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Eliminate pressure by multiplying the radial equation by $k_z$ and subtracting $k_x$ times the vertical equation. For $k_z\ne 0$, incompressibility gives ${\tilde{u}}_z=-(k_x/k_z){\tilde{u}}_x$, and the pressure-free relation is $s(k_x^2+k_z^2){\tilde{u}}_x/k_z=2\Omega k_z{\tilde{u}}_y$. Substitution in the azimuthal equation yieldsThus the vertical shear instability occurs when $2q{k}_x{k}_z>k_z^2$, or $2q(k_x/k_z)>1$ for $k_z\ne 0$. Otherwise the nonzero modes oscillate; equality is marginal in the exponential-growth sense. With $q>0$, growing disturbances have $k_x/k_z>1/(2q)$ and are strongly inclined in wavevector space.

Put $R=k_x/k_z$. Maximizing $(2qR-1)/(1+R^2)$ gives $q{R}^2-R-q=0$. For $q>0$, the unstable maximizing root isMore precisely its bracket is $1-q^2/2+O(q^4)$, and $R_{\max }\sim q^{-1}$. This is the maximum growth rate of the vertical shear instability. For negative $q$, the maximum uses the corresponding negative root and is asymptotically $\Omega |q|$. At $q=0$ there is no exponentially growing mode. The undivided formula handles $k_z=0$: $s^2=0$, not an exponential instability; an allowed vertical velocity can instead force a secular azimuthal change. The excluded $\mathbf{k}=0$ is a spatially uniform disturbance and does not belong to the stipulated nonconstant one-phase family.