By the absence of arbitrage principle, all options are non-negative. (If this were not the case an option buyer would receive cash upfront if say a buyer purchased a call at a negative price). Hence
American Call.
The relation must hold true. Otherwise, ignoring transaction costs, an investor can sell the stock short, and buy the arbitraged in-the-money call strike $K < S$ (usually significantly cheaper than the stock) and then immediately or anytime before expiration $T$ buy the stock at the lower strike price $K$, to close the short position. The arbitrage profit will be the difference in stock prices less the call price.
SELL: – Stock $S_{mkt}$ + Call (strike $K < S_{mkt}$)
BUY: Stock $S_{K}$
PROFIT: -Stock $S_{K}$ – (- $S_{mkt}$ + Call (strike $K < S_{mkt}$) )
= $S_{mkt}$ – Call – Stock $S_{K}$
In regular notation,
Profit = $S – K – C^A$

Snapshot at time t to expiration of an American call. Point S* is optimal exercise price where the intrinsic value broken line is tangential to the call price curve.
We see that as the stock price increases so does the call price in a linear fashion beyond $S_*$. For $S \geq S_*$, exercise is optimal and so this is known as the stopping region.
Consider a stock $S$ currently trading at $80$.
Intrinsic + time value | price at $t$ is time value alone. Intrinsic value is zero |
|
price at $t$ is time value alone. Intrinsic value is zero | Intrinsic + time value |
|
American Put
The relation must hold true. Otherwise, ignoring transaction costs, an investor can buy the stock, and the arbitraged in-the-money put strike $K > S$ (usually significantly cheaper than the stock) and then immediately sell the stock at the higher strike price $K$, making a profit of the difference in stock prices, less the put price.
BUY: Stock $S_{mkt}$ + Put (strike $K > S_{mkt}$)
SELL: Stock $S_{K}$
PROFIT: Stock $S_{K}$ – ($S_{mkt}$ + Put (strike $K> S_{mkt}$) )
=Stock $S_{K}$ – $S_{mkt}$ – Put
In regular notation,
Profit = $ K – S – P^A$

Time t to expiration of an American put. Point S* is optimal exercise price where the intrinsic value broken line is tangential to the put price curve.
\begin{align}& \dfrac{\partial V^A}{\partial t} + \dfrac{1}{2}\sigma^2 S^2 \dfrac{\partial^2 V^A}{\partial S^2} + (r – q) S \dfrac{\partial V^A}{\partial S} – rV^A = 0 \quad (t < T) \tag{1}\\
\\& V(S,t) > F(S, t)\end{align}
Corollary~ 8.5.3. The price of an American call on an asset not paying a
dividend is the same as the price of the European all on the same asset with the same expirationThe idea behind Corollary 8.5.3 is that the discounted process $e^{-rt}(S (t) –
K)^+$ is a submartingale under $\widetilde{\mathbb{P}}$ and hence tends to rise. Therefore, it is optimal to wait until expiration before deciding whether to exercise. There are two factors that contribute to the submartingale property for $e^{- rt}(S (t )-K )^+ $.One is the discounting of the strike. In fact, $e^{- rt}(S(t )- K)$ (without the $^+$) is a submartingale because $e^{- rt}S(t)$ is a martingale under the risk-neutral measure and $e^{- rt}K$ increases as $t$ increases (throughout this chapter, we assume strictly positive interest rate $r$). When we reinstate the $^+$, we are taking a convex function of a submartingale and, because of Jensen’s inequality, this reinforces the upward trend.
The previous argument does not apply to the American put, whose discounted intrinsic value $e^{-rt}(K – S(t))$ (without the +) is a supermartingale $(e^{-rt}K$ falls and $- e^{- rt}S (t )$ is a martingale). Jensen’s inequality creates an upward trend that competes with this supermartingale property, and the analysis becomes complicated.
If the underlying asset pays a dividend, the case considered in the next subsection, the argument above no longer applies to the American call. In this case, $e^{- rt}S(t )$ is a supermartingale and tends to fall because of the dividend outflow.
If $S = S_*$, we obtain the boundary condition for the Black Scholes problem
$$P^A(S_{*}(t)) = \text{max} (K – S_{*}(t), 0) \tag{2}$$
The put will never be exercised when its exercise value is zero. We need to satisfy the final boundary condition
$$P^A(S, T) = \text{max} (K – S, 0) \tag{3}$$
which implies that the value at $T$ is simply its payoff it has not been exercised prior to $T$. However, equations $1$ and $2$ above do not suffice for solving the problem of determining a unique value for optimal exercise $S_*$. Making $P^A(S, t) \equiv V_{\text{put}}(S;S^*) $ as large as possible is the origin of the third equation known as the smooth pasting condition.
Such that $$\dfrac{\partial P^A }{\partial S} (S_*(t), t) = -1 \tag{4}$$
Equations $1$ to $4$ above constitute a free boundary problem. The determination of the optimal exercise boundary $S_∗(t)$ as function of $t$ is part of the problem, and is what makes it non-linear.
In the case of a Perpetual American Option the option is never exercised (unlike the finite maturity American option that can be exercised at anytime). This eliminates the dependency on time $t$ in the Black Scholes model and reduces the free boundary problem to one with an explicit solution.
Finite expiration American Puts are beyond the scope of the course but are discussed in literature[ref]Steven E. Shreve. Stochastic Calculus for Finance II, Continuous-Time Models. Chapter 8.4[/ref]