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In order to find the stability of a nonlinear system of differential equations (in the real plane) we need to show that the eigenvalues of the linearized system are all negative. Can someone explain to me why finding the trace to be negative and the determinant to be positive is enough for this purpose? I thought the previous statement only applies to $2 \times 2$ systems?

Thanks.

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  • $\begingroup$ @Ian You are right! I was just thinking about that - I found that if the eigenvalues are say, $3,-2,-2$, then the product (determinant) is positive, but the trace is negative, but not all the eigenvalues are negative! What's wrong here? $\endgroup$
    – user85362
    Commented Jun 22, 2014 at 18:46

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So this will not hold in dimension higher than 2. For example, in odd dimension, when all eigenvalues are negative, the determinant is also negative. In even dimension $>2$, although this is possible unlike in odd dimension, these conditions are still not sufficient. For example, in dimension 4, one can have eigenvalues of $-10,1,2,-1$ and still satisfy both conditions.

In nonlinear 2 dimensional systems, this does hold. Those two conditions allow you to check that the eigenvalues of the Jacobian have strictly negative real part, which is all you need in a nonlinear system in a neighborhood of an equilibrium point.

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  • $\begingroup$ OK. Can you help me with a specific question about a research paper? I am trying to interpret something related to what you said. The link is here: pubs.sciepub.com/ajeid/2/1/1, and the information I am targeting is in the section $3.$ Stability of Disease-Free Equilibrium. $\endgroup$
    – user85362
    Commented Jun 22, 2014 at 18:56
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    $\begingroup$ So the determinant will have to be positive for asymptotic stability, because the system is even dimensional. Evidently they have shown that when the determinant is positive, the spectral radius is $<1$. If they have also shown that there is an eigenvalue whose real part is $<-1$, then they're done in terms of showing all eigenvalues have negative real part. $\endgroup$
    – Ian
    Commented Jun 22, 2014 at 19:01
  • $\begingroup$ So you say that "in even dimension $>2$, ... these conditions are still not sufficient". Then what are the sufficient conditions? $\endgroup$
    – user85362
    Commented Jun 22, 2014 at 19:10
  • $\begingroup$ Just the eigenvalue condition. Any other sufficient condition you might come up with is really just providing a way to bound the eigenvalues when you don't exactly know what they are. $\endgroup$
    – Ian
    Commented Jun 22, 2014 at 19:12

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