# Introduction

The bottom mounted cylinder is one of the simplest scattering problems in water waves. It is the basis for many more complicated scattering problems in which it is desired to keep the scatterers as simple as possible. The theory can be found in many books. The derivation here is related to Cylindrical Eigenfunction Expansion. We begin with the Frequency Domain Problem.

# Boundary Problem for a Fixed Body

We assume small amplitude so that we can linearise all the equations (see Linear and Second-Order Wave Theory). We also assume that Frequency Domain Problem with frequency $\omega$ and we assume that all variables are proportional to $\exp(-\mathrm{i}\omega t)\,$ The water motion is represented by a velocity potential which is denoted by $\phi\,$ so that

$\Phi(\mathbf{x},t) = \mathrm{Re} \left\{\phi(\mathbf{x},\omega)e^{-\mathrm{i} \omega t}\right\}.$

The coordinate system is the standard Cartesian coordinate system with the $z-$axis pointing vertically up. The water surface is at $z=0$ and the region of interest is $-h\lt z\lt 0$. There is a body which occupies the region $\Omega$ and we denote the wetted surface of the body by $\partial\Omega$ We denote $\mathbf{r}=(x,y)$ as the horizontal coordinate in two or three dimensions respectively and the Cartesian system we denote by $\mathbf{x}$. We assume that the bottom surface is of constant depth at $z=-h$.

The Standard Linear Wave Scattering Problem in Finite Depth for a fixed body is

\begin{align} \Delta\phi &=0, &-h\lt z\lt 0,\,\,\mathbf{x} \in \Omega \\ \partial_z\phi &= 0, &z=-h, \\ \partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align}

(note that the last expression can be obtained from combining the expressions:

\begin{align} \partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\ \mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align}

where $\alpha = \omega^2/g \,$) The body boundary condition for a rigid body is just

$\partial_{n}\phi=0,\ \ \mathbf{x}\in\partial\Omega_{\mathrm{B}},$

The equation is subject to some radiation conditions at infinity. We assume the following. $\phi^{\mathrm{I}}\,$ is a plane wave travelling in the $x$ direction,

$\phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \,$

where $A$ is the wave amplitude (in potential) $\mathrm{i} k$ is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form $\exp(-\mathrm{i}\omega t)$) and

$\phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h}$

In three-dimensions the Sommerfeld Radiation Condition is

$\sqrt{|\mathbf{r}|}\left( \frac{\partial}{\partial|\mathbf{r}|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}$

where $\phi^{\mathrm{{I}}}$ is the incident potential.

If we have a problem in which all the scatterers are of constant cross sections so that

$\partial\Omega = \partial\hat{\Omega} \times z\in[-h,0]$

where $\partial\hat{\Omega}$ is a function only of $x,y$ i.e. the boundary of the scattering bodies is uniform with respect to depth. We can remove the depth dependence separation of variables and obtain that the dependence on depth is given by

$\phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \bar{\phi}(x,y)$

Since $\phi$ satisfies Laplace's Equation, then $\Phi$ satisfies Helmholtz's Equation

$\nabla^2 \bar{\phi} + k^2 \bar{\phi} = 0$

in the region not occupied by the scatterers.

# Equation for a Cylinder

A cylinder of radius $a$ (which we will suppose is at the centre of our cylindrical coordinate system) occupies the region $r\lt a$. On the boundary of the cylinder we assume that the normal derivative vanishes.

We can now use Separation of Variables by writing Helmholtz's Equation in cylindrical polar coordinates,

## Separation of Variable for the $r$ and $\theta$ coordinates

For the solution of

$\frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial Y}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 Y}{\partial \theta^2} = k_m^2 Y(r,\theta),$

we use the separation

$\,\!Y(r,\theta) =: R(r) \Theta(\theta).$

Substituting this into Laplace's equation yields

$\frac{r^2}{R(r)} \left[ \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d}r} \right) - k_m^2 R(r) \right] = - \frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d} \theta^2} = \eta^2,$

where the separation constant $\eta$ must be an integer, say $\nu$, in order for the potential to be continuous. $\Theta (\theta)$ can therefore be expressed as

$\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.$

We also obtain the following expression

$r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d} R}{\mathrm{d} r} \right) - (\nu^2 + k_m^2 r^2) R(r) = 0, \quad \nu \in \mathbb{Z}.$

Substituting $\tilde{r}:=k_m r$ and writing $\tilde{R} (\tilde{r}) := R(\tilde{r}/k_m) = R(r)$, this can be rewritten as

$\tilde{r}^2 \frac{\mathrm{d}^2 \tilde{R}}{\mathrm{d} \tilde{r}^2} + \tilde{r} \frac{\mathrm{d} \tilde{R}}{\mathrm{d} \tilde{r}} - (\nu^2 + \tilde{r}^2)\, \tilde{R} = 0, \quad \nu \in \mathbb{Z},$

which is the modified version of Bessel's equation. Substituting back, the general solution is given by

$R(r) = D_\nu \, I_\nu(k_m r) + E_\nu \, K_\nu(k_m r),\ \nu \in \mathbb{Z},$

where $I_\nu$ and $K_\nu$ are the modified Bessel functions of the first and second kind, respectively, of order $\nu$.

Note that $K_\nu (\mathrm{i} x) = \pi / 2\,\, \mathrm{i}^{\nu+1} H_\nu^{(2)}(x)$ with $H_\nu^{(2)}$ denoting the Hankel function of the second kind of order $\nu$. Also, $I_\nu$ does not satisfy the Sommerfeld Radiation Condition since it becomes unbounded for increasing real argument and it represents incoming waves.

The solution is

$\phi (r,\theta) = \sum_{\nu = - \infty}^{\infty} \left[ D_{\nu} I_\nu (k_0 r) + E_{\nu} K_\nu (k_0 r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},$

Note that in many cases a symmetry argument is used to express the complex exponentials in terms of sine and cosines. This follows from a symmetry in the incident potential.

# Boundary condition at $r = a$

The boundary condition at $r = a$ is that the normal derivative vanishes. If this is substituted into the above equation we obtain

$D_{\nu} I^{\prime}_\nu (k_0 a) + E_{\nu} K^{\prime}_\nu (k_0 a) = 0,$

so that

$E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)},$

and we often assume that the incident wave is a plane wave travelling in the positive $x$ direction, i.e.

$\phi^{i} (r,\theta) = e^{k_0 x} = \sum_{\nu = - \infty}^{\infty} I_\nu (k_0 r) \mathrm{e}^{\mathrm{i} \nu \theta},$

We therefore have the total potential as

$\phi (r,\theta) = \sum_{\nu = - \infty}^{\infty} \left[ I_\nu (k_0 r) - \frac{ I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)} K_\nu (k_0 r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}.$