Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 2 / 1 / c

Write $Y={\sum }_{n\ge 0}{b}_n{y}_n$ in the complete Sturm-Liouville eigenfunction expansion. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The Sturm-Liouville operator has the symmetry identitybecause both functions obey the homogeneous Dirichlet boundary conditions. Consequentlyso that $({\lambda }_m-\alpha )b_m=a_m$. The stipulated exclusion of every eigenvalue makes all denominators nonzero, and givesThis is the resolvent operator applied to $f$ in the weighted space. The same projection shows uniqueness: a homogeneous solution would have $({\lambda }_m-\alpha )b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.

Only two coefficients in the Sturm-Liouville eigenfunction expansion are nonzero: $a_0=2$ and $a_1=2-\alpha$. Since $\alpha \ne 2$, the result of part (i) simplifies toUsing Parseval's identity, the prescribed weighted norm becomesThus $1-\alpha =\pm 2$, so $\alpha =-1$ or $3$. Positivity selectsThe standing nonresonance assumption still applies: this answer requires that $3$ is not one of the unspecified higher eigenvalues. Their actual values are not supplied. If a higher eigenvalue were $3$, the stated exclusion would fail and the data would not have an admissible $\alpha$ under that exclusion.