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I'm studying the stability of steady states by means of the eigenvalues of $J$.

So far the criteria is this:

  • All eigenvalues $\gt 0 \implies$ unstable

  • All eigenvalues $\lt 0 \implies$ stable.

  • In 2D: one eigenvalue $\gt 0$, and another $\lt 0 \implies$ saddle.

  • All eigenvalues $\leq 0 \implies$ critical (stability cannot be analyzed using this approximation).

But what about the following cases:

  • Saddle points in $n$-dim (how do they look like?), can an eigenvalue be zero?
  • Eigenvalues $\geq 0$, how do I classify these?
Daniel Donnelly
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Cate
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  • Firstly, use $\LaTeX$ and next, put in what you've tried, what you mean by what they look like. See my edits to your post, and also define the variables in your post such as $J$. :) – Daniel Donnelly Dec 14 '21 at 05:31
  • Additionally, you always want to put in all 5 tags, in order to increase your post's visibility. Please see the two I added, to be sure that they're appropriate. Also, stability of what etc? Please, show more equations! – Daniel Donnelly Dec 14 '21 at 05:40

1 Answers1

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I'm assuming you want to analyze the local stability of a fixed point of a nonlinear system using linear approximation. First of all we need to clear some confusion. The fixed point is unstable if at least one eigenvalue is at the right-half plane, i.e. its real part is positive. An unstable fixed point is either a source (all eigenvalues are at RHP) or a saddle (some eigenvalues are at LHP and at least one of them is at RHP).

If a FP is a source, then all initial conditions around the FP go to infinity. If a FP is a saddle some initial conditions go to the FP, but if you select an initial condition randomly, there is $0$ probability of selecting such a point (so "almost all" of the initial conditions go to infinity). Imagine trying to randomly select a point on a line in 2D space.

A saddle in higher dimensions work exactly the same way, for example in 3D space it is possible starting on a plane go to the FP but starting from any other point go to infinity (in this case 2 of the eigenvalues are at LHP and one of them is at RHP).

If at least one eigenvalue is on the imaginary axis and others are at LHP nothing can be concluded with this approximation. In the $\geq 0$ case you mentioned if any of the eigenvalues are at LHP the LP is unstable. If all eigenvalues are on the imaginary axis again nothing can be concluded.

obareey
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  • Thanks I got everything except the last paragraph. Why are you taking about the LHP in the $\geq 0$ case. Wouldn't it be the RHP? And why having one eigenvalue in the imaginary axis and the others in LHP result in nothing can be concluded. Wouldn't it be a stable focus? – Cate Dec 16 '21 at 10:40