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Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode $k$, the Stokes temperature-slaving operator maps $\theta$ to $w={\mathcal{W}}_R\theta$, where $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $w=w^{\prime \prime }=0$. The temperature evolution has linear operator ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$ and bilinear map $\mathcal{N}(\varphi ,\chi )=-{\mathbf{u}}_R[\varphi ]\cdot \nabla \chi$. Under the homogeneous thermal Dirichlet boundary conditions, ${\mathcal{L}}_{R_c}$ is self-adjoint. Normalize its critical eigenfunction as $g(z)=\sin \pi z$ and set $s=k_c^2+{\pi }^2$; the critical vertical velocity is $f=sg$.

At order $ϵ$, the critical eigenfunction equation gives ${\mathcal{L}}_{R_c}(g{e}^{i{k}_{c}x})=0$. At order ${ϵ}^2$, the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to $|A{|}^2$ and, in a general vertical-mode calculation, a second harmonic proportional to $A^2$. These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.

At order ${ϵ}^3$, the method of multiple scales produces the slow derivative $A_{\tau }$, the detuning term $R_2{\partial }_R{\mathcal{L}}_R$, and the two cross-advection terms involving first- and second-order fields. Project the $e^{i{k}_{c}x}$ component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by $⟨g,g{⟩}_z$. The detuning supplies $\mu A$ with $\mu \propto R_2$; interactions of horizontal wavenumbers $2k_c$ and $-k_c$ permit $\beta A^{\ast }$ with $\beta \propto \gamma$; self-interaction through the slaved mean and second harmonic supplies $-\lambda |A{|}^{2}A$. Other products have the wrong horizontal wavenumber. Reflection $x\mapsto -x$ permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes $\beta =0$ at this order. To see this, write a positive second-harmonic forcing component as $(w,\theta )=(W(z),\Theta (z))e^{2i{k}_{c}x}$, incorporating the cosine's factor $1/2$. Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor $A^{\ast }$,The integral vanishes because $g=0$ at both plates, even though $\Theta (0)$ is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times $\gamma$; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.

The same normalization makes the remaining coefficients explicit. Since ${\mathcal{W}}_{R}g=(R{k}_c^2/s^2)g$, $\mu =R_2{k}_c^2/s^2$. The quadratic second harmonic cancels for $f=sg$, while the uniform correction is ${\theta }_{20}=-s|A{|}^2\sin (2\pi z)/(2\pi )$. Projecting $-w_1{\partial }_z{\theta }_{20}$ gives $-s^2|A{|}^{2}A/2$. Thus for the literal model and this temperature normalization,A generic nonzero $\beta$ would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.

Write $A=x+iy$. Then $x_{\tau }=(\mu +\beta -\lambda (x^2+y^2))x$ and $y_{\tau }=(\mu -\beta -\lambda (x^2+y^2))y$. These are a gradient flow for $V=-\frac{1}{2}(\mu +\beta )x^2-\frac{1}{2}(\mu -\beta )y^2+\frac{\lambda }{4}(x^2+y^2{)}^2$, so local minima give stable equilibrium points. At the origin the two eigenvalues are $\mu +\beta$ and $\mu -\beta$. The origin has exponential asymptotic stability if $\mu <-|\beta |$, retains asymptotic stability with algebraic decay at $\mu =-|\beta |$, and is unstable if $\mu >-|\beta |$. At equality, ${\rho }^2=x^2+y^2$ obeys $({\rho }^2{)}_{\tau }\le -2\lambda {\rho }^4$, since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.

For $\beta >0$ the stable nonzero equilibrium points are real; for $\beta <0$ they are imaginary:The real branch has Jacobian matrix eigenvalues $-2(\mu +\beta ),-2\beta$; the imaginary branch has $-2(\mu -\beta ),2\beta$. The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when $\beta \ne 0$.

For $\beta =0$, the origin is stable for $\mu \le 0$, with algebraic decay at zero. If $\mu >0$, the circle $|A{|}^2=\mu /\lambda$ is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero-$\beta$ branches instead exhibit phase locking to one of two phases separated by $\pi$.

Assume $R^2=1-q^2>0$. The symbol $R$ in this part is the convection roll amplitude, not the Rayleigh number of the previous part. Set $r=R^2$ for the algebra. The periodic boundary condition requires $qL\in 2\pi \mathbb{Z}$ and modulation wavenumbers $lL\in 2\pi \mathbb{Z}$. The conserved-field equation preserves the imposed zero mean of $B$, since the integral of an $X$ derivative over a period vanishes.

Equating the $e^{ilX+\lambda T}$ coefficients after linearization of the convection amplitude coupled to a conserved field givesThe conjugate sideband is needed because the variation of $|A{|}^{2}A$ couples a perturbation to its complex conjugate. Put $S=a_1+a_2$, $D=a_1-a_2$. The linear operator on $(S,D,b)$ isIts characteristic polynomial, expressed conveniently without expanding every coefficient, isAt $l=0$ it has roots $-2r,0,0$. The phase root is neutral; the uniform $B$ root is excluded by the prescribed zero-flux-perturbation mean. Nonzero arbitrarily long-wave modes remain allowed on a sufficiently large domain. For fixed $r>0$, the two roots tending to zero have $\lambda =l^2\nu +O(l^4)$ at simple limiting roots. Dividing the polynomial by $2l^4$ yieldsRepeated limiting roots may require a further expansion, but do not affect the strict instability criterion. If $(\mu -\sigma )r+2\sigma q^2>0$, the constant term is negative and the two real roots have opposite signs. The positive root gives a growing long-wave sideband instability. Using $r=1-q^2$ givesThis is sufficient for instability when a sufficiently small nonzero allowed $l$ is available. It is not an unconditional instability theorem for every fixed finite period. For example $q=0$, $\mu =2$, $\sigma =1$, $L=1$ satisfy the inequality, but the nonzero allowed $l$ obey $l^2\ge 4{\pi }^2$. Their roots are $-l^2$ and $-l^2-1\pm \sqrt{1+4l^2}$, all negative. The prescribed zero mean removes the uniform flux root and the remaining uniform phase is merely neutral. The corrected conclusion is the long-wave instability with a conserved mean field, with domain size and mode availability stated.

Consider a horizontal fluid layer under the Boussinesq approximation heated from below, rotating uniformly about its vertical axis. To make the linear formulas definite, take depth one, stress-free impermeable fixed-temperature plates, a horizontally infinite or sufficiently large periodic domain, and negligible centrifugal-buoyancy modifications. Rigid plates change the vertical eigenfunctions and thresholds and require their own boundary value problem; the explicit formulas below use the free-slip model. Let $P=\nu /{\kappa }_T$ be the Prandtl number, $\mathrm{Ta}=(2\Omega d^2/\nu {)}^2$ the Taylor number, and $R$ the Rayleigh number. Time is again measured on the thermal diffusion time.

The linearized rotating Rayleigh-Bénard convection equations about conduction areThe Coriolis force does no work on the fluid, but it couples vertical motion to vertical vorticity and thereby changes the balance of buoyancy and viscous damping. On the plates require $w=\theta =0$ and ${\partial }_z{u}_x={\partial }_z{u}_y=0$. Write $\zeta =(\nabla \times \mathbf{u}{)}_z$. Taking a curl and a double curl removes pressure and givesA normal mode with horizontal wavenumber $k>0$, vertical index $n\ge 1$ and complex growth rate $s$ has $w=W\sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$, $\theta =\Theta \sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$ and $\zeta =Z\cos (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$. Put $a=k^2+n^2{\pi }^2$. The three equations become $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, and $(s+a)\Theta =W$. Their determinant gives the rotating-convection growth-rate polynomialThis cubic includes viscous and thermal modes as well as the convective instability. It avoids excluding a root by division by $s/P+a$. Linear stability means every root has negative real part; stationary onset has $s=0$, while overstability has $s=\pm i\omega$ with nonzero angular frequency.

Setting $s=0$ gives the stationary neutral curve of rotating convectionFor oscillatory convection, set $s=i\omega$ and equate real and imaginary parts. Eliminating $R$ yieldsThe oscillatory neutral curve of rotating convection is physically admissible only when ${\omega }^2>0$. Thus $P<1$ and sufficiently rapid rotation are necessary; a formal minimum with ${\omega }^2\le 0$ is not a Hopf bifurcation. Convection begins at the smaller of the stationary minimum and the admissible oscillatory minimum. The angular frequency approaches zero where a fixed-wave-number stationary and oscillatory threshold meet; that is a double-zero limit of the cubic, requiring a different slow-time reduction.

For wavenumber selection in rotating convection, the lowest vertical index is $n=1$. To compare indices, set $k^2=n^2{\pi }^{2}y$ at fixed $y>0$. Both neutral thresholds have a positive term proportional to $n^4$ and a rotation term independent of $n$, while the admissible angular frequency squared decreases with $n$. Their minima therefore cannot improve on $n=1$. Set $t=k^2$, $m={\pi }^2$. Minimizing each neutral curve givesFor the latter equation also check ${\omega }^2>0$, or minimize over the admissible set instead. Without rotation, $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At rapid rotation, $k_s\sim (\mathrm{Ta}{\pi }^2/2{)}^{1/6}$ and $R_{s,c}\sim 3(\mathrm{Ta}{\pi }^2/2{)}^{2/3}$: rotation narrows the horizontal convection rolls and raises their threshold. In a finite box only discrete wavenumbers are permitted, so the minimum must be taken over those allowed modes.

The rapid-rotation comparison has $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Its equality gives $8P^4-P-1=0$, with positive root about $0.67660$. Below this value, sufficiently rapid rotation can make oscillatory convection the primary instability; low Prandtl number allows inertial motions to interact with the thermal field before viscosity damps them. This estimate describes the selected neutral-curve comparison in the stated free-slip asymptotic model, rather than claiming every $P<1$ has oscillatory primary onset at every rotation rate.

Linear selection does not determine the nonlinear planform or its stability. A weakly nonlinear expansion with a solvability condition gives amplitude equations. A stationary convection roll amplitude has the Landau amplitude equation $A_T=rA-g|A{|}^{2}A$; a supercritical branch requires $g>0$ and has $|A{|}^2=r/g$, whereas $g<0$ requires higher-order saturation and permits subcritical behaviour. The coefficient depends on the physical parameters and boundary conditions, and should not be assumed positive merely because the linear threshold is known.

A nonzero-frequency Hopf bifurcation supplies oppositely travelling convection roll amplitudes $Z_{+},Z_{-}$. At a fixed orientation, a cubic normal form iswith complex $g,h$ and the understood symmetry exchanging propagation directions. For $r>0$ and $g_r=\mathrm{Re}g>0$, a travelling-roll branch has one nonzero amplitude, of squared modulus $r/g_r$; it is stable to the opposite travelling amplitude when $h_r=\mathrm{Re}h>g_r$. A standing-roll branch has equal squared moduli $r/(g_r+h_r)$, and is stable within this pair when $g_r>h_r$ and $g_r+h_r>0$. The imaginary parts shift angular frequencies. These conditions concern amplitude perturbations in that reduced subspace; orientation and modulation modes still have to be tested. This is the competition of travelling and standing convection rolls.

Near simultaneous stationary and oscillatory thresholds, a codimension-two bifurcation requires retaining both types of mode. After selecting an orientation and fixing a steady spatial phase, a schematic nonresonant steady–Hopf mode interaction isHere $a$ is real, $z$ is the chosen complex oscillatory amplitude, $g_s,h_s$ are real, and $g_o,h_o$ may be complex. Assume $r_s>0$, $g_s>0$ for a radially stable pure steady branch, and $r_o>0$, $g_{o,r}=\mathrm{Re}(g_o)>0$ for a radially stable pure oscillatory branch. Steady convection rolls suppress the oscillatory mode if $r_o-h_{o,r}{r}_s/g_s<0$, where $h_{o,r}=\mathrm{Re}(h_o)$; oscillatory convection rolls suppress the steady mode if $r_s-h_s{r}_o/g_{o,r}<0$. These are transverse tests for branches whose existence and radial stability have already been checked. Mixed states have positive intensities $u=a^2$, $v=|z{|}^2$ solving $g_{s}u+h_{s}v=r_s$, $h_{o,r}u+g_{o,r}v=r_o$. If $D=g_s{g}_{o,r}-h_s{h}_{o,r}\ne 0$, then $u=(r_s{g}_{o,r}-h_s{r}_o)/D$, $v=(g_s{r}_o-h_{o,r}{r}_s)/D$. For positive $u,v$ and the positive self-saturation coefficients above, the intensity Jacobian matrix has negative trace and determinant $4uvD$, so the mixed state is attracting in intensities when $D>0$ and is a saddle equilibrium when $D<0$; phase and other-mode perturbations remain separate tests. Depending on the cross-couplings, one obtains coexistence or competition and bistability; degeneracies or resonances require extra terms. A full travelling/standing-wave competition must retain both $Z_{\pm }$, not just this illustrative one-mode $z$. If the Hopf angular frequency tends to zero, averaging over fast oscillations fails and a double-zero normal form must keep the corresponding two slow variables. A crossing of global minima at distinct wavenumbers is instead an interaction of distinct modes, not automatically the same double-zero problem.

Weakly nonlinear convection rolls can also lose stability spatially. The real Ginzburg–Landau equation $A_T=rA+\xi A_{XX}-g|A{|}^{2}A$, with $r,\xi ,g>0$, has detuned convection rolls with $|A{|}^2=(r-\xi Q^2)/g$. Linear phase modulation gives diffusion coefficient $\xi (r-3\xi Q^2)/(r-\xi Q^2)$, so the Eckhaus instability excludes $Q^2>r/(3\xi )$ even though convection rolls exist up to $Q^2<r/\xi$. Transverse bending and mean-flow couplings impose additional restrictions in the actual rotating layer; the scalar equation is a local longitudinal example, not its complete stability theory.

Most distinctively, rotation breaks mirror symmetry between competing convection roll orientations. For two sets of convection rolls at relative angle $\vartheta$, write ${\dot{A}}_2=r{A}_2-g_0|A_2{|}^2{A}_2-g(\vartheta )|A_1{|}^2{A}_2$ and interchange indices with $\vartheta \mapsto -\vartheta$ for the other equation. The established convection roll $|A_1{|}^2=r/g_0$ is unstable to the new orientation when its linear growth rate $r[1-g(\vartheta )/g_0]$ is positive. The inequality need not be symmetric under $\vartheta \mapsto -\vartheta$ because the imposed rotation supplies handedness. This is the Küppers–Lortz instability: sufficiently rapid rotation can destabilize steady convection rolls immediately above their stationary onset to oblique convection rolls of another orientation. Successive replacements can produce time-dependent convection roll switching or heteroclinic cycles, instead of a stable single convection roll pattern. It requires three-dimensional perturbations even when the original convection roll solution is independent of its axial coordinate.

At finite Prandtl number and stress-free plates, nearly parallel convection rolls can couple resonantly to a slowly damped large-scale Eulerian mean flow. This small-angle instability of rotating convection rolls makes a regular two-roll expansion nonuniform as the angle tends to zero; one must retain the mean-flow mode. It should not be identified automatically with the finite-angle Küppers–Lortz instability or assigned a universal threshold from a one-amplitude equation. Consequently the onset type and selected scale follow from the admissible neutral curves, while persistent convection roll patterns require a separate nonlinear and sideband stability calculation.

With periodic or sufficiently decaying boundary conditions, integration by parts givesAssume $\kappa \ge 0$ for a stable quadratic free energy; the diffusive smoothing discussed below requires $\kappa >0$. In a periodic box of volume $V$, use the unitary convention ${\varphi }_{\mathbf{q}}=V^{-1/2}{\int }_V\varphi (\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}d\mathbf{r}$. A real field has ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. Since the Fourier transform of a derivative converts ${\nabla }^2$ to $-q^2$, each Fourier mode obeysIt is an Ornstein-Uhlenbeck process. Multiplication by the integrating factor $e^{r(q)t}$ givesThe noise integral is understood as an Itô integral with a deterministic kernel; writing $\eta ds$ is shorthand for the corresponding Wiener increment. Direct differentiation in the mild stochastic sense checks both the equation and the initial value. The stated Kronecker-delta noise covariance corresponds to this finite-volume normalization. Unlike a conserved order parameter, the zero wavenumber mode decays at the nonzero rate $\Gamma a$.

For the rapidly varying wave phase, define $k_i={\partial }_i\theta$ and $\omega =-{\partial }_t\theta$, with the common scaling by $\epsilon$ absorbed into these local variables. A packet moves with group velocity $c_{g,i}=\partial \Omega /\partial k_i$. Along that trajectory, $d/dt={\partial }_t+c_g\cdot {\nabla }_x$. Compatibility gives ${\partial }_t{k}_i=-{\partial }_i\omega$ and ${\partial }_i{k}_j={\partial }_j{k}_i$. Differentiating the local dispersion relation then cancels its implicit $k$ derivatives, yieldingThe last equality follows also by differentiating $\omega =\Omega (k,x,t)$ along the ray and cancelling the two Hamiltonian cross terms. The partial derivatives of $\Omega$ hold its other arguments fixed.

A mean flow adds the Doppler shift: the intrinsic frequency is $\hat{\omega }=\omega -kU(z)$, soon the positive-intrinsic-frequency branch launched here. The medium is stationary and independent of $x$, so $\omega$ and $k$ are constant. Initially $m=0,U=0$, hence $\omega =N$. Moreover $\dot{m}=-k{U}^{\prime }<0$, and then $\dot{z}=-Nkm/(k^2+m^2{)}^{3/2}>0$. The initial vertical group velocity is zero, but $m$ immediately becomes negative and the ray accelerates upwards.

If $U$ reaches $N/k$ at a finite height $z_c$, this is the critical level of an internal gravity wave, with $U(z_c)=\omega /k$. Below it,Assuming $0<U^{\prime }(z_c)<\infty$, put $\delta =z_c-z$. Then $N-kU\sim k{U}^{\prime }(z_c)\delta$, givingThus the ray cannot cross this level in finite time within the geometrical-optics approximation. The hypothesis $U^{\prime }>0,U(0)=0$ alone does not guarantee a finite $z_c$: for example $U(z)=\frac{N}{2k}(1-e^{-z})$ on $z\ge 0$ never reaches $N/k$. Existence of a finite critical level is an additional assumption. Very close to one, diverging wavenumber can invalidate the inviscid, linear, slowly varying approximation; the result is the formal ray prediction.

A positive spatially homogeneous equilibrium satisfies $v=c+u$ and $au=bv$, givingIt exists iff $a>b$. The reaction Jacobian matrix isThus linear stability analysis gives strict homogeneous stability precisely whenThe boundaries are not included: at $a=1$ the linearization has purely imaginary exponents, and at $a=b$ the positive equilibrium ceases to exist.

For a mode $\cos (kx)$, put $q=k^2$. The perturbation matrix is $J-\mathrm{diag}(q,dq)$. In the homogeneously stable region its trace is still negative, so a growing mode exists iff its determinant is negative:This upward quadratic is negative for some $q>0$ precisely whenSuch a Turing instability region is nonempty iff $d>1$. The unstable band isFor a finite domain only the admissible boundary-condition wavenumbers in this band grow. The region shown below describes a continuous wavenumber spectrum.

At onset the minimum of $D$ is zero and $q_c=u_{\ast }(d-a)/(2d)$. On the threshold $b=4d{a}^2/(d+a{)}^2$, substitution gives $u_{\ast }=4dac/(d-a{)}^2$. HenceThis last expression is the threshold wavenumber; away from onset the fastest-growing mode need not coincide with the minimum of the determinant.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

For $-\beta /{\ell }^2<V\le 0$, there is one positive stationary wavenumber $k_0\in (0,\ell ]$ satisfying $\beta (k_0^2-{\ell }^2)/(k_0^2+{\ell }^2{)}^2=V$, and its negative. Since ${\omega }^{\prime \prime }(k_0)>0$, their paired stationary-phase contribution isAt $V=0$ the finite stationary points are $\pm \ell$ and are nondegenerate, so the same formula applies, assuming sufficient decay to control the large-wavenumber tail.

For $V>\beta /(8{\ell }^2)$ or $V<-\beta /{\ell }^2$, there are no real stationary points: no group-velocity ray reaches those slopes. Integration by parts gives much smaller contributions; for a Schwartz amplitude they decay faster than every inverse power of $t$ at a fixed such $V$. For less regular data the rate depends on the amplitude, and the printed even function assumption alone does not specify it. At the excluded endpoint values of the group velocity stationary points coalesce and an ordinary $t^{-1/2}$ formula fails, typically requiring an Airy-type transition rather than the expressions above.

Substitution of $e^{i(kx-\omega t)}$ givesThe phase velocity is negative for all nonzero wavenumbers, so crests travel left. The group velocity is negative for $|k|<\ell$, zero at $|k|=\ell$, and positive for $|k|>\ell$. Its minimum is $-\beta /{\ell }^2$ at zero and maximum $\beta /(8{\ell }^2)$ at $|k|=\sqrt{3}\ell$; it tends to zero from above at infinity. The odd dispersion relation has extrema at $k=\pm \ell$, with $\omega (\ell )=-\beta /(2\ell )$.

The Fourier solution is $\phi (x,t)={\int }_{\mathbb{R}}A(k)e^{i[kx-\omega (k)t]}dk$. The stationary-phase statements require enough amplitude regularity and decay, for example $A$ smooth and rapidly decreasing; realness and evenness alone do not guarantee a convergent integral or a stationary-phase expansion. For $x=Vt$, set ${\Phi }_V(k)=kV-\omega (k)$ and solve ${\omega }^{\prime }(k)=V$ implicitly.

For an axisymmetric vertical mode of a shearing sheet in this Boussinesq approximation, the perturbations have no horizontal gradients, so background-shear advection vanishes. The linear Coriolis acceleration and perturbation advection of the background shear givewhile thermal conduction and incompressibility giveHere $\beta$ is the thermal diffusion rate of a Fourier mode, not the cooling exponent in Question 2. Since $k\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $P^{\prime }=0$ for this nonzero-wavenumber amplitude.

The three remaining amplitudes satisfyA nontrivial solution requires the determinant to vanish, giving the convective overstability dispersion relationNo division by $s$ or $s+\beta$ was needed, so the thermal energy mode of a shearing sheet is retained. The notation $N^2$ denotes a signed buoyancy frequency squared and can be negative for an adverse stratification; it must not be restricted to the square of a real positive frequency.

Use superscripts $T$ and $D$ for the thin-upper-layer and two-deep-layer cases. The cutoff is identical, whileThus confinement suppresses the Rayleigh-Taylor instability growth strongly, but moves the most amplified disturbance to a somewhat larger wavenumber. The cutoff depends on the gravity-capillary balance, whereas growth depends also on the inertia of the moving layers.

Take $x$ in the injection direction and $y$ along the flat interface. The specified $U$ must be a Darcy velocity, so the pore-fluid velocity and unperturbed interface speed are $U/\varphi$. Write the perturbed interface as $x=X(t)+\eta (y,t)$, with $\dot{X}=U/\varphi$, and choose the normal from fluid 1 into fluid 2. The background pressure gradients from Darcy law are ${\partial }_x{p}_i^{(0)}=-{\mu }_{i}U/k$.

By incompressibility and uniform permeability of a porous medium, pressure perturbations are harmonic functions. A transverse Fourier mode with positive wavenumber $\alpha$ has the decaying formsThe kinematic boundary condition and equal normal Darcy flux implyWithout capillarity the pressure is continuous at the displaced interface. Its linear perturbation is thereforeConsequently the planar viscous-fingering dispersion relation reduces toThis is the Saffman–Taylor instability: less viscous fluid advances preferentially through protrusions. Without a short-wave regularizing mechanism the idealized rate has no finite maximum.

For the prescribed apparent capillary pressure, the chosen normal is $\mathbf{n}=(1,-{\eta }_y)/\sqrt{1+{\eta }_y^2}$, so $\nabla \cdot \mathbf{n}=-{\eta }_{yy}+O({\eta }^2)={\alpha }^2\eta +O({\eta }^2)$. Subtract the flat-front pressure jump $\gamma (1+\beta X){\kappa }_0$. Linearization of the jump at $x=X+\eta$ givesThere is no first-order product of the surface-tension perturbation and the perturbed macroscopic curvature. ThusUsing the capillary number convention $\mathrm{Ca}={\mu }_{2}U/\gamma$ and ${\mu }_2/({\mu }_1+{\mu }_2)=(1+M)/2$, this isIf $x$ is measured from the unperturbed front and $\gamma$ denotes its local apparent tension, set $X=0$ to obtain the usual fixed-coefficient expression. For a physically fixed spatial gradient, $X=Ut/\varphi$ and the displayed rate is an instantaneous, generally time-dependent growth rate; the amplitude solves $\dot{a}=\sigma (\alpha ;X(t))a$, rather than one constant exponential law for the entire displacement. The parameter $\beta$ has dimensions of inverse length; the absolute tension gradient is $\gamma \beta$.

Assume $1+\beta X>0$, as required for positive apparent surface tension, and define $A=({\mu }_2-{\mu }_1)U-k\gamma \beta {\kappa }_0$. If $A>0$, unstable modes satisfy $0<{\alpha }^2<A/[k\gamma (1+\beta X)]$. Differentiating the cubic dispersion relation gives the fastest-growing modeWhen $A\le 0$, every nonzero mode decays, with a neutral zero-wavenumber translation in the infinite-domain limit. For $\beta {\kappa }_0>0$, surface-tension-gradient stabilization of viscous fingering suppresses the instability entirely at positive injection speeds satisfyingIf $\beta {\kappa }_0\le 0$, no positive injection speed eliminates every long-wave unstable mode in an unbounded interface. The prescribed wetting variation is treated as an effective normal pressure law for porous-media flow.