Two Semi-Infinite Elastic Plates of Identical Properties - Revision history

Meylan at 05:36, 10 May 2010

2010-05-10T05:36:02Z

← Older revision		Revision as of 05:36, 10 May 2010
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			{{incomplete pages}}

	= Introduction =		= Introduction =

Meylan: /* Solution using the Free-Surface Green Function for a Floating Elastic Plate */

2007-11-10T07:05:42Z

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

@@ Line 40: / Line 40: @@
 </math></center>
 for <math>\psi = G</math> with z = 0, we obtain
 <center><math>
 -\frac{1}{\alpha}\phi^{\rm In}_n =

Akohout: /* Solution using the Free-Surface Green Function for a Floating Elastic Plate */

2007-11-09T01:32:27Z

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

← Older revision		Revision as of 01:32, 9 November 2007
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	<center><math>		<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~~[\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{\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}}		= -\frac{1}{\alpha}\phi_{n}^{\mathrm{In}}
	</math></center>		</math></center>

Meylan: /* Expression for the source functions */

2007-11-08T08:25:42Z

Expression for the source functions

← Older revision		Revision as of 08:25, 8 November 2007
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	\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\|},		\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>		</math></center>
			It can easily be shown that these functions are symmetric or antisymmetric about <math>x = 0</math>.

	~~-----~~

	~~It can easily be shown that <math>\chi_s</math> and <math>\psi_s</math> are symmetric about <math>x = 0</math> and~~
	~~<math>\chi_a</math> and <math>\psi_a</math> are antisymmetric about <math>x = 0</math>.~~
	We therefore have the following equations		We therefore have the following equations
	<center><math>		<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\|}		\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]		\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]		-k_n[\partial_{x^\prime}^3\phi_n]\right]
	</math></center>		</math></center>

	<center><math>		<center><math>
	\partial_x\phi_n(x) = \partial_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\|}		\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]		\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]		- sgn(x)k_n^2[\partial_{x^\prime}^3\phi_n]\right]
	</math></center>		</math></center>

	<center><math>		<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\|}		\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]		\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]		+ k_n^3[\partial_{x^\prime}^3\phi_n]\right]
	</math></center>		</math></center>

	<center><math>		<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\|}		\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]		\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]		- sgn(x)k_n^4[\partial_{x^\prime}^3\phi_n]\right]
	</math></center>		</math></center>

Meylan: /* break */

2007-11-08T08:20:45Z

break

← Older revision		Revision as of 08:20, 8 November 2007
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	-----		-----

	~~=break=~~
	~~<b> We know that its wrong below here </b>~~

	~~----~~
	~~Well I'm not sure. Should we be taking <math>\partial_{x^\prime}\phi_n</math> or <math>\partial_x\phi_n</math> ?~~

	It can easily be shown that <math>\chi_s</math> and <math>\psi_s</math> are symmetric about <math>x = 0</math> and		It can easily be shown that <math>\chi_s</math> and <math>\psi_s</math> are symmetric about <math>x = 0</math> and
	<math>\chi_a</math> and <math>\psi_a</math> are antisymmetric about <math>x = 0</math>.		<math>\chi_a</math> and <math>\psi_a</math> are antisymmetric about <math>x = 0</math>.
	~~Note, this notation is different from that used in [[Evans and Porter 2005]].~~		We therefore have the following equations

	<center><math>		<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\|}		\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]		\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]		-k_n[\partial_{x^\prime}^3\phi_n]\right]

Meylan: /* Expression for the source functions */

2007-11-08T08:17:20Z

Expression for the source functions

← Older revision		Revision as of 08:17, 8 November 2007
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	It follows that		It follows that
	<center><math>		<center><math>
	G_n(x,x^\prime)\|_{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}\|},		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>		</math></center>

	We require the following functions		We require the following functions
	<center><math>		<center><math>
	~~\chi_s(x) =~~ \beta G_n(x,x^\prime)\|_{x^\prime=0,z=0} = i\beta\sum_{n=-2}^\infty\frac{k_n\sin^2{(k_n H)}}{2\alpha C(k_n)}e^{-k_n\|x\|},		\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>		</math></center>

	<center><math>		<center><math>
	~~\chi_a(x) =~~ \beta\partial_{x^\prime} G_n \|_{x^\prime=0,z=0}= i\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} 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>		</math></center>

	<center><math>		<center><math>
	~~\psi_s(x) =~~ \beta\partial_{x^\prime}^2 G_n \|_{x^\prime=0,z=0}= -i\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}^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>		</math></center>

	<center><math>		<center><math>
	~~\psi_a(x)=~~\beta\partial_{x^\prime}^3 G_n\|_{x^\prime=0,z=0} = i\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\|},		\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>		</math></center>

	-----		-----

	=break=		=break=
	<b> We know that its wrong below here </b>		<b> We know that its wrong below here </b>

Meylan: /* Expression for the source functions */

2007-11-08T08:12:25Z

Expression for the source functions

← Older revision		Revision as of 08:12, 8 November 2007
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	The [[Free-Surface Green Function for a Floating Elastic Plate]] is given by		The [[Free-Surface Green Function for a Floating Elastic Plate]] is given by
	<center><math>		<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}\|},		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>		</math></center>

	It follows that		It follows that
	<center><math>		<center><math>
	~~\chi_s(x) =~~ G_n(x,x^\prime)\|_{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}\|},		G_n(x,x^\prime)\|_{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>		</math></center>

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	-----		-----
			=break=
	<b> We know that its wrong below here </b>		<b> We know that its wrong below here </b>

Meylan: /* Solution using the Free-Surface Green Function for a Floating Elastic Plate */

2007-11-08T08:08:19Z

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

← Older revision		Revision as of 08:08, 8 November 2007
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	no real point because the boundary conditions are given in terms of <math>\phi_n</math> since this represents		no real point because the boundary conditions are given in terms of <math>\phi_n</math> since this represents
	the displacement.		the displacement.
			This equation is solved by applying the edge conditions at <math>x</math> = 0 and z = 0


	~~The correct~~ equation is
	~~<center><math>~~
	~~\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></center>~~

	~~which can be~~ solved by applying the edge conditions at <math>x</math> = 0 and z = 0
	<center>		<center>
	<math>		<math>

Meylan: /* Solution using the Free-Surface Green Function for a Floating Elastic Plate */

2007-11-08T08:06:37Z

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

← Older revision		Revision as of 08:06, 8 November 2007
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	for <math>\psi = G</math> with z = 0, we obtain		for <math>\psi = G</math> with z = 0, we obtain
	<center><math>		<center><math>
	-\phi^{\rm In}_n =		-\frac{1}{\alpha}\phi^{\rm In}_n =
	\int_{-\infty}^{\infty}\left(		\int_{-\infty}^{\infty}\left(
	G_{n}\left( x,x^{\prime }\right) \phi \left( x		G_{n}\left( x,x^{\prime }\right) \phi \left( x
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	_{n}\left( x^{\prime }\right) \right) dx^{\prime }		_{n}\left( x^{\prime }\right) \right) dx^{\prime }
	</math></center>		</math></center>
	where the vertical integrals at the ends have given the contribution <math>\~~phi_n~~^{In}</math>.		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		We then substitute for <math>\phi</math> and obtain
	<center><math>		<center><math>
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	\gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right)		\gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right)
	-G\left( x,x^{\prime }\right) \phi		-G\left( x,x^{\prime }\right) \phi
	_{n}\left( x^{\prime }\right) \right) dx^{\prime } = -\~~phi~~^{\rm In}_n		_{n}\left( x^{\prime }\right) \right) dx^{\prime } = -\frac{1}{\alpha}\phi_{n}^{\mathrm{In}}
	</math></center>		</math></center>


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

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

	and obtain		and obtain
			<center>
	<math>		<math>
	\int_{-\infty}^{\infty}\left\{ \frac{1}{\alpha}		\int_{-\infty}^{\infty}\left\{ \frac{1}{\alpha}
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	\phi_{n}\left( x^{\prime }\right)		\phi_{n}\left( x^{\prime }\right)
	dx^\prime		dx^\prime
	</math>		</math></center>

	<center><math>		<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[\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{\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		= -\frac{1}{\alpha}\phi_{n}^{\mathrm{In}}
	</math></center>		</math></center>

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	<center><math>		<center><math>
	\phi_{n}\left( x\right)		\phi_{n}\left( x\right)
	= \beta		= \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]		\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)		+ \partial_{x^\prime} G_n [\partial_{x^\prime}^2\phi_n] - G_n [\partial_{x^\prime}^3\phi_n]\right)
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	the displacement.		the displacement.

	~~We are missing the incident wave which comes into the equations through the boundary conditions at infinity.~~

	The correct equation is		The correct equation is

Meylan: /* Solution using the Free-Surface Green Function for a Floating Elastic Plate */

2007-11-08T08:01:04Z

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

← Older revision		Revision as of 08:01, 8 November 2007
Line 41:		Line 41:
	for <math>\psi = G</math> with z = 0, we obtain		for <math>\psi = G</math> with z = 0, we obtain
	<center><math>		<center><math>
	0 =		-\phi^{\rm In}_n =
	\int_{-\infty}^{\infty}\left(		\int_{-\infty}^{\infty}\left(
	G_{n}\left( x,x^{\prime }\right) \phi \left( x		G_{n}\left( x,x^{\prime }\right) \phi \left( x
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	_{n}\left( x^{\prime }\right) \right) dx^{\prime }		_{n}\left( x^{\prime }\right) \right) dx^{\prime }
	</math></center>		</math></center>
			where the vertical integrals at the ends have given the contribution <math>\phi_n^{In}</math>.
	We then substitute <math>~~G =~~ \phi</math> to obtain		We then substitute for <math>\phi</math> and obtain
	<center><math>		<center><math>
	\int_{-\infty}^{\infty}\left(		\int_{-\infty}^{\infty}\left(
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	\gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right)		\gamma\alpha + 1\right)\phi_{n}\left( x^{\prime }\right)
	-G\left( x,x^{\prime }\right) \phi		-G\left( x,x^{\prime }\right) \phi
	_{n}\left( x^{\prime }\right) \right) dx^{\prime } = 0		_{n}\left( x^{\prime }\right) \right) dx^{\prime } = -\phi^{\rm In}_n
	</math></center>		</math></center>
	~~plus the vertical integrals at the ends (which will give a contribution <math>\phi_n^{In}</math>).~~

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