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The leapfrog method for the ODE $y' = f(t,y)$ is derived like the forward and backward Euler methods, except that centered differencing is used. This yields the formula $$\frac{y_{i+1}-y_{i-1}}{2h}=f(t_i,y_i)$$ Show that this is an explicit linear two-step method that is second order accurate and does not belong to the Adams or BDF families.

So we can get the relation that $y_i'=\frac{y_{i+1}-y_{i-1}}{2h}$. But I am not sure what game I should play with this to derive that it is explicit linear two-step. I.e. do I look for a interpolating polynomial or do I play around with Taylor series and plug this in to show it is second order accurate? Any help is appreciated. Thank you!

Lutterbach
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  • This is also known as the two-step Nyström or the central Euler method. For exhaustive analysis see https://math.stackexchange.com/q/3067795/115115, https://math.stackexchange.com/q/3074952/115115, https://math.stackexchange.com/q/3330910/115115 and the sources mentioned there. – Lutz Lehmann Nov 21 '19 at 16:31
  • Explicit: $y_{i+1}$ occurs exactly once, and can be easily isolated. Two-step: Only data from $t_i$ and $t_{i-1}$ is used in one step of the method. The true question is to show that it is second order accurate, at least locally, which follows from the error order of the central difference quotient. – Lutz Lehmann Nov 21 '19 at 16:39

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You can use Taylor's formula to get the local truncation error, which is the error in the approximation of $y(t_{i+1})$ by applying the method with exact values for $y_i$ and $y_{i-1}$. \begin{align*} \tilde y_{i+1}-y(t_{i+1})= & y(t_i-h)+2h y'(t_i)-y(t_i+h)\\ = & \left(y(t_i)-hy'(t_i)+\frac{h^2}{2}y''(t_i) + O(h^3)\right)+2hy'(t_i)-\\ & -\left(y(t_i)+hy'(t_i)+\frac{h^2}{2}y''(t_i)+O(h^3)\right) \\ = & O(h^3) \end{align*}

A local truncation error of order 3 will yield a global error of order 2.

PierreCarre
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