= Solution
Take $x$ downslope and $z$ normal to the plane. To leading order in the <lubrication approximation>, <Newtonian gravity> gives the <hydrostatic pressure> $p=p_{\mathrm{atm}}+\rho g(h-z)$. The downslope <Stokes equation> is therefore $\mu u_{zz}=\rho g(h_x-\theta)$. Imposing the <no-slip boundary condition> at $z=0$ and the <stress-free boundary condition> at $z=h$ gives
$$
u=\frac g\nu(\theta-h_x)\left(hz-\frac{z^2}{2}\right),\qquad q=\int_0^h u\,dz=\frac g{3\nu}h^3(\theta-h_x).
$$
Here $\nu=\mu/\rho$ is the <kinematic viscosity>; $u$ is the downslope <velocity>. Applying <mass conservation> to this <lubrication gravity-current flux>,
$$
\boxed{h_t=\frac g{3\nu}\partial_x[h^3(h_x-\theta)]}.
$$
For typical thickness $H=\hat h$ and length $L=\hat x$, the transverse <velocity> is $O(UH/L)$ by <incompressibility>. Both convective terms are then $O(U^2/L)$, while transverse viscous acceleration is $O(\nu U/H^2)$. Thus the <inertia criterion for an inclined viscous film> is $UH^2/(\nu L)\ll1$, in addition to $H/L\ll1$. With evolution time $T\sim L/U$ the same condition controls unsteady inertia. If a faster time is independently imposed, also require $H^2/(\nu T)\ll1$. Estimating $U$ from the <velocity> profile gives the two requested limits:
$$
\boxed{H\ll\theta L:\quad U\sim\frac{g\theta H^2}{\nu},\quad \frac{g\theta H^4}{\nu^2L}\ll1;\qquad H\gg\theta L:\quad U\sim\frac{gH^3}{\nu L},\quad \frac{gH^5}{\nu^2L^2}\ll1.}
$$
At long times, $h\sim A/x_N$ and the ratio of hydrostatic-gradient forcing to downslope <Newtonian gravity> in the bulk is $h_x/\theta=O(A/(\theta x_N^2))\ll1$. Put $K=g\theta/(3\nu)$; the bulk <conservation law> is $h_t+(Kh^3)_x=0$. The <finite-volume gravity-driven thin-film current> scales as $x\sim t^{1/3}$, $h\sim t^{-1/3}$. More explicitly, writing $h=t^{-1/3}f(\xi)$, $\xi=x/t^{1/3}$, gives
$$
-\frac13(f+\xi f')+K(f^3)'=0,\qquad Kf^3-\frac13\xi f=0.
$$
The <constant of integration> is zero for the zero-flux dry tail. Consequently
$$
\boxed{h(x,t)\simeq\sqrt{\frac{x}{3Kt}}=\sqrt{\frac{\nu x}{g\theta t}},\qquad 0<x<x_N(t)}.
$$
The reduced bulk solution is zero outside this interval. Its thin nose region is resolved below. Integrating to impose the conserved area,
$$
A=\int_0^{x_N}\sqrt{\frac{x}{3Kt}}\,dx=\frac{2x_N^{3/2}}{3\sqrt{3Kt}},\qquad \boxed{x_N^3=\frac{27}{4}KA^2t=\frac{9g\theta A^2t}{4\nu}}.
$$
The bulk front height is $h_N=3A/(2x_N)$, with $h_N^2=x_N/(3Kt)$. Differentiation of the front law gives
$$
\boxed{\dot x_N=\frac{x_N}{3t}=Kh_N^2=\frac{g\theta h_N^2}{3\nu}}.
$$
Equivalently, this is the <Rankine-Hugoniot condition> for the bulk <shock wave>, whose speed is flux divided by height.
In moving coordinate $y=x-x_N(t)$, the full <thin-film equation> becomes $h_t|_y-\dot x_N h_y+q_y=0$. A nose thickness $h_N$ and slope $\theta$ give axial scale $\ell_N=h_N/\theta$. Since $\ell_N/x_N=O(A/(\theta x_N^2))\ll1$, the explicit time derivative is smaller than translation by that ratio. The <travelling front of a gravity-driven thin film> is therefore quasisteady. Integrate once with dry-tip flux zero to obtain $q=\dot x_N h$, or
$$
h_y=\theta\left(1-\frac{h_N^2}{h^2}\right).
$$
With $f=h/h_N$, separation of variables and the convention $y=0$ at $f=0$ give
$$
\boxed{\frac{\theta y}{h_N}=f-\operatorname{atanh}f=f+\frac12\log\frac{1-f}{1+f},\qquad 0\le f<1}.
$$
The nose approaches $h_N$ exponentially far behind its tip, then falls to zero with a vertical tangent in this small-slope model. The requested sketch shows this monotone profile:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-74-nose.png]
{title=Quasisteady gravity-current nose, with a dry tip and an asymptotically uniform upstream film}
Expanding the <inverse hyperbolic tangent>, $f-\operatorname{atanh}f=-f^3/3+O(f^5)$, so $h\sim(-3\theta h_N^2y)^{1/3}$. Its slope has magnitude $\theta/f^2$ near the tip. The <small-slope breakdown at an inclined-current nose> occurs at $f=O(\sqrt\theta)$, which gives
$$
\boxed{h=O(\sqrt\theta h_N),\qquad |y|=O(\sqrt\theta h_N)}.
$$
Finally, <surface tension> produces a pressure-gradient scale $\gamma h_N/\ell_N^3$, whereas downslope <Newtonian gravity> gives $\rho g\theta$. Their ratio is $\gamma\theta^2/(\rho g h_N^2)$. Thus <surface tension> changes the principal nose length when
$$
\boxed{h_N\sim\theta\sqrt{\frac{\gamma}{\rho g}}}.
$$
This is a bulk-nose crossover; capillarity and contact-line physics can already affect a smaller region very close to the dry tip even when the bulk nose is <Newtonian gravity> dominated.
Back to article page
