= Dirichlet eigenfunctions of the Chebyshev differential operator
{c}
{title2=$y_n(x)=\sqrt{2/\pi}\sin(n\arccos x)$}
The <Chebyshev differential equation> $-(1-x^2)y''+xy'=\lambda y$ on $-1<x<1$, with zero endpoint values, has the <Sturm-Liouville form>
$$
-\frac d{dx}\left[\sqrt{1-x^2}\,y'\right]=\lambda\frac{y}{\sqrt{1-x^2}}.
$$
Thus $p=\sqrt{1-x^2}$ and the weight defining the <weighted inner product> is $w=(1-x^2)^{-1/2}$. The substitution $x=\cos\theta$ maps the problem to $-Y''=\lambda Y$ on $0<\theta<\pi$ with homogeneous <Dirichlet boundary conditions>. It also transforms $w(x)\,dx$ into $-d\theta$. Consequently the normalized <eigenfunctions> are
$$
\boxed{y_n(x)=\sqrt{\frac2\pi}\sin(n\arccos x),\qquad\lambda_n=n^2,\quad n=1,2,\ldots.}
$$
Their <weighted inner product> is the ordinary sine <inner product> on $[0,\pi]$. Equivalently, $y_n=\sqrt{2/\pi}\sqrt{1-x^2}\,U_{n-1}(x)$, where $U_{n-1}$ is a <Chebyshev polynomial of the second kind>. These are not the cosine solutions $T_n$, which have different endpoint values. The original endpoints are singular because $p$ vanishes and $w$ diverges there, but the transformed Dirichlet problem specifies the self-adjoint realization. The eigenfunctions vanish at both endpoints and their quasi-derivatives $py_n'$ have finite limits, so the boundary form vanishes.
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