= Solution
Write the perturbation as $\eta e^{ikx+st}$ and use the same factor for all velocity and pressure amplitudes. At the upper interface, the linearized <kinematic boundary condition>, zero tangential traction, and normal-stress balance are
$$
v=s\eta,
\qquad
u_y+v_x=0,
\qquad
-p+2\mu v_y
=\left(\frac{3V}{h_0^4}-\gamma k^2\right)\eta
\quad(y=h_0).
$$
The first term in the normal stress is the linearization of the attractive <disjoining pressure> $V/h^3$, while the second is the stabilizing <capillary pressure>. Symmetry about $y=0$ makes $u$ even and $v$ odd, and supplies the lower-interface conditions.
For a two-dimensional Fourier mode, the <Papkovich–Neuber representation> is equivalently expressed by the odd <Biharmonic stream function for planar Stokes flow>
$$
\psi(y)=A\sinh(ky)+By\cosh(ky),
\qquad
u=\psi_y,
\qquad
v=-ik\psi.
$$
The tangential-stress condition at $y=h_0$, with $K=kh_0$, gives
$$
Ak\sinh K+B(\sinh K+K\cosh K)=0,
$$
whereas the kinematic condition gives $iB\sinh K=s\eta$. The associated normal traction is
$$
-p+2\mu v_y
=\mu k s\eta\frac{2K+\sinh2K}{\sinh^2K}.
$$
Equating this with the linearized interfacial traction yields the <dispersion relation>
$$
\boxed{
s=\frac{3V}{\mu h_0^3}
\frac{(1-\Gamma K^2)\sinh^2K}
{K(2K+\sinh2K)},
\qquad
\Gamma=\frac{\gamma h_0^2}{3V}.}
$$
This is the <Van der Waals rupture instability of a viscous sheet>.
For $\Gamma=0$, the growth rate is positive, starts from $3V/(4\mu h_0^3)$ as $K\to0$, and decreases to zero like $3V/(2\mu h_0^3K)$ as $K\to\infty$. For $\Gamma=1$, it has the same long-wave limit, vanishes at $K=1$, and is negative for $K>1$: <surface tension> damps wavelengths shorter than the cutoff. Long waves feel the attractive interaction but require coherent flow over a large distance; at short wavelengths viscous resistance suppresses the clean-film instability, while capillarity adds direct decay. A finite film size, <Navier-Stokes equation>[fluid inertia], surrounding-fluid stresses, gravity, surface viscosity, and failure of the continuum <disjoining pressure> law can shift the observable most unstable wavelength.
During a growing thin spot, interfacial flow stretches the surface and dilutes its <surfactant>, thereby increasing the local surface tension above $\gamma_0$. Adjacent less-stretched regions retain more surfactant and lower tension. The resulting <Marangoni effect>[surface-tension gradient] pulls toward the thin spot and opposes the outward flow that drives thinning. This is <surfactant stabilization of film rupture>; in the strong limit the surfaces behave almost as immobile boundaries.
For strong surfactant and $\Gamma\gg1$, instability requires $K=O(\Gamma^{-1/2})$. The <Taylor expansion>
$$
\frac{\sinh2K-2K}{4K\cosh^2K}
=\frac{K^2}{3}+O(K^4)
$$
reduces the supplied relation to
$$
s\sim\frac{V}{\mu h_0^3}K^2(1-\Gamma K^2).
$$
It is maximal at
$$
\boxed{K_{\max}^2=\frac1{2\Gamma},
\qquad
s_{\max}=\frac{V}{4\mu h_0^3\Gamma}
=\frac{3V^2}{4\mu\gamma_0h_0^5}.}
$$
Thus the characteristic rupture time is $s_{\max}^{-1}=4\mu\gamma_0h_0^5/(3V^2)$. Surfactant greatly extends the life of a soap bubble while its film is moderately thick, but the $h_0^{-5}$ growth-rate dependence predicts rapid final rupture after drainage has made the film sufficiently thin.
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