There are two necessary qualifications to the printed formula. Critical points satisfy $g^{\prime }=0$, not $g^{\prime }=1$. At a point with $g^{\prime }=1$, the local degree is one and the printed summand is zero. In addition, $\log \Vert g^{\prime }(0)\Vert$ is finite only when $g^{\prime }(0)\ne 0$; constant maps and origin-critical maps need separate treatment.

These are genuine issues even for otherwise ordinary disc maps. The map $g(z)=(z-1/2{)}^2/4$ maps the unit disc into itself and has $g^{\prime }(0)\ne 0$. Its derivative never equals one in the disc, so the printed count is zero, but it has a ramification point at $1/2$ of multiplicity one. For $R>1/2$ the correct identity has the additional term $\log (2R)$, proving that the displayed identity with the literal printed count is false. Also $g(z)=z^2$ shows the undefined basepoint logarithm in an origin-critical example.

Prove the intended identity first for a nonconstant map with $g^{\prime }(0)\ne 0$. Let $H(z)=2|g^{\prime }(z)|/(1-|g(z){|}^2)$ be its hyperbolic derivative density and $u=\log H$. Away from derivative zeros, $\log |g^{\prime }|$ is harmonic, and direct differentiation givesAt a zero $a$ of $g^{\prime }$ of multiplicity $m_a$, factor $g^{\prime }=(z-a{)}^{m_a}h$ with $h(a)\ne 0$. Its logarithm contains $m_a\log |z-a|$, whose distributional Laplacian is $2\pi m_a{\delta }_a$. ThusThe local map degree is $m_a+1$, so this is the required excess-degree weighting.

For a smooth function, differentiating its circular mean and applying the divergence theorem gives ${\overline{u}}^{\prime }(r)=(2\pi r{)}^{-1}{\int }_{|z|<r}\Delta udA$. Integrating in $r$ gives the logarithmic Green-Jensen formula. Factoring the isolated logarithmic singularities extends it to the present $u$:Substitution yields the critical-point Jensen identity for the hyperbolic derivativewhere $N_{\mathrm{crit}}={\sum }_{|a|<R:g^{\prime }(a)=0}{m}_a\log (R/|a|)$. This is the requested formula with the critical-point typo corrected. Initially choose a circle containing no ramification point on its boundary; continuity and integrability of logarithmic boundary singularities extend the result to the remaining radii.

By Schwarz-Pick theorem, $H(z)\le 2/(1-|z{|}^2)$. ThereforeThe normalized hyperbolic distance is $\rho (R)=\log ((1+R)/(1-R))$, andConsequentlyOne can take $C_1=1$ and $C_2=\log (2/H(0))\ge 0$. Since the corrected count is nonnegative in the noncritical-basepoint case, this also proves the growth bound for $T_{\mathbb{D}}$ with the literal printed, identically zero count, though not its false identity.

For completeness, if $g^{\prime }$ vanishes to order $m\ge 1$ at zero, put $c={\lim }_{z\to 0}H(z)/|z{|}^m>0$ andApply the same formula to $\log H-m\log |z|$, which is regular at zero. It givesThe origin contribution is a regularized Jensen term, not the undefined $\log (R/0)$. If one instead omits the origin term, the upper bound gains $-m\log R$; this is bounded for $R\ge 1/2$, while the characteristic and off-origin count are bounded on any smaller compact disc. Thus a bound $C_1\rho +C_2$ still holds for that nonnegative-count convention after enlarging $C_2$. For a constant map $H=0$ and $T_{\mathbb{D}}=0$, so its characteristic growth bound is trivial; the logarithmic identity and a finite critical-divisor count are not defined. These qualifications resolve all maps allowed by the original opening sentence without asserting an undefined equality.

If $\varphi$ is identically zero there are no exceptions. Otherwise choose $x_0$ with $\varphi (x_0)>0$ and put $I_{+}=\{x\ge x_0:{\varphi }^{\prime }(x)>\varphi (x{)}^2\}$. Monotonicity implies $\varphi (x)>0$ thereafter. On this set, $1<{\varphi }^{\prime }/{\varphi }^2$, whenceAdding the finite initial interval $[0,x_0]$ provesThis derivative growth outside a finite-measure set controls exceptional radii in Nevanlinna estimates. For example let $T_s$ be the smooth spherical characteristic and $A(r)=r{T}_s^{\prime }(r)$ its increasing area-counting function. Apply the lemma to $T_s+1$ and $A+1$: outside the union of two finite-measure sets,Their logarithms are consequently $O(\log r+\log (T_s+1))$. Such derivative/area bounds in estimates from the Poisson-Jensen formula give the Nevanlinna logarithmic derivative lemma, $m(r,f^{\prime }/f)=O({\log }^{+}{T}_f+\log r)$ outside a finite-length set. The second main theorem uses that estimate as its error term. The elementary growth lemma controls that error and its exceptional set; it is not itself a substitute for the analytic estimates.

We use the truncated second main theorem stated in the preceding solution, together with the fact that a transcendental meromorphic function has $T_f(r)/\log r\to \infty$. Here is a justification of the latter fact. If there are infinitely many poles, fixing any arbitrarily large finite pole count gives $N(r;\infty )\ge A\log r-O_A(1)$. If there are only finitely many poles, multiply by a polynomial $p$ removing them to obtain a transcendental entire function $h=pf$. Its Taylor series has nonzero coefficients of arbitrarily high degrees $n$. Cauchy's coefficient estimate gives $\log M_h(r/2)\ge n\log r-O_n(1)$; the subharmonic function Poisson estimate gives $\log M_h(r/2)\le 3m(r,h)$. Since $T_h\le T_f+O(\log r)$, letting $n$ be arbitrarily large proves the claimed ratio limit. Thus the second-theorem error is $o(T_f)$ along large nonexceptional radii.

Let $N_{\mathrm{simple}}(r;a)$ count just the simple $a$-points. A multiple $a$-point of degree $m\ge 2$ contributes one to the truncated count and at least two to the full count; a simple point contributes one to each. Hence, with the same positive logarithmic weights for large radii,If each of the five values had only finitely many simple preimages, each simple counting function would be $O(\log r)$. Summing and using the first main theorem would give ${\sum }_{j=1}^5\overline{N}(r;a_j)\le (5/2)T_f+O(\log r)$. But the second theorem requires $3T_f\le \sum \overline{N}+o(T_f)$, a contradiction. This proves the stronger result five values force infinitely many simple preimages:In particular more than one simple zero is necessary, but more than one target with simple zeros is not necessary.

For the last distinction, take a nonsingular Weierstrass elliptic function scaled so that ${\wp }^{\prime 2}=4\wp (\wp -1)(\wp +1)$; it has branch values $-1,0,1,\infty$. These are the four totally ramified Weierstrass values: finite branch preimages have local degree two, and all poles are double. SetThe first four targets are the images of those branch values under the Möbius transformation. Their preimages remain double; at a lattice pole, for example, $f(z)\sim (z-z_{\ast }{)}^2$. The last target corresponds to $\wp =3$, where ${\wp }^{\prime 2}=96\ne 0$, so every such preimage is simple. The degree-two torus map attains that regular value, and periodicity produces infinitely many preimages in the plane. This nonconstant elliptic function is transcendental because it has infinitely many poles. Exactly one of the five selected finite targets therefore has simple preimages.

The growth order of a meromorphic function isThe order of the integrated count is defined analogously using ${\log }^{+}N(R;a)$, with an identically zero count assigned order zero. Constants have function order zero. These are growth orders, not local zero or pole orders. For nonconstant $f$, the first main theorem gives $\rho (N(\cdot ;a))\le \rho (f)$ for every target.

Use the truncated Nevanlinna second main theorem in the following standard form: for distinct fixed targets $a_1,\dots ,a_q$,outside an exceptional set of radii of finite linear measure. For finite order the error is $O(\log R)$; replacing each truncated count by its full count preserves the inequality.

Suppose $0<\rho (f)<\infty$ and three targets have counting orders below $\rho (f)$. Choose $\mu$ strictly between the maximum of their three orders and $\rho (f)$. Each full count is at most $R^{\mu }$ eventually. The theorem with $q=3$ then gives $T_f(R)\le C{R}^{\mu }$ for sufficiently large radii outside the exceptional set. Its tail measure is eventually less than one, so every interval $[R,R+1]$ contains a nonexceptional radius $R^{\prime }$. Monotonicity of the characteristic yields $T_f(R)\le T_f(R^{\prime })\le C(R+1{)}^{\mu }$ for all large $R$. This contradicts its order. If $\rho (f)=0$, all the counting orders are already zero. Hence the counting-order exceptions for a finite-order meromorphic function satisfyThe exponential has order one but omits both zero and infinity, so those two counts are zero; this realizes two exceptions.

For a Möbius map $U(w)=(aw+b)/(cw+d)$ with $ad-bc\ne 0$, translation and multiplication by a nonzero constant change $T$ by $O(1)$, as follows directly from the proximity inequalities and unchanged pole multiplicities. The first main theorem also gives $T(1/h)=T(h)+O(1)$. If $c=0$, $U$ is affine. If $c\ne 0$, write $U(w)=a/c+K/(cw+d)$ with $K\ne 0$ and use these three elementary operations. Thus $T_{U\circ f}=T_f+O(1)$, proving

For the sum, a pole of $f_1+f_2$ has order no larger than the sum of the two pole orders, andTherefore $T_{f_1+f_2}\le T_{f_1}+T_{f_2}+O(1)$ andIf $\rho (f_1)>\rho (f_2)$, apply the same inequality to $f_1=(f_1+f_2)-f_2$. It forces $\rho (f_1)\le \max (\rho (f_1+f_2),\rho (f_2))$, hence the sum's order equals $\rho (f_1)$. The reversed case is identical. This reasoning also handles an infinite larger order. For equal orders cancellation can be strict: $f_1=e^z$, $f_2=-e^z$ both have order one, but their sum is zero of order zero.

For a nonconstant meromorphic map, write $T_f(R)=m(R,f)+N(R;\infty )$ in the usual normalization, where $m=(2\pi {)}^{-1}\int {\log }^{+}|f(R{e}^{i\theta })|d\theta$. Let $N(R;a)$ count $a$-points with their local degrees, and let $\overline{N}(R;a)$ count each distinct point once, using the same integrated logarithmic weights. At infinity the local degree is the pole order. The Nevanlinna deficiency and Nevanlinna ramification index areThe latter is an asymptotic excess-multiplicity index, not the integer local ramification degree. The Nevanlinna first main theorem gives $N(R;a)\le T_f(R)+O(1)$ and identifies deficiency with the lower limiting normalized proximity. The target $a$ ranges over the Riemann sphere, not just the real line.

The exponential is entire and has no poles. ConsequentlyThus $c=1/\pi$ in this normalization.

To obtain the rational-composition growth, represent the nonconstant degree-$d$ sphere map by relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit sphere in ${\mathbb{C}}^2$ has a positive minimum and finite maximum: simultaneous vanishing would contradict relative primeness on the projective line. Homogeneity therefore gives constants $0<A\le B<\infty$ withApply this to the entire nonvanishing lift $(e^z,1)$. The logarithmic joint norm differs from $d\log \Vert (e^z,1)\Vert$ by a bounded function. Its circular mean minus the value at zero is the spherical characteristic. It differs from the ordinary characteristic by $O(1)$: Jensen's formula applied to the entire denominator counts its zeros, whilecontrols the proximity replacement. If the denominator vanishes at zero, factor its zero and use its leading coefficient in Jensen's formula; the discrepancy is still a basepoint-dependent constant. This proves the rational composition law for the Nevanlinna characteristic here and givesIf $U$ is constant, $d=0$ and its characteristic is bounded instead.

For a simultaneous nonzero deficiency and ramification example choosea degree-three polynomial composed with the exponential. Its characteristic is $3R/\pi +O(1)$. Its only zeros are $2\pi in$, $n\in \mathbb{Z}$, all of multiplicity two. The number of distinct such points in a large disc is $R/\pi +O(1)$, so integration of their count givesThe origin contribution is handled by its usual logarithmic counting term. ThereforeThis is simultaneous deficiency and ramification for an exponential polynomial: the missing exponential value contributes a deficit, while the attained value is attained with double multiplicity.

Use the curvature-minus-one normalization of the hyperbolic metric:The associated distance is $d(z,w)=2\mathrm{artanh}|(z-w)/(1-\overline{w}z)|$. A connected Riemann surface has its complete canonical Poincare metric on a Riemann surface precisely when its universal cover is conformally the unit disc. By the uniformization theorem, the other simply connected covering types are the complex plane and Riemann sphere. Disc automorphisms preserve $\lambda (z)|dz|$, as direct differentiation of $(z-a)/(1-\overline{a}z)$ shows; therefore the metric descends through the deck group. This is the uniformization notion of hyperbolicity, not the different potential-theoretic Green-function classification.

First prove the needed Schwarz lemma. If $F:\mathbb{D}\to \mathbb{D}$ is holomorphic and $F(0)=0$, then $F(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle and $r\uparrow 1$ give $|F(z)|\le |z|$ and $|F^{\prime }(0)|\le 1$.

Write ${\psi }_a(z)=(z-a)/(1-\overline{a}z)$. Conjugating $f$ by the source and target disc automorphisms, ${\psi }_{f(a)}\circ f\circ {\psi }_a^{-1}$ fixes zero. Applying the derivative bound givesIntegrate along any piecewise smooth path and then take the infimum of lengths: $d(f(z),f(w))\le d(z,w)$. This proves the Schwarz-Pick theorem. Contraction means nonexpansion, not a uniform Lipschitz constant strictly smaller than one; automorphisms preserve all distances.

If both $R$ and $S$ have disc universal covers, lift $g\circ {\pi }_R$ to $\tilde{g}:\mathbb{D}\to \mathbb{D}$ through ${\pi }_S$. The disc is simply connected, so the lift exists, and it is holomorphic because the covering maps are local biholomorphisms. Both covering maps are local hyperbolic isometries. The disc contraction therefore descends to $g^{\ast }d{s}_S\le d{s}_R$ and, by the path-length argument, $d_S(g(p),g(q))\le d_R(p,q)$. If only $S$ is known to have a hyperbolic metric, a nonconstant $g$ forces $R$ to have one too: otherwise lift from the plane or sphere universal cover of $R$ to the disc cover of $S$. A bounded entire function is constant by the Liouville theorem, and a holomorphic map from the compact sphere into the disc is constant by the maximum modulus principle. Thus a nonconstant map into a hyperbolic surface is distance-nonincreasing between the canonical hyperbolic metrics; a nonhyperbolic source admits only constant such maps.

For the prescribed point, Schwarz-Pick theorem gives $|{\psi }_{w_0}(f(z))|\le |{\psi }_{z_0}(z)|$. Thereforeis holomorphic after filling the removable singularity at $z_0$, and $|h|\le 1$. Inverting the target automorphism provesConversely every holomorphic $h$ with this bound gives an admissible disc map: $|{\psi }_{z_0}(z)h(z)|<1$, its inverse target automorphism remains in the disc, and the denominator cannot vanish.

Fix $z$ and put $r=|{\psi }_{z_0}(z)|<1$. The possible intermediate values fill $|v|\le r$, since constant choices $h\equiv c$, $|c|\le 1$, attain them all. Thus the one-point value region of a holomorphic disc map is $|{\psi }_{w_0}(w)|\le r$. Completing the square in $|w-w_0{|}^2\le r^2|1-{\overline{w}}_{0}w{|}^2$ gives its Euclidean centre and radius:This closed disc lies strictly inside the unit disc because $r<1$. For $z=z_0$ it degenerates to the single point $w_0$. Its boundary is attainable by unimodular constant $h$; requiring $f$ to map into the open disc does not remove that boundary.

Let $f$ be holomorphic on a neighborhood of the closure of a bounded region with piecewise smooth positively oriented boundary, with no boundary zeros and with $f$ not identically zero. The argument principle stateswhere $m_a$ is the zero multiplicity. More generally a closed contour avoiding the zeros gives ${\sum }_a{m}_a\mathrm{Ind}(\Gamma ,a)$, and the integral is the winding number of $f\circ \Gamma$ about zero.

To prove it, factor near a zero $a$ as $f(z)=(z-a{)}^{m_a}h(z)$ with $h(a)\ne 0$. Then $f^{\prime }/f=m_a/(z-a)+h^{\prime }/h$. Remove small disjoint circles about the finitely many zeros in the region. On the remainder $f^{\prime }/f$ is holomorphic, so the Cauchy integral theorem reduces the boundary integral to the sum of the small-circle integrals. Each is $2\pi i{m}_a$ because $h^{\prime }/h$ is holomorphic on its small disc. This proves the formula. Along a contour, the imaginary part of $d\log f$ is the change in a continuous argument, giving the winding interpretation and the indexed version.

Now choose a closed disc $\overline{D(z_0,r)}\subset \Omega$ on which $z_0$ is the only zero of the nonconstant limit $f$. Its boundary has $\min |f|>0$. Local uniform convergence gives $\max |f_n-f|<\min |f|$ there for all sufficiently large $n$. The contour functions $f+t(f_n-f)$, $0\le t\le 1$, have no boundary zero. Their winding numbers are constant under this homotopy, so the argument principle shows that $f_n$ has exactly the same positive number $m$ of zeros, counted with multiplicity, inside the disc. This is the Rouche theorem argument, here derived from the contour count.

Choose a zero $z_n$ in that disc. For every $0<ϵ<r$, the compact annulus $ϵ\le |z-z_0|\le r$ contains no zero of $f$, so uniform convergence makes $f_n$ nonzero on it eventually. Therefore all of the chosen zeros eventually lie within $ϵ$ of $z_0$. ConsequentlyThe maps $f_n$ are eventually not identically zero, since they converge at any fixed point where $f$ is nonzero; thus their zero counts above are legitimate.

A double zero need not persist as a double zero. On the unit disc take $f(z)=z^2$, $f_n(z)=z^2+1/n$. Its zeros $\pm i/\sqrt{n}$ converge to zero but are simple, while $f_n^{\prime }$ vanishes only at zero and $f_n(0)=1/n\ne 0$. Hence the additional simultaneous derivative condition cannot always be imposed. Local uniform convergence preserves total nearby multiplicity, not its concentration at a single point.

Take rings and modules to be unital. Let $J_{\ell }$ denote the intersection of the maximal left ideals. For each nonzero vector $v$ of a simple module $M$ on the left, the map $R\to M$, $r\mapsto rv$, is onto and its kernel is a maximal left ideal. Consequently $J_{\ell }$ annihilates every simple module on the left. Conversely, annihilating $R/L$ forces an element to belong to the maximal left ideal $L$. Thuswhich shows that $J_{\ell }$ is a two-sided ideal.

For $j\in J_{\ell }$ and $r\in R$, the left ideal $R(1-rj)$ cannot be proper: a containing maximal left ideal would also contain $rj$ and hence $1$. Therefore $u(1-rj)=1$ for some $u\in R$. Here $u=1+urj$ also differs from $1$ by an element of $J_{\ell }$, so it too has a left inverse $v$. Multiplying the first equality by $v$ gives $1-rj=v$; hence $(1-rj)u=1$. We have proved that $1-rj$ is a unit, not just left invertible.

The identityis verified by multiplying on either side, using $(1-ab{)}^{-1}(1-ab)=(1-ab)(1-ab{)}^{-1}=1$. It implies that $1-jr$ is a unit as well. If $j$ were outside a maximal right ideal $I$, then $I+jR=R$, so $1-jr\in I$ for some $r$. A proper right ideal cannot contain a unit, giving a contradiction. Thus $J_{\ell }$ belongs to every maximal right ideal. Applying the same argument to the opposite ring proves the reverse inclusion. This proves the Jacobson radical is independent of handedness:For the zero ring the intersections are interpreted as the whole ring, so the equality remains valid.

If a two-sided ideal $N$ satisfies $N^m=0$, then for $x\in N$ and $r\in R$, $(rx{)}^m=0$. The finite geometric series ${\sum }_{i=0}^{m-1}(rx{)}^i$ inverts $1-rx$. The unit criterion for the Jacobson radical, also proved above by the maximal-ideal argument, gives $N\subseteq J(R)$. This assertion concerns a nilpotent ideal; arbitrary nilpotent elements of a noncommutative ring need not lie in its Jacobson radical.

For the formal power series ring over the p-adic integers, a series $f={\sum }_{i\ge 0}{a}_i{t}^i$ is a unit exactly when $a_0$ is a unit of ${\mathbb{Z}}_p$. Necessity follows by comparing constant coefficients. If $a_0$ is invertible, construct the inverse recursively by $b_0=a_0^{-1}$ and $b_m=-a_0^{-1}{\sum }_{i=1}^m{a}_i{b}_{m-i}$. Therefore the nonunits are exactly the series whose constant coefficient lies in $p{\mathbb{Z}}_p$. They form the unique maximal ideal $(p,t)$, and the quotient is ${\mathbb{F}}_p$. Hence

Use the steady slender-plume model implicit in the requested height-only similarity. Put $\theta =T-T_0$ and introduce permeability $K$, dynamic viscosity $\mu ={\rho }_0\nu$, fluid heat capacity per volume $C_h={\rho }_0{c}_p$, and effective transverse thermal diffusivity $\kappa$. These material constants are needed to give dimensional fluxes. Take $u,w$ to be Darcy velocities, so the thermal transport coefficient for the steady equation is $\kappa =k_e/C_h$ when $k_e$ is the effective conductivity of the saturated medium.

The Darcy law with gravity isThe ambient is hydrostatic with $p_{a,z}=-{\rho }_{0}g$. A slender plume's pressure correction has vertical gradient smaller than the buoyancy term by the squared width-to-height ratio. Indeed, incompressibility gives $u/w$ of order width/height, and the horizontal Darcy relation makes the pressure correction of order $\mu w{\text{width}}^2/(K\text{height})$. Its vertical derivative has the additional inverse-height factor. Thus the leading vertical Darcy balance isThe proportionality is to temperature excess $T-T_0$, not absolute $T$; the velocity vanishes in the unheated ambient. This is the local buoyancy/drag balance of a porous thermal plume, rather than a free-fluid inertial momentum balance.

The governing continuity equation and steady advection-diffusion equation, with axial conduction neglected in the slender high-Rayleigh number limit, areChoose the streamfunction convention $u=-{\psi }_z$, $w={\psi }_x$. It enforces continuity identically and givesA change to the opposite streamfunction convention changes the sign of $\psi$, not the physical velocity or temperature.

The upward convected excess heat flux per unit source length isThis is the quantity denoted $F$ in this question; it differs from Question 2's buoyancy-flux symbol. In conservative form the heat equation is ${\partial }_x(u\theta )+{\partial }_z(w\theta )=\kappa {\theta }_{xx}$. Integrate across the plume:The temperature anomaly and its derivative vanish far from the plume; lateral entrainment brings in ambient-temperature fluid and thus no excess heat. There is no heat source or sink above the line source. Transverse conduction redistributes this heat without removing its total, while axial conduction is negligible in the stated approximation. In an unsteady plume a heat-storage term would have to be retained, so the convective flux alone would not in general be independent of height.

Let $\mathcal{J}=\beta F_h/C_h=\int w^{2}dx>0$. It is constant because $w=\beta \theta$. To obtain similarity exponents without guessing a profile, write the width as $\delta \propto z^a$ and central velocity as $W\propto z^b$. The fixed heat flux requires $W^2\delta$ constant, hence $2b+a=0$. Balancing vertical advection, including its continuity-required horizontal component, against transverse conduction gives $W{\delta }^2\propto \kappa z$, hence $b+2a=1$. Thus $a=2/3$ and $b=-1/3$, and the streamfunction scale $W\delta$ is proportional to $z^{1/3}$. A suitable similarity form isThe volume flux per unit span is $Q=\psi (+\infty )-\psi (-\infty )$ and therefore grows as $z^{1/3}$; it tends to zero at the ideal source. Thus the zero-source-volume requirement fixes the absence of an additive source-length scale, not a finite temperature at the singular point.

Substitution into the streamfunction equation givesChoose the scale normalization $AB=6\kappa$ and $f(+\infty )=1$. Reflection symmetry makes $f$ odd, so the relevant conditions are $f(0)=0$, $f^{\prime \prime }(0)=0$, $f^{\prime }(\infty )=0$, with finite $f(\infty )$; the heat input fixes the remaining dimensional scale. The nonlinear similarity equation becomesIntegrate once. Since $f^{\prime },f^{\prime \prime }\to 0$ in the ambient, the integration constant is zero, giving $f^{\prime \prime }+2f{f}^{\prime }=0$. A second integration gives $f^{\prime }+f^2=C$; the normalized far limit fixes $C=1$. Separation with $f(0)=0$ now yields the complete positive-plume profileIt satisfies the symmetry and decay conditions; the solution of the first-order initial-value equation is unique. The physical heat strength is determined using ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^4\eta d\eta =4/3$, obtained by substituting $t=\tanh \eta$:Equivalently, defineThe hyperbolic-secant porous plume profile and its velocity field are thenIn particular $W{b}^2=6\kappa z$. The far horizontal velocities are inward, $u(\pm \infty ,z)=\mp Wb/(3z)$, supplying the entrained ambient volume. They are consistent with $Q^{\prime }=2Wb/(3z)$.

Finally use specific Darcy-flux moments, per unit source length, for the requested flux relationships. The volume flux, kinematic momentum flux and buoyancy flux areThus the physical mass flux is ${\rho }_{0}Q(z)$, and the conventional density-weighted Darcy second moment is ${\rho }_{0}M$. If intrinsic pore-velocity momentum flux is required instead, a uniform porosity $\phi$ gives ${\rho }_{0}M/\phi$; keeping this convention distinct does not change the displayed Darcy moments. These are the conserved momentum flux of a porous thermal plume relationships. Constancy of $M$ follows here from proportionality of temperature and velocity plus heat conservation, not from ignoring the drag on the solid matrix in a total momentum balance.

The ideal source has zero limiting volume flux but divergent central speed and temperature. Consequently the Boussinesq approximation and slender-plume limit describe the region outside a finite source, where $|\alpha \theta |\ll 1$ and $b/z\ll 1$. Indeed $Wz/\kappa =6(z/b{)}^2\gg 1$ is the local high-Rayleigh condition. These limits improve downstream; they do not make the singular similarity field a literal finite-temperature solution at $x=z=0$.

Let ${\rho }_r$ be a constant reference density close to the ambient density, and let ${\rho }_p$ denote the mean plume density. The Boussinesq approximation permits full-cross-section kinematic fluxesThus $Q$ is the specific mass flux, equal to volume flux at this order; $M$ is the specific vertical momentum flux; and $F$ is the buoyancy flux. Physical mass and momentum fluxes are ${\rho }_{r}Q$ and ${\rho }_{r}M$. These definitions include the full circular area. If factors of $\pi$ are removed in a course's flux convention, divide $Q,M,F$ by $\pi$ throughout rather than mixing normalizations. The PDF's source density is ${\rho }_0$, distinct from ambient ${\rho }_a(z)$; the TeX changes the source symbol incorrectly.

For the top-hat plume model, let the plume radius be $r(z)$ and let $w,b$ be uniform inside it. ThenThe Batchelor entrainment hypothesis supplies the necessary closure: the inward edge velocity is $v_e={\alpha }_{e}w$, with constant entrainment coefficient ${\alpha }_e>0$. Entrainment across a cylindrical slice gives $Q^{\prime }=2\pi r{\alpha }_{e}w=2{\alpha }_e\sqrt{\pi }\sqrt{M}$. The entrained ambient has negligible upward momentum. The vertical buoyancy force on the slice is $\pi r^{2}b=F/w$, hence $M^{\prime }=FQ/M$.

For the buoyancy balance, let $R={\int }_A{\rho }_{p}wdA$ be the transported density scalar. Entrainment gives $R^{\prime }={\rho }_a{Q}^{\prime }$. Differentiate $F=g({\rho }_{a}Q-R)/{\rho }_r$ to get $F^{\prime }=g{\rho }_a^{\prime }Q/{\rho }_r$. Therefore the closed flux equations areHere $N$ is the buoyancy frequency when the ambient is stably stratified. These are the Boussinesq top-hat plume in a stratified ambient equations in the explicitly stated full-area convention. Stable stratification consumes positive buoyancy flux as the plume rises. The top-hat assumption alone would not determine $Q^{\prime }$ without an entrainment law.

For uniform ambient density, $F=F_0$ is constant. The source volume $Q_0$ becomes negligible when accumulated entrainment ${\int }_0^{z}c\sqrt{M(s)}ds$ is much greater than $Q_0$. In a jet-like region, $M\simeq M_0$, so the criterion is $z\gg Q_0/(c\sqrt{M_0})$. In a buoyancy-dominated far field with $Q\simeq C_Q{z}^{5/3}$, it is $z\gg (Q_0/C_Q{)}^{3/5}$, with $C_Q$ determined below. These are the two source-volume lengths of a turbulent plume. They are not the jet length, which measures the persistence of imposed source momentum. A nonzero source volume can also appear as a plume virtual origin; it is not valid to discard it at every distance simply because the source is geometrically small.

For $Q_0=M_0=0$, $F_0>0$, seek $Q=C_Q{z}^{5/3}$, $M=C_M{z}^{4/3}$. The two flux equations give $C_Q=(3c/5)\sqrt{C_M}$ and $C_M^{3/2}=9c{F}_0/20$. Thus the pure-plume solution isFor completeness its directly measurable top-hat fields areThis is an axisymmetric pure plume. In this regime the explicit source-volume criterion isFor $Q_0=F_0=0$, $M_0>0$, the buoyancy vanishes and momentum is constant. The pure-jet solution isThis is the jet limit of the top-hat plume model. Both ideal zero-volume sources have singular speeds at $z=0$; they model regions outside a finite source where entrainment has already made the source-volume correction small.

When $M_0$ and $F_0$ are both positive, use the jet solution near the source and the pure-plume solution far above it, with an appropriate virtual-origin correction. To quantify this transition, eliminate $z$:The conserved quantity is the plume flux-balance invariant. With negligible $Q_0$, comparison of its two terms and $Q\simeq c\sqrt{M_0}z$ gives a transition height of order $M_0^{3/4}/(\sqrt{c}\sqrt{F_0})$. The exact connection, including finite $Q_0$, is the finite-source forced-plume quadraturewith $M$ then recovered from the invariant. A positive integrand gives the physical increasing-volume branch. As $Q$ grows, the source constant is negligible relative to $Q^2$, proving attraction to pure plume similarity. One matches the two asymptotic regimes or uses this quadrature; the jet and plume solutions cannot be linearly superposed because their governing flux equations are nonlinear.