= Solution
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 gives
$$
\sigma_{zz}=-\rho g(h-z),\qquad \sigma_{xx}=-\rho g(h-z)+4\mu u_x.
$$
The 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> $\partial_x(4\mu Hu_x)$ and the gravitational driving force $-\rho gHh_x$. 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> yield
$$
\boxed{4\mu\partial_x(Hu_x)-\rho gHh_x-\lambda\mu\frac ul=0,\qquad H_t+\partial_x(Hu)=0.}
$$
Since $b$ is fixed, $H_t=h_t$. No accumulation or ablation is included.
At the <grounding line>, the first boundary condition is <ice-sheet flotation>:
$$
\boxed{\rho H_G=\rho_wb_G,\qquad H_G=\frac{\rho_w}{\rho}\alpha x_G.}
$$
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 $\rho gH_G^2/2$, while the <water> contribution is $\rho_wgb_G^2/2$. On using <ice-sheet flotation>, their difference is $\rho g'H_G^2/2$, with $g'=g(1-\rho/\rho_w)>0$. Equivalently, shelf force balance integrates to $4\mu Hu_x-\rho g'H^2/2=\text{constant}$; absence of a buttressing force sets this constant to zero. Thus the <unbuttressed Newtonian grounding-line stress> condition is
$$
\boxed{4\mu H_Gu_{x,G}=\frac12\rho g'H_G^2,\qquad u_{x,G}=\frac{g'H_G}{8\nu},\qquad \nu=\mu/\rho.}
$$
Here $\nu$ 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 relation
$$
\boxed{u=-\frac l\lambda\frac g\nu Hh_x,\qquad h_t=\frac l\lambda\frac g\nu\partial_x(H^2h_x).}
$$
The 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>:
$$
H_t+H_x\dot x_G=\alpha\frac{\rho_w}{\rho}\dot x_G,\qquad \left(\alpha\frac{\rho_w}{\rho}-H_x\right)\dot x_G=H_t=-uH_x-Hu_x.
$$
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>:
$$
\boxed{\left(\alpha\frac{\rho_w}{\rho}-H_x\right)\dot x_G=\frac l\lambda\frac g\nu Hh_x(h_x+\alpha)-\frac{g'}{8\nu}H^2\quad\text{at }x=x_G(t).}
$$
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 gives
$$
h_x=-\frac{\lambda\nu q_0}{lgH^2},\qquad H_x=\alpha-\frac{\lambda\nu q_0}{lgH^2}.
$$
Differentiating $Hu=q_0$ and using $u_x=g'H/(8\nu)$ at the <grounding line> gives
$$
\frac{q_0}{H}H_x+\frac{g'H^2}{8\nu}=0,\qquad \frac{g'H^5}{8\nu}=q_0\left(\frac{\lambda\nu q_0}{lg}-\alpha H^2\right).
$$
Put $\beta=l/\lambda$, $\mathcal H=H/\beta$ 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>:
$$
\boxed{\frac{g'}{8g}\mathcal H^5=\mathcal Q(\mathcal Q-\alpha\mathcal H^2).}
$$
For $q_0>0$ and a bed deepening seaward ($\alpha\ge0$), the polynomial $(g'/8g)\mathcal H^5+\alpha\mathcal Q\mathcal H^2-\mathcal Q^2$ 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.
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