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| r} \left( r \frac{\partial Y}{\partial r} \right) + \frac{1}{r^2} | | r} \left( r \frac{\partial Y}{\partial r} \right) + \frac{1}{r^2} |
| \frac{\partial^2 Y}{\partial \theta^2} \right] = - \frac{1}{Z(z)} | | \frac{\partial^2 Y}{\partial \theta^2} \right] = - \frac{1}{Z(z)} |
− | \frac{\mathrm{d}^2 Z}{\mathrm{d} z^2} = \eta^2. | + | \frac{\mathrm{d}^2 Z}{\mathrm{d} z^2} = k^2. |
| </math> | | </math> |
| </center> | | </center> |
− | The possible separation constants <math>\eta</math> will be determined by the | + | The possible separation constants <math>k</math> will be determined by the |
| free surface condition and the bed condition. | | free surface condition and the bed condition. |
− |
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− | {{separation of variables for a free surface}}
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− |
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− | For the solution of
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− | <center>
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− | <math>
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− | \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial
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− | Y}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 Y}{\partial
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− | \theta^2} = k_m^2 Y(r,\theta),
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− | </math>
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− | </center>
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− | another separation will be used,
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− | <center>
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− | <math>
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− | Y(r,\theta) =: R(r) \Theta(\theta).
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− | </math>
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− | </center>
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− | Substituting this into Laplace's equation yields
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− | <center>
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− | <math>
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− | \frac{r^2}{R(r)} \left[ \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d}r} \left( r
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− | \frac{\mathrm{d} R}{\mathrm{d}r} \right) - k_m^2 R(r) \right] = -
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− | \frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d}
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− | \theta^2} = \eta^2,
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− | </math>
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− | </center>
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− | where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
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− | in order for the potential to be continuous. <math>\Theta
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− | (\theta)</math> can therefore be expressed as
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− | <center>
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− | <math>
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− | \Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
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− | </math>
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− | </center>
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− | We also obtain the following expression
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− | <center>
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− | <math>
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− | r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d}
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− | R}{\mathrm{d} r} \right) - (\nu^2 + k_m^2 r^2) R(r) = 0, \quad \nu \in
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− | \mathbb{Z}.
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− | </math>
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− | </center>
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− | Substituting <math>\tilde{r}:=k_m r</math> and writing <math>\tilde{R} (\tilde{r}) :=
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− | R(\tilde{r}/k_m) = R(r)</math>, this can be rewritten as
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− | <center>
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− | <math>
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− | \tilde{r}^2 \frac{\mathrm{d}^2 \tilde{R}}{\mathrm{d} \tilde{r}^2}
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− | + \tilde{r} \frac{\mathrm{d} \tilde{R}}{\mathrm{d} \tilde{r}}
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− | - (\nu^2 + \tilde{r}^2)\, \tilde{R} = 0, \quad \nu \in \mathbb{Z},
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− | </math>
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− | </center>
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− | which is the modified version of Bessel's equation. Substituting back,
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− | the general solution is given by
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− | <center>
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− | <math>
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− | R(r) = D \, I_\nu(k_m r) + E \, K_\nu(k_m r), \quad m \in
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− | \mathbb{N},\ \nu \in \mathbb{Z},
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− | </math>
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− | </center>
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− | where <math>I_\nu</math> and <math>K_\nu</math> are the modified
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− | [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first
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− | and second kind, respectively, of order <math>\nu</math>.
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− |
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− | The potential <math>\phi</math> can thus be expressed in local cylindrical
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− | coordinates as
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− | <center>
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− | <math>
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− | \phi (r,\theta,z) = \sum_{m = 0}^{\infty} \phi_m(z) \sum_{\nu = -
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− | \infty}^{\infty} \left[ D_{m\nu} I_\nu (k_m r) + E_{m\nu} K_\nu (k_m
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− | r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},
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− | </math>
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− | </center>
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− |
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− |
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− | Note that <math>K_\nu (-\mathrm{i} x) = \pi / 2\,\,
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− | \mathrm{i}^{\nu+1} H_\nu^{(1)}(x)</math> with <math>H_\nu^{(1)}</math> denoting
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− | the Hankel function of the first kind of order <math>\nu</math>.
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− |
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− |
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− | Also, <math>I_\nu</math> does not satisfy the [[Sommerfeld Radiation Condition]]
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− | since it becomes unbounded for increasing real argument. These
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− | solution represents incoming waves.
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The solution of the problem for the potential in finite water depth
can be found by a separation ansatz,
[math]\displaystyle{
\phi (r,\theta,z) =: Y(r,\theta) Z(z).\,
}[/math]
Substituting this into the equation for [math]\displaystyle{ \phi }[/math] yields
[math]\displaystyle{
\frac{1}{Y(r,\theta)} \left[ \frac{1}{r} \frac{\partial}{\partial
r} \left( r \frac{\partial Y}{\partial r} \right) + \frac{1}{r^2}
\frac{\partial^2 Y}{\partial \theta^2} \right] = - \frac{1}{Z(z)}
\frac{\mathrm{d}^2 Z}{\mathrm{d} z^2} = k^2.
}[/math]
The possible separation constants [math]\displaystyle{ k }[/math] will be determined by the
free surface condition and the bed condition.