Ordinary points, regular singular points and irregular singular points

JamesGoh
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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 ?

thanks in advance
 
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JamesGoh said:
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 ?
Yes, it is.

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 ?
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.

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

thanks in advance
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".
 
HallsofIvy said:
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.

You mean \lim (x- x_0)p(x) for the last quote right ?
 
Yes, I managed to mess up a couple of formulas in that!
 
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