Difference between Lyapunov and uniform stability

4.2k Views Asked by At

I'm trying to understand the difference between general Lyapunov stability and uniform stability.

I understand, that Lyapunov s. is the situation, when for every $\epsilon > 0$, there exists $\delta > 0$, such that when the solution starts within a distance $\delta$ from the equilibrium point, it remains within a distance $\epsilon$ from it forever.

But I don't get, what's the difference with uniform stability at all. I've found some info on this site, but still I don't get it.

Could you, please, explain it to me in some intuitive way?

2

There are 2 best solutions below

1
On BEST ANSWER

The concept of uniform stability is mainly defined for non-autonomous systems, i.e. the systems of the form $$\dot x = f(t,x),$$ but Lyapunov stability or what is often called just stability is defined for both autonomous systems and non-autonomous systems.

We first assume that the system is non-autonomous and let $x_0(t)$ be solution of $\dot x = f(t,x)$ starting at $x_0(t_0)$. Then we have the following definitions [Khalil, Hassan K., Nonlinear systems., Upper Saddle River, NJ: Prentice Hall. xv, 750 p. (2002). ZBL1003.34002.]:

  • $\textbf{Lyapunov stability (stability)}$: for each $\varepsilon>0$, there is $\delta=\delta(\mathbf{\epsilon,t_0})>0$ such that $$\|x(t_0)-x_0(t_0)\|<\delta~~\Rightarrow ~~\|x(t)-x_0(t)\|<\epsilon, ~~\forall t\geq t_0\geq0$$
  • $\textbf{Uniform stability}$: for each $\varepsilon>0$, there is $\delta=\delta(\mathbf{\epsilon})>0$, $\textbf{independent of}$ $\mathbf{t_0}$, such that $$\|x(t_0)-x_0(t_0)\|<\delta~~\Rightarrow ~~\|x(t)-x_0(t)\|<\epsilon, ~~\forall t\geq t_0\geq0$$

Note that the only difference is in the definition of $\delta$, where for uniform stability, $\delta$ is independent of $t_0$.

For autonomous system, $\dot x = f(x)$, uniform stability is the same as Lyapunov stability because $f$ is not explicitly a function of $t$.

9
On

Let us consider the following nonlinear system $$\dot x = f(t,x),\ x(t_0)=x_0,$$ and assume that there is a unique solution starting from $x_0$ at time $t_0$. Let $\phi(t,t_0,x_0)$ be that solution.

Let $x^*$ be an equilibrium for this system, i.e. $f(t,x^*)=0$ for all $t\ge t_0$. Then, we have the following definitions

  • The equilibrium point $x^*$ is stable if for each $\varepsilon>0$, there is $\delta=\delta(\mathbf{\epsilon,t_0})>0$ such that $$\|x_0-x^*\|<\delta~~\Rightarrow ~~\|\phi(t,t_0,x_0)-x^*\|<\epsilon, ~~\forall t\geq t_0\geq0$$

  • The equilibrium point $x^*$ is uniformly stable if for each $\varepsilon>0$, there is $\delta=\delta(\mathbf{\epsilon})>0$ such that $$\|x_0-x^*\|<\delta~~\Rightarrow ~~\|\phi(t,t_0,x_0)-x^*\|<\epsilon, ~~\forall t\geq t_0\geq0$$

Note that the only difference is in the definition of $\delta$, where for uniform stability, $\delta$ is independent of $t_0$.