 Solution

Assume $\gamma >0$ and zero initial cavity amplitude. Taking Laplace transforms givesEliminating the mode yields the passive cavity input-output transfer functionIts unique pole is $s=-\gamma /2$, so the cavity's internal mode decays and its causal input-output response is stable. A nonzero initial amplitude would add $\sqrt{\gamma }a(0)/(s+\gamma /2)$ to the output, a decaying transient.

For a precise Nyquist stability criterion interpretation of this open cavity, rewrite the mode equation as an integrator with negative state feedback. Its return ratio is $L(s)=(\gamma /2)/s$, and its characteristic equation is $1+L(s)=0$. The frequency locus $L(i\omega )=-i\gamma /(2\omega )$ lies on the imaginary axis. The Nyquist indentation excluding the pole at the origin maps to a large semicircle in the right half-plane; the whole contour has no winding around $-1$. There are no right-half-plane open-loop poles within that indented contour, so there are no unstable characteristic zeros. There is also no zero-frequency characteristic root, since the physical characteristic polynomial is $s+\gamma /2$. This recovers asymptotic stability.

Do not use the plant $G$ as a return ratio without specifying an additional feedback connection. Indeedand its locus passes through $-1$ at $\omega =0$. That would indicate a marginal mode for a new unity-negative-feedback loop with characteristic factor $1+G$, but not instability of the original cavity. Part (e) supplies the actual return ratio $\alpha G$ for the beam-splitter network.