 Solution

In the thin-current approximation, vertical momentum balance is hydrostatic: $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglect inertia and radial viscous derivatives compared with the vertical ones. The radial Stokes equation is $\mu u_{zz}=p_r=\rho g{h}_r$, with no slip at $z=0$ and zero tangential stress $u_z=0$ at $z=h$. Integration givesIts depth-integrated radial flux is $Q_r={\int }_0^{h}udz=-\rho g{h}^3{h}_r/(3\mu )$. Axisymmetric mass conservation $h_t+r^{-1}{\partial }_r(r{Q}_r)=0$ therefore givesThe remaining conditions are conserved volume $2\pi {\int }_0^{R(t)}rhdr=V$, regularity and zero radial flux at the center, $h(R(t),t)=0$, zero thickness outside the front, and the initially concentrated volume at the origin. The advancing edge obeys $R^{\prime }={\lim }_{r\uparrow R}{Q}_r/h$ when this limit exists. One must not impose a finite zero slope at this compact-support edge.

Put $h=t^{a}F(r{t}^{-b})$ with dimensional factors to be restored. Constant volume gives $a+2b=0$, while balancing the equation gives $a-1=4a-2b$. Hence $a=-1/4,b=1/8$. In normalized similarity variables $\eta =r/(\beta V^{3}t{)}^{1/8}$ and $h=V^{1/4}(\beta t{)}^{-1/4}F(\eta )$, the equation becomesMultiplying by $\eta$ and integrating from the regular center yields $\eta F^3{F}^{\prime }=-{\eta }^{2}F/8$. Where $F>0$, this integrates to $F^3=3({\eta }_0^2-{\eta }^2)/16$. Restoring physical variables giveswith zero thickness outside. Write $h=A(t)(1-r^2/R^2{)}^{1/3}$. Its volume is $3\pi A{R}^2/4$, so $A=4V/(3\pi R^2)$. Combining this with $A^3=3R^2/(16\beta t)$ givesThe center height scales as $t^{-1/4}$, while the front scales as $t^{1/8}$. The ideal profile has a singular slope at the edge; microscopic contact-line physics is excluded from the lubrication model, which applies to the sufficiently spread bulk current.