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Removing a Singularity for a Second Order ODE

  1. Oct 24, 2012 #1
    I was given the following equation to solve:

    x^2*y'' + x*y' + k^2*x^2*y = 0
    B.C. y'(0)=0, y(1)=0

    where k is just some constant.

    I am having a hard time removing the singularity created by the boundary condition at y' and not aware of a method how. Any advice would be greatly appreciated.
     
  2. jcsd
  3. Oct 26, 2012 #2

    haruspex

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    For x nonzero, you can obviously simplify the equation. When x is zero, there simply is no information from the DE. So any solution of the simplified equation is a valid solution for the original.
    Please show your working so far.
     
  4. Oct 27, 2012 #3

    LCKurtz

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    Not sure what you mean by removing the singularity. You have a regular singular point at ##x=0##. Have you looked for a solution of the form$$
    y=\sum_{n=0}^\infty x^{n+r}\hbox{?}$$
     
  5. Oct 29, 2012 #4
    Hi !

    it is a Bessel ODE:
    y = c1*J0(k*x) +c2*Y0(k*x)
    The condition y'(0)=0 implies c2=0, hence y(x) = c1*J0(k*x)
    J(k)=0 only in case of some particular values of k.
    So, generally, the condition y(1)=0 implies c1=0, hence y(x)=0 in general.
    But, y(x)=c1*J0(k*x) in case of some particular values of k.
     
  6. Nov 2, 2012 #5
    Thanks for all the help so far.
    I'm just trying to express the Bessel function in a way that satisfies the boundary conditions. So far I have written it out to:
    1- 1/4*k^2*x^2 + (.25*k^2*x^2)^2/4 - (1/4*k^2*x^2)^3/
    36 + (.25*x^2*k^2)^4/24^2

    Using equation 78 off this link http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
    My problem is that I can't get it to meet the boundary conditions. If I could take out the 1 and set "k" equal to about 5.3, it fits... but I think that's cheating.
     
  7. Nov 2, 2012 #6

    haruspex

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    Does this help:
    Let ri be the ith +ve root of J1(x). So it also satisfies J0'(x) = 0.
    Consider Y(x) = Ʃ aiJ0(rix)
    This satisfies Y'(1) = 0.
    If we can find ai s.t. Ʃairi2 = k2Ʃ ai then I believe Y satisfies the original equation, and Y'(0) = Ʃ airiJ0'(0) = 0
     
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