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\noindent
Regular solution $p = m$, has recursion solution
\[ (n + m)^2 a_n + a_{n - 2} - m^2 a_n = 0 \]
\[ \implies a_n = \frac{-1}{n(n + 2m)}a_{n - 2} \]
Put $n \to 2n'$
\[ \implies a_{2n'} = \frac{-1}{4n'(n' + m)}a_{2n'
- 2} \]
so stepping up from $a_0$ we have (dropping
primes)
\[ a_{2n} = \frac{(-1)^n}{2^{2n} n! (n + m)(n + m
- 1) \cdots (m + 1)} a_0 \]
Take $a_0 = \frac{1}{2^m m!}$ (convention) to find
the \emph{Bessel function of the first kind}:
\[ J_m(z) = \left( \frac{z}{2} \right)^m \sum_{n =
0}^\infty \frac{(-1)^n}{n! (n + m)!} \left(
\frac{z}{2} \right)^{2n} \tag{3.2} \]
Exercise: Use $y = \sqrt{z} R$ in Bessel equation
(3.27) to find
\[ y'' + y \left( 1 + \frac{1}{4z} -
\frac{m^2}{z^"} \right) = 0 \]
So as $z \to \infty$, $y'' = -y$ so we have
solutions
\[ R = \frac{1}{\sqrt{z}} (A \cos z + B \sin z) .\]
Also works for $m = \gamma$ (non-integer) if $(n
+ m)! \to \Gamma(n + m + 1)$. Second solution with
$p = -m$ (integer) is the Neumann function (Bessel
function of the second kind)
\[ Y_m(z) = \lim_{\gamma \to m} \frac{J_\gamma(z)
\cos(\gamma \pi) - J_{-\gamma}(z)}{\sin
\gamma\pi} \]
Exercise$^*$: Use (3.28) to show that
\[ \dfrac{}{z} (z^m J_m(z)) = z^m J_{m - 1}(z) \]
and hence
\[ J_m'(z) + \frac{m}{z} J_m(z) = J_{m - 1}(z)
\tag{3.29} \]
Repeat with $z^{-m}$ to find \emph{recursion
relations}.
\[ J_{m - 1}(z) + J_{m + 1}(z) = \frac{2m}{z}
J_m(z) \tag{3.30} \]
\[ J_{m - 1}(z) - J_{m + 1}(z) = 2J'_m(z) \]
Asymptotic behaviour $J_m(z), Y_m(z)$:
\begin{itemize}
    \item Small $z \to 0$, $J_0(z) \to 1$, $J_m(z)
        \to \frac{1}{m!} \left( \frac{z}{2}
        \right)^m$, $m > 0$.
        \[ Y_0(z) \to \frac{2}{\pi}\ln \left(
        \frac{z}{2} \right), \quad Y_m(z) \to
        -\frac{(m - 1)!}{\pi} \left( \frac{2}{z}
        \right)^m \tag{3.31} \]
        ($Y_m$ is divergent as $z \to 0$)
    \item Large $z \to \infty$: oscillatory
        solutions:
        \[ J_m(z) \approx \sqrt{\frac{2}{\pi z}}
        \cos \left( z - \frac{m\pi}{2} -
        \frac{\pi}{4} \right) \tag{3.32} \]
        \[ Y_m(z) \approx \sqrt{\frac{2}{\pi z}}
        \sin \left( z - \frac{m\pi}{2} -
        \frac{\pi}{4} \right) \]
\end{itemize}

\subsubsection*{Zeros of Bessel function $J_m(z)$}

Define $j_{mn}$ to be $n$-th zero,
\[ J_m(j{mn}) = 0 \quad (z > 0) \]
From (3.32) this occurs when (approximately)
\[ \cos \left( z - \frac{m\pi}{2} - \frac{\pi}{4}
\right) = 0 \]
i.e. $z - \frac{m\pi}{2} - \frac{\pi}{4} = n\pi -
\frac{\pi}{2}$ (modal point). So zero at
\[ z \approx n\pi + \frac{m\pi}{2} - \frac{\pi}{4}
\equiv \tilde{j}_{mn} \tag{3.33} \]
(Accuracy: $\frac{j_{mn} - \tilde{j}_{mn}}{j_{mn}}
< \frac{0.1}{n}$, for $n > \frac{m^2}{2}$
(non-examinable)) \\
For $J_0(z)$ actual values are
\[ j_{01} = 2.405, \quad j_{02} = 5.520, \quad
j_{03} = 8.653 \]
\[ j_{0n} \approx n\pi - \frac{\pi}{4} \]
(precision $\sim \frac{1\%}{n}$).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/d0319d92551111ed.png}
\end{center}

\subsection{2D Wave equation (continued):
Vibrating drum}

From section 3.8, radial solutions to (3.26)
become
\[ R_m(z) = R_m(\sqrt{\lambda} r) = A
J_m(\sqrt{\lambda} r) + B Y_m(\sqrt{\lambda} r) \]
Impose boundary conditions:
\begin{itemize}
    \item Regularity at $r = 0 \implies B = 0$ by
        (3.31)
    \item Unit disk $r = 1$ with $R = 0$ implies
        \[ J_m(\sqrt{\lambda}) = 0 \]
        But these zeros occur at
        \[ j_{mn} (\approx \tilde{j}_{mn} = n\pi +
        \frac{m\pi}{2} - \frac{\pi}{4}) \]
        so our eigenvalues must be
        \[ \lambda_{mn} = j_{mn}^2 \tag{3.34} \]
\end{itemize}
With the polar mode (3.26) the spatial solution
is
\[ V_{mn}(r, \theta) = H_m(\theta)
R_{mn}(\sqrt{\lambda_{mn}} r) = (A_{mn} \cos
m\theta + B_{mn} \sin m\theta) J_m (j_{mn} r)
\tag{3.35} \]
The temporal solution to (3.21) $\ddot{T} =
-\lambda c^2 T$ are $T_{mn}(t) = \cos (j_{mn} ct)$
and $\sin(j_{mn} ct)$. For our linear homogeneous
PDE (3.17)we can sum together to obtain general
solution (noting the special case for $m = 0$):
\begin{align*}
    u(r, \theta, t)
    &= \sum_{n = 1}^\infty J_0 (j_{mn} r) (A_{0n}
    \cos (j_{0n} ct) + C_{0n} \sin (j_{0n} ct)) \\
    &+
    \sum_{m = 1}^\infty \sum_{n = 1}^\infty
    J_m(j_{mn} r) (A_{nm} \cos m\theta + B_{nm}
    \sin m\theta) \cos (j_{mn} ct) \\
    &+ \sum_{m = 1}^\infty \sum_{n = 1}^\infty J_m
    (j_{mn} r) (C_{mn} \cos m\theta + D_{mn} \sin
    m\theta) \sin (j_{nm} ct) \tag{3.36}
\end{align*}
Now impose \emph{initial conditions} (3.19) at $t
= 0$
\[ u(r, \theta, 0) = \phi(r, \theta) = \sum_{m =
0}^\infty \sum_{n = 1}^\infty J_m(j_{mn} r) \times
(A_{nn} \cos m\theta + B_{nn} \sin m\theta)
\tag{3.37} \]
\[ \pfrac{u}{t}(r, \theta, 0) = \psi(r, \theta) =
\sum_{m = 0}^\infty \sum_{n = 1}^\infty j_{mn} c
J_m (j_{mn} r) \times (C_{mn} \cos m\theta +
D_{mn} \sin m\theta) \]
Orthogonality: Find coefficients by multiplying by
$J_m$, $\cos$, $\sin$ and exploit orthogonality
(1.1-3) and Example sheet 1, Q8.
\begin{align*}
    \int_0^1 J_m(j_{mn} r) J_m(j_{m k} r) r \dd r
    &= \half [J'_m (j_{mn})]^2 \delta_{nk}
    \tag{3.28} \\
    &= \half [J_{m + 1}(j_{mn})]^2 \delta_{nk}
    && \text{by recursion} \tag{3.29}
\end{align*}
Now integrate to obtain $A_{mn}$
\[ \int_0^{2\pi} \dd \theta \cos p\theta \int_0^1
r \dd r J_{pq}(j_{pq} r) \phi(r, \theta) =
\frac{\pi}{2} [J_{p + 1}(j_{pq})]^2 A_{pq} \]
Exercises: Find $B$, $C$, $D$.

\begin{example*}
    Initial radial profile.
    \[ u(r, \theta, 0) = \phi(r) = 1 - r^2 \]
    \[ \implies m = 0, \quad B_{mn} = 0, A_{mn} =
    0, m \neq n \]
    \[ \pfrac{u}{t}(r, \theta, 0) = 0 \]
    \[ \implies C_{mn} = D_{mn} = 0 \]
    We need to find:
    \begin{align*}
        A_{0n}
        &= \frac{2}{J_1(j_{0n})^2} \int_0^1
        J_0(j_{0n} r) (1 - r^2) r \dd r \\
        &= \frac{2}{J_1(j_{0n})^2}
        \frac{J_2(j_{0n})}{j_{0n}^2} \\
        &\approx \frac{J_2(j_{0n})}{n}
    \end{align*}
    as $n \to \infty$. (Exercise$^*$ using
    (3.29-30)).
\end{example*}
