|Nov28-06, 02:49 PM||#1|
singular points in 3-dim space
For a linearized system I have eigenvalues [tex]\lambda_1, \lambda_2 = a \pm bi \;(a>0)[/tex] and [tex]\lambda_3 < 0 [/tex],
then it should be an unstable spiral point. As [tex]t \to +\infty[/tex] the trajectory will lie in the plane which is parallel with the plane spanned by eigenvectors [tex]v_1,v_2[/tex] corresponding to [tex]\lambda_1, \lambda_2[/tex].
Right? I am just not very sure.
|Nov29-06, 07:07 AM||#2|
Yes, that is correct.
|Nov29-06, 08:33 AM||#3|
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