Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 321 / 2

Let $K={\rho }_0|\mathbf{u}{|}^2/2$ be the kinetic energy per unit volume. Dot the momentum equation with ${\rho }_0\mathbf{u}$. The Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Incompressibility turns the advection and pressure terms into divergences, yieldingThus $\mathbf{F}=(K+P)\mathbf{u}$ is the energy flux and $S={\rho }_0{\Omega }^2(3x-2qz)u_x$ is the source in the requested kinetic-energy form.

To decide about conservation of energy, absorb the ordinary radial shearing-sheet tidal potential ${\Phi }_t=-3{\Omega }^2{x}^2/2$ into the density. Its material derivative cancels the $3x$ work term:For $q=0$ this is the usual conservative Jacobi energy in a shearing sheet, subject to vanishing boundary energy flux. For $q\ne 0$, the extra body force has nonzero curl, $\nabla \times [-2{\Omega }^{2}qz{\mathbf{e}}_x]=-2{\Omega }^{2}q{\mathbf{e}}_y$, so no scalar potential can absorb it. A potential $2{\Omega }^{2}qxz$ would necessarily introduce a vertical force $-2{\Omega }^{2}qx{\mathbf{e}}_z$, absent from the model. Therefore this vertical-shear model does not generically conserve the standard total mechanical energy: its maintained shear/irradiation background can supply or remove energy. This is the energy balance of the incompressible vertical-shear model.

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.

Set $\mathbf{k}=(k_x,0,k_z)\ne 0$. Incompressibility gives $(\mathbf{k}\cdot \tilde{\mathbf{u}})f^{\prime }(\xi )e^{st}=0$. Since $f$ is nonconstant and differentiable, it follows that $\mathbf{k}\cdot \tilde{\mathbf{u}}=0$; thereforeThe background has no advection of a $y$-independent perturbation. Its only remaining cross-advection term is $({\mathbf{u}}^{\prime }\cdot \nabla )\mathbf{U}=(-3\Omega u_x^{\prime }/2+q\Omega u_z^{\prime }){\mathbf{e}}_y$. The velocity/time terms have spatial factor $f$, while $\nabla P^{\prime }=\tilde{P}\mathbf{k}{g}^{\prime }(\xi )e^{st}$. For nonzero pressure amplitude, $g^{\prime }$ must therefore be a constant multiple of $f$; absorbing that multiple into $\tilde{P}$ gives $g^{\prime }=f$. An additive spatial constant in $g$ is a pressure gauge. If $\tilde{P}=0$, there is no such restriction on the unused pressure profile.

Matching coefficients givesBecause the quadratic perturbation term vanishes identically, these amplitudes give an exact single-phase incompressible perturbation of a shear flow, not merely a first-order truncation. Any sufficiently differentiable profile with $g^{\prime }=f$ works locally; boundary conditions may restrict the profiles in a particular domain.

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.