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Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

The field $\varphi$ is a compositional order parameter in a binary fluid mixture. Its spatial integral is conserved in a closed sample, so advection and diffusion enter a continuity equation; the material derivative follows the incompressible fluid. The diffusive current $\mathbf{J}=-M\nabla \mu$, with positive order-parameter mobility $M$, moves material down the chemical potential gradient. The chemical potential of a composition field is the functional derivative of the symmetric quartic Landau free energyso $\mu =a\varphi +b{\varphi }^3-\kappa {\nabla }^2\varphi$. Positive $b,\kappa$ stabilize large amplitudes and penalize sharp interfaces; $a<0$ permits phase separation.

The momentum equation is the Navier-Stokes equation with dynamic viscosity $\eta$, constant mass density $\rho$, and the Korteweg force $-\varphi \nabla \mu$ arising from composition stresses. The pressure enforces incompressible flow, $\nabla \cdot \mathbf{v}=0$, and can absorb pure gradient contributions to that thermodynamic force. For example, $-\varphi \nabla \mu =\mu \nabla \varphi -\nabla (\varphi \mu )$ gives an equivalent force after redefining the pressure. These equations constitute deterministic Model H dynamics.

Their dissipative signs can be checked by the energy dissipation identity for Model H. Under periodic or no-flux, no-work boundary conditions, composition advection and capillary work cancel between $F$ and the kinetic energy, givingThus diffusion and viscosity dissipate energy, while the reversible advective coupling transfers it between composition and motion. Thermal noises are omitted at this deterministic hydrodynamic level.

Write $h(t)$ for the height of the filling-box first front above the floor and $P(h)$ for the upward plume volume flux crossing it. Neglect the plume's horizontal area compared with $A$, use the Boussinesq approximation, and measure buoyancy relative to the initially unmodified ambient. The first front separates fluid already modified by plume discharge from the unmodified fluid below; the entire upper region need not be uniformly mixed during an unventilated filling box model transient.

A finite injected volume flux $Q$ cannot enter a hermetically sealed, fixed-volume room under the incompressible flow assumption. If $Q>0$, an equal overflow or exhaust is needed. Taking that exhaust at the ceiling, volume conservation of the region above the front givesThe source volume flux is subtracted because only entrained room fluid is removed from the initially unmodified region. The printed pure plume law has $P(0)=0$, so it cannot be an exact finite-$Q$ source law down to the physical floor. In the far field where $Q\ll P(h)$, or for a heat source whose injected volume flux is negligible, it givesHere $t=0$ is when the plume discharge first reaches the ceiling. Retaining $Q$ while using the far-field law gives $-\dot{h}=[\lambda B^{1/3}{h}^{5/3}-Q]/A$ only while that law remains a valid approximation; it must not be extrapolated to predict negative plume entrainment. A plume virtual origin below the floor is one way to match a nonzero source volume flux.

For displacement ventilation, let $F$ be the total ceiling exhaust, $S$ the total buoyant-source volume flux, and $L=F-S$ the separate ambient inflow into the lower layer. Volume conservation and the upper-layer buoyancy flux balance giveThe ambient inflow has zero reference buoyancy. In the ideal two-layer closure the upper layer is uniformly mixed with reduced gravity $b>0$, and the lower layer has zero reference buoyancy. The plume supplies the upper layer while lighter upper-layer air is removed at the ceiling. An interface is driven downward when $P(h)>F$ and upward when $P(h)<F$; since $P^{\prime }(h)>0$, its equilibrium is stable.

In the negligible-source-volume pure plume limit, identifying $F=Q_v$ gives the intended one-source answerThe prescribed inequality makes $h_1/H\ll 1$, so both layers fit in the room. It does not, by itself, justify neglecting a finite source $Q$: the pure plume source region must also be small compared with $h_1$, normally requiring $Q\ll Q_v$. If $Q_v$ denotes additional background ventilation rather than total exhaust, replace $Q_v$ in both formulas by $Q+Q_v$. If it is total exhaust, the separate ambient inlet carries $Q_v-Q$. These conventions must not be mixed.

For $n$ independent equal axisymmetric pure plumes, adding their volume fluxes givesThus the conventional multiple-plume displacement ventilation calculation with negligible source volume flux and separate ambient ventilation givesSplitting the buoyancy flux increases total fluid entrainment, so the same ventilation balances the plumes at a lower interface. The global buoyancy flux is unchanged, which leaves the upper-layer reduced gravity unchanged. Independence requires sufficiently separated sources and plumes that do not merge below the interface.

There is, however, a substantive problem with the literal source-volume specification in this paragraph of the PDF. It puts all $Q_v$ through the buoyant sources, so $S=F=Q_v$ and $L=0$. For every positive interface height, entraining plumes haveThere is no fresh ambient inflow to replace the lower-layer fluid being entrained. Consequently the stated positive steady two-layer configuration does not follow from those literal hypotheses. This is no steady displacement layer without ambient supply, not a small correction to the $n^{-2/5}$ scaling. For example, a matched plume virtual origin givesWhen $S=F$, this yields $h_{\ast }=0$. The unshifted answer $h_n$ equals the omitted $z_0$ in this case, so neglecting the source region cannot be justified. Also each source has reference buoyancy $(B/n)/(Q_v/n)=B/Q_v$, already equal to the proposed upper-layer value; a steady positive plume cannot entrain zero-buoyancy lower fluid and still deliver that same value. The conventional answer is valid for a corrected arrangement with negligible-volume buoyancy sources and a separate ambient inlet, or with source flux $S<Q_v$ and the remaining $Q_v-S$ supplied directly to the lower layer. A real room may have other circulation or continuous stratification, but that requires a different model.

For the intended displacement ventilation model, reducing the exhaust to $F=Q_v/2$ makes the new equilibriumThe interface initially descends since its old plume volume flux exceeds the new exhaust. In ventilated filling-box relaxation, writing $V_{\ast }=A(H-h_{\ast })$ and taking the linearization of the two balances above givesAs $P_n^{\prime }(h_{\ast })=5F/(3h_{\ast })$, the two decay times areA dimensional analysis estimate of the full adjustment time is thereforeThe interface adjusts on the shorter lower-layer scale; replacing the buoyancy stored throughout the upper layer is slower. Exact equilibrium is approached asymptotically, and a specified small relative tolerance adds a logarithmic factor. For the literal all-inflow-through-sources arrangement there is no positive two-layer equilibrium to return to; $2AH/Q_v$ is only a flushing-scale estimate for a different, whole-room adjustment.

Expanding the kinetic helicity conservation law separates advection from the other flux:Thus material conservation of kinetic helicity density holds precisely whenAn especially useful sufficient pair of conditions is incompressible flow, $\nabla \cdot \mathbf{u}=0$, and constancy of $w-u^2/2$ along integral curves of the vorticity, $\mathbf{\omega }\cdot \nabla (w-u^2/2)=0$. Together these conditions make both terms vanish without cancellation. They are not necessary individually, because the two terms in the displayed balance can cancel. An irrotational flow is a trivial conserved case with $q=0$.

For contrast, combining the balance with mass conservation givesConsequently constancy of $w-u^2/2$ along integral curves of the vorticity makes $q/\rho$ an advected scalar even in compressible flow; it does not generally make $q$ itself constant.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.

The velocity field is incompressible becauseWith the convention $u_x={\psi }_y$ and $u_y=-{\psi }_x$, a stream function isThe vorticity is vertical:For a steady inviscid incompressible flow, the vorticity equation is $(\mathbf{u}\cdot \nabla )\mathbf{\omega }=(\mathbf{\omega }\cdot \nabla )\mathbf{u}$. Here the right side is zero because the flow is independent of $z$, while the left side is $2(\mathbf{u}\cdot \nabla \psi ){\mathbf{e}}_z=0$ because velocity is tangent to the level sets of $\psi$. Thus the equation is satisfied.

The streamlines are the level curves $\sin x\sin y=\text{constant}$. The lines $x=\pi$ and $y=\pi$, together with the boundary, are separatrices; each of the four cells contains closed nested curves around a centre at $(\pi /2,\pi /2)$, $(3\pi /2,\pi /2)$, $(\pi /2,3\pi /2)$, or $(3\pi /2,3\pi /2)$.

For a velocity potential, $\mathbf{u}=\nabla \varphi$. The incompressible flow condition $\nabla \cdot \mathbf{u}=0$ therefore becomes Laplace equation ${\nabla }^2\varphi =0$. The Laplace equation in polar coordinates givesso positivity of $\gamma$ and $\lambda$ requiresThe velocity components areThe no-penetration boundary condition on both walls says $u_{\theta }=0$ at $\theta =0$ and $\theta =\alpha$, hence $\lambda \alpha =n\pi$. If $n\ge 2$, $\sin (n\pi \theta /\alpha )$ changes sign inside the wedge. The stated sign condition therefore selects $n=1$, and

Choose $x$ tangential to the solid boundary and locally along the outer inviscid streamlines, and let $y$ be the inward normal coordinate. Write $u$ and $v$ for the corresponding tangential and normal velocity field components, $U(x)$ for the outer tangential velocity, and $\nu$ for the kinematic viscosity.

For a layer of streamwise scale $L$ and thickness $\delta \ll L$, incompressible flow givesso $u=O(U)$ and $v=O(U\delta /L)$. The normal Navier-Stokes equation then gives $p_y=0$ to leading order, while the outer Euler equations giveIn the tangential Navier-Stokes equation, streamwise viscous diffusion is smaller than normal diffusion by $O((\delta /L{)}^2)$. Retaining the leading inertial and normal-diffusion terms produces the Prandtl boundary-layer equationIt is supplemented by $u=v=0$ at a stationary wall and $u\to U(x)$ on matching to the outer flow.

Writing $R=x^2+y^2$, direct differentiation givesso $\nabla \cdot \mathbf{u}=0$. Likewise,so the two-dimensional vorticity ${\partial }_{x}v-{\partial }_{y}u$ vanishes. The flow is therefore incompressible and irrotational away from the origin.

In spherical symmetry, the velocity is radial and constant over each sphere. HenceApply the divergence theorem to the fluid shell between radii $R_1$ and $R_2$. Since the flow is incompressible,Thus $Q$ is independent of $R$ and is a function of time alone. This is the flux law for spherically symmetric incompressible radial flow.

With ${\mathbf{u}}_0=0$ and ${\theta }_0=-z$, the incompressibility condition and temperature equation hold because ${\nabla }^{2}z=0$. The momentum equation is satisfied by the hydrostatic pressurebecause $\nabla p_0=\sigma Ra{\theta }_0\hat{\mathbf{z}}$. Thus this is the conductive basic state.

Let the unperturbed interface rise at pore velocity $U/\varphi$, and write its displacement asIn each fluid, Darcy's law and incompressible flow implyThe decaying pressure perturbations are proportional to $e^{-\alpha |z|}$. Continuity of normal velocity, the kinematic boundary condition $\varphi {\eta }_t=u_n^{\prime }$, and continuity of pressure giveThus a less mobile displaced fluid, ${\mu }_2>{\mu }_1$, and a denser fluid above a lighter one both drive the Saffman–Taylor instability.

For immiscible fluids, the Young–Laplace equation adds the pressure jump $-\gamma {\eta }_{xx}$. The dispersion relation becomeswhereIf $A>0$, the unstable band is $0<\alpha <\sqrt{A/C}$ and differentiation gives

Define the signed characteristic buoyancy velocitySince ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$ and $Ca={\mu }_{2}U/\gamma$,This expression applies when the quantity in the final parentheses is positive.

Every growth curve starts at the origin. For equal densities its initial slope is proportional to $M$; for ${\rho }_2>{\rho }_1$ buoyancy shifts that slope upward, while for ${\rho }_2<{\rho }_1$ it shifts it downward. When $A>0$, the curve rises to one positive maximum and then crosses zero before its stabilizing $-{\alpha }^3$ capillary tail. When $A\le 0$, every nonzero wavenumber decays.

For a prescribed nonzero wavenumber, neutral stability requiresorIn the quasistatic limit $U\to 0$, viscosity contrast disappears and this reduces to the capillary Rayleigh-Taylor instability threshold ${\rho }_2-{\rho }_1=\gamma {\alpha }^2/g$. Without surface tension, neutral stability in that limit simply requires equal densities.

The definitions through the Stokes stream function giveso incompressibility is automatic.

Balancing inertia $A^2/r^5$ with viscosity $\nu U/{\delta }^2$ givesThe streamfunction scale is $rU\delta \sim (A\nu r{)}^{1/2}$, motivating

The divergence isThus the flow is incompressible.

Axisymmetry and incompressibility giveThus $r{u}_r=C$. Finiteness at the origin forces $C=0$, so

Choose the polar axis along $\mathbf{d}$ and write $d=|\mathbf{d}|$. Sincethe velocity components areTheir divergence isThus the flow is incompressible for $r\ne 0$.

Let $u=ps{i}_z$ and $w=-ps{i}_x$, so mass conservation for the two-dimensional incompressible flow is automatic. Write $b$ for buoyancy and introduce the diffusive operatorsThe Linearized Boussinesq equations areTaking the curl of the momentum conservation equations eliminates the pressure and givesApply $D_{\kappa }$ and use $D_{\kappa }b=N^2{\psi }_x$. Since the constant-coefficient linear partial differential operators commute,This is the viscous-diffusive internal gravity wave equation.

A radially symmetric velocity potential has $u_r={\partial }_r\varphi$. The incompressible flow condition in two-dimensional polar coordinates isso $r{u}_r=A(t)$. The kinematic boundary condition at the bubble surface gives $u_r(a,t)=\dot{a}$, henceAt the outer interface, $\dot{b}=u_r(b,t)=a\dot{a}/b$. ThereforeandThe area $\pi (b^2-a^2)$ of the incompressible liquid is conserved. These formulas are the radially symmetric incompressible flow in a planar annulus.