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How to prove derivative of Airy function

  1. Jun 26, 2010 #1
    The Airy function is defined as follows:

    [tex]
    \textrm{Ai}(x) = \frac{1}{\pi}\int\limits_0^{\infty} \cos\big(\frac{1}{3}t^3 + xt\big)dt
    [/tex]

    It seems pretty probable, that its derivative is this:

    [tex]
    \textrm{Ai}'(x) = -\frac{1}{\pi}\int\limits_0^{\infty} t\sin\big(\frac{1}{3}t^3 + xt\big)dt
    [/tex]

    I'm interested to know how this result is proven. It's not as simple as it might seem on quick glance. Only thing that's trivial is this:

    [tex]
    \textrm{Ai}'(x) = \lim_{\Delta x\to 0}\frac{1}{\pi}\int\limits_0^{\infty}
    \frac{\cos\big(\frac{1}{3}t^3 + (x+\Delta x)t\big) - \cos\big(\frac{1}{3}t^3 + xt\big)}{\Delta x}dt
    = -\lim_{\Delta x\to 0}\frac{1}{\pi}\int\limits_0^{\infty} t\sin\big(\frac{1}{3}t^3 + (x+\xi_{t,\Delta x})t\big) dt
    [/tex]

    Where [itex]|\xi_{t,\Delta x}|\leq |\Delta x|[/itex]. So the non-trivial task is to prove that the order of limit and integration can be changed. The integral does not converge absolutely, but instead relies on cancellation of accelerating oscillations. If we have no better knowledge about [itex]\xi_{t,\Delta x}[/itex], it could behave in such manner that it starts picking more positive parts of the integrand than negative parts, and have surprising effect on the value of the integral.


    EDIT 1:

    At first I thought it would be simple thing to prove that

    [tex]
    \textrm{Ai}''(x) = x\textrm{Ai}(x)
    [/tex]

    after it has been shown that the derivative and integral can be commuted, but actually that's not the only problem. If I assume that I can commute [itex]D_x^2[/itex] with the integral, I get the following:

    [tex]
    \textrm{Ai}''(x) = -\frac{1}{\pi}\int\limits_0^{\infty} t^2\cos\big(\frac{1}{3}t^3 + xt\big)dt
    = -\frac{1}{\pi}\int\limits_0^{\infty}\Big( D_t\sin\big(\frac{1}{3}t^3 + xt\big) \;-\; x\cos\big(\frac{1}{3}t^3 + xt\big)\Big)dt
    [/tex]

    But [itex]\lim_{t\to\infty}\sin(\frac{1}{3}t^3 + xt)[/itex] does not exist and is not zero, so actually we can't conclude that the integral of the derivative would be zero. So how do you compute the second derivative then?

    EDIT 2:

    hmhmh... I just noticed that

    [tex]
    \int\limits_0^{\infty} t^{\alpha} \cos\big(\frac{1}{3}t^3 + xt\big)dt = \frac{1}{3}\int\limits_0^{\infty} u^{(\alpha - 2)/3} \cos\big(\frac{1}{3}u + xu^{1/3}\big) du
    [/tex]

    seems to converge only when [itex]-1 < \alpha < 2[/itex], and diverges for [itex]2 \leq \alpha[/itex].
     
    Last edited: Jun 26, 2010
  2. jcsd
  3. Jun 27, 2010 #2
    Use some complex analysis. I.e, use a variant of Jordan's lemma to rotate the contour in the complex plane so that the integrand converges exponentially. The problem then becomes trivial.
     
  4. Jun 27, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is generally true that
    [tex]\frac{d}{dx}\int_a^b f(x,t)dt= \int_a^b \frac{\partial f}{\partial x} dt[/tex]

    Are you allowed to use that or is that part of what you are trying to prove?
     
  5. Jun 29, 2010 #4
    An application of that rule could only give us a formal result, because it is not clear that the right hand side of your expression exists in this case. If the derivative were dominated by some integrable function, we could apply your result, but we can't do that here.

    However, if we first rotate the contour slightly (using a Jordan's lemma type argument) then our integrand converges exponentially and we can apply your result. :)
     
  6. Aug 27, 2010 #5
    [tex]
    \textrm{Ai}(x) = \frac{1}{\pi}\int\limits_0^{\infty} \cos\big(\frac{1}{3}t^3 + xt\big) dt
    = \frac{1}{2\pi} \int\limits_0^{\infty} e^{i(\frac{1}{3}t^3 + xt)}dt
    \;+\;\frac{1}{2\pi} \int\limits_0^{\infty} e^{-i(\frac{1}{3}t^3 + xt)}dt
    [/tex]

    So... the domains of integration should be rotated counter clockwise for the first integrand, and clockwise for the second integrand. I'll fix some [itex]0<\theta<\frac{\pi}{3}[/itex], and then examine counter clockwise integration domain

    [tex]
    [0,R]\;\cup\; \{Re^{i\phi}\;|\;0\leq\phi\leq\theta\}\;\cup\; \{re^{i\theta}\;|\;0\leq r\leq R\}
    [/tex]

    for the first integrand.

    With fixed [itex]\theta[/itex], it is possible to find [itex]a>0[/itex] so that [itex]a\phi \leq \sin(\phi)[/itex] for all [itex]0\leq\phi\leq 3\theta[/itex]. The integral over the arc can be approximated above as follows.

    [tex]
    \Big| \int\limits_0^{\theta} e^{i(\frac{1}{3}R^3e^{3i\phi} + xRe^{i\phi})} Rie^{i\phi} d\phi\Big|
    \leq R\int\limits_0^{\theta} e^{-\frac{1}{3}R^3\sin(3\phi) - xR\sin(\phi)}d\phi
    \leq R\int\limits_0^{\theta}e^{(-aR^3 + |x|R)\phi} d\phi
    [/tex]
    [tex]
    = \frac{1}{aR^2 - |x|}\Big(1 - e^{(-aR^3+|x|R)\theta}\Big) \underset{R\to\infty}{\to} 0
    [/tex]

    This means that

    [tex]
    \textrm{Ai}(x) = \frac{1}{2\pi} \int\limits_0^{\infty} e^{i(\frac{1}{3}t^3 e^{3i\theta} + xte^{i\theta})} e^{i\theta} dt
    \;+\;\frac{1}{2\pi} \int\limits_0^{\infty} e^{-i(\frac{1}{3}t^3e^{-3i\theta} + xte^{-i\theta})} e^{-i\theta} dt
    [/tex]
    [tex]
    =\frac{1}{\pi}\textrm{Re}\Big( \int\limits_0^{\infty}
    e^{\frac{1}{3}t^3(-\sin(3\theta)+i\cos(3\theta)) +xt(-\sin(\theta) + i\cos(\theta)) + i\theta} dt\Big)
    [/tex]

    The rest is clear.


    -------------

    Life is lot easier for those who simply believe Wikipedia: http://en.wikipedia.org/wiki/Airy_function

    *sigh!*
     
    Last edited: Aug 27, 2010
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