ordinary points, regular singular points and irregular singular points

JamesGoh

Say we have an ODE

[itex]\frac{d^{2}x}{d^{2}y}+ p(x)\frac{dx}{dy}+q(x)y=0[/itex]

Now, we introduce a point of interest [itex]x_{0}[/itex]

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

Now let's do some multiplication with [itex]x_{0}[/itex]
still being
the point of interest

[itex](x-x_{0})p(x)[/itex] (1)

and

[itex](x-x_{0})^{2}q(x)[/itex] (2)

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

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

thanks in advance

HallsofIvy

Yes, it is.

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

Yes.

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

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

In any other situation, [itex]x_0[/itex] is an "irregular singular" point.

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

JamesGoh

You mean [itex]\lim (x- x_0)p(x)[/itex] for the last quote right ?

HallsofIvy

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