Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.6 STOKES’ EQUATION AND AIRY INTEGRALS

the integrand. Consider the contour integral

y(z)=

∫b

a

f(t) exp(zt)dt, (25.33)

in whicha,bandf(t) are all yet to be chosen. Note that the contour is in the

complext-plane and that the path fromatobcan be distorted as required so

long as no poles of the integrand are trapped between an original path and its

distortion.

Substitution of (25.33) into (25.32) yields

∫b

a

t^2 f(t) exp(zt)dt=

∫b

a

zf(t) exp(zt)dt

=[f(t) exp(zt)]ba−

∫b

a

df(t)

dt

exp(zt)dt.

If we could choose the limitsaandbso that the end-point contributions vanish

then Stokes’ equation would be satisfied by (25.33), providedf(t) satisfies

df(t)

dt

+t^2 f(t)=0 ⇒ f(t)=Aexp(−^13 t^3 ), (25.34)

whereAis any constant.

To make the end-point contributions vanish we must chooseaandbsuch that

exp(−^13 t^3 +zt) = 0 for both values oft. This can only happen if|a|→∞and

|b|→∞and, even then, only if Re (t^3 ) is positive. This condition is satisfied if

2 nπ−^12 π<3arg(t)< 2 nπ+^12 πfor some integern.

Thusaandbmust each be at infinity in one of the three shaded areas shown in

figure 25.10, but clearly not in the same area as this would lead to a zero value

for the contour integral. This leaves three contours (markedC 1 ,C 2 andC 3 in the

figure) that start and end in different sectors. However, only two of them give rise

to independent integrals since the pathC 2 +C 3 is equivalent to (can be distorted

into) the pathC 1.

The two integral functions given particular names are

Ai(z)=

1

2 πi

∫

C 1

exp(−^13 t^3 +zt)dt (25.35)

and

Bi(z)=

1

2 π

∫

C 2

exp(−^13 t^3 +zt)dt−

1

2 π

∫

C 3

exp(−^13 t^3 +zt)dt.

(25.36)

Stokes’ equation is unchanged if the independent variable is changed fromz

toζ,whereζ=exp(2πi/3)z≡Ωz. This is also true for the repeated change

z→Ωζ=Ω^2 z. The same changes of variable, rotations of the complex plane

through 2π/3or4π/3, carry the three contoursC 1 ,C 2 andC 3 into each other,