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For large $\alpha$, the evanescent interaction between the two interfaces is proportional to $E=e^{-2\alpha }$. Ignoring it first gives isolated interfacial gravity waves with phase speedsThe lower wave propagating positively relative to its local current and the upper wave propagating negatively have equal laboratory speed whenTheir common speed is zero. This is counterpropagating wave instability: the interfaces communicate through their decaying velocity fields, and near resonance their wave phases can lock while extracting energy from the mean shear.

The instability interval has asymptotic endpoints $J_{\pm }=2\alpha [1\pm E+O(E^2)]$ and widthNear exact resonance the interaction is weak but destabilizing: at $q=1$, $C_{-}^2=2-\sqrt{4+E^2}=-E^2/4+O(E^4)$, so $\mathrm{Im}C\sim E/2$. Thus the resonant band narrows and growth weakens exponentially with separation measured in decay lengths. The resonance here is between two density interfaces in globally uniform shear, not a resonance modified by additional vorticity jumps.

Take $k>0$ without loss of generality, so $\alpha =kh/2>0$, and require decay as $Z\to \pm \infty$. The base velocity is $\tilde{U}=Z$ everywhere; there is no jump in its derivative. Let $C=\tilde{c}$ and $E=e^{-2\alpha }$. A decaying representation that already solves the layer equations isIt is continuous at both interfaces. At $Z=-1$ the value is $A+BE$ and the derivative jump is $-2\alpha A$; at $Z=1$ they are $B+AE$ and $-2\alpha B$. The density anomaly drops by one across each interface. The jump conditions for stratified inviscid shear flow therefore reduce toFor nonreal $C$ the displacement denominators do not vanish. Neutral limiting values are obtained by continuation, rather than by dividing by zero at an interface.

Set $q=J/(2\alpha )$. A nonzero mode requires the determinant to vanish:Expansion gives the dispersion relation for two density interfaces in uniform shear,Uniform shear throughout the exterior is important: replacing it with constant exterior velocities would introduce vorticity jumps and give a different dispersion relation.

Sphere 2 is now force-free and torque-free. Its translation and spin follow the incident rigid-motion components, so it emits neither a Stokeslet nor a rotlet. Its first disturbance is a stresslet from the incident rate of strain, of magnitude $\mu a^3(a|\mathbf{V}|/R^2)$. Its return velocity at sphere 1 is consequently $O(a^4|\mathbf{V}|/R^4)$:More precisely, that leading correction is $-(15/4){ϵ}^4(\mathbf{V}\cdot \mathbf{n})\mathbf{n}$. It vanishes for purely transverse forcing, for which the first translational correction is of order ${ϵ}^6$.

The first-return spin requires an additional cancellation. The leading incident strain at sphere 2 isIt is axisymmetric about the line of centers, so the reflected stresslet has zero vorticity at sphere 1 on that axis. A naive vorticity estimate of order $(|\mathbf{V}|/a){ϵ}^5$ therefore has zero coefficient. The finite-size correction to the incident strain and the next force-free scattering multipoles are two powers of $ϵ$ smaller; these generically giveThis suppressed spin from a force-free distant sphere is also consistent with the reciprocal torque-to-self-translation coupling. Purely longitudinal forcing gives zero spin exactly by axial symmetry. Distinguishing the stresslet's on-axis curl from its generic off-axis scale is essential here.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

Taking the curl of the Brinkman equation eliminates pressure and the constant matrix velocity. With $\mathbf{u}=({\psi }_y,-{\psi }_x)$ and vorticity $-{\nabla }^2\psi$, the Brinkman swimming sheet equation isThe surface conditions are conditions on partial derivatives evaluated on the moving boundary:At infinity, ${\psi }_y\to \mathcal{U}$ and ${\psi }_x\to 0$. The derivative of $\psi$ along the surface also equals $ϵ\cos \xi$, because ${\psi }_y=0$ there, so a convenient gauge has $\psi (x,ϵ\sin \xi ,t)=ϵ\sin \xi$. Decaying perturbations and the force-free condition determine the swimming solution.

For a flat impermeable free surface with zero tangential stress, the image of a parallel stresslet is a stresslet of the same strength and parallel orientation at the reflected point. Reflection makes the normal velocity odd and tangential velocity even across the plane, enforcing zero normal flow and zero tangential shear. This is the free-surface image of a force dipole; it is not the no-slip-wall image system.

The cell at $z=-h$ lies a distance $2h$ vertically below its image. Since $\mathbf{e}\cdot \mathbf{r}=0$ there, its induced vertical velocity is $S/(32\pi \mu h^2)$ towards the surface. There is no image-induced tangential velocity, vorticity or tangent-normal strain at this point, so a parallel orientation remains parallel in this idealization. Therefore the finite-time free-surface approach of a point stresslet isThe trajectory combines its parallel self-propulsion with this normal drift. Finite-time contact is the point-model prediction for $S>0$, carried over from part (b). A puller with $S<0$ moves away instead. For a finite cell, the far-field approximation fails before $h$ reaches zero; surface deformation and near-contact physics can change the final encounter.

Take $\mathbf{e}={\mathbf{e}}_x$ and the surface normal ${\mathbf{e}}_z$. Under reflection in the free surface, polar vectors transform by $M=\mathrm{diag}(1,1,-1)$ and an axial torque transforms by $\det (M)M$. A parallel torque therefore reverses sign, while the separation direction ${\mathbf{e}}_x$ does not. The free-surface image of a rotlet dipole is a dipole of strength $-D$ at $z=+h$.

The observation point at the cell is directly below the image, so $\mathbf{e}\cdot \mathbf{r}=0$ and the image velocity vanishes. A symmetry proof of the vanishing normal drift is also useful: reflection in the vertical plane containing $\mathbf{e}$ reverses the axial dipole moment, preserves the normal component of a polar velocity, and leaves the geometry unchanged. Linearity would make that same component change sign with $D$, forcing it to be zero. Thus this singularity alone produces no attraction or repulsion.

Write $X,Y,Z$ relative to the image, so the cell is at $(0,0,-2h)$. The image field isAlthough its value at the cell is zero, its velocity gradient is not. In particular, the surface-induced yaw of a rotlet dipole hasA spherical torque-free body rotates with the local fluid angular velocity ${\Omega }_z={\omega }_z/2$. Its swimming direction therefore turns in the surface plane and, at fixed height and speed, it traces a circle with radius $V_{\mathrm{swim}}/|{\Omega }_z|$. The sign of $D$ sets the sense of turning in this convention. An elongated swimmer also responds to strain; its rotation coefficient depends on its shape and need not equal one half of the vorticity. When the stresslet is included, its changing height produces a changing turning rate rather than an exactly fixed-radius circle.

Let $\mathbf{n}=\mathbf{X}/R$, $\zeta =6\pi \mu a$, and letUse the method of reflections for Stokes flow. At the fixed sphere, the first sphere produces the incident Stokeslet velocity ${\mathbf{u}}_{\infty }(0)=\mathsf{A}\mathbf{V}$ to leading order. In Faxén translation law, set the second sphere's translational velocity to zero. The applied holding force is thereforeThe first sphere's finite-radius potential dipole adds $O(V{a}^3/R^3)$ to the incident velocity, as does the Laplacian term in Faxén translation law at the fixed sphere. Multiplying by $\zeta$ gives the stated next correction. There is no intermediate $O(\mu V{a}^3/R^2)$ term. This is the holding force and torque for a sphere in a distant Stokeslet.

The vorticity of a Stokeslet at displacement $\mathbf{r}$ is $\mathbf{F}\times \mathbf{r}/(4\pi \mu r^3)$. At the fixed sphere, $\mathbf{r}=-\mathbf{X}$, soSet its angular velocity to zero in Faxén rotation law. The applied holding couple isThe displayed sign is the external couple needed to oppose the ambient rotation, rather than the hydrodynamic couple on the sphere.

The leading reflected flow at the first sphere is $\mathsf{G}(\mathbf{X}){\mathbf{F}}_2=-{\mathsf{A}}^2\mathbf{V}$. Apply Faxén translation law there with its prescribed force $\mathbf{F}=\zeta \mathbf{V}$. Since $(I+\mathbf{nn}{)}^2=I+3\mathbf{nn}$, the mobility correction from a fixed distant sphere isThe next correction comes from finite-radius terms in the incident/reflected flow and in Faxén translation law, together with the fixed sphere's induced stresslet and holding-couple rotlet. Each gives $O(V{a}^4/R^4)$ at the first sphere. The symbol $\mathbf{V}$ is its isolated-sphere velocity scale, not its actual velocity in the two-sphere problem.

The first sphere has no applied couple. Its angular velocity follows from Faxén rotation law and the reflected Stokeslet:The component of ${\mathbf{F}}_2$ parallel to $\mathbf{X}$ contributes no vorticity. The next reflected rotlet and stresslet give the stated error order.