Eigenfunction Matching for a Circular Dock

Introduction

We show here a solution for a dock on Finite Depth water.

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}=\alpha\phi, \,\, z=0,\,r\gt a, }[/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

Separation of variables

We now separate variables, noting that since the problem has circular symmetry we can write the potential as

[math]\displaystyle{ \phi(r,\theta,z)=\zeta(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta} }[/math]

Applying Laplace's equation we obtain

[math]\displaystyle{ \zeta_{zz}+\mu^{2}\zeta=0 }[/math]

so that:

[math]\displaystyle{ \zeta=\cos\mu(z+H) }[/math]

where the separation constant [math]\displaystyle{ \mu^{2} }[/math] must satisfy the Dispersion Relation for a Free Surface

[math]\displaystyle{ k\tan\left( kh\right) =-\alpha,\quad r\gt a }[/math]

and

[math]\displaystyle{ \kappa\tan(\kappa h)=0,\quad r\lt a }[/math]

Note that we have set [math]\displaystyle{ \mu=k }[/math] under the free surface and [math]\displaystyle{ \mu=\kappa }[/math] under the dock. We denote the positive imaginary solution of the Dispersion Relation for a Free Surface by [math]\displaystyle{ k_{0} }[/math] and the positive real solutions by [math]\displaystyle{ k_{m} }[/math], [math]\displaystyle{ m\geq1 }[/math]. The solutions of second equation will be denoted by [math]\displaystyle{ \kappa_{m} = m\pi/h }[/math], [math]\displaystyle{ m\geq 0 }[/math].

We define

[math]\displaystyle{ \phi_{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 and

[math]\displaystyle{ \psi_{m}\left( z\right) = \cos\kappa_{m}(z+h),\quad m\geq 0 }[/math]

as the vertical eigenfunction of the potential in the dock covered region. For later reference, we note that:

[math]\displaystyle{ \int\nolimits_{-h}^{0}\phi_{m}(z)\phi_{n}(z) d z=A_{m}\delta_{mn} }[/math]

where

[math]\displaystyle{ A_{m}=\frac{1}{2}\left( \frac{\cos k_{m}h\sin k_{m}h+k_{m}h}{k_{m}\cos ^{2}k_{m}h}\right) }[/math]

and

[math]\displaystyle{ \int\nolimits_{-h}^{0}\phi_{n}(z)\psi_{m}(z) d z=B_{mn} }[/math]

where

[math]\displaystyle{ B_{mn}=\frac{k_{n}\sin k_{n}h\cos\kappa_{m}h-\kappa_{m}\cos k_{n}h\sin \kappa_{m}h}{\left( \cos k_{n}h\cos\kappa_{m}h\right) \left( k_{n} ^{2}-\kappa_{m}^{2}\right) } }[/math]

and

[math]\displaystyle{ \int\nolimits_{-h}^{0}\psi_{m}(z)\psi_{n}(z) d z=C_{m}\delta_{mn} }[/math]

where

[math]\displaystyle{ C_{m}=\frac{1}{2}h,\quad,m\neq 0 \quad \mathrm{and} \quad C_0 = h }[/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}}+\mu^{2}\right) \rho_{n}=0 }[/math]

where [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ k_{m} }[/math] or [math]\displaystyle{ \kappa_{m}, }[/math] depending on whether [math]\displaystyle{ r }[/math] is greater or less than [math]\displaystyle{ a }[/math]. We can convert this equation to the standard form by substituting [math]\displaystyle{ y=\mu r }[/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). 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]. 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}\sum_{m=0}^{\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}} =\frac{A}{i\sqrt{\alpha}}e^{k_{0}x}\phi_{0}\left( z\right) =\sum\limits_{n=-\infty}^{\infty}e_{n}I_{n}(k_{0}r)\phi_{0}\left(z\right) e^{i n \theta} }[/math]

where [math]\displaystyle{ e_{n}=A/\left(i\sqrt{\alpha}\right) }[/math] (we retain the dependence on [math]\displaystyle{ n }[/math] for situations where the incident potential might take another form).

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{ e_{n}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) =\sum_{m=0}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) }[/math]

and

[math]\displaystyle{ e_{n}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) =\sum_{m=0}^{\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{ e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l} =\sum_{m=0}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \,\,\,(8) }[/math]

and

[math]\displaystyle{ e_{n}k_{0}I_{n}^{\prime}(k_{0}a)A_{0}\delta_{0l}+a_{ln}k_{l}K_{n}^{\prime }(k_{l}a)A_{l} =\sum_{m=0}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m} a)B_{ml} }[/math]

The first equation can be solved for the open water coefficients [math]\displaystyle{ a_{mn} }[/math]

[math]\displaystyle{ a_{ln}=-e_{n}\frac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum _{m=0}^{\infty}b_{mn}\frac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}} }[/math]

and if this is substituted into the second equation we obtain

[math]\displaystyle{ \left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\frac{K_{n}^{\prime}(k_{0} a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l} =\sum_{m=0}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m} a)-k_{l}\frac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa _{m}a)\right) B_{ml}b_{mn} }[/math]

for each [math]\displaystyle{ n }[/math]. This gives the required equations to solve for the coefficients of the water velocity potential in the plate covered region

Numerical Solution

To solve the system of equations (10) together with the boundary conditions (6 and 7) we set the upper limit of [math]\displaystyle{ l }[/math] to be [math]\displaystyle{ M }[/math]. We also set the angular expansion to be from [math]\displaystyle{ n=-N }[/math] to [math]\displaystyle{ N }[/math]. This gives us

[math]\displaystyle{ \phi(r,\theta,z)=\sum_{n=-N}^{N}\sum_{m=0}^{M}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=-N}^{N}\sum_{m=0}^{M}b_{mn}I_{n}(\kappa _{m}r)e^{i n\theta}\psi_{m}(z), \;\;r\lt a }[/math]