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Thread: ODE with cross product

  1. #1
    Forum Admin topsquark's Avatar
    Jan 2006
    Wellsville, NY

    ODE with cross product

    This is a post by Amiram.

    Quote Originally Posted by Amiram View Post
    Hello, as I was studying quantum dynamics I ran into this simple looking differential equation for the spin of a particle:

    dX/dt=w^X , where w is the axis around which the spin processes (all in R3 of course). X is in fact also a function of space but let us assume that X(t,{qi})= T(t)*Q({qi}) where i=1,2,3 for any 3D coordinate system. From the cross product's distributive property we can deduce to a problem which X is a function of time only At first I tried solve the equations with pure algebra but the system had too many free variables. what I did managed to do is to find a private solution for w=(1,1,1) which is T(t)=e^-t(-2t-3,1,2+2t).

    I'd be glad if someone could help with the general case, thank you.
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  2. #2
    Super Member Rebesques's Avatar
    Jul 2005
    My house.

    Re: ODE with cross product

    If I'm reading this right, the simplified equation is $\displaystyle X'(t)=w\times X(t)$, yes?

    Then, by introducing the matrix $\displaystyle M=\left[ \begin{array}{ccc} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1& 0 \end{array} \right]$

    the system becomes $\displaystyle X'(t)=MX(t)$, which has the general solution $\displaystyle X(t)=c\exp(Mt)$, c a constant.
    Thanks from topsquark
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