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QUESTION

$Y(x,t)$ satisfies the wave equation $$Y_{tt}=c^2Y_{xx},\ t&gt;0,\
0&lt;x&lt;a,$$ with $$Y(x,0)=\frac{\beta x}{a},\ \ Y_t(x,0)=0,\ \
Y(0,t)=0,\ \ Y_x(a,t)=0.$$
\begin{description}

\item[(a)]
Show that $$y(x,s)={\cal{L}}Y(x,t)=\frac{\beta x}{a}-\frac{\beta c
\sinh \frac {sx}{c}}{s^2a\cosh \frac{sa}{c}}$$

\item[(b)]
Use the complex inversion formula to find $Y(x,t)$.

\end{description}

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ANSWER

\begin{description}

\item[(a)]
 $s^2Y-sY(x,0)-Y_t(x,0)=c^2Y_{xx}\\
 Y_{xx}-\frac{s^2}{c^2}Y=-\frac{s}{c^2}\frac{\beta x}{a}\\
 Y=A(s)e^{\frac{s}{c}x}+B(s)e^{-\frac{s}{c}x}+\frac{\beta x}{as}\\
 Y(0,t)=0\Rightarrow A(s)+B(s)=0\\
 Y_x(a,t)=0\Rightarrow
 A(s)\frac{s}{c}\left(e^{\frac{s}{c}a}+e^{-\frac{s}{c}a}+\frac{\beta}{as}\right)=0\\
 \Rightarrow Y(x,s)=\frac{\beta x}{as}-\frac{\beta
 c}{as^2}\frac{\sinh \frac{sx}{c}}{\cosh \frac{sa}{c}}$

\item[(b)]
 Simple pole at $s=0$
 \begin{eqnarray*}
 Ye^{st}&amp;=&amp;\left(\frac{\beta x}{as}-\frac{\beta
 c}{as^2}\frac{\frac{sx}{c}+O(s^3)}{1+O(s^2)}\right)(1+st+O(st))\\
 &amp;=&amp;\frac{\beta x}{as}-\frac{\beta c}{as^2}\frac{sx}{c}+O(1)\\
 &amp;\Rightarrow&amp;\textrm{Res}(0)=0
 \end{eqnarray*}

 Simple poles at $s_n=(2n+1)\frac{c\pi i}{2a}$, use l'H\^opital's rule.

 \begin{eqnarray*}
 \textrm{Res}(Ye^{st},s_n)&amp;=&amp;\lim_{s \to s_n}\frac{\beta
 c}{a}\frac{e^{st} \sinh \frac{sx}{c}}{s^2 \frac{a}{c}\sinh
 \frac{a}{c}\sinh \frac{sa}{c}+2s \cosh \frac{sa}{c}}\\
 &amp;=&amp;\frac{\beta c^2}{a^2}\frac{e^{s_nt}\sinh \frac{s_n
 x}{c}}{s_n^2(-1)^n}\\
 Y(x,t)&amp;=&amp;\sum_{n=1}^\infty \textrm{Res}(s_n)+\textrm{Res}(-s_n)\\
 &amp;=&amp;\sum_{n=1}^\infty\frac{8\beta}{\pi^2}\frac{(-1)^n}{(2n+1)^2}\sin
 \frac{(2n+1)\pi x}{2a}\cos \frac{(2n+1)\pi c t}{2a}
\end{eqnarray*}
\end{description}

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