Here I have a system of nonlinear differential equations:

$ (M+2m)\ddot{x} + m(l_1 \ddot{\theta}_1\cos\theta_1 - l_1\dot{\theta}_1^2\sin\theta_1) + m(l_2\ddot{\theta}_2\cos\theta_2-l_2\dot{\theta}_2^2\sin\theta_2) = F $

$ l_1\ddot{\theta}_1 + \ddot{x}\cos\theta_1 - g\sin\theta_1 = 0 $

$ l_2\ddot{\theta}_2 + \ddot{x}\cos\theta_2 - g\sin\theta_2 = 0 $

States are defined by me like this:

$x_1 = \theta_1 , x_2 = \dot{x_1} = \dot{\theta}_1,x_3 = \theta_2, x_4 = \dot{x_3} = \dot{\theta}_2,x_5 = x, x_6 = \dot{x_5} = \dot{x}$

I mean actually what I need to do is to linearize this system about all the states are equal to $0$. But I cannot find the $\cdots$ places below

$ \dot{x_1} = x_2 \\ \dot{x_2} = \cdots \\ \dot{x_3} = x_4 \\ \dot{x_4} = \cdots \\ \dot{x_5} = x_6 \\ \dot{x_6} = \cdots \\ $

How can I find $\dot{x_2},\dot{x_4},\dot{x_6}$?

Thanks

  • First you need to solve your system for $\ddot x,\,\ddot \theta_1$ and $\ddot \theta_2$. After that you can find what you are looking for. – Artem Oct 29 '12 at 18:37
  • I will use Cramer's rule, is this ok for the system of nonlinear differential equations? I mean for $\ddot{x}$... I have done this... I will post it again, will you look at? Thanks by the way. – kemal acikgoz Oct 29 '12 at 18:42
  • $ \ddot{x} = (Fl_1*l_2 + l_1^2*l_2*mx_2^2*sin(x_1) + l_1*l_2^2*mx_4^2*sin(x_3) - gl_1*l_2*mcos(x_1)*sin(x_1) - gl_1*l_2*mcos(x_3)*sin(x_3))/(Ml_1*l_2 + 2*l_1*l_2*m - l_1*l_2*mcos(x_1)^2 - l_1*l_2*mcos(x_3)^2) $ – kemal acikgoz Oct 29 '12 at 18:45
  • Your system is linear with respect to the second order derivatives. And I do not think the Cramer's rule is the best option in this situation. – Artem Oct 29 '12 at 18:47
  • ok, what else I can do? I found $\ddot{x}$ by using Cramer's rule as above. – kemal acikgoz Oct 29 '12 at 18:49
up vote 1 down vote accepted

First, put your system

$$ (M+2m)\ddot{x} + m(l_1 \ddot{\theta}_1\cos\theta_1 - l_1\dot{\theta}_1^2\sin\theta_1) + m(l_2\ddot{\theta}_2\cos\theta_2-l_2\dot{\theta}_2^2\sin\theta_2) = F $$

$$ l_1\ddot{\theta}_1 + \ddot{x}\cos\theta_1 - g\sin\theta_1 = 0 $$

$$ l_2\ddot{\theta}_2 + \ddot{x}\cos\theta_2 - g\sin\theta_2 = 0 $$

into a matrix formulation

$$ \underbrace{\begin{bmatrix} M+2m & m l_1\cos\theta_1 & ml_2\cos \theta_2\\ \cos \theta_1 & l_1 & 0\\ \cos \theta_2 & 0 & l_2 \end{bmatrix}}_{\boldsymbol{M}(\theta_1,\theta_2)} \dfrac{d}{dt} \begin{bmatrix} \dot{x}\\ \dot{\theta_1}\\ \dot{\theta_2}\\ \end{bmatrix}= \underbrace{\begin{bmatrix}ml_1\sin \theta_1\dot{\theta}^2_1+ml_2\sin\theta_2\dot{\theta}^2_2\\g\sin\theta_1\\g\sin\theta_2 \end{bmatrix}}_{\boldsymbol{f}(\theta_1,\theta_2,\dot{\theta_1},\dot{\theta_2})}+ \begin{bmatrix}1\\0\\0 \end{bmatrix}F. $$

by inverting the mass matrix $\boldsymbol{M}(\theta_1,\theta_2)$, this can be done with many computer algebra systems (e.g. Maple, Python, Mathematica, MATLAB symbolic toolbox) we obtain:

$$\dfrac{d}{dt}x = \dot{x}$$ $$\dfrac{d}{dt}\theta_1 = \dot{\theta}_1$$ $$\dfrac{d}{dt}\theta_2 = \dot{\theta}_2$$ $$\dfrac{d}{dt}\begin{bmatrix}\dot{x}\\\dot{\theta}_1\\\dot{\theta}_2\end{bmatrix}=\boldsymbol{M}(\theta_1,\theta_2)^{-1}\boldsymbol{f}(\theta_1,\theta_2,\dot{\theta_1},\dot{\theta_2})+\boldsymbol{M}(\theta_1,\theta_2)^{-1}\begin{bmatrix}1\\0\\0 \end{bmatrix}F$$

  • Thanks a lot! :) – kemal acikgoz Feb 12 at 11:50
  • 2
    No problem, I know it is a late answer but I couldn't resist to answer it because there were no answer :). – MrYouMath Feb 12 at 12:42

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