Bottom Mounted Cylinder - Revision history

Meylan at 09:33, 20 October 2009

2009-10-20T09:33:12Z

← Older revision		Revision as of 09:33, 20 October 2009
Line 1:		Line 1:
			{{incomplete pages}}

	= Introduction =		= Introduction =

Meylan: /* Boundary condition at $r = a$ */

2008-08-25T22:30:31Z

Boundary condition at <math>r = a</math>

← Older revision		Revision as of 22:30, 25 August 2008
Line 46:		Line 46:
	equation we obtain		equation we obtain
	<center>		<center>
	<math> D_{\nu} I^{\prime}_\nu (k_0 a) + E_{\nu} K^{\prime}_\nu (k_0 a) = 0 </math>		<math> D_{\nu} I^{\prime}_\nu (k_0 a) + E_{\nu} K^{\prime}_\nu (k_0 a) = 0, </math>
	</center>		</center>
	so that		so that
	<center>		<center>
	<math> E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)}</math>		<math> E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)},</math>
	</center>		</center>
	and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>		and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>
Line 65:		Line 65:
	\phi (r,\theta) = \sum_{\nu = -		\phi (r,\theta) = \sum_{\nu = -
	\infty}^{\infty} \left[ I_\nu (k_0 r) - \frac{ I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)} K_\nu (k_0		\infty}^{\infty} \left[ I_\nu (k_0 r) - \frac{ I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)} K_\nu (k_0
	r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},		r) \right] \mathrm{e}^{\mathrm{i} \nu \theta}.
	</math>		</math>
	</center>		</center>
	[[Category:Problems with Cylindrical Symmetry]]		[[Category:Problems with Cylindrical Symmetry]]

Meylan at 22:29, 25 August 2008

2008-08-25T22:29:19Z

← Older revision		Revision as of 22:29, 25 August 2008
Line 29:		Line 29:
	{{separation of variables for the r and theta coordinates}}		{{separation of variables for the r and theta coordinates}}

	~~</center>~~
	The solution is		The solution is
	<center>		<center>

Meylan: /* Equation for a Cylinder */

2008-08-25T22:27:45Z

Equation for a Cylinder

← Older revision		Revision as of 22:27, 25 August 2008
Line 22:		Line 22:
	A cylinder of radius <math>a</math> (which we will suppose is at the centre		A cylinder of radius <math>a</math> (which we will suppose is at the centre
	of our cylindrical coordinate system) occupies the		of our cylindrical coordinate system) occupies the
	region <math> r>a </math>.		region <math> r<a </math>. On the boundary of the cylinder we assume that
			the normal derivative vanishes.

	We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,		We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,

Meylan: /* Boundary condition at $r = a$ */

2008-08-25T09:14:22Z

Boundary condition at <math>r = a</math>

← Older revision		Revision as of 09:14, 25 August 2008
Line 56:		Line 56:
	<center>		<center>
	<math>		<math>
	\phi^{i} (r,\theta) = e^{kx} = \sum_{\nu = -		\phi^{i} (r,\theta) = e^{k_0 x} = \sum_{\nu = -
	\infty}^{\infty} I_\nu (k_0 r) \mathrm{e}^{\mathrm{i} \nu \theta},		\infty}^{\infty} I_\nu (k_0 r) \mathrm{e}^{\mathrm{i} \nu \theta},
	</math>		</math>

Meylan: /* Boundary condition at $r = a$ */

2008-08-25T09:13:52Z

Boundary condition at <math>r = a</math>

← Older revision		Revision as of 09:13, 25 August 2008
Line 46:		Line 46:
	equation we obtain		equation we obtain
	<center>		<center>
	<math> D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 </math>		<math> D_{\nu} I^{\prime}_\nu (k_0 a) + E_{\nu} K^{\prime}_\nu (k_0 a) = 0 </math>
	</center>		</center>
	so that		so that
	<center>		<center>
	<math> E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k a)}{K^{\prime}_\nu (k a)}</math>		<math> E_{\nu} = -\frac{D_{\nu} I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)}</math>
	</center>		</center>
	and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>		and we often assume that the incident wave is a plane wave travelling in the positive <math>x</math>
Line 57:		Line 57:
	<math>		<math>
	\phi^{i} (r,\theta) = e^{kx} = \sum_{\nu = -		\phi^{i} (r,\theta) = e^{kx} = \sum_{\nu = -
	\infty}^{\infty} I_\nu (k r) \mathrm{e}^{\mathrm{i} \nu \theta},		\infty}^{\infty} I_\nu (k_0 r) \mathrm{e}^{\mathrm{i} \nu \theta},
	</math>		</math>
	</center>		</center>
Line 64:		Line 64:
	<math>		<math>
	\phi (r,\theta) = \sum_{\nu = -		\phi (r,\theta) = \sum_{\nu = -
	\infty}^{\infty} \left[ I_\nu (k r) - \frac{ I^{\prime}_\nu (k a)}{K^{\prime}_\nu (k a)} K_\nu (k		\infty}^{\infty} \left[ I_\nu (k_0 r) - \frac{ I^{\prime}_\nu (k_0 a)}{K^{\prime}_\nu (k_0 a)} K_\nu (k_0
	r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},		r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},
	</math>		</math>
	</center>		</center>
	[[Category:Problems with Cylindrical Symmetry]]		[[Category:Problems with Cylindrical Symmetry]]

Meylan: /* Boundary condition at $r = a$ */

2008-08-25T09:12:08Z

Boundary condition at <math>r = a</math>

← Older revision		Revision as of 09:12, 25 August 2008
Line 43:		Line 43:
	= Boundary condition at <math>r = a</math> =		= Boundary condition at <math>r = a</math> =

	The boundary condition at <math>r = a</math> is ~~now~~ substituted into the above		The boundary condition at <math>r = a</math> is that the normal derivative vanishes. If this is substituted into the above
	equation ~~to give~~		equation we obtain
	<center>		<center>
	<math> D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 </math>		<math> D_{\nu} I^{\prime}_\nu (k a) + E_{\nu} K^{\prime}_\nu (k a) = 0 </math>

Meylan: /* Equation for a Cylinder */

2008-08-25T09:09:40Z

Equation for a Cylinder

← Older revision		Revision as of 09:09, 25 August 2008
Line 27:		Line 27:

	{{separation of variables for the r and theta coordinates}}		{{separation of variables for the r and theta coordinates}}
	~~<center>~~
	~~<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial~~
	~~\phi}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2~~
	~~\phi}{\partial \theta^2} - k^2 \phi = 0, \quad r>a,~~
	~~</math>~~
	</center>		</center>
	The ~~separate~~ solution is		The solution is
	<center>		<center>
	<math>		<math>
	\phi (r,\theta) = \sum_{\nu = -		\phi (r,\theta) = \sum_{\nu = -
	\infty}^{\infty} \left[ D_{\nu} I_\nu (k r) + E_{\nu} K_\nu (k		\infty}^{\infty} \left[ D_{\nu} I_\nu (k_0 r) + E_{\nu} K_\nu (k_0
	r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},		r) \right] \mathrm{e}^{\mathrm{i} \nu \theta},
	</math>		</math>
	</center>		</center>
	~~(this follows exactly as for [[Cylindrical Eigenfunction Expansion]]).~~ Note that in many		Note that in many
	cases a symmetry argument is used to express the complex exponentials in terms		cases a symmetry argument is used to express the complex exponentials in terms
	of sine and cosines. This follows from a symmetry in the incident potential.		of sine and cosines. This follows from a symmetry in the incident potential.

Meylan: /* Equation for a Cylinder */

2008-08-25T09:04:48Z

Equation for a Cylinder

← Older revision		Revision as of 09:04, 25 August 2008
Line 26:		Line 26:
	We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,		We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,

	{{separation of variables for the r and theta ~~directions~~}}		{{separation of variables for the r and theta coordinates}}
	<center>		<center>
	<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial		<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial

Meylan: /* Equation for a Cylinder */

2008-08-25T09:03:57Z

Equation for a Cylinder

← Older revision		Revision as of 09:03, 25 August 2008
Line 26:		Line 26:
	We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,		We can now use [[Separation of Variables]] by writing [[Helmholtz's Equation]] in cylindrical polar coordinates,

	{{separation of variables in the r and theta directions}}		{{separation of variables for the r and theta directions}}
	<center>		<center>
	<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial		<math> \frac{1}{r} \frac{\partial}{\partial r} \left( r \frac{\partial

Bottom Mounted Cylinder - Revision history

Meylan at 09:33, 20 October 2009

Meylan: /* Boundary condition at r = a */

Meylan at 22:29, 25 August 2008

Meylan: /* Equation for a Cylinder */

Meylan: /* Boundary condition at r = a */

Meylan: /* Boundary condition at r = a */

Meylan: /* Boundary condition at r = a */

Meylan: /* Equation for a Cylinder */

Meylan: /* Equation for a Cylinder */

Meylan: /* Equation for a Cylinder */

Meylan: /* Boundary condition at $r = a$ */

Meylan: /* Boundary condition at $r = a$ */

Meylan: /* Boundary condition at $r = a$ */

Meylan: /* Boundary condition at $r = a$ */