How to solve Legendre's differential equation without power series assumption?

Question

Legendre's differential equation $\,(1-x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+\ell(\ell+1)y=0\, $ is usually solved in most text-books either by assuming a power series solution or by Frobenius method.Are their other ways to solve the same?

Answer 1

There is another way and after some clever guessing it will give us the Rodrigues' formula $$P_n(x) = \frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$ where $f^{(n)}$ is symbolic for the $n$th derivative of $f$

Looking at $$[(1-x^2)y']' = -n(n+1)y$$ and after having found polynomial solutions for some values of $n$, but not finding a general polynomial expression, one could try $y=F^{(n)}$; after all the factor n(n+1) might result from taking a second derivative of $x^{n+1}$, and maybe, just maybe, it will all fit.

In order to profit the most of the factor $(1-x^2)$ in subsequent evaluations, let us try $F(x) = [(x^2-1)^n]^{(n)}$ and see whether we can come close or not.

Starting with $$(x^2-1) \cdot [(x^2-1)^n]^{(1)} = (x^2-1) \cdot n(x^2-1)^{(n-1)} \cdot 2x = 2nx(x^2-1)^{n}$$ and taking left and right the $(n+1)$st derivative, we find that $$(x^2-1)[(x^2-1)^n]^{(n+2)} + \binom{n+1}{1}\cdot 2x \cdot [(x^2-1)^n]^{(n+1)} + \binom{n+1}{2}\cdot 2 \cdot [(x^2-1)^n]^{(n)} \\ = 2nx \cdot [(x^2-1)^n]^{(n+1)} + \binom{n+1}{1}\cdot 2n \cdot [(x^2-1)^n]^{(n)}$$ or $$(x^2-1)[(x^2-1)^n]^{(n+2)} + (2x(n+1)-2nx)[(x^2-1)^n]^{(n+1)} + (n(n+1) - 2n(n+1)) [(x^2-1)^n]^{(n)} = 0$$ which leads to $$(x^2-1)F''(x) + 2xF'(x) - n(n+1)F(x) = 0$$ or, equivalently $$[(1-x^2)F'(x)]' + n(n+1)F(x) = 0$$ which miraculously does not require any further adjustments, and shows that $F(x) = [(x^2-1)^n]^{(n)}$ is a solution of the Legendre differential equation; this is a multiple of the Legendre polynomial $P_n(x)$.

But, there is a drawback: the Frobenius power series method also leads to the other set of solutions, the non-polynomials $Q_n(x)$. And it is almost impossible to think of these solutions when plugging in some expression into the Legendre equation. See About the Legendre differential equation for more information on $Q_n$.