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The contact set is the region where the obstacle inequality $w\ge f$ is active. On its complement the obstacle supplies no reaction; its boundary is a free boundary, determined along with the unknown solution. A point can belong to the contact set while its contact force is zero, as for a taut horizontal string just touching an obstacle at one point. Thus equality of the gap alone does not imply strictly positive pressure.

A free boundary problem determines both a field and part of its domain boundary. A glacier terminus, for example, separates positive ice thickness from bare bed. Mass conservation, nonnegative thickness, a rule for ice ablation only where ice exists, and suitable interface or entropy solution conditions are needed; a formal zero of one characteristic branch alone need not give the selected physical boundary.

Use the usual nondegenerate positive-rate case $r>0$, $\sigma >0$, $K>0$. Define the Black-Scholes equationThe American value dominates immediate exercise. Its discounted value is a supermartingale under the risk-neutral measure, so $\mathcal{L}V\le 0$, with equality where it is optimal to continue. The obstacle problem isEquivalently $\max (\mathcal{L}V,g-V)=0$.

For the put, write the exercise boundary as $S=b(t)$. On $S>b(t)$ solve $\mathcal{L}V=0$; on $0<S\le b(t)$ set $V=K-S$. Value matching and smooth pasting requireThe remaining conditions are $V(t,0)=K$, $V(t,S)\to 0$ as $S\to \infty$, and the terminal contingent claim payoff, with $b(T-)=K$ in this positive-rate setting. The boundary is an unknown to be found with the price, which is what makes this a free boundary problem.

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 givesFor 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,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 isThe 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 requiresEquivalently 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 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 $L_{\xi }=1-2\xi$ givesBoth 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 $z=\xi +(1-2\xi )y$ givesThis 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 $H\ge 0$, 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 $n>0$.