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A critical level is a height at which the intrinsic frequency of an internal gravity wave vanishes. For finite positive buoyancy frequency, its local vertical wavenumber diverges as $\hat{\omega }\to 0$. The inviscid Taylor–Goldstein equation is singular there; viscosity, diffusion or nonlinear wave breaking can regularize the resulting small scales.

For positive background velocity $U=U_0{e}^{Mz}$ and buoyancy frequency $N=N_0{e}^{Mz}$, a stationary nonzero-horizontal-wavenumber Taylor–Goldstein equation reduces to the displayed constant-coefficient equation. Its structures are oscillatory, affine or exponential according to the sign of $N_0^2/U_0^2-M^2-k^2$. The disturbance Laplacian is a constant times its streamfunction, but its buoyancy anomaly is $N_0^2{e}^{Mz}\varphi /U_0$. Therefore vorticity self-advection vanishes while buoyancy self-advection produces a nonzero second harmonic for $M\ne 0$. These nontrivial real linear modes are not exact finite-amplitude solutions without additional fields. For $M=0$, both polarizations are constant and these nonlinear self-interactions vanish.

The zero-phase velocity ansatz $w=(\mathrm{sech}z{)}^k(\tanh z{)}^{1-k}$ solves the Taylor–Goldstein equation away from its critical level of an internal gravity wave exactly when $J=k(1-k)$. Let $y=\tanh z$ and $a=1-k$. Then $w^{\prime }/w=a/y-y$ and $w^{\prime \prime }/w=a(a-1)/y^2-a-1+2y^2$, so the differential-equation residual divided by $w$ is $[J-k(1-k)](y^{-2}-1)$. The curve has maximum $1/4$ at $k=1/2$. The endpoint $k=0$ does not decay at infinity; fractional powers at the critical level require a branch and regularity qualification.

After removing the hydrostatic background pressure, the two-dimensional ideal Boussinesq equations arewith the buoyancy equation ${\sigma }_t+u{\sigma }_x+w{\sigma }_z+N^2(z)w=0$. The last equation follows by materially advecting total buoyancy $B_0(z)+\sigma$ with $B_0^{\prime }(z)=N^2(z)$. Take the streamfunction conventionso the $y$-component of vorticity is $q=u_z-w_x={\nabla }^2\psi$. Taking ${\partial }_z$ of horizontal momentum minus ${\partial }_x$ of vertical momentum cancels pressure. Incompressibility combines the remaining nonlinear terms into advection of $q$, giving $q_t+u{q}_x+w{q}_z=-{\sigma }_x$. Hence the full equations areThere is no vortex stretching in this genuinely two-dimensional velocity field.

Write $\psi ={\psi }_0(z)+\varphi$, with ${\psi }_0^{\prime }=\overline{u}=U(z)$ and background buoyancy anomaly zero. The Linearized Boussinesq equations becomeFor a stationary disturbance $\varphi =\hat{\psi }(z)e^{ikx}$ with real $k\ne 0$ and $U>0$, the second equation gives $\hat{\sigma }=(N^2/U)\hat{\psi }$. The first therefore reduces to the stationary Taylor–Goldstein equationThe restriction $k\ne 0$ is used when cancelling $ik$; a purely horizontally uniform alteration of the background is not this wave problem.

For the exponential-profile stationary stratified waves, defineSince $U^{\prime \prime }/U=M^2$ and $N/U=N_0/U_0$, all coefficients are constant. The complete vertical solution set isShifting the origin of $z$ simply changes the two constants. Vertical oscillation requires $N_0/U_0>\sqrt{M^2+k^2}$; equality gives an affine structure; below that value the structure is evanescent or growing, with admissibility determined by the physical boundaries.

The Rayleigh quotient with one Neumann endpoint must retain its upper boundary term. On a finite interval with $D=z_2-z_1>0$, choose a nonzero real vertical structure, so the denominator is strictly positive. Multiplication of the differential equation by $\hat{\psi }$ and integration by parts givesWith only ${\hat{\psi }}^{\prime }(z_1)=0$, it equals $k^2$ precisely when $\hat{\psi }(z_2){\hat{\psi }}^{\prime }(z_2)=0$. A lower Neumann boundary condition alone does not ensure this.

For $s>0$, the lower condition selects $\hat{\psi }=A\cos [\sqrt{s}(z-z_1)]$. Equality of the quotient requires $\sqrt{s}D=n\pi /2$ for a positive integer $n$: even $n$ gives an upper Neumann condition, odd $n$ an upper Dirichlet boundary condition. Thuswhich additionally restricts $N_0^2/U_0^2>M^2+n^2{\pi }^2/(4D^2)$. If $s=0$, the lower condition leaves a constant structure; it has quotient $C=k^2$, possible for a genuine wave only when $C>0$. If $s<0$, the lower condition leaves $A\cosh [\sqrt{-s}(z-z_1)]$, whose upper product is positive and whose quotient is strictly less than $k^2$. On an infinite interval, none of these nonzero lower-Neumann structures is square-integrable, so the displayed ratio of integrals is not a defined finite-integral quotient there. Ratios of truncated integrals would be a different limiting prescription.

For every mode above, including the affine and exponential cases, the disturbance satisfiesThis statement concerns the disturbance, not the sum with the background streamfunction; the latter generally has a different Laplacian eigenvalue. Its buoyancy polarization isThus no nonzero mode has a constant buoyancy-to-streamfunction ratio when $M\ne 0$.

The nonlinear exactness test for a stratified streamfunction mode now distinguishes the two equations. The disturbance's vorticity self-advection vanishes because its Laplacian is proportional to itself. But its buoyancy self-advection isFor a nonzero real sinusoidal disturbance and $M>0$, this is a nonzero second harmonic somewhere. Consequently none of the nontrivial modes with the stipulated positive $M$ is by itself an exact finite-amplitude solution of both nonlinear equations. Additional harmonics or mean corrections would be needed. In the limiting case $M=0$, $U,N,\alpha$ are constant; both self-interactions vanish, and these same single-wavenumber structures, or their superpositions with the same $C$, do solve the full ideal equations wherever their boundary conditions are admissible.

Use the streamfunction convention $u={\psi }_z$, $w=-{\psi }_x$, and take $v=0$. With $J(A,B)=A_x{B}_z-A_z{B}_x$, the material derivative is $D={\partial }_t-J(\psi ,\cdot )$. This convention differs from the horizontal quasi-geostrophic streamfunction convention used below. The nonlinear Boussinesq equations areThe $y$ component of vorticity is $\zeta =u_z-w_x={\nabla }^2\psi$. Differentiate the first momentum equation in $z$ and subtract the $x$ derivative of the second. The pressure terms cancel; the derivatives of the advecting velocity cancel by $u_x+w_z=0$. Thus $D\zeta =-{\sigma }_x$, orConservation of total buoyancy, with background gradient $N^2(z)$, givesBoth equations retain the nonlinear advection terms.

Write $\psi =\bar{\psi }(z)+\varphi$, with ${\bar{\psi }}_z=U(z)=\bar{u}(z)$, and take the background buoyancy anomaly to be zero. The linear equations for $(\varphi ,\sigma )$ areFor a normal mode $\varphi =\hat{\psi }(z)e^{ik(x-ct)}$, the second equation gives $\hat{\sigma }=N^2\hat{\psi }/(U-c)$. Substitute this into the first to obtain the Taylor–Goldstein equationHere $k\ne 0$ and the calculation is initially away from critical levels $U=c$. The first term is the restoring effect of stable buoyancy on vertically displaced parcels. The second comes from advection of the background vorticity gradient $U^{\prime \prime }$: a displacement creates a vorticity anomaly, whose induced velocity acts back on the displacement. This is the shear-flow counterpart of a Rossby wave restoring mechanism; its sign can also permit instability, so it is not invariably stabilizing.

Rigid boundaries imply $w=-{\varphi }_x=0$, hence $\hat{\psi }(0)=\hat{\psi }(\pi H)=0$. For a real, regular normal mode, multiply the Taylor–Goldstein equation by $\hat{\psi }$, integrate, and integrate its second derivative by parts. The endpoint term vanishes, leavingFor a smooth real dispersion branch, differentiate this identity with respect to $k$. The stated stationarity of $I$ removes the term involving $d\hat{\psi }/dk$, so $2k=I_{c}dc/dk$. Direct differentiation at fixed $\hat{\psi }$ givesSince the frequency is $kc$, the group velocity is $d(kc)/dk$, and thereforeThis is variational group velocity of a Taylor-Goldstein mode. Singular critical levels or a vanishing denominator require separate limiting analysis; no stationarity proof is needed here.

For the sinusoidal profiles, choose $c=0$. In the interior,The profile ratios extend to the endpoints in this reduced equation. Thus $n=1$ always gives a nonzero real $k$, with $k_1^2=N_0^2/U_0^2$. For this normal mode,The denominator in the group velocity expression is $2H(2N_0^2/U_0^3+1/(U_0{H}^2))$. It follows thatAlthough $U$ vanishes at the endpoints, this weighted integral is finite; the vanishing of ${\hat{\psi }}_1$ supplies the required cancellation. The formula is the limiting derivative approached from the regular real branch with $c<0$, before any interior critical level is encountered.

If instead $N=N_0$ and $U=U_0$ are constant, still take $c=0$. The same vertical eigenfunctions satisfy the Taylor–Goldstein equation, now withThere is no value of $n\ge 1$ guaranteed to give real $k$ for arbitrary positive parameters: if $N_{0}H<U_0$, all these squared wavenumbers are negative. A nonzero real $k$ exists exactly when $N_{0}H>U_0$; equality gives only the degenerate $n=1$, $k=0$ limit. For a real branch the same group velocity calculation gives $c_g=k_n^2{U}_0^3/N_0^2$.

To test exact nonlinear validity, take a real normal mode $\varphi =A\sin (nz/H)\cos (kx)$ and its corresponding buoyancy perturbation $\sigma =a(z)\varphi$, where $a=N^2/(U-c)$. Becausethe quadratic vorticity self-advection is zero: $J(\varphi ,{\nabla }^2\varphi )=0$. The quadratic buoyancy self-advection, however, isThis nonlinear exactness test for a stratified streamfunction mode distinguishes the cases. For constant $N,U$, $a=N_0^2/U_0$ is constant, so both quadratic terms vanish. The background-mode terms already satisfy the linear equations, hence the constant-profile modes are exact finite-amplitude solutions of the ideal Boussinesq equations. Pressure can be recovered from the momentum equations: the vorticity equation supplies their compatibility condition in this simply connected strip.

For the sinusoidal profiles, $a=(N_0^2/U_0)\sin (z/H)$ has a nonzero derivative in the interior. For every nonzero $A$ and nonzero real $k$, $a^{\prime }\varphi {\varphi }_x$ is not identically zero; it produces a second horizontal harmonic that is absent from the proposed solution. Thus none of these nontrivial sinusoidal-profile modes is an exact finite-amplitude solution. A $k=0$ field is merely an added horizontal shear with no vertical motion and is a degenerate exception to the wave calculation, not a propagating member of it.

Let $V=U-c$ and suppose a growing normal mode has $\mathrm{Im}c=c_i>0$, taking $\alpha >0$ without loss of generality. Since $V$ never vanishes on the real channel, choose a continuous square-root branch and write $\varphi =V^{1/2}H$. Direct differentiation givesSubstitution into the Taylor–Goldstein equation and division by $V^{1/2}$ givesThus the required transformation isThe impermeable walls give $\varphi (\pm 1)=0$, hence $H(\pm 1)=0$. Multiply by $\overline{H}$, integrate across the channel, and use integration by parts. With the boundary term zero, the resulting power-transformed Taylor–Goldstein energy identity isBoth $U^{\prime \prime }$ and $J$ are real. Since $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part, divided by $-c_i$, isIf $J\ge U^{\prime 2}/4$ everywhere, the integrand is nonnegative and the first two terms have positive integral for any nonzero mode. This contradicts the identity. Therefore instability requiresThis is the necessary instability condition of the Miles–Howard theorem. Where $U^{\prime }\ne 0$, it says that the gradient Richardson number $J/U^{\prime 2}$ must be below $1/4$ somewhere. It is not a sufficient condition for instability, and the conclusion concerns exponential normal-mode growth, not exclusion of the algebraic lift-up effect in part (a). The algebra uses the given nondimensional $J$; in dimensional notation its counterpart is the squared buoyancy frequency.

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.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

Take a common constant buoyancy frequency $N$ and put $k=k_T>0$. Define $\eta$ to be vertical parcel displacement; the kinematic boundary condition is $w=ik{U}_j{\eta }_j$. In either uniform region the exact Taylor–Goldstein equation givesNo WKB approximation is needed. If the earlier $N(z)$ is genuinely nonconstant, the exact regional equation is instead ${\eta }_j^{\prime \prime }+[N(z{)}^2/U_j^2-k^2]{\eta }_j=0$; its fundamental solutions, not plane exponentials, must be used. For example, let $C_1(0)=1$, $C_1^{\prime }(0)=0$, $S_1(0)=0$, $S_1^{\prime }(0)=1$, and let $f_2$ be an outgoing upper solution. Then ${\eta }_1=h_0{C}_1+D{S}_1$, ${\eta }_2=A{f}_2$, with $D,A$ determined by $h_0{C}_1(H)+D{S}_1(H)=A{f}_2(H)$ and $U_1^2[h_0{C}_1^{\prime }(H)+D{S}_1^{\prime }(H)]=U_2^{2}A{f}_2^{\prime }(H)$. This gives the exact arbitrary-profile field implicitly, while a simple angle-only coefficient requires further stratification assumptions.

Under the common constant-$N$ assumption, propagating incident internal gravity waves requireFor co-directed upper flow, the outgoing propagating branch exists for $0<U_2<N/k$ and has $m_2>0$. The radiation condition chooses $m_j>0$ because the laboratory vertical group velocity is positive on the negative intrinsic frequency branch. Define $\tan {\theta }_j=m_j/k$, so $\cos {\theta }_j=k{U}_j/N$.

At the velocity jump, the material-interface jump conditions for stratified inviscid shear flow require continuity of displacement and of fluid pressure. Vertical velocity itself is generally discontinuous: $w_j=ik{U}_j\eta$. In terms of displacement the pressure is $p_j=U_j^2{\eta }_j^{\prime }$. The displacement impedance for an internal wave at a velocity jump is $Z_j=U_j^2{m}_j$. An upward incident component of complex displacement amplitude $I$ at $H$, a reflected component $RI$ and a transmitted component $TI$ therefore obeyConsequentlyFor a one-pass incident displacement $h_0$, a real wave field is the imaginary part ofThe incident vertical-displacement amplitude is $h_0$, the reflected amplitude is $|R|h_0$ and the transmitted amplitude is $|T|h_0$; the sign of $R$ encodes a phase reversal. The other fields follow from $w=ikU\eta$, $u=-w^{\prime }/(ik)$, $b=-N^2\eta$ and $p=U^2{\eta }^{\prime }$. If a perfectly reflecting terrain boundary is enforced after the return wave arrives, replace the incident coefficient $h_0$ by $h_0/[1+R{e}^{2i{m}_{1}H}]$. The one-pass amplitudes use the permitted neglect of repeated terrain reflections.

For common $N$, $Z_j=(N^2/k)\cos {\theta }_j\sin {\theta }_j$. The physical vertical-displacement transmission coefficient is thereforeThe printed coefficient has a different denominator and cannot be obtained for this standard displacement convention. For example, when $U_2=U_1$, there is no interface scattering and the vertical amplitude stays $h_0$, whereas the printed expression gives $h_0/\cos {\theta }_1$. A total displacement magnitude would instead be the vertical amplitude divided by $\cos {\theta }_2$; with unequal velocities it still does not give the printed expression. The transmission formula needs an explicit correction or a different, presently unspecified amplitude/interface model.

For $U_2>N/k$, choose $m_2=i\kappa$ with $\kappa =(k^2-N^2/U_2^2{)}^{1/2}$: the transmitted response decays as $e^{-\kappa (z-H)}$ and the same impedance equations give a complex $R,T$, with $|R|=1$. At $U_2=0$ the steady equation is singular. Oppositely directed upper flow formally supports propagating waves when $0<|U_2|<N/k$, but upward radiation then requires $m_2<0$ and signed $Z_2$; the positive-angle co-directed formula is not valid there. A real velocity jump can also have Kelvin-Helmholtz instability, which is excluded from this formal steady scattering calculation.

Use the Boussinesq approximation with a constant inertial reference mass density ${\rho }_0$, keeping the background mass density gradient in buoyancy. Write $V(z)=\overline{U}(z)-c$ and $P=\hat{p}/{\rho }_0$. For $k\ne 0$, the linearized normal mode equations areThe reference mass density is chosen close to the background values, whose relative variation must be small. Hydrostatic background pressure satisfies ${\overline{p}}^{\prime }=-g\overline{\rho }$.

From incompressibility, $\hat{u}=i{\hat{w}}^{\prime }/k$. The horizontal momentum equation and the density equation then giveSubstitute these into the vertical momentum equation and multiply by $ik$:For $V\ne 0$, rearrangement gives the Taylor–Goldstein equationThe buoyancy frequency $N$ is real for stable stratification, where ${\overline{\rho }}^{\prime }<0$. Points where $\overline{U}=c$ are critical levels of an internal gravity wave; the derivation there is understood through a limiting or piecewise formulation. The printed density gradient is ${\overline{\rho }}^{\prime }$, not the pressure gradient accidentally substituted in the local TeX.

Let $\hat{\eta }$ be the displacement of a material interface and use $[f{]}_{-}^{+}=f_{+}-f_{-}$. On each side, the kinematic boundary condition isThere is one displaced material interface, so its displacement is continuous. Thus the first jump condition isIn particular, vertical velocity need not be continuous when the background velocity jumps.

With no surface tension, pressure continuity applies on the displaced interface, rather than at its undisplaced height. Linearizing gives $[\hat{p}+\hat{\eta }{\overline{p}}^{\prime }{]}_{-}^{+}=0$. Since ${\overline{p}}^{\prime }=-g\overline{\rho }$, this becomes $[\hat{p}-g\overline{\rho }\hat{\eta }{]}_{-}^{+}=0$. The horizontal momentum calculation in part (a) givesSubstitute this and $\hat{\eta }=\hat{w}/[ik(\overline{U}-c)]$ to obtain the second jump condition:These jump conditions for stratified inviscid shear flow use one-sided values of ${\overline{U}}^{\prime }$ and apply to jumps of mass density, vorticity, or velocity. If $\overline{U}$ is continuous, the first condition reduces to continuity of $\hat{w}$; a jump of vorticity generally still prevents continuity of ${\hat{w}}^{\prime }$. This derivation avoids multiplying singular distributional derivatives of a discontinuous velocity in the Taylor–Goldstein equation.

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.

Set $y=\tanh z$ and $a=1-k$, so $dy/dz=1-y^2$ and $\hat{w}=(1-y^2{)}^{k/2}{y}^a$. Away from $y=0$, the logarithmic derivative givesFor $c=0$, ${\overline{U}}^{\prime \prime }/\overline{U}=-2(1-y^2)$ and $N^2/{\overline{U}}^2=J(1-y^2)/y^2$. The Taylor–Goldstein equation residual divided by $\hat{w}$ consequently simplifies toThus the neutral mode of the equal-width Hazel model requiresThe qualifier “solution” needs care at the critical level of an internal gravity wave $z=0$. For $0<k<1$, $\hat{w}\sim z^{1-k}$ and ${\hat{w}}^{\prime }\sim (1-k)z^{-k}$: it is a solution separately on either side, not a classical smooth eigenfunction across the critical level of an internal gravity wave. A branch for negative $\tanh z$ is also needed. For example, a limit from $c_i>0$ assigns $(\tanh z{)}^{1-k}=|\tanh z{|}^{1-k}{e}^{-i\pi (1-k)}$ on $z<0$.