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Reset $t=0$ when gravity is switched off. The subsequent overdamped relaxation of a clamped--free filament obeys ${\zeta }_{\perp }{h}_t=-B{h}_{xxxx}$ with the same homogeneous endpoint conditions. Separate a normal mode as $h=\varphi (x/L)e^{-\Gamma t}$. Writing $s=x/L$ givesThe two clamp conditions reduce the general hyperbolic/trigonometric solution toThe two free-end conditions have matrixIts determinant is $2(1+\cosh \alpha \cos \alpha )$. Hence the clamped--free bending mode values satisfyThere is no zero mode: at zero eigenvalue $\varphi$ is cubic, and the four clamp/free conditions force all coefficients to vanish. With the following normalization a mode isThe fourth-order operator is self-adjoint and positive under these endpoint conditions, since $\int f{g}^{\prime \prime \prime \prime }=\int f^{\prime \prime }{g}^{\prime \prime }$. Its orthogonal eigenfunction expansion givesFor the previously attained steady profile, the first coefficient is nonzero: $h_s$ is positive and the first mode can be chosen positive in the interior. Two applications of integration by parts, using both functions' endpoint conditions, also giveThus the dominant long-time answer, including its shape and amplitude, isThe next mode decays much faster. For other initial shapes with an accidentally zero first-mode projection, the lowest mode with $c_n\ne 0$ determines the asymptotic behaviour. The decay time has the correct units and exhibits the strong $L^4$ dependence characteristic of bending opposed by local viscous drag.

Let $\varphi$ be the velocity potential. The linearized deep-water equations are ${\nabla }^2\varphi =0$ for $z<0$, ${\eta }_t={\varphi }_z$ and ${\varphi }_t+g\eta =0$ at $z=0$, with zero normal velocity at the four side walls. The normal modes arewhere $m,n\ge 0$ are not both zero. The free-surface conditions give the dispersion relation for a deep-water gravity wave:

The initial displacement is the $(m,n)=(3,4)$ mode, so $k=5\pi /a$. Release from rest selects a cosine in time:Velocity decays as $e^{kz}$, so it falls to $1/e$ of its surface value at depth $a/(5\pi )$.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.

For the growing normal mode in two deep layers,It vanishes at the endpoints of the unstable band and has its unique interior maximum where $A-3\gamma k^2=0$. Using $k_c=\sqrt{A/\gamma }$ givesThis is the maximum exponential growth rate, not the real oscillation frequency.

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

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.