Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading something on Bessel functions of first-second kind as the solutions to the Bessel diff. equation, and the difference, according to the text, is whether the function is singular or not in the origin. What does that exactly mean for a diff. equation of the form:


to be singular?

Specifically, I wnat to write the solution of a cylindrical symmetrical wave as bessel functions.

share|improve this question
The coefficient of $y_{xx}$ is the culprit. It becomes zero when $x=0$. This is not nice, to suddenly lose the highest derivative in the equation. –  user53153 Feb 27 '13 at 20:25
@5pm Then if the term of the second derivative is $x^2y_{xx}$, is it always singular? For example here: mathworld.wolfram.com/BesselFunctionoftheFirstKind.html the first term is like that and they say: "which are non-singular at the origin". –  MyUserIsThis Feb 27 '13 at 20:45
Read carefully: "solutions to the Bessel differential equation ... which are nonsingular at the origin." Bessel functions are defined as solutions that do not blow up at $x=0$. This is actually related to the singularity of the equation: most solutions of the equation blow up when reaching $x=0$. Bessel functions are special solutions which do not do this. –  user53153 Feb 27 '13 at 20:54
@5pm So I have to know the solution, or at least if it diverges in the origin (or is singular), before being able solve it? –  MyUserIsThis Feb 27 '13 at 20:58
Depends on what you want. If you want a cylindrical symmetric wave, then the solution should not blow up at zero. –  user53153 Feb 27 '13 at 21:12
show 3 more comments

1 Answer

up vote 1 down vote accepted

A linear ODE has a singular point when the coefficient of its highest derivative turns to $0$. Such points are further classified as regular and irregular. References:

  1. Regular singular point
  2. Frobenius method
share|improve this answer
Thank you for your answer. –  MyUserIsThis Feb 27 '13 at 21:46
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.