= Solution
Write $s=R^2$. A steady <complex amplitude> obeys $(\mu-s+i\Lambda)A=-1$, so
$$
\boxed{A=\frac{s-\mu+i\Lambda}{(\mu-s)^2+\Lambda^2},\qquad s[(\mu-s)^2+\Lambda^2]=1.}
$$
Thus every positive intensity root gives exactly one steady phase, and $s=0$ is impossible. The cubic intensity <polynomial> is $F(s,\mu)=s^3-2\mu s^2+(\mu^2+\Lambda^2)s-1$. Its positive steady branch can be parameterized completely as
$$
\mu_\pm(s)=s\pm\sqrt{s^{-1}-\Lambda^2},\qquad 0<s\leq\Lambda^{-2},\qquad R=\sqrt s.
$$
The two formulae join smoothly as a single curve at the maximum radius $R=\Lambda^{-1}$; this joining point is not a <saddle-node bifurcation>. A turning point with respect to $\mu$ satisfies $F_s=0$. Subtracting the <equilibrium> and <derivative> conditions gives
$$
\mu-s=\frac1{2s^2},\qquad
\boxed{\Lambda^2=f(s):=\frac1s-\frac1{4s^4},\qquad \mu=s+\frac1{2s^2}.}
$$
These points lie on $\mu_+$. Since $f'(s)=(1-s^3)/s^5$, its maximum is $f(1)=3/4$. Therefore \b[there is exactly one distinct steady state for every $\mu$ when $\Lambda\geq\sqrt3/2$]. One can also see uniqueness directly: $\mu_-$ is increasing and $\mu_+$ is decreasing when no turning points exist, and their respective ranges meet only at their common endpoint. At the equality there is a <cusp bifurcation> with a triple intensity root at $s=1,\mu=3/2$.
When $0<\Lambda<\sqrt3/2$, the two turning intensities satisfy $s_-<1<s_+$. Define $\mu_L=s_-+1/(2s_-^2)$ and $\mu_U=s_++1/(2s_+^2)$; the curve has a local minimum at the former and a local maximum at the latter, with $\mu_L<\mu_U$. There are three distinct steady intensities for $\mu_L<\mu<\mu_U$, two distinct ones at an endpoint, and one otherwise. The <bifurcation diagrams> below show the folded small-$\Lambda$ curves and the unfurled large-$\Lambda$ curve, together with the <linear stability> calculated next. Both ends of the full branch have $R\to0$ as $|\mu|\to\infty$, while its middle reaches the finite maximum $\Lambda^{-1}$.
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