Stokes equation and Airy integrals

APPLICATIONS OF COMPLEX VARIABLES
25.6 Stokes’ equation and Airy integrals
Much of the analysis of situations occurring in physics and engineering is concerned with what happens at a boundary within or surrounding a physical system.
Sometimes the existence of a boundary imposes conditions on the behaviour of
variables describing the state of the system; obvious examples include the zero
displacement at its end-points of an anchored vibrating string and the zero
potential contour that must coincide with a grounded electrical conductor.
More subtle are the effects at internal boundaries, where the same non-vanishing
variable has to describe the situation on either side of the boundary but its
behaviour is quantitatively, or even qualitatively, different in the two regions. In
this section we will study an equation, Stokes’ equation, whose solutions have
this latter property; as well as solutions written as series in the usual way, we will
find others expressed as complex integrals.
The Stokes’ equation can be written in several forms, e.g.
d2 y
+ λxy = 0;
dx2
d2 y
+ xy = 0;
dx2
d2 y
= xy.
dx2
We will adopt the last of these, but write it as
d2 y
= zy
dz 2
(25.32)
to emphasis that its complex solutions are valid for a complex independent
variable z, though this also means that particular care has to be exercised when
examining their behaviour in different parts of the complex z-plane. The other
forms of Stokes’ equation can all be reduced to that of (25.32) by suitable
(complex) scaling of the independent variable.
25.6.1 The solutions of Stokes’ equation
It will be immediately apparent that, even for z restricted to be real and denoted
by x, the behaviour of the solutions to (25.32) will change markedly as x passes
through x = 0. For positive x they will have similar characteristics to the solutions
of y = k 2 y, where k is real; these have monotonic exponential forms, either
increasing or decreasing. On the other hand, when x is negative the solutions
will be similar to those of y + k 2 y = 0, i.e. oscillatory functions of x. This is
just the sort of behaviour shown by the wavefunction describing light diffracted
by a sharp edge or by the quantum wavefunction describing a particle near to
the boundary of a region which it is classically forbidden to enter on energy
grounds. Other examples could be taken from the propagation of electromagnetic
radiation in an ion plasma or wave-guide.
Let us examine in a bit more detail the behaviour of plots of possible solutions
y(z) of Stokes’ equation in the region near z = 0 and, in particular, what may
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25.6 STOKES’ EQUATION AND AIRY INTEGRALS
(b)
y
(c)
(a)
(a)
(c)
z
(b)
Figure 25.9 Behaviour of the solutions y(z) of Stokes’ equation near z = 0
for various values of λ = −y (0). (a) with λ small, (b) with λ large and (c) with
λ appropriate to the Airy function Ai(z).
happen in the region z > 0. For definiteness and ease of illustration (see figure
25.9), let us suppose that both y and z, and hence the derivatives of y, are real and
that y(0) is positive; if it were negative, our conclusions would not be changed
since equation (25.32) is invariant under y(z) → −y(z). The only difference would
be that all plots of y(z) would be reflected in the z-axis.
We first note that d2 y/dx2 , and hence also the curvature of the plot, has the
same sign as z, i.e. it has positive curvature when z > 0, for so long as y(z)
remains positive there. What will happen to the plot for z > 0 therefore depends
crucially on the value of y (0). If this slope is positive or only slightly negative
the positive curvature will carry the plot, either immediately or ultimately, further
away from the z-axis. On the other hand, if y (0) is negative but sufficiently large
in magnitude, the plot will cross the y = 0 line; if this happens the sign of the
curvature reverses and again the plot will be carried ever further from the z-axis,
only this time towards large negative values.
Between these two extremes it seems at least plausible that there is a particular
negative value of y (0) that leads to a plot that approaches the z-axis asymptotically, never crosses it (and so always has positive curvature), and has a slope
that, whilst always negative, tends to zero in magnitude. There is such a solution, known as Ai(z), whose properties we will examine further in the following
subsections. The three cases are illustrated in figure 25.9.
The behaviour of the solutions of (25.32) in the region z < 0 is more straight889
APPLICATIONS OF COMPLEX VARIABLES
forward, in that, whatever the sign of y at any particular point z, the curvature
always has the opposite sign. Consequently the curve always bends towards the
z-axis, crosses it, and then bends towards the axis again. Thus the curve exhibits
oscillatory behaviour. Furthermore, as −z increases, the curvature for any given
|y| gets larger; as a consequence, the oscillations become increasingly more rapid
and their amplitude decreases.
25.6.2 Series solution of Stokes’ equation
Obtaining a series solution of Stokes’ equation presents no particular difficulty
when the methods of chapter 16 are used. The equation, written in the form
d2 y
− zy = 0,
dz 2
has no singular points except at z = ∞. Every other point in the z-plane is
an ordinary point and so two linearly independent series expansions about it
(formally with indicial values σ = 0 and σ = 1) can be found. Those about z = 0
∞
n
n+1
. The corresponding recurrence relations
take the forms ∞
0 an z and
0 bn z
are
(n + 3)(n + 2)an+3 = an
and (n + 4)(n + 3)bn+3 = bn ,
and the two series (with a0 = b0 = 1) take the forms
z6
z3
+
+ ··· ,
(3)(2) (6)(5)(3)(2)
z7
z4
+
+ ··· .
y2 (z) = z +
(4)(3) (7)(6)(4)(3)
y1 (z) = 1 +
The ratios of successive terms for the two series are thus
z3
an+3 z n+3
=
n
an z
(n + 3)(n + 2)
and
bn+3 z n+4
z3
.
=
n+1
bn z
(n + 4)(n + 3)
It follows from the ratio test that both series are absolutely convergent for all z.
A similar argument shows that the series for their derivatives are also absolutely
convergent for all z. Any solution of the Stokes’ equation is representable as a
superposition of the two series and so is analytic for all finite z; it is therefore an
integral function with its only singularity at infinity.
25.6.3 Contour integral solutions
We now move on to another form of solution of the Stokes’ equation (25.32), one
that takes the form of a contour integral in which z appears as a parameter in
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25.6 STOKES’ EQUATION AND AIRY INTEGRALS
the integrand. Consider the contour integral
b
y(z) =
f(t) exp(zt) dt,
(25.33)
a
in which a, b and f(t) are all yet to be chosen. Note that the contour is in the
complex t-plane and that the path from a to b can 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
b
t2 f(t) exp(zt) dt =
z f(t) exp(zt) dt
a
a
= [ f(t) exp(zt) ] ba −
b
a
df(t)
exp(zt) dt.
dt
If we could choose the limits a and b so that the end-point contributions vanish
then Stokes’ equation would be satisfied by (25.33), provided f(t) satisfies
df(t)
+ t2 f(t) = 0 ⇒ f(t) = A exp(− 13 t3 ),
(25.34)
dt
where A is any constant.
To make the end-point contributions vanish we must choose a and b such that
exp(− 13 t3 + zt) = 0 for both values of t. This can only happen if |a| → ∞ and
|b| → ∞ and, even then, only if Re (t3 ) is positive. This condition is satisfied if
2nπ − 12 π < 3 arg(t) < 2nπ + 12 π for some integer n.
Thus a and b must 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 (marked C1 , C2 and C3 in the
figure) that start and end in different sectors. However, only two of them give rise
to independent integrals since the path C2 + C3 is equivalent to (can be distorted
into) the path C1 .
The two integral functions given particular names are
1
exp(− 13 t3 + zt) dt
(25.35)
Ai(z) =
2πi C1
and
Bi(z) =
1
2π
C2
exp(− 13 t3 + zt) dt −
1
2π
C3
exp(− 13 t3 + zt) dt.
(25.36)
Stokes’ equation is unchanged if the independent variable is changed from z
to ζ, where ζ = exp(2πi/3)z ≡ Ωz. This is also true for the repeated change
z → Ωζ = Ω2 z. The same changes of variable, rotations of the complex plane
through 2π/3 or 4π/3, carry the three contours C1 , C2 and C3 into each other,
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APPLICATIONS OF COMPLEX VARIABLES
Im t
C3
C1
Re t
C2
Figure 25.10 The contours used in the complex t-plane to define the functions
Ai(z) and Bi(z).
though sometimes the sense of traversal is reversed. Consequently there are
relationships connecting Ai and Bi when the rotated variables are used as their
arguments. As two examples,
Ai(z) + ΩAi(Ωz) + Ω2 Ai(Ω2 z) = 0,
(25.37)
Bi(z) = i[ Ω2 Ai(Ω2 z) − ΩAi(Ωz) ] = e−πi/6 Ai(ze−2πi/3 ) + eπi/6 Ai(ze2πi/3 ).
(25.38)
Since the only requirements for the integral paths is that they start and end in
the correct sectors, we can distort path C1 so that it lies on the imaginary axis
for virtually its whole length and just to the left of the axis at its two ends. This
enables us to obtain an alternative expression for Ai(z), as follows.
Setting t = is, where s is real and −∞ < s < ∞, converts the integral representation of Ai(z) to
∞
1
exp[ i( 13 s3 + zs) ] ds.
Ai(z) =
2π −∞
Now, the exponent in this integral is an odd function of s and so the imaginary
part of the integrand contributes nothing to the integral. What is left is therefore
1 ∞
Ai(z) =
cos( 13 s3 + zs) ds.
(25.39)
π 0
This form shows explicitly that when z is real, so is Ai(z).
This same representation can also be used to justify the association of the
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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 real z > 0 as |z| → ∞. As discussed in subsection
25.6.1, all solutions except the one called Ai(z) tend to ±∞ as z (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) of Stokes’
equation is a less obvious one. However, it is made on the basis of its behaviour for
negative real values of z. 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 −z increases. The solution Bi(z) is chosen to be
the particular function that exhibits the same behaviour as Ai(z) except that it is
in quadrature with Ai, i.e. it is π/2 out of phase with it. Specifically, as x → −∞,
2|x|3/2
π
1
+
sin
,
3
4
2πx1/4
2|x|3/2
π
1
+
cos
.
Bi(x) ∼ √
3
4
2πx1/4
Ai(x) ∼ √
(25.40)
(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) and Bi(z) for real z are shown in figure 25.11.
By choosing a suitable contour for C1 in (25.35), express Ai(0) in terms of the gamma
function.
With z set equal to zero, (25.35) takes the form
Ai(0) =
1
2πi
C1
exp(− 13 t3 ) 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.
Here we consider C1 as made up of two straight-line segments: one along the line
arg t = 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 arg t = 2π/3, thus ending in the
correct final sector. On each, we set 13 t3 = s, where s is real and positive on both lines.
Then dt = e4πi/3 (3s)−2/3 ds on the first segment and dt = e2πi/3 (3s)−2/3 ds on the second.
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APPLICATIONS OF COMPLEX VARIABLES
1
0.5
−10
z
−5
−0.5
Figure 25.11 The functions Ai(z) (full line) and Bi(z) (broken line) for real
z.
Then we have
Ai(0) =
1
2πi
0
e−s e4πi/3 (3s)−2/3 ds +
∞
1
2πi
∞
e−s e2πi/3 (3s)−2/3 ds
0
∞
3−2/3
e−s (−e4πi/3 + e2πi/3 )s−2/3 ds
2πi 0
√ −2/3 ∞
3i 3
=
e−s s−2/3 ds
2πi
0
=
=
3−1/6 1
Γ( 3 ),
2π
where we have used the standard integral defining the gamma function in the last line. Finally in this subsection we should mention that the Airy functions and their
derivatives are closely related to Bessel functions of orders ± 13 and ± 23 and that
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