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Linear Differential Equation,Legendre's Equation, sturm liouville - eigenvalues/eigenfunction

Consider the linear differential operator: $$ L = \frac{1}{4}(1+x^2)\frac{d^2}{dx^2}+\frac{1}{2}x(1+x^2)\frac{d}{dx}+a $$ acting on functions defined in $-1 \le x \le 1$ and vanishing at the endpoints of the interval.

(a) Is $L$ Hermitian?
(b) Determine the weight function necessary to make $L$ Hermitian.
(c) Show explicitly that $$ \int_{-1}^{1}V^*(x)W(x)Lu(x)dx = \int_{-1}^{1}(LV)^*W(x)u(x)dx $$ and thereby determine the condition on 'a'.
(d) Change variables to $$ x= \tan\left(\frac{\Theta}{2}\right) $$ Find $2$ even eigenfunctions $f_1(x)$ and $f_2(x)$ of the diferential equation $$ Lu=\lambda u. $$

It's my first time posting question, so wasn't sure how to type the differential equation.

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  • $\begingroup$ Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. $\endgroup$ – José Carlos Santos Nov 25 '18 at 14:21
  • $\begingroup$ I know, I am new to this and still practicing. The equations weren't coming the way I wanted them to. $\endgroup$ – cisko Nov 25 '18 at 15:11
  • $\begingroup$ You can take a look at how I edited your question. That will help you learn MathJax for the next time. $\endgroup$ – DisintegratingByParts Nov 25 '18 at 20:33
  • $\begingroup$ Thank you so much $\endgroup$ – cisko Nov 26 '18 at 21:02
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Note that this is not Legendre's equation; Legendre's equation is singular at $x=\pm 1$ because it has $1-x^2$ instead of $1+x^2$. Your operator may be written in selfadjoint form as $$ Lf = \frac{1}{4}((1+x^2)f')'+af $$

This operator is symmetric on $[-1,1]$ with respect to weight function $1$, assuming you impose the stated conditions $f(-1)=f(1)=0$. That is, if $f,g$ vanish at the endpoints, then $$ \int_{-1}^{1}\{(Lf)g-fLg\}dx =0. $$ This follows from the fact that $f,g$ vanish at $\pm 1$, and from the Lagrange identity: $$ (Lf)g-f(Lg) = \frac{d}{dx}\left(\frac{1}{4}(1+x^2)(f'g-fg')\right) $$ There is a standard trick to get rid of the weight, such as $1+x^2$. In $$ Lf = \frac{1}{4}((1+x^2)f')'+af $$ let $$ f(x) = g(\int \frac{1}{1+x^2}dx)=g(\tan^{-1}x). $$ Then $$ Lf = \frac{1}{4}\frac{1}{1+x^2}g''(x)+ag $$ Solving $Lf=\lambda f$ requires solving $$ g'' = 4(\lambda -a)(1+x^2)g. $$ It is natural to try $g(x)=\sum_{n=0}^{\infty}a_n x^{2n}$: $$ \sum_{n=1}^{\infty}(2n)(2n-1)a_{n}x^{2n-2}=4(\lambda-a)\sum_{n=0}^{\infty}a_n x^{2n}+4\sum_{n=0}^{\infty}a_n x^{2n+2}. $$ This gives a 3-term recursion relation, which leads to 2 independent solutions, both of which are even.

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