I don't understand how the solution of 'D2x-x*t=0'can be
x(t,x)?=(1/pi)int(cos((1/3)*t^3+x*t),t,0,inf).That is to say, what is the x doing on the right side of the equation. Or how can x=f(x,t)?
replied to: georgesimpso
Replied to: I don't understand how the solution of 'D2x-x*t=0'can be
x(t,x)?=(1/pi)int(cos((1/3)*t^3+x*t),t,0,inf).That is...
Well, you're dealing with Airy's ODE
d^2x/dt^2 - xt = 0, or x'' - tx = 0.
Pair of linearly independent of this ODE are the Airy functions Ai(t) and Bi(t). You're asking for an integral representation of the former one. So either x(t,x) or x=f(x,t) don't make sense. Indeed such an integral representation for Ai(x) is given by the integral
x(t) = Ai(t) = (1/pi)int_0^infty cos(u^3/3 + ut)du,
where you easily see that the solution x(t)=Ai(t) depends only on the independent variable t (of course you know that "u" is a "dumb" variable).
Well, I'm writing your solution by using LaTeX, so I explain you those symbols
pi = 3.14159...
int stands for integration
_0 and ^infty = infinite are the limits of integration
cos = cosine function
_ = underline to indicate the lower limit of int.
^ = is used for the upper limit of int. but also stands for power symbol, e.g t^3.
Hope all of this helps.
perucho
replied to: perucho
Replied to: Well, you're dealing with Airy's ODE
d^2x/dt^2 - xt = 0,...
Hi,
I had a couple of mistakes, sorry.
By "linearly independent" I meant "linearly independent solutions"
On the fifth row I put Ai(x), but correct is Ai(t).
No big deal but anyway.
