Template:Separation of variables in cylindrical coordinates in 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 the Dispersion Relation for a Free Surface

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

This equation 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 for a Free Surface

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

again it suffices to consider only the positive root of this equation. 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 for a Free Surface gives the required relation between the radian frequency and the wavenumber.

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]

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 \, 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} }[/math] [math]\displaystyle{ + \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) = 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] [math]\displaystyle{ + \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 Sommerfeld Radiation Condition 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]

(where we have set the parameters [math]\displaystyle{ F_m }[/math] so that our vertical eigenfunctions are unity at the free surface [math]\displaystyle{ z=0 }[/math]). The two terms describe the propagating and the decaying wavefields respectively.

We can write this expression in compact notation as

[math]\displaystyle{ \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]

where

[math]\displaystyle{ f_m(z) = \frac{\cos k_m (z+d)}{\cos k_m d}. }[/math]