Codex @codex  0

Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. Incompressible flow with axial strain rate $E$ givesIts radial Stokes equation gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial Newtonian fluid stress tensor component is ${\sigma }_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner Young–Laplace equation conditions have opposite curvature signs:Subtracting and eliminating $C$ givesIn particular, the boxed pressure is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas pressures, $C=-\gamma R_i{R}_e/(2\mu D)$, so surface tension draws both interfaces inward in addition to the imposed extension.

The kinematic boundary condition is ${\dot{R}}_j=-E{R}_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, givesThe cancellation of $C$ in the area equation is mass conservation: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal pressures, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu )$. Since $S=A/D$,Closure occurs at $D=\sqrt{A}=\sqrt{3}b$, soThis capillary collapse of an annular viscous cylinder has finite closure time for $\gamma >0$ and $b>0$. For $\gamma =0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a boundary condition on the subsequent solid cylinder.

Use $A=R_e^2-R_i^2$, $S=R_e+R_i$, $D=R_e-R_i$, and $Q=Aw=A(0)w_0$. Locally replace $E$ in the preceding annular viscous extension calculation by $w^{\prime }(z)$. Its leading pressure isThe axial liquid stress relative to ambient pressure, together with the two circumferential surface tension forces, transmits a cut forceIt would be incorrect simply to differentiate this and retain an extra gas-pressure term. The sloping inner gas interface also exerts an axial pressure force $-\pi (p_i-p_e)(R_i^2{)}^{\prime }$ per unit height, after subtracting the common ambient pressure contribution. For constant gas pressures these terms cancel. The resulting capillary tensile force of a hollow slender thread therefore givesThus the printed axial equation holds even for unequal constant gas pressures; their influence remains in the radial evolution.

For equal pressures, the kinematic boundary condition from the previous calculation becomes $w{D}^{\prime }=-w^{\prime }D/2+\gamma /(2\mu )$. Multiplying by the appropriate integrating factor givesAt closure $R_i=0$, so $D=\sqrt{A}$ and $D\sqrt{w}=\sqrt{Q}$. HenceThe right side is positive for a genuine annulus. On the gravity-driven quadratic family from the previous part, $w^{1/2}$ grows linearly with height and this integral diverges logarithmically. It must then reach the finite positive threshold at a finite height. This explains the expected capillary closure of a falling hollow thread on that freely falling branch without requiring its full coupled profile.

There is a real boundary-data limitation to that expectation. Neither the printed nozzle speed nor the local equations alone guarantee finite closure for every imposed nozzle tension. To see this within the same leading model, define $T=3\mu Q{w}^{\prime }/w+\gamma S$, so $T^{\prime }=-\rho gQ/w$. For equal pressures and $w^{\prime }>0$, both radii decrease and $S\le S_0$. Ifthen a bootstrap gives $T>T_0/2$ and $w\ge w_0{e}^{cz}$: the maximum accumulated weight is smaller than the assumed tension margin. Consequently the closure integral, including its prefactor, is at most $\gamma /(\mu c{w}_0)$. If this is smaller than $\sqrt{A_0}-D_0$, closure never occurs at finite height. For example, in consistent units choose $\mu =\rho g=w_0=1$, $R_i(0)=\lambda$, $R_e(0)=2\lambda$, $\gamma =\lambda$, $T_0=100{\lambda }^2$. Then $c=47/9$, the weight bound is $27{\lambda }^2/47<50{\lambda }^2$, and $9\lambda /47<(\sqrt{3}-1)\lambda$. Small $\lambda$ makes the initial geometry slender. Thus finite closure is the expected freely falling branch behaviour, not a theorem following from nozzle speed alone. If $\gamma =0$ there is no capillary closure by this mechanism.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

Take $z$ upwards and write $\zeta =z+Wt$, so that the unperturbed interface is $\zeta =0$, with carbon dioxide above and water below. Use the sharp-interface, incompressible flow version of Darcy's law, with constant porosity and permeability of a porous medium, no capillary pressure, and effectively unbounded layers. The distinction between Darcy flux and interface speed matters: the base flux is $q_z=-\varphi W$, since the printed $W$ is the speed of the interface. Thus the base pressure gradients satisfyThe mass density ordering is stabilizing, whereas injection of the less viscous fluid into the more viscous fluid produces the Saffman–Taylor instability.

For linear stability analysis, displace the interface by $\eta =\hat{\eta }{e}^{\sigma t+i\alpha x}$. Taking the divergence of Darcy's law shows that the perturbation pressure satisfies the Laplace equation. Decay away from the interface gives, for $a_{\alpha }=|\alpha |>0$,The kinematic boundary condition and continuity of normal Darcy flux giveMeanwhile pressure continuity, evaluated on the displaced interface, givesEliminating $A,B$ yields the buoyancy-modified Darcy fingering dispersion relation:Every nonzero wavenumber decays when $W<W_c$, grows when $W>W_c$, and is neutral at equality, whereThe zero wavenumber is a uniform displacement and is neutral. Without a short-scale regularization such as surface tension, this model has no fastest-growing finite wavenumber: on the unstable side its growth rate increases without bound as $|\alpha |$ increases. Finite layer depths would change the dispersion relation; none are specified here.

Taking $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a year of $365.25$ days, the data giveThe numerical paragraph supplies no porosity, so it cannot determine a unique numerical interface speed. If $W$ were instead intended to mean Darcy velocity, the numerical critical value would be $103.2\mathrm{m}\mathrm{y}{\mathrm{r}}^{-1}$, and the interface would move at $W/\varphi$. These two velocity conventions must not be conflated.

Write $Q=U_j{x}_j$ and use repeated-index summation. Direct differentiation givesTracing yields ${\partial }_i{u}_i=0$. Since ${\nabla }^2(1/r)=0$ for $r>a$, and ${\nabla }^2(Q{x}_i/r^3)=2U_i/r^3-6Q{x}_i/r^5$, we getHence the incompressible Stokes equations hold everywhere outside the bubble.

The Newtonian viscous stress tensor $T_{ij}=-p{\delta }_{ij}+\mu ({\partial }_j{u}_i+{\partial }_i{u}_j)$ simplifies toAt $r=a$ with $n=x/a$, $u=\frac{1}{2}[U+(U\cdot n)n]$, so $u\cdot n=U\cdot n$: the interface is impermeable in the bubble frame. Also $Tn=-(3\mu /a)(U\cdot n)n$ is purely normal, so the tangential traction vanishes, appropriate to a clean bubble with negligible internal viscosity. The tangential fluid velocity need not equal the bubble's velocity. At infinity $u,p\to 0$. The calculation assumes the usual spherical, negligible-deformation limit; exact normal-stress balance includes interior pressure, surface tension and any buoyancy sustaining the motion.

Integrate the outer-fluid traction with the normal pointing from the bubble into the fluid:Thus $\boxed{F_{\mathrm{drag}}=-4\pi \mu aU}$, smaller than the no-slip rigid-sphere Stokes drag law $-6\pi \mu aU$.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

Let $z$ measure distance normal to the plane, with the wall at $z=0$ and the free surface at $z=h$. Take density $\rho$, dynamic viscosity $\mu$, and gravitational acceleration $g$. In lubrication theory, normal momentum balance gives hydrostatic pressure while inertia and streamwise viscous derivatives are smaller. With negligible surface tension and negligible air shear,Apply the no-slip boundary condition $u=v=0$ at $z=0$, and zero leading tangential stress $u_z=v_z=0$ at $z=h$. The velocity profiles areDepth integration gives the volume flux, with $K=\rho g/(3\mu )$:Combining incompressibility with the wall and surface kinematic conditions gives conservation of mass, $h_t+{\partial }_x{q}_x+{\partial }_y{q}_y=0$. Therefore the two-dimensional version of the gravity-driven thin film on an incline satisfiesDirect downslope gravity supplies the advective term, while gradients of the hydrostatic pressure spread the film along the surface. At $\alpha =0$ this reduces to nonlinear gravitational spreading in a horizontal film.

The interface is material: fluid particles on $F=z-\eta (x,t)=0$ remain there. Its material derivative therefore vanishes, giving $w_j={\eta }_t+u_j{\eta }_x=D_j\eta /Dt$ on the interface. Since $w_j={\partial }_z{\varphi }_j$, this is the kinematic boundary condition for a free-surface graph on each side. The material derivatives use the respective tangential fluid velocities; they enforce the same moving normal boundary.

The Unsteady Bernoulli equation gives $p_j+{\rho }_j[{\partial }_t{\varphi }_j+|\nabla {\varphi }_j{|}^2/2+gz]=C_j(t)$. Absorb the spatially constant functions $C_j$ into the gauges of the velocity potentials. Evaluating at $z=\eta$ and subtracting the equations yields the stated dynamic boundary condition for an inviscid interface. Surface tension supplies the pressure jump, rather than pressure continuity. Thus the first condition enforces material motion, and the second enforces the Bernoulli pressure balance.

Take $x$ in the injection direction and $y$ along the flat interface. The specified $U$ must be a Darcy velocity, so the pore-fluid velocity and unperturbed interface speed are $U/\varphi$. Write the perturbed interface as $x=X(t)+\eta (y,t)$, with $\dot{X}=U/\varphi$, and choose the normal from fluid 1 into fluid 2. The background pressure gradients from Darcy law are ${\partial }_x{p}_i^{(0)}=-{\mu }_{i}U/k$.

By incompressibility and uniform permeability of a porous medium, pressure perturbations are harmonic functions. A transverse Fourier mode with positive wavenumber $\alpha$ has the decaying formsThe kinematic boundary condition and equal normal Darcy flux implyWithout capillarity the pressure is continuous at the displaced interface. Its linear perturbation is thereforeConsequently the planar viscous-fingering dispersion relation reduces toThis is the Saffman–Taylor instability: less viscous fluid advances preferentially through protrusions. Without a short-wave regularizing mechanism the idealized rate has no finite maximum.

For the prescribed apparent capillary pressure, the chosen normal is $\mathbf{n}=(1,-{\eta }_y)/\sqrt{1+{\eta }_y^2}$, so $\nabla \cdot \mathbf{n}=-{\eta }_{yy}+O({\eta }^2)={\alpha }^2\eta +O({\eta }^2)$. Subtract the flat-front pressure jump $\gamma (1+\beta X){\kappa }_0$. Linearization of the jump at $x=X+\eta$ givesThere is no first-order product of the surface-tension perturbation and the perturbed macroscopic curvature. ThusUsing the capillary number convention $\mathrm{Ca}={\mu }_{2}U/\gamma$ and ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$, this isIf $x$ is measured from the unperturbed front and $\gamma$ denotes its local apparent tension, set $X=0$ to obtain the usual fixed-coefficient expression. For a physically fixed spatial gradient, $X=Ut/\varphi$ and the displayed rate is an instantaneous, generally time-dependent growth rate; the amplitude solves $\dot{a}=\sigma (\alpha ;X(t))a$, rather than one constant exponential law for the entire displacement. The parameter $\beta$ has dimensions of inverse length; the absolute tension gradient is $\gamma \beta$.

Assume $1+\beta X>0$, as required for positive apparent surface tension, and define $A=({\mu }_2-{\mu }_1)U-k\gamma \beta {\kappa }_0$. If $A>0$, unstable modes satisfy $0<{\alpha }^2<A/[k\gamma (1+\beta X)]$. Differentiating the cubic dispersion relation gives the fastest-growing modeWhen $A\le 0$, every nonzero mode decays, with a neutral zero-wavenumber translation in the infinite-domain limit. For $\beta {\kappa }_0>0$, surface-tension-gradient stabilization of viscous fingering suppresses the instability entirely at positive injection speeds satisfyingIf $\beta {\kappa }_0\le 0$, no positive injection speed eliminates every long-wave unstable mode in an unbounded interface. The prescribed wetting variation is treated as an effective normal pressure law for porous-media flow.

Each broad face supplies surface tension $\gamma$ pulling the rounded hole edge into the remaining sheet. Their resultant is $2\gamma$ per unit circumference. At the inner boundary the fluid's outward normal is $-{\mathbf{e}}_r$, so the boundary traction is outward in the radial direction when $h{\sigma }_{rr}=-2\gamma$. The right panel of the preceding diagram shows these two capillary pulls.

With uniform thickness and $p_{\mathrm{ext}}=0$, the axisymmetric viscous-sheet stretching equations reduce toThis Euler-Cauchy equation gives $u=Ar+B/r$. The fixed outer rim imposes $u(R_0)=0$, hence $B=-A{R}_0^2$. Because $(ru{)}_r/r=2A$, conservation of mass gives $\dot{h}=-2Ah$, independent of $r$. Thus a uniform sheet remains uniform.

The radial stress is ${\sigma }_{rr}=6\mu A-2\mu B/r^2$. Its value at the hole edge determinesThe edge is material, so $\dot{R}=u(R)$ andNeglecting the initially tiny hole's volume, conservation of mass gives $\pi h(R_0^2-R^2)=\pi h_0{R}_0^2$, or $h=h_0/(1-x^2)$ with $x=R/R_0$. Therefore the capillary growth of a hole in a viscous sheet obeysFor a finite initial hole $x_i$, replace $h_0$ in the denominator by $h_0(1-x_i^2)$. A nonzero seed is needed: the exact initial condition $x(0)=0$ gives the stationary solution of this differential equation. For a small positive seed, $x$ initially grows exponentially at rate $\gamma /(\mu h_0)$.

Let $\hat{\eta }$ be the displacement amplitude of an interface. Its kinematic boundary condition is $ik(U-c)\hat{\eta }=\hat{w}$, on either side, so the first of the jump conditions for stratified inviscid shear flow isFor zero surface tension, pressure is continuous at the displaced interface. Since the background hydrostatic pressure has derivative $-g\rho$, linearization gives $[\hat{p}-g\rho \hat{\eta }{]}_{-}^{+}=0$. Using $\hat{p}={\rho }_0[(U-c){\hat{w}}^{\prime }-U^{\prime }\hat{w}]/(ik)$ and the kinematic condition givesThis derivation also works when $U$ jumps: the displacement is shared, while the vertical velocity need not be. It avoids ambiguous products of distributions at a discontinuous velocity profile.

If the lower density ${\rho }_{-}$ exceeds the upper density ${\rho }_{+}$, stable density stratification adds a gravitational restoring term. Including surface tension $\gamma$, the same calculation has the schematic dispersion relationThe gravity term dominates the shear term as $K\to 0$ and therefore suppresses sufficiently long waves. The capillary term dominates as $K\to \infty$ and suppresses sufficiently short waves. Together they reduce the Kelvin-Helmholtz instability growth rate and can leave only a finite unstable band, or eliminate instability if the restoring effects are strong enough.

More surfactant is encountered on the side toward which the far-field concentration $\Gamma$ increases. Adsorption lowers the surface tension there, so the resulting Marangoni stress drives interfacial flow toward the cleaner, higher-tension side. The reaction force propels the bubble along the concentration gradient; this is chemophoresis.

The bulk concentration obeys the advection-diffusion equationAt $r=a$,equates the outward bulk diffusive flux to minus the net adsorption rate: $C>b\Gamma$ favors desorption into the bulk, while $C<b\Gamma$ favors adsorption onto the interface.

With advection neglected, $\Gamma$ is harmonic. Rotational symmetry about $\mathbf{G}$ and the imposed far-field gradient select the dipolar formWhen $kab\ll D$, the boundary flux is smaller than the characteristic diffusive flux, so the leading boundary condition is ${\partial }_r\Gamma =0$ at $r=a$. This gives $B=a^3/2$, and hence

For an interface with unit normal directed from the bubble into the exterior liquid, the interfacial stress balance with variable surface tension may be writtenwith signs tied to the stated curvature convention. Put $C^{\prime }=A\mathbf{G}\cdot \mathbf{n}$ and $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$. The constant part of the normal traction is balanced by the uniform bubble pressure. Since $\kappa =2/a$ andthe remaining exterior traction isIts resultant vanishes becauseand therefore the integrals of the two terms cancel. This is required because the bubble and its interfacial stresses exert no external body force on the combined bubble–fluid system.

The traction is a first spherical harmonic, so the decaying, force-free Stokes flow has no Stokeslet and is generated by the indicated Papkovich–Neuber representation. Comparing the supplied tractionwith the capillary traction givesIn the convention for these potentials, the normal velocity at $r=a$ is $(\beta /\mu )\mathbf{G}\cdot \mathbf{n}$. The kinematic boundary condition for a translating sphere is $\mathbf{u}\cdot \mathbf{n}=\mathbf{U}\cdot \mathbf{n}$, so

Linearize the surface transport equation about $C=C_0$ by writing $C=C_0+C^{\prime }$ and neglecting products of small perturbations. The tangential velocity relative to the translating bubble obtained from the same potential isUsing the identities supplied in the question,The bulk result gives ${\Gamma }^{\prime }=\frac{3}{2}a\mathbf{G}\cdot \mathbf{n}$ on the surface. The linearized equationthen yieldsThus

Increasing $k$ strengthens exchange with the imposed bulk gradient, so $A$ and $U$ increase toward a saturation value. Increasing $D_s$ smooths surface-concentration differences and decreases $U$. Increasing $C_0$ strengthens advective redistribution of the background surfactant; the resulting feedback opposes the imposed dipole, so $U$ decreases.

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.

Surfactants accumulate at an interface and orient their two chemically distinct ends toward their preferred fluid components. The relaxation found in part c lowers the coefficient of the gradient energy, and hence generally lowers the surface tension. If the reduction makes $\tilde{\kappa }<0$, the positive fourth-gradient term proportional to $\gamma$ can stabilize structure at a nonzero wavevector, producing the modulated correlations characteristic of a microemulsion.

The long-wave approximation makes $u$ independent of $z$ at leading order. Mass conservation and symmetry about $z=0$ then giveFor an incompressible Newtonian fluid,The leading normal-stress balance on either surface isSince ${\sigma }_{xx}-{\sigma }_{zz}=4\mu u_x$, it follows that