= Solution
Write $A(x)=x-x^2$, $q=1/(n+2)$ and $p=(n+1)/(n+2)$. Integrating the steady <mass conservation> equation from the head gives
$$
\boxed{H_0(x)=[(n+2)A(x)]^q,\qquad0<x<1.}
$$
For the time-dependent first-order equation $H_t+H^{n+1}H_x=1-2x$, <method of characteristics> gives $\dot x=H^{n+1}$ and $\dot H=1-2x$. Consequently $d[H^{n+2}/(n+2)-A(x)]/dt=0$. On a <characteristic> starting at $(\xi,0)$ with zero thickness,
$$
H^{n+2}=(n+2)[A(x)-A(\xi)],\qquad t=\int_\xi^x\frac{dx'}{[(n+2)(A(x')-A(\xi))]^p}.
$$
On a <characteristic> emitted by the boundary $x=0$, the invariant is zero and $H=H_0(x)$. The travel time of the one emitted at time zero is
$$
\boxed{f(x)=(n+2)^{-p}\int_0^x[z(1-z)]^{-p}\,dz=(n+2)^{-p}B_x(q,q).}
$$
The <incomplete beta function> is finite at both endpoints because $0<p<1$ for the physical exponents $n>0$. Before <characteristic> crossings, the initial-data branch and boundary-data branch have the displayed forms in the source, with the appropriate emission time $t-f(x)$ for the latter.
<Glacier characteristic admissibility> gives a substantive domain restriction. Factoring the invariant gives $A(x)-A(\xi)=(x-\xi)(1-x-\xi)$. A positive forward branch requires
$$
\boxed{0\leq\xi<x<1-\xi,\qquad0\leq\xi<1/2.}
$$
Equivalently it is the image of this parameter domain under $(\xi,x)\mapsto(x,t(\xi,x))$, restricted further to regions where that map is single-valued. <Characteristics> cannot start with positive thickness growth at $\xi>1/2$, where the prescribed initial rate $H_t=1-2\xi$ is negative. This is why \b[the printed piecewise expression is a formal <characteristic> construction, not a global nonnegative solution from a bare bed]. For example for $x>1/2$, putting $L_\xi=1-2\xi$ gives
$$
t(\xi,x)=(n+2)^{-p}L_\xi^{-n/(n+2)}B_{(x-\xi)/L_\xi}(q,q)\geq f(x).
$$
Both factors increase with $\xi$ there, so no positive initial-data branch exists at that $x$ for $0<t<f(x)$, contrary to a literal global reading of the first case. Compression can also make <characteristics> intersect; <entropy solutions> then require <shock waves>.
One can nevertheless give the exact zero-thickness endpoint locus of these formal <characteristics>. Let $K=f(1)=(n+2)^{-p}B(q,q)$. At $x=1-\xi$ the change of variable $z=\xi+(1-2\xi)y$ gives
$$
t_{\rm end}(\xi)=K(1-2\xi)^{-n/(n+2)},\qquad\boxed{x_{\rm end}(t)=\frac12\left[1+\left(\frac Kt\right)^{(n+2)/n}\right],\quad t\geq K.}
$$
This is the formal snout locus inferred from the given <characteristic> family; it retreats from $1$ towards $1/2$. It is not justified as the physical terminus of an accumulating glacier formed from zero initial <ice>. In particular it says nothing about a positive advancing snout at $t<K$. A physical <free boundary problem> must specify no ablation on a bare bed, maintain $H\geq0$, and select an <entropy solution> or retain the small slope regularization. Those additional conditions are absent from the printed first-order construction. Thus a unique global physical snout trajectory cannot be claimed from that construction alone. The steady endpoint remains $x=1$. The degenerate $n=0$ limit would instead have the same endpoint time $K$ for every initial <characteristic>, so the inverted formula presumes $n>0$.
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