# Regular and irregular points

In the following 2 examples, I am trying to find all the singular points of the given equations and determine whether each one is regular or irregular

I can determine the singular points of the equations but I am having some trouble determining if they are regular or irregular.

1) $$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$

The singular points of the equation are $x=\pm 1, 0$

The answer is that $0$ is irregular and $\pm 1$ regular but I am not sure why

2) $$xy'' + (1-x)y' +xy =0$$

The singular point is $0$ but why is this considered regular?

Do I have to consider the $p (x)$ term in each equation?

• The coefficient of y" should be just x – jh123 Nov 12 '17 at 16:48

The definition of wikipedia :

For the ordinary differential equation: $$f''(x)+p_{1}(x)f'(x)+p_{0}(x)f(x)=0.\,$$

• Point $a$ is an ordinary point when functions $p_1(x)$ and $p_0(x)$ are analytic at $x = a$.
• Point $a$ is a regular singular point if $p_1(x)$ has a pole up to order $1$ at $x = a$ and $p_0$ has a pole of order up to $2$ at $x = a$.
• Otherwise, point a is an irregular singular point.

For your example 1) $$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$

we divide by the coefficient function of $y''$ to obtain:

$$y'' + \frac {2}{x^3 (1-x^2)}y'+\dfrac{4}{x^2 (1-x^2)}y =0.$$

The singular points are $x=0$ and $x=\pm 1$. From the definition, we see that $x=0$ is an irregular singular point because it is a third order pole. $x=\pm 1$ are order $1$ poles of the coefficient function of $y'$ and also order $1$ poles of the coefficient function of $y$. Hence, we know that these are regular singular points.

Can you do the next example?

• What is meant by pole? – jh123 Nov 12 '17 at 16:37
• pole = sigular point – MrYouMath Nov 12 '17 at 16:40
• I am getting a little confused, when I plug I'm $\pm 1$ after we divided by the coefficient doesn't the second and third term go to 0? So how is it order 1 poles of the coedeficient y' and y – jh123 Nov 12 '17 at 16:45
• Note that $x^3(1-x^2)=x^3(1-x)^1(1+x)^1$ and $x^2(1-x^2)=x^2(1-x)^1(1+x)^1$. The powers are indicating the order of the pole/singular point. – MrYouMath Nov 12 '17 at 16:48
• So why is the pole order of p when x=0 3? – jh123 Nov 12 '17 at 16:52

Consider the first of your equations:

$$x^2 (1-x^2)y'' + \frac {2}{x}y'+4y =0$$

Reducing it to standard form, the coefficients of $y',y$ are $p(x)=\dfrac{2}{x^3(1-x^2)},q(x)=\dfrac 4{x^2(1-x^3)}$ respectively.

The functions $p,q$ each have poles on $x=0,\pm 1$, so they are singularities, or singular points.

Definition: Now, we call a singular point $a$ to be regular if either $p$ or $q$ diverges as $x\to a$ but $\lim\limits_{x\to a}(x-a)p(x)$ and $\lim\limits_{x\to a}(x-a)^2q(x)$ remain finite. Otherwise, we call it irregular.

This definition is equivalent to saying that $p$ has a pole of order at most $1$ at $x=a$ and $q$ has a pole of order at most $2$ at $x=a$ for $a$ to be regular, otherwise it's irregular

Now, notice that at $x=\pm 1$, $p$ has a pole of order $1\leq 1$ and $q$ has a pole of order $1\leq 2$, so $x=\pm 1$ is regular.

But at $x=0$, $p$ has a pole of order $3\gt 1$, so $x=0$ is irregular.

Here's a head start for you:

The coefficients of $y',y$ after reduction to standard form are:

$$p(x)=\frac{1-x}x^2\quad\textrm{and}\quad q(x)=\frac 1x$$

Clearly, the only pole is at $x=0$ for both $p,q$. What is the order of the pole in $p,q$ ? Compute and compare with the definition.

• What us meant by pole? – jh123 Nov 12 '17 at 16:38
• @jh123, A pole is just another name for "singular point". – Prasun Biswas Nov 12 '17 at 16:43
• I'm a little confused when you say p has a pole order of 1 and q has a pole order of 1? What is meant by the order? – jh123 Nov 12 '17 at 16:47
• @jh123, Basically, a singular point (pole) $x=a$ of $f(x)$ is of order $n$ if it is the least positive integer such that $\lim\limits_{x\to a}x^nf(x)$ is finite. – Prasun Biswas Nov 12 '17 at 16:48
• This is equivalent to the multiplicity of the singular point (akin to the multiplicity of a root in a polynomial). – Prasun Biswas Nov 12 '17 at 16:49