# Steepest descent vs. stationary phase method

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1. Feb 16, 2015

### blue_leaf77

Up to this point I have got a grasp of some basics of "steepest descent method" to evaluate the integral of a complex exponential function $f(z) = \exp(A(x,y))\exp(iB(x,y))$. Using this method the original integration path is modified in such a way that it passes through its saddle points, assuming this function is analytic everywhere. The modified path coincides with the line on which the magnitude of $f(z)$, namely $\exp(A(x,y))$, is changing most rapidly.

But then I also saw there is another method to calculate the integral of the same function, "stationary phase method". I haven't gone through the related literature though, but judging from the name I suspect both methods are identical. This is because in steepest descent method one follows the line of the most rapid change of $A(x,y)$, which is also the line of constant phase (constant $B(x,y)$), for them being analytic. Now my intuition says that stationary phase means constant phase, so does this mean that in "stationary phase method" one also follows the same path as that in "steepest descent method"? If so, what is the difference between both methods? And in which cases one method is better than the other?

2. Feb 21, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Feb 21, 2015

### mathwonk

this question is too vague for me, and suggests you have not done enough work yourself even to understand the terms you are asking about.

4. Feb 22, 2015

### blue_leaf77

Ok let me explain it in a another way. One of applications of complex analysis finds itself in computing the integral of this form $\int f(z) dz$ where $f(z) = \exp(A(z))\exp(iB(z))$, assuming the integrand has no singularities then we can modify the integration path in such a way that it passes its saddle points. The result is an approximate and is called steepest decent method, because the integration path is modified to follow points along which the amplitude of $f(z)$ changes most rapidly (either while ascending or descending).
Now there is another seemingly similar method to calculate the same integral, called stationary phase method. To be honest I haven't gone through this topic because in my literature the related discussion is long enough. But judging from the name it says "stationary phase", I am thinking that in this method the integration path is defined so that it follows points on the complex plane where the phase of $f(z)$ is stationary or constant. But according to theorem in complex analysis, lines of constant phase coincides with line of fastest change in amplitude, the latter is precisely what one deals with when using steepest descent method described above. Up to now this is just my guess, and my question is if what I'm thinking there was right? If not, in other words if both methods are actually different, would someone please explain what the difference is?
I hope this time my problem can be understood easier.

5. Feb 28, 2015

### jasonRF

It is difficult to respond with too much information since you haven't really defined things - are A and B real-valued? So I will define my own terms in order to state how I think of this.

Consider an integral $I(x) = \int dz \, g(z) \, \exp\left( x h(z) \right)$. The steepest descent paths end up coinciding with paths along which $h(z)$ has a constant imaginary part. Thus we can essentially used Laplace's method - a great method because the complete asymptotic expansion is determined by arbitrarily short segments of the contour.

If you instead go along paths for which the real part of $h(z)$ is constant, things are different. If there is a location where $h_i^\prime (z)=0$, where $h_i(z)$ is the imaginary part of $h(z)$ then the leading order asymptotic behaviour is determined by an arbitrary short segment of the contour in the vicinity of that point (a point of stationary phase), but higher order terms in general may depend on behavior over the entire contour. It is easy to understand why: away from the stationary phase point the integrand oscillates so rapidly that adjacent segments cancel to a large degree to leading order.

I use stationary phase a lot - primarily for determining leading order behaviour of Fourier integrals of one sort or another that occur in signal processing / linear system analysis. I almost never need more than the leading order behavior, as I am looking for insight, not many digits of precision. Also, the stationary phase formula is simple to derive once you understand the idea - simple enough that is just takes a couple minutes to rederive whenever I need it. Multivariate versions of the stationary phase formula are hard to derive but easy to use. However, the stationary phase approach does not work with all Fourier-type integrals (eg when there are no stationary points over the integration interval), so then you have to resort to steepest descents.

I learned this stuff while taking a radiowave propagation course, so I have a heuristic understanding only and likely do not know all of the precise caveats. In any case you should read up on this a little more yourself. In particular, stationary phase is much much simpler to understand than steepest descents and 30 minutes would likely be enough for you to learn this.

jason

Last edited: Feb 28, 2015
6. Feb 28, 2015

### blue_leaf77

Ah now it's clear, so in stationary phase method one simply swaps the role between real and imaginary part of $h(z)$ from that in steepest descent, right. I was previously confused in thinking that they are identical with the reason I already mentioned in my previous comment.
In steepest descent, the integration path is approximated as only to be done around saddle points, while in stationary phase around the points of stationary phase. Both of them lead to Gaussian type integration.

Thanks a lot Jason.