Legendre differential equation

The Legendre differential equation is the second order ordinary differential equation (ODE) which can be written as:
 * $$(1-x^2)d^2y/dx^2-2xdy/dx+l(l+1)y=0\,$$

which when rearranged to:
 * $${d \over dx }[(1 - x^2){dy \over dx }]+l(l+1)y=0\,$$ is called Legendre differential equation of order $$l$$, where the quantity $$l$$ is a constant.


 * $$Ly=0\,$$

where $$L\,$$ is the Legendre operator:
 * $$L={d \over dx }[(1 - x^2){d \over dx }]+l(l+1)\,$$

In principle, $$l$$ can be any number, but it is usually an integer.

We use the Frobenius method to solve the equation in the region $$|x|\leq 1$$. We start by setting the parameter p in Frobenius method zero.


 * $$y= \sum_{n=0}^{\infty}a_n x^n$$,
 * $$y' = \sum_{n=0}^{\infty}n a_n x^{n-1}$$,
 * $$y'' = \sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}$$.

Substituting these terms into the original equation, one obtains


 * {| border="0" cellpadding="0" cellspacing="0"

- 2x\sum_{n=0}^{\infty}n a_n x^{n-1} + l(l+1)\sum_{n=0}^{\infty}a_n x^n$$ + \sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}$$ + \sum_{n=-2}^{\infty}(n+2)(n+1) a_{n+2} x^n$$
 * $$0=Ly\,$$
 * $$ = \big(1-x^2)y'' -2xy'+l(l+1)y$$
 * $$=(1-x^2)\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}
 * $$=(1-x^2)\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}
 * $$=(1-x^2)\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}
 * $$=(1-x^2)\sum_{n=0}^{\infty}n(n-1) a_n x^{n-2}
 * $$=\sum_{n=0}^{\infty}\left[-n(n-1)-2n+l(l+1)\right] a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[-n(n-1)-2n+l(l+1)\right] a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[-n(n-1)-2n+l(l+1)\right] a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[l^2-n^2+l-n\right]a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[l^2-n^2+l-n\right]a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[l^2-n^2+l-n\right]a_n x^n
 * $$=\sum_{n=0}^{\infty}\left[(l+n+1)(l-n)a_n + (n+2)(n+1)a_{n+2}\right]x^n$$.
 * }
 * $$=\sum_{n=0}^{\infty}\left[(l+n+1)(l-n)a_n + (n+2)(n+1)a_{n+2}\right]x^n$$.
 * }
 * }

Thus


 * $$ a_2 = -{l(l+1) \over 2} a_0$$,

and in general,


 * $$ a_{n+2} = -{(l+n+1)(l-n) \over (n+2)(n+1)}a_n $$.

This series converges when


 * $$\lim_{n \to \infty}\left|{a_{n+2}x^{n+2} \over a_nx^n}\right|<1$$.

Therefore the series solution has to be cut by choosing:


 * $$ n =l \mbox { or } n = -(l+1)\,$$.

The series cut in specific integers $$l$$ and $$l+1$$ produce polynomials called Legendre polynomials.