<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/cb5f877b-6eaf-432e-be39-2e1361e0de6c/Legendres_equation.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/cb5f877b-6eaf-432e-be39-2e1361e0de6c/Legendres_equation.png" width="40px" /> Legendre’s equation is given by
$$ (1-x^2)\frac{\text d^2 y}{\text d x^2} -2x \frac{\text dy}{\text dx}+l(l+1)y=0 $$
where $y(x)$ is defined in the range $-1\le x \le 1$
</aside>
We assume a solution of the form
$$ \begin{aligned} y&=\sum^\infin_{n=0}a_n x^n \\ y'&=\sum^\infin_{n=1}n a_n x^{n-1} \\ y''&=\sum^\infin_{n=2} n(n-1) a_n x^{n-2}
\end{aligned} $$
We substitute these into Legendre’s equation
$$ \begin{aligned} \sum^\infin_{n=2} [n(n-1) a_n x^{n-2}-n(n-1) a_n x^n]-2\sum^\infin_{n=1} na_n x^n+l(l+1) \sum^\infin_{n=0}a_n x^n=0 \end{aligned} $$
Lets write all in terms of powers of $x^n$
$$ \begin{aligned} \sum^\infin_{n=0} (n+2)(n+1) a_{n+2}x^n-\sum^\infin_{n=1}n(n-1)a_nx^n -2\sum^\infin_{n=1}na_nx^n+l(l+1) \sum^\infin_{n=0}a_n x^n=0 \end{aligned} $$
if we equate poweres of $x$ for $n=0$ and $n=1$ we get
$$ \begin{aligned} n&=0: \\ n&=1: \end{aligned} \qquad \begin{aligned} 2a_2+l(l+1)a_0&=0 \\ 6a_3-2a_1+l(l+1)a_1&=0 \end{aligned} $$
Equating coefficients for $n\ge 2$ gives
$$ (n+1)(n+2)a_{n+2}-n(n-1)a_n-2na_n+l(l+1)a_n=0 $$
which can be re-written as
$$ a_{n+2}=\frac{n(n+1)-l(l+1)}{(n+1)(n+2)} a_n $$
Thus we have a solution to the Legendre equation
$$ \begin{aligned} y=&a_0 \left [ 1-l(l+1) \frac{x^2}{2!}+(l-2)l(l+1)(l+3)\frac{x^4}{4!}+\ldots \right ] \\ &+ a_1\left [ x-(l-1)(l+2)\frac{x^3}{3!}+(l-3)(l-1)(l+2)(l+4)\frac{x^5}{5!}+\ldots \right ]
\end{aligned} $$
🗒️ Note: in most applications $x$ is in the domain $-1 \le x \le 1$
However in both cases the convergence ratio is violated
$$ \left | \frac{a_{n+2}}{a_n}x^2\right |<1 $$
The divergence of the series at $x=\pm 1$ is a limitation which can be overcome when $l$ is a non-negative integer and we find $n=l$
if $l$ is a non-negative integer then:
when $n=l, \; a_{l+1}=0\times a_l$ so $a_{l+2}=a_{l+4}=a_{l+6}=0$ and $a_{l+1}, a_{l+3} \ne 0$ unless $a_0=0$ when n $l=$ odd
Similarly $a_{l+1}, a_{l+3}\ne 0$ unless $a_1=0$ when $l=$ even
Thus when $l=$ even we get
$$ y=a_0+ a_2 x^2 + a_4 x^4 + \ldots + a_l x^l $$
and when $l=$ odd
$$ y=a_1 x+a_3 x^3+ a_5 x^5 +\ldots + a_l x^l $$
each series terminates at $a_l x^l$ as there are no terms past $n=l$
when $y(1)=1$ these solutions are known as the Legendre polynomials $(P_l(x))$
We solve using the solution to Legendre’s equation
| $l$ | $y(x)$ | Legendre Polynomial | |
|---|---|---|---|
| $0$ | $y=a_0$ | $P_0(x)=1$ | |
| $1$ | $y=a_1x$ | $P_1(x)=x$ | |
| $2$ | $y=a_0+a_2 x^2$ | $P_2(x)=\frac{3x^2-1}{2}$ | |
$P_2(x)$ was obtained from
$$ y(1)=a_0-3 a_0 x^2 =a_0(1-3)=1 $$
and hence $a_0=-\frac 12$ which gives
$$ P_2(x)=\frac{3x^2-1}{2} $$
🗒️ Note: this corresponds to solving an eigenvalue equation
$$ \hat L^2 P_l(x)=\lambda P_l(x) $$
where $\lambda=l(l+1)$ and $\hat L^2=-(1-x^2)\frac{\text d^2}{\text d x^2}+2 \frac{\text d}{\text dx}$.
Here $P_l$ are eigenfunctions of $\hat L^2$ subject to the boundary condition that they are convergent at $x=\pm 1$
The boundary conditions led to the discrete eigenvalues $l(l+1)$
This bears comparison to the equation
$$ \frac{\text d^2 y}{\text dx^2}+\lambda y=0 $$
which with boundary conditions $y(0)=y(l)=0$ posses a solution with eigenvalues $\lambda=k^2 \pi^2 /l^2$ with characteristic eigenfunctions
$$ \phi_k=\sin \left ( \frac{k\pi x}{l} \right ) $$
The operator $\frac{\text d^2}{\text dx^2}$ has eigenfunctions above and these form a orthonormal basis for expanding a general function
$$ f(x)=\sum^\infin_{k=0}C_k \sin \left ( \frac{k\pi x}{l}\right ) \qquad (0<x<l) $$
with
$$ C_k=\frac{2}{l}\int^l_0 f(x) \sin \left ( \frac{k\pi x}{l} \right )\,\text dx $$
🗒️ Note: this forms part of the general theory of linear operator in the context of ODE’s especially the Sturm-Liouville theory
We want to prove that over the interval $-1\le x \le 1$ we have