Template:Cylindrical equations: Difference between revisions

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\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} + \frac{\partial^2 \phi}{\partial z^2} = 0,
\phi}{\partial \theta^2} + \frac{\partial^2 \phi}{\partial z^2} = 0,
\quad (r,\theta,z) \in \mathbb{R}_{>0} \, \times \ ]- \pi, \pi]
\quad (r,\theta,z) \in \Omega
\times  \mathbb{R}_{<0}, </math>
</center>
<center>
<math>\frac{\partial \phi}{\partial z} - \alpha \phi = 0, \quad
(r,\theta,z) \in \mathbb{R}_{>0}\,
\times \, ]\!- \pi, \pi]  \times  \{ 0 \},</math>
</center>
as well as
<center>
<math>
\frac{\partial \phi}{\partial z} = 0, \quad (r,\theta,z) \in
\mathbb{R}_{>0}\, \times \,]\!- \pi, \pi] \times \{ -d \},
</math>
</math>
</center>
</center>
in the case of constant finite water depth <math>d</math> and
<center>
<center>
<math>
<math>\frac{\partial \phi}{\partial z} = \alpha \phi , \quad
\sup \big\{ \, |\phi| \ \big| \ (r,\theta,z) \in \mathbb{R}_{>0}\,
z=0
\times \, ]\!- \pi, \pi] \times \mathbb{R}_{<0} \,\big\} < \infty
</math>
</math>
</center>
</center>
in the case of infinite water depth. Moreover, the [[Sommerfeld Radiation Condition]]
<center>
<center>
<math>
<math>
\lim_{r \rightarrow \infty} \sqrt{r} \, \Big(
\frac{\partial \phi}{\partial z} = 0, z=-h
\frac{\partial}{\partial r} - \mathrm{i} k \Big) \phi = 0
</math>
</math>
</center>
</center>
with the wavenumber <math>k</math> also applies.

Latest revision as of 07:53, 25 August 2008

The problem for the complex water velocity potential in suitable non-dimensionalised 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, \quad (r,\theta,z) \in \Omega }[/math]

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

[math]\displaystyle{ \frac{\partial \phi}{\partial z} = 0, z=-h }[/math]

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