Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 8A / Solution

At zero an ordinary point requires $p,q$ to be analytic. A regular singular point requires $zp$ and $z^{2}q$ to extend analytically to zero, with at least one of $p,q$ failing to be analytic if “singular” excludes ordinary points.

Put $\zeta =1/z$ and $v(\zeta )=w(1/\zeta )$. Differentiation gives $w^{\prime }=-{\zeta }^2{v}^{\prime }$ and $w^{\prime \prime }={\zeta }^4{v}^{\prime \prime }+2{\zeta }^3{v}^{\prime }$, so the transformed equation isThe point at infinity is ordinary precisely when both transformed coefficients are analytic at $\zeta =0$. Solving these conditions for the original coefficients giveswith $P,Q$ analytic near zero; the sign can be absorbed into the arbitrary analytic $P$.

If the only singular point on the sphere is the regular singularity at zero, then $zp(z)$ and $z^{2}q(z)$ are entire. The condition at infinity makes the first tend to $2$ and the second to $0$. By Liouville theorem they are respectively the constants $2$ and $0$. Thus the unique equation iswith general solution $w=A+B/z$. Zero is genuinely regular singular and infinity is ordinary. If “no other singular points” were restricted to finite points only, the general coefficients would instead be $p=P(z)/z$ and $q=Q(z)/z^2$ with entire $P,Q$; the preceding question about infinity indicates the spherical interpretation.