How Does the Airy Function Solve the Given PDE?

[Qyzren]

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 ∞).


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

 

[mighty2000]

!



Thank you for your post. Your approach so far seems correct, but there are a few small mistakes that I would like to point out. First, in your expression for U(w,t), it should be e^(-iw³t) instead of e^(-iw³t). This will affect the final result.

Next, when you interchange the order of integration, you should also change the limits of integration for w to be from 0 to ∞, since the Airy function is only defined for positive values of ω.

Finally, to get the desired form of u(x,t), you need to use the identity for the Fourier transform of the product of two functions. This will give you the integral in the form of the Airy function, with the appropriate scaling factor of (3t)^(-1/3).

I hope this helps you complete your solution. Good luck with your research!