I have the following problem:

Suppose a function $f\in C(\mathbb{R}^{n+1}\to\mathbb{R}^n)$ is such that $f(t,\textbf0)=0$ and $$\textbf x^*f(t,\textbf x)\le0\quad\text{for all }(t,\textbf x)\in\mathbb{R}^{n+1}, $$ where $\textbf x^*$ stands for the conjugate transposition of the vector $\textbf x$.

I am asked to prove that the equilibrium solution $\textbf x(t)=0$ of $\dot{\textbf x}=f(t,\textbf x)$ is uniformly stable, that is,

$$\forall\epsilon>0,\ \exists\delta>0\quad\text{such that}\quad |x(t_0)|<\delta\implies|x(t)|<\epsilon\quad\forall t\ge t_0. $$ I have no idea how to start. Thanks in advance!


$$ (|\mathbf{x}(t)|^2)'=2\,\mathbf{x}^*(t)\ \mathbf{x}'(t)=2\,\mathbf{x}^*(t)\ f(t,\mathbf{x}(t))\le0. $$ This shows that $|\mathbf{x}(t)|$ is decreasing.

  • $\begingroup$ Why not enough? Letting $ \delta $ be equal to $ \epsilon $, the stability condition is satisfied, isn't it? $\endgroup$ – Mohsen Shahriari Mar 30 '17 at 11:11
  • $\begingroup$ You are right. I misunderstood the definition of uniform stability, and thought it was asymptotic stability. I will delete that part of the answer. $\endgroup$ – Julián Aguirre Mar 30 '17 at 12:14

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