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Use a time Laplace transform and a spatial Green function. The first-order transport term must be retained: the Airy resolvent kernel for ${\partial }_x^3+p$ alone would solve a different equation. The governing linear partial differential equation is the linear dispersive Stokes equation, distinct from the viscous-fluid Stokes equation.

For the time interval of interest, extend the given boundary signal by zero after $T$ and putThe possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting causal Green function produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.

Writing $Q(x,p)$ for the time Laplace transform, the Laplace transform of a derivative givesThe corresponding characteristic roots are the three roots of a polynomialFor $\mathrm{Re}p>0$, label the unique root of negative real part by $r_{-}(p)$, and the two roots of positive real part by $r_1(p),r_2(p)$. This spatial root splitting for the dispersive Stokes resolvent is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive real parts. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_{-}(p)$ is a nonzero holomorphic function.

Set $d_j=3r_j^2+1$, including $j=-$. Define the Stokes resolvent kernel with advectionThe sign on the negative half-line is essential. To verify the Green function, use the partial fraction decompositionIts expansion at infinity yields the reciprocal-polynomial root identitiesConsequently $R_p$ and $R_p^{\prime }$ are continuous at zero, and $R_p^{\prime \prime }$ has jump one. Each piece solves the homogeneous ordinary differential equation, so the jump conditions for a third-order Green function imply, in the sense of distributional derivatives,Both pieces decay in their respective spatial directions.

Now defineIt satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous linear ordinary differential equation; spatial decay leaves only $e^{r_{-}x}$. Imposing the Neumann boundary condition fixes its coefficient, giving the half-line Stokes solution with a boundary derivativeNo division by a vanishing root is involved, since $r_{-}=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:Thus introduce the completely explicit initial-data kernelFor any upward Bromwich contour $\mathrm{Re}p=\sigma >0$ to the right of any transform singularities, the requested integral representation isThe contour integral is understood in the standard Bromwich inversion formula sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic characteristic roots; there are no unknown Dirichlet or second-derivative boundary traces.

The transformed equation verifies the linear partial differential equation and the initial term, while taking the spatial derivative gives $Q_x(0,p)=G(p)$. Bromwich inversion formula therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the Laplace transform time-shift rule. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0^{\prime }(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.

The drift makes the natural measure weighted. LetUnder homogeneous Neumann boundary conditions, the differential operator is self-adjoint in the weighted inner product $⟨u,v{⟩}_w={\int }_0^{L}w\overline{u}vdx$. Indeed integration by parts giveswhen the derivative boundary terms vanish. In particular its eigenvalues are nonpositive and it has the constant stationary eigenfunction. We first construct the weighted Neumann heat kernel with constant drift, then incorporate the two prescribed coordinate derivatives.

Put $b=\beta /2$. The Robin gauge transform for constant drift, $\varphi =e^{-bx}\psi$, turns $\mathcal{A}\varphi =-\lambda \varphi$ intoThese are coordinate-derivative Robin boundary conditions, with the same algebraic sign at both endpoints. The left condition for an oscillatory solution of wavenumber $k$ gives $\psi =\cos (kx)+(b/k)\sin (kx)$. The right condition then reduces to $(k+b^2/k)\sin (kL)=0$, giving $k_n=n\pi /L$. There is also a separate mode $\lambda =0$, $\psi =e^{bx}$, which must not be discarded by restricting to oscillatory $k$. For $0\le \lambda <b^2$, write $k=i\eta$: the endpoint determinant reduces to $\lambda \sinh (\eta L)=0$, leaving only that zero mode. At $\lambda =b^2$, the linear solution $\psi =A+Bx$ satisfies both endpoint conditions only if it vanishes, because $b>0$. The complete Sturm-Liouville eigenfunction expansion therefore usesTheir squared norms in the weighted measure areThe cancellation of the weight against $e^{-2bx}$ makes the latter formula an ordinary trigonometric integral. DefineFor $t>0$ this kernel is smooth in the interior. It integrates initial values against $w(y)dy$, not against unweighted Lebesgue measure. It is symmetric in $x,y$ as a weighted kernel, even though the drift operator does not look symmetric in an unweighted inner product.

To determine the boundary forcing, set $a_n(t)=⟨{\varphi }_n,q(\cdot ,t){⟩}_w/N_n$. Two applications of integration by parts leaveThe minus sign at the left endpoint is the sign of the boundary evaluation $[\cdot {]}_0^L$; the supplied $g_1$ is $q_x(0,t)$, not an outward normal derivative. Solving this linear differential equation by an integrating factor and recombining the modes gives the requested space-time integral representation:This is the Neumann boundary-forcing formula. The zero mode also gives the useful consistency checkOrdinary unweighted mass instead obeys $d{\int }_0^{L}qdx/dt=h_1-g_1+\beta [q(L,t)-q(0,t)]$. Replacing the weighted measure by $dy$ or deleting the constant mode would therefore give an incorrect solution.

There is a subtle point when checking nonzero derivative data: every homogeneous eigenfunction has zero endpoint derivative, but differentiating the boundary-forcing sum and its time integral term by term at an endpoint is not valid. The kernel is singular as $t-s\downarrow 0$, and the prescribed derivatives are interior limits of the full solution. An explicit Laplace transform form verifies those limits without this interchange and gives an alternative purely contour-integral version.

For $\mathrm{Re}p>0$, set $\kappa =\sqrt{p+b^2}$ using the principal square root, and defineBoth solve $(p-\mathcal{A})f=0$, with $u_p(0)=v_p(L)=1$, $u_p^{\prime }(0)=v_p^{\prime }(L)=0$. Their weighted Wronskian is the nonzero constantHence the resolvent kernel for Neumann advection-diffusion on an interval isThe first derivative in $x$ has jump $-e^{-\beta y}$, giving $(p-{\mathcal{A}}_x){\mathcal{R}}_p=e^{-\beta y}{\delta }_y$ in the sense of distributional derivatives. If $G(p)={\int }_0^T{e}^{-ps}{g}_1(s)ds$ and $H(p)={\int }_0^T{e}^{-ps}{h}_1(s)ds$, an equivalent transformed solution isIn fact $v_p^{\prime }(0)=-D_p$ and $u_p^{\prime }(L)=e^{-\beta L}{D}_p$, so $Q_x(0,p)=G(p)$ and $Q_x(L,p)=H(p)$ exactly. The inverse Bromwich contour integral of $e^{pt}Q(x,p)/(2\pi i)$ is an alternative final answer. Its poles at $p=0$ and $p=-{\lambda }_n$ reproduce the constant and decaying modes of $K_{\beta }$, respectively; apparent square-root singularities in the normalized hyperbolic factors are removable. This also independently checks the kernel normalization and boundary signs.

As in question 1, the printed dot on $q_0$ denotes its spatial derivative: the two corner conditions are $q_0^{\prime }(0)=g_1(0)$ and $q_0^{\prime }(L)=h_1(0)$. The initial eigenfunction expansion converges to $q_0$ in the weighted space and, for the stated smooth compatible data, has the appropriate classical initial and boundary limits. Differences of solutions with zero data satisfy $\frac{1}{2}d\Vert q{\Vert }_w^2/dt=-{\int }_0^{L}w|q_x{|}^{2}dx$, proving uniqueness in the usual regular solution class. In the limit $\beta \to 0$, $N_0\to L$, ${\varphi }_n\to \cos (n\pi x/L)$, and the ordinary Neumann heat kernel on an interval and its boundary signs are recovered.

Let $F(x,z,t)=z-\eta (x,t)$. The interface is a material surface for both fluids: its material derivative vanishes on $F=0$. With $u_j={\partial }_x{\varphi }_j$ and $w_j={\partial }_z{\varphi }_j$, this givesThus the kinematic boundary condition says that particles stay on the moving interface. The tangential velocities can differ; each side has its own material derivative. Replacing both derivatives by one background speed would be incorrect.

The dynamic boundary condition for an inviscid interface is pressure continuity, because the fluids have no viscous normal stress and there is no surface tension. The unsteady Bernoulli equation in each layer of irrotational flow readsEquating these pressures at $z=\eta$ gives the required dynamic boundary condition for an inviscid interface. Choose the potential gauges so that ${\rho }_1{C}_1={\rho }_2{C}_2$, or set both constants to zero relative to a common reference pressure. For example, ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ removes the distinct uniform-stream Bernoulli constants without changing either velocity. Material-interface kinematics and pressure continuity supply the two matching conditions, with the Bernoulli gauge understood.

For an even velocity potential, take ${\Phi }_I=C\cosh (kz)$ and the decaying exterior amplitudes equal at the upper and lower interfaces. At $z=L$, kinematics gives $A=ic{B}_{+}$ and $C\sinh (kL)=i(V-c)B_{+}$. The dynamic condition becomes $-cA=(V-c)C\cosh (kL)$. ThereforeWriting $s=\tanh (kL)$, its two roots areThe interior vertical velocity is odd, so $B_{-}=-B_{+}$; this is a varicose mode of a planar jet. As with the odd potential, $k>0$, finite $L>0$ and $V\ne 0$ give a growing root. The parity of the potential and the parity of the interface displacements are opposite because the vertical velocity is its derivative.

Write ${\varphi }^{\prime }=\Phi (z)e^{ik(x-ct)}$ and let $B_{\pm }$ be the displacement amplitudes at $z=\pm L$. Use subscripts $I$ for the jet interior and $O$ for the adjoining exterior. Each interface is a material vortex sheet, so its linearized kinematic boundary conditions areThe equal mass densities make the hydrostatic displacement terms cancel. The dynamic boundary condition for an inviscid interface therefore givesThe division form assumes $c\ne 0,V$; the preceding undivided equations remain the proper conditions in those exceptional cases. Potentials themselves need not be continuous, and their vertical derivatives need not be equal: the same material displacement is advected by different tangential base velocities. At infinity the exterior potentials decay. These displacement and pressure conditions, rather than continuity of $\Phi$ and ${\Phi }^{\prime }$, determine the equal-density top-hat planar-jet dispersion relation.

For the equal-width Hazel model with stable density stratification $J\ge 0$,The Miles–Howard theorem therefore excludes exponentially growing regular normal modes for $J\ge 1/4$. The candidate neutral modes of the equal-width Hazel model instead lie onThe parabola touches the sufficient-stability threshold at its maximum. There is no contradiction: neutrality has $c_i=0$, whereas the proof in part (a) assumes $c_i>0$. Moreover, the $k=1/2$ profile has a singular critical-layer derivative and logarithmically divergent horizontal kinetic energy; it is not a regular growing mode satisfying the proof's hypotheses.

For $J<1/4$, the local gradient Richardson number condition permits instability but does not establish it. Substitution of a neutral ansatz alone also does not determine on which side of the curve unstable eigenvalues lie. It identifies the formal neutral curve; concluding a full stability boundary requires additional continuation analysis of the eigenvalues. The nondecaying $k=0$ endpoint and the smooth $k=1,J=0$ endpoint have the distinct qualifications described above.

For this nonlinear ansatz, take $k$ real as well as $U,\mu$ real. Then $|e^{ik(x-Ut)}|=1$, and the advective time derivative vanishes. Substitution into the nonlinear Ginzburg-Landau equation givesThe stipulated positive-amplitude Ginzburg-Landau plane wave therefore hasThe strict inequality $\mu >k^2$ guarantees that this is real and positive. The zero solution of the nonlinear Ginzburg-Landau equation also exists but is not the required positive wave. Although complex $k$ is useful in the linear impulse analysis, it cannot generally be carried into this constant-amplitude nonlinear ansatz: its spatially varying modulus would make the cubic term carry a different spatial factor.

The temporal integration by parts terms areThe terminal states are free and the objective has no explicit terminal-velocity dependence. ThereforeThe factor two follows from the displayed objective without a $1/2$ prefactor. The initial scalar field perturbation is fixed, so $\delta \theta (0)=0$ and there is no independently prescribed adjoint-scalar condition at $t=0$. Its value there is obtained by backward integration. Independent variation of the initial velocity state gives ${\mathbf{u}}^{†}(0)={\mathbf{u}}_0^{†}$.

Variation of the candidate ${\mathbf{u}}_0$ in the energy-augmented functional gives the fixed-energy initial-condition optimality condition. For the usual divergence-free no-slip control space $H$, write $P_H$ for its orthogonal projection, the Leray-Helmholtz projection in the usual incompressible velocity space. This is the standard formal control condition when pressure is determined by incompressible flow and normal momentum. If the independently imposed direct pressure flux is retained as an additional constraint, the admissible control variations must also satisfy its compatibility conditions along the trajectory; they need not fill the usual space $H$. The standard projection formula alone does not prove stationarity for that more restricted problem. ThenFor the natural solenoidal adjoint space, $P_H{\mathbf{u}}^{†}(0)={\mathbf{u}}^{†}(0)$. Equivalently, its component tangent to the sphere in a normed vector space of fixed kinetic energy vanishes:Here $E_0>0$ is needed for the sphere in a normed vector space to be regular and for division by $2E_0$. The multiplier is real and can have either sign; normalizing the initial adjoint with a prescribed positive sign is not a general necessary condition for minimization. If $E_0=0$, the only feasible initial velocity is zero and this tangent formula for the sphere in a normed vector space is inapplicable. First-order stationarity alone also allows maxima or saddles; a local minimum requires the appropriate nonnegative constrained second variation.

The spatial boundary terms vanish with periodic adjoints in $x,z$, homogeneous no-slip boundary conditions ${\mathbf{u}}^{†}=0$ at $y=\pm 1$, and ${\partial }_y{\theta }^{†}=0$ there. To see this, velocity variations vanish at the wall while their normal derivatives need not, so the viscous boundary term forces the adjoint velocity to vanish. Scalar variations have zero normal derivative but free values, forcing the adjoint scalar's normal derivative to vanish. Pressure variation gives adjoint incompressible flow; no independent terminal or initial datum is assigned to $p^{†}$. Its additive time-dependent constant may be fixed by a zero spatial mean.

In particular, a homogeneous Neumann condition for the direct pressure does not by variational transposition require ${\partial }_y{p}^{†}=0$. For a smooth adjoint solution, the wall-normal adjoint momentum equation instead suppliesat the flat walls, because the direct scalar normal derivative is zero there. A zero adjoint-pressure derivative would be an additional compatibility choice, valid only when this right-hand side vanishes.

There are also literal direct-data compatibility qualifications in the printed setup. The initial total scalar field is $\overline{\theta }=-\mathrm{erf}(30y)$, whose wall derivative is $-60e^{-900}/\sqrt{\pi }\ne 0$. It is extraordinarily small but mathematically not zero. Thus exact initial scalar data and exact zero wall flux are incompatible for a classical solution smooth at $t=0$. One may use a parabolic mild solution of an abstract Cauchy problem with the boundary condition enforced for $t>0$, or replace the initial profile by an exactly compatible smooth one; these are conventions, not an equality $e^{-900}=0$. The reference scalar field is also not a stationary profile of the diffusion equation at finite $Pe$: the full $\Delta \Theta$ term must be retained.

The pressure compatibility at a no-slip wall is similarly important. Direct normal momentum gives ${\partial }_{y}p=\nu \Delta u_y-R{i}_B\theta$. The separately imposed zero pressure derivative requires this right side to vanish. It is not automatic for arbitrary fixed-energy initial velocities. For example, the divergence-free no-slip field generated by the periodic stream function $S=A(1-y^2{)}^2\sin (\alpha x)$ with $\alpha =\pi /L_x>0$ has $u_x={\partial }_{y}S$, $u_y=-{\partial }_{x}S$ and $\Delta u_y{|}_{y=\pm 1}=-8A\alpha \cos (\alpha x)\ne 0$, while $\theta (0)=0$. With $A\ne 0$ adjusted to any positive energy, no classical solution smooth up to the initial wall can satisfy that extra pressure condition. The displayed adjoints are the formal necessary equations along admissible smooth trajectories; a well-posed physical formulation normally determines pressure from incompressible flow and normal momentum, rather than imposing independent homogeneous pressure flux for every control.

Introduce specific heat capacity $c_p$, latent heat per unit mass $L_f$, and $\Delta T=T_m-T_s>0$. Write the prescribed convective heat flux as $F=K(\overline{T}-T_m)$, whereThere is a dimensional convention to specify: with $Q$ measured in $\mathrm{W}\mathrm{k}{\mathrm{g}}^{-1}$ and dimensionless $\lambda$, this printed expression makes $K$ a heat transfer coefficient only if $\mu$ is kinematic viscosity. If $\mu$ denotes dynamic viscosity, the numerator needs ${\rho }_0^{2}g\alpha Q$ instead. The question says only “viscosity”; use the consistent kinematic interpretation, or equivalently the corrected dynamic formula. The following balances apply to either convention once $K$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

For a thin lava lake crust with a linear temperature profile, Fourier's law gives the outward conductive flux $k\Delta T/a$. The Stefan condition balances it against upward thermal convection and freezing, while the well-mixed liquid loses the convective flux and gains radiogenic heating. To leading order in $a/H$ the radiogenically heated lava-lake crust model isThe fixed surface temperature already incorporates radiative cooling, so no extra radiative loss is added to the interior balance. The linear conductive profile is interpreted as a quasi-steady approximation, so sensible-heat storage within the crust is not evolved. Thinness alone does not establish this approximation: rapid thermal adjustment in the crust is also needed, for example a small Stefan number during the initial conduction-driven growth. Replacing the liquid depth by $H-a$ and retaining the moving liquid-volume enthalpy would introduce thin-crust corrections: the requested constant-temperature branch refers to the leading model above. A transient thermal profile in a thick or newly nucleating crust would require the full Stefan problem, rather than this linear-profile closure.

Using the thermal diffusion time $H^2/\kappa$, defineThe three independent dimensionless parameters are the convective-conductive coefficient $B$, the heating parameter $R$, and the Stefan number $S$. For positive $B,R,S$ the evolution equations becomewhere prime denotes $d/d\tau$. Temperature has the exact solutionConsequently the unique constant-temperature initial condition isAt this temperature, the convective flux equals the column's radiogenic heat production. Starting with no crust, the constant-temperature lava-crust growth law is obtained by separating $h^{\prime }=S(1/h-R)$:The crust grows towards $h_{\ast }=1/R$, or $a_{\ast }=k\Delta T/({\rho }_{0}QH)$, where conduction carries all the radiogenic input. The logarithm makes reaching this equilibrium take infinite time. For an initial thickness $h_0>0$, subtract $[-R{h}_0-\log (1-R{h}_0)]/(S{R}^2)$ on the same side of the equilibrium; the corresponding antiderivative uses $\log |1-Rh|$ on either branch. The thin-crust description through equilibrium requires $R\gg 1$.

For general initial temperature, the full early and intermediate history is specified by the exact temperature above and one scalar ordinary differential equation. A particularly convenient integral formulation for an initially bare surface isThis removes the infinite initial derivative of $h$. The right-hand side points into $y>0$ at zero, and is decreasing in $y$ when ${\theta }_0\ge 0$; two nonnegative solutions therefore cannot separate, which gives uniqueness. Its solution, together with the temperature formula, determines the entire model evolution and can be computed without a phase-boundary search. The early expansion follows by substitution:The dimensional leading crust thickness is $a\sim \sqrt{2k\Delta Tt/({\rho }_0{L}_f)}$. Thus hotter liquid reduces the first correction to conductive crust growth, whereas colder liquid permits more rapid thickening.

If ${\theta }_0>{\theta }_{\ast }$, the temperature decreases and the instantaneous stationary lava-crust thickness $h_e(\tau )=1/[B\theta (\tau )]$ increases towards $h_{\ast }$. Starting from zero, the crust always remains below this rising curve: at a first contact $h^{\prime }=0$ but $h_e^{\prime }>0$, so it cannot cross from below. Hence hot-start crust grows monotonically. It also stays thinner than the constant-temperature solution: the early expansion puts it below, and at a proposed first crossing their derivative difference is $-SB({\theta }_0-{\theta }_{\ast })e^{-B\tau }<0$, preventing crossing from below. If $0\le {\theta }_0<{\theta }_{\ast }$, the temperature increases and the instantaneous stationary lava-crust thickness decreases. The crust stays thicker than the constant-temperature solution by scalar comparison, but it may cross the decreasing $h_e$, reach a maximum and subsequently melt back towards $h_{\ast }$. Such overshoot is possible, not inevitable; the two relevant relaxation rates are $B$ and $S{R}^2$. The numerical plot uses $B=4$, $R=20$, $S=0.1$, so crust adjustment is faster than temperature relaxation and the cool-start example does overshoot:

The curves use ${\theta }_0/{\theta }_{\ast }=0.5,1,1.5$ and initially $h=0$. They solve the complete reduced time-dependent equations, rather than replacing the moving stationary thickness by the actual crust. At late times, set $h=h_{\ast }+v$, $\delta ={\theta }_0-{\theta }_{\ast }$; linearization gives $v^{\prime }=-S{R}^{2}v-SB\delta e^{-B\tau }$. For $S{R}^2\ne B$,At equal rates the forced term is $-SB\delta \tau e^{-B\tau }$. In particular, slow warming with $B<S{R}^2$ approaches equilibrium from a thicker crust, explaining the plotted remelting. These long-time expressions describe the linearized response; nonlinear corrections can have their own faster exponential rates. All predictions are restricted to $a/H\ll 1$, nonnegative liquid superheat, and the assumed vigorous-convection flux law. If crust growth violates those conditions, the reduced model must be replaced rather than extrapolated to total freezing.

Let $z$ increase upwards, so the bed is at $z=-b(x)$, the surface is at $z=h(x,t)$, and $H=h+b$. In the shallow-shelf approximation the incompressible flow has nearly depth-uniform horizontal speed $u$ and vertical strain $w_z=-u_x$. The Newtonian fluid stress tensor, together with the hydrostatic approximation, then givesThe factor four includes both horizontal extension and the pressure correction required by vertical compression. Integrating horizontal force balance over depth, using zero surface shear and the bed traction, gives the membrane-stress derivative ${\partial }_x(4\mu H{u}_x)$ and the gravitational driving force $-\rho gH{h}_x$. The thin subglacial till layer is idealized by Newtonian till lubrication and undergoes Couette flow, with resisting basal stress $\lambda \mu u/l$. Hence force balance and mass conservation yieldSince $b$ is fixed, $H_t=h_t$. No accumulation or ablation is included.

At the grounding line, the first boundary condition is ice-sheet flotation:For an unbuttressed ice shelf, the second is the extensional stress required to balance the difference between the integrated ice and seawater hydrostatic pressures. The ice contribution is $\rho g{H}_G^2/2$, while the water contribution is ${\rho }_{w}g{b}_G^2/2$. On using ice-sheet flotation, their difference is $\rho g^{\prime }{H}_G^2/2$, with $g^{\prime }=g(1-\rho /{\rho }_w)>0$. Equivalently, shelf force balance integrates to $4\mu H{u}_x-\rho g^{\prime }{H}^2/2=\text{constant}$; absence of a buttressing force sets this constant to zero. Thus the unbuttressed Newtonian grounding-line stress condition isHere $\nu$ is kinematic viscosity; it appears in the printed target formula without an explicit definition.

The ratio of longitudinal stress divergence to till drag is of order $4lH/(\lambda L^2)$. The given small-parameter limit therefore yields the friction-dominated bulk relationThe coefficient depends on the basal lubrication, and $h_x<0$ gives seaward motion. This is a bulk reduction: longitudinal stress has been neglected in the differential equation but its boundary traction remains specified. A complete uniformly valid approximation close to the grounding line could require a membrane-stress boundary layer. In the remainder, use the pointwise friction closure and retain the shelf stress condition as the boundary data, as requested in the paper.

Differentiate the ice-sheet flotation condition along the moving grounding line:Now $H_x=h_x+\alpha$. Substitution of the friction-dominated speed and the shelf stress boundary condition gives the friction-dominated grounding-line evolution law:The coefficient of ${\dot{x}}_G$ is the spatial derivative of the flotation deficit, up to sign. If it vanishes, the implicit equation remains the correct compatibility condition but division to determine a finite speed is invalid; tangential contact needs separate analysis.

In a steady state, volume flux per unit width is constant: $Hu=q_0$. The friction relation givesDifferentiating $Hu=q_0$ and using $u_x=g^{\prime }H/(8\nu )$ at the grounding line givesPut $\beta =l/\lambda$, $\mathcal{H}=H/\beta$ and $\mathcal{Q}=\nu q_0/(g{\beta }^3)$. Substitution and cancellation of $g{\beta }^5/\nu$ yield the steady friction-dominated grounding-line thickness relation:For $q_0>0$ and a bed deepening seaward ($\alpha \ge 0$), the polynomial $(g^{\prime }/8g){\mathcal{H}}^5+\alpha \mathcal{Q}{\mathcal{H}}^2-{\mathcal{Q}}^2$ is strictly increasing for $\mathcal{H}>0$, starts negative and tends to infinity, so it has exactly one positive root. At that root $\mathcal{Q}>\alpha {\mathcal{H}}^2$ and $H_x<0$, as required by steady outward extension. The relation belongs to the specified friction closure; it is not a flux law for arbitrary till rheology or arbitrary membrane-stress matching.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.