Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / 7

Multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The common $h$ terms cancel, and integration by parts gives the Lagrange identity for a Sturm-Liouville operator:At each endpoint the two pairs $(y_n^{\prime },y_n)$ and $(y_m^{\prime },y_m)$ satisfy the same nontrivial homogeneous linear equation. They therefore lie on the same one-dimensional line, and their determinant vanishes. The boundary expression is zero. Since the eigenvalues are distinct, division gives the required orthogonality of Sturm-Liouville eigenfunctions:For complex-valued eigenfunctions of a real self-adjoint problem one may choose real eigenfunctions, or use the conjugated version as a weighted inner product.

Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda {\int }_1^b{y}^{2}dx/x={\int }_1^{b}x(y^{\prime }{)}^{2}dx>0$ for a nonzero real eigenfunction; hence $\lambda >0$. Using the printed change of variables $\xi =\lambda \ln x$, write $y(x)=Y(\xi )$. Then $(x{y}^{\prime }{)}^{\prime }={\lambda }^2{Y}_{\xi \xi }/x$, so $Y_{\xi \xi }+Y/\lambda =0$. The first endpoint gives $Y=A\sin (\xi /\sqrt{\lambda })$, and the other gives $\sin (\sqrt{\lambda }\ln b)=0$. ThusThis is the logarithmic change of variable in a Sturm-Liouville problem. The measure $dx/x=d(\ln x)$ gives squared weighted norm $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.