Discretising the equation

Replace the derivatives in the Black-Scholes equation with differences on a grid of spot values \(S_j = j\,\Delta S\) and times \(t_n = n\,\Delta t\):

$$ \frac{\partial V}{\partial S} \approx \frac{V_{j+1} - V_{j-1}}{2\Delta S}, \qquad \frac{\partial^2 V}{\partial S^2} \approx \frac{V_{j+1} - 2V_j + V_{j-1}}{\Delta S^2} $$

The value at expiry is the payoff. Stepping backwards in time, each level is found from the one after it. Boundary conditions pin the edges: a call is worthless at \(S = 0\) and worth \(S e^{-q\tau} - K e^{-r\tau}\) for very large \(S\).

Three schemes

Write the spatial operator as \(L\). The theta-scheme is

$$ V^{n} - \theta\, L V^{n} = V^{n+1} + (1 - \theta)\, L V^{n+1} $$

• Explicit, \(\theta = 0\). Each new value is a weighted sum of three old ones. No equations to solve, but stable only when \(\Delta t \lesssim 1 / (\sigma^2 M^2)\) for \(M\) spot steps. Fine grids need enormous numbers of time steps.
• Implicit, \(\theta = 1\). A tridiagonal system every step, solved in linear time by the Thomas algorithm. Stable for any step sizes; first-order accurate in time.
• Crank-Nicolson, \(\theta = \tfrac{1}{2}\). The average of the two. Stable and second-order accurate in time. The default choice, though the kink in the payoff can cause small oscillations near the strike.

American exercise

After each time step, replace every grid value by the larger of itself and the payoff. This projection is simple and works well; more careful methods solve the linear complementarity problem exactly at each step.

What you get

Unlike a tree or a simulation, a grid solve returns the option value at every spot and every time at once. Delta and Gamma are differences of neighbouring grid values, Theta the difference between two time levels. The heatmap on the solver page is that whole surface.