Generalized Eigenfunction Expansion for Water Waves for a Fixed Body

Introduction

The generalized eigenfunction method goes back to the work of P ovzner 1953 and Ikebe 1060. The generalized eigenfunction method has been applied to water-wave problems by \cite{friedman_shinbrot67,hazard_lenoir02,JFM02,

 hazard_loret07,hazard_meylan07}.  

The theory of generalised eigenfunction is described in Hazard and Lenoir 2002 and Meylan 2002 for the case of a rigid body in water of infinite depth and for a thin plate on water of shallow draft respectively. We will present here the theory for a rigid body in water of finite depth. rms of this solutions (which we call the generalised eigenfunctions) because they solve for

The generalized eigenfunction method is based on an inner product in which the evolution operator is self-adjoint. It follows from the self-adjointness that we can expand the solution in the eigenfunctions of the operator. These eigenfunctions are nothing more than the single-frequency solutions. The main difficulty is that the eigenfunctions are associated with a continuous spectrum, and this requires that they be carefully normalized. Once this is done, we can derive simple expressions which allow the time-domain problem to be solved in terms of the single-frequency solutions. The mathematical ideas are discussed in detail in Hazard and L oret 2007.

We consider a two-dimensional fluid domain of constant depth, which contains a finite number of fixed bodies of arbitrary geometry. We denote the fluid domain by [math]\displaystyle{ \Omega }[/math], the boundary of the fluid domain which touches the fixed bodies by [math]\displaystyle{ \partial\Omega }[/math], and the free surface by [math]\displaystyle{ F. }[/math] The [math]\displaystyle{ x }[/math] and [math]\displaystyle{ z }[/math] coordinates are such that [math]\displaystyle{ x }[/math] is pointing in the horizontal direction and [math]\displaystyle{ z }[/math] is pointing in the vertical upwards direction (we denote [math]\displaystyle{ \mathbf{x}=\left( x,z\right) ). }[/math] The free surface is at [math]\displaystyle{ z=0 }[/math] and the sea floor is at [math]\displaystyle{ z=-h }[/math]. The fluid motion is described by a velocity potential [math]\displaystyle{ \Phi }[/math] and free surface by [math]\displaystyle{ \zeta }[/math].

The equations of motion in the time domain are Laplace's equation through out the fluid

[math]\displaystyle{ \Delta\Phi\left( \mathbf{x,}t\right) =0,\ \ \mathbf{x}\in\Omega. }[/math]

At the bottom surface we have no flow

[math]\displaystyle{ \partial_{n}\Phi=0,\ \ z=-h. }[/math]

At the free surface we have the kinematic condition

[math]\displaystyle{ \partial_{t}\zeta=\partial_{n}\Phi,\ \ z=0,\ x\in \partial\Omega_{F}, }[/math]

and the dynamic condition (the linearized Bernoulli equation)

[math]\displaystyle{ \partial_{t}\Phi = -g\zeta ,\ \ z=0,\ x\in \partial\Omega_{F}. }[/math]

The body boundary condition for a fixed body is

[math]\displaystyle{ \partial_{n}\Phi=0,\ \ \mathbf{x}\in\partial\Omega, }[/math]

The initial conditions are

[math]\displaystyle{ \left.\zeta\right|_{t=0} = \zeta_0(x)\,\,\,\mathrm{and}\,\,\, \left.\partial_t\zeta\right|_{t=0} = \partial_t\zeta_0(x). }[/math]

Single frequency equations

The single frequency solution is based on the assumption that all time-dependence is given by [math]\displaystyle{ \mathrm{e}^{-\mathrm{i}\omega t} }[/math] and that the system is excited by an incident wave. We can then write

[math]\displaystyle{ \Phi\left( \mathbf{x},t\right) ={\Phi}_\kappa\left( \mathbf{x},\omega\right) \mathrm{e}^{-\mathrm{i}\omega t},\ \ \ =and= \ \ \ \zeta\left( x,t\right) ={\zeta}_\kappa\left( x,\omega\right) \mathrm{e}^{-\mathrm{i}\omega t}, }[/math]

where [math]\displaystyle{ \kappa=1 }[/math] for waves excited by an incident wave from the left and [math]\displaystyle{ -1 }[/math] for waves incident from the right. Under these assumptions, equations (1) to (2)\ become

[math]\displaystyle{ \Delta{\Phi}_\kappa\left( \mathbf{x,}\omega\right) =0,\ \ \mathbf{x}\in \Omega, }[/math] [math]\displaystyle{ \partial_{n}{\Phi}_\kappa=0,\ \ \mathbf{x}\in\partial\Omega, }[/math] [math]\displaystyle{ \partial_{n}{\Phi}_\kappa=0,\ \ z=-h, }[/math] [math]\displaystyle{ -\mathrm{i}\omega{\zeta}_\kappa=\partial_{n}{\Phi}_\kappa,\ \ z=0,\ x\in F, }[/math] [math]\displaystyle{ {\zeta}_\kappa = \mathrm{i}\omega{\Phi}_\kappa,\ \ z=0,\ x\in F. }[/math]

We must also specify radiations conditions, which are given by

[math]\displaystyle{ {\Phi}_1=\frac{1}{\mathrm{i} \omega} \left( \mathrm{e}^{\mathrm{i} kx} + R_1 e^{-\mathrm{i} k x} \right)\frac{\cosh k\left( z+h\right) }{\cosh kh}, \,\,\,\mathrm{as}\,\,x\to-\infty, }[/math] [math]\displaystyle{ {\Phi}_1=\frac{1}{\mathrm{i} \omega} T_1\mathrm{e}^{\mathrm{i} kx}\frac{\cosh k\left( z+h\right) }{\cosh kh}, \,\,\,\mathrm{as}\,\,x\to\infty, }[/math] [math]\displaystyle{ {\Phi}_{-1}=\frac{1}{\mathrm{i} \omega} T_{-1}\mathrm{e}^{-\mathrm{i} kx} \frac{\cosh k\left( z+h\right) }{\cosh kh}, \,\,\,\mathrm{as}\,\,x\to\infty, }[/math] [math]\displaystyle{ {\Phi}_{-1}=\frac{1}{\mathrm{i} \omega} \left( \mathrm{e}^{-\mathrm{i} kx} + R_{-1} e^{\mathrm{i} k x} \right)\frac{\cosh k\left( z+h\right) }{\cosh kh}, \,\,\,\mathrm{as}\,\,x\to-\infty. }[/math]

Note that [math]\displaystyle{ R_{\kappa} }[/math] and [math]\displaystyle{ T_{\kappa} }[/math] are the reflection and transmission coefficients respectively and that we have normalized so that the amplitude (in displacement) is unity. The wavenumber [math]\displaystyle{ k }[/math] is the positive real solution of the dispersion equation </math>k\tanh

kh=\omega^{2}[math]\displaystyle{ , and we will consider both }[/math]k(\omega)[math]\displaystyle{ and }[/math]\omega(k)[math]\displaystyle{ in what follows as required. The solution of the single-frequency equation may be computationally challenging and for the generalized eigenfunction expansion the major numerical work is to determine the single-frequency solutions. == Time domain calculations == The solution in the frequency domain can be used to construct the solution in the time domain. This is well-known for the case of a plane incident wave which is initially far from the body, and in this case the solution can be calculated straightforwardly using the standard Fourier transform. However, when we consider an initial displacement which is non-zero around the bodies, we cannot express the time-dependent solution in terms of the single frequency solutions by a standard Fourier transform. We begin with the equations in the time domain subject to the initial conditions given by Denoting the potential at the surface by \lt center\gt \lt math\gt \phi(x,t)=\left.\Phi\left(\mathbf{x},t\right)\right|_{z=0}, }[/math]

we introduce the operator [math]\displaystyle{ \mathbf{G} }[/math] which maps the surface potential to the potential throughout the fluid domain. The operator [math]\displaystyle{ \mathbf{G}\psi }[/math] is found by solving

[math]\displaystyle{ \begin{matrix} \Delta\Psi\left( \mathbf{x}\right) & = 0,\ \ \mathbf{x}\in\Omega,\\ \partial_{n}\Psi & = 0,\ \ \mathbf{x}\in\partial\Omega,\\ \partial_{n}\Psi & = 0,\ \ z=-h,\\ \Psi & =\psi,\ \ z = 0,\ x\in F, \end{matrix} }[/math]

and is defined by [math]\displaystyle{ \mathbf{G}\psi=\Psi. }[/math] The operator </math>\partial

_{n}\mathbf{G}[math]\displaystyle{ , which maps the surface potential to the normal derivative of potential at the surface (called the Dirichlet-to-Neumann map) is given by \lt center\gt \lt math\gt \partial_{n}\mathbf{G}\psi=\left. \partial_{n}\Psi\right\vert _{z=0}. }[/math]

Therefore equations (1) to (2) can be written as

[math]\displaystyle{ \partial_{t}^2 \zeta + \partial_{n}\mathbf{G} \zeta = 0. }[/math]

where we can recover the potential using the operator [math]\displaystyle{ \mathbf{G} }[/math]. The evolution operator [math]\displaystyle{ \partial_{n}\mathbf{G} }[/math] is symmetric in the Hilbert space given by the following inner product

[math]\displaystyle{ \left\langle \zeta,\eta \right\rangle _{\mathcal{H}}= \int_{F}\zeta \eta ^{*}\,\mathrm{d} x, }[/math]

where [math]\displaystyle{ ^{*} }[/math] denotes complex conjugate, and we assume that this symmetry implies that the operator is self-adjoint. We can prove the symmetry by using Green's second identity

[math]\displaystyle{ \begin{matrix} \int_{F}\left( \partial_{n}\mathbf{G}\zeta \right) \left( \eta\right) ^{*}\,\mathrm{d} x &=\int_{F}\left( \partial_{n}\mathbf{G}\partial_t\phi \right) \left( \partial_t\psi\right) ^{*}\,\mathrm{d} x \\ &=\int_{F}\left( \partial_t\phi \right) \left( \partial_{n}\mathbf{G} \partial_t\psi\right) ^{*}\,\mathrm{d} x \\ &=\int_{F}\left( \zeta \right) \left( \partial_{n}\mathbf{G}\eta\right) ^{*}\,\mathrm{d} x, \end{matrix} }[/math]

where [math]\displaystyle{ \phi }[/math] and [math]\displaystyle{ \psi }[/math] are the surface potentials associated with [math]\displaystyle{ \zeta }[/math] and [math]\displaystyle{ \eta }[/math] respectively.

==Eigenfunctions of [math]\displaystyle{ \partial_{n==\mathbf{G} }[/math]}

The eigenfunctions of [math]\displaystyle{ \partial_{n}\mathbf{G} }[/math] satisfy

[math]\displaystyle{ \partial_{n}\mathbf{G} \zeta = \omega^2 \zeta. }[/math]

Equation (8) is nothing more than equations (5) to (6). This means that to solve for the eigenfunctions of [math]\displaystyle{ \partial_{n}\mathbf{G} }[/math] we need to solve the frequency-domain equations, and the radian frequency [math]\displaystyle{ \omega }[/math] is exactly the eigenvalue. To actually calculate the eigenfunctions of [math]\displaystyle{ \partial_{n}\mathbf{G} }[/math] we need to specify the incident wave potential, and for each frequency we have two eigenfunctions (waves incident from the left and from the right). It is possible for there to exist point spectra for this operator which correspond to the existence of a trapped mode mciver96, but this case is not discussed here. Note that the presence of a trapped mode requires that the generalized eigenfunction expansion we derive must be modified.

Normalization of the Eigenfunctions

The eigenfunctions of [math]\displaystyle{ \partial_{n}\mathbf{G} }[/math] (with eigenvalue [math]\displaystyle{ \omega }[/math]) are denoted by [math]\displaystyle{ \zeta_{\kappa}(x,k\left( \omega\right) ) }[/math]. As mentioned previously, determining [math]\displaystyle{ \zeta_\kappa }[/math] is the major computation of the generalized eigenfunction method, but we simply assume that they are known. We know that the eigenfunctions are orthogonal for different [math]\displaystyle{ \omega }[/math] (from the self-adjointness of [math]\displaystyle{ \partial_n\mathbf{G} }[/math]), and that the waves incident from the left and right with the same [math]\displaystyle{ \omega }[/math] are orthogonal from the identity

[math]\displaystyle{ R_1 T_{-1}^{*} + R_{-1}^{*} T_{1}=0, }[/math]

Mei 1989. It therefore follows that

[math]\displaystyle{ \left\langle \left( {\zeta}_{\kappa}(x,k\left( \omega_{1}\right) )\right) ,{\zeta}_{\kappa^{\prime}}(x,k\left( \omega_{2}\right) )\right\rangle _{\mathcal{H} }=\Lambda_{n}\left( \omega_{1}\right) \delta\left( \omega_{1}-\omega _{2}\right) \delta_{\kappa\kappa^{\prime}}, }[/math]

but we need to determine the normalizing function </math>\Lambda_{n}\left(

 \omega_{n}\right)[math]\displaystyle{ . This is achieved by using the result that the
eigenfunctions satisfy the same normalizing condition with and without
the scatterers present.  This result, the proof of which is quite
technical, is well-known and has been shown for many different
situations. The original proof was for Schr\"odinger's equation and was
due to [[povzner53,ikebe60]]. A proof for the case of Helmholtz
equation was given by [[wilcox75]].  Recently the proof was given
for water waves by [[hazard_lenoir02,hazard_loret07]].

Since the eigenfunctions satisfy the same normalizing condition with
and without the scatterers, we normalize with the scatterers absent.
This means that the eigenfunctions are simply the incident waves, and
the free surface \lt math\gt F }[/math] is the entire axis. This allows us to derive

[math]\displaystyle{ \begin{matrix} \left\langle \left( {\zeta}_{\kappa}(x,k\left( \omega_{1}\right) )\right) ,{\zeta}_{\kappa^{\prime}}(x,k\left( \omega_{2}\right) )\right\rangle _{\mathcal{H}} &= \int_{\mathbb{R}}\left( e^{\kappa\mathrm{i} k_{1} x}\right) \left( e^{\kappa^{\prime}\mathrm{i} k_{2}x}\right) ^{*}\,\mathrm{d} x \\ & =2\pi \delta_{\kappa\kappa^{\prime}}\delta\left( k_{1}-k_{2}\right) \\ & =2\pi\delta_{\kappa\kappa^{\prime}}\delta\left( \omega_{1}-\omega_{2}\right) \left. \frac{d\omega}{dk}\right\vert _{\omega=\omega_{1}}. \end{matrix} }[/math]

This result allows us to calculate the time-dependent solution in the eigenfunctions (or single-frequency solutions).

Expansion in Eigenfunctions

We expand the solution for the displacement in the time domain as

[math]\displaystyle{ \zeta\left( x,t\right) =\int_{\mathbb{R}^{+}} \sum_{\kappa\in\left\{ -1,1\right\}} \left\{ f_{\kappa}\left( \omega\right) \cos(\omega t)+ g_{\kappa}\left( \omega\right) \frac{\sin(\omega t)}{\omega}\right\} \zeta_{\kappa}(x,k) d\omega, (9) }[/math]

where [math]\displaystyle{ f_\kappa }[/math] and [math]\displaystyle{ g_\kappa }[/math] will be determined from the initial conditions. Note that here, and in subsequent equations, we are assuming that [math]\displaystyle{ k }[/math] is a function of [math]\displaystyle{ \omega }[/math] or that [math]\displaystyle{ \omega }[/math] is a function of [math]\displaystyle{ k }[/math] as required. If we take the inner product with respect to the eigenfunctions [math]\displaystyle{ \zeta_\kappa }[/math] we obtain

[math]\displaystyle{ \left\langle \zeta_0\left( x\right) ,\zeta_{\kappa}(x,k)\right\rangle _{\mathcal{H}}=2\pi f_{\kappa}\left( \omega\right) \frac{d\omega}{dk}, }[/math]

and

[math]\displaystyle{ \left\langle v_0\left( x\right) ,\zeta_{\kappa}(x,k)\right\rangle _{\mathcal{H}}=2\pi g_{\kappa}\left( \omega\right) \frac{d\omega}{dk}. }[/math]

We can therefore write equation \eqref{expansion}, changing the variable of integration to [math]\displaystyle{ k }[/math], as \begin{multline} \zeta\left( x,t\right)

=\frac{1}{2\pi}\int_{\mathbb{R}^{+}}  \sum_{\kappa\in\left\{  -1,1\right\}}

\Big\{ \left\langle \zeta_0\left( x\right)

,\zeta_{\kappa}(x,k)\right\rangle _{\mathcal{H}}

\cos(\omega t)\\ + \left\langle v_0\left( x\right)

,\zeta_{\kappa}(x,k)\right\rangle _{\mathcal{H}}

\frac{\sin(\omega t)}{\omega}\Big\} \zeta_{\kappa}(x,k) dk, (10) \end{multline}

If we take the case when [math]\displaystyle{ v_0( x) =0 }[/math] and write the integral given by the inner product explicitly, we obtain

[math]\displaystyle{ \zeta\left( x,t\right) =\int_{\mathbb{R}^{+}}\Big\{ \sum_{\kappa\in\left\{ -1,1\right\} }\left( \frac{1}{2\pi}\int_{F}\zeta_0\left( x^{\prime}\right) \zeta_{\kappa}(x^{\prime},k) ^{*}\,\mathrm{d} x^{\prime}\right) \zeta_{\kappa}(x,k)\Big\}\cos(\omega t)dk.(11) }[/math]

An identity linking waves from the left and right

A consequence of the requirement that the displacement be real, if the initial displacement and initial derivative of displacement is real, is that

[math]\displaystyle{ \sum_{\kappa\in\left\{ -1,1\right\}} \left\langle \zeta_0\left( x\right) ,\zeta_{\kappa}(x,k)\right\rangle _{\mathcal{H}} \zeta_{\kappa}(x,k), }[/math]

must be purely real. This can only be true if

[math]\displaystyle{ (13) \Im \left\{ \zeta_{1}(x^\prime,k)^{*} \zeta_{1}(x,k) \right\} = - \Im \left\{ \zeta_{-1}(x^\prime,k)^{*} \zeta_{-1}(x,k) \right\}, \,\,\,x,x^{\prime} \in F. }[/math]

To the best of the author's knowledge, this result has not appeared previously.

Results

We present here results for a very simple geometry, so that frequency domain calculations are straightforward. We consider a pair of rigid plates of negligible thickness with the boundary of the structure given by

[math]\displaystyle{ \partial\Omega = \{-L_2\leq x \leq -L_1, z=-d\}\cup \{L_1\leq x \leq L_2, z=-d\}. }[/math]

The method used to solve the equations is based on the matched eigenfunction expansion method. Full details and computer code can be found on meylan_wikiwaves and the problem is also discussed in linton_evans91. The initial displacement is given by

[math]\displaystyle{ \zeta_0(x) = \exp(-4x^2)\,\,\,\mathrm{and}\,\,\,v_0(x)=0. (14) }[/math]

We fix the water depth [math]\displaystyle{ h=-2 }[/math], the dock lengths are [math]\displaystyle{ L_1=1 }[/math] and [math]\displaystyle{ L_2=1.75 }[/math]. We consider the case when [math]\displaystyle{ d=0 }[/math] (Figure 15), [math]\displaystyle{ d=0.1 }[/math] (Figure 16), and [math]\displaystyle{ d=0.2 }[/math] (Figure 17). Note that the dock is also shown as a dark line in the figures, but this is only shown for illustrative purposes and has no relationship with the free surface. The solutions are also shown in Movie 1 and Movie 4 and Movie 5. Further solutions with different dock lengths and submergences are illustrated in Movies 2, 3, and 6. These figures show interesting resonant effects, which are strongly a function of the depth of submergence.

\begin{figure} \begin{center} \psfrag{x}[][][0.8]{[math]\displaystyle{ x }[/math]} \psfrag{xi}[][][0.8]{[math]\displaystyle{ \zeta }[/math]} \psfrag{t = 0}[][][0.8]{[math]\displaystyle{ t = 0 }[/math]} \psfrag{t = 25}[][][0.8]{[math]\displaystyle{ t = 25 }[/math]} \psfrag{t = 5}[][][0.8]{[math]\displaystyle{ t = 5 }[/math]} \psfrag{t = 30}[][][0.8]{[math]\displaystyle{ t = 30 }[/math]} \psfrag{t = 10}[][][0.8]{[math]\displaystyle{ t = 10 }[/math]} \psfrag{t = 35}[][][0.8]{[math]\displaystyle{ t = 35 }[/math]} \psfrag{t = 15}[][][0.8]{[math]\displaystyle{ t = 15 }[/math]} \psfrag{t = 40}[][][0.8]{[math]\displaystyle{ t = 40 }[/math]} \psfrag{t = 20}[][][0.8]{[math]\displaystyle{ t = 20 }[/math]} \includegraphics[width=12cm]{figures/large_dock} \end{center} \caption{The surface displacement for the initial conditions given

 by equation \eqref{initial_figure} for the times shown.  The water
 depth is [math]\displaystyle{ h=-2 }[/math], and the dock lengths are [math]\displaystyle{ L_1=1 }[/math] and [math]\displaystyle{ L_2=1.75 }[/math]. The
 dock submergence is [math]\displaystyle{ d=0 }[/math]. The dock
 position is shown by the dark line for illustrative purposes only. }

(15) \end{figure}

\begin{figure} \begin{center} \psfrag{x}[][][0.8]{[math]\displaystyle{ x }[/math]} \psfrag{xi}[][][0.8]{[math]\displaystyle{ \zeta }[/math]} \psfrag{t = 0}[][][0.8]{[math]\displaystyle{ t = 0 }[/math]} \psfrag{t = 25}[][][0.8]{[math]\displaystyle{ t = 25 }[/math]} \psfrag{t = 5}[][][0.8]{[math]\displaystyle{ t = 5 }[/math]} \psfrag{t = 30}[][][0.8]{[math]\displaystyle{ t = 30 }[/math]} \psfrag{t = 10}[][][0.8]{[math]\displaystyle{ t = 10 }[/math]} \psfrag{t = 35}[][][0.8]{[math]\displaystyle{ t = 35 }[/math]} \psfrag{t = 15}[][][0.8]{[math]\displaystyle{ t = 15 }[/math]} \psfrag{t = 40}[][][0.8]{[math]\displaystyle{ t = 40 }[/math]} \psfrag{t = 20}[][][0.8]{[math]\displaystyle{ t = 20 }[/math]} \includegraphics[width=12cm]{figures/submerged_large_dockd0p1} \end{center} \caption{As in Figure 15, except that the dock

 submergence is [math]\displaystyle{ d=0.1 }[/math].}

(16) \end{figure}

\begin{figure} \begin{center} \psfrag{x}[][][0.8]{[math]\displaystyle{ x }[/math]} \psfrag{xi}[][][0.8]{[math]\displaystyle{ \zeta }[/math]} \psfrag{t = 0}[][][0.8]{[math]\displaystyle{ t = 0 }[/math]} \psfrag{t = 25}[][][0.8]{[math]\displaystyle{ t = 25 }[/math]} \psfrag{t = 5}[][][0.8]{[math]\displaystyle{ t = 5 }[/math]} \psfrag{t = 30}[][][0.8]{[math]\displaystyle{ t = 30 }[/math]} \psfrag{t = 10}[][][0.8]{[math]\displaystyle{ t = 10 }[/math]} \psfrag{t = 35}[][][0.8]{[math]\displaystyle{ t = 35 }[/math]} \psfrag{t = 15}[][][0.8]{[math]\displaystyle{ t = 15 }[/math]} \psfrag{t = 40}[][][0.8]{[math]\displaystyle{ t = 40 }[/math]} \psfrag{t = 20}[][][0.8]{[math]\displaystyle{ t = 20 }[/math]} \includegraphics[width=12cm]{figures/submerged_large_dockd0p2} \end{center} \caption{As in Figure 15, except that the dock

 submergence is [math]\displaystyle{ d=0.2 }[/math].}

(17) \end{figure}

Solution for twin submerged docks

We give a brief account of the solution for a pair of submerged docks in the frequency domain. This is based on the solution given in linton_evans91. Further details and computer code can be found on meylan_wikiwaves. While we have concentrated on expressing the solution in terms of the displacements, the eigenfunctions [math]\displaystyle{ \zeta_\kappa }[/math] cannot be found without solving for the potential throughout the fluid domain. The boundary value problem can be expressed as

[math]\displaystyle{ \Delta\Phi_\kappa=0, \,\, -h\lt z\lt 0, }[/math] [math]\displaystyle{ \partial_n\Phi_{\kappa}=0, \,\, z=-h, }[/math] [math]\displaystyle{ \partial_n\Phi_\kappa=\omega^2\Phi_\kappa, \,\, z=0, }[/math] [math]\displaystyle{ \partial_z\phi=0, \,\, z=-d,\,-L_2\lt x\lt -L_1,\,\,\mathrm{and}\,\,L_1\lt x\lt L_2, }[/math]

subject to an incident plane wave coming from negative infinity. We use the eigenfunction matching method coupled with symmetry, which allows us decompose the solutions into an symmetric ([math]\displaystyle{ \Phi_{s} }[/math]) and an anti-symmetric solution ([math]\displaystyle{ \Phi_{a} }[/math]). Details of this method can be found in linton_mciver01. It is interesting to note that these solutions are orthogonal with respect to the inner product given by equation \eqref{inner_product} and we could use these in our generalized eigenfunction expansion. However, this decomposition is only useful when the body geometry has symmetry, in which case we are reduced to having only to solve for the potential in a half space. The solutions for waves incident from the left and right can be found as

[math]\displaystyle{ \Phi_1 = \frac{1}{2}\left(\Phi_{s} + \Phi_{a}\right), \,\,\, \mathrm{and} \,\,\, \Phi_{-1} = \frac{1}{2}\left(\Phi_{s} - \Phi_{a}\right). }[/math]

The symmetric solution can be written as

[math]\displaystyle{ \Phi_{s}(x,z)= \left\{ \begin{matrix}[c]{l} { \frac{1}{\mathrm{i} \omega}} e^{-k_{0}^h (x+L)}\phi_{0}^h\left( z\right) + \sum_{m=0}^{\infty}a_{m}^{s}e^{k_{m}^h x}\phi_{m}^h(z), \;\;x\lt -L_2 \\ \sum_{m=0}^{\infty}b_{m}^{s} e^{-k_{m}^d (x+L)}\phi_{m}^d(z) + \sum_{m=0}^{\infty}c_{m}^{s} e^{k_{m}^d (x-L)}\phi_{m}^d(z) , \;\;-d\lt z\lt 0,\,-L_2\lt x\lt -L_1, \\ d_0^{s}\frac{x+L_1}{L_2-L_1} + \sum_{m=1}^{\infty}d_{m}^{s} e^{\kappa_{m} (x+L)}\psi_{m}(z) + e_0^{s}\frac{x+L_2}{L_2-L_1} \\ \quad\quad + \sum_{m=1}^{\infty}e_{m}^{s} e^{-\kappa_{m} (x-L)}\psi_{m}(z) , \;\;-h\lt z\lt -d,\,-L_2\lt x\lt -L_1, \\ \sum_{m=0}^{\infty}f_{m}^{s}\frac{\cosh(k_{m}^h x)}{\cosh(k_m^h L)}\phi_{m}^h(z), \;\;-L_1\lt x\lt 0, \end{matrix} \right. }[/math]

where [math]\displaystyle{ k_n^l }[/math] are the roots of the dispersion relation for a free surface

[math]\displaystyle{ k \tan(kl) = -\omega^2. }[/math]

We denote the positive imaginary solutions by [math]\displaystyle{ k_{0}^l }[/math] and the positive real solutions by [math]\displaystyle{ k_{m}^l }[/math], [math]\displaystyle{ m\geq1 }[/math] (ordered with increasing imaginary part) and [math]\displaystyle{ \kappa_{m}=m\pi/(h-d) }[/math]. We define

[math]\displaystyle{ \phi_{m}^l\left( z\right) = \frac{\cos k_{m}(z+l)}{\cos k_{m}l},\quad m\geq0, }[/math]

as the vertical eigenfunction of the potential in the open water regions 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 region. The anti-symmetric solution is the same, except that the last term is a ratio of [math]\displaystyle{ \sinh }[/math] rather than [math]\displaystyle{ \cosh }[/math]. We solve by truncating the expansions and matching the potential and the [math]\displaystyle{ x }[/math] derivative of the potential at the boundaries [math]\displaystyle{ x=-L_2 }[/math] and [math]\displaystyle{ x=-L_1 }[/math].

Generalised Eigenfunction Expansion for a Elastic Plate on Shallow Water