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 ∞).

Attempt:

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