Eigenfunction Matching for a Circular Floating Elastic Plate

We present a solution for a

circular thin plate of shallow draft on water of finite depth subject to linear wave forcing of a single frequency. The solution, which is given in a closed form, is based on decomposing the solution into angular eigenfunctions. The coefficients in the expansion are then found by matching the potential and its derivative at the plate edge and imposing the free edge conditions for the plate. The matching is accomplished by taking the inner product with respect to the vertical eigenfunctions for the free surface. The equations that are derived are transformed so that the final system of equations involves only the unknowns under the plate. Solutions are presented and compared to the results of \cite{JGR02} who presented a solution for a plate of arbitrary geometry.}

={ INTRODUCTION+

The problem of a linear floating thin plate of shallow draft subject to wave forcing is a standard problem in hydroelasticity which can be used to model a range of physical systems. In two dimensions many different solution methods exist. In shallow water a solution was presented by \cite{stoker}, for finite depth solutions have been presented by \cite{jgrfloe1d} and \cite{Newman_deform} amongst others. In three dimensions a solution for a circular plate on shallow water was presented by \cite{Miloh00} and general solution methods have been presented by \cite{jgrfloecirc}, \cite{JGR02} and \cite{Kashiwagibspline} amongst others. However, with the exception of \cite{Miloh00}, these solutions were based on the free surface Green's function and were highly numerical. Even the solution presented by \cite{jgrfloecirc} for a circular plate only exploited the circular geometry to calculate the modes of vibration of the free plate.

The only three dimensional solution for a thin plate of shallow draft that is not based on a highly numerical me\-thod was presented by \cite{Miloh00}. The exact same solution method was also independently derived by \cite{Tsubogo01}. Their solution was for the case of shallow water and a circular plate. The circular geometry allows separation of variables in the angular direction so that the solution may be found by de-coupling the solutions for each angular eigenfunction. Once this has been accomplished the solution for each angular direction can be found by solving a linear system of four equations. The four equations were derived by matching the potential and its derivative and by imposing the two boundary conditions at the edge of the plate.

We present here an extension of the method of \cite{Miloh00} to the case where the water depth is finite. In this case we can still solve for each angular eigenfunction separately and we match the potential and its derivative and impose the boundary conditions at the plate edge. However, we must match the potential not at a point but throughout the water depth. This matching is accomplished by taking the inner product with respect to the vertical eigenfunctions which satisfy the free surface condition. We present results for the method which are compared to the results of \cite{JGR02}.

GOVERNING EQUATIONS

We begin with the equations for the plate-water system in non-dimensional form as the problem is so well-known. The derivation and non-dimensionalisation is discussed in detail in \cite{JGR02}. We non-dimensionalise the spatial variables with respect to a length parameter [math]\displaystyle{ L\lt math\gt (for example, \lt math\gt L\lt math\gt may be derived from the area of the plate or \lt math\gt L\lt math\gt may be the characteristic length \lt math\gt \left( D/\rho g\right) ^{1/4}\lt math\gt , where \lt math\gt D\lt math\gt is the rigidity constant of the plate, \lt math\gt \rho\lt math\gt is the density of the water and \lt math\gt g\lt math\gt is the gravitational constant) and the time variables with respect to \lt math\gt \sqrt{L/g}\lt math\gt . We assume that all motions are time harmonic with radian frequency \lt math\gt \omega=\sqrt{\alpha}\lt math\gt so that the velocity potential of the water, \lt math\gt \bar{\phi}(x,t)\lt math\gt , can be expressed as the real part of a complex quantity \lt math\gt \phi\lt math\gt : \lt math\gt \bar{\phi}(\mathbf{x},t)=\Re \{\phi(\mathbf{x})\e^{-\i\sqrt{\alpha }t}\} \label{time} }[/math] We will use a cylindrical coordinate system, [math]\displaystyle{ \mathbf{x}=(r,\theta,z)\lt math\gt , assumed to have its origin at the centre of the circular plate which has radius \lt math\gt a\lt math\gt . The water is assumed to have constant finite depth \lt math\gt H\lt math\gt and the \lt math\gt z\lt math\gt -direction points vertically upward with the water surface at \lt math\gt z=0\lt math\gt and the sea floor at \lt math\gt z=-H\lt math\gt . The boundary value problem can therefore be expressed as: \lt math\gt \left. \begin{array} [c]{cc} \Delta\phi=0, & -H\lt z\lt 0,\\ \phi_{z}=0, & z=-H,\\ \phi_{z}=\alpha\phi, & z=0,\,r\gt a,\\ (\beta\Delta^{2}+1-\alpha\gamma)\phi_{z}=\alpha\phi, & z=0,\,r\lt a \end{array} \right\} \label{bvp_plate} }[/math] where the constants [math]\displaystyle{ \beta\lt math\gt and \lt math\gt \gamma\lt math\gt are given by: \lt math\gt \beta=\frac{D}{\rho\,L^{4}g}, \quad \gamma=\frac{\rho_{i}h}{\rho\,L} }[/math] and [math]\displaystyle{ \rho_{i}\lt math\gt is the density of the plate. We must also apply the edge conditions for the plate and the radiation condition as \lt math\gt r\rightarrow\infty\lt math\gt . The subscript \lt math\gt z\lt math\gt denotes the derivative in \lt math\gt z\lt math\gt -direction. =SOLUTION METHOD= == Separation of variables= We now separate variables, noting that since the problem has circular symmetry we can write the potential as: \lt math\gt \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}\lt math\gt must satisfy the standard dispersion equations: \lt math\gt k\tan\left( kH\right) =-\alpha,\quad r\gt a \label{eq_k} }[/math] and: [math]\displaystyle{ \kappa\tan(\kappa H)=\frac{-\alpha}{\beta\kappa^{4}+1-\alpha\gamma},\quad r\lt a \label{eq_kappa} }[/math] Note that we have set [math]\displaystyle{ \mu=k\lt math\gt under the free surface and \lt math\gt \mu=\kappa\lt math\gt under the plate. The dispersion equations are discussed in detail in \cite{FoxandSquire}. We denote the positive imaginary solution of (\ref{eq_k}) by \lt math\gt k_{0}\lt math\gt and the positive real solutions by \lt math\gt k_{m}\lt math\gt , \lt math\gt m\geq1\lt math\gt . The solutions of (\ref{eq_kappa}) will be denoted by \lt math\gt \kappa_{m}\lt math\gt , \lt math\gt m\geq-2\lt math\gt . The fully complex solutions with positive imaginary part are \lt math\gt \kappa_{-2}\lt math\gt and \lt math\gt \kappa_{-1}\lt math\gt (where \lt math\gt \kappa_{-1}=\overline{\kappa_{-2}}\lt math\gt ), the negative imaginary solution is \lt math\gt \kappa_{0}\lt math\gt and the positive real solutions are \lt math\gt \kappa_{m}\lt math\gt , \lt math\gt m\geq1\lt math\gt . We define: \lt math\gt \phi_{m}\left( z\right) =\dfrac{\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) =\dfrac{\cos\kappa_{m}(z+H)}{\cos\kappa_{m}H},\quad m\geq-2 }[/math] as the vertical eigenfunction of the potential in the plate 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}=\dfrac{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]

We now solve for the function [math]\displaystyle{ \rho_{n}(r)\lt math\gt . Using Laplace's equation in polar coordinates we obtain: \lt math\gt \dfrac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r} \dfrac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left( \frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0 }[/math] where [math]\displaystyle{ \mu\lt math\gt is \lt math\gt k_{m}\lt math\gt or \lt math\gt \kappa_{m},\lt math\gt depending on whether \lt math\gt r\lt math\gt is greater or less than \lt math\gt a\lt math\gt . We can convert this equation to the standard form by substituting \lt math\gt y=\mu r\lt math\gt to obtain: \lt math\gt y^{2}\dfrac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\dfrac{\mathrm{d}\rho_{n} }{\mathrm{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\lt math\gt , \lt math\gt I_{n}(y)\lt math\gt and \lt math\gt K_{n}(y)\lt math\gt \cite[see][]{abr_ste}. Since the solution must be bounded we know that under the plate the solution will be a linear combination of \lt math\gt I_{n}(y)\lt math\gt while outside the plate the solution will be a linear combination of \lt math\gt K_{n}(y)\lt math\gt . Therefore the potential can be expanded as: \lt math\gt \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=-2}^{\infty}b_{mn} I_{n}(\kappa_{m}r)\e^{\i n\theta}\psi_{m}(z), \;\;r\lt a }[/math] where [math]\displaystyle{ a_{mn}\lt math\gt and \lt math\gt b_{mn}\lt math\gt 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 \lt math\gt A\lt math\gt in displacement travelling in the positive \lt math\gt x\lt math\gt -direction. Following \cite{Miloh00} the incident potential can therefore be written as: \begin{align} \phi^{\mathrm{I}} & =\dfrac{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} \end{align} where \lt math\gt e_{n}=A/\left( \i\sqrt{\alpha}\right)\lt math\gt (we retain the dependence on \lt math\gt n\lt math\gt for situations where the incident potential might take another form). ==Boundary conditions== The boundary conditions for the plate also have to be considered. The vertical force and bending moment must vanish, which can be written as \cite[following][]{Miloh00}: \lt math\gt \left[ \bar{\Delta}-\frac{1-\nu}{r}\left( \frac{\partial}{\partial r} +\frac{1}{r}\frac{\partial^{2}}{\partial\theta^{2}}\right) \right] w=0 \label{plate_boundary_condition_1} }[/math] and: [math]\displaystyle{ \left[ \frac{\partial}{\partial r}\bar{\Delta}-\frac{1-\nu}{r^{2}}\left( \frac{\partial}{\partial r}+\frac{1}{r}\right) \frac{\partial^{2}} {\partial\theta^{2}}\right] w=0 \label{plate_boundary_condition_2} }[/math] where [math]\displaystyle{ w\lt math\gt is the time-independent surface displacement, \lt math\gt \nu\lt math\gt is Poisson's ratio, and \lt math\gt \bar{\Delta}\lt math\gt is the polar coordinate Laplacian: \lt math\gt \bar{\Delta}=\frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{r^{2}}\frac{\partial^{2}}{\partial\theta^{2}} }[/math]

= Displacement of the plate

The surface displacement and the water velocity potential at the water surface are linked through the kinematic boundary condition: [math]\displaystyle{ \phi_{z}=-\i\sqrt{\alpha}w\quad\text{at }z=0 }[/math] >From equations (\ref{bvp_plate}) the potential and the surface displacement are therefore related by: [math]\displaystyle{ w=\i\sqrt{\alpha}\phi,\quad r\gt a }[/math] and: [math]\displaystyle{ (\beta\bar{\Delta}^{2}+1-\alpha\gamma)w=\i\sqrt{\alpha}\phi,\quad r\lt a }[/math] The surface displacement can also be expanded in eigenfunctions as: [math]\displaystyle{ w(r,\theta)=\sum_{n=-\infty}^{\infty}\sum_{m=0}^{\infty}\i\sqrt{\alpha} a_{mn}K_{n}(k_{m}r)\e^{\i n\theta},\;\;r\gt a }[/math] and: \begin{align} &w(r,\theta)=\\ \notag &\sum_{n=-\infty}^{\infty}\sum_{m=-2}^{\infty}\i\sqrt{\alpha}(\beta\kappa _{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}I_{n}(\kappa_{m}r)\e^{\i n\theta},\; r<a \end{align} using the fact that: [math]\displaystyle{ \bar{\Delta}\left( I_{n}(\kappa_{m}r)\e^{\i n\theta}\right) =\kappa_{m} ^{2}I_{n}(\kappa_{m}r)\e^{\i n\theta} \label{laplacian_besseli} }[/math]

An infinite dimensional system of equations

The boundary conditions (\ref{plate_boundary_condition_1}) and (\ref{plate_boundary_condition_2}) can be expressed in terms of the potential using (\ref{laplacian_besseli}). Since the angular modes are uncoupled the conditions apply to each mode, giving: \begin{align}\label{plate1} &\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn} \times\\\notag &\left( \kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left( \kappa _{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right) \right) =0 \end{align} and: \begin{align} &\sum_{m=-2}^{\infty}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn} \times\label{plate2}\\\notag &\left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2} }\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa _{m}a)\right) \right)\\\notag & =0 \end{align} 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\lt math\gt have to be equal. Again we know that this must be true for each angle and we obtain: \begin{align} 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) \\\notag & =\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)\psi_{m}(z) \end{align} and: \begin{align} 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=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m}a)\psi _{m}(z)\nonumber \end{align} for each \lt math\gt n\lt math\gt . We solve these equations by multiplying both equations by \lt math\gt \phi_{l}(z)\lt math\gt and integrating from \lt math\gt -H\lt math\gt to \lt math\gt 0\lt math\gt to obtain: \begin{align} e_{n}I_{n}(k_{0}a)A_{0}\delta_{0l}+a_{ln}K_{n}(k_{l}a)A_{l}\label{cont_pot} =\sum_{m=-2}^{\infty}b_{mn}I_{n}(\kappa_{m}a)B_{ml} \nonumber \end{align} and: \begin{align} 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}\label{cont_derivative_pot}\\ & =\sum_{m=-2}^{\infty}b_{mn}\kappa_{m}I_{n}^{\prime}(\kappa_{m} a)B_{ml} \nonumber \end{align} Equation (\ref{cont_pot}) can be solved for the open water coefficients \lt math\gt a_{mn}\lt math\gt : \lt math\gt a_{ln}=-e_{n}\dfrac{I_{n}(k_{0}a)}{K_{n}(k_{0}a)}\delta_{0l}+\sum _{m=-2}^{\infty}b_{mn}\dfrac{I_{n}(\kappa_{m}a)B_{ml}}{K_{n}(k_{l}a)A_{l}} }[/math] which can then be substituted into equation (\ref{cont_derivative_pot}) to give us: \begin{align} & \left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\dfrac{K_{n}^{\prime}(k_{0} a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l} \label{final_equation}\\ &\quad =\sum_{m=-2}^{\infty}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m} a)-k_{l}\dfrac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa _{m}a)\right) B_{ml}b_{mn}\nonumber \end{align} for each [math]\displaystyle{ n\lt math\gt . Together with equations (\ref{plate1}) and (\ref{plate2}) equation (\ref{final_equation}) 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 (\ref{final_equation}) together with the boundary conditions we set the upper limit of \lt math\gt l\lt math\gt to be \lt math\gt M\lt math\gt . We also set the angular expansion to be from \lt math\gt n=-N\lt math\gt to \lt math\gt N\lt math\gt . This gives us: \lt math\gt \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=-2}^{M}b_{mn}I_{n}(\kappa _{m}r)\e^{\i n\theta}\psi_{m}(z), \;\;r\lt a }[/math] Since [math]\displaystyle{ l\lt math\gt is an integer with \lt math\gt 0\leq l\leq M\lt math\gt this leads to a system of \lt math\gt M+1\lt math\gt equations. The number of unknowns is \lt math\gt M+3\lt math\gt and the two extra equations are obtained from the boundary conditions for the free plate (\ref{plate1}) and (\ref{plate2}). The equations to be solved for each \lt math\gt n\lt math\gt are: \begin{align} & \left( k_{0}I_{n}^{\prime}(k_{0}a)-k_{0}\dfrac{K_{n}^{\prime}(k_{0} a)}{K_{n}(k_{0}a)}I_{n}(k_{0}a)\right) e_{n}A_{0}\delta_{0l}\\ &\quad =\sum_{m=-2}^{M}\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)-k_{l} \dfrac{K_{n}^{\prime}(k_{l}a)}{K_{n}(k_{l}a)}I_{n}(\kappa_{m}a)\right) B_{ml}b_{mn} \nonumber \end{align} \begin{align} &\sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}\times\\\notag & \left( \kappa_{m}^{2}I_{n}(\kappa_{m}a)-\frac{1-\nu}{a}\left( \kappa _{m}I_{n}^{\prime}(\kappa_{m}a)-\frac{n^{2}}{a}I_{n}(\kappa_{m}a)\right) \right) =0 \end{align} and: \begin{align} &\sum_{m=-2}^{M}(\beta\kappa_{m}^{4}+1-\alpha\gamma)^{-1}b_{mn}\times\\\notag & \left( \kappa_{m}^{3}I_{n}^{\prime}(\kappa_{m}a)+n^{2}\frac{1-\nu}{a^{2} }\left( \kappa_{m}I_{n}^{\prime}(\kappa_{m}a)+\frac{1}{a}I_{n}(\kappa _{m}a)\right) \right) =0 \end{align} It should be noted that the solutions for positive and negative \lt math\gt n\lt math\gt are identical so that they do not both need to be calculated. There are some minor simplifications which are a consequence of this which are discussed in more detail in \cite{Miloh00}. =THE SHALLOW WATER THEORY OF ZILMAN AND MILOH= The shallow water theory of \cite{Miloh00} can be recovered by simply setting the depth shallow enough that the shallow water theory is valid and setting \lt math\gt M=0\lt math\gt . If the shallow water theory is valid then the first three roots of the dispersion equation for the ice will be exactly the same roots found in the shallow water theory by solving the polynomial equation. The system of equations has four unknowns (three under the plate and one in the open water) exactly as for the theory of \cite{Miloh00}. We do not present any comparison with the results of \cite{Miloh00} because they applied their theory in a situation (wavelength 200, water depth 20) where the shallow water approximation is not valid. It should be noted however that our equations become identical to those of \cite{Miloh00} when \lt math\gt M=0\lt math\gt and the water depth is chosen so that the shallow water approximation is valid. =NUMERICAL RESULTS= We present solutions for a plate of radius \lt math\gt a=100\lt math\gt . The wavelength is \lt math\gt \lambda=50\lt math\gt (recall that \lt math\gt \alpha=2\pi/\lambda\tanh\left( 2\pi H/\lambda\right)\lt math\gt ), \lt math\gt \beta=10^{5}\lt math\gt and \lt math\gt \gamma=0\lt math\gt . The incident wave is of unit amplitude. We begin with some convergence results, first of all fixing the number of roots of the dispersion equation \lt math\gt M=8\lt math\gt and varying the number of terms in the angular expansion \lt math\gt N\lt math\gt . Fig.~\ref{convergence_n} shows the real part of the displacement. The number of points in the angular expansion is \lt math\gt N=2\lt math\gt (a), \lt math\gt 4\lt math\gt (b), \lt math\gt 8\lt math\gt (c), and \lt math\gt 16\lt math\gt (d). The depth \lt math\gt H=25\lt math\gt . For this situation it follows that we only require \lt math\gt N=8\lt math\gt for an accurate solution which means we only need to solve \lt math\gt 9\lt math\gt systems of equations. Now we fix the number of points in the angular expansion is \lt math\gt N=16\lt math\gt \ and vary the number of roots of the dispersion equation \lt math\gt M\lt math\gt . Fig.~\ref{convergence_roots} shows the real part of the displacement. The number of roots of the dispersion equation is \lt math\gt M=0\lt math\gt (a), \lt math\gt 2\lt math\gt (b), \lt math\gt 4\lt math\gt (c), and \lt math\gt 8\lt math\gt (d). The depth \lt math\gt H=25\lt math\gt . It follows that we only require \lt math\gt M=2\lt math\gt for an accurate solution which means that we only need to solve a \lt math\gt 5\times 5\lt math\gt system of equations. This shows how efficient this closed form solution is. We can trivially compare our results with those of \cite{Miloh00} and they are in agreement. We compare with the results presented in \cite{JGR02} for an arbitrary shaped plate modified to compute the solution for finite depth. The circle is represented in this scheme by square panels which are arranged to, as nearly as possible, form a circular shape. Fig.~\ref{comparisionh25} and fig.~\ref{comparisionh1} show the real part (a and c) and imaginary part (b and d) of the displacement for depth \lt math\gt H=25\lt math\gt and \lt math\gt H=1\lt math\gt respectively. The number of points in the angular expansion is \lt math\gt N=16\lt math\gt . The number of roots of the dispersion equation is \lt math\gt M=8\lt math\gt . Plots (a) and (b) are calculated using the circular plate method described here. Plots (c) and (d) are calculated using an arbitrary shaped plate method, with the panels shown being the actual panels used in the calculation. We see the expected agreement between the two methods. Finally tables 1 and 2 show the values of the coefficients \lt math\gt b_{mn}\lt math\gt for the case \lt math\gt H=25\lt math\gt . The very rapid decay of the higher evanescent modes is apparent which explains the rapid convergence in \lt math\gt M\lt math\gt shown in fig.~\ref{convergence_roots}. It is important to realise also that the expansion is in the potential whereas we have shown the displacement in the figures. }[/math]