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The conductive state is the no-flow equilibrium in which heat conduction alone maintains a linear temperature profile across a layer with fixed boundary temperatures. Its hydrostatic pressure follows by integrating $P_b^{\prime }=-g{\rho }_b$, with mass density obtained from the equation of state. Under the Boussinesq approximation, deviations from this state are the variables used in the linear Rayleigh-Bénard convection problem.

A convection roll is a circulating cell pattern whose ideal velocity and temperature fields are periodic across a chosen horizontal direction and independent of the coordinate along the convection roll axis. In a two-dimensional stream function representation, its velocity field is $(-{\psi }_z,0,{\psi }_x)$. Rayleigh-Bénard convection can select stationary or oscillatory convection rolls; a two-dimensional convection roll may be unstable to three-dimensional disturbances, a different convection roll orientation, or slow amplitude and phase modulations. Existence of a saturated Landau amplitude equation branch does not alone establish its stability.

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 ${\mathrm{Ra}}_c=27{\pi }^4/4$.

The Rayleigh number compares destabilizing thermal buoyancy with viscous and thermal diffusion:The conductive heat flux is $k\Delta T/h$, and the Nusselt number is the actual upward heat flux divided by this value. The four-thirds convective heat-transfer law is the empirical large-Rayleigh closure $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$. It givesThe layer depth cancels: thin thermal boundary layers control the resistance while convection mixes the interior. The dimensionless coefficient $C$ depends on the boundary and material-property regime; the two boundary-layer temperature drops can be absorbed into its convention. This is not a universal law for all Rayleigh-Bénard convection. Below convective onset the stable conductive solution instead has $\mathrm{Nu}=1$, and an impulsive initial conductive transient is not described by the steady large-Rayleigh closure.

Write $\beta$ for the thermal expansion coefficient, $\nu$ for kinematic viscosity, $\kappa$ for thermal diffusivity, and $k_T=\rho c_p\kappa$ for thermal conductivity. For a temperature difference $\Delta >0$ across depth $h$,The four-thirds convective heat-transfer law posits $\mathrm{Nu}=C{\mathrm{Ra}}^{1/3}$ at high Rayleigh number, and thereforeThe layer depth cancels. A useful physical justification is a mixed interior with thin conductive boundary layers of thickness $\delta$, each maintained near a fixed critical local Rayleigh number: $g\beta {\Delta }_b{\delta }^3/(\nu \kappa )\sim {\mathrm{Ra}}_{\mathrm{crit}}$ and $q\sim k_T{\Delta }_b/\delta$. Eliminating $\delta$ gives the exponent $4/3$. Whether $\Delta$ denotes the whole-layer contrast or the drop across one boundary layer changes the coefficient, not the exponent.

Assume Boussinesq approximation, constant material properties, a laterally extensive statistically turbulent Rayleigh-Bénard convection layer, negligible side-wall heat transfer, no rotation or imposed shear, approximately uniform bulk temperature, and the same specified Prandtl number regime when choosing $C$. The closure is empirical and regime dependent; high Rayleigh number alone does not make it exact or extend it to the impulsive conductive initial transient.

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.

Use the layer depth $d$, the thermal diffusion time $d^2/\kappa$, velocity $\kappa /d$, and the imposed temperature difference $\Delta T$ as scales. Let $\theta$ be the temperature departure from the conductive profile, $\mathbf{u}$ the velocity, $w=\mathbf{u}\cdot \hat{\mathbf{z}}$, and let pressure absorb hydrostatic terms. The dimensionless Boussinesq equations for Rayleigh-Bénard convection areThe Prandtl number and Rayleigh number are $\sigma =\nu /\kappa$ and $R=g{\alpha }_T\Delta T{d}^3/(\nu \kappa )$, with kinematic viscosity $\nu$, thermal diffusivity $\kappa$ and coefficient of thermal expansion ${\alpha }_T$. At $z=0,1$, the stress-free boundary condition and perfectly conducting thermal boundary condition give $w=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and $\theta =0$.

For the fundamental vertical normal mode, take $w=W\sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$ and $\theta =\vartheta \sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Here $k=|\mathbf{k}|>0$ is the horizontal wavenumber. With $s=k^2+{\pi }^2$, projection onto incompressible flow eliminates pressure and givesThus the stress-free convection growth-rate polynomial is $(\lambda +\sigma s)(\lambda +s)=\sigma R{k}^2/s$.

For the infinite-Prandtl number limit, the finite thermal growth rate isThe other, viscously damped root tends to $-\sigma s$. For $\sigma =1$, both roots are explicit:The plus root determines instability. A higher vertical harmonic replaces ${\pi }^2$ by $n^2{\pi }^2$; the fundamental is the most favorable one.

For the large-$R$ maximum in the infinite-Prandtl number limit, put $x=k^2$ and $p={\pi }^2$. Differentiating gives $d{\lambda }_{\infty }/dx=R(p-x)/(p+x{)}^3-1$. The unique maximum for sufficiently large $R$ therefore satisfies $R(p-x)=(p+x{)}^3$, whenceThis is fastest growth in infinite-Prandtl convection. The large-$R$ expansion is taken after the infinite-Prandtl number limit; at finite $\sigma$, neglecting inertia additionally requires $|\lambda |\ll \sigma s$. It selects the fastest growing mode far above onset, rather than the neutral-curve minimum $k^2={\pi }^2/2$.

DefineThe first equation and the bounded inverse of the Stokes operator show that $u\in L^{\infty }(0,T;D(A))$ and thatin $L^{\infty }(0,T;H)$. In fact, the strong convergence from part (i) and convergence of the spectral projections implyIn three dimensions $D(A)\subset H^2\hookrightarrow L^{\infty }$, so $u_m{\theta }_m\to u\theta$ in $L^1(0,T;L^2)$. Because $\nabla \cdot u_m=0$,and the nonlinear term consequently converges in distributions and in the required weak $L^2(0,T;H_{\mathrm{per}}^{-1})$ sense. The linear terms pass by weak convergence, while ${\Pi }_m$ tends strongly to the identity. Hencein $L^2(0,T;H_{\mathrm{per}}^{-1})$. Together with part (i), this proves existence of a global weak solution of the Rayleigh-Bénard convection system on every finite interval.