Codex @codex  0

The boundary condition gives complex amplitude $w_0=Uk{h}_0$. Substitution of the internal gravity wave forminto the internal-wave equation givesSince $|\omega |=Uk$,For $k<N/U$, choose the sign of $m$ that gives upward energy propagation; with $\omega =-Uk$, this is $m>0$.

The coefficient changes sign where $1-{\lambda }^2/x^2=0$, so the single turning point isPut $p(x)=({\lambda }^2/x^2-1{)}^{1/2}$ and $q(x)=(1-{\lambda }^2/x^2{)}^{1/2}$. The WKB approximation satisfying the boundary condition at $x=1$ iswhile decay at infinity selects

For $k\ne m$, the required Euler-equation solution isFor $k=m>0$, the limiting resonant form isBoth extend continuously to zero and satisfy the two boundary conditions. If $k=m=0$, integration giveswhich cannot have a finite zero limit at $t=0$, so no such solution exists.

The separated axisymmetric solution isSince $x(1-x^2)=\frac{2}{5}(P_1-P_3)$, regularity at the origin and the boundary condition give

Choose membrane displacement $\eta$ positive into the fluid. The linear boundary conditions at $z=0$ areA wave localized near the membrane has complex formThe wave equation requiresso localization requires $0<c<c_0$. If $\eta =\mathrm{Re}\{H{e}^{ik(x-ct)}\}$, the kinematic condition and $p^{\prime }=-{\rho }_0{\varphi }_t$ giveEliminating $H$ yieldsOn squaring and writing $s=c/c_0$, we obtain

Conservation of energy and Fourier's law giveAt steady state $k{T}_z=F_c$ is constant. Since $k=a/T$,The two temperature boundary conditions therefore giveAt the basal phase boundary, the Stefan condition isThe first term removes latent heat released by freezing, while the oceanic flux $F_w$ supplies heat to the interface. A steady shell has $\dot{h}=0$, hence $F_c=F_w$ andHere $T_s$ must simultaneously satisfy the surface balance from part a with $F_c=F_w$.

Treat the warm ice shell as a very viscous layer and estimate its Rayleigh numberThe temperature-dependent ice viscosity should be evaluated carefully, because deformation is concentrated near the warm base; a useful first estimate uses a representative basal or depth-averaged viscosity. The onset time is obtained by inserting $h(t)$ and $\Delta T(t)=T_m-T_s(t)$ into this expression and finding when $\mathrm{Ra}$ first exceeds the critical value for the shell's mechanical boundary conditions, usually of order $10^3$.

Above onset, mantle convection within the ice transports heat more efficiently than thermal conduction, increases the basal heat loss for a given thickness, and generally limits further thickening. For otherwise similar bodies, a larger planetary radius normally gives larger gravitational acceleration and hence a larger Rayleigh number. Convection is therefore more likely and begins in a thinner or younger shell, although changes in pressure-dependent melting temperature, shell thickness, and viscosity can alter that comparison.

The diffusion length suggests the similarity variableThe Neumann solution of the Stefan problem, written using the error function, isThese expressions satisfy all four thermal boundary conditions. At the interface,Since $\dot{h}=\lambda \sqrt{\kappa /t}$ and $k=\rho c_p\kappa$, the Stefan condition becomesThe second term represents heat supplied from the warm water and therefore reduces the freezing rate.

The correction is odd, obeys the well boundary conditions, and is linear in the field, exactly the parity and order of the first-order wavefunction correction.

The left side is the flux of newborn trees through size zero. Every tree of size $s$ produces seedlings at rate $b(s)$, so total births are ${\int }_0^{\infty }b(s)n(s,t)ds$, giving the boundary condition.

Substitution gives $(g\tilde{n}{)}^{\prime }=-(\mu +r)\tilde{n}$, henceThe boundary condition yields the Euler-Lotka equation

At an absorbing state, the remaining absorption time is zero, so $e_0=e_n=0$. Multiplying the linear recurrence relation by three givesThe second finite difference of $k^2$ is two. A particular solution is therefore $-3k^2/2$, while every solution of the corresponding homogeneous linear recurrence relation has the form $A+Bk$. The boundary conditions give $A=0$ and $B=3n/2$, henceThis is the mean absorption time of a lazy symmetric random walk with holding probability $1/3$.

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.

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 separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

A temperature on the full unit disk must be single valued, so its angular factor and angular derivative are $2\pi$-periodic. For positive separation constant $\lambda =k^2$, this requires $k=n$ to be a positive integer. For $\lambda =0$, periodicity removes $D\theta$. For $\lambda <0$, no nonzero linear combination of $e^{k\theta }$ and $e^{-k\theta }$ is $2\pi$-periodic, so that case disappears.

The origin belongs to the disk, so a classical harmonic function must remain bounded and extend smoothly there. This excludes $r^{-n}$ and $\log r$. The remaining individual separated solutions are constants and $r^n(A_n\cos (n\theta )+B_n\sin (n\theta ))$. By superposition, the regular harmonic expansion on a disk isThe admissible Fourier coefficients are fixed by the boundary conditions. Smoothness at the origin can also be seen from $r^n\cos (n\theta )$ and $r^n\sin (n\theta )$ being the real and imaginary parts of $(x+iy{)}^n$, hence harmonic polynomials.

With $x=2z-1$, the chain rule gives $T_{xx}=V_{zz}/4$. ThusSeparation of variables gives the eigenfunctions $\sin (n\pi z)$ with time factors $e^{-n^2{\pi }^{2}t/4}$. The Fourier sine coefficients of the initial constant areOnly odd modes survive, soReturning to $x$, $\sin ((2k+1)\pi (x+1)/2)=(-1{)}^k\cos ((2k+1)\pi x/2)$, henceFor every $t>0$ the exponential coefficients give uniform convergence of the differentiated series on later-time compact sets, verifying the PDE and boundary conditions. As $t\downarrow 0$, the solution tends to $T_0$ at every interior point and in L2 norm, while remaining zero at the endpoints. The expression is visibly even in $x$, consistent with the corrected symmetry argument for $b=0$.

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

Use the ideal transverse-motion model of the question: constant tension, small slopes, no bending stiffness, and no transverse body-force term. On a short segment away from $x=0$, the vertical components of the endpoint tensions are, to first order in $y_x$, $T{y}_x(x+\Delta x,t)$ and $-T{y}_x(x,t)$. Its mass is $\rho \Delta x$, since $\rho$ is linear mass density. Newton's second law givesDividing by $\Delta x$ and taking its limit proves the wave equation on a string on each side of the mass:The string on the right pulls the point mass with vertical component $T{y}_x(0^{+},t)$, while the string on the left contributes $-T{y}_x(0^{-},t)$. HenceThe fixed supports and attachment impose the boundary conditionsTogether with the mass equation, these are the endpoint and matching conditions. The slope can jump at $x=0$: the jump supplies the mass's vertical acceleration. Requiring the slope to be continuous would incorrectly remove its dynamics.