Consider $V=\mathbb{R}^d\times\mathbb{R}^n$ with coordinate $x^T=[\theta^T,\sigma^T]$. I have an ODE of the form: $\dot{x}=F(x)$, where $F$ is assumed to be sufficiently smooth.

Suppose that there's some $\delta>0$ so that all trajectories of the ODE satisfy the inequality $$ \frac{d}{dt}(x^Tx)=\frac{d}{dt}(\theta^T\theta+\sigma^T\sigma)\le-\delta \theta ^T \theta $$ Barbalat's lemma allows us to conclude that:

  1. $\theta\to0$ as $t\to\infty$
  2. There exists some $C>0$ such that $\sigma^T\sigma\to C$

I want to prove that in this case, $\theta \to 0$ exponentially fast, with exponent depending on $\delta$. If $F$ was an affine function, $F(x)=Ax+b$ then one can prove this result by writing $x(0)$ as a linear combination of generalized eigenvectors for $A$. Is there a well-known result for the general, nonlinear case?

  • 1
    $\begingroup$ Under your hypotheses, it is straightforward to show that $e^{\delta t} \| \theta_t \|^2 \le \|x_0\|^2 + \delta \int_0^t e^{\delta s} \| \sigma_s \|^2 ds$. Hence, $\| \theta_t \|^2 \le e^{-\delta t} \|x_0\|^2 + \delta t e^{-\delta t} C + \delta \int_0^t e^{\delta (s-t) } | \| \sigma_s \|^2 - C | ds$ where $C$ is the one from Barbalat's lemma. It seems like this latter term is the one that dictates the rate of convergence. $\endgroup$ – Nawaf Bou-Rabee Jan 6 '18 at 19:14

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