Free-Surface Green Function for a Floating Elastic Plate

This is a special version of the free-surface Green function which applied when the Floating Elastic Plate boundary condition applies at the free-surface

Two Dimensions

The Green function [math]\displaystyle{ G(x,z) }[/math] representing outgoing waves as [math]\displaystyle{ |x|\rightarrow \infty }[/math] satisfies

[math]\displaystyle{ \nabla^2 G = 0, -h\lt z\lt 0, }[/math] [math]\displaystyle{ \frac{\partial G}{\partial z} =0, z=-h, }[/math] [math]\displaystyle{ {\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G} = \delta(x), z=0, }[/math]

(note we are only considering singularities at the free surface).

This problem can be solved to give

[math]\displaystyle{ G(x,z) = -i\sum_{n=-2}^\infty\frac{\sin{(k_n h)}\cos{(k_n(z+h))}}{2\alpha C(k_n)}e^{-k_n|x|}, }[/math]

where

[math]\displaystyle{ C(k_n)=\frac{1}{2}\left(h + \frac{(5\beta k(n)^4 + 1 - \alpha\gamma)\sin^2{(k(n)h)}}{\alpha}\right), }[/math]

and [math]\displaystyle{ k_n }[/math] are the solutions of the Dispersion Relation for a Floating Elastic Plate,

[math]\displaystyle{ \beta k^5 \sin(kH) - k \left(1 - \alpha \gamma \right) \sin(kH) = -\alpha \cos(kH) \, }[/math]

with [math]\displaystyle{ n=-1,-2 }[/math] corresponding to the complex solutions with positive real part, [math]\displaystyle{ n=0 }[/math] corresponding to the imaginary solution with positive imaginary part and [math]\displaystyle{ n\gt 0 }[/math] corresponding to the real solutions with positive real part. We can also write the Green function as

[math]\displaystyle{ G\left( r,z\right) = -\omega\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{q_n\sinh q_nH}\exp\left( \mathrm{i}q_nr\right) \cosh q_n\left( z+H\right) . }[/math]

where [math]\displaystyle{ q_n=ik_n }[/math] and we have used a different non-dimensionalisation so that the Dispersion Relation for a Floating Elastic Plate is written as

[math]\displaystyle{ -q^5 \sinh(kH) - k \left(1 - m\omega^2 \right) \sinh(kH) = -\omega^2 \cosh(kH) \, }[/math]

and

[math]\displaystyle{ R\left( q_n\right) =\frac{\omega^{2}q_n}{\omega^{2}\left( 5q_n^{4}+u\right) +H\left[ \left( q_n^{5}+uq\right) ^{2}-\omega^{4}\right] }. }[/math]

Three Dimensions

[math]\displaystyle{ \phi_{=P= }\left( r,z\right) =-\frac{\omega}{2}\sum_{q\in K^{ }}\frac{R\left( q\right) }{\sinh qH}H_{0}^{\left( 1\right) }\left( qr\right) \cosh q\left( z+H\right) }[/math]

Derivation of the Green function

We present here a derivation of the Green function for both the two and three dimensional problem. The devivation uses the Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Equation. This derivation is based on the PhD thesis of Hyuck Chung.

Previous Work

Radially symmetric waves in a thin plate without the hydrodynamic support, have studied by Sneddon 1945 using the two dimensional radial Fourier transform. This integral transform is commonly called the Hankel transform

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\frac{1}{2\pi}\int_{0}^{\infty}w\left( r\right) rJ_{0}\left( \gamma r\right) dr. }[/math]

A complete analytical solution for deflection of an infinite plate due to a localized static load was derived by Wyman 1950 by choosing the bounded solutions of

[math]\displaystyle{ w_{rr}+\frac{1}{r}w_{r}+\mathrm{i}w=0 (3-81) }[/math]

whose solutions satisfy the radial form of the (non-dimensional) thin plate equation

[math]\displaystyle{ \left[ \frac{\partial^{2}}{\partial r^{2}}+\frac{1}{r}\frac{\partial }{\partial r}\right] ^{2}w+w=0. }[/math]

If we consider a plate floating on water we have to to solve Laplace's equation, and thus the application of the Fourier transform is not as straightforward as in the two examples cited (Sneddon Sneddon45 and Wyman Wyman50). Response of an infinite thin elastic plate under fluid loading is studied in a series of papers by Crighton Crighton72, Crighton77, Crighton79. In the following sections, the deflection of a plate is expressed by the infinite number of modes that exist in the vibrating plate floating on the water, instead of a finite number of modes that represent the free oscillation of an elastic plate without restraints. The fact that waves in an incompressible fluid can be expressed by an infinite series of natural modes is shown by John fritz, fritz2, each mode being a Hankel function of the first kind. It is known that the oscillation of an infinite plate can also be expressed in terms of Hankel functions. Kheisin, in 1967 (Kheisin67 chapter IV), studied the same problem solved in this chapter. Kheisin derived the same dispersion equation for the physical variables, and then studied the properties of the inverse transform for the simple shallow water and static load cases, which are approximated versions of the solutions reported here.

Mathematical model

In terms of non-dimensional variables the Floating Elastic Plate equation becomes

[math]\displaystyle{ \left[ \nabla^{4}-m\omega^{2}+1\right] w+\mathrm{i}\omega \phi=p_{=a= }. }[/math]

plus Laplace's equation for the velocity potential and the bottom condition

[math]\displaystyle{ \nabla^{2}\phi=0 }[/math]

for [math]\displaystyle{ -\infty\lt x,y\lt \infty, }[/math] [math]\displaystyle{ -H\lt z\lt 0 }[/math] and

[math]\displaystyle{ \phi_{z}=0\quad \mathrm{at} \quad \;z=-H. }[/math]

The system of equations from together with the Sommerfeld Radiation Condition form the BVP, which we will solve for [math]\displaystyle{ w }[/math] and [math]\displaystyle{ \phi }[/math] for a given [math]\displaystyle{ \omega }[/math].

Spatial Fourier transform

We solve the system of equations using the Fourier Transform in [math]\displaystyle{ \left( x,y\right) }[/math]-plane for three-dimensions and with a Fourier Transform in [math]\displaystyle{ x }[/math] for two dimensions. We choose the Fourier transform with respect to [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] defined as

[math]\displaystyle{ \hat{\phi}\left( \alpha,k,z\right) =\int_{-\infty}^{\infty}\int_{-\infty }^{\infty}\phi\left( x,y,z\right) e^{\mathrm{i}\left( \alpha x+ky\right) }dxdy }[/math]

and the inverse Fourier transform defined as

[math]\displaystyle{ \phi\left( x,y,z\right) =\frac{1}{4\pi^{2}}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\hat{\phi}\left( \alpha,k,z\right) e^{-\mathrm{i}\left( \alpha x+ky\right) }d\alpha dk. }[/math]

For the one-dimensional case, the definitions are

[math]\displaystyle{ \hat{\phi}\left( \alpha,z\right) =\int_{-\infty}^{\infty}\phi\left( x,z\right) e^{\mathrm{i}\alpha x}dx, }[/math] [math]\displaystyle{ \phi\left( x,z\right) =\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{\phi }\left( \alpha,z\right) e^{-\mathrm{i}\alpha x}d\alpha. }[/math]

We denote the spatial Fourier transform by using a hat over [math]\displaystyle{ w }[/math] and [math]\displaystyle{ \phi }[/math].

The Fourier transform of both sides of the Laplace's equation becomes an ordinary differential equation with respect to [math]\displaystyle{ z }[/math],

[math]\displaystyle{ \frac{\partial^{2}\hat{\phi}}{\partial z^{2}}\left( \alpha,k,z\right) -\left( \alpha^{2}+k^{2}\right) \hat{\phi}\left( \alpha,k,z\right) =0 }[/math]

with a solution

[math]\displaystyle{ \hat{\phi}\left( \alpha,k,z\right) =A\left( \gamma\right) e^{\gamma z}+B\left( \gamma\right) e^{-\gamma z} }[/math]

where [math]\displaystyle{ \gamma^{2}=\alpha^{2}+k^{2} }[/math] and [math]\displaystyle{ A }[/math], [math]\displaystyle{ B }[/math] are unknown coefficients to be determined by the boundary condition. We find that [math]\displaystyle{ \hat{\phi} }[/math] is a function only of the magnitude of the Fourier transform variables, [math]\displaystyle{ \gamma=\left\| \mathbf{\gamma}\right\| =\left| \left| \left( \alpha,k\right) \right| \right| }[/math], hence we may now denote [math]\displaystyle{ \hat{\phi }\left( \alpha,k,z\right) }[/math] by [math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) }[/math]. We can reach the same conclusion using the fact that [math]\displaystyle{ w\left( x,y\right) }[/math] and [math]\displaystyle{ \phi\left( x,y,z\right) }[/math] are functions of [math]\displaystyle{ r=\sqrt{x^{2}+y^{2}} }[/math] thus the Fourier transforms must be functions of [math]\displaystyle{ \gamma=\sqrt{\alpha^{2}+k^{2}} }[/math].

We find the unknown coefficients [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] from the Fourier transformed ocean floor condition that [math]\displaystyle{ \left. \hat{\phi}_{z}\right| _{z=-H}=0 }[/math] to be [math]\displaystyle{ A\left( \gamma\right) =Ce^{\gamma H} }[/math], and [math]\displaystyle{ B\left( \gamma\right) =Ce^{-\gamma H} }[/math]. Thus, we obtain the depth dependence of the Fourier transform of the potential

[math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) =\hat{\phi}\left( \gamma,0\right) \frac{\cosh\gamma\left( z+H\right) }{\cosh\gamma H}. }[/math]

At the surface, [math]\displaystyle{ z=0 }[/math], differentiating both sides with respect to [math]\displaystyle{ z }[/math] the vertical component of the velocity is

[math]\displaystyle{ \hat{\phi}_{z}\left( \gamma,0\right) =\hat{\phi}\left( \gamma,0\right) \gamma\tanh\gamma H. }[/math]

Using this relationship to substitute for [math]\displaystyle{ \hat{\phi}_{z} }[/math] in the non-dimensional form of the kinematic condition, we find that

[math]\displaystyle{ \hat{\phi}\left( \gamma,0\right) =\frac{\mathrm{i}\omega\hat {w}\left( \gamma\right) }{\gamma\tanh\gamma H}. }[/math]

Thus, once we find [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] then we can find [math]\displaystyle{ \hat{\phi}\left( \gamma,z\right) }[/math].

The Fourier transform of the plate equation is also an algebraic equation in the parameter [math]\displaystyle{ \gamma }[/math]

[math]\displaystyle{ \left( \gamma^{4}+1-m\omega^{2}\right) \hat{w}+\mathrm{i}\omega \hat{\phi}=1. }[/math]

Hence, we have the single algebraic equation for each spatial Fourier component of [math]\displaystyle{ w }[/math]

[math]\displaystyle{ \left( \gamma^{4}+1-m\omega^{2}-\frac{\omega^{2}}{\gamma\tanh\gamma H}\right) \hat{w}=1. }[/math]

Notice that we have used the fact that the Fourier transform of the delta function is [math]\displaystyle{ 1 }[/math].

We find that the spatial Fourier transform of the displacement of the ice sheet for the localized forcing, both point and line, is

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\frac{1}{d\left( \gamma,\omega\right) } }[/math]

where

[math]\displaystyle{ d\left( \gamma,\omega\right) =\gamma^{4}+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}. }[/math]

Where [math]\displaystyle{ d\left( \gamma,\omega\right) }[/math]\ in the Dispersion Relation for a Floating Elastic Plate function, and the associated algebraic equation [math]\displaystyle{ d\left( \gamma ,\omega\right) =0 }[/math] the dispersion equation for waves propagating through an ice sheet. This dispersion relation (and the Fourier transform was derived by Kheisin 1967 chapter IV, and Fox and Squire 1994.

Inverse Fourier Transform

Our task now is to derive the inverse Fourier transform). We notice that the roots of the Dispersion Relation for a Floating Elastic Plate for a fixed [math]\displaystyle{ \omega }[/math], are the poles of the function [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math], which are necessary for calculation of integrals involved in the inverse Fourier transform. There are a pair of two real roots [math]\displaystyle{ \pm q_0 }[/math], an infinite number of pure imaginary roots [math]\displaystyle{ \left\{ \pm q_{n}\right\} {n=1,2,...} }[/math], ([math]\displaystyle{ \Im q_{n}\gt 0 }[/math])plus four complex roots (in all physical situations) occur at plus and minus complex-conjugate pairs [math]\displaystyle{ \pm q_{-1} }[/math] and [math]\displaystyle{ \pm q_{-2}^{} }[/math] with [math]\displaystyle{ Im \left( q_{-1,-2}\right) \gt 0 }[/math]).

We note that the dispersion equation represents a relationship between the spatial wavenumbers [math]\displaystyle{ \gamma }[/math] and the radial frequency [math]\displaystyle{ \omega }[/math], which is how the name "dispersion" came about.

We may also solve for the velocity potential at the surface of the water

[math]\displaystyle{ \hat{\phi}\left( \gamma,0\right) =\frac{\mathrm{i}\omega w} {\gamma\tanh\gamma H}=\frac{\mathrm{i}\omega}{\gamma\tanh\left( \gamma H\right) d\left( \gamma,\omega\right) }, (3-10) }[/math]

which is also a function of [math]\displaystyle{ \gamma }[/math] only and has the same poles as [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] does.

The Fourier transform of the differential equations can be interpreted as wave-like forcing of the ice sheet, i.e.,

[math]\displaystyle{ p_{=a= }\left( x,y,t\right) =\hat{p}_{=a= }\left( \alpha ,k,\omega\right) e^{\mathrm{i}\left( \omega t-\alpha x-ky\right) } }[/math]

where [math]\displaystyle{ \hat{p}_{=a= }\left( \alpha,k,\omega\right) }[/math] is the amplitude of the wave-like forcing or a Fourier transform of the wave-like force which we set to be one. Therefore, superposition or the inverse Fourier transform of the solution under this wave-like force represents the response of a plate to a localized force expressed by the delta function in Eqn.~((3-23)). Since the waves are radially symmetric, [math]\displaystyle{ p_{=a= } }[/math] can be written as a function of [math]\displaystyle{ r }[/math],

[math]\displaystyle{ p_{=a= }\left( r,t\right) =\hat{p}_{=a= }\left( \gamma ,\omega\right) J_{0}\left( \gamma r\right) e^{\mathrm{i}\omega t} }[/math]

where [math]\displaystyle{ \hat{p}_{=a= }\left( \gamma,\omega\right) }[/math] is the Fourier component of radially symmetric wave-like function.

The inverse Fourier transform (sec:3)

We calculate the displacement [math]\displaystyle{ w\left( x,y\right) }[/math] by performing the two dimensional inverse Fourier transform of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] in Eqn.~((3-8)) and, since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is radially symmetric, the inverse transform may be written (Bracewell Bracewell65)

[math]\displaystyle{ w_{=P= }\left( r\right) =\frac{1}{2\pi}\int_{0}^{\infty}\hat{w}\left( \gamma\right) \gamma J_{0}\left( \gamma r\right) d\gamma (3-11) }[/math]

where [math]\displaystyle{ J_{0} }[/math] is Bessel function and [math]\displaystyle{ r }[/math] is the distance from the point of forcing. The response to line forcing is given by the inverse Fourier transform of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] in the [math]\displaystyle{ x }[/math]-axis and since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is an even function, this is

[math]\displaystyle{ w_{=L= }\left( r\right) =\frac{1}{\pi}\int_{0}^{\infty}\hat{w}\left( \gamma\right) \cos\left( \gamma r\right) d\gamma (3-12) }[/math]

where again [math]\displaystyle{ r=\left| x\right| }[/math] is the distance from the line of forcing. Note that the factors [math]\displaystyle{ 1/2\pi }[/math] and [math]\displaystyle{ 1/\pi }[/math] result from the form of the Fourier transform in two and one dimensional spaces that we defined in Eqn.~((3-22)) and Eqn.~((3-36)).

The integrals of Eqn.~((3-11)) and Eqn.~((3-12)) are calculated using the singularities of [math]\displaystyle{ \hat{w} }[/math], i.e., the roots of the dispersion equation. First, since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is an even fractional function that equals zero when [math]\displaystyle{ \gamma=0 }[/math] and is bounded in the whole plane except in regions around its poles, we find that [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] can be expressed using the Mittag-Leffler expansion in section 3.2. Pairing each pole [math]\displaystyle{ q }[/math] with its negative counterpart [math]\displaystyle{ -q }[/math], gives

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{q\in K^{}}\frac{2qR\left( q\right) }{\gamma^{2}-q^{2}} (3-50) }[/math]

where [math]\displaystyle{ R\left( q\right) }[/math] is the residue of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] at [math]\displaystyle{ \gamma=q }[/math]. We denoted the set of roots of the dispersion equation with positive imaginary part and a positive real root by [math]\displaystyle{ K^{} }[/math]. Thus, this set is [math]\displaystyle{ K^{}=\left\{ q_{=T= },q_{=D= },-q_{=D= }^{},\mathrm{i}q_{1},\mathrm{i}q_{2},\mathrm{i} q_{3},\cdots\right\} }[/math]. Note that the rest of the roots of the dispersion equation are the negative of the values in [math]\displaystyle{ K^{} }[/math]. By substituting this expansion into the integrals in Eqn.~((3-11)) and Eqn.~((3-12)), we are able to perform the integration and write each result as a summation over the roots [math]\displaystyle{ q\in K^{} }[/math].

\begin{figure}[tbh] \begin{center} \includegraphics[height=5.0918cm,width=7.5037cm]{polyres.eps} \caption{Graphs of [math]\displaystyle{ \left| R\left( \mathrm{i}q\right) \right| }[/math] as function of real number [math]\displaystyle{ q }[/math], with [math]\displaystyle{ \tanh }[/math]-functions (solid line) and polynomial expression (dashed line).}

(fig:3-31)

\end{center} \end{figure}

The residues [math]\displaystyle{ R\left( q\right) }[/math] can be calculated using the usual formula. Since each of the poles of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is simple, the residue [math]\displaystyle{ R\left( q\right) }[/math] at a pole [math]\displaystyle{ q }[/math] can be found using the expression

[math]\displaystyle{ \begin{matrix} R\left( q\right) & =\left[ \left. \frac{d}{d\gamma}d\left( \gamma ,\omega\right) \right| _{\gamma=q}\right] ^{-1}\\ & =\left[ 4q^{3}+\omega^{2}\left( \frac{qH+\tanh qH-qH\tanh^{2}qH} {q^{2}\tanh^{2}qH}\right) \right] ^{-1}. (3-16) \end{matrix} }[/math]

As each pole [math]\displaystyle{ q }[/math] is a root of the dispersion equation, we may substitute [math]\displaystyle{ \tanh qH=\omega^{2}/\left( q^{5}+uq\right) }[/math], where for brevity we have defined [math]\displaystyle{ u=\left( 1-m\omega^{2}\right) }[/math]. The residue may then be given as the rational function of the pole

[math]\displaystyle{ R\left( q\right) =\frac{\omega^{2}q}{\omega^{2}\left( 5q^{4}+u\right) +H\left[ \left( q^{5}+uq\right) ^{2}-\omega^{4}\right] }. (3-17) }[/math]

This form avoids calculation of the hyperbolic tangent which becomes small at the imaginary roots. This causes numerical round-off problems and rapid variation in computing Eqn.~((3-16)) since [math]\displaystyle{ q_{n}H }[/math] tends to [math]\displaystyle{ n\pi }[/math] as [math]\displaystyle{ n }[/math] becomes large, which makes [math]\displaystyle{ \tan q_{n}H }[/math] become small. Fig.~((fig:3-31)) shows the graphs of the two expressions, Eqn.~((3-16)) and Eqn.~((3-17)) of the residue for imaginary argument. The imaginary roots [math]\displaystyle{ \mathrm{i}q_{1},\mathrm{i}q_{2},\mathrm{i}q_{3},... }[/math] are located where the two curves in Fig.~((fig:3-31)) coincide (the spiky parts of the solid curve). There are two such points at a spike and the root is the one on the left. Eqn.~((3-16)), from the direct calculation, is a rapidly varying function near the roots [math]\displaystyle{ \left\{ \mathrm{i} q_{n}\right\} _{n=1}^{\infty} }[/math], hence a small numerical error in the values of the roots will result in a large error in the residue. In contrast, Eqn.~((3-17)) gives us a smooth function, and the resulting calculation of the residue is stable.

Using the identities (Abramowitz and Stegun Stegun formula 11.4.44 with [math]\displaystyle{ \nu=0 }[/math], [math]\displaystyle{ \mu=0 }[/math], [math]\displaystyle{ z=-\mathrm{i}q }[/math] and [math]\displaystyle{ a=r }[/math])

[math]\displaystyle{ \int_{0}^{\infty}\frac{\gamma}{\gamma^{2}-q^{2}}J_{0}\left( \gamma r\right) d\gamma=K_{0}\left( -\mathrm{i}qr\right) (3-24) }[/math]

for [math]\displaystyle{ Im q\gt 0 }[/math], [math]\displaystyle{ r\gt 0 }[/math], where [math]\displaystyle{ K_{0} }[/math] is a modified Bessel function and an identity between the modified Bessel function, we notice that [math]\displaystyle{ q_{=T= } }[/math] term of Eqn.~((3-50)) may pose a problem. However, considering that [math]\displaystyle{ q_{=T= }=\lim_{\varepsilon\searrow0}\left( q_{=T= }+\mathrm{i}\varepsilon\right) }[/math], we are able to apply Eqn.~((3-24)) to all terms of Eqn.~((3-50)). Alternatively, we may put an additional imaginary term [math]\displaystyle{ \mathrm{i}\beta\omega }[/math], [math]\displaystyle{ \beta\gt 0 }[/math], representing damping, to the dispersion equation

[math]\displaystyle{ \gamma^{4}+\mathrm{i}\beta\omega+1-m\omega^{2}-\frac{\omega^{2} }{\gamma\tanh\gamma H}. }[/math]

Every element in [math]\displaystyle{ K^{} }[/math] has positive imaginary part. Addition of the damping term proportional to [math]\displaystyle{ w_{t} }[/math] ensures that the inverse transform satisfies the radiation condition. We omit the damping term here for the sake of algebraic simplicity.

We can calculate the integral of the inverse Fourier transform of [math]\displaystyle{ \hat{w} }[/math] and, using the identity between [math]\displaystyle{ K_{0} }[/math] and Hankel function of the first kind (Abramowitz and Stegun Stegun formula 9.6.4), [math]\displaystyle{ K_{0}\left( \zeta\right) =\frac{\mathrm{i}\pi}{2}H_{0}^{\left( 1\right) }\left( \mathrm{i}\zeta\right) }[/math] for [math]\displaystyle{ Re \zeta\geq0 }[/math], the displacement for point forcing may be written in the equivalent forms

[math]\displaystyle{ \begin{matrix} w_{=P= }\left( r\right) & =\frac{1}{\pi}\sum_{q\in K^{} }qR\left( q\right) K_{0}\left( -\mathrm{i}qr\right) (3-13)\\ & =\frac{\mathrm{i}}{2}\sum_{q\in K^{}}qR\left( q\right) H_{0}^{\left( 1\right) }\left( qr\right) (3-14) \end{matrix} }[/math]

where the subscript P of [math]\displaystyle{ w_{=P= } }[/math] indicates the response to a point load.

The identity (Erdelyi bateman formula 1.2 (11) with [math]\displaystyle{ x=\gamma }[/math], [math]\displaystyle{ y=r }[/math], and [math]\displaystyle{ a=-\mathrm{i}q }[/math])

[math]\displaystyle{ \int_{0}^{\infty}\frac{\cos\left( \gamma r\right) }{\gamma^{2}-q^{2}} d\gamma=-\frac{\pi}{\mathrm{i}2q}\exp\left( \mathrm{i}qr\right) (3-25) }[/math]

for [math]\displaystyle{ Im q\gt 0 }[/math], [math]\displaystyle{ r\gt 0 }[/math]. The above integration gives the surface displacement for line forcing

[math]\displaystyle{ w_{=L= }\left( r\right) =\mathrm{i}\sum_{q\in K^{} }R\left( q\right) \exp\left( \mathrm{i}qr\right) (3-15) }[/math]

where the subscript L of [math]\displaystyle{ w_{=L= } }[/math] indicates the response to a line load.

Modal expansion of the solutions

We have shown that the fundamental solution for finite water depth [math]\displaystyle{ H }[/math] is found by first finding the roots in the upper-half plane, [math]\displaystyle{ K^{} }[/math], of the dispersion Eqn.~((3-9)). Numerical calculation is achieved by truncating the sum after some finite number of roots. Straightforward computer code (in MatLab) to find the roots has been given by Fox and Chung report. After calculating the residue for each root given in Eqn.~((3-17)), the sum in Eqn.~((3-14)) can be calculated by separating it into

[math]\displaystyle{ w_{=P= }\left( r\right) =\frac{\mathrm{i}}{2}q_{=T= }R_{=T= }H_{0}^{\left( 1\right) }\left( q_{=T= }r\right) -Im \left[ q_{=D= }R_{=D= }H_{0}^{\left( 1\right) }\left( q_{=D= }r\right) \right] +\frac{1}{\pi}\sum_{n=1}^{\infty }\mathrm{i}q_{n}R_{n}K_{0}\left( q_{n}r\right) (3-18) }[/math]

We used the identities [math]\displaystyle{ -\mathrm{i}\left( -q_{=D= }\right) =\left( -\mathrm{i}q_{=D= }\right) ^{}\lt math\gt , }[/math]R\left( q^{ }\right) =\left( R\left( q\right) \right) ^{}[math]\displaystyle{ and }[/math]H_{0}^{\left( 1\right) }\left( -\zeta^{}\right) =-\left( H_{0}^{\left( 1\right) }\left( \zeta\right) \right) ^{}</math>, and denote the residues for the poles in [math]\displaystyle{ K^{} }[/math] by [math]\displaystyle{ R_{=T= }=R\left( q_{=T= }\right) }[/math], [math]\displaystyle{ R_{=D= }=R\left( q_{=D= }\right) }[/math], and [math]\displaystyle{ R_{n}=R\left( \mathrm{i}q_{n}\right) \lt math\gt , }[/math]n\in\mathbf{N}</math>, respectively. Note that [math]\displaystyle{ R\left( -\gamma_{=D= }^{}\right) \rightarrow-R_{=D= }^{} }[/math] as [math]\displaystyle{ \beta\rightarrow0 }[/math]. Since, from Eqn.~((3-17)), [math]\displaystyle{ \left| qR\left( q\right) \right| \propto\left| q\right| ^{-8}\lt math\gt as }[/math]\left| q\right| </math> increases, the sum may be terminated after a relatively small number of terms without significant error, as shown in Fig.~((fig:rsd)). We also notice that the smaller the depth [math]\displaystyle{ H }[/math], the fewer number of roots we need to achieve the desired accuracy, since for a given [math]\displaystyle{ n }[/math], [math]\displaystyle{ q_{n} }[/math] increases as [math]\displaystyle{ H }[/math] becomes small.

\begin{figure}[tbh] \begin{center} \includegraphics[height=7.15cm,width=7.3126cm]{rsd.eps} \caption{Log-log plot of the residues corresponding to the evanescent modes. +, * and [math]\displaystyle{ \diamondsuit }[/math] indicate for [math]\displaystyle{ \omega=0.1,1.0,10 }[/math] respectively. The water depth is [math]\displaystyle{ H=20\pi }[/math] (deep water).}

(fig:rsd)

\end{center} \end{figure}

The surface displacement for line forcing is given by the sum

[math]\displaystyle{ w_{=L= }\left( r\right) =\mathrm{i}R_{=T= }\exp\left( \mathrm{i}q_{=T= }r\right) -2Im \left[ R_{=D= }\exp\left( \mathrm{i}q_{=D= }r\right) \right] +\mathrm{i}\sum_{n=1}^{\infty}R_{n}\exp\left( -q_{n}r\right) . (3-19) }[/math]

We have written the solutions ((3-18)) and ((3-19)) in terms of the travelling, damped travelling and evanescent mode in order to emphasize the behavior of each mode.

We consider here the energy flux due to the wave produced by the force at [math]\displaystyle{ r=0 }[/math]. From the equation of motion ((2-41)) and [math]\displaystyle{ \mathbf{v}=\nabla\phi }[/math], the energy crossing a surface [math]\displaystyle{ S }[/math] in a time period [math]\displaystyle{ \left[ t,t+T\right] }[/math] is

[math]\displaystyle{ \rho\int_{t}^{t+T}\int_{S}\phi_{t}\left( x,y,z,t\right) \phi_{n}\left( x,y,z,t\right) d\sigma dt. }[/math]

For time-harmonic waves, i.e., [math]\displaystyle{ \phi=Re \left[ \phi\exp\left( i\omega t\right) \right] \lt math\gt , and }[/math]T=2\pi/\omega</math>, the above equation becomes

[math]\displaystyle{ -i2\rho\pi\int_{S}\left[ \phi^{}\phi_{n}-\phi\phi_{n}^{}\right] d\sigma=2\rho\piIm \int_{S}\phi^{}\phi_{n}d\sigma . (3-52) }[/math]

We notice that if [math]\displaystyle{ S }[/math] is the closed surface of domain [math]\displaystyle{ \mathcal{V} }[/math], then from Green's theorem, we have

[math]\displaystyle{ \int_{S}\left[ \phi^{}\phi_{n}-\phi\phi_{n}^{}\right] d\sigma =\int_{\mathcal{V}}\left[ \phi^{}\nabla^{2}\phi-\phi\nabla^{2}\phi^{ }\right] dV=0, (3-82) }[/math]

which states that there is no net energy propagation to infinity, i.e., the law of energy conservation. Hence, only the travelling mode [math]\displaystyle{ H_{0}^{\left( 1\right) }\left( q_{=T= }r\right) \lt math\gt for the point load and }[/math]\exp\left( iq_{=T= }r\right) </math> for the line load carry the energy to infinity since the rest of the solution decays exponentially. Note that the decaying rate of the Hankel function of a real variable is [math]\displaystyle{ 1/\sqrt{r} }[/math] and the integral on the cylindrical surface gives [math]\displaystyle{ d\sigma=rdrd\theta dz }[/math], thus we have non-zero energy flux for the travelling mode of the point load response.

The evanescent modes in Eqn.~((3-19)) and ((3-18)) are always real since [math]\displaystyle{ \mathrm{i}R_{n} }[/math] in Eqn.~((3-18)) and Eqn.~((3-19)) is real for each [math]\displaystyle{ n=1,2,\cdots }[/math]. Hence, the only imaginary term in the responses due to point or line load that give non-zero in Eqn.~((3-52)) is the coefficient of the travelling wave. This corresponds to the travelling waves being the only modes that carry energy away from the load. The damped-travelling and evanescent modes contribute motion that is in phase with the forcing, while the travelling mode has a component in quadrature to the forcing.

The velocity potential in the water can be found using Eqn.~((3-5)) to Eqn.~((3-6)), giving

[math]\displaystyle{ \phi_{=P= }\left( r,z\right) =-\frac{\omega}{2}\sum_{q\in K^{ }}\frac{R\left( q\right) }{\sinh qH}H_{0}^{\left( 1\right) }\left( qr\right) \cosh q\left( z+H\right) }[/math]

for the point load and

[math]\displaystyle{ \phi_{=L= }\left( r,z\right) =-\omega\sum_{q\in K^{}} \frac{R\left( q\right) }{q\sinh qH}\exp\left( \mathrm{i}qr\right) \cosh q\left( z+H\right) . }[/math]

for the line load.

Note that the zeros of [math]\displaystyle{ d\left( \gamma\right) \gamma\tanh\gamma H }[/math]\ are the same as those of the dispersion equation. Thus the singularities of [math]\displaystyle{ \hat {\phi}\lt math\gt and }[/math]\hat{w}[math]\displaystyle{ are the same. The term }[/math]\sinh qH</math> is close to zero for most imaginary roots so these expressions are not directly suitable for numerical computation. The substitution following Eqn.~((3-16)) may be used to give computationally stable expressions.

Each term of [math]\displaystyle{ w_{=P= } }[/math] and [math]\displaystyle{ w_{=L= } }[/math] in Eqn.~((3-18)) and Eqn.~((3-19)) is a natural mode of a floating ice sheet whose wavenumber is a root of the dispersion equation. We showed that the mode of [math]\displaystyle{ q_{=T= } }[/math], the travelling mode carries the wave energy outwards, and the modes of [math]\displaystyle{ q_{=D= } }[/math] and [math]\displaystyle{ \mathrm{i}q_{n} }[/math], [math]\displaystyle{ n=1,2,... }[/math], the damped travelling and evanescent modes, are exponentially decaying.

====Summary of the analytic structure of [math]\displaystyle{ \hat{w====\left( \gamma\right) }[/math]}

Here we summarize and clarify the analytic properties of the complex valued function [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] that have been discussed so far.

It may be worth reminding ourselves that the Mittag-Leffler expansion of [math]\displaystyle{ \hat{w} }[/math] given by Eqn.~((ap-20)) has an extra term [math]\displaystyle{ \hat{w}\left( 0\right) =0 }[/math] and is unique. Hence, even though a function such as [math]\displaystyle{ \hat{w}\left( \gamma\right) +1 }[/math] has the same singularities and residues, the resulting series expansion will be different. This uniqueness of the expansion guaranties the uniqueness of the solution derived using the inverse Fourier transform of the series expansion form.

We consider the question of whether we can reconstruct the Fourier transform [math]\displaystyle{ \hat{w} }[/math] from the modified formula of residues given by Eqn.~((3-17)) and the positions of the roots given by the dispersion equation [math]\displaystyle{ d\left( \gamma\right) =0 }[/math]. In addition to the residues and poles, we require that [math]\displaystyle{ \hat{w}\left( 0\right) =0 }[/math], to uniquely reconstruct [math]\displaystyle{ \hat{w} }[/math]. The answer to the question is yes and no depending on how the formulae are used. It follows that the function [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] for [math]\displaystyle{ \gamma\in C }[/math] can be reconstructed from [math]\displaystyle{ q\in K^{} }[/math] and [math]\displaystyle{ R\left( q\right) }[/math] (either Eqn.~((3-16)) or Eqn.~((3-17))) as

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{q\in K^{}}\frac{2qR\left( q\right) }{\gamma^{2}-q^{2}}. }[/math]

However, we find the following

[math]\displaystyle{ \hat{w}\left( \gamma\right) \neq R\left( \gamma\right) \sum_{q\in K^{}}\frac{2q}{\gamma^{2}-q^{2}}. (3-80) }[/math]

We note that near [math]\displaystyle{ \gamma=q\in K^{} }[/math] the left and the right hand sides share the same analytic property, i.e.,

[math]\displaystyle{ \int_{C_{\varepsilon}\left( q\right) }\hat{w}\left( \gamma\right) d\gamma=\int_{C_{\varepsilon}\left( q\right) }R\left( \gamma\right) \sum_{q\in K^{}}\frac{2q}{\gamma^{2}-q^{2}}\,d\gamma }[/math]

where [math]\displaystyle{ C_{\varepsilon}\left( q\right) }[/math] is a circular contour of small radius [math]\displaystyle{ \varepsilon }[/math] around a pole [math]\displaystyle{ q }[/math]. The reason for ((3-80)) is because the function [math]\displaystyle{ R\left( \gamma\right) }[/math] defined by Eqn.~((3-16)) and Eqn.~((3-17)) introduce zeros and singularities of their own, which change the analytic properties of the right hand side of Eqn.~((3-80)).

It may seem trivial that the formulae of the residues Eqn.~((3-16)) and Eqn.~((3-17)) are valid only at the pole [math]\displaystyle{ \gamma=q }[/math], nevertheless we have confirmed Eqn.~((3-80)).