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From what I understood: The series solution of an ODE is found using Frobenius Method. For the Legendre's equation:

$\displaystyle (1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y=0 $

The solution steps start with $y=x^m(a_0+a_1x+a_2x^2+....) $

The initial requirement for using Frobenius method is that $x=0$ should be a regular singularity. Singularity is defined as the situation when the coefficient of $\frac{d^2y}{dx^2}$ is zero. Regularity requires differentiability of the other two coefficients.

My doubt: Having $x=0$ is not a singularity here. So, how can we start with $y=x^m(a_0+a_1x+a_2x^2+....) $ here ?

What did I miss here ? Please advise.

square_one
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  • Fröbenius will give you the solution about the two singular regular points of the ODE, which are $x_0 = \pm 1$. The radii of convergence for both solutions are left to you to determine, but maybe setting $z = x-x_0$ can be useful here... – Dmoreno Sep 10 '14 at 15:51

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In this case, $m=0$ and Frobenius' method leads to substituting $y=\sum_{n=0}^{\infty} a_nx^n$ into the differential equation. $x=0$ is not a singular point, hence $m=0$; if $x=0$ would have been a singular point of order $m$, then $m>0$.

On one hand, this leads to one set of solutions, the well-known Legendre polynomials $y=P_n(x)$. The other, non-terminating series solutions is commonly denoted as $y=Q_n(x)$ and these are illustrated in About the Legendre differential equation.

The non-terminating series will have a convergence radius equal to the distance of $x=0$ to the nearest singular point. There are two singular points here, $x=1$ and $x=-1$, so the radius of convergence is 1, and the series will thus (at least) converge on the interval $(-1,1)$.

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Maestro13
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