Template:Separation of variables in cylindrical coordinates - Revision history

Meylan at 09:00, 27 September 2008

2008-09-27T09:00:25Z

← Older revision		Revision as of 09:00, 27 September 2008
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	== Separation for Cylindrical Coordinates ==		=== Separation for Cylindrical Coordinates ===

	We now separate variables, noting that since the problem has		We now separate variables, noting that since the problem has

Meylan: /* Separation for Cylindrical Coordinates */

2008-08-26T05:54:50Z

Separation for Cylindrical Coordinates

← Older revision		Revision as of 05:54, 26 August 2008
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	\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}		\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
	\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(		\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
	\frac{n^{2}}{r^{2}}+k{2}\right) \rho_{n}=0.		\frac{n^{2}}{r^{2}}+k^{2}\right) \rho_{n}=0.
	</math>		</math>
	</center>		</center>

Meylan: /* Separation for Cylindrical Coordinates */

2008-08-26T05:52:35Z

Separation for Cylindrical Coordinates

← Older revision		Revision as of 05:52, 26 August 2008
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	<center>		<center>
	<math>		<math>
	\rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)		\rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)\,
	</math>		</math>
	</center>		</center>
	for some constants <math>C_1</math> and <math>C_2</math>		for some constants <math>C_1</math> and <math>C_2</math>

Meylan: /* Separation for Cylindrical Coordinates */

2008-08-26T05:52:18Z

Separation for Cylindrical Coordinates

← Older revision		Revision as of 05:52, 26 August 2008
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	<center>		<center>
	<math>		<math>
	\phi(r,\theta,z)=Z(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}		\phi(r,\theta,z)=\frac{\cos k(z+h)}{\cos kh}\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}
	</math>		</math>
	</center>		</center>
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	\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}		\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
	\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(		\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
	\frac{n^{2}}{r^{2}}+~~\mu^~~{2}\right) \rho_{n}=0		\frac{n^{2}}{r^{2}}+k{2}\right) \rho_{n}=0.
	</math>		</math>
	</center>		</center>
	~~where <math>\mu</math> is <math>k_{m}</math> or~~		We can convert this equation to the
	~~<math>\kappa_{m},</math> depending on whether <math>r</math> is~~		standard form by substituting <math>y=k r</math> (provided that
	~~greater or less than <math>a</math>.~~ We can convert this equation to the
	standard form by substituting <math>y=~~\mu~~ r</math> (provided that
	<math>\mu\neq 0</math>to obtain		<math>\mu\neq 0</math>to obtain
	<center>		<center>
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	modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and		modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
	<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]).		<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]).

			Therefore
			<center>
			<math>
			\rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)
			</math>
			</center>
			for some constants <math>C_1</math> and <math>C_2</math>

Meylan at 05:20, 26 August 2008

2008-08-26T05:20:30Z

← Older revision		Revision as of 05:20, 26 August 2008
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			== Separation for Cylindrical Coordinates ==

	We now separate variables, noting that since the problem has		We now separate variables, noting that since the problem has
	circular symmetry we can write the potential as		circular symmetry we can write the potential as
	<center>		<center>
	<math>		<math>
	\phi(r,\theta,z)=~~\zeta~~(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}		\phi(r,\theta,z)=Z(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}
	</math>		</math>
	</center>		</center>

Meylan at 23:49, 8 August 2008

2008-08-08T23:49:06Z

New page

We now separate variables, noting that since the problem has
circular symmetry we can write the potential as
<center>
<math>
\phi(r,\theta,z)=\zeta(z)\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta}
</math>
</center>
We now solve for the function <math>\rho_{n}(r)</math>.
Using Laplace's equation in polar coordinates we obtain
<center>
<math>
\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}r^{2}}+\frac{1}{r}
\frac{\mathrm{d}\rho_{n}}{\mathrm{d}r}-\left(
\frac{n^{2}}{r^{2}}+\mu^{2}\right) \rho_{n}=0
</math>
</center>
where <math>\mu</math> is <math>k_{m}</math> or
<math>\kappa_{m},</math> depending on whether <math>r</math> is
greater or less than <math>a</math>. We can convert this equation to the
standard form by substituting <math>y=\mu r</math> (provided that
<math>\mu\neq 0</math>to obtain
<center>
<math>
y^{2}\frac{\mathrm{d}^{2}\rho_{n}}{\mathrm{d}y^{2}}+y\frac{\mathrm{d}\rho_{n}
}{\rm{d}y}-(n^{2}+y^{2})\rho_{n}=0
</math>
</center>
The solution of this equation is a linear combination of the
modified Bessel functions of order <math>n</math>, <math>I_{n}(y)</math> and
<math>K_{n}(y)</math> ([[Abramowitz and Stegun 1964]]).