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Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the shock frame, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to \infty$ in the Rankine-Hugoniot conditions for a perfect gas givesThe downstream laboratory velocity is $D-D/r=2D/(\gamma +1)$, which distinguishes the gas speed from the front speed.

For the planar blast-wave energy scaling of a self-similar blast wave, integration of the total energy density over the shocked interval $0<z<Z$ gives the energy per unit area on this side:The similarity solution makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot{Z}=(E_{+}/({\rho }_{1}I){)}^{1/2}{Z}^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,If $E$ denotes the one-sided energy, $C=(9/(4I){)}^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_{+}=E/2$ and $C=(9/(8I){)}^{1/3}$. In either convention the requested scaling is $Z=C(E/{\rho }_1{)}^{1/3}{t}^{2/3}$. The constant depends on the similarity profiles and the energy convention; energy conservation determines the exponent without solving those profiles. The Strong-shock Rankine-Hugoniot conditions additionally fix $f(1)=(\gamma +1)/(\gamma -1)$ and $g(1)=h(1)=2/(\gamma +1)$.

Assume $D_0,n_0,Q>0$. The density-dependent diffusion coefficient makes this a cubic porous medium equation, since its two-dimensional form is $n_t=[D_0/(3n_0^2)]\Delta (n^3)$. The mass-preserving similarity solution is a radial cubic-diffusion source profile. At fixed $r$, the proposed scaled density givesSeparation is possible whenwhere $\gamma$ is a constant. The resulting ordinary differential equation isRecognize the second term as $\gamma ({\xi }^{2}F{)}^{\prime }$. Regularity and zero radial flux at the origin set the integration constant to zero:In the region $F>0$, this reduces to $F{F}^{\prime }=-\gamma \xi$, giving $F^2=1-\gamma {\xi }^2$ from the central normalization. For a nonnegative profile whose front is precisely at $\xi =1$, the first zero fixes $\gamma =1$. Continuity of the density excludes truncating a positive value at the front. HenceHere integrating the scale equation gives ${\lambda }^6=6D_{0}t/r_0^2+\text{constant}$, and the point-source initial condition requires $\lambda (0)=0$.

Use the planar area element, not the one-dimensional area under the drawn profile. Mass conservation givesThus the scale and the complete density arewhere $[u{]}_{+}=\max (u,0)$. The factors of $n_0$ are necessary on dimensional grounds; the length scale is not a function of $Q$ alone.

The front and peak obeyEach radial profile starts with horizontal tangent at the origin, decreases to a square-root edge, and is identically zero beyond $R(t)$. Later profiles are wider and lower. This is finite propagation in porous-medium diffusion, because the diffusion coefficient vanishes at zero density. The sketch uses dimensionless time $\tau =6D_{0}t/r_0^2$; it preserves the radial integral $2\pi \int rndr$, not the unweighted area under each curve.

At the front $F^{\prime }$ diverges, so the profile is a weak solution, not a globally smooth classical solution. Nevertheless $F^2{F}^{\prime }=-\xi \sqrt{1-{\xi }^2}$ tends to zero there. The density and flux both match their zero exterior values, so extending the interior solution by zero creates no spurious boundary source in the conservation law. The behavior at the origin is regular because $F^{\prime }(0)=0$.

Finally, the point release is recovered as a distributional initial condition. For any continuous compactly supported test function $\phi$ on the plane, the total density remains $Q$ and all its support lies in the shrinking disk $r\le R(t)$. ThereforeThus $n(\cdot ,t)\to Q{\delta }^{(2)}$ as $t\downarrow 0$, in the sense of the Dirac delta function.

Take $\Delta \rho >0$ and $\tau >0$, with $x$ pointing along the imposed shear stress, and put $G=\Delta \rho g$. Relative to the upper fluid's hydrostatic reference, the hydrostatic pressure in the current is $p=G(h-z)$. Thus the horizontal pressure gradient is $G\nabla h$. The lubrication theory momentum balance, lower no-slip boundary condition, and imposed upper shear stress giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe two terms have opposite roles: imposed shear stress carries fluid downstream, whereas the hydrostatic pressure gradient spreads it from thick regions towards thin ones.

For the approximation, a representative horizontal velocity is $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis condition controls viscous normal stress, vertical viscous corrections and the normal projection of shear at a slightly tilted interface relative to $GH$. With it and the previous inertia bounds, vertical inertia is small as well. Time variations here are on the transport timescale $L/U$; separately imposed rapid forcing would need its own unsteady inertia bound. A small ordinary Reynolds number is a stronger sufficient restriction, but the reduced inertia ratios above are what the thin-layer momentum balance directly requires.

In the steady far-downstream regime, cross-stream derivatives dominate gravity-driven spreading. Dropping the smaller downstream gravity flux leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, with $h=0$ outside. The squared height is a parabolic Barenblatt solution; the height cross-section is a semicircular profile after rescaling its axes. The cross-stream volume flux vanishes at the edges even though the height slope becomes singular there. The profile describes the outer lubrication theory region, not a resolved microscopic front.

Near the point source the two horizontal dimensions are comparable, say $\ell$. Upstream spreading arrests where outward gravity-driven volume flux balances downstream shear-driven volume flux:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. The same horizontal scale is obtained by setting $y_N\sim x$ in the downstream similarity solution, confirming where that solution fails.

For the line source the steady volume flux per unit width is $q=A{h}^2-D{h}^3{h}_x$. On the downstream constant-height branch, $q=Q_{2d}$ gives $h_d^2=Q_{2d}/A$. Upstream there is no net flux through the finite nose, so $q=0$. Thus $h{h}_x=A/D$, and continuity of height at the source givesThe line-source upstream reach under imposed shear is thereforeIn the upstream part, shear-driven and gravity-driven volume fluxes cancel. The volume flux per unit width jumps by exactly $Q_{2d}$ at the source. As in the point-source profile, the square-root nose is a formal outer solution with an unresolved steep front; its finite upstream reach follows from flux balance, not from assuming a small slope all the way to the nose.

Use a planar saline Stefan problem with upward coordinate $z$, the initial contact at $z=0$, and phase boundaries $z=a(t)$ and $z=b(t)$ enclosing pure ice, where $a<b$. Fresh liquid occupies $z>b$ and brine occupies $z<a$. We assume negligible bulk flow, equal constant mass density $\rho$, specific heat capacity $c_p$ and thermal diffusivity $\kappa$ in all regions, and zero salt content and salt transport in the ice. Equal densities remove phase-change volume flow; suppressing convection isolates the molecular-transport mechanism. Different material properties would change the numerical coefficients. Set $K=\rho c_p\kappa$ for the common thermal conductivity and $\Delta T=T_m-T_{\infty }>0$. Denote the brine salt diffusion coefficient by $D$.

Use a linear ice liquidus, $T_L(C)=T_m-mC$, with $m>0$ and $C$ the salt mass fraction. Assume the initial brine is a stable liquid, $T_{\infty }>T_L(C_0)$, and all concentrations considered are below the eutectic composition. Thus $R=m{C}_0/\Delta T>1$. These assumptions exclude independent nucleation in a supercooled brine or an eutectic system event, neither of which is specified by the initial data alone. Start with a negligible seed of ice and use local phase equilibrium without interfacial kinetics or curvature. Temperature differences may be in Celsius, since only differences enter this calculation.

WriteThe paper calls $S$ a Stefan number; it is the latent-to-sensible heat ratio, reciprocal to another common Stefan number convention. At the lower phase boundary, define $T_i=T_L(C_i)$ and letThe liquidus relation is $\vartheta =Rc$.

A similarity solution gives an explicit calculation of the two positions. Putand use $u=z/(2\sqrt{\kappa t})$, $v=z/(2\sqrt{Dt})$. The temperatures and salinities areThese error function and complementary error function profiles satisfy the heat equation and the salt diffusion equation, their far-field conditions, and the interfacial temperature conditions. The fresh-water temperature is constant because the interface and the far field are both at $T_m$.

Salt conservation at the moving lower phase boundary requiresFor upward motion this is dilution by melting salt-free ice, rather than salt rejection by freezing brine. Substitution gives the dilution function for a melting saline Stefan front:For $A>0$, $0<c<1$, so the interfacial brine is fresher than the remote liquid.

The Stefan condition at the upper boundary is $\rho L\dot{b}=K{T}_{s,z}(b)$, since the fresh liquid has no temperature gradient. At the lower boundary it isThese signs follow from the jump in enthalpy: the upper front freezes liquid while a positive $\dot{a}$ melts solid. With $E=\mathrm{erf}B-\mathrm{erf}\delta$, they reduce toTogether with $\delta =ϵA$, these equations determine $A,B,\vartheta$, and hence both positions, without discarding the salt-diffusion correction.

For the stated large latent-to-sensible heat ratio, the layer is thin relative to the thermal diffusion length. Expanding the thermal equations for $B,\delta \ll 1$ givesConsequently a useful leading calculation iswhere the positive $A$ is obtained fromThe left side decreases with $A\ge 0$ while the right side increases, and their values at zero and infinity guarantee a unique positive root for large $S$. Thus this leading calculation includes the translation of the lower boundary as well as the increasing thickness.

If the scale separation also obeys $ϵSA\ll 1$, the familiar simpler result iswithThis last simplification needs the logarithmic refinement $ϵS\sqrt{\log (RS)}\ll 1$. The algebraic ordering $ϵ\ll S^{-1}$ alone should not be used to discard an arbitrarily large logarithmic correction; the preceding coupled equations remain the appropriate calculation when that refinement is unavailable. This distinction is captured by large latent heat in a freezing and melting ice layer.

The illustration uses $S=40$, $ϵ=2.5\times 10^{-4}$ and $R=2$. Solving the full equations gives $A=1.73707$, $B=0.0143053$ and $\vartheta =0.0158766$. It shows the ice layer between freshwater and cold brine, with both fronts advancing upward and a strongly diluted lower liquid boundary.

The phase diagram shows the spatial path from remote brine to the lower interface: temperature first changes over $\sqrt{\kappa t}$ at almost unchanged salinity, then salinity changes over its much smaller diffusion scale at nearly constant temperature. The liquid path ends on the liquidus at $(C_i,T_i)$; the solid has $C=0$ and its temperature rises from $T_i$ to $T_m$. The eutectic position is schematic, with $C_E=2C_0$ in the illustration, and is not used in the calculation. A point initially in the fresh layer freezes when $b$ reaches it, cools in the ice, and later melts when $a$ reaches it. The inset indicates this temporal path in the opposite direction through the solid branch and into the liquid.

The mechanism is simultaneous upper freezing and lower melting, with net growth of the ice thickness. Cold brine accepts most of the latent heat released by upper freezing. Salt reaching the lower contact lowers its equilibrium melting temperature, and melting freshens that liquid until its liquidus is close to $T_m$. The heat delivered through the ice supplies the much smaller melting demand there. The remote brine remains liquid even though it is below the pure-water melting temperature, because it lies above its own saline liquidus. Thus “cold” does not by itself decide which phase is stable. These conclusions apply while the layers remain effectively deep and the stated no-convection, pure-ice and phase-equilibrium assumptions hold.

Use $F=F_s>0$ and constant entrainment parameter $\alpha >0$. The printed integrals are normalized by $\pi$, so a top-hat plume model has $Q=b^{2}w$, $M=b^2{w}^2$ and $F=b^{2}w{g}^{\prime }$; the physical area is still $\pi b^2$. Keeping this normalization avoids spurious factors of $\pi$ in the flux equations.

Eliminate height between the two flux balances. The plume flux-balance invariant follows fromsoFor positive fluxes, both $M$ and $Q$ increase. In particular $Q^{\prime }\ge 2\alpha \sqrt{M_s}>0$, so $Q\to \infty$ as $z\to \infty$. Therefore $C/(\lambda Q^2)\to 0$, whatever the finite source imbalance, and $M\sim {\lambda }^{2/5}{Q}^{4/5}$. Integrating $Q^{\prime }\sim 2\alpha {\lambda }^{1/5}{Q}^{2/5}$ gives the attracting similarity solutionThis proves attraction to pure plume similarity for physical positive source fluxes. Sources with a vanishing initial volume or momentum flux enter the positive regime immediately and can be obtained by the corresponding limiting solutions; a point buoyancy source has the same similarity form with zero virtual-origin offset. The statement assumes positive buoyancy flux: a jet with $F=0$ is a different asymptotic problem.

Define the plume balance parameterPure plume balance means $\Gamma =1$ at each height, or equivalently $C=0$: momentum, buoyancy and entrainment have their similarity relation, even when the source has finite radius. A forced plume has $\Gamma <1$, meaning excess momentum relative to this balance; a lazy plume has $\Gamma >1$, meaning insufficient momentum. The invariant ensures that the sign of the imbalance is preserved while $\Gamma \to 1$. In particular, a strictly positive source remains pure at every height precisely whenFor this source the exact solution is the same similarity form with $z$ replaced by $Z=z+z_0$. Writing $a=(9\alpha F_s/10{)}^{1/3}$,Thus the radius growth of an axisymmetric pure plume is $b^{\prime }=6\alpha /5$ and the plume virtual origin lies at $-z_0$, with $z_0=5b_s/(6\alpha )$. Equivalently, in terms of top-hat source properties, pure plume balance is $g_s^{\prime }{b}_s/w_s^2=8\alpha /5$.

For the stated source-to-source geometry, the relevant horizontal distance is $L_s$. Consequently the edge criterion is $b_m=L_s$, rather than the radius at which two expanding edges first touch. Since $b(z)=b_s+6\alpha z/5$,Here $b_s=Q_s/\sqrt{M_s}={\lambda }^{-1/5}{Q}_s^{3/5}$. A positive merging height requires $L_s>b_s$; equality puts it at the source, while an already wider source satisfies the geometric condition from the start. The no-interaction premise extrapolates independent plumes even though their edges would first touch earlier, when each radius is $L_s/2$.

Use $M^{5/2}=\lambda Q^2$ with $Q=b^{2}w$ and $M=b^2{w}^2$ to get $F_s=(8\alpha /5)b{w}^3$. The individual plume properties at the specified height are thereforeThe source volume flux affects the height through the virtual origin, but it does not affect these properties at a prescribed radius in an exactly pure plume.

For the stipulated merger of two equal pure plumes, the new radius is $b_c=\sqrt{2}{b}_m$. This factor is outside $b_m$, as required by the area. Since speed and reduced gravity are held fixed, the merged fluxes are $Q_c=2Q_m$, $M_c=2M_m$ and $F_c=2F_s$. Their balance parameter isThe immediate combined plume is lazy. The same result follows from $\Gamma =5g^{\prime }b/(8\alpha w^2)$: enlarging the radius at fixed $g^{\prime },w$ increases this parameter.

At fixed $w_m,g_m^{\prime }$, the radius required for pure plume balance isThat is half the stipulated combined area $2\pi L_s^2$. This fixed-speed pure-plume merger and flux conservation comparison exposes the limitation of demanding immediate pure balance with unchanged speed and reduced gravity: reducing the area to this value would also halve all three merged fluxes. A conservative merger must instead adjust its properties and begin out of pure balance. Under the given constant-entrainment equations, the resulting lazy plume subsequently approaches the attracting similarity solution with total buoyancy flux $2F_s$.

Seek a nonzero-buoyancy, separable similarity solution of the unsteady Boussinesq triangular-profile line-plume balances, using powers of distance and time. Let $\tau =t+t_{\ast }>0$, allowing a time origin, and initially take distance $z$ from the virtual origin. Balancing the powers in the momentum and buoyancy equations and the entrainment balance gives the formOne way to recover the exponents is to let $Q\propto z^a{\tau }^p$ and $M\propto z^b{\tau }^q$. Matching the spatial powers in ${\partial }_t(Q^2/M)$, $Q_z$ and $M/Q$ gives $a=2,b=3$; matching the time powers gives $p=-1,q=-2$. The momentum equation then gives the powers of $F$. In the resulting solution, $Q^2/M$ is time-independent, so the storage term in the first equation vanishes exactly.

Direct substitution yields the coefficient equationsFor a rising buoyant solution $C>0$, the last equation gives $B=2A/3$. The first then gives $A=\alpha {\rho }_0$, and the second gives $C=B$. The separable decaying line-plume similarity is thereforeReconstructing the physical fields confirmsFor example, ${\partial }_t(FQ/M)=-2\alpha {\rho }_0{z}^2/{\tau }^3$ and $F_z=2\alpha {\rho }_0{z}^2/{\tau }^3$ cancel; the momentum balance similarly gives $-\alpha {\rho }_0{z}^2/{\tau }^2+2\alpha {\rho }_0{z}^2/{\tau }^2=FQ/M$.

The unshifted form uses $t_{\ast }=0$ and is defined for $t>0$, with a singular zero-time limit. To give a finite solution for all $t\ge 0$, choose $t_{\ast }>0$. The equations also permit replacing $z$ by $z+z_{\ast }>0$, a plume virtual origin; at a finite physical source distance this family has flux histories proportional to ${\tau }^{-1},{\tau }^{-2},{\tau }^{-3}$. The Boussinesq approximation requires $(z+z_{\ast })/(g{\tau }^2)\ll 1$ in the region modelled. This is a decaying similarity family; its time and virtual origins require source or initial data. The PDF supplies neither, so a unique forced-startup history cannot be selected from it. In particular, the singular unshifted field should not be presented as a regular solution at $t=0$.

For the finite-mass similarity solution above, the total disk mass isSubstitute $\xi =1/(\tau \sqrt{x})$, so $x={\tau }^{-2}{\xi }^{-2}$ and $x^{-3/2}|dx|=2\tau d\xi$. This givesThe requested linear increase of mass is therefore correct. It is supplied by the infinite-radius boundary, rather than by a source-free isolated finite disk. The inward mass accretion rate iswhich tends to zero at the origin and to $3\pi C{\bar{\nu }}_0{\Sigma }_0$ at infinity.

For Keplerian rotation, the specific angular momentum is $\ell =\sqrt{G{M}_{\star }r}$. The total disk angular momentum would beThis integral diverges at $\xi =0$, corresponding to infinite radius. There is no finite constant total angular momentum for the nonzero printed similarity solution. The seemingly time-independent last expression is an infinite quantity; it cannot prove a finite conservation law.

The problem remains visible under a cutoff. With $r\le r_{0}R$, write $J_R=J_{\ast }{E}_1(1/(\tau \sqrt{R}))$, where $J_{\ast }=4\pi C{\Sigma }_0\sqrt{G{M}_{\star }}{r}_0^{5/2}$ and $E_1$ is the exponential integral. ThenEven the finite change after subtracting a common logarithmic divergence is not zero. Direct integration of the disk diffusion equation gives the boundary identityfor this profile the outer expression has a nonzero limit, giving the same finite change rate. An inner zero torque alone does not impose zero outer angular-momentum flux.

More generally, the printed family $A+B{e}^{-\xi }$ can have finite angular momentum at both ends only if $A=0$ and $A+B=0$, hence it is trivial. Thus the request for a nonzero finite constant $J_d$ is incompatible with the stated viscosity exponent, similarity ansatz and infinite radial domain. A physical finite-domain or altered similarity problem would need different boundary data or assumptions; silently assigning a finite value to the divergent integral does not repair it.

Let $g=G(\xi )$ with $\xi ={\tau }^{-1}{x}^{-1/2}$. The derivatives of this similarity solution areThe first-derivative terms cancel in the dimensionless diffusion equation, leaving $G^{\prime \prime }+G^{\prime }=0$ for $\xi >0$. ThusAt the inner boundary $x\to 0$, $\xi \to \infty$. Interpreting the named variable as vanishing there sets $A=0$. Even if only the true viscous torque in an accretion disk $x^{1/2}g$ is required to vanish, a nonzero $A$ would give infinite total mass near $r=0$; finite mass again selects $A=0$. With nonnegative surface density, choose $B=C>0$; its value fixes the initial mass normalization. ThereforeThe surface density vanishes exponentially at the origin and has an $x^{-5/2}$ tail at infinity. Its peak occurs at $x_{\max }=1/(25{\tau }^2)$, with ${\Sigma }_{\max }/{\Sigma }_0=C(5\tau {)}^5{e}^{-5}$. The peak moves inward and grows as time increases.

The plot uses $C=1$ and shows the actual surface density, with logarithmic radius to resolve both peaks. The later profile is larger at every fixed radius. Its total mass is finite; its total angular momentum is not, as the next calculation establishes.

Write $d_1={\rho }_0-{\rho }_1>0$, $d_2={\rho }_2-{\rho }_0>0$, and choose the common undisturbed interface pressure $P$. The hydrostatic pressure is $p_1=P-{\rho }_{1}gz$ in the upper fluid and $p_2=P-{\rho }_{2}gz$ in the lower fluid. Continuity of hydrostatic pressure across both current interfaces givesThus $h_1=d_{2}h/(d_1+d_2)$ and $h_2=-d_{1}h/(d_1+d_2)$. This isostatic depth partition of an interfacial current expresses local vertical buoyancy balance: a displaced column with an imbalance would move vertically until the partition was restored. With the hydrostatic pressure $P-{\rho }_{0}gz$ subtracted, the pressure within the current is $\Delta \rho gh$, whereThe equivalent body force is outward for a profile decreasing with $r$. Its magnitude is $\Delta \rho g|h_r|$; the derivative without an absolute value is a signed quantity.

For an ambient-controlled interfacial plug current, integrating the body force through depth gives radial force per area $O(\Delta \rho g{H}^2/R)$. The ambient Stokes flow varies over distance $R$, exerting traction $O(\mu u/R)$, so $u\sim \Delta \rho g{H}^2/\mu$. The current's internal shear supports the ambient traction with a velocity difference $\Delta u\sim (\mu u/R)H/(\lambda \mu )$; $\Delta u/u\sim H/(\lambda R)\ll 1$ requires $\lambda \gg H/R$. The force per area from its in-plane viscous extension is $O(\lambda \mu uH/R^2)$, negligible against ambient traction when $\lambda \ll R/H$. Together these give the viscosity window for an interfacial plug currentFixed volume gives $V\sim H{R}^2$, and the spreading velocity has the scale $\dot{R}$. Thus $\dot{R}\sim \Delta \rho g{V}^2/(\mu R^4)$ andChoosing $A$ so that $R=A{t}^{1/5}$ exactly, the similarity solution has $h=(V/A^2)t^{-2/5}F(\eta )$, $\eta =r/(A{t}^{1/5})$, and $u=A{t}^{-4/5}U(\eta )$. In radial conservation $h_t+r^{-1}(rhu{)}_r=0$, substitution gives $(\eta FU{)}^{\prime }=({\eta }^{2}F{)}^{\prime }/5$. Regularity and no source at the centre remove the integration constant. Therefore the radial plug velocity is linear:

For the disc-traction analogy for an interfacial current, subtract the imposed straining flow from the stationary-disc solution. The perturbation has velocity $-Er{\mathbf{e}}_r$ on the disc and zero far-field velocity. Reverse its sign: the resulting Stokes flow is driven by a radially expanding disc with velocity $Er{\mathbf{e}}_r$ in otherwise quiescent fluid. The background straining flow contributes no radial shear traction on $z=0$, so the given disc traction changes sign. This step uses Linearity of Stokes flow, Uniqueness of Stokes flow, and the no-slip boundary condition. The thin plug current has the same leading boundary velocity and hence the same ambient resistance. Each face contributes $-8\mu Er/[\pi \sqrt{R^2-r^2}]$; both faces must be counted. With $E=1/(5t)$, radial force balance isIntegrate with $h(R,t)=0$. The interfacial plug-current similarity profile isVolume integration gives $V=2\pi {\int }_0^{R}rhdr=4\pi H{R}^2/5$. Eliminating $H$ gives the full prefactor:The formal edge slope diverges, so a narrow edge region lies outside the thin-current approximation. Its traction and volume contributions are integrable, and this is a leading thin-current solution rather than a uniformly valid edge description.