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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-

  • 0
    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.2012-11-16
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    You may get the integral representions of the two type of the Legendre functions in such as http://people.math.sfu.ca/~cbm/aands/page_335.htm.2012-11-21

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