Helmholtz's Equation - Revision history

Mike smith: spelling fix

2013-04-27T21:03:06Z

spelling fix

← Older revision		Revision as of 21:03, 27 April 2013
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	</math>		</math>
	</center>		</center>
	where <math>\epsilon = 1,1/2 \ \mbox{or} \ 0</math>, depending on whether we are exterior, on the boundary or in the interior of the domain (respectively), and the fundamental solution for the ~~Helmholz~~ Equation (which incorporates Sommerfeld Radiation conditions) is given by		where <math>\epsilon = 1,1/2 \ \mbox{or} \ 0</math>, depending on whether we are exterior, on the boundary or in the interior of the domain (respectively), and the fundamental solution for the Helmholtz Equation (which incorporates Sommerfeld Radiation conditions) is given by
	<math>G(\|\mathbf{x} - \mathbf{x}^\prime)\|) = \frac{i}{4} H_{0}^{(1)}(k \|\mathbf{x} - \mathbf{x}^\prime)\|).\,</math>		<math>G(\|\mathbf{x} - \mathbf{x}^\prime)\|) = \frac{i}{4} H_{0}^{(1)}(k \|\mathbf{x} - \mathbf{x}^\prime)\|).\,</math>

Mike smith: consistency

2010-07-16T03:29:39Z

consistency

← Older revision		Revision as of 03:29, 16 July 2010
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	<center>		<center>
	<math>		<math>
	R(r) = B \, J_\nu(~~k_m~~ r) + C \, H^{(1)}_\nu(~~k_m~~ r),\ \nu \in \mathbb{Z},		R(r) = B \, J_\nu(k r) + C \, H^{(1)}_\nu(k r),\ \nu \in \mathbb{Z},
	</math>		</math>
	</center>		</center>

Mike smith: missing i/4

2010-06-21T00:26:30Z

missing i/4

← Older revision		Revision as of 00:26, 21 June 2010
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	<center>		<center>
	<math>		<math>
	\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0		\frac{1}{2}\phi(\mathbf{x}) = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
	(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)\phi(\mathbf{x^{\prime}})		(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)\phi(\mathbf{x^{\prime}})
	\mathrm{d} S^{\prime}.		\mathrm{d} S^{\prime}.
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	<center>		<center>
	<math>		<math>
	\frac{1}{2}\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma} = \phi^{\mathrm{I}}(\mathbf{x}) + \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0		\frac{1}{2}\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma} = \phi^{\mathrm{I}}(\mathbf{x}) + \frac{i}{4} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
	(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{ \prime}}		(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)\sum_{n=-N}^{N} a_n e^{\mathrm{i} n \gamma^{ \prime}}
	\mathrm{d} S^{\prime}.		\mathrm{d} S^{\prime}.
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	\mathrm{d} S		\mathrm{d} S
	= \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma}		= \int_{\partial\Omega} \phi^{\mathrm{I}}(\mathbf{x})e^{\mathrm{i} m \gamma}
	\mathrm{d} S +		\mathrm{d} S + \frac{i}{4}
	\sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0		\sum_{n=-N}^{N} a_n \int_{\partial\Omega} \int_{\partial\Omega} \partial_{n^{\prime}} H^{(1)}_0
	(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)e^{\mathrm{i} n \gamma^{\prime}}		(k\|\mathbf{x} - \mathbf{x^{\prime}}\|)e^{\mathrm{i} n \gamma^{\prime}}

Mike smith: /* Solution for a Circle */

2010-06-16T00:14:49Z

Solution for a Circle

← Older revision		Revision as of 00:14, 16 June 2010
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	</math>		</math>
	</center>		</center>
	which tells us that providing we know the form of the incident wave, we can compute the <math>D_\nu \,</math> coefficients and ~~compute~~ the potential throughout the circle. It is possible to ~~equate~~ plane ~~waves and~~ cylindrical waves using the [http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion Jacobi-Anger Identity].		which tells us that providing we know the form of the incident wave, we can compute the <math>D_\nu \,</math> coefficients and ultimately determine the potential throughout the circle. It is possible to expand a plane wave in terms of cylindrical waves using the [http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion Jacobi-Anger Identity].

	== Solution for an arbitrary scatterer ==		== Solution for an arbitrary scatterer ==

Mike smith: /* Solution for a Circle */

2010-06-16T00:13:19Z

Solution for a Circle

← Older revision		Revision as of 00:13, 16 June 2010
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	where <math>J_\nu \,</math> denotes a Bessel function		where <math>J_\nu \,</math> denotes a Bessel function
	of the first kind and <math>H^{(1)}_\nu \,</math>		of the first kind and <math>H^{(1)}_\nu \,</math>
	denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). ~~Note that the first term represents~~		denotes a Hankel functions of order <math>\nu</math> (see [http://en.wikipedia.org/wiki/Bessel_function Bessel functions] for more information ). The choice of which
	~~incoming waves and the second term represents the scattered wave.~~ The choice of which
	Hankel function depends on whether we have positive or negative exponential time dependence.		Hankel function depends on whether we have positive or negative exponential time dependence.
	The potential outside the circle can therefore be written as		The potential outside the circle can therefore be written as
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	</math>		</math>
	</center>		</center>
			Note that the first term represents the incident wave
			(incoming wave) and the second term represents the scattered wave. In other words, we say that <math>\phi = \phi^{\mathrm{I}}+\phi^{\mathrm{S}} \,</math>, where
			<center>
			<math>
			\phi^{\mathrm{I}} (r,\theta)= \sum_{\nu = -
			\infty}^{\infty} D_{\nu} J_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta},
			</math>
			</center>
			<center>
			<math>
			\phi^{\mathrm{S}} (r,\theta)= \sum_{\nu = -
			\infty}^{\infty} E_{\nu} H^{(1)}_\nu (k
			r) \mathrm{e}^{\mathrm{i} \nu \theta}.
			</math>
			</center>
	We consider the case where we have Neumann boundary condition on the circle. Therefore		We consider the case where we have Neumann boundary condition on the circle. Therefore
	we have <math>\partial_n\phi=0</math> at <math>r=a \,</math>. This allows us to obtain		we have <math>\partial_n\phi=0</math> at <math>r=a \,</math>. This allows us to obtain
	<center>		<center>
	<math>		<math>
	E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)}.		E_{\nu} = - \frac{D_{\nu} J^{\prime}_\nu (k a)}{ H^{(1)\prime}_\nu (ka)},
	</math>		</math>
	</center>		</center>
			which tells us that providing we know the form of the incident wave, we can compute the <math>D_\nu \,</math> coefficients and compute the potential throughout the circle. It is possible to equate plane waves and cylindrical waves using the [http://en.wikipedia.org/wiki/Jacobi%E2%80%93Anger_expansion Jacobi-Anger Identity].

	== Solution for an arbitrary scatterer ==		== Solution for an arbitrary scatterer ==

Mike smith: /* Solution for a Circle */

2010-06-15T22:06:18Z

Solution for a Circle

← Older revision		Revision as of 22:06, 15 June 2010
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	We consider the case where we have Neumann boundary condition on the circle. Therefore		We consider the case where we have Neumann boundary condition on the circle. Therefore
	we have <math>\partial_n\phi=0</math> at <math>r=a \,</math>. ~~We can therefore~~ obtain		we have <math>\partial_n\phi=0</math> at <math>r=a \,</math>. This allows us to obtain
	<center>		<center>
	<math>		<math>

Mike smith: /* Solution for an arbitrary scatterer */

2010-06-15T22:04:09Z

Solution for an arbitrary scatterer

← Older revision		Revision as of 22:04, 15 June 2010
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	</center>		</center>

	We solve this equation by the Galerkin method using a Fourier series as the basis. We parameterise the curve <math>\partial\Omega</math> by <math>\mathbf{s}(\gamma)</math> where <math>-\pi \leq \gamma \leq \pi</math>. We write the potential on the boundary ~~are~~		We solve this equation by the Galerkin method using a Fourier series as the basis. We parameterise the curve <math>\partial\Omega</math> by <math>\mathbf{s}(\gamma)</math> where <math>-\pi \leq \gamma \leq \pi</math>. We write the potential on the boundary as
	<center>		<center>
	<math>		<math>

Mike smith: /* Solution for an arbitrary scatterer */

2010-06-15T22:03:47Z

Solution for an arbitrary scatterer

← Older revision		Revision as of 22:03, 15 June 2010
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	<math>G(\|\mathbf{x} - \mathbf{x}^\prime)\|) = \frac{i}{4} H_{0}^{(1)}(k \|\mathbf{x} - \mathbf{x}^\prime)\|).\,</math>		<math>G(\|\mathbf{x} - \mathbf{x}^\prime)\|) = \frac{i}{4} H_{0}^{(1)}(k \|\mathbf{x} - \mathbf{x}^\prime)\|).\,</math>

	It we consider again Neumann boundary conditions <math>\partial_{n^\prime}\phi(\mathbf{x}) = 0</math> and restrict ourselves to the boundary we obtain the following integral equation		If we consider again Neumann boundary conditions <math>\partial_{n^\prime}\phi(\mathbf{x}) = 0</math> and restrict ourselves to the boundary we obtain the following integral equation
	<center>		<center>
	<math>		<math>
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	</math>		</math>
	</center>		</center>

	==External Links==		==External Links==
	[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]		[http://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz Equation on Wikipedia]

	[[Category:Linear Water-Wave Theory]]		[[Category:Linear Water-Wave Theory]]

Mike smith at 00:29, 10 June 2010

2010-06-10T00:29:31Z

Show changes

Meylan: /* Solution for a Circle */

2010-06-09T08:08:17Z

Solution for a Circle

← Older revision		Revision as of 08:08, 9 June 2010
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	where		where
	[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind		[http://en.wikipedia.org/wiki/Bessel_function Bessel functions] of the first kind
	and Hankel functions of order <math>\nu</math>.		and Hankel functions of order <math>\nu</math>. Note that the first represent
			incoming waves and the second represent the scattered wave. The choice of which
			Hankel function depends on whether we have positive or negative exponential time dependence.
	The potential outside the circle can therefore be written as		The potential outside the circle can therefore be written as
	<center>		<center>