Introduction

Legendre Equation plays a vital role in many problems of mathematical Physics and in the theory of quadratures (as applied to Numerical Integration).

DEFINITION 9.4.1   The equation

$$(1 - x^2) y^{\prime\prime} - 2 x y^\prime + p (p+1) y = 0, \; -1 < x < 1$$ (9.4.1)

where $p \in {\mathbb{R}},$ is called a LEGENDRE EQUATION of order $p.$

Equation (9.4.1) was studied by Legendre and hence the name Legendre Equation.

Equation (9.4.1) may be rewritten as

$$y^{\prime\prime} - \frac{2 x}{(1 - x^2)} y^\prime + \frac{p (p+1)}{(1 - x^2)} y = 0.$$

The functions $$\frac{ 2 x}{1 - x^2}$$ and $$\frac{p(p+1)}{1 - x^2}$$ are analytic around $x_0 = 0$ (since they have power series expressions with centre at $x_0 = 0$ and with $R = 1$ as the radius of convergence). By Theorem 9.3.1, a solution $y$ of (9.4.1) admits a power series solution (with centre at $x_0 = 0$ ) with radius of convergence $R = 1.$ Let us assume that $y = \sum\limits_{k=0}^\infty a_k x^k$ is a solution of (9.4.1). We have to find the value of $a_k$ 's. Substituting the expression for

$$y^\prime = \sum\limits_{k=0}^\infty k a_k x^{k-1} \; {\mbox{ and }} \; y^{\prime\prime} = \sum\limits_{k=0}^\infty k(k-1) a_k x^{k-2}$$

in Equation (9.4.1), we get

$$\sum_{k=0}^\infty \left\{ (k+1)(k+2) a_{k+2} + a_k (p-k)(p+k+1) \right\} x^k = 0.$$

Hence, for $k = 0, 1, 2, \ldots$

$$a_{k+2} = - \frac{(p-k)(p+k+1)}{(k+1)(k+2)} a_k.$$

It now follows that

$$\begin{array}{lll} a_2 &= - \frac{p(p+1)}{2!}a_0, & \hspace{... ..._1 \\ &= (-1)^2 \frac{p(p-2)(p+1)(p+3)}{4!} a_0, & \end{array}$$

etc. In general,

$$a_{2m} = (-1)^m \frac{p(p-2)\cdots(p-2m+2) (p+1) (p+3) \cdots (p+2m -1)}{(2m)!} a_0$$

and

$$a_{2m+1} = (-1)^m \frac{(p-1)(p-3)\cdots(p-2m+1) (p+2) (p+4) \cdots (p+2m)}{(2m+1)!} a_1.$$

It turns out that both $a_0$ and $a_1$ are arbitrary. So, by choosing $a_0 = 1, a_1 = 0$ and $a_0 = 0, a_1 = 1$ in the above expressions, we have the following two solutions of the Legendre Equation (9.4.1), namely,

$$y_1 = 1 - \frac{p(p+1)}{2!} x^2 + \cdots + (-1)^m \frac{(p-2m+2)\cdots (p+2m-1)}{(2m)!} x^{2m} + \cdots$$ (9.4.2)

and

$$y_2 = x - \frac{(p-1)(p+2)}{3!} x^3 + \cdots + (-1)^m \frac{(p-2m+1) \cdots (p+2m)}{(2m+1)!} x^{2m+1} + \cdots.$$ (9.4.3)

Remark 9.4.2   $y_1$ and $y_2$ are two linearly independent solutions of the Legendre Equation (9.4.1). It now follows that the general solution of (9.4.1) is

$$y = c_1 y_1 + c_2 y_2$$ (9.4.4)

where $c_1$ and $c_2$ are arbitrary real numbers.