= Solution
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 $p_1=P-\rho_1gz$ in the upper fluid and $p_2=P-\rho_2gz$ in the lower fluid. Continuity of <hydrostatic pressure> across both current interfaces gives
$$
p_0=P-\rho_0gz+d_1gh_1=P-\rho_0gz-d_2gh_2,\qquad \boxed{d_1h_1=-d_2h_2.}
$$
Thus $h_1=d_2h/(d_1+d_2)$ and $h_2=-d_1h/(d_1+d_2)$. 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> $P-\rho_0gz$ subtracted, the pressure within the current is $\Delta\rho gh$, where
$$
\boxed{\Delta\rho=\frac{(\rho_0-\rho_1)(\rho_2-\rho_0)}{\rho_2-\rho_1},\qquad\mathbf f=-\Delta\rho g h_r\mathbf e_r.}
$$
The 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 $O(\Delta\rho gH^2/R)$. The ambient <Stokes flow> varies over distance $R$, exerting <traction> $O(\mu u/R)$, so \b[$u\sim\Delta\rho gH^2/\mu$]. The current's internal shear supports the ambient <traction> with a velocity difference $\Delta u\sim(\mu u/R)H/(\lambda\mu)$; $\Delta u/u\sim H/(\lambda R)\ll1$ 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 current>
$$
\boxed{H/R\ll\lambda\ll R/H.}
$$
Fixed volume gives $V\sim HR^2$, and the spreading velocity has the scale $\dot R$. Thus $\dot R\sim\Delta\rho gV^2/(\mu R^4)$ and
$$
\boxed{R\sim At^{1/5},\qquad A\sim(\Delta\rho gV^2/\mu)^{1/5}.}
$$
Choosing $A$ so that $R=At^{1/5}$ exactly, the <similarity solution> has $h=(V/A^2)t^{-2/5}F(\eta)$, $\eta=r/(At^{1/5})$, and $u=At^{-4/5}U(\eta)$. In radial conservation $h_t+r^{-1}(rhu)_r=0$, substitution gives $(\eta FU)'=(\eta^2F)'/5$. Regularity and no source at the centre remove the integration constant. Therefore \b[the radial plug velocity is linear]:
$$
\boxed{U(\eta)=\eta/5,\qquad u=r/(5t).}
$$
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 is
$$
-\Delta\rho g h h_r=\frac{16\mu Er}{\pi\sqrt{R^2-r^2}},\qquad\boxed{h h_r=-\frac{16\mu r}{5\pi\Delta\rho gt\sqrt{R^2-r^2}}.}
$$
Integrate with $h(R,t)=0$. The <interfacial plug-current similarity profile> is
$$
\boxed{h^2=\frac{32\mu\sqrt{R^2-r^2}}{5\pi\Delta\rho gt},\qquad h=H(1-r^2/R^2)^{1/4},\quad H^2=\frac{32\mu R}{5\pi\Delta\rho gt}.}
$$
Volume integration gives $V=2\pi\int_0^Rrh\,dr=4\pi HR^2/5$. Eliminating $H$ gives the full prefactor:
$$
\boxed{R(t)=\left(\frac{125\Delta\rho gV^2t}{512\pi\mu}\right)^{1/5}.}
$$
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.
For a <diffusion-controlled solute gravity current>, excess solute mass, not enriched-fluid volume, is conserved. At late times the effective <mass density> excess is proportional to solute concentration, so $\Delta\rho(t)R^2H\sim\Delta\rho_0V_0$, and $H\sim(Dt)^{1/2}$. The same ambient resistance scaling gives
$$
\dot R\sim\frac{\Delta\rho_0gV_0}{\mu}\frac{(Dt)^{1/2}}{R^2},\qquad\boxed{R(t)\sim\left(\frac{\Delta\rho_0gV_0}{\mu}\right)^{1/3}D^{1/6}t^{1/2}.}
$$
This scaling does not determine a universal numerical prefactor. The <mass diffusivity> $D$ controls the growing depth, while dilution weakens buoyancy; assuming a fixed current volume would incorrectly retain the $t^{1/5}$ law. The thin-layer requirement becomes $(\mu D/(\Delta\rho_0gV_0))^{1/3}\ll1$; when only dissolved solute changes the upper fluid, its <viscosity> is of order the ambient <viscosity>, consistent with the plug-current window.
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