Mar 23, 2012 #1 Poirot1 245 0 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.
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.
Mar 23, 2012 #2 chisigma Gold Member MHB 1,628 0 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 $\chi$ $\sigma$
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 $\chi$ $\sigma$