American Put Free-Boundary Conditions in the Black–Scholes Model

Summary

The document asks how to state the Black–Scholes free-boundary problem for an American put. It contrasts two proposed condition sets involving the exercise region, continuation region, terminal payoff, behavior at large stock prices, and value and delta matching at the exercise boundary. The central modeling idea is that the option equals intrinsic value below the optimal exercise boundary, while above it the value satisfies the Black–Scholes equation; value and slope meet smoothly at the boundary.

The post does not provide a worked derivation or a definitive corrected formulation. Its equations appear to contain notation or completeness issues, so readers should distinguish conditions imposed in the continuation region from value matching, smooth pasting, terminal payoff, and far-field behavior. The question about whether the boundary lies below the strike is also raised but not answered. Boundary behavior can depend on model assumptions and time conventions, so the listed conditions should be checked against a standard American-option formulation before implementation.

Key ideas

• The American put problem divides the stock-price domain into exercise and continuation regions.
• In the continuation region, the option value satisfies the Black–Scholes differential equation.
• At the exercise boundary, value matching and smooth pasting connect the two regions.
• The terminal payoff and large-stock boundary condition help complete the pricing problem.
• The source poses questions about the conditions but does not resolve them.

Tags

# Some confusion on american put pde


# Some confusion on american put pde












Suppose $$L(v) = \dfrac{\partial v}{\partial t} + rS\dfrac{\partial v}{\partial S} + \dfrac{1}{2}\sigma^2S^2$\dfrac{\partial^2 v}{\partial S^2} -rv$$ is Black-Scholes operator.

`First version` is $$P = K - S, L(P)<0\quad 0\leq S< B(t).$$ $$P>K - S,\ L(P) =0\quad S> B(t)$$ $$P = \max\{K - S, 0 \},\ \dfrac{\partial P}{\partial S} = -1\quad S = B(t).$$

`Second version`: $$P = \max\{K - S\}\quad 0\leq S< B(t).$$ $$L(P) = 0,\quad S>B(t)$$ $$P(S,T) = \max(K-S,0),\quad \lim\limits_{S\rightarrow\infty}P = 0;$$ $$P = K-S,\ \dfrac{\partial P}{\partial S} = -1\quad S = B(t),\ B(T) = K$$

Could any one tell the lack or repeat conditions in each version? Give me a correct version.

By the way, do we need the condition $B(t) \leq K？$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.