# Ordinary points, regular singular points and irregular singular points

1. Aug 27, 2011

### JamesGoh

Say we have an ODE

$\frac{d^{2}x}{d^{2}y}+ p(x)\frac{dx}{dy}+q(x)y=0$

Now, we introduce a point of interest $x_{0}$

If p(x) and q(x) remain finite at at $x_{0}$
is $x_{0}$
considered as an
ordinary point ?

Now let's do some multiplication with $x_{0}$
still being
the point of interest

$(x-x_{0})p(x)$ (1)

and

$(x-x_{0})^{2}q(x)$ (2)

If (1) and (2) remain finite, is $x_{0}$
considered as a regular singular point ?

Otherwise if (1) and (2) are undefined, is $x_{0}$
an irregular singular point ?

2. Aug 27, 2011

### HallsofIvy

Staff Emeritus
Yes, it is.

Well, that depends. You started with the equation
$$\frac{d^2y}{dx^2}+ p(x)\frac{dy}{dx}+ q(x)y= 0$$
Multiplying the second derivative by $x- x_0$ would be the same as having
$$\frac{d^2y}{dx^2}+ \frac{p(x)}{x- x_0}\frac{dy}{dx}+ \frac{q(x)}{x- x_0}y= 0$$
Whether $x_0$ is a "regular singular point" or not now depends upon the limits of those two fractions as x goes to $x_0$. IF p(x) and q(x) were 0 at $x= x_0$, then $x_0$ might still be an ordinary point.

Yes.

Given the differential equation
$$\frac{d^2y}{dx^2}+ p(x)\frac{dy}{dx}+ q(x)y= 0$$
If $\lim_{x\to x_0}p(x)$ and $\lim_{x\to x_0} q(x)$ exist, then $x_0$ is an "ordinary" point.

If those do not exist but $\lim (x- x_0)(x- x_0)p(x)$ and $\lim(x-x_0)^2q(x)$ exist, then $x_0$ is a "regular singular" point.

In any other situation, $x_0$ is an "irregular singular" point.

It might be helpful to remember that the "Euler-Lagrange" type equation,
$$(x- x_0)^2\frac{d^2y}{dx^2}+ (x- x_0)\frac{dy}{dx}+ y= 0$$
has $x_0$ as a "regular singular point".

3. Aug 27, 2011

### JamesGoh

You mean $\lim (x- x_0)p(x)$ for the last quote right ?

4. Aug 29, 2011

### HallsofIvy

Staff Emeritus
Yes, I managed to mess up a couple of formulas in that!