Ordinary points, regular singular points and irregular singular points

[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 ?

thanks in advance

 

[HallsofIvy]

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".

 

[JamesGoh]

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 ?

 

[HallsofIvy]

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

 

[blue_raver22]

I would like to clarify and provide a more complete understanding of the terms mentioned in this content. In the context of ordinary differential equations (ODEs), there are three types of points that can arise: ordinary points, regular singular points, and irregular singular points.

An ordinary point, also known as a regular point, is a point in the domain of an ODE where both the coefficients p(x) and q(x) are finite and well-behaved. This means that the equation is well-defined and can be solved using standard methods. In the given ODE, if p(x) and q(x) remain finite at x_{0}, then x_{0} can be considered as an ordinary point.

On the other hand, a regular singular point is a point where both p(x) and q(x) become infinite or undefined, but their product remains finite. In other words, if the expressions (1) and (2) in the given ODE remain finite at x_{0}, then x_{0} can be considered as a regular singular point. These points require special techniques to solve the ODE, such as the method of Frobenius.

Lastly, an irregular singular point is a point where either p(x) or q(x) (or both) become infinite or undefined. In this case, neither (1) nor (2) will remain finite at x_{0}. These points pose a significant challenge in solving the ODE, and their behavior can be unpredictable.

In summary, the type of point at x_{0} in the given ODE can be determined by examining the behavior of p(x) and q(x) at that point. If both remain finite, it is an ordinary point; if their product remains finite, it is a regular singular point; and if either becomes infinite or undefined, it is an irregular singular point. I hope this explanation helps to clarify the concept of ordinary, regular singular, and irregular singular points in ODEs.