Consider the non-autonomous nonlinear dynamical system $$ \dot{x}=f(t,x) $$ It is a known result (Theorem 4.13 in Khalil (Nonlinear Systems)) that if the linearization $$ \dot{x}=A(t)x $$ where $A(t)=\frac{\partial f}{\partial x}(t,x) \rvert_{x=0}$, is such that the origin is an exponentially stable equilibrium point, then the origin is an exponentially stable equilibrium point for the nonlinear system. I am wondering whether an analogous result holds in the unstable case, i.e, if the linearized system is unstable and with diverging solutions then the nonlinear system is also unstable. Is anyone aware of such a result? I have not been able to find it in the literature or on the standard nonlinear systems textbooks.
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As usual, eigenvalues of real part $0$ can make things complicated. Consider the system
$$ \eqalign{\dot{x} &= y \cr \dot{y} &= -x^3 \cr}$$ The linearized system (with an eigenvalue $0$ of algebraic multiplicity $2$) is unstable, but the nonlinear system is stable.
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$\begingroup$ Hi, I am aware of this result. However, in the autonomous case if the linearization has some eigenvalues with positive real part you can conclude that the nonlinear system is unstable. I am looking for an analogous result in the non-autonomous case. I have slightly edited the question to stress that in my particular problem, I happen to know that the solutions of the linearized non-autonomous system diverge. $\endgroup$ Commented Nov 30, 2021 at 4:27