Eigenfunction Matching for a Circular Dock
Introduction
We show here a solution for a dock on Finite Depth water, which is circular. This is the three-dimensional analog of the Eigenfunction Matching for a Semi-Infinite Dock.
Governing Equations
We begin with the Frequency Domain Problem. We will use a cylindrical coordinate system, [math]\displaystyle{ (r,\theta,z) }[/math], assumed to have its origin at the centre of the circular plate which has radius [math]\displaystyle{ a }[/math]. The water is assumed to have constant finite depth [math]\displaystyle{ h }[/math] and the [math]\displaystyle{ z }[/math]-direction points vertically upward with the water surface at [math]\displaystyle{ z=0 }[/math] and the sea floor at [math]\displaystyle{ z=-h }[/math]. The boundary value problem can therefore be expressed as
[math]\displaystyle{ \Delta\phi=0, \,\, -h\lt z\lt 0, }[/math]
[math]\displaystyle{ \phi_{z}=0, \,\, z=-h, }[/math]
[math]\displaystyle{ \phi_{z}=0, \,\, z=0,\,r\lt a }[/math]
We must also apply the Sommerfeld Radiation Condition as [math]\displaystyle{ r\rightarrow\infty }[/math]. The subscript [math]\displaystyle{ z }[/math] denotes the derivative in [math]\displaystyle{ z }[/math]-direction.
Solution Method
We use separation of variables in the two regions, [math]\displaystyle{ r\lt a }[/math] and [math]\displaystyle{ r\gt a }[/math].
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]
and
[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:
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]
where
[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 Variables for a Dock
The separation of variables equation for a floating dock
[math]\displaystyle{ Z^{\prime\prime} + k^2 Z =0, }[/math]
subject to the boundary conditions
[math]\displaystyle{ Z^{\prime} (-h) = 0, }[/math]
and
[math]\displaystyle{ Z^{\prime} (0) = 0. }[/math]
The solution is [math]\displaystyle{ k=\kappa_{m}= \frac{m\pi}{h} \, }[/math], [math]\displaystyle{ m\geq 0 }[/math] and
[math]\displaystyle{ Z = \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad m\geq 0. }[/math]
We note that
[math]\displaystyle{ \int\nolimits_{-h}^{0}\psi_{m}(z)\psi_{n}(z) \mathrm{d} z=C_{m}\delta_{mn}, }[/math]
where
[math]\displaystyle{ C_{m} = \begin{cases} h,\quad m=0 \\ \frac{1}{2}h,\,\,\,m\neq 0 \end{cases} }[/math]
Inner product between free surface and dock modes
[math]\displaystyle{ \int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) \mathrm{d} z=B_{mn} }[/math]
where
Separation for Cylindrical Coordinates
We now separate variables, noting that since the problem has circular symmetry we can write the potential as
[math]\displaystyle{ \phi(r,\theta,z)=\frac{\cos k(z+h)}{\cos kh}\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta} }[/math]
We now solve for the function [math]\displaystyle{ \rho_{n}(r) }[/math]. Using Laplace's equation in polar coordinates we obtain
[math]\displaystyle{ \frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r} \frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left( \frac{n^{2}}{r^{2}}+k^{2}\right) \rho_{n}=0. }[/math]
We can convert this equation to the standard form by substituting [math]\displaystyle{ y=k r }[/math] (provided that [math]\displaystyle{ \mu\neq 0 }[/math]to obtain
[math]\displaystyle{ y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n} }{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0 }[/math]
The solution of this equation is a linear combination of the modified Bessel functions of order [math]\displaystyle{ n }[/math], [math]\displaystyle{ I_{n}(y) }[/math] and [math]\displaystyle{ K_{n}(y) }[/math] (Abramowitz and Stegun 1964).
Therefore
[math]\displaystyle{ \rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)\, }[/math]
for some constants [math]\displaystyle{ C_1 }[/math] and [math]\displaystyle{ C_2 }[/math]
Since the solution must be bounded we know that under the plate the solution will be a linear combination of [math]\displaystyle{ I_{n}(y) }[/math] while outside the plate the solution will be a linear combination of [math]\displaystyle{ K_{n}(y) }[/math]. The case [math]\displaystyle{ \kappa_0 =0 }[/math] is a special case and the solution under the dock is [math]\displaystyle{ (r/a)^{|n|} }[/math]. Therefore the potential can be expanded as
[math]\displaystyle{ \phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}a_{mn}K_{n} (k_{m}r)e^{i n\theta}\phi_{m}(z), \;\;r\gt a }[/math]
and
[math]\displaystyle{ \phi(r,\theta,z)=\sum_{n=-\infty}^{\infty}b_{0n}(r/a)^{|n|} e^{i n\theta}\psi_{0}(z)+ \sum_{n=-\infty}^{\infty}\sum_{m=1}^{\infty}b_{mn} I_{n}(\kappa_{m}r)e^{i n\theta}\psi_{m}(z), \;\;r\lt a }[/math]
where [math]\displaystyle{ a_{mn} }[/math] and [math]\displaystyle{ b_{mn} }[/math] are the coefficients of the potential in the open water and the plate covered region respectively.
Incident potential
The incident potential is a wave of amplitude [math]\displaystyle{ A }[/math] in displacement travelling in the positive [math]\displaystyle{ x }[/math]-direction. The incident potential can therefore be written as
[math]\displaystyle{ \phi^{\mathrm{I}} =e^{k_{0}x}\phi_{0}\left( z\right) =\sum\limits_{n=-\infty}^{\infty} I_{n}(k_{0}r)\phi_{0}\left(z\right) e^{i n \theta} }[/math]
An infinite dimensional system of equations
The potential and its derivative must be continuous across the transition from open water to the plate covered region. Therefore, the potentials and their derivatives at [math]\displaystyle{ r=a }[/math] have to be equal for each angle and we obtain
[math]\displaystyle{ I_{n}(k_{0}a)\phi_{0}\left( z\right) + \sum_{m=0}^{\infty} a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right) = b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) }[/math]
and
[math]\displaystyle{ k_{0}I_{n}^{\prime}(k_{0}a)\phi_{0}\left( z\right) +\sum _{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left( z\right) =b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty} b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi _{m}(z) }[/math]
for each [math]\displaystyle{ n }[/math]. We solve these equations by multiplying both equations by [math]\displaystyle{ \phi_{l}(z) }[/math] and integrating from [math]\displaystyle{ -h }[/math] to [math]\displaystyle{ 0 }[/math] to obtain:
[math]\displaystyle{ I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l} =b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} }[/math]
and
[math]\displaystyle{ k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime }(k_{l}a)A_{l} = b_{0n}B_{0l}\frac{|n|}{a} + \sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml} }[/math]
Numerical Solution
To solve the system of equations we set the upper limit of [math]\displaystyle{ l }[/math] to be [math]\displaystyle{ M }[/math].
A Simple Method To Calculate The Diffraction Transfer Matrix For The Case Of A Circular Dock
Let's consider an incident wave whose potential has the following expression
Such an incident potential is found in the Kagemoto and Yue Interaction Theory, where it can be written as the sum of an ambient incident potential and the scattered potentials of the other bodies, which are interpretated as incident potentials for the studied body.
We can apply the same eigenfunction matching that previously, considering the potential and its normal derivative continuous at [math]\displaystyle{ r=a }[/math]. Thus the potential and its normal derivative expressed at each side of this value of the radius have to be equal. We obtain the relationships
[math]\displaystyle{ \sum_{m=0}^{\infty} D_{mn} I_n (k_m a) \phi_m(z) + \sum_{m=0}^{\infty} a_{mn} K_{n}(k_{m}a)\phi_{m}\left( z\right) = b_{0n} \psi_{0}(z) +\sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) }[/math]
and
[math]\displaystyle{ \sum_{m=0}^{\infty} D_{mn} k_m I'_n (k_m a) \phi_m(z) +\sum _{m=0}^{\infty} a_{mn}k_{m}K_{n}^{\prime}(k_{m}a)\phi_{m}\left( z\right) =b_{0n} \frac{|n|}{a} \psi_{0}(z) +\sum_{m=1}^{\infty} b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi _{m}(z) }[/math]
for each [math]\displaystyle{ n }[/math]. We solve these equations with the same method that before, by multiplying both equations by [math]\displaystyle{ \phi_{l}(z) }[/math] and integrating from [math]\displaystyle{ -H }[/math] to [math]\displaystyle{ 0 }[/math] to obtain:
[math]\displaystyle{ D_{ln} I_n (k_l a) A_l+a_{ln}K_{n}(k_{l}a)A_{l} =b_{0n}B_{0l} + \sum_{m=1}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml}, \ \ \ (1) }[/math]
and
[math]\displaystyle{ D_{ln} k_l I'_n (k_l a) A_l+a_{ln}k_{l}K_{n}^{\prime }(k_{l}a)A_{l} = b_{0n}B_{0l}\frac{|n|}{a} + \sum_{m=1}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)B_{ml}, \ \ \ (2) }[/math]
The Diffraction Transfer Matrix maps the coefficients of the incident wave with the coefficients of the scattered wave within the open water domain. The relation which links these two coefficients can be written as follows
From the equations (1) and (2) we can write a linear system of equation, limiting the number of modes of the dispersion equation to [math]\displaystyle{ N }[/math] real ones
[math]\displaystyle{ \begin{bmatrix} \begin{bmatrix} -A_0 K_n(k_0 a)&0 \quad \cdots&0\\ 0&&\\ \vdots&-A_l K_n(k_l a)&\vdots\\ &&0\\ 0&\cdots \quad 0 & -A_N K_n(k_N a) \end{bmatrix} & \begin{bmatrix} B_{00}&I_n(\kappa_{1}a) B_{10}&\cdots&I_n(\kappa_{N}a) B_{N0}\\ B_{01}&I_n(\kappa_{1}a) B_{11}&&\\ \vdots&\vdots&&\vdots\\ &&&\\ B_{0N}&I_n(\kappa_{1}a) B_{1N}&\cdots&I_n(\kappa_{N}a) B_{NN} \end{bmatrix} \\ \begin{bmatrix} -k_0 K'_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&-k_l K'_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 & -k_N K'_n(k_N a) A_N \end{bmatrix} & \begin{bmatrix} B_{00}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{10}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{N0}\\ B_{01}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{11}&&\\ \vdots&\vdots&&\vdots\\ &&&\\ B_{0N}\frac{|n|}{a}&\kappa_{1} I'_n(\kappa_{1}a) B_{1N}&\cdots&\kappa_{N} I'_n(\kappa_{N}a) B_{NN} \end{bmatrix} \end{bmatrix} \begin{bmatrix} a_{0n} \\ \\ \vdots \\ \\ a_{Nn} \\ b_{0n}\\ \\ \vdots \\ \\ b_{Nn} \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} I_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&I_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 & I_n(k_N a) A_N \end{bmatrix} \\ \begin{bmatrix} k_0 I'_n(k_0 a) A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&k_l I'_n(k_l a) A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 &k_N I'_n(k_N a) A_N \end{bmatrix} \end{bmatrix} \begin{bmatrix} D_{0n} \\ \\ \vdots \\ \\ D_{Nn} \\ D_{0n}\\ \\ \vdots \\ \\ D_{Nn} \end{bmatrix} }[/math]
for each [math]\displaystyle{ n }[/math].
Therefore we can find a Diffraction Transfer Matrix for each [math]\displaystyle{ n }[/math], by setting
Then we solve the linear system defined previously, so that we can find the coefficients [math]\displaystyle{ (a_{ln})_{l \in [0, N]} }[/math] for each [math]\displaystyle{ i }[/math]. This vector represents exactly the [math]\displaystyle{ i^{th} }[/math] column of the Diffraction Transfer Matrix, [math]\displaystyle{ n }[/math] being set.
This method permits to obtain the matrix which links the coefficients of the incident and scattered potential in the free water domain. Applying this for each [math]\displaystyle{ n }[/math], we finally obtain a 3-dimensional matrix for the Diffraction Transfer Matrix.
Matlab Code
A program to calculate the coefficients for circular dock problems can be found here circle_dock_matching_one_n.m Note that this problem solves only for a single n.
Additional code
This program requires
