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For a steady point-source shear-driven viscous gravity current far downstream, $h\ll y_N\ll x$ permits dropping the downstream gravity flux. With $w=h^2$, the equation is $w_x=D(w^2{)}_{yy}/(4A)$, a quadratic porous medium equation. Conserving $A\int wdy=Q$ yields $y_N=(3\mu \Delta \rho gQx/{\tau }^2{)}^{1/3}$ and $h^2=\tau (y_N^2-y^2)/(2\Delta \rho gx)$ inside the current. The central height decays as $x^{-1/6}$. Near the source, balancing shear and gravity gives an upstream reach of order $\sqrt{\mu \Delta \rho gQ}/\tau$, whose numerical coefficient is not determined by scaling.

For a steady line-source shear-driven viscous gravity current with downstream constant thickness, $h_d^2=2\mu Q_{2d}/\tau$. Zero upstream volume flux per unit width gives $h{h}_x=3\tau /(2\Delta \rho g)$, so $h^2=h_d^2+3\tau x/(\Delta \rho g)$ until it vanishes at $x=-x_N$. Unlike the point-source scaling estimate, this reduced line-source model fixes the exact coefficient $2/3$. The square-root nose has an unresolved steep edge.

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.

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 giveIts depth-integrated volume flux isDefine $D=G/(3\mu )$ and $A=\tau /(2\mu )$. The continuity equation gives the shear-driven viscous gravity current equationThe 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 $U\sim \tau H/\mu +G{H}^3/(\mu L)$, with vertical velocity $UH/L$. Small slopes require $H/L\ll 1$. Horizontal fluid inertia relative to vertical viscous resistance is ${\rho }_{c}U{H}^2/(\mu L)$, where ${\rho }_c=\rho +\Delta \rho$. Hence sufficient small parameters, expressed without an unknown velocity, areFor the hydrostatic pressure and normal-stress approximation also require $\mu U/(GHL)\ll 1$. Its gravity-driven part is already $(H/L{)}^2$; its shear-driven part addsThis 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 leavesWrite $Y=y_N(x)$ and let $h_c$ be a typical central thickness. Balance gives $Y^2\sim (D/A)h_c^{2}x$, while conservation of the source volume flux gives $Q\sim A{h}_c^{2}Y$. Therefore the downstream similarity of a shear-driven gravity current hasThe omitted downstream gravity flux relative to the imposed shear flux is $D{h}_c^2/(Ax)\sim (Y/x{)}^2\ll 1$, so this approximation is self-consistent in the stated regime.

For the full profile put $w=h^2$. Then $h^3{h}_y=(w^2{)}_y/4$ andThis is a porous medium equation with downstream distance as its evolution coordinate. Conservation of $\int wdy=Q/A$ and a similarity solution $w=x^{-1/3}f(\eta )$, $\eta =y/x^{1/3}$, givewhere symmetry gives zero integration constant at $\eta =0$. Inside the positive region, $f^{\prime }=-2A\eta /(3D)$, so $f=(A/(3D))({\eta }_N^2-{\eta }^2)$. Requiring the total downstream volume flux to be $Q$ fixes the coefficient:Consequentlyfor $|y|<y_N$, 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:Eliminating $h_{\ast }$ givesThis is a scaling estimate, not a determination of a numerical prefactor. The associated depth is $h_{\ast }\sim (\mu Q/(\Delta \rho g){)}^{1/4}$. 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 $q=A{h}^2-D{h}^3{h}_x$. 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 givesThe line-source upstream reach under imposed shear is thereforeIn 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.