 Solution

Differentiate $b=Q/\sqrt{M}$ and use both balances:Substituting $Q^2=8\alpha M^{5/2}/(5B_0)+c{Q}_0^2$ givesFor $c<0$, the radius grows faster than the pure-plume slope $6\alpha /5$ and its growth rate decreases toward that slope as $M$ increases. For $c=0$, it grows at that slope exactly. For $0<c<1$, it grows more slowly, possibly initially contracting, and its slope increases toward $6\alpha /5$. Since $M^{\prime }=B_{0}Q/M>0$ and $M\to \infty$, all branches satisfy $db/dz\to 6\alpha /5$ as $z\to \infty$.

Initial necking requiresThus a sufficiently lazy plume initially narrows as buoyancy accelerates its fluid faster than fluid entrainment supplies additional volume. It still has increasing volume and momentum fluxes; shrinking radius does not imply volume loss. At equality the initial slope is zero, and for $c>3/5$ the slope later becomes positive once $M^{5/2}=5c{B}_0{Q}_0^2/(12\alpha )$. This gives a finite minimum radius before the eventual pure plume regime.