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[SOLVED] Singular Points

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Poirot

Banned
Feb 15, 2012
250
I have computed the singular points of Chebyshev equation to be x= 1, -1. What is the best way to find whether they are regular? Thanks.
 

chisigma

Well-known member
Feb 13, 2012
1,704
The Chebisheff DE is...

$\displaystyle y^{\ ''} - \frac{x}{1-x^{2}}\ y^{\ '} + \frac{\alpha^{2}}{1-x^{2}}\ y= y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y=0$ (1)

If $x_{0}$ is a singularity of p(x) and q(x) and both the limits...

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})\ p(x)$

$\displaystyle \lim_{x \rightarrow x_{0}} (x-x_{0})^{2}\ q(x)$ (2)

... exist finite, then $x_{0}$ is a regular singular point. You can verify that $x_{0}=1$ and $x_{0}=-1$ are both regular singular points...

Kind regards

$\chi$ $\sigma$
 
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Poirot

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Feb 15, 2012
250
Thanks