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For thickness $h(r,t)$, radial velocity $u(r,t)$ and external pressure $p_{\mathrm{ext}}$, the leading Newtonian fluid stress tensor and incompressibility giveRadial force balance on an annular sector includes the inward projection of hoop traction and external pressure on the sloping broad faces. Together with conservation of mass it yields

A spherical clean bubble with negligible internal dynamic viscosity has zero tangential traction and permits tangential slip. In the creeping-flow, negligible-deformation limit its drag is $-4\pi \mu a\mathbf{U}$, versus $-6\pi \mu a\mathbf{U}$ for a no-slip rigid sphere.

A sphere of radius $a$ at gap $aϵ$ translates at $U{\mathbf{e}}_x$ and rotates at $\omega {\mathbf{e}}_y$ above a stationary wall in a right-handed coordinate system. At leading order its gap is $h=aϵ+r^2/(2a)$. In its translating frame the lower and upper tangential velocities are $-U{\mathbf{e}}_x$ and $-a\omega {\mathbf{e}}_x$. The Reynolds lubrication equation givesIntegrating the upper-boundary traction, including pressure on the sloping surface, and its moment about the sphere centre gives the sphere-on-fluid force and torqueThe off-diagonal symmetry is required by the Lorentz reciprocal theorem for Stokes flow. A torque-free sphere therefore has $a\omega =U/4$ and leading drag $3\pi \mu aU\log (1/ϵ)$. Compare the rigid-wall terms of Bertin et al., equations (4.4)–(4.5); those forces act on the sphere and have the opposite sign.

Choose $x$ downslope along the rigid plane and $z$ normal to it into the fluid, with the wall at $z=0$ and the flat free surface at $z=h$. The gravitational body force per unit volume has components $(\rho g\sin \alpha ,-\rho g\cos \alpha )$. The wind's tangential traction on the top surface is $-S$ in the $x$ direction; the atmospheric normal traction is $-p_{\mathrm{atm}}$ in the $z$ direction. At the wall there is no-slip boundary condition, and the wall supplies the opposing normal force and a tangential traction whose sign depends on the wind strength.

The velocity is parallel to the plane, $(u(z),0)$. Its velocity profile is a concave quadratic, with $u(0)=0$ and negative slope at the free surface when $S>0$. The diagram uses moderate wind, for which the flow is still everywhere downslope. Stronger wind can reverse the upper portion, or the entire moving layer; the later cases quantify this.

For an inclined viscous film with opposing surface shear, the steady parallel velocity automatically satisfies the continuity equation and the advective acceleration vanishes. The two components of the Navier-Stokes equations reduce toThe uniform thickness and constant atmospheric pressure make $p_x=0$. The boundary conditions areThese are respectively no-slip boundary condition, imposed tangential traction on the surface with outward normal $+{\mathbf{e}}_z$, and the normal stress condition on the flat free surface. The normal viscous stress is zero because the normal velocity vanishes. Integrating givesIn particular $\mu u^{\prime }(0)=\rho gh\sin \alpha -S$ is the downslope tangential traction exerted by the fluid on the wall. The wall's tangential traction on the fluid has the opposite sign, because the outward normal of the fluid at its bottom is $-{\mathbf{e}}_z$. This fixes the potentially confusing wall sign convention.

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 $\gamma /(\mu h_0)$.

For the stated axisymmetric flow, the diagonal components of the rate-of-strain tensor areTheir sum is $u_r+u/r+w_z=0$, the incompressibility condition. In the leading thin-sheet approximation, vanishing tangential traction makes $u$ independent of $z$. The normal stress boundary condition is ${\sigma }_{zz}=-p_{\mathrm{ext}}$, so the Newtonian fluid stress tensor givesThereforeFor the small annular sector, the inner and outer radial faces contribute $2\delta \theta {\partial }_r(rh{\sigma }_{rr})\delta r$ in the radial direction. The two azimuthal faces contribute $-2\delta \theta h{\sigma }_{\theta \theta }\delta r$: their hoop tractions have inward radial components. The combined radial force of the external pressure on the sloping upper and lower surfaces is $2\delta \theta r{p}_{\mathrm{ext}}{h}_r\delta r$.

Neglecting inertia, force balance is thusSubstituting the two stresses cancels the terms involving $p_{\mathrm{ext}}{h}_r$ and gives the axisymmetric viscous-sheet stretching equations:Finally, conservation of mass in the sector gives