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For constant diffusivity, the trial Rayleigh quotient simplifies toIts derivative is $1-2/q^2$, so its only positive critical point is $q=\sqrt{2}$. The second derivative is $4/q^3>0$; also the quotient diverges as $q\downarrow 0$ or $q\to \infty$. Therefore this is the global optimum within the trial family:The two requested profiles areThe initial radial graph decreases from one at the centre to zero at the rim, has horizontal tangent at the centre, and is concave downward for $0<r<1$. The central temperature decreases exponentially from one to zero, with initial time derivative $-(3+2\sqrt{2})$. These are trial profiles, not exact solutions of the full heat equation. Although the radial trial function is not twice differentiable at the centre, it has finite weighted energy and belongs to the variational trial class used by the Rayleigh-Ritz variational principle.

Write a nonzero separated solution as $T=X(x)Y(y)S(t)$. Division of the heat equation by this product givesThe two spatial factors solve independent constant-coefficient eigenvalue problems. Zero endpoint values in $x$ requireThe insulating conditions requireThe zero cosine mode is the nonzero constant function and must be retained; there is no nonzero zero-index sine mode. The temporal equation is $S^{\prime }=-\lambda {\pi }^2(n^2/a^2+m^2/b^2)S$. Superposing the complete sine-cosine basis yieldsThe constants are fixed by expansion of the initial profile. Orthogonality gives sine norm $a/2$; the cosine norm is $b$ at $m=0$ and $b/2$ otherwise. This is the heat equation on a rectangle with two insulated sides.

Seek a nonzero separated solution $\psi =T(t)R(r)\Theta (\theta )$. Substitution into the heat equation and division by the product givesThe left side depends only on time, so set it to $-k^2$. Multiplication by $r^2$ then separates the angular dependence, with separation constant $\ell (\ell +1)$. The three ordinary differential equations areThe temporal solution is $T=T_0{e}^{-\kappa k^{2}t}$. With $q=\cos \theta$, the angular equation is the Legendre differential equation; regularity at both angular endpoints selects $\ell =0,1,2,\dots$ and $\Theta \propto P_{\ell }(q)$, the Legendre polynomial. The radial equation has the Spherical Bessel functions $j_{\ell }(kr)$ and $y_{\ell }(kr)$ for $k>0$. Regularity at $r=0$, if imposed, excludes the singular second solution. For $k=0$ the radial powers are $r^{\ell }$ and $r^{-\ell -1}$. Without specified radial boundary conditions, no discrete set of $k$ is selected; $-k^2$ is a convention for the temporal separation constant.

For spherical symmetry the angular mode is $\ell =0$. Put $u(t,r)=r\psi (t,r)$. Direct differentiation givesso the radial heat equation in three dimensions becomes the ordinary one-dimensional heat equation $u_t=\kappa u_{rr}$. Since $r_{\pm }>0$, the zero endpoint temperatures become $u(t,r_{\pm })=0$.

Write $L=r_{+}-r_{-}$. The two initial sine modes already satisfy these Dirichlet boundary conditions; their squared wave numbers are $(\pi /L{)}^2$ and $(2\pi /L{)}^2$. Each therefore evolves independently with its corresponding exponential decay. Dividing back by $r$ givesAt $t=0$ the initial profile is recovered, and each sine is zero at both endpoints. The first mode decays four times more slowly in its exponential rate than the second. This is the two-mode instance of the spherical-shell heat equation via radial rescaling; uniqueness follows, for example, by applying the heat equation energy identity to the difference of two solutions.

The exponential slab heating solution separates three useful controls: the heat source, the conductive path and the boundary temperature. Reduce deposited power, improve thermal conduction, shorten paths to a cooled surface, or lower the actual surface temperature. Each proposal has to preserve the shield's radiation-protection function.

Reduce incident photon intensity, intercept part of the beam in a separately cooled upstream shield, or distribute the deposited power over a larger cooled region. These reduce the relevant volumetric heat generation rate or local heat load. Intermittent exposure can reduce average heating, but the peak then requires a transient heat equation calculation; a smaller average alone does not establish a safe pulse maximum.

Increase the effective thermal conductivity, for example through a suitable material or a well-bonded conductive path to a heat sink. Reduce contact thermal resistance in any added path. Material substitution must also satisfy shielding, mechanical and environmental constraints.

Shorten the maximum conduction distance by introducing cooled internal surfaces or dividing the shield into thinner, separately cooled layers. Reducing the thickness of a single slab lowers its temperature rise at fixed $q_0$ and $\mu$, but also reduces attenuation; it is not automatically an acceptable shield modification. A layered arrangement can preserve the required total shielding while changing the thermal boundary-value problem.

Lower the cooling-air temperature, improve forced convection by appropriate flow and surface design, enlarge the heat-exchange area, or use a more effective coolant where compatible. These can lower the actual boundary temperature when the external cooling is modeled. In the stipulated model $T_0$ is already prescribed: increasing the external heat transfer coefficient while holding $T_0$ fixed leaves the calculated interior profile unchanged. It helps physically by reducing the surface-to-coolant temperature difference under a heat load.

Do not change the attenuation coefficient independently of the heat-source normalization without explaining the comparison. At fixed incident photon energy flux, the entrance deposition $q_0$ generally changes with absorption properties. A larger $\mu$ with artificially fixed $q_0$ is a different experiment. Similarly, increasing total slab heat capacity delays heating but does not lower the steady state maximum for an unchanged source and unchanged thermal boundary conditions.

Let ${\delta }_{xx}{U}_{ij}=(U_{i+1,j}-2U_{ij}+U_{i-1,j})/(\Delta x{)}^2$, and define ${\delta }_{yy}$ analogously. The Crank-Nicolson diffusion scheme isUse $U_{ij}^0=u_0(x_i,y_j)$ and zero values on every boundary node for every time level. This is the theta method with $\theta =1/2$ applied to the spatial finite difference method for the heat equation.

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces toMultiplying by $\rho$ and integrating givesA finite temperature and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+H{a}^2/(4\kappa )$. ThereforeThis is the steady temperature of a uniformly heated cylindrical wire. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4{\pi }^2{a}^2\kappa \sigma )$. The outward heat flux is $J_Q(\rho )=-\kappa T^{\prime }(\rho )=H\rho /2$, so the power crossing a cylindrical surface per unit length, $2\pi \rho J_Q(\rho )=H\pi {\rho }^2$, equals the heating generated inside it.

Far from the ridge, $T(z,\infty )=T_0+\Delta T(1-z/a)$. Transients have decayed and the heat equation reduces to $T_{zz}=0$. A finite basal heat supply sustains this gradient. Thermal thickness, seafloor depth and surface heat flux therefore approach finite limits rather than continuing the laws of the half-space cooling model indefinitely.

Write $h=\Delta x$, $k=\Delta t$ and ${\delta }_{xx}{u}_i=(u_{i+1}-2u_i+u_{i-1})/h^2$. Insert a smooth exact heat equation solution into the Crank-Nicolson method. Expanding about the midpoint $(x_i,t_n+k/2)$ givesThus the residual, defined as left side minus right side, isSince $u_t=D{u}_{xx}$ implies $u_{ttt}=D{u}_{xxtt}$,This proves second-order numerical consistency in both space and time. About the old time level, uncombined Taylor remainders can include $O(k^3)$; the midpoint expression displays the symmetric leading errors most cleanly. Multiplying by $k$ gives the one-step defect $O(k^3+k{h}^2)$. Stable evolution with suitable initial and boundary data then gives global error $O(k^2+h^2)$; consistency alone does not prove global convergence.

Use the forward spatial transform $\hat{f}(k)=\int e^{-ikx}f(x)dx$ and inverse $f(x)=\int e^{ikx}\hat{f}(k)dk/(2\pi )$. The retarded transformed Green function obeysIt vanishes for negative $\tau$. Continuity at zero excludes an unwanted ${\delta }^{\prime }$ term; integrating across zero then fixes $G_k^{\prime }(0^{+})=c^2$. Define $\gamma =\alpha c^2$ and $d_k=\sqrt{{\gamma }^2-c^2{k}^2}$. The two homogeneous exponents are $-\gamma \pm d_k$. Imposing the two matching conditions gives the Fourier-mode Green function of the telegraph equation,For an explicit time-Fourier route, use $G_k(\tau )=\int e^{-i\omega \tau }{\mathcal{G}}_k(\omega )d\omega /(2\pi )$. ThenFor $k\ne 0$, both poles lie below the real axis. Closing above for $\tau <0$ gives zero; closing below for $\tau >0$ is clockwise, giving $-i$ times the residue sum. Since ${\mathcal{G}}_k=-c^2/[(\omega -{\omega }_{+})(\omega -{\omega }_{-})]$, this iswhich reduces to the same damped sine or hyperbolic sine above. The $k=0$ pole on the real axis is interpreted by the retarded limiting prescription, or directly by the initial-value matching above.

For $|k|<\alpha c$ the response is overdamped and nonoscillatory. For $|k|>\alpha c$, put ${\Omega }_k=\sqrt{c^2{k}^2-{\gamma }^2}$; analytic continuation gives $G_k=\Theta (\tau )c^2{e}^{-\gamma \tau }\sin ({\Omega }_k\tau )/{\Omega }_k$, a real damped oscillation. At equality the finite limit is $\Theta (\tau )c^2\tau e^{-\gamma \tau }$.

All nonzero Fourier modes decay. The uniform mode instead has $G_0=(1-e^{-2\gamma \tau })/(2\alpha )$ for positive time, leaving a constant temperature increment. At small wavenumber the slow decay exponent is $-k^2/(2\alpha )+O(k^4)$, so the diffusion limit of the damped heat-transmission equation has diffusivity $1/(2\alpha )$. The full hyperbolic equation has finite propagation speed $c$, unlike the limiting parabolic heat equation.

Assume a stationary temperature profile, spatially uniform mass density and radioactive heat generation rate, constant thermal conductivity, isotropic solid thermal conduction, and blackbody emission into effectively cold surroundings. The volumetric source isTotal radiogenic power equals the surface blackbody radiation loss. The Stefan–Boltzmann law givesFor steady radiogenic heating of a conducting sphere, the interior heat equation isRegularity at the centre removes the $1/r^2$ integration term and yields $dT/dr=-q_{v}r/(3\kappa )$. Matching $T(R)=T_S$ givesThe high central temperature is compatible with a cold surface because finite thermal conductivity requires a large temperature gradient to carry the accumulated interior heat. A nonzero background radiation temperature $T_{\mathrm{bg}}$ replaces $T_S^4$ by $T_S^4-T_{\mathrm{bg}}^4$ in the surface balance. The calculation neglects convection, stored transient heat, changing radioactive abundances, phase changes and any substantial external irradiation.

Write $\Delta T=T_0-T_B$, $\Delta C=C_0-C_B$ and assume these scales are nonzero. The two reference liquidus relations imply $\Delta T=-m\Delta C$. For the similarity variable $\eta =z/(2\sqrt{\kappa t})$ and front $h=2\lambda \sqrt{\kappa t}$,Thus a temperature profile $T=T_{\ast }+\Delta T\theta (\eta )$ has $T_t=-\Delta T\eta {\theta }^{\prime }/(2t)$ and $\kappa T_{zz}=\Delta T{\theta }^{\prime \prime }/(4t)$. Each heat equation becomes ${\theta }^{\prime \prime }+2\eta {\theta }^{\prime }=0$. The solute diffusion equation becomes ${\mathcal{C}}^{l\prime \prime }+2Le\eta {\mathcal{C}}^{l\prime }=0$, withThis Lewis number compares thermal and solutal diffusivities. Inside the mushy layer, $T^s-T_B=-m(C^s-C_B)$ gives $\boxed{{\mathcal{C}}^s={\theta }^s}$. The wall and far-field conditions become ${\theta }^s(0)=0$ and ${\theta }^l,{\mathcal{C}}^l\to 0$ as $\eta \to \infty$.

At the moving phase boundary, equality of temperatures gives ${\theta }^s={\theta }^l+(T_{\infty }-T_B)/\Delta T$. Equality of concentrations gives ${\mathcal{C}}^s={\mathcal{C}}^l+1$. Therefore the dimensionless parameters with the convention used here areNeither concentration nor temperature offsets should be dropped when enforcing continuity. In particular ${\mathcal{R}}_{\theta }$ can be negative. If $k=\rho c_p\kappa$, then $S=L/[c_p\Delta T]$: this is the inverse Stefan number, although the paper calls $S$ its Stefan number.

To check every factor of two in the flux conditions, divide the thermal Stefan condition by $k\Delta T/(2\sqrt{\kappa t})$, and the solute balance by $D\Delta C/(2\sqrt{\kappa t})$. This givesFinally $T_z^l=-m{C}_z^l$ and $-m\Delta C=\Delta T$ imply $\boxed{{\theta }^{l\prime }={\mathcal{C}}^{l\prime }}$. This is marginal equilibrium at a mush-liquid boundary. Together these calculations produce all the rescaled differential equations and all three interface balances.

The heat equation and its homogeneous Dirichlet boundary conditions are linear, so superposition of the separated modes givesThe Fourier sine basis represents the initial profile on the interior of the interval. At $t=0$ its Fourier sine series must equal $x(1-x)$. Multiplying by $\sin (m\pi x/L)$, integrating from $0$ to $L$, and using orthogonality givesTherefore the required Fourier sine coefficients areFor every $t>0$, the exponential factors ensure convergence of the differentiated series, and every term satisfies the partial differential equation and the boundary conditions. The Fourier sine series recovers the smooth initial profile at every $0<x<L$ as $t\downarrow 0$.

The scan explicitly prints $x(1-x)$ on $(0,L)$; it must not silently become $x(L-x)$. For $L\ne 1$, the interior profile tends to $L(1-L)\ne 0$ at the right endpoint, whereas the prescribed boundary condition is zero for $t>0$. The solution is therefore not continuous at the corner $(L,0)$, although it satisfies every condition on the printed open ranges. A solution continuous on the closed space-time domain with these data would require $L=1$.^[ref]

Substitute the proposed Fourier sine series into the heat equation:Projecting onto each sine using orthogonality gives $T_n^{\prime }=-D{k}_n^2{T}_n$ for every mode with $b_n\ne 0$. The initial condition fixes $T_n(0)=1$, since $b_n$ already describes the initial amplitude. HenceFor a mode with $b_n=0$, the value of $T_n$ is immaterial and the same convention is convenient. The positive thermal diffusivity $D$ damps every mode, with higher frequencies decaying faster. For square-integrable initial data the damping justifies termwise differentiation for $t>0$ and recovery of the initial data in L2 norm convergence as $t\downarrow 0$.

The left panel illustrates the Fourier series in part (b). The right panel is a self-generated example with $D=L=1$ and ${\Theta }_0(x)=\sin (\pi x)+\frac{1}{2}\sin (3\pi x)$, showing the two normal modes decaying at rates ${\pi }^2$ and $9{\pi }^2$.

For $t>0$, fixed zero endpoint temperatures are homogeneous Dirichlet boundary conditions:An initial condition is also needed to determine a particular solution of the heat equation.

For $T_n=A\cos (k_{n}x)e^{-\kappa k_n^{2}t}$, direct differentiation again gives ${\partial }_t{T}_n=-\kappa k_n^2{T}_n$ and ${\partial }_x^2{T}_n=-k_n^2{T}_n$. Thus it satisfies the heat equation. Moreover,vanishes at both endpoints for integer $n$, proving the homogeneous Neumann boundary conditions. Now $n=0$ gives the nonzero constant mode $T=A$, which does not decay. It represents conserved mean temperature in an insulated plate.

Put $k_n=n\pi /L$ and ${\lambda }_n=\kappa k_n^2$. Direct differentiation of $T_n=A\sin (k_{n}x)e^{-{\lambda }_{n}t}$ givesso ${\partial }_t{T}_n=\kappa {\partial }_x^2{T}_n$, the heat equation. At $x=0$, the sine factor vanishes; at $x=L$, it is $\sin (n\pi )=0$ for integer $n$. Thus the homogeneous Dirichlet boundary conditions hold. The nonzero, nonredundant spatial modes use $n=1,2,\dots$; $n=0$ is the zero function and negative $n$ only change the amplitude sign.

The reference to part “(c)(i)” in the source is circular; the relevant modes are those verified in part (b)(i). Linearity of the heat equation and its homogeneous Dirichlet boundary conditions allows superposition of those modes:Match the initial uniform interior temperature $T_0$ using the orthogonality of the Fourier sine series:Thus $b_n=0$ for even integers $n$ and $b_n=4T_0/(n\pi )$ for odd integers $n$, givingFor every $t>0$, the exponential factor makes the series and the derivatives needed in the heat equation converge uniformly on later time intervals. At $t=0$, the Fourier sine series equals $T_0$ at interior points but zero at the endpoints: the instantaneous cooling boundary condition is incompatible with the pre-immersion endpoint values. This corner mismatch is physically the idealization of an instantaneous bath contact and does not invalidate the solution for $t>0$.

Write $U(x,\tau )=T(x,t_f+\tau )$ after removal, with $\tau \ge 0$. Neglecting heat loss to air means thermal insulation, rather than an end to internal thermal conduction. The evolution and conditions areHere $T(x,t_f)$ is the nonuniform bath-cooled profile derived in part (c). At the switching instant it need not have zero endpoint derivatives: the sudden replacement of Dirichlet boundary conditions by Neumann boundary conditions can produce another initial-boundary corner incompatibility. The insulated heat equation nevertheless recovers the given profile and satisfies its new boundary conditions for $\tau >0$.

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 $\kappa >0$.