\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\newcommand{\lin}{\int_0^\infty}
\parindent=0pt
\begin{document}


{\bf Question}

The equation for the radial displacement $u(r,t)$ of a vibrating
circular membrane of radius $\ell$ is $$ {{\partial^2u}\over
{\partial r^2}}+{1 \over r} {{\partial u} \over {\partial r}}-{1
\over {c^2}} {{\partial^2u}\over {\partial t^2}}=0 $$ together
with the boundary condition $u(\ell, t)=0$. If the membrane
satisfies the initial conditions $$ u(r,0)=0, \qquad {{\partial
u}\over {\partial r}}(r,0)=v_0 $$ derive the solution $$
u(r,t)={{2\ell v_0} \over c}\sum_{n=1}^\infty {{J_0(\gamma_n
r/\ell)\sin(\gamma_n ct/\ell)} \over {\gamma_n^2J_1(\gamma_n)}} $$
where $\gamma_n$ is the $n$-th positive root of $J_0(x)$.



\vspace{.5in}

{\bf Answer}

$$ \frac{X''}{X} = \frac{1}{k}\frac{T''}{T} = \lambda$$
$$\Rightarrow X'' - \lambda X = 0$$

$\ds X(0) = 0 $ and $\ds X(l) = 0$ $$\Rightarrow \lambda =
-\frac{n^2\pi^2}{l^2}$$

Therefore \begin{eqnarray*} x_n(x) & = & A_n \sin \left(\frac{n\pi
x}{l}\right) \\ T' & = & -\frac{n^2\pi^2k}{l^2} T \\ T_n(t) & = &
B_n e^{-\frac{n^2\pi^2kt}{l^2}} \end{eqnarray*}

Thus \begin{eqnarray*}u(x,t) & = & \sum_{n = 1}^\infty Q_n
\sin\left( \frac{n \pi x}{l} \right) e ^{-\frac{n^2\pi^2kt}{l^2}}
\\ f(x) & = &\sum_{n = 1}^\infty Q_n \sin\left( \frac{n \pi x}{l}
\right) \end{eqnarray*}

If $\ds f(x) = 3 \sin \left(\frac{4\pi x}{l} \right)$ then $Q_4 =
3$ and $ Q_n = 0$ otherwise.  So $$ u(x,t) = 3 \sin \left( \frac{4
\pi x}{l} \right) e^{-\frac{16\pi^2 kt}{l^2}}$$

If $\ds l = 1 $ then $\ds f(x) = \sum_{n - 1}^\infty Q_n \sin
(n\pi x)$.  But $\ds f(x) = \sin(2 \pi x) - 2 \sin (5 \pi x)$

$\Rightarrow Q_2 = 1 \hspace{.1in} Q_5 = -2$ and $ Q_n = 0$
otherwise. So $$ u(x,t) =  \sin (2\pi x) e^{-400 \pi^2t} - 2 \sin
(5\pi x) e^{-2500 \pi^2t}$$


\end{document}