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Two equal stable density jumps of size $\Delta \rho /2$ at $z=\pm h/2$ lie in the global linear shear flow $U=\Delta Uz/h$. Decaying normal modes haveThe jump conditions for stratified inviscid shear flow give the determinant equationWriting $\alpha =kh/2$, $J=g\Delta \rho h/({\rho }_0\Delta U^2)$ and $\tilde{c}=2c/\Delta U$, this isThe two roots for ${\tilde{c}}^2$ are real; one is negative exactly when $2\alpha /(1+e^{-2\alpha })<J<2\alpha /(1-e^{-2\alpha })$. At large $\alpha$, this narrow band centres on $J=2\alpha$, where the isolated counterpropagating interfacial gravity waves have the same zero laboratory speed. This realizes counterpropagating wave instability.

A linear shear flow has velocity that varies linearly in a transverse coordinate, such as $\mathbf{u}=-Sx{\mathbf{e}}_y$. Its velocity gradient and shear rate are spatially constant.

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.

For a Keplerian shearing sheet, the shear rate is $S=3\Omega /2$ and the shearing-sheet tidal potential isA uniform steady solution isThe Coriolis acceleration of this linear shear flow balances the radial tidal acceleration. The only nonzero background component of the viscous stress tensor that matters is $T_{xy}=-\nu ({\Sigma }_0){\Sigma }_{0}S$, which is spatially constant. Hence $\nabla \cdot \mathbf{T}=0$: a local uniform patch has no stress gradient or torque divergence to drive an accretion flow.

Let $D/Dt={\partial }_t-Sx{\partial }_y$ be the material derivative along the background linear shear flow ${\mathbf{u}}_0=-Sx{\mathbf{e}}_y$. Linearizing about constant $\Sigma$ givesThe coefficient $2\Omega -S$ includes the contribution $\mathbf{v}\cdot \nabla {\mathbf{u}}_0=-S{v}_x{\mathbf{e}}_y$ from perturbation advection of the background shear.

A shear flow has a velocity field varying across adjacent fluid layers, giving a nonzero shear rate. A parallel example is $\mathbf{u}=U(z){\mathbf{e}}_x$, with vertical shear $U^{\prime }$. Linear shear flow has constant shear, while a shear layer concentrates the velocity change in a narrow region.

Use a linear shear flow to see why straight pathlines can still carry local spin. Faster layers pull past slower ones, and a small symmetric paddle can turn clockwise while its center travels straight ahead.

For the planar simple shear flow $\mathbf{u}=(\gamma y,0,0)$ with constant signed shear rate $\gamma$, direct differentiation givesWith $\gamma =1{\mathrm{s}}^{-1}$, these are $-1{\mathrm{s}}^{-1}$ and $-0.5{\mathrm{s}}^{-1}$: both signs indicate clockwise orientation under the right-hand rule. Yet a particle initially at $(x_0,y_0)$ follows the straight pathline $x(t)=x_0+\gamma y_{0}t$, $y(t)=y_0$. The curl measures differences of velocity between neighboring particles, rather than bending of one particle's pathline.

The velocity gradient has $L_{xy}=\gamma$ as its only nonzero entry. Its spin tensor supplies clockwise rotation, while its rate-of-strain tensor supplies simultaneous deformation. For a material line making angle $\alpha$ to the horizontal, the equation $\dot{\delta \mathbf{x}}=L\delta \mathbf{x}$ gives $\dot{\alpha }=-\gamma {\sin }^2\alpha$. Initially horizontal and vertical material lines therefore have angular velocities $0$ and $-\gamma$; their mean is $-\gamma /2$. At $t=0.5\mathrm{s}$ with $\gamma =1{\mathrm{s}}^{-1}$, an initially vertical displacement becomes $(0.5,1)$ times its original length and has angle $\arctan 2\approx 63.435^{\circ }$, a clockwise change of $26.565^{\circ }$. The plotted cross deforms; an actual paddle's response also depends on its shape and interaction with the fluid. The exact local spin statement concerns the spin tensor, rather than every tracer's orientation.