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Following a parcel. Let $\mathbf{X}(t)$ be a fluid pathline, with $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The multivariable chain rule gives the material derivative,The first term measures change at fixed position; the advection term accounts for the parcel moving through spatial gradients. The derivative is also called the Lagrangian derivative.

The forces on a parcel. A material volume $dV$ has mass $\rho dV$. The pressure forces on its boundary sum to $-\nabla pdV$, while the gravitational acceleration is $-\nabla \psi$. Newton's second law therefore gives the Euler equations for an inviscid fluid,This neglects viscosity and other body forces. The left side is the parcel's acceleration, not just the time derivative at a fixed spatial location.

Bernoulli conservation. The claimed streamline result needs steady flow and a barotropic fluid, $p=p(\rho )$, or an equivalent enthalpy relation along each isentropic streamline. These hypotheses are necessary; the unsteady Euler equation alone does not imply the stated conservation law. Define the barotropic enthalpy functionDot the steady Euler equation with $\mathbf{u}$, using $\mathbf{u}\cdot [(\mathbf{u}\cdot \nabla )\mathbf{u}]=\mathbf{u}\cdot \nabla (u^2/2)$. This givesThus the Bernoulli function isEvery term has units of energy per mass, ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Different streamlines can have different constants.

The spherical accretion equation. Write the inward speed as $u(r)>0$, so the velocity vector is $-u(r)\hat{\mathbf{r}}$ and the positive mass accretion rate isAssume a prescribed central potential $\psi =-GM/r$ and negligible gas self-gravity. The steady radial Euler equation isFor a barotropic equation of state, $dp/dr=c_s^{2}d\rho /dr$, with $c_s^2=dp/d\rho$. Eliminate the density derivative to obtainA smooth sonic point requires the right side to vanish where $u=c_s$, henceFor a nonisothermal gas this is an implicit relation involving the local sound speed. In the isothermal case $c_s$ is constant and this is the Bondi sonic point.

The isothermal integral. For $p=c_s^2\rho$, choose the enthalpy reference so that $h_b=c_s^2\ln (\rho /{\rho }_{\infty })$. The reservoir is at rest at infinity and $\psi (\infty )=0$, making its Bernoulli constant zero. ThusorThe logarithm has a dimensionless argument. Constant temperature during compression generally requires heat removal, so this is Isothermal Bondi accretion, rather than adiabatic accretion with $\gamma >1$.

The critical accretion rate. Select the smooth reservoir-connected transonic branch. Evaluate the last relation at $u_s=c_s$ and $GM/r_s=2c_s^2$:Mass conservation then gives the isothermal Bondi accretion rate,The units are $\mathrm{kg}{\mathrm{s}}^{-1}$. Equivalently $\dot{M}=4\pi \lambda G^2{M}^2{\rho }_{\infty }/c_s^3$, with $\lambda =e^{3/2}/4\simeq 1.12$.

The inner-boundary selection of isothermal Bondi flow is important: a standard absorbing accretor inside the sonic radius selects this critical solution. The steady equations and ${\rho }_{\infty }$ alone also admit subcritical subsonic branches supported by a different inner boundary; they do not by themselves fix a unique rate. The usual answer presumes transonic capture.

Following a fluid element. Let $\mathbf{X}(t)$ be a pathline, satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The material derivative, also called the Lagrangian time derivative, is the rate observed while following that element. The multivariable chain rule givesThe partial derivative holds position fixed; the advection term accounts for motion through a spatially varying field. For a vector field, the same formula applies component by component in fixed Cartesian axes.

Momentum balance. For an inviscid flow, pressure is the only surface stress. A small material volume $dV$ has net pressure force $-\nabla pdV$, obtained by the divergence theorem. Its mass is $\rho dV$ and its material acceleration is $D\mathbf{u}/Dt$. Applying Newton's second law gives the Euler equations for an inviscid fluid,This is the local momentum conservation equation when viscosity and physical body forces are absent. A prescribed body force per unit mass would add $\rho \mathbf{f}$ on the right.

Changing to rotating coordinates. In a rotating frame with constant angular velocity $\mathbf{\Omega }$, the lab acceleration is the rotating-frame material acceleration plus $2\mathbf{\Omega }\times \mathbf{u}+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. ThereforeFor uniform rotation,With constant ${\rho }_0$, this gives the Euler equation in a uniformly rotating frame,The term subtracted from $p$ is the equilibrium centrifugal contribution. Dividing the bracket by ${\rho }_0$ defines a reduced pressure $\Pi =p/{\rho }_0-|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. The Coriolis acceleration remains a velocity-dependent term.

Linear disturbances. Linearize about rest in the rotating coordinates, with the equilibrium part of $\Pi$ removed. The advective acceleration is quadratic in disturbance amplitude, so the linearized Euler equations and incompressible flow condition areTaking the curl removes the pressure gradient. The identitygives the inertial-wave vorticity equation,

Plane-wave frequencies. Let $\mathbf{k}\ne 0$ and use the given plane wave convention. Incompressible flow requires $\mathbf{k}\cdot {\mathbf{u}}_0=0$, and the vorticity equation givesFor a nonzero frequency, apply $\mathbf{k}\times$ again, using $\mathbf{k}\times (\mathbf{k}\times {\mathbf{u}}_0)=-k^2{\mathbf{u}}_0$. It follows thatThis is the inertial-wave dispersion relation. If $\theta$ is the angle of the wavevector to the rotation axis, the positive angular frequency is $2|\Omega \cos \theta |$. It is largest, $2|\Omega |$, for a parallel wavevector, decreases to zero for a perpendicular one, and is independent of the magnitude of the wavenumber at fixed direction. The two signs describe the two oscillatory polarization branches. When $k_z=0$, the zero-frequency result is obtained directly from the linear equations rather than by dividing by $\omega$; it describes the nonoscillating balanced limit.

The observed value is about 4 ppb higher than the local chemical–dry deposition balance predicts. Entrainment from the ozone-richer free troposphere adds a term $(w_e/H)({\chi }_{\mathrm{FT}}-{\chi }_{\mathrm{BL}})$; horizontal Advection can likewise import ozone or longer-lived precursors from other regions. Time dependence, including the day–night cycle of photodissociation and changes in mixing depth, is also omitted. The approximately four-day model adjustment time means that air arriving at the measurement location need not have attained the assumed local steady state. These transport and history terms can raise ozone without changing the given local chemical constants.

Take $u>0$. Despite the source's terminology, $P(x)$ is a dimensionless conditional bound-state probability at phase $x$, not a spatially normalized probability density; the phase-average bound fraction is ${\Delta }^{-1}{\int }_0^{\Delta }Pdx$. The given equation has advection and binding reactions but no diffusion term. Large $N$ permits the prescribed steady $u$ approximation.

For spatially gated myosin attachment, define $A=e^{-k_1{x}_1/u}$, $B=e^{-k_2(\Delta -x_2)/u}$, $p_0=P(0)$ and $p_b=P(x_1)$. Continuity and periodicity give $p_b=1-(1-p_0)A$, $p_0=B{p}_b$, hence $p_b=(1-A)/(1-AB)$. The full periodic steady solution isThe requested limit makes $B\ll 1$, so $p_0\simeq 0$ and $p_b\simeq 1-e^{-k_1{x}_1/u}$. The curve rises exponentially over the binding window, remains flat to $x_2$, then decays over length $u/k_2$ to an exponentially small value before the next period. The piecewise expression is continuous at both switching points and at the period boundary.

Taking the imaginary part of the complex potential gives the streamfunctionThe corresponding velocity field isPut $\zeta =z{e}^{i\alpha /2}=\xi +i\eta$. Then $w=\beta {\zeta }^2$ and $\psi =2\beta \xi \eta$. Thus the streamlines are the previous rectangular hyperbolas rotated clockwise by $\alpha /2$. Their separatrices have angles $\theta =-\alpha /2+n\pi /2$; the extension axis is $\theta =-\alpha /2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.

For pure advection, a marked fluid element obeysConsequently the dye circle becomes an ellipse with semiaxes $R{e}^{2\beta t}$ and $R{e}^{-2\beta t}$. Its area isStretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map Jacobian determinant is one, as required by incompressibility. Negligible diffusion is needed to regard the sharp dye boundary as a material curve.

Let ${\mathbf{U}}_h=\Gamma z{\mathbf{u}}_h(x,y)$ denote the actual horizontal velocity field. With constant nonzero Coriolis parameter $f_0$, geostrophic balance and hydrostatic balance, using pressure per reference density $\pi$, giveDifferentiate the first equation in $z$ to obtain thermal-wind balance:Taking the dot product with ${\mathbf{U}}_h$ provesThus the horizontal velocity is tangent to the horizontal buoyancy contours at every height. For a fluid whose thermal equation of state makes horizontal buoyancy contours coincide with horizontal temperature contours, there is no horizontal advection across those contours. In a dry terrestrial atmosphere, use potential temperature as the materially conserved thermal variable. A salinity contribution to buoyancy would require separate treatment before interpreting the result as a statement about temperature.

The usual statement about net meridional heat transport also follows explicitly. Introduce a horizontal streamfunction $\Psi$ such that ${\mathbf{u}}_h=(-{\Psi }_y,{\Psi }_x)$. Integration of thermal-wind balance givesand henceFor a zonally periodic function $\Psi$, or vanishing endpoint contributions, the zonal integral is zero. Likewise, the anomalous eddy buoyancy flux isThe required correlation between meridional velocity and thermal anomaly is absent. A nonzero vertical phase tilt, as in the next part, permits that correlation.

Qualification of the wording. The result does not say that the local heat flux vector $b{\mathbf{U}}_h$ vanishes, or that an arbitrary open-domain integral vanishes without boundary conditions. For a concrete counterexample to that literal reading, take ${\mathbf{u}}_h=(0,1)$ and $b=f_0\Gamma x+N^{2}z$. These satisfy the displayed thermal-wind balance, but $vb=\Gamma z(f_0\Gamma x+N^{2}z)$ is generally nonzero. What the hypotheses alone guarantee is zero transport across horizontal buoyancy contours; the vanishing zonally integrated heat transport uses the stated endpoint condition.

The quasi-geostrophic potential-vorticity equation is the leading balanced form of potential-vorticity conservation for an inviscid rotating fluid with stable density stratification, small Rossby number and an adiabatic process for each fluid element. Its parent conservation law is Ertel potential vorticity conservation. Horizontal geostrophic balance, vertical hydrostatic balance, and the beta-plane approximation reduce it to advection of one scalar by the leading horizontal geostrophic flow.

One can see its origin by combining the vertical vorticity equation with the thermal equation. Set $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_h$, ${\mathbf{u}}_g=(-{\psi }_y,{\psi }_x)$, and ${\zeta }_g={\nabla }_h^2\psi$. The quasi-geostrophic approximation givesHere $b$ is the buoyancy perturbation relative to the reference stable density stratification, $N=N(z)$ its buoyancy frequency, and $w$ the small vertical velocity. The stretching of absolute vorticity in the first equation is offset by the change of stratification represented by the second. More precisely,The last dot product is zero because thermal-wind balance makes ${\partial }_z{\mathbf{u}}_g$ perpendicular to ${\nabla }_{h}b$. Adding therefore yieldsThe constant $f_0$ has no effect on the tendency, but its inclusion makes the connection with absolute vorticity clear. Vertical advection of $q$ is beyond the retained order; the vertical velocity has already supplied the stretching and thermodynamic terms.

For a simple balanced weather forecast, the essential cycle is diagnose, advect, invert:

This is an elliptic boundary value problem, not a local algebraic diagnosis. Its solution supplies the next geostrophic flow and, through $b=f_0{\psi }_z$, the next thermal field. Repeat the cycle to move the balanced weather systems forward.

Interior potential vorticity alone is insufficient: boundary thermal anomalies can induce flow even with zero interior anomaly, as in the Eady model. The method filters fast inertia-gravity waves and is useful for large-scale extratropical weather, while tropical dynamics, convection, moisture, friction and diabatic heating require additional physics. If vertical motion is wanted, the quasi-geostrophic omega equation supplies a further diagnostic calculation.

On a beta plane, $f=f_0+\beta y$, the preceding vorticity equation isIn the broad ocean interior, assume a statistically steady flow, weak advection of relative vorticity, and negligible friction compared with planetary-vorticity advection. The constant-depth, constant-density assumptions of 2(a) exclude topographic and baroclinic vorticity generation contributions. More precisely, require $|{\xi }_t|,|\mathbf{u}\cdot \nabla \xi |,|r\xi |\ll |\beta v|$. Then Sverdrup balance isSmall Rossby number and slow forcing help justify these conditions in the interior; they do not justify them inside an intense western boundary current. For fluctuations, the mean eddy contribution to relative-vorticity advection must also be negligible.

The depth-integrated northward volume flux per unit zonal distance is $V=Hv$. For a basin $0<x<L$ and wind stress independent of $x$, the northward interior volume flux isA negative value denotes southward transport. The cancellation of $H$ is why this interior volume flux is insensitive to the chosen constant depth. In a closed, steady basin the net volume flux across a full latitude section is zero: a boundary return must compensate $T$. Thus the displayed formula cannot represent the total closed-basin transport including that return.

For a concrete circulation sketch, use a northern-hemisphere subtropical basin $0<y<B$ andThis represents westward trade winds at the southern edge and eastward westerlies at the northern edge. The negative wind stress curl drives southward interior flow. Define an ocean transport streamfunction by $Hu=-{\Psi }_y$, $Hv={\Psi }_x$ and impose no normal flow at the eastern wall through $\Psi (L,y)=0$. Integrating Sverdrup balance westward givesSince $V=0$ at $y=0,B$, this also gives no normal flow at the north and south walls. Interior water moves westward in the southern part, southward across the basin, and eastward in the northern part. A northward western boundary current completes the clockwise ocean gyre.

Pure Sverdrup balance cannot satisfy the western impermeability condition as well: ${\Psi }_I(0,y)=-LV(y)$ varies along that wall. Indeed imposing no normal flow on both east and west walls would require their constant streamfunction values to agree in this simply connected closed basin, whereas their difference here is the nonzero $T(y)$. The shaded return in the sketch represents a boundary region where the interior approximation breaks down. It is supplied by friction in 2(c), rather than inferred from Sverdrup balance alone. A southern-hemisphere subtropical ocean gyre has the reflected, anticlockwise circulation.

For an order-of-magnitude estimate, use ${\tau }_0=0.1\mathrm{N}{\mathrm{m}}^{-2}$, ${\rho }_0=10^3\mathrm{kg}{\mathrm{m}}^{-3}$, $\beta =2\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$, $L=5\times 10^6\mathrm{m}$ and a wind stress variation scale ${\ell }_{\tau }=10^6\mathrm{m}$. ThenHere $1\mathrm{Sv}=10^6{\mathrm{m}}^3{\mathrm{s}}^{-1}$ is a Sverdrup. The sinusoidal example has maximum $|{\tau }_{s,y}|=\pi {\tau }_0/B$; taking $B=\pi \times 10^6\mathrm{m}$ makes it agree with the chosen variation scale. Other reasonable basin widths and wind profiles change the coefficient, but retain a transport of order tens of Sverdrups.

Let $V_b$ be the northward speed in the western boundary current, whose zonal width $\delta$ is much less than its meridional scale. Its dominant relative vorticity is $\xi \simeq v_x\sim V_b/\delta$. Neglecting local wind stress curl and relative-vorticity advection in the narrow return region, the linear stationary vorticity equation becomesConsequentlyThis decay into the interior from the western wall explains why the Stommel boundary layer lies on the west. The sign also matters: a northward return has $v_x<0$, so its negative relative vorticity supplies a positive drag contribution $-r\xi$ to balance $\beta v$. Here $r$ has units ${\mathrm{s}}^{-1}$; no extra factor of $f_0$ belongs in $r/\beta$ with the paper's definition of linear drag.

For example, at fixed $y$ the leading correction to the interior ocean transport streamfunction isIt cancels the western value of ${\Psi }_I$ and carries ${\int }_0^{\infty }H{v}_{B}dx=-LV=-T$. Exponentially small eastern-wall corrections and the turning regions near the northern and southern walls do not affect the leading width estimate.

Using one year $=3.156\times 10^7\mathrm{s}$, the representative $\beta$ of 2(b), and $H=10^3\mathrm{m}$ givesTo return the $25\mathrm{Sv}$ interior volume flux, conservation of mass requiresFor the exponential profile this is the boundary speed associated with its effective transport width $\delta$. It is much narrower and faster than the familiar large-scale western boundary currents, whose energetic cores span tens of kilometres and have speeds of order $1\mathrm{m}{\mathrm{s}}^{-1}$. The simplified constant-depth linear drag model with a one-year decay time therefore does not give a realistic quantitative boundary structure. Lateral viscosity, eddies, stratification and inertia can all affect a real current.

The corresponding relative vorticity scale isThus relative vorticity cannot be discarded relative to planetary vorticity in this predicted return. Even a more realistic $1\mathrm{m}{\mathrm{s}}^{-1}$ across $50\mathrm{km}$ gives $|\xi |\sim 2\times 10^{-5}{\mathrm{s}}^{-1}$, an appreciable fraction of $|f_0|$.

Magnitude alone does not prove that its material advection is large: a perfectly parallel current $v=v(x)$, $u=0$ has $\mathbf{u}\cdot \nabla \xi =0$ exactly. For a gyre return varying over meridional scale $B$, however, incompressibility gives $u\sim \delta V_b/B$, and both generic contributions to $\mathbf{u}\cdot \nabla \xi$ scale as $V_b^2/(B\delta )$. Relative to $\beta V_b$, this isAbsent special cancellations, the assumed neglect of relative-vorticity advection is therefore not self-consistent for these parameters. Its natural inertial width is instead ${\delta }_I\sim \sqrt{V_I/\beta }$, where $V_I=|T|/(HL)\simeq 5\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$; this gives about $16\mathrm{km}$. That comparison diagnoses the missing inertia rather than supplying a complete nonlinear boundary-current solution.

Far downslope let the typical height be $H(X)$ and the cross-slope half-width be $W(X)$. Slenderness gives $W\ll X$ and slowly varying longitudinal thickness. In a steady porous gravity current, the longitudinal spreading term is smaller than advection by the ratio $H/X$, while longitudinal diffusion relative to transverse diffusion is of order $(W/X{)}^2$. Thus the leading balance isIntegrating across the finite current, with zero flux through its dry side edges, makes the leading downslope volume flux constant:Balancing the two differential terms givesEliminate $H\propto W^{-1/\beta }$ to find the far-downslope width of a constant-flux porous current:The typical height is correspondingly $H\propto X^{-1/(2\beta +1)}$. For every $\beta >0$, the width exponent is less than one half, so $W/X\to 0$ and $H/X\to 0$, consistent with the discarded longitudinal terms. The porosity exponent $\alpha$ drops out because it enters storage, which has zero time derivative in this steady problem.

The cell conservation equation has the form ${\bar{n}}_{\bar{t}}=-\bar{\nabla }\cdot \bar{\mathbf{J}}$, withThe first term is advection by the suspension's fluid velocity, the second is active swimming at fixed speed $V_c$ in direction $\mathbf{p}$, and the last is an isotropic diffusive flux due to random swimming. Here $D$ has units length squared per time, $\bar{n}$ is cell number per volume, and the divergence measures local loss by transport. No birth or death term is included.

For a spherical cell of radius $a_c$, its excess buoyancy force is $(4\pi a_c^3/3)\Delta \rho g$. The Stokes drag law gives passive settling speed $V_{\mathrm{set}}=2a_c^2\Delta \rho g/(9\mu )$. Neglecting individual settling requires $V_{\mathrm{set}}\ll V_c$, not exact neutral buoyancy. A dilute suspension can still generate appreciable collective buoyancy because many small excess masses contribute to the bulk density.

Use the half-rate convention for the gyrotactic reorientation time $B$, required by the coefficients printed in the question:The first torque tends to point the bottom-heavy cell upward; the second rotates a sphere with half the fluid vorticity. This is a quasistatic torque balance, assuming orientation relaxation is fast enough to follow the local flow. Keeping orientation dynamics instead would put $d\mathbf{p}/d\bar{t}$ on the left. The convention with aligning rate $1/B$ uses a reorientation parameter twice this $B$. Thus the factor of two is a definition, not a physical discrepancy. Close to the upward state the quasistatic gyrotactic balance with half-rate reorientation gives ${\mathbf{p}}_{\perp }=B\bar{\mathbf{\omega }}\times {\mathbf{e}}_z$.

For a dilute Newtonian suspension with the Boussinesq approximation, the other equations areThe background-water hydrostatic pressure is absorbed in $\bar{p}$. The cell volume is $\upsilon$, density is $\rho$, cell density is $\rho +\Delta \rho$, dynamic viscosity is $\mu =\rho \nu$, and $\upsilon \bar{n}\ll 1$ is the volume fraction. Extra cell mass produces the downward buoyancy term. Constant viscosity and no explicit swimming stress are part of this closure. Rigid impermeable plates impose no-slip boundary condition and zero normal total cell flux.

At rest, the stable cell direction is upward and the steady vertical flux is $V_c\bar{n}-D{\bar{n}}_{\bar{z}}$. It is constant in depth by conservation and zero at an impermeable plate, soThe pressure gradient balances its excess weight. The mean-concentration formula follows by averaging over $0\le \bar{z}\le h$ and is continuous at $\beta =0$.

Use $\mathbf{x}=\bar{\mathbf{x}}/h$, $t=D\bar{t}/h^2$, $\mathbf{u}=h\bar{\mathbf{u}}/D$, $n=\bar{n}/{\bar{N}}_0$, and $p=h^2\bar{p}/(\rho \nu D)$. ThenwhereThe dimensionless base concentration is $n_0=e^{\beta z}$.

Linearize with the horizontal Fourier mode $e^{\sigma t+ilx+imy}$ and $k^2=l^2+m^2$. Write the first-order orientation perturbation as $\mathbf{p}={\mathbf{e}}_z+ϵ{\mathbf{p}}_1+\cdots$. Unit length gives $p_{1z}=0$, while torque balance gives ${\mathbf{p}}_{1\perp }=G{\mathbf{\omega }}_1\times {\mathbf{e}}_z$. Incompressibility impliesLet $\mathcal{L}=d^2/d{z}^2-k^2$. The perturbation of fluid advection is $n_0^{\prime }W=\beta n_{0}W$, and swimming contributes $\beta N^{\prime }+\beta n_0\nabla \cdot {\mathbf{p}}_1$. Cell conservation therefore givesorThis explains the destabilizing concentration-advection term and the gyrotactic redistribution term with their signs.

To eliminate pressure from the fluid equations, take their divergence, giving $\mathcal{L}{p}_1=-(R/\beta )N^{\prime }$. Apply $\mathcal{L}$ to the vertical equation $S{c}^{-1}\sigma W=-p_1^{\prime }+\mathcal{L}W-(R/\beta )N$. The concentration second derivatives cancel, leavingNo-slip conditions give $W=0$ and $W^{\prime }=0$ at each plate; for nonzero $k$, the latter follows from the vanishing horizontal velocity and incompressibility. Zero perturbed normal cell flux givesThe normal orientation correction is zero at this order, and the fluid velocity is zero at the plate, so neither adds another term to this concentration condition.

Now use the stated distinguished scalings $k=\beta \alpha$, $\sigma ={\beta }^2{\sigma }_2$, $G={\beta }^{-1}{G}_{-1}$, $N=N_0+\beta N_1+{\beta }^2{N}_2+\cdots$, $W=\beta W_1+{\beta }^2{W}_2+\cdots$, and $R=R_0+\beta R_1+\cdots$. Here $\alpha$ is a scaled wavenumber, unrelated to Question 2's shear rate. At order one, the concentration equation and boundary conditions give $N_0^{\prime \prime }=0$, $N_0^{\prime }=0$ at both endpoints, so $N_0$ is constant. Absorb its nonzero value into the perturbation amplitude and chooseAt order $\beta$, the fluid equation is $W_1^{(4)}=-R_0{\alpha }^2$ with $W_1=W_1^{\prime }=0$ at both ends. Its unique solution isA homogeneous difference is cubic and the four clamped conditions make it zero.

The concentration equation at order $\beta$ givesso $N_1=z-G_{-1}{W}_1+C_1$, where $C_1$ fixes the amplitude normalization. At order ${\beta }^2$ it givesThe boundary conditions expand to $N_2^{\prime }=N_1$ at each endpoint. Integrating this equation from zero to one cancels $[N_2^{\prime }{]}_0^1$ against $[N_1{]}_0^1$. Also $[W_2^{\prime }{]}_0^1=0$ andThus the required solvability condition isSince ${\int }_0^1{z}^2(1-z{)}^{2}dz=1/30$, the clamped-quartic solvability in shallow bioconvection gives

For nonzero long horizontal waves, $R_0<720$ gives decay and $R_0>720$ gives growth; the leading neutral threshold is $R_0=720$. The leading instability is stationary and independent of $G_{-1}$ and $Sc$. The exactly uniform horizontal mode $\alpha =0$ has zero growth at this order because total cell number is conserved, so it does not determine the threshold. Upward swimming produces a dense upper layer; the unstable disturbance is its buoyancy-driven overturning, or bioconvection. Higher orders are needed for wavelength selection and finite-depth threshold corrections.

With fixed material parameters and mean concentration, $R\sim h^4{V}_c\upsilon \Delta \rho g{\bar{N}}_{00}/(\rho \nu D^2)$ for small $\beta$, so a sufficiently thin chamber is stable to this branch. Formally its threshold corresponds to $h_c\sim [720\rho \nu D^2/(V_c\upsilon \Delta \rho g{\bar{N}}_{00}){]}^{1/4}$ when the weak-stratification and distinguished-gyrotaxis assumptions apply. In particular, the assumed $G=O({\beta }^{-1})$ is a specified asymptotic analysis; at fixed $B,D,V_c$ and decreasing $h$, one instead has $G=O({\beta }^{-2})$. The leading result should not be asserted unchanged in that different ordering without redoing the expansion.

The first equation is vertical hydrostatic equilibrium: the upward pressure gradient balances the central mass's vertical gravitational force. For $|z|\ll r$, the point-mass acceleration is $-GMz/(r^2+z^2{)}^{3/2}\simeq -{\Omega }_K^{2}z$, where Keplerian rotation gives ${\Omega }_K^2=GM/r^3$. The disc's self-gravity and vertical inertia are neglected.

The second equation is local thermal balance. The divergence of the vertical radiative flux equals viscous dissipation. In the convention of this alpha disk, $\nu =\alpha p/(\rho \Omega )$, and the dominant Keplerian shear has $rd\Omega /dr=-3\Omega /2$. ThereforeThe radiative flux increases from zero at the symmetric midplane to a positive outgoing value at the upper surface. Radial heat transport, thermal advection and rapid storage of thermal energy are excluded from this local balance.

The third equation is radiative diffusion, equivalentlyIt applies to optically thick material, with $\kappa$ the opacity per unit mass and $\sigma$ the Stefan-Boltzmann constant. Here constant electron-scattering opacity represents Thomson scattering; the formula uses a diffusive, nearly thermal radiation field. Positive upward flux requires temperature to decrease upwards.

The final equation is the ideal gas pressure law, with mean molecular weight ${\mu }_m$ in proton-mass units. Writing $\mathcal{R}=k/({\mu }_m{m}_p)$ gives $p=\mathcal{R}\rho T$. The pressure is gas pressure because radiation pressure is assumed negligible. The four equations express vertical support, local heating balance, diffusive cooling and the gas equation of state.

Estimate characteristic bulk terms. Put $ϵ=H/r\ll 1$ and assume smooth radial variation on scale $r$, secular evolution, and $\alpha \lesssim 1$. Vertical support gives $c_s^2\sim p/\rho \sim {\Omega }^2{H}^2$, while the alpha disk prescription and slow accretion implyThese velocity estimates follow from the viscous radial evolution scale and mass conservation. They are estimates for a slowly evolving thin disc, not a claim that geometric thinness excludes independently forced rapid vertical motion.

For vertical balance, the exact central force expands asAt $z\sim H$, the neglected correction has magnitude $\rho {\Omega }^{2}H{ϵ}^2$, compared with the retained force $\rho {\Omega }^{2}H$. Radial pressure support similarly changes the rotation from its Keplerian value byVertical meridional inertia scales as $u_r{u}_z/r$ or $u_z^2/H$, both $O({\alpha }^2{ϵ}^4{\Omega }^{2}H)$. Meridional viscous accelerations can scale as $\nu u_z/H^2$ or $\nu u_r/(rH)$, both $O({\alpha }^2{ϵ}^2{\Omega }^{2}H)$. Thus the gravity correction and possible viscous correction giveSome omitted terms are smaller than this leading order.