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One needs to show why solutions for the system $$x'=\left[\begin{matrix} 0&-1&0 \\ 0&-2&0 \\ -1&2&-1\end{matrix}\right]x$$ are Lypunov or asymptotically stable/unstable. Most probably we would need to check eigenvalues real part to determine stability. However $\lambda_1=0, \lambda_2=-1, \lambda_3=-2$, which makes the whole process a little more difficult, especially when it comes to check the former type of stability (eigenvalues are nonpositive and that implies solutions are Lyapunov stable).

But how do I know if they are asymptotically stable? Shoould I find strictly decreasing Lyapunov function?

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Zero eigenvalue means $x'$ is zero on some line. (Specifically, the line $x_1+x_3=0=x_2$.) What happens to solutions that begin on that line? They stay where they started. So much for asymptotic stability.

In general, an asymptotically stable equilibrium must be an isolated equilibrium: since nearby points are drawn to it, they can't be equilibria too.

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  • $\begingroup$ Yes, that makes sense. And what would happen if, let say, $\lambda_1=i, \lambda_2=-i, \lambda_3=-2$ ? What do such eigenvalues tell us about the solution? $\endgroup$ – Jules Sep 20 '14 at 11:00
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    $\begingroup$ @Jules Wrote down the general solution: it will involve periodic terms. Therefore, the equilibrium is not asymptotically stable. $\endgroup$ – user147263 Sep 20 '14 at 15:36

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