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For the unstratified local vertical shear instability, maximize $s^2/{\Omega }^2=(2qR-1)/(1+R^2)$ over $R=k_x/k_z$. Its derivative vanishes when $q{R}^2-R-q=0$. For $q>0$ the maximum uses $R=(1+\sqrt{1+4q^2})/(2q)$, and substitution gives the displayed expression. For small $q$, $R\sim q^{-1}$ and $s_{\max }=\Omega q(1-q^2/2+O(q^4))$. Changing the sign of the shear changes the preferred sign of $R$ but gives the same maximum magnitude; there is no exponential growth at $q=0$.

Write the proposed background as $\mathbf{U}=U(x,z){\mathbf{e}}_y$, $U=-3\Omega x/2+q\Omega z$. It satisfies incompressibility, and $\mathbf{U}\cdot \nabla \mathbf{U}=U{\partial }_y\mathbf{U}=0$. A constant pressure has zero gradient. The Coriolis acceleration is $-2\Omega {\mathbf{e}}_z\times \mathbf{U}=2\Omega U{\mathbf{e}}_x=-(3x-2qz){\Omega }^2{\mathbf{e}}_x$, which exactly cancels the prescribed body force. Thus it is a steady state.

The background is ordinary Keplerian radial shear together with vertical shear: ${\partial }_{x}U=-3\Omega /2$ and ${\partial }_{z}U=q\Omega$. At a given radius, different heights have different azimuthal velocities. The small $q$ in this question measures vertical shear; it is distinct from the orbital shear parameter, whose radial value here remains $3/2$. The vertical shear instability draws on this height-dependent differential rotation.

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.