Diagnosing an Unstable Explicit Black–Scholes Grid

An explicit finite-difference step uses the old price curve to advance its diffusion and curvature. On a fixed spatial grid, a large time step can amplify small grid-scale errors at every update. Refine the time step, or use backward Euler or Crank–Nicolson with suitable treatment of the payoff kink and a convergence check.

Track what the explicit update does to an alternating error

For the diffusion equation $U_\tau=D U_{xx}$ , the standard explicit update uses $\lambda=D\Delta\tau/\Delta x^2$ :

U_j^{n+1}=(1-2\lambda)U_j^n+\lambda(U_{j-1}^n+U_{j+1}^n).

For the alternating error

e_j=(-1)^j

, each neighbour equals

-e_j

. One update multiplies the error by

1-4\lambda

. Its absolute value exceeds one when

\lambda>1/2

. Fourier modes at lower frequencies impose no tighter bound for this simple stencil. The Black–Scholes log-price equation has diffusion coefficient

D=\sigma^2/2

, along with drift and discounting, so this heat-equation analysis explains the scaling. Derive or check the constraint for the actual discretisation used.

On a uniform spot grid $S_j=j\Delta S$ , the explicit central Black–Scholes update has diagonal coefficient $1-\Delta\tau(\sigma^2j^2+r)$ . Requiring all coefficients to be nonnegative is a useful sufficient monotonicity check for this stencil with nonnegative $r$ . Also check the off-diagonal coefficients, especially at small $j$ where central differencing of the drift can create a negative weight.

Insert $e_{j-1}=e_{j+1}=-e_j$ into the update to get a multiplier of $1-4\lambda$ . If its magnitude exceeds one, even a small alternating perturbation grows each time you step. This explains the wiggles that appear before the prices become obviously wrong. In Black–Scholes the diffusion coefficient increases with $S_j^2$ , so the largest spot indices can be the most restrictive. Halving the spatial step at fixed upper spot typically calls for about four times as many time steps in the diffusion-dominated part of an explicit grid.

Calculate the limit on one stock grid

Let maturity be one year, $r=5\%$ , $\sigma=25\%$ , and $S_{\max}=400$ with 200 spot intervals. The highest interior index is $199$ , so keeping its diagonal coefficient nonnegative requires $\Delta\tau\le1/(0.25^2(199)^2+0.05)\approx0.000404$ . The off-diagonal weights are nonnegative for these inputs. Using 4,000 time steps gives a diagonal weight of about $0.381$ at that node. Using 1,000 time steps gives a weight near $-1.475$ . Inspect the entire price curve, including the far right, because an instability seeded there can grow long before the price at spot $100$ appears visibly wrong.

Use the options pricer to spot-check your explicit-grid European prices after refining the time step.

Track what the explicit update does to an alternating error

Calculate the limit on one stock grid

References and next steps