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For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

For an inviscid fluid of constant density, with irrotational velocity $\mathbf{u}=\nabla \Phi$ and conservative body-force potential $\Psi$, the Unsteady Bernoulli equation isIt holds throughout a connected irrotational fluid region, up to the time-dependent gauge of $\Phi$. Indeed the Euler equations and $\nabla \times \mathbf{u}=0$ imply that the gradient of the left side is zero. In the bubble problem there is no body-force contribution compatible with the assumed spherical symmetry.

Spherical incompressibility gives ${\partial }_r(r^2{u}_r)=0$. The moving interface imposes $u_r(R,t)=\dot{R}$, soThis potential tends to zero at infinity, where the velocity tends to zero and $C(t)=p_{\infty }/\rho$. Differentiate $\Phi$ at fixed spatial $r$, not along the moving interface: ${\Phi }_t=-(2R{\dot{R}}^2+R^2\ddot{R})/r$. At $r=R$, pressure continuity without surface tension gives $p=p_{\infty }-\Delta p$. Bernoulli therefore yields $-2{\dot{R}}^2-R\ddot{R}+{\dot{R}}^2/2-\Delta p/\rho =0$, or the Rayleigh equation for an inviscid spherical bubble

Integrating the kinetic energy over the exterior fluid givesDifferentiate and use the radius equation:With $R(0)=R_0$ and $\dot{R}(0)=0$, the pressure-work energy balance for a spherical bubble integrates toorThe collapsing branch takes the negative square root for $\dot{R}$. The energy gain is exactly the pressure difference times the lost bubble volume, providing a check on the sign and numerical factor.

Let $\mathbf{n}$ point from fluid $-$ to fluid $+$, $P_s=I-\mathbf{nn}$, ${\nabla }_s=P_s\nabla$, and $\mathcal{K}={\nabla }_s\cdot \mathbf{n}$. A clean material interface without phase change has continuous velocity, with normal interface speed $V_n={\mathbf{u}}_{\pm }\cdot \mathbf{n}$. Its surface stress is $\gamma P_s$. The surface divergence of that tensor is ${\nabla }_s\gamma -\gamma \mathcal{K}\mathbf{n}$. Balancing forces on a small interfacial patch therefore gives the interfacial stress balance with variable surface tensionThe normal condition is $\mathbf{n}\cdot ({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=\gamma \mathcal{K}$; the tangential condition is $P_s({\mathbf{\sigma }}_{+}-{\mathbf{\sigma }}_{-})\mathbf{n}=-{\nabla }_s\gamma$. On a stationary sphere with outward normal these reduce to $p_{-}-p_{+}=2\gamma /a$ when viscous normal stresses vanish. This fixes the sign convention for the Marangoni stress rather than leaving the direction ambiguous.

For the bubble take negligible internal dynamic viscosity, no contaminants, and initially the standard effectively insulating gas-bubble approximation. With heat advection neglected, the exterior temperature satisfies Laplace's equation, approaches $T_0+\mathbf{H}\cdot \mathbf{r}$ remotely, and has zero normal heat flux at $r=a$. The temperature dipole around an insulating sphere givesThe zero-flux condition follows by differentiating the radial factor. Let ${\gamma }_T=(d\gamma /dT{)}_{T_0}$. To first order in the weak gradient,

In the laboratory frame the force-free bubble can be described by the harmonic Papkovich–Neuber representation potentialsThey give constant exterior pressure and the decaying source dipole fieldAt the surface $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so the normal kinematic boundary condition holds for translation. In coordinates aligned with $\mathbf{U}$, $u_r=a^{3}U\cos \theta /r^3$ and $u_{\theta }=a^{3}U\sin \theta /(2r^3)$. Direct differentiation givesEquivalently, the surface tangential traction is $(3\mu /a)P_s\mathbf{U}$. The tangential stress condition therefore requiresFor the usual negative ${\gamma }_T$, the bubble moves toward hotter liquid. The normal condition is consistent as well: its viscous dipole is $-6\mu \mathbf{U}\cdot \mathbf{n}/a=3{\gamma }_T\mathbf{H}\cdot \mathbf{n}$, exactly the dipolar part of $2{\gamma }_s/a$. The constant part fixes $p_{\mathrm{gas}}-p_{\infty }=2{\gamma }_0/a$. There is no Stokeslet term, hence no net applied force or drag imbalance.

The thermal assumption matters. The problem gives the exterior thermal diffusivity but no gas thermal conductivity. For an inviscid bubble with conductivity ratio $K=k_{\mathrm{gas}}/k_{\mathrm{liquid}}$, temperature and normal heat-flux continuity instead giveso the same stress calculation givesThe insulating-bubble result uses $K\ll 1$. If instead an undisturbed surface temperature gradient is imposed, corresponding to $K=1$, the coefficient is $1/3$ rather than $1/2$. Exterior thermal diffusivity alone does not select between those thermal boundary conditions. This dependence is explicit rather than silently treating a missing gas property as a prescribed value.

The small parameters can now be stated precisely. With $U_{\ast }=a|{\gamma }_T||\mathbf{H}|/(2\mu )$ and $\nu =\mu /\rho$, requireSmall Reynolds number justifies Stokes flow; small Péclet number allows quasi-static heat conduction without advective distortion; small capillary number and fractional surface-tension variation support a nearly spherical interface and a weak perturbation of its base surface tension. These last two conditions are proportional for this velocity scale but describe different approximations. Linearizing $\gamma (T)$ also needs $a|\mathbf{H}||{\gamma }_{TT}/{\gamma }_T|\ll 1$ if its derivative varies appreciably. Negligible internal viscosity and the insulating specialization require the corresponding gas-to-liquid viscosity and conductivity ratios to be small. Gravity has been excluded; if restored, its buoyancy-induced migration must be small compared with $U_{\ast }$, not merely small enough to avoid deformation.

Finally, surfactants change both the constitutive surface tension and the surface stress dynamics. The surface flow driven by the thermal Marangoni effect carries insoluble surfactant toward the high-tension side, lowering its tension and producing an opposing concentration-driven gradient. When surface diffusion or exchange is slow, this tends to immobilize the interface, suppressing the clean-bubble thermocapillary migration of an insulating bubble, sometimes nearly stopping it. If the surfactant stays uniformly equilibrated, it can instead change ${\gamma }_T$ without a large opposing gradient. Its concentration, surface diffusivity and adsorption kinetics must be supplied to compute a replacement speed; no universal numerical reduction factor follows merely from its presence.

Let $z$ measure height above the substrate, and write $\mu =\rho \nu$. In the lubrication approximation, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:The no-slip boundary condition gives $u(0)=0$; negligible air viscosity and constant surface tension give $u_z(h)=0$. Integrating twice and then through the depth yieldsThe free-surface kinematic boundary condition and incompressibility imply $h_t+q_x=0$. ThereforeThe cross-sectional area $A=\int hdx$ is conserved when the endpoint volume flux vanishes.

The relevant small groups follow from the same balances. Put ${ϵ}_h=H_0/\ell$ and $U_{\ast }=g{H}_0^3/(3\nu \ell )$. Small ${ϵ}_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O({ϵ}_h^2)$. The ratio of horizontal inertia to vertical viscous stress isThus a small full-length Reynolds number is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_{\ast })$ must likewise be small; on the spreading timescale $t_{\ast }=\ell /U_{\ast }$ it equals the displayed group. If surface tension is retained, its pressure-gradient correction relative to gravity isThe Bond number here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/{\ell }^2)$ whereas the hydrostatic pressure is $O(\rho g{H}_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant kinematic viscosity.

To find the similarity solution, put $K=g/(3\nu )$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the time derivative against the nonlinear diffusion gives ${\ell }^5\sim K{A}^{3}t$. A normalized ansatz isSubstitution gives $(f^3{f}^{\prime }{)}^{\prime }=-(f+\xi f^{\prime })/5$. For the symmetric zero-flux profile, one integration gives $f^3{f}^{\prime }=-\xi f/5$, and on its positive support a second integration givesThus the solution is a compactly supported Barenblatt solution of the porous medium equation. Let $\ell (t)$ denote its half-width. In dimensional variables,Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell \sin \theta$,Writing the remaining integral as $I$, the final widths areThe height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The volume flux tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported weak solution of the reduced equation. The diverging edge slope means the lubrication approximation is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.

Take $x$ downslope and $z$ normal to the plane. To leading order in the lubrication approximation, Newtonian gravity gives the hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. The downslope Stokes equation is therefore $\mu u_{zz}=\rho g(h_x-\theta )$. Imposing the no-slip boundary condition at $z=0$ and the stress-free boundary condition at $z=h$ givesHere $\nu =\mu /\rho$ is the kinematic viscosity; $u$ is the downslope velocity. Applying mass conservation to this lubrication gravity-current flux,For typical thickness $H=\hat{h}$ and length $L=\hat{x}$, the transverse velocity is $O(UH/L)$ by incompressibility. Both convective terms are then $O(U^2/L)$, while transverse viscous acceleration is $O(\nu U/H^2)$. Thus the inertia criterion for an inclined viscous film is $U{H}^2/(\nu L)\ll 1$, in addition to $H/L\ll 1$. With evolution time $T\sim L/U$ the same condition controls unsteady inertia. If a faster time is independently imposed, also require $H^2/(\nu T)\ll 1$. Estimating $U$ from the velocity profile gives the two requested limits:

At long times, $h\sim A/x_N$ and the ratio of hydrostatic-gradient forcing to downslope Newtonian gravity in the bulk is $h_x/\theta =O(A/(\theta x_N^2))\ll 1$. Put $K=g\theta /(3\nu )$; the bulk conservation law is $h_t+(K{h}^3{)}_x=0$. The finite-volume gravity-driven thin-film current scales as $x\sim t^{1/3}$, $h\sim t^{-1/3}$. More explicitly, writing $h=t^{-1/3}f(\xi )$, $\xi =x/t^{1/3}$, givesThe constant of integration is zero for the zero-flux dry tail. ConsequentlyThe reduced bulk solution is zero outside this interval. Its thin nose region is resolved below. Integrating to impose the conserved area,The bulk front height is $h_N=3A/(2x_N)$, with $h_N^2=x_N/(3Kt)$. Differentiation of the front law givesEquivalently, this is the Rankine-Hugoniot condition for the bulk shock wave, whose speed is flux divided by height.

In moving coordinate $y=x-x_N(t)$, the full thin-film equation becomes $h_t{|}_y-{\dot{x}}_N{h}_y+q_y=0$. A nose thickness $h_N$ and slope $\theta$ give axial scale ${\ell }_N=h_N/\theta$. Since ${\ell }_N/x_N=O(A/(\theta x_N^2))\ll 1$, the explicit time derivative is smaller than translation by that ratio. The travelling front of a gravity-driven thin film is therefore quasisteady. Integrate once with dry-tip flux zero to obtain $q={\dot{x}}_{N}h$, orWith $f=h/h_N$, separation of variables and the convention $y=0$ at $f=0$ giveThe nose approaches $h_N$ exponentially far behind its tip, then falls to zero with a vertical tangent in this small-slope model. The requested sketch shows this monotone profile:

Expanding the inverse hyperbolic tangent, $f-\mathrm{atanh}f=-f^3/3+O(f^5)$, so $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. Its slope has magnitude $\theta /f^2$ near the tip. The small-slope breakdown at an inclined-current nose occurs at $f=O(\sqrt{\theta })$, which givesFinally, surface tension produces a pressure-gradient scale $\gamma h_N/{\ell }_N^3$, whereas downslope Newtonian gravity gives $\rho g\theta$. Their ratio is $\gamma {\theta }^2/(\rho g{h}_N^2)$. Thus surface tension changes the principal nose length whenThis is a bulk-nose crossover; capillarity and contact-line physics can already affect a smaller region very close to the dry tip even when the bulk nose is Newtonian gravity dominated.

Use incompressibility, negligible viscosity and surface tension, and an initially irrotational flow. Take the undisturbed free surface at $z=0$ and bed at $z=-H$. The velocity potential $\phi$ satisfies Laplace equation and ${\phi }_z(-H)=0$. For a Fourier mode, writeThe linear kinematic boundary condition and linear Bernoulli equation at the free surface are ${\eta }_t={\phi }_z$ and ${\phi }_t+g\eta =0$. Eliminating $A,{\eta }_a$ gives the surface gravity wave dispersion relationThe rightward phase velocity and group velocity areThe phase velocity transports a crest; the group velocity transports a narrow-band envelope and its leading-order energy. For $kH\gg 1$, $\omega \sim \sqrt{gk}$ and $c_g\sim c_p/2\sim \sqrt{g/k}/2$: deep-water surface gravity waves are dispersive. For $kH\ll 1$, $\omega \sim k\sqrt{gH}$ and $c_g\sim c_p\sim \sqrt{gH}$, the nondispersive shallow-water dispersion relation. The boundary velocity controls the excited amplitude; it does not change the linear dispersion relation.

For potential flow $\mathbf{u}=\nabla \phi$ in an inviscid fluid of constant density under the conservative force per unit mass $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by inserting the velocity potential into the Euler equations and integrating the resulting spatial gradient. A time-only change of the velocity potential changes $C(t)$ without altering the velocity field.

For spherical motion in an unbounded liquid, incompressibility gives $r^2{u}_r=A(t)$; the kinematic condition $u_r(R)=\dot{R}$ gives $A=R^2\dot{R}$. Choose $\phi =-R^2\dot{R}/r$, vanishing at infinity. To obtain the displayed bubble relation, take no spatial body-force contribution, neglect viscosity and surface tension, and equate the liquid-side interface pressure to $p_b$. With the far-field pressure $p_{\infty }$, Bernoulli equation evaluated at the surface uses the partial derivative at fixed $r$:Hence the Rayleigh equation for an inviscid spherical bubble is

The exterior kinetic energy isDifferentiating it and using $\dot{V}=4\pi R^2\dot{R}$ gives the pressure-work energy balance for a spherical bubble:This derivation also holds at a turning point $\dot{R}=0$, without dividing by $\dot{R}$.

For constant $p_{\infty }$ and the stated inverse-volume bubble pressure, integrate this exact time derivative:The relation is valid on physically accessible portions of the motion, where the resulting kinetic energy is nonnegative. In particular $\log q\le q-1$ makes the bracket nonpositive, as expected for the restoring pressure about $V_0$.

For $\alpha /\lambda \ll 1$, follow the unstable root near the clean-film value $L=\lambda s(1-s)$. Write $\beta =L+\Delta$ in $(\beta -L)(\beta +S)+3LS/4=0$. At first order in $S$, $L\Delta +3LS/4=0$, givingThis expansion is uniform near the maximizing wavenumber. Differentiation gives $s_{\ast }=1/2-3\alpha /(8\lambda )+O((\alpha /\lambda {)}^2)$. Evaluating the growth there yieldsThe change in maximizing wavenumber only affects the maximum at second order. The first-order decrease therefore has exactly the requested coefficient, relative to the clean-film rate with its surface tension replaced by $\bar{\sigma }$.

Without surfactant, capillary pressure drives a surface flow towards a growing bulge. Advection compresses the insoluble surfactant in converging parts of that motion, increasing concentration and lowering surface tension. The resulting gradient supplies a tangential Marangoni stress that opposes the surface motion. This additional resistance slows capillary-driven redistribution of liquid even after the base-tension decrease has already been accounted for.

In the strong-elasticity limit, concentration gradients suppress tangential interface motion. The surface behaves as effectively immobile at leading order. Setting $u_s\approx 0$ in the velocity formula gives ${\sigma }_x\approx h{p}_x/2$, henceone-quarter of the clean shear-free mobility $h^3/(3\mu )$. This mobility reduction explains the quarter-rate limit; it slows the unstable capillary mode without removing the long-wave unstable band.

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 givesAt 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 becomeSet $s=k^2{a}^2$, $\lambda =h_0^3\bar{\sigma }/(3\mu a^4)$ and $\alpha =h_{0}A{\Gamma }_0/(\mu a^2)$. 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 isThis proves the dispersion relation before its lettered deductions. The physical assumption $\bar{\sigma }>0$ is needed for the claimed capillary instability band.

Let $n$ point out of the drop, and use $-$ for the interior and $+$ for the exterior. The interfacial velocity is continuous. Define $\kappa ={\nabla }_s\cdot n$; for a sphere of radius $a$ it is $2/a$. With constant surface tension, the interfacial stress balance with variable surface tension reduces toUse the exterior-viscosity kernels to define $S[t]=\int J(y-x)t(x)d{S}_x$ and $D[u]=\int u(x)K(y-x)n(x)d{S}_x$. The interior viscosity is $\lambda \mu$, so its velocity kernel is $J/\lambda$, while its stress kernel $K$ is unchanged. The interior boundary equation isFor the decaying exterior flow, the inner boundary's fluid-domain normal is $-n$, soAdd these equations and use the traction jump. The required capillary boundary integral equation for a viscous drop isThe formula uses the drop-outward normal consistently in both integrals. If $\lambda =1$, the double-layer term cancels. A spherical drop of constant curvature has zero velocity: a constant Laplace-pressure jump is balanced without motion. A prescribed ambient flow would add its known far-boundary contribution; none is present for the surface-tension-driven flow here.

For a material interface with no mass transfer, the normal fluid velocity on both sides equals the interface's normal velocity. For ordinary viscous fluids without interfacial slip, tangential velocities also agree. The stress tensors $T_i=-p_{i}I+{\mu }_i(\nabla {\mathbf{u}}_i+\nabla {\mathbf{u}}_i^T)$ have continuous tangential traction when surface tension is uniform; their normal traction jump balances surface-tension curvature. Nonuniform surface tension instead supplies a tangential Marangoni stress. These conditions express conservation of mass, no slip between the contacting fluids and force balance at the interface. For a flat interface of constant tension the full traction is continuous.

Choose $x$ down the plane and $y$ perpendicular outward from it. A steady parallel solution has velocity $(u(y),0)$ and equationsFor a uniform free layer, $p_x=0$, $u(0)=0$, $u^{\prime }(h)=0$ and $p(h)=p_{\mathrm{atm}}$. Integration givesIts flux per unit cross-slope width is

For the two-layer falling film of equal-density viscous fluids let $H=(1+\alpha )h$ and $A=\rho g\sin \theta /\mu$. Equal density gives the same continuous hydrostatic pressure in both, $p=p_{\mathrm{atm}}+\rho g\cos \theta (H-y)$. The wall condition is $u_1(0)=0$, the upper condition $u_2^{\prime }(H)=0$, and at $h$ we require $u_1=u_2$, $\mu u_1^{\prime }=\beta \mu u_2^{\prime }$. Integrating the two momentum equations with these conditions givesThe shear stress transmitted to the lower layer is the upper layer's downslope weight per unit area, $\rho g\alpha h\sin \theta$, obtained by integrating the upper momentum balance from its stress-free surface. It depends on thickness but not upper viscosity. Changing $\beta$ changes the upper shear rate needed to transmit that stress, while the lower profile depends on $\alpha$ alone.

For a slender viscous bubble, approximate the interface normal by the radial direction and its curvature by $1/a$. Axial curvature and axial derivatives of the leading external motion are smaller by the slenderness factors. The exterior leading Stokes flow is locally radial. Incompressibility gives ${\partial }_r(r{u}_r)=0$, while the kinematic boundary condition gives $u_r(a)=a_t$ to leading order, henceThe radial viscous Laplacian of this $1/r$ velocity vanishes, so exterior pressure is constant to leading order; set its reference value to zero. Its radial normal stress at the interface is $2\mu {\partial }_r{u}_r{|}_a=-2\mu a_t/a$. Inner viscous normal stress is smaller by the viscosity ratio $\lambda$, and the bubble's leading normal stress is $-P$. The interfacial stress balance with variable surface tension, here with constant $\gamma$, consequently givesThese are local leading-order relations: the outer axial flow and tangential traction enter at higher slenderness order. The sign also has a useful check: a uniform cylinder with no imposed excess pressure contracts, $a_t=-\gamma /(2\mu )$, under surface tension.