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Direct differentiation of the velocity field gives $(\mathbf{u}\cdot \nabla )\mathbf{u}=(-{\Omega }^{2}x,-{\Omega }^{2}y,0)$. Since $u^2={\Omega }^2(x^2+y^2)$, this equals $\nabla (-u^2/2)$ as required.

In cylindrical coordinates, steady rigid rotation has $u_r=u_z=0$, $u_{\theta }=\Omega r$. The radial and vertical Euler equations for constant-density fluid are $p_r=\rho {\Omega }^{2}r$ and $p_z=-\rho g$, hence $p=\rho ({\Omega }^2{r}^2/2-gz)+C$. Constant atmospheric pressure on the free surface gives $z_s(r)=H+{\Omega }^2{r}^2/(2g)$. Conservation of fluid volume determines $H$:ThereforeThe lowest point is the axis, where $z_s(0)=h-{\Omega }^2{a}^2/(4g)$. The free surface first meets the bottom whenThis assumes no overflow and uses the full disk of wetted bottom until first contact. Once a dry central region forms, the volume constraint and geometry must be modified.

An isotropic simple fluid has local stress determined by the deformation history of the material element. A viscometric flow is one whose relative deformation history is that of steady simple shear flow, up to rigid changes of frame. At a given element the shear rate is constant along its history, although it can differ between elements. This includes the locally simple-shear kinematics used in standard viscometric measurements.

Use the component-first velocity gradient $K_{ij}=\partial v_i/\partial x_j$. The first two Rivlin-Ericksen tensors areIn the axes of a steady shear $v_x=sy$, the relative deformation gradient over a lag $q$ is $F=I+qK$, since $K^2=0$. The material history is determined by $s$, and its tensor symmetries exclude $xz$ and $yz$ stress components. The remaining symmetric stress consists of an isotropic part, one shear component, and two independent normal-stress differences. The relevant matrices areThus $I,A_1,A_1^2,A_2$ span the allowed stress components. Material isotropy and reversal of the shear show that their scalar coefficients depend on $|s|=(A_1:A_1/2{)}^{1/2}$. Define the viscometric functions bySolving for the matrix coefficients givesThis is an exact steady-viscometric representation, rather than an assumption that the fluid is everywhere a second-order fluid. The shear viscosity $\mu$ measures the shear resistance, while ${\psi }_1$ and ${\psi }_2$ measure the first and second normal-stress differences generated by shear. Their zero-rate values are understood as limits when these exist. An isotropic pressure changes neither normal difference.

For a flow in the lubrication approximation with gap $H$, longitudinal scale $\ell \gg H$ and speed $U$, the shear-rate scale is $U/H$. The viscous shear force per unit volume scales as $\mu U/H^2$. Quadratic viscoelastic normal stresses scale as ${\psi }_i{U}^2/H^2$; their longitudinal gradients, and the comparable mixed-component corrections, contribute forces of scale ${\psi }_i{U}^2/(H^2\ell )$. Therefore the normal-stress smallness in lubrication criterion isEquivalently, the thin-gap aspect ratio must multiply the shear-based elastic parameter to give a small quantity: $(H/\ell )|{\psi }_i|\dot{\gamma }/\mu \ll 1$. Normal stresses need not themselves be smaller than the shear stress; it is their effect on the leading force balance that matters. If the flow changes on an additional short time scale $T$, the $D_t{A}_1$ term also requires $|{\psi }_1|/(\mu T)\ll 1$ for its neglect. Material coefficients should be evaluated at the characteristic shear rate.

Now take positive cylinder rotation, measure $\theta$ counterclockwise from the rightward horizontal, and let $y=r-a$ increase outward through the film. The tangent has upward component $\cos \theta$, so the tangential gravitational acceleration is $-g\cos \theta$. To leading order in $h/a$, neglect inertia and higher-order hydrostatic or capillary pressure gradients, and write $U=\Omega a$. Tangential force balance and the shear-free free surface givehenceThe wall shear stress is $-\rho gh\cos \theta$. It opposes rotation on the uphill side and assists it on the downhill side.

For a power-law fluid, ${\sigma }_{\theta y}=k|u_y{|}^{n-1}{u}_y$. Its inversion must retain the sign:Integrating with the no-slip boundary condition $u(0)=U$ yieldsThe volume flux per unit axial length is $Q={\int }_0^{h}u(y)dy$. The integrated bracket is $[(n+1)/(2n+1)]h^{(2n+1)/n}$, so the power-law coating on a rotating cylinder hasAt $\cos \theta =0$ the gravitational term vanishes by continuity. Steady mass conservation makes $Q$ independent of $\theta$.

On the uphill half, put $D=(\rho g\cos \theta /k{)}^{1/n}$ and $m=(2n+1)/n$. The local flux is $Q(h)=Uh-(D/m)h^m$, withIt has a unique maximum atA positive coating must carry the same steady flux at every angle. The tightest local bound occurs at $\cos \theta =1$, not on the downhill side, where the flux increases monotonically with thickness. ConsequentlyFor $0<Q<Q_{\max }$ the thinner positive root on the uphill side connects through $\cos \theta =0$ to the unique downhill root, giving a periodic leading-order coating profile. At the limiting flux the two uphill roots meet at $\theta =0$; it is the maximum flux of this lubrication model, with a possible limiting corner resolved by neglected higher-order effects. For $n=1$ the check is $Q_{\max }=(2/3)U\sqrt{kU/(\rho g)}$. All these conclusions require the resulting film to remain thin enough for the lubrication approximation and for the neglected normal stresses and inertia to satisfy their scaling restrictions.

For steady inviscid flow at atmospheric pressure, Bernoulli's equation along a freely falling streamline gives $u^2/2+gz=u_0^2/2+gH$. The downward speed is thereforeFor the incompressible jet, mass conservation fixes its volume flux: $\pi R^{2}u=\pi a^2{u}_0$. ConsequentlyFar from impact, the inviscid falling jet and radial impact film is thin and nearly horizontal. Let $v$ be its radial speed and $h$ its thickness. The same volume flux now crosses a cylindrical surface of area $2\pi rh$, giving $v=a^2{u}_0/(2rh)$. At the free surface, atmospheric pressure and Bernoulli's equation give $v^2=u_0^2+2g(H-h)$. Eliminate $v$ to obtainThis description assumes a steady incompressible jet, negligible surface tension, and negligible dissipation between the tube and the thin far-field film. It is a free-surface relation, not an assertion that pressure is atmospheric throughout the film or that the impact region itself is thin.

In the inviscid incompressible irrotational model, write the total velocity as $(U+{\varphi }_x,{\varphi }_y)$. The perturbation potential obeys Laplace's equation in $-h<y<\eta (x,t)$. The complete boundary conditions, neglecting surface tension and taking the air pressure constant, areThe first surface condition expresses that the free surface is material; the second is the Bernoulli equation with atmospheric pressure and the constant background kinetic energy absorbed into the potential's time gauge.

At first order, evaluation is on $y=0$ and products of perturbations vanish. HenceFor a mode $e^{i(\omega t-kx)}$, the bottom condition selects $\varphi =C\cosh [k(y+h)]e^{i(\omega t-kx)}$. Let $\Omega =\omega -kU$. The linear surface conditions areEliminating $C$ gives the dispersion relationIt is the intrinsic gravity-wave frequency squared, with the uniform current supplying the Doppler shift. For either sign of real $k$, $k\tanh (kh)$ is nonnegative.

Use incompressibility, negligible viscosity and surface tension, and an initially irrotational flow. Take the undisturbed free surface at $z=0$ and bed at $z=-H$. The velocity potential $\phi$ satisfies Laplace equation and ${\phi }_z(-H)=0$. For a Fourier mode, writeThe linear kinematic boundary condition and linear Bernoulli equation at the free surface are ${\eta }_t={\phi }_z$ and ${\phi }_t+g\eta =0$. Eliminating $A,{\eta }_a$ gives the surface gravity wave dispersion relationThe rightward phase velocity and group velocity areThe phase velocity transports a crest; the group velocity transports a narrow-band envelope and its leading-order energy. For $kH\gg 1$, $\omega \sim \sqrt{gk}$ and $c_g\sim c_p/2\sim \sqrt{g/k}/2$: deep-water surface gravity waves are dispersive. For $kH\ll 1$, $\omega \sim k\sqrt{gH}$ and $c_g\sim c_p\sim \sqrt{gH}$, the nondispersive shallow-water dispersion relation. The boundary velocity controls the excited amplitude; it does not change the linear dispersion relation.

Set $a=2\pi G{\rho }_0$. The axisymmetric Poisson equation for Newtonian gravity integrates to $(R{\Phi }_R{)}^{\prime }=4\pi G{\rho }_{0}R$. Regularity on the axis fixes the integration constant, so ${\Phi }_R=aR$. Radial force balance in the rotating fluid givesWith the stated central pressure, $p_c={\rho }_0{a}^2/(4k^2)$, this is a perfect square:The free surface lies at the first zero of pressure. For the source's positive rotation parameter, $R_0=\sqrt{2\pi G{\rho }_0}/k$; allowing either rotation direction requires $|k|$ in the denominator. The inward effective gravity is ${\Phi }_R-R{\Omega }^2=R(a-k^2{R}^2)$, so $\boxed{g_{\mathrm{eff}}(R_0)=0}$. In particular $p_R(R_0)=0$. This is the critically rotating self-gravitating cylinder.

For a normal mode proportional to $e^{i\omega t+im\varphi }$, the background material derivative becomes $i\sigma$, where $\sigma =\omega +m\Omega (R)$. Linearizing the centrifugal term gives $-2\Omega u_{\varphi }$ in the radial equation. In the azimuthal equation the background velocity gradient and cylindrical geometric term together give $(2\Omega +R{\Omega }^{\prime })u_R=3\Omega u_R$. Since the interior mass density is constant, write $W=p^{\prime }/{\rho }_0+{\Phi }^{\prime }$. The two momentum equations and incompressible flow condition are consequentlyFor $m\ne 0$, the last two equations express the other amplitudes in terms of $u=u_R$:Differentiate the second expression using $R{\Omega }^{\prime }=\Omega$ and $R{\sigma }^{\prime }=m\Omega$, then substitute into the radial momentum equation. Before any division by $\sigma$, the result isAway from corotation this gives the required ordinary differential equationAt corotation the original coupled equations, or the undivided equation, must be used. A regular axisymmetric radial velocity would vanish by the incompressible flow condition; the requested dipolar calculation has $m=1$.

For $m=1$, use the amplitude normalization $u=A(\omega +kR)$, which remains useful when $\omega$ tends to zero. Substitution into the undivided equation gives zero identically. For nonzero $\omega$ this is the source's $1+kR/\omega$ after choosing $A=1/\omega$. The other amplitudes areThese formulas also satisfy the three original equations directly. By the regularity assumption in the question, this is the interior velocity family to which the free surface conditions must be applied.

The gravitational potential of a displaced cylindrical interface supplies an essential part of those conditions. For surface fluid displacement $\eta e^{im\varphi }$, the bulk mass density perturbation vanishes but the displaced mass density step contributes ${\rho }_0\eta \delta (R-R_0)$. The interior and exterior gravitational potential amplitudes solve Laplace's equation, are proportional to $R^m$ and $R^{-m}$ for $m\ge 1$, and are continuous at the surface. Integrating the Poisson equation for Newtonian gravity across the mass density step givesSince the two radial derivatives at the surface are $-m{\Phi }_s^{\prime }/R_0$ and $m{\Phi }_s^{\prime }/R_0$, respectively,It would be incorrect to set the entire potential perturbation to zero merely because the bulk mass density is unchanged.

Let ${\Omega }_s=k{R}_0$, so $a={\Omega }_s^2$. Zero fluid Lagrangian perturbation of surface pressure gives $p_s^{\prime }+\eta p_R(R_0)=0$, hence $p_s^{\prime }=0$. The surface momentum potential and kinematic condition for $m=1$ are thereforeIf $\omega +{\Omega }_s\ne 0$, the kinematic condition fixes $\eta =-iA$. The same value follows if the radial fluid Lagrangian displacement $u_R/[i(\omega +kR)]=-iA$ is required to extend continuously to the free surface, including a possible surface-corotation limit. Comparing the gravitational condition with the interior value of $W_s$ then givesfor a nontrivial perturbation. These dipolar perturbations of a critically rotating cylinder describe a displacement of the whole cylinder's axis. Translating an isolated self-gravitating system gives another equilibrium, so there is no external restoring force and the translation is neutral. A uniform drift of the centre also follows from invariance under a Galilean transformation, explaining the double zero of the frequency equation. The singular $1/\omega$ in the originally chosen velocity normalization does not obstruct the finite-amplitude limiting solution.

There is an admissibility qualification to the frequency conclusion as printed. Regular interior fluid Eulerian perturbation of velocity alone does not justify cancelling the surface factor $\omega +{\Omega }_s$. Eliminating $\eta$ without that cancellation instead givesThe additional surface-corotation case $\omega =-{\Omega }_s$ formally satisfies every displayed velocity equation and the usual linearized surface conditions withHere $u_R(R_0)=p^{\prime }(R_0)=0$, and the velocity is finite on the axis. However, its interior radial fluid Lagrangian displacement tends to $-iA$, whereas the undisplaced surface has $\eta =0$. Thus a regular single-frequency fluid Lagrangian displacement continuous up to the surface excludes this degenerate branch. The intended ${\omega }^2=0$ conclusion holds with that regular-displacement condition, or with surface corotation excluded; the printed velocity regularity assumption by itself leaves this formal exception.

Take the undisturbed free surface as $z=0$, the bottom as $z=-h$, and the walls as $x,y=0,a$. For an incompressible flow which is an irrotational flow, write $\mathbf{u}=\nabla \varphi$ with a velocity potential satisfying ${\nabla }^2\varphi =0$. Impermeability gives ${\varphi }_z=0$ at the bottom, ${\varphi }_x=0$ at the $x$ walls, and ${\varphi }_y=0$ at the $y$ walls.

For the actual surface $z=\eta (x,y,t)$, the exact kinematic boundary condition for a free-surface graph isThe Unsteady Bernoulli equation, with atmospheric pressure at the surface and its time-dependent constant absorbed into $\varphi$, givesFor small disturbances about fluid at rest, $\varphi$ and $\eta$ are first order. Products of first-order quantities and Taylor corrections from evaluating at $z=\eta$ are second order. The linear surface conditions are thereforeor, after eliminating the elevation,

The separated normal modes of surface gravity waves in a rectangular tank havewhere $m,n$ are nonnegative integers not both zero. The horizontal factors satisfy the wall conditions, and the vertical factor satisfies both the Laplace equation and the bottom condition. At the top, ${\varphi }_z=k_{mn}\tanh (k_{mn}h)\varphi$; substitution into the surface condition givesThe elevation isReal physical standing waves are the real parts or linear combinations of these complex modes. The uniform $m=n=0$ mode has zero frequency and would change the mean depth; conservation of the fixed water volume excludes it from the disturbances. Modes with equal $m^2+n^2$ are degenerate and can be superposed.

For the shallow-water approximation, let the vertical scale be much smaller than the horizontal scale, so vertical acceleration is negligible and pressure is hydrostatic. Assume constant density, a flat impermeable bottom, a free surface under constant atmospheric pressure, and a horizontally moving column with no leading-order vertical shear. Neglect friction. Taking the bottom at $z=0$, hydrostatic pressure is $p=p_a+\rho g(h-z)$, so its horizontal gradient is independent of $z$. The horizontal momentum equations are thereforeIntegrating incompressibility from the bottom to the free surface, using zero bottom vertical velocity and the surface kinematic condition $w=h_t+u{h}_x+v{h}_y$, givesTogether these are the nonrotating shallow water equations.

Rotation with angular velocity $f\hat{\mathbf{z}}/2$ adds Coriolis force terms $-fv$ and $+fu$ to the left sides of the $u$ and $v$ equations, respectively. The centrifugal potential is absorbed into the background pressure and effective gravity. ThusWrite $\zeta =v_x-u_y$ and $d=u_x+v_y$. Taking the horizontal curl, including the derivatives of the advecting velocity, yieldsFor example the curl of the Coriolis force contribution is $fd$, and the curl of nonlinear advection supplies $\zeta d$. Since $q_a=f+\zeta$ and $D_{H}h=-hd$, division givesThis is exact shallow-water potential vorticity conservation within the rotating shallow water equations.

For the quasi-geostrophic approximation, take a horizontal scale $L$, speed $U$, mean depth $H$, and slow time $L/U$. Require small Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$ and small surface departure $\eta =h-H$. Introduce the Rossby deformation radius $L_R=\sqrt{gH}/|f|$ and Burger number $\mathrm{Bu}=L_R^2/L^2=O(1)$. Geostrophic balance gives $g\eta /L\sim |f|U$, hence $\eta /H\sim \mathrm{Ro}/\mathrm{Bu}\ll 1$. The Froude number is $U/\sqrt{gH}=\mathrm{Ro}/\sqrt{\mathrm{Bu}}\ll 1$. These assumptions select slow balanced motion, with fast gravity-wave transients absent at leading order.

Define the quasi-geostrophic streamfunction by $\psi =g\eta /f$. The leading geostrophic flow is $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, which is horizontally nondivergent. Expanding the exact shallow-water potential vorticity givesThe correction is smaller than $f/H$ by $O(\mathrm{Ro})$. Advecting it by the leading geostrophic flow, while the constant leading term has zero gradient, givesHere potential-vorticity conservation means that $Q$ is carried unchanged by the leading horizontal trajectories. Potential-vorticity inversion means solving the elliptic equation $({\nabla }_H^2-L_R^{-2})\psi =Q-f$, with appropriate boundary or decay conditions, to recover $\psi$ and thus the velocity. Conservation evolves the scalar field; inversion obtains its velocity field.

For a mean current $U(y)=-{\bar{\psi }}_y$, substitute $Q=\bar{Q}+Q^{\prime }$ and $\psi =\bar{\psi }+{\psi }^{\prime }$ into potential-vorticity conservation. Retaining first-order terms givesLet $\Delta Q=\bar{Q}(0^{+})-\bar{Q}(0^{-})>0$. Differentiating the mean inversion relation gives ${\bar{Q}}_y=-U^{\prime \prime }+L_R^{-2}U$. Away from $y=0$ this is zero, so decay at both infinities gives $U=U_0{e}^{-|y|/L_R}$. A velocity jump would introduce a derivative of a delta function, absent from ${\bar{Q}}_y$, so $U$ is continuous. Integrating through zero now gives $-[U^{\prime }]=\Delta Q$, and therefore

For the regular interfacial normal mode ${\psi }^{\prime }=\varphi (y)e^{ik(x-ct)}$, the linear equation readsIn each region, choose zero perturbation potential vorticity, giving ${\varphi }^{\prime \prime }-(k^2+L_R^{-2})\varphi =0$. This supplies a regular solution even if $U=c$ at isolated locations away from the interface. Continuity and decay require $\varphi =A{e}^{-\gamma |y|}$, where $\gamma =\sqrt{k^2+L_R^{-2}}$. Integrating the equation across zero givesSince $[{\varphi }^{\prime }]=-2\gamma A$, the nonzero normal mode hasThis potential-vorticity step edge wave is an interfacial Rossby wave. Displacing the potential vorticity jump creates alternating anomalies; their inverted velocity displaces the interface further with an intrinsic propagation opposite to the positive mean current. The mechanism requires a nonzero jump, or more generally a gradient, of the mean potential vorticity. It does not require planetary variation of $f$. Reversing the jump reverses the intrinsic propagation; setting the jump to zero removes this interfacial restoring mechanism.

Substitute the Killing-flow four-acceleration and stationary pressure into the relativistic Euler equation. It givesFor the equation of state of blackbody radiation, $p=\rho /3$, this becomes $d\rho =-4\rho d\log N$. The barotropic hydrostatic first integral therefore gives, on each connected nonvacuum fluid region,This is the stationary radiation density law. Combining it with the Stefan–Boltzmann law $\rho \propto T^4$ also gives the Tolman–Ehrenfest relation $TN=\mathrm{constant}$.

At a regular finite free surface, a smooth Killing vector field has finite norm. Hence $C/N^4$ cannot approach zero there. More globally, its norm has a finite upper bound on a compact regular fluid closure, which bounds the density away from zero whenever $C>0$. If the timelike norm instead tends to zero, the density diverges. Thus a radiation fluid has no regular free surface: a finite radiation-filled cavity needs a material wall supporting nonzero pressure, rather than an unsupported radiation-vacuum interface.

A precise stratification–rotation analogy uses constant $N$, ideal fluid, and independence of one horizontal coordinate. For the stratified system take ${\partial }_y=0$, $v=0$, and allow prescribed forces $(F_x,0,F_z)$. The possibly nonlinear Boussinesq equations areFor the unstratified system choose right-handed coordinates $(X,Y,Z)=(z,y,-x)$, rotation $2\mathbf{\Omega }=N{\mathbf{e}}_Z$, and ${\partial }_Y=0$. Its rotating Euler equations, with the static gravitational and centrifugal contributions absorbed into the background pressure, arewhere $D_R={\partial }_t+U{\partial }_X+W{\partial }_Z$. The identification isThen $D_R=D$, $P_X=p_z$, and $P_Z=-p_x$, proving the component-by-component equivalence. In particular, buoyancy perturbation maps to transverse rotating velocity, and stratified horizontal velocity maps to rotation-axis velocity. The mapping permits arbitrary prescribed forces in the two-dimensional plane, but no transverse force: a transverse force in the rotating problem would create a source in the stratified buoyancy equation. A transverse stratified force would also invalidate $v=0$. Constant $N$ is essential to this constant-rotation equivalence.

A physical free surface introduces extra structure beyond the interior equations. After subtracting hydrostatic pressure, its zero total pressure condition gives $p=g\eta$ to first order, together with the kinematic condition ${\eta }_t=w$ for the stratified vertical displacement. The mapping takes this normal velocity into $U$, whereas an actual rotating free surface normal to its physical vertical $Z$ has normal velocity $W$ and a displacement along $Z$. The image of the stratified horizontal surface is a vertical surface in the rotating coordinates. Consequently the physical free-surface problems are not analogous, although deliberately mapped rigid boundaries or mathematical boundary data can be.

For the first prescribed force, the steady linear stratified equations are satisfied byReal parts are understood. The horizontal pressure gradient balances the force; the vertical pressure gradient is balanced by buoyancy; and $w=0$ satisfies buoyancy conservation and incompressible flow. Thus neither $u(z)$ nor $b(z)$ is fixed by the steady equations alone.

Initial rest and switching on in the unbounded fluid remove this ambiguity by causal pressure adjustment by internal waves. Work with the $e^{imz}$ coefficient and let $\hat{F}(q)$ be the horizontal Fourier transform of $f^{\prime }(x)$. For an abrupt switch at time zero, elimination in the linear equations givesAt $q=0$ the bracket cancels the apparent singularity at every finite time. The persistent pressure term is the symmetric antiderivative of the force: the inverse transform of $-i\hat{F}/q$, interpreted as a Cauchy principal value, is $f(x)-\epsilon /2$. The oscillatory part propagates away. To see why its zero-frequency part does not leave an arbitrary offset, note that $\Omega (q)\sim cq$ with $c=N/m$; its long-wave contribution consists of oppositely moving fronts. At fixed $x$, the two far-field values of $f-\epsilon /2$ are $-\epsilon /2$ and $+\epsilon /2$, whose average is zero. Nonzero horizontal wavenumbers give dispersive transients that decay locally. Smooth finite-duration switching changes the transient, not this selected steady limit. ThereforeA spatially constant pressure gauge is immaterial. The long-wave group velocity $c=N/m$ gives the adjustment scale: the elapsed time since switching must satisfy $Nt\gg 1$ and $(N/m)t\gg |x|+x_1$, and exceed the switching duration. These are asymptotic conditions, not an assertion of exact equilibration after a finite time; the full dispersion relation allows tails. The same long-wave limit gives $u_{\infty }=\epsilon m{e}^{imz}/(2N)$ if desired. Equal-amplitude outward fronts leave opposite pressure offsets and the same velocity on their two sides.

For the force depending instead on $y$, any steady linear horizontal momentum balance requires $p_x=f^{\prime }(x)e^{i\ell y}$ and $p_y=0$. Their mixed derivatives disagree wherever $f^{\prime }\ne 0$, since ${\partial }_y{p}_x=i\ell f^{\prime }(x)e^{i\ell y}$ but ${\partial }_x{p}_y=0$. Hence no steady linear velocity field can satisfy the equations, irrespective of whether it is $x$-independent. This statement concerns the linearized problem; nonlinear inertial balances were excluded by the assumptions.

For an unsteady solution put $\mathbf{u}=(-t{\varphi }_y,t{\varphi }_x,0)$ and take $\sigma =0$. Its horizontal divergence is zero. The pressure derivatives needed in momentum balance areTheir compatibility is exactlyThus the Poisson equation supplies an actual pressure and the proposed velocity solves both momentum equations; the vertical momentum and buoyancy equations hold because $w=\sigma =0$. This flow grows linearly in time, so it remains a valid small-disturbance prediction only while its velocities are small.

As $x_1\to 0$ with total force $\epsilon$ fixed, $f^{\prime }\to \epsilon \delta (x)$. Set $\varphi =\Phi (x)e^{i\ell y}$. Away from the source ${\Phi }^{\prime \prime }-{\ell }^2\Phi =0$, and decay at both ends requires $\Phi =A{e}^{-\ell |x|}$. Integrating across the source gives the jump ${\Phi }^{\prime }(0^{+})-{\Phi }^{\prime }(0^{-})=-i\ell \epsilon$. Since this jump is $-2\ell A$, $A=i\epsilon /2$, soThis verifies the distributional source as well as the decay condition.

The corresponding rotating problem is not expected to reproduce this evanescent accelerating flow. Its forcing varies in $Y$, precisely the coordinate whose independence was required in the stratification–rotation analogy. Moreover, $x$ maps to the rotation-axis coordinate $-Z$. At small Rossby number, the Taylor–Proudman theorem gives ${\partial }_Z{\mathbf{u}}_R=0$ in the unforced exterior regions. Such nonzero columns cannot decay with $|Z|$, hence cannot reproduce an $e^{-\ell |x|}$ tail. There is also a direct unsteady obstruction to copying the proposed flow: in a homogeneous fluid rotating about the mapped $x$ axis, a mode with nonzero axial wavenumber has inertial wave frequency of magnitude $N|q|/\sqrt{q^2+{\ell }^2}$, rather than the zero-frequency horizontal vortical response of the stratified problem. The steady or slowly varying exterior rotating response must be columnar or accompanied by outgoing inertial waves; the decaying stratified solution is outside the partial analogy.

The force scale is $F\sim \rho g{a}^3$. At the distant free surface, the Stokeslet's pressure and viscous normal-stress scales are $F/d^2$. Gravity restores a displaced surface by a pressure change $\rho gh$, soThus $h/d=O((a/d{)}^3)\ll 1$, consistent with treating the boundary as approximately flat and impermeable at leading order. This uses the stipulated gravity-dominated restoring balance.

The liquid-air interface has negligible tangential traction. The stress-free image of a normal Stokeslet places an opposite point force at the reflection point, a distance $2d$ above the bubble centre. Across the flat surface the combined normal velocity is odd and the tangential velocity is even, so the normal velocity and tangential stress vanish there. A same-radius inviscid drop of density $2\rho$ has downward excess weight $4\pi \rho g{a}^3/3$, exactly opposite the bubble's upward buoyancy. Its exterior Stokeslet therefore supplies this image, giving the asserted full-space comparison to leading order in $a/d$.

The image's incident axial velocity at the bubble isTo first reflection, the bubble translates with this locally uniform incident velocity plus its isolated rise speed $F/(4\pi \mu a)$. Hence the bubble rise near a distant stress-free free surface isOnly the first relative correction is retained; the image's spatial variation and additional reflections affect higher terms.

The customary capillary-controlled small-deformation condition is a small capillary number,equivalently a small Bond number $\mathrm{Bo}=\rho g{a}^2/\gamma$. This is a sufficient condition, not a necessary condition for the special isolated spherical solution in part (b). More precisely, the image's strain is $O(F/(\mu d^2))$, so its deformation-producing normal stress compared with $\gamma /a$ is $O(\mathrm{Bo}(a/d{)}^2)$. When exploiting the exact isolated solution, that smaller combination controls the leading boundary-induced distortion. If capillarity at the distant free surface is also retained, the gravity-dominated deflection scaling requires $\gamma /(\rho g{d}^2)\ll 1$; $a\ll \sqrt{\gamma /(\rho g)}\ll d$ allows both the usual spherical-bubble criterion and gravity-dominated surface restoration.