 Solution

At $e_{\max }=\sqrt{B}$, torque balance givesThe magnitude of the right-hand side must not exceed one. This fixes the dissipative offset of the resonant-argument libration centre.

For $C>0$, put $S=5A/[C{a}^2(5p-q)B^{q/2}]$. The two torque-balanced phases are ${\varphi }_0=-\arcsin S$ and ${\varphi }_0=\pi +\arcsin S$, modulo $2\pi$. In the usual weak-perturbation pendulum approximation, $\delta \dot{\varphi }\simeq -3(p+q)n\delta a/(2a)$ and $\delta \dot{a}\simeq -2(p+q)Ca{B}^{q/2}\cos {\varphi }_0\delta \varphi$. Restoring libration thus selects $\cos {\varphi }_0<0$, givingIt approaches $\pi$ when drag is weak compared with resonant torque. For $C<0$, the corresponding restoring branch has $\cos {\varphi }_0>0$ and approaches zero. The sine equation is the direct result independent of this leading-order stability approximation.