Show that root $x\equiv 0$ of $\dfrac{dx}{dt}=F(t,x)$ is uniformly stable (uniformly asymptotically stable)

Question

I have a problem:

For the system of equations: $$\bf \dfrac{dx}{dt}=F(t,x) \tag 1$$ where $F$ is continuous in $I \times D \subset\mathbb{R}\times \mathbb{R}^n$ and $F(t,0)\equiv0$, $F(t+\omega,x)=F(t,x), \omega >0$, it means that $F$ is periodic function. Prove that, if root $x\equiv 0$ of (1) is stable (asymptotically stable) then it is uniformly stable (uniformly asymptotically stable).

• Uniformly Stable: If any given $\epsilon >0$, $\exists \delta=\delta(\epsilon)>0$: $$\|x(t_0)\|< \delta \implies \|x(t)\|<\epsilon, \forall t\ge t_0 >0$$

• I have thought about my problem, I used the definition uniformly stable/ uniformly asymptotically stable.

But I'm having trouble when I try to find a solution to the problem, and I still have no solution. Any help will be appreciated. Thanks!

Answer 1

Here's the solution by José L. Massera

1/ $x \equiv 0$ is uniformly stable.

• We assume that If $x \equiv 0$ is not uniformly stable, there would be an $\epsilon >0$ and sequences $\{x_{0n}\}$, $\lim_{n \to \infty}x_{0n}=0,t_{0n},t_n,t_{0n} \le t_n$, such that:$$\left \|F(t_n,t_{0n},x_{0n}) \right \| \ge \epsilon$$

• Assuming that the period is $1$ , let $\tau_{0n}=t_{0n}-\left \lfloor t_{0n} \right \rfloor \le 1$, $\tau_n=t_n-\left \lfloor t_{0n} \right \rfloor \ge \tau_{0n} \ge 0;$

and we have

$$\left \|F(\tau_n,\tau_{0n},x_{0n}) \right \| =\left \|F(t_n,t_{0n},x_{0n}) \right \| \ge \epsilon; (1)$$ - But $x_{1n}=F(0,\tau_{0n},x_{0n})$ exists for $n$ sufficiently large and tends to zero and we would have $$\left \|F(\tau_n,0,x_{1n}) \right \|=\left \|F(\tau_n,\tau_{0n},x_{0n}) \right \| \ge \epsilon; (2) $$ against the stability assumption. $\blacksquare$

I don't understand (1) and (2). I think we must use If $x(t)$ is solution then $y(t)=x(t+\omega)$ too.

If you understand this solution, you can explain to me? anyone? Thanks!

Answer 2

Here's the solution (my professor):

$\blacktriangle $. Prove that $x\equiv 0$ is Uniformly Stable:

• Because $x\equiv 0$ is stable, so $\forall \varepsilon >0, \forall t_0 \ge 0, \exists \eta (t_0,\varepsilon ) >0 $ such that $\left \| x_0 \right \|<\eta$ then $$\left \| x(t) \right \|<\varepsilon ,\forall t \ge t_0 \ge 0$$

• Because $x(t;\tau,u)$ is continuous (follow $(\tau,u)$), so: $\exists \delta_1>0: \left | \tau-t_0 \right |<\delta_1,\left \| u-x_0 \right \|<\delta_1, \forall t_0,\tau \in \left [ 0,\omega \right ]$ then $$\left \| x(t;\tau,u) -x(t;t_0,x_0)\right \|<\dfrac{\eta}{2}, \forall t \in \left [ 0,\omega \right ]$$

• Because $x(t;t_0,x_0)$ is continuous (follow $t$), so $\exists \delta_2>0$ such that $$\left | t-t_0 \right |<\delta_2 \implies \left \| x(t;t_0,x_0)-x_0 \right \|<\eta$$

• We take $\delta(t_0)=\min \left \{ \delta_1,\delta_2,\eta \right \}$.

Whence, $\forall t' \in \left ( t_0-\delta(t_0),t_0+\delta(t_0) \right ), \left \| x_0 \right \|<\delta(t_0)$ we consider solution $x(t;t',x_0)$ of $(1)$: $\left \| x(t_0;t',x_0) \right \|\le \left \|x(t_0;t',x_0)-x(t_0;t_0,x_0) \right \|+ \left \|x(t_0;t_0,x_0) \right \|<\dfrac{\eta}{2}+\dfrac{\eta}{2}=\eta$ then $$ \left \| x(t;t_0,x(t_0;t',x_0)) \right \|<\epsilon$$

• Therefore, $\forall s \in \left [ 0,\omega \right ],\exists \delta (s)>0$ such that $\forall t_0 \in \left [ s-\delta(s),s+\delta(s) \right ], \forall x_0: \left \| x_0 \right \|<\delta$ then $$\left \| x(t;t_0,x_0) \right \|<\epsilon, \forall t\ge t_0$$

• Thus, we have $$\bigcup_{s \in \left [ 0,\omega \right ]}\left ( s-\delta(s),s+\delta(s) \right )\supset \left [ 0,\omega \right ]$$ Hence, $\exists s_1,\ldots, s_m$ such that $$\bigcup_{k=\overline{1,m}}\left ( s_k-\delta(s_k),s_k+\delta(s_k) \right )\supset \left [ 0,\omega \right ]$$

• We let $\delta :=\min_{k=\overline{1,m}} \delta (s_k)>0$. We consider any solution $x(t;t_0,x_0)$ of $(1)$: $$\left \| x_0 \right \|< \delta, \exists m \in \Bbb Z:\tau=t_0+m\omega\in \left [ 0, \omega \right ]$$

• By taking $y(t)=x(t-m\omega) $ is a solution of $(1)$. Whence, $$\left \| y(\tau) \right \|=\left \| x(\tau-m\omega) \right \|=\left \| x_0 \right \|< \delta \implies \left \| y(t) \right \|<\epsilon, \forall t\ge \tau \implies \left \| x(t) \right \|<\epsilon, \forall t\ge t_0 \blacksquare $$

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$\blacktriangle $ But how can we show that $x\equiv 0$ is Uniformly asymptotically Stable?

Any help will be appreciated! Thanks!