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It is well known that for a system of ODEs, $\dot{\boldsymbol{y}} = \boldsymbol{Ay}$, the global stable equlibrium point is given by the eigenvector that correponds to the largest eigenvalue of $\boldsymbol{A}$, provided that the matrix $\boldsymbol{A}$ is primitive. This is according to the Perron-Frobenius theorem.

Then my question is that for a system $\dot{\boldsymbol{y}} = \boldsymbol{Ay} + \boldsymbol{B}\boldsymbol{y}\boldsymbol{y}^{T}$, do we have a similar theory for the global stability as that for $\dot{\boldsymbol{y}} = \boldsymbol{Ay}$?

We know that if we have $\boldsymbol{y}^{\ast}$ such that $\boldsymbol{0} = \boldsymbol{Ay}^{\ast} + \boldsymbol{B}\boldsymbol{y}^{\ast}\boldsymbol{y}^{\ast T}$, and the eigenvalues of the Jacobian matrix $J(\boldsymbol{y}^{\ast})$ are negative, we can conclude $\boldsymbol{y}^{\ast}$ is locally stable, but how about the global stablibility?

If all the entries of $\boldsymbol{y}(0)$ are between $0$ and $1$, do things change?

Thank you guys for helpful discussions!

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  • $\begingroup$ It is not globally stable in general. Consider $1\times1$ matrices - if $B$ is positive large $y$ will run off to infinity, while for $B$ negative large negative $y$ will do the same. $\endgroup$
    – user25199
    Commented Nov 25, 2013 at 10:30

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In the case of $1\times 1$ matrices, and $b\neq 0$, your equation is $y'=ay+by^2$, which is never globally stable.

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  • $\begingroup$ Thank you. But local stability is possible? Is there anyway to estimate the domain of attraction, i.e., the bounday within which the trajectory starts and will then approach the eqlibirum point? $\endgroup$
    – Jeff
    Commented Nov 25, 2013 at 16:01
  • $\begingroup$ Local stability is possible. And you stated yourself the condition of local stability in your question. Concerning the size of the domain of attraction, you can obtain various estimates: roughly speaking if the eigenvalues of $A$ are sufficiently negative and the size of $B$ is sufficiently small you can get the estimate. Look to any book which has "stability" in its title. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 25, 2013 at 18:22
  • $\begingroup$ Thank you for the helpful discussion. If all the entries of $\boldsymbol{y}$ are between $0$ and $1$, do things change? $\endgroup$
    – Jeff
    Commented Nov 26, 2013 at 11:47
  • $\begingroup$ What do you mean "between $0$ and $1$" ? If the initial condition is between $0$ and $1$, then, as time goes $y$ may escape from this interval. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 26, 2013 at 14:19

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