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For $k_y=0$ and $k_z\ne 0$, pressure elimination gives ${\dot{v}}_x=2\Omega \chi v_y-\nu k^2{v}_x$ and ${\dot{v}}_y=-2(\Omega -A)v_x-\nu k^2{v}_y$, where $\chi =k_z^2/k^2$. The inviscid frequency is the radial epicyclic frequency times $\sqrt{\chi }$. With $w=v_x/\sqrt{\chi }$, $(\Omega -A)|w{|}^2+\Omega |v_y{|}^2$ is conserved when viscosity vanishes. In a Keplerian flow it bounds kinetic-energy amplification by four, whereas with zero rotation the same system has the algebraic lift-up effect.

// Target: turbulence.bigb

For axisymmetric disturbances, $k_y=0$, the wavevector is fixed, so there is no winding/swing mechanism. Take $k_z\ne 0$ and set $\chi =k_z^2/k^2$. Eliminating pressure with incompressibility givesThe axisymmetric incompressible epicyclic wave has inviscid frequencywhere ${\kappa }_r$ is the radial epicyclic frequency.

In a Keplerian flow $A=3\Omega /4$ and ${\kappa }_r=\Omega$. The axisymmetric disturbances therefore execute stable inertial/epicyclic oscillations, with viscous amplitude decay $e^{-\nu k^{2}t}$. Rotation opposes sustained radial displacement through angular-momentum conservation. There can be bounded exchanges of kinetic energy between components, but no large Reynolds-number-dependent amplification of this axisymmetric family. Indeed, with $w=v_x/\sqrt{\chi }$, inviscid motion conservesSince the physical velocity norm is $|w{|}^2+|v_y{|}^2$, Keplerian energy amplification is at most $\Omega /(\Omega -A)=4$; viscosity can only reduce that bound. A pure initial azimuthal disturbance can attain the inviscid factor four after a quarter epicyclic period. By contrast, the two-dimensional nonaxisymmetric Orr mechanism remains available in a Keplerian flow and has the ${\mathrm{Re}}^{2/3}$ gain derived above.

In non-rotating shear, $\Omega =0$ while $A$ remains nonzero. The axisymmetric solution isCross-stream motion transports the basic shear and generates an azimuthal streak: this is the lift-up effect. Inviscid azimuthal velocity grows linearly and its energy quadratically, rather than oscillating. With viscosity, the gain for a suitable initial cross-stream disturbance scales as $[A/(\nu k^2){]}^2$ at large Reynolds number, appreciably stronger than the two-dimensional ${\mathrm{Re}}^{2/3}$ scaling. For instance, $C_y=0$ and fixed $\chi >0$ give the leading optimized gain $4\chi A^2/(e^2{\nu }^2{k}^4)$ at $\tau \simeq 1/(\nu k^2)$. Neither mechanism is an exponentially growing hydrodynamic normal mode. Rotation bounds axisymmetric lift-up in Keplerian shear; without rotation, lift-up gives strong algebraic transient growth. If $k_z=0$ as well as $k_y=0$, incompressibility forces $v_x=0$ and removes this coupling altogether.

Use incompressible Euler equations with pressure divided by the constant reference density. Writing $L={\partial }_t+U(y){\partial }_x$, their linearization about the parallel shear flow isTaking the divergence requires the commutator $[{\partial }_y,L]=U^{\prime }{\partial }_x$. Both this commutator and the divergence of $U^{\prime }v$ contribute, so ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying the Laplacian to $Lv=-p_y$ givesHence the wall-normal velocity equation isTaking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise equation cancels pressure and yields the inviscid Squire equation,This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm 1,t)=0$. They do not impose a viscous no-slip condition on the tangential perturbation velocity.

For a Fourier mode, set $k^2={\alpha }^2+{\beta }^2$, $D=d/dy$ and $q=(D^2-k^2)\tilde{v}$. The equations becomeFor Couette flow $U=\lambda y$, $U^{\prime \prime }=0$, so $q(y,t)=e^{-i\alpha \lambda yt}{q}_0(y)$ with $q_0=(D^2-k^2){\tilde{v}}_0$. Invert this using the Dirichlet Green function of $D^2-k^2$ on $[-1,1]$. If $y_{<}=\min (y,s)$ and $y_{>}=\max (y,s)$, it isIt vanishes at both walls and its first derivative has jump one at $y=s$. ThusAt $t=0$, uniqueness of the Dirichlet inversion recovers the compatible initial velocity. This inviscid Couette initial-value Green function directly solves the initial-value problem without a Laplace transform.

The Squire equation is now a first-order forced equation in time. Its integrating factor givesFor nonzero $\alpha$, the velocity formula contains shearing phases and the inviscid Couette continuous spectrum. Velocity cancellation can occur even though the transported $q$ has constant magnitude; it must not be confused with exponential modal instability.

For the separate case $\alpha =0$, $\beta \ne 0$, the Dirichlet operator $D^2-{\beta }^2$ is invertible and $q_t=0$. Hence $\tilde{v}={\tilde{v}}_0$ andSince $\tilde{\eta }=i\beta \tilde{u}$ in this case, $\tilde{u}={\tilde{u}}_0-\lambda t{\tilde{v}}_0$. This is the lift-up effect: a transverse disturbance transports base-flow momentum and creates a streamwise perturbation growing linearly in time, with perturbation kinetic energy generically growing as $t^2$. The growth is algebraic and is possible without an exponentially unstable normal mode. For nonzero initial normal velocity and nonzero shear, nonlinear effects eventually limit the validity of the linearized calculation. The same streamwise-independent result holds with $\lambda$ replaced by $U^{\prime }(y)$ for a general parallel profile.

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.