Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Showing Integral form satisfies the Airy function

  1. Dec 10, 2009 #1
    Hey All,

    Can someone please give me the gist of how to show that the integral form of the Airy function for real inputs:
    [tex]
    Ai(x) = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,
    [/tex]

    satisfies the Airy Differential Equation: y'' - xy = 0

    I tried differentiating twice wrt to the x variable (assuming I could just bring it inside the integration) and then subbing back into the ODE but that failed.

    Regards,
    Thrillhouse
     
  2. jcsd
  3. Dec 18, 2009 #2
    Assume that
    [tex]y = \frac{1}{\pi} \int_0^\infty \cos\left(\tfrac13t^3 + xt\right)\, dt,[/tex]

    and the operations integrate and differentiate can be interchange (??), I obtain

    [tex]y''-xy = -\frac{1}{\pi} \int_0^\infty (t^2+x)\cos\left(\tfrac13t^3 + xt\right)\, dt[/tex]

    Why is RHS identically zero ? :cry:
     
    Last edited: Dec 18, 2009
  4. Dec 18, 2009 #3
    make a change of variables:

    [tex]u=\frac{1}{3}t^{3}+xt => du=(t^{2}+x)dt [/tex]

    you will simply have:

    [tex]-\frac{1}{\pi}\int^{\infty}_{0}cos(u)du[/tex]

    that technically doesn't converge to zero. So that's the most far you can get
     
  5. Dec 19, 2009 #4
    surely the positive and negative components of the cos function will add up to zero when you integrate ?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook