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For the steady positive-velocity branch, conservation of buoyancy flux gives $bhu=F_0$. Substituting into the other two balances givesUse a power-law ansatz $h=C{x}^p$, $u=D{x}^q$. The entrainment balance forces $p=1$ and $C(1+q)=\alpha$. The momentum powers are $x^{2q}$, $x^{-q}$ and $x^{-q}$, so the nonzero inertial branch has $q=0$. Therefore the constant-speed entraining gravity current on a slope isIts positive speed requires $\tan \theta >\alpha /2$ for $\theta <\pi /2$. In the shallow, strictly hyperbolic range this meansThe same steady integral balances have a formal vertical-plume limit at $\theta =\pi /2$, although their time-dependent system then loses strict hyperbolicity. At the lower threshold the assumed positive-flux, finite-speed solution fails.

The line source is singular: $h,uh$ and the momentum flux tend to zero, but $b$ diverges as $x\downarrow 0$. Thus this is a downstream ideal-source solution, valid only far enough away that $b/g\ll 1$ and the Boussinesq approximation and top-hat plume model hold. It does not describe the immediate source region.

Seek a steady downslope solution with $u>0$ constant. The volume conservation equation reduces to $h_x=E$, so the constant-speed entraining gravity current on a slope hasThe conserved buoyancy flux per unit width is $B=u{g}^{\prime }h$, givingSubstitute these expressions into the momentum equation. Its advective term is $E{u}^2$ and its hydrostatic term is $\frac{1}{2}E(B/u)\cos \theta$. The balance is thereforewhich givesThe assumed slope inequality makes ${\beta }_{\ast }>0$. The increasing thickness dilutes reduced gravity while leaving $g^{\prime }h$ and the buoyancy flux constant. The denominator includes both bottom drag and the momentum needed to accelerate newly entrained ambient fluid.

For slowly depositing particles, $B_x=-v_{s}B/(uh)$ and local adjustment gives $u\sim {\beta }_{\ast }{B}^{1/3}$ with the constant-speed entraining gravity current on a slope coefficient. At leading order $h\sim h_0+E(x-x_0)$ and $B^{1/3}=B_o^{1/3}-[v_s/(3{\beta }_{\ast }E)]\ln [1+E(x-x_0)/h_0]$. Its validity requires settling slow compared with the current and, for small $E$, compared with entrainment.