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Thread: Parabolic cylinder functions

  1. #1
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    Parabolic cylinder functions

    Hi everyone! I have a differential equation of the form

    $\displaystyle \frac{d^2 f}{d x^2} + \lambda^2(1 + x^2)f(x) = 0.$

    The solutions to this equation are the parabolic cylinder functions. Using the definition as stated in Wikipedia, the even and odd solutions are $\displaystyle y_1(i \lambda/2; \sqrt{2 i \lambda}x)$ and $\displaystyle y_2(i \lambda/2; \sqrt{2 i \lambda}x)$, respectively. For real, nonzero values of $\displaystyle \lambda$, these solutions are oscillating with decreasing amplitude and period. I am interested in calculating the value of

    $\displaystyle \int_0^\infty dx\:y_1(i\lambda/2; \sqrt{2 i\lambda}x)$

    as a function of $\displaystyle \lambda$. With the normalization as used in the Wikipedia article, this value is a positive and finite constant (when $\displaystyle \lambda \neq 0$). I found numerically for $\displaystyle \lambda = 1$ that this value is $\displaystyle \approx 0.57$. Can anyone help me with this problem?

    Kind regards,
    Simon
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  2. #2
    Junior Member
    Joined
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    Re: Parabolic cylinder functions

    I have made some progress, but the problem is not yet solved. Doing some further numerical studies, I found the value of the integral for a set of $\displaystyle \lambda$. For $\displaystyle \lambda \ll 1$ the value seems to be $\displaystyle \approx 0.85 \lambda^{-0.32}$ (dashed curve in the figure).

    Parabolic cylinder functions-lambda.png

    Edit: Never mind about the fit (dashed curve) =P. For even smaller $\displaystyle \lambda$ the integral curve has the same curvature in the log-log plot, so that the proportionality constant (0.85) and the exponent (-0.32) continuously increases.
    Last edited by sitho; Jul 17th 2014 at 03:15 AM.
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