 Solution

Write $s=\tilde{z}$ and use full positive coefficients, $\tilde{Q}=A{s}^{\theta }$, $\tilde{M}=D{s}^{\mu }$, $\tilde{B}=C{s}^{\psi }$. Comparing exponents in the three plume equations givesSolving these equations yieldsThe coefficient equations are $\theta A=\sqrt{D}$, $\mu D=CA/D$ and $\psi C=-{\lambda }^{\beta }A$. ThusFor a rising plume with $Q,M,B>0$, the first equation requires $\theta >0$, the momentum equation $\mu >0$, and the decreasing-buoyancy equation $\psi <0$. Together these requireThey are the physical plume similarity in power-law stratification branches, not positive-plume solutions for arbitrary powers. The endpoint coefficients are singular; the endpoints cannot be included by simply substituting into these formulas. At the prescribed positive source coordinate $s_s=1/\lambda$, the matched source buoyancy isfinite and positive for every finite $\lambda >0$ in the open range. In contrast, the same branch has divergent buoyancy at the mathematical origin $s=0$. A source at positive height makes the branch physically usable. Merely saying a power is finite at a nonzero coordinate would not itself determine this range; positivity and the plume balances are essential.

At large $s$, $Q$ and $M$ increase without bound, $B$ decreases to zero, radius $Q/\sqrt{\pi M}$ grows proportionally to height, and mean upward velocity $M/Q\propto s^{1+\beta /2}$ decreases but remains positive. The integrated buoyancy loss is finite because $\theta +\beta =\psi -1<-1$. The weakening ambient stratification never makes this matched plume negatively buoyant at a finite height; its upward momentum continues to grow even while its speed decreases through entrainment. Consequently its maximum rise height is infinite in this ideal model. This particular exact branch requires its three source fluxes to match the displayed coefficients; it is not a proof that every source in the same ambient escapes.