Free-Surface Green Function

Equations for the Green function

The Free-Surface Green function is a function which satisfies the following equation (in Finite Depth) [math]\displaystyle{ \mathbf{x}=(x,y,z) }[/math] and [math]\displaystyle{ \mathbf{\xi}=(a,b,c) }[/math]

[math]\displaystyle{ \nabla_{\mathbf{x}}^{2}G(\mathbf{x},\mathbf{\xi})=\delta(\mathbf{x}-\mathbf{\xi}), \, -\infty\lt z\lt 0 }[/math]

[math]\displaystyle{ \frac{\partial G}{\partial z}=0, \, z=h, }[/math]

[math]\displaystyle{ \frac{\partial G}{\partial z} = k_{\infty}\phi,\,z\in\Gamma_s, }[/math]

[math]\displaystyle{ \frac{\partial G}{\partial z} = L\phi, \, z\in\Gamma_w. }[/math]

where [math]\displaystyle{ k_{\infty} }[/math] is the wavenumber in Infinite Depth which is given by [math]\displaystyle{ k_{\infty}=\omega^2/g }[/math] where [math]\displaystyle{ g }[/math] is gravity.

Two Dimensional Representations

Many expressions for the Green function have been given. In two dimensions it can be written as

Three Dimensional Representations

Finite Depth

The most important representation of the finite depth free surface Green function is the eigenfunction expansion given by John 1950

[math]\displaystyle{ G(\mathbf{x};\mathbf{\xi}) = \frac{i}{2} \, \frac{\alpha^2-k^2}{(\alpha^2-k^2)d-\alpha}\, \cosh k(z+d)\, \cosh k(c+d) \, H_0^{(1)}(k r) + \frac{1}{\pi} \sum_{m=1}^{\infty} \frac{k_m^2+\alpha^2}{(k_m^2+\alpha^2)d-\alpha}\, \cos k_m(z+d)\, \cos k_m(c+d) \, K_0(k_m r), }[/math]

Infinite Depth

In three dimensions and infinite depth the Green function [math]\displaystyle{ G }[/math], for [math]\displaystyle{ r\gt 0 }[/math], was given by {\em Havelock\/} \cite{havelock55} as

Let [math]\displaystyle{ (r,\theta) }[/math] be spherical coordinates such that

[math]\displaystyle{ x - a = r \cos \theta,\, }[/math]

[math]\displaystyle{ y - b = r \sin \theta,\, }[/math]

and let [math]\displaystyle{ R_0 }[/math] and [math]\displaystyle{ R_1 }[/math] denote the distance from the source point [math]\displaystyle{ \mathbf{\xi} = (a,b,c) }[/math] and the distance from the mirror source point [math]\displaystyle{ \bar{\mathbf{\xi}} = (a,b,-c) }[/math] respectively, [math]\displaystyle{ R_0^2 = (x-a)^2 + (y-b)^2 + (z-c)^2 }[/math] and [math]\displaystyle{ R_1^2 = (x-a)^2 + (y-b)^2 + (z+c)^2 }[/math].

[math]\displaystyle{ G(\mathbf{x};\mathbf{\xi}) = \frac{i \alpha}{2} e^{\alpha (z+c)} \, H_0^{(1)}(\alpha r) + \frac{1}{4 \pi R_0} + \frac{1}{4 \pi R_1} - \frac{1}{\pi^2} \int\limits_{0}^{\infty} \frac{\alpha}{\eta^2 + \alpha^2} \big( \alpha \cos \eta (z+c) - \eta \sin \eta (z+c) \big) K_0(\eta r) d\eta, }[/math]

where [math]\displaystyle{ H^{(1)}_0 }[/math] and [math]\displaystyle{ K_0 }[/math] denote the Hankel function of the first kind and the modified Bessel function of the second kind, both of order zero as defined in {\em Abramowitz \& Stegun} \cite{abr_ste}. This Green function will be referred to as {\em Havelock\/}'s Green function. It should be noted that {\em Havelock\/}'s Green function can also be written in the following closely related form,

[math]\displaystyle{ G(\mathbf{x};\mathbf{\xi}) = \frac{i \alpha}{2} e^{\alpha (z+c)} \, H_0^{(1)}(\alpha r) + \frac{1}{4 \pi R_0} + \frac{1}{2 \pi^2} \int\limits_{0}^{\infty} \frac{(\eta^2 - \alpha^2) \cos \eta (z+c) + 2 \eta \alpha \sin \eta (z+c)}{\eta^2 + \alpha^2} K_0(\eta r) d\eta }[/math]

\cite{linton01}. An equivalent representation is due to {\em Kim} \cite{kim65} for [math]\displaystyle{ r\gt 0 }[/math], although implicitly given in the work of {\em Havelock} \cite{havelock55}, and is given by

[math]\displaystyle{ G(\mathbf{x};\mathbf{\xi}) = \frac{1}{4 \pi R_0} + \frac{1}{4 \pi R_1} - \frac{\alpha}{4} e^{\alpha (z+c)} \Big(\mathbf{H}_0(\alpha r) + Y_0(\alpha r) - 2i J_0 (\alpha r) + \frac{2}{\pi} \int\limits_{z+c}^0 \frac{e^{-\alpha \eta}}{\sqrt{r^2 + \eta^2}} d\eta \Big), }[/math]

where [math]\displaystyle{ J_0 }[/math] and [math]\displaystyle{ Y_0 }[/math] are the Bessel functions of order zero of the first and second kind and [math]\displaystyle{ \mathbf{H}_0 }[/math] is the Struve function of order zero.

The expression due to Peter and Meylan 2004 is

[math]\displaystyle{ G(\mathbf{x};\mathbf{\xi}) = \frac{i \alpha}{2} e^{\alpha (z+c)} H_0^{(1)}(\alpha r) + \frac{1}{\pi^2} \int\limits_0^{\infty} \Big( \cos \eta z + \frac{\alpha}{\eta} \sin \eta z \Big) \frac{\eta^2}{\eta^2+\alpha^2} \Big( \cos \eta c + \frac{\alpha}{\eta} \sin \eta c \Big) K_0(\eta r) d\eta. }[/math]