I am confused by the definition of a controllable pair (A,B) from "Linear Systems" by Chen. Definition of a controllable linear system

But can we get convergence in continuous case in finite time? Say a scalar system: $\frac{dx}{dt}=ax+bu$, ($b\neq0$ to be controllable). Assign control $u=-kx, (k>0)$, the state $x$ will only converge to the origin exponentially fast, but will never reach zero. In discrete time case, since we only consider the state at discrete time points, we can control the state to zero in finite time.

Thanks a lot for your answers!

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    $\begingroup$ Controllability does not mean that a reached state can be maintained, merely that any state can be reached. $\endgroup$
    – Cesareo
    Commented Jun 8, 2018 at 15:07
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    $\begingroup$ @Cesareo: I don't think the OP means this. I think the problem is just the equilibrium point (the origin of a linear system). Zixi (great question +1): I think controllability does not only constrain itself to linear feedbacks. You could use bang-bang like nonlinear controllers that would drive the state to zero in finite time (e.g. see Feldbaum's n-interval theorem). $\endgroup$
    – MrYouMath
    Commented Jun 8, 2018 at 15:12
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    $\begingroup$ A controllable system does not have to be stable to initial conditions.A system is controllable when any state can be reached with adequate input. $\endgroup$
    – Cesareo
    Commented Jun 8, 2018 at 15:16
  • $\begingroup$ In the following papers "Finite Time Controllers" and "Finite Time Differential Equations" by V. T. Haimo are studied 1st and 2nd order scalar autonomous ODEs that achieve solutions of finite duration, where is explained uniqueness of solutions is not hold since is required for the ODE to have at least one point in time where is not-Lipschitz. With this, no Linear ODE could have a solution that reach zero in finite time and stays there forever after. As counter example x'=-sgn(x)sqrt(|x|) , x(0)=1 could stand the solution x(t)=1/4*(1-t/2+|1-t/2|)^2 $\endgroup$
    – Joako
    Commented May 20, 2022 at 11:07

1 Answer 1


Assume, without loss of generality, that the initial condition $x(0) > 0$. Now assign the control

$$ u = r - k x $$

where $r$ is the reference value for the control system. Now set

$$ r = -1. $$

Then $x(t)$ will converge exponentially fast to $-1$. Because the initial condition was assumed to be positive, $x(t)$ will eventually cross the zero line. Due to continuity, there will be a time $t^*$ at which $x(t^*) = 0$ holds exactly. Therefore, the origin can be reached exactly.

The same holds analogously for negative initial conditions.

Notice that the definition just states that $x(t)$ is transfered to a final state, not that this state can actually be maintained, as mentioned by Cesareo.

  • $\begingroup$ Thank you @SampleTime for the detailed answer. Yes, I think my understanding of "transfer from initial state to a final state" was wrong, as I was interpreting that as "stabilize at the final state". $\endgroup$
    – Zixi
    Commented Jun 8, 2018 at 19:03

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