Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 2 / 1 / a

Two integrations by parts give the boundary form of the Sturm-Liouville operator:The interior terms cancel because $p,q$ are real. Thus the required endpoint condition isFor a regular interval, a standard self-adjoint domain is specified by one real homogeneous Robin boundary condition at each endpoint:where neither pair of real constants is identically zero. At each end, these conditions force the two boundary data vectors of $f,g$ to be proportional, making the displayed form zero. Dirichlet boundary conditions and Neumann boundary conditions are the cases $B=0$ and $A=0$. Coupled conditions also work, for example $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$, whose two endpoint contributions cancel. Self-adjointness uses the maximal common domain with these conditions, not an arbitrary smaller symmetric restriction.

The weighted inner product must be nondegenerate. The usual interpretation is $w>0$ almost everywhere, allowing isolated endpoint zeros as in part (b), with the operator defined on its appropriate weighted domain. Merely allowing a weight that is zero on an entire interval would not by itself define the stated operator or a positive-definite inner product on ordinary functions.

Let $L\varphi =\lambda \varphi$ with $\varphi \ne 0$ in the self-adjoint domain. By the boundary identity from part (i),Since its squared weighted norm is positive, $\boxed{\lambda \in \mathbb{R}}$.

For $L{\varphi }_n={\lambda }_n{\varphi }_n$ and $L{\varphi }_m={\lambda }_m{\varphi }_m$, the same identity givesThus ${\lambda }_n\ne {\lambda }_m$ impliesThis is orthogonality of Sturm-Liouville eigenfunctions. Degenerate eigenspaces can be given an orthonormal basis, although vectors in a degenerate eigenspace are not automatically orthogonal.

Choose a complete orthonormal system of eigenfunctions ${\varphi }_n$ for the self-adjoint problem. Expand $f={\sum }_n{f}_n{\varphi }_n$ and $y={\sum }_n{c}_n{\varphi }_n$, withSubstitution into $Ly=f$ gives ${\lambda }_n{c}_n=f_n$. Since no eigenvalue is zero, the Sturm-Liouville eigenfunction expansion isFor eigenfunctions not normalized to one, divide each term additionally by $⟨{\varphi }_n,{\varphi }_n⟩$. The expansion is understood in the weighted function space and, with sufficient data regularity, in the operator domain. A zero eigenvalue would instead require its corresponding source coefficient to vanish and would leave an arbitrary homogeneous component.