MAT244-2013S > Quiz 4

Q4--day section--problem 2

**Victor Ivrii**:

--- Quote from: Alexander Jankowski on March 24, 2013, 06:10:31 PM ---Oh, okay. I was confused because the origin is a saddle point (the eigenvalues of the coefficient matrix are $\lambda = Â±4$) and there is no apparent basin of attraction. I should note that the textbook discusses separatrices only in that context. Anyway, I updated the solution again and plotted more level curves for completion.

--- End quote ---

Sure: unstable equilibrium points do not have basins of attractions as almost all trajectories go away. However if a linear system has a matrix with $n_+$ positive and $n_-$ negative eigenvalues (but not $0$ so $n_++n_-=n$) ) then the whole space breaks into direct sum of linear subspaces $V_+ \oplus V_-$ where all trajectories in $V_\pm$ tend to $0$ as $t\to \pm \infty$ and $V_+$, $V_-$ are unstable and stable subspaces respectively.

Similar picture but with "curved" $W_\pm$ (called manifolds or (hyper)surfaces) instead of $V_pm$ happens for nonlinear systems. As $n=2$ we have $n_\pm=1$ and both of them are lines (separatrices) but for $n=3$ one of them is a surface and another a line and there are more possibilities as $n\ge 4$.

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