Simplifying an integral expression involving Airy functions

In summary, the conversation discusses the search for an analytical expression for a specific integral involving Airy functions. The speaker mentions their inability to find the desired formula in a published article and their current approach of numerical integration. Another speaker suggests reaching out to the authors of the article for advice. The conversation then shifts to discussing possible solutions and methods, with one speaker proposing a potential solution based on the results in the article. The other speaker is able to successfully derive the desired formula and thanks the first speaker for the inspiration.
  • #1
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1
Hello,
Is there some way to express the following integral in terms of some simpler functions?

[itex]f(x,s) = \int^{\infty}_{-\infty} dk\, e^{-ks} \text{Ai}(-k) \text{Ai}(x-k) [/itex]

where the parameter [itex]s \in (0,1) [/itex] and the coordinate [itex]x \in (-\infty,+\infty) [/itex]

The best I can come up with is to integrate numerically, but it takes time to get a good resolution :(
 
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  • #2
I made a google search, and came up with the following article by Vallee, Soares, de Izarra: "An integral representation for the product of Airy functions", published in ZAMP (1997), pages 156-160
On the following preview image, eq. (12), at the bottom of preview, seems sufficiently close, for some parameter choices, to be worthwhile for you to take a closer look at (the actual results are NOT, unfortunately, part of the preview).
http://link.springer.com/article/10.1007/PL00001464#page-1

In any case, even if that article does not contain what you seek, perhaps the authors might give you valuable advice?

Best of luck!
 
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  • #3
Thanks for your effort! I downloaded the full article and it does not contain the formula I am looking for :( There is a similar one when the 's' in the exponent is imaginary, and they give some general strategy to calculate those integrals, but I need to work it all out by myself and I suspect that it's going to take me a whole lot of time :(
 
  • #4
Irid said:
Thanks for your effort! I downloaded the full article and it does not contain the formula I am looking for :( There is a similar one when the 's' in the exponent is imaginary, and they give some general strategy to calculate those integrals, but I need to work it all out by myself and I suspect that it's going to take me a whole lot of time :(
Not a bull's eye, then?
Okay, I hope you have zoomed in a bit closer on your problem..:smile:
 
  • #5
well I'm integrating the thing numerically, but when x->0, the analytical result is a dirac delta function, but in terms of numerical integration i have to include very high k values to smooth out unphysical oscillations, which is heavy duty on my pc :(
 
  • #6
Aah, so you are to get that Dirac delta perversity, then?
No wonder why that conventionalist minded computer grumbles and writhes in pain.

Couldn't you, in order to avoid that particular numerical problem, try to devise some clever asymptotic matching scheme? (I'm not at all sure it will work, it's more than a decade since I did these sorts of things, so my suggestion might prove silly or impractical)
 
  • #7
arildno said:
Aah, so you are to get that Dirac delta perversity, then?
No wonder why that conventionalist minded computer grumbles and writhes in pain.

Couldn't you, in order to avoid that particular numerical problem, try to devise some clever asymptotic matching scheme? (I'm not at all sure it will work, it's more than a decade since I did these sorts of things, so my suggestion might prove silly or impractical)

yeah I'm trying that right now. Going to adjust my range of k-integration according to the value of s I'm considering, should do the trick. But it would still be cool to have an analytical expression...
 
  • #8
In my experience, functions that behave like beasts in some region of the domain are rarely tamed by analytical means.
Anyways, I would suggest that you pop off an e-mail to one of the authors of the 1997 article, asking if they happen to know of any work done on your specific problem. There's a jungle out there, also in the realm of published materials.
 
  • #9
Using the results in that paper, I believe I have derived

[itex]f(y,s) = {\exp\left[{s^4-6s^2y-3y^2\over12s}\right]\over2\sqrt{\pi}\sqrt{s}}[/itex]

Start with eq.(20), but replace the factor of Ai(x) with exp(-sx), and take z=0. Then I used Mathematica to do the x integral in eq.(21). Mathematica would only do it if I shifted and rescaled x so that the argument of the Airy function was just -x. Then it gave a horrible mess, but using FullSimplify followed by PowerExpand gave a simple result, and then the xi integral was simple, giving the result above.

No promises that I did it right!

EDIT: actually, if we replace s by i*s, then that x integral is just the Fourier transform of Ai(-x). But then the result is analytic in s in some domain which includes imaginary s. So I didn't need Mathematica to do that integral.
 
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  • #10
Avodyne said:
Using the results in that paper, I believe I have derived

[itex]f(y,s) = {\exp\left[{s^4-6s^2y-3y^2\over12s}\right]\over2\sqrt{\pi}\sqrt{s}}[/itex]

Start with eq.(20), but replace the factor of Ai(x) with exp(-sx), and take z=0. Then I used Mathematica to do the x integral in eq.(21). Mathematica would only do it if I shifted and rescaled x so that the argument of the Airy function was just -x. Then it gave a horrible mess, but using FullSimplify followed by PowerExpand gave a simple result, and then the xi integral was simple, giving the result above.

No promises that I did it right!

EDIT: actually, if we replace s by i*s, then that x integral is just the Fourier transform of Ai(-x). But then the result is analytic in s in some domain which includes imaginary s. So I didn't need Mathematica to do that integral.

Wow, this is correct! Could you do the same for this integral as well?

[itex]g(x,s) = \int^{\infty}_{-\infty} dk\, e^{-ks} \text{Ai}(x-k)[/itex]
 
  • #11
Nice that the paper was useful after all!
(Slapping my own back, :approve:)
 
  • #12
woops, never mind, found it myself:

[itex] g(x,s) \propto \exp(-s^3/3-sx) [/itex]

Anyway, thanks for the inspiration
 
  • #13
arildno said:
Nice that the paper was useful after all!
(Slapping my own back, :approve:)

well I guess that paper is way beyond my own level, that's why I have a hard time applying it...
 
  • #14
Irid said:
well I guess that paper is way beyond my own level, that's why I have a hard time applying it...
I am too rusty to follow that paper myself, so kudos to Avodyne who could.
 
  • #15
I've realized that you can do this integral without using the results in the paper. The key is to let [itex]s[/itex] be imaginary, and then analytically continue to real [itex]s[/itex] later.

Start with

[itex]f(x,-iz) = \int^{\infty}_{-\infty} dk\, e^{ikz} \text{Ai}(-k) \text{Ai}(x-k) [/itex]

and substitute in

[itex]\text{Ai}(-k) = {1\over2\pi}\int^{\infty}_{-\infty} dt\, \exp\bigl[it^3\!/3 + i(-k)t\bigr] [/itex]

[itex]\text{Ai}(x-k) = {1\over2\pi}\int^{\infty}_{-\infty} du\, \exp\bigl[iu^3\!/3 + i(x-k)u\bigr] [/itex]

Then do the integral over [itex]k[/itex], which yields [itex]2\pi\delta(z-t-u)[/itex]. Then integrate over [itex]u[/itex], which (because of the delta function) just replaces [itex]u[/itex] with [itex]z-t[/itex] everywhere. Now we have

[itex]f(x,-iz) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[i\bigl(zt^2-(z^2+x)t + z^3\!/3+zx\bigr)\bigr][/itex]

At this point, let [itex]z=is[/itex]. Now we have

[itex]f(x,s) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[-st^2+i(s^2-x)t + s^3\!/3-sx\bigr][/itex]

This is a simple gaussian integral which yields the result I gave above.
 
  • #16
Avodyne said:
I've realized that you can do this integral without using the results in the paper. The key is to let [itex]s[/itex] be imaginary, and then analytically continue to real [itex]s[/itex] later.

Start with

[itex]f(x,-iz) = \int^{\infty}_{-\infty} dk\, e^{ikz} \text{Ai}(-k) \text{Ai}(x-k) [/itex]

and substitute in

[itex]\text{Ai}(-k) = {1\over2\pi}\int^{\infty}_{-\infty} dt\, \exp\bigl[it^3\!/3 + i(-k)t\bigr] [/itex]

[itex]\text{Ai}(x-k) = {1\over2\pi}\int^{\infty}_{-\infty} du\, \exp\bigl[iu^3\!/3 + i(x-k)u\bigr] [/itex]

Then do the integral over [itex]k[/itex], which yields [itex]2\pi\delta(z-t-u)[/itex]. Then integrate over [itex]u[/itex], which (because of the delta function) just replaces [itex]u[/itex] with [itex]z-t[/itex] everywhere. Now we have

[itex]f(x,-iz) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[i\bigl(zt^2-(z^2+x)t + z^3\!/3+zx\bigr)\bigr][/itex]

At this point, let [itex]z=is[/itex]. Now we have

[itex]f(x,s) = {1\over2\pi}\int^{\infty}_{-\infty} dt\,\exp\bigl[-st^2+i(s^2-x)t + s^3\!/3-sx\bigr][/itex]

This is a simple gaussian integral which yields the result I gave above.
Yep, now that you've shown it, it seems easy :D
 

1. What are Airy functions and why are they important in integral expressions?

Airy functions are special mathematical functions that arise in many areas of science, including physics, engineering, and mathematics. They are important in integral expressions because they can help solve certain types of differential equations and are often used to model physical phenomena.

2. How do you simplify an integral expression involving Airy functions?

Simplifying an integral expression involving Airy functions typically involves using various mathematical techniques, such as integration by parts, substitution, and trigonometric identities. It may also involve using properties of Airy functions, such as their asymptotic behavior and recurrence relations.

3. Can an integral expression involving Airy functions be expressed in closed form?

It depends on the specific integral expression. In some cases, it is possible to express the integral in terms of simpler mathematical functions, while in other cases an exact closed form solution may not exist. In these cases, numerical methods may be used to approximate the integral.

4. Are there any special cases or properties to consider when simplifying an integral expression involving Airy functions?

Yes, there are a few special cases and properties to consider when simplifying an integral expression involving Airy functions. For example, the integral may involve complex numbers, in which case techniques for handling complex numbers may be necessary. Additionally, the integral may involve limits of integration that approach infinity, in which case asymptotic properties of Airy functions may be useful.

5. What are some applications of simplifying integral expressions involving Airy functions?

Simplifying integral expressions involving Airy functions has many practical applications in various fields of science and engineering. Some examples include modeling the behavior of optical fibers and studying the flow of fluids in porous media. It can also be used in solving boundary value problems in physics and engineering.

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