# Perpetual Option Paying Chooser Option

A perpetual option solves the ODE $$rSV_S+\frac{1}{2}\sigma^2S^2V_{SS}-rV=0$$ The general solution is $$V(S)=aS+bS^{\gamma}$$ where $$\gamma=-\frac{2r}{\sigma^2}<0$$.

For an American put option with payoff $$K-S$$, we find $$a=0$$ because we require $$V(S)=0$$ as $$S\to\infty$$. We find the free boundary (exercise point, $$S^*$$) and the remaining free parameter ($$b$$) by value-matching and smooth-pasting of $$V(S)$$ with the payoff $$K-S$$ at $$S=S^*$$, that is \begin{align} b(S^*)^{\gamma} &= K-S^* \\ b\gamma(S^*)^{\gamma-1} &= -1 \end{align} The option value is then \begin{align} V(S)=\begin{cases} K-S &if\; S

Question: What happens if the option pays $$\max(K_1-S,K_2-2S)=K_2-2S+\max(S+K_1-K_2,0)$$ instead of $$K-S$$? The payoff now resembles a chooser option (between two puts). We still require $$a=0$$ such that $$V(S)=0$$ as $$S\to\infty$$. But how to proceed? I don't think it's as simple as finding $$b$$ and $$S^*$$ by solving value-matching and smooth-pasting condition and setting \begin{align} V(S)=\begin{cases} \max(K_1-S,K_2-2S) &if\; S