# Wave Scattering By A Vertical Circular Cylinder

Wave and Wave Body Interactions
Current Chapter Wave Scattering By A Vertical Circular Cylinder
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## Introduction

This important flow accepts a closed-form analytical solution for arbitrary values of the wavelength $\displaystyle{ \lambda\, }$. This was shown to be the case by McCamy and Fuchs 1954 using separation of variables.

## Problem

The incident potential is given as

$\displaystyle{ \Phi_I = \mathrm{Re} \left\{\phi_I e^{i\omega t} \right \} \, }$
$\displaystyle{ \phi_I = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cosh k h} e^{-ikx} }$

Let the diffraction potential be

$\displaystyle{ \phi_7 = \frac{i g A}{\omega} \frac{\cosh k(z+h)}{\cos k h} \psi(x,y) }$

For $\displaystyle{ \phi_7\, }$ to satisfy the 3D Laplace equation, it is easy to show that $\displaystyle{ \psi\, }$ must satisfy the Helmholtz equation:

$\displaystyle{ \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + k^2 \right) \psi = 0\, }$

In polar coordinates:

$\displaystyle{ \begin{Bmatrix} x=R\cos\theta \\ y=R\sin\theta \end{Bmatrix} ; \quad \psi(R,\theta) }$

The Helmholtz equation takes the form:

$\displaystyle{ \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} + \frac{1}{R^2} \frac{\partial^2}{\partial\theta^2} + k^2 \right) \psi = 0 \, }$

On the cylinder:

$\displaystyle{ \frac{\partial\phi_7}{\partial n} = - \frac{\partial\phi_I}{\partial n} \, }$

or

$\displaystyle{ \frac{\partial\psi}{\partial R} = - \frac{\partial}{\partial R} \left( e^{-ikx} \right) = -\frac{\partial}{\partial R} \left( e^{-ikR\cos\theta} \right) }$

Here we make use of the familiar identity:

$\displaystyle{ e^{-ikR\cos\theta} = \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \cos m \theta }$
$\displaystyle{ \epsilon_m = \begin{Bmatrix} 1, & m = 0 \\ 2(-i)^m, & m \gt 0 \end{Bmatrix} }$

## Solution

Try:

$\displaystyle{ \psi(R,\theta) = \sum_{m=0}^{\infty} A_m F_m ( k R ) \cos m \theta \, }$

Upon substitution in Helmholtz's equation we obtain:

$\displaystyle{ \left( \frac{\partial^2}{\partial R^2} + \frac{1}{R} \frac{\partial}{\partial R} - \frac{m^2}{R^2} + k^2 \right) F_m ( k R ) = 0 }$

This is the Bessel equation of order m accepting as solutions linear combinations of the Bessel functions

$\displaystyle{ \begin{Bmatrix} J_m ( k R ) \\ Y_m ( k R ) \end{Bmatrix} }$

The proper linear combination in the present problem is suggested by the radiation condition that $\displaystyle{ \psi\, }$ must satisfy:

As $\displaystyle{ R \to \infty\, }$:

$\displaystyle{ \psi(R,\theta) \sim e^{-ikR + i\omega t} \, }$

Also as $\displaystyle{ R \to \infty\, }$:

$\displaystyle{ J_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \cos \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) }$
$\displaystyle{ Y_m ( k R ) \sim \left( \frac{2}{\pi k R} \right)^{1/2} \sin \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right) }$

Hence the Hankel function:

$\displaystyle{ H_m^{(2)} ( k R ) = J_m ( k R ) - i Y_m ( k R ) \, }$
$\displaystyle{ \sim \left( \frac{2}{\pi k R} \right)^{1/2} e^{-i \left( k R - \frac{1}{2} m \pi - \frac{\pi}{4} \right)} }$

Satisfies the far field condition required by $\displaystyle{ \psi(R,\theta) \, }$. So we set:

$\displaystyle{ \psi(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m A_m H_m^{(2)} ( k R ) \cos m \theta }$

with the constants $\displaystyle{ A_m \, }$ to be determined. The cylinder condition requires:

$\displaystyle{ \left. \frac{\partial\psi}{\partial R} \right|_{R=a} = - \frac{\partial}{\partial R} \sum_{m=0}^{\infty} \epsilon_m J_m ( k R ) \left.\cos m \theta \right|_{r=a} }$

It follows that:

$\displaystyle{ A_m {H_m^{(2)}}^' (k a) = - J_m^' (k a) \, }$

or:

$\displaystyle{ A_m = - \frac{J_m^' ( k a ) }{{H_m^{(2)}}^' (k a)} \, }$

where $\displaystyle{ (')\, }$ denotes derivatives with respect to the argument. The solution for the total velocity potential follows in the form

$\displaystyle{ (\psi+x)(r,\theta) = \sum_{m=0}^{\infty} \epsilon_m \left[ J_m (k R) - \frac{J_m^'(k a)}{{H_m^{(2)}}^'(k a)} H_m^{(2)} (k a) \right] \cos m \theta }$

And the total original potential follows:

$\displaystyle{ \phi = \phi_I + \phi_7 = \frac{i g A}{\omega} \frac{\cosh k (z+h)}{\cosh k h } (\psi+x) (r,\theta) }$

In the limit as $\displaystyle{ h \to \infty\,, \quad \frac{\cosh k (z+h)}{k h} \to e^{k z} \, }$ and the series expansion solution survives.

The total complex potential, incident and scattered, was derived above.

The hydrodynamic pressure follows from Bernoulli:

$\displaystyle{ P = \mathrm{Re} \left\{ \mathbf{P} e^{i\omega t} \right\} \, }$
$\displaystyle{ \mathbf{P} = - i\omega \rho \left( \phi_I + \phi_7 \right) \, }$

## Surge exciting force

The surge exciting force is given by

$\displaystyle{ X_1 = \iint_{S_B} P n_1 \mathrm{d}S = \mathrm{Re} \left\{ \mathbf{X}_1 e^{i\omega t} \right\} }$
$\displaystyle{ \mathbf{X}_1 = \rho \int_{-\infty}^0 \mathrm{d}z \int_0^{2\pi} a \mathrm{d}\theta \left( - i \omega \frac{i g A}{\omega} \right) e^{k z} n_1 (\psi + x)_{R=a} }$

Simple algebra in this case of water of infinite depth leads to the expression.