Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 1 / i

For the weighted inner product, the factor $w$ cancels the $w^{-1}$ in the Sturm-Liouville operator. Assuming the coefficients are real, two applications of integration by parts giveThis is the boundary form in the Lagrange identity for a Sturm-Liouville operator. The potential terms cancel; if $q$ were complex, an additional integral of $(q-q^{\ast })f^{\ast }g$ would remain, so suitable real coefficients are part of formal self-adjointness. The endpoint restrictions in part (a) eliminate the displayed form for every pair of functions in the domain.

The necessary boundary identity for a self-adjoint operator isAt regular endpoints, a standard full family of separated self-adjoint conditions iswith real coefficients and neither pair $({\alpha }_a,{\beta }_a)$, $({\alpha }_b,{\beta }_b)$ zero. For example, if ${\beta }_a\ne 0$, both functions obey $p{f}^{\prime }=-{\alpha }_{a}f/{\beta }_a$ at $a$, so $p{f}^{\prime \ast }g-p{f}^{\ast }{g}^{\prime }=0$ there. The cases ${\beta }_a=0$ and the endpoint $b$ follow directly. These include Dirichlet boundary conditions, Neumann boundary conditions and Robin boundary conditions.

Separate vanishing at each endpoint is sufficient but not necessary. Coupled conditions such as $y(b)=e^{i\chi }y(a)$ and $p(b)y^{\prime }(b)=e^{i\chi }p(a)y^{\prime }(a)$, with real $\chi$, make the two endpoint forms equal and hence their difference zero. More generally the allowed endpoint data must be a maximal subspace on which the boundary form vanishes; this ensures that the adjoint has the same domain. Merely imposing extra endpoint restrictions until the boundary term vanishes proves symmetry, but can overrestrict the domain and fail to give self-adjointness.

Let $L{y}_n={\lambda }_n{y}_n$. By self-adjointness,Taking $m=n$ first shows that each eigenvalue is real: $⟨y_n,y_n{⟩}_w>0$ for a nonzero eigenfunction, since $w>0$. ThereforeFor distinct eigenvalues this proves orthogonality of Sturm-Liouville eigenfunctions:The weight is essential: this need not be orthogonality under the unweighted integral.