Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2011 / Paper 2 / 1

For two real eigenfunctions, multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The $q$ terms cancel. The resulting Lagrange identity for a Sturm-Liouville operator isIntegrating first over $[ϵ,R]$ and then taking the endpoint limits givesThe needed assumptions are that the improper integral exists and that the displayed boundary expression tends to zero at both endpoints. Bounded $y_m,y_n$ and their first derivatives are sufficient because $p\to 0$, but are stronger than necessary. Merely knowing that $p\to 0$ does not control arbitrarily growing eigenfunctions. With the vanishing boundary form and distinct eigenvalues, division by ${\lambda }_n-{\lambda }_m$ proves orthogonality of Sturm-Liouville eigenfunctions:For complex eigenfunctions in a real self-adjoint differential operator problem, use $\overline{y_m}$ instead, giving the corresponding weighted inner product.

Multiplying the particular equation by $e^{-x}$ puts it in Sturm-Liouville form:ConsequentlyIts polynomial eigenfunctions are the Laguerre polynomials. Products of these polynomials, with or without one derivative, grow only algebraically at infinity. The exponential in $p$ therefore makes the boundary form vanish there; its factor $x$ makes it vanish at zero. Their orthogonality is thus

For $\lambda =1$, substitution of $Ax+B$ gives $A+B=0$. For $\lambda =2$, substitution of $A{x}^2+Bx+C$ gives $4A+B=0$ and $B+2C=0$. Choosing $y_n(0)=1$, which fixes the otherwise arbitrary nonzero multiplicative constants, yieldsTheir product is $1-3x+\frac{5}{2}{x}^2-\frac{1}{2}{x}^3$. Using the Gamma function identity ${\int }_0^{\infty }{x}^k{e}^{-x}dx=k!$, also obtainable by repeated integration by parts, verifies the required orthogonality directly: