= Three-level laser rate equation
{title2=$\dot u=r(1-u)-u-C(2u-1)s$}
Consider an original reduced model with a fixed number $N$ of active emitters. Let $u=N_2/N$ be the upper lasing-state fraction, $1-u=N_1/N$ the ground-state fraction, and $s=n/N$ the number of cavity <photons> per emitter. Assume the pump state empties rapidly, so its population can be neglected. Measure time by $x=t/\tau$, where $\tau$ is the upper state's spontaneous lifetime. Let $r=W\tau$ be the effective pump rate, $C>0$ the dimensionless stimulated-coupling coefficient, $K>0$ the photon-loss rate in these units, and $0\leq\beta\leq1$ the fraction of spontaneous decays feeding this optical <normal mode>. The <rate equations> are
$$
\frac{du}{dx}=r(1-u)-u-C(2u-1)s,
\qquad
\frac{ds}{dx}=\bigl[C(2u-1)-K\bigr]s+\beta u.
$$
Each term has a physical role. Pumping transfers emitters out of the ground state; <spontaneous emission> removes upper-state population. The net stimulated term includes emission from fraction $u$ and absorption by fraction $1-u$, giving their difference $2u-1$. A stimulated transition removes one stored excitation and creates one cavity <photon>, or does the reverse. The factor $\beta$ accounts for spontaneous emission into the selected mode without pretending that all <spontaneous emission> enters it.
For negligible cavity light, the upper-state equation is linear. A constant pump applied to $u(0)=0$ gives
$$
u(x)=\frac{r}{r+1}\left(1-e^{-(r+1)x}\right),\qquad
u_\infty=\frac{r}{r+1}.
$$
Thus this model first permits <population inversion> at $r>1$. Positive <optical gain> is still insufficient for an oscillator: the gain must exceed the complete <laser threshold>, including photon loss. The small-signal photon growth coefficient changes sign when
$$
C(2u-1)=K,\qquad
u_{\rm th}=\frac{1+K/C}{2},\qquad
\boxed{r_{\rm th}=\frac{C+K}{C-K}\quad(C>K).}
$$
If $C\leq K$, even complete inversion cannot exceed the assumed loss in this model. For chosen $C=100$ and $K=20$, the threshold population is $u_{\rm th}=0.6$ and the pump threshold is $r_{\rm th}=1.5$. At $r=2$, the unsaturated population first reaches $0.6$ at $x=(\log10)/3\simeq0.7675$. These numbers are stipulated examples, not ruby material parameters.
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