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For unbiased one-dimensional diffusion, the mean squared displacement is $⟨x^2(t)⟩=2Dt$. With $t=n\tau$, compare this to the mutant expression:Therefore the mutant diffusion coefficient is

Take $x=0$ at chamber A and $x=L$ at chamber B. At steady state, the diffusion equation with constant positive diffusion coefficient givesThe two Dirichlet boundary conditions give $A=c_0$ and $B=-c_0/L$. ThusThe concentration gradient is constant, $dc/dx=-c_0/L$. By Fick's first law, the diffusive flux is $J=-Ddc/dx=D{c}_0/L$, directed from A towards B. The profile is linear independently of $D$, while the flux and the time needed to approach it depend on $D$.

This result assumes no significant bulk flow, chemoattractant loss to the walls, degradation or cellular uptake. Experimentally, a uniform-concentration control with matched reservoir concentrations tests for directional cell migration caused by flow or apparatus asymmetry; reversing the reservoir concentrations tests whether the directional response follows the chemical gradient. Keeping both reservoirs well mixed and measuring the concentration profile checks the assumed boundary concentrations and whether uptake has distorted it. These controls distinguish the assumed passive diffusion field from other directional cues; the derived profile alone does not establish the mechanism of chemotaxis.

For a stationary film of uniform thickness, thermal conduction gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:Consequently the surface temperature isThis is the quasistatic temperature of a conductively cooled film, with a weak-cooling expansion when $\alpha h/\kappa \ll 1$.

Let the thickness and longitudinal scales be $h_{\ast },L$, with $ϵ=h_{\ast }/L\ll 1$. Incompressibility gives transverse speed of order $ϵU$, and film evolution has the kinematic time scale $L/U$. For the temperature variation scale $\Theta$, longitudinal diffusion divided by transverse diffusion is $h_{\ast }^2/L^2={ϵ}^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse diffusion isThus ${ϵ}^2\ll 1$ and $Pe\ll 1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_{\ast }^2/(\kappa t_{\mathrm{evol}})$ to be small.

Define the positive surface-tension gradient coefficientThe weak-cooling approximation gives ${\gamma }_s\simeq {\gamma }_0+Bh$, so the Marangoni stress is ${\gamma }_{s,x}=B{h}_x$. The Young–Laplace equation gives the leading liquid pressure $p=p_a-{\gamma }_0{h}_{xx}$, with constant ambient pressure. In lubrication theory, $p_z=0$ and $\mu u_{zz}=p_x$. The no-slip boundary condition at the wall and the tangential surface condition $\mu u_z(h)={\gamma }_{s,x}$ giveHence the volume flux isConservation of mass gives the thermocapillary thin-film equationThe replacement of surface tension by ${\gamma }_0$ in the capillary term needs $B{h}_{\ast }/{\gamma }_0\ll 1$. For comparable capillary and thermocapillary fluxes,so the fractional tension change is only $O({ϵ}^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution ${\gamma }_x{h}_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|{\gamma }^{\prime }|\Delta T/{\gamma }_0$ were allowed to grow without bound.

The capillary term smooths the film: a crest has negative curvature and higher liquid pressure, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher surface tension; Marangoni stress pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.

Use $H=h/\hat{h}$, $X=ϵx/\hat{h}$, and $\tau =t/\hat{t}$. The two dimensionless flux-divergence coefficients are $B{ϵ}^2\hat{t}/(2\mu )$ and ${\gamma }_0{ϵ}^4\hat{t}/(3\mu \hat{h})$. Setting both equal to one gives the balanced scales for a thermocapillary film:These choices yield $H_{\tau }+(H^2{H}_X{)}_X+(H^3{H}_{XXX}{)}_X=0$. The predicted $ϵ$ must be small for the assumed long-wave approximation.

Linearize about $H=1$ using a normal mode of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the dispersion relation isModes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbersThe physical fastest wavenumber is $ϵ/(\sqrt{2}\hat{h})$.

For a positive steady film with zero volume flux, $H^2{H}_X+H^3{H}_{XXX}=0$. Divide by $H^3$ and integrate once:Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the zero-flux thermocapillary film profiles relationThus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.

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.

Take $\kappa >0$ and initially $\omega \ne 0$. Write $m_j(y,t)={\int }_{\mathbb{R}}{x}^j\chi (x,y,t)dx$, and set $k=\pi /L$, $a=\kappa k^2$. The advection-diffusion equation is ${\chi }_t+u{\chi }_x=\kappa ({\chi }_{xx}+{\chi }_{yy})$. Integration by parts in the unbounded coordinate gives the longitudinal moment hierarchy for shear transportAll three moments obey Neumann boundary conditions at the walls. The initial values are $m_0=L$, $m_1=0$, $m_2=L^3/12$. These manipulations require finite moments and vanishing boundary terms at $x=\pm \infty$; diffusion of the compact initial distribution supplies the required decay for positive time.

The zeroth moment remains exactly $L$. Since $\sin (ky)$ is a Neumann eigenfunction, put $m_1=Lg(t)\sin (ky)$. Then $g^{\prime }+ag=U\cos (\omega t)$ and $g(0)=0$, givingFor the second moment, $2u{m}_1=LUg\cos (\omega t)[1-\cos (2ky)]$. Thus $m_2=A(t)+B(t)\cos (2ky)$, withThe equation for $B$ has positive damping and bounded forcing, so $B$ remains bounded. In the equation for $A$, the decaying term in $g$ contributes only a bounded accumulated transient. Over a complete nonzero-frequency cycle, $⟨{\cos }^2(\omega t){⟩}_t=1/2$ and $⟨\sin (\omega t)\cos (\omega t){⟩}_t=0$. ConsequentlyThis proves oscillating shear dispersion in an insulating channel, including the local-in-$y$ leading term, not just a cross-channel average. The instantaneous derivative retains bounded periodic oscillations; its cycle-averaged slope is $2K{m}_0$. Thus $2Kt{m}_0$ is the leading second moment, rather than a literal rate proportional to $t$.

The asymptotic statement holds for fixed positive diffusivity and fixed nonzero frequency, after times long compared with $a^{-1}$ and the oscillation period. If $|\omega |L^2/\kappa \ll 1$, transverse diffusion adjusts rapidly to the changing shear flow, andThis is the cycle average of the instantaneous steady-shear enhancement $U^2{\cos }^2(\omega t)/(2a)$. At exactly $\omega =0$, there is no cycle average: $g\to U/a$ and $K=\kappa +U^2/(2a)$. Hence the zero-frequency and long-time limits do not commute. In the intermediate range $a^{-1}\ll t\ll |\omega {|}^{-1}$, the shear is approximately steady at its initial amplitude and that latter enhancement is appropriate.

Let ${\sigma }_v^2=⟨v(0{)}^2⟩$, assumed finite. The displacement is $\Delta X(t)={\int }_0^{t}v(s)ds$. By stationarity, its two-time covariance is $⟨v(s)v(r)⟩={\sigma }_v^2\rho (s-r)$, with $\rho$ even for a real scalar process. Integrating over the two triangles of the square gives the Taylor turbulent dispersion identityA sufficient condition for ordinary long-time variance growth is ${\int }_0^{\infty }|\rho (s)|ds<\infty$ with ${\int }_0^{\infty }\rho (s)ds>0$. The dominated convergence theorem then givesMore generally the necessary asymptotic condition for a finite positive variance-based diffusion coefficient is convergence of ${\sigma }_v^2{\int }_0^t(1-s/t)\rho (s)ds$ to that coefficient. Absolute integrability is sufficient, not necessary. A zero integral does not give nondegenerate ordinary diffusion. Moreover a second-moment calculation alone does not establish a normal distribution or an entire diffusive scaling limit; those require additional probabilistic assumptions.

For the specified positive autocorrelation, direct integration gives, when $\alpha \ne 1,2$,Thus the power-law velocity-correlation dispersion has the following regimes. If $\alpha >1$, the Lagrangian integral time is $1/(\alpha -1)$ and ordinary variance growth holds, with $D={\sigma }_v^2/(\alpha -1)$. At $\alpha =2$ the exact expression is $2{\sigma }_v^2[t-\log (1+t)]$. If $\alpha =1$,For $0<\alpha <1$,The last two regimes spread faster than ordinary diffusion and have no finite long-time diffusion coefficient. Their slowly decaying velocity memory is responsible. At short times all these correlations instead give the ballistic law $⟨\Delta X^2⟩\sim {\sigma }_v^2{t}^2$.

Use a Poincaré map sampled once per forcing period. A hyperbolic periodic trajectory has stable manifolds containing parcels that approach it under forward iteration and unstable manifolds containing parcels that approach it under backward iteration. In the integrable limit the relevant branches coincide along a heteroclinic orbit and form a barrier between neighboring cells.

A transverse crossing in the perturbed map means that this barrier has split: a parcel on the intersection belongs to both asymptotic manifolds, and adjoining branches enclose fluid lobes transported across the former boundary. Iteration stretches and folds those lobes. In the connected heteroclinic network this creates complicated intercell transport and chaotic advection. The crossing is between invariant curves at the same forcing phase; it does not mean that different trajectories cross at one time in violation of uniqueness.

The incompressible flow gives an area-preserving map, so lobe exchange preserves parcel area. Without diffusion, an advected passive scalar also keeps its value on each trajectory. Therefore manifold intersections can generate fine filaments and efficient advective exchange but do not by themselves produce molecular homogenization, nor prove that every region of the flow becomes chaotic.

On $y=0$, the unperturbed trajectory satisfies $\dot{x}=\sin x$. Put $a_0=\log \tan (X_0/2)$. Integration with $x(0)=X_0$ givesIt runs from $(0,0)$ to $(\pi ,0)$, so it is the appropriate heteroclinic orbit. From the perturbed stream function,On the chosen connection, $\nabla {\psi }_0=(0,-\sin x)$; hence the scalar product in the heteroclinic Melnikov function for a periodic planar flow is ${\sin }^{2}x\sin (\omega t)$. ThereforeSet $s=\tau -t_0+a_0$. The odd term ${\mathrm{sech}}^{2}s\sin (\omega s)$ integrates to zero, leaving the supplied even Fourier integral multiplied by $\sin [\omega (t_0-a_0)]$. ThusFor every fixed $\omega \ne 0$ the amplitude is nonzero and the phase zeros are simple: ${\partial }_{t_0}M$ there is $\pm \pi {\omega }^2/\sinh (\pi \omega /2)\ne 0$. At fixed phase, varying $X_0$ also gives simple zeros because $d{a}_0/d{X}_0=1/\sin X_0$ is nonzero for $0<X_0<\pi$. The Melnikov splitting of a periodically perturbed cellular flow therefore produces transverse manifold intersections at sufficiently small perturbation and fluid-lobe exchange between adjacent cells. The unperturbed separatrix is no longer a material transport barrier.

At high frequency the leading splitting amplitude is asymptotic to $2\pi |\omega |e^{-\pi |\omega |/2}$, so fast forcing has an exponentially weak first-order effect on this connection. This is a fixed-frequency, small-$ϵ$ result; an exponentially small leading term need not dominate higher orders in a simultaneous high-frequency limit. At exactly $\omega =0$ the perturbation itself vanishes and $M=0$, so the nonzero-frequency conclusion does not apply. Manifold splitting demonstrates intercell chaotic advection locally, not a proof of uniform mixing or ordinary diffusion throughout the entire domain.

A density deficit in a flux tube is not by itself an instability calculation. For example, a straight magnetized tube in otherwise field-free gas at the same temperature has lateral balance $p_i+B_i^2/(2{\mu }_0)=p_e$. The perfect gas law then gives ${\rho }_i<{\rho }_e$, so the tube rises unless some other force balances its weight deficit. This configuration simply lacks magnetostatic equilibrium. A magnetic buoyancy instability, in contrast, starts from a force-balanced atmosphere and asks whether a small allowed displacement grows. A balanced horizontal magnetic layer can be stable or unstable; the presence of magnetic pressure alone does not decide this.

Assume a smooth equilibrium, with downward gravitational acceleration $g>0$, and write $\mathcal{R}$ for the gas constant per unit mass. The ambient total pressure and force balance areWe use ideal magnetic flux freezing on the displacement time scale and rapid lateral pressure equilibration. A bodily displacement perpendicular to a straight horizontal tube does not change its length or bend its field. If its area is $\mathcal{A}$, conservation of magnetic flux and mass per unit length gives $B\mathcal{A}=\mathrm{constant}$ and $\rho \mathcal{A}=\mathrm{constant}$. Thus the tube preservesEfficient heat exchange fixes its temperature to the ambient temperature at its new height, not to its original temperature. This is the thermal assumption that distinguishes the requested criterion from the adiabatic one.

Take a tube initially at $z_0$ and let $f_0=B_0(z_0)/\rho (z_0)$. After a small upward displacement $\xi$, let ${\rho }_i(z)$ be the density it would have in lateral balance at $z=z_0+\xi$. Its preserved $f_0$ and imposed local temperature giveAt the initial position ${\rho }_i=\rho$. Differentiate this relation at $z_0$:For the ambient atmosphere the same differentiation must also differentiate $f(z)=B_0(z)/\rho (z)$. It givesSubtracting isolates the actual displaced density contrast:The pressure-gradient force on the tube balances the weight of the displaced ambient gas, so its leading vertical buoyancy acceleration isThe coefficient multiplying $\xi$ is positive precisely when the tube becomes lighter on upward displacement. Thus the isothermal interchange criterion for magnetic buoyancy isWhere $B_0>0$, this is simply $d(B_0/\rho )/dz<0$, or $d\log B_0/dz<d\log \rho /dz$. The squared form is independent of the field's choice of direction. Equality is neutral at this order, and the reversed inequality is restoring for these displacements. A layer with an interval satisfying the strict inequality admits local growing interchange disturbances under the assumed pressure/thermal equilibration. The calculation does not address field-bending undular modes, finite heat-transfer times or magnetic diffusion; those can change the criterion. In particular, the localized interchange criterion for a magnetized atmosphere with an adiabatic gas response is not interchangeable with this fast-thermal-exchange result.

Magnetic flux concentration and flux expulsion describe redistribution of an existing field, not necessarily creation of net flux. For a weak field in a prescribed conducting flow, the resistive induction equation isIts ideal part combines transport, stretching and compression:Thus a converging flow can intensify a component, and stretching can increase magnetic energy even if the signed net flux is zero. At large magnetic Reynolds number $\mathrm{Rm}=UL/\eta$, diffusion is initially slow on the eddy scale, so magnetic flux freezing carries field lines with the flow. These statements do not imply that an ideal closed eddy can simply lose its initially enclosed flux.

The missing step is the production of small scales by differential circulation. In a two-dimensional incompressible eddy, the Cartesian magnetic flux function obeysDifferent streamlines have different circulation rates. An initially smooth field is wound into fine folds; diffusion and reconnection then smooth the variation of $A$ inside the eddy. As $A$ becomes nearly uniform there, its gradient and hence the interior magnetic field become small. The surviving flux lies mainly near the circulation boundary and adjacent separatrices, where stronger gradients remain. This is flux expulsion. With exactly zero diffusivity, field topology and the flux of material loops remain frozen, so a relaxed flux-free interior is not obtained by pure advection alone. Rigid rotation also fails to generate the differential winding responsible for this process.

The phase mixing in magnetic flux expulsion estimate makes the enhanced role of small diffusion explicit. If the angular-velocity gradient is of order $U/L^2$, a wound radial wavenumber grows as $k_r\sim Ut/L^2$. The accumulated damping exponent is of orderIt becomes order one atThis is much shorter than the large-scale diffusion time $(L/U)\mathrm{Rm}$ at large $\mathrm{Rm}$, but longer than one turnover time. It is a kinematic differential-shear estimate, not a universal time for every turbulent or rapidly evolving convection cell.

A related concentration mechanism occurs at converging flow boundaries. For a locally vertical bundle, conservation of flux gives $B\mathcal{A}\simeq \mathrm{constant}$, so reducing its cross-section increases $B$. In a simple converging boundary region with strain rate $U/L$, balancing strain and transverse diffusion givesThis is the thickness estimate for that local strain-diffusion model; it need not be the thickness or energy scaling of every closed-eddy flux-expulsion problem. Stronger field eventually exerts Lorentz force feedback and changes the flow. The equipartition magnetic field $B_{\mathrm{eq}}=\sqrt{{\mu }_0\rho }U$ is a useful indication that the kinematic assumption can fail, not a universal upper bound on concentrated fields.

At the Sun's photosphere, hot upflows form solar granules, and their spreading horizontal motion transports weak vertical field toward cooler intergranular lanes and downflows. A thin-layer illustration, neglecting vertical transport terms, is ${\partial }_t{B}_z+{\nabla }_h\cdot ({\mathbf{u}}_h{B}_z)=0$. It gives $D_h{B}_z/Dt=-B_z{\nabla }_h\cdot {\mathbf{u}}_h$: horizontal convergence intensifies the vertical flux density. This surface transport is related to the redistribution by convective eddies, but is not identical to the closed-streamline mechanism in the preceding planar example. Larger-scale supergranulation also gathers magnetic elements toward its boundaries, helping organize the solar magnetic network.

Concentration can be further intensified by compressible thermal dynamics. In the thin-tube picture of convective collapse, an unstable downward flow drains gas, reduces internal pressure and allows contraction at approximately conserved flux. Lateral balance $p_i+B^2/(2{\mu }_0)\simeq p_e$ then permits field stronger than the local kinetic equipartition estimate. This evacuation mechanism is excluded by a strict Boussinesq approximation. It is not a description of every evolving surface concentration: gas pressure, dynamic stresses and magnetic curvature can all matter when thin-tube equilibrium fails. The resulting strong, intermittent magnetic elements are associated with intergranular structure; their radiative appearance also depends on thermal structure and opacity. The organized fields of sunspots require a broader account of flux supply and magnetic feedback, rather than attribution to weak-field expulsion alone.

Thus convection redistributes and concentrates flux, finite diffusion enables classical expulsion, and nonlinear magnetic and thermal feedback determines the strong-field outcome. An expulsion argument alone neither supplies the original solar field nor establishes a self-sustaining solar dynamo.

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

Put $a=\kappa m^2$, with $\kappa >0$ and $m\ne 0$. The relevant homogenization of a periodic advection-diffusion equation limit is a long horizontal wavelength observed on its diffusive time scale. Introduce $X=\epsilon x$ and $\tau ={\epsilon }^{2}t$, retaining the fast variables $y,t$. Since the velocity field is independent of $x$, no fast horizontal cell variable is needed. Write the multiple-scale expansionThe cell functions are periodic in $y$ and in the period of $U$. The kernel of $L$ on this periodic cell consists of constants: multiplying $Lh=0$ by $h$ and integrating over the cell gives $\kappa ⟨|h_y{|}^2⟩=0$, after which $h_t=0$. Thus the leading concentration really is independent of the fast variables. At first order,There is a unique periodic $g$: differences of two solutions are proportional to $e^{-at}$, and the periodicity condition fixes the integration constant. Equivalently,At second order the advection-diffusion equation givesIntegration over the fast cell annihilates the left side. Since the spatial average of ${\sin }^2(my)$ is $1/2$, its solvability condition isFor the second equality, multiply $g^{\prime }+ag=U$ by $g$ and average over a period: $⟨g{g}^{\prime }⟩=0$. In particular the enhancement of diffusivity is nonnegative. Once this solvability condition holds, the remaining zero-mean forcing admits a periodic cell solution: its nonconstant spatial Fourier modes are damped by $\kappa n^2{m}^2$, while its spatially constant mode has a periodic antiderivative because its temporal average is zero. This also explains the relaxation of initial transverse inhomogeneity. The first corrector has zero spatial mean, so $\bar{\chi }=C+O({\epsilon }^2)$.

This derivation describes the large-scale, long-time limit, after times comparable to the period and to $a^{-1}$. It does not give an exact closed equation for the spatial average of arbitrary finite-time data: that average contains a correlation between the velocity field and concentration. In physical coordinates the leading long-wavelength concentration therefore obeys ${\bar{\chi }}_t={\kappa }_e{\bar{\chi }}_{xx}$ asymptotically. The cell expansion, its bounded periodic correctors, and the decay of transverse cell transients justify this effective diffusivity of a periodic sinusoidal shear.

For constant $U_0$, the periodic cell solution is $g=U_0/a$. ConsequentlyThis is Taylor dispersion: molecular diffusion moves particles between different horizontal velocities and turns their accumulated horizontal displacement into large-scale diffusion.

For the impulsive flow, $g$ jumps by $U_1$ at each kick and decays exponentially between kicks. If $G$ denotes its value just after a kick, periodicity requires $G=e^{-aT}G+U_1$. Thus, within one period,Using the mean-square expression, which has no ambiguous product of a Dirac delta with a discontinuous function, givesThe impulsive model can also be obtained by shrinking smooth pulses of fixed integrated strength. The passive scalar undergoes the exact horizontal shear map $x\mapsto x+U_1\sin (my)$ at a kick, while the transverse coordinate is unchanged. The cell solution above is the corresponding limit of the smooth cell solutions. Using a one-sided value in $Ug$ instead would miss the contribution from the jump of $g^2$.

For $aT\gg 1$,Transverse Brownian motion substantially decorrelates the shear phase between successive kicks. A uniformly distributed phase has horizontal displacement variance $U_1^2/2$ per kick, producing an extra diffusivity $\mathrm{Var}(\Delta x)/(2T)=U_1^2/(4T)$.