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At leading order in the lubrication approximation, surface arclength is horizontal distance. An insoluble surfactant is transported by the surface velocity, without bulk exchange or diffusion. Conservation of insoluble surfactant on a moving interface therefore becomesThe local hydrostatic pressure and capillary normal stress giveSolve $\mu u_{zz}=p_x$ with no-slip boundary condition at $z=0$ and Marangoni stress $\mu u_z(h)={\gamma }_x=-A{C}_x$. The velocity, surface velocity and volume flux per unit width areMass conservation, $h_t+q_x=0$, yields the thin-film equations with insoluble surfactant:and the corresponding surfactant equation isThe hydrostatic and capillary terms thus affect both the film flux and the surface transport, with different coefficients.

For finite-mass Marangoni spreading on a liquid film, first neglect both pressure-gradient terms. If the spread has size $\ell$, then $C\sim M/\ell$ and $C_x\sim M/{\ell }^2$, giving surface speed $u_s\sim AM{h}_0/(\mu {\ell }^2)$. Equating $\ell /t$ to this speed givesSet $\ell =(AM{h}_{0}t/\mu {)}^{1/3}$ exactly as a similarity scale, and writeOn the positive half of the pool, substitution produces two ordinary differential equations:Symmetry fixes the flux integration constant in the second equation to zero, giving $H{\Gamma }^{\prime }=-\eta /3$ wherever $\Gamma >0$. Substitution into the first equation gives $\eta H^{\prime }=H$. Thus the linear similarity profiles for surfactant spreading have the formThe concentration vanishes at the moving front. The two integral constraints are total surfactant mass and conservation of the fluid volume relative to the undisturbed layer:The first gives $k={\eta }_N^2/3$; the second gives $k{\eta }_N=2$. Therefore ${\eta }_N^3=6$. In physical variables, the solution isOutside the pool the leading outer solution has $h=h_0$, $C=0$. In particular, $h(x_{N-},t)=2h_0$: the subscript means the one-sided limit just behind the front. As a consistency check, the surface speed there is $x_N/(3t)={\dot{x}}_N$ and the fluid jump satisfies the moving-front mass conservation condition. This ideal outer solution has a height jump from $2h_0$ to $h_0$.

That jump cannot persist in a physical interface with nonzero gravity or surface tension. Across a smoothing region of width $\Delta$, $h$ changes by $O(h_0)$. The Marangoni flux scales as $AM{h}_0^2/(\mu x_N^2)$. At $g=0$, its balance with the capillary flux ${\gamma }_0{h}_0^4/(\mu {\Delta }^3)$ gives capillary smoothing of a Marangoni front:Here the surface tension approaches ${\gamma }_0$ at the surfactant-free edge. The assumed concentration-gradient scale is $M/x_N^2$ even inside the smoothing region.

When gravity dominates capillarity, the hydrostatic flux is $\rho g{h}_0^4/(\mu \Delta )$. The same balance gives gravity smoothing of a Marangoni front:The gravity-dominated condition is ${\Delta }^2\gg {\gamma }_0/(\rho g)$, the square of the capillary length. The printed condition omits $\rho$; the expression above restores dimensional consistency.

This width grows faster than $x_N$. It becomes comparable with the pool size whenThe time is a scaling estimate, so numerical factors cannot be fixed by the width balance. For $t\gg t^{\ast }$, the sharp-front similarity profile no longer describes the film: hydrostatic pressure levels the layer over the whole pool, and its thickness tends towards $h_0$ with an increasingly small depression beneath the surfactant. The surface can still spread by Marangoni stress while a hydrostatic-pressure-driven return flow nearly cancels the net film flux. In this gravity-levelled surfactant film, $q\simeq 0$ gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$, so the fractional height variation is of order $AM/(\rho g{h}_0^2{x}_N)\ll 1$.

Apply volume conservation, mass conservation and vertical momentum flux balance to a slice of the triangular-profile line plume. Ambient ingestion contributes $2e$ of volume and $2{\rho }_{0}e$ of mass per unit height and source length. The ambient is quiescent, so it supplies no leading vertical momentum. The integrated driving force is the stored buoyancy $\mathcal{B}$. Using the quantities from part (a), the non-Boussinesq triangular-profile line-plume balances areTogether with Batchelor entrainment, these are three equations for $W,\rho ,b$. They neglect viscous boundary stresses and streamwise pressure-force corrections within the integral-plume approximation.

Multiplying volume conservation by $g{\rho }_0$ and subtracting $g$ times mass conservation cancels the ambient sources exactly, givingThis is buoyancy conservation in the homogeneous ambient; it is a consequence of the first two balances, not an additional independent equation.

Now take the Boussinesq approximation in inertia and entrainment, retaining the small density deficit in buoyancy. Then $Q={\rho }_{0}bW$, $M=(2/3){\rho }_{0}b{W}^2$ and $F=(2/3)gbW({\rho }_0-\rho )$. ConsequentlyThe mass source is $2\alpha {\rho }_{0}W=3\alpha {\rho }_{0}M/Q$. Substitution into the mass conservation, momentum and buoyancy balances gives the unsteady Boussinesq triangular-profile line-plume balancesThese use density-weighted mass and momentum fluxes. Removing ${\rho }_0$ from some flux definitions while retaining it in the source would mix incompatible conventions.

Let $e$ be specific internal energy, $Q$ the heat supplied per unit volume per unit time, and $E=\rho (e+|\mathbf{u}{|}^2/2)$. For an inviscid perfect gas in a prescribed Newtonian gravitational potential $\Phi$, the fluid total-energy equation isThere is no gravitational term if no external body is present. For a time-independent $\Phi$, adding potential energy gives the equivalent conservative form ${\partial }_t(E+\rho \Phi )+\nabla \cdot [(E+p+\rho \Phi )\mathbf{u}]=Q$. Subtracting the kinetic-energy equation and using mass conservation givesFor a perfect gas with constant specific-heat ratio $\gamma$, $e=p/[(\gamma -1)\rho ]$. The isothermal equation of state therefore makes $e=c_s^2/(\gamma -1)$ constant. Hence the required heat supply and heat loss areCompression requires cooling; expansion requires heating. The isothermal sound speed $c_s$ differs from the adiabatic sound speed $\sqrt{\gamma }{c}_s$.

Define the isothermal Mach numbers by ${\mathcal{M}}_i=u_i/c_s$. Orient the normal to a stationary isothermal shock along the flow, so the mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2>0$. Tangential velocity is continuous for this planar hydrodynamic shock. Normal momentum conservation givesFor a genuine discontinuity $u_1\ne u_2$, factorization gives $u_1{u}_2=c_s^2$. Thus the isothermal shock jump relations areAcross a thin shock, gravitational potential and tangential kinetic energy are unchanged. The specific enthalpy $h=\gamma c_s^2/(\gamma -1)$ is the same on both sides, so the energy removed per unit area per unit time isPositive net cooling requires $u_1>u_2$, equivalently ${\rho }_2>{\rho }_1$ and ${\mathcal{M}}_1>1>{\mathcal{M}}_2$. Cooling across an isothermal shock permits only compression shocks; the formal expansion discontinuity requires energy supplied by the surroundings. The continuous state $u_1=u_2$ is not a shock.

For steady radial flow, mass conservation gives $r^2\rho u_r=\text{constant}$. Differentiating it and eliminating the mass density gradient from radial momentum balance givesA regular sonic point has $|u_r|=c_s$, so the right-hand side must vanish there as well. ThereforeThis is the Bondi sonic point for accretion; the same critical radius appears in an isothermal Parker wind. A general steady solution need not pass through a sonic point, but a smooth transonic one must obey both critical conditions.

Put $x=r/r_s$ and $\mathcal{M}=|u_r|/c_s$ on a branch whose flow direction is fixed. The radial equation becomesIntegrating yields the integrated isothermal Bondi flow relation:For a transonic branch, $x=\mathcal{M}=1$ gives $C=-3/2$. Expansion about the critical point gives $(\mathcal{M}-1{)}^2=(x-1{)}^2+O((x-1{)}^3)$, with slope $-1$ for inflow supplied by gas at rest at infinity and slope $+1$ for a transonic outflow.

For Isothermal Bondi accretion, integrate radial momentum balance once more, using $u_r\to 0$ and $\rho \to {\rho }_{\infty }$ at infinity:At the Bondi sonic point, $GM/r_s=2c_s^2$ and $u_r^2=c_s^2$, so ${\rho }_s=e^{3/2}{\rho }_{\infty }$. The positive inward isothermal Bondi accretion rate is consequentlyThe critical mass density and rate are fixed by the regular transonic solution and the specified reservoir at infinity; an arbitrary subsonic solution does not share this accretion rate.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

Use the steady Boussinesq top-hat plume in a stratified ambient, with negligible ambient vertical momentum and no externally added heat or particles. The Batchelor entrainment hypothesis supplies inward velocity $\alpha w$ across circumference $2\pi b$. Volume conservation gives $\pi Q_z=2\pi b\alpha w$, and $bw=\sqrt{M}$, henceThe momentum flux increases by the integrated buoyancy force. Writing $g^{\prime }=g\Delta \rho /{\rho }_o$, the force per reference density and per $\pi$ is $g^{\prime }{b}^2$. Since $F=g^{\prime }Q$ and $b^2=Q^2/M$, this givesFor the buoyancy flux, entrained material has the local environmental mass density. Let ${\rho }_p$ be the plume's averaged density. Its scalar mass conservation is $({\rho }_{p}Q{)}_z={\rho }_a{Q}_z$. Since $F=(g/{\rho }_o)({\rho }_a-{\rho }_p)Q$, differentiating cancels both entrainment terms and givesIn stable density stratification, ${\rho }_a^{\prime }<0$ and a rising plume loses buoyancy flux. These are kinematic fluxes with the common factor $\pi {\rho }_o$ removed, so no additional factor of $\pi$ belongs in the three differential equations.

For identical suspended particles, reduced gravity is proportional to particle volume fraction. If particles settle through the bed at speed $v_s$ and are not resuspended, the downward particle deposition flux removes a buoyancy-equivalent amount $v_s{g}^{\prime }$ per unit bed area. The depth-integrated scalar mass conservation equation isFor a steady particle-laden gravity current, $B_x=-v_{s}B/(uh)$. If $v_s/u\ll 1$, with fixed positive $E$ and drag/slope coefficients, the locally adjusted leading solution of the momentum and volume equations is $u\sim {\beta }_{\ast }{B}^{1/3}$ and $h\sim h_0+E(x-x_0)$. The weak-settling attenuation of a sloping gravity current then givesand henceFor an inlet at $x_0=0$ this has the requested logarithmic form. If the inlet is genuinely at $x_0\ne 0$, the argument must contain $x-x_0$ in order to satisfy the specified inlet condition.

The speed factor must be the ${\beta }_{\ast }$ derived in the previous part. The printed $\beta$ omits the factor one-half multiplying $E\cos \theta$ and uses an undefined uppercase $C$ instead of the governing equation's $c$. Taking $C=c$ still leaves a nonzero leading momentum residual $-\frac{1}{2}E(B/u)\cos \theta$ for the printed speed; this discrepancy cannot be repaired by a correction of order $v_s/u$. With the given hydrostatic momentum equation, the logarithmic law is valid with ${\beta }_{\ast }$, not the printed speed factor.

A controlled local approximation also needs $v_s/(Eu)\ll 1$ if $E$ is allowed to become small. Indeed the exact volume balance is $h_x=E-h{u}_x/u$; under $u\sim {\beta }_{\ast }{B}^{1/3}$ its second term is $v_s/(3u)$. The neglected depth and acceleration corrections become significant as $B$ approaches zero.

Take $z$ upward and $u_s>0$ downward. The quadratic hindered-settling flux is $j(\varphi )=-u_s\varphi (1-\varphi )$. Particle mass conservation gives the kinematic sedimentation equationon a smooth characteristic curve. Integrate the scalar conservation law across a moving discontinuity: $V[\varphi ]=[j]$. The Rankine-Hugoniot condition gives the settling shock speedwhere $L$ is the lower side and $R$ the upper side. Since $j^{\prime \prime }=2u_s>0$, a compressive shock has ${\varphi }_L>{\varphi }_R$ and its neighboring characteristic curves point into it.

Put $a={\varphi }_1$, so ${\varphi }_2=2a$, with $0<a<1/4$. The two-layer sedimentation with a compression shock hasThese respectively separate upper suspension from clear fluid, lower suspension from upper suspension, and deposit from lower suspension. Their initial positions are $H,z_m,0$. The characteristic curves in the two suspended states have slopes $u_s(2a-1)$ and $u_s(4a-1)$; clear-fluid and deposit characteristic curves have slopes $-u_s$ and $u_s$.

For $z_m=H/2$, $V_2$ and $V_3$ meet first. Their intersection givesThe potential meeting of the upper two fronts would take $(H-z_m)/(2u_{s}a)$, which is later in the stated range. The new front separates particle volume fraction one from particle volume fraction $a$, soIt follows $z=h_{c1}+u_{s}a(t-t_{c1})$ until it meets $z=H-u_s(1-a)t$. The final mass-conserving intersection isFor $z_m=H/2$, these are $H(1-3a/2)/[u_s(1-a)]$ and $3aH/2$.

The PDF prints a plus sign before $a{z}_m/H$ in the stopping-time numerator. That sign contradicts both the top-front trajectory and particle mass conservation; the correct sign is minus, as derived here. Once all particles are deposited, the remaining $1$-to-$0$ front is stationary because both limiting fluxes vanish.

The first panel shows the two-front merger and the final intersection, with straight characteristic curves in each suspended layer. The second panel shows the tuned three-front intersection.

Retain $a={\varphi }_1$, ${\varphi }_2=2a$ and the three settling shock speeds. A triple-shock sedimentation point requires the meeting of $V_2$ with $V_3$ to coincide with the meeting of $V_1$ with $V_2$. HenceSolving, and using the deposit-front position $2u_{s}at$, givesThe common height also equals $a(H+z_m)$, independently checking mass conservation. The second panel of the compression-shock diagram shows $V_1,V_2,V_3$ meeting at this point. For $t>t_c$ the suspension is exhausted: deposit below $h_c$ and clear fluid above it are separated by a stationary shock, with zero flux on both sides.

Take $z$ upward from the horizontal bed and measure $x$ away from an ice divide. In the shallow-ice approximation, the hydrostatic pressure is $p=p_a+\rho g(h-z)$ and the horizontal balance for a Newtonian fluid isThe upper stress-free boundary condition gives $u_z(h)=0$. The resisting basal traction is $-\mu \beta u_b$, so the fluid-side shear stress satisfies $\mu u_z(0)=\mu \beta u_b$. This is a Navier slip boundary condition with slip length ${\ell }_s=1/\beta$. Integrating twice givesThe first term is the uniform basal sliding contribution, while the quadratic term is internal viscous deformation. Both are positive where the thickness decreases downstream.

The shallow-ice flux with linear basal drag isNo ice accumulation or ice ablation appears. For a symmetric glacier, work on one half, $0\le x\le x_N(t)$, with $q(0,t)=0$, $h(x_N,t)=0$ and no outgoing volume flux. Its conserved half-volume per unit width is $V_g={\int }_0^{x_N}hdx$; the full glacier has volume $2V_g$. This specifies the volume convention rather than silently supplying an unspecified value. For a general finite release on the whole line, the late similarity solution is centred on its conserved centre of mass and uses half its total volume for $V_g$.

The internal-shear and linear basal drag law mobilities are equal at $H_{\ast }=3/\beta$. Suitable vertical, horizontal and temporal scales areWrite $h=H_{\ast }\mathcal{H}$, $x=L_{\ast }X$, $t=T_{\ast }\tau$. ThenThe initial typical thickness is $\mathcal{H}\simeq 10$, the initial extent is of order $L_{\ast }/10$, and its shear-controlled spreading time is of order $T_{\ast }/10^5$. The shallow-ice approximation also requires small aspect ratio: in particular the initial thickness and extent must satisfy approximately $100H_{\ast }/L_{\ast }\ll 1$. The equations model the broad shallow bulk; the very tip can require physics beyond that approximation.

The two limiting equations are porous medium equations of the form ${\mathcal{H}}_{\tau }=({\mathcal{H}}^n{\mathcal{H}}_X{)}_X$, with $n=3$ for a thick shear-dominated glacier and $n=2$ for a thin sliding-dominated glacier. Volume conservation and balance of the time derivative give exponent $\alpha =1/(n+2)$. Set $\mathcal{H}={\tau }^{-\alpha }f(\eta )$, $\eta =X{\tau }^{-\alpha }$. One integration, using zero volume flux at the ice divide, gives $f^n{f}^{\prime }=-\alpha \eta f$, henceThe planar volume-conserving nonlinear-diffusion similarity is thereforeThe Beta function fixes its constants:Consequently the explicit early shear and late sliding limits areTheir centre depths are $A_3{\tau }^{-1/5}$ with $A_3\simeq 0.842252$, and $A_2{\tau }^{-1/4}$ with $A_2=\sqrt{2/\pi }$. Thus the early extent grows as $t^{1/5}$ and the late extent as $t^{1/4}$. These are asymptotic regime profiles; they solve the respective limiting equations, not the full sum of both mobilities. For arbitrary finite-width initial data they are appropriate after the corresponding spatial relaxation, with a virtual time origin depending on that data.

The shear-to-slip transition in a shallow ice current occurs at thickness of order $H_{\ast }$. Using equality at the centre of the early similarity solution givesIf the initial profile is already approximated by that early similarity solution and its centre depth is 10, the virtual age is ${\tau }_0=A_3^5/10^5\simeq 4.2385\times 10^{-6}$, and the elapsed estimate is ${\tau }_{\mathrm{tr}}-{\tau }_0$. Other reasonable definitions of typical thickness change the order-one coefficient; the initial thickness alone does not uniquely determine an exact transition time or an exact initial profile. The robust nondimensional estimate is a transition time of order one in the $T_{\ast }$ scale.

Finally, let a local nose translate steadily at positive speed $v$ and write $s=x_N-x\ge 0$. This is a local traveling nose of a glacier with linear basal drag; the speed of the globally spreading glacier changes slowly with time. The local mass conservation equation integrates to $q=vh$, since both $q$ and $h$ vanish at the front. Therefore

The right side is strictly increasing, giving a unique positive thickness at every $s>0$. With $S=X_N-X$ and $c=d{X}_N/d\tau$, the same relation reads $cS={\mathcal{H}}^3/3+{\mathcal{H}}^2/2$. Its limiting profiles areThe square-root tip reflects basal sliding; the cube-root thicker region reflects internal viscous deformation. Even during the early thick regime, the very nose is thin and has the square-root inner form. At late times that sliding balance governs nearly the whole profile. The dimensionless crossover thickness is $\mathcal{H}=1$ and its distance is $S=5/(6c)$; the early thick cube-root outer profile matches this smaller sliding region. The diverging geometric slope at the idealized tip also marks the local limit of the long-wave shallow-ice approximation.

For constant-density incompressible inviscid flow with conservative body force, Euler equations for an inviscid fluid give ${\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}=-\nabla (p/\rho +\Phi )$. Using $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$, take the curl to obtain ${\partial }_t\mathbf{\omega }=\nabla \times (\mathbf{u}\times \mathbf{\omega })$. Since both divergences vanish, the vector identity expands this toWithout incompressibility or the constant-density/barotropic hypothesis, extra compression or baroclinic terms are present; the printed equation uses these flow assumptions.

Let $\kappa (t)=\dot{h}/h$, with $h>0$. Axisymmetric mass conservation is $r^{-1}{\partial }_r(r{u}_r)+{\partial }_z{u}_z=0$. If $u_r$ is independent of $z$, then $u_z$ is affine in $z$. The impermeable moving-plate conditions $u_z(r,\pm h,t)=\pm \dot{h}$ force $u_z=\kappa z$, and integrating the radial equation gives $u_r=-\kappa r/2+C(t)/r$. Regularity at the axis excludes $C(t)/r$. The meridional part of the axisymmetric inviscid flow between moving parallel plates is thereforeFor no swirl, $u_{\theta }=0$ completes the field. Continuity alone does not determine azimuthal motion; this must be specified separately. More generally, prescribe a smooth regular initial profile $U(r)$ independent of $z$. The azimuthal part of Euler equations for an inviscid fluid isso conservation of $r{u}_{\theta }$ along radial particle paths gives $u_{\theta }(r,t)=\lambda U(\lambda r)$, where $\lambda =\sqrt{h/h_0}$. Together with the meridional components above, this gives a family of complete fields, with a suitable pressure determined by the radial and vertical equations. The field is therefore not unique without rotational data.

With the stated solid-body rotation, $u_{\theta }=\Omega (t)r$ and $\mathbf{\omega }=2\Omega {\mathbf{e}}_z$. The vorticity equation gives $2\dot{\Omega }=2\Omega {\partial }_z{u}_z=2\Omega \kappa$, henceFor a fluid particle, $\dot{r}=-\kappa r/2$, so $r^{2}h$ is constant and $r^2\Omega$ is constant. This is conservation of angular momentum about the axis: radial spreading when the plates approach decreases the angular velocity. The meridional and rotating fields satisfy the inviscid equations with an appropriate pressure; no tangential no-slip condition is imposed on inviscid plates.