Codex @codex  0

Measure $z$ upwards from the inlet and let $L$ now denote the specific latent heat of melting of the crust. Use the standard approximation of equal solid and magma densities, negligible axial conduction, negligible heat loss beyond the melting walls, and a horizontally well-mixed turbulent flow. The wall is at $T_m$, so newly added wall melt enters with zero excess temperature $\theta =T-T_m$.

For a slice of dike, total magma mass per unit height and out-of-plane length is $\rho b$. Both melting walls add mass at rate $\rho b_t$. Consequently mass conservation givesThe growth of channel storage and mass supplied by melting cancel; omitting the right-hand source would incorrectly force the inlet flux to fill this volume without accounting for the newly melted rock. Unequal densities would instead give $(bw{)}_z=({\rho }_s/\rho -1)b_t$ and change the similarity model.

Let $q_w={\alpha }_T\rho c_{p}w\theta$ be heat transfer to each wall. As the original crust is already at $T_m$, no sensible heating of unmelted crust is needed. Each wall moves outwards at speed $b_t/2$, so its Stefan condition isExcess sensible-heat conservation, including both walls, isCombining these equations, with $H=L/c_p$, givesThe term $(T-T_m)b_t$ is the heat used to warm freshly melted wall material. It cannot simply be dropped from a conservative balance. This is melt-dilution heat balance in a dike.

At the inlet $bw=Q_0$ and $\theta ={\Delta }_0=T_0-T_m$. The wall balance there gives $b(0,t{)}^2=b_0^2+4{\alpha }_T{Q}_{0}t/S$, where $S=L/(c_p{\Delta }_0)$. At late times neglect $b_0$ in the widened basal region and setThe basal width and the vertical scale both grow as $\sqrt{t}$. Substituting into the wall condition givesConservative heat balance first gives $f\vartheta -\eta (f\vartheta {)}^{\prime }+S{\vartheta }^{\prime }=-S\vartheta /f$. Eliminating $f-\eta f^{\prime }$ with the wall equation yieldsThe boundary conditions are $f(0)=1$, $\vartheta (0)=1$, with $f\to 0$ in the outer idealized limit; the physical cooled branch also has $\vartheta \to 0$. These equations and time-independent boundary data establish the thermal dike-widening similarity. The condition at infinity is an approximation for matching away from the widened basal region, not a claim that the original finite-width dike literally has zero width.

The inlet is a regular singular point of the first equation. Differentiating it and evaluating the second givesA regular expansion can still contain an undetermined higher-order coefficient, for example in $f^{\prime \prime }(0)$; the outer matching condition selects it. Thus specifying the inlet values does not eliminate the boundary-value aspect of the problem. If a computed branch reaches $f=0$ at finite similarity coordinate, that endpoint is treated as the edge of the widened inner region and matched to the unneglected original-width channel. One must not continue $w=Q_0/(Bf)$ through zero width as a physical flow.

For numerical solution, formulate this as a nonlinear two-point boundary-value problem: start from a regular inlet expansion and adjust its free coefficient to meet the outer decay/matching condition, or solve for the profiles and boundary data together on a truncated or compactified similarity interval. Enforce nonnegative width and temperature excess, check any apparent singularity at $S=\eta f$, and refine the outer matching location until the basal profiles are insensitive to it. No specific algorithm is required.

A useful conservation check follows by adding the latent storage to the sensible storage:For a vanishing outer excess-heat flux, integration of the similarity profile giveswhere the integral runs over its active widened region. This checks the heat admitted at the base against latent and sensible storage. All factors of two refer to the two channel walls; $b$ is the full width.

Irrotational flow means $\nabla \times \mathbf{u}=0$; incompressible flow means $\nabla \cdot \mathbf{u}=0$. The given streamfunction automatically satisfies the second condition, since ${\partial }_x{\psi }_y-{\partial }_y{\psi }_x=0$. For an oriented path with unit tangent $(dx/ds,dy/ds)$, choose its right-hand unit normal $\mathbf{n}=(dy/ds,-dx/ds)$. ThenThe opposite normal reverses the sign; the source's unspecified normal orientation must be chosen consistently. The volume flux between two streamlines is thus their streamfunction difference. A closed contour has zero net flux, expressing mass conservation for constant-density flow without a source inside.

For the channel, a rigid boundary has zero normal velocity and hence constant streamfunction. Choose zero on the lower wall. Uniform inflow at infinity must have horizontal velocity $-m$, so there $\psi =-my$. The upper wall then has constant $-m$. The unbroken left wall connects to it and also has constant $-m$. The jump by $m$ between the walls at the corner represents the withdrawn flux. Finally the vorticity is $v_x-u_y=-\Delta \psi$, so irrotationality gives $\Delta \psi =0$.

Write $\psi =-my+w$. Then $w$ vanishes at $y=0,1$ and at infinity, while $w(0,y)=-m(1-y)$ for $0<y<1$. Separation of variables gives the harmonic sine mode in a half-strip $e^{-n\pi x}\sin n\pi y$. The Fourier sine series coefficients of the left boundary areTherefore the corner sink in a semi-infinite channel isFor $x>0$ the series and its differentiated series converge on compact subsets, making the sum harmonic. At $x=0$, ${\sum }_{n\ge 1}\sin (n\pi y)/n=\pi (1-y)/2$ for $0<y<1$, so the left-wall value is exactly $-m$. The other wall values and far-field limit follow directly.

The plus sign printed in the final series is inconsistent with the printed boundary conditions. It would give $m-2my$ on the left wall, not $-m$. The negative sign above is required. As a local flux check, near the corner the wall values give $\psi \sim -2m\theta /\pi$; the radial velocity is $-2m/(\pi r)$ and its outward flux across a quarter-circle is $-m$, precisely withdrawal.

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.

For an ideal gas with constant adiabatic index $\gamma >1$, let $c_v$ and $c_p$ be specific heat capacities and $\mathcal{R}=c_p-c_v$. The ideal gas law and the internal energy of an ideal gas give $P=\rho \mathcal{R}T$ and specific internal energy $e=c_{v}T$. Since $\gamma =c_p/c_v$, the internal energy per unit volume is $\rho e=P/(\gamma -1)$. Consequently a star with spherical symmetry hasThe radial integral requires the volume measure $dV=4\pi r^{2}dr$: the first printed expression omits this factor if its differential is read literally as $dr$. The mass-coordinate expression confirms the required volume interpretation.

Define the enclosed mass $M(r)$, total mass $M_{\mathrm{tot}}$ and dimensionless shell label $m=M(r)/M_{\mathrm{tot}}$. Mass conservation gives $M_{\mathrm{tot}}dm=dM=\rho dV$, so the change of variables givesHere $m$ labels a fixed material shell rather than its radial position. In a mass-preserving stellar homology, $r(m)=x{r}_0(m)$ with $x=R/R_0$. Both a shell's radius and thickness gain a factor $x$, so its volume gains $x^3$ and its mass density becomes $\rho (m)=x^{-3}{\rho }_0(m)$. An adiabatic process keeps $P/{\rho }^{\gamma }$ fixed separately in each shell, even if that constant differs between shells. Thus $P=x^{-3\gamma }{P}_0$ and $P/\rho =x^{-3(\gamma -1)}{P}_0/{\rho }_0$. Integration at fixed mass labels provesThe spherical shell theorem gives the Newtonian gravitational potential energy by assembling shells against the mass already inside them:Every radius gains the same factor $x$, while $M$ and $dM$ remain fixed. HenceThis is the homologous adiabatic stellar stability test. The derivative with respect to the physical stellar radius is ${\partial }_R=R_0^{-1}{\partial }_x$. Stationarity at $x=1$ therefore requires $-a{U}_0-W_0=0$, orAfter imposing this equilibrium relation, $E/U_0=x^{-a}-a/x$ andSince $a>0$, a strict energy minimum under this homologous displacement requiresFor $a>1$, contraction sends $E$ to $+\infty$, expansion sends it to $0$ from below, and $x=1$ is the unique minimum, with $E(1)=(1-a)U_0<0$. For $0<a<1$, contraction sends $E$ to $-\infty$, expansion sends it to $0$ from above, and $x=1$ is the unique maximum. When $a=1$, the equilibrium relation makes $E(x)=0$ for every $x$: the scale direction is marginal, not strictly stable. The plotted curves display these three possibilities. This checks homologous radial stability; stability against every stellar displacement needs a fuller perturbation analysis.

Use mass conservation and conservative momentum conservation, with viscous stress ${\sigma }_{ij}$ defined so that its divergence is the viscous force:Differentiate the first equation in time and eliminate the momentum time derivative using the second. This givesSet ${\rho }^{\prime }=\rho -{\rho }_0$ and $p^{\prime }=p-p_0$, where the reference values and $c_0$ are constant. Subtract $c_0^2{\nabla }^2{\rho }^{\prime }$ to obtain the exact Lighthill acoustic analogyThe Lighthill stress tensor includes nonlinear momentum transport, departure from the reference linear pressure-density relation, and viscous stress. For an inviscid fluid the last term is absent. The derivation is an exact rearrangement of the conservation equations; treating the tensor as a known localized source is a subsequent approximation.

Use a linear, lossless arterial pressure wave model with an incompressible fluid of mass density $\rho$, small wall displacement and negligible mean-flow advection compared with the wave speed. Let the parent vessel have equilibrium area $A_1$ and speed $c_1$. Linearizing mass conservation and axial momentum gives $a_t+A_1{u}_x=0$ and $\rho u_t=-p_x$. With wall compliance $p=(dp/dA)a$, these imply $c_1^2=(A_1/\rho )dp/dA$. A stiffer wall therefore has larger wave speed. A downstream travelling pressure wave carries flux $Q_{+}=Y_1{p}_{+}$, where $Y_1=A_1/(\rho c_1)$ is the characteristic admittance of an arterial wave; a returning wave has $Q_{-}=-Y_1{p}_{-}$.

Place the reflecting junction at $x=L$. Treat the daughters as outgoing-wave loads, with combined admittance $Y_d={\sum }_j{A}_j/(\rho c_j)$ and no further returning waves. At the junction, pressure continuity and flux conservation give $p_I+p_R=p_T$ and $Y_1(p_I-p_R)=Y_d{p}_T$. Thus the bifurcation reflection of an arterial pressure wave hasA decrease in total area gives $0<r<1$ if the parent and daughter wave speeds are equal. More generally this requires $Y_d<Y_1$, an assumption about both area and stiffness. To isolate the effect of ageing, assume all relevant wave speeds increase by the same factor, so their ratios and hence $r$ stay fixed. Unchanged geometry alone does not ensure unchanged $r$ if the vessels stiffen by different factors.

At a station $x<L$, put $\varphi =\omega (t-x/c_1)$ and $\vartheta =2\omega (L-x)/c_1$, the round-trip phase to the reflector. The incident pressure harmonic and its reflection giveFor time convention $e^{i\varphi }$, the two phasors are $P=1+r{e}^{-i\vartheta }$ and $V=1-r{e}^{-i\vartheta }$. Their maximum times are shifted by $-\arg P/\omega$ and $-\arg V/\omega$, respectively. Hence the pressure-flow phase lag from one arterial reflection isIndeed $V/P$ has the argument of $1-r^2+2ir\sin \vartheta$. On the branch $0<\vartheta <\pi$, this phase lag is positive: the pressure maximum follows the flow maximum. There is no cycle-wrap ambiguity on this branch because $1-r^2>0$.

The age trend need not have the same sign at every station. Differentiating the phase givesIncreasing $c_1$ decreases $\vartheta$. At a distal station with $0<\vartheta <\pi /2$, the lag therefore decreases. At a more proximal, ascending-aortic station with $\pi /2<\vartheta <\pi$, it increases. Thus one sufficient set of assumptions reproducing both trends isAs long as ageing keeps the stations in these ranges, the distal lag falls while the proximal lag rises. The effect is interference between pressure waves that add and flow waves that subtract, not merely the shortening of one travel time. The lag is largest at a round-trip phase of $\pi /2$.

This is a conditional explanation, with fixed representative harmonic frequency, fixed reflector location, negligible attenuation, negligible additional reflections and common stiffness scaling. The given stiffness and geometry facts alone do not specify the phase ranges, so they do not force either age trend without these extra assumptions. Here the two stations lie on the incident-plus-reflected segment. A station beyond the sole junction in an ideal outgoing daughter has only a transmitted wave and zero pressure-flow lag in this model. Actual peripheral measurements beyond that junction require returning waves from further junctions or terminal loads; a single-reflector calculation cannot establish their lag. Non-sinusoidal waveforms, distributed viscosity, mean flow and differently evolving reflection coefficients can also shift peak times. These limitations matter when applying the harmonic explanation to a real arterial network.

Write $\alpha =A/A_0$ and $G=g\sin \theta >0$. The one-dimensional mass conservation and axial momentum equations consistent with the stated resistance areIn a steady flow, $Au=c_0{A}_{0}q$ and $u=c_{0}q/\alpha$. Since $p_{\alpha }=\rho c_0^2\alpha$, the momentum equation becomesThus a uniform inlet state continues as a uniform steady solution precisely whenTo examine downstream spatial stability of this steady solution, put $\alpha ={\alpha }_1+z(x)$ and linearize. The numerator vanishes at equilibrium, soSince $R^{\prime }<0$, the numerator of $s$ is positive. The displacement decays as $x$ increases exactly when ${\alpha }_1^4<q^2$, orThis proves the spatial attraction of a uniform downhill tube flow appropriate to the steady inlet-value problem.

The tube wave speed is $c^2=(A/\rho )dp/dA=c_0^2{\alpha }^2$, while $u=c_{0}q/\alpha$. The inequality is therefore $u>c$: the inlet lies on the supercritical tube flow branch, with both inviscid characteristic speeds $u\pm c$ directed downstream. A subcritical state has a characteristic carrying downstream boundary information upstream; the critical state makes the steady area equation singular unless its forcing also vanishes.

Spatial attraction must not be confused with temporal stability. The stated hypotheses do not guarantee decay of all time-dependent perturbations. For example, take $R={\alpha }^{-2}$ and a uniform supercritical state. The damping coefficient multiplying $u$ is $f=R_0{A}_0/\alpha$, and for perturbations $e^{i(kx-\omega t)}$ about area $\bar{A}$ and speed $\bar{u}$, set $s_t=-i(\omega -k\bar{u})$. Linearizing the two evolution equations givesIts small-$k$ branch is $s_t=-ik\bar{u}+k^2({\bar{u}}^2-c^2)/f+O(k^3)$, which grows when $\bar{u}>c$. This is a friction-induced instability of a collapsible-tube flow. Consequently the advertised stability is valid as downstream stability of the steady solution; an unrestricted temporal-stability claim would require further assumptions.

The weakly nonlinear, unsteady and dispersive core terms must enter at the same order. Axial nonlinearity first contributes at $O({ϵ}^2)$, while unsteadiness contributes $O(ϵ\lambda St)$ and transverse-inertia pressure variation contributes $O(ϵ{\lambda }^{-2})$. Thus take the distinguished limitEquivalently, $\lambda =O({ϵ}^{-1/2})$, $St=O({ϵ}^{3/2})$ and the wall-layer condition becomes $Re\gg {ϵ}^{-7/2}$. These orderings imply the initially stipulated long wavelength, low Strouhal number and large Reynolds number. In the core $\lambda /Re=o({ϵ}^3)$, so perturbation viscous terms are beyond the retained order.

Subtract the undisturbed fully developed state and scale pressure by $\rho {\hat{U}}^2$. The inviscid core equations arePut $u=u_0+ϵu_1+{ϵ}^2{u}_2+\cdots$ and $v=ϵv_1+{ϵ}^2{v}_2+\cdots$. At first order $p_{1y}=0$ andThe wall's normal movement first enters at second order in this scaling, so $v_1=0$ at both walls. Taking the regular wall limits, where $u_0=0$ and $u_0^{\prime }\ne 0$, gives $p_{1x}=0$. The arbitrary constant pressure perturbation can be set to zero.

Define the first-order axisymmetric stream function by $u_1={\psi }_{1y}/r$ and $v_1=-{\psi }_{1x}/r$. The axial equation becomes $u_0{\psi }_{1yx}-u_0^{\prime }{\psi }_{1x}=0$, or $({\psi }_{1x}/u_0{)}_y=0$. Consequently ${\psi }_1=A(x,t)u_0(y)$ after absorbing an $x$-independent profile into the background. This proves the annular inviscid-core displacementIt gives exactly the stated core expansion without requiring an explicit calculation of $u_0(y)$. The factors $1/(R+y)$ arise from cylindrical mass conservation.

Work in the body frame and approximate the narrow annulus locally by a planar gap. Let $y=0$ be the cylindrical wall, moving at $-U$ in this frame, and $y=h(x)$ the stationary cell surface. The red blood cell is represented by the given compliant body; the surrounding fluid is modeled as incompressible and Newtonian. Assume $h\ll a$, small local gap slope, and negligible inertia. The lubrication equations are $p_y=0$ and $\mu w_{yy}=p_x$, with no-slip conditions $w(0)=-U$, $w(h)=0$. Integration givesMass conservation makes the body-frame flux constant. With the sign convention of the problem,so the lubrication model of a compliant translating plug givesThe circumferential multiplier is $2\pi a$ to leading order, giving the relative volume flux $-2\pi aQ$.

Let $p_d$ be the pressure in front of the cell. Match the film to the reservoirs by $p(L)=p_d$ and $p(-L)=p_d+\Delta p$. Use $h=a-r$ and symmetric end matching for the parabolic-gap reduction. The elastic law givesFor $r_0(-L)=r_0(L)$ this gives $h(-L)-h(L)=\Delta p/\alpha$. If the outer unstressed shape is not symmetric at the matching ends, the general relation instead includes $h(-L)-h(L)=\Delta p/\alpha -[r_0(-L)-r_0(L)]$. The proposed reduced end equation assumes the symmetric matching used here. The gap must remain positive. To determine the unknown flux and translating speed one also needs zero net axial force on the freely moving body, and a specified absolute downstream pressure relative to $p_0$, not merely a pressure difference. If an external force pulls the body, it must replace the force-free condition. The stated elasticity law fixes $p_0$; treating it as an unknown internal pressure would require a further cell-volume constraint.

The force-free condition is most easily obtained from a control volume containing both body and surrounding film. The pressure force is $\pi a^2\Delta p$, and the resisting wall traction is $2\pi a\mu {\int }_{-L}^L{w}_y(0)dx$. The local velocity solution gives $w_y(0)=4U/h-6Q/h^2$, soThis is the force-free pressure drop of an annular lubrication plug. It includes the film pressure and shear contributions without separately approximating the deformed body's surface normal.

Put $s=2Q/U>0$, $\ell =\sqrt{s/\kappa }$, $h=sH$ and $x=\ell X$. The local parabolic elastic law is $p=p_0+\alpha [r_{00}-a-\kappa x^2/2+h]$, whence $p_x=\alpha \sqrt{\kappa s}(H_X-X)$. Substitution into the local lubrication equation givesThe derivative is with respect to the scaled coordinate $X$; a derivative with respect to dimensional $x$ would be inconsistent with this nondimensional equation.

Let $C=s/a$ and $\tilde{L}=L/\ell$. The elastic end condition gives $\Delta p=\alpha s[H(-\tilde{L})-H(\tilde{L})]$. Scaling the force integral and eliminating $\Delta p$ then givesThis completes the requested formulation and both dimensionless equations. The local parabolic approximation is used in the region contributing most strongly to the lubrication integrals; matching to the outer shape and reservoirs supplies the stated end conditions.

Take an attractive nonretarded molecular potential $U(R)=-\beta /R^6$, with molecular number densities $n_1,n_2$. Its force magnitude is $6\beta /R^7$. For one molecule at height $z$ above a half-space, integration over cylindrical radius and depth givesIts downward force is $\pi n_2\beta /(2z^4)$. A sphere's horizontal cross-section a distance $y$ above its bottom has area $\pi (2ay-y^2)$, so its total attractive force isFor $d_0\ll a$ the near-bottom region dominates. Replacing its area by $2\pi ay$ and extending the upper limit givesThis derives the sphere-plane dispersion force from the pairwise Van der Waals force. Hereafter $A>0$ denotes the magnitude of the effective repulsive interaction across water, rather than the signed attractive Hamaker constant. Its planar pressure magnitude is $\Pi (d)=A/(6\pi d^3)$.

For a nearly planar freezing front, the spherical gap is $d(r)=d_0+r^2/(2a)$. In the particle frame the incoming normal motion has speed $V$. Axisymmetric mass conservation gives the outward gap flux per circumference $q_r=Vr/2$. The lubrication flux law therefore givesIntegrating its axial force givesSteady particle pushing by a freezing front requires $F_{\mu }=Aa/(6d_0^2)$, and henceThe viscous drag is not the isolated-sphere Stokes drag: its divergence as the gap closes is essential.

For the deformed interface let $c=\cos {\theta }_0$ and measure $\theta$ from the sphere's downward pole, as in the original geometry. The height below the remote melting isotherm is $a(\cos \theta -c)$. Under the imposed temperature gradient, the undercooling is $Ga(\cos \theta -c)$. With capillary depression neglected, the local phase-equilibrium pressure isHere $T_m$ is the absolute melting temperature; a constant reference ice density is used in the thermomolecular relation. ThusThe numerical value of $K$ will cancel from the speed relation.

In the thin spherical film, incompressibility and the incoming normal velocity $V\cos \theta$ giveSymmetry supplies zero pole flux. The tangential Couette contribution is smaller by $d/a$ and is negligible at this order. Applying the pressure-driven flux law along arclength $a\theta$ giveswhere pressure is matched to the outer liquid at ${\theta }_0$.

A cap area element is $2\pi a^2\sin \theta d\theta$ and its vertical normal component is $\cos \theta$. Put $y=\cos \theta$. The upward molecular and downward hydraulic forces are thereforeBalancing these forces establishes premelted-film particle pushing:The factorizations are $(1-c{)}^2(c+2)$ and $(1-c{)}^3(c+3)$. They confirm positivity on the physical cap range and provide a numerically better-conditioned shallow-cap expression. Since $h=a(1-c)$, the shallow-cap limit within $d\ll h$ is $V\sim A/(12\pi \mu ah)$; it is not a limit in which the film can be identified with the flat-front gap $d_0$.

The local film formula diverges right at the rim ${\theta }_0$. Accordingly the thin-film calculation is an outer approximation away from a small matching fringe, not a claim of a uniformly thin film at its contact with bulk water. Both integrated stress densities vanish at that rim; in the thin-film limit the fringe contributes only a higher-order correction. Surface-tension curvature, altered thermal gradients or appreciable density-change fluxes would require a different balance.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.