In control theory, we talk about the direct and indirect Lyapunov methods, applied to the stability analysis of nonlinear systems.

There are systems that are locally unstable for specific operating points, but that are globally stable.

What i would like to know is if there is any system which is locally stable, for any linearized point, but is globally unstable.

In other words, I would like to know if a system that is guaranteed to be locally stable, thorough linearization, for any point, can be said to be globally stable as well.

In the case it is positive, I would like to know if uniform stability and exponential stability holds as well.

  • 6
    $\begingroup$ You are really referring to the Markus-Yamabe conjecture. It holds in dimension two but fails in general in bigger dimensions. See for example en.wikipedia.org/wiki/Markus-Yamabe_conjecture. On the other hand, you may want to have a look at the Kalman conjecture, which is a special case of the Markus-Yamabe conjecture and holds in more generality (in particular it holds in dimension 3, as far as I remember with a proof by Barabanov, somewhat cryptic but considered correct). $\endgroup$ – John B Nov 17 '17 at 21:49

The answer is no. A system might be unstable for some initial conditions even if its linearization is Hurwitz for all $\boldsymbol{x}$.

An example is provided in this lecture, page 4 (Nonlinear Systems and Control — Spring 2015: State-Dependent Riccati Equation Method). Let

$$ \begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \end{bmatrix} = \begin{bmatrix} \frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} & 1 + \sqrt{x_1^2 + x_2^2} \\ 0 & \frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} $$

with $\epsilon > 0.$ The eigenvalues are given by the matrix entries $a_{11}$ and $a_{22}$, which are obviously negative for any $\boldsymbol{x} \in \mathbb{R}^2$. Thus $A(\boldsymbol{x})$ is Hurwitz for all $\boldsymbol{x} \in \mathbb{R}^2$.

However, take $\epsilon = 1$ and solve the ODE numerically, using for example the Runge-Kutta method, with initial conditions $\boldsymbol{x}_0 = \begin{bmatrix}1 & 1\end{bmatrix}^T$. You will observe that $x_1 \rightarrow \infty$ and $x_2 \rightarrow \text{const}$.

  • 1
    $\begingroup$ But $A(x)$ is not the Jacobian of this system, so you can't make any conclusion from its eigenvalues. If one does calculate the true Jacobian you will see that for almost half of the state space the linearisation will be unstable. $\endgroup$ – Kwin van der Veen Nov 17 '17 at 19:38
  • $\begingroup$ This is a nonlinear system system, not a linearization of a system as pointed out by van der Veen. $\endgroup$ – ITA Nov 17 '17 at 22:57
  • $\begingroup$ But there should exist some nonlinear system $\dot{\boldsymbol{x}} = f(\boldsymbol{x})$ such that the above $A(\boldsymbol{x})$ is the Jacobian of $f$, right? $\endgroup$ – SampleTime Nov 18 '17 at 10:06
  • $\begingroup$ @SampleTime I am not sure about this namely $\frac{\partial\,f_2}{\partial\,x_1}=0$, which would imply that $f_2$ is not a function of $x_1$. However $\frac{\partial\,f_2}{\partial\,x_2}$ is a function of $x_1$, which would suggest the opposite. $\endgroup$ – Kwin van der Veen Nov 18 '17 at 19:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.