How can I solve a system of second order ODE's numerically?

[peterbone]

Hello. First post here.

I'm trying to write a program (from scratch) to simulate a double inverted pendulum (a cart with 2 pendulums one on top of the other). The system is modeled with a system of 3 second order ODE's, which I need to solve numerically using Runge Kutta. I know how to solve a system of first order ODE's numerically but not a system of second order ODE's. The equations are shown in this paper (there's no point in me re-writing them here):

http://www.tf.uni-kiel.de/etech/ART/paper/2001/ieee_cca_isic_zhong.pdf
(equations 4 to 6)

So can anyone tell me how to go about solving this initial value problem numerically? I have looked in many books but can only find examples of systems of first order equations and single second order equations.

Thanks

Peter Bone

 

[HallsofIvy]

Each second order differential equation is equivalent to two first order equations so you could write this system as six first order equations.

 

[peterbone]

Thanks, but I don't know how to go about reducing the order of coupled differential equations because the 3 unkown variables x, theta1 and theta2 all appear in the same equations.

 

[arildno]

Suppose you've got a second order diff.eq system:
\frac{d^{2}\vec{Y}}{dt^{2}}=\vec{F}(y_{1},..y_{n},\dot{y}_{1},...,\dot{y}_{n},t), \vec{Y}(t)=(y_{1}(t),...,y_{n}(t)),\dot{y}_{m}\equiv\frac{dy_{m}}{dt},1\leq{m}\leq{n}; m,n\in\mathbb{N}

Now, define:
\vec{X}(t)=(x_{1}(t),...,x_{n}(t),...,x_{2n}(t))
with:
x_{i}=y_{i}, x_{n+i}=\frac{dy_{i}}{dt}=\frac{dx_{i}}{dt}, 1\leq{i}\leq{n}
Thus, we may form the 1-order differential system of 2n equations:
\frac{d\vec{X}}{dt}=\vec{G}(\vec{X},t)
where:
G_{i}(\vec{X},t)=x_{n+i}, 1\leq{i}\leq{n}
G_{i}(\vec{X},t)=F_{i-n}(\vec{X},t), n<i\leq{2n}

 

[peterbone]

Thankyou, that was helpful.
I also found this site which explains the whole process of simulating a single inverted pendulum and includes the reduction stage.
http://drewk.net/projects/ipendulum/ipendulum.html
I should be able to use the same method for the double inverted pendulum.