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\begin{center}

{\bf Solutions of Laplace's Equation and others in Spherical
Co-ordinates}

\end{center}

${}$

$\tri^2V=0 \hspace{1.3in}$ Laplace's Equation

$\ds\tri^2V=\frac{1}{c^2}\frac{\p^2V}{\p t^2} \hspace{0.9in}$ Wave
Equation

$\ds\tri^2V=\frac{1}{K}\frac{\p V}{\p t} \hspace{0.95in}$
Diffusion Equation

$\tri^2\psi+\{l-v(x,y,z)\}\psi=0 \hspace{0.1in}$ Wave Mechanics
Equation

$\ds\tri^2=\frac{1}{r^2}\frac{\p}{\p r}r^2\frac{\p}{\p r}+
\frac{1}{r^2\sin\theta}\frac{\p}{\p\theta}\sin\theta\frac{\p}{\p\theta}
+\frac{1}{r^2\sin^2\theta}\frac{\p^2}{\p\phi^2}$

${}$

{\bf Axially Symmetric Solutions of $\tri^2V=0$}

$\ds\frac{1}{R}=\frac{1}{|\vec{r}-\vec{r_0}|}$ is a solution of
Laplaces equation in the coordinates $x,y,z$.


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Consider $\ds\frac{1}{r}=
\frac{1}{(r^2+c^2-2cr\cos\theta)^\frac{1}{2}}=
\frac{1}{(x^2+y^2+(z-c)^2)^\frac{1}{2}}$

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We have $\ds\tri^2\left(\frac{1}{R}\right)=0$ and so
$\ds\frac{\p^n}{\p c^n}\tri^2\frac{1}{r}=0$ or
$\ds\tri^2\frac{\p^n}{\p c^n}\frac{1}{R}=0$

i.e. $\ds\frac{1}{n!}\frac{\p^n}{\p c^n}\left(\frac{1}{R}\right)$
is a solution of Laplaces equation and in particular

$\ds\frac{1}{n!}\left[\frac{\p^n}{\p
c^n}\left(\frac{1}{R}\right)\right]_{c=0}$ is a solution.

${}$

[N.B. $\frac{\p}{\p c}(\frac{1}{R})=-\frac{\p}{\p z}(\frac{1}{R})$
and so $[\frac{\p^n}{\p c^n}(\frac{1}{r})]_{c=0}=
(-1)^n\frac{\p^n}{\p z^n}\frac{1}{r}$ i.e. the above solution can
be written $\frac{(-1)^n}{n!}\frac{\p^n}{\p c^n}\frac{1}{r}]$

${}$

$\ds\frac{1}{R}=
\frac{1}{[(r-ce^{i\theta})(r-ce^{-i\theta})]^\frac{1}{2}}$ and
$(r-ce^{i\theta})^{-\frac{1}{2}}$ has a power series expansion in
powers of $c$, which is absolutely convergent for
$\ds\left|\frac{ce^{i\theta}}{r}\right|<1$  i.e. for
$\ds\frac{|c|}{r}<1$ when $\theta$ is real.

Similarly for $(r-ce^{-i\theta})^{-\frac{1}{2}}$.

Therefore $\ds\frac{1}{R}=
\frac{1}{(r-ce^{i\theta})^\frac{1}{2}(r-ce^{-i\theta})^\frac{1}{2}}$
has a power series expansion in $c$ which is also convergent for
$\ds\frac{|c|}{r}<1$ ($\theta$ real) and the coefficient of $c^n$
is $\ds\frac{1}{n!}\left[\frac{\p^n}{\p
c^n}\frac{1}{R}\right]_{c=0}$.

Therefore the coefficient of $c^n$ in the above expansion of
$\frac{1}{R}$ in powers of $c$ is a solution of Laplace's equation

$\ds\frac{1}{R}= \frac{1}{r\left[1-\frac{2c}{r}\cos\theta+
\frac{c^2}{r^2}\right]^\frac{1}{2}}$ and so $\ds\frac{1}{R}=
\frac{1}{r}\sum_{n=0}^\infty\left(\frac{c}{r}\right)^nP_n(\cos\theta)$
for $\ds\frac{|c|}{r}<1$ and $\theta$ real.

Thus $\ds\frac{P_n(\cos\theta)}{r^{n+1}}$ is a solution of
Laplace's equation and is axially symmetric.  Similarly, by
considering the expansion of $\frac{1}{R}$ for positive powers of
$\frac{1}{c}$ with $|c|>r, \,\, \theta$ real, we find that
$r^nP_n(\cos\theta)$ is also an axially symmetric solution.

${}$

{\bf Alternative Argument}

If $V$ is a solution of $\tri^2V=0$, homogeneous and of degree
$n$, then $\ds\frac{V}{r^{2n+1}}$ is also a solution of
$\tri^2V=0$ of degree $-(n+1)$.

{\bf Proof}

$\tri^2Vr^m=r^m\tri^2V+V\tri^2r^m+2\vec{\tri}V\vec{\tri}r^m=
r^m\tri^2V+Vm(m+1)r^{m-2}+2mr^{m-1}\frac{\p V}{\p r}$

and since $V$ is homogeneous of degree $n$, $\ds r\frac{\p V}{\p
r}=nV$.  So if $\tri^2V=0$, then

$\tri^2Vr^m=r^{m-2}[m(m+1)+2mn]V=0$ if $m=0,-2n-1$.

From $\ds\frac{1}{(r^2+c^2-2rc\cos\theta)^\frac{1}{2}}=
\frac{1}{r}\sum_{n=0}^\infty\left(\frac{c}{r}\right)^n
P_n(\cos\theta) \hfill (|c|<r, \,\, \theta$ real)

Putting $r=1, \,\, c=h, \,\, \cos\theta=\mu$, we have the
definition of the $P_n$'s.

$\ds\frac{1}{(1-2\mu h+h^2)\frac{1}{2}}=\sum_{n=0}^\infty
h^nP_n(\mu) \hfill (|h|<1, \mu$ real $-1\leq\mu\leq1)$

This expansion is valid for all $h$ and $\mu$, where
$|h|<|\mu\pm(\mu^2-1)^\frac{1}{2}|$, since

$(\mu+(\mu^2-1)\frac{1}{2}-h)(\mu-(\mu^2-1)^\frac{1}{2}-h)=1-2\mu
h+h^2$

${}$

{\bf Properties of ${\bf P_n(\mu)}$}

\begin{itemize}
\item[1)]
$P_n(\mu)$ is a polynomial in $\mu$ of degree $n$, in alternate
powers $n, n-2, \cdots 1$ or $0$.  i.e.$P_{2n}(\mu)$ contains even
powers and is an even function.  $P_{2n+1}(\mu)$ contains odd
powers and is an odd function.

\item[2)]
$P_n(1)=1 \hspace{0.3in} P_n(-1)=(-1)^n$

\item[3)]
$|P_n(\mu)|\leq1 \hspace{0.3in} -1\leq\mu\leq1$

\item[4)]
Legendre's Equation

$P_n(\mu)$ is a solution of
$\ds\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu}w+n(n+1)w=0$

\item[5)]
Orthogonal Properties

$\ds\int_{-1}^1P_n(\mu)P_m(\mu)d\mu=\left\{\begin{array}{cc}
0&m\not=n\\ \frac{2}{2n+1}&m=n\end{array}\right.$

\item[6)]
Rodriguez's Formula

$\ds P_n(\mu)=\frac{1}{2^nn!}\frac{d^n}{d\mu^n}(\mu^2-1)^n$

\item[7)]
Recurrence Formulae

$(n+1)P_{n+1}(\mu)-(2n+1)\mu P_n(\mu)+nP_{n-1}(\mu)=0$

$P'_{n+1}(\mu)-P'_{n-1}(\mu)=(2n+1)P_n(\mu)$

$(\mu^2-1)P_n'(\mu)=n[\mu P_n(\mu)-P_{n-1}(\mu)]$

\end{itemize}

{\bf Proofs}

\begin{itemize}
\item[1)]
Write $c_0=1$, $\ds c_r=\frac{1.3\cdots 2r-1}{2.4\cdots 2r}=
\frac{\Gamma\left(r+\frac{1}{2}\right)}
{r!\Gamma\left(\frac{1}{2}\right)}$

Then $\ds(1-z)^{-\frac{1}{2}}=\sum_0^\infty c_rz^r \hspace{0.2in}
|z|<1$

$\ds\frac{1}{(1+h^2-2\mu
h)^\frac{1}{2}}=\frac{1}{[(1-he^{i\theta})(1-he^{-i\theta})]^\frac{1}{2}}$

where $\cos\theta=\mu \hspace{0.2in} (\theta$ real if
$-1\leq\mu\leq1$, but we don't assume this).

So $\ds\frac{1}{(1+h^2-2\mu h)^\frac{1}{2}}=\sum_0^\infty c_r
h^re^{ir\theta}\sum_0^\infty c_sh^se^{is\theta} \hfill
|h|<|\mu\pm(\mu^2-1)^\frac{1}{2}|$

$P_n(\mu)$= coefficient of $h^n$ on RHS

$=c_nc_0e^{in\theta}+ c_{n-1}c_1e^{i(n-1)\theta}e^{-i\theta}+
c_{n-2}c_2e^{i(n-2)\theta}e^{-i2\theta}+\cdots+
c_0c_ne^{-in\theta}$

$=c_0c_n[e^{in\theta}+e^{-in\theta}]+
c_1c_{n-1}[e^{i(n-2)\theta}+e^{-i(n-2)\theta}]+\cdots$

$\hspace{0.5in}+ \left\{\begin{array}{ll} \frac{c_n^2}{2} & n {\rm
\ even\ }\\
\frac{c_{n-1}}{2}{c_{n+1}}{2}(e^{i\theta}+e^{-i\theta}) & n {\rm \
odd\ }\end{array}\right.$

$=2c_0c_n\cos n\theta+2c_1c_{n-1}\cos(n-2)\theta+
\cdots+\left\{\begin{array}{l} \frac{c_n^2}{2}\\
\frac{c_{n-1}}{2}{c_{n+1}}{2}\cos\theta \end{array}\right.$

$\cos n\theta=$ polynomial in $\cos\theta$ of degree $n$, in
alternate powers

$n, n-2, \cdots 0$ or 1 for odd and even.

Therefore $P_n(\mu)$ is a polynomial in $\mu$ of degree $n$, in
alternate powers $n, n-2 \cdots$


\item[2)]
Putting $\mu=1$ we have $\ds\frac{1}{1-h}=\sum_0^\infty
h^nP_n(1)$, therefore $P_n(1)=1$

Since $P_n(-\mu)=(-1)^nP_n(\mu) \hspace{0.2in} P_n(-1)=(-1)^n$

Values of $P_0, P_1, P_2, P_3$:

$\ds\frac{1}{[1+(h^2-2\mu h)]^\frac{1}{2}}=\sum_0^\infty c_r(2\mu
h-h^2)^r$

$=1+hc_1(2\mu-h)+h^2c_2(4\mu^2-4\mu h+h^2)+h^3c_3(8\mu^3+\cdots)$

Therefore

$P_0(\mu)=1$

$P_1(\mu)=2c_1\mu=\mu$

$P_2(\mu)=-c_1+4c_2\mu^2=-\frac{1}{2}+\frac{4.1.3}{2.4}\mu^2=
\frac{3\mu^2-1}{2}$

$P_3(\mu)=\frac{5}{2}\mu^3-\frac{3}{2}\mu$


\item[3)]
From $P_n(\mu)=2c_0c_n\cos n\theta+\cdots$

$|P_n(\mu)|\leq2c_0c_n|\cos n\theta|+\cdots$

As $c_0c_1\cdots$ are all positive.  If $-1\leq\mu\leq1
\hspace{0.2in} \mu\cos\theta$ is real and $|\cos n\theta|\leq1$.
Therefore $|P_n(\mu)|\leq2c_0c_n+\cdots=P_n(1)=1$


\item[4)]
Legendre's equation

We have the result that $r^nP_n(\cos\theta)$ is a solution of
$\tri^2V=0$ in spherical polar co-ordinates.

Therefore $\ds\frac{1}{r^2}\frac{\p}{\p r}\left\{r^2\frac{\p}{\p
r}r^nP_n(\cos\theta)\right\}+\frac{1}{r^2\sin\theta}
\frac{\p}{\p\theta}\sin\theta\frac{\p}{\p\theta}r^n
P_n(\cos\theta)=0$

Therefore $\ds n(n+1)r^{n-2}P_n(\mu)+
r^{n+2}\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu}P_n{\mu}=0$

Therefore $P_n(\mu)$ satisfies
$\ds\frac{d}{d\mu}(1-\mu^2)\frac{dw}{d\mu}+n(n+1)w=0$

[N.B.  This equation has solutions linearly independent of
$P_n(\mu)$ since it is of the second order.  These solutions are
unbounded at $\mu\pm1$ corresponding to $\theta=0$ or $\pi$ (i.e.
the 2-axis).

\item[5)]
Orthogonal Property

\begin{itemize}
\item[i)]
$\ds\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu}P_n(\mu)+
n(n+1)P_n(\mu)=0$

\item[ii)]

$\ds\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu}P_m(\mu)+
m(m+1)P_m(\mu)=0$
\end{itemize}

$(i)P_m(\mu)-(ii)P_n(\mu)$ gives

$\ds\frac{d}{d\mu}(1-\mu^2)\left\{P_m\frac{d}{d\mu}P_n-
P_n\frac{d}{d\mu}P_m\right\}+ [n(n+1)-m(m+1)]P_mP_n=0$

So $\ds (n-m)(n+m+1)\int_{-1}^1P_mP_nd\mu+
[(1-\mu^2)(P_mP_n'-P_nP_m')]_{-1}^1=0$

Therefore $\ds\int_{-1}^1P_mP_nd\mu=0 \hspace{0.3in} m\not=n$

${}$

Value of $\ds\int_{-1}^1P_n^2(\mu)d\mu$

$\ds\frac{1}{(1-2\mu h+h^2)}=\left[\sum_0^\infty
h^nP_n(\mu)\right]^2$

$\ds\int_{-1}^1\frac{d\mu}{1-2\mu h+h^2}=
\int_{-1}^1\left[\sum_0^\infty h^nP_n(\mu)\right]^2d\mu$

LHS=$\ds\left[-\frac{1}{2h}\log(1-2\mu
h+h^2)\right]_{-1}^1=\frac{1}{2h}\log\frac{(1+h)^2}{(1-h)^2}$

$\ds\hspace{0.2in}=\frac{1}{h}\log\frac{1+h}{1-h}=
2\sum_0^\infty\frac{h^{2n}}{2n+1}$

RHS=$\ds\int_{-1}^1\sum_{n=0}^\infty h^nP_n(\mu)\sum_{m=0}^\infty
h^mP_m(\mu)d\mu$

$\ds\hspace{0.2in}=\sum_{n=0}^\infty h^n\int_{-1}^1P_n(\mu)
\sum_{m=0}^\infty h^mP_m(\mu)d\mu$

$\ds\hspace{0.2in}=\sum_{n=0}^\infty h^n\sum_{m=0}^\infty h^m
\int_{-1}^1P_n(\mu)P_m(\mu)d\mu$

$\ds\hspace{0.2in}=\sum_{n=0}^\infty h^{2n}
\int_{-1}^1P_n^2(\mu)d\mu \hspace{0.2in}$ therefore
$\ds\int_{-1}^1P_n^2(\mu)d\mu=\frac{2}{2n+1}$


\item[6)]
Rodriguez's Formula

$P_m$ is perpendicular to $P_n, \hspace{0.1in} m\not=n,
\hspace{0.1in} P_n(1)=1, \hspace{0.1in} P_n(\mu)$ is of degree
$n$. Define $F(\mu)$ of degree $2n$ such that
$F^{(n)}(\mu)=P_n(\mu)$

$F(1)=F'(1)=\cdots=F'(n-1)=0$.

In fact $F(\mu)=
\frac{1}{(n-1)!}\int_1^\mu(\mu-\lambda)^{n-1}P_n(\lambda)d\lambda$

\begin{itemize}
\item[i)]
$F(\mu)$ has a zero of order $n$ at $\mu=+1$

\item[ii)]
We show from the orthogonal properties that $F(\mu)$ has a zero of
order $n$ at $\mu=-1$.
\end{itemize}

Assuming this we have

$F(\mu)=(\mu-1)^n(\mu+1)^n*$ poly. of degree 0

$=c(\mu-1)^n(\mu+1)^n=c(\mu^2-1)^n$

Therefore $\ds P_n(\mu)=c\frac{d^n}{d\mu^n}(\mu^2-1)^n$

To find $c$ let $\mu=1$

$\ds1=c\left[\frac{d^n}{d\mu^n}(\mu-1)^n(\mu+1)^n\right]_{\mu=1}=
cn!2^n$

Therefore $\ds c=\frac{1}{n!2^n}$ using Leibniz theorem.

${}$

Proof of (ii)

Since $P_0,P_1\cdots$ are linearly independent polynomials, any
polynomial $f(\mu)$ of degree $r$ can be expressed uniquely as
$c_0P_0+c_1P_1+\cdots+c_rP_r$.

$P_n$ is perpendicular to $P_r, \hspace{0.1in} r<n$, therefore
$P_n$ is perpendicular to any polynomial of degree $r<n$.  In
particular $P_n$ is perpendicular to $(1+\mu)^r, \hspace{0.1in}
r<n$.

i.e. $\ds\int_{-1}^1P_n(\mu)(1+\mu)^rd\mu=0, \hspace{0.2in} r<n$

i.e. $\ds\int_{-1}^1F^{(n)}(\mu)(1+\mu)^rd\mu=0,\hspace{0.2in}r<n$

Denote this by $I_{n,r}$, where $I_{n,r}=0$ for $0<r<n$

$\ds I_{n,r}=-r\int_{-1}^1F^{(n-1)}(\mu)(1+\mu)^{r-1}d\mu$

Therefore $I_{n-1,r-1}=0$ for $1<r<n$

By $r$ integrations by parts $\ds\int_{-1}^1F^{(n-r)}(\mu)d\mu=0$

$[F^{(n-r-1)}(\mu)]_{-1}^1=0 \hspace{0.1in}
F^{(n-r-1)}(1)=F^{(n-r-1)}(-1)=0 \hspace{0.1in} r=0,1,\cdots n-1$

Hence (ii) follows.

${}$

Suppose $f(\mu)$ has derivatives of all orders in $[-1,1]$.

$\ds\int_{-1}^1f(\mu)P_n(\mu)d\mu=
\frac{1}{2^nn!}\int_{-1}^1f(\mu)D^n(\mu-1)^nd\mu$

$\ds\hspace{0.2in}=\frac{1}{2^nn!}
\int_{-1}^1[-f'(\mu)]D^{n-1}(\mu^2-1)^nd\mu+0$

$\ds\hspace{0.2in}=\frac{(-1)^n}{2^nn!}
\int_{-1}^1f^{(n)}(\mu)(\mu^2-1)d\mu=\frac{1}{2^nn!}
\int_{-1}^1(1-\mu^2)^nf^{(n)}(\mu)d\mu$

e.g. $\ds\int_{-1}^1
P_n^2(\mu)=\frac{1}{2^nn!}\int_{-1}^1(1-\mu^2)^nD^nP_n(\mu)d\mu=
\frac{2}{2n+1}$


\item[7)]
Recurrence Formulae

We have $\ds G(\mu,h)\equiv(1-2\mu h+h^2)^{-\frac{1}{2}}=
\sum_0^\infty h^nP_n(\mu)$

(a) $\ds\frac{\p G}{\p h}=\frac{\mu-h}{(1-2\mu h+h^2)}G$

i.e. $\ds(1-2\mu h+h^2)\sum_1^\infty nP_n(\mu)h^{n-1}=
(\mu-h)\sum_0^\infty h^nP_n(\mu)$

Equating coefficients of $h^n$ on each side

$(n+1)P_{n+1}(\mu)-2\mu nP_n(\mu)+(n-1)P_{n-1}(\mu)=\mu
P_n(\mu)-P_{n-1}(\mu) \hfill (i)$

$2P_2(\mu)-2\mu P_1(\mu)=\mu P_1(\mu)-P_0(\mu) \hfill (ii)$

$P_1(\mu)=\mu P_0(\mu) \hfill (iii)$

(i) gives $(m+1)P_{m+1}(\mu)-(2n+1)\mu P_n(\mu)+nP_{n-1}(\mu)=0$

(b) $\ds\frac{\p g}{\p\mu}=\frac{h}{(1-2\mu h+h^2)}G
\hspace{0.3in} \frac{\p G}{\p h}=\frac{\mu-h}{(1-2\mu h+h^2)}G$

$\ds\left(\frac{1}{h}-h\right)\frac{\p G}{\p\mu}-2h\frac{\p G}{\p
h}=\frac{1-h^2-2h(\mu-h)}{(1-2\mu h+h^2)}G=G$

Therefore $\ds\left(\frac{1}{h}-h\right)\frac{\p
G}{\p\mu}=\left(2h\frac{\p}{\p h}+1\right)G$

$\ds\left(\frac{1}{h}-h\right)\sum_1^\infty
h^nP_n'(\mu)=\sum_0^\infty(2n+1)h^nP_n(\mu)$

therefore $P_{n+1}'(\mu)-P_{n-1}'(\mu)=(2n+1)P_n(\mu)$

$P_2'(\mu)=3P_1(\mu)$

This formula gives

$\ds\int_1^{\mu}P_n(\lambda)d\lambda=
\frac{1}{2n+1}\left\{P_n+1(\mu)-P_{n-1}(\mu)\right\}$

(c) $\ds\frac{\mu^2-1}{h}\frac{\p G}{\p\mu}-(\mu-h)\frac{\p G}{\p
h}=-G$

Therefore $\ds(\mu^2-1)P_n'(\mu)=n(\mu P_n(\mu)-P_{n-1}(\mu))$

$\ds\hspace{0.2in}=
n\left\{\frac{(n+1)P_{n+1}(\mu)+nP_{n-1}(\mu)}{2n+1}-
P_{n-1}(\mu)\right\}$

$\ds\hspace{0.2in}=\frac{n(n+1)}{2n+1}\{P_{n+1}(\mu)-P_{n-1}(\mu)\}$

Differentiating the above gives

$\ds\frac{d}{d\mu}(\mu^2-1)\frac{dP_n(\mu)}{d\mu}=
\frac{n(n+1)}{2n+1}\{P_{n+1}'(\mu)-P_{n-1}'(\mu)\}=n(n+1)P_n(\mu)$

which is Legendre's equation.

[Note that $\ds\frac{\p}{\p\mu}(\mu^2-1)\frac{\p
G}{\p\mu}=h\frac{\p}{\p h}\left(h\frac{\p}{\p h}+1\right)G$ and
this leads to the differential equation.]

${}$

Zeros of $P_n(\mu)$

$\ds P_n(\mu)=\frac{1}{2^nn!}\frac{d^n}{d\mu^n}(\mu^2-1)^n$

$(\mu^2-1)^n$ has $n$ zeros at $-1$ and $n$ zeros at $+1$.

therfore $\ds\frac{d}{d\mu}(\mu^2-1)^n$ has $n-1$ zeros at $-1$,
$n-1$ zeros at $+1$ and therefore one (say $\mu$) in $(-1,1)$, as
it has $2n-1$ altogether.

Continuing this process $\ds\frac{d^n}{d\mu^n}(\mu^2-1)^n$ has $n$
zeros in $-1<\mu<1$, all simple.

\end{itemize}

${}$

{\bf Axially Symmetric Potentials (in spherical co-ordinates)}


DIAGRAM


Let $U$ be a solution of $\tri^2U=0$, existing in $a\leq r\leq b$
and axially symmetric about $Oz$.

If $(r,\theta,\phi)$ are spherical polar co-ordinates, where
$\theta=0$ and $\theta=\pi$ is the same $z$-axis, then $U$ has the
form $\ds\sum_0^\infty
\left(A_nr^n+\frac{B_n}{r^{n+1}}\right)P_n(\cos\theta)$

on $\theta=0$ this becomes $\ds
U(r,0)=\sum_0^\infty\left(A_nr^n+\frac{B_n}{r^{n+1}}\right)$.

Conversely if $U(r,0)$ has this form and exists in $a\leq r\leq b$
then

$\ds U(r,\theta)=\sum_0^\infty
\left(A_nr^n+\frac{B_n}{r^{n+1}}\right)P_n(\cos\theta)$

${}$

Example


DIAGRAM



$\ds U=\int\frac{dS}{|\vec{r}-\vec{r_0}|}$

where $\vec{r}$ is the position vector of the field point, and
where $\vec{r_0}$ is the position vector of a point on the disc,
and the integral is taken over the disc with boundary $r=C, \,\,
\theta=\alpha$, referred to $Oz$.


DIAGRAM


$\ds U(r,0)=2\pi\int_)^a\frac{pdp}{(p^2+h^2)^\frac{1}{2}}=
2\pi[\sqrt{a^2+h^2}-|h|]=2\pi(R-h)$

$h=c\cos\alpha-r$

Therefore
$U(r,0)=2\pi[(r^2+c^2-2rc\cos\alpha)^\frac{1}{2}-(c\cos\alpha-r)]$

$(r^2+c^2-2rc\cos\alpha)^\frac{1}{2}=
(r^2+c^2-2rc\cos\alpha)(r^2+c^2-2rc\cos\alpha)^{\frac{1}{2}}$

$\hspace{0.5in}=(r^2+c^2-2rc\cos\alpha) \left\{\begin{array}{ll}
\ds\sum_{n=0}^\infty\frac{c^n}{r^{n+1}}P_n(\cos\alpha) & r<c\\
\ds\sum_{n=0}^\infty\frac{r^n}{c^{n+1}}P_n(\cos\alpha) &
r>c\end{array}\right.$

${}$

For $r>c$, putting $\lambda=\cos\alpha$,

$\ds\frac{(r^2+c^2-2rc\cos\alpha)^\frac{1}{2}}{r}=
\left(1+\frac{c^2}{r^2}-2\lambda\frac{e}{r}\right)
\sum_0^\infty\left(\frac{c}{r}\right)^nP_n(\lambda)$

$\ds\hspace{0.2in}=P_0(\lambda)+\frac{c}{r}\{P_1(\lambda)-2\lambda
P_0(\lambda)\}+\sum_{r=2}^\infty\left(\frac{c}{r}\right)^n
\{P_n(\lambda)-2\lambda P_{n-1}(\lambda)+P_{n-2}(\lambda)\}$

Therefore $(r^2+c^2-2rc\cos\alpha)^\frac{1}{2}-(r-c\lambda)$

$\ds\hspace{0.2in}=r\left(1+\frac{c}{R}(-\lambda)+\sum_{n=2}^\infty
\{P_n(\lambda)-2\lambda P_{n-1}(\lambda)+P_{n-2}(\lambda)\}\right)
-(r-c\lambda)$

$\hfill (r>c\Rightarrow r>c\lambda)$

$\ds\hspace{0.2in}=r\sum_{n=2}^\infty\frac{c^n}{r^n}F_n(\lambda),
\hspace{0.3in} F_n(\lambda)=P_n(\lambda)-2\lambda
P_{n-1}(\lambda)+P_{n-2}(\lambda)$

Therefore $\ds\frac{U(r,0)}{2\pi}=
\sum_{n=2}^\infty\frac{c^n}{r^{n-1}}F_n(\lambda)=
\sum_{n=0}^\infty\frac{c^{n+2}}{r^{n+1}}F_{n+2}(\lambda) \hfill
r>c$

Therefore $\ds\frac{U(r,0)}{2\pi}=
\sum_{n=0}^\infty\frac{c^{n+2}}{r^{n+1}}
P_n(\cos\theta)F_{n+2}(\cos\alpha)$

[For large $r$, RHS$\approx\frac{c^2}{r}F_2(\cos\alpha)$

$F_2(\cos\alpha)=P_2(\cos\alpha)-2\cos\alpha
P_1(\cos\alpha)+P_0(\cos\alpha)$

$\ds\hspace{0.5in}=\frac{3\cos^2\alpha-1}{2}-2\cos^2\alpha+1=
\frac{1}{2}-\frac{1}{2}\cos^2\alpha=\frac{1}{2}\sin^2\alpha$

Therefore $\ds U(r,\theta)\sim\frac{\pi
c^2\sin^2\alpha}{r}=\frac{\pi a^2}{r}$ as $r\to\infty$]

$F_{n+2}(\lambda)=P_{n+2}-2\lambda P_{n+1}+P_n$

$\ds\hspace{0.6in}=P_{n+2}+P_n-2
\left\{\frac{(n+2)P_{n+2}+(n+1)P_n}{2n+3}\right\}$

$\ds\hspace{0.6in}=\frac{-P_{n+2}+P_n}{2n+3}=
\frac{(1-\lambda^2)P_{n+1}'(\lambda)}{(n+1)(n+2)}$

${}$

{\bf Example of Boundary Problem}

To find a potential $V$ existing in $0\leq r\leq a$ such that
$V+U=0$ on $r=a$ where $U(r,\theta)$ is the potential considered
above.

$V(r,\theta)$ must be of the form
$\ds\sum_{n=0}^\infty\frac{r^n}{a^n}A_nP_n(\cos\theta)$

Hence we require $\ds\sum_0^\infty
A_nP_n(\cos\theta)+U(a,\theta)=0$

$\ds\frac{U(a,\theta)}{2\pi}=\sum_{n=0}^\infty\frac{c^{n+2}}{a^{n+1}}
P_n(\cos\theta)F_{n+2}(\lambda)$

Therefore $\ds\frac{V}{2\pi}=-\sum_0^\infty
r^n\frac{c^{n+2}}{a^{2n+1}}P_n(\cos\theta)F_{n+2}(\lambda)$.

${}$

{\bf Definition - Solid Harmonic of degree ${\bf n}$}

If $f(x,y,z)$ is a polynomial in $x,y,z$ homogeneous and of degree
$n$, and if $\tri^2f=0$, then $f$ is said to be a solid harmonic
of degree $n$.

Example: $1; \,\, x,y,z; \,\, yz,zx,xy,z^2-x^2,z^2-y^2,$ etc.

${}$

{\bf Definition - Surface Harmonic of degree ${\bf n}$}

If $f(x,y,z)=r^nS_n({\bf u})$ or $r^nS_n(\theta,\phi)$ where $S_n$
depends only on the unit vector ${\bf u}$ along the position
vector, or on the spherical polar angles $\theta,\phi,\,\,\, S_n$
is called a surface harmonic of degree $n$.

${}$

{\bf Differential Equation satisfied by ${\bf S_n}$}

Substitute $f=r^nS_n$ in $\tri^2f=0$.

$\ds\left\{\frac{1}{r^2}\frac{\p}{\p r}r^2\frac{\p}{\p r}+
\frac{1}{r^2\sin\theta}\frac{\p}{\p\theta}\sin\theta\frac{d}{d\theta}+
\frac{1}{r^2\sin^2\theta}\frac{\p^2}{\p\phi^2}\right\}f=0$

Therefore
$\ds\left\{\frac{1}{\sin\theta}\frac{\p}{\p\theta}\sin\theta\frac{d}{d\theta}+
\frac{1}{\sin^2\theta}\frac{\p^2}{\p\phi^2}+n(n+1)\right\}S_n=0
\hfill (1)$

This equation admits solutions of the form $s(\theta)e^{\pm
im\phi}, \,\, m$ constant, where

$\ds\frac{1}{\sin\theta}\frac{d}{d\theta}\sin\theta\frac{ds}{d\theta}+
\left(n(n+1)-\frac{m^2}{\sin^2\theta}\right)s=0$

Putting $\cos\theta=\mu$, so
$\ds\frac{1}{\sin\theta}\frac{d}{d\theta}=-\frac{d}{d\mu}$ this
becomes

$\ds\frac{d}{d\mu}(1-\mu^2)\frac{ds}{d\mu}+
\left(n(n-1)-\frac{m^2}{1-\mu^2}\right)s=0 \hfill (2)$

Equation (2) is called Legendre's associated equation.  For $m=0$
it reduces to Legendre's equation.  In this case $S_n$ is
independent of $\phi$, i.e. is axially symmetric.  One solution is
$P_n(\mu)$.

${}$

{\bf The number of linearly independent Surface Harmonics of
degree ${\bf n}$ is ${\bf 2n+1}$}

$f$ can always be written

$\ds f=\phi_n(x,y)+\frac{z}{1!}\phi_{n-1}(x,y)+
\frac{z^2}{2!}\phi_{n-2}(x,y)+\cdots$

where $\phi_r$ is a homogeneous polynomial in $x,y$ of degree $r$.

$\ds\tri^2f=\tri_1^2+\frac{\p^2}{\p z^2}=
(\tri_1^2\phi_n+\phi_{n-2})+\frac{z}{1!}(\tri_1^2\phi_{n-1}+\phi_{n-3})$

$\ds\hspace{0.7in}+\cdots+
\frac{z^{n-3}}{(n-3)!}(\tri_1^2\phi_3+\phi_1)+
\frac{z^{n-2}}{(n-2)!}(\tri_1^2\phi_2+\phi_0)$

Since this must vanish identically

$\tri_1^2\phi_n+\phi_{n-2}=0 \hspace{0.7in}
\tri_1^2\phi_{n-1}+\phi_{n-3}=0$

$\tri_1^2\phi_{n-2}+\phi_{n-4}=0\cdots \hspace{0.5in}
\tri_1^2\phi_{n-3}+\phi_{n-5}=0\cdots$

Therefore $\phi_n, \,\, \phi_{n-1}$ are arbitrary polynomials in
$x,y$ of degrees $n$ and $n-1$, and for the others we have

$\phi_{n-2r}=(-1)^r(\tri_1^2)^r\phi_n$

$\phi_{n-2r-1}=(-1)^r(\tri_1^2)^r\phi_{n-1}$

Therefore

$\ds f=\phi_n-\frac{z^2}{2!}\tri_1^2\phi_n+
\frac{z^4}{4!}(\tri_1^2)^2\phi_n-\cdots+\frac{z}{1!}\phi_{n-1}-
\frac{z^3}{3!}(\tri_1^2)\phi_{n-1}+
\frac{z^5}{5!}(\tri_1^2)^2\phi_{n-1}-\cdots$

where both series must terminate.

$\phi_n$ can have any one of the forms $x^n, x^{n-1}y,\cdots y^n$.
There are $n+1$ of these, and they are linearly independent.

$\phi_{n-1}$ can have any one of the forms $x^{n-1},
x^{n-2}y,\cdots y^{n-1}$. There are $n$ of these, and they are
linearly independent.

Therefore the total number of forms is $2n+1$ and the
corresponding $f$'s are linearly independent.

${}$

{\bf Associated Legendre Functions [Ferrer's definition]}

$\ds P_n^m(\mu)=(1-\mu^2)^\frac{m}{2}\frac{d^m}{d\mu^m}P_n(\mu)=
(1-\mu^2)^\frac{m}{2}\frac{1}{2^nn!}\frac{d^{m+n}}{d\mu^{m+n}}(\mu^2-1)^n$

is the associated Legendre function of the first kind of degree
$n$, order $m$.  There are $n+1$ such functions for $m=0,1,2\cdots
n$.

We show

\begin{itemize}
\item[i)]
$r^ne^{\pm mi\phi}P_n^m(\cos\theta)=$ polynomial in $x,y,z$ of
degree $n$.

\item[ii)]
$\tri^2r^ne^{\pm mi\phi}P_n^m(\cos\theta)=0$

${}$

\item[i)]
$P_n^m(\cos\theta)=\sin^m\theta$ [Poly. in
$\cos\theta;\cos^{n-m}\theta\cdots\left\{\begin{array}{cl}
\cos\theta & n-m {\rm\ odd}\\ 1 & n-m {\rm\
even}\end{array}\right.$

$r^ne^{im\phi}P_n^m\cos\theta=(r\sin\theta)^me^{im\phi}r^{n-m}[\cdots]$

$\hspace{0.1in}=(x+iy)^m$ [Poly in
$z,r^2;z^{n-m},z^{n-m-2}r^2\cdots \left\{\begin{array}{cl}
zr^{n-m-1} & n-m {\rm\ odd}\\ r^{n-m} & n-m {\rm\
even}\end{array}\right.$

$\hspace{0.2in}= {\rm poly\ in\ } x,y,z,\sin\theta,\,\,\,
r^2=x^2+y^2+z^2$

\item[ii)]
To show $\ds \tri^2r^ne^{\pm
im\phi}(1-\mu^2)^\frac{m}{2}\frac{d^m}{d\mu^m}P_n(\mu)=0$

This is so if $\ds e^{\pm
im\phi}(1-\mu^2)^\frac{m}{2}\frac{d^m}{d\mu^m}P_n(\mu)$

satisfies $\ds\left[\frac{\p}{\p\mu}(1-\mu^2)\frac{\p}{\p\mu}+
\left(n(n+1)+\frac{1}{1-\mu^2}\right)\frac{ap^2}{\p\phi^2}\right](\,\,)=0$

i.e. if $(1-\mu^2)^\frac{m}{2}D^mP_n(\mu)$ satisfies

$\ds L(m;w)=\left[\frac{d}{d\mu}(1-\mu^2)\frac{d}{d\mu}+
n(n+1)-\frac{m^2}{1-\mu^2}\right]w=0$

This is know as Legendre's Associated equation.

Now

$\ds L(m:(1-\mu^2)^\frac{m}{2}W)=
\frac{d}{d\mu}\left\{(1-\mu^2)^{\frac{m}{2}+1}\frac{dW}{d\mu}-
m\mu(1-\mu^2)^\frac{m}{2}W\right\}$

$\ds\hspace{0.5in}+\left\{n(n+1)-\frac{m^2}{1-\mu^2}\right\}
(1-\mu^2)^\frac{m}{2}W$

$\ds=(1-\mu^2)^{\frac{m}{2}+1}\frac{d^2W}{d\mu^2}+
(-\mu(m+2)-m\mu)(1-\mu^2)^\frac{m}{2}\frac{dW}{d\mu}$

$\ds\hspace{0.4in}-
mW((1-\mu^2)^\frac{m}{2}-m\mu^2(1-\mu^2)^{\frac{m}{2}-1})$

$\ds\hspace{0.6in}+
\left\{n(n+1)-\frac{m^2}{1-\mu^2}\right\}(1-\mu^2)^\frac{m}{2}W$

$=(1-\mu^2)^\frac{m}{2}L_1(m:W)$

${}$

where $L_1(m:W)$

$\ds\hspace{0.2in}=\left\{(1-\mu^2)\frac{d^2}{d\mu^2}-
2(m+1)|mu\frac{d}{d\mu}+n(n+1)-m(m+1)\right\}W$

${}$

We must now show that

$L_1(m:D^mP_n(\mu))=0$

Since $L_1(m;W)=0 \Rightarrow DL_1(m:W)=0$ we get

$[(1-\mu^2)D^3+(-2\mu D^2-2(m+1)\mu D^2)$

$\hspace{0.5in}+n(n+1)-m(m+1)-2(m+1)D]W=0$

i.e. $[(1-\mu^2)D^2-2\mu(m+2)D+n(n+1)-(m+1)(m+2)]DW=0$

i.e. $L_1(m+1;DW)=0$

i.e. $L_1(m;W)=0 \Rightarrow L_1(m+1;DW)=0$

i.e. $L_1(0,W)=0 \Rightarrow L_1(m;D^mW)=0$

$L_1(0;P_n(\mu))=[(1-\mu^2)D^2-2\mu D+n(n+1)]P_n(\mu)=0$

Therefore $L_1(m;D^mP_n(\mu))=0$ as required.
\end{itemize}

${}$

{\bf General Surface Harmonic of degree ${\bf n}$}

Giving $m$ the values $0,1,\cdots n$ in $r^ne^{\pm
mi\phi}P_n^m(\cos\theta)$ we have

$r^nP_n(\cos\theta), \,\,\, r^ne^{\pm i\phi}P_n'(\cos\theta),
\cdots r^ne^{\pm ni\phi}P_n^n(\cos\theta)$.

These are $2n+1$ in number and are linearly independent (from the
orthogonality of $1, \,\, e^{\pm i\phi} \cdots$ over
$0\leq\phi\leq2\pi$).

Therefore

$\ds S_n=A_0P_n(\mu)+\sum_{m=1}^n
(C_me^{mi\phi}+C_m'e^{-mi\phi})P_n^m(\mu)$

$\ds\hspace{0.2in}=A_0P_n(\mu)+\sum_{m=1}^n(A_m\cos m\phi+B_m\sin
m\phi)P_n^m(\mu)$

${}$

{\bf Solutions of Legendre's equation when ${\bf n}\not=$ integer}



\end{document}