Difference between revisions of "Template:Separation of variables in cylindrical coordinates"

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== Separation for Cylindrical Coordinates ==
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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
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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>

Latest revision as of 09:00, 27 September 2008

Separation for Cylindrical Coordinates

We now separate variables, noting that since the problem has circular symmetry we can write the potential as

[math]\displaystyle{ \phi(r,\theta,z)=\frac{\cos k(z+h)}{\cos kh}\sum_{n=-\infty}^{\infty}\rho_{n}(r)e^{i n \theta} }[/math]

We now solve for the function [math]\displaystyle{ \rho_{n}(r) }[/math]. Using Laplace's equation in polar coordinates we obtain

[math]\displaystyle{ \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}}+k^{2}\right) \rho_{n}=0. }[/math]

We can convert this equation to the standard form by substituting [math]\displaystyle{ y=k r }[/math] (provided that [math]\displaystyle{ \mu\neq 0 }[/math]to obtain

[math]\displaystyle{ 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]

The solution of this equation is a linear combination of the modified Bessel functions of order [math]\displaystyle{ n }[/math], [math]\displaystyle{ I_{n}(y) }[/math] and [math]\displaystyle{ K_{n}(y) }[/math] (Abramowitz and Stegun 1964).

Therefore

[math]\displaystyle{ \rho_n(r) = C_1 I_{n}(kr) + C_2 K_{n}(kr)\, }[/math]

for some constants [math]\displaystyle{ C_1 }[/math] and [math]\displaystyle{ C_2 }[/math]