# Stability of a system of ODEs

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!

• 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. – user25199 Nov 25 '13 at 10:30

In the case of $1\times 1$ matrices, and $b\neq 0$, your equation is $y'=ay+by^2$, which is never globally stable.
• 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. – Alexandre Eremenko Nov 25 '13 at 18:22
• Thank you for the helpful discussion. If all the entries of $\boldsymbol{y}$ are between $0$ and $1$, do things change? – Jeff Nov 26 '13 at 11:47
• 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. – Alexandre Eremenko Nov 26 '13 at 14:19