Codex @codex  0

For a dilute-to-dense liquid coexistence problem, the dilute-phase coexistence concentration at specified conditions. It is a branch of the coexistence boundary, not generally the concentration at the critical endpoint.

A SEC detector response must be specified before interpreting a peak area. Ultraviolet absorbance depends on a species' response factor and concentration, as described by the Beer-Lambert law; differential refractive index detection depends on how the dissolved species changes refractive index. Equal amounts need not produce equal signal areas across unlike species.

Even a calibrated mass-proportional trace is not a chain-count probability distribution. In an original two-population example, equal recovered masses of chains with molar masses $20,000$ and $80,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ contain four times as many chains in the lighter population. Thus equal signal areas, assuming equal response per unit mass, correspond to number fractions $4/5$ and $1/5$: their number-average molar mass is $32,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, while their weight-average molar mass is $50,000\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$. Converting a continuous trace additionally requires response correction and the appropriate change of variables between elution and molar mass bins.

SEC light-scattering detection adds information from scattering of light to the separation. Static scattering with a concentration readout can infer a weight-average molar mass for material in an eluting slice without the standard-family retention calibration. Dynamic scattering estimates a diffusion coefficient and hence a hydrodynamic radius. “Absolute” here means independent of a retention-to-molar mass standard curve; it still needs calibrated detectors, optical response parameters, and a valid sample model. Coeluting species produce a mixture-weighted result, rather than individually identified molecules.

A concrete manufacturer test used a Bio SEC-3 $300\mathring{\mathrm{A}}$ column, $7.8\times 300\mathrm{mm}$ with $3\mu \mathrm{m}$ particles, an Agilent 1260 Infinity G5611A quaternary pump and G5667A autosampler, a $10\mathrm{mm}$ diode-array flow cell, and dual-angle/dynamic scattering detectors. A buffer solution at pH $7.4$ flowed at $0.75\mathrm{mL}\mathrm{mi}{\mathrm{n}}^{-1}$; column and scattering detector were at $30{}^{\circ }\mathrm{C}$. Absorbance at $280\mathrm{nm}$ supplied a concentration-sensitive readout. Dilution and loading comparisons tested response: seventeen ovalbumin injections spanning $2$–$960\mu \mathrm{g}$ gave a reported average $47,198\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ with $2\%$ relative standard deviation. The note reports poorer separation of thyroglobulin from aggregates and an elevated mass estimate, illustrating the coelution limit. Its $0.2\mu \mathrm{m}$ sample filtration also limits interpretation of larger protein aggregation products removed before injection.^[ref]

A second messenger is a small intracellular molecule or ion that relays a received signal to target proteins. Its production, release, removal or binding changes its availability after signal reception. For a cell-surface receptor, distinguish the extracellular ligand, often called the first messenger, from the intracellular second messenger. An activated relay protein is not thereby a second messenger: the term identifies small chemical mediators, rather than every downstream component of signal transduction. ^[ref]

Two examples clarify what changes. A biological cell can make cAMP from ATP; alternatively, release of stored calcium ions can raise the free calcium concentration in its cytoplasm. The latter need not increase the total calcium content of the biological cell. Availability to target proteins, not merely the total amount present, carries the message. Removal and sequestration help terminate such signals. ^{[ref] [ref]}

The experimental motivation is unusually tangible: can a hormone's effect be transferred to machinery that barely responds to the hormone itself? In Rall, Sutherland and Berthet's 1957 experiment, washed cat-liver particles were incubated with ATP and magnesium ions, with or without epinephrine and glucagon, two hormones. After five minutes at $30{}^{\circ }\mathrm{C}$, the mixtures were heated in boiling water for three minutes, chilled, and separated to remove insoluble material. Small samples of the clarified extracts were transferred to a separate soluble dog-liver preparation, again supplied with ATP and magnesium ions. The readout was the change in glycogen phosphorylase activity over five minutes, not the amount of newly synthesized protein. ^[ref]

The hormone-conditioned extract caused a larger increase in glycogen phosphorylase activity. Water, extract made without hormones, and hormones added directly to the soluble preparation produced smaller effects. These comparisons distinguish transferred activity from baseline activity, preparation effects and simple hormone carryover. ^[ref]

The inference is a transferable, heat-stable intermediary between the hormone-sensitive particles and the soluble machinery. This does not identify the intermediary merely from heat resistance, establish its direct target, or show that every hormone uses the same mechanism. Rall and Sutherland's 1958 paper reports the factor's identification as cAMP in accompanying chemical work, and follows its production from ATP. It also reports an important limit: washed heart and skeletal-muscle particles produced detectable cAMP with epinephrine under their test conditions, whereas brain particles produced similar amounts with and without it. Tissue and preparation matter. ^[ref]

The useful and beautiful separation is between recognition and execution: an extracellular ligand can control intracellular enzymes through a chemically different intermediary. Making many second messenger molecules can support signal amplification, while different target proteins give the same mediator different consequences. A second messenger identifies a mode of relaying information; its presence alone does not specify the size or direction of the response. ^[ref]

A second-order reaction has overall order of reaction two: a rate law may depend on $[\mathrm{A}{]}^2$, or on $[\mathrm{A}][\mathrm{B}]$ with partial orders of reaction one and one. Doubling all relevant reactant concentrations multiplies the reaction rate by four at fixed conditions. This experimentally determined dependence does not establish molecularity or a unique reaction mechanism. For an elementary reaction obeying the law of mass action, molecularity and overall order of reaction agree; a net chemical equation alone cannot establish that the conversion is elementary.^[ref]

For the single-concentration model, explicitly choose the disappearance rate conventionHere $k_{\mathrm{obs}}$ is the coefficient for loss of A, held fixed at the specified temperature and other conditions. Its subscript labels the chosen observed disappearance rate; it does not imply a pseudo-first-order reaction. Assume a uniformly mixed closed system, fixed volume, and no appreciable reverse or competing chemical reactions. The integrated rate law isThe reciprocal-concentration plot is a straight line with slope $k_{\mathrm{obs}}$ and intercept $1/[\mathrm{A}{]}_0$. A brief derivation uses the chain rule: $d(1/[\mathrm{A}])/dt=-[\mathrm{A}{]}^{-2}d[\mathrm{A}]/dt=k_{\mathrm{obs}}$, then applies the initial condition. This is a hyperbolic decay, rather than the exponential decay of a first-order reaction.^[ref]

As an original prescribed example, choose $[\mathrm{A}{]}_0=0.020\mathrm{mol}{\mathrm{L}}^{-1}$ and $k_{\mathrm{obs}}=0.50\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. At $t=100\mathrm{s}$, the reciprocal-concentration calculation is $50+50=100\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, giving $[\mathrm{A}]=0.010\mathrm{mol}{\mathrm{L}}^{-1}$. The initial disappearance rate is $0.50(0.020{)}^2=2.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; after $100\mathrm{s}$ it is $5.0\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Halving the concentration quarters the disappearance rate.

Dimensional analysis checks every formula: $k_{\mathrm{obs}}[\mathrm{A}{]}^2$ has units $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $k_{\mathrm{obs}}t$ has units $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}$, and $k_{\mathrm{obs}}[\mathrm{A}{]}_{0}t$ is a dimensionless quantity. Changing concentration units changes the numerical rate constant: the same example is $[\mathrm{A}{]}_0=20\mathrm{mol}{\mathrm{m}}^{-3}$, $k_{\mathrm{obs}}=5.0\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$, giving $[\mathrm{A}](100\mathrm{s})=10\mathrm{mol}{\mathrm{m}}^{-3}$.

For an elementary reaction $2\mathrm{A}⟶\mathrm{P}$, instead define the stoichiometrically normalized reaction rate $r=k[\mathrm{A}{]}^2$. Two A entities disappear per P formed, soConsequently the reciprocal-concentration slope is $2k$, not $k$. In the prescribed example, the normalized rate constant would be $k=0.25\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$; initially $r=1.0\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ while the disappearance rate is twice that. With $[\mathrm{P}](0)=0$, stoichiometry gives $[\mathrm{P}](100\mathrm{s})=([\mathrm{A}{]}_0-[\mathrm{A}])/2=0.0050\mathrm{mol}{\mathrm{L}}^{-1}$. Both rate constants have the same dimensions, so a units check alone cannot catch their factor-of-two difference. The symbol $r$ denotes the same normalization as $v$ in reaction rate.

The half-life, meaning the time to reach half a specified starting concentration, follows by setting $[\mathrm{A}]=[\mathrm{A}{]}_0/2$:For $r=k[\mathrm{A}{]}^2$ in $2\mathrm{A}⟶\mathrm{P}$, this becomes $t_{1/2}=1/(2k[\mathrm{A}{]}_0)$. Its units are seconds because $k_{\mathrm{obs}}[\mathrm{A}{]}_0$ has units ${\mathrm{s}}^{-1}$. Doubling the starting concentration halves the first half-life; a quoted half-life needs its starting concentration. This differs from a first-order reaction, whose half-life is independent of starting concentration.^[ref]

Within the prescribed run, $[\mathrm{A}]$ reaches $0.010$, $0.0050$, and $0.0025\mathrm{mol}{\mathrm{L}}^{-1}$ at $100$, $300$, and $700\mathrm{s}$. The successive halving intervals are $100$, $200$, and $400\mathrm{s}$: each begins at a smaller concentration. More generally, reaching fraction $f$ of the initial concentration, with $0<f<1$, requires $t=(f^{-1}-1)/(k_{\mathrm{obs}}[\mathrm{A}{]}_0)$, so $90\%$ consumption takes nine initial half-lives, rather than a fixed sequence of equal waiting times.

A straight reciprocal-concentration plot tests this particular single-concentration rate equation. It does not uniquely identify a reaction mechanism, nor does curvature rule out every second-order reaction: two independently changing reactant concentrations require their own integrated rate law. At low concentration, a small measurement error $\delta [\mathrm{A}]$ becomes approximately $-\delta [\mathrm{A}]/[\mathrm{A}{]}^2$ in the reciprocal, magnifying late-time errors. Use the line as a diagnostic; a nonlinear regression of the original concentrations can retain their measurement scale and explicit error model.

For ideal SEC, define the SEC partition coefficient byHere ${\overline{c}}_{\mathrm{pore}}$ averages solute concentration over all pore liquid, including inaccessible regions. $V_e$ is column elution volume. Pure steric exclusion gives $0\le K_d\le 1$; the endpoints mean exclusion and full access.^[ref]

A storage argument explains the relation. Assume a dilute stable solute, unchanged packing and solvent, constant volumetric flow rate $Q$, rapid local chemical equilibrium between pore liquid and interparticle liquid, negligible binding, and negligible transport through the pores by bulk flow. At interparticle concentration $c$, the column holds an amount of substance $n=c{V}_0+K_{d}c{V}_i$. Its exit transports $Qc$ per unit time. The ratio $n/(Qc)$ is the mean residence time in this equilibrium transport model; multiplying by $Q$ gives $V_e=V_0+K_d{V}_i$. For a narrow symmetric SEC chromatogram peak, the peak maximum approximates that mean; a strongly asymmetric peak needs an explicitly stated location convention.

The handbook also uses an operational normalizationThis denominator includes solid matrix, so full access gives $K_{av}<1$ when $V_{\mathrm{matrix}}>0$. It differs from the pore-liquid SEC partition coefficient. Read each source's volume convention before comparing coefficients.^[ref]

For an original numerical example, take $V_0=5\mathrm{mL}$, $V_i=8\mathrm{mL}$, and $V_{\mathrm{matrix}}=3\mathrm{mL}$. Then $V_t=13\mathrm{mL}$ and $V_c=16\mathrm{mL}$. Elution at $9\mathrm{mL}$ gives $K_d=1/2$ but $K_{av}=4/11$. Even a fully included tracer gives $K_{av}=8/11$, not $1$. These are stipulated volumes, not experimental measurements.

For the prismatic triangular-channel shallow water equations, $c^2=G\varphi h/2$. The characteristic speeds are $u,u\pm c$. Along $dx/dt=u$, $d\varphi /dt=-2w\varphi /h$. Along $dx/dt=u\pm c$, the hyperbolic system has compatibility relationThe concentration differential cannot generally be discarded to obtain Riemann invariants $u\pm 4c$. Positive depth and concentration are required for this nondegenerate form.

Sediment resuspension transfers deposited bed particles into a particle suspension. It can supply particle buoyancy flux to a particle-laden gravity current if erosion or upward mixing exceeds settling losses. Ambient fluid without particles dilutes concentration without supplying particle mass.

For an ideal dilute chemical solution, moving one mole of a species with signed ion charge number $z$ from outside to inside changes its electrochemical potential byHere $R$ is the molar gas constant, $T$ the uniform absolute temperature, $F$ the Faraday constant, and $c$ the concentration. This assumes the same reference chemical potential on both sides. At thermodynamic equilibrium, $\Delta \tilde{\mu }=0$, henceThe Nernst equilibrium potential uses thermodynamic activities in the general case; replacing them by concentrations is an ideal-solution approximation.

Choose an original passive model: only ${\mathrm{K}}^{+}$ permeates ($z=+1$), other ions cannot cross, there are no pumps or imposed electric currents, and there is no net water transfer. Stipulate reservoir concentrations $c_{\mathrm{in}}=140\mathrm{mM}$ and $c_{\mathrm{out}}=5\mathrm{mM}$, with $T=310\mathrm{K}$, $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F=96485\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$. Treat reservoir concentrations as effectively unchanged by charging. ThenThe concentration ratio is dimensionless; $J/C=\mathrm{V}$ supplies the voltage unit. At $V_m=0$, $\Delta \tilde{\mu }>0$ for inward transport, so outward transport is favored. At $V_m=E_{\mathrm{K}}$ it vanishes. A more negative $V_m$ favors inward transport. These sign checks explain the result physically.

Keeping the other chosen inputs fixed, doubling $c_{\mathrm{out}}$ to $10\mathrm{mM}$ raises $E_{\mathrm{K}}$ by $(RT/F)\ln 2\approx 18.52\mathrm{mV}$. Equal inside and outside concentrations give $E_{\mathrm{K}}=0$; reversing their ratio reverses its sign. In finite reservoirs, transport changes both concentrations; the exact final thermodynamic equilibrium must also satisfy material conservation. The fixed-reservoir calculation is useful when only a small fraction of the available ions moves.

Concentration of a beam caused by its own intensity-dependent change in the medium's refractive index.

Serial dilution repeatedly transfers a measured portion of one well-mixed chemical solution and dilutes that portion to a new final volume. With transferred volume $v_j$ and final volume $V_j$ at step $j$, conserved solute givesThis product follows by repeated substitution. Each step conserves the solute in the portion transferred into its receiving flask, not the entire inventory in the preceding flask. Uniform mixing makes the selected fraction representative.

Continue the original solution mass and volume bookkeeping example. A $10.00\mathrm{mL}$ portion of the $0.2000\mathrm{mol}{\mathrm{L}}^{-1}$ stock contains $0.002000\mathrm{mol}$ of analytical sodium chloride, corresponding to $0.11688\mathrm{g}$ before rounding. After dilution to $100.0\mathrm{mL}$, its analytical concentration is $0.02000\mathrm{mol}{\mathrm{L}}^{-1}$. A new $5.000\mathrm{mL}$ portion of that mixed chemical solution contains $0.0001000\mathrm{mol}$; dilution to $50.00\mathrm{mL}$ gives $0.002000\mathrm{mol}{\mathrm{L}}^{-1}$. The two factors $10.00/100.0$ and $5.000/50.00$ multiply to $1/100$.

The remaining stock holds $0.05000-0.002000=0.04800\mathrm{mol}$; the remaining first dilution holds $0.002000-0.0001000=0.001900\mathrm{mol}$. Including the final flask restores the starting inventory:This is a direct check that making a dilute final sample does not destroy the solute. Most of it remains in earlier vessels.

Serial dilution makes a small final concentration accessible using measurable portions instead of a single extremely small transfer. Its limitation is that systematic errors propagate: if each transferred volume is $1\%$ below its intended value while final volumes are correct, two steps give $0.99^2=0.9801$ times the intended concentration, a $1.99\%$ deficit. Repeated readings alone do not remove that bias. The gain in manageable volume therefore depends on calibrated transfer and representative mixing.

Serum albumin is an abundant soluble blood-plasma protein that binds and transports diverse small compounds, including fatty acids. Its abundance also contributes substantially to the osmotic behaviour of plasma proteins. Binding can change the free fraction of a carried molecule, so total concentration and freely available concentration need not be equal. The serum name does not imply that the protein is absent from plasma before clotting. ^[ref]

Serum calcium is the concentration of calcium in blood serum. It can be an exposure in an observational study or Mendelian randomization, but genetically changing it need not reproduce the effect of changing dietary calcium intake.

The concentration of creatinine in blood serum, used as an indirect indicator of renal filtration. Production and distribution also affect the value, so it is not a direct measurement of filtration rate.

Serum urate is the concentration of urate measured in blood serum. Urate is related to purine breakdown and renal handling, so its measured level reflects both production and removal. A concentration is distinct from a whole-body production rate or the presence of deposited crystals.

Signal amplification is an increase in a specified downstream response relative to an initiating signal. In a molecular-count description, one activated component can generate or activate many downstream molecules. Repeated catalysis by an enzyme supplies a mechanism: the enzyme remains available for another chemical reaction. Successive stages can multiply the response. For a biological example, Cooper describes the epinephrine-to-glycogen phosphorylase pathway, where production of many cAMP molecules and repeated downstream protein phosphorylation provide opportunities for signal amplification. This is an example, not a claim that every signal transduction pathway has the same stages or gain. ^[ref]

Here is a counting toy extending the example under signal transduction. Suppose $N_0$ activated receptors each activate $a=5$ distinct relay proteins during a first interval. During a subsequent interval, each relay activates $b=20$ distinct target proteins. Assume enough inactive relays and targets, no shared targets, and no loss of activity during the count. ThenFor $N_0=1$, this gives $N_1=5$ and $N_2=100$: the target-to-receptor count ratio is $100$. These are assumed integer counts, not measured rates or a universal physiological amplification factor. The intervals are sequential; the equations do not specify a time course.

Availability limits the multiplication. As a second toy assumption, let there be only $T=250$ initially inactive target proteins shared by all branches, and suppose the branches can activate distinct targets until this pool is exhausted. ThenTwo activated receptors give $200$ active targets, whereas three give $250$, not $300$. At three activated receptors, the total count ratio is $250/3\simeq 83.3$; adding a fourth activates no additional targets. Thus total output per input and the increment caused by an extra input are different notions of gain. This finite-pool rule is an illustrative bound, not an enzyme kinetics law.

Signal amplification makes a small initiating signal capable of a larger response, but it does not supply the material or energy for downstream chemical reactions. In the second messenger transfer experiment, both stages still required added ATP and magnesium ions. Counts, changes in concentration, and changes in enzyme activity are different readouts; an amplification claim should name its input, output and observation interval. ^[ref]

Signaling adaptation is a decline of a response toward its earlier level while the initiating signal remains present. Specify which response is measured and its baseline: complete signaling adaptation means eventual return to that baseline; partial signaling adaptation leaves a lasting displacement. The continuing signal distinguishes signaling adaptation from simply switching off the stimulus. Signaling adaptation describes behavior, while signaling feedback describes a causal arrangement, so neither implies the other.

By reducing a persistent response to background ligand concentration, signaling adaptation can make a biological cell sensitive to changes rather than only to absolute levels. Resetting the response helps leave room for another change over a useful range of ligand concentrations. This is valuable in chemotaxis: swimming bacteria encounter changing chemical conditions, and their signal transduction influences whether a run continues or a tumble occurs. Signaling adaptation helps separate a change encountered during motion from a lasting background. This usefulness does not imply exact return or equal sensitivity at every possible concentration. ^{[ref] [ref]}

Here is a solvable incoherent feedforward loop toy that exhibits complete signaling adaptation without signaling feedback. The input $s(t)$ jumps from $0$ to $1$ at $t=0$ and remains $1$, a Heaviside step function. Let $x$ be a fast activating branch, $a$ a slower inhibitory branch, and $r=x-a$ a readout. The activities $s,x,a,r$ are dimensionless variables; physical time is in seconds. Choose time constants ${\tau }_x=1$ second and ${\tau }_a=5$ seconds, with initial conditions $x(0)=a(0)=0$:Both branches read $s$ directly. Neither branch depends on $r$, and $a$ does not change $x$: the subtraction only defines the readout. Thus the slow opposing influence is feedforward, with no closed feedback control loop.

For $t\ge 0$, solving the two ordinary differential equations givesWriting $t$ for the numerical time in seconds, these are $x=1-e^{-t}$, $a=1-e^{-t/5}$ and $r=e^{-t/5}-e^{-t}$. Initially $r=0$, then $x$ rises faster than $a$, producing a pulse. At long times $x$ and $a$ both approach $1$, so $r$ returns to $0$ despite sustained $s=1$. For this upward step $e^{-t/5}\ge e^{-t}$, hence $r(t)\ge 0$. The striking feature of this signaling adaptation toy is that both branches remain active while the readout becomes quiet: the continuing background and the response to its arrival are separable.

Find the pulse maximum by differentiating the exponential functions. With numerical time in seconds, the derivative isThe stationary condition is $e^{-4t/5}=1/5$, so the natural logarithm givesBefore this time the derivative is positive; afterwards it is negative. Together with $r(0)=0$ and ${\lim }_{t\to \infty }r(t)=0$, this proves that it is the unique pulse maximum. As another numerical check, at $t=10$ seconds $x\simeq 0.999955$, $a\simeq 0.864665$ and $r\simeq 0.135290$: the slow branch is still catching up.

The subtraction $r=x-a$ is a mathematical readout, not a claim that an actual chemical concentration equals this difference for arbitrary inputs. Nonnegativity was established only for the specified upward step; other input histories can give a negative signed readout. Complete signaling adaptation here also relies on equal eventual branch gains: changing the readout to $r=x-0.9a$ leaves $r_{\ast }=0.1$ for $s=1$. This toy illustrates cancellation and different time constants; it does not establish molecular implementation or robustness to arbitrary parameter changes. Ma and colleagues' 2009 study separately analyzed biochemical mathematical models with negative feedback and incoherent feedforward loops, identifying additional conditions for signaling adaptation in those models. ^[ref]

A historical connection to feedback control comes from Yi, Huang, Simon and Doyle's 2000 analysis of a bacterial chemotaxis model. They identified integral action in its regulation of receptor activity: accumulated deviations alter a slower adjustment, which in turn changes the measured activity. At a stationary state, an adjustment that integrates a nonzero error cannot remain constant, explaining how zero error can follow. Their conclusion depended on stated biochemical assumptions and stable behavior; it was not a claim that every biological cell uses that mechanism. A transient pulse alone, as the incoherent feedforward loop toy shows, cannot identify the underlying signaling feedback architecture. ^[ref]

For $r=-\lambda >0$ and vanishing solid-to-liquid diffusivity ratio, $H(r)=\sqrt{\pi }r{e}^{r^2}\mathrm{erfc}(-r)$ gives the interfacial liquid concentration. It increases from zero to $C_s$ as $r$ increases. Its derivative is $C_s{H}^{\prime }/(1+H{)}^2$, which decreases from $\sqrt{\pi }{C}_s$ to zero. For a direct proof, write $A=H/r=\sqrt{\pi }{e}^{r^2}\mathrm{erfc}(-r)$. Then $A^{\prime }=2rA+2$, so $A\ge \sqrt{\pi }+2r$. Differentiation gives $2H^{\prime 2}-(1+H)H^{\prime \prime }=2[(1+r^2+2r^4)A^2+r(4r^2-1)A+2r^2-2]>0$: the bracket increases with $A$, and substitution of its lower bound gives only positive polynomial coefficients, including constant $\pi -2$. Thus the concentration is strictly concave as a function of melting speed parameter $r$. Combining this with a linearized thermal balance produces one fold on the melting branch and, with the freezing branch, an interval of three possible interface speeds.

Start at sodium electrochemical equilibrium with $n_0$ internal ions. Exporting $n$ ions changes the membrane potential by $-ne/C$ and the internal number density to $(n_0-n)/V$. Subtracting the initially balanced chemical and electrical contributions gives the displayed incremental cost. An initial zero potential further implies $n_0=c_{o}V$ for an ideal fixed external reservoir. Nonzero initial equilibrium potential permits unequal concentrations, so that latter equality is an additional assumption.

Define the displayed quantity as the free energy available per unit positive charge for sodium entry, with $V_m={\psi }_i-{\psi }_o$. Its negative is the signed inward electrochemical cost per charge. A reversal of the transport direction reverses the sign. At equilibrium it vanishes without requiring equal concentrations.