Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 330 / 2

Use the steady Boussinesq top-hat plume in a stratified ambient, with negligible ambient vertical momentum and no externally added heat or particles. The Batchelor entrainment hypothesis supplies inward velocity $\alpha w$ across circumference $2\pi b$. Volume conservation gives $\pi Q_z=2\pi b\alpha w$, and $bw=\sqrt{M}$, henceThe momentum flux increases by the integrated buoyancy force. Writing $g^{\prime }=g\Delta \rho /{\rho }_o$, the force per reference density and per $\pi$ is $g^{\prime }{b}^2$. Since $F=g^{\prime }Q$ and $b^2=Q^2/M$, this givesFor the buoyancy flux, entrained material has the local environmental mass density. Let ${\rho }_p$ be the plume's averaged density. Its scalar mass conservation is $({\rho }_{p}Q{)}_z={\rho }_a{Q}_z$. Since $F=(g/{\rho }_o)({\rho }_a-{\rho }_p)Q$, differentiating cancels both entrainment terms and givesIn stable density stratification, ${\rho }_a^{\prime }<0$ and a rising plume loses buoyancy flux. These are kinematic fluxes with the common factor $\pi {\rho }_o$ removed, so no additional factor of $\pi$ belongs in the three differential equations.

Define $K=N_s^2{z}_s^{-\beta }>0$. The buoyancy frequency definition gives $F_z=-KQ{z}^{\beta }$. The powers in a plume similarity in power-law stratification satisfySolving givesA rising positive-buoyancy turbulent plume entrains, so $Q_z>0$; its momentum flux increases, so $M_z>0$; its buoyancy flux decreases in the stable ambient, so $F_z<0$. With positive coefficients, this requires $q>0$, $m>0$ and $f<0$. The strongest restrictions areIn this interval $1<q<5/3$, $0<m<4/3$ and $-2<f<0$.

Coefficient matching makes the positive similarity solution explicit:soIt is an ideal scaling for $z>0$, often a far-field description. As $z\to 0$, the prescribed ambient stratification and this model's buoyancy flux are singular; a finite physical source would require a regularized near-source region.

Take logarithmic height $\xi =\ln (z/z_{\ast })$ with an arbitrary fixed length $z_{\ast }$, so $d/d\xi =zd/dz$. This is dimensionally meaningful and has the same derivatives as the abbreviated $\ln z$. For example,The exponent vanishes and the coefficient is $q$, by the plume similarity in power-law stratification. Likewise the momentum coefficient is $m$ and the buoyancy coefficient is $f$. The logarithmic-height plume equations are thereforeThe reference similarity solution becomes the equilibrium $(1,1,1)$ of an autonomous differential equation. Normalizing each flux by its own height-dependent similarity value is essential; simply differentiating the unnormalized fluxes would miss the three subtraction terms.

Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains $-2m{M}^{\prime }$, since the reciprocal of $\hat{M}$ contributes another $-M^{\prime }$.

For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval $q,m>0$ and $f<0$, $P(0)<0$ while $P(\lambda )\to +\infty$ as $\lambda \to +\infty$. There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as $e^{\lambda \xi }\propto z^{\lambda }$; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.

At the upper endpoint,The limiting polynomial is $\lambda (\lambda +1)(\lambda +10/3)$, reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with $\lambda \sim -2f/3$. The two negative modes decay as $z^{-1}$ and $z^{-10/3}$; the remaining mode is neutral at the limiting linear system. Its eigenvector is $(1,2,3)$ in relative-flux coordinates.

The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with $Q\propto F^{1/3}{z}^{5/3}$ and $M\propto F^{2/3}{z}^{4/3}$. A small fractional change in $F$ produces changes in $Q,M$ in the ratio $1:2:3$. It is retained rather than damped out.

There is a qualification at the exact endpoint. For fixed nonzero $K$ and $\beta =-8/3$, taking $Q\propto z^{5/3}$ gives $F_z\propto -z^{-1}$, not zero. The flux coefficients above also diverge as $f\to 0$. Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.