Cylindrical Eigenfunction Expansion

The problem for the complex water velocity potential in 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 \mathbb{R}_{\gt 0} \, \times \ ]- \pi, \pi] \times \mathbb{R}_{\lt 0}, }[/math]

[math]\displaystyle{ \frac{\partial \phi}{\partial z} - \alpha \phi = 0, \quad (r,\theta,z) \in \mathbb{R}_{\gt 0}\, \times \, ]\!- \pi, \pi] \times \{ 0 \}, }[/math]

as well as

[math]\displaystyle{ \frac{\partial \phi}{\partial z} = 0, \quad (r,\theta,z) \in \mathbb{R}_{\gt 0}\, \times \,]\!- \pi, \pi] \times \{ -d \}, }[/math]

in the case of constant finite water depth [math]\displaystyle{ d }[/math] and

[math]\displaystyle{ \sup \big\{ \, |\phi| \ \big| \ (r,\theta,z) \in \mathbb{R}_{\gt 0}\, \times \, ]\!- \pi, \pi] \times \mathbb{R}_{\lt 0} \,\big\} \lt \infty }[/math]

in the case of infinite water depth. Moreover, the radiation condition

[math]\displaystyle{ \lim_{r \rightarrow \infty} \sqrt{r} \, \Big( \frac{\partial}{\partial r} - \mathrm{i} k \Big) \phi = 0 }[/math]

with the wavenumber [math]\displaystyle{ k }[/math] also applies.

The case of water of finite depth

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{ \pi }[/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} = \eta^2. }[/math]

The possible separation constants [math]\displaystyle{ \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]\displaystyle{ Z(z) }[/math] can be written as

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

since [math]\displaystyle{ \eta = 0 }[/math] is not an eigenvalue. To satisfy the bed condition, [math]\displaystyle{ G }[/math] must be [math]\displaystyle{ 0 }[/math]. [math]\displaystyle{ Z(z) }[/math] satisfies the free surface condition, provided the separation constants [math]\displaystyle{ \eta }[/math] are roots of the equation

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

or, equivalently, if they satisfy

[math]\displaystyle{ \alpha + \eta \tan \eta d = 0. }[/math]

This equation, also called dispersion relation, has an infinite number of real roots, denoted by [math]\displaystyle{ k_m }[/math] and [math]\displaystyle{ -k_m }[/math] ([math]\displaystyle{ 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]\displaystyle{ k_0 }[/math]. Writing [math]\displaystyle{ k_0 = - \mathrm{i} k }[/math], [math]\displaystyle{ k }[/math] is the (positive) root of the dispersion relation

[math]\displaystyle{ \alpha = k \tanh k d, }[/math]

again it suffices to consider only the positive root. The solutions can therefore be written as

[math]\displaystyle{ Z_m(z) = F_m \cos \big( k_m (z+d) \big), \quad m \geq 0. }[/math]

It follows that [math]\displaystyle{ k }[/math] is the previously introduced wavenumber and the dispersion relation gives the required relation to the radian frequency.

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]

another separation will be used,

[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]

Equation (\ref{pot_cyl_rt2}) also yields

[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 \, I_\nu(k_m r) + E \, K_\nu(k_m r), \quad m \in \mathbb{N},\ \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].

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} 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]

where [math]\displaystyle{ Z_m(z) }[/math] is given by equation \eqref{sol_Z_fin}. Substituting [math]\displaystyle{ Z_m }[/math] back as well as noting that [math]\displaystyle{ k_0=-\mathrm{i} k }[/math] yields

[math]\displaystyle{ \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} + \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]

Noting that [math]\displaystyle{ \cos \mathrm{i} x = \cosh x }[/math] is an even function and the relations [math]\displaystyle{ I_\nu(-\mathrm{i} x) = (-\mathrm{i})^{\nu} J_\nu(x) }[/math] where [math]\displaystyle{ J_\nu }[/math] is the Bessel function of the first kind of order [math]\displaystyle{ \nu }[/math] and [math]\displaystyle{ K_\nu (-\mathrm{i} x) = \pi / 2\,\, \mathrm{i}^{\nu+1} H_\nu^{(1)}(x) }[/math] with [math]\displaystyle{ H_\nu^{(1)} }[/math] denoting the Hankel function of the first kind of order [math]\displaystyle{ \nu }[/math], it follows that

[math]\displaystyle{ \phi (r,\theta,z) = \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} + \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]

However, [math]\displaystyle{ J_\nu }[/math] does not satisfy the radiation condition \eqref{water_rad} and neither does [math]\displaystyle{ 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]\displaystyle{ D_{m\nu}'=0 }[/math] for all [math]\displaystyle{ m,\nu }[/math]. Therefore, the eigenfunction expansion of the water velocity potential in cylindrical outgoing waves with coefficients [math]\displaystyle{ A_{m\nu} }[/math] is given by

[math]\displaystyle{ \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]

The two terms describe the propagating and the decaying wavefields respectively.

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} + \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]