Past exam of the differential equations 2 course of the University of Oxford / 2023 / 2 / b / ii

First take the usual endpoint-regular interpretation $f\in C([-1,1])$. Integrating the divergence form shows that $F=(1-x^2)y^{\prime }$ has finite endpoint limits. A nonzero limit would make $y^{\prime }$ proportional to $1/(1\mp x)$ near that endpoint, forcing a logarithmically unbounded $y$. Thus boundedness requires $F(-1)=F(1)=0$, and consequentlyThis is orthogonality to the constant eigenfunction $P_0$, hence the Fredholm alternative condition. It is also sufficient under the stated endpoint regularity: putIf $M=\max |f|$, the zero-mean condition gives $|F(s)|\le M\min (1+s,1-s)$. Hence the last integrand is bounded, so $y$ extends continuously to both endpoints. Any two bounded solutions differ by a constant: their difference solves $((1-x^2)z^{\prime }{)}^{\prime }=0$, whose other solution is $\frac{1}{2}\log ((1+x)/(1-x))$.

For the requested eigenfunction expansion, setThe Legendre polynomials span all polynomials, so their completeness follows from the Weierstrass approximation theorem. Projecting $\mathcal{L}y=f$ onto each eigenfunction gives $-k(k+1)y_k=f_k$ for $k\ge 1$. The zero mode is free, while $f_0=0$. Thus the full family isThe formula is unchanged by rescaling any $P_k$. For the usual normalization $P_k(1)=1$, the facts $|P_k(x)|\le 1$ on $[-1,1]$ and $N_k=2/(2k+1)$, together with the Cauchy-Schwarz inequality, bound each summand by a constant times $k^{-3/2}$. The series therefore converges absolutely and uniformly. Its identification with the preceding integral solution avoids unjustified termwise second differentiation.

There is a source ambiguity: continuity only on $(-1,1)$ does not supply the endpoint control used above. Even for integrable forcing, zero mean alone is insufficient. For example, letThis $f$ is continuous in the interior, absolutely integrable, and has zero mean because $F(\pm 1)=0$. Yet every candidate with zero endpoint values of $(1-x^2)y^{\prime }$ haswhose integral diverges logarithmically in $H$ at each endpoint. Thus no bounded solution exists. For a merely interior-continuous $f$, the exact criterion is that some constant $B$ makesbounded as $x$ approaches both endpoints. The Bounded solvability for the Legendre operator criterion above covers the intended regular forcing and makes this endpoint limitation explicit.