Eigenfunction Matching for a Semi-Infinite Floating Elastic Plate

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Contents

Introduction

We show here a solution for a semi-infinite Floating Elastic Plate on Finite Depth. The problem was solved by Fox and Squire 1994 but the solution method here is slightly different. The simpler theory for a Dock describes many of the ideas here in more detail.

Wave scattering by a submerged semi-infinite elastic plate

Equations

We consider the problem of small-amplitude waves which are incident on a semi-infinite floating elastic plate occupying water surface for x > 0. The submergence of the plate is considered negligible. We assume that the problem is invariant in the y direction, although we allow the waves to be incident from an angle. We also assume that the plate edges are free to move at each boundary, although other boundary conditions could easily be considered using the methods of solution presented here. We begin with the Frequency Domain Problem for a semi-infinite Floating Elastic Plates in the non-dimensional form of Tayler 1986 (Dispersion Relation for a Floating Elastic Plate). We also assume that the waves are normally incident (incidence at an angle will be discussed later).

\Delta \phi = 0, \;\;\; -h < z \leq 0,
\partial_z \phi = 0, \;\;\; z = - h,
\partial_z\phi=\alpha\phi, \,\, z=0,\,x<0,
\beta \partial_x^4\partial_z \phi - \left( \gamma\alpha - 1 \right) \partial_z \phi = \alpha\phi, \;\; z = 0, \;\;\; x \geq 0,

where α = ω2, β and γ are the stiffness and mass constant for the plate respectively. The free edge conditions at the edge of the plate imply

\partial_x^3 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,
\partial_x^2 \partial_z\phi = 0, \;\; z = 0, \;\;\; x = 0,

Method of solution

We use separation of variables in the two regions, x < 0 and x > 0.

We express the potential as

\phi(x,z) = X(x)Z(z)\,

and then Laplace's equation becomes

\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2

Separation of variables for a free surface

We use separation of variables

We express the potential as

\phi(x,z) = X(x)Z(z)\,

and then Laplace's equation becomes

\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2

The separation of variables equation for deriving free surface eigenfunctions is as follows:

Z^{\prime\prime} + k^2 Z =0.

subject to the boundary conditions

Z^{\prime}(-h) = 0

and

Z^{\prime}(0) = \alpha Z(0)

We can then use the boundary condition at z=-h \, to write

Z = \frac{\cos k(z+h)}{\cos kh}

where we have chosen the value of the coefficent so we have unit value at z = 0. The boundary condition at the free surface (z=0 \,) gives rise to:

k\tan\left( kh\right) =-\alpha \,

which is the Dispersion Relation for a Free Surface

We denote the negative imaginary solution of this equation by k_{0} \, and the positive real solutions by k_{m} \,, m\geq1. We define

\phi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

\int\nolimits_{-h}^{0}\phi_{m}(z)\phi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}

where

A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right).

Separation of variables under the Plate

Z^{\prime\prime} + \kappa^2 Z =0.

subject to the boundary conditions

Z^{\prime}(-h) = 0

and

\left(\beta \kappa^4 + 1 - \alpha\gamma\right)Z^{\prime}(0) = \alpha Z(0)

(the first term comes from the beam eigenvalue problem, where \partial_x^4 X = \kappa^4 X). We then use the boundary condition at z=-h \, to write

Z = \frac{\cos \kappa(z+h)}{\cos \kappa h}

The boundary condition at the free surface (z = 0) is the Dispersion Relation for a Floating Elastic Plate

\kappa \tan{(\kappa h)}= -\frac{\alpha}{\beta \kappa^{4} + 1 - \alpha\gamma}

Solving for \kappa \, gives a pure imaginary root with positive imaginary part, two complex roots (two complex conjugate paired roots with positive imaginary part in most physical situations), an infinite number of positive real roots which approach {n\pi}/{h} \, as n approaches infinity, and also the negative of all these roots (Dispersion Relation for a Floating Elastic Plate) . We denote the two complex roots with positive imaginary part by \kappa_{-2} \, and \kappa_{-1} \,, the purely imaginary root with positive imaginary part by \kappa_{0} \, and the real roots with positive imaginary part by \kappa_{n} \, for n a positive integer. The imaginary root with positive imaginary part corresponds to a reflected travelling mode propagating along the x axis. The complex roots with positive imaginary parts correspond to damped reflected travelling modes and the real roots correspond to reflected evanescent modes.

Inner product between free surface and elastic plate modes

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

where

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

Incident potential

The incident potential is a wave of amplitude A in displacement travelling in the positive x-direction. The incident potential can therefore be written as

\phi^{\mathrm{I}} =e^{-k_{0}x}\phi_{0}\left( z\right)

An infinite dimensional system of equations

The potential and its derivative must be continuous across the transition from open water to the plate covered region. Therefore, the potentials and their derivatives at x = 0 have to be equal. We also truncate the sum at N being careful that we have two extra modes on the plate covered region to satisfy the edge conditions. We obtain

\phi_{0}\left( z\right) + \sum_{m=0}^{N} a_{m} \phi_{m}\left( z\right) =\sum_{m=-2}^{N}b_{m}\psi_{m}(z)

and

-k_{0}\phi_{0}\left( z\right) +\sum _{m=0}^{N} a_{m}k_{m}\phi_{m}\left( z\right) =-\sum_{m=-2}^{N}b_{m}\kappa_{m}\psi _{m}(z)

for each N. We solve these equations by multiplying both equations by \phi_{l}(z)\, and integrating from − h to 0 to obtain:

A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{m=-2}^{N}b_{m}B_{ml},\,\,0 \leq l \leq N

and

-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l =-\sum_{m=-2}^{N}b_{m}\kappa_{m}B_{ml},\,\,0 \leq l \leq N

If we multiply the first equation by kl and subtract the second equation we obtain

(k_{0}+k_l)A_{0}\delta_{0l} =\sum_{m=-2}^{N}b_{m}(k_l + \kappa_{m})B_{ml}

Finally, we need to apply the conditions at the edge of the plate to give us two further equations,

\partial_x^2\partial_z\phi = \sum_{m=-2}^{N}b_{m} \kappa_m^3 \tan\kappa_m h = 0

and

\partial_x^3\partial_z\phi = - \sum_{m=-2}^{N}b_{m} \kappa_m^4 \tan\kappa_m h = 0

Numerical Solution

To solve the system of equations previously defined we set the upper limit of l to be N, as stated before. In terms of matrix, we obtain

\begin{bmatrix} \begin{bmatrix} A_0&0 \quad \cdots&0\\ 0&&\\ \vdots&A_l&\vdots\\ &&0\\ 0&\cdots \quad 0 &A_N \end{bmatrix} & \begin{bmatrix} -B_{-2,0}&-B_{-1,0}\\ &\\ \vdots&\vdots\\ &\\ -B_{-2,N}&-B_{-1,N} \end{bmatrix} & \begin{bmatrix} -B_{0,0}&\cdots&-B_{0,N}\\ &&\\ \vdots&-B_{m,l}&\vdots\\ &&\\ -B_{N,0}&\cdots&-B_{N,N} \end{bmatrix} \\ \\ \begin{bmatrix} 0&&\cdots&&0\\ 0&&\cdots&&0 \end{bmatrix} & \begin{bmatrix} \kappa_{-2}^3\tan\kappa_{-2}h&\kappa_{-1}^3\tan\kappa_{-1}h\\ \kappa_{-2}^4\tan\kappa_{-2}h&\kappa_{-1}^4\tan\kappa_{-1}h \end{bmatrix} & \begin{bmatrix} \kappa_0^3\tan\kappa_0h&\cdots&\kappa_N^3\tan\kappa_Nh\\ \kappa_0^4\tan\kappa_0h&\cdots&\kappa_N^4\tan\kappa_Nh \end{bmatrix} \\ \\ \begin{bmatrix} 0&&\cdots&&0\\ &&&&\\ \vdots&&\ddots&&\vdots\\ &&&&\\ 0&&\cdots&&0 \end{bmatrix} & \begin{bmatrix} (k_0+\kappa_{-2})B_{-2,0}&(k_0+\kappa_{-1})B_{-1,0}\\ &\\ \vdots&\vdots\\ &\\ (k_N+\kappa_{-2})B_{-2,N}&(k_N+\kappa_{-1})B_{-1,N} \end{bmatrix} & \begin{bmatrix} (k_0 + \kappa_0) \, B_{0,0}&\cdots&(k_N + \kappa_{0}) \, B_{0,N}\\ &&\\ \vdots&(k_l + \kappa_{m}) \, B_{m,l}&\vdots\\ &&\\ (k_0 + \kappa_N) \, B_{N,0}&\cdots&(k_N + \kappa_{N}) \, B_{N,N}\\ \end{bmatrix} \end{bmatrix} \begin{bmatrix} \begin{bmatrix} a_{0} \\ \\ \vdots \\ \\ a_N \end{bmatrix} \\ \\ \begin{bmatrix} b_{-2}\\ b_{-1} \end{bmatrix} \\ \\ \begin{bmatrix} b_{0}\\ \\ \vdots \\ \\ b_N \end{bmatrix} \end{bmatrix} = \begin{bmatrix} \begin{bmatrix} - A_{0} \\ 0 \\ \vdots \\ \\ 0 \end{bmatrix} \\ \\ \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \\ \begin{bmatrix} 2k_{0}A_{0} \\ 0 \\ \vdots \\ \\ 0 \end{bmatrix} \end{bmatrix}

We then simply need to solve the linear system of equations. Note that we can solve this equation for bn first and then solve for an

Waves Incident at an Angle

We can consider the case of Waves Incident at an Angle θ. When a wave in incident at an angle θ we have the wavenumber in the y direction is ky = sinθk0 where k0 is as defined previously (note that ky is imaginary).

This means that the potential is now of the form \phi(x,y,z)=e^{k_y y}\phi(x,z) so that when we separate variables we obtain

k^2 = k_x^2 + k_y^2

where k is the separation constant calculated without an incident angle.

It is shown that the potential can be expanded as

\phi(x,z)=e^{-\hat{k}_0x}\phi_0(z)+\sum_{m=0}^{\infty}a_{m}e^{\hat{k}_{m}x}\phi_{m}(z), \;\;x<0

and

\phi(x,z)=\sum_{m=-2}^{\infty}b_{m} e^{-\hat{\kappa}_{m}x}\psi_{m}(z), \;\;x>0

where \hat{k}_{m} = \sqrt{k_m^2 - k_y^2} and \hat{\kappa}_{m} = \sqrt{\kappa_m^2 - k_y^2} where we always take the positive real root or the root with positive imaginary part.

The edge conditions are also different and are

\left(\frac{\partial^3}{\partial x^3} + (2 - \nu)k^2_y\frac{\partial}{\partial x}\right) \frac{\partial\phi}{\partial z}= 0,

and

\left(\frac{\partial^2}{\partial x^2} + \nu k^2_y\right)\frac{\partial\phi}{\partial z} = 0,

where ν is Poisons ratio.

We can expend these edge conditions, which respectively gives

-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^3+\hat{\kappa}_m k_y^2(2-\nu))\tan \kappa_m h=0

and

-\sum_{m=-2}^{\infty}b_{m}\kappa_m (\hat{\kappa}_m^2+k_y^2\nu))\tan \kappa_m h=0

The equations are derived almost identically to those above and we obtain

A_{0}\delta_{0l}+a_{l}A_{l} =\sum_{m=-2}^{\infty}b_{m}B_{ml}

and

-\hat{k_{0}}A_{0}\delta_{0l}+a_{l}\hat{k}_{l}A_l =-\sum_{m=-2}^{\infty}b_{m}\hat{\kappa}_{m}B_{ml}

and these are solved exactly as before.

Energy Balance

We present a derivation of the energy balance here and also refer to the derivation Energy Balance for Two Elastic Plates

Based on the method used in Evans and Davies 1968, a check can be made to ensure the solutions energy balance. The energy balance equation is derived by applying Green's theorem to φ and its conjugate. The domain of integration is shown in the figure on the right. We assume that the angle is sufficiently small that we do not get total reflection.

A diagram depicting the area Ω which is bounded by the rectangle \partial\Omega. The rectangle \partial\Omega is bounded by -h\leq z \leq0 and -\infty\leq x \leq \infty

Applying Green's theorem to φ and its conjugate φ * gives

{ \iint_\Omega\left(\phi^*\nabla^2\phi - \phi\nabla^2\phi^* \right)\mathrm{d}x\mathrm{d}z = \int_{\partial\Omega}\left(\phi^*\frac{\partial\phi}{\partial n} - \phi\frac{\partial\phi^*}{\partial n} \right)\mathrm{d}l },

where n denotes the outward plane normal to the boundary and l denotes the plane parallel to the boundary. As φ and φ * satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to

\Im\int_\mathcal{S}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}l = 0,

The contributions from the vertical ends are

\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz = \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right) \left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz
= \frac{\hat{k}_0}{i} A_0 \left(1 - |a_0^2| \right)

and

\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz = \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\kappa}_0 \left(b_0^{*} e^{\hat{\kappa}_0 x} \right) \left(b_0 e^{-\hat{\kappa}_0 x}\right)\psi_0(z)^2 dz
= -\frac{\hat{\kappa}_0}{i} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz

The contribution from the surface integral under the plate is

\Im\int_{0}^{\infty}\phi^*\frac{\partial\phi}{\partial n} dx = \Im\int_{0}^{\infty} 1/\alpha \left(\beta \partial_x^4 \frac{\partial\phi^{*}}{\partial n} - (\gamma\alpha - 1 ) \frac{\partial\phi^{*}}{\partial n} \right) \frac{\partial\phi}{\partial n} dx
= \frac{2\beta}{\alpha} \frac{\hat{\kappa}_0^{3}}{i} |b_0^2| \partial_z\psi_0(0)^2

where we have integrated by parts and used the condition at the ends of the plate.

The energy balance is therefore

{\hat{k}_0} A_0 |a_0^2| + {\hat{\kappa}_0} |b_0^2| \int_{-h}^{0} \psi_0(z)^2 dz - \frac{2\beta}{\alpha}{ \hat{\kappa}_0^{3}} |b_0^2| \partial_z\psi_0(0)^2 = {\hat{k}_0} A_0


Note that this formula is only valid for angles \theta \in [-\theta_0, \theta_0], where θ0 is defined by \sin \theta_0 =\frac{\kappa_0}{k_0}. For large angles there is total reflection and then | R | 2 = 1

Matlab Code

A program to calculate the coefficients for the semi-infinite dock problems can be found here semiinfinite_plate.m

Additional code

This program requires

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