---
tag: 99879Z5Y1K
kind: example
name: Chandrasekhar's H-function
msc: [45G10, 85A25, "45E10"]
instance_of: [BSCYBP6AS6]
uses: [EX9YYRMAJA, KKT4Y0TBRZ, BSCYBP6AS6]
sources:
  - url: https://en.wikipedia.org/wiki/Chandrasekhar%27s_H-function
same_as:
  wikidata: Q48996454
  wikipedia: Chandrasekhar's H-function
---

## Example

Let $\Psi$ be an even polynomial in $\mu$, the characteristic function, with $\Psi \ge 0$ on $[0, 1]$ and $\int_0^1 \Psi(\mu)\, d\mu \le \frac{1}{2}$.
*Chandrasekhar's H-function* is the solution $H$ on $[0, 1]$ of
$$H(\mu) = 1 + \mu\, H(\mu) \int_0^1 \frac{\Psi(\mu')\, H(\mu')}{\mu + \mu'}\, d\mu'.$$
One takes the solution that is analytic in $\mu$ in the right half-plane and continuous on $[0, 1]$.
The simplest case is isotropic scattering, $\Psi = \frac{\omega_0}{2}$ with albedo $0 < \omega_0 \le 1$.
The unknown appears in a product with itself, so this is a [[nonlinear integral equation|BSCYBP6AS6]].

Check of a moment identity.
Let $I = \int_0^1 \Psi(\mu) H(\mu)\, d\mu$.
Multiply the equation by $\Psi(\mu)$ and integrate over $[0, 1]$.
The double integral $\int_0^1 \int_0^1 \frac{\mu}{\mu + \mu'}\, \Psi(\mu) H(\mu)\, \Psi(\mu') H(\mu')\, d\mu'\, d\mu$ does not change when $\mu$ and $\mu'$ swap, so it equals half the integral with $\frac{\mu + \mu'}{\mu + \mu'} = 1$, that is $\frac{1}{2} I^2$.
So $I = \int_0^1 \Psi\, d\mu + \frac{1}{2} I^2$, and the physical root is
$$\int_0^1 \Psi(\mu) H(\mu)\, d\mu = 1 - \sqrt{1 - 2 \int_0^1 \Psi(\mu)\, d\mu}.$$
In the conservative case $\Psi = \frac{1}{2}$ this gives $\int_0^1 H(\mu)\, d\mu = 2$.
Also $H(0) = 1$, directly from the equation.

## Remarks

The function $H$ is tied to a Wiener–Hopf factorization.
For $z$ off the segment $[-1, 1]$, with the analytic continuation of $H$, one has
$$\frac{1}{H(z)\, H(-z)} = 1 - 2 z^2 \int_0^1 \frac{\Psi(\mu)}{z^2 - \mu^2}\, d\mu.$$
The right side is the dispersion function of the [[radiative transfer equation|EX9YYRMAJA]] in a half-space; $H(-z)$ is the factor holomorphic for $\operatorname{Re} z < 0$ and $H(z)$ the factor holomorphic for $\operatorname{Re} z > 0$.
As $z \to \infty$, both sides tend to $1 - 2\int_0^1 \Psi\, d\mu$, consistent with $H(\infty) = 1/(1 - I)$ from the equation.
In the conservative case, $H$ gives the exact angular distribution of the light leaving a grey atmosphere, the companion of the [[Milne problem|KKT4Y0TBRZ]].

## Explanations

### L1 human

Light that enters a thick cloud or a stellar atmosphere is scattered many times before some of it comes back out.
Chandrasekhar showed that the outgoing light is described by a single function $H$ of the cosine $\mu$ of the exit angle.
This function solves an equation in which it appears twice, once outside and once inside an integral.
Simple integral identities, such as $\int_0^1 H\, d\mu = 2$ for pure scattering, follow by symmetry without solving the equation.