On the Global Uniform Asymptotic Stability of Time-Varying Systems

Abstract

This work is motivated by the problem of practical stability of nonlinear time-varying system

$$\begin{aligned} \dot{x}(t)&= f(t,x(t)),\\ x(t_0)&= x_0, \end{aligned}$$

where $t\in \mathbb R , x\in \mathbb R ^n$, and $f:\mathbb R \times \mathbb R ^n \longrightarrow \mathbb R ^n$ is continuous in $t$ and locally Lipschitz in $x.$ We give some sufficient conditions to guarantee practical uniform asymptotic stability. An invariance principle is given when the origin is not an equilibrium point. The main result of this paper is illustrated by an example in three dimensional.

Appendix

Proof of Lemma 1

We apply the Ascoli–Arzel lemma twice (see [17]): first to extract a subsequence such that $(w_i)_{i}$ is uniformly convergent and second to extract further another subsequence such that $(\dot{w_i})_{i}$ is uniformly convergent. Then we obtain

$$\begin{aligned} \displaystyle \lim _{i \longrightarrow +\infty }\frac{dw_i}{dt}=\frac{d}{dt}\displaystyle \lim _{i \longrightarrow + \infty }w_i. \end{aligned}$$

$1.$ We verify that $(w_i)_{i}$ is uniformly bounded and equicontinuous. The uniformly boundedness comes from $\Vert w_i\Vert _{\infty }< M.$ The equicontinuity comes from the uniformly boundedness of the first derivative. Indeed, since $\Vert w_i\Vert _{\infty }\le M$, the closed ball $\overline{B}_{M}$ is compact and $f(t,\cdot )$ is continuous on $\overline{B}_{M}$, there exists a constant $A$ such that

$$\begin{aligned} \Vert f(t,x)\Vert \le A, \quad \forall (t,x)\in \mathbb R \times \overline{B}_{M}. \end{aligned}$$

Then

$$\begin{aligned} \Vert \dot{w}_i(t)\Vert =\Vert f(t+t_i,w_i(t))\Vert \le A, \quad \forall i\in I, \quad \forall t\in [a,b]. \end{aligned}$$

Thus, $(\dot{w_i})_{i}$ is relatively compact and we can extract a subsequence that we denote also by $(\dot{w_i})_{i}$, which is uniformly convergent to a function $w\in \mathcal C ^{0}([a,b];D).$

$2.$ We prove that $(\dot{w}_{i})_{i}$ is relatively compact. We have already proved the uniform boundedness $\Vert \dot{w}_i\Vert _{\infty }\le A.$ For the equicontinuity, we use both the uniform continuity in $t$ and uniform local Lipschitz continuity in $x$ of $f.$ Let $L_{M}$ be the uniform Lipschitz constant corresponding to the compact set $\overline{B}_{M}.$ Then $\Vert \dot{w}_{i}(t_1)-\dot{w}_{i}(t_2)\Vert =\Vert f(t_i+t_1,w_i(t_1))-f(t_i+t_2,w_i(t_2)) \le \Vert f(t_i+t_1,w_i(t_1))-f(t_i+t_2,w_i(t_1))\Vert +\Vert f(t_i+t_2,w_i(t_1))-f(t_i+t_2,w_i(t_2))\Vert .$ Let $\varepsilon > 0$ be arbitrarily. Then we choose $\delta _{1}$ such that

$$\begin{aligned} \Vert f(s_1,x)-f(s_2,x)\Vert <\frac{\varepsilon }{2}, \quad \forall |s_1-s_2|<\delta _{1}, \quad \forall x \in \overline{B}_{M}. \end{aligned}$$

On the one hand,

$$\begin{aligned} \displaystyle \Vert f(t_i+t_2,w_i(t_1))-f(t_i+t_2,w_i(t_2))\Vert \le L_{M}\Vert w_i(t_1)-w_i(t_2)\Vert \le L_{M}A |t_1-t_2|. \end{aligned}$$

Then we choose

$$\begin{aligned} \delta =\min \left(\delta _1,\displaystyle \frac{\varepsilon }{2L_{M}A}\right). \end{aligned}$$

It follows that the left-hand side from the above inequality is also bounded by $\displaystyle {\varepsilon }/{2}$ for any $t_1, t_2$ with $|t_1-t_2|<\delta .$ So,

$$\begin{aligned} \Vert \dot{w}_{i}(t_1)-\dot{w}_{i}(t_2)\Vert <\frac{\varepsilon }{2} +\frac{\varepsilon }{2}=\varepsilon \end{aligned}$$

for any $i$ and $t_1,t_2 \in [a,b], |t_1-t_2|<\delta .$ We can now extract a second subsequence from $(\dot{w_i})_{i}$ which is also uniformly convergent and this ends the proof of lemma.$\square$

Proof of Lemma 3

Let $\tau >0$ be an arbitrary time interval. Let $(t_{k })_{k}$ be the sequence that renders $x^{*}$ a w-limit point for the trajectory $x(t;t_0,x_0).$ Then if we denote by $x_k=x(t_k),$ we have

$$\begin{aligned} \displaystyle \lim _{k \rightarrow +\infty }x_{k}=x^{*}. \end{aligned}$$

Consider the following sequence of solutions:

$$\begin{aligned} \displaystyle w_k:[0,\tau ]\longrightarrow D, w_k(t)=x(t+t_k;t_k,x^{*}). \end{aligned}$$

We have chosen $x_0, t_0$ such that all these functions are bounded by $M$, it means that $\Vert w_k\Vert _{\infty }<M.$ We have $w_k$ not in $B_r, w_k(0)=x^{*}$ and $V(w_k(t))\le V(x^{*}).$ Let us denote by

$$\begin{aligned} \displaystyle y_k^{t}=x(t+t_k), \quad \forall 0\le t\le \tau , \end{aligned}$$

and let $L_M$ be the Lipschitz constant of $f$ on the compact $\overline{B}_M.$ Then

$$\begin{aligned} \Vert y_k^{t}-w_k(t)\Vert \le \displaystyle \Vert y_k^{0}-w_k(0)\Vert +L_M\int _{0}^{t}\Vert w_k(s)-y_k^{s}\Vert ds. \end{aligned}$$

Thus, by applying Gronwall–Bellman’s lemma, we obtain

$$\begin{aligned} \displaystyle \Vert y_k^{t}-w_k(t)\Vert \le \displaystyle \Vert x_k-x^{*}\Vert e^{L_Mt}. \end{aligned}$$

Since

$$\begin{aligned} \displaystyle \lim _{k \rightarrow +\infty }x_{k}=x^{*}, \end{aligned}$$

we get

$$\begin{aligned} \displaystyle \lim _{k \rightarrow +\infty }\Vert y_k^{t}-w_k(t)\Vert =0. \end{aligned}$$

We have $V(x^{*})=\displaystyle \lim \nolimits _{t\rightarrow +\infty }V(x(t))$ and $V$ is nonincreasing on trajectories in $D\backslash B_r$ which implies that

$$\begin{aligned} V(y_k^{t})>V(x^{*}), \end{aligned}$$

and also

$$\begin{aligned} \displaystyle \lim _{k \rightarrow +\infty }V(y_k^{t})=V(x^{*})=\displaystyle \lim _{k \rightarrow + \infty }V(w_k^{t}). \end{aligned}$$

Since $(w_k)_{k}$ is uniformly bounded, we apply Lemma 1 and we obtain a subsequence uniformly convergent to a function $\displaystyle w\in \mathcal C ^{1}([0,\tau ];(D\cup \overline{B}_{M})\backslash B_r).$ Obviously,

$$\begin{aligned} V(w(t))=V(x^{*}), \quad \forall 0\le t \le \tau . \end{aligned}$$

Thus, $W(w(t))=0$ and $w(t)\in E.$ Since $f$ is continuous in $(t,x),$ we obtain that $w$ is an integral curve of $f(t_{*},\cdot ),$ it means that $\dot{w}(t)=f(t_{*},w(t))$ for $0\le t\le \tau$ and any $t_{*}.$ In particular, for $t_{*}=t+t_k,$ we get $w(t)$ is a solution of the same equation as $w_k(t)$ and $w(0)=w_k(0)=x^{*}.$ By the uniqueness of the solution, they must coincide. Then

$$\begin{aligned} x(t+t_k;t_k,x^{*})\in E, \quad 0\le t \le \tau . \end{aligned}$$

But $\tau$ was arbitrarily, thus $x(t;t_0,x^{*})\in E$ for any $t,$ and then $x^{*}\in N$. $\square$