Legendre operator's eigenvalues and eigenfunctions

Question

Consider the Lengendre operator which is a Sturm-Liouville operator defined in [-1,1] give as: $$ \begin{equation} \mathcal{L} u=-\frac{d}{d x}\left[\left(1-x^{2}\right) \frac{d u}{d x}\right]=-\left(1-x^{2}\right) \frac{d^{2} u}{d x^{2}}+2 x \frac{d u}{d x} \end{equation} $$ Where $u\in \mathcal{C}^\infty [-1,1] $. I want to find the operator's eigenvalues and eigenfunctions. So I'm trying to solve next equation using Frobenius method: $$ \begin{equation} -\left(1-x^{2}\right) \frac{d^{2} u_{n}(x)}{d x^{2}}+2 x \frac{d u_{n}(x)}{d x}=\lambda_{n} u_{n}(x) \end{equation} $$ I got that $u=u_1 + u_2$, where $u_1=\sum_{k=0}^{\infty} C_k\ t^{k+1}$ and $ u_2=\sum_{k=0}^{\infty} b_k\ t^{k}+d\sum_{k=0}^{\infty} C_k\ t^{k+1}\cdot log(t)$. I've tried to sustitute in the equation but I'm not able to reach to the solution, which is $\lambda_n =n(n+1)$ and $u_n=p_n$, which are the Legendre polynomials.

Answer 1

There are $2$ independent solutions of $\mathcal{L}u=\lambda u$ for all $\lambda\in\mathbb{C}$. Take $\lambda=0$ for example. The solutions are found by solving directly: $$ \frac{d}{dx}\left((1-x^2)\frac{du}{dx}\right) = 0 $$ $$ (1-x^2)\frac{du}{dx} = C $$ $$ \frac{du}{dx} = \frac{C}{1-x^2}= \frac{C}{2}\left(\frac{1}{1-x}+\frac{1}{1+x}\right) $$ $$ u = \frac{C}{2}\ln\frac{1+x}{1-x}+D $$ The classical Legendre polynomials are uniquely determined by the requirement that the solutions remain bounded near $x=\pm 1$. Then you get solutions only when $\lambda=n(n+1)$ for $n=0,1,2,3,\cdots$, and these are the classical Legendre polynomials. $\lambda=0$ is one such value of $\lambda$, and, as you see, one solution is bounded--namely, the constant function $u=1$, which is the solution where $C=0$ and $D=1$. The solution where $C=1$ and $D=0$ is not bounded near $\pm 1$.