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QUESTION

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\item[(a)]
Show that ${\cal H}_0\left(\frac{1}{r}\right)=\frac{1}{a}$ by
using the fact that the Hankel transform is its own inverse.

\item[(b)]
Solve the cylindrical wave equation $$\frac{1}{r}\frac{\partial
}{\partial r}\left(r\frac{\partial \phi}{\partial
r}\right)=\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}$$ with
the boundary conditions $$\phi(r,0)=\frac{1}{r},\ \phi_t(r,0)=0,$$
$$\phi\to 0\textrm{ as either }r\to \infty \textrm{or
}t\to\infty,\ \phi_r\to0\textrm{ as }r\to \infty.$$

You can leave your final answer as an (inverse Hankel transform)
integral.

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ANSWER

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\item[(a)]
$${\cal H}_0\left(\frac{1}{r}\right)=\int_0^\infty\frac{1}{r}{\cal
J}_0(\alpha r)r\,dr=\int_0^\infty {\cal J}_0(\alpha
r)\,dr=\frac{1}{\alpha}\int_0^\infty {\cal
J}_0(x)\,dx=\frac{k}{\alpha}$$ for some constant $k$. But as the
Hankel transform is self inverse, $k=1$.

\item[(b)]
$$\frac{1}{r}(r\phi_r)_r=\frac{1}{c^2}\phi_{tt},\
\phi(r,0)=\frac{1}{r},\ \phi_t(r,0)=0,$$ \\ $ \phi\to 0$ as $r\to
\infty$ or $t \to \infty,\ phi_r\to 0$ as $r\to \infty$

Take a Hankel transform of order zero with respect to $r$
$$-\alpha^2\Phi(\alpha,t)=\frac{1}{c^2}\Phi_{tt}(\alpha, t)$$

\begin{eqnarray*}
\Phi(\alpha,t)&=&A(\alpha)\cos c\alpha t+B(\alpha)\sin c\alpha t\\
\phi_t(r,0)&=&0 \forall r\Rightarrow\Phi_t(\alpha,0)=0\ \textrm{
forall } \alpha \Rightarrow B(\alpha)=0\\
\phi(r,0)&=&\frac{1}{R}\Rightarrow
\Phi(\alpha,0)=\frac{1}{\alpha}\Rightarrow
A(\alpha)=\frac{1}{\alpha}
\end{eqnarray*}

Inverse Hankel transform
$$\phi(r,t)=\int_0^\infty\frac{1}{\alpha}\cos c\alpha t {\cal
J}_0(\alpha r)\alpha \,d\alpha=\int_0^\infty \cos c \alpha t {\cal
J}_0(\alpha r)\,d\alpha$$
 Could look this up as a cosine
transform.

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