# A question about the definition of stability of a critical point

From an introductory chapter to differential equations:

A critical point, say, $x_0$, of the differential equation $$\frac{dx}{dt}=f(x)\tag{1}$$ is called stable if given $\varepsilon > 0$, there is a $\delta > 0$, such that for all $t \geq t_0,$ $\lVert x(t)−x_0(t)\rVert< \varepsilon$, whenever $\lVert x(t_0) − x_0(t_0) \rVert< \delta,$ where $x(t)$ is a solution of (1).

I understand that "for all $t \ge t_0,$ $\lVert x(t)−x_0(t)\rVert< \varepsilon$" is a way of writing that the value of $x(t)$ is not going away from $x_0$. But why is the "whenever $\lVert x(t_0) − x_0(t_0) \rVert< \delta$" useful?

• Are you sure it is not $t\ge t_c$? Jul 30, 2013 at 16:17

The delta part is necessary because it establishes that $x_0$ is a critical point if the subsequent condition holds, no matter how near or far the critical point is away from the solution. That's what makes it critical. It could be super close or super far away. All we know is that if the downstream solutions are always relatively bounded, then the point is a critical point.

• If we set time to $t_0$ we will see, that initial condition should suffice to both inequalities: $\lVert x(0)-x_0 \rVert < \delta$ and $\lVert x(0) - x_0 \rVert < \varepsilon$, so $\delta < \varepsilon$. The best explanation for this condition would be an example that shows some special behaviour of trajectories with respect to initial neighbourhood. Jul 30, 2013 at 16:08

Yes, the condition "whenever $\| x(t_0)−x_0(t_0) \|<\delta$" is useful.

A critical point, say $x_0$, is stable means that if a point $x(t_0)$ is near $x_0$ at $t_0$, then for all $t \geq t_0$, the corresponding solutions $x(t)$ and $x_0(t)$ starting from $x(t_0)$ and $x_0(t_0)$ at $t$ are near too.

In mathematical language, the definition is

"A critical point, say, $x_0$, of the differential equation $\frac{dx}{dt}=f(x)(1)$ is called stable if given $\varepsilon>0$, there is a $\delta>0$, such that for all $t \geq t_0$,$\| x(t)−x_0(t) \| <\varepsilon$, whenever $\| x(t_0)−x_0(t_0) \| < \delta$, where $x(t)$ is a solution of $(1)$."

• Awww ok, I did not get the being near at t0 part. It makes sense. Thanks. Jul 30, 2013 at 15:46

I've mentioned that there's some peculiarity behind the definition of Lyapunov stability, especially the part that was asked by topic starter. Consider the following system:

This is just a sample of dynamics, but you can extend it to the whole plane. And of course there'll be a critical point inside this family of closed curves.

So, you look at picture; what associations do you have? Trajectories don't go to infinity, that's good, and they have a behaviour that resembles "staying in the neighbourhood of critical point", something stable. It's good definitions try to catch the most important examples which behaviour we want describe. I think that you agree that behaviour in this example should be called stable too (like an example of conservative harmonic oscillator $x'' = -\omega^2 x$).

Now, if we try to remove the part that you was asking about, we'll find that our new definition don't capture this case: some trajectories that begin in $\epsilon$-ball go out of it. And that's bad: definition don't work. But look, if we have two neighbourhoods (as in original definition), we can capture this example: for every neighbourhood in this example you can know the maximal distance along all trajectories from origin. And here begins the $\epsilon-\delta$ interplay.