobian matrix 
\begin{equation*}
\Im =\left( 
\begin{array}{cc}
\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ 
\frac{\partial x}{\partial \theta } & \frac{\partial y}{\partial \theta }
\end{array}
\right) =\left( 
\begin{array}{cc}
\cos \theta & \sin \theta \\ 
-r\sin \theta & r\cos \theta
\end{array}
\right)
\end{equation*}
with the inverse matrix 
\begin{equation*}
\Im ^{-1}=\left( 
\begin{array}{cc}
\frac{\partial r}{\partial x} & \frac{\partial \theta }{\partial x} \\ 
\frac{\partial r}{\partial y} & \frac{\partial \theta }{\partial y}
\end{array}
\right) =\left( 
\begin{array}{cc}
\cos \theta & \frac{-\sin \theta }{r} \\ 
\sin \theta & \frac{\cos \theta }{r}
\end{array}
\right) .
\end{equation*}
So by the chain rule we have 
\begin{eqnarray*}
\frac{\partial }{\partial x} &=&\cos \theta \frac{\partial }{\partial r}-%
\frac{\sin \theta }{r}\frac{\partial }{\partial \theta }, \\
\frac{\partial }{\partial y} &=&\sin \theta \frac{\partial }{\partial r}-%
\frac{\cos \theta }{r}\frac{\partial }{\partial \theta }.
\end{eqnarray*}
These operators are squared to give 
\begin{eqnarray*}
\frac{\partial ^{2}}{\partial x^{2}} &=&\left[ \cos \theta \frac{\partial }{%
\partial r}-\frac{\sin \theta }{r}\frac{\partial }{\partial \theta }\right]
^{2} \\
&=&\cos ^{2}\theta \frac{\partial ^{2}}{\partial r^{2}}-2\left( \frac{\sin
\theta \cos \theta }{r}\right) \frac{\partial ^{2}}{\partial r\partial
\theta }+\frac{\sin ^{2}\theta }{r^{2}}\frac{\partial ^{2}}{\partial \theta
^{2}} \\
&&+2\left( \frac{\sin \theta \cos \theta }{r^{2}}\right) \frac{\partial }{%
\partial \theta }+\frac{\sin ^{2}\theta }{r}\frac{\partial }{\partial r}
\end{eqnarray*}

\begin{eqnarray*}
\frac{\partial ^{2}}{\partial y^{2}} &=&\left[ \sin \theta \frac{\partial }{%
\partial r}+\frac{\cos \theta }{r}\frac{\partial }{\partial \theta }\right]
^{2} \\
&=&\sin ^{2}\theta \frac{\partial ^{2}}{\partial r^{2}}+2\left( \frac{\sin
\theta \cos \theta }{r}\right) \frac{\partial ^{2}}{\partial r\partial
\theta }+\frac{\cos ^{2}\theta }{r^{2}}\frac{\partial ^{2}}{\partial \theta
^{2}} \\
&&-2\left( \frac{\sin \theta \cos \theta }{r^{2}}\right) \frac{\partial }{%
\partial \theta }+\frac{\cos ^{2}\theta }{r}\frac{\partial }{\partial r}.
\end{eqnarray*}
Adding these operators, we get

\begin{eqnarray*}
\triangle &=&\frac{\partial ^{2}}{\partial x^{2}}+\frac{\partial ^{2}}{%
\partial y^{2}} \\
&=&\frac{\partial ^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial }{\partial
r}+\frac{1}{r^{2}}\frac{\partial ^{2}}{\partial \theta ^{2}}.
\end{eqnarray*}
It is also natural to look for special panharmonic functions that themselves
are rotationally invariant. In two dimensions this means that we use polar
coordinates $\left( r,\theta \right) $ and look for solutions depending only
on $r.$ Thus we have 
\begin{equation*}
u_{rr}+\frac{1}{r}u_{r}=\mu ^{2}u(r)
\end{equation*}
if $u$ does not depend on $\theta .$ This ordinary differential equation is
known as the modified Bessel equation of zero order. Since this can be
written in the form

\begin{equation*}
r^{2}u_{rr}+ru_{r}+\left( \mu ir\right) ^{2}u=0,
\end{equation*}
and the complete solution of this equation can be written in the form 
\begin{equation*}
u(r)=aI_{0}(\mu r)+bK_{0}(\mu r),
\end{equation*}
where 
\begin{equation*}
I_{0}(x)=1+\frac{\left( \frac{x}{2}\right) ^{2}}{1!1!}+\frac{\left( \frac{x}{%
2}\right) ^{4}}{2!2!}+\frac{\left( \frac{x}{2}\right) ^{6}}{3!3!}+.....
\end{equation*}
and 
\begin{equation*}
K_{0}\left( x\right) =I_{0}(x)\int_{x}^{\infty }\frac{dx}{xI_{0}^{2}(x)}.
\end{equation*}
The function $K_{0}(x)$ is the modified Bessel function of the second kind.
It follows from $I_{0}$ and $K_{0\text{ }}$ in above, we have 
\begin{equation*}
K_{0}\left( x\right) \approx \log x^{-1}\text{ as }x\longmapsto 0^{+},
\end{equation*}
but this is not interesting because of $K_{0}(x)\longrightarrow \infty ,$ if 
$x\longrightarrow 0^{+}$, therefore choose $b=0,$ so $u(r)=aI_{0}(\mu r),$
where $a$ is a constant.

\end{document}
