Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 1

Multiplying by $x^2$ makes the first two terms a total derivative:The resulting Sturm-Liouville form isEquivalently, an integrating multiplier $\mu$ must satisfy $(\mu (1-1/x){)}^{\prime }=\mu (2/x-1/x^2)$, which reduces to ${\mu }^{\prime }/\mu =2/x$ and gives $\mu =x^2$ up to a constant. The coefficient $p$ is positive for $1<x<R$ and vanishes linearly at the left endpoint; the weight $w$ is positive throughout. The boundedness condition removes the logarithmic local branch, giving the no-log boundary condition at a linearly vanishing Sturm-Liouville endpoint.

The appropriate weighted inner product and Sturm-Liouville operator areFor two functions obeying the endpoint conditions, integration by parts gives the Lagrange identity for a Sturm-Liouville operatorAt $R$, both $u$ and $v$ vanish. At $1$, the factor $x(x-1)$ tends to zero while the functions and their first derivatives remain bounded. Thus the boundary form vanishes at both ends.

This gives the regular self-adjoint differential operator realization: its singular-end condition is ${\lim }_{x\downarrow 1}x(x-1)v^{\prime }=0$, together with $v(R)=0$. The bounded classical eigenfunctions satisfy exactly this no-log choice. More precisely the operator domain is the maximal differential-operator domain in the weighted space with these separated endpoint conditions; the preceding calculation verifies its self-adjoint boundary form, rather than merely the differential expression's formal symmetry.

Let $y\ne 0$ be an eigenfunction, allowing complex values for the reality argument. The self-adjoint differential operator identity and integration by parts giveThe boundary term $[x(x-1)\overline{y}{y}^{\prime }{]}_1^R$ vanishes for the same reasons as in part ii. The denominator is real and positive; the numerator is real and strictly positive because $l\ge 1$ and $y\ne 0$. Hence the Rayleigh quotient isThe positivity uses a nondegenerate interval $R>1$.

For real eigenfunctions $y_i,y_j$, multiply their Sturm-Liouville problems crosswise and subtract. Integration givesAt $R$, both eigenfunctions vanish, while at $1$ their bounded values and bounded first derivatives multiply a factor tending to zero. The right side is zero. Distinct eigenvalues therefore give the orthogonality of Sturm-Liouville eigenfunctionsFor complex eigenfunctions, replace $y_i$ by $\overline{y_i}$, in accordance with the weighted inner product.