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Clockwise or counterclockwise? (linear system phase portrait)

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Given a 2x2 matrix A with entries a,b,c,d (real) with complex eigenvalues I would like to know how to find out whether the solutions to the linear system are clockwise or counterclockwise. (Some kind of inequality between a,b,c,d).

    2. Relevant equations

    3. The attempt at a solution

    I tried looking at the signs of each component of the derivative. It seems to me that clockwise means x1' > 0 and x2' < 0 for x1,x2 > 0 as long as x1 or x2 are not too small. Then im not sure...
  2. jcsd
  3. Feb 13, 2007 #2


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    Science Advisor

    The simplest way is to see what the matrix does to (1, 0) and (0, 1).

    For example, if the problem is
    [tex]\frac{dX}{dt}= \left( \begin{array}{cc}0 & -1\\ 1 & 0\end{array}\right)X[/tex]
    [tex]\left( \begin{array}{cc}0 & -1\\ 1 & 0\end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right)= \left(\begin{array}{c}0 \\ 1\end{array}\right)[/tex]
    That's counter-clockwise rotation. The matrix
    [tex]\left(\begin{array}{cc}0 & 1\\ -1 & 0\end{array}\right)[/tex]
    has exactly the same eigenvalues but is clockwise rotation.
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