= Solution
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> give
$$
\mathbf u_{\parallel}(z)=-\frac{G}{2\mu}z(2h-z)\nabla h+\frac{\tau}{\mu}z\mathbf e_x.
$$
Its depth-integrated <volume flux> is
$$
\mathbf q=-\frac{G}{3\mu}h^3\nabla h+\frac{\tau}{2\mu}h^2\mathbf e_x.
$$
Define $D=G/(3\mu)$ and $A=\tau/(2\mu)$. The <continuity equation> gives the <shear-driven viscous gravity current> equation
$$
\boxed{h_t+A\partial_x(h^2)=D\nabla\cdot(h^3\nabla h).}
$$
The 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 $U\sim\tau H/\mu+GH^3/(\mu L)$, with vertical <velocity> $UH/L$. Small slopes require $H/L\ll1$. Horizontal fluid inertia relative to vertical viscous resistance is $\rho_c UH^2/(\mu L)$, where $\rho_c=\rho+\Delta\rho$. Hence sufficient small parameters, expressed without an unknown <velocity>, are
$$
\boxed{\frac HL\ll1,\qquad
\frac{\rho_c\tau H^3}{\mu^2L}\ll1,\qquad
\frac{\rho_c GH^5}{\mu^2L^2}\ll1.}
$$
For the <hydrostatic pressure> and normal-stress approximation also require $\mu U/(GHL)\ll1$. Its gravity-driven part is already $(H/L)^2$; its shear-driven part adds
$$
\boxed{\frac{\tau}{GL}\ll1.}
$$
This 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 leaves
$$
A(h^2)_x=D(h^3h_y)_y.
$$
Write $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim(D/A)h_c^2x$, while conservation of the source <volume flux> gives $Q\sim Ah_c^2Y$. Therefore the <downstream similarity of a shear-driven gravity current> has
$$
\boxed{Y\sim\left(\frac{DQx}{A^2}\right)^{1/3},\qquad h_c\propto x^{-1/6}.}
$$
The omitted downstream gravity flux relative to the imposed shear flux is $Dh_c^2/(Ax)\sim(Y/x)^2\ll1$, so this approximation is self-consistent in the stated regime.
For the full profile put $w=h^2$. Then $h^3h_y=(w^2)_y/4$ and
$$
w_x=\frac{D}{4A}(w^2)_{yy}.
$$
This is a <porous medium equation> with downstream distance as its evolution coordinate. Conservation of $\int w\,dy=Q/A$ and a <similarity solution> $w=x^{-1/3}f(\eta)$, $\eta=y/x^{1/3}$, give
$$
-\frac A3(f+\eta f')=\frac D4(f^2)'',\qquad
-\frac A3\eta f=\frac D2ff',
$$
where symmetry gives zero integration constant at $\eta=0$. Inside the positive region, $f'= -2A\eta/(3D)$, so $f=(A/(3D))(\eta_N^2-\eta^2)$. Requiring the total downstream <volume flux> to be $Q$ fixes the coefficient:
$$
Q=A\int_{-Y}^Yh^2\,dy=\frac{4A^2Y^3}{9Dx}.
$$
Consequently
$$
\boxed{y_N(x)=\left(\frac{3\mu\Delta\rho gQ}{\tau^2}x\right)^{1/3},\qquad
h(x,y)=\left[\frac{\tau}{2\Delta\rho gx}\bigl(y_N(x)^2-y^2\bigr)\right]^{1/2}}
$$
for $|y|<y_N$, 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>:
$$
\frac{Dh_*^4}{\ell}\sim Ah_*^2,\qquad Q\sim Ah_*^2\ell.
$$
Eliminating $h_*$ gives
$$
\boxed{x_N\sim\frac{\sqrt{DQ}}A\sim\frac{\sqrt{\mu\Delta\rho gQ}}{\tau}.}
$$
This is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_*\sim(\mu Q/(\Delta\rho g))^{1/4}$. 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 $q=Ah^2-Dh^3h_x$. 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 gives
$$
h^2(x)=h_d^2+\frac{2A}{D}x=\frac{2\mu Q_{2d}}{\tau}+\frac{3\tau}{\Delta\rho g}x.
$$
The <line-source upstream reach under imposed shear> is therefore
$$
\boxed{x_N=\frac{DQ_{2d}}{2A^2}=\frac{2\mu\Delta\rho gQ_{2d}}{3\tau^2},\qquad
h(x)=\begin{cases}
0,&x\leq-x_N,\\
\sqrt{\dfrac{3\tau}{\Delta\rho g}(x+x_N)},&-x_N<x<0,\\
\sqrt{\dfrac{2\mu Q_{2d}}\tau},&x>0.
\end{cases}}
$$
In 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.
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