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In the frame of a planar interface moving at speed $V$, the steady liquid temperature obeys $-V{T}_z=\kappa T_{zz}$ and approaches $T_{\infty }$. ThusThe Stefan condition, with negligible solid-side heat flux, givesso $T_i-T_{\infty }=L/c_p=S\Delta T$. The kinetic undercooling law then giveswhich requires $S<1$ for advancing solidification.

Scale length with $\kappa /V$, time with $\kappa /V^2$, and temperature with $\Delta T$. The base liquid temperature is ${\Theta }_0=S{e}^{-z}$. For an interface displacement $\zeta e^{i\alpha x+\sigma t}$, write the thermal perturbation asThe heat equation givesLinearizing the Stefan condition giveswhile the kinetic law with stabilizing curvature undercooling givesEliminating $A$ gives the general implicit dispersion relation

At marginal stability $\sigma =0$,and the dispersion relation becomesSince ${\alpha }^2=\lambda (\lambda -1)$ on this curve, the right-hand side is simply $\Gamma {\lambda }^2$. A positive cutoff wavenumber therefore exists exactly whenThe unstable band is $0<\alpha <{\alpha }_c$, whereA sketch of $\Gamma {\lambda }^2$ against the constant $S/(1-S)$ shows no crossing above $\alpha =0$ below threshold and one cutoff above it. As $S\uparrow 1$, ${\lambda }_c$ and ${\alpha }_c$ diverge like $(1-S{)}^{-1/2}$, so the range of the morphological instability of a kinetically limited solidification front becomes unbounded.

Take the Fourier transform in $x$. Since $\widetilde{{\theta }_{xx}}=-k^2\tilde{\theta }$, the transformed heat equation isThereforeThe given transform pair and the convolution theorem yield the heat-kernel solution

When the heat equation is viewed as a second-order equation, its principal symbol contains only the spatial second derivatives:up to an irrelevant sign. It vanishes on $dt$, so the initial plane is characteristic in this second-order sense. The equation remains a well-posed first-order evolution equation in time; these are different notions of order.

Fix $t>0$ and define $w(s,y)=u(t-s,|y|)$ for $0\le s\le t$. The assumed smooth extension and the Neumann boundary condition at zero make $w$ a $C^{1,2}$ function. The heat equation givesThe Itô formula therefore makes $w(s,B_s)$ a local martingale. Stop first when $|B|$ leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on $[0,t]$ give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.

Let $\rho =t\wedge {\tau }_0\wedge {\tau }_1$ and apply the Itô formula to $u(t-s,B_s)$ for $0\le s\le \rho$. The heat equation cancels the drift, so the stopped process is a bounded martingale. The optional sampling theorem for a supermartingale gives $u(t,x)={\mathbb{E}}_x[u(t-\rho ,B_{\rho })]$. On the three mutually exclusive terminal events, the initial and Dirichlet boundary conditions identify this value asThis is the probabilistic representation of the heat equation with time-dependent Dirichlet data.

Let $X=x/ϵ$, $a(X)=1/d(X)$, and seekwith periodic correctors. The leading cell equation makes $u_0$ independent of $X$. At the next order, continuity of microscopic heat flux givesAveraging over one period and using $⟨u_{1X}⟩=0$ yieldsThus the periodic homogenization of a diffusion equation has effective diffusivity $1/\overline{d}$:

For the sawtooth profile, the two triangular areas giveThe homogenized problem is therefore the unit-diffusivity heat equation. Its half-line step solution isIt has the required initial and boundary limits, and direct differentiation verifies $u_t=u_{xx}$.

Take $z$ vertically downward from the cold surface. The ice occupies $0<z<h(t)$ and the water occupies $z>h(t)$. The temperature satisfies the heat equationin both phases, withThe remaining interfacial condition is the Stefan conditionA schematic should show the cold surface, the descending phase boundary, the ice temperature rising from $T_b$ to $T_m$, and the water temperature tending from $T_m$ to $T_{\infty }$.

Conservation of thermal energy and Fourier's law give the heat equationsThe density times the specific heat capacity, ${\rho }_i{c}_i$, is the volumetric heat capacity, and $k_i$ is the thermal conductivity.

The condition $T_1(-L,t)=0$ holds the left exterior face at a fixed zero reference temperature; $T_2(L,t)=0$ does the same for the right exterior face. The condition $T_1(0_{-},t)=T_2(0_{+},t)$ expresses perfect thermal contact, with no temperature jump at the interface. Finally,expresses continuity of the signed heat flux in the positive $x$ direction: the interface neither stores nor produces heat.

For $\omega \ne 0$, substitution of the separated solutions into the heat equations givesThe zero exterior boundary values are incorporated byTo obtain a eigenvalue equation for heat conduction in two joined slabs, write $u=\omega {\alpha }_{1}L$ and $v=\omega {\alpha }_{2}L$. Continuity of temperature and heat flux at zero gives the homogeneous linear systemA nonzero amplitude pair exists precisely when its determinant vanishes, equivalentlyIf $\cos u\cos v\ne 0$, division by $k_1{k}_2{\alpha }_1{\alpha }_2\cos u\cos v$ gives the printed transcendental equationThe printed necessity claim needs the qualification that both tangents are defined. If just one of $\cos u,\cos v$ vanishes, $D(\omega )\ne 0$. If both cosines vanish, however, $D(\omega )=0$ and legitimate normal modes remain. Their temperature at the thermal interface need not be zero; the temperature condition fixes the amplitude ratio, while each heat flux is zero there.

For a concrete counterexample, take identical slabs: ${\rho }_1={\rho }_2=\rho$, $c_1=c_2=c$, $k_1=k_2=k$, and $\alpha =\sqrt{\rho c/k}$. Thenwith each formula restricted to its own slab, satisfies both heat equations and all four boundary conditions. But $\omega \alpha L=\pi /2$, so neither tangent is defined. This directly disproves the unqualified printed assertion. The determinant condition retains these missing normal modes.

Finally, $\omega =0$ must be checked before dividing by it. Then $G_1=A(x+L)$, $G_2=B(L-x)$, and interface continuity gives $A=B$, while heat flux continuity gives $k_{1}A=-k_{2}B$. Positivity of the thermal conductivities forces $A=B=0$. Thus $\boxed{\omega =0\text{ gives only the trivial solution}.}$ In the determinant formula, its apparent zero at $\omega =0$ is spurious because the chosen sine basis degenerates there.

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.

A steady state has $T_s^{\prime \prime }=0$, so $T_s=a+bx$. The fixed temperature gives $T_s(0)=T_0$, and thermal insulation gives $T_s^{\prime }(L)=0$. ThusSubtracting this steady state produces the homogeneous mixed boundary conditionsThe partial differential equation and boundary conditions apply for $t>0$. The initial condition is attained as $t\downarrow 0$. Its slope at $x=L$ is $-T_0/L$, so it does not satisfy the insulated-end condition at the instant $t=0$. The resulting heat equation solution is smooth for $t>0$ and approaches the initial temperature continuously; its spatial derivative need not be continuous at $(L,0)$.

For any fixed subregion $\Omega$, with no internal heat generation, conservation of energy says that the rate of stored thermal energy equals the incoming heat flux:The minus sign is needed because $\mathbf{q}\cdot \mathbf{n}>0$ denotes outgoing heat flux. Substitute Fourier's law $\mathbf{q}=-k\nabla T$ and apply the divergence theorem:Here differentiation under the integral uses a fixed region and sufficient regularity of $T$. Since the identity holds for every $\Omega$, continuity of the integrands gives the local heat equationFor spatially varying thermal conductivity, the product rule gives $\nabla \cdot (k\nabla T)=k{\nabla }^{2}T+\nabla k\cdot \nabla T$; the second term cannot generally be discarded.

Let $A$ be the constant cross-sectional area, and suppose that the temperature is uniform on each cross-section. Consider a slice $[a,b]$. Its stored thermal energy is $\rho cA{\int }_a^{b}T(x,t)dx$. Since the lateral surface is insulated and there is no internal heat generation, conservation of energy giveswhere $q$ is the heat flux in the positive $x$ direction. By Fourier's law, $q=-k{T}_x$, soThe slice is arbitrary; therefore the local heat equation isThe constant $\kappa$ is the thermal diffusivity. The cancellation of $A$ uses a rod of uniform cross-section; the one-dimensional model assumes no transverse temperature variation.

Take a bar of constant cross-sectional area $A$, thermally insulated along its sides, with no internal heat production. Fourier's law says that the heat flux in the positive $x$ direction isThe thermal conductivity $k(x)$ is positive. On a small slice $[x,x+\Delta x]$, conservation of energy givesHere $\rho c_p$ is the volumetric heat capacity, the product of mass density and specific heat capacity. Divide by $A\Delta x$, take $\Delta x\to 0$, and use $\rho c_p=1$. The resulting heat equation isFor differentiable $k$, this is $T_t=k{T}_{xx}+k^{\prime }{T}_x$; retaining the divergence form also expresses conservation of energy correctly across an interface where $k$ jumps.

Write $T=T_s+T_u$. The transient satisfies the heat equation and homogeneous mixed boundary conditionsThe left condition is a Neumann boundary condition and the right one a Dirichlet boundary condition. Separation of variables, $T_u=X(x)G(t)$, givesThe condition $X^{\prime }(0)=0$ removes the sine term. The condition $X(2\ell )=0$ then selectsThere is no nonzero mode with $\mu =0$, because a linear $X$ satisfying both conditions is zero. There is no growing mode either: integration by parts gives ${\int }_0^{2\ell }(X^{\prime }{)}^{2}dx={\mu }^2{\int }_0^{2\ell }{X}^{2}dx$, so every nonzero mode has ${\mu }^2>0$.

These Sturm-Liouville eigenfunctions have orthogonality and squared normExpand the initial transient in this half-integer Fourier cosine series. Its coefficients areThe integration formula follows by integration by parts; the cosine at the endpoint is zero. Thus the half-integer cosine expansion of a linear profile givesAt $t=0$ the Fourier series reproduces $-T_s$ everywhere, including the endpoints; its $n^{-2}$ coefficient decay gives uniform convergence. For every $t>0$ the exponential factors allow termwise differentiation and verify the heat equation and both boundary conditions. The sudden switch-on of nonzero heat flux is incompatible with the spatial derivative of the constant initial temperature at the corner $(0,0)$. Accordingly, the positive-time boundary derivative must not be inferred by differentiating the initial Fourier series term by term at that corner.

Inward heat flux at the left end points in the positive $x$ direction, so the boundary conditions areFor uniform thermal conductivity, the steady heat equation gives $T_s^{\prime \prime }=0$. The heat flux is constant and equal to $Q$, so $T_s^{\prime }=-Q/k$. Imposing the right-end temperature gives

In a steady state the heat equation says $(k{T}_s^{\prime }{)}^{\prime }=0$, so the positive-$x$ heat flux $q=-k{T}_s^{\prime }$ is the same constant $Q$ in both materials. Assume perfect thermal contact: temperature is continuous at $x=\ell$, and conservation of energy requires continuous heat flux there. Integrating from the right endpoint givesThe two formulas agree at the interface. Their slopes differ, but $-k_1{T}_s^{\prime }({\ell }^{-})=-k_2{T}_s^{\prime }({\ell }^{+})=Q$. The total temperature drop is $Q\ell (1/k_1+1/k_2)$, the sum of the two layers' contributions in steady heat conduction through layers.

Put $\tau =t+b$ and $q=x+a$. For $\theta =f(t)e^{-q^2/(4\tau )}$, partial differentiation givesThe heat equation therefore requires $f^{\prime }=-f/(2\tau )$, giving $f=C/\sqrt{|t+b|}$ on an interval not containing $-b$. A decaying spatial Gaussian function uses $t+b>0$, so $f=C/\sqrt{t+b}$ there.

The initial profile is a difference of two translated Gaussian functions. Set $b=1/4$, $a=-2$ and $a=2$, with respective constants $C=1/2$ and $C=-1/2$. The superposition principle givesThis is an odd pair of Gaussian profiles under diffusion, obtained from Gaussian initial data for the heat equation. It is odd in $x$, vanishes at zero, is positive for $x>0$ and negative for $x<0$. The two lobes broaden and decrease in height, and $\theta (x,t)\to 0$ at every fixed $x$ as $t\to \infty$. Their extrema move; they do not remain exactly at $\pm 2$.

The last two terms in the preceding result equal $v_s$. ThereforeThe left-hand side of the original partial differential equation is consequently $v_{tt}-v_s$, givingIn these variables it has the form of a heat equation, with logarithmic radial coordinate $s$ playing the evolution variable and angular coordinate $t$ playing the spatial variable. If $u$ is single-valued around the whole punctured plane, then $v(s,t+2\pi )=v(s,t)$ as well; that periodicity is a global coordinate condition, not an extra local transformation term.

For a nonzero separated solution $U=X(x)T(t)$, separation of variables gives $X^{\prime \prime }/X=T^{\prime }/T=-\lambda$. The zero Dirichlet boundary conditions force $X=\sin nx$, $\lambda =n^2$, with integers $n\ge 1$; the time factors are $e^{-n^{2}t}$. Expanding the initial profile in this Fourier sine series requires only the two already displayed spatial modes. ThusBoth terms individually solve the heat equation, vanish at the endpoints and give the required initial profile at $t=0$. The initial function is also $4{\sin }^{3}x\ge 0$ on the bar; the later solution stays nonnegative and tends to zero as heat leaves through the fixed-temperature ends.