 Solution

For identical suspended particles, reduced gravity is proportional to particle volume fraction. If particles settle through the bed at speed $v_s$ and are not resuspended, the downward particle deposition flux removes a buoyancy-equivalent amount $v_s{g}^{\prime }$ per unit bed area. The depth-integrated scalar mass conservation equation isFor a steady particle-laden gravity current, $B_x=-v_{s}B/(uh)$. If $v_s/u\ll 1$, with fixed positive $E$ and drag/slope coefficients, the locally adjusted leading solution of the momentum and volume equations is $u\sim {\beta }_{\ast }{B}^{1/3}$ and $h\sim h_0+E(x-x_0)$. The weak-settling attenuation of a sloping gravity current then givesand henceFor an inlet at $x_0=0$ this has the requested logarithmic form. If the inlet is genuinely at $x_0\ne 0$, the argument must contain $x-x_0$ in order to satisfy the specified inlet condition.

The speed factor must be the ${\beta }_{\ast }$ derived in the previous part. The printed $\beta$ omits the factor one-half multiplying $E\cos \theta$ and uses an undefined uppercase $C$ instead of the governing equation's $c$. Taking $C=c$ still leaves a nonzero leading momentum residual $-\frac{1}{2}E(B/u)\cos \theta$ for the printed speed; this discrepancy cannot be repaired by a correction of order $v_s/u$. With the given hydrostatic momentum equation, the logarithmic law is valid with ${\beta }_{\ast }$, not the printed speed factor.

A controlled local approximation also needs $v_s/(Eu)\ll 1$ if $E$ is allowed to become small. Indeed the exact volume balance is $h_x=E-h{u}_x/u$; under $u\sim {\beta }_{\ast }{B}^{1/3}$ its second term is $v_s/(3u)$. The neglected depth and acceleration corrections become significant as $B$ approaches zero.