MHB Are the Singular Points of the Chebyshev Equation Regular?

Poirot1
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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.
 
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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

$\chi$ $\sigma$
 
Thanks
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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