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Another question dealing with Frobenius method

  1. Feb 8, 2005 #1
    This is another question I have trouble proving:

    Suppose the coefficients of the equation: w'' + p(z)w' + q(z)w = 0 are analytic and single-valued in a punctured neighborhood of the origin. Suppose it is known that the function w(z) = f(z) ln z is a solution, where f is analytic and single-valued in a punctured neighborhood. Deduce that f is also a solution.

    Thanks for your help.
     
  2. jcsd
  3. Feb 8, 2005 #2

    Hurkyl

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    Can't you just make that substitution into the DE?
     
  4. Feb 8, 2005 #3
    When I make the substitution it becomes very messy and it is not easy to see that f is a solution. There are terms without the natural log which I can't see how they would vanish.
     
  5. Feb 8, 2005 #4

    Hurkyl

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    Actually, it's clearly not true. Take p(z) = q(z) = 0. Try the solution w(z) = 1.
     
  6. Feb 8, 2005 #5
    Hmmm, it's probably because p and q are required to have poles at z = 0. p(z) = q(z) = 0 is analytic at z = 0.
     
  7. Feb 8, 2005 #6

    Hurkyl

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    The nice thing about being everywhere analytic is that you're also analytic on a punctured neighborhood of the origin. :smile: Whether they're analytic at the origin or not is irrelevant.

    If you're still not convinced, try making your own differential equation whose coefficients are singular at the origin. Mine had 1/z as a solution.
     
  8. Feb 8, 2005 #7
    Yes I agree with you now. Thanks.
     
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