Differential Equation with Vectors

[nburo]

Hello everyone, I have the following differential equation :

\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)]

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

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

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

This type of math is too high for me. If you guys know a way to solve it for \theta(t) 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.

Thanks.

 

[djeitnstine]

Try decomposing it into its scalar components first.

Then you get a system of 3 diffeq's

 

[nburo]

Thanks, but ok, le's say I decomposed it in 3 differential equations. Here I'll have :

\frac{d}{dt}\theta_1(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )
\frac{d}{dt}\theta_2(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )
\frac{d}{dt}\theta_3(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )

Let's say I use Maple to solve, what do I have to do next?

 

[Marin]

Hi there!

Why not trying to simplify the equation at first:

\theta\times(\theta\times\omega)=\theta(\theta\omega)-\omega ||\theta||^2)

plugging it in the equation gives:

\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)

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

\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)


\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)

observe that \theta\times\theta=0 and (\theta\omega) is a scalar

\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)

now, simplfying a little gives:

\theta\times\dot\theta=\theta\times\omega-\frac{1}{2}\theta\times(\theta\times\omega)+(1-\frac{||\theta||}{2}\cot\frac{||\theta||}{2})(-\theta\times\omega)

and using \omega\times\theta=-\theta\times\omega

\theta\times\dot\theta=\theta\times\omega-\frac{1}{2}\theta\times(\theta\times\omega)-\theta\times\omega-\frac{||\theta||}{2}\cot\frac{||\theta||}{2}\theta\times\omega

or

-2\theta\times\dot\theta=\theta\times(\theta\times\omega)+||\theta||\cot\frac{||\theta||}{2}\theta\times\omega

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.

 

[matematikawan]

Thanks, but ok, le's say I decomposed it in 3 differential equations. Here I'll have :

\frac{d}{dt}\theta_1(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )
\frac{d}{dt}\theta_2(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )
\frac{d}{dt}\theta_3(t) = F(\theta_1(t), \theta_2(t), \theta_3(t) )

Let's say I use Maple to solve, what do I have to do next?

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?

 

[nburo]

Wow thanks guys! Your responses are very interesting and useful! Unfortunately, I don't really know much about the Runge-Kutta method =/

 

[Feldoh]

Wow thanks guys! Your responses are very interesting and useful! Unfortunately, I don't really know much about the Runge-Kutta method =/

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.