= Solution
Let $z$ measure height above the substrate, and write $\mu=\rho\nu$. In the <lubrication approximation>, the horizontal scale $\ell$ is much larger than the typical height $H_0$. Vertical momentum balance is hydrostatic to leading order, while the horizontal balance retains the dominant vertical shear:
$$
p=p_{\rm atm}+\rho g(h-z),\qquad \mu u_{zz}=p_x=\rho g h_x.
$$
The <no-slip boundary condition> gives $u(0)=0$; negligible air viscosity and constant <surface tension> give $u_z(h)=0$. Integrating twice and then through the depth yields
$$
u(x,z,t)=\frac{g h_x}{\nu}\left(\frac{z^2}{2}-hz\right),\qquad
q=\int_0^hu\,dz=-\frac{g}{3\nu}h^3h_x.
$$
The free-surface <kinematic boundary condition> and <incompressibility> imply $h_t+q_x=0$. Therefore
$$
\boxed{h_t=\frac{g}{3\nu}(h^3h_x)_x.}
$$
The cross-sectional area $A=\int h\,dx$ is conserved when the endpoint <volume flux> vanishes.
The relevant small groups follow from the same balances. Put $\epsilon_h=H_0/\ell$ and $U_*=gH_0^3/(3\nu\ell)$. Small $\epsilon_h$ suppresses horizontal viscous derivatives and makes vertical viscous corrections to hydrostatic pressure smaller by $O(\epsilon_h^2)$. The ratio of horizontal inertia to vertical viscous stress is
$$
\boxed{\frac{U_*H_0^2}{\nu\ell}=\epsilon_h\mathrm{Re}_{H_0}
=\epsilon_h^2\mathrm{Re}_{\ell}\ll1.}
$$
Thus a small full-length <Reynolds number> is sufficient but not necessary: thin-layer geometry reduces the inertial ratio. The unsteady viscous ratio $H_0^2/(\nu t_*)$ must likewise be small; on the spreading timescale $t_*=\ell/U_*$ it equals the displayed group. If <surface tension> is retained, its pressure-gradient correction relative to gravity is
$$
\boxed{\frac{\gamma}{\rho g\ell^2}=\mathrm{Bo}_{\ell}^{-1}\ll1.}
$$
The <Bond number> here uses the horizontal scale, because the capillary pressure is $O(\gamma H_0/\ell^2)$ whereas the hydrostatic pressure is $O(\rho gH_0)$. We also neglect air shear and evaporation and assume a Newtonian fluid with constant <kinematic viscosity>.
To find the <similarity solution>, put $K=g/(3\nu)$. Conservation of $A$ gives $h\sim A/\ell$, and balancing the <time derivative> against the nonlinear diffusion gives $\ell^5\sim KA^3t$. A normalized ansatz is
$$
h=\left(\frac{A^2}{Kt}\right)^{1/5}f(\xi),\qquad
\xi=\frac{x-x_c}{(KA^3t)^{1/5}},\qquad \int f\,d\xi=1.
$$
Substitution gives $(f^3f')'=-(f+\xi f')/5$. For the symmetric zero-flux profile, one integration gives $f^3f'=-\xi f/5$, and on its positive support a second integration gives
$$
f^3=C-3\xi^2/10.
$$
Thus the solution is a compactly supported <Barenblatt solution> of the <porous medium equation>. Let $\ell(t)$ denote its half-width. In dimensional variables,
$$
\boxed{h(x,t)=\left[\frac{9\nu}{10gt}\{\ell(t)^2-(x-x_c)^2\}\right]_+^{1/3}.}
$$
Here the positive-part notation sets the height to zero outside $|x-x_c|<\ell$. The area normalization fixes the coefficient, not merely the exponent. With $x-x_c=\ell\sin\theta$,
$$
A=2\left(\frac{9\nu}{10gt}\right)^{1/3}\ell^{5/3}
\int_0^{\pi/2}\cos^{5/3}\theta\,d\theta.
$$
Writing the remaining integral as $I$, the final widths are
$$
\boxed{\ell(t)=\left(\frac{5gA^3t}{36\nu I^3}\right)^{1/5},\qquad
\text{full drop width}=2\left(\frac{5gA^3t}{36\nu I^3}\right)^{1/5}.}
$$
The height falls like $t^{-1/5}$ and the width grows like $t^{1/5}$. The <volume flux> tends to zero at the front, even though the profile slope diverges; this gives a valid compactly supported <weak solution> of the reduced equation. The diverging edge slope means the <lubrication approximation> is not uniform at a microscopic contact line. Surface-tension or contact-line physics supplies an inner regularization without changing the gravity-dominated bulk similarity law in its validity regime. This is an exact source-type solution for $t>0$; an arbitrary finite initial drop approaches an appropriate late-time profile rather than being exactly represented by this singular initial profile.
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