Consider the singular Sturm-liouville problem:
$$(1-x^2)y''-2xy'+\lambda y=((1-x^2)y')'+\lambda y=0, \quad y(0)=0, \lim_{x\to 1}y(x)<\infty $$
Show that the eigenvalues of this problem are $\lambda_{n}=2n(2n+1)(2n+2)$, $n=0,1,2,...$ and the corresponding eigenfunctions are the odd Legendre polynomials $P_{2n+1}(x)$
I'm trying to solve this Sturm-Liouville problem. But I haven't gotten a solution yet. How can I solve this problem? I need some help or a way to solve this problem. I would like to know a solution. Any help is appreciated
The Legendre differential equation is singular at $x=\pm 1$. But you are working on $[0,1)$; so your problem is regular at $0$ and singular at $x=1$. Your problem is posed as $$ Lf=\lambda f, \; 0 \le x < 1, \\ f(0)=0, \lim_{x\rightarrow 1}f(x) \mbox{ is finite.} $$ The condition at $x=1$ is only satisfied by the Legendre polynomials $P_n$, which satisfy $LP_n=n(n+1)P_n$. For all other $\lambda$, $f(x)$ is unbounded near $x=1$. The even order Legendre polynomials are not $0$ at $x=0$, but the odd order Legendre polynomials do vanish at $x=0$. That's why the eigenvalues are $\lambda_n=(2n+1)(2n+2)$ for $n=0,1,2,3,\cdots$. I'm not sure how you came up with $2n(2n+1)(2n+2)$; one of us is wrong about that.