Legendre Differential equation (#1) : A friendly introduction

February 26, 2017February 26, 2017 / 153armstrong

In this series of posts about Legendre differential equation, I would like to de-construct the differential equation down to the very bones. The motivation for this series is to put all that I know about the LDE in one place and also maybe help someone as a result.

The Legendre differential equation is the following:

$(1-x^2)y^{''} -2xy^{'} + l(l+1)y = 0$

where $y^{'} = \frac{dy}{dx}$ and $y^{''} = \frac{d^{2}y}{dx}$

We will find solutions for this differential equation using the power series expansion i.e
$y = \sum\limits_{n=0}^{\infty} a_n x^n$

$y^{'} = \sum\limits_{n=0}^{\infty} na_n x^{n-1}$

$y^{''} = \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

We will plug in these expressions for the derivatives into the differential equation.

$l(l+1)y = l(l+1)\sum\limits_{n=0}^{\infty} a_n x^n$ – (i)

$-2xy^{'} = -2\sum\limits_{n=0}^{\infty} na_n x^{n}$ – (ii)

$(1-x^2)y^{''} = (1-x^2)\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

$= \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2} - \sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n}$ – (iii)

** Note: Begin

$\sum\limits_{n=0}^{\infty} n(n-1)a_n x^{n-2}$

Let’s take $\lambda = n-2$ .
As n -> $0$ . , $\lambda$ -> $-2$ .
As n -> $\infty$ , $\lambda$ -> $\infty$ .

$\sum\limits_{\lambda = -2}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}$

$= 0 + 0 + \sum\limits_{\lambda = 0}^{\infty} (\lambda+2)(\lambda+1)a_n x^{\lambda}$

Again performing a change of variables from $\lambda$ to n.

$= \sum\limits_{n= 0}^{\infty} (n+2)(n+1)a_n x^{n}$

** Note: End

(iii) can now be written as follows.

$\sum\limits_{n=0}^{\infty} x^n \left((n+1)(n+2)a_{n+2} - n(n-1)a_n \right)$ – (iv)

(i)+(ii)+(iv).

$\sum\limits_{n=0}^{\infty} x^n \left((n+2)(n+1)a_{n+2} - (l(l+1)-n(n+1))a_n \right)$

x = 0 is a trivial solution and therefore we get the indicial equation:

$(n+2)(n+1)a_{n+2} - (l(l+1)-n(n+1))a_n = 0$

$(n+2)(n+1)a_{n+2} = (l^2 - n^2 + l - n)a_n = 0$

$(n+2)(n+1)a_{n+2} = ((l-n)(l+n)+ l - n)a_n = 0$

$(n+2)(n+1)a_{n+2} = (l-n)(l+n+1)a_n = 0$

We get the following recursion relation on the coefficients of the power series expansion.

$a_{n+2} = a_n \frac{(l+n+1)(l-n)}{(n+1)(n+2)}$

Next post: What do these coefficients mean ?

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