# What are the ordinary and singular points of the first order diff. equation?

Consider a first order differential equation. What do ordinary and singular points mean? What do they represent? (I cannot understand their formal definitions so please explain with examples. Thank you.)

UPDATE

After some research I noticed something. Consider the direction field of the $x\dot{y}=y-1$: I am pretty sure that an ordinary point should have one and only one (existence and uniqueness) line element (arrows in the picture).

Therefore all points except P(0,1) are ordinary points. On the other hand, P(0,1) is part of more than one integral curve and there are different line elements for each curve. This point violates uniqueness condition.

I think this point is a singular point. However, I don't know if it is regular or irregular. Please correct me if I am wrong and tell me if P(0,1) is irregular or not.

• I do not normally suggest other math sites but take a look at paul's online math help site it goes through a nice explanation of both terms. Aug 12, 2014 at 8:20
• @Chinny84 Most of the explanations in the internet are about second order equations and their series solution. They are confusing for me. Therefore, I want to learn their meaning for the first order equations.
– user137035
Aug 12, 2014 at 8:24
• en.wikipedia.org/wiki/Regular_singular_point The idea is the same, a point is ordinary when the coefficient are analytic around it. Aug 12, 2014 at 8:28
• @Lolman That explanation does not make sense for me. I have to admit that I am a little dumb about maths.
– user137035
Aug 12, 2014 at 8:34
• Do you know what an analytic function is? Do you know what a Taylor series is? Do you know what is an holomorphic or meromorphic function? It is not about being dumb it is about not knowing the definitions of the thing you use. Aug 12, 2014 at 9:39

We read from the wikipedia arcticle http://en.wikipedia.org/wiki/Regular_singular_point.

"Consider an ordinary linear differential equation of n-th order: $$\sum_{i=0}^n p_i(z)f^{(i)}(z)=0$$ with $p_i$ meromorphic functions and $p_n(z)=1$."

A meromorphic function is a function that is holomorphic on $\mathbb{C}\setminus D$ where $D$ is a discrete set. And holomorphic or analytic at a point $x_0$ means equivalently: it admit complex derivative at $x_0$, its Taylor series converges to the function in a neighbourhood of $x_0$. What happens in $D$? Those points are called poles and the function goes to infinity like $\frac{1}{z^n}$ for some $n$, and that $n$ is the order of the pole in $x_0$.

Now given an ordinary linear differential equation of order n, a point is $\textbf{ordinary}$ if $p_i$ are analytic around $x_0$, it is a $\textbf{regular singular point}$ if $p_{n-i}$ has a pole of order at most $i$ in $x_0$, and it is a $\textbf{irregular point}$ if none of the above holds.

• Thank you for your efforts, now I have an idea about the formal definitions. But what I really seek is the consequences. Examples can give me much more insight. I want to be able to say things like for $\dot{y}=y+1$ all points in the phase plane are ordinary because .... or this point in $\dot{y}=(y+1)/x$ is a singular point because ...
– user137035
Aug 12, 2014 at 10:35
• I updated my question accordingly.
– user137035
Aug 13, 2014 at 6:10

Following definitions are taken from Tenenbaum's ODE book. Consider a linear differential equation:

$$y^{(n)}+F_{n-1}(x)y^{(n-1)}+\dots+F_0(x)y=Q(x)$$

A point $x=x_0$ is called an ordinary point of the linear differential equation if each function $(F_0, F_1, \dots, Q)$ is analytic at $x=x_0$.

By analytic, we mean that there is a Taylor series for the function in a neighborhood of $x_0$. This can be checked via Taylor's theorem and the series should be convergent.

A point $x=x_0$ is called a singularity if one or more of the functions $(F_0, F_1, \dots, Q)$ is not analytic at $x=x_0$.

The definition of regular and irregular singularities is given only for the second order equations:

$$y^"+F_1(x)y'+F_0(x)y=0$$

If the multiplication of $F_1(x)$ by $(x-x_0)$ and of $F_0(x)$ by $(x-x_0)^2$ result in functions each of which is analytic at $x=x_0$, then the $x_0$ is called regular singular point. Otherwise it is called irregular singular point.

The example at the above question can be written as below:

$$y'-\frac 1 x y=-\frac 1 x$$

Since there is no Taylor series for $\frac 1 x$ at $x_0=0$. This point is a singularity. We cannot tell whether it is irregular or not because it is not a second order equation.

The scope of this answer is only limited with Tenenbaum's book and what I understand from it. I think Lolman's answer is much broader but you should have a higher level of knowledge to understand it.