---
tag: WWE9H2948V
kind: definition
name: Method of steepest descent
msc: [41A60, "30E15"]
defines: Q5H9VYFK98
uses: [0NAHXR1FRH, 06ZGPYTWK8, 9G7K8DBT9Y, 0JW6MDHAZD]
sources:
  - url: https://en.wikipedia.org/wiki/Method_of_steepest_descent
---

## Definition

Let $f$ and $g$ be [[holomorphic functions|0NAHXR1FRH]] on a domain $\Omega \subseteq \C$, let $C$ be a [[contour|06ZGPYTWK8]] in $\Omega$, and let
$$I(\lambda) = \int_C g(z)\, e^{\lambda f(z)}\,dz, \qquad \lambda > 0.$$
A simple [[saddle point|0JW6MDHAZD]] of $f$ is a point $z_0$ with $f'(z_0) = 0$ and $f''(z_0) \ne 0$.
A *steepest descent path* through $z_0$ is a path through $z_0$ on which $\operatorname{Im} f$ is constant.
Along it, $\operatorname{Re} f$ must decrease strictly as one moves away from $z_0$.
The *method of steepest descent* finds the behaviour of $I(\lambda)$ as $\lambda \to \infty$ in three steps:

1. by Cauchy's theorem, replace $C$ by a contour with the same ends, or ending in the same valleys of $\operatorname{Re} f$ at infinity.
   The new contour consists of steepest descent paths through saddle points of $f$ and of pieces where $\operatorname{Re} f$ is smaller;
2. on each steepest descent path, where $e^{\lambda f}$ has constant phase, apply [[Laplace's method|9G7K8DBT9Y]];
3. keep the contributions of the saddle points with the largest value of $\operatorname{Re} f$.

## Remarks

Let $z_0$ be a simple saddle with $g(z_0) \ne 0$, and let $\theta = \frac12 \big(\pi - \arg f''(z_0)\big)$.
The angle $\theta$, defined modulo $\pi$, is the direction in which the path crosses $z_0$, taken with the orientation of the contour.
The contribution of $z_0$ is
$$g(z_0)\, e^{\lambda f(z_0)}\, e^{i\theta} \sqrt{\frac{2\pi}{\lambda \abs{f''(z_0)}}}\,\big(1 + o(1)\big).$$
Indeed $f(z_0 + r e^{i\theta}) \approx f(z_0) + \frac12 f''(z_0) r^2 e^{2i\theta}$, and $f''(z_0) e^{2i\theta}$ is real and negative.

Example: $I(\lambda) = \int_{\R} e^{i\lambda x^2}\,dx$, with $f(z) = i z^2$ and $g = 1$.
The only saddle is $z_0 = 0$ with $f''(0) = 2i$, so $\theta = \pi/4$.
On the line $z = r e^{i\pi/4}$ one has $f(z) = -r^2$, which is real and decreasing in $\abs{r}$, so this line is the steepest descent path.
Rotating the contour to it gives exactly $I(\lambda) = e^{i\pi/4} \sqrt{\pi/\lambda}$.

Non-example: for $f(z) = z^2$ the real axis passes through the saddle $0$, but $\operatorname{Re} f = x^2$ increases away from $0$ along it.
So it is a path of steepest ascent; the steepest descent path is the imaginary axis.