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Math Help - Zeros of Euler's equation (y''+(k/x^2)y=0)

  1. #1
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    Zeros of Euler's equation, y''+(k/x^2)y=0

    "Show that every nontrivial solution of y''+\frac{k}{x^2}y=0 has an infinite number of positive zeros if k>1/4 and only finitely many positive zeros if k\le 1/4."

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    I set y=x^M = e^{M \log{x}} (for some constant M), differentiated twice and put it back into the equation, which gives M=\frac{1\pm \sqrt{1-4k}}{2}. So, y_1 = x^{\frac{1}{2} (1+\sqrt{1-4k}) and y_2 = x^{\frac{1}{2} (1-\sqrt{1-4k}) solves y''+\frac{k}{x^2}y=0.

    The Wronskian seems to be identially nonzero, so then every solution of y''+\frac{k}{x^2}y=0 can be written as y = C_1 x^{\frac{1}{2} (1+\sqrt{1-4k})} + C_2 x^{\frac{1}{2} (1-\sqrt{1-4k})}.

    The "finitely many positive zeros if k\le 1/4" part follows, but I'm not sure about the "infinite number of positive zeros of k>1/4" part. Obviously the exponents are complex numbers that avoid the real and imaginary axes in that case.

    Any other approaches?
    Last edited by Combinatus; March 28th 2012 at 12:21 PM. Reason: Fixed a typo
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  2. #2
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    Re: Zeros of Euler's equation (y''+(k/x^2)y=0)

    I solved it, I think. y = C_1 x^{\frac{1}{2} (1+\sqrt{1-4k})} + C_2 x^{\frac{1}{2} (1-\sqrt{1-4k})} could be written as the composition of sines and cosines of (the monotonic function) \ln x when k > 1/4 with a preceding (monotonic) factor e^{\ln \sqrt{x}}, so y should have infinitely many solutions in that case.

    It turns out that k=1/4 makes the Wronskian vanish, so with a bit of effort, y_3 = x^{1/2} and y_4 = x^{1/2} \ln x can be found, which can be shown to be linearly independent solutions in this case. The general solution when k=1/4, then, has finitely many zeros in x>0.

    I think this was somehow a stupid and overly lengthy approach to the problem, though, so suggestions for alternatives are welcome!
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