Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ordinary points, regular singular points and irregular singular points

  1. Aug 27, 2011 #1
    Say we have an ODE

    [itex]\frac{d^{2}x}{d^{2}y}+ p(x)\frac{dx}{dy}+q(x)y=0[/itex]

    Now, we introduce a point of interest [itex]x_{0}[/itex]

    If p(x) and q(x) remain finite at at [itex]x_{0}[/itex]
    is [itex]x_{0}[/itex]
    considered as an
    ordinary point ?

    Now let's do some multiplication with [itex]x_{0}[/itex]
    still being
    the point of interest

    [itex](x-x_{0})p(x)[/itex] (1)

    and

    [itex](x-x_{0})^{2}q(x)[/itex] (2)

    If (1) and (2) remain finite, is [itex]x_{0}[/itex]
    considered as a regular singular point ?

    Otherwise if (1) and (2) are undefined, is [itex]x_{0}[/itex]
    an irregular singular point ?

    thanks in advance
     
  2. jcsd
  3. Aug 27, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, it is.

    Well, that depends. You started with the equation
    [tex]\frac{d^2y}{dx^2}+ p(x)\frac{dy}{dx}+ q(x)y= 0[/tex]
    Multiplying the second derivative by [itex]x- x_0[/itex] would be the same as having
    [tex]\frac{d^2y}{dx^2}+ \frac{p(x)}{x- x_0}\frac{dy}{dx}+ \frac{q(x)}{x- x_0}y= 0[/tex]
    Whether [itex]x_0[/itex] is a "regular singular point" or not now depends upon the limits of those two fractions as x goes to [itex]x_0[/itex]. IF p(x) and q(x) were 0 at [itex]x= x_0[/itex], then [itex]x_0[/itex] might still be an ordinary point.

    Yes.

    Given the differential equation
    [tex]\frac{d^2y}{dx^2}+ p(x)\frac{dy}{dx}+ q(x)y= 0[/tex]
    If [itex]\lim_{x\to x_0}p(x)[/itex] and [itex]\lim_{x\to x_0} q(x)[/itex] exist, then [itex]x_0[/itex] is an "ordinary" point.

    If those do not exist but [itex]\lim (x- x_0)(x- x_0)p(x)[/itex] and [itex]\lim(x-x_0)^2q(x)[/itex] exist, then [itex]x_0[/itex] is a "regular singular" point.

    In any other situation, [itex]x_0[/itex] is an "irregular singular" point.

    It might be helpful to remember that the "Euler-Lagrange" type equation,
    [tex](x- x_0)^2\frac{d^2y}{dx^2}+ (x- x_0)\frac{dy}{dx}+ y= 0[/tex]
    has [itex]x_0[/itex] as a "regular singular point".
     
  4. Aug 27, 2011 #3
    You mean [itex]\lim (x- x_0)p(x)[/itex] for the last quote right ?
     
  5. Aug 29, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, I managed to mess up a couple of formulas in that!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ordinary points, regular singular points and irregular singular points
Loading...