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Use pressure per unit constant density, $\pi$, with the gravitational potential and centrifugal potential absorbed into it. Strict geostrophic balance is the exact balancewith material acceleration omitted, for a homogeneous incompressible flow. Writing $f=2\Omega \ne 0$, its components are $fv={\pi }_x$, $fu=-{\pi }_y$ and ${\pi }_z=0$. Differentiating the first two equations with respect to $z$ gives $u_z=v_z=0$. Alternatively, the vector calculus identityshows, on taking the curl of the balance, that $(\mathbf{\Omega }\cdot \nabla )\mathbf{u}=0$. Thus $w_z=0$ too, proving the Taylor–Proudman theorem:The physical pressure still includes its background hydrostatic pressure; it is the reduced pressure $\pi$ that is independent of $z$. Between horizontal impermeable boundaries, the vertically constant $w$ is zero.

For boundaries $z=z_b(x,y)$ and $z=z_t(x,y)$, impermeability condition gives $w_b={\mathbf{u}}_H\cdot {\nabla }_H{z}_b$ and $w_t={\mathbf{u}}_H\cdot {\nabla }_H{z}_t$. A nearly columnar flow crossing contours of $H=z_t-z_b$ must therefore have $w_t-w_b\simeq {\mathbf{u}}_H\cdot {\nabla }_{H}H\ne 0$. Exact columnarity must give way to small corrections from ageostrophic flow. The leading horizontal geostrophic flow can remain nearly independent of $z$, and the leading reduced pressure has ${\pi }_z=0$, but $w_z$ is now nonzero and exact three-dimensional strict geostrophic balance no longer holds. For small slopes, the dominant vertical vorticity equation iswhere $q=v_x-u_y$ is relative vorticity. Equivalently, $D[(q+f)/H]/Dt\simeq 0$. Crossing geostrophic contours stretches or squashes columns: increasing $H$ produces cyclonic relative vorticity, decreasing $H$ produces anticyclonic relative vorticity. The full vorticity equation also contains tilting terms, neglected at this columnar order.

For the circular shallowing, fluid arriving from upstream has potential vorticity $f/h$. Its height is $h(1-ϵ)$ inside, so conservation givesThis uses the stated column-stretching description and initially open streamlines. To obtain the leading disturbance velocity field, use the conventional streamfunction definition $u=-{\Psi }_y$, $v={\Psi }_x$, and write $\Psi =-Uy+\psi (r)$. Then $q={\nabla }_H^2\psi =r^{-1}(r{\psi }_r{)}_r$. Regularity at $r=0$ gives ${\psi }_r=-fϵr/2$ inside. Outside, ${\psi }_r=B/r$; matching tangential velocity at $r=a$ fixes $B=-fϵa^2/2$. Choose the irrelevant additive constant so thatBoth $\psi$ and ${\psi }_r$ match at $a$, so there is no extra vorticity sheet on the rim. This is the circular topographic anticyclone. The radial expression specifies the disturbance; the complete streamfunction is $\Psi =-Uy+\psi$, since a radial function alone cannot represent the required uniform far-field current. Reversing the streamfunction sign convention reverses the displayed proportionality constant, but leaves the physical velocity unchanged.

Differentiation gives the full relative velocity field in the rotating frame:away from the narrow transition across the change of depth. The disturbance turns clockwise for $fϵ>0$ and decays as $1/r$. The horizontal field is the leading small-$ϵ$ columnar inversion: the exact variable-depth transport across a sharp rim needs higher-order corrections, rather than an exactly two-dimensional velocity continuous across a literal vertical step.

Take $U>0$ and $ϵ>0$. The greatest opposing zonal disturbance speed is $Ca$, attained at $(0,-a)$. If $Ca<U$, then $u>0$ everywhere and every particle advances in $x$, precluding closed streamlines. If $Ca>U$, the interior stagnation point $(0,-U/C)$ is a center, since the interior contours are circles about that point. The exterior stagnation point $(0,-C{a}^2/U)$ is a saddle: linearization there has real eigenvalues of opposite sign. Closed contours around the center are bounded by a separatrix through the saddle. The two points meet at the rim at onset, giving the closed-streamline threshold for a circular topographic anticyclone:At this threshold $U/(fa)=ϵ/2\ll 1$, consistent with small Rossby number. Equality is the onset, not a finite-area closed cell. Beyond onset, upstream data alone no longer determine the potential vorticity of trapped fluid; the threshold follows from the continuation of the supplied open-flow model.

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.

Let $\mathbf{n}=\mathbf{R}/R$. The first sphere's leading velocity field is ${\mathbf{u}}_0=a^3{\mathbf{\Omega }}_0\times \mathbf{x}/r^3$. Taking its curl givesThe second sphere is torque-free, soIts own force-free and torque-free motion emits no Stokeslet or rotlet; further reflections are smaller at large separation. In particular it spins in the opposite direction for a primary rotation perpendicular to the line of centers, and in the same direction with twice that magnitude for a parallel rotation.

Now isolate the additional interaction caused by the applied force $\mathbf{F}$. Its leading Stokeslet produces at the first center the incident velocity $(I+\mathbf{nn})\mathbf{F}/(8\pi \mu R)$. The Stokes drag law makes the increment of the force required to hold that sphere fixedThe sign opposes the incident velocity. This holding force acts on the fluid and generates a reflected Stokeslet. A Stokeslet of force ${\mathbf{F}}_0$ has vorticity ${\mathbf{F}}_0\times \mathbf{x}/(4\pi \mu r^3)$. Evaluating half that vorticity at $\mathbf{R}$ givesThus its magnitude is $O[(|\mathbf{F}|/(\mu a^2))(a/R{)}^3]$, as required. This is forced-sphere rotation reflected from a held sphere. Longitudinal forcing has zero leading spin by symmetry. Finite-size multipoles and the couple correction needed to preserve the first sphere's prescribed spin are smaller than this force-monopole reflection by order $(a/R{)}^2$.

The basic quasi-geostrophic streamfunction $\bar{\psi }=-\Lambda yz$ gives $\bar{u}=\Lambda z$, $\bar{v}=0$ and constant quasi-geostrophic potential vorticity $\bar{Q}=f$. DefineBecause the background has no potential-vorticity gradient, the linear potential-vorticity equation is simplyIts characteristics have constant $y,z,x-\Lambda zt$, so the general solution is $q^{\prime }=\mathcal{F}(y,z,x-\Lambda zt)$. In particular, $q^{\prime }=F(y,z)G(x-\Lambda zt)$ is the requested arbitrary separated family; one product is not the most general initial field.

Linearizing the vertical velocity requires retaining advection of ${\bar{\psi }}_z=-\Lambda y$ by $v^{\prime }={\psi }_x^{\prime }$. ThereforeThe impermeable boundary at $z=0$ imposesThis is conservation of boundary buoyancy, not merely a condition that ${\psi }_z^{\prime }$ vanish.

For zero interior potential vorticity, write ${\psi }^{\prime }=\mathrm{Re}\{\varphi (z,t)e^{i(kx+ly)}\}$. With $K_h=\sqrt{k^2+l^2}>0$, stratified quasi-geostrophic inversion givesThe decaying solution isNo time dependence has been assumed to obtain this vertical structure. Applying the boundary condition gives $-\mu \dot{A}-i\Lambda kA=0$, henceThus the Eady edge wave travels eastward with horizontal phase velocity $\omega /k=\Lambda /\mu$ for $k\ne 0$.

The propagation mechanism can be made explicit using boundary parcels. Thermal wind gives the background buoyancy gradient ${\bar{b}}_y=f{\bar{\psi }}_{zy}=-f\Lambda$. A small meridional parcel displacement $d$ produces boundary buoyancy perturbation $b^{\prime }=-d{\bar{b}}_y=f\Lambda d$. Meanwhile the decaying potential-vorticity inversion gives $b^{\prime }=f{\psi }_z^{\prime }=-f\mu {\psi }^{\prime }$ at the boundary. Consequently ${\psi }^{\prime }=-\Lambda d/\mu$ there, and the parcel equation $d_t=v^{\prime }={\psi }_x^{\prime }$ becomesA displaced buoyancy contour therefore induces a velocity field that advances that displacement eastward. No interior potential-vorticity gradient is needed because the boundary buoyancy supplies the invertibility data.

Vortex stretching is what sets the induced flow's vertical reach. In the interior, the zero-potential vorticity condition requiresBoundary isopycnal displacement decreases with height as $e^{-\mu z}$, so fluid columns are stretched or compressed over depth ${\mu }^{-1}=f/(N{K}_h)$. Conservation of potential vorticity makes this stretching balance the relative vorticity. The resulting inversion factor $b^{\prime }=-f\mu {\psi }^{\prime }$ gives the propagation speed: stronger density stratification reduces penetration and speed, while larger $f$ increases them. Zero interior potential vorticity therefore does not imply zero interior flow.

For imposed boundary vertical velocity, retain the same decaying structure and writeThe boundary relation becomes the forced Eady edge wave equationThe integrating factor $e^{i\omega t}$ gives, for ${\omega }_0\ne \omega$,The arbitrary homogeneous coefficient $B$ sets the initial disturbance. With $A(0)=0$, this is $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$.

At resonance, ${\omega }_0=\omega$, integration instead yieldsAn initially undisturbed fluid has ${\psi }^{\prime }=Ct{e}^{-\mu z}\cos (kx+ly-\omega t)$. The imposed pumping reinforces the same boundary normal mode every cycle, with no dissipative loss, so the linear amplitude grows secularly. The divergence describes resonance of the undamped linear model; sufficiently large wave amplitude invalidates the assumed small-disturbance approximation.

Put $a=\kappa m^2$, with $\kappa >0$ and $m\ne 0$. The relevant homogenization of a periodic advection-diffusion equation limit is a long horizontal wavelength observed on its diffusive time scale. Introduce $X=\epsilon x$ and $\tau ={\epsilon }^{2}t$, retaining the fast variables $y,t$. Since the velocity field is independent of $x$, no fast horizontal cell variable is needed. Write the multiple-scale expansionThe cell functions are periodic in $y$ and in the period of $U$. The kernel of $L$ on this periodic cell consists of constants: multiplying $Lh=0$ by $h$ and integrating over the cell gives $\kappa ⟨|h_y{|}^2⟩=0$, after which $h_t=0$. Thus the leading concentration really is independent of the fast variables. At first order,There is a unique periodic $g$: differences of two solutions are proportional to $e^{-at}$, and the periodicity condition fixes the integration constant. Equivalently,At second order the advection-diffusion equation givesIntegration over the fast cell annihilates the left side. Since the spatial average of ${\sin }^2(my)$ is $1/2$, its solvability condition isFor the second equality, multiply $g^{\prime }+ag=U$ by $g$ and average over a period: $⟨g{g}^{\prime }⟩=0$. In particular the enhancement of diffusivity is nonnegative. Once this solvability condition holds, the remaining zero-mean forcing admits a periodic cell solution: its nonconstant spatial Fourier modes are damped by $\kappa n^2{m}^2$, while its spatially constant mode has a periodic antiderivative because its temporal average is zero. This also explains the relaxation of initial transverse inhomogeneity. The first corrector has zero spatial mean, so $\bar{\chi }=C+O({\epsilon }^2)$.

This derivation describes the large-scale, long-time limit, after times comparable to the period and to $a^{-1}$. It does not give an exact closed equation for the spatial average of arbitrary finite-time data: that average contains a correlation between the velocity field and concentration. In physical coordinates the leading long-wavelength concentration therefore obeys ${\bar{\chi }}_t={\kappa }_e{\bar{\chi }}_{xx}$ asymptotically. The cell expansion, its bounded periodic correctors, and the decay of transverse cell transients justify this effective diffusivity of a periodic sinusoidal shear.

For constant $U_0$, the periodic cell solution is $g=U_0/a$. ConsequentlyThis is Taylor dispersion: molecular diffusion moves particles between different horizontal velocities and turns their accumulated horizontal displacement into large-scale diffusion.

For the impulsive flow, $g$ jumps by $U_1$ at each kick and decays exponentially between kicks. If $G$ denotes its value just after a kick, periodicity requires $G=e^{-aT}G+U_1$. Thus, within one period,Using the mean-square expression, which has no ambiguous product of a Dirac delta with a discontinuous function, givesThe impulsive model can also be obtained by shrinking smooth pulses of fixed integrated strength. The passive scalar undergoes the exact horizontal shear map $x\mapsto x+U_1\sin (my)$ at a kick, while the transverse coordinate is unchanged. The cell solution above is the corresponding limit of the smooth cell solutions. Using a one-sided value in $Ug$ instead would miss the contribution from the jump of $g^2$.

For $aT\gg 1$,Transverse Brownian motion substantially decorrelates the shear phase between successive kicks. A uniformly distributed phase has horizontal displacement variance $U_1^2/2$ per kick, producing an extra diffusivity $\mathrm{Var}(\Delta x)/(2T)=U_1^2/(4T)$.

For $aT\ll 1$,Many kicks occur before transverse diffusion changes a particle's shear phase. Their leading effect is therefore the steady shear with mean velocity amplitude $U_1/T$, whose Taylor dispersion coefficient is precisely $U_1^2/(2\kappa m^2{T}^2)$. The divergence as $\kappa \to 0$ describes a very long decorrelation time; it does not imply infinite finite-time spreading. At $\kappa =0$ the phase never changes and, for a uniform phase, $n$ kicks give variance $n^2{U}_1^2/2$. The spreading is then ballistic transport, rather than diffusion. The long-time diffusive limit must precede the zero-diffusivity limit, with $t\gg a^{-1}$.

A large molecular Schmidt number $\mathrm{Sc}=\nu /\kappa$ separates the velocity's smallest smooth length scale from the scalar's smaller diffusive length scale. Follow a Lagrangian trajectory $X(t)$ and measure displacement $x$ from it. Within a sufficiently small scalar packet, Taylor expansion of the smooth velocity field givesThe moving frame removes uniform sweeping, leaving a linear random velocity gradient tensor. Thus the local scalar equation is the advection-diffusion equation with velocity $Cx$, and averaging over realizations of $C$ gives the corresponding ensemble average. The approximation requires the packet to lie within the velocity's locally smooth region. Its thin, diffusive direction can satisfy this condition even when its long direction eventually becomes too large for one global linearization; applications to a spatially varying flow then use local packet descriptions. The statistical conclusions below refer to the linear-flow model and the usual mixing assumptions, not to every conceivable trace-free random matrix process.

Write $Q=\int \chi dA$ and assume sufficiently rapid decay to discard boundary terms. Integration by parts givesFor $J_{ij}=\int x_i{x}_j\chi dA$, integrate the advective term once and the diffusive term twice. Since ${\partial }_k(x_i{x}_j)={\delta }_{ki}{x}_j+{\delta }_{kj}{x}_i$ and $\Delta (x_i{x}_j)=2{\delta }_{ij}$, one obtainsThese are mass conservation and the evolution of the unnormalized second moment tensor. The first moment obeys $\dot{M}=CM$, so a centered packet stays centered.

For a positive Gaussian scalar packet in a linear incompressible flow, $B$ must be a positive-definite matrix. Diagonalizing its symmetric matrix and applying the Gaussian integral givesThe factor $Q^2$ in the last identity matters because $J$ has not been divided by the scalar mass. Substitution into the advection-diffusion equation also givesThese identities agree with the second moment equation and prove that the Gaussian family is preserved. For an initially negative packet the same analysis applies after replacing $\chi$ by $-\chi$ and taking its positive mass.

To see the roles of stretching and diffusion explicitly, let the deformation gradient $F$ solve $\dot{F}=CF$, $F(0)=I$. Incompressibility gives $\det F=1$. Differentiating $F^{-1}J{F}^{-T}$ integrates the moment equation:Without diffusion, $J=FJ(0)F^T$ and $\det J$ is constant. In a statistically stationary, mixing flow with positive Lyapunov exponent $\lambda$, the two stretching exponents are $\lambda$ and $-\lambda$. Writing the ordered eigenvalues of $J$ as $e^{2{\rho }_1},e^{2{\rho }_2}$, their typical initially nondiffusive evolution isThe fluctuation scale presumes the usual central limit theorem for accumulated strain; it is not implied by incompressibility alone. During this stage the packet becomes long and thin but keeps its area and peak concentration.

In the instantaneous principal-axis basis, write $\tilde{C}$ for the rotated velocity gradient tensor. Differentiation of an eigenvalue givesThe formulas hold through a repeated eigenvalue by continuity of their sum. The compressed physical variance is $e^{2{\rho }_2}/Q$. Its contraction is arrested when it is of order $\kappa /\lambda$, the square of the Batchelor scalar microscale ${\ell }_B\sim \sqrt{\kappa /\lambda }$. If its initial length is $L\gg {\ell }_B$, the crossover occurs atThereafter ${\rho }_2$ has stationary fluctuations associated with the strain and the diffusive floor, while ${\rho }_1$ continues to grow typically at rate $\lambda$. Under sufficiently fast mixing, the central part of the distribution of ${\rho }_1$ is approximately Gaussian with mean $\lambda t+O(1)$ and variance $D_{\rho }t+o(t)$ for a strain-dependent coefficient $D_{\rho }$. This central approximation does not determine far tails, which control high concentration moments; those require a full large deviation principle. The two logarithmic widths need not be independent. The area now increases and $f$ decreases typically as $e^{-\lambda t}$. This is diffusion arrest of Gaussian packet compression.

For example, a constant principal strain $C=\mathrm{diag}(\lambda ,-\lambda )$ gives the exact physical variancesThe thin variance approaches its diffusive floor, the long variance grows exponentially, and the mass-to-area relation gives the quoted typical amplitude decay. A random strain additionally makes rare histories important for concentration moments.

For a precise statistical formulation define the logarithmic area increaseThe Gaussian identities give the exact relation $f(t)=f(0)e^{-H_t}$; henceAssume the long-time strain statistics give a large deviation principle for $H_t/t$ on $s\ge 0$, with convex good rate function $I(s)$, $I(\lambda )=0$. At exponential accuracy the probability of $H_t/t$ near $s$ is $e^{-tI(s)}$. Combining that cost with the amplitude factor and applying the Laplace principle for probability measures gives the large-deviation decay rates of a Gaussian scalar packetThe initial amplitude only changes a time-independent prefactor. One can see the minimization directly by dividing the possible area-growth rates into small intervals: each contributes its probability cost times $e^{-\mu ts}$, and the interval with least total exponential cost dominates. An interval around a minimizing $s$ gives the matching lower bound. The nonnegative area increase bounds the exponential weight by one, so unusually large positive growth cannot produce a divergent weighted tail. Using the area rather than only ${\rho }_1$ also avoids discarding possibly important fluctuations of the compressed direction.

Suppose $I$ is strictly convex and differentiable, with its minimum at $\lambda >0$ and finite negative right derivative at zero. DefineFor $0<\mu <{\mu }_{\ast }$ the minimizing rate $s_{\mu }$ is positive and uniquely solvesFor $\mu \ge {\mu }_{\ast }$, convexity puts the minimum at the endpoint $s=0$, givingHigh moments are dominated by rare histories with little net area growth, in which a packet keeps a substantial concentration. Their probability cost is $I(0)$, independent of the moment order. A finite threshold is conditional: if $I^{\prime }(0+)=-\infty$, the minimizer can approach zero without reaching it at any finite $\mu$.

Below the threshold, differentiation of the minimizing expression cancels the terms involving $s_{\mu }^{\prime }$ because $I^{\prime }(s_{\mu })+\mu =0$. Thus ${\gamma }_{\mu }^{\prime }=s_{\mu }$, andStrict negativity uses a nondegenerate rate function with a unique zero at $\lambda$: for positive $\mu$, $s_{\mu }<\lambda$ and $I(s_{\mu })>0$. For deterministic strain the ratio can instead be constant. Near the typical rate, if $I(s)\simeq (s-\lambda {)}^2/(2D)$ with $D>0$, then ${\gamma }_{\mu }\simeq \lambda \mu -D{\mu }^2/2$ for small $\mu$. If this quadratic expression holds all the way to zero, it gives the illustrative full resultThe quadratic form is an example, not a universal law of turbulent strain.

A thin scalar packet is stretched until molecular diffusion broadens its compressed direction; conservation of its scalar mass then converts area growth into dilution. Ensemble averages sample different stretching histories, producing different decay rates for different moments and saturation when weakly stretched histories dominate. The present moment calculation is specifically for the centered Gaussian packet and averaging over the flow; signed random initial fields with overlapping packets can also have cancellations and require their initial-field statistics.

Finally, positive exponential decay is a statistical conclusion requiring the strain assumptions above, not a consequence of the local differential equation by itself. For $C\equiv 0$ and initially isotropic physical variance $L^2$, the exact solution has $J=Q(L^2+2\kappa t)I$ and $f=Q/[2\pi (L^2+2\kappa t)]$. Its moments decay algebraically as $t^{-\mu }$. Thus the general equations for the exponential rates are the rate-function minimization above; numerical rates and a finite saturation threshold require the long-time statistics of the flow.

With polar unit vectors ${\mathbf{e}}_r=(\cos \theta ,\sin \theta ,0)$ and ${\mathbf{e}}_{\theta }=(-\sin \theta ,\cos \theta ,0)$, the second term is $\Gamma {\mathbf{e}}_{\theta }/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero divergence and zero vorticity away from the axis. Its singular axis carries a line vortex; adding the constant velocity field gives a uniform flow plus that line vortex.

The counterclockwise circulation around a positively oriented loop is $\oint \mathbf{u}\cdot d\mathbf{r}$. For a radius-$R$ circle, the uniform flow contributes zero and the line vortex contributesAny loop winding once around the axis has the same circulation, by Stokes theorem in the nonsingular intervening region.

On $C_R$, $\mathbf{n}={\mathbf{e}}_r$, $\mathbf{u}\cdot \mathbf{n}=U\cos \theta$, and $dl=Rd\theta$. ThusIn fact this equality holds for every $R>0$ for the stated field, since the remaining vector integral is $(0,\pi ,0)$.

This is the advective part of the momentum flux. To obtain the force, include pressure. For steady irrotational flow, the Bernoulli equation givesThe constant terms integrate to zero against $\mathbf{n}$; the remaining pressure contribution is $\oint p\mathbf{n}dl=\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The total outward momentum flux plus pressure term is therefore $\rho \mathbf{\Gamma }\times \mathbf{U}$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, isThis is the Kutta–Joukowski theorem, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional aerofoil, faster-decaying far-field terms do not change this limit. The displayed advective integral supplies half the lift; pressure supplies the other half.

For potential flow $\mathbf{u}=\nabla \phi$ in an inviscid fluid of constant density under the conservative force per unit mass $-\nabla \chi$, the unsteady Bernoulli equation isIt follows by inserting the velocity potential into the Euler equations and integrating the resulting spatial gradient. A time-only change of the velocity potential changes $C(t)$ without altering the velocity field.

For spherical motion in an unbounded liquid, incompressibility gives $r^2{u}_r=A(t)$; the kinematic condition $u_r(R)=\dot{R}$ gives $A=R^2\dot{R}$. Choose $\phi =-R^2\dot{R}/r$, vanishing at infinity. To obtain the displayed bubble relation, take no spatial body-force contribution, neglect viscosity and surface tension, and equate the liquid-side interface pressure to $p_b$. With the far-field pressure $p_{\infty }$, Bernoulli equation evaluated at the surface uses the partial derivative at fixed $r$:Hence the Rayleigh equation for an inviscid spherical bubble is

The exterior kinetic energy isDifferentiating it and using $\dot{V}=4\pi R^2\dot{R}$ gives the pressure-work energy balance for a spherical bubble:This derivation also holds at a turning point $\dot{R}=0$, without dividing by $\dot{R}$.

For constant $p_{\infty }$ and the stated inverse-volume bubble pressure, integrate this exact time derivative:The relation is valid on physically accessible portions of the motion, where the resulting kinetic energy is nonnegative. In particular $\log q\le q-1$ makes the bracket nonpositive, as expected for the restoring pressure about $V_0$.

A particle pathline follows the time-dependent velocity field, so its coordinates satisfy $\dot{x}-x=\sin t$ and $\dot{y}=-y$. The second equation and the initial data give $y=e^{-t}$. Using the integrating factor $e^{-t}$ in the first,Therefore the complete pathline isIf $x_0+\frac{1}{2}\ne 0$, its first coordinate grows without bound in magnitude. If $x_0=-\frac{1}{2}$, the first coordinate remains bounded and periodic while $y\to 0$. The unique release coordinate giving a bounded particle path is $x_0=-1/2$.

For a two-dimensional incompressible flow $\mathbf{u}=(u,v)$, the continuity equation is $u_x+v_y=0$. Locally, or globally on a simply connected fluid region, this permits a streamfunction withIndeed, the one-form $-vdx+udy$ is closed by incompressibility and hence is locally the differential of $\psi$. Conversely these expressions automatically give zero divergence by equality of mixed derivatives.

A streamline at a fixed time is tangent to the instantaneous velocity field. Along it,Thus streamlines are connected curves on which the streamfunction is constant; wherever $\nabla \psi \ne 0$, the level curves have exactly the required tangent direction. At a stagnation point, the tangent is degenerate and distinct branches of a level set need not form a single streamline. For unsteady flow this argument fixes time: it does not identify streamlines with particle pathlines.

Take a positive initial separation $0<l_0\ll \eta$. In the dissipation range, the velocity field is smooth, so the difference of the two Lagrangian trajectories obeys the linearized separation equation $\dot{\mathbf{l}}\simeq (\nabla \mathbf{u})\mathbf{l}$. Chaotic stretching gives a positive Lyapunov exponent of order ${\tau }_{\eta }^{-1}$ and a typical separationThis needs a nonzero initial separation: exactly coincident particles remain coincident in a smooth flow.

Once $l$ lies in the inertial range, the Kolmogorov 1941 theory gives a relative speed of order $(ϵl{)}^{1/3}$. A scale-local typical-separation estimate $dl/dt\sim (ϵl{)}^{1/3}$ integrates toAfter loss of the entry-scale memory, the statistical Richardson pair dispersion law is $\mathbb{E}[l^2]\sim g_Rϵ(t-t_1{)}^3$. Thus the characteristic separation grows as $(t-t_1{)}^{3/2}$ while $\eta \ll l\ll L$. This is a statistical scaling closure, rather than a deterministic equation for every pair.

For $l\gg L$, the particle velocities become approximately independent. If their Lagrangian velocity autocorrelations have a finite Lagrangian integral time $T_L\sim L/U$, the diffusive large-scale pair dispersion has effective relative diffusivity of order $UL$. After a further velocity-decorrelation transient,This is the long-time mechanism of Taylor turbulent dispersion applied to the difference of two decorrelated trajectories. The root mean square separation eventually loses memory of the $L^2$ term.

Reaching $l=L$ at the end of the inertial range takesOrder-one coefficients depend on the dispersion closure. The initial exponential stage can take a long time if $l_0$ is extremely small, but the inertial-range stage is of order one outer eddy turnover time.

The three-regime turbulent pair-separation model in the sketch matches the regimes continuously for illustration. Its coefficients are schematic; the universal claims here are the scaling powers under the stated stretching, locality and decorrelation assumptions.

A passive scalar $\theta$ is transported by the velocity field without feeding back on it:Take its mean to be zero, and define the scalar dissipation rate by the half-scalar variance convention $\chi =\kappa ⟨|\nabla \theta {|}^2⟩$. In a statistically equilibrated energy cascade of scalar fluctuations, this is also the flux of half-scalar variance toward small scales.

The Kolmogorov two-thirds law gives a typical velocity increment $\delta u_r\sim (ϵr{)}^{1/3}$ and hence eddy turnover time ${\tau }_r\sim r/\delta u_r\sim {ϵ}^{-1/3}{r}^{2/3}$. Assuming local scalar transfer on this same time scale, $\chi \sim ⟨(\Delta \theta {)}^2⟩/{\tau }_r$. The Obukhov-Corrsin theory therefore gives the scalar analogue:This applies to separations at which both direct forcing and molecular diffusion are negligible and the transporting velocity increments lie in the inertial range. For very large or very small ratios $\nu /\kappa$, the scalar and velocity cutoff scales differ, so this common range must be checked. Defining $\chi$ as dissipation of the full scalar variance instead simply changes the convention for $C_{\theta }$.

Write $\mathbf{X}=ϵ\mathbf{x}$, $\mathbf{k}={\nabla }_{\mathbf{X}}\theta$, and denote the slow density and velocity field amplitudes by $a_{\rho }$ and ${\mathbf{a}}_u$. Acting on the rapid phase, ${\partial }_t+\mathbf{U}\cdot \nabla$ gives $i\Omega$, where $\Omega =\omega -\mathbf{U}\cdot \mathbf{k}$. Spatial differentiation gives $-i\mathbf{k}$, while derivatives of the amplitudes and the mean shear flow are $O(ϵ)$. The leading continuity equation and linearized Euler equations therefore becomeEliminating ${\mathbf{a}}_u$ on the nonzero acoustic branch gives the acoustic eikonal equation in a shear flow:The mean-shear term is small here because $\nabla \mathbf{U}=O(ϵ)$; retaining advection does not require retaining that gradient in the leading phase equation. The positive intrinsic-frequency acoustic dispersion relation is $\omega =U{k}_x+c_0|\mathbf{k}|$, with group velocity ${\mathbf{v}}_g=U{\mathbf{e}}_x+c_0\mathbf{k}/|\mathbf{k}|$.