Energy Balance for Two Elastic Plates - Revision history

Meylan at 09:38, 20 October 2009

2009-10-20T09:38:29Z

← Older revision		Revision as of 09:38, 20 October 2009
Line 1:		Line 1:
			{{pages to be deleted}}

	== Introduction ==		== Introduction ==

Meylan at 08:29, 12 March 2009

2009-03-12T08:29:10Z

← Older revision		Revision as of 08:29, 12 March 2009
Line 1:		Line 1:
	= Introduction =		== Introduction ==

	We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies		We derive here a condition which must be satisfied for energy balance for two semi-infinite elastic plates. The solution applies
Line 25:		Line 25:
	(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).		(for details of this notation see [[Eigenfunction Matching Method for Floating Elastic Plates]]).

	= Equations =		== Equations ==

	Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.		Based on the method used in [[Evans and Davies 1968]], a check can be made to ensure the solutions of the floating plate problem are in energy balance.
Line 64:		Line 64:
	where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big\|_{z=-h}=0</math>).		where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big\|_{z=-h}=0</math>).

	= Expanding <math>\mathbf{\xi_1}</math> =		== Expanding <math>\mathbf{\xi_1}</math> ==

	Near <math>x=-\infty</math>, we approximate <math>\phi</math> by		Near <math>x=-\infty</math>, we approximate <math>\phi</math> by
Line 87:		Line 87:
	where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.		where <math>R_1(0)^*</math> is the conjugate of <math>R_1(0)</math>.

	= Expanding <math>\mathbf{\xi_2}</math> =		== Expanding <math>\mathbf{\xi_2}</math> ==

	Near <math>x=\infty</math>, we approximate <math>\phi</math> by		Near <math>x=\infty</math>, we approximate <math>\phi</math> by
Line 116:		Line 116:
	otherwise <math>\xi_2=0</math>.		otherwise <math>\xi_2=0</math>.

	= Expanding <math>\mathbf{\xi_3}</math> =		== Expanding <math>\mathbf{\xi_3}</math> ==

	The ice-covered boundary condition for the [[Floating Elastic Plate]] gives		The ice-covered boundary condition for the [[Floating Elastic Plate]] gives
Line 239:		Line 239:
	</center>		</center>

	= Solving the Energy Balance Equation =		== Solving the Energy Balance Equation ==

	Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain		Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain

Meylan: /* Equations */

2009-03-12T08:27:37Z

Equations

← Older revision		Revision as of 08:27, 12 March 2009
Line 42:		Line 42:
	As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to		As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
	<center><math>		<center><math>
	\Im\int_\~~mathcal{S~~}\phi\frac{\partial\phi^*}{\partial n} \mathrm{d}l = 0,		\Im\int_{\partial\Omega}\phi\frac{\partial\phi^*}{\partial n} \mathrm{d}l = 0,
	</math></center>		</math></center>
	Expanding gives		Expanding gives

Meylan: /* Equations */

2009-03-12T08:26:55Z

Equations

← Older revision		Revision as of 08:26, 12 March 2009
Line 32:		Line 32:
	The domain of integration is shown in the figure on the right.		The domain of integration is shown in the figure on the right.

	~~[[Image:energy_schematic.jpg\|thumb\|right\|300px\|A diagram depicting the area <math>\mathcal~~{~~U}</math> which is bounded by the rectangle <math>\mathcal{S}</math>.]~~		{{energy_region_plates}}
	{~~A diagram depicting the area <math>\mathcal{U~~}~~</math> which is bounded by the rectangle <math>\mathcal{S}</math>.~~
	~~The rectangle <math>\mathcal{S~~}~~</math> is bounded by <math> -h\leq z \leq0</math> and <math>-\infty\leq x \leq \infty</math>]]~~

	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives

Meylan: /* Equations */

2009-03-12T08:07:43Z

Equations

← Older revision		Revision as of 08:07, 12 March 2009
Line 38:		Line 38:
	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
	<center><math>		<center><math>
	{ \~~int\int_~~\~~mathcal{U~~}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)\mathrm{d}x\mathrm{d}z		{ \iint_{\Omega}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)\mathrm{d}x\mathrm{d}z
	= \int_\~~mathcal{S~~}(\phi\frac{\partial\phi^}{\partial n} - \phi^\frac{\partial\phi}{\partial n})\mathrm{d}l },		= \int_{\partial\Omega}(\phi\frac{\partial\phi^}{\partial n} - \phi^\frac{\partial\phi}{\partial n})\mathrm{d}l },
	</math></center>		</math></center>
	where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.		where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.

Adi Kurniawan: /* Expanding $\mathbf{\xi_1}$ */

2009-03-12T05:55:45Z

Expanding <math>\mathbf{\xi_1}</math>

← Older revision		Revision as of 05:55, 12 March 2009
Line 78:		Line 78:
	\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}		\xi_1 & = & {-\Im\int_{-h}^0 \left[\left(e^{-i\kappa^{I}_1(x)}+ R_1(0)e^{i\kappa^{I}_1(x)}\right)\right.}
	{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }		{ \left(i\kappa^{I}_1e^{i\kappa^{I}_1(x)} - i\kappa^{I}_1R_1(0)^*e^{-i\kappa^{I}_1(x)}\right) }
	{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]dz, }\\ \\		{ \left.\left(\frac{\cosh^2{(k^I_1(z+h))}}{\cosh^2{(k^I_1h)}}\right)\right]\mathrm{d}z, }\\ \\
	& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - \|R_1(0)\|^2\right)}{2\cosh^2{(k^I_1h)}}		& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - \|R_1(0)\|^2\right)}{2\cosh^2{(k^I_1h)}}
	\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)dz\right] ,}\\ \\		\int_{-h}^0 \left(\cosh{(2k^I_1(z+h))+1}\right)\mathrm{d}z\right] ,}\\ \\
	& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - \|R_1(0)\|^2\right)}{2\cosh^2{(k^I_1h)}}		& = & { - \Im\left[\frac{i\kappa^{I}_1\left(1 - \|R_1(0)\|^2\right)}{2\cosh^2{(k^I_1h)}}
	\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\		\left[\frac{1}{2k^I_1}\sinh{(2k^I_1(z+h))} + z\right]_{-h}^0\right], }\\ \\

Adi Kurniawan: /* Equations */

2009-03-12T05:54:35Z

Equations

← Older revision		Revision as of 05:54, 12 March 2009
Line 38:		Line 38:
	Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives		Applying Green's theorem to <math>\phi</math> and its conjugate <math>\phi^*</math> gives
	<center><math>		<center><math>
	{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)~~dxdz~~		{ \int\int_\mathcal{U}(\phi\nabla^2\phi^* - \phi^*\nabla^2\phi)\mathrm{d}x\mathrm{d}z
	= \int_\mathcal{S}(\phi\frac{\partial\phi^}{\partial n} - \phi^\frac{\partial\phi}{\partial n})dl },		= \int_\mathcal{S}(\phi\frac{\partial\phi^}{\partial n} - \phi^\frac{\partial\phi}{\partial n})\mathrm{d}l },
	</math></center>		</math></center>
	where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.		where <math>n</math> denotes the outward plane normal to the boundary and <math>l</math> denotes the plane parallel to the boundary.
	As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to		As <math>\phi</math> and <math>\phi^*</math> satisfy the Laplace's equation, the left hand side of the Green theorem equation vanishes so that it reduces to
	<center><math>		<center><math>
	\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} dl = 0,		\Im\int_\mathcal{S}\phi\frac{\partial\phi^*}{\partial n} \mathrm{d}l = 0,
	</math></center>		</math></center>
	Expanding gives		Expanding gives
Line 53:		Line 53:
	<center><math>		<center><math>
	\xi_1 =		\xi_1 =
	-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big\|_{x=-\infty}dz },		-{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big\|_{x=-\infty}\mathrm{d}z },
	</math></center>		</math></center>
	<center><math>		<center><math>
	\xi_2 =		\xi_2 =
	{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big\|_{x=\infty}dz },		{ \Im\int_{-h}^0(\phi\frac{\partial\phi^*}{\partial x})\big\|_{x=\infty}\mathrm{d}z },
	</math></center>		</math></center>
	and		and
	<center><math>		<center><math>
	\xi_3 =		\xi_3 =
	{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big\|_{z=0}dx } ,		{ \Im\int_{-\infty}^{\infty}(\phi\frac{\partial\phi^*}{\partial z})\big\|_{z=0}\mathrm{d}x } ,
	</math></center>		</math></center>
	where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big\|_{z=-h}=0</math>).		where <math>\Im</math> denotes the imaginary part. The bottom domain composant is obviously equal to zero because of the no-influx condition over the seabed (<math>\frac{\partial\phi^*}{\partial z}\big\|_{z=-h}=0</math>).

Fmontiel: /* Expanding $\mathbf{\xi_2}$ */

2008-08-27T00:15:59Z

Expanding <math>\mathbf{\xi_2}</math>

← Older revision		Revision as of 00:15, 27 August 2008
Line 105:		Line 105:
	\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.		\frac{\partial\phi}{\partial x} \approx -i\kappa^I_{\Lambda}T_\Lambda(0)e^{-i\kappa^I_{\Lambda}(x)}\frac{\cosh{(k^I_\Lambda(z+h))}}{\cosh{(k^I_\Lambda h)}}.
	</math></center>		</math></center>
	We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=~~\sqrt{~~\frac{k_{\Lambda}(0)}{k_{1}(0)}}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.		We know that <math>\kappa_\Lambda(0)=\sqrt{k_{\Lambda}(0)^2-(ik_y)^2}</math>, where <math>k_y</math> is real and depends on the incident angle <math>\theta</math>. When <math>\theta</math> becomes greater than a certain angle <math>\theta_0</math> defined by <math>\sin \theta_0=\frac{k_{\Lambda}(0)}{k_{1}(0)}</math>, <math>\kappa_\Lambda(0)</math> becomes real so the potential becomes real as well. Thus the imaginary part is equal to 0. In that case, we have <math>\xi_2=0</math>.
	Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,		Therefore, if <math>\theta \in [-\theta_0, \theta_0]</math>,
	<center><math>		<center><math>

Fmontiel: /* Solving the Energy Balance Equation */

2008-08-26T23:43:11Z

Solving the Energy Balance Equation

← Older revision		Revision as of 23:43, 26 August 2008
Line 276:		Line 276:
	\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)		\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
	</math></center>		</math></center>
	For the case of greater angles, there are no terms depending on \|T_{\Lambda}(0)\|^2, so we obtain <math>D=0</math> and <math>\|R_{1}(0)\|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.		For the case of greater angles, there are no terms depending on <math>\|T_{\Lambda}(0)\|^2</math>, so we obtain <math>D=0</math> and <math>\|R_{1}(0)\|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.

	[[Category:Floating Elastic Plate]]		[[Category:Floating Elastic Plate]]

Fmontiel: /* Solving the Energy Balance Equation */

2008-08-26T23:42:33Z

Solving the Energy Balance Equation

← Older revision		Revision as of 23:42, 26 August 2008
Line 243:		Line 243:
	= Solving the Energy Balance Equation =		= Solving the Energy Balance Equation =

	Pulling it all together, we finally obtain		Pulling it all together for the case <math>\theta \in [-\theta_0, \theta_0]</math>, we finally obtain
	<center><math>		<center><math>
	\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)		\frac{\beta_\Lambda}{\alpha}\left[2\kappa^I_{\Lambda}(k^I_\Lambda)^2((\kappa^I_{\Lambda})^2 +k_y^2)
Line 276:		Line 276:
	\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)		\frac{1}{2}{\sinh{(2k^I_1h)}}+k^I_1h}\right)
	</math></center>		</math></center>
			For the case of greater angles, there are no terms depending on \|T_{\Lambda}(0)\|^2, so we obtain <math>D=0</math> and <math>\|R_{1}(0)\|^2 = 1</math>, which is the case of the total reflexion. No energy is transmitted in the <math>x>0</math> region.

	[[Category:Floating Elastic Plate]]		[[Category:Floating Elastic Plate]]

Energy Balance for Two Elastic Plates - Revision history

Meylan at 09:38, 20 October 2009

Meylan at 08:29, 12 March 2009

Meylan: /* Equations */

Meylan: /* Equations */

Meylan: /* Equations */

Adi Kurniawan: /* Expanding \mathbf{\xi_1} */

Adi Kurniawan: /* Equations */

Fmontiel: /* Expanding \mathbf{\xi_2} */

Fmontiel: /* Solving the Energy Balance Equation */

Fmontiel: /* Solving the Energy Balance Equation */

Adi Kurniawan: /* Expanding $\mathbf{\xi_1}$ */

Fmontiel: /* Expanding $\mathbf{\xi_2}$ */