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Integrating 2nd order ODE using midpoint rule |
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| Apr28-12, 04:43 AM | #1 |
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Integrating 2nd order ODE using midpoint rule
Hi
I am trying to integrate Newtons equations for my system [tex] a = \frac{F}{m} = \frac{d^2x}{dt^2} [/tex] This is only for the first coordinate of the particle. I wish to do it for y and z as well, but let us just work with x for now to make it simple. The force in the x-direction depends on the velocity in the x-direction, vx, and the y- and z-coordinate. In other words [tex] F=F(v_x, y, z) [/tex] Now, I wish to solve this equation, and I have currently implemented an Euler method. This is how I iterate [tex] v_{n+1} = v_n + dt\cdot a(v_{x,n},y_n,z_n) \\ x_{n+1} = x_{n} + dt\cdot v_{n} [/tex] I now want to improve the error, and use a 2nd order Runge-Kutta method, i.e. the midpoint rule as briefly summarized here: http://www.efunda.com/math/num_ode/num_ode.cfm I am not quite sure how to do this. In the link they say that now I should generally write [tex] y_{n+1} = y_{n} + dt\cdot f(x_n + dt/2, y_n + k_1/2) [/tex] where [tex] k_1 = dt\cdot f(x_n, y_n). [/tex] This is where my confusion arises: What does [itex]f(x_n + dt/2, y_n + k_1/2)[/itex] correspond to for me? I would really appreciate a hint or two with this. Best, Niles. |
| Apr29-12, 03:47 PM | #2 |
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Here you have it explained:
Computational physics page 292, "13.4 More on finite difference methods, Runge-Kutta methods" |
| May4-12, 11:33 AM | #3 |
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Thanks!
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