Past exam of the differential equations and applications course of the University of Oxford / 2010 / A / 3

Normalize the Laguerre differential equation as $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, where $P(x)=1/x-1$ and $Q(x)=n/x$. At $x=0$, the products $xP=1-x$ and $x^{2}Q=nx$ are analytic. Thus zero is a regular singular point, with indicial equationAt $x=1$ the coefficients $P,Q$ are analytic, so it is an ordinary point. No singular-point indicial equation is needed there.

For infinity put $z=1/x$ and $Y(z)=y(1/z)$. Then the equation becomesAt $z=0$, the coefficient $z(z^{-1}+z^{-2})=1+1/z$ is not analytic. Therefore infinity is an irregular singular point, including when $n=0$. A terminating polynomial solution does not change the classification of the differential equation.

Substitute $y={\sum }_{k\ge 0}{a}_k{x}^k$ and compare the coefficient of $x^k$. The two derivative terms give $(k+1{)}^2{a}_{k+1}$, while the remaining terms give $(n-k)a_k$. ThusIterating the recurrence givesIndeed the recurrence first vanishes at $k=n$ and then remains zero. Consequently the Laguerre polynomial iswith exact degree $n$ and leading coefficient $(-1{)}^n/n!$. This is the regular power-series branch normalized at zero; the second branch at the repeated indicial root is not an additional analytic solution with this normalization.

Assume first that $n\ge m$. Substituting Rodrigues' formula for $L_n$ givesHere integration by parts was performed $n$ times. Each boundary term vanishes: at infinity a polynomial times $e^{-x}$ tends to zero; at zero every derivative of $x^n{e}^{-x}$ of order less than $n$ vanishes. If $n>m$, the derivative $L_m^{(n)}$ is zero, proving orthogonality. The case $m>n$ follows by exchanging the two factors.

For $m=n$, the leading coefficient found above gives $L_n^{(n)}=(-1{)}^n$. ThereforeThe same conclusion includes $n=0$ directly. Thus the full Laguerre orthonormality relation is