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Take $z$ positive upwards, so the source's sign convention makes $p$ a downward applied load. The small-slope tension contribution on a short interval is $T{w}^{\prime \prime }$. Let $r\ge 0$ be the upward contact reaction. Force balance gives $T{w}^{\prime \prime }-p+r=0$. Nonpenetration is $w\ge f$, and reaction is absent where $w>f$. ThereforeThe contact boundary must have continuous $w$ to avoid a broken string, and continuous slope in this smooth-contact model with no concentrated point force. Integrating force balance over a vanishing interval around a contact endpoint gives $T[w^{\prime }]=0$ when the load and reaction are locally bounded. This is the one-dimensional obstacle problem, with its reaction acting as a nonnegative constraint force.

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.

An American option allows exercise at a stopping time $\tau \in [t,T]$. For its exercise financial payoff $H$, the pricing problem isThe discounted value is the Snell envelope of discounted exercise payoffs. Define the pricing operatorIn the continuation region the Black-Scholes equation holds, $\mathcal{L}V=0$. In the exercise region $V=H$, while the ability to wait and the supermartingale property give $\mathcal{L}V\le 0$. Together these conditions are the obstacle problemThus an unknown exercise boundary must be found alongside the value. At a regular boundary in the nondegenerate Itô diffusion, value matching is $V=H$, and smooth pasting is $V_S=H^{\prime }$ where the financial payoff is differentiable. A finite difference method or a binomial options pricing model steps backward taking the larger of continuation and immediate exercise values at each node.

For a non-dividend-paying American call option with $r\ge 0$, no early exercise of a call without dividends follows directly from the European lower boundAt every candidate exercise time, keeping the European call has value at least the immediate exercise financial payoff, so early exercise cannot improve the value; the American and European prices coincide. The first inequality follows from Jensen inequality or put-call parity and nonnegative put value. With $r>0$ an American put option can benefit from early receipt of its strike at sufficiently low stock prices, so its exercise boundary is generally nontrivial and its value is at least the European put value. Dividends can make call exercise worthwhile, and negative interest invalidates the stated no-early-exercise call argument.

Put ${\mathcal{L}}_{r}V(s)=\frac{1}{2}{\sigma }^2{s}^2{V}^{\prime \prime }(s)+rs{V}^{\prime }(s)$ and define the nonnegative reserve rate $a(s)=rV(s)-{\mathcal{L}}_{r}V(s)$. The obstacle problem gives $V\ge g$ and $a\ge 0$. The American-option superhedge with a funded reserve invests the local surplus in the bond rather than consuming it.

For initial wealth $x\ge V(S_0)$, setand choose the stock and bond holdingsThus $D_t\ge 0$ and $F_t\ge V(S_t)\ge g(S_t)$, pathwise at every time. The Itô formula under the original drift givesAdding these equations yieldsThis is a self-financing strategy, and its nonnegative wealth makes it an admissible trading strategy. Continuity of the stock and local regularity of $V$ ensure local integrability of the holdings. The construction does not require $\mu =r$.

For a classical solution, the usual Itô formula applies directly. The smooth fit solution below is $C^1$ and piecewise $C^2$, with locally absolutely continuous first derivative. The generalized Itô formula applies with its almost-everywhere second derivative; the absence of a derivative jump means no boundary local time of a semimartingale term. This is the usual regularity interpretation of the perpetual American option obstacle equation.

Write $\alpha =2r/{\sigma }^2\in (0,1)$. In a continuation interval, the Euler differential equation has power solutions $s$ and $s^{-\alpha }$: substitution of $s^{\beta }$ gives $(\beta -1)(\beta +\alpha )=0$. A decreasing bounded value uses the second solution.

Let $b$ be the exercise boundary. Value matching and smooth fit require $A{b}^{-\alpha }=(1+b{)}^{-1}$ and $-\alpha A{b}^{-\alpha -1}=-(1+b{)}^{-2}$. Dividing gives $b/(1+b)=\alpha$. ConsequentlyThis is a perpetual reciprocal-payoff American option.

To verify the obstacle inequality, observe that $s^{\alpha }/(1+s)$ increases up to $b$ and decreases afterwards, because its logarithmic derivative is $\alpha /s-1/(1+s)$. Therefore $V(s)\ge (1+s{)}^{-1}$ for $s>b$. In the continuation region $({\mathcal{L}}_r-r)V=0$. In the exercise region,The quadratic is convex. At the two endpoints of $[0,b]$ it equals $-\alpha /2$ and $-\alpha /[2(1-\alpha )]$, respectively, so it is negative throughout that interval. Thus the obstacle problem is satisfied on both regions, with value matching and smooth fit at $b$. The second derivative has a jump at $b$; the equation is understood piecewise and in the generalized Itô sense described in part (a).

For optimality, use the Risk-neutral measure for the Black-Scholes model, under which $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. The Itô formula makes $e^{-rt}V(S_t)$ a nonnegative supermartingale, so every exercise time $\tau$ satisfiesBefore hitting the exercise region, its drift vanishes. Since $0\le V\le 1$, the stopped process is a bounded martingale. Apply the optional sampling theorem at ${\tau }_{\ast }\wedge T$ and let $T\to \infty$. The contribution from $\{{\tau }_{\ast }>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds forIf $S_0\le b$, exercise immediately; otherwise wait for the first down-crossing. If that time is infinite, the discounted payout is zero. This proves both the value and the optimal stopping policy, and part (a) supplies its superhedge.

For payoff $g(s)=(1+s{)}^{-1}$ and $\alpha =2r/{\sigma }^2\in (0,1)$, the exercise boundary is $b=\alpha /(1-\alpha )$. The value equals $g$ below $b$ and $(1+b{)}^{-1}(s/b{)}^{-\alpha }$ above it. Value matching, smooth fit, and the obstacle problem verify the first down-crossing policy.