Template:Removing the depth dependence - Revision history

Administrator at 00:04, 9 June 2010

2010-06-09T00:04:09Z

← Older revision		Revision as of 00:04, 9 June 2010
Line 11:		Line 11:
	<center>		<center>
	<math>		<math>
	\phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \~~Phi~~(x,y)		\phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \bar{\phi}(x,y)
	</math>		</math>
	</center>		</center>
	Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]		Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]
	<center>		<center>
	<math>\nabla^2 \~~Phi~~ + k^2 \~~Phi~~ = 0 </math>		<math>\nabla^2 \bar{\phi} + k^2 \bar{\phi} = 0 </math>
	</center>		</center>
	in the region not occupied by the scatterers.		in the region not occupied by the scatterers.

Meylan at 22:58, 29 April 2010

2010-04-29T22:58:08Z

← Older revision		Revision as of 22:58, 29 April 2010
Line 11:		Line 11:
	<center>		<center>
	<math>		<math>
	\phi(x,y,z) = \frac{\~~cos~~ \big( ~~k_0~~ (z+h) \big)}{\~~cos~~(~~k_0~~ h)} \Phi(x,y)		\phi(x,y,z) = \frac{\cosh \big( k (z+h) \big)}{\cosh(k h)} \Phi(x,y)
	</math>		</math>
	</center>		</center>
	Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]		Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]
	<center>		<center>
	<math>\nabla^2 \Phi + ~~k_0~~^2 \Phi = 0 </math>		<math>\nabla^2 \Phi + k^2 \Phi = 0 </math>
	</center>		</center>
	in the region not occupied by the scatterers. ~~Not that this is not the standard way to write [[Helmholtz's Equation]]~~		in the region not occupied by the scatterers.
	~~because <math>k_0</math> is the pure imaginary, and it is more normal to write~~
	~~<center>~~
	~~<math>\nabla^2 \Phi - k^2 \Phi = 0 </math>~~
	~~</center>~~
	~~where <math>k=-ik_0.</math>~~

Adi Kurniawan at 02:38, 25 August 2008

2008-08-25T02:38:07Z

← Older revision		Revision as of 02:38, 25 August 2008
Line 18:		Line 18:
	<math>\nabla^2 \Phi + k_0^2 \Phi = 0 </math>		<math>\nabla^2 \Phi + k_0^2 \Phi = 0 </math>
	</center>		</center>
	in the region not occupied by the scatterers. Not that this is not the standard ~~why~~ to write [[Helmholtz's Equation]]		in the region not occupied by the scatterers. Not that this is not the standard way to write [[Helmholtz's Equation]]
	because <math>k_0</math> is the pure imaginary, and it is more normal to write		because <math>k_0</math> is the pure imaginary, and it is more normal to write
	<center>		<center>

Meylan at 08:55, 24 August 2008

2008-08-24T08:55:18Z

New page

If we have a problem in which all the scatterers are of constant cross sections so
that
<center>
<math>\partial\Omega = \partial\hat{\Omega} \times z\in[-h,0]
</math>
</center>
where <math>\partial\hat{\Omega} </math> is a function only of <math>x,y</math>
i.e. the boundary of the scattering bodies is uniform with respect to depth.
We can remove the depth dependence [http://en.wikipedia.org/wiki/Separation_of_Variables separation of variables]
and obtain that the dependence on depth is given by
<center>
<math>
\phi(x,y,z) = \frac{\cos \big( k_0 (z+h) \big)}{\cos(k_0 h)} \Phi(x,y)
</math>
</center>
Since <math>\phi</math> satisfies [[Laplace's Equation]], then <math>\Phi</math> satisfies [[Helmholtz's Equation]]
<center>
<math>\nabla^2 \Phi + k_0^2 \Phi = 0 </math>
</center>
in the region not occupied by the scatterers. Not that this is not the standard why to write [[Helmholtz's Equation]]
because <math>k_0</math> is the pure imaginary, and it is more normal to write
<center>
<math>\nabla^2 \Phi - k^2 \Phi = 0 </math>
</center>
where <math>k=-ik_0.</math>