Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 73 / 2

The variable $\psi$ is a rotating-flow pressure potential including the conservative body-force and centrifugal potentials. For constant mass density,up to an additive function of time, where $\Phi$ is any imposed conservative potential. The last term is the centrifugal potential; its negative gradient is the centrifugal acceleration. The Coriolis force remains explicit in the equation because it is velocity-dependent and cannot be absorbed into a scalar position potential.

The further constraint is incompressible flow:For the specified shear, ${\mathbf{u}}_0\cdot \nabla {\mathbf{u}}_0=0$ and ${\nabla }^2{\mathbf{u}}_0=0$. Its Coriolis acceleration is $4\Omega Ax{\mathbf{e}}_x$, so the corresponding total basic pressure potential can be chosen as ${\psi }_0=-2\Omega A{x}^2$. This supplies the force balance required for the basic state used below.

Let the perturbation phase be $\vartheta =\mathbf{k}(t)\cdot \mathbf{r}$ and write $\psi ={\psi }_0+\mathrm{Re}[\Pi (t)e^{i\vartheta }]$. Incompressible flow imposes $\mathbf{k}\cdot \mathbf{v}=0$. Applying the background advective derivative to the phase givesCancel this position-dependent phase by choosingThe perturbation acting on the basic flow gives $({\mathbf{u}}^{\prime }\cdot \nabla ){\mathbf{u}}_0=-2A{u}_x^{\prime }{\mathbf{e}}_y$. Its self-advection vanishes exactly: the physical velocity ${\mathbf{u}}^{\prime }$ is perpendicular to $\mathbf{k}$, while every spatial derivative of ${\mathbf{u}}^{\prime }$ is proportional to a component of $\mathbf{k}$. Thus ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }=0$, including the products between the complex mode and its conjugate. Substituting into the full Navier-Stokes equation therefore gives an exact incompressible shearing wave, with no small-amplitude approximation:In components these equations areDifferentiating the transversality constraint closes the pressure:The coefficient $4A$ includes both the shear term in the momentum equation and $\dot{\mathbf{k}}\cdot \mathbf{v}$; treating the wavevector as fixed would miss one contribution. Initial data with $\mathbf{k}(t_0)\cdot \mathbf{v}(t_0)=0$ remain transverse under this closed evolution. These shearing waves are exact in an unbounded local shear flow or with compatible shearing-periodic boundaries; arbitrary rigid boundary conditions need not admit a single such wave.

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 a fixed $\mathrm{Re}\gg 1$, optimize the gain over the initial wave tilt and a later observation time. From the energy profile,Its stationary points satisfyThere are two negative roots in the high-Reynolds-number regime: an early energy minimum and a later maximum. Write them as $T_{-}$ and $T_{+}$. Balancing the large-$|T|$ terms for the early root and the small-$|T|$ terms for the late root givesThus the optimum disturbance starts as a sufficiently tightly wound leading wave and reaches its peak slightly before $k_x=0$; finite viscosity shifts the peak away from the inviscid swing.

Normalize its initial energy at $T_{-}$. The exact maximum ratio isThe prefactor is asymptotic to $(2\mathrm{Re}{)}^{2/3}$, while $g(T_{-})/\mathrm{Re}\to -2/3$ and $g(T_{+})/\mathrm{Re}\to 0$. Therefore optimal transient amplification of a viscous shearing wave givesThe time between the optimum initial tilt and the peak is approximately $(2\mathrm{Re}{)}^{1/3}/(2A)$. Without viscosity, ever larger initial tilts would allow unbounded gain; viscosity damps them before they can swing, selecting the finite optimum. The coefficient above uses the paper's Reynolds number based on $A$, not one based on the full shear rate $2A$.

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.