Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 1 / b

To obtain Sturm-Liouville form, multiply the equation by $(1-x^2{)}^{-1/2}$. The derivative identitygivesThe weight determines the weighted inner product on $(-1,1)$. It is integrable but singular at the endpoints, where $p$ also vanishes. Thus this is a singular-endpoint Sturm-Liouville problem, not a regular problem with strictly positive endpoint values of $p$. The transformation in part (ii) specifies its admissible Dirichlet realization; these are the Dirichlet eigenfunctions of the Chebyshev differential operator.

Let $Y(\theta )=y(\cos \theta )$ for $0\le \theta \le \pi$. The chain rule givesSubstitution cancels the first-derivative terms and leaves $-Y^{\prime \prime }=n^{2}Y$. The endpoint conditions become $Y(0)=Y(\pi )=0$. For $n\ne 0$, the first condition selects $Y=A\sin (n\theta )$, and the second requires $\sin (n\pi )=0$. Thus a nonzero solution requires an integer $n$. The case $n=0$ gives $Y=A+B\theta$ and both conditions force it to vanish. Negative integers give the same eigenvalues and eigenfunctions up to sign, so take $n=1,2,\dots$.

For vector normalization, use $w(x)dx=-d\theta$:Choosing a positive real normalization constant givesHere $U_{n-1}$ is the Chebyshev polynomial of the second kind. The factors $\sqrt{1-x^2}$ make both endpoint values zero. The derivatives need not stay finite there; instead $p{y}_n^{\prime }=-n\sqrt{2/\pi }\cos (n\theta )$ has finite endpoint limits, ensuring that the relevant boundary form vanishes.

Using the normalized eigenfunctions and the same change of variables, their weighted inner product isFor distinct positive integers $m,n$, the product-to-sum identity givesbecause each integrated sine has integer-multiple-of-$\pi$ endpoints. For $m=n$, the integral is one by the normalization already calculated. HenceThis verifies orthonormality explicitly, with ${\delta }_{mn}$ the Kronecker delta.