# Legendre differential equation

Legendre differential equation is the second order ordinary differential equation (ODE) which can be written as:

${\displaystyle (1-x^{2})d^{2}y/dx^{2}-2xdy/dx+l(l+1)y=0\,}$

which can be written as:

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

where ${\displaystyle L\,}$ is the Legendre operator:

${\displaystyle L={d \over dx}[(1-x^{2}){d \over dx}]+l(l+1)\,}$

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 ${\displaystyle |x|\leq 1}$. We start by setting the parameter p in Frobenius method zero.

${\displaystyle y=\sum _{n=0}^{\infty }a_{n}x^{n}}$,
${\displaystyle y'=\sum _{n=0}^{\infty }na_{n}x^{n-1}}$,
${\displaystyle y''=\sum _{n=0}^{\infty }n(n-1)a_{n}x^{n-2}}$.

Substituting these terms into the original equation, one obtains

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

Thus

${\displaystyle a_{2}=-{l(l+1) \over 2}a_{0}}$,

and in general,

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

This series converges when

${\displaystyle \lim _{n\to \infty }\left|{a_{n+2}x^{n+2} \over a_{n}x^{n}}\right|<1}$.

Therefore the series solution has to be cut by choosing:

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

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