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Pde involving airy function

  1. Oct 19, 2007 #1
    pde involving airy function!!

    If u(x,t) satisfies ∂u/∂t + ∂³u/∂x³ = 0, with u(x,0) = f(x), and u, ∂u/∂x, ∂²u/∂x² -> 0 as |x| -> ∞, use Fourier transform methods to show that u(x,t) = (3t)^(-1/3) ∫f(y) Ai[(x-y)/((3t)^(-1/3))] dy (integral from -∞ to ∞), where Ai(x) is the Airy function, for which Ai(x) = 1/π ∫cos(ω³/3 + ωx) dω (integral from 0 to ∞).

    all my integrals are from now on are -∞ to ∞
    F{u(x,t)} = U(ω,t) = 1/(2π) ∫u(x,t) e^[iωx] dx
    F{u_t} = ∂U/∂t(ω,t)
    F{u_xxx} = (-iω)³U(ω,t)

    => ∂U/∂t + iω³U = 0
    ∂U/U = -iω³∂t
    (i'm going to use w for ω for more convience)
    U(w,t) = c(w)*e^[-iw³t] where c(w) arb fn of w.
    => u(x,t) = ∫c(w) e^[-iw³t] e^[-iwx] dw
    The IC gives
    u(x,0) = f(x) = ∫c(w) e^[-iwx] dw
    => c(w) = F{f(x)} = 1/(2π) ∫f(x)e^[iwx] dx
    u(x,t) = ∫1/(2π) ∫f(y) e^[iwy] dy e^[-iw³t] e^[-iwx] dw
    interchanging order of integration
    u(x,t) = 1/(2π)∫f(y)[∫e^[-iw(x-y)] e^[-iw³t] dw] dy

    and i'm stuck...
    did i make a mistake somewhere? if not how do i get it to the form they want?
    ie u(x,t) = (3t)^(-1/3) ∫f(y) Ai[(x-y)/((3t)^(-1/3))] dy

    Thank you for your help
  2. jcsd
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