series solutions to 2nd order ODEs i am very confused about how to tell whether a point is ordinary or regular singular. i know the definitions but think that I am physically doing it wrong. Do you sub the point into p @ q and see if you get zero, or see if they are equal? can you please explain in the most basic manner the step used.
 A: Let's look at the definitions: given a differential equation
$$ y'' + p(x) y' + q(x) = 0, $$
(the leading coefficient must be $1$ to do this: if it isn't, divide by it!) the point $x=a$ is


*

*An ordinary point if $p(x)$ and $q(x)$ are regular at $x=a$ (continuous or bounded in a neighbourhood is good enough).

*A regular singular point if not an ordinary point and $(x-a)p(x)$ and $(x-a)^2 q(x)$ are bounded in a neighbourhood of $x=a$

*An irregular singular point if neither of these is true.
To check which one we have, normally the best way is first to check if $p(a)$ and $q(a)$ exist. If they do, it's an ordinary point. If not, find $\lim_{x \to a} (x-a)p(x)$ and $\lim_{x \to a} (x-a)^2q(x)$. If these exist, it is a regular singular point. If they don't, it's an irregular singular point.
It may be possible to spot what sort of singularities $p$ and $q$ have without needing to take the limit in simple cases.

Examples:


*

*$y'' + y'\sin{x}+y\cos{x} = 0$


$\sin{x}$ and $\cos{x}$ are analytic at every point, so every point is an ordinary point.


*$y'' + \frac{1}{x}y'+\frac{1}{x}y = 0$


$p(x) = 1/x$, $q(x)=1/x$. $p(0)$ and $q(0)$ are undefined, so $0$ is not a regular point. $xp(x) \to 1$ as $x \to 0$ and $x^2 q(x) \to 0$ as $x \to 0$, so $x=0$ is a regular singular point. Elsewhere $p(x)$ and $q(x)$ are defined, so we have ordinary points.


*$y'' + \frac{1}{x^2}y' + y = 0$


$p(x) = 1/x^2$. $\lim_{x \to 0}xp(x)$ does not exist, so $x=0$ is an irregular singular point.


*$ y'' + y\cot{x} = 0 $
$p(x)=0$, $q(x)=\cot{x}$. Since $\cos{x}/\sin{x} \approx 1/x$ as $x \to 0$, $x^2q(x) \to 0$ as $x \to 0$, so $0$ is a regular singular point. $\cot{x}$ has singularities whenever $\sin{x}=0$, so at $n\pi$. $\cot{(x+n\pi)} = \cot{x}$, so every singularity looks like the one at $x=0$. Hence $x=n\pi$ are regular singular points.



You may wonder why there is the distinction between types of singular points. Suppose $a=0$. Suppose that  $y=x^{\sigma}u$, where $u$ is analytic at $a$ (has a power series expansion $a_0+a_1x+a_2x^2+\dotsb$). Then
$$ y' = \sigma x^{\sigma-1} u + x^{\sigma} u' \\
 y'' = \sigma(\sigma-1) x^{\sigma-2}u+2\sigma x^{\sigma-1}u' + x^{\sigma} u'', $$
so the differential equation becomes
$$ ( x^2u'' + 2\sigma x u' + \sigma(\sigma-1) u + xp(x) (\sigma u+xu') + x^2q(x) u )x^{\sigma-2} = 0 $$
In particular, everything but $p$ and $q$ has lowest term constant in the vicinity of $x=0$. To avoid having terms with negative powers in the bracket, which won't necessarily cancel out, $xp(x)$ needs to have an expansion in nonnegative powers, and likewise $q(x)$. Hence a solution of the form $x^{\sigma} (a_0+a_1x+\dotsb)$ will only definitely work if $0$ is a regular singular point.
