Difference between a uniform stable and a uniform attractive solution

Question

Consider the following time-varying system:

$\dot{x} = f(x,t)$

The solution of this system which starts from the point $x_0$ at time $t_0 \geq 0$ is denoted as $x(t;x_0,t_0)$ with $x(t_0;x_0,t_0)=x_0$. The solution $x(t;x_0,t_0)$ of above system is:

1. Uniform stable, if for each $\epsilon >0$ there exists a $\delta(\epsilon)>0$ such that: $|\tilde{x}_0-x_0|<\delta \Rightarrow |x(t;\tilde{x}_0,t_0)-x(t;x_0,t_0)|<\epsilon, \forall t \geq t_0$
2. Uniform attractive, if there exist an $r>0$ and, for each $\epsilon >0$, a $T(\epsilon)>0$ such that: $|\tilde{x}_0-x_0|<r \Rightarrow |x(t;\tilde{x}_0,t_0)-x(t;x_0,t_0)|<\epsilon, \forall t \geq t_0+T$

I find it very difficult to understand the differences between these definitions.

Answer 1

Let us interpret the initial datum $\tilde{x}_0$ as a perturbation of $x_0$. The "uniform stable" condition means that, if the perturbation is small enough, then the solution with perturbed data stays near the original solution at all times.

On the other hand, the "uniform attractive" condition means that there exists an attractive region, namely the ball $B(x_0, r)$. Every time the perturbed datum lies in this region, the corresponding solution is asymptotically close to the original. This means that, if we go far enough in the future, the two solutions are as close as we want.

Note that the "uniform stable" condition does not imply that the solution with perturbed data is asymptotically close to the original. An example that might be illuminating is a simple harmonic oscillator $$ \begin{cases} \dot{v}+\omega^2 x=0 \\ \dot{x}=v. \end{cases} $$ Here $x$ can be interpreted as the position and $v$ as the velocity of a mass point attached to a spring. Clearly, the initial condition $(x_0, v_0)=(0,0)$ gives rise to the stationary solution $(x, v)=(0,0)$, corresponding to the point being motionless. But a small perturbation $(\tilde{x}_0, \tilde{v}_0)$ will give rise to the oscillating solution $$ (\tilde{x}(t), \tilde{v}(t))=\left( \tilde{x}_0\cos(\omega t)+\frac{\tilde{v}_0 \sin(\omega t)}{\omega} , -\omega \tilde{x}_0 \sin(\omega t)+\tilde{v}_0 \cos(\omega t)\right), $$ which is close to $(0,0)$ if $\sqrt{ \tilde{x}_0^2+\tilde{v}_0^2 }$ is small enough, but is not asymptotic to $(0,0)$ since it keeps oscillating.

Therefore the initial datum $(0,0)$ is for this system uniform stable but not uniform attractive.

Answer 2

Your notion of uniform attractive is also known as asymptotic stability. The notion of uniform stability is often called just the stability (or Lyapunov stability). In my experience, the notations in italics are much more widespread.

The difference between these stabilities lies in the behaviour of the solutions. In the case of stability, your solution changes by at most a constant for a sufficiently small change of initial data.

In the case of asymptotic stability, if the change of initial data is suffciently small, then the distance between the solutions tends to zero (some sort of convergence).

You can try to look at the system

$$\dot x(t)=\begin{pmatrix}0&1\\-1&0\end{pmatrix}x(t)$$ The zero solution is stable (in fact, you can prove that $\|x(t)\| = \|x(0)\|$), but not asymptotically.

However, in the case of $$\dot x(t)=\begin{pmatrix}-2&1\\0&-2\end{pmatrix}x(t)$$ you have asymptotical stability of zero solution. Whatever the initial data, the solution converges to zero as $t\to\infty$.