Proving the Airy Equation with an Improper Integral

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Homework Help Overview

The discussion revolves around proving that the improper integral \( \frac{1}{\pi} \int_0^\infty \cos\left(\frac{1}{3}t^3 + xt\right) dt \) satisfies the Airy equation \( y'' - xy' = 0 \). Participants explore the implications of differentiating under the integral sign and the convergence of the integral involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss differentiating the integral and the challenges of convergence. Questions arise about the use of an integrating factor and its implications for the proof. There is also uncertainty about the mathematical justification for introducing convergence factors.

Discussion Status

The discussion is active, with participants providing insights and clarifications regarding the convergence of integrals and the use of factors to aid in proving the equation. There is no explicit consensus, but several productive lines of reasoning are being explored.

Contextual Notes

Participants note that the problem is considered challenging and is intended for extra credit, indicating a level of complexity that may be beyond the typical coursework for some students involved.

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Homework Statement



So the Airy equation states that y''-xy'=0. My problem is to proof that the improper integral
1/[tex]\pi[/tex] [tex]\int[/tex]cos(1/3*t^3+x*t) from 0 to [tex]\infty[/tex] satisfies this equation.

I've tried differentiating under the integral sign, but all I've gotten is the integrand to be
y''+ty = -1/pi [tex]\int[/tex] cos(1/3*t^3+x*t)(t^2+x) from zero to infinity. Naturally I do u substitution, but my final answer comes out to be -sin(infinity)/Pi

What should I be doing?

Homework Equations





The Attempt at a Solution

 
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After the substitution, you have

[tex]y''-xy=\int_0^\infty \cos u\,du[/tex]

That integral doesn't converge because the integrand doesn't vanish as [itex]u\rightarrow\infty[/itex]. The trick is to introduce an integrating factor [itex]e^{-\lambda u}[/itex] and then take the limit as [itex]\lambda\rightarrow 0^+[/itex].
 
I'm sorry, but i don't quite understand what I do from there. If the integrand is cos u e^-k*u
it'll converge, but do i find k and why you take the limit as k go to 0 from the right?

this problem is supposed to be over the heads of students in our class b/c its extra credit and I'm a freshman in a sophomore class, so I don't know if I'm missing anything.
 
No, you don't find k; you're taking a limit as k goes to 0. The idea is that e-ku will go to 1 as k goes to 0, so the integrand is again just cos u. You have to approach from the right so that -ku<0, otherwise e-ku would go to positive infinity when you plug in the upper limit of the integral.
 
ahhh, my answer works out to zero now. thank you.
 
The only thing I don't get is, if I'm trying to do a proof of y''+xy=0 for a given equation, why am i allowed to put an integrating factor into the equation?? Wouldn't that show I was unable to prove what I wanted to prove?
 
I should have said "convergence factor," not "integrating factor."

Honestly, I don't know the mathematical justification for when such factors are allowed and what their requirements are. It's just a technique that I occasionally ran across in physics, but the mathematical theory behind it was never explained.
 
I'll make sure to read up on convergence factors once my tests are over. Thanks for the help!
 

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