The Method of Steepest Descent and Stationary Phase

Lecture 46

The repeated encounter with complex integrals such as

$$H_\nu (\rho )=\int^{\alpha_2}_{\alpha_1} e^{i\rho\cos\alpha} e^{i\nu\alpha} d\alpha~~,$$

especially when $\rho\gg 1$ , demands that we have at our disposal a systematic method for evaluating, at least approximately, integrals of the type

$$I(\rho )= \int^B_A X(z)e^{\rho f(z)} dz \,;~~\qquad~~1\ll\rho\,.$$ (548)

This is an integral in the complex $z$ -plane along a curve which starts at $A$ and terminates at $B$ . The exponential is a rapidly changing function because $1\ll\rho$ . The function $X(z)$ , 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 $\rho$ .

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 $A$ and $B$ can be quite arbitrary provided that

$$f(z) = f(x+iy)=u(x,y)+iv(x,y)$$

is analytic, i.e., all its derivatives exist. This is usually, if not always, the case. Analyticity of $f(z) = f(x+iy)$ is equivalent to

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial (iy)}$$

$$\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x} = \frac{1} {i}~\frac{\partial u}{\partial y} +\frac{\partial v}{\partial y}\,,$$

which yields the Cauchy-Riemann equations

$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\,;~~ \frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x}\,.$$

They imply

$$\left[\frac{\partial^2}{\partial x^2} +\frac{\partial^2}{\partial y^2}\right] \left\{\begin{array}{c} u\\ v\end{array}\right\} = 0$$

or

$$\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2} =0\,,$$

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

Let $z_0$ be an extremum of $f$ , i.e.,

$$\frac{\partial f}{\partial x} = 0~\textrm{and}~\frac{\partial f}{\partial y} = 0,~~ \textrm{or~equivalently}~~\frac{\partial f}{\partial z} = 0\,.$$

At such a critical point, $f$ has neither a maximum nor a minimum, it has a saddle point instead, because $\frac{\partial^2 f}{\partial x^2} = -\frac{\partial^2 f}{\partial y^2}$ prevents $f$ from having a maximum or minimum anywhere.

Figure 5.17: Critical points of the function $f(z)=i\rho \cos z$ . The solid contours in the shaded regions (``valleys'') are the isograms of $Re~f$ which are below zero elevation, while the dashed contours in the unshaded regions (``mountain ranges'') are the isograms of $Re~f$ which are above zero elevation. Each solid dot is located in a mountain pass which connects two valleys by a path of steepest ascent and descent.

Example:

$$f(z) = i\cos z = i\cos x\cosh y +\sin x\sinh y$$

$$f'(z) = -i\sin z$$

$$z_0 = n\pi\,,~~n=0\,,~\pm 1\,,~\pm 2 ~~~~~\textrm{(location~of critical~points)}$$

The integrand of $I(\rho )$ is

$$X(z) e^{\rho f(z)} = X(z) e^{\rho u(x,y)+i\rho v(x,y)}$$

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

$$u(A) = -\infty$$

$$u(B) = -\infty\,.$$

This means that, in the example, points $A$ and $B$ 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 $z_0$ in such a way that

$$Ref(z_0) = \textrm{maximun~along~the~path}$$

and that the rate at which $Ref(z_0)$ decreases along either direction away from $z_0$ as rapidly as possible.

One suspects that the integral

$$\int^B_A X(z) e^{\rho u(x,y)+i\rho v(x,y)}dz$$

gets its major contribution along this path through $z_0$ . A possible objection against such a suspicion is that along this path the integrand

$$X(z)e^{\rho f(z)}=X(z) e^{\rho u(x,y)} e^{i\rho v(x,y)}$$

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

$$e^{i\rho v(x,y)}\,.$$

As a consequence, one might think that the value of the integral would average to approximately zero and make its evaluation through $z_0$ 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 $u(x,y)$ changes must rapidly, namely,

$$\vec\nabla u =\left(\frac{\partial u}{\partial x}\,,~\frac{\partial u}{\partial y}\right)~~~(\textrm{\lq\lq gradient~of~}u\textrm{''})$$

is also the direction along which

$$v=v(x(\tau ),y(\tau ))$$

is constant; indeed,

$$\frac{dv}{d\tau} = \frac{dx}{d\tau} \frac{\partial v}{\partial x}+ \frac{dy}{d\tau}\frac{\partial v}{\partial y}$$

$$= \frac{\partial u}{\partial x}~\frac{\partial v}{\partial x} +\frac{\partial u}{\partial y}~\frac{\partial v}{\partial y}$$

$$= \frac{\partial v}{\partial y}~\frac{\partial v}{\partial x}-\frac{\partial v}{\partial x} ~\frac{\partial v}{\partial y} =0\,.$$

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

Figure 5.18: Path of steepest ascent and descent through the critical point $z_0$ of the function $f(z)$ . The dotted lines are the isograms of $Im~f~(=v(x,y))$ , the locus of points where the phase of the integrand $e^{\rho f}$ is constant. Perpendicular to these are the solid lines. They are the isograms of $Re~f~(=u(x,y))$ . The heavily dotted directed line through $z_0$ is the integration path of stationary phase and steepest descent.

The important conclusion is, therefore, this:

$$e^{i\rho v(x,y)}$$

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

We now expand $f(z)$ in the neighborhood of the critical point $z_0$ :

$$f(z) = f(z_0)+\frac{1}{2} (z-z_0)^2 f''(z_0)+\frac{1}{3!}(z-z_0)^3 f'''(z_0) +\cdots$$

$$= f(z_0)+\frac{1}{2} (z-z_0)^2 e^{i\delta_0}\vert f''(z_0)\vert +\cdots$$ (549)

Here $\delta_0$ is the phase of $f''(z_0)$ . We are assuming that the third and higher derivative terms make a negligible contribution in controlling the asymptotic behavior of

$$e^{\rho f(z)}\,.$$

This is a good assumption provided the second derivative of $f(z)$ does not vanish at $z_0$ ,

$$f''(z_0)\not= 0\,.$$

Assuming that this is the case, we now must choose the integration path through $z_0$ . The linear part of this path is

$$z = z_0+e^{i\phi}\tau\,.$$ (550)

so that

$$dz = e^{i\phi} d\tau\,.$$ (551)

Here $\tau$ is the path parameter and $e^{i\phi}$ 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 $z_0$ the function $f(z)$ makes the integrand

$$e^{\rho f(z)}$$

rise as fast as possible to a maximum at $z_0$ and subsequently makes that integrand decrease as rapidly as possible. Such a path is exhibited in Figure 5.18. Along this path the function $f(z)$ has the form

$$f(z) = f(z_0)-\frac{1}{2} \tau^2\vert f''(z_0)\vert +\cdots$$

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

$$(z-z_0)^2 e^{i\delta_0}\equiv e^{2i\phi}\tau^2 e^{i\delta_0} = -\tau^2\,.$$

This condition determines the angle $\phi$ of the integration path.

$$e^{2i\phi}e^{i\delta_0} = -1~(=e^{\pm i\pi})\,.$$ (552)

The path itself is

$$z-z_0=e^{-i\delta_0/2}e^{\pm i\pi /2}\tau\,.$$

The $(\pm )$ ambiguity expresses the fact that the integration may proceed into the forward direction or the backward direction. The two directions obviously differ by $\pi$ radians. The ambiguity is resolved by the fact that the integral $\int^B_A\cdots dz$ has its integration path along a direct path from $A$ over the critical point $z_0$ to $B$ . For example, the complex integrals for the Hankel functions

$$H_\nu (\rho ) = \frac{e^{-i\pi\nu /2}}{\pi} \int_A^B e^{i\rho\cos z+i\nu z} dz$$

have the integrand $e^{i\rho\cos z}$ whose critical points are located at $z_0=0,\pm\pi ,\cdots$ , 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

$$dz = e^{i(\pm\pi /2-\delta_0/2)}d\tau$$

can only be resolved by drawing a global picture in which the direct, and hence directed, integration path connecting $A\to z_0\to B$ 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 $-\tau_1 \le \tau \le \tau_1$ centered around the saddle point:

$$I(\rho ) = e^{\rho f(z_0)}\int^{\tau_1}_{-\tau_1} X(z_0+e^{i\phi}... ...}{2} \tau^2 \vert f''(z_0)\vert } d\tau\left.\frac{dz}{d\tau}\right\vert _{z_0}$$ (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

$$\tau^2_1 \vert f''\vert\gg\tau^3_1\vert f'''\vert ~~~\textrm{or~ equivalently}~~~\tau_1\ll\left\vert\frac{f''}{f'''}\right\vert ~.$$ (554)

On the other hand, at first glance one would think that $\tau_1$ 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 $\pm\tau_1$ , even when $\tau_1$ is ìsmallî, i.e. satisfies Eq.(5.54). This is because the localized nature of the exponential is controlled by the positive parameter $\rho$ . To make the value of the integral independent of $\tau_1$ , this parameter must be so large that

$$\tau_1\gg\frac{2}{\sqrt{\rho\vert f''(z_0)\vert}}\left( =\begin{a... ...of~neighborhood}\\ \textrm{where~integrand~is~non-neglegible}\end{array}\right)$$ (555)

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

$$\left\vert\frac{f''}{f'''}\right\vert \gg \tau_1\gg\frac{2}{\sqrt{\rho\vert f''(z_0)\vert}} ~~.$$

This chain of inequalities reconciles the two seemingly contradictory demands. The more the two length scales $\vert f^{''}\vert / \vert f^{'''}\vert$ and $2/ \sqrt{\rho\vert f''(z_0)\vert}$ 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 $X$ in a Taylor series and obtain

$$I(\rho) = e^{\rho f(z_0)}\sum_{n=0}^\infty \left.\frac{d^n X}{dz^n}\right\v... ...}{2} \tau^2 \vert f''(z_0)\vert } d\tau\left.\frac{dz}{d\tau}\right\vert _{z_0}$$

$$= \frac{e^{\rho f(z_0)}}{\sqrt{\rho\vert f''(z_0)\vert}} \left.\fra... ...vert f''\vert}}_{-\tau_1\sqrt{\rho\vert f''\vert}} t^n e^{-\frac{1}{2} t^2 } dt$$

One can simplify this expression in two ways:

First of all, it is permissible to replace the integration limits by $t=\pm\infty$ whenever

$$\tau_1\sqrt{\rho\vert f''\vert}\gg \sqrt{n} \ge 1~~.$$

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

$$\frac{1}{n!}\int_{-\infty}^\infty t^n e^{-t^2 /2}~dt =\sqrt{... ...xtrm{odd}\\ 1&n=0\\ \frac{1}{2^m m!}&n=2m \end{array}\right.$$

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

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

$$\frac{e^{2i\phi}}{ \vert f^{''}(z_0) \vert } = \frac{-e^{-i\delta_0}}{\vert f^{''}(z_0) \vert } \equiv \frac{-1}{ f^{''}(z_0) }~~.$$

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

$$\boxed{ \int^B_A X(z)e^{\rho f(z)} dz = \sqrt{ \frac{2\pi}{\rho }... ...ht\vert _{z_0} \frac{(-1)^m}{m!} \left( \frac{1}{2\rho f^{''}(z_0)} \right)^m }$$ (556)

Here $N$ is the mandatory truncation integer, and

$$[-f^{''}(z_0)]^{1/2}=\vert f^{''}(z_0)\vert^{1/2}e^{-i\phi}~(= \vert f^{''}(z_0)\vert^{1/2}e^{i\delta_0/2} e^{\pm i\pi /2})$$

is that root which has the phase factor $e^{-i\phi}$ whose angle $\phi$ points along the integration path through the critical point $z_0$ .

Example: Evaluate

$$H^{(1)}_\nu = \frac{e^{-i\pi\nu /2}}{\pi}\int_{C_1} e^{i\rho\cos z} e^{i\nu z}dz$$

$$= \frac{e^{-i\pi\nu /2}}{\pi}\int_{C_1}e^{\rho f(z)}X(z)dz\,.$$ (557)

to second order accuracy in $1/\rho$ . Here

$$X(z) = e^{i\nu z}$$

$$f(z) = i\cos z$$

$$f'(z) = -i\sin z$$

$$f^{''}(z) = -i\cos z$$

The critical points determined by $f'(z)=0$ are

$$z=0,\pm\pi ,\cdots~.$$

The integration limits of $H^{(1)}_\nu$ in the complex $z-$ plane are indicated in Figure 5.3. They dictate that the most direct path of steepest descent passes through the critical point

$$z_0=0~~.$$

Consequently,

$$f^{''}(z_0) = -i=e^{-i\pi /2}$$

$$X(z_0) = 1$$

$$X^{''}(z_0) = -\nu^2$$

The phase angle $\phi$ of the integration path $z-z_0=\tau e^{i\phi}$ is determined by the condition that

$$(z-z_0)^2f^{''}(z_0)=-\tau^2 \vert f^{''}(z_0)\vert~~.$$

Consequently, Eq.(5.52) becomes

$$e^{2i\phi}e^{-i\pi/2} = -1~(=e^{\pm i\pi})$$

or

$$e^{i\phi}=\pm e^{-i\pi/4}$$

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,

$$e^{i\phi}=e^{-i\pi/4}$$

Thus

$$[-f^{''}(z_0)]^{1/2}=e^{i\pi/4}~~.$$

It follows that the large $\rho$ expansion of Eq.(5.57) is

$$H^{(1)}_\nu (\rho ) = \frac{e^{-i\pi \nu/2}}{\pi} \sqrt{ \frac{2\pi}{\rho }}e^{i\rho}e^... .../4}\left[1-\nu^2\frac{-1}{1} \left(\frac{1}{2\rho(-i)}\right)^1 +\cdots \right]$$

$$= \sqrt{ \frac{2}{\pi\rho }}e^{i\rho-i(\nu+1/2)\frac{\pi}{2}} \left[1+i\frac{\nu^2}{2\rho}+\cdots \right]$$ (558)

Exercise 55.1 (STEEPEST DESCENT)

(a)

Using the method of steepest descent FIND an asymptotic expression for $H^{(2)}_\nu (\rho )$ and for $J_\nu (\rho )$ when $\nu<<\rho$ .

(b)

The gamma function $\Gamma(w+1)$ which for $Re~w>-1$ is represented by

$$\Gamma(w+1)=\int^{\infty}_0e^{-\tau}\tau^wd\tau$$

Using the steepest descent approach, FIND an asymptotic expression for $\Gamma(w+1)$ when $Re~w>>1$ . Why does't it work? Try again by substituting $wz$ for $\tau$ , and obtaining

$$\Gamma(w+1)=w^{w+1}\int^{\infty}_0 e^{-wz}z^wdz=w^{w+1}\int^{\infty}_0e^{w(\ln z-z)} dz$$