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 / ii

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.