= Solution
For the steady positive-velocity branch, conservation of <buoyancy flux> gives $bhu=F_0$. Substituting into the other two balances gives
$$
(hu)'=\alpha u,\qquad
\left(hu^2+\frac{F_0h}{2u}\cos\theta\right)'=\frac{F_0}{u}\sin\theta.
$$
Use a <power-law ansatz> $h=Cx^p$, $u=Dx^q$. 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> is
$$
\boxed{h=\alpha x,\qquad
u=\left[\frac{F_0}{\alpha}\left(\sin\theta-\frac{\alpha}{2}\cos\theta\right)\right]^{1/3},\qquad
\rho-\rho_0=\frac{\rho_0F_0}{g\alpha ux}.}
$$
Its positive speed requires $\tan\theta>\alpha/2$ for $\theta<\pi/2$. In the shallow, strictly hyperbolic range this means
$$
\boxed{\arctan(\alpha/2)<\theta<\pi/2,\qquad \alpha\ll1.}
$$
The 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: $h,u h$ and the momentum flux tend to zero, but $b$ diverges as $x\downarrow0$. Thus this is a downstream ideal-source solution, valid only far enough away that $b/g\ll1$ and the <Boussinesq approximation> and <top-hat plume model> hold. It does not describe the immediate source region.
Back to article page
