Milne problem

Example

Let $E_1(x) = \int_1^\infty e^{-x\mu}\, \mu^{-1}\, d\mu$ for $x > 0$ be the exponential integral, and more generally $E_n(x) = \int_1^\infty e^{-x\mu}\, \mu^{-n}\, d\mu$. The Milne problem asks for a nonzero function $J$ on $[0, \infty)$ of at most linear growth with $$J(\tau) = \frac{1}{2} \int_0^\infty E_1(\abs{\tau - t})\, J(t)\, dt, \quad \tau \ge 0.$$ This is a homogeneous Wiener–Hopf equation with kernel $k(x) = \frac{1}{2} E_1(\abs{x})$. In astrophysics $\tau$ is the optical depth in a plane-parallel grey atmosphere in radiative equilibrium, and $J$ is the mean intensity.

Check of the main features. The kernel has total mass $1$: $\int_{\R} \frac{1}{2} E_1(\abs{x})\, dx = \int_1^\infty \mu^{-2}\, d\mu = 1$. So on the whole line every linear function is a solution, and the cut at $\tau = 0$ makes the problem nontrivial. Using $E_n' = -E_{n-1}$ for $n \ge 2$, one computes for $\tau \ge 0$ $$\frac{1}{2} \int_0^\infty E_1(\abs{\tau - t})\, dt = 1 - \frac{1}{2} E_2(\tau), \qquad \frac{1}{2} \int_0^\infty E_1(\abs{\tau - t})\, t\, dt = \tau + \frac{1}{2} E_3(\tau).$$ Hence the Eddington guess $J = \tau + \frac{2}{3}$ gives a right side that differs from $J$ by $\frac{1}{2} E_3(\tau) - \frac{1}{3} E_2(\tau)$. This error tends to $0$ as $\tau \to \infty$, but at $\tau = 0$ it is $\frac{1}{4} - \frac{1}{3} = -\frac{1}{12}$, since $E_2(0) = 1$ and $E_3(0) = \frac{1}{2}$.