## Appendix B Threshold Gain Reduction and Lasing Frequency Shift  
Under feedback from both sides of the laser cavity, new lasing conditions are found which subsequently result in a shift of the laser threshold gain and lasing frequency with respect to the case without feedback, in which:  
\[\kappa_{j}=0.\] (A13)  
In this case, the amplitude condition from Equation (A11) becomes:  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\] (A14)  
Using the expansion:  
\[\ln(1+x)\simeq x,\] (A15)  
the threshold gain reduction due to feedback can be found by computing the ratio between Equations (A11) and (A14):  
\[1 = \frac{\rho_{2}\rho_{1}[1+\kappa_{2}t_{\rm cav}\cos(\phi_{2})][1+ \kappa_{1}t_{\rm cav}\cos(\phi_{1})]e^{\big{(}\Gamma g_{\rm FB}-\alpha\big{)} L}}{\rho_{2}\rho_{1}e^{\Gamma g_{\rm int}-\alpha L}}\] (A16) \[= [1+\kappa_{2}t_{\rm cav}\cos(\phi_{2})][1+\kappa_{1}t_{\rm cav} \cos(\phi_{1})]e^{\Gamma\big{(}g_{\rm FB}-g_{\rm ret}\big{)}L}\] \[\stackrel{{\rm(2)}}{{\Leftrightarrow}}G_{\rm FB}-G_ {\rm ref}\approx-2\kappa_{2}\cos(\phi_{2})-2\kappa_{1}\cos(\phi_{1}).\]  
The relation between the right hand terms determines the threshold gain reduction, as discussed in Section 2.1. The phase lasing condition from Equation (A12) yields the lasing frequency shift equation. Consider the following definitions related to the effective refractive index [52]:  
\[n \equiv n^{\prime}+in^{\prime\prime}\] (A17a) \[n_{\rm g} \equiv n+\omega\frac{\partial n}{\partial\omega}\] (A17b) \[n^{\prime\prime} \equiv -\frac{cG}{2\omega v_{\rm g}},\] (A17c)\[\alpha_{{}_{\rm H}}\equiv\Delta n^{\prime}/\Delta n^{\prime\prime}\] (A17d) \[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]0 \[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]1 (A17e)  
where \(n_{\rm g}\) is the group refractive index. To find the lasing frequency shift equation, calculating the change in \(\beta\) is first needed:  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]2 \[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]3 (A18)  
with which:  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]4 (A19)  
Furthermore, using:  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]5 (A20a) \[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]6 (A20b)  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]7 (A20c)  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]8 (A20d)  
the following can be computed:  
\[1\stackrel{{\rm(A13)}}{{=}}\rho_{2}\rho_{1}e^{\big{(}\Gamma g_{ \rm th}-\alpha\big{)}L}.\]9 \[\ln(1+x)\simeq x,\]0 (A21) \[\ln(1+x)\simeq x,\]1 \[\ln(1+x)\simeq x,\]2  
Finally, using the phase condition in Equation (A12), and assuming without generality loss that \(m=0\), it is possible to compute:  
\[\ln(1+x)\simeq x,\]3 \[\ln(1+x)\simeq x,\]4 (A22) \[\delta\omega\stackrel{{\eqref{eq:A20a}}}{{=}}- \gamma_{{}_{\rm H}}[\