Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 7E (Further Complex Methods)

The point $z_0$ is an ordinary point when both $p$ and $q$ are holomorphic at $z_0$.

The point $z_0$ is a regular singular point when it is singular butextend holomorphically to $z_0$.

A singular point which is not regular singular is an irregular singular point: at least one of $(z-z_0)p(z)$ and $(z-z_0{)}^{2}q(z)$ fails to extend holomorphically to $z_0$.

Put $t=1/z$ and $y(t)=w(1/t)$. Sincethe transformed equation is

By the transformed equation, infinity is ordinary precisely whenare holomorphic at $t=0$. Equivalently, as convergent Laurent expansions near infinity,

By the regular singular point at infinity criterion, infinity is regular singular precisely whenare holomorphic at $t=0$, but the ordinary-point conditions fail. Equivalently,with convergent expansions in $1/z$, excluding the ordinary case.

Infinity is irregular singular when at least one offails to be holomorphic as a function of $1/z$ at $1/z=0$.

Suppose there were no singular point in the Riemann sphere. Then every finite point would be ordinary, so $p$ and $q$ would be entire functions. Ordinary behavior at infinity would force $p(z)=2/z+O(z^{-2})$ and $q(z)=O(z^{-4})$. In particular $p(z)\to 0$, so Liouville theorem would give $p\equiv 0$, contradicting the required leading term $2/z$. Thus infinity cannot also be ordinary, and every such differential equation has at least one singular point in $\overline{\mathbb{C}}$.