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I want to show a scholar example of a system such that its linearization is not controllable but the system can be stabilized with nonlinear feedback. I am thinking about this one $$ \begin{aligned} \dot{x}_1 &= x_2^3, \\ \dot{x}_2 &= u, \end{aligned} $$ where $x_1$, $x_2$, and $u$ are scalars, and the goal is to drive the system to the origin.

What is the simplest control that you would propose for this system?

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    $\begingroup$ What have you tried yourself, for example have you tried backstepping? Also wouldn't $\dot{x}=-x$ satisfy your description, since it is not controllable, but is stable? And how do you define "simplest control"? $\endgroup$
    – Kwin van der Veen
    Nov 4, 2019 at 15:02
  • $\begingroup$ I have tried myself to design this example :) My answer was $u=-x_2^3-x_2-x_1$, but as shown by @SampleTime, the term $x_2^3$ can be omitted. Normally we do not discuss controllability for autonomous systems. $\endgroup$
    – Arastas
    Nov 4, 2019 at 15:41
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    $\begingroup$ You could interpret it as $\dot{x}=-x+a\,u$ with $a=0$, otherwise another simple example would be the system $\dot{x}=x\,u$, for which $u=-x^2$ would be a "simple" stabilizing controller. $\endgroup$
    – Kwin van der Veen
    Nov 4, 2019 at 15:50
  • $\begingroup$ Thanks, $\dot{x}=xu$ is a nice example! $\endgroup$
    – Arastas
    Nov 4, 2019 at 15:53

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Even though the linearization is not controllable, the nonlinear system can still be stabilized with linear feedback. I propose the control law

$$ u = -x_1 - x_2 \tag{1} $$

which leads to the closed loop dynamics

$$ \begin{align} \dot{x}_1 &= x_2^3 \\ \dot{x}_2 &= -x_1 - x_2 \end{align} \tag{2} $$

Take the Lyapunov function

$$ V(x) = x_1^2 + 2 x_1 x_2 + x_2^2 + \frac{1}{2} x_2^4 $$

It is easy to show that $V$ has a unique minimum at $(0, 0)$ so it is positive definite. The derivative is

$$ \dot{V}(x) = -2 (x_1 + x_2)^2 $$

which is negative semi-definite (zero along the $x_1 = -x_2$ line). If we insert that into $(2)$, we have $\dot{x}_2 = 0$ but $\dot{x}_1 = -x_1^3$, so no solution can stay in the set $\dot{V}(x) = 0$ except $x_1 = x_2 = 0$.

So, by LaSalle, the system is globally asymptotically stabilized by the linear feedback $(1)$.

This is probably also the "simplest" stabilizing control law (linear feedback with both gains being 1), but that depends on your definition of simple.

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    $\begingroup$ Thanks! Note also that it is sufficient to consider $V(x) = x_1^2 + 0.5 x_2^4$ yeilding $\dot{V} = -2x_2^2$, and then the same invariance argument works. $\endgroup$
    – Arastas
    Nov 4, 2019 at 15:38
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    $\begingroup$ Your answer destroys my example. :) I was looking for an example that cannot be stabilized by linear feedback. :) $\endgroup$
    – Arastas
    Nov 4, 2019 at 15:39
  • $\begingroup$ @Arastas Sorry about that :) I would also be interested whether such systems even exist (I guess so, but I haven't seen an example yet). $\endgroup$
    – SampleTime
    Nov 4, 2019 at 15:42
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    $\begingroup$ I do not believe there exists a linear stabilizing control law for $\dot{x}=x\,u$, namely $u=-x^2$ does stabilize it. $\endgroup$
    – Kwin van der Veen
    Nov 4, 2019 at 15:53
  • $\begingroup$ @KwinvanderVeen That is indeed a nice example! $\endgroup$
    – SampleTime
    Nov 4, 2019 at 15:55

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