Pass to $B={\overline{A}}^{\Vert \cdot \Vert }$. Closed invariant subspaces and intertwining maps are unchanged by this closure, and $BH$ is still dense. Because $B$ is a star-algebra, every closed invariant subspace reduces it.

Every nonzero nondegenerate compact-operator representation contains an irreducible reducing subspace. Indeed choose a nonzero positive compact element $b$; an isolated nonzero eigenvalue gives a nonzero finite-rank spectral projection $e\in B$ by continuous functional calculus, using a function vanishing at zero. The finite-dimensional corner $eBe$ has a minimal nonzero projection $p$, with $pBp=\mathbb{C}p$. Choose a unit vector $v\in pH$ and set $K=\overline{Bv}$. Then $pK=\mathbb{C}v$, since $pbv=pbpv$ is scalar times $v$. If $Q$ is the projection onto a reducing subspace of $K$, it commutes with $B$, and $Qv=pQv\in \mathbb{C}v$. Its projection property makes $Qv$ either zero or $v$. Cyclicity then makes $Q=0$ or $I_K$. Thus $K$ is irreducible.

Take a maximal orthogonal family of such irreducible reducing subspaces, by the axiom of choice. If their orthogonal complement were nonzero, its restricted representation would be nondegenerate: a vector annihilated by all restricted $B$ would, by taking adjoints, be orthogonal to $BH$ and hence zero. The image of $B$ on this complement is a norm-closed C-star algebra of compact operators, so the preceding construction gives another irreducible summand, contradicting maximality. The family therefore exhausts $H$.

Finally, suppose one irreducible equivalence class occurred infinitely often. Choose some $b\in B$ acting nontrivially on that class, and a unit vector $v$ with $\Vert bv\Vert >0$. Transport $v$ by unitary intertwiners into distinct equivalent summands. These vectors are orthonormal; their images under $b$ are orthogonal and all have the same nonzero norm. That contradicts compactness of $b$. The decomposition has finite multiplicity for every irreducible type, as claimed in nondegenerate representations by compact operators.

The Von Neumann double commutant theorem states that for a unital star-subalgebra $A\subseteq B(H)$,The commutant of an operator algebra consists of operators commuting with every element. Each commutation equation is closed in the weak operator topology, so $A^{\prime \prime }$ is weakly closed and contains both indicated closures.

Let $T\in A^{\prime \prime }$ and fix ${\xi }_1,\dots ,{\xi }_m\in H$. On $H^m$ let $A$ act diagonally and put $\xi =({\xi }_1,\dots ,{\xi }_m)$. The subspace $K=\overline{\{(a{\xi }_1,\dots ,a{\xi }_m):a\in A\}}$ reduces the diagonal star-algebra. Its orthogonal projection $P$ therefore commutes with every diagonal $a$. Writing $P$ as an $m\times m$ operator matrix shows that each entry belongs to $A^{\prime }$. Since $T\in A^{\prime \prime }$, the diagonal $T$ commutes with $P$. Unitality gives $\xi \in K$, so $P\xi =\xi$ and $P(T\xi )=T\xi$. Thus $T\xi \in K$. By definition of $K$, for any $\epsilon >0$ a single $a\in A$ satisfies ${\sum }_j\Vert (T-a){\xi }_j{\Vert }^2<{\epsilon }^2$. These are precisely the finite-vector neighbourhoods of the strong operator topology. Hence $T$ belongs to the strong closure, completing the proof. Unitality, or the corresponding nondegeneracy hypothesis, is essential to the theorem's stated form.

Apply part (a) to the compact positive operator $T^{\ast }T$. Its positive square root $B=|T|$ is defined by multiplying each eigenvector of eigenvalue $\lambda$ by $\sqrt{\lambda }$ and vanishing on the kernel. The eigenvalues tend to zero, so $B$ is compact. For every $x$, $\Vert Tx{\Vert }^2=⟨T^{\ast }Tx,x⟩=\Vert Bx{\Vert }^2$.

Consequently $U(Bx)=Tx$ is a well-defined isometry on $\mathrm{ran}B$. Extend it to the closure of that range and set it to zero on its orthogonal complement. This gives a partial isometry with initial space $(\ker T{)}^{\perp }$, final space $\overline{\mathrm{ran}T}$, andFor uniqueness, the positive factor in any such polar decomposition of a bounded operator satisfies $B^2=T^{\ast }T$. The positive-square-root uniqueness proved in 4(b) fixes $B$. The relation $U(Bx)=Tx$ then fixes $U$ on the dense subset $\mathrm{ran}B$ of its initial space and the kernel convention fixes it elsewhere. This proves uniqueness with the canonical initial-space condition. Without that condition the word “unique” would be false: $T=0$ can be written $U\cdot 0$ with many different unitary $U$.

The spectral theorem for compact self-adjoint operators says that a compact self-adjoint $A$ has an orthogonal decompositionwith real nonzero eigenvalues of finite multiplicity. Only finitely many eigenvalues, counting multiplicity, have absolute value at least any given $\epsilon >0$. Thus the nonzero eigenvalues form a finite list or a sequence tending to zero, and $A={\sum }_{\lambda \ne 0}\lambda P_{\lambda }$ in operator norm. The kernel may have arbitrary Hilbert dimension.

For existence of an extremal eigenvector, self-adjointness gives $\Vert A\Vert ={\sup }_{\Vert x\Vert =1}|⟨Ax,x⟩|$. One elementary proof of the nontrivial inequality uses polarization: if the quadratic form is bounded in absolute value by $m\Vert x{\Vert }^2$, its polarized real part is bounded by $m$ on two unit vectors; adjusting a phase bounds $|⟨Ax,y⟩|$ by $m$. Choose unit $x_j$ with quadratic values tending to $\lambda =\Vert A\Vert$ or $-\Vert A\Vert$. ThenIf $A\ne 0$, compactness supplies a convergent subsequence of $A{x}_j$, and the last estimate supplies a convergent unit subsequence of $x_j$, with eigenvalue $\lambda$.

Eigenvectors for distinct eigenvalues are orthogonal, since $A=A^{\ast }$. An infinite orthonormal sequence in one nonzero eigenspace would violate compactness. More generally an infinite orthonormal family with eigenvalues bounded away from zero has images separated in norm, again impossible. Take all nonzero eigenspaces and let $M$ be their closed span. Its complement reduces $A$; if the restriction there were nonzero, the extremal-eigenvector argument would produce another nonzero eigenvector, a contradiction. Thus $M^{\perp }=\ker A$. The finite truncations of the eigenvalue sum have remainder norm tending to zero, proving the asserted norm expansion.

On the Hardy space of the circle $H^2(\mathbb{T})\subset L^2(\mathbb{T})$, let $P_{+}$ be the projection onto nonnegative Fourier modes and $T_f=P_{+}{M}_f{|}_{H^2}$. The Toeplitz index theorem for continuous symbols states: for continuous $f$, $T_f$ is Fredholm exactly when $f$ never vanishes, and thenwith counterclockwise winding positive.

First, $T_f{T}_g-T_{fg}=-P_{+}{M}_f(I-P_{+})M_g{P}_{+}$ is compact. For Laurent-polynomial symbols the negative-mode crossing involves only finitely many modes and is finite rank. Uniform approximation and $\Vert T_f\Vert \le \Vert f{\Vert }_{\infty }$ extend this to continuous symbols. If $f$ is nonvanishing, $T_{1/f}$ is therefore a two-sided parametrix, and part (b) gives the Fredholm property.

Writing $n=\mathrm{wind}(f,0)$, a continuous argument along the parametrized circle increases by $2\pi n$. Removing this increase gives a periodic continuous logarithm $h$ of $z^{-n}f(z)$, so $f=z^n{e}^h$. The symbols $z^n{e}^{th}$ give a continuous Fredholm path from $z^n$ to $f$. Part (c) makes their indices equal. For $n\ge 0$, $T_{z^n}$ is the $n$th unilateral shift, with zero kernel and $n$-dimensional cokernel. For negative $n$ it is the corresponding backward shift, with kernel dimension $-n$ and zero cokernel. In both cases the index is $-n$.

For necessity, if $f(\zeta )=0$ use the normalized Hardy kernelsThey have norm one and converge weakly to zero, as their inner products with each Fourier basis vector tend to zero. Their squared moduli are Poisson kernels concentrating at $\zeta$, so continuity gives $\Vert f{k}_{r,\zeta }{\Vert }_2\to 0$ and hence $\Vert T_f{k}_{r,\zeta }{\Vert }_2\to 0$. A Fredholm parametrix would imply $k_{r,\zeta }=B{T}_f{k}_{r,\zeta }+K{k}_{r,\zeta }\to 0$ in norm, because a compact operator takes this weakly null bounded family to zero in norm. This contradicts unit norm, proving the converse.

Fix a Fredholm operator $T$. Split its domain as $X\oplus N=(\ker T{)}^{\perp }\oplus \ker T$ and its target as $Y\oplus M=\mathrm{ran}T\oplus \ker T^{\ast }$. For a sufficiently small norm perturbation $R$, writewhere $A:X\to Y$ remains invertible by the Neumann series. Invertible triangular row and column operations reduce this matrix toThe second block maps the finite-dimensional space $N$ into $M$. Hence the perturbed operator has closed range, finite kernel and cokernel, andsince the rank of that finite block cancels from the difference. Thus the Fredholm set is open and its index is locally constant, hence norm continuous.

If $S,T$ are Fredholm, their product is Fredholm by multiplying their invertible images in the Calkin algebra. For index additivity use the following finite-dimensional exact sequence:Here $q$ is reduction modulo $\mathrm{ran}T$, the induced $S$-map sends $[x]$ to $[Sx]$, and the last map is the quotient by the larger range $\mathrm{ran}S$. These maps are well-defined. For example, $\ker q=\ker S\cap \mathrm{ran}T$ is the image of the preceding map. If $Sx\in \mathrm{ran}ST$, subtracting a suitable $Ty$ makes $x-Ty\in \ker S$, proving exactness at $\mathrm{coker}T$. The other positions follow directly from the definitions. Alternating dimensions in this exact sequence gives

For a Fredholm operator $T$, define $B$ to invert $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ and to vanish on the range complement. The bounded inverse theorem makes $B$ bounded, andThe errors are finite rank, so $T$ is invertible in the Calkin algebra.

Conversely suppose $BT=I-K$ and $TB=I-L$ for bounded $B$ and compact $K,L$. On $\ker T$, $K$ is the identity. An infinite-dimensional closed kernel would contain an orthonormal sequence, whose images under $K$ have no convergent subsequence, contradicting compactness. Thus the kernel is finite-dimensional.

There is a positive lower bound for $T$ on $(\ker T{)}^{\perp }$. Otherwise choose unit vectors $x_j$ there with $T{x}_j\to 0$. Since $x_j=BT{x}_j+K{x}_j$, compactness gives a norm-convergent subsequence with unit limit $x$. The limit lies in the kernel complement and has $Tx=0$, a contradiction. This lower bound proves closed range: if $T{x}_j$ converges, remove each kernel component; the remaining $x_j$ form a Cauchy sequence and converge to a preimage of the limit.

Finally adjointing $TB=I-L$ gives $B^{\ast }{T}^{\ast }=I-L^{\ast }$. The same kernel argument makes $\ker T^{\ast }$ finite-dimensional. Therefore $T$ is Fredholm. $\boxed{T\text{ is Fredholm}⟺[T]\text{ is invertible modulo compacts}}$, the Atkinson theorem. The finite-dimensional case also satisfies the parametrix formulation directly, though its quotient algebra is the zero algebra.

A bounded operator $T:H\to H$ is a Fredholm operator when its range is closed and both $\ker T$ and $\mathrm{ran}(T{)}^{\perp }=\ker T^{\ast }$ have finite dimension. Its Fredholm index isFor index zero, choose an isomorphism $J:\ker T\to \ker T^{\ast }$ and extend it by zero on $(\ker T{)}^{\perp }$. The restricted map $T:(\ker T{)}^{\perp }\to \mathrm{ran}T$ is a bounded isomorphism, by the bounded inverse theorem. The finite-rank operator $J$ fills exactly the missing kernel-to-cokernel block. Thus $T+J$ is invertible and $T=(T+J)-J$ is an invertible operator plus a compact operator.

Conversely suppose $T=A+K$ with $A$ invertible and $K$ compact. The Atkinson theorem proved in part (b) makes every $A+tK$, $0\le t\le 1$, Fredholm: its quotient image is always $[A]$. The block argument in part (c) proves local constancy of the Fredholm index, so its index is constant along this interval and equals $\mathrm{ind}A=0$. This invokes the actual parametrix and block proofs below, not an unproved perturbation assertion. The claimed equivalence holds.

A hydraulic control of a zero-PV rotating channel is a critical section separating the upstream branch, where long-wave disturbances can propagate upstream, from the downstream branch, where they cannot. In the steady specific-energy description it is a stationary point of $E$ with respect to depth at fixed volume flux and local width. The control therefore fixes the discharge for a given upstream Bernoulli function and channel geometry; an arbitrary downstream condition cannot generally be imposed independently on the upstream reservoir.

Evaluate all quantities at $x=0$ and keep $Q,b$ fixed while differentiating. With $\Delta =fb$, the preceding branch givesAt the hydraulic control of a zero-PV rotating channel, this derivative vanishes. Using $h_1-h_2=\Delta v/g$ reduces it to $u_{1}v=g{h}_2$, where $v=u_1+\Delta /2$. Equivalently,Combining this with $Q=b{h}_{1}v-b\Delta v^2/(2g)$ cancels the quadratic term and leavesAt control $u_{1c}=\sqrt{g{h}_1+{\Delta }^2/16}-3\Delta /4$ and $h_{2c}=u_{1c}{v}_c/g$. Both walls must remain wet. Since $v_c>0$, the second depth is positive precisely when $u_{1c}>0$, givingEquality is the limiting case $u_{1c}=h_{2c}=0$, with the second wall just dry. For a wider channel, a dry lateral region requires a different solution and the two-wet-wall hydraulic control of a zero-PV rotating channel formula cannot be continued. The channel must also remain sufficiently long and slowly varying for the transverse geostrophic balance approximation to hold. As a check, $f\to 0$ gives $Q_c=b\sqrt{g}{h}_1^{3/2}$, the familiar rectangular-channel critical discharge.

Put $\Delta =fb$ and $v=u_1+\Delta /2$, the mean of the two wall velocities. The zero-potential-vorticity rotating channel flow has $h_2=h_1-\Delta v/g$. Substitution into the volume flux formula givesThe resulting quadratic equation has two roots, $v=g{h}_1/\Delta \pm \sqrt{g^2{h}_1^2/{\Delta }^2-2gQ/(b^{2}f)}$. The positive-depth condition $h_2>0$ selects the minus sign. ThusThe discarded sign would make $h_2<0$. On the retained branch $h_2=\sqrt{h_1^2-2fQ/g}$, which also shows the admissibility condition $h_1^2\ge 2fQ/g$.

The depth-based specific energy at the first wall isBecause ${\partial }_y(h+u^2/(2g))=0$, this same specific energy holds everywhere across the channel at fixed $x$. Its sum with the bed elevation is the Bernoulli function divided by $g$. Zero shallow-water potential vorticity makes that Bernoulli function constant throughout the connected steady channel flow, soConsequently the depth-based specific energy decreases as the bed rises and increases as it falls. It is constant along a level-bed channel, even if the width varies; that width variation instead changes the wall depths and velocities compatible with the conserved volume flux and Bernoulli function.

Material conservation of shallow-water potential vorticity carries $P=0$ from the reservoir into the channel. At leading order in a long, slowly varying channel, $v$ and $v_x$ are negligible, so $\zeta \simeq -u_y$. Zero absolute vorticity therefore implies $u_y=f$. Integration from $y=0$, followed by cross-channel geostrophic balance, gives the zero-potential-vorticity rotating channel flow:In particular, $u_2=u_1+fb$ and $h_2=h_1-(fb/g)(u_1+fb/2)$. The velocity increases towards the shallower wall. Differentiating the depth-based specific energy confirmsThis cross-channel constancy is also the $P=0$ consequence of $\nabla B=-P\nabla \psi$.

For a reservoir of finite depth, exact $P=0$ requires relative vorticity $\zeta =-f$, an anticyclonic rotation cancelling the planetary vorticity. A reservoir at rest in the rotating frame has $\zeta =0$ and $P=f/h$, so it does not have exactly zero shallow-water potential vorticity. A very deep, nearly quiescent reservoir can nevertheless supply the approximation $P\simeq 0$ because $f/h$ is small. The idealization must describe vanishing absolute vorticity or this deep-reservoir limit, rather than ordinary finite-depth rest in the rotating frame.

Take $y$ from the first wall to the second, with $f>0$. Negligible transverse velocity and cross-channel geostrophic balance give $fu=-g{\partial }_y(h+H)=-g{h}_y$, since the bed elevation $H(x)$ is independent of $y$. Therefore the volume flux isFor positive downstream volume flux, the layer is deeper at the first wall: $h_1>h_2$. This calculation does not require zero shallow-water potential vorticity.

Here $h_j$ must denote the wall depth above the local bed. If the surface heights are instead measured from a fixed vertical datum, write them as ${\eta }_j$ and use $h_j={\eta }_j-H(x)$, giving $Q=g[({\eta }_1-H{)}^2-({\eta }_2-H{)}^2]/(2f)$. The compact formula thus uses a bed-relative meaning of the wall heights.

Let $\mathbf{U}=(u,v)$, $\zeta =v_x-u_y$ and $D/Dt={\partial }_t+\mathbf{U}\cdot \nabla$. The inviscid shallow water equations areTaking the vertical curl of the momentum equation, with constant $f$, gives $D\zeta /Dt=-(f+\zeta )\nabla \cdot \mathbf{U}$. Combining this with the depth equation provesThus shallow-water potential vorticity is materially conserved: changes in absolute vorticity are exactly compensated by stretching or compression of the layer depth.

The relation to a Bernoulli function presupposes steady flow. In that case define $B=g(h+H)+\frac{1}{2}|\mathbf{U}{|}^2$. The vector identity $(\mathbf{U}\cdot \nabla )\mathbf{U}=\nabla (\frac{1}{2}|\mathbf{U}{|}^2)+\zeta \hat{\mathbf{z}}\times \mathbf{U}$ givesSteady volume conservation allows a transport stream function with $h\mathbf{U}=({\psi }_y,-{\psi }_x)$. Hence $\hat{\mathbf{z}}\times (h\mathbf{U})=\nabla \psi$, soIn particular, $\mathbf{U}\cdot \nabla B=0$: the Bernoulli function is constant along each streamline. Where the flow permits a single-valued dependence on the stream function, the displayed derivative describes how that constant changes between neighbouring streamlines. For an unsteady flow, shallow-water potential vorticity remains conserved but the stated steady Bernoulli function relation does not generally hold.

In the Boussinesq approximation, put $\gamma =K{W}^2\sqrt{g/{\rho }_0}$. The buoyancy-driven turbulent density diffusion equation gives, wherever $q={\bar{\rho }}_z>0$,The coefficient outside the bracket is positive, so the local density gradient increases precisely whenAt a point where $q=0$, use the original degenerate diffusion equation rather than dividing by $\sqrt{q}$.

For the proposed self-similar solution, $\bar{\rho }={\rho }_0[1+F(t)G(z)]$ and $F(0)=1$. Substitution into ${\bar{\rho }}_t={\partial }_z{F}_{\rho }$ givesAn odd power profile $G(z)=C\mathrm{sgn}(z)|z{|}^p$ has right-hand spatial power $(3p-5)/2$. Matching it to $p$ requires $p=5$. Thus choose $G=C{z}^5$ with $C>0$, an odd, increasing profile. Since $G^{\prime }=5C{z}^4$, its flux is proportional to $z^6$ on both sides of the midplane. The resulting quintic density-profile similarity in an unstable tube satisfiesIntegration gives $F^{-1/2}=1-At/2$. ConsequentlyAlthough mixing transports density downwards, the local density gradient can grow because spatial variations of the flux redistribute density. In this example the downward flux increases towards each end, driving the formal amplification of the odd profile.

This self-similar solution deliberately ignores the end conditions: its nonzero end flux cannot satisfy impermeable ends. Moreover, the Boussinesq approximation requires $FC|z{|}^5\ll 1$, which fails before the formal divergence. A closed tube has only finite available potential energy, so it cannot sustain indefinitely growing turbulent kinetic energy or the assumed instantaneous flux law. End effects, changing eddy statistics and decay of the unstable stratification must eventually replace this local model. The conflicting endpoint labels in the source do not enter the calculation because end conditions have been expressly excluded.

Write $q={\partial }_z\bar{\rho }>0$ for the unstable, horizontally averaged density gradient. A turbulent displacement over a mixing length of order $W$ encounters a density contrast $\Delta \rho \sim Wq$. Balancing its buoyancy acceleration with the inertial acceleration givesA turbulent diffusivity for density is therefore $D_T=K{u}_{\ast }W=K{W}^2\sqrt{gq/\bar{\rho }}$, absorbing dimensionless factors into $K>0$. The upward density flux is $J_{\rho }=⟨w^{\prime }{\rho }^{\prime }⟩=-D_{T}q$: turbulent exchange carries denser water downwards and lighter water upwards. Defining $F_{\rho }=-J_{\rho }$ as the positive downward flux yields the buoyancy-driven turbulent density diffusion lawThis is a flux per unit horizontal area, with units of mass per area per time. The total density transport across a horizontal section is $W^2{F}_{\rho }$.

The assumptions are established high-Reynolds number turbulence, width-limited eddies, an instantaneous local balance between buoyancy and inertia, approximately constant $K$, negligible molecular diffusion, and zero mean vertical volume flux. The last condition follows from incompressibility and impermeable sidewalls in a closed tube. Density variations and the eddy statistics must also vary slowly enough for a local mixing-length closure to be meaningful.

Horizontal averaging of mass conservation, with upward $z$, gives ${\bar{\rho }}_t=-{\partial }_z{J}_{\rho }={\partial }_z{F}_{\rho }$. Differentiating once more supplies the requested evolution of the density gradient:Under the Boussinesq approximation, replace $\bar{\rho }$ in the inertial prefactor by the constant reference density ${\rho }_0$; retain the changing density gradient inside the flux.

Let $E(k)$ be the one-dimensional turbulent energy spectrum, normalized so that $\frac{1}{2}⟨|{\mathbf{u}}^{\prime }{|}^2⟩={\int }_0^{\infty }E(k)dk$. Large energy-containing eddies receive energy from the forcing. In the inertial range, nonlinear transfer passes this energy towards smaller scales at a mean rate $ϵ$ per unit mass. Dimensional analysis of Kolmogorov 1941 theory givesThe dissipation range begins near the inverse of the Kolmogorov length scale, where kinematic viscosity removes the transferred kinetic energy. The large-scale spectrum depends on the forcing; the rising low-$k$ branch in this sketch is illustrative, rather than a universal power law.

The tube has neither homogeneous nor isotropic large-scale turbulence: gravity selects a vertical direction, its walls constrain the motion, and its density gradient varies with height. The width $W$ bounds the transverse size of the energy-containing eddies. At sufficiently high Reynolds number, much smaller eddies may become approximately locally isotropic and exhibit an inertial range, even though the full tube flow does not satisfy the homogeneous-isotropic assumptions.

To obtain the Reynolds stress, write $u_i=U_i+u_i^{\prime }$ and $p=P+p^{\prime }$, with $⟨u_i^{\prime }⟩=0$. Averaging the incompressible Navier-Stokes equations uses $⟨u_i{u}_j⟩=U_i{U}_j+⟨u_i^{\prime }{u}_j^{\prime }⟩$ and gives the Reynolds-averaged momentum equationHere the final term is the mean buoyancy force, with the hydrostatic reference removed. Thus averaging the quadratic momentum flux introduces the covariance $⟨u_i^{\prime }{u}_j^{\prime }⟩$; multiplied by ${\rho }_0$, it has the dimensions of a stress. Its divergence transfers momentum between the mean flow and the fluctuations.

An eddy viscosity model represents this Reynolds stress byThe isotropic term can be absorbed into the mean pressure, while the deviatoric term gives a diffusion of mean momentum with turbulent viscosity ${\nu }_T\sim u_{\ast }\ell$, where $u_{\ast }$ and $\ell$ are representative turbulent speed and mixing length. Unlike kinematic viscosity, this coefficient depends on the flow. The closure assumes a local relation between stress and mean strain; anisotropic, buoyancy-driven large eddies may require a more detailed Reynolds stress model.

For a very deep release, $z/H_p\ll 1$ over the rise, so bubble expansion is negligible. The neutral-height equation reduces toThusIf the stopping thermal is also much larger than its initial size, the leading form isThe first expression does not require this extra entrainment-dominated limit.

Near the neutral height the quasi-steady speed tends to zero. Momentum can produce some overshoot and subsequent adjustment or oscillation. The water spreads horizontally as a buoyant intrusion near its equilibrium density level. Bubble slip then becomes important relative to the slowing liquid, so bubbles can separate and continue rising toward the surface, reducing the residual liquid's buoyancy. Loss of bubbles and further mixing alter the final intrusion level; the retained-bubble model no longer applies indefinitely after stopping.

For constant buoyancy frequency in the Boussinesq approximation, the ambient density isConservation of liquid mass during fluid entrainment gives $d(V\rho )/dz={\rho }_a(z)V^{\prime }(z)$, with $\rho (0)={\rho }_0$. Integrating by parts,Since $a=a_0+\alpha z$ and $V\propto a^3$,Hence the water density within the thermal isThe entrained water is denser than its new surroundings because it retains a mixture of fluid drawn from lower levels.

The bulk density is $\rho (1-\varphi )$. Dropping products of the small density contrast with $\varphi$, its net reduced gravity isThe first term is the expanding-bubble contribution and the second is the restoring effect of stratification. In the quasi-steady Froude model, ascent stops when $g^{\prime }=0$. The neutral height of a bubbly thermal in a stratified fluid therefore satisfiesThis is a fifth-degree polynomial, with leading coefficient $-{\alpha }^4/H_p$ at finite release pressure. The relevant height is the first positive zero reached while the thermal is still below the surface, dilute and within the Boussinesq range. Depending on parameters there may be no such zero before the surface, or more than one mathematical zero; a later root beyond a first stopping point is not another continuous ascent solution. The model's stopping height is a neutral-buoyancy estimate, not a claim that inertial overshoot is absent.

For a high-Reynolds number thermal, buoyancy force scales as ${\rho }_0{g}^{\prime }{a}^3$ and inertial resistance scales as ${\rho }_0{W}^2{a}^2$. Their balance gives $W^2\propto g^{\prime }a$. It is therefore natural to write a diameter-based Froude number closureThe order-one coefficient $F$ accounts for the thermal's shape, entrainment and flow structure. The main resistance is inertial pressure or form drag and the momentum needed to entrain and accelerate surrounding water, rather than molecular skin friction. Added mass matters when the thermal accelerates, so the Froude closure is a quasi-steady approximation.

In a homogeneous ocean all of the leading buoyancy comes from the bubbles. Parts (a) and (b) giveSubstituting into the speed law yieldsFor $z\gg a_0/\alpha$, this becomesThe square root is essential both physically and dimensionally. This approximation still requires dilute bubbles and retained gas; it is not a valid prediction of infinite speed where the pressure approximation approaches zero.

The Boussinesq approximation retains small density differences only in the gravitational buoyancy force and uses a constant reference density in inertia and volume conservation. The entrainment coefficient $\alpha$ specifies the inward velocity of ambient fluid relative to the thermal boundary, $u_e=\alpha W$. Turbulent interface motions engulf ambient water and mix it into the coherent rising volume.

For a spherical buoyant thermal, $V=4\pi a^3/3$. To leading order in the dilute gas fraction, expansion of the gas does not dominate the total-volume balance, so fluid entrainment givesBut $dV/dt=4\pi a^{2}da/dt$ and $dz/dt=W$. Therefore, wherever the thermal is rising,This growth law for a dilute bubbly thermal is independent of stratification under the constant-$\alpha$ model. Stratification changes the speed and eventual stopping height; it does not enter this volume-per-distance relation. In real flows the entrainment coefficient itself can vary with stratification and flow structure.