I am trying to understand the method of steepest descent (complex integral). I looked in some complex analysis books and also on Wikipedia, but I still don't understand the methodology of approximating such integrals nor the name of this method. Could someone give me a step by step example and a way to visualize it?


1 Answer 1


I will give you a conceptual explanation; I trust you can find the computational details elsewhere once you understand the idea. We are often interested in the behavior of quantities of the type $I(\lambda)=\int_\gamma e^{\lambda f(z)}\,\mathrm{d}z$, where $\gamma$ is a curve in the plane, $\lambda$ is a (real) parameter and $f$ is a comple-valued analytic function. The steppest descent method uses the freedom we have in deforming $\gamma$ between its endpoints using Cauchy's theorem, together with some basic observations about the nature of the integrand $e^{\lambda f(z)}$.

Writing $\lambda f(z)$ as $\lambda(\Re f +i \Im f)$, we see that the integrand will be much larger around the points $z$ on $\gamma$ where $\Re f$ attains a maximum than elsewhere. In fact, it will turn out that as $\lambda$ grows, the main contribution to the integral will come from arbitrarily small neighborhoods of maxima of $\Re f$ (and hence of $|e^{\lambda f(z)}|$. This observation is also the basis of Laplace's method, with which you may already be familiar. In Laplace's method, we are dealing with an integral $J(\lambda)=\int_{[a,b]} e^{\lambda \phi(x)}\,\mathrm{d}x$, where $\phi$ is now a real function. Suppose $\phi$ has a maximum at an interior point $x_0 \in (a,b)$. Then $\phi'(x_0)=0$, and moreover only a vanishingly small neighborhood of $x_0$ contributes to the asymptotic behavior. This is explained clearly in the Wikipedia article.

Returning to $I(\lambda)$, we want to use the same insight that Laplace's method is based on in the complex case, except that now we need to measure the size of $e^{\lambda f}$ in terms of its modulus. Only a small neighborhood of the maximum of $|e^{\lambda f}|=e^{\lambda \Re f}$ along $\gamma$ matters. The additional twist is that now we have the freedom to choose $\gamma$ by contour deformation: only the endpoints are fixed. The condition that $|e^{\lambda f(z_0)}|$ be a maximum along a curve for some point $z_0$ can be understood geometrically. You can draw a 3D plot of the modulus of an analytic function (you can see such plots on Wikipedia). Such plots look like smooth landscapes. If you observe a modulus plot for $e^{\lambda f(z)}$, you will see that there are only two types of points: ordinary points, at which $f'(z)\neq 0$, and saddle points, where $f'(z)=0$. $e^{\lambda f}$ being an analytic function, $|e^{\lambda f(z)}|$ cannot have a local maximum (this is the maximum modulus principle): we never see isolated peaks in the modulus landscape. At ordinary points, there will be a single direction such that $|e^f|$ increases fastest. At saddle points there will be two or more (depending on the order of vanishing of $(e^{\lambda f})'$).

The method of steepest descent is based on the observation that it is advantageous to deform the contour of integration $\gamma$, as far as possible, so as to travel along directions where $\Re f$ increases or decreases monotonically. This allows us to isolate the main contribution to the integral to the neighborhood of such points. For example, if $z_0$ is an endpoint of $\gamma$, and we can deform $\gamma$ such that $\Re f$ decreases as we move along $\gamma$ towards the other endpoint, say $z_1$. Then as $\lambda$ gets large, the contributions from points away from $z_0$ disappear, and we find: $$I(\lambda) \sim \frac{e^{\lambda f(z_0)}}{\lambda f'(z_0)}.$$

The most interesting case is when we deform $\gamma$ to pass through a saddle point $z_0$ of $f$, also orienting $\gamma$ such that the modulus $|e^{\lambda f(z)}|$ decreases as we move away from the saddle point (choosing a path of ``descent'' away from $z_0$). To see an example of such a contour, look at the picture of the $$w=x^2-y^2$$ plot on the Wikipedia article for saddle point linked above. Good contours $\gamma$ in this case would pass through the point $(0,0)$ and then go downwards (in a direction of decreasing $w$) as they move away from the origin. (This is just for illustration. This is not the modulus of an analytic function, but exponentiating will give you the modulus plot for $e^{z^2}$.) If we choose $\gamma$ such that 1. it goes through a saddle point, 2. the real part of $f$ (and hence $|e^{\lambda f}|$) decreases as we move away from the saddle point and towards the endpoints of $\gamma$, then the maximum of $|e^{\lambda f(z)}|$ on our contour will occur at $z_0$, and we can evaluate the integral $I(\lambda)$ approximately using by Taylor expanding $f$ around $z_0$, just like in the case of Laplace's method. We can now expand in a suitable neighborhood of $z_0$: $$f(z) = f(z_0)+\dfrac{1}{2}f''(z_0)(z-z_0)^2+\epsilon.$$ Assuming $f''(z_0)\neq 0$. The expansion is quadratic because $f'(z_0)=0$. For large $\lambda$, the resulting integral will be asymptotically equivalent to a Gaussian with a phase factor. Notice that we can easily identify saddle points by the equation $f'(z_0)=0$.

In some applications, it is useful to use specific contours $\gamma$. Here I have basically assumed that the endpoints of $\gamma$ lie in valleys on both sides of a saddle point, so that the exact choice of contour is not important. Contours through $z_0$ that are exactly colinear with the gradient of $\Re f$ will be the contours of ``steppest descent''. By the Cauchy-Riemann equations, the gradient of $\Im f$ is perpendicular to such contours, and so the phase is constant. In simple situations you can solve equations for these contours, and often you can solve approximately, which can help you guess a good $\gamma$ in cases where it matters.

  • $\begingroup$ Thanks, your answer is very clear and illuminating. $\endgroup$
    – bill
    Commented Sep 5, 2012 at 1:00
  • $\begingroup$ Great answer! I have a question though (which I suspect is pretty elementary): How did you get the approximate result for $I(λ)$ from above of the example with the surface $w$? $\endgroup$ Commented Jul 30, 2018 at 22:27

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