The Leap-Frog Method

If we expand both and as before (1.28) becomes

(1.29) | |||

(1.30) |

from which all terms up to cancel so that the method is clearly 2nd order accurate. Note in passing that using (1.28) 2 consecutive values of are required in order to calculate the next one: and are required to calculate . Hence 2 boundary conditions are required, even though (1.28) was derived from a 1st order differential equation. This so-called

We analyse this equation by writing and to obtain

which has the solutions

The product of the 2 solutions is equal to the constant in the quadratic equation, i.e. . Since the 2 solutions are different, one of them always has magnitude . Since for a small random error it is impossible to guarantee that there will be no contribution with this contribution will tend to dominate as the equation is iterated. Hence the method is unstable. There is an important exception to this instability: when the partial derivative is purely imaginary (but not when it has some general complex value), the quantity under the square root in (1.33) can be negative and both 's have modulus unity. Hence, for the case of oscillation (1.1c) where , the algorithm is just stable, as long as

The stability properties of the leap-frog method are summarised below

Decay | Growth | Oscillation |

unstable | unstable |