Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 13A (Methods)

The Lagrange identity for this Sturm-Liouville theory operator isIntegrating from zero to one, the boundary term vanishes because both $y$ and $y_0$ vanish at both endpoints. Since $L(y_0;{\lambda }_0)=0$ and $L(y;{\lambda }_0)=f$, it follows thatThis is the solvability condition at a Sturm-Liouville eigenvalue.

Because the parameter enters through $\lambda wy$,Use this as $f$ in part (a). The necessary orthogonality condition isWith $y=\epsilon y_0+{\epsilon }^2{y}_1$ and $\mu =O(1)$,ConsequentlyThe normalization makes the second integral one. Divide by ${\epsilon }^{m+1}$ to obtainThis is the leading nonlinear eigenvalue shift in a Sturm-Liouville problem.