Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 14A / b

Multiply the equation by $x^2$ and combine the derivative terms:Thus the self-adjoint form isThe Dirichlet boundary conditions remain at $1$ and $e$, and the weight $w=x$ is positive on this interval, as required for the Sturm-Liouville problem.

Set $s=\log x$ and $Y(s)=y(e^s)$, so $0\le s\le 1$. By the chain rule, the equation becomes $Y^{\prime \prime }+2Y^{\prime }+(1+\lambda )Y=0$. The substitution $Y=e^{-s}v$ reduces it further toMultiplication by $\overline{v}$ and integration by parts gives $\lambda {\int }_0^1|v{|}^2={\int }_0^1|v^{\prime }{|}^2>0$ for a nonzero solution, ruling out zero and negative eigenvalues. For $\lambda =k^2>0$, the first boundary condition gives $v=A\sin (ks)$, and the second gives $k=n\pi$, $n=1,2,\dots$. Thus all eigenpairs arewith arbitrary nonzero scalar multiples of each eigenfunction. These are the complete list since all solutions of the constant-coefficient differential equation have been considered; the transformed sine system is also complete in the corresponding weighted Hilbert space.

For the weight $w=x$, substitution $x=e^s$ givesThe product-to-sum formula makes this zero when $m\ne n$ and $1/2$ when $m=n$. ThereforeIn particular the eigenfunctions are orthogonal, and multiplying each by $\sqrt{2}$ gives an orthonormal basis for the weighted space.