# Introduction

We present here the solution for case of a single crack with waves incident from normal. The solution method is derived from Squire and Dixon 2001 and Evans and Porter 2005.

# Governing Equations

We consider the entire free surface to be occupied by a Floating Elastic Plate with a single discontinuity at $x=0$. The equations are the following

$\nabla^2 \phi = 0, -H\ltz\lt0,$
$\frac{\partial \phi}{\partial z} =0, z=-H,$
${\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial \phi}{\partial z} - \alpha \phi} = 0, z=0, x\neq 0.$

# Solution using the Free-Surface Green Function for a Floating Elastic Plate

We then use Green's second identity If φ and ψ are both twice continuously differentiable on U, then

$\int_U \left( \psi \nabla^2 \varphi - \varphi \nabla^2 \psi\right)\, dV = \oint_{\partial U} \left( \psi {\partial \varphi \over \partial n} - \varphi {\partial \psi \over \partial n}\right)\, dS$

If we then substitiute the Free-Surface Green Function for a Floating Elastic Plate which satisfies the following equations (plus the Sommerfeld Radiation Condition far from the body)

$\nabla^2 G = 0, -H\ltz\lt0,$
$\frac{\partial G}{\partial z} =0, z=-H,$
${\left( \beta \frac{\partial^4}{\partial x^4} - \gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G} = \delta(x), z=0,$

for $\psi = G$ with z = 0, we obtain

$0 = \lim_{x^{\prime}\to\infty}\int_{z^{\prime}=-H}^{0}$
$G_{n}\left( x,x^{\prime,z}\right) \phi \left( x ^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime }$
$\int_{-\infty}^{\infty}\left( G_{n}\left( x,x^{\prime }\right) \phi \left( x ^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime }$
$-\frac{1}{\alpha}\phi^{\rm In}_n = \int_{-\infty}^{\infty}\left( G_{n}\left( x,x^{\prime }\right) \phi \left( x ^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime }$

where the vertical integrals at the ends have given the contribution $\phi_{n}^{\mathrm{In}}$. We then substitute for $\phi$ and obtain

$\int_{-\infty}^{\infty}\left( G_{n}\left( x,x^{\prime }\right) \frac{1}{\alpha} \left( \beta \partial_{x^\prime}^4 - \gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right) -G\left( x,x^{\prime }\right) \phi _{n}\left( x^{\prime }\right) \right) dx^{\prime } = -\frac{1}{\alpha}\phi_{n}^{\mathrm{In}}$

We now integrate by parts remembering that $\phi_n$ is continuous everywhere except at $x^\prime = 0$ and obtain

$\int_{-\infty}^{\infty}\left\{ \frac{1}{\alpha} \left( \beta \partial_{x^\prime}^4 - \gamma\alpha + 1\right)G_{n}\left( x,x^{\prime }\right) - G\left( x,x^{\prime }\right)\right\} \phi_{n}\left( x^{\prime }\right) dx^\prime$
$+ \frac{\beta}{\alpha}\left(-\partial_{x^\prime}^3G_n(x,0)[\phi_n] + \partial_{x^\prime}^2 G_n(x,0)\partial_{x^\prime}[\phi_n] - \partial_{x^\prime} G_n(x,0)\partial_{x^\prime}^2[\phi_n] + G(x,0)\partial_{x^\prime}^3[\phi_n] \right) = -\frac{1}{\alpha}\phi_{n}^{\mathrm{In}}$

where [] denotes the jump in the function at $x^{\prime}=0$.

The integral can be simplified using the delta function property of the Green function to give us

$\phi_{n}\left( x\right) = \phi_{n}^{\mathrm{In}} + \beta \left(\partial_{x^\prime}^3 G_n [\phi_n] - \partial_{x^\prime}^2 G_n [\partial_{x^\prime}\phi_n] + \partial_{x^\prime} G_n [\partial_{x^\prime}^2\phi_n] - G_n [\partial_{x^\prime}^3\phi_n]\right)$

We can write the equation in terms of $\phi$ as was done by Porter and Evans 2005 but there is no real point because the boundary conditions are given in terms of $\phi_n$ since this represents the displacement. This equation is solved by applying the edge conditions at $x$ = 0 and z = 0

$\partial_x^2\phi_n=0,\,\,\, {\rm and}\,\,\,\, \partial_x^3\phi_n=0.$

# Expression for the source functions

The Free-Surface Green Function for a Floating Elastic Plate is given by

$G(x,x^{\prime},z) = -\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$

It follows that

$G_n(x,x^\prime)|_{z=0} = \sum_{n=-2}^\infty\frac{k_n\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$

We require the following functions

$\beta G_n(x,x^\prime)|_{x^\prime=0,z=0} = \beta\sum_{n=-2}^\infty\frac{k_n\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x|},$
$\beta\partial_{x^\prime} G_n |_{x^\prime=0,z=0}= \beta \sgn(x)\sum_{n=-2}^\infty\frac{k_n^2\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x|},$
$\beta\partial_{x^\prime}^2 G_n |_{x^\prime=0,z=0}= \beta \sum_{n=-2}^\infty\frac{k_n^3\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x|},$
$\beta\partial_{x^\prime}^3 G_n|_{x^\prime=0,z=0} = \beta \sgn(x)\sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x|},$

It can easily be shown that these functions are symmetric or antisymmetric about $x = 0$.

We therefore have the following equations

$\phi_n(x) = \phi_n^{In}+ \frac{\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[-sgn(x)k_n^4[\phi_n] + k_n^3[\partial_{x^\prime}\phi_n] - sgn(x)k_n^2[\partial_{x^\prime}^2\phi_n] -k_n[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x\phi_n(x) = \partial_x\phi_n^{In}+ \frac{\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[k_n^5[\phi_n] - sgn(x)k_n^4[\partial_{x^\prime}\phi_n] + k_n^3[\partial_{x^\prime}^2\phi_n] - sgn(x)k_n^2[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x^2\phi_n(x) = \partial_x^2\phi_n^{In}+ \frac{\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[-sgn(x)k_n^6[\phi_n] + k_n^5[\partial_{x^\prime}\phi_n] - sgn(x)k_n^4[\partial_{x^\prime}^2\phi_n] + k_n^3[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x^3\phi_n(x) = \partial_x^3\phi_n^{In}+ \frac{\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[k_n^7[\phi_n] - sgn(x)k_n^6[\partial_{x^\prime}\phi_n] + k_n^5[\partial_{x^\prime}^2\phi_n] - sgn(x)k_n^4[\partial_{x^\prime}^3\phi_n]\right]$

where

$\phi^{In} = e^{-ik_0x}\frac{\cos(k_0(z+H))}{\cos(k_0H)}$
$\phi_n^{In} = -k_0e^{-ik_0x}\tan(k_0H)$
$\partial_x\phi_n^{In} = ik_0^2e^{-ik_0x}\tan(k_0H)$
$\partial_x^2\phi_n^{In} = k_0^3e^{-ik_0x}\tan(k_0H)$
$\partial_x^3\phi_n^{In} = -ik_0^4e^{-ik_0x}\tan(k_0H)$

Mike, in the code, I actually use the -ve of these. Any thoughts?

# Formulation and solution

Alison - what is your definition of $\phi_{n}^{\mathrm{In}}$. I think this explains the difference. You can calculate it so there is unit incidence in $\phi$ or $\phi_n$. You have to be careful about what is R and T.

The equation for the normal derivitive of the surface potential is given by

$\phi_{n}\left( x\right) = \phi_{n}^{\mathrm{In}}\left( x\right) + \psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n] + \chi_a[\partial_{x^\prime}^2\phi_n] - \chi_s [\partial_{x^\prime}^3\phi_n]$

The jump coefficients $[\phi_n], [\partial_{x^\prime}\phi_n], [\partial_{x^\prime}^2\phi_n]$ and $[\partial_{x^\prime}^3\phi_n]$ can be found by applying the edge conditions

$\frac{\partial^2}{\partial x^2}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0,\,\,\, {\rm and}\,\,\,\, \frac{\partial^3}{\partial x^3}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0.$

which immediately imply that $[\partial_{x^\prime}^2\phi_n]$ and $[\partial_{x^\prime}^3\phi_n]$ are zero so that $\phi$ becomes

$\phi_{n}\left( x\right) = \phi_{n}^{\mathrm{In}}\left( x\right) + \psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]$

We are now left with two unknowns which can be solved using the same edge conditions which imply

$\partial_x^2\phi_n(x) = \partial_x^2\phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[-sgn(x)k_n^6[\phi_n] - k_n^5[\partial_{x^\prime}\phi_n] \right] = 0$

My code multiples by i i.e.

$\partial_x^2\phi_n(x) = \partial_x^2\phi_n^{In}+ \frac{\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{k_n|x|} \left[sgn(x)k_n^6[\phi_n] + k_n^5[\partial_{x^\prime}\phi_n] \right] = 0$

Any thoughts?

$\partial_x^3\phi_n(x) = \partial_x^3\phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[k_n^7[\phi_n] + sgn(x)k_n^6[\partial_{x^\prime}\phi_n] \right] = 0$

The jump conditions $[\phi_n]$ and $[\partial_{x^\prime}\phi_n]$ can be solved simultaneously.

The reflection and transmission coefficients, $R$ and $T$ can be found by taking the limit of $\phi_n$ as $x\rightarrow\pm\infty$ to obtain

$R = \frac{i\beta\sin^2(k_0h)}{2\alpha C(k_0)}\left[k_0^4[\phi_n] - k_0^3[\partial_{x^\prime}\phi_n]\right]$

My code uses

$R = -\frac{\beta\sin(k_0h)}{2\alpha k_0C(k_0)}\left[k_0^4[\phi_n] + k_0^3[\partial_x\phi_n]\right]$

and

$T= 1 - \frac{i\beta\sin^2(k_0h)}{2\alpha C(k_0)}\left[k_0^4[\phi_n] + k_0^3[\partial_{x^\prime}\phi_n]\right]$

but Even and Porter give:

$R = \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] + k_0^3[\partial_x\phi_n]\right]$

and

$T= 1 - \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] - k_0^3[\partial_x\phi_n]\right]$

Mike, can you explain where I may have gone wrong? One big difference I have made is take the limits of $\phi_n$ rather than $\phi$.

They are wrong here. They say P and Q are the jump but in fact one is the negative of the jump (my theory - you might like to check it)

# Other boundary conditions

More complicated boundary conditions can be treated using this formulation.

Previously, we considered plates with free edges ie with a zero bending moment and zero shear force at each edge. Here, we consider that each plate be connected by a series of flexural rotational springs (stiffness denoted by $S_r$) and vertical linear springs (stiffness denoted by $S_l$). The edge conditions become:

$\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right)$
$\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right)$
$\frac{\partial^2\phi_n^+(x)}{\partial x^2} = \frac{S_r}{\beta}\left( \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} \right)$
$\frac{\partial^2\phi_n^-(x)}{\partial x^2} = \frac{S_r}{\beta}\left( \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} \right)$

where x = 0. These edge conditions imply that $\frac{\partial^2\phi_n^+(x)}{\partial x^2} = \frac{\partial^2\phi_n^+(x)}{\partial x^2}$ and $\frac{\partial^3\phi_n^+(x)}{\partial x^3} = \frac{\partial^3\phi_n^-(x)}{\partial x^3}$ which implie $[\partial_x^2\phi_n] = 0$ and $[\partial_x^3\phi_n] = 0$. $\phi_n$ can now be re-expressed by

$\phi_{n}\left( x\right) = \phi_{n}^{\mathrm{In}}\left( x\right) + \psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]$

Now $\phi_n^+(x) - \phi_n^-(x)$ can be expressed as $[\phi]$ and $\frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x}$ can be expressed as $[\partial_x\phi_n]$ so that our edge conditions become

$[\phi_n] = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3} \right)$
$= -s_l \left[ \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left(k_n^7[\phi_n] +sgn(x) k_n^6[\partial_{x^\prime}\phi_n] \right) \right]$
$[\partial_x\phi_n] = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)$
$= s_r \left[ \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left(-sgn(x)k_n^6[\phi_n] - k_n^5[\partial_{x^\prime}\phi_n] \right)\right]$

where $s_l = \beta/S_l$ and $s_r = \beta/S_r$.

$[\phi_n]$ and $[\partial_x\phi_n]$ can be solved simultaneously.

# Evans and Porter

Substituting (1) into (2) gives

$\psi_s(x,z)= { -\frac{\beta}{\alpha} \sum_{n=-2}^\infty \frac{g_n\cos{(k_n(z+h))}}{2k_{xn}C_n}e^{k_n|x|} }, \psi_a(x,z)= { {\rm sgn}(x) i\frac{\beta}{\alpha}\sum_{n=-2}^\infty \frac{g_n'\cos{(k_n(z+h))}}{2k_{xn}C_n}e^{k_n|x|}},$

where

$g_n = ik_n^3 \sin{k_n h},\,\,\,\, g'_n= -k_n^4 \sin{k_n h}.$

We then express the solution to the problem as a linear combination of the incident wave and pairs of source functions at each crack,

$\phi(x,z) = e^{-k_0 x}\frac{\cos(k_0(z+h))}{\cos(k_0h)} + (P\psi_s(x,z) + Q\psi_a(x,z))\,\,\,(3)$

where $P$ and $Q$ are coefficients to be solved which represent the jump in the gradient and elevation respectively of the plates across the crack $x = a_j$. The coefficients $P$ and $Q$ are found by applying the edge conditions and to the $z$ derivative of $\phi$ at $z=0$,

$\frac{\partial^2}{\partial x^2}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0,\,\,\, {\rm and}\,\,\,\, \frac{\partial^3}{\partial x^3}\left. \frac{\partial \phi}{\partial z}\right|_{x=0,z=0}=0.$

The reflection and transmission coefficients, $R$ and $T$ can be found from (3) by taking the limits as $x\rightarrow\pm\infty$ to obtain

$R = {- \frac{\beta}{\alpha} (g'_0Q + ig_0P)}$

and

$T= 1 + {\frac{\beta}{\alpha}(g'_0Q - ig_0P)}$

# Other boundary conditions

More complicated boundary conditions can be treated using this formulation.

Previously, we considered plates with free edges ie with a zero bending moment and zero shear force at each edge. Here, we consider that each plate be connected by a series of flexural rotational springs (stiffness denoted by $S_r$) and vertical linear springs (stiffness denoted by $S_l$). The edge conditions become:

$\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right)$
$\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right)$
$\frac{\partial^2\phi_n^+(x)}{\partial x^2} = \frac{S_r}{\beta}\left( \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} \right)$
$\frac{\partial^2\phi_n^-(x)}{\partial x^2} = \frac{S_r}{\beta}\left( \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} \right)$

where x = 0.

Since $(\phi_n^+(x) - \phi_n^-(x)$ represents the jump in elevation i.e. Q and $\frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x}$ represents the jump in the gradient i.e. P, the above can be re expressed as

$Q = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3} \right)$
$P = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)$

where $s_l = \beta/S_l$ and $s_r = \beta/S_r$. Since $\partial_x^3 \psi_a$ is symmetric, the jump about x = 0 is zero i.e. Q = 0 and

$Q = - s_l \left( \partial_x^3 \phi_n^{In} + P \partial_x^3\psi_{sn} \right)$
$= - s_l \left( k_0k_{x0}^2\tan(k_0h) + P \frac{\beta}{\alpha}sgn(x)\sum_{n=-2}^\infty\frac{g_nk_nk_{xn}^2\sin(k_nh)}{2C_n} \right)$

Similarly, $\partial_x^2 \psi_s$ is symmetric and hence the jump about x = 0 is zero i.e. P = 0

$P = s_r \left( \partial_x^2 \phi_n^{In} + Q\partial_x^2\psi_{an} \right)$
$= s_r \left( -k_0k_{x0}^2\tan(k_0h) - Qsgn(x)i\frac{\beta}{\alpha} \sum\frac{g_n^'k_nk_{xn}\sin(k_nh)}{2C_n} \right)$

P and Q can be solved simultaneously.

Taking the limit values for $s_r$ and $s_l$, we can also model for a hinge-connector where $s_l=\infty$ and $s_r = 0$. ie

$\phi_n^-(x) = \phi_n^+(x)$
$\frac{\partial^3\phi_n^-(x)}{\partial x^3} = \frac{\partial^3\phi_n^+(x)}{\partial x^3}$
$\frac{\partial^2\phi_n^+(x)}{\partial x^2} = 0$
$\frac{\partial^2\phi_n^-(x)}{\partial x^2} = 0$

This implies that the jump in the elevation, Q, is zero. Hence, we only have one unknown to solve, P. We use the following boundary condition to solve for P:

$\frac{\partial^2\phi_n(x)}{\partial x^2} = 0$

which gives

$P = -\frac{\partial_x^2\phi_n^{In}(x)}{\partial_x^2\psi_{sn}(x)}$
$P = \frac{2\alpha k_{x0}k_0\tan (k_0h)}{\beta}\sum_{n=-2}^\infty\frac{C_n}{k_nk_{kn}g_n\sin(k_nh)}$

# Solution

$G(x,x^{\prime},z) = -i\sum_{n=-2}^\infty\frac{\sin{(k_n H)}\cos{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$
$G(x,x^\prime,z)_n = i\sum_{n=-2}^\infty\frac{k_n\sin{(k_n H)}\sin{(k_n(z+H))}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$
$G(x,x^\prime)_n|_{z=0} = i\sum_{n=-2}^\infty\frac{k_n\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$

$\partial_{x^\prime} G_n = sgn(x-x^\prime)i\sum_{n=-2}^\infty\frac{k_n^2\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$
$\partial_{x^\prime}^2 G_n = i\sum_{n=-2}^\infty\frac{k_n^3\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$
$\partial_{x^\prime}^3 G_n = sgn(x-x^\prime)i\sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x-x^{\prime}|},$
$\partial_{x^\prime}^3 G_n|_{x^\prime=0} = sgn(x)i\sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n|x|},$

$\phi_n(x) = \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[sgn(x)k_n^4[\phi_n] - k_n^3[\partial_{x^\prime}\phi_n] + sgn(x)k_n^2[\partial_{x^\prime}^2\phi_n] -k_n[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x\phi_n(x) = \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[-k_n^5[\phi_n] + sgn(x)k_n^4[\partial_{x^\prime}\phi_n] -k_n^3[\partial_{x^\prime}^2\phi_n] + sgn(x)k_n^2[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x^2\phi_n(x) = \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[sgn(x)k_n^6[\phi_n] - k_n^5[\partial_{x^\prime}\phi_n] + sgn(x)k_n^4[\partial_{x^\prime}^2\phi_n] - k_n^3[\partial_{x^\prime}^3\phi_n]\right]$
$\partial_x^3\phi_n(x) = \phi_n^{In}+ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}e^{-k_n|x|} \left[-k_n^7[\phi_n] + sgn(x)k_n^6[\partial_{x^\prime}\phi_n] - k_n^5[\partial_{x^\prime}^2\phi_n] + sgn(x)k_n^4[\partial_{x^\prime}^3\phi_n]\right]$

We now use the boundary conditions to solve for the jump conditions.

$\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[ k_n^4[\phi_n] + k_n^2[\partial_{x^\prime}^2\phi_n] \right] = 0$
$\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[ k_n^6[\partial_{x^\prime}\phi_n] + k_n^4[\partial_{x^\prime}^3\phi_n] \right]= 0$
$\phi_n^{In} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[-k_n^6[\phi_n] - k_n^5[\partial_{x^\prime}\phi_n] - k_n^4[\partial_{x^\prime}^2\phi_n] - k_n^3[\partial_{x^\prime}^3\phi_n]\right] = 0$
$\phi_n^{In} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left[k_n^6[\phi_n] - k_n^5[\partial_{x^\prime}\phi_n] + k_n^4[\partial_{x^\prime}^2\phi_n] - k_n^3[\partial_{x^\prime}^3\phi_n]\right] = 0$

Mike, I have assumed that sgn(x) for the left plate is -ve and visa versa. I'm not sure if this is correct.