= Solution
Let $v_g\simeq c/n_r$ be the light's <group velocity> in the fibre, with unit mode overlap and a uniform transverse mode area. The <fibre-laser gain-volume rate equations> follow because the upper-level <density> is defined per doped volume, whereas the <photon> <density> is defined per full cavity volume. <Stimulated emission> therefore removes upper-level ions at rate $v_g\sigma_{21}N_2n$ per doped volume, but supplies cavity <photons> at the volume-averaged rate $fv_g\sigma_{21}N_2n$. Thus
$$
\dot N_2=S-v_g\sigma_{21}N_2n-\frac{N_2}{\tau_2},\qquad
\dot n=\left(fv_g\sigma_{21}N_2-\frac1{\tau_c}\right)n.
$$
At threshold the stimulated <photon> growth rate balances the <photon> loss rate, giving
$$
\boxed{N_{\mathrm{th}}=\frac1{fv_g\sigma_{21}\tau_c}\simeq\frac{n_r}{fc\sigma_{21}\tau_c}.}
$$
Using this relation yields the stated equations, including the essential $1/f$ in the ion-depletion term:
$$
\dot N_2=S-\frac1f\frac{N_2}{N_{\mathrm{th}}}\frac n{\tau_c}-\frac{N_2}{\tau_2},\qquad
\dot n=\left(\frac{N_2}{N_{\mathrm{th}}}-1\right)\frac n{\tau_c}.
$$
Above threshold, the nonzero stationary <photon> <density> requires $N_{2,\mathrm{ss}}=N_{\mathrm{th}}$. The remaining stationary equation then gives
$$
\boxed{r=\frac{S\tau_2}{N_{\mathrm{th}}}=\frac S{S_{\mathrm{th}}},\qquad
n_{\mathrm{ss}}=(r-1)f\frac{\tau_c}{\tau_2}N_{\mathrm{th}},\qquad S_{\mathrm{th}}=\frac{N_{\mathrm{th}}}{\tau_2}.}
$$
The parameter $r$ is the pump rate in units of its lasing-threshold value. The above-threshold branch requires $r>1$; below threshold the ideal stationary solution is $n=0$, $N_2=S\tau_2$. <Spontaneous emission> into the lasing mode has been omitted in this deterministic model, so an exactly zero initial field would need a seed or noise to start, even when the zero-field branch is unstable.
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