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# The Method of Steepest Descent and Stationary Phase

Lecture 46

The repeated encounter with complex integrals such as

especially when , demands that we have at our disposal a systematic method for evaluating, at least approximately, integrals of the type

 (548)

This is an integral in the complex -plane along a curve which starts at and terminates at . The exponential is a rapidly changing function because . The function , by contrast, is a slowly varying function. The success of the method hinges on the following circumstance: the dominant contribution to the integral comes from only a small segment of the integration contour, and the accuracy of that dominant contribution improves with increasing .

The value of the integral depends obviously on the behavior of the integrand along the integration path. However, the Cauchy-Goursat theorem implies that the integration path between the fixed limits and can be quite arbitrary provided that

is analytic, i.e., all its derivatives exist. This is usually, if not always, the case. Analyticity of is equivalent to

which yields the Cauchy-Riemann equations

They imply

or

i.e., is harmonic''. (Nota bene: a harmonic function need not be analytic.)

Let be an extremum of , i.e.,

At such a critical point, has neither a maximum nor a minimum, it has a saddle point instead, because prevents from having a maximum or minimum anywhere.

Example:

The integrand of is

and the integration path is assumed to start and end where this integrand vanishes, i.e., where

This means that, in the example, points and would lie in different shaded strips in Figure 5.17.

The integration path between these end points can be deformed without changing the value of the integral. The method of steepest descent takes advantage of this fact by deforming the integration path so that it goes through the critical point in such a way that

and that the rate at which decreases along either direction away from as rapidly as possible.

One suspects that the integral

gets its major contribution along this path through . A possible objection against such a suspicion is that along this path the integrand

might oscillate very rapidly. One might blame such a behaviour on the phase factor

As a consequence, one might think that the value of the integral would average to approximately zero and make its evaluation through not give the dominant contribution to the total integral. Fortunately this can never happen. Remarkably enough, the opposite is the case, the path of steepest ascent and descent is also the path of stationary phase. In other words, the direction along which changes must rapidly, namely,

is also the direction along which

is constant; indeed,

Thus is constant along the direction of the gradient of . In still other words, the level surfaces of and are perpendicular to each other, a direct consequence of the Cauchy-Riemann equations.

The important conclusion is, therefore, this:

has constant phase along the direction of . It is clear that if were not tangent to the line of constant phase, then the method of steepest descent would not work.

We now expand in the neighborhood of the critical point :

 (549)

Here is the phase of . We are assuming that the third and higher derivative terms make a negligible contribution in controlling the asymptotic behavior of

This is a good assumption provided the second derivative of does not vanish at ,

Assuming that this is the case, we now must choose the integration path through . The linear part of this path is
 (550)

so that

 (551)

Here is the path parameter and controls the direction of the path. Now comes the important step: We choose the direction of the path so that in the process of passing through the function makes the integrand

rise as fast as possible to a maximum at and subsequently makes that integrand decrease as rapidly as possible. Such a path is exhibited in Figure 5.18. Along this path the function has the form

This form must coincide with Eq.(5.49) along the path. Consequently,

This condition determines the angle of the integration path.
 (552)

The path itself is

The ambiguity expresses the fact that the integration may proceed into the forward direction or the backward direction. The two directions obviously differ by radians. The ambiguity is resolved by the fact that the integral has its integration path along a direct path from over the critical point to . For example, the complex integrals for the Hankel functions

have the integrand whose critical points are located at , as in Figure . A cursory inspection of this integrand reveals quite readily through which of these critical points the directed integration path must pass.

In general, the ambiguity in

can only be resolved by drawing a global picture in which the direct, and hence directed, integration path connecting is exhibited.

After the global ambiguity has been settled, the evaluation of the integral becomes straightforward. The integral, Eq.(5.48), is approximated by restricting the integration to the path segment centered around the saddle point:

 (553)

The accuracy of this approximation is determined by two seemingly irreconcilable demands. On one hand we are neglecting cubical (and higher) order terms in the exponential, and this is permitted only if

 (554)

On the other hand, at first glance one would think that would have to be large enough in order not to miss any contributions to the to-be evaluated integral. However, there is no conflict. The highly localized nature of the gaussian guarantees that the integral be independent of its limts , even when is ìsmallî, i.e. satisfies Eq.(5.54). This is because the localized nature of the exponential is controlled by the positive parameter . To make the value of the integral independent of , this parameter must be so large that

 (555)

Comparing Eq.(5.55) with (5.54), one finds

This chain of inequalities reconciles the two seemingly contradictory demands. The more the two length scales and differ from each other the better the chain of inequalities can be satisfied., and the greater the accuracy with which the given integral Eq.(5.48) gets approximated by Eq.(5.53).

Moving forward, expand the slowly varying function in a Taylor series and obtain

One can simplify this expression in two ways:

First of all, it is permissible to replace the integration limits by whenever

Under this condition the integral may be replaced by its limiting value,

It is obvious that the inequality is violated for sufficiently large . However, this will not happen if the Taylor series representation of can be truncated without compromising the accuracy with which is to be represented.

Secondly, one may apply Eqs.(5.52) and (5.49) to

 (556)

With these two simplifications the steepest descent evaluation of the contour integral Eq.(5.48) yields the following series in inverse powers of :

 (557)

Here is the mandatory truncation integer, and

is that root which has the phase factor whose angle points along the integration path through the critical point .

Example: Evaluate

 (558)

to second order accuracy in . Here

The critical points determined by are

The integration limits of in the complex plane are indicated in Figure 5.3. They dictate that the most direct path of steepest descent passes through the critical point

Consequently,

The phase angle of the integration path is determined by the condition that

Consequently, Eq.(5.52) becomes

or

The fact that the path goes from the second to the fourth quadrant (as in Figure 5.18) requires that one choose the upper sign,

Thus, in light of Eq.(5.56, namely

one has

This is because51 the square root of a polar representation is unique52It follows that the large expansion of Eq.(5.58) is
 (559)

Exercise 55.1 (STEEPEST DESCENT)
(a)
Using the method of steepest descent FIND an asymptotic expression for and for when .
(b)
The gamma function which for is represented by

Using the steepest descent approach, FIND an asymptotic expression for when . Why does't it work? Try again by substituting for , and obtaining

#### Footnotes

... because51
See the discussion surrounding Figures 4.10 and 4.11 on page .
... unique52
Indeed, another polar representation, namely,

would have given the wrong result

Next: Boundary Value Problems in Up: Special Function Theory Previous: More Properties of Hankel   Contents   Index
Ulrich Gerlach 2010-12-09