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I'm confused about this statement from Wikipedia:

Controllability does not mean that a reached state can be maintained, merely that any state can be reached.

What is an example of a system that is controllable, but has a state that is not maintainable?

Here is my confusion: suppose such a state $x$ exists. If I can reach any state from any other state (controllability), why can't I maintain that state by applying the appropriate control input to go from any $x+\epsilon$ back to $x$?

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The system, $$\begin{aligned} \dot{x} &= y, \\ \dot{y} &= u \end{aligned}$$ is controllable. Thus one can drive the system to the state $(x, y) = (1, 1).$ This state cannot be maintained.

What you will find when constructing the control action that reaches that state is that the system is simultaneously forced to leave that state once it reaches there, because there is an inherent "delay" in how the system responds to your input.

Intuitively, you can think of a car with a target state being a position-velocity pair $(p, v)$ and control action being acceleration. You can target some non-zero position $p$ with non-zero velocity $v$, but, if your velocity is non-zero, you can't possibly stay at that position $p$ without instantaneously changing velocity (i.e. infinite acceleration) which is not possible.

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  • $\begingroup$ Perfect example, thank you. As a quick follow-up, is there any formal notion for controllability, where all reachable states can also be maintained? (I think this is different from stabilizability) $\endgroup$ Commented Dec 24, 2024 at 19:27
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    $\begingroup$ @dkv the set of equilibrium points contains all the states that can be maintained. $\endgroup$ Commented Dec 24, 2024 at 19:39
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    $\begingroup$ @dkv, A set is controlled-invariant if there exists a control action that keeps you on that set if you start there. So you'd ask for a target set-point to be controlled-invariant. With that, you can use your control action to reach the point, then switch to the maintaining action to render the point invariant (i.e. make it an equilibrium). This still won't guarantee continuous control action. I'm not aware of a definition that combines the two into one. $\endgroup$ Commented Dec 24, 2024 at 20:51
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    $\begingroup$ Making a note for myself (and anyone else reading who needs things spelled out ;)) on the comment by @KBS. This the the set of all (x*, u*) such that x*=Ax*+Bu*, which can be found by solving [(I-A); -B][x*; u*]=0 $\endgroup$ Commented Dec 28, 2024 at 1:08

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