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\message{
Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics,
The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of
Physics, The University of Cincinnati, Cincinnati, OH 45221-0011). This
material may not be reproduced for profit, modified or published in any
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\title{Boundary Value Problems in Electrostatics II}
\author{Friedrich Wilhelm Bessel\\(1784 - 1846)}
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\bdo
\maketitle
\tableofcontents
\pagebreak
\large
Although this is a new chapter, we continue to do things begun in the
previous chapter. In particular, the first topic is the separation of
variable method in spherical polar coordinates.
\section{Laplace Equation in Spherical Coordinates}
The Laplacian operator in spherical coordinates is
\beq
\lap=\frac1r\spd{}{r}r+\frac1{r^2\sin\th}\pde{}{\th}\sin\th\pde{}{\th}
+\frac1{r^2\sin^2\th}\spd{}{\ph}.
\eeq
This is also a coordinate system in which it is possible to find a solution
in the form of a product of three functions of a single variable each:
\beq
\Ph(r,\th,\ph)=R(r)P(\th)Q(\ph)=U(r)P(\th)Q(\ph)/r.
\eeq
Operate on $\Ph$ with $\lap$, and set the result equal to zero to find
\beq
\frac{PQ}{r}\sde{U}{r}+\frac{UQ}{r^2\sin\th}\der{}{\th}\lep\sin\th
\der{P}{\th}\rip+\frac{UP}{r^3\sin^2\th}\sde{Q}{\ph}=0
\eeq
Multiply by $r^3\sin^2\th/UPQ$ to find
\beq
\frac{r^2\sin^2\th}{U}\sde{U}{r}+\frac{\sin\th}{P}\der{}{\th}\lep\sin\th
\der{P}{\th}\rip+\frac1Q\sde{Q}{\ph}=0.
\eeq
The first two terms are independent of $\ph$ while the third depends only
on this variable. Thus the third must be a constant as must the sum of the
first two; the first of these conditions is
\beq
\frac1Q\sde{Q}{\ph}=C \hbox{ or } \sde{Q}{\ph}=CQ,
\eeq
from which it follows that $Q\sim e^{\sqrt{C}\ph}$.
Now, a change in $\ph$
by $2\pi$ corresponds to no change whatsoever in spatial position;
therefore, we must have $Q(\ph+2\pi)=Q(\ph)$ because a function describing
a measurable quantity must be a single-valued function of position. Hence
we can conclude that $\sqrt{C}=im$ where $m$ is an integer so that $e^{im2
\pi}=1$. Thus $C=-m^2$, and $Q(\ph)\rightarrow Q_m(\ph)=e^{im\ph}$, with
$m=0,\pm1,\pm2,$... . We recognize that the functions $Q_m$ can be used to
construct a Fourier series and are a complete orthogonal set on the
interval $\ph_0\le\ph\le\ph+2\pi$.
Returning now to Laplace's equation, \eq{4}, and
using $-m^2$ for $\frac1Q\sde{Q}{\ph}$, we find
\beq
\frac{r^2\sin^2\th}{U}\sde{U}{r}+\frac{\sth}{P}\der{}{\th}\lep\sth\der{P}{\th}
\rip-m^2=0
\eeq
or
\beq
\frac{r^2}{U}\sde{U}{r}+\frac1{P\sth}\der{}{\th}\lep\sth\der{P}{\th}\rip
-\frac1{\sin^2\th}m^2=0.
\eeq
In this expression we recognize that the first term depends only on $r$ and
next two, only on $\th$, so we as usual conclude that each term must be
separately a constant and that the constants must add to zero. The first
equation extracted by this device is
\beq
\sde{U}{r}=\frac{A}{r^2}U
\eeq
where $A$ is the constant. It is a standard convention to write $A$ as
$l(l+1)$ which is still quite general if $l$ is allowed to be complex. Thus
the preceding equation becomes
\beq
\sde{U}{r}=\frac{l(l+1)}{r^2}U.
\eeq
The solutions of this ordinary, second-order, linear, differential equation
are two in number and are $U\sim r^{l+1}$ and $U\sim1/r^l$. Before
commenting further on that, let us go on to the equation for $P(\th)$.
\subsection{Legendre Equation and Polynomials}
Substitution of $l(l+1)$ for the first term in \eq{7} produces
\beq
\frac1\sth\der{}{\th}\lep\sth\der{P}{\th}\rip+\lep l(l+1)
-\frac{m^2}{\sin^2\th}\rip P=0.
\eeq
This is the {\em generalized Legendre Equation}; it is commonly written in
terms of a different variable, namely $u\equiv\cth$. Then one has
\beq
\der{}{\th}\lep\sth\der{P}{\th}\rip=\der{u}{\th}\der{}{u}\lep\sqrt{1-u^2}
\der{u}{\th}\der{P}{u}\rip=-\sqrt{1-u^2}\der{}{u}\lep-(1-u^2)\der{P}{u}\rip
\eeq
and
\beq
l(l+1)-\frac{m^2}{\sin^2\th}=l(l+1)-\frac{m^2}{1-u^2};
\eeq
hence,
\beq
\der{}{u}\lep(1-u^2)\der{P}{u}\rip+\lep l(l+1)-\frac{m^2}{1-u^2}\rip P=0
\eeq
is the form of the generalized Legendre equation using $u$ as the variable.
The interval of interest to us is $0\le\th\le\pi$ which is $-1\le u\le1$.
We shall discuss first the special case of $m=0$ which corresponds to
$Q(\ph)=1$, or a system for which $\Ph(\x)$ is independent of $\ph$; we
shall call such a potential {\em azimuthally invariant}; there are many
interesting systems which are more or less of this type. The equation for
$P$ is
\beq
\der{}{u}\lep(1-u^2)\der{P}{u}\rip+l(l+1)P=0,
\eeq
which is called the {\em Legendre equation}.
A standard procedure for solving this equation (and other similar
second-order differential equations) is to assume that the solution can be
written as a power series. Then there must be a smallest power $\al$ in the
series, so we can write
\beq
P(u)=u^{\al}\sum_{j=0}^\infty a_ju^j=\sum_{j=0}^\infty a_ju^{j+\al}
\eeq
from which we may evaluate the derivatives as
\beq
\der{P}{u}=\sum_{j=0}^\infty(j+\al)a_ju^{j+\al-1},
\eeq
\beq
\sde{P}{u}=\sum_{j=0}^\infty(j+\al)(j+\al-1)a_ju^{j+\al-2},
\eeq
and
\beq
-\der{}{u}u^2\der{P}{u}=-\sum_{j=0}^\infty(j+\al)(j+\al+1)a_ju^{j+\al}.
\eeq
Substitution into the Legendre equation gives
\beq
\sum_{j=0}^\infty\leb(\al+j)(\al+j-1) u^{\al+j-2}a_j-(\al+j)(\al+j+1)
u^{\al+j}a_j+l(l+1)u^{\al+j}a_j\rib=0,
\eeq
or, if we shift the zero of $j$ in each term so as to isolate individual
powers of $u$,
\beqa
\al(\al-1)u^{\al-2}a_0+\al(\al+1)a_1u^{\al-1}+\nonumber \\
\sum_{j=0}^\infty\leb(\al+j+2)(\al+j+1)a_{j+2}+\lep l(l+1)
-(\al+j)(\al+j+1)\rip a_j\rib u^{\al+j}=0.
\eeqa
The only way that this power series can vanish for all $u$ on the interval
is to have the coefficient of each power of $u$ vanish separately. Thus
I will list the coefficients:
\begin{center}
\begin{tabular}{||l|l||} \hline\hline
$j$ & Coefficient of $u^{\al+j}$\\ \hline\hline
$-2$ & $a_0\al(\al-1)$ \\ \hline
$-1$ & $a_1\al(\al+1)$ \\ \hline
$j\ge 0$ & $\leb(\al+j+2)(\al+j+1)a_{j+2}+\lep l(l+1)
-(\al+j)(\al+j+1)\rip a_j\rib$ \\ \hline\hline
\end{tabular}
\end{center}
The
coefficient of the leading (smallest) power $j=-2$ is
zero if $\al=0$ or 1; $a_0=0$ is not an option because by definition the
first term in the expansion has a nonvanishing coefficient.
Thus we find at this juncture two possible allowed values of $\al$.
\beq
\al=0,1
\eeq
In order that the
coefficient of the next power $j=-1$ of $u$ vanish we
must have either $\al=0$ or $a_1=0$ (we can also have both); we can't have
$\al=-1$ because of our first condition. Finally, the condition that
the coefficient of $u^{\al+j}$ vanish for $j\ge0$ is
\beq
a_{j+2}=\frac{(\al+j)(\al+j+1)-l(l+1)}{(\al+j+1)(\al+j+2)}a_j.
\eeq
This is call the {\em{ recurrence relation}}.
Consider this relation when $j$ is very large, much larger than 1. In this
limit it simplifies to the statement that $a_{j+2}=a_j(1+{\cal O}(1/j))$
which will produce a power series (for large powers) in the form of a sum
of terms proportional to
$u^{2j}$, all with the same coefficient. At $u\rightarrow1$, this sum will
not converge, and so $P$ is singular at $u=1$. This is not an allowed
behavior for a solution to the Laplace equation, so we cannot have such a
function representing the potential. What must therefore happen is that the
series terminates which means that there must be some $j$ such that $a_j
\ne0$ while $a_{j+2}=0$. Examining \eq{22}, we see that this $j$ is such
that
\beq
(\al+j)(\al+j+1)-l(l+1)=0.
\eeq
This condition requires that $\al+j=l$ which is a condition on $l$; since $
\al$ is 0 or 1, and $j$ is a non-negative integer, we see that $l$ must be
an integer equal to or larger than $\al$.
\beq
l\in{\cal Z}\;\;\;l\ge\al
\eeq
Now, our recurrence relation
gives us $a_{j+2}$ from $a_j$; hence, starting from $a_0$, we can get only
the even coefficients $a_j$, $j$ even, and starting from $a_1$, we
get the odd coefficients. Lets consider the odd series and the
even series separately. First consider the {\bf even series}.
Since at termination of the series $\al+j=l$, we see that $l$ is
even when $\al=0$ and $l$ is odd when $\al=1$. Thus the even series
terminates when
\beq
\al=0\;{\rm and}\;l\;{\rm even}
\eeq
\[
\al=1\;{\rm and}\;l\;{\rm odd}
\]
By similar arguments applied to the {\bf odd series}, we can see
that it terminates when
\beq
\al=0\;{\rm and}\;l\;{\rm odd}
\eeq
\[
\al=1\;{\rm and}\;l\;{\rm even}
\]
Since $l$ cannot be both odd and even, we can only have an even or an
odd series (they are actually equivalent). The other must be zero.
Since by convention, we choose $a_0\ne 0$, it must be that the odd
series vanishes, and thus $a_1=0$.
Remembering that $l$
is odd when $\al=1$ and is even when $\al=0$, we see that the solutions are
polynomials of degree $l$. They are known as {\em Legendre polynomials}. It
is easy to generate a few of them, aside from normalization, starting from
$l=0$ and using the recurrence relation. As for normalization, they
are traditionally chosen to be such that $P(1)=1$. Let us add a subscript
$l$ to $P$ to designate the particular Legendre polynomial. The first few
are
\begin{center}
\begin{tabular}{||l|l||} \hline\hline
$l$ & $P_l(u)$ \\ \hline\hline
$0$ & $P_0(u)=1$ \\\hline
$1$ & $P_1(u)=u $ \\\hline
$2$ & $P_2(u)=\frac32u^2-\frac12 $\\\hline
$3$ & $P_3(u)=\frac52u^3-\frac32u $\\\hline
$4$ & $P_4(u)=\frac{35}8u^4-\frac{15}4u^2+\frac38 $\\\hline\hline
\end{tabular}
\end{center}
There are other ways to generate the Legendre polynomials. For example, one
has {\em Rodrigues' formula} which is
\beq
P_l(u)=\frac1{2^ll!}\frac{d^l}{du^l}(u^2-1)^l;
\eeq
it is easy to see that this generates a polynomial of degree $l$; one may
show that it is a solution to the Legendre equation by direct substitution
into that equation. Thus, it must be the Legendre polynomial (one should
also check normalization). If one expands the factor $(u^2-1)^l$ in
Rodrigues' formula using the binomial expansion and then takes the
derivatives, she/he will find that
\beq
P_l(u)=\frac{(-1)^l}{2^l}\sum_{n\ge l/2}^l\frac{(-1)^n(2n)!}{n!(l-n)!(2n-l)
!}u^{2n-l}
\eeq
Another way of generating $P_l(u)$ is via the generating function
\beq
T(u,x)=(1-2ux+x^2)^{-1/2}=\sum_{l=0}^\infty x^lP_l(u).
\eeq
If one takes $l$ derivatives of this function with respect to $x$ and then
sets $x=0$, the result is $l!P_l(u)$.
The Legendre functions have many properties that we will need to make use
of from time to time. For summaries of these, see \eg, the section on
Legendre functions in Abramowitz and Stegun starting on p.~332 and also the
section on orthogonal polynomials starting on p.~771. Here we summarize
some of the most significant properties. First, orthogonality and
normalization. Consider the integral
\beq
\inoou P_l(u)P_{l'}(u)=\frac1{2^{l+l'}l!l'!}\inoou\frac{d^l}{du^l}(u^2-1)^l
\frac{d^{l'}}{du^{l'}}(u^2-1)^{l'}.
\eeq
Suppose, without loss of generality, that $l'\ge l$ and start by
integrating by parts,
\beqa
\inoou P_l(u)P_{l'}(u)=\nonumber \\
\frac1{2^{l+l'}l!l'!}\leb\left.\frac{d^l}{du^l}(u^2-1)^l
\frac{d^{l'-1}}{du^{l'-1}}(u^2-1)^{l'}\ril_{-1}^1-\inoou\frac{d^{l+1}}{du^{l
+1}}(u^2-1)^l\frac{d^{l'-1}}{du^{l'-1}}(u^2-1)^{l'}\rib.
\eeqa
The first term in brackets vanishes because $(u^2-1)$ is zero at the end
points of the interval. Continuing from the right-hand side, integrate in
like fashion $l'-1$ more times. The result is
\beq
\inoou P_l(u)P_{l'}(u)=(-1)^{l'}\frac1{2^{l+l'}l!l'!}\inoou
\frac{d^{l+l'}(u^2-1)^l}{du^{l+l'}}(u^2-1)^{l'}.
\eeq
Now, $d^{l+l'}(u^2-1)^l/du^{l+l'}=0$ if $l'>l$, and so the integral is zero
in this case. If $l'=l$, we have
\beqa
\inoou P_l(u)P_l(u)=(-1)^l\frac{(2l)!}{2^{2l}(l!)^2}\inoou(u^2-1)^l
\nonumber \\
=(-1)^l\frac{(2l)!}{2^{2l}(l!)^2}2(-1)^l\frac{(2l)!!}{(2l+1)!!}
=2\frac{(2^ll!)^2(2l-1)!!}{2^{2l}(l!)^2(2l+1)!!}=\frac{2}{2l+1}.
\eeqa
Thus we have derived the relation
\beq
\inoou P_l(u)P_{l'}(u)=\frac2{2l+1}\de_{ll'}
\eeq
which expresses the orthogonality and normalization of the Legendre
polynomials.
Consider next recurrence relations. These provide, among other things, a
good way to generate values of Legendre polynomials on computers. A number
of recurrence relations can be derived using Rodrigues' formula and the
Legendre equation. Consider, for example,
\beqa
\der{P_{l+1}}{u}=\frac1{2^{l+1}(l+1)!}\frac{d^{l+2}}{du^{l+2}}(u^2-1)^{l+1}
\nonumber \\
=\frac{l+1}{2^l(l+1)!}\frac{d^{l+1}}{du^{l+1}}\lep(u^2-1)^lu\rip\nonumber
\\ =\frac1{2^ll!}\frac{d^l}{du^l}\leb(u^2-1)^l+2lu^2(u^2-1)^{l-1}\rib
\nonumber \\
=\frac1{2^ll!}\frac{d^l}{du^l}\leb(u^2-1)^l+2l(u^2-1)^l+2l(u^2-1)^{l-1}
\rib \nonumber \\
=(2l+1)P_l(u)+\frac{dP_{l-1}(u)}{du},
\eeqa
or
\beq
\der{P_{l+1}(u)}{u}-(2l+1)P_l(u)-\der{P_{l-1}(u)}{u}=0.
\eeq
From this relation and the Legendre equation
\beq
\frac{d}{du}\lep(1-u^2)\der{P_l}{u}\rip+l(l+1)P_l=0
\eeq
one may derive additional standard recurrence relations for the Legendre
polynomials. Several of these are
\beqa
(l+1)P_{l+1}-u(2l+1)P_l+lP_{l-1}=0 \nonumber \\
(1-u^2)\der{P_l}{u}+luP_l-lP_{l-1}=0 \nonumber \\
\der{P_{l+1}}{u}-u\der{P_l}{u}-(l+1)P_l=0.
\eeqa
These may be used to advantage in numerous applications such as doing
integrals of products of two Legendre polynomials and a power of $u$.
\subsection{Solution of Boundary Value Problems with Azimuthal Symmetry}
Using what we have learned in the previous two sections, we are now in a
position to construct a general solution to the Laplace equation in
spherical coordinates under conditions of azimuthal invariance, that is,
when $\Ph(\x)$ is independent of $\ph$. The most general form that a
solution can have is
\beq
\Ph(r,\th)=\sum_{l=0}^\infty\lep A_lr^l+B_lr^{-(l+1)}\rip P_l(\cos\th).
\eeq
The Legendre polynomials form a complete set on the interval $-1\le u\le1$
or $0\le\cos\th\le\pi$. Thus any specified $\ph$-independent potential on a
spherical surface can be expressed as a sum of $P_l$'s. If the volume in
which a solution is to be found includes the origin, then none of the terms
$\sim r^{-(l+1)}$ can be included in the sum as they are singular at the
origin, and the potential will not be singular there. Similarly, if the
volume extends to $r\rightarrow\infty$, then no terms $\sim r^l$ are
allowed. In the former case, the conclusion is that $B_l=0$ for all $l$,
and in the latter case, all $A_l=0$.
We consider now some examples.
\subsubsection{Example: A Sphere With a Specified Potential}
An isolated sphere of radius $a$ is centered at the origin. By unspecified
means, the potential on its surface is maintained at
\[
\Ph(a,\theta,\phi)=V_o \cos^3(\theta)
\]
where $\theta$ is the polar angle. Find $\Ph(r,\theta,\phi)$ for
all $r>a$.
This problem is azimuthally symmetric. Thus, in general,
\[
\Ph(r,\th)=\sum_{l=0}^\infty\lep A_lr^l+B_lr^{-(l+1)}\rip P_l(\cos\th).
\]
Since our volume contains all $r>a$, physics demands that $A_l=0$
for all $l$. The constants $B_l$ are then determined by matching the
terms in the series the boundary condition on the surface of the sphere.
Recall that
\medskip
\begin{center}
\begin{tabular}{||l|l||}\hline\hline
$P_0(x)$ & $1$ \\ \hline
$P_1(x)$ & $x$ \\ \hline
$P_2(x)$ & $\frac12(3x^2-1)$ \\ \hline
$P_3(x)$ & $\frac12(5x^3-3x)$ \\ \hline
$P_4(x)$ & $\frac18(35x^4-30x^2+3)$ \\ \hline
$P_5(x)$ & $\frac18(63x^5-70x^3+15x)$ \\ \hline\hline
\end{tabular}
\end{center}
So that,
\[
\Ph(a,\theta,\phi)=V_o \cos^3(\theta) =
V_o\left( \frac{2}{5} P_3(\cos(\theta)) +
\frac{3}{5}P_1(\cos(\theta))\right)
\]
Thus
\[
\Ph(r,\theta,\phi)=
V_o\left( \frac{2}{5} \left(\frac{a}{r}\right)^4 P_3(\cos(\theta)) +
\frac{3}{5} \left(\frac{a}{r}\right)^2 P_1(\cos(\theta))\right).
\]
\subsubsection{Example: Hemispheres of Opposite Potential}
For the first, suppose that we need to solve
the Laplace equation inside of a sphere of radius $a$ given that on the
surface, the potential is specified as follows:
\beq
\Ph(a,\th)=\lec\barr{cc} V, & 0\le\th\le\pi/2 \\
-V, & \pi/2\le\th\le\pi. \ear\right.
\eeq
\centerline{\psfig{figure=fig1.ps,height=1.8in,width=7.0in}}
\noindent Then the expansion must take the form
\beq
\Ph(r,\th)=V\sum_{l=0}^\infty A_l\lep\frac{r}{a}\rip^lP_l(\cos\th).
\eeq
Notice the introduction of the factor $V$ on the right-hand side, along
with the use of the powers of $a$ in the sum. These are included for
convenience. The scale of the potential and hence the size of a leading term
in the sum is set by $V$ which also gives the correct dimensions to the
terms in the sum; it is thus natural to put this factor in each term. The
powers of $a$ are included for the same reasons; $r$ is of order $a$ and
has the same dimensions so that leading coefficients $A_l$ are of order
unity and have dimension unity.
On the spherical surface, we have
\beq
\Ph(a,\th)=V\sum_{l=0}^\infty A_lP_l(\cos\th).
\eeq
In order to find a given coefficient $A_n$, we multiply this equation by
$P_n(\cos\th)$ and integrate over $\cos\th$, recalling that $d\cos\th=-
\sin\th d\th$. Making use of the orthogonality and normalization of the
Legendre polynomials, we find
\beq
\int_0^\pi d\th\Ph(a,\th)P_n(\cos\th)\sin\th=V\sum_{l=0}^\infty A_l\lep
\frac2{2l+1}\rip\de_{ln}=V\lep\frac{2}{2n+1}\rip A_n,
\eeq
or
\beqa
A_n=\frac{2n+1}{2}\leb\int_0^{\pi/2}d\th P_n(\cos\th)\sin\th-
\int_{\pi/2}^\pi d\th P_n(\cos\th)\sin\th\rib \nonumber \\
=\frac{2n+1}2\leb\int_0^1duP_n(u)-\int_{-1}^0duP_n(u)\rib
\eeqa
Now use the inversion property of the Legendre polynomials, $P_n(u)=(-1)
^nP_n(-u)$ to conclude that
\beq
A_n=\frac{2n+1}2\leb\int_0^1duP_n(u)-(-1)^n\int_0^1duP_n(u)\rib=
\lec\barr{cl} 0 & \hbox{$n$ even} \\
(2n+1)\int_0^1duP_n(u) & \hbox{$n$ odd.} \ear\right.
\eeq
To complete the integral for the case of odd $n$ we use a recurrence
relation
\beq
\der{P_{n+1}}{u}=(2n+1)P_n +\der{P_{n-1}}{u}
\eeq
so that:
\beq
A_n=(2n+1)\int_0^1duP_n(u)=\int_0^1du\leb\der{P_{n+1}}{u}-\der{P_{n-1}}{u}
\rib=P_{n-1}(0)-P_{n+1}(0)
\eeq
where we make use of the fact that $P_n(1)=1$, independent of $n$.
Further, for even $l$,
\beq
P_l(0)=(-1)^{l/2}\frac{(l-1)!!}{l!!}
\eeq
where the
``double factorial'' sign means $l!!=l(l-2)(l-4)...(2 \hbox{ or }1)$.
Hence
\beqa
P_{n-1}(0)-P_{n+1}(0)=(-1)^{(n-1)/2}\leb\frac{(n-2)!!}{(n-1)!!}+\frac{n!!}{(n+
1)!!}\rib \nonumber \\
=(-1)^{(n-1)/2}\frac{(n-2)!!}{(n+1)!!}(n+1-n)=(-1)^{(n-1)/2}\frac
{(n-2)!!}{(n+1)!!}(2n+1).
\eeqa
Now set $n=2m+1$, $m=0$,1,2,..., and have
\beq
\Ph(r,\th)=V\sum_{m=0}^\infty B_m\lep\frac{r}{a}\rip^{2m+1}P_{2m+1}(\cth)
\eeq
where
\beq
B_m=(-1)^m\frac{(2m-1)!!}{(2m+2)!!}(4m+3).
\eeq
The first few terms in the expansion are
\beq
\Ph(r,\th)=\frac32V\lec\frac r a P_1(\cth)-\frac7{12}\lep\frac r a \rip^3P_3
+\frac{11}{24}\lep\frac r a\rip^5P_5-\frac{25}{64}\lep\frac ra\rip^7P_7+...
\ric
\eeq
\subsubsection{Example: Potential of an Isolated Charge}
Another method of finding the coefficients in the expansion makes use of
the fact that the expansion is unique. If, for example, we are able to find
the potential at fixed $\th=\th_0$ for all $r$,
\beq
\Ph(\th_0,r)=g(r)=\sum_{l=0}^\infty\lep A_lr^l+B_lr^{-(l+1)}\rip P_l(\cos\th_0)
\eeq
then we can infer the form of the
expansion by expanding $g(r)$ in powers of $r$ and
recognizing that the coefficient of $r^l$ {\em must} be $P_l(\cos\th_0)$ times a
coefficient $A_l$ while the coefficient of $(1/r)^{-(l+1)}$ must be $B_lP_l
(\cos\th_0)$. The most convenient value of $\th_0$ to use is certainly 0 or
$\pi$ since we know immediately the value of $P_l(\cth)$ in these instances.
Consider the following specific example: Suppose that there is a charge $q$
at a position $\x=a\zh$
\centerline{\psfig{figure=isol_charge.ps,height=2.0in,width=2.0in}}
\noindent in which case we know that the potential is
\beq
\Ph(r,\th)=\frac{q}{\sqrt{r^2+a^2-2ar\cos\th}}.
\eeq
For $\th=0$, we have simply $\phx=q/|r-a|$. At $ra$ yields
\beq
\Ph(r,\th)=\frac qr\sum_{l=0}^\infty\lep\frac ar\rip^lP_l(\cth)
\;\;\;\;r>a\,.
\eeq
There are two points that are worth making in connection with these
expansions. First, as stated earlier, there is a generating function $T(u,x
)$ for the Legendre polynomials; see \eq{29}. We have just derived it; that
is, it is $\Ph$, \eq{54}, equal to the sum in \eq{56}.
Also, we have obtained a convenient and useful expansion for the potential
of a point charge; in more general notation, we have derived
\beq
\xxpi=\sum_{l=0}^\infty\frac{r_<^l}{r_>^{l+1}}P_l(\cos\ga)
\eeq
where $\ga$ is the angle between $\x$ and $\xp$ while $r_<$ ($r_>$) is the
smaller (larger) of $|\x|$ and $|\xp|$.
\subsection{Behavior of Fields in Conical Holes and Near Sharp Points}
The field in the vicinity of the apex of a cone-shaped tip or depression
can also be investigated using the separation of variables method in
spherical coordinates. The solution for the potential is of the form, for
$r$ small enough,
\beq
\Ph(r,\th)\sim r^{\nu}P_\nu(\cth)
\eeq
where $P_\nu(u)$ is a solution to the Legendre equation
\beq
\der{}{u}(1-u^2)\der{P_\nu}{u}+\nu(\nu+1)P_\nu=0
\eeq
with $\nu$ to be determined.
\centerline{\psfig{figure=fig2.ps,height=3.5in,width=6.0in}}
\noindent For the geometry shown, the solution must be well-behaved as $\th
\rightarrow0$, or $u=\cth\rightarrow1$, but not necessarily as
$\th\rightarrow\pi$ or
$u=-1$. Introduce the variable $y\equiv\frac12(1-u)$ or $u=1-2y$; then
\eq{60} becomes
\beq
-\frac12\der{}{y}\lep1-(1-2y)^2\rip\lep-\frac12\der{P_\nu}{y}\rip+
\nu(\nu+1)P_\nu=0
\eeq
or
\beq
\der{}{y}\lep y(1-y)\der{P_\nu}{y}\rip+\nu(\nu+1)P_\nu=0.
\eeq
Let us look once again for a solution in the form of a power series
expansion,
\beq
P_\nu=y^\al\sum_{j=0}^\infty a_jy^j,
\eeq
with $0\le y\le y_0\le1$.
Then
\beq
\der{P_\nu}{y}=\sum_{j=0}^\infty(\al+j)a_jy^{\al+j-1},
\eeq
\beq
y(1-y)\der{P_\nu}{y}=\sum_{j=0}^\infty(\al+j)a_j(y^{\al+j}-y^{\al+j+1}),
\eeq
and
\beq
\der{}{y}\lep y(1-y)\der{P_\nu}{y}\rip=\sum_{j=0}^\infty a_j\leb(\al+j)y^{\al
+j-1}-(\al+j+1)y^{\al+j}\rib(\al+j).
\eeq
Now combine these equations to find
\beq
\sum_{j=0}^\infty\leb a_j(\al+j)^2y^{\al+j-1}+a_j\lep(\al+j)(\al+j+1)+\nu(
\nu+1)\rip y^{\al+j}\rib=0
\eeq
or, isolating individual powers of $y$,
\beq
a_0\al^2y^{\al-1}=0
\eeq
which implies that $\al=0$, and
\beq
a_{j+1}=a_j\frac{j(j+1)-\nu(\nu+1)}{(j+1)^2}.
\eeq
If one lets $\nu=l$, a non-negative integer, the result is just the
Legendre polynomials (no surprise), viewed as functions of $y$. More
generally, for any real $\nu>0$, one finds that the solutions are {\em
Legendre functions of the first kind of order $\nu$}.
\beq
P_\nu=\sum_{j=0}^\infty a_j(\nu)y^j,
\eeq
These are
well-behaved (that is, they are not singular) functions of $y$ for
$y<1$ corresponding to $u>-1$ and are singular at $y=1$.
For $1>\nu>0$, $P_
\nu(y)$ has a single zero; for $2>\nu>1$, $P_\nu(y)$ has two zeroes, etc.
This is important because if we have a cone of half-angle $\be$ with
equipotential surfaces, we need $\nu$ to be such that
$P_\nu\lep(1-\cos\be)/2\rip=0$.
There will thus be a sequence of allowed values of $\nu$, which we
designate by $\nu_k$, $k=1$, 2, 3,..., which are such that $y_\be\equiv
\frac12(1-\cos\be)\equiv k^{th} $ zero of $P_\nu$.
The general solution at finite values of $r$, and including the point $r=0$
, is
\beq
\Ph(r,\th)=\sum_{k=1}^\infty A_kr^{\nu_k}P_{\nu_k}(\cos\th).
\eeq
For small $r$, the leading term is the one with the smallest power of $r$,
that is the $k=1$ term. Hence we may approximate the sum sufficiently
close to the origin by its leading term
\beq
\Ph(r,\th)\approx Ar^{\nu_1}P_{\nu_1}(\cos\th).
\eeq
The dominant contribution to the electric field in this region comes from
this term; we have, by the usual $\Ex=-\grad\Phx$,
\beq
E_r=\der{\Ph}{r}=-\nu_1Ar^{\nu_1-1}P_{\nu_1}(\cos\th)
\eeq
and
\beq
E_\th=-\frac{1}{r}\der{\Ph}{\th}=
A\sth\, r^{\nu_1-1}\left.\der{P_{\nu_1}(u)}{u}\ril_{\cth}
\eeq
The behavior of $\nu_1$ as a function of $\be$ is shown below.\vspace{3.5in} For
$\be$ less than about $0.8\pi$, one has\footnote{This relation comes from
study of the properties of the Legendre functions.} $\nu_1\approx\frac{2.405}
{\be}-\frac12$, while for $\be$ larger than about the same number,
$\nu_1\approx\leb2\ln\lep\frac2{\pi-\be}\rip\rib^{-1}.$ As $\be\rightarrow
\pi$, $\nu_1\rightarrow0$ and so $E_r\sim E_\th\sim1/r$ in this limit. The
enhancement of a field near \eg a lightning rod is thus $\sim(R/\de)$ if
$R$ is the size of the system and $\de$ is the radius of curvature of the
tip of the rod. Recall that in two dimensions we found an enhancement of
order $(R/\de)^{1/2}$ The enhancement is much more pronounced in three
dimensions; a three dimensional tip is a much sharper thing than an edge.
\subsection{Associated Legendre Polynomials; Spherical Harmonics}
Let us now return to the more general case of a solution to Laplace's
equation (i.e. a potential) which depends on
the azimuthal angle $\ph$. Then we must have the functions of $\ph$
$e^{im\ph}$, or, equivalently, $\sin m\ph$ and $\cos m\ph$, and the
differential equation we have to face on the space of $\th$ is
\beq
\der{}{u}(1-u^2)\der{P}{u}+\lep l(l+1)-\frac{m^2}{1-u^2}\rip P=0.
\eeq
The solutions are not finite polynomials in $u$ in general but can be
expressed as infinite power series. They are only ``well-behaved''
on the interval $-1\le u\le1$ when $l\ge|m|$, with $l$ an integer.
Then there is just one well-behaved solution which is known as the {\em
associated Legendre function of degree $l$ and order $m$}. For $m\ge0$, the
associated Legendre function can be written in terms of the Legendre
polynomial of the same degree as
\beq
P^m_l(u)=(-1)^m(1-u^2)^{m/2}\frac{d^m}{dx^m}\lep P_l(u)\rip;
\eeq
one can read all about this in Abramowitz and Stegun on pages 332 to 353.
Making use of Rodrigues' formula for the Legendre polynomials, we see that
\beq
P_l^m(u)=(-1)^m\frac{(1-u^2)^{m/2}}{2^ll!}\frac{d^{l+m}}{du^{l+m}}
[(u^2-1)^l].
\eeq
This last formula is also valid for negative $m$ \footnote{That's in part a
matter of definition.}; comparing the two cases,
one may see that
\beq
P^{-m}_l(u)=(-1)^m\frac{(l-m)!}{(l+m)!}P_l^m(u).
\eeq
As for the Legendre polynomials, there is a generating function for the
associated Legendre functions as well as a variety of recurrence relations.
For example, a recurrence relation in degree is given by
\beq
(2l+1)uP_l^m(u)=(l-m+1)P_{l+1}^m(u)+(l+m)P_{l-1}^m(u)
\eeq
and one in order is
\beq
P_l^{m+1}+\frac{2mu}{\sqrt{1-u^2}}P_l^m(u)+(l-m+1)(l+m)P_l^{m-1}(u)=0
\eeq
Out of all of this, what is of importance to us is that the product
\beq
(Ar^l+Br^{-l-1})P^m_l(\cth)e^{im\ph}
\eeq
is a solution of the Laplace equation and that the set of functions $e^{im\ph}
P^m_l(\cth)$ with $l=0,1,2$,..., and $m=-l,-l+1,$...$l-1,l$ form a complete
orthogonal set on the two-dimensional domain $0\le\th\le\pi$ and $0\le\ph
\le 2\pi$. As usual, completeness is difficult to demonstrate but
orthogonality is quite straightforward using the formulae we have already
written down. Consider the integral
\beq
I=\int_0^{2\pi}d\ph\int_0^{\pi}d\th\sth e^{-im\ph}P_l^m(\cth)e^{im'\ph}P_{l'}
^{m'}(\cth)=2\pi\de_{mm'}\int_{-1}^1duP_l^m(u)P_{l'}^m(u)
\eeq
Assume $l'\ge l$, $m\ge0$, and write $P_{l'}^m$ in terms of $P_{l'}^{-m}$:
\beqa
I=2\pi\de_{mm'}(-1)^m\frac{(l'+m)!}{(l'-m)!}\int_{-1}^1duP_{l'}^{-m}(u)
P_l^m(u) \nonumber \\
=2\pi\de_{mm'}(-1)^m\frac{(l'+m)!}{(l-m)!}\frac1{2^{l+l'}l!l'!}\int_{-1}^1
du\frac{d^{l'-m}}{du^{l'-m}}(u^2-1)^{l'}\frac{d^{l+m}}{du^{l+m}}(u^2-1)^l
\nonumber \\
=2\pi\de_{mm'}(-1)^{l'}\frac{(l'+m)!}{(l'-m)!}\frac1{2^{l+l'}l!l'!}
\int_{-1}^1du(u^2-1)^{l'}\frac{d^{l+l'}}{du^{l+l'}}(u^2-1)^l \nonumber \\
=2\pi\de_{ll'}\de_{mm'}\frac{(l+m)!(2l)!(2l)!!}{(l-m)!2^{2l}(l!)^2(2l+1)!!
}2=\frac{4\pi}{2l+1}\de_{ll'}\de_{mm'}\frac{(l+m)!}{(l-m)!}.
\eeqa
Thus we may construct an orthonormal set of functions on the surface of the
unit sphere; these are called {\em spherical harmonics} and are defined as
\beq
\ylm\equiv\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}
P_l^m(\cth)e^{im\ph},
\eeq
with $m=-l,-l+1,...,l-1,m$ and $l=0,1,2,...$. These functions have the
property that
\beq
\ylms=(-1)^mY_{l,-m}(\th,\ph).
\eeq
The condition of orthonormality is
\beq
\int_0^{2\pi}d\ph\int_0^\pi\sth d\th\ylms\ylmp=\de_{ll'}\de_{mm'}.
\eeq
The completeness relation (not derived as usual) is
\beq
\sum_{l=0}^\infty\sum_{m=-l}^m\ylmsp\ylm=\de(\ph-\ph')\de(\cth-\cos\th').
\eeq
A general function $g(\th,\ph)$ is expanded in terms of the spherical
harmonics as
\beq
g(\th,\ph)=\sum_{l,m}A_{lm}\ylm
\eeq
with
\beq
A_{lm}=\int_0^{2\pi}d\ph\int_0^\pi d\th\sth g(\th,\ph)\ylms.
\eeq
\subsection{The Addition Theorem}
In applications we will occasionally have need to know the function $P_l(
\cos\ga)$ where $\ga$ is the angle between two vectors $\x$ and $\xp$; it
will prove to be useful to be able to write this function in terms of the
variables $\th$, $\ph$, $\th'$, and $\ph'$.
\centerline{\psfig{figure=fig3.ps,height=2.5in,width=5.0in}}
\noindent It must be possible to do so in
terms of any complete sets of functions of these variables such as the
spherical harmonics. In fact, the expansion is
\beq
P_l(\cos\ga)=\frac{4\pi}{2l+1}\sum_{m=-l}^l\ylmsp\ylm.
\eeq
We shall derive this expression as an example of the use of spherical
harmonics and their properties. First, let us set up a second coordinate
system rotated relative to the original one in such a way that its polar
axis lies along the direction of $\xp$. In this system, the vector $\x$ has
components $(r_R,\th_R,\ph_R)$.
\centerline{\psfig{figure=fig4.ps,height=2.5in,width=5.0in}}
\noindent Further, $\th_R=\ga$. Next, we may regard
$P_l(\cos\ga)$ as a function of $\th$ and $\ph$ for fixed $\th'$ and $\ph'$
and so can certainly expand it as
\beq
P_l(\cos\ga)=\sum_{l'=0}^\infty\sum_{m'=-l}^l A_{l'm'}(\th',\ph')\ylmp.
\eeq
Similarly, in terms of spherical harmonics whose arguments are coordinates in
the rotated system, it is easy to see that
\beq
P_l(\cos\ga)=\sqrt{\frac{4\pi}{2l+1}} Y_{l,0}(\th_R,\ph_R).
\eeq
Now, the spherical harmonics satisfy the differential equation
\beq
\lap\ylm+\frac{l(l+1)}{r^2}\ylm=0
\eeq
and they also satisfy this equation with variables $\th_R$, $\ph_R$. But
the Laplacian operator $\lap=\grad\cdot\grad$ is a scalar object which
is invariant under coordinate rotations which means we can write it in
the unrotated frame while
writing the spherical harmonic in the rotated frame:
\beq
\lap Y_{l,m}(\th_R,\ph_R)+\frac{l(l+1)}{r^2}Y_{l,m}(\th_R,\ph_R)=0.
\eeq
Now recall that
\beq
Y_{l,0}(\th_R,\ph_R)=\sqrt{\frac{2l+1}{4\pi}} P_l(\cos\ga)=
\sqrt{\frac{2l+1}{4\pi}}
\sum_{l'=0}^\infty\sum_{m'=-l'}^{l'} A_{l'm'}(\th',\ph')\ylmp.
\eeq
If we plug this into the differential equation above, we obtain:
\beq
\sum_{l'=0}^\infty\sum_{m'=-l'}^{l'} A_{l'm'}(\th',\ph')[\lap\ylmp+
\frac{l(l+1)}{r^2}\ylmp]=0
\eeq
or
\beq
\sum_{l'=0}^\infty\sum_{m'=-l'}^{l'} A_{l'm'}(\th',\ph')\leb-\frac{l'(l'+1)
}{r^2}+\frac{l(l+1)}{r^2}\rib\ylmp=0.
\eeq
This equation can be true only if $l=l'$ {\bf or} if $A_{l'm'}=0$. Thus we
have demonstrated that $A_{l'm'}=0$ for $l'\ne l$, and the expansion of
$P_l$ reduces to
\beq
P_l(\cos\ga)=\sum_{m=-l}^lA_{lm}(\th',\ph')\ylm.
\eeq
The coefficients in this expansion are found in the usual way for an orthogonal
function expansion,
\beq
A_{lm}=\inom P_l(\cos\ga)\ylms=\inom_R P_l(\cos\th_R)\ylms.
\eeq
Following the same line of reasoning, we may express $\sqrt{4\pi/(2l+1)}
\ylms$ as a sum of the form
\beq
\sqrt{\frac{4\pi}{2l+1}}\ylms=\sum_{m'=-l}^l B_{lm'}(m)Y_{lm'}(\th_R,\ph_R)
\eeq
where $B_{l0}$ in particular is
\beqa
B_{l0}(m)=\int d\Om_R\sqrt{\frac{4\pi}{2l+1}}\ylms Y_{l,0}^*(\th_R,\ph_R)
\nonumber \\ =\int d\Om_R\ylms P_l(\cos\th_R)\equiv A_{lm}.
\eeqa
However, from \eq{76}, it is clear that when $u=1$ $P_l^m(u)=0$ when
$m\ne 0$,and $P_l^m(u)=P_l(u)$ when $m= 0$, thus
\beq
\sqrt{\frac{4\pi}{2l+1}}\left.\ylms\ril_{\th_R=0}=B_{l0}(m)\sqrt{\frac{2l+
1}{4\pi}}P_l(1)=B_{l0}(m)\sqrt{\frac{2l+1}{4\pi}},
\eeq
so \beq
A_{lm}=B_{l0}(m)=\frac{4\pi}{2l+1}\left.\ylms\ril_{\th_R=0}=\ylmsp
\frac{4\pi}{2l+1},
\eeq
where the last step follow since when $\th_R=0$, $\x=\xp$.
Thus,
\beq
P_l(\cos\ga)=\frac{4\pi}{2l+1}\sum_{m=-l}^l\ylmsp\ylm.
\eeq
Thus ends our demonstration of the spherical harmonic addition theorem.
An application if this theorem is that we can write $\xxpi$ as an expansion
in spherical harmonics:
\beq
\xxpi=\sum_{l=0}^\infty\frac{r_<^l}{r_>^{l+1}}P_l(\cos\ga)=
\sum_{l=0}^\infty\lec\frac{4\pi}{2l+1}\frac{r_<^l}{r_>^{l+1}}\sum_{m=-l}^l
\leb\ylmsp\ylm\rib\ric.
\eeq
This expansion is often useful when faced with common integrals in
electrostatics such as as $\int d^3x'\rhxp/|\x-\xp|$.
\subsection{Expansion of the Green's Function in Spherical Harmonics}
More as an example of the use of the addition theorem than anything else,
let us devise an expansion for the Dirichlet Green's function for the
region V bounded by $r=a$ and $r=b$, $a**^{l+1}}\sum_{m=-
l}^l\ylmsp\ylm \nonumber \\
+\sum_{l=0}^\infty\sum_{m=-l}^l\lep A_{lm}\frac{r^l}{b^{l+1}}+B_{lm}
\frac{a^l}{r^{l+1}}\rip\ylm
\eeqa
where $A_{lm}$ and $B_{lm}$ can be functions of $\xp$. The first term on
the right-hand side is $\xxpi$ and the second is a general solution of the
Laplace equation. The coefficients are determined by requiring that the
boundary conditions on $\gxxp$ are satisfied ($G(\x,\xp)=0$ for $\x$ on
one of the two bounding spherical surfaces). At $r=a$ ($r_<=r=a$ $r_>=r'$),
we have
\beq
0=\sum_{l,m}\lec\frac{4\pi}{2l+1}\frac{a^l}{r'^{l+1}}\ylmsp+A_{lm}
\frac{a^l}{b^{l+1}}+B_{lm}\frac1a\ric\ylm=0,
\eeq from which it follows, using the orthogonality of the spherical
harmonics, that
\beq
\frac{4\pi}{2l+1}\frac{a^l}{r'^{l+1}}\ylmsp+A_{lm}\frac{a^l}{b^{l+1}}+
B_{lm}\frac1a=0.
\eeq
By similar means applied at $r=b$ ($r_<=r'$ $r_>=r=b$), one may show that
\beq
\frac{4\pi}{2l+1}\frac{r'^l}{b^{l+1}}\ylmsp+A_{lm}\frac1b+B_{lm}
\frac{a^l}{b^{l+1}}=0.
\eeq
These present us with two linear equations that may be solved for $A_{lm}$
and $B_{lm}$; the solutions are
\beq
A_{lm}=\frac{4\pi}{(2l+1)}\ylmsp
\left.\lep\frac{a^{2l+1}}{b^lr'^{l+1}}-
\frac{r'^l}{b^l}\rip\ris\leb1-\lep\frac ab\rip^{2l+1}\rib
\eeq
and
\beq
B_{lm}=\frac{4\pi}{2l+1}\ylmsp\left.\lep\frac{a^{l+1}r'^l}{b^{2l+1}}-\frac{a^{l+
1}}{r'^{l+1}}\rip\ris\leb1-\lep\frac ab\rip^{2l+1}\rib.
\eeq
From these and the expansion \eq{107}, we find that the Green's function is
\beqa
\gxxp=\sum_{l,m}\frac{4\pi}{(2l+1)\leb1-\lep\frac ab\rip^{2l+1}\rib}\ylmsp
\ylm \nonumber \\
\lec\leb1-\lep\frac ab\rip^{2l+1}\rib\frac{r_<^l}{r_>^{l+1}}+\lep\frac ab
\rip^{2l+1}\frac{r^l}{r'^{l+1}}-\frac{r'^lr^l}{b^{2l+1}}+\lep
\frac ab\rip^{2l+1}\frac{r'^l}{r^{l+1}}-\frac{a^{2l+1}}{r^{l+1}r'^{l+1}}\ric.
\eeqa
This result, if we can call it that, can be written in a somewhat more
compact form by factoring the quantity $\{...\}$. Suppose that $r_>=r'$ and
$r_<=r$; then
\beqa
\{...\}=\lep r^l-\frac{a^{2l+1}}{r^{l+1}}\rip\lep\frac1{r'^{l+1}}-
\frac{r'^l}{b^{2l+1}}\rip \nonumber \\
\equiv\lep r_<^l-\frac{a^{2l+1}}{r_<^{l+1}}\rip\lep\frac1{r_>^{l+1}}-
\frac{r_>^l}{b^{2l+1}}\rip.
\eeqa
If, on the other hand, $r_>=r$ and $r_<=r'$, then
\beq
\{...\}=\lep r'^l-\frac{a^{2l+1}}{r'^{l+1}}\rip\lep\frac1{r^{l+1}}-
\frac{r^l}{b^{2l+1}}\rip \nonumber \\
\equiv\lep r_<^l-\frac{a^{2l+1}}{r_<^{l+1}}\rip\lep\frac1{r_>^{l+1}}
-\frac{r_>^l}{b^{2l+1}}\rip.
\eeq
Comparing these results, we see that we may in general write
\beq
\gxxp=\sum_{l,m}\frac{4\pi/(2l+1)}{1-\lep\frac ab\rip^{2l+1}}\ylmsp
\ylm\lep r_<^l-\frac{a^{2l+1}}{r_<^{l+1}}\rip\lep\frac1{r_>^{l+1}}
-\frac{r_>^l}{b^{2l+1}}\rip.
\eeq
Notice that the Green's functions for the interior of a sphere of radius
$b$ and for the exterior of a sphere of radius $a$ are easily obtained by
taking the limits $a\rightarrow0$ and $b\rightarrow\infty$, respectively.
In the former case, for example, one finds
\beqa
\gxxp=\sum_{l,m}\frac{4\pi}{2l+1}\ylmsp\ylm\lep\frac{r_<^l}{r_>^{l+1}}-
\frac{r_<^lr_>^l}{b^{2l+1}}\rip \nonumber \\
=\xxpi-\sum_{l,m}\frac{4\pi}{2l+1}\ylmsp\ylm\frac b{r'}\frac{r^l}{(b^2/r')^{l+
1}} \nonumber \\
=\xxpi-\frac{b/r'}{|\x-\x'_R|}
\eeqa
where $\x'_R=(b^2/r',\th',\ph')$ in spherical coordinates.
\section{Laplace Equation in Cylindrical Coordinates; Bessel Functions}
In cylindrical coordinates the Laplacian is
\beq
\lap=\spd{}{\rh}+\frac1\rh\pde{}{\rh}+\frac1{\rh^2}\spd{}{\ph}+\spd{}{z}.
\eeq
We once again look for solutions of the Laplace equation in the form of
products of functions of a single variable,
\beq
\Phx=R(\rh)Q(\ph)Z(z);
\eeq
Following the usual procedure (substitute into the Laplace equation; divide
by appropriate functions to obtain terms which appear to depend on a single
variable; argue that such terms must be constants; etc.), we wind up with
the following three ordinary differential equations:
\beq
\sde Q\ph+\nu^2Q=0,\hbox{ $\nu=0,\pm1,\pm2,$...,}
\eeq
\beq
\sde Zz-k^2Z=0,
\eeq
and
\beq
\sde R\rh+\frac1\rh\der R\rh+\lep k^2-\frac{\nu^2}{\rh^2}\rip R=0
\eeq
where the value of $k$ is yet to be determined, and $\nu$ is determined as
indicated by the same argument as in the case of spherical coordinates; the
functions $Q(\ph)$ are the same as in spherical coordinates also, $Q(\ph)
\sim e^{i\nu\ph}$.
The choice of $k$ is specified by the sort of boundary conditions one has.
One could imagine having to satisfy quite arbitrary conditions on an end
face $z=c$ where $c$ is constant; alternatively, one may have to fit some
function on a side wall $\rh=c$. In the former case, one wants to have
functions of $\rh$ which form a complete set on an appropriate interval of
$\rh$; and in the latter case, one wants functions of $z$ to form a
complete set on some interval of $z$; in both cases we will need a complete
set of functions of $\ph$, which we have. Now, looking at the equations for
$R$ and $Z$, we can see that the latter function in particular is going to
be simple exponentials of $kz$; for $k$ real, these do not form a complete
set; for $k$ imaginary, they are sines and cosines and can form a complete
set. We may not recognize it yet, but a similar thing happens to $R$; for
$k$ imaginary, it is roughly exponential in character and we cannot get a
complete set of functions in this way. But for $k$ real, the functions $R$
are oscillatory (although not sines and cosines) and can form a complete
set.
\begin{center}
\begin{tabular}{||l|l|l||}\hline
$k$ & $Z(z)$ & $R(\rho)$ \\ \hline\hline
real & incomplete ($e^{\pm kz}$) & complete (oscillatory) \\ \hline
imaginary & complete ($e^{i\pm |k|z}$) & incomplete \\ \hline\hline
\end{tabular}
\end{center}
The functions of $z$ in either case ($k$ real or imaginary) are familiar to
us and do not require further discussion. The functions of $R$, although
probably known to all of us at least vaguely, are much less familiar so we
will spend some time presenting their most important, to us, properties.
Let's start by defining a dimension-free variable $x=k\rh$. Then \eq{122}
becomes
\beq
k^2\sde{R}{x}+k^2\frac1x\der Rx+k^2\lep1-\frac{\nu^2}{x^2}\rip R=0
\eeq
or
\beq
\sde Rx+\frac1x\der Rx+\lep1-\frac{\nu^2}{x^2}\rip R=0
\eeq
which is {\em Bessel's Equation}. Its solutions are {\em Bessel functions of
order $\nu$}. In our particular case, $\nu$ is an integer, although this
need not be true in general. If $k$ is imaginary, then $x$ is imaginary, so
we must deal with Bessel functions of imaginary argument; viewed as
functions of a real variable $|x|$, these are known as {\em modified Bessel
functions}.
For a given $\nu$, there are two linearly independent solutions of Bessel's
equation. Their choice is somewhat arbitrary since any linear combination
of them is also a solution. One possible and common way to choose them is
as follows:
\beq
J_\nu(x)=\lep\frac x2\rip^\nu\sum_{j=0}^\infty\frac{(-1)^j(x/2)^{2j}}
{j!\Ga(j+\nu+1)}
\eeq
and
\beq
J_{-\nu}(x)=\lep\frac x2\rip^{-\nu}\sum_{j=0}^\infty\frac{(-1)^j(x/2)^{2j}}
{j!\Ga(j-\nu+1)}
\eeq
where
\beq
\Ga(z)\equiv\int_0^\infty dt\,t^{z-1}e^{-t}
\eeq
is the {\em gamma function} which, for $z$ a real, positive integer $n$, is
$\Ga(n)=(n-1)!$. For $z=0$ or a negative integer, it is singular.
The two Bessel functions introduced above are linearly independent
solutions of Bessel's equation so long as $\nu$ is not an integer. It is
easy to verify that they are solutions by direct substitution into the
differential equation. If, however, $\nu$ is an integer, they become
identical (aside from a possible sign difference) and so do not provide us
with everything we need in this case, which is the important one for us.
Another function, which is a solution and which is linearly independent of
either of the two solutions introduced above (taken one at a time) is given
by
\beq
N_\nu(x)\equiv\frac{\jnx\cos(\nu\pi)-J_{-\nu}(x)}{\sin(\nu\pi)}.
\eeq
This is the {\em Neumann function}; it is also called the {Bessel function
of the second kind of order $\nu$}. \footnote{Why does this work?
Consider the limit $\nu\to m$ by La Hopital's rule. To formally show that
$N_m$ and $J_m$ are independent, one must calculate
the Wronskian and show that $W[N_m,J_m]\ne 0$.Note that in Abramowitz
and Stegun's book, this function is written as $Y_\nu(x)$; see p.~358.}
For $\nu=n$, a non-negative integer, the Neumann function has a series
representation which is
\beqa
N_n(x)\equiv -\frac1\pi\lep\frac x2\rip^{-n}\sum_{j=0}^{n-1}
\frac{(n-j-1)!}{j!}\lep \frac x2\rip^{2j}+\frac2\pi\ln(x/2)J_n(x) \nonumber
\\-\frac1\pi\lep\frac x2\rip^n\sum_{j=0}^\infty\lec\leb\ps(j+1)+\ps(n+j+1)
\rib\frac{(x/2)^{2j}(-1)^j}{j!(n+j)!}\ric
\eeqa
where $\ps(y)=d(\ln\Ga(y))/dy$ is known as the {\em digamma or psi
function}.
Finally, for some purposes it is more useful to use {\em Bessel functions
of the third kind}, also called {\em Hankel functions}; these are given by
\beq
H_\nu^{(1)}=\jnx+i\nnx
\eeq
and
\beq
H_\nu^{(2)}=\jnx-i\nnx.
\eeq
It is not easy to see what are the properties of the various kinds of
Bessel functions from the expansions we have written down so far. As it
turns out, their behavior is really quite simple; many of the important
features are laid bare by their behavior at small and large arguments. For
$x<<1$ and real, non-negative $\nu$, one finds
\beq
\jnx=\frac1{\Ga(\nu+1)}\lep\frac x2\rip^\nu[1-O(x^2)]
\eeq
and
\beq
\nnx=\lec\barr{cc}
\frac2\pi[\ln(x/2)+0.5772+...] & \nu=0 \\
-\frac{\Ga(\nu)}{\pi}\lep\frac2x\rip^\nu & \nu\ne0 \ear\right.
\eeq
For $x>>>1,\;\nu$,
\beq
\jnx\sim\sqrt{\frac2{\pi x}}\cos\lep x-\frac{\nu\pi}{2}-\frac\pi4\rip
\eeq
and
\beq
\nnx\sim\sqrt{\frac2{\pi x}}\sin\lep x-\frac{\nu\pi}2-\frac\pi4\rip\,.
\eeq
\centerline{\psfig{figure=bessels.ps,height=3.0in,width=4.0in}}
Notice in particular that as $x\rightarrow0$, the Bessel functions of the
first kind are well-behaved (finite) whereas the Neumann functions are
singular; $N_0(x)$ has only a logarithmic singularity while the
higher-order functions are progressively more singular. At large (real)
argument, on the other hand, both $\jnx$ and $\nnx$ are finite and
oscillatory. Hence, the Bessel functions (by which we mean the $\jnx$) of
non-negative order are
allowable as solutions of the Laplace equation at all values of $\rh$; the
Neumann functions, on the other hand, are not allowable on a domain which
includes the point $\rh=0$.
Bessel functions of all kinds satisfy certain recurrence relations. It is a
straightforward if tedious matter to show by direct substitution of the
series expansions that they obey the following:
\beq
\Om_{\nu+1}-\frac{2\nu}{x}\Om_\nu+\Om_{\nu-1}=0
\eeq
and
\beq
\Om_{\nu+1}+2\der{\Om_\nu}{x}-\Om_{\nu-1}=0.
\eeq
By taking the sum and difference of these relations, we find also
\beq
\Om_{\nu\pm1}=\frac\nu x\Om_\nu\mp\der{\Om_\nu}{x}.
\eeq
These are valid for all three kinds of Bessel functions.
The Bessel function $\jnx$ can form a complete orthogonal set on an interval
$0\le x\le x_0$ in much the same way as the sine function $\sin(n\pi x/x_0)$
does (Note that $x_0$ is a zero of the sine function.) Similarly, let us
denote the $n^{th}$ zero of $\jnx$ by $x_{\nu n}$ and then form the
functions $J_\nu(x_{\nu n}y)$, with $n=1$,2,... . Then it turns out that
for fixed $\nu$, these functions provide a complete orthogonal set on the
interval $0\le y\le1$. As usual, we shall not demonstrate completeness.
Orthogonality can be demonstrated by making use of the Bessel equation and
recurrence relations. It is a useful exercise to do so. Start from the
Bessel equation for $\jxy$,
\beq
\frac1y\der{}{y}\lep y\der{\jxy}{y}\rip+\lep x^2-\frac{\nu^2}{y^2}\rip\jxy=0
\eeq
Multiply this equation by $y\jxpy$ and integrate from 0 to 1:
\beq
\int_0^1dy\lec\jxpy\der{}{y}\lep y\der\jxy y\rip+y\jxpy\lep\x^2-\frac
{\nu^2}{y^2}\rip\jxy\ric=0
\eeq
or if we integrate the first term by parts:
\beqa
\jxpy y\left.\der\jxy y\ril_0^1-\int_0^1dy\der\jxpy y y\der\jxy y +
\nonumber \\
\int_0^1dyy\jxpy\jxy\lep\x^2-\frac{\nu^2}{y^2}\rip=0
\eeqa
Similarly, if we start from the differential equation for $\jxpy$ and
perform the same manipulations, we find the same equation with $x$ and $x'$
interchanged. Subtract the second equation from the first to find
\beq
J_\nu(x')x\left.\der{J_\nu(u)}{u}\ril_x-J_\nu(x)x'\left.\der{J_\nu(u)}
{u}\ril_{x'}+(x^2-x'^2)\int_0^1ydy\jxy\jxpy=0
\eeq
If we let $x=\xnn$ and $x'=\xnnp$, two distinct zeros of the Bessel
function, then the integrated terms vanish and we may conclude that the
Bessel functions $\jnxy$ and $\jnxpy$ are orthogonal when integrated over
$y$ from 0 to one, provided a factor of $y$ is included in the integrand.
We still have to determine normalization in the case $n=n'$. In the
preceding equation, let $x'=\xnn$ and rearrange the terms to have
\beq
-J_\nu(x)\xnn\left.\der{J_\nu(u)}{u}\ril_{\xnn}=-\int_0^1dy\,y(x^2-\xnn^2)
\jnxy\jxy,
\eeq
or
\beq
\int_0^1dy\,y\jnxy\jxy=\frac{J_\nu(x)\xnn(dJ_\nu(u)/du)|_{\xnn}}{x^2-\xnn^2}.
\eeq
Use L'H\^opital's Rule to evaluate the limit of this expression as $x
\rightarrow\xnn$:
\beq
\int_0^1dy\,y[\jnxy]^2=\frac12\left.\der{J_\nu(x)}{x}\ril_{\xnn}\left.
\der{J_\nu(u)}{u}\ril_{\xnn}=\frac12[J_\nu'(\xnn)]^2,
\eeq
where the prime ' denotes a derivative with respect to argument.
Now employ the recurrence relation
\beq
xJ_\nu'(x)=\nu J_\nu(x)-xJ_{\nu+1}(x)
\eeq
to find $J_\nu'(\xnn)=-J_{\nu+1}(\xnn)$, from which the normalization
integral becomes
\beq
\int_0^1dy\,y[J_\nu(\xnn y)]^2=\frac12[J_{\nu+1}(\xnn)]^2
\eeq
The expansion of an arbitrary function of $\rh$ on the interval $0\le\rh\le
a$ may be written as
\beq
f(\rh)=\sum_{n=1}^\infty A_nJ_\nu(\rh\xnn/a)
\eeq
with coefficients which may be determined from the orthonormalization
properties of the basis functions as
\beq
A_n=\frac{\int_0^a\rh d\rh f(\rh)J_\nu(\rh\xnn/a)}{\frac{a^2}{2}
\leb J_{\nu+1}(\xnn)\rib^2}.
\eeq
This type of expansion is termed a {\em Fourier-Bessel series}. The
completeness relation for the basis functions is
\beq
\sum_{n=1}^\infty\frac{J_\nu(\rh\xnn/a)J_\nu(\rh'\xnn/a)}{(a^2/2)[J_{\nu+1}
(\xnn)]^2}=\de(\rh^2/2-\rh'^2/2)\equiv\frac1\rh\de(\rh-\rh').
\eeq
It is also of importance to consider the case of imaginary $k$, $k=i\ka$
with real $\ka$. Then the functions of $z$ are oscillatory, being of the
form $Z(z)\sim e^{\pm i\ka z}$, and the functions of $\rh$ will be Bessel
functions of imaginary argument, \eg $J_\nu(i\ka\rh)$. For given $\nu^2$,
there are two linearly independent solutions which are conventionally
chosen to be $J_\nu$ and $H_\nu^{(1)}$, the reason being that they have
particularly simple behaviors at large and small arguments. Let us
introduce the {\em modified Bessel functions} $I_\nu(x)$ and $K_\nu(x)$,
\beqa
I_\nu(x)\equiv i^{-\nu}J_\nu(ix) \\
K_\nu(x)\equiv\frac\pi 2i^{\nu+1}H_\nu^{(1)}(ix).
\eeqa
These have the forms at small argument, $x<<1$,
\beq
I_\nu(x)=\frac1{\Ga(\nu+1)}\lep\frac x2\rip^\nu
\eeq
and
\beq
K_\nu(x)=\lec\barr{cc} -\ln(x/2)+0.5772+... & \nu=0 \\
\frac{\Ga(\nu)}2\lep\frac2x\rip^\nu & \nu\ne0 \ear \right.
\eeq
while at large argument, $x>>>1,\nu$
\beq
I_\nu(x)=\frac1{\sqrt{2\pi x}}e^x\leb1+O\lep\frac1x\rip\rib
\eeq
and
\beq
K_\nu(x)=\sqrt{\frac{\pi}{2x}}e^{-x}\leb1+O\lep\frac1x\rip\rib.
\eeq
\centerline{\psfig{figure=bessels2.ps,height=2.5in,width=4.0in}}
\noindent From these equations we see that $I_\nu(x)$ is well-behaved
for $x<\infty$,
corresponding to $\rh<\infty$, while $K_\nu(x)$ is well-behaved for
$x>0$ or $\rh>0$, from which we can decide which function(s) to use in
expanding any given potential problem.
Let's look at some examples of expansions in cylindrical coordinates.
\subsection{Example I}
Consider a charge-free right-circular cylinder bounded by S given by
$\rh=a$, $z=0$, and $z=c$. Let $\Phx$ be zero on S except for the top face
$z=c$ where $\Ph(\rh,\ph,c)=V(\rh,\ph)$ with $V$ given.
\centerline{\psfig{figure=cyl1.ps,height=2.5in,width=5.0in}}
For this distribution of boundary potential, we need a complete set of
functions of the space $0\le\ph\le 2\pi$ and $0\le\rh\le a$. Thus we
take ($k$ real) $Z(z)$ to be damped exponentials ($\sinh$ and $\cosh$)
and $R(\rho)$ to be ordinary Bessels functions.
\beq
\Ph\sim e^{im\ph}\lep A J_m(k\rho) + B N_m(k\rho)
\rip \lep \cos(kz) \pm \sin(kz) \rip
\eeq
It will be
convenient if each of these functions is equal to zero when $\rh=a$ and
also if each one is zero when $z=0$. With just a little thought,
\begin{enumerate}
\item No Neumann functions $N_m$ since they diverge at $\rho=0$
\item No $\cosh$ since it is finite at $z=0$, and hence would not
satisfy the B.C.
\item Since $J_m$ and $J_{-m}$ are not independent, use $J_{|m|}$.
\end{enumerate}
we realize
that we want to use the hyperbolic sine function of $z$ and the Bessel
function of the first kind for $R$. Our expansion is thus of the form
\beq
\Ph(\rh,\ph,z)=\sum_{m=-\infty}^\infty\sum_{n=1}^\infty A_{mn}e^{im\ph}
J_{|m|}(x_{mn}\rh/a)\sinh(x_{mn}z/a)
\eeq
where $x_{mn}$ is the $n^{th}$ zero of $J_{|m|}(x)$. Each term in the sum is
itself a solution of the Laplace equation; each one satisfies the boundary
conditions on $z=0$ and $\rh=a$, and, for given $n$, we have a complete set
of functions of $\ph$ while for given $m$, we have a complete set of
functions of $\rh$.
The coefficients in the expansion are determined from the condition that
$\Ph$ reduce to the given potential $V$ on the top face of the cylinder.
Making use of the orthogonality of the basis functions of both $\ph$ and
$\rh$, we have
\beqa
\int_0^a\rh d\rh\int_0^{2\pi}d\ph V(\rh,\ph)J_{|m|}(x_{mn}\rh/a)e^{-im\ph}
\\ \nonumber=A_{mn}2\pi(a^2/2)[J_{|m|+1}(x_{mn})]^2\sinh(x_{mn}c/a)
\eeqa
or
\beq
A_{mn}=\frac{\int_0^{2\pi}d\ph\int_0^a\rh d\rh e^{-im\ph}J_{|m|}(x_{mn}\rh/
a)V(\rh,\ph)}{\pi a^2\sinh(x_{mn}c/a)[J_{|m|+1}(x_{mn})]^2}.
\eeq
For any given function $V$, one may now attempt to complete the integrals.
\subsection{Example II}
Consider the same geometry as in the first example but now with boundary
condition $\Ph=0$ on the constant-$z$ faces and some given value $V(\ph,z)$
on the surface at $\rh=a$.
\centerline{\psfig{figure=cyl2.ps,height=3.0in,width=5.0in}}
For this system we need a complete set of functions on the domain $0\le z
\le c$ and $0\le\ph\le 2\pi$ which means picking $k$ imaginary, $k=i\ka$.
The appropriate functions of $z$ are $\sin$ and $\cos$, and the appropriate
functions of $\rho$ are the modified Bessels Functions.
\beq
\Ph\sim e^{i m\ph}\lep \sin(kz) \pm \cos(kz) \rip
\lep A I(k\rho) + B K(k\rho) \rip
\eeq
\begin{enumerate}
\item We may eliminate the $K$ modified Bessels functions since they diverge
when $\rho\to 0$.
\item Since $I_m$ is not independent of $I_{-m}$, we use $I_{|m|}$.
\item The $\cos$ function of $z$ cannot be zero at both $z=0$ and $z=c$,
and so may be eliminated.
\item Take $k=n\pi/c$ so that $\sin(n\pi z/c)|_{z=c}=0$
\end{enumerate}
Thus the expansion is
\beq
\Ph(\rh,\ph,z)=\sum_{m=-\infty}^\infty\sum_{n=1}^\infty A_{mn}e^{im\ph}
\sin(n\pi z/c)I_{|m|}(n\pi\rh/c)
\eeq
with coefficients given by
\beq
A_{mn}=\frac{\int_0^c dz\int_0^{2\pi}d\ph e^{-im\ph}\sin(n\pi z/c)V(\ph,z)
}{\pi cI_{|m|}(n\pi a/c)}
\eeq
\subsection{B.V.P. on Large Cylinders}
By applying the same considerations, one may solve other boundary-value
problems on cylinders. A case of special interest, and requiring special
treatment, is one in which $a\rightarrow\infty$; then $k_{\nu n}\equiv\xnn/a$
becomes a continuous variable and instead of a Fourier-Bessel series, we
come up with an integral. The orthogonality condition is
\beq
\int_0^\infty xdxJ_m(kx)J_m(k'x)=\frac1k\de(k-k')
\eeq
and the completeness relation is the same, with different names for the
variables,
\beq
\int_0^\infty kdkJ_m(kx)J_m(kx')=\frac1x\de(x-x').
\eeq
To see how this comes to be, consider the completeness relation on a finite
interval,
\beq
\sum_{n=1}^\infty\frac{J_m(x_{mn}\rh/a)J_m(x_{mn}\rh'/a)}{(a^2/2)[J_{m+1}(
x_{mn})]^2}=\frac1\rh\de(\rh-\rh')
\eeq
and then let $a\rightarrow\infty$, defining $x_{mn}/a$ as $k$ while noting
that the interval between roots of the Bessel function at large argument is
$\pi$. Also, the asymptotic form of the Bessel function, valid at large
argument, is
\beqa
J_{m+1}(x_{mn})\sim\sqrt{\frac2{\pi x_{mn}}}\cos[n\pi+(m-1/2)\pi/2-(m+1)\pi
/2-\pi/4] \nonumber \\
=\sqrt{\frac2{\pi x_{mn}}}\cos[(n-1)\pi]=-(-1)^n\sqrt{\frac2{\pi
x_{mn}}}.
\eeqa
Using this substitution in the closure relation for the finite interval and
taking the limit of large $a$, one finds that the sum becomes the integral,
\eq{165}. Of course, this is not a rigorous derivation because the
asymptotic expression is not arbitrarily accurate for all roots.
\subsection{Green's Function Expansion in Cylindrical Coordinates}
We can expand the Dirichlet Green's function in cylindrical coordinates in
much the same manner as we did in spherical coordinates. We shall go
through the derivation to expose a somewhat different approach from what we
employed in the latter case. We consider a domain between two infinitely
long right-circular cylindrical surfaces. Then $G$ must vanish on these
surfaces and it must also satisfy a Poisson equation,
\beqa
\lap\gxxp=-4\pi\de(\rh^2/2-\rh'^2/2)\de(\ph-\ph')\de(z-z')\nonumber \\
=-\frac{4\pi}{\rh}\de(\rh-\rh')\de(\ph-\ph')\de(z-z').
\eeqa
Let us write the delta functions of $\ph$ and $z$ using closure relations,
\beq
\de(z-z')=\frac1{2\pi}\int_{-\infty}^\infty dke^{ik(z-z')}=\frac1\pi\int_0
^\infty dk\cos[k(z-z')]
\eeq
and
\beq
\de(\ph-\ph')=\frac1{2\pi}\sum_{m=-\infty}^\infty e^{im(\ph-\ph')}.
\eeq
Similarly, expand the $\ph$-dependence and $z$-dependence of $G$ using the
same basis functions,
\beq
\gxxp=\sum_{m=-\infty}^\infty\int_0^\infty\frac{dk}{2\pi^2}g_m(k,\rh,\rh')
e^{im(\ph-\ph')}\cos[k(z-z')].
\eeq
Now operate on this expansion with the Laplacian using cylindrical
coordinates:
\beqa
\lap\gxxp=\sum_{m=-\infty}^\infty\int_0^\infty\frac{dk}{2\pi^2}
\lep\sde{}{\rh}+\frac1\rh\der{}{\rh}-\leb\frac{m^2}{\rh^2}+k^2\rib\rip g_m
e^{im(\ph-\ph')}\cos[k(z-z')]\nonumber\\=-\frac{4\pi}{\rh}\de(\rh-\rh')
\sum_{m=-\infty}^\infty\int_0^\infty\frac{dk}{2\pi^2}e^{im(\ph-\ph')}
\cos[k(z-z')].
\eeqa
Multiply by members of the basis sets, \ie $e^{-im'(\ph-\ph')}$ and
$\cos[k'(z-z')]$ and integrate over the appropriate intervals of $\ph-\ph'$
and $z-z'$ to find a differential equation for $g_m$,
\beq
\sde{g_m}{\rh}+\frac1\rh\der{g_m}{\rh}-\lep k^2+\frac{m^2}{\rh^2}\rip g_m
=-\frac{4\pi}{\rh}\de(\rh-\rh').
\eeq
For $\rh\ne\rh'$, this is Bessel's equation with solutions (viewed as
functions or $\rh$) which are Bessel functions of imaginary argument, or,
as we have described, modified Bessel functions of argument $k\rh$.
Because of the delta function inhomogeneous term, the solution for
$\rh<\rh'$ is different from the solution for $\rh>\rh'$. Hence we may
write that, for $\rh<\rh'$,
\beq
g_m(k,\rh,\rh')=A_<(\rh')K_m(k\rh)+B_<(\rh')I_m(k\rh),
\eeq
and, for $\rh>\rh'$,
\beq
g_m(k,\rh,\rh')=A_>(\rh')K_m(k\rh)+B_>(\rh')I_m(k\rh).
\eeq
The various coefficients are functions of $\rh'$ and must in fact be linear
combinations of $K_m(\rh')$ and $I_m(\rh')$ because $\gxxp$ is also a
solution of $\nabla'^2\gxxp=0$ when $\rh\ne\rh'$; another way to see this same
point is to recall that $\gxxp=G(\xp,\x)$. Finally, the coefficients are
further constrained by the condition that the Green's function must vanish
when $\rh$ becomes equal to the radius of either the inner or outer
cylinder.
Let us at this point restrict our attention to a special (and simple)
limiting case which is the infinite space. The radius of the inner cylinder is
0 and that of the outer one becomes infinite in this limit. Then we have to
have a function $g_m$ which remains finite as $\rh\rightarrow0$ which can
only be $I_m(k\rh)$; also, we must have $g_m$ vanish as $\rh\rightarrow
\infty$, which can only be $K_m(k\rh)$. Thus we have
\beq
g_m(k,\rh,\rh')=\lec\barr{cc}
A_<(\rh')I_m(k\rh) & \rh<\rh' \\
A_>(\rh')K_m(k\rh) & \rh>\rh'. \ear\right.
\eeq
The symmetry condition on $G$ tells us that $A_<(\rh')=AK_m(k\rh')$ while
$A_>(\rh')=AI_m(k\rh')$. All of these conditions are included in the
statement
\beq
g_m(k,\rh,\rh')=AI_m(k\rh_<)K_m(k\rh_>)
\eeq
where $\rh_<$ ($\rh_>$) is the smaller (larger) of $\rh$ and $\rh'$.
The remaining constant in the determination of $g_m$ can be found from the
required normalization of $G$. Let us integrate \eq{173} across the point
$\rh=\rh'$:
\beq
\int_{\rh'-\ep}^{\rh'+\ep} d\rho \lep\sde{}{\rh}+\frac1\rh\der{}{\rh}-
\leb k^2+\frac{m^2}{\rh^2}\rib\rip g_m=-\frac{4\pi}{\rh'}
\eeq
If we take the limit of $\ep\rightarrow0$ and realize that $g_m$ is
continuous while its first derivative is not, we find that this equation
gives
\beq
\lim_{\ep\rightarrow0}\left.\der{g_m}{\rh}\ril_{\rh'-\ep}^{\rh'+\ep}
=A\leb I_m(k\rh')k\left.\der{K_m(x)}{x}\ril_{k\rh'}-K_m(k\rh')k\left.
\der{I_m(x)}{x}\ril_{k\rh'}\rib=-\frac{4\pi}{\rh'},
\eeq
or
\beq
A[I_m(x)K_m'(x)-K_m(x)I_m'(x)]=-\frac{4\pi}{x};
\eeq
the primes denote derivatives with respect to the argument $x$.
The quantity in [...] here is the {\em Wronskian} of $I_m$ and $K_m$. One
may learn by consulting, \eg the section on Bessel functions in Abramowitz
and Stegun, that Bessel functions have simple Wronskians:
\beq
I_m(x)K_m'(x)-K_m(x)I_m'(x)\equiv W[I_m(x),K_m(x)]=-\frac1x.
\eeq
Comparison of the two preceding equations leads one to conclude that
$A=4\pi$. Hence our expansion of $\gxxp$, which is just $1/|\x-\xp|$,
is
\beq
\gxxp=\xxpi=\frac2\pi\sum_{m=-\infty}^\infty\int_0^\infty dke^{im(\ph-\ph')
}\cos[k(z-z')]I_m(k\rh_<)K_m(k\rh_>)
\eeq
which may also be written entirely in terms of real functions as
\beqa
\xxpi=\frac4\pi\int_0^\infty dk\cos[k(z-z')]\nonumber \\
\times\lec\frac12I_0(k\rh_<)K_0(k\rh_
>)+\sum_{m=1}^\infty\cos[m(\ph-\ph')]I_m(k\rh_<)K_m(k\rh_>)\ric.
\eeqa
This turns out to be a useful expansion of $1/|\x-\xp|$; it also provides a
starting point for the derivation of some other equally useful expansions.
For example, if we let $\xp=0$, then $\rh_<=0$ and all $I_m$ vanish except
for $m=0$, while $1/|\x-\xp|=1/|\x|=1/\sqrt{\rh^2+z^2}$, so we find, using also
$I_0(0)=1$,
\beq
\frac1{\sqrt{\rh^2+z^2}}=\frac2\pi\int_0^\infty dk\cos(kz)K_0(k\rh).
\eeq
Other useful identities may be obtained.
\edo
**