Cylindrical Eigenfunction Expansion

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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]


[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]


[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]