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Consider steady flow of a Newtonian fluid in incompressible flow with constant dynamic viscosity $\mu >0$ between parallel plates at $z=0$ and $z=H$. The bottom plate is stationary and obeys the Navier slip boundary condition with slip length $b$. The top plate moves at velocity $U>0$ and obeys the no-slip boundary condition. There is no streamwise pressure gradient or body force. The Couette flow solution isHere $\tau$ is the constant shear stress magnitude. The lower wall's Navier slip boundary condition holds exactly because $b{u}^{\prime }(0)=bU/(H+b)=u(0)$. At the top, $u(H)=U$ is exactly the prescribed no-slip boundary condition; no finite upper-wall slip length is imposed in this example.

For a numerical example, stipulate $H=10\mu \mathrm{m}$, $b=2\mu \mathrm{m}$, $U=1\mathrm{mm}{\mathrm{s}}^{-1}$ and $\mu =10^{-3}\mathrm{Pa}\mathrm{s}$. The bottom velocity is $0.167\mathrm{mm}{\mathrm{s}}^{-1}$ and the shear stress is $0.0833\mathrm{Pa}$, compared with $0.100\mathrm{Pa}$ when both plates obey the no-slip boundary condition. The extrapolated profile reaches zero at $z=-2\mu \mathrm{m}$.

For fixed $H$, $U$ and $\mu$, $b\to 0$ recovers $u=Uz/H$. As $b\to \infty$, $u\to U$ throughout the gap and $\tau \to 0$: the moving upper wall selects the limiting uniform velocity, while the lower wall approaches a stress-free boundary condition. The dynamic viscosity stays positive; the disappearing shear stress comes from the disappearing velocity gradient.

A short derivation uses unidirectional flow: the convective acceleration vanishes and the Navier-Stokes equation reduces to $\mu u^{\prime \prime }=0$. Write $u=Az+B$. The bottom condition gives $B=bA$ and the top condition gives $A(H+b)=U$.

In nonrotating Rayleigh-Bénard convection with fixed-temperature, stress-free boundary conditions, the stress-free convection growth-rate polynomial has real roots on the destabilizing side. Consequently the first instability is stationary, rather than oscillatory. Minimizing the free-slip convection neutral curve selects ${\lambda }_c=\pi /\sqrt{2}$ and ${\mathrm{Ra}}_c=27{\pi }^4/4$.

For nonrotating convection under the Boussinesq approximation with fixed-temperature impermeable stress-free boundary conditions, a marginal vertical mode $\sin (n\pi z)$ has neutral Rayleigh number $(k^2+n^2{\pi }^2{)}^3/k^2$. The minimum is at $n=1$, $k=\pi /\sqrt{2}$, and $R_c=27{\pi }^4/4$. These follow by eliminating the temperature from $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $(D^2-k^2)\theta =-w$, then minimizing over $k$.

For a unit-depth Boussinesq layer with stress-free boundary conditions, fixed boundary temperatures, imposed field along $y$, and horizontal magnetic perturbations at the boundaries, put $a_h^2=k^2+m^2$ and $\Lambda =a_h^2+{\pi }^2$. Poloidal vertical velocity, temperature and vertical field proportional to $e^{st+ikx+imy}\sin \pi z$ have amplitudes satisfyingThe determinant gives the stated dispersion relation. Here $\sigma$ is the Prandtl number, $\zeta$ the magnetic-to-thermal diffusivity ratio, $Q$ the Chandrasekhar number and $R$ the Rayleigh number. The steady neutral curve is $R={\Lambda }^3/a_h^2+Q{m}^2\Lambda /a_h^2$.

// Target: fluid-mechanics.bigb

For a thin viscous gravity current with hydrostatic pressure $p=\rho g(h-z)$, a no-slip boundary condition at its base and a stress-free boundary condition at its top, horizontal momentum balance gives $u=(g/\nu )h_x(z^2/2-hz)$. Integrating across the depth gives the volume flux $q=-(g/3\nu )h^3{h}_x$. Horizontal spreading follows $h_t+q_x=0$ if the substrate is impermeable.

Let $H$ be the typical thickness, ${\ell }_r$ the radial variation length, $U$ the horizontal speed, and $W$ the vertical speed. For a thin smooth profile, $\epsilon =H/{\ell }_r\ll 1$. The axisymmetric continuity equation isAway from singular boundary regions, balancing its terms gives $U/{\ell }_r\sim W/H$, hence $\boxed{W/U\sim \epsilon \ll 1}$. Near $r=0$, regularity gives $v_r=O(r)$ and the same conclusion follows without a divergent $v_r/r$ term.

Vertical viscous derivatives dominate horizontal ones by ${\epsilon }^{-2}$. The ratio of convective acceleration to vertical viscous acceleration isFor thickness-evolution time $T$, the local inertial ratio is $H^2/(\nu T)$. When both ratios are small, lubrication theory reduces the Navier-Stokes equation toThinness must therefore accompany a slow enough motion; it is not a justification for dropping arbitrarily rapid imposed acceleration. The small vertical velocity also makes the leading vertical balance hydrostatic balance, $p_z=-\rho g$. The neglected normal viscous stress at the interface is smaller than the characteristic lubrication pressure by order ${\epsilon }^2$.

Use the source's curvature sign: $p_{\mathrm{air}}-p_{\mathrm{liquid}}=-\gamma \kappa$. The Young–Laplace equation and atmospheric pressure $p_0$ thus give $p(r,h,t)=p_0+\gamma \kappa$. Integrating the hydrostatic balance yieldsConsequently $p_r=\gamma {\kappa }_r+\rho g{h}_r$, independent of $z$ at this order. The radial Stokes equation and boundary conditions areThe first is the no-slip boundary condition; the second is the leading stress-free boundary condition, since the atmosphere has negligible dynamic viscosity and the surface tension is constant. Integrating twice givesThis is the horizontal velocity only. A smaller vertical velocity is generally required by the continuity equation.

Differentiating the axisymmetric stream function givesThe normal kinematic boundary condition fixes $C=Ua/2$, soBoth velocity components vanish at infinity, and the axisymmetric stream function is regular at the polar axes. The tangential component of the strain-rate tensor vanishes everywhere:Thus the stress-free boundary condition is satisfied. The vorticity is ${\omega }_{\varphi }=2C\sin \theta /r^2=Ua\sin \theta /r^2$, and the supplied dipolar pressure is consistent with the Stokes equation.

Use the Newtonian fluid stress tensor $\mathbf{\sigma }=-p\mathbf{I}+2\eta \mathbf{e}$. At the bubble surface,The vertical force exerted by the liquid is the integral of the vertical traction:Since the integral is $2/3$, the drag is downward and has magnitudeThe constant pressure integrates to zero force. The pressure perturbation contributes $4\pi \eta aU/3$ and the normal viscous stress contributes $8\pi \eta aU/3$; there is no tangential traction contribution.

For the free-interface qualification in part (b), the calculated normal traction at $\theta =0$ and $\theta =\pi$ differs by $6\eta U/a$. It cannot equal a single uniform gas pressure if every restoring effect is discarded. Restoring the external-minus-internal variation of hydrostatic pressure supplies a term $({\rho }_w-{\rho }_a)ga\cos \theta$. It cancels the dynamic normal traction at the terminal velocity $U=({\rho }_w-{\rho }_a)g{a}^2/(3\eta )$, consistent with a spherical clean bubble without surface tension. This distinguishes the drag calculation from the extra conditions needed for a freely rising bubble.

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.

Let $\delta =H/L\ll 1$. Incompressibility makes the transverse velocity $O(\delta U)$. In the axial Navier-Stokes equation, transverse viscous stress scales as $\mu U/H^2$, longitudinal viscous stress as $\mu U/L^2$, and convective inertia as $\rho U^2/L$. Thus lubrication theory requiresUnsteady inertia is also negligible if $\rho H^2/(\mu T)\ll 1$; this coincides with the preceding condition when $T=L/U$. The transverse pressure variation is smaller than this pressure scale by $O({\delta }^2)$, and viscous normal stress at the free surface is similarly smaller. Consequently pressure is uniform across the film to leading order, while axial viscous shear balances its axial gradient. These estimates justify the lubrication approximation with small surface slopes and a slowly varying rigid wall.

Write the physical cylindrical-film thickness as $d=ϵ\bar{h}$, so that $\bar{h}=O(a)$, and use $\bar{z},\bar{t}$ for dimensional coordinates. For a surface of radius $a+d$, its outward mean curvature, with the convention giving cylindrical curvature $1/a$, isThe Young–Laplace equation gives $p=p_{\mathrm{ext}}+\gamma \kappa$. With local normal coordinate $y$ measured from the rigid cylinder, the axial velocity solves $\mu u_{yy}=p_{\bar{z}}$, with no-slip boundary condition at $y=0$ and stress-free boundary condition at $y=d$. ThereforeThe factors of circumference in axisymmetric mass conservation cancel at leading thin-film order, giving $d_{\bar{t}}+q_{\bar{z}}=0$. IntroduceThis produces the capillary instability of a thin film on a rigid cylinder governed byThe curvature term $h_z$ destabilizes long waves; the axial-curvature term $h_{zzz}$ smooths short waves.

Linearize about $h_0$ using $h=h_0+\eta e^{st+ikz}$ with small $\eta$. The linear equation is $s+h_0^3(-k^2+k^4)=0$, soThe growth rate is positive for $0<|k|<1$, zero at $k=0,\pm 1$, and negative for $|k|>1$. Its derivative is $2h_0^{3}k(1-2k^2)$, givingThe requested sketch is even in $k$:

For the Rayleigh–Plateau instability of an entire liquid cylinder in the purely viscous limit with negligible exterior viscosity, the fastest mode tends to infinite wavelength. The rigid-wall coating instead selects a finite wavelength. The difference comes from the flow resistance: a coating must shear across its thickness, so its long-wave growth vanishes as $k^2$; a free liquid cylinder can undergo nearly plug-like axial extension. For example, the long-wave free-cylinder balances ${\eta }_t+(a/2)u_z=0$ and $3\mu u_{zz}=-\gamma ({\eta }_z/a^2+{\eta }_{zzz})$ give $s\sim \gamma (1-a^2{k}^2)/(6\mu a)$, which does not vanish as $k\to 0$. This comparison is for the creeping-flow, negligible-exterior-viscosity limit; finite inertia or exterior viscosity changes the wavelength selection. The full free-cylinder dispersion relation is discussed in Plateau-Rayleigh Instability of a Cylinder of Viscous Liquid.

The collar is a quasisteady sliding collar on a coated cylinder. It must slowly grow as well as translate: an exact travelling wave would obey $q-ch=\text{constant}$, whereas its stationary uniform far films have $q=0$ and would require $-c{h}_{+}=-c{h}_{-}$. Thus the stipulated unequal films require the stated slow evolution of the collar amplitude. In its $h=O(1)$ core, $c\ll 1$ and the volume flux is small, so the leading pressure gradient vanishes: $h^{\prime }+h^{\prime \prime \prime }=0$. Integration gives a constant plus sine and cosine terms. Choose the collar centre at its maximum to remove the sine term, and parameterize its edge height as $B$. ThenThe core is a local capillary equilibrium. In the moving frame, swept-up volume minus deposited volume gives the growth rate $c(h_{+}-h_{-})$. Its leading excess volume is $2\pi A$, and for constant $B$,Thus the printed rate is the rate of collar volume, or an order-of-magnitude amplitude rate; interpreted as an exact equation for $\dot{A}$, it lacks $1/(2\pi )$. This slow change is smaller than the translational core terms by the small far-film thickness and does not alter the leading local matching below.

Near either edge $x_f=\pm \pi$, let $h_f$ denote the adjacent film thickness, $h_{-}$ at the rear and $h_{+}$ at the front. Across the short transition region the collar's slow growth is higher order, so local mass conservation integrates to $q=c(h-h_f)$. The axial third derivative dominates the first derivative in the capillary flux, givingFor collar-to-film capillary matching, chooseThe two sides become $c{h}_f{H}^3{H}^{\prime \prime \prime }$ and $c{h}_f(H-1)$ respectively, and henceThe resulting transition length is small compared with the core length: the matching relations below imply $h_f=O(c^{2/3}/A)$, so $h_f/c^{1/3}=O(c^{1/3}/A)$ for an order-one core amplitude.

To count admissible branches, put $H=1+\zeta$ and linearize. The characteristic equation is $r^3=1$, with rootsAt the rear, decay as $X\to -\infty$ permits only $\zeta \sim D{e}^X$. For the nontrivial branch growing into a positive collar, $D>0$; translation rescales $D$, leaving one orbit. This branch has positive $H^{\prime },H^{\prime \prime },H^{\prime \prime \prime }$. It exists for all forward $X$, since $0<H^{\prime \prime \prime }=(H-1)/H^3\le 4/27$ for $H\ge 1$, and once $H^{\prime \prime }>0$, $H$ grows at least quadratically. Then $H^{\prime \prime \prime }=O(X^{-4})$, so successive integrations give a quadratic asymptote with a finite constant term; translation removes the linear term. The constant solution is not a rear-to-collar match. On the downward branch, $H<1$ gives $H^{\prime \prime \prime }<0$ while $H^{\prime }$ and $H^{\prime \prime }$ are initially negative; the slope remains negative and $H$ reaches zero, so it cannot match a positive collar. Uniqueness here refers to the positive, nontrivial rear matching branch, modulo translations.

At the front, decay as $X\to +\infty$ permits the two real amplitudes inTranslation removes one degree of freedom, leaving one parameter $\lambda$. These oscillatory branches account for the small capillary ripples ahead of the collar. This local stable manifold count gives the family of front branches; admissible global branches must also remain positive and reach a parabolic core asymptote. It does not, by itself, prove the global uniqueness of the matching parameter supplied in the question.

Write the rear and front core asymptotes, after translation removes their linear terms, as $a_{-}{X}^2+b_{-}$ and $a_{+}(\lambda )X^2+b_{+}(\lambda )$. Expanding the outer collar near either edge giveswhereas the inner expansion givesMatching the constant and quadratic coefficients therefore yieldsEliminate $h_{-}/h_{+}$ to obtain $a_{+}(\lambda )b_{+}(\lambda )=a_{-}{b}_{-}$. Using the specified selected root ${\lambda }_0$ and writing $\alpha =a_{+}({\lambda }_0)$, $\beta =b_{+}({\lambda }_0)$,The physical branch has $\alpha >a_{-}$, as required by $h_{-}<h_{+}$. No numerical values of these universal matching coefficients are needed for the requested relations.

Finally, the energy source is decreasing capillary surface energy, not Newtonian gravity. This can be checked through the volume-constrained capillary energy of a cylindrical coating. The exact liquid volume per axial length is $\pi [(a+d{)}^2-a^2]$, and the free-surface area is $2\pi \int (a+d)\sqrt{1+d_{\bar{z}}^2}d\bar{z}$. Expanding the latter and eliminating $\int d$ with fixed exact volume gives the nonconstant quadratic partApart from its positive dimensional factor, this is the surface energy $\mathcal{E}=\frac{1}{2}\int (h_z^2-h^2)dz$. Its variational derivative is $-h-h_{zz}$, and the thin-film equation is a conserved gradient flow. With periodic or zero-energy-flux boundaries, integration by parts givesFor an isolated collar between unequal infinite films, use the corresponding local energy balance, including far-film fluxes and subtracting background energies; the unsubtracted total energy is infinite. Collecting film into the growing collar lowers surface energy, which supplies the viscous dissipation sustaining its slow motion.

The Boussinesq approximation assumes that the fractional mass density variations are small. Replace density by a constant reference value in inertia, pressure scaling and mass conservation, but retain its small anomaly in the gravitational force: the large hydrostatic force otherwise cancels, leaving precisely the buoyancy-driving term. For temperature-driven motion, write $\rho ={\rho }_r[1-\alpha (T-T_r)]$ with $|\alpha (T-T_r)|\ll 1$. The resulting velocity is an incompressible flow; compressional heating and acoustic effects are omitted in the shallow, low-compressibility regime.

For this convection problem, use the strong Boussinesq approximation in the constant-property sense: the thermal expansion coefficient, kinematic viscosity, thermal diffusivity and heat capacity are also independent of temperature over the layer. This additional assumption makes the conduction profile linear and the perturbation coefficients constant. This usage of “strong” is the convection convention described in courses.physics.ucsd.edu/2021/Spring/physics218c/Spiegel%20Convection.pdf . Some stratified-flow literature uses “strong” instead to emphasize a globally constant reference density; that condition is also satisfied here. The equations below specify precisely which assumptions are being used.

Introduce the thermal diffusivity $\kappa$ and positive thermal expansion coefficient $\alpha$, which are necessary material data even though the question names only $\nu$ explicitly. Measure $z$ upward from the bottom. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$ with zero velocity. Let $\theta =T-T_b$ and let $\pi$ be perturbation pressure divided by the reference density, after subtracting the hydrostatic base pressure. Linearizing givesThe sign in the last equation is positive because upward motion carries warmer base-state fluid into cooler surroundings. Taking the divergence of momentum gives ${\nabla }^2\pi =g\alpha {\theta }_z$. Apply a Laplacian to its vertical component and subtract the vertical derivative of that pressure identity. This eliminates pressure:Apply the temperature operator to obtain the vertical-velocity equationThis is the pressure- and temperature-eliminated linear stability analysis equation for Rayleigh-Bénard convection.

At each boundary, impermeability gives $w=0$, the stress-free boundary condition gives $w_{zz}=0$, and perfect thermal conduction gives $\theta =0$. To obtain the second condition, zero tangential traction gives ${\partial }_z{u}_h+{\nabla }_{h}w=0$; since $w$ vanishes identically along the boundary, differentiate horizontal incompressibility to get $w_{zz}=0$. For the fully eliminated velocity equation, $\theta =0$ and the preceding momentum relation also imply $w_{zzzz}=0$ there. Thus suitable formulations areor, for the sixth-order vertical equation alone, $w=w_{zz}=w_{zzzz}=0$ at both boundaries. Fixed boundary temperature means a zero perturbation there, not zero perturbation heat flux.

Use normal modes $w=W\sin (n\pi z/d)e^{\sigma t+i\mathbf{k}\cdot {\mathbf{x}}_h}$, $n\ge 1$, $k=|\mathbf{k}|>0$. Put $K^2=k^2+n^2{\pi }^2/d^2$. Substitution yields the stress-free convection growth-rate polynomialAt stationary marginal stability, $\sigma =0$. In terms of the Rayleigh number $\mathrm{Ra}=g\alpha \Delta T{d}^3/(\nu \kappa )$ and dimensionless horizontal wave number $a=kd$, the free-slip convection neutral curve isWriting $y=a^2$ and differentiating gives $d\log {\mathrm{Ra}}_n/dy=3/(y+n^2{\pi }^2)-1/y$. Its zero is $a^2=n^2{\pi }^2/2$, with minimum $27n^4{\pi }^4/4$. The smallest vertical mode is $n=1$, so the critical temperature difference isThe first unstable horizontal wavelength is $2\sqrt{2}d$. To check that an oscillatory mode cannot destabilize earlier, expand the growth-rate equation: its linear coefficient is $(\nu +\kappa )K^2>0$, and its constant term changes sign exactly at the stationary neutral curve. For positive temperature difference the discriminant is $(\nu -\kappa {)}^2{K}^4+4g\alpha \Delta T{k}^2/(d{K}^2)>0$, so the crossing root is real. Below the global critical value every convection mode decays; at criticality the selected mode is marginal; above it some modes grow and eventually require nonlinear advection, producing convective heat transport. This is exchange of stabilities in stress-free convection.

Strictly, the linearized equations remain the correct small-amplitude equations on either side of onset. What fails above $\Delta T_{\max }$ is stability of the conductive state and long-time validity of a small-perturbation description, not the algebra of linearization. The threshold also presumes that the required small density variations and constant-property approximation hold at that temperature difference.

Put $a_h^2=k^2+m^2>0$ and $\Lambda =a_h^2+{\pi }^2$. The plane-layer poloidal-toroidal decomposition supplies the assumed poloidal fields asSince $w=-{\Delta }_h\varphi$ and $b_z=-{\Delta }_h\chi$, potentials proportional to $e^{st+ikx+imy}\sin \pi z$ have the required vertical components. Their horizontal components are proportional to $\cos \pi z$, satisfying the stress-free boundary conditions and the horizontal-field boundary conditions. The double-curl representation automatically enforces incompressibility and the solenoidal magnetic-field constraint.

Let $W,\Theta ,C$ be the amplitudes of $w,\theta ,b_z$. The temperature and resistive induction equations immediately giveEliminate the total-pressure perturbation by applying $\Delta$ to the vertical momentum equation and subtracting ${\partial }_z$ times its divergence. The resulting vertical equation isIn particular, $imC=-m^{2}W/(s+\zeta \Lambda )$ shows explicitly that magnetic tension opposes the displacement.

Taking the determinant of these three amplitude equations gives the horizontal-field magnetoconvection dispersion relation:This cubic polynomial includes all roots of the assumed poloidal coupled system; using its determinant form avoids discarding possible roots by dividing by $s+\Lambda$ or $s+\zeta \Lambda$. As a check, $Q{m}^2=0$ factors off magnetic diffusion and leaves the standard stress-free Rayleigh-Bénard convection growth equation.

Take the upward direction as ${\mathbf{e}}_z$, with polar angle $\theta$ measured from it. The liquid is quiescent at infinity. On a translating clean inviscid bubble, the kinematic boundary condition is $(\mathbf{u}-U{\mathbf{e}}_z)\cdot \mathbf{n}=0$, and the stress-free boundary condition is zero tangential traction. The normal traction must balance uniform gas pressure and any constant-curvature surface-tension jump.

For a vertical Stokeslet of strength $F$, the radial velocity is $u_r=F\cos \theta /(4\pi \mu r)$. Its traction on a concentric sphere is purely normal, so zero tangential traction is automatic. The kinematic condition at $r=a$ givesThe singularity is only a device for generating the exterior solution; there is no liquid point force inside the actual bubble. The integrated dynamic traction on the bubble is $-F{\mathbf{e}}_z$, giving the clean-bubble Stokes drag. The upward buoyancy is $4\pi \rho g{a}^3/3$, hence force balance gives

The normal stress also shows that surface tension is not required for this particular spherical solution. Let the ambient hydrostatic pressure at the bubble centre be $p_c$, so on its surface $p_h=p_c-\rho ga\cos \theta$. Adding the Stokeslet's dynamic normal traction giveswhere the terminal buoyancy value of $F$ cancels the dipole. This is the spherical clean bubble without surface tension balance: an initially spherical bubble can remain spherical in this ideal steady Stokes solution even when surface tension is zero. With uniform surface tension $\gamma$, choose gas pressure $p_b=p_c+2\gamma /a$; with $\gamma =0$, $p_b=p_c$. This proves compatibility of the spherical shape, not stability of arbitrary deformations or validity when inertia matters.

Use the flat-surface Boussinesq approximation, with small thermal expansion, constant properties, negligible surface deformation, and constant surface tension. The top is impermeable and shear-free, while the solid walls are no-slip. Temperature-dependent surface tension would introduce a separate Marangoni stress, which is not included in this buoyancy-driven model. The stated time scale requires $\kappa$ to mean thermal diffusivity; if the material property given is a dimensional thermal conductivity $k$, use $\kappa =k/({\rho }_0{c}_p)$.

Take $z$ positive downwards and define the nondimensional streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. Velocity is scaled by $\kappa /L$, streamfunction by $\kappa$, and temperature is measured from the cold-end surface value. Define the Prandtl number and horizontal-length Rayleigh number bywhere ${\alpha }_T$ is the thermal expansion coefficient. Put $J(\psi ,f)={\psi }_z{f}_x-{\psi }_x{f}_z$, and let ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the curl of the Boussinesq equations eliminates pressure and gives the full nonlinear vorticity-temperature systemThe buoyancy sign follows from the downward coordinate: warmer fluid has body acceleration $-g{\alpha }_T\Delta TT{\mathbf{e}}_z$.

Choose the same streamfunction constant, zero, on the connected boundary. The mechanical and thermal boundary conditions areThe upper stress-free boundary condition reduces to ${\psi }_{zz}=0$, since $\psi =0$ there already gives ${\psi }_{xx}=0$. An arbitrary additive constant in $T$ changes only the reference hydrostatic pressure.

In steady infinite-Prandtl-number convection, momentum inertia disappears but thermal advection remains:For a shallow lake, with fixed $\mathrm{Ra}$ as $ϵ\to 0$, the leading interior temperature is $T_0=x$. This is an outer solution: it does not satisfy the insulated sidewall derivative right at $x=0,1$. The leading global conductive solution, which resolves those end regions, isIt is harmonic, has $T_c(x,0)=x$, and satisfies the insulated sides and bottom. Away from sidewall layers of horizontal width $O(ϵ)$ it reduces to $x$ with exponentially small end corrections. These layers also turn the flow and enforce the sidewall no-slip boundary condition.

In the shallow interior the dominant momentum derivatives are vertical, so ${\psi }_{0,zzzz}=-\mathrm{Ra}$. Integrating with the surface and bottom conditions gives the shallow horizontal convection with an insulated bottom solution, with $y=z/ϵ$:Surface flow is toward the colder end, with a deeper return flow. The depth-integrated volume flux is zero because ${\psi }_0(ϵ)={\psi }_0(0)=0$. The bulk expressions apply away from the turning regions, not unchanged on the vertical walls. A useful small-advection condition in the end layers is $\mathrm{Ra}{ϵ}^4\ll 1$, automatic for fixed Rayleigh number in the stated thin limit.

The heat flux in these units is $\mathbf{q}=(uT-T_x,wT-T_z)$. For the leading bulk temperature, the horizontal transport through a vertical section per unit out-of-plane width isThus leading advective horizontal transport vanishes: both layers have the same temperature at a given $x$, and their opposing volume fluxes cancel. The leading conductive transport is from the hot end toward the cold end. In dimensional form it is $-{\rho }_0{c}_p\kappa \Delta Tϵ=-k\Delta Tϵ$ per unit width. Heat enters through the warmer portion of the maintained upper surface and leaves through its colder portion, with the conductive end adjustments particularly important near the two insulated sidewalls. It cannot enter through the insulated sides or bottom.

For the advection correction in shallow horizontal convection, set $T=x+\Theta (z)$ in the bulk. Since $w_0=0$ and $T_x=1$, the steady heat equation gives ${\Theta }_{zz}=u_0={\psi }_{0,z}$. The conditions $\Theta (0)=0$ and ${\Theta }_z(ϵ)=0$ imply ${\Theta }_z={\psi }_0$ and henceFor example the bottom bulk temperature is $T(x,ϵ)=x-\mathrm{Ra}{ϵ}^5/320$. The correction makes the returning deeper fluid slightly colder. Its horizontal advective heat transport isThus the transport including this bulk correction is $\boxed{{\mathcal{H}}_x=-ϵ-19{\mathrm{Ra}}^2{ϵ}^9/1451520}$, both terms directed toward the colder end.

Vertical transport must distinguish local flux from net transport. In the leading interior $w_0=T_{0,z}=0$, whereas the temperature correction gives a local downward conductive flux $q_z=-{\Theta }_z=-{\psi }_0>0$. The end circulation gives an upward advective transport: with impermeable sidewalls and $T_0\simeq x$, integration by parts gives ${\int }_0^{1}w{T}_{0}dx=-{\int }_0^{1}x{\psi }_{x}dx={\int }_0^1\psi dx\simeq {\psi }_0(z)<0$. It compensates that downward conduction. More generally, steady heat conservation and insulated sides give $d[{\int }_0^1{q}_{z}dx]/dz=0$, and the insulated impermeable bottom sets the constant to zero. ThereforeThere is no net vertical heat transport through the lake, despite its local recirculating heat flux and its nonzero horizontal heat transport.