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Prove that y(x) has finitely many positive zero

  1. Dec 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Given $$y''+e^{-x}y=0. \qquad (*)$$ Let ##y(x)## be any nontrivial solution of ##(*)##, show that y has finitely many positive zeros.
    Hint: Consider ##z''+\frac{C}{x^4}z=0## where ##C>0## is sufficiently large, which has a solution ##z(x)=x\sin \frac{\sqrt C}{x}##.

    2. Relevant equations
    1. (Sturm's Comparison Theorem)
    Let y(x) and z(x) be nontrivial solutions of
    $$y'' + q(x)y = 0$$
    and
    $$z'' + r(x)z = 0, $$
    where q(x) and r(x) are positive functions such that q(x) > r(x). Then y(x) vanishes at least once between any two successive zeros of z(x).
    2. Moreover, if there exists a constant ##m>0## such that ##q(x) \geq m^2## for all x, then y(x) has infinitely many zeros.

    3. The attempt at a solution
    My guess is ##C## must be big enough s.t ##\frac{C}{x^4}>e^{-x}##, so we can make use of the theorem. I also think that a proof by contradiction (i.e. suppose y has infinitely many positive zeros) may be useful, but don't know how to proceed further from here.
    Could anyone shed some light on it please?
     
  2. jcsd
  3. Dec 25, 2013 #2

    mfb

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    Your attempt looks good so far.

    "Then y(x) vanishes at least once between any two successive zeros of z(x). "

    As you use q(x) > r(x), q(r) will be your C/x^4 and r(x) will be e^(-x).
    Therefore, your contradiction starts with "assume ##z'' + r(x)z = 0## has infinitely many zeroes [...]"
    And then use the known solution for the other case to get a contradiction.


    That is a nice theorem.
     
  4. Dec 25, 2013 #3

    haruspex

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    Right, but noting that y and z in the problem statement have reversed roles from the y and z in the theorem statement.
    Also, it doesn't quite get you there because sin(1/x) has infinitely many positive zeroes. You might need to split the range of x into (0, ε) and [ε, ∞).
     
  5. Dec 26, 2013 #4
    Thank you for the help, but doesn't sin(1/x) have infinitely many zeros in [ε, ∞)?
     
  6. Dec 26, 2013 #5

    haruspex

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    I don't think so. Can you find any in (1/π, ∞)?
     
  7. Dec 26, 2013 #6
    Oh that's enlightening, thank you!
     
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