Difference between revisions of "Cylindrical Eigenfunction Expansion"

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 +
{{incomplete pages}}
 +
 
= Introduction =
 
= Introduction =
  
 
There are any situations where we want to expand the three-dimensional linear water wave
 
There are any situations where we want to expand the three-dimensional linear water wave
 
solution in cylindrical co-ordinates. For example, scattering from a  
 
solution in cylindrical co-ordinates. For example, scattering from a  
[[Bottom Mounted Cylinder]] or scattering from a [[Circular Elastic Plate]]. In these cases it is easy to find
+
[[Bottom Mounted Cylinder]] or scattering from a [[Circular Floating Elastic Plate]]. In these cases it is easy to find
 
the solution by an expansion in the cylindrical eigenfunctions. If the depth dependence can be
 
the solution by an expansion in the cylindrical eigenfunctions. If the depth dependence can be
 
removed the solution reduces to a two dimensional problem (see [[Removing The Depth Dependence]]). While
 
removed the solution reduces to a two dimensional problem (see [[Removing The Depth Dependence]]). While
Line 11: Line 13:
 
= Outine of the theory =  
 
= Outine of the theory =  
  
 +
{{cylindrical equations}}
 +
{{sommerfeld radiation condition three dimensions}}
  
The problem for the complex water velocity potential in suitable non-dimensionalised
+
{{separation of variables in cylindrical coordinates in finite depth}}
cylindrical coordinates, <math>\phi (r,\theta,z)</math>, is given by
 
<center>
 
<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 \theta^2} + \frac{\partial^2 \phi}{\partial z^2} = 0,
 
\quad (r,\theta,z) \in \mathbb{R}_{>0} \, \times \ ]- \pi, \pi]
 
\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>
 
</center>
 
in the case of constant finite water depth <math>d</math> and
 
<center>
 
<math>
 
\sup \big\{ \, |\phi| \ \big| \ (r,\theta,z) \in \mathbb{R}_{>0}\,
 
\times \, ]\!- \pi, \pi] \times \mathbb{R}_{<0} \,\big\} < \infty
 
</math>
 
</center>
 
in the case of infinite water depth. Moreover, the [[Sommerfeld Radiation Condition]]
 
<center>
 
<math>
 
\lim_{r \rightarrow \infty} \sqrt{r} \, \Big(
 
\frac{\partial}{\partial r} - \mathrm{i} k \Big) \phi = 0
 
</math>
 
</center>
 
with the wavenumber <math>k</math> also applies.
 
  
== The case of water of finite depth ==
+
{{separation of variables for a free surface}}
 
 
The solution of the problem for the potential in finite water depth
 
can be found by a separation ansatz,
 
<center>
 
<math>
 
\phi (r,\theta,z) =: Y(r,\theta) Z(z).\,
 
</math>
 
</center>
 
Substituting this into the equation for <math>\pi</math> yields
 
<center>
 
<math>
 
\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} = \eta^2.
 
</math>
 
</center>
 
The possible separation constants <math>\eta</math> will be determined by the
 
free surface condition and the bed condition.
 
 
 
In the setting of water of finite depth, the general solution 
 
<math>Z(z)</math> can be written as
 
<center>
 
<math>
 
Z(z) = F \cos \big( \eta (z+d) \big) + G \sin \big( \eta (z+d) \big),
 
\quad \eta \in \mathbb{C} \backslash \{ 0 \},
 
</math>
 
</center>
 
since <math>\eta = 0</math> is not an eigenvalue.
 
To satisfy the bed condition, <math>G</math> must be <math>0</math>.
 
<math>Z(z)</math> satisfies the free surface condition, provided the separation
 
constants <math>\eta</math> are roots of the equation
 
<center>
 
<math>
 
- F \eta \sin \big( \eta (z+d) \big) - \alpha F \cos \big( \eta (z+d)
 
  \big) = 0, \quad z=0,
 
</math>
 
</center>
 
or, equivalently, if they satisfy the [[Dispersion Relation for a Free Surface]]
 
<center><math>
 
\alpha + \eta \tan \eta d = 0\,.
 
</math></center>
 
This equation has an
 
infinite number of real roots, denoted by <math>k_m</math> and <math>-k_m</math> (<math>m \geq
 
1</math>), but the negative roots produce the same eigenfunctions as the
 
positive ones and will therefore not be considered. It also has a pair of purely imaginary roots which
 
will be denoted by <math>k_0</math>. Writing <math>k_0 = - \mathrm{i} k</math>, <math>k</math> is the
 
(positive) root of the [[Dispersion Relation for a Free Surface]]
 
<center><math>
 
\alpha = k \tanh k d,\,
 
</math></center>
 
again it suffices to consider only the positive root of this equation. The solutions can
 
therefore be written as
 
<center>
 
<math>
 
Z_m(z) = F_m \cos \big( k_m (z+d) \big), \quad m \geq 0.
 
</math>
 
</center>
 
It follows that <math>k</math> is the previously introduced wavenumber and the [[Dispersion Relation for a Free Surface]]
 
gives the required relation between the radian frequency and the wavenumber.
 
  
For the solution of
+
{{separation of variables for the r and theta coordinates}}
<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>
 
Equation (\ref{pot_cyl_rt2}) also yields
 
<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
 
The potential <math>\phi</math> can thus be expressed in local cylindrical
Line 175: Line 26:
 
<center>
 
<center>
 
<math>
 
<math>
\phi (r,\theta,z) = \sum_{m = 0}^{\infty} Z_m(z) \sum_{\nu = -
+
\phi (r,\theta,z) = \sum_{m = 0}^{\infty} \phi_m(z) \sum_{\nu = -
 
\infty}^{\infty} \left[ D_{m\nu} I_\nu (k_m r) + E_{m\nu} K_\nu (k_m
 
\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},  
 
r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},  
 
</math>
 
</math>
 
</center>
 
</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>
 
  
 
== The case of infinitely deep water ==  
 
== The case of infinitely deep water ==  
Line 256: Line 44:
  
 
As above, the general solution can be represented as
 
As above, the general solution can be represented as
 
+
<center>
 
<math>
 
<math>
 
Z(z) = F \mathrm{e}^{\mathrm{i} \eta z} + G \mathrm{e}^{- \mathrm{i} \eta z}, \quad \eta \in \mathbb{C}
 
Z(z) = F \mathrm{e}^{\mathrm{i} \eta z} + G \mathrm{e}^{- \mathrm{i} \eta z}, \quad \eta \in \mathbb{C}
 
\backslash \{0\}.
 
\backslash \{0\}.
 
</math>
 
</math>
 
+
</center>
 
Assuming <math>\eta</math> has got a positive
 
Assuming <math>\eta</math> has got a positive
 
imaginary part, in order to satisfy the depth condition, <math>F<math> must be
 
imaginary part, in order to satisfy the depth condition, <math>F<math> must be
 
zero. <math>Z(z)</math> then satisfies the free surface condition if <math>\eta</math> is a root of
 
zero. <math>Z(z)</math> then satisfies the free surface condition if <math>\eta</math> is a root of
+
<center>
 
<math>
 
<math>
 
-G \mathrm{i} \eta \mathrm{e}^{-\mathrm{i} \eta z} - \alpha G \mathrm{e}^{-\mathrm{i} \eta z} = 0, \quad  z=0,
 
-G \mathrm{i} \eta \mathrm{e}^{-\mathrm{i} \eta z} - \alpha G \mathrm{e}^{-\mathrm{i} \eta z} = 0, \quad  z=0,
 
</math>
 
</math>
 
+
</center>
 
which yields the dispersion relation  
 
which yields the dispersion relation  
 
+
<center>
 
<math>
 
<math>
 
\eta = - \mathrm{i} \alpha.
 
\eta = - \mathrm{i} \alpha.
 
</math>
 
</math>
 
+
</center>
 
Therefore, <math>\eta</math> must even be purely imaginary. For <math>\Im \eta < 0</math>,  
 
Therefore, <math>\eta</math> must even be purely imaginary. For <math>\Im \eta < 0</math>,  
 
this is also obtained, but with a minus sign in front of
 
this is also obtained, but with a minus sign in front of
 
<math>\eta</math>. However, this yields the same solution. One solution can
 
<math>\eta</math>. However, this yields the same solution. One solution can
 
therefore be written as
 
therefore be written as
 
+
<center>
 
<math>
 
<math>
 
Z(z) = G \mathrm{e}^{\alpha z}.
 
Z(z) = G \mathrm{e}^{\alpha z}.
 
</math>
 
</math>
 
+
</center>
 
Now, <math>\eta</math> is assumed real. In this case, it is convenient to write
 
Now, <math>\eta</math> is assumed real. In this case, it is convenient to write
 
the general solution in terms of cosine and sine,
 
the general solution in terms of cosine and sine,
 
+
<center>
 
<math>
 
<math>
 
Z(z) = F \cos(\eta z) + G \sin(\eta z), \quad \eta \in \mathbb{R}
 
Z(z) = F \cos(\eta z) + G \sin(\eta z), \quad \eta \in \mathbb{R}
 
\backslash \{0\}.
 
\backslash \{0\}.
 
</math>
 
</math>
 
+
</center>
 
This solution satisfies the depth condition automatically.
 
This solution satisfies the depth condition automatically.
 
Making use of the free surface condition, it follows that
 
Making use of the free surface condition, it follows that
 
+
<center>
 
<math>
 
<math>
 
(-\eta F - \alpha G) \sin (\eta z) + (\eta G - \alpha F) \cos(\eta z)
 
(-\eta F - \alpha G) \sin (\eta z) + (\eta G - \alpha F) \cos(\eta z)
 
= 0, \quad z=0,
 
= 0, \quad z=0,
 
</math>
 
</math>
 
+
</center>
 
which can be solved for <math>G</math>,
 
which can be solved for <math>G</math>,
 
+
<center>
 
<math>
 
<math>
 
G = \frac{\alpha}{\eta} F.
 
G = \frac{\alpha}{\eta} F.
 
</math>
 
</math>
 
+
</center>
 
Substituting this back gives
 
Substituting this back gives
 
+
<center>
 
<math>
 
<math>
 
Z(z) = F \big( \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z)
 
Z(z) = F \big( \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z)
 
\big) , \quad \eta \in \mathbb{R} \backslash \{0\}.  
 
\big) , \quad \eta \in \mathbb{R} \backslash \{0\}.  
 
</math>
 
</math>
 
+
</center>
 
Obviously, a negative value of <math>\eta</math> produces the same
 
Obviously, a negative value of <math>\eta</math> produces the same
 
eigenfunction as the positive one. Therefore, only positive ones are
 
eigenfunction as the positive one. Therefore, only positive ones are
 
considered, leading to the definition
 
considered, leading to the definition
 
+
<center>
 
<math>
 
<math>
 
\psi(z,\eta) := \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z), \quad
 
\psi(z,\eta) := \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z), \quad
 
(z,\eta) \in \mathbb{R}_{\leq0} \times \mathbb{R}_{>0}.
 
(z,\eta) \in \mathbb{R}_{\leq0} \times \mathbb{R}_{>0}.
 
</math>
 
</math>
 
+
</center>
 
This gives the vertical eigenfunctions in infinite depth.
 
This gives the vertical eigenfunctions in infinite depth.
  
Line 345: Line 133:
 
expansion of the water velocity potential into cylindrical outgoing
 
expansion of the water velocity potential into cylindrical outgoing
 
waves in water of infinite depth,  
 
waves in water of infinite depth,  
 
+
<center>
 
<math>
 
<math>
 
\phi (r,\theta,z) =  \mathrm{e}^{\alpha z} \sum_{\nu = -
 
\phi (r,\theta,z) =  \mathrm{e}^{\alpha z} \sum_{\nu = -
Line 353: Line 141:
 
\theta} \mathrm{d}\eta.
 
\theta} \mathrm{d}\eta.
 
</math>
 
</math>
 +
</center>
  
 
= Example: Expansion of a plane wave =  
 
= Example: Expansion of a plane wave =  
Line 391: Line 180:
 
== Finite depth ==
 
== Finite depth ==
  
It is easily seen that this is the same as in the infinite-depth case except that we need to replace <math>\mathrm{e}^{ \alpha z}</math> by <math>\frac{\cosh(k (z+d))}{\cosh kd}</math> and <math>\alpha</math> by <math>k</math>. Therefore, we have
+
It is easily seen that this is the same as in the infinite-depth case except that we need to replace <math>\mathrm{e}^{ \alpha z}</math> by <math>\frac{\cosh(k (z+h))}{\cosh kh}</math> and <math>\alpha</math> by <math>k</math>. Therefore, we have
  
 
<center><math>
 
<center><math>
 
\phi^{\mathrm{In}}(r_l,\theta_l,z) = A \frac{g}{\omega} \, e^{\mathrm{i} k (O_x^l
 
\phi^{\mathrm{In}}(r_l,\theta_l,z) = A \frac{g}{\omega} \, e^{\mathrm{i} k (O_x^l
\cos \chi + O_x^l \sin \chi)} \, \frac{\cosh(k (z+d))}{\cosh kd}
+
\cos \chi + O_x^l \sin \chi)} \, \frac{\cosh(k (z+h))}{\cosh kh}
 
\sum_{\mu = -\infty}^{\infty} \mathrm{e}^{\mathrm{i}\mu (\pi/2 -  \chi)}
 
\sum_{\mu = -\infty}^{\infty} \mathrm{e}^{\mathrm{i}\mu (\pi/2 -  \chi)}
 
J_\mu(k r_l) \mathrm{e}^{\mathrm{i}\mu \theta_l}.
 
J_\mu(k r_l) \mathrm{e}^{\mathrm{i}\mu \theta_l}.
Line 401: Line 190:
  
 
[[Category:Problems with Cylindrical Symmetry]]
 
[[Category:Problems with Cylindrical Symmetry]]
 +
[[Category:Eigenfunction Matching Method]]

Latest revision as of 09:34, 20 October 2009


Introduction

There are any situations where we want to expand the three-dimensional linear water wave solution in cylindrical co-ordinates. For example, scattering from a Bottom Mounted Cylinder or scattering from a Circular Floating Elastic Plate. In these cases it is easy to find the solution by an expansion in the cylindrical eigenfunctions. If the depth dependence can be removed the solution reduces to a two dimensional problem (see Removing The Depth Dependence). While the theory here does apply in this two dimensional situtation, the theory is presented here for the fully three dimensional (depth dependent) case. We begin by assuming the Frequency Domain Problem.

Outine of the theory

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]

In three-dimensions the Sommerfeld Radiation Condition is

[math]\displaystyle{ \sqrt{|\mathbf{r}|}\left( \frac{\partial}{\partial|\mathbf{r}|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.} }[/math]

where [math]\displaystyle{ \phi^{\mathrm{{I}}} }[/math] is the incident potential.

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.

Separation of variables for a free surface

We use separation of variables

We express the potential as

[math]\displaystyle{ \phi(x,z) = X(x)Z(z)\, }[/math]

and then Laplace's equation becomes

[math]\displaystyle{ \frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2 }[/math]

The separation of variables equation for deriving free surface eigenfunctions is as follows:

[math]\displaystyle{ Z^{\prime\prime} + k^2 Z =0. }[/math]

subject to the boundary conditions

[math]\displaystyle{ Z^{\prime}(-h) = 0 }[/math]

and

[math]\displaystyle{ Z^{\prime}(0) = \alpha Z(0) }[/math]

We can then use the boundary condition at [math]\displaystyle{ z=-h \, }[/math] to write

[math]\displaystyle{ Z = \frac{\cos k(z+h)}{\cos kh} }[/math]

where we have chosen the value of the coefficent so we have unit value at [math]\displaystyle{ z=0 }[/math]. The boundary condition at the free surface ([math]\displaystyle{ z=0 \, }[/math]) gives rise to:

[math]\displaystyle{ k\tan\left( kh\right) =-\alpha \, }[/math]

which is the Dispersion Relation for a Free Surface

The above equation is a transcendental equation. If we solve for all roots in the complex plane we find that the first root is a pair of imaginary roots. We denote the imaginary solutions of this equation by [math]\displaystyle{ k_{0}=\pm ik \, }[/math] and the positive real solutions by [math]\displaystyle{ k_{m} \, }[/math], [math]\displaystyle{ m\geq1 }[/math]. The [math]\displaystyle{ k \, }[/math] of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

[math]\displaystyle{ \cos ix = \cosh x, \quad \sin ix = i\sinh x, }[/math]

to arrive at the dispersion relation

[math]\displaystyle{ \alpha = k\tanh kh. }[/math]

We note that for a specified frequency [math]\displaystyle{ \omega \, }[/math] the equation determines the wavenumber [math]\displaystyle{ k \, }[/math].

Finally we define the function [math]\displaystyle{ Z(z) \, }[/math] as

[math]\displaystyle{ \chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0 }[/math]

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

[math]\displaystyle{ \int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn} }[/math]

where

[math]\displaystyle{ A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right). }[/math]

Separation of Variable for the [math]\displaystyle{ r }[/math] and [math]\displaystyle{ \theta }[/math] coordinates

For the solution of

[math]\displaystyle{ \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]

we use the separation

[math]\displaystyle{ \,\!Y(r,\theta) =: R(r) \Theta(\theta). }[/math]

Substituting this into Laplace's equation yields

[math]\displaystyle{ \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]

where the separation constant [math]\displaystyle{ \eta }[/math] must be an integer, say [math]\displaystyle{ \nu }[/math], in order for the potential to be continuous. [math]\displaystyle{ \Theta (\theta) }[/math] can therefore be expressed as

[math]\displaystyle{ \Theta (\theta) = C \, \mathrm{e}^{\mathrm{i} \nu \theta}, \quad \nu \in \mathbb{Z}. }[/math]

We also obtain the following expression

[math]\displaystyle{ 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]

Substituting [math]\displaystyle{ \tilde{r}:=k_m r }[/math] and writing [math]\displaystyle{ \tilde{R} (\tilde{r}) := R(\tilde{r}/k_m) = R(r) }[/math], this can be rewritten as

[math]\displaystyle{ \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]

which is the modified version of Bessel's equation. Substituting back, the general solution is given by

[math]\displaystyle{ R(r) = D_\nu \, I_\nu(k_m r) + E_\nu \, K_\nu(k_m r),\ \nu \in \mathbb{Z}, }[/math]

where [math]\displaystyle{ I_\nu }[/math] and [math]\displaystyle{ K_\nu }[/math] are the modified Bessel functions of the first and second kind, respectively, of order [math]\displaystyle{ \nu }[/math].

Note that [math]\displaystyle{ K_\nu (\mathrm{i} x) = \pi / 2\,\, \mathrm{i}^{\nu+1} H_\nu^{(2)}(x) }[/math] with [math]\displaystyle{ H_\nu^{(2)} }[/math] denoting the Hankel function of the second kind of order [math]\displaystyle{ \nu }[/math]. Also, [math]\displaystyle{ I_\nu }[/math] does not satisfy the Sommerfeld Radiation Condition since it becomes unbounded for increasing real argument and it represents incoming waves.

The potential [math]\displaystyle{ \phi }[/math] can thus be expressed in local cylindrical coordinates as

[math]\displaystyle{ \phi (r,\theta,z) = \sum_{m = 0}^{\infty} \phi_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]

The case of infinitely deep water

A solution will be developed for the same setting as before but under the assumption of water of infinite depth. As in the previous section, Laplace's equation must be solved in cylindrical coordinates satisfying the free surface and the radiation condition. However, instead of the bed condition, the water velocity potential is also required to satisfy the depth condition. Therefore, [math]\displaystyle{ Z(z) }[/math] must be solved for satisfying the depth condition. It will turn out that in the case of infinitely deep water an uncountable amount of separation constants [math]\displaystyle{ \eta }[/math] is valid.

As above, the general solution can be represented as

[math]\displaystyle{ Z(z) = F \mathrm{e}^{\mathrm{i} \eta z} + G \mathrm{e}^{- \mathrm{i} \eta z}, \quad \eta \in \mathbb{C} \backslash \{0\}. }[/math]

Assuming [math]\displaystyle{ \eta }[/math] has got a positive imaginary part, in order to satisfy the depth condition, [math]\displaystyle{ F\lt math\gt must be zero. \lt math\gt Z(z) }[/math] then satisfies the free surface condition if [math]\displaystyle{ \eta }[/math] is a root of

[math]\displaystyle{ -G \mathrm{i} \eta \mathrm{e}^{-\mathrm{i} \eta z} - \alpha G \mathrm{e}^{-\mathrm{i} \eta z} = 0, \quad z=0, }[/math]

which yields the dispersion relation

[math]\displaystyle{ \eta = - \mathrm{i} \alpha. }[/math]

Therefore, [math]\displaystyle{ \eta }[/math] must even be purely imaginary. For [math]\displaystyle{ \Im \eta \lt 0 }[/math], this is also obtained, but with a minus sign in front of [math]\displaystyle{ \eta }[/math]. However, this yields the same solution. One solution can therefore be written as

[math]\displaystyle{ Z(z) = G \mathrm{e}^{\alpha z}. }[/math]

Now, [math]\displaystyle{ \eta }[/math] is assumed real. In this case, it is convenient to write the general solution in terms of cosine and sine,

[math]\displaystyle{ Z(z) = F \cos(\eta z) + G \sin(\eta z), \quad \eta \in \mathbb{R} \backslash \{0\}. }[/math]

This solution satisfies the depth condition automatically. Making use of the free surface condition, it follows that

[math]\displaystyle{ (-\eta F - \alpha G) \sin (\eta z) + (\eta G - \alpha F) \cos(\eta z) = 0, \quad z=0, }[/math]

which can be solved for [math]\displaystyle{ G }[/math],

[math]\displaystyle{ G = \frac{\alpha}{\eta} F. }[/math]

Substituting this back gives

[math]\displaystyle{ Z(z) = F \big( \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z) \big) , \quad \eta \in \mathbb{R} \backslash \{0\}. }[/math]

Obviously, a negative value of [math]\displaystyle{ \eta }[/math] produces the same eigenfunction as the positive one. Therefore, only positive ones are considered, leading to the definition

[math]\displaystyle{ \psi(z,\eta) := \cos(\eta z) + \frac{\alpha}{\eta} \sin(\eta z), \quad (z,\eta) \in \mathbb{R}_{\leq0} \times \mathbb{R}_{\gt 0}. }[/math]

This gives the vertical eigenfunctions in infinite depth.

For the radial and angular coordinate the same separation can be used as in the finite depth case so that the general solution of problem can be written as

[math]\displaystyle{ \phi (r,\theta,z) = \mathrm{e}^{\alpha z} \sum_{\nu = - \infty}^{\infty} \left[ E_\nu (-\mathrm{i} \alpha) I_\nu (-\mathrm{i} \alpha r) + F_{\nu} (-\mathrm{i} \alpha) K_\nu (-\mathrm{i} \alpha r) \right] \mathrm{e}^{\mathrm{i} \nu \theta} }[/math]
[math]\displaystyle{ + \int\limits_0^{\infty} \psi (z,\eta) \sum_{\nu = - \infty}^{\infty} \left[ E_\nu I_\nu (\eta r) + F_{\nu} (\eta) K_\nu (\eta r) \right] \mathrm{e}^{\mathrm{i} \nu \theta} \mathrm{d}\eta. }[/math]

Making use of the radiation condition as well as the relations of the Bessel functions in the same way as in the finite depth case, this can be rewritten as the eigenfunction expansion of the water velocity potential into cylindrical outgoing waves in water of infinite depth,

[math]\displaystyle{ \phi (r,\theta,z) = \mathrm{e}^{\alpha z} \sum_{\nu = - \infty}^{\infty} A_{\nu} (\mathrm{i} \alpha) H_\nu^{(1)} (\alpha r) \mathrm{e}^{\mathrm{i} \nu \theta} + \int\limits_0^{\infty} \psi (z,\eta) \sum_{\nu = - \infty}^{\infty} A_{\nu} (\eta) K_\nu (\eta r) \mathrm{e}^{\mathrm{i} \nu \theta} \mathrm{d}\eta. }[/math]

Example: Expansion of a plane wave

Infinite depth

In Cartesian coordinates centred at the origin, the wavefield due to a plane incident wave travelling in the direction making an angle [math]\displaystyle{ \chi }[/math] with the [math]\displaystyle{ x }[/math]-axis is given by

[math]\displaystyle{ \phi^{\mathrm{In}}(x,y,z) = A \frac{g}{\omega} \, \mathrm{e}^{\mathrm{i}\alpha (x \cos \chi + y \sin \chi)+ \alpha z}, }[/math]

where [math]\displaystyle{ A }[/math] is the amplitude (in displacement). We want to express the ambient wavefield in the eigenfunction expansion of an incoming wave in the local coordinates of a body whose mean-centre position is [math]\displaystyle{ O = (O_x,O_y) }[/math]. The ambient wave can be represented in an eigenfunction expansion centred at the origin as

[math]\displaystyle{ \phi^{\mathrm{In}}(x,y,z) = A \frac{g}{\omega} \mathrm{e}^{ \alpha z} \sum_{\mu = -\infty}^{\infty} \mathrm{e}^{\mathrm{i}\mu (\pi/2 - \theta + \chi)} J_\mu(\alpha r) }[/math]

(cf. Linton and McIver 2001, p. 169). Since the local coordinates of the body, that is [math]\displaystyle{ (r_l,\theta_l,z) }[/math], are centred at its mean-centre position, a phase factor has to be defined which accounts for the position from the origin. Including this phase factor, the ambient wavefield at body is given by

[math]\displaystyle{ \phi^{\mathrm{In}}(r_l,\theta_l,z) = A \frac{g}{\omega} \, e^{\mathrm{i}\alpha (O_x^l \cos \chi + O_x^l \sin \chi)} \, \mathrm{e}^{\alpha z} \sum_{\mu = -\infty}^{\infty} \mathrm{e}^{\mathrm{i}\mu (\pi/2 - \chi)} J_\mu(\alpha r_l) \mathrm{e}^{\mathrm{i}\mu \theta_l}. }[/math]


Finite depth

It is easily seen that this is the same as in the infinite-depth case except that we need to replace [math]\displaystyle{ \mathrm{e}^{ \alpha z} }[/math] by [math]\displaystyle{ \frac{\cosh(k (z+h))}{\cosh kh} }[/math] and [math]\displaystyle{ \alpha }[/math] by [math]\displaystyle{ k }[/math]. Therefore, we have

[math]\displaystyle{ \phi^{\mathrm{In}}(r_l,\theta_l,z) = A \frac{g}{\omega} \, e^{\mathrm{i} k (O_x^l \cos \chi + O_x^l \sin \chi)} \, \frac{\cosh(k (z+h))}{\cosh kh} \sum_{\mu = -\infty}^{\infty} \mathrm{e}^{\mathrm{i}\mu (\pi/2 - \chi)} J_\mu(k r_l) \mathrm{e}^{\mathrm{i}\mu \theta_l}. }[/math]