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A normal mode $x\propto e^{-i\omega t}$ has eigenvalue $\lambda =-i\omega$ of $A$. Setting $d_{\omega }=\omega -{\omega }_1$, the characteristic equation isThusOne branch then has positive imaginary frequency and hence positive exponential growth rate. If $4ac<b^2$, both frequencies are real and the distinct-eigenvalue system has bounded oscillatory motion. At equality there is no positive exponential growth rate, but a nontrivial Jordan block can cause algebraic growth, so exponential neutrality must not be confused with Lyapunov stability. The parameter $d$ listed in the PDF does not appear in the printed matrix and has no effect on this calculation.

The central stratified layer supports upward and downward internal gravity waves, whose counterpropagating components form a standing wave. The unstratified regions support evanescent waves, and the upper disturbance decays as $e^{-k(z-2H)}$. Consequently the central layer is a stratified internal-wave guide, rather than a source of propagating energy at infinity.

The largest response occurs at its trapped normal mode frequencies. They can be specified without solving the forced problem. Set the boundary forcing to zero to find a free normal mode. The lower solution is proportional to $\sinh (kz)$, so continuity of fluid pressure and vertical velocity givesIn the central layer writeThe upper Robin boundary condition then gives the exact trapped-mode conditionThe left side increases strictly from zero to infinity, so there is one positive root for each $n$; the frequencies accumulate at zero. This is constructive phase matching after reflection at both ends. The phase shifts from the evanescent waves matter: simply imposing integer half-wavelengths across the stratified layer is generally incorrect.

In the ideal inviscid model, exact resonant forcing has no bounded steady harmonic solution. The undamped normal mode grows secularly under sustained forcing. Weak viscosity or other losses would produce large finite peaks near the displayed frequencies; the linear approximation eventually fails if the disturbance becomes too large.

The illustrated free normal mode has an oscillatory central region and evanescent wave tails. Its lower tail reaches the fixed zero-displacement boundary; its upper tail decays to infinity.

Eliminate the transport amplitudes in the linear stability analysis. For a general bed shear response $\mathcal{T}(k)$, the dispersion relation isThis is the most specific result available without adding the missing hydrodynamic closure. With $\mathcal{T}(k)={\tau }_{0}k(A+iB)$, setMultiplication by $1-ix$ gives the growth rate and migration velocityThe packed-bed volume fraction cancels. For a real physical closure the negative wavenumber response is the complex conjugate; the displayed $k>0$ formulas cover the independent normal modes.

If $A>0$, $B>0$ are treated as constants, the upstream shear stress phase lead promotes growth, while the inclined-bed sediment threshold reduces it and the finite saturation length delays sediment transport. There is an unstable band precisely when $b>0$: $0<k<b/(a{L}_{\mathrm{sat}})$. Within this band $c>0$, so the growing bedforms migrate downstream. The fastest normal mode satisfiesThis last maximization assumes constant $A,B$; if the hydrodynamic coefficients vary with wavenumber, their derivatives also enter the selection condition. When $b\le 0$ and $a>0$, all nonzero normal modes decay.

Just above the sediment entrainment threshold,Thus the common regime $B<1/{\mu }_s$ is stable near onset: the slope penalty overwhelms the hydrodynamic phase lead. This conclusion is conditional on that inequality, not universal. For constant $B>0$, instability begins only when ${\tau }_0>{\tau }_{\mathrm{th},0}/({\mu }_{s}B)$ as well as ${\tau }_0>{\tau }_{\mathrm{th},0}$. The transport sensitivity $C=\chi \gamma {\Delta }^{\gamma -1}$ vanishes at onset for $\gamma >1$, is finite for $\gamma =1$, and is singular for $0<\gamma <1$. Exactly at threshold, the positive-part transport law needs separate treatment; the above linearization with perturbations small compared with $\Delta$ is unavailable.

Far above the sediment entrainment threshold, $b\simeq {\tau }_{0}B$. If $A,B>0$, the unstable cutoff tends to $B/(A{L}_{\mathrm{sat}})$, and the dominant wavelength is set by the saturation length. Both growth rate and migration velocity scale with $C{\tau }_0\sim \chi \gamma {\tau }_0^{\gamma }$, apart from their length factors. Their numerical values and any detailed dependence on wavenumber still require a flow closure.

For example, prescribing uniform shear stress independently of the bed gives $\mathcal{T}=0$ andThis equally admissible closure illustrates why instability cannot be asserted from the printed sediment equations alone.

The plots use constant $a>0$ and three illustrative ratios $b/a$. They show how the sign of the shear phase lead minus the slope correction determines stability; they are not numerical predictions for unspecified flow conditions.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

The kinetic energy of the three unit masses is $\frac{1}{2}{\sum }_i{\dot{x}}_i^2$. Expanding the potential givesThus the small-oscillation mass and stiffness matrices arewhere $J$ is the all-ones matrix.

The vector $(1,1,1{)}^T$ is an eigenvector of the stiffness matrix with eigenvalue $1$. Every vector orthogonal to it has coordinate sum zero and is an eigenvector with eigenvalue $4$. Since $T=I$, the generalized eigenvalue problem for small oscillations gives ${\omega }^2$ equal to these eigenvalues. One orthonormal choice of normal modes isThe last two may be replaced by any orthonormal basis of the sum-zero plane because the frequency is degenerate. Hence the normal frequencies are $\boxed{1,2,2}$.

The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.

The lowest allowed vertical wavenumber, $k=\pi /(2H)$, is the last to be stabilized as $|B_z|$ increases. Therefore the finite-thickness magnetorotational instability criterion isThere is also a vertically uniform velocity mode with zero magnetic perturbation; for the usual orbitally stable regime $q<2$, it is just stable epicyclic motion. For $q>2$, that uniform mode is already hydrodynamically unstable, independently of the magnetic criterion. At $q=2$ it is marginal. The criterion above concerns the magnetic modes.

Let $V=U-c$ and take $k>0$ without loss of generality. The Boussinesq approximation replaces density by ${\rho }_0$ in inertia and pressure acceleration, while retaining the density perturbation in buoyancy. Linearizing about the hydrostatic equilibrium and substituting the normal mode givesThe last two equations give $\hat{u}=i{\hat{w}}^{\prime }/k$ and $\hat{\rho }=i{\rho }^{\prime }\hat{w}/(kV)$. The horizontal momentum equation givesSubstituting in the vertical momentum equation yields the Taylor–Goldstein equation:The buoyancy frequency is real for stable density stratification. The impermeable walls impose $\hat{w}(\pm L)=0$.

Choose $a=1/2$ in the power-transformed Taylor–Goldstein energy identity and let $Q=N^2-(U^{\prime }{)}^2/4$. The result isBecause $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part givesIf $4N^2>(U^{\prime }{)}^2$ everywhere, the integral is strictly positive for every nonzero eigenfunction, so a mode with $c_i\ne 0$ is impossible. Thus the flow is neutrally stable to these inviscid normal modes. This is the Miles–Howard theorem, expressed as a lower bound $\mathrm{Ri}>1/4$ on the gradient Richardson number. It excludes exponential modal growth; it does not by itself exclude transient growth.

For a possibly unstable normal mode, $c_i\ne 0$ ensures $V=U-c$ never vanishes, so choose one continuous branch of $V^a$. Substitute $\hat{w}=V^{a}q$ into the Taylor–Goldstein equation and collect terms:Multiplication by $V^{2a}$ puts its first two terms in divergence form:Multiply by the complex conjugate $\overline{q}$ and apply integration by parts. Since $q(\pm L)=0$, the boundary term vanishes and the power-transformed Taylor–Goldstein energy identity isThe weight is $V^{2a}$, not $|V{|}^{2a}$: preserving its complex phase is essential for the subsequent stability proofs.

Put $y={\tilde{c}}^2$ and $s=J/(2\alpha )>0$. The dispersion relation for two density interfaces in uniform shear isIts discriminant is $16s+4s^2{e}^{-4\alpha }>0$, so both roots $y_{\pm }$ are real. Their sum is positive; therefore one is negative precisely when their product is negative. That condition isSolving for $J$ givesIn this interval $y_{-}<0$, so $\tilde{c}=i\sqrt{-y_{-}}$ is the growing normal mode. At either endpoint the corresponding root is zero and the mode is neutral.

Linearizing about the conductive state givesThe perturbations vanish at $z=0,1$. For a normal mode with vertical dependence $\sin (n\pi z)$, put $q=k^2+n^2{\pi }^2$. The two modal equations are $-q\hat{\psi }=-ikR\hat{\theta }$ and $s\hat{\theta }+ik\hat{\psi }=-q\hat{\theta }$. Eliminating $\hat{\psi }$ givesNeutrality requires $R=(k^2+n^2{\pi }^2{)}^2/k^2$. The arithmetic-geometric mean inequality bounds this below by $4n^2{\pi }^2$, with equality at $k=n\pi$. The first unstable vertical mode is therefore $n=1$, givingThe wavelength statement applies at the first onset. At a general larger $R$, maximizing the growth rate within a fixed vertical mode gives $k^2=n\pi \sqrt{R}-n^2{\pi }^2$, so its preferred wavelength need not remain exactly two. This is the onset of convection in a horizontal Darcy layer.

Put $k_a=k$ ahead of the front and $k_b=k+\Delta k$ behind. With viscosity $\mu$, Darcy's law gives the planar pressure gradients $p_{a0}^{\prime }=-\mu U/k_a$ and $p_{b0}^{\prime }=-\mu U/k_b$. Write the displacement as $\eta e^{\sigma t+i\alpha y}$ with $\alpha >0$, and take the unperturbed interface to be $x=0$ in its translating frame.

The liquid flux is divergence-free away from the interface, so the pressure perturbations are harmonic. The decaying normal modes areExpanding pressure continuity at the displaced interface and matching normal Darcy flux givesHenceThe common normal-flux perturbation is $\delta u_n=-k_b\alpha B/\mu =\alpha U\Delta k\eta /(k_a+k_b)$. The local enthalpy balance converts it into a front-speed perturbation, $\sigma \eta =\delta u_n/(1+S\Delta \varphi )$. Thus the melting-front instability due to permeability contrast hasFor a signed transverse wavenumber, replace $\alpha$ by $|\alpha |$.

Take $f_0>0$ for definiteness and define $K=(k^2+l^2{)}^{1/2}>0$. Thermal-wind balance gives $U_z=-b_y/f_0=\Lambda$, and the lower boundary condition fixes $U=\Lambda z$. The basic interior quasi-geostrophic potential vorticity has no gradient, so a boundary-supported normal mode ${\psi }^{\prime }=\mathrm{Re}[\varphi (z)e^{i(kx+ly-\omega t)}]$ has zero interior potential-vorticity perturbation. Its vertical equation isDecay upward selects $\varphi =A{e}^{-{\mu }_{0}z}$.

The buoyancy perturbation is $b^{\prime }=f_0{\psi }_z^{\prime }$. At a rigid boundary $w^{\prime }=0$, linearized material conservation of buoyancy givesAt $z=0$, this is $c{\varphi }^{\prime }(0)+\Lambda \varphi (0)=0$, where $c=\omega /k$ for $k\ne 0$. The Eady edge wave therefore hasIt is an exponentially trapped boundary-buoyancy wave, not a vertically propagating interior mode. The formula for $\omega$ extends to $k=0$, where the perturbation is stationary.

For the two-region problem, absence of a density jump at the common material interface requires continuity of buoyancy:Thus its basic slope is $a=h_y=f_0({\Lambda }_1-{\Lambda }_2)/(N_1^2-N_2^2)$. For a nondegenerate stratification contrast, its variation over horizontal scale $L$ is $aL/D=O(\mathrm{Ro}/\mathrm{Bu})$, which is small in the usual quasi-geostrophic approximation with Burgers number of order one. The matching conditions can therefore be applied at the constant reference level $z=0$ to leading order. An exceptionally small stratification contrast would require checking this flat-interface ordering separately.

Pressure and normal velocity must be continuous because there is no membrane or singular interfacial force. Since base buoyancy is continuous, expanding pressure continuity at the displaced interface gives simply ${\varphi }_1(0)={\varphi }_2(0)$. The base horizontal velocity is also continuous there; call it $U_I$. Decay away from the interface then givesLinearized buoyancy conservation in each region givesMatching normal velocity gives $w_1^{\prime }=w_2^{\prime }$: the common slope correction to normal velocity cancels because pressure continuity also makes the horizontal velocity perturbation continuous. HenceThe quasi-geostrophic wave at a stratification interface has intrinsic phase velocityAs $N_1/N_2\to \infty$ with $N_2$ and the shears bounded, the lower region becomes effectively rigid andrecovering the upper-fluid Eady edge wave. The matching calculation describes regular interface-supported modes; arbitrary interior potential-vorticity initial disturbances can additionally be advected by the shear.

Assume $\beta >0$, $c=\sqrt{gH}$, and use complex normal mode amplitudes. The zonal momentum and continuity equations giveWhen ${\omega }^2\ne k^2{c}^2$, solving these two linear equations gives

For $\hat{v}=0$, a nonzero solution requires $\omega =\pm kc$ and $\hat{\eta }=\pm c\hat{u}/g$. The remaining momentum equation gives ${\hat{u}}^{\prime }=\mp \beta y\hat{u}/c$. Only the positive sign decays at both infinities, so the equatorial Kelvin wave isIts phase and group velocities are both eastward with speed $c$.

For nonzero meridional velocity, substitute the eliminated amplitudes into $-i\omega \hat{v}+\beta y\hat{u}=-g{\hat{\eta }}^{\prime }$. The result isSet $Y=y\sqrt{\beta /c}$. The decaying Hermite polynomial modes have $\hat{v}=V_n{H}_n(Y)e^{-Y^2/2}$, up to normalization, and their eigenvalue gives the equatorial shallow-water dispersion relation

For $n=0$, the quadratic in $k$ factors asThe admissible Yanai wave root is $k=\omega /c-\beta /\omega$. With $\hat{v}(0)=1$, substitution and cancellation giveThese expressions satisfy the original equations even at the isolated point where the two algebraic roots coincide and the earlier elimination formula has a zero denominator. The other root is not an independent admissible branch: its generic singular reconstruction fails the trapping conditions.

In the zonal momentum equation, the acceleration term is $-i\omega \hat{u}=({\omega }^{2}y/c)\hat{v}$, the Coriolis term is $-\beta y\hat{v}$, and the pressure term is $-ikg\hat{\eta }=k\omega y\hat{v}$. When ${\omega }^2\ll \beta c$, acceleration is small and the leading balance is between Coriolis acceleration and the pressure gradient. When ${\omega }^2\gg \beta c$, acceleration balances pressure and the Coriolis term is small; the branch tends to an eastward gravity wave.

For $n\ge 1$, the fixed-frequency zonal roots areThe radicand is nonnegative exactly when $s={\omega }^2/(\beta c)$ satisfies $s^2-(2n+1)s+1/4\ge 0$. This gives the frequency gap for higher equatorial wave modes:

For the forcing frequency $\omega =\frac{1}{2}\sqrt{\beta c}$, we have $s=1/4$. Already the $n=1$ low-frequency upper bound is $3/2-\sqrt{2}<1/4$, and that bound decreases with $n$, while the high-frequency bound increases. Hence no $n\ge 1$ mode propagates. The available real-wavenumber branches areFor the equatorial Kelvin wave, $c_{g,K}=c$. For the Yanai wave, differentiating $k=\omega /c-\beta /\omega$ givesThus both propagating responses are detected east of the localized forcing, provided it has nonzero projection onto their meridional structures. The Yanai phase travels westward at this frequency, but its wave packet and energy travel eastward. Higher modes contribute only an evanescent response near the forcing.

By separation of variables, the axisymmetric normal modes satisfying regularity at $r=0$ and the fixed-edge condition at $r=1$ arewhere $J_0({\gamma }_k)=0$. The zero initial displacement removes all cosine terms, soDifferentiating at $t=0$ gives the weighted Fourier-Bessel expansionMultiply by $r{J}_0({\gamma }_{\ell }r)$ and integrate from zero to one. The orthogonality and norm from part (i) giveUsingthe remaining integral isHenceand therefore

At $z=0$, the two free-surface conditions for a normal mode giveEliminating $\hat{\zeta }$ yieldsFor the vertical factor found in part (b),Thus the gravity-wave dispersion relation iswhere

Let each mass be $m$ and each spring constant be $k$, with the $x_i$ denoting longitudinal displacements from equilibrium. The Lagrangian isThe inverse coordinate transformation isSubstitution givesThus the Euler-Lagrange equations separate into a symmetric pair,and an antisymmetric pair,