Eigenfunction Matching for a Submerged Semi-Infinite Dock - Revision history

Meylan at 23:59, 16 October 2009

2009-10-16T23:59:47Z

← Older revision		Revision as of 23:59, 16 October 2009
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			{{complete pages}}

	== Introduction ==		== Introduction ==

Mike smith at 02:10, 9 August 2009

2009-08-09T02:10:33Z

← Older revision		Revision as of 02:10, 9 August 2009
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	<center>		<center>
	<math>		<math>
	\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}^{h}(z), \;\;x<0		\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}^{h}x}\phi_{m}^{h}(z), \;\;x<0
	</math>		</math>
	</center>		</center>

Mike smith: consistent notation

2009-08-09T02:10:21Z

consistent notation

← Older revision		Revision as of 02:10, 9 August 2009
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	<math>		<math>
	\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}		\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}
	a_{m} \phi_{m}\left( z\right)		a_{m} \phi_{m}^{h}\left( z\right)
	=\sum_{m=0}^{\infty}b_{m}\chi_{m}(z)		=\sum_{m=0}^{\infty}b_{m}\chi_{m}(z)
	</math>		</math>
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	<center>		<center>
	<math>		<math>
	-k_{0}\phi_{0}\left( z\right) +\sum		-k_{0}^{h}\phi_{0}\left( z\right) +\sum
	_{m=0}^{\infty} a_{m}k_{m}\phi_{m}\left( z\right)		_{m=0}^{\infty} a_{m}k_{m}^{h}\phi_{m}^{h}\left( z\right)
	=-\sum_{m=0}^{\infty}b_{m}\mu_{m}\chi		=-\sum_{m=0}^{\infty}b_{m}\mu_{m}\chi
	_{m}(z)		_{m}(z)
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	for each <math>n</math>.		for each <math>n</math>.
	We solve these equations by multiplying both equations by		We solve these equations by multiplying both equations by
	<math>\phi_{l}(z)</math> and integrating from <math>-h</math> to <math>0</math> to obtain:		<math> \phi_{l}^{h}(z) \,</math> and integrating from <math>-h</math> to <math>0</math> to obtain:
	<center>		<center>
	<math>		<math>
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	<center>		<center>
	<math>		<math>
	-k_{0}A_{0}\delta_{0l}+a_{l}k_{l}A_l		-k_{0}^{h}A_{0}\delta_{0l}+a_{l}k_{l}^{h}A_l
	=-\sum_{m=0}^{\infty}b_{m}\mu_{m}B_{ml}		=-\sum_{m=0}^{\infty}b_{m}\mu_{m}B_{ml}
	</math>		</math>

Mike smith: spelling error, consistent notation, removed scattered ref

2009-08-09T02:06:11Z

spelling error, consistent notation, removed scattered ref

← Older revision		Revision as of 02:06, 9 August 2009
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	We need to apply some boundary conditions at plus and minus infinity,		We need to apply some boundary conditions at plus and minus infinity,
	where are essentially ~~the~~ the solution cannot grow. This means that we		where are essentially that the solution cannot grow. This means that we
	only have the positive (or ~~negaitve~~) roots of the dispersion equation.		only have the positive (or negative) roots of the dispersion equation.
	However, it does not help us with the purely imaginary root. Here we		However, it does not help us with the purely imaginary root. Here we
	must use a different condition, essentially identifying one solution		must use a different condition, essentially identifying one solution
	as the incoming wave and the other as the outgoing wave.		as the incoming wave and the other as the outgoing wave.

	Therefore the ~~scattered~~ potential (without the incident wave, which will		Therefore the potential (without the incident wave, which will
	be added later) can		be added later) can
	be expanded as		be expanded as
	<center>		<center>
	<math>		<math>
	\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}(z), \;\;x<0		\phi(x,z)=\sum_{m=0}^{\infty}a_{m}e^{k_{m}x}\phi_{m}^{h}(z), \;\;x<0
	</math>		</math>
	</center>		</center>

Mike smith at 23:25, 8 August 2009

2009-08-08T23:25:47Z

← Older revision		Revision as of 23:25, 8 August 2009
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	{{separation of variables for a dock}}		{{separation of variables for a dock}}
			We now adjust for the fact that the dock is submerged.
	{{separation of variables for a submerged dock}}		{{separation of variables for a submerged dock}}

Mike smith: /* Solution Method */

2009-08-08T23:03:19Z

Solution Method

← Older revision		Revision as of 23:03, 8 August 2009
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	== Solution Method ==		== Solution Method ==

	We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the three regions, <math>x<0</math>		We use [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables] in the three regions: {<math>x<0 \,</math>},
	<math>-d<z<0,\,\,x>0</math>, and <math>-h<z<-d,\,\,x>0</math>. The first two regions use the free-surface eigenfunction		{<math>-d<z<0,\,\,x>0</math> }, and {<math>-h<z<-d,\,\,x>0</math> }. The first two regions use the free-surface eigenfunction
	and the third uses the dock eigenfunctions.		and the third uses the dock eigenfunctions.

Mike smith: /* Governing Equations */

2009-08-08T22:58:47Z

Governing Equations

← Older revision		Revision as of 22:58, 8 August 2009
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	</math>		</math>
	</center>		</center>
	<center><math>		<center>
			<math>
	\partial_z\phi=\alpha\phi, \,\, z=0,		\partial_z\phi=\alpha\phi, \,\, z=0,
	</math></center>		</math>
			</center>
	<center>		<center>
	<math>		<math>

Adi Kurniawan: /* Energy Balance */

2009-02-24T07:29:09Z

Energy Balance

← Older revision		Revision as of 07:29, 24 February 2009
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	<center><math>		<center><math>
	\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz		\lim_{x\to-\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}z
	= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)		= \lim_{x\to-\infty} \Im\int_{-h}^{0} \hat{k}_0 \left(e^{\hat{k}_0 x} + a_0^{*} e^{-\hat{k}_0 x} \right)
	\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 dz		\left(e^{-\hat{k}_0 x} - a_0 e^{\hat{k}_0 x} \right)\phi_0(z)^2 \mathrm{d}z
	</math></center>		</math></center>
	<center><math>		<center><math>
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	and		and
	<center><math>		<center><math>
	\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} dz		\lim_{x\to\infty} \Im\int_{-h}^{0}\phi^*\frac{\partial\phi}{\partial n} \mathrm{d}z
	= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\mu}_0 \left(b_0^{*} e^{\hat{\mu}_0 x} \right)		= \lim_{x\to\infty} \Im\int_{-h}^{0} -\hat{\mu}_0 \left(b_0^{*} e^{\hat{\mu}_0 x} \right)
	\left(b_0 e^{-\hat{\mu}_0 x}\right)\chi_0(z)^2 dz		\left(b_0 e^{-\hat{\mu}_0 x}\right)\chi_0(z)^2 \mathrm{d}z
	</math></center>		</math></center>
	<center><math>		<center><math>
	= - \frac{\hat{\mu}_0}{i} \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz		= - \frac{\hat{\mu}_0}{i} \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 \mathrm{d}z
	</math></center>		</math></center>
	Therefore the energy balance can be written as		Therefore the energy balance can be written as
	<center><math>		<center><math>
	\hat{k}_0 A_0 \|a_0^2\| + \hat{\mu}_0 \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz = \hat{k}_0 A_0		\hat{k}_0 A_0 \|a_0^2\| + \hat{\mu}_0 \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 \mathrm{d}z = \hat{k}_0 A_0
	</math></center>		</math></center>
	Note that the second integral above is actually only from <math>-d</math> to 0.		Note that the second integral above is actually only from <math>-d</math> to 0.

Meylan: /* Energy Balance */

2008-10-28T01:14:37Z

Energy Balance

← Older revision		Revision as of 01:14, 28 October 2008
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	</math></center>		</math></center>
	<center><math>		<center><math>
	= \hat{k}_0 A_0 \left(1 - \|a_0^2\| \right)		= \frac{\hat{k}_0}{i} A_0 \left(1 - \|a_0^2\| \right)
	</math></center>		</math></center>
	and		and
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	</math></center>		</math></center>
	<center><math>		<center><math>
	= -\hat{\mu}_0 \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz		= - \frac{\hat{\mu}_0}{i} \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz
	</math></center>		</math></center>
	Therefore the energy balance can be written as		Therefore the energy balance can be written as

Meylan: /* Energy Balance */

2008-10-28T00:41:24Z

Energy Balance

← Older revision		Revision as of 00:41, 28 October 2008
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	Therefore the energy balance can be written as		Therefore the energy balance can be written as
	<center><math>		<center><math>
	\hat{k}_0 A_0 \|a_0^2\| + \hat{\mu}_0 \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz = \hat{k}_0		\hat{k}_0 A_0 \|a_0^2\| + \hat{\mu}_0 \|b_0^2\| \int_{-h}^{0} \chi_0(z)^2 dz = \hat{k}_0 A_0
	</math></center>		</math></center>
	Note that the second integral above is actually only from <math>-d</math> to 0.		Note that the second integral above is actually only from <math>-d</math> to 0.