= Solution
For
$$
y''+P(z)y'+Q(z)y=0,
$$
a finite point $z_0$ is a <regular singular point> when $(z-z_0)P(z)$ and $(z-z_0)^2Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.
For the displayed <Bessel differential equation>,
$$
P(z)=\frac1z,
\qquad
Q(z)=1-\frac1{4z^2}.
$$
At $z=0$, the functions $zP(z)=1$ and $z^2Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomes
$$
Y''+\frac1wY'
+\left(\frac1{w^4}-\frac1{4w^2}\right)Y=0.
$$
Here $w^2Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an <irregular singular point>. Thus
$$
\boxed{z=0\text{ is regular singular, while }z=\infty\text{ is irregular singular}.}
$$
Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:
$$
z^2y''+zy'+\left(z^2-\frac14\right)y
=z^{3/2}(f''+f).
$$
Hence $f''+f=0$, and on any domain carrying a branch of $\sqrt z$ two linearly independent solutions are
$$
\boxed{y_1(z)=\frac{\cos z}{\sqrt z},
\qquad
y_2(z)=\frac{\sin z}{\sqrt z}.}
$$
Near the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two <Frobenius method>[Frobenius exponents]. At infinity their combinations are $z^{-1/2}e^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.
Back to article page
