Legendre's Equation

$$(1-x^2)y'' -2xy' + p(p+1)y = 0, \tag{11}$$

Is called the Legendre's Differential equation of order $p$ .
Notice that both $P$ and $Q$ are analytic functions in the neighborhood of $x_{0} = 0$ .
Therefore, for $- 1 < x < 1$ , we can expand and write a series solution as $$y = \sum a_{m}x^m$$
Substituting and evaluating coefficients, we get

\begin{aligned} a_{2} & = - \frac{(p) (p + 1)}{2} a_{0} \\ a_{3} & = - \frac{(p - 1) (p + 2)}{6} a_{1} \end{aligned}

and the general coefficient of $x^{m}$ will give recurrence $$a_{m+2} = - \frac{(p+m)(p+m+1)}{(m+2)(m+1)}a_{m}$$
Reducing it to order of $a_{0}$ , $$\begin{align} a_{2m} &= (-1)^{m} \frac{(p -2m + 2)\dots(p-4)(p-2)p(p+1)(p+3)(p+5)\dots(p+2m-1)}{(2m)!}a_{0} \ \ \ \ a_{2m+1} &= (-1)^{m} \frac{(p-2m+1)\dots(p-3)(p-1)(p+2)(p+4)\dots(p+2m)}{(2m+1)!}a_{1} \end{align}$$
And we finally have the solution as $$\begin{align} y &= \sum a_{m}x^m \ \ &= a_{0}\left( 1-\frac{(p)(p+1)}{2!}x^2 + \frac{(p-2)p(p+1)(p+3)}{4!}x^4 \dots + a_{2m}x^{2m} \right) \ &+ a_{1}\left( x - \frac{(p-1)(p+2)}{3!}x^3 + \frac{(p-3)(p-1)(p+2)(p+4)}{5!} x^5 \dots + a_{2m+1}x^{2m+1}\right) \ \ &= a_{0}y_{1}(x) + a_{1}y_{2}(x) \end{align}$$
Notice that for $p = n$ , a non-negative integer coefficient, the coefficient $a_{2 m} = 0$ (provided $n$ is even) for $2 m > n$ and $a_{2 m + 1} = 0$ (provided $n$ is odd) for $2 m + 1 > n$
In the former case, $y_{1}$ is polynomial of degree $n$ , $y_{2}$ is infinite series, and it is the opposite in the latter.

Thus solution is always going to be of form of one polynomial and one infinite series.

Choosing a specific value for constants ( $a_{0}$ when $n$ is even and $a_{1}$ when $n$ is odd) corresponding to polynomial solutions, we get the solutions of Legendre's equation as polynomials known as "Legendre's Polynomials", $P_{n} (x)$

Legendre's Polynomials

Definition

A polynomial solution denoted by $P_{n} (x)$ of degree $n$ of $(11)$ which satisfies $P_{n} (1) = 1$ is called the Legendre polynomial of degree $n$ .

A few Legendre's Polynomials are as follows $$\begin{align} P_{0}(x) &= 1, & P_{1}(x) &= x \ P_{2}(x) &= \frac{1}{2}(3x^2 - 1), & P_{3}(x) &= \frac{1}{2}(5x^3 - 3x) \ \ P_{4}(x) &= \frac{3}{8}\left( 1-10x^2 + \frac{35}{3}x^4 \right)\ \ P_{n}(x) &= \frac{1}{2^nn!} \frac{d^n}{dx^n}(x^2-1)^n \
\end{align}$$
In general, we have the recurrence relation $$(n+1)P_{n+1}(x) = (2n+1)xP_{n}(x) - nP_{n-1}(x)$$

Properties of Legendre's polynomials

Generating function: The function $G (t, x)$ given by $$G(t,x) = \frac{1}{\sqrt{ 1 - 2xt + t^2 }}$$is called the generating function of the polynomials. It can be shown that for small $t$ , $$\frac{1}{\sqrt{ 1-2xt + t^2 }} = \sum_{n=0}^\infty P_{n}(x)t^n$$
Orthogonality : $$\int_{-1}^1 P_{m}(x)P_{n}(x)dx = \begin{cases}0, &\text{if } m \neq n \ \ \frac{2}{2n+1}, &\text{if }m=n\end{cases}$$
Fourier-Legendre Series : By using orthogonality, any piecewise continuous function in $- 1 < x < 1$ can be expressed in terms of Legendre's Polynomials as $$f(x) \sim \sum_{n=0}^\infty c_{n}P_{n}(x)$$where $$c_{n} = \frac{2n+ {1}}{2}\int_{-1}^1 f(x)P_{n}(x)dx$$