Singular points in 3-dim space

firenze
Messages
5
Reaction score
0
For a linearized system I have eigenvalues \lambda_1, \lambda_2 = a \pm bi \;(a>0) and \lambda_3 < 0,
then it should be an unstable spiral point. As t \to +\infty the trajectory will lie in the plane which is parallel with the plane spanned by eigenvectors v_1,v_2 corresponding to \lambda_1, \lambda_2.

Right? I am just not very sure.
 
Physics news on Phys.org
Yes, that is correct.
 
Thanks:smile:
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top