Template:Frequency domain equations for a floating plate - Revision history

Meylan at 23:01, 12 November 2025

2025-11-12T23:01:52Z

← Older revision		Revision as of 23:01, 12 November 2025
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	\Delta \phi &= 0, &-h < z \leq 0 \\		\Delta \phi &= 0, &-h < z \leq 0 \\
	\partial_z \phi &= 0, &z = - h \\		\partial_z \phi &= 0, &z = - h \\
	\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha/g}\phi, &z = 0 \\		\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{g \alpha}\phi, &z = 0 \\
	-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .		-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .
	\end{align}		\end{align}
	</math></center>		</math></center>

Meylan at 23:01, 12 November 2025

2025-11-12T23:01:33Z

← Older revision		Revision as of 23:01, 12 November 2025
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	\Delta \phi &= 0, &-h < z \leq 0 \\		\Delta \phi &= 0, &-h < z \leq 0 \\
	\partial_z \phi &= 0, &z = - h \\		\partial_z \phi &= 0, &z = - h \\
	\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha}\phi, &z = 0 \\		\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha/g}\phi, &z = 0 \\
	-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .		-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .
	\end{align}		\end{align}
	</math></center>		</math></center>

Meylan at 22:55, 12 November 2025

2025-11-12T22:55:04Z

← Older revision		Revision as of 22:55, 12 November 2025
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	\begin{align}		\begin{align}
	-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\		-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\
	\rho g\zeta - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \~~eta~~ -\omega^2 \rho_i h \zeta, &z=0 \\		-\rho g\zeta - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \zeta -\omega^2 \rho_i h \zeta, &z=0 \\
	\Delta \phi &= 0, &-h<z<0 \\		\Delta \phi &= 0, &-h<z<0 \\
	\partial_z \phi &= 0, &z=-h,		\partial_z \phi &= 0, &z=-h,

Adi Kurniawan at 11:03, 6 November 2010

2010-11-06T11:03:02Z

← Older revision		Revision as of 11:03, 6 November 2010
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	<math>\exp (-\mathrm{i}\omega t)\,</math> the equations become		<math>\exp (-\mathrm{i}\omega t)\,</math> the equations become
	<center><math>		<center><math>
	-\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; ~~</math></center>~~		\begin{align}
	~~<center><math>~~ \rho g\zeta - \mathrm{i}\omega\rho \phi		-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 \\
	= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0~~; </math></center>~~		\rho g\zeta - \mathrm{i}\omega\rho \phi &= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, &z=0 \\
	~~<center><math>~~		\Delta \phi &= 0, &-h<z<0 \\
	\Delta \phi = 0,~~\,\,~~-h<z<0		\partial_z \phi &= 0, &z=-h,
	~~</math></center>~~		\end{align}
	~~<center><math>~~
	\partial_z \phi = 0,~~\,\,~~z=-h,
	</math></center>		</math></center>
	where		where
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	<math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain		<math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain
	<center><math>		<center><math>
	\Delta \phi = 0, ~~\;\;\;~~ -h < z \leq 0,		\begin{align}
	~~</math></center>~~		\Delta \phi &= 0, &-h < z \leq 0 \\
	~~<center><math>~~		\partial_z \phi &= 0, &z = - h \\
	\partial_z \phi = 0, ~~\;\;\;~~ z = - h,		\beta \partial_x^4 \zeta + \left( 1 - \gamma\alpha \right) \zeta &= -\mathrm{i} \sqrt{\alpha}\phi, &z = 0 \\
	~~</math></center>~~		-\mathrm{i}\omega\zeta &= \partial_z\phi , &z=0 .
	~~<center><math>~~		\end{align}
	\beta \partial_x^4 \zeta
	+ \left( 1 - \gamma\alpha \right) \zeta = -\mathrm{i} \sqrt{\alpha}\phi, \;\;
	z = 0.
	</math></center>		</math></center>
	~~<center><math>~~
	~~-\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>~~

Meylan at 10:39, 28 April 2010

2010-04-28T10:39:25Z

← Older revision		Revision as of 10:39, 28 April 2010
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	If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to		If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to
	<math>\exp (\mathrm{i}\omega t)\,</math> the equations become		<math>\exp (-\mathrm{i}\omega t)\,</math> the equations become
	<center><math>		<center><math>
	\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>		-\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>
	<center><math> \rho g\zeta + \mathrm{i}\omega\rho \phi		<center><math> \rho g\zeta - \mathrm{i}\omega\rho \phi
	= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>		= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>
	<center><math>		<center><math>
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	<center><math>		<center><math>
	\beta \partial_x^4 \zeta		\beta \partial_x^4 \zeta
	+ \left( 1 - \gamma\alpha \right) \zeta = \mathrm{i} \sqrt{\alpha}\phi, \;\;		+ \left( 1 - \gamma\alpha \right) \zeta = -\mathrm{i} \sqrt{\alpha}\phi, \;\;
	z = 0.		z = 0.
	</math></center>		</math></center>
	<center><math>		<center><math>
	\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>		-\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>

Meylan at 02:01, 30 November 2009

2009-11-30T02:01:27Z

← Older revision		Revision as of 02:01, 30 November 2009
Line 27:		Line 27:
	z = 0.		z = 0.
	</math></center>		</math></center>
			<center><math>
			\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>

Meylan at 02:00, 30 November 2009

2009-11-30T02:00:41Z

← Older revision		Revision as of 02:00, 30 November 2009
Line 1:		Line 1:
	If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to		If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to
	<math>\exp (i\omega t)\,</math> the equations become		<math>\exp (\mathrm{i}\omega t)\,</math> the equations become
	<center><math>		<center><math>
	i\omega\zeta = \partial_z\phi , \ z=0; </math></center>		\mathrm{i}\omega\zeta = \partial_z\phi , \ z=0; </math></center>
	<center><math> \rho g\zeta + i\omega\rho \phi		<center><math> \rho g\zeta + \mathrm{i}\omega\rho \phi
	= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>		= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>
	<center><math>		<center><math>

Meylan at 02:00, 30 November 2009

2009-11-30T02:00:14Z

← Older revision		Revision as of 02:00, 30 November 2009
Line 23:		Line 23:
	</math></center>		</math></center>
	<center><math>		<center><math>
	\beta \partial_x^4\~~partial_z \phi~~		\beta \partial_x^4 \zeta
	- \left( \gamma\alpha ~~- 1~~ \right) \~~partial_z~~ \~~phi +~~ \alpha\phi ~~= 0~~, \;\;		+ \left( 1 - \gamma\alpha \right) \zeta = \mathrm{i} \sqrt{\alpha}\phi, \;\;
	z = 0.		z = 0.
	</math></center>		</math></center>

Meylan at 01:28, 3 April 2009

2009-04-03T01:28:20Z

New page

If we make the assumption of [[Frequency Domain Problem]] that everything is proportional to
<math>\exp (i\omega t)\,</math> the equations become
<center><math>
i\omega\zeta = \partial_z\phi , \ z=0; </math></center>
<center><math> \rho g\zeta + i\omega\rho \phi
= D \partial_x^4 \eta -\omega^2 \rho_i h \zeta, \ z=0; </math></center>
<center><math>
\Delta \phi = 0,\,\,-h<z<0
</math></center>
<center><math>
\partial_z \phi = 0,\,\,z=-h,
</math></center>
where
<math>\zeta</math> is the surface displacement and <math>\phi</math> is the velocity potential in the frequency domain.

These equations can be simplified by defining <math>\alpha = \omega^2/g</math>,
<math>\beta = D/\rho g</math> and <math>\gamma = \rho_i h/\rho</math> to obtain
<center><math>
\Delta \phi = 0, \;\;\; -h < z \leq 0,
</math></center>
<center><math>
\partial_z \phi = 0, \;\;\; z = - h,
</math></center>
<center><math>
\beta \partial_x^4\partial_z \phi
- \left( \gamma\alpha - 1 \right) \partial_z \phi + \alpha\phi = 0, \;\;
z = 0.
</math></center>