# solution of Lagrange differential equation are square integrable

I was recently posing myself this question. Given the Lagrange DE $$[(1-x^2)u']'+\lambda u=0,$$ where $\lambda$ is a real parameter and $x\in[-1,1]$, it is well known that, if $\lambda=n(n+1)$ for some integer $n$, then we get the Legendre polynomials as solutions of the DE.

However, if we consider a general parameter $\lambda$ and we consider the solution $u=u_\lambda$ which solves the DE with that particular parameter, then it is true that $u\in L^2([-1,1])$? Moreover, do we still have some boundary conditions like $$\lim_{x\to\pm 1}(1-x^2)u(x)=0?$$

Thanks for your attention. Best regards,

-Guido-

-
This is called the Legendre DE, not the Lagrange DE. And the space of solutions is two-dimensional, so even for integer $n$ there is a second fundamental solution which is not a polynomial. – Lukas Geyer Nov 16 '12 at 16:15
You may get the integral representions of the two type of the Legendre functions in such as people.math.sfu.ca/~cbm/aands/page_335.htm. – doraemonpaul Nov 21 '12 at 9:24

I am not sure about the behavior at $\pm 1$, but the solutions to these differential equations are called Legendre functions of the first and second kind. There is a lot of literature about them, and they are implemented in Mathematica and Maple.