Hankel trasformation of acoustic wave equation We consider a simplified version of acoustic wave equation
\begin{equation}
\frac{\partial^2 p}{\partial r^2}+\frac{1}{r}\frac{\partial p}{\partial r}+\frac{\partial^2 p}{\partial z^2}+k^2 p=\frac{1}{r} \delta(r) \delta(z-z_0)
\end{equation}
where p is a complex quantity, defined by $P=p \exp(-i\omega t)$, $\omega$ is the circular frequency, $k=\omega/c$ is the wavenumber, c is the sound speed, z is the depth below the ocean surface, r is the radius. The formulation is in cylindrical coordinates because the ocean may be treated as an cylindrical waveguide over relatively large distances. The article (which I am studying) says that by applying the Hankel transformation we obtain
\begin{equation}
\frac{\partial^2 q}{\partial z^2}+(k^2-h^2) q=\delta(z-z_0)
\end{equation}
I do not know the Hankel transformation. We accept suggestions for understanding the transition between the two mathematical formulas. 
Thank you very much.
 A: The Hankel transform takes the form
$$\hat{f}(\rho) = 2 \pi \int_0^{\infty} dr \, r J_0(\rho r) \, f(r)$$
One may derive this from the two-dimensional Fourier transform, assuming that the functions are radially symmetric (i.e., $f(x,y)=f(r)$, as follows:
$$\begin{align}\hat{f}(u,v) &= \int_{-\infty}^{\infty} dx \, \int_{-\infty}^{\infty} dy \, e^{i (u x+v y)} f(x,y)\\ &= \int_0^{\infty} dr \, r f(r) \int_0^{2 \pi} d\theta \, e^{i \rho r \cos{(\theta-\phi)}} \end{align}$$
where $u=\rho \cos{\phi}$, etc.  The Bessel function comes from that last inner integral.
The inverse is 
$$f(r) = \int_0^{\infty} d\rho \, \rho J_0(\rho r) \, \hat{f}(\rho)$$
Now, the Bessel function obeys the following relation:
$$\left (\frac{d^2}{dr^2} + \frac{1}{r} \frac{d}{dr} \right ) J_0(\rho r) = -\rho^2 J_0(\rho r)$$
and we also have
$$\frac{\delta(r)}{r} = \int_0^{\infty} d\rho \, \rho J_0(\rho r)$$
So that, assuming that $q$ is the Hankel transform of $p$, we find that, from the differential equation,
$$\int_0^{\infty} d\rho \, \rho J_0(\rho r) \, \left [\frac{d^2 q}{dz^2} + (k^2-\rho^2) q - \delta(z-z_0) \right ] = 0$$
Note that the $h^2$ in your equation is the radial frequency variable in what I have derived.  As the integral is zero identically, the quantity inside the brackets is zero.  Thus, the second equation.
