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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)


    [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


    User Avatar
    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.


    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


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    Science Advisor

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