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

The relevant weighted inner product is $⟨u,v{⟩}_w={\int }_0^{1}w(x)u(x)v(x)dx$. The normalized eigenfunctions of the Sturm-Liouville problem have the orthonormality propertyHere ${\delta }_{nm}$ is the Kronecker delta: the integral is one for $n=m$ and zero otherwise. The weight is $w$, rather than $p$ or the unweighted measure.

Projecting the Sturm-Liouville eigenfunction expansion onto $y_m$ using the weighted inner product givesFor the partial sum $f_N={\sum }_{n=0}^N{a}_n{y}_n$, orthonormality therefore givesFor $f$ of finite weighted norm, the given completeness means that $f_N\to f$ in that norm. The triangle inequality implies $|\Vert f_N{\Vert }_w-\Vert f{\Vert }_w|\le \Vert f_N-f{\Vert }_w\to 0$. Taking the limit of the finite identity proves Parseval's identity:This argument justifies the limiting step without assuming that a double infinite series can be integrated term by term. As usual, the expansion is understood for square-integrable functions with respect to the weighted measure; the boundary and smoothness assumptions of the regular Sturm-Liouville problem provide the usual stronger interpretations when needed.

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.