Difference between revisions of "Template:Separation of variables in cylindrical coordinates in finite depth"

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</math>
 
</math>
 
</center>
 
</center>
Substituting this into the equation for <math>\pi</math> yields
+
Substituting this into the equation for <math>\phi</math> yields
 
<center>
 
<center>
 
<math>
 
<math>
Line 12: Line 12:
 
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.
 
{{separation of variables for a free surface}}
 
 
For the solution of
 
<center>
 
<math>
 
\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} = k_m^2 Y(r,\theta),
 
</math>
 
</center>
 
another separation will be used,
 
<center>
 
<math>
 
Y(r,\theta) =: R(r) \Theta(\theta).
 
</math>
 
</center>
 
Substituting this into Laplace's equation yields
 
<center>
 
<math>
 
\frac{r^2}{R(r)} \left[ \frac{1}{r} \frac{\mathrm{d}}{\mathrm{d}r} \left( r
 
\frac{\mathrm{d} R}{\mathrm{d}r} \right) - k_m^2 R(r) \right] = -
 
\frac{1}{\Theta (\theta)} \frac{\mathrm{d}^2 \Theta}{\mathrm{d}
 
\theta^2} = \eta^2,
 
</math>
 
</center>
 
where the separation constant <math>\eta</math> must be an integer, say <math>\nu</math>,
 
in order for the potential to be continuous. <math>\Theta
 
(\theta)</math> can therefore be expressed as
 
<center>
 
<math>
 
\Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}.
 
</math>
 
</center>
 
We also obtain the following expression
 
<center>
 
<math>
 
r \frac{\mathrm{d}}{\mathrm{d}r} \left( r \frac{\mathrm{d}
 
R}{\mathrm{d} r} \right) - (\nu^2 + k_m^2 r^2) R(r) = 0, \quad \nu \in
 
\mathbb{Z}.
 
</math>
 
</center>
 
Substituting <math>\tilde{r}:=k_m r</math> and writing <math>\tilde{R} (\tilde{r}) :=
 
R(\tilde{r}/k_m) = R(r)</math>, this can be rewritten as
 
<center>
 
<math>
 
\tilde{r}^2 \frac{\mathrm{d}^2 \tilde{R}}{\mathrm{d} \tilde{r}^2}
 
+ \tilde{r} \frac{\mathrm{d} \tilde{R}}{\mathrm{d} \tilde{r}}
 
- (\nu^2 + \tilde{r}^2)\, \tilde{R} = 0, \quad \nu \in \mathbb{Z},
 
</math>
 
</center>
 
which is the modified version of Bessel's equation. Substituting back,
 
the general solution is given by
 
<center>
 
<math>
 
R(r) = D \, I_\nu(k_m r) + E \, K_\nu(k_m r), \quad m \in
 
\mathbb{N},\ \nu \in \mathbb{Z},
 
</math>
 
</center>
 
where <math>I_\nu</math> and <math>K_\nu</math> are the modified
 
[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first
 
and second kind, respectively, of order <math>\nu</math>.
 
 
The potential <math>\phi</math> can thus be expressed in local cylindrical
 
coordinates as
 
<center>
 
<math>
 
\phi (r,\theta,z) = \sum_{m = 0}^{\infty} Z_m(z) \sum_{\nu = -
 
\infty}^{\infty} \left[ D_{m\nu} I_\nu (k_m r) + E_{m\nu} K_\nu (k_m
 
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},
 
</math>
 
</center>
 
where <math>Z_m(z)</math> is given by equation \eqref{sol_Z_fin}. Substituting <math>Z_m</math>
 
back as well as noting that <math>k_0=-\mathrm{i} k</math> yields
 
 
<center><math>
 
\phi (r,\theta,z)
 
= F_0\cos(-\mathrm{i} k (z+d)) \sum_{\nu = - \infty}^{\infty}
 
\left[ D_{0\nu} I_\nu (-\mathrm{i} k r) + E_{0\nu} K_\nu (-\mathrm{i} k r)\right]
 
\mathrm{e}^{\mathrm{i} \nu \theta}
 
</math></center>
 
<center><math>
 
+ \sum_{m = 1}^{\infty} F_m\cos(k_m(z+d)) \sum_{\nu = -
 
\infty}^{\infty} \left[ D_{m\nu} I_\nu (k_m r) + E_{m\nu} K_\nu (k_m
 
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}.
 
</math></center>
 
 
Noting that <math>\cos \mathrm{i} x = \cosh x</math> is an even function and the
 
relations <math>I_\nu(-\mathrm{i} x) = (-\mathrm{i})^{\nu} J_\nu(x)</math> where <math>J_\nu</math> is the Bessel
 
function of the first kind of order <math>\nu</math> and <math>K_\nu (-\mathrm{i} x) = \pi / 2\,\,
 
\mathrm{i}^{\nu+1} H_\nu^{(1)}(x)</math> with <math>H_\nu^{(1)}</math> denoting
 
the Hankel function of the first kind of order <math>\nu</math>, it follows that
 
 
<center><math>
 
\phi (r,\theta,z)
 
= F_0\cosh(k (z+d)) \sum_{\nu = - \infty}^{\infty}
 
\left[ D_{0\nu}' J_\nu (k r) + E_{0\nu}' H_\nu^{(1)} (k r)\right]
 
\mathrm{e}^{\mathrm{i} \nu \theta}
 
</math></center>
 
<center><math>
 
+ \sum_{m = 1}^{\infty} F_m \cos(k_m(z+d)) \sum_{\nu = -
 
\infty}^{\infty} \left[ D_{m\nu}' I_\nu (k_m r) + E_{m\nu}' K_\nu (k_m
 
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}.
 
</math></center>
 
 
However, <math>J_\nu</math> does not satisfy the [[Sommerfeld Radiation Condition]]
 
and neither does <math>I_\nu</math>
 
since it becomes unbounded for increasing real argument. These
 
two solutions represent incoming waves which will also be
 
required later.
 
 
Therefore, the solution of the problem requires <math>D_{m\nu}'=0</math>
 
for all <math>m,\nu</math>. Therefore, the
 
eigenfunction expansion of the water velocity potential in
 
cylindrical outgoing waves with coefficients <math>A_{m\nu}</math> is given by
 
<center><math>
 
\phi (r,\theta,z) = \frac{\cosh(k (z+d))}{\cosh kd} \sum_{\nu = -
 
\infty}^{\infty} A_{0\nu} H_\nu^{(1)} (k r) \mathrm{e}^{\mathrm{i} \nu \theta} + \sum_{m = 1}^{\infty} \frac{\cos(k_m(z+d))}{\cos k_m d}
 
\sum_{\nu = - \infty}^{\infty} A_{m\nu} K_\nu (k_m r) \mathrm{e}^{\mathrm{i} \nu \theta}.
 
</math></center>
 
(where we have set the parameters <math>F_m</math> so that our vertical
 
eigenfunctions are unity at the free surface <math>z=0</math>).
 
The two terms describe the propagating and the decaying wavefields
 
respectively.
 
 
We can write this expression in compact notation as
 
 
<center><math>
 
\phi (r,\theta,z) =  \sum_{m = 0}^{\infty} f_m(z)
 
\sum_{\nu = - \infty}^{\infty} A_{m\nu} K_\nu (k_m r) \mathrm{e}^{\mathrm{i} \nu \theta}.
 
</math></center>
 
where
 
<center><math>
 
f_m(z) = \frac{\cos k_m (z+d)}{\cos k_m d}.
 
</math></center>
 

Latest revision as of 05:16, 26 August 2008

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.