 Solution

Write $G=\Delta \rho g>0$ and take $\tau >0$ in the positive $x$ direction. Removing the ambient hydrostatic pressure leaves excess pressure $G(h-z)$, so the horizontal pressure gradient in the dense gravity current is $G\nabla h$. The lubrication approximation givesThe no-slip boundary condition at $z=0$ and the imposed upper shear stress require $u(0)=v(0)=0$, $\mu u_z(h)=\tau$ and $\mu v_z(h)=0$. Integration yieldsConsequently the depth-integrated volume flux isUsing conservation of mass, $h_t+\nabla \cdot \mathbf{q}=0$, proves the evolution equation for a shear-driven viscous gravity current:The gravity term smooths thickness gradients, while the imposed shear transports the current downstream.

The approximations need more than a small ordinary Reynolds number. With thickness $H$, horizontal scale $L$, dense-fluid density ${\rho }_c=\rho +\Delta \rho$ and typical horizontal speed $V$, sufficient small parameters areThe first neglects horizontal viscous derivatives relative to vertical ones. The second compares convective inertia with vertical viscous stress: it is the reduced Reynolds number, ${\mathrm{Re}}_H(H/L)$, rather than a requirement that ${\mathrm{Re}}_H$ itself be small. The third keeps viscous normal stress small relative to the excess hydrostatic pressure $GH$. Here $V$ is of order $\tau H/\mu +G{H}^3/(\mu L)$, so this last condition requires $\tau /(GL)\ll 1$ together with $(H/L{)}^2\ll 1$. Explicit sufficient inertia conditions are ${\rho }_c\tau H^3/({\mu }^{2}L)\ll 1$ and ${\rho }_{c}G{H}^5/({\mu }^2{L}^2)\ll 1$. If an independently imposed unsteady time scale is $T$, also require ${\rho }_c{H}^2/(\mu T)\ll 1$; for $T\sim L/V$ this is the same reduced-inertia condition.

Set $A=\tau /(2\mu )$ and $D=G/(3\mu )$. Far downstream, the transverse scale $Y=y_N$ is much smaller than $x$, so the downstream gravity-flux divergence is smaller than the transverse one by order $(Y/x{)}^2$. For the steady flow, away from the source,The latter expresses conservation of the prescribed point-source volume flux. In scaling form, $Q\sim A{H}^{2}Y$ and $A{H}^2/x\sim D{H}^4/Y^2$. Eliminating $H$ gives $Y^3\sim DQx/A^2$, proving $y_N\propto x^{1/3}$ and $H\propto x^{-1/6}$.

For the exact downstream similarity solution, put $w=h^2$. Since $h^3{h}_y=(w^2{)}_y/4$, the reduced equation is the porous medium equationSeek $w=x^{-1/3}F(\eta )$ with $\eta =y/x^{1/3}$. Substitution givesIntegrate once, using symmetry and zero transverse flux at $\eta =0$, to obtain $-A\eta F/3=DF{F}^{\prime }/2$. Where $F>0$, this gives $F^{\prime }=-2A\eta /(3D)$, hence $F=A({\eta }_N^2-{\eta }^2)/(3D)$. Vanishing thickness fixes its support. The volume-flux condition then readsThus the downstream profile and half-width arewith $h=0$ outside that interval. This is the downstream similarity of a shear-driven gravity current. The square-root edge is the formal outer thin-layer solution; arbitrarily close to that edge its slope is not small, so the local front requires a separate description. This does not change the downstream scaling and flux calculation.

Near the source the two horizontal scales are comparable, say $\ell$, so the far-downstream reduction cannot locate the upstream nose. In the upstream direction the downstream shear flux $A{H}^2$ must balance the upstream gravity flux of order $D{H}^4/\ell$. Hence $\ell \sim (D/A)H^2$. The total discharge through a region of width $\ell$ is $Q\sim A{H}^2\ell$, giving $H^4\sim Q/D$. Combining these estimates gives the upstream reach scaleThe shear advects liquid toward positive $x$, while the hydrostatic thickness gradient drives liquid back toward negative $x$; their balance stops further upstream spreading. A numerical prefactor requires the full two-dimensional near-source free-boundary solution and cannot be fixed by these scaling arguments alone.