Floating Elastic Plates of Identical Properties - Revision history

Meylan: /* Solution */

2007-06-06T14:11:06Z

Solution

← Older revision		Revision as of 14:11, 6 June 2007
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	</math></center>		</math></center>
	<center><math>		<center><math>
	\chi_a(x-x^\prime) = \beta\partial_x G_z = -sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^2\sin^2{(k_n H)}}{C(k_n)}e^{-k_n\|x-x^\prime\|},		\chi_a(x-x^\prime) = \beta\partial_x G_z = -\sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^2\sin^2{(k_n H)}}{C(k_n)}e^{-k_n\|x-x^\prime\|},
	</math></center>		</math></center>
	<center><math>		<center><math>
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	</math></center>		</math></center>
	<center><math>		<center><math>
	\psi_a(x-x^\prime) =\beta\partial_x^3 G_z = -sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{C(k_n)}e^{-k_n\|x-x^\prime\|}		\psi_a(x-x^\prime) =\beta\partial_x^3 G_z = -\sgn(x-x^\prime)\frac{i\beta}{2\alpha} \sum_{n=-2}^\infty\frac{k_n^4\sin^2{(k_n H)}}{C(k_n)}e^{-k_n\|x-x^\prime\|}
	</math></center>		</math></center>

Akohout: /* More complicated boundary conditions */

2007-05-31T05:11:36Z

More complicated boundary conditions

← Older revision		Revision as of 05:11, 31 May 2007
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	<center><math>		<center><math>
	k_0^3e^{-k_0x^\prime} = \left(\frac{S_l}{\beta} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^7\right)[\phi_z]		k_0^3e^{-k_0x^\prime} = \left(\frac{S_l}{\beta} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^7\right)[\phi_z]
	+ \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left(sgn(x-x^\prime)k_n^6\right)\right)[\partial_{x}\phi_z]		- \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left(sgn(x-x^\prime)k_n^6\right)\right)[\partial_{x}\phi_z]
	</math></center>		</math></center>

Meylan: /* Integral at the top */

2007-05-30T07:06:58Z

Integral at the top

← Older revision		Revision as of 07:06, 30 May 2007
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	The final integral is		The final integral is
	<center><math>		<center><math>
	~~0 =~~ -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx		-\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx
	</math></center>		</math></center>

	At z=0, the z variable disappears to give		At z=0, the z variable disappears to give
	<center><math>		<center><math>
	~~0 =~~ -\int_{-\infty}^\infty \left( G(x,x^\prime) \phi_z(x) - \phi(x) G_z(x,x^\prime)\right)\, dx		-\int_{-\infty}^\infty \left( G(x,x^\prime) \phi_z(x) - \phi(x) G_z(x,x^\prime)\right)\, dx
	</math></center>		</math></center>

	We then substitute ~~<math>G = \phi</math> to remove~~ <math>\phi</math> and obtain		We then substitute for <math>\phi</math> and obtain
	<center><math>		<center><math>
	\int_{-\infty}^{\infty}\left(		\int_{-\infty}^{\infty}\left(

Meylan: /* Integral and the top */

2007-05-30T07:04:21Z

Integral and the top

← Older revision		Revision as of 07:04, 30 May 2007
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	0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx		0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx
	</math></center>		</math></center>

	~~== Integral and the top ==~~

	At z=0, the z variable disappears to give		At z=0, the z variable disappears to give

Meylan: /* Integral at the bottom */

2007-05-30T07:03:27Z

Integral at the bottom

← Older revision		Revision as of 07:03, 30 May 2007
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	</math></center>		</math></center>

	== Integral at the top		== Integral at the top ==

	The final integral is		The final integral is

Meylan: /* Integral and the bottom */

2007-05-30T07:02:52Z

Integral and the bottom

← Older revision		Revision as of 07:02, 30 May 2007
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	---------		---------

	== Integral ~~and~~ the bottom ==		== Integral at the bottom ==
	Also, ~~0ur~~ governing equations imply <math>G {\partial \phi \over \partial z}\|_{z=-h} = 0 </math> and <math>\phi {\partial G \over \partial z}\|_{z=-h} = 0</math> so that,		Also, our governing equations imply <math>G {\partial \phi \over \partial z}\|_{z=-h} = 0 </math> and <math>\phi {\partial G \over \partial z}\|_{z=-h} = 0</math> so that,
	<center><math>		<center><math>
	\int_{-\infty}^\infty \left( G {\partial \phi \over \partial z}\|_{z=-h} - \phi {\partial G \over \partial z}\|_{z=-h}\right)\, dx =0		\int_{-\infty}^\infty \left( G {\partial \phi \over \partial z}\|_{z=-h} - \phi {\partial G \over \partial z}\|_{z=-h}\right)\, dx =0
	</math></center>		</math></center>
	~~and we are left with~~
			== Integral at the top

			The final integral is
	<center><math>		<center><math>
	0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx		0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx

Akohout: /* More complicated boundary conditions */

2007-05-30T03:10:21Z

More complicated boundary conditions

← Older revision		Revision as of 03:10, 30 May 2007
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	We now have two unknowns which can be solved simultaneously using the following two edge conditions		We now have two unknowns which can be solved simultaneously using the following two edge conditions
	<center><math>		<center><math>
	\frac{S_r}{\beta}[\phi_z] = \partial_x^2\phi_z		\frac{S_r}{\beta}[\partial_x\phi_z] = \partial_x^2\phi_z
	</math></center>		</math></center>
	<center><math>		<center><math>
	\frac{S_r}{\beta}[\phi_z] = k_0^2e^{-k_0x^\prime} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		\frac{S_r}{\beta}[\partial_x\phi_z] = k_0^2e^{-k_0x^\prime} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)		\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)
	</math></center>		</math></center>
	<center><math>		<center><math>
	-k_0^2e^{-k_0x^\prime} = -\frac{S_r}{\beta}[\phi_z] + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		-k_0^2e^{-k_0x^\prime} = -\frac{S_r}{\beta}[\partial_x\phi_z] + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)		\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)
	</math></center>		</math></center>
	<center><math>		<center><math>
	-k_0^2e^{-k_0x^\prime} = \left(~~-\frac{S_r}{\beta} +~~ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{C(k_n)}\left(\sgn(x-x^\prime)k_n^6\right)\right)[\phi_z]		-k_0^2e^{-k_0x^\prime} = \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{C(k_n)}\left(\sgn(x-x^\prime)k_n^6\right)\right)[\phi_z]
	+ \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^5\right)[\partial_{x}\phi_z]		+ \left(-\frac{S_r}{\beta} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^5\right)[\partial_{x}\phi_z]
	</math></center>		</math></center>
	and		and
	<center><math>		<center><math>
	-\frac{S_l}{\beta}[~~\partial_x~~\phi_z] = \partial_x^3\phi_z(x)		-\frac{S_l}{\beta}[\phi_z] = \partial_x^3\phi_z(x)
	</math></center>		</math></center>
	<center><math>		<center><math>
	-\frac{S_l}{\beta}[~~\partial_x~~\phi_z] = -k_0^3e^{-k_0x^\prime} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		-\frac{S_l}{\beta}[\phi_z] = -k_0^3e^{-k_0x^\prime} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)		\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)
	</math></center>		</math></center>
	<center><math>		<center><math>
	k_0^3e^{-k_0x^\prime} = \frac{S_l}{\beta}[~~\partial_x~~\phi_z] - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		k_0^3e^{-k_0x^\prime} = \frac{S_l}{\beta}[\phi_z] - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)		\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)
	</math></center>		</math></center>
	<center><math>		<center><math>
	k_0^3e^{-k_0x^\prime} = \left(-\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^7\right)[\phi_z]		k_0^3e^{-k_0x^\prime} = \left(\frac{S_l}{\beta} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^7\right)[\phi_z]
	+ \left(~~\frac{S_l}{\beta} -~~ \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left(sgn(x-x^\prime)k_n^6\right)\right)[\partial_{x}\phi_z]		+ \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left(sgn(x-x^\prime)k_n^6\right)\right)[\partial_{x}\phi_z]
	</math></center>		</math></center>

Akohout: /* More complicated boundary conditions */

2007-05-30T02:57:39Z

More complicated boundary conditions

← Older revision		Revision as of 02:57, 30 May 2007
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	These edge conditions imply <math>\frac{\partial^2\phi_n^+(x^\prime)}{\partial x^2} = \frac{\partial^2\phi_n^-(x^\prime)}{\partial x^2} </math> and <math>\frac{\partial^3\phi_n^+(x^\prime)}{\partial x^3} = \frac{\partial^3\phi_n^-(x^\prime)}{\partial x^3} </math> which imply <math>[\partial_x^2\phi_n] = 0</math> and <math>[\partial_x^3\phi_n] = 0 </math>.		These edge conditions imply <math>\frac{\partial^2\phi_n^+(x^\prime)}{\partial x^2} = \frac{\partial^2\phi_n^-(x^\prime)}{\partial x^2} </math> and <math>\frac{\partial^3\phi_n^+(x^\prime)}{\partial x^3} = \frac{\partial^3\phi_n^-(x^\prime)}{\partial x^3} </math> which imply <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		<math> \phi_n </math> can again be expressed by
	<center><math>		<center><math>
	\phi_{n}\left( x\right)		\phi_{n}\left( x\right)
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	</math></center>		</math></center>
	<center><math>		<center><math>
	= k_0^2e^{-k_0x^\prime} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		\frac{S_r}{\beta}[\phi_z] = k_0^2e^{-k_0x^\prime} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)		\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)
			</math></center>
			<center><math>
			-k_0^2e^{-k_0x^\prime} = -\frac{S_r}{\beta}[\phi_z] + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
			\left(sgn(x-x^\prime)k_n^6[\phi_z] + k_n^5[\partial_{x}\phi_z] \right)
			</math></center>
			<center><math>
			-k_0^2e^{-k_0x^\prime} = \left(-\frac{S_r}{\beta} + \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{C(k_n)}\left(\sgn(x-x^\prime)k_n^6\right)\right)[\phi_z]
			+ \left(\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^5\right)[\partial_{x}\phi_z]
	</math></center>		</math></center>
	and		and
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	</math></center>		</math></center>
	<center><math>		<center><math>
	= -k_0^3e^{-k_0x^\prime} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}		-\frac{S_l}{\beta}[\partial_x\phi_z] = -k_0^3e^{-k_0x^\prime} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
	\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)		\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)
	</math></center>		</math></center>
			<center><math>
	~~where~~ <math>~~s_l~~ = \beta/~~S_l~~</~~math~~> ~~and~~ <math>~~s_r~~ = \beta/~~S_r~~</~~math~~>.		k_0^3e^{-k_0x^\prime} = \frac{S_l}{\beta}[\partial_x\phi_z] - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)}
			\left(k_n^7[\phi_z] + sgn(x-x^\prime)k_n^6[\partial_{x}\phi_z] \right)
			</math></center>
			<center><math>
			k_0^3e^{-k_0x^\prime} = \left(-\frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} k_n^7\right)[\phi_z]
			+ \left(\frac{S_l}{\beta} - \frac{i\beta}{2\alpha}\sum_{n=-2}^\infty \frac{\sin^2{(k_n H)}}{ C(k_n)} \left(sgn(x-x^\prime)k_n^6\right)\right)[\partial_{x}\phi_z]
			</math></center>

	= A Set of Floating Elastic Plates of Identical Properties =		= A Set of Floating Elastic Plates of Identical Properties =

Meylan: /* Evaluation of the integrals */

2007-05-29T22:24:09Z

Evaluation of the integrals

← Older revision		Revision as of 22:24, 29 May 2007
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	\phi = A(e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))}		\phi = A(e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))}
	</math></center>		</math></center>
	and ~~<math>x\to\infty</math>~~		and
	<center><math>		<center><math>
	\phi_x = Ak_0(-e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))}		\phi_x = Ak_0(-e^{-k_0x} + R e^{k_0x})\cos{(k_n(z+H))}
	</math></center>		</math></center>
	where <math>R</math> and <math>T</math> and reflection and transmission coefficient.		where <math>R</math> and <math>T</math> and reflection and transmission coefficient.
	For ~~large~~ x		For <math>x\to\infty</math>
	<center><math>		<center><math>
	\phi = AT e^{-k_0x}\cos{(k_n(z+H))}		\phi = AT e^{-k_0x}\cos{(k_n(z+H))}
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	</math></center>		</math></center>

	~~Now,~~		== Integral at the right end ==

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

	and		== Integral and the left end ==

	<math>		<math>
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	---------		---------

			== Integral and the bottom ==
	Also, 0ur governing equations imply <math>G {\partial \phi \over \partial z}\|_{z=-h} = 0 </math> and <math>\phi {\partial G \over \partial z}\|_{z=-h} = 0</math> so that,		Also, 0ur governing equations imply <math>G {\partial \phi \over \partial z}\|_{z=-h} = 0 </math> and <math>\phi {\partial G \over \partial z}\|_{z=-h} = 0</math> so that,
	<center><math>		<center><math>
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	0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx		0 = -\int_{-\infty}^\infty \left( G(x,x^\prime,z) \phi_z(x,z)\|_{z=0} - \phi(x,z) G_z(x,x^\prime,z) \|_{z=0}\right)\, dx
	</math></center>		</math></center>

			== Integral and the top ==

	At z=0, the z variable disappears to give		At z=0, the z variable disappears to give

Meylan: /* Evaluation of the integrals */

2007-05-29T22:16:55Z

Evaluation of the integrals

← Older revision		Revision as of 22:16, 29 May 2007
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	where we take the limit as N goes to infinity. We have to evaluate four integrals, the integral at the top, the integral at the bottom and the integrals at either end.		where we take the limit as N goes to infinity. We have to evaluate four integrals, the integral at the top, the integral at the bottom and the integrals at either end.

	~~= Evaluation of the integrals =~~

	= Evaluation of the integrals =		= Evaluation of the integrals =