Given a system of equations:

\begin{align} &f''(x) = -a \cdot \sin(f(x))\\ &f(0) = b\\ &f'(0) = c \end{align}

$a, b, c, dt, N$ are arbitrary parameters.

How to get a values of $f(0), f(dt), f(2dt) ... f(N)$. I am stuck with the non-linearity of the right part of the first equation.

I will be very glad, if someone will show me an implementation of an algorithm calculating this.

  • $\begingroup$ I am stuck with the non-linearity of the right part of the first equation. Does that comment mean that you are able to numerically solve the system for the linear case? $\endgroup$ – nicoguaro Apr 24 at 18:34
  • $\begingroup$ Yes, i can build a system of linear equations for each $f_i$ and build a matrix, which can be solved via Gaussian method. $\endgroup$ – Constantor Apr 24 at 18:40
  • $\begingroup$ It's better to be consistent in using x or t for the independent variable. $\endgroup$ – Maxim Umansky Apr 25 at 4:53
  • $\begingroup$ Since this is an initial-value problem, you can solve it numerically using the Euler method, for example. $\endgroup$ – Christoph Apr 25 at 5:17

With the functions $y_1 := f$, $y_2 := f'$, $\pmb{y} := (y_1,y_2)^{\top}$, we obtain an initial-value problem with an autonomous first-order system: $$ \pmb{y}' = \left( \begin{array}{c} y_2\\ -a \sin(y_1) \end{array} \right) =: \pmb{f}(\pmb{y}), \quad \pmb{y}(0) = \left( \begin{array}{c} b\\ c \end{array} \right) =: \pmb{y}_0. $$ We now choose the Euler method for the numerical solution: $$ \pmb{y}_i = \pmb{y}_{i-1} + h \pmb{f}(\pmb{y}_{i-1}), $$ $i = 1, 2, \dots$. This yields approximations $\pmb{y}_i \simeq \pmb{y}(x_i)$ at the positions $x_i = ih$.

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    $\begingroup$ Euler is of course just one simple possibility. For the conservation of energy it might be advantageous to use a symplectic method such as Verlet integration. $\endgroup$ – Christoph Apr 25 at 8:14
  • $\begingroup$ Thank you very much! But I know that Euler's method is not accuarate enough. Is there a way with better accuracy? $\endgroup$ – Constantor Apr 25 at 12:38
  • $\begingroup$ You could use a higher-order Runge-Kutta method instead of Euler, and/or reduce the step size $h$. $\endgroup$ – Christoph Apr 25 at 13:34
  • $\begingroup$ Higher order (explicit) RK does not imply a smaller step size. A-Stability has to be taken into account $\endgroup$ – VoB Apr 25 at 17:19

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