Difference between revisions of "Cylindrical Eigenfunction Expansion"

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Line 15: Line 15:
 
<math>
 
<math>
 
\frac{\partial \phi}{\partial z} = 0, (r,\theta,z) \in
 
\frac{\partial \phi}{\partial z} = 0, (r,\theta,z) \in
\mathds{R}_{>0}\, \times \,]\!- \pi, \pi] \times \{ -d \},
+
\mathbb{R}_{>0}\, \times \,]\!- \pi, \pi] \times \{ -d \},
 
</math>
 
</math>
  
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<math>
 
<math>
 
\sup \big\{ \, \abs{\phi} \ \big| \ (r,\theta,z) \in \mathbb{R}_{>0}\,
 
\sup \big\{ \, \abs{\phi} \ \big| \ (r,\theta,z) \in \mathbb{R}_{>0}\,
\times \, ]\!- \pi, \pi] \times \mathds{R}_{<0} \,\big\} < \infty
+
\times \, ]\!- \pi, \pi] \times \mathbb{R}_{<0} \,\big\} < \infty
 
</math>
 
</math>
  

Revision as of 09:16, 20 April 2006

The problem for the potential in cylindrical coordinates, [math]\displaystyle{ \phi (r,\theta,z) }[/math], is given by

[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} + \frac{\partial^2 \phi}{\partial z^2} = 0, && (r,\theta,z) \in \mathbb{R}_{\gt 0} \, \times \ ]- \pi, \pi] \times \mathds{R}_{\lt 0}, }[/math]

[math]\displaystyle{ \frac{\partial \phi}{\partial z} - \alpha \phi = 0, (r,\theta,z) \in \mathds{R}_{\gt 0}\, \times \, ]\!- \pi, \pi] \times \{ 0 \}, }[/math]

as well as

[math]\displaystyle{ \frac{\partial \phi}{\partial z} = 0, (r,\theta,z) \in \mathbb{R}_{\gt 0}\, \times \,]\!- \pi, \pi] \times \{ -d \}, }[/math]

in the case of constant finite depth $d$ and

[math]\displaystyle{ \sup \big\{ \, \abs{\phi} \ \big| \ (r,\theta,z) \in \mathbb{R}_{\gt 0}\, \times \, ]\!- \pi, \pi] \times \mathbb{R}_{\lt 0} \,\big\} \lt \infty }[/math]

in the case of infinite depth. Moreover, the radiation condition

[math]\displaystyle{ \lim_{r \rightarrow \infty} \sqrt{r} \, \Big( \frac{\partial}{\partial r} - \i k \Big) \phi = 0 }[/math]

with the wavenumber $k$ also applies.