## ordinary points, regular singular points and irregular singular points

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
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity

Recognitions:
Gold Member
Science Advisor
Staff Emeritus
 Quote by 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 ?
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".

 Quote by HallsofIvy 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 ?

Recognitions:
Gold Member
Science Advisor
Staff Emeritus

## ordinary points, regular singular points and irregular singular points

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

 Similar discussions for: ordinary points, regular singular points and irregular singular points Thread Forum Replies Calculus & Beyond Homework 2 Calculus & Beyond Homework 1 Differential Equations 3 Differential Equations 1 Calculus & Beyond Homework 0