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I have two questions that are related to stability in the sense of Lyapunov.

  1. Is a system with multiple poles on the imaginary axis (e.g. double pole at $z=0$ or double pole at $z=i$) unstable in the sense of Lyapunov? From the example of $y''=0$ I would think that this linear system is unstable.

  2. When using the Linearization method from Lyapunov for investigating the stability of a nonlinear system, I know that if the linear system is asymptotically stable in the equilibrium point, then the equilibrium point of the nonlinear system is also asymptotically stable. If the linear system is unstable at the equilibrium point, then the equilibrium point of the nonlinear system is also unstable. It is said that for the case in which the linear system is marginally stable at the equilibrium point, then the linearization method is indecisive. Does the case from question 1 belong to the indecisive case, or would it imply that the nonlinear system is unstable at the equilibrium point?

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  • $\begingroup$ Lyapunov certifies that the origin is globally asymptotically stable (GAS). Eigenvalues on the imaginary axis make GAS impossible. One has only marginal internal stability. However, one does not have BIBO stability, as a sinusoidal input leads to resonance and the output blows up. $\endgroup$ – Rodrigo de Azevedo Aug 17 '17 at 16:06
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I think I have an answer to the first equation. The example $y''=0$ can be transferred into state space form:

$\dot{y}_1=y_2 \text{ and } \dot{y}_2=0.$

The solution is given by $\mathbf{y}(t)=\begin{bmatrix} y_1,y_2\end{bmatrix}^T=\begin{bmatrix} at+b,a\end{bmatrix}^T$. Now, If we refer to Lyapunov's definition of stability, we have to check if for any given $R>0$ we can find $r>0$ such that for $|| \mathbf{y}(0)||<r$ the solution $\mathbf{y}(t)$ stays bounded for all time $t$. Hence, we must try to find a $r$ such that for a given $R$ we can guarantee $||\mathbf{y}(t)||<R$. Using the euclidian norm, we see that $||\mathbf{y}(t)||=\sqrt{(at+b)^2+a^2}$. This expression cannot be bounded by any finite $R$, hence the origin is an unstable equilibrium point. As the system is a linear system we can conclude that the system is unstable.

  1. I think I also have an answer to this problem. Let us look at the ODE $\ddot{x}+b(\dot{x})+c(x)=0$ in state-space representation: $$x=x_1$$ $$\dot{x}_1=x_2$$ $$\dot{x}_2=-b(x_2)-c(x_1).$$

The functions $b$ and $c$ always have the same sign as the argument, hence $x_2b(x_2)\geq 0$ and $x_1c(x_1)\geq 0$. Now define the Lyapunov candidate function for the origin:

$$V(x_1,x_2)=\frac{1}{2}x^2_2+\int_{0}^{x_1}c(\bar{x})d\bar{x}.$$

Because of $x_1c(x_1)\geq 0$ it is obvious that $V$ is positive definit. Now, let us investigate the derivative:

$$\dot{V}(x_1,x_2)=x_2\dot{x}_2+c(x_1)\dot{x}_1=x_2\left[-b(x_2)-c(x_1)\right] + c(x_1)x_2=-x_2b(x_2)\leq 0.$$

The result shows that the equilibrium point in the origin is at least stable because $\dot{V}$ is negative semidefinite. The linearized equation $\ddot{x}=0$ is unstable in the sense of Lyapunov as we ruled out in question 1. Hence, even if the linearized equation is unstable like in the 1st question the linearized equations are indecisive.

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  • $\begingroup$ 1. Not necessarily. Consider the system $\dot x=0$, $\dot y=0$. $\endgroup$ – AVK Oct 3 '17 at 8:08

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