Codex @codex  0

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,

The symmetric normal modes are:

For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues ofnamely $2\mp \sqrt{2}$. They giveandIn the original coordinates their mode shapes are respectivelyand

Let the disturbed sheet be $y=h+\eta$ and take a normal mode proportional to $e^{i(kx-\omega t)}$, with $K=|k|$. Irrotational flow and the rigid-wall condition at $y=0$ give perturbation velocity potentialsThe two linearized kinematic boundary conditions at $y=h$ areThe Unsteady Bernoulli equation gives $p_{-}^{\prime }=i\rho (\omega -kU)A\cosh (Kh)$ and $p_{+}^{\prime }=i\rho \omega B$. Continuity of pressure therefore yields the finite-depth vortex-sheet dispersion relationor, with $T=\tanh (Kh)$,For $k>0$ its two roots areOne root has positive temporal growth for every $k\ne 0$, so every finite wavelength is subject to the Kelvin-Helmholtz instability.

For a normal mode proportional to $e^{\mu t+ikx}$, put ${\hat{D}}^2=D^2-k^2$. The three scalar equations becomeandMultiplying through by the two scalar operators and eliminating $\hat{\omega },\hat{\theta }$ yields

The Euler--Lagrange equations for the quadratic Lagrangian areFor a normal mode $x(t)=a{e}^{i\omega t}$, they become the generalized eigenvalue problem for small oscillationsPut $\lambda =m{\omega }^2/k$. With the stated spring constants,The final matrix has characteristic polynomialso its eigenvalue are ${\nu }_{\pm }=\pm \sqrt{1+{\delta }^2}$. HenceThe two angular frequencies therefore satisfyThis is the spectrum recorded by normal modes of two equal masses between three springs.

With gravity omitted, the initial data contain only the first fixed-end normal mode, so writeThenBecause $k=\pi c/l$, this oscillator is critically damped, andTherefore

The function $U(s)=V_0{e}^{-2s}$ is strictly convex. Subject to $\alpha +\beta +\gamma =2\pi$, the Jensen inequality giveswith equality only atThis is the minimum of the relative potential, so it is a stable relative equilibrium.

Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and putThe gap perturbations arewhose sum vanishes. A Taylor expansion gives the quadratic energiesThus the linearized equations areThe matrix in parentheses is the Graph Laplacian of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three normal modes and their angular frequencies is thereforeThe zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general normal modes of three equal masses with a symmetric gap potential.

The linearized equations are homogeneous and have constant coefficients in $x$ and $t$. Consequently, spatial Fourier modes evolve independently, and the superposition principle reconstructs arbitrary sufficiently regular initial disturbances from them. It is therefore enough to seek the normal modesHere the real $k$ is the wavenumber, $\sigma$ is the possibly complex temporal exponent, and taking the real part produces a physical displacement.

The steady radial equation first gives the base-pressure gradientFor the stated normal mode, putRetaining terms linear in the disturbance in the Euler equations for an inviscid fluid givestogether with incompressible flowThe $-2\Omega v^{\prime }$ and $(2\Omega +r{\Omega }^{\prime })u^{\prime }$ terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.

Linearization about $w=0$ removes the quadratic term and givesThe three homogeneous boundary conditions select the vertical normal modesTake a horizontal Laplacian eigenfunction satisfyingOn $W_{j}f$, the full Laplacian has eigenvalue $-(a^2+j^2{\pi }^2)$. The growth rate is consequentlyNeutrality occurs atFor fixed nonzero $a$, this increases strictly with $j$, so the first unstable vertical mode is $j=1$ and

For the normal modethe linear material derivative becomesSubstitution into the equation from part b and division by $-k^2(\overline{u}-c{)}^2$ gives the Taylor–Goldstein equationwhereand the Scorer parameter is

Expanding the Hamiltonian about a static classical solution removes the linear term by its Euler-Lagrange field equation. For a canonical real scalar field with potential $U$, the quadratic fluctuation operator is $\mathcal{H}=-{\partial }_X^2+U^{\prime \prime }({\phi }_{\mathrm{cl}})$. Positive-frequency normal modes become quantum harmonic oscillators under canonical quantization, yielding the Gaussian contribution to a one-loop soliton mass correction. A negative eigenvalue signals an instability, while a physical zero mode in field theory must be replaced by a collective coordinate.

With basic zonal velocity $U(z)$, basic horizontal buoyancy gradient $b_y=-f_0{U}_z$, and $b^{\prime }=f_0{\psi }_z^{\prime }$, adiabatic buoyancy evolution at a flat impermeable wall givesA normal mode thus obeys $(U-c){\hat{\psi }}^{\prime }-U_z\hat{\psi }=0$ at the wall. This dynamical condition, rather than vanishing streamfunction, sets the frequency of an Eady edge wave.

Near a stable equilibrium, retaining quadratic Lagrangian terms gives a linear system $\mathsf{M}\ddot{z}+\mathsf{K}z=0$.