Prove that $x \equiv 0$ of $\dot{x}(t)=a(t)x$ is Uniformly Asymptotically Stable

Question

I have a problem:

Consider the scalar equation: $$\dot{x}(t)=a(t)x \tag{I}$$ where $a(t) \in C(\mathbb{R}^+)$.

Prove that $x \equiv 0$ of $(I)$ is Uniformly Asymptotically Stable iff $$\forall M>0, \exists T>0, \forall t_0 \ge 0: \int_{t_0}^{t}a(s)\mathrm{d}s<-M, \forall t \ge t_0+T$$

Here's my sketch:

• The first, since the equation $\dot{x}(t)=a(t)x $, we have $$x(t)=x_0 \cdot \exp \left (\int_{t_0}^{t}a(s)\mathrm{d}s \right)$$
• The second, $\forall M>0, \exists T>0, \forall t_0 \ge 0: \int_{t_0}^{t}a(s)\mathrm{d}s<-M \iff \exp \left (\int_{t_0}^{t}a(s)\mathrm{d}s \right) < e^{-M} $

Thus, $$\left|x(t) \right|=\left|x_0 \cdot \exp \left (\int_{t_0}^{t}a(s)\mathrm{d}s \right) \right| \le \left|x_0 \right| \cdot \exp \left (\int_{t_0}^{t}a(s)\mathrm{d}s \right) \le \left|x_0 \right| e^{-M}$$

• Now, I have stuck when I'm trying to show that $x \equiv 0$ of $(I)$ is Uniformly Stable and Uniformly convergent, we know

$1/$ Uniformly convergent:

$\forall \epsilon >0,\exists \delta_1>0, \exists T=T(\epsilon)>0, \text{s.t}: \|x(t_0)\|<\delta_1 \implies \|x(t)\|< \epsilon, \forall t \ge t_0 +T$.

$2/$ Uniformly Stable:

$\forall \epsilon >0,\exists \delta = \delta (\epsilon)>0, \text{s.t}: \|x(t_0)\|<\delta \implies \|x(t)\|< \epsilon, \forall t \ge t_0 \ge 0$.

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I choose $\epsilon = \left|x_0 \right| e^{-M}>0$. But How do we find $T=T(\epsilon)=???$

Whence, I still have no solution.

Any help will be appreciated! Thanks!

Answer 1

Disclaimer: The definition you linked says the "equilibrium $x_e=0$ at $t_0$ of $\dot{x}=f(x,t)$" which seems to imply that we are only interested in a particular fixed initial time $t_0$. I'm just going to roll with that: In the answer I assume that $t_0$ is some fixed initial time. However, it seems a bit strange given that the terminology "uniformly stable" and "uniformly convergent" which, in my experience, usually involves making statements for all possible times $t_0$ (see, for example, the book by Khalil).

To get the if we first show that

$$\forall M>0, \exists T>0:\int_{t_0}^{t}a(s)\mathrm{d}s<-M, \forall t \ge t_0+T\quad\quad(*)$$

implies that $0$ is uniformly asymptotically stable. Then, to get the only if it is enough to show that $0$ being uniformly convergent implies $(*)$.

To show that $(*)$ implies that $0$ is uniformly asymptotically stable we have show that $(*)$ implies that $0$ is both uniformly convergent and uniformly stable.

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Uniformly convergent is easy. Fix any $\varepsilon>0$. Then, require that there exists a $\delta>0$ and $T>0$ such that

$$|x(t_0)|\leq \delta\Rightarrow |x(t)|\leq \varepsilon\quad\forall t\geq t_0+T.$$

Fix any $M>0$. From your sketch we have that there exists a $T>0$ such that

$$\left|x(t) \right|\leq \left|x(t_0) \right| e^{-M}$$

for all $t\geq t_0+T$. Picking $\delta=\varepsilon e^M$ then gives us the desired $|x(t)|\leq\varepsilon$ for all $t\geq t_0+T$.

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For uniform stability: Again pick any $M>0$ and let $T>0$ be such that

$$|x(t)|\leq |x(t_0)|e^{-M} \quad (**)$$

for all $t\geq t_0+T$. Since $a$ is continuous, it is easy to see that

$$\int_{t_0}^{t_0+T}a(s)ds\leq \alpha$$

for some $\alpha\in\mathbb{R}$. Then, for any $t\in[t_0,t_0+T]$,

$$|x(t)|=|x(t_0)|\left|exp\left(\int_{t_0}^{t_0+T}a(s)ds\right)\right|\leq |x(t_0)|e^\alpha. \quad (***)$$

So, if we set $\delta=\min\{\varepsilon e^M,\varepsilon e^{-\alpha}\}$, then $(**)$ and $(***)$ gives the desired $|x(t)|\leq \varepsilon$ for all $t\geq t_0$.

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Lastly, for the only if: By uniform convergence you have that for any $x(t_0)$

$$\lim_{t\to\infty} x(t)=\lim_{t\to\infty} x(t_0)\exp\left(\int_{t_0}^{t}a(s)ds\right)=0\Leftrightarrow \lim_{t\to\infty} \int_{t_0}^{t}a(s)ds=0$$

which is equivalent to $(*)$.

Answer 2

Here's another solution:

1/ We'll show that $x \equiv 0$ is Uniformly Stable iff $$\int_{t_0}^{t}a(s)\mathrm{d}s \le M, \forall t \ge t_0 \ge 0$$

• Proof (Sufficiency).

We assume that $x \equiv 0$ is Uniformly Stable. $\forall \epsilon >0, \forall t_0 \ge 0 ,\exists \delta=\delta(\epsilon)>0 $ such that $$\|x_0\|< \delta \implies \|x(t)\| < \epsilon, \forall t \ge t_0 \ge 0$$ Taking $t_0=0,x_0=\dfrac{\delta}{2}, \epsilon =1$. So: $$\|x(t;0;x_0)\|=|x_0|\exp\left(\int_{0}^{t}a(s)\mathrm{d}s \right) <\epsilon =1$$ Therefore, $\int_{0}^{t}a(s)\mathrm{d}s< M:=\log\left(\dfrac{2}{\delta} \right)$.

• Proof (Necesity).

We assume that $\int_{t_0}^{t}a(s)\mathrm{d}s \le M, \forall t \ge t_0 \ge 0 $

We have $\forall \epsilon >0, \forall t_0 \ge 0 ,\exists \delta=\delta(\epsilon)>0 $ such that $\|x_0\|< \delta$.

We consider $\|x(t)\| \le \epsilon:=\delta \exp \left(-\int_{0}^{t_0}a(s)\mathrm{d}s \right) e^M, \forall t \ge t_0 \ge 0$

2/ We'll show that $x \equiv 0$ is Asymptotically Stable iff $$\lim_{t \to \infty}\int_{t_0}^{t}a(s)\mathrm{d}s =-\infty $$

3/ Since $(1)$ and $(2)$, we have $x \equiv 0$ is Uniformly Asymptotically Stable if and only if $$\forall M>0, \exists T>0, \forall t_0 \ge 0: \int_{t_0}^{t}a(s)\mathrm{d}s<-M, \forall t \ge t_0+T$$.