= Solution
The <beam splitter> relation and the cavity output equation give
$$
b_0=\beta b_{
m in}-\alpha(\sqrt\gamma\,a+b_0),
\qquad (1+\alpha)b_0=\beta b_{
m in}-\alpha\sqrt\gamma\,a.
$$
For $\alpha\ne-1$, solve this scalar algebraic relation and substitute back:
$$
\boxed{b_0=\frac\beta{1+\alpha}b_{
m in}-\frac{\alpha\sqrt\gamma}{1+\alpha}a,\qquad
b_1=\frac\beta{1+\alpha}b_{
m in}+\frac{\sqrt\gamma}{1+\alpha}a.}
$$
The cavity equation becomes
$$
\boxed{\dot a=\frac{\gamma(\alpha-1)}{2(1+\alpha)}a-\frac{\beta\sqrt\gamma}{1+\alpha}b_{
m in}.}
$$
Finally, use the other beam-splitter output and $\alpha^2+\beta^2=1$:
$$
b_2=\alpha b_{
m in}+\beta b_1
=\left(\alpha+\frac{\beta^2}{1+\alpha}\right)b_{
m in}
+\frac{\beta\sqrt\gamma}{1+\alpha}a
=\boxed{b_{
m in}+\frac{\beta\sqrt\gamma}{1+\alpha}a}.
$$
These derivations retain the printed input $b_{
m in}$ in both beam-splitter equations; the converted TeX changes one of these subscripts. Define $\gamma_{
m eff}=\gamma(1-\alpha)/(1+\alpha)$ and $c=\beta\sqrt\gamma/(1+\alpha)$, so $\dot a=-\gamma_{
m eff}a/2-cb_{
m in}$ and $b_2=b_{
m in}+ca$. In the nondegenerate range $|\alpha|<1$, $\gamma_{
m eff}>0$ and $c^2=\gamma_{
m eff}$, the expected passive input-output relation. This is <instantaneous coherent cavity feedback through a beam splitter>; it uses the zero-delay interconnection assumed by the equations.
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