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For the proposed plane wave, define the positive squared effective wavenumberThe horizontal Laplacian contributes $-(k^2+l^2){\psi }^{\prime }$ and the vertical stretching operator contributes $-(f_0^2/N^2)m^2{\psi }^{\prime }$. Substitution in the Linear Rossby-wave equation therefore givesThe intrinsic phase velocity is westward for positive $\beta$. A stationary disturbance with $\omega =0$ and $k\ne 0$ must have its westward intrinsic phase velocity exactly cancelled by the eastward background current. Solving for the squared vertical wavenumber yieldsFor Earth, with northward $y$, $\beta =2\Omega \cos \varphi /a>0$ in both hemispheres away from the poles. Thus an oscillatory vertical structure requiresThe background flow must be eastward but not too strong for the chosen horizontal scale. Long horizontal wavelengths have the largest allowed speed. An easterly current, $U<0$, gives $m^2<0$ and an evanescent wave; so does an eastward current exceeding the upper bound. At equality $m=0$ and the vertical group velocity vanishes. A zonally uniform disturbance, $k=0$, is a degenerate stationary pattern with no vertical Rossby wave energy propagation, so it cannot be obtained by cancelling $k$ in the preceding calculation.

A real $m$ alone does not select upward propagation. Differentiate the dispersion relation at fixed $k,l$ to obtain the vertical group velocity:The outgoing radiation condition for a surface source requiresFor the convention $k>0$, take $m>0$. The stationary phase lines then obey $dx/dz=-m/k<0$: they tilt westward with height while wave energy travels upward. The real conjugate component $(-k,-l,-m)$ describes the same physical upward-propagating wave.

The figure suggests a winter extratropical path from the lower terrestrial atmosphere into the stratosphere, where the mean zonal flow is eastward. The summer stratosphere has westward flow and blocks such stationary Rossby waves in this idealization. A zero-wind surface is a stationary-wave critical layer in a shear flow; it also bounds the winter westerlies aloft near the upper part of the diagram. For a slowly varying real wind profile, local propagation needs $m^2(z)>0$ along the path, sufficient surface forcing, and no intervening absorbing critical layer in a shear flow. A layer where the eastward wind becomes too strong is a turning barrier for the given horizontal wavenumbers. The uniform-current result is a local guide, not a complete prediction through a sheared spherical terrestrial atmosphere.

For scale, at $60^{\circ }$ latitude $\beta \simeq 1.14\times 10^{-11}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$. A planetary zonal wavenumber $s$ has local $k=s/(a\cos \varphi )$. Taking $l=0$ gives the most permissive limitThe resulting limits are approximately $116$, $29$ and $13\mathrm{m}{\mathrm{s}}^{-1}$ for $s=1,2,3$. Therefore the long planetary waves are favored; the strong core of the winter vortex can exclude shorter ones even though it is westerly. Nonzero meridional wavenumber lowers these limits further. A compressible log-pressure coordinate model would also contain a density-scale-height correction; it is absent from the equation supplied here and should not be silently inserted.

Interchanging the Northern and Southern Hemispheres changes the sign of $f_0$, but only $f_0^2$ enters the dispersion relation and vertical group velocity. With the same eastward $x$ and northward $y$, $\beta$ stays positive. Thus the propagation criterion is the same in both hemispheres: it selects the winter westerlies, irrespective of which hemisphere is winter. Actual amplitudes can differ because mountains, land-sea contrasts, forcing spectra and mean winds differ between hemispheres; those source differences do not reverse this propagation criterion.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

Let $\pi$ be pressure divided by the reference density and $b$ the buoyancy perturbation. The linear, inviscid Boussinesq equations about $U(z)$ areFor a stationary Fourier component $e^{ikx}$, incompressibility gives $u=i{w}^{\prime }/k$. The horizontal momentum equation then gives $\pi =i(U^{\prime }w-U{w}^{\prime })/k$, while buoyancy conservation gives $b=i{N}^{2}w/(kU)$. Substitute these into the vertical momentum equation to obtain the stationary Taylor–Goldstein equation:Hence its local WKB approximation isFor $U^{\prime \prime }$ negligible, an outgoing upward wave has $mU/k>0$. In particular $m>0$ for $k,U>0$. In an upper evanescent region choose the branch with $\mathrm{Im}m>0$ in $e^{i\int mdz}$ so the solution decays upward. These choices express the radiation condition; a reflected wave has the opposite real branch and downward group velocity.

There are two different limiting levels. A critical layer in a shear flow occurs at $U=0$, where the intrinsic frequency vanishes, $|m|\to \infty$, and upward propagation slows while the vertical wavelength contracts. Dissipation or nonlinear effects can absorb the wave and transfer its momentum to the mean flow. For a stable, slowly varying critical layer with Richardson number $\mathrm{Ri}=N_0^2/(U^{\prime }{)}^2>1/4$, the usual causal inviscid continuation suppresses transmission; it is not the same as reflection at $m=0$. When the Richardson number is small, a general absorption-only conclusion needs extra stability assumptions.

Reflection instead occurs at the turning level of a stationary internal wave $k|U|=N_0$, beyond which $m^2<0$. It is sometimes called a critical reflection height, but the wind there is nonzero. Assuming $0<k{U}_0<N_0$, the specified profiles giveThe decreasing-wind profile never encounters $m=0$ before its zero-wind layer. Both profiles have $U^{\prime \prime }=0$; a well-separated WKB description requires $N_{0}H/U_0\gg 1$, and a turning-layer transition is still required for the increasing profile.

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.

A vertically propagating stationary topographic internal gravity wave meets a turning level when its squared vertical wavenumber changes from positive to negative. The stationary internal-wave WKB solution then fails locally. A simple zero is resolved by an Airy function transition connecting incident and reflected waves below to a decaying evanescent wave above. For negligible wind curvature the condition is $|kU(z_t)|=N(z_t)$, distinct from the zero-intrinsic-frequency critical layer in a shear flow.