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The cell conservation equation equates the change of cell number density $n$ to minus the divergence of its total flux. The flux $n\mathbf{u}$ is advection by the water, $n{V}_s⟨\mathbf{p}⟩$ is the mean swimming flux, and $-\mathbf{D}\cdot \nabla n$ is the dispersal flux. The tensor $\mathbf{D}$ describes effective translational diffusion, whereas $D_r$ in part (i) describes rotational diffusion; they have different dimensions. This conservation law neglects birth, death and sources at the walls.

The equation $\nabla \cdot \mathbf{u}=0$ expresses incompressible flow. In the momentum equation, $\rho$ is the reference suspension density, $\rho [{\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}]$ is material acceleration times density, $-\nabla p_e$ is the pressure force after subtracting the pure-water hydrostatic pressure, and $\mu {\nabla }^2\mathbf{u}$ is the viscous force for constant dynamic viscosity $\mu$ in a Newtonian fluid. Since $n$ counts cells per volume, $g^{\prime }$ in $-g^{\prime }n{\mathbf{e}}_z$ is excess downward weight per cell; it is not an acceleration. The small cell-fluid density difference permits the Boussinesq approximation: retain that difference in buoyancy while using constant $\rho$ elsewhere. Diluteness permits neglect of cell-cell interactions, changes of viscosity, and active cell stresses. Any separate settling speed is assumed negligible or already incorporated in the specified mean drift. The light strength and swimming speed are uniform.

Consider the intended long-channel bulk approximation. At the sidewalls impose zero normal cell flux and the no-slip boundary condition. The closed ends require zero total fluid flux through each horizontal section, so ${\int }_{-a}^{a}wdx=0$; localized turning regions are omitted from the bulk equations. The mean concentration is ${\int }_{-a}^{a}ndx=2a{n}_0$. The final paragraph below distinguishes these bulk conditions from exact closed-end cell conservation.

For $\mathbf{u}=w(x){\mathbf{e}}_z$, the vorticity is $-w^{\prime }(x){\mathbf{e}}_y$. Hence $\epsilon =-w^{\prime }/(2D_r)$, and the leading horizontal mean swimming direction is $K_0{\mathbf{e}}_x$. With $\mathbf{D}=DI$ and $n=n(x)$, steady cell conservation equation says that $J_x=V_s{K}_{0}n-D{n}^{\prime }$ is constant. Its wall value is zero, so $n^{\prime }=\beta n$ with $\beta =V_s{K}_0/D$. The phototactic concentration in a vertical channel is thereforeThis result neglects the $O({\epsilon }^2)$ correction to horizontal swimming. For positive phototaxis, $\beta >0$ and cells accumulate towards the illuminated wall $x=a$.

The steady inertial terms vanish for $w=w(x)$, since $\mathbf{u}\cdot \nabla =w{\partial }_z$. The horizontal momentum equation gives ${\partial }_x{p}_e=0$. The vertical momentum equation then requires a constant excess pressure gradient $G={\partial }_z{p}_e$ and gives $\mu w^{\prime \prime }=G+g^{\prime }{n}_1{e}^{\beta x}$. Twice integrating producesSet $b=a\beta$. The two no-slip boundary conditions give $A=-C\sinh b/a$ and $B=-C\cosh b-G{a}^2/(2\mu )$. The zero fluid-flux condition follows by integrating this expression:Consequently the remaining constants areThe resulting velocity can also be written without separate polynomial constants asHere $G$ is the zero-flux pressure gradient in a cell-driven channel. It is negative for $b>0$: pressure decreases upwards, creating the upward pressure force that balances the excess cell weight and drives the return flow. Its magnitude is determined by zero net fluid transport, not by prescribing pressure independently. In the zero-light limit, the continuous result is $n=n_0$, $G=-g^{\prime }{n}_0$ and $w=0$.

For the requested large-$b$ sketch, cells occupy a boundary layer of thickness ${\beta }^{-1}$ near $x=a$. Put $\xi =x/a$ and $W=g^{\prime }{n}_0{a}^2/(\mu b)$. The exact dimensionless profile above isAway from the illuminated-wall layer it tends to $(1+\xi )(1-3\xi )/2$. There is broad upward return flow for $-1<\xi <1/3$ and downward flow on the illuminated side. The outer maximum is $2W/3$ at $\xi =-1/3$. Near $\xi =1$, the exponential term is approximately $2e^{-b(1-\xi )}$ and restores the no-slip boundary condition from the negative outer velocity to zero at the wall. The finite-$b$ profile shown below retains that layer. The small-vorticity assumption must still be enforced, for example by sufficiently weak excess weight or a sufficiently dilute suspension; $b\gg 1$ alone does not ensure $|w^{\prime }|/(2D_r)\ll 1$.

There is a genuine global qualification to the bulk construction. For an exactly steady channel with impermeable ends as well as impermeable sidewalls, the closed-channel cell-flux constraint also requires $\int J_{z}dx=0$ at every height. In this bulk ansatz the first-order orientation closure givessince integration by parts, $n^{\prime }=\beta n$ and $w(\pm a)=0$ imply $\int n{w}^{\prime }=-\beta \int nw$. Moreover, multiply $\mu w^{\prime \prime }=G+g^{\prime }n$ by $w$ and integrate. The wall values and zero fluid flux givefor a nontrivial profile. Thus zero axial cell flux generally fails unless the extra factor $1+V_s\beta K_1/(2D_r)$ vanishes, or the flow is trivial. A complete steady closed-channel solution generically needs axial concentration variation. The explicit profile solves the stated bulk equations, sidewall conditions, concentration normalization and zero fluid-flux condition; it is not, for arbitrary parameters, an exact globally closed steady cell distribution. Ignoring end regions gives the intended bulk approximation, but does not by itself prove the stronger global existence claim.

Equations and conventions. Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. With reduced pressure $\Pi =p/{\rho }_0+gz$, the nonrotating ideal Boussinesq equations areDensity departures are small compared with ${\rho }_0$, so they can be neglected in inertia and mass continuity but retained in the gravitational force. The reference stratification must also change density only weakly over the depth of interest. incompressible flow is already assumed; the Boussinesq approximation additionally restricts the size of the density variations, rather than following automatically from incompressible flow. An ideal fluid here has neither viscosity nor buoyancy diffusion.

A stationary reference profile $\bar{\sigma }(z)$ has buoyancy frequencyStable stratification has $N^2>0$: an upward-displaced parcel retains its old buoyancy and has negative buoyancy relative to its new surroundings.

Vorticity equation and linearization. For motion in the $(x,z)$ plane, choose the streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. The $y$-component of vorticity is then $u_z-w_x={\nabla }^2\psi$, where ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the $y$-component of the curl of momentum givesThe minus sign in the buoyancy curl follows from $(\nabla \times \sigma \hat{z}{)}_y=-{\sigma }_x$.

Write $\psi =\bar{\psi }(z)+{\psi }^{\prime }$ with ${\bar{\psi }}_z=\bar{u}(z)$ and $\sigma =\bar{\sigma }(z)+{\sigma }^{\prime }$. The linear material derivative is $D_t={\partial }_t+\bar{u}{\partial }_x$. The perturbation equations areThe shear-curvature term arises because $w^{\prime }$ advects the basic vorticity ${\bar{u}}_z$. Apply $D_t$ to the first equation and use the second; $D_t$ commutes with ${\partial }_x$ and does not differentiate the height-only coefficients. Therefore

For a stationary nonzero horizontal wavenumber $k$, the buoyancy equation gives $\hat{\sigma }=N^2\hat{\psi }/\bar{u}$. Substitution into the vorticity equation givesHere ${\ell }^2$ is the Scorer parameter. Where $m^2>0$, the vertical structure is oscillatory; where $m^2<0$, it is evanescent. A rigid flat lower boundary gives $\hat{\psi }(0)=0$ for $k\ne 0$. In an upper evanescent region the growing solution is excluded by boundedness. When propagation persists to infinity, an appropriate radiation condition distinguishes outgoing from incoming energy for a forced problem.

If $\bar{u}=U_0+Sz$ with $S>0$, its curvature term vanishes. Bounded $N^2$ implies ${\ell }^2=N^2/{\bar{u}}^2\to 0$ aloft. Thus every fixed nonzero $k$ eventually lies in an evanescent region. If stratification is sufficiently strong near the bottom, a lower oscillatory region is separated from the upper evanescent region by a turning level of a stationary internal wave. Waves reflect at that level and at the rigid lower boundary, giving vertically trapped modes for the horizontal wavenumbers that satisfy both conditions. In a slowly varying example the phase condition is approximately ${\int }_0^{z_t}mdz=(n-1/4)\pi$, from a lower node and the turning-point connection; the key point is trapping by the decrease in the Scorer parameter, not a physical lid.

Rigid channel modes. With constant $U=\bar{u}>0$ and $N$, the vertical nodes require $m_n=n\pi /H$, $n=1,2,\dots$. Stationarity then requiresThe specified strict interval admits positive $k_n^2$ for exactly $n=1,2$. Choosing $k_1,k_2>0$, every real superposition of these nonzero-wavenumber modes isThe two independent complex constants encode the amplitudes and horizontal phases; negative wavenumbers are supplied by complex conjugation, not additional independent constants.

Exact finite amplitude. Put $\lambda =N^2/U^2$. Both modes and every superposition satisfy ${\nabla }^2{\psi }^{\prime }=-\lambda {\psi }^{\prime }$. For the total streamfunction $\Psi =Uz+{\psi }^{\prime }$, the total buoyancy is $\sigma =N^{2}z+(N^2/U){\psi }^{\prime }=(N^2/U)\Psi$. A stationary velocity $({\Psi }_z,-{\Psi }_x)$ is tangent to its streamfunction contours, so $D\sigma /Dt=0$ exactly. Moreover the nonlinear self-advection of ${\nabla }^2{\psi }^{\prime }$ vanishes because it is proportional to ${\psi }^{\prime }$ itself. The remaining vorticity equation iswhich is an identity. Both rigid boundaries remain impermeable, and the curl-free residual in momentum determines a pressure. Thus the displayed two-mode family consists of exact finite-amplitude Boussinesq solutions, with no small-amplitude restriction on the constants in these equations. Very large amplitudes can overturn the stratification or leave the physical Boussinesq regime; exact existence does not establish stability.

The two-constant statement concerns wave disturbances with $k\ne 0$. If $k=0$ is literally included, arbitrary stationary horizontal shear adjustments ${\psi }^{\prime }=F(z)$ and height-only buoyancy changes are also possible. They are changes of the basic state and are not constrained by the wave equation obtained after division by $k$. Excluding that mean-flow sector is necessary for the intended count.

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.

Let ${\rho }_r$ be a constant reference density close to the ambient density, and let ${\rho }_p$ denote the mean plume density. The Boussinesq approximation permits full-cross-section kinematic fluxesThus $Q$ is the specific mass flux, equal to volume flux at this order; $M$ is the specific vertical momentum flux; and $F$ is the buoyancy flux. Physical mass and momentum fluxes are ${\rho }_{r}Q$ and ${\rho }_{r}M$. These definitions include the full circular area. If factors of $\pi$ are removed in a course's flux convention, divide $Q,M,F$ by $\pi$ throughout rather than mixing normalizations. The PDF's source density is ${\rho }_0$, distinct from ambient ${\rho }_a(z)$; the TeX changes the source symbol incorrectly.

For the top-hat plume model, let the plume radius be $r(z)$ and let $w,b$ be uniform inside it. ThenThe Batchelor entrainment hypothesis supplies the necessary closure: the inward edge velocity is $v_e={\alpha }_{e}w$, with constant entrainment coefficient ${\alpha }_e>0$. Entrainment across a cylindrical slice gives $Q^{\prime }=2\pi r{\alpha }_{e}w=2{\alpha }_e\sqrt{\pi }\sqrt{M}$. The entrained ambient has negligible upward momentum. The vertical buoyancy force on the slice is $\pi r^{2}b=F/w$, hence $M^{\prime }=FQ/M$.

For the buoyancy balance, let $R={\int }_A{\rho }_{p}wdA$ be the transported density scalar. Entrainment gives $R^{\prime }={\rho }_a{Q}^{\prime }$. Differentiate $F=g({\rho }_{a}Q-R)/{\rho }_r$ to get $F^{\prime }=g{\rho }_a^{\prime }Q/{\rho }_r$. Therefore the closed flux equations areHere $N$ is the buoyancy frequency when the ambient is stably stratified. These are the Boussinesq top-hat plume in a stratified ambient equations in the explicitly stated full-area convention. Stable stratification consumes positive buoyancy flux as the plume rises. The top-hat assumption alone would not determine $Q^{\prime }$ without an entrainment law.

For uniform ambient density, $F=F_0$ is constant. The source volume $Q_0$ becomes negligible when accumulated entrainment ${\int }_0^{z}c\sqrt{M(s)}ds$ is much greater than $Q_0$. In a jet-like region, $M\simeq M_0$, so the criterion is $z\gg Q_0/(c\sqrt{M_0})$. In a buoyancy-dominated far field with $Q\simeq C_Q{z}^{5/3}$, it is $z\gg (Q_0/C_Q{)}^{3/5}$, with $C_Q$ determined below. These are the two source-volume lengths of a turbulent plume. They are not the jet length, which measures the persistence of imposed source momentum. A nonzero source volume can also appear as a plume virtual origin; it is not valid to discard it at every distance simply because the source is geometrically small.

For $Q_0=M_0=0$, $F_0>0$, seek $Q=C_Q{z}^{5/3}$, $M=C_M{z}^{4/3}$. The two flux equations give $C_Q=(3c/5)\sqrt{C_M}$ and $C_M^{3/2}=9c{F}_0/20$. Thus the pure-plume solution isFor completeness its directly measurable top-hat fields areThis is an axisymmetric pure plume. In this regime the explicit source-volume criterion isFor $Q_0=F_0=0$, $M_0>0$, the buoyancy vanishes and momentum is constant. The pure-jet solution isThis is the jet limit of the top-hat plume model. Both ideal zero-volume sources have singular speeds at $z=0$; they model regions outside a finite source where entrainment has already made the source-volume correction small.

When $M_0$ and $F_0$ are both positive, use the jet solution near the source and the pure-plume solution far above it, with an appropriate virtual-origin correction. To quantify this transition, eliminate $z$:The conserved quantity is the plume flux-balance invariant. With negligible $Q_0$, comparison of its two terms and $Q\simeq c\sqrt{M_0}z$ gives a transition height of order $M_0^{3/4}/(\sqrt{c}\sqrt{F_0})$. The exact connection, including finite $Q_0$, is the finite-source forced-plume quadraturewith $M$ then recovered from the invariant. A positive integrand gives the physical increasing-volume branch. As $Q$ grows, the source constant is negligible relative to $Q^2$, proving attraction to pure plume similarity. One matches the two asymptotic regimes or uses this quadrature; the jet and plume solutions cannot be linearly superposed because their governing flux equations are nonlinear.

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Let ${\rho }_0$ be the reference mass density, $\nu$ the kinematic viscosity, $\kappa$ the thermal diffusivity, $\eta$ the magnetic diffusivity, and ${\alpha }_T$ the coefficient of thermal expansion. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$. For the Boussinesq approximation, density variation is retained only in buoyancy. Linearizing about this state and the imposed magnetic field $B_0\hat{\mathbf{y}}$ givesHere $w=u_z$ and $p_{\ast }=p^{\prime }+B_0{b}_y/{\mu }_0$ is the perturbation of magnetohydrodynamic total pressure. The gradient part of the Lorentz force has been absorbed into $p_{\ast }$; the remaining term is magnetic tension. The sign in the temperature equation follows from $d{T}_b/dz=-\Delta T/d$.

Use the thermal-diffusion scaling of a convection layer: length $d$, time $d^2/\kappa$, velocity $\kappa /d$, temperature perturbation $\Delta T$, field perturbation $B_0$, and total-pressure perturbation ${\rho }_0\nu \kappa /d^2$. Dropping dimensionless-variable decorations giveswithThese are respectively the Rayleigh number, Chandrasekhar number, Prandtl number, and magnetic-to-thermal diffusivity ratio. In particular, the magnetic coefficient is $\zeta Q$, not $Q$ alone, in these thermal-time units. At $z=0,1$, the boundary conditions are $\theta =w=b_z=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and ${\partial }_z{b}_x={\partial }_z{b}_y=0$.

Let ${\rho }_0$ be a constant reference density and define total buoyancy acceleration by $\sigma =-g(\rho -{\rho }_0)/{\rho }_0$. Absorb the reference hydrostatic pressure into a kinematic pressure $p$. The ideal nonrotating Boussinesq equations areFor a resting density profile ${\rho }_b(z)$, the background satisfies $p_b^{\prime }(z)={\sigma }_b(z)$ and the squared buoyancy frequency isStable stratification has $N^2>0$. The Boussinesq approximation requires fractional density variations to be small over the domain and parcel excursions, so density can be replaced by ${\rho }_0$ in inertia while its small variation is retained in the gravitational force. For an approximately uniform density gradient over height $H$, this requires $N^{2}H/g\ll 1$. The stipulated incompressibility alone does not justify neglecting density variation in inertia.

Put $b=\sigma -{\sigma }_b$ and $p^{\prime }=p-p_b$. To linear order the two-dimensional internal gravity wave equations areThe exact impermeability condition on $z=\eta (x,t)$ is the kinematic boundary conditionTaylor expansion about $z=0$ permits $w(x,0,t)={\eta }_t$ at leading order if the boundary slope and its displacement relative to the vertical wave scale are small: $|kϵ|\ll 1$ and $|mϵ|\ll 1$. The second condition controls both the omitted evaluation correction $\eta w_z$ and the product $u{\eta }_x$, since continuity gives $u\sim (m/k)w$. Hence for the harmonic forcing,The propagating calculation assumes $k\ne 0$ and $0<|\omega |<N$; a time-independent boundary does no oscillatory work on fluid initially at rest. If $k=0$ but $\omega \ne 0$, incompressibility forces the vertical velocity to be independent of height, so the unbounded column has no finite-energy radiating internal-wave solution. The propagating wave-generation problem therefore needs the stated nonzero horizontal wavenumber.

For a plane mode with phase $\theta =kx+mz-\omega t$, let $U,W,P,B$ be its complex amplitudes. The linear equations giveSubstituting into $-i\omega W=-imP+B$ gives the dispersion relationNo waves enter from above. The appropriate radiation condition therefore selects upward vertical group velocity, not upward phase propagation:The boundary-forced internal gravity wave satisfying the boundary condition isThe displacement is $ϵ\sin \theta$, consistent with $b=-N^2$ times the displacement. For a right-moving boundary choose $k>0$, $\omega >0$ without loss of generality. Then $m<0$, and the constant-phase lines of an internal gravity wave have slope $dz/dx=-k/m>0$: the crests rise toward the right. Their acute angle with the horizontal isThe group velocity is along these crests in the upward-right direction, whereas the wavevector and phase propagation point downward-right.

The physical pressure perturbation is ${\rho }_0{p}^{\prime }$. Pressure work at the lower boundary, to second order in amplitude, is its period-mean pressure times vertical boundary velocity. Static hydrostatic terms average to zero over a cycle. ThereforeAs an independent energy check, the wave's mean kinetic-plus-buoyancy energy density is ${\rho }_0{ϵ}^2{N}^2/2$, and multiplication by $c_{gz}$ gives exactly this upward energy flux.

For the exact horizontal mean-momentum equation, incompressibility writes the horizontal momentum equation asAverage over a horizontal period. Require the pressure and velocity fluxes to be periodic, or more generally their horizontal derivative averages to vanish; in particular there must be no imposed mean pressure gradient. ThenHorizontal averaging also makes ${\partial }_z\overline{w}=0$; the zero mean boundary displacement rate gives $\overline{w}=0$. These conditions exclude an independent mean mass flux or horizontal force. The mean equation applies in the fluid, with flattening of the small moving boundary understood at the order used above.