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In the nonrelativistic regime,The assumption $E-\mu \gg k_{B}T$ puts the Fermi-Dirac distribution in its Maxwell-Boltzmann limit, soTherefore the nonrelativistic Maxwell--Boltzmann number density is

For $p+e^{-}\leftrightarrow H+\gamma$, chemical equilibrium and the vanishing photon chemical potential implyApplying the number-density formula to each massive species and eliminating the chemical potentials givesUse charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_e{g}_p/g_H$. With the Hydrogen binding energywe obtain the Saha ionization equationThe assumptions are thermal equilibrium and chemical equilibrium before cosmological recombination, a nonrelativistic nondegenerate gas, zero photon chemical potential, charge neutrality, negligible proton-electron mass correction in the translational prefactor, and the stated degeneracy-factor approximation.

Fora finite point $z_0$ is an ordinary point exactly when $p$ and $q$ are holomorphic there. It is a regular singular point exactly whenare holomorphic at $z_0$.

Put $\zeta =1/z$ and $W(\zeta )=w(1/\zeta )$. Direct differentiation givesso the transformed equation isConsequently $z=\infty$ is ordinary precisely whenare holomorphic at $\zeta =0$. It is a regular singular point at infinity precisely when $zp(z)$ and $z^{2}q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the Laurent series of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hencefor constants $a,b$. The equation is a Cauchy-Euler differential equation. Its indicial equation isFor distinct roots ${\lambda }_1,{\lambda }_2$, the general solution on a domain with a chosen logarithm branch isFor a repeated root $\lambda$, the general solution of an Euler-Cauchy equation is

Finally require infinity to be ordinary. In the transformed equation its coefficients becomeBoth are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is

With $s=0$, the equilibria satisfyBesides $g=0$, the two positive equilibria arewhere $g_{-}{g}_{+}=1$. Sincethe nonzero equilibrium relation $k=(1+g^2)/g$ givesThe fixed point stability for an autonomous differential equation therefore shows that $g=0$ and $g_{+}$ are stable, while $g_{-}$ is unstable. The graph of $f(g,0)$ starts at zero with negative slope, crosses upward at $g_{-}$, crosses downward at $g_{+}$, and tends to $-\infty$ as $g\to \infty$.

For a constant input $s$, write the equilibrium equation asThe low stable equilibrium and the intervening unstable equilibrium coalesce in a saddle-node bifurcation. More precisely, let $g_c$ be the first positive solution ofand defineEquivalently, $f(g_c,s_c)=0$ and ${\partial }_{g}f(g_c,s_c)=0$, with $g_c$ on the low-concentration branch. If $s_1>s_c(k)$, that branch no longer exists. Holding the input long enough carries the trajectory into the basin of attraction of the high state. When the input returns to zero, the concentration converges toThis is the saturating autocatalytic switch.

For $k\gg 1$, the threshold occurs at $g=O(k^{-1})$, soThe approximate double-root conditions areThus $g_c\simeq 1/(2k)$ andHence the constant in $s_c(k)\simeq C{k}^{-1}$ is $\boxed{C=1/4}$, as recorded by the strong-autocatalysis switching threshold.