$(u,\mathrm{A}v)=\int_{-1}^{1}u^*(x)\frac{d}{dx}[(x^2-1)\frac{d}{dx}v(x)]dx=(\mathrm{A}u,v)$

Question

I want to demonstrate that the following operator is a self-adjoint operator in $L^2(-1,1)$ $$(u,\mathrm{A}v)$$ $$\mathrm{A} =u^*(x)\frac{d}{dx}(x^2-1)\frac{d}{dx}v(x)$$

$$w=a+ib$$ $$w^*=a-ib$$

I want the proof of the following equality step by step. $$(u,\mathrm{A}v)=\int_{-1}^{1}u^*(x)\frac{d}{dx}[(x^2-1)\frac{d}{dx}v(x)]dx=(\mathrm{A}u,v)$$ I'm stuck at the beginning. $$(u,\mathrm{A}v)=\int_{-1}^{1}u^*(x)\frac{d}{dx}[(x^2-1)\frac{d}{dx}v(x)]dx=\int_{-1}^{1}u^*(x)\, 2x\, \,v''(x)\, dx$$

Thank you so much for the generous help.

Answer 1

Integrating by parts, with $U=u^*$ and $V=(x^2-1)\frac{dv(x)}{dx}$ reveals

$$\begin{align} (u,Av)&=\int_{-1}^{1}u^*(x)\frac{d}{dx}\left((x^2-1)\frac{d}{dx}v(x)\right)dx\\\\ &=-\int_{-1}^1 \left(\frac{d}{dx}u^*(x)\right)\left((x^2-1)\frac{dv(x)}{dx}\right)\,dx\\\\ &=-\int_{-1}^1 \left((x^2-1)\frac{d}{dx}u^*(x)\right)\left(\frac{dv(x)}{dx}\right)\,dx\tag 1\\\\ \end{align}$$

Then, integrating by parts the right-hand side of $(1)$ with U=$(x^2-1)\frac{d}{dx}u^*(x)$ and $V=v(x)$ yields

$$(u,Av)=\int_{-1}^1 v(x)\frac{d}{dx}\left((x^2-1)\frac{d}{dx}u(x)\right)dx=(Au,v)$$

as was to be shown!