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For stable density jumps, $J>0$ and $q=J/(2\alpha )>0$. Let $y=C^2$. The two roots of the dispersion relation for two density interfaces in uniform shear areThey are real, and $y_{+}>0$. An exponentially growing normal mode exists precisely when $y_{-}<0$, giving a pair of imaginary phase speeds; the member with positive imaginary part grows in $e^{ik(x-ct)}$ for $k>0$.

Since $y_{+}{y}_{-}=(1-q{)}^2-q^2{E}^2$, this condition is $|1-q|<qE$. Solving both inequalities yieldsAt either endpoint $y_{-}=0$, so the exponential growth rate vanishes. Outside the open band both squared phase speeds are nonnegative and the interfacial modes have real frequencies. This is a modal instability test for the stably stratified configuration; unstable density inversions would require a separate analysis. Within the band the dimensional growth rate is $k\Delta U\sqrt{-y_{-}}/2$.

Substitution of a normal mode gives the dispersion relationFor real $k$, $\sigma$ is the temporal growth rate. Since $\beta \ne 0$, its supremum is finite exactly whenFor $\beta <0$, arbitrarily short wavelengths grow arbitrarily fast. For $\beta >0$, the finite maximum temporal growth rate isThe Briggs-Bers criterion starts the inverse temporal Laplace transform above all temporal singularities and then deforms its contour downward while following the spatial roots. A finite growth bound supplies such an initial contour and a causal, high-frequency-controlled Green function. Unbounded temporal growth prevents that standard construction.

For absolute wave-packet instability, a candidate spatial pinch point must satisfy $D=0$, $D_k=0$, and $\mathrm{Im}\omega >0$. HeregivesThus candidate growing saddles require $\gamma <0$ at $k_0$, or $\gamma <{\alpha }^2/(4\beta )$ at $k_{\pm }$. Together with $\beta >0$, existence of at least one such candidate requires $\gamma <{\alpha }^2/(4\beta )$.

A growing double root is not sufficient: the roots must pinch the spatial inversion contour from opposite sides. Collisions of branches originating in the same spatial half-plane do not obstruct the relevant deformation. This distinction is part of the Briggs-Bers criterion; it is stated, for example, in the primary study doi.org/10.1017/jfm.2016.195.

An explicit false spatial saddle in quartic dispersion is $\alpha =-2$, $\beta =1$, $\gamma =1/2$. Its candidates $k=\pm i$ have $\omega =i/2$, but all real modes have $\sigma =-k^4-2k^2-1/2<0$. Each imaginary collision joins two branches in the same half-plane; neither is a relevant pinch. For $\omega =i\Omega$, the spatial roots obey $k^2=-1\pm \sqrt{1/2-\Omega }$, making those same-half-plane collisions transparent as $\Omega \downarrow 1/2$.

For this particular real, even dispersion relation one can also establish the actual threshold directly. At the origin its impulse Green function isLaplace method selects the real maximum, and gives a positive prefactor times $t^{-1/2}{e}^{{\sigma }_{\max }t}$ because $\alpha \ne 0$. Therefore the actual absolute wave-packet instability condition isThese are sufficient for this model as well as necessary. They follow after identifying relevant real saddles; the earlier algebraic double-root test alone lacks the pinch information. Equality is marginal, not exponential absolute growth.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

Write the horizontal velocity amplitudes as $(u,v)$ and magnetic amplitudes as $(b_x,b_y)$. The perturbation is horizontally uniform, divergence-free, and has no vertical velocity, so the unperturbed density and pressure are consistent at linear order. In the Keplerian shearing sheet, the horizontal components of the linearized ideal magnetohydrodynamic equations areUse the undivided equations at a zero of $F$. If $F^{\prime \prime }=-k^{2}hF$, the magnetic-force coefficient becomes $ik{v}_A^2$; the stratification cancels. The ideal magnetohydrodynamic induction equation similarly reduces toThe azimuthal induction term is field stretching by the background differential rotation. The four amplitudes obey the homogeneous systemIts determinant must vanish for a nonzero normal mode. Put $A=v_A^2{k}^2$. For nonzero $s$, elimination first gives $(s^2+A)u-2\Omega sv=0$ and $s(s^2+A)v+\Omega (s^2-3A)u/2=0$, whose solvability condition is $(s^2+A{)}^2+{\Omega }^2(s^2-3A)=0$. The original determinant extends the same polynomial to marginal $s=0$. ThusThis is the ideal magnetorotational dispersion relation with the midplane Alfvén speed $v_A=B_0/\sqrt{4\pi {\rho }_0}$. The vertical structure enters through the admissible eigenvalues $k$, rather than through a different horizontal dispersion polynomial. No division by a vanishing growth rate is required in the determinant derivation.

Keep the prescribed outer slip velocity fixed when perturbing the unsteady Prandtl equation. Cancelling the base equation and retaining terms linear in the disturbance gives the linearized unsteady Prandtl equationThe no-slip boundary condition supplies $\tilde{u}=\tilde{v}=0$ at the wall; the fixed outer velocity supplies $\tilde{u}\to 0$ at infinity. There is no prescribed zero normal-velocity disturbance at infinity: a finite displacement-related value is permitted.

With the frozen parallel profile $U=U(y)$, $V=0$, the coefficients are invariant under translations in $x$ and $t$. Normal modes therefore separate those variables. For the mode convention used here the equations areEliminate $u=-v^{\prime }/(i\alpha )$ to obtain the Prandtl normal-mode equationwith a bounded far-field $v$. Requiring $v(\infty )=0$ would incorrectly eliminate the proposed outer profile. For real $U$ and real $\alpha$, complex conjugation of the equation and its boundary conditions replaces $(\alpha ,c,v)$ by $(-\alpha ,c^{\ast },v^{\ast })$. This proves the stated eigenvalue symmetry. The paired modes have the same temporal growth rate $\alpha \mathrm{Im}c$, so choosing positive wavenumber loses no real physical disturbance.

Here the prescribed parallel profile is a local frozen-coefficient model. A nontrivial arbitrary $U(y)$ with $V=0$ and constant outer slip is not generally an exact steady solution of the unforced unsteady Prandtl equation; its base equation would require $U^{\prime \prime }=0$. The following stability calculation uses the simplifying local model stipulated for the mode analysis.

Set $c_0=U_c=U(y_c)$. Away from the critical point the dominant equation is $(U-c_0)v_0^{\prime }-U^{\prime }{v}_0=0$, whose solutions are multiples of $U-c_0$. Choosing the coefficient to be zero below $y_c$ and one above it satisfies the wall and tangential far-field conditions. The choice $c_0=U_c$ makes $v_0$ continuous across the joining point; $U^{\prime }(y_c)=0$ also makes its first derivative continuous. Its second derivative jumps. Thus it is a valid leading outer solution away from $y_c$, but viscosity must smooth the join in a critical layer in a shear flow.

Near the join, $U-c_0\sim (y-y_c{)}^2/2$. If the layer width is $\ell$, matching gives $v=O({\ell }^2)$. The advection side of the mode equation scales as $\alpha {\ell }^3$ and $v^{\prime \prime \prime }$ scales as ${\ell }^{-1}$. Balance gives ${\ell }^4\sim {\alpha }^{-1}=\epsilon /k$. Equivalently, the eigenvalue correction ${\epsilon }^{1/2}{c}_1$ balances $(y-y_c{)}^2$, so the quarter-power critical layer hasSubstitution, retaining the leading terms and cancelling their common factor, gives

At order ${\epsilon }^{1/2}$ outside the layer,A solution consistent with the chosen lower branch and the wall conditions is $v_1=0$ below the join. Above it a particular solution is $v_1=-c_1$; a multiple of $U-c_0$ represents an arbitrary amplitude renormalization and may be set to zero. Matching consequently requiresThe corresponding derivatives match as $w_Y\sim Y$, $w_{YY}\to 1$ on the positive side and as zero on the negative side. The matching statements require that growing homogeneous corrections be absent.

To symmetrize these different end conditions and remove $k$ and $i$, chooseThe polynomial $A(z)=z^2/2-C$ is itself a solution of the transformed homogeneous equation. Direct substitution, using $a^4=ik$, givesThe derivative is third order, as required by the original mode equation. Initially the two ends are along the rotated rays $z=aY$ with real $Y$. Continuing them to the real $z$ axis is compatible with the asymptotic end conditions: the decreasing homogeneous correction behaves exponentially as $\exp [-z^2/(2\sqrt{2})]$ and still decreases throughout the rotation from angle $\pi /8$ to zero. Thus the supplied real-axis spectral normalization uses the same recessive conditions; a complex coordinate change should not be mistaken for a real stretching alone.

For each supplied real eigenvalue $C_n=(4n+7)/\sqrt{2}$, the phase-speed correction isThe mode factor has temporal magnitude $e^{\alpha \mathrm{Im}ct}$. ThereforeAmong the listed indices, $n=-2$ gives positive growth $\gamma \sim \sqrt{\alpha }/2$; $n=-1,1,2,\dots$ give decay. Hence the frozen non-monotone flow has high-wavenumber instability of a non-monotone Prandtl layer, with arbitrarily rapid linear growth as the positive wavenumber increases. The normal-mode calculation demonstrates linear instability; it is not by itself a claim about the final nonlinear state.

Use the fluid displacement $\mathbf{\xi }$, with velocity perturbation $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Denote Eulerian perturbations by $\delta$ and the corresponding Lagrangian pressure perturbation by ${\Delta }_{L}p=\delta p+\mathbf{\xi }\cdot \nabla p$. For adiabatic perturbations, the linearized mass conservation, momentum, adiabatic equation of state and Poisson equation areThese self-gravitating adiabatic displacement equations retain the perturbation of the star's own Newtonian gravitational potential. In the uniform-density interior, $\delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, $\nabla \Phi ={\omega }_d^2\mathbf{r}$, and $\nabla p=-\rho {\omega }_d^2\mathbf{r}$. The dynamical frequency of a uniform-density star ${\omega }_d=(GM/R^3{)}^{1/2}$ sets the natural timescale. For a normal mode with time factor $e^{-i\omega t}$, replace ${\partial }_t^2$ by $-{\omega }^2$. The requested interior analysis needs no surface or exterior matching conditions.

The azimuthal derivative of a cylindrical vector Fourier mode must include basis rotation: the azimuthal dependence refers to the components in the rotating cylindrical basis. Since ${\partial }_{\varphi }\hat{\mathbf{s}}=\hat{\mathbf{\varphi }}$ and ${\partial }_{\varphi }\hat{\mathbf{\varphi }}=-\hat{\mathbf{s}}$, a vector amplitude $\mathbf{v}\propto e^{im\varphi }$ obeysAlso $\mathbf{B}=(J/2)(-y,x,0)$ in Cartesian coordinates, so $(\mathbf{v}\cdot \nabla )\mathbf{B}=(J/2)\hat{\mathbf{z}}\times \mathbf{v}$. Consequentlyand the basis-rotation terms cancel in the ideal magnetohydrodynamic induction equation:Here the printed $J$ is the coefficient in the specified magnetic field: $\nabla \times \mathbf{B}=J\hat{\mathbf{z}}$. If $J_{\mathrm{phys}}$ denotes SI electric current density, then $J={\mu }_0{J}_{\mathrm{phys}}$. The field and all subsequent printed coefficients are mutually consistent with this normalization.

For a normal mode with $\sigma \ne 0$, induction gives $\mathbf{b}=imJ\mathbf{u}/(2\sigma )$. Substitution into the momentum equation, including the Coriolis force, gives the reduced velocity equationThus the coefficients are the printed $\lambda$ and $\nu$. Let $C=J^2/({\mu }_0\rho )$. Inserting the allowed relation $\lambda =i\delta \nu$ and multiplying by $\sigma$ yieldsHence the uniform-current rotating magnetohydrodynamic wave dispersion isThe radicand is real. For $|m|\ge 2$, $2m\delta -m^2<0$ because $|\delta |<1$, so both roots are purely imaginary. For $m=0$ the roots of the quadratic are also imaginary; any zero-frequency case must be checked in the original equations, since the elimination divided by $\sigma$. A positive real part requiresFor integer $m$ this is possible only when $|m|=1$ and $m\delta >1/2$. Taking the conventional representative $m\ge 0$ gives the printed exception $m=1$, with $\delta >1/2$ and sufficiently large $J$. Literally allowing negative integers also gives $m=-1$, $\delta <-1/2$; complex conjugation maps $(m,\delta ,\sigma )$ to $(-m,-\delta ,\overline{\sigma })$, so these describe the conjugate real disturbance. Purely imaginary roots describe neutral oscillatory modes; a repeated root at threshold does not itself prove boundedness of arbitrary initial perturbations.

For an axisymmetric vertical mode of a shearing sheet with real $k\ne 0$, the two divergence constraints give $ik{v}_z=ik{b}_z=0$. The vertical momentum equation then yields $ikp=0$. Thus all three amplitudes vanish:Define the magnetic amplitudes $a_i=b_i/\sqrt{4\pi {\rho }_0}$ and the signed vertical Alfvén velocity $v_A=B_0/\sqrt{4\pi {\rho }_0}$. Write $A=k^2{v}_A^2$. The remaining equations for the normal mode areFor a growing or oscillatory mode with $\sigma \ne 0$, eliminating $a_x,a_y,\vartheta$ givesA nonzero velocity requires the determinant to vanish. After removing its factor $\sigma$, the dynamical dispersion relation isThis is the radially stratified magnetorotational dispersion relation, with $A=k^2{B}_0^2/(4\pi {\rho }_0)$. Keeping the original five-amplitude system instead gives characteristic polynomial $\sigma$ times this quartic. There is also a stationary balanced normal mode; division by $\sigma$ excludes it but loses no exponentially growing mode. At $k=0$ the divergence argument for vanishing vertical components does not apply, so that spatially uniform case must be treated separately.

For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

The potential-vorticity gradients in a meridional two-layer current are important here: besides ${\bar{q}}_{n,y}=\beta$, one has ${\bar{q}}_{1,x}=-FV$ and ${\bar{q}}_{2,x}=FV$. They must be retained when linearizing, even though the basic relative vorticity vanishes.

Let $K^2=k^2+l^2$ and $A=K^2+F$. For a normal mode the two-layer quasi-geostrophic potential vorticity amplitudes areBecause the basic velocities are northward $(V,0)$ in the two layers, their advection frequencies are $\sigma -lV$ and $\sigma$. The uniform forcing has no perturbation, $W^{\prime }=0$. The linear equations are thereforeIn particular, a perturbation's zonal velocity advects the basic interface-induced zonal potential-vorticity gradient. Substitution of the plane wave givesA nonzero disturbance exists exactly when the determinant vanishes. The requested relation for linear stability of a meridional two-layer current isEquivalently, with $\mathcal{B}=K^2(K^2+2F)$,Its discriminant isFor $K>0$, exponential baroclinic instability occurs when this discriminant is negative; otherwise the two frequencies are real. This also displays the stabilizing contribution of the planetary vorticity gradient when $k\ne 0$. As a check, $V=0$ gives the uncoupled barotropic mode and baroclinic mode frequencies $-\beta k/K^2$ and $-\beta k/(K^2+2F)$.

Put $S=M^2/f_0$. The thermal wind is $\bar{U}(z)=-Sz$, and the basic potential vorticity is spatially constant: $\bar{q}=2f_0$ when the full reference buoyancy is included, or $f_0$ when that reference contribution has been subtracted. Either way ${\nabla }_h\bar{q}=0$.

For a smooth normal mode, the linear quasi-geostrophic potential-vorticity equation becomesTake $k\ne 0$ and $K>0$, as required for a nontrivial wave with the printed phase-speed parameterization. Except possibly at an isolated critical level of a shear-flow wave, this implies $\hat{q}=0$; regularity then extends that condition through the isolated level. HenceDecay at infinity excludes the growing exponential, giving the decaying vertical structure of a semi-infinite Eady edge waveThe amplitude $C$ is arbitrary. The semi-infinite Eady model supports a wave trapped at its lower boundary with penetration depth $m^{-1}$. The zero-horizontal-wavenumber case has no nonzero decaying solution of this homogeneous vertical equation. Singular neutral interior potential vorticity sheets belong to a different continuous-spectrum class; they are not the smooth decaying edge-wave eigenfunction requested here.

The PDF contains the request for this vertical structure and uses the phase factor $e^{i(kx+ly-kct)}$. The TeX aid omits that request and also inserts a duplicate factor without $c$. The solution uses the original PDF form throughout.

For an exponentially growing normal mode with $k\ne 0$, $\mathrm{Im}(kc)>0$. Such a complex $c$ cannot equal the real basic velocity anywhere, so the interior equation forces $\hat{q}=0$ at every height. Decay then fixes the unique vertical structure $C{e}^{-mz}$, and the lower-boundary condition fixes $c=-S/m$, which is real. This contradiction proves the absence of exponential instability in the semi-infinite Eady model:For $k=0$, there is no advective frequency in the linear interior equation, so an exponentially growing mode would again have $q^{\prime }=0$. The wall buoyancy equation then forces ${\psi }_z^{\prime }(0)=0$, which the nonzero decaying exponential cannot satisfy. Thus these modes do not provide a missing growing branch. This conclusion rules out exponential eigenmode growth, not every possible transient response or singular neutral potential vorticity disturbance.

With a second rigid boundary at $z=D$, the same uniform-shear flow is the Eady model. Both boundaries carry buoyancy perturbations and support Boundary Rossby waves. Their intrinsic propagation directions oppose one another relative to their respective local basic flows. At suitable wavelengths their interacting fields can phase-lock and release basic-state potential energy, producing baroclinic instability. The single-boundary system lacks that second interacting boundary wave.

For a direct comparison, write $\Lambda ={\bar{U}}_z=-S$ and $m=NK/|f_0|$. The vertical solution between the two boundaries is $A\cosh (mz)+B\sinh (mz)$, and their buoyancy conditions areEliminating $A,B$ gives the finite-depth Eady dispersion relationThe bracket is negative for $0<mD<2.399\dots$, so those modes with $k\ne 0$ have a growing and a decaying member. At short wavelengths the two edge waves interact weakly and the frequencies are real. In the semi-infinite limit, the lower-wave root tends to $c=\Lambda /m=-S/m$, recovering the neutral result above.

For a normal mode write $x_j=A_j{e}^{i\omega t}$. The generalized eigenvalue problem isThe three independent eigenvectors and corresponding squared frequencies areThe last two have zero mass-weighted displacement and are orthogonal in the mass inner product. The lowest frequency is zero: it represents a rigid translation with both springs unchanged. Its general motion includes uniform centre-of-mass velocity, not just a constant displacement. Superpositions of the two oscillatory normal modes and this translation give the general solution.