Mathematical Methods for Physics and Engineering : A Comprehensive Guide

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.