Let's consider the following nonlinear system,

$$\dot{x}_1=x_2^3+u$$ $$\dot{x}_2=-u$$ $$y=x_1$$

Taking the derivative of $y$,

$$\dot{y}=\dot{x}_1=x_2^3+u \qquad \text{Equation 1}$$

Let $f_1(x) = x_2^3$ and $v=\dot{y}$ then Equation 1 becomes

$$u=v-f_1(x) \qquad \text{Equation 2}$$

Let $v=\dot{y}_d-e$ where $e=y_d-y$ and $y_d$ is a continuous differentiable trajectory. Then isolating input $u$ in Equation 2 yields,

$$u = \dot{y}_d-e-f_1(x)$$

or equivalently,

$$u=\dot{y}_d-e-x_2^3 \qquad \text{Equation 3}$$

Defining the internal state $\eta$ as $$\eta=\dot{x}_2$$ How can I show that the internal dynamics is unstable?

  • $\begingroup$ When you substitute the definition of $e$ back in you get $u=\dot{y}-x_2^3=x_2^3+u-x_2^3=u$ so it is unclear to me what this accomplishes. $\endgroup$ – Kwin van der Veen Oct 22 '18 at 19:57
  • $\begingroup$ Do you know about control Lyapunov functions? What do you mean by "internal dynamics"? What is the initial condition of the system? $\endgroup$ – Mortified Through Math Oct 23 '18 at 14:28
  • $\begingroup$ If you're interested in the internal dynamics, then make $u=0$, ignore the output and see how the state evolves when starting from a nonzero initial state. $\endgroup$ – Rodrigo de Azevedo Oct 27 '18 at 20:10

Substituting Equation 3 into Equation 1,

$$\dot{y} = \dot{y}_d-e$$

Noting that $\dot{e} = \dot{y}_d - \dot{y}$, we are left with

$$\dot{e}+e=0\qquad \text{Equation 4}$$

Substituting $-u=\dot{x}_2$ into Equation 3 we get,

$$-\dot{x}_2+x_2^3=\dot{y}_d-e\qquad \text{Equation 5}$$

In the view of facts that e is guaranteed to be bounded by Equation 4 and $\dot{y}_d$ is assured to be bounded, we have the following condition

$$|\dot{y}_d-e|\leq D$$

where D is a positive constant. We can conclude from Equation 5 that

$$|x_2|\leq D^{1/3}$$

Then, when

$$\dot{x}_2<0 \Rightarrow x_2<-D^{1/3}$$

and when

$$\dot{x}_2>0 \Rightarrow x_2>D^{1/3}$$

According to these two Equations, the system has negative velocity at negative position and positive velocity at positive position, respectively. This result proves that the internal state ($\eta$) of $\dot{x}_2$ is indeed unstable.


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