Difference between revisions of "Bottom Mounted Cylinder"

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= Removing the depth dependence =
 
 
We can [[Removing The Depth Dependence | remove the depth dependence]] and the equation reduces to the [[Helmholtz's Equation]]
 
in a two dimensional domain (i.e. there is no dependence on <math>z</math>
 
 
<math> \nabla^2\phi + k^2 0, \quad r>a, </math>
 
 
<math> \frac{\partial \phi}{\partial r} = 0, \quad r = a. </math>
 
 
where <math>k</math> is the positive imaginary root of the [[Dispersion Equation]]. We can now separate
 
variables by writing Laplace's equation in cylindrical polar coordinates.
 
 
<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial
 
\phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2
 
\phi}{\partial \theta^2} + k^2 = 0, \quad r>a,
 
</math>
 
 
The separate solution is
 
 
<math>
 
\phi^{i} (r,\theta) =  \sum_{\nu = -
 
\infty}^{\infty} D_{\nu} I_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta},
 
</math>
 
  
 
and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>
 
and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>

Revision as of 08:57, 8 August 2006

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.

Theory

A cylinder of radius [math]\displaystyle{ a }[/math] (which we will suppose is at the centre of a cylindrical coordinate system) extends from the free surface to the bottom of the fluid

The problem is Laplace's equation through out the fluid,

[math]\displaystyle{ \nabla^2\phi = 0, \quad r\gt a, }[/math]

the linearised free surface condition

[math]\displaystyle{ \frac{\partial \phi}{\partial z} - \alpha \phi = 0, \quad r\gt a, \quad z =0 }[/math]

and the non flow conditions through the bottom boundary at [math]\displaystyle{ z=d }[/math]

[math]\displaystyle{ \frac{\partial \phi}{\partial z} = 0, \quad z = d, }[/math]

and the cylinder

[math]\displaystyle{ \frac{\partial \phi}{\partial r} = 0, \quad r = a. }[/math]

Removing the depth dependence

We can remove the depth dependence and the equation reduces to the Helmholtz's Equation in a two dimensional domain (i.e. there is no dependence on [math]\displaystyle{ z }[/math]

[math]\displaystyle{ \nabla^2\phi - k^2 0, \quad r\gt a, }[/math]

[math]\displaystyle{ \frac{\partial \phi}{\partial r} = 0, \quad r = a. }[/math]

where [math]\displaystyle{ k }[/math] is the positive imaginary root of the Dispersion Relation for a Free Surface

[math]\displaystyle{ \alpha + k \tan k d = 0. }[/math]

We can now separate variables by writing Laplace's equation in cylindrical polar coordinates.

[math]\displaystyle{ \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial \phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} - k^2 = 0, \quad r\gt a, }[/math]

The separate solution is

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

(this follows exactly as for Cylindrical Eigenfunction Expansion). Note that in many cases a symmetry arguement 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 [math]\displaystyle{ r = a }[/math]

The boundary condition at [math]\displaystyle{ r = a }[/math] is now substituted into the above equation to give

[math]\displaystyle{ D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 }[/math]

so that

[math]\displaystyle{ E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k a)}{K^{\prime}_\nu (k a)} }[/math]




and we often assume that the incident wave is a plane wave travelling in the positive [math]\displaystyle{ x }[/math] direction, i.e.

[math]\displaystyle{ \phi^{i} (r,\theta) = e^{kx} = \sum_{\nu = - \infty}^{\infty} I_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta}, }[/math]

We therefore have the total potential is

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