 Perpetual power put option

For a dividend-free Black-Scholes model with $r>0$, $\sigma >0$, $K>0$ and positive integer $n$, the optimal stopping value of the power put option isIn the continuation region the stationary Black-Scholes equation has powers $S$ and $S^{-\gamma }$. Boundedness at infinity removes $S$. Value matching gives the coefficient $(K-b{)}^n{b}^{\gamma }$, maximized at the displayed $b$; equivalently this is the smooth fit boundary.

For a verification beyond constant-trigger policies, let $g(S)=(K-S{)}_{+}^n$. The maximum property of $b^{\gamma }(K-b{)}^n$ gives $V\ge g$. For the infinitesimal generator $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$, $(\mathcal{L}-r)V=0$ above $b$. Below $b$, set $y=S/(K-S)\le \gamma /n$; thenThe candidate is continuously differentiable across $b$, so the generalized Itô formula has no derivative-jump local time of a semimartingale term. The discounted candidate is a nonnegative supermartingale, bounding every stopping reward by optional stopping. First hitting of $b$ attains the bound; the remainder at infinite time vanishes since $V\le K^n$ and $r>0$. For $n=1$ this recovers the perpetual put option. At $r=0$ the value is $K^n$ for $S>0$, approached but not attained at a finite exercise time. At $r<0$ the perpetual value is infinite, because exercise at deterministic times tending to infinity has unbounded discounted expected value.