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Coupled 2nd Order DE: Convert to system of 1st Order

  1. Apr 30, 2013 #1
    It's been a while since I've played with systems of ODEs, and I seem to have forgotten some of the tricks. As an example, I have two coupled nonlinear DE that I want to convert to a system of four 1st order nonlinear DE. But, the normal way of making variable substitutions is not working of me.

    Two 2nd order nonlinear DE:
    ##\ddot{\alpha} + \ddot{\beta}cos(\alpha-\beta)+\dot{\beta}^2sin(\alpha-\beta)=\tau_{1}##
    ##\ddot{\beta} + \ddot{\alpha}cos(\alpha-\beta)-\dot{\alpha}^2sin(\alpha-\beta)=\tau_{2}##

    If I substitute variables ##x_1=\alpha##, ##x_2=\dot{\alpha}##,##x_3=\beta##,##x_4=\dot{\beta}##, like my initial reaction was, I get:

    ##\dot{x_1}=x_2##
    ##\dot{x_2}=-\dot{x_4}cos(x_1-x_3)-x_4^2sin(x_1-x_3)+\tau_1##
    ##\dot{x_3}=x_4##
    ##\dot{x_4}=-\dot{x_2}cos(x_1-x_3)+x_2^2sin(x_1-x_3)+\tau_2##

    which isn't quite right, since it is not in the form ##\dot{\vec{\textbf{x}}}=\vec{\textbf{f}}(t,\vec{\textbf{x}})##

    I'm trying to get it a form where I can run it through ODE45 or a similar solver.

    Refresh my memory! :)


    EDIT! I believe I figured it out. I did do it right, just implementing it in Matlab had me confused. But, all I have to do in my function is use the old values for ##\dot{x_2}## and ##\dot{x_4}##. Seeing those derivatives in the rhs just threw me off. Is this the right way to go about it?
     
    Last edited: Apr 30, 2013
  2. jcsd
  3. Apr 30, 2013 #2

    AlephZero

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    You can write your equations as
    $$\begin{bmatrix} 1 & \cos(\alpha - \beta) \\ \cos(\alpha - \beta) & 1 \end{bmatrix}
    \begin{bmatrix} \ddot\alpha \\ \ddot\beta \end{bmatrix}
    = \begin{bmatrix} -\dot\beta^2 \sin(\alpha - \beta) + r_1 \\
    \dot\alpha^2 \sin(\alpha - \beta) + r_2\end{bmatrix}$$
    So
    $$\begin{bmatrix} \ddot\alpha \\ \ddot\beta \end{bmatrix}
    = \begin{bmatrix} 1 & \cos(\alpha - \beta) \\ \cos(\alpha - \beta) & 1 \end{bmatrix}^{-1}
    \begin{bmatrix} -\dot\beta^2 \sin(\alpha - \beta) + r_1 \\
    \dot\alpha^2 \sin(\alpha - \beta) + r_2\end{bmatrix}$$

    The matrix is singular if ##\alpha = \beta##, but in that case you only have one differential equation not two.
     
  4. May 1, 2013 #3
    That makes sense, but is that the only way to implement it into Matlab (ode45)? I thought I was onto something (see my edit in my original post), but now I'm thinking it won't work.
     
  5. May 1, 2013 #4
    Move everything with overdots to the left hand sides. Then re-express the left hand side as a matrix times the column vector of the dotted first derivatives. So the equation is

    Mxdot=f(x,t)

    In your subroutine for evaluating derivatives, first calculate the right hand side vector. Then premultiply it by the inverse of M. Some equation solvers accept the more general inputs as M and f(x,t), and do the inversion for you. I'm not sure whether the equation solver you are using will do this.

    Chet
     
  6. May 1, 2013 #5
    Thanks, Chet. Turns out I was able to keep it in my original form (system of implicit DEs) and use Matlabs ode15i solver. I believe it does a similar operation by inverting the matrix....but it just saves me a lot of work! It wouldn't be too bad, but when the system I have has 8 state variables, it gets messy really fast.

    Anyways, looks like ode15i was the solution I was looking for.
     
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