# Legendre's Equation, sturm liouville - eigenvalues/eigenfunction

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.

• Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Nov 25 '18 at 14:21
• I know, I am new to this and still practicing. The equations weren't coming the way I wanted them to. – cisko Nov 25 '18 at 15:11
• You can take a look at how I edited your question. That will help you learn MathJax for the next time. – DisintegratingByParts Nov 25 '18 at 20:33
• Thank you so much – cisko Nov 26 '18 at 21:02

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.