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Consider the differential equation

$x^{\prime}=f(t,x),{\qquad}t\geq0.\tag{1}$

Definition 1. If there exists a constant $x^{\ast}\in\mathbb{R}$ such that $f(t,x^{\ast})=0$ for all $t\geq0$, then $x^{\ast}$ is said to be the equilibrium solution of (1).

Definition 2. A solution $x$ of (1) is uniformly stable if for each $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)>0$ such that if $x$ is a solution of (1) and $|x(s)-x^{\ast}|<\delta$ for some $s\geq0$, then $|x(t)−x^{\ast}|<\varepsilon$ for all $t{\geq}s$.

Do you think replacing last two of the strict inequalities ($<$) in Definition 2 with weaker ones ($\leq$) gives an equivalent definition?

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  • $\begingroup$ For homogeneous linear equations, the two of these definitions are equivalent, which works for me. $\endgroup$ – bkarpuz Oct 3 '15 at 8:31
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Let us consider the homogeneous linear equation

$x^{\prime}=p(t)x,{\qquad}t\geq0.\tag{2}$

Clearly, for this equation $x^{\ast}=0$. Further, for any given $\lambda\in\mathbb{R}\backslash\{0\}$, $x$ is a solution of (2) if and only if $\lambda{}x$ is a solution of (2) since (2) is linear.

We say that (2) satisfies the property (P) if for each $\varepsilon>0$ there exists $\delta=\delta(\varepsilon)>0$ such that if $x$ is a solution of (2) and $|x(s)|\leq\delta$ for some $s\geq0$, then $|x(t)|\leq\varepsilon$ for all $t{\geq}s$.

Obviously, if (2) is uniformly stable, then it satisfies (P).

Now, let (2) satisfy (P). Let be $\varepsilon>0$. From (P), we can find $\delta>0$ such that $|{\lambda}x(s)|\leq\delta$ for some $s\geq0$ implies $|{\lambda}x(t)|\leq\varepsilon$ for all $t{\geq}s$, where $\lambda>1$. Then, $|x(s)|<\delta$ for some $s\geq0$ implies $|x(t)|<\varepsilon$ for all $t{\geq}s$. That is, (2) is uniformly stable.

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