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MathGroup Archive 2003

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Airy's Gi(x) function; asymptotic matching and asymptotic limits

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43783] Airy's Gi(x) function; asymptotic matching and asymptotic limits
  • From: "Curt Fischer" <crf3 at po.cwru.edu>
  • Date: Sat, 4 Oct 2003 02:05:01 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Group,

Recently I had to solve the differential equation y''[x] - x y[x] == -1,
with one known boundary condition y'[0]==0.

The general solution is conveniently written as C[1] AiryAi[x] + C[2]
AiryBi[x] + Pi airyGi[x].  One of the constants can be solved for with
respect to the other

Question 1:

Mathematica does not have the airyGi[x] function built-in.  It returns the
solution
\!\(1\/2\ \((2\ \@3\ AiryAi[x]\ C[2] + 2\ AiryBi[x]\ C[2] -
2\ x\^2\ Hypergeometric0F1[4\/3,
x\^3\/9]\ HypergeometricPFQ[{1\/3}, {2\/3, 4\/3}, x\^3\/9] +
x\^2\ Hypergeometric0F1[2\/3,
x\^3\/9]\ HypergeometricPFQ[{2\/3}, {4\/3, 5\/3}, x\^3\/9])\)\)

which is a big messy expression involving AiryAi[x], AiryBi[x], and
hypergeometric functions.  Does anyone know how I can relate this
hypergeometric stuff is equal to airyGi[x] == Integrate[Sin[t^3 + z t]
dz,{t,0,infinity}] ?

Question 2: When I solved my problem analytically, I was interested in
evaluating the unknown integration constant by asymptotic matching to
another function which I knew.  This worked great on paper, but Mathematica
could not take the limit of

2\ x\^2\ Hypergeometric0F1[4\/3,
x\^3\/9]\ HypergeometricPFQ[{1\/3}, {2\/3, 4\/3}, x\^3\/9] +
x\^2\ Hypergeometric0F1[2\/3,
x\^3\/9]\ HypergeometricPFQ[{2\/3}, {4\/3, 5\/3}, x\^3\/9])\)\)

Is there a way to evaluate this limit in Mathematica?  Also, in general is
there anyway to get the an "asymptotic limit" of a function in Mathematica?
For example, airyGi[x] -> 1/(Pi x) for large x.  Is there any way to elicit
this type of info about a function from Mathematica?

(See Abramowitz and Stegun, 1974, Handbook of Mathematical Functions, for
this and other info on Airy functions.)

thanks for any help anyone can provide,



Curt Fischer





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