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Landau theory expands a thermodynamic potential in an order parameter, constrained by symmetry. For a scalar with sign-reversal symmetry, a useful free-energy density isHere $h$ is the conjugate field. At $h=0$, minimizing rather than merely finding stationary points givesThe symmetry-related minima represent spontaneous symmetry breaking, as for a simple uniaxial ferromagnet with magnetization as order parameter. An infinitesimal field selects one sign. Above $T_c$, the susceptibility is $1/[a(T-T_c)]$; below it the stable-branch susceptibility is $1/[2a(T_c-T)]$. Minimization lowers $f$ by $-a^2(T-T_c{)}^2/(4b)$ below $T_c$, producing a heat-capacity density jump $T_c{a}^2/(2b)$ with continuous entropy.

A negative quartic coefficient requires a stabilizing $+c{\eta }^6/6$ with $c>0$. It can produce a first-order transition before the quadratic coefficient changes sign: coexistence occurs at $a(T-T_c)=3b^2/(16c)$ and ${\eta }^2=-3b/(4c)$. Odd terms are allowed when the assumed symmetry is absent. Landau theory predicts mean-field exponents; spatial fluctuations can change critical behaviour, especially close to a critical point in low dimensions. A gradient term extends the model to nonuniform order parameters but does not by itself supply dynamical equations.

A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by temperature alone. A cavity with a small aperture approximates this ideal. Photon number is not conserved in thermal equilibrium, so its chemical potential is zero. Combining the Bose occupation with the two-polarization electromagnetic mode density gives the spectral energy densityChanging variable to $x=\hslash \omega /(k_{B}T)$ shows directly why the total energy density is proportional to $T^4$:Three powers of T arise from the number of thermally populated modes and one from their characteristic energy. Isotropic radiation incident on a surface has outward hemispherical flux $cu/4$, givingThe factor 1/4 is the hemispherical angular average; multiplying energy density by c alone would overcount. The spectrum's frequency scale grows with T. Classical equipartition over all modes would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency modes. Real materials can have wavelength-dependent emissivity, and a body's net radiative exchange also includes absorbed environmental radiation.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

At zero applied field, time-reversal symmetry interchanges $M$ and $-M$ without changing the free energy, so the analytic expansion contains only even powers. The term $-{\mu }_{0}HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability requireFor $T>T_c$ at small field, the cubic term is higher order and $M\simeq {\mu }_{0}H/a(T)$. Thus the susceptibility as defined in the question isThe three requested free-energy sketches and their minima are given separately above.

For the plotted isotherms, divide the equation of state by nonzero M and rearrange:This is an Arrott plot: its slope determines b, and the isotherm through the origin identifies the mean-field Curie temperature.^[ref] Under the paper's stated uniform-field Landau model, use H as supplied on its axis.

The ordinate unit is $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2=10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times 10^6{\mathrm{A}}^2{\mathrm{m}}^{-2}$. ConsequentlyApproximate horizontal intercepts and the corresponding $a={\mu }_0(H/M{)}_{M^2=0}$ are:

The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha (T-T_c)$ across the four read-offs gives approximately $\alpha =4.8\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ andInterpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.

Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the free-energy density relative to the omitted smooth background:The equilibrium excess heat-capacity density is $c_H=-T{\partial }_T^2{F}_{\min }=T{\alpha }^2/(2b)$ below $T_c$ and zero above it. ThereforeThe all-isotherm linear fit gives about $1.5\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha \simeq 6.9\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and a jump near $3\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so a jump of order a few $10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ is supported; finer precision is not. This is per volume because F is a density; no sample volume is given to convert it to a total heat capacity.

If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $c{M}^6/6$ with $c>0$. At zero field the coexistence conditions giveThus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible metastability. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

For a single quantum state, the equilibrium occupations areBoth reduce to $e^{(\mu -ϵ)/(k_{B}T)}$ when $ϵ-\mu \gg k_{B}T$.

For one three-dimensional quadratic branch, the density of states per unit volume isWith ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, changing integration variable to $x=ϵ/(k_{B}T)$ givesA degeneracy factor multiplies both densities if additional equivalent branches are present. The first expression is the noncondensed density: a macroscopically occupied ground state must be counted separately.

Bose-Einstein condensation occurs when the excited states can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total density,For field-induced triplon condensation, write $\Delta =\nu n-\eta (H-H_g)$, so $ϵ(k)={\hslash }^2{k}^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic chemical potential is zero, the Bose distribution requires $\Delta \ge 0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.

For the triplon condensation field scaling, the critical density at onset is $n_T=\zeta (3/2)/{\lambda }_T^3$. ThereforeThe low-temperature phase boundary rises from $(T,H)=(0,H_g)$; the condensed phase lies on the higher-field side.

For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the condensate density in a shifted-parabola triplon model, added condensate density keeps the minimum at zero: $\nu (n_0+n_T)=\eta (H-H_g)$. At fixed temperature this giveswith $n_0=0$ below onset. At zero temperature it becomes $n_0=\eta (H-H_g)/\nu$ within that model.

This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed phase. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{{ϵ}_k^0({ϵ}_k^0+2g{n}_0)}$, with ${ϵ}_k^0={\hslash }^2{k}^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. The gap closes and the condensate density grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model. A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.^[ref]

Use $v$ for molar volume to distinguish it from the fluctuating parcel volume below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded volume, while $a>0$ models the reduction in pressure from attractive interactions.

Below the critical temperature, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,The endpoints are liquid and vapour molar volumes; intermediate mean volumes correspond to phase mixtures with changing proportions. Above the critical temperature, liquid-like and gas-like states connect continuously without a coexistence plateau.

In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the critical point. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the critical temperature, where compression alone cannot cross a liquid-vapour transition. The model has no solid phase.

The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.

The whole cylinder's volume is fixed. For the volume fluctuations of a small fluid parcel, interpret the small quantity as a material parcel containing $s$ moles, whose volume can fluctuate against the much larger remainder. Its mean volume at the critical density is $\bar{V}=s{v}_c=3sb$. At fixed temperature its availability is $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. The mean-field variance diverges as $(T-T_c{)}^{-1}$, so its critical exponent is $1$; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric volume. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the critical point.

Use the number-density virial expansion, following the density versus molar second virial convention, with $n=N/V$:Thus $B_2$ is the coefficient of the first density correction and has units of volume. With molar density instead, the molar coefficient is $N_A{B}_2$. For a classical pair potential $u(r)$,Attractive separations contribute negatively and short-range repulsion contributes positively. For inert-gas interactions the attractive contribution dominates at sufficiently low temperature, $B_2$ crosses zero at the Boyle temperature, and repulsion dominates above it.

The soft-core high-temperature second virial coefficient illustrates why the far-high-temperature tail depends on the model. A hard-core estimate $B_2\simeq b-a/(k_{B}T)$ tends to positive excluded volume $b$. A soft repulsive potential, such as the Lennard-Jones potential in the original diagram, allows a positive maximum followed by decay toward zero. The sign change is the main attraction-versus-repulsion feature; neither sketch should be interpreted as measured data for a specified gas. This classical, dilute-gas sketch omits low-temperature quantum-exchange corrections.^[ref]

Brownian motion is the irregular motion of a mesoscopic particle caused by thermal molecular impacts. A colloidal sphere in a fluid provides a physical example. The random force competes with dissipative drag; averaging many trajectories produces diffusion even though each trajectory fluctuates.

A free-particle Langevin equation with thermal Langevin force normalization isThe noise has zero mean, and its strength is linked to drag by the fluctuation-dissipation theorem. The velocity relaxes on the time scale $\tau =m/\zeta$. At long times $t\gg \tau$, the Einstein diffusion relation isFor a slowly moving no-slip sphere of radius $a$ in a continuum fluid of dynamic viscosity $\eta$, Stokes drag law gives $\zeta =6\pi \eta a$. The particle probability density obeys ${\partial }_{t}P=D{\nabla }^{2}P$; a localized initial position develops a Gaussian profile. The ballistic-to-diffusive Brownian displacement changes character: at $t\ll \tau$, equilibrium velocities instead give ballistic mean-square displacement $3(k_{B}T/m)t^2$.

The drag-noise model assumes a classical bath with a sufficiently short correlation time; a mathematical white-noise trajectory does not resolve individual molecular collisions. Tracking nonliving colloidal particles measures diffusion and tests the connection between thermal fluctuations and dissipation.^[ref]

Landau theory expands the free energy in a small order parameter and keeps terms allowed by the system's symmetries. For a scalar order parameter $q$ with sign-reversal symmetry in Landau theory $q\mapsto -q$, a useful free-energy density isAt zero conjugate field $h$, the quartic Landau equilibrium branches obtained by minimizing areThe single minimum becomes two symmetry-related minima: the order parameter turns on continuously and a chosen equilibrium phase spontaneously breaks the symmetry. At the critical temperature the quadratic restoring term vanishes. The mean-field exponent is $q\propto (T_c-T{)}^{1/2}$; the susceptibility above $T_c$ is $\partial q/\partial h=1/[a(T-T_c)]$.

More general allowed terms describe other transitions. A negative quartic term, stabilized by a positive sixth-order term, can give a first-order transition with a discontinuous order parameter and metastable minima. Without the sign-reversal symmetry, a cubic term may also be allowed. The expansion is phenomenological: its coefficients depend on microscopic physics. Minimizing a uniform polynomial neglects important spatial fluctuations, particularly near a critical point, so its critical exponents need not be exact.^[ref]

Measure energies from the single-particle ground state, and assume the three-dimensional ideal Bose gas in the large-particle-number limit. The harmonic-trap chemical-potential pinning discussed here uses a chemical potential satisfying ${\mu }_{\mathrm{ch}}\le 0$. On cooling, it approaches zero; below the condensation temperature it remains at zero to leading thermodynamic order. Above that temperature it is negative. The detailed negative-chemical-potential branch shape is determined by the density of states; the figure uses the harmonic trap analysed below. The finite-size condensate chemical potential stays slightly below zero rather than being exactly pinned there.

A Bose-Einstein condensate has macroscopic occupation of a single quantum state. A dilute atomic gas needs low temperature and high phase-space density while remaining metastable against ordinary liquid or solid formation. Laser cooling followed by evaporative cooling of atoms in a weak trap can achieve the required regime. For the trapped ideal gas the quantitative threshold is derived below; a uniform ideal gas has $n{\lambda }_T^3=\zeta (3/2)$ at onset. An experimental signature is a narrow low-momentum component appearing within the broader thermal distribution after time-of-flight expansion, as observed in magnetically trapped rubidium-87.^[ref]

Spinless cubic-box wave-vector counting. Impose periodic boundary conditions on a cube of side $L$. The allowed vectors are $\mathbf{k}=(2\pi /L)(n_x,n_y,n_z)$ with integer components. Each state occupies $(2\pi /L{)}^3$ in wave-vector space. For a spinless particle, counting the spherical shell givesThere is no spin multiplicity. This is continuum shell counting; a finite box has discrete levels. Using $ϵ={\hslash }^2{k}^2/(2m)$ gives the free-particle density of states per unit energy and volume

Semiclassical density of states in a central potential. At position $\mathbf{r}$, the available kinetic energy is $ϵ-U(r)$. The local-density approximation integrates the preceding free-particle result over the classically allowed region:This form assumes the allowed radial region is the connected interval $0\le r\le r_{ϵ}$. A nonmonotonic potential would require integrating all allowed shells.

For the isotropic harmonic trap, $r_{ϵ}=\sqrt{2ϵ/(m{\omega }^2)}$. Substituting $r=r_{ϵ}\sin \vartheta$ givesConsequentlyThe exact oscillator ground energy is $3\hslash \omega /2$ above the potential minimum. The harmonic-trap energy-zero convention subtracts that constant, while the displayed density of states is the leading semiclassical continuum expression, neglecting the finite-level corrections produced by that shift. The discrete ground state must be counted separately.

Condensate population. Below the transition, use ${\mu }_{\mathrm{ch}}\simeq 0$ for the excited states. The saturation of harmonic-trap excited states follows from the supplied integral:where $\zeta$ is the Riemann zeta function and $\zeta (3)\simeq 1.202$. Setting $N_{\mathrm{ex}}(T_0)=N$ gives the harmonic-trap critical-temperature prefactor:Above $T_0$ there is no macroscopic ground-state fraction in this approximation. For completeness, above the transition the quantum gas fugacity $z=e^{{\mu }_{\mathrm{ch}}/(k_{B}T)}<1$ satisfies $N=(k_{B}T/\hslash \omega {)}^3{\mathrm{Li}}_3(z)$; the polylogarithm ${\mathrm{Li}}_3$ therefore defines the plotted negative branch. A finite trapped gas has rounded behaviour, weak interactions shift the threshold, and the continuum thermal population is inaccurate when $k_{B}T$ is comparable to or below $\hslash \omega$. The power law is intended for the large-$N$ condensed regime, not an exact discrete-spectrum formula at arbitrarily small temperature.^[ref]

For the ideal dilute Maxwell-Boltzmann gas implicit in a single-particle partition function, indistinguishability gives $Z_N=Z_1^N/N!$. With Stirling's approximation,Differentiate at fixed $T,V$:This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.

Equilibrium constraints and the Gibbs minimum. The open-system equilibrium contacts allow exchange of heat, volume and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' chemical potentials across the contacts. These conditions follow by maximizing total entropy under conserved total energy, volume and species amounts.

The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir entropy change givesFor candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus the Fixed-particle-number Gibbs minimum selects stable equilibrium at fixed $T,p,N$, with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.

The simple-compressible-system first law givesFor a homogeneous extensive single-component phase, the Single-component Gibbs Euler relation follows from $U=TS-pV+{\mu }_{\mathrm{ch}}N$, yielding $G={\mu }_{\mathrm{ch}}N$. For a mixture, the corresponding identities are $G={\sum }_i{\mu }_i{N}_i$ and $dG=-SdT+Vdp+{\sum }_i{\mu }_{i}d{N}_i$. Surface effects or other nonextensive contributions require additional terms.^[ref]

Reaction equilibrium. Let $d\xi$ be an allowed reaction extent for the forward process, so $d{N}_A=-d\xi$ and $d{N}_B=2d\xi$. At fixed $T,p$,For an interior equilibrium allowing both reaction directions, the coefficient vanishes:At fixed $T,V$, minimizing $F$ gives the same condition. The reaction formula-unit conservation gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This conservation law replaces fixed total particle number for the reacting mixture.

Using the stipulated single-particle functions,The number-based dissociation equilibrium ratio is thereforeThis number-based quantity has a factor of volume; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on temperature alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.

Define the dissociated fraction of original A formula units by $\alpha =N_B/(2M)$. Then $N_A=M(1-\alpha )$ and $N_B=2M\alpha$, soThe dissociation fraction versus conserved density follows because the left side increases monotonically from zero to infinity on $0<\alpha <1$. At fixed $V,T$, increasing $M$ decreases $q$, hence decreases dissociation. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high density the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.

The four dimensional equations express mass conservation, momentum balance with pressure, gravity and viscous stress, heat advection and conduction, and a linear thermal-expansion equation of state. For heating from below, suitable plate conditions are $\mathbf{u}=0$ and $T=T_0+\Delta T$ at $z=0$, and $\mathbf{u}=0$, $T=T_0$ at $z=d$. No slip and thermal insulation on the sidewall provide one suitable lateral completion; prescribed lateral temperatures would be another specified thermal setup.

Let $\kappa =k/({\rho }_0{c}_p)$ be thermal diffusivity and $\nu =\mu /{\rho }_0$ kinematic viscosity. Use $d$, $d^2/\kappa$, $\kappa /d$ and $\Delta T$ as length, time, velocity and temperature scales. Subtract the static reference pressure $p_H$ with $\nabla p_H=-{\rho }_{0}g\mathbf{k}$ and scale its remainder by $\mu \kappa /d^2$. ThenDensity becomes ${\rho }_0(1-BT)$. Substitution and division by the corresponding diffusion/viscous scales giveIn the Boussinesq small-density-contrast limit retain buoyancy while neglecting the other $B$ factors. At infinite Prandtl number, after any fast initial adjustment, the equations areThe velocity is instantaneously constrained by a viscous elliptic balance; the temperature still evolves.

Take the divergence of the vector Poisson equation. Derivatives commute, and divergence of a curl is zero, so $\Delta (\nabla \cdot \mathbf{A})=0$. Its boundary value is zero, hence the maximum principle gives $\nabla \cdot \mathbf{A}=0$ throughout $D$. The vector identity now gives $\mathrm{curl}\mathrm{curl}\mathbf{A}=\mathrm{curl}\mathbf{u}$, orThe layer domain is simply connected because its horizontal section is simply connected. For a curl-free field on such a domain, define $\chi (x)$ as its line integral from a fixed base point to $x$. Contractibility of loops and Stokes theorem make the integral path independent, and differentiation gives its gradient equal to the field. Thus $\mathrm{curl}\mathbf{A}-\mathbf{u}=\nabla \chi$.

Incompressibility then gives $\Delta \chi =0$, while the prescribed normal curl gives ${\partial }_n\chi =0$. Green's identity yieldsHence $\chi$ is constant on the connected domain and $\boxed{\mathrm{curl}\mathbf{A}=\mathbf{u}}$. The extra condition $\mathbf{A}\cdot \mathbf{n}=0$ belongs to the stated potential boundary problem; the reconstruction argument above uses the divergence and normal-curl conditions. On a multiply connected domain a curl-free field need not be a globally single-valued gradient.

Finally $\mathrm{curl}(T\mathbf{k})=(T_y,-T_x,0)$. Taking the curl of the infinite-Prandtl momentum balance and using $\mathrm{curl}\mathbf{u}=-\Delta \mathbf{A}$ givesThe derivation is an instance of velocity reconstruction from a mixed-boundary vector Poisson problem.

Define ${\chi }_1$ to be one in phase 1 and zero in phase 2, including its dependence on both transverse coordinates even though the averaged fields depend only on $x,t$. With no interphase mass transfer, its distributional phase-mass equation is ${\partial }_t(\rho {\chi }_1)+\nabla \cdot (\rho \mathbf{v}{\chi }_1)=0$. At a pipe cross section $\Sigma$ of area $A$, setThus ${\rho }_1$ is an intrinsic phase density and $u_1$ is the density-weighted mean axial velocity. This weighting is necessary when density varies within the phase; it becomes the ordinary intrinsic mean for an incompressible phase of constant density.

Integrate over the cross section. The transverse divergence contributes no boundary mass flux at the impermeable pipe wall, and the material interface has no mass leakage. The section area is constant, soFor phase 2 use ${\chi }_2=1-{\chi }_1$ and fraction $1-\alpha$, with the analogous intrinsic density and density-weighted velocity:The intrinsic quantities are defined where their phase fraction is nonzero. These are the intrinsic phase averages in a two-fluid pipe model.

For constant phase density, let $u_1$ be the intrinsic mean axial speed and defineThe profile coefficient measures the momentum flux relative to a uniform velocity profile with the same mean. Cauchy-Schwarz gives $D_1\ge 1$ where the mean is nonzero, with equality for a uniform phase profile. At zero mean it is safer to retain the second moment itself rather than divide by $u_1^2$.

In the stated common-pressure, no-extra-force closure, the averaged conservative momentum equation isFor constant $D_1$, expand its derivatives and use ${\alpha }_t+u_1{\alpha }_x+\alpha (u_1{)}_x=0$. The terms combine asThereforeThis is the phase momentum profile coefficient correction. It accounts for the second moment of velocity, not merely a rescaling of the convective speed. A variable coefficient would also produce ${\rho }_1\alpha u_1^2(D_1{)}_x$.

For $x>1$, write the proposed density as $x^{-\alpha }(c-1/x)$. Since the first factor is positive, nonnegativity for all such $x$ requires $c\ge 1$; if $c<1$, values sufficiently close to one make it negative. With $c\ge 1$, the positive tail is integrable only if $\alpha >1$. In that case normalization requiresImposing $c\ge 1$ now gives ${\alpha }^2-\alpha -1\ge 0$. Combining this with $\alpha >1$ yieldsConversely, these conditions ensure a nonnegative, integrable function with integral one, so they are sufficient as well as necessary. The endpoint is allowed: the density vanishes at the left-hand boundary limit, not on its whole support.

After bath cooling followed by insulated equilibration, the remaining mean temperature is conserved: integrating the insulated heat equation givesFor constant density, specific heat capacity and cross-section, this is conservation of energy. Its Fourier cosine series has a constant mode equal to the initial integral average; every nonconstant mode decays. Therefore the eventual uniform temperature isFor each mode with an odd integer index in part (c), $L^{-1}{\int }_0^L\sin ((2j+1)\pi x/L)dx=2/[(2j+1)\pi ]$. Integrating the bath-cooled series yieldsFor $t_f>0$, uniform convergence justifies termwise integration; the $t_f=0$ result also follows by dominated convergence of the summable positive coefficients. As a check, ${\sum }_{j\ge 0}(2j+1{)}^{-2}={\pi }^2/8$ gives $T_f=T_0$ for zero bath time, and $T_f\to 0$ as $t_f\to \infty$. More bath time strictly lowers the retained average temperature when $T_0>0$ and $\kappa >0$.