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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.