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For $k_z=v_z=0$, define the complex vertical vorticity amplitudeTake the curl of the amplitude equations, including ${\dot{k}}_x=2A{k}_y$. Incompressibility cancels the pressure and rotation terms, givingEquivalently, the uniform basic absolute vorticity has no gradient for the perturbation to advect. The general two-dimensional viscous shearing wave solution iswhere $Z_0$ is an arbitrary complex constant and $k_x(t_0)$ is an arbitrary real initial radial wavenumber. The perturbation pressure follows from part (ii), or explicitly $\Pi =(4A{k}_y^2-2\Omega k^2)Z/k^4$.

For $A\ne 0$, shift the time origin to the swing $k_x=0$ and put $T=k_x/k_y=2At$, $T_0=2A{t}_0$ and $\mathrm{Re}=A/(\nu k_y^2)$. ThenThe spatially averaged perturbation kinetic-energy mass density is $E=\rho ⟨|{\mathbf{u}}^{\prime }{|}^2⟩/2=\rho |\mathbf{v}{|}^2/4$, and $|\mathbf{v}{|}^2=|Z{|}^2/k^2$. ConsequentlyThe proportionality constant depends on the initial data; the exponential profile is not a claim of infinite physical growth from the remote past. For $A>0$, leading waves with $T<0$ can gain velocity amplitude as their wavevector magnitude decreases, before viscosity and later winding damp them. This is the Orr mechanism, and it is independent of uniform rotation for these two-dimensional disturbances.

If $A=0$, the wavevector is constant and $\mathbf{v}(t)=\mathbf{v}(t_0)e^{-\nu k^2(t-t_0)}$, so there is no shear amplification. The formulas with the physical-time integral remain valid, although $T=2At$ is no longer a useful coordinate. The high-positive-Reynolds-number case in the next part takes $A>0$.

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.