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-4
10
-5
10
-6
10
-7
10
-8
10
0 5 10 15 20 25 30
j

Figure 3. Stieltjes function with =1/10. Solid-with-x™s: Absolute value of the abso-
lute error in the partial sum of the asymptotic series, up to and including aj where
j is the abscissa. Dashed-with-circles: The result of Euler acceleration. The terms
up to and including the optimum order, here Nopt ( ) = 9, are unweighted. Terms of
degree j > Nopt are multiplied by the appropriate Euler weight factors as described
in the text. The circle above j = 15 is thus the sum of nine unweighted and six
Euler-weighted terms.


Qualitatively, the numerator resembles a Gaussian centered on t =
1/ . The heart of the “steepest descent” method for evaluating integrals
is to (i) rewrite the rapidly varying part of the integral as an exponential
(ii) make a change of variable so that this exponential is equal to the
Gaussian function exp(’z 2 / ) and expand dt/dz, multiplied by the
slowly varying part of the integral (here 1/(1 + t(z), in powers of z.
Since this method is described in Sec. 11 below, the details will be
omitted here. The lowest order is identical with the lowest order Euler
approximation.
W. G. C. Boyd (no relation) has developed systematic methods for
integrals that are Stieltjes functions, a class that includes the Stielt-
jes function as a special case[77, 78, 79, 80]. The simpler treatment
described here is based on Olver™s monograph[249] and forty-year old
articles by Rosser[273, 274].




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21
Exponential Asymptotics

1



µ=1/5
0.8


0.6
Integrand




0.4


0.2

µ=1/80
0
0 0.5 1 1.5 2
T

Figure 4. Integrand of the integral ENoptimum ( ), which is the error in the regular
asymptotic series truncated at the N “th term, as a function of T ≡ t for =
1/5, 1/10, 1/20, 1/40, 1/80 in order of increasing narrowness.


6. A Linear Di¬erential Equation



Our second example is the linear problem

uxx ’ u = ’f (x)
2
(23)

on the in¬nite interval x ∈ [’∞, ∞] subject to the conditions that
both |u(x)|, |f (x)| ’ 0 as |x| ’ ∞ where the subscripts denote second
di¬erentiation with respect to x, f (x) is a known forcing function, and
u(x) is the unknown. This problem is a prototype for boundary layers
in the sense that the term multiplying the highest derivative formally
vanishes in the limit ’ 0, but it has been simpli¬ed further by omit-
ting boundaries. The divergence, however, is not eliminated when the
boundaries are.
At ¬rst, this linear boundary value problem seems very di¬erent
from the Stieltjes integral. However, Eq. (23) is solved without approx-
imation by the Fourier integral
∞ F (k)
u(x) = exp(ikx)dk (24)
1 + 2 k2
’∞




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22 John P. Boyd

where F (k) is the Fourier transform of the forcing function:

1
F (k) = f (x) exp(’ikx)dx (25)
2π ’∞

The Fourier integral (24) is very similar in form to the Stieltjes
function. To be sure, the range of integration is now in¬nite rather
than semi-in¬nite and the exponential has a complex argument. The
similarity is crucial, however: for both the Stieljes integral and the
Fourier integral, expanding the denominator of the integrand in powers
of generates an asymptotic series. In both cases, the series is divergent
because the expansion of the denominator has only a ¬nite radius of
convergence whereas the range of integration is unbounded.
The asymptotic solution to (23) may be derived by either of two
routes. One is to expand 1/(1 + 2 k 2 ) as a series in and then recall
that the product of F (k) with (’k 2 ) is the transform of the second
derivative of f(x) for any f(x). The second route is to use the method of
multiple scales. If we assume that the solution u(x) varies only on the
same “slow” O(1) length scale as f (x), and not on the “fast” O(1/ )
scale of the homogeneous solutions of the di¬erential equation, then the
second derivative may be neglected to lowest order to give the solution
u(x) ≈ f (x). This is called the “outer” solution in the language of
matched asymptotic expansions.) Expanding u(x) as a series of even
powers of and continuing this reasoning to higher order gives
∞ 2j f
2d
u(x) ∼ (26)
dx2j
j=0

This di¬erential equation seems to have little connection to our pre-
vious example, but this is a mirage. For the special case
4
f (x) = (27)
1 + x2
the Fourier transform F (k) = 2 exp(’ | k | ) . Using the partial fraction
expansion 1/(1 + 2 k 2 ) = (1/2){1/(1 ’ i k) + 1/(1 + i k)}, one can
show that the solution to (23) is
1 i i
S’
u(x; ) = +S
1 + ix 1 + ix 1 + ix
1 i i
S’
+ +S (28)
1 ’ ix 1 ’ ix 1 ’ ix
where S( ) is the Stieltjes function. At x = 0, the solution simpli¬es
to u(0) = 2{S(i ) + S(’i )}. The odd powers of cancel, but the even




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23
Exponential Asymptotics

powers reinforce to give

u(0) ∼ 4 2j
(2j)! (’1)j (29)
j=0

There is nothing special about the Lorentzian function (or x = 0),
however. As explained at greater length in [61] and [69], the exponential
decay of a Fourier transform with wavenumber k is generic if f (x) is
free of singularities for real x. The factorial growth of the power series
coe¬cients with j, explicit in (29), is typical of the general multiple
scale series (26) for all x for most forcing functions f (x).
To obtain the optimal truncation, apply the identity 1/(1 + z) =
N N +1 /(1 + z) for all z and any positive integer N to
j
j=0 (’z) + (’z)
the integral (24) with z = 2 k 2 to obtain, without approximation,

N ∞
2j f k 2(N +1) F (k)
2d N +1 2(N +1)
u= + (’1) exp(ikx)dk (30)
dx2j 1 + 2 k2
’∞
j=0

The N -th order asymptotic approximation is to neglect the integral. For
large N , the error integral in Eq. (30) can be approximatedly evaluated
by steepest descent (Sec. 11 below). The optimal truncation is obtained
by choosing N so as to minimize this error integral for a given . It is
not possible to proceed further without speci¬c information about the
transform F (k). If, however, one knows that

F (k) ∼ A exp(’µ|k|) as |k| ’ ∞ (31)

for some positive constant µ where A denotes factors that vary alge-
braically rather than exponentially with wavenumber, then independent
of A (to lowest order), the optimal truncation as estimated by steepest
descent is
µ
Nopt ( ) ∼ ’ 1, << 1 (32)
2
and the error in the “superasymptotic” approximation is

Nopt 2j f
2d µ
u(x; ) ’ ¤ A exp ’ , << 1 (33)
dx2j
j=0


where A denotes factors that vary algebraically with , i. e., slowly
compared to the exponential, in the limit of small .
In textbooks on perturbation theory, the di¬erential equation (23) is
most commonly used to illustrate the method of matched asymptotic




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24 John P. Boyd

expansions. The multiple scales series (26) is the interior or “outer”
solution. To satisfy the boundary conditions
u(’1) = u(1) = 0 (34)
it is necessary to add “inner” solutions which are functions of the “fast”
variable X = x/ . For (23), the exact solution is
u(x; ) = up (x; ) + a exp(’[x + 1]/ ) + b exp([x ’ 1]/ ) (35)
where up (x; ), the particular solution, is the solution to the same prob-
lem on the in¬nite interval, already described above, and
’up (’1; ) + e’2/ up (1; ) ’up (1; ) + e’2/ up (’1; )
a= , b= (36)
1 ’ exp(’4/ ) 1 ’ exp(’4/ )
The “inner” expansion is just the perturbative approximation to the
exponentials in (35). The matched asymptotics solution is completed
by matching the inner and outer expansions together, term-by-term.
It is important to note that for the ¬nite domain x ∈ [’1, 1], it is
perfectly reasonable to choose a function like g(x) = x4 /(1 + x2 ), which
is unbounded as |x| ’ ∞ and therefore lacks a well-behaved Fourier
transform. However, the hyperasymptotic method can be extended to
such cases by de¬ning the function f in the Fourier integral to be
1
f (x) ≡ g(x) {erf(»[x ’ 2]) ’ erf( »[x + 2])} (37)
2
If the constant » is large, the multiplier of g di¬ers from 1 by an expo-
nentially small amount on the interval x ∈ [’1, 1] so that f ≈ g on the
¬nite domain. The modi¬ed function f , unlike g, decays exponentially
with |x| as |x| ’ ∞ so that it has a well-behaved Fourier transform.
We can therefore proceed exactly as before with f used to generate the
“outer” approximation in the form of a Fourier transform. For exam-
ple, for the particular case g = x4 /(1 + x2 ), the poles at x = ±i imply
that F (k) decays as exp(’|k|) so that the optimal truncation and error
bound are the same as for the Lorentzian forcing, f = 4/(1 + x2 ).
Since asymptotic matching is needed only because of the boundaries
(and boundary layers), it is natural to assume that the inner expansion
is the villain, responsible for the divergence of the matched asymptotic
expansions. This is only half-true. In the perturbative scheme,
a ∼ ’ up (’1; ); b ∼ ’ up (1; ) (38)
to all orders in with an error which is O(exp(’2/ ))). The boundary
layers have indeed enforced a minimum error below which the ordi-
nary perturbative scheme cannot go, but it depends on the separation



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