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Differential Equation with Vectors

  1. Aug 6, 2009 #1
    Hello everyone, I have the following differential equation :

    [tex]\dot{\theta}(t) = \omega(t) - \frac{1}{2}\theta(t)\times\omega(t) + \frac{1}{\Vert\theta(t)\Vert^2}\left(1-\frac{\Vert\theta(t)\Vert}{2}cot\frac{\Vert\theta(t)\Vert}{2}\right)\theta(t)\times[\theta(t)\times\omega(t)][/tex]

    where [tex]\omega(t)[/tex] is a known 3D vector with the form :

    [tex]\omega(t) = [a_1+b_1*t , a_2+b_2*t , a_3+b_3*t ][/tex]

    also, [tex]\theta(t)[/tex] is the unknown 3D vector that we need to find. Its norm is [tex]\Vert\theta(t)\Vert[/tex].

    This type of math is too high for me. If you guys know a way to solve it for [tex]\theta(t)[/tex] with Maple, Mathlab or Scilab, I'd be happy, because anyone of those seems to explain how to solve an equation that contains vectors. I think a solver would be more appropriate because the equation is quite "big", if you know what I mean.

  2. jcsd
  3. Aug 6, 2009 #2


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    Gold Member

    Try decomposing it into its scalar components first.

    Then you get a system of 3 diffeq's
  4. Aug 6, 2009 #3
    Thanks, but ok, le's say I decomposed it in 3 differential equations. Here I'll have :

    [tex]\frac{d}{dt}\theta_1(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )[/tex]
    [tex]\frac{d}{dt}\theta_2(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )[/tex]
    [tex]\frac{d}{dt}\theta_3(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )[/tex]

    Let's say I use Maple to solve, what do I have to do next?
  5. Aug 18, 2009 #4
    Hi there!

    Why not trying to simplify the equation at first:

    [tex]\theta\times(\theta\times\omega)=\theta(\theta\omega)-\omega ||\theta||^2)[/tex]

    plugging it in the equation gives:

    [tex]\dot\theta=\omega-\frac{1}{2}\theta\times\omega+\frac{1}{||\theta||^2}(1-\frac{||\theta||}{2}\cot\frac{||\theta||}{2})(\theta(\theta\omega)-\omega ||\theta||^2)[/tex]

    now, let's get rid of some terms, let's left-operate with x theta:

    [tex]\theta\times\dot\theta=\theta\times\omega-\frac{1}{2}\theta\times(\theta\times\omega)+\theta\times\frac{1}{||\theta||^2}(1-\frac{||\theta||}{2}\cot\frac{||\theta||}{2})(\theta(\theta\omega)-\omega ||\theta||^2)[/tex]

    [tex]\theta\times\dot\theta=\theta\times\omega-\frac{1}{2}\theta\times(\theta\times\omega)+\frac{1}{||\theta||^2}(1-\frac{||\theta||}{2}\cot\frac{||\theta||}{2})(\theta\times\theta(\theta\omega)-\theta\times\omega ||\theta||^2)[/tex]

    observe that [tex]\theta\times\theta=0[/tex] and (\theta\omega) is a scalar

    [tex]\theta\times\dot\theta=\theta\times\omega-\frac{1}{2}\theta\times(\theta\times\omega)+\frac{1}{||\theta||^2}(1-\frac{||\theta||}{2}\cot\frac{||\theta||}{2})(-\theta\times\omega ||\theta||^2)[/tex]

    now, simplfying a little gives:


    and using [tex]\omega\times\theta=-\theta\times\omega[/tex]




    Well, to me it looks a tiny little bit better, the inhomogenety is at least not there any more , but the equation remains nonlinear which makes it hardly solvable analytically.
  6. Aug 20, 2009 #5

    Marin had taken all the trouble to show that probably there is no analytical solution. So I suggestion is that we go for numerical solution. Much easier. How about using Runge-Kutta method?
  7. Aug 20, 2009 #6
    Wow thanks guys! Your responses are very interesting and useful! Unfortunately, I don't really know much about the Runge-Kutta method =/
  8. Aug 20, 2009 #7
    That's all right there are a lot of programs that will do Runge-Kutta 4th order method for you. I know in particular that Matlab has ode45() which is pretty easy to use if you have matlab.
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