Past exam of the options I course of the University of Oxford / 2004 / G

Multiply the equation for $y_i$ by $y_j$ and the equation for $y_j$ by $y_i$, then subtract. The $q$ terms cancel and the derivative terms combine into a Wronskian boundary derivative:Integrating over the interval eliminates the boundary term because both eigenfunctions vanish at both endpoints. Since their eigenvalues differ,This is orthogonality in a weighted inner product in the regular Sturm-Liouville setting; the endpoint values must justify the integration by parts.

If $\lambda =0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda <0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda =k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive integer $n$. Thus $\boxed{{\lambda }_{\min }={\pi }^2}$.

The admissible trial function $y=\sin \pi x$ has quotientTherefore the minimum, or infimum before its attainment is established, is $\boxed{\le {\pi }^2}$ as requested. The sharp Poincaré inequality supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also ${\pi }^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.