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For a steady upward plume with prescribed $N^2=-G^2<0$, the kinematic plume fluxes satisfyThe positive sign in the last equation expresses the gain of buoyancy flux as the plume rises through the prescribed unstable ambient. Seek a positive power law solution $Q=q{z}^p$, $M=m{z}^r$, $F=f{z}^s$, with $z>0$. Matching powers givesThe last relation gives $s=p+1$ and the first $r=2p-2$. Substitution into the middle relation yields $4p-5=2p+1$, henceThe amplitude equations areFrom the first relation $m=9q^2/(4{\alpha }^2)$, while the last two give $m^2=G^2{q}^2/16$. Taking positive amplitudes gives $m=Gq/4$, and consequentlyReconstructing the radius, speed and reduced gravity gives the top-hat plume in constant unstable stratification:As a direct check, $M_z={\alpha }^2{G}^2{z}^3/9=FQ/M$ and $F_z={\alpha }^2{G}^3{z}^3/9=G^{2}Q$, while $Q_z={\alpha }^{2}G{z}^2/3=2\alpha \sqrt{M}$.

The corresponding ambient gradient is ${\rho }_{0z}={\rho }_r{G}^2/g$. Thus $\rho (z)={\rho }_0(z)-{\rho }_r{G}^{2}z/(4g)$; for a linear ambient this is ${\rho }_0(0)+3{\rho }_r{G}^{2}z/(4g)$. The plume is locally lighter than the ambient even though both densities increase upward in this prescribed unstable stratification.

The coordinate $z$ can be measured from a virtual origin. This exact power law branch has fluxes tending to zero there and describes buoyancy generated by the maintained unstable ambient; it does not impose arbitrary nonzero source fluxes at that origin. The given idealization holds the ambient quiescent despite its static instability. Finally, the Boussinesq approximation limits the modeled height range to small relative density changes, here $G^{2}z/g\ll 1$. Within these assumptions, the displayed positive branch proves the requested existence.