Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3 / 15D / Solution

For the real Sturm-Liouville problem, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. Integration by parts givesThe boundary term vanishes because both functions are zero at both endpoints. Distinct eigenvalues therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so

Let $c_n={\int }_0^{1}w{y}_n^{2}dx>0$. For $y=\sum a_n{y}_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality givesFor finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the Rayleigh quotient is a weighted average of the eigenvalues, soEquality occurs precisely when only a lowest-eigenvalue component is present.

Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy ${\dot{a}}_n=-{\lambda }_n{a}_n$. Alternatively differentiate its energy directly and use the preceding inequality:Thus the squared weighted norm is at most its initial value times $e^{-2{\lambda }_{1}t}$. This describes decay when ${\lambda }_1>0$; positivity of $p,w$ alone does not require ${\lambda }_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.