Diffraction Transfer Matrix

Introduction

The diffraction transfer matrix relates the incident and scattered potential in Cylindrical Eigenfunction Expansion. The simplest problem is that of a Bottom Mounted Cylinder. Here we present the theory for bodies of arbitrary geometry. While Kagemoto and Yue 1986 presented theory for bodies of arbitrary shape, they did not explain how to actually obtain the diffraction transfer matrices for bodies which did not have an axisymmetric geometry. This step was performed by Goo and Yoshida 1990 who came up with an explicit method to calculate the diffraction transfer matrices for bodies of arbitrary geometry in the case of finite depth. Utilising a Green's function they used the standard method of transforming the single diffraction boundary-value problem to an integral equation for the source strength distribution function over the immersed surface of the body. However, the representation of the scattered potential which is obtained using this method is not automatically given in the cylindrical eigenfunction expansion. To obtain such cylindrical eigenfunction expansions of the potential Goo and Yoshida 1990 used the representation of the free surface finite depth Green's function given by Black 1975 and Fenton 1978. Their representation of the Green's function was based on applying Graf's addition theorem to the eigenfunction representation of the free surface finite depth Green's function given by John 1950. Their representation allowed the scattered potential to be represented in the eigenfunction expansion with the cylindrical coordinate system fixed at the point of the water surface above the mean centre position of the body. The theory is extended to Infinite Depth in Diffraction Transfer Matrix for Infinite Depth

Eigenfunction expansion of the potential

The scattered potential of a body $\Delta_j$ can be expanded in the Cylindrical Eigenfunction Expansion,

$\phi_j^\mathrm{S} (r_j,\theta_j,z) = \sum_{m=0}^{\infty} f_m(z) \sum_{\mu = - \infty}^{\infty} A_{m \mu}^j K_\mu (k_m r_j) \mathrm{e}^{\mathrm{i}\mu \theta_j},$

with discrete coefficients $A_{m \mu}^j$, where

$f_m(z) = \frac{\cos k_m (z+H)}{\cos k_m H}.$

The incident potential upon body $\Delta_j$ can be also be expanded in regular cylindrical eigenfunctions,

$\phi_j^\mathrm{I} (r_j,\theta_j,z) = \sum_{n=0}^{\infty} f_n(z) \sum_{\nu = - \infty}^{\infty} D_{n\nu}^j I_\nu (k_n r_j) \mathrm{e}^{\mathrm{i}\nu \theta_j},$

with discrete coefficients $D_{n\nu}^j$. In these expansions, $I_\nu$ and $K_\nu$ denote the modified Bessel functions of the first and second kind, respectively, both of order $\nu$.

Note that the term for $m =0$ or $n=0$ corresponds to the propagating modes while the terms for $m\geq 1$ ($n\geq 1$) correspond to the evanescent modes.

Calculation of the diffraction transfer matrix for bodies of arbitrary geometry

The scattered and incident potential can therefore be related by a diffraction transfer operator acting in the following way,

$A_{m \mu}^j = \sum_{n=0}^{\infty} \sum_{\nu = -\infty}^{\infty} B_{m n \mu \nu}^j D_{n\nu}^j.$

Before we can apply the interaction theory we require the diffraction transfer matrices $\mathbf{B}^j$ which relate the incident and the scattered potential for a body $\Delta_j$ in isolation. The elements of the diffraction transfer matrix, $({\mathbf B}^j)_{pq}$, are the coefficients of the $p$th partial wave of the scattered potential due to a single unit-amplitude incident wave of mode $q$ upon $\Delta_j$.

It should be noted that, instead of using the source strength distribution function, it is also possible to consider an integral equation for the total potential and calculate the elements of the diffraction transfer matrix from the solution of this integral equation. An outline of this method for water of finite depth is given by Kashiwagi 2000. We will present here a derivation of the diffraction transfer matrices for the case infinite depth based on a solution for the source strength distribution function. However, an equivalent derivation would be possible based on the solution for the total velocity potential.

The Free-Surface Green Function for Finite Depth in cylindrical polar coordinates

$G(r,\theta,z;s,\varphi,c)= \frac{1}{\pi} \sum_{m=0}^{\infty} \frac{k_m^2+\alpha^2}{H(k_m^2+\alpha^2)-\alpha}\, \cos k_m(z+H) \cos k_m(c+H) \sum_{\nu=-\infty}^{\infty} K_\nu(k_m r) I_\nu(k_m s) \mathrm{e}^{\mathrm{i}\nu (\theta - \varphi)},$

given by Black 1975 and Fenton 1978 is used. The elements of ${\mathbf B}^j$ are therefore given by

$({\mathbf B}^j)_{pq} = \frac{1}{\pi} \frac{(k_m^2+\alpha^2)\cos^2 k_mH}{H(k_m^2+\alpha^2)-\alpha} \int\limits_{\Gamma_j} \cos k_m(c+H) I_p(\eta s) \mathrm{e}^{-\mathrm{i}p \varphi} \varsigma_q^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}$

where $\varsigma_q^j(\mathbf{\zeta})$ is the source strength distribution due to an incident potential of mode $q$ of the form

$\phi_q^{\mathrm{I}}(s,\varphi,c) = \frac{\cos k_m(c+H)}{\cos k_m H} I_q (k_m s) \mathrm{e}^{\mathrm{i}q \varphi}$

We assume that we have represented the scattered potential in terms of the source strength distribution $\varsigma^j$ so that the scattered potential can be written as

$\phi_j^\mathrm{S}(\mathbf{y}) = \int\limits_{\Gamma_j} G (\mathbf{y},\mathbf{\zeta}) \, \varsigma^j (\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}, \quad \mathbf{y} \in D,$

where $D$ is the volume occupied by the water and $\Gamma_j$ is the immersed surface of body $\Delta_j$. The source strength distribution function $\varsigma^j$ can be found by solving an integral equation. The integral equation is described in Wehausen and Laitone 1960 and in Green Function Solution Method.

Calculation of the coefficients $A_{m \mu}^j$, using a source strength distribution

The idea is to represent the scattered potential for the body $\Delta_j$ in terms of a source strength distribution $\varsigma^j$, as we already did in the previous section

$\phi_j^\mathrm{S}(r_j,\theta_j,z) = \int\limits_{\Gamma_j} G (r_j,\theta_j,z;s,\varphi,c) \, \varsigma^j (s,\varphi,c) \mathrm{d}\sigma(s,\varphi,c), \quad \mathbf{y} \in D,$

where $D$ is the volume occupied by the water and $\Gamma_j$ is the immersed surface of body $\Delta_j$. As we will use the Cylindrical Eigenfunction Expansion, we express all the variables in a cylindrical coordinate system. The expression of the Free-Surface Green Function has already been expanded in cylindrical coordinates in the previous section, so we obtain the expression of the scattered potential as follow

$\phi_j^\mathrm{S}(r_j,\theta_j,z) = \int\limits_{\Gamma_j} \Big[ \frac{1}{\pi} \sum_{m=0}^{\infty} \frac{k_m^2+\alpha^2}{H(k_m^2+\alpha^2)-\alpha}\, \cos k_m(z+H) \cos k_m(c+H) \sum_{\mu=-\infty}^{\infty} K_\mu(k_m r_j) I_\mu(k_m s) \mathrm{e}^{\mathrm{i}\mu (\theta_j - \varphi)} \Big] \varsigma^j (s,\varphi,c) \mathrm{d}\sigma(s,\varphi,c)$

We can rearrange this expression in order to obtain a Cylindrical Eigenfunction Expansion.

$\phi_j^\mathrm{S}(r_j,\theta_j,z) = \sum_{m=0}^{\infty} \frac{\cos k_m(z+H)}{\cos k_m H} \sum_{\mu=-\infty}^{\infty} \Big[ \frac{1}{\pi} \frac{(k_m^2+\alpha^2) \cos k_m H}{H(k_m^2+\alpha^2)-\alpha} \int\limits_{\Gamma_j} \cos k_m(c+H) I_\mu(k_m s) \mathrm{e}^{-\mathrm{i}\mu \varphi} \varsigma^j (s,\varphi,c) \mathrm{d}\sigma(s,\varphi,c) \Big] K_\mu(k_m r_j) \mathrm{e}^{\mathrm{i}\mu \theta_j}$

An eigenfunction matching method permits us to identify the coefficients of the scattered potential expression

$A_{m \mu}^j = \frac{1}{\pi} \frac{(k_m^2+\alpha^2) \cos k_m H}{H(k_m^2+\alpha^2)-\alpha} \int\limits_{\Gamma_j} \cos k_m(c+H) I_\mu(k_m s) \mathrm{e}^{-\mathrm{i}\mu \varphi} \varsigma^j (s,\varphi,c) \mathrm{d}\sigma(s,\varphi,c)$

The diffraction transfer matrix of rotated bodies

For a non-axisymmetric body, a rotation about the mean centre position in the $(x,y)$-plane will result in a different diffraction transfer matrix. We will show how the diffraction transfer matrix of a body rotated by an angle $\beta$ can be easily calculated from the diffraction transfer matrix of the non-rotated body. The rotation of the body influences the form of the elements of the diffraction transfer matrices in two ways. Firstly, the angular dependence in the integral over the immersed surface of the body is altered and, secondly, the source strength distribution function is different if the body is rotated. However, the source strength distribution function of the rotated body can be obtained by calculating the response of the non-rotated body due to rotated incident potentials. It will be shown that the additional angular dependence can be easily factored out of the elements of the diffraction transfer matrix.

The additional angular dependence caused by the rotation of the incident potential can be factored out of the normal derivative of the incident potential such that

$\frac{\partial \phi_{q\beta}^{\mathrm{I}}}{\partial n} = \frac{\partial \phi_{q}^{\mathrm{I}}}{\partial n} \mathrm{e}^{\mathrm{i}q \beta},$

where $\phi_{q\beta}^{\mathrm{I}}$ is the rotated incident potential. Since the integral equation for the determination of the source strength distribution function is linear, the source strength distribution function due to the rotated incident potential is thus just given by

$\varsigma_{q\beta}^j = \varsigma_q^j \, \mathrm{e}^{\mathrm{i}q \beta}.$
$({\mathbf B}^j)_{pq} = \frac{1}{\pi} \frac{(k_m^2+\alpha^2)\cos^2 k_md}{d(k_m^2+\alpha^2)-\alpha} \int\limits_{\Gamma_j} \cos k_m(c+d) I_p(\eta s) \mathrm{e}^{-\mathrm{i}p \varphi} \varsigma_q^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}$

This is also the source strength distribution function of the rotated body due to the standard incident modes.

The elements of the diffraction transfer matrix $\mathbf{B}^j$ are given by equations (B_elem). Keeping in mind that the body is rotated by the angle $\beta$, the elements of the diffraction transfer matrix of the rotated body are given by

$({\mathbf B}^j_\beta)_{pq} = \frac{1}{\pi} \frac{(k_m^2+\alpha^2)\cos^2 k_md}{d(k_m^2+\alpha^2)-\alpha} \int\limits_{\Gamma_j} \cos k_m(c+d) I_p(\eta s) \mathrm{e}^{-\mathrm{i}p (\varphi+\beta)} \varsigma_{q\beta}^j(\mathbf{\zeta}) \mathrm{d}\sigma_\mathbf{\zeta}$

Thus the additional angular dependence caused by the rotation of the body can be factored out of the elements of the diffraction transfer matrix. The elements of the diffraction transfer matrix corresponding to the body rotated by the angle $\beta$, $\mathbf{B}^j_\beta$, are given by

$(\mathbf{B}^j_\beta)_{pq} = (\mathbf{B}^j)_{pq} \, \mathrm{e}^{\mathrm{i}(q-p) \beta}.$