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.

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.