This question is generated from Example 6.3 in Slotine and Li "Applied Nonlinear Control", Prentice Hall, 1991

A nonlinear system

$$ \begin{bmatrix}\dot{x_1}\\\dot{x_2}\end{bmatrix}=\begin{bmatrix}{x_2}^3+u\\u\end{bmatrix}$$ $$ y=x_1 $$ as the control objective to make $y$ track $y_d$, and the tracking error is $e=y-y_d$.

It is, $\dot{y}=\dot{x_1}$ and choosing the control law

$$ u=-{x_2}^3-e+\dot{y_d} $$ gives $$ \dot{e}+e=0 $$ which is stable and converging to zero as the time $t\rightarrow\infty$ (or in other words, the error equation has one pole in -1).

The same control input $u$ applies also to the second equation of the nonlinear system, representing the internal dynamics. With the choice of $u$ as above, $\dot{x_2}=u$ yields $$ \dot{x_2}+{x_2}^3=\dot{y_d}-e. $$

With the choice of a bounded $\dot{y_d}$ and $e$ (bounded since $\dot{e}+e=0$), then it is $$ |\dot{y_d}-e|\leq D, $$ being $D$ a positive constant.

From now on my question starts, as I would like to know what are the math steps to derive the following conclusion:

the example concludes that $|x_2|\leq D^{1/3}$ (i.e. $x_2$ is bounded too!), since $\dot{x_2}<0$ when $x_2>D^{1/3}$, and $\dot{x_2}>0$ when $x_2<D^{1/3}$. Can someone explain how to derive these inequalities?

The example demonstrates that the chosen control law for $u$, i.e. $u=-{x_2}^3-e+\dot{y_d}$ -chosen from the first state equation- does not causes the second state $x_2$ to become unbounded.


1 Answer 1


Let us rewrite your equation as $\dot{x}_2 = -x_2^3 + d(t)$, where $|d(t)|\le D$ for all $t$. If $x_2$ is positive, then $\dot{x}_2$ is negative for all $x_2>D^{1/3}$. If $x_2$ is negative, then $\dot{x}_2$ is positive for all $x_2<-D^{1/3}$. It implies that the set $|x_2^3|\le D$ is invariant and attractive: if the initial condition $x_2(t_0)$ belongs to the set, then the trajectory reamins in the set. If the initial condition $x_2(t_0)$ is outside the set, then it will converge to the set.

Thus, $x_2(t)$ is bounded. However, the claim that $|x_2|\le D^{1/3}$ is valid only if you know for sure that the initial condition satisfies this inequality, or if you consider it as $t \to \infty$.

  • $\begingroup$ Thank you @Arastas, you made it very clear. So, if I understand well, we are trying to ensure that $x_2$ is always bounded. So we state the condition for $\dot{x_2}>0$ and $\dot{x_2}<0$ (as I assume we don't know $\dot{x_2}$ itself, but we try to estimate what happens in each case)... overall this approach is looking at what happens to $x_2$ when the control law chosen based on controlling $x_1$ is applied $\endgroup$
    – Lello
    Commented Nov 17, 2019 at 22:11
  • 1
    $\begingroup$ Yes, you are right. There are also more general ways to study it, i.e., the Lyapunov functions. But for the scalar case is more simple. $\endgroup$
    – Arastas
    Commented Nov 18, 2019 at 9:17

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