So I have been studying control theory, and I have found different authors use different definitions for the concept of accessibility. In particular, I find that with one definition it is trivial to prove that accessibility is a necessary condition for controllability, and with the other one...not so much.

Let me introduce some concepts first. We'll be dealing with a differential equation on a smooth $n$-manifold $M$ of the form \begin{equation}\label{controlsys} \dot{x} = f(x,u(t)), \end{equation} where $u(t)$ is a time-dependent map from the nonnegative reals $\mathbb{R}^+$ to a constraint set $\Omega \subset \mathbb{R}^m$; $u$ is the control.

Define the reachable sets: Given $x_0 \in M$, we define $R(x_0,t)$ to be the set of all $x \in M$ for which there exists an admissible control $u$ such that there is a trajectory of the control system with $x(0)=x_0, x(t) = x$. The reachable set from $x_0$ at time $T$ is defined to be \begin{equation} R_T(x_0) = \bigcup_{0 \leq t \leq T} R(x_0,t). \end{equation}

Now here come the two definitions of accessibility I have found.

  1. The control system on $M$ is said to be accessible from $p \in M$ if for every $T>0$, the reachable set $R_T(p)$ contains a nonempty open set.
  2. The control system on $M$ is said to be accessible from $p \in M$ if for some $T>0$, the reachable set $R_T(p)$ contains a nonempty open set.

Since my definition of controllability is simply: for all $p \in M$ there exists a $T>0$ such that $R_T(p) = M$ , it is clear that controllability implies accessibility from every point $p$ according to the second definition.

However, having seen how many papers use either one of the two definitions of accessibility above nonchalantly, I believe (1) and (2) are equivalent. Intuitively it kinda makes sense (it'd be very weird for $R_T(p)$ to go from having an empty interior to a nonempty interior all of a sudden as we vary $T$--and accessibility after time $T$ implies accessibility for all later times, of course), but I'm stuck trying to prove it. I also couldn't find any authoritative references discussing the distinction, which I thought was weird.

Can you please help me prove these 2 definitions are equivalent? Thanks!

  • $\begingroup$ The most helpful thing to do in the light of the answer below is to link to one or more of the papers where you suspect that the equivalence of the two notions has been used. Then we can talk about the the equivalence in the context of those papers. $\endgroup$ Jun 1, 2021 at 6:55
  • $\begingroup$ Sure, let me link a couple: References using definition (2) tend to be older ones, such as the ones mentioned in encyclopediaofmath.org/wiki/Accessibility. References using definition (1) tend to be more recent ones, such as christophkawan.de/img/Introduction_Control_Theory.pdf $\endgroup$ Jun 5, 2021 at 22:19
  • $\begingroup$ Thanks, I'll let you know why those statements work. $\endgroup$ Jun 6, 2021 at 7:04

1 Answer 1


They're not equivalent. Take $\ M=\mathbb{R}^2\ $, $\ \Omega=[0,2]^2\ $ and $\ f\ $ to be the function defined by $$ f(x,u)=\cases{(1,0)&if $\ x_1<1$\\ x+((u_1x_1-1,u_2x_2-1))&if $\ x_1\ge1\ $. } $$ For this control system $$ R_T((0,1))=\cases{ [0,T]\times\{1\}\ &for $\ 0<T\le1$\\ \big[0,\frac{2+e^{3(T-1)}}{3}\big]\times\big[1,\frac{2+e^{3(T-1)}}{3}\big]&for $\ T>1\ $, } $$ which contains an open subset of $\ M\ $ for $\ T>1\ $ but not for $\ 0<T\le0\ $. This system is therefor accessible from $\ (0,1)\ $ according to the second definition but not according to the first.

  • 1
    $\begingroup$ I think there's something wrong with this-- it seems you have computed $R(0,T)$ (that is, the points we can reach in the manifold at exactly time $T$) instead of $R_T(0)$ (that is, the points we can reach in the manifold for times $t \leq T$). In fact, reachable sets should form a chain under inclusion for increasing T, and the proposed formula doesn't satisfy that. Playing around with this control system gives me reachable sets with non-empty interior right from the get-go. $\endgroup$ Jun 5, 2021 at 22:12
  • $\begingroup$ Yes, you're right. I misread the definition of $\ R\ $. I've now modified my counterxample by adding an extra dimension to the state space. As long as the differential equation has a unique solution for any (sufficiently regular) $\ u(t)\ $, I don't think there exists a counterxample with a one-dimensional state space. Thank you for picking up the error. $\endgroup$ Jun 5, 2021 at 23:21
  • $\begingroup$ Yup, probably because these $R_T(x_0)$ sets are path-connected, which implies they are intervals on $\mathbb{R}$. This is a sweet counterexample extending that 1D intuition you had. Thank you! $\endgroup$ Jun 6, 2021 at 3:02

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