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## Main Question or Discussion Point

Consider an ODE of general form:

[itex]\frac{d}{dt}y(t) = f(y(t),t)[/itex]

with an initial condition

[itex]y(t_0) = y_0 [/itex].

We create an approximate solution by an intitialization step

[itex] v := f(y_0,t_0) [/itex]

[itex] t:= t_0 [/itex]

[itex] y:= y_0 [/itex]

and iteration of the following modified leapfrog step

[itex] \tau := h/2 [/itex]

[itex] t += \tau [/itex]

[itex] y += \tau v [/itex]

[itex] v = 2 f(y,t) -v[/itex]

[itex] y += \tau v [/itex]

[itex] t += \tau [/itex]

which increases t by h.

Thus one has to evaluate the right-hand side of the differential equation only

once per step. The method turns out to be second order and very robust.

Did anybody find this method mentioned in the literature or is it even well known

in some communities? Please tell me.

More on the method can be found in

http://www.ma.utexas.edu/mp_arc/c/08/08-197.pdf

[itex]\frac{d}{dt}y(t) = f(y(t),t)[/itex]

with an initial condition

[itex]y(t_0) = y_0 [/itex].

We create an approximate solution by an intitialization step

[itex] v := f(y_0,t_0) [/itex]

[itex] t:= t_0 [/itex]

[itex] y:= y_0 [/itex]

and iteration of the following modified leapfrog step

[itex] \tau := h/2 [/itex]

[itex] t += \tau [/itex]

[itex] y += \tau v [/itex]

[itex] v = 2 f(y,t) -v[/itex]

[itex] y += \tau v [/itex]

[itex] t += \tau [/itex]

which increases t by h.

Thus one has to evaluate the right-hand side of the differential equation only

once per step. The method turns out to be second order and very robust.

Did anybody find this method mentioned in the literature or is it even well known

in some communities? Please tell me.

More on the method can be found in

http://www.ma.utexas.edu/mp_arc/c/08/08-197.pdf

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