Difference between revisions of "Bottom Mounted Cylinder"

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The problem is Laplace's equation through out the fluid,
 
The problem is Laplace's equation through out the fluid,
 
+
<center>
 
<math> \nabla^2\phi = 0, \quad r>a, </math>
 
<math> \nabla^2\phi = 0, \quad r>a, </math>
 
+
</center>
 
the linearised free surface condition  
 
the linearised free surface condition  
 
+
<center>
 
<math>\frac{\partial \phi}{\partial z} - \alpha \phi = 0, \quad r>a, \quad z =0 </math>
 
<math>\frac{\partial \phi}{\partial z} - \alpha \phi = 0, \quad r>a, \quad z =0 </math>
 
+
</center>
 
and the non flow conditions through the bottom boundary at <math>z=d</math>
 
and the non flow conditions through the bottom boundary at <math>z=d</math>
 
+
<center>
 
<math> \frac{\partial \phi}{\partial z} = 0, \quad z = d, </math>
 
<math> \frac{\partial \phi}{\partial z} = 0, \quad z = d, </math>
 
+
</center>
 
and the cylinder
 
and the cylinder
 
+
<center>
 
<math> \frac{\partial \phi}{\partial r} = 0, \quad r = a. </math>
 
<math> \frac{\partial \phi}{\partial r} = 0, \quad r = a. </math>
 +
</center>
  
 
= Removing the depth dependence =
 
= Removing the depth dependence =
Line 31: Line 32:
 
We can [[Removing the Depth Dependence| remove the depth dependence]] and the equation reduces to the [[Helmholtz's Equation]]  
 
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>
 
in a two dimensional domain (i.e. there is no dependence on <math>z</math>
 
+
<center>
 
<math> \nabla^2\phi - k^2 0, \quad r>a, </math>
 
<math> \nabla^2\phi - k^2 0, \quad r>a, </math>
 
+
</center>
 +
<center>
 
<math> \frac{\partial \phi}{\partial r} = 0, \quad r = a. </math>
 
<math> \frac{\partial \phi}{\partial r} = 0, \quad r = a. </math>
 
+
<center>
 
where <math>k</math> is the positive imaginary root of the [[Dispersion Relation for a Free Surface]]
 
where <math>k</math> is the positive imaginary root of the [[Dispersion Relation for a Free Surface]]
 
+
<center>
 
<math>
 
<math>
 
\alpha + k \tan k d = 0.
 
\alpha + k \tan k d = 0.
 
</math>
 
</math>
 
+
</center>
 
We can now separate
 
We can now separate
 
variables by writing Laplace's equation in cylindrical polar coordinates.
 
variables by writing Laplace's equation in cylindrical polar coordinates.
 
+
<center>
 
<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial
 
<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 r} \right) + \frac{1}{r^2} \frac{\partial^2
 
\phi}{\partial \theta^2} - k^2 = 0, \quad r>a,
 
\phi}{\partial \theta^2} - k^2 = 0, \quad r>a,
 
  </math>
 
  </math>
 
+
</center>
 
The separate solution is  
 
The separate solution is  
 
+
<center>
 
<math>
 
<math>
 
\phi (r,\theta) =  \sum_{\nu = -
 
\phi (r,\theta) =  \sum_{\nu = -
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r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},  
 
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},  
 
</math>
 
</math>
 
+
</center>
 
(this follows exactly as for [[Cylindrical Eigenfunction Expansion]]). Note that in many
 
(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
 
cases a symmetry arguement is used to express the complex exponentials in terms
Line 66: Line 68:
 
The boundary condition at <math>r = a</math> is now substituted into the above
 
The boundary condition at <math>r = a</math> is now substituted into the above
 
equation to give
 
equation to give
 
+
<center>
 
<math> D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 </math>
 
<math> D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 </math>
 
+
</center>
 
so that  
 
so that  
 
+
<center>
 
<math> E_{\nu}  = -\frac{D_{\nu} I^{\prime}_\nu (k a)}{K^{\prime}_\nu (k a)}</math>
 
<math> E_{\nu}  = -\frac{D_{\nu} I^{\prime}_\nu (k a)}{K^{\prime}_\nu (k a)}</math>
 
+
</center>
 
 
 
 
 
 
 
 
 
 
 
 
 
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>
 
direction, i.e.
 
direction, i.e.

Revision as of 09:00, 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]