Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 14A (Methods)

Let $Q_k=P^{(k)}$. Differentiating Legendre's differential equation $k$ times and using the Leibniz rule givesThe induction step follows by differentiating this equation once: the derivative of the coefficient $-2(k+1)x$ contributes the additional term $-2(k+1)Q_k^{\prime }$. ThusFor fixed $\lambda$, this is negative for every sufficiently large $k$.

Multiplication by $(1-x^2{)}^k$ puts the equation in self-adjoint form:Multiply by $Q_k$ and integrate over $[-1,1]$. Regularity and the vanishing factor at both endpoints remove the boundary term in integration by parts, leavingIf $Q_k$ is not identically zero, the integral on the right without ${\lambda }_k$ is positive, whereas the left side is nonnegative. Therefore

Part (a) makes ${\lambda }_k<0$ for all sufficiently large $k$, while part (b) would make ${\lambda }_k\ge 0$ whenever $P^{(k)}$ were nonzero. Hence some derivative of $P$ vanishes identically, so $P$ is a polynomial.

Let its degree be $n$. Then $P^{(n)}$ is a nonzero constant. Substituting $k=n$ into the differentiated equation annihilates both derivative terms and givesThus ${\lambda }_n=0$, andThe nonzero regular solutions are therefore constant multiples of the Legendre polynomial of degree $n$.