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]\displaystyle{ x=0 }[/math].
The equations are the following
[math]\displaystyle{
\nabla^2 \phi = 0, -H\lt z\lt 0,
}[/math]
[math]\displaystyle{
\frac{\partial \phi}{\partial z} =0, z=-H,
}[/math]
[math]\displaystyle{
{\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]
We then use Green's second identity
If φ and ψ are both twice continuously differentiable on U, then
[math]\displaystyle{ \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]
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)
[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]
for [math]\displaystyle{ \psi = G }[/math] with z = 0, we obtain
[math]\displaystyle{
0 = \lim_{x^{\prime}\to\infty}\int_{z^{\prime}=-H}^{0}
}[/math]
[math]\displaystyle{
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]
[math]\displaystyle{
\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]
[math]\displaystyle{
-\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]
where the vertical integrals at the ends have given the contribution [math]\displaystyle{ \phi_{n}^{\mathrm{In}} }[/math].
We then substitute for [math]\displaystyle{ \phi }[/math] and obtain
[math]\displaystyle{
\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]
We now integrate by parts remembering that [math]\displaystyle{ \phi_n }[/math] is continuous everywhere except at [math]\displaystyle{ x^\prime = 0 }[/math]
and obtain
[math]\displaystyle{
\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]
[math]\displaystyle{
+ \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]
where [] denotes the jump in the function at [math]\displaystyle{ x^{\prime}=0 }[/math].
The integral can be simplified using the delta function property of the Green function
to give us
[math]\displaystyle{
\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]
We can write the equation in terms of [math]\displaystyle{ \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]\displaystyle{ \phi_n }[/math] since this represents
the displacement.
This equation is solved by applying the edge conditions at [math]\displaystyle{ x }[/math] = 0 and z = 0
[math]\displaystyle{
\partial_x^2\phi_n=0,\,\,\,
{\rm and}\,\,\,\,
\partial_x^3\phi_n=0.
}[/math]
Expression for the source functions
The Free-Surface Green Function for a Floating Elastic Plate is given by
[math]\displaystyle{
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]
It follows that
[math]\displaystyle{
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]
We require the following functions
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
It can easily be shown that these functions are symmetric or antisymmetric about [math]\displaystyle{ x = 0 }[/math].
We therefore have the following equations
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
where
[math]\displaystyle{
\phi^{In} = e^{-ik_0x}\frac{\cos(k_0(z+H))}{\cos(k_0H)}
}[/math]
[math]\displaystyle{
\phi_n^{In} = -k_0e^{-ik_0x}\tan(k_0H)
}[/math]
[math]\displaystyle{
\partial_x\phi_n^{In} = ik_0^2e^{-ik_0x}\tan(k_0H)
}[/math]
[math]\displaystyle{
\partial_x^2\phi_n^{In} = k_0^3e^{-ik_0x}\tan(k_0H)
}[/math]
[math]\displaystyle{
\partial_x^3\phi_n^{In} = -ik_0^4e^{-ik_0x}\tan(k_0H)
}[/math]
Mike, in the code, I actually use the -ve of these. Any thoughts?
Formulation and solution
Alison - what is your definition of [math]\displaystyle{ \phi_{n}^{\mathrm{In}} }[/math]. I think this explains the difference. You can calculate it so there is unit incidence in [math]\displaystyle{ \phi }[/math] or [math]\displaystyle{ \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
[math]\displaystyle{
\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]
The jump coefficients [math]\displaystyle{ [\phi_n], [\partial_{x^\prime}\phi_n], [\partial_{x^\prime}^2\phi_n] }[/math] and [math]\displaystyle{ [\partial_{x^\prime}^3\phi_n] }[/math] can be found by applying the edge conditions
[math]\displaystyle{
\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]
which immediately imply that [math]\displaystyle{ [\partial_{x^\prime}^2\phi_n] }[/math] and [math]\displaystyle{ [\partial_{x^\prime}^3\phi_n] }[/math] are zero so that [math]\displaystyle{ \phi }[/math] becomes
[math]\displaystyle{
\phi_{n}\left( x\right)
= \phi_{n}^{\mathrm{In}}\left( x\right) +
\psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]
}[/math]
We are now left with two unknowns which can be solved using the same edge conditions which imply
[math]\displaystyle{
\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]
My code multiples by i i.e.
[math]\displaystyle{
\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]
Any thoughts?
[math]\displaystyle{
\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]
The jump conditions [math]\displaystyle{ [\phi_n] }[/math] and [math]\displaystyle{ [\partial_{x^\prime}\phi_n] }[/math] can be solved simultaneously.
The reflection and transmission coefficients, [math]\displaystyle{ R }[/math] and [math]\displaystyle{ T }[/math] can be found by taking the limit of [math]\displaystyle{ \phi_n }[/math] as
[math]\displaystyle{ x\rightarrow\pm\infty }[/math] to obtain
[math]\displaystyle{
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]
My code uses
[math]\displaystyle{
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]
and
[math]\displaystyle{
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]
but Even and Porter give:
[math]\displaystyle{
R = \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] + k_0^3[\partial_x\phi_n]\right]
}[/math]
and
[math]\displaystyle{
T= 1 - \frac{\beta\sin(k_0h)}{\alpha}\left[k_0^4[\phi_n] - k_0^3[\partial_x\phi_n]\right]
}[/math]
Mike, can you explain where I may have gone wrong? One big difference I have made is take the limits of [math]\displaystyle{ \phi_n }[/math] rather than [math]\displaystyle{ \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]\displaystyle{ S_r }[/math]) and vertical linear springs (stiffness denoted by [math]\displaystyle{ S_l }[/math]).
The edge conditions become:
[math]\displaystyle{
\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right)
}[/math]
[math]\displaystyle{
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right)
}[/math]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
where x = 0.
These edge conditions imply that [math]\displaystyle{ \frac{\partial^2\phi_n^+(x)}{\partial x^2} = \frac{\partial^2\phi_n^+(x)}{\partial x^2} }[/math] and [math]\displaystyle{ \frac{\partial^3\phi_n^+(x)}{\partial x^3} = \frac{\partial^3\phi_n^-(x)}{\partial x^3} }[/math] which implie [math]\displaystyle{ [\partial_x^2\phi_n] = 0 }[/math] and [math]\displaystyle{ [\partial_x^3\phi_n] = 0 }[/math]. [math]\displaystyle{ \phi_n }[/math] can now be re-expressed by
[math]\displaystyle{
\phi_{n}\left( x\right)
= \phi_{n}^{\mathrm{In}}\left( x\right) +
\psi_a [\phi_n] - \psi_s [\partial_{x^\prime}\phi_n]
}[/math]
Now [math]\displaystyle{ \phi_n^+(x) - \phi_n^-(x) }[/math] can be expressed as [math]\displaystyle{ [\phi] }[/math] and [math]\displaystyle{ \frac{\partial\phi_n^+(x)}{\partial x} - \frac{\partial\phi_n^-(x)}{\partial x} }[/math] can be expressed as [math]\displaystyle{ [\partial_x\phi_n] }[/math] so that our edge conditions become
[math]\displaystyle{
[\phi_n] = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3} \right)
}[/math]
[math]\displaystyle{
= -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]
[math]\displaystyle{
[\partial_x\phi_n] = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)
}[/math]
[math]\displaystyle{
= 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]
where [math]\displaystyle{ s_l = \beta/S_l }[/math] and [math]\displaystyle{ s_r = \beta/S_r }[/math].
[math]\displaystyle{ [\phi_n] }[/math] and [math]\displaystyle{ [\partial_x\phi_n] }[/math] can be solved simultaneously.
Evans and Porter
Substituting (1) into (2) gives
[math]\displaystyle{
\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]
where
[math]\displaystyle{
g_n = ik_n^3 \sin{k_n h},\,\,\,\,
g'_n= -k_n^4 \sin{k_n h}.
}[/math]
We then express the solution to the problem as a linear combination of the
incident wave and pairs of source functions at each crack,
[math]\displaystyle{
\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]
where [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math] are coefficients to be solved which represent the jump in the gradient
and elevation respectively of the plates across the crack [math]\displaystyle{ x = a_j }[/math].
The coefficients [math]\displaystyle{ P }[/math] and [math]\displaystyle{ Q }[/math] are found by applying the edge conditions and to
the [math]\displaystyle{ z }[/math] derivative of [math]\displaystyle{ \phi }[/math] at [math]\displaystyle{ z=0 }[/math],
[math]\displaystyle{
\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]
The reflection and transmission coefficients, [math]\displaystyle{ R }[/math] and [math]\displaystyle{ T }[/math] can be found from (3)
by taking the limits as [math]\displaystyle{ x\rightarrow\pm\infty }[/math] to obtain
[math]\displaystyle{
R = {- \frac{\beta}{\alpha}
(g'_0Q + ig_0P)}
}[/math]
and
[math]\displaystyle{
T= 1 + {\frac{\beta}{\alpha}(g'_0Q - ig_0P)}
}[/math]
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]\displaystyle{ S_r }[/math]) and vertical linear springs (stiffness denoted by [math]\displaystyle{ S_l }[/math]).
The edge conditions become:
[math]\displaystyle{
\frac{\partial^3\phi_n^+(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( \phi_n^+(x) - \phi_n^-(x)\right)
}[/math]
[math]\displaystyle{
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = -\frac{S_l}{\beta}\left( (\phi_n^+(x) - \phi_n^-(x)\right)
}[/math]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
where x = 0.
Since [math]\displaystyle{ (\phi_n^+(x) - \phi_n^-(x) }[/math] represents the jump in elevation i.e. Q and [math]\displaystyle{ \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
[math]\displaystyle{
Q = -s_l \left( \frac{\partial^3\phi_n(x)}{\partial x^3} \right)
}[/math]
[math]\displaystyle{
P = s_r \left( \frac{\partial^2\phi_n(x)}{\partial x^2} \right)
}[/math]
where [math]\displaystyle{ s_l = \beta/S_l }[/math] and [math]\displaystyle{ s_r = \beta/S_r }[/math].
Since [math]\displaystyle{ \partial_x^3 \psi_a }[/math] is symmetric, the jump about x = 0 is zero i.e. Q = 0 and
[math]\displaystyle{
Q = - s_l \left( \partial_x^3 \phi_n^{In} + P \partial_x^3\psi_{sn} \right)
}[/math]
[math]\displaystyle{
= - 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]
Similarly, [math]\displaystyle{ \partial_x^2 \psi_s }[/math] is symmetric and hence the jump about x = 0 is zero i.e. P = 0
[math]\displaystyle{
P = s_r \left( \partial_x^2 \phi_n^{In} + Q\partial_x^2\psi_{an} \right)
}[/math]
[math]\displaystyle{
= 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]
P and Q can be solved simultaneously.
Taking the limit values for [math]\displaystyle{ s_r }[/math] and [math]\displaystyle{ s_l }[/math], we can also model for a hinge-connector where [math]\displaystyle{ s_l=\infty }[/math] and [math]\displaystyle{ s_r = 0 }[/math].
ie
[math]\displaystyle{
\phi_n^-(x) = \phi_n^+(x)
}[/math]
[math]\displaystyle{
\frac{\partial^3\phi_n^-(x)}{\partial x^3} = \frac{\partial^3\phi_n^+(x)}{\partial x^3}
}[/math]
[math]\displaystyle{
\frac{\partial^2\phi_n^+(x)}{\partial x^2} = 0
}[/math]
[math]\displaystyle{
\frac{\partial^2\phi_n^-(x)}{\partial x^2} = 0
}[/math]
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:
[math]\displaystyle{
\frac{\partial^2\phi_n(x)}{\partial x^2} = 0
}[/math]
which gives
[math]\displaystyle{
P = -\frac{\partial_x^2\phi_n^{In}(x)}{\partial_x^2\psi_{sn}(x)}
}[/math]
[math]\displaystyle{
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]
Solution
[math]\displaystyle{
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]
[math]\displaystyle{
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]
[math]\displaystyle{
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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
We now use the boundary conditions to solve for the jump conditions.
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
[math]\displaystyle{
\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]
Mike, I have assumed that sgn(x) for the left plate is -ve and visa versa. I'm not sure if this is correct.
