Airy functional equation.

  • #1
MathematicalPhysicist
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Main Question or Discussion Point

I want to show that: [tex]Ai(x)+jAi(jx)+j^2Ai(j^2x)=0[/tex], where:
[tex]Ai(x)=\int_{-i\infty}^{i\infty}e^{xz-z^3/3}dz[/tex] and [tex]j=e^{2i\pi/3}[/tex], so far I got that I need to show that:
[tex]e^{zx}+je^{jxz}+j^2e^{xzj^2}=0[/tex] but didn't succeed in doing so.

Any hints?
 

Answers and Replies

  • #3
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I doubt your proposition is correct, but you could try the following:

1) Prove that j^2 is the conjugate of j using Euler's formula

2) write down j=a+ib

3) plug in the condition and do some algebra, using the trig. definitions of sin and cos

You will find (if my calculations are correct) that the number is indeed real. I couldn't show that it vanishes though.

4) use a calculator and see, if it's really so ;)


Another thing that comes up in my mind is to try reproducing the differential equation of the Airy fnct by the equation you have, but it's definitely not easy


Btw, your definition of the Airy fnct. is a little bit different from the one in Wikipedia

Good luck!

marin
 
  • #4
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I am using the textbook by Olivier Vallee and Manuel Soares called: "Airy functions and applications to physics", the identity I am trying to prove is in page 6, equation 2.3, so it's not my proposition.
:smile:
 
  • #5
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hmmm, I guess it must be correct, since you've found it in a book... Unfortunately I have no access to this book :(

Here's something interesting, but I am not sure if it's mathematically correct, since I haven't done complex analysis yet:

Consider the equation:

[tex]e^{zx}+je^{jzx}+j^2e^{j^2zx}=:f(j)[/tex]

As mentioned above, j^2 is the complex conjugate of j. We know that the complex roots of an equation come up as pairs in the real polynomial functions. So I looked for this function.

Now if we consider f(j) as a function of a complex variable, but not over the field C but over the reals R, then all the term e-to-the are equal to 1, since e^{ix} has magnitude 1.**

So the equation simplifies to:

1+j+j^2=0

which solutions are exactly j and j^2.

For j our solution coincides with the given one. But what about j^2?! It turns out that if you plug in the equation j^2 instead of j, the equation does not change since j^4=j :) ! So the symmetry is preserved.


** I am not sure if I am allowed to do so, maybe some of you can discuss it further and say whether and why it's right/wrong.


So much from me for now,

MathematicalPhysicist, if you get the solution, I would be very interested in it :)
 
  • #6
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Ok, I found it in another textbook of Olver called Asymptotic and Special analysis edition 1974, page 88.

Cheers, mate.
(-:
 
  • #7
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ok, could you paste it here, or type it, since I don't have access to this book.

(or if it's too long, just sketch it :))

thanks, marin
 
  • #8
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It's in page 55, sorry for the mistake. Anyway, also in the book it's only a sketch, I'll PM you with the book.
 

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