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\begin{document}
\begin{center}
{\bf Bessel Functions}

\end{center}

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{\bf Vibrations of a Membrane}

${}$

The governing equation for the displacement $w(x, y, t)$from the
equilibrium position (the plane $z=0$) is

$\ds \hspace{2in} \nabla_1^2w = \frac{1}{c^2}\frac{\pl ^2w}{\pl
t^2} \hfill(1)$ $$\nabla_1^2 = \frac{\pl^2}{\pl x^2} +
\frac{\pl^2}{\pl y^2}$$

[ The assumptions are:-
\begin{itemize}
\item[(i)] That the action across the element $Delta S$ is a force
$T \Delta S$ perpendicular to $\Delta S$, where $T' \to T$ as
$\Delta S \to 0$
\item[(ii)] the displacement $w$ of any point of the membrane is
purely transverse.
\item[(iii)] that $\ds \left[\left( \frac{\pl w}{\pl x}\right)^2 +
\left( \frac{\pl w}{\pl y} \right)^2 \right]^\frac{1}{2} $ is
small. \hfill]
\end{itemize}

It can be shown that the tension is isotropic at a point at a
point (i.e. independent of the orientation of $\Delta S$),
independently of (ii) and (iii), and for equilibrium or motion
from (ii) it can be shown that $T$ is uniform over the membrane
and $c = T$/ mass/unit area.  We assume $c^2$=constant.  The usual
boundary condition for a finite membrane) is that $w=0$ on the
boundary.

${}$

{\bf Simple Harmonic Vibrations}

${}$

$\ds \hspace{1.75in} w(x,y,t) = W(x,y) \cos (wt + \epsilon)$
\hfill(2)

Then

$\ds \hspace{1.5in} \nabla_1^2W + k^2W =0 \hspace{1in} k^2 =
\frac{w^2}{c^2}$ \hfill(3)

For a circular membrane (complete or annular) we use plane polar
coordinates $r, \, \theta$ $$ \nabla_1^2 = \frac{1}{r}
\frac{\pl}{\pl r} r \frac{\pl}{\pl r} + \frac{1}{r^2}
\frac{\pl^2}{\pl \theta^2}$$ Therefore $$\frac{1}{r} \frac{\pl
}{\pl r} r \frac{\pl W}{\pl r} + \frac{1}{r^2} \frac{\pl^2W}{\pl
\theta^2} + k^2 W =0$$ This is separable.  i.e. we can find
solutions of the form $\ds W(r \theta) = F(r) G(\theta)$ by
substitution we have $$ \frac{1}{F} \frac{1}{r} \frac{d}{dr}
\left( r \frac{dF}{dr} \right) + k^2 + \frac{1}{r^2} \frac{1}{G}
\frac{d^2G}{d\theta^2} =0$$ Therefore

$\ds \hspace{2in} \frac{1}{G}
\frac{d^2G}{d\theta^2}=$constant$=-n^2$\hfill(4)

Therefore $$ G(\theta) = A \cos n \theta + B \sin n \theta$$

$\ds \hspace{1.5in} r \frac{d}{dr} \left( r \frac{dF}{dr} \right)
+ (k^2r^2 - n^2)F =0$ \hfill(5)

${}$

${}$

[Note that $\frac{d}{dr}(r \frac{dF}{dr}) + (k^2r -
\frac{n^2}{r})F =0$ is the self adjoint for ?????????]

${}$

In (5) write $kr = x$ (not the co-ordinate)

$\ds \hspace{1.5in} x \frac{d}{dx} \left(x \frac{dF}{dx}\right) +
(x^2 - n^2)F =0$ \hfill(6)

This is {\bf Bessel's Equation of order ${\bf n}$}.

${}$

The solution of the original equation $\nabla^2 W+ k^2 W=0$ must
be periodic in $\theta$ of period $2\pi$, for otherwise $W$ would
not be  a one valued function of position.  Therefore $n$ is an
integer.

${}$

{\bf Series solution for ${\bf F}$}

${}$

We assume $\ds F \sum_{m=0}^\infty a_mx^{m+c}$ where $c$ is to be
found.

$\ds \left[ \left( x \frac{d}{dx} \right)^2 - n^2 \right]x^{(m+c)}
= [(m+c)^2 - n^2]x^{m+c}$

$\ds \left[ \left( x \frac{d}{dx} \right)^2 - n^2 \right] F =
\sum_{m=0}^\infty a_m[(m+c)^2 - n^2]x^{m+c}$

$\ds  x^2 F = \sum_{m=0}^\infty a_mx^{n+c+2} = \sum_{m=2}^\infty
a_{m-2}x^{m+c}$

Hence we require $\ds \sum_{m=0}^\infty a_m [(m+c)^2 - n^2]
x^{m+c} + \sum_{m=2}^\infty a_{m-2}x^{m+c} =0$

this is true if

$\ds a_0(c^2 - n^2)=0$ Indicial equation

$\ds a_1(\overline{c+1}^2 - n^2)=0$

$a_m(\overline{m+c}^2 - n^2) + a_{m-2} \hspace{.2in} m=2, 3,
\ldots$

Since $a_0 \not=0, \, c = \pm n$

In the second, putting $c=n$ $$a_1(2n+1)=0$$

In the third, putting $c=n, \, m = 3, 5, \ldots$ $$a_3 3(2n+3) +
a_1 = 0 \cdots$$

We suppose $n \not= -\frac{1}{2} , -\frac{3}{2}, \ldots$

Then $a_1 = a_3 = \ldots =0$

For $m = 2, 4, 6, \ldots$ in the third relation

$\ds a_2 2(2 + 2n) + a_0 \Rightarrow a_2 = \frac{-a_0}{2^2 \cdot
1(n+1)}$

$\ds a_4 4(4 + 2n) + a_2 \Rightarrow a_4 = \frac{a_0}{2^4 \cdot
1\cdot 2(n+1)(n+2)}$

$\ds a_{2m} = \frac{(-1)^ma_0}{2^{2m} m!(n+1)\ldots(n+m)}$

Hence we have one solution (taking $c=n$)

$\ds a_0 x^n \sum_{m=0}^\infty \frac{(-1)^m}{m!(n+1)\ldots(n+m)}
\left( \frac{x}{2} \right)^{2m}$

The Bessel function $J_n(x)$ is defined by taking $\ds a_0 =
\frac{1}{2^n \Gamma(n+1)}$

\begin{eqnarray*} J_n(x) & = & \left( \frac{x}{2} \right)^n
\frac{1}{\Gamma(n+1)} \sum_{m=0}^\infty
\frac{(-1)^m}{m!(n+1)\ldots(n+m)}\left( \frac{x}{2}\right)^{2m} \\
& = & \sum_{m=1}^\infty \frac{(-1)^m}{m!\Gamma(m+n+1)}\left(
\frac{x}{2}\right)^{2m} \end{eqnarray*}

The series converges for all $x$, by the ratio test.  So $J_n(x)$
is an integral function of $x$.  If $L_n$ is the operation in
Bessels equation $L_{-n}(y) = L_n(y)$ since n appears as $n^2$

$L_n[J_n(x)] =0$ for all n

Therefore $L_{-n} [j_{-n}(x)]=0$ i.e. $L_nJ_{-n}(x)]=0$

Hence $J_{-n} (x) $ is a solution.  $$J_{-n}(x) =
\sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m-n+1}\left(
\frac{x}{2}\right)^{-n+2m}$$ When $n$ is a non-negative integer
$\Gamma(m+n+1)$ is infinite from

$ m = 0, 1, 2, \ldots , n-1$ since $\Gamma(z)$ has poles at $z = 0
, -1, -2, \ldots$  Therefore $$J_{-n}(x) = \sum_{m=n}^\infty
\frac{(-1)^m}{m!\Gamma(m-n+1}\left( \frac{x}{2}\right)^{-n+2m}$$


Write $m = n+k$ $$J_{-n}(x) = \sum_{k=o}^\infty
\frac{(-1)^k}{(n+k)!k!}\left( \frac{x}{2}\right)^{2k+n} = (-i)^n
J_n(x)$$

${}$

{\bf Wronskin of $J_n$ and $J_{-n}$}

${}$

$$W = \left| \begin{array}{cc} J_n & J_{-n} \\ J'_n & J'_{-n}
\end{array} \right|$$

${}$

we have $\ds \frac{d}{dx} \left( x \frac{d}{dx} j_{\pm n}\right) +
\left(x - \frac{n^2}{x}\right) J_{\pm n}=0$

Therefore $\ds \frac{d}{dx} \left\{ x [J'_n  J_{-n} - J_n J'_{-n}]
\right\} = 0$

i.e. $\ds x[J'_n  J_{-n} - J_n J'_{-n}] = C$ (constant).

$\ds C = \lim_{x \to 0} x[J'_n  J_{-n} - J_n J'_{-n}]$
\begin{eqnarray*} J_n & = & \frac{\left( \frac{x}{2}
\right)^n}{r(n+1)} \left[ 1 -
\frac{\left(\frac{x}{2}\right)^2}{1!(n+1)} + \cdots \right] \\
J_{-n} & = & \frac{\left( \frac{x}{2} \right)^{-n}}{r(-n+1)}
\left[ 1 - \frac{\left(\frac{x}{2}\right)^2}{1!(-n+1)} + \cdots
\right]
\\ xJ'_n & = & \frac{n\left( \frac{x}{2}
\right)^n}{r(n+1)} \left[ 1 + 0x^2 \right] \\ xJ'_{-n} & = &
\frac{-n\left( \frac{x}{2} \right)^{-n}}{r(-n+1)} \left[ 1 + 0x^2
\right] \end{eqnarray*}

Therefore $\ds J_{-n} \cdot x J'_n - xJ'_{-n} J_n =
\frac{2n}{\Gamma(1+n)\Gamma(1-m)}[1+0x^2]$

${}$

Therefore $\ds C = \frac{2n}{\Gamma(1+n)\Gamma(1-m)} =
\frac{2}{\Gamma (1-n) \Gamma(n)} = \frac{2\sin n\pi}{\pi}$

${}$

Therefore $\ds J_{-n} \cdot x J'_n - xJ'_{-n} J_n =\frac{2}{x}
\frac{\sin n \pi}{\pi}$

Thus $J_n$ and $J_{-n}$ are linearly independent when $n$ is not
an integer.

and $J_n$ and $J_{-n}$ are linearly dependent when $n$ is an
integer.

${}$

{\bf Definition of the second solution [Weber]}

${}$

$$Y_n(x) = \frac{\cos n \pi J_n - J_{-n}}{\sin n \pi}$$

when $n$ tends to an integer the numerator tends to 0 nd the
denominator tens to 0.  Hence when $m$ is an integer we define
$\ds Y_m(x) = \lim_{n\to m} Y_n(x)$ Therefore \begin{eqnarray*}
Y_m(x) = \left. \frac{\frac{\pl}{\pl n} [\cos n \pi J_n -
J_{-n}]}{\frac{\pl}{\pl n} \sin n \pi} \right\}_{n=m} & = &
\left.\frac{\pi\sin n \pi J_n  + \cos n \pi \frac{\pl}{\pl n} -
\frac{\pl}{\pl n} J_{-n}}{\pi \cos \pi n}\right\}_{n=m} \\ & = &
\frac{1}{\pi} \left[\frac{\pl}{\pl n} (J_n) - (-1)^m
\frac{\pl}{\pl n} (J_{-n})\right]_{n=m}
\end{eqnarray*}

In particular $\ds Y_0(x) =
\frac{1}{\pi}\left[\left(\frac{\pl}{\pl n} J_n\right)_{n=0} -
\left(-\frac{\pl}{\pl n} J_n\right)_{n=0} \right] = \frac{2}{\pi}
\left[ \frac{\pl}{\pl n} J_n\right]_{n=0}$


The functions of the second kind of order $n$.  They are unbounded
at $x=0$, for fro the relation between $J_n$ and and other
solution of Bessel's equations, say $Y_n(x)$, we have

$\ds x[Y_nJ'_n - Y'_nJ_n] = C$

$\ds Y_nJ'_N - Y' _n J_n = \frac{c}{x}$

${}$

If $c \not=0$ then $Y_n$ and $Y'_n$ can not exist at x=0.  Hence
the general solution of Bessels equations is

either $A_1J_n(x) + B_1J_{-n}(x)$ (n not an integer)

or $A_1J_n(x) + B_1Y_{n}(x)$ (all cases)

A solution bounded at $x=0$ is necessarily $AJ_n(x)$

${}$

{\bf Returning to the membrane problem}

${}$

\begin{enumerate}
\item Complete membrane ($0 \leq r \leq a$).  Since $W$ must be
bounded at $r=0$, we have $F(r) = AJ_n(kr)$

and $W(r\theta) = AJ_n(kr)\cos(n\theta+\varepsilon)$ (n integer
$\geq 0$)

Since $W=0$ on $r=a$.  $AJ_n(ka)=0$

$A=0$ is trivial therefore $J_n(ka)=0$.

This is an equation for the eigenvalues $k_1^2, \, k_2^2, \ldots$.
Hence for a given $n$ the values of $k$ are given by $ka =
j_{n1},\, j_{n2}, \ldots$  where $j_{n1},\, j_{n2}, \ldots$ are
the positive zeros of $J_n(x)$.  The allowed frequencies
(frequencies of normal modes) are $\ds \frac{w}{a2pi}, \, \, w =
kc$  Therefore $$w = \frac{c}{a}(j_{n1},\, j_{n2}, \ldots)
\hspace{.2in} n = 0, 1, \ldots$$

\item Annular Membrane $b \leq r \leq a$.

In this case $r=0$ is not in the \lq \lq physical space \rq \rq.
Therefore we must take $F(r) = AJ_n(kr) +BY_n(kr)$

$F(a) =0 \, \, \, \, F(b) = 0$ give:- \begin{eqnarray*}  AJ_n(ka)
+BY_n(ka) & = & 0 \\  AJ_n(kb) +BY_n(kb) & = & 0 \end{eqnarray*}
Hence for non-trivial A, B $$\left| \begin{array}{cc} AJ_n(ka) &
BY_n(ka)  \\  AJ_n(kb) & BY_n(kb) \end{array} \right|=0$$
\end{enumerate}

${}$

\underline{ Sketch of $J_0\,  J_1\, J_2$}

${}$

$\ds J_0 = 1 = \left( \frac{x}{2}\right)^2 \cdot \frac{1}{(1!)^2}
+ \left( \frac{x}{4}\right)^4 \frac{1}{(2!)^2} + \cdots$

$\ds J_1 = \left( \frac{x}{2} \right) -\left( \frac{x}{2}\right)^3
\frac{1}{1!2!} + \left( \frac{x}{2} \right)^5 \frac{1}{2!3!}$

Note also $J'_0(x) = -J_1(x)$

${}$

PICTURE

${}$

The asymptotic formula for $J_n(x)$ is $\ds J_n(x) \sim \left(
\frac{2}{\pi x} \right)^\frac{1}{2}\cos\left( x - \frac{\pi}{4} +
n\frac{\pi}{2}\right)$

${}$

${}$

{\bf Orthogonal and normal Properties of ${\bf J_0\, \, (j_mx)}$}

${}$

where $j1, \, j_2\ldots$ are the zeros of $j_0(x)$.

We show that \begin{eqnarray*} \int_0^1xJ_0(j_mx)J_0(j_px) & = & 0
\hspace{.7in} p \not=m \\ & = & \frac{1}{2} J_1^2(j_m)
\hspace{.2in} p=m \end{eqnarray*} the functions are orthogonal to
weight function $x$ over $[a,b]$.

$\ds \frac{d}{dx} x \frac{d}{dx} J_0(\alpha x) +
\alpha^2xJ_0(\alpha x) =0$

$\ds \frac{d}{dx} x \frac{d}{dx}J_0(\beta x) + \beta^2xJ_0(\beta
x)=0$

$\ds \frac{d}{dx} x \left[J_0(\beta x)\frac{d}{dx}J_0(\alpha x) -
J_0(\alpha x)\frac{d}{dx}J_0(\beta x) \right] = (\alpha^2 -
\beta^2)x J_0(\alpha x) J_0(\beta x)$

Therefore $\ds \left.x\left[J_0(\beta x)\frac{d}{dx}J_0(\alpha x)
- J_0(\alpha x)\frac{d}{dx}J_0(\beta x) \right]\right\}_0^1 $

$I(\alpha^2 - \beta^2) \int_0^1 J_0(\alpha x) J_0(\beta x) \, dx$

i.e. \begin{eqnarray*} (\alpha^2 - \beta^2) \int_0^1 x J_0(\alpha
x) - J_0(\beta x) \, dx & = & J_0(\alpha)\beta J'_0(\beta) -
J_0(\beta) \alpha J'_0(\alpha) \hspace{.4in} (1)\\ & = &
J_0(\beta) J_1(\alpha) - J_0(\alpha) \beta J_1(\beta)
\end{eqnarray*}

If $\alpha = j_m \hspace{.2in} \beta = j_p \hspace{.2in} m \not=p$
$$ \int_0^1 x J_0(j_mx) J_0(j_px) dx =0$$
\begin{eqnarray*} \int_0^1xJ_0^2(\alpha x) & = &  \lim_{\beta \to \alpha}
\frac{-\alpha
 J_0(\beta) J'_0(\alpha) + \beta J_0(\alpha) J_0'
\beta}{\alpha^2 - \beta^2} \\ & = & \left[ \frac{\frac{\pl}{\pl
\beta} {\rm Numerator}}{\frac{\pl}{\pl \beta} {\rm
Denominator}}\right]_{\beta=\alpha} \\ & = & \left[
\frac{-j'_0(\beta) \cdot \alpha J'_0(\alpha) \frac{\pl}{\pl \beta}
(\beta J'_0\beta)}{-2\beta} \right]_{\beta = \alpha} \\ & = &
-\frac{\alpha J_0^{'2} (\alpha) - \alpha J_0^2(\alpha)}{-2\alpha}
\left[ \frac{\pl}{\pl\beta} (\beta J'_0(\beta)) + \beta J_0(\beta)
=0\right] \end{eqnarray*}

Therefore $\ds \int_0^1 x J_o^2(\alpha x) = \frac{1}{2}
J_1^2(\alpha) + J_0^2(\alpha) \hspace{.3in} [J_0' = -J_1]$ \hfill
(2)

Therefore then $\alpha = j_m$

$\ds \int_0^1 x J_0(j_m x) J_0(j_p x) \, dx = \frac{1}{2}
J_1^2(j_m) \delta _{mp}$

It also follows that if $f'_1, \, f'_2 \, , \ldots$ are the zeros
of $J'_0(x)$ then $$ \int_0^1 x J_0(j'_mx)J_0(j'_px) \, dx =
\frac{1}{2} J_0^2 (j_m) \delta _{mp}$$

\newpage

{\bf Special Cases of ${\bf \ds \int_0^1 J_0(\alpha_n x)f(x) dx}$}


${}$

In (1) take $\alpha = \alpha\-n$ ($a$ zero) $\beta \not= \alpha_m
\, \, \, m = 1,2, \ldots$

Then $\ds \int_0^1 xJ_0 (\alpha_n x)J_0(\beta x) \, dx =
\frac{\alpha_n}{\alpha_n^2 - \beta ^2}
J_1(\alpha_n)J_0(\beta)$\hfill (3)

In this put $\beta =0$ and $J_0(0) =1$ and so $\ds
\int_0^1xJ_0(\alpha_n x) \, dx =
\frac{J_1(\alpha_n)}{\alpha_n}$\hfill (4)

(4) can also be obtained as follows:

$\ds \frac{d}{dx} x \frac{d}{dx}J_0(\alpha_n x) = -\alpha_n^2
xJ_0(\alpha_n x)$

Therefore \begin{eqnarray*}-\alpha_n^2 \int_0^1 x J_0(\alpha_n x)
dx & = & \left[x \frac{d}{dx} J_0(\alpha_n x)\right]_0^1 \\ & = &
a_n J'_0(\alpha_n) \\ & = & -\alpha_n J_1(\alpha_n)
\end{eqnarray*} Next we consider $$I_k = \int_0^1 x J_0(\alpha
x)x^k\,dx$$

We have

$\ds \hspace{1in} \frac{d}{dx} x \frac{d}{dx} J_0(\alpha x) =
-\alpha^2xJ_0(\alpha x)$\hfill(i)

$\ds\hspace{1in} \frac{d}{dx} x \frac{d}{dx} x^k = k^2x^{k-1}$
\hfill(ii)

Therefore $$ \frac{d}{dx} x \left\{ x^k \frac{d}{dx} J_0(\alpha x)
- J_0(\alpha x) \frac{d}{dx} x^k\right\} = -J_0(\alpha x) \{ x
\alpha^2 x^k + k^2x^{k-1}\}$$

Therefore by integration over [0,1] $$-\alpha^2I_k - k^2I_{k-2} =
\alpha J'_0(\alpha) - k J_0(\alpha)$$


Therefore $I_0\, I_2, \ldots$ can be found in terms of
$J_0(\alpha)$ and $J_1(\alpha)$.  $I_1$ and $I_�$ can be found in
terms of the \lq \lq Sturve function \rq \rq $$\rightarrow
\int_0^1J_0(\alpha x) dx q J_0(\alpha) J_1(\alpha)$$


\newpage

{\bf Formal Fourier - Bessel  Explanations}

${}$

Assume that $f(x)$, defined in [0,1], possesses and expansion
$$f(x) = \sum_{n=1}^\infty A_n J_0(\alpha_n x)$$ Then
\begin{eqnarray*} \int_0^1 x J_0 (\alpha_m x) f(x) dx & = &  \sum_{n=1}^\infty
A_m \int_0^1 xJ_0(\alpha_n x) J_0(\alpha_m x) dx \\ & = & A_m
\int_0^1 x J_0^2 (\alpha_m x) \, dx \\ & = & A_m
\frac{J_1^2(\alpha _m)}{2} \\ A_m & = & \frac{2}{J_1^2(\alpha_m) }
\int_0^1 x J_0 (\alpha _m) f(x) \, dx
\end{eqnarray*}

${}$

{\bf Initial and Boundary Problem for the Vibrating Membrane}

${}$

We have, for the displacement $w(r,t)$ in radially symmetric
vibrations $$\frac{1}{r} \frac{\pl}{\pl r} \left( r \frac{\pl
w}{\pl r} \right) = \frac{1}{c^2} \frac{\pl^2w}{\pl t^2}
\hspace{.5in} 0 \leq r \leq a$$ $w(0,t) $ exists and $w(a,t) =0$

Also $\ds w(r,0) = f(r) \hspace{.3in} 0 \leq r \leq a
\hspace{.5in} \frac{\pl w} {\pl t} (r,0) =0 \hspace{.3in} 0 \leq r
\leq a$

${}$

Choose $a$ - unit of length and choose unit on time so that $c=1$.
Also replace $r$ by $x$, giving $$\frac{1}{x} \frac{\pl}{\pl x}
x\frac{ \pl w}{\pl x} = \frac{\pl^2w}{\pl t^2}$$ $w(0,t) $ exists
and $w(1,t) =0$

Also $\ds w(x,0) = f(x) \hspace{.5in} \frac{\pl w} {\pl t} (x,0)
=0$

${}$

Assume $\ds w(x,t) = \sum_{n=1}^\infty J_0(\al_nx)\phi_n(t)$

This satisfies the boundary conditions $$\frac{1}{2} J_1^2 (\al
_n) \phi_n(t) = \int_0^1 x w(x,t) J_0(\al_n x) dx$$
\begin{eqnarray*} \frac{1}{2} J_1^2(\al_n)\ddot\phi_n(t) & = &
\int_0^1 xJ_0(\al _n x) \frac{\pl^2w}{\pl t^2}(xt) \, dx \\& = &
\int_0^1 J_0(\al_n x) \frac{\pl}{\pl x} \left( x \frac{\pl w}{\pl
x} \right) \, dx  \\ & = & \left[J_0(\al_n x) x \frac{\pl w}{\pl
x} \right]_0^1 - \int_0^1 x \frac{\pl w}{\pl x} \frac{d}{dx}
J_0(\al _nx) dx \\ & = & 0 + \left[ -w x \frac{d}{dx} J_0(\al _n
x) \right]_0^1 + \int_0^1 w \frac{d}{dx} x \frac{d}{dx} J_0(\al _n
x) \, dx \\ & = & 0 + \int_0^1 w  [-\al_n^2 x J_0(\al_n x)]\, dx
\\ & = & -\al^2 \cdot \frac{1}{2} J_1^2(\al_n) \phi_n(t)
\end{eqnarray*} Therefore $$\ddot \phi(t) + \al_n^2 \phi(t) =0$$
and $$\phi_n(t) = A_n \cos(\al_n(t) + B_n \sin(\al_n t))$$
$\begin{array}{ll} w(x,0) =f(x) \hspace{.2in} & 0 \leq x \leq 1 \\
\ds \frac{\pl}{\pl t} w(x.0) =0 & 0 \leq x \leq i\end{array}$

Therefore \begin{eqnarray*} f(x) & = & \sum_0^\infty J_0(\al_n x)
\phi_n(0) \\ 0 & = & \sum_9^\infty J_0(\al_n x) \dot \phi_n(0) \\
\dot \phi_n = 0, \hspace{.75in} \phi_n(0) \cdot \frac{1}{2}
J_1^2(\al_n) & = & \int_0^1 xJ_0(\al_n x) f(x) \, dx \\ {\rm i.e.\
}B_n=0 \hspace{1.5in} A_n & = & \frac{2}{J_1^2(\al _n)} \int_0^1
xJ_0(\al_n x) f(x) \, dx \end{eqnarray*} Hence we have the series
solution for $w(x,t)$ $$w(x,t) = \sum_1^\infty J_0(\al _n x) \cos
\al_nt \frac{2}{J_1^2(\al_n)} \int_0^1 y J_0(\al_n y) f(y) \, dy$$

\newpage

{\bf Alternative procedure}

${}$

Solutions of the differential equation bounded at $x=0$

are $J_0(kx)[A\cos kt + \sin kt]$

This satisfies the boundary conditions $w(1,t) =0$ if $J_0(k) =0$
thus

$k = \al_1, \, \al_2 , \ldots$

The solution also satisfies $\ds \frac{\pl w}{\pl t} (x0) =0$ if
$B_n=0$.

Formally the series $\ds \sum_{n=1}^\infty A_n J_0(\al_n x) \cos
\al_n t$ satisfies both boundary conditions and the initial
conditions on $\ds \frac{\pl w}{\pl t}$.  Therefore as
$w(x_0)=f(x)$ thus

$f(x) = \sum A_n J_0(\al_n x)$  Therefore $$A_n \cdot \frac{1}{2}
J_1^2(\al_n) = \int_0^1 xJ_0(\al_n x) f(x) \, dx.$$

${}$

\underline{Example}

$$f(x) = 1- \frac{J_0(kx)}{J_0(k)}$$ $k$ real and $J_0(k) \not= 0$
\begin{eqnarray*} \int_0^1 x J_0(\al_n x)J_0(kx) dx & = &
\frac{J_0(k) \al_nJ_1(\al_n) - J_0(\al_n) kJ_1(k)}{\al_n^2 - k^2}
\\ & = & J_0(k) \frac{\al_nJ_1(\al_n)}{\al_n^2 - k^2} \\ k=0 {\rm
\ gives\ \ } \int_0^1 x J_0(\al_n x) dx & = &
\frac{J_1(\al_n)}{\al_n} \\ \int_0^1 x J_0(\al_n x) f(x) dx & = &
J_1(\al_n) \left\{ \frac{1}{\al_n} - \frac{\al_n}{\al_n^2 -
k^2}\right\} \\ & = & \frac{k^2 J_1(\al_n)}{\al_n(k^2 - \al_n^2)}
\end{eqnarray*} Hence in this case: $$ w(x,t) = \sum_{n=1}^\infty
J_0(\al_n x) \cos \al_n t\cdot
\frac{2k^2}{\al_n(k^2-\al_n^2)}\frac{1}{J_1(\al_n)}$$ Since $\al_n
= O(n)$ for large $N$, and $\ds J_1(\al_n) = O\left(
\frac{1}{\al_n^\frac{1}{2}}\right) = O\left(
\frac{1}{n^\frac{1}{2}} \right)$

Then the coefficient of $J_0(\al_n x) \cos(\al_n t)$ is $\ds
O\left(\frac{1}{n^\frac{5}{2}}\right)$

\newpage

{\bf Solution of a linear differential equation by definite
integral (or a contour integral)}

${}$

{\bf Preliminary Remarks}

${}$

Consider the differential equation $$x \phi (D) y + \psi (D) y=0$$
where \begin{eqnarray*} \phi(p) & = & a_0 p^n + a_1 p^{n-1} +
\ldots +a_n \\ \psi(p) & = & b_0p^m + b_1p^{m-1} + \ldots + b_m
\end{eqnarray*} Seek a solution $$y = \int_a^b e^{px} K(p) dp
h\space{.2in}{\rm ot} \hspace{.2in} y = \int_C e^{px} K(p) dp$$
where $K(p)$ is to be found and $a, b$ (or $C$) are also to be
found (and are independent of $x$.)

Then \begin{eqnarray*} \phi(D) y & = & \int_a^b \phi(D) e^{px}
K(p) dp \\ & = & \int_a^b \phi(p) e^{px} K(p) dp \\ \psi(D) y & =
& \int_a^b \psi(p) e^{px} K(p) dp \\ {\rm then\ }\hspace{.75in} x
\phi(D) y & = & \int_a^b x e^{px} \phi(p)K(p) dp \\ & = & \int_a^b
\frac{d}{dp} (e^{px}) \phi(p) K(p) dp \\ & = & \left[ e^{px}
\phi(p) K(p) \right]_a^b - \int_a^b e^{px} \frac{d}{dp} (\phi(p)
K(p) ) dp
\end{eqnarray*} Hence $$x \phi(D) y + \psi(D) y = \left[ e^{px}
\phi(p) K(p)\right]_a^b + \int_a^b e^{px} \left\{ \psi(p) K(p) -
\frac{d}{dp} \phi(p) K(p) dp\right\} dp$$

\begin{itemize}
\item[(i)] Choose $K(p)$ so that the integrand is zero.
i.e.$$\frac{d}{dp} \{ \phi(p) K(p) \} = \psi(p)K(p) =
\frac{\psi(p)}{\phi(p)} \phi(p) K(p)$$ therefore $$\phi(p) k(p) =
C \exp\left\{\int^p \frac{\psi(q)}{\phi(q)} dq\right\}$$

\underline{Note:} if all the zeros of $\phi$ are simple

$\ds \frac{\psi}{\phi} = \sum_1^n \frac{a_r}{q-q_r} \Rightarrow
\int^p\frac{\psi}{\phi} = \log(p - p_r)^{a_r}$

Therefore $\ds K_r = \prod_1^n (p-p_r)^{a_{r-1}}$

\item[(ii)] when $K(p)$ is known we choose $a$ and $b$ (or $C$) so
that

$\ds \left[ e^{px} K(p) \phi(p)\right]_a^b =0$
\end{itemize}

${}$

${}$

{\bf Consider Bessel's equation on order n}

${}$

$$\left\{ \left( x \frac{d}{dx} \right)^2 + x^2 - n^2\right\} y
=0$$ $$x^2 \left\{ \frac{d^2y}{dx^2} + y\right\} + x\frac{dy}{dx}
- n^2y =0$$ Substitute $y = x^nz$ \begin{eqnarray*} x\frac{dy}{dx}
& = &  x^n \left\{ x \frac{d}{dx} + n \right\} \\ \left( x
\frac{d}{dx} \right)^2 y & = & x^n \left\{ x \frac{d}{dx} +n
\right]^2 z \\ & = & x^n \left\{ \left( x \frac{d}{dx}\right)^2 +
2n x \frac{d}{dx} + n^2 \right\}z \\ & =& x^n \left\{ x^2
\frac{d^2}{dx^2} + (2n+1) x \frac{d}{dx} + n^2 \right\} \\ \left\{
\left( x \frac{d}{dx}\right)^2 + x^2 - n^2\right\}y & = & x^n
\left\{ x^2 \frac{d^2}{dx^2} + (2n+1) x \frac{d}{dx} + x^2\right\}
z \\ & = & x^{n+1} \left\{ x \left( \frac{d^2}{dx^2} _1 \right) +
(2n+1) \frac{d}{dx} \right\} z \end{eqnarray*}  Hence $y$
satisfies Bessel's equation if $z$ satisfies $$\left\{ x \left(
\frac{d^2}{dx^2} +1\right) + (2n+1) \frac{d}{dx} \right\} z =0$$
Consider a solution for $z$ of the form $$\int_a^b e^{itx} K(t)
dt$$ Then

$\ds \left\{ x \left( \frac{d^2}{dx^2} +1\right) + (2n+1)
\frac{d}{dx} \right\} \int_a^b e^{itx} K(t) dt$
\begin{eqnarray*}\hspace{.1in} & = & \int_a^b \{ x(1-t^2) + (2n+1)it \} K(t)
e^{itx} dt \\ & = & \left[ \frac{1-t^2}{i} e^{itx} K(t)
\right]_a^b + \int_a^b e^{itx} \left[ -\frac{d}{dt} \left\{
\frac{1-t^2}{i}K(t) \right\} + it (2n+1)K(t) \right] dt
\end{eqnarray*} Hence choose $K(t)$ so that
$$\frac{d}{dt}(1-t^2)K(t) = -(2n-1) t K(t)$$ $$K(t) = c(1-t^2)^{n
- \frac{1}{2}}$$ hence a solution of Bessel's equation is $$ y =
x^n \int_a^b e^{itx} (1-t^2)^{n - \frac{1}{2}} dt$$ if $a$ and $b$
are chosen so that $$[(1-t^1)^{n+\frac{1}{2}}e^{itx}]_a^b =0$$
Suppose $n>-\frac{1}{2}$ and also $x$ is real and positive

[There is no difficulty if $x=z$ is complex]

${}$

{\bf Admissible Pairs of limits}

${}$

(i) $(-1, +1)$

(ii) $(-1, -1+i\infty)$

(iii) $(+1, +1+i\infty)$

PICTURE

${}$

These integrals must be linearly dependent since that are
solutions of a second order equation.  With proper specification
of $(1-t^2)^{n-\frac{1}{2}}$ the relation is 1 = (ii) - (iii)

Consider $$\int e^{itx}(1 -t^2)^{n-\frac{1}{2}} dt$$ round the
contour shown.

PICTURE

We choose that branch of $(1-t2)^{n-\frac{1}{2}}$ which is real
and positive on $AA'$.  i.e. $\arg(1-t)=0, \, \arg(1+t)=0$ on
$AA'$.  AS $t$ passes from $a$ to $b$ round $AB \arg(1-t)$
decreases by $\frac{\pi}{2}$; as $t$ passes from $A'$ to $B'$
round $A'B' \arg(1+t)$ increases by$\frac{\pi}{2}$.  Since the
integrand $e^{itx}(1-t^2)^{n-\frac{1}{2}}$ is now one-valued and
regular on and inside the countour, by Cauchy's Theorem we have
$$\int_{\{A'A\}} = \int_{\{A'A\}}+ \int_{\{B'C'\}}+
\int_{\{CB\}}+\int_{\{C'C\}}+\int_{\{A'B'\}}+\int_{\{BA\}}$$ We
show \begin{itemize} \item[(a)] $\ds \lim_{h \to \infty}
\int_{\{C'C\}} =0$ \item[(b)] $\ds \lim_{\epsilon \to \infty}
\int_{\{A'B'\}} = \lim_{\epsilon\to \infty}\int_{\{BA\}}=0$
\end{itemize} We shall then have $$\int_{-1}^1
e^{itx}(1-t^2)^{n-\frac{1}{2}} dt = \int_{-1}^{-1_i\infty}
e^{itx}(1-t^2)^{n-\frac{1}{2}} - \int_{1}^{1+i\infty}
e^{itx}(1-t^2)^{n-\frac{1}{2}}$$  i.e. $$(i) = (ii) - (iii) $$
since the limits if the three integrals exist as
$\epsilon\to\infty$ if $n> -\frac{1}{2}$ and as $h \to \infty.$ on
$CC'$, $|1 - t^2| =PX \cdot PX' \leq C'X \cdot CX' = h^2 +4$

$|e^{itx}| = |e^{ix(u+ih)}| = e^{-x+h}$

Therefore $\left| \int_{CC'} \right| \leq e^{-xa} (h^2
+4)^{n-\frac{1}{2}} \cdot 2 \to 0$ as $h \to \infty$

on $AB$ we have $t-1 = \epsilon e^{t\theta}$  Therefore $|t-1|
^{n-\frac{1}{2}} = \epsilon^{n - \frac{1}{2}}$

$e^{itx} (t^2-1)^{n-\frac{1}{2}}$ is bounded in the neighbourhood
of $t=1$ with bound $M$ say.

Therefore $\ds |e^{itx} (t^2-1(^{n-\frac{1}{2}}| \leq
M\epsilon^{n-\frac{1}{2}}$

Thus $\ds \left| \int_{AB}\right| \leq M\epsilon^{n-\frac{1}{2}}
\cdot \frac{\pi}{2} \epsilon \frac{M \pi}{2} e^{n-\frac{1}{2}} \to
0 $ as $\epsilon\to 0$

Similarly $\ds \int_{B'A'} \to 0$ as $\epsilon \to 0$ ($n >
-\frac{1}{2}$)

\newpage

Hence we have the following solutions of Bessel's equation

\begin{itemize}
\item[(i)] $\ds x^n \int_0^1 e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $

\item[(ii)] $\ds x^n \int_{-1}^{-1+i\infty}e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $

\item[(iii)] $\ds x^n \int_{+1}^{+1+i\infty}e^{itx}(1 - t^2)^{n-\frac{1}{2}} dt $
\end{itemize}

${}$

${}$

{\bf Series Expansions of (i)}

${}$

$$(i) = x^n \int_{-1}^1 (i-t^2)^{n-\frac{1}{2}} \sum_0^\infty
\frac{(itx)^m}{m!} dt$$ the series for all $x, \, t$ absolutely
and uniformly.
\begin{eqnarray*} (i) & = & x^n \sum_{m=0}^\infty
\frac{(ix)^m}{m!} \int_{-1}^1 t^m(1 - t^2)^{n-\frac{1}{2}} dt \\
\int_{-1}^1 t^m(1 - t^2)^{n-\frac{1}{2}} dt & = & 0 {\rm if\ m\
is\ odd} \\ \int_{-1}^1 t^{2m}(1 - t^2)^{n-\frac{1}{2}} dt & = & 2
\int_{0}^1 t^{2m}(1 - t^2)^{n-\frac{1}{2}} dt \\ & = & 2
\int_{0}^1 u^{m}(1 - u)^{n-\frac{1}{2}}\cdot \frac{1}{2}
u^\frac{1}{2} du \\ & = & \int_{0}^1 u^{m-\frac{1}{2}}(1 -
u)^{n-\frac{1}{2}} du \\ & = & \frac{\Gamma (n+\frac{1}{2}) \Gamma
(m+\frac{1}{2})}{\Gamma (m+n+1)} \\ \Rightarrow (i) & = &x^n
\sum_{m=0}^\infty \frac{(ix)^{2m}}{(2m)!}\frac{\gamma
(n+\frac{1}{2}) \Gamma (m+\frac{1}{2})}{\Gamma (m+n+1)} \\
\frac{\Gamma (m+\frac{1}{2})}{(2m)!} & = & \frac{\Gamma
(\frac{1}{2} )\frac{1}{2} \frac{3}{2} \cdots m - \frac{1}{2}}{1
\cdot 2 \cdot 3 \cdots (2m-1)2m} \\ & = & \frac{\Gamma
(\frac{1}{2})}{2^{2m} \cdot m!} \\ \Rightarrow (i) & = & \Gamma
(n+\frac{1}{2}) \Gamma \left(\frac{1}{2}\right) x^n
\sum_{m=0}^\infty \frac{(-1)^m \left(\frac{x}{2}\right)^{2m}}{m!
\Gamma(m+n+1)} \\ & = & 2^n \Gamma (n+\frac{1}{2}) \Gamma
\left(\frac{1}{2}\right) J_n(x) \Rightarrow J_n(x) \\ & = &
\frac{1}{2^n \Gamma (n+\frac{1}{2}) \Gamma
\left(\frac{1}{2}\right)} \times (i) \end{eqnarray*}

${}$

{\bf Hanbel Functions of order n (Bessel functions of the third
kind)}

${}$

\underline{Definition}
\begin{eqnarray*} H_n^{(1)} (x) & = & -2 \left( \frac{1}{2^n
\Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2}) }\right) \cdot (iii)
\\H_n^{(2)} (x) & = & 2 \left( \frac{1}{2^n
\Gamma (\frac{1}{2}) \Gamma (n + \frac{1}{2}) }\right) \cdot (ii)
\end{eqnarray*} Then $$J_n(x) = \frac{1}{2} (H_n^{(1)}(x) +
H_n^{(2)}(x))$$ we also define $$Y_n(x) = \frac{1}{2i}
(H_n^{(1)}(x) - H_n^{(2)}(x))$$

${}$

{\bf Alternative integral representation of ${\bf H_n^{(1)}(x)}$
and $\bf{ H_n^{(1)}(x)}$}

${}$

In the integral representation for $H_n^{(1)}(x)$, $\arg (1-t) =
-\frac{\pi}{2}$ Therefore we write $(1-t) = \eta e^{-\frac{\pi
i}{2} (= -i\eta}$.  Where $\eta$ goes from 0 to $\infty$ through
real values as $t$ goes from 1 to $1+i\infty$. Thus $$ (1-t)^{n -
\frac{1}{2}} e^{-\frac{\pi i}{2}( n - \frac{1}{2})}$$ Also
\begin{eqnarray*} (1+t) & = & 2 - (1-t) = 2 - \eta e^{-\frac{\pi
i}{2}} = 2(1 + \frac{i \eta}{2}) \\ \Rightarrow (1+t)^{n -
\frac{1}{2}} & = & 2 ^{n - \frac{1}{2}}( 1 + \frac{in}{2}) ^{n -
\frac{1}{2}} \\ \Rightarrow (1-t)^{n - \frac{1}{2}} & = & 2^{n -
\frac{1}{2}} \eta ^{n - \frac{1}{2}} e^{-\frac{\pi i}{2}(n -
\frac{1}{2})} \left( 1 + \frac{i\eta}{2} \right)^{n - \frac{1}{2}}
\\ \\ e^{itx} & = & e^{ix ( 1+ i\eta)} = e^{ix} e^{-x\eta} \\ dt &
= & -e^{-\frac{\pi i}{2} d \eta} \\ \Rightarrow e^{itx}(1 -
t^2)^{n - \frac{1}{2}} dt & = & 2^{n - \frac{1}{2}} e^{i(x -
\frac{n\pi}{2} + \frac{pi}{4})} e^{-x\eta} \eta^{n - \frac{1}{2}}
\left( 1 + \frac{i\eta}{2}\right)^{n - \frac{1}{2}} d\eta \\
\Rightarrow H_n^{(1)} (x) &  = & \frac{2^\frac{1}{2}x^ne^{i(x -
\frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (\frac{1}{2}) \Gamma (n +
\frac{1}{2})} \int_0^\infty e^{-x\eta}\eta ^{n -
\frac{1}{2}}\left( 1 + \frac{i\eta}{2}\right)^{n - \frac{1}{2}}
d\eta \hspace{.2in}(I) \end{eqnarray*}

In the integral for $H_n^{(2)} (x) \arg(t+1) = \frac{\pi}{2}$
Therefore $(t+1) = \eta e^\frac{\pi i}{2} = i\eta$ where $\eta$
goes from 0 to $\infty$ as $t$ goes from -1 to $+1+i\infty$ we get
similarly $$H_n^{(2)} (x)   =  \frac{2^\frac{1}{2}x^ne^{-i(x -
\frac{n\pi}{2} - \frac{\pi}{4})}}{\Gamma (\frac{1}{2}) \Gamma (n +
\frac{1}{2})} \int_0^\infty e^{-x\eta}\eta ^{n -
\frac{1}{2}}\left( 1 - \frac{i\eta}{2}\right)^{n - \frac{1}{2}}
d\eta \hspace{.2in}(I) $$

[Note: when $x$ is real and positive, $H_n^{(1)} (x)$, $H_n^{(2)}
(x)$ are complex conjugates.  Therefore $\frac{1}{2}[H_n^{(1)} (x)
+ H_n^{(2)}(x)]$ is real, so is $\frac{1}{2i}[H_n^{(1)} (x) -
H_n^{(2)}(x)]$.]

${}$

Finally substitute $\eta x = u$ in both integrals.  For real and
positive $u$ goes from 0 to $\infty$ as $\eta$ goes from 0 to
$\infty$

$$H_n^{(1)} (x)   = \left( \frac{2}{x}\right)^\frac{1}{2}
\frac{1}{\Gamma(\frac{1}{2})} \frac{e^{i(x - \frac{n\pi}{2} -
\frac{\pi}{4})}}{\Gamma (n+\frac{1}{2}) } \int_0^\infty e^{-u}u
^{n - \frac{1}{2}}\left( 1 + \frac{iu}{2x}\right)^{n -
\frac{1}{2}} du  $$

$$H_n^{(2)} (x)   = \left( \frac{2}{x}\right)^\frac{1}{2}
\frac{1}{\Gamma(\frac{1}{2})} \frac{e^{-i(x - \frac{n\pi}{2} -
\frac{\pi}{4})}}{\Gamma (n+\frac{1}{2}) } \int_0^\infty e^{-u}u
^{n - \frac{1}{2}}\left( 1 - \frac{iu}{2x}\right)^{n -
\frac{1}{2}} du  $$


${}$

${}$

{\bf Asymptotic Expansions of ${\bf \ds \int_0^\infty e^{-u}u ^{n
- \frac{1}{2}}\left( 1 \pm  \frac{iu}{2x}\right)^{n - \frac{1}{2}}
du }$}

${}$

We consider the case $n=0$ i.e.$$ \int_0^\infty e^{-u}u ^{ -
\frac{1}{2}}\left( 1 \pm  \frac{iu}{2x}\right)^{ - \frac{1}{2}}
du$$ We apply Taylor's formula $$f(t) = \sum_{r=0}^{n-1}
\frac{f^{(r)}(0) t^r}{r!} + \frac{1}{(n-1!)} \int_0^t (t-s)^{n-1}
f^{(n)} (s) ds$$ to the function $\ds (1 -t)^{-\frac{1}{2}}$ this
gives $$ (1 - t)^{-\frac{1}{2}} = \sum_{r=0}^{n-1}
\frac{\frac{1}{2} \cdot \frac{3}{2} \cdots r - \frac{1}{2}}{r!}t^r
+\frac{\frac{1}{2} \cdot \frac{3}{2} \cdots n -
\frac{1}{2}}{(n-1)!} \int_0^t(t-s)^{n-1} (1-s)^{-n - \frac{1}{2}}
ds $$ the last term is $$ \frac{\frac{1}{2} \cdot \frac{3}{2}
\cdots n - \frac{1}{2}}{(n-1)!} t^n \int_0^1 (1-v)^{n-1}(1 -
tv)^{-n - \frac{1}{2}} dv$$ writing $\ds t = \frac{u}{2ix}$
$$\left( 1 - \frac{u}{2ix}\right)^{-\frac{1}{2}} =
\sum_{r=0}^{n-1} \frac{\frac{1}{2} \frac{3}{2} \cdots n -
\frac{1}{2}}{r!} \frac{u^n}{(2ix)^n} + r_n \left( \frac{u}{x}
\right) $$ $$ r_n\left( \frac{u}{x} \right) = \frac{\frac{1}{2}
\frac{3}{2} \cdots n - \frac{1}{2}}{(n-1)!} \frac{u^r}{(2ix)^r}
\int_0^1 (1 - v)^{n-1} \left( 1 - \frac{u}{2ix} v \right) ^{-n -
\frac{1}{2}} dv $$

If $x$ is real and positive $$\left| 1 - \frac{u}{2ix}v \right| =
\left( 1 + \frac{u^2v^2}{4x^2} \right)^\frac{1}{2} \geq 1$$
Therefore \begin{eqnarray*} \left| r_n \left( \frac{u}{x} \right)
\right| & \leq &  \frac{\frac{1}{2} \frac{3}{2} \cdots n -
\frac{1}{2}}{n!} \frac{u^n}{(2x)^n} \int_0^1 (1 - v)^{n-1} \cdot 1
dv \\ & = &\frac{\frac{1}{2} \frac{3}{2} \cdots n -
\frac{1}{2}}{n!} \frac{u^n}{(2x)^n} \end{eqnarray*} Hence

$$ \int_0 ^\infty e^{-u} u ^{-\frac{1}{2}} \left( 1 -
\frac{u}{2ix} \right)^{-\frac{1}{2}} du = \sum_{r=0}^{n-1}
\frac{\frac{1}{2} \frac{3}{2} \cdots r - \frac{1}{2}}{r!}
\frac{1}{(2ix)^r} \int_0^\infty e^{-u} y^{r - \frac{1}{2}} du +
R_n(x)$$

$$R_n(x) = \int_0^\infty e^{-u} u^{-\frac{1}{2}} r_n\left(
\frac{u}{x} \right) du $$

$$ \int_0^\infty e^{-u} u^{r-\frac{1}{2}}du = \Gamma (r+
\frac{1}{2} = \Gamma \left( \frac{1}{2} \right) \frac{1}{2} \cdot
\frac{3}{2} \cdots r - \frac{1}{2}$$

Also \begin{eqnarray*} |R_n(x)| & \leq &  \int_0^\infty e^{-u}
u^{n - \frac{1}{2}}  \left| r_n \frac{u}{x} \right| du \\ & \leq &
\frac{\frac{1}{2} \cdot \frac{3}{2} \cdots n - \frac{1}{2}}{n!}
\frac{1}{(2x)^n} \int_0^\infty e^{-u} u^{n-\frac{1}{2}} du
\end{eqnarray*}

Therefore $$ \int_0^\infty e^{-u} u^{-\frac[1}{2} \left( 1 -
\frac{u}{2ix} \right)^{-\frac{1}{2}} du = \Gamma \left(
\frac{1}{2} \right) \sum_{r=0} ^{n-1} \frac{[\frac{1}{2}
\frac{3}{2} \cdots r - \frac{1}{2}]^2}{r!} \frac{1}{(2ix)^r} +
\bar R_n \bar x$$

Where $\ds |\bar R_n| \leq \frac{[\frac{1}{2} \cdot \frac{3}{2}
\cdots n - \frac{1}{2}]^2}{n!} \frac{1}{(2x)^n} \cdot \lim_{x \to
\infty} x^{n-1} =0$

There the series is the asymptotic expansion of the left hand
side.

$\left({\rm In\ fact\ } R_n = 0\left( \frac{1}{x^n}\right)\right)$

\newpage

{\bf Divergence of the Infinite series}

${}$

The D'Alernbert ratio is $$\left| \frac{(n+\frac{1}{2})^2}{n}
\cdot \frac{1}{2ix}\right| = \frac{1}{2x}
\frac{(n+\frac{1}{2})^2}{n} $$ which tends to infity for and $x$.

${}$

{\bf Asymptotic Expansion of $\bf H_0^{(1)}(x) \, H_0^{(2)}(x)$}

$$H_0^{(1)}(x) \sim \left( \frac{2}{\pi
x}\right)^\frac{1}{2}e^{i(x - \frac{\pi}{4})} \sum_{r=0}^\infty
\frac{\left[ \frac{1}{2} \frac{3}{2} \cdots r -
\frac{1}{2}\right]^2 }{r!} \frac{1}{(2ix)^r}$$

$$H_0^{(2)}(x) \sim \left( \frac{2}{\pi
x}\right)^\frac{1}{2}e^{-i(x - \frac{\pi}{4})} \sum_{r=0}^\infty
\frac{\left[ \frac{1}{2} \frac{3}{2} \cdots r -
\frac{1}{2}\right]^2 }{r!} \frac{1}{(2ix)^r}$$

where the remainder after the term in $\ds \frac{1}{x^{n-1}}$ has
modulas $\leq | {\rm term\ in\ } \frac{1}{x^n}|$

Write

$\ds A(x)  =  \frac{1}{2} \frac{1}{\Gamma (\frac{1}{2})} \left[
\int_0^\infty e^{-u} u^{-\frac{1}{2}}\left(1 -
\frac{u}{2ix}\right)^{-\frac{1}{2}}du+\int_0^\infty e^{-u}
u^{-\frac{1}{2}} \left( 1+
\frac{u}{2ix}\right)^{-\frac{1}{2}}du\right] $

$\ds B(x)  =  \frac{1}{2} \frac{1}{\Gamma (\frac{1}{2})} \left[
\int_0^\infty e^{-u} u^{-\frac{1}{2}}\left(1 -
\frac{u}{2ix}\right)^{-\frac{1}{2}}du-\int_0^\infty e^{-u}
u^{-\frac{1}{2}} \left( 1+
\frac{u}{2ix}\right)^{-\frac{1}{2}}du\right] $

$\ds H_0^{(1)}(x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2}
e^{i(x - \frac{\pi}{4})} [A(x) + iB(x)]$

$\ds H_0^{(2)}(x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2}
e^{-i(x - \frac{\pi}{4})} [A(x) - iB(x)]$

\begin{eqnarray*} J_0(x) & = &  \frac{1}{2}\left( H_0^{(1)}(x) +
H_0^{(2)}(x)\right)\\ & = &  \left( \frac{2}{\pi
x}\right)^\frac{1}{2} \left[A(x) \cos \left( x - \frac{\pi}{4}
\right) -  B(x) \sin \left( x - \frac{\pi}{4}
\right)\right]\end{eqnarray*}

\begin{eqnarray*} Y_0(x) & = &  \frac{1}{2i}\left( H_0^{(1)}(x) -
H_0^{(2)}(x)\right] \\ & = &  \left( \frac{2}{\pi
x}\right)^\frac{1}{2} \left[A(x) \sin \left( x - \frac{\pi}{4}
\right) -  B(x) \cos \left( x - \frac{\pi}{4}
\right)\right]\end{eqnarray*}

The general Bessel function of zero order is

$\ds A_1 J_0(x) + B_1Y_0(x) = C(\cos x J_0() + \sin x Y_0(x))$

From the definitions of $A(x)$ and $B(x)$

$\ds A(x) \sim \sum_{r=0}^\infty \frac{[\frac{1}{2} \frac{3}{2}
\cdots 2r - \frac{1}{2}]^2}{(2r)!} \frac{(-1)^r}{(2x)^{2r}}$

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$\ds B(x) \sim \sum_{r=0}^\infty \frac{[\frac{1}{2} \frac{3}{2}
\cdots 2r + \frac{1}{2}]^2}{(2r+1)!}
\frac{(-1)^{r+1}}{(2x)^{2r+1}}$

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{\bf Zeros of a Bessel Function of zero order}

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The zeros are given by $\ds \cot (x - \frac{\pi}{4} - \al) =
\frac{B(x)}{A(x)}$

$$A(x) = 1+O\left( \frac{1}{x^2}\right) \hspace{.4in} B(x) =
-\frac{1}{8x} + O\left(\frac{1}{x^3}\right)$$ Therefore
$$\frac{A(x)}{B(x)} = -\frac{1}{8x} + O\left(
\frac{1}{x^3}\right)$$ Therefore for large $x$, the zeros are
approximately given by $\cot (x - \frac{\pi}{4} - \al) =0$ i.e.
$\ds x - \frac{\pi}{4} -\al = \left( k + \frac{1}{2} \right) \pi$
k = large integer.

$\ds x = \al + \left( k + \frac{3}{4}\right) \pi$

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{\bf Asymptotic Expansions of ${\bf H_0^{(1)}(x) \, \, \& \, \,
H_0^{(2)}(x)}$}


$\ds H_0^{(1)}(x) \sim\left( \frac{2}{\pi x}
\right)^\frac{1}{2}e^{i(x - \frac{n\pi}{2} - \frac{\pi}{4})}
\sum_{m=0}^\infty \frac{(-1)^m(m,n)}{(2ix)^m}$

$\ds H_0^{(2)}(x) \sim\left( \frac{2}{\pi x}
\right)^\frac{1}{2}e^{-i(x - \frac{n\pi}{2} - \frac{\pi}{4})}
\sum_{m=0}^\infty \frac{(-1)^m(m,n)}{(2ix)^m}$

where $(0,n) =1.$

$\ds (m,n) = \frac{(4n^2 - 1^2)(4n^2 - 3^2) \cdots (4n^2 - (2m -
1)^2)}{2^{2m} m!}$

There expansion are only useful when $x >> n$

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{\bf Bessel Functions of order $\bf {(k+\frac{1}{2}) \, k = 0, 1,
\cdots}$}

We have $$ J_n(x) = \sum_{m=0}^\infty \frac{(-1)^m \left(
\frac{x}{2} \right) ^{n+2m}}{m! \Gamma (n+m+1)}$$

$$ J_\frac{1}{2} (x) = \left(\frac{2}{\pi x}\right)^\frac{1}{2}
\sin x \hspace{.3in}  J_{-\frac{1}{2}} (x) = \left(\frac{2}{\pi
x}\right)^\frac{1}{2} \cos x$$

$\ds J_\frac{1}{2} (x) =
\sum_{m=0}^\infty\left(\frac{x}{2}\right)^\frac{1}{2}\frac{(-1)^m
x^{2m}}{2^{2m}m! \Gamma (m+\frac{3}{2})}$

\begin{eqnarray*} 2^{2m}m! \Gamma (m+\frac{3}{2})  & = &  2^{2m
}m! \Gamma (\frac{3}{2}) \frac{3}{2} \frac{5}{2} \cdots m +
\frac{1}{2} \\ & = & \Gamma (\frac{3}{2}) 2 \cdot 4 \cdots 2m 3
\cdot 5 \cdots (2m+1)  \\ & = & (2m+1)! \Gamma (\frac{3}{2}) \\ &
= & (2m+1)! \frac{1}{2} \Gamma(\frac{1}{2}) \\ \Rightarrow
J_\frac{1}{2} (x) & = & \left( \frac{x}{2} \right)^\frac{1}{2}
\frac{1}{\frac{1}{2} \Gamma \frac{1}{2}}
\sum_{m=0}^\infty\frac{(-1)^m x^{2m}}{(2m+1)!}  \\ &= & \left(
\frac{2x}{\pi} \right)^\frac{1}{2} \frac{\sin x}{x}\\ &  = &
\left( \frac{2}{\pi x} \right)^\frac{1}{2} \sin x \end{eqnarray*}

Similarly $$J_{-\frac{1}{2}} (x) = \left( \frac{2}{\pi x}
\right)^\frac{1}{2} \cos x$$

\begin{eqnarray*} H_\frac{1}{2}^{(1)} & = & -i e^{ix} \left(
\frac{2}{\pi x}\right)^\frac{1}{2} \hspace{,2in}
H_{-\frac{1}{2}}^{(2)} = i e^{-ix} \left( \frac{2}{\pi
x}\right)^\frac{1}{2} \\ H_{k + \frac{1}{2}}^{(1)} (x) & = & -2
\left( \frac{x}{2} \right)^{k+ \frac{1}{2} } \frac{1}{\Gamma
(\frac{1}{2}\Gamma(k+1)} \int_1^{1+i\infty} e^{itx} (1 - t^2)^k dt
\\ & = &-2 \left( \frac{x}{2} \right)^{k+\frac{1}{2}} \frac{1}{ \Gamma
(\frac{1}{2})k!} \left( 1+ \frac{d^2}{dx^2}
\right)\int_1^{1+i\infty} e^{itx}  dt \\ & =& 2 \left( \frac{x}{2}
\right)^{k+\frac{1}{2}} \frac{1}{ \Gamma (\frac{1}{2})k!} \left(
1+ \frac{d^2}{dx^2} \right)\frac{e^{itx}}{x} \\ & = &
\frac{e^x}{x^\frac{1}{2}} \{ {\rm Polynimial\ in\ } \frac{1}{x},
{\rm degree\ } k \} \end{eqnarray*} The functions $\ds H_{k +
\frac{1}{2}}^{(1)} (x), \, H_{k + \frac{1}{2}}^{(2)} (x)$ are
called spherical Bessel functions.  They arise in solution of the
wave equation in spherical Polar coordinates.

\newpage

{\bf Radially Progressive Waves in two dimensions}

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We had for the membrane $$ \nabla_1^2 w = \frac{1}{c^2}
\frac{\pl^2w}{\pl t^2} \hspace{.3in} \nabla_1^2 = \frac{\pl^2}{\pl
x^2} +\frac{\pl^2}{\pl y^2}$$ and found solutions (in the case of
radial symmetry) $$w = [AJ_0(kr) + BY_0(kr)]\cos (wt + \epsilon)
\hspace{.2in} k = \frac{w}{c}$$ assuming the form $$w = f(r)
e^{iwt}$$ (real parts to be taken eventually)  we find similar
form $$w = [A_1H_0^{(1)} (kr) + A_2H_0^{(2)}(kr)]e^{iwt}$$ since
$$H_0^{(1)}(kr) \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2}
e^{i(kr - \frac{\pi}{4})} {\rm \ \ as\ } r \to \infty$$
$$H_0^{(2)}(kr) \sim \left( \frac{2}{\pi k r}\right)^\frac{1}{2}
e^{i(kr - \frac{\pi}{4})} {\rm \ \ as\ } r \to \infty$$ we get
$$H_0^{(1)}(kr) e^{iwt} \sim \left( \frac{2}{\pi k
r}\right)^\frac{1}{2} e^{i(w (t+ \frac{r}{c}) - \frac{\pi}{4})}$$
$$H_0^{(2)}(kr) e^{iwt} \sim \left( \frac{2}{\pi k
r}\right)^\frac{1}{2} e^{i(w (t- \frac{r}{c}) - \frac{\pi}{4})}$$

The first repents a wave converging to the origin with velocity
$c$, the second a wane diverging from the origin with velocity
$c$.






\end{document}
