Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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25.6 STOKES’ EQUATION AND AIRY INTEGRALS


contour integral (25.35) with the particular solution of Stokes’ equation that


decays monotonically to zero for realz>0as|z|→∞. As discussed in subsection


25.6.1, all solutions except the one called Ai(z)tendto±∞asz(real) takes on


increasingly large positive values and so their asymptotic forms reflect this. In a


worked example in subsection 25.8.2 we use the method of steepest descents (a


saddle-point method) to show that the function defined by (25.39) has exactly


the characteristic asymptotic property expected of Ai(z) (see page 911). It follows


that it is the same function as Ai(z), up to a real multiplicative constant.


The choice of definition (25.36) as the other named solution Bi(z)ofStokes’

equation is a less obvious one. However, it is made on the basis of its behaviour for


negative real values ofz. As discussed earlier, Ai(z) oscillates almost sinusoidally


in this region, except for a relatively slow increase in frequency and an even


slower decrease in amplitude as−zincreases. The solution Bi(z) is chosen to be


the particular function that exhibits the same behaviour as Ai(z)exceptthatitis


in quadrature with Ai, i.e. it isπ/2 out of phase with it. Specifically, asx→−∞,


Ai(x)∼

1

2 πx^1 /^4

sin

(
2 |x|^3 /^2
3

+

π
4

)
, (25.40)

Bi(x)∼

1

2 πx^1 /^4

cos

(
2 |x|^3 /^2
3

+

π
4

)

. (25.41)


There is a close parallel between this choice and that of taking sine and cosine


functions as the basic independent solutions of the simple harmonic oscillator


equation. Plots of Ai(z)andBi(z)forrealzare shown in figure 25.11.


By choosing a suitable contour forC 1 in (25.35), expressAi(0)in terms of the gamma
function.

Withzset equal to zero, (25.35) takes the form


Ai(0) =

1


2 πi


C 1

exp(−^13 t^3 )dt.

We again use the freedom to choose the specific line of the contour so as to make the
actual integration as simple as possible.
HereweconsiderC 1 as made up of two straight-line segments: one along the line
argt=4π/3, starting at infinity in the correct sector and ending at the origin; the other
starting at the origin and going to infinity along the line argt=2π/3, thus ending in the
correct final sector. On each, we set^13 t^3 =s,wheresis real and positive on both lines.
Thendt=e^4 πi/^3 (3s)−^2 /^3 dson the first segment anddt=e^2 πi/^3 (3s)−^2 /^3 dson the second.

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