= Solution
Let the positive undisturbed <surface tension> be $\bar\sigma=\sigma_0-A\Gamma_0>0$. An <insoluble surfactant> perturbation $g$ produces $\sigma_1=-Ag$. The tangential <Marangoni stress> changes the surface condition to $\mu u_z(h)=\sigma_x$. Solving the <lubrication theory> momentum equation now gives
$$
u=\frac{p_x}{2\mu}(z^2-2hz)+\frac{\sigma_x}{\mu}z,\qquad
q=-\frac{h^3}{3\mu}p_x+\frac{h^2}{2\mu}\sigma_x,\qquad
u_s=-\frac{h^2}{2\mu}p_x+\frac h\mu\sigma_x.
$$
At leading order, <conservation of insoluble surfactant on a moving interface> reduces to $\Gamma_t+(\Gamma u_s)_x=0$. Corrections from the changing circumference are smaller by $h_0/a$. Likewise the normal-<pressure> contribution $-\sigma_1/a$ gives a coupling smaller by $h_0/a$ than the retained tangential <Marangoni stress>, so it is omitted at the same order.
For thickness and concentration amplitudes $H,g$ proportional to $e^{\beta t+ikx}$, put $D=a^{-2}-k^2$ and use $p_1=-\bar\sigma DH$. The two conservation equations become
$$
\begin{aligned}
\left(\beta-\frac{h_0^3\bar\sigma}{3\mu}k^2D\right)H+\frac{h_0^2A}{2\mu}k^2g&=0,\\
-\frac{\Gamma_0h_0^2\bar\sigma}{2\mu}k^2D\,H+\left(\beta+\frac{h_0A\Gamma_0}{\mu}k^2\right)g&=0.
\end{aligned}
$$
Set $s=k^2a^2$, $\lambda=h_0^3\bar\sigma/(3\mu a^4)$ and $\alpha=h_0A\Gamma_0/(\mu a^2)$. The uncoupled capillary growth and elastic relaxation rates are $L=\lambda s(1-s)$ and $S=\alpha s$. The off-diagonal product is $3LS/4$, so the <determinant> is $(\beta-L)(\beta+S)+3LS/4$. Thus the <surfactant stabilization of an annular liquid lining> relation is
$$
\boxed{\beta^2+\beta s[\lambda(s-1)+\alpha]+\frac{\alpha\lambda}{4}s^2(s-1)=0.}
$$
This proves the <dispersion relation> before its lettered deductions. The physical assumption $\bar\sigma>0$ is needed for the claimed capillary instability band.
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