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With carbon-ion thermal energy $U=M{k}_B{T}_c/(8m_u)$ and luminosity $L=A(M/M_{\odot })T_c^{7/2}$, the time between two core temperatures is $k_B{M}_{\odot }(T_f^{-5/2}-T_i^{-5/2})/(20m_{u}A)$. Late cooling is slow in this approximation. Extrapolating the formula does not prove a categorical observational absence of ultracool remnants, since atmosphere, crystallization and heat-capacity physics alter the simple model.

An ideal solution is a chemical solution whose component chemical potentials depend on composition throughHere $n_i$ is an amount of substance, $x_i$ its mole fraction, $R$ the molar gas constant, and ${\mu }_i^{\ast }$ the pure-component liquid reference at the same temperature $T$ and pressure $P$. This model applies to every component across the composition range; it is stronger than an approximation valid only near a pure solvent.^[ref]

With this pure-liquid standard state convention, the thermodynamic activity satisfies ${\mu }_i={\mu }_i^{\ast }+RT\ln a_i$, so an ideal solution has $a_i=x_i$, or activity coefficient ${\gamma }_i=a_i/x_i=1$.^[ref] This statement concerns the liquid. Assuming that a coexisting vapor is an ideal gas mixture is an additional approximation. Its mole fractions $y_i$ generally differ from the liquid mole fractions $x_i$.

For an original two-component example, stipulate $T=300\mathrm{K}$ and pure-component vapor pressures $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. A and B are invented distinguishable components; these numbers are chosen model inputs, not material data. Combine an ideal solution with an ideal gas vapor, neglect the pure-liquid reference's pressure correction, and exclude other gases. At liquid mole fractions $x_A=0.4$ and $x_B=0.6$, Raoult's law and addition of partial pressures giveThus the coexisting vapor contains about $72.73\%$ A by amount of substance, although the liquid contains only $40\%$. The total pressure is neither pure-component vapor pressure, and $x_A$ cannot be substituted for $y_A$. This enrichment is the basis of distillation.

The usefulness of the ideal solution is that one composition law connects chemical potentials, vapor-liquid equilibrium, entropy of mixing and Gibbs free energy of mixing. Its liquid components can interact strongly: the defining condition is their composition dependence, rather than the absence of all interactions as in an ideal gas model.

The Mixture measurements and the Gibbs-Duhem constraint article describes the inspected acetone–chloroform preparation, apparatus, paired readouts, controls and sampling limits.^[ref]

That existing apparatus account supplies the experimental controls and limitations. Its refractive index calibration yields mass fractions, which must be converted to mole fractions. For a measured component-A mass fraction $w_A$ and component molar masses $M_A,M_B$,Apply this conversion separately to the flask sample and to completely condensed vapor to obtain $x_A$ and $y_A$. Comparing $y_{i}P$ with $x_i{p}_i^{\ast }(T)$ tests Raoult's law at the measured temperature; using the changing temperature at each composition is essential. The chosen A/B values above are unrelated to the protocol's substances.

The more general diagnostic, retaining an ideal gas vapor and negligible pure-liquid pressure correction, isThis follows by replacing $\ln x_i$ with $\ln (x_i{\gamma }_i)$ in the chemical potential subtraction used for Raoult's law. A reproducible discrepancy from ${\gamma }_i=1$ can expose failure of the ideal solution approximation, but interpreting it requires checking measurement uncertainty, representative paired sampling and the ideal gas approximation. Apparent agreement within uncertainty supports an approximation over the tested conditions; it cannot establish exact ideality at every composition.

Positive or negative deviations of component partial pressures require activity coefficients different from one under these assumptions. An azeotrope is also absent from the chosen unequal-vapor pressure model: the explicit inequality in ideal binary vapor composition gives $y_A>x_A$ throughout $0<x_A<1$. An interior equality would require $p_A^{\ast }=p_B^{\ast }$; in that degenerate case every composition has $y_A=x_A$, rather than an isolated enrichment reversal. Neither an azeotrope nor liquid phase separation should be inferred from the ideal curves.

Consider a neutral monoprotic acid $\mathrm{HA}$ dissolved in water, with no added salts or other acids. Assume dilute ideal thermodynamic activities, fixed temperature, water thermodynamic activity one, and negligible self-ionization of water. Write $C_T$ for its analytical concentration, $c=C_T/c^{\circ }$ with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$, and $\alpha$ for its weak acid dissociation fraction. Conservation of acid units and electric charge givesHere ${\mathrm{A}}^{-}$ is the conjugate base, and ${\mathrm{H}}^{+}$ is conventional shorthand for hydrated hydrogen ions, represented by hydronium in this model. Substitution into the dimensionless acid dissociation constant givesSolving this quadratic equation and rejecting its negative root of a polynomial gives an exact result within the stated model:The second form avoids subtracting nearly equal numbers. Unlike the approximation $\alpha \approx \sqrt{K_a/c}$, it retains depletion of neutral weak acid and always satisfies $0<\alpha <1$ for positive $K_a,c$. The broader weak acid mass and charge balances restore self-ionization of water when that omitted contribution matters.

For an original worked example, stipulate $K_a=10^{-5}$ without identifying a real substance. At $C_T=10^{-2}\mathrm{mol}{\mathrm{L}}^{-1}$, the weak acid dissociation fraction is approximately $0.0311267$, or $3.11267\%$, and $[{\mathrm{H}}_3{\mathrm{O}}^{+}]=3.11267\times 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$. A hundredfold dilution to $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ raises the fraction to $0.270156$, or $27.0156\%$, but lowers $[{\mathrm{H}}_3{\mathrm{O}}^{+}]$ to $2.70156\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$. A completely ionized strong acid at $C_T=10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$ would have $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\approx 10^{-4}\mathrm{mol}{\mathrm{L}}^{-1}$: less than the more concentrated weak acid in the first calculation, despite its greater converted fraction.

The directions of both changes follow without relying on the plot. Put $x=[{\mathrm{H}}_3{\mathrm{O}}^{+}]/c^{\circ }=c\alpha$. Rearranging the chemical equilibrium relation givesThus smaller $c$ forces smaller $x$, which in turn makes $\alpha$ larger. The derivative proves that total hydronium concentration and converted fraction move in opposite directions during dilution. In this model, adding solvent also makes additional acid units ionize at fixed total acid amount of substance, even though their ions are spread over more volume. That distinction explains why “more ionized” alone does not predict greater electrical conductivity.

Do not extend the no-background model to unlimited dilution: its prediction $[{\mathrm{H}}_3{\mathrm{O}}^{+}]\to 0$ omits water's own ions. The weak acid mass and charge balances include that background. Added salts, other acids or bases, and nonideal activity coefficients also change the relation between analytical concentration, converted fraction and hydronium concentration.

Facing a companion and facing the source of light are different relations. The Moon keeps its near side of the Moon approximately toward Earth, while the direction to the distant Sun is nearly fixed during the idealized orbit shown below. A fixed surface patch therefore turns toward, across and away from sunlight. The far side of the Moon has day and night too; its name does not mean permanent darkness. ^[ref].

If the Sun direction has fixed angle $\alpha$ and the marked surface patch has outward normal vector at angle $\theta$, its direct illumination is proportional to $\max (0,\cos (\theta -\alpha ))$. With synchronous rotation, $\theta =f+\pi$, so the patch goes through day and night even though it always faces Earth. In reality the Earth-Moon system advances around the Sun: the lunar spin and orbital cycle relative to distant stars is about $27.3$ Earth days, while the lunar phase cycle has a synodic period of about $29.5$ Earth days. These numbers describe the actual system, not the fixed-sunlight cartoon. ^[ref].

For a synchronously rotating exoplanet whose orbital companion is itself the illuminating star, the direction to the star turns with the orbit. On an ideal circular orbit, with perpendicular spin axis and no additional oscillations, this direction is fixed relative to the surface, giving persistent day and night hemispheres. Day-night heat redistribution then helps determine the nightside temperature; permanent lack of direct starlight does not imply zero temperature. Orbital eccentricity and departures from this ideal spin geometry modify that simple picture.

An electric lamp whose current heats a filament until it emits visible thermal radiation. Joule heating converts electrical input into thermal energy and radiation. Its electrical resistance changes with temperature, so one fixed resistor value need not describe different operating voltages.

Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

Consider an explicitly stipulated dilute independent spin-one-half two-level system with zero applied magnetic field. Each constituent has $J_z=\pm \hslash /2$ and $m_z=\gamma J_z$, with constant signed gyromagnetic ratio $\gamma$. Ignore interactions and the specimen's own magnetic field. At positive temperature $T$, thermal equilibrium with the rotating lattice gives energy levels $E_{\pm }=\mp \hslash \Omega /2$. Define the dimensionless variable $x=\hslash \Omega /(2k_{B}T)$, where $\hslash$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. The partition function and Boltzmann distribution giveTaking the expected value giveswhere $n$ is number density. This is the rotational version of the independent-moment magnetic response. The angular momentum aligns with positive rotation even when the magnetic moment points oppositely. The probability ratio $p_{+}/p_{-}=e^{2x}$ provides a direct population check of the sign.

For $|x|\ll 1$, expand the hyperbolic tangent to obtainThe slope against $B_{\Omega }$ is positive for either sign of $\gamma$, while the slope against $\Omega$ has the sign of $\gamma$. At fixed $T>0$, reversing $\Omega$ reverses both means, and $\Omega =0$ gives zero. At fixed nonzero $\Omega$, the formal $T\to 0^{+}$ limit saturates $|⟨J_z⟩|$ at $\hslash /2$; the linear approximation must not be extrapolated there. At $T\to \infty$, the response vanishes. These limits describe the toy canonical ensemble, whose equilibration is assumed.

For a numerical example, stipulate $n=10^{28}{\mathrm{m}}^{-3}$, $T=300\mathrm{K}$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and rotation frequency $f=1000\mathrm{Hz}$. With $\Omega =2\pi f$, $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}J/K$, the model givesThese are teaching inputs and calculated outputs, not measured material values. The corresponding internal angular momentum density is $M_z/\gamma =+4.22\times 10^{-17}\mathrm{J}s/{\mathrm{m}}^3$. Keeping a rotor at prescribed $\Omega$ while its internal angular momentum changes requires the drive to supply the appropriate angular impulse. In an isolated body the mechanical part instead changes oppositely, as in magnetomechanical angular momentum balance. There is no creation of net angular momentum by choosing a rotating frame.

For $N$ distinguishable noninteracting spins, let a fraction $\alpha$ occupy the level $+ϵ$ and the remainder occupy $-ϵ$. The Stirling formula givesThe relation $1/T=\partial S/\partial E$ yieldsPositive temperature corresponds to $0\le \alpha <1/2$; a population inversion $1/2<\alpha \le 1$ has negative temperature.

The helium-4 characteristic-property table gives a normal boiling point of 4.224 K and a critical temperature of 5.2 K.

Indium antimonide is a narrow-gap semiconductor formed from indium and antimony.

A reversible heat pump maintaining $T_h$ against an external $T_c<T_h$ has heating coefficient of performance $T_h/(T_h-T_c)$. If its maximum work input is ${\dot{W}}_{\max }$ and the building loses heat at rate $\lambda (T_h-T_c)$, maintaining equilibrium requires $\lambda (T_h-T_c{)}^2\le {\dot{W}}_{\max }{T}_h$. This yields the displayed minimum external temperature, provided it lies in the physical positive-temperature range. If the rating instead specifies maximum delivered heat, the bound is $T_h-T_c\le {\dot{Q}}_{h,\max }/\lambda$; a heat-output rating cannot be substituted for a work-input rating in the Carnot formula.

In thermal-time units, taking the Prandtl number large removes momentum inertia at leading order. The Stokes flow momentum equation then determines velocity instantaneously from the temperature perturbation, while the nonlinear heat equation retains its time derivative. Small Prandtl number does not generally give this model. An exact contrasting shear in a free-slip layer is $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$: it decays slowly as the Prandtl number tends to zero but is excluded by the instantaneous Stokes equation.

Infrared is electromagnetic radiation with wavelengths longer than the red end of ordinarily visible light. A useful named convention is the CIE range $780\mathrm{nm}$ to $1\mathrm{mm}$; the approximate visible range $400$–$700\mathrm{nm}$ used in some teaching texts puts the boundary nearer $700\mathrm{nm}$. ^[ref]. Neither number is a sharp physical cutoff of human vision. The CIE divisions IR-A, IR-B and IR-C are respectively $0.78$–$1.4\mu \mathrm{m}$, $1.4$–$3\mu \mathrm{m}$ and $3$–$1000\mu \mathrm{m}$; the boundaries of near-, middle- and far-infrared vary between applications. ^[ref].

Use vacuum wavelength $\lambda$ when comparing these ranges. With frequency $\nu$, speed of light $c$, Planck constant $h$ and photon energy $E$,Thus the CIE infrared interval corresponds approximately to $3.00\times 10^{11}$–$3.84\times 10^{14}\mathrm{Hz}$ and $1.99\times 10^{-22}$–$2.55\times 10^{-19}\mathrm{J}$ per photon, with the smaller energy at the longer wavelength. A $940\mathrm{nm}$ infrared photon has $E\simeq 2.11\times 10^{-19}\mathrm{J}$, whereas a $254\mathrm{nm}$ ultraviolet photon has $E\simeq 7.82\times 10^{-19}\mathrm{J}$. These are calculations from $hc/\lambda$, not measurements of a particular source. A television remote is a familiar approximately $940\mathrm{nm}$ source; its information is carried by variations in the emitted radiant flux. ^[ref].

Detection and visual perception answer different questions. A photodetector can respond to infrared even when an observer perceives no light; a thermometer detects an increase in temperature after photon absorption. Conversely, a small readout can reflect a small spectral responsivity, rather than absence of incident electromagnetic radiation. Images made from infrared readouts use display colors chosen by the imaging system; those colors are not the colors of the incoming infrared.

The important microscopic mechanisms also differ. In infrared spectroscopy, photon absorption often increases the energy of a normal mode in which chemical bonds stretch or bend. The corresponding optical spectrum helps distinguish molecules by their motions. For the usual electric dipole mechanism, the motion must change the electric dipole moment; matching the frequency alone is not sufficient. ^[ref]. Electronic processes can also emit or absorb infrared, so it is not exclusively radiation from molecular motion. ^[ref]; ^[ref].

Infrared is especially useful for observing thermal radiation from objects near everyday temperatures. For an ideal $300\mathrm{K}$ blackbody, Wien's displacement law gives a wavelength peak near $9.66\mu \mathrm{m}$. A thermal camera's readout nevertheless depends on emissivity, reflected surroundings and transmission along the path, as well as temperature. ^[ref]. A source emitting infrared need not be hot: an infrared remote is useful precisely because a photodetector can distinguish its signal from a background. Visible light and ultraviolet can also cause heating when absorbed; “infrared” names a wavelength range, not a unique form of heat.

Herschel's 1800 discovery made the distinction between sight and detection experimentally concrete. His report describes a optical prism spreading sunlight across a white, ruled receiving surface, with three thermometers at different positions near the end of the visible optical spectrum. He observed heating beyond the red edge where no visible color appeared. The result established that sunlight contained invisible radiation less strongly refracted by that optical prism than the visible red; it did not measure a photon energy or establish today's infrared boundaries. ^[ref]; ^[ref].

A documented modern version by Caltech/IPAC uses a glass optical prism, white paper in a shaded box and three blackened liquid-filled thermometers. Initial readings in shade provide each thermometer's baseline; subsequent readings compare blue, yellow and just beyond red after the bulbs have had time to warm. The measured quantity is the rise in temperature, not a “temperature of a color”. The lesson reports the expected largest rise beyond visible red and specifically discusses matching the bulb coatings and thermometer calibration. Clouds and changes in the projected optical spectrum affect the comparison. ^[ref].

For a stronger modern interpretation, an additional shaded thermometer would monitor ambient drift, exchanging bulbs between positions would test differences between thermometers, and blocking the incident beam would distinguish beam-dependent heating from warmed supports or air. These are suggested controls, not claims about Herschel's historical procedure. Thermal conduction through the mounting and convection can change the readout. A broad bulb also samples a range of wavelengths; the optical prism's transmission and optical dispersion, the bulb's absorption and heat loss all affect the position of the largest rise. In particular, the result does not show that an infrared photon carries more energy than a visible photon.

A simple analytical illustration makes those limitations explicit. Let $\theta =T-T_{\mathrm{amb}}$ be the bulb's excess temperature, $C$ its heat capacity, $G$ its effective heat-loss coefficient in $\mathrm{W}{\mathrm{K}}^{-1}$, and $P_{\mathrm{abs}}$ the absorbed power. Assume one uniform bulb temperature, constant ambient temperature, constant $C,G$, and losses linear in $\theta$. Thenfor a beam switched on at $t=0$ with $\theta (0)=0$. After switching the beam off at $t=t_s$, $\theta (t)=\theta (t_s)e^{-(t-t_s)/\tau }$. The exponential function follows by solving this linear differential equation. The steady excess temperature measures $P_{\mathrm{abs}}/G$; it is not a direct measurement of the optical spectrum. With illustrative $C=0.020\mathrm{J}{\mathrm{K}}^{-1}$ and $G=0.0010\mathrm{W}{\mathrm{K}}^{-1}$, $\tau =20\mathrm{s}$. Absorbed powers of $1.0$ and $0.5\mathrm{mW}$ give steady rises of $1.0$ and $0.5\mathrm{K}$, regardless of which wavelengths supplied that absorbed power.

The upper schematic represents the logic of the optical prism experiment; its light ray positions are not a calibrated wavelength scale. The lower curves use the illustrative heat-balance model, with the beam blocked at $60\mathrm{s}$; they are not historical or modern measured data. The beauty of the experiment is that a simple change of detector extends observation beyond what the eye can perceive.

The derivative of outgoing spectral radiance with respect to a pressure-resolved absorber abundance defines a kernel once the vertical measure is fixed. In an optically thin column with $k\propto p$ and $B/B_0=p/p_0$, $K_p=[k_0{B}_0/(p_{0}gM)]p(p/p_0-1)$. It is negative inside the atmosphere and zero at both endpoints. Its largest magnitude lies at $p_0/2$, whereas the logarithmic-pressure kernel $p{K}_p$ peaks in magnitude at $2p_0/3$. A uniformly perturbed abundance instead has one integrated derivative, not a pressure location.

A preparation obtained by allowing medicinal material to stand in a solvent for a specified time, at the prescribed temperature, then straining or filtering.

On $0<x<L$, $t>0$, an initial condition $u(x,0)=g(x)$ gives the starting profile. For the heat equation, Dirichlet boundary conditions $u(0,t)=a(t)$, $u(L,t)=b(t)$ can model ends held at prescribed temperatures. Alternatively, zero Neumann boundary conditions describe insulated ends in the constant-conductivity model. These choices can produce different evolutions from the same initial condition.

For a continuous classical solution reaching the corners, the Dirichlet boundary data must satisfy $g(0)=a(0)$ and $g(L)=b(0)$; higher smoothness can require further compatibility. In dimensionless variables, a complete worked heat equation problem isIts exact classical solution is $u(x,t)=x+e^{-{\pi }^{2}t}\sin (\pi x)$. The straight profile $x$ is the equilibrium; the extra bump decays. Substituting verifies the equation and every datum, and the energy estimate proves uniqueness among smooth solutions with these data.

For the transport equation with $c>0$ on the same interval, prescribe an inflow transport boundary condition only at $x=0$. Backtracking a characteristic curve gives $g(x-ct)$ if $x\ge ct$, and $a(t-x/c)$ otherwise. Values at $x=L$ are determined by this formula; independently prescribing them can make the problem inconsistent.^[ref]

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

Instantaneous interface kinetics is the limit of negligible kinetic undercooling at a moving solidification interface. Its temperature is set by local liquidus equilibrium and curvature-induced melting-temperature depression, without a further velocity-dependent attachment term. It does not imply that diffusion or heat transport is instantaneous.