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For a slowly varying parallel shear flow $U(y){\mathbf{e}}_x$, the leading WKB method neglects mean-velocity gradients while retaining local advection. The continuity equation and linearized Euler equations then give the displayed acoustic dispersion relation. On the positive intrinsic-frequency branch, $\omega =U{k}_x+c_0|\mathbf{k}|$ and the group velocity is $U{\mathbf{e}}_x+c_0\mathbf{k}/|\mathbf{k}|$. A wavefront normal is parallel to $\mathbf{k}$; the acoustic energy-ray direction generally differs because of the advection term.

For an incompressible shear flow $U(z){\mathbf{e}}_x$ with uniform magnetic field $B{\mathbf{e}}_x$, define $S=\sigma +k_{x}U$ for time dependence $e^{i\sigma t}$ and $\zeta =i{k}_{x}v-i{k}_{y}u$. Eliminating the magnetic perturbations from the horizontal momentum equations gives the displayed product for $S\ne 0$. Away from the Alfvén wave resonance, it yields $\zeta =k_y{U}^{\prime }w/S$. At resonance one must retain the original system: for constant $U$, $k_y=0$, $S^2=k_x^2{B}^2/\rho$, any smooth transverse amplitude $v(z)$ with $b_y=k_{x}Bv/S$ and all other amplitudes zero is an Alfvén wave. Its nonzero vorticity disproves the unrestricted zero-vorticity reduction. Exponentially growing modes with real background coefficients cannot encounter these real-frequency resonances.

For nonresonant incompressible normal modes of a uniform-density shear flow with aligned uniform magnetic field, let $S=\sigma +k_{x}U(z)$, $k^2=k_x^2+k_y^2$ and $f=w/S$. The normal displacement is $f/i$. The linearized horizontal equations and incompressibility give $i{k}^2{p}_1=\rho (S{w}^{\prime }-S^{\prime }w)=\rho S^2{f}^{\prime }$. Substituting the ideal magnetohydrodynamic induction equation into normal momentum gives the displayed conservative equation. Across a zero-thickness shear layer, continuity of $f$ and of $(\rho S^2-k_x^2{B}^2)f^{\prime }$ supplies the interface conditions. They express common normal displacement and continuous perturbed magnetohydrodynamic total pressure, avoiding undefined products of distributional shear gradients.

For a stratified parallel shear flow, the gradient Richardson number compares the squared buoyancy frequency with squared vertical shear. The Miles–Howard theorem excludes exponentially growing inviscid normal modes when this ratio is at least $1/4$ everywhere. Where $U^{\prime }=0$, the equivalent criterion is written directly as $N^2\ge (U^{\prime }{)}^2/4$.

The Hazel model is a smooth stratified shear flow whose base velocity has a hyperbolic-tangent profile and whose squared buoyancy frequency has a hyperbolic-secant-squared profile. The equal-width nondimensional example is $U=\tanh z$ and $N^2=J{\mathrm{sech}}^{2}z$. For $J\ge 0$, its gradient Richardson number is $J{\cosh }^{2}z$, with minimum $J$.

For a passive scalar in a planar shear flow $u(y,t)$, let $m_j(y,t)={\int }_{\mathbb{R}}{x}^j\chi (x,y,t)dx$. Integrating the advection-diffusion equation by parts in $x$ gives the displayed hierarchy for $j\ge 0$, with negative-index terms omitted. Sufficient decay and finite moments justify the integrations. Insulating walls give Neumann boundary conditions for every moment. Solving the first two equations then determines the long-time growth of the second moment without solving the full scalar field.

The energy of the residual flow and magnetic field in a maintained shear flow has source $S=-a{T}_{xy}$, where $T_{xy}=\rho v_x{v}_y-B_x{B}_y/{\mu }_0$. The kinetic and magnetic stresses can exchange energy with the imposed background. The source vanishes when their off-diagonal contributions cancel.

The Orr-Sommerfeld equation governs the wall-normal velocity of normal mode disturbances to an incompressible viscous parallel shear flow. Here $D=d/dy$, ${\kappa }^2={\alpha }^2+{\beta }^2$, and $Re$ is the Reynolds number. It follows by eliminating pressure from the linearized Navier-Stokes equations. At a rigid no-slip wall, $\hat{v}=D\hat{v}=0$. In the inviscid limit it reduces to the Rayleigh equation for inviscid shear flow.

Dense water formed in marginal seas spills over sills and descends the continental slope as a marginal-sea overflow. Strong shear flow generates turbulent mixing and entrainment of surrounding, lighter water. The descending current's volume flux grows while its density anomaly is diluted, modifying the depth at which it joins the deep circulation and the temperature and salinity of the exported water. Thus the deep volume flux can substantially exceed the source overflow volume flux; it cannot be inferred from northern surface cooling alone.^[ref]

Take $\kappa >0$ and initially $\omega \ne 0$. Write $m_j(y,t)={\int }_{\mathbb{R}}{x}^j\chi (x,y,t)dx$, and set $k=\pi /L$, $a=\kappa k^2$. The advection-diffusion equation is ${\chi }_t+u{\chi }_x=\kappa ({\chi }_{xx}+{\chi }_{yy})$. Integration by parts in the unbounded coordinate gives the longitudinal moment hierarchy for shear transportAll three moments obey Neumann boundary conditions at the walls. The initial values are $m_0=L$, $m_1=0$, $m_2=L^3/12$. These manipulations require finite moments and vanishing boundary terms at $x=\pm \infty$; diffusion of the compact initial distribution supplies the required decay for positive time.

The zeroth moment remains exactly $L$. Since $\sin (ky)$ is a Neumann eigenfunction, put $m_1=Lg(t)\sin (ky)$. Then $g^{\prime }+ag=U\cos (\omega t)$ and $g(0)=0$, givingFor the second moment, $2u{m}_1=LUg\cos (\omega t)[1-\cos (2ky)]$. Thus $m_2=A(t)+B(t)\cos (2ky)$, withThe equation for $B$ has positive damping and bounded forcing, so $B$ remains bounded. In the equation for $A$, the decaying term in $g$ contributes only a bounded accumulated transient. Over a complete nonzero-frequency cycle, $⟨{\cos }^2(\omega t){⟩}_t=1/2$ and $⟨\sin (\omega t)\cos (\omega t){⟩}_t=0$. ConsequentlyThis proves oscillating shear dispersion in an insulating channel, including the local-in-$y$ leading term, not just a cross-channel average. The instantaneous derivative retains bounded periodic oscillations; its cycle-averaged slope is $2K{m}_0$. Thus $2Kt{m}_0$ is the leading second moment, rather than a literal rate proportional to $t$.

The asymptotic statement holds for fixed positive diffusivity and fixed nonzero frequency, after times long compared with $a^{-1}$ and the oscillation period. If $|\omega |L^2/\kappa \ll 1$, transverse diffusion adjusts rapidly to the changing shear flow, andThis is the cycle average of the instantaneous steady-shear enhancement $U^2{\cos }^2(\omega t)/(2a)$. At exactly $\omega =0$, there is no cycle average: $g\to U/a$ and $K=\kappa +U^2/(2a)$. Hence the zero-frequency and long-time limits do not commute. In the intermediate range $a^{-1}\ll t\ll |\omega {|}^{-1}$, the shear is approximately steady at its initial amplitude and that latter enhancement is appropriate.

Write $a=\Lambda T$. The deformation gradient of one interval of shear flow is $S(a)=\left(\begin{array}{cc}1& a\\ 0& 1\end{array}\right)$. An initial unit material line at angle $\theta$ therefore becomes $(\cos \theta +a\sin \theta ,\sin \theta )$. Its orientation, including the correct quadrant, isFor an unoriented line the angle is taken modulo $\pi$. If $\Lambda =0$, there is no shear flow: every ${\lambda }_N=1$ and $\mu =0$. The fluctuation and maximizing calculations below concern nonzero shear.

To quantify the renovating shear flow, use the intended isotropic model: each shear orientation is independent and uniform modulo $\pi$. Independence by itself does not specify a probability law and cannot fix the answer. Under the uniform law, conditional on every earlier deformation, the relative angle to the next shear is again uniform. Thus the logarithmic increments ${\xi }_n=\log \ell ({\theta }_n)$ are independent and identically distributed, even though the absolute orientation of the stretched material line need not be uniform. Since $\log {\lambda }_N={\sum }_{n=1}^N{\xi }_n$, only the one-step angular mean is needed.

Write ${\ell }^2=A+B\cos (2\theta -\delta )$, where $A=1+a^2/2$, $B=|a|\sqrt{1+a^2/4}$ and $A^2-B^2=1$. The supplied angular logarithmic integral, with parameter $B/A$, givesTherefore the mean logarithmic stretching rate isThis is the mean logarithmic stretching in an isotropic renovating shear, and the strong law of large numbers gives the same almost-sure Lyapunov exponent. Taking $\Lambda >0$ without loss for this statistic puts it in the requested form $\Lambda F(\Lambda T)$.

For $b=|\Lambda |T$, $F(b)=\log (1+b^2/4)/(2b)$. At small $b$, $\mu ={\Lambda }^{2}T/8+O({\Lambda }^4{T}^3)$; at large $b$, $\mu \sim \log (b/2)/T$, tending to zero when $T$ increases at fixed shear rate. The maximizing nonzero $b$ solvesThe difference $2(1-z^{-1})-\log z$ rises from zero up to $z=2$ and then strictly decreases to $-\infty$, so there is exactly one root beyond $2$. Numerically,Very rapid renovation gives little deformation per interval and cancels the first-order mean stretch. Long intervals provide strong deformation but progressively align a material line with the shear direction, so one interval stretches it only algebraically in $T$. Intermediate persistence combines substantial deformation with frequent reorientation, producing the finite maximum.

For the distribution, let $\gamma =\mathrm{arsinh}(|a|/2)$, so the singular stretches of $S(a)$ are $e^{\pm \gamma }$. In the right-singular-vector basis, ${\ell }^2=e^{2\gamma }{\cos }^2\phi +e^{-2\gamma }{\sin }^2\phi$. Put $q=\tanh \gamma =|a|/\sqrt{a^2+4}$. Factoring this expression and expanding its logarithm givesThe variance follows from angular orthogonality. These bounded independent increments implyAccordingly the central part of the large-$N$ stretch distribution is approximately described by a log-normal distribution, with density $[\lambda \sqrt{2\pi Nv}{]}^{-1}\exp [-(\log \lambda -Nr{)}^2/(2Nv)]$. This central limit theorem is not an assertion about arbitrarily extreme tails: the exact stretch has support inside $e^{-N\gamma }\le {\lambda }_N\le e^{N\gamma }$.

For $|a|\ll 1$, the log-stretch fluctuations in an isotropic renovating shear haveTo leading order the mean and variance of $\log {\lambda }_N$ both equal $N{a}^2/8$. Typical stretch is $\exp (N{a}^2/8)$, while its mean grows as $\exp (3N{a}^2/16)$. These moment statements can be justified directly rather than extrapolating the central Gaussian approximation: expansion of ${\ell }^p$ givesfor fixed $p$. In particular $⟨{\lambda }_N^2⟩=(1+a^2/2{)}^N$ exactly. The leading small-$a$ log-normal approximation is especially precise in the many-small-renovations scaling with $N{a}^2$ fixed and $N{a}^4\to 0$.

If the orientation law is not isotropic these formulas need not hold. For example, repeating a fixed shear direction gives $S(a{)}^N=S(Na)$, only algebraic stretch, and zero asymptotic logarithmic rate. The associated degenerate direction variables are independent, illustrating why the stated independence condition alone is insufficient to select the isotropic result.

Let $p$ be kinematic pressure, with the resting hydrostatic part subtracted, and let $\sigma$ be the buoyancy perturbation about background buoyancy $N^{2}z$. The nonrotating inviscid Boussinesq equations, with no buoyancy diffusion, areHere the third equation is mass conservation for incompressible flow, and the last expresses buoyancy conservation following a parcel. Linearization about rest removes the products of disturbances. For a plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$, the resulting Linearized Boussinesq equations becomeWith $K=(k^2+m^2{)}^{1/2}$ and $k\ne 0$, the first and third equations give $\hat{p}=-m\omega \hat{w}/k^2$. Substitution in vertical momentum and buoyancy gives the dispersion relationSince the velocity is the time derivative of the displacement field, $\hat{\mathbf{u}}=-i\omega \hat{\mathbf{\xi }}$. Thus the complete buoyancy polarization of a plane internal gravity wave isBoth velocity and displacement are perpendicular to the wavevector; pressure and velocity are in quadrature with buoyancy and displacement.

For upward/rightward phase velocity, choose $k,m>0$ and $\omega =Nk/K$. Differentiating this dispersion relation gives the group velocity, while the vector phase velocity points along the wavevector:Their dot product is zero and ${\mathbf{c}}_p+{\mathbf{c}}_g=(N/K,0)$. Draw ${\mathbf{c}}_g$ from the tip of ${\mathbf{c}}_p$: the right-triangle geometry of internal-wave velocities has horizontal hypotenuse $N/K$. This hypotenuse is distinct from the horizontal phase velocity $\omega /k$.

At an instant with real complex amplitude $\hat{\sigma }=A$, write $\varphi =kx+mz-\omega t$. The spatial fields areThe constant-phase lines of an internal gravity wave slope down to the right, with $dz/dx=-k/m$. Buoyancy maxima lie a quarter wavelength from pressure maxima; the oscillating displacement and velocity run along the phase lines. The group velocity points down and right along them, despite upward/rightward motion of the wave phase.

Let $P=\overline{{\xi }_{x}u+{\zeta }_{x}w}$, using horizontal averaging over a wavelength. This is the displacement pseudomomentum of an internal gravity wave, whose sign convention is opposite to $k\overline{E}/\omega$. For complex coefficients, $\overline{ab}=\frac{1}{2}\mathrm{Re}(\hat{a}{\hat{b}}^{\ast })$. The above wave polarization consequently givesSince $c_{gz}=-\omega m/K^2$, these obey $\overline{{\zeta }_{x}p}=c_{gz}P$. With weak kinematic viscosity, each leading monochromatic velocity component satisfies ${\nabla }^{2}u=-K^{2}u$, ${\nabla }^{2}w=-K^{2}w$. The dissipative term is therefore $-\nu K^{2}P$, so the supplied averaged balance reduces toFor constant $k,m,N$ and a steady wave envelope, $c_{gz}{P}_z=-\nu K^{2}P$. As $P$ is proportional to squared wave amplitude, viscous attenuation of an internal gravity wave givesHere $c_{gz}<0$, so amplitude decreases below a source at $z_s$: the source is above the receiving level. The same conclusion follows directly because downward group velocity carries energy from the source downward. Upward phase velocity does not locate the source.

For a slowly varying mean shear flow $U(z)$, the local intrinsic frequency is $\hat{\omega }=\omega -kU(z)$. A local WKB approximation uses ${\hat{\omega }}^2=N^2{k}^2/(k^2+m(z{)}^2)$, with varying vertical wavenumber and group velocity. The appropriate transported quantity is wave action, or horizontal wave pseudomomentum $kE/\hat{\omega }$; its vertical flux has viscous loss, and the changing polarization factors must be retained. In particular, $P/|\hat{\sigma }{|}^2$ is no longer constant.

Near a critical level of an internal gravity wave, $\hat{\omega }\to 0$, so $|m|\sim N|k|/|\hat{\omega }|$ and $|c_{gz}|\sim {\hat{\omega }}^2/(N|k|)$. The damping per vertical distance, of order $\nu K^2/|c_{gz}|$, grows rapidly. The waves are strongly attenuated over a narrow altitude range and transfer momentum to the mean flow there or before reaching the critical level. The inviscid WKB approximation eventually fails; viscosity and possibly internal-wave breaking regulate the small scales.

Use $S(z)=\sigma +k_{x}U(z)$, $k^2=k_x^2+k_y^2>0$ and primes for $z$ derivatives. Time dependence is $e^{i\sigma t}$, so exponential growth means $\mathrm{Im}\sigma <0$. The constant mass density and incompressible velocity giveAll magnetic factors below use the source convention ${\mu }_0=1$.

Linearizing the ideal magnetohydrodynamic momentum equation, the advective perturbation is $iS{\mathbf{u}}_1+U^{\prime }w{\mathbf{e}}_x$. The magnetic force is $(\nabla \times \mathbf{b})\times B{\mathbf{e}}_x$, whose $x$ component vanishes. Thus the three component equations areIn particular the vertical equation has the pressure-gradient and magnetic signs printed in the PDF.

The incompressible ideal magnetohydrodynamic induction equation can be written $D\mathbf{B}/Dt=(\mathbf{B}\cdot \nabla )\mathbf{u}$. Linearizing it givesFor $S\ne 0$, these areThe $U^{\prime }{b}_z$ term is essential: a perturbed vertical magnetic component is stretched by the background shear flow. Differentiating these amplitudes and using incompressibility also verifies $i{k}_x{b}_x+i{k}_y{b}_y+b_z^{\prime }=0$.

For completeness, substitution into all three momentum equations giveswhere $\zeta =i{k}_{x}v-i{k}_{y}u$ is the vertical vorticity amplitude. Multiplying the $y$ equation by $k_x$ and subtracting $k_y$ times the $x$ equation eliminates pressure:Therefore, away from $\rho S^2=k_x^2{B}^2$,This is the nonresonant reduction relevant to the instability calculation. Division at the Alfvén wave resonance would be invalid. For a concrete counterexample to an unrestricted reading, take constant $U$, $k_y=0$, $S=\pm k_{x}B/\sqrt{\rho }$, $v=a(z)$ and $b_y=k_{x}Ba(z)/S$, with all other amplitudes zero. Every smooth $a(z)$ solves the original linear system, but $\zeta =i{k}_{x}a(z)$ need not vanish. These transverse resonant waves do not drive $w$. A genuinely growing mode in a real background has complex $S$ and cannot satisfy either $S=0$ or $S^2=k_x^2{B}^2/\rho$ on the real $z$ axis. This explains exactly when the requested reductions may be used.

On that nonresonant branch, multiply the horizontal equations by $k_x$ and $k_y$ and add. Incompressibility gives $k_{x}u+k_{y}v=i{w}^{\prime }$, soCombining incompressibility with vorticity, multiply its first equation by $k_x$ and subtract $k_y$ times the definition of $\zeta$. This givesConsequentlySubstitute this and the derivative of the displayed pressure relation into vertical momentum. Multiplication by $k^2/i$ givesExpanding the second derivative once gives exactly the requested vertical-velocity equation:

To take the discontinuous-layer limit without multiplying distributions, introduce $f=w/S$. It is proportional to the normal displacement: the material kinematic condition is $w=iS\eta$, so $f=i\eta$. Since $S^{\prime }=k_x{U}^{\prime }$, $S{w}^{\prime }-S^{\prime }w=S^2{f}^{\prime }$. The same equation becomes the displacement equation for a parallel magnetic shear flowThe physical interface condition is continuity of $f$, because both fluids move the same interface. For constant $U_j$ on each side, away from resonance the coefficient is constant and nonzero, so $f^{\prime \prime }-k^{2}f=0$. Requiring decay at infinity and matching $f(0)$ yieldsThe exponent is $z$, as in the actual PDF; the converted TeX's $x$ there is a transcription error.

Integrate the conservative displacement equation across a layer of thickness $2\epsilon$. Its undifferentiated term has an integral tending to zero, while the derivative givesEquivalently, the perturbed magnetohydrodynamic total pressure is $p_1+B{b}_x=(\rho S^2-k_x^2{B}^2)f^{\prime }/(i{k}^2)$ and is continuous. With $f_{+}^{\prime }=-kA$ and $f_{-}^{\prime }=kA$, a nontrivial amplitude requiresThis is also the rigorous thin-layer integration of the original vertical equation, after combining the derivative terms before taking the limit.

Put $\overline{U}=(U_1+U_2)/2$ and $\Delta U=U_2-U_1$. Completing the square gives the dispersion relationFor $(\Delta U{)}^2<4B^2/\rho$ both interface frequencies are real, so there is no exponentially growing interface mode. Above that threshold, the growing sign has growth rate $|k_x|\sqrt{(\Delta U{)}^2/4-B^2/\rho }$. This is magnetic stabilization of an equal-density vortex sheet: magnetic tension from bending the field overcomes the kinetic shear driving Kelvin-Helmholtz instability. Equality is marginal and the strict stable condition avoids its double-root degeneracy. The case $k_x=0$ has no streamwise shear drive and has the neutral interface frequency $\sigma =0$; the nonzero-frequency elimination should be interpreted by the original equations or a limiting displacement formulation in that case. The horizontal $k_y$ affects the exponential decay length through $k$, but not the stated threshold.

Write $\mathbf{X}=ϵ\mathbf{x}$, $\mathbf{k}={\nabla }_{\mathbf{X}}\theta$, and denote the slow density and velocity field amplitudes by $a_{\rho }$ and ${\mathbf{a}}_u$. Acting on the rapid phase, ${\partial }_t+\mathbf{U}\cdot \nabla$ gives $i\Omega$, where $\Omega =\omega -\mathbf{U}\cdot \mathbf{k}$. Spatial differentiation gives $-i\mathbf{k}$, while derivatives of the amplitudes and the mean shear flow are $O(ϵ)$. The leading continuity equation and linearized Euler equations therefore becomeEliminating ${\mathbf{a}}_u$ on the nonzero acoustic branch gives the acoustic eikonal equation in a shear flow:The mean-shear term is small here because $\nabla \mathbf{U}=O(ϵ)$; retaining advection does not require retaining that gradient in the leading phase equation. The positive intrinsic-frequency acoustic dispersion relation is $\omega =U{k}_x+c_0|\mathbf{k}|$, with group velocity ${\mathbf{v}}_g=U{\mathbf{e}}_x+c_0\mathbf{k}/|\mathbf{k}|$.

In a stationary wall boundary layer, horizontal homogeneity preserves $\omega$ and $k_x$ along a Hamiltonian ray-tracing equations trajectory. For a two-dimensional ray,Suppose $U$ increases from zero at the wall to a positive free-stream value. For a downstream-directed wave normal, $k_x>0$, increasing $U$ decreases $\Omega$ and $k_y$. The wave normal bends towards the wall, and the ray becomes more nearly parallel to it. If $U$ reaches $\omega /k_x-c_0$, then $k_y=0$: the ray turns and returns towards the wall rather than reaching the free stream. With launch angle ${\beta }_0$ above the wall and free-stream Mach number $M_{\infty }=U_{\infty }/c_0$, the escape condition isFor upstream-directed wave normals, $k_x<0$, $\Omega$ and $|k_y|$ increase through the layer; there is no such downstream turning barrier on this branch. A wavefront normal is parallel to $\mathbf{k}$, while the energy ray follows the group velocity: advection makes their directions different. In particular, a wall-normal wave normal with $k_x=0$ can still have a ray drifting downstream. The ordinary WKB method needs a local turning-point continuation at $k_y=0$; the ray sketch depicts that qualitative continuation.

Use $D={\partial }_t+w{\partial }_z$, $w=v_z$, and a prime for ${\partial }_z$. For the plane-parallel magnetohydrodynamic flow with imposed shear, $\nabla \cdot \mathbf{u}=w^{\prime }$, whileThe solenoidal magnetic field constraint gives $B_z^{\prime }=0$. The $z$ component of the MHD induction equation then gives ${\partial }_t{B}_z=0$. Thus $B_z$ is constant in both space and time.

Use the magnetic tension and magnetic pressure decompositionThe continuity equation and momentum equation now reduce toThe transverse components of the MHD induction equation areThe terms involving $a$ respectively accelerate the flow across the imposed shear flow and wind the transverse magnetic field. The closure is the isothermal equation of state $p=c_s^2\rho$, as printed in the PDF.

The logarithmic density term is the barotropic fluid energy for $p=c_s^2\rho$, rather than the thermodynamic heat content of an isolated gas. DefineFor any scalar $q$, the continuity equation implies ${\partial }_t(\rho q)+{\partial }_z(\rho wq)=\rho Dq$. Therefore $De=-c_s^2{w}^{\prime }$, $D\Phi =wg$, and dotting the momentum equations with $\rho \mathbf{v}$ givesThe MHD induction equation gives the complementary magnetic energy balanceAdding these two identities, and collecting the product derivatives, yields the barotropic magnetic energy equationHere $E=E_h+E_B$. The two stress terms in $S$ describe magnetohydrodynamic shear work: the maintained background shear flow can supply energy to, or remove energy from, the perturbation flow and magnetic field. The sign depends on the off-diagonal total stress; this is why the energy excluding the background shear flow is not generally conserved. Changing ${\rho }_0$ merely adds a constant multiple of the conserved density to $E$ and the corresponding mass flux to $F$, leaving $S$ unchanged.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.