Regular singular point

Remarks

For $n = 2$, written $y'' + p(z) y' + q(z) y = 0$, condition 2 says that $(z - z_0) p(z)$ and $(z - z_0)^2 q(z)$ are analytic at $z_0$. For $n = 2$, the point $\infty$ is ordinary or regular singular if and only if $z\, p(z)$ and $z^2 q(z)$ are analytic at $\infty$.

Example: the Cauchy–Euler equation $z^2 y'' + a z y' + b y = 0$ with $(a, b) \ne (0, 0)$ has $p = \frac{a}{z}$ and $q = \frac{b}{z^2}$. So $0$ is singular, while $z p = a$ and $z^2 q = b$ are constant; hence $0$ is a regular singular point.

Non-example: for $z^2 y'' + y' = 0$, one has $p = \frac{1}{z^2}$, and $z p = \frac{1}{z}$ is not analytic at $0$. So $0$ is singular but not regular singular; it is an irregular singular point.

Fuchs proved a characterisation by growth. A singular point is regular singular if and only if every solution grows at most like a power of $\abs{z - z_0}^{-1}$ as $z \to z_0$ in each sector with vertex $z_0$.