Introduction
We present here the solution of Evans and Porter 2005 for the simple
case of a single crack with waves incident from normal (they also considered multiple cracks
and waves incident from different angles).
The solution of Evans and Porter 2005 expresses the potential
[math]\displaystyle{ \phi }[/math] in terms of a linear combination of the incident wave and certain source functions located at the crack.
Along with satisfying the field and boundary conditions, these source functions satisfy the jump conditions in the displacements and gradients across the crack.
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 }[/math] with z = 0, we obtain
[math]\displaystyle{
0 =
\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]
We then substitute [math]\displaystyle{ G = \phi }[/math] to 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 } = 0
}[/math]
plus the vertical integrals at the ends (which will give a contribution). [math]\displaystyle{ \phi_n^{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] so that
[math]\displaystyle{
\int_{-\infty}^\infty(\partial_{x^\prime}^4\phi_n)G_n dx^\prime = \int_{-\infty}^0(\partial_{x^\prime}^4\phi_n)G_n dx^\prime + \int_0^\infty(\partial_{x^\prime}^4\phi_n)G_n dx^\prime
}[/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[\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)
=0
}[/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)
= \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.
We are missing the incident wave which comes into the equations through the boundary conditions at infinity.
The correct equation is
[math]\displaystyle{
\phi_{n}\left( x\right)
= \phi_{n}^{\mathrm{In}}\left( x\right) +
\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]
which can be 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 velocity potential
[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]
Consequently, the source functions for a single crack at [math]\displaystyle{ x=0 }[/math] can be defined as
[math]\displaystyle{
\psi_s(x,z)= \beta\chi_{xx}(x,z),\,\,\,
\psi_a(x,z)= \beta\chi_{xxx}(x,z),\,\,\,(2)
}[/math]
It can easily be shown that [math]\displaystyle{ \psi_s }[/math] is symmetric about [math]\displaystyle{ x = 0 }[/math] and
[math]\displaystyle{ \psi_a }[/math] is antisymmetric about [math]\displaystyle{ x = 0 }[/math].
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.
