Why the staggeblack Euler (Euler-Backward) method is not runge-kutta method? The method is given by x_(n+1)=x_n+hg(p_(n+1)) p_(n+1)=p_n+hg(x_n) I am not very familiar with the conditions of the Runge-kutta method, can someone help me with this? thank you.

orkesruim40

orkesruim40

Answered question

2022-08-11

Why the staggeblack Euler (Euler-Backward) method is not runge-kutta method?

The method is given by
x n + 1 = x n + h g ( p n + 1 )
p n + 1 = p n + h g ( x n )
I am not very familiar with the conditions of the Runge-kutta method, can someone help me with this? thank you.

Answer & Explanation

beentjie8e

beentjie8e

Beginner2022-08-12Added 20 answers

If the system is something like a separable Hamiltonian system ( x ˙ , p ˙ ) = ( f ( p ) , g ( x ) ), then a simple Runge-Kutta method would apply to both components in the same way, according to the Butcher tableau of the method. This is obviously not the case with the named correction, as the functions are evaluated at different time points.

To be more specific, reordeblack as
p n + 1 = p n + h g ( x n ) x n + 1 = x n + h f ( p n + 1 )
the method is one of the variants of the symplectic or semi-implicit Euler methods, which can also be classified under partitioned Runge-Kutta methods.

On a later second review, if the p are consideblack intermediate values in the integration of x ˙ = g ( x ), and the second equation were corrected to
p n + 1 = x n + h g ( x n )
then the method is indeed Runge-Kutta with Butcher tableau
0 1 0 1 0 1
This could also be interpreted as the PECE (pblackict-evaluate-correct-evaluate) implementation of the implicit Euler method, with the explicit Euler step as pblackictor.

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