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Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

Write $\mathbf{u}={\mathbf{u}}^{\prime }$ and $\theta =-z+{\theta }^{\prime }$. Dropping quadratic perturbation terms gives the Linearized Boussinesq equationsandThe fixed temperatures give ${\theta }^{\prime }=0$ at $z=0,1$. Impermeable stress-free boundary conditions give

For each Volvox colony, the excess gravitational force isFor a normal Stokeslet a distance $R$ below a flat stress-free interface, the image is an oppositely directed Stokeslet a distance $R$ above it. The real force creates no lateral velocity at the other colony because both centres have the same height. Their separation from the image is $(x,-2R)$, however, so each colony moves toward the other with speedBoth colonies move, and therefore

DefineThe dimensionless dynamics areThis gradient flow descends an even attractive potential with minimum $V(0)=-1$ and $V\to 0^{-}$ as $|\xi |\to \infty$.

For $\xi \gg 1$, $d\xi /d\tau \simeq -1/{\xi }^2$, so the approach from ${\xi }_0=x_0/(2R)$ takes ${\tau }_c\simeq {\xi }_0^3/3$. Restoring dimensions,With water viscosity $\mu \simeq 10^{-3}\mathrm{Pa}\mathrm{s}$, ${\rho }_w\simeq 10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $ϵ=0.03$, $R=200\mu \mathrm{m}$, and $x_0=5R$, this givesBecause $x_0/(2R)=2.5$ is only moderately large, this is a far-field estimate rather than the exact collision time.

Let $a$ denote the bubble size, $U$ its speed, and $\mathrm{Re}=Ua/\nu \gg 1$ the Reynolds number. Outside a thin boundary layer, inertia dominates and the flow is an irrotational flow with velocity scale $U$ and strain scale $U/a$. Its viscous dissipation has orderin three dimensions.

The boundary-layer thickness isFor a clean bubble the stress-free boundary condition prescribes tangential stress rather than tangential velocity. The outer flow already has tangential strain of order $U/a$, so cancelling that stress requires a velocity correction of onlynot a correction of order $U$. Its gradient remains $O(U/a)$, and its boundary-layer dissipation iswhich is asymptotically smaller.

In steady translation, the drag power $DU$ balances viscous dissipation. The leading estimate therefore follows from the known outer flow alone:in three dimensions. This is the dissipation estimate for high-Reynolds-number bubble drag.

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

Let $d(z,t)$ be the local water-film thickness. To first order in the interface amplitudes,The lubrication theory flux down the vertical surface, with a no-slip boundary condition at the ice and a stress-free boundary condition at the water-air interface, isFor the unperturbed film $p_z=0$, so

The linearized Young–Laplace equation gives the capillary-pressure perturbationBecause the prescribed volume flux is uniform, its first-order perturbation must vanish. Linearization of the flux law givesThus, with $\Gamma =\gamma /(3\rho g)$,and henceThe complex amplitude ratio records the phase shift caused by surface tension in the long-wave approximation.

Let the unit normal $\mathbf{n}$ point from the fluid toward the free surface, let $\mathbf{t}$ lie along the surface, and write the swimmer direction asReflecting the swimmer position across the plane and reflecting its orientation toconstructs the free-surface image of a force dipole. At the surface the two dipoles have equal tangential velocity and opposite normal velocity. Their sum therefore satisfies the no-penetration boundary condition; tangential velocity is even across the plane and normal velocity is odd, so the tangential traction vanishes and the stress-free boundary condition is also satisfied.

The vector from the image to the swimmer is $\mathbf{r}=-2h\mathbf{n}$, for which $({\mathbf{p}}^{\ast }\cdot \mathbf{r}{)}^2=4h^2{\sin }^2\theta$. Evaluating the image force-dipole flow at the swimmer givesThere is no tangential image velocity at the swimmer in this point-dipole approximation. With the convention that $\mathcal{P}>0$ is an extensile pusher microswimmer, a nearly parallel pusher is attracted toward the surface, whereas a nearly parallel contractile puller microswimmer is repelled. For $|\sin \theta |>1/\sqrt{3}$ the normal drift reverses because the image samples the axial rather than equatorial part of the dipolar flow.

The shallow-ice approximation makes the pressure hydrostatic pressure,Horizontal Stokes flow balance and the stress-free boundary condition at $z=h$ givesoThe basal shear stress and the stated basal sliding law are thereforeIntegrating once more and imposing $u(0)=u_b$ yieldsFor the right half of an ice cap, $h_x<0$, so both basal sliding and internal deformation carry ice away from the ice divide.

For a fixed-temperature layer with stress-free boundary conditions, a vertical Fourier sine series has $a^2={\lambda }^2+n^2{\pi }^2$, where $\lambda$ is horizontal wavenumber. The coupled vertical velocity and temperature amplitudes satisfy the displayed quadratic for their growth rate $\sigma$. Its roots are real for positive Rayleigh number, while both roots have negative real parts for negative Rayleigh number. Setting $\sigma =0$ gives the free-slip convection neutral curve.