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 (a solution is given below) 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{ r=|x| }[/math], [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{ G\left( r,z\right) =-\frac{\omega}{2}\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{\sinh q_nH}H_{0}^{\left( 1\right) }\left( q_n r\right) \cosh q_n\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

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 1949 and John 1950 and discussed in Cylindrical Eigenfunction Expansion Each mode being a Hankel function of the first kind in the horizontal (in the three-dimensional case) and a hyperbolic function in the vertical. Kheisin 1967 chapter IV, studied the same problem solved in this chapter and derived the same dispersion equation for the physical variables. He studied the properties of the inverse transform only for the simple shallow water and static load cases, which are approximated versions of the full solutions which is presented 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_n. }[/math]

where [math]\displaystyle{ p_n }[/math] is the pressure and [math]\displaystyle{ n }[/math] is the dimension. We assume that [math]\displaystyle{ p_2 = \delta(x) }[/math] and that [math]\displaystyle{ p_3 = \delta(x) \delta(y) }[/math]. [math]\displaystyle{ w }[/math] is the displacement of the surface which is related to the velocity potential by the equation [math]\displaystyle{ \frac{\partial\phi}{\partial z} = \frac{\partial w}{\partial t} }[/math] We also have 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 of it) 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{ \pm q_{n},\,n\in\mathbb{N} }[/math], ([math]\displaystyle{ \mathrm{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{ \mathrm{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.

We calculate the displacement [math]\displaystyle{ w\left( x,y\right) }[/math] using the inverse Fourier transform of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math]. Since [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] is radially symmetric, the inverse transform may be written (Bracewell 1965)

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

for the three dimensional problem, where [math]\displaystyle{ J_{0} }[/math] is Bessel function and [math]\displaystyle{ r }[/math] is the distance from the point of forcing. For the two-dimensional problem 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\left( r\right) =\frac{1}{\pi}\int_{0}^{\infty}\hat{w}\left( \gamma\right) \cos\left( \gamma r\right) d\gamma }[/math]

where again [math]\displaystyle{ r=\left| x\right| }[/math]. 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.

The integrals are calculated using the Mittag-Leffler Expansion for the Floating Elastic Plate Dispersion Relation where we show that

[math]\displaystyle{ \hat{w}\left( \gamma\right) =\sum_{n=-2}^{\infty}\frac{2q_nR\left( q_n\right) }{\gamma^{2}-q_n^{2}} }[/math]

where [math]\displaystyle{ R\left( q_n\right) }[/math] is the residue of [math]\displaystyle{ \hat{w}\left( \gamma\right) }[/math] at [math]\displaystyle{ \gamma=q_n }[/math] given by

[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_n\right) ^{2}-\omega^{4}\right] }. }[/math]

.

Using the identities Abramowitz and Stegun 1964 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) }[/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 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 1964 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 three-dimensions may be written in the equivalent forms

[math]\displaystyle{ w\left( r\right) =\frac{1}{\pi}\sum_{n=-2}^{\infty} q_n R\left( q_n\right) K_{0}\left( -\mathrm{i}q_n r\right) }[/math] [math]\displaystyle{ =\frac{\mathrm{i}}{2}\sum_{n=-2}^{\infty}q_n R\left( q_n\right) H_{0}^{\left( 1\right) }\left( q_n r\right). }[/math]

In two-dimensions the identity (Erdelyi et. al.1954 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) }[/math]

for [math]\displaystyle{ Im q\gt 0 }[/math], [math]\displaystyle{ r\gt 0 }[/math] can be used. This gives us

[math]\displaystyle{ w\left( r\right) =\mathrm{i}\sum_{n=-2}^{\infty} R\left( q\right) \exp\left( \mathrm{i}qr\right). }[/math]

Solution for the Velocity Potential

The velocity potential in the water can be found in two-dimensions

[math]\displaystyle{ \phi\left( r,z\right) =-\frac{\omega}{2}\sum_{n=-2}^{\infty} \frac{R\left( q_n\right) }{\sinh q_nH}H_{0}^{\left( 1\right) }\left( q_nr\right) \cosh q_n\left( z+H\right) }[/math]

and for three-dimensions

[math]\displaystyle{ \phi\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]