Setting Finite-Difference Boundary Conditions for a European Put

For a non-dividend-paying European put under Black–Scholes, use the terminal payoff

P(0,S)=(K-S)^+

when the time variable is remaining maturity

\tau

. At spot zero, set

P(\tau,0)=Ke^{-r\tau}

. At a sufficiently high truncation point

S_{\max}

, use

P(\tau,S_{\max})\approx0

and check whether moving the boundary materially changes the prices you need.

Derive each edge from the payoff

With $\tau=T-t$ , write the risk-neutral value as

P(\tau,S)=e^{-r\tau}\mathbb{E}^{\mathbb Q}[(K-SX)^+], \qquad X=\exp\bigl((r-\tfrac12\sigma^2)\tau+\sigma\sqrt{\tau}Z\bigr),

where

Z

is standard normal. Substituting

S=0

gives the exact lower boundary. Since

X>0

almost surely,

(K-SX)^+\to0

pointwise as

S\to\infty

. The payoff is bounded by

K

, so dominated convergence gives the zero limit. With finite

S_{\max}

, the value at the boundary generally remains positive.

On a spot grid, suppose a backward-Euler row next to the left edge contains $aP_0^{n+1}+bP_1^{n+1}+cP_2^{n+1}$ on the left side. Substituting the known boundary moves $aP_0^{n+1}$ to the right side with a minus sign. The boundary is a known value at the new time layer, whereas the right side also contains known values from the previous time layer.

S=0

, a stock in this multiplicative model stays at zero, so the put payoff is the certain payment

K

at maturity. Today's value is therefore the discounted strike for that layer's remaining time. At high spot the put payoff tends to zero, but the limit is attained only as spot tends to infinity. Using zero at finite

S_{\max}

is a numerical approximation. Compare the price near the strike after moving

S_{\max}

farther out while keeping comparable local grid spacing. That separates the boundary choice from how finely you resolve the payoff kink.

Take $K=100$ , $r=4\%$ , $\sigma=25\%$ , and a maximum remaining maturity of nine months. At the payoff layer the zero-spot value is $100$ . At the nine-month layer it is $100e^{-0.04(0.75)}\approx97.04$ . Set the upper edge at $350$ . The analytic put value there is about $10^{-8}$ in this example, so zero is an accurate approximation at that edge. At spot $85$ , the analytic price is about $15.29$ . Compare that value to the solver’s output and repeat with a higher upper boundary while keeping spot step fixed. Holding the spot step fixed helps separate boundary truncation from grid resolution.

It helps to picture the time-marching array. At the initial layer of remaining maturity zero, the left node is the strike and each interior node is the immediate exercise payoff. After one step the left node is the discounted strike for one step of time. Its value changes again at every later layer. In an implicit solver, the new-layer boundary contribution moves to the right-hand side of the linear system. Using the old-layer value there would solve a slightly different boundary problem.

Derive each edge from the payoff

References and next steps