View source for Two Semi-Infinite Elastic Plates of Identical Properties

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= 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 <math>x=0</math>. 
The equations are the following
<center><math>
\nabla^2 \phi = 0, -H<z<0, 
</math></center>
<center><math>
\frac{\partial \phi}{\partial z} =0, z=-H,
</math></center>
<center><math>
{\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.
</math></center>

= Solution using the [[Free-Surface Green Function for a Floating Elastic Plate]] =

We then use [http://en.wikipedia.org/wiki/Green's_identities Green's second identity]
If &phi; and &psi; are both twice continuously differentiable on ''U'', then
<center><math> \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 
</math></center>
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)
<center><math>
\nabla^2 G = 0, -H<z<0, 
</math></center>
<center><math>
\frac{\partial G}{\partial z} =0, z=-H,
</math></center>
<center><math>
{\left( \beta \frac{\partial^4}{\partial x^4} - 
\gamma\alpha + 1\right)\frac{\partial G}{\partial z} - \alpha G}
 = \delta(x), z=0,
</math></center>
for <math>\psi = G</math> with z = 0, we obtain 
<center><math>
0 = \lim_{x^{\prime}\to\infty}\int_{z^{\prime}=-H}^{0}
</math></center>
<center><math>
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 }
</math></center>
<center><math>
\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 }
</math></center>
<center><math>
-\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 }
</math></center>
where the vertical integrals at the ends have given the contribution <math>\phi_{n}^{\mathrm{In}}</math>.
We then substitute for <math>\phi</math> and obtain
<center><math>
\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}}
</math></center>


We now integrate by parts remembering that <math>\phi_n</math> is continuous everywhere except at <math>x^\prime = 0</math>
and obtain
<center>
<math>
\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
</math></center>

<center><math> 
+ \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}}
</math></center>

where [] denotes the jump in the function at <math>x^{\prime}=0</math>.

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

<center><math>
\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)
</math></center>

We can write the equation in terms of <math>\phi</math> as was done by [[Porter and Evans 2005]] but there is 
no real point because the boundary conditions are given in terms of <math>\phi_n</math> since this represents
the displacement.
This equation is solved by applying the edge conditions at <math>x</math> = 0 and z = 0
<center>
<math>
\partial_x^2\phi_n=0,\,\,\,
{\rm and}\,\,\,\,
\partial_x^3\phi_n=0.
</math>
</center>

= Expression for the source functions = 

The [[Free-Surface Green Function for a Floating Elastic Plate]] is given by 
<center><math>
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}|},
</math></center>

It follows that 
<center><math>
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}|},
</math></center> 

We require the following functions 
<center><math>
\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|},
</math></center> 

<center><math>
\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|},
</math></center> 

<center><math>
\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|},
</math></center> 

<center><math>
\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|},
</math></center> 
It can easily be shown that these functions are symmetric or antisymmetric about <math>x = 0</math>.

We therefore have the following equations
<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>



where
<center><math>
\phi^{In} = e^{-ik_0x}\frac{\cos(k_0(z+H))}{\cos(k_0H)}
</math></center>
<center><math>
\phi_n^{In} = -k_0e^{-ik_0x}\tan(k_0H)
</math></center>
<center><math>
\partial_x\phi_n^{In} = ik_0^2e^{-ik_0x}\tan(k_0H)
</math></center>
<center><math>
\partial_x^2\phi_n^{In} = k_0^3e^{-ik_0x}\tan(k_0H)
</math></center>
<center><math>
\partial_x^3\phi_n^{In} = -ik_0^4e^{-ik_0x}\tan(k_0H)
</math></center>

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

= Formulation and solution= 

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

The equation for the normal derivitive of the surface potential is given by 
<center><math>
\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]
</math></center>

The jump coefficients <math>[\phi_n], [\partial_{x^\prime}\phi_n], [\partial_{x^\prime}^2\phi_n]</math> and <math>[\partial_{x^\prime}^3\phi_n]</math> can be found by applying the edge conditions
<center><math>
\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.
</math></center>
which immediately imply that <math>[\partial_{x^\prime}^2\phi_n]</math> and <math>[\partial_{x^\prime}^3\phi_n]</math> are zero so that <math>\phi</math> becomes 
<center><math>
\phi_{n}\left( x\right)
 = \phi_{n}^{\mathrm{In}}\left( x\right) +
\psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]
</math></center>

We are now left with two unknowns which can be solved using the same edge conditions which imply
<center><math>
\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
</math></center>
My code multiples by i i.e. 
<center><math>
\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
</math></center>
Any thoughts?

<center><math>
\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
</math></center>

The jump conditions <math>[\phi_n]</math> and <math>[\partial_{x^\prime}\phi_n]</math> can be solved simultaneously.
  
The reflection and transmission coefficients, <math>R</math> and <math>T</math> can be found by taking the limit of <math>\phi_n</math> as 
<math>x\rightarrow\pm\infty</math> to obtain
<center>
<math> 
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]
</math>
</center>
My code uses
<center>
<math> 
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]
</math>
</center>
and 
<center>
<math>
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]
</math>
</center>

but Even and Porter give:
<center>
<math> 
R = \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] + k_0^3[\partial_x\phi_n]\right]
</math>
</center>
and 
<center>
<math>
T= 1 - \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] - k_0^3[\partial_x\phi_n]\right]
</math>
</center>

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

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 <math>S_r</math>) and vertical linear springs (stiffness denoted by <math>S_l</math>). 
The edge conditions become:

<center><math>
\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right) 
</math></center>

<center><math>
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right) 
</math></center>

<center><math>
\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) 
</math></center>

<center><math>
\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)
</math></center>

where x = 0. 
These edge conditions imply that <math>\frac{\partial^2\phi_n^+(x)}{\partial x^2} = \frac{\partial^2\phi_n^+(x)}{\partial x^2} </math> and <math>\frac{\partial^3\phi_n^+(x)}{\partial x^3} = \frac{\partial^3\phi_n^-(x)}{\partial x^3} </math> which implie <math>[\partial_x^2\phi_n] = 0</math>  and <math>[\partial_x^3\phi_n] = 0 </math>. <math> \phi_n </math> can now be re-expressed by
<center><math>
\phi_{n}\left( x\right)
 = \phi_{n}^{\mathrm{In}}\left( x\right) +
\psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]
</math></center>

Now <math> \phi_n^+(x) - \phi_n^-(x)</math>  can be expressed as <math>[\phi]</math> and <math> \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} </math> can be expressed as <math>[\partial_x\phi_n]</math> so that our edge conditions become
<center><math>
[\phi_n] = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3}  \right)
</math></center>
<center><math>
= -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]
</math></center>

<center><math>
[\partial_x\phi_n] = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)
</math></center>
<center><math>
= 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]
</math></center>

where <math>s_l = \beta/S_l</math> and  <math>s_r = \beta/S_r</math>.

<math> [\phi_n] </math> and <math> [\partial_x\phi_n] </math> can be solved simultaneously. 

= Evans and Porter = 
Substituting (1) into (2) gives
<center>
<math>
\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|}},
</math>
</center>
where
<center>
<math>
g_n = ik_n^3 \sin{k_n h},\,\,\,\,
g'_n= -k_n^4 \sin{k_n h}.
</math>
</center>
We then express the solution to the problem as a linear combination of the 
incident wave and pairs of source functions at each crack,
<center>
<math>
\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)
</math>
</center>
where <math>P</math> and <math>Q</math> are coefficients to be solved which represent the jump in the gradient 
and elevation respectively of the plates across the crack <math>x = a_j</math>. 
The coefficients <math>P</math> and <math>Q</math> are found by applying the edge conditions  and to
the <math>z</math> derivative of <math>\phi</math> at <math>z=0</math>,
<center>
<math>
\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.
</math>
</center>
The reflection and transmission coefficients, <math>R</math> and <math>T</math> can be found from (3) 
by taking the limits as <math>x\rightarrow\pm\infty</math> to obtain
<center>
<math> 
R = {- \frac{\beta}{\alpha}
(g'_0Q + ig_0P)}
</math>
</center>
and 
<center>
<math>
T= 1 + {\frac{\beta}{\alpha}(g'_0Q - ig_0P)}
</math>
</center>

= 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 <math>S_r</math>) and vertical linear springs (stiffness denoted by <math>S_l</math>). 
The edge conditions become:

<center><math>
\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right) 
</math></center>

<center><math>
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right) 
</math></center>

<center><math>
\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) 
</math></center>

<center><math>
\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)
</math></center>

where x = 0. 

Since <math> (\phi_n^+(x) - \phi_n^-(x)</math> represents the jump in elevation i.e. Q and <math> \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} </math> represents the jump in the gradient i.e. P, the above can be re expressed as
<center><math>
Q = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3}  \right)
</math></center>

<center><math>
P = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)
</math></center>

where <math>s_l = \beta/S_l</math> and  <math>s_r = \beta/S_r</math>.
Since <math> \partial_x^3 \psi_a </math> is symmetric, the jump about x = 0 is zero i.e. Q = 0 and
<center><math>
Q = - s_l \left( \partial_x^3 \phi_n^{In} + P \partial_x^3\psi_{sn} \right)
</math></center>
<center><math>
= - 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)
</math></center>

Similarly, <math> \partial_x^2 \psi_s </math> is symmetric and hence the jump about x = 0 is zero i.e. P = 0 
<center><math>
P = s_r \left(  \partial_x^2 \phi_n^{In} + Q\partial_x^2\psi_{an} \right)
</math></center>
<center><math>
= 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)
</math></center>

P and Q can be solved simultaneously. 

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

<center><math>
\phi_n^-(x) = \phi_n^+(x) 
</math></center>

<center><math>
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = \frac{\partial^3\phi_n^+(x)}{\partial x^3}
</math></center>

<center><math>
\frac{\partial^2\phi_n^+(x)}{\partial x^2} = 0 
</math></center>

<center><math>
\frac{\partial^2\phi_n^-(x)}{\partial x^2}  = 0 
</math></center>

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:
<center><math>
\frac{\partial^2\phi_n(x)}{\partial x^2} = 0 
</math></center>

which gives
<center><math>
P = -\frac{\partial_x^2\phi_n^{In}(x)}{\partial_x^2\psi_{sn}(x)}
</math></center>
<center><math>
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)}
</math></center>

= Solution =

<center><math>
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}|},
</math></center>

<center><math>
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}|},
</math></center> 

<center><math>
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}|},
</math></center> 


<center><math>
\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}|},
</math></center> 

<center><math>
\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}|},
</math></center> 

<center><math>
\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}|},
</math></center> 

<center><math>
\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|},
</math></center> 


<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>

<center><math>
\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]
</math></center>


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

<center><math>
\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
</math></center>

<center><math>
\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
</math></center>

<center><math>
\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
</math></center>

<center><math>
\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
</math></center>

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

[[Category:Floating Elastic Plate]]