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Take pressure positive in compression, outward normal on the upper interface, and curvature $\kappa \simeq -h_{xx}$. The external-minus-internal traction convention gives the base pressure $p_0=V/h_0^3$. For the symmetric Fourier perturbation, the linearized upper-surface conditions areIsotropic base stress cancels all terms due merely to rotating the normal. Velocity $u$ is even and $v$ odd in $y$.

Use the same Papkovich–Neuber representation as in Q1, with $\mathbf{\Phi }=(0,\varphi )$ and symmetric harmonic potentialsSuppress the common exponential. They giveThe zero tangential stress is $A{h}_0\cosh K+B\sinh K=0$, where $K=k{h}_0>0$. Hence $B=-A{h}_0\coth K$, $v(h_0)=A\sinh K/2$ and the normal stress becomes $\mu kA(\cosh K+K/\sinh K)$. Eliminating $A$ using the kinematic condition gives the Van der Waals rupture instability of a viscous sheet rateAttraction destabilizes thickness variations; curvature pressure opposes them.

For the clean-sheet long-wave rupture plateau, writing $s_0=3V/(\mu h_0^3)$, the rate is $s/s_0\to 1/4$. Without capillarity, the curve decreases toward $1/(2K)$ at large $K$; with $\Gamma =1$ it changes sign at $K=1$ and behaves as $-K/2$. Thus short waves are weakly unstable without surface tension, but stabilized by it. The largest clean-sheet rate is approached at long wavelength. The strictly uniform $k=0$ thickness change is not an admissible fixed-volume Fourier mode; this is the limit of nonzero waves. Stress-free surfaces permit extensional, nearly plug-like motion whose mobility compensates the weak long-wave pressure gradient. Unsteady inertia eventually matters in that limit; finite sheet extent, drainage, gravity, surface elasticity/viscosity, surfactant transport and molecular-scale corrections can also modify wavelength selection.

For the comparison with surfactant, keep the same equilibrium tension ${\gamma }_0$. Flow away from a thinning neck stretches the surface, diluting its surfactant and raising tension there. Convergence into thicker regions concentrates surfactant and lowers tension. The resulting Marangoni stress points back toward the neck, opposing the draining surface motion. It therefore reduces mobility and the rupture rate relative to the clean constant-${\gamma }_0$ sheet. This mechanism needs surface transport to sustain concentration gradients; instantaneous concentration equilibration would remove it.

For the supplied strong-surfactant rate, unstable waves satisfy $K<{\Gamma }^{-1/2}$. When $\Gamma \gg 1$, all of them have $K\ll 1$. Expanding its mobility factor givesMaximizing in $K^2$ yields the strong-surfactant long-wave rupture maximumHere $\gamma$ is the equilibrium tension ${\gamma }_0$ in the soap-film comparison. Surface immobilization removes the clean long-wave plateau and selects a finite long wavelength. The growth timescale is lengthened by a factor about $3\Gamma$ compared with the clean maximum. A linear rupture-time estimate is $s_{\max }^{-1}\log (h_0/|{\eta }_{\mathrm{initial}}|)$, not a universal bubble lifetime: drainage and changing thickness, initial defects and nonlinear rupture also matter.

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 areThe 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 flowThe tangential-stress condition at $y=h_0$, with $K=k{h}_0$, giveswhereas the kinematic condition gives $iB\sinh K=s\eta$. The associated normal traction isEquating this with the linearized interfacial traction yields the dispersion relationThis 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 $K\to 0$, and decreases to zero like $3V/(2\mu h_0^{3}K)$ 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, 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 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 $\Gamma \gg 1$, instability requires $K=O({\Gamma }^{-1/2})$. The Taylor expansionreduces the supplied relation toIt is maximal atThus the characteristic rupture time is $s_{\max }^{-1}=4\mu {\gamma }_0{h}_0^5/(3V^2)$. 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.