= Solution
The <Newtonian shallow-ice flux> on the flat bed is
$$
q=-\frac{gh^3}{3\nu}h_x,
$$
so <mass conservation> gives
$$
\boxed{
h_t=\frac g{3\nu}(h^3h_x)_x+A\quad(0<x<x_s),
\qquad
h_t=\frac g{3\nu}(h^3h_x)_x-A\quad(x_s<x<x_N).}
$$
At the <ice divide>, symmetry gives $h_x(0,t)=q(0,t)=0$. At the terminus, $h(x_N,t)=0$ and the moving-front mass balance applies; in a steady state $q(x_N)=0$. At the <snowline>, $h(x_s,t)=h_s$ and both $h$ and $q$ are continuous, hence $h_x$ is continuous because $h_s>0$.
In a steady state, $q_x=A$ in the accumulation region and $q_x=-A$ in the ablation region. Therefore
$$
q=Ax\quad(0<x<x_s),
\qquad
q=A(x_N-x)\quad(x_s<x<x_N).
$$
Flux continuity gives $x_N=2x_s$. Integrating the flux law gives the piecewise shape
$$
\boxed{
h^4=h_0^4-\frac{6\nu A}{g}x^2
\quad(0\leq x\leq x_s),}
$$
$$
\boxed{
h^4=\frac{6\nu A}{g}(x_N-x)^2
\quad(x_s\leq x\leq x_N).}
$$
Matching $h(x_s)=h_s$ yields
$$
\boxed{x_s=\left(\frac g{6\nu A}\right)^{1/2}h_s^2,
\qquad x_N=2x_s,
\qquad h_0=2^{1/4}h_s.}
$$
The largest surface slope is of order $\sqrt{\nu A/g}/h_s$, so the <shallow-ice approximation> requires
$$
\boxed{A\ll\frac{gh_s^2}{\nu}.}
$$
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